In order to extend the local solution of the initial-boundary value problem (8)–(11) globally in time, we shall establish a differential inequality which provides us the uniform estimates of solutions for both time and the Mach number. Suppose that \((\sigma ,\mathbf {u})\) is the local solution to the initial-boundary value problem (8)–(11) in \(\Omega \times (0,T)\), for \(0< T<\infty \). Moreover, we assume that \(1/c\leq \rho =1+\epsilon \sigma \leq c\) for some constant \(c>1\). Then the viscosity coefficients can be estimated as follows: \(1/c^{\alpha }\le \mu (\rho )=\rho ^{\alpha }\le c^{\alpha }\) and \(1/c^{\beta }\le \lambda (\rho )=\rho ^{\beta }\le c^{\beta }\).
\(L^{2}\) estimate
Lemma 3.1
For the solution to (8)–(11), we have
$$\begin{aligned}& \frac{d}{dt} \bigl\Vert \bigl(\sqrt{p'(\rho )}\sigma , \sqrt{\rho }\mathbf {u}\bigr) \bigr\Vert _{L^{2}}^{2}+ \gamma _{1} \Vert \mathbf {u}\Vert _{H^{1}}^{2} \\& \quad \le \epsilon \Vert \sigma _{t} \Vert _{L^{2}}^{2}+C \Vert \sigma \Vert _{H^{1}}^{2}\bigl( \Vert \mathbf {u}\Vert _{H^{2}}^{2}+ \Vert \sigma \Vert _{H^{1}}^{2}\bigr), \end{aligned}$$
where \(\gamma _{1}\) is a positive constant independent of ϵ.
Proof
We integrate the product of (8) and \(p'(\rho )\sigma \) to get
$$\begin{aligned}& \frac{1}{2}\frac{d}{dt} \bigl\Vert \sqrt{p'( \rho )}\sigma \bigr\Vert _{L^{2}}^{2}+ \frac{1}{\epsilon } \int _{\Omega }p'(\rho )\operatorname{div} \mathbf {u}\sigma \,dx \\& \quad = - \int _{\Omega }p'(\rho )\sigma \operatorname{div}( \sigma \mathbf {u}) \,dx+\frac{1}{2} \int _{\Omega }p''(\rho )\epsilon \sigma _{t}\sigma ^{2} \,dx \\& \quad \leq \epsilon \Vert \sigma _{t} \Vert _{L^{2}}^{2}+ \eta \Vert \mathbf {u}\Vert _{H^{1}}^{2}+C_{\eta } \Vert \sigma \Vert _{H^{1}}^{4}. \end{aligned}$$
Due to the boundary conditions (11) and Lemma 2.3, we have
$$\begin{aligned}& - \int _{\Omega }\bigl(2\mu (\rho )\operatorname {div}\bigl( \mathrm{D}(\mathbf {u})\bigr)+\lambda ( \rho )\nabla \operatorname{div} \mathbf {u}\bigr)\cdot \mathbf {u}\,dx \\& \quad = - \int _{\Omega }\bigl(\bigl(2\mu (\rho )+\lambda (\rho )\bigr) \nabla \operatorname {div}\mathbf {u}-\mu (\rho )\operatorname {curl}\operatorname {curl}\mathbf {u}\bigr)\cdot \mathbf {u}\,dx \\& \quad = \int _{\Omega }\bigl(\bigl(2\mu (\rho )+\lambda (\rho )\bigr) \vert \operatorname {div}\mathbf {u}\vert ^{2}+ \mu (\rho ) \vert \operatorname {curl}\mathbf {u}\vert ^{2}\bigr) \,dx \\& \qquad {} + \int _{\Omega }\bigl[\nabla \bigl(2\mu (\rho )+\lambda (\rho ) \bigr)\operatorname {div}\mathbf {u}+ \nabla \bigl(\mu (\rho )\bigr)\times \operatorname {curl}\mathbf {u}\bigr] \cdot \mathbf {u}\,dx \\& \quad \geq \iota _{0} \Vert \mathbf {u}\Vert _{H^{1}}^{2}+ \int _{\Omega }\bigl[(\nabla \bigl(2\mu ( \rho )+\lambda (\rho ) \bigr)\operatorname {div}\mathbf {u}+\nabla \bigl(\mu (\rho )\bigr) \operatorname {curl}\mathbf {u}\bigr]\cdot \mathbf {u}\,dx. \end{aligned}$$
Integrating the product of (9) and u, we get
$$\begin{aligned}& \frac{1}{2}\frac{d}{dt} \Vert \sqrt{\rho } \mathbf {u}\Vert _{L^{2}}^{2}+ \frac{1}{\epsilon } \int _{\Omega }p'(\rho )\mathbf {u}\cdot \nabla \sigma \,dx + \bigl\Vert \sqrt{\bigl(2\mu (\rho )+\lambda (\rho )\bigr)}\operatorname {div}\mathbf {u}\bigr\Vert _{L^{2}}^{2}+ \bigl\Vert \sqrt{\mu (\rho )} \operatorname {curl}\mathbf {u}\bigr\Vert _{L^{2}}^{2} \\& \quad = - \int _{\Omega }\bigl[\nabla \bigl(2\mu (\rho )+\lambda ( \rho )\bigr)\operatorname {div}\mathbf {u}+\nabla \bigl(\mu (\rho )\bigr)\times \operatorname {curl}\mathbf {u}\bigr]\cdot \mathbf {u}\,dx \\& \qquad {} + \int _{\Omega }\bigl[\nabla \bigl(2\mu (\rho )\bigr)\cdot \mathrm{D}(\mathbf {u})+ \nabla \bigl(\lambda (\rho )\bigr) \operatorname {div}\mathbf {u}\bigr]\cdot \mathbf {u}\,dx \\& \quad \leq \eta \Vert \mathbf {u}\Vert _{H^{1}}^{2}+C_{\eta } \Vert \mathbf {u}\Vert _{H^{2}}^{2} \Vert \sigma \Vert _{H^{1}}^{2}. \end{aligned}$$
Using (11) again and integration by parts, we have
$$\begin{aligned}& \frac{1}{\epsilon } \int _{\Omega }p'(\rho )\operatorname{div} \mathbf {u}\sigma \,dx +\frac{1}{\epsilon } \int _{\Omega }p'(\rho )\mathbf {u}\cdot \nabla \sigma \,dx \\& \quad =- \int _{\Omega }p''(\rho )\nabla \sigma \cdot \mathbf {u}\sigma \,dx \leq \eta \Vert \mathbf {u}\Vert _{H^{1}}^{2}+C_{\eta } \Vert \sigma \Vert _{H^{1}}^{4}. \end{aligned}$$
(19)
Summing up the above equalities and choosing η small enough, we get the lemma. □
Estimates of first order derivatives
Lemma 3.2
For the solution to (8)–(11), we have
$$ \begin{aligned} &\frac{1}{2} \frac{d}{dt} \bigl[ \bigl\Vert \sqrt{2\mu (\rho )+\lambda (\rho )} \operatorname {div}\mathbf {u}\bigr\Vert _{L^{2}}^{2} + \bigl\Vert \sqrt{\mu (\rho )} \operatorname {curl}\mathbf {u}\bigr\Vert _{L^{2}}^{2} \bigr] \\ &\qquad {}+ \frac{d}{dt} \int _{\Omega }\rho \mathbf {u}_{t}\cdot \mathbf {u}\,dx + \frac{1}{2} \bigl\Vert \sqrt{p'(\rho )}\sigma _{t} \bigr\Vert _{L^{2}}^{2} \\ &\quad \le C \Vert \mathbf {u}_{t} \Vert _{H^{1}}^{2}+ \eta \Vert \mathbf {u}\Vert _{H^{2}}^{2} \\ &\qquad {}+C_{\eta } \bigl[ \Vert \mathbf {u}\Vert _{H^{2}}^{2}\bigl( \Vert \mathbf {u}\Vert _{H^{2}}^{2}+ \Vert \sigma _{t} \Vert _{H^{1}}^{2}+ \Vert \sigma \Vert _{H^{2}}^{2}\bigr)+ \Vert \sigma \Vert _{H^{2}}^{2}\bigl( \Vert \mathbf {u}_{t} \Vert _{H^{1}}^{2}+ \Vert \sigma _{t} \Vert _{H^{1}}^{2}\bigr) \bigr], \end{aligned} $$
where η is to be determined later.
Proof
First, by differentiating (9) with respect to t, we have
$$ \begin{aligned}[b] &(\rho \mathbf {u}_{t})_{t}+ \epsilon \sigma _{t} \mathbf {u}\cdot \nabla \mathbf {u}+\rho (\mathbf {u}_{t}\cdot \nabla \mathbf {u}+\mathbf {u}\cdot \nabla \mathbf {u}_{t})+p''(\rho )\sigma _{t} \nabla \sigma +\frac{1}{\epsilon }p'( \rho )\nabla \sigma _{t} \\ &\quad =\operatorname{div}\bigl(2\mu (\rho ) D(\mathbf {u}_{t})\bigr)+ \nabla \bigl( \lambda (\rho ) \operatorname{div} \mathbf {u}_{t}\bigr) \\ &\qquad {}+\operatorname{div}\bigl(2\mu '(\rho )\epsilon \sigma _{t}D( \mathbf {u})\bigr)+ \nabla \bigl(\lambda '(\rho )\epsilon \sigma _{t} \operatorname{div} \mathbf {u}\bigr). \end{aligned} $$
(20)
Multiplying (20) by u in \(L^{2}\), integrating by parts and using the boundary conditions (11), we deduce that
$$\begin{aligned} &\frac{1}{2}\frac{d}{dt} \bigl[ \bigl\Vert \sqrt{2\mu ( \rho )+\lambda (\rho )} \operatorname {div}\mathbf {u}\bigr\| _{L^{2}}^{2} + \bigl\Vert \sqrt{\mu (\rho )}\operatorname {curl}\mathbf {u}\bigr\| _{L^{2}}^{2} \bigr] \\ &\qquad {}+\frac{d}{dt} \int _{\Omega }\rho \mathbf {u}_{t}\mathbf {u}\,dx+ \frac{1}{\epsilon } \int _{\Omega }p'(\rho )\nabla \sigma _{t}\cdot \mathbf {u}\,dx \\ &\quad = \int _{\Omega }\rho \mathbf {u}_{t}^{2} \,dx- \int _{\Omega }\rho _{t}( \mathbf {u}\cdot \nabla ) \mathbf {u}\cdot \mathbf {u}\,dx- \int _{\Omega }\rho ( \mathbf {u}_{t}\cdot \nabla \mathbf {u}+ \mathbf {u}\cdot \nabla \mathbf {u}_{t})\cdot \mathbf {u}\,dx \\ &\qquad {}- \int _{\Omega }p''(\rho )\sigma _{t}\nabla \sigma \cdot \mathbf {u}\,dx+ \int _{\Omega }\operatorname{div}\bigl(2\mu '(\rho ) \epsilon \sigma _{t}D(\mathbf {u})\bigr)+ \nabla \bigl(\lambda '(\rho )\epsilon \sigma _{t} \operatorname{div} \mathbf {u}\bigr) \cdot \mathbf {u}\,dx \\ &\qquad {}- \int _{\Omega }\nabla \bigl(2\mu (\rho )+\lambda (\rho )\bigr) \cdot \mathbf {u}\operatorname {div}\mathbf {u}_{t} \,dx- \int _{\Omega }\nabla \bigl(\mu (\rho )\bigr)\times \operatorname {curl}\mathbf {u}_{t}\cdot \mathbf {u}\,dx \\ &\qquad {}+\frac{1}{2} \int _{\Omega }\partial _{t}\bigl(2\mu (\rho )+\lambda (\rho )\bigr) \vert \operatorname {div}\mathbf {u}\vert ^{2} \,dx + \frac{1}{2} \int _{\Omega }\partial _{t}\bigl(\mu ( \rho )\bigr) \vert \operatorname {curl}\mathbf {u}\vert ^{2} \,dx \\ &\quad \leq C \Vert \mathbf {u}_{t} \Vert _{L^{2}}^{2}+ \eta \Vert \mathbf {u}\Vert _{H^{2}}^{2} \\ &\qquad {}+C_{\eta } \bigl[ \Vert \mathbf {u}\Vert _{H^{2}}^{2}\bigl( \Vert \mathbf {u}\Vert _{H^{2}}^{2}+ \Vert \sigma _{t} \Vert _{H^{1}}^{2}+ \Vert \sigma \Vert _{H^{2}}^{2}\bigr) + \Vert \sigma \Vert _{H^{2}}^{2}\bigl( \Vert \mathbf {u}_{t} \Vert _{H^{1}}^{2}+ \Vert \sigma _{t} \Vert _{H^{1}}^{2}\bigr)\bigr]. \end{aligned}$$
We multiply (8) by \(p'(\rho )\sigma _{t}\), integrate by parts and use the boundary conditions (11) again to infer that
$$ \begin{aligned} & \bigl\Vert \sqrt{p'( \rho )\sigma _{t}} \bigr\Vert _{L^{2}}^{2} - \frac{1}{\epsilon } \int _{\Omega }p'(\rho )\nabla \sigma _{t}\cdot \mathbf {u}\,dx \\ &\quad =- \int _{\Omega }p'(\rho )\sigma _{t} \operatorname {div}(\sigma \mathbf {u}) \,dx+ \int _{\Omega }\nabla \bigl( p'(\rho )\bigr)\cdot \mathbf {u}\sigma _{t} \,dx \\ &\quad \leq \eta \Vert \sigma _{t} \Vert _{L^{2}}^{2}+C_{\eta } \Vert \sigma \Vert _{H^{1}}^{2} \Vert \mathbf {u}\Vert _{H^{1}}^{2}. \end{aligned} $$
Summing up the above estimates, we obtain the above lemma. □
Lemma 3.3
For the solution to (8)–(11), we have
$$ \begin{aligned} &\frac{d}{dt} \Vert \nabla \sigma \Vert _{L^{2}}^{2}+ \bigl\Vert \sqrt{p'(1)^{-1}}\sqrt{2 \mu (\rho ) +\lambda (\rho )} \nabla \operatorname{div} \mathbf {u}\bigr\Vert _{L^{2}}^{2} \\ &\quad \le C \Vert \mathbf {u}_{t} \Vert _{L^{2}}^{2}+ \eta \Vert \mathbf {u}\Vert _{H^{1}}^{2} +C_{\eta } \Vert \sigma \Vert _{H^{2}}^{4}+C \Vert \mathbf {u}\Vert _{H^{2}}^{2} \Vert \sigma \Vert _{H^{2}}^{2},\quad 0< \eta < 1, \end{aligned} $$
where η is to be determined later.
Proof
Applying ∇ to (8), multiplying the resulting equation by ∇σ, integrating in \(L^{2}\), we obtain
$$ \begin{aligned} &\frac{1}{2}\frac{d}{dt} \Vert \nabla \sigma \Vert _{L^{2}}^{2} + \frac{1}{\epsilon } \int _{\Omega }\nabla \operatorname{div} \mathbf {u}\cdot \nabla \sigma \,dx \\ &\quad =- \int _{\Omega }\bigl((\mathbf {u}\cdot \nabla )\nabla \sigma +\nabla \mathbf {u}\nabla \sigma +\nabla \sigma \operatorname{div} \mathbf {u}+\sigma \nabla \operatorname{div} \mathbf {u}\bigr)\nabla \sigma \,dx \\ &\quad \le \eta \bigl( \Vert \mathbf {u}\Vert _{H^{1}}^{2}+ \Vert \nabla \operatorname {div}\mathbf {u}\Vert _{L^{2}}^{2}\bigr) +C_{\eta } \Vert \sigma \Vert _{H^{2}}^{4}. \end{aligned} $$
Now, we apply \(\langle \text{(9)},p'(\rho )^{-1}\nabla \operatorname{div} \mathbf {u}\rangle \) to derive that
$$\begin{aligned}& \bigl\Vert \sqrt{p'(\rho )^{-1}}\sqrt{2\mu (\rho )+\lambda (\rho )}\nabla \operatorname{div} \mathbf {u}\bigr\Vert _{L^{2}}^{2} -\frac{1}{\epsilon } \int _{\Omega }\nabla \operatorname{div} \mathbf {u}\cdot \nabla \sigma \,dx \\& \quad = \int _{\Omega }\rho (\mathbf {u}_{t}+\mathbf {u}\cdot \nabla \mathbf {u})\cdot p'( \rho )^{-1}\nabla \operatorname{div} \mathbf {u}\,dx \\& \qquad {} + \int _{\Omega }p'(\rho )^{-1}\mu (\rho )\operatorname {curl}\operatorname {curl}\mathbf {u}\cdot \nabla \operatorname{div} \mathbf {u}\,dx \\& \qquad {} - \int _{\Omega }\bigl[2\nabla \mu (\rho )\cdot D(\mathbf {u})+\nabla \lambda (\rho )\cdot \operatorname {div}\mathbf {u}\bigr]\cdot p'(\rho )^{-1}\nabla \operatorname{div} \mathbf {u}\,dx \\& \quad \le \eta \Vert \nabla \operatorname {div}\mathbf {u}\Vert _{L^{2}}^{2} +C_{\eta }\bigl( \Vert \mathbf {u}_{t} \Vert _{L^{2}}^{2}+ \Vert \mathbf {u}\Vert _{H^{2}}^{2}( \Vert \mathbf {u}\Vert _{H^{2}}^{2}+ \Vert \sigma \Vert _{H^{2}}^{2})\bigr), \end{aligned}$$
where with the aid of \(\operatorname {curl}\nabla =0\) and \(\operatorname {curl}\mathbf {u}\times \mathbf {n}|_{\partial \Omega }=0\),
$$\begin{aligned}& \int _{\Omega }p'(\rho )^{-1}\mu (\rho )\operatorname {curl}\operatorname {curl}\mathbf {u}\cdot \nabla \operatorname{div} \mathbf {u}\,dx \\& \quad =- \int _{\Omega }\nabla \bigl[p'(\rho )^{-1}\mu (\rho )\bigr]\times \operatorname {curl}\mathbf {u}\cdot \nabla \operatorname{div} \mathbf {u}\,dx \\& \qquad {} + \int _{\partial \Omega }p'(\rho )^{-1}\mu (\rho ) (\mathbf {n}\times \operatorname {curl}\mathbf {u})\cdot \nabla \operatorname{div} \mathbf {u}\,dS \\& \quad \le \eta \Vert \nabla \operatorname {div}\mathbf {u}\Vert _{L^{2}}^{2} +C_{\eta } \Vert \mathbf {u}\Vert _{H^{2}}^{2} \Vert \sigma \Vert _{H^{2}}^{2}. \end{aligned}$$
Putting the above estimates together, we get this lemma. □
Lemma 3.4
For the solution to (8)–(11), we have
$$ \begin{aligned}[b] & \frac{d}{dt}\bigl( \bigl\Vert \sqrt{p'(1)}\sigma _{t} \bigr\Vert _{L^{2}}^{2}+ \Vert \sqrt{\rho } \mathbf {u}_{t} \Vert _{L^{2}}^{2} \bigr)+\gamma _{2} \Vert \mathbf {u}_{t} \Vert _{H^{1}}^{2} \\ &\quad \le \eta \Vert \mathbf {u}\Vert _{H^{1}}^{2}+C_{\eta } \Vert \sigma _{t} \Vert _{H^{1}}^{4} \\ &\qquad {} +C\bigl[ \Vert \sigma _{t} \Vert _{H^{1}}^{2} \bigl( \Vert \mathbf {u}\Vert _{H^{2}}^{2}+ \Vert \sigma \Vert _{H^{2}}^{2}+ \Vert \mathbf {u}_{t} \Vert _{H^{2}}^{2}\bigr) + \Vert \mathbf {u}_{t} \Vert _{H^{1}}^{2} \bigl\Vert (\mathbf {u},\sigma ) \bigr\Vert _{H^{2}}^{2}+ \Vert \sigma \Vert _{H^{2}}^{4} \bigr], \end{aligned} $$
(21)
where \(0<\eta <1\) is to be determined later, and \(\gamma _{2}\) is a positive constant independent of ϵ.
Proof
Applying \(\partial _{t}\) to (8), multiplying the resulting equation by \(p'(1)\sigma _{t}\), integrating in \(L^{2}\), we get
$$ \begin{aligned} &\frac{1}{2} \frac{d}{dt} \bigl\Vert \sqrt{p'(1)}\sigma _{t} \bigr\Vert _{L^{2}}^{2} + \frac{p'(1)}{\epsilon } \int _{\Omega }\sigma _{t}\operatorname {div}\mathbf {u}_{t} \,dx \\ &\quad = -p'(1) \int _{\Omega }(\mathbf {u}\cdot \nabla \sigma _{t}+\mathbf {u}_{t} \cdot \nabla \sigma +\sigma _{t}\operatorname{div} \mathbf {u}+\sigma \operatorname{div} \mathbf {u}_{t})\sigma _{t}\,dx \\ &\quad \le \eta \bigl( \Vert \mathbf {u}_{t} \Vert _{H^{1}}^{2}+ \Vert \mathbf {u}\Vert _{H^{1}}^{2}\bigr)+C_{\eta } \bigl( \Vert \sigma \Vert _{H^{1}}^{4} + \Vert \sigma _{t} \Vert _{H^{1}}^{4}\bigr). \end{aligned} $$
Applying \(\partial _{t}\) to (9), we have
$$\begin{aligned} &\rho (\partial _{tt} \mathbf {u}+ \mathbf {u}\cdot \nabla \mathbf {u}_{t})+ \frac{1}{\epsilon }p'(1)\nabla \sigma _{t} \\ &\quad = \operatorname {div}\bigl(2\mu (\rho ) \mathrm{D}(\mathbf {u}_{t})\bigr)+\nabla \bigl(\lambda ( \rho )\operatorname{div} \mathbf {u}_{t}\bigr) \\ &\qquad {}-\rho _{t}(\mathbf {u}_{t}+ \mathbf {u}\cdot \nabla \mathbf {u}) - \rho \mathbf {u}_{t} \cdot \nabla \mathbf {u}+\partial _{t}\biggl[ \frac{p'(1)-p'(1+\epsilon \sigma )}{\epsilon }\nabla \sigma \biggr] \\ &\qquad {}+\operatorname {div}\bigl(2\partial _{t}\mu (\rho ) \mathrm{D}(\mathbf {u})\bigr)+ \nabla \bigl( \partial _{t}\lambda (\rho )\operatorname{div} \mathbf {u}\bigr). \end{aligned}$$
(22)
Taking \(\langle \text{(22)},\mathbf {u}_{t}\rangle \) and using the boundary conditions (11), we find that
$$\begin{aligned}& \frac{1}{2}\frac{d}{dt} \Vert \sqrt{\rho }\mathbf {u}_{t} \Vert _{L^{2}}^{2}+ \bigl\Vert \sqrt{2 \mu (\rho )+\lambda (\rho )}\operatorname {div}(\mathbf {u}_{t}) \bigr\Vert _{L^{2}}^{2} \\& \qquad {}+ \bigl\Vert \sqrt{\mu (\rho )}\operatorname {curl}\mathbf {u}_{t} \bigr\Vert _{L^{2}}^{2} + \frac{p'(1)}{\epsilon } \int _{\Omega }\nabla \sigma _{t}\cdot \mathbf {u}_{t} \,dx \\& \quad = \int _{\Omega } \biggl[\frac{p'(1)-p'(1+\epsilon \sigma )}{\epsilon } \nabla \sigma \biggr]_{t}\cdot \mathbf {u}_{t}\,dx - \int _{\Omega }\bigl[\epsilon \sigma _{t} (\mathbf {u}\cdot \nabla ) \mathbf {u}+\rho (\mathbf {u}_{t}\cdot \nabla ) \mathbf {u}\bigr] \cdot \mathbf {u}_{t}\,dx \\& \qquad {} - \int _{\Omega }\bigl[\nabla \bigl(2\mu (\rho )+\lambda (\rho ) \bigr)\operatorname {div}\mathbf {u}_{t}+ \nabla \bigl(\mu (\rho )\bigr)\times \operatorname {curl}\mathbf {u}_{t}\bigr]\cdot \mathbf {u}_{t} \,dx \\& \qquad {} - \int _{\Omega }\partial _{t}\bigl(2\mu (\rho )+\lambda (\rho )\bigr) \vert \operatorname {div}\mathbf {u}_{t} \vert ^{2}+ \partial _{t}\bigl(\mu (\rho )\bigr) \vert \operatorname {curl}\mathbf {u}_{t} \vert ^{2}\,dx \\& \qquad {} + \int _{\Omega }\bigl[\operatorname {div}\bigl(2\partial _{t}\mu ( \rho ) \mathrm{D}(\mathbf {u})\bigr)+ \nabla \bigl(\partial _{t}\lambda ( \rho )\operatorname{div} \mathbf {u}\bigr)\bigr]\cdot \mathbf {u}_{t} \,dx \\& \quad \le \eta \Vert \mathbf {u}_{t} \Vert _{H^{1}}^{2} +C_{\eta } \bigl( \Vert \sigma _{t} \Vert _{H^{1}}^{2}\bigl( \bigl\Vert (\mathbf {u},\sigma ) \bigr\Vert _{H^{2}}^{2} + \Vert \mathbf {u}_{t} \Vert _{H^{2}}^{2}\bigr)+ \Vert \mathbf {u}_{t} \Vert _{H^{1}}^{2} \bigl\Vert (\mathbf {u},\sigma ) \bigr\Vert _{H^{2}}^{2} \bigr). \end{aligned}$$
Hence, by choosing η appropriately small and using Korn’s inequality, we obtain the estimate (21). □
Next, we estimate the vorticity of the velocity, which is denoted by \(\omega =\operatorname{curl} \mathbf {u}\). By virtue of (8) and (9), it is easy to see that ω satisfies the following systems:
$$ \begin{aligned} &\rho \omega _{t}+\rho \mathbf {u}\cdot \nabla \omega -\mu (\rho ) \triangle \omega =g, \\ &\omega \times \mathbf {n}=0 \quad \mbox{on } \partial \Omega , \end{aligned} $$
(23)
where
$$ \begin{aligned} g={}&{-}\rho \omega \operatorname{div} \mathbf {u}- \frac{\epsilon }{\rho }\nabla \sigma \times \bigl(\operatorname {div}\bigl(\mu (\rho )\nabla \mathbf {u}\bigr)+\nabla \bigl[\lambda (\rho )\operatorname{div} \mathbf {u}\bigr]\bigr) \\ &{}+\operatorname{curl}\bigl(\mu (\rho )\bigr)\Delta \mathbf {u}+\bigl(\nabla \mu ( \rho )\bigr)\cdot \nabla \operatorname{curl} \mathbf {u}. \end{aligned} $$
Then we have the following.
Lemma 3.5
$$ \frac{d}{dt} \Vert \sqrt{\rho }\omega \Vert _{L^{2}}^{2}+ \bigl\Vert \sqrt{\mu (\rho )} \operatorname {curl}\omega \bigr\Vert _{L^{2}}^{2} \le \eta \Vert \omega \Vert _{L^{2}}^{2}+C_{\eta } \Vert \mathbf {u}\Vert _{H^{2}}^{2}\bigl( \Vert \mathbf {u}\Vert _{H^{2}}^{2}+ \Vert \sigma \Vert _{H^{2}}^{2} \bigr), $$
(24)
where \(0<\eta <1\) is a positive constant which is to be determined.
Proof
Multiplying (23)1 by ω, with the aid of the boundary condition (23)2 we infer that
$$ \frac{1}{2}\frac{d}{dt} \Vert \sqrt{\rho }\omega \Vert _{L^{2}}^{2}+ \bigl\Vert \sqrt{\mu ( \rho )}\operatorname {curl}\omega \bigr\Vert _{L^{2}}^{2} = \int _{\Omega }g\cdot \omega \,dx- \int _{\Omega }\nabla \mu (\rho )\times \operatorname {curl}\omega \cdot \omega \,dx, $$
(25)
where
$$ \begin{aligned} - \int _{\Omega }\mu (\rho )\triangle \omega \cdot \omega \,dx&= \int _{\Omega }\mu (\rho )\operatorname {curl}\operatorname {curl}\omega \cdot \omega \,dx \\ &= \int _{\Omega }\mu (\rho ) \vert \operatorname {curl}\omega \vert ^{2} \,dx+ \int _{\Omega }\nabla \bigl(\mu (\rho )\bigr)\times \operatorname {curl}\omega \cdot \omega \,dx \\ &\quad {} + \int _{\partial \Omega }\mu (\rho ) (\mathbf {n}\times \operatorname {curl}\mathbf {u})\cdot \omega \,dS. \end{aligned} $$
With the aid of Lemma 2.2, it is easy to verify that
$$ \begin{aligned} \int _{\Omega }g\omega \,dx & \le \eta \Vert \omega \Vert _{H^{1}}^{2}+C_{\eta } \Vert \mathbf {u}\Vert _{H^{2}}^{2}\bigl( \Vert \mathbf {u}\Vert _{H^{2}}^{2} + \Vert \sigma \Vert _{H^{2}}^{2} \bigr) \\ & \le \eta \bigl( \Vert \omega \Vert _{L^{2}}^{2}+ \Vert \operatorname {curl}\omega \Vert _{L^{2}}^{2} \bigr)+C_{\eta } \Vert \mathbf {u}\Vert _{H^{2}}^{2} \bigl( \Vert \mathbf {u}\Vert _{H^{2}}^{2} + \Vert \sigma \Vert _{H^{2}}^{2}\bigr). \end{aligned} $$
Using Holder’s inequality and Young’s inequality, we have
$$ \int _{\Omega }\nabla \mu (\rho )\times \operatorname {curl}\omega \cdot \omega \,dx \le \eta \Vert \operatorname {curl}\omega \Vert _{L^{2}}^{2} +C_{\eta } \Vert \sigma \Vert _{H^{2}}^{2} \Vert \mathbf {u}\Vert _{H^{2}}^{2}. $$
Inserting the above two inequalities into (25) and choosing η appropriately small, we get the above lemma. □
Definition 3.1
Now, defining two functions:
$$ \begin{aligned} & \Psi _{1}(t):= \int _{\Omega }\rho \mathbf {u}\cdot \mathbf {u}_{t} \,dx + \bigl\Vert ( \sigma _{t},\mathbf {u}_{t}) \bigr\Vert _{L^{2}}^{2} + \bigl\Vert (\sigma ,\mathbf {u}) \bigr\Vert _{H^{1}}^{2}, \\ & \Phi _{1}(t):= \Vert \sigma _{t} \Vert _{L^{2}}^{2}+ \Vert \mathbf {u}\Vert _{H^{1}}^{2}+ \bigl\Vert (\operatorname {curl}\operatorname {curl}\mathbf {u},\nabla \operatorname {div}\mathbf {u}) \bigr\Vert _{L^{2}}^{2} + \Vert \mathbf {u}_{t} \Vert _{H^{1}}^{2}. \end{aligned} $$
(26)
we conclude from Lemmas 3.1–3.5 that, for small ϵ, there is a positive constant \(C_{1}\), such that
$$\begin{aligned} \frac{d}{dt}\Psi _{1}(t)+\Phi _{1}(t) \le& C_{1}\bigl( \Vert \sigma _{t} \Vert _{H^{1}}^{2} \bigl\Vert (\mathbf {u}_{t},\sigma _{t}) \bigr\Vert _{H^{1}}^{2} + \Vert \sigma \Vert _{H^{2}}^{2}\bigl( \Vert \sigma \Vert _{H^{2}}^{2}+ \Vert \sigma _{t} \Vert _{H^{1}}^{2}\bigr) \\ &{} + \Vert \mathbf {u}\Vert _{H^{2}}^{2}\bigl( \bigl\Vert (\mathbf {u}_{t}, \sigma _{t}) \bigr\Vert _{H^{1}}^{2}+ \bigl\Vert (\sigma ,\mathbf {u}) \bigr\Vert _{H^{2}}^{2}\bigr)\bigr). \end{aligned}$$
(27)
Boundedness of second order derivatives
First, we show the following lemma.
Lemma 3.6
For the solution to (8)–(11), we have
$$ \begin{aligned}[b] &\frac{d}{dt} \bigl\Vert \sqrt{2\mu (\rho )+\lambda (\rho )}\nabla \operatorname{div} \mathbf {u}\bigr\Vert _{L^{2}}^{2}-2\frac{d}{dt} \int _{\Omega }\rho \mathbf {u}_{t} \cdot \nabla \operatorname{div} \mathbf {u}\,dx+ \bigl\Vert \sqrt{p'(1)}\nabla \sigma _{t} \bigr\Vert _{L^{2}}^{2} \\ &\quad \le \eta \Vert \nabla \operatorname{div} \mathbf {u}_{t} \Vert _{L^{2}}^{2} +C_{\eta } \Vert \mathbf {u}_{t} \Vert _{L^{2}}^{2}+C \Vert \nabla \operatorname{div} \mathbf {u}\Vert _{L^{2}}^{2} \\ &\qquad {} +C\bigl[ \Vert \sigma \Vert _{H^{2}}^{2} \bigl\Vert (\sigma _{t},\mathbf {u}_{t}) \bigr\Vert _{H^{1}}^{2} + \Vert \mathbf {u}\Vert _{H^{3}}^{2} \bigl( \bigl\Vert (\mathbf {u},\sigma ) \bigr\Vert _{H^{2}}^{2}+ \bigl\Vert (\sigma _{t}, \mathbf {u}_{t}) \bigr\Vert _{H^{1}}^{2}\bigr)\bigr], \end{aligned} $$
(28)
where \(0<\eta <1\) is a small positive constant which is to be determined.
Proof
Differentiating (9) with respect to t, we have
$$\begin{aligned}& \rho \bigl(\mathbf {u}_{tt}+ (\mathbf {u}\cdot \nabla ) \mathbf {u}_{t}\bigr)+ \frac{1}{\epsilon }p'(1)\nabla \sigma _{t} \\& \quad = \bigl(2\mu (\rho )+\lambda (\rho )\bigr)\nabla \operatorname{div} \mathbf {u}_{t}-\mu (\rho ) \operatorname {curl}\operatorname{curl} \mathbf {u}_{t} \\& \qquad {} -\rho _{t}\bigl(\mathbf {u}_{t}+ (\mathbf {u}\cdot \nabla ) \mathbf {u}\bigr) -\rho \mathbf {u}_{t}\cdot \nabla \mathbf {u}+\biggl[ \frac{p'(1)-p'(1+\epsilon \sigma )}{\epsilon }\nabla \sigma \biggr]_{t} \\& \qquad {} +\bigl(2\mu (\rho )+\lambda (\rho )\bigr)_{t}\nabla \operatorname{div} \mathbf {u}-\mu (\rho )_{t} \operatorname {curl}\operatorname{curl} \mathbf {u} \\& \qquad {} +2\nabla \mu (\rho )\cdot D(u_{t})+\operatorname {div}\mathbf {u}_{t} \nabla \lambda (\rho ) +2\nabla \bigl(\mu (\rho )\bigr)_{t}\cdot D(u)+ \operatorname {div}\mathbf {u}\nabla \bigl(\lambda ( \rho )\bigr)_{t}. \end{aligned}$$
(29)
Multiplying (29) by \(\nabla \operatorname{div} \mathbf {u}\) and integrating in \(L^{2}\), we get
$$\begin{aligned}& \frac{1}{2}\frac{d}{dt} \bigl\Vert \sqrt{2 \mu (\rho )+\lambda (\rho )}\nabla \operatorname{div} \mathbf {u}\bigr\Vert _{L^{2}}^{2} -\frac{d}{dt} \int _{\Omega }\rho \mathbf {u}_{t} \cdot \nabla \operatorname{div} \mathbf {u}\,dx -\frac{p'(1)}{\epsilon } \int _{\Omega }\nabla \sigma _{t}\cdot \nabla \operatorname{div} \mathbf {u}\,dx \\& \quad = \int _{\Omega } \biggl[ \biggl( \frac{p'(1+\epsilon \sigma )-p'(1)}{\epsilon }\nabla \sigma \biggr)_{t} + \epsilon \sigma _{t}\mathbf {u}\cdot \nabla \mathbf {u}+\rho \bigl((\mathbf {u}_{t} \cdot \nabla ) \mathbf {u}+(\mathbf {u}\cdot \nabla ) \mathbf {u}_{t}\bigr) \biggr]\cdot \nabla \operatorname{div} \mathbf {u}\,dx \\& \qquad {} - \int _{\Omega }\rho \mathbf {u}_{t}\cdot \nabla \operatorname{div} \mathbf {u}_{t}\,dx \\& \qquad {} + \int _{\Omega }\biggl[-\frac{1}{2}\bigl(2\mu (\rho )+ \lambda (\rho )\bigr)_{t}\nabla \operatorname{div} \mathbf {u}+\mu (\rho )_{t}\operatorname {curl}\operatorname{curl} \mathbf {u}\biggr]\cdot \nabla \operatorname{div} \mathbf {u}\,dx \\& \qquad {} + \int _{\Omega }\mu (\rho )\operatorname {curl}\operatorname {curl}\mathbf {u}_{t} \cdot \nabla \operatorname {div}\mathbf {u}\,dx \\& \qquad {} - \int _{\Omega }\bigl[2\nabla \mu (\rho )\cdot D(u_{t})+ \nabla \lambda ( \rho )\cdot \nabla \mathbf {u}_{t} +2\nabla \bigl(\mu ( \rho )\bigr)_{t}\cdot D(u)+ \nabla \bigl(\lambda (\rho ) \bigr)_{t}\cdot \nabla \mathbf {u}\bigr] \\& \qquad {}\cdot \nabla \operatorname {div}\mathbf {u}\,dx \\& \quad \le \eta \bigl( \bigl\Vert (\nabla \operatorname{div} \mathbf {u}_{t}, \nabla \operatorname{div} \mathbf {u}, \nabla \sigma _{t}) \bigr\Vert _{L^{2}}^{2}\bigr) +C_{\eta }\bigl( \Vert \mathbf {u}_{t} \Vert _{L^{2}}^{2}+ \Vert \nabla \operatorname {div}\mathbf {u}\Vert _{L^{2}}^{2}\bigr) \\& \qquad {} +C_{\eta }\bigl[ \Vert \sigma \Vert _{H^{2}}^{2} \bigl\Vert (\mathbf {u}_{t},\sigma _{t}) \bigr\Vert _{H^{1}}^{2} + \Vert \mathbf {u}\Vert _{H^{3}}^{2} \bigl( \bigl\Vert (\mathbf {u}_{t},\sigma _{t}) \bigr\Vert _{H^{1}}^{2}+ \Vert \mathbf {u}\Vert _{H^{2}}^{2} \bigr)\bigr], \end{aligned}$$
where we have used the following estimate:
$$\begin{aligned}& \int _{\Omega }\mu (\rho )\operatorname {curl}\operatorname {curl}\mathbf {u}_{t} \cdot \nabla \operatorname {div}\mathbf {u}\,dx \\& \quad \le \biggl\vert \int _{\partial \Omega }\mu (\rho ) (\mathbf {n}\times \operatorname {curl}\mathbf {u}_{t}) \cdot \nabla \operatorname {div}\mathbf {u}\,dS \biggr\vert + \biggl\vert \int _{\Omega }\nabla \mu (\rho )\times \operatorname {curl}\mathbf {u}_{t}\cdot \nabla \operatorname {div}\mathbf {u}\,dx \biggr\vert \\& \quad \leq \eta \Vert \nabla \operatorname {div}\mathbf {u}\Vert _{L^{2}}^{2} +C_{\eta } \Vert \mathbf {u}_{t} \Vert _{H^{1}}^{2} \Vert \sigma \Vert _{H^{2}}^{2}. \end{aligned}$$
Similarly, we take \(\langle \nabla \text{(8)},p'(1)\nabla \sigma _{t}\rangle \) to infer that
$$\begin{aligned}& \bigl\Vert \sqrt{p'(1)}\nabla \sigma _{t} \bigr\Vert _{L^{2}}^{2}+ \frac{p'(1)}{\epsilon } \int _{\Omega }\nabla \sigma _{t}\cdot \nabla \operatorname{div} \mathbf {u}\,dx \\& \quad = -p'(1) \int _{\Omega }\bigl((\mathbf {u}\cdot \nabla )\nabla \sigma +\nabla \mathbf {u}\nabla \sigma +\nabla \sigma \operatorname{div} \mathbf {u}+\sigma \nabla \operatorname{div} \mathbf {u}\bigr)\cdot \nabla \sigma _{t}\,dx \\& \quad \le \eta \Vert \nabla \sigma _{t} \Vert _{L^{2}}^{2}+C_{\eta } \Vert \mathbf {u}\Vert _{H^{2}}^{2} \Vert \sigma \Vert _{H^{2}}^{2}. \end{aligned}$$
Summing up the above inequalities together and choosing η small, we get the above estimate. □
Lemma 3.7
We have
$$\begin{aligned}& \frac{1}{2}\frac{d}{dt} \bigl\Vert \nabla ^{2}\sigma \bigr\Vert _{L^{2}}^{2} + \bigl\Vert \sqrt{p'( \rho )^{-1}}\sqrt{2\mu (\rho )+ \lambda (\rho )}\nabla ^{2}\operatorname{div} \mathbf {u}\bigr\Vert _{L^{2}}^{2} \\& \quad \le \eta \Vert \mathbf {u}\Vert _{H^{3}}^{2}+C_{\eta } \Vert \sigma \Vert _{H^{2}}^{4} +C\bigl( \bigl\Vert \nabla ^{2}\operatorname{curl} \mathbf {u}\bigr\Vert _{L^{2}}^{2}+ \Vert \mathbf {u}_{t} \Vert _{H^{1}}^{2}\bigr) \\& \qquad {} +C\bigl( \Vert \mathbf {u}\Vert _{H^{2}}^{4}+ \Vert \sigma \Vert _{H^{2}}^{2}( \Vert \mathbf {u}_{t} \Vert _{H^{1}}^{2} + \Vert \mathbf {u}\Vert _{H^{3}}^{2})\bigr), \end{aligned}$$
(30)
where \(0<\eta <1\) is a small positive constant which is to be determined.
Proof
We differentiate (8) twice with respect to x to have
$$ \nabla ^{2}\sigma _{t}+\mathbf {u}\cdot \nabla \bigl(\nabla ^{2}\sigma \bigr)+2 \nabla \mathbf {u}\cdot \nabla ( \nabla \sigma ) +\nabla ^{2}\mathbf {u}\cdot \nabla \sigma +\nabla ^{2}(\sigma \operatorname{div} \mathbf {u})+\frac{1}{\epsilon } \nabla ^{2}\operatorname{div} \mathbf {u}=0. $$
(31)
Taking \(\langle \text{(31)},p'(1)\nabla ^{2}\sigma \rangle \), we obtain
$$ \begin{aligned} &\frac{1}{2}\frac{d}{dt} \bigl\Vert \sqrt{p'(1)}\nabla ^{2}\sigma \bigr\Vert _{L^{2}}^{2}+ \frac{p'(1)}{\epsilon } \int _{\Omega }\nabla ^{2}\operatorname{div} \mathbf {u}\nabla ^{2} \sigma \,dx \\ &\quad =-p'(1) \int _{\Omega }\bigl[\bigl(\mathbf {u}\cdot \nabla \bigl(\nabla ^{2}\sigma \bigr)+2 \nabla \mathbf {u}\cdot \nabla (\nabla \sigma ) + \nabla ^{2}\mathbf {u}\cdot \nabla \sigma +\nabla ^{2}(\sigma \operatorname{div} \mathbf {u})\bigr)\bigr]\nabla ^{2} \sigma \,dx \\ &\quad \le \eta \Vert \mathbf {u}\Vert _{H^{3}}^{2}+C_{\eta } \Vert \sigma \Vert _{H^{2}}^{4}. \end{aligned} $$
Then we apply ∇ to (9) to get
$$\begin{aligned}& \bigl(2\mu (\rho )+\lambda (\rho )\bigr)\nabla ^{2} \operatorname{div} \mathbf {u}-\mu ( \rho )\nabla \operatorname {curl}\operatorname{curl} \mathbf {u}- \frac{1}{\epsilon }p'(1)\nabla ^{2} \sigma \\& \quad =\nabla \biggl[\frac{p'(1+\epsilon \sigma )-p'(1)}{\epsilon }\nabla \sigma \biggr] +\rho \bigl(\nabla \mathbf {u}_{t}+\nabla \mathbf {u}\cdot \nabla \mathbf {u}+\mathbf {u}\cdot \nabla ^{2}\mathbf {u}\bigr) +\epsilon \nabla \sigma ( \mathbf {u}_{t}+ \mathbf {u}\cdot \nabla \mathbf {u}) \\& \qquad {} -\nabla \bigl[2\nabla \mu (\rho )\cdot D(u)+\nabla \bigl(\lambda ( \rho ) \bigr)\operatorname {div}\mathbf {u}\bigr] \\& \qquad {}- \nabla \bigl(2\mu (\rho )+\lambda (\rho )\bigr) \nabla \operatorname{div} \mathbf {u}+\nabla \mu (\rho )\operatorname {curl}\operatorname{curl} \mathbf {u}, \end{aligned}$$
which, by multiplying \(\nabla ^{2}\operatorname{div} \mathbf {u}\) in \(L^{2}\), gives
$$ \begin{aligned} & \bigl\Vert \sqrt{2\mu (\rho )+\lambda (\rho )}\nabla ^{2}\operatorname{div} \mathbf {u}\bigr\Vert _{L^{2}}^{2} -\frac{p'(1)}{\epsilon } \int _{\Omega }\nabla ^{2}\operatorname{div} \mathbf {u}\nabla ^{2}\sigma \,dx \\ &\quad \le \eta \bigl\Vert \nabla ^{2}\operatorname{div} \mathbf {u}\bigr\Vert _{L^{2}}^{2} +C_{\eta }\bigl( \bigl\Vert \nabla ^{2}\operatorname{curl} \mathbf {u}\bigr\Vert _{L^{2}}^{2}+ \Vert \mathbf {u}_{t} \Vert _{H^{1}}^{2}\bigr) \\ &\qquad {} +C_{\eta } \bigl[ \Vert \sigma \Vert _{H^{2}}^{2} \bigl( \Vert \sigma \Vert _{H^{2}}^{2}+ \Vert \mathbf {u}\Vert _{H^{3}}^{2}+ \Vert \mathbf {u}_{t} \Vert _{H^{1}}^{2}\bigr)+ \Vert \mathbf {u}\Vert _{H^{2}}^{4} \bigr], \end{aligned} $$
where we have used the fact that \(\|\nabla \operatorname {curl}\operatorname{curl} \mathbf {u}\|_{L^{2}}\le \|\nabla ^{2} \operatorname{curl} \mathbf {u}\|_{L^{2}}\).
Summing up the above two inequalities together and choosing η suitably small, we get the estimate (30). □
Lemma 3.8
For the solution to (8)–(11), we have
$$\begin{aligned}& \frac{d}{dt}\bigl( \Vert \operatorname{div} \mathbf {u}_{t} \Vert _{L^{2}}^{2}+ \bigl\Vert \sqrt{\rho ^{-1}p'(1)} \nabla \sigma _{t} \bigr\Vert _{L^{2}}^{2}\bigr) + \bigl\Vert \sqrt{\rho ^{-1}}\sqrt{2\mu ( \rho )+\lambda (\rho )}\nabla \operatorname{div} \mathbf {u}_{t} \bigr\Vert _{L^{2}}^{2} \\& \quad \le \eta \bigl( \Vert \mathbf {u}_{t} \Vert _{H^{2}}^{2}+ \Vert \mathbf {u}\Vert _{H^{3}}^{2}\bigr)+C \Vert \nabla \operatorname {curl}\mathbf {u}_{t} \Vert _{L^{2}}^{2} +C_{\eta } \bigl( \Vert \sigma \Vert _{H^{2}}^{2} \Vert \mathbf {u}_{t} \Vert _{H^{2}}^{2}+ \Vert \mathbf {u}_{t} \Vert _{H^{1}}^{2} \Vert \mathbf {u}\Vert _{H^{2}}^{2} \\& \qquad {} + \Vert \mathbf {u}\Vert _{H^{2}}^{4}+ \Vert \sigma _{t} \Vert _{H^{1}}^{2}( \Vert \mathbf {u}_{t} \Vert _{H^{2}}^{2}+ \Vert \mathbf {u}\Vert _{H^{3}}^{2}+ \Vert \sigma \Vert _{H^{2}}^{2}+ \Vert \sigma _{t} \Vert _{H^{1}}^{2})\bigr), \quad 0< \eta < 1, \end{aligned}$$
where η is a small positive constant which is to be determined.
Proof
It is obvious that \(\mathbf {u}_{tt}\cdot \mathrm{n}|_{\partial \Omega }=0\), thus
$$ \int _{\Omega }\mathbf {u}_{tt}\nabla \operatorname {div}\mathbf {u}_{t} \,dx=-\frac{1}{2} \frac{d}{dt} \int _{\Omega } \vert \operatorname {div}\mathbf {u}_{t} \vert ^{2} \,dx. $$
We take \(\langle \text{(29)},\rho ^{-1}\nabla \operatorname{div} \mathbf {u}_{t}\rangle \) and \(\langle \partial _{t}\nabla \text{(8)},\rho ^{-1}p'(1)\nabla \sigma _{t} \rangle \), summing up the resulting equations to obtain the above lemma. Here we use the following estimate:
$$\begin{aligned}& \frac{1}{2} \int _{\Omega }\partial _{t}\bigl(\rho ^{-1}p'(1)\bigr) \vert \nabla \sigma _{t} \vert ^{2} \,dx - \int _{\Omega }\rho ^{-1}p'(1)\nabla \sigma _{t} \cdot \mathbf {u}\cdot \nabla ^{2}\sigma _{t} \,dx \\& \quad = \frac{1}{2} \int _{\Omega }\partial _{t}\bigl(\rho ^{-1}p'(1)\bigr) \vert \nabla \sigma _{t} \vert ^{2} \,dx +\frac{1}{2} \int _{\Omega }\operatorname {div}\bigl(\rho ^{-1}p'(1) \mathbf {u}\bigr) \vert \nabla \sigma _{t} \vert ^{2} \,dx \\& \quad = -\frac{1}{2} \int _{\Omega }\rho _{t}\rho ^{-2}p'(1) \vert \nabla \sigma _{t} \vert ^{2} \,dx + \frac{1}{2} \int _{\Omega }\rho ^{-2}p'(1)\operatorname {div}( \rho \mathbf {u}) \vert \nabla \sigma _{t} \vert ^{2} \,dx \\& \quad = -\frac{1}{2} \int _{\Omega }\bigl[\rho _{t}+\operatorname {div}(\rho \mathbf {u}) \bigr]\rho ^{-2}p'(1) \vert \nabla \sigma _{t} \vert ^{2} \,dx \\& \quad = 0. \end{aligned}$$
□
Next, we estimate the derivatives of curlu.
Lemma 3.9
$$\begin{aligned}& \frac{d}{dt} \Vert \sqrt{\rho }\omega _{t} \Vert _{L^{2}}^{2}+\gamma _{3} \Vert \operatorname {curl}\omega _{t} \Vert _{L^{2}}^{2} \\& \quad \le \eta \Vert \omega _{t} \Vert _{L^{2}}^{2}+C_{\eta } \bigl(\bigl( \Vert \sigma _{t} \Vert _{H^{1}}^{2}+ \Vert \sigma \Vert _{H^{2}}^{2}\bigr) \bigl( \Vert \mathbf {u}_{t} \Vert _{H^{1}}^{2}+ \Vert \mathbf {u}\Vert _{H^{3}}^{2}\bigr)+ \Vert \mathbf {u}_{t} \Vert _{H^{1}}^{2} \Vert \mathbf {u}\Vert _{H^{3}}^{2}\bigr) \\& \qquad {}+C \Vert \sigma \Vert _{H^{2}}^{4} \Vert \mathbf {u}\Vert _{H^{2}}^{2}. \end{aligned}$$
where \(\gamma _{3}>0\) is a positive constant and \(0<\eta <1\) is a small positive constant which is to be determined.
Proof
Firstly, we apply \(\partial _{t}\) to (23)1 to see that
$$ \rho (\omega _{tt}+ \mathbf {u}\cdot \nabla \omega _{t})-\mu (\rho ) \triangle \omega _{t}=h, $$
(32)
where
$$ h:=-\epsilon \sigma _{t}(\omega _{t}+\mathbf {u}\cdot \nabla \omega )- \rho \mathbf {u}_{t}\nabla \omega +\partial _{t}\bigl(\mu (\rho )\bigr)\triangle \omega +g_{t}, $$
with
$$ \begin{aligned} \vert g_{t} \vert \le{} &C\bigl( \epsilon \vert \sigma _{t} \vert \vert \nabla u \vert ^{2}+ \vert \nabla u_{t} \vert \vert \nabla u \vert +\epsilon ^{2} \vert \sigma _{t} \vert \vert \nabla \sigma \vert \bigl( \vert \nabla \sigma \vert \vert \nabla \mathbf {u}\vert + \bigl\vert \nabla ^{2} u \bigr\vert \bigr) \\ &{}+ \vert \nabla \sigma _{t} \vert \bigl( \vert \nabla \sigma \vert \vert \nabla \mathbf {u}\vert + \bigl\vert \nabla ^{2} u \bigr\vert \bigr) +\epsilon \vert \nabla \sigma \vert \bigl( \vert \nabla \sigma _{t} \vert \vert \nabla \mathbf {u}\vert + \epsilon \vert \nabla \sigma \vert \vert \nabla u_{t} \vert + \bigl\vert \nabla ^{2} u_{t} \bigr\vert \bigr) \\ &{}+\epsilon \bigl( \vert \nabla \sigma _{t} \vert \bigl\vert \nabla ^{2} u \bigr\vert + \vert \nabla \sigma \vert \bigl\vert \nabla ^{2} u_{t} \bigr\vert \bigr)\bigr). \end{aligned} $$
Obviously, the boundary condition for (32) reads
$$ \omega _{t}\times \mathbf {n}=0\quad \mbox{on } \partial \Omega . $$
(33)
Therefore, by virtue of (33) and integration by parts,
$$\begin{aligned}& - \int _{\Omega }\mu (\rho )\Delta \omega _{t}\cdot \omega _{t} \,dx \\& \quad = \int _{\Omega }\mu (\rho )\operatorname {curl}\operatorname {curl}\omega _{t} \cdot \omega _{t} \,dx \\& \quad = \int _{\Omega }\mu (\rho ) \vert \operatorname {curl}\omega _{t} \vert ^{2} \,dx+ \int _{\Omega }\nabla \mu (\rho )\times \operatorname {curl}\omega _{t}\cdot \omega _{t} \,dx + \int _{\partial \Omega }\mu (\rho )\operatorname {curl}\omega _{t}(\omega _{t} \times \mathbf{n}) \,dS. \end{aligned}$$
Multiplying (32) by \(\omega _{t}\) in \(L^{2}\), we obtain
$$ \begin{aligned}[b] &\frac{1}{2} \frac{d}{dt} \Vert \sqrt{\rho }\omega _{t} \Vert _{L^{2}}^{2}+ \bigl\Vert \sqrt{ \mu (\rho )}\operatorname {curl}\omega _{t} \bigr\Vert _{L^{2}}^{2} \\ &\quad = \int _{\Omega }h\cdot \omega _{t} \,dx - \int _{\Omega }\nabla \mu ( \rho )\times \nabla \omega _{t}\cdot \omega _{t} \,dx \\ &\quad \le \Vert \omega _{t} \Vert _{H^{1}}\bigl[ \Vert g_{t} \Vert _{L^{2}}+ \Vert \sigma _{t} \Vert _{H^{1}}\bigl( \Vert \mathbf {u}_{t} \Vert _{H^{2}}+ \Vert \mathbf {u}\Vert _{H^{2}} \Vert \mathbf {u}\Vert _{H^{3}}\bigr) \\ &\qquad {}+ \Vert \mathbf {u}_{t} \Vert _{H^{1}} \Vert \mathbf {u}\Vert _{H^{3}}+ \Vert \sigma _{t} \Vert _{H^{1}} \Vert \mathbf {u}\Vert _{H^{3}}+ \Vert \sigma \Vert _{H^{2}} \Vert \mathbf {u}_{t} \Vert _{H^{2}}\bigr]. \end{aligned} $$
(34)
□
With the aid of Lemma 2.2 and (33), we have
$$ \Vert \omega _{t} \Vert _{H^{1}}^{2}\leq C \bigl( \Vert \operatorname {curl}\omega _{t} \Vert _{L^{2}}^{2}+ \Vert \omega _{t} \Vert _{L^{2}}^{2}\bigr). $$
Using Young’s inequality and the above equality, we obtain the following lemma.
Lemma 3.10
For the solution to (8)–(11), we have
$$ \begin{aligned} &\frac{d}{dt} \bigl\Vert \sqrt{\mu (\rho )}\nabla \omega \bigr\Vert _{L^{2}}^{2}+ \Vert \sqrt{ \rho }\omega _{t} \Vert _{L^{2}}^{2} +\frac{1}{20} \bigl\Vert \sqrt{\mu (\rho )} \triangle \omega \bigr\Vert _{L^{2}}^{2} \\ &\quad \le \eta \bigl( \Vert \omega _{t} \Vert _{L^{2}}^{2}+ \Vert \operatorname {curl}\omega \Vert _{L^{2}}^{2}\bigr) +C_{\eta } \bigl( \Vert \mathbf {u}\Vert _{H^{3}}^{2} \bigl( \Vert \mathbf {u}\Vert _{H^{2}}^{2}+ \Vert \sigma \Vert _{H^{2}}^{2}+ \Vert \sigma _{t} \Vert _{H^{1}}^{2}\bigr) + \Vert \mathbf {u}_{t} \Vert _{H^{2}}^{2} \Vert \sigma \Vert _{H^{2}}^{2} \bigr), \end{aligned} $$
where \(0<\eta <1\) is a small positive constant which is to be determined.
Proof
We take \(\langle \text{(23)}_{1},\omega _{t}-\delta \triangle \omega \rangle \) (in which δ is a positive constant to be determined later) to get
$$\begin{aligned}& \frac{1}{2}\frac{d}{dt} \bigl\Vert \sqrt{\delta \rho + \mu (\rho )}\operatorname {curl}\omega \bigr\Vert _{L^{2}}^{2} + \Vert \sqrt{\rho }\omega _{t} \Vert _{L^{2}}^{2}+ \delta \bigl\Vert \sqrt{\mu (\rho )}\triangle \omega \bigr\Vert _{L^{2}}^{2} \\& \quad = \int _{\Omega }g\cdot (\omega _{t}-\delta \triangle \omega ) \,dx + \frac{1}{2} \int _{\Omega }\partial _{t}\bigl(\delta \rho +\mu ( \rho )\bigr) \vert \operatorname {curl}\omega \vert ^{2}\,dx \\& \qquad {} + \int _{\Omega }\nabla \bigl(\delta \rho +\mu (\rho )\bigr)\times \operatorname {curl}\omega \cdot \omega _{t} \,dx+ \int _{\Omega }\rho \mathbf {u}\cdot \nabla \omega \cdot (\omega _{t}-\delta \triangle \omega ) \,dx \\& \quad \le \eta \bigl( \Vert \omega _{t} \Vert _{L^{2}}^{2}+ \Vert \triangle \omega \Vert _{L^{2}}^{2}\bigr) +C_{\eta }\bigl( \Vert \mathbf {u}\Vert _{H^{2}}^{4}+ \Vert \sigma \Vert _{H^{2}}^{2}\bigl( \Vert \mathbf {u}\Vert _{H^{3}}^{2}+ \Vert \mathbf {u}_{t} \Vert _{H^{1}}^{2}\bigr)\bigr) \\& \qquad {} +\eta \Vert \operatorname {curl}\omega \Vert _{L^{2}}^{2}+C_{\eta } \bigl( \Vert \sigma _{t} \Vert _{H^{1}}^{2} \Vert \mathbf {u}\Vert _{H^{3}}^{2} + \Vert \sigma \Vert _{H^{2}}^{2} \Vert \mathbf {u}_{t} \Vert _{H^{2}}^{2}\bigr), \end{aligned}$$
where we use the following estimate:
$$\begin{aligned}& - \int _{\Omega }\bigl(\delta \rho +\mu (\rho )\bigr)\Delta \omega \cdot \omega _{t} \,dx \\& \quad = \int _{\Omega }\bigl(\delta \rho +\mu (\rho )\bigr)\operatorname {curl}\operatorname {curl}\omega \cdot \omega _{t} \,dx \\& \quad = \int _{\Omega }\bigl(\delta \rho +\mu (\rho )\bigr)\operatorname {curl}\omega \cdot \operatorname {curl}\omega _{t} \,dx+ \int _{\Omega }\nabla \bigl(\delta \rho +\mu ( \rho )\bigr)\times \operatorname {curl}\omega \cdot \omega _{t} \,dx \\& \qquad {} + \int _{\partial \Omega }\bigl(\delta \rho +\mu (\rho )\bigr)\operatorname {curl}\omega \cdot (\omega _{t}\times \mathbf {n}) \,dS \\& \quad = \frac{1}{2}\frac{d}{dt} \bigl\Vert \sqrt{\delta \rho +\mu (\rho )}\operatorname {curl}\omega \bigr\Vert ^{2}_{L^{2}} - \frac{1}{2} \int _{\Omega }\partial _{t}\bigl(\delta \rho +\mu ( \rho )\bigr) \vert \operatorname {curl}\omega \vert ^{2}\,dx \\& \qquad {} + \int _{\Omega }\nabla \bigl(\delta \rho +\mu (\rho )\bigr)\times \operatorname {curl}\omega \cdot \omega _{t} \,dx. \end{aligned}$$
Thus, we choose δ and η suitably small to conclude the lemma.
In order to close the estimates, we have to estimate \(\|\sigma \|_{H^{2}}\). To this end, we obtain from the continuity equation (8) and the boundary condition \(\mathbf {u}\cdot \mathbf {n}|_{\partial _{\Omega }}=0\)
$$ \frac{d}{dt} \int _{\Omega }\sigma \,dx=- \int _{\partial \Omega }\biggl( \sigma +\frac{1}{\epsilon }\biggr)\mathbf {u}\cdot \mathbf {n}\,dS=0, $$
thus
$$ \int _{\Omega }\sigma \,dx= \int _{\Omega }\sigma _{0} \,dx=0. $$
From Eqs. (9) and Poincare’s inequality, we have
$$\begin{aligned} \Vert \sigma \Vert _{H^{2}}^{2}&\le C \Vert \nabla \sigma \Vert _{H^{1}}^{2} \\ &\le C\epsilon ^{2}\bigl( \Vert \mathbf {u}_{t} \Vert _{H^{1}}^{2}+ \Vert \mathbf {u}\Vert _{H^{3}}^{2} \bigr) \\ &\quad {}+C\epsilon \bigl[ \Vert \sigma \Vert _{H^{2}}^{2} \bigl( \Vert \mathbf {u}_{t} \Vert _{H^{1}}^{2}+ \Vert \mathbf {u}\Vert _{H^{3}}^{2}+ \Vert \mathbf {u}\Vert _{H^{2}}^{4}\bigr)+ \Vert \mathbf {u}\Vert _{H^{2}}^{4}\bigr]. \end{aligned}$$
(35)
In addition, in order to control the terms \(\|\mathbf {u}_{t}\|_{H^{2}}\) and \(\|\mathbf {u}\|_{H^{3}}\), we use the following fact which is obtained from Lemmas 2.1–2.2 and the boundary condition (11):
$$\begin{aligned}& \Vert \mathbf {u}\Vert _{H^{3}} \leq C\bigl( \Vert \operatorname {div}u \Vert _{H^{2}}+ \Vert \operatorname {curl}\mathbf {u}\Vert _{H^{2}}+ \Vert \mathbf {u}\Vert _{H^{2}}\bigr), \\& \Vert \operatorname {curl}\mathbf {u}\Vert _{H^{2}} \leq C\bigl( \Vert \operatorname {curl}\operatorname {curl}\mathbf {u}\Vert _{H^{1}}+ \Vert \operatorname {curl}\mathbf {u}\Vert _{H^{1}}\bigr), \\& \Vert \operatorname {curl}\operatorname {curl}\mathbf {u}\Vert _{H^{1}} \leq C\bigl( \Vert \triangle \operatorname {curl}u \Vert _{L^{2}}+ \Vert \operatorname {curl}\operatorname {curl}\mathbf {u}\times \mathbf {n}\Vert _{H^{1/2}( \partial \Omega )}+ \Vert \operatorname {curl}\mathbf {u}\Vert _{L^{2}}\bigr), \\& \Vert \mathbf {u}_{t} \Vert _{H^{2}} \leq C\bigl( \Vert \operatorname {div}u_{t} \Vert _{H^{1}}+ \Vert \operatorname {curl}\mathbf {u}_{t} \Vert _{H^{1}}+ \Vert \mathbf {u}_{t} \Vert _{H^{1}}\bigr), \\& \Vert \operatorname {curl}\mathbf {u}_{t} \Vert _{H^{1}} \leq C\bigl( \Vert \operatorname {curl}\operatorname {curl}\mathbf {u}_{t} \Vert _{L^{2}}+ \Vert \operatorname {curl}\mathbf {u}_{t} \Vert _{L^{2}}\bigr), \end{aligned}$$
(36)
where the estimate of the term \(\|\operatorname {curl}\operatorname {curl}\mathbf {u}\times \mathbf {n}\|_{H^{1/2}(\partial \Omega )}\) is crucial for the proof. We can estimate it by the strategy in [22] as follows: In order to derive an estimate near the boundary, we firstly construct the local coordinates by the isothermal coordinates \(\lambda (\psi ,\varphi )\), where \(\lambda (\psi ,\varphi )\) satisfies
$$ \lambda _{\psi }\cdot \lambda _{\psi }>0, \qquad \lambda _{\varphi }\cdot \lambda _{\varphi }>0 \quad \mbox{and}\quad \lambda _{\psi }\cdot \lambda _{\varphi }=0. $$
We cover the boundary ∂Ω by a finite number of bounded open sets \(W^{k}\subset \mathcal{R}^{3}\), \(k=1,2,\ldots,L\), such that, for any \(x \in W^{k}\cap \Omega \),
$$ x = \lambda ^{k}(\psi ,\varphi )+rn\bigl(\lambda ^{k}( \psi ,\varphi )\bigr) = \Lambda ^{k}(\psi ,\varphi ,r), $$
where \(\lambda ^{k}(\psi ,\varphi )\) is the isothermal coordinate and n is the unit outer normal to ∂Ω. For simplicity, we will omit the superscript k in each \(W^{k}\). Then we construct the orthonormal system corresponding to the local coordinates by
$$ e_{1} = \frac{\lambda _{\psi }}{ \vert \lambda _{\psi } \vert },\qquad e_{2} = \frac{\lambda _{\varphi }}{ \vert \lambda _{\varphi } \vert },\qquad e_{3} = n(\lambda ) = e_{1} \times e_{2}. $$
By direct calculations, we can use the fact that \(J \in C^{2}\) and
$$ \begin{aligned} J &= \det \operatorname{Jac}\Lambda = (\Lambda _{\psi }\times \Lambda _{\varphi })\cdot e_{3} \\ &= \vert \lambda _{\psi } \vert \vert \lambda _{\varphi } \vert +r\bigl( \vert \lambda _{\psi } \vert n_{\varphi }\cdot e_{2}+ \vert \lambda _{\varphi } \vert n_{\psi } \cdot e_{1}\bigr)+r^{2}\bigl[(n_{\psi }\cdot e_{1}) (n_{\varphi }\cdot e_{2})-(n_{\psi } \cdot e_{2}) (n_{\varphi }\cdot e_{1})\bigr] > 0, \end{aligned} $$
for sufficiently small \(r>0\). Furthermore, we can derive some other relations:
$$\begin{aligned}& \operatorname{Jac}\bigl(\Lambda ^{-1}\bigr) = (\operatorname{Jac}\Lambda )^{-1}, \\& \bigl[\nabla \bigl(\Lambda ^{-1}\bigr)^{1}\bigr]\circ \Lambda = \frac{1}{J}(\Lambda _{\psi }\times e_{3}), \\& \bigl[\nabla \bigl(\Lambda ^{-1}\bigr)^{2}\bigr]\circ \Lambda = \frac{1}{J}(e_{3} \times \Lambda _{\varphi }), \\& \bigl[\nabla \bigl(\Lambda ^{-1}\bigr)^{3}\bigr]\circ \Lambda = \frac{1}{J}(\Lambda _{\varphi }\times \Lambda _{\psi }), \end{aligned}$$
where the notation ‘∘’ is the composite of operators. Set \(y:=(y_{1},y_{2},y_{3}):=(\psi ,\varphi ,r)\), \(a_{ij} = ((\operatorname{Jac}\Lambda )^{-1})_{ij}\). Then \(n = (a_{31},a_{32},a_{33})\), the tangential directions \(\tau _{i} = (a_{i1},a_{i2},a_{i3})\) (\(i=1,2\)), and
$$ a_{ij}a_{3j}=0,\quad \mbox{for } i=1,2. $$
We denote by \(D_{i}\) the partial derivative with respect to \(y_{i}\) in local coordinates. To be precise, \(D_{3}\) is the normal derivative and \(D_{i}\) for \(i=1,2\) are the tangential derivatives in the original coordinates. Moreover, we have
$$ \partial _{x_{j}}=a_{kj}D_{k}. $$
Next, we denote the vorticity near the boundary as \(\widetilde{w}:=(\widetilde{w}_{1},\widetilde{w}_{2},\widetilde{w}_{3})^{t}:=w(t, \Lambda (y))\). So we get
$$ \begin{aligned} \operatorname{curl}w\cdot n ={}& (a_{k2}D_{k} \widetilde{w}_{3}-a_{k3}D_{k} \widetilde{w}_{2},a_{k3}D_{k} \widetilde{w}_{1}-a_{k1}D_{k} \widetilde{w}_{3},a_{k1}D_{k} \widetilde{w}_{2}-a_{k2}D_{k} \widetilde{w}_{1}) \\ &{}\cdot (a_{31},a_{32},a_{33}) \\ ={}& \bigl[(a_{32}a_{13}-a_{33}a_{12})D_{1}+(a_{32}a_{23}-a_{33}a_{22})D_{2} \bigr] \widetilde{w}_{1} \\ &{}+\bigl[(a_{33}a_{11}-a_{31}a_{13})D_{1}+(a_{33}a_{21}-a_{31}a_{23})D_{2} \bigr] \widetilde{w}_{2} \\ &{}+\bigl[(a_{31}a_{12}-a_{32}a_{11})D_{1}+(a_{31}a_{22}-a_{32}a_{21})D_{2} \bigr] \widetilde{w}_{3} \\ ={}& \sum_{i=1}^{2}(n\times \tau _{i})\cdot D_{i}\widetilde{w} \\ ={}& \sum_{i=1}^{2}\bigl(D_{i} \bigl((n\times \tau _{i})\cdot \widetilde{w}\bigr)-D_{i}(n \times \tau _{i})\cdot \widetilde{w}\bigr) \\ ={}&\sum_{i=1}^{2}\bigl(D_{i} \bigl((n\times \widetilde{w})\cdot \tau _{i}\bigr)-D_{i}(n \times \tau _{i})\cdot \widetilde{w}\bigr). \end{aligned} $$
Thus, with the boundary condition (11) we get the estimate
$$ \begin{aligned}[b] \Vert \operatorname{curl}w\cdot n \Vert _{H^{\frac{1}{2}}(\partial \Omega )} &\leq C \sum_{i=1}^{3} \Vert w_{j} \Vert _{H^{\frac{1}{2}}(\partial \Omega )} \\ &\le C \Vert u \Vert _{H^{2}}. \end{aligned} $$
(37)
□
Definition 3.2
We define
$$ \Psi =\sum_{i=1}^{2} \Psi _{i}(t),\qquad \Phi =\sum_{i=1}^{2} \Phi _{i}(t), $$
(38)
where \(\Psi _{1}(t)\) and \(\Phi _{1}(t)\) are defined by (26), and
$$\begin{aligned}& \begin{aligned} \Psi _{2}(t)&= \Vert \nabla \operatorname{div} \mathbf {u}\Vert _{L^{2}}^{2}-2 \int _{\Omega }\rho \mathbf {u}_{t}\cdot \nabla \operatorname{div} \mathbf {u}\,dx + \bigl\Vert \nabla ^{2}\sigma \bigr\Vert _{L^{2}}^{2} + \Vert \operatorname{div} \mathbf {u}_{t} \Vert _{L^{2}}^{2} \\ &\quad {} + \Vert \nabla \sigma _{t} \Vert _{L^{2}}^{2} + \Vert \nabla \operatorname{curl} \mathbf {u}\Vert _{L^{2}}^{2}+ \Vert \operatorname{curl} \mathbf {u}_{t} \Vert _{L^{2}}^{2}, \end{aligned} \\& \begin{aligned} \Phi _{2}(t)& =\bigl( \bigl\Vert \nabla ^{2} \operatorname{div} \mathbf {u}\bigr\Vert _{L^{2}}^{2}+ \Vert \nabla \operatorname{div} \mathbf {u}_{t} \Vert _{L^{2}}^{2} \bigr) + \Vert \nabla \sigma _{t} \Vert _{L^{2}}^{2}+ \Vert \nabla \operatorname{curl} \mathbf {u}_{t} \Vert _{L^{2}}^{2} \\ &\quad {} + \bigl\Vert \nabla ^{2}\operatorname{curl} \mathbf {u}\bigr\Vert _{L^{2}}^{2}+ \Vert \sigma \Vert _{H^{2}}^{2}. \end{aligned} \end{aligned}$$
Combining Lemmas 3.1–3.10 with the estimates (35)–(37) and choosing a suitable constant C, and small enough constants ϵ and η, we finally conclude that
$$ \frac{d}{dt}\Psi (t)+\Phi (t)\le c_{0}\Phi (t) \bigl(\Psi (t)+\Psi ^{2}(t) \bigr), $$
(39)
where \(c_{0}\geq 1\) is a constant independent of ϵ.
Now, employing (39), and following the analysis in [25], we obtain the following uniform estimate.
Lemma 3.11
(Uniform estimate)
Let \(\Omega \subset \mathcal{R}^{3}\) be a simply connected, bounded domain with smooth boundary ∂Ω. Let \((u,\sigma )\) be a solution to (8)–(11) in \(\Omega \times (0,T)\) with \(c^{-1} \le 1+\epsilon \sigma \le c\) for some \(c>1\), \(\forall (x,t)\in \Omega \times (0,T)\), \(\epsilon \in (0,\epsilon _{1}]\). Suppose that
$$ \Psi (0) \le \beta /(2c_{0}), \quad \beta \in \biggl(0,\frac{1}{2}\biggr]. $$
Then we have
$$ \Psi (t) \le \beta /(2c_{0}),\quad t\in [0,T]. $$