设(σ , u φ )∈X M T ])为方程组(10)的局部解，将方程组(1)变形如下。
(10) $\left\{ {\begin{array}{*{20}{l}}{{\sigma _t} + \bar \rho {\mathop{\rm div}\nolimits} {\boldsymbol{u}} = {g_1}}\\{{{\boldsymbol{u}}_t} - \frac{2}{{\bar \rho }}{\mathop{\rm div}\nolimits} [\nu (\phi ){\boldsymbol{D}}({\boldsymbol{u}})] - \frac{1}{{\bar \rho }}\mathit{\boldsymbol{ \boldsymbol{\nabla}}} [(\nu (\phi ) + }\\{\quad \quad \lambda (\phi )){\mathop{\rm div}\nolimits} {\boldsymbol{u}}] + \frac{{{p^\prime }(\bar \rho )}}{{\bar \rho }}\mathit{\boldsymbol{ \boldsymbol{\nabla}}} \sigma + \frac{1}{{\bar \rho }}\mathit{\boldsymbol{ \boldsymbol{\nabla}}} \phi \mathit{\boldsymbol{ \boldsymbol{\Delta}}} \phi = {g_2}}\\{\rho {\phi _t} + \rho {\boldsymbol{u}} \cdot \mathit{\boldsymbol{ \boldsymbol{\nabla}}} \phi = \mathit{\boldsymbol{ \boldsymbol{\Delta}}} \mu }\\{\rho \mu = \rho \left( {{\phi ^3} - \phi } \right) - \mathit{\boldsymbol{ \boldsymbol{\Delta}}} \phi }\end{array}} \right.$
$\left\{\begin{aligned}g_1= & -\operatorname{div}(\sigma \boldsymbol{u}) \\g_2= & -(\boldsymbol{u} \cdot \mathit{\boldsymbol{ \boldsymbol{\nabla}}}) \boldsymbol{u}+h_1(\sigma) \mathit{\boldsymbol{ \boldsymbol{\nabla}}} \sigma-2 h_2(\sigma) \cdot \\& \operatorname{div}[ \nu(\phi) \boldsymbol{D}(\boldsymbol{u})]-h_2(\sigma) \mathit{\boldsymbol{ \boldsymbol{\nabla}}}[(\nu(\phi)+ \\& \lambda(\phi)) \operatorname{div} \boldsymbol{u}-\mathit{\boldsymbol{ \boldsymbol{\nabla}}} \phi \mathit{\boldsymbol{ \boldsymbol{\Delta}}} \phi]\end{aligned}\right.$
其中，${h_1}(\sigma ) = \frac{{{p^\prime }(\bar \rho )}}{{\bar \rho }} - \frac{{{p^\prime }(\rho )}}{\rho }, {h_2}(\sigma ) = \frac{1}{{\bar \rho }} - \frac{1}{\rho }$
由解空间(9)，结合Sobolev嵌入定理可得, 存在M 0 ＞0足够小，使得$\forall$ M ＜M 0 ，从而有
(11) $0<\frac{\bar{\rho}}{2} \leqslant \rho(x, t) \leqslant 2 \bar{\rho}$
命题2 假设(σ 0 , u 0 )∈H 3 , φ 0 2 -1∈L 2 , $\mathit{\boldsymbol{ \boldsymbol{\nabla}}} \phi_0$ H 2 ，则存在一个只依赖于初值和T 的常数C ，使得
(12) $\begin{aligned}& \|(\sigma, \boldsymbol{u})(t)\|_{H^3}^2+\|\mathit{\boldsymbol{ \boldsymbol{\nabla}}} \phi(t)\|_{H^2}^2+ \\& \left\|\phi^2(t)-1\right\|_{L^2}^2+\int_0^t\left(\|\sigma\|_{H^3}^2+\|\mathit{\boldsymbol{ \boldsymbol{\nabla}}} \boldsymbol{u}\|_{H^3}^2+\right. \\& \left.\|\mathit{\boldsymbol{ \boldsymbol{\nabla}}} \phi\|_{H^4}^2\right) \mathrm{d} \tau \leqslant C\left(\left\|\left(\sigma_0, \boldsymbol{u}_0\right)\right\|_{H^3}^2+\left\|\mathit{\boldsymbol{ \boldsymbol{\nabla}}} \phi_0\right\|_{H^2}^2+\right. \\& \left.\left\|\phi_0^2-1\right\|_{L^2}^2\right)\end{aligned}$
引理1 若(σ , u φ )∈X M T ])是方程(10)的局部解，则
(13) $\begin{array}{r}\left\|\left(\sigma, \boldsymbol{u}, \phi^2-1, \mathit{\boldsymbol{ \boldsymbol{\nabla}}} \phi\right)(t)\right\|^2+\int_0^t \|(\mathit{\boldsymbol{ \boldsymbol{\nabla}}} \mu, \mathit{\boldsymbol{ \boldsymbol{\nabla}}} \boldsymbol{u}, \\\mathit{\boldsymbol{ \boldsymbol{\Delta}}} \phi, \mathit{\boldsymbol{ \boldsymbol{\nabla}}} \phi)\left\|^2 \mathrm{~d} \tau \leqslant\right\|\left(\sigma, \boldsymbol{u}, \phi^2-1, \mathit{\boldsymbol{ \boldsymbol{\nabla}}} \phi\right)(0) \|^2\end{array}$
(14) $\|\phi\|_{L^{\infty}} \leqslant C$
(15) $G(\rho)=\rho \int_{\bar{\rho}}^\rho \frac{p(\eta)-p(\bar{\rho})}{\eta^2} \mathrm{~d} \eta, \rho>0$
(16) $\begin{gathered}\rho G^{\prime}(\rho)=G(\rho)+(p(\rho)-p(\bar{\rho})), \rho G^{\prime \prime}(\rho)=p^{\prime}(\rho) \\G(\rho)_t+\operatorname{div}(G(\rho) \boldsymbol{u})+(p(\rho)-p(\bar{\rho})) \operatorname{div} \boldsymbol{u}=0\end{gathered}$
方程组(1)中第二式乘以u
(17) $\begin{aligned}& \quad \frac{\mathrm{d}}{\mathrm{d} t} \int\left(\frac{1}{2} \rho \boldsymbol{u}^2+G(\rho)\right) \mathrm{d} x+\frac{1}{2} \int \nu(\phi) \cdot\\& \left|\mathit{\boldsymbol{ \boldsymbol{\nabla}}} \boldsymbol{u}+\mathit{\boldsymbol{ \boldsymbol{\nabla}}}^{\mathrm{T}} \boldsymbol{u}\right|^2 \mathrm{~d} \boldsymbol{x}+\int \lambda(\phi)|\operatorname{div} \boldsymbol{u}|^2 \mathrm{~d} \boldsymbol{x}+\int \boldsymbol{u} \cdot \mathit{\boldsymbol{ \boldsymbol{\nabla}}} \phi \mathit{\boldsymbol{ \boldsymbol{\Delta}}} \phi \cdot \\& \mathrm{d} \boldsymbol{x}=0\end{aligned}$
方程组(10)中第三式乘以μ , 并运用(10)中第四式，可得
(18) $\begin{aligned}& \frac{1}{2} \frac{\mathrm{d}}{\mathrm{d} t} \int|\mathit{\boldsymbol{ \boldsymbol{\nabla}}} \phi|^2 \mathrm{~d} \boldsymbol{x}+\frac{1}{4} \frac{\mathrm{d}}{\mathrm{d} t} \int \rho\left(\phi^2-1\right)^2 \mathrm{~d} \boldsymbol{x}+ \\& \int|\mathit{\boldsymbol{ \boldsymbol{\nabla}}} \mu|^2 \mathrm{~d} \boldsymbol{x}=-\int \mathit{\boldsymbol{ \boldsymbol{\nabla}}}(\boldsymbol{u} \cdot \mathit{\boldsymbol{ \boldsymbol{\nabla}}} \phi) \mathit{\boldsymbol{ \boldsymbol{\nabla}}} \phi \mathrm{d} \boldsymbol{x}\end{aligned}$
(19) $\begin{aligned}& \frac{\mathrm{d}}{\mathrm{d} t} \int\left(\frac{1}{2} \rho \boldsymbol{u}^2+G(\rho)+\frac{1}{2}|\mathit{\boldsymbol{ \boldsymbol{\nabla}}} \phi|^2+\frac{\rho}{4} \cdot\right. \\& \left.\left(\phi^2-1\right)^2\right) \mathrm{d} \boldsymbol{x}+\frac{\nu_0}{2}\|\mathit{\boldsymbol{ \boldsymbol{\nabla}}} \boldsymbol{u}\|^2+\|\mathit{\boldsymbol{ \boldsymbol{\nabla}}} \mu\|^2 \leqslant 0\end{aligned}$
(20) $c_{\bar{\rho}}(\rho-\bar{\rho})^2 \leqslant G(\rho) \leqslant C_{\bar{\rho}}(\rho-\bar{\rho})^2$
其中，$c_{\bar{\rho}}$ $C_{\bar{\rho}}$ $\bar{\rho}$
将式(19)在[0, T ]上积分，结合式(20)得
(21) $\begin{aligned}& \left\|\left(\sigma, \boldsymbol{u}, \phi^2-1, \mathit{\boldsymbol{ \boldsymbol{\nabla}}} \phi\right)\right\|^2+\int_0^t\|(\mathit{\boldsymbol{ \boldsymbol{\nabla}}} \mu, \mathit{\boldsymbol{ \boldsymbol{\nabla}}} \boldsymbol{u})\|^2 \mathrm{~d} \tau \leqslant \\& \left\|\left(\sigma_0, \boldsymbol{u}_0, \phi_0^2-1, \mathit{\boldsymbol{ \boldsymbol{\nabla}}} \phi_0\right)\right\|^2\end{aligned}$
$\begin{aligned}& \left\|\phi^2-1\right\|_{L^{\infty}} \leqslant C\left\|\phi^2-1\right\|_{H^2}=C\left(\left\|\phi^2-1\right\|_{L^2}+\right. \\& \left.\|\phi \mathit{\boldsymbol{ \boldsymbol{\nabla}}} \phi\|_{L^2}+\left\|\mathit{\boldsymbol{ \boldsymbol{\nabla}}}^2\left(\phi^2-1\right)\right\|_{L^2}\right) \leqslant M\end{aligned}$
在方程组(10)中第四式两边乘以－Δ φ , 关于x
(22) $\begin{array}{r}\|\mathit{\boldsymbol{ \boldsymbol{\Delta}}} \phi\|^2+\int \rho\left(3 \phi^2-1\right)|\mathit{\boldsymbol{ \boldsymbol{\nabla}}} \phi|^2 \mathrm{~d} \boldsymbol{x}= \\-\int \rho \mu \mathit{\boldsymbol{ \boldsymbol{\Delta}}} \phi \mathrm{d} \boldsymbol{x}-\int\left(\phi^2-1\right) \phi \mathit{\boldsymbol{ \boldsymbol{\nabla}}} \phi \cdot \mathit{\boldsymbol{ \boldsymbol{\nabla}}} \sigma \mathrm{d} \boldsymbol{x}\end{array}$
$\begin{aligned}& \int \rho\left(3 \phi^2-1\right)|\mathit{\boldsymbol{ \boldsymbol{\nabla}}} \phi|^2 \mathrm{~d} \boldsymbol{x} \geqslant \frac{\bar{\rho} m_0}{2}\|\mathit{\boldsymbol{ \boldsymbol{\nabla}}} \phi\|^2- \\& \int \rho \mu \mathit{\boldsymbol{ \boldsymbol{\Delta}}} \phi \mathrm{d} \boldsymbol{x} \leqslant 2 \bar{\rho}\|\mathit{\boldsymbol{ \boldsymbol{\nabla}}} \mu\|\|\mathit{\boldsymbol{ \boldsymbol{\nabla}}} \phi\| \leqslant \frac{\varepsilon}{4}\|\mathit{\boldsymbol{ \boldsymbol{\nabla}}} \phi\|^2+ \\& \frac{4 \bar{\rho}^2}{\varepsilon}\|\mathit{\boldsymbol{ \boldsymbol{\nabla}}} \mu\|^2 \int\left(\phi^2-1\right) \phi \mathit{\boldsymbol{ \boldsymbol{\nabla}}} \phi \cdot \mathit{\boldsymbol{ \boldsymbol{\nabla}}} \sigma \mathrm{d} \boldsymbol{x} \leqslant\left\|\left(\phi^2-1\right)\right\|_{L^6} \cdot \\& \|\phi \mathit{\boldsymbol{ \boldsymbol{\nabla}}} \phi\|\|\mathit{\boldsymbol{ \boldsymbol{\nabla}}} \sigma\|_{L^3} \leqslant M\|\mathit{\boldsymbol{ \boldsymbol{\nabla}}} \phi\|^2\end{aligned}$
(23) $\|\mathit{\boldsymbol{ \boldsymbol{\Delta}}} \phi\|^2+\|\mathit{\boldsymbol{ \boldsymbol{\nabla}}} \phi\|^2 \leqslant\|\mathit{\boldsymbol{ \boldsymbol{\nabla}}} \mu\|^2$
结合式(23)、(21)得到式(13), 引理1得证。
引理2 若(σ , u φ )∈X M T ])是方程(10)的局部解，则
(24) $\begin{aligned}& \quad \frac{\mathrm{d}}{\mathrm{d} t}\left\|\mathit{\boldsymbol{ \boldsymbol{\nabla}}}^{k+1} \phi\right\|^2+\frac{1}{4 \bar{\rho}^2}\left\|\mathit{\boldsymbol{ \boldsymbol{\nabla}}}^{k+3} \phi\right\|^2 \leqslant M \cdot\\& \left(\left\|\mathit{\boldsymbol{ \boldsymbol{\nabla}}}^{k+1} \sigma\right\|^2+\left\|\mathit{\boldsymbol{ \boldsymbol{\nabla}}}^{k+2} \phi\right\|^2+\left\|\mathit{\boldsymbol{ \boldsymbol{\nabla}}}^{k+2} \boldsymbol{u}\right\|^2\right), k=0, \\& 1, 2\end{aligned}$
证明：将方程组(10)中第四式代入(10)中第三式，可得
(25) $\begin{aligned}& \phi_t+\boldsymbol{u} \cdot \mathit{\boldsymbol{ \boldsymbol{\nabla}}} \phi+\frac{1}{\rho} \mathit{\boldsymbol{ \boldsymbol{\Delta}}}\left(\frac{\mathit{\boldsymbol{ \boldsymbol{\Delta}}} \phi}{\rho}\right)-\frac{3 \phi^2-1}{\rho} \mathit{\boldsymbol{ \boldsymbol{\Delta}}} \phi=\frac{6 \phi}{\rho} . \\& (\mathit{\boldsymbol{ \boldsymbol{\nabla}}} \phi)^2\end{aligned}$
对式(25)求$\mathit{\boldsymbol{ \boldsymbol{\nabla}}}$ k $-\mathit{\boldsymbol{ \boldsymbol{\Delta}}} \mathit{\boldsymbol{ \boldsymbol{\nabla}}}^k \phi$
(26) $\begin{aligned}& \quad \frac{1}{2} \frac{\mathrm{d}}{\mathrm{d} t} \int\left|\mathit{\boldsymbol{ \boldsymbol{\nabla}}}^{k+1} \phi\right|^2 \mathrm{~d} \boldsymbol{x}+\int\left|\mathit{\boldsymbol{ \boldsymbol{\nabla}}}\left(\frac{\mathit{\boldsymbol{ \boldsymbol{\nabla}}}^k \mathit{\boldsymbol{ \boldsymbol{\Delta}}} \phi}{\rho}\right)\right|^2 \mathrm{~d} \boldsymbol{x}- \\& \int \mathit{\boldsymbol{ \boldsymbol{\nabla}}}^k(\boldsymbol{u} \cdot \mathit{\boldsymbol{ \boldsymbol{\nabla}}} \phi) \mathit{\boldsymbol{ \boldsymbol{\Delta}}} \mathit{\boldsymbol{ \boldsymbol{\nabla}}}^k \phi \mathrm{d} \boldsymbol{x}=\sum\limits_{1 \leqslant l \leqslant k} C_k^l \int \mathit{\boldsymbol{ \boldsymbol{\nabla}}}^l\left(\frac{1}{\rho}\right) \mathit{\boldsymbol{ \boldsymbol{\nabla}}}^{k-l} \cdot \\& \mathit{\boldsymbol{ \boldsymbol{\Delta}}}\left(\frac{\mathit{\boldsymbol{ \boldsymbol{\Delta}}} \phi}{\rho}\right) \mathit{\boldsymbol{ \boldsymbol{\nabla}}}^k \mathit{\boldsymbol{ \boldsymbol{\Delta}}} \phi \mathrm{d} \boldsymbol{x}+\sum\limits_{1 \leqslant l \leqslant k} C_k^l \int \mathit{\boldsymbol{ \boldsymbol{\Delta}}}\left[\mathit{\boldsymbol{ \boldsymbol{\nabla}}}^l\left(\frac{1}{\rho}\right) \mathit{\boldsymbol{ \boldsymbol{\nabla}}}^{k-l} \mathit{\boldsymbol{ \boldsymbol{\Delta}}} \phi\right] . \\& \frac{\mathit{\boldsymbol{ \boldsymbol{\nabla}}}^k \mathit{\boldsymbol{ \boldsymbol{\Delta}}} \phi}{\rho} \mathrm{d} \boldsymbol{x}-\sum\limits_{1 \leqslant l \leqslant k} C_k^l \int \mathit{\boldsymbol{ \boldsymbol{\nabla}}}^l\left(\frac{3 \phi^2-1}{\rho}\right) \mathit{\boldsymbol{ \boldsymbol{\nabla}}}^{k-l} \mathit{\boldsymbol{ \boldsymbol{\Delta}}} \phi \mathit{\boldsymbol{ \boldsymbol{\nabla}}}^k \mathit{\boldsymbol{ \boldsymbol{\Delta}}} \phi \mathrm{d} \boldsymbol{x}- \\& \int \frac{3 \phi^2-1}{\rho}\left|\mathit{\boldsymbol{ \boldsymbol{\nabla}}}^k \mathit{\boldsymbol{ \boldsymbol{\Delta}}} \phi\right|^2 \mathrm{~d} \boldsymbol{x}-\int \mathit{\boldsymbol{ \boldsymbol{\nabla}}}^k\left(\frac{6 \phi}{\rho}(\mathit{\boldsymbol{ \boldsymbol{\nabla}}} \phi)^2\right) \mathit{\boldsymbol{ \boldsymbol{\Delta}}} \mathit{\boldsymbol{ \boldsymbol{\nabla}}}^k \phi \mathrm{d} \boldsymbol{x}\end{aligned}$
$\begin{array}{*{20}{l}}{\int {{{\left| {\mathit{\boldsymbol{ \boldsymbol{\nabla}}} \left( {\frac{{{\mathit{\boldsymbol{ \boldsymbol{\nabla}}} ^k}\mathit{\boldsymbol{ \boldsymbol{\Delta}}} \phi }}{\rho }} \right)} \right|}^2}} {\rm{d}}\mathit{\boldsymbol{x}} = \int {\frac{1}{{{\rho ^2}}}} {{\left| {{\mathit{\boldsymbol{ \boldsymbol{\nabla}}} ^{k + 1}}\mathit{\boldsymbol{ \boldsymbol{\Delta}}} \phi } \right|}^2}{\rm{d}}\mathit{\boldsymbol{x}} + }\\{\int {\frac{{|\mathit{\boldsymbol{ \boldsymbol{\nabla}}} \rho {|^2}}}{{{\rho ^4}}}} {{\left| {{\mathit{\boldsymbol{ \boldsymbol{\nabla}}} ^k}\mathit{\boldsymbol{ \boldsymbol{\Delta}}} \phi } \right|}^2}{\rm{d}}\mathit{\boldsymbol{x}} - \int {\frac{{3{\phi ^2} - 1}}{\rho }} {{\left| {{\mathit{\boldsymbol{ \boldsymbol{\nabla}}} ^k}\mathit{\boldsymbol{ \boldsymbol{\Delta}}} \phi } \right|}^2}{\rm{d}}\mathit{\boldsymbol{x}} \le }\\{ - \int {\frac{{{\phi ^2} - 1}}{\rho }} {{\left| {{\mathit{\boldsymbol{ \boldsymbol{\nabla}}} ^k}\mathit{\boldsymbol{ \boldsymbol{\Delta}}} \phi } \right|}^2}{\rm{d}}\mathit{\boldsymbol{x}}}\end{array}$
(27) $\begin{aligned}& \quad \frac{1}{2} \frac{\mathrm{d}}{\mathrm{d} t}\left\|\mathit{\boldsymbol{ \boldsymbol{\nabla}}}^{k+1} \phi\right\|^2+\frac{1}{4 \bar{\rho}^2}\left\|\mathit{\boldsymbol{ \boldsymbol{\nabla}}}^{k+1} \mathit{\boldsymbol{ \boldsymbol{\Delta}}} \phi\right\|^2 \leqslant I_1^i, i= \\& 1, 2, \cdots, 6\end{aligned}$
根据Hölder不等式和Gagliardo-Nirenberg不等式对I 1 i (i =1, 2，…，6)进行估计。
$\begin{aligned}& I_1^1=\int \frac{\phi^2-1}{\rho}\left|\mathit{\boldsymbol{ \boldsymbol{\nabla}}}^k \mathit{\boldsymbol{ \boldsymbol{\Delta}}} \phi\right|^2 \mathrm{~d} \boldsymbol{x} \leqslant\left\|\phi^2-1\right\|_{L^{\infty}} \cdot \\& \left\|\mathit{\boldsymbol{ \boldsymbol{\nabla}}}^k \mathit{\boldsymbol{ \boldsymbol{\Delta}}} \phi\right\|^2 \leqslant M\left\|\mathit{\boldsymbol{ \boldsymbol{\nabla}}}^{k+2} \phi\right\|^2\end{aligned}$
$\begin{aligned}& I_1^2=\sum\limits_{0 \leqslant l \leqslant k} C_k^l \int \mathit{\boldsymbol{ \boldsymbol{\nabla}}}^l \boldsymbol{u} \cdot \mathit{\boldsymbol{ \boldsymbol{\nabla}}}^{k-l+1} \phi \mathit{\boldsymbol{ \boldsymbol{\Delta}}} \mathit{\boldsymbol{ \boldsymbol{\nabla}}}^k \phi \mathrm{d} \boldsymbol{x} \leqslant \sum\limits_{0 \leqslant l \leqslant k} \| \mathit{\boldsymbol{ \boldsymbol{\nabla}}}^l \boldsymbol{u} \cdot \\& \mathit{\boldsymbol{ \boldsymbol{\nabla}}}^{k-l+1} \phi\|\| \mathit{\boldsymbol{ \boldsymbol{\nabla}}}^{k+2} \phi \|\end{aligned}$
当$l \leqslant\left[\frac{k+1}{2}\right]$
$\begin{aligned}& \left\|\mathit{\boldsymbol{ \boldsymbol{\nabla}}}^l \boldsymbol{u} \cdot \mathit{\boldsymbol{ \boldsymbol{\nabla}}}^{k-l+1} \phi\right\| \leqslant\left\|\mathit{\boldsymbol{ \boldsymbol{\nabla}}}^l \boldsymbol{u}\right\|_{L^3}\left\|\mathit{\boldsymbol{ \boldsymbol{\nabla}}}^{k-l+1} \phi\right\|_{L^6} \leqslant \\& \left\|\mathit{\boldsymbol{ \boldsymbol{\nabla}}}^\alpha \boldsymbol{u}\right\|^{1-\frac{l}{k+1}}\left\|\mathit{\boldsymbol{ \boldsymbol{\nabla}}}^{k+2} \boldsymbol{u}\right\|^{\frac{l}{k+1}}\|\mathit{\boldsymbol{ \boldsymbol{\nabla}}} \phi\|^{\frac{l}{k+1}}\left\|\mathit{\boldsymbol{ \boldsymbol{\nabla}}}^{k+2} \phi\right\|^{1-\frac{l}{k+1}} \leqslant \\& M\left\|\mathit{\boldsymbol{ \boldsymbol{\nabla}}}^{k+2} \boldsymbol{u}\right\|^{\frac{l}{k+1}}\left\|\mathit{\boldsymbol{ \boldsymbol{\nabla}}}^{k+2} \phi\right\|^{1-\frac{l}{k+1}} \leqslant M\left(\left\|\mathit{\boldsymbol{ \boldsymbol{\nabla}}}^{k+2} \boldsymbol{u}\right\|+\right. \\& \left.\left\|\mathit{\boldsymbol{ \boldsymbol{\nabla}}}^{k+2} \phi\right\|\right)\end{aligned}$
由$\frac{{l - 1}}{3} = \left( {\frac{\alpha }{3} - \frac{1}{2}} \right)\left( {1 - \frac{l}{{k + 1}}} \right) + \left( {\frac{{k + 2}}{3} - \frac{1}{2}} \right){\rm{\cdot}}\frac{l}{{k + 1}}$ $\alpha=\frac{1}{2}-\frac{l}{2(k+1-l)} \in\left[0, \frac{1}{2}\right]$
当$\left[\frac{k+1}{2}\right]+1 \leqslant l \leqslant k$
$\begin{aligned}& \left\|\mathit{\boldsymbol{ \boldsymbol{\nabla}}}^l \boldsymbol{u} \cdot \mathit{\boldsymbol{ \boldsymbol{\nabla}}}^{k-l+1} \phi\right\| \leqslant\left\|\mathit{\boldsymbol{ \boldsymbol{\nabla}}}^l \boldsymbol{u}\right\|_{L^6}\left\|\mathit{\boldsymbol{ \boldsymbol{\nabla}}}^{k-l+1} \phi\right\|_{L^3} \leqslant \\& \|\boldsymbol{u}\|^{1-\frac{l+1}{k+2}}\left\|\mathit{\boldsymbol{ \boldsymbol{\nabla}}}^{k+2} \boldsymbol{u}\right\|^{\frac{l+1}{k+2}}\left\|\mathit{\boldsymbol{ \boldsymbol{\nabla}}}^\alpha \phi\right\|^{\frac{l+1}{k+2}}\left\|\mathit{\boldsymbol{ \boldsymbol{\nabla}}}^{k+2} \phi\right\|^{1-\frac{l+1}{k+2}} \leqslant \\& M\left\|\mathit{\boldsymbol{ \boldsymbol{\nabla}}}^{k+2} \boldsymbol{u}\right\|^{\frac{l+1}{k+2}}\left\|\mathit{\boldsymbol{ \boldsymbol{\nabla}}}^{k+2} \phi\right\|^{1-\frac{l+1}{k+2}} \leqslant M\left(\left\|\mathit{\boldsymbol{ \boldsymbol{\nabla}}}^{k+2} \boldsymbol{u}\right\|+\right. \\& \left.\left\|\mathit{\boldsymbol{ \boldsymbol{\nabla}}}^{k+2} \phi\right\|\right)\end{aligned}$
由$\frac{{k - l}}{3} = \left( {\frac{\alpha }{3} - \frac{1}{2}} \right)\frac{{l + 1}}{{k + 2}} + \left( {\frac{{k + 2}}{3} - \frac{1}{2}} \right)(1 - \left. {\frac{{l + 1}}{{k + 2}}} \right)$
可确定$\alpha=\frac{k+2}{2(l+1)} \in\left[\frac{1}{2}, 1\right]$
$I_1^2 \leqslant M\left(\left\|\mathit{\boldsymbol{ \boldsymbol{\nabla}}}^{k+2} \boldsymbol{u}\right\|^2+\left\|\mathit{\boldsymbol{ \boldsymbol{\nabla}}}^{k+2} \phi\right\|^2\right)$
$\begin{aligned}\;\;\;I_1^3 & =\sum\limits_{1 \leqslant l \leqslant k} C_k^l \int \mathit{\boldsymbol{ \boldsymbol{\nabla}}}^l\left(\frac{1}{\rho}\right) \mathit{\boldsymbol{ \boldsymbol{\nabla}}}^{k-l} \mathit{\boldsymbol{ \boldsymbol{\Delta}}}\left(\frac{\mathit{\boldsymbol{ \boldsymbol{\Delta}}} \phi}{\rho}\right) \mathit{\boldsymbol{ \boldsymbol{\nabla}}}^k \mathit{\boldsymbol{ \boldsymbol{\Delta}}} \phi \mathrm{d} \boldsymbol{x} \leqslant \\\sum\limits_{1 \leqslant l} & \left(\left\|\mathit{\boldsymbol{ \boldsymbol{\nabla}}}^l\left(\frac{1}{\rho}\right) \mathit{\boldsymbol{ \boldsymbol{\nabla}}}^{k-l+2}\left(\frac{\mathit{\boldsymbol{ \boldsymbol{\Delta}}} \phi}{\rho}\right)\right\|\right)\left\|\mathit{\boldsymbol{ \boldsymbol{\nabla}}}^{k+2} \phi\right\|_{L^2}\end{aligned}$
$\begin{aligned}& \left\|\mathit{\boldsymbol{ \boldsymbol{\nabla}}}^l\left(\frac{1}{\rho}\right) \mathit{\boldsymbol{ \boldsymbol{\nabla}}}^{k-l+2}\left(\frac{\mathit{\boldsymbol{ \boldsymbol{\nabla}}} \phi}{\rho}\right)\right\| \leqslant\left\|\mathit{\boldsymbol{ \boldsymbol{\nabla}}}^l(\sigma)\right\|_{L^6} \cdot \\& \left\|\mathit{\boldsymbol{ \boldsymbol{\nabla}}}^{k-l+2}\left(\frac{\mathit{\boldsymbol{ \boldsymbol{\Delta}}} \phi}{\rho}\right)\right\|_{L^3} \leqslant\left\|\mathit{\boldsymbol{ \boldsymbol{\nabla}}}^l(\sigma)\right\|_{L^6}\left(\|\left(\mathit{\boldsymbol{ \boldsymbol{\nabla}}}^{k-l+2} \mathit{\boldsymbol{ \boldsymbol{\Delta}}} \phi\right) \cdot\right. \\& \left.\frac{1}{\rho}+\mathit{\boldsymbol{ \boldsymbol{\nabla}}}^{k-l+2}\left(\frac{1}{\rho}\right) \mathit{\boldsymbol{ \boldsymbol{\Delta}}} \phi \|_{L^3}\right) \leqslant\left\|\mathit{\boldsymbol{ \boldsymbol{\nabla}}}^l(\sigma)\right\|_{L^6} \cdot \\& \left(\left\|\left(\mathit{\boldsymbol{ \boldsymbol{\nabla}}}^{k-l+4} \phi\right)\right\|_{L^3}\left\|\frac{1}{\rho}\right\|_{L^{\infty}}+\left\|\mathit{\boldsymbol{ \boldsymbol{\nabla}}}^{k-l+2}\left(\frac{1}{\rho}\right)\right\|_{L^3} \cdot\right. \\& \left.\|\mathit{\boldsymbol{ \boldsymbol{\Delta}}} \phi\|_{L^{\infty}}\right) \leqslant M\left(\left\|\mathit{\boldsymbol{ \boldsymbol{\nabla}}}^{k+1} \sigma\right\|+\left\|\mathit{\boldsymbol{ \boldsymbol{\nabla}}}^{k+2} \phi\right\|\right)\end{aligned}$
故$I_1^3 \leqslant M\left(\left\|\mathit{\boldsymbol{ \boldsymbol{\nabla}}}^{k+1} \sigma\right\|^2+\left\|\mathit{\boldsymbol{ \boldsymbol{\nabla}}}^{k+2} \phi\right\|^2\right)$
$\begin{aligned}& I_1^4=-\sum\limits_{1 \leqslant l \leqslant k} C_k^l \int \mathit{\boldsymbol{ \boldsymbol{\nabla}}}\left[\mathit{\boldsymbol{ \boldsymbol{\nabla}}}^l\left(\frac{1}{\rho}\right) \mathit{\boldsymbol{ \boldsymbol{\nabla}}}^{k-l} \mathit{\boldsymbol{ \boldsymbol{\Delta}}} \phi\right] . \\& \mathit{\boldsymbol{ \boldsymbol{\nabla}}}\left(\frac{\mathit{\boldsymbol{ \boldsymbol{\nabla}}}^k \mathit{\boldsymbol{ \boldsymbol{\Delta}}} \phi}{\rho}\right) \mathrm{d} \boldsymbol{x} \leqslant \sum\limits_{1 \leqslant l \leqslant k}\left(\left\|\mathit{\boldsymbol{ \boldsymbol{\nabla}}}^l\left(\frac{1}{\rho}\right)\right\|_{L^3} \cdot\right. \\& \left.\left\|\mathit{\boldsymbol{ \boldsymbol{\nabla}}}^{k-l+3} \phi\right\|_{L^6}+\left\|\mathit{\boldsymbol{ \boldsymbol{\nabla}}}^{l+1}\left(\frac{1}{\rho}\right)\right\|_{L^3}\left\|\mathit{\boldsymbol{ \boldsymbol{\nabla}}}^{k-l+2} \phi\right\|_{L^6}\right) \times \\& \left(\left\|\mathit{\boldsymbol{ \boldsymbol{\nabla}}}^{k+3} \phi\right\|+\left\|\mathit{\boldsymbol{ \boldsymbol{\nabla}}}^{k+2} \phi\right\|\left\|\mathit{\boldsymbol{ \boldsymbol{\nabla}}}\left(\frac{1}{\rho}\right)\right\|_{L^{\infty}}\right) \leqslant \\& \sum\limits_{1 \leqslant l \leqslant k}\left(\left\|\mathit{\boldsymbol{ \boldsymbol{\nabla}}}^l \sigma\right\|_{L^3}\left\|\mathit{\boldsymbol{ \boldsymbol{\nabla}}}^{k-l+3} \phi\right\|_{L^6}+\left\|\mathit{\boldsymbol{ \boldsymbol{\nabla}}}^{l+1} \sigma\right\|_{L^3} \cdot\right. \\& \left.\left\|\mathit{\boldsymbol{ \boldsymbol{\nabla}}}^{k-l+2} \phi\right\|_{L^6}\right)\left(\left\|\mathit{\boldsymbol{ \boldsymbol{\nabla}}}^{k+3} \phi\right\|+\left\|\mathit{\boldsymbol{ \boldsymbol{\nabla}}}^{k+2} \sigma\right\|\right) \leqslant \\& M \sum\limits_{1 \leqslant l \leqslant k}\left(\left\|\mathit{\boldsymbol{ \boldsymbol{\nabla}}}^{k+1} \sigma\right\|+\left\|\mathit{\boldsymbol{ \boldsymbol{\nabla}}}^{k+3} \phi\right\|\right)\left(\left\|\mathit{\boldsymbol{ \boldsymbol{\nabla}}}^{k+3} \phi\right\|+\right. \\& \left.\left\|\mathit{\boldsymbol{ \boldsymbol{\nabla}}}^{k+2} \phi\right\|\right) \leqslant M\left(\left\|\mathit{\boldsymbol{ \boldsymbol{\nabla}}}^{k+1} \sigma\right\|^2+\left\|\mathit{\boldsymbol{ \boldsymbol{\nabla}}}^{k+2} \phi\right\|^2+\right. \\& \left.\left\|\mathit{\boldsymbol{ \boldsymbol{\nabla}}}^{k+3} \phi\right\|^2\right)\end{aligned}$
$\begin{aligned}& I_1^5=\sum\limits_{1 \leqslant l \leqslant k} C_k^l \int \mathit{\boldsymbol{ \boldsymbol{\nabla}}}^l\left(\frac{3 \phi^2-1}{\rho}\right) \mathit{\boldsymbol{ \boldsymbol{\nabla}}}^{k-l} \mathit{\boldsymbol{ \boldsymbol{\Delta}}} \phi \mathit{\boldsymbol{ \boldsymbol{\nabla}}}^k \mathit{\boldsymbol{ \boldsymbol{\Delta}}} \phi \mathrm{d} \boldsymbol{x} \leqslant \\& \sum\limits_{1 \leqslant l \leqslant k}\left\|\mathit{\boldsymbol{ \boldsymbol{\nabla}}}^l\left(\frac{3 \phi^2-1}{\rho}\right) \mathit{\boldsymbol{ \boldsymbol{\nabla}}}^{k-l+2} \phi\right\|\left\|\mathit{\boldsymbol{ \boldsymbol{\nabla}}}^{k+2} \phi\right\| \cdot \\& \left\|\mathit{\boldsymbol{ \boldsymbol{\nabla}}}^l\left(\frac{3 \phi^2-1}{\rho}\right) \mathit{\boldsymbol{ \boldsymbol{\nabla}}}^{k-l+2} \phi\right\| \leqslant\left(\left\|\mathit{\boldsymbol{ \boldsymbol{\nabla}}}^l \sigma\right\|_{L^3}+\left\|\mathit{\boldsymbol{ \boldsymbol{\nabla}}}^l \phi\right\|_{L^3}\right) \cdot\\& \left\|\mathit{\boldsymbol{ \boldsymbol{\nabla}}}^{k-l+2} \phi\right\|_{L^6} \leqslant\left(\left\|\mathit{\boldsymbol{ \boldsymbol{\nabla}}}^\alpha \sigma\right\|^{1-\frac{l}{k+2}}\left\|\mathit{\boldsymbol{ \boldsymbol{\nabla}}}^{k+1} \sigma\right\|^{\frac{l}{k+2}}+\right. \\& \left.\left\|\mathit{\boldsymbol{ \boldsymbol{\nabla}}}^\alpha \phi\right\|^{1-\frac{l}{k+2}}\left\|\mathit{\boldsymbol{ \boldsymbol{\nabla}}}^{k+2} \phi\right\|^{\frac{l}{k+2}}\right) \quad \times \quad\|\mathit{\boldsymbol{ \boldsymbol{\nabla}}} \phi\|^{\frac{l}{k+2}} \cdot \\& \left\|\mathit{\boldsymbol{ \boldsymbol{\nabla}}}^{k+2} \phi\right\|^{1-\frac{l}{k+2}} \leqslant M\left(\left\|\mathit{\boldsymbol{ \boldsymbol{\nabla}}}^{k+1} \sigma\right\|+\left\|\mathit{\boldsymbol{ \boldsymbol{\nabla}}}^{k+2} \phi\right\|\right)\end{aligned}$
$\begin{aligned}& \quad \frac{l}{3}-\frac{1}{3}=\left(\frac{\alpha}{3}-\frac{1}{2}\right)\left(1-\frac{l}{k+2}\right)+\left(\frac{k+1}{3}-\right. \\& \left.\frac{1}{2}\right) \frac{l}{k+2}\end{aligned}$
$I_1^5 \leqslant M\left(\left\|\mathit{\boldsymbol{ \boldsymbol{\nabla}}}^{k+1} \sigma\right\|^2+\left\|\mathit{\boldsymbol{ \boldsymbol{\nabla}}}^{k+2} \phi\right\|^2\right)$
$\begin{aligned}& I_1^6=\sum\limits_{0 \leqslant l \leqslant k} C_k^l \int \mathit{\boldsymbol{ \boldsymbol{\nabla}}}^l\left(\frac{6 \phi}{\rho}\right) \mathit{\boldsymbol{ \boldsymbol{\nabla}}}^{k-l}(\mathit{\boldsymbol{ \boldsymbol{\nabla}}} \phi) 2 \mathit{\boldsymbol{ \boldsymbol{\Delta}}} \mathit{\boldsymbol{ \boldsymbol{\nabla}}}^k \phi \mathrm{d} \boldsymbol{x} \leqslant \\& \sum\limits_{1 \leqslant l \leqslant k}\left\|\mathit{\boldsymbol{ \boldsymbol{\nabla}}}^l\left(\frac{6 \phi}{\rho}\right) \mathit{\boldsymbol{ \boldsymbol{\nabla}}}^{k-l}(\mathit{\boldsymbol{ \boldsymbol{\nabla}}} \phi)^2\right\|\left\|\mathit{\boldsymbol{ \boldsymbol{\nabla}}}^{k+2} \phi\right\| \| \mathit{\boldsymbol{ \boldsymbol{\nabla}}}^l\left(\frac{6 \phi}{\rho}\right) . \\& \mathit{\boldsymbol{ \boldsymbol{\nabla}}}^{k-l}(\mathit{\boldsymbol{ \boldsymbol{\nabla}}} \phi)^2 \| \leqslant\left(\left\|\mathit{\boldsymbol{ \boldsymbol{\nabla}}}^l \sigma\right\|_{L^3}+\left\|\mathit{\boldsymbol{ \boldsymbol{\nabla}}}^l \phi\right\|_{L^3}\right) . \\& \left\|\mathit{\boldsymbol{ \boldsymbol{\nabla}}}^{k-l+1} \phi\right\|_{L^6}\|\mathit{\boldsymbol{ \boldsymbol{\nabla}}} \phi\|_{L^{\infty}} \leqslant\left(\left\|\mathit{\boldsymbol{ \boldsymbol{\nabla}}}^l \sigma\right\|_{L^3}+\right. \\& \left.\left\|\mathit{\boldsymbol{ \boldsymbol{\nabla}}}^l \phi\right\|_{L^3}\right)\left\|\mathit{\boldsymbol{ \boldsymbol{\nabla}}}^{k-l+1} \phi\right\|_{L^6} \leqslant M\left(\left\|\mathit{\boldsymbol{ \boldsymbol{\nabla}}}^{k+1} \sigma\right\|+\right. \\& \left.\left\|\mathit{\boldsymbol{ \boldsymbol{\nabla}}}^{k+2} \phi\right\|\right)\end{aligned}$
$I_1^6 \leqslant M\left(\left\|\mathit{\boldsymbol{ \boldsymbol{\nabla}}}^{k+1} \sigma\right\|^2+\left\|\mathit{\boldsymbol{ \boldsymbol{\nabla}}}^{k+2} \phi\right\|^2\right)$
把I 1 i (i =1, 2, …, 6)的估计式代入式(27)，可得式(24)，引理2得证。
引理3 若(σ , u φ )∈X M T ])是方程(10)的局部解，则
(28) $\begin{aligned}& \frac{\mathrm{d}}{\mathrm{d} t}\left(\left\|\mathit{\boldsymbol{ \boldsymbol{\nabla}}}^k \boldsymbol{u}\right\|+\frac{p^{\prime}(\bar{\rho})}{\bar{\rho}^2}\left\|\mathit{\boldsymbol{ \boldsymbol{\nabla}}}^k \sigma\right\|^2\right)+\frac{\nu_0}{\bar{\rho}} \cdot \\& \left\|\mathit{\boldsymbol{ \boldsymbol{\nabla}}}^{k+1} \boldsymbol{u}\right\|^2 \leqslant M\left(\left\|\mathit{\boldsymbol{ \boldsymbol{\nabla}}}^k \sigma\right\|^2+\left\|\mathit{\boldsymbol{ \boldsymbol{\nabla}}}^{k+2} \phi\right\|^2\right), k=0, \\& 1, 2, \cdots, 3\end{aligned}$
证明：对式(10)求$\mathit{\boldsymbol{ \boldsymbol{\nabla}}}^k$
$\mathit{\boldsymbol{ \boldsymbol{\nabla}}}^k \sigma_t+\bar{\rho} \operatorname{div} \mathit{\boldsymbol{ \boldsymbol{\nabla}}}^k \boldsymbol{u}=\mathit{\boldsymbol{ \boldsymbol{\nabla}}}^k g_1$
(29) $\begin{aligned}& \mathit{\boldsymbol{ \boldsymbol{\nabla}}}^k \boldsymbol{u}_t-\frac{2}{\bar{\rho}} \operatorname{div}\left(\nu(\phi) \boldsymbol{D}\left(\mathit{\boldsymbol{ \boldsymbol{\nabla}}}^k \boldsymbol{u}\right)\right)-\frac{1}{\bar{\rho}} \mathit{\boldsymbol{ \boldsymbol{\nabla}}}\{[\boldsymbol{\nu}(\phi)+ \\& \left.\lambda(\phi)] \operatorname{div} \mathit{\boldsymbol{ \boldsymbol{\nabla}}}^k \boldsymbol{u}\right\}+\frac{p^{\prime}(\bar{\rho})}{\bar{\rho}} \mathit{\boldsymbol{ \boldsymbol{\nabla}}}^k \mathit{\boldsymbol{ \boldsymbol{\nabla}}} \sigma+\frac{1}{\bar{\rho}} \mathit{\boldsymbol{ \boldsymbol{\nabla}}}^k(\mathit{\boldsymbol{ \boldsymbol{\nabla}}} \phi \mathit{\boldsymbol{ \boldsymbol{\Delta}}} \phi)= \\& \mathit{\boldsymbol{ \boldsymbol{\nabla}}}^k g_2+\frac{2}{\bar{\rho}} \operatorname{div}\left\{\mathit{\boldsymbol{ \boldsymbol{\nabla}}}^k[\nu(\phi) \boldsymbol{D}(\boldsymbol{u})]-\nu(\phi) \boldsymbol{D}\left(\mathit{\boldsymbol{ \boldsymbol{\nabla}}}^k \boldsymbol{u}\right)\right\}+ \\& \frac{1}{\bar{\rho}} \mathit{\boldsymbol{ \boldsymbol{\nabla}}}\left\{\mathit{\boldsymbol{ \boldsymbol{\nabla}}}^k[\nu(\phi)+\lambda(\phi)] \operatorname{div} \boldsymbol{u}-[\nu(\phi)+\lambda(\phi)] \cdot\right. \\& \left.\operatorname{div} \mathit{\boldsymbol{ \boldsymbol{\nabla}}}^k \boldsymbol{u}\right\}\end{aligned}$
对式(29)的第二式乘以$\mathit{\boldsymbol{ \boldsymbol{\nabla}}}^k \boldsymbol{u}$
(30) $\begin{aligned}& \frac{1}{2} \frac{\mathrm{d}}{\mathrm{d} t}\left(\left\|\mathit{\boldsymbol{ \boldsymbol{\nabla}}}^k \boldsymbol{u}\right\|^2+\frac{p^{\prime}(\bar{\rho})}{\bar{\rho}^2}\left\|\mathit{\boldsymbol{ \boldsymbol{\nabla}}}^k \sigma\right\|^2\right)+\frac{1}{\bar{\rho}} \cdot \\& \int \nu(\phi)\left|\boldsymbol{D}\left(\mathit{\boldsymbol{ \boldsymbol{\nabla}}}^k \boldsymbol{u}\right)\right|^2 \mathrm{~d} \boldsymbol{x}+\frac{1}{\bar{\rho}} \int[\nu(\phi)+\lambda(\phi)] \cdot \\& \left|\operatorname{div} \mathit{\boldsymbol{ \boldsymbol{\nabla}}}^k \boldsymbol{u}\right|^2 \mathrm{~d} \boldsymbol{x}=I_2^1+I_2^2+I_2^3+I_2^4+I_2^5+I_2^6+I_2^7\end{aligned}$
$\begin{aligned}& I_2^1=\int \mathit{\boldsymbol{ \boldsymbol{\nabla}}}^k \operatorname{div}(\sigma \boldsymbol{u}) \mathit{\boldsymbol{ \boldsymbol{\nabla}}}^k \sigma \mathrm{d} \boldsymbol{x} \leqslant \sum\limits_{0 \leqslant l \leqslant k+1} \| \mathit{\boldsymbol{ \boldsymbol{\nabla}}}^l \sigma \cdot \\& \mathit{\boldsymbol{ \boldsymbol{\nabla}}}^{k-l+1} \boldsymbol{u}\|\| \mathit{\boldsymbol{ \boldsymbol{\nabla}}}^k \sigma \|\end{aligned}$
当$l \leqslant\left[\frac{k}{2}\right]$
$\begin{aligned}& \left\|\mathit{\boldsymbol{ \boldsymbol{\nabla}}}^l \sigma \cdot \mathit{\boldsymbol{ \boldsymbol{\nabla}}}^{k-l+1} \boldsymbol{u}\right\| \leqslant\left\|\mathit{\boldsymbol{ \boldsymbol{\nabla}}}^l \boldsymbol{\sigma}\right\|_{L^3}\left\|\mathit{\boldsymbol{ \boldsymbol{\nabla}}}^{k-l+1} \boldsymbol{u}\right\|_{L^6} \leqslant \\& \left\|\mathit{\boldsymbol{ \boldsymbol{\nabla}}}^\alpha \sigma\right\|^{1-\frac{l-1}{k}}\left\|\mathit{\boldsymbol{ \boldsymbol{\nabla}}}^k \sigma\right\|^{\frac{l-1}{k}}\|\mathit{\boldsymbol{ \boldsymbol{\nabla}}} \boldsymbol{u}\|^{\frac{l-1}{k}}\left\|\mathit{\boldsymbol{ \boldsymbol{\nabla}}}^{k+1} \boldsymbol{u}\right\|^{1-\frac{l-1}{k}} \leqslant \\& M\left(\left\|\mathit{\boldsymbol{ \boldsymbol{\nabla}}}^k \sigma\right\|+\left\|\mathit{\boldsymbol{ \boldsymbol{\nabla}}}^{k+1} \boldsymbol{u}\right\|\right)\end{aligned}$
其中α 由$\frac{{l - 1}}{3} = \left( {\frac{\alpha }{3} - \frac{1}{2}} \right)\left( {1 - \frac{{l - 1}}{k}} \right) + \left( {\frac{k}{3} - \frac{1}{2}} \right) \cdot \frac{{l - 1}}{k}$
当$l \geqslant\left[\frac{k}{2}\right]+1$
$\begin{aligned}& \left\|\mathit{\boldsymbol{ \boldsymbol{\nabla}}}^l \sigma \cdot \mathit{\boldsymbol{ \boldsymbol{\nabla}}}^{k-l+1} \boldsymbol{u}\right\| \leqslant\left\|\mathit{\boldsymbol{ \boldsymbol{\nabla}}}^l \sigma\right\|_{L^6}\left\|\mathit{\boldsymbol{ \boldsymbol{\nabla}}}^{k-l+1} \phi\right\|_{L^3} \leqslant \\& \|\mathit{\boldsymbol{ \boldsymbol{\nabla}}} \sigma\|^{1-\frac{l-1}{k}}\left\|\mathit{\boldsymbol{ \boldsymbol{\nabla}}}^k \sigma\right\|^{\frac{l+1}{k}}\left\|\mathit{\boldsymbol{ \boldsymbol{\nabla}}}^\alpha \boldsymbol{u}\right\|^{\frac{l+1}{k}}\left\|\mathit{\boldsymbol{ \boldsymbol{\nabla}}}^{k+1} \boldsymbol{u}\right\|^{1-\frac{l+1}{k}} \leqslant \\& M\left(\left\|\mathit{\boldsymbol{ \boldsymbol{\nabla}}}^k \sigma\right\|+\left\|\mathit{\boldsymbol{ \boldsymbol{\nabla}}}^{k+1} \boldsymbol{u}\right\|\right)\end{aligned}$
其中α 由$\frac{{k - l}}{3} = \left( {\frac{\alpha }{3} - \frac{1}{2}} \right)\left( {\frac{{l + 1}}{k}} \right) + \left( {\frac{{k + 1}}{3} - \frac{1}{2}} \right) \cdot \left( {1 - \frac{{l + 1}}{k}} \right)$
故$I_2^1 \leqslant M\left(\left\|\mathit{\boldsymbol{ \boldsymbol{\nabla}}}^k \sigma\right\|^2+\left\|\mathit{\boldsymbol{ \boldsymbol{\nabla}}}^{k+1} \boldsymbol{u}\right\|^2\right)$
$\begin{array}{*{20}{l}} {I_2^2 = \int {{\nabla ^k}} \left[ { - (\boldsymbol{u},\nabla )\boldsymbol{u} + {h_1}(\sigma )\nabla \sigma } \right]{\nabla ^k}\boldsymbol{u}{\text{d}}\boldsymbol{x} = } \\ { - \int {{\nabla ^{k - 1}}} \left[ { - (\boldsymbol{u},\nabla )\boldsymbol{u} + {h_1}(\sigma )\nabla \sigma } \right]{\nabla ^{k + 1}}\boldsymbol{u}{\text{d}}\boldsymbol{x} \leqslant } \\ {\left( {\parallel \boldsymbol{u}{\parallel _{{L^3}}}{{\left\| {{\nabla ^k}\boldsymbol{u}} \right\|}_{{L^6}}} + {{\left\| {{\nabla ^{k - 1}}\boldsymbol{u}} \right\|}_{{L^6}}}\parallel \nabla \boldsymbol{u}{\parallel _{{L^3}}}} \right) \cdot } \\ {\left\| {{\nabla ^{k + 1}}{\mathbf{u}}} \right\| + \left( {{{\left\| {{h_1}(\sigma )} \right\|}_{{L^3}}}{{\left\| {{\nabla ^k}\sigma } \right\|}_{{L^6}}} + \parallel \nabla \sigma {\parallel _{{L^6}}} \cdot } \right.} \\ {\left. {{{\left\| {{\nabla ^{k - 1}}{h_1}(\sigma )} \right\|}_{{L^3}}}} \right)\left\| {{\nabla ^{k + 1}}\boldsymbol{u}} \right\| \leqslant M\left( {\left\| {{\nabla ^{k + 1}}\boldsymbol{u}} \right\| + } \right.} \\ {\left. {\left\| {{\nabla ^k}\sigma } \right\|} \right)\left\| {{\nabla ^{k + 1}}\boldsymbol{u}} \right\| \leqslant M\left( {{{\left\| {{\nabla ^k}\sigma } \right\|}^2} + {{\left\| {{\nabla ^{k + 1}}\boldsymbol{u}} \right\|}^2}} \right)} \end{array}$
$\begin{aligned}& I_2^3=\int \mathit{\boldsymbol{ \boldsymbol{\nabla}}}^{k-1}\left\{2 h_2(\sigma) \operatorname{div}[\nu(\phi) \boldsymbol{D}(\boldsymbol{u})]\right\} \mathit{\boldsymbol{ \boldsymbol{\nabla}}}^{k+1} \boldsymbol{u} \mathrm{d} \boldsymbol{x}+ \\& \int \mathit{\boldsymbol{ \boldsymbol{\nabla}}}^{k-1}\left\{h_2(\sigma) \mathit{\boldsymbol{ \boldsymbol{\nabla}}}[(\nu(\phi)+\lambda(\phi)) \operatorname{div} \boldsymbol{u}]\right\} \mathit{\boldsymbol{ \boldsymbol{\nabla}}}^{k+1} \boldsymbol{u} \mathrm{d} \boldsymbol{x}= \\& \sum\limits_{0 \leqslant l \leqslant k-1} C_{k-1}^l \int \mathit{\boldsymbol{ \boldsymbol{\nabla}}}^l h_2(\sigma) \operatorname{div} \mathit{\boldsymbol{ \boldsymbol{\nabla}}}^{k-l-1}(\nu(\phi) \boldsymbol{D}(\boldsymbol{u})) \mathit{\boldsymbol{ \boldsymbol{\nabla}}}^{k+1} \boldsymbol{u} \cdot \\& \mathrm{d} \boldsymbol{x}+\sum\limits_{0 \leqslant l \leqslant k-1} C_{k-1}^l \int \mathit{\boldsymbol{ \boldsymbol{\nabla}}}^l h_2(\sigma) \mathit{\boldsymbol{ \boldsymbol{\nabla}}}^{k-l}((\nu(\phi)+\lambda(\phi)) \cdot \\& \operatorname{div} \boldsymbol{u}) \mathit{\boldsymbol{ \boldsymbol{\nabla}}}^{k+1} \boldsymbol{u} \mathrm{d} \boldsymbol{x}\end{aligned}$
$\begin{aligned}& \sum\limits_{0 \leqslant l \leqslant k-1} C_{k-1}^l \int \mathit{\boldsymbol{ \boldsymbol{\nabla}}}^l h_2(\sigma) \operatorname{div} \mathit{\boldsymbol{ \boldsymbol{\nabla}}}^{k-l-1}(\nu(\phi) \boldsymbol{D}(\boldsymbol{u})) \cdot \\& \mathit{\boldsymbol{ \boldsymbol{\nabla}}}^{k+1} \boldsymbol{u} \mathrm{d} \boldsymbol{x} \leqslant \sum\limits_{0 \leqslant l \leqslant k-1}\left\|\mathit{\boldsymbol{ \boldsymbol{\nabla}}}^l h_2(\sigma) \mathit{\boldsymbol{ \boldsymbol{\nabla}}}^{k-l}(\nu(\phi) \boldsymbol{D}(\boldsymbol{u}))\right\| \cdot \\& \left\|\mathit{\boldsymbol{ \boldsymbol{\nabla}}}^{k+1} \boldsymbol{u}\right\| \leqslant \sum\limits_{0 \leqslant l \leqslant k-1}\left\|\mathit{\boldsymbol{ \boldsymbol{\nabla}}}^l \sigma\right\|_{L^3}\left(\left\|\mathit{\boldsymbol{ \boldsymbol{\nabla}}}^{k-l+1} \boldsymbol{u}\right\|_{L^6} \cdot\right. \\& \left.\|\nu(\phi)\|_{L^{\infty}}+\|\mathit{\boldsymbol{ \boldsymbol{\nabla}}} \boldsymbol{u}\|_{L^{\infty}}\left\|\mathit{\boldsymbol{ \boldsymbol{\nabla}}}^{k-l} \phi\right\|_{L^6}\right)\left\|\mathit{\boldsymbol{ \boldsymbol{\nabla}}}^{k+1} \boldsymbol{u}\right\| \leqslant \\& \sum\limits_{0 \leqslant l \leqslant k-1}\left\|\mathit{\boldsymbol{ \boldsymbol{\nabla}}}^l \sigma\right\|_{L^3}\left\|\mathit{\boldsymbol{ \boldsymbol{\nabla}}}^{k-l+1} \boldsymbol{u}\right\|_{L^6}\left\|\mathit{\boldsymbol{ \boldsymbol{\nabla}}}^{k+1} \boldsymbol{u}\right\|+ \\& \sum\limits_{0 \leqslant l \leqslant k-1}\left\|\mathit{\boldsymbol{ \boldsymbol{\nabla}}}^l \sigma\right\|_{L^3}\left\|\mathit{\boldsymbol{ \boldsymbol{\nabla}}}^{k-l} \phi\right\|_{L^6}\left\|\mathit{\boldsymbol{ \boldsymbol{\nabla}}}^{k+1} \boldsymbol{u}\right\| \leqslant \\& M\left(\left\|\mathit{\boldsymbol{ \boldsymbol{\nabla}}}^{k+2} \phi\right\|^2+\left\|\mathit{\boldsymbol{ \boldsymbol{\nabla}}}^{k+1} \boldsymbol{u}\right\|^2\right)\end{aligned}$
$\begin{aligned}& \sum\limits_{0 \leqslant l \leqslant k-1} C_{k-1}^l \int \mathit{\boldsymbol{ \boldsymbol{\nabla}}}^l h_2(\sigma) \mathit{\boldsymbol{ \boldsymbol{\nabla}}}^{k-l}((\nu(\phi)+\lambda(\phi)) \cdot\\& \operatorname{div} \boldsymbol{u}) \mathit{\boldsymbol{ \boldsymbol{\nabla}}}^{k+1} \boldsymbol{u} \mathrm{d} \boldsymbol{x} \\&\end{aligned}$
$I_2^3 \leqslant M\left(\left\|\mathit{\boldsymbol{ \boldsymbol{\nabla}}}^{k+2} \phi\right\|^2+\left\|\mathit{\boldsymbol{ \boldsymbol{\nabla}}}^{k+1} \boldsymbol{u}\right\|^2\right)$
$I_2^4=\int \mathit{\boldsymbol{ \boldsymbol{\nabla}}}^k(\mathit{\boldsymbol{ \boldsymbol{\nabla}}} \phi \mathit{\boldsymbol{ \boldsymbol{\Delta}}} \phi) \mathit{\boldsymbol{ \boldsymbol{\nabla}}}^k \boldsymbol{u} \mathrm{d} \boldsymbol{x}$
当k =0时，$I_2^4 = \int \mathit{\boldsymbol{ \boldsymbol{\nabla}}} \phi \mathit{\boldsymbol{ \boldsymbol{\Delta}}} \phi {\rm{\cdot}}\mathit{\boldsymbol{u}}{\rm{d}}\mathit{\boldsymbol{x}} \le \parallel \mathit{\boldsymbol{ \boldsymbol{\nabla}}} \phi {\parallel _{{L^3}}}\parallel \mathit{\boldsymbol{ \boldsymbol{\Delta}}} \phi \parallel {\rm{\cdot}}\parallel {\boldsymbol{u}}{\parallel _{{L^6}}} \le \parallel \mathit{\boldsymbol{ \boldsymbol{\nabla}}} \phi {\parallel ^{\frac{1}{2}}}{\left\| {{\mathit{\boldsymbol{ \boldsymbol{\nabla}}} ^2}\phi } \right\|^{\frac{1}{2}}}\left\| {{\mathit{\boldsymbol{ \boldsymbol{\nabla}}} ^2}\phi } \right\|\parallel \mathit{\boldsymbol{ \boldsymbol{\nabla}}} \mathit{\boldsymbol{u}}\parallel \le M\left( {{{\left\| {{\mathit{\boldsymbol{ \boldsymbol{\nabla}}} ^2}\phi } \right\|}^2} + \parallel \mathit{\boldsymbol{ \boldsymbol{\nabla}}} \mathit{\boldsymbol{u}}{\parallel ^2}} \right)$
当k =1时，$I_2^4 = \int \mathit{\boldsymbol{ \boldsymbol{\nabla}}} (\mathit{\boldsymbol{ \boldsymbol{\nabla}}} \phi \mathit{\boldsymbol{ \boldsymbol{\Delta}}} \phi ) \cdot \mathit{\boldsymbol{ \boldsymbol{\nabla}}} \mathit{\boldsymbol{u}}{\rm{d}}\mathit{\boldsymbol{x}} = - \int {(\mathit{\boldsymbol{ \boldsymbol{\nabla}}} \phi \mathit{\boldsymbol{ \boldsymbol{\Delta}}} \phi )} \cdot {\mathit{\boldsymbol{ \boldsymbol{\nabla}}} ^2}\mathit{\boldsymbol{u}}{\rm{d}}\mathit{\boldsymbol{x}} \le \parallel \mathit{\boldsymbol{ \boldsymbol{\nabla}}} \phi {\parallel _{{L^3}}}\parallel \mathit{\boldsymbol{ \boldsymbol{\Delta}}} \phi {\parallel _{{L^6}}}\left\| {{\mathit{\boldsymbol{ \boldsymbol{\nabla}}} ^2}\mathit{\boldsymbol{u}}} \right\| \le \parallel \mathit{\boldsymbol{ \boldsymbol{\nabla}}} \phi {\parallel ^{\frac{1}{2}}}{\left\| {{\mathit{\boldsymbol{ \boldsymbol{\nabla}}} ^2}\phi } \right\|^{\frac{1}{2}}}\left\| {{\mathit{\boldsymbol{ \boldsymbol{\nabla}}} ^3}\phi } \right\|\left\| {{\mathit{\boldsymbol{ \boldsymbol{\nabla}}} ^2}\mathit{\boldsymbol{u}}} \right\| \le M \cdot \left( {{{\left\| {{\mathit{\boldsymbol{ \boldsymbol{\nabla}}} ^2}\phi } \right\|}^2} + \parallel \mathit{\boldsymbol{ \boldsymbol{\nabla}}} \mathit{\boldsymbol{u}}{\parallel ^2}} \right)$
k ≥2时，$I_2^4 = \int {{\mathit{\boldsymbol{ \boldsymbol{\nabla}}} ^k}} (\mathit{\boldsymbol{ \boldsymbol{\nabla}}} \phi \mathit{\boldsymbol{ \boldsymbol{\Delta}}} \phi ) \cdot {\mathit{\boldsymbol{ \boldsymbol{\nabla}}} ^k}\mathit{\boldsymbol{u}}{\rm{d}}\mathit{\boldsymbol{x}} = - \int {{\mathit{\boldsymbol{ \boldsymbol{\nabla}}} ^{k - 1}}} \cdot (\mathit{\boldsymbol{ \boldsymbol{\nabla}}} \phi \mathit{\boldsymbol{ \boldsymbol{\Delta}}} \phi ) \cdot {\mathit{\boldsymbol{ \boldsymbol{\nabla}}} ^{k + 1}}\mathit{\boldsymbol{u}}{\rm{d}}\mathit{\boldsymbol{x}} = - \sum\limits_{0 \le l \le k - 1} {C_{k - 1}^l} \int {{\mathit{\boldsymbol{ \boldsymbol{\nabla}}} ^{l + 1}}} \phi \cdot {\mathit{\boldsymbol{ \boldsymbol{\nabla}}} ^{k - l + 1}}\phi \cdot {\mathit{\boldsymbol{ \boldsymbol{\nabla}}} ^{k + 1}}\mathit{\boldsymbol{u}}{\rm{d}}\mathit{\boldsymbol{x}} \le \sum\limits_{0 \le l \le k - 1} {\left\| {{\mathit{\boldsymbol{ \boldsymbol{\nabla}}} ^{l + 1}}\phi \cdot {\mathit{\boldsymbol{ \boldsymbol{\nabla}}} ^{k - l + 1}}\phi } \right\|} \cdot \left\| {{\mathit{\boldsymbol{ \boldsymbol{\nabla}}} ^{k + 1}}\mathit{\boldsymbol{u}}} \right\|$
$\begin{aligned}& \left\|\mathit{\boldsymbol{ \boldsymbol{\nabla}}}^{l+1} \phi \mathit{\boldsymbol{ \boldsymbol{\nabla}}}^{k-l+1} \phi\right\| \leqslant\left\|\mathit{\boldsymbol{ \boldsymbol{\nabla}}}^{l+1} \phi\right\|_{L^3} \cdot \\& \left\|\mathit{\boldsymbol{ \boldsymbol{\nabla}}}^{k-l+1} \phi\right\|_{L^6} \leqslant\left\|\mathit{\boldsymbol{ \boldsymbol{\nabla}}}^\alpha \phi\right\|^{1-\frac{l}{k}}\left\|\mathit{\boldsymbol{ \boldsymbol{\nabla}}}^{k+2} \phi\right\|^{\frac{l}{k}} . \\& \left\|\mathit{\boldsymbol{ \boldsymbol{\nabla}}}^2 \phi\right\|^{\frac{l}{k}}\left\|\mathit{\boldsymbol{ \boldsymbol{\nabla}}}^{k+2} \phi\right\|^{1-\frac{l}{k}} \leqslant M\left\|\mathit{\boldsymbol{ \boldsymbol{\nabla}}}^{k+2} \phi\right\|\end{aligned}$
其中α 由$\frac{l}{3} = \left( {\frac{\alpha }{3} - \frac{1}{2}} \right)\left( {1 - \frac{l}{k}} \right) + \left( {\frac{{k + 2}}{3} - \frac{1}{2}} \right){\rm{\cdot}}\frac{l}{k}$
$I_2^4 \leqslant M\left(\left\|\mathit{\boldsymbol{ \boldsymbol{\nabla}}}^{k+2} \phi\right\|^2+\left\|\mathit{\boldsymbol{ \boldsymbol{\nabla}}}^{k+1} \boldsymbol{u}\right\|^2\right)$
$\begin{array}{*{20}{l}} {I_2^5 = \sum\limits_{0 \leqslant l \leqslant k} {C_k^l} \int {{\nabla ^l}} \nu (\phi )\boldsymbol{D}\left( {{\nabla ^{k - l}}\boldsymbol{u}} \right):\nabla {\nabla ^k}\boldsymbol{u}{\text{d}}\boldsymbol{x} \leqslant } \\ {\sum\limits_{0 \leqslant l \leqslant k} {{{\left\| {{\nabla ^l}\phi } \right\|}_{{L^3}}}} \left\| {{\nabla ^{k + 1}}\boldsymbol{u}} \right\|{{\left\| {{\nabla ^{k - l + 1}}\boldsymbol{u}} \right\|}_{{L^6}}}} \end{array}$
$\begin{aligned}& \left\|\mathit{\boldsymbol{ \boldsymbol{\nabla}}}^l \phi\right\|_{L^3}\left\|\mathit{\boldsymbol{ \boldsymbol{\nabla}}}^{k-l+1} \boldsymbol{u}\right\|_{L^6} \leqslant\left\|\mathit{\boldsymbol{ \boldsymbol{\nabla}}}^\alpha \phi\right\|^{1-\frac{l-1}{k}} \cdot \\& \left\|\mathit{\boldsymbol{ \boldsymbol{\nabla}}}^{k+2} \phi\right\|^{\frac{l-1}{k}}\|\mathit{\boldsymbol{ \boldsymbol{\nabla}}} \boldsymbol{u}\|^{\frac{l-1}{k}}\left\|\mathit{\boldsymbol{ \boldsymbol{\nabla}}}^{k+1} \boldsymbol{u}\right\|^{1-\frac{l-1}{k}} \leqslant M \cdot \\& \left(\left\|\mathit{\boldsymbol{ \boldsymbol{\nabla}}}^{k+2} \phi\right\|+\left\|\mathit{\boldsymbol{ \boldsymbol{\nabla}}}^{k+1} \boldsymbol{u}\right\|\right)\end{aligned}$
其中α 由$\frac{{l - 1}}{3} = \left( {\frac{\alpha }{3} - \frac{1}{2}} \right)\left( {1 - \frac{{l - 1}}{k}} \right) + \left( {\frac{{k + 2}}{3} - } \right.\left. {\frac{1}{2}} \right)\frac{{l - 1}}{k}$
故$I_2^5 \leqslant M\left(\left\|\mathit{\boldsymbol{ \boldsymbol{\nabla}}}^{k+2} \phi\right\|^2+\left\|\mathit{\boldsymbol{ \boldsymbol{\nabla}}}^{k+1} \boldsymbol{u}\right\|^2\right)$
$\begin{aligned}& I_2^6=\frac{1}{\bar{\rho}} \sum\limits_{1 \leqslant l \leqslant k} C_k^l \int \mathit{\boldsymbol{ \boldsymbol{\nabla}}}^l(\nu(\phi)+\lambda(\phi)) \operatorname{div} \mathit{\boldsymbol{ \boldsymbol{\nabla}}}^{k-l} \cdot \\& \boldsymbol{u} \operatorname{div} \mathit{\boldsymbol{ \boldsymbol{\nabla}}}^k \boldsymbol{u} \mathrm{d} \boldsymbol{x} \leqslant M\left\|\mathit{\boldsymbol{ \boldsymbol{\nabla}}}^{k+1} \boldsymbol{u}\right\|^2\end{aligned}$
$\begin{aligned}& I_2^7=\int \mathit{\boldsymbol{ \boldsymbol{\nabla}}}^k\left(h_2(\sigma) \mathit{\boldsymbol{ \boldsymbol{\nabla}}} \phi \mathit{\boldsymbol{ \boldsymbol{\Delta}}} \phi\right) \cdot \mathit{\boldsymbol{ \boldsymbol{\nabla}}}^k \boldsymbol{u} \mathrm{d} \boldsymbol{x} \leqslant M\left(\| \mathit{\boldsymbol{ \boldsymbol{\nabla}}}^{k+2} \cdot\right. \\& \left.\phi\left\|^2+\right\| \mathit{\boldsymbol{ \boldsymbol{\nabla}}}^{k+1} \boldsymbol{u} \|^2\right)\end{aligned}$
将$I_2^i(i=1, 2, \cdots, 7)$
引理4 若(σ , u φ )∈X M T ])是方程(10)的局部解，则
(31) $\begin{aligned}& \quad \frac{\mathrm{d}}{\mathrm{d} t} \int \mathit{\boldsymbol{ \boldsymbol{\nabla}}}^k \boldsymbol{u} \cdot \mathit{\boldsymbol{ \boldsymbol{\nabla}}}^{k+1} \sigma \mathrm{d} \boldsymbol{x}+\frac{p^{\prime}(\bar{\rho})}{2 \bar{\rho}}\left\|\mathit{\boldsymbol{ \boldsymbol{\nabla}}}^{k+1} \sigma\right\|^2 \leqslant \\& M\left(\left\|\mathit{\boldsymbol{ \boldsymbol{\nabla}}}^{k+2} \boldsymbol{u}\right\|^2+\left\|\mathit{\boldsymbol{ \boldsymbol{\nabla}}}^{k+3} \phi\right\|^2\right)+\left\|\mathit{\boldsymbol{ \boldsymbol{\nabla}}}^{k+1} \boldsymbol{u}\right\|^2, k= \\& 0, 1, 2\end{aligned}$
证明：对方程组(29)的第二式乘以$\mathit{\boldsymbol{ \boldsymbol{\nabla}}}^{k+1} \sigma$
(32) $\begin{array}{*{20}{l}} {\frac{{\text{d}}}{{{\text{d}}t}}\int {{\nabla ^k}} \boldsymbol{u} \cdot {\nabla ^{k + 1}}\sigma {\text{d}}\boldsymbol{x} + \frac{{{p^\prime }(\bar \rho )}}{{\bar \rho }}{{\left\| {{\nabla ^{k + 1}}\sigma } \right\|}^2} - \bar \rho \cdot } \\ {\int {{{\left( {\operatorname{div} {\nabla ^k}\boldsymbol{u}} \right)}^2}} {\text{d}}\boldsymbol{x} \leqslant I_3^1 + I_3^2 + I_3^3 + I_3^4 + I_3^5 + I_3^6 + I_3^7} \end{array}$
$\begin{aligned}& I_3^1=\int \mathit{\boldsymbol{ \boldsymbol{\nabla}}}^k \operatorname{div}(\boldsymbol{u} \sigma) \operatorname{div} \mathit{\boldsymbol{ \boldsymbol{\nabla}}}^k \boldsymbol{u} \mathrm{d} \boldsymbol{x} \leqslant\left\|\mathit{\boldsymbol{ \boldsymbol{\nabla}}}^{k+1} \boldsymbol{u}\right\| \cdot \\& \left\|\mathit{\boldsymbol{ \boldsymbol{\nabla}}}^{k+1}(\boldsymbol{u} \sigma)\right\| \leqslant\left(\|\boldsymbol{\sigma}\|_{L^{\infty}}\left\|\mathit{\boldsymbol{ \boldsymbol{\nabla}}}^{k+1} \boldsymbol{u}\right\|+\left\|\mathit{\boldsymbol{ \boldsymbol{\nabla}}}^{k+1} \sigma\right\| \cdot\right. \\& \left.\|\boldsymbol{u}\|_{L^{\infty}}\right)\left\|\mathit{\boldsymbol{ \boldsymbol{\nabla}}}^{k+1} \boldsymbol{u}\right\| \leqslant M\left(\left\|\mathit{\boldsymbol{ \boldsymbol{\nabla}}}^{k+1} \sigma\right\|^2+\right. \\& \left.\left\|\mathit{\boldsymbol{ \boldsymbol{\nabla}}}^{k+1} \boldsymbol{u}\right\|^2\right)\end{aligned}$
$\begin{aligned}& I_3^2=\int \mathit{\boldsymbol{ \boldsymbol{\nabla}}}^k\left[-(\boldsymbol{u}, \mathit{\boldsymbol{ \boldsymbol{\nabla}}}) \boldsymbol{u}+h_1(\sigma) \mathit{\boldsymbol{ \boldsymbol{\nabla}}} \boldsymbol{\sigma}\right] \cdot \mathit{\boldsymbol{ \boldsymbol{\nabla}}}^{k+1} \sigma \mathrm{d} \boldsymbol{x}= \\& \int \mathit{\boldsymbol{ \boldsymbol{\nabla}}}^k[-(\boldsymbol{u}, \mathit{\boldsymbol{ \boldsymbol{\nabla}}}) \boldsymbol{u}] \cdot \mathit{\boldsymbol{ \boldsymbol{\nabla}}}^{k+1} \sigma \mathrm{d} \boldsymbol{x}+\int \mathit{\boldsymbol{ \boldsymbol{\nabla}}}^k\left[h_1(\sigma) \mathit{\boldsymbol{ \boldsymbol{\nabla}}} \sigma\right] \cdot \\& \mathit{\boldsymbol{ \boldsymbol{\nabla}}}^{k+1} \sigma \mathrm{d} \boldsymbol{x}\end{aligned}$
$\begin{aligned}& \quad \int \mathit{\boldsymbol{ \boldsymbol{\nabla}}}^k[-(\boldsymbol{u}, \mathit{\boldsymbol{ \boldsymbol{\nabla}}}) \boldsymbol{\mu}] \cdot \mathit{\boldsymbol{ \boldsymbol{\nabla}}}^{k+1} \sigma \mathrm{d} \boldsymbol{x} \leqslant \sum\limits_{0 \leqslant l \leqslant k} C_k^l \int \mathit{\boldsymbol{ \boldsymbol{\nabla}}}^l \boldsymbol{u} \cdot \\& \mathit{\boldsymbol{ \boldsymbol{\nabla}}}^{k-l+1} \boldsymbol{u} \cdot \mathit{\boldsymbol{ \boldsymbol{\nabla}}}^{k+1} \sigma \mathrm{d} \boldsymbol{x} \leqslant M \sum\limits_{0 \leqslant l \leqslant k}\left\|\mathit{\boldsymbol{ \boldsymbol{\nabla}}}^l \boldsymbol{u} \cdot \mathit{\boldsymbol{ \boldsymbol{\nabla}}}^{k-l+1} \boldsymbol{u}\right\| \cdot \\& \left\|\mathit{\boldsymbol{ \boldsymbol{\nabla}}}^{k+1} \boldsymbol{\sigma}\right\|\end{aligned}$
$\begin{aligned}& \left\|\mathit{\boldsymbol{ \boldsymbol{\nabla}}}^l \boldsymbol{u} \cdot \mathit{\boldsymbol{ \boldsymbol{\nabla}}}^{k-l+1} \boldsymbol{u}\right\| \leqslant\left\|\mathit{\boldsymbol{ \boldsymbol{\nabla}}}^l \boldsymbol{u}\right\|_{L^3}\left\|\mathit{\boldsymbol{ \boldsymbol{\nabla}}}^{k-l+1} \boldsymbol{u}\right\|_{L^6} \leqslant\\& \left\|\mathit{\boldsymbol{ \boldsymbol{\nabla}}}^\alpha \boldsymbol{u}\right\|^{1-\frac{l}{k+1}}\left\|\mathit{\boldsymbol{ \boldsymbol{\nabla}}}^{k+2} \boldsymbol{u}\right\|^{\frac{l}{k+1}}\|\mathit{\boldsymbol{ \boldsymbol{\nabla}}} \boldsymbol{u}\|^{\frac{l}{k+1}} \| \mathit{\boldsymbol{ \boldsymbol{\nabla}}}^{k+2} \cdot\\& \boldsymbol{u}\left\|^{1-\frac{l}{k+1}} \leqslant M\right\| \mathit{\boldsymbol{ \boldsymbol{\nabla}}}^{k+2} \boldsymbol{u} \|\end{aligned}$
其中α 由$\frac{{l - 1}}{3} = \left( {\frac{\alpha }{3} - \frac{1}{2}} \right)\left( {1 - \frac{l}{{k + 1}}} \right) + \left( {\frac{{k + 2}}{3} - } \right.\left. {\frac{1}{2}} \right)\frac{l}{{k + 1}}$
$\int \mathit{\boldsymbol{ \boldsymbol{\nabla}}}^k\left[h_1(\sigma) \mathit{\boldsymbol{ \boldsymbol{\nabla}}} \sigma\right] \cdot \mathit{\boldsymbol{ \boldsymbol{\nabla}}}^{k+1} \sigma \mathrm{d} \boldsymbol{x} \leqslant M\left\|\mathit{\boldsymbol{ \boldsymbol{\nabla}}}^{k+1} \sigma\right\|^2$ $I_3^2 \leqslant M\left(\left\|\mathit{\boldsymbol{ \boldsymbol{\nabla}}}^{k+1} \sigma\right\|^2+\left\|\mathit{\boldsymbol{ \boldsymbol{\nabla}}}^{k+2} \boldsymbol{u}\right\|^2\right)$
$\begin{aligned}& I_3^3=-\int \mathit{\boldsymbol{ \boldsymbol{\nabla}}}^k\left\{2 h_2(\sigma) \operatorname{div}[\nu(\phi) \boldsymbol{D}(\boldsymbol{u})]\right\} \mathit{\boldsymbol{ \boldsymbol{\nabla}}}^{k+1} \cdot \\& \sigma \mathrm{d} \boldsymbol{x}-\int \mathit{\boldsymbol{ \boldsymbol{\nabla}}}^k\left\{h_2(\sigma) \mathit{\boldsymbol{ \boldsymbol{\nabla}}}[(\nu(\phi)+\lambda(\phi)) \operatorname{div} \boldsymbol{u}]\right\} \mathit{\boldsymbol{ \boldsymbol{\nabla}}}^{k+1} \sigma \cdot \\& \mathrm{d} \boldsymbol{x}=-\sum\limits_{0 \leqslant l \leqslant k} C_k^l \int \mathit{\boldsymbol{ \boldsymbol{\nabla}}}^l h_2(\sigma) \operatorname{div} \mathit{\boldsymbol{ \boldsymbol{\nabla}}}^{k-l+1}[\nu(\phi) \boldsymbol{D}(\boldsymbol{u})] \cdot \\& \mathit{\boldsymbol{ \boldsymbol{\nabla}}}^{k+1} \sigma \mathrm{d} \boldsymbol{x}-\sum\limits_{0 \leqslant l \leqslant k} C_{k-1}^l \int \mathit{\boldsymbol{ \boldsymbol{\nabla}}}^l h_2(\sigma) \mathit{\boldsymbol{ \boldsymbol{\nabla}}}^{k-l}[(\nu(\phi)+ \\& \lambda(\phi)) \operatorname{div} \boldsymbol{u}] \cdot \mathit{\boldsymbol{ \boldsymbol{\nabla}}}^{k+1} \sigma \mathrm{d} \boldsymbol{x} \leqslant M\left(\left\|\mathit{\boldsymbol{ \boldsymbol{\nabla}}}^{k+1} \sigma\right\|^2+\right. \\& \left.\left\|\mathit{\boldsymbol{ \boldsymbol{\nabla}}}^{k+2} \boldsymbol{u}\right\|^2+\left\|\mathit{\boldsymbol{ \boldsymbol{\nabla}}}^{k+3} \phi\right\|^2\right)\end{aligned}$
$\begin{aligned}& I_3^4=-\frac{1}{\bar{\rho}} \int \mathit{\boldsymbol{ \boldsymbol{\nabla}}}^k(\mathit{\boldsymbol{ \boldsymbol{\nabla}}} \phi \mathit{\boldsymbol{ \boldsymbol{\Delta}}} \phi) \cdot \mathit{\boldsymbol{ \boldsymbol{\nabla}}}^{k+1} \sigma \mathrm{d} \boldsymbol{x} \leqslant \\& -\sum\limits_{0 \leqslant l \leqslant k} C_k^l \int \mathit{\boldsymbol{ \boldsymbol{\nabla}}}^{l+1} \phi \mathit{\boldsymbol{ \boldsymbol{\nabla}}}^{k-l} \mathit{\boldsymbol{ \boldsymbol{\Delta}}} \phi \cdot \mathit{\boldsymbol{ \boldsymbol{\nabla}}}^{k+1} \sigma \mathrm{d} \boldsymbol{x} \leqslant \sum\limits_{0 \leqslant l \leqslant k} \| \mathit{\boldsymbol{ \boldsymbol{\nabla}}}^{l+1} \phi \cdot \\& \mathit{\boldsymbol{ \boldsymbol{\nabla}}}^{k-l+2} \phi\|\| \mathit{\boldsymbol{ \boldsymbol{\nabla}}}^{k+1} \sigma \|\end{aligned}$
$\begin{aligned}& \left\|\mathit{\boldsymbol{ \boldsymbol{\nabla}}}^{l+1} \phi \cdot \mathit{\boldsymbol{ \boldsymbol{\nabla}}}^{k-l+2} \phi\right\| \leqslant\left\|\mathit{\boldsymbol{ \boldsymbol{\nabla}}}^{l+1} \phi\right\|_{L^3} \cdot \\& \left\|\mathit{\boldsymbol{ \boldsymbol{\nabla}}}^{k-l+2} \phi\right\|_{L^6} \leqslant\left\|\mathit{\boldsymbol{ \boldsymbol{\nabla}}}^\alpha \phi\right\|^{1-\frac{l+1}{k+2}} \mid\left\|\mathit{\boldsymbol{ \boldsymbol{\nabla}}}^{k+3} \phi\right\|^{\frac{l+1}{k+2}} \cdot \\& \left\|\mathit{\boldsymbol{ \boldsymbol{\nabla}}}^2 \phi\right\|^{\frac{l+1}{k+2}}\left\|\mathit{\boldsymbol{ \boldsymbol{\nabla}}}^{k+3} \phi\right\|^{1-\frac{l+1}{k+2}} \leqslant M\left\|\mathit{\boldsymbol{ \boldsymbol{\nabla}}}^{k+3} \phi\right\|\end{aligned}$
其中α 由$\frac{l}{3} = \left( {\frac{\alpha }{3} - \frac{1}{2}} \right)\left( {1 - \frac{{l + 1}}{{k + 2}}} \right) + \left( {\frac{{k + 3}}{3} - } \right.\left. {\frac{1}{2}} \right)\frac{{l + 1}}{{k + 2}}$
$I_3^4 \leqslant M\left(\left\|\mathit{\boldsymbol{ \boldsymbol{\nabla}}}^{k+3} \phi\right\|^2+\left\|\mathit{\boldsymbol{ \boldsymbol{\nabla}}}^{k+1} \sigma\right\|^2\right)$
$\begin{aligned}& I_3^5=\frac{2}{\bar{\rho}} \int \operatorname{div} \mathit{\boldsymbol{ \boldsymbol{\nabla}}}^k(\nu(\phi) \boldsymbol{D}(\boldsymbol{u})) \cdot \mathit{\boldsymbol{ \boldsymbol{\nabla}}}^{k+1} \sigma \mathrm{d} \boldsymbol{x} \leqslant \\& \int \operatorname{div} \mathit{\boldsymbol{ \boldsymbol{\nabla}}}^k(\nu(\phi) \boldsymbol{D}(\boldsymbol{u})) \cdot \mathit{\boldsymbol{ \boldsymbol{\nabla}}}^{k+1} \sigma \mathrm{d} \boldsymbol{x}=\sum\limits_{0 \leqslant l \leqslant k} C_{k+1}^l \int\left(\mathit{\boldsymbol{ \boldsymbol{\nabla}}}^l \cdot\right. \\& \left.\nu(\phi) \operatorname{div} \boldsymbol{D}\left(\mathit{\boldsymbol{ \boldsymbol{\nabla}}}^{k-l} \boldsymbol{u}\right)+\mathit{\boldsymbol{ \boldsymbol{\nabla}}}^{l+1} \nu(\phi) \boldsymbol{D}\left(\mathit{\boldsymbol{ \boldsymbol{\nabla}}}^{k-l} \boldsymbol{u}\right)\right) \cdot \\& \mathit{\boldsymbol{ \boldsymbol{\nabla}}}^{k+1} \sigma \mathrm{d} \boldsymbol{x} \leqslant \sum\limits_{1 \leqslant l \leqslant k}\left(\left\|\mathit{\boldsymbol{ \boldsymbol{\nabla}}}^l \phi\right\|_{L^3} \quad\left\|\mathit{\boldsymbol{ \boldsymbol{\nabla}}}^{k-l+2} \boldsymbol{u}\right\|_{L^6}+\right. \\& \left.\left\|\mathit{\boldsymbol{ \boldsymbol{\nabla}}}^{l+1} \phi\right\|_{L^3}\left\|\mathit{\boldsymbol{ \boldsymbol{\nabla}}}^{k-l+1} \boldsymbol{u}\right\|_{L^6}\right)\left\|\mathit{\boldsymbol{ \boldsymbol{\nabla}}}^{k+1} \sigma\right\| \leqslant M\left(\| \mathit{\boldsymbol{ \boldsymbol{\nabla}}}^{k+2} \cdot\right. \\& \left.\boldsymbol{u}\left\|^2+\right\| \mathit{\boldsymbol{ \boldsymbol{\nabla}}}^{k+1} \sigma \|^2\right)\end{aligned}$
$\begin{aligned}& I_3^6=\frac{1}{\bar{\rho}} \int \mathit{\boldsymbol{ \boldsymbol{\nabla}}}^{k+1}\{[\nu(\phi)+\lambda(\phi)] \operatorname{div} \boldsymbol{u}\} \cdot \mathit{\boldsymbol{ \boldsymbol{\nabla}}}^{k+1} \sigma \mathrm{d} \boldsymbol{x} \leqslant \\& \sum\limits_{0 \leqslant l \leqslant k} C_{k+1}^l \int \mathit{\boldsymbol{ \boldsymbol{\nabla}}}^l[\nu(\phi)+\lambda(\phi)] \cdot \operatorname{div} \mathit{\boldsymbol{ \boldsymbol{\nabla}}}^{k-l} \boldsymbol{u} \cdot \mathit{\boldsymbol{ \boldsymbol{\nabla}}}^{k+1} \sigma \mathrm{d} \boldsymbol{x} \leqslant \\& \sum\limits_{0 \leqslant l \leqslant k}\left\|\mathit{\boldsymbol{ \boldsymbol{\nabla}}}^l \phi\right\|_{L^3}\left\|\mathit{\boldsymbol{ \boldsymbol{\nabla}}}^{k-l+1} \boldsymbol{u}\right\|_{L^6}\left\|\mathit{\boldsymbol{ \boldsymbol{\nabla}}}^{k+1} \sigma\right\|\end{aligned}$
$\begin{aligned}& \left\|\mathit{\boldsymbol{ \boldsymbol{\nabla}}}^l \phi\right\|_{L^3}\left\|\mathit{\boldsymbol{ \boldsymbol{\nabla}}}^{k-l+1} \boldsymbol{u}\right\|_{L^6} \leqslant\left\|\mathit{\boldsymbol{ \boldsymbol{\nabla}}}^\alpha \phi\right\|^{1-\frac{l}{k}} \cdot \\& \left\|\mathit{\boldsymbol{ \boldsymbol{\nabla}}}^{k+3} \phi\right\|^{\frac{l}{k}}\|\mathit{\boldsymbol{ \boldsymbol{\nabla}}} \boldsymbol{u}\|^{\frac{l}{k}}\left\|\mathit{\boldsymbol{ \boldsymbol{\nabla}}}^{k+2} \boldsymbol{u}\right\|^{1-\frac{l}{k}} \leqslant M\left(\| \mathit{\boldsymbol{ \boldsymbol{\nabla}}}^{k+3} \cdot\right. \\& \left.\phi\|+\| \mathit{\boldsymbol{ \boldsymbol{\nabla}}}^{k+2} \boldsymbol{u} \|\right)\end{aligned}$
其中α 由$\frac{{l - 1}}{3} = \left( {\frac{\alpha }{3} - \frac{1}{2}} \right)\left( {1 - \frac{l}{k}} \right) + \left( {\frac{{k + 3}}{3} - } \right.\left. {\frac{1}{2}} \right)\frac{l}{k}$
$\begin{aligned}& I_3^6 \leqslant M\left(\left\|\mathit{\boldsymbol{ \boldsymbol{\nabla}}}^{k+3} \phi\right\|^2+\left\|\mathit{\boldsymbol{ \boldsymbol{\nabla}}}^{k+1} \sigma\right\|^2+\right. \\& \left.\left\|\mathit{\boldsymbol{ \boldsymbol{\nabla}}}^{k+2} \boldsymbol{u}\right\|^2\right)\end{aligned}$
$\begin{aligned}& I_3^7=\int \mathit{\boldsymbol{ \boldsymbol{\nabla}}}^k\left[h_2(\sigma) \mathit{\boldsymbol{ \boldsymbol{\nabla}}} \phi \mathit{\boldsymbol{ \boldsymbol{\Delta}}} \phi\right] \cdot \mathit{\boldsymbol{ \boldsymbol{\nabla}}}^{k+1} \sigma \mathrm{d} \boldsymbol{x} \leqslant M\left(\| \mathit{\boldsymbol{ \boldsymbol{\nabla}}}^{k+3} \cdot\right. \\& \left.\phi\left\|^2+\right\| \mathit{\boldsymbol{ \boldsymbol{\nabla}}}^{k+1} \sigma \|^2\right)\end{aligned}$
将$I_3^i(i=1, 2, \cdots, 7)$
