偏微分方程

中山大学数学学院2019学年秋季学期数学与应用数学专业《偏微分方程》课程网站

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第八周作业

本次作业题包括习题2.5第3题,习题3.1第6、9题和习题3.2第2题。

习题2.5

3

设$\varphi\in L^1(\mathbb{R}^3)\cap C(\mathbb{R}^3)$有界,则以下问题

$\begin{cases}\frac{\partial u}{\partial t}=a^2(\frac{\partial^2 u}{\partial x^2}+\frac{\partial^2 u}{\partial y^2}+\frac{\partial^2 u}{\partial z^2})&,(x,y,z,t)\in \mathbb{R}^3\times (0,+\infty)\\u(x,y,z,0)=\varphi (x,y,z)&,(x,y,z)\in\mathbb{R}^3\end{cases}$

有惟一有界经典解$u(x,t)=\frac{1}{(2a\sqrt{\pi t})^3}\int^\infty_{-\infty}\int^\infty_{-\infty}\int^\infty_{-\infty}\varphi (\xi,\eta,\zeta)e^{-\frac{(x-\xi)^2+(y-\eta)^2+(z-\zeta)^2}{4a^2t}}\mathrm{d}\xi\mathrm{d}\eta\mathrm{d}\zeta$,从而$\vert u(x,t)\vert \leq\frac{1}{(2a\sqrt{\pi t})^3}\int^\infty_{-\infty}\int^\infty_{-\infty}\int^\infty_{-\infty}\vert\varphi (\xi,\eta,\zeta)\vert\mathrm{d}\xi\mathrm{d}\eta\mathrm{d}\zeta\leq\frac{\Vert\varphi\Vert_{L^1(\mathbb{R}^3)}}{(2a\sqrt{\pi})^3}t^{-\frac{3}{2}}$

习题3.1

6

若初边值问题有形如$u(x,y)=X(x)Y(y)$的非零解,则$X’‘(x)Y(y)+X(x)Y’‘(y)=0$,即$\frac{X’‘(x)}{X(x)}=-\frac{Y’‘(y)}{Y(y)}$为与$x,y$无关的常数,记为$-\lambda$。由边界条件$X(0)=X(a)=0$

  • 当$\lambda\lt 0$,$X(x)=Ce^{\sqrt{-\lambda}x}+De^{-\sqrt{-\lambda}x}$(其中$C$、$D$为待定常数),于是$0=X(0)=C+D$而$0=X(a)=Ce^{\sqrt{-\lambda}a}+De^{-\sqrt{-\lambda}a}$,解出$C=D=0$从而$X$恒为零,与$u$不恒为零的假设矛盾。
  • 当$\lambda=0$,$X(x)=Cx+D$(其中$C$、$D$为待定常数),于是$0=X(0)=D$而$0=X(a)=Ca$,从而$C=0$,$X$恒为零,与$u$不恒为零的假设矛盾。
  • 当$\lambda\gt 0$,$X(x)=C\cos({\sqrt{\lambda}x})+D\sin({\sqrt{\lambda}x})$(其中$C$、$D$为待定常数),于是$0=X(0)=C$而$0=X(a)=D\sin({\sqrt{\lambda}a})$,于是存在正整数$k$使$\sqrt{\lambda}a=k\pi$,$X(x)=D\sin({\frac{k\pi x}{a}})$。此时,$Y(y)=Ae^{\sqrt{\lambda}y}+Be^{-\sqrt{\lambda}y}=Ae^{\frac{k\pi y}{a}}+Be^{-\frac{k\pi y}{a}}$(其中$A$、$B$为待定常数),由边界条件$0=Y(b)$得$B=-e^{2\frac{k\pi b}{a}}A$

于是得到$u(x,y)=AD(e^{\frac{k\pi y}{a}}-e^{2\frac{k\pi b}{a}}e^{-\frac{k\pi y}{a}})\sin({\frac{k\pi x}{a}})$,特别地$u(x,0)=AD(1-e^{2\frac{k\pi b}{a}})\sin({\frac{k\pi x}{a}})$,与边值条件$u(x,0)=\sin({\frac{\pi x}{a}})$对比知取$k=1$,$AD=\frac{1}{1-e^{2\frac{\pi b}{a}}}$即可。也就是说,$u(x,y)=\frac{e^{\frac{\pi y}{a}}-e^{2\frac{\pi b}{a}}e^{-\frac{\pi y}{a}}}{1-e^{2\frac{\pi b}{a}}}\sin({\frac{\pi x}{a}})$,容易验证它是问题的经典解。

9

对任何$u,v\in V, t\in\mathbb{R}$,成立$u+tv\in V$,并且

$\begin{align}J(u+tv)&=\iiint_{\Omega}\frac{1}{2}((\frac{\partial (u+tv)}{\partial x})^2+(\frac{\partial (u+tv)}{\partial y})^2+(\frac{\partial (u+tv)}{\partial z})^2)\mathrm{d}x\mathrm{d}y\mathrm{d}z+\iint_{\Gamma}(\frac{1}{2}\sigma (u+tv)^2-g(u+tv))\mathrm{d}S\\&=\iiint_{\Omega}\frac{1}{2}((\frac{\partial u}{\partial x})^2+(\frac{\partial u}{\partial y})^2+(\frac{\partial u}{\partial z})^2)\mathrm{d}x\mathrm{d}y\mathrm{d}z+\iint_{\Gamma}(\frac{1}{2}\sigma u^2-gu)\mathrm{d}S+t(\iiint_{\Omega}(\frac{\partial u}{\partial x}\frac{\partial v}{\partial x}+\frac{\partial u}{\partial y}\frac{\partial v}{\partial y}+\frac{\partial u}{\partial z}\frac{\partial v}{\partial z})\mathrm{d}x\mathrm{d}y\mathrm{d}z+\iint_{\Gamma}(\sigma uv-gv)\mathrm{d}S)+t^2(\iiint_{\Omega}\frac{1}{2}((\frac{\partial v}{\partial x})^2+(\frac{\partial v}{\partial y})^2+(\frac{\partial v}{\partial z})^2)\mathrm{d}x\mathrm{d}y\mathrm{d}z+\iint_{\Gamma}(\frac{1}{2}\sigma v^2)\mathrm{d}S)\\&=J(u)+t(-\iiint_{\Omega}v\Delta u\mathrm{d}x\mathrm{d}y\mathrm{d}z+\iint_{\Gamma}(v\frac{\partial u}{\partial\vec{n}}+\sigma uv-gv)\mathrm{d}S)+t^2(\iiint_{\Omega}\frac{1}{2}((\frac{\partial v}{\partial x})^2+(\frac{\partial v}{\partial y})^2+(\frac{\partial v}{\partial z})^2)\mathrm{d}x\mathrm{d}y\mathrm{d}z+\iint_{\Gamma}(\frac{1}{2}\sigma v^2)\mathrm{d}S)\end{align}$

因此,若$u\in V$使$J(u)=\min_{v\in V}J(v)$,则对任何$v\in V$成立$0=\frac{\mathrm{d}}{\mathrm{d}t}\vert_{t=0}(J(u+tv))=-\iiint_{\Omega}v\Delta u\mathrm{d}x\mathrm{d}y\mathrm{d}z+\iint_{\Gamma}(v\frac{\partial u}{\partial\vec{n}}+\sigma uv-gv)\mathrm{d}S$。于是对任何$M_0\in\Omega$,若$\Delta u(M_0)\neq 0$,则存在$M_0$的某邻域$B_{2r}(M_0)\subseteq\Omega$使$\Delta u$在$B_{2r}(M_0)$中与$\Delta u(M_0)$同号,可取$v\in V$使$v(M)\begin{cases}=1&, M\in B_{r}(M_0)\\0&, M\notin B_{2r}(M_0)\\\geq 0&, 其它情况\end{cases}\geq 0$,于是$\iiint_{\Omega}v\Delta u\mathrm{d}x\mathrm{d}y\mathrm{d}z+\iint_{\Gamma}(v\frac{\partial u}{\partial\vec{n}}+\sigma uv-gv)\mathrm{d}S=\iiint_{\Omega}v\Delta u\mathrm{d}x\mathrm{d}y\mathrm{d}z\neq 0$矛盾,所以$\Delta u(M_0)=0$。进而对任何$v\in V$成立$0=\iint_{\Gamma}(\frac{\partial u}{\partial\vec{n}}+\sigma u-g)v\mathrm{d}S$。于是对任何$M_0\in\Gamma$,$\frac{\partial u}{\partial\vec{n}}(M_0)+\sigma u(M_0)=g(M_0)$。这说明$u$是以下问题的经典解:

$\begin{cases}\Delta u(M)= 0&,M\in\Omega\\\frac{\partial u}{\partial\vec{n}}(M)+\sigma u(M)=g(M)&,M\in\Gamma\end{cases}$

反之,若$u\in V$为上述初边值问题的经典解,则对任何$v\in V$,

$\begin{align}J(v)&=J(u+(v-u))\\&=J(u)+(-\iiint_{\Omega}(v-u)\Delta u\mathrm{d}x\mathrm{d}y\mathrm{d}z+\iint_{\Gamma}(v-u)(\frac{\partial u}{\partial\vec{n}}+\sigma u-g)\mathrm{d}S)+(\iiint_{\Omega}\frac{1}{2}((\frac{\partial (v-u)}{\partial x})^2+(\frac{\partial (v-u)}{\partial y})^2+(\frac{\partial (v-u)}{\partial z})^2)\mathrm{d}x\mathrm{d}y\mathrm{d}z+\iint_{\Gamma}(\frac{1}{2}\sigma (v-u)^2)\mathrm{d}S)\\&=J(u)+(\iiint_{\Omega}\frac{1}{2}((\frac{\partial (v-u)}{\partial x})^2+(\frac{\partial (v-u)}{\partial y})^2+(\frac{\partial (v-u)}{\partial z})^2)\mathrm{d}x\mathrm{d}y\mathrm{d}z+\iint_{\Gamma}(\frac{1}{2}\sigma (v-u)^2)\mathrm{d}S)\\&\geq J(u)\end{align}$

因此,$J(u)=\min_{v\in V}J(v)$。

习题3.2

2

由调和函数的均值性质,$u(0,0)=\frac{1}{2\pi}\int_{\partial B((0,0),1)}u\mathrm{d}s=\frac{1}{2\pi}\int^{2\pi}_{0}\sin (\theta)\mathrm{d}\theta=0$

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