Lipschitz Stability for a Transport Coefficient Inverse Problem of a Linearly Coupled Korteweg-de Vries System

Wu Bin,, Chen Qun

 基金资助: 国家自然科学基金.  11601240

 Fund supported: the NSFC.  11601240

Abstract

This paper concerns an inverse problem of determining two spatially varying transport coefficients simultaneously in a linearly coupled Korteweg-de Vries (KdV) system with the first order terms. To obtain the stability result for the inverse problem with only one internal measurement data, we first prove a Carleman estimate including only one local integral for this coupled KdV system. Based on this Carleman estimate, we then obtain Lipschitz stability for the inverse problem under some priori information.

Keywords： Carleman estimate ; Coupled KdV system ; Lipschitz stability ; Coefficient inverse problem

Wu Bin, Chen Qun. Lipschitz Stability for a Transport Coefficient Inverse Problem of a Linearly Coupled Korteweg-de Vries System. Acta Mathematica Scientia[J], 2021, 41(3): 740-761 doi:

1 引言

$T>0$, $L>0$, $\Omega: = (0, L) $$\Omega_T: = \Omega\times (0, T) . 本文考虑一类带有一阶导数项耦合的线性Korteweg-de Vires (KdV) 方程组 \begin{eqnarray} \left\{\begin{array}{ll} u_t+ u_{xxx}+ \left(M_1(x)u\right)_{x}+k_1v_x = 0, &(x, t)\in \Omega_T, \\ v_t+ v_{xxx}+\left(M_2(x)v\right)_{x}+k_2u_x = 0, &(x, t)\in \Omega_T, \\ u(0, t) = g_1(t), \ u(L, t) = g_2(t), \ u_x(L, t) = g_3(t), &t\in (0, T), \\ v(0, t) = h_1(t), \ v(L, t) = h_2(t), \ v_x(L, t) = h_3(t), &t\in (0, T), \\ u(x, 0) = u_0(x), \quad v(x, 0) = v_0(x), &x\in\Omega, \end{array}\right. \end{eqnarray} 其中常数 k_1 , k_2\in {{\Bbb R}} 表示色散系数, 仅依赖于 x\in\Omega 的输运系数 M_1(x) , M_2(x) 表示相应线性波的速度. 方程组(1.1) 可以看作是包含两个耦合长波的KdV方程组[13] \begin{eqnarray} \left\{\begin{array}{ll} u_t+ \lambda_1u_{xxx}+ \alpha_1 u u_x+\delta u_x+k_1v_x = 0, &(x, t)\in \Omega_T, \\ v_t+\lambda_2 v_{xxx}+\alpha_2 v v_x-\delta v_x+k_2u_x = 0, &(x, t)\in \Omega_T \end{array}\right. \end{eqnarray} 的线性化形式. 这个方程组是用以描述分层流体中两个长重力内波的强相互作用. 对于更多的物理细节, 参见文献[12, 16]. 本文的目的是研究方程组(1.1) 中输运系数的反演问题. 更确切地说, 令 \omega\subset \Omega$$ \Omega$的一个适当子区域, $t_0$是某给定时刻, 本文希望通过下列测量数据

$\begin{eqnarray} v|_{\omega_T}\quad {\rm{及}}\quad (u, v)(x, t_0), \quad x\in\Omega, \ \ t_0 = \frac{T}{2} \end{eqnarray}$

2 适定性

$\begin{eqnarray} \left\{\begin{array}{ll} u_t+ u_{xxx}+ \left(M_1(x)u\right)_{x}+k_1v_x = 0, &(x, t)\in \Omega_T, \\ v_t+ v_{xxx}+\left(M_2(x)v\right)_{x}+k_2u_x = 0, &(x, t)\in \Omega_T, \\ u(0, t) = g_1(t), \ u(L, t) = g_2(t), \ u_{x}(L, t) = g_3(t), &t\in (0, T), \\ v(0, t) = h_1(t), \ v(L, t) = h_2(t), \ v_{x}(L, t) = h_3(t), &t\in (0, T), \\ u(x, 0) = u_0(x), \ v(x, 0) = v_0(x), &x\in\Omega. \end{array}\right. \end{eqnarray}$

$\begin{eqnarray} \left\{\begin{array}{ll} \varrho_t+ \varrho_{xxx}+ \left(M_1(x)\varrho\right)_{x}+k_1\vartheta_x = F, &(x, t)\in \Omega_T, \\ \vartheta_t+\vartheta_{xxx}+\left(M_2(x)\vartheta\right)_{x}+k_2\varrho_x = G, &(x, t)\in \Omega_T, \\ \varrho(0, t) = \varrho(L, t) = \varrho_{x}(L, t) = 0, &t\in (0, T), \\ \vartheta(0, t) = \vartheta(L, t) = \vartheta_{x}(L, t) = 0, &t\in (0, T), \\ \varrho(x, 0) = \varrho_0(x), \quad \vartheta(x, 0) = \vartheta_0(x), &x\in\Omega, \end{array}\right. \end{eqnarray}$

我们应用不动点方法证明方程组(2.2) 的解的存在性和唯一性. 为此, 定义函数集

$(\hat\varrho, \hat\vartheta)\in {{\cal B}}_R$. 考虑

$\begin{eqnarray} \left\{\begin{array}{ll} \varrho_t+ \varrho_{xxx} = -\left(M_1(x)\hat\varrho\right)_{x}-k_1\hat\vartheta_x+F, &(x, t)\in \Omega_T, \\ \vartheta_t+\vartheta_{xxx} = -(M_2(x)\hat\vartheta)_{x}-k_2\hat\varrho_x+G, &(x, t)\in \Omega_T, \\ \varrho(0, t) = \varrho(L, t) = \varrho_{x}(L, t) = 0, &t\in (0, T), \\ \vartheta(0, t) = \vartheta(L, t) = \vartheta_{x}(L, t) = 0, &t\in (0, T), \\ \varrho(x, 0) = \varrho_0(x), \ \ \vartheta(x, 0) = \vartheta_0(x), &x\in\Omega. \end{array}\right. \end{eqnarray}$

3 卡勒曼估计

$\begin{eqnarray} \left\{\begin{array}{ll} p_t+ p_{xxx}+(M_1(x)p)_x+k_1q_x = f_1, &(x, t)\in \Omega_T, \\ q_t+ q_{xxx}+(M_2(x)q)_{x}+k_2p_x = f_2, &(x, t)\in \Omega_T, \end{array}\right. \end{eqnarray}$

$$$\left\{\begin{array}{ll}p(0, t) = p(L, t) = p_{x}(L, t) = 0, &t\in (0, T), \\ q(0, t) = q(L, t) = q_{x}(L, t) = 0, &t\in (0, T).\end{array}\right.$$$

$$$\varphi(x, t) = \frac{e^{\lambda d(x)}-e^{2\lambda\|d\|_{C(\overline{\Omega})}}}{t(T-t)}, \quad\phi(x, t) = \frac{e^{\lambda d(x)}}{t(T-t)},$$$

$\begin{eqnarray} &&{{\cal I}}(p_x)+{{\cal I}}(q_x)+s{{\cal I}}(p)+s{{\cal I}}(q)+{\cal B} (p_x)|_{x = 0}+{\cal B} (q_x)|_{x = 0}\\ &\leq& C\int_{\Omega_T}s\phi(|f_1|^2+|f_{1, x}|^2+|f_2|^2+|f_{2, x}|^2)e^{2s\varphi}{\rm d}x{\rm d}t+Cs^{35}e^{Cs}\|q\|^2_{L^2(\omega_T)} \end{eqnarray}$

3.1 带非齐次Neumann边界KdV方程的卡勒曼估计

$\begin{eqnarray} y(L, t) = 0, \ y_{xx}(0, t) = \theta(t), \ y_{xx}(L, t) = 0, \quad t\in (0, T) \end{eqnarray}$

$$${\cal L}[y]: = y_t+ y_{xxx}, \quad (x, t)\in\Omega_T$$$

$\begin{eqnarray} ({\cal L}_1[Y], {\cal L}_2[Y]) = L(Y)+N(Y)+B(Y), \end{eqnarray}$

$\begin{eqnarray} \left\{\begin{array}{l} \varphi_x = \lambda d_x \phi, \quad \varphi_{xx} = \left(\lambda^2d_x^2+\lambda d_{xx}\right)\phi, \quad |\varphi_t|\leq Ce^{C\lambda}\phi^2, \\ |\varphi_{xxx}|+|\varphi_{xxxx}|\leq C\lambda^4\phi, \quad |\varphi_{xt}|\leq C\lambda\phi^2.\end{array}\right. \end{eqnarray}$

$\begin{eqnarray} L(Y)+N(Y) &\geq &\int_{{\Omega_T}}s\lambda^2\phi|y_{xx}|^2e^{2s\varphi}{\rm d}x{\rm d}t+\int_{\Omega_T}s^3\lambda^4\phi^3|y_x|^2e^{2s\varphi}{\rm d}x{\rm d}t+\int_{\Omega_T}s^5\lambda^6\phi^5|y|^2e^{2s\varphi}{\rm d}x{\rm d}t\\ &&-C\sum\limits_{x = \hat x}\int_0^T(s^2\lambda^4\phi^3|y_x|^2e^{2s\varphi}+s^4\lambda^6\phi^5|y|^2e^{2s\varphi})|_{\hat x = 0, L}{\rm d}t. \end{eqnarray}$

$\begin{eqnarray} \left\{\begin{array}{l} Y|_{x = 0, L} = (ye^{s\varphi})|_{x = 0, L}, \quad Y_x|_{x = 0, L} = (y_xe^{s\varphi}+s\varphi_xye^{s\varphi})|_{ x = 0, L}, \\ Y_{xx}|_{x = 0, L} = (y_{xx}e^{s\varphi}+2s\varphi_xy_xe^{s\varphi}+(s\varphi_{xx}+s^2\varphi_x^2)ye^{s\varphi})|_{ x = 0, L}. \end{array}\right. \end{eqnarray}$

$\begin{eqnarray} B(Y)-I_{11}^3&\geq &\int_0^T\left(s|\varphi_x||Y_{xx}|^2+s^3|\varphi_x|^3|Y_x|^2+s^5|\varphi_x|^5|Y|^2\right)|_{x = 0}{\rm d}t\\ &&-C\int_0^T\left(s|\varphi_x||Y_{xx}|^2+s^3|\varphi_x|^3|Y_x|^2+s^5|\varphi_x|^5|Y|^2\right)|_{x = L}{\rm d}t\\ &\geq &\frac{1}{4}\int_0^T\left(s^3\lambda^3\phi^3|y_x|^2e^{2s\varphi}+s^5\lambda^5\phi^5|y|^2e^{2s\varphi}\right)|_{x = 0}{\rm d}t\\ &&-C\int_0^T \left(s^3\lambda^3\phi^3|y_x|^2e^{2s\varphi}\right)|_{x = L}{\rm d}t. \end{eqnarray}$

$\begin{eqnarray} I_{11}^3& = &-3\int_0^T\left[s\varphi_x(y_x+s\varphi_x y)(y_t+s\varphi_t y)e^{2s\varphi}\right]_{x = 0}^{x = L}{\rm d}t\\ & = &\int_0^T\left(-\frac{3}{2}s^2\left(\varphi_x^2e^{2s\varphi}\right)_t|y|^2+3s^2\varphi_x\varphi_tyy_xe^{2s\varphi}+3s^3\varphi_x^2\varphi_t |y|^2e^{2s\varphi}\right)|_{x = 0}{\rm d}t\\ &&+\int_0^T(3s\varphi_xy_xy_t e^{2s\varphi})|_{x = 0}{\rm d}t\\ &\geq &-Ce^{C\lambda}\int_0^t \left(s^3\lambda^2\phi^4|y|^2e^{2s\varphi}\right)|_{ x = 0}{\rm d}t-Ce^{C\lambda}\int_0^t \left(s\phi^2|y_x|^2e^{2s\varphi}\right)|_{ x = 0}{\rm d}t\\ &&+3\int_0^T\left(s\varphi_xy_xy_t e^{2s\varphi}\right)|_{x = 0}{\rm d}t. \end{eqnarray}$

$\begin{eqnarray} &&-\int_0^T\left[(s\varphi_x e^{2s\varphi})|_{x = 0}y_xy_t\right]_{x = 0}^{x = L}{\rm d}t\\ & = &-\int_{\Omega_T}(s\varphi_xe^{2s\varphi})|_{x = 0}y_{xt}y_x{\rm d}x{\rm d}t-\int_{\Omega_T}(s\varphi_xe^{2s\varphi})|_{x = 0}y_{xx}(-y_{xxx}+{{\cal L}}[y]){\rm d}x{\rm d}t\\ & = &\frac{1}{2}\int_{\Omega_T}\left((s\varphi_xe^{2s\varphi})|_{x = 0}\right)_t|y_x|^2{\rm d}x{\rm d}t+\frac{1}{2}\int_{0}^T\left[(s\varphi_xe^{2s\varphi})|_{x = 0}|y_{xx}|^2\right]_{x = 0}^{x = L}{\rm d}t\\ &&-\int_{\Omega_T}(s\varphi_xe^{2s\varphi})|_{x = 0}y_{xx}{{\cal L}}[y]{\rm d}x{\rm d}t. \end{eqnarray}$

$\begin{eqnarray} \int_0^T\left(s\varphi_xy_xy_t e^{2s\varphi}\right)|_{x = 0}{\rm d}t &\geq &-Ce^{C\lambda}\int_{\Omega_T}s^2\lambda\phi^3|y_x|^2e^{2s\varphi}{\rm d}x{\rm d}t -C\int_0^T\left(s\lambda\phi|\theta|^2e^{2s\varphi}\right)|_{x = 0}{\rm d}t \\ &&-C\int_{\Omega_T}s\lambda\phi|y_{xx}|^2e^{2s\varphi}{\rm d}x{\rm d}t-C\int_{\Omega_T}s\lambda\phi |{\mathcal L}[y]|^2e^{2s\varphi}{\rm d}x{\rm d}t. \end{eqnarray}$

$\begin{eqnarray} B(Y) &\geq&\frac{1}{8}\int_0^T\left(s^3\lambda^3\phi^3|y_x|^2e^{2s\varphi}+s^5\lambda^5\phi^5|y|^2e^{2s\varphi}\right)|_{x = 0}{\rm d}t-C\int_0^T \left(s^3\lambda^3\phi^3|y_x|^2e^{2s\varphi}\right)|_{x = L}{\rm d}t\\ &&-C\int_0^T\left(s\lambda\phi|\theta|^2e^{2s\varphi}\right)|_{x = 0}{\rm d}t-C\int_{\Omega_T}s^3\lambda\phi^3|y_x|^2e^{2s\varphi}{\rm d}x{\rm d}t\\ &&-C\int_{\Omega_T}s\lambda\phi|y_{xx}|^2e^{2s\varphi}{\rm d}x{\rm d}t-C\int_{\Omega_T}s\lambda\phi |{\mathcal L}[y]|^2e^{2s\varphi}{\rm d}x{\rm d}t. \end{eqnarray}$

$\begin{eqnarray} {{\cal I}}(z)\leq C\int_{\Omega_T}|{\cal L}[z]|^2e^{2s\varphi}{\rm d}x{\rm d}t+Cs^{-1}{\cal I} (z_x) \end{eqnarray}$

$\begin{eqnarray} \left\{ \begin{array}{ll}{\cal L}[z]: = z_t+ z_{xxx}, &(x, t)\in\Omega_T, \\ z(0, t) = z(L, t) = z_x(L) = 0, & t\in (0, T) \end{array} \right. \end{eqnarray}$

$z\in L^2(0, T;H^3(\Omega))$都成立.

为简单记, 这里我们沿用引理3.2中的记号. 令$Z = ze^{s\varphi}$. 因为估计式(3.19) 不涉及边界条件, 所以(3.19) 式对于$Z$仍然成立, 即

$\begin{eqnarray} L(Z)+N(Z) & \geq & \int_{{\Omega_T}}s\lambda^2\phi|z_{xx}|^2e^{2s\varphi}{\rm d}x{\rm d}t+\int_{\Omega_T}s^3\lambda^4\phi^3|z_x|^2e^{2s\varphi}{\rm d}x{\rm d}t+\int_{\Omega_T}s^5\lambda^6\phi^5|z|^2e^{2s\varphi}{\rm d}x{\rm d}t\\ & &-C\sum\limits_{x = \hat x}\int_0^T(s^2\lambda^4\phi^3|z_x|^2e^{2s\varphi}+s^4\lambda^6\phi^5|z|^2e^{2s\varphi})|_{\hat x = 0, L}{\rm d}t. \end{eqnarray}$

$\begin{eqnarray} B(Z)-I_{11}^3&\geq &\int_0^T\left(s|\varphi_x||Z_{xx}|^2+s^3|\varphi_x|^3|Z_x|^2+s^5|\varphi_x|^5|Z|^2\right)|_{x = 0}{\rm d}t\\ &&-C\int_0^T\left(s|\varphi_x||Z_{xx}|^2+s^3|\varphi_x|^3|Z_x|^2+s^5|\varphi_x|^5|Z|^2\right)|_{x = L}{\rm d}t\\ &\geq &\int_0^T\left(s^3\lambda^3\phi^3|z_x|^2e^{2s\varphi}\right)|_{x = 0}{\rm d}t-C\int_0^T \left(s\lambda\phi|z_{xx}|^2e^{2s\varphi}\right)|_{x = L}{\rm d}t. \end{eqnarray}$

$\begin{eqnarray} B(Z)\geq \int_0^T\left(s^3\lambda^3\phi^3|z_x|^2e^{2s\varphi}\right)|_{x = 0}{\rm d}t-C\int_0^T \left(s\lambda\phi|z_{xx}|^2e^{2s\varphi}\right)|_{x = L}{\rm d}t. \end{eqnarray}$

$\begin{eqnarray} ({\cal L}_1[Z], {\cal L}_2[Z]) &\geq &\int_{{\Omega_T}}s\lambda^2\phi|z_{xx}|^2e^{2s\varphi}{\rm d}x{\rm d}t +\int_{\Omega_T}s^3\lambda^4\phi^3|z_x|^2e^{2s\varphi}{\rm d}x{\rm d}t \\ &&+\int_{\Omega_T}s^5\lambda^6\phi^5|z|^2e^{2s\varphi}{\rm d}x{\rm d}t-C\int_{0}^T\left(s\lambda\phi|z_{xx}|^2e^{2s\varphi}\right)|_{x = L}{\rm d}t. \end{eqnarray}$

$\begin{eqnarray} \|{\mathcal L}_1[Z]\|^2_{L^2(\Omega_T)}+\|{\mathcal L}_2[Z]\|^2_{L^2(\Omega_T)}+2({\cal L}_1[Z], {\cal L}_2[Z]) = \|e^{s\varphi}{\cal L}[z]+{\mathcal R}[Z]\|_{L^2(\Omega_T)}^2, \end{eqnarray}$

$$${{\cal I}}(z)\leq C\int_{\Omega_T}|{\cal L}[z]|^2e^{2s\varphi}{\rm d}x{\rm d}t+C\int_{0}^T\left(s\lambda\phi|z_{xx}|^2e^{2s\varphi}\right)|_{x = L}{\rm d}t.$$$

$\begin{eqnarray} \int_{0}^T\left(s\lambda\rho_1\phi|z_{xx}|^2e^{2s\varphi}\right)|_{x = L}{\rm d}t & = &2\int_{\Omega_T}s\lambda\rho_1\phi z_{xx}z_{xxx}e^{2s\varphi}{\rm d}x{\rm d}t+\int_{\Omega_T}s\lambda\left(\rho_1\phi e^{2s\varphi}\right)_x |z_{xx}|^2{\rm d}x{\rm d}t\\ &\leq& Cs^{-1}{\cal I} (z_x). \end{eqnarray}$

$\begin{eqnarray} \left\{\begin{array}{ll} p_{x, t}+ p_{x, xxx}+(M_1(x)p)_{xx}+k_1q_{xx} = f_{1, x}, &(x, t)\in \Omega_T, \\ q_{x, t}+ q_{x, xxx}+(M_2(x)q)_{xx}+k_2p_{xx} = f_{2, x}, &(x, t)\in \Omega_T, \\ p_{x}(L, t) = 0, \ p_{x, xx}(0, t) = \theta_1(t), \ p_{x, xx}(L, t) = 0, &t\in (0, T), \\ q_{x}(L, t) = 0, \ q_{x, xx}(0, t) = \theta_2(t), \ q_{x, xx}(L, t) = 0, &t\in (0, T), \end{array}\right. \end{eqnarray}$

$$${{\cal I}}(p)+{{\cal I}}(q) \leq C\int_{\Omega_T}(|p|^2+|p_x|^2+|q|^2+|q_x|^2+|f_1|^2+|f_2|^2)e^{2s\varphi}{\rm d}x{\rm d}t+Cs^{-1}({\cal I} (p_x)+{\cal I} (q_x)).$$$

$\begin{eqnarray} \int_0^T(s\lambda\phi(|f_1|^2+|f_2|^2)e^{2s\varphi})|_{x = 0}{\rm d}t & = &\int_0^T(s\lambda\phi e^{2s\varphi})|_{x = 0}{\rm d}t\left(\left|\int_\Omega f_{1, x} dx\right|^2+\left|\int_\Omega f_{2, x} dx\right|^2\right)\\ &\leq& \int_0^T(s\lambda\phi e^{2s\varphi})|_{x = 0}\int_{\Omega}(|f_{1, x}|^2+|f_{2, x}|^2){\rm d}x{\rm d}t\\ &\leq &C\int_{\Omega_T}s\lambda\phi(|f_{1, x}|^2+|f_{2, x}|^2)e^{2s\varphi}{\rm d}x{\rm d}t. \end{eqnarray}$

$\begin{eqnarray} &&{{\cal I}}(p_x)+{{\cal I}}(q_x)+s{\cal I} (p)+s{\cal I} (q)+{\cal B} (p_x)|_{x = 0}+{\cal B} (q_x)|_{x = 0}\\ &\leq &C\int_{\Omega_T}s\lambda\phi(|f_1|^2+|f_{1, x}|^2+|f_2|^2+|f_{2, x}|^2)e^{2s\varphi}{\rm d}x{\rm d}t\\ &&+C\int_{{\omega^{(3)}_T}}s^9\lambda^6\phi^9(|p_x|^2+|q_x|^2)e^{2s\varphi}{\rm d}x{\rm d}t. \end{eqnarray}$

$\begin{eqnarray} {\cal N} [\zeta] = A(x)\zeta_x(x)+B(x)\zeta(x), \quad x\in\Omega \end{eqnarray}$

$\begin{eqnarray} |A(x)(x-x_0)|>0, \quad x\in\Omega, \end{eqnarray}$

$\begin{eqnarray} \left\{\begin{array}{ll} \overline u_{t}+ \overline u_{xxx}+ \left(M_1(x)\overline u\right)_{x}+k_1\overline v_{x} = -({\overline M}_1(x)\tilde u)_x, &(x, t)\in \Omega_T, \\ \overline v_{t}+ \overline v_{xxx}+ \left(M_2(x)\overline v\right)_{x}+k_2\overline u_{x} = -({\overline M}_2(x)\tilde v)_x, &(x, t)\in \Omega_T, \\ \overline u (0, t) = \overline u(L, t) = \overline u_{x}(L, t) = 0, &t\in (0, T), \\ \overline v(0, t) = \overline v(L, t) = \overline v_{x}(L, t) = 0, &t\in (0, T). \end{array}\right. \end{eqnarray}$

$\begin{eqnarray} & &\int_{\Omega}s\left(|{\overline M}_{1, xx}(x)|^2+|{\overline M}_{1, x}(x)|^2+|{\overline M}_{1}(x)|^2\right)e^{2s\varphi(x, t_0)}{\rm d}x\\ &\leq&{\cal I} (\overline u_{xt})+{\cal I} ( {\overline u}_{xtt})+{\cal I} ( {\overline u}_{t})+{\cal I} ( {\overline u}_{tt})+{\cal I} (\overline v_x)+{\cal I} ( {\overline v}_{xt})+Cse^{Cs}\|{\overline u}(\cdot, t_0)\|^2_{H^4(\Omega)}\\ & \leq &C\sum\limits_{i = 1}^2\int_{\Omega_T}s\left(|{\overline M}_{i, xx}(x)|^2+|{\overline M}_{i, x}(x)|^2+|{\overline M}_i(x)|^2\right)e^{2s\varphi}{\rm d}x{\rm d}t\\ & &+Cs^{35}e^{Cs}\|\overline v\|_{H^2(0, T;L^2(\omega_T))}^2+Cse^{Cs}\|{\overline u}(\cdot, t_0)\|^2_{H^4(\Omega)}. \end{eqnarray}$

$\begin{eqnarray} &&\int_{\Omega}s\left(|{\overline M}_{2, xx}(x)|^2+|{\overline M}_{2, x}(x)|^2+|{\overline M}_{2}(x)|^2\right)e^{2s\varphi(x, t_0)}{\rm d}x\\ &\leq &C\sum\limits_{i = 1}^2\int_{\Omega_T}s\left(|{\overline M}_{i, xx}(x)|^2+|{\overline M}_{i, x}(x)|^2+|{\overline M}_i(x)|^2\right)e^{2s\varphi}{\rm d}x{\rm d}t\\ &&+Cs^{35}e^{Cs}\|\overline v\|_{H^2(0, T;L^2(\omega_T))}^2+Cse^{Cs}\|{\overline v}(\cdot, t_0)\|^2_{H^4(\Omega)}. \end{eqnarray}$

$\begin{eqnarray} &&\sum\limits_{i = 1}^2\int_{\Omega}s\left(|{\overline M}_{i, xx}(x)|^2+|{\overline M}_{i, x}(x)|^2+|{\overline M}_{i}(x)|^2\right)e^{2s\varphi(x, t_0)}{\rm d}x\\ &\leq &C\sum\limits_{i = 1}^2\int_{\Omega_T}s\left(|{\overline M}_{i, xx}(x)|^2+|{\overline M}_{i, x}(x)|^2+|{\overline M}_i(x)|^2\right)e^{2s\varphi}{\rm d}x{\rm d}t\\ & &+Cs^{35}e^{Cs}\|\overline v\|_{H^2(0, T;L^2(\omega_T))}^2+Cse^{Cs}\|{\overline u}(\cdot, t_0)\|^2_{H^4(\Omega)}+Cse^{Cs}\|{\overline v}(\cdot, t_0)\|^2_{H^4(\Omega)}. \end{eqnarray}$

$D: = \frac{4}{T^2} \min\limits_{(x, t)\in\Omega_T}\frac{e^{2\lambda\|d\|_{C(\overline \Omega)}}-e^{\lambda d(x)}}{t(T-t)}>0$.$s\rightarrow \infty$时, 我们有

$\begin{eqnarray} \sup\limits_{x\in\Omega}\left|\int_0^T e^{2s(\varphi(x, t)-\varphi(x, t_0))}{\rm d}t\right|\leq \int_0^T e^{-2sD\left(t-\frac{T}{2}\right)^2}{\rm d}t\leq \int_{-\infty}^{+\infty} e^{-2sD t^2}{\rm d}t = o(1). \end{eqnarray}$

$\begin{eqnarray} &&\sum\limits_{i = 1}^2\int_{\Omega_T}\left(|{\overline M}_{i, xx}(x)|^2+|{\overline M}_{i, x}(x)|^2+|{\overline M}_i(x)|^2\right)e^{2s\varphi}{\rm d}x{\rm d}t\\ & = &\sum\limits_{i = 1}^2\int_{\Omega}\left(|{\overline M}_{i, xx}(x)|^2+|{\overline M}_{i, x}(x)|^2+|{\overline M}_i(x)|^2\right)e^{2s\varphi(x, t_0)}{\rm d}x\int_0^T e^{2s(\varphi(x, t)-\varphi(x, t_0))}{\rm d}t\\ &\leq &\epsilon \sum\limits_{i = 1}^2\int_{\Omega}\left(|{\overline M}_{i, xx}(x)|^2+|{\overline M}_{i, x}(x)|^2+|{\overline M}_i(x)|^2\right)e^{2s\varphi(x, t_0)}{\rm d}x. \end{eqnarray}$

$\begin{eqnarray} &&(s- \epsilon C s)\sum\limits_{i = 1}^2\int_{\Omega_T}\left(|{\overline M}_{i, xx}(x)|^2+|{\overline M}_{i, x}(x)|^2+|{\overline M}_i(x)|^2\right)e^{2s\varphi(x, t_0)}{\rm d}x{\rm d}t\\ &\leq& Cs^{35}e^{Cs}\|\overline v\|_{H^2(0, T;L^2(\omega_T))}^2+Cse^{Cs}\|{\overline u}(\cdot, t_0)\|^2_{H^4(\Omega)}+Cse^{Cs}\|{\overline v}(\cdot, t_0)\|^2_{H^4(\Omega)}. \end{eqnarray}$

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