第四周作业

本次作业题包括习题1.6第3、5题。

习题1.6

3

令$E_1(t)=\iint_{\Omega_{t}}(u_t^2+a^2(u_x^2+u_y^2))\mathrm{d}x\mathrm{d}y$，则

$\begin{align}\frac{\mathrm{d}E_1(t)}{\mathrm{d}t}&=\frac{\mathrm{d}}{\mathrm{d}t}(\iint_{\Omega_t}(u_{t}^2+a^2(u_x^2+u_y^2))\mathrm{d}x\mathrm{d}y)\\&=\iint_{\Omega_t}(2u_{t}u_{tt}+a^2(2u_xu_{xt}+2u_yu_{yt}))\mathrm{d}x\mathrm{d}y-a\int_{\Gamma_t}(u_t^2+a^2(u_x^2+u_y^2))\mathrm{d}s\\&=\iint_{\Omega_t}(2u_{t}(a^2(u_{xx}+u_{yy})+f)+a^2(2u_xu_{xt}+2u_yu_{yt}))\mathrm{d}x\mathrm{d}y-a\int_{\Gamma_t}(u_t^2+a^2(u_x^2+u_y^2))\mathrm{d}s\\&=2a^2\iint_{\Omega_t}(\frac{\partial}{\partial x}(u_{t}u_x)+\frac{\partial}{\partial y}(u_{t}u_y))\mathrm{d}x\mathrm{d}y+\iint_{\Omega_t}2u_{t}f\mathrm{d}x\mathrm{d}y-a\int_{\Gamma_t}(u_t^2+a^2(u_x^2+u_y^2))\mathrm{d}s\\&=2a^2\int_{\Gamma_t}(u_{t}u_x\cos (\vec{n},x)+u_{t}u_y\cos (\vec{n},y))\mathrm{d}x\mathrm{d}y+\iint_{\Omega_t}2u_{t}f\mathrm{d}x\mathrm{d}y-a\int_{\Gamma_t}(u_t^2+a^2(u_x^2+u_y^2))\mathrm{d}s\\&=-a\int_{\Gamma_t}((u_t\cos (\vec{n},x)-au_x)^2+(u_t\cos (\vec{n},y)-au_y)^2)\mathrm{d}s+\iint_{\Omega_t}2u_{t}f\mathrm{d}x\mathrm{d}y\\&\leq \iint_{\Omega_t}u_{t}^2\mathrm{d}x\mathrm{d}y+\iint_{\Omega_t}f^2\mathrm{d}x\mathrm{d}y\end{align}$

也就是说，$\frac{\mathrm{d}e^{-t}E_1(t)}{\mathrm{d}t}\leq e^{-t}\iint_{\Omega_t}f^2\mathrm{d}x\mathrm{d}y$，积分得$e^{-t}E_1(t)-E_1(0)\leq \int^{t}_{0}e^{-\tau}\iint_{\Omega_t}f^2\mathrm{d}x\mathrm{d}y\mathrm{d}\tau$，$E_1(t)\leq e^t E_1(0)+\int^{t}_{0}e^{t-\tau}\iint_{\Omega_t}f^2\mathrm{d}x\mathrm{d}y\mathrm{d}\tau$。

令$E_0(t)=\iint_{\Omega_{t}}u^2\mathrm{d}x\mathrm{d}y$，则

$\begin{align}\frac{\mathrm{d}E_0(t)}{\mathrm{d}t}&=\frac{\mathrm{d}}{\mathrm{d}t}(\iint_{\Omega_t}u^2\mathrm{d}x\mathrm{d}y)\\&=\iint_{\Omega_t}2uu_{t}\mathrm{d}x\mathrm{d}y-a\int_{\Gamma_t}u^2\mathrm{d}s\\&\leq \iint_{\Omega_t}(u^2+u_{t}^2)\mathrm{d}x\mathrm{d}y\\&\leq E_0(t)+E_1(t)\end{align}$

也就是说，$\frac{\mathrm{d}e^{-t}E_0(t)}{\mathrm{d}t}\leq e^{-t}E_1(t)$，积分得$e^{-t}E_0(t)-E_0(0)\leq \int^{t}_{0}e^{-\tau}E_1(\tau)\mathrm{d}\tau$，$E_0(t)\leq e^t E_0(0)+\int^{t}_{0}e^{t-\tau}E_1(\tau)\mathrm{d}\tau\leq e^t E_0(0)+e^{t}\int^{t}_{0}(E_1(0)+\Vert f\Vert_{L^2(K)}^2)\mathrm{d}\tau\leq e^t E_0(0)+e^{t}t(E_1(0)+\Vert f\Vert_{L^2(K)}^2)$，$\Vert u\Vert_{L^2(K)}^2=\int^{\frac{R}{a}}_{0}E_0(t)\mathrm{d}t\leq e^{\frac{R}{a}} E_0(0)+e^{\frac{R}{a}}\frac{R}{a}(E_1(0)+\Vert f\Vert_{L^2(K)}^2)$。

特别地，若$\frac{\partial^2u_1}{\partial t^2}-a^2(\frac{\partial^2u_1}{\partial x^2}+\frac{\partial^2u_1}{\partial y^2})=f_1$而$\frac{\partial^2u_2}{\partial t^2}-a^2(\frac{\partial^2u_2}{\partial x^2}+\frac{\partial^2u_2}{\partial y^2})=f_2$，$u_1\vert_{t=0}=u_2\vert_{t=0}$且$\frac{\partial u_1}{\partial t}\vert_{t=0}=\frac{\partial u_2}{\partial t}\vert_{t=0}$，即有$\Vert u_2-u_1\Vert_{L^2(K)}^2\leq e^{\frac{R}{a}}\frac{R}{a}\Vert f_2-f_1\Vert_{L^2(K)}^2$。

5

$\begin{align}\frac{\mathrm{d}E(t)}{\mathrm{d}t}&=\frac{\mathrm{d}}{\mathrm{d}t}(\iint_{\Omega}(u_{t}^2+a^2(u_x^2+u_y^2))\mathrm{d}x\mathrm{d}y+a^2\int_{\Gamma}\sigma u^2\mathrm{d}s)\\&=\iint_{\Omega}(2u_{t}u_{tt}+a^2(2u_xu_{xt}+2u_yu_{yt}))\mathrm{d}x\mathrm{d}y+a^2\int_{\Gamma}2\sigma uu_{t}\mathrm{d}s\\&=2a^2\iint_{\Omega}(u_{t}(u_{xx}+u_{yy})+u_xu_{xt}+u_yu_{yt})\mathrm{d}x\mathrm{d}y+2a^2\int_{\Gamma}\sigma uu_{t}\mathrm{d}s\\&=2a^2\iint_{\Omega}(\frac{\partial}{\partial x}(u_{t}u_x)+\frac{\partial}{\partial y}(u_{t}u_y))\mathrm{d}x\mathrm{d}y+2a^2\int_{\Gamma}\sigma uu_{t}\mathrm{d}s\\&=2a^2(\int_{\Gamma}(u_{t}u_x\cos(\vec{n},x)+u_{t}u_y\cos(\vec{n},y))\mathrm{d}s+\int_{\Gamma}\sigma uu_{t}\mathrm{d}s)\\&=2a^2(\int_{\Gamma}u_t\frac{\partial u}{\partial \vec{n}}\mathrm{d}s+\int_{\Gamma}\sigma uu_{t}\mathrm{d}s)\\&=2a^2\int_{\Gamma}u_t(\frac{\partial u}{\partial \vec{n}}+\sigma u)\mathrm{d}s\\&=0\end{align}$

这表明$E$关于$t$为常数。若$u_1,u_2$为方程的解，令$u=u_2-u_1$，则

$\begin{cases}u_{tt}-a^2(u_{xx}+u_{yy})=0&,t>0,(x,y)\in\Omega\\u(x,y,0)=0,u_{t}(x,y,0)=0 &,(x,y)\in\Omega\\(\frac{\partial u}{\partial \vec{n}}+\sigma u)\vert_{\Gamma}=0&,(x,y)\in\Gamma\end{cases}$

注意到$t=0$时有$u_t=u_x=u_y=0=u$。因此对$t>0$都有$\iint_{\Omega}(u_{t}^2+a^2(u_x^2+u_y^2))\mathrm{d}x\mathrm{d}y+a^2\int_{\Gamma}\sigma u^2\mathrm{d}s=E(t)=E(0)=0$，由各被积函数非负知它们恒为零。注意到对所有$t>0,(x,y)\in\Omega$，$u(x,y,t)=u(x,y,t)-u(x,y,0)=\int^t_{0}\frac{\partial}{\partial s}u(x,y,s)\mathrm{d}s=0$，所以$u_1=u_2$。