I am not sure how to prove the following. (P591 Problem 4(b),Problem Set 12.10 "Advanced Engineering Mathmatics" by Erwin Kreyszig 10th ed.)
Assuming that termwise differentiation is permissible, show that a solution to Laplace Equation in the disk r<R, and satisfying the boundary condition of $u(R,\theta)=f(\theta)$$ \ $(R and $ f(\theta)$ is given) ,is $$ u(r,\theta)=a_0+ \sum_{n=1}^{\infty} [ a_n(\frac{r}{R})^n \cos{n\theta}+b_n (\frac{r}{R})^n \sin{n \theta} ]$$ $a_n, b_n$ are Fourier coefficients of f.
I have already proved that $$ u_n=r^n \cos{n\theta} , u_n=r^n \sin{n\theta} $$ are solution to Laplace Equation $$ \nabla^2 u= u_{rr}+\frac{1}{r} u_r + \frac{1}{r^2} u_{\theta \theta} =0$$
In particular, how I can prove the case of $\frac{r}{R}$ in addition to the case of $r$ is the solution to Laplace Equation and why termwise differentiation becomes a condition. Thank you for your kind explanation in advance.
I consulted with Bing and Bing gave me the following derivation as I understand it.
1:We know $u_n=r^n \cos{n \theta}, u_n=r^n \sin{n \theta}$ are solutions to $\nabla^2 u =0$.
2:Dirichlet boudary condition is defined at the boundary of domain of circle of radius R. $u(R,\theta)=f(\theta) $.
3:For solution $u(r,\theta)$ to satisfy Dirichlet boundary condition $u(R,\theta)=f(\theta)$ at all $r \leq R$, radius component r must be "Normalized" by dividing R.
4:Normalized solutions are $u(r,\theta)=(\frac{r}{R})^n \cos{n\theta}$ and $u(r,\theta)=(\frac{r}{R})^n \sin{n\theta}$. Since both solutions are the solution to linear equation $\nabla^2 u=0$, by the princilple of superposition, we get general solution $$ u(r,\theta)=a_0+\sum_{n=1}^{\infty} [ a_n (\frac{r}{R})^n \cos{n\theta}+b_n (\frac{r}{R})^n \sin{n\theta}]$$ Because of normalization, $(\frac{r}{R})^n$ always is less than 1 and including the case $r=0$, solution $u(r,\theta)$ is always finite for all $r \leq R$ with good behavier.
5:Particular Solution
By Dirichlet boundary condition $$ u(R,\theta)=a_0+\sum_{n=1}^{\infty} [ a_n \cos{n\theta} + b_n \sin{n\theta} ]$$ $ u(R,\theta)=f(\theta) $ is satified if we define $a_0,a_n$ are fourier cosine coefficient and $b_n$ is fourier sine coefficient.
I cannot find problem in this derivation given by Artificial Intelligence, Bing. "Normalization" is still not fully understood by me, though. Bing is getting clever day by day but still not fully trustworthy. Also I do not find "termwise differentiation is permissible" is a requirement for this proposition. Any comments are welcome.