Appendix A: Spherically Symmetric Equilibria of the Full Liquid/Gas Model
In this appendix, we prove Proposition 4.1. That is, we show that (4.3) is the unique regular spherically symmetric equilibrium solution to the system (2.1)–(2.4) under the radiation condition (4.2) for \(T_l\).
Proof of Proposition 4.1
We consider the full liquid/gas model (2.1)–(2.4) and prove Proposition 4.1. Steady-state solutions of (2.1)–(2.4) solve
For the spherically symmetric case, (A.1b) we have \({\textbf{v}}_{l,*}(x) = v_{l,*}(r) \frac{x}{r}\). Therefore, \(0={\textrm{div}}{\textbf{v}}_{l,*}= \partial _r v_{l,*}(r) + (2/r)v_{l,*}(r)\) and hence
$$\begin{aligned} \begin{aligned} \frac{1}{r^2}\partial _r(r^2 v_{l,*}(r)) = 0,\quad r\ge R_*. \end{aligned}\end{aligned}$$
(A.4)
Therefore,
$$\begin{aligned} \begin{aligned} v_{l,*}(r) = \frac{a}{r^2},\quad r\ge R_*, \end{aligned}\end{aligned}$$
(A.5)
for constant a. But the boundary condition (A.3a) implies \(v_{l,*}(R_*) = 0\). So \(a=0\), and thus \(v_{l,*}\equiv 0\). Therefore, (A.1a) becomes \(\nabla p_{l,*} = 0\). So the pressure \(p_{l,*}\) is a constant and equal to its value at infinity, \(p_{\infty ,*}\).
For the gas velocity \(v_{g,*}\), (A.2a) becomes
$$\begin{aligned} \frac{1}{r^2} \partial _r(\rho _{g,*} v_{g,*})= 0,\quad 0\le r\le R_*, \end{aligned}$$
which implies \(\rho _{g,*} v_{g,*}\) is a constant. Again, the boundary condition (A.3a) implies \(v_{g,*}(R_*) = 0\). So \(\rho _{g,*} v_{g,*}\equiv 0\). But \(\rho _{g,*}\ne 0\) since otherwise \(s_*\) is singular in (A.2e). Therefore, \({\textbf{v}}_{g,*}\equiv {\textbf{0}}\) and thus (A.2b) becomes \(\nabla p_{g,*} = 0\). So \(p_{g,*}\) is a constant. Moreover, by \(v_{l,*}=v_{g,*}\equiv 0\), (A.3b) yields \(-p_{l,*} + p_{g,*} = \frac{2\sigma }{R_*}\). So \(p_{g,*} = p_{\infty ,*} + \frac{2\sigma }{R_*}\).
For the equations of the temperatures, due to \(v_{l,*}=0\) (A.1c) becomes \(\Delta T_{l,*} = 0\) in \({\mathbb {R}}^3{\setminus } B_{R_*}\), or, in spherical coordinates,
$$\begin{aligned} \frac{1}{r^2} \partial _r(r^2 \partial _r T_{l,*}) = 0,\quad r\ge R_*, \end{aligned}$$
which implies
$$\begin{aligned} \partial _r T_{l,*}=\frac{a_1}{r^2},\quad r\ge R_* \end{aligned}$$
for some constant \(a_1\). Integrating over \((r,\infty )\) we get
$$\begin{aligned} T_{l,*}(r) = T_\infty - \frac{a_1}{r},\quad r\ge R_*. \end{aligned}$$
By the radiation condition (4.2), \(T_{l,*}(r) = T_\infty + o(r^{-1})\) as \(r\rightarrow \infty \). This gives \(a_1=0\) and thus \(T_{l,*}\equiv T_\infty \). On the other hand, (A.2c) becomes \(\Delta T_{g,*} = 0\) in \(B_{R_*}\) since \(v_{g,*}=0\). Since \(T_{g,*}\) is regular, \(T_{g,*}\equiv T_{g,*}(R_*)=T_\infty \) by the maximum principle.
For the gas density \(\rho _{g,*}\), by (A.2d) \(\rho _{g,*} = \frac{p_{g,*}}{{\mathscr {R}}_gT_\infty } = \frac{1}{{\mathscr {R}}_gT_\infty } \left( p_{\infty ,*}+\frac{2\sigma }{R_*} \right) \). Due to the conservation of mass (7.1),
$$\begin{aligned} M:= \int _{B_{R_0}} \rho _0 = \lim _{t\rightarrow \infty } \int _{B_{R(t)}} \rho _g(\cdot ,t) = \frac{4\pi }{3} \rho _{g,*} R_*^3, \end{aligned}$$
where \((\rho _0(x),R_0)\), \(\rho _0(x)\ge 0, R_0>0\), is the initial data. Therefore, the steady state \((\rho _{g,*},R_*)\) can be obtained by solving
In particular, the stationary radius \(R_*\) satisfies the cubic equation
$$\begin{aligned} \begin{aligned} p_{\infty ,*} R_*^3 + 2\sigma R_*^2 - \frac{3{\mathscr {R}}_gT_\infty M}{4\pi } = 0. \end{aligned}\end{aligned}$$
(A.7)
It is readily seen that for any \(M>0\), the cubic equation has a unique positive root \(R_*[M]\). This proves Proposition 4.1. \(\square \)
Appendix B: Derivation of the Reduced System for \(\rho (r,t)\) and R(t): Proof of Proposition 5.1
Considering the uniformity of the pressure \(p_g\) in (3.2b), we can eliminate \(T_g\) by plugging (3.2d) into (3.2c) and deduce
$$\begin{aligned} \begin{aligned} \partial _t s + {\textbf{v}}_g\cdot \nabla s = \kappa _g \Delta \left( \frac{1}{\rho _g} \right) . \end{aligned}\end{aligned}$$
(B.1)
Plugging (3.2e) into the left hand side of (B.1) and using (3.2b), we have
$$\begin{aligned} \begin{aligned} c_v \left\{ \frac{\partial _t p_g}{p_g} - \frac{\gamma }{\rho _g} \left[ \partial _t\rho _g + {\textbf{v}}_g\cdot \nabla \rho _g \right] \right\} = \kappa _g \Delta \left( \frac{1}{\rho _g} \right) . \end{aligned}\end{aligned}$$
(B.2)
Using (3.2a) in (B.2), we obtain
$$\begin{aligned} \begin{aligned} c_v \left\{ \frac{\partial _t p_g}{p_g} + \gamma {\textrm{div}}{\textbf{v}}_g \right\} = \kappa _g \Delta \left( \frac{1}{\rho _g} \right) , \end{aligned}\end{aligned}$$
(B.3)
Therefore, the system (3.2) is reduced to
Expanding the term \({\textrm{div}}(\rho _g{\textbf{v}}_g)\) in (B.4a) and substituting \({\textrm{div}}{\textbf{v}}_g\) using (B.4b), and using the elementary identity
$$\begin{aligned} \begin{aligned} \rho _g \Delta \left( \frac{1}{\rho _g} \right) = -\Delta \log \rho _g + \frac{|\nabla \rho _g|^2}{\rho _g^2}, \end{aligned}\end{aligned}$$
(B.5)
we get
$$\begin{aligned} \begin{aligned} \partial _t\rho _g = \frac{\kappa }{\gamma c_v} \Delta \log \rho _g - \frac{\kappa }{\gamma c_v} \frac{|\nabla \rho _g|^2}{\rho _g^2} - {\textbf{v}}_g\cdot \nabla \rho _g + \frac{\partial _tp_g}{\gamma p_g} \rho _g. \end{aligned}\end{aligned}$$
(B.6)
Assuming the bubble is a sphere \(B_{R(t)}\) and solutions are spherically symmetric, and recalling we denoted the radial components of the gas and liquid velocity by \(v_g(r,t)\) and \(v_l(r,t)\), respectively, the systems (3.1) and (B.4) become
and
and the boundary condition (3.3) becomes
The liquid velocity and pressure \((v_l,p_l)\) can be directly solved in terms of R(t), \(\dot{R}(t)\), the liquid pressure \(p_l(R(t),t)\) on the bubble wall, and the far-field liquid pressure \(p_\infty (t):= p_l(r=\infty ,t)\). In fact, the incompressibility condition (B.7b) and the kinematic boundary condition (B.9a) imply
$$\begin{aligned} \begin{aligned} v_l(r,t) = \frac{(R(t))^2\dot{R}(t)}{r^2},\quad r\ge R(t),\ t>0. \end{aligned}\end{aligned}$$
(B.10)
Plugging Equation (B.10) into Equation (B.7a), we have
$$\begin{aligned} \begin{aligned} \frac{2R\dot{R}^2 {+} R^2 {\ddot{R}}}{r^2} {=} \nu _l\left( \frac{2R^2\dot{R}}{r^4} {-} 2\frac{R^2\dot{R}}{r^4} \right) {+} 2 \frac{R^4\dot{R}^2}{r^5} {-} \frac{1}{\rho _l} \partial _rp_l,\quad r\ge R(t),\ t>0. \end{aligned}\nonumber \\\end{aligned}$$
(B.11)
Note that the diffusion term in (B.11) vanishes. So the reduction using spherical symmetry assumption also works for Euler equation, i.e., we can take \(\nu _l=0\) in (3.1a). Integrating Equation (B.11) over \(r>R(t)\), we deduce
$$\begin{aligned} \begin{aligned} p_l(r,t) = p_\infty (t) + \rho _l \left( \frac{2R(t) (\dot{R}(t))^2 + (R(t))^2{\ddot{R}}(t)}{r} - \frac{(R(t))^4(\dot{R}(t))^2}{2r^4} \right) ,\quad r\ge R(t),\ t>0. \end{aligned}\nonumber \\\end{aligned}$$
(B.12)
In particular, on the boundary the liquid pressure is
$$\begin{aligned} p_l(R(t),t) = p_\infty (t) + \rho _l\left( \frac{3}{2}\dot{R}^2 + R{\ddot{R}} \right) ,\quad t>0. \end{aligned}$$
Moreover, (B.10) implies \(\partial _rv_l(r,t) = -2(R(t))^2\dot{R}(t)/r^3\) so that
$$\begin{aligned} \partial _rv_l(R(t),t) = -2\, \frac{\dot{R}(t)}{R(t)}. \end{aligned}$$
This implies
$$\begin{aligned} \begin{aligned} R{\ddot{R}} + \frac{3}{2} \dot{R}^2 = \frac{1}{\rho _l} \left( p_g(t) - p_\infty (t) - \frac{2\sigma }{R} -4\mu _l \frac{\dot{R}}{R} \right) ,\quad t>0, \end{aligned}\end{aligned}$$
(B.13)
where the Young–Laplace boundary condition (B.9b) has been used.
For the gas dynamics in the bubble, by integrating (B.8b) in r, the radial component of the gas velocity \(v_g\) can be expressed in terms of \(\rho _g(r,t)\), \(\partial _r\rho _g(r,t)\), \(p_g(t)\), and \(\partial _tp_g(t)\). To be more precise,
$$\begin{aligned} \begin{aligned} v_g(r,t) = \frac{\kappa }{\gamma c_v} \partial _r\left( \frac{1}{\rho _g(r,t)} \right) - \frac{\partial _tp_g(t)}{p_g(t)}\frac{r}{3\gamma },\quad 0\le r\le R(t),\ t>0. \end{aligned}\nonumber \\\end{aligned}$$
(B.14)
Using (B.14) we can eliminate \({\textbf{v}}_g\) in (B.6) and obtain
$$\begin{aligned} \begin{aligned} \partial _t\rho _g = \frac{\kappa }{\gamma c_v} \Delta _r\log \rho _g + \frac{\partial _t p_g}{3\gamma p_g} r \partial _r\rho _g + \frac{\partial _t p_g}{\gamma p_g} \rho _g,\quad 0\le r\le R(t),\ t>0, \end{aligned}\nonumber \\\end{aligned}$$
(B.15)
where \(\Delta _r f = \frac{1}{r^2}\partial _r(r^2\partial _rf)\) for spherically symmetric functions f. From the boundary condition (B.9c) for the gas temperature and the equation of state (3.2d),
$$\begin{aligned} \begin{aligned} p_g(t) = {\mathscr {R}}_g\rho _g(R(t),t) T_\infty . \end{aligned}\end{aligned}$$
(B.16)
Taking time derivative of (B.16) we obtain
$$\begin{aligned} \begin{aligned} \frac{\partial _t p_g}{p_g} = \frac{\partial _t\rho _g(R(t),t)}{\rho _g(R(t),t)} + \frac{\partial _r\rho _g(R(t),t)}{\rho _g(R(t),t)} \dot{R}(t). \end{aligned}\end{aligned}$$
(B.17)
Evaluating (B.14) at \(r=R(t)\) and using the kinematic boundary condition (B.9a) it follows that
$$\begin{aligned} \begin{aligned} \dot{R}(t) = -\frac{\kappa }{\gamma c_v} \frac{\partial _r\rho _g(R(t),t)}{\left( \rho _g(R(t),t) \right) ^2} - \frac{R(t)}{3\gamma } \frac{\partial _t p_g}{p_g}. \end{aligned}\end{aligned}$$
(B.18)
For the boundary data for the gas density, we use (B.16) and (B.13) to deduce
$$\begin{aligned} \begin{aligned} \rho _g(R(t),t) = \frac{1}{{\mathscr {R}}_gT_\infty } \left[ p_\infty + \frac{2\sigma }{R} + \rho _l\left( R{\ddot{R}} + \frac{3}{2} (\dot{R})^2 \right) \right] . \end{aligned}\end{aligned}$$
(B.19)
Collecting the results (B.15), (B.16), (B.18), (B.19), we conclude that, under the spherical symmetry assumption, the system (3.1)–(3.3) is reduced to a system of \((\rho (r,t),R(t))\):
$$\begin{aligned} \partial _t\rho= & {} \frac{\kappa }{\gamma c_v} \Delta _r\log \rho + \frac{\partial _t p}{3\gamma p} r \partial _r\rho + \frac{\partial _t p}{\gamma p} \rho ,\quad 0\le r\le R(t),\ t>0, \end{aligned}$$
(B.20)
$$\begin{aligned} p(t)= & {} {\mathscr {R}}_gT_\infty \rho (R(t),t),\quad t>0, \end{aligned}$$
(B.21)
$$\begin{aligned} \dot{R}(t)= & {} -\frac{\kappa }{\gamma c_v} \frac{\partial _r\rho (R(t),t)}{\left( \rho (R(t),t) \right) ^2} - \frac{R(t)}{3\gamma } \frac{\partial _t p}{p},\quad t>0, \end{aligned}$$
(B.22)
$$\begin{aligned} \rho (R(t),t)= & {} \frac{1}{{\mathscr {R}}_gT_\infty } \left[ p_\infty + \frac{2\sigma }{R} + 4\mu _l \frac{\dot{R}}{R} + \rho _l\left( R{\ddot{R}} + \frac{3}{2} (\dot{R})^2 \right) \right] , \quad t>0,\nonumber \\ \end{aligned}$$
(B.23)
where \(\rho \equiv \rho _g\), \(p\equiv p_g\), \(\kappa =\kappa _g\). This is the reduced system (5.1a)–(5.1c).
Appendix C: A Perspective on Coercive Energy Estimate of Biro–Velázquez, and an Extension
In this appendix, we prove Theorem 7.5, which extends the coercivity estimate of Biro-Velázquez to the case where \(p_\infty -p_{\infty ,*}\) is small in norm.
Proof of Theorem 7.5
Let us recall the total energy
$$\begin{aligned} {\mathcal {E}}_{\textrm{total}} = FE + KE_l + U_{g-l} + PV_{p_\infty }, \end{aligned}$$
where
The energy is a functional of state variables, which are defined on a deforming regime, \(B_{R}\). We fix the region to be \(B_1\) by setting \(x=Ry\), where \(y\in B_1\). Defining \({\overline{\rho }}(y)=\rho (Rr)\) and using the constitutive relation \(p = {\mathscr {R}}_gT_\infty \rho (R) = {\mathscr {R}}_gT_\infty {\overline{\rho }}(1)\) we have that
$$\begin{aligned} FE= & {} \frac{4\pi c_v T_\infty }{3} {\overline{\rho }}(1) R^3 - c_v T_\infty M_0 \log ({\mathscr {R}}_gT_\infty ) - c_v T_\infty M_0 \log {\overline{\rho }}(1)\\{} & {} +\, c_v\gamma T_\infty R^3 \int _{B_1} {\overline{\rho }} \log {\overline{\rho }}. \end{aligned}$$
Thus, \({\mathcal {E}}_{\textrm{total}} \) is a functional of \(({\overline{\rho }}, R,\dot{R})\):
$$\begin{aligned} \begin{aligned} {\mathcal {E}}_{\textrm{total}}&= {\mathcal {E}}_{\textrm{total}} [{\overline{\rho }}, R,\dot{R}]\\&= \frac{4\pi c_v T_\infty }{3} {\overline{\rho }}(1)R^3 - c_vT_\infty M_0\log ({\mathscr {R}}_gT_\infty ) - c_vT_\infty M_0 \log {\overline{\rho }}(1)\\&\quad + c_v\gamma T_\infty R^3 \int _{B_1} {\overline{\rho }} \log {\overline{\rho }}\\&\quad + 2\pi \rho _lR^3\dot{R}^2 + 4\pi \sigma R^2 + \frac{4\pi }{3}R^3p_\infty . \end{aligned} \end{aligned}$$
We set \({\overline{\rho }} = \rho _* + \varrho \), \(R = R_* + {\mathcal {R}}\) and expand the total energy \({\mathcal {E}}_{\textrm{total}}[{\overline{\rho }}, R,\dot{R}]\) at \((\rho _*,R_*,\dot{R}_*=0)\) along the mass preserving hypersurface \(M_0 = \text {Mass}[\rho ,R]\):
$$\begin{aligned} \begin{aligned}&{\mathcal {E}}_{\textrm{total}} [{\overline{\rho }},R,\dot{R}] = {\mathcal {E}}_{*} + d{\mathcal {E}}_{*}[ \varrho , {\mathcal {R}}, \dot{{\mathcal {R}}}]+ \frac{1}{2} d^2{\mathcal {E}}_{*}[ \varrho , {\mathcal {R}}, \dot{{\mathcal {R}}}] + O(|( \varrho , {\mathcal {R}}, \dot{{\mathcal {R}}})|^3),\quad \textrm{where}\\&{\mathcal {E}}_{*}={\mathcal {E}}_{\textrm{total}} [\rho _*,R_*,\dot{R}_*=0]\\&d{\mathcal {E}}_{*}[ \varrho , {\mathcal {R}}, \dot{{\mathcal {R}}}]= \frac{\textrm{d}}{\textrm{d}\varepsilon }\Big |_{\varepsilon =0} {\mathcal {E}}_{\textrm{total}} (\rho _*,R_*+\varepsilon {\mathcal {R}},\varepsilon \dot{{\mathcal {R}}})\\&d^2{\mathcal {E}}_{*}[ \varrho , {\mathcal {R}}, \dot{{\mathcal {R}}}]= \frac{\textrm{d}^2}{\textrm{d}\varepsilon ^2}\Big |_{\varepsilon =0} {\mathcal {E}}_{\textrm{total}} (\rho _*,R_*+\varepsilon {\mathcal {R}},\varepsilon \dot{{\mathcal {R}}})\ \end{aligned} \end{aligned}$$
(C.2)
To expand along the mass preserving hypersurface, we first use \(M_0 = \int _{B_R} \rho \, \textrm{d}x\) to rewrite
$$\begin{aligned} \begin{aligned} R^3 \int _{B_1} {\overline{\rho }} \log {\overline{\rho }}&= \int _{B_R} \rho \log \rho = \int _{B_R} \rho \log \rho _* + \int _{B_R} \rho \log \frac{\rho }{\rho _*}\\&= \log \rho _*\int _{B_R} \rho + \int _{B_R} \rho + \int _{B_R} \rho \left( \log \frac{\rho }{\rho _*} - 1 \right) \\&= M_0\log \rho _* + M_0 + R^3 \int _{B_1} {\overline{\rho }} \left( \log \frac{{\overline{\rho }}}{\rho _*} - 1 \right) , \end{aligned} \end{aligned}$$
giving the following expression for the total energy:
$$\begin{aligned} \begin{aligned} {\mathcal {E}}_{\textrm{total}} [{\overline{\rho }}, R,\dot{R}]&= \frac{4\pi c_v T_\infty }{3} {\overline{\rho }}(1,t)R^3 - c_vT_\infty M_0\log ({\mathscr {R}}_gT_\infty ) - c_vT_\infty M_0 \log {\overline{\rho }}(1,t) \\&\quad + c_v\gamma T_\infty M_0\log \rho _* + c_v\gamma T_\infty M_0 + c_v\gamma T_\infty R^3 \int _{B_1} {\overline{\rho }} \left( \log \frac{{\overline{\rho }}}{\rho _*} - 1 \right) \\&\quad + 2\pi \rho _lR^3\dot{R}^2 + 4\pi \sigma R^2 + \frac{4\pi }{3}R^3 p_\infty . \end{aligned} \end{aligned}$$
To expand the logarithmic terms we note that for \(z_*\ne 0\) and \(|z-z_*|<\frac{1}{2} |z_*|\):
$$\begin{aligned} \Big | \Big ( z\left( \log \left( \frac{z}{z_*}\right) - 1 \Big ) - \Big (-z_* + \frac{1}{2z_*} (z-z_*)^2 \right) \Big |&\le \frac{2}{|z_*|}|z-z_*|^3 \end{aligned}$$
(C.3)
$$\begin{aligned} \Big | \log z - \left( \log z_* + \frac{1}{z_*}(z-z_*)- \frac{1}{2 z_*^2} (z-z_*)^2 \right) \Big |&\le \frac{2}{3|z_*|^3}|z-z_*|^3 \end{aligned}$$
(C.4)
Applying (C.3) and (C.4) we have
$$\begin{aligned} \begin{aligned} {\mathcal {E}}_{\textrm{total}} [{\overline{\rho }}, R,\dot{R}]&= \frac{4\pi c_v T_\infty }{3} {\overline{\rho }}(1,t)R^3 - c_vT_\infty M_0\log ({\mathscr {R}}_gT_\infty )\\&\quad - c_vT_\infty M_0\left( \log \rho _* + \frac{1}{\rho _*} \varrho (1)- \frac{1}{2 \rho _*^2} \varrho (1)^2\right) + O\left( \varrho (1)^3\right) \\&\quad + c_v\gamma T_\infty M_0\log \rho _* + c_v\gamma T_\infty M_0\\&\quad + c_v\gamma T_\infty R^3\left( -\frac{4\pi }{3}\rho _* +\frac{1}{2\rho _*}\int _{B_1} \varrho ^2 \right) + O\left( R^3\int _{B_1}| \varrho |^3 \right) \\&\quad + 2\pi \rho _lR^3\dot{R}^2 + 4\pi \sigma R^2 + \frac{4\pi }{3}R^3 p_\infty . \end{aligned} \end{aligned}$$
Rearranging and simplifying gives
$$\begin{aligned} {\mathcal {E}}_{\textrm{total}} [{\overline{\rho }}, R,\dot{R}]&= - c_vT_\infty M_0\log ({\mathscr {R}}_gT_\infty ) + c_v(\gamma -1) T_\infty M_0\log \rho _* + c_v\gamma T_\infty M_0 \nonumber \\&\quad +\, \frac{4\pi c_v T_\infty }{3} {\overline{\rho }}(1)R^3 + c_vT_\infty M_0\left( -\frac{1}{\rho _*} \varrho (1) + \frac{1}{2 \rho _*^2} \varrho (1)^2\right) + O\left( \varrho (1)^3\right) \nonumber \\&\quad + c_v\gamma T_\infty R^3\left( -\frac{4\pi }{3}\rho _* +\frac{1}{2\rho _*}\int _{B_1} \varrho ^2 \right) + O\left( R^3\int _{B_1} | \varrho |^3 \right) \nonumber \\&\quad + 2\pi \rho _lR^3\dot{R}^2 + 4\pi \sigma R^2 + \frac{4\pi }{3}R^3 p_\infty . \end{aligned}$$
(C.5)
Verification that \(d{\mathcal {E}}_*[\varrho ,{\mathcal {R}},\dot{{\mathcal {R}}}]=0\) when \(p_\infty = p_{\infty ,*}\). Starting with (C.5) we calculate:
$$\begin{aligned} d{\mathcal {E}}_*[ \varrho ,{\mathcal {R}},\dot{{\mathcal {R}}}]&= \frac{4\pi c_v T_\infty }{3}\left( 3\rho _*R_*^2{\mathcal {R}} +R_*^3 \varrho (1)\right) -\frac{c_vT_\infty }{\rho _*}\left( \frac{4\pi }{3}\rho _*R_*^3\right) \varrho (1) \nonumber \\&\quad -4\pi c_v\gamma T_\infty R_*^2\rho _*{\mathcal {R}}+8\pi \sigma R_*{\mathcal {R}}+4\pi R_*^2 p_\infty {\mathcal {R}}\nonumber \\&= 4\pi R_*^2\left( c_vT_\infty \rho _*(1-\gamma )+ \frac{2\sigma }{R_*}+p_\infty \right) {\mathcal {R}}\nonumber \\&= 4\pi R_*^2\left( -{\mathscr {R}}_gT_\infty \rho _*+ \frac{2\sigma }{R_*}+p_\infty \right) {\mathcal {R}}\qquad \left[ \gamma -1=\frac{{\mathscr {R}}_g}{c_v}\ \text {by (2.3)}\right] \nonumber \\&= 4\pi R_*^2 {\mathcal {P}}_\infty {\mathcal {R}} \qquad \left[ \text {by}\, (4.14b) \right] , \end{aligned}$$
(C.6)
where \({\mathcal {P}}_\infty = p_\infty - p_{\infty ,*}\). It is readily to see that \(d{\mathcal {E}}_*[ \varrho ,{\mathcal {R}},\dot{{\mathcal {R}}}]=0\) when \(p_\infty = p_{\infty ,*}\).
Computation of \(\frac{1}{2}d^2{\mathcal {E}}[ \varrho , {\mathcal {R}}, \dot{{\mathcal {R}}}]\).
From (C.5) we compute the quadratic terms:
$$\begin{aligned} \frac{1}{2}d^2{\mathcal {E}}[ \varrho , {\mathcal {R}}, \dot{{\mathcal {R}}}]&= 4\pi c_v T_\infty \left( \rho _*R_* {\mathcal {R}}^2 + R_*^2 \varrho (1) {\mathcal {R}}\right) + \frac{c_v T_\infty M_0}{2\rho _*^2} \varrho (1)^2 \nonumber \\&\quad + \frac{c_v\gamma T_\infty R_*^3}{2\rho _*} \int _{B_1} \varrho ^2 - 4\pi c_v\gamma \rho _* T_\infty R_* {\mathcal {R}}^2\nonumber \\&\quad + 2\pi \rho _l R_*^3 \dot{{\mathcal {R}}}^2 + 4\pi \sigma {\mathcal {R}}^2 + 4\pi R_* p_\infty {\mathcal {R}}^2. \end{aligned}$$
(C.7)
Next, using that the perturbed bubble is assumed to have mass equal to \(M_0=\textrm{Mass}(\rho _*,R_*)\), we express the cross-term just above in terms of a quadratic expression in \(\varrho \) as follows:
$$\begin{aligned} M_0=R^3 \int _{B_1}{\overline{\rho }}&= (R_*+{\mathcal {R}})^3 \int _{B_1}\left( \rho _*+ \varrho \right) \\&= M_0 + 4\pi R_*^2 \rho _* {\mathcal {R}} + R_*^3 \int _{B_1} \varrho + O\Big ({\mathcal {R}}^2 + \left( \int _{B_1} \varrho \right) ^2 \Big ) \end{aligned}$$
and therefore
$$\begin{aligned} {\mathcal {R}} = -\frac{R_*}{4\pi \rho _*}\int _{B_1} \varrho + O\Big ({\mathcal {R}}^2 + \left( \int _{B_1} \varrho \right) ^2 \Big ). \end{aligned}$$
(C.8)
Substitution of (C.8) into (C.7) we obtain a leading expression entirely in terms of the perturbed density \( \varrho \). We list the various terms that we rewrite exclusively in terms of \( \varrho \):
$$\begin{aligned} 4\pi c_v T_\infty \rho _*R_* {\mathcal {R}}^2&= \frac{c_v T_\infty \rho _* R_*^3}{4\pi \rho _*^2} \left( \int _{B_1} \varrho \right) ^2 + O\left( |{\mathcal {R}}|^3 + \left( \int _{B_1} | \varrho |\right) ^3 \right) \\ 4\pi c_v T_\infty R_*^2 \varrho (1) {\mathcal {R}}&= -\frac{c_v T_\infty R_*^3}{\rho _*}\ \varrho (1) \int _{B_1} \varrho + O\left( |{\mathcal {R}}|^3 + | \varrho (1)|^3 + \left( \int _{B_1} | \varrho |\right) ^3 \right) \\ - 4\pi c_v\gamma \rho _* T_\infty R_* {\mathcal {R}}^2&= -\frac{c_v\gamma T_\infty R_*^3}{4\pi \rho _*} \left( \int _{B_1} \varrho \right) ^2 + O\left( |{\mathcal {R}}|^3 + \left( \int _{B_1} | \varrho |\right) ^3 \right) \\ 4\pi \left( \sigma + R_* p_\infty \right) {\mathcal {R}}^2&= \frac{1}{4\pi \rho _*^2} \left( \frac{\sigma }{R_*} + p_\infty \right) R_*^3\left( \int _{B_1} \varrho \right) ^2 + O\left( |{\mathcal {R}}|^3 + \left( \int _{B_1} | \varrho |\right) ^3 \right) . \end{aligned}$$
Inserting these expressions into (C.7), we obtain
$$\begin{aligned} \frac{1}{2}d^2{\mathcal {E}}[ \varrho , {\mathcal {R}}, \dot{{\mathcal {R}}}]&= \frac{c_v T_\infty M_0}{2\rho _*^2} \varrho (1)^2 + \frac{c_v\gamma T_\infty R_*^3}{2\rho _*} \int _{B_1} \varrho ^2 + 2\pi \rho _l R_*^3 \dot{{\mathcal {R}}}^2 \nonumber \\&\quad + \frac{R_*^3}{4\pi \rho _*^2}\left( c_v(1-\gamma )T_\infty \rho _* + \frac{\sigma }{R_*}+p_\infty \right) \left( \int _{B_1} \varrho \right) ^2 - \frac{c_v T_\infty R_*^3}{\rho _*} \varrho (1) \int _{B_1} \varrho \nonumber \\&\quad + O\left( |{\mathcal {R}}|^3 + | \varrho (1)|^3 +\left( \int _{B_1} | \varrho |\right) ^3 \right) \end{aligned}$$
(C.9)
The coefficient of the fourth term on the right of (C.9) can be simplified using the relation \(1-\gamma = -{\mathscr {R}}_g/c_v\) and the relation \({\mathscr {R}}_gT_\infty \rho _* = p_{\infty ,*} + 2\sigma /R_*\) between the equilibrium density and bubble radius:
$$\begin{aligned} c_v(1-\gamma )T_\infty \rho _* + \frac{\sigma }{R_*}+p_\infty = -{\mathscr {R}}_gT_\infty \rho _* + \frac{\sigma }{R_*}+p_\infty =-\frac{\sigma }{R_*} + {\mathcal {P}}_\infty , \end{aligned}$$
where \({\mathcal {P}}_\infty = p_\infty - p_{\infty ,*}\). Thus,
$$\begin{aligned} \frac{1}{2}d^2{\mathcal {E}}[ \varrho , {\mathcal {R}}, \dot{{\mathcal {R}}}]&= \frac{c_v T_\infty M_0}{2\rho _*^2} \varrho (1)^2 + \frac{c_v\gamma T_\infty R_*^3}{2\rho _*} \int _{B_1} \varrho ^2 + 2\pi \rho _l R_*^3 \dot{{\mathcal {R}}}^2 \nonumber \\&\quad {-} \frac{\sigma R_*^2}{4\pi \rho _*^2}\left( \int _{B_1} \varrho \right) ^2 {+} \frac{R_*^3}{4\pi \rho _*^2} {\mathcal {P}}_\infty \left( \int _{B_1} \varrho \right) ^2 {-} \frac{c_v T_\infty R_*^3}{\rho _*} \varrho (1) \int _{B_1} \varrho \nonumber \\&\quad + O\left( |{\mathcal {R}}|^3 + | \varrho (1)|^3 +\left( \int _{B_1} | \varrho |\right) ^3 \right) \end{aligned}$$
(C.10)
Using that \(M_0=\rho _* R_*^3 |B_1|\), we may rewrite (C.10) as
$$\begin{aligned}&\frac{1}{2}d^2{\mathcal {E}}[ \varrho , {\mathcal {R}}, \dot{{\mathcal {R}}}] \nonumber \\&\quad = \frac{c_v T_\infty R_*^3 |B_1|}{2} \left( \frac{ \varrho (1)^2}{\rho _*} - 2\frac{ \varrho (1)}{\rho _*} \frac{1}{|B_1|}\int _{B_1} \varrho \right) + \frac{c_v\gamma T_\infty R_*^3 }{2\rho _*} \int _{B_1} \varrho ^2 + 2\pi \rho _l R_*^3 \dot{{\mathcal {R}}}^2 \nonumber \\&\qquad - \frac{\sigma R_*^2}{4\pi \rho _*^2}\left( \int _{B_1} \varrho \right) ^2 + \frac{R_*^3}{4\pi \rho _*^2} {\mathcal {P}}_\infty \left( \int _{B_1} \varrho \right) ^2 + O\left( |{\mathcal {R}}|^3 + | \varrho (1)|^3 +\left( \int _{B_1} | \varrho |\right) ^3 \right) \end{aligned}$$
(C.11)
or
$$\begin{aligned} \frac{1}{2}d^2{\mathcal {E}}[ \varrho , {\mathcal {R}}, \dot{{\mathcal {R}}}]&= \frac{c_v T_\infty R_*^3 |B_1|}{2\rho _*} \left( \varrho (1)- \frac{1}{|B_1|}\int _{B_1} \varrho \right) ^2 + \frac{c_v\gamma T_\infty R_*^3}{2\rho _*} \int _{B_1} \varrho ^2 + 2\pi \rho _l R_*^3 \dot{{\mathcal {R}}}^2 \nonumber \\&\quad - \left[ \frac{\sigma R_*^2}{4\pi \rho _*^2} + \frac{c_vT_\infty R_*^3}{2 \rho _*|B_1|} \right] \left( \int _{B_1} \varrho \right) ^2 + \frac{R_*^3}{4\pi \rho _*^2} {\mathcal {P}}_\infty \left( \int _{B_1} \varrho \right) ^2 \nonumber \\&\quad + O\left( |{\mathcal {R}}|^3 + | \varrho (1)|^3 +\left( \int _{B_1} | \varrho |\right) ^3 \right) \end{aligned}$$
(C.12)
By the Cauchy-Schwarz inequality \(\left( \int _{B_1} \varrho \right) ^2\le |B_1| \int _{B_1} \varrho ^2 \) and therefore
$$\begin{aligned} \frac{1}{2}d^2{\mathcal {E}}[ \varrho , {\mathcal {R}}, \dot{{\mathcal {R}}}]&\ge \frac{c_v T_\infty R_*^3 |B_1|}{2\rho _*} \left( \varrho (1)- \frac{1}{|B_1|}\int _{B_1} \varrho \right) ^2 + 2\pi \rho _l R_*^3 \dot{{\mathcal {R}}}^2 \nonumber \\&\quad + \left( \frac{c_v\gamma T_\infty R_*^3}{2\rho _*}- \left[ \frac{\sigma R_*^2|B_1|}{4\pi \rho _*^2} + \frac{c_vT_\infty R_*^3}{2 \rho _*} \right] \right) \int _{B_1} \varrho ^2 + \frac{R_*^3}{4\pi \rho _*^2}{\mathcal {P}}_\infty \left( \int _{B_1} \varrho \right) ^2 \nonumber \\&\quad + O\left( |{\mathcal {R}}|^3 + | \varrho (1)|^3 +\left( \int _{B_1} | \varrho |\right) ^3 \right) \end{aligned}$$
(C.13)
Finally, we find for the constant in (C.13) that
$$\begin{aligned} \frac{c_v\gamma T_\infty R_*^3}{ 2\rho _*}- \left[ \frac{\sigma R_*^2|B_1|}{4\pi \rho _*^2} + \frac{c_vT_\infty R_*^3}{2 \rho _*} \right] = \frac{R_*^3}{ \rho _*^2} \left( \frac{p_{\infty ,*}}{2} + \frac{2\sigma }{3R_*}\right) . \end{aligned}$$
(C.14)
This follows, yet again, from the relations \(\gamma - 1= {\mathscr {R}}_g/c_v\) and \({\mathscr {R}}_gT_\infty \rho _* = p_{\infty ,*} + 2\sigma /R_*\). For the term involving \({\mathcal {P}}_\infty \), using the Cauchy-Schwarz inequality \(\left( \int _{B_1} \varrho \right) ^2\le |B_1| \int _{B_1} \varrho ^2 \),
$$\begin{aligned} {\mathcal {P}}_\infty \left( \int _{B_1} \varrho \right) ^2 \ge -|{\mathcal {P}}_\infty | |B_1| \int _{B_1} \varrho ^2. \end{aligned}$$
Summarizing
$$\begin{aligned} {\mathcal {E}}_{\textrm{total}} - {\mathcal {E}}_*&\ge \frac{c_v T_\infty R_*^3 |B_1|}{2\rho _*} \left( \varrho (1)- \frac{1}{|B_1|}\int _{B_1} \varrho \right) ^2\nonumber \\&\quad + 2\pi \rho _l R_*^3 \dot{{\mathcal {R}}}^2 + \frac{R_*^3}{\rho _*^2} \left( \frac{p_{\infty ,*}}{2} + \frac{2\sigma }{3R_*}\right) \int _{B_1} \varrho ^2 \nonumber \\&\quad - 4\pi R_*^2 |{\mathcal {P}}_\infty | |{\mathcal {R}}| - \frac{R_*^3}{4\pi \rho _*^2} |{\mathcal {P}}_\infty | |B_1| \int _{B_1} \varrho ^2\nonumber \\&\quad + O\left( |{\mathcal {R}}|^3 + | \varrho (1)|^3 +\left( \int _{B_1} | \varrho |\right) ^3 \right) , \end{aligned}$$
(C.15)
where all explicit terms are non-negative except for the terms involving \({\mathcal {P}}_\infty \) which can be made small since \(|{\mathcal {P}}_\infty | = |p_\infty - p_{\infty ,*}|\le \delta _0\).
We now conclude the proof by bounding the error term in (C.15) from above by a sufficiently small constant times \(\int _{B_1} \varrho ^2\).
For the fourth term on the right hand side of (C.15), using (C.8), in terms of the perturbed density \( \varrho \),
$$\begin{aligned} - 4\pi R_*^2 |{\mathcal {P}}_\infty | |{\mathcal {R}}| \ge - \frac{R_*^3}{\rho _*} |{\mathcal {P}}_\infty | \int _{B_1} | \varrho | - C_0|{\mathcal {P}}_\infty | \left( \int _{B_1} \varrho \right) ^2 \end{aligned}$$
for some constant \(C_0>0\). Since \(| \varrho |\le \delta _0\le 1\) and \(|{\mathcal {P}}_\infty | = |p_\infty - p_{\infty ,*}|\le \delta _0\), by the Cauchy-Schwarz inequality \(\left( \int _{B_1} \varrho \right) ^2\le |B_1| \int _{B_1} \varrho ^2 \)
$$\begin{aligned} - 4\pi R_*^2 |{\mathcal {P}}_\infty | |{\mathcal {R}}| \ge - \frac{R_*^3}{\rho _*} |{\mathcal {P}}_\infty | \int _{B_1} \varrho ^2 - C_0|{\mathcal {P}}_\infty | |B_1| \int _{B_1} \varrho ^2 \ge - C_1 \delta _0 \int _{B_1} \varrho ^2 \end{aligned}$$
for some constant \(C_1>0\).
Now we estimate the cubic term in the third line on the right hand side of (C.15). Since \(M_0 = \text {Mass}[\rho ,R]\),
$$\begin{aligned} \int _{B_R} (\rho - \rho _*)\, \textrm{d}x = \int _{B_R} \varrho \, \textrm{d}x = M_0 - \frac{4\pi R^3}{3} \rho _* = \frac{4\pi R_*^3}{3} \rho _* - \frac{4\pi R^3}{3} \rho _*, \end{aligned}$$
or
$$\begin{aligned} \begin{aligned} \frac{4\pi \rho _*}{3} (R^3 - R_*^3) = -\int _{B_R} (\rho -\rho _*)\, \textrm{d}x, \end{aligned}\end{aligned}$$
(C.16)
which implies
$$\begin{aligned} \begin{aligned} |{\mathcal {R}}|&= |R - R_*| \le \frac{3}{4\pi \rho _*(R^2+RR_*+R_*^2)} |B_R|^{\frac{1}{2}} \left( \int _{B_R} |\rho -\rho _*|^2\, \textrm{d}x \right) ^{\frac{1}{2}} \\&\le C_2 \left( \int _{B_R} |\rho -\rho _*|^2\, \textrm{d}x \right) ^{\frac{1}{2}}, \end{aligned}\end{aligned}$$
(C.17)
where \(C_2>0\) depends only on \(\nu , M_0, T_\infty \). We now control \(| \varrho (1,t)|^3\) by the first and the third terms on the right hand side of (C.15). Indeed,
$$\begin{aligned} \begin{aligned} | \varrho (1)|&= \left| \left( \varrho (1) - \frac{1}{|B_1|} \int _{B_1} \varrho \right) + \frac{1}{|B_1|} \int _{B_1} \varrho \, \right| \\&\le \left| \varrho (1) - \frac{1}{|B_1|} \int _{B_1} \varrho \, \right| + \frac{1}{|B_1|^{\frac{1}{2}}} \left( \int _{B_1} \varrho ^2 \right) ^{\frac{1}{2}}\\&\le C_3\left\{ \left| \varrho (1) - \frac{1}{|B_1|} \int _{B_1} \varrho \, \right| + \left( \int _{B_1} \varrho ^2 \right) ^{\frac{1}{2}} \right\} \end{aligned} \end{aligned}$$
for some \(C_3>0\) depending only on \(\nu , M_0, T_\infty \). Since \(| \varrho (1)| = |\rho (R) - \rho _*| \le \delta _0\),
$$\begin{aligned} \begin{aligned} | \varrho (1)|^3 = | \varrho (1)| |\varrho (1)|^2&\le \delta _0 C_3 \left\{ \left| \varrho (1) - \frac{1}{|B_1|} \int _{B_1} \varrho \, \right| + \left( \int _{B_1} \varrho ^2 \right) ^{\frac{1}{2}} \right\} ^2\\&\le 2 \delta _0 C_3 \left| \varrho (1) - \frac{1}{|B_1|} \int _{B_1} \varrho \, \right| ^2 + 2\delta _0 C_3 \int _{B_1} \varrho ^2. \end{aligned}\end{aligned}$$
(C.18)
Using (C.17) and (C.18), one has
$$\begin{aligned} \begin{aligned}&O\left( |{\mathcal {R}}|^3 + | \varrho (1)|^3 +\left( \int _{B_1} | \varrho |\right) ^3 \right) \\&\quad \ge - C \left( \int _{B_R} (\rho -\rho _*)^2 \right) ^{\frac{3}{2}} - C\delta _0 \left| \varrho (1) - \frac{1}{|B_1|} \int _{B_1} \varrho \, \right| ^2 - C\delta _0 \int _{B_1} \varrho ^2 - C \int _{B_1}| \varrho |^3 \end{aligned} \end{aligned}$$
for some \(C>0\) depending only on \(\nu , M_0, T_\infty \).
Consequently, using \(|{\mathcal {P}}_\infty | = |p_\infty - p_{\infty ,*}|\le \delta _0\), (C.15) can be further computed as
$$\begin{aligned} \begin{aligned} {\mathcal {E}}_{\textrm{total}} - {\mathcal {E}}_*&\ge - C_1 \delta _0 \int _{B_1} \varrho ^2 + \left( \frac{c_v T_\infty R_*^3 |B_1|}{2\rho _*} - C\delta _0 \right) \left( \varrho (1) - \frac{1}{|B_1|} \int _{B_1} \varrho \right) ^2 \\&\quad + \frac{R_*^3}{\rho _*}\left( \frac{p_{\infty ,*}}{2} + \frac{2\sigma }{3R_*} \right) \int _{B_1} \varrho ^2\\&\quad - C\delta _0 |B_R|^{\frac{1}{2}} \int _{B_R} (\rho -\rho _*)^2 \,\textrm{d}x - 2C\delta _0 \int _{B_1} \varrho ^2 - \delta _0 \frac{R_*^3}{4\pi \rho _*} |B_1| \int _{B_1} \varrho ^2\\&\ge \Theta \left( \int _{B_R} (\rho -\rho _*)^2 \right) \end{aligned} \end{aligned}$$
for some constant \(\Theta >0\), provided \(\delta _0>0\) is sufficiently small. Note that we’ve used \(\int _{B_1} \varrho ^2\, \textrm{d}y = R^{-3}\int _{B_R}(\rho - \rho _*)^2\, \textrm{d}x\) in which \(R^{-3}\ge \nu ^3\) above. This completes the proof of Theorem 7.5.
Appendix D: An Interpolation Lemma
Lemma D.1
Let \(\Omega \) be a bounded Lipschitz domain in \({\mathbb {R}}^n\), \(k< m\), and \(0<\gamma \le 1\). For \(u\in C^\infty (\Omega )\),
$$\begin{aligned} \left\| \nabla ^k u \right\| _{L^\infty (\Omega )} \le C_1 \left\| u \right\| _{L^p(\Omega )}^\lambda \left\| u \right\| _{C^{m,\gamma }(\Omega )}^{1-\lambda } + C_2 \left\| u \right\| _{L^s(\Omega )} \end{aligned}$$
for arbitrary \(s\ge 1\), where \(-\frac{k}{n} = \frac{\lambda }{p} - (1-\lambda ) \frac{m}{n}\), and the constants \(C_1\), \(C_2\) depend on the domain \(\Omega \) and on s in addition to the other parameters.
Proof
By Gagliardo–Nirenberg interpolation inequality,
$$\begin{aligned} \left\| \nabla ^k u \right\| _{L^\infty (\Omega )} \le C_1 \left\| u \right\| _{L^p(\Omega )}^{\lambda } \left\| \nabla ^m u \right\| _{L^\infty (\Omega )}^{1-\lambda } + C_2 \left\| u \right\| _{L^s(\Omega )} \end{aligned}$$
for arbitrary \(s\ge 1\), where
$$\begin{aligned} 0 = \frac{k}{n} - \frac{m}{n}(1-\lambda ) + \frac{\lambda }{p}, \end{aligned}$$
and the constants \(C_1\), \(C_2\) depend on the domain \(\Omega \) and on s in addition to the other parameters. The lemma then follows since \(\left\| \nabla ^m u \right\| _{L^\infty (\Omega )} \le \left\| u \right\| _{C^{m,\gamma }(\Omega )}\). \(\quad \square \)
Appendix E: Estimate of the Exponential Decay Rate \(\beta \) in the Linearized System
In this appendix, we prove parts (2) and (3) of Theorem 9.3. In particular, we investigate the location of the roots of the meromorphic function \(Q(\tau )\) defined in (9.20) which corresponds to the spectrum of the linear operator \({\mathcal {L}}\).
Lemma E.1
There exists a negative upper bound for the real parts of the roots of the meromorphic function \(Q(\tau )\) in (9.20). More precisely, there exists \(\beta >0\) such that \(\xi <-\beta \) for all roots \(\tau =\xi +i\eta \) of \(Q(\tau )\). The constant \(\beta \) can be chosen as
$$\begin{aligned} \begin{aligned} \beta&= \min \Bigg \{ \left( 1 - \sqrt{\dfrac{1-\frac{1}{\gamma }}{\frac{3p_{\infty ,*}R_*+6\sigma }{2p_{\infty ,*}R_*+6\sigma } - \frac{1}{\gamma }}} \right) \pi ^2\overline{\kappa },\, \sqrt{ \frac{{\mathscr {R}}_gT_\infty \rho _*}{\rho _lR_*^2} },\\&\quad \frac{2\mu _l}{\rho _lR_*^2} + {{\mathbbm {1}}}_{\Delta \le 0}\, \frac{{\mathscr {R}}_gT_\infty \rho _*}{\pi ^4\overline{\kappa }\rho _lR_*^2} \left( 1-\frac{1}{\gamma } \right) \left[ \frac{\pi ^4}{90} + O\left( \left( \frac{1}{\pi ^2\overline{\kappa }}\sqrt{ \frac{{\mathscr {R}}_gT_\infty \rho _*}{\rho _lR_*^2} } \right) ^{3/2} \right) \right] \\&\quad - {{\mathbbm {1}}}_{\Delta >0}\, \frac{\sqrt{\Delta }}{2\rho _lR_*} \Bigg \}, \end{aligned}\nonumber \\\end{aligned}$$
(E.1)
in which
$$\begin{aligned} \Delta := \left( \frac{4\mu _l}{R_*} \right) ^2 - 8\rho _l{\mathscr {R}}_gT_\infty \rho _*. \end{aligned}$$
Proof of Lemma E.1
Let \(\tau = \xi + i\eta \) be a root of \(Q(\tau )\), i.e., \(Q(\tau )=0\). Plugging \(\tau = \xi + i\eta \), \(\xi \in {\mathbb {R}},\eta \in {\mathbb {R}}\), into (9.20), we have
$$\begin{aligned} Q(\xi + i\eta ) = \frac{1}{{\mathscr {R}}_gT_\infty } \left( \** _1 + iH_1 \right) \left( \** _2 + iH_2 \right) + 4\pi \,\frac{\rho _*}{R_*}, \end{aligned}$$
where
$$\begin{aligned} \begin{aligned} \** _1&= \frac{4\pi }{3\gamma } + \frac{8(\gamma -1)}{\pi \gamma }\sum _{j=1}^\infty \frac{\pi ^2\overline{\kappa }\left( \pi ^2\overline{\kappa }j^2 + \xi \right) }{\left( \pi ^2\overline{\kappa }j^2 + \xi \right) ^2 + \eta ^2},\\ H_1&= - \frac{8(\gamma -1)}{\pi \gamma } \sum _{j=1}^\infty \frac{\pi ^2\overline{\kappa }\eta }{\left( \pi ^2\overline{\kappa }j^2 + \xi \right) ^2 + \eta ^2},\\ \** _2&= \rho _lR_*\left( \xi ^2 - \eta ^2 \right) + \frac{4\mu _l}{R_*}\, \xi - \frac{2\sigma }{R_*^2},\\ H_2&= \rho _lR_*(2\xi \eta ) + \frac{4\mu _l}{R_*}\, \eta . \end{aligned}\end{aligned}$$
(E.2)
Setting real and imaginary parts of Q equal to zero, we obtain
$$\begin{aligned} \begin{aligned} \text {real part: }&\frac{1}{{\mathscr {R}}_gT_\infty } \left( \** _1\** _2 - H_1H_2 \right) + 4\pi \, \frac{\rho _*}{R_*} = 0,\\ \text {imaginary part: }&\frac{1}{{\mathscr {R}}_gT_\infty } \left( \** _1H_2 + H_1\** _2 \right) = 0. \end{aligned}\end{aligned}$$
(E.3)
The real part in (E.3) reads
$$\begin{aligned} \begin{aligned} 0&= \frac{1}{{\mathscr {R}}_gT_\infty } \bigg [ \left( \frac{4\pi }{3\gamma } + \frac{8(\gamma -1)}{\pi \gamma }\sum _{j=1}^\infty \frac{\pi ^2\overline{\kappa }\left( \pi ^2\overline{\kappa }j^2 + \xi \right) }{\left( \pi ^2\overline{\kappa }j^2 + \xi \right) ^2 + \eta ^2} \right) \left( \rho _lR_*\left( \xi ^2-\eta ^2 \right) + \frac{4\mu _l}{R_*}\, \xi - \frac{2\sigma }{R_*^2} \right) \\&\quad + \frac{8(\gamma -1)}{\pi \gamma } \sum _{j=1}^\infty \frac{\pi ^2\overline{\kappa }\eta ^2}{\left( \pi ^2\overline{\kappa }j^2 + \xi \right) ^2 + \eta ^2 } \left( \rho _lR_*(2\xi ) + \frac{4\mu _l}{R_*} \right) \bigg ] + 4\pi \, \frac{\rho _*}{R_*}\\&= \frac{1}{{\mathscr {R}}_gT_\infty } \bigg [ \left( \frac{4\pi }{3\gamma } + \frac{8(\gamma -1)}{\pi \gamma }\sum _{j=1}^\infty \frac{\pi ^4\overline{\kappa }^2j^2}{\left( \pi ^2\overline{\kappa }j^2 + \xi \right) ^2 + \eta ^2} \right) \left( \rho _lR_*\left( \xi ^2-\eta ^2 \right) + \frac{4\mu _l}{R_*}\, \xi - \frac{2\sigma }{R_*^2} \right) \\&\quad + \frac{8(\gamma -1)}{\pi \gamma } \sum _{j=1}^\infty \frac{\pi ^2\overline{\kappa }}{\left( \pi ^2\overline{\kappa }j^2 + \xi \right) ^2 + \eta ^2 } \left( \rho _lR_*\xi (\xi ^2+\eta ^2) + \frac{4\mu _l}{R_*}(\xi ^2+\eta ^2) - \frac{2\sigma }{R_*^2}\, \xi \right) \bigg ] + 4\pi \, \frac{\rho _*}{R_*}. \end{aligned}\end{aligned}$$
(E.4)
When \(\eta \ne 0\), the imaginary part in (E.3) reads
$$\begin{aligned} 0&= \left( \frac{4\pi }{3\gamma } + \frac{8(\gamma -1)}{\pi \gamma } \sum _{j=1}^\infty \frac{\pi ^2\overline{\kappa }\left( \pi ^2\overline{\kappa }j^2 + \xi \right) }{\left( \pi ^2\overline{\kappa }j^2 + \xi \right) ^2 + \eta ^2} \right) \left( \rho _lR_*(2\xi ) + \frac{4\mu _l}{R_*} \right) \nonumber \\&\quad - \frac{8(\gamma -1)}{\pi \gamma } \sum _{j=1}^\infty \frac{\pi ^2\overline{\kappa }}{\left( \pi ^2\overline{\kappa }j^2 + \xi \right) ^2 + \eta ^2 } \left( \rho _lR_*(\xi ^2 - \eta ^2) + \frac{4\mu _l}{R_*}\, \xi - \frac{2\sigma }{R_*^2} \right) \nonumber \\&= \left( \frac{4\pi }{3\gamma } + \frac{8(\gamma -1)}{\pi \gamma } \sum _{j=1}^\infty \frac{\pi ^4\overline{\kappa }^2 j^2}{\left( \pi ^2\overline{\kappa }j^2 + \xi \right) ^2 + \eta ^2} \right) \left( \rho _lR_*(2\xi ) + \frac{4\mu _l}{R_*} \right) \nonumber \\&\quad + \frac{8(\gamma -1)}{\pi \gamma } \sum _{j=1}^\infty \frac{\pi ^2\overline{\kappa }}{\left( \pi ^2\overline{\kappa }j^2 + \xi \right) ^2 + \eta ^2 } \left( \rho _lR_*(\xi ^2 + \eta ^2) + \frac{2\sigma }{R_*^2} \right) . \end{aligned}$$
(E.5)
For \(\eta \ne 0\), the equation (E.5) implies
$$\begin{aligned} \begin{aligned} \rho _lR_*(2\xi ) + \frac{4\mu _l}{R_*} < 0, \end{aligned}\end{aligned}$$
(E.6)
which gives
$$\begin{aligned} \begin{aligned} \xi < - \frac{2\mu _l}{\rho _lR_*^2} \le 0,\qquad \eta \ne 0. \end{aligned}\end{aligned}$$
(E.7)
The equation (E.5) also implies
$$\begin{aligned} \begin{aligned}&\frac{8(\gamma -1)}{\pi \gamma } \sum _{j=1}^\infty \frac{\pi ^2\overline{\kappa }}{\left( \pi ^2\overline{\kappa }j^2 + \xi \right) ^2 + \eta ^2 }\\&\quad = - \left( \frac{4\pi }{3\gamma } + \frac{8(\gamma -1)}{\pi \gamma } \sum _{j=1}^\infty \frac{\pi ^4\overline{\kappa }^2 j^2}{\left( \pi ^2\overline{\kappa }j^2 + \xi \right) ^2 + \eta ^2} \right) \dfrac{\rho _lR_*(2\xi ) + \frac{4\mu _l}{R_*}}{\rho _lR_*(\xi ^2+\eta ^2) + \frac{2\sigma }{R_*^2}}. \end{aligned}\end{aligned}$$
(E.8)
Plugging (E.8) into the real part (E.4), we derive
$$\begin{aligned} \begin{aligned} 0&= \frac{1}{{\mathscr {R}}_gT_\infty } \left( \frac{4\pi }{3\gamma } + \frac{8(\gamma -1)}{\pi \gamma }\sum _{j=1}^\infty \frac{\pi ^4\overline{\kappa }^2j^2}{\left( \pi ^2\overline{\kappa }j^2 + \xi \right) ^2 + \eta ^2} \right) \Bigg [ \rho _lR_*\left( \xi ^2-\eta ^2 \right) + \frac{4\mu _l}{R_*}\, \xi - \frac{2\sigma }{R_*^2} \\&\quad - \dfrac{\left( \rho _lR_*(2\xi ) + \frac{4\mu _l}{R_*} \right) \left( \rho _lR_*\xi (\xi ^2+\eta ^2) + \frac{4\mu _l}{R_*}(\xi ^2+\eta ^2) - \frac{2\sigma }{R_*^2}\, \xi \right) }{\rho _lR_*(\xi ^2+\eta ^2) + \frac{2\sigma }{R_*^2}} \Bigg ] + 4\pi \, \frac{\rho _*}{R_*}\\&=: \frac{1}{{\mathscr {R}}_gT_\infty } \left( \frac{4\pi }{3\gamma } + \frac{8(\gamma -1)}{\pi \gamma }\sum _{j=1}^\infty \frac{\pi ^4\overline{\kappa }^2j^2}{\left( \pi ^2\overline{\kappa }j^2 + \xi \right) ^2 + \eta ^2} \right) \Lambda + 4\pi \, \frac{\rho _*}{R_*}, \end{aligned}\end{aligned}$$
(E.9)
where \(\Lambda \) is the square bracket on the right hand side of the first equation. A straightforward calculation shows that
$$\begin{aligned} \begin{aligned}&\left( \rho _lR_*(\xi ^2+\eta ^2) + \frac{2\sigma }{R_*^2} \right) \Lambda \\&\quad = -\rho _l^2R_*^2 (\xi ^2 + \eta ^2)^2 - \frac{4\mu _l}{R_*} \left( \rho _lR_*(2\xi ) + \frac{4\mu _l}{R_*} \right) (\xi ^2 + \eta ^2)\\&\qquad + \frac{2\sigma }{R_*^2} \left( 2\rho _lR_*\xi ^2 + 2\,\frac{4\mu _l}{R_*}\, \xi - 2\rho _lR_*\eta ^2 \right) - \left( \frac{2\sigma }{R_*^2} \right) ^2\\&\quad > -\rho _l^2R_*^2 (\xi ^2 + \eta ^2)^2 - \frac{2\sigma }{R_*^2}\, 2\rho _lR_*(\xi ^2+\eta ^2) - \left( \frac{2\sigma }{R_*^2} \right) ^2\\&\quad = -\left( \rho _lR_*(\xi ^2+\eta ^2) + \frac{2\sigma }{R_*^2} \right) ^2, \end{aligned} \end{aligned}$$
where we’ve used (E.6) and so \(2\rho _lR_*\xi ^2 + 2\,\frac{4\mu _l}{R_*}\, \xi = \xi \left( \rho _lR_*(2\xi ) + 2\,\frac{4\mu _l}{R_*} \right) > -2\rho _lR_*\xi ^2\) in the last inequality. This implies
$$\begin{aligned} \begin{aligned} \Lambda > -\left( \rho _lR_*(\xi ^2+\eta ^2) + \frac{2\sigma }{R_*^2} \right) . \end{aligned}\end{aligned}$$
(E.10)
Using (E.10) in (E.9), we get
$$\begin{aligned} \begin{aligned} 0&> - \frac{1}{{\mathscr {R}}_gT_\infty } \left( \frac{4\pi }{3\gamma } + \frac{8(\gamma -1)}{\pi \gamma }\sum _{j=1}^\infty \frac{\pi ^4\overline{\kappa }^2j^2}{\left( \pi ^2\overline{\kappa }j^2 + \xi \right) ^2 + \eta ^2} \right) \left( \rho _lR_*(\xi ^2+\eta ^2) + \frac{2\sigma }{R_*^2} \right) \\&\quad + 4\pi \, \frac{\rho _*}{R_*}. \end{aligned}\end{aligned}$$
(E.11)
Suppose
$$\begin{aligned} \begin{aligned} \xi \ge -\theta \pi ^2\overline{\kappa }, \end{aligned}\end{aligned}$$
(E.12)
where \(\theta \in (0,1)\), to be chosen. Then \(\xi \ge -\theta \pi ^2\overline{\kappa }j^2\) for all \(j=1,2,\ldots \). We further assume that
$$\begin{aligned} \begin{aligned} \xi \ge - \sqrt{\frac{2p_{\infty ,*}}{\rho _lR_*^2} + \frac{4\sigma }{\rho _lR_*^3} - \eta ^2}, \end{aligned}\end{aligned}$$
(E.13)
provided \(\eta ^2 \le \frac{2p_{\infty ,*}}{\rho _lR_*^2} + \frac{4\sigma }{\rho _lR_*^3}\), so that \(\rho _lR_*(\xi ^2+\eta ^2) + \frac{2\sigma }{R_*^2} \le \frac{2p_{\infty ,*}}{R_*} + \frac{6\sigma }{R_*^2}\). Then (E.11) gives
$$\begin{aligned} \begin{aligned} 0&> - \frac{1}{{\mathscr {R}}_gT_\infty } \left( \frac{4\pi }{3\gamma } + \frac{8(\gamma -1)}{\pi \gamma }\, \frac{1}{(1-\theta )^2} \sum _{j=1}^\infty \frac{1}{j^2} \right) \left( \frac{2p_{\infty ,*}}{R_*} + \frac{6\sigma }{R_*^2} \right) + 4\pi \, \frac{\rho _*}{R_*}. \end{aligned} \end{aligned}$$
Using \(\sum _{j=1}^\infty j^{-2} = \pi ^2/6\), one has
$$\begin{aligned} 3{\mathscr {R}}_gT_\infty \, \frac{\rho _*}{R_*} < \left[ \frac{1}{\gamma }+ \left( 1-\frac{1}{\gamma } \right) \frac{1}{(1-\theta )^2} \right] \left( \frac{2p_{\infty ,*}}{R_*} + \frac{6\sigma }{R_*^2} \right) , \end{aligned}$$
or, equivalently, using \({\mathscr {R}}_gT_\infty \rho _* = p_{\infty ,*} + 2\sigma /R_*\),
$$\begin{aligned} \theta > 1 - \sqrt{\dfrac{1-\frac{1}{\gamma }}{\frac{3p_{\infty ,*}R_*+6\sigma }{2p_{\infty ,*}R_*+6\sigma } - \frac{1}{\gamma }}}. \end{aligned}$$
We simply choose
$$\begin{aligned} \theta = 1 - \sqrt{\dfrac{1-\frac{1}{\gamma }}{\frac{3p_{\infty ,*}R_*+6\sigma }{2p_{\infty ,*}R_*+6\sigma } - \frac{1}{\gamma }}} \in (0,1) \end{aligned}$$
to reach a contradiction to (E.12) and (E.13). Therefore, we have for \(\eta \in {\mathbb {R}}\) with \(\eta \ne 0\) and \(\eta ^2\le \frac{2p_{\infty ,*}}{\rho _lR_*^2} + \frac{4\sigma }{\rho _lR_*^3}\) that
$$\begin{aligned} \xi < - \min \left\{ \left( 1 - \sqrt{\dfrac{1-\frac{1}{\gamma }}{\frac{3p_{\infty ,*}R_*+6\sigma }{2p_{\infty ,*}R_*+6\sigma } - \frac{1}{\gamma }}} \right) \pi ^2\overline{\kappa },\, \sqrt{\frac{2p_{\infty ,*}}{\rho _lR_*^2} + \frac{4\sigma }{\rho _lR_*^3} - \eta ^2} \right\} . \end{aligned}$$
Combining (E.7), for \(\eta \in {\mathbb {R}}\) with \(\eta \ne 0\) and \(\eta ^2\le \frac{p_{\infty ,*}}{\rho _lR_*^2} + \frac{2\sigma }{\rho _lR_*^3} \) we have
$$\begin{aligned} \begin{aligned} \xi < - \max \left\{ \frac{2\mu _l}{\rho _lR_*^2},\, \min \left\{ \left( 1 - \sqrt{\dfrac{1-\frac{1}{\gamma }}{\frac{3p_{\infty ,*}R_*+6\sigma }{2p_{\infty ,*}R_*+6\sigma } - \frac{1}{\gamma }}} \right) \pi ^2\overline{\kappa },\, \sqrt{ \frac{p_{\infty ,*}}{\rho _lR_*^2} + \frac{2\sigma }{\rho _lR_*^3} } \right\} \right\} . \end{aligned}\nonumber \\\end{aligned}$$
(E.14)
Now, we consider the case \(\eta ^2 > \frac{p_{\infty ,*}}{\rho _lR_*^2} + \frac{2\sigma }{\rho _lR_*^3} \). Since \(\eta \ne 0\), the imaginary part in (E.3) gives the identity \(\** _1 = - \frac{H_1}{H_2}\, \** _2\). Using this identity in the real part in (E.3), we derive
$$\begin{aligned} 4\pi \, \frac{\rho _*}{R_*}\, {\mathscr {R}}_gT_\infty H_2 = H_1\left( \** _2^2 + H_2^2 \right) , \end{aligned}$$
which implies
$$\begin{aligned} \begin{aligned}&4\pi \, \frac{\rho _*}{R_*}\, {\mathscr {R}}_gT_\infty \left( \rho _lR_*(2\xi ) + \frac{4\mu _l}{R_*} \right) \\&\quad = -\frac{8(\gamma -1)}{\pi \gamma } \sum _{j=1}^\infty \frac{\pi ^2\overline{\kappa }}{\left( \pi ^2\overline{\kappa }j^2 + \xi \right) ^2 + \eta ^2}\, \left( \** _2^2 + H_2^2 \right) . \end{aligned}\end{aligned}$$
(E.15)
To find a positive lower bound for \(\** _2^2 + H_2^2\), note that
$$\begin{aligned} \** _2^2 + H_2^2 = \left| \rho _lR_*\tau ^2 + \frac{4\mu _l}{R_*}\, \tau - \frac{2\sigma }{R_*^2} \right| ^2 = \rho _l^2R_*^2 \left| \tau - \tau _+ \right| ^2 \left| \tau - \tau _- \right| ^2, \end{aligned}$$
where
$$\begin{aligned} \tau _{\pm } = \dfrac{-\dfrac{4\mu _l}{R_*} \pm \sqrt{\left( \dfrac{4\mu _l}{R_*} \right) ^2 + 4\rho _lR_*\, \dfrac{2\sigma }{R_*^2}}}{2\rho _lR_*} \end{aligned}$$
are on the real axis. By the triangular inequality \(|\tau - \tau _{\pm }| > |\eta |\), and so
$$\begin{aligned} \** _2^2 + H_2^2 > \rho _l^2 R_*^2 \eta ^4. \end{aligned}$$
Therefore, (E.15) yields
$$\begin{aligned} 4\pi \, \frac{\rho _*}{R_*}\, {\mathscr {R}}_gT_\infty \left( \rho _lR_*(2\xi ) + \frac{4\mu _l}{R_*} \right) < -\frac{8(\gamma -1)}{\pi \gamma } \sum _{j=1}^\infty \frac{\pi ^2\overline{\kappa }}{\pi ^4\overline{\kappa }^2 j^4 + \eta ^2}\, \rho _l^2R_*^2\eta ^4. \end{aligned}$$
Since \(\eta ^2 > \frac{p_{\infty ,*}}{\rho _lR_*^2} + \frac{2\sigma }{\rho _lR_*^3}\) and \(\frac{a^2}{\pi ^4\overline{\kappa }^2 j^4 + a}\) is increasing in a for \(a = \eta ^2 \ge 0\), we further derive
$$\begin{aligned} \begin{aligned}&4\pi \, \frac{\rho _*}{R_*}\, {\mathscr {R}}_gT_\infty \left( \rho _lR_*(2\xi ) + \frac{4\mu _l}{R_*} \right) \\&\quad < -\frac{8(\gamma -1)}{\pi \gamma } \sum _{j=1}^\infty \frac{\pi ^2\overline{\kappa }}{\pi ^4\overline{\kappa }^2 j^4 + \left( \frac{p_{\infty ,*}}{\rho _lR_*^2} + \frac{2\sigma }{\rho _lR_*^3} \right) }\, \rho _l^2R_*^2 \left( \frac{p_{\infty ,*}}{\rho _lR_*^2} + \frac{2\sigma }{\rho _lR_*^3} \right) ^2\\&\quad = -\frac{8(\gamma -1)}{\pi \gamma }\, \frac{1}{\pi ^2\overline{\kappa }} \sum _{j=1}^\infty \frac{1}{j^4 + B^2}\, \rho _l^2R_*^2 \left( \frac{p_{\infty ,*}}{\rho _lR_*^2} + \frac{2\sigma }{\rho _lR_*^3} \right) ^2, \end{aligned}\end{aligned}$$
(E.16)
where
$$\begin{aligned} B:= \frac{1}{\pi ^2\overline{\kappa }}\sqrt{ \frac{p_{\infty ,*}}{\rho _lR_*^2} + \frac{2\sigma }{\rho _lR_*^3}}. \end{aligned}$$
Using
$$\begin{aligned} \begin{aligned} \sum _{j=1}^\infty \frac{1}{j^4 + B^2}&= \frac{e^{\frac{i\pi }{4}}\pi \cot \left( e^{\frac{i\pi }{4}}\pi \sqrt{B} \right) + e^{\frac{3i\pi }{4}}\pi \cot \left( e^{\frac{3i\pi }{4}}\pi \sqrt{B} \right) }{4B^{3/2}} - \frac{1}{2B^2}\\&= \dfrac{\frac{2\pi }{\sqrt{2}} \left( \frac{\cot \left( \pi \sqrt{B/2} \right) {{\,\textrm{csch}\,}}^2\left( \pi \sqrt{B/2} \right) + \csc ^2\left( \pi \sqrt{B/2} \right) \coth \left( \pi \sqrt{B/2} \right) }{\cot ^2\left( \pi \sqrt{B/2} \right) + \coth ^2\left( \pi \sqrt{B/2} \right) } \right) }{4B^{3/2}} - \frac{1}{2B^2}\\&= \dfrac{\frac{2\pi }{\sqrt{2}} \left( \frac{1}{\pi \sqrt{B/2}} + \frac{4}{45} \left( \pi \sqrt{B/2} \right) ^3 + O\left( \left( \pi \sqrt{B/2} \right) ^6 \right) \right) }{4B^{\frac{3}{2}}} - \frac{1}{2B^2}\\&= \frac{\pi ^4}{90} + O\left( B^{3/2} \right) \ \text { for }B\ll 1, \end{aligned} \end{aligned}$$
we have from (E.16) that
$$\begin{aligned} \begin{aligned}&4\pi \, \frac{\rho _*}{R_*}\, {\mathscr {R}}_gT_\infty \left( \rho _lR_*(2\xi ) + \frac{4\mu _l}{R_*} \right) \\&\quad < -\frac{8(\gamma -1)}{\pi \gamma }\, \frac{1}{\pi ^2\overline{\kappa }} \left( \frac{\pi ^4}{90} + O\left( \left( \frac{1}{\pi ^2\overline{\kappa }}\sqrt{ \frac{p_{\infty ,*}}{\rho _lR_*^2} + \frac{2\sigma }{\rho _lR_*^3} } \right) ^{3/2} \right) \right) \rho _l^2R_*^2 \left( \frac{p_{\infty ,*}}{\rho _lR_*^2} + \frac{2\sigma }{\rho _lR_*^3} \right) ^2. \end{aligned} \end{aligned}$$
Consequently, we have for \(\eta ^2 > \frac{p_{\infty ,*}}{\rho _lR_*^2} + \frac{2\sigma }{\rho _lR_*^3} \) that
$$\begin{aligned} \begin{aligned} \xi < -\frac{2\mu _l}{\rho _lR_*^2} - \frac{{\mathscr {R}}_gT_\infty \rho _*}{\pi ^4\overline{\kappa }\rho _lR_*^2} \left( 1-\frac{1}{\gamma } \right) \left( \frac{\pi ^4}{90} + O\left( \left( \frac{1}{\pi ^2\overline{\kappa }}\sqrt{\frac{p_{\infty ,*}}{\rho _lR_*^2} + \frac{2\sigma }{\rho _lR_*^3}} \right) ^{3/2} \right) \right) , \end{aligned}\nonumber \\\end{aligned}$$
(E.17)
where \({\mathscr {R}}_gT_\infty \rho _* = p_{\infty ,*} + 2\sigma /R_*\) is used.
It remains to consider the case \(\eta =0\). We first show that \(\xi <0\). Suppose for the sake of contradiction that \(\xi \ge 0\) then
$$\begin{aligned} \begin{aligned} Q(\xi )&= \frac{1}{{\mathscr {R}}_gT_\infty } \left( \frac{4\pi }{3\gamma } + \frac{8(\gamma -1)}{\pi \gamma } \sum _{j=1}^\infty \frac{\pi ^2\overline{\kappa }}{\pi ^2\overline{\kappa }j^2 + \xi } \right) \left( \rho _lR_*\xi ^2 + \frac{4\mu _l}{R_*}\, \xi \right) \\&\quad - \frac{2\sigma }{{\mathscr {R}}_gT_\infty R_*^2} \left( \frac{4\pi }{3\gamma } + \frac{8(\gamma -1)}{\pi \gamma } \sum _{j=1}^\infty \frac{\pi ^2\overline{\kappa }}{\pi ^2\overline{\kappa }j^2 + \xi } \right) + 4\pi \,\frac{\rho _*}{R_*}, \end{aligned} \end{aligned}$$
where the second line is greater than
$$\begin{aligned} - \frac{2\sigma }{{\mathscr {R}}_gT_\infty R_*^2} \left( \frac{4\pi }{3\gamma } + \frac{8(\gamma -1)}{\pi \gamma } \sum _{j=1}^\infty \frac{\pi ^2\overline{\kappa }}{\pi ^2\overline{\kappa }j^2} \right) + 4\pi \,\frac{\rho _*}{R_*} = \frac{8\pi \rho _*}{3R_*} + \frac{4\pi p_{\infty ,*}}{3{\mathscr {R}}_gT_\infty R_*} > 0. \end{aligned}$$
Here the identities \(\sum _{j=1}^\infty j^{-2} = \frac{\pi ^2}{6}\) and \({\mathscr {R}}_gT_\infty \rho _* = p_{\infty ,*} + \frac{2\sigma }{R_*}\) are used. This yields \(Q(\xi )>0\), which contradicts to the fact that \(Q(\tau )=0\). Thus, we have \(\xi <0\) for \(\eta =0\).
Now we search for a negative upper bound for \(\xi \) when \(\eta =0\). Suppose
$$\begin{aligned} \begin{aligned} \xi \ge -\theta _0\pi ^2\overline{\kappa }, \end{aligned}\end{aligned}$$
(E.18)
where \(0<\theta _0<1\) to be chosen. Suppose further that
$$\begin{aligned} \begin{aligned} \xi> \frac{-\frac{4\mu _l}{R_*} + \sqrt{\Delta }}{2\rho _lR_*}\quad \text { if }\Delta := \left( \frac{4\mu _l}{R_*} \right) ^2 - 4\rho _lR_* \left( \frac{2p_{\infty ,*}}{R_*} + \frac{4\sigma }{R_*^2} \right) >0 \end{aligned}\end{aligned}$$
(E.19)
such that
$$\begin{aligned} \begin{aligned} \rho _lR_*\xi ^2 + \frac{4\mu _l}{R_*}\, \xi - \frac{2\sigma }{R_*^2} \ge -\frac{2p_{\infty ,*}}{R_*} - \frac{6\sigma }{R_*^2}. \end{aligned}\end{aligned}$$
(E.20)
Note that the inequality (E.20) always holds when \(\Delta \le 0\). Then
$$\begin{aligned} \begin{aligned} 0 = Q(\xi )&= \frac{1}{{\mathscr {R}}_gT_\infty } \left( \frac{4\pi }{3\gamma } + \frac{8(\gamma -1)}{\pi \gamma } \sum _{j=1}^\infty \frac{\pi ^2\overline{\kappa }}{\pi ^2\overline{\kappa }j^2 + \xi } \right) \left( \rho _lR_*\xi ^2 + \frac{4\mu _l}{R_*}\, \xi - \frac{2\sigma }{R_*^2} \right) \\&\quad + 4\pi \,\frac{\rho _*}{R_*}\\&> - \frac{1}{{\mathscr {R}}_gT_\infty } \left( \frac{4\pi }{3\gamma } + \frac{8(\gamma -1)}{\pi \gamma (1-\theta _0)} \sum _{j=1}^\infty \frac{\pi ^2\overline{\kappa }}{\pi ^2\overline{\kappa }j^2} \right) \left( \frac{2p_{\infty ,*}}{R_*} + \frac{6\sigma }{R_*^2} \right) + 4\pi \,\frac{\rho _*}{R_*}\\&= - \frac{1}{{\mathscr {R}}_gT_\infty } \left( \frac{4\pi }{3\gamma } + \frac{4\pi }{3} \left( 1-\frac{1}{\gamma } \right) \frac{1}{1-\theta _0} \right) \left( \frac{2p_{\infty ,*}}{R_*} + \frac{6\sigma }{R_*^2} \right) + 4\pi \,\frac{\rho _*}{R_*} \end{aligned} \end{aligned}$$
So
$$\begin{aligned} 4\pi {\mathscr {R}}_gT_\infty \, \frac{\rho _*}{R_*} < \left( \frac{4\pi }{3\gamma } + \frac{4\pi }{3} \left( 1-\frac{1}{\gamma } \right) \frac{1}{1-\theta _0} \right) \left( \frac{2p_{\infty ,*}}{R_*} + \frac{6\sigma }{R_*^2} \right) , \end{aligned}$$
or, equivalently,
$$\begin{aligned} \theta _0 > 1 - \dfrac{1-\dfrac{1}{\gamma }}{\dfrac{3{\mathscr {R}}_gT_\infty \, \frac{\rho _*}{R_*}}{\frac{2p_{\infty ,*}}{R_*} + \frac{6\sigma }{R_*^2}} - \dfrac{1}{\gamma } } = 1 - \dfrac{1-\dfrac{1}{\gamma }}{ \frac{3p_{\infty ,*}R_*+6\sigma }{2p_{\infty ,*}R_*+6\sigma } - \dfrac{1}{\gamma } }, \end{aligned}$$
where \({\mathscr {R}}_gT_\infty \rho _* = p_{\infty ,*} + \frac{2\sigma }{R_*}\) has been used in the last equation. We then choose
$$\begin{aligned} \theta _0 = 1 - \dfrac{1-\dfrac{1}{\gamma }}{ \dfrac{3p_{\infty ,*}R_*+6\sigma }{2p_{\infty ,*}R_*+6\sigma } - \dfrac{1}{\gamma } } \in (0,1) \end{aligned}$$
to reach a contradiction to (E.18) and (E.19). Hence we derive for \(\eta =0\)
$$\begin{aligned} \begin{aligned} \xi < {\left\{ \begin{array}{ll} - \min \left\{ \left( 1 - \dfrac{1-\dfrac{1}{\gamma }}{ \dfrac{3p_{\infty ,*}R_*+6\sigma }{2p_{\infty ,*}R_*+6\sigma } - \dfrac{1}{\gamma } } \right) \pi ^2\overline{\kappa },\, \dfrac{ \dfrac{4\mu _l}{R_*} - \sqrt{\Delta }}{2\rho _lR_*} \right\} &{}\ \text { if }\Delta >0,\\ - \left( 1 - \dfrac{1-\dfrac{1}{\gamma }}{ \dfrac{3p_{\infty ,*}R_*+6\sigma }{2p_{\infty ,*}R_*+6\sigma } - \dfrac{1}{\gamma } } \right) \pi ^2\overline{\kappa }&{}\ \text { if }\Delta \le 0. \end{array}\right. } \end{aligned}\end{aligned}$$
(E.21)
Combining the upper bounds (E.14), (E.17), and (E.21) for different cases of \(\eta \) and using the identity \({\mathscr {R}}_gT_\infty \rho _* = p_{\infty ,*} + \frac{2\sigma }{R_*}\), the lemma follows. \(\quad \square \)
Appendix F: Rate of Convergence of Slow Solutions Approaching to Center Manifold for a Class of Fully Nonlinear Autonomous Systems
As mentioned in the paragraph below Theorem 6.7, there are several obstacles preventing us from direct applying center manifold theorem to prove the exponential decay in nonlinear bubble oscillations. One of which is the quasilinear character of the problem (9.5). For this purpose, we develop in this appendix a geometric theory for a class of fully nonlinear autonomous systems which covers the quasilinear system (9.5).
We study a larger class of fully nonlinear autonomous systems of the form \(\dot{\textbf{w}} = {\mathcal {L}}{\textbf{w}} + {\mathcal {N}}({\textbf{w}}, \dot{\textbf{w}})\) that includes the quasilinear autonomous system (9.5) of our concern. Assuming that the solution of such equation converges toward a given center manifold and that the time derivative of the solution is sufficiently small for all time, we prove that the convergence rate is exponential. The proof is an adaptation of [10, Sections 2.4 and 6.3], where the existence and stability of center manifold for semilinear equations are established.
Setup of the fully nonlinear autonomous system, assumptions on the solution, and the center manifold. Let Z be a Banach space with norm \(\left\| \,\cdot \, \right\| \). We consider the evolution equation
$$\begin{aligned} \begin{aligned} \dot{\textbf{w}} = {\mathcal {L}}{\textbf{w}} + {\mathcal {N}}({\textbf{w}}, \dot{\textbf{w}}),\quad {\textbf{w}}(0) \in Z, \end{aligned}\end{aligned}$$
(F.1)
where \({\mathcal {N}}({\textbf{w}},\textbf{p}):Z\times Z\rightarrow Z\) has a uniformly continuous second derivative with \({\mathcal {N}}({\textbf{0}},{\textbf{p}}) = {\textbf{0}}\) and \(\partial _{({\textbf{w}}, {\textbf{p}})}{\mathcal {N}}({\textbf{0}}, {\textbf{0}}) = {\textbf{O}}\).
Assume
(i) \(Z = X\,\oplus \, Y\) where X is finite dimensional and Y is closed.
(ii) X is \({\mathcal {L}}\)-invariant and that if \({\mathcal {A}}:= {\mathcal {L}}|_X\), then the real parts of the eigenvalues of \({\mathcal {A}}\) are all zeros.
(iii) Y is \(e^{{\mathcal {L}} t}\)-invariant. Let \(Q_1\) be a projection on X (not necessarily along Y) and \(Q_2:= I - Q_1\). For some positive constants b, c,
$$\begin{aligned} \begin{aligned} \left\| e^{{\mathcal {L}} t} Q_2 \right\| \le ce^{-bt},\quad t\ge 0. \end{aligned}\end{aligned}$$
(F.2)
Let \({\textbf{w}}\) be a solution of (F.1). Decompose \({\textbf{w}}\) into \({\textbf{w}} = {\textbf{x}} + {\textbf{y}}\) where \(\textbf{x} = Q_1{\textbf{w}}\) and \({\textbf{y}} = Q_2{\textbf{w}}\). Let \({\mathcal {B}} = Q_2{\mathcal {L}}\). Then equation (9.12) can be written as
$$\begin{aligned} \begin{aligned} \dot{\textbf{x}}&= {\mathcal {A}}{\textbf{x}} + f({\textbf{x}}, {\textbf{y}}, \dot{\textbf{x}}, \dot{\textbf{y}}),\\ \dot{\textbf{y}}&= {\mathcal {B}}{\textbf{y}} + g({\textbf{x}}, {\textbf{y}}, \dot{\textbf{x}}, \dot{\textbf{y}}), \end{aligned}\end{aligned}$$
(F.3)
where
$$\begin{aligned} f({\textbf{x}}, {\textbf{y}}, \dot{\textbf{x}}, \dot{\textbf{y}}) = Q_1{\mathcal {N}}({\textbf{x}}+{\textbf{y}}, \dot{\textbf{x}}+\dot{\textbf{y}}),\quad g({\textbf{x}}, {\textbf{y}}, \dot{\textbf{x}}, \dot{\textbf{y}}) = Q_2{\mathcal {N}}({\textbf{x}}+{\textbf{y}}, \dot{\textbf{x}}+\dot{\textbf{y}}). \end{aligned}$$
A curve \({\textbf{y}} = h({\textbf{x}})\), defined for \(|{\textbf{x}}|\) small, is said to be an invariant manifold for (F.3) if the solution \(({\textbf{x}}(t),\textbf{y}(t))\) of (F.3) through \((\textbf{x}(0),h({\textbf{x}}(0)))\) satisfies \({\textbf{y}}(t) = h({\textbf{x}}(t))\). A center manifold for (F.3) is an invariant manifold that is tangent to X space at the origin.
By the assumption on the nonlinearity \({\mathcal {N}}(\textbf{w},\dot{\textbf{w}})\), there exists a continuous function \(k(\varepsilon )\) with \(k(0)=0\) such that
$$\begin{aligned} \begin{aligned} \left\| f({\textbf{x}}, {\textbf{y}}, \dot{\textbf{x}}, \dot{\textbf{y}}) \right\| + \left\| g({\textbf{x}}, {\textbf{y}}, \dot{\textbf{x}}, \dot{\textbf{y}}) \right\|&\le \varepsilon k(\varepsilon ),\\ \left\| f({\textbf{x}}, {\textbf{y}}, \dot{\textbf{x}}, \dot{\textbf{y}}) - f({\textbf{x}}', {\textbf{y}}', \dot{\textbf{x}}', \dot{\textbf{y}}') \right\|&\le k(\varepsilon ) \left[ \left\| \textbf{x} - {\textbf{x}}' \right\| + \left\| {\textbf{y}} - {\textbf{y}}' \right\| + \left\| \dot{\textbf{x}} - \dot{\textbf{x}}' \right\| + \left\| \dot{\textbf{y}} - \dot{\textbf{y}}' \right\| \right] ,\\ \left\| g({\textbf{x}}, {\textbf{y}}, \dot{\textbf{x}}, \dot{\textbf{y}}) - g({\textbf{x}}', {\textbf{y}}', \dot{\textbf{x}}', \dot{\textbf{y}}') \right\|&\le k(\varepsilon ) \left[ \left\| \textbf{x} - {\textbf{x}}' \right\| + \left\| {\textbf{y}} - {\textbf{y}}' \right\| + \left\| \dot{\textbf{x}} - \dot{\textbf{x}}' \right\| + \left\| \dot{\textbf{y}} - \dot{\textbf{y}}' \right\| \right] , \end{aligned}\nonumber \\\end{aligned}$$
(F.4)
for all \(\textbf{x}, {\textbf{x}}'\in X\), \({\textbf{y}}, {\textbf{y}}'\in Y\) and all \(\dot{\textbf{x}}, \dot{\textbf{x}}', \dot{\textbf{y}}, \dot{\textbf{y}}' \in Z\) with \(\left\| ({\textbf{x}}, {\textbf{y}}) \right\| , \left\| (\dot{\textbf{x}}, \dot{\textbf{y}}) \right\| , \left\| ({\textbf{x}}',\textbf{y}') \right\| , \left\| (\dot{\textbf{x}}', \dot{\textbf{y}}') \right\| < \varepsilon \).
Let \({\mathcal {M}}\) be a center manifold for (F.3) given by \({\textbf{y}} = h({\textbf{x}})\). If we substitute \({\textbf{y}}(t)=h({\textbf{x}}(t))\) into (F.3) we obtain
$$\begin{aligned} \begin{aligned} h'({\textbf{x}}) \left[ {\mathcal {A}}{\textbf{x}} + f({\textbf{x}}, h({\textbf{x}}), \dot{\textbf{x}}, h'({\textbf{x}})\dot{\textbf{x}}) \right] = \mathcal Bh({\textbf{x}}) + g({\textbf{x}}, h({\mathbf{x)}}, \dot{\textbf{x}}, h'({\textbf{x}})\dot{\textbf{x}}). \end{aligned}\end{aligned}$$
(F.5)
The equation on the center manifold is given by
$$\begin{aligned} \begin{aligned} \dot{\textbf{u}} = {\mathcal {A}}{\textbf{u}} + f({\textbf{u}}, h({\textbf{u}}), \dot{\textbf{u}}, h'({\textbf{u}})\dot{\textbf{u}}). \end{aligned}\end{aligned}$$
(F.6)
We assume that \({\textbf{w}}(t)\) converges to some element in \({\mathcal {M}}\), as \(t\rightarrow \infty \), and that \(\sup _{t\ge 0} \left\| \dot{\textbf{w}}(t) \right\| \) is sufficiently small.
Rate of convergence to the center manifold. The following lemma describes that the trajectory shadows the center manifold and corresponds to [10, Lemma 2.4.1].
Lemma F.1
Let \(({\textbf{x}}(t), {\textbf{y}}(t))\) be a solution of (F.3) with \(\left\| ({\textbf{x}}(0), \textbf{y}(0)) \right\| \) and \(\left\| (\dot{\textbf{x}}(t), \dot{\textbf{y}}(t)) \right\| \), for all \(t\ge 0\), sufficiently small. Then there exist positive \(C_1\) and \(\beta _1\) such that
$$\begin{aligned} \left\| {\textbf{y}}(t) - h({\textbf{x}}(t)) \right\| \le C_1 e^{-\beta _1 t} \left\| \textbf{y}(0) - h({\textbf{x}}(0)) \right\| \end{aligned}$$
for all \(t\ge 0\).
Proof
Let \(({\textbf{x}}(t),{\textbf{y}}(t))\) be a solution of (F.3) with \(({\textbf{x}}(0),{\textbf{y}}(0))\) sufficiently small. Let \({\textbf{z}}(t) = {\textbf{y}}(t) - h({\textbf{x}}(t))\), then
$$\begin{aligned} \begin{aligned} \dot{\textbf{z}} = {\mathcal {B}}{\textbf{z}} + {\mathcal {R}}({\textbf{x}},\textbf{z},\dot{\textbf{x}},\dot{\textbf{z}}) \end{aligned}\end{aligned}$$
(F.7)
where
$$\begin{aligned} \begin{aligned} {\mathcal {R}}({\textbf{x}},{\textbf{z}},\dot{\textbf{x}},\dot{\textbf{z}})&= h'(\textbf{x}) \left[ f({\textbf{x}}, h({\textbf{x}}), \dot{\textbf{x}}, h'({\textbf{x}})\dot{\textbf{x}}) - f({\textbf{x}},{\textbf{z}}+h({\textbf{x}}), \dot{\textbf{x}}, \dot{\textbf{z}} + h'(\textbf{x})\dot{\textbf{x}}) \right] \\&\quad + g({\textbf{x}},{\textbf{z}}+h({\textbf{x}}), \dot{\textbf{x}}, \dot{\textbf{z}} + h'({\textbf{x}})\dot{\textbf{x}}) - g({\textbf{x}}, h({\textbf{x}}), \dot{\textbf{x}}, h'({\textbf{x}})\dot{\textbf{x}}). \end{aligned}\end{aligned}$$
(F.8)
Using the hypotheses of f and g and the bounds on h,
$$\begin{aligned} \begin{aligned} \left\| {\mathcal {R}}({\textbf{x}},{\textbf{z}},\dot{\textbf{x}},\dot{\textbf{z}}) \right\|&\le \left\| h'({\textbf{x}}) \right\| \Big ( \left\| f({\textbf{x}}, h({\textbf{x}}), \dot{\textbf{x}}, h'({\textbf{x}})\dot{\textbf{x}}) - f({\textbf{x}},{\textbf{z}}+h({\textbf{x}}), \dot{\textbf{x}}, h'({\textbf{x}})\dot{\textbf{x}}) \right\| \\&\quad + \left\| f({\textbf{x}}, {\textbf{z}} + h({\textbf{x}}), \dot{\textbf{x}}, h'({\textbf{x}})\dot{\textbf{x}}) - f({\textbf{x}},{\textbf{z}}+h({\textbf{x}}), \dot{\textbf{x}}, \dot{\textbf{z}} + h'({\textbf{x}})\dot{\textbf{x}}) \right\| \Big )\\&\quad + \left\| g({\textbf{x}},{\textbf{z}}+h({\textbf{x}}), \dot{\textbf{x}}, \dot{\textbf{z}} + h'({\textbf{x}})\dot{\textbf{x}}) - g({\textbf{x}}, h({\textbf{x}}), \dot{\textbf{x}}, \dot{\textbf{z}} + h'({\textbf{x}})\dot{\textbf{x}}) \right\| \\&\quad + \left\| g({\textbf{x}}, h({\textbf{x}}), \dot{\textbf{x}}, \dot{\textbf{z}} + h'({\textbf{x}})\dot{\textbf{x}}) - g({\textbf{x}}, h({\textbf{x}}), \dot{\textbf{x}}, h'({\textbf{x}})\dot{\textbf{x}}) \right\| \\&\le \delta (\varepsilon ) \left[ \left\| \textbf{z} \right\| + \left\| \dot{\textbf{z}} \right\| \right] , \end{aligned}\end{aligned}$$
(F.9)
if \(\left\| \textbf{z} \right\| , \left\| \dot{\textbf{z}} \right\| < \varepsilon \), for some continuous function \(\delta (\varepsilon )\) with \(\delta (0)=0\). Using (F.2) we obtain, from (F.7),
$$\begin{aligned} \begin{aligned} \left\| {\textbf{z}}(t) \right\| \le ce^{-bt} \left\| {\textbf{z}}(0) \right\| + c\delta (\varepsilon ) \int _0^t e^{-b(t-s)} \left[ \left\| \mathbf{z(s)} \right\| + \left\| \dot{\textbf{z}}(s) \right\| \right] \textrm{d}s. \end{aligned}\end{aligned}$$
(F.10)
Using (F.9) in (F.7) one has
$$\begin{aligned} \left\| \dot{\textbf{z}} \right\| \le \left\| {\mathcal {B}} \right\| \left\| \textbf{z} \right\| + \left\| {\mathcal {R}}({\textbf{x}},{\textbf{z}},\dot{\textbf{x}},\dot{\textbf{z}}) \right\| \le \left\| {\mathcal {B}} \right\| \left\| \textbf{z} \right\| + \delta (\varepsilon )\left[ \left\| \textbf{z} \right\| + \left\| \dot{\textbf{z}} \right\| \right] , \end{aligned}$$
and so
$$\begin{aligned} \begin{aligned} \left\| \dot{\textbf{z}} \right\| \le C_0 \left\| \textbf{z} \right\| ,\qquad C_0 = \left( 1-\delta (\varepsilon ) \right) ^{-1} \left( \left\| {\mathcal {B}} \right\| + \delta (\varepsilon ) \right) . \end{aligned}\end{aligned}$$
(F.11)
Therefore, (F.10) yields
$$\begin{aligned} \begin{aligned} e^{bt} \left\| {\textbf{z}}(t) \right\|&\le c \left\| {\textbf{z}}(0) \right\| + c\delta (\varepsilon ) \left( 1 + C_0 \right) \int _0^t e^{bs} \left\| \mathbf{z(s)} \right\| \textrm{d}s \end{aligned}\end{aligned}$$
(F.12)
By Gronwall’s lemma,
$$\begin{aligned} e^{bt} \left\| {\textbf{z}}(t) \right\| \le c \left\| {\textbf{z}}(0) \right\| e^{c\delta (\varepsilon ) \left( 1 + C_0 \right) t }. \end{aligned}$$
The lemma follows. \(\square \)
Proposition F.2
Suppose that the zero solution of (F.6) is Lyapunov stable. Let \(({\textbf{x}}(t), {\textbf{y}}(t))\) be a solution of (F.3). There exists \(\varepsilon >0\) such that if \(\left\| ({\textbf{x}}(0), {\textbf{y}}(0)) \right\| < \varepsilon \) and if \(\left\| (\dot{\textbf{x}}(t), \dot{\textbf{y}}(t)) \right\| < \varepsilon \) for all \(t\ge 0\), then there exists a solution \({\textbf{u}}(t)\) of (F.6) such that as \(t\rightarrow \infty \),
$$\begin{aligned} \begin{aligned} {\textbf{x}}(t)&= {\textbf{u}}(t) + O(e^{-b_1 t}),\\ {\textbf{y}}(t)&= h({\textbf{u}}(t)) + O(e^{-b_1 t}), \end{aligned}\end{aligned}$$
(F.13)
where \(b_1 = \min (b, \beta _1)\), b and \(\beta _1\) are given in the assumption (F.2) and in Lemma F.1, respectively.
Proof
The proof is based on that of [10, Theorem 2.4.2]. Let \(({\textbf{x}}(t), {\textbf{y}}(t))\) be a solution of (F.3). Since the zero solution of (F.6) is Lyapunov stable, solutions \({\textbf{u}}(t)\) of (F.6) are Lyapunov stable if \({\textbf{u}}(0)\) is sufficiently small. Let \({\textbf{u}}(t)\) be a solution of (F.6) with \({\textbf{u}}(0)\) sufficiently small. Let \({\textbf{z}}(t) = {\textbf{y}}(t) - h({\textbf{x}}(t))\), \(\varvec{\phi }(t) = {\textbf{x}}(t) - {\textbf{u}}(t)\). Then
$$\begin{aligned} \dot{\textbf{z}}&= {\mathcal {B}}{\textbf{z}} + {\mathcal {R}}(\varvec{\phi } + {\textbf{u}},\textbf{z},\dot{\varvec{\phi }} + \dot{\textbf{u}},\dot{\textbf{z}}), \end{aligned}$$
(F.14a)
$$\begin{aligned} \dot{\varvec{\phi }}&= {\mathcal {A}}\varvec{\phi } + {\mathcal {V}}(\varvec{\phi },\textbf{z},\dot{\varvec{\phi }},\dot{\textbf{z}}), \end{aligned}$$
(F.14b)
where \({\mathcal {R}}\) is defined in (F.8) and
$$\begin{aligned} \begin{aligned} {\mathcal {V}}(\varvec{\phi },{\textbf {z}},\dot{\varvec{\phi }},\dot{{\textbf {z}}})&= f({\textbf {u}} {+} \varvec{\phi }, {{\textbf {z}}} + h({{\textbf {u}}} {+} \varvec{\phi }), \dot{{\textbf {u}}} {+} \dot{\varvec{\phi }}, \dot{{\textbf {z}}} + h'({\textbf {u}} {+} \varvec{\phi }) (\dot{{\textbf {u}}} {+} \dot{\varvec{\phi }}))\\ {}&\quad - f({{\textbf {u}}}, h({{\textbf {u}}}), \dot{{\textbf {u}}}, h'({{\textbf {u}}})\dot{{\textbf {u}}}). \end{aligned} \end{aligned}$$
We now formulate (F.14a)–(F.14b) as a fixed point problem. Let \({\mathscr {X}}\) be the set of continuous differentiable functions \(\varvec{\phi }:[0,\infty )\rightarrow X\) of (F.14b) with \(\left\| \varvec{\phi }(t) e^{at} \right\| \le 1\) and \(\left\| \dot{\varvec{\phi }}(t) e^{at} \right\| \le a\) for all \(t\ge 0\), where \(a=b/2\) in which b is defined in (F.2). We define the norm \(\left\| \varvec{\phi } \right\| _{{\mathscr {X}}} = \sup \{\left\| \varvec{\phi }(t) e^{at} \right\| + \left\| \dot{\varvec{\phi }}(t) e^{at} \right\| : t\ge 0\}\). By the assumption on the operator \({\mathcal {A}}\), we can decompose \({\mathcal {A}}\) into \({\mathcal {A}} = {\mathcal {A}}_1 + {\mathcal {A}}_2\) where
$$\begin{aligned} \begin{aligned} \left\| e^{{\mathcal {A}}_1 t}{\textbf{x}} \right\| = \left\| {\textbf{x}} \right\| ,\qquad \left\| {\mathcal {A}}_2{\textbf{x}} \right\| \le (b/4) \left\| {\textbf{x}} \right\| , \end{aligned}\end{aligned}$$
(F.15)
where b is defined by (F.2). Then (F.14b) can be written as
$$\begin{aligned} \dot{\varvec{\phi }} = {\mathcal {A}}_1\varvec{\phi } + \left[ {\mathcal {A}}_2\varvec{\phi } + {\mathcal {V}}(\varvec{\phi },\textbf{z},\dot{\varvec{\phi }},\dot{\textbf{z}}) \right] , \end{aligned}$$
and \(\varvec{\phi }(\infty ) = 0\) for \(\varvec{\phi }\in {\mathscr {X}}\). Let \({\textbf{z}}(t)\) be a given solution of (F.14a). By Duhamel’s formula, a solution \(\varvec{\phi }\in {\mathscr {X}}\) of (F.14b) must satisfy
$$\begin{aligned} \varvec{\phi }(t) = -\int _t^\infty e^{{\mathcal {A}}_1(t-s)} \left[ {\mathcal {A}}_2\varvec{\phi }(s) + {\mathcal {V}}(\varvec{\phi }(s),\textbf{z}(s),\dot{\varvec{\phi }}(s),\dot{\textbf{z}}(s)) \right] \textrm{d}s. \end{aligned}$$
Thus, a solution \(\varvec{\phi }\in {\mathscr {X}}\) of (F.14b) is a fixed point of the map** T that defined by
$$\begin{aligned} \begin{aligned} (T\varvec{\phi })(t) = -\int _t^\infty e^{{\mathcal {A}}_1(t-s)} \left[ {\mathcal {A}}_2\varvec{\phi }(s) + {\mathcal {V}}(\varvec{\phi }(s),\textbf{z}(s),\dot{\varvec{\phi }}(s),\dot{\textbf{z}}(s)) \right] \textrm{d}s. \end{aligned}\end{aligned}$$
(F.16)
Using the bounds on f, g, h, and the fact that \({\mathcal {R}}({\textbf{x}},{\textbf{0}},\dot{\textbf{x}},{\textbf{0}}) = {\textbf{0}}\) and \({\mathcal {V}}({\textbf{0}}, {\textbf{0}}, {\textbf{0}}, {\textbf{0}}) = {\textbf{0}}\), there is a continuous function \(k(\varepsilon )\) with \(k(0)=0\) such that if \(\varvec{\phi }_1,\varvec{\phi }_2\in X\), \({\textbf{z}}_1, \textbf{z}_2\in Y\), and \(\dot{\varvec{\phi }}_1,\dot{\varvec{\phi }}_2, \dot{\textbf{z}}_1, \dot{\textbf{z}}_2\in Z\) with \(\left\| (\varvec{\phi }_i, {\textbf{z}}_i) \right\| \), \(\left\| (\dot{\varvec{\phi }}_i, \dot{\textbf{z}}_i) \right\| < \varepsilon \), \(i=1,2\), then
$$\begin{aligned} \begin{aligned}&\left\| {\mathcal {R}}(\varvec{\phi }_1,\textbf{z}_1,\dot{\varvec{\phi }}_1,\dot{\textbf{z}}_1) - {\mathcal {R}}(\varvec{\phi }_2,\textbf{z}_2,\dot{\varvec{\phi }}_2,\dot{\textbf{z}}_2) \right\| \\&\quad \le k(\varepsilon ) \left[ \left\| ({\textbf{z}}_1, \dot{\textbf{z}}_1) \right\| \left( \left\| \varvec{\phi }_1 - \varvec{\phi }_2 \right\| + \left\| \dot{\varvec{\phi }}_1 - \dot{\varvec{\phi }}_2 \right\| \right) + \left\| {\textbf{z}}_1 - {\textbf{z}}_2 \right\| + \left\| \dot{\textbf{z}}_1 - \dot{\textbf{z}}_2 \right\| \right] ,\\&\left\| {\mathcal {V}}(\varvec{\phi }_1,\textbf{z}_1,\dot{\varvec{\phi }}_1,\dot{\textbf{z}}_1) - {\mathcal {V}}(\varvec{\phi }_2,\textbf{z}_2,\dot{\varvec{\phi }}_2,\dot{\textbf{z}}_2) \right\| \le k(\varepsilon ) \left[ \left\| {\textbf{z}}_1 - {\textbf{z}}_2 \right\| + \left\| \dot{\textbf{z}}_1 - \dot{\textbf{z}}_2 \right\| + \left\| \varvec{\phi }_1 - \varvec{\phi }_2 \right\| + \left\| \dot{\varvec{\phi }}_1 - \dot{\varvec{\phi }}_2 \right\| \right] . \end{aligned}\end{aligned}$$
(F.17)
By the same argument as in the proof of Lemma F.1, one has
$$\begin{aligned} \begin{aligned} \left\| {\textbf{z}}(t) \right\| \le C_1 \left\| {\textbf{z}}(0) \right\| e^{-\beta _1 t}, \end{aligned}\end{aligned}$$
(F.18)
where
$$\begin{aligned} \beta _1 = b - ck(\varepsilon ) \left( 1 + C_0 \right) ,\qquad C_0 = \left( 1-k(\varepsilon ) \right) ^{-1} \left( \left\| {\mathcal {B}} \right\| + k(\varepsilon ) \right) . \end{aligned}$$
Using (F.15), we obtain, from (F.16), that
$$\begin{aligned} \begin{aligned} \left\| T\varvec{\phi }(t) \right\|&\le \int _t^\infty \frac{b}{4} |\varvec{\phi }(s)|\, \textrm{d}s + \int _t^\infty \left\| {\mathcal {V}}(\varvec{\phi }(s),\textbf{z}(s),\dot{\varvec{\phi }}(s),\dot{\textbf{z}}(s)) \right\| \textrm{d}s\\&= \frac{a}{4} \int _t^\infty \left\| \varvec{\phi }(s) \right\| \textrm{d}s + \int _t^\infty \left\| {\mathcal {V}}(\varvec{\phi }(s),\textbf{z}(s),\dot{\varvec{\phi }}(s),\dot{\textbf{z}}(s)) - {\mathcal {V}}({\textbf{0}}, {\textbf{0}}, {\textbf{0}}, {\textbf{0}}) \right\| \textrm{d}s\\&\le \frac{e^{-at}}{2} + k(\varepsilon ) \int _t^\infty \left[ \left\| \varvec{\phi }(s) \right\| + \left\| {\textbf{z}}(s) \right\| + \left\| \dot{\varvec{\phi }}(s) \right\| + \left\| \dot{\textbf{z}}(s) \right\| \right] \textrm{d}s, \end{aligned}\nonumber \\\end{aligned}$$
(F.19)
where we’ve used \({\mathcal {V}}({\textbf{0}}, {\textbf{0}}, {\textbf{0}}, {\textbf{0}}) = \textbf{0}\) in the second equation and (F.17) in the last inequality. In view of (F.11) in the proof of Lemma F.1 and (F.18),
$$\begin{aligned} \begin{aligned} \left\| {\textbf{z}}(s) \right\| + \left\| \dot{\textbf{z}}(s) \right\| \le \left( 1 + C_0 \right) \left\| {\textbf{z}}(s) \right\| \le \left( 1 + C_0 \right) C_1 \left\| {\textbf{z}}(0) \right\| e^{-\beta _1 s} =: C_2\, e^{-\beta _1 s}. \end{aligned}\nonumber \\\end{aligned}$$
(F.20)
Together with the hypothesis \(\varvec{\phi }\in {\mathscr {X}}\), we derive
$$\begin{aligned} \left\| T\varvec{\phi }(t) \right\| \le \frac{e^{-at}}{2} + k(\varepsilon ) \int _t^\infty \left[ (1+a) e^{-as} + C_2 e^{-\beta _1 s} \right] \textrm{d}s \le e^{-at} \end{aligned}$$
for \(\varepsilon \) sufficiently small such that \(\beta _1 = b - ck(\varepsilon ) \left( 1 + C_0 \right) \ge b/2 = a\) and \(k(\varepsilon )\le \min \{(1+a)^{-1}, C_2^{-1}\}/ 2\). To estimate \(\frac{d}{dt} T\varvec{\phi }\), we compute
$$\begin{aligned} \begin{aligned} \left\| \frac{d}{dt} T\varvec{\phi }(t) \right\|&= \left\| A_2\varvec{\phi }(t) + V(\varvec{\phi }(t),\textbf{z}(t),\dot{\varvec{\phi }}(t),\dot{\textbf{z}}(t)) \right\| \\&\le \frac{a}{2} \left\| \varvec{\phi }(t) \right\| + k(\varepsilon ) \left( \left\| \varvec{\phi }(t) \right\| + \left\| {\textbf{z}}(t) \right\| + \left\| \dot{\varvec{\phi }}(t) \right\| + \left\| \dot{\textbf{z}}(t) \right\| \right) \\&\le \left( \frac{a}{2} + k(\varepsilon ) (1+a) \right) e^{-at} + k(\varepsilon ) C_2 e^{-\beta _1t} \le ae^{-at} \end{aligned} \end{aligned}$$
for \(\varepsilon \) sufficiently small. This proves T maps \({\mathscr {X}}\) into \({\mathscr {X}}\).
We now show that T is a contraction on \({\mathscr {X}}\). Let \(\varvec{\phi }_1, \varvec{\phi }_2\in {\mathscr {X}}\) and let \({\textbf{z}}_1, {\textbf{z}}_2\) be the corresponding solutions of (F.14a) with \({\textbf{z}}_i(0) = {\textbf{z}}_0\), \(i=1,2\). We first estimate \({\textbf{v}}(t) = {\textbf{z}}_1(t) - {\textbf{z}}_2(t)\). From (F.14a) and (F.17),
$$\begin{aligned} \begin{aligned} \left\| {\textbf{v}}(t) \right\|&\le ck(\varepsilon ) \int _0^t e^{-b(t-s)} \left[ \left\| (\textbf{z}_1(s), \dot{\textbf{z}}_1(s)) \right\| \left( \left\| \varvec{\phi }_1(s) - \varvec{\phi }_2(s) \right\| + \left\| \dot{\varvec{\phi }}_1(s) - \dot{\varvec{\phi }}_2(s) \right\| \right) + \left\| {\textbf{v}}(s) \right\| + \left\| \dot{\textbf{v}}(s) \right\| \right] \textrm{d}s. \end{aligned}\nonumber \\\end{aligned}$$
(F.21)
Since
$$\begin{aligned} \dot{\textbf{v}} = {\mathcal {B}}{\textbf{v}} + {\mathcal {R}}(\varvec{\phi }_1 + {\textbf{u}},\textbf{z}_1,\dot{\varvec{\phi }}_1 + \dot{\textbf{u}},\dot{\textbf{z}}_1) - {\mathcal {R}}(\varvec{\phi }_2 + {\textbf{u}},\textbf{z}_2,\dot{\varvec{\phi }}_2 + \dot{\textbf{u}},\dot{\textbf{z}}_2), \end{aligned}$$
using (F.17), we get
$$\begin{aligned} \left\| \dot{\textbf{v}} \right\| \le \left\| {\mathcal {B}} \right\| \left\| v \right\| + k(\varepsilon ) \left[ \left\| ({\textbf{z}}_1, \dot{\textbf{z}}_1) \right\| \left( \left\| \varvec{\phi }_1 - \varvec{\phi }_2 \right\| + \left\| \dot{\varvec{\phi }}_1 - \dot{\varvec{\phi }}_2 \right\| \right) + \left\| {\textbf{v}} \right\| + \left\| \dot{\textbf{v}} \right\| \right] . \end{aligned}$$
So
$$\begin{aligned} \left\| \dot{\textbf{v}} \right\| \le (1 - k(\varepsilon ))^{-1} \left[ (\left\| {\mathcal {B}} \right\| + k(\varepsilon )) \left\| \textbf{v} \right\| + k(\varepsilon ) \left\| ({\textbf{z}}_1, \dot{\textbf{z}}_1) \right\| \left( \left\| \varvec{\phi }_1 - \varvec{\phi }_2 \right\| + \left\| \dot{\varvec{\phi }}_1 - \dot{\varvec{\phi }}_2 \right\| \right) \right] , \end{aligned}$$
implying
$$\begin{aligned} \left\| \dot{\textbf{v}}(s) \right\| \le C_3 \left\| {\textbf{v}}(s) \right\| + k_1(\varepsilon ) \left\| ({\textbf{z}}_1, \dot{\textbf{z}}_1) \right\| \left\| \varvec{\phi }_ 1 - \varvec{\phi }_2 \right\| _{{\mathscr {X}}} e^{-as},\qquad k_1(0) = 0, \end{aligned}$$
for some \(C_3>0\). Together with \(\left\| ({\textbf{z}}_1(s), \dot{\textbf{z}}_1(s)) \right\| \le C_2 e^{-\beta _1s}\), which is followed by (F.20), (F.21) implies
$$\begin{aligned} \begin{aligned} \left\| {\textbf{v}}(t) \right\|&\le ck(\varepsilon ) \int _0^t e^{-b(t-s)} \left[ \left( 1 + k_1(\varepsilon ) \right) C_2 e^{-\beta _1s} \left\| \varvec{\phi }_ 1 - \varvec{\phi }_2 \right\| _{{\mathscr {X}}} e^{-as} + (1+C_3) \left\| \textbf{v}(s) \right\| \right] ds\\&\le C_4 k(\varepsilon ) \left\| \varvec{\phi }_ 1 - \varvec{\phi }_2 \right\| _{{\mathscr {X}}} e^{-bs} + c k(\varepsilon )(1+C_3) \int _0^t e^{-b(t-s)} \left\| {\textbf{v}}(s) \right\| \textrm{d}s \end{aligned} \end{aligned}$$
for \(\varepsilon \) sufficiently small, where \(C_4>0\) is a constant. By Gronwall’s lemma,
$$\begin{aligned} \begin{aligned} \left\| {\textbf{v}}(t) \right\| \le C_4 k(\varepsilon ) \left\| \varvec{\phi }_ 1 - \varvec{\phi }_2 \right\| _{{\mathscr {X}}} e^{-\beta _2 t},\qquad \beta _2 = b - c k(\varepsilon )(1+C_3). \end{aligned}\end{aligned}$$
(F.22)
Using (F.15) and (F.22),
$$\begin{aligned} \begin{aligned}&\left\| T\varvec{\phi }_1(t) - T\varvec{\phi }_2(t) \right\| \\&\ \ \le \int _t^\infty \left[ \frac{a}{2} \left\| \varvec{\phi }_1(s) - \varvec{\phi }_2(s) \right\| + \left\| {\mathcal {R}}(\varvec{\phi }_1(s),{\textbf{z}}_1(s), \dot{\varvec{\phi }}_1(s), \dot{\textbf{z}}_1(s)) - {\mathcal {R}}(\varvec{\phi }_2(s),{\textbf{z}}_2(s), \dot{\varvec{\phi }}_2(s), \dot{\textbf{z}}_2(s) ) \right\| \right] \\&\ \ \le \frac{a}{2} \int _t^\infty \left\| \varvec{\phi }_ 1 - \varvec{\phi }_2 \right\| _{{\mathscr {X}}} e^{-as}\, \textrm{d}s\\&\ \ \quad + k(\varepsilon ) \int _t^\infty \left[ \left\| ({\textbf{z}}_1(s), \dot{\textbf{z}}_1(s)) \right\| \left( \left\| \varvec{\phi }_1(s) - \varvec{\phi }_2(s) \right\| + \left\| \dot{\varvec{\phi }}_1(s) - \dot{\varvec{\phi }}_2(s) \right\| \right) + \left\| {\textbf{v}}(s) \right\| + \left\| \dot{\textbf{v}}(s) \right\| \right] \textrm{d}s\\&\ \ \le \frac{1}{2} \left\| \varvec{\phi }_ 1 - \varvec{\phi }_2 \right\| _{{\mathscr {X}}}\\&\ \ \quad + k(\varepsilon ) \int _t^\infty \left[ (1+k_1(\varepsilon ))C_2 e^{-\beta _1s} \left\| \varvec{\phi }_ 1 - \varvec{\phi }_2 \right\| _{{\mathscr {X}}} e^{-as} + (1+C_3)C_4 k(\varepsilon ) \left\| \varvec{\phi }_ 1 - \varvec{\phi }_2 \right\| _{{\mathscr {X}}} e^{-\beta _2 s} \right] \textrm{d}s\\&\ \ \le \alpha \left\| \varvec{\phi }_ 1 - \varvec{\phi }_2 \right\| _{{\mathscr {X}}},\qquad \alpha <1, \end{aligned} \end{aligned}$$
for \(\alpha \) sufficiently small. This shows that T is a contraction and, hence, has a unique fixed point.
Note that \(T = T_{\textbf{z}}\) and the above analysis proves that \(T_\textbf{z}\) has a unique fixed point in \({\mathscr {X}}\) provided \({\textbf{z}}\) and \(\dot{\textbf{z}}\) are sufficiently small. Denote \(\varvec{\phi }\in {\mathscr {X}}\) the fixed point of the contraction \(T_{\textbf{z}}\), where \({\textbf{z}} = {\textbf{y}} - h({\textbf{x}})\). Let \({\textbf{u}}(0) = {\textbf{x}}(0) - \varvec{\phi }(0)\). Then let \(\textbf{u}(t)\) be the solution of (F.6) evolving from \(\textbf{u}(0)\). Then \(({\textbf{x}}(t), {\textbf{y}}(t))\) can be decomposed as in \({\textbf{x}}(t) = {\textbf{u}}(t) + \varvec{\phi }(t)\) and \({\textbf{y}} = h({\textbf{x}}(t)) + {\textbf{z}}(t)\). In view of the fact that \(\varvec{\phi }\in {\mathscr {X}}\) and Lemma F.1, the asymptotic limits in (F.13) follow, which completing the proof of Proposition F.2. \(\quad \square \)
Appendix G: Asymptotic Expansion of the Local Center Manifold
In this appendix, we check the expression of local center manifold in Lemma 9.6 by asymptotic expansion.
If we substitute \({\textbf{y}} = h({\textbf{x}})\) into (9.32b), we see that the center manifold can be obtained by solving
$$\begin{aligned} h'({\textbf{x}}) \left[ Q_1 {\mathcal {N}}({\textbf{x}} + h({\textbf{x}}), \dot{\textbf{x}} + h'({\textbf{x}})\dot{\textbf{x}}) \right] = {\mathcal {L}} h({\textbf{x}}) + Q_2 {\mathcal {N}}({\textbf{x}} + h({\mathbf{x)}}, \dot{\textbf{x}} + h'(\textbf{x})\dot{\textbf{x}}). \end{aligned}$$
We want to show the manifold of equilibria \({\mathcal {M}}_*\) is a center manifold by showing it satisfies the above equation. On the manifold of equilibria \({\mathcal {M}}_*\), \(\dot{\textbf{x}} = 0\). So we need to check
$$\begin{aligned} \begin{aligned} h'({\textbf{x}}) \left[ Q_1 {\mathcal {N}}({\textbf{x}} + h({\textbf{x}}), {\textbf{0}}) \right] = {\mathcal {L}} h({\textbf{x}}) + Q_2 {\mathcal {N}}({\textbf{x}} + h({\mathbf{x)}}, {\textbf{0}}). \end{aligned}\end{aligned}$$
(G.1)
Note that
$$\begin{aligned} {\mathcal {N}}({\textbf{x}} + h({\textbf{x}}), {\textbf{0}})= & {} {\mathcal {N}}^0(\textbf{x} + h({\textbf{x}})) = \begin{bmatrix} 0\\ 0\\ -\frac{{\mathscr {R}}_gT_\infty }{\rho _lR_*}\, H^0({\textbf{x}} + h({\textbf{x}}))\\ 0\\ 0\\ \vdots \end{bmatrix}\\= & {} \begin{bmatrix} 0\\ 0\\ -\frac{2\sigma }{\rho _lR_*^3} \frac{(\alpha +R_{**}(\alpha )-R_*)^2}{\alpha +R_{**}(\alpha )}\\ 0\\ 0\\ \vdots \end{bmatrix}. \end{aligned}$$
So \(Q_1{\mathcal {N}}({\textbf{x}} + h({\textbf{x}}), {\textbf{0}}) = {\textbf{0}}\) and \({\mathcal {N}}({\textbf{x}} + h({\textbf{x}}), {\textbf{0}}) = {\mathcal {N}}(\textbf{x} + h({\textbf{x}}), {\textbf{0}})\). It suffices to check \({\mathcal {L}} h({\textbf{x}}) + Q_2 {\mathcal {N}}({\textbf{x}} + h({\mathbf{x)}}, {\textbf{0}}) = \textbf{0}\) in (G.1). Taylor expand \(h({\textbf{x}}) = h(\alpha {\textbf{b}})\) at \(\alpha =0\), one gets
$$\begin{aligned} h({\textbf{x}}) = \begin{bmatrix} \rho _{**}(\alpha ) - \rho _*\\ R_{**}(\alpha ) - R_*\\ 0\\ 0\\ 0\\ \vdots \end{bmatrix} = \begin{bmatrix} \rho _{**,1} \alpha + \rho _{**,2} \alpha ^2 + \rho _{**,3} \alpha ^3 + \cdots \\ R_{**,1} \alpha + R_{**,2} \alpha ^2 + R_{**,3} \alpha ^3 + \cdots \\ 0\\ 0\\ 0\\ \vdots \end{bmatrix}, \end{aligned}$$
where \(\rho _{**,m} = (m!)^{-1} \rho _{**}^{(m)}(0)\) and \(R_{**,m} = (m!)^{-1} R_{**}^{(m)}(0)\) in which \(\rho _{**}^{(m)}(0)\) and \(R_{**}^{(m)}(0)\) are the mth derivatives of \(\rho _{**}\) and \(R_{**}\) at \(\alpha =0\), respectively. Then \({\mathcal {L}} h({\textbf{x}}) + {\mathcal {N}}({\textbf{x}} + h({\mathbf{x)}}, {\textbf{0}}) = {\textbf{0}}\) is equivalent to
$$\begin{aligned} \begin{aligned} \sum _{m=1}^\infty \left( \frac{{\mathscr {R}}_gT_\infty }{\rho _lR_*} \rho _{**,m} + \frac{2\sigma }{\rho _lR_*^3} R_{**,m} \right) \alpha ^m - \frac{2\sigma }{\rho _lR_*^3} \frac{\left( \alpha +R_{**}(\alpha )-R_* \right) ^2}{\alpha +R_{**}(\alpha )} = 0. \end{aligned}\nonumber \\\end{aligned}$$
(G.2)
For \(m=1\), we differentiate (9.34) with respect to \(\alpha \), evaluate at \(\alpha =0\) and use \(R_{**}(0) = R_*\), we get \({\mathscr {R}}_gT_\infty \rho _{**}'(0) = -(2\sigma /R_*^2) R_{**}'(0)\). So we have
$$\begin{aligned} \frac{{\mathscr {R}}_gT_\infty }{\rho _lR_*} \rho _{**,1} + \frac{2\sigma }{\rho _lR_*^3} R_{**,1} = 0. \end{aligned}$$
For \(m=2\), we first expand
$$\begin{aligned} \frac{\left( \alpha +R_{**}(\alpha )-R_* \right) ^2}{\alpha +R_{**}(\alpha )} = \frac{( 1+R_{**}'(0) )^2}{R_*} \alpha ^2 + \cdots . \end{aligned}$$
Then the coefficient of \(\alpha ^2\) in (G.2) is
$$\begin{aligned} \begin{aligned} \frac{{\mathscr {R}}_gT_\infty }{\rho _lR_*} \rho _{**,2} + \frac{2\sigma }{\rho _lR_*^3} R_{**,2} - \frac{2\sigma }{\rho _lR_*^4} \left( 1+R_{**,1} \right) ^2 . \end{aligned}\end{aligned}$$
(G.3)
By differentiating (9.34) with respect to \(\alpha \) twice, evaluating at \(\alpha =0\) and using \(R_{**}(0) = R_*\), we have \({\mathscr {R}}_gT_\infty \rho _{**}''(0) = (4\sigma /R_*^3) (R_{**}'(0))^2 - (2\sigma /R_*^2) R_{**}''(0)\). So
$$\begin{aligned} \frac{{\mathscr {R}}_gT_\infty }{\rho _lR_*} \rho _{**,2} + \frac{2\sigma }{\rho _lR_*^3} R_{**,2} = (2!)^{-1} \frac{4\sigma }{\rho _l R_*^4} (R_{**,1})^2. \end{aligned}$$
In order to have the term (G.3) for \(m=2\) vanishing, we require
$$\begin{aligned} 0 = \frac{2\sigma }{\rho _lR_*^4} R_{**,1}^2 - \frac{2\sigma }{\rho _lR_*^4} (1 + R_{**,1})^2, \end{aligned}$$
whose solution is \(R_{**,1} = -1/2\).
For \(m=3\), we further expand, using \(R_{**,1} = -1/2\), that
$$\begin{aligned} \frac{\left( \alpha +R_{**}(\alpha )-R_* \right) ^2}{\alpha +R_{**}(\alpha )} = \frac{( 1+R_{**,1} )^2}{R_*} \alpha ^2 + \left( \frac{1}{2R_*} R_{**}''(0) - \frac{1}{8R_*^2} \right) \alpha ^3 + \cdots \end{aligned}$$
Then the coefficient of \(\alpha ^3\) in (G.2) is
$$\begin{aligned} \begin{aligned} \frac{{\mathscr {R}}_gT_\infty }{\rho _lR_*} \rho _{**,3} + \frac{2\sigma }{\rho _lR_*^3} R_{**,3} - \frac{2\sigma }{\rho _lR_*^3} \left( \frac{1}{2R_*} R_{**}''(0) - \frac{1}{8R_*^2} \right) . \end{aligned}\end{aligned}$$
(G.4)
By differentiating (9.34) with respect to \(\alpha \) three times, evaluating at \(\alpha =0\) and using \(R_{**}(0) = R_*\), we get \({\mathscr {R}}_gT_\infty \rho _{**}'''(0) = -(12\sigma /R_*^4)(R_{**}'(0))^3 + (12\sigma /R_*^3)R_{**}'(0)R_{**}''(0) -(2\sigma /R_*^2) R_{**}'''(0)\). So we have
$$\begin{aligned} \frac{{\mathscr {R}}_gT_\infty }{\rho _lR_*} \rho _{**,3} = -\frac{2\sigma }{\rho _lR_*^5} R_{**,1}^3 + \frac{4\sigma }{\rho _lR_*^4} R_{**,1} R_{**,2} - \frac{2\sigma }{\rho _lR_*^3} R_{**,3}. \end{aligned}$$
In order to have the term (G.4) for \(m=3\) vanishing, we require
$$\begin{aligned} 0= & {} -\frac{2\sigma }{\rho _lR_*^5} R_{**,1}^3 + \frac{4\sigma }{\rho _lR_*^4} R_{**,1} R_{**,2} - \frac{2\sigma }{\rho _lR_*^3} R_{**,3} + \frac{2\sigma }{\rho _lR_*^3} R_{**,3} \\{} & {} - \frac{\sigma }{\rho _lR_*^3} \left( \frac{1}{R_*} R_{**}''(0) - \frac{1}{4R_*^2} \right) , \end{aligned}$$
where the third term cancels the fourth. Using \(R_{**,1}=-1/2\) and \(R_{**}''(0) = 2R_{**,2}\),
$$\begin{aligned} 0 = \frac{\sigma }{4\rho _lR_*^5} - \frac{2\sigma }{\rho _lR_*^4} R_{**,2} - \frac{2\sigma }{\rho _lR_*^4} R_{**,2} + \frac{\sigma }{4\rho _lR_*^5} \end{aligned}$$
for which the solution is \(R_{**,2} = 1/(8R_*)\). Thus,
$$\begin{aligned} R_{**}(\alpha ) = R_* - \frac{1}{2} \alpha + \frac{1}{8R_*} \alpha ^2 + \cdots = \frac{-\alpha + \sqrt{\alpha ^2 + 4R_{*}^2}}{2}, \end{aligned}$$
which coincides with the center manifold expression in (9.34).
