A Decoupled Unconditionally Stable Numerical Scheme for the Cahn–Hilliard–Hele-Shaw System

Abstract

We propose a novel decoupled unconditionally stable numerical scheme for the simulation of two-phase flow in a Hele-Shaw cell which is governed by the Cahn–Hilliard–Hele-Shaw system (CHHS) with variable viscosity. The temporal discretization of the Cahn–Hilliard equation is based on a convex-splitting of the associated energy functional. Moreover, the capillary forcing term in the Darcy equation is separated from the pressure gradient at the time discrete level by using an operator-splitting strategy. Thus the computation of the nonlinear Cahn–Hilliard equation is completely decoupled from the update of pressure. Finally, a pressure-stabilization technique is used in the update of pressure so that at each time step one only needs to solve a Poisson equation with constant coefficient. We show that the scheme is unconditionally stable. Numerical results are presented to demonstrate the accuracy and efficiency of our scheme.

Appendix

Proof of Proposition 1

Taking the test function $v_h=1$ in Eq. (3.13) gives

$$\begin{aligned} \int _{{\varOmega }} \phi _h^{n+1}\, dx= \int _{{\varOmega }} \phi _h^{n}\, dx. \end{aligned}$$

We define the average $a:=\int _{{\varOmega }} \phi _h^n\, dx/ \int _{{\varOmega }}1\, dx$.

Thanks to the decoupling of the pressure equation from the Cahn–Hilliard equation, we only need to show the existence and uniqueness of solutions to Eqs. (3.13) and (3.14). For convenience, we first look for solutions in $M_h\, \times \, M_h$, i.e., seek $\{\phi _h, \mu _h\} \in M_h \times M_h$ such that

$$\begin{aligned}&\left( \phi _h-\phi _h^n, v_h\right) +k(\tilde{m}(\phi _h^n)\nabla \mu _h, \nabla v_h)+k\left( \frac{\phi _h^n}{12\eta (\phi _h^n)}\nabla p_h^n, \nabla v_h\right) =0, \forall v_h \in M_h, \end{aligned}$$

(6.1)

$$\begin{aligned}&\epsilon ^2 \left( \nabla \phi _h, \nabla \varphi _h\right) +\left( (\phi _h+a)^3-\phi _h^n, \varphi _h\right) -(\mu _h, \varphi _h)=0, \quad \forall \varphi _h \in M_h, \end{aligned}$$

(6.2)

with a new mobility function $\tilde{m}(\phi _h^n)=\frac{1}{Pe}\big [m(\phi _h^n)+\frac{\gamma }{\epsilon } \frac{(\phi ^n_h)^2}{12\eta (\phi _h^n)}\big ]$. It follows from the boundedness assumption (1.3) that

$$\begin{aligned} 0<\frac{m_1}{Pe} \le \tilde{m}(\phi _h^n) \le C(||\phi _h^n||_{L^\infty }) \end{aligned}$$

(6.3)

Note also that $M_h$ is a Hilbert space with the inner product $(u_h, v_h)_{M_h}=(\nabla u_h, \nabla v_h)$ and induced norm $|| u_h ||_{M_h}^2=(u_h, u_h)_{M_h}$, thanks to the Poincare inequality.

Claim: Eq. (6.2) defines a bounded and continuous solution operator $\phi _h(\mu _h): M_h \rightarrow M_h$.

Indeed, given $\mu _h \in M_h$, Eq. (6.2) is the Euler–Lagrange equation of the following strict convex minimization problem

$$\begin{aligned} \min _{\phi _h \in M_h} J(\phi _h):=\min _{\phi _h \in M_h} \int _{{\varOmega }}\frac{\epsilon ^2}{2}|\nabla \phi _h|^2+ \frac{1}{4}(\phi _h+a)^4- (\mu _h+\phi _h^n) \phi _h\, dx. \end{aligned}$$

Thus the theory of calculus of variation implies the unique existence of $\phi _h \in M_h$ such that Eq. (6.2) holds.

For continuity of the operator $\phi _h(\mu _h)$, let us assume that $\phi _h^{(1)}$ and $\phi _h^{(2)}$ are solutions to Eq. (6.2) with source terms $\mu _h^{(1)}$ and $\mu _h^{(2)}$, respectively. Subtract the two equations satisfied by $\phi _h^{(1)}$ and $\phi _h^{(2)}$ and take the test function $\varphi _h=\phi _h^{(1)}-\phi _h^{(2)}$

$$\begin{aligned}&\epsilon ^2||\phi _h^{(1)}-\phi _h^{2}||_{M_h}^2+\left( (\phi _h^{(1)}+a)^3-(\phi _h^{(2)}+a)^3, \phi _h^{(1)}-\phi _h^{(2)}\right) \\&\quad =(\mu _h^{(1)}-\mu _h^{(2)}, \phi _h^{(1)}-\phi _h^{(2)}) \\&\quad \le ||\mu _h^{(1)}-\mu _h^{(2)}||_{L^2}||\phi _h^{(1)}-\phi _h^{(2)}||_{L^2}\\&\quad \le C||\mu _h^{(1)}-\mu _h^{(2)}||_{M_h}||\phi _h^{(1)}-\phi _h^{(2)}||_{M_h}. \end{aligned}$$

The monotonicity of the function $f(x)=x^3$ implies (cf. [29])

$$\begin{aligned} \left( (\phi _h^{(1)}+a)^3-(\phi _h^{(2)}+a)^3, (\phi _h^{(1)}-a)-(\phi _h^{(2)}-a)\right) \ge 0. \end{aligned}$$

(6.4)

The continuity of $\phi _h(\mu _h)$ is thus proved.

To prove the boundedness of $\phi _h(\mu _h)$, we take the test function $\varphi _h=\phi _h$ in Eq. (6.2)

$$\begin{aligned} \epsilon ^2||\phi _h||_{M_h}^2+ \int _{{\varOmega }} (\phi _h+a)^4\, dx = a \int _{{\varOmega }} (\phi _h+a)^3\, dx +(\mu _h+\phi _h^n, \phi _h) \end{aligned}$$

(6.5)

Applying the Young’s inequality $bc \le \frac{b^p}{p}+\frac{c^q}{q}$ with $p=\frac{4}{3}$ and $q=4$, one obtains

$$\begin{aligned} a\int _{{\varOmega }} (\phi _h+a)^3\, dx \le \int _{{\varOmega }}\frac{3}{4}(\phi _h+a)^4+\frac{a^4}{4} |{\varOmega }|. \end{aligned}$$

(6.6)

It follows, by Cauchy–Schwartz inequality and Poincare’s inequality, that

$$\begin{aligned} ||\phi _h||_{M_h} \le C(|a|, |{\varOmega }|, ||\phi _h^n||_{L^2})(1+||\mu _h||_{M_h}). \end{aligned}$$

Now we define an operator $T: M_h \rightarrow M_h$ such that for any $v_h \in M_h$

$$\begin{aligned} \left( T(\mu _h), v_h \right) _{M_h}:= \left( \phi _h(\mu _h)-\phi _h^n, v_h\right) +k(\tilde{m}(\phi _h^n)\nabla \mu _h, \nabla v_h)+k\left( \frac{\phi _h^n}{12\eta (\phi _h^n)}\nabla p_h^n, \nabla v_h\right) , \end{aligned}$$

where $\phi _h(\mu _h)$ is the solution operator defined via Eq. (6.2).

It is clear that $T(\mu _h)$ is continuous in $M_h$, in view of the continuity of $\phi _h(\mu _h)$. Further, the boundedness of $\phi _h(\mu _h)$ implies that

$$\begin{aligned}&\Big |\left( T(\mu _h), v_h \right) _{M_h} \Big | \le C(||\phi _h^n||_{L^\infty }, ||\phi _h^n||_{L^2}, ||\nabla p_h^n||_{L^2})(1+||\mu _h||_{M_h})||v_h||_{M_h}, \end{aligned}$$

from which the boundedness of T follows.

Next, we show that T is coercive in the sense that $\frac{\left( T(\mu _h), \mu _h \right) _{M_h}}{||\mu _h||_{M_h}} \rightarrow \infty$, as $||\mu _h||_{M_h} \rightarrow \infty$. One has, by using Eq. (6.2) with $\varphi _h=\phi _h$

$$\begin{aligned}&\left( T(\mu _h), \mu _h \right) _{M_h} \\&\quad = \left( \phi _h(\mu _h)-\phi _h^n, \mu _h\right) +k(\tilde{m}(\phi _h^n)\nabla \mu _h, \nabla \mu _h)+k\left( \frac{\phi _h^n}{12\eta (\phi _h^n)}\nabla p_h^n, \nabla \mu _h\right) \\&\quad \ge k\frac{m_1}{Pe}||\mu _h||_{M_h}^2+\epsilon ^2 ||\nabla \phi _h||_{L^2}^2 + \int _{{\varOmega }}(\phi _h+a)^4\, dx - a \int _{{\varOmega }} (\phi _h+a)^3\, dx \\&\quad \quad -(\phi _h^n, \mu _h)+k\left( \frac{\phi _h^n}{12\eta (\phi _h^n)}\nabla p_h^n, \nabla \mu _h\right) \\&\quad \ge k\frac{m_1}{Pe}||\mu _h||_{M_h}^2-C||\phi _h^n||_{L^2}||\mu _h||_{M_h}-C(||\phi _h||_{L^\infty })||\nabla p_h^n||_{L^2}||\mu _h||_{M_h}-C \\&\quad \ge C||\mu _h||_{M_h}^2-C(||\phi _h^n||_{L^2}, ||\phi _h^n||_{L^\infty }, ||\nabla p_h^n||_{L^2}), \end{aligned}$$

where one has utilized the inequality (6.6). The coercivity of T is thus justified.

The operator T is also strictly monotone, since Eqns. (6.1) and (6.2) yield

$$\begin{aligned}&\left( T(\mu _h)-T(v_h), \mu _h-v_h \right) _{M_h} \\&\quad = \left( \phi _h(\mu _h)-\phi _h(v_h), \mu _h-v_h \right) +k\left( \tilde{m}(\phi _h^n)\nabla (\mu _h-v_h), \nabla (\mu _h-v_h)\right) \\&\quad \ge k\frac{m_1}{Pe}||\mu _h-v_h||_{M_h}^2+\epsilon ^2\left( \nabla (\phi _h(\mu _h)-\phi _h(v_h)), \nabla (\phi _h(\mu _h)-\phi _h(v_h))\right) \\&\qquad +\left( (\phi _h(\mu _h)+a)^3-(\phi _h(v_h)+a)^3, \phi _h(\mu _h)-\phi _h(v_h)\right) \\&\quad \ge 0, \end{aligned}$$

with equality if and only if $\mu _h=v_h$. Here the inequality (6.4) is applied again.

The Browder–Minty Lemma ([37], p. 557, Theorem 26.A.) implies that there is a unique $\mu _h \in M_h$ such that $T(\mu _h)=0$. That is, there exists a unique pair $\{\phi _h, \mu _h\} \in M_h \times M_h$ such that Eqs. (6.1) and (6.2) hold. Now one defines $\phi _h^{n+1}=\phi _h+a$ and $\mu _h^{n+1}=\mu _h+\frac{1}{|{\varOmega }|} \int _{{\varOmega }}(\phi _h^{n+1})^3-\phi _h^n\, dx$. Then $\{\phi _h^{n+1}, \mu _h^{n+1}\} \in Y_h \times Y_h$ is the unique solution to Eqs. (3.13) and (3.14). The proof of Proposition 1 is now complete. $\square$