求解扇型域上的Dirichlet问题
{ Δ 2 u = 0 , 1 < r < e , 0 < θ < π 2 u ∣ r = 1 = u ∣ r = e = 0 u ∣ θ = 0 = 0 , u ∣ θ = π 2 = g ( r ) (15) \begin{cases} \Delta_2u=0, \quad 1<r<e,0<\theta<\frac{\pi}{2} \\ u|_{r=1}=u|_{r=e}=0 \\ u|_{\theta=0}=0, \quad u|_{\theta=\frac{\pi}{2}}=g(r) \end{cases} \tag{15} ??????Δ2?u=0,1<r<e,0<θ<2π?u∣r=1?=u∣r=e?=0u∣θ=0?=0,u∣θ=2π??=g(r)?(15)
这里， ( r , θ ) (r,\theta) (r,θ)为极坐标，e为自然对数的底数。
解：极坐标下，方程为
r 2 ? 2 u ? r 2 + r ? u ? r + ? 2 u ? θ 2 = 0 r^2\frac{\partial^2u}{\partial r^2}+r\frac{\partial u}{\partial r}+\frac{\partial^2u}{\partial \theta^2}=0 r2?r2?2u?+r?r?u?+?θ2?2u?=0
设 u ( r , θ ) = R ( r ) Θ ( θ ) u(r,\theta)=R(r)\Theta(\theta) u(r,θ)=R(r)Θ(θ)，代入方程和关于r的齐次边界问题，分离变量得固有值问题
{ r 2 R ′ ′ ( r ) + r R ′ ( r ) + λ R ( r ) = 0 , 1 < r < e R ( 1 ) = R ( e ) = 0 (16) \begin{cases} r^2R''(r)+rR'(r)+\lambda R(r)=0,\quad 1<r<e \\ R(1)=R(e)=0 \end{cases} \tag{16} {
r2R′′(r)+rR′(r)+λR(r)=0,1<r<eR(1)=R(e)=0?(16)
及常微分方程
Θ ′ ′ ( θ ) ? λ Θ ( θ ) = 0 \Theta''(\theta)-\lambda \Theta(\theta)=0 Θ′′(θ)?λΘ(θ)=0
固有值问题中的方程是变系数二阶线性方程，可化为 S ? L S-L S?L型
[ r R ′ ( r ) ] ′ + λ 1 r R ( r ) = 0 [rR'(r)]'+\lambda\frac{1}{r}R(r)=0 [rR′(r)]′+λr1?R(r)=0
这里， k ( r ) = r , q ( r ) ≡ 0 , ρ ( r ) = 1 r k(r)=r,q(r)\equiv 0,\rho(r)=\frac{1}{r} k(r)=r,q(r)≡0,ρ(r)=r1?。由S-L定理知 λ > 0 \lambda >0 λ>0
R ( r ) R(r) R(r)的方程时欧拉（Euler）方程，作变量代换 r = e t r=e^t r=et，记 y ( t ) = R ( e t ) y(t)=R(e^t) y(t)=R(et)，则变系数方程的固有值问题（16）式转化为最简S-L型方程固有值问题
{ y ′ ′ ( t ) + λ y = 0 , 0 < t < 1 y ( 0 ) = y ( 1 ) = 0 \begin{cases} y''(t)+\lambda y=0,\quad 0<t<1 \\ y(0)=y(1)=0 \end{cases} {
y′′(t)+λy=0,0<t<1y(0)=y(1)=0?
即得固有值
λ n = ( n π ) 2 , n = 1 , 2 , ? \lambda_n=(n\pi)^2,\quad n=1,2,\cdots λn?=(nπ)2,n=1,2,?
及固有函数
y n ( t ) = s i n n π t y_n(t)=sin\,n\pi t yn?(t)=sinnπt
亦即
R n ( r ) = s i n ( n π l n r ) R_n(r)=sin(n\pi\,lnr) Rn?(r)=sin(nπlnr)
相应地
Θ n ( θ ) = A n c h n π θ + B n s h n π θ \Theta_n(\theta)=A_nch\,n\pi\theta+B_nsh\,n\pi\theta Θn?(θ)=An?chnπθ+Bn?shnπθ
设
u ( r , θ ) = ∑ n = 1 + ∞ ( A n c h n π θ + B n s h n π θ ) s i n ( n π l n r ) u(r,\theta)=\sum_{n=1}^{+\infty}(A_nch\,n\pi\theta+B_nsh\,n\pi\theta)sin(n\pi lnr) u(r,θ)=n=1∑+∞?(An?chnπθ+Bn?shnπθ)sin(nπlnr)
代入关于 θ \theta θ的边界条件，得
u ∣ θ = 0 = ∑ n = 1 + ∞ A n s i n ( n π l n r ) = 0 u ∣ θ = π 2 = ∑ n = 1 + ∞ B n s h n π 2 2 s i n ( n π l n r ) = g ( r ) u|_{\theta=0}=\sum_{n=1}^{+\infty}A_nsin(n\pi lnr)=0 \\ u|_{\theta=\frac{\pi}{2}}=\sum_{n=1}^{+\infty}B_nsh\frac{n\pi^2}{2}sin(n\pi lnr)=g(r) u∣θ=0?=n=1∑+∞?An?sin(nπlnr)=0u∣θ=2π??=n=1∑+∞?Bn?sh2nπ2?sin(nπlnr)=g(r)
这里， { s i n ( n π l n r ) , n = 1 , 2 , ? } \{sin(n\pi lnr),n=1,2,\cdots \} {
sin(nπlnr),n=1,2,?}是区间 [ 1 , e ] [1,e] [1,e]上加权函数 ρ ( r ) = 1 r \rho(r)=\frac{1}{r} ρ(r)=r1?的完备正交函数系，故
A n = 0 B n = 1 s h n π 2 2 ? ∫ 1 e g ( r ) s i n ( n π l n r ) 1 r d r ∫ 1 e s i n 2 ( n π l n r ) 1 r d r A_n=0 \\ B_n=\frac{1}{sh\frac{n\pi^2}{2}}·\frac{\int_1^eg(r)sin(n\pi lnr)\frac{1}{r}dr}{\int_1^esin^2(n\pi lnr)\frac{1}{r}dr} An?=0Bn?=sh2nπ2?1??∫1e?sin2(nπlnr)r1?dr∫1e?g(r)sin(nπlnr)r1?dr?