令$\Omega\subset \mathbb{R} ^3$是带有光滑边界$\partial \Omega:=\Gamma$的有界区域, $Q_T:=\Omega\times (0, T)$, $\Sigma_T=\Gamma\times (0, T)$. $\eta$表示在$\Omega$的边界$\Gamma$上的单位外法方向.我们考虑一类带有非奇异主部系数矩阵$(a_{ij})_{1\leq i, j\leq 2}$的强耦合微分方程组

$ \begin{eqnarray}\label{2-1.1} \left\{\begin{array}{l} \partial^2_{t}{u_1}-a_{11}\Delta {u_1}-a_{12}\Delta u_2=f_1, \quad (x, t)\in Q_T, \\ \partial^2_{t}{u_2}-a_{21}\Delta {u_1}-a_{22}\Delta u_2=f_2, \quad (x, t)\in Q_T, \end{array}\right. \end{eqnarray} $

(1.1)

这里$a_{ij}$是常数.本文所讨论的方程组(1.1)不仅仅局限于双曲类型.当矩阵$(a_{ij})_{1\leq i, j\leq 2}$有一个正的和一个负的特征值时, 该系统是一个双曲椭圆型方程组.本文研究这样一个强耦合微分方程组的卡勒曼估计及其反源问题的稳定性.

双曲型方程组(1.1)已被广泛应用于工程和物理领域^{[9, 13-14, 40]}.例如, 一维的Ⅱ型热弹性模型^[33]

$ \begin{eqnarray}\label{3.1.2}\left\{\begin{array}{l} \rho u_{tt}=(\lambda+2\mu+p)u_{xx}+m\alpha_{xt}, \\ c\alpha_{tt}=k\alpha_{xx}+mu_{xt}, \end{array}\right. \end{eqnarray} $

(1.2)

这里所有的物理参数$\rho, \lambda, \mu, p, m, c, k$都是常数.令$u_t=\hat u$, $\alpha_x=\hat \alpha$.则方程组(1.2)可以改写为类似于方程组(1.1)的新形式

$ \begin{eqnarray}\label{3.1.3}\left\{\begin{array}{l} \hat u_{tt}=\left(\frac{\lambda+2\mu+p}{\rho}+\frac{m^2}{\rho c}\right)\hat u_{xx}+\frac{mk}{\rho c}\hat\alpha_{xx}, \\[3mm] \hat\alpha_{tt}=\frac{k}{c}\hat\alpha_{xx}+\frac{m}{c}\hat u_{xx}. \end{array}\right. \end{eqnarray} $

(1.3)

卡勒曼估计是一类与微分算子有关的加权能量估计.它可以用于证明偏微分方程系数反问题的稳定性与唯一性.在文献[2]中, Bukhgeim与Klibanov提出了一种基于卡勒曼估计去解决反问题稳定性的有效方法.这种方法的应用也可见于后续的文献[5, 18, 21, 26-27, 31, 36].从那时起, 卡勒曼估计被许多研究者广泛的应用于证明偏微分方程系数反问题的稳定性和唯一性, 参见文献[3, 22, 28, 30, 36-37].关于最近的Bukhgeim-Klibanov方法的应用, 可以参见文献[24].对于卡勒曼估计的其他应用, 参见文献[12, 16, 20, 23, 25, 32, 34, 38-39]等.

最近, 卡勒曼估计被用于解决一些耦合微分方程组的系数反问题, 如Cristofol, Gaitan和Ramoul^[19], Benabdallah, Cristofol, Gaitan和Yamamoto^[1], Isakov和Kim^[36], Bellassoued和Yamamoto^[17], Paolo和Daniel^[4], Fan, Jiang和Nakamura^[11], Wu和Liu^[6].这些文章中, 作者通过将每个方程所对应的卡勒曼估计相加来消除耦合项所带来的困难.然而, 这种方法并不适用于强耦合微分方程组(1.1), 因为强耦合项不能被两个方程所对应的卡勒曼估计相加所得的左边项所吸收.对于强耦合方程组, 与上述方法不同, Wu和Liu^[7]将双曲系统对角线化, 证明了卡勒曼估计并应用于热弹性系统的反源问题. Bellassoued和Yamamoto^[15]研究了Biot方程的系数反问题, 他们证明了带有变系数的强耦合双曲系统的卡勒曼估计.此外, Fan和Chen^[10]建立了一个$2\times2$强耦合抛物系统的卡勒曼估计.这些文章都是应用将主部系数矩阵对角化的方法.更确切地说, 就是用矩阵$P$将主部系数矩阵$(a_{ij})_{1\leq i, j\leq 2}$对角化, 使得$P^{-1}(u_1, u_2)$满足一个弱耦合系统, 再应用已有的单个方程的卡勒曼估计相加得到需要证明的方程组的卡勒曼估计.在这些情况下, 矩阵$(a_{ij})_{1\leq i, j\leq 2}$是具有两个正特征值的正定矩阵.与这些文章相比, 我们在$a_{ij}$更弱的条件下证明了类似的卡勒曼估计，此时我们只需要$a_{ij}, 1\leq i, j\leq 2$是常数并且满足${\rm det}(a_{ij})_{1\leq i, j\leq 2}\neq 0$, 参见条件(A1).

在本文中, 我们用$t$和$x=(x_1, x_2, x_3)$分别代表时间和空间变量, 对于函数$v$, 我们令$\partial_tv=\frac{\partial v}{\partial t}$, $\nabla v=\nabla_x v=(\partial_{x_1} v, \partial_{x_2} v, \partial_{x_3} v)$, $\nabla_{x, t} v=\left(\nabla v, v_t\right)$和$|\nabla_{x, t} v|^2=|\nabla v|^2+|v_t|^2$.对于固定点$x_0\in\mathbb{R} ^3\setminus\bar\Omega$, 我们引入函数$\varphi$: $\Omega\times (0, T)\rightarrow \mathbb{R} $, 定义如下

$ \begin{eqnarray}\label{2-1.6} \varphi(x, t)=|x-x_0|^2-\theta t^2, \quad (x, t)\in \Omega\times (0, T), \end{eqnarray} $

(1.4)

其中$\theta$是正参数.定义两个常数

$ \begin{eqnarray*}d_0=\inf\limits_{x\in\overline\Omega}|x-x_0|^2 , A: d_1=\sup\limits_{x\in\overline\Omega}|x-x_0|^2.\end{eqnarray*} $

此外, 对于$\delta>0$, 我们定义集合

$ \begin{eqnarray*}Q(\delta):=\{(x, t)\in \Omega\times (0, \infty):\varphi(x, t)>\delta\}.\end{eqnarray*} $

易知$\partial Q(\delta)=\Sigma(\delta)\cup\Sigma_1(\delta)\cup (\Omega\times\{0\})$, 这里

$ \Sigma(\delta):=\left\{(x, t):t\in(0, t_\delta(x)), t_\delta(x):=\left(\frac{|x-x_0|^2-\delta}{\theta}\right)^{1/2}, x\in\Gamma\right\}, $

(1.5)

$ \Sigma_1(\delta):=\left\{(x, t)\in \Omega\times (0, \infty):\varphi(x, t)=\delta\right\}. $

(1.6)

此外我们用$C$表示某个正常数, 其值在不同的不等式中会有所变化.

现在我们给出定理1.1中所需的条件

(A1)  $a_{ij}, 1\leq i, j\leq 2$是正常数且满足

$ \begin{eqnarray} \det(a_{ij})_{1\leq i, j\leq 2}\neq 0, \quad a_{12}a_{21}>0. \end{eqnarray} $

(1.7)

这篇文章的主要结果是以下对于强耦合微分方程组(1.1)的全局卡勒曼估计.

定理1.1  设(A1)成立, $f_i\in L^2(Q_T), i=1, 2$.则对于任意$\delta > 0$, 存在正常数$\theta_1=\theta_1(\Omega, T, x_0, a_{ij})$, $s_1=s_1(\Omega, T, x_0, a_{ij})$和$C=C(\Omega, T, x_0, a_{ij})$满足对所有的$\theta\leq\theta_1$和$s\geq s_1$, 成立以下估计

$ \begin{eqnarray}\label{1-1.6} &&\sum\limits_{i=1}^2\int_{Q(\delta)}\left(s|\nabla_{x, t}u_i|^2+s^3|u_i|^2\right){\rm e}^{2s\varphi}{\rm d}x{\rm d}t\nonumber\\ &\leq& C\sum\limits_{i=1}^2\int_{Q(\delta)}|f_i|^2{\rm e}^{2s\varphi}{\rm d}x{\rm d}t+C\sum\limits_{i=1}^2\int_{\Sigma(\delta)}\left(s|\nabla_{x, t}u_i|^2+s^3|u_i|^2\right){\rm e}^{2s\varphi}{\rm d}S{\rm d}t\nonumber\\ &&+Cs^3{\rm e}^{4\delta s}\sum\limits_{i=1}^2\|u_i\|^2_{H^1(Q(\delta))}, \end{eqnarray} $

(1.8)

这里$(u_1, u_2)\in \left(H^2(Q_T)\right)^2$满足(1.1)式及$u_1(x, 0)=u_2(x, 0)=0$, $x\in \Omega$.

注1.1  此定理中矩阵$(a_{ij})_{1\leq i, j\leq 2}$的条件比文献[7, 10, 15]中的条件要弱, 那里要求矩阵$(a_{ij})_{1\leq i, j\leq 2}$是带有两个正特征值的正定矩阵.

作为定理1.1的应用, 我们考虑对于一个强耦合微分方程组的反源问题.考虑

$ \begin{equation} \label{1.7}\partial^2_{t}{u_1}-a_{11}\Delta {u_1}-a_{12}\Delta u_2=\sigma_1(x, t)p_1(x), \quad (x, t)\in Q_T, \end{equation} $

(1.9)

$ \begin{equation} \label{1.8}\partial^2_{t}{u_2}-a_{21}\Delta {u_1}-a_{22}\Delta u_2=\sigma_2(x, t)p_2(x), \quad (x, t)\in Q_T, \end{equation} $

(1.10)

$ \begin{equation} \label{110}u_1(x, 0)=u_2(x, 0)=0, \quad \partial_t u_1(x, 0)=\partial_t u_2(x, 0)=0, \quad x\in\Omega, \end{equation} $

(1.11)

其中$\sigma_i$是两个给定函数, $p_i$是两个未知源.

反源问题  通过观测数据

$ \begin{eqnarray*} \left.{u_i}\right.|_{\Sigma(\delta)}\ \ \mbox{及}\ \ \left.\frac{\partial u_i}{\partial\eta} \right|_{\Sigma(\delta)}, \quad i=1, 2, \end{eqnarray*} $

确定在(1.9)-(1.11)式中的函数$p_1$和$p_2$, 这里$\Sigma(\delta)$是由(1.5)式所定义的$\Sigma_T$的开集.

我们作以下假设

(A2)  $p_1, p_2\in L^2(\Omega)$, $u_1, u_2\in H^3(Q_T)$则存在正常数$M$使得

$ \begin{eqnarray} \sum\limits_{i=1}^2\left(\|p_i\|^2_{L^2(\Omega)}+\|u_i\|^2_{H^3(Q_T)}\right)\leq M; \end{eqnarray} $

(1.12)

(A3)  $T$满足

$ \begin{eqnarray}\label{112} T>\frac{\sup\limits_{x\in\overline{\Omega}}|x-x_0|}{\sqrt{\theta}}; \end{eqnarray} $

(1.13)

(A4)  $\sigma_1, \sigma_2\in H^1(Q_T)$满足

$ \begin{eqnarray} \label{1.12}|\sigma_i(x, 0)|>0, \quad x\in\overline\Omega, \ i=1, 2. \end{eqnarray} $

(1.14)

定理1.2  令$\theta>0$且充分小和$\delta\in (0, \frac{d_0}{2})$.则在假设(A1)-(A4)下, 存在两个常数$C=C(\Omega, T, x_0, \delta, \sigma_0, a_{ij}, M, \theta)>0$和$\kappa=\kappa(\Omega, T, x_0, \delta, \sigma_0, a_{ij}, M, \theta)\in (0, 1)$使得

$ \begin{eqnarray}\label{3.1.13} \sum\limits_{i=1}^2\|p_i\|_{L^2(\Omega)}\leq C\sum\limits_{i=1}^2\left(\|u_i\|_{H^1(\Sigma(\delta))}+\|\partial_t u_i\|_{H^1(\Sigma(\delta))}\right)^\kappa, \end{eqnarray} $

(1.15)

这里$(u_1, u_2)$满足(1.9)-(1.11)式.

注1.2  作为定理1.2的直接推论，可以得到反源问题的唯一性.

注1.3  (A3)表示观测时间不应该太小.

本文的其余部分安排如下:在第二节, 我们给出了带非奇异主部系数矩阵的强耦合微分方程组的逐点卡勒曼估计.第三节证明方程组(1.1)的全局卡勒曼估计, 即定理1.1.在最后一节, 我们给出了关于反源问题的Hölder稳定性, 即定理1.2的证明.

本节我们建立以下强耦合微分方程组

$ \begin{equation}\label{2-2.5} \left\{\begin{array}{l} U_{tt}-a\Delta U-b\Delta V=F, \quad (x, t)\in Q_T, \\ V_{tt}-c\Delta U-d\Delta V=G, \quad (x, t)\in Q_T \end{array}\right. \end{equation} $

(2.1)

的逐点卡勒曼估计, 这里的系数$a, b, c, d$是正常数且满足

$ \begin{equation}\label{2-2.6} {\rm det}\Bigg(\begin{array}{cc} a~~&b\\ c~~&d\end{array}\Bigg)=ad-bc\neq 0, \quad bc>0. \end{equation} $

(2.2)

引理2.1  设$F, G\in L^2(\overline{Q_T})$, $U, V\in H^2(\overline{Q_T})$满足(2.1)和(2.2)式.则对于任意的$\delta>0$, 存在正常数$\theta=\theta_0(a, b, c, d, \Omega, T)>0$, $s=s_0(a, b, c, d, \Omega, T)>0$和$C=C(a, b, c, d, \Omega, T)>0$使得当$\theta\leq \theta_0$和$s\geq s_0$时有

$ \begin{equation}\label{2-2.7} (|F|^2+|G|^2){\rm e}^{2s\varphi}\geq Cs(|\nabla_{x, t}U|^2+|\nabla_{x, t}V|^2+s^2U^2+s^2V^2){\rm e}^{2s\varphi}+\nabla\cdot X+Y_t, \end{equation} $

(2.3)

其中$(x, t)\in Q(\delta)$.这里的函数$X=X(U, V)$和$Y=Y(U, V)$满足

$ \begin{equation}\label{2-2.8} |X(U, V)|+|Y(U, V)|\leq Cs\left(|\nabla_{x, t} U|^2+|\nabla_{x, t} V|^2+s^2U^2+s^2V^2\right). \end{equation} $

(2.4)

另外, 若$(U, V)(x, 0)=(0, 0)$或$(U_t, V_t)(x, 0)=(0, 0)$, 我们有

$ \begin{equation}\label{2-2.9} Y(x, 0)=0. \end{equation} $

(2.5)

证  令

$ \begin{equation}\label{2-2.10} W(x, t)=U(x, t)+m, V(x, t), \quad (x, t)\in Q_T, \end{equation} $

(2.6)

其中常数$m$为

$ m=\frac{d-a}{2c}. $

那么$(W, V)$满足

$ \begin{eqnarray}\label{2-2.11}\left\{\begin{array}{ll} {\cal L}_1[W, V]:=W_{tt}-\nu \Delta W-\tau_1\Delta V=F+mG, &\quad(x, t)\in Q_T, \\ {\cal L}_2[W, V]:=V_{tt}-\nu \Delta V-\tau_2\Delta W=G, &\quad(x, t)\in Q_T, \end{array}\right. \end{eqnarray} $

(2.7)

这里

$ \nu=a+cm, \quad \tau_1=b+cm^2, \quad \tau_2=c. $

由于$\nu, \tau_1, \tau_2>0$, 根据(2.2)式我们有

$ \begin{equation}\label{2-2.12} \nu^2-\tau_1\tau_2\neq 0, \quad{\rm i.e.}\quad \left|\nu-\sqrt{\tau_1\tau_2}\right|>0. \end{equation} $

(2.8)

令

$ \begin{equation} \tilde W=W {\rm e}^{s\varphi}, \quad \tilde V=V {\rm e}^{s\varphi}, \end{equation} $

(2.9)

则

$ W_{tt}=\left[\tilde W_{tt}+4s\theta t\tilde W_t+4s^2\left(\theta^2 t^2+O_1\left(\frac{1}{s}\right)\right)\tilde W\right]{\rm e}^{-s\varphi}, $

$ \Delta W=\left[\Delta \tilde W-4s(x-x_0)\cdot\nabla\tilde W+4s^2\left(|x-x_0|^2+O_2\left(\frac{1}{s}\right)\right)\tilde W\right]{\rm e}^{-s\varphi}. $

对于$V$我们有类似的等式.记

$ O_1\left(\frac{1}{s}\right)= \frac{\theta}{2s}, \quad O_2\left(\frac{1}{s}\right)=-\frac{3}{2s}. $

我们有

$ \begin{equation}\label{2-2.14} \left(\sigma{\cal L}_1[W, V]\right)^2{\rm e}^{2s\varphi}=\left(I_{11}+I_{12}+I_{13}+I_{14}\right)^2, \end{equation} $

(2.10)

其中

$ \begin{eqnarray*} I_{11}&=&\sigma\tilde W_{tt}-\sigma\nu\Delta \tilde W-4 \sigma \nu s^2\left(|x-x_0|^2-\frac{1}{\nu}\theta^2t^2+O_3\left(\frac{1}{s}\right)\right)\tilde W\nonumber\\ &&-\sigma\tau_1\Delta \tilde V-4\sigma\tau_1 s^2\left(|x-x_0|^2+O_2\left(\frac{1}{s}\right)\right)\tilde V, \\ \end{eqnarray*} $

$ I_{12}=4\sigma \theta st \tilde W_t, \quad I_{13}=4\sigma\nu s(x-x_0)\cdot\nabla\tilde W, \quad I_{14}=4\sigma\tau_1 s(x-x_0)\cdot\nabla\tilde V, $

这里

$ O_3\left(\frac{1}{s}\right)=-\frac{1}{\nu} O_1\left(\frac{1}{s}\right)+O_2\left(\frac{1}{s}\right). $

选取常数$\sigma>0$, 使得

$ \begin{eqnarray}\label{2-2.15}\sigma^2=\frac{\tau_2}{\tau_1}.\end{eqnarray} $

(2.11)

由(2.10)式, 可得

$ \begin{eqnarray}\label{2-2.16} \left(\sigma{\cal L}_1[W, V]\right)^2{\rm e}^{2s\varphi}\geq I_{11}^2+2I_{11}I_{12}+2I_{11}I_{13}+2I_{11}I_{14}. \end{eqnarray} $

(2.12)

下面我们分别估计不等式(2.12)右边的四项. $I_{11}^2$可改写为

$ \begin{eqnarray*} I_{11}^2&=&\frac{1}{2}(I_{11}-\sigma\nu\rho s\tilde W+\sigma\nu\rho s\tilde W)^2+\frac{1}{2}(I_{11}-12\sigma\tau_1 s\tilde V+12\sigma\tau_1 s\tilde V)^2\\ &=&J_1+J_2, \end{eqnarray*} $

其中$\rho$是待定的正常数.对于$I_{11}^2$中的$J_1$, 我们有

$ \begin{eqnarray*} J_1&\geq& \sigma\nu\rho s\tilde W I_{11}-\sigma^2\nu^2\rho^2 s^2\tilde W^2\nonumber\\ &=&(\sigma^2\nu\rho s\tilde W\tilde W_t)_t-\sigma^2\nu\rho s\tilde W_t^2-\nabla\cdot(\sigma^2\nu^2\rho s\tilde W\nabla\tilde W)+\sigma^2\nu^2\rho s|\nabla\tilde W|^2\nonumber\\ &&-4\sigma^2\nu^2\rho s^3\left(|x-x_0|^2-\frac{1}{\nu}\theta^2t^2+O_3\left(\frac{1}{s}\right)\right)\tilde W^2-\sigma^2\nu^2\rho^2s^2\tilde W^2\nonumber\\ &&-\sigma^2\tau_1\nu\rho s\left[\Delta \tilde V+4 s^2\left(|x-x_0|^2+O_2\left(\frac{1}{s}\right)\right)\tilde V\right]\tilde W. \end{eqnarray*} $

类似地

$ \begin{eqnarray*} J_2&\geq& 12\sigma\tau_1 s\tilde V I_{11}-12^2\sigma^2\tau_1^2s^2\tilde V^2\nonumber\\ &=&12\sigma^2\tau_1s\left[\tilde W_{tt}-\nu\Delta \tilde W-4\nu s^2\left(|x-x_0|^2-\frac{1}{\nu}\theta^2t^2+O_3\left(\frac{1}{s}\right)\right)\tilde W\right]\tilde V\nonumber\\ &&-\nabla\cdot(12\sigma^2\tau_1^2s \tilde V\nabla\tilde V)+12\sigma^2\tau_1^2s|\nabla\tilde V|^2-12^2\sigma^2\tau_1^2s^2\tilde V^2\nonumber\\ &&-48\sigma^2\tau_1^2 s^3\left(|x-x_0|^2+O_2\left(\frac{1}{s}\right)\right)\tilde V^2. \end{eqnarray*} $

将上述两个不等式相加, 得

$ \begin{eqnarray}\label{2-2.17} I_{11}^2&\geq&-\sigma^2\nu\rho s\tilde W_t^2+\sigma^2\nu^2\rho s|\nabla\tilde W|^2-4\sigma^2\nu^2\rho s^3\left(|x-x_0|^2-\frac{1}{\nu}\theta^2t^2+O_4\left(\frac{1}{s}\right)\right)\tilde W^2\nonumber\\ &&+12\sigma^2\tau_1^2s|\nabla\tilde V|^2-48\sigma^2\tau_1^2 s^3\left(|x-x_0|^2+O_5\left(\frac{1}{s}\right)\right)\tilde V^2\nonumber\\ &&+\nabla \cdot \hat X_1+(\hat Y_1)_t+\hat Z_1, \end{eqnarray} $

(2.13)

其中

$ O_4\left(\frac{1}{s}\right)=O_3\left(\frac{1}{s}\right)+\frac{\rho}{4s}, \quad O_5\left(\frac{1}{s}\right)=O_2\left(\frac{1}{s}\right)+\frac{12^2}{48s} $

及

$ \hat X_1=-\sigma^2\nu^2\rho s\tilde W\nabla\tilde W-12\sigma^2\tau_1^2s \tilde V\nabla\tilde V, \quad \hat Y_1=\sigma^2\nu\rho s\tilde W\tilde W_t, $

$ \begin{eqnarray*} \hat Z_1&=&-\sigma^2\tau_1\nu\rho s\left[\Delta \tilde V+4 s^2\left(|x-x_0|^2+O_2\left(\frac{1}{s}\right)\right)\tilde V\right]\tilde W\nonumber\\ &&+12\sigma^2\tau_1s\left[\tilde W_{tt}-\nu\Delta \tilde W-4\nu s^2\left(|x-x_0|^2-\frac{1}{\nu}\theta^2t^2+O_3\left(\frac{1}{s}\right)\right)\tilde W\right]\tilde V. \end{eqnarray*} $

通过类似的计算, 可得第二项$2I_{11}I_{12}$的估计

$ \begin{eqnarray}\label{2-2.18} 2I_{11}I_{12}&=&\left(4\sigma^2\theta st\tilde W_t^2\right)_t-4\sigma^2\theta s\tilde W_t^2-\nabla\cdot(8\sigma^2\nu\theta st\tilde W_t\nabla \tilde W)+\left(4\sigma^2\nu \theta st|\nabla \tilde W|^2\right)_t\nonumber\\ &&-4\sigma^2\nu \theta s|\nabla\tilde W|^2-\left[16\sigma^2\nu \theta s^3\left(t|x-x_0|^2-\frac{1}{\nu}\theta^2t^3+tO_3\left(\frac{1}{s}\right)\right)\tilde W^2\right]_t\nonumber\\ &&+16\sigma^2\nu\theta s^3\left[|x-x_0|^2-\frac{3}{\nu}\theta^2t^2+O_3\left(\frac{1}{s}\right)\right]\tilde W^2-8\sigma^2\tau_1\theta st\tilde W_t\Delta \tilde V\nonumber\\ &&-32\sigma^2\tau_1\theta s^3 t\left(|x-x_0|^2+O_2\left(\frac{1}{s}\right)\right)\tilde W_t\tilde V\nonumber\\ &=&-4\sigma^2 \theta s\tilde W_t^2-4\sigma^2\nu \theta s|\nabla \tilde W|^2+16\sigma^2\nu \theta s^3\left[|x-x_0|^2-\frac{3}{\nu}\theta^2t^2+O_3\left(\frac{1}{s}\right)\right]\tilde W^2\nonumber\\ &&+\nabla\cdot \hat X_2+(\hat Y_2)_t+\hat Z_2, \end{eqnarray} $

(2.14)

其中

$ \hat X_2=-8\sigma^2\nu \theta st\tilde W_t\nabla \tilde W, $

$ \hat Y_2=4\sigma^2\theta st\tilde W_t^2+4\sigma^2\nu \theta st|\nabla \tilde W|^2-16\sigma^2\nu \theta s^3\left(t|x-x_0|^2-\frac{1}{\nu}\theta^2t^3+tO_3\left(\frac{1}{s}\right)\right)\tilde W^2, $

$ \hat Z_2=-8\sigma^2\tau_1\theta st\tilde W_t\Delta \tilde V-32\sigma^2\tau_1\theta s^3 t\left(|x-x_0|^2+O_2\left(\frac{1}{s}\right)\right)\tilde W_t\tilde V. $

同理, 第三项$2I_{11}I_{13}$为

$ \begin{eqnarray}\label{2-2.19} 2I_{11}I_{13}&=&\left(8\sigma^2\nu s(x-x_0)\cdot\nabla \tilde W\tilde W_t\right)_t-4\sigma^2\nu s(x-x_0)\cdot \nabla\left(\tilde W_t^2\right)\nonumber\\ &&-\nabla\cdot(8\sigma^2\nu^2 s(x-x_0)\cdot\nabla\tilde W\nabla\tilde W)+\nabla(8\sigma^2\nu^2 s(x-x_0)\cdot\nabla\tilde W)\cdot\nabla\tilde W\nonumber\\ &&-16\sigma^2\nu^2 s^3\left( |x-x_0|^2-\frac{1}{\nu}\theta^2 t^2+O_3\left(\frac{1}{s}\right)\right)(x-x_0)\cdot\nabla(\tilde W^2)\nonumber\\ &&-8\sigma^2\tau_1\nu s(x-x_0)\cdot\nabla\tilde W\left[\Delta \tilde V+4s^2\left(|x-x_0|^2+O_2\left(\frac{1}{s}\right)\right)\tilde V\right]\nonumber\\ &=&16\sigma^2\nu^2 s^3\left[5|x-x_0|^2-\frac{3}{\nu}\theta^2t^2+3O_3\left(\frac{1}{s}\right)\right]\tilde W^2\nonumber\\ &&+12\sigma^2\nu s \tilde W_t^2-4\sigma^2\nu^2 s |\nabla \tilde W|^2+\nabla \cdot \hat X_3+(\hat Y_3)_t+\hat Z_3, \end{eqnarray} $

(2.15)

其中

$ \begin{eqnarray*} \hat X_3&=&-4\sigma^2\nu s(x-x_0)\tilde W_t^2-8\sigma^2\nu^2 s(x-x_0)\cdot\nabla \tilde W\nabla \tilde W+4\sigma^2\nu^2 s(x-x_0)|\nabla \tilde W|^2\\ &&-16\sigma^2\nu^2 s^3\left(|x-x_0|^2-\frac{1}{\nu}\theta^2 t^2+O_3\left(\frac{1}{s}\right)\right)(x-x_0)\tilde W^2, \end{eqnarray*} $

$ \hat Y_3=8\sigma^2\nu s(x-x_0)\cdot\nabla\tilde W\tilde W_t, $

$ \hat Z_3=-8\sigma^2\tau_1\nu s(x-x_0)\cdot\nabla\tilde W\left[\Delta \tilde V+4s^2\left(|x-x_0|^2+O_2\left(\frac{1}{s}\right)\right)\tilde V\right]. $

在(2.15)式, 我们利用了

$ \begin{eqnarray}\label{2-219} \nabla[(x-x_0)\cdot\nabla\tilde W]\cdot\nabla\tilde W&=&(x_i-x^0_i)_{x_j}\tilde W_{x_i}\tilde W_{x_j}+(x_i-x^0_i)\tilde W_{x_ix_j}\tilde W_{x_j}\nonumber\\ &=&-\frac{1}{2}|\nabla\tilde W|^2+\frac{1}{2}\nabla\cdot\left[(x-x_0)|\nabla\tilde W|^2\right], \end{eqnarray} $

(2.16)

这里$x_0=(x^0_1, x^0_2, x^0_3)$.为简便计, 我们省略了符号$\sum$.最后, 对于$2I_{11}I_{14}$, 我们有

$ \begin{eqnarray}\label{2-2.20} 2I_{11}I_{14}&=&-\nabla\cdot(8\sigma^2\tau_1^2s(x-x_0)\cdot\nabla \tilde V\nabla\tilde V)+\nabla(8\sigma^2\tau_1^2s(x-x_0)\cdot\nabla \tilde V)\cdot\nabla\tilde V\nonumber\\ &&-16\sigma^2\tau_1^2s^3\left(|x-x_0|^2+O_2\left(\frac{1}{s}\right)\right)(x-x_0)\cdot\nabla(\tilde V^2)\nonumber\\ &&+8\sigma^2\tau_1s(x-x_0)\cdot\nabla\tilde V\left[\tilde W_{tt}-\nu\Delta \tilde W-4\nu s^2\left(|x-x_0|^2-\frac{1}{\nu}\theta^2t^2+O_3\left(\frac{1}{s}\right)\right)\tilde W\right]\nonumber\\ &=&16\sigma^2\tau_1^2 s^3\left[5|x-x_0|^2+3O_2\left(\frac{1}{s}\right)\right]\tilde V^2-4\sigma^2\tau_1^2s|\nabla\tilde V|^2+\nabla\cdot \hat X_4+\hat Z_4, \end{eqnarray} $

(2.17)

其中

$ \begin{eqnarray*} \hat X_4&=&-8\sigma^2\tau_1^2s(x-x_0)\cdot\nabla \tilde V\nabla\tilde V+4\sigma^2\tau_1^2s(x-x_0)|\nabla\tilde V|^2\\ &&-16\sigma^2\tau_1^2s^3\left(|x-x_0|^2+O_2\left(\frac{1}{s}\right)\right)(x-x_0)\tilde V^2, \end{eqnarray*} $

$ \hat Z_4=8\sigma^2\tau_1s(x-x_0)\cdot\nabla\tilde V\left[\tilde W_{tt}-\nu\Delta \tilde W-4\nu s^2\left(|x-x_0|^2-\frac{1}{\nu}\theta^2t^2+O_3\left(\frac{1}{s}\right)\right)\tilde W\right]. $

这里$\nabla\left[(x-x_0)\cdot\nabla \tilde V\right]\cdot\nabla\tilde V$的计算类似于(2.16)式.

现将估计(2.13)-(2.15)和(2.17)式代入(2.12)式, 我们得到

$ \begin{eqnarray}\label{2-2.21} & &\left(\sigma{\cal L}_1[W, V]\right)^2{\rm e}^{2s\varphi}\nonumber\\ &\geq& 4s\left(3\sigma^2\nu -\sigma^2\theta-\frac{1}{4}\sigma^2\nu\rho\right)\tilde W_t^2+4s\left(-\sigma^2\nu^2+\frac{1}{4}\sigma^2\nu^2\rho-\sigma^2\nu\theta\right)|\nabla \tilde W|^2\nonumber\\ &&+16\sigma^2s^3\left[\left(\nu\theta+5\nu^2-\frac{1}{4}\nu^2\rho\right)|x-x_0|^2-\left(3\theta-\frac{1}{4}\nu\rho+3\nu\right)\theta^2t^2 +O_6\left(\frac{1}{s}\right)\right]\tilde W^2\nonumber\\ &&+16\sigma^2\tau_1^2s^3\left(2|x-x_0|^2+O_7\left(\frac{1}{s}\right)\right)\tilde V^2+8\sigma^2\tau_1^2s|\nabla\tilde V|^2+\nabla\cdot \hat X+\hat Y_t+\hat Z, \end{eqnarray} $

(2.18)

其中

$ \begin{eqnarray*} O_6\left(\frac{1}{s}\right)=-\frac{\nu^2\rho}{4} O_4\left(\frac{1}{s}\right)+\nu\theta O_3\left(\frac{1}{s}\right)+3\nu^2 O_3\left(\frac{1}{s}\right), O_7\left(\frac{1}{s}\right)=3O_2\left(\frac{1}{s}\right)-3O_5\left(\frac{1}{s}\right), \end{eqnarray*} $

及

$ \hat X=\sum\limits_{i=1}^4\hat X_i, \quad \hat Y=\sum\limits_{i=1}^3\hat Y_i, \quad \hat Z=\sum\limits_{i=1}^4 \hat Z_i. $

此外, 对于$\hat X_i$和$\hat Y_i$, 若$\tilde W(x, 0)=0$或$\tilde W_t(x, 0)=0$, 我们有

$ \begin{equation}\label{2-2.22} |\hat X|+|\hat Y|\leq Cs\left(|\nabla_{x, t}\tilde W|^2+s^2\tilde W^2+|\nabla \tilde V|^2+s^2\tilde V^2\right) \end{equation} $

(2.19)

及

$ \begin{equation}\label{2-2.23} \hat Y(x, 0)=0, \quad x\in \Omega.\end{equation} $

(2.20)

现在研究算子${\cal L}_2[W, V]$的估计.由于${\cal L}_1[W, V]$和${\cal L}_2[W, V]$结构相似, 在(2.18)式中取$\sigma=1$, 用$\tau_2$取代$\tau_1$, 并用$\tilde W$代替$\tilde V$, 可得

$ \begin{eqnarray}\label{2-2.24} &&\left({\cal L}_2[W, V]\right)^2{\rm e}^{2s\varphi}\nonumber\\ &\geq& 4s\left(3\nu -\theta-\frac{1}{4}\nu\rho\right)\tilde V_t^2+4s\left(-\nu^2+\frac{1}{4}\nu^2\rho-\nu\theta\right)|\nabla \tilde V|^2\nonumber\\ &&+16s^3\left[\left(\nu\theta+5\nu^2-\frac{1}{4}\nu^2\rho\right)|x-x_0|^2-\left(3\theta-\frac{1}{4}\nu\rho+3\nu\right)\theta^2t^2+O_6\left(\frac{1}{s}\right)\right]\tilde V^2\nonumber\\ &&+16\tau^2_2s^3\left(2|x-x_0|^2+O_7\left(\frac{1}{s}\right)\right)\tilde W^2+8\tau_2^2s|\nabla\tilde W|^2+\nabla\cdot \tilde X+\tilde Y_t+\tilde Z, \end{eqnarray} $

(2.21)

其中

$ \tilde X=\sum\limits_{i=1}^4\tilde X_i, \quad \tilde Y=\sum\limits_{i=1}^3\tilde Y_i, \quad \tilde Z=\sum\limits_{i=1}^4 \tilde Z_i $

满足

$ \begin{equation}\label{2-2.25} |\tilde X|+|\tilde Y|\leq Cs\left(|\nabla_{x, t}\tilde V|^2+s^2\tilde V^2+|\nabla \tilde W|^2+s^2\tilde W^2\right) \end{equation} $

(2.22)

及

$ \begin{equation}\label{2-2.26}\tilde Y(x, 0)=0, \quad x\in\Omega, \end{equation} $

(2.23)

若$\tilde V(x, 0)=0$或$\tilde V_t(x, 0)=0$.因为$\hat Z+\tilde Z$的计算在下面的证明中尤其重要性, 所以我们在这里列出$\tilde Z_i$的表达式

$ \begin{eqnarray*} \tilde Z_1&=&-\tau_2\nu\rho s\left[\Delta \tilde W+4 s^2\left(|x-x_0|^2+O_2\left(\frac{1}{s}\right)\right)\tilde W\right]\\%\tilde \nonumber\\ & &+12\tau_2s\left[\tilde V_{tt}-\nu\Delta \tilde V-4\nu s^2\left(|x-x_0|^2-\frac{1}{\nu}\theta^2t^2+O_3\left(\frac{1}{s}\right)\right)\tilde V\right]\tilde W, \end{eqnarray*} $

$ \tilde Z_2=-8\tau_2\theta st\tilde V_t \Delta \tilde W-32\tau_2\theta s^3t\left(|x-x_0|^2+O_2\left(\frac{1}{s}\right)\right)\tilde V_t\tilde W, \nonumber $

$ \tilde Z_3=-8\tau_2\nu s(x-x_0)\cdot\nabla\tilde V\left[\Delta \tilde W+4s^2\left(|x-x_0|^2+O_2\left(\frac{1}{s}\right)\right)\tilde W\right], \nonumber $

$ \tilde Z_4=8\tau_2s(x-x_0)\cdot\nabla\tilde W\left[\tilde V_{tt}-\nu\Delta \tilde V-4\nu s^2\left(|x-x_0|^2-\frac{1}{\nu}\theta^2t^2+O_3\left(\frac{1}{s}\right)\right)\tilde V\right]. $

$\tilde X_i$和$\tilde Y_i$具有类似于$\hat X_i$和$\hat Y_i$的形式.

从(2.18)和(2.21)式, 可知

$ \begin{eqnarray}\label{2-2.27} &&\left(\sigma{\cal L}_1[W, V]\right)^2{\rm e}^{2s\varphi}+\left({\cal L}_2[W, V]\right)^2{\rm e}^{2s\varphi}\nonumber\\ &\geq& 4s\left(3\sigma^2\nu -\sigma^2\theta-\frac{1}{4}\sigma^2\nu\rho\right)\tilde W_t^2+4s\left(-\sigma^2\nu^2+\frac{1}{4}\sigma^2\nu^2\rho-\sigma^2\nu\theta+2\tau_2^2\right)|\nabla\tilde W|^2\nonumber\\ &&+16s^3\left\{\left(\sigma^2\nu\theta+5\sigma^2\nu^2-\frac{1}{4}\sigma^2\nu^2\rho+2\tau_2^2\right)|x-x_0|^2-\left(3\sigma^2\theta-\frac{1}{4}\sigma^2\nu\rho+3\sigma^2\nu\right)\theta^2t^2\right.\nonumber\\ &&\left.+O_8\left(\frac{1}{s}\right)\right\}\tilde W^2+4s\left(3\nu -\theta-\frac{1}{4}\nu\rho\right)\tilde V_t^2+4s\left(-\nu^2+\frac{1}{4}\nu^2\rho-\nu\theta+2\sigma^2\tau_1^2\right)|\nabla \tilde V|^2\nonumber\\ &&+16s^3\left[\left(\nu\theta+5\nu^2-\frac{1}{4}\nu^2\rho+2\sigma^2\tau_1^2\right)|x-x_0|^2-\left(3\theta-\frac{1}{4}\nu\rho+3\nu\right)\theta^2t^2+O_9\left(\frac{1}{s}\right)\right] \nonumber\\ &&\times \tilde V^2+\nabla\cdot(\hat X+\tilde X)+(\hat Y+\tilde Y)_t+(\hat Z+\tilde Z), \end{eqnarray} $

(2.24)

其中

$ O_8\left(\frac{1}{s}\right)=\sigma^2 O_6\left(\frac{1}{s}\right)+\tau_2^2O_7\left(\frac{1}{s}\right), $

$ O_9\left(\frac{1}{s}\right)=O_6\left(\frac{1}{s}\right)+\sigma^2\tau_1^2O_7\left(\frac{1}{s}\right). $

现在, 我们来寻求$\hat Z+\tilde Z$的下界.注意到

$ \sigma^2\tau_1=\tau_2, $

我们有

$ \begin{eqnarray*} \hat Z_1&=&-\nabla\cdot(\tau_2\nu\rho s\tilde W\nabla\tilde V)+\tau_2\nu\rho s\nabla\tilde W\cdot\nabla\tilde V-4\tau_2\nu\rho s^3\left(|x-x_0|^2+O_2\left(\frac{1}{s}\right)\right)\tilde W\tilde V\nonumber\\ &&+(12\tau_2s\tilde W_t\tilde V)_t-12\tau_2 s\tilde W_t\tilde V_t-\nabla\cdot(12\tau_2\nu s\tilde V\nabla\tilde W)+12\tau_2\nu s\nabla\tilde V\cdot\nabla\tilde W\\ &&-48\tau_2\nu s^3\left(|x-x_0|^2-\frac{1}{\nu}\theta^2t^2+O_3\left(\frac{1}{s}\right)\right)\tilde W\tilde V. \end{eqnarray*} $

$\tilde Z_1$的计算类似于$\hat Z_1$, 即

$ \begin{eqnarray*} \tilde Z_1&=&-\nabla\cdot(\tau_2\nu\rho s\tilde V\nabla\tilde W)+\tau_2\nu\rho s\nabla\tilde V\cdot\nabla\tilde W-4\tau_2\nu\rho s^3\left(|x-x_0|^2+O_2\left(\frac{1}{s}\right)\right)\tilde V\tilde W\nonumber\\ &&+(12\tau_2s\tilde V_t\tilde W)_t-12\tau_2 s\tilde V_t\tilde W_t-\nabla\cdot(12\tau_2\nu s\tilde W\nabla\tilde V)+12\tau_2\nu s\nabla\tilde W\cdot\nabla\tilde V\\ &&-48\tau_2\nu s^3\left(|x-x_0|^2-\frac{1}{\nu}\theta^2t^2+O_3\left(\frac{1}{s}\right)\right)\tilde W\tilde V. \end{eqnarray*} $

因此

$ \begin{eqnarray}\label{2-2.28} \hat Z_1+\tilde Z_1&=&\nabla\cdot X_1+(Y_1)_t-8\tau_2\nu\rho s^3\left(|x-x_0|^2+O_2\left(\frac{1}{s}\right)\right)\tilde V\tilde W\nonumber\\ &&+2\tau_2\nu\rho s\nabla\tilde W\cdot\nabla\tilde V-24\tau_2 s\tilde W_t\tilde V_t+24\tau_2\nu s\nabla\tilde V\cdot\nabla\tilde W\nonumber\\ &&-96\tau_2\nu s^3\left(|x-x_0|^2-\frac{1}{\nu}\theta^2t^2+O_3\left(\frac{1}{s}\right)\right)\tilde V\tilde W, \end{eqnarray} $

(2.25)

其中

$ X_1=-\tau_2\nu\rho s(\tilde V\nabla\tilde W+\tilde W\nabla\tilde V)-12\tau_2\nu s(\tilde V\nabla\tilde W+\tilde W\nabla\tilde V), $

$ Y_1=12\tau_2s\tilde V\tilde W_t+12\tau_2s\tilde W\tilde V_t. $

对于$\hat Z_2$和$\tilde Z_2$, 我们有

$ \begin{eqnarray*} \hat Z_2&=&-\nabla\cdot(8\tau_2\theta st\tilde W_t\nabla \tilde V)+8\tau_2\theta st\nabla\tilde W_t\cdot\nabla \tilde V-\left[32\tau_2 \theta s^3t\left(|x-x_0|^2+O_2\left(\frac{1}{s}\right)\right)\tilde W\tilde V\right]_t\\ &&+32\tau_2\theta s^3t\left(|x-x_0|^2+O_2\left(\frac{1}{s}\right)\right)\tilde W\tilde V_t+32\tau_2\theta s^3\left(|x-x_0|^2+O_2\left(\frac{1}{s}\right)\right)\tilde W\tilde V\\ &=&-\nabla\cdot(8\tau_2\theta st\tilde W_t\nabla \tilde V)+(8\tau_2\theta st\nabla\tilde W\cdot\nabla\tilde V)_t-8\tau_2\theta st\nabla\tilde W\cdot\nabla\tilde V_t-8\tau_2\theta s\nabla\tilde W\cdot\nabla\tilde V\\ &&-\left[32\tau_2 \theta s^3t\left(|x-x_0|^2+O_2\left(\frac{1}{s}\right)\right)\tilde W\tilde V\right]_t+32\tau_2\theta s^3t\left(|x-x_0|^2+O_2\left(\frac{1}{s}\right)\right)\tilde W\tilde V_t\\ &&+32\tau_2\theta s^3\left(|x-x_0|^2+O_2\left(\frac{1}{s}\right)\right)\tilde W\tilde V \end{eqnarray*} $

和

$ \tilde Z_2=-\nabla\cdot(8\tau_2\theta st\tilde V_t\nabla\tilde W)+8\tau_2\theta st\nabla\tilde V_t\cdot\nabla\tilde W-32\tau_2\theta s^3t\left(|x-x_0|^2+O_2\left(\frac{1}{s}\right)\right)\tilde V_t\tilde W. $

因此

$ \begin{eqnarray}\label{2-2.29} \hat Z_2+\tilde Z_2=\nabla \cdot X_2+(Y_2)_t-8\tau_2\theta s\nabla\tilde W\cdot\nabla\tilde V+32\tau_2\theta s^3\left(|x-x_0|^2+O_2\left(\frac{1}{s}\right)\right)\tilde W\tilde V, \end{eqnarray} $

(2.26)

其中

$ X_2=-8\tau_2\theta st\tilde W_t\nabla\tilde V-8\tau_2\theta st\tilde V_t\nabla \tilde W, $

$ Y_2=8\tau_2\theta st\nabla\tilde W\cdot\nabla\tilde V-32\tau_2 \theta s^3t\left(|x-x_0|^2+O_2\left(\frac{1}{s}\right)\right)\tilde W\tilde V. $

通过直接的计算, 有

$ \begin{eqnarray*} &&-8\sigma^2\tau_1\nu s(x-x_0)\cdot\nabla \tilde W\Delta \tilde V\\ &=&-8\tau_2\nu s\left\{\left[(x_i-x^0_i)\tilde W_{x_i}\tilde V_{x_j}\right]_{x_j}-(x_i-x^0_i)_{x_j}\tilde W_{x_i}\tilde V_{x_j}-(x_i-x^0_i)\tilde W_{x_ix_j}\tilde V_{x_j}\right\}\\ &=&-8\tau_2\nu s\left\{\left[(x_i-x^0_i)\tilde W_{x_i}\tilde V_{x_j}\right]_{x_j}-\tilde W_{x_j}\tilde V_{x_j}-(x_i-x^0_i)\left[\left(\tilde W_{x_j}\tilde V_{x_j}\right)_{x_i}-\tilde W_{x_j}\tilde V_{x_ix_j}\right]\right\}\\ &=&-8\tau_2\nu s\left\{\left[(x_i-x^0_i)\tilde W_{x_i}\tilde V_{x_j}\right]_{x_j}+2\tilde W_{x_j}\tilde V_{x_j}-\left[(x_i-x_i^0)\tilde W_{x_j}\tilde V_{x_j}\right]_{x_i}+(x_i-x_i^0)\tilde W_{x_j}\tilde V_{x_ix_j}\right\} \end{eqnarray*} $

和

$ \begin{eqnarray*} &&-8\tau_2\nu s(x-x_0)\cdot\nabla \tilde V\Delta \tilde W\\ &=&-8\tau_2\nu s\left\{\left[(x_i-x^0_i)\tilde V_{x_i}\tilde W_{x_j}\right]_{x_j}-\tilde V_{x_j}\tilde W_{x_j}-(x_i-x_i^0)\tilde V_{x_ix_j}\tilde W_{x_j}\right\}. \end{eqnarray*} $

此外

$ \begin{eqnarray*} &&-32\sigma^2\tau_1\nu s^3\left(|x-x_0|^2+O_2\left(\frac{1}{s}\right)\right)(x-x_0)\cdot\nabla\tilde W\tilde V\\ &=&-\nabla\cdot\left[32\tau_2\nu s^3\left(|x-x_0|^2+O_2\left(\frac{1}{s}\right)\right)(x-x_0)\tilde W\tilde V\right]\\ &&+32\tau_2\nu s^3\left(|x-x_0|^2+O_2\left(\frac{1}{s}\right)\right)(x-x_0)\cdot\nabla\tilde V \tilde W\\ &&+32\tau_2\nu s^3\left(5|x-x_0|^2+3O_2\left(\frac{1}{s}\right)\right)\tilde W\tilde V. \end{eqnarray*} $

因此, 我们有

$ \begin{eqnarray}\label{2-2.30} \hat Z_3+\tilde Z_3=\nabla\cdot X_3-8\tau_2\nu s\nabla\tilde W\cdot\nabla \tilde V+32\tau_2\nu s^3\left(5|x-x_0|^2+3O_2\left(\frac{1}{s}\right)\right)\tilde W\tilde V, \end{eqnarray} $

(2.27)

其中

$ \begin{eqnarray*} X_3&=&-8\tau_2\nu s\left[(x-x_0)\cdot\nabla \tilde W\nabla\tilde V+(x-x_0)\cdot\nabla\tilde V\nabla\tilde W-(x-x_0)\nabla\tilde W\cdot\nabla\tilde V\right]\\ &&-32\tau_2\nu s^3\left(|x-x_0|^2+O_2\left(\frac{1}{s}\right)\right)(x-x_0)\tilde W\tilde V. \end{eqnarray*} $

最后, 通过类似于(2.27)式的计算和

$ \begin{eqnarray*} &&8\sigma^2\tau_1s(x-x_0)\cdot\nabla\tilde V\tilde W_{tt}\\ &=&[8\tau_2s(x-x_0)\cdot\nabla\tilde V\tilde W_t]_t-8\tau_2s(x-x_0)\cdot\nabla\tilde V_t\tilde W_t\\ &=&[8\tau_2s(x-x_0)\cdot\nabla\tilde V\tilde W_t]_t-\nabla\cdot[8\tau_2s(x-x_0)\tilde V_t\tilde W_t]+24\tau_2s\tilde V_t\tilde W_t\\ &&+[8\tau_2s(x-x_0)\cdot\nabla\tilde W\tilde V_t]_t-8\tau_2s(x-x_0)\cdot\nabla\tilde W\tilde V_{tt}, \end{eqnarray*} $

我们得到

$ \begin{eqnarray}\label{2-2.31} \hat Z_4+\tilde Z_4&=&\nabla\cdot X_4+(Y_3)_t+24\tau_2s\tilde W_t\tilde V_t-8\tau_2\nu s\nabla\tilde W\cdot\nabla\tilde V\nonumber\\ &&+32\tau_2\nu s^3\left[5|x-x_0|^2-\frac{3}{\nu}\theta^2t^2+3O_3\left(\frac{1}{s}\right)\right]\tilde W\tilde V, \end{eqnarray} $

(2.28)

其中

$ \begin{eqnarray*} X_4&=&-8\tau_2\nu s\left[(x-x_0)\cdot\nabla \tilde W\nabla\tilde V+(x-x_0)\cdot\nabla\tilde V\nabla\tilde W-(x-x_0)\nabla\tilde V\cdot\nabla\tilde W\right]\\ &&-32\tau_2\nu s^3\left(|x-x_0|^2-\frac{1}{\nu}\theta^2t^2+O_3\left(\frac{1}{s}\right)\right)(x-x_0)\tilde W\tilde V-8\tau_2s(x-x_0)\tilde W_t\tilde V_t, \\ &&Y_3=8\tau_2s\left[(x-x_0)\cdot\nabla\tilde W\tilde V_t+(x-x_0)\cdot\nabla\tilde V\tilde W_t\right]. \end{eqnarray*} $

从(2.25)-(2.28)式, 我们得到

$ \begin{eqnarray}\label{2-2.32} \hat Z+\tilde Z&=&\nabla\cdot\left(\sum\limits_{i=1}^4 X_i\right)+\left(\sum\limits_{i=1}^3Y_i\right)_t-4s\left(2\tau_2\theta-\frac{1}{2}\tau_2\nu\rho-2\tau_2\nu\right)\nabla\tilde W\cdot\nabla\tilde V\nonumber\\ & &-16s^3\left[\left(\frac{1}{2}\tau_2\nu\rho-2\tau_2\theta-14\tau_2\nu\right)|x-x_0|^2+O_{10}\left(\frac{1}{s}\right)\right]\tilde V\tilde W, \end{eqnarray} $

(2.29)

其中

$ O_{10}\left(\frac{1}{s}\right)=\frac{1}{2}\tau_2\nu\rho O_{2}\left(\frac{1}{s}\right)-2\tau_2\theta O_{2}\left(\frac{1}{s}\right)-6\tau_2\nu O_{2}\left(\frac{1}{s}\right) $

及

$ \begin{eqnarray}\label{2-2.33} \left|\sum\limits_{i=1}^4X_i\right|+\left|\sum\limits_{i=1}^3Y_i\right|\leq Cs\left(|\nabla_{x, t}\tilde W|^2+s^2\tilde W^2+|\nabla_{x, t} \tilde V|^2+s^2\tilde V^2\right). \end{eqnarray} $

(2.30)

此外, 若$(\tilde W, \tilde V)(x, 0)=(0, 0)$或$(\tilde W_t, \tilde V_t)(x, 0)=(0, 0)$, 有

$ \begin{eqnarray}\label{2-2.34} \sum\limits_{i=1}^3Y_i(x, 0)=0, \quad x\in\Omega. \end{eqnarray} $

(2.31)

为了在(2.24)式中有$\tilde W_t^2, \tilde V_t^2$和$|\nabla\tilde W|^2, |\nabla\tilde V|^2$的下界, 对于充分小的$\theta$我们首先选择

$ \rho=12-\frac{4\theta}{\nu}-\epsilon_0, $

这里$\epsilon_0$是待定的正常数.然后将(2.29)式代入(2.24)式, 并利用(2.11)式, 我们有

$ \begin{eqnarray}\label{2-2.35} &&\left(\sigma{\cal L}_1[W, V]\right)^2{\rm e}^{2s\varphi}+\left({\cal L}_2[W, V]\right)^2{\rm e}^{2s\varphi}\nonumber\\ &\geq& \epsilon_0\nu\sigma^2 s\tilde W_t^2+\epsilon_0\nu s\tilde V_t^2+4s\left(2\nu^2+2\sigma^2\tau_1^2-2\nu\theta-\frac{1}{4}\epsilon_0\nu^2\right)(\sigma^2|\nabla\tilde W|^2+|\nabla \tilde V|^2)\nonumber\\ &&+16s^3\left[\left(2\nu^2+2\sigma^2\tau_1^2+2\nu\theta+\frac{1}{4}\epsilon_0\nu^2\right)|x-x_0|^2-\left(4\theta+\frac{1}{4}\epsilon_0\nu\right)\theta^2 t^2\right](\sigma^2\tilde W^2+\tilde V^2)\nonumber\\ &&-4s\left(-8\tau_2\nu+4\tau_2\theta+\frac{1}{2}\epsilon_0\tau_2\nu\right)\nabla\tilde W\cdot\nabla\tilde V-16s^3\left(-8\tau_2\nu-4\tau_2\theta-\frac{1}{2}\epsilon_0\tau_2\nu\right)\nonumber\\ &&\times|x-x_0|^2\tilde W\tilde V-O_{11}\left(\frac{1}{s}\right)(\tilde W^2+\tilde V^2)+\nabla\cdot X+Y_t, \end{eqnarray} $

(2.32)

其中

$ O_{11}\left(\frac{1}{s}\right)=16s^3\left(\left|O_8\left(\frac{1}{s}\right)\right|+\left|O_9\left(\frac{1}{s}\right)\right|+\left|O_{10}\left(\frac{1}{s}\right)\right|\right) $

及

$ X=\hat X+\tilde X+\sum\limits_{i=1}^4 X_i, \quad Y=\hat Y+\tilde Y+\sum\limits_{i=1}^3Y_i. $

由Young不等式, 我们有

$ \begin{eqnarray*} &&\left(-8\tau_2\nu+4\tau_2\theta+\frac{1}{2}{\epsilon_0\tau_2\nu}\right)\nabla\tilde W\cdot\nabla\tilde V\\ &\leq& \sigma\left(4\tau_2\nu+2\tau_2\theta+\frac{1}{4}{\epsilon_0\tau_2\nu}\right)|\nabla\tilde W|^2+\frac{1}{\sigma}\left(4\tau_2\nu+2\tau_2\theta+\frac{1}{4}{\epsilon_0\tau_2\nu}\right)|\nabla\tilde V|^2 \end{eqnarray*} $

和

$ \begin{eqnarray*} & &\left(-8\tau_2\nu-4\tau_2\theta-\frac{1}{2}{\epsilon_0\tau_2\nu}\right)\tilde W\tilde V\\ &\leq& \sigma\left(4\tau_2\nu+2\tau_2\theta+\frac{1}{4}\epsilon_0\tau_2\nu\right)\tilde W^2+\frac{1}{\sigma}\left(4\tau_2\nu+2\tau_2\theta+\frac{1}{4}\epsilon_0\tau_2\nu\right)\tilde V^2. \end{eqnarray*} $

那么

$ \begin{eqnarray}\label{2-2.37} & &4s\left(2\nu^2+2\sigma^2\tau_1^2-2\nu\theta-\frac{1}{4}{\epsilon_0\nu^2}\right)(\sigma^2|\nabla\tilde W|^2+|\nabla \tilde V|^2)\nonumber\\ &&-4s\left(-8\tau_2\nu+4\tau_2\theta+\frac{1}{2}{\epsilon_0\tau_2\nu}\right)\nabla\tilde W\cdot\nabla\tilde V\nonumber\\ &\geq&4s\left(2\nu^2+2\sigma^2\tau_1^2-\frac{4\tau_2\nu}{\sigma}-2\nu\theta-\frac{2\tau_2\theta}{\sigma} -\frac{\epsilon_0\tau_2\nu}{4\sigma}-\frac{\epsilon_0\nu^2}{4}\right)(\sigma^2|\nabla\tilde W|^2+|\nabla \tilde V|^2)\nonumber\\ &\geq&4s\left[2\left(\nu-\sqrt{\tau_1\tau_2}\right)^2-\left(2\nu+\frac{2\tau_2}{\sigma}\right)\theta -\frac{\epsilon_0\tau_2\nu}{4\sigma}-\frac{\epsilon_0\nu^2}{4}\right](\sigma^2|\nabla\tilde W|^2+|\nabla \tilde V|^2) \end{eqnarray} $

(2.33)

和

$ \begin{eqnarray}\label{2-2.38} &&16s^3\left[(2\nu^2+2\sigma^2\tau_1^2+2\nu\theta+\frac{1}{4}{\epsilon_0\nu^2}\right)|x-x_0|^2-\left(4\theta+\frac{1}{4}{\epsilon_0\nu}\right)\theta^2 t^2](\sigma^2\tilde W^2+\tilde V^2)\nonumber\\ &&-16s^3(-8\tau_2\nu-4\tau_2\theta-\frac{1}{2}{\epsilon_0\tau_2\nu})|x-x_0|^2\tilde W\tilde V\nonumber\\ &\geq&16s^3\left[\left(2\nu^2+2\sigma^2\tau_1^2-\frac{4\tau_2\nu}{\sigma}+2\nu\theta+\frac{\epsilon_0\nu^2}{4} -\frac{2\tau_2\theta}{\sigma}-\frac{\epsilon_0\tau_2\nu}{4\sigma}\right)|x-x_0|^2\right.\nonumber\\ &&\left.-\left(4\theta+\frac{\epsilon_0\nu}{4}\right)\theta^2 t^2\right](\sigma^2\tilde W^2+\tilde V^2)\nonumber\\ &\geq&16s^3\left\{\left[2\left(\nu-\sqrt{\tau_1\tau_2}\right)^2-\frac{2\tau_2\theta}{\sigma} -\frac{\epsilon_0\tau_2\nu}{4\sigma}\right]|x-x_0|^2-\left(4\theta+\frac{\epsilon_0\nu}{4}\right)\theta^2 t^2\right\}\nonumber\\ &&\times(\sigma^2\tilde W^2+\tilde V^2). \end{eqnarray} $

(2.34)

选择充分小的仅依赖于$\nu, \tau_1$及$\tau_2$的$\theta_0>0$和$\epsilon_0>0$, 使得对于任意的$(x, t)\in Q(\delta)$, 成立

$ \begin{eqnarray} \label{2-2.39}&& 2\left(\nu-\sqrt{\tau_1\tau_2}\right)^2-\left(2\nu+\frac{2\tau_2}{\sigma}\right)\theta-\frac{\epsilon_0\tau_2\nu}{4\sigma}-\frac{\epsilon_0\nu^2}{4}\nonumber\\ &\geq& \left(\nu-\sqrt{\tau_1\tau_2}\right)^2-\frac{\epsilon_0\tau_2\nu}{4\sigma}-\frac{\epsilon_0\nu^2}{2}\geq \frac{1}{2}\left(\nu-\sqrt{\tau_1\tau_2}\right)^2 \end{eqnarray} $

(2.35)

及

$ \begin{eqnarray} \label{2-2.40}&& \left[2\left(\nu-\sqrt{\tau_1\tau_2}\right)^2-\frac{2\tau_2\theta}{\sigma}-\frac{\epsilon_0\tau_2\nu}{4\sigma}\right]|x-x_0|^2-\left(4\theta+\frac{\epsilon_0\nu}{4}\right)\theta^2 t^2\nonumber\\ &\geq&\frac{1}{2}\left(\nu-\sqrt{\tau_1\tau_2}\right)^2|x-x_0|^2+\frac{1}{2}\left(\nu-\sqrt{\tau_1\tau_2}\right)^2\left(|x-x_0|^2-\theta t^2\right)\geq \frac{1}{2}\left(\nu-\sqrt{\tau_1\tau_2}\right)^2d_0. \end{eqnarray} $

(2.36)

那么, 由(2.8)式和不等式(2.32)-(2.36), 我们有对所有的$s\geq s_0$, 成立

$ \begin{eqnarray}\label{2-2.41} &&\left(\sigma{\cal L}_1[W, V]\right)^2{\rm e}^{2s\varphi}+\left({\cal L}_2[W, V]\right)^2{\rm e}^{2s\varphi}\nonumber\\ &\geq& Cs\left(|\nabla_{x, t} \tilde W|^2+|\nabla_{x, t} \tilde V|^2+s^2\tilde W^2+s^2\tilde V^2\right)+\nabla \cdot X+Y_t, \end{eqnarray} $

(2.37)

其中$s_0$是充分大的正常数.这里$s_0$和$C$依赖于$\nu, \tau_1$, $\tau_2$和$a, b, c, d$.利用$(W, V)=(\tilde W, \tilde V) {\rm e}^{-s\varphi}$, 我们有

$ \begin{eqnarray}\label{2-2.42} &&\left(\sigma{\cal L}_1[W, V]\right)^2{\rm e}^{2s\varphi}+\left({\cal L}_2[W, V]\right)^2{\rm e}^{2s\varphi}\nonumber\\ &\geq& Cs\left(|\nabla_{x, t} W|^2+|\nabla_{x, t} V|^2+s^2W^2+s^2V^2\right){\rm e}^{2s\varphi}+\nabla \cdot X+Y_t. \end{eqnarray} $

(2.38)

另一方面, 有

$ \begin{eqnarray}\label{2-2.43} \left({\cal L}_1[W, V]\right)^2{\rm e}^{2s\varphi} +\left({\cal L}_2[W, V]\right)^2{\rm e}^{2s\varphi} &=&\left( \left|F+mG\right|^2+|G|^2\right){\rm e}^{2s\varphi}\nonumber\\ &\leq& \left[2|F|^2+\left(2m^2+1\right)|G|^2\right]{\rm e}^{2s\varphi}. \end{eqnarray} $

(2.39)

因此, 由(2.38)和(2.39)式, 我们得到(2.3)式.此外, 联立(2.19), (2.20), (2.22), (2.23), (2.30)和(2.31)式, 我们有(2.4)和(2.5)式.证毕.

注2.1  我们不能直接应用现有的卡勒曼估计得到所需要证明的估计.确切地说, 第一个方程中的$\Delta V$项不能被第二个方程的卡勒曼估计所吸收.同时, 因为矩阵$\Bigg(\begin{array}{cc} a \quad b\\ c \quad d\end{array} \Bigg)$可能没有两个不同的特征值, 那就无法找到矩阵$P$将主部系数矩阵对角化.因此我们无法使用文献[7, 13]中的对角化方法.我们的证明是将整个系统当作一个整体, 参考Klibanov^[27]对单个微分方程逐点卡勒曼估计的证明方法来建立卡勒曼估计.这里的主要困难来自于涉及$\tilde W$和$\tilde V$的强耦合项.

首先引入定义在$Q_T$上的截断函数$\chi$

$ \begin{eqnarray} \chi_1(x, t)=\left\{\begin{array}{ll} 1, &(x, t)\in Q(2\delta), \\ 0, &(x, t)\in Q(\delta)\setminus Q(3/2\delta). \end{array}\right. \end{eqnarray} $

(3.1)

那么$(v_1, v_2):=(\chi_1 u_1, \chi_1 u_2)$满足

$ \begin{eqnarray} \left\{\begin{array}{l} \partial^2_{t}{v_1}-a_{11}\Delta {v_1}-a_{12}\Delta v_2=\chi_1f_1+F_1(\chi_1, u_1, u_2), \quad (x, t)\in Q_T, \\ \partial^2_{t}{v_2}-a_{21}\Delta {v_1}-a_{22}\Delta v_2=\chi_1f_2+F_2(\chi_1, u_1, u_2), \quad (x, t)\in Q_T, \end{array}\right. \end{eqnarray} $

(3.2)

其中

$ \begin{eqnarray} \sum\limits_{i=1}^2|F_i(\chi_1, u_1, u_2)|\leq C\left(|\partial_t^2\chi_1|+|\nabla_{x, t}\chi_1|+|\Delta\chi_1|\right)\sum\limits_{i=1}^2\left(|\nabla_{x, t}u_i|+|u_i|\right). \end{eqnarray} $

(3.3)

从引理2.1.可知, 存在$\theta_1=\theta_1(\Omega, T, x_0, a_{ij})$, $s_1=s_1(\Omega, T, x_0, a_{ij})$和$C=C(\Omega, T, x_0, a_{ij})$, 使得对$\theta\leq \theta_1$和$s\geq s_1$, 有

$ \begin{eqnarray}\label{2.44} s\sum\limits_{i=1}^2|\nabla_{x, t}v_i|^2{\rm e}^{2s\varphi}+s^3\sum\limits_{i=1}^2|v_i|^2{\rm e}^{2s\varphi}+\nabla\cdot X+Y_t\leq C\sum\limits_{i=1}^2\left(|f_i|^2+|F_i|^2\right){\rm e}^{2s\varphi}, \ \end{eqnarray} $

(3.4)

其中$(x, t)\in Q(\delta)$, 这里的函数$X=X(v_1, v_2)$和$Y=Y(v_1, v_2)$满足

$ \begin{eqnarray}\label{2.45} |X(v_1, v_2)|+|Y(v_1, v_2)| \leq Cs\sum\limits_{i=1}^2\left(|\nabla_{x, t} v_i|^2+s^2|v_i|^2\right) \leq Cs\sum\limits_{i=1}^2\left(|\nabla_{x, t} u_i|^2+s^2|u_i|^2\right). \end{eqnarray} $

(3.5)

从$\chi_1$的定义, 我们可知$(v_1, v_2)(x, t)=0$, $(x, t)\in \Sigma_1(\delta)$.此外, 由$(u_1, u_2)(x, 0)=0$, $x\in\Omega$可知$(v_1, v_2)(x, 0)=0$, $x\in\Omega$.在$Q(\delta)$上对(3.4)式积分, 并利用(3.5)式, 我们得到对任意的$\theta\leq \theta_1$和$s\geq s_1$有

$ \begin{eqnarray}\label{2-2.45} &&\sum\limits_{i=1}^2\int_{Q(\delta)}\left(s|\nabla_{x, t}v_i|^2+s^3|v_i|^2\right){\rm e}^{2s\varphi}{\rm d}x{\rm d}t\nonumber\\ &\leq& C\sum\limits_{i=1}^2\int_{Q(\delta)}\left(|f_i|^2+|F_i|^2\right){\rm e}^{2s\varphi}{\rm d}x{\rm d}t+C\sum\limits_{i=1}^2\int_{\Sigma(\delta)}\left(s|\nabla_{x, t}u_i|^2+s^3|u_i|^2\right){\rm e}^{2s\varphi}{\rm d}S{\rm d}t. \end{eqnarray} $

(3.6)

利用$v(x, t)=u(x, t)$, $(x, t)\in Q(2\delta)$及$\varphi(x, t)\leq 2\delta$, $(x, t)\in Q(\delta)\setminus Q(2\delta)$, 我们进一步有

$ \begin{eqnarray}\label{2-2.46} &&\sum\limits_{i=1}^2\int_{Q(\delta)}\left(s|\nabla_{x, t}u_i|^2+s^3|u_i|^2\right){\rm e}^{2s\varphi}{\rm d}x{\rm d}t\nonumber\\ &=&\sum\limits_{i=1}^2\int_{Q(2\delta)}\left(s|\nabla_{x, t}v_i|^2+s^3|v_i|^2\right){\rm e}^{2s\varphi}{\rm d}x{\rm d}t+\sum\limits_{i=1}^2\int_{Q(\delta)\setminus Q(2\delta)}\left(s|\nabla_{x, t}u_i|^2+s^3|u_i|^2\right){\rm e}^{2s\varphi}{\rm d}x{\rm d}t\nonumber\\ & \leq&\sum\limits_{i=1}^2\int_{Q(2\delta)}\left(s|\nabla_{x, t}v_i|^2+s^3|v_i|^2\right){\rm e}^{2s\varphi}{\rm d}x{\rm d}t+Cs^3{\rm e}^{4\delta s}\|u_i\|^2_{H^1(Q(\delta))}. \end{eqnarray} $

(3.7)

由${\rm Supp}(\nabla_{x, t}\chi_1)\subset Q(\delta)\setminus Q(2\delta)$可知

$ \begin{eqnarray}\label{2-2.47} \sum\limits_{i=1}^2\int_{Q(\delta)}|F_i|^2{\rm e}^{2s\varphi}{\rm d}x{\rm d}t\leq C{\rm e}^{4\delta s}\sum\limits_{i=1}^2\|u_i\|^2_{H^1(Q(\delta))}. \end{eqnarray} $

(3.8)

最后将估计(3.6)和(3.8)式代入(3.7)式可得(1.8)式.

本节中我们利用Imanuvilov和Yamamoto^[29]或者Cavaterra, Lorenzi和Yamamoto^[8]的方法证明反源问题的Hölder稳定性.这种方法是Bukhgeim-Klibanov在文献[2]中提出的方法的一种改进.对Bukhgeim-Klibanov方法的最新应用, 也可参阅文献[24].

固定$\theta$充分小且满足$0<\theta<\theta_1$.我们首先证明下面的两个引理.

引理4.1  存在常数$C>0$使得

$ \begin{eqnarray}\label{2.4.1} &&\sum\limits_{i=1}^2\int_{Q(\delta)}\left(s^3|\partial_t u_i|^2+s|\nabla_{x, t}\partial_t u_i|^2\right){\rm e}^{2s\varphi}{\rm d}x{\rm d}t\\ &\leq& Cs^3{\rm e}^{Cs}\sum\limits_{i=1}^2\left(\|u_i\|^2_{H^1(\Sigma(\delta))}+\|\partial_t u_i\|^2_{H^1(\Sigma(\delta))}\right)\nonumber\\ &&+C\sum\limits_{i=1}^2\int_{Q(\delta)}|p_i|^2{\rm e}^{2s\varphi}{\rm d}x{\rm d}t+Cs^3{\rm e}^{4\delta s}M. \end{eqnarray} $

(4.1)

证  易知对于$i=0, 1$有

$ \begin{eqnarray} \label{2.1.7}\left\{\begin{array}{ll}\partial^2_{t}{\partial_t^iu_1}-a_{11}\Delta {\partial_t^iu_1}-a_{12}\Delta \partial_t^i u_2=\partial_t^i \sigma_1(x, t)p_1(x), &(x, t)\in Q_T, \\ \partial^2_{t}{\partial_t^i u_2}-a_{21}\Delta {\partial_t^i u_1}-a_{22}\Delta \partial_t^i u_2=\partial_t^i\sigma_2(x, t)p_2(x), &(x, t)\in Q_T, \\ \partial_t^i u_1(x, 0)=\partial_t^i u_2(x, 0)=0, &x\in\Omega.\end{array}\right. \end{eqnarray} $

(4.2)

对(4.2)式应用定理1.1, 则对充分大的$s$, 我们有

$ \begin{eqnarray} \label{242} &&\sum\limits_{i=1}^2\int_{Q(\delta)}\left(s^3|\partial_t u_i|^2+s|\nabla_{x, t}\partial_t u_i|^2\right){\rm e}^{2s\varphi}{\rm d}x{\rm d}t\\ &\leq& Cs^3{\rm e}^{Cs}\sum\limits_{i=1}^2\left(\|u_i\|^2_{H^1(\Sigma(\delta))}+\|\partial_t u_i\|^2_{H^1(\Sigma(\delta))}\right)\nonumber\\ &&+C\sum\limits_{i=1}^2\int_{Q(\delta)}|p_i|^2{\rm e}^{2s\varphi}{\rm d}x{\rm d}t+Cs^3{\rm e}^{4\delta s}\sum\limits_{i=1}^2\|u_i\|^2_{H^2(Q(\delta))}. \end{eqnarray} $

(4.3)

由(1.13)式, 我们可得$\theta T^2>|x-x_0|^2-\delta$, $x\in\Omega$.同时$(x, t)\in Q(\delta)$隐含$x\in\Omega$和$\theta t^2<|x-x_0|^2-\delta$.因此$0<t<T$和$Q(\delta)\subset Q_T$.进一步,

$ \begin{eqnarray}\label{244} \|u_i\|^2_{H^2(Q(\delta))}\leq M. \end{eqnarray} $

(4.4)

最后, 将(4.4)式代入(4.3)式可得估计(4.1).证毕.

引理4.2  存在常数$C>0$使得

$ \begin{eqnarray}\label{1.4.4} &&\sum\limits_{i=1}^2\int_\Omega|p_i(x)|^2{\rm e}^{2s\varphi(x, 0)}{\rm d}x\\ &\leq& Cs^2{\rm e}^{Cs}\sum\limits_{i=1}^2\left(\|u_i\|^2_{H^1(\Sigma(\delta))}+\|\partial_tu_i\|^2_{H^1(\Sigma(\delta))}\right) +\sum\limits_{i=1}^2\int_{Q(\delta)}|p_i|^2{\rm e}^{2s\varphi}{\rm d}x{\rm d}t\\ &&+C\sum\limits_{i=1}^2\int_{Q(\delta)}\left(s^3|\partial_t u_i|^2+s|\nabla_{x, t}\partial_t u_i|^2\right){\rm e}^{2s\varphi}{\rm d}x{\rm d}t. \end{eqnarray} $

(4.5)

证  令$T_0:=\min\limits_{x\in\Omega}t_\delta(x)$.由$\delta\in \left(0, \frac{d_0}{2}\right)$可知$T_0=\min\limits_{x\in\Omega}\sqrt{\frac{{|x-x_0|^2-\delta}}{\beta}}>\sqrt{\frac{d_0}{2\beta}}>0$.引入截断函数$\chi_2$满足

$ \left\{\begin{array}{ll} 0\leq\chi_2(t)\leq 1,&-\infty < t< \infty, \\ \chi_2(t)=1, &t\leq T_0/4, \\ \chi_2(t)=0,&t\geq 3T_0/{4}. \end{array}\right. $

令

$ u(x, t)=\chi_2(t)\partial_tu_1(x, t){\rm e}^{s\varphi(x, t)}, \ \ v(x, t)=\chi_2(t)\partial_tu_2(x, t){\rm e}^{s\varphi(x, t)}. $

通过直接的计算, 我们有

$ \begin{equation}\label{4.1} \partial^2_{t}{u}-a_{11}\Delta {u}-a_{12}\Delta v=\left({\cal N}_1[u_1]-a_{11}{\cal N}_2[u_1]-a_{12}{\cal N}_2[u_2]+\chi_2\sigma_1p_1\right){\rm e}^{s\varphi}, \end{equation} $

(4.6)

$ \begin{equation} \label{4.2}\partial^2_{t}{v}-a_{21}\Delta {u}-a_{22}\Delta v=\left({\cal N}_1[u_2]-a_{21}{\cal N}_2[u_1]-a_{22}{\cal N}_2[u_2]+\chi_2\sigma_2p_2\right){\rm e}^{s\varphi}, \end{equation} $

(4.7)

其中

$ {\cal N}_1[w]= (\chi_2)_{tt} {w}_t+s \chi_2 \left(\varphi_{tt}+s\varphi_t^2\right){w}_t+ 2(\chi_2)_t {w}_{tt}+ 2s(\chi_2)_t\varphi_t {w}_t+2s\chi_2 \varphi_t{w}_{tt}, $

$ {\cal N}_2[w]= s \chi_2 \left(\Delta\varphi+s|\nabla\varphi|^2\right){w}_t+2s\chi_2 \nabla\varphi\cdot\nabla{w}_t, $

满足

$ \begin{equation}\label{4.3} \left|{\cal N}_1[w]\right|+\left|{\cal N}_2[w]\right|\leq C\left(s^2|w_t|+s|\nabla_{x, t}w_t|\right). \end{equation} $

(4.8)

用$-a_{21}u_t$和$-a_{12}v_t$分别乘以(4.6)和(4.7)式, 再在$Q(\delta)$上积分, 然后相加, 可得

$ \begin{eqnarray}\label{4.4} &&-\int_{Q(\delta)}\left[\left(a_{21}u_t^2\right)_t+\left(a_{12}v_t^2\right)_t\right]{\rm d}x{\rm d}t\\ &=&-2\int_{Q(\delta)}\left(a_{11}a_{21}\Delta u u_t+a_{22} a_{12}\Delta v v_t\right){\rm d}x{\rm d}t -2\int_{Q(\delta)}\left(a_{12}a_{21}\Delta v u_t+a_{21}a_{12}\Delta u v_t\right){\rm d}x{\rm d}t\\ &&-2\int_{Q(\delta)}\left(a_{21}\chi_2\sigma_1 p_1 u_t+a_{12}\chi_2\sigma_2 p_2v_t\right){\rm e}^{s\varphi}{\rm d}x{\rm d}t\nonumber\\ &&-2\int_{Q(\delta)}\left(a_{21}{\cal L}_1[u_1, u_2]u_t+a_{12}{\cal L}_2[u_1, u_2]v_t\right){\rm e}^{s\varphi}{\rm d}x{\rm d}t\\ &:=&K_1+K_2+K_3+K_4, \end{eqnarray} $

(4.9)

其中

$ {\cal L}_1[u_1, u_2]={\cal N}_1[u_1]-a_{11}{\cal N}_2[u_1]-a_{12}{\cal N}_2[u_2], \\ {\cal L}_2[u_1, u_2]={\cal N}_1[u_2]-a_{21}{\cal N}_2[u_1]-a_{22}{\cal N}_2[u_2]. $

接下来我们逐一地估计$K_i, i=1, \cdots, 4$.从$\chi_2$和$T_0$的定义, 我们有$\nabla_{x, t}u(x, t_\delta(x))=\nabla_{x, t}v(x, t_\delta(x))=0$, $x\in\Omega$及$u_t(x, t)=v_t(x, t)=0$, $(x, t)\in\Sigma_1(\delta)$.另外, 由(1.11)式我们有$\nabla u(x, 0)=\nabla v(x, 0)=0$, $x\in\Omega$.再结合$Q(\delta)=\Omega\times(0, t_\delta(x))$和

$ \begin{equation}\label{410}\left\{\begin{array}{l} |u_t|+|v_t|\leq \sum\limits_{i=1}^2\left(s|\partial_tu_i|+|\partial_t^2u_i|\right){\rm e}^{s\varphi}, \\ |\nabla u|+|\nabla v|\leq \sum\limits_{i=1}^2\left(s|\partial_t u_i|+|\nabla \partial_tu_i|\right){\rm e}^{s\varphi}, \end{array}\right. \end{equation} $

(4.10)

我们有

$ \begin{eqnarray}\label{4.5} K_1&=&\int_\Omega \int_0^{t_\delta(x)}\left[\left(a_{11}a_{21}|\nabla u|^2\right)_t+\left(a_{22}a_{12}|\nabla v|^2\right)_t\right]{\rm d}t{\rm d}x\nonumber\\ &&-2\int_{\Sigma(\delta)}\left(a_{11}a_{21}u_t\nabla u\cdot\eta+a_{22}a_{12}v_t\nabla v\cdot\eta\right){\rm d}S{\rm d}t\nonumber\\ &=&\int_\Omega \left(a_{11}a_{21}|\nabla u|^2+a_{22}a_{12}|\nabla v|^2\right)_{t=0}^{t=t_\delta(x)}{\rm d}x\nonumber\\ &&-2\int_{\Sigma(\delta)}\left(a_{11}a_{21}u_t\nabla u\cdot\eta+a_{22}a_{12}v_t\nabla v\cdot\eta\right){\rm d}S{\rm d}t\nonumber\\ &\leq&Cs^2{\rm e}^{Cs}\sum\limits_{i=1}^2\left(\|u_i\|_{H^1(\Sigma(\delta))}^2+\|\partial_t u_i\|_{H^1(\Sigma(\delta))}^2\right). \end{eqnarray} $

(4.11)

对于$K_2$, 利用分部积分可得

$ \begin{eqnarray}\label{4.6} K_2&=&2a_{12}a_{21}\left(\int_{Q(\delta)}\nabla v\cdot \nabla u_t{\rm d}x{\rm d}t+\int_{Q(\delta)}\nabla u\cdot\nabla v_t{\rm d}x{\rm d}t\right)\nonumber\\ &&-2a_{12}a_{21}\int_{\Sigma(\delta)}\left(u_t\nabla v\cdot\eta+v_t\nabla u\cdot\eta\right){\rm d}S{\rm d}t\nonumber\\ &\leq& Cs^2{\rm e}^{Cs}\sum\limits_{i=1}^2\left(\|u_i\|_{H^1(\Sigma(\delta))}^2+\|\partial_t u_i\|_{H^1(\Sigma(\delta))}^2\right). \end{eqnarray} $

(4.12)

应用Hölder不等式, 有

$ \begin{eqnarray}\label{4.7} K_3\leq C\sum\limits_{i=1}^2\int_{Q(\delta)}|p_i|^2{\rm e}^{2s\varphi}{\rm d}x{\rm d}t+C\sum\limits_{i=1}^2\int_{Q(\delta)}\left(|\partial_t^2u_i|^2+s^2|\partial_t u_i|^2\right){\rm e}^{2s\varphi}{\rm d}x{\rm d}t. \end{eqnarray} $

(4.13)

由(4.8)和(4.10)式可知

$ \begin{eqnarray*} &&\left|{\cal L}_1[u_1, u_2] u_t\right|+\left|{\cal L}_2[u_1, u_2] v_t\right|\nonumber\\ &\leq& C\left(s^2\sum\limits_{i=1}^2|\partial_tu_i|+s\sum\limits_{i=1}^2|\nabla_{x, t}\partial_tu_i|\right)\left(s\sum\limits_{i=1}^2|\partial_tu_i|+\sum\limits_{i=1}^2|\partial_t^2u_i|\right){\rm e}^{s\varphi}\nonumber\\ &\leq& C\left(s^3\sum\limits_{i=1}^2|\partial_tu_i|^2+s|\nabla_{x, t}\partial_t u_i|^2\right){\rm e}^{s\varphi}, \end{eqnarray*} $

这意味着

$ \begin{eqnarray}\label{4.9} K_4\leq C\sum\limits_{i=1}^2\int_{Q(\delta)}\left(s^3|\partial_t u_i|^2+s|\nabla_{x, t}\partial_t u_i|^2\right){\rm e}^{2s\varphi}{\rm d}x{\rm d}t. \end{eqnarray} $

(4.14)

现将(4.11)-(4.14式代入(4.9)式, 我们得到

$ \begin{eqnarray}\label{4.10} &&-\int_{Q(\delta)}\left[a_{21}\left(u_t^2\right)_t+a_{12}\left(v_t^2\right)_t\right]{\rm d}x{\rm d}t\nonumber\\ &\leq&Cs^2{\rm e}^{Cs}\sum\limits_{i=1}^2\left(\|u_i\|_{H^1(\Sigma(\delta))}^2+\|\partial_t u_i\|_{H^1(\Sigma(\delta))}^2\right)+C\sum\limits_{i=1}^2\int_{Q_T}|p_i|^2{\rm e}^{2s\varphi}{\rm d}x{\rm d}t\nonumber\\ &&+C\sum\limits_{i=1}^2\int_{Q(\delta)}\left(s^3|\partial_t u_i|^2+s|\nabla_{x, t}\partial_t u_i|^2\right){\rm e}^{2s\varphi}{\rm d}x{\rm d}t. \end{eqnarray} $

(4.15)

另一方面, 利用(1.9)-(1.11)式和$\chi_2$的定义, 我们得到

$ \begin{equation} \label{4.11}u_t(x, 0)=\chi_2(0)\partial_t^2u_1(x, 0){\rm e}^{s\varphi(x, 0)}=\sigma_1(x, 0)p_1(x){\rm e}^{s\varphi(x, 0)}, \end{equation} $

(4.16)

$ \begin{equation} \label{4.12}v_t(x, 0)=\chi_2(0)\partial_t^2v_1(x, 0){\rm e}^{s\varphi(x, 0)}=\sigma_2(x, 0)p_2(x){\rm e}^{s\varphi(x, 0)}. \end{equation} $

(4.17)

然后, 由(4.16)-(4.17)式和${\rm Supp}(\chi_2)\subset (T_0/4, 3T_0/4)$, 我们进一步有

$ \begin{eqnarray}\label{4.13} &&-\int_{Q(\delta)}\left[a_{21}\left(u_t^2\right)_t+a_{12}\left(v_t^2\right)_t\right]{\rm d}x{\rm d}t\\ &=&-\int_{\Omega}\int_0^{t_\delta(x)}\left[a_{21}\left(u_t^2\right)_t+a_{12}\left(v_t^2\right)_t\right]{\rm d}x{\rm d}t\nonumber\\ &=&\int_\Omega\left(a_{21}\left|\sigma_1(x, 0)p_1(x)\right|^2+ a_{12}\left|\sigma_2(x, 0)p_2(x)\right|^2\right){\rm e}^{2s\varphi(x, 0)}{\rm d}x{\rm d}t. \end{eqnarray} $

(4.18)

最后, 由(1.14), (4.15)和(4.18)式可得估计(4.5)式.证毕.

定理1.2的证明  由引理4.1和引理4.2可知, 对充分大的$s$我们有

$ \begin{eqnarray}\label{4.17} \sum\limits_{i=1}^2\int_\Omega|p_i(x)|^2{\rm e}^{2s\varphi(x, 0)}{\rm d}x &\leq& Cs^3{\rm e}^{Cs}\sum\limits_{i=1}^2\left(\|u_i\|_{H^1(\Sigma(\delta))}^2+\|\partial_t u_i\|_{H^1(\Sigma(\delta))}^2\right)\nonumber\\ &&+C\sum\limits_{i=1}^2\int_{Q(\delta)}|p_i|^2{\rm e}^{2s\varphi}{\rm d}x{\rm d}t+Cs^3{\rm e}^{4\delta s}M. \end{eqnarray} $

(4.19)

此外, 有

$ \begin{eqnarray}\label{4.18} \sum\limits_{i=1}^2\int_{Q(\delta)}|p_i|^2{\rm e}^{2s\varphi}{\rm d}x{\rm d}t&=&\sum\limits_{i=1}^2\int_{\Omega}|p_i(x)|^2{\rm e}^{2s\varphi(x, 0)}{\rm d}x\int_0^{t_{\delta}(x)} {\rm e}^{-2s\theta t^2}{\rm d}t\nonumber\\ &\leq& C\frac{1}{\sqrt{\theta}}\frac{1}{\sqrt{s}}\sum\limits_{i=1}^2\int_{\Omega}|p_i(x)|^2{\rm e}^{2s\varphi(x, 0)}{\rm d}x. \end{eqnarray} $

(4.20)

若取$s$充分大, (4.19)式右边第二项能够被左边项吸收.因此, 由(4.19)和(4.20)式可得

$ \begin{eqnarray} \sum\limits_{i=1}^2\|p_i\|^2_{L^2(\Omega)}{\rm e}^{2sd_0}\leq Cs^3{\rm e}^{Cs}\sum\limits_{i=1}^2\left(\|u_i\|_{H^1(\Sigma(\delta))}^2+\|\partial_t u_i\|_{H^1(\Sigma(\delta))}^2\right)+Cs^3{\rm e}^{4\delta}M, \end{eqnarray} $

(4.21)

也就是, 当$s_2$是充分大的常数且$s>s_2$时, 有

$ \begin{eqnarray}\label{4.20} \sum\limits_{i=1}^2\|p_i\|^2_{L^2(\Omega)}\leq C{\rm e}^{Cs}\sum\limits_{i=1}^2\left(\|u_i\|_{H^1(\Sigma(\delta))}^2+\|\partial_t u_i\|_{H^1(\Sigma(\delta))}^2\right)+C{\rm e}^{-(d_0-2\delta) s}M. \end{eqnarray} $

(4.22)

记$s:=s+s_2$, 并用$C{\rm e}^{Cs_2}$代替$C$, 则(4.22)式对所有的$s\geq 0$成立.最后, 对$s$极小化(4.22)式的右边项, 我们得到(1.15)式对于$\kappa=\frac{(d_0-2\delta)}{C+(d_0-2\delta)}$成立.证毕.