第三周作业

本次作业题包括习题1.4第4、5题和题1.5第3题。

习题1.4

4

因解轴对称，初始条件中$\varphi,\psi$也轴对称，即可以看作$r$的函数。利用极坐标，可记$(x,y)=(r\cos\theta_0,r\sin\theta_0)$，则$(x+s\cos\theta)^2+(y+s\sin\theta)^2=x^2+y^2+s^2+2rs(\cos\theta\cos\theta_0+\sin\theta\sin\theta_0)=r^2+s^2+2rs\cos(\theta-\theta_0)$，于是

$\begin{align}u(r,t)=u(x,y,t)&=\frac{1}{2\pi a}(\frac{\partial}{\partial t}(\int^{at}_{0}\int^{2\pi}_{0}\frac{\varphi(x+s\cos\theta,y+s\sin\theta)}{\sqrt{(at)^2-s^2}}s\mathrm{d}\theta\mathrm{d}s)+\int^{at}_{0}\int^{2\pi}_{0}\frac{\psi(x+s\cos\theta,y+s\sin\theta)}{\sqrt{(at)^2-s^2}}s\mathrm{d}\theta\mathrm{d}s) \\&=\frac{1}{2\pi a}(\frac{\partial}{\partial t}(\int^{at}_{0}\int^{2\pi}_{0}\frac{\varphi(\sqrt{r^2+s^2+2rs\cos(\theta-\theta_0)}}{\sqrt{(at)^2-s^2}}s\mathrm{d}\theta\mathrm{d}s)+\int^{at}_{0}\int^{2\pi}_{0}\frac{\psi(\sqrt{r^2+s^2+2rs\cos(\theta-\theta_0)}}{\sqrt{(at)^2-s^2}}s\mathrm{d}\theta\mathrm{d}s)\\&=\frac{1}{2\pi a}(\frac{\partial}{\partial t}(\int^{at}_{0}\int^{2\pi}_{0}\frac{\varphi(\sqrt{r^2+s^2+2rs\cos(\theta)}}{\sqrt{(at)^2-s^2}}s\mathrm{d}\theta\mathrm{d}s)+\int^{at}_{0}\int^{2\pi}_{0}\frac{\psi(\sqrt{r^2+s^2+2rs\cos(\theta)}}{\sqrt{(at)^2-s^2}}s\mathrm{d}\theta\mathrm{d}s)\end{align}$

5

令$v(x,y,z,t)=e^{\frac{cz}{a}}u(x,y,t)$，则$v$满足：

$\begin{cases}v_{tt}=a^2(v_{xx}+v_{yy}+v_{zz}) & , x,y,z\in \mathbb{R}, t\in (0,+\infty)\\v(x,y,z,0)=e^{\frac{cz}{a}}\varphi (x,y), v_t(x,y,z,0)=e^{\frac{cz}{a}}\psi (x,y)& , x,y,z\in \mathbb{R}\end{cases}$

由定理4.1，

$\begin{align}v(x,y,z,t)&=\frac{\partial}{\partial t}(\frac{1}{4\pi a^2 t}\iint_{S((x,y,z),at)} e^{\frac{c\zeta}{a}}\varphi (\xi,\eta)\mathrm{d}S)+\frac{1}{4\pi a^2 t}\iint_{S((x,y,z),at)} e^{\frac{c\zeta}{a}}\psi (\xi,\eta)\mathrm{d}S\\&=\frac{\partial}{\partial t}(\frac{1}{4\pi a}\iint_{B((x,y),at)} \frac{(e^{\frac{c(z+ \sqrt{(at)^2-(\xi -x)^2-(\eta -y)^2}}{a}}+e^{\frac{c(z- \sqrt{(at)^2-(\xi -x)^2-(\eta -y)^2}}{a}})\varphi (\xi,\eta)}{\sqrt{(at)^2-(\xi -x)^2-(\eta -y)^2}}\mathrm{d}\xi\mathrm{d}\eta)+\frac{1}{4\pi a}\iint_{B((x,y),at)} \frac{(e^{\frac{c(z+ \sqrt{(at)^2-(\xi -x)^2-(\eta -y)^2}}{a}}+e^{\frac{c(z- \sqrt{(at)^2-(\xi -x)^2-(\eta -y)^2}}{a}})\psi (\xi,\eta)}{\sqrt{(at)^2-(\xi -x)^2-(\eta -y)^2}}\mathrm{d}\xi\mathrm{d}\eta\\&=\frac{\partial}{\partial t}(\frac{1}{2\pi a}\int^{2\pi}_{0}\int^{at}_{0} \frac{e^{\frac{cz}{a}}\cosh (\frac{c}{a}\sqrt{(at)^2-r^2})\varphi (x+r\cos\theta,y+r\sin\theta)}{\sqrt{(at)^2-r^2}}r\mathrm{d}r\mathrm{d}\theta)+\frac{1}{2\pi a}\int^{2\pi}_{0}\int^{at}_{0} \frac{e^{\frac{cz}{a}}\cosh (\frac{c}{a}\sqrt{(at)^2-r^2})\psi (x+r\cos\theta,y+r\sin\theta)}{\sqrt{(at)^2-r^2}}r\mathrm{d}r\mathrm{d}\theta\end{align}$

，从而$u(x,y,t)=e^{-\frac{cz}{a}}v(x,y,z,t)=\frac{\partial}{\partial t}(\frac{1}{2\pi a}\int^{2\pi}_{0}\int^{at}_{0} \frac{\cosh (\frac{c}{a}\sqrt{(at)^2-r^2})\varphi (x+r\cos\theta,y+r\sin\theta)}{\sqrt{(at)^2-r^2}}r\mathrm{d}r\mathrm{d}\theta)+\frac{1}{2\pi a}\int^{2\pi}_{0}\int^{at}_{0} \frac{\cosh (\frac{c}{a}\sqrt{(at)^2-r^2})\psi (x+r\cos\theta,y+r\sin\theta)}{\sqrt{(at)^2-r^2}}r\mathrm{d}r\mathrm{d}\theta$。

习题1.5

3

因$\varphi,\psi$紧支，存在$R,M>0$使$\mathrm{supp}(\varphi),\mathrm{supp}(\psi)\subseteq B(O,R)$，$\vert\varphi\vert,\vert\psi\vert\leq M$。记$P=(x_0,y_0)$，当$t>\frac{1}{a}\max\{\overline{OP}+R,\vert\overline{OP}-R\vert\}$

$\begin{align}\vert u(x_0,y_0,t)\vert &=\vert \frac{1}{2\pi a}(\frac{\partial}{\partial t}(\int^{at}_{0}\int^{2\pi}_{0}\frac{\varphi(x_0+r\cos\theta,y_0+r\sin\theta)}{\sqrt{(at)^2-r^2}}r\mathrm{d}\theta\mathrm{d}r)+\int^{at}_{0}\int^{2\pi}_{0}\frac{\psi(x_0+r\cos\theta,y_0+r\sin\theta)}{\sqrt{(at)^2-r^2}}r\mathrm{d}\theta\mathrm{d}r)\vert \\&=\vert \frac{1}{2\pi a}(\frac{\partial}{\partial t}(\int^{\overline{OP}+R}_{\max\{0,\overline{OP}-R\}}\int^{2\pi}_{0}\frac{\varphi(x_0+r\cos\theta,y_0+r\sin\theta)}{\sqrt{(at)^2-r^2}}r\mathrm{d}\theta\mathrm{d}r)+\int^{\overline{OP}+R}_{\max\{0,\overline{OP}-R\}}\int^{2\pi}_{0}\frac{\psi(x_0+r\cos\theta,y_0+r\sin\theta)}{\sqrt{(at)^2-r^2}}r\mathrm{d}\theta\mathrm{d}r)\vert \\&=\vert \frac{1}{2\pi a}(\int^{\overline{OP}+R}_{\max\{0,\overline{OP}-R\}}\int^{2\pi}_{0}\frac{-a^2t\varphi(x_0+r\cos\theta,y_0+r\sin\theta)}{\sqrt{((at)^2-r^2)^3}}r\mathrm{d}\theta\mathrm{d}r+\int^{\overline{OP}+R}_{\max\{0,\overline{OP}-R\}}\int^{2\pi}_{0}\frac{\psi(x_0+r\cos\theta,y_0+r\sin\theta)}{\sqrt{(at)^2-r^2}}r\mathrm{d}\theta\mathrm{d}r)\vert \\&\leq \frac{1}{2\pi a}(\int^{\overline{OP}+R}_{\max\{0,\overline{OP}-R\}}\int^{2\pi}_{0}\frac{a^2t\vert\varphi(x_0+r\cos\theta,y_0+r\sin\theta)\vert}{\sqrt{((at)^2-r^2)^3}}r\mathrm{d}\theta\mathrm{d}r+\int^{\overline{OP}+R}_{\max\{0,\overline{OP}-R\}}\int^{2\pi}_{0}\frac{\vert\psi(x_0+r\cos\theta,y_0+r\sin\theta)\vert}{\sqrt{(at)^2-r^2}}r\mathrm{d}\theta\mathrm{d}r)\\&\leq \frac{1}{2\pi a}(\frac{(2R)(2\pi)(\overline{OP}+R)a^2tM}{\sqrt{((at)^2-(\overline{OP}+R)^2)^3}}+\frac{(2R)(2\pi)(\overline{OP}+R)M}{\sqrt{(at)^2-(\overline{OP}+R)^2}})\end{align}$

因此，$\displaystyle\lim_{t\to +\infty}u(x_0,y_0,t)=0$。