1 Introduction and the Main Result

1.1 State-of-the-Art

In this paper we study the problem of charged liquid drop from rigorous mathematical point of view. In the model the two effecting forces are the surface tension, which prefers to keep the drop spherical, and the repulsive electrostatic force, which both act on the boundary of the drop. The problem is well-known and goes back to Rayleigh [48] who studied the linear stability of the sphere and showed that the sphere becomes unstable when the total electric charge is above a given threshold. When the total electric charge is above this Rayleigh threshold, the drop begins to elongate and may eventually form a conical singularity at the tip with a certain opening angle. Such singularities are called Taylor cones due to the work by Taylor [52] and the numerical and experimental evidence suggest that the charged drop typically forms such a singularity [24, 45, 52, 61]. In this paper our goal is to study the well-posedness of the problem and the regularity of the solution. We refer to [46] and [30] for an introduction to the topic.

The static problem of charged liquid drop can be seen as a nonlocal isoperimetric problem and it has been studied from the point of view of Calculus of Variations in recent years [28, 29, 37, 46]. The main issue is that the associated minimization problem, formulated in the framework of Calculus of Variations, is not well-posed, in the sense that the problem does not have a minimizer [28, 46]. Even more surprising is that the results in [28, 46] show that even if the total electric charge is below the Rayleigh threshold, the sphere is not a local minimizer of the associated energy. This means that the electrostatic term is not lower order with respect to the surface tension, which makes the problem mathematically challenging. In order to make the variational problem well-posed one may restrict the problem to convex sets [29] or regularize the functional by adding a curvature term [30] which could lead to the existence of minimizer as the result in [27] suggests.

Here we study this problem from the point of view of fluid-dynamics, which is the framework studied e.g. in [24], where the authors derive the PDE system in the irrotational case (see also [5]). Indeed, as it is observed in [24] the problem is by nature evolutionary, where the drop deforms as a function of time given by the Euler equations with the surface of the drop being the free boundary, which law of motions is coupled with the system which we give in (1.3) below. The problem can thus be seen as the Euler equations for incompressible fluids with free boundary with an additional term given by the electric field. The Euler equations with free boundary without the electric field has been studied rather extensively in recent years. We give only a brief overview on this challenging problem below and refer to [12, 44] for more detailed introduction to the topic. Regarding the problem with electric field we mention the recent works by Yang [59, 60] and Wang-Yang [55], where the authors study the case of the water-wave problem. We also mention the work [20], where the authors study the Stokes flow associated with the charged liquid drop near the sphere and show that under smallness assumption the flow is well defined.

We stress that in our case it is crucial to include the surface tension in the model since otherwise the problem might be ill-posed. For the problem without the electric field one may study the Euler equations also without the surface tension, when one assumes the so called Rayleigh-Taylor sign condition [17], which one should not confuse with the Rayleigh threshold mentioned above. For the water-wave problem the well-posedness is proven by Wu [56, 57] and the general case is due to Lindblad [40], see also [3, 38]. Concerning the problem with surface tension, which is closer to ours, the short time existence of solution in the irrotational case for starshaped sets is due to Beyer-Günther [6, 7] and the general case is proven by Coutand-Shkoller [11]. We also mention the earlier works concerning the well-posedness of the problem in the planar case [2, 34, 58]. The works that are closest to ours are Shatah-Zheng [50] and Schweizer [49], where the authors prove regularity estimates for the free boundary Euler equations with surface tension. Our work is also inspired by Masmoudi-Rousset [44], where the authors prove similar estimates for the Euler equations without the surface tension.

As usual with geometric evolution equations, the Euler equations with free boundary may develop singularities in finite time. In [13] the authors construct an example where the equations develop singularities where the drop changes its topology. We stress that in the absence of the electric field, we do not expect the flow to develop conical singularities predicted by Taylor [52], where both the curvature and the velocity become singular. Indeed, Taylor cones are special type of singularities as the evolving sets \(\Omega _t\) do not change their topology, but only lose their regularity. We also point out that the analysis in [52] does not give a rigorous mathematical proof for the fact that the Taylor cone is a singularity of the associated flow. Indeed, since there is no monotonicity formula for the Euler equations, similar to the one by Huisken [33, 43] for the mean curvature flow, there is little hope to have general classification of the singularities at the moment. We refer to [21] and [23] which both study critical points of energy functionals, which are very much similar to ours, with conical or cusp-like singularities. Then again, as we interpret the Taylor cone as a singularity of the flow given by the Euler equations (1.3), it is not clear why the singularity is a critical point of the potential energy.

1.2 Statement of the Main Theorem

We study the motion of an incompressible charged drop in vacuum in \(\mathbb {R}^3\) and denote the fluid domain by \(\Omega _t\). We assume that we have an initial smooth and compact set \(\Omega _0 \subset \mathbb {R}^3\) and a smooth initial velocity field \(v_0: \Omega _0 \rightarrow \mathbb {R}^3\) which evolve to a smooth family of sets and vector fields \((\Omega _t, v(\cdot , t))_{t \in [0,T)}\). The total energy is given by

$$\begin{aligned} J_t(\Omega _t, v) = \frac{1}{2} \int _{\Omega _t} |v(x,t)|^2 \, dx + \mathcal {H}^2(\partial \Omega _t) + \frac{Q}{\text {Cap}(\Omega _t)}, \end{aligned}$$
(1.1)

where \(Q>0\) and \(\text {Cap}(\Omega )\) is the electrostatic capacity given by

$$\begin{aligned} \text {Cap}(\Omega ): = \inf \big \{ \int _{\mathbb {R}^3}\frac{1}{2}|\nabla u|^2 \, dx: \, u(x) \ge 1 \,\, \text {for all } \, x \in \Omega \,, u \in {\dot{H}}^1(\mathbb {R}^3) \big \}, \end{aligned}$$

and by \(\mathcal {H}^2(\partial \Omega )\) we denote the two dimensional Hausdorff measure of the set \(\partial \Omega \). Define the norm \(\Vert u\Vert _{{\dot{H}}^1(\mathbb {R}^3)} = \Vert \nabla u\Vert _{L^2(\mathbb {R}^3)} + \Vert u\Vert _{L^6(\mathbb {R}^3)}\). We denote the capacitary potential as \(U_\Omega \in {\dot{H}}^1(\mathbb {R}^3)\) which is the function for which \(\text {Cap}(\Omega ) = \int _{\mathbb {R}^3} \frac{1}{2}|\nabla U_\Omega |^2 \, dx\). This is equivalent to say that \(U_\Omega \in {\dot{H}}^1(\mathbb {R}^3)\) satisfies

$$\begin{aligned} {\left\{ \begin{array}{ll} -\Delta U_\Omega = 0 \qquad \text {in } \, \mathbb {R}^3 \setminus {\overline{\Omega }} \\ U_\Omega = 1 \qquad \text {on } \, {\overline{\Omega }}. \end{array}\right. } \end{aligned}$$
(1.2)

We denote the mean curvature of \(\Sigma _t= \partial \Omega _t\) by \(H_{\Sigma _t}\), which for us is the sum of the principal curvatures given by orientation via the outer normal \(\nu _{\Sigma _t}\). With this convention convex sets have positive mean curvature. As usual we denote the material derivative of a vector field F by

$$\begin{aligned} \mathcal {D}_t F = \partial _t F + (v\cdot \nabla )F. \end{aligned}$$

The equations of motion are given by the Euler equations with free boundary (for the derivation see [24])

$$\begin{aligned} {\left\{ \begin{array}{ll} \mathcal {D}_t v + \nabla p =0 \qquad \text {in } \, \Omega _t \\ \textrm{div}\,v=0 \qquad \text {in } \, \Omega _t \\ v_n = V_t \qquad \text {on } \, \Sigma _t= \partial \Omega _t \\ p= H_{\Sigma _t} - \frac{Q}{2 C_t^2} |\nabla U_{\Omega _t}|^2 \qquad \text {on } \Sigma _t, \end{array}\right. } \end{aligned}$$
(1.3)

where \(C_t = \text {Cap}(\Omega _t) \), \(V_t\) is the normal velocity, \(v_n = v \cdot \nu \) and p is the pressure. We say that the system (1.3) has a smooth solution in time-interval (0, T) with initial data \((\Omega _0, v_0)\), if there is a family of \(C^\infty \)-diffeomorphisms \((\Phi _t)_{t \in [0,T)}\), which depend smoothly on t, such that \(\Phi _0 = id\) and \(\Phi _t(\Omega _0) = \Omega _t\), the functions \(v(t, \Phi (t,x))\) and \(p(t, \Phi (t,x))\) are smooth and the equations hold in the classical sense. Moreover we require that \(v(t, \Phi (t,\cdot )) \rightarrow v_0\) as \(t \rightarrow 0\), where \(v_0: \Omega _0 \rightarrow \mathbb {R}^3\) is the initial velocity field. When the total electric charge is zero, i.e. \(Q= 0\), the system reduces to the more familiar Euler equations with free boundary with surface tension. We stress that formally it may seem that the term given by the electric field in the pressure is of lower order than the curvature. However, even for Lipschitz domains this naive intuition fails as we will observe in the beginning of Sect. 3.

The characteristic property of the solution of (1.3) is the conservation of the energy (1.1), i.e.,

$$\begin{aligned} \frac{d}{dt} J(\Omega _t, v) = 0, \end{aligned}$$

which follows from straightforward calculation. Therefore one could guess, and we will prove this in our main theorem, that assuming that the flow given by the system (1.3) does not develop singularities, then it preserves the regularity of the initial data \((\Omega _0, v_0)\) at least in sense of certain Sobolev norm. In particular, we point out that, unlike the mean curvature flow [43], the flow given by the system (1.3) is not smoothing.

We parametrize the moving boundary \(\Sigma _t = \partial \Omega _t\) by using a fixed reference surface \(\Gamma \) which we assume to be smooth and compact. We use the height function parametrization which means that for every t we associate the function \(h(\cdot , t): \Gamma \rightarrow \mathbb {R}\) with the moving boundary \(\Sigma _t\) as

$$\begin{aligned} \Sigma _t = \{x+ h(x,t)\nu _{\Gamma }(x): x \in \Gamma \}. \end{aligned}$$

We assume that \(\Gamma \) satisfies the interior and exterior ball condition with radius \(\eta >0\) and note that \(\eta \) is not necessarily small. For example, from application point of view a relevant case is when the initial set is star-shaped in which case it is natural to choose the reference manifold to be a sphere in which case \(\eta \) is its radius. It is clear that the height-function parametrization is well defined as long as

$$\begin{aligned} \sup _{t \in [0,T)} \Vert h(\cdot ,t)\Vert _{L^\infty (\Gamma ) }< \eta . \end{aligned}$$

Therefore we define the quantity

$$\begin{aligned} \sigma _T := \eta - \sup _{t \in [0,T)} \Vert h(\cdot ,t)\Vert _{L^\infty (\Gamma ) } \end{aligned}$$
(1.4)

and the above condition reads as \(\sigma _T >0\).

As in [44, 49, 50] we note that we do not consider the existence in this paper. Instead, as in [49] we assume that the following qualitative short time existence result holds.

Throughout the paper we assume that for every smooth initial set and smooth initial velocity field the system (1.3) yhas a smooth solution which exists a short interval of time.

Since we will prove a priori estimates, we expect the existence to follow from an argument in the spirit of [51].

In this paper we are interested in finding a priori estimates which guarantee that the system (1.3) does not develop singularities. To this aim we fix a small \(\alpha >0\) and define

$$\begin{aligned} \Lambda _T:= \sup _{t \in [0,T)} \left( \Vert h(\cdot ,t)\Vert _{C^{1,\alpha }(\Gamma )} + \Vert \nabla v(\cdot ,t) \Vert _{L^\infty (\Omega _t)} + \Vert v_n(\cdot ,t) \Vert _{H^2(\Sigma _t) } \right) . \end{aligned}$$
(1.5)

We note that \(\alpha \) can be any positive number. We could also replace the \(C^{1,\alpha }\)-norm by the \(C^{1,\text {Dini}}\)-norm, but we choose to work with Hölder norms as the problem is already technically involved. Our goal is to show that if the quantity \(\Lambda _T\) is bounded then the flow can be extended beyond time T and is smooth if the initial data is smooth. We will prove this in a quantitative way and define an energy quantity of order \(l \ge 1\) as

$$\begin{aligned} {\hat{E}}_l(t) := \sum _{k = 0}^l\Vert \mathcal {D}_t^{l+1-k} v\Vert _{H^{\frac{3}{2}k}(\Omega _t)}^2 + \Vert v(\cdot , t)\Vert _{H^{\lfloor \frac{3}{2}(l+1) \rfloor }(\Omega _t)}^2, \end{aligned}$$
(1.6)

where \(\lfloor \frac{3}{2}(l+1) \rfloor \) denotes the integer part of \(\frac{3}{2}(l+1)\). We define the Hilbert space for half-integers \(H^{\frac{3}{2}k}(\Omega _t)\) via extension in Sect. 2. In the last term we use a Hilbert space of integer order since it simplifies the calculations. The fact that the boundedness of \({\tilde{E}}_l(t)\) for every l implies the smoothness of the flow will be clear from the results in Sect. 8. Indeed, we first show that the bound on \({\hat{E}}_l(t) \) implies a bound for the pressure p. By the a priori estimate we know that the fluid domain remains \(C^{1,\alpha }\)-regular. We use this and estimates for harmonic functions to conclude that the bound on the pressure implies bound on the curvature (see Lemma 5.2), which then gives the regularity of the fluid domain \(\Omega _t\).

Our main result reads as follows. Recall that we assume that the reference surface \(\Gamma \subset \mathbb {R}^3\) satisfies the interior and exterior ball condition with radius \(\eta \).

MainTheorem

Assume that \(\Omega _0\) is a smooth initial set which boundary satisfies \(\partial \Omega _0 = \{x+ h_0(x)\nu _{\Gamma }(x): x \in \Gamma \}\) with \( \Vert h_0\Vert _{L^\infty (\Gamma )}< \eta \) and let \(v_0 \in C^{\infty }(\Omega _0; \mathbb {R}^3)\) be the initial velocity field. Assume that the system (1.3) has a smooth solution in time-interval [0, T) and the parametrization satisfies

$$\begin{aligned} \Lambda _T \le M \quad \text {and} \quad \sigma _T \ge \frac{1}{M} \end{aligned}$$
(1.7)

for some \(M>0\), where \(\sigma _T\) is defined in (1.4) and \(\Lambda _T\) in (1.5). Then for every \(l \in \mathbb {N}\) there is a constant \(C_l\), which depends on \(M, l, {\hat{E}}_l(0) \), and on T if \(T>1\), such that the flow satisfies

$$\begin{aligned} \sup _{0< t < T} {\hat{E}}_l(t) \le C_l, \end{aligned}$$

where \({\hat{E}}_l(t)\) is defined in (1.6). In particular, the system (1.3) does not develop singularity at time T, but remains quantitatively smooth.

Moreover, there are \(T_0 >0 \) and M, which depend on \(\sigma _0\), i.e., \(\sigma _t\) at \(t=0\), \(\Vert H_{\Sigma _0}\Vert _{H^2(\Sigma _0)}\), \(\Vert v\Vert _{H^3(\Omega _0)}\) and the \(C^{1,\alpha }\)-norm of \(h_0\), such that the a priori estimates (1.7) hold for M up to time \({\hat{T}} = \min \{ T, T_0\}\).

Let us make a few comments on the Main Theorem. First, from the point of view of the shape of the drop, the result says that if the parametrization of the flow remains \(C^{1,\alpha }\)-regular then the flow does not develop singularities. We expect this to be optimal in the sense that, we cannot relax the \(C^{1,\alpha }\)-regularity to Lipschitz regularity as the flow may create conical singularities as discussed before.

From the point of view of the velocity, the assumption on Lipschitz regularity of v, which is stronger than the boundedness of the \(\textrm{curl}\,v\), is in the spirit of the Beale-Kato-Majda criterion and thus natural in the theory of the Euler equations [41]. Indeed, in the case when the drop does not chance its shape, i.e., \(\Omega _t = \Omega _0\) the condition (1.7) reduces to

$$\begin{aligned} \sup _{t \in [0,T)} \Vert \nabla v (\cdot ,t)\Vert _{L^{\infty }(\Omega _0)} < \infty , \end{aligned}$$

which guarantees that the equations do not develop singularities by standard results for the Euler-equations [41]. Whether one may remove this condition is beyond our reach at the moment as the gradient level estimates are a fundamental problem in the theory of the Euler equations without the free boundary. The condition on the \(H^2\)-integrability of the normal component of the velocity v on the other hand is related to the fact that the boundary \(\Sigma _t\) is moving. We do not expect this to be optimal but again this problem is too involved for us to solve at the moment. Our main contribution to the problem is to find the optimal sufficient condition for the shape of the drop which guarantee that the flow is well-defined and provide the regularity estimates of all order l. For a drop without surface tension similar type of estimate is proven by Ginsberg [26] with an a priori assumption on the uniform curvature bound.

Finally the last statement of the Main Theorem says that the first statement is not empty, i.e., that the a priori estimates stay bounded up to time \(T_0\), which depends on the initial data by requiring that \(\Vert H_{\Sigma _0}\Vert _{H^2(\Sigma _0)}\) and \(\Vert v_0\Vert _{H^3(\Omega _0)}\) are bounded. We also note that since the regularity estimates in Main Theorem are quantitative, the result can be applied for non-smooth initial data by standard approximation. We note that all quantities in the paper depend of course on the chosen reference surface \(\Gamma \) even if it is not explicitly mentioned.

1.3 Overview of the Proof and the Structure of the Paper

As the paper is long we give a brief overview of the proof of the Main Theorem and of the structure of the paper. The proof is based on energy estimates and to that aim we define the energy functional of order \(l \ge 1\) as

$$\begin{aligned} \begin{aligned} {\mathcal {E}}_l(t) =\frac{1}{2} \int _{\Omega _t}&|\mathcal {D}_t^{l+1} v|^2 \, dx + \frac{1}{2}\int _{\Sigma _t} |\nabla _\tau (\mathcal {D}_t^l v \cdot \nu )|^2 \, d \mathcal {H}^2 \\&- \frac{Q}{2C_t^2} \int _{\Omega _t^c} |\nabla (\partial _t^{l+1} U_{\Omega _t}) |^2 \, dx + \int _{\Omega _t} |\nabla ^{\lfloor \frac{1}{2}(3l+1) \rfloor }( \textrm{curl}\,v)|^2 \, dx, \end{aligned} \end{aligned}$$

which is similar to the quantity in [49] defined on graphs. Here v is the velocity, \(\mathcal {D}_t^l v\) is the material derivative of order l and \(\lfloor \frac{1}{2}(3\,l+1) \rfloor \) denotes the largest integer smaller than \(\frac{1}{2}(3l+1)\). Note that we need an additional term involving the time derivative of the capacitary potential \(U_{\Omega _t}\), as it appears as a high order term in the linearization of the pressure (see Lemma 4.7). This additional term causes problems as it is not immediately clear why the energy is positive or even bounded from below. We also define the associated energy quantity, where we include the spatial regularity

$$\begin{aligned} E_l(t) = \sum _{k=0}^{l} \Vert \mathcal {D}_t^{l+1-k}v\Vert _{H^{\frac{3}{2} k}(\Omega _t)}^2 + \Vert v\Vert _{H^{\lfloor \frac{3}{2}(1+1)\rfloor }(\Omega _t)}^2+ \Vert \mathcal {D}_t^l v \cdot \nu \Vert _{H^1(\Sigma _t)}^2+1. \end{aligned}$$

Note that this quantity takes into account the natural scaling of the system (1.3), where time scales of order \(\frac{3}{2}\) with respect to space as observed e.g. in [50].

We prove high order energy estimates by first showing that if \(\Lambda _T, \sigma _T\) satisfy (1.7) then for all \(t<T\) it holds

$$\begin{aligned} \frac{d}{dt} {\mathcal {E}}_l(t) \le C_l E_l(t) \end{aligned}$$
(1.8)

for \(l \ge 2\). The novelty of (1.8) is that the RHS has linear and not polynomial dependence on \(E_l(t)\), which is crucial in order to show that the flow remains smooth as long as (1.7) holds. This makes the proof technically challenging as we need to estimate all nonlinear error terms in \(\frac{d}{dt} {\mathcal {E}}_l(t)\) in an optimal way. We complete the argument by proving

$$\begin{aligned} E_l(t) \le C_l({\tilde{C}}_l + {\mathcal {E}}_l(t)) \end{aligned}$$
(1.9)

which holds for \(l \ge 1\). The inequalities (1.8) and (1.9) then imply the energy estimates for \(l\ge 2\) and the quantitative \(C^\infty \)-regularity of the flow.

We prove (1.8) and (1.9) by an induction argument over l, where the constants depend on \(\sup _{t<T} E_{l-1}(t)\) which is bounded by the previous step. Therefore the first challenge is to start the argument and to bound \(E_1(t)\). The issue is that (1.8) does not hold for \(l=1\). Instead, we show a weaker estimate

$$\begin{aligned} \frac{d}{dt} {\mathcal {E}}_1(t) \le C_1 (1+ \Vert p\Vert _{H^2(\Omega _t)}^2) E_1(t), \end{aligned}$$
(1.10)

which we expect to be sharp. Therefore in order to start the induction argument we use an ad-hoc argument to show

$$\begin{aligned} \int _0^T \Vert p\Vert _{H^2(\Omega _t)}^2 \, dt \le C. \end{aligned}$$
(1.11)

The inequalities (1.9), (1.10) and (1.11) then imply the first order energy estimate.

We show (1.11) by studying the function

$$\begin{aligned} \Phi (t) = -\int _{\Sigma _t} p \, \Delta _{\Sigma _t} v_n \, d \mathcal {H}^2, \end{aligned}$$

where p is the pressure and \(\Delta _{\Sigma _t}\) the Laplace-Beltrami operator, and prove that it holds

$$\begin{aligned} \frac{d}{dt}\Phi (t) \le - \frac{1}{3} \Vert p\Vert _{H^2(\Omega _t)}^2 + \text {lower order terms}. \end{aligned}$$

We show that the a priori estimates (1.7) imply that \(\Phi \) is bounded and thus we obtain (1.11) by integrating the above inequality over (0, T). We point out that the low order energy estimate is the most challenging part of the proof as we have to work with domains with low regularity. We need rather deep results from differential geometry, boundary regularity for harmonic functions and elliptic regularity in order to overcome this problem. Let us finally outline the structure of the paper.

In Sect. 2 we introduce our notation. Due to the presence of the surface tension, the problem is geometrically involved and we need notation and tools from differential geometry to overcome these issues. We also define the function spaces that we need which include the Hilbert spaces in the domain \(H^{k}(\Omega )\) and on the boundary \(H^{k}(\Sigma )\) for half-integers \(k =0, \frac{1}{2}, 1, \dots \). We also recall functional inequalities such as interpolation inequality and Kato-Ponce inequality.

In Sect. 3 we prove div-curl type estimates in order to transform the high order energy estimates into regularity for the shape and the velocity. We first recall the result from [10] and prove its lower order version in Theorem 3.6. In Theorem 3.9 we prove sharp boundary regularity estimates for harmonic functions with Dirichlet boundary data by using methods from [22]. We believe that these two results are of independent interest.

In Sect. 4 we derive commutation formulas as in [50] and formula for the material derivatives of the pressure on the moving boundary. These formulas include four different error terms which we bound in Sect. 5. All the error terms have different structure and therefore we need to treat them one by one, which makes the Sect. 5 long. The further difficulty is due to the fact that the time and space derivatives have different scaling.

The core of the proof of the Main Theorem is in the next three sections. In Sect. 6 we prove (1.11), in Sect. 7 we prove (1.8) and (1.10), and in Sect. 8 we prove (1.9). The short final section then contains the proof of the Main Theorem.

2 Notation and Preliminary Results

In this section we introduce our notation and recall some basic results on function spaces and geometric inequalities. Many of these results are well-known for experts but we include them since they might be difficult to find, while some results we did not find at all in the existing literature. Throughout the paper C denotes a large constant, which value may change from line to line.

We first introduce notation related to Riemannian geometry. As an introduction to the topic we refer to [39]. We will always deal with compact hypersurfaces \(\Sigma \subset \mathbb {R}^3\), which then can be seen as boundaries of sets \(\Omega \), i.e., \(\partial \Omega = \Sigma \). We denote its outer unit normal by \(\nu _{\Omega }\) and denote it sometimes by \( \nu _{\Sigma }\) or merely \(\nu \) when its meaning is obvious from the context. We use the outward orientation and denote the second fundamental form by \(B_\Sigma \) and the mean curvature by \(H_\Sigma \), which is defined as the sum of the principal curvatures. Again we write simply B and H when the meaning is clear from the context. We note that we use the convention in our notation that \(\Sigma = \partial \Omega \) denotes a generic surface, \(\Sigma _t= \partial \Omega _t\) denotes the evolving surface given by the equations (1.3) and \(\Gamma = \partial G\) is our reference surface which we introduce later. We note that the constants in the paper will depend on the chosen reference surface. We take this for granted and do not mention it in the statements.

Since \(\Sigma \) is embedded in \(\mathbb {R}^{3}\) it has natural metric g induced by the Euclidian metric. Then \((\Sigma , g)\) is a Riemannian manifold and we denote the inner product on each tangent space \(X, Y \in T_x \Sigma \) by \(\langle X, Y \rangle \), which we may write in local coordinates as

$$\begin{aligned} \langle X,Y \rangle = g(X,Y) = g_{ij} X^iY^j. \end{aligned}$$

We extend the inner product in a natural way for tensors. We denote smooth vector fields on \(\Sigma \) by \({\mathscr {T}}(\Sigma )\) and by a slight abuse of notation we denote smooth kth order tensor fields on \(\Sigma \) by \({\mathscr {T}}^k(\Sigma )\). We write \(X^i\) for vectors and \(Z_i\) for covectors in local coordinates.

We denote the Riemannian connection on \(\Sigma \) by \({\bar{\nabla }}\) and recall that for a function \(u \in C^\infty (\Sigma )\) the covariant derivative \({\bar{\nabla }} u \) is a 1-tensor field defined for \(X \in {\mathscr {T}}( \Sigma )\) as

$$\begin{aligned} {\bar{\nabla }} u(X) = {\bar{\nabla }}_X u = X u, \end{aligned}$$

i.e., the derivative of u in the direction of X. The covariant derivative of a smooth k-tensor field \(F \in {\mathscr {T}}^k( \Sigma )\), denoted by \({\bar{\nabla }} F\), is a \((k+1)\)-tensor field and we have the following recursive formula for \( Y_1, \dots , Y_k, X \in {\mathscr {T}}( \Sigma )\)

$$\begin{aligned} {\bar{\nabla }} F(Y_1, \dots , Y_k, X) = ({\bar{\nabla }}_X F)(Y_1, \dots , Y_k), \end{aligned}$$

where

$$\begin{aligned} ({\bar{\nabla }}_X F)(Y_1, \dots , Y_k) = X F(Y_1, \dots , Y_k) - \sum _{i=1}^k F(Y_1, \dots , {\bar{\nabla }}_X Y_i,\dots , Y_k). \end{aligned}$$

Here \({\bar{\nabla }}_X Y\) is the covariant derivative of Y in the direction of X (see [39]) and since \({\bar{\nabla }}\) is the Riemannian connection it holds \({\bar{\nabla }}_X Y = {\bar{\nabla }}_Y X + [X,Y]\) for every \(X, Y \in {\mathscr {T}}( \Sigma )\). We denote the kth order covariant derivative of a function u on \(\Sigma \) by \({\bar{\nabla }}^k u \in {\mathscr {T}}^k( \Sigma )\). The notation \({\bar{\nabla }}_{i_1} \cdots {\bar{\nabla }}_{i_k} u\) means a coefficient of \({\bar{\nabla }}^k u\) in local coordinates. We may raise the index of \({\bar{\nabla }}_i u\) by using the inverse of the metric tensor \(g^{ij}\) as \({\bar{\nabla }}^i u = g^{ij}{\bar{\nabla }}_j u\). We denote the divergence of a vector field \(X \in {\mathscr {T}}(\Sigma )\) by \(\textrm{div}\,_{\Sigma } X\) and the Laplace-Beltrami operator for a function \(u: \Sigma \rightarrow \mathbb {R}\) by \(\Delta _\Sigma u\). We recall that by the divergence theorem

$$\begin{aligned} \int _{\Sigma } \textrm{div}\,_{\Sigma } X \, d \mathcal {H}^2 = 0. \end{aligned}$$

We will first fix our reference surface which we denote by \(\Gamma \) which is a boundary of a smooth, compact set G, i.e., \(\Gamma = \partial G\). Since G is smooth it satisfies the interior and exterior ball condition with radius \(\eta \), and we denote the tubular neighborhood of \(\Gamma \) by \({\mathcal {N}}_\eta (\Gamma )\) which is defined as

$$\begin{aligned} {\mathcal {N}}_\eta (\Gamma ) = \{ x \in \mathbb {R}^3: \text {dist}(x,\Gamma ) < \eta \}. \end{aligned}$$

Then the map \(\Psi : \Gamma \times (-\eta , \eta ) \rightarrow {\mathcal {N}}_\eta (\Gamma )\) defined as \(\Psi (x,s) = x+ s \nu _{\Gamma }(x)\) is a diffeomorphism. We say that a hypersurface \(\Sigma \), or a domain \(\Omega \) with \(\partial \Omega = \Sigma \), is \(C^{1,\alpha }(\Gamma )\)-regular for some small \(\alpha >0\), when it can be written as

$$\begin{aligned} \Sigma = \{ x + h(x) \nu _{\Gamma }(x): x \in \Gamma \}, \end{aligned}$$

for a \(C^{1,\alpha }(\Gamma )\)-regular function \(h: \Gamma \rightarrow \mathbb {R}\) with \(\Vert h\Vert _{L^\infty } < \eta \). In particular, all \(C^{1,\alpha }(\Gamma )\)-regular sets are diffeomorphic. We say that a set \(\Sigma \) is uniformly \(C^{1,\alpha }(\Gamma )\)-regular if the height-function satisfies \(\Vert h\Vert _{C^{1,\alpha }(\Gamma )} \le C\) and \(\Vert h\Vert _{L^\infty } \le c \eta \) for constants C and \(c <1\). Finally we say that \(\Sigma \) is uniformly \(C^1\)-regular if \(\Vert h\Vert _{C^{1}(\Gamma )} \le C\).

Let us next fix our notation in the ambient space \(\mathbb {R}^3\). We denote the kth order differential of a vector field \(F: \mathbb {R}^3 \rightarrow \mathbb {R}^m\) by \(\nabla ^k F\), the divergence of \(F: \mathbb {R}^3 \rightarrow \mathbb {R}^3\) by \(\textrm{div}\,F\) and the Laplace operator in \(\mathbb {R}^3\) by \(\Delta \). The notation \((\nabla F)^T\) stands for the transpose of \(\nabla F\). When we restrict \(F: \mathbb {R}^3 \rightarrow \mathbb {R}^3\) on \(\Sigma \), we define its normal and tangential part as

$$\begin{aligned} F_n:= F \cdot \nu _\Sigma \qquad \text {and} \qquad F_\tau = F - F_n \, \nu _\Sigma . \end{aligned}$$

We use the notation \(x\cdot y\) for the inner product of two vectors in \(\mathbb {R}^n\).

Since \(\Sigma \) is a smooth hypersurface we may extend every function and vector field defined on \(\Sigma \) to \(\mathbb {R}^3\). We may thus define a tangential differential of a vector field \(F: \Sigma \rightarrow \mathbb {R}^m\) by

$$\begin{aligned} \nabla _\tau F = \nabla F - (\nabla F \nu _\Sigma ) \otimes \nu _\Sigma \end{aligned}$$

where we have extended F to \(\mathbb {R}^3\). We may then extend the definition of \(\textrm{div}\,_\Sigma \) to fields \(F: \Sigma \rightarrow \mathbb {R}^3\) by \(\textrm{div}\,_\Sigma F = \text {Tr}(\nabla _\tau F)\) and the divergence theorem generalizes to

$$\begin{aligned} \int _{\Sigma } \textrm{div}\,_{\Sigma } F \, d \mathcal {H}^2 = \int _{\Sigma } H_\Sigma (F \cdot \nu _\Sigma ) \, d \mathcal {H}^2. \end{aligned}$$

We note that the tangential gradient of \(u \in C^\infty (\Sigma )\) is equivalent to its covariant derivative in the sense that for every vector field \(X \in {\mathscr {T}}(\Sigma )\) we find a vector field \(\tilde{X}: \Sigma \rightarrow \mathbb {R}^{3}\) which satisfies \(\tilde{X}\cdot \nu _\Sigma = 0\) and

$$\begin{aligned} {\bar{\nabla }}_X u = \nabla _\tau u \cdot \tilde{X}. \end{aligned}$$

Let us comment briefly on the notation related to the equations (1.3). We denote the derivative with respect to time by \(\partial _t F\) and the material derivative as

$$\begin{aligned} \mathcal {D}_t F:= \partial _t F +(v \cdot \nabla ) F. \end{aligned}$$

The material derivative does not commute with the spatial derivative and we denote the commutation

$$\begin{aligned} {[}\mathcal {D}_t,\nabla ] u = \mathcal {D}_t\nabla u - \nabla \mathcal {D}_t u. \end{aligned}$$

We denote by \(U_\Omega \) the capacitary potential defined in (1.2) and denote \(U_t = U_{\Omega _t}\), \(H_t = H_{\Sigma _t}\) etc...when the meaning is clear from the context. To shorten further the notation we denote

$$\begin{aligned} Q(t): = \frac{Q}{(\text {Cap}(\Omega _t))^2}. \end{aligned}$$
(2.1)

We may thus write the pressure in (1.3) as

$$\begin{aligned} p = H_t - \frac{Q(t)}{2} |\nabla U_t|^2. \end{aligned}$$

Let us next fix the notation for the function spaces. We define the Sobolev space \(W^{l,p}(\Sigma )\) in a standard way for \(p \in [1,\infty ]\), see e.g. [4], denote the Hilbert space \(H^l(\Sigma ) = W^{l,2}(\Sigma )\) and define the associated norm for \(u \in W^{l,p}(\Sigma )\) as

$$\begin{aligned} \Vert u\Vert _{W^{l,p}(\Sigma )}^p = \sum _{k = 0}^l \int _\Sigma |{\bar{\nabla }}^k u|^p\, d \mathcal {H}^2 \end{aligned}$$

and for \(p = \infty \)

$$\begin{aligned} \Vert u\Vert _{W^{l,\infty }(\Sigma )} = \sum _{k = 0}^l \sup _{x \in \Sigma } |{\bar{\nabla }}^k u|. \end{aligned}$$

We often denote \(\Vert u\Vert _{C^0(\Sigma )} = \Vert u\Vert _{L^\infty (\Sigma )} = \sup _{x \in \Sigma } |u(x)|\) for continuous function \(u: \Sigma \rightarrow \mathbb {R}\) and \(\Vert u\Vert _{C^{m}(\Sigma )} = \Vert u\Vert _{W^{m,\infty }(\Sigma )}\). We define the Hölder norm of a continuous function \(u: \Sigma \rightarrow \mathbb {R}\) by

$$\begin{aligned} \Vert u\Vert _{C^\alpha (\Sigma )} = \Vert u\Vert _{L^\infty (\Sigma )} + \sup _{\begin{array}{c} x \ne y \\ x,y \in \Sigma \end{array}} \frac{|u(y) - u(x)|}{|y-x|^\alpha }. \end{aligned}$$

We define the Hölder norm for a tensor field \(F \in {\mathscr {T}}^k(\Sigma )\) as in [36]

$$\begin{aligned} \Vert F\Vert _{C^\alpha (\Sigma )} = \sup \{ \Vert F(X_1, \dots , X_k) \Vert _{C^\alpha (\Sigma )}: X_i \in {\mathscr {T}}(\Sigma ) \, \text { with } \, \Vert X_i \Vert _{C^1(\Sigma )} \le 1\}. \end{aligned}$$

Finally we define the \(H^{-1}(\Sigma )\)-norm by duality, i.e.,

$$\begin{aligned} \Vert u \Vert _{H^{-1}(\Sigma )}:= \sup \Big \{ \int _{\Sigma } u \, g \, d\mathcal {H}^2: \Vert g\Vert _{H^{1}(\Sigma )} \le 1 \Big \}. \end{aligned}$$

For functions defined in the domain \(u:\Omega \rightarrow \mathbb {R}\) we define the Sobolev space \(W^{l,p}(\Omega )\) as functions which have kth order weak derivative in \(\Omega \) and the corresponding norm is bounded

$$\begin{aligned} \Vert u\Vert _{W^{l,p}(\Omega )}^p:= \sum _{k=0}^l \int _{\Omega } |\nabla ^k u|^p \, dx <\infty . \end{aligned}$$

As before we denote the Hilbert space as \(H^l(\Omega ) = W^{l,2}(\Omega )\) and define \(H^{-1}(\Omega )\) by duality. Finally given an index vector \(\alpha = (\alpha )_{i=1}^k \in \mathbb {N}^k\) we define its norm by

$$\begin{aligned} |\alpha | = \sum _{i =1}^k \alpha _i. \end{aligned}$$

Throughout the paper we use the notation \(S \star T\) from [32, 42] to denote a tensor formed by contraction on some indexes of tensors S and T, using the coefficients of the metric tensor \(g_{ij}\) if S and T are defined on the boundary \(\Sigma \). We also use the convention that \({\bar{\nabla }}^k u \star {\bar{\nabla }}^l v\) denotes contraction of some indexes of tensors \({\bar{\nabla }}^{i} u\) and \( {\bar{\nabla }}^{j} v\) for any \(i \le k\) and \(j \le l\). In other words, we include also the lower order covariant derivatives.

Following the notation from [54], we first introduce the real interpolation method and then the interpolation spaces. Let X and Y be Banach spaces endowed respectively with the norm \(\Vert \cdot \Vert _X\) and \(\Vert \cdot \Vert _Y\). The couple (XY) is said to be an interpolation couple if both X and Y are embedded in a Hausdorff topological vector space V. In this case we have that \(X\cap Y\) endowed with the norm \(\Vert v\Vert _{X\cap Y}=\max \{\Vert v\Vert _X, \Vert v\Vert _Y\}\) is a Banach space. Moreover, we also have that \(X+Y=\{z=x+y,\, x\in X,\,y\in Y \}\) endowed with the norm

$$\begin{aligned} \Vert z\Vert _{X+Y}= \inf _{x\in X,\, y \in Y}\{ \Vert x\Vert _X+\Vert y\Vert _Y, \, z=x+y\} \end{aligned}$$

is a Banach space and it is immediate to check that

$$\begin{aligned} X\cap Y\subset X,Y\subset X+ Y. \end{aligned}$$

For \(z \in X+Y\) and \(t>0\) we introduce the K functional

$$\begin{aligned} K(t,z,X,Y)=\inf _{x\in X,\, y\in Y} \{\Vert x\Vert _X+t\Vert y\Vert _Y,\, x+y=z\}. \end{aligned}$$

For \(\theta \in (0,1)\), \(p\in [1,\infty )\) and \(z\in X+Y\) we let

$$\begin{aligned} \Vert z\Vert ^p_{\theta , p}= \int _{0}^\infty \left( \frac{K(t,z,X,Y)}{t^{\theta }}\right) ^p \frac{dt}{t} \end{aligned}$$

and define

$$\begin{aligned} (X,Y)_{\theta , p}=\{z\in X+Y: \Vert z\Vert _{\theta ,p}<\infty \}. \end{aligned}$$

We note that \((X,Y)_{\theta ,p}\) is a Banach space.

Finally, we recall that if \((X_1,X_2)\) and \((Y_1,Y_2)\) are interpolation couples and \({\mathcal {F}}: X_1+X_2 \rightarrow Y_1+Y_2\) is a linear operator which is bounded \(X_i \rightarrow Y_i\) by \(M_i\). Then for \(\theta \in (0,1)\) and \(p\in [1,\infty )\) the operator

$$\begin{aligned} {\mathcal {F}}: (X_1,X_2)_{\theta , p} \rightarrow (Y_1,Y_2)_{\theta , p} \end{aligned}$$
(2.2)

is bounded and we may estimate its norm by \(M_1^{1-\theta }M_2^\theta \).

2.1 Half-Integer Sobolev Spaces

Before giving the definition of half-integer Sobolev space in a domain, we exploit the extension properties of Sobolev functions. Throughout the paper we assume that the boundary \(\Sigma = \partial \Omega \) is uniformly \(C^{1,\alpha }(\Gamma )\)-regular and thus it is \(H^1\) extension domain, i.e., there is a linear operator \(T:H^{1}(\Omega )\rightarrow H^{1} (\mathbb {R}^3)\) such that

$$\begin{aligned} \Vert T(u)\Vert _{H^{1} (\mathbb {R}^3)} \le C\Vert u\Vert _{H^{1}(\Omega )}. \end{aligned}$$

We refer to [9] for the study of Sobolev spaces under Lipschitz-regularity and the references therein.

We need more regularity for the boundary for higher order Sobolev extension \(m \ge 2\), although we do not need the optimal condition. Instead, we assume the following for the second fundamental form

figure a

which guarantees that we may extend a given function \(u \in H^{m}(\Omega )\) to the whole space. Note that for \(m \ge 4\) the condition (\(\hbox {H}_{m}\)) is implied by \(\Vert B_\Sigma \Vert _{H^{m-2}(\Sigma )}\le C_m \) by the Sobolev-embedding, which agrees with the assumption e.g. in [10]. In the following we do not specify that a given quantity depends on the constant \(C_m\), but take it for granted when we refer to the condition (\(\hbox {H}_{m}\)).

Even if there are many results for extensions of Sobolev functions in the literature, the condition (\(\hbox {H}_{m}\)) is too weak to apply them. To this aim we need the following result.

Proposition 2.1

Let \(m \in \mathbb {N}\), with \(m\ge 2\), and let \(\Omega \) be a smooth domain which is uniformly \(C^{1,\alpha }(\Gamma )\)-regular and satisfies (\(\hbox {H}_{m}\)). Then there is an extension operator \(T:H^{m}(\Omega )\rightarrow H_0^{m} (\mathbb {R}^3)\) such that

$$\begin{aligned} \Vert T(u)\Vert _{H^m(\mathbb {R}^3)}\le C\Vert u\Vert _{H^m(\Omega )}. \end{aligned}$$

Proof

Let \(x_0\in \partial \Omega \). There exist \(\delta >0\) and a diffeomorphism \(\Psi : \mathbb {R}^3 \rightarrow \mathbb {R}^3 \) such that \(\Psi ^{-1}( \Omega \cap B_\delta (x_0))= B_1^+ = B_1 \cap \mathbb {R}_+^3\). We note that the \(C^{1,\alpha }\) regularity of \(\partial \Omega \) and (\(\hbox {H}_{m}\)) imply that we may choose the diffeomorphism such that it satisfies \(\Vert \Psi \Vert _{C^{1,\alpha }(\mathbb {R}^3)} \le C\) and

$$\begin{aligned} \Vert \nabla ^2 \Psi \Vert _{L^4(\mathbb {R}^3)}\le C \quad \text { if }\, m=2,\qquad \Vert \Psi \Vert _{H^{m}(\mathbb {R}^3)}\le C \quad \text { if } \, m>2. \end{aligned}$$
(2.3)

For \(u: \Omega \rightarrow \mathbb {R}\) smooth we let \(u'=u\circ \Psi \). We may extend \(u'\) to a function \(T(u') \in H^m (B_1)\) such that

$$\begin{aligned} \Vert T(u')\Vert _{H^m(B_1)}\le C\Vert u'\Vert _{H^m(B_1^+)}. \end{aligned}$$

The construction of \(T(u') \in H^m (B_1)\) is classical but we recall it for the reader’s convenience. We define

$$\begin{aligned} T(u')(x',x_n)= {\left\{ \begin{array}{ll} u'(x',x_n) \,\,\,\,\ {} &{}x_n\ge 0 \\ {\sum }_{j=1}^{m+1} \lambda _j u'(x', -j^{-1}x_n )\,\,\,\,\, &{}x_n<0 \end{array}\right. } \end{aligned}$$

where \(\lambda =(\lambda _1,\dots ,\lambda _{m+1})\) solve the system

$$\begin{aligned} {\left\{ \begin{array}{ll} {\sum }_{j}\lambda _j=1 \\ {\sum } _{j}(-j)^{-1}\lambda _j=1 \\ \vdots \\ {\sum }_{j} (-j)^{-m}\lambda _j=1. \end{array}\right. } \end{aligned}$$

This system, known as Vandermonde system, has a unique solution, hence \(Tu'\) is well defined. Finally we define the extension operator as

$$\begin{aligned} T(u):= T(u') \circ \Psi ^{-1}. \end{aligned}$$

Let us show that T is a bounded operator.

It is straightforward to check that

$$\begin{aligned} \Vert Tu'\Vert _{H^m(B_1)}\le C\Vert u'\Vert _{H^m(B_1^+)}. \end{aligned}$$

Let us then show that

$$\begin{aligned} \Vert u'\Vert _{H^m(B_1^+)}\le C \Vert u\Vert _{H^m(\Omega )}. \end{aligned}$$
(2.4)

We first note that

$$\begin{aligned} \nabla u'= \nabla u \star \nabla \Psi \end{aligned}$$

and for \(m\ge 2\)

$$\begin{aligned} \nabla ^m u'= \sum _{|\alpha |\le m-1 } \nabla ^{1+\alpha _1}\Psi \star \cdots \star \nabla ^{1+\alpha _m} \Psi \star \nabla ^{1+\alpha _{m+1}} u. \end{aligned}$$

For \(m=2\) we have then by Hölder’s inequality, by (2.3) and by the Sobolev embedding

$$\begin{aligned} \Vert \nabla ^2 u'\Vert _{L^2(B_1)} \le C \Vert \Psi \Vert _{C^1(\mathbb {R}^3)}\Vert \nabla ^2 u\Vert _{L^2(\Omega )} + \Vert \nabla ^2 \Psi \Vert _{L^4(\mathbb {R}^3)}\Vert \nabla u\Vert _{L^4(\Omega )} \le C\Vert u\Vert _{H^2(\Omega )}. \end{aligned}$$

To treat the case \(m\ge 3\) we first observe that by Sobolev embedding it holds

$$\begin{aligned} \Vert \nabla ^{m-2}u\Vert _{L^\infty (B_1^+)}\le \Vert u\Vert _{H^m(\Omega )} \quad \text {and} \quad \Vert \nabla ^{m-2}\Psi \Vert _{L^\infty (B_1^+)}\le \Vert \Psi \Vert _{H^m(\mathbb {R}^3)}. \end{aligned}$$

Hence for \(m \ge 3\) we have by Hölder’s inequality, by (2.3) and by the Sobolev embedding

$$\begin{aligned} \begin{aligned} \Vert \nabla ^m u'\Vert _{L^2(B_1)}&\le C\big ((1+\Vert \Psi \Vert _{H^m(\mathbb {R}^3)}^m)\Vert u\Vert _{H^m(\Omega )} + \Vert \nabla ^2 \Psi \Vert _{L^4(\mathbb {R}^3)}\Vert \nabla ^{m-1} u\Vert _{L^4(\Omega )}\\&\,\,\,\,\,\,\,+\Vert \nabla ^{m-1} \Psi \Vert _{L^4(\mathbb {R}^3)}\Vert \nabla ^{2} u\Vert _{L^4(\Omega )} \big )\\&\le C (1+\Vert \Psi \Vert _{H^m(\mathbb {R}^3)}^m)\Vert u\Vert _{H^m(\Omega )}\\&\le C \Vert u\Vert _{H^m(\Omega )}. \end{aligned} \end{aligned}$$

Thus we have (2.4).

Similarly we show that

$$\begin{aligned} \Vert T(u') \circ \Psi ^{-1}\Vert _{H^m(\Omega \cap B_\delta (x_0))} \le C\Vert Tu'\Vert _{H^m(B_1)}. \end{aligned}$$

The claim then follows from standard covering argument. \(\square \)

Throughout the paper we will refer to the operator T as the canonical extension operator or simply as the extension operator. We define the half-integer Sobolev space in the domain \(\Omega \) using the canonical extension operator.

Definition 2.2

We say that a function \(u\in L^2(\Omega )\) is in \(H^\frac{1}{2}(\Omega )\) if

$$\begin{aligned} \Vert u\Vert ^2_{H^\frac{1}{2}(\Omega )}:= \Vert u\Vert ^2_{L^2(\Omega )}+\int _{\mathbb {R}^3}\int _{\mathbb {R}^3}\frac{|Tu(x)-Tu(y)|^2}{|x-y|^3}\,dxdy<\infty . \end{aligned}$$

For \(k\ge 1\) we say that \(u\in H^{k+\frac{1}{2}}(\Omega )\) if \(u\in H^k(\Omega )\) and

$$\begin{aligned} \Vert u\Vert _{H^{k+\frac{1}{2}}(\Omega )}:=\Vert u\Vert _{H^k(\Omega )}+ \Vert T(\nabla ^k u)\Vert _{H^\frac{1}{2}(\Omega )} < \infty . \end{aligned}$$

Finally we define the \(H^{-\frac{1}{2}}(\Omega )\)-norm by duality.

We define \(H^\frac{1}{2}_\star (\Omega )\) as the space of functions via interpolation such that \(u \in H^\frac{1}{2}_\star (\Omega )\) if \(T(u) \in (L^2(\mathbb {R}^3),H^1(\mathbb {R}^3))_{\frac{1}{2},2}\) and endow it with the norm

$$\begin{aligned} \Vert u\Vert _{H_\star ^\frac{1}{2}(\Omega )}^2:= \Vert u\Vert _{L^2(\Omega )}^2+ \int _0^\infty \left( \frac{ K(t,T(u),L^2(\mathbb {R}^3),H^1(\mathbb {R}^3))}{t^{1/2}}\right) ^2\, \frac{dt}{t}. \end{aligned}$$
(2.5)

This gives an equivalent definition for the half-integer Sobolev space.

Proposition 2.3

Let \(m \in \mathbb {N}\), with \(m\ge 2\), and let \(\Omega \) be a smooth domain which is uniformly \(C^{1,\alpha }(\Gamma )\)-regular and satisfies (\(\hbox {H}_{m}\)). The norms in Definition 2.2 and (2.5) are equivalent, i.e.,

$$\begin{aligned} \Vert u\Vert _{H^\frac{1}{2}(\Omega )}\simeq \Vert u\Vert _{H_\star ^\frac{1}{2}(\Omega )}. \end{aligned}$$

We do not give the details of the proof, but only refer to [9] and mention that it follows from the fact that \(H^\frac{1}{2}(\mathbb {R}^3)=(L^2(\mathbb {R}^3),H^1(\mathbb {R}^3))_{\frac{1}{2},2}\), see [54].

2.2 Half-Integer Sobolev Spaces on a Surfaces

We begin by defining the space \(H^\frac{1}{2}(\Sigma )\). Again there are many ways to do this. We choose the definition via harmonic extension.

Definition 2.4

Let \(\Sigma = \partial \Omega \) be uniformly \(C^{1,\alpha }(\Gamma )\)-regular. We say that \(u\in H^\frac{1}{2}(\Sigma )\) if \(u\in L^2(\Sigma )\) and

$$\begin{aligned} \Vert u\Vert _{H^\frac{1}{2}(\Sigma )}=\Vert u\Vert _{L^2(\Sigma )}+\inf \{\Vert \nabla v\Vert _{L^2(\Omega )}: \,v-u \in H^{1}_0 (\Omega )\} <\infty . \end{aligned}$$

We define the space \(H^{-\frac{1}{2}}(\Sigma )\) and its norm by duality.

By standard theory the \(C^{1,\alpha }(\Gamma )\)-regularity of \(\Sigma \) ensures that Definition 2.4 is equivalent to the definition via Gagliardo seminorm

$$\begin{aligned} \Vert u\Vert _{H^\frac{1}{2}(\Sigma )}^2 \simeq \Vert u\Vert _{L^2(\Sigma )}^2 +\int _\Sigma \int _\Sigma \frac{|u(x)-u(y)|^2}{|x-y|^3}\, d\mathcal {H}^2_xd\mathcal {H}^2_y. \end{aligned}$$

Moreover, this norm is also equivalent to the norm obtained via interpolation. Indeed, let us define the interpolation space (see beginning of Sect. 2)

$$\begin{aligned} H^\frac{1}{2}_\star (\Sigma )=(L^2(\Sigma ),H^1(\Sigma ))_{\frac{1}{2},2}. \end{aligned}$$

Let us show that

$$\begin{aligned} \Vert u\Vert _{H^\frac{1}{2}(\Sigma )} \simeq \Vert u\Vert _{H_\star ^\frac{1}{2}(\Sigma )}. \end{aligned}$$
(2.6)

Due to the non-local nature of the problem we give the proof of (2.6) in detail.

We fix a small \(\delta >0\) and cover \(\Sigma \) with finitely many balls of radius \(\delta \) centered at \(x_i \in \Sigma \), i.e.,

$$\begin{aligned} \Sigma \subset \bigcup _{i=1}^N B_\delta (x_i). \end{aligned}$$

Since \(\Sigma \) is \(C^{1,\alpha }(\Gamma )\), there are \(C^{1,\alpha }\)-regular functions \(\phi _i\) such that \(\Sigma \cap B_{2\delta } (x_i)\) is contained in the graph of \(\phi _i\) for every \(i =1, \dots , N\), when \(\delta \) is small enough. Let \(\{\eta _i\}_{i=1,\dots , N}\) be a partition of unity subordinated to the open covering \(B_{\delta }(x_i)\). Then it holds

$$\begin{aligned} \Vert u\Vert _{H^\frac{1}{2}(\Sigma )}\le \sum _{i=1}^N\Vert \eta _i u\Vert _{H^\frac{1}{2}(\Sigma )}. \end{aligned}$$

Let us fix \(i =1, \dots , N\) and by rotating and translating the coordinates we may assume that \(x_i = 0\) and \(\Sigma \cap B_{2\delta } \subset \{ (x',\phi _i(x')): x' \in \mathbb {R}^2\}\) with \(\phi _i(0) = 0\) and \(\nabla \phi _i(0) = 0\). Denote \(u_i= \eta _i u\) and \(v_i(x') = u_i(x', \phi _i(x'))\). Note that then \(v_i: \mathbb {R}^2 \rightarrow \mathbb {R}\), \(\text {supp}\, v_i \subset B_{\delta '} \subset \mathbb {R}^2\) and \(\text {supp}\, u_i \subset B_{\delta } \subset \mathbb {R}^3\) with \(\delta /2 \le \delta ' \le \delta \). Therefore we deduce by the \(C^{1,\alpha }\)-regularity of \(\phi _i\) that

$$\begin{aligned} \begin{aligned}&\int _{\Sigma }\int _{\Sigma }\frac{|u_i(x)-u_i(y)|^2}{|x-y|^3}\, d\mathcal {H}^2_x\,d\mathcal {H}^2_y\\&= \int _{\Sigma \cap B_{2\delta } }\int _{\Sigma \cap B_{2\delta }}\frac{|u_i(x)-u_i(y)|^2}{|x-y|^3}\, d\mathcal {H}_x\,d\mathcal {H}_y + 2\int _{\Sigma \setminus B_{2\delta } }\int _{\Sigma \cap B_{2\delta }}\frac{|u_i(x)-u_i(y)|^2}{|x-y|^3}\, d\mathcal {H}_x\,d\mathcal {H}_y \\&\le C\int _{B_{2\delta '}}\int _{B_{2\delta '}} \frac{|u_i(x',\phi _i(x'))-u_i(y',\phi _i(y'))|^2}{(|x'-y'|^2 +(\phi _i(x')-\phi _i(y')^2)^{3/2}}\, dx'\,dy' +C \Vert u\Vert _{L^2(\Sigma )}^2 \\&\le C\Vert v_i\Vert _{H^\frac{1}{2}(\mathbb {R}^2)}^2 + C\Vert u\Vert _{L^2(\Sigma )}^2\\&\le C\Vert v_i\Vert _{H_{\star }^\frac{1}{2}(\mathbb {R}^2)}^2+ C\Vert u\Vert _{L^2(\Sigma )}^2. \end{aligned} \end{aligned}$$

This implies \(\Vert u_i\Vert _{H^{\frac{1}{2}}(\Sigma )} \le C\Vert v_i\Vert _{H_{\star }^\frac{1}{2}(\mathbb {R}^2))}+ C\Vert u\Vert _{L^2(\Sigma )}\). Let us denote by \(L_0^2(B_{2\delta '})\) and \(H_0^1(B_{2\delta '})\) for functions \(f \in L^2(\mathbb {R}^2)\), and respectively \(f \in H^1(\mathbb {R}^2)\), with \(\text {supp} f \subset B_{2\delta '}\). Denote also \(\Psi : B_{2\delta '} \rightarrow \Psi (B_{2\delta '}) \subset \mathbb {R}^3\), \(\Psi (x') = (x', \phi _i(x'))\). We may estimate

$$\begin{aligned} \begin{aligned}&K(t,v_i, L^2(\mathbb {R}^2),H^1(\mathbb {R}^2))\\&\quad =\inf _{f+g=v_i}\{ \Vert f\Vert _{L^2(\mathbb {R}^2)}+ t \Vert g\Vert _{H^1(\mathbb {R}^2)},\, f \in L^2(\mathbb {R}^2),\, g\in H^1(\mathbb {R}^2) \}\\&\quad \le \inf _{f+g=v_i}\{ \Vert f\Vert _{L^2(\mathbb {R}^2)}+ t \Vert g\Vert _{H^1(\mathbb {R}^2)},\, f \in L_0^2(B_{2\delta '}),\, g\in H_0^1(B_{2\delta '}) \} \\&\quad \le C\inf _{f+ g = v_i}\{ \Vert f \circ \Psi ^{-1}\Vert _{L^2(\Sigma \cap B_{2\delta })}+ t \Vert g\circ \Psi ^{-1}\Vert _{H^1(\Sigma \cap B_{2\delta })},\, f \in L_0^2(B_{2\delta '}),\, g\in H_0^1(B_{2\delta '}) \} \\&\quad \le C\inf _{{\tilde{f}}+ {\tilde{g}} = u_i}\{ \Vert {\tilde{f}} \Vert _{L^2(\Sigma \cap B_{2\delta })}+ t \Vert {\tilde{g}}\Vert _{H^1(\Sigma \cap B_{2\delta })},\,{\tilde{f}} \in L_0^2(\Sigma \cap B_{2\delta }),\, {\tilde{g}}\in H_0^1(\Sigma \cap B_{2\delta }) \} \\&\quad =C \, K(t,u_i, L_0^2(\Sigma \cap B_{2\delta }),H_0^1(\Sigma \cap B_{2\delta })). \end{aligned} \end{aligned}$$

Since \(\text {supp} \, u_i \subset \Sigma \cap B_{\delta }\), it is easy to see that

$$\begin{aligned} K(t,u_i, L_0^2(\Sigma \cap B_{2\delta }),H_0^1(\Sigma \cap B_{2\delta })) \le C\, K(t,u_i, L^2(\Sigma ),H^1(\Sigma )). \end{aligned}$$

Therefore, we deduce

$$\begin{aligned} \Vert v_i\Vert _{H_{\star }^\frac{1}{2}(\mathbb {R}^2)}\le C\Vert u_i\Vert _{H_{\star }^\frac{1}{2}(\Sigma )} \le C\Vert u\Vert _{H_{\star }^\frac{1}{2}(\Sigma )}. \end{aligned}$$

Repeating the argument for every \(i =1, \dots , N\) yields

$$\begin{aligned} \Vert u\Vert _{H^\frac{1}{2}(\Sigma )}\le C\Vert u\Vert _{H_{\star }^\frac{1}{2}(\Sigma )}. \end{aligned}$$

The opposite inequality can be proved in a similar way.

We also note that we may interpolate between \(H^{\frac{1}{2}}(\Sigma )\) and its dual and obtain by using [9, Theorem 4.1] and by a localization argument that

$$\begin{aligned} (H^{-\frac{1}{2}}(\Sigma ), H^{\frac{1}{2}}(\Sigma ))_{\frac{1}{2}, 2} = L^2(\Sigma ). \end{aligned}$$

In order to define higher order half-integer Sobolev spaces on the boundary we use the fact that in our setting the boundary \(\partial \Omega = \Sigma \) is given by the parametrization \(\Psi _{\Sigma }: \Gamma \rightarrow \Sigma \), \(\Psi _{\Sigma }(x) = x + h(x)\nu _\Gamma (x)\), where \(\Gamma \) is the reference surface. In our case \(\Gamma \) is the boundary of a smooth set G and the map \(\Psi : \Gamma \times (-\eta , \eta ) \rightarrow {\mathcal {N}}_\eta (\Gamma )\), \(\Psi (x,s) = x + s \nu _{\Gamma }(x)\), is a diffeomorphism. Here \({\mathcal {N}}_\eta (\Gamma )\) is the tubular neighborhood of \(\Gamma \). Therefore the projection map \(\pi _\Gamma : {\mathcal {N}}_\eta (\Gamma ) \rightarrow \Gamma \) is well defined as

$$\begin{aligned} \pi _\Gamma (y) = x \qquad \text {where } \, y = x + s \nu _{\Gamma }(x) \,\,\, \text {for some } \, \, s \in (- \eta , \eta ). \end{aligned}$$
(2.7)

We extend \(\pi _\Gamma \) to whole \(\mathbb {R}^3\) and thus we may extend a given function \(u: \Gamma \rightarrow \mathbb {R}\) to \(\mathbb {R}^3\) as \((u \circ \pi _\Gamma ): \mathbb {R}^3 \rightarrow \mathbb {R}\). In particular, the kth order derivative \(\nabla ^k (u \circ \pi _\Gamma )(x)\) is well defined for all \(x \in \Gamma \), and for \(x \in \Gamma \) the function \(s \mapsto (u \circ \pi _\Gamma )(x+ s\nu _{\Gamma }(x))\) is constant for |s| small. We use this extension to define the half-integer Sobolev norm on the reference surface.

Definition 2.5

For \(m \ge 2\) we say that \(u \in H^{m-\frac{1}{2}}(\Gamma )\) if \(u \in H^{m-1}(\Gamma )\) and the norm

$$\begin{aligned} \Vert u \Vert _{H^{m-\frac{1}{2}}(\Gamma )}:= \Vert \nabla ^{m-1} (u \circ \pi _\Gamma )\Vert _{H^{\frac{1}{2}}(\Gamma )}+ \Vert u\Vert _{H^{m-1}(\Gamma )} \end{aligned}$$

is bounded.

We define the half-integer Sobolev spaces on \(\Sigma \) by map** a function \(u \in C^\infty (\Sigma )\) back to \(\Gamma \) by using the parametrization \(\Psi _{\Sigma }: \Gamma \rightarrow \Sigma \), \(\Psi _{\Sigma }(x) = x + h(x)\nu _{\Gamma }(x)\). Let us fix \(m \ge 2\) and recall that

$$\begin{aligned} {\bar{\nabla }}^m (u\circ \Psi _\Sigma )= \sum _{|\alpha |\le m-1}{\bar{\nabla }}^{1+\alpha _1}\Psi _\Sigma \star \cdots \star {\bar{\nabla }}^{1+\alpha _{k}}\Psi _\Sigma \star {\bar{\nabla }}^{1+\alpha _{m+1}} u. \end{aligned}$$

If \(\Sigma \) satisfies (\(\hbox {H}_{m}\)), then arguing as in the proof of Proposition 2.1 we deduce

$$\begin{aligned} \Vert u\Vert _{H^m(\Sigma )} \simeq \Vert u\circ \Psi _\Sigma \Vert _{H^{m}(\Gamma )}. \end{aligned}$$

Based on this we define the half-integer Sobolev space of order \(m - 1/2\) on \(\Sigma \) in the following way.

Definition 2.6

Let \(m\ge 2\) be an integer and assume \(\Sigma \) is \(C^{1,\alpha }(\Gamma )\)-regular. We say that u is in the space \(H^{m-\frac{1}{2}}(\Sigma )\) if \((u\circ \Psi _\Sigma ) \in H^{m-\frac{1}{2}}(\Gamma )\) and define the norm as

$$\begin{aligned} \Vert u\Vert _{H^{m-\frac{1}{2}}(\Sigma )}:= \Vert u\circ \Psi _\Sigma \Vert _{H^{m-\frac{1}{2}}(\Gamma )}, \end{aligned}$$

where \(\Psi _{\Sigma }: \Gamma \rightarrow \Sigma \) is the parametrization \(\Psi _{\Sigma }(x) = x + h(x)\nu _{\Gamma }(x)\).

We define the space \(H^{m-\frac{1}{2}}_\star (\Sigma )\) via interpolation as the functions \(u\in H^{m-1}(\Sigma )\) such that

$$\begin{aligned} \Vert u\Vert _{H_\star ^{m-\frac{1}{2}}(\Sigma )}^2:= \Vert u\Vert _{H^{m-1}(\Sigma )}^2+ \int _{0}^\infty \left( \frac{K(t,u,H^{m-1}(\Sigma ),H^m(\Sigma ))}{t^{1/2}} \right) ^2\, \frac{dt}{t} <\infty . \end{aligned}$$
(2.8)

We note that if \(\Sigma \) satisfies the assumption (\(\hbox {H}_{m}\)) for \(m\ge 2\) the norm \(H^{m-\frac{1}{2}}(\Sigma )\) in Definition 2.6 is equivalent with the interpolation norm \(\Vert u\Vert _{H_\star ^{m-\frac{1}{2}}(\Sigma )}\) in (2.8). We state this in the next proposition. The proof is similar to the argument for (2.6) and we omit it.

Proposition 2.7

Let \(m\ge 2\) and assume that \(\Sigma \) is uniformly \(C^{1,\alpha }(\Gamma )\)-regular and satisfies the assumption (\(\hbox {H}_{m}\)). Then it holds

$$\begin{aligned} H^{m-\frac{1}{2}}(\Sigma )=H_\star ^{m-\frac{1}{2}}(\Sigma ) \qquad \text {and} \qquad \Vert u\Vert _{H^{m-\frac{1}{2}}(\Sigma )} \simeq \Vert u\Vert _{H_\star ^{m-\frac{1}{2}}(\Sigma )}. \end{aligned}$$

2.3 Geometric Preliminaries

We begin by recalling basic results from differential geometry. We define the Riemann curvature tensor \(R \in {\mathscr {T}}^4(\Sigma )\) [39, 43] via interchange of covariant derivatives of a vector field \(Y^i\) and a covector field \(Z_i\) as

$$\begin{aligned} \begin{aligned}&{\bar{\nabla }}_i {\bar{\nabla }}_j Y^s - {\bar{\nabla }}_j {\bar{\nabla }}_i Y^s = R_{ijkl} g^{ks} Y^l,\\&{\bar{\nabla }}_i {\bar{\nabla }}_j Z_k - {\bar{\nabla }}_j {\bar{\nabla }}_i Z_k = R_{ijkl} g^{ls} Z_s, \end{aligned} \end{aligned}$$
(2.9)

where we have used the Einstein summation convention. We may write the Riemann tensor in local coordinates by using the second fundamental form B as

$$\begin{aligned} R_{ijkl} = B_{ik}B_{jl} - B_{il}B_{jk}. \end{aligned}$$
(2.10)

We will also need the Simon’s identity which reads as

$$\begin{aligned} \Delta _{\Sigma } B_{ij} = {\bar{\nabla }}_i {\bar{\nabla }}_j H + H B_{il} g^{ls} B_{sj} - |B|^2 B_{ij}. \end{aligned}$$
(2.11)

Let us recall that the interpolation inequality holds for smooth compact n-dimensional hypersurface \(\Sigma \subset \mathbb {R}^{n+1}\), see e.g. [4],

$$\begin{aligned} \Vert {\bar{\nabla }}^k u \Vert _{L^p(\Sigma )} \le C_{\Sigma } \Vert u \Vert _{W^{l,r}(\Sigma )}^\theta \Vert u \Vert _{L^q(\Sigma )}^{(1-\theta )}, \end{aligned}$$
(2.12)

where

$$\begin{aligned} \frac{1}{p} = \frac{k}{n} + \theta \left( \frac{1}{r} - \frac{l}{n} \right) + \frac{1}{q}(1- \theta ). \end{aligned}$$

In particular, (2.12) holds on the reference surface \(\Gamma \subset \mathbb {R}^3\) and in \(\mathbb {R}^{n}\) for functions with compact support \(\text {supp} \, u \subset B_R\).

In order to have the interpolation inequality for a general surface \(\Sigma \subset \mathbb {R}^{n+1}\) with control on the constant \(C_{\Sigma }\), we use the result in [42], which states that once the mean curvature \(H_\Sigma \) satisfies the bound \(\Vert H_{\Sigma }\Vert _{L^{n+\delta }(\Sigma )} \le C\), then the above interpolation inequality holds on \(\Sigma \) with uniform bound on the constant. We state this for our purpose, where \(\Sigma \) is 2-dimensional surfaces that is uniformly \(C^{1,\alpha }(\Gamma )\)-regular and satisfies the bound \(\Vert B_\Sigma \Vert _{L^4(\Sigma )} \le C\). The reason for the \(L^4\)-curvature bound will be clear from the results in Sect. 6. The following interpolation inequality follows from [42, Proposition 6.5].

Proposition 2.8

Assume \(\Sigma \subset \mathbb {R}^3\) is a compact 2-dimensional hypersurface which is uniformly \(C^{1,\alpha }(\Gamma )\)-regular and satisfies the bound \(\Vert B_\Sigma \Vert _{L^4(\Sigma )} \le M\). Then for integers kl, \(0 \le k <l\) and numbers \(p, r \in [1,\infty )\) and \(q \in [1,\infty ]\) we have for all tensor fields T that

$$\begin{aligned} \Vert {\bar{\nabla }}^k T \Vert _{L^p(\Sigma )} \le C \Vert T \Vert _{W^{l,r}(\Sigma )}^\theta \Vert T \Vert _{L^q(\Sigma )}^{(1-\theta )}, \end{aligned}$$

where p and \(\theta \in [0,1]\) are given by

$$\begin{aligned} \frac{1}{p} = \frac{k}{2} + \theta \left( \frac{1}{r} - \frac{l}{2} \right) + \frac{1}{q}(1- \theta ). \end{aligned}$$

The constant C depends on Mkplrq.

In particular, we have the Sobolev embedding, i.e., for \(p \in [1,n)\) it holds \(\Vert u\Vert _{L^{p^*}(\Sigma )} \le C \Vert u\Vert _{W^{1,p}(\Sigma )}\) with \(p^* = \frac{np}{n-p}\), for \(p = n\) it holds \(\Vert u\Vert _{L^{q}(\Sigma )} \le C \Vert u\Vert _{W^{1,p}(\Sigma )}\) for all \(q <\infty \) and for \(p >n\) it holds \(\Vert u\Vert _{C^{\alpha }(\Sigma )}\le C \Vert u\Vert _{W^{1,p}(\Sigma )}\) for \(\alpha = 1- \frac{n}{p}\).

There is a danger for confusion in terminology when we use interpolation of function spaces and interpolation inequality. We use the term ’interpolation argument’, when we interpolate between two function spaces, and ’interpolation inequality’ or merely ’interpolation’ when we refer to Proposition 2.8.

Let \(\Sigma = \partial \Omega \subset \mathbb {R}^3\) be a compact hypersurface in \(\mathbb {R}^3\) such that \(\Sigma = \partial \Omega \) which is \(C^{1,\alpha }(\Gamma )\)-regular. Then the Sobolev embedding extends to half-integers, i.e., it holds

$$\begin{aligned} \Vert u\Vert _{L^p(\Sigma )}\le C \Vert u\Vert _{H^{\frac{1}{2}}(\Sigma )}, \qquad \text {for } \, p \le 4 \end{aligned}$$

and

$$\begin{aligned} \Vert u\Vert _{L^p(\Omega )}\le C \Vert u\Vert _{H^{\frac{1}{2}}(\Omega )}, \qquad \text {for } \, p \le 3. \end{aligned}$$

We need the above interpolation inequality also for half-integers and for functions defined in \(\Omega \). To this aim we need to assume that \(\Sigma \) satisfies the condition (\(\hbox {H}_{m}\)).

Corollary 2.9

Let \(m \in \mathbb {N}\) and \(\Sigma \subset \mathbb {R}^3\) is compact 2-dimensional hypersurface which is uniformly \(C^{1,\alpha }(\Gamma )\)-regular such that \(\Sigma = \partial \Omega \) and satisfies the condition (\(\hbox {H}_{m}\)). Then for all half-integers k and l with \(k< l \le m\) and for \(q \in [1,\infty ]\) it holds

$$\begin{aligned} \Vert u\Vert _{H^{k}(\Sigma )}\le C \Vert u\Vert _{H^{l}(\Sigma )}^\theta \Vert u\Vert _{L^q(\Sigma )}^{1-\theta }, \end{aligned}$$

where \(\theta \in [0,1]\) is given by

$$\begin{aligned} 1 = k - \theta (l - 1) + \frac{2}{q}(1- \theta ). \end{aligned}$$

In addition, it holds

$$\begin{aligned} \Vert u\Vert _{H^{k}(\Omega )}\le C \Vert u\Vert _{H^{l}(\Omega )}^\theta \Vert u\Vert _{L^q(\Omega )}^{1-\theta }, \end{aligned}$$

where \(\theta \in [0,1]\) is given by

$$\begin{aligned} \frac{1}{2} = \frac{k}{3} + \theta \left( \frac{1}{2} - \frac{l}{3} \right) + \frac{1}{q}(1- \theta ). \end{aligned}$$

Moreover, the inequality (2.12) holds on \(\Omega \subset \mathbb {R}^3\) with \(r=2\) and integers \(k <l \le m\). The constants depends on mq and on the \(C^{1,\alpha }\)-norm of the heightfunction.

Proof

We sketch the proof only for the first claim when \(k={\tilde{k}} -\frac{1}{2}\) for \({\tilde{k}} \in \mathbb {N}\) and l is an integer. By Proposition 2.7 and by the classical interpolation theory stated in (2.2) we have

$$\begin{aligned} \Vert u\Vert _{H^{k}(\Sigma )} \le C\Vert u\Vert _{H_\star ^{k}(\Sigma )} \le C\Vert u\Vert _{H^{{\tilde{k}}}(\Sigma )}^\frac{1}{2}\Vert u\Vert _{H^{{\tilde{k}}-1}(\Sigma )}^\frac{1}{2}. \end{aligned}$$

Proposition 2.8 yields

$$\begin{aligned} \Vert u\Vert _{H^{{\tilde{k}}}(\Sigma )} \le C \Vert u\Vert _{H^{l}(E)}^{\theta _1} \Vert u\Vert _{L^q(E)}^{1-\theta _1}, \end{aligned}$$

where \(\theta _1\) is given by \(1 = {\tilde{k}} - \theta _1 (l - 1) + \frac{2}{q}(1- \theta _1)\), and

$$\begin{aligned} \Vert u\Vert _{H^{{\tilde{k}}-1}(\Sigma )} \le C \Vert u\Vert _{H^{l}(E)}^{\theta _2} \Vert u\Vert _{L^q(E)}^{1-\theta _2}, \end{aligned}$$

where \(\theta _2\) is given by \(1 = ({\tilde{k}} -1) - \theta _2 (l - 1) + \frac{2}{q}(1- \theta _2)\). This implies the claim. The case when l is half-integer follows from the same argument. Finally the second interpolation inequality follows by extending u to whole \(\mathbb {R}^3\), where the inequality is well-known, and using Proposition 2.1. \(\square \)

2.4 Functional and Geometric Inequalities

We begin by recalling the extension of the interpolation inequality (2.12), or the Gagliardo-Nirenberg inequality, in \(\mathbb {R}^n\) for fractional Sobolev spaces [8]. We state the result in the setting that we need, where for all \(f \in C_0^\infty (B_R)\) it holds

$$\begin{aligned} \Vert f\Vert _{W^{s,p}(B_R)} \le C \Vert f\Vert _{W^{s_1,p_1}(B_R)}^\theta \Vert f\Vert _{L^{p_2}(B_R)}^{1-\theta }, \end{aligned}$$
(2.13)

for \(0 \le s \le s_1\), \(p_2 \in (1,\infty )\) and \(\theta \in (0,1)\) which satisfy

$$\begin{aligned} s = \theta s_1 \qquad \text {and}\qquad \frac{1}{p} = \frac{\theta }{p_1} + \frac{1-\theta }{p_2}. \end{aligned}$$

Next we recall the Kato-Ponce inequality, or the fractional Leibniz rule, in \(\mathbb {R}^n\) which is proven e.g. in [31]. We may define the norm \(\Vert f\Vert _{W^{k, p}(\mathbb {R}^n)}\) for half-integer \(k \ge 0\) and \(p \in (1,\infty )\) by using Bessel potentials \(\langle D \rangle ^k\) as

$$\begin{aligned} \Vert f\Vert _{W^{k, p}(\mathbb {R}^n)} = \Vert \langle D \rangle ^k f\Vert _{L^p(\mathbb {R}^n)}. \end{aligned}$$

The Kato-Ponce inequality, in the form we are interested in, states that for \(f,g \in C_0^\infty (\mathbb {R}^n)\) and for numbers \(2 \le p_1, q_2 < \infty \) and \(2\le p_2, q_1 \le \infty \) with

$$\begin{aligned} \frac{1}{p_1} + \frac{1}{q_1} = \frac{1}{p_2} + \frac{1}{q_2} = \frac{1}{2} \end{aligned}$$
(2.14)

it holds

$$\begin{aligned} \Vert fg\Vert _{H^k(\mathbb {R}^n)}\le C\Vert f\Vert _{W^{k,p_1}(\mathbb {R}^n)}\Vert g\Vert _{L^{q_1}(\mathbb {R}^n)}+C \Vert f\Vert _{L^{p_2}(\mathbb {R}^n)}\Vert g\Vert _{W^{k,q_2}(\mathbb {R}^n)}. \end{aligned}$$
(2.15)

We need the following generalization of the Kato-Ponce inequality both on the boundary \(\Sigma \) and in the domain \(\Omega \).

Proposition 2.10

Let \(m\ge 1\) be an integer and assume \(\Sigma \) is uniformly \(C^{1,\alpha }(\Gamma )\)-regular and satisfies the condition (\(\hbox {H}_{m}\)). Then for all half-integers \(k \le m\) it holds

$$\begin{aligned} \Vert fg\Vert _{H^k(\Sigma )}\le C\Vert f\Vert _{H^{k}(\Sigma )}\Vert g\Vert _{L^{\infty }(\Sigma )}+C\Vert f\Vert _{L^{\infty }(\Sigma )}\Vert g\Vert _{H^{k}(\Sigma )} \end{aligned}$$

and

$$\begin{aligned} \Vert fg\Vert _{H^k(\Omega )}\le C\Vert f\Vert _{H^{k}(\Omega )}\Vert g\Vert _{L^{\infty }(\Omega )}+C\Vert f\Vert _{L^{\infty }(\Omega )}\Vert g\Vert _{H^{k}(\Omega )}. \end{aligned}$$

Moreover, assume that \(\Vert B\Vert _{L^4} \le M\) and let \(k \in \mathbb {N}\). Then for \(p_1,p_2, q_1,q_2 \in [2,\infty ]\) with \(p_1, q_2 <\infty \) which satisfies (2.14) it holds

$$\begin{aligned} \Vert fg\Vert _{H^k(\Sigma )}\le C\Vert f\Vert _{W^{k,p_1}(\Sigma )}\Vert g\Vert _{L^{q_1}(\Sigma )}+C\Vert f\Vert _{L^{p_2}(\Sigma )}\Vert g\Vert _{W^{k,q_2}(\Sigma )}. \end{aligned}$$

The constants depend on \(M, m, k, p_1,p_2, q_1,q_2\) and on the \(C^{1,\alpha }\)-norm of the heightfunction.

Proof

The second inequality follows immediately from the property of the extension operator given by Proposition 2.1 and by the classical Kato-Ponce inequality (2.15), see e.g. [10]. Also the first inequality follows from a similar localization argument as we used in (2.6).

We prove the third inequality, since we will use the argument also later. First by Leibniz formula we may write

$$\begin{aligned} {\bar{\nabla }}^k (fg) = \sum _{i+j=k} {\bar{\nabla }}^i f \star {\bar{\nabla }}^j g. \end{aligned}$$

The claim thus follows once we prove

$$\begin{aligned} \sum _{i+j=k} \Vert {\bar{\nabla }}^i f \star {\bar{\nabla }}^j g \Vert _{L^2(\Sigma )} \le C\Vert f\Vert _{W^{k,p_1}(\Sigma )}\Vert g\Vert _{L^{q_1}(\Sigma )}+\Vert f\Vert _{L^{p_2}(\Sigma )}\Vert g\Vert _{W^{k,q_2}(\Sigma )}. \end{aligned}$$
(2.16)

To this aim we use Hölder’s inequality as

$$\begin{aligned} \sum _{i+j=k} \Vert {\bar{\nabla }}^i f \star {\bar{\nabla }}^j g \Vert _{L^2(\Sigma )} \le \sum _{i+j=k} \Vert {\bar{\nabla }}^i f\Vert _{L^4(\Sigma )} \Vert {\bar{\nabla }}^j g \Vert _{L^4(\Sigma )}. \end{aligned}$$

By interpolation inequality in Proposition 2.8 we have

$$\begin{aligned} \Vert {\bar{\nabla }}^{i} f\Vert _{L^4(\Sigma )} \le C\Vert f\Vert _{W^{k,p_1}(\Sigma )}^{\theta _i} \Vert f\Vert _{L^{p_2}(\Sigma )}^{1-\theta _i} \qquad \text { with }\qquad \theta _i= \frac{\frac{i}{2}-\frac{1}{4} + \frac{1}{p_2}}{\frac{k}{2} + \frac{1}{p_2} - \frac{1}{p_1}} \end{aligned}$$

and recalling that \(\frac{1}{p_1} + \frac{1}{q_1} = \frac{1}{p_2} + \frac{1}{q_2} = \frac{1}{2}\) we have

$$\begin{aligned} \Vert {\bar{\nabla }}^{j} g\Vert _{L^4(\Sigma )} \le C\Vert g\Vert _{W^{k,q_2}(\Sigma )}^{\theta _j} \Vert g\Vert _{L^{q_1}(\Sigma )}^{1-\theta _j} \qquad \text { with }\qquad \theta _j= \frac{\frac{j}{2}-\frac{1}{4} + \frac{1}{q_1}}{\frac{k}{2} + \frac{1}{p_2} - \frac{1}{p_1}}. \end{aligned}$$

In particular, \(i+j = k\) implies \( \theta _i + \theta _j =1\). Therefore we have by Young’s inequality \( a^{\theta _i} b^{\theta _j} \le \theta _i a+ \theta _j b \le a+b \) that

$$\begin{aligned} \begin{aligned} \sum _{i+j=k} \Vert {\bar{\nabla }}^i f \star {\bar{\nabla }}^j g \Vert _{L^2}&\le C \Vert f\Vert _{L^{p_2}}\Vert g\Vert _{L^{q_1}}\sum _{i+j=k} \Vert f\Vert _{W^{k,p_1}}^{\theta _i} \Vert f\Vert _{L^{p_2}}^{-\theta _i}\Vert g\Vert _{W^{k,q_2}}^{\theta _j} \Vert g\Vert _{L^{q_1}}^{-\theta _j}\\&\le C \Vert f\Vert _{L^{p_2}}\Vert g\Vert _{L^{q_1}} \left( \frac{\Vert f\Vert _{W^{k,p_1}}}{\Vert f\Vert _{L^{p_2}}} + \frac{\Vert g\Vert _{W^{k,q_2}}}{\Vert g\Vert _{L^{q_1}}} \right) \end{aligned} \end{aligned}$$

and the claim follows. \(\square \)

We remark that we do not generalize the last inequality in Proposition 2.10 for half-integers k since we do not define the space \(W^{k,p}(\Sigma )\) for \(p \ne 2\), when k is not an integer. However, under the assumption of Proposition 2.10, we obtain a weaker version which reads as follows

$$\begin{aligned} \Vert fg\Vert _{H^{\frac{1}{2}}(\Sigma )}\le C \Vert f\Vert _{H^{\frac{1}{2}}(\Sigma )}\Vert g\Vert _{L^{\infty }(\Sigma )}+\Vert f\Vert _{L^{p}(\Sigma )}\Vert g\Vert _{W^{1,q}(\Sigma )}, \end{aligned}$$
(2.17)

for \(\frac{1}{p} + \frac{1}{q} = \frac{1}{2}\). Again, since the proof is similar to the argument we used in (2.6) we leave it for the reader, but refer to [15, Lemma 4.3] for the proof of the case \(p=q=4\). In particular, when g is Lipschitz, we may estimate the product simply by

$$\begin{aligned} \Vert fg\Vert _{H^{\frac{1}{2}}(\Sigma )} \le C\Vert f\Vert _{H^{\frac{1}{2}}(\Sigma )}\Vert g\Vert _{C^{1}(\Sigma )}. \end{aligned}$$

Next we recall (see e.g. [22]) that it holds \(\Vert u\Vert _{H^{k+2}(\Sigma )} \le C_{\Sigma }(\Vert \Delta _\Sigma u\Vert _{H^{k}(\Sigma )} + \Vert u \Vert _{L^2(\Sigma )})\). However, the constant depends on the curvature of \(\Sigma \) and we need to quantify this dependence.

Proposition 2.11

Assume that \(\Sigma \) is uniformly \(C^{1,\alpha }(\Gamma )\)-regular and satisfies \(\Vert B_\Sigma \Vert _{L^4} \le M\). For all \(\varepsilon >0\) there exists a constant \(C_\varepsilon \) such that for \(k = 0, \frac{1}{2}, 1\) it holds

$$\begin{aligned} \Vert u\Vert _{H^{k+2}(\Sigma )} \le (1+\varepsilon )\Vert \Delta _\Sigma u\Vert _{H^{k}(\Sigma )} + C_{\varepsilon } \Vert u \Vert _{L^2(\Sigma )}. \end{aligned}$$

Let m be an integer with \(m \ge 3\) and assume that \(\Sigma \) satisfies in addition the condition (\(\hbox {H}_{m}\)). Then for every half-integer \(2 \le k \le m\) it holds

$$\begin{aligned} \Vert u\Vert _{H^{k}(\Sigma )} \le (1+\varepsilon )\Vert \Delta _\Sigma u\Vert _{H^{k-2}(\Sigma )} + C_{\varepsilon } \Vert u \Vert _{L^2(\Sigma )}. \end{aligned}$$

The constant \(C_\varepsilon \) depends on \(\varepsilon , M, m\) and on the \(C^{1,\alpha }\)-norm of the heightfunction.

Proof

The case \(k = 0\) follows from [15, Lemma 4.11] but we give the proof for the reader’s convenience. We recall that the Riemann tensor R satisfies by (2.10) \(|R|\le C|B|^2\) and deduce by [22, Remark 2.4] (see also [4]) that

$$\begin{aligned} \Vert {\bar{\nabla }}^2 u\Vert _{L^{2}(\Sigma )}^2 \le \Vert \Delta _\Sigma u\Vert _{L^{2}(\Sigma )}^2 + C \int _\Sigma |B|^2 |{\bar{\nabla }} u|^2\, d \mathcal {H}^2. \end{aligned}$$

By Proposition 2.8 there is \(\theta \in (0,1)\) such that

$$\begin{aligned} \int _\Sigma |B|^2 |{\bar{\nabla }} u|^2\, d \mathcal {H}^2 \le \Vert B\Vert _{L^4}^2 \Vert {\bar{\nabla }} u\Vert _{L^{4}}^2 \le C \Vert u\Vert _{H^{2}}^{2\theta } \Vert u\Vert _{L^{2}}^{2(1-\theta )} \le \varepsilon \Vert u\Vert _{H^{2}} + C_{\varepsilon }\Vert u\Vert _{L^{2}}^2. \end{aligned}$$

This implies the claim for \(k =0\).

For the case \(k=1\) we use (2.9) and integration by parts

$$\begin{aligned} \begin{aligned} \Vert {\bar{\nabla }} \Delta _\Sigma u\Vert _{L^{2}(\Sigma )}^2&= \int _{\Sigma } {\bar{\nabla }}^k {\bar{\nabla }}_i{\bar{\nabla }}^i u \, {\bar{\nabla }}_k {\bar{\nabla }}^j{\bar{\nabla }}_j u \, d \mathcal {H}^2 \\&= \int _{\Sigma } {\bar{\nabla }}_i {\bar{\nabla }}^k {\bar{\nabla }}^i u \, {\bar{\nabla }}_k {\bar{\nabla }}^j{\bar{\nabla }}_j u \, d \mathcal {H}^2 + \int _{\Sigma } (R \star {\bar{\nabla }} u \star {\bar{\nabla }}^3 u) \, d \mathcal {H}^2\\&\ge -\int _{\Sigma } {\bar{\nabla }}^k {\bar{\nabla }}^i u \, {\bar{\nabla }}_i {\bar{\nabla }}_k {\bar{\nabla }}^j{\bar{\nabla }}_j u \, d \mathcal {H}^2 - \Vert R \star {\bar{\nabla }} u\Vert _{L^2}\, \Vert {\bar{\nabla }}^3 u\Vert _{L^2}. \end{aligned} \end{aligned}$$

As before we have by Proposition 2.8 and by \(|R|\le C|B|^2\) that

$$\begin{aligned} \Vert R \star {\bar{\nabla }} u\Vert _{L^2} \le C\Vert B\Vert _{L^4}^2\Vert {\bar{\nabla }} u\Vert _{L^\infty } \le \varepsilon \Vert u\Vert _{H^3} + C_{\varepsilon } \Vert u\Vert _{L^2}. \end{aligned}$$
(2.18)

We proceed by using (2.9) and by integrating by parts

$$\begin{aligned} \begin{aligned} -\int _{\Sigma }&{\bar{\nabla }}^k {\bar{\nabla }}^i u \, {\bar{\nabla }}_i {\bar{\nabla }}_k {\bar{\nabla }}^j{\bar{\nabla }}_j u \, d \mathcal {H}^2 \\&\ge -\int _{\Sigma } {\bar{\nabla }}^k {\bar{\nabla }}^i u \, {\bar{\nabla }}_k {\bar{\nabla }}_i {\bar{\nabla }}^j{\bar{\nabla }}_j u \, d \mathcal {H}^2 + \int _{\Sigma } {\bar{\nabla }}^2 u \star R \star {\bar{\nabla }}^2 u\, d \mathcal {H}^2 \\&\ge -\int _{\Sigma } {\bar{\nabla }}^k {\bar{\nabla }}^i u \, {\bar{\nabla }}_k {\bar{\nabla }}^j {\bar{\nabla }}_i {\bar{\nabla }}_j u \, d \mathcal {H}^2 - \int _{\Sigma } {\bar{\nabla }}^k {\bar{\nabla }}^i u \, {\bar{\nabla }}_k \bar{[} {\bar{\nabla }}_i\nabla ^j - {\bar{\nabla }}_j\nabla ^i] {\bar{\nabla }}_j u \, d \mathcal {H}^2\\&\,\,\,\,\,\,\,\,- C\Vert B\Vert _{L^4}^2 \Vert {\bar{\nabla }}^2 u\Vert _{L^4}^2\\&\ge \int _{\Sigma } {\bar{\nabla }}^j {\bar{\nabla }}^k {\bar{\nabla }}^i u \, {\bar{\nabla }}_k {\bar{\nabla }}_i {\bar{\nabla }}_j u \, d \mathcal {H}^2 + \int _{\Sigma } {\bar{\nabla }}_k {\bar{\nabla }}^k {\bar{\nabla }}^i u \, \bar{[} {\bar{\nabla }}_i\nabla ^j - {\bar{\nabla }}_j\nabla ^i] {\bar{\nabla }}_j u \, d \mathcal {H}^2\\&\,\,\,\,\,\,\,\,- C\Vert B\Vert _{L^4}^2 \Vert {\bar{\nabla }}^2 u\Vert _{L^4}^2\\&\ge \Vert {\bar{\nabla }}^3 u\Vert _{L^2}^2 - C\Vert B\Vert _{L^4}^2 \Vert {\bar{\nabla }}^2 u\Vert _{L^4}^2 - \Vert R \star {\bar{\nabla }} u\Vert _{L^2}\, \Vert {\bar{\nabla }}^3 u\Vert _{L^2}. \end{aligned} \end{aligned}$$

The inequality for \(k =1\) then follows from (2.18) and from Proposition 2.8 which yields

$$\begin{aligned} \Vert {\bar{\nabla }}^2 u\Vert _{L^4} \le \varepsilon \Vert u\Vert _{H^3} + C_\varepsilon \Vert u\Vert _{L^2}. \end{aligned}$$

The case \(k=1/2\) follows from the previous two estimates and Proposition 2.7 with standard interpolation argument which we briefly sketch here for the reader’s convenience. We define a linear operator \({\mathcal {F}}: {\tilde{H}}^{k}(\Sigma ) \rightarrow {\tilde{H}}^{k+2}(\Sigma )\) such that \({\mathcal {F}}(g) = u\), where u is the solution of

$$\begin{aligned} \Delta _\Sigma u = g \qquad \text {on } \, \Sigma , \end{aligned}$$

and \({\tilde{H}}^k(\Sigma ) = \{ f \in H^k(\Sigma ): \int _{\Sigma } f \, d \mathcal {H}^2 = 0\}\). The operator \({\mathcal {F}}\) is well-defined and by the previous estimates it satisfies

$$\begin{aligned} \Vert {\mathcal {F}}\Vert _\mathcal {L(L^2,H^2)} \le C \qquad \text {and} \qquad \Vert {\mathcal {F}}\Vert _\mathcal {L(H^1,H^3)} \le C \end{aligned}$$

By the interpolation theory discussed in (2.2) it holds

$$\begin{aligned} \Vert {\mathcal {F}}(g)\Vert _{H_\star ^{\frac{5}{2}}(\Sigma )} \le C \Vert g\Vert _{H_\star ^{\frac{1}{2}}(\Sigma )}. \end{aligned}$$

Proposition 2.7 then yields

$$\begin{aligned} \Vert {\mathcal {F}}(g)\Vert _{H^{\frac{5}{2}}(\Sigma )} \le C \Vert g\Vert _{H^{\frac{1}{2}}(\Sigma )}. \end{aligned}$$

We apply this to \({\tilde{u}} = u - {\bar{u}}\), where and the claim follows.

The argument for higher m and k is similar and we merely sketch it. Let k be an integer with \(2 \le k \le m\). Using (2.9) and arguing as above we obtain after long but straightforward calculations that

$$\begin{aligned} \Vert {\bar{\nabla }}^{k} u\Vert _{L^{2}(\Sigma )}^2 \le \Vert {\bar{\nabla }}^{k-2} \Delta _\Sigma u\Vert _{L^{2}(\Sigma )}^2 + C \sum _{\alpha + \beta \le k-2}\Vert {\bar{\nabla }}^\alpha R \star {\bar{\nabla }}^{1+\beta } u \Vert _{L^2(\Sigma )}^2. \end{aligned}$$

Then by (2.10), (2.16), Proposition 2.8 and by the assumption \(\Vert B\Vert _{L^\infty }, \Vert B\Vert _{H^{k-2}}\le C\) we have

$$\begin{aligned} \begin{aligned} \sum _{\alpha + \beta \le k-2}\Vert {\bar{\nabla }}^\alpha (R \star {\bar{\nabla }}^{1+\beta } u) \Vert _{L^2(\Sigma )}&\le C \Vert B\Vert _{L^\infty }^2 \Vert u\Vert _{H^{k-1}} + C \Vert B\Vert _{L^\infty } \Vert B\Vert _{H^{k-2}} \Vert {\bar{\nabla }} u\Vert _{L^\infty }\\&\le \varepsilon \Vert u\Vert _{H^k(\Sigma )} + C_{\varepsilon } \Vert u\Vert _{L^2(\Sigma )}. \end{aligned} \end{aligned}$$

This yields the claim for integers \(2 \le k \le l\).

If \(k \le m-\frac{1}{2}\) is an half-integer but not an integer, then we may use the previous argument for integer \(l = k + \frac{1}{2} \le m\) and deduce

$$\begin{aligned} \Vert u\Vert _{H^{l}} \le (1+\varepsilon )\Vert \Delta _\Sigma u\Vert _{H^{l-2}(\Sigma )} + C_{\varepsilon } \Vert u \Vert _{L^2(\Sigma )}. \end{aligned}$$

The same holds for \(l-1\). Hence, the claim follows by Proposition 2.7 and by the same interpolation argument we used above. \(\square \)

By using the previous proposition and the Simon’s identity (2.11) we deduce that we may bound the second fundamental form by the mean curvature.

Proposition 2.12

Assume that \(\Sigma \) is uniformly \(C^{1, \alpha }(\Gamma )\)-regular. Then for every \(p \in (1,\infty )\) it holds

$$\begin{aligned} \Vert B_\Sigma \Vert _{L^p(\Sigma )} \le C(1+ \Vert H_\Sigma \Vert _{L^p(\Sigma )}). \end{aligned}$$

If in addition \(\Vert B_\Sigma \Vert _{L^4(\Sigma )} \le M\), then for \(k =\frac{1}{2}, 1, 2\) it holds

$$\begin{aligned} \Vert B_\Sigma \Vert _{H^{k}(\Sigma )} \le C(1+ \Vert H_\Sigma \Vert _{H^{k}(\Sigma )}). \end{aligned}$$

Finally let \(m\ge 3\) be an integer and assume that \(\Sigma \) satisfies in addition the condition (\(\hbox {H}_{m}\)) for m. Then the above estimate holds for all half-integers \(k\le m\). The constants depend on Mpm and on the \(C^{1,\alpha }\)-norm of the heightfunction.

Proof

The first claim follows from standard Calderon-Zygmund estimate [25] and we omit it. Let us proof the second claim for \(k=\frac{1}{2}\). We recall the geometric fact

$$\begin{aligned} \Delta _\Sigma x_i = - H_\Sigma \nu _i, \end{aligned}$$

where \(x_i = x \cdot e_i\) and \(\nu _i = \nu _\Sigma \cdot e_i\). Then we have by Proposition 2.11 and (2.17)

$$\begin{aligned} \begin{aligned} \Vert B_\Sigma \Vert _{H^{\frac{1}{2}}(\Sigma )}&\le C\sum _{i=1}^3(1+ \Vert \nabla ^2_\Sigma x_i \Vert _{H^{\frac{1}{2}}(\Sigma )}) \le \sum _{i=1}^3 C(1+ \Vert \Delta _\Sigma x_i \Vert _{H^{\frac{1}{2}}(\Sigma )}) \\&= \sum _{i=1}^3 C(1+ \Vert H_\Sigma \nu _i\Vert _{H^{\frac{1}{2}}(\Sigma )}) \\&\le C(1+ \Vert H_\Sigma \Vert _{H^{\frac{1}{2}}(\Sigma )} + \Vert H_\Sigma \Vert _{L^{4}(\Sigma )} \Vert \nu _\Sigma \Vert _{W^{1,4}(\Sigma )}) \le C(1+ \Vert H_\Sigma \Vert _{H^{\frac{1}{2}}(\Sigma )}). \end{aligned} \end{aligned}$$

The argument for \(k=1\) is similar.

In the case \(k =2\) we use the Simon’s identity (2.11) to deduce

$$\begin{aligned} \Vert \Delta _\Sigma B\Vert _{L^2(\Sigma )}^2 \le \Vert {\bar{\nabla }}^{2} H\Vert _{L^2(\Sigma )}^2 + C \Vert B\Vert _{L^6(\Sigma )}^6. \end{aligned}$$

Proposition 2.11 yields \(\Vert B\Vert _{H^2(\Sigma )} \le 2\Vert \Delta _\Sigma B\Vert _{L^{2}(\Sigma )} +C\). The claim then follows from interpolation inequality (Proposition 2.8)

$$\begin{aligned} \Vert B\Vert _{L^6(\Sigma )}^6 \le \Vert B\Vert _{H^2(\Sigma )}^{\frac{2}{3}} \Vert B\Vert _{L^4(\Sigma )}^{\frac{16}{3}} \le \varepsilon \Vert B\Vert _{H^2(\Sigma )} + C_\varepsilon . \end{aligned}$$

Let us then fix \(m\ge 3\), assume that \(\Sigma \) satisfies the condition (\(\hbox {H}_{m}\)) for m and let \(k \le m\). We use the Simon’s identity (2.11) and Proposition 2.10 to deduce

$$\begin{aligned} \begin{aligned} \Vert \Delta _\Sigma B\Vert _{H^{k-2}(\Sigma )}&\le \Vert H\Vert _{H^k(\Sigma )} + C \Vert B\star B \star B\Vert _{H^{k-2}(\Sigma )}\\&\le \Vert H\Vert _{H^k(\Sigma )} + C \Vert B\Vert _{L^\infty (\Sigma )}^2 \Vert B\Vert _{H^{k-2}(\Sigma )}\\&\le \Vert H\Vert _{H^k(\Sigma )} + \varepsilon \Vert B\Vert _{H^{k}(\Sigma )} + C_\varepsilon , \end{aligned} \end{aligned}$$

where the last inequality follows from \(\Vert B\Vert _{L^\infty }\le C\) and from interpolation. The claim then follows from Proposition 2.11. \(\square \)

Note that by the definition of the space \(\Vert \cdot \Vert _{H^{k}(\Sigma )}\) in Definition 2.6 it is not yet clear if it holds

$$\begin{aligned} \Vert \nabla _\tau u\Vert _{H^{k-1}(\Sigma )} \le C \Vert u\Vert _{H^{k}(\Sigma )} \end{aligned}$$

when k is not an integer. We conclude this by section by proving this in the following technical lemma.

Lemma 2.13

Let m be an integer with \(m \ge 3\) and assume that \(\Sigma \) is uniformly \(C^{1, \alpha }(\Gamma )\)-regular and satisfies the condition (\(\hbox {H}_{m}\)). Then it holds

$$\begin{aligned} \Vert \nabla _\tau u \Vert _{H^{m-\frac{3}{2}}(\Sigma )} \le C \Vert u \Vert _{H^{m-\frac{1}{2}}(\Sigma )}. \end{aligned}$$

Proof

Let us denote \({\tilde{u}} = u \circ \Psi : \Gamma \rightarrow \mathbb {R}\) and in order to simplify the notation denote the extension given by the projection in (2.7) \(({\tilde{u}} \circ \pi _\Gamma )\) simply by \({\tilde{u}}\). We observe that there is a matrix field \(A(x) = A(x,h, {\bar{\nabla }} h)\) such that

$$\begin{aligned} (\nabla _\tau u \circ \Psi )(x) = A(x) \nabla {\tilde{u}}(x) \qquad \text {for } \, x \in \Gamma . \end{aligned}$$

Therefore we have by Definition 2.6 and by Proposition 2.10

$$\begin{aligned} \begin{aligned} \Vert \nabla _\tau u \Vert _{H^{m-\frac{3}{2}}(\Sigma )}&= \Vert \nabla _\tau u \circ \Psi \Vert _{H^{m-\frac{3}{2}}(\Gamma )} = \Vert A \, \nabla {\tilde{u}} \Vert _{H^{m-\frac{3}{2}}(\Gamma )} \\&\le C\Vert A\Vert _{L^\infty } \, \Vert \nabla {\tilde{u}} \Vert _{H^{m-\frac{3}{2}}(\Gamma )} + C \Vert A\Vert _{H^{m-\frac{3}{2}}(\Gamma )} \, \Vert \nabla {\tilde{u}} \Vert _{L^\infty }. \end{aligned} \end{aligned}$$

The assumption \(\Vert B\Vert _{L^\infty (\Sigma )},\Vert B\Vert _{H^{m-2}(\Sigma )}\le C\) implies for the height function \(\Vert h\Vert _{C^2(\Gamma )}\le C\) and \( \Vert h\Vert _{H^m(\Gamma )} \le C\) and therefore \(\Vert A\Vert _{L^{\infty }(\Gamma )}, \Vert A\Vert _{H^{m-1}(\Gamma )} \le C\). Moreover, since \(m\ge 3\) the Sobolev embedding yields \(\Vert \nabla {\tilde{u}}\Vert _{L^\infty (\Gamma )} \le C\Vert \nabla {\tilde{u}}\Vert _{H^{m-\frac{3}{2}}(\Gamma )}\). Therefore we have

$$\begin{aligned} \Vert \nabla _\tau u \Vert _{H^{m-\frac{3}{2}}(\Sigma )} \le C \Vert \nabla {\tilde{u}} \Vert _{H^{m-\frac{3}{2}}(\Gamma )} \le C\Vert {\tilde{u}} \Vert _{H^{m-\frac{1}{2}}(\Gamma )} = C \Vert u \Vert _{H^{m-\frac{1}{2}}(\Sigma )}. \end{aligned}$$

\(\square \)

3 Elliptic Estimates for Vector Fields and Functions

In this section we recall some known and provide some new div-curl type estimates for vector fields in the domain, i.e., \(F: \Omega \rightarrow \mathbb {R}^3\). We will need estimates where we control the norm \(\Vert F\Vert _{H^k(\Omega )}\) by the \(\textrm{div}\,F\), \(\textrm{curl}\,F\) in \(\Omega \) and with \(F_n\) on the boundary \(\Sigma \). The main result of the section is Theorem 3.6 where we prove this estimate for \(k=1\) and require the boundary merely to satisfy \(\Vert B_\Sigma \Vert _{L^4} \le C\). We do not expect the \(L^4\)-integrability to be the optimal condition. However, related to this we note that we may construct a cone \(\Omega \subset \mathbb {R}^3\) and a harmonic function \(u: \Omega \rightarrow \mathbb {R}\) with zero Neumann boundary data \(\partial _\nu u = 0\) arguing as in [21, Section 3], such that u can be written in spherical coordinates as \(u(\rho ,\theta ) = \sqrt{\rho } f(\theta )\) for a smooth functions f. In particular, \( u \notin H^2(\Omega \cap B_R)\) and therefore we may deduce that a necessary condition for the curvature is at least \(\Vert B_\Sigma \Vert _{L^2} \le C\) for Lemma 3.5 and Theorem 3.6 to hold.

We will also prove boundary regularity estimates for harmonic functions in Theorem 3.9, which quantify the boundary regularity of the harmonic functions with respect to the regularity of the boundary. We note that in Theorem 3.9 it is crucial to assume that the boundary is uniformly \(C^{1,\alpha }(\Gamma )\)-regular. Indeed, the statement does not hold for Lipschitz domains.

3.1 Regularity Estimates for Vector Fields

We begin this section by recalling the following result which is essentially from [10] (see also [53]). Recall that we define

$$\begin{aligned} \textrm{curl}\,F = \nabla F - (\nabla F)^T. \end{aligned}$$

Throughout the section we assume that \(\Omega \) is connected, but its boundary \(\Sigma = \partial \Omega \) may have many components.

Theorem 3.1

Let \(l \ge 2\) be an integer and let \(\Omega \) be a domain such that \(\Sigma = \partial \Omega \) is uniformly \(C^{1,\alpha }(\Gamma )\)-regular and \(\Vert B_{\Sigma }\Vert _{H^{\frac{3}{2}l-1}(\Sigma )}\le M\). Then there exists a constant C, which depends on Ml and on the \(C^{1,\alpha }\)-norm of the heightfunction, such that for all smooth vector fields \(F:\Omega \rightarrow \mathbb {R}^3\) and every half-integers \(1\le k \le \frac{3}{2} l\) it holds

$$\begin{aligned} \Vert F\Vert _{H^{k}(\Omega )} \le C( \Vert F_n\Vert _{H^{k-\frac{1}{2}}(\Sigma )}+ \Vert F\Vert _{L^2( \Omega )} +\Vert \textrm{div}\,F\Vert _{H^{k-1}( \Omega )}+\Vert \textrm{curl}\,F\Vert _{H^{k-1}} ). \end{aligned}$$

Moreover, for \(k = \lfloor \frac{3}{2}(l+1)\rfloor \) it holds

$$\begin{aligned} \Vert F\Vert _{H^{k}( \Omega )} \le C( \Vert \nabla _\tau F_n\Vert _{H^{k-\frac{3}{2}}(\Sigma )}+ (1+\Vert B_\Sigma \Vert _{H^{\frac{3}{2} l}}) \Vert F\Vert _{L^\infty } +\Vert \textrm{div}\,F\Vert _{H^{k-1}( \Omega )}+\Vert \textrm{curl}\,F\Vert _{H^{k-1}} ). \end{aligned}$$

Proof

We first note that the assumption \(\Vert B\Vert _{H^{\frac{3}{2}l-1}(\Sigma )}\le M\) implies that \(\Sigma \) satisfies the condition (\(\hbox {H}_{m}\)) for \(m =\lfloor \frac{3}{2} l + 1\rfloor \ge 4\). We use [10, Theorem 1.3] to deduce

$$\begin{aligned} \Vert F\Vert _{H^{k}(\Omega )} \le C( \Vert \nabla _\tau F \cdot \nu \Vert _{H^{k-\frac{3}{2}}(\Sigma )}+ \Vert F\Vert _{L^2(\Omega )} +\Vert \textrm{div}\,F\Vert _{H^{k-1}(\Omega )}+\Vert \textrm{curl}\,F\Vert _{H^{k-1}(\Omega )} ) \end{aligned}$$

for all \(k \le \lfloor \frac{3}{2}(l+1)\rfloor \). We write \(\nabla _\tau F \cdot \nu = \nabla _\tau F_n + F \star B\) and use Proposition 2.10 to obtain

$$\begin{aligned} \Vert \nabla _\tau F \cdot \nu \Vert _{H^{k-\frac{3}{2}}(\Sigma )} \le \Vert \nabla _\tau F_n \Vert _{H^{k-\frac{3}{2}}(\Sigma )} + C\Vert F\Vert _{L^\infty } \Vert B\Vert _{H^{k-\frac{3}{2}}(\Sigma )} + C\Vert B\Vert _{L^\infty } \Vert F\Vert _{H^{k-\frac{3}{2}}(\Sigma )}. \end{aligned}$$

Interpolation inequality yields

$$\begin{aligned} \Vert F\Vert _{H^{k-\frac{3}{2}}(\Sigma )} \le \Vert F\Vert _{H^{k-1}(\Omega )} \le \varepsilon \Vert F\Vert _{H^{k}(\Omega )} + C_{\varepsilon }\Vert F\Vert _{L^{\infty }(\Omega )}. \end{aligned}$$

Thus we have the second inequality. The first one follows from the fact that for \(k \le \frac{3}{2} l\) it holds \(\Vert B\Vert _{H^{k-\frac{3}{2}}(\Sigma )} \le M\) by the assumption. \(\square \)

We combine Proposition 2.11 and Theorem 3.1 and obtain the following inequality which is suitable to our purpose.

Proposition 3.2

Let l and \(\Omega \) be as in Theorem 3.1. Then for all smooth vector fields \(F:\Omega \rightarrow \mathbb {R}^3\) and every half-integer \(\frac{5}{2} \le k \le \frac{3}{2} l\) it holds

$$\begin{aligned} \Vert F\Vert _{H^{k}(\Omega )} \le C( \Vert \Delta _\Sigma F_n\Vert _{H^{k-\frac{5}{2}}(\Sigma )}+ \Vert F\Vert _{L^2(\Omega )} +\Vert \textrm{div}\,F\Vert _{H^{k-1}(\Omega )}+\Vert \textrm{curl}\,F\Vert _{H^{k-1}(\Omega )} ). \end{aligned}$$

Moreover, for \(k = \lfloor \frac{3}{2}(l+1)\rfloor \) it holds

$$\begin{aligned} \Vert F\Vert _{H^{k}(\Omega )} \le C( \Vert \Delta _\Sigma F_n\Vert _{H^{k-\frac{5}{2}}(\Sigma )}+ (1+\Vert B\Vert _{H^{\frac{3}{2} l}}) \Vert F\Vert _{L^\infty } +\Vert \textrm{div}\,F\Vert _{H^{k-1}(\Omega )}+\Vert \textrm{curl}\,F\Vert _{H^{k-1}(\Omega )} ). \end{aligned}$$

Proof

Recall that \(\Vert B\Vert _{H^{\frac{3}{2} l-1}(\Sigma )}\le M\) implies that \(\Sigma \) satisfies the condition (\(\hbox {H}_{m}\)) for \(m =\lfloor \frac{3}{2} l + 1\rfloor \ge 4\). The first inequality then follows from Theorem 3.1 and Proposition 2.11.

Let us then prove the last inequality. We have by Proposition 2.11 that

$$\begin{aligned} \Vert \nabla _\tau F_n\Vert _{H^{k-\frac{3}{2}}(\Sigma )} \le C(\Vert \Delta _{\Sigma } \nabla _\tau F_n\Vert _{H^{k-\frac{7}{2}}(\Sigma )} + \Vert F_n\Vert _{L^2(\Sigma )}). \end{aligned}$$

We use the commutation formula (2.9) for the tangential gradient of \(u: \Sigma \rightarrow \mathbb {R}\) and obtain

$$\begin{aligned} \Delta _{\Sigma } (\nabla _\tau u) = \nabla _\tau (\Delta _\Sigma u) + (B \star B) \star \nabla _\tau u. \end{aligned}$$

Therefore we have by Lemma 2.13

$$\begin{aligned} \begin{aligned} \Vert \Delta _{\Sigma } \nabla _\tau F_n\Vert _{H^{k-\frac{7}{2}}(\Sigma )}&\le C\left( \Vert \nabla _\tau (\Delta _\Sigma F_n) \Vert _{H^{k-\frac{7}{2}}(\Sigma )} + \Vert (B \star B) \star \nabla _\tau F_n \Vert _{H^{k-\frac{7}{2}}(\Sigma )} \right) \\&\le C ( \Vert \Delta _\Sigma F_n \Vert _{H^{k-\frac{5}{2}}(\Sigma )} + \Vert (B \star B) \star \nabla _\tau F_n \Vert _{H^{k-\frac{7}{2}}(\Sigma )} ). \end{aligned} \end{aligned}$$

Recall that \(k = \lfloor \frac{3}{2}(l+1)\rfloor \). We have by Proposition 2.10, by Lemma 2.13 and by the assumption \(\Vert B\Vert _{H^{\frac{3}{2} l -1}(\Sigma )} \le C\) that

$$\begin{aligned} \begin{aligned} \Vert (B \star B) \star \nabla _\tau F_n \Vert _{H^{k-\frac{7}{2}}(\Sigma )}&\le C\left( \Vert B\Vert _{L^\infty }^2 \Vert \nabla _\tau F_n \Vert _{H^{k-\frac{7}{2}}} + \Vert B\Vert _{L^\infty } \Vert B\Vert _{H^{k-\frac{7}{2}}} \Vert \nabla _\tau F_n\Vert _{L^\infty } \right) \\&\le C (\Vert F_n \Vert _{H^{k-\frac{5}{2}}(\Sigma )} + \Vert \nabla _\tau F_n\Vert _{L^\infty (\Sigma )} ). \end{aligned} \end{aligned}$$

Finally we have by the Sobolev embedding and by Corollary 2.9

$$\begin{aligned} \Vert \nabla _\tau F_n\Vert _{L^\infty (\Sigma )} +\Vert F_n \Vert _{H^{k-\frac{5}{2}}(\Sigma )} \le \varepsilon \Vert \nabla _\tau F_n\Vert _{H^{k-\frac{3}{2}}(\Sigma )} + \varepsilon \Vert F\Vert _{H^k(\Omega )}+ C_{\varepsilon } \Vert F\Vert _{L^\infty }. \end{aligned}$$

The second inequality then follows from Theorem 3.1 and combining the above inequalities. \(\square \)

Proposition 3.2 provides the inequality we need when we have the bound \(\Vert B_\Sigma \Vert _{H^{\frac{3}{2}l -1}(\Sigma )} \le C\) for \(l \ge 2\). When \(l=1\) the above bound reduces to \(\Vert B_\Sigma \Vert _{H^{\frac{1}{2}}(\Sigma )}\), which is the bound that we are able to prove in Sect. 6, but is not enough to apply the results from [10, 53]. Note that by the Sobolev embedding this implies \(\Vert B_{\Sigma }\Vert _{L^{4}(\Sigma )} \le C\). We need to work more in order to prove the first inequality in Theorem 3.1 under the assumption \(\Vert B_{\Sigma }\Vert _{L^{4}(\Sigma )} \le C\).

We begin by recalling the following Reilly’s type identity for vector fields. First, if \(\psi :\Omega \rightarrow \mathbb {R}^3\) is a smooth divergence free vector field such that \(\psi \cdot \nu = 0\) on \(\Sigma \) then it holds

$$\begin{aligned} \Vert \nabla \psi \Vert _{L^2(\Omega )}^2 = \frac{1}{2}\Vert \textrm{curl}\,\psi \Vert _{L^2(\Omega )}^2 - \int _{\Sigma } \langle B_\Sigma \, \psi , \psi \rangle \, d \mathcal {H}^2. \end{aligned}$$
(3.1)

Second, if \(u: \Omega \rightarrow \mathbb {R}\) is a smooth function then it holds

$$\begin{aligned} \begin{aligned} \Vert \nabla ^2 u\Vert _{L^2(\Omega )}^2 =&\Vert \Delta u\Vert _{L^2(\Omega )}^2 -2\int _{\Sigma } \Delta _\Sigma u \, \partial _\nu u \, d \mathcal {H}^2 \\&- \int _{\Sigma } \langle B_\Sigma {\bar{\nabla }} u, {\bar{\nabla }} u \rangle \, d \mathcal {H}^2 - \int _{\Sigma } H_\Sigma (\partial _\nu u)^2 \, d \mathcal {H}^2. \end{aligned} \end{aligned}$$
(3.2)

We give the calculations for (3.1) and (3.2) for the reader’s convenience. First, for a generic smooth vector field \(F: \Omega \rightarrow \mathbb {R}^3\) it holds

$$\begin{aligned} \begin{aligned} \int _{\Omega } |\textrm{div}\,F|^2 + \frac{1}{2} |\text {curl}\, F|^2\,dx&= \sum _{i,j=1}^3\int _{\Omega } (\partial _i F_j)^2\,dx + \sum _{i,j=1}^3\int _{\Omega } (\partial _i F_i\partial _j F_j - \partial _i F_j \partial _j F_i) \,dx\\&= \Vert \nabla F\Vert _{L^2(\Omega )}^2 + \sum _{i,j=1}^3\int _{\Omega } (\partial _i F_i\partial _j F_j - \partial _i F_j \partial _j F_i) \,dx. \end{aligned} \end{aligned}$$

By using divergence theorem twice we obtain

$$\begin{aligned} \begin{aligned} \int _{\Omega } \partial _i F_i\partial _j F_j\, dx&= - \int _{\Omega } F_i \partial _i \partial _j F_j\, dx + \int _{\Sigma } \partial _j F_j \, F_i \nu _i,\ d \mathcal {H}^2\\&= \int _{\Omega } \partial _j F_i \partial _i F_j\, dx + \int _{\Sigma } \partial _j F_j \, F_i \nu _i,\ d \mathcal {H}^2 - \int _{\Sigma } F_i \partial _i F_j \nu _j \ d \mathcal {H}^2. \end{aligned} \end{aligned}$$

Combining the two above equalities yield

$$\begin{aligned} \begin{aligned} \Vert \nabla F\Vert _{L^2(\Omega )}^2&= \Vert \textrm{div}\,F\Vert _{L^2(\Omega )}^2 + \frac{1}{2}\Vert \text {curl}\, F\Vert _{L^2(\Omega )}^2\\&+ \int _{\Sigma } (\nabla F \, F) \cdot \nu \ d \mathcal {H}^2 - \int _{\Sigma } \textrm{div}\,F \, F_n \, d \mathcal {H}^2 . \end{aligned} \end{aligned}$$
(3.3)

Assume now that \(\psi \) is a divergence free vector field such that \(\psi \cdot \nu = 0\) on \(\Sigma \). Since \(\psi \) is a tangent field on \(\Sigma \) we have \(\nabla _\psi (\psi \cdot \nu )=0\) and thus \( \nabla \psi \, \psi \cdot \nu = -\langle {\bar{\nabla }}_\psi \nu ,\psi \rangle = -\langle B_\Sigma \psi ,\psi \rangle \). Therefore the equality (3.1) follows from (3.3) and from \(F_n = \psi \cdot \nu = 0\).

To obtain (3.2) we apply (3.3) for \(F = \nabla u\) and deduce

$$\begin{aligned} \Vert \nabla ^2 u\Vert _{L^2(\Omega )}^2 = \Vert \Delta u\Vert _{L^2(\Omega )}^2 + \int _{\Sigma } ((\nabla ^2 u \, \nabla u) \cdot \nu -\Delta u \, \partial _\nu u) \, d \mathcal {H}^2. \end{aligned}$$

We write \(\nabla u = \nabla _\tau u + \partial _\nu u \, \nu \) and observe

$$\begin{aligned} \begin{aligned} (\nabla ^2 u \, \nabla u )\cdot \nu&= (\nabla ^2 u \, \nabla _\tau u) \cdot \nu + (\nabla ^2 u \, \nu )\cdot \nu \,\partial _\nu u \\&=\nabla _\tau (\partial _\nu u) \cdot \nabla _\tau u - \langle B_\Sigma \, {\bar{\nabla }} u, {\bar{\nabla }} u \rangle + (\nabla ^2 u \, \nu ) \cdot \nu \,\partial _\nu u. \end{aligned} \end{aligned}$$

Again by divergence theorem

$$\begin{aligned} \int _{\Sigma } \nabla _\tau (\partial _\nu u) \cdot \nabla _\tau u \, d \mathcal {H}^2 = - \int _{\Sigma } \Delta _\Sigma u \, \partial _\nu u\, d \mathcal {H}^2. \end{aligned}$$

The equality (3.2) then follows from

$$\begin{aligned} \Delta _\Sigma u = \Delta u - (\nabla ^2 u \, \nu )\cdot \nu - H_\Sigma \, \partial _\nu u. \end{aligned}$$
(3.4)

We remark that it is crucial in Theorem 3.1 that the boundary term on the RHS has only the normal component of the vector field. The next lemma is a generalization of [35] and it essentially states that we may control the vector field on the boundary by its normal or its tangential component.

Lemma 3.3

Let \(\Omega \subset \mathbb {R}^3\) with \(\Sigma =\partial \Omega \) be uniformly \(C^{1,\alpha }(\Gamma )\)-regular. Then for all vector fields \(F \in {\dot{H}}^1(\Omega ; \mathbb {R}^3)\) it holds

$$\begin{aligned} \Vert F\Vert _{L^2(\Sigma )}^2 \le C \left( \Vert F_n \Vert _{L^2(\Sigma )}^2 + \Vert F\Vert _{L^2(\Omega )}^2 +\Vert \textrm{div}\,F\Vert _{L^2(\Omega )}^2 +\Vert \textrm{curl}\,F\Vert _{L^2(\Omega )}^2 \right) \end{aligned}$$

and

$$\begin{aligned} \Vert F\Vert _{L^2(\Sigma )}^2 \le C \left( \Vert F_\tau \Vert _{L^2(\Sigma )}^2 + \Vert F\Vert _{L^2(\Omega )}^2 +\Vert \textrm{div}\,F\Vert _{L^2(\Omega )}^2 +\Vert \textrm{curl}\,F\Vert _{L^2(\Omega )}^2 \right) , \end{aligned}$$

where \(F_n = F\cdot \nu \) and \(F_\tau = F - F_n \, \nu \). Here \(F \in {\dot{H}}^1(\Omega ; \mathbb {R}^3)\) means that \(\Vert F\Vert _{{\dot{H}}^1(\Omega )} = \Vert \nabla F \Vert _{L^2(\Omega )} + \Vert F \Vert _{L^6(\Omega )} < \infty \). Note that \(\Omega \) may be unbounded, but its boundary is compact.

Proof

We only consider the case when \(\Omega \) is bounded. Let us first assume that \(\Omega \) is uniformly star shaped with respect to the origin, i.e., we have that there exists a constant \(c_0>0\) such that \(x \cdot \nu _\Omega \ge c_0\) for all \(x \in \Sigma \). We claim that the following identity holds

$$\begin{aligned} \textrm{div}\,\left( |F|^2x- 2 (F\cdot x) F \right) = |F|^2 -2 \textrm{curl}\,F (F \cdot x) -2 \textrm{div}\,F \, (F \cdot x). \end{aligned}$$
(3.5)

Indeed, this follows from the following straightforward computation, where we denote the Dirac delta by \(\delta _i^j\),

$$\begin{aligned} \begin{aligned} \textrm{div}\,( |F|^2x-&2 (F\cdot x) F ) = \sum _{i,j =1}^3 \partial _i \left( F_j^2 x_i - 2x_jF_j F_i\right) \\&= 3|F|^2 + \sum _{i,j =1}^3 2x_i\partial _iF_jF_j - 2 \delta _i^j F_jF_i - 2 x_j \partial _iF_j F_i-2 x_j F_j\partial _iF_i \\&= |F|^2 -2 ( F\cdot x) \textrm{div}\,F -2\sum _{i,j =1}^3 (\partial _jF_i-\partial _i F_j )F_j x_i. \end{aligned} \end{aligned}$$

Thus we integrate (3.5) to find

$$\begin{aligned} \int _{\Sigma } \big ((x \cdot \nu )|F|^2 -2 F_n (F \cdot x)\big )\, d\mathcal {H}^2 = \int _{\Omega } |F|^2 - 2\textrm{curl}\,F (F \cdot x) -2 \textrm{div}\,F (F \cdot x) \,dx. \end{aligned}$$

Note that \(|F|^2 = |F_\tau |^2 + F_n^2\) and \((F \cdot x) = ( x \cdot \nu ) F_n + (F_\tau \cdot x)\). Therefore we have the equality

$$\begin{aligned} \begin{aligned} \int _{\Sigma } \big (-( x \cdot \nu ) F_n^2&+ ( x \cdot \nu ) |F_\tau |^2 -2 F_n (F_\tau \cdot x) \big )\, d\mathcal {H}^2\\&=\int _{\Omega } |F|^2 + 2\textrm{curl}\,F (F \cdot x) -2 \textrm{div}\,F (F \cdot x) \,dx. \end{aligned} \end{aligned}$$

We use the fact that \(x \cdot \nu \ge c_0\) on \(\Sigma \) and obtain the first claim by re-organizing the terms in above and estimating \(| F_n (F_\tau \cdot x)| \le \varepsilon |F_\tau |^2 + C_e F_n^2 \)

$$\begin{aligned} \begin{aligned} c_0\int _{\Sigma } |F_\tau |^2\, d\mathcal {H}^2 \le&\int _{\Sigma } (\varepsilon |F_\tau |^2 + C_\varepsilon |F_n|^2) \, d\mathcal {H}^2 \\&+ C(\Vert F\Vert _{L^2(\Omega )}^2 + \Vert \textrm{curl}\,F\Vert _{L^2(\Omega )}^2 + \Vert \textrm{div}\,F\Vert _{L^2(\Omega )}^2). \end{aligned} \end{aligned}$$

This yields the first inequality. The second follows from similar argument.

To prove the general case, i.e. when \(\Omega \) is not starshaped, we use a localization argument which is similar to [1]. \(\square \)

Remark 3.4

We observe that the proof gives us a slightly stronger estimate. Indeed, we may improve the second inequality in Lemma 3.3 as

$$\begin{aligned} \Vert F\Vert _{L^2(\Sigma )}^2 \le C \left( \Vert F_\tau \Vert _{L^2(\Sigma )}^2 + \Vert F\Vert _{L^2(\Omega )}^2 +\Vert F\textrm{div}\,F\Vert _{L^1(\Omega )} +\Vert F\textrm{curl}\,F\Vert _{L^1(\Omega )})\right) . \end{aligned}$$

In order to estimate \(\Vert \nabla F\Vert _{H^1(\Omega )}\) we first consider the case when F is curl-free, i.e., \(F = \nabla u\). Since we define the norm \(\Vert \partial _\nu u\Vert _{H^{\frac{1}{2}}(\Sigma )}\) via the harmonic extension, we prove the next lemma using standard results from harmonic analysis instead of localizing and flattening the boundary.

Lemma 3.5

Assume that \(\Omega \), with \(\Sigma = \partial \Omega \), is uniformly \(C^{1,\alpha }(\Gamma )\)-regular and \(\Vert B_\Sigma \Vert _{L^4} \le M\) and \(u:\Omega \rightarrow \mathbb {R}\) is a smooth function. There exists a constant C, depending on M and the \(C^{1,\alpha }\)-norm of the heightfunction, such that it holds

$$\begin{aligned} \Vert u\Vert _{H^2(\Omega )} \le C\left( \Vert \partial _\nu u\Vert _{H^{\frac{1}{2}}(\Sigma )} + \Vert u\Vert _{L^2(\Omega )} + \Vert \Delta u\Vert _{L^2(\Omega )} \right) . \end{aligned}$$

The reverse also holds \(\Vert \partial _\nu u\Vert _{H^{\frac{1}{2}}(\Sigma )} \le C \Vert u\Vert _{H^2(\Omega )}\).

Proof

We have by (3.2) and (3.4) that

$$\begin{aligned} \Vert \nabla ^2 u\Vert _{L^2(\Omega )}^2 \le \Vert \Delta u\Vert _{L^2(\Omega )}^2 + 2 \int _{\Sigma } ((\nabla ^2 u\, \nu )\cdot \nu - \Delta u) \, \partial _\nu u \, d \mathcal {H}^2 + C \int _{\Sigma } |B_\Sigma | \, |\nabla u|^2\, d \mathcal {H}^2. \end{aligned}$$

To estimate the last terms, we use the interpolation inequality (Proposition 2.8), Lemma 3.3 and the assumption \(\Vert B\Vert _{L^4} \le C\) and have for some \(\theta \in (0,1)\)

$$\begin{aligned} \begin{aligned} \int _{\Sigma } |B_\Sigma | \, |\nabla u|^2\, d \mathcal {H}^2&\le C \Vert B\Vert _{L^4(\Sigma )} \Vert \nabla u \Vert _{L^{\frac{8}{3}}(\Sigma )}^2 \le \Vert \nabla u \Vert _{H^{\frac{1}{2}}(\Sigma )}^{2\theta } \Vert \nabla u \Vert _{L^{2}(\Sigma )}^{2(1 -\theta )}\\&\le \varepsilon \Vert u \Vert _{H^{2}(\Omega )}^{2} + C_\varepsilon \Vert \nabla u \Vert _{L^{2}(\Sigma )}^{2}\\&\le \varepsilon \Vert u \Vert _{H^{2}(\Omega )}^{2} + C_\varepsilon (\Vert \partial _\nu u\Vert _{L^2(\Sigma )}^2 + \Vert \nabla u \Vert _{L^{2}(\Omega )}^{2} + \Vert \Delta u\Vert _{L^2(\Omega )}^2 )\\&\le \varepsilon \Vert u \Vert _{H^{2}(\Omega )}^{2} + C_\varepsilon (\Vert \partial _\nu u\Vert _{L^2(\Sigma )}^2 + \Vert u \Vert _{L^{2}(\Omega )}^{2} + \Vert \Delta u\Vert _{L^2(\Omega )}^2 ). \end{aligned} \end{aligned}$$
(3.6)

Let us then show that

$$\begin{aligned} \int _{\Sigma } ((\nabla ^2 u\, \nu )\cdot \nu - \Delta u) \, \partial _\nu u \, d \mathcal {H}^2 \le \varepsilon \Vert u \Vert _{H^{2}(\Omega )}^{2} + C_\varepsilon ( \Vert \partial _\nu u\Vert _{H^{\frac{1}{2}}(\Sigma )}^2 + \Vert u \Vert _{L^{2}(\Omega )}^{2} + \Vert \Delta u\Vert _{L^2(\Omega )}^2 ). \end{aligned}$$
(3.7)

To this aim we denote the harmonic extension of \(\nu \) by \({\tilde{\nu }}\) and denote \(f = \Delta u\). Then we have

$$\begin{aligned} \begin{aligned} \int _{\Sigma } (\nabla ^2 u\, \nu )\cdot \nu \, \partial _\nu u \, d \mathcal {H}^2&= \int _{\Sigma } \big (\partial _\nu (\nabla u \cdot {\tilde{\nu }}) -(\partial _\nu {\tilde{\nu }} \cdot \nabla u) \big )\,(\nabla u \cdot {\tilde{\nu }})\, d \mathcal {H}^2 \\&\le \int _{\Sigma } \partial _\nu (\nabla u \cdot {\tilde{\nu }})\,(\nabla u \cdot {\tilde{\nu }})\, d \mathcal {H}^2 + C\Vert \partial _\nu {\tilde{\nu }}\Vert _{L^4(\Sigma )} \Vert \nabla u \Vert _{L^{\frac{8}{3}}(\Sigma )}^2. \end{aligned} \end{aligned}$$

We argue as in (3.6) and obtain

$$\begin{aligned} \Vert \nabla u \Vert _{L^{\frac{8}{3}}(\Sigma )}^2 \le \varepsilon \Vert u \Vert _{H^{2}(\Omega )}^{2} + C_\varepsilon (\Vert \partial _\nu u\Vert _{L^2(\Sigma )}^2 + \Vert u \Vert _{L^{2}(\Omega )}^{2} + \Vert f\Vert _{L^2(\Omega )}^2 ). \end{aligned}$$

Next we use the result from [19] for harmonic functions \(\varphi :\Omega \rightarrow \mathbb {R}\) in \(C^{1,\alpha }\)-domains which states that

$$\begin{aligned} \Vert \partial _\nu \varphi \Vert _{L^p(\Sigma )} \le C_p \Vert \nabla _\tau \varphi \Vert _{L^p(\Sigma )} \qquad \text {for }\, p \in (1,\infty ). \end{aligned}$$

We use this for \({\tilde{\nu }}\) component-wise, use the fact that on \(\Sigma \) it holds \({\tilde{\nu }} = \nu \) and obtain

$$\begin{aligned} \Vert \nabla {\tilde{\nu }}\Vert _{L^4(\Sigma )} \le C\Vert \nabla _\tau {\tilde{\nu }}\Vert _{L^4(\Sigma )} \le \Vert B \Vert _{L^4(\Sigma )}\le C. \end{aligned}$$
(3.8)

Therefore we have

$$\begin{aligned} \begin{aligned} \int _{\Sigma } (\nabla ^2 u\, \nu \cdot ) \nu \, \partial _\nu u \, d \mathcal {H}^2 \le&\int _{\Sigma } \partial _\nu (\nabla u \cdot {\tilde{\nu }})\,(\nabla u \cdot {\tilde{\nu }})\, d \mathcal {H}^2 \\&+ \varepsilon \Vert u \Vert _{H^{2}(\Omega )}^{2} + C_\varepsilon (\Vert \partial _\nu u\Vert _{L^2(\Sigma )}^2 + \Vert u \Vert _{L^{2}(\Omega )}^{2} + \Vert f\Vert _{L^2(\Omega )}^2 ). \end{aligned} \end{aligned}$$
(3.9)

Let us denote \({\tilde{u}} = (\nabla u \cdot {\tilde{\nu }})\) for short and let v be the harmonic extension of \({\tilde{u}}\) to \(\Omega \). Let us show that v is close to \({\tilde{u}}\), i.e., we show that

$$\begin{aligned} \Vert \nabla ({\tilde{u}} - v) \Vert _{L^2(\Omega )} \le \varepsilon \Vert \nabla ^2 u \Vert _{L^2(\Omega )} + C_\varepsilon (\Vert f\Vert _{L^2(\Omega )} + \Vert {\tilde{u}} - v\Vert _{L^2(\Omega )}) . \end{aligned}$$
(3.10)

To this aim we calculate (recall that \(f= \Delta u\))

$$\begin{aligned} \Delta {\tilde{u}} = \nabla f \cdot {\tilde{\nu }} + 2 \nabla ^2 u : \nabla {\tilde{\nu }}. \end{aligned}$$
(3.11)

This implies by integration by parts

$$\begin{aligned} \begin{aligned} \Vert \nabla&({\tilde{u}} - v) \Vert _{L^2(\Omega )}^2 = -\int _{\Omega } \Delta {\tilde{u}} ({\tilde{u}} - v)\, dx = \int _{\Omega }( \nabla f \cdot {\tilde{\nu }} + 2 \nabla ^2 u: \nabla {\tilde{\nu }}) ({\tilde{u}} - v )\, dx \\&= \int _{\Omega } f \textrm{div}\,\, (({\tilde{u}} - v) \, {\tilde{\nu }}) - 2( \nabla ^2 u: \nabla {\tilde{\nu }}) ({\tilde{u}} - v )\, dx \\&\le C \Vert \nabla ({\tilde{u}} - v) \Vert _{L^2(\Omega )}\Vert f\Vert _{L^2(\Omega )} + C (\Vert \nabla ^2 u \Vert _{L^2(\Omega )}+ \Vert f\Vert _{L^2(\Omega )}) \Vert \nabla {\tilde{\nu }} \Vert _{L^{4}(\Omega )}\Vert ({\tilde{u}} - v) \Vert _{L^{4}(\Omega )}. \end{aligned} \end{aligned}$$

By standard estimates from harmonic analysis [16] and by (3.8) it holds

$$\begin{aligned} \Vert \nabla {\tilde{\nu }} \Vert _{L^{4}(\Omega )} \le C \Vert \nabla {\tilde{\nu }} \Vert _{L^{4}(\Sigma )} \le C. \end{aligned}$$
(3.12)

On the other hand we have by Hölder’s inequality and by Sobolev embedding (recall that \({\tilde{u}} - v = 0\) on \(\Sigma \))

$$\begin{aligned} \Vert ({\tilde{u}} - v) \Vert _{L^{4}(\Omega )} \le \Vert ({\tilde{u}} - v) \Vert _{L^6(\Omega )}^{\frac{1}{2}} \Vert {\tilde{u}} - v\Vert _{L^2(\Omega )}^{\frac{1}{2}} \le C \Vert \nabla ({\tilde{u}} - v)\Vert _{L^2(\Omega )}^{\frac{1}{2}} \Vert {\tilde{u}} - v\Vert _{L^2(\Omega )}^{\frac{1}{2}}. \end{aligned}$$

Therefore by combining the previous inequalities we obtain (3.10) by Young’s inequality.

We proceed by using (3.11) and by integrating by parts

$$\begin{aligned} \begin{aligned} \int _{\Sigma }&\partial _\nu {\tilde{u}}\,{\tilde{u}} \, d \mathcal {H}^2 =\int _{\Sigma } \partial _\nu {\tilde{u}}\,v \, d \mathcal {H}^2 = \int _{\Omega } (\nabla {\tilde{u}} \cdot \nabla v + \Delta {\tilde{u}} \, v) \, dx \\&\le 2\Vert \nabla v\Vert _{L^2(\Omega )}^2 + 2\Vert \nabla ({\tilde{u}}- v)\Vert _{L^2(\Omega )}^2 + \int _\Omega (\nabla f \cdot {\tilde{\nu }} + 2 \nabla ^2 u: \nabla {\tilde{\nu }}) \, v\, dx \\&= 2\Vert \nabla v\Vert _{L^2(\Omega )}^2 + 2\Vert \nabla ({\tilde{u}} - v)\Vert _{L^2(\Omega )}^2 + \int _{\Sigma } f \, v \, ({\tilde{\nu }} \cdot \nu ) \, d \mathcal {H}^2\\&\,\,\,\,\,\,\,\,\,\,\,\,\,\,+ \int _{\Omega } -f \textrm{div}\,\, (v \, {\tilde{\nu }}) + 2(\nabla ^2 u: \nabla {\tilde{\nu }}) v \, dx. \end{aligned} \end{aligned}$$

Recall that \({\tilde{\nu }} = \nu \) and \(v = {\tilde{u}}\) on \(\Sigma \). Therefore we obtain by the above inequality, by \(\Vert v\Vert _{L^4(\Omega )} \le C \Vert v\Vert _{H^1(\Omega )}\), (3.10) and (3.12) that

$$\begin{aligned} \begin{aligned} \int _{\Sigma } \partial _\nu {\tilde{u}}\,{\tilde{u}} - f \, {\tilde{u}} \, d \mathcal {H}^2&\le \varepsilon \Vert \nabla ^2 u \Vert _{L^2(\Omega )} \\&\,\,\,\,\,\,\,\,\,\,\,+ C_\varepsilon (\Vert \nabla \nu \Vert _{L^4(\Omega )}\Vert v\Vert _{L^4(\Omega )}+ \Vert v\Vert _{H^1(\Omega )}^2 +\Vert f\Vert _{L^2(\Omega )}^2 + \Vert {\tilde{u}} - v\Vert _{L^2(\Omega )}^2) \\&\le \varepsilon \Vert \nabla ^2 u \Vert _{L^2(\Omega )} + C_\varepsilon (\Vert v\Vert _{H^1(\Omega )}^2 +\Vert f\Vert _{L^2(\Omega )}^2 + \Vert {\tilde{u}} - v\Vert _{L^2(\Omega )}^2). \end{aligned} \end{aligned}$$

The inequality (3.7) then follows from the above and (3.9) together with

$$\begin{aligned} \Vert v\Vert _{H^1(\Omega )} = \Vert \partial _\nu u\Vert _{H^{\frac{1}{2}}(\Sigma )}, \qquad \Vert {\tilde{u}} - v\Vert _{L^2(\Omega )}^2 \le \varepsilon \Vert \nabla ^2 u \Vert _{L^2(\Omega )} + C_\varepsilon \Vert u\Vert _{L^2(\Omega )}^2 + \Vert v\Vert _{L^2(\Omega )}^2, \end{aligned}$$

and by recalling that \({\tilde{u}} = (\nabla u \cdot {\tilde{\nu }}) = \partial _\nu u\) on \(\Sigma \) and \(f = \Delta u\). This yields the first claim. The second inequality follows from reversing the previous calculations. \(\square \)

We state our lower order version of Theorem 3.1.

Theorem 3.6

Assume that \(\Omega \), with \(\Sigma = \partial \Omega \), is uniformly \(C^{1,\alpha }(\Gamma )\)-regular and \(\Vert B_\Sigma \Vert _{L^4(\Sigma )} \le M\). There exists a constant C, depending on M and the \(C^{1,\alpha }\)-norm of the heightfunction, such that for all vector fields \(F \in H^1(\Omega ;\mathbb {R}^3)\) it holds

$$\begin{aligned} \Vert F\Vert _{H^1(\Omega )} \le M( \Vert F_n\Vert _{H^\frac{1}{2}(\Sigma )} + \Vert F\Vert _{L^2(\Omega )} +\Vert \textrm{div}\,F\Vert _{L^2(\Omega )}+\Vert \textrm{curl}\,F\Vert _{L^2(\Omega )} ). \end{aligned}$$

Proof

By approximation argument we may assume that F and \(\Omega \) are smooth. We use the Helmholtz-Hodge decomposition and write \(F=\nabla \phi +\psi \) where \(\phi \) is the unique solution of the Neumann problem

$$\begin{aligned} {\left\{ \begin{array}{ll} \Delta \phi =\textrm{div}\,F\qquad &{}x\in \Omega \\ \partial _\nu \phi = F_n &{}x\in \Sigma \end{array}\right. } \end{aligned}$$

with zero average and \(\psi \) solves

$$\begin{aligned} {\left\{ \begin{array}{ll} \textrm{curl}\,\psi = \textrm{curl}\,F \qquad &{}x\in \Omega \\ \textrm{div}\,\psi =0 &{}x\in \Omega \\ \psi \cdot \nu = 0 &{}x\in \Sigma . \end{array}\right. } \end{aligned}$$

We also note that \(\nabla \phi \) and \(\psi \) are orthogonal in \(L^2(\Omega )\) and thus

$$\begin{aligned} \int _{\Omega } |\nabla \phi |^2 +|\psi |^2dx =\int _{\Omega }|F|^2 dx. \end{aligned}$$

For \(\phi \) we have by Lemma 3.5 that

$$\begin{aligned} \begin{aligned} \Vert \phi \Vert _{H^2(\Omega )}&\le C( \Vert \partial _\nu \phi \Vert _{H^\frac{1}{2}(\Sigma )} + \Vert \nabla \phi \Vert _{L^2(\Omega )} +\Vert \Delta \phi \Vert _{L^2(\Omega )})\\&\le C( \Vert F_n\Vert _{H^\frac{1}{2}(\Sigma )} + \Vert F\Vert _{L^2(\Omega )} +\Vert \textrm{div}\,F\Vert _{L^2(\Omega )}). \end{aligned} \end{aligned}$$
(3.13)

For \(\psi \) we have by (3.1)

$$\begin{aligned} \Vert \nabla \psi \Vert _{L^2(\Omega )}^2\, dx= \frac{1}{2}\Vert \text {curl}\, \psi \Vert _{L^2(\Omega )}^2 - \int _{\Sigma } \langle B_\Sigma \, \psi , \psi \rangle \, d\mathcal {H}^2. \end{aligned}$$

We use the assumption \(\Vert B_\Sigma \Vert _{L^4} \le C\), Hölder’s inequality and interpolation inequality to deduce

$$\begin{aligned} \begin{aligned} -\int _{\Sigma } \langle B_\Sigma \, \psi , \psi \rangle \, d\mathcal {H}^2&\le \Vert B_{\Sigma }\Vert _{L^4(\Sigma )}\Vert \psi \Vert _{L^{\frac{8}{3}}(\Sigma )}^2 \\&\le \varepsilon \Vert \psi \Vert _{H^{\frac{1}{2}}(\Sigma )}^2 + C_\varepsilon \Vert \psi \Vert _{L^2(\Omega )}^2 \le \varepsilon \Vert \nabla \psi \Vert _{L^{2}(\Omega )}^2 + C_\varepsilon \Vert F\Vert _{L^2(\Omega )}^2. \end{aligned} \end{aligned}$$

Thus we have

$$\begin{aligned} \Vert \nabla \psi \Vert _{L^2(\Omega )} \le C(\Vert \textrm{curl}\,F\Vert _{L^2(\Omega )} + \Vert F\Vert _{L^2(\Omega )}) \end{aligned}$$

This together with (3.13) yields the claim. \(\square \)

We proceed by using Theorem 3.6 to control the higher order norms \(\Vert F\Vert _{H^2(\Omega )}\) and \(\Vert F\Vert _{H^3(\Omega )}\). We do not need the sharp dependence on the curvature for these estimates and we state the result in a form that is suitable for us. We also treat the case \(\Vert F\Vert _{H^{\frac{3}{2}}(\Omega )}\), but only for curl-free vector fields. In this case we need the ’sharp’ curvature dependence but this time we have non-optimal dependence on the divergence.

Lemma 3.7

Assume that \(\Omega \), with \(\Sigma = \partial \Omega \), is uniformly \(C^{1,\alpha }(\Gamma )\)-regular and \(\Vert B_\Sigma \Vert _{H^{\frac{1}{2}}(\Sigma )} \le M\). Then for all vector fields \(F \in H^3(\Omega ;\mathbb {R}^3)\) it holds

$$\begin{aligned} \Vert F\Vert _{H^3(\Omega )} \le C \big (\Vert \Delta _\Sigma F_n\Vert _{H^\frac{1}{2}(\Sigma )} + (1+\Vert H_\Sigma \Vert _{H^{2}(\Sigma )})\Vert F\Vert _{L^{\infty }(\Omega )} +\Vert \textrm{div}\,F\Vert _{H^2(\Omega )}+\Vert \textrm{curl}\,F\Vert _{H^2(\Omega )} \big ) \end{aligned}$$

and

$$\begin{aligned} \Vert F\Vert _{H^2(\Omega )} \le C\big (\Vert \Delta _\Sigma F_n\Vert _{H^\frac{1}{2}(\Sigma )} + \Vert F\Vert _{L^{\infty }(\Omega )} +\Vert \textrm{div}\,F\Vert _{H^1(\Omega )}+\Vert \textrm{curl}\,F\Vert _{H^1(\Omega )} \big ) \end{aligned}$$

for some constant C, depending on M and the \(C^{1,\alpha }\)-norm of the heightfunction. Moreover, if \(F = \nabla u\) then it holds

$$\begin{aligned} \Vert \nabla u\Vert _{H^{\frac{3}{2}}(\Omega )} \le C(\Vert \partial _\nu u \Vert _{H^1(\Sigma )} + \Vert u\Vert _{L^2(\Omega )} +\Vert \Delta u\Vert _{H^1(\Omega )}) \end{aligned}$$

and for \(k = \frac{1}{2}, 1\) it holds

$$\begin{aligned} \Vert u\Vert _{H^{k}(\Sigma )} \le C \Vert u\Vert _{H^{k+\frac{1}{2}}(\Omega )}. \end{aligned}$$

Proof

By approximation argument we may assume that F and \(\Omega \) are smooth. Note also that by Sobolev embedding \(\Vert B_\Sigma \Vert _{L^4(\Sigma )} \le \Vert B_\Sigma \Vert _{H^{\frac{1}{2}}(\Sigma )} \le M\).

Let \({\tilde{\nu }}\) be the harmonic extension of the normal field \(\nu \) to \(\Omega \). Let us define the vector fields \(\tau _i = e_i - ({\tilde{\nu }} \cdot e_i) {\tilde{\nu }}\) for \(i =1,2,3\), where \(\{e_i\}_i\) is a coordinate basis of \(\mathbb {R}^3\). For ij we define a vector field \(F_{ij}:\Omega \rightarrow \mathbb {R}^3\) as \((F_{ij})_k = \nabla ^2 F_k \tau _i \cdot \tau _j \). We apply Theorem 3.6 for \(F_{ij}\) and obtain

$$\begin{aligned} \Vert \nabla F_{ij}\Vert _{L^2(\Omega )} \le C( \Vert F_{ij} \cdot \nu \Vert _{H^\frac{1}{2}(\Sigma )} + \Vert F_{ij}\Vert _{L^2(\Omega )} +\Vert \textrm{div}\,F_{ij}\Vert _{L^2(\Omega )}+\Vert \textrm{curl}\,F_{ij}\Vert _{L^2(\Omega )} ). \end{aligned}$$

Recall that (3.12) implies \(\Vert \nabla {\tilde{\nu }}\Vert _{L^4(\Omega )} \le C\). Moreover by maximum principle it holds \(\Vert {\tilde{\nu }} \Vert _{L^\infty (\Omega )} \le C\). Therefore

$$\begin{aligned} \begin{aligned} \Vert \textrm{div}\,F_{ij}\Vert _{L^2(\Omega )}&\le C\Vert \textrm{div}\,F\Vert _{H^2(\Omega )} + C\Vert \nabla ^2 F \Vert _{L^4(\Omega )} \Vert \nabla \nu \Vert _{L^4(\Omega )} \\&\le C\Vert \textrm{div}\,F\Vert _{H^2(\Omega )} + C \Vert \nabla ^2 F \Vert _{L^4(\Omega )}. \end{aligned} \end{aligned}$$

By interpolation we have \( \Vert \nabla ^2 F \Vert _{L^4(\Omega )} \le \varepsilon \Vert F\Vert _{H^3(\Omega )} + C_{\varepsilon } \Vert F\Vert _{L^2(\Omega )}\) and thus

$$\begin{aligned} \Vert F_{ij}\Vert _{L^2(\Omega )} + \Vert \textrm{div}\,F_{ij}\Vert _{L^2(\Omega )} \le \varepsilon \Vert F \Vert _{H^3(\Omega )} + C_\varepsilon (\Vert \textrm{div}\,F\Vert _{H^2(\Omega )}+ \Vert F\Vert _{L^2(\Omega )}). \end{aligned}$$

By a similar argument

$$\begin{aligned} \Vert \textrm{curl}\,F_{ij}\Vert _{L^2(\Omega )} \le \varepsilon \Vert F \Vert _{H^3(\Omega )} + C_\varepsilon (\Vert \textrm{curl}\,F\Vert _{H^2(\Omega )}+ \Vert F\Vert _{L^2(\Omega )}) \end{aligned}$$

and

$$\begin{aligned} \Vert \nabla F_{ij}\Vert _{L^2(\Omega )}^2 \ge \sum _{k,l=1}^3 \Vert \nabla ^2 \nabla _l F_k \, \tau _i \cdot \tau _j \Vert _{L^2(\Omega )}^2 - \varepsilon \Vert F \Vert _{H^3(\Omega )}^2 - C_\varepsilon \Vert F\Vert _{L^2(\Omega )}^2. \end{aligned}$$

Let us fix a point \(x \in \Omega \) and estimate the norm

$$\begin{aligned} \sum _{i,j,k,l=1}^3|\nabla ^2 \nabla _l F_k(x) \, \tau _i \cdot \tau _j|^2. \end{aligned}$$

First we observe that the above quantity does not depend on the choice of the coordinates in \(\mathbb {R}^3\). Let us choose the coordinates such that \( {\tilde{\nu }}(x) \cdot e_i = 0\) for \(i=1,2\). Then we have

$$\begin{aligned} \sum _{i,j,k,l=1}^3 |\nabla ^2 \nabla _l F_k(x) \, \tau _i \cdot \tau _j |^2 \ge \sum _{k,l=1}^3 \sum _{i,j=1}^2 | \nabla _i\nabla _j\nabla _l F_k(x)|^2. \end{aligned}$$

By a simple combinatorial argument we deduce

$$\begin{aligned} \begin{aligned}&\sum _{i,j,k,l=1}^3 | \nabla _i\nabla _j\nabla _l F_k(x)|^2\\&\le C\sum _{k,l=1}^3 \sum _{i,j=1}^2 | \nabla _i\nabla _j\nabla _l F_k(x)|^2 + C |\nabla ^2 \textrm{div}\,F(x)|^2 + C |\nabla ^2 \textrm{curl}\,F(x)|^2. \end{aligned} \end{aligned}$$

By applying the above argument for every x we have

$$\begin{aligned} \sum _{i,j,k,l=1}^3 \Vert \nabla F_{ij}\Vert _{L^2(\Omega )}^2 \ge c_0 \Vert \nabla ^3 F\Vert _{L^2(\Omega )}^2 - C(\Vert \textrm{div}\,F\Vert _{H^2(\Omega )}^2+ \Vert \textrm{curl}\,F\Vert _{H^2(\Omega )}^2+ \Vert F\Vert _{L^2(\Omega )}^2). \end{aligned}$$

Combing all the previous estimates we obtain

$$\begin{aligned} \Vert \nabla ^3 F\Vert _{L^2(\Omega )} \le \sum _{i,j=1}^3 C(\Vert F_{ij} \cdot \nu \Vert _{H^\frac{1}{2}(\Sigma )} + \Vert F\Vert _{L^2(\Omega )} +\Vert \textrm{div}\,F\Vert _{H^2(\Omega )}+ \Vert \textrm{curl}\,F\Vert _{H^2(\Omega )}). \end{aligned}$$

The first inequality follows once we show

$$\begin{aligned} \sum _{i,j=1}^3 \Vert F_{ij}\cdot \nu \Vert _{H^\frac{1}{2}(\Sigma )} \le C\Vert \Delta _\Sigma F_n \Vert _{H^\frac{1}{2}(\Sigma )}+ \varepsilon \Vert F\Vert _{H^3(\Omega )} +C_\varepsilon (1 + \Vert H_\Sigma \Vert _{H^{2}})\Vert F\Vert _{L^{\infty }(\Sigma )}. \end{aligned}$$
(3.14)

To this aim we first note that on \(\Sigma \) it holds \(\tau _i = e_i - (\nu \cdot e_i)\nu \) and therefore \(\tau _i\) is tangential on \(\Sigma \). Thus we have

$$\begin{aligned} (\nabla ^2 F_n) \tau _i \cdot \tau _j = F_{ij} \cdot \nu + \nabla _\tau F \star B_\Sigma + F \star \nabla _\tau B_\Sigma . \end{aligned}$$
(3.15)

We use Proposition 2.10 and get

$$\begin{aligned} \Vert \nabla _\tau F \star B_\Sigma \Vert _{H^{\frac{1}{2}}(\Sigma )} \le C\Vert \nabla F\Vert _{H^{\frac{1}{2}}(\Sigma )} \Vert B_\Sigma \Vert _{L^{\infty }(\Sigma )} + C\Vert \nabla _\tau F\Vert _{L^{\infty }(\Sigma )} \Vert B_\Sigma \Vert _{H^{\frac{1}{2}}(\Sigma )}. \end{aligned}$$

Recall that \(\Vert B_\Sigma \Vert _{H^{\frac{1}{2}}(\Sigma )} \le C\). By interpolation we have

$$\begin{aligned} \Vert \nabla _\tau F\Vert _{L^{\infty }(\Sigma )} \le \varepsilon \Vert F\Vert _{H^3(\Omega )} + C_\varepsilon \Vert F\Vert _{L^{\infty }}. \end{aligned}$$

Moreover, by Sobolev embedding, Proposition 2.8 and by Proposition 2.12 we have for \(\theta < \frac{1}{2}\)

$$\begin{aligned} \Vert B_\Sigma \Vert _{L^{\infty }(\Sigma )} \le C\Vert B_\Sigma \Vert _{W^{1, \frac{7}{3}}(\Sigma )} \le C\Vert B_\Sigma \Vert _{H^{2}(\Sigma )}^{\theta }\Vert B_\Sigma \Vert _{L^{4}(\Sigma )}^{1 - \theta }\le C(1+ \Vert H_\Sigma \Vert _{H^{2}(\Sigma )}^{\theta }). \end{aligned}$$

On the other hand, Corollary 2.9 implies

$$\begin{aligned} \Vert \nabla F\Vert _{H^{\frac{1}{2}}(\Sigma )}\le C \Vert \nabla F\Vert _{H^{1}(\Omega )}\le C\Vert F\Vert _{H^3(\Omega )}^{\frac{1}{2}} \Vert \nabla F\Vert _{L^2(\Omega )}^{\frac{1}{2}}. \end{aligned}$$

Therefore we have by the above estimates and by Young’s inequality

$$\begin{aligned} \begin{aligned} \Vert \nabla _\tau F \star B_\Sigma \Vert _{H^{\frac{1}{2}}(\Sigma )}&\le \varepsilon \Vert F\Vert _{H^3(\Omega )} + C_{\varepsilon }\Vert F\Vert _{L^\infty } + C_{\varepsilon }(1+ \Vert H_\Sigma \Vert _{H^{2}(\Sigma )}^{\theta })\Vert \nabla F\Vert _{L^2(\Omega )}\\&\le \varepsilon \Vert F\Vert _{H^3(\Omega )} + C_\varepsilon (1+ \Vert H_\Sigma \Vert _{H^{2}(\Sigma )}) \Vert F\Vert _{L^\infty (\Omega )}, \end{aligned} \end{aligned}$$

where the last inequality follows from interpolation.

Let us then bound the last term in (3.15). We have by Proposition 2.10

$$\begin{aligned} \begin{aligned} \Vert F \star \nabla _\tau B_\Sigma \Vert _{H^{\frac{1}{2}}(\Sigma )}&\le \Vert F \star \nabla _\tau B_\Sigma \Vert _{H^{1}(\Sigma )} \\&\le C \Vert F\Vert _{L^\infty (\Sigma )}\Vert {\bar{\nabla }} B_\Sigma \Vert _{H^{1}(\Sigma )} + C \Vert F\Vert _{W^{1,3}(\Sigma )} \Vert {\bar{\nabla }} B_\Sigma \Vert _{L^{6}(\Sigma )}. \end{aligned} \end{aligned}$$

Proposition 2.12 yields \(\Vert B_\Sigma \Vert _{H^{2}(\Sigma )} \le C(1+ \Vert H_\Sigma \Vert _{H^{2}(\Sigma )})\). Interpolation implies

$$\begin{aligned} \Vert {\bar{\nabla }} B_\Sigma \Vert _{L^{6}(\Sigma )}\le C \Vert B_\Sigma \Vert _{H^{2}(\Sigma )}^{\theta } \Vert B_\Sigma \Vert _{L^{2}(\Sigma )}^{1-\theta } \end{aligned}$$

and

$$\begin{aligned} \Vert F\Vert _{W^{1,3}(\Sigma )} \le \Vert F\Vert _{H^{3}(\Omega )}^{1-\theta } \Vert F\Vert _{L^{\infty }(\Omega )}^{\theta } \end{aligned}$$

for some \(\theta \in (0,1)\). Therefore we have by (3.15)

$$\begin{aligned} \Vert F_{ij} \cdot \nu \Vert _{H^\frac{1}{2}(\Sigma )} \le \Vert {\bar{\nabla }}^2 F_n \Vert _{H^\frac{1}{2}(\Sigma )} + \varepsilon \Vert F\Vert _{H^3(E)} +C_\varepsilon (1 + \Vert H_\Sigma \Vert _{H^{2}(\Sigma )})\Vert F\Vert _{L^{\infty }(\Omega )}. \end{aligned}$$

The inequality (3.14) then follows from Proposition 2.11 as

$$\begin{aligned} \Vert {\bar{\nabla }}^2 F_n \Vert _{H^\frac{1}{2}(\Sigma )} \le 2\Vert \Delta _\Sigma F_n \Vert _{H^\frac{1}{2}(\Sigma )} + C_\varepsilon \Vert F_n\Vert _{L^2(\Sigma )}. \end{aligned}$$

The second inequality follows from a similar argument.

Let us next prove the last part of the statement, i.e. the inequalities when \(F=\nabla u\). Let u be a solution of the Neumann boundary problem

$$\begin{aligned} {\left\{ \begin{array}{ll} \Delta u =f \qquad &{} \text {in } \Omega \\ \partial _\nu u = g &{}\text {on } \Sigma , \end{array}\right. } \end{aligned}$$

where \(\int _{\Sigma }g = \int _{\Omega } f\) and \(\int _\Omega u \, dx = 0\). First, clearly \(\Vert u\Vert _{H^{\frac{1}{2}}(\Sigma )} \le C \Vert u\Vert _{H^1(\Omega )}\). By the equation and by divergence theorem

$$\begin{aligned} \int _\Omega |\nabla u|^2 \, dx + \int _\Omega u \, f \, dx = \int _{\Sigma } u \, \partial _\nu u \, d \mathcal {H}^2 \le \Vert u\Vert _{H^{\frac{1}{2}}(\Sigma )} \Vert g\Vert _{H^{-\frac{1}{2}}(\Sigma )} \le C\Vert u\Vert _{H^1(\Omega )} \Vert g\Vert _{H^{-\frac{1}{2}}(\Sigma )}. \end{aligned}$$

Therefore by

$$\begin{aligned} \left| \int _\Omega u \, f \, dx \right| \le \Vert u\Vert _{L^{2}(\Omega )} \Vert f\Vert _{L^{2}(\Omega )} \end{aligned}$$

and by Poincaré inequality \(\Vert u\Vert _{L^{2}(\Omega )} \le C \Vert \nabla u\Vert _{L^{2}(\Omega )}\) we have

$$\begin{aligned} \Vert u\Vert _{H^1(\Omega )} \le C(\Vert g\Vert _{H^{-\frac{1}{2}}(\Sigma )} + \Vert f\Vert _{L^{2}(\Omega )}). \end{aligned}$$

On the other hand Lemma 3.5 implies

$$\begin{aligned} \begin{aligned} \Vert u\Vert _{H^{2}(\Omega )}&\le C(\Vert g\Vert _{H^{\frac{1}{2}}(\Sigma )} + \Vert u\Vert _{L^2} + \Vert f\Vert _{L^{2}(\Omega )})\\&\le C(\Vert g\Vert _{H^{\frac{1}{2}}(\Sigma )} + \Vert f\Vert _{L^{2}(\Omega )}). \end{aligned} \end{aligned}$$

We use the two above inequalities and standard interpolation argument to deduce

$$\begin{aligned} \Vert u\Vert _{H^{3/2}(\Omega )} \le C(\Vert g\Vert _{L^2(\Sigma )} + \Vert f\Vert _{L^2(\Omega )}). \end{aligned}$$
(3.16)

We proceed by applying (3.16) for \(u_{x_i} = \nabla u \cdot e_i - c_i\), for \(i = 1,2,3\), , and obtain

$$\begin{aligned} \Vert \nabla u\Vert _{H^{3/2}(\Omega )} \le C(\Vert \partial _\nu (\nabla u)\Vert _{L^2(\Sigma )}+ \Vert f\Vert _{H^1(\Omega )}). \end{aligned}$$
(3.17)

In order to treat the first term on the RHS we let \({\tilde{\nu }}\) be the harmonic extension of \(\nu \) to \(\Omega \). We write \(\nabla u = \nabla _\tau u + (\partial _\nu u)\, \nu \) and have

$$\begin{aligned} \partial _\nu (\nabla u) = \nabla _\tau (\partial _\nu u) + \partial _\nu (\nabla u \cdot {\tilde{\nu }}) \nu + \nabla {\tilde{\nu }} \star \nabla u. \end{aligned}$$

Recall that we have by maximum principle \(\Vert {\tilde{\nu }}\Vert _{L^\infty } \le C\) and by (3.12) \(\Vert \nabla {\tilde{\nu }}\Vert _{L^4(\Omega )} \le C\). We argue as in (3.6) and obtain

$$\begin{aligned} \Vert \nabla {\tilde{\nu }} \star \nabla u\Vert _{L^2(\Sigma )} \le \Vert \nabla {\tilde{\nu }} \Vert _{L^4(\Sigma )}\, \Vert \nabla u\Vert _{L^{4}(\Sigma )} \le C\Vert u\Vert _{H^2(\Omega )}. \end{aligned}$$

We use Remark 3.4 for \(F = \nabla (\nabla u \cdot {\tilde{\nu }})\) and (3.11) and have

$$\begin{aligned} \begin{aligned} \Vert \partial _\nu (\nabla u \cdot {\tilde{\nu }})\Vert _{L^2(\Sigma )}^2&\le C(\Vert \nabla _\tau (\nabla u \cdot {\tilde{\nu }})\Vert _{L^2(\Sigma )}^2 + \Vert \nabla (\nabla u \cdot {\tilde{\nu }}) \, \Delta (\nabla u \cdot {\tilde{\nu }})\Vert _{L^1(\Omega )} + \Vert \nabla u \cdot {\tilde{\nu }}\Vert _{H^1(\Omega )}^2)\\&\le C(\Vert \partial _\nu u \Vert _{H^1(\Sigma )}^2 + \Vert u\Vert _{H^2(\Omega )}^2 + \Vert f\Vert _{H^1(\Omega )}^2+ \Vert \nabla u \cdot {\tilde{\nu }}\Vert _{H^1(\Omega )}^2)\\&\,\,\,\,\,\,\,+ C(\Vert \nabla ^2 u \star \nabla ^2 u \star \nabla {\tilde{\nu }}\Vert _{L^1(\Omega )} + \Vert \nabla ^2 u \star \nabla u \star \nabla {\tilde{\nu }} \star \nabla {\tilde{\nu }}\Vert _{L^1(\Omega )}). \end{aligned} \end{aligned}$$

First, we obtain by using the previous estimates

$$\begin{aligned} \Vert \nabla u \cdot {\tilde{\nu }}\Vert _{H^1(\Omega )}^2 \le C \Vert u \Vert _{H^2(\Omega )}^2. \end{aligned}$$

We bound the second last term by Hölder’s inequality and by the Sobolev embedding

$$\begin{aligned} \Vert \nabla ^2 u \star \nabla ^2 u \star \nabla {\tilde{\nu }}\Vert _{L^1(\Omega )} \le C\Vert \nabla {\tilde{\nu }}\Vert _{L^4(\Omega )}\Vert \nabla ^2 u\Vert _{L^{\frac{8}{3}}(\Omega )}^2 \le \varepsilon \Vert \nabla u\Vert _{H^{\frac{3}{2}}(\Omega )}^2 + C_{\varepsilon }\Vert u\Vert _{H^2(\Omega )}^2. \end{aligned}$$

Similarly we estimate the last term

$$\begin{aligned} \begin{aligned} \Vert \nabla ^2 u \star \nabla u \star \nabla {\tilde{\nu }} \star \nabla {\tilde{\nu }}\Vert _{L^1(\Omega )}&\le C\Vert \nabla {\tilde{\nu }}\Vert _{L^4(\Omega )}^2 \Vert \nabla ^2 u\Vert _{L^{\frac{8}{3}}(\Omega )} \Vert \nabla u\Vert _{L^8(\Omega )}\\&\le \varepsilon \Vert \nabla u\Vert _{H^{\frac{3}{2}}(\Omega )}^2 + C_{\varepsilon }\Vert u\Vert _{H^2(\Omega )}^2. \end{aligned} \end{aligned}$$

Therefore we have

$$\begin{aligned} \Vert \partial _\nu (\nabla u)\Vert _{L^2(\Sigma )} \le \varepsilon \Vert \nabla u \Vert _{H^{\frac{3}{2}}(\Omega )} + C_\varepsilon (\Vert \partial _\nu u \Vert _{H^1(\Sigma )}+ \Vert u\Vert _{H^2(\Omega )} +\Vert f\Vert _{H^1(\Omega )}). \end{aligned}$$

Recall that we have

$$\begin{aligned} \Vert u\Vert _{H^2(\Omega )} \le C(\Vert g\Vert _{H^{\frac{1}{2}}(\Sigma )} + \Vert f\Vert _{L^2(\Omega )}). \end{aligned}$$

Therefore the third inequality follows from (3.17).

For the last inequality we recall that while the Trace operator is not bounded \(T: H^{\frac{1}{2}}(\mathbb {R}^3) \rightarrow L^{2}(\mathbb {R}^2)\) it is bounded as \(T: H^{\frac{3}{2}}(\mathbb {R}^3) \rightarrow H^{1}(\mathbb {R}^2)\). We prove the statement by localization argument similar to the one in the proof of Proposition 2.10 and we only give the sketch of the proof.

We cover \(\Sigma \) with balls of radius \(\delta \), \(B_{\delta }(x_i), i = 1, \dots , N\) such that the set \(\Sigma \cap B_{2\delta }(x_i)\) is contained in the graph of \(\phi _i\) and \(\Omega \) is above the graph. We denote the partition of unity by \(\eta _i\). Let us fix i and we may assume that \(x_i = 0\) and \(\phi _i(0) = \nabla \phi _i(0) = 0\). By the regularity assumptions it holds \(\Vert \phi _i\Vert _{C^{1,\alpha }(\mathbb {R}^2)}, \Vert \phi _i\Vert _{W^{2,4}(\mathbb {R}^2)} \le C\). We define \(u_i(x) = \eta _i(x)u_i(x)\) and \(v_i(x',x_3)=u_i(x', x_3+\phi _i(x'))\) for \(x_3 \ge 0\) and extend \(v_i\) to \(\mathbb {R}^3\) by the extension operator. Then we have by the Trace Theorem

$$\begin{aligned} \Vert u_i\Vert _{H^1(\Sigma \cap B_\delta )} \le C\Vert v_i\Vert _{H^1(\mathbb {R}^2)} \le C \Vert v_i\Vert _{H^{\frac{3}{2}}(\mathbb {R}^3)}. \end{aligned}$$

Recall that the assumption \(\Vert B_\Sigma \Vert _{L^4} \le C\) guarantees that \(\Omega \) is an \(H^2\)-extension domain. Therefore it holds \(\Vert v_i\Vert _{H^{\frac{3}{2}}(\mathbb {R}^3)} \le C\Vert u\Vert _{H^{\frac{3}{2}}(\Omega )}\) and the last inequality follows. \(\square \)

3.2 Regularity Estimates for Functions

In this subsection we prove regularity estimates for functions \(u: \Omega \rightarrow \mathbb {R}\) defined as a solution of the Dirichlet problem

$$\begin{aligned} {\left\{ \begin{array}{ll} \Delta u= f \qquad &{} x\in \Omega \\ u=g \qquad &{}x\in \Sigma \end{array}\right. } \end{aligned}$$
(3.18)

We first consider the case when \(g = 0\) and improve in this case the third inequality in Lemma 3.7. Here we assume that the boundary has the regularity \(\Vert B_\Sigma \Vert _{H^{\frac{1}{2}}(\Sigma )} \le C\). Note that by the Sobolev embedding this implies \(\Vert B_\Sigma \Vert _{L^{4}(\Sigma )} \le C\).

Proposition 3.8

Assume \(\Omega \), with \(\Sigma = \partial \Omega \), is uniformly \(C^{1,\alpha }(\Gamma )\)-regular and \(\Vert B_\Sigma \Vert _{H^{\frac{1}{2}}(\Sigma )} \le M\). There exists a constant C, depending on M and the \(C^{1,\alpha }\)-norm of the heightfunction, such that the solution of the problem (3.18) with zero Dircihlet boundary datum, i.e., \(g = 0\) satisfies

$$\begin{aligned} \Vert \partial _\nu u\Vert _{H^1(\Sigma )} +\Vert \nabla u\Vert _{H^{\frac{3}{2}}(\Omega )} \le C\Vert f\Vert _{H^\frac{1}{2}(\Omega )}. \end{aligned}$$

Proof

First we note that since \(u = 0\) on \(\Sigma \) then by (3.2) we have

$$\begin{aligned} \Vert \nabla ^2 u\Vert _{L^2(\Omega )}^2 =\Vert f\Vert _{L^2(\Omega )}^2 - \int _{\Sigma } H_\Sigma |\partial _\nu u|^2\, d \mathcal {H}^2. \end{aligned}$$

By (3.6) it holds

$$\begin{aligned} -\int _{\Sigma } H_\Sigma |\partial _\nu u|^2\, d \mathcal {H}^2 \le \varepsilon \Vert u\Vert _{H^2(\Omega )}^2 + C_\varepsilon \Vert \nabla u\Vert _{L^2(\Sigma )}^2. \end{aligned}$$

We apply Lemma 3.3 for \(F = \nabla u\) and recall that \(u=0\) on \(\Sigma \) to deduce

$$\begin{aligned} \begin{aligned} \Vert \nabla u\Vert _{L^2(\Sigma )}&\le C(\Vert \nabla _\tau u\Vert _{L^2(\Sigma )} + \Vert u\Vert _{H^1(\Omega )} + \Vert f\Vert _{L^2(\Omega )})\\&\le \varepsilon \Vert u\Vert _{H^2(\Omega )} + C_\varepsilon (\Vert u\Vert _{L^2(\Omega )} + \Vert f\Vert _{L^2(\Omega )}). \end{aligned} \end{aligned}$$

Therefore we have

$$\begin{aligned} \Vert \nabla ^2 u\Vert _{L^2(\Omega )}^2 \le C(\Vert u\Vert _{L^2(\Omega )} + \Vert f\Vert _{L^2(\Omega )}). \end{aligned}$$

We bound \(\Vert u\Vert _{L^2(\Omega )}\) simply by multiplying the equation (3.18) by u and integrating by parts \(\Vert \nabla u\Vert _{L^2(\Omega )}^2 \le \Vert f\Vert _{L^2(\Omega )}\Vert u\Vert _{L^2(\Omega )}\). Poincaré inequality then implies \(\Vert u\Vert _{L^2(\Omega )} \le C \Vert f\Vert _{L^2(\Omega )}\) and we have

$$\begin{aligned} \Vert u\Vert _{H^2(\Omega )} \le C \Vert f\Vert _{L^2(\Omega )}. \end{aligned}$$
(3.19)

Let \({\tilde{\nu }}\) be the harmonic extension of the normal field and let us define \(\tau _i = e_i - \langle e_i,{\tilde{\nu }} \rangle {\tilde{\nu }}\) as in the proof of Lemma 3.7. Define \(u_i = \nabla u \cdot \tau _i\). Observe that \(u_i = 0\) on \(\Sigma \) and apply (3.19) to deduce

$$\begin{aligned} \Vert \nabla ^2 u_i\Vert _{L^2(\Omega )} \le C \Vert \Delta u_i \Vert _{L^2(\Omega )}. \end{aligned}$$
(3.20)

We have (recall \(\Delta u = f\))

$$\begin{aligned} \Delta u_i = \nabla f \star \tau _i + \nabla ^2 u \star \nabla {\tilde{\nu }} + \nabla u \star \nabla {\tilde{\nu }}\star \nabla {\tilde{\nu }}. \end{aligned}$$

Arguing similarly as in the proof of Lemma 3.5 and using (3.19) yields

$$\begin{aligned} \Vert \Delta u_i \Vert _{L^2(\Omega )} \le \varepsilon \Vert u\Vert _{H^3(\Omega )} +C\Vert f\Vert _{H^1(\Omega )} \end{aligned}$$
(3.21)

Let us then treat the LHS of (3.20). We have (recall that \(\tau _i = e_i - \langle e_i, {\tilde{\nu }} \rangle {\tilde{\nu }}\))

$$\begin{aligned} \nabla _{j}\nabla _{k} u_i = \nabla (\nabla _{j}\nabla _{k} u) \cdot \tau _i + \nabla ^2 u \star \nabla {\tilde{\nu }} + \nabla u \star \nabla {\tilde{\nu }} \star \nabla {\tilde{\nu }} + \nabla u \star \nabla ^2 {\tilde{\nu }}. \end{aligned}$$

Therefore arguing as in the proof of Lemma 3.5, we obtain

$$\begin{aligned} \begin{aligned} \Vert \nabla ^2 u_i\Vert _{H^2(\Omega )} \ge \sum _{i,j,k =1}^3&\Vert \nabla (\nabla _{j}\nabla _{k} u) \cdot \tau _i \Vert _{L^2(\Omega )} \\&- \varepsilon \Vert u\Vert _{H^3(\Omega )} - C_\varepsilon \Vert f\Vert _{H^1(\Omega )} - C\Vert \nabla u\Vert _{L^\infty (\Omega )}\Vert \nabla ^2 {\tilde{\nu }}\Vert _{L^2(\Omega )}. \end{aligned} \end{aligned}$$
(3.22)

Let us fix a point \(x \in \Omega \) and as in the proof of Lemma 3.7 we may assume that \({\tilde{\nu }}(x) \cdot e_i =0\) for \(i =1,2\). Then it is easy to see that

$$\begin{aligned} \begin{aligned} \sum _{i,j,k =1}^3 |\nabla (\nabla _{j}\nabla _{k} u(x)) \cdot \tau _i |^2&\ge \sum _{i=1}^2\sum _{j,k =1}^3 |\langle \nabla \nabla _j \nabla _k u(x), \tau _i \rangle |^2\\&\ge c\sum _{i,j,k =1}^3 | \nabla _i \nabla _j \nabla _k u(x)|^2 - C |\nabla \Delta u(x)|^2. \end{aligned} \end{aligned}$$

This together with (3.22) yields

$$\begin{aligned} \Vert u_i\Vert _{H^2(\Omega )} \ge c \Vert \nabla ^3 u\Vert _{L^2(\Omega )} - C \Vert f\Vert _{H^1(\Omega )}- C\Vert \nabla u\Vert _{L^\infty }\Vert \nabla ^2 {\tilde{\nu }}\Vert _{L^2(\Omega )}. \end{aligned}$$
(3.23)

We proceed by recalling that \({\tilde{\nu }}\) is the harmonic extension of \(\nu \). We claim that it holds

$$\begin{aligned} \Vert \nabla ^2 {\tilde{\nu }}\Vert _{L^2(\Omega )} \le C. \end{aligned}$$
(3.24)

Indeed, this follows from already familiar argument and we only give its outline. Define \(\tau _i = e_i - \langle e_i,{\tilde{\nu }} \rangle {\tilde{\nu }}\) as in the proof of Lemma 3.7 and let \(u_{ij} = \langle \nabla {\tilde{\nu }} \, \tau _i \tau _j \rangle \). Then it holds \(u_{ij} = \langle B_\Sigma \tau _i, \tau _j \rangle \) on \(\Sigma \) and therefore by the assumptions it holds \(\Vert u_{ij}\Vert _{H^{\frac{1}{2}}(\Sigma )} \le C\). Arguing as in the proof of Lemma 3.5 we deduce

$$\begin{aligned} \Vert \nabla u_{ij}\Vert _{L^2(\Omega )}^2 \le \Vert u_{ij}\Vert _{H^{\frac{1}{2}}(\Sigma )}^2 + \varepsilon \Vert \nabla ^2 {\tilde{\nu }}\Vert _{L^2(\Omega )}^2 +C_\varepsilon . \end{aligned}$$

By applying this to every \(i,j, =1, 2,3\) and arguing as above we obtain (3.24).

We have by interpolation inequality in Corollary 2.9

$$\begin{aligned} \Vert \nabla u\Vert _{L^\infty (\Omega )} \le C \Vert \nabla ^2 u\Vert _{L^{4}(\Omega )} \le C\Vert \nabla ^3 u\Vert _{L^2(\Omega )}^{\frac{3}{4}}\Vert \nabla ^2 u\Vert _{L^2(\Omega )}^{\frac{1}{4}}. \end{aligned}$$

Therefore by Young’s inequality and by (3.19)

$$\begin{aligned} \begin{aligned} \Vert \nabla u\Vert _{L^\infty (\Omega )}\Vert \nabla ^2 {\tilde{\nu }}\Vert _{L^2(\Omega )}&\le \varepsilon \Vert \nabla ^3 u\Vert _{L^2(\Omega )} +C_\varepsilon \Vert \nabla ^2 u\Vert _{L^2(\Omega )}\\&\le \varepsilon \Vert \nabla ^3 u\Vert _{L^2(\Omega )}+ C_\varepsilon \Vert f\Vert _{L^2(\Omega )}. \end{aligned} \end{aligned}$$

Hence, (3.20), (3.21) and (3.23) imply

$$\begin{aligned} \Vert u\Vert _{H^3(\Omega )} \le C \Vert f\Vert _{H^1(\Omega )}. \end{aligned}$$
(3.25)

We set \(\mathcal F\) to be the linear operator such that it associates f with the unique solution u of the problem (3.18). Then we have by (3.19) and (3.25)

$$\begin{aligned} \Vert {\mathcal {F}}\Vert _\mathcal {L(L^2,H^2)} \le C \qquad \text {and} \qquad \Vert {\mathcal {F}}\Vert _\mathcal {L(H^1,H^3)} \le C. \end{aligned}$$

Then we have the inequality

$$\begin{aligned} \Vert \nabla u\Vert _{H^{\frac{3}{2}}(\Omega )} \le C \Vert f\Vert _{H^{\frac{1}{2}}(\Omega )} \end{aligned}$$
(3.26)

by standard interpolation theory.

We need yet to bound \(\Vert \partial _\nu u\Vert _{H^1(\Sigma )}\). To this aim we extend \(\nabla u\) to \(\mathbb {R}^3\) by T such that

$$\begin{aligned} \Vert T(\nabla u)\Vert _{H^{\frac{3}{2}}(\mathbb {R}^3)} \le C \Vert \nabla u\Vert _{H^{\frac{3}{2}}(\Omega )}. \end{aligned}$$

Let us denote \(U = T(\nabla u)\). Let \({\tilde{\nu }}\) be the Harmonic extension of \(\nu \) as before, which we may also extend to \(\mathbb {R}^3\). We note that we may assume that the extensions have support in \(B_R\). We have by Lemma 3.7

$$\begin{aligned} \Vert \nabla u \cdot \nu \Vert _{H^1(\Sigma )} \le C\Vert U \cdot {\tilde{\nu }}\Vert _{H^{\frac{3}{2}}(\Omega )} \le C \Vert U \cdot {\tilde{\nu }}\Vert _{H^{\frac{3}{2}}(\mathbb {R}^3)}. \end{aligned}$$

The Kato-Ponce inequality (2.15) with \(p_2 =8, q_2=8/3\) yields

$$\begin{aligned} \Vert U \cdot {\tilde{\nu }}\Vert _{H^{\frac{3}{2}}(\mathbb {R}^3)} \le C \Vert U \Vert _{H^{\frac{3}{2}}(\mathbb {R}^3)} \Vert {\tilde{\nu }} \Vert _{L^\infty (\mathbb {R}^3)}+ C \Vert U \Vert _{L^{8}(\mathbb {R}^3)}\Vert {\tilde{\nu }}\Vert _{W^{\frac{3}{2},\frac{8}{3}}(\mathbb {R}^3)}. \end{aligned}$$

We have \(\Vert {\tilde{\nu }} \Vert _{L^\infty (\mathbb {R}^3)}\le C\) and by the Sobolev embedding \(\Vert U \Vert _{L^{8}(\mathbb {R}^3)} \le C\Vert U \Vert _{H^{\frac{3}{2}}(\mathbb {R}^3)}\). We use (2.13) to deduce that

$$\begin{aligned} \Vert \nabla {\tilde{\nu }}\Vert _{W^{\frac{1}{2},\frac{8}{3}}(\mathbb {R}^3)} \le C \Vert \nabla {\tilde{\nu }}\Vert _{H^{1}(\mathbb {R}^3)}^{\frac{1}{2}} \Vert \nabla {\tilde{\nu }}\Vert _{L^{4}(\mathbb {R}^3)}^{\frac{1}{2}}. \end{aligned}$$

By (3.12) we have

$$\begin{aligned} \Vert \nabla {\tilde{\nu }}\Vert _{L^{4}(\mathbb {R}^3)} \le \Vert {\tilde{\nabla }} \nu \Vert _{L^{4}(\Omega )} \le C \end{aligned}$$

and by (3.24)

$$\begin{aligned} \Vert \nabla {\tilde{\nu }}\Vert _{H^{1}(\mathbb {R}^3)} \le C \Vert \nabla ^2 {\tilde{\nu }}\Vert _{L^{2}(\Omega )} \le C. \end{aligned}$$

Therefore by combining the previous inequalities we have

$$\begin{aligned} \Vert \nabla u \cdot \nu \Vert _{H^1(\Sigma )} \le C \Vert U \cdot {\tilde{\nu }}\Vert _{H^{\frac{3}{2}}(\mathbb {R}^3)} \le C\Vert U \Vert _{H^{\frac{3}{2}}(\mathbb {R}^3)} \le C\Vert \nabla u \Vert _{H^{\frac{3}{2}}(\Omega )}. \end{aligned}$$

The result then follows from (3.26). \(\square \)

We conclude this section by proving the sharp boundary regularity estimate for the Dirichlet problem. The proof follows the argument in [22, Theoreom 4.1], with the difference that here we have Dirichlet boundary datum, instead of the zero Neumann case.

Theorem 3.9

Assume \(\Omega \), with \(\Sigma = \partial \Omega \), is uniformly \(C^{1,\alpha }(\Gamma )\)-regular and satisfies (\(\hbox {H}_{m}\)) for \(m \ge 2\). Let \(u \in {\dot{H}}^1(\Omega ^c)\) be the solution of

$$\begin{aligned} {\left\{ \begin{array}{ll} \Delta u = 0 \qquad &{}x\in \Omega ^c \\ u=g &{}x\in \Sigma . \end{array}\right. } \end{aligned}$$
(3.27)

Then for all integers \(0\le k \le m-1\) it holds

$$\begin{aligned} \Vert \nabla ^k u\Vert _{H^\frac{1}{2}(\Sigma )} \le C (1+ \Vert B_{\Sigma }\Vert _{H^{k-1}(\Sigma )} +\Vert g \Vert _{H^{k+\frac{1}{2}}(\Sigma )}) \end{aligned}$$
(3.28)

for some constant C, depending on m and on the \(C^{1,\alpha }\)-norm of the heightfunction. Moreover, if g is constant then the above holds for all \(k \in \mathbb {N}\).

Proof

Step 1: Flattening the boundary. Since \(\Sigma \) is \(C^{1,\alpha }(\Gamma )\), for any \(x\in \Sigma \) we find \(\delta >0\) such that after rotating and translating the coordinates

$$\begin{aligned} \Omega ^c\cap B_\delta = \{(x', x_3):\, x_3>\phi (x') \} \end{aligned}$$

with \(\phi \in C^{1,\alpha }(B_\delta ) \), \(\phi (0)=0\) and \(\nabla \phi (0)= 0\). Consider the diffeomorfism \(\Psi :\Omega ^c\cap B_\delta \rightarrow B_\delta ^+\) \(\Psi (x',x_3)\rightarrow (x',x_3-\phi (x'))\) and let \(v:= u\circ \Psi ^{-1}\) and \(w:= g\circ \Psi ^{-1}\). Let us extend g by its harmonic extension, denote it by \({\tilde{g}}\), and thus \(w = {\tilde{g}}\circ \Psi ^{-1}\) is defined in \(B_{\delta }^+\). By standard calculations we deduce that v is the solution of

$$\begin{aligned} {\left\{ \begin{array}{ll} \textrm{div}\,(A_\phi \nabla v) = 0 \qquad &{}x\in B^+_\delta \\ v=w &{}x_3=0, \end{array}\right. } \end{aligned}$$
(3.29)

where \(A_\phi \) is symmetric matrix which can be written as \(A_\phi = I + {\tilde{A}}(\nabla \phi )\) where \({\tilde{A}}(\nabla \phi (x)) = 0\) if \(\nabla \phi (x)= 0\). In particular, by choosing \(\delta \) small enough \(A_\phi \) is postitive definite. In weak form (3.29) reads as

$$\begin{aligned} \int _{B_\delta ^+} A_\phi \nabla v \cdot \nabla \varphi \, dx= 0 \end{aligned}$$

for all \(\varphi \in C_0^\infty (B_\delta ^+)\).

Let k be an integer as in the statement. Let us differentiate the equation (3.29) k times in tangential directions. To this aim let us fix an index vector \(\gamma =(\gamma _1,\gamma _2,0)\) with \(\gamma _1+\gamma _2=k \), and denote \({\bar{v}} = \nabla ^\gamma v\) and \({\bar{w}}= \nabla ^\gamma w\). Then \({\bar{v}}\) is the solution of

$$\begin{aligned} {\left\{ \begin{array}{ll} \textrm{div}\,(A_\phi \nabla {\bar{v}}) = -{\sum }_{{\tilde{\alpha }},\beta } \textrm{div}\,(\nabla ^{{\tilde{\alpha }}} A_\phi \nabla \nabla ^{\beta } v) \qquad &{}x\in B^+_\delta \\ {\bar{v}}={\bar{w}} &{}x_3=0 . \end{array}\right. } \end{aligned}$$
(3.30)

with \({\tilde{\alpha }}=({\tilde{\alpha }}_1,{\tilde{\alpha }}_2,0)\), \(\beta =(\beta _1,\beta _2,0)\), \(|\beta |\le k-1\) and \(|{\tilde{\alpha }}|+|\beta |\le k\). In the weak form this reads as

$$\begin{aligned} \int _{B_\delta ^+} A_\phi \nabla {\bar{v}} \cdot \nabla \varphi \, dx= - \sum _{{\tilde{\alpha }},\beta } \int _{B_\delta ^+} (\nabla ^{{\tilde{\alpha }}} A_\phi \nabla \nabla ^\beta v ) \cdot \nabla \varphi \, dx \end{aligned}$$

for all \(\varphi \in C_0^\infty (B_\delta ^+)\).

Step 2: Choice of the test function that has zero boundary value Let \(\zeta \in C_0^\infty (B_\delta ^+)\) be a smooth cut-off function such that \(\zeta (x) =1\) for \(|x|\le \frac{\delta }{2}\) and \(0\le \zeta \le 1\). We choose a test function \(\varphi =({\bar{v}}- {\bar{w}} )\zeta ^2\), which has zero boundary value. With this choice we have

$$\begin{aligned} \begin{aligned}&\int _{B_\delta ^+} (A_\phi \nabla {\bar{v}} \cdot \nabla {\bar{v}} )\,\zeta ^2 dx=\int _{B_\delta ^+} ( A_\phi \nabla {\bar{v}} \cdot \nabla {\bar{w}} ) \zeta ^2 \, dx + \int _{B_\delta ^+} ( A_\phi \nabla {\bar{v}} \cdot \nabla \zeta )\, ({\bar{w}} -{\bar{v}}) \zeta dx\\&- \sum _{{\tilde{\alpha }},\beta }\int _{B_\delta ^+} (\nabla ^{{\tilde{\alpha }}} A_\phi \nabla \nabla ^\beta v \cdot \nabla ({\bar{v}}- {\bar{w}} )) \zeta ^2\, dx -2\sum _{{\tilde{\alpha }},\beta }\int _{B_\delta ^+} (\nabla ^{{\tilde{\alpha }}} A_\phi \nabla \nabla ^\beta v \cdot \nabla \zeta ) ({\bar{v}}- {\bar{w}} )\zeta \, dx\\&=I_1+I_2+I_3+I_4. \end{aligned} \end{aligned}$$

By the assumption \(\phi \in C^{1,\alpha }\) it holds \(\Vert A_\phi \Vert _{L^\infty } \le C\). Thus we may bound the first two terms as

$$\begin{aligned} I_1+I_2\le C\Vert \nabla {\bar{v}}\Vert _{L^2(B_\delta ^+)}( \Vert \nabla {\bar{w}}\Vert _{L^2(B_\delta ^+)} +\Vert {\bar{v}} - {\bar{w}}\Vert _{L^2(B^+_\delta )}). \end{aligned}$$

The term \(I_3\) is more difficult to treat. Note first since \(A_\phi \) is of the form \(I + {\tilde{A}}(\nabla \phi )\) we have a point-wise bound by the Leibniz rule

$$\begin{aligned} \sum _{{\tilde{\alpha }},\beta }|\nabla ^{{\tilde{\alpha }}} A_\phi |^2||\nabla \nabla ^\beta v|^2 \le C\sum _{\begin{array}{c} |\alpha | +|\beta |\le k\\ |\beta |\le k-1 \end{array}}(1+ |\nabla ^{\alpha _1} \nabla \phi |^2\dots |\nabla ^{\alpha _{k}}\nabla \phi |^2 ) |\nabla \nabla ^{\beta } v|^2. \end{aligned}$$

Hence, we obtain by Hölder’s inequality

$$\begin{aligned} \begin{aligned} I_3 \le&C\Vert \nabla ({\bar{v}}-{\bar{w}})\Vert _{L^2(B^+_\delta )} \\&\cdot \sum _{\begin{array}{c} |\alpha | +|\beta |\le k\\ |\beta |\le k-1 \end{array}}\Big (1+ \Vert \nabla ^{\alpha _1} \nabla \phi \Vert _{L^{\frac{2k}{\alpha _1}}(B_\delta ^+)}\dots \Vert \nabla ^{\alpha _{k}}\nabla \phi \Vert _{L^{\frac{2k}{ \alpha _k}}(B^+_\delta )} \Big )\Vert \nabla \nabla ^{\beta } v\Vert _{L^{\frac{2k}{|\beta |}}(B^+_\delta )}. \end{aligned} \end{aligned}$$

We use interpolation inequality to estimate

$$\begin{aligned} \Vert \nabla ^{\alpha _i} \nabla \phi \Vert _{L^{\frac{2k}{\alpha _1}}(B_\delta ^+)} \le \Vert \nabla \phi \Vert _{H^{k}(B_\delta ^+)}^{\frac{\alpha _i}{k} } \Vert \nabla \phi \Vert ^{1-\frac{\alpha _i}{k}}_{L^\infty (B_\delta ^+)}. \end{aligned}$$

Also by interpolation we have

$$\begin{aligned} \Vert \nabla \nabla ^\beta v \Vert _{L^{\frac{2k}{|\beta |}} (B_\delta ^+)}\le \Vert v\Vert _{H^{k+1}(B_\delta ^+)}^{\frac{|\beta |}{k}}\Vert \nabla v\Vert _{L^\infty (B_\delta ^+)}^{1-\frac{|\beta |}{k}} \end{aligned}$$

and \(|\beta |\le k-1\). Since \(\Sigma \) is \(C^{1,\alpha }\)-regular, we have by Schauder estimates [25] that \(\nabla v \in C^{0,\alpha }(B_\delta ^+)\). Note that \(\sum _i \frac{\alpha _i}{k} \le \frac{k-|\beta |}{k} \) and \(\frac{|\beta |}{k} <1\). Therefore by Young’s inequality we deduce

$$\begin{aligned} \begin{aligned} |I_3|\le&C \Vert \nabla ({\bar{v}}-{\bar{w}})\Vert _{L^2(B^+_\delta )}( 1+ \Vert \nabla \phi \Vert _{H^k(B_\delta ^+)}^{\frac{k-|\beta |}{k}}) \Vert v\Vert _{H^{k+1}(B_\delta ^+)}^{\frac{|\beta |}{k}}\\ \le&\varepsilon \Vert \nabla ({\bar{v}}-{\bar{w}})\Vert _{L^2(B^+_\delta )}^2 + \varepsilon \Vert v\Vert _{H^{k+1}(B_\delta ^+)}^2 + C_\varepsilon (1+ \Vert \nabla \phi \Vert _{H^k(B_\delta ^+)}^2). \end{aligned} \end{aligned}$$

We bound the last term \(I_4\) similarly.

Finally we collect the previous estimates, use the ellipticity of the matrix \(A_\phi \) and the definition of \({\bar{w}}\) and obtain

$$\begin{aligned} \Vert \nabla {\bar{v}}\Vert _{L^2(B^+_{\delta /2})}^2 \le 4\varepsilon \Vert v\Vert _{H^{k+1}(B_\delta ^+)}^2 + C(1+ \Vert \phi \Vert _{H^{k+1}(B_\delta ^+)}^2 + \Vert w\Vert _{H^{k+1}(B_\delta ^+)}^2). \end{aligned}$$

Summing over all the multi index of the type \((\gamma _1,\gamma _2)\) we have the control over the horizontal derivatives. To estimate the vertical derivatives, we use the equation in the strong form as in [22], and obtain

$$\begin{aligned} \Vert v\Vert _{H^{k+1}(B^+_{\delta /2})}^2 \le C\varepsilon \Vert v\Vert _{H^{k+1}(B_\delta ^+)}^2 + C(1+ \Vert \phi \Vert _{H^{k+1}(B_\delta ^+)}^2 + \Vert w\Vert _{H^{k+1}(B_\delta ^+)}^2). \end{aligned}$$
(3.31)

Step 3: Going back to the original function. We need to go back to the original function u. The argument is similar to [22] and we merely sketch it. We note that arguing as in [22, Thorem 4.1] we may control

$$\begin{aligned} \Vert \phi \Vert _{H^{k+1}(B_\delta ^+)} \le C(1+ \Vert B_{\Sigma }\Vert _{H^{k-1}(\Sigma )}) \end{aligned}$$

for all \(k \in \mathbb {N}\). Recall that \({\tilde{g}}\) is the harmonic extension of g. Using the assumption that the curvature satisfies the condition (\(\hbox {H}_{m}\)) for m, we may deduce, arguing as in the proof of Proposition 2.1, that for \(k \le m-1\) it holds

$$\begin{aligned} \Vert w\Vert _{H^{k+1}(B_\delta ^+)} \le C\Vert {\tilde{g}}\Vert _{H^{k+1}(\Omega ^c \cap B_{\delta })} \le C \Vert g\Vert _{H^{k+\frac{1}{2}}(\Sigma )}. \end{aligned}$$

Obviously if g is constant the above inequality is trivial.

Fix \(\sigma \) small such that \(\cup _{x\in \Sigma }B_\delta (x)\) covers \(\mathcal N_\delta =\{x\in \Omega ^c: d(x,\Omega )\le \delta \}\) and \(\sigma _1<\sigma _2<\sigma \). By compactness we may choose a finite family of balls covering \(\mathcal N_\delta \). Choosing \(\varepsilon \) small enough we have by (3.31) and by the above inequalities

$$\begin{aligned} \Vert u \Vert _{H^{k+1}(\mathcal N_{\sigma _1})}^2 \le C (\Vert u \Vert _{H^{k+1}(\mathcal N_{\sigma _2}\setminus \mathcal N_{\sigma _1})}^2 + 1+ \Vert B_{\Sigma }\Vert ^2_{H^{k-1}(\Sigma )} +\Vert g \Vert ^2_{H^{k+\frac{1}{2}}(\Sigma )}). \end{aligned}$$

Since u is harmonic, the interior regularity yields

$$\begin{aligned} \Vert u \Vert _{H^{k+1}(\mathcal N_{\sigma _2}\setminus \mathcal N_{\sigma _1})}\le C \Vert u\Vert _{L^2(\mathcal N_{\sigma _2})}. \end{aligned}$$

By standard estimates from harmonic analysis [16] it holds for R large

$$\begin{aligned} \Vert u\Vert _{L^2(N_{{\sigma }_2})} \le \Vert u\Vert _{L^2(\Omega ^c\cap B_R)} \le C(\Vert u\Vert _{L^2(\Sigma )} + \Vert u\Vert _{L^2(\partial B_R)} ) \le C(1 + \Vert g\Vert _{L^2(\Sigma )} ). \end{aligned}$$

Therefore we have

$$\begin{aligned} \Vert u \Vert _{H^{k+1}(\mathcal N_{\sigma _1})} \le C \left( 1+ \Vert B_{\Sigma }\Vert _{H^{k-1}(\Sigma )} +\Vert g \Vert _{H^{k+\frac{1}{2}}(\Sigma )}\right) . \end{aligned}$$

The claim follows from

$$\begin{aligned} \Vert \nabla ^k u\Vert _{H^\frac{1}{2}(\Sigma )} \le C \Vert \nabla ^{k+1}u \Vert ^2_{L^2(\mathcal N_{\sigma _1})}. \end{aligned}$$

\(\square \)

4 Useful Formulas

In this section we focus on the equations (1.3) and assume that the family of sets \((\Omega _t)_{t \in (0,T)}\) and velocities \(v(\cdot ,t)\) are solution of (1.3). We derive a general formula for the commutators of the material derivative of high order \(\mathcal {D}_t^k\) with spatial derivatives. We apply this to calculate \([\mathcal {D}_t^k,\nabla ] v\) and \([\mathcal {D}_t^k,\nabla ] p\), which will produce two types of error terms, (4.13) and (4.14), defined in the fluid domain \(\Omega _t\). We will also calculate the formula for \(\mathcal {D}_t^k p\) on the moving boundary \(\Sigma _t\) in Lemma 4.7, which includes third type of error term defined in (4.26). The precise structures of these error terms are complicated and we only need to trace the order of the derivatives that appear. Therefore we effectively use the notation from [32]

$$\begin{aligned} \nabla ^k f \star \nabla ^l g \end{aligned}$$

to denote a contraction of some indexes of tensors \(\nabla ^i f\) and \(\nabla ^j g\) for \(i \le k\) and \(j \le l\). Note that we include the lower order derivatives.

We begin by recalling the following formulas from [50]

$$\begin{aligned}{} & {} {[}\mathcal {D}_t,\nabla ] f = \mathcal {D}_t \nabla f - \nabla \mathcal {D}_t f = - \nabla v^T\nabla f, \end{aligned}$$
(4.1)
$$\begin{aligned}{} & {} {[}\mathcal {D}_t, \nabla _\tau ] f = - (\nabla _\tau v)^T \nabla _\tau f \end{aligned}$$
(4.2)

and

$$\begin{aligned} {[}\mathcal {D}_t, \Delta _\Sigma ] f = \nabla _\tau ^2 f \star \nabla v - \nabla _\tau f \cdot \Delta _\Sigma v + B \star \nabla v \star \nabla _\tau f. \end{aligned}$$
(4.3)

Let us also recall the material derivative of the normal field. We use the shorthand notation \(\nu = \nu _{\Sigma _t}\), \(B= B_{\Sigma _t}\) and \(v_n = v \cdot \nu \). We have by [50]

$$\begin{aligned} \mathcal {D}_t \nu = - (\nabla _\tau v)^T \nu . \end{aligned}$$
(4.4)

Since \(\nabla _\tau v_n = \nabla _\tau v^T \nu + B v_\tau \), we may write (4.4) as

$$\begin{aligned} \mathcal {D}_t \nu = - \nabla _\tau v_n+ B v_\tau . \end{aligned}$$
(4.5)

We need higher order versions of the commutation formula (4.1), i.e., for

$$\begin{aligned} {[}\mathcal {D}_t^l, \nabla ^k] f = \mathcal {D}_t^l \nabla ^k f -\nabla ^k \mathcal {D}_t^l f. \end{aligned}$$

Recall the definition of the norm of an index vector \(\alpha = (\alpha )_{i=1}^k \in \mathbb {N}^k\)

$$\begin{aligned} |\alpha | = \sum _{i =1}^k \alpha _i \end{aligned}$$

and note that we include zero in the set of natural numbers \(\mathbb {N}\).

Lemma 4.1

For \(l, k \in \mathbb {N}\) with \(l,k \ge 1\) it holds

$$\begin{aligned} {[}\mathcal {D}_t^l, \nabla ^k] f = \sum _{\begin{array}{c} |\alpha | \le k-1 \\ |\beta | \le l-1 \end{array}} \nabla ^{1+\alpha _1} \mathcal {D}_t^{\beta _1} v \star \cdots \star \nabla ^{1+\alpha _l} \mathcal {D}_t^{\beta _l} v \star \nabla ^{1+\alpha _{l+1}} \mathcal {D}_t^{\beta _{l+1}}f. \end{aligned}$$

Proof

Let us first assume \(l=1\) and prove

$$\begin{aligned} \mathcal {D}_t\nabla ^k f -\nabla ^k \mathcal {D}_t f = \sum _{|\alpha | \le k-1} \nabla ^{1+\alpha _1} v \star \nabla ^{1+\alpha _{2}} f. \end{aligned}$$
(4.6)

We argue by induction over k and observe immediately that the case \(k=1\) follows from (4.1). Assume that (4.6) holds for \(k-1\) and note that by (4.1) we have

$$\begin{aligned} \mathcal {D}_t\nabla ^k f = \mathcal {D}_t \, \nabla (\nabla ^{k-1} f) = \nabla \, \mathcal {D}_t (\nabla ^{k-1} f) + \nabla v \star (\nabla ^{k} f). \end{aligned}$$

By induction assumption we have

$$\begin{aligned} \begin{aligned} \nabla \, \mathcal {D}_t (\nabla ^{k-1} f)&= \nabla \big (\nabla ^{k-1} \mathcal {D}_t f + \sum _{|\alpha | \le k-2} \nabla ^{1+\alpha _1} v \star \nabla ^{1+\alpha _{2}} f\big ) \\ {}&= \nabla ^k \mathcal {D}_t f + \sum _{|\alpha | \le k-1} \nabla ^{1+\alpha _1} v \star \nabla ^{1+\alpha _{2}} f. \end{aligned} \end{aligned}$$

This yields the claim for \(l=1\).

The proof for \(l\ge 1\) follows from a similar induction argument. Assume that the claim holds for \(l-1\) and note that

$$\begin{aligned} \mathcal {D}_t^l \nabla ^k f= \nabla ^k \mathcal {D}_t^l f +\mathcal {D}_t ([ \mathcal {D}_t^{l-1},\nabla ^k ] f) + [\mathcal {D}_t, \nabla ^k] (\mathcal {D}_t^{l-1} f). \end{aligned}$$

By the first claim we have

$$\begin{aligned} {[}\mathcal {D}_t, \nabla ^k] (\mathcal {D}_t^{l-1} f) = \sum _{|\alpha | \le k-1 } \nabla ^{1+\alpha _1} v \star \nabla ^{1+\alpha _2} \mathcal {D}_t^{l-1}f. \end{aligned}$$

On the other hand, by the induction assumption we have

$$\begin{aligned} \begin{aligned} \mathcal {D}_t[\mathcal {D}_t^{l-1},\nabla ^k] f = \mathcal {D}_t \sum _{\begin{array}{c} |\alpha | \le k-1 \\ |\beta | \le l-2 \end{array}} \nabla ^{1+\alpha _1} \mathcal {D}_t^{\beta _1} v \star \cdots \star \nabla ^{1+\alpha _{l-1}} \mathcal {D}_t^{\beta _{l-1}} v \star \nabla ^{1+\alpha _{l}} \mathcal {D}_t^{\beta _{l}}f. \end{aligned} \end{aligned}$$

We use the Leibniz rule and the first claim to deduce that

$$\begin{aligned} \begin{aligned} \mathcal {D}_t\nabla ^{1+\alpha _1} \mathcal {D}_t^{\beta _1} v=\nabla ^{1+\alpha _1} \mathcal {D}_t^{1+\beta _1} v+ \sum _{|{\tilde{\alpha }}| \le \alpha _1} \nabla ^{1+{\tilde{\alpha }}_1} v \star \nabla ^{1+{\tilde{\alpha }}_{2}} \mathcal {D}_t^{\beta _1}v. \end{aligned} \end{aligned}$$

Similar formula holds also for \(\mathcal {D}_t\nabla ^{1+\alpha _l} \mathcal {D}_t^{\beta _l} f\). Hence, we obtain the claim. \(\square \)

Let us next prove higher order commutation formulas for (4.2) and a formula for \(\mathcal {D}_t^l \nu \) and \(\mathcal {D}_t B\). Below \(a_{\beta }(\nu )\) and \(a_{\alpha ,\beta }(\nu , B)\) denote bounded coefficients which depend on \(\nu \) and on \(\nu \) and B respectively.

Lemma 4.2

For \(l \ge 1\) it holds

$$\begin{aligned} {[}\mathcal {D}_t^l, \nabla _\tau ] f = \sum _{ |\beta | \le l-1} a_{\beta }(\nu ) \nabla \mathcal {D}_t^{\beta _1} v \star \cdots \star \nabla \mathcal {D}_t^{\beta _l} v \star \nabla _\tau \mathcal {D}_t^{\beta _{l+1}}f \end{aligned}$$
(4.7)

and

$$\begin{aligned} {[}\mathcal {D}_t^l, \nabla ^2_\tau ] f = \sum _{\begin{array}{c} |\alpha |\le 1 \\ |\beta | \le l-1 \end{array} } a_{\alpha , \beta }(\nu , B) \nabla ^{1+\alpha _1} \mathcal {D}_t^{\beta _1} v \star \cdots \star \nabla ^{1+\alpha _{l}} \mathcal {D}_t^{\beta _{l}} v \star \nabla _\tau ^{1+ \alpha _{l+1}} \mathcal {D}_t^{\beta _{l+1}}f. \end{aligned}$$
(4.8)

Moreover we have

$$\begin{aligned} \mathcal {D}_t^l \nu = \sum _{ |\beta |\le l-1} a_{\beta }(\nu ) \,\nabla \mathcal {D}_t^{\beta _1} v \star \cdots \star \nabla \mathcal {D}_t^{\beta _l} v \end{aligned}$$
(4.9)

and

$$\begin{aligned} \mathcal {D}_t^l B = \sum _{\begin{array}{c} |\alpha | \le 1\\ |\beta | \le l-1 \end{array}} a_{\alpha , \beta }(\nu , B) \nabla ^{ 1+\alpha _1} \mathcal {D}_t^{\beta _i} v \star \cdots \star \nabla ^{1+\alpha _{l+1}} \mathcal {D}_t^{\beta _{l+1}} v . \end{aligned}$$
(4.10)

Proof

Let us first prove (4.9). First, the claim holds for \(l=1\) by (4.4). Let us assume that (4.9) holds for \(l-1\). Then

$$\begin{aligned} \mathcal {D}_t^{l} \nu = \mathcal {D}_t \sum _{ |\beta |\le l-2} a_{\beta }(\nu ) \,\nabla \mathcal {D}_t^{\beta _1} v \star \cdots \star \nabla \mathcal {D}_t^{\beta _{l-1}} v. \end{aligned}$$

By (4.4) it holds \(\mathcal {D}_t a_{\beta }(\nu ) = {\tilde{a}}_{\beta }(\nu ) \nabla v\) and by (4.1) we have

$$\begin{aligned} \mathcal {D}_t \nabla \mathcal {D}_t^{\beta _i} v =\nabla \mathcal {D}_t^{\beta _i+1} v + \nabla v \star \nabla \mathcal {D}_t^{\beta _i} v. \end{aligned}$$

Thus we deduce

$$\begin{aligned} \mathcal {D}_t \sum _{|\beta | \le l-2 } a_{\beta }(\nu ) \, \nabla \mathcal {D}_t^{\beta _1} v \star \cdots \star \nabla \mathcal {D}_t^{\beta _{l-1}} v = \sum _{|\beta | \le l-1} {\tilde{a}}_{\beta }(\nu ) \,\nabla \mathcal {D}_t^{\beta _1} v \star \cdots \star \nabla \mathcal {D}_t^{\beta _{l}} v \end{aligned}$$

which implies (4.9).

Let us next prove (4.7). By (4.2) the claim holds for \(l=1\). Let us assume that the claim holds for \(l-1\). Then

$$\begin{aligned} \mathcal {D}_t^{l} \nabla _\tau f = \mathcal {D}_t \nabla _\tau \mathcal {D}_t^{l-1}f+\mathcal {D}_t \sum _{|\beta | \le l-2} a_{\beta }(\nu ) \nabla \mathcal {D}_t^{\beta _1} v \star \cdots \star \nabla \mathcal {D}_t^{\beta _{l-1}} v \star \nabla _\tau \mathcal {D}_t^{\beta _{l}}f. \end{aligned}$$

As before we have by (4.4) \(\mathcal {D}_t a_{\beta }(\nu ) = {\tilde{a}}_{\beta }(\nu ) \nabla v\) and by (4.2)

$$\begin{aligned} \mathcal {D}_t \nabla _\tau \mathcal {D}_t^{\beta _i} f = \nabla _\tau \mathcal {D}^{\beta _i+1}_t f + a(\nu ) \nabla v \star \nabla _\tau \mathcal {D}_t^{\beta _i} f. \end{aligned}$$

Therefore we obtain by Leibniz rule

$$\begin{aligned} \mathcal {D}_t^{l} \nabla _\tau f = \nabla _\tau \mathcal {D}^{l}_t f + \sum _{ |\beta | \le l-1} {\tilde{a}}_{\beta }(\nu ) \nabla \mathcal {D}_t^{\beta _1} v \star \cdots \star \nabla \mathcal {D}_t^{\beta _{l}} v \star \nabla _\tau \mathcal {D}_t^{\beta _{l}}f \end{aligned}$$

and (4.7) follows.

We notice next that (4.10) follows from \(B = \nabla _\tau \nu \) and by combining (4.7) with (4.9). Finally we obtain (4.8) by first applying (4.7) as

$$\begin{aligned} \begin{aligned} \mathcal {D}_t^l \nabla _\tau ^2 f&= \nabla _\tau (\mathcal {D}_t^l \nabla _\tau f) + \sum _{ |\beta | \le l-1} a_{\beta }(\nu ) \nabla \mathcal {D}_t^{\beta _1} v \star \cdots \star \nabla \mathcal {D}_t^{\beta _l} v \star \nabla _\tau \mathcal {D}_t^{\beta _{l+1}} \nabla _\tau f. \end{aligned} \end{aligned}$$

The claim then follows by differentiating (4.7). \(\square \)

Remark 4.3

By Lemma 4.2 we have in particular that

$$\begin{aligned} \mathcal {D}_t^l v_n = \sum _{ |\beta | \le l} a_{ \beta }(\nu ) \,\nabla \mathcal {D}_t^{\beta _1} v \star \cdots \star \nabla \mathcal {D}_t^{\beta _l} v \star D_t^{\beta _{l+1}} v, \end{aligned}$$

where \(\beta _i\le l-1\) for \(i \le l\). Moreover, since we may write the Laplace-Beltrami operator as \(\Delta _\Sigma f = \text {Tr}(\nabla _\tau ^2 f)\) then Lemma 4.2 yields

$$\begin{aligned} \mathcal {D}_t^l \Delta _\Sigma f = \Delta _\Sigma \mathcal {D}_t^l f + \sum _{\begin{array}{c} |\alpha | \le 1 \\ |\beta | \le l-1 \end{array}} a_{\alpha , \beta }(\nu , B) \nabla ^{1+\alpha _1} \mathcal {D}_t^{\beta _1} v \star \cdots \star \nabla ^{1+\alpha _{l}} \mathcal {D}_t^{\beta _{l}} v \star \nabla _\tau ^{1+ \alpha _{l+1}} \mathcal {D}_t^{\beta _{l+1}}f. \end{aligned}$$

Let us next derive formulas for the divergence and the curl of the vector field \(\mathcal {D}_t^l v\). Recall that by (1.3) we have \(\textrm{div}\,v = 0\) which then implies

$$\begin{aligned} - \Delta p= \textrm{div}\,(\mathcal {D}_t v) = \text {Tr}((\nabla v)^2) = \textrm{div}\,\textrm{div}\,(v \otimes v). \end{aligned}$$
(4.11)

For the curl we have \(\textrm{curl}\,(D_t v) = 0\) and \(\omega = \textrm{curl}\,v = \nabla v - \nabla v^T \) satisfies (see e.g. [50])

$$\begin{aligned} \mathcal {D}_t\omega = -\nabla v^T \omega - \omega \nabla v. \end{aligned}$$
(4.12)

We will derive formulas for \(\textrm{div}\,\mathcal {D}_t^l v\) and \(\textrm{curl}\,\mathcal {D}_t^l v\) below by using (4.11), (4.12) and the commutation formula in Lemma 4.1. To this aim we introduce two type of error functions. The first type we denote by \(R_{\textrm{div}\,}^l\), which stands for any function which can be written in the form

$$\begin{aligned} R^{l}_{\textrm{div}\,}=\sum _{|\beta |\le l} a_{ \beta }(\nabla v) \nabla \mathcal {D}_t^{\beta _1} v \star \cdots \star \nabla \mathcal {D}_t^{\beta _l}v, \end{aligned}$$
(4.13)

for \(l \ge 0\). We also use the convention that the indexes are ordered as \(\beta _1 \ge \beta _2 \ge \dots \ge \beta _l\). The second type of error function is slightly more general and it can be written in the form

$$\begin{aligned} R_{bulk}^l = \sum _{|\alpha |\le 1, |\beta | \le l} a_{\alpha , \beta }(\nabla v) \nabla \mathcal {D}_t^{\beta _1} v \star \cdots \star \nabla \mathcal {D}_t^{\beta _{l}} v \star \nabla ^{\alpha _1} \mathcal {D}_t^{\alpha _2 + \beta _{l+1}} v, \end{aligned}$$
(4.14)

for \(l \ge 0\). Note that \(R_{bulk}^l\) has one higher order term compared to \( R^{l}_{\textrm{div}\,}\). In particular, all functions of type \( R^{l}_{\textrm{div}\,}\) are contained in \(R_{bulk}^l\). The reason for introducing these two notations is that we need to estimate them in different norms. We will do this in the next section. Note that using Lemma 4.1 and \(-\nabla p = \mathcal {D}_t v\) we deduce that

$$\begin{aligned}{}[\mathcal {D}_t^{l+1},\nabla ]p = R_{bulk}^l. \end{aligned}$$
(4.15)

Lemma 4.4

Let \(l\ge 1\) and denote \(\omega = \textrm{curl}\,v\). Then it holds

$$\begin{aligned} \mathcal {D}_t \nabla ^{l} \omega = \nabla v\star \nabla ^{l} \omega + \sum _{|\alpha |\le l}\nabla ^{1+\alpha _1} v \star \nabla ^{1+\alpha _{2}}v. \end{aligned}$$

Moreover, \(\textrm{curl}\,\mathcal {D}_t^l v \) and \(\textrm{div}\,\mathcal {D}_t^l v \) can be written in the form

$$\begin{aligned} \textrm{curl}\,\mathcal {D}_t^l v = R_{\textrm{div}\,}^{l-1}\qquad \text {and} \qquad \textrm{div}\,\mathcal {D}_t^l v = R_{\textrm{div}\,}^{l-1}. \end{aligned}$$

We may also write the divergence of \(\mathcal {D}_t^{l+1} v\) as

$$\begin{aligned} \textrm{div}\,\mathcal {D}_t^{l+1} v= \textrm{div}\,\textrm{div}\,( v \otimes \mathcal {D}_t^{l} v) + \textrm{div}\,R^{l-1}_{bulk}. \end{aligned}$$

Proof

The first claim is an immediate consequence of Lemma 4.1 and (4.12). The second claim follows from Lemma 4.1 and from \(\textrm{curl}\,(\mathcal {D}_t v) = 0\). Similarly the third claim follows from Lemma 4.1 and \(\textrm{div}\,v = 0\).

Let us then prove the last claim. We begin by proving two useful identities. First, we claim that for a vector field F it holds

$$\begin{aligned}{}[\mathcal {D}_t,\textrm{div}\,]F = -\textrm{div}\,(\nabla v F). \end{aligned}$$
(4.16)

Indeed, since \(\textrm{div}\,v = 0\) we have

$$\begin{aligned} \mathcal {D}_t\textrm{div}\,F- \textrm{div}\,\mathcal {D}_t F =\sum _{i,j=1}^3 v_i\partial _i (\partial _jF_j)- \partial _i(\partial _jF_i v_j)=- \sum _{i,j=1}^3 \partial _i v_j \partial _j F_i= -\textrm{div}\,(\nabla v F). \end{aligned}$$

The second identity follows also from \(\textrm{div}\,v = 0\) and we may write it

$$\begin{aligned} \textrm{div}\,\textrm{div}\,( v\otimes \mathcal {D}_t^lv ) =\textrm{div}\,(\nabla \mathcal {D}_t^lv \, v). \end{aligned}$$
(4.17)

Let us prove the last claim in the case \(l = 1\). We use (4.1), (4.11), (4.16), (4.17) and the definition of \(R_{bulk}\) in (4.14) to deduce

$$\begin{aligned} \begin{aligned} \textrm{div}\,\mathcal {D}_t^{2}v&= \mathcal {D}_t \textrm{div}\,\mathcal {D}_t v - [\mathcal {D}_t,\textrm{div}\,] \mathcal {D}_t v \\&= \mathcal {D}_t \textrm{div}\,( \nabla v\, v ) + \textrm{div}\,(\nabla v \, \mathcal {D}_t v)\\&= \textrm{div}\,( \mathcal {D}_t ( \nabla v\, v ) ) -\textrm{div}\,(\nabla v \, \nabla v \, v)+ \textrm{div}\,(R_{bulk}^0)\\&= \textrm{div}\,( \nabla \mathcal {D}_t v\, v ) ) + \textrm{div}\,(R_{bulk}^0)\\&= \textrm{div}\,\textrm{div}\,(v \otimes \mathcal {D}_t v) ) + \textrm{div}\,(R_{bulk}^0). \end{aligned} \end{aligned}$$

Let us assume that the claim holds for \(l -1\). We argue as before and obtain by (4.16), (4.17) and by the induction assumption

$$\begin{aligned} \begin{aligned} \textrm{div}\,\mathcal {D}_t^{l+1}v&= \mathcal {D}_t \textrm{div}\,\mathcal {D}_t^{l} v - [\mathcal {D}_t, \textrm{div}\,] \mathcal {D}_t^{l} v \\&= \mathcal {D}_t \textrm{div}\,\left( \nabla \mathcal {D}_t^{l-1} v\, v + R_{\textrm{div}\,}^{l-2} \right) + \textrm{div}\,(\nabla v \, \mathcal {D}_t^{l} v). \end{aligned} \end{aligned}$$

We use (4.16), (4.1) and the definition of \(R_{bulk}^{l-1}\) in (4.14) and obtain

$$\begin{aligned} \begin{aligned} \mathcal {D}_t \textrm{div}\,&\left( \nabla \mathcal {D}_t^{l-1} v\, v\right) = \textrm{div}\,\left( \mathcal {D}_t( \nabla \mathcal {D}_t^{l-1} v\, v)\right) + [\mathcal {D}_t,\textrm{div}\,](\nabla \mathcal {D}_t^{l-1} v\, v) \\&=\textrm{div}\,(\nabla \mathcal {D}_t^{l} v \, v) + \textrm{div}\,\left( \nabla \mathcal {D}_t^{l-1} v \star \nabla v \star v + \nabla \mathcal {D}_t^{l-1} v \star \mathcal {D}_t v \right) \\&= \textrm{div}\,(\nabla \mathcal {D}_t^{l} v \, v) + \textrm{div}\,R_{bulk}^{l-1}\\&= \textrm{div}\,\textrm{div}\,(v \otimes \mathcal {D}_t^{l} v) + \textrm{div}\,R_{bulk}^{l-1}. \end{aligned} \end{aligned}$$

Similarly we have

$$\begin{aligned} \mathcal {D}_t \textrm{div}\,R_{\textrm{div}\,}^{l-2} = \textrm{div}\,R_{\textrm{div}\,}^{l-1} \qquad \text {and} \qquad \textrm{div}\,(\nabla v \, \mathcal {D}_t^{l} v) = \textrm{div}\,R_{\textrm{div}\,}^{l-1} \end{aligned}$$

and the claim follows. \(\square \)

Let us then turn our focus on the pressure. By (4.11) and (1.3) we have that p is a solution of

$$\begin{aligned} {\left\{ \begin{array}{ll} -\Delta p = \textrm{div}\,\textrm{div}\,(v \otimes v), \quad \text {in }\, \Omega _t, \\ p = H - \frac{Q(t)}{2} |\nabla U|^2, \quad \text {on }\, \Sigma _t, \end{array}\right. } \end{aligned}$$

where Q(t) is a real valued function of time defined in (2.1) as

$$\begin{aligned} Q(t) = \frac{Q}{(\text {Cap}(\Omega _t))^2}, \end{aligned}$$

\(U = U_{\Omega _t}\) is the capacitary potential and \(H = H_{\Sigma _t}\) is the mean curvature.

We need to derive the equation for \(\mathcal {D}_t^l p\). We obtain the equation for \(\mathcal {D}_t^l p\) in the bulk from Lemma 4.4.

Remark 4.5

By Lemma 4.4 and by (4.15) the function \(-\Delta \mathcal {D}_t^{l+1}p\) can be written as

$$\begin{aligned} \begin{aligned} - \Delta \mathcal {D}_t^{l+1}p&= -\textrm{div}\,\mathcal {D}_t^{l+1}\nabla p+ \textrm{div}\,[\mathcal {D}_t^{l+1},\nabla ]p = \textrm{div}\,\mathcal {D}_t^{l+2} v + \textrm{div}\,R_{bulk}^{l} \\&= \textrm{div}\,\textrm{div}\,(v \otimes \mathcal {D}_t^{l+1}v) + \textrm{div}\,( R_{bulk}^{l}). \end{aligned} \end{aligned}$$

To find the formula for \(\mathcal {D}_t^l p\) on the boundary \(\Sigma _t\) is more challenging. To that aim we first need to study the capacitary potential U. We introduce an error term which appears when we deal with the capacity term on the boundary, i.e., for \(l \ge 0 \) we denote by \(R_U^l\) as functions on \(\Sigma _t\), which can be written in the form

$$\begin{aligned} R_U^l = \sum _{\begin{array}{c} |\alpha | + |\beta | \le l+1\\ |\beta | \le l \end{array}} a_{\alpha ,\beta }(v) \mathcal {D}_t^{\beta _1} v \star \cdots \star \mathcal {D}_t^{\beta _{l}} v \star \nabla ^{1+ \alpha _1} \partial _t^{\alpha _2} U. \end{aligned}$$
(4.18)

We note that v is defined in \(\Omega _t\) while U is defined in \(\Omega _t^c\), but they are both well-defined on the boundary \(\Sigma _t\). We have the following formulas for U on \(\Sigma _t\).

Lemma 4.6

Let \(l \ge 1\). Then on \(\Sigma _t\) it holds

$$\begin{aligned} \mathcal {D}_t^l \nabla U = R_U^{l-1} \end{aligned}$$

and

$$\begin{aligned} \begin{aligned} \mathcal {D}_t^{l+1} \nabla U =&\nabla \partial _t^{l+1} U + \nabla ^2 U \, \mathcal {D}_t^l v\\&+ \sum _{\begin{array}{c} \alpha + |\beta | + \gamma \le l+1\\ |\beta |\le l-1, \gamma \le l \end{array}} a_{\alpha , \beta ,\gamma }(v) \mathcal {D}_t^{\beta _1} v \star \cdots \star \mathcal {D}_t^{\beta _{l-1}} v \star \nabla ^{1+ \alpha } \partial _t^{\gamma } U. \end{aligned} \end{aligned}$$

Moreover we have the following formula for \(\partial _t^{l+1} U\)

$$\begin{aligned} \partial _t^{l+1} U = - \partial _{\nu } U \, (\mathcal {D}_t^{l} v \cdot \nu ) + R_U^{l-1} \qquad \text {on } \, \Sigma _t. \end{aligned}$$

Proof

The proof of the first statement is straightforward. Note that

$$\begin{aligned} \mathcal {D}_t \nabla U = \nabla \partial _t U + \nabla ^2 U v \end{aligned}$$

and

$$\begin{aligned} \mathcal {D}_t^2 \nabla U = \nabla \partial _t^2 U + \nabla ^2 U \mathcal {D}_t v + \nabla ^3 U\star v \star v + \nabla ^2 \partial _t U \star v. \end{aligned}$$

Thus the first claim holds for \(l =1,2\) and the second for \(l=1\). The general case \(l \ge 1\) follows by an induction argument.

For the third claim we recall that the potential satisfies \(U = 1\) on \(\Sigma _t\). Therefore it holds \(\mathcal {D}_t U = 0\) on \(\Sigma _t\) which we write as

$$\begin{aligned} \partial _t U = - ( \nabla U \cdot v ). \end{aligned}$$

Differentiating this yields

$$\begin{aligned} \mathcal {D}_t^{l} \partial _t U = - (\nabla U \cdot \mathcal {D}_t^{l} v ) + \sum _{\begin{array}{c} i+j = l\\ i \le l-1 \end{array}} \mathcal {D}_t^{i} v \star \mathcal {D}_t^{j} \nabla U. \end{aligned}$$

By the first claim we have \(\mathcal {D}_t^{j} \nabla U = R_U^{j-1}\) and thus by the definition of \(R_U^{j-1}\) in (4.18) we may write

$$\begin{aligned} \mathcal {D}_t^{l} \partial _t U = - (\nabla U \cdot \mathcal {D}_t^{l} v ) + \sum _{\begin{array}{c} |\alpha | + |\beta | \le l\\ |\beta | \le l-1 \end{array}} a_{\alpha ,\beta }(v) \mathcal {D}_t^{\beta _1} v \star \cdots \star \mathcal {D}_t^{\beta _{l-1}} v \star \nabla ^{1+ \alpha _1} \partial _t^{\alpha _2} U. \end{aligned}$$

It also holds

$$\begin{aligned} \mathcal {D}_t^{l} \partial _t U = \partial _t^{l+1} U + \sum _{\begin{array}{c} |\alpha | + |\beta | \le l\\ |\beta | \le l-1 \end{array}} a_{\alpha ,\beta }(v) \mathcal {D}_t^{\beta _1} v \star \cdots \star \mathcal {D}_t^{\beta _{l-1}} v \star \nabla ^{1+ \alpha _1} \partial _t^{\alpha _2} U. \end{aligned}$$

Since U is constant on \(\Sigma _t\) it holds \(\nabla U = \partial _{\nu } U \, \nu \). This implies the third claim. \(\square \)

We conclude this section by deriving a formula for \(\mathcal {D}_t^{l+1} p\). Recall that

$$\begin{aligned} p = H - \frac{Q(t)}{2} |\nabla U|^2 \qquad \text {on } \, \Sigma , \end{aligned}$$
(4.19)

where \(Q(t)\) is defined in (2.1). It is well known that (e.g. [15])

$$\begin{aligned} \mathcal {D}_t H = - \Delta _\Sigma v_n - |B|^2 v_n + \nabla _\tau H\cdot v \end{aligned}$$
(4.20)

where \(v_n = v \cdot \nu \). Using the geometric fact

$$\begin{aligned} \Delta _\Sigma \nu = -|B|^2 \nu + \nabla _\tau H \end{aligned}$$
(4.21)

and (4.19) we obtain the formula

$$\begin{aligned} \mathcal {D}_t p = - \Delta _\Sigma v \cdot \nu - 2 B : \nabla _\tau v - Q(t)( \mathcal {D}_t \nabla U \cdot \nabla U) -\frac{Q'(t)}{2} |\nabla U|^2 . \end{aligned}$$
(4.22)

We may write (4.22) in a different form. Indeed we use \(\nabla U = \partial _\nu U \, \nu = - |\nabla U|\nu \) and obtain

$$\begin{aligned} \begin{aligned} - \mathcal {D}_t \nabla U \cdot \nabla U&= -(\nabla \partial _t U \cdot \nabla U) - (\nabla ^2 U \nu \cdot v) \, \partial _\nu U\\&= -(\nabla \partial _t U \cdot \nabla U) + ( \nabla ^2 U \nu \cdot \nu ) \, v_n\, |\nabla U| - ( \nabla ^2 U \nabla U \cdot v_\tau ). \end{aligned} \end{aligned}$$

We notice that

$$\begin{aligned} (\nabla ^2 U \nabla U \cdot v_\tau ) = \frac{1}{2} ( \nabla _\tau |\nabla U|^2 \cdot v). \end{aligned}$$
(4.23)

Moreover, we recall that U is harmonic in \(\Omega _t^c\) and constant on \(\Sigma _t\). Therefore it holds by (3.4)

$$\begin{aligned} 0 = \Delta U = \overbrace{\Delta _{\tau } U}^{=0} + (\nabla ^2 U\nu \cdot \nu ) + H \partial _{\nu } U = (\nabla ^2 U\nu \cdot \nu ) - H|\nabla U|. \end{aligned}$$
(4.24)

Thus we have by (4.19), (4.20), (4.23), (4.24) that

$$\begin{aligned} \mathcal {D}_t p = - \Delta _\Sigma v_n - |B|^2 v_n - Q(t)\,\big ( \partial _\nu U \, (\partial _\nu \partial _t U) - H |\nabla U|^2 v_n\big ) + \langle \nabla _\tau p, v\rangle - Q'(t) \frac{|\nabla U|^2}{2}. \end{aligned}$$
(4.25)

In the next lemma we find a suitable expression for \(\mathcal {D}_t^{l} p\) for \(l \ge 1\). Again we will have an error term which in this case is defined on the boundary \(\Sigma _t\) and is more complicated than the previous ones. We define the error term \(R_p^l\) for \(l \ge 1\) as

$$\begin{aligned} R_{p}^l = R_{I}^l + R_{II}^l + R_{III}^l. \end{aligned}$$
(4.26)

where

$$\begin{aligned} \begin{aligned}&R_{I}^l = - (|B|^2 - Q(t) H\, |\nabla U|^2 ) (\mathcal {D}_t^{l}v \cdot \nu ) + (\nabla _\tau p \cdot \mathcal {D}_t^{l}v ), \\&R_{II}^l = \sum _{|\alpha |\le 1, \, |\beta |\le l-1}a_{\alpha , \beta ,\gamma }(B) \nabla ^{1+\alpha _1} \mathcal {D}_t^{\beta _1} v \star \cdots \star \nabla ^{1+\alpha _{l+1}} \mathcal {D}_t^{\beta _{l+1}} v \qquad \text {and} \\&R_{III}^l =\sum _{\begin{array}{c} |\alpha | + |\beta | + |\gamma |\le l+1\\ |\beta |\le l-1, \gamma _i \le l \end{array}} a_{\alpha , \beta ,\gamma , Q}(v) \mathcal {D}_t^{\beta _1} v \star \cdots \star \mathcal {D}_t^{\beta _{l-1}} v \star \nabla ^{1+\alpha _1} \partial _t^{\gamma _1} U \star \nabla ^{1+\alpha _1} \partial _t^{\gamma _2} U , \end{aligned} \end{aligned}$$
(4.27)

where the coefficients \(a_{\alpha , \beta ,\gamma ,Q}(v)\) depend on v and on the derivatives \(Q^{(k)}(t)\) for \(k \le l+1\). Above \(a_{\alpha ,\beta , \gamma }(B)\) means that the coefficient depends on the second fundamental form. For \(l=1\) we need to quantify this dependence in which case \(R_{II}^1\) reads as

$$\begin{aligned} R_{II}^1 = a_1(\nu , \nabla v) \star B + a_2(\nu , \nabla v) \star \nabla ^2 v. \end{aligned}$$
(4.28)

The reason why \(R_p^l\) has three terms is that \(R_{II}^l\) contains the error terms arising from the surface tension and \(R_{III}^l\) from the capacity. The first term \(R_{I}^l\) is separate merely from notational reasons as it contains the highest order material derivatives.

Lemma 4.7

For \(l \ge 2\) It holds

$$\begin{aligned} \mathcal {D}_t^{l} p = -\Delta _\Sigma (\mathcal {D}_t^{l-1} v \cdot \nu ) - Q(t) \, \partial _{\nu } U \, (\partial _{\nu } \partial _t^{l} U) + R_p^{l-1} \end{aligned}$$

on \(\Sigma _t\), where \(Q(t)\) is defined in (2.1).

Proof

We first claim that it holds

$$\begin{aligned} \begin{aligned} \mathcal {D}_t^{l} p&= - (\Delta _\Sigma \mathcal {D}_t^{l-1} v) \cdot \nu - 2 B : \nabla _\tau (\mathcal {D}_t^{l-1} v) \\&- Q(t) ( \nabla \partial _t^{l} U \cdot \nabla U ) - Q(t) ( \nabla ^2 U \nabla U \cdot \mathcal {D}_t^{l-1} v) + R_{II}^{l-1} + R_{III}^{l-1}. \end{aligned} \end{aligned}$$
(4.29)

To obtain the claim (4.29) for \(l=2\) we first recall that by (4.3) we have

$$\begin{aligned}{}[\mathcal {D}_t,\Delta _{\Sigma }] v = a_1(\nu , \nabla v) \star B + a_2(\nu , \nabla v) \star \nabla ^2 v, \end{aligned}$$

and that (4.4) implies \(\mathcal {D}_t \nu = -(\nabla _\tau v)^T \nu \) and (4.2) implies \([\mathcal {D}_t, \nabla _\tau ] v= a(\nu ) \nabla v \star \nabla v\). We use (4.2) and (4.4) to obtain

$$\begin{aligned} \begin{aligned} \mathcal {D}_t B =\mathcal {D}_t (\nabla _\tau \nu )&= \nabla _\Sigma (\mathcal {D}_t \nu ) + [\mathcal {D}_t, \nabla _\tau ] \nu \\&= -\nabla _\tau ((\nabla _\tau v)^T \nu ) + a_1(\nu , \nabla v) \star B \\&= a_1(\nu , \nabla v) \star B + a_2(\nu , \nabla v ) \star \nabla ^2 v. \end{aligned} \end{aligned}$$

We differentiate (4.22) and use the above identities and have

$$\begin{aligned} \begin{aligned} \mathcal {D}_t^{2} p = - (\Delta _\Sigma \mathcal {D}_t v) \cdot \nu - 2 B: \nabla _\tau (\mathcal {D}_t v) -\mathcal {D}_t \Big (Q(t) ( \mathcal {D}_t \nabla U \cdot \nabla U ) +\frac{Q'(t)}{2} |\nabla U|^2 \Big ) + R_{II}^1. \end{aligned} \end{aligned}$$

Lemma 4.6 yields

$$\begin{aligned} \begin{aligned} \mathcal {D}_t (\mathcal {D}_t \nabla U \cdot \nabla U)&= (\mathcal {D}_t^2 \nabla U \cdot \nabla U)+ (\mathcal {D}_t \nabla U \cdot \mathcal {D}_t \nabla U)\\&= ( \nabla \partial _t^{2} U \cdot \nabla U ) + (\nabla ^2 U \nabla U \cdot \, \mathcal {D}_t v ) \\&\,\,\,\,\,\,\,\,\,\,+ \sum _{\begin{array}{c} |\alpha | + |\beta | \le 2,\\ \beta _i \le 1 \end{array}} a_{\alpha ,\beta }(v) \nabla ^{1+\alpha _1} \partial _t^{\beta _1} U \star \nabla ^{1+\alpha _2} \partial _t^{\beta _2} U. \end{aligned} \end{aligned}$$

We may embed the rest of the terms to \(R_{III}^1\). This implies (4.29) for \(l=2\).

To obtain the claim (4.29) for \(l\ge 2\) we differentiate (4.22) \((l-1)\)-times. Since the argument is similar to the case \(l=2\), we only highlight the most subtle steps. To identify the error terms we recall that by Lemma 4.2 we have for \(i \le l-1\)

$$\begin{aligned}{} & {} \mathcal {D}_t^i \nu = \sum _{|\beta | \le i-1} a_{\beta }(\nu )\nabla \mathcal {D}_t^{\beta _1} v \star \cdots \star \nabla \mathcal {D}_t^{\beta _i} v, \\{} & {} {[}\mathcal {D}_t^i, \nabla _\Sigma ] v = \sum _{|\beta | \le i-1} a_{\beta }(\nu )\nabla \mathcal {D}_t^{\beta _1} v \star \cdots \star \nabla \mathcal {D}_t^{\beta _{i+1}} v, \end{aligned}$$

and by Remark 4.3

$$\begin{aligned} {[}\mathcal {D}_t^i,\Delta _\Sigma ] v = \sum _{\begin{array}{c} |\alpha | \le 1 \\ |\beta | \le i-1 \end{array}} a_{\alpha , \beta }( B) \nabla ^{1+\alpha _1} \mathcal {D}_t^{\beta _1} v \star \cdots \star \nabla ^{1+\alpha _{i}} \mathcal {D}_t^{\beta _{i}} v \star \nabla _\tau ^{1+ \alpha _{i+1}} \mathcal {D}_t^{\beta _{i+1}}v. \end{aligned}$$

In order to treat the capacitary terms we first observe that

$$\begin{aligned} \mathcal {D}_t^{l-1} \langle \mathcal {D}_t \nabla U, \nabla U\rangle = \langle \mathcal {D}_t^{l} \nabla U, \nabla U \rangle + \sum _{\begin{array}{c} i+j\le l\\ i,j \le l-1 \end{array}} \mathcal {D}_t^{i} \nabla U \star \mathcal {D}_t^{i}\nabla U \end{aligned}$$

and then use Lemma 4.6 to deduce

$$\begin{aligned} \begin{aligned} \mathcal {D}_t^{l-1}&(\mathcal {D}_t \nabla U \cdot \nabla U) = ( \nabla \partial _t^{l} U \cdot \nabla U ) + ( \nabla ^2 U \nabla U \cdot \mathcal {D}_t^{l-1} v) \\&+ \sum _{\begin{array}{c} |\alpha | + |\beta | +|\gamma | \le l,\\ |\beta | \le l-2, \, \gamma _i \le l-1 \end{array}} a_{\alpha ,\beta ,\gamma }(v) \mathcal {D}_t^{\beta _1} v \star \cdots \star \mathcal {D}_t^{\beta _{l-1}} \star \nabla ^{1+\alpha _1} \partial _t^{\gamma _1} U \star \nabla ^{1+\alpha _2} \partial _t^{\gamma _2} U. \end{aligned} \end{aligned}$$

This implies (4.29).

We proceed by calculating and by using (4.21)

$$\begin{aligned} \begin{aligned} \Delta _\Sigma (\mathcal {D}_t^{l-1} v \cdot \nu )&= (\Delta _\Sigma \mathcal {D}_t^{l-1} v) \cdot \nu + 2 B: \nabla _\tau (\mathcal {D}_t^{l-1} v) + ( \Delta _\Sigma \nu ) \cdot (\mathcal {D}_t^{l-1} v)\\&= (\Delta _\Sigma \mathcal {D}_t^{l-1} v) \cdot \nu + 2 B: \nabla _\tau (\mathcal {D}_t^{l-1} v) -|B|^2 (\mathcal {D}_t^{l-1} v \cdot \nu ) + ( \nabla _\tau H \cdot \mathcal {D}_t^{l-1} v). \end{aligned} \end{aligned}$$

Moreover, we recall that \(\nabla U = \partial _{\nu } U \nu \) and that (4.24) implies \(( \nabla ^2 U \nu \cdot \nu ) = H \partial _\nu U \). Therefore we have

$$\begin{aligned} \begin{aligned} \langle \nabla ^2 U \nabla U, \mathcal {D}_t^{l-1} v \rangle&= \langle \nabla ^2 U \nu , \nu \rangle \, \partial _{\nu } U \, (\mathcal {D}_t^{l-1} v \cdot \nu ) + \langle \nabla ^2 U \nabla U, (\mathcal {D}_t^{l-1} v)_\tau \rangle \\&= H |\nabla U|^2 (\mathcal {D}_t^{l-1} v \cdot \nu ) + ( \nabla _\tau \frac{|\nabla U|^2}{2} \cdot \mathcal {D}_t^{l-1} v). \end{aligned} \end{aligned}$$

By combining the previous identities with (4.29) implies

$$\begin{aligned} \begin{aligned} \mathcal {D}_t^l p&= - \Delta _\Sigma (\mathcal {D}_t^{l-1} v \cdot \nu ) - Q(t) \, \partial _{\nu } U \, (\partial _{\nu } \partial _t^{l} U) \\&-(|B|^2 - Q(t)\, H|\nabla U|^2) (\mathcal {D}_t^{l-1} v \cdot \nu ) + \big (\nabla _\tau (H - Q(t)\frac{|\nabla U|^2}{2} )\cdot \mathcal {D}_t^{l-1} v\big )+ R_p^{l-1}. \end{aligned} \end{aligned}$$

Hence, the claim follows from (4.19). \(\square \)

5 Estimation of the Error Terms

In the previous section we introduced four error terms \(R_{\textrm{div}\,}^l, R_{bulk}^l, R_U^l\) and \(R_p^l\), which will appear later in the proof of the Main Theorem. These are nonlinear and characterized by their order \(l \ge 0\) and their precise forms can be found in (4.13), (4.14) (4.18) and (4.26) respectively. The first two terms \(R_{\textrm{div}\,}^l\) and \(R_{bulk}^l\) are defined in the fluid domain and appear already in the case when the shape of the drop does not change. The term \(R_U^l\) is due to the nonlinearity of the capacitary term. The term \(R_p^l\) is due to the nonlinear behavior of the pressure on the moving boundary and it is by far the most difficult to treat.

In this section our goal is to estimate these error terms by the energy quantity of order \(l \in \ \mathbb {N}\) which we define as

$$\begin{aligned} E_l(t) := \sum _{k=0}^{l} \Vert \mathcal {D}_t^{l+1-k}v\Vert _{H^{\frac{3}{2} k}(\Omega _t)}^2 + \Vert v\Vert _{H^{\lfloor \frac{3}{2}(1+1)\rfloor }(\Omega _t)}^2+ \Vert \mathcal {D}_t^l v \cdot \nu \Vert _{H^1(\Sigma _t)}^2+1. \end{aligned}$$
(5.1)

The most difficult is to estimate the lowest order terms \(R_{div}^1, \dots \), i.e., the case \(l=1\), and we treat it separately. The difficulty of the case \(l=1\) makes the arguments in this section long and cumbersome.

As we explained in the introduction, the proof of the Main Theorem is by induction argument, where we assume that we have the bound \(E_{l-1}(t) \le C\) and then use this to bound \(E_l(t)\). We begin this section by proving that the bound \(E_{l-1}(t) \le C\) for \(l \ge 2\) implies

$$\begin{aligned} \Vert B\Vert _{H^{\frac{3}{2}l -1}(\Sigma _t)} \le M(C). \end{aligned}$$

This will guarantee that every step improves the regularity of the flow. Perhaps the most challenging part is to start the argument and we show in Sect. 6 that the a priori bounds (1.7) imply the following estimate on the pressure

$$\begin{aligned} \Vert p\Vert _{H^1(\Omega _t)} \le C \end{aligned}$$

for all \(t \le T\). We will show that this implies the following curvature bound

$$\begin{aligned} \Vert B\Vert _{H^{\frac{1}{2}}(\Sigma _t)} \le C, \end{aligned}$$

which in particular implies the bound \(\Vert B\Vert _{L^{4}(\Sigma _t)} \le C\).

We notice that the above curvature bounds ensure that \(\Sigma _t\) satisfies the condition (\(\hbox {H}_{m}\)) for \(m = \lfloor \frac{3}{2}l\rfloor +1\) when \(l \ge 2\) and \(m=2\) for \(l =1\). This means that the results from Sect. 2 such as Proposition 2.1, Proposition 2.7, Corollary 2.9, Proposition 2.10, Proposition 2.11 and Proposition 2.12 hold for all \(k \le m\). We take this for granted in the calculations throughout this section without further mention in order to make the presentation less heavy.

We begin by estimating the capacitary potential U by the pressure via the identity (4.19). We note that in the next lemma the a priori \(C^{1,\alpha }\)-bound for the boundary \(\Sigma _t= \partial \Omega _t\) is crucial.

Lemma 5.1

Let \(l\ge 1\) and assume that \(\Sigma _t\) is uniformly \(C^{1,\alpha }(\Gamma )\)-regular and satisfies the condition \(\Vert B\Vert _{L^4(\Sigma _t)} \le M\) when \(l = 1\) and \(\Vert B\Vert _{H^{\frac{3}{2} l- 1}(\Sigma _t)} \le M\) when \(l \ge 2\). Let U be the capacitary potential defined in (1.2). There exists a constant C, depending on Ml and on the \(C^{1,\alpha }\)-norm of the heightfunction, such that

$$\begin{aligned} \Vert \nabla ^{1+k} U \Vert _{H^{\frac{1}{2}}(\Sigma _t)} \le C (1+ \Vert p \Vert _{H^{k}(\Sigma _t)} ) \qquad \text {on } \, \Sigma _t\end{aligned}$$

for all integers \(k \le \frac{3}{2}l +\frac{1}{2}\).

Proof

Let us note that the assumptions on the curvature imply that \(\Sigma _t\) satisfies the condition (\(\hbox {H}_{m}\)) for \(m = \lfloor \frac{3}{2}l\rfloor +1\). In particular, the condition \(k \le \frac{3}{2}l +\frac{1}{2}\) implies \(k \le m\).

Let us prove the claim by induction over k and consider first the case \(k = 0\). This follows immediately from Theorem 3.9 and from \(\Vert B\Vert _{L^4(\Sigma _t)} \le C\) as

$$\begin{aligned} \Vert \nabla U\Vert _{H^{\frac{1}{2}}(\Sigma _t)} \le C(1+ \Vert B\Vert _{L^2(\Sigma _t)}) \le C. \end{aligned}$$

Let us then fix k and assume that the claim holds for \(k-1\). Since U is constant on \(\Sigma _t\), Theorem 3.9 implies

$$\begin{aligned} \Vert \nabla ^{1+k} U\Vert _{H^{\frac{1}{2}}(\Sigma _t)} \le C(1 + \Vert B\Vert _{H^k(\Sigma _t)}). \end{aligned}$$

By Proposition 2.12 we have

$$\begin{aligned} \Vert B\Vert _{H^k(\Sigma _t)} \le C(1+ \Vert H\Vert _{H^k(\Sigma _t)}) \le C(1+\Vert p\Vert _{H^k(\Sigma _t)}+\Vert |\nabla U|^2\Vert _{H^k(\Sigma _t)}). \end{aligned}$$

Proposition 2.10 yields

$$\begin{aligned} \Vert |\nabla U|^2\Vert _{H^k(\Sigma _t)} \le C\Vert \nabla U\Vert _{L^\infty (\Sigma _t)} \Vert \nabla U\Vert _{H^k(\Sigma _t)}. \end{aligned}$$

Since \(\Omega _t\) is uniformly \(C^{1,\alpha }\)-regular we have by Schauder estimates \(\Vert U\Vert _{C^{1,\alpha }(\Sigma _t)} \le C\). By combining the previous inequalities we obtain

$$\begin{aligned} \Vert \nabla ^{1+k} U\Vert _{H^{\frac{1}{2}}(\Sigma _t)} \le C(1 + \Vert p\Vert _{H^k(\Sigma _t)} + \Vert \nabla U\Vert _{H^k(\Sigma _t)}) \end{aligned}$$
(5.2)

We claim next that under the assumptions of the lemma, it holds for every smooth function \(u: \Omega _t \rightarrow \mathbb {R}\) and for all \(k \le m\)

$$\begin{aligned} \Vert \nabla u\Vert _{H^k(\Sigma _t)} \le C(\Vert u\Vert _{L^2(\Sigma _t)} + \Vert \nabla ^{1+k} u\Vert _{L^{2}(\Sigma _t)}). \end{aligned}$$
(5.3)

Indeed, for \(k =1\) we have \(\Vert \nabla u\Vert _{H^1(\Sigma _t)} \le \Vert u\Vert _{L^2(\Sigma _t)} + \Vert \nabla ^2 u\Vert _{L^2(\Sigma _t)}\), while for \(k=2\) the assumption \(\Vert B\Vert _{L^4} \le C\) and the Sobolev embedding imply

$$\begin{aligned} \begin{aligned} \Vert \nabla u\Vert _{H^2(\Sigma _t)}&\le \Vert u\Vert _{L^2(\Sigma _t)}+ \Vert \nabla ^3 u\Vert _{L^2(\Sigma _t)} + \Vert B\star \nabla ^2 u \Vert _{L^2(\Sigma _t)}\\&\le \Vert u\Vert _{L^2(\Sigma _t)} + \Vert \nabla ^3 u\Vert _{L^2(\Sigma _t)} + \Vert B\Vert _{L^4(\Sigma _t)} \Vert \nabla ^2 u \Vert _{L^4(\Sigma _t)}\\&\le C(\Vert u\Vert _{L^2(\Sigma _t)}+\Vert \nabla ^3 u\Vert _{L^2(\Sigma _t)}). \end{aligned} \end{aligned}$$

The case \(k \ge 3\) follows from the same argument. We will take (5.3) for granted from now on.

We obtain by (5.2) and (5.3) that

$$\begin{aligned} \Vert \nabla ^{1+k} U\Vert _{H^{\frac{1}{2}}(\Sigma _t)} \le C(1 + \Vert p\Vert _{H^k(\Sigma _t)} + \Vert \nabla ^{1+k} U\Vert _{L^{2}(\Sigma _t)}). \end{aligned}$$

We deduce by Lemma 3.3, by interpolation and by the induction assumption (that the claim holds for \(k-1\))

$$\begin{aligned} \begin{aligned} \Vert \nabla ^{1+k} U\Vert _{L^{2}(\Sigma _t)}&\le C(1+\Vert \nabla ^{k} U\Vert _{H^{1}(\Sigma _t)}) \le \varepsilon \Vert \nabla ^{k} U\Vert _{H^{\frac{3}{2}}(\Sigma _t)} +C_\varepsilon (1+\Vert \nabla ^{k} U\Vert _{L^{2}(\Sigma _t)})\\&\le \varepsilon \Vert \nabla ^{1+k} U\Vert _{H^{\frac{1}{2}}(\Sigma _t)} + C_\varepsilon (1+\Vert p\Vert _{H^{k-1}(\Sigma _t)}). \end{aligned} \end{aligned}$$

Thus by choosing \(\varepsilon \) small enough we obtain the claim. \(\square \)

From Lemma 5.1 we deduce that an estimate on the pressure implies bound on the curvature. The statement follows from the proof of Lemma 5.1 and we leave the proof for the reader.

Lemma 5.2

Let \(l \ge 1\) and assume that \(\Sigma _t\) is uniformly \(C^{1,\alpha }(\Gamma )\)-regular and for \(l=1\) it holds \(\Vert p\Vert _{H^1(\Omega _t)} \le M\) and for \(l \ge 2 \) it holds \(E_{l-1}(t) \le M\). In the case \(l=1\) we have

$$\begin{aligned} \Vert B\Vert _{H^{\frac{1}{2}}(\Sigma _t)} \le C \end{aligned}$$

and \(\Vert B\Vert _{L^{4}(\Sigma _t)} \le C\). In the case \(l \ge 2\) we have

$$\begin{aligned} \Vert B\Vert _{H^{\frac{3}{2} l- 1}(\Sigma _t)} \le C. \end{aligned}$$

Moreover for \(l \ge 1\) we have

$$\begin{aligned} \Vert B \Vert _{H^{k}(\Sigma _t)} \le M(1+ \Vert p \Vert _{H^{k}(\Sigma _t)} ) \end{aligned}$$

for integers \(k \le \frac{3}{2}l + \frac{1}{2}\). The constants depend on Ml and on the \(C^{1,\alpha }\)-norm of the heightfunction.

From now on we will assume that, in addition to (1.7), we have for \(l=1\) the estimate \(\Vert p\Vert _{H^1(\Omega _t)} \le C\) and for \(l \ge 2\) \(E_{l-1}(t)\le C\). By Lemma 5.2 these imply curvature bounds that we mentioned at the beginning of the section.

We begin to estimate the error terms and we begin with \(R_{\textrm{div}\,}^l\) defined in (4.13).

Lemma 5.3

Consider \(R_{\textrm{div}\,}^l\) defined in (4.13). Assume that (1.7) holds and \(\Vert p\Vert _{H^1(\Omega _t)} \le M\). Then we have

$$\begin{aligned} \Vert R_{\textrm{div}\,}^1\Vert _{H^\frac{1}{2}(\Omega _t)}^2 \le C(1+ \Vert p\Vert _{H^2(\Omega _t)}^2)E_1(t) \end{aligned}$$

for \(C= C(M)\).

Let \(l \ge 2\) and assume also that \(E_{l-1}(t)\le M\). Then there exists \(C = C(M,l)\), such that

$$\begin{aligned} \Vert R_{\textrm{div}\,}^l\Vert _{H^{\frac{1}{2}}(\Omega _t)}^2 \le C E_l(t) \end{aligned}$$
(5.4)

and for integers \(1 \le k \le l\) and every \(\varepsilon >0\) it holds

$$\begin{aligned} \Vert R_{\textrm{div}\,}^{l-k}\Vert _{H^{\frac{3}{2} k -1}(\Omega _t)}^2 \le \varepsilon E_{l}(t) + C_\varepsilon \end{aligned}$$
(5.5)

for some \(C_\varepsilon = C_\varepsilon (M,l,\varepsilon )\).

Proof

For \(l=1\) we have by the definition of \(R_{\textrm{div}\,}^1 \) (4.13) that

$$\begin{aligned} R_{\textrm{div}\,}^1 = a(\nabla v) \star \nabla \mathcal {D}_t v, \end{aligned}$$

where a is smooth. Note that in this case \(E_1(t)\), defined in (5.1), reads as

$$\begin{aligned} E_1(t) = \Vert \mathcal {D}_t^{2} v\Vert _{L^{2}(\Omega _t)}^2 + \Vert \mathcal {D}_t v\Vert _{H^{\frac{3}{2}}(\Omega _t)}^2 + \Vert v\Vert _{H^3(\Omega _t)}^2+ \Vert \mathcal {D}_t v \cdot \nu \Vert _{H^1(\Sigma _t)}^2+1. \end{aligned}$$

Since \(\Vert B\Vert _{L^4} \le C\), we may extend \(\nabla v\) and \(\nabla \mathcal {D}_t v\) to \(\mathbb {R}^3\), denote the extensions \(F_v\) and \(G_{v_t} \) respectively, such that the extensions satisfy

$$\begin{aligned} \Vert F_v\Vert _{L^\infty (\mathbb {R}^3)} \le C \Vert \nabla v\Vert _{L^\infty (\Omega _t)},\qquad \Vert F_v\Vert _{H^2(\mathbb {R}^3)} \le C \Vert \nabla v\Vert _{H^2(\Omega _t)} \end{aligned}$$

and

$$\begin{aligned} \Vert G_{v_t}\Vert _{L^2(\mathbb {R}^3)} \le C \Vert \nabla \mathcal {D}_t v\Vert _{L^2(\Omega _t)}, \qquad \Vert G_{v_t}\Vert _{H^{\frac{1}{2}}(\mathbb {R}^3)} \le C \Vert \mathcal {D}_t v\Vert _{H^{\frac{3}{2}}(\Omega _t)}. \end{aligned}$$

Moreover, since \(\Omega _t\) is bounded we may assume that \(F_{v}, G_{v_t} \in C_0^\infty (B_R)\).

We use the Kato-Ponce inequality (2.15) in \(\mathbb {R}^3\) with \(p_1 = 2, q_1 = \infty \), \(p_2 = \frac{12}{5}\) and \(q_2= 12\) to deduce

$$\begin{aligned} \begin{aligned} \Vert R_{\textrm{div}\,}^1\Vert _{H^{\frac{1}{2}}(\Omega _t)}&\le C \Vert \nabla \mathcal {D}_t v\star a(\nabla v)\Vert _{H^{\frac{1}{2}}(\Omega _t)} \le C\Vert G_{v_t} \star a(F_v)\Vert _{H^{\frac{1}{2}}(\mathbb {R}^3)} \\&\le C \Vert G_{v_t}\Vert _{H^{\frac{1}{2}}(\mathbb {R}^3)} \Vert F_v\Vert _{L^\infty (\mathbb {R}^3)} + C \Vert G_{v_t}\Vert _{L^{\frac{12}{5}}(\mathbb {R}^3)}\Vert F_v\Vert _{W^{\frac{1}{2}, 12}(\mathbb {R}^3)}. \end{aligned} \end{aligned}$$

Since \(\Vert F_v\Vert _{L^\infty } \le C\) we have

$$\begin{aligned} \Vert G_{v_t}\Vert _{H^{\frac{1}{2}}(\mathbb {R}^3)} \Vert F_v\Vert _{L^\infty (\mathbb {R}^3)} \le C \Vert \mathcal {D}_t v\Vert _{H^{\frac{3}{2}}(\Omega _t)} \le C E_1(t)^{\frac{1}{2}}. \end{aligned}$$

We have by using the Sobolev embedding \(\Vert u\Vert _{L^p(B_R)} \le C \Vert u\Vert _{H^s(B_R)} = C \Vert u\Vert _{W^{s,2}(B_R)}\), for \(p = \frac{6}{3-2\,s}\) and \(s=\frac{1}{4}\), and by the general Gagliardo-Nirenberg inequality (2.13) that

$$\begin{aligned} \Vert G_{v_t}\Vert _{L^{\frac{12}{5}}(\mathbb {R}^3)} \le C\Vert G_{v_t}\Vert _{H^{\frac{1}{4}}(\mathbb {R}^3)} \le C \Vert G_{v_t}\Vert _{H^{\frac{1}{2}}(\mathbb {R}^3)}^{\frac{1}{2}}\Vert G_{v_t}\Vert _{L^{2}(\mathbb {R}^3)}^{\frac{1}{2}}. \end{aligned}$$

By (2.13) we also have

$$\begin{aligned} \Vert F_{v}\Vert _{W^{\frac{1}{2}, 12}(\mathbb {R}^3)} \le C \Vert F_{v}\Vert _{W^{1,6}(\mathbb {R}^3)}^{\frac{1}{2}}\Vert F_{v}\Vert _{L^{\infty }(\mathbb {R}^3)}^{\frac{1}{2}} \le C \Vert F_{v}\Vert _{H^{2}(\mathbb {R}^3)}^{\frac{1}{2}}\Vert F_{v}\Vert _{L^{\infty }(\mathbb {R}^3)}^{\frac{1}{2}}. \end{aligned}$$

Therefore by \(\Vert F_{v}\Vert _{L^{\infty }(\mathbb {R}^3)} \le C\), \(\Vert F_{v}\Vert _{H^{2}(\mathbb {R}^3)} \le C\Vert \nabla v\Vert _{H^2(\Omega _t)}\le C\Vert v\Vert _{H^3(\Omega _t)}\) and

$$\begin{aligned} \Vert G_{v_t}\Vert _{L^{2}(\mathbb {R}^3)} \le C \Vert \nabla \mathcal {D}_t v \Vert _{L^2(\Omega _t)} = C \Vert p\Vert _{H^2(\Omega _t)} \end{aligned}$$

we have

$$\begin{aligned} \Vert G_{v_t}\Vert _{L^{\frac{12}{5}}(\mathbb {R}^3)}\Vert F_{v}\Vert _{W^{\frac{1}{2}, 12}(\mathbb {R}^3)} \le C \Vert p\Vert _{H^2(\Omega _t)}^{\frac{1}{2}} \Vert \mathcal {D}_t v\Vert _{H^{\frac{3}{2}}(\Omega _t)}^{\frac{1}{2}} \Vert F_{v}\Vert _{H^{2}(\mathbb {R}^3)}^{\frac{1}{2}} \le C \Vert p\Vert _{H^2(\Omega _t)}^{\frac{1}{2}} E_1(t)^{\frac{1}{2}}. \end{aligned}$$

This implies the first inequality.

Let \(l\ge 2\). In order to estimate the product (4.13) we use Proposition 2.10 to deduce

$$\begin{aligned} \Vert R^l_{\textrm{div}\,}\Vert _{H^\frac{1}{2}(\Omega _t)} \le \sum _{|\beta |\le l}\Vert \nabla \mathcal {D}^{\beta _1}_t v\Vert _{H^\frac{1}{2}(\Omega _t)} \prod _{i=2}^l\Vert \nabla \mathcal {D}_t^{\beta _i} v\Vert _{L^\infty (\Omega _t)}, \end{aligned}$$

where we use the convention that \(\beta _1\ge \beta _2 \ge \dots \ge \beta _l \). By Recalling the definition of \(E_{l-1}(t)\) in (5.1), by the assumption \(E_{l-1}(t)\le C\) and by Sobolev embedding it holds for \(\beta _i \le l-2\)

$$\begin{aligned} \Vert \nabla \mathcal {D}^{\beta _i}_t v\Vert _{L^{\infty }(\Omega _t)}^2 \le C \Vert \mathcal {D}_t^{\beta _i}v\Vert _{H^3(\Omega _t)}^2 \le \sum _{k=0}^{l-1} \Vert \mathcal {D}_t^{l-k} v\Vert _{H^{\frac{3}{2} k}(\Omega _t)}^2 \le C E_{l-1}(t) \le C. \end{aligned}$$

For future purpose we also note that by the same argument we have

$$\begin{aligned} \Vert \nabla ^{1+ \alpha }\mathcal {D}^{\beta }_t v\Vert _{L^{\infty }(\Omega _t)}^2 \le C E_{l-1}(t) \quad \text {for } \, \alpha + \beta \le l-2. \end{aligned}$$
(5.6)

Moreover, by the same argument it holds

$$\begin{aligned} \Vert \nabla \mathcal {D}_t^{l-1}v\Vert ^2_{L^\infty (\Omega _t)}\le C E_l(t). \end{aligned}$$

We also have

$$\begin{aligned} \Vert \nabla \mathcal {D}_t^{l-1}v\Vert _{H^\frac{1}{2}(\Omega _t)}^2\le C\Vert \mathcal {D}_t^{l-1}v\Vert _{H^\frac{3}{2}(\Omega _t)}^2 \le CE_{l-1}(t)\le C. \end{aligned}$$

Recall that by the definition of \(R_{\textrm{div}\,}^l\) above, the norm of the index is \(|\beta | \le l\). Therefore since \(l \ge 2\) it holds \(\beta _i \le l-1\) for \(i \ge 2\) and \(\beta _i \le l-2\) for \(i \ge 3\). Thus we conclude by the above estimates that

$$\begin{aligned} \Vert R^l_{\textrm{div}\,}\Vert _{H^\frac{1}{2}(\Omega _t)} \le C(1+ \Vert \nabla \mathcal {D}^l_t v\Vert _{H^\frac{1}{2}(\Omega _t)} +\Vert \nabla \mathcal {D}_t^{l-1}v\Vert _{H^\frac{1}{2}(\Omega _t)} \Vert \mathcal {D}_t^{l-1} v\Vert _{L^\infty (\Omega _t)}) \le C E_{l}(t)^{\frac{1}{2}}, \end{aligned}$$

which implies (5.4).

The proof of (5.5) follows from similar argument and we merely sketch it. For \(k =l\) the statement is trivial. For \(1 \le k\le l-1\) we recall that

$$\begin{aligned} R^{l-k}_{\textrm{div}\,}=\sum _{|\beta |\le l-k} a_\beta (\nabla v) \nabla \mathcal {D}_t^{\beta _1} v \star \cdots \star \nabla \mathcal {D}_t^{\beta _{l-k}}v. \end{aligned}$$

First, if \(k = 1\) then by applying the previous estimate for \(l-1\) we obtain

$$\begin{aligned} \Vert R^{l-1}_{\textrm{div}\,}\Vert _{H^{\frac{1}{2}}(\Sigma _t)}^2 \le C(1 + \Vert p\Vert _{H^2(\Omega _t)}^2) E_{l-1}(t). \end{aligned}$$

But now the condition \(E_{l-1}(t)\le C\) yields

$$\begin{aligned} \Vert p\Vert _{H^2(\Omega _t)}^2 \le \Vert \mathcal {D}_t v\Vert _{H^1(\Omega _t)}^2 \le C E_{l-1}(t) \le C. \end{aligned}$$

This implies the inequality for \(k = 1\).

Assume \(2 \le k \le l-1\). We apply Proposition 2.10 to bound

$$\begin{aligned} \Vert R^{l-k}_{\textrm{div}\,}\Vert _{H^{\frac{3}{2}k -1}(\Sigma _t)} \le \sum _{|\beta |\le l-k} \Vert \nabla \mathcal {D}_t^{\beta _1} v \Vert _{L^\infty (\Sigma _t)} \cdots \Vert \nabla \mathcal {D}_t^{\beta _{l-k-1}} v \Vert _{L^\infty (\Sigma _t)} \Vert \nabla \mathcal {D}_t^{\beta _{l-k}}v\Vert _{H^{\frac{3}{2}k -1}(\Sigma _t)}. \end{aligned}$$

Since \(k \ge 2\) then \(\beta _i \le l-2\) for all i. Therefore by (5.6) we have \(\Vert \nabla \mathcal {D}_t^{\beta _i}v\Vert _{L^\infty (\Omega _t)}\le C\) for all i. Moreover since \(\beta _i \le l-k\) it holds by the Trace Theorem and by interpolation

$$\begin{aligned} \begin{aligned} \Vert \nabla \mathcal {D}_t^{\beta _i} v\Vert _{H^{\frac{3}{2}k -1}(\Sigma _t)}^2&\le C\Vert \mathcal {D}_t^{\beta _i} v\Vert _{H^{\frac{3}{2}k+1}(\Omega _t)}^2 \\&\le \varepsilon \Vert \mathcal {D}_t^{\beta _i} v\Vert _{H^{\frac{3}{2}(k+1)}(\Omega _t)}^2 + C_\varepsilon \Vert \mathcal {D}_t^{\beta _i} v\Vert _{H^{\frac{3}{2}k}(\Omega _t)}^2\\&\le \varepsilon E_l(t) + C_\varepsilon E_{l-1}(t) \le \varepsilon E_l(t) + C_\varepsilon . \end{aligned} \end{aligned}$$

Hence, we have (5.5). \(\square \)

We proceed to bound the \(L^2\)-norm of \(R_{bulk}^l\), which is defined in (4.14), in the fluid domain \(\Omega _t\). Formally \(R_{bulk}^l\) is of order 1/2 higher than \(R^{l}_{\textrm{div}\,}\), and therefore this bound is of the same order than the previous lemma.

Lemma 5.4

Consider \(R_{bulk}^l\) defined in (4.14). Assume that (1.7) holds and \(\Vert p\Vert _{H^1(\Omega _t)} \le M\). There exists \(C = C(M)\) such that

$$\begin{aligned} \Vert R_{bulk}^1 \Vert _{L^2(\Omega _t)}^2 \le C (1+ \Vert p\Vert _{H^2(\Omega _t)}^2)E_1(t). \end{aligned}$$

Let \(l \ge 2\) and assume also that \(E_{l-1}(t)\le M\). There exists \(C=C (M,l)\) such that

$$\begin{aligned} \Vert R_{bulk}^l \Vert _{L^2(\Omega _t)}^2 \le C E_l(t) \end{aligned}$$

and for integers \(1\le k \le l-1\) and for \(\varepsilon >0 \) it holds

$$\begin{aligned} \Vert R_{bulk}^{l-k} \Vert _{H^{\frac{3}{2}(k-1)}(\Omega _t)}^2 \le \varepsilon E_l(t) + C_\varepsilon \end{aligned}$$

for some constant \(C_\varepsilon = C_\varepsilon (M,l,\varepsilon )\).

Proof

By (4.14) and the uniform bound on \(\nabla v\) given by (1.7) we have a pointwise estimate

$$\begin{aligned} |R_{bulk}^l| \le C \sum _{|\alpha |\le 1, |\beta | \le l} |\nabla \mathcal {D}_t^{\beta _1} v | \cdots |\nabla \mathcal {D}_t^{\beta _{l}} v | \, | \nabla ^{\alpha _1} \mathcal {D}_t^{\alpha _2 + \beta _{l+1}} v|. \end{aligned}$$

Let us first consider the case \(l=1\). Then we have by the above inequality, by \(\mathcal {D}_t v = -\nabla p\) and by ignoring the terms of the form \(|\nabla v |\), as they are uniformly bounded, and obtain a pointwise bound

$$\begin{aligned} |R_{bulk}^1| \le C \left( 1+ |\mathcal {D}_t^{2} v| + | \nabla \mathcal {D}_t v||\nabla p|\right) . \end{aligned}$$

Therefore we have by Hölder’s inequality and by the Sobolev embedding

$$\begin{aligned} \begin{aligned} \Vert R_{bulk}^1 \Vert _{L^2(\Omega _t)}^2&\le C(\Vert \mathcal {D}_t^{2} v\Vert _{L^2(\Omega _t)}^2 + \Vert \nabla p \Vert _{L^6(\Omega _t)}^2 \Vert \nabla \mathcal {D}_t v \Vert _{L^3(\Omega _t)}^2 )\\&\le C(1+ \Vert p\Vert _{H^2(\Omega _t)}^2)(1+ \Vert \mathcal {D}_t^2 v\Vert _{L^2(\Omega _t)}^2 + \Vert \mathcal {D}_t v\Vert _{H^{3/2}(\Omega _t)}^2)\\&\le C(1+ \Vert p\Vert _{H^2(\Omega _t)}^2)(1+ E_1(t)). \end{aligned} \end{aligned}$$

This implies the claim for \(l =1\).

Let us then treat the case \(l \ge 2\). Let us assume that the first l indexes are ordered as \(\beta _1 \ge \beta _2 \ge \dots \ge \beta _{l}\). As before we ignore all the terms in above which are uniformly bounded by the a priori assumption and by the assumption \(E_{l-1}(t) \le C\). Recall first that by (5.6) it holds

$$\begin{aligned} \Vert \nabla \mathcal {D}_t^{\beta _i} v\Vert _{L^\infty } \le C \quad \text {when } \quad \beta _i \le l-2. \end{aligned}$$

Recall that it holds \(|\beta | \le l\). Therefore, if \(\beta _{l+1} \ge l-1\) then \(\beta _1 \le 1\) and \(\beta _i =0\) for \(i \ge 2\). When \(\beta _{l+1} = l-2\) then the only possible other none-zero indexes are when \(\beta _1 = 2\) or when \(\beta _1 = \beta _2 = 1\). Finally when \(l \ge 3\) and \(\beta _{l+1} \le l-3\) then \(\nabla ^{\alpha _1}\mathcal {D}_t^{\alpha _2+\beta _{l+1}} v\) is itself uniformly bounded and the only nontrivial terms are given by the indexes \(\beta _1 = l-1\) and \(\beta _2 = 1\). Hence, we have a pointwise bound which we write by relabeling the indexes as

$$\begin{aligned} \begin{aligned}&|R_{bulk}^l| \le C \sum _{|\alpha |\le 1} |\nabla ^{\alpha _1} \mathcal {D}_t^{\alpha _2 + l} v | +C\sum _{|\alpha |\le 1, \beta _i \le l-1}|\nabla \mathcal {D}_t^{\beta _1} v| |\nabla ^{\alpha _1} \mathcal {D}_t^{\alpha _2 + \beta _2} v | \\&\quad + C \sum _{|\alpha | \le 1}(|\nabla \mathcal {D}_t^2 v | +|\nabla \mathcal {D}_t v |^2 ) | \nabla ^{\alpha _1} \mathcal {D}_t^{\alpha _2 + l-2} v| +C(|\nabla \mathcal {D}_t^{l} v| + |\nabla \mathcal {D}_t^{l-1} v||\nabla \mathcal {D}_t v|) |\nabla \mathcal {D}_t v| \end{aligned} \end{aligned}$$

Then by Hölder’s inequality we deduce

$$\begin{aligned} \begin{aligned} \Vert R_{bulk}^l\Vert _{L^2(\Omega _t)}^2&\le C \sum _{|\alpha |\le 1} \Vert \nabla ^{\alpha _1} \mathcal {D}_t^{\alpha _2 +l} v\Vert _{L^2}^2 +C \sum _{|\alpha | \le 1, \beta _i \le l-1} \Vert \nabla \mathcal {D}_t^{\beta _1} v\Vert _{L^3}^2 \Vert \nabla ^{\alpha _1} \mathcal {D}_t^{\alpha _2 +\beta _2} v \Vert _{L^6}^2\\&+ C \sum _{|\alpha | \le 1} ( \Vert \nabla \mathcal {D}_t^2 v\Vert _{L^3}^2 + \Vert \nabla \mathcal {D}_t v\Vert _{L^6}^4 ) \Vert \nabla ^{\alpha _1} \mathcal {D}_t^{\alpha _2 + l-2} v\Vert _{L^6}^2 \\&+ C(\Vert \nabla \mathcal {D}_t^l v\Vert _{L^3}^2 + \Vert \nabla \mathcal {D}_t^{l-1} v\Vert _{L^6}^2\Vert \nabla \mathcal {D}_t v\Vert _{L^6}^2) \Vert \mathcal {D}_t v\Vert _{L^6}^2. \end{aligned} \end{aligned}$$
(5.7)

We bound the first term on RHS of (5.7) for \(\alpha _1+ \alpha _2 \le 1\)

$$\begin{aligned} \Vert \nabla ^{\alpha _1} \mathcal {D}_t^{\alpha _2 +l} v\Vert _{L^2(\Omega _t)}^2 \le \Vert \mathcal {D}_t^{l+1} v \Vert _{L^2(\Omega _t)}^2 + \Vert \mathcal {D}_t^{l} v \Vert _{H^{3/2}(\Omega _t)}^2 \le E_l(t). \end{aligned}$$

We claim that the next term with the \(L^3\)-norm, \(\Vert \nabla \mathcal {D}_t^{\beta _1} v \Vert _{L^3}\), is bounded. Indeed, we use the Sobolev embedding, the induction assumption and the fact that \(\beta _1 \le l-1\) and have

$$\begin{aligned} \Vert \nabla \mathcal {D}_t^{\beta _1} v \Vert _{L^3(\Omega _t)}^2 \le \Vert \mathcal {D}_t^{\beta _1} v \Vert _{H^{\frac{3}{2}}(\Omega _t)}^2 \le \sum _{k=0}^l \Vert \mathcal {D}_t^{l -k} v\Vert _{H^{\frac{3}{2} k}(\Omega _t)}^2 \le E_{l-1} (t) \le C. \end{aligned}$$

We bound the coupling term with \(\alpha _1 + \alpha _2 \le 1\) and \(\beta _2 \le l-1\) by

$$\begin{aligned} \Vert \nabla ^{\alpha _1} \mathcal {D}_t^{\alpha _2 +\beta _2} v\Vert _{L^6(\Omega _t)}^2 \le C\Vert \mathcal {D}_t^{\alpha _2+\beta _2} v\Vert _{H^{1+\alpha _1}(\Omega _t)}^2 \le C \sum _{k=0}^l \Vert \mathcal {D}_t^{l+1 -k} v\Vert _{H^{\frac{3}{2} k}(\Omega _t)}^2 \le C E_l(t). \end{aligned}$$

We proceed to the next row in (5.7) and for \(\alpha _1 + \alpha _2 \le 1\) we have

$$\begin{aligned} \Vert \nabla ^{\alpha _1} \mathcal {D}_t^{\alpha _2 +l-2} v \Vert _{L^6(\Omega _t)}^2 \le \Vert \mathcal {D}_t^{\alpha _2 +l-2} v \Vert _{H^{1+\alpha _1}(\Omega _t)}^2 \le C \sum _{k=0}^l \Vert \mathcal {D}_t^{l -k} v\Vert _{H^{\frac{3}{2} k}(\Omega _t)}^2 \le C E_{l-1}(t) \le C. \end{aligned}$$

We also have

$$\begin{aligned} \Vert \nabla \mathcal {D}_t^2 v\Vert _{L^3(\Omega _t)}^2 \le \Vert \mathcal {D}_t^{2} v \Vert _{H^{3/2}(\Omega _t)}^2 \le C E_2(t) \le C E_l(t) \end{aligned}$$

since \(l \ge 2\). Moreover, by interpolation it holds

$$\begin{aligned} \Vert \nabla \mathcal {D}_t v\Vert _{L^6} \le C\Vert \nabla \mathcal {D}_t v \Vert _{H^2}^{\frac{1}{2}} \Vert \nabla \mathcal {D}_t v\Vert _{L^2}^{\frac{1}{2}} \le C E_2(t)^{\frac{1}{4}} E_1(t)^{\frac{1}{4}} \le C E_l(t)^{\frac{1}{4}}, \end{aligned}$$

when \(l \ge 2\). Hence, the second row in (5.7) is bounded by \(E_l(t)\).

We are left with the last row in (5.7). We bound the first term by

$$\begin{aligned} \Vert \nabla \mathcal {D}_t^l v\Vert _{L^3(\Omega _t)}^2 \le \Vert \nabla \mathcal {D}_t^l v\Vert _{H^{\frac{1}{2}}(\Omega _t)}^2\le \Vert \mathcal {D}_t^l v\Vert _{H^{\frac{3}{2}}(\Omega _t)}^2 \le E_l(t) \end{aligned}$$

and the last by \(\Vert \mathcal {D}_t v\Vert _{L^6(\Omega _t)}^2 \le \Vert \mathcal {D}_t v\Vert _{H^1(\Omega _t)}^2 \le C E_1(t) \le C \). Finally we treat the two remaining terms by the same argument. Indeed for \(\beta \le l-1\) we have by interpolation as before

$$\begin{aligned} \Vert \nabla \mathcal {D}_t^{\beta } v\Vert _{L^6(\Omega _t)} \le C\Vert \nabla \mathcal {D}_t^{\beta } v \Vert _{H^2(\Omega _t)}^{\frac{1}{2}} \Vert \nabla \mathcal {D}_t^{\beta } v\Vert _{L^2(\Omega _t)}^{\frac{1}{2}} \le C E_l(t)^{\frac{1}{4}} E_{l-1}(t)^{\frac{1}{4}} \le C E_l(t)^{\frac{1}{4}}. \end{aligned}$$

Hence, we have

$$\begin{aligned} \Vert R_{bulk}^l\Vert _{L^2(\Omega _t)}^2 \le CE_l(t). \end{aligned}$$

We are left with the last inequality. For \(k=1\) the claim follows by applying the previous inequality with \(l-1\). Let us then assume \(l\ge 3\) and \(2 \le k\le l-1\). By definition of \(R_{bulk}^{l-k}\) in (4.14) it holds \(|\beta | \le l-k \le l-2\). Therefore (5.6) implies

$$\begin{aligned} \Vert \nabla \mathcal {D}_t^{\beta _i} v\Vert _{L^\infty (\Omega _t)} \le \Vert \mathcal {D}_t^{\beta _i} v\Vert _{H^3(\Omega _t)}\le C \end{aligned}$$

for all i. Therefore by Proposition 2.10 and by relabeling the indexes we have

$$\begin{aligned} \begin{aligned} \Vert R_{bulk}^{l-k}\Vert _{H^{\frac{3}{2}(k-1)}(\Omega _t)} \le C\sum _{\begin{array}{c} {|\beta |\le l-k}\\ {|\alpha |\le 1} \end{array}}&\Vert \nabla ^{\alpha _1}\mathcal {D}_t^{\alpha _2+\beta _{2}}v\Vert _{H^{\frac{3}{2}(k-1)}(\Omega _t)} \\&+ \Vert \nabla \mathcal {D}_t^{\beta _{1}}v\Vert _{H^{\frac{3}{2}(k-1)}(\Omega _t)}\Vert \nabla ^{\alpha _1}\mathcal {D}_t^{\alpha _2+\beta _{2}}v\Vert _{L^\infty (\Omega _t)}. \end{aligned} \end{aligned}$$

Since \(\beta _2 \le l-k\), we may estimate the first term on RHS as

$$\begin{aligned} \Vert \nabla ^{\alpha _1}\mathcal {D}_t^{\alpha _2+\beta _{2}}v\Vert _{H^{\frac{3}{2}(k-1)}(\Omega _t)}^2 \le C \sum _{i=0}^{l-1} \Vert \mathcal {D}_t^{l-i}v\Vert _{H^{\frac{3}{2}i}(\Omega _t)}^2\le CE_{l-1}(t) \le C. \end{aligned}$$

We estimate the second similarly by using \(\beta _1 \le l-k\)

$$\begin{aligned} \Vert \nabla \mathcal {D}_t^{\beta _{1}}v\Vert _{H^{\frac{3}{2}(k-1)}(\Omega _t)}^2 \le \Vert \mathcal {D}_t^{\beta _{1}}v\Vert _{H^{\frac{3}{2}k}(\Omega _t)}^2 \le CE_{l-1}(t) \le C. \end{aligned}$$

Finally we bound the last term by the Sobolev embedding, by \(\beta _{2}\le l-k\), \(\alpha _1 + \alpha _2 \le 1\) and by interpolation

$$\begin{aligned} \begin{aligned} \Vert \nabla ^{\alpha _1}\mathcal {D}_t^{\alpha _2+\beta _{2}}v\Vert _{L^\infty (\Omega _t)}^2&\le C\Vert \mathcal {D}_t^{\alpha _2+\beta _{2}}v\Vert _{H^{2+\alpha _1}(\Omega _t)}^2 \\&\le \varepsilon \Vert \mathcal {D}_t^{\alpha _2+\beta _{2}}v\Vert _{H^{\frac{3}{2}(2+\alpha _1)}(\Omega _t)}^2 + C_\varepsilon \Vert \mathcal {D}_t^{\alpha _2+\beta _{2}}v\Vert _{L^{2}(\Omega _t)}^2\\&\le \varepsilon E_l(t) + C_\varepsilon E_{l-1}(t) \le \varepsilon E_l(t) + C_\varepsilon . \end{aligned} \end{aligned}$$

Thus we have

$$\begin{aligned} \Vert R_{bulk}^{l-k}\Vert _{H^{\frac{3}{2}(k-1)}(\Omega _t)}^2 \le \varepsilon E_l(t) + C_\varepsilon . \end{aligned}$$

\(\square \)

The two previous error bounds in Lemma 5.3 and Lemma 5.4 are similar in the sense that they only involve the material derivatives of the velocity field. We proceed to the error terms which involve the time derivatives of the capacity potential. Note that \(\partial _t^k U\) for all k is a harmonic function in \(\Omega _t^c\) but not constant on \(\Sigma _t\). We will use again Theorem 3.9 together with Lemma 4.6 which gives the formula for \(\partial _t^k U\) on the boundary \(\Sigma _t\).

We first prove a generic bound which will be useful when we bound the pressure.

Lemma 5.5

Let \(l\ge 2\) and assume that (1.7) and the condition \(E_{l-1}(t) \le M\) hold. Let \(\alpha , \beta \ge 0\) be integers. When \(\alpha + \beta \le l\), it holds

$$\begin{aligned} \Vert \nabla ^{1+\alpha }\partial _t^{\beta } U \Vert _{H^{\frac{\alpha }{2}+ \frac{1}{2}}(\Sigma _t)}^2 \le \varepsilon E_l(t) +C_\varepsilon , \end{aligned}$$
(5.8)

for \(C_\varepsilon = C_\varepsilon (M, l, \varepsilon )\). On the other hand, when \(\alpha + \beta \le l+1\) and \(\beta \le l\) then

$$\begin{aligned} \Vert \nabla ^{1+\alpha }\partial _t^\beta U \Vert _{H^{\frac{1}{2}}(\Sigma _t)}^2 \le C E_l(t) \end{aligned}$$
(5.9)

for \(C = C_\varepsilon (M, l)\).

Proof

Instead of (5.8) we prove in fact a slightly stronger result, namely

$$\begin{aligned} \begin{aligned}&\Vert \nabla ^{1+\alpha }\partial _t^{\beta } U \Vert _{H^{\frac{\alpha }{2} + \frac{1}{2} }(\Sigma _t)}^2 \le \varepsilon E_l(t) +C_\varepsilon \quad \text {when } \, \alpha \, \text {is even and } \\&\Vert \nabla ^{1+\alpha }\partial _t^{\beta } U \Vert _{H^{\frac{\alpha }{2} + 1}(\Sigma _t)}^2 \le C E_l(t) \quad \text {when } \, \alpha \, \text { is odd}. \end{aligned} \end{aligned}$$
(5.10)

The inequality (5.8) then follows from (5.10) by interpolation.

We prove (5.10) by induction over \(\beta \) and consider first the case \(\beta = 0\). Note that then \(\alpha \le l\). Let us first consider the case when \(\alpha \) is even. Then by Lemma 5.1 and by interpolation we have

$$\begin{aligned} \begin{aligned} \Vert \nabla ^{1+\alpha } U\Vert _{H^{\frac{\alpha }{2}+\frac{1}{2}}(\Sigma _t)}&\le C(1+ \Vert p\Vert _{H^{\frac{3}{2}\alpha }(\Sigma _t)}) \le \varepsilon \Vert p\Vert _{H^{\frac{3}{2}\alpha +\frac{1}{2}}(\Sigma _t)} + C_\varepsilon \Vert p\Vert _{L^2(\Sigma _t)}\\&\le \varepsilon \Vert p\Vert _{H^{\frac{3}{2}\alpha +\frac{1}{2}}(\Sigma _t)} + C_\varepsilon . \end{aligned} \end{aligned}$$

We use Lemma 3.7, \(-\nabla p = \mathcal {D}_t v\), \(\alpha \le l\) and the definition of \(E_l(t)\) to estimate

$$\begin{aligned} \Vert p\Vert _{H^{\frac{3}{2}\alpha +\frac{1}{2}}(\Sigma _t)}^2 \le C \Vert p\Vert _{H^{\frac{3}{2}\alpha +1}(\Omega _t)}^2 \le C(1+ \Vert \mathcal {D}_t v \Vert _{H^{\frac{3l}{2}}(\Omega _t)}^2) \le C E_l(t). \end{aligned}$$

This implies (5.10) when \(\alpha \) is even. When \(\alpha \) is odd we have again by Lemma 5.1, Lemma 3.7 and by the definition of \(E_l\) that

$$\begin{aligned} \Vert \nabla ^{1+\alpha } U\Vert _{H^{\frac{\alpha }{2}+1}(\Sigma _t)}^2 \le C (1+ \Vert p\Vert _{H^{\frac{3\alpha }{2} +\frac{1}{2}}(\Sigma _t)}^2) \le C (1+ \Vert p\Vert _{H^{\frac{3l}{2} +1}(\Omega _t)}^2) \le CE_{l}(t). \end{aligned}$$

Let us assume that (5.10) holds for \(\beta \le k-1\) for \(2 \le k \le l\). In particular, this implies that (5.8) holds for \(\beta \le k-1\). Let us consider only the case when \(\alpha \) is even, since the argument for \(\alpha \) odd is similar. We first observe that since \(\alpha \le l-k \le l-1\) then by the Trace Theorem

$$\begin{aligned} \Vert p\Vert _{H^{\frac{3}{2}\alpha }(\Sigma _t)}^2 \le C(1+\Vert \nabla p \Vert _{H^{\frac{3}{2}\alpha }(\Omega _t)}^2) \le C(1+\Vert \mathcal {D}_t v \Vert _{H^{\frac{3}{2}(l-1)}(\Omega _t)}^2) \le C E_{l-1}(t) \le C. \end{aligned}$$
(5.11)

We have then by Theorem 3.9, Lemma 5.2 and by (5.11) that

$$\begin{aligned} \begin{aligned} \Vert \nabla ^{1+\alpha } \partial _t^k U \Vert _{H^{ \frac{\alpha }{2}+\frac{1}{2}}(\Sigma _t)}&\le C(1+ \Vert p\Vert _{H^{\frac{3}{2}\alpha }(\Sigma _t)} + \Vert \partial _t^k U\Vert _{H^{ \frac{3}{2}\alpha +\frac{3}{2}}(\Sigma _t)})\\&\le C(1+ \Vert \partial _t^k U\Vert _{H^{ \frac{3}{2}(1+\alpha )}(\Sigma _t)}). \end{aligned} \end{aligned}$$

We use the expression of \(\partial _t^k U\) from Lemma 4.6 and Proposition 2.10 to estimate the last term in above as follows

$$\begin{aligned} \begin{aligned} \Vert \partial _t ^{k} U\Vert _{H^{\frac{3}{2}(1+\alpha )}(\Sigma _t)} \le&\Vert \nabla U\Vert _{H^{\frac{3}{2}(1+\alpha )}(\Sigma _t)} \Vert \mathcal {D}_t^{k-1}v \Vert _{L^\infty (\Sigma _t)} \\&+ \Vert \nabla U\Vert _{L^{\infty }(\Sigma _t)}\Vert \mathcal {D}_t^{k-1}v\Vert _{H^{\frac{3}{2}(1+\alpha )}(\Sigma _t)}+\Vert R^{k-2}_U\Vert _{H^{\frac{3}{2}(1+\alpha )}(\Sigma _t)}. \end{aligned} \end{aligned}$$

To estimate the first term on RHS let us first assume that \(k \le l-1\). Then by (5.6) \(\Vert \mathcal {D}_t^{k-1}v \Vert _{L^\infty (\Sigma _t)} \le C\) and since \(\alpha \le l-1\) we have by (5.8) for \(\beta = 0\) that

$$\begin{aligned} \Vert \nabla U\Vert _{H^{\frac{3}{2}(1+\alpha )}(\Sigma _t)}^2 \le \varepsilon E_l(t) + C_\varepsilon . \end{aligned}$$

On the other hand, when \(k = l\) then \(\alpha = 0\). Thus again by (5.8) it holds

$$\begin{aligned} \Vert \nabla U\Vert _{H^{\frac{3}{2}}(\Sigma _t)}^2 \le CE_{1}(t) \le CE_{l-1}(t) \le C. \end{aligned}$$

By the Sobolev embedding, by interpolation, by the Trace Theorem and by \(\alpha + k \le l\) we have

$$\begin{aligned} \begin{aligned} \Vert \mathcal {D}_t^{k-1}v \Vert _{L^\infty (\Sigma _t)}^2&\le C \Vert \mathcal {D}_t^{k-1}v\Vert _{H^{\frac{3}{2}(1+\alpha )}(\Sigma _t)}^2 \\&\le \varepsilon \Vert \mathcal {D}_t^{k-1}v\Vert _{H^{ \frac{3}{2}\alpha +2}(\Sigma _t)}^2+ C_\varepsilon \Vert \mathcal {D}_t^{k-1}v\Vert _{L^2(\Sigma _t)}^2\\&\le C\varepsilon \Vert \mathcal {D}_t^{k-1}v\Vert _{H^{ \frac{3}{2}(\alpha +2)}(\Omega _t)}^2 + C_\varepsilon E_{l-1}(t) \le C\varepsilon E_{l}(t) + C_\varepsilon . \end{aligned} \end{aligned}$$

Hence, we need yet to prove

$$\begin{aligned} \Vert R^{k-2}_U\Vert _{H^{\frac{3}{2}(1+\alpha )}(\Sigma _t)}^2 \le \varepsilon E_l(t) +C_\varepsilon , \end{aligned}$$
(5.12)

where \(k \le l\).

By the definition in (4.18) and by the Kato-Ponce inequality (Proposition 2.10) we may estimate

$$\begin{aligned} \begin{aligned} \Vert R^{k-2}_U\Vert _{H^{\frac{3}{2}(1+\alpha )}(\Sigma _t)} \le&C\sum _{\begin{array}{c} |\gamma |+|\beta | \le k-1\\ |\beta |\le k-2 \end{array}}\Big ( \Vert \mathcal {D}_t^{\beta _1}v \Vert _{L^\infty } \cdots \Vert \mathcal {D}_t^{\beta _{k-2}}v \Vert _{L^\infty }\Vert \nabla ^{1+ \gamma _1} \partial _t^{\gamma _2} U\Vert _{H^{\frac{3}{2}(1+\alpha )}(\Sigma _t)}\\&\,\,\,\,\,\,\,+ \Vert \mathcal {D}_t^{\beta _1}v \Vert _{H^{\frac{3}{2}(1+\alpha )}(\Sigma _t)} \cdots \Vert \mathcal {D}_t^{\beta _{k-2}}v \Vert _{L^\infty }\Vert \nabla ^{1+ \gamma _1} \partial _t^{\gamma _2} U\Vert _{L^{\infty }(\Sigma _t)} \Big ). \end{aligned} \end{aligned}$$
(5.13)

Since \(|\beta | \le k-2 \le l-2\), (5.6) implies \(\Vert \mathcal {D}_t^{\beta _1}v \Vert _{L^\infty } \le C\) for all i. Note that \(\alpha +k \le l\) and \(\beta _1 \le k-2\) implies \(\alpha + \beta _1 \le l-2\). Therefore we have by the Trace Theorem and by the definition of \(E_{l-1}(t)\)

$$\begin{aligned} \Vert \mathcal {D}_t^{\beta _1}v \Vert _{H^{\frac{3}{2}(1+\alpha )}(\Sigma _t)}^2 \le C\Vert \mathcal {D}_t^{\beta _1}v \Vert _{H^{ \frac{3}{2}(\alpha +2)}(\Omega _t)}^2 \le CE_{l-1}(t) \le C. \end{aligned}$$

We bound the both capacitary terms by the Sobolev embedding and by the induction assumption, which states that (5.8) holds for \(\beta \le k-1\). Indeed we have by \(\gamma _1 + \alpha \le |\gamma | + \alpha \le k-1 + \alpha \le l-1\) and \(\gamma _2 \le k-1\) that

$$\begin{aligned} \begin{aligned} \Vert \nabla ^{1+ \gamma _1} \partial _t^{\gamma _2} U\Vert _{L^{\infty }(\Sigma _t)}^2&\le C \Vert \nabla ^{1+ \gamma _1} \partial _t^{\gamma _2} U\Vert _{H^{\frac{3}{2}(1+\alpha )}(\Sigma _t)}^2\\&\le C(1+\Vert \nabla ^{1+ (1+ \alpha + \gamma _1)} \partial _t^{\gamma _2} U\Vert _{H^{\frac{1}{2}(1+\alpha + \gamma _1)}(\Sigma _t)}^2)\\&\le \varepsilon E_l(t) + C_\varepsilon . \end{aligned} \end{aligned}$$
(5.14)

Hence, we have (5.12) when \(\alpha \) is even.

Let us then prove (5.9). We notice that Theorem 3.9 and (5.8) imply (5.9) when \(2 \le \alpha \le l\). On the other Lemma 5.1 implies (5.9) when \(\alpha = l+1\). (Note that the assumption \(\alpha \le \frac{3}{2} l\) is satisfied for all \(\alpha \le l+1\) since \(l \ge 2\).) We need thus to consider the case \(\alpha =1\) and \(\beta = l\). For this the argument is similar than before and we only sketch it.

By Theorem 3.9 and by (5.11) we have

$$\begin{aligned} \begin{aligned} \Vert \nabla ^{2} \partial _t^l U \Vert _{H^{\frac{1}{2}}(\Sigma _t)}&\le C(1+ \Vert p\Vert _{H^{1}(\Sigma _t)} + \Vert \partial _t^l U\Vert _{H^{\frac{5}{2}}(\Sigma _t)})\\&\le C(1+ \Vert \partial _t^l U\Vert _{H^{\frac{5}{2}}(\Sigma _t)}). \end{aligned} \end{aligned}$$

We use the expression of \(\partial _t^l U\) from Lemma 4.6 and Proposition 2.10 to estimate the last term in above as follows

$$\begin{aligned} \begin{aligned} \Vert \partial _t ^{l} U\Vert _{H^{\frac{5}{2}}(\Sigma _t)}&\le \Vert \nabla U\Vert _{H^{\frac{5}{2}}(\Sigma _t)} \Vert \mathcal {D}_t^{l-1}v \Vert _{L^\infty (\Sigma _t)} \\&+ \Vert \nabla U\Vert _{L^{\infty }(\Sigma _t)}\Vert \mathcal {D}_t^{l-1}v\Vert _{H^{\frac{5}{2}}(\Sigma _t)}+\Vert R^{l-2}_U\Vert _{H^{\frac{5}{2}}(\Sigma _t)}. \end{aligned} \end{aligned}$$

By Lemma 5.1 and Lemma 3.7 we have

$$\begin{aligned} \begin{aligned} \Vert \nabla U\Vert _{H^{\frac{5}{2}}(\Sigma _t)}^2 \le C(1+ \Vert p\Vert _{H^2(\Sigma _t)}^2) \le C(1+ \Vert \nabla p\Vert _{H^{\frac{3}{2}}(\Omega _t)}^2) \le CE_{1}(t) \le C. \end{aligned} \end{aligned}$$

Recall also that \(\Vert \nabla U\Vert _{L^\infty } \le C\). Sobolev embedding and Trace Theorem yield

$$\begin{aligned} \Vert \mathcal {D}_t^{l-1}v\Vert _{L^{\infty }(\Sigma _t)}^2\le C\Vert \mathcal {D}_t^{l-1}v\Vert _{H^{2}(\Sigma _t)}^2 \le C\Vert \mathcal {D}_t^{l-1}v\Vert _{H^{3}(\Omega _t)}^2 \le C E_l(t). \end{aligned}$$

Therefore we need yet to estimate \( \Vert R^{l-2}_U\Vert _{H^{\frac{5}{2}}(\Sigma _t)} \).

Arguing as in (5.13) we obtain

$$\begin{aligned} \begin{aligned} \Vert R^{l-2}_U\Vert _{H^{\frac{5}{2}}(\Sigma _t)} \le&\sum _{\begin{array}{c} |\gamma |+|\beta | \le l-1\\ |\beta |\le l-2 \end{array}}\Big ( \Vert \mathcal {D}_t^{\beta _1}v \Vert _{L^\infty } \cdots \Vert \mathcal {D}_t^{\beta _{l-2}}v \Vert _{L^\infty }\Vert \nabla ^{1+ \gamma _1} \partial _t^{\gamma _2} U\Vert _{H^{\frac{5}{2}}(\Sigma _t)}\\&\,\,\,\,\,\,\,+ \Vert \mathcal {D}_t^{\beta _1}v \Vert _{H^{\frac{5}{2}}(\Sigma _t)} \cdots \Vert \mathcal {D}_t^{\beta _{l-2}}v \Vert _{L^\infty }\Vert \nabla ^{1+ \gamma _1} \partial _t^{\gamma _2} U\Vert _{L^{\infty }(\Sigma _t)} \Big ). \end{aligned} \end{aligned}$$

Arguing as before we deduce \(\Vert \mathcal {D}_t^{\beta _i}v \Vert _{L^\infty } \le C\) for all i and

$$\begin{aligned} \Vert \mathcal {D}_t^{\beta _1}v \Vert _{H^{\frac{5}{2}}(\Sigma _t)}^2 \le \Vert \mathcal {D}_t^{\beta _1}v \Vert _{H^{3}(\Omega _t)}^2 \le E_{l-1}(t) \le C. \end{aligned}$$

Finally we use the fact that (5.9) holds for \(\beta \le l-1\) and \(\gamma _1 + \gamma _2 \le l-1\) to conclude

$$\begin{aligned} \begin{aligned} \Vert \nabla ^{1+ \gamma _1} \partial _t^{\gamma _2} U\Vert _{L^{\infty }(\Sigma _t)}^2&\le C \Vert \nabla ^{1+ \gamma _1} \partial _t^{\gamma _2} U\Vert _{H^{\frac{5}{2}}(\Sigma _t)}^2\\&\le C(1+ \Vert \nabla ^{1+ (2+\gamma _1)} \partial _t^{\gamma _2} U\Vert _{H^{\frac{1}{2}}(\Sigma _t)}^2) \le CE_l(t). \end{aligned} \end{aligned}$$

This concludes the proof. \(\square \)

We need also the following bound on the capacitary potential and for the error term \(R_U^l\), defined in (4.18), associated with it. In the first statement of the following lemma we need to relax the usual assumption on the quantity (1.5) being bounded to assume that the set \(\Omega _t\) is uniformly \(C^{1,\alpha }(\Gamma )\)-regular and that the velocity satisfies \(\Vert v\Vert _{W^{1,4}(\Sigma _t)} \le C\). The point is that we need the following estimate when we do not have the Lipschitz bound on v. This does not complicate the proof and will be useful later.

Lemma 5.6

Consider \(R_U^l\) defined in (4.18). Assume that \(\Sigma _t\) is uniformly \(C^{1,\alpha }(\Gamma )\)-regular and satisfies

$$\begin{aligned} \Vert p\Vert _{H^1(\Omega _t)} + \Vert v\Vert _{W^{1,4}(\Sigma _t)} \le M. \end{aligned}$$

There exists C such that

$$\begin{aligned} \Vert \nabla \partial _t^2 U\Vert _{L^2(\Sigma _t)}^2 + \Vert R_U^1\Vert _{L^2(\Sigma _t)}^2 \le C(1+ \Vert p\Vert _{H^2(\Omega _t)}^2)E_1(t). \end{aligned}$$
(5.15)

Let \(l \ge 2\) and assume that \(E_{l-1}(t) \le M\). Then it holds

$$\begin{aligned} \Vert \nabla \partial _t^{l+1} U\Vert _{L^2(\Sigma _t)}^2 + \Vert R_U^l\Vert _{L^2(\Sigma _t)}^2 \le C E_l(t). \end{aligned}$$
(5.16)

The constants depend on Ml and on the \(C^{1,\alpha }\)-norm of the heightfunction.

Proof

This time we only prove (5.15) since (5.16) follows from similar argument. Note also that the \(C^{1,\alpha }\)-bound on \(\Sigma _t\) implies \(C^{1,\alpha }\)-bound on U. Recall also that \(\Vert p\Vert _{H^1(\Omega _t)} \le C\) implies \(\Vert B\Vert _{L^{4}(\Sigma _t)} \le \Vert B\Vert _{H^{\frac{1}{2}}(\Sigma _t)} \le C\) by Lemma 5.2. We begin by noticing that, since \(\partial _t^{2} U\) is harmonic, it holds by Lemma 3.3

$$\begin{aligned} \Vert \nabla \partial _t^{2} U \Vert _{L^2(\Sigma _t)}\le C(1+ \Vert \partial _t^{2} U \Vert _{H^1(\Sigma _t)}). \end{aligned}$$

Then by Lemma 4.6 we have

$$\begin{aligned} \Vert \partial _t^{2} U \Vert _{H^1(\Sigma _t)}\le C(1+ \Vert \partial _\nu U (\mathcal {D}_t v \cdot \nu ) \Vert _{H^1(\Sigma _t)} + \Vert R_U^0\Vert _{H^1(\Sigma _t)}), \end{aligned}$$

where

$$\begin{aligned} R_U^0 = \sum _{|\alpha |\le 1} a_{\alpha }(v) \nabla ^{1+\alpha _1} \partial _t^{\alpha _2} U. \end{aligned}$$

Since \(\Vert \nabla U\Vert _{L^{\infty }(\Sigma _t)} \le C\), we may estimate by Proposition 2.10, \(\Vert B\Vert _{L^4} \le C\) and by the Sobolev embedding

$$\begin{aligned} \begin{aligned}&\Vert \partial _t^{2} U \Vert _{H^1(\Sigma _t)}\\&\quad \le C+ C \Vert \mathcal {D}_t v \cdot \nu \Vert _{H^1(\Sigma _t)} + C(\Vert \nabla ^2 U\Vert _{L^4(\Sigma _t)}+ \Vert B\Vert _{L^4(\Sigma _t)}) \Vert \mathcal {D}_t v \cdot \nu \Vert _{L^4(\Sigma _t)} + \Vert R_U^0\Vert _{H^1(\Sigma _t)}\\&\quad \le C+ C(1+ \Vert \nabla ^2 U\Vert _{L^4(\Sigma _t)}) \Vert \mathcal {D}_t v \cdot \nu \Vert _{H^1(\Sigma _t)} + \Vert R_U^0\Vert _{H^1(\Sigma _t)}. \end{aligned} \end{aligned}$$
(5.17)

We have by the Sobolev embedding and by Lemma 5.1

$$\begin{aligned} \Vert \nabla ^2 U\Vert _{L^4(\Sigma _t)} \le \Vert \nabla ^2 U\Vert _{H^{1/2}(\Sigma _t)} \le C(1+ \Vert p\Vert _{H^1(\Sigma _t)}). \end{aligned}$$
(5.18)

Since \(\Vert \mathcal {D}_t v \cdot \nu \Vert _{H^1(\Sigma _t)}^2 \le E_1(t)\), we need yet to show that \(\Vert R_U^0\Vert _{H^1(\Sigma _t)}^2 \le CE_1(t)\).

Since \(\Vert v\Vert _{L^\infty (\Sigma _t)} \le C \Vert v\Vert _{W^{1,4}(\Sigma _t)} \le C\) we have by the Sobolev embedding

$$\begin{aligned} \begin{aligned} \Vert R_U^0\Vert _{H^1(\Sigma _t)}&\le C \sum _{|\alpha |\le 1} \Vert v\Vert _{L^\infty } \Vert \nabla ^{1+\alpha _1} \partial _t^{\alpha _2} U\Vert _{H^1(\Sigma _t)} + C \sum _{|\alpha |\le 1} \Vert v\Vert _{W^{1,4}} \Vert \nabla ^{1+\alpha _1} \partial _t^{\alpha _2} U\Vert _{L^4(\Sigma _t)}\\&\le C(1 + \Vert \nabla ^{2} U\Vert _{H^1(\Sigma _t)} + \Vert \nabla \partial _t U\Vert _{H^1(\Sigma _t)}). \end{aligned} \end{aligned}$$

Lemma 5.1 and Lemma 3.7 yield

$$\begin{aligned} \begin{aligned} \Vert \nabla ^3 U\Vert _{H^{\frac{1}{2}}(\Sigma _t)}^2 \le C(1+ \Vert p\Vert _{H^2(\Sigma _t)}^2) \le C(1+ \Vert \nabla p\Vert _{H^{\frac{3}{2}}(\Omega _t)}^2) \le CE_{1}(t). \end{aligned} \end{aligned}$$
(5.19)

We bound \(\Vert \nabla ^2 \partial _t U\Vert _{H^{\frac{1}{2}}(\Sigma _t)}\) with a similar argument and thus we only sketch it. First, we recall that it holds \(\partial _t U = - \nabla U \cdot v\) on \(\Sigma _t\). We use Theorem 3.9, Lemma 5.2, Proposition 2.10 and (5.19) to deduce

$$\begin{aligned} \begin{aligned} \Vert \nabla ^2 \partial _t U\Vert _{H^{\frac{1}{2}}(\Sigma _t)}^2&\le C(1 + \Vert p\Vert _{H^1(\Sigma _t)}^2 + \Vert \nabla U \cdot v\Vert _{H^{\frac{5}{2}}(\Sigma _t)}^2)\\&\le C( 1+ \Vert p\Vert _{H^1(\Sigma _t)}^2 + \Vert \nabla ^3 U\Vert _{H^{\frac{1}{2}}(\Sigma _t)}^2 + \Vert v\Vert _{H^3(\Omega _t)}^2)\\&\le CE_1(t). \end{aligned} \end{aligned}$$
(5.20)

We thus deduce by (5.17), (5.18), (5.19) and (5.20) that

$$\begin{aligned} \Vert \nabla \partial _t^{2} U \Vert _{L^2(\Sigma _t)}^2 \le C(1+ \Vert p\Vert _{H^2(\Omega _t)}^2)E_1(t). \end{aligned}$$
(5.21)

We are left with

$$\begin{aligned} \Vert R_U^1\Vert _{L^2(\Sigma _t)}^2 \le C(1+ \Vert p\Vert _{H^2(\Omega _t)}^2)E_1(t). \end{aligned}$$
(5.22)

To this aim we recall the definition of \(R_U^1\) in (4.18)

$$\begin{aligned} R_U^1 = \sum _{|\alpha |+\beta \le 2, \,\beta \le 1} a_{\alpha , \beta } \mathcal {D}_t^\beta v \star \nabla ^{1+ \alpha _1} \partial _t^{\alpha _2} U. \end{aligned}$$

We use Hölder’s inequality as

$$\begin{aligned} \Vert R_U^1\Vert _{L^2(\Sigma _t)} \le C \big (\sum _{|\alpha |\le 2} \Vert \nabla ^{1+ \alpha _1} \partial _t^{\alpha _2} U\Vert _{L^2(\Sigma _t)} + \Vert \mathcal {D}_t v\Vert _{L^4(\Sigma _t)} \sum _{|\alpha |\le 1} \Vert \nabla ^{1+ \alpha _1} \partial _t^{\alpha _2} U\Vert _{L^4(\Sigma _t)}\big ). \end{aligned}$$

We have by (5.19), (5.20) and (5.21)

$$\begin{aligned} \sum _{|\alpha |\le 2} \Vert \nabla ^{1+ \alpha _1} \partial _t^{\alpha _2} U\Vert _{L^2(\Sigma _t)}^2 \le C(1+ \Vert p\Vert _{H^2(\Omega _t)}^2)E_1(t). \end{aligned}$$

By the Sobolev embedding and \(-\nabla p = \mathcal {D}_t v\) it holds

$$\begin{aligned} \Vert \mathcal {D}_t v\Vert _{L^4(\Sigma _t)} \le \Vert \mathcal {D}_t v\Vert _{H^1(\Omega _t)} \le C \Vert p\Vert _{H^2(\Omega _t)}. \end{aligned}$$

Moreover (5.19) and (5.20) imply for \(\alpha _1 + \alpha _2 \le 1\)

$$\begin{aligned} \Vert \nabla ^{1+ \alpha _1} \partial _t^{\alpha _2} U\Vert _{L^4(\Sigma _t)}^2 \le \Vert \nabla ^{1+ \alpha _1} \partial _t^{\alpha _2} U\Vert _{H^{\frac{1}{2}}(\Sigma _t)}^2 \le CE_1(t). \end{aligned}$$

Hence we have (5.22). \(\square \)

Finally we need to bound the error term \(R_p^l\), defined in (4.26), which is associated with the pressure. This term is the most difficult to treat and it turns out that the lower order case \(l=1\) is the most challenging to deal with. Therefore we state it as an own lemma.

Lemma 5.7

Let \(R_p^l\) be as defined in (4.26). Assume that (1.7) holds and \(\Vert p\Vert _{H^1(\Omega _t)}\le M\). Then it holds

$$\begin{aligned} \Vert R_p^1\Vert _{H^{\frac{1}{2}}(\Sigma _t)}^2 \le C(1+ \Vert p\Vert _{H^2(\Omega _t)}^2) E_1(t). \end{aligned}$$

for some constant \(C=C(M)\).

Proof

Let us begin by recalling that by the definition of \(R_p^1\) in (4.26), (4.27) (4.28) we may write

$$\begin{aligned} \begin{aligned} R_p^1 =&- |\nabla _\tau p|^2 - (|B|^2 - Q(t) H |\nabla U|^2) \, \partial _\nu p + a_1(\nu , \nabla v) \star B + a_2(\nu , \nabla v) \star \nabla ^2 v \\&+ \sum _{|\alpha |+ |\gamma | \le 2,\, \gamma _i \le 1} a_{\alpha ,\gamma , Q}(v) \nabla ^{1+\alpha _1} \partial _t^{\gamma _1} U \star \nabla ^{1+\alpha _2} \partial _t^{\gamma _2} U. \end{aligned} \end{aligned}$$
(5.23)

We first recall it holds \(\Vert U\Vert _{C^{1,\alpha }(\Sigma _t)} \le C\). We may bound the curvature by Sobolev embedding and interpolation as

$$\begin{aligned} \Vert B\Vert _{C^\alpha (\Sigma _t)} \le C (1+\Vert p\Vert _{C^\alpha (\Sigma _t)}) \le C(1+\Vert {\bar{\nabla }} p\Vert _{L^{\frac{11}{5}}(\Sigma _t)}) \le C(1+ \Vert p\Vert _{H^{2}(\Sigma _t)}^{\theta } \Vert p\Vert _{L^{4}(\Sigma _t)}^{1-\theta }), \end{aligned}$$

for \(\theta < \frac{4}{9}\). Recall that \( \Vert p\Vert _{H^1(\Omega _t)} \le C\) implies \(\Vert p\Vert _{L^{4}(\Sigma _t)}, \Vert B\Vert _{L^4} \le C\). By the Sobolev embedding and by Lemma 3.7 we have

$$\begin{aligned} \begin{aligned} \Vert p\Vert _{H^{2}(\Sigma _t)}^2 \le C\Vert \nabla p\Vert _{H^{1}(\Sigma _t)}^2 \le C \Vert \nabla p\Vert _{H^{\frac{3}{2}}(\Omega _t)}^2 \le C E_{1}(t). \end{aligned} \end{aligned}$$
(5.24)

Therefore we obtain

$$\begin{aligned} \Vert B\Vert _{C^\alpha (\Sigma _t)}^2 \le CE_1(t)^{\theta } \qquad \text {for } \, \theta < \frac{4}{9}. \end{aligned}$$
(5.25)

We may also bound the curvature simply as

$$\begin{aligned} \Vert B\Vert _{L^\infty (\Sigma _t)} \le \Vert B\Vert _{C^\alpha (\Sigma _t)} \le C (1+\Vert p\Vert _{C^\alpha (\Sigma _t)}) \le C(1+ \Vert p\Vert _{H^2(\Omega _t)}). \end{aligned}$$
(5.26)

Let us bound the first term in (5.23) which is of the highest order. We observe that by interpolation it holds

$$\begin{aligned} \Vert \nabla p\Vert _{L^{8}(\Sigma _t)} \le C\Vert \nabla p\Vert _{H^{1}(\Sigma _t)}^{\frac{1}{2}} \Vert \nabla p\Vert _{L^{4}(\Sigma _t)}^{\frac{1}{2}}. \end{aligned}$$

By using this, the Sobolev embedding and (2.17) we have

$$\begin{aligned} \begin{aligned} \Vert |\nabla _\tau p|^2 \Vert _{H^{1/2}(\Sigma _t)}&\le C\Vert |\nabla p|^2 \Vert _{H^{\frac{1}{2}}(\Sigma _t)} + C \Vert \nu \Vert _{W^{1,4}(\Sigma _t)} \Vert |\nabla p|^2\Vert _{L^{4}(\Sigma _t)} \\&\le C\Vert |\nabla p|^2 \Vert _{H^{1}(\Omega _t)} + C\Vert B\Vert _{L^4(\Sigma _t)} \Vert \nabla p\Vert _{L^{8}(\Sigma _t)}^2 \\&\le C\Vert \nabla p \Vert _{L^{6}(\Omega _t)} \Vert \nabla ^2 p \Vert _{L^{3}(\Omega _t)} + C \Vert \nabla p\Vert _{L^{4}(\Sigma _t)}\Vert \nabla p\Vert _{H^{1}(\Sigma _t)} \\&\le C\Vert p \Vert _{H^{2}(\Omega _t)} \Vert \nabla p \Vert _{H^{\frac{3}{2}}(\Omega _t)} \\&\le C (1+ \Vert p \Vert _{H^{2}(\Omega _t)}) E_1(t)^{\frac{1}{2}}. \end{aligned} \end{aligned}$$

This gives bound for the first term.

In order to bound the next term in (5.23) we let \({\tilde{\nu }}\) and \({\tilde{B}}\) be the harmonic extensions of the normal \(\nu \) and of the second fundamental form B to \(\Omega _t\). By maximum principle \(\Vert {\tilde{B}}\Vert _{L^\infty (\Omega _t)} \le \Vert B\Vert _{L^\infty (\Sigma _t)}\), while by standard results from harmonic analysis [16] it holds

$$\begin{aligned} \Vert {\tilde{B}}\Vert _{W^{1,3}(\Omega _t)} \le C\Vert B\Vert _{W^{1,3}(\Sigma _t)} \end{aligned}$$

and by (3.12) \(\Vert {\tilde{\nu }}\Vert _{W^{1,4}(\Omega _t)} \le C\). Then we have

$$\begin{aligned} \Vert \nabla p \cdot {\tilde{\nu }} \Vert _{H^{1}(\Omega _t)} \le C\Vert p \Vert _{H^{2}(\Omega _t)} + C \Vert {\tilde{\nu }}\Vert _{W^{1,4}(\Omega _t)} \Vert p\Vert _{W^{1,4}(\Omega _t)}\le C\Vert p \Vert _{H^{2}(\Omega _t)}. \end{aligned}$$
(5.27)

We have by (5.25), (5.27) and by the Sobolev embedding

$$\begin{aligned} \begin{aligned} \Vert |B|^2 \,&\partial _\nu p \Vert _{H^{1/2}(\Sigma _t)}^2 \le C(1+\Vert \nabla (|{\tilde{B}}|^2 \, (\nabla p \cdot {\tilde{\nu }}))\Vert _{L^{2}(\Omega _t)}^2\\&\le C(1+ \Vert B\Vert _{L^\infty (\Sigma _t)}^4\Vert \nabla p \cdot {\tilde{\nu }} \Vert _{H^1(\Omega _t)}^2 + \Vert B\Vert _{L^\infty (\Sigma _t)}^2\Vert \nabla {\tilde{B}}\Vert _{L^3(\Omega _t)}^2 \Vert \nabla p\Vert _{L^6(\Omega _t)}^2)\\&\le C(1+ \Vert p\Vert _{H^2(\Omega _t)}^2) E_1(t)^{2\theta } + E_1(t)^{\theta } \Vert B\Vert _{W^{1,3}(\Sigma _t)}^2\Vert p\Vert _{H^2(\Omega _t)}^2), \end{aligned} \end{aligned}$$

for \(\theta < \frac{4}{9}\). By interpolation in Proposition 2.8, by Lemma 5.2 and (5.24) we have

$$\begin{aligned} \Vert B\Vert _{W^{1,3}(\Sigma _t)} \le C\Vert B\Vert _{H^2(\Sigma _t)}^{\frac{5}{9}}\Vert B\Vert _{L^4(\Sigma _t)}^{\frac{4}{9}} \le C(1+ \Vert p\Vert _{H^2(\Sigma _t)}^{\frac{5}{9}})\le C E_1(t)^{\frac{1}{2} \cdot \frac{5}{9}}. \end{aligned}$$
(5.28)

Therefore, since \(\theta < \frac{4}{9}\), we may bound the second term as

$$\begin{aligned} \Vert |B|^2 \,\partial _\nu p \Vert _{H^{1/2}(\Sigma _t)}^2 \le C(1+ \Vert p\Vert _{H^2(\Omega _t)}^2)E_1(t). \end{aligned}$$

By the same argument we also have

$$\begin{aligned} \Vert H |\nabla U|^2 \partial _\nu p \Vert _{H^{1/2}(\Sigma _t)}^2 \le C(1+ \Vert p\Vert _{H^2(\Omega _t)}^2)E_1(t). \end{aligned}$$
(5.29)

Indeed, the same calculations as above lead to

$$\begin{aligned} \Vert H |\nabla U|^2 \partial _\nu p \Vert _{H^{1/2}(\Sigma _t)}^2 \le C(1+ \Vert p\Vert _{H^2(\Omega _t)}^2) E_1(t)^{\frac{4}{9}}(\Vert B\Vert _{W^{1,3}(\Sigma _t)}^2+ \Vert |\nabla U|^2\Vert _{W^{1,3}(\Sigma _t)}^2). \end{aligned}$$

Recall that (5.28) yields \(\Vert B\Vert _{W^{1,3}(\Sigma _t)}^2\le CE_1(t)^{\frac{5}{9}}\). By Lemma 5.1 we deduce

$$\begin{aligned} \begin{aligned} \Vert |\nabla U|^2\Vert _{H^2(\Sigma _t)}&\le C\Vert \nabla U\Vert _{L^\infty }\Vert \nabla U\Vert _{H^2(\Sigma _t)} \\&\le C(1+ \Vert \nabla ^2 U\Vert _{H^{\frac{1}{2}}(\Sigma _t)})\le C(1+ \Vert p\Vert _{H^2(\Sigma _t)}) \le CE_1(t)^{\frac{1}{2}}. \end{aligned} \end{aligned}$$

Hence, by interpolation

$$\begin{aligned} \Vert |\nabla U|^2\Vert _{W^{1,3}(\Sigma _t)} \le C\Vert |\nabla U|^2 \Vert _{H^2(\Sigma _t)}^{\frac{1}{3}} \Vert |\nabla U|^2\Vert _{L^\infty (\Sigma _t)}^{\frac{2}{3}} \le C E_1(t)^{\frac{1}{2} \cdot \frac{1}{3}} \end{aligned}$$

and (5.29) follows.

The term \(a_1(\nu , \nabla v) \star B \) is easy to bound and leave the details for the reader. Also the term \(a_2(\nu , \nabla v) \, \nabla ^2 v \) is not difficult and we merely point out that by interpolation

$$\begin{aligned} \Vert \nabla ^2 v\Vert _{L^4(\Omega _t)} \le C \Vert v\Vert _{H^3(\Omega _t)}^{1/2} \Vert \nabla v\Vert _{L^\infty (\Omega _t)}^{1/2} \le C \Vert v\Vert _{H^3(\Omega _t)}^{1/2}. \end{aligned}$$

Thus we have by (5.26)

$$\begin{aligned} \begin{aligned} \Vert a_1(\nu , \nabla v) \, \nabla ^2 v\Vert _{H^{\frac{1}{2}}(\Sigma _t)}&\le \Vert a_1(\nu , \nabla v) \, \nabla ^2 v\Vert _{H^{1}(\Omega _t)}\\&\le C\Vert \nabla ^3 v\Vert _{H^3(\Omega _t)} + C \Vert \nabla ^2 v\Vert _{L^4(\Omega _t)}^2 +C \Vert B\Vert _{C^\alpha (\Sigma _t)} \Vert \nabla ^2 v\Vert _{L^2(\Omega _t)}\\&\le C(1 + \Vert p \Vert _{H^2(\Omega _t)}) \Vert v \Vert _{H^3(\Omega _t)}. \end{aligned} \end{aligned}$$

Before we treat the last term in (5.23) we need to show that the coefficients \(a_{\alpha ,\gamma , Q}\) are bounded. To this aim we need to show that \(Q^{(1)}= Q'(t), Q^{(2)} = Q''(t)\), where \(Q(t)\) is defined in (2.1), are bounded since \(a_{\alpha , \gamma ,Q}\) depend smoothly on them. It is clear that it is enough to show that the first and second derivative of \(\text {Cap}(\Omega _t)\) are bounded. It is easy to see, and in fact we already calculated, that

$$\begin{aligned} \frac{d}{dt} \text {Cap}(\Omega _t) = -\frac{1}{2} \int _{\Sigma _t} |\nabla U|^2 v_n d\mathcal {H}^2. \end{aligned}$$

This is clearly bounded. We calculate further and obtain

$$\begin{aligned} \frac{d^2}{dt^2} \text {Cap}(\Omega _t) = -\frac{1}{2} \int _{\Sigma _t} H_{\Sigma _t}|\nabla U|^2 v_n d\mathcal {H}^2 - \int _{\Sigma _t} (\nabla \partial _t U \cdot \nabla U) v_n + |\nabla U|^2 \partial _t (v_n)\, d\mathcal {H}^2. \end{aligned}$$
(5.30)

The first term on RHS is clearly bounded. For the second term on RHS in (5.30) we note that since U is constant on \(\Sigma _t\) we have \(\nabla U= -|\nabla U| \nu \). Therefore

$$\begin{aligned} |\int _{\Sigma _t} \nabla \partial _t U \cdot \nabla U v_n d\mathcal {H}^2| =|\int _{\Sigma _t}|\nabla U| v_n \nabla \partial _t U \cdot \nu d\mathcal {H}^2| \le \Vert \nabla U v_n \Vert _{H^\frac{1}{2}(\Sigma _t)}\Vert \nabla \partial _t U \cdot \nu \Vert _{H^{-\frac{1}{2}}(\Sigma )}. \end{aligned}$$

We note that we may use the Kato Ponce inequality (Proposition 2.10) and Lemma 5.1 to deduce

$$\begin{aligned} \Vert \nabla U v_n \Vert _{H^\frac{1}{2}(\Sigma _t)} \le \Vert \nabla U \Vert _{H^\frac{1}{2}(\Sigma _t)} \Vert v_n\Vert _{L^\infty (\Sigma _t)}+\Vert v_n \Vert _{H^\frac{1}{2}(\Sigma _t)}\Vert \nabla U \Vert _{L^\infty (\Sigma _t)} \le C. \end{aligned}$$

Next we let \({\tilde{U}}_t\) the harmonic extension of \(\partial _t U\) in \(\Omega _t\) and note that for any \( \phi \in H^\frac{1}{2}(\Sigma _t)\) it holds

$$\begin{aligned} \begin{aligned} \int _{\Sigma _t} \phi \nabla \partial _t U \cdot \nu \, d\mathcal {H}^2 = \int _{\Omega _t} \textrm{div}\,(\phi \nabla {\tilde{U}}_t ) \,dx&\le \Vert \nabla \phi \Vert _{L^2(\Omega _t)} \Vert \nabla \partial _t {\tilde{U}}\Vert _{L^2(\Omega _t)}\\&\le \Vert \phi \Vert _{H^\frac{1}{2}(\Sigma _t)} \Vert \partial _t U\Vert _{H^\frac{1}{2}(\Sigma _t)}. \end{aligned} \end{aligned}$$

This and \(\partial _t U = - \nabla U \cdot v\) imply

$$\begin{aligned} \Vert \nabla \partial _t U \cdot \nu \Vert _{H^{-\frac{1}{2}}(\Sigma _t)} \le \Vert \partial _t U\Vert _{H^\frac{1}{2}(\Sigma _t)}= \Vert \partial _\nu U v_n\Vert _{H^\frac{1}{2}(\Sigma _t)} \le C. \end{aligned}$$

Let us estimate the last term in (5.30). We note that

$$\begin{aligned} \partial _t(v_n) = \mathcal {D}_t v \cdot \nu + a(\nabla v)\star B_{\Sigma _t}. \end{aligned}$$

By (4.11) it holds \(\textrm{div}\,\mathcal {D}_t v=- \text {Tr}((\nabla v)^2)\). Therefore we estimate

$$\begin{aligned} \begin{aligned} |\int _{\Sigma _t} |\nabla U|^2 \partial _t v_n\, d\mathcal {H}^2|&\le C + \int _{\Sigma _t} |\nabla U|^2\mathcal {D}_t v\cdot \nu d\mathcal {H}^2\\&\le C(1 + \Vert \textrm{div}\,\mathcal {D}_t v\Vert _{L^2(\Omega _t)} + \Vert \nabla U\Vert _{H^{\frac{1}{2}}(\Sigma _t)} \Vert \mathcal {D}_t v\Vert _{L^2(\Omega _t)}) \le C. \end{aligned} \end{aligned}$$

Thus we have \(|Q^{(2)}(t)| \le C\) and the coefficients \(a_{\alpha ,\gamma , Q}\) are bounded.

Let us treat the last term in (5.23). We may assume that \(\alpha _1+\gamma _1 \ge \alpha _2+\gamma _2\) and assume first that \(\alpha _1 + \gamma _1=2\) (in which case \(\alpha _2 = \gamma _2 = 0\)). This means that either \(\alpha _1 =2, \gamma _1 = 0\) or \(\alpha _1 =1, \gamma _1 = 1\). Therefore we have by (5.19) and (5.20)

$$\begin{aligned} \Vert \nabla ^{1+ \alpha _1}\partial _t^{\gamma _1} U \Vert _{H^{\frac{1}{2}}(\Sigma _t)}^2 \le CE_1(t). \end{aligned}$$
(5.31)

We extend v to the complement \(\Omega _t^c\) such that it remains Lipschitz. Recall that U (and \(\partial _t U\)) is harmonic in \(\Omega _t^c\). Since \(\Omega _t\) is bounded we may choose a large ball such that \(\Omega _t \subset B_{R/2}\) and \(\Vert U\Vert _{H^{\frac{1}{2}}(\Sigma _t)}\simeq \Vert \nabla U\Vert _{L^{2}(B_R {\setminus } \Omega _t)}\). Then by (5.18), (5.31) and by the Sobolev embedding it holds

$$\begin{aligned} \begin{aligned} \Vert a_{\alpha ,\gamma }(v)&\nabla ^{1+ \alpha _1}\partial _t^{\gamma _1} U \star \nabla U \Vert _{H^{\frac{1}{2}}(\Sigma _t)} \le C \Vert a_{\alpha ,\gamma }(v) \nabla ^{1+ \alpha _1}\partial _t^{\gamma _1} U \star \nabla U \Vert _{H^{1}(B_R \setminus \Omega _t)} \\&\le C(1+\Vert \nabla ^{2+ \alpha _1}\partial _t^{\gamma _1} U \Vert _{L^{2}(B_R \setminus \Omega _t)} +\Vert \nabla ^{1+ \alpha _1}\partial _t^{\gamma _1} U\Vert _{L^4(B_R \setminus \Omega _t)} \Vert \nabla ^{2} U\Vert _{L^4(B_R \setminus \Omega _t)}) \\&\le C(1+ \Vert \nabla ^2 U\Vert _{H^{\frac{1}{2}}(\Sigma _t)})(1+ \Vert \nabla ^{1+ \alpha _1}\partial _t^{\gamma _1} U \Vert _{H^{\frac{1}{2}}(\Sigma _t)}) \\&\le (1+ \Vert p\Vert _{H^{1}(\Sigma _t)})E_1(t)^{\frac{1}{2}}. \end{aligned} \end{aligned}$$

We are left with the last term (5.23) in the case when \(\alpha _i + \gamma _i \le 1\) for \(i = 1,2\). We estimate this by the Kato-Ponce inequality (Proposition 2.10) and the Sobolev embedding as

$$\begin{aligned} \begin{aligned} \sum _{\alpha _i + \gamma _i\le 1, \, i =1,2}&\Vert a_{\alpha ,\gamma }(v) \nabla ^{1+\alpha _1} \partial _t^{\gamma _1} U \star \nabla ^{1+\alpha _2} \partial _t^{\gamma _2} U\Vert _{H^{\frac{1}{2}}(\Sigma _t)} \\&\le C \sum _{\alpha + \gamma \le 1 } \Vert \nabla ^{1+\alpha } \partial _t^{\gamma } U\Vert _{L^{\infty }(\Sigma _t)} \sum _{\alpha + \gamma \le 1} \Vert \nabla ^{1+\alpha } \partial _t^{\gamma } U\Vert _{H^{\frac{1}{2}}(\Sigma _t)} \\&\le C \sum _{\alpha + \gamma \le 1 } \Vert \nabla ^{1+\alpha } \partial _t^{\gamma } U\Vert _{H^{\frac{3}{2}}(\Sigma _t)} \sum _{\alpha + \gamma \le 1} \Vert \nabla ^{1+\alpha } \partial _t^{\gamma } U\Vert _{H^{\frac{1}{2}}(\Sigma _t)}. \end{aligned} \end{aligned}$$

We bound the first term in the last row by (5.31)

$$\begin{aligned} \Vert \nabla ^{1+\alpha } \partial _t^{\gamma } U\Vert _{H^{\frac{3}{2}}(\Sigma _t)}^2 \le CE_1(t). \end{aligned}$$

We bound the last term in the last row when \(\alpha =1\) and \(\gamma = 0\) by (5.18) as before \(\Vert \nabla ^{2} U\Vert _{H^{\frac{1}{2}}(\Sigma _t)} \le C(1 + \Vert p\Vert _{H^1(\Sigma _t)})\). We need yet to prove

$$\begin{aligned} \Vert \nabla \partial _t U\Vert _{H^{\frac{1}{2}}(\Sigma _t)} \le C(1 + \Vert p\Vert _{H^1(\Sigma _t)}) \end{aligned}$$
(5.32)

to conclude the proof. We use the fact that \(\partial _t U\) is harmonic and \(\partial _t U = \partial _\nu U \, v_n\) and therefore by Theorem 3.9 we deduce

$$\begin{aligned} \Vert \nabla \partial _t U\Vert _{H^{\frac{1}{2}}(\Sigma _t)} \le C(1 + \Vert \partial _\nu U \, v_n\Vert _{H^{3/2}(\Sigma _t)}). \end{aligned}$$

We recall that \(\Vert B\Vert _{L^4(\Sigma _t)} \le C \) and \(\Vert B\Vert _{H^1(\Sigma _t)} \le C(1+ \Vert p\Vert _{H^1(\Sigma _t)})\). Therefore we deduce by by Proposition 2.10, the a priori bound \(\Vert v_n\Vert _{H^2(\Sigma _t)} \le C\) in (1.7) and by (5.18) that

$$\begin{aligned} \begin{aligned} \Vert \partial _\nu U \, v_n\Vert _{H^{3/2}(\Sigma _t)}&\le C(\Vert v_n\Vert _{H^{\frac{3}{2}}(\Sigma _t)} + \Vert \nabla U\Vert _{H^{\frac{3}{2}}(\Sigma _t)} + \Vert \nu \Vert _{H^{\frac{3}{2}}(\Sigma _t)}) \\&\le C(1+ \Vert v_n\Vert _{H^{2}(\Sigma _t)} + \Vert \nabla ^2 U\Vert _{H^{\frac{1}{2}}(\Sigma _t)} + \Vert B\Vert _{H^{1}(\Sigma _t)})\\&\le C(1+ \Vert p\Vert _{H^1(\Sigma _t)}). \end{aligned} \end{aligned}$$

Hence, we have (5.32) and the claim follows. \(\square \)

We conclude this section with the higher order version of Lemma 5.7.

Lemma 5.8

Let \(l \ge 2\) and consider \(R_p^l\) defined in (4.26). Assume that (1.7) holds and \(E_{l-1}(t) \le M\). There exists \(C = C(M,l)\) such that

$$\begin{aligned} \Vert R_p^l\Vert _{H^{\frac{1}{2}}(\Sigma _t)}^2 \le C E_l(t) \end{aligned}$$
(5.33)

and for integers \(1 \le k \le l-1\) and \(\varepsilon >0\) it holds

$$\begin{aligned} \Vert R_p^{l-k}\Vert _{H^{\frac{3}{2}k-1}(\Sigma _t)}^2 \le \varepsilon E_l(t) +C_\varepsilon \end{aligned}$$
(5.34)

for some constant \(C_\varepsilon =C(M,l\varepsilon )\).

Proof

Let us first prove (5.33). We begin by showing

$$\begin{aligned} \Vert R_I^l\Vert _{H^{\frac{1}{2}}(\Sigma _t)}^2 \le CE_l(t), \end{aligned}$$
(5.35)

where

$$\begin{aligned} R_{I}^l = - (|B|^2 + Q(t)\, H\, |\nabla U|^2 ) (\mathcal {D}_t^{l}v \cdot \nu ) + \langle \nabla _\tau p, \mathcal {D}_t^{l}v \rangle , \end{aligned}$$

here \(Q(t)\) is defined in (2.1). Let us first recall that \(E_{1}(t)\le C\) implies \(\Vert B\Vert _{H^{2}(\Sigma _t)} \le C\) and \(\Vert p\Vert _{H^2(\Sigma _t)} \le C\). By Lemma 5.1 this implies \(\Vert \nabla ^3 U\Vert _{H^{\frac{1}{2}}(\Sigma _t)} \le C\). In particular, \(\Vert B\Vert _{L^\infty } \le C\) and \(\Vert \nabla ^2 U \Vert _{L^\infty } \le C\). Therefore we may bound by Sobolev embedding and by Proposition 2.10

$$\begin{aligned} \begin{aligned} \Vert (|B|^2 + Q(t)\,H\, |\nabla U|^2 )&(\mathcal {D}_t^{l}v \cdot \nu )\Vert _{H^{\frac{1}{2}}(\Sigma _t)}^2 \le \Vert (|B|^2 + Q(t)\,H\, |\nabla U|^2 ) (\mathcal {D}_t^{l}v \cdot \nu )\Vert _{H^{1}(\Sigma _t)}^2 \\&\le C(1+ \Vert B\Vert _{W^{1,4}(\Sigma _t)}^2 \Vert \mathcal {D}_t^{l}v \cdot \nu \Vert _{L^4(\Sigma _t)}^2 + \Vert \mathcal {D}_t^{l}v\cdot \nu \Vert _{H^1(\Sigma _t)}^2)\\&\le C(1+ \Vert B\Vert _{H^2(\Sigma _t)}^2) \Vert \mathcal {D}_t^{l}v \cdot \nu \Vert _{H^1(\Sigma _t)}^2 \\&\le C E_l(t). \end{aligned} \end{aligned}$$

In order bound \(\Vert \nabla _\tau p \cdot \mathcal {D}_t^{l}v\Vert _{H^{\frac{1}{2}}(\Sigma _t)}\) we observe that by the curvature bound \(\Vert B\Vert _{L^\infty }\le C\), by \(-\nabla p = \mathcal {D}_t v\) and by the Sobolev embeddings \(\Vert u\Vert _{L^3(\Omega _t)} \le C\Vert u\Vert _{H^{\frac{1}{2}}(\Omega _t)}\) and \(\Vert u\Vert _{L^6(\Omega _t)} \le C\Vert u\Vert _{H^{1}(\Omega _t)}\) we have

$$\begin{aligned} \begin{aligned} \Vert \nabla _\tau p \cdot \mathcal {D}_t^{l}v\Vert _{H^{\frac{1}{2}}(\Sigma _t)}&\le C\Vert \mathcal {D}_t v \cdot \mathcal {D}_t^{l}v\Vert _{H^{1}(\Omega _t)}\\&\le C(1+ \Vert \nabla \mathcal {D}_t v \cdot \mathcal {D}_t^{l}v\Vert _{L^{2}(\Omega _t)} + \Vert \mathcal {D}_t v \cdot \nabla \mathcal {D}_t^{l}v\Vert _{L^{2}(\Omega _t)})\\&\le C(1 + \Vert \nabla \mathcal {D}_t v\Vert _{L^{3}(\Omega _t)} \Vert \mathcal {D}_t^{l}v\Vert _{L^{6}(\Omega _t)} + \Vert \mathcal {D}_t v\Vert _{L^{6}(\Omega _t)} \Vert \nabla \mathcal {D}_t^{l}v\Vert _{L^{3}(\Omega _t)})\\&\le C(1 + \Vert \mathcal {D}_t v\Vert _{H^{\frac{3}{2}}(\Omega _t)} \Vert \mathcal {D}_t^{l}v\Vert _{H^{1}(\Omega _t)} + \Vert \mathcal {D}_t v\Vert _{H^{1}(\Omega _t)} \Vert \mathcal {D}_t^{l}v\Vert _{H^{\frac{3}{2}}(\Omega _t)}). \end{aligned} \end{aligned}$$

By definition of \(E_l(t)\) it holds

$$\begin{aligned} \Vert \mathcal {D}_t v\Vert _{H^{\frac{3}{2}}(\Omega _t)}^2 \le E_1(t) \le E_{l-1}(t) \le C \quad \text {and} \quad \Vert \mathcal {D}_t^{l}v\Vert _{H^{\frac{3}{2}}(\Omega _t)}^2\le E_{l}(t). \end{aligned}$$

Therefore we have \(\Vert \nabla _\tau p \cdot \mathcal {D}_t^{l}v\Vert _{H^{\frac{1}{2}}(\Sigma _t)}^2 \le C E_{l}(t) \) and (5.35) follows.

Let us next show

$$\begin{aligned} \Vert R_{II}^l\Vert _{H^{\frac{1}{2}}(\Sigma _t)}^2 \le CE_l(t), \end{aligned}$$
(5.36)

where

$$\begin{aligned} R_{II}^l = \sum _{|\alpha |\le 1, \, |\beta |\le l-1}a_{\alpha , \beta }(B) \overbrace{\nabla ^{1+\alpha _1} \mathcal {D}_t^{\beta _1} v \star \cdots \star \nabla ^{1+\alpha _{l+1}} \mathcal {D}_t^{\beta _{l+1}} v}^{=: R_{\alpha , \beta }(v)}. \end{aligned}$$

We first observe that we may ignore the coefficients \(a_{\alpha , \beta }(B) \). Indeed, we may extend B to \(\Omega _t\), call the extension \({\tilde{B}}\), such that \(\Vert {\tilde{B}}\Vert _{H^2(\Omega _t)} \le C\). Then by the above notation

$$\begin{aligned} \begin{aligned} \Vert R_{II}^l\Vert _{H^{\frac{1}{2}}(\Sigma _t)}&\le C \Vert R_{II}^l\Vert _{H^{1}(\Omega _t)} \le C(1+ \sum _{\alpha , \beta } \Vert \nabla (\nabla a_{\alpha , \beta }({\tilde{B}}) \star R_{\alpha , \beta }(v))\Vert _{L^2(\Omega _t)}\\&\le C(1+ \sum _{\alpha , \beta } \Vert \nabla {\tilde{B}} \star R_{\alpha , \beta }(v)\Vert _{L^2(\Omega _t)} + \Vert \nabla R_{\alpha , \beta }(v)\Vert _{L^2(\Omega _t)}) \\&\le C(1+ \sum _{\alpha , \beta } \Vert \nabla {\tilde{B}}\Vert _{L^4(\Omega _t)} \Vert R_{\alpha , \beta }(v)\Vert _{L^4(\Omega _t)} + \Vert \nabla R_{\alpha , \beta }(v)\Vert _{L^2(\Omega _t)}) \\&\le C(1+ \sum _{\alpha , \beta } \Vert \nabla R_{\alpha , \beta }(v)\Vert _{L^2(\Omega _t)}) \\&\le C\big (1 + \sum _{|\alpha |\le 2, \, |\beta |\le l-1} \Vert \nabla ^{1+\alpha _1} \mathcal {D}_t^{\beta _1} v \star \cdots \star \nabla ^{1+\alpha _{l+1}} \mathcal {D}_t^{\beta _{l+1}} v\Vert _{L^2(\Omega _t)}\big ). \end{aligned} \end{aligned}$$
(5.37)

By the assumption \(E_{l-1}(t) \le C\) and by (5.6) we deduce \(\Vert \nabla ^{1+ \alpha _i} \mathcal {D}_t^{\beta _i} v\Vert _{L^\infty } \le C\) for \(\alpha _i + \beta _i \le l-2\). We note also that by \(|\alpha | \le 2\) and \(|\beta |\le l-1\) it follows that \(|\alpha | + |\beta | \le l+1\). We ignore all the terms in the last row of (5.37) which indexes satisfy \(\alpha _i + \beta _i \le l-2\) as these are uniformly bounded. For the rest of the terms we use Hölder’s inequality and relabel the indexes (note that below we assume \(\alpha \le 2\) and \(\beta \le l-1\))

$$\begin{aligned} \begin{aligned} \sum _{|\alpha |\le 2, \, |\beta |\le l-1}&\Vert \nabla ^{1+\alpha _1} \mathcal {D}_t^{\beta _1} v \star \cdots \star \nabla ^{1+\alpha _{l+1}} \mathcal {D}_t^{\beta _{l+1}} v\Vert _{L^2(\Omega _t)} \\&\le C\sum _{\alpha \le 2, \beta \le l-1} \Vert \nabla ^{1 + \alpha } \mathcal {D}_t^{\beta } v\Vert _{L^2}^2 + \sum _{\alpha + \beta = l} \Vert \nabla ^{1 + \alpha } \mathcal {D}_t^{\beta } v\Vert _{L^6}^2 \cdot \sum _{\alpha + \beta = 1} \Vert \nabla ^{1 + \alpha } \mathcal {D}_t^{\beta } v\Vert _{L^3}^2 \\&\,\,\,\,\,\,\,\,+ \sum _{\alpha + \beta = l-1} \Vert \nabla ^{1 + \alpha } \mathcal {D}_t^{\beta } v\Vert _{L^3}^2 \cdot \sum _{\alpha + \beta = 2} \Vert \nabla ^{1 + \alpha } \mathcal {D}_t^{\beta } v\Vert _{L^6}^2 + \sum _{\alpha + \beta \le l-1} \Vert \nabla ^{1 + \alpha } \mathcal {D}_t^{\beta } v\Vert _{L^6}^6 . \end{aligned} \end{aligned}$$
(5.38)

To bound the first term on the RHS of (5.38) we simply note that for \(\beta \le l-1\) and \(\alpha \le 2\) it holds

$$\begin{aligned} \Vert \nabla ^{1+\alpha } \mathcal {D}_t^{\beta } v\Vert _{L^2(\Omega _t)}^2 \le C\Vert \mathcal {D}_t^{\beta } v\Vert _{H^3(\Omega _t)}^2 \le C E_{l}(t). \end{aligned}$$

For the second and the third terms we have first for \(\alpha + \beta \le l\) and \(\beta \le l-1\) (which include the case \(\alpha + \beta = 2\) as \(l \ge 2\)) that

$$\begin{aligned} \Vert \nabla ^{1 + \alpha } \mathcal {D}_t^{\beta } v\Vert _{L^6(\Omega _t)}^2 \le C\Vert \mathcal {D}_t^{\beta } v\Vert _{H^{l -\beta +2}(\Omega _t)}^2 \le C \Vert \mathcal {D}_t^{l+1 -(l+1+\beta )} v\Vert _{H^{\frac{3}{2}(l+1+\beta )}(\Omega _t)}^2 \le C E_{l}(t). \end{aligned}$$

For \(\alpha + \beta \le l-1\) (which includes the case \(\alpha + \beta = 1\)) we deduce

$$\begin{aligned} \Vert \nabla ^{1 + \alpha } \mathcal {D}_t^{\beta } v\Vert _{L^3(\Omega _t)}^2 \le C \Vert \mathcal {D}_t^{\beta }v\Vert _{H^{\frac{1}{2} +(l- \beta )}(\Omega _t)}^2 \le C E_{l-1}(t) \le C. \end{aligned}$$
(5.39)

For the last term we interpolate in the fluid domain \(\Omega _t \subset \mathbb {R}^3\) for \(\alpha + \beta \le l-1\) as

$$\begin{aligned} \begin{aligned} \Vert \nabla ^{1 + \alpha } \mathcal {D}_t^{\beta } v\Vert _{L^6(\Omega _t)}&\le \Vert \nabla ^{1+\alpha } \mathcal {D}_t^{\beta } v\Vert _{H^2(\Omega _t)}^{1/3} \Vert \nabla ^{1 + \alpha } \mathcal {D}_t^{\beta } v\Vert _{L^3(\Omega _t)}^{2/3}\le C E_l(t)^{1/6} \Vert \nabla ^{1 + \alpha } \mathcal {D}_t^{\beta } v\Vert _{L^3}^{2/3}. \end{aligned} \end{aligned}$$

By (5.39) we have \(\Vert \nabla ^{1 + \alpha } \mathcal {D}_t^{\beta } v\Vert _{L^3}\le C\) and thus

$$\begin{aligned} \Vert \nabla ^{1 + \alpha } \mathcal {D}_t^{\beta } v\Vert _{L^6}^6 \le C E_l(t). \end{aligned}$$

By combing the previous estimates with (5.37) and (5.38) we obtain

$$\begin{aligned} \Vert R_{II}^l\Vert _{H^{1/2}(\Sigma _t)}^2 \le C(1+\Vert \nabla R_{II}^l\Vert _{L^{2}(\Omega _t)}^2) \le C E_l(t), \end{aligned}$$

and (5.36) follows.

We are left to prove

$$\begin{aligned} \Vert R_{III}^l\Vert _{H^{\frac{1}{2}}(\Sigma _t)}^2 \le CE_l(t), \end{aligned}$$
(5.40)

where

$$\begin{aligned} R_{III}^l = \sum _{\begin{array}{c} |\alpha | + |\beta | + |\gamma |\le l+1\\ |\beta |\le l-1, \gamma _i \le l \end{array}} a_{\alpha , \beta ,\gamma ,Q}(v) \mathcal {D}_t^{\beta _1} v \star \cdots \star \mathcal {D}_t^{\beta _{l-1}} v \star \nabla ^{1+\alpha _1} \partial _t^{\gamma _1} U \star \nabla ^{1+\alpha _1} \partial _t^{\gamma _2} U \end{aligned}$$

and the coefficients \(a_{\alpha ,\beta ,\gamma ,Q}\) depend on the time derivatives of \(Q(t)\) up to order \(l+1\). Recall that \(Q(t)\) is defined in (2.1)

This time we will not give the argument which proves the boundedness of \(Q(t)^{(l+1)} = \frac{d^{l+1}}{dt^{l+1}} Q(t)\) as it simpler than the rest of the proof and is similar to the argument in (5.30).

Recall that by (5.6) it holds \(\Vert \mathcal {D}_t^{\beta _i} v\Vert _{L^\infty } \le C\) for \(\beta _i \le l-2\). On the other hand for \(\alpha + \gamma \le l-1\) we have by Lemma 5.5

$$\begin{aligned} \begin{aligned} \Vert \nabla ^{1+\alpha }\partial _t^\gamma U\Vert _{L^{\infty }(\Sigma _t)}^2 \le C\Vert \nabla ^{1+\alpha }\partial _t^\gamma U\Vert _{H^{\frac{3}{2}}(\Sigma _t)}^2\le C(1+\Vert \nabla ^{2+\alpha }\partial _t^\gamma U\Vert _{H^{\frac{1}{2}}(\Sigma _t)}^2) \le CE_{l-1}(t) \le C. \end{aligned} \end{aligned}$$
(5.41)

Similarly we have

$$\begin{aligned} \Vert \mathcal {D}_t^{\beta } v\Vert _{H^{\frac{1}{2}}(\Sigma _t)} \le C \quad \text {for }\, \beta \le l-1\quad \text {and} \quad \Vert \nabla ^{1+\alpha }\partial _t^\gamma U\Vert _{H^{\frac{1}{2}}(\Sigma _t)} \le C \quad \text {for} \, \alpha +\gamma \le l. \end{aligned}$$
(5.42)

Therefore we may ignore all the terms with indexes which satisfy \(\beta _i \le l-2\) and \(\alpha _i + \gamma _i \le l-1\). Recall that we assume \(\beta _1 \ge \dots \ge \beta _{l-1}\). Therefore we may estimate by Proposition 2.10, (5.41) and by (5.42)

$$\begin{aligned} \begin{aligned}&\Vert R_{III}^l\Vert _{H^{\frac{1}{2}}(\Sigma _t)} \\&\quad \le \sum _{\alpha + \gamma \le l}C\big (\Vert \mathcal {D}_t^{l-1} v\Vert _{L^{\infty }(\Sigma _t)}\Vert \nabla ^{1+\alpha }\partial _t^\gamma U\Vert _{H^{\frac{1}{2}}(\Sigma _t)} + \Vert \mathcal {D}_t^{l-1} v\Vert _{H^{\frac{1}{2}}(\Sigma _t)}\Vert \nabla ^{1+\alpha }\partial _t^\gamma U\Vert _{L^{\infty }(\Sigma _t)}\big )\\&\,\,\,\,\,\,\,\,\,\,\,+C \sum _{\alpha + \gamma \le l+1, \gamma \le l}\Vert \nabla ^{1+\alpha }\partial _t^\gamma U\Vert _{H^{\frac{1}{2}}(\Sigma _t)}\\&\quad \le C\big (\Vert \mathcal {D}_t^{l-1} v\Vert _{L^{\infty }(\Sigma _t)} + \sum _{\alpha + \gamma \le l}\Vert \nabla ^{1+\alpha }\partial _t^\gamma U\Vert _{L^{\infty }(\Sigma _t)} + \sum _{\begin{array}{c} \alpha + \gamma \le l+1 \\ \gamma \le l \end{array}}\Vert \nabla ^{1+\alpha }\partial _t^\gamma U\Vert _{H^{\frac{1}{2}}(\Sigma _t)} \big ). \end{aligned} \end{aligned}$$

We estimate the first term in the last row by Sobolev embedding

$$\begin{aligned} \Vert \mathcal {D}_t^{l-1} v\Vert _{L^{\infty }(\Sigma _t)}^2 \le C \Vert \mathcal {D}_t^{l-1} v\Vert _{H^{3}(\Omega _t)}^2 \le CE_{l}(t). \end{aligned}$$

For the second we use Sobolev embedding and Lemma 5.5 and obtain for \(\alpha +\gamma \le l\)

$$\begin{aligned} \Vert \nabla ^{1+\alpha }\partial _t^\gamma U\Vert _{L^{\infty }(\Sigma _t)}^2 \le C(1+\Vert \nabla ^{2+\alpha }\partial _t^\gamma U\Vert _{H^{\frac{1}{2}}(\Sigma _t)}^2 )\le CE_l(t). \end{aligned}$$

The same argument also implies

$$\begin{aligned} \Vert \nabla ^{1+\alpha }\partial _t^\gamma U\Vert _{H^{\frac{1}{2}}(\Sigma _t)}^2 \le CE_{l}(t) \end{aligned}$$

for \(\alpha + \gamma \le l+1\) and \(\gamma \le l\). Hence we obtain (5.40) and this concludes the proof of (5.33).

Let us then prove (5.34). Let us first treat the first term in the definition of \(R_{p}^{l-k}\) and bound \(\Vert R_I^{l-k}\Vert _{H^{\frac{3}{2}k-1}(\Sigma _t)}\). We first observe that the case \(k=1\) follows from (5.33). Let us then assume \(k \ge 2\). By the Sobolev embedding it holds \(\Vert u\Vert _{L^\infty (\Sigma )} \le C \Vert u\Vert _{L^{\frac{3}{2}k-1}(\Sigma )}\). We use this and the Kato-Ponce inequality in Proposition 2.10 to deduce that

$$\begin{aligned} \begin{aligned}&\Vert R_I^{l-k}\Vert _{H^{\frac{3}{2}k-1}(\Sigma _t)} \\&\quad \le C(1+ \Vert B\Vert _{H^{\frac{3}{2}k-1}(\Sigma _t)}^{2} + \Vert \nabla U\Vert _{H^{\frac{3}{2}k-1}(\Sigma _t)}^{2}+ \Vert \mathcal {D}_t^{l-k} v\Vert _{H^{\frac{3}{2}k-1}(\Sigma _t)}^{2} + \Vert \nabla p \Vert _{H^{\frac{3}{2}k-1}(\Sigma _t)}^{2}). \end{aligned} \end{aligned}$$

Let us show that all the terms on RHS are bounded.

To this aim we first recall that the bound \(E_{l-1}(t) \le C\) implies \(\Vert B\Vert _{H^{\frac{3}{2} l -1}(\Sigma _t)} \le C\). Since \(k \le l-1\) we deduce \(\Vert B\Vert _{H^{\frac{3}{2}k-1}(\Sigma _t)} \le C\). Lemma 5.1 implies \(\Vert \nabla U\Vert _{H^{\frac{3}{2}k-1}(\Sigma _t)}^2 \le C E_{l-1}(t) \le C\). The condition \(k \le l-1\) and the Trace Theorem also yields

$$\begin{aligned} \Vert \mathcal {D}_t^{l-k} v\Vert _{H^{\frac{3}{2}k-1}(\Sigma _t)}^2 \le C\Vert \mathcal {D}_t^{l-k} v\Vert _{H^{\frac{3}{2}k}(\Omega _t)}^2 \le C E_{l-1}(t) \le C. \end{aligned}$$

Similarly we deduce by \(-\nabla p = \mathcal {D}_t v\) that \( \Vert \nabla p \Vert _{H^{\frac{3}{2}k-1}(\Sigma _t)}\le C\). Hence, we have

$$\begin{aligned} \Vert R_I^{l-k}\Vert _{H^{\frac{3}{2}k-1}(\Sigma _t)}\le C. \end{aligned}$$

Let us then bound \(\Vert R_{II}^{l-k}\Vert _{H^{\frac{3}{2}k-1}(\Sigma _t)}\). As before, Proposition 2.10 and the Sobolev embedding yield

$$\begin{aligned} \Vert R_{II}^{l-k}\Vert _{H^{\frac{3}{2}k-1}(\Sigma _t)} \le \sum _{\alpha \le 1, \beta \le l-k-1}C\left( 1+ \Vert a_{\alpha ,\beta }(B)\Vert _{H^{\frac{3}{2}k-1}(\Sigma _t)}^2 + \Vert \nabla ^{1+\alpha } \mathcal {D}_t^{\beta } v\Vert _{H^{\frac{3}{2}k-1}(\Sigma _t)}^q\right) \end{aligned}$$

for \(q \ge 1\). Recall that \(\Vert B\Vert _{H^{\frac{3}{2}k-1}(\Sigma _t)} \le C\) and \(k \ge 2\). Therefore \(\Vert B\Vert _{L^\infty } \le C \) and thus \(\Vert a_{\alpha ,\beta }(B)\Vert _{H^{\frac{3}{2}k-1}(\Sigma _t)} \le C\). On the other hand by \(\alpha \le 1\), \(\beta \le l - (k+1)\) and by Lemma 3.7 we deduce

$$\begin{aligned} \Vert \nabla ^{1+\alpha } \mathcal {D}_t^{\beta } v\Vert _{H^{\frac{3}{2}k-1}(\Sigma _t)}^2 \le C \Vert \mathcal {D}_t^{\beta } v\Vert _{H^{\frac{3}{2}(k+1)}(\Omega _t)}^2 \le C \sum _{i =0}^{l-1} \Vert \mathcal {D}_t^{l-i} v\Vert _{H^{\frac{3}{2}i}(\Omega _t)}^2 \le C E_{l-1}(t) \le C. \end{aligned}$$
(5.43)

Hence, we have \(\Vert R_{II}^{l-k}\Vert _{H^{\frac{3}{2}k-1}(\Sigma _t)} \le C\).

Let us finally treat \(R_{III}^{l-k}\). We note that it holds \(\Vert v\Vert _{H^{\frac{3}{2}k-1}(\Sigma _t)}^2 \le CE_{l-1}(t) \le C\) and therefore we may ignore the coefficients \(a_{\alpha ,\beta ,\gamma , Q}(v)\). Then by Proposition 2.10 and by the Sobolev embedding we have

$$\begin{aligned} \begin{aligned}&\!\!\!\! \Vert R_{III}^{l-k}\Vert _{H^{\frac{3}{2}k-1}(\Sigma _t)} \\&\le C\left( 1+ \sum _{\beta \le l-(k+1)} \Vert \mathcal {D}_t^{\beta } v\Vert _{H^{\frac{3}{2}k-1}(\Sigma _t)}^{q} \cdot \cdot \sum _{\begin{array}{c} |\alpha | + |\gamma | \le l-k+1, \\ |\gamma | \le l-k \end{array}} \Vert \nabla ^{1+\alpha _1}\partial _t^{\gamma _1} U\Vert _{H^{\frac{3}{2}k-1}(\Sigma _t)}\Vert \nabla ^{1+\alpha _2}\partial _t^{\gamma _2} U\Vert _{H^{\frac{3}{2}k-1}(\Sigma _t)}\right) \end{aligned} \end{aligned}$$
(5.44)

for \(q \ge 1\). By (5.43) we have \(\Vert \mathcal {D}_t^{\beta } v\Vert _{H^{\frac{3}{2}k-1}(\Sigma _t)} \le C\) for all \(\beta \le l-(k+1)\). To bound the last term we may assume that \(\alpha _1 + \gamma _1 \ge \alpha _1 + \gamma _1 \). If \(\alpha _1 + \gamma _1 = l-k +1\) then necessarily \(\alpha _2 + \gamma _2 \le l-k\). Therefore by Lemma 5.5

$$\begin{aligned} \Vert \nabla ^{1+\alpha _1}\partial _t^{\gamma _1} U\Vert _{H^{\frac{3}{2}k-1}(\Sigma _t)}^2 \le C\left( 1+ \Vert \nabla ^{1+(\alpha _1 +k-1)}\partial _t^{\gamma _1} U\Vert _{H^{\frac{k-1}{2} + \frac{1}{2}}(\Sigma _t)}^2\right) \le \varepsilon E_l(t) + C_\varepsilon \end{aligned}$$

and

$$\begin{aligned} \Vert \nabla ^{1+\alpha _2}\partial _t^{\gamma _2} U\Vert _{H^{\frac{3}{2}k-1}(\Sigma _t)}^2 \le C\left( 1+ \Vert \nabla ^{1+(\alpha _2+k-1) }\partial _t^{\gamma _2} U\Vert _{H^{\frac{k-1}{2} + \frac{1}{2}}(\Sigma _t)}^2\right) \le CE_{l-1}(t) \le C. \end{aligned}$$

Therefore we deduce by (5.44) that

$$\begin{aligned} \Vert R_{III}^{l-k}\Vert _{H^{\frac{3}{2}k-1}(\Sigma _t)}^2 \le \varepsilon E_l(t) + C_\varepsilon . \end{aligned}$$

This concludes the proof of (5.34). \(\square \)

6 First Regularity Estimates

In this section we prove our first regularity estimates for the solution of (1.3). We assume that the solution satisfies the a priori estimates (1.7), i.e., \(\Lambda _T <\infty \) and \(\sigma _T>0\), where \(\Lambda _T\) and \(\sigma _T\) are defined in (1.5) and (1.4). We recall that

$$\begin{aligned} \Lambda _T:= \sup _{t \in (0,T]} \left( \Vert h(\cdot ,t)\Vert _{C^{1,\alpha }(\Sigma _t)} + \Vert \nabla v(\cdot ,t) \Vert _{L^\infty (\Omega _t)} + \Vert v_n(\cdot ,t) \Vert _{H^2(\Sigma _t) } \right) . \end{aligned}$$

In particular, bound on \(\Lambda _T\) does not imply curvature bounds, and thus we need to be careful as we may not use e.g. the interpolation results from Proposition 2.8. Our goal in this section is to show that the a priori estimates (1.7) imply the following bounds for the pressure

$$\begin{aligned} \sup _{t \le T}\Vert p\Vert _{H^1(\Omega _t)} \le C \qquad \text {and} \qquad \int _0^T \Vert p\Vert _{H^2(\Omega _t)}^2 \, dt \le C. \end{aligned}$$

The first bound above is important as it implies \(\Vert B_{\Sigma }\Vert _{L^4(\Sigma _t)} \le C\), which is crucial e.g. for the interpolation inequality in Proposition 2.8 to hold. The second estimate is important for the first order energy estimate which we prove in the next section in Proposition 7.1.

Let us begin by stating regularity estimates that we have by the a priori estimate. First, recall once again that by the uniform \(C^{1,\alpha }(\Gamma )\)-bound we have for the capacitary potential U that \(\Vert U\Vert _{C^{1,\alpha }({\bar{\Omega }}_t^c)} \le C\). Let us prove the following estimates for the second fundamental form and for the capcacitary potential.

Lemma 6.1

Assume that (1.7) holds for \(T>0\). Then for all \(t < T\) we have

$$\begin{aligned} \Vert B\Vert _{L^4(\Sigma _t)}^4 \le \varepsilon \Vert p\Vert _{H^1(\Sigma _t)}^2 + C_{\varepsilon } \end{aligned}$$

for \(C= C(M, \varepsilon )\) and

$$\begin{aligned} \Vert B\Vert _{H^1(\Sigma _t)} + \Vert \nabla ^2 U \Vert _{H^{\frac{1}{2}}(\Sigma _t)}\le C(1+ \Vert p\Vert _{H^1(\Sigma _t)}) \end{aligned}$$

for \(C= C(M)\).

Proof

We denote the height-function by \(h = h(\cdot ,t)\). Then by standard calculations (see e.g. [22, 43]) we may write the second fundamental form on \(\Sigma \) as \(B = a(h,{\bar{\nabla }} h) {\bar{\nabla }}^2\,h\). Therefore we may bound

$$\begin{aligned} \Vert B\Vert _{L^4(\Sigma _t)}^4 \le C(1+\Vert {\bar{\nabla }}^2 h\Vert _{L^4(\Gamma )}^4) \quad \text {and} \quad \Vert {\bar{\nabla }}^3 h\Vert _{L^2(\Gamma )}^2 \le C(1+ \Vert {\bar{\nabla }} B \Vert _{L^2(\Sigma _t)}^2 + \Vert {\bar{\nabla }}^2 h\Vert _{L^4(\Gamma )}^4). \end{aligned}$$

We use interpolation on \(\Gamma \) as

$$\begin{aligned} \Vert {\bar{\nabla }}^2 h\Vert _{L^4(\Gamma )} \le C \Vert {\bar{\nabla }}^3 h\Vert _{L^2(\Gamma )}^{\theta } \Vert h\Vert _{C^{1,\alpha }(\Gamma )}^{1-\theta } \le C \Vert {\bar{\nabla }}^3 h\Vert _{L^2(\Gamma )}^{\theta } \end{aligned}$$

for \(\theta < 1/2\). This implies by Young’s inequality \(\Vert {\bar{\nabla }}^2\,h\Vert _{L^4(\Gamma )}^4 \le \varepsilon \Vert {\bar{\nabla }}^3\,h\Vert _{L^2(\Gamma )}^2 + C_\varepsilon \). Thus by choosing \(\varepsilon \) small we obtain

$$\begin{aligned} \Vert {\bar{\nabla }}^3 h\Vert _{L^2(\Gamma )}^2 \le C(1+ \Vert {\bar{\nabla }} B \Vert _{L^2(\Sigma _t)}^2) \end{aligned}$$

and

$$\begin{aligned} \Vert B\Vert _{L^4(\Sigma _t)}^4 \le \varepsilon \Vert {\bar{\nabla }}^3 h\Vert _{L^2(\Gamma )}^2 + C_\varepsilon . \end{aligned}$$

By the Simon’s identity (2.11) we deduce

$$\begin{aligned} \Vert {\bar{\nabla }} B \Vert _{L^2(\Sigma _t)}^2 \le \Vert {\bar{\nabla }} H \Vert _{L^2(\Sigma _t)}^2 + C \Vert B\Vert _{L^4(\Sigma _t)}^4. \end{aligned}$$

Therefore we have

$$\begin{aligned} \Vert B\Vert _{L^4(\Sigma _t)}^4 \le \varepsilon \Vert H \Vert _{H^1(\Sigma _t)}^2 + C_\varepsilon \end{aligned}$$
(6.1)

and

$$\begin{aligned} \Vert B \Vert _{H^1(\Sigma _t)} \le C(1+\Vert H \Vert _{H^1(\Sigma _t)}). \end{aligned}$$
(6.2)

Let us consider the capacitary potential U. Let us show that even though we may not use Proposition 2.8, the \(C^{1,\alpha }(\Gamma )\)-regularity still implies the following weak interpolation inequality

$$\begin{aligned} \Vert \nabla U \Vert _{H^1(\Sigma _t)} \le \varepsilon \Vert \nabla ^2 U \Vert _{H^{\frac{1}{2}}(\Sigma _t)} + C_\varepsilon \Vert \nabla U \Vert _{L^{2}(\Sigma _t)}\le \varepsilon \Vert \nabla ^2 U \Vert _{H^{\frac{1}{2}}(\Sigma _t)} + C_\varepsilon . \end{aligned}$$
(6.3)

In order to prove (6.3) we first observe that the \(C^{1,\alpha }(\Gamma )\)-regularity of \(\Sigma _t\) implies the following inequalities for \(p \in (1,2)\) and \(u,v \in C^\infty (\Sigma _t)\)

$$\begin{aligned} \Vert u\Vert _{L^2(\Sigma _t)} \le \varepsilon \Vert u\Vert _{H^{\frac{1}{2}}(\Sigma _t)} + C_\varepsilon \Vert u\Vert _{L^p(\Sigma _t)} \quad \text {and} \quad \Vert \nabla _\tau v\Vert _{L^p(\Sigma _t)} \le \delta \Vert v\Vert _{H^1(\Sigma _t)} + C_\delta \Vert v\Vert _{L^2(\Sigma _t)}. \end{aligned}$$

We apply these for \(u = \nabla ^2 U\) and \(v = \nabla U\) and have

$$\begin{aligned} \begin{aligned}&\Vert \nabla ^2 U\Vert _{L^2(\Sigma _t)} \le \varepsilon \Vert \nabla ^2 U\Vert _{H^{\frac{1}{2}}(\Sigma _t)} + C_\varepsilon \Vert \nabla ^2 U\Vert _{L^p(\Sigma _t)} \quad \text {and} \\&\Vert \nabla _\tau \nabla U\Vert _{L^p(\Sigma _t)} \le \delta \Vert \nabla U\Vert _{H^1(\Sigma _t)} + C_\delta \Vert \nabla U\Vert _{L^2(\Sigma _t)}. \end{aligned} \end{aligned}$$

Since \(\partial _{x_i} U\) is harmonic and \(\Sigma _t\) is \(C^{1,\alpha }(\Gamma )\)-regular we have by [19] that

$$\begin{aligned} \Vert \nabla ^2 U\Vert _{L^p(\Sigma _t)} \le C( \Vert \nabla _\tau \nabla U\Vert _{L^p(\Sigma _t)} + \Vert \nabla U\Vert _{L^2(\Omega _t)}). \end{aligned}$$

Therefore by first choosing \(\varepsilon \) small and then \(\delta \) even smaller we obtain (6.3).

By \(p = H - \frac{Q(t)}{2} |\nabla U|^2\), where \(Q(t)\) is defined in (2.1), we may estimate

$$\begin{aligned} \Vert H\Vert _{H^{1}(\Sigma _t)} \le C( \Vert p\Vert _{H^1(\Sigma _t)} + \Vert \nabla U \Vert _{H^1(\Sigma _t)}). \end{aligned}$$

Then by (6.3) we have

$$\begin{aligned} \Vert H\Vert _{H^{1}(\Sigma _t)} \le C_\varepsilon (1+ \Vert p\Vert _{H^1(\Sigma _t)}) + \varepsilon \Vert \nabla ^2 U \Vert _{H^{\frac{1}{2}}(\Sigma _t)}. \end{aligned}$$

We use Theorem 3.9 and (6.2) and have

$$\begin{aligned} \Vert \nabla ^2 U \Vert _{H^{\frac{1}{2}}(\Sigma _t)} \le C( 1 + \Vert B\Vert _{H^1(\Sigma _t)}) \le C(1+\Vert H \Vert _{H^1(\Sigma _t)}). \end{aligned}$$

Therefore we deduce from the two above inequalities that

$$\begin{aligned} \Vert H\Vert _{H^{1}(\Sigma _t)} \le C(1+ \Vert p\Vert _{H^1(\Sigma _t)}). \end{aligned}$$

The claim follows from this together with (6.1) and (6.2). \(\square \)

Let us proceed to the following regularity estimate.

Lemma 6.2

Assume that the a priori estimates (1.7) hold for \(T>0\). Then

$$\begin{aligned} \sup _{t \in [0,T]} \Vert p\Vert _{L^2(\Sigma _t)}^2 + \int _0^T \Vert p\Vert _{H^1(\Sigma _t)}^2 \, dt \le C(1+ T), \end{aligned}$$

for \(C = C(M)\).

Proof

The idea is to consider the following function

$$\begin{aligned} \Psi (t):= \int _{\Sigma _t} p\, ( \nabla v \, \nu )\cdot \nu + \varepsilon p^2 \, d \mathcal {H}^2, \end{aligned}$$

where the choice of \(\varepsilon \) will be clear later. First, we observe that under the a priori estimates (1.7) v is uniformly Lipschitz and therefore \(\Psi \) is bounded from below by

$$\begin{aligned} \Psi (t) \ge - C\Vert p(\cdot , t)\Vert _{L^1(\Sigma _t)} + \varepsilon \Vert p(\cdot , t)\Vert _{L^2(\Sigma _t)}^2 \ge -C_\varepsilon + \frac{\varepsilon }{2}\Vert p(\cdot , t)\Vert _{L^2(\Sigma _t)}^2, \end{aligned}$$
(6.4)

where the last inequality follows from the Young’s inequality, i.e., \(\Vert p(\cdot , t)\Vert _{L^1(\Sigma _t)} \le \frac{\varepsilon }{2} \Vert p(\cdot , t)\Vert _{L^2(\Sigma _t)}^2 + C_\varepsilon \) and \(C_\varepsilon \) is a large constant that depends on \(\varepsilon \). By differentiating and using the a priori estimates (1.7) we obtain

$$\begin{aligned} \begin{aligned} \frac{d}{dt}&\int _{\Sigma _t} p\, ( \nabla v \, \nu ) \cdot \nu \, d \mathcal {H}^2 \\&= \int _{\Sigma _t} p \, (\nabla v \, \nu )\cdot \nu \textrm{div}\,_\tau v \, d \mathcal {H}^2 + \int _{\Sigma _t} \mathcal {D}_t p \, (\nabla v \, \nu ) \cdot \nu \, d \mathcal {H}^2 + \int _{\Sigma _t} p\, \mathcal {D}_t ((\nabla v \, \nu ) \cdot \nu ) \, d \mathcal {H}^2\\&\le C_\varepsilon + \varepsilon \int _{\Sigma _t} |\mathcal {D}_t p |^2 \, d \mathcal {H}^2 + \int _{\Sigma _t} p\, \mathcal {D}_t ((\nabla v \, \nu ) \cdot \nu ) \, d \mathcal {H}^2. \end{aligned} \end{aligned}$$
(6.5)

We estimate the second last term in (6.5) by (4.25) and (1.7) and have

$$\begin{aligned} \int _{\Sigma _t} |\mathcal {D}_t p |^2 \, d \mathcal {H}^2 \le C (1+ \Vert p\Vert _{H^1(\Sigma _t)}^2 + \Vert B\Vert _{L^4(\Sigma _t)}^4 + \Vert \nabla \partial _t U \Vert _{L^2(\Sigma _t)}^2). \end{aligned}$$

Lemma 6.1 yields \(\Vert B\Vert _{L^4(\Sigma _t)}^4 \le C(1+ \Vert p\Vert _{H^1(\Sigma _t)}^2)\). On the other hand, by Lemma 3.3 it holds \(\Vert \nabla \partial _t U \Vert _{\Sigma _t}^2 \le C\Vert \partial _t U \Vert _{H^1(\Sigma _t)}^2\). Since \(\partial _t U = - \nabla U \cdot v\), we may estimate by Theorem 3.9 and by Lemma 6.1

$$\begin{aligned} \begin{aligned} \Vert \nabla \partial _t U \Vert _{L^2(\Sigma _t)}^2&\le C(1+\Vert {\bar{\nabla }} \partial _t U \Vert _{L^2(\Sigma _t)}^2) \le C(1+ \Vert {\bar{\nabla }} (\nabla U \cdot v) \Vert _{L^2(\Sigma _t)}^2) \le C(1+ \Vert \nabla ^2 U\Vert _{L^2(\Sigma _t)}^2)\\ {}&\le C(1+ \Vert \nabla ^2 U\Vert _{H^{1/2}(\Sigma _t)}^2) \le C(1+ \Vert p\Vert _{H^1(\Sigma _t)}^2). \end{aligned} \end{aligned}$$

Therefore we may bound

$$\begin{aligned} \int _{\Sigma _t} |\mathcal {D}_t p |^2 \, d \mathcal {H}^2 \le C (1+ \Vert p\Vert _{H^1(\Sigma _t)}^2 ). \end{aligned}$$
(6.6)

Let us treat the last in term in (6.5). First, we have by (4.1), (4.4) and (1.7) that

$$\begin{aligned} \begin{aligned} \int _{\Sigma _t} p\, \mathcal {D}_t ( (\nabla v \, \nu ) \cdot \nu ) \, d \mathcal {H}^2&\le \int _{\Sigma _t} p\, ((\nabla \mathcal {D}_t v \, \nu ) \cdot \nu ) \, d \mathcal {H}^2 + \varepsilon \Vert p\Vert _{L^2(\Sigma _t)}^2 + C_\varepsilon \\&= - \int _{\Sigma _t} p\, ( \nabla ^2 p \, \nu )\cdot \nu ) \, d \mathcal {H}^2 + \varepsilon \Vert p\Vert _{L^2(\Sigma _t)}^2 + C_\varepsilon . \end{aligned} \end{aligned}$$

We use (4.11) to estimate \(|\Delta p| \le C\Vert \nabla v\Vert _{L^\infty }^2 \le C \) and recall that by (3.4) it holds \((\nabla ^2 p \, \nu )\cdot \nu = \Delta p - \Delta _\Sigma p - H \partial _\nu p\) to deduce

$$\begin{aligned} \begin{aligned} -\int _{\Sigma _t} p\, ( \nabla ^2 p \, \nu )\cdot \nu \, d \mathcal {H}^2&\le C + \int _{\Sigma _t} p\, \Delta _\Sigma p \, d \mathcal {H}^2 + \int _{\Sigma _t} H p\, \partial _\nu p \, d \mathcal {H}^2\\&\le C - \int _{\Sigma _t} |{\bar{\nabla }} p |^2 \, d \mathcal {H}^2 + \int _{\Sigma _t} (\varepsilon |\partial _\nu p|^2 + C_\varepsilon (1+ |H|^4)) \, d \mathcal {H}^2, \end{aligned} \end{aligned}$$

where in the last inequality we have used \(p = H - \frac{Q(t)}{2}|\nabla U|^2\), where \(Q(t)\) is defined in (2.1). Lemma 3.3 yields \(\Vert \partial _\nu p\Vert _{L^2(\Sigma _t)} \le C(1 + \Vert p\Vert _{H^1(\Sigma _t)})\) while Lemma 6.1 implies \(\Vert H\Vert _{L^4}^4 \le \delta \Vert p\Vert _{H^1(\Sigma _t)}^2 + C_{\delta }\). Therefore by first choosing \(\varepsilon \) small and then \(\delta \) even smaller, we obtain

$$\begin{aligned} -\int _{\Sigma _t} p\, (\nabla ^2 p \, \nu )\cdot \nu \, d \mathcal {H}^2 \le - \Vert {\bar{\nabla }} p \Vert _{L^2(\Sigma _t)}^2 + \varepsilon \Vert p\Vert _{H^1(\Sigma _t)}^2 + C_\varepsilon . \end{aligned}$$

By direct calculation and by using (6.6) we have

$$\begin{aligned} \frac{d}{dt} \int _{\Sigma _t} p^2 \, d \mathcal {H}^2 \le C (1+ \Vert p\Vert _{H^1(\Sigma _t)}^2 ). \end{aligned}$$

Combining the two above inequalities with (6.5) and (6.6) we conclude

$$\begin{aligned} \frac{d}{dt} \Psi (t) \le - \frac{1}{2}\Vert {\bar{\nabla }} p \Vert _{L^2(\Sigma _t)}^2 + \varepsilon \Vert p\Vert _{L^2(\Sigma _t)}^2 + C_\varepsilon , \end{aligned}$$
(6.7)

when \(\varepsilon \) is small. Finally we use Lemma 6.1 to estimate

$$\begin{aligned} \Vert p\Vert _{L^2(\Sigma _t)}^2 \le C(1+\Vert H \Vert _{L^2(\Sigma _t)}^2) \le \Vert B\Vert _{L^4(\Sigma _t)}^4 + C \le \varepsilon \Vert p\Vert _{H^1(\Sigma _t)}^2 + C_\varepsilon . \end{aligned}$$

This yields \(\Vert p\Vert _{L^2(\Sigma _t)}^2 \le 2 \Vert {\bar{\nabla }} p \Vert _{L^2(\Sigma _t)}^2 + C_\varepsilon \) when \(\varepsilon \) is small. Thus we may estimate (6.7)

$$\begin{aligned} \frac{d}{dt} \Psi (t) \le - \frac{1}{4}\Vert {\bar{\nabla }} p \Vert _{L^2(\Sigma _t)}^2 + C_\varepsilon , \end{aligned}$$

when \(\varepsilon \) is small. The conclusion follows by integrating the above over [0, T] and using (6.4). \(\square \)

We proceed to higher order regularity estimate which is uniform in time.

Proposition 6.3

Assume that the a priori estimates (1.7) hold for \(T>0\). Then

$$\begin{aligned} \sup _{t \in (0,T]} \Vert \nabla p\Vert _{L^2(\Omega _t)}^2 \le e^{C(1+T)}(1+ \Vert \nabla p\Vert _{L^2(\Omega _0)}^2) \end{aligned}$$

for \(C = C(M)\).

Proof

We differentiate

$$\begin{aligned} \begin{aligned} \frac{d}{dt} \frac{1}{2} \int _{\Omega _t} |\nabla p|^2 \, dx&=\frac{1}{2} \int _{\Omega _t} |\nabla p|^2 \overbrace{\textrm{div}\,(v)}^{=0}\, dx + \int _{\Omega _t} ( \mathcal {D}_t \nabla p \cdot \nabla p) \, dx \\&= \int _{\Omega _t} (\nabla \mathcal {D}_t p \cdot \nabla p) \, dx + \int _{\Omega _t} ([\mathcal {D}_t,\nabla ] p \cdot \nabla p) \, dx. \end{aligned} \end{aligned}$$

By (4.1) and \(\Vert \nabla v\Vert _{L^\infty }\le C\) we have a pointwise estimate \(|[\mathcal {D}_t,\nabla ] p| \le C |\nabla v| |\nabla p| \le C|\nabla p|\). Therefore we deduce

$$\begin{aligned} \begin{aligned} \frac{d}{dt} \frac{1}{2} \int _{\Omega _t} |\nabla p|^2 \, dx&\le \int _{\Omega _t} ( \nabla \mathcal {D}_t p \cdot \nabla p ) \, dx + C \Vert \nabla p\Vert _{L^2(\Omega _t)}^2 \\&= \int _{\Omega _t} \textrm{div}\,( \mathcal {D}_t p \, \nabla p) \, dx - \int _{\Omega _t} \mathcal {D}_t p \, \Delta p \, dx + C \Vert \nabla p\Vert _{L^2(\Omega _t)}^2 \\&= \int _{\Sigma _t} \mathcal {D}_t p \, \partial _\nu p \, dx - \int _{\Omega _t} \mathcal {D}_t p \, \Delta p \, dx + C \Vert \nabla p\Vert _{L^2(\Omega _t)}^2\\&\le \Vert \mathcal {D}_t p\Vert _{L^2(\Sigma _t)}^2 + \Vert \partial _\nu p\Vert _{L^2(\Sigma _t)}^2 - \int _{\Omega _t} \mathcal {D}_t p \, \Delta p \, dx + C \Vert \nabla p\Vert _{L^2(\Omega _t)}^2. \end{aligned} \end{aligned}$$
(6.8)

We have by (6.6) that \(\Vert \mathcal {D}_t p\Vert _{L^2(\Sigma _t)}^2 \le C(1 + \Vert p\Vert _{H^1(\Sigma _t)}^2)\) and by Lemma 3.3

$$\begin{aligned} \Vert \partial _\nu p\Vert _{L^2(\Sigma _t)}^2 \le C(\Vert p\Vert _{H^1(\Sigma _t)}^2 + \Vert \Delta p\Vert _{L^2(\Omega _t)}^2) \le C(1 + \Vert p\Vert _{H^1(\Sigma _t)}^2). \end{aligned}$$

We are left with the second last term in (6.8).

To that aim let \(u: \Omega _t \rightarrow \mathbb {R}\) be the solution of

$$\begin{aligned} {\left\{ \begin{array}{ll} -\Delta u = \Delta p \quad \text {in }\, \Omega _t\\ u= 0 \quad \text {on }\, \Sigma _t. \end{array}\right. } \end{aligned}$$

Then it holds

$$\begin{aligned} -\int _{\Omega _t} \mathcal {D}_t p \, \Delta p \, dx = \int _{\Omega _t} \mathcal {D}_t p \, \Delta u \, dx = \int _{\Omega _t} \Delta \mathcal {D}_t p \, u \, dx + \int _{\Sigma _t} \mathcal {D}_t p \, \partial _\nu u \, d \mathcal {H}^2. \end{aligned}$$

Since \(|\Delta u| = |\Delta p| \le C\) and \(u =0 \) on \(\Sigma _t\) it holds \(\Vert u\Vert _{H^1(\Omega _t)} \le C\) and by Lemma 3.3 we deduce \(\Vert \nabla u\Vert _{L^2(\Sigma _t)} \le C\). We may bound the last term on the RHS by (6.6)

$$\begin{aligned} \int _{\Sigma _t} \mathcal {D}_t p \, \partial _\nu u \, d \mathcal {H}^2 \le \Vert \mathcal {D}_t p\Vert _{L^2(\Sigma _t)}^2 + \Vert \partial _\nu u\Vert _{L^2(\Sigma _t)}^2 \le C(1 + \Vert p\Vert _{H^1(\Sigma _t)}^2). \end{aligned}$$

We are thus left with the second last term.

We have by (3.2), by Lemma 6.1, by Sobolev embedding and by interpolation that

$$\begin{aligned} \begin{aligned} \Vert \nabla ^2 u\Vert _{L^2(\Omega _t)}^2&\le C + \int _{\Sigma _t}|H_{\Sigma _t}| |\nabla u|^2 \, d \mathcal {H}^2 \le C + C_\varepsilon \Vert H_{\Sigma _t}\Vert _{L^3(\Sigma _t)}^3 + \varepsilon \Vert \nabla u \Vert _{L^{3}(\Sigma _t)}^{3}\\&\le C_\varepsilon (1+ \Vert p\Vert _{L^2(\Sigma _t)}^2) + \varepsilon \Vert \nabla u \Vert _{L^{4}(\Sigma _t)}^{2} \Vert \nabla u\Vert _{L^2(\Sigma _t)}\\&\le C_\varepsilon (1+ \Vert p\Vert _{L^2(\Sigma _t)}^2) + C \varepsilon \Vert \nabla ^2 u \Vert _{L^{2}(\Omega _t)}^{2}. \end{aligned} \end{aligned}$$

Therefore it holds

$$\begin{aligned} \Vert u\Vert _{H^2(\Omega _t)}^2 \le C(1+\Vert p\Vert _{H^1(\Sigma _t)}^2). \end{aligned}$$

By Remark 4.5 we have

$$\begin{aligned} \Delta \mathcal {D}_t p = \textrm{div}\,\textrm{div}\,(v \otimes \nabla p) + \textrm{div}\,( R_{bulk}^0). \end{aligned}$$

Therefore by integrating by parts

$$\begin{aligned} \begin{aligned} \int _{\Omega _t} \Delta \mathcal {D}_t p \, u \, dx&= \int _{\Omega _t} ( v \otimes \nabla p): \nabla ^2 u \, dx + \int _{\Omega _t} R_{bulk}^0 \star \nabla u \, dx - \int _{\Sigma _t} (\nabla p \cdot \nu ) (\nabla u \cdot v) \, d \mathcal {H}^2 \\&\le C(1+\Vert p\Vert _{H^1(\Sigma _t)}^2 + \Vert \nabla p\Vert _{L^2(\Omega _t)}^2). \end{aligned} \end{aligned}$$

We deduce by (6.8) and by the above estimates that

$$\begin{aligned} \frac{d}{dt} \frac{1}{2} \Vert \nabla p \Vert _{L^2(\Omega _t)}^2 \le C(1+ \Vert p\Vert _{H^1(\Sigma _t)}^2 + \Vert \nabla p \Vert _{L^2(\Omega _t)}^2 ). \end{aligned}$$

This implies

$$\begin{aligned} \frac{d}{dt} \log (1 + \Vert \nabla p \Vert _{L^2(\Omega _t)}^2)\le C(1+ \Vert p\Vert _{H^1(\Sigma _t)}^2) \end{aligned}$$

and the claim follows from Lemma 6.2. \(\square \)

An important consequence of Proposition 6.3 is that by Lemma 5.2 we have the following bound for the curvature

$$\begin{aligned} \Vert B\Vert _{L^4(\Sigma _t)} + \Vert B\Vert _{H^{\frac{1}{2}}(\Sigma _t)}\le C. \end{aligned}$$
(6.9)

This means that from now on we may use the general interpolation inequality from Proposition 2.8.

At the end of this section we improve Lemma 6.2. We recall the definition of the energy quantity \(E_1(t)\) from (5.1)

$$\begin{aligned} E_1(t) = \Vert \mathcal {D}_t^2 v\Vert _{L^{2}(\Omega _t)}^2 + \Vert \nabla p\Vert _{H^{\frac{3}{2}}(\Omega _t)}^2 + \Vert v\Vert _{H^{3}(\Omega _t)}^2 + \Vert \partial _\nu p\Vert _{H^1(\Sigma _t)}^2 +1. \end{aligned}$$

In particular, \(E_1(0)\) denotes the above quantity at time \(t = 0\). It is clear that

$$\begin{aligned} \Vert p\Vert _{H^1(\Omega _t)}^2 \le E_1(t). \end{aligned}$$

Lemma 6.4

Assume that the a priori estimates (1.7) hold for \(T>0\). Then

$$\begin{aligned} \int _0^T \Vert p\Vert _{H^2(\Omega _t)}^2 \, dt \le C, \end{aligned}$$

where the constant C depends on MT and on \(E_1(0)\).

Proof

The proof is similar to Lemma 6.2. This time we differentiate

$$\begin{aligned} \Phi (t):= -\int _{\Sigma _t} p \Delta _{\Sigma _t} v_n\, d \mathcal {H}^2. \end{aligned}$$

Note that by the a priori estimates (1.7) and by Proposition 6.3, \(\Phi \) is uniformly bounded on [0, T]. Note also that by Proposition 6.3 it holds

$$\begin{aligned} \sup _{t < T} \Vert p\Vert _{H^1(\Omega _t)}^2 \le C, \end{aligned}$$

where the constant depends on \(T, \Lambda _T\) and on \(E_1(0)\).

We calculate as in (6.5) by using (6.6) and \(\Vert v_n\Vert _{H^2(\Sigma _t)} \le C\) that

$$\begin{aligned} \begin{aligned} \frac{d}{dt} \Phi (t)&\le C + \Vert \mathcal {D}_t p\Vert _{L^2(\Sigma _t)}^2 -\int _{\Sigma _t} p\, (\mathcal {D}_t \Delta _{\Sigma _t} v_n)\, d \mathcal {H}^2\\&\le C(1+ \Vert p\Vert _{H^1(\Sigma _t)}^2) -\int _{\Sigma _t} p\, (\mathcal {D}_t \Delta _{\Sigma _t} v_n)\, d \mathcal {H}^2. \end{aligned} \end{aligned}$$

To bound the last term we recall that by (4.3) it holds

$$\begin{aligned} (\mathcal {D}_t \Delta _{\Sigma _t} v_n) = \Delta _{\Sigma _t} (\mathcal {D}_t v_n) + \nabla _\tau ^2 v_n \star \nabla v - \nabla _\tau v_n \cdot \Delta _{\Sigma _t} v + B \star \nabla v\star \nabla _\tau v_n. \end{aligned}$$

Therefore by \(\Vert \nabla v\Vert _{L^\infty } +\Vert v_n\Vert _{H^2(\Sigma _t)} \le C\), by \(\Vert \nabla _\tau v_n\Vert _{L^4(\Sigma _t)} \le \Vert v_n\Vert _{H^2(\Sigma _t)} \) and by (6.9) it holds

$$\begin{aligned} \begin{aligned}&-\int _{\Sigma _t} p\, (\mathcal {D}_t \Delta _{\Sigma _t} v_n)\, d \mathcal {H}^2 \le - \int _{\Sigma _t} p\, \Delta _{\Sigma _t} (\mathcal {D}_t v_n)\, d \mathcal {H}^2 +\int _{\Sigma _t} p\, \nabla _\tau v_n \cdot \Delta _{\Sigma _t} v\, d \mathcal {H}^2 \\&\,\,\,\,\,\,\,\, + C(1+ \Vert p\Vert _{L^2(\Sigma _t)}\Vert B\Vert _{L^4(\Sigma _t)}\Vert {\bar{\nabla }} v_n\Vert _{L^4(\Sigma _t)}) \\&\le -\int _{\Sigma _t} p\, \Delta _{\Sigma _t} (\mathcal {D}_t v \cdot \nu )\, d \mathcal {H}^2 - \int _{\Sigma _t} p\, \Delta _{\Sigma _t} (\mathcal {D}_t \nu \cdot v)\, d \mathcal {H}^2 - \int _{\Sigma _t} (\nabla _\tau ( p\, \nabla _\tau v_n) \cdot \nabla _\tau v) \, d \mathcal {H}^2 + C . \end{aligned} \end{aligned}$$
(6.10)

We may write the first term on RHS in (6.10) by the formula (3.2), \(\mathcal {D}_t v = - \nabla p\) and by the estimate (6.9)

$$\begin{aligned} \begin{aligned} -\int _{\Sigma _t} p\, \Delta _{\Sigma _t} (\mathcal {D}_t v \cdot \nu )\, d \mathcal {H}^2&= \int _{\Sigma _t} \Delta _{\Sigma _t} p\, \partial _\nu p\, d \mathcal {H}^2\\&\le -\frac{1}{2} \int _{\Omega _t} (|\nabla ^2 p|^2-|\Delta p|^2) \, dx + C\int _{\Sigma _t}|B||\nabla p|^2\, d \mathcal {H}^2\\&\le -\frac{1}{2} \int _{\Omega _t} |\nabla ^2 p|^2 \, dx + C+ C_\varepsilon \Vert B\Vert _{L^4}+ \varepsilon \Vert \nabla p\Vert _{L^{\frac{8}{3}}(\Sigma _t)}^2\\&\le -\frac{1}{2} \int _{\Omega _t} |\nabla ^2 p|^2 \, dx + C_\varepsilon + \varepsilon \Vert \nabla p\Vert _{L^{\frac{8}{3}}(\Sigma _t)}^2. \end{aligned} \end{aligned}$$

By the Sobolev embedding it holds \(\Vert \nabla p\Vert _{L^{\frac{8}{3}}(\Sigma _t)}^2 \le C\Vert \nabla p\Vert _{H^{1/2}(\Sigma _t)}^2 \le C(1+ \Vert \nabla ^2 p\Vert ^2_{L^2(\Omega _t)}).\) Therefore by choosing \(\varepsilon >0\) small we deduce

$$\begin{aligned} -\int _{\Sigma _t} p\, \Delta _{\Sigma _t} (\mathcal {D}_t v \cdot \nu )\, d \mathcal {H}^2 \le -\frac{1}{3} \int _{\Omega _t} |\nabla ^2 p|^2 \, dx + C_\varepsilon . \end{aligned}$$

We bound the third term on RHS in (6.10) simply by \(\Vert \nabla v\Vert _{L^\infty } +\Vert v_n\Vert _{H^2(\Sigma _t)} \le C\) as

$$\begin{aligned} -\int _{\Sigma _t} \nabla _\tau ( p\, \nabla _\tau v_n) \cdot \nabla _\tau v\, d \mathcal {H}^2 \le C(1+ \Vert p\Vert _{H^1(\Sigma _t)}^2 ). \end{aligned}$$

For the remaining term in (6.10) we recall (4.5) which states \(\mathcal {D}_t \nu = - \nabla _\tau v_n + B v_\tau \). Therefore we obtain by Lemma 6.1 and by \(\Vert \nabla v\Vert _{L^\infty } +\Vert v_n\Vert _{H^2(\Sigma _t)} \le C\) that

$$\begin{aligned} \begin{aligned} -\int _{\Sigma _t} p\, \Delta _{\Sigma _t} (\mathcal {D}_t \nu \cdot v)\, d \mathcal {H}^2&= \int _{\Sigma _t} \langle {\bar{\nabla }} p, {\bar{\nabla }}(\mathcal {D}_t \nu \cdot v)\rangle \, d \mathcal {H}^2 \\&\le C(1+ \Vert p\Vert _{H^1(\Sigma _t)}^2 + \Vert B \Vert _{H^1(\Sigma _t)}^2 + \Vert B\Vert _{L^4(\Sigma _t)}^4)\\&\le C(1+ \Vert p\Vert _{H^1(\Sigma _t)}^2). \end{aligned} \end{aligned}$$

Hence, we have

$$\begin{aligned} \frac{d}{dt} \Phi (t) \le - \frac{1}{3} \int _{\Omega _t} |\nabla ^2 p|^2 \, dx + C(1+ \Vert p\Vert _{H^1(\Sigma _t)}^2) \end{aligned}$$

and the claim follows from Lemma 6.2. \(\square \)

7 Energy Estimates

As we mentioned before, the fundamental property of the solution of (1.3) is the conservation of the energy (1.1). In this section we define high order energy functions and show that their derivatives are controlled by the quantity (5.1) of the same order. This will be the first step in proving that the high order energy quantities remain bounded along the flow.

We define the energy of order \(l \ge 1\) as

$$\begin{aligned} \begin{aligned} {\mathcal {E}}_l(t) =\frac{1}{2} \int _{\Omega _t}&|\mathcal {D}_t^{l+1} v|^2 \, dx + \frac{1}{2}\int _{\Sigma _t} |\nabla _\tau (\mathcal {D}_t^l v \cdot \nu )|^2 \, d \mathcal {H}^2 \\&- \frac{Q(t)}{2} \int _{\Omega _t^c} |\nabla (\partial _t^{l+1} U) |^2 \, dx + \int _{\Omega _t} |\nabla ^{\lfloor \frac{1}{2}(3l+1) \rfloor }\omega |^2 \, dx, \end{aligned} \end{aligned}$$
(7.1)

where \(\lfloor \frac{1}{2}(3l+1) \rfloor \) is the integer part of \(\frac{1}{2}(3l+1)\), \(\omega \) is the curl of v defined as

$$\begin{aligned} \omega = \textrm{curl}\,v= \nabla v - \nabla v^T \end{aligned}$$

and \(Q(t)\) is defined in (2.1).

In this section we calculate \(\frac{d}{dt}{\mathcal {E}}_l\) and estimate it in terms of the \(E_l(t)\), which we recall is defined in (5.1) as

$$\begin{aligned} E_l(t) = \sum _{k=0}^{l} \Vert \mathcal {D}_t^{l+1-k}v\Vert _{H^{\frac{3}{2} k}(\Omega _t)}^2 + \Vert v\Vert _{H^{\lfloor \frac{3}{2}(1+1)\rfloor }(\Omega _t)}^2+ \Vert \mathcal {D}_t^l v \cdot \nu \Vert _{H^1(\Sigma _t)}^2+1. \end{aligned}$$

In particular, it holds \({\mathcal {E}}_l(t) \le C E_l(t)\). We state the main results of this section below and prove them later.

Proposition 7.1

Assume that the a priori estimates (1.7) hold for \(T>0\). Then for all \(t <T\) it holds

$$\begin{aligned} \frac{d}{dt} {\mathcal {E}}_1(t) \le C(1 + \Vert p\Vert _{H^2(\Omega _t)}^2)E_1(t), \end{aligned}$$

where the constant depends on MT and \(E_1(0)\), i.e., \(E_1(t)\) at time \(t=0\).

Proposition 7.2

Let \(l \ge 2\) and assume that (1.7) and \(E_{l-1}(t) \le M\) hold for all \(t \in [0,T)\). Then all \(t <T\) it holds

$$\begin{aligned} \frac{d}{dt} {\mathcal {E}}_l(t) \le CE_l(t) \end{aligned}$$

where the constant depends on MlT and on \(\sup _{t < T} E_{l-1}(t)\).

The proof of the both above energy estimates is based on the calculations of the differential of \({\mathcal {E}}_l(t)\), which we state first for all \(l \ge 1\). In the proof of Proposition 7.1 and Proposition 7.2 we then need to estimate the remainder terms by the quantity \(E_l(t)\).

We begin by differentiating the first term of \({\mathcal {E}}_l(t)\) in (7.1) and obtain by \(\textrm{div}\,v = 0\), by (4.15) and by the definition of \(E_l(t)\) that

$$\begin{aligned} \begin{aligned}&\frac{d}{dt} \frac{1}{2} \int _{\Omega _t}|\mathcal {D}_t^{l+1} v|^2 \, dx = \int _{\Omega _t} ( \mathcal {D}_t^{l+2} v \cdot \mathcal {D}_t^{l+1} v) \, dx\\&\quad = - \int _{\Omega _t} (\nabla \mathcal {D}_t^{l+1} p \cdot \mathcal {D}_t^{l+1} v ) \, dx - \int _{\Omega _t} ([\mathcal {D}_t^{l+1},\nabla ] p \cdot \mathcal {D}_t^{l+1} v ) \, dx \\&\quad \le - \int _{\Omega _t} (\nabla \mathcal {D}_t^{l+1} p \cdot \mathcal {D}_t^{l+1} v) \, dx + \Vert \mathcal {D}_t^{l+1} v \Vert _{L^2(\Omega _t)}^2 + \Vert [\mathcal {D}_t^{l+1},\nabla ] p \Vert _{L^2(\Omega _t)}^2 \\&\quad = - \int _{\Omega _t} \textrm{div}\,(\mathcal {D}_t^{l+1} p \, \mathcal {D}_t^{l+1} v) \, dx + \int _{\Omega _t} \mathcal {D}_t^{l+1} p \, \textrm{div}\,(\mathcal {D}_t^{l+1} v) \, dx + E_l(t) + \Vert R_{bulk}^l \Vert _{L^{2}(\Omega _t)}^2\\&\quad = - \int _{\Sigma _t} \mathcal {D}_t^{l+1} p \, (\mathcal {D}_t^{l+1} v \cdot \nu ) \, d\mathcal {H}^2 + E_l(t) + \Vert R_{bulk}^l \Vert _{L^{2}(\Omega _t)}^2 + \int _{\Omega _t} \mathcal {D}_t^{l+1} p \, \textrm{div}\,(\mathcal {D}_t^{l+1} v) \, dx. \end{aligned} \end{aligned}$$
(7.2)

Next we differentiate the second term in the energy and obtain by \(\Vert \nabla v\Vert _{L^\infty } \le C\) and (4.2)

$$\begin{aligned}{} & {} \frac{d}{dt} \frac{1}{2} \int _{\Sigma _t} |\nabla _\tau (\mathcal {D}_t^{l} v \cdot \nu )|^2\, d\mathcal {H}^2 \nonumber \\{} & {} \quad =\int _{\Sigma _t} ( \mathcal {D}_t \nabla _\tau (\mathcal {D}_t^{l} v \cdot \nu ) \cdot \nabla _\tau (\mathcal {D}_t^{l} v \cdot \nu )) \, d\mathcal {H}^2 + \frac{1}{2} \int _{\Sigma _t} |\nabla _\tau (\mathcal {D}_t^{l} v \cdot \nu )|^2 (\textrm{div}\,_\tau v) \, d\mathcal {H}^2 \nonumber \\{} & {} \quad \le \int _{\Sigma _t} (\nabla _\tau \mathcal {D}_t (\mathcal {D}_t^{l} v \cdot \nu ) \cdot \nabla _\tau (\mathcal {D}_t^{l} v \cdot \nu )) \, d\mathcal {H}^2 + C \Vert \mathcal {D}_t^{l} v \cdot \nu \Vert _{H^1(\Sigma _t)}^2\nonumber \\{} & {} \quad = -\int _{\Sigma _t} (\Delta _{\Sigma _t} (\mathcal {D}_t^{l} v \cdot \nu )) ( \mathcal {D}_t (\mathcal {D}_t^{l} v \cdot \nu ))\, d\mathcal {H}^2 +C E_l(t) \\{} & {} \quad = -\int _{\Sigma _t} (\Delta _{\Sigma _t} (\mathcal {D}_t^{l} v \cdot \nu )) ( \mathcal {D}_t^{l+1} v \cdot \nu )\, d\mathcal {H}^2 \nonumber \\{} & {} \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,+ \int _{\Sigma _t} ( \nabla _{\tau } (\mathcal {D}_t^{l} v \cdot \nu ) \cdot \nabla _\tau ( \mathcal {D}_t^{l} v \cdot \mathcal {D}_t \nu )) \rangle \, d\mathcal {H}^2 +C E_l(t)\nonumber \\{} & {} \quad \le -\int _{\Sigma _t} (\Delta _{\Sigma _t} (\mathcal {D}_t^{l} v \cdot \nu )) ( \mathcal {D}_t^{l+1} v \cdot \nu )\, d\mathcal {H}^2 +C E_l(t) + \Vert \mathcal {D}_t^{l} v \cdot \mathcal {D}_t \nu \Vert _{H^1(\Sigma _t)}^2.\nonumber \end{aligned}$$
(7.3)

We differentiate the third term, use the fact that \(\partial _t^{l+1} U\) is harmonic and Lemma 4.6 and have (recall that it holds \(\Vert \nabla \partial _t^{l+1} U\Vert _{L^2(\Omega _t^c)} \le C\Vert \nabla \partial _t^{l+1} U\Vert _{L^2(\Sigma _t)}\))

$$\begin{aligned}{} & {} \frac{d}{dt} -\frac{Q(t)}{2} \int _{\Omega _t^c} |\nabla \partial _t^{l+1} U|^2\, dx = - Q(t)\int _{\Omega _t^c} \langle \nabla \partial _t^{l+2} U,\nabla \partial _t^{l+1} U \rangle \, dx \nonumber \\{} & {} \,\,\,\,\,\,\,\,+ \frac{Q(t)}{2} \int _{\Sigma _t} |\nabla \partial _t^{l+1} U|^2 v_n \, d\mathcal {H}^2 -\frac{Q'(t)}{2} \int _{\Omega _t^c} |\nabla \partial _t^{l+1} U|^2\, dx \nonumber \\{} & {} \le Q(t)\int _{\Sigma _t} \partial _t^{l+2} U \, (\partial _{\nu } \partial _t^{l+1} U)\, d\mathcal {H}^2 + C\Vert \nabla \partial _t^{l+1} U\Vert _{L^2(\Sigma _t)}^2 \nonumber \\{} & {} = - Q(t)\int _{\Sigma _t} (\partial _{\nu } U (\mathcal {D}_t^{l+1} v \cdot \nu ) +R_U^{l})\, (\partial _{\nu } \partial _t^{l+1} U)\, d\mathcal {H}^2 + C\Vert \nabla \partial _t^{l+1} U\Vert _{L^2(\Sigma _t)}^2 \nonumber \\{} & {} \le -Q(t)\int _{\Sigma _t} (\partial _{\nu } U \, \partial _{\nu } \partial _t^{l+1} U) (\mathcal {D}_t^{l+1} v \cdot \nu ) \, d\mathcal {H}^2 + \Vert R_U^{l}\Vert _{L^2(\Sigma _t)}^2 +C\Vert \nabla \partial _t^{l+1} U\Vert _{L^2(\Sigma _t)}^2, \end{aligned}$$
(7.4)

where \(R_U^l\) in the remainder term defined in (4.18) and \(Q'(t) = \frac{d}{dt }Q(t)\), where \(Q(t)\) is defined in (2.1). Recall that in the proof of Lemma 5.7 we proved that \(Q'(t)\) and \(Q''(t)\) are bounded, see (5.30).

Finally, we differentiate the fourth term involving the curl. To that aim we denote \(\lambda _l:= \lfloor \frac{1}{2}(3l+1)\rfloor \). We have by Lemma 4.4 and by (2.16)

$$\begin{aligned} \begin{aligned} \frac{d}{dt}\int _{\Omega _t} |\nabla ^{\lambda _l}\omega |^2\, dx&= \int _{\Omega _t} \nabla v \star \nabla ^{\lambda _l} \omega \star \nabla ^{\lambda _l}\omega \, dx + C\Vert \nabla ^{\lambda _l}\omega \Vert _{L^2(\Omega _t)}^2 \\&\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,+ \sum _{|\alpha |\le \lambda _l}\Vert \nabla ^{1+\alpha _1} v \star \nabla ^{1+\alpha _2} v \Vert _{L^2(\Omega _t)}^2 \\&\le C\Vert \nabla ^{\lambda _l}\omega \Vert _{L^2(\Omega _t)}^2 + C\Vert \nabla v\Vert _{L^\infty (\Omega _t)}^2 \Vert \nabla v\Vert _{H^{\lambda _l}(\Omega _t)}^2 \\&\le C\Vert \nabla ^{\lambda _l}\omega \Vert _{L^2(\Omega _t)}^2 + C\Vert v\Vert _{H^{\lfloor \frac{3}{2}(l+1)\rfloor }(\Omega _t)}^2 \le C E_{l}(t). \end{aligned} \end{aligned}$$
(7.5)

Let us next show that the highest order terms in (7.2), (7.3) and (7.4) cancel out. Indeed, this follows from Lemma 4.7 which states that

$$\begin{aligned} \mathcal {D}_t^{l+1}p = -\Delta _{\Sigma _t} (\mathcal {D}_t^l v \cdot \nu ) - Q(t) \, \partial _{\nu } U \, (\partial _{\nu } \partial _t^{l+1} U) + R_p^{l}, \end{aligned}$$

where \(R_p^l\) denotes the error term defined in (4.26) and \(Q(t)\) is defined in (2.1). Note that we may estimate

$$\begin{aligned} \big |\int _{\Sigma _t} R_p^{l} (\mathcal {D}_t^{l+1} v \cdot \nu ) \, d \mathcal {H}^2 \big | \le \Vert R_p^l\Vert _{H^{\frac{1}{2}}(\Sigma _t)}^2 + \Vert \mathcal {D}_t^{l+1} v \cdot \nu \Vert _{H^{-\frac{1}{2}}(\Sigma _t)}^2. \end{aligned}$$

By divergence theorem and by Lemma 4.4 we have

$$\begin{aligned} \Vert \mathcal {D}_t^{l+1} v \cdot \nu \Vert _{H^{-\frac{1}{2}}(\Sigma _t)}^2 \le C\Vert \mathcal {D}_t^{l+1} v\Vert _{L^2(\Omega _t)}^2 + \Vert R_{\textrm{div}\,}^l\Vert _{L^2(\Omega _t)}^2 \le CE_l(t) + \Vert R_{\textrm{div}\,}^l\Vert _{L^2(\Omega _t)}^2. \end{aligned}$$

Therefore we have by (7.2), (7.3), (7.4) and (7.5) that

$$\begin{aligned} \begin{aligned} \frac{d}{dt} {\mathcal {E}}_l(t)&\le C E_l(t) + \Vert R_{bulk}^l\Vert _{L^2(\Omega _t)}^2 + \Vert R_{U}^l\Vert _{L^2(\Sigma _t)}^2 +\Vert R_{\textrm{div}\,}^l\Vert _{L^2(\Omega _t)}^2+ \Vert R_p^l\Vert _{H^{\frac{1}{2}}(\Sigma _t)}^2\\&\,\,\,\,+ C \Vert \nabla \partial _t^{l+1} U\Vert _{L^2(\Sigma _t)}^2 +\Vert \mathcal {D}_t^l v \cdot \mathcal {D}_t \nu \Vert _{H^1(\Sigma _t)}^2 + \int _{\Omega _t} \mathcal {D}_t^{l+1} p \, \textrm{div}\,(\mathcal {D}_t^{l+1} v) \, dx. \end{aligned} \end{aligned}$$
(7.6)

We need thus to bound the remainder terms in (7.6). As we mentioned before, we prove the energy bounds by induction such that we bound \(E_l(t)\) when we know that \(E_{l-1}(t)\) is already bounded. The difficulty is to bound the first order quantity \(E_1(t)\) and we do this separately in Proposition 7.1.

Proof of Proposition 7.1

First we recall that Proposition 6.3 implies

$$\begin{aligned} \sup _{t \in (0,T)} \Vert \nabla p\Vert _{L^2(\Omega _t)}^2 \le e^{CT}(1+\Vert \nabla p\Vert _{L^2(\Omega _0)}^2) \le C, \end{aligned}$$

where the constant C on the RHS depends on \(T, \Lambda _T\) defined in (1.7) and on \(E_1(0)\), which is the energy quantity \(E_1(t)\) at time \(t=0\). Then we have the bound (6.9) which we recall is

$$\begin{aligned} \Vert B\Vert _{L^4(\Sigma _t)} + \Vert B\Vert _{H^{\frac{1}{2}}(\Sigma _t)} \le C \end{aligned}$$

for all \(t <T\).

We recall the estimate (7.6) for \(l=1\) and use Lemma 5.3, Lemma 5.4, estimate (5.15) from Lemma 5.6 and Lemma 5.7 to deduce

$$\begin{aligned} \frac{d}{dt} {\mathcal {E}}_1(t) \le C(1+ \Vert p\Vert _{H^2(\Omega _t)}^2) E_1(t) + \Vert \mathcal {D}_t v \cdot \mathcal {D}_t \nu \Vert _{H^1(\Sigma _t)}^2 + \int _{\Omega _t} \mathcal {D}_t^{2} p \, \textrm{div}\,(\mathcal {D}_t^{2} v) \, dx. \end{aligned}$$

Hence, we need to bound the two last terms. The second last term is easy to treat and we merely claim that it holds

$$\begin{aligned} \Vert \mathcal {D}_t v \cdot \mathcal {D}_t \nu \Vert _{H^1(\Sigma _t)}^2 \le C(1+ \Vert p\Vert _{H^2(\Omega _t)}^2) E_1(t). \end{aligned}$$

Indeed, this follows from \(\mathcal {D}_t v = -\nabla p\), \(\mathcal {D}_t \nu = -(\nabla _\tau v)^T \nu \) from (4.4) and from Proposition 2.10. We leave the details for the reader. The last term is challenging and we will prove that

$$\begin{aligned} \int _{\Omega _t} \mathcal {D}_t^{2} p \, \textrm{div}\,(\mathcal {D}_t^{2} v) \, dx \le C(1+ \Vert p\Vert _{H^2(\Omega _t)}^2) E_1(t). \end{aligned}$$
(7.7)

The argument is similar than in the proof of Proposition 6.3. The idea is to use the fact that the term \(\textrm{div}\,(\mathcal {D}_t^{2} v)\) is lower order due to the fact that \(\textrm{div}\,v = 0\). Indeed, we have by Lemma 4.4 that \(\textrm{div}\,(\mathcal {D}_t^{2} v) = R^{1}_{\textrm{div}\,}\), where \(R^{1}_{\textrm{div}\,}\) is defined in (4.13) and is lower order than \(\mathcal {D}_t^{2} v\). To this aim let u be a solution of

$$\begin{aligned} {\left\{ \begin{array}{ll} &{}-\Delta u = \textrm{div}\,(\mathcal {D}_t^{2} v), \quad \text {in } \, \Omega _t\\ &{}\,\,\,\,\,\,\,u = 0 \qquad \text {on }\, \Sigma _t. \end{array}\right. } \end{aligned}$$

We have by integration by parts

$$\begin{aligned} \int _{\Omega _t} \mathcal {D}_t^{2} p \, \textrm{div}\,(\mathcal {D}_t^{2} v) \, dx = -\int _{\Omega _t} \mathcal {D}_t^{2} p \, \Delta u \, dx = -\int _{\Omega _t} \Delta \mathcal {D}_t^{2} p \, u \, dx - \int _{\Sigma _t} \mathcal {D}_t^2 p\, \partial _\nu u\, d \mathcal {H}^2. \end{aligned}$$

We use Remark 4.5 and write

$$\begin{aligned} -\Delta \mathcal {D}_t^{2} p = \textrm{div}\,\textrm{div}\,(v \otimes \mathcal {D}_t^2 v) + \textrm{div}\,(R_{bulk}^1). \end{aligned}$$

By integration by parts we deduce

$$\begin{aligned} \begin{aligned} -\int _{\Omega _t} \Delta \mathcal {D}_t^{2} p \, u \, dx =&\int _{\Omega _t} (v \otimes \mathcal {D}_t^2 v): \nabla ^2 u \, dx + \int _{\Omega _t} R_{bulk}^1 \star \nabla u \, dx \\&- \int _{\Sigma _t} (\mathcal {D}_t^2 v \cdot \nu ) (\nabla u \cdot v) \, d \mathcal {H}^2. \end{aligned} \end{aligned}$$
(7.8)

We use divergence theorem, the definition of \(E_1(t)\), \(\textrm{div}\,(\mathcal {D}_t^{2} v) = R^{1}_{\textrm{div}\,}\) and Lemma 5.3 for the last term

$$\begin{aligned} \begin{aligned} - \int _{\Sigma _t} (\mathcal {D}_t^2 v \cdot \nu ) (\nabla u \cdot v) \, d \mathcal {H}^2&= - \int _{\Omega _t} \textrm{div}\,\big ( (\nabla u \cdot v)\mathcal {D}_t^2 v\big )\\&\le CE_1(t) + \Vert u\Vert _{H^2(\Omega _t)}^2 +C\Vert R^{1}_{\textrm{div}\,}\Vert _{L^2(\Omega _t)}^2\\&\le C(1+\Vert p\Vert _{H^2(\Omega _t)}^2)E_1(t) + \Vert u\Vert _{H^2(\Omega _t)}^2. \end{aligned} \end{aligned}$$

Since \(-\Delta u = \textrm{div}\,(D_t^2 v)\), we may use the inequality (3.19) and Lemma 5.3 to obtain

$$\begin{aligned} \Vert u\Vert _{H^2(\Omega _t)}^2 \le C\Vert R^{1}_{\textrm{div}\,}\Vert _{L^2(\Omega _t)}^2 \le C(1+\Vert p\Vert _{H^2(\Omega _t)}^2)E_1(t). \end{aligned}$$

Therefore we have by (7.8), by the definition of \(E_1(t)\) and by Lemma 5.4 that

$$\begin{aligned} \begin{aligned} \int _{\Omega _t} \mathcal {D}_t^{2} p \, \textrm{div}\,(\mathcal {D}_t^{2} v) \, dx \le C(1+\Vert p\Vert _{H^2(\Omega _t)}^2)E_1(t) - \int _{\Sigma _t} \mathcal {D}_t^2 p\, \partial _\nu u\, d \mathcal {H}^2. \end{aligned} \end{aligned}$$
(7.9)

We proceed by using Lemma 4.7 to write

$$\begin{aligned} \mathcal {D}_t^2 p = - \Delta _{\Sigma _t} (\mathcal {D}_t v \cdot \nu ) - Q(t) \nabla U \cdot \nabla \partial _t^2 U + R_p^1, \end{aligned}$$

where \(Q(t)\) is defined in (2.1). We integrate by parts on \(\Sigma _t\) the term

$$\begin{aligned} - \int _{\Sigma _t}\Delta _{\Sigma _t} (\mathcal {D}_t v \cdot \nu ) \partial _\nu u\, d \mathcal {H}^2 = \int _{\Sigma _t}\langle {\bar{\nabla }} (\mathcal {D}_t v \cdot \nu ), {\bar{\nabla }} \partial _\nu u \rangle \, d \mathcal {H}^2 \end{aligned}$$

and deduce

$$\begin{aligned} \begin{aligned} -\int _{\Sigma _t} \mathcal {D}_t^2 p\, \partial _\nu u\, d \mathcal {H}^2 \le&\Vert \mathcal {D}_t v \cdot \nu \Vert _{H^1(\Sigma _t)}^2 + \Vert \partial _\nu u\Vert _{H^1(\Sigma _t)}^2 \\&+ C\Vert \nabla \partial _t^2 U\Vert _{L^{2}(\Sigma _t)}^2 + C\Vert R_p^1\Vert _{L^{2}(\Sigma _t)}^2. \end{aligned} \end{aligned}$$

Lemma 5.7 and (5.21) imply

$$\begin{aligned} \Vert \nabla \partial _t^2 U\Vert _{L^{2}(\Sigma _t)}^2 + \Vert R_p^1\Vert _{H^{\frac{1}{2}}(\Sigma _t)}^2 \le C(1+\Vert p\Vert _{H^2(\Omega _t)}^2)E_1(t). \end{aligned}$$

We use Proposition 3.8, Lemma 4.4 and Lemma 5.3 to deduce

$$\begin{aligned} \begin{aligned} \Vert \partial _\nu u\Vert _{H^1(\Sigma _t)}^2&\le C\Vert \textrm{div}\,(\mathcal {D}_t^2 v)\Vert _{H^{\frac{1}{2}}(\Omega _t)}^2 \le C\Vert R_{\textrm{div}\,}\Vert _{H^{\frac{1}{2}}(\Omega _t)}^2 \\&\le C(1+\Vert p\Vert _{H^2(\Omega _t)}^2)E_1(t). \end{aligned} \end{aligned}$$

Therefore since \(\Vert \mathcal {D}_t v \cdot \nu \Vert _{H^1(\Sigma _t)}^2 \le E_1(t)\) we obtain by combining the previous estimates

$$\begin{aligned} \int _{\Sigma _t} \mathcal {D}_t^2 p\, \partial _\nu u\, d \mathcal {H}^2 \le C(1+\Vert p\Vert _{H^2(\Omega _t)}^2)E_1(t). \end{aligned}$$

Hence, (7.8) implies (7.7) which concludes the proof. \(\square \)

By a similar argument we prove the higher order case. We begin again with (7.6) and we need to bound the remainder terms.

Proof of Proposition 7.2

The assumption \(E_{l-1}(t) \le C\) and Lemma 5.2 imply the curvature bound \(\Vert B\Vert _{H^{\frac{3}{2}l-1}(\Sigma _t)} \le C\). Thus we may use the estimate (7.6), Lemma 5.3, Lemma 5.4, estimate (5.16) from Lemma 5.6 and Lemma 5.7 to deduce

$$\begin{aligned} \frac{d}{dt} {\mathcal {E}}_l(t) \le C E_l(t) + \Vert \mathcal {D}_t^l v \cdot \mathcal {D}_t \nu \Vert _{H^1(\Sigma _t)}^2 + \int _{\Omega _t} \mathcal {D}_t^{l+1} p \, \textrm{div}\,(\mathcal {D}_t^{l+1} v) \, dx. \end{aligned}$$

Hence, we need to bound the two last terms. Again the second last term is easy to treat and we merely sketch it. We have by (4.4) \(\mathcal {D}_t \nu = - (\nabla _\tau v)^T \nu \) and have by Proposition 2.10

$$\begin{aligned} \Vert \mathcal {D}_t^l v \cdot \mathcal {D}_t \nu \Vert _{H^1(\Sigma _t)}^2 \le C \Vert \mathcal {D}_t \nu \Vert _{L^\infty (\Sigma _t)}^2\Vert \mathcal {D}_t^l v\Vert _{H^1(\Sigma _t)}^2 + C \Vert \mathcal {D}_t \nu \Vert _{W^{1,4}(\Sigma _t)}^2\Vert \mathcal {D}_t^l v\Vert _{L^4(\Sigma _t)}^2 \le C E_l(t). \end{aligned}$$

We treat the last term similarly as in the previous proof and claim that it holds

$$\begin{aligned} \int _{\Omega _t} \mathcal {D}_t^{l+1} p \, \textrm{div}\,(\mathcal {D}_t^{l+1} v) \, dx \le C E_l(t). \end{aligned}$$
(7.10)

Since the argument is almost the same as with (7.7) we only sketch it. We let u be the solution of

$$\begin{aligned} {\left\{ \begin{array}{ll} &{}-\Delta u = \textrm{div}\,(\mathcal {D}_t^{l+1} v) \quad \text {in } \, \Omega _t,\\ &{}\,\,\,\,\,\,\,u = 0 \qquad \text {on }\, \Sigma _t. \end{array}\right. } \end{aligned}$$

Again by integration by parts and by using Remark 4.5, Lemma 5.3 and Lemma 5.4 we obtain the higher order version of (7.9) which reads as

$$\begin{aligned} \int _{\Omega _t} \mathcal {D}_t^{l+1} p \, \textrm{div}\,(\mathcal {D}_t^{l+1} v) \, dx \le C E_l(t) - \int _{\Sigma _t} \mathcal {D}_t^{l+1}p \, \partial _\nu u \, d \mathcal {H}^2. \end{aligned}$$

Lemma 4.7 yields

$$\begin{aligned} \mathcal {D}_t^{l+1}p = -\Delta _{\Sigma _t}(\mathcal {D}_t^l v \cdot \nu ) - Q(t)\nabla U \cdot \nabla \partial _t^{l+1} U + R_p^l, \end{aligned}$$

where \(Q(t)\) is defined in (2.1). By integration by parts on \(\Sigma _t\) we have

$$\begin{aligned} \begin{aligned} - \int _{\Sigma _t} \mathcal {D}_t^{l+1}p \, \partial _\nu u \, d \mathcal {H}^2 \le&\Vert \mathcal {D}_t^l v \cdot \nu \Vert _{H^1(\Sigma _t)}^2 + \Vert \partial _\nu u\Vert _{H^1(\Sigma _t)}^2 \\&+ C\Vert \nabla \partial _t^{l+1} U\Vert _{L^{2}(\Sigma _t)}^2 + C\Vert R_p^l\Vert _{L^{2}(\Sigma _t)}^2. \end{aligned} \end{aligned}$$

Note that \(\Vert \mathcal {D}_t^l v \cdot \nu \Vert _{H^1(\Sigma _t)}^2 \le E_l(t)\). By Lemma 5.6 and Lemma 5.8 we have

$$\begin{aligned} \Vert \nabla \partial _t^{l+1} U\Vert _{H^{\frac{1}{2}}(\Sigma _t)}^2 + \Vert R_p^l\Vert _{H^{\frac{1}{2}}(\Sigma _t)}^2 \le CE_l(t). \end{aligned}$$

Finally by Proposition 3.8, by the curvature bound \(\Vert B\Vert _{H^2(\Sigma _t)} \le \Vert B\Vert _{H^{\frac{3}{2}l-1}(\Sigma _t)} \le C\), by Lemma 4.4 and by Lemma 5.3 we have

$$\begin{aligned} \Vert \partial _\nu u\Vert _{H^{1}(\Sigma _t)}^2 \le C\Vert \textrm{div}\,(\mathcal {D}_t^{l+1} v)\Vert _{H^{\frac{1}{2}}(\Omega _t)}^2 \le C \Vert R_{\textrm{div}\,}^l\Vert _{H^{\frac{1}{2}}(\Omega _t)}^2 \le C E_l(t). \end{aligned}$$

This proves (7.10) and concludes the proof. \(\square \)

8 Higher Regularity Estimates

Let us recall the definition of the energies \(E_l(t)\) and \({\mathcal {E}}_l(t)\) for \(l \ge 1\) in (5.1) and (7.1) respectively. In the previous section we proved energy estimates where we control the derivative of \({\mathcal {E}}_l(t)\) by \(E_l(t)\). In this section we complete the estimate and prove that the energy \({\mathcal {E}}_l(t)\) in fact controls \(E_l(t)\). This, together with the results in the previous section, will give us control for \({\mathcal {E}}_l(t)\) and implies the regularity of the flow.

Note that the energy \({\mathcal {E}}_l(t)\) is defined in (7.1) as

$$\begin{aligned} \begin{aligned} {\mathcal {E}}_l(t) =\frac{1}{2} \int _{\Omega _t}&|\mathcal {D}_t^{l+1} v|^2 \, dx + \frac{1}{2}\int _{\Sigma _t} |\nabla _\tau (\mathcal {D}_t^l v \cdot \nu )|^2 \, d \mathcal {H}^2 \\&- \frac{Q(t)}{2} \int _{\Omega _t^c} |\nabla (\partial _t^{l+1} U) |^2 \, dx + \int _{\Omega _t} |\nabla ^{\lfloor \frac{1}{2}(3l+1) \rfloor }\textrm{curl}\,v |^2 \, dx, \end{aligned} \end{aligned}$$

where \(Q(t)\) is defined in (2.1). Since \(Qt \ge 0\), the energy has one negative term and we define its positive part as

$$\begin{aligned} {\mathcal {E}}_l^+(t) := 1+ \frac{1}{2}\Vert \mathcal {D}_t^{l+1}v\Vert _{L^{2}(\Omega _t)}^2 + \Vert \textrm{curl}\,v\Vert _{H^{\lfloor \frac{3}{2}l+\frac{1}{2}\rfloor }(\Omega _t)}^2+ \frac{1}{2} \Vert \mathcal {D}_t^l v \cdot \nu \Vert _{H^1(\Sigma _t)}^2. \end{aligned}$$
(8.1)

Then it holds

$$\begin{aligned} c \, {\mathcal {E}}_l^+(t) \le {\mathcal {E}}_l(t) + \frac{Q(t)}{2} \int _{\Omega _t^c} |\nabla (\partial _t^{l+1} U)|^2 \, dx+ 1, \end{aligned}$$

for \(c>0\).

The first main result of this section states that the energy \( {\mathcal {E}}_1(t)\) controls \(E_1(t)\). For later purpose we need this bound when the boundary \(\Sigma _t\) is \(C^{1,\alpha }\)-regular but the velocity field is only bounded in \(W^{1,4}\). This makes the statement slightly heavy.

Proposition 8.1

Assume that \(\Sigma _t\) is uniformly \(C^{1,\alpha }(\Gamma )\)-regular and the pressure and the velocity satisfies

$$\begin{aligned} \Vert p\Vert _{H^1(\Omega _t)} + \Vert v\Vert _{W^{1,4}(\Sigma _t)} + \Vert v\Vert _{W^{1,4}(\Omega _t)} \le M. \end{aligned}$$

Then there are constants C and \(C_0\) such that

$$\begin{aligned} E_1(t) \le C(C_0 + {\mathcal {E}}_1(t)). \end{aligned}$$

The constants depend on \(\sigma _t\) defined in (1.4), the \(C^{1,\alpha }\)-norm of the heightfunction and on M.

Before the proof we remark that if the a priori estimates (1.7) hold for \(T>0\), then the above assumptions hold for constants \(C, C_0\), which depend on \(T, \Lambda _T, \sigma _T\) and \(E_1(0)\). Indeed, then by Proposition 6.3 it holds

$$\begin{aligned} \Vert p\Vert _{H^1(\Omega _t)} \le C. \end{aligned}$$

Proof

We first recall that by Lemma 5.2 we have

$$\begin{aligned} \Vert B\Vert _{L^4(\Sigma _t)} + \Vert B\Vert _{H^{\frac{1}{2}}(\Sigma _t)} \le C\Vert p\Vert _{H^1(\Omega _t)}\le C. \end{aligned}$$

The claim follows once we prove that for any \(\varepsilon >0\) it holds

$$\begin{aligned} {\mathcal {E}}_1^+(t) \le {\mathcal {E}}_1(t) + \varepsilon E_1(t) +C_\varepsilon \end{aligned}$$
(8.2)

and

$$\begin{aligned} E_1(t) \le C {\mathcal {E}}_1^+(t). \end{aligned}$$
(8.3)

Let us first prove (8.2). Since \(\partial _t^2 U\) is harmonic in \(\Omega _t^c\) it holds

$$\begin{aligned} \int _{\Omega _t^c} |\nabla \partial _t^2 U|^2 \, dx \le C \Vert \partial _t^2 U\Vert _{H^{\frac{1}{2}}(\Sigma _t)}^2. \end{aligned}$$

We use interpolation (Corollary 2.9) to deduce

$$\begin{aligned} \Vert \partial _t^2 U\Vert _{H^{\frac{1}{2}}(\Sigma _t)} \le C\Vert \partial _t^2 U\Vert _{H^1(\Sigma _t)}^{\frac{1}{2}}\Vert \partial _t^2 U\Vert _{L^{2}(\Sigma _t)}^{\frac{1}{2}}. \end{aligned}$$
(8.4)

By (5.21) it holds

$$\begin{aligned} \Vert \partial _t^2 U\Vert _{H^1(\Sigma _t)} \le C(1 + \Vert p\Vert _{H^1(\Sigma _t)})E_1(t)^{\frac{1}{2}}. \end{aligned}$$
(8.5)

In order to estimate \(\Vert \partial _t^2 U\Vert _{L^{2}(\Sigma _t)}\) we use Lemma 4.6 and \(\Vert \nabla U\Vert _{L^\infty }, \Vert v\Vert _{L^\infty } \le C\) and have

$$\begin{aligned} \begin{aligned} \Vert \partial _t^2 U\Vert _{L^{2}(\Sigma _t)}&\le C \Vert \mathcal {D}_t v\Vert _{L^{2}(\Sigma _t)} + C\sum _{|\alpha | \le 1}\Vert \nabla ^{1+\alpha _1}\partial _t^{\alpha _2} U\Vert _{L^{2}(\Sigma _t)}\\&\le C\Vert p\Vert _{H^1(\Sigma _t)} + C(1+ \Vert \nabla ^{2}U\Vert _{L^{2}(\Sigma _t)} + \Vert \nabla \partial _t U\Vert _{L^{2}(\Sigma _t)}). \end{aligned} \end{aligned}$$

We have by (5.18) \( \Vert \nabla ^{2}U\Vert _{L^{2}(\Sigma _t)} \le C(1+ \Vert p\Vert _{H^1(\Sigma _t)})\). We use Lemma 3.3 and \(\partial _t U = - \nabla U \cdot v\) to deduce

$$\begin{aligned} \begin{aligned} \Vert \nabla \partial _t U\Vert _{L^{2}(\Sigma _t)}\le C(1+\Vert \nabla _\tau \partial _t U\Vert _{L^{2}(\Sigma _t)})&\le C\Vert v\Vert _{L^\infty } \Vert \nabla ^2 U\Vert _{L^{2}(\Sigma _t)} + C \Vert \nabla U\Vert _{L^\infty } \Vert v\Vert _{H^1(\Sigma _t)} \\&\le C(1+ \Vert \nabla ^2 U\Vert _{L^{2}(\Sigma _t)}) \le C(1+ \Vert p\Vert _{H^1(\Sigma _t)}). \end{aligned} \end{aligned}$$

Therefore by (8.4), (8.5), \(\Vert p\Vert _{L^4(\Sigma _t)}\le \Vert p\Vert _{H^1(\Omega _t)} \le C\) and by interpolation we obtain

$$\begin{aligned} \begin{aligned} \Vert \partial _t^2 U\Vert _{H^{\frac{1}{2}}(\Sigma _t)}^2&\le C + C\Vert p\Vert _{H^1(\Sigma _t)}^2 E_1(t)^{\frac{1}{2}} \\&\le C + C\Vert p\Vert _{H^2(\Sigma _t)}^{\frac{2}{3}}\Vert p\Vert _{L^4(\Sigma _t)}^{\frac{4}{3}} E_1(t)^{\frac{1}{2}}\\&\le \varepsilon (\Vert p\Vert _{H^2(\Sigma _t)}^2 + E_1(t)) + C_\varepsilon . \end{aligned} \end{aligned}$$

Lemma 3.7 yields \(\Vert p\Vert _{H^2(\Sigma _t)}^2 \le C\Vert \nabla p\Vert _{H^{\frac{3}{2}}(\Omega _t)}^2 \le C E_1(t)\) and (8.2) follows.

Let us then prove (8.3). Recall that it holds

$$\begin{aligned} E_1(t) \le 2 {\mathcal {E}}_{1}^+(t) + \Vert v\Vert _{H^3(\Omega _t)}^2 + \Vert \mathcal {D}_t v\Vert _{H^{\frac{3}{2}}(\Omega _t)}^2. \end{aligned}$$
(8.6)

By (4.11) we have \( -\Delta p = \text {Tr}((\nabla v)^2)\). We use the third inequality in Lemma 3.7 and \(\Vert {\bar{\nabla }} \partial _\nu p\Vert _{L^{2}(\Sigma _t)}^2 \le 2 {\mathcal {E}}_{1}^+(t)\) and have by interpolation

$$\begin{aligned} \begin{aligned} \Vert \mathcal {D}_t v\Vert _{H^{\frac{3}{2}}(\Omega _t)}^2&= \Vert \nabla p\Vert _{H^{\frac{3}{2}}(\Omega _t)}^2 \le C (\Vert \partial _\nu p\Vert _{H^{1}(\Sigma _t)}^2 + \Vert p\Vert _{L^2(\Omega _t)}^2 + \Vert \Delta p\Vert _{H^1(\Omega _t)}^2) \\&\le C (\Vert \partial _\nu p\Vert _{H^{1}(\Sigma _t)}^2 + \Vert p\Vert _{L^2(\Omega _t)}^2 + \Vert \nabla ^2 v \Vert _{L^4(\Omega _t)}^2 \Vert \nabla v\Vert _{L^4(\Omega _t)}^2 +1 ) \\&\le C_\varepsilon {\mathcal {E}}_1^+(t) + \varepsilon \Vert v\Vert _{H^3(\Omega _t)}^2. \end{aligned} \end{aligned}$$
(8.7)

We proceed estimating \(\Vert v\Vert _{H^3(\Omega _t)}\). By the first inequality in Lemma 3.7, by \(\Vert v\Vert _{L^\infty (\Omega _t)} \le \Vert v\Vert _{W^{1,4}(\Omega _t)}\le C\) and by Lemma 5.2 we have

$$\begin{aligned} \begin{aligned} \Vert v\Vert _{H^3(\Omega _t)}^2&\le C\big (\Vert \Delta _{\Sigma _t} v_n\Vert _{H^{\frac{1}{2}}(\Sigma _t)}^2 + (1+ \Vert H_{\Sigma _t}\Vert _{H^2(\Sigma _t)}^2)\Vert v\Vert _{L^\infty }^2 + \Vert \textrm{curl}\,v \Vert _{H^1(\Omega _t)}^2\big ) \\&\le C\big (\Vert \Delta _{\Sigma _t} v_n\Vert _{H^{\frac{1}{2}}(\Sigma _t)}^2 + \Vert p\Vert _{H^2(\Sigma _t)}^2 + {\mathcal {E}}_1^+(t)\big ). \end{aligned} \end{aligned}$$

By the fourth inequality in Lemma 3.7 and by (8.7) we have

$$\begin{aligned} \Vert p\Vert _{H^2(\Sigma _t)}^2 \le C(\Vert p\Vert ^2_{L^2(\Sigma _t)} +\Vert \nabla p\Vert _{H^{\frac{3}{2}}(\Omega _t)}^2) \le C_\varepsilon {\mathcal {E}}_1^+(t) +\varepsilon \Vert v\Vert _{H^3(\Omega _t)}^2. \end{aligned}$$

Therefore by choosing \(\varepsilon \) small enough we deduce

$$\begin{aligned} \Vert v\Vert _{H^3(\Omega _t)}^2 \le C\Vert \Delta _{\Sigma _t} v_n\Vert _{H^{\frac{1}{2}}(\Sigma _t)}^2 + C{\mathcal {E}}_1^+(t) \end{aligned}$$
(8.8)

In order to control \(\Vert \Delta _{\Sigma _t} v_n\Vert _{H^{\frac{1}{2}}(\Sigma _t)}\) we use (4.25) which states

$$\begin{aligned} \mathcal {D}_t p = -\Delta _{\Sigma _t} v_n - Q(t) \nabla U \cdot \nabla \partial _t U + R_p^0, \end{aligned}$$
(8.9)

where \(Q(t)\) is defined in (2.1) and

$$\begin{aligned} R_p^0 = -(|B|^2 -Q(t)\, H|\nabla U|^2)v_n + (\nabla p \cdot v_\tau ) -\frac{Q'(t)}{2} |\nabla U|^2. \end{aligned}$$

Therefore we have

$$\begin{aligned} \Vert \Delta _{\Sigma _t} v_n\Vert _{H^{\frac{1}{2}}(\Sigma _t)} \le \Vert \mathcal {D}_t p\Vert _{H^{\frac{1}{2}}(\Sigma _t)} + \Vert \nabla U \cdot \nabla \partial _t U\Vert _{H^{\frac{1}{2}}(\Sigma _t)} + \Vert R_p^0\Vert _{H^{\frac{1}{2}}(\Sigma _t)}. \end{aligned}$$

We estimate the first term on RHS as

$$\begin{aligned} \begin{aligned} \Vert \mathcal {D}_t p\Vert _{H^{\frac{1}{2}}(\Sigma _t)}^2&\le C(1+ \Vert \nabla \mathcal {D}_t p\Vert _{L^{2}(\Omega _t)}^2) \\&\le C(1+ \Vert \mathcal {D}_t \nabla p\Vert _{L^{2}(\Omega _t)}^2 + \Vert [\mathcal {D}_t, \nabla ] p\Vert _{L^{2}(\Omega _t)}^2)\\&\le C(1+ \Vert \mathcal {D}_t^2 v\Vert _{L^{2}(\Omega _t)}^2 + \Vert \nabla v\Vert _{L^{4}(\Omega _t)}^2\Vert \nabla p\Vert _{L^{4}(\Omega _t)}^2) \le C {\mathcal {E}}_1^+(t). \end{aligned} \end{aligned}$$

By an already familiar argument we get

$$\begin{aligned} \Vert \nabla U \cdot \nabla \partial _t U\Vert _{H^{\frac{1}{2}}(\Sigma _t)}^2 + \Vert R_p^0\Vert _{H^{\frac{1}{2}}(\Sigma _t)}^2 \le \varepsilon E_1(t) + C_{\varepsilon }. \end{aligned}$$

We leave the details for the reader. Combing the previous three inequalities yield

$$\begin{aligned} \Vert \Delta _{\Sigma _t} v_n\Vert _{H^{\frac{1}{2}}(\Sigma _t)}^2 \le C_{\varepsilon } {\mathcal {E}}_1^+(t) + \varepsilon E_1(t). \end{aligned}$$

By combining (8.6), (8.7), (8.8) with the above inequality and by choosing \(\varepsilon \) small enough imply

$$\begin{aligned} E_1(t) \le C{\mathcal {E}}_1^+(t) \end{aligned}$$

and the claim (8.3) follows. \(\square \)

Proposition 8.1 implies that the bound on \(\textrm{curl}\,v\) and \( \mathcal {D}_t^2 v\) in the fluid domain and on \(\mathcal {D}_t v\) on the boundary imply the bound on v and \(\mathcal {D}_t v\) in the domain. In the next lemma we show the converse for the initial set \(t=0\), i.e., the bound on v in the domain and on the mean curvature \(H_{\Sigma _0}\) imply that \(E_1(0)\) is bounded.

Lemma 8.2

Assume that \(\Omega _0\) is a smooth set such that \(\Vert h_0\Vert _{L^\infty (\Sigma )} < \eta \). Then it holds

$$\begin{aligned} E_1(0) \le C_0, \end{aligned}$$

for a constant \(C_0\) which depends on \(\sigma _0 = \eta - \Vert h_0\Vert _{L^\infty (\Sigma )}\), \(\Vert v\Vert _{H^3(\Omega _0)}\) on \(\Vert H_{\Sigma _0}\Vert _{H^2(\Sigma _0)}\) and on \(\Vert h_0\Vert _{C^{1,\alpha }}\).

Proof

The bound \(\Vert H_{\Sigma _0}\Vert _{H^2(\Sigma _0)} \le C\) and Proposition 2.12 imply \(\Vert B_{\Sigma _0}\Vert _{H^2(\Sigma _0)} \le C\). Then we obtain by Theorem 3.9 that \(\Vert \nabla ^3 U\Vert _{H^{\frac{1}{2}}(\Sigma _0)} \le C\). Hence, we have

$$\begin{aligned} \Vert p\Vert _{H^2(\Sigma _0)} \le C. \end{aligned}$$
(8.10)

Let us show that

$$\begin{aligned} \Vert \partial _\nu p\Vert _{H^1(\Sigma _0)} \le C. \end{aligned}$$
(8.11)

Let \({\tilde{\nu }}\) be the harmonic extension of the normal field. Note that since

$$\begin{aligned} \Vert B\Vert _{C^{\alpha }(\Sigma _0)} \le C\Vert B_{\Sigma _0}\Vert _{H^2(\Sigma _0)} \le C, \end{aligned}$$

then by standard elliptic regularity theory [25] we deduce that \(\Vert \nabla {\tilde{\nu }} \Vert _{C^{\alpha }(\Sigma _0)} \le C\). Then (4.11), \(\nabla \Delta p = \nabla ^2 v \star \nabla v\) and \(\Vert v\Vert _{H^3(\Omega _0)}\le C\) imply that

$$\begin{aligned} \begin{aligned} \Vert \Delta (\nabla p \cdot {\tilde{\nu }})\Vert _{L^2(\Omega _0)}&\le C\Vert \nabla ^2 v \star \nabla v\Vert _{L^2(\Omega _0)} + C\Vert \nabla ^2 p \star \nabla {\tilde{\nu }}\Vert _{L^2(\Omega _0)} \\&\le C(1+ \Vert p\Vert _{H^2(\Omega _0)}). \end{aligned} \end{aligned}$$

Lemma 3.5 together with interpolation yields

$$\begin{aligned} \Vert p\Vert _{H^2(\Omega _0)} \le C(1+ \Vert \partial _\nu p\Vert _{H^{\frac{1}{2}}(\Sigma _0)}) \le \varepsilon \Vert \partial _\nu p\Vert _{H^1(\Sigma _0)} + C_\varepsilon . \end{aligned}$$

Therefore by combing the two estimates with Lemma 3.3 we obtain

$$\begin{aligned} \Vert \partial _\nu p\Vert _{H^1(\Sigma _0)} = \Vert \nabla p \cdot {\tilde{\nu }}\Vert _{H^1(\Sigma _0)} \le C_\varepsilon (1+ \Vert \partial _\nu (\nabla p \cdot {\tilde{\nu }})\Vert _{L^2(\Sigma _0)}) + \varepsilon \Vert \partial _\nu p\Vert _{H^1(\Sigma _0)}. \end{aligned}$$

Choosing \(\varepsilon \) small yields

$$\begin{aligned} \Vert \partial _\nu p\Vert _{H^1(\Sigma _0)} \le C(1+ \Vert \partial _\nu (\nabla p \cdot {\tilde{\nu }})\Vert _{L^2(\Sigma _0)}). \end{aligned}$$

Note that by \(\Vert \nabla {\tilde{\nu }} \Vert _{C^{\alpha }(\Sigma _0)} \le C\), Lemma 3.3 and by (8.10) we have

$$\begin{aligned} \begin{aligned} \Vert \partial _\nu (\nabla p \cdot {\tilde{\nu }})\Vert _{L^2(\Sigma _0)}&\le C( \Vert (\nabla ^2 p \, \nu \cdot \nu ) \Vert _{L^2(\Sigma _0)} + C\Vert \nabla p\Vert _{L^2(\Sigma _0)})\\&\le C \Vert (\nabla ^2 p \, \nu \cdot \nu ) \Vert _{L^2(\Sigma _0)} + C(1 + \Vert p\Vert _{H^1(\Sigma _0)})\\&\le C(1+ \Vert (\nabla ^2 p \, \nu \cdot \nu ) \Vert _{L^2(\Sigma _0)}). \end{aligned} \end{aligned}$$

Therefore since

$$\begin{aligned} \Delta _{\Sigma _0} p = \Delta p - (\nabla ^2 p \, \nu \cdot \nu ) - H_{\Sigma _0} \partial _\nu p \end{aligned}$$

we obtain by (8.10) and by \(\Vert \partial _\nu p\Vert _{L^2(\Sigma _t)} \le C(1+ \Vert p\Vert _{H^1(\Sigma _t)}) \le C\) that

$$\begin{aligned} \Vert (\nabla ^2 p \, \nu \cdot \nu )\Vert _{L^2(\Sigma _0)} \le C( 1+ \Vert p\Vert _{H^2(\Sigma _0)})\le C. \end{aligned}$$

Thus we have (8.11) by the three inequalities above.

We estimate \(\Vert \nabla p\Vert _{H^{\frac{3}{2}}(\Omega _0)}\) similarly. We use (8.11) and Lemma 3.7 to estimate

$$\begin{aligned} \Vert \nabla p\Vert _{H^{\frac{3}{2}}(\Omega _0)} \le C(\Vert \partial _\nu p\Vert _{H^1(\Sigma _0)} + \Vert p \Vert _{L^2(\Omega _0)} + \Vert \Delta p \Vert _{H^1(\Omega _0)}) \le C. \end{aligned}$$
(8.12)

In order to show that \(\Vert \mathcal {D}_t^2 v\Vert _{L^2(\Omega _0)}\) is bounded we first observe that by (4.1) and by (8.12) we have

$$\begin{aligned} \Vert \mathcal {D}_t^2 v\Vert _{L^2(\Omega _0)} \le \Vert \nabla \mathcal {D}_t p\Vert _{L^2(\Omega _0)} + \Vert \nabla v \star \nabla p\Vert _{L^2(\Omega _0)} \le \Vert \nabla \mathcal {D}_t p\Vert _{L^2(\Omega _0)} + C. \end{aligned}$$

Recall that we define the \(H^{\frac{1}{2}}(\Sigma _0)\)-norm using harmonic extension. Then it holds

$$\begin{aligned} \Vert \nabla \mathcal {D}_t p\Vert _{L^2(\Omega _0)} \le C(\Vert \mathcal {D}_t p\Vert _{H^{\frac{1}{2}}(\Sigma _0)} +\Vert \mathcal {D}_t p\Vert _{L^2(\Omega _0)} + \Vert \Delta \mathcal {D}_t p\Vert _{L^2(\Omega _0)}). \end{aligned}$$

Note that it holds \(\Vert \mathcal {D}_t p\Vert _{L^2(\Omega _0)} \le C \Vert \mathcal {D}_t p\Vert _{H^{\frac{1}{2}}(\Sigma _0)}\). By Remark 4.5 and Lemma 4.4 we have

$$\begin{aligned} \begin{aligned} \Vert \Delta \mathcal {D}_t p\Vert _{L^2(\Omega _0)}&\le C\Vert R_{\textrm{div}\,}^1\Vert _{L^2(\Omega _0)} + C\Vert R_{bulk}^0\Vert _{H^1(\Omega _0)} \\&\le C(1+ \Vert p\Vert _{H^2(\Omega _0)}+ \Vert v\Vert _{H^2(\Omega _0)} )\le C. \end{aligned} \end{aligned}$$

We proceed by using (4.22) to write

$$\begin{aligned} \mathcal {D}_t p = - \Delta _{\Sigma _0} v \cdot \nu -2 B:\nabla _\tau v - Q(0)(D_t \nabla U \cdot \nabla U) - \frac{Q'(0)}{2} |\nabla U|^2. \end{aligned}$$

We only bound the first term on RHS as the others are lower order. By the \(C^{1,\alpha }\)-regularity of \(\nu \) we immediately estimate

$$\begin{aligned} \Vert \Delta _{\Sigma _0} v \cdot \nu \Vert _{H^{\frac{1}{2}}(\Sigma _0)} \le C \Vert \Delta _{\Sigma _0} v \Vert _{H^{\frac{1}{2}}(\Sigma _0)} \le C\Vert v\Vert _{H^3(\Omega _0)}. \end{aligned}$$

This concludes the proof. \(\square \)

Let us next prove the higher order version of Proposition 8.1.

Proposition 8.3

Let \(l \ge 2\) and assume that (1.7) and \(E_{l-1}(t) \le M\) hold for all \(t \in [0,T)\). Then there are constants C and \(C_0\) such that

$$\begin{aligned} E_l(t) \le C( C_0+ {\mathcal {E}}_l(t)), \end{aligned}$$

where the constants C and \(C_0\) depend on Ml and T.

Proof

We recall that by the definition of \(E_l(t)\) in (5.1), \({\mathcal {E}}_l(t)\) in (7.1) and of \({\mathcal {E}}_l^+(t)\) it holds

$$\begin{aligned} c\, {\mathcal {E}}_l^+(t)\le {\mathcal {E}}_l(t) + \frac{Q(t)}{2} \int _{\Omega _t^c}|\nabla (\partial _t^{l+1} U)|^2 \, dx + 1 \end{aligned}$$

for \(c>0\), \(Q(t)\) defined in (2.1), and

$$\begin{aligned} E_{l}(t) \le 2 {\mathcal {E}}_{l}^+(t) + \sum _{k=1}^{l} \Vert \mathcal {D}_t^{l+1-k} v\Vert _{H^{\frac{3}{2} k}(\Omega _t)}^2 + \Vert v \Vert _{H^{\lfloor \frac{3}{2}(l +1)\rfloor }(\Omega _t)}^2. \end{aligned}$$
(8.13)

The claim follows once we prove that for any \(\varepsilon >0\) it holds

$$\begin{aligned} {\mathcal {E}}_l^+(t) \le {\mathcal {E}}_l(t) + \varepsilon E_l(t) +C_\varepsilon \end{aligned}$$
(8.14)

and

$$\begin{aligned} E_l(t) \le C {\mathcal {E}}_l^+(t). \end{aligned}$$
(8.15)

In order to prove (8.14) we use the fact that \(\partial ^{l+1} U\) is harmonic in \(\Omega _t^c\), interpolation (Corollary 2.9) and Lemma 5.6 and have

$$\begin{aligned} \begin{aligned} \int _{\Omega _t^c} |\nabla \partial _t^{l+1} U|^2 \, dx&\le C \Vert \partial _t^{l+1} U\Vert _{H^{\frac{1}{2}}(\Sigma _t)}^2 \le C \Vert \partial _t^{l+1} U\Vert _{H^{1}(\Sigma _t)}\Vert \partial _t^{l+1} U\Vert _{L^{2}(\Sigma _t)}\\&\le C E_l(t)^{\frac{1}{2}} \Vert \partial _t^{l+1} U\Vert _{L^{2}(\Sigma _t)} \le \varepsilon _1 E_l(t) + C_{\varepsilon _1} \Vert \partial _t^{l+1} U\Vert _{L^{2}(\Sigma _t)}^2. \end{aligned} \end{aligned}$$

We use Lemma 4.6, \(\Vert \nabla U\Vert _{L^\infty } \le C\), Lemma 5.6 and the assumption \(E_{l-1}(t) \le {\tilde{C}}\) to deduce

$$\begin{aligned} \Vert \partial _t^{l+1} U\Vert _{L^{2}(\Sigma _t)} \le \Vert \nabla U \cdot \mathcal {D}_t^l v\Vert _{L^{2}(\Sigma _t)} + \Vert R_U^{l-1}\Vert _{L^{2}(\Sigma _t)} \le C\Vert \mathcal {D}_t^l v\Vert _{L^{2}(\Sigma _t)} + C. \end{aligned}$$

By the Trace Theorem, by interpolation (Corollary 2.9) and by the definition of \(E_l(t)\) it holds

$$\begin{aligned} \begin{aligned} \Vert \mathcal {D}_t^l v\Vert _{L^{2}(\Sigma _t)}^2&\le C \Vert \mathcal {D}_t^l v\Vert _{H^{1}(\Omega _t)}^2\le C\Vert \mathcal {D}_t^l v\Vert _{H^{\frac{3}{2}}(\Omega _t)}^{\frac{4}{3}} \Vert \mathcal {D}_t^l v\Vert _{L^{2}(\Omega _t)}^{\frac{2}{3}} \\&\le C E_l(t)^{\frac{2}{3}} E_{l-1}(t)^{\frac{1}{3}} \le \varepsilon _2 E_l(t) + C_{\varepsilon _2}. \end{aligned} \end{aligned}$$

By choosing first \(\varepsilon _1\) and then \(\varepsilon _2\) small implies (8.14).

Let us then prove (8.15). By (8.13) we have to bound \(\Vert \mathcal {D}_t^{l+1-k} v\Vert _{H^{\frac{3}{2} k} (\Omega _t)}\) for all \(k = 1, \dots , l\) and \(\Vert v\Vert _{H^{\lfloor \frac{3}{2}(l+1)\rfloor }(\Omega _t)}\). We claim first that it holds

$$\begin{aligned} \Vert \mathcal {D}_t^{l} v\Vert _{H^{\frac{3}{2}}(\Omega _t)}^2 \le C {\mathcal {E}}_{l}^+(t). \end{aligned}$$
(8.16)

Indeed, by Theorem 3.1, Lemma 4.4 and Lemma 5.3 it holds

$$\begin{aligned} \begin{aligned} \Vert&\mathcal {D}_t^{l} v\Vert _{H^{\frac{3}{2}}(\Omega _t)}^2 \\&\le C(\Vert (\mathcal {D}_t^{l} v\cdot \nu )\Vert _{H^{1}(\Sigma _t)}^2 + \Vert \mathcal {D}_t^l v\Vert _{L^{2}(\Omega _t)}^2 + \Vert \textrm{div}\,\mathcal {D}_t^l v\Vert _{H^{\frac{1}{2}}(\Omega _t)}^2 + \Vert \textrm{curl}\,\mathcal {D}_t^l v\Vert _{H^{\frac{1}{2}}(\Omega _t)}^2)\\&\le C({\mathcal {E}}_l^+(t) + E_{l-1}(t) + \Vert R_{\textrm{div}\,}^{l-1} \Vert _{H^{\frac{1}{2}}(\Omega _t)}^2 )\\&\le C({\mathcal {E}}_l^+(t) + E_{l-1}(t)) \le C{\mathcal {E}}_l^+(t) \end{aligned} \end{aligned}$$

and (8.16) follows.

Next we claim that for \(2 \le k \le l\) it holds

$$\begin{aligned} \Vert \mathcal {D}_t^{l+1-k} v\Vert _{H^{\frac{3}{2}k}(\Omega _t)}^2 \le C \Vert \mathcal {D}_t^{l+3 -k} v \Vert _{H^{\frac{3}{2}k-3}(\Omega _t)}^2 + \varepsilon E_l(t) + C_{\varepsilon }. \end{aligned}$$
(8.17)

This inequality means that two derivatives in time implies regularity for three derivatives in space. We first use Proposition 3.2, Lemma 4.4 and Lemma 5.3 to deduce

$$\begin{aligned} \begin{aligned} \Vert \mathcal {D}_t^{l+1-k}&v\Vert _{H^{\frac{3}{2}k}(\Omega _t)}^2 \\&\le C \big (\Vert \Delta _\Sigma (\mathcal {D}_t^{l+1-k} v \cdot \nu )\Vert _{H^{\frac{3}{2}k-\frac{5}{2}}(\Sigma _t)}^2 + \Vert \mathcal {D}_t^{l+1-k} v\Vert _{L^2(\Omega _t)}^2 \\&\,\,\,\,\,\,\,+ \Vert \textrm{div}\,(\mathcal {D}_t^{l+1-k} v)\Vert _{H^{\frac{3}{2}k -1}(\Omega _t)}^2 + \Vert \textrm{curl}\,(\mathcal {D}_t^{l+1-k} v)\Vert _{H^{\frac{3}{2}k -1}(\Omega _t)}^2\big )\\&\le C\big ( \Vert \Delta _\Sigma (\mathcal {D}_t^{l+1-k} v \cdot \nu )\Vert _{H^{\frac{3}{2}k-\frac{5}{2}}(\Sigma _t)}^2 + E_{l-1}(t) + \Vert R_{\textrm{div}\,}^{l-k}\Vert _{H^{\frac{3}{2}k -1}(\Omega _t)}^2\big )\\&\le C \Vert \Delta _\Sigma (\mathcal {D}_t^{l+1-k} v \cdot \nu )\Vert _{H^{\frac{3}{2}k-\frac{5}{2}}(\Sigma _t)}^2 + \varepsilon E_{l} + C_\varepsilon . \end{aligned} \end{aligned}$$
(8.18)

We proceed by using Lemma 4.7 to write

$$\begin{aligned} \mathcal {D}_t^{l+2 -k} p = - \Delta _\Sigma (\mathcal {D}_t^{l+1-k} v \cdot \nu ) - Q(t)(\nabla U \cdot \nabla \partial _t^{l+2-k} U) + R_p^{l+1-k}. \end{aligned}$$
(8.19)

Lemma 5.8 yields

$$\begin{aligned} \Vert R_p^{l+1-k}\Vert _{H^{\frac{3}{2} k -\frac{5}{2}}(\Sigma _t)}^2 = \Vert R_p^{l-(k-1)}\Vert _{H^{\frac{3}{2} (k-1) -1}(\Sigma _t)}^2 \le \varepsilon E_l(t) + C_{\varepsilon }. \end{aligned}$$
(8.20)

Next we claim that

$$\begin{aligned} \Vert (\nabla U \cdot \nabla \partial _t^{l+2-k} U)\Vert _{H^{\frac{3}{2}k-\frac{5}{2}}(\Sigma _t)}^2 \le \varepsilon E_l(t) + C_\varepsilon . \end{aligned}$$
(8.21)

If \(k =2\) then we use the fact that by the assumption \(\Vert B\Vert _{H^2(\Sigma _t)} \le \Vert B\Vert _{H^{\frac{3}{2}l-1}(\Sigma _t)} \le C\) and by Theorem 3.9 the function U is uniformly \(C^{2,\alpha }\)-regular. Therefore we have by Lemma 5.5

$$\begin{aligned} \Vert (\nabla U \cdot \nabla \partial _t^{l} U)\Vert _{H^{\frac{1}{2}}(\Sigma _t)}^2 \le C\Vert \nabla \partial _t^{l} U\Vert _{H^{\frac{1}{2}}(\Sigma _t)}^2\le \varepsilon E_{l}(t) + C_{\varepsilon }. \end{aligned}$$

If \(k \ge 3\) then \(2 \le \frac{3}{2}k-\frac{5}{2} \le \lfloor \frac{3}{2} l \rfloor -2\). We have by Proposition 2.10, by the Sobolev embedding, by Lemma 5.1 and by Lemma 5.5 that

$$\begin{aligned} \begin{aligned} \Vert (\nabla U \cdot \nabla \partial _t^{l+2-k}&U)\Vert _{H^{\frac{3}{2}k-\frac{5}{2}}(\Sigma _t)}^2 \le C\Vert \nabla U\Vert _{L^\infty (\Sigma _t)}^2 \Vert \nabla \partial _t^{l+2-k} U\Vert _{H^{\frac{3}{2}k-\frac{5}{2}}(\Sigma _t)}^2 \\&\,\,\,\,\,\,\,\,\,\,\,\,+\Vert \nabla U\Vert _{H^{\frac{3}{2}k-\frac{5}{2}}(\Sigma _t)}^2 \Vert \nabla \partial _t^{l+2-k} U\Vert _{L^\infty (\Sigma _t)}^2\\&\le C(1+ \Vert \nabla U\Vert _{H^{\frac{3}{2}k-\frac{5}{2}}(\Sigma _t)}^2 )\Vert \nabla \partial _t^{l+2-k} U\Vert _{H^{\frac{3}{2}k-\frac{5}{2}}(\Sigma _t)}^2 \\&\le C(1+ \Vert p\Vert _{H^{\lfloor \frac{3}{2} l \rfloor -2}(\Sigma _t)}^2) (\varepsilon E_{l}(t) + C_\varepsilon ). \end{aligned} \end{aligned}$$

Hence, (8.21) follows from the Trace Theorem as

$$\begin{aligned} \Vert p\Vert _{H^{\lfloor \frac{3}{2} l \rfloor -2}(\Sigma _t)}^2 \le C(1+\Vert \nabla p\Vert _{H^{\frac{3}{2} l -2}(\Omega _t)}^2) \le CE_{l-1}(t) \le C. \end{aligned}$$

We deduce by (8.18), (8.19), (8.20), (8.21), Lemma 3.7 and (4.15) that

$$\begin{aligned} \begin{aligned}&\Vert \mathcal {D}_t^{l+1-k} v\Vert _{H^{\frac{3}{2}k}(\Omega _t)}^2 \\&\quad \le C\Vert \mathcal {D}_t^{l+2 -k} p\Vert _{H^{\frac{3}{2}k-\frac{5}{2}}(\Sigma _t)}^2 + \varepsilon E_l(t) + C_{\varepsilon }\\&\quad \le C\Vert \nabla \mathcal {D}_t^{l+2 -k} p\Vert _{H^{\frac{3}{2}k-3}(\Omega _t)}^2 + \varepsilon E_l(t) + C_{\varepsilon }\\&\quad \le C\Vert \mathcal {D}_t^{l+2 -k} \nabla p\Vert _{H^{\frac{3}{2}k-3}(\Omega _t)}^2 +C\Vert [\mathcal {D}_t^{l+2 -k},\nabla ] p\Vert _{H^{\frac{3}{2}k-3}(\Omega _t)}^2 + \varepsilon E_l(t) + C_{\varepsilon }\\&\quad \le C\Vert \mathcal {D}_t^{l+3 -k} v \Vert _{H^{\frac{3}{2}k-3}(\Omega _t)}^2 +C\Vert R_{bulk}^{l+1-k} \Vert _{H^{\frac{3}{2}k-3}(\Omega _t)}^2 + \varepsilon E_l(t) + C_{\varepsilon }. \end{aligned} \end{aligned}$$

Lemma 5.3 implies

$$\begin{aligned} \Vert R_{bulk}^{l+1-k} \Vert _{H^{\frac{3}{2}k-3}(\Omega _t)}^2 \le \Vert R_{bulk}^{l-(k-1)} \Vert _{H^{\frac{3}{2}(k-1)-1}(\Omega _t)}^2 \le \varepsilon E_l(t) + C_{\varepsilon } \end{aligned}$$

and the estimate (8.17) follows.

Let us then prove

$$\begin{aligned} \Vert v \Vert _{H^{\lfloor \frac{3}{2}(l +1)\rfloor }(\Omega _t)}^2 \le C\Vert \mathcal {D}_t^2 v\Vert _{H^{\frac{3}{2}(l-1)}(\Omega _t)}^2+ C \Vert \mathcal {D}_t v\Vert _{H^{\frac{3}{2}l}(\Omega _t)}^2 + \varepsilon E_{l}(t) + C_\varepsilon {\mathcal {E}}_l^+(t). \end{aligned}$$
(8.22)

We denote \(\lambda _l = \lfloor \frac{3}{2}(l +1)\rfloor -1\) and use the second inequality in Proposition 3.2 and have

$$\begin{aligned} \Vert v \Vert _{H^{\lfloor \frac{3}{2}(l +1)\rfloor }(\Omega _t)}^2 \le C(1+ \Vert \Delta _{\Sigma _t} v_n\Vert _{H^{\lambda _l -\frac{3}{2}}(\Sigma _t)}^2 +\Vert B\Vert _{H^{\frac{3}{2}l}(\Sigma _t)}^2+ \Vert \textrm{curl}\,v \Vert _{H^{\lambda _l}(\Omega _t)}^2). \end{aligned}$$

By the definition of \({\mathcal {E}}_l^+(t)\) in (8.1) it holds \(\Vert \textrm{curl}\,v \Vert _{H^{\lambda _l}(\Omega _t)}^2 \le {\mathcal {E}}_l^+(t)\). Lemma 5.2 and Trace Theorem yield

$$\begin{aligned} \begin{aligned} \Vert B\Vert _{H^{\frac{3}{2}l}(\Sigma _t)}^2&\le C\left( 1+ \Vert p\Vert _{H^{\lfloor \frac{3}{2}l +\frac{1}{2}\rfloor }(\Sigma _t)}^2\right) \\&\le C\left( 1+ \Vert \nabla p\Vert _{H^{\frac{3}{2}l}(\Omega _t)}^2) = C (1+\Vert \mathcal {D}_t v\Vert _{H^{\frac{3}{2}l}(\Omega _t)}^2\right) . \end{aligned} \end{aligned}$$

We treat the term \(\Vert \Delta _{\Sigma _t} v_n\Vert _{H^{\lambda _l -\frac{3}{2}}(\Sigma _t)}\) by using (8.9) and have

$$\begin{aligned} \Vert \Delta _{\Sigma _t} v_n\Vert _{H^{\lambda _l -\frac{3}{2}}(\Sigma _t)} \le C\Vert \mathcal {D}_t p\Vert _{H^{\lambda _l -\frac{3}{2}}(\Sigma _t)} + C\Vert \nabla U \cdot \nabla \partial _t U\Vert _{H^{\frac{3}{2}l -1}(\Sigma _t)} + \Vert R_p^0\Vert _{H^{\frac{3}{2}l -1}(\Sigma _t)}. \end{aligned}$$

By Lemma 5.5 we have

$$\begin{aligned} \Vert \nabla \partial _t U\Vert _{H^{\frac{3}{2}l -1}(\Sigma _t)}^2 \le \varepsilon E_l + C_\varepsilon \end{aligned}$$

and

$$\begin{aligned} \Vert \nabla U\Vert _{H^{\frac{3}{2}l -1}(\Sigma _t)}^2 \le CE_{l-1}(t) \le C. \end{aligned}$$

Therefore we have by Proposition 2.10 and by the Sobolev embedding

$$\begin{aligned} \Vert \nabla U \cdot \nabla \partial _t U\Vert _{H^{\frac{3}{2}l -1}(\Sigma _t)}^2 \le C\Vert \nabla U\Vert _{H^{\frac{3}{2}l -1}(\Sigma _t)}^2 \Vert \nabla \partial _t U\Vert _{H^{\frac{3}{2}l -1}(\Sigma _t)}^2 \le \varepsilon E_l + C_\varepsilon . \end{aligned}$$

Similarly we obtain

$$\begin{aligned} \Vert R_p^0\Vert _{H^{\frac{3}{2}l -1}(\Sigma _t)}^2 \le \varepsilon E_l + C_\varepsilon . \end{aligned}$$

We leave the details for the reader. Therefore we have by arguing as before

$$\begin{aligned} \begin{aligned} \Vert v \Vert _{H^{\lfloor \frac{3}{2}(l +1)\rfloor }(\Omega _t)}^2&\le C\Vert \mathcal {D}_t p\Vert _{H^{\lambda _l -\frac{3}{2}}(\Sigma _t)} + \varepsilon E_l + C_\varepsilon {\mathcal {E}}_l^+(t)\\&\le C\Vert \nabla \mathcal {D}_t p\Vert _{H^{\lambda _l -2}(\Omega _t)}^2 + \varepsilon E_l + C_\varepsilon {\mathcal {E}}_l^+(t)\\&\le C\Vert \mathcal {D}_t \nabla p\Vert _{H^{\lambda _l -2}(\Omega _t)}^2 + \Vert [\mathcal {D}_t, \nabla ] p \Vert _{H^{\lambda _l -2}(\Omega _t)}^2 + \varepsilon E_l + C_\varepsilon {\mathcal {E}}_l^+(t)\\&\le C\Vert \mathcal {D}_t^2 v \Vert _{H^{\lambda _l -2}(\Omega _t)}^2 + \Vert \nabla v \star \nabla p\Vert _{H^{\lambda _l -2}(\Omega _t)}^2 + \varepsilon E_l + C_\varepsilon {\mathcal {E}}_l^+(t). \end{aligned} \end{aligned}$$

Note that \(\lambda _l -2 \le \frac{3}{2}(l-1)\) and \(\lambda _l -1 \le \lfloor \frac{3}{2} l\rfloor \). Thus by the definition of \(E_{l-1}(t)\) it holds

$$\begin{aligned} \Vert \nabla p\Vert _{H^{\lambda _l -2}(\Omega _t)}^2 + \Vert \nabla v\Vert _{H^{\lambda _l -2}(\Omega _t)}^2 \le \Vert \mathcal {D}_t v \Vert _{H^{\frac{3}{2}(l-1)}(\Omega _t)}^2 + \Vert v\Vert _{H^{\lfloor \frac{3}{2} l\rfloor }(\Omega _t)}^2 \le C E_{l-1}(t)\le C. \end{aligned}$$

Proposition 2.10, the assumption \( \Vert \nabla v \Vert _{L^\infty (\Omega _t)} \le C\), and the Sobolev embedding then imply

$$\begin{aligned} \Vert \nabla v \star \nabla p\Vert _{H^{\lambda _l -2}(\Omega _t)}^2 \le C E_{l-1}^2(t)\le C \end{aligned}$$

and the inequality (8.22) follows.

We deduce by (8.16), (8.22) and by using (8.17) an iterative way that

$$\begin{aligned} \sum _{k=1}^{l} \Vert \mathcal {D}_t^{l+1-k} v\Vert _{H^{\frac{3}{2} k}(\Omega _t)}^2 + \Vert v \Vert _{H^{\lfloor \frac{3}{2}(l +1)\rfloor }(\Omega _t)}^2 \le C_\varepsilon {\mathcal {E}}_{l}^+(t) + \varepsilon E_{l}(t). \end{aligned}$$

Thus we obtain (8.15) by using the above inequality and (8.13). \(\square \)

9 Proof of the Main Theorem

In this short section we collect the results from Sects. 6, 7 and 8 and prove the Main Theorem. The proof is fairly straightforward, and the only delicate part is to show that the a priori estimates (1.7) hold for a short time.

Proof of the Main Theorem

Let us assume that the quantities \(\Lambda _T\) and \(\sigma _T\), which are defined in (1.5) and (1.4) respectively, satisfy \(\Lambda _T \le M \) and \(\sigma _T\ge \frac{1}{M}\) for \(T>0\). We show that this implies the bound

$$\begin{aligned} E_l(t) \le C_l \qquad \text {for all }\, t \le T \end{aligned}$$
(9.1)

for every positive integer l, where the constant \(C_l\) depends on lTM and on \(E_l(0)\). Here the dependence on T means that if \(T<1\), then the constant \(C_l\) may be chosen to be independent of T. The estimate (9.1) is crucial as it quantifies the smoothness of the flow under the assumption that the a priori estimates are bounded.

We obtain first by Lemma 6.4 that

$$\begin{aligned} \int _0^T \Vert p\Vert _{H^2(\Omega _t)}^2 \, dt \le {\tilde{C}}, \end{aligned}$$
(9.2)

where \({\tilde{C}}\) depends on TM and on \(E_1(0)\). Proposition 7.1 and Proposition 8.1 in turn imply

$$\begin{aligned} \begin{aligned} \frac{d}{dt} {\mathcal {E}}_1(t)&\le C(1+ \Vert p\Vert _{H^2(\Omega _t)}^2)E_1(t)\\&\le C(1+ \Vert p\Vert _{H^2(\Omega _t)}^2)(C_0 + {\mathcal {E}}_1(t)) \end{aligned} \end{aligned}$$
(9.3)

for all \(t \le T\). In particular, the quantity \(C_0 + {\mathcal {E}}_1(t)\) is positive. Therefore we obtain by integrating over (0, T) and using (9.2) that

$$\begin{aligned} C_0 + {\mathcal {E}}_1(t) \le {\hat{C}}(C_0 + {\mathcal {E}}_1(0)) \end{aligned}$$

for all \(t \le T\). By using Proposition 8.1 again we have

$$\begin{aligned} E_1(t) \le C(C_0 + {\mathcal {E}}_1(t)) \le C {\hat{C}}(C_0 + {\mathcal {E}}_1(0)) \le C_1, \end{aligned}$$

where the constant \(C_1 \) depends on MT and \(E_1(0)\).

We may then use Proposition 7.2 and Proposition 8.3 in an inductive way and deduce that if \(E_{l-1}(t) \le C_{l-1}\) for \(t \le T\) then it holds

$$\begin{aligned} \frac{d}{dt} {\mathcal {E}}_l(t) \le CE_l(t) \le C(C_0 + {\mathcal {E}}_l(t)). \end{aligned}$$

By integrating we deduce

$$\begin{aligned} C_0 + {\mathcal {E}}_l(t) \le (C_0 + {\mathcal {E}}_l(0))e^{C T} \end{aligned}$$

and using Proposition 8.3 again we have

$$\begin{aligned} E_l(t) \le C(C_0 + {\mathcal {E}}_l(0))e^{C T} \le C_l, \end{aligned}$$
(9.4)

where the constant \(C_l\) depends on lTM and on \(E_{l}(0)\). Note that we obtain (9.4) under the assumption \(E_{l-1}(t) \le C_{l-1}\) for \(t \le T\) and thus an induction argument implies that (9.4) holds for all l for a constant which depends on TlM and on \(E_l(0)\). Therefore we have (9.1).

Let us then prove the last claim, i.e., that the a priori estimates (1.7) hold for M for a short time

$$\begin{aligned} T_0 \ge c_0 \end{aligned}$$
(9.5)

for a positive constant \(c_0\) which depends on \(\Vert H_{\Sigma _0}\Vert _{H^2(\Sigma )}\), \(\Vert v\Vert _{H^3(\Omega _0)}\) and on \(\sigma _0\).

To this aim we define the quantity

$$\begin{aligned} \lambda _t:= \Vert \nabla p\Vert _{L^2(\Omega _t)}^2 + \Vert B\Vert _{L^4(\Sigma _t)}^4 + \Vert \nabla v\Vert _{L^4(\Sigma _t)}^4 + \Vert \nabla v\Vert _{L^4(\Omega _t)}^4 +1, \end{aligned}$$

where p is the pressure and v the velocity field. Let us also denote by

$$\begin{aligned} \delta (t): = d_{{\mathcal {H}}}(\Omega _t, \Omega _0) \end{aligned}$$

the Hausdorff distance between the sets \(\Omega _t\) and \(\Omega _0\). The point is that if we would know that it holds \(\lambda _t \le 2\lambda _0\) and \(\delta (t) \le \varepsilon _0\), where \(\lambda _0\) is the value at time \(t=0\) and \(\varepsilon _0\) is a small number, then we have by the curvature bound and by standard argument from regularity theory (e.g. by Allard regularity theory) that \(\Sigma _t\) is uniformly \(C^{1,\alpha }(\Gamma )\)-regular. We choose the number \(\varepsilon _0\) such that it depends also on \(\sigma _0\) so that \(\delta (t) \le \varepsilon _0\) implies \(\sigma _t \ge \frac{\sigma _0}{2}\). Moreover, by Proposition 8.1 we deduce that there are constants C and \(C_0\) such that

$$\begin{aligned} E_1(t) \le C(C_0 + {\mathcal {E}}_1(t)). \end{aligned}$$
(9.6)

Let us then define \(T_0 \in (0, T]\) to be the largest number such that

$$\begin{aligned} \begin{aligned} \sup _{t \le T_0} \lambda _t \le 2 \lambda _0, \quad \sup _{t \le T_0 } \delta (t) \le \varepsilon _0 \quad \text {and} \quad \sup _{t \le T_0 } {\mathcal {E}}_1(t) \le C_0 + {\mathcal {E}}_1(0), \end{aligned} \end{aligned}$$

where \(C_0\) is the constant in (9.6). We note that the last condition together with (9.6) implies that

$$\begin{aligned} E_1(t) \le C(C_0 + {\mathcal {E}}_1(t))\le C(2C_0 + {\mathcal {E}}_1(0)) \le {\tilde{C}} E_1(0), \end{aligned}$$
(9.7)

for \(t \le T_0\). It is also easy to see that for \(\Lambda _{T}\) defined in (1.5) it holds \(\Lambda _{T}^2 \le C \sup _{t \le T} E_1(t)\). This means that (9.7) ensures that the a priori estimates (1.7) hold for the time interval \([0,T_0]\). Therefore it is enough to show that \(T_0 \ge c_0\). We may assume that \(T_0 < \min \{T,1\}\) since otherwise the claim is trivially true.

If \(T_0 < \min \{T,1\}\) then at least in one of the three conditions in the definition of \(T_0\) we have an equality. Assume that \(\lambda _{T_0} = 2 \lambda _0\). Note that by (9.7) it holds \(E_1(t) \le {\tilde{C}} E_1(0)\) for all \(t \le T_0\). We remark that it holds

$$\begin{aligned} \Vert B\Vert _{L^\infty (\Sigma _t)}^2 + \Vert \nabla v\Vert _{L^\infty (\Omega _t)}^2 \le CE_1(t). \end{aligned}$$

Moreover, by using the formula (4.10) in Lemma 4.2 we obtain

$$\begin{aligned} \Vert \mathcal {D}_t \nabla p\Vert _{L^2(\Omega _t)} + \Vert \mathcal {D}_t B\Vert _{L^2(\Sigma _t)} + \Vert \mathcal {D}_t \nabla v\Vert _{L^2(\Sigma _t)} \le CE_1(t). \end{aligned}$$

We leave the details for the reader. Therefore by a straightforward calculation we deduce that for some \(q \ge 1\) it holds

$$\begin{aligned} \frac{d}{dt} \lambda _t \le C E_1(t)^q \le C E_1(0)^q \end{aligned}$$

where the last inequality follows from (9.7). By integrating the above over \((0,T_0)\) and using \(\lambda _{T_0} = 2 \lambda _0\) we obtain

$$\begin{aligned} \lambda _0 \le C E_1(0)^q T_0. \end{aligned}$$

Since \(\lambda _0 \ge 1\) we have \(T_0 \ge c_0\), for a constant that depends on \(E_1(0)\) and \(\sigma _0\).

We argue similarly if we have an equality in the third condition in the definition of \(T_0\), i.e., \( {\mathcal {E}}_1(T_0) = C_0 + {\mathcal {E}}_1(0)\). Indeed, then by the definition of \(E_1(t)\) we have that

$$\begin{aligned} \Vert p\Vert _{H^2(\Omega _t)}^2 \le E_1(t). \end{aligned}$$

Therefore we obtain by (9.3) and (9.7) that

$$\begin{aligned} \frac{d}{dt} {\mathcal {E}}_1(t) \le C(1+ \Vert p\Vert _{H^2(\Omega _t)}^2)E_1(t) \le C, \end{aligned}$$

where the constant C depends on \(E_1(0)\) and on \(\sigma _0\). We integrate the above over \((0,T_0)\) and obtain

$$\begin{aligned} C_0 = {\mathcal {E}}_1(T_0) - {\mathcal {E}}_1(0) \le CT_0. \end{aligned}$$

Thus we have again \(T_0 \ge c_0\).

Finally assume that it holds \( \delta (T_0) = \varepsilon _0\). By definition the flow gives a diffeomorphism \(\Phi _{T_0}: \Sigma _0 \rightarrow \Sigma _{T_0}\). We note that the velocity is uniformly bounded by the Sobolev embedding and by (9.7)

$$\begin{aligned} \Vert v\Vert _{L^\infty (\Omega _t)}^2 \le CE_1(t) \le C E_1(0). \end{aligned}$$

Therefore we have by the fundamental Theorem of Calculus that for every \(x \in \Sigma _0\) it holds

$$\begin{aligned} |\Phi _{T_0}(x) - x| \le \int _0^{T_0} \Vert v\Vert _{L^\infty }\, dt \le C T_0. \end{aligned}$$

Since \(\sup _{x \in \Sigma _0}|\Phi _{T_0}(x) - x|\ge \delta (T_0) \ge \varepsilon _0\), we again have \(T_0 \ge c_0\).

We have thus obtained (9.5) for a constant \(c_0\) which depends on \(\sigma _0\) and \(E_1(0)\). By Lemma 8.2 we deduce that \(c_0\) in fact depends on \(\sigma _0, \Vert v\Vert _{H^3(\Omega _0)}\), \(\Vert H_{\Sigma _0}\Vert _{H^2(\Sigma _0)}\) and on \(\Vert h_0\Vert _{C^{1,\alpha }(\Gamma )}\). This concludes the proof of the second claim. \(\square \)