# Laplace's Equation in Polar Coordinates with an Inhomogeneous Boundary Condition

I've looked at other similar problems to this on the site, but haven't found anything useful for what I'm dealing with.

I'm trying to solve Laplace's equation in polar coordinates with the following conditions:

$$1\le r\le 2$$ $$u(1,\theta)=1+\cos(2\theta)$$ $$u(2,\theta)=0$$

When $$0$$ is an allowed value for $$r$$, I was shown that the general solution is:

$$u(r,\theta)=A_0/2 \ + \sum_{n=1}^\infty \ r^n[A_n\cos(n\theta)+B_n\sin(n\theta)]$$

But I also know that in general, the solutions for $$R(r)$$ are: $$R(r)=r^n$$ $$R(r)=r^{-n}$$ $$R(r)=ln(r)$$

So what I tried as my general solution for this problem is:

$$u(r,\theta)=A_0/2 \ + \sum_{n=1}^\infty \ [r^n+r^{-n}+ln(r)][A_n\cos(n\theta)+B_n\sin(n\theta)]$$

The first boundary condition was simple enough. I got that $$A_0=1, \ \ \ \ \ \ A_2=1, \ \ \ \ \ A_n=0 \ (n\neq2), \ \ \ \ \ B_n=0$$

And now I am left with the following monster to solve:

$$0=1+\cos(2\theta)\sum_{n=1}(2^{n}+2^{-n}+ln(2))$$

I'm not too sure how to proceed. I'm thinking it will have to be an othogonal expansion to get the sum equal to $$-\sec(2\theta)$$, but that seems rather disgusting. Also, I do not have weights on my linearly independent functions which can be solved for.

• Great first post and welcome to Mathematics Stack Exchange! A quick tour will enhance your experience. – dantopa Feb 8 '19 at 1:53

Separate variables for the Laplace equation in polar coordinates: $$0=\nabla^2 (R(r)\Theta(\theta))=\frac{1}{r}(rR'(r))'\Theta(\theta)+\frac{1}{r^2}R(r)\Theta''(\theta) \\ r(rR')'\frac{1}{R}=\lambda = -\frac{\Theta''}{\Theta}$$ $$\lambda=n^2$$ for periodicity in $$\theta$$, with $$\Theta_n(\theta)=Ae^{in\theta}+Be^{-in\theta}$$ and corresponding $$\lambda_n=n^2$$ and $$r(rR_n')'=n^2R_n$$ has solutions $$R_0 = C_0+D_0\ln(r), \\ R_n = C_nr^n+D_nr^{-n},\;\; n=1,2,3,\cdots.$$ In order to match $$u(1,\theta)=1+\cos(2\theta)$$, $$u(2,\theta)=0$$
$$u(r,\theta)=(1+D_0\ln(r))+(C_2r^2+D_2/r^2)\cos(2\theta)$$ where $$1+D_0\ln(2)=0,\;\; 4C_2+D_2/4=0$$ Therefore, $$u(r,\theta)=\left(1-\frac{\ln(r)}{\ln(2)}\right)+C_2(r^2-16/r^2)\cos(2\theta)$$ $$u(2,\theta)=0$$ is easily verified. And $$u(1,\theta)=1+\cos(2\theta)$$ holds if $$C_2=-1/15$$. Therefore, the desired solution is $$u(r,\theta)=\left(1-\frac{\ln(r)}{\ln(2)}\right)-\frac{1}{15}(r^2-16/r^2)\cos(2\theta).$$
• @Austin : The $\cos(n\theta)$ and $\sin(m\theta)$ are mutually orthogonal, which happens because of the Sturm-Liouville periodic problem in $\theta$. – DisintegratingByParts Feb 9 '19 at 0:53