Solving Laplace Equation in polar coordinates using Separation of Variables

Question

I'm trying to solve Laplace's Equation in Polar Coordinates with a given condition but I think I am stuck.

Here is the problem: $$\Delta u=0, (0 \leq r<0, 0 \leq \theta < 2 \pi)$$ $$u(a, \theta)= 2\sin ^2(2\theta)$$

We can assume that the function needs to be regular when $r \rightarrow 0$ and that it is periodic in $2 \pi$. I did my separation of variables and I chose my separation constant in order to get the sin and cos combination for the solution in $\theta$.

However when I arrive at the step where I need to determine my constants using the orthogonality of the sin and cos functions I find that they are both zero.

Here is what I did:

By using $u(r, \theta) = R(r)T(\theta)$ and plugging in Laplace's Equation we have: $$R''T + \frac{1}{r}R'T + \frac{1}{r^2}RT''=0$$ $$r^2\frac{R''}{R}+r\frac{R'}{R} = -\frac{T''}{T} = \lambda$$

Choosing $\lambda=k^2$ and using the periodicity conditions:

$$T(\theta)=A_k\sin(kx)+B_k\cos(kx) , k=1,2,3,...$$

Then using a power rule: $R=r^{\alpha}$

$$R(r) = r^k+r^{-k}$$

since we want the solution to be regular we drop the $r^{-k}$ term and we have:

$$u(r, \theta)=\sum^{\infty}_{k=1}r^k(A_k\sin(kx)+B_k\cos(kx))$$

Then using the orthogonality of the cosinus and sinus functions and my condition I find that $A_k=B_k=0$.

Thank you !

Answer 1

Firstly, you forgot the case $\lambda=0$.

Secondly, note that $2 \sin^2 (2 \theta) = 1 - \cos(4 \theta)$, and $\cos (k\theta)$ is not orthogonal to $\cos(4\theta)$ if $k = 4$. So somewhere in your calculations you may have happened to implicitly assume that $k \ne 4$?