数学物理方程 第9次作业

Chasse_neige

7.计算下列积分

(a)

$${\int }_{-1}^1{\mathrm{P}}_k^{\prime }(x){\mathrm{P}}_l(x)\mathrm{d}x$$

当 $k\le l$ 时，利用 Legendre 多项式的正交性，积分为 $0$

当 $k>l$ 时

$$\begin{array}{c}{\int }_{-1}^1{\mathrm{P}}_k^{\prime }(x){\mathrm{P}}_l(x)\mathrm{d}x={P_k(x)P_l(x)|}_{-1}^1-{\int }_{-1}^1{P}_k(x)P_l^{\prime }(x)\mathrm{d}x\\ =P_k(1)P_l(1)-P_k(-1)P_l(-1)=1-(-1{)}^{k+l}\end{array}$$

(b)

$${\int }_{-1}^1{\displaystyle \frac{{\mathrm{P}}_l(x)}{(1-2xt+t^2{)}^{3/2}}}\mathrm{d}x$$

利用生成函数，得到

$$\frac{\partial }{\partial x}\frac{1}{\sqrt{1-2xt+t^2}}=\sum _l{P}_l^{\prime }(x)t^l$$

所以

$$\frac{t}{(1-2xt+t^2{)}^{\frac{3}{2}}}=\sum _l{P}_l^{\prime }(x)t^l$$

所以

$${\int }_{-1}^1{\displaystyle \frac{{\mathrm{P}}_l(x)}{(1-2xt+t^2{)}^{3/2}}}\mathrm{d}x={\int }_{-1}^1\sum _{k=1}^{\mathrm{\infty }}{P}_k^{\prime }(x)P_l(x)t^{k-1}\mathrm{d}x$$

带入上一问的结果，直接得到

$${\int }_{-1}^1{\displaystyle \frac{{\mathrm{P}}_l(x)}{(1-2xt+t^2{)}^{3/2}}}\mathrm{d}x=\sum _{k=l+1}^{\mathrm{\infty }}(1-(-1{)}^{k+l})t^{k-1}=2t^l\frac{1}{1-t^2}=\frac{2t^l}{1-t^2}$$

8.一半径为 $a$ 的半球, 球面温度为 $u_0$, 底面温度为 $0$. 求半球内的温度分布.

方程以及边界条件为

$$\begin{array}{c}{\mathrm{\nabla }}^{2}u=0\\ u{|}_{r=a}=u_0,u{|}_{\theta =\frac{\pi }{2}}=0\end{array}$$

由于问题关于 $z$ 轴是轴对称的，所以解可以写成径向函数以及 Legendre 多项式的叠加

$$u(r,\theta )=\sum _l(A_l{r}^l+B_l{r}^{-l-1})P_l(\cos \theta )$$

由原点处的有界性，得到

$$B_l=0$$

所以

$$u(r,\theta )=\sum _l{A}_l{r}^l{P}_l(\cos \theta )$$

利用边界条件 $u(r,\frac{\pi }{2})=0$，得到解仅仅包含奇次项

$$u(r,\theta )=\sum _l{A}_{2l+1}{r}^{2l+1}{P}_{2l+1}(\cos \theta )$$

再利用正交性待定系数

$$u_0=u(a,\theta )=\sum _l{A}_{2l+1}{a}^{2l+1}{P}_{2l+1}(\cos \theta )$$$${\int }_0^{\frac{\pi }{2}}{u}_0{P}_{2l+1}(\cos \theta )\sin \theta \mathrm{d}\theta -{\int }_{\frac{\pi }{2}}^{\pi }{u}_0{P}_{2l+1}(\cos \theta )\sin \theta \mathrm{d}\theta =A_{2l+1}{a}^{2l+1}{\int }_0^{\pi }{P}_{2l+1}^2(\cos \theta )\sin \theta \mathrm{d}\theta$$

注意这个积分已经把问题进行了延拓至整个球

$$A_{2l+1}=\frac{4l+3}{2a^{2l+1}}({\int }_0^1{u}_0{P}_{2l+1}(x)\mathrm{d}x-{\int }_{-1}^0{u}_0{P}_{2l+1}(x)\mathrm{d}x)=(-1{)}^l\frac{1}{a^{2l+1}}\frac{(2l)!(4l+3)}{2^{2l+1}l!(l+1)!}$$

所以温度分布为

$$u(r,\theta )=\sum _{l=0}^{\mathrm{\infty }}(-1{)}^l\frac{r^{2l+1}}{a^{2l+1}}\frac{(2l)!(4l+3)}{2^{2l+1}l!(l+1)!}{P}_{2l+1}(\cos \theta )$$

9.一半径为 $b$ 的接地导体球壳, 内部放一均匀带电圆环, 环的半径为 $a$, 电量为 $Q$, 环心与球心重合. 试求球壳内电势分布.

取圆环轴线为 $z$ 轴，直接猜测解为

$$\varphi (r,\theta )=\{\begin{array}{l}\sum _l{A}_{2l}{r}^{2l}{P}_{2l}(\cos \theta )(0\le r\le a)\\ \sum _l(B_{2l}{r}^{2l}+C_{2l}{r}^{-2l-1})P_{2l}(\cos \theta )(a\le r\le b)\end{array}$$

的形式，利用边界条件待定系数

对于 $0\le r\le a$

$$r^2{\frac{\partial u(r,\frac{\pi }{2})}{\partial r}|}_{r=a-}^{r=a+}=-\frac{Q}{2\pi {ϵ}_0}\delta (\theta -\frac{\pi }{2})$$$$(2l{a}^{2l+1}(B_{2l}-A_{2l})-(2l+1)C_{2l}{a}^{-2l})\frac{2}{4l+1}=-\frac{Q}{2\pi {ϵ}_0}\frac{(-1{)}^l(2l)!}{2^{2l}(l!{)}^2}$$

再利用电势连续的条件

$$A_{2l}{a}^{2l}=B_{2l}{a}^{2l}+C_{2l}{a}^{-2l-1}$$

对于 $a\le r\le b$

$$u(b,\theta )=0$$

所以

$$B_{2l}{b}^{2l}+C_{2l}{b}^{-2l-1}=0$$

解得

$$\begin{array}{c}{A}_{2l}=\frac{Q}{4\pi {ϵ}_0}{P}_{2n}(0)(\frac{1}{a^{2l+1}}-\frac{a^{2l}}{b^{4l+1}})\\ B_{2l}=-\frac{Q}{4\pi {ϵ}_0}{P}_{2n}(0)\frac{a^{2l}}{b^{4l+1}}\\ C_{2l}=\frac{Q}{4\pi {ϵ}_0}{P}_{2n}(0)a^{2l}\end{array}$$

最终得到

$$u(r,\theta )=\{\begin{array}{ll}{\displaystyle \frac{Q}{4\pi {ϵ}_0}}\sum _{n=0}^{\mathrm{\infty }}\frac{(-1{)}^n(2n)!}{2^{2n}(n!{)}^2}{P}_{2n}(\cos \theta )r^{2n}({\displaystyle \frac{1}{a^{2n+1}}}-{\displaystyle \frac{a^{2n}}{b^{4n+1}}})& (0\le r<a)\\ {\displaystyle \frac{Q}{4\pi {ϵ}_0}}\sum _{n=0}^{\mathrm{\infty }}\frac{(-1{)}^n(2n)!}{2^{2n}(n!{)}^2}{P}_{2n}(\cos \theta )({\displaystyle \frac{a^{2n}}{r^{2n+1}}}-{\displaystyle \frac{a^{2n}{r}^{2n}}{b^{4n+1}}})& (a<r<b)\end{array}$$