# polar Laplace equation solution:

Question: $\frac{\partial^2 u}{\partial r^2} + \frac{1}{r}\frac{\partial u}{\partial r} + \frac{1}{r^2}\frac{\partial^2 u}{\partial {\theta}^2} = 0,\:\:\:\: 0\leq r \leq 3 \:\:\:\: -\pi \leq \theta \leq \pi$

with the boundary condition : $u(3,\theta) = 2 + \theta$ and the periodicity conditions : $u(r,\pi) = u(r,-\pi)$ and $u_{\theta}(r,\pi) = u_{\theta}(r,-\pi)$ :

My solution:

I assumed $u(r,\theta) = \psi(\theta)R(r)$

I solved for $\psi(\theta)$ and $R(r)$ and got $R(r)=c r^n$ and $\psi(\theta) = Asin(\lambda \theta) + Bcos(\lambda \theta)$

Solving the two periodicity condition just told me that $\lambda$ ( the separation constant) is $n$

My final answer is $u(r,\theta) = 2 + \sum\limits_{n=1}^\infty \frac{2}{n}(\frac{r}{3})^n (-1)^{n+1} sin(n\theta) + \frac{2}{3^n\pi n^2}((-1)^n - 1)cos(n\theta)$

But the final correct solution is : $u(r,\theta) = 2 + \sum\limits_{n=1}^\infty \frac{2}{n}(\frac{r}{3})^n (-1)^{n+1} sin(n\theta)$

so obviously the cosine coefficient (B) is $0$ according to one of the conditions, but I don't know how or where that happens.

• ${\rm u}\left(r = 3^{-},\theta\right) - 2 = \theta$ is an odd function of $\theta$. This symmetry vanishes out the $B_{n}$ coefficients since $\cos\left(n\theta\right)$ is an even function of $\theta$. – Felix Marin Jan 12 '14 at 20:43

Also, $${\rm u}\pars{r,\theta} = 2 - 2\Im\overbrace{\sum_{n = 1}^{\infty}{\pars{-1}^{n} \over n}\,z^{n}} ^{\ds{\equiv\ \varphi\pars{z}}}\quad\mbox{where}\quad z = {r \over 3}\,\expo{\ic \theta}$$ $$\varphi'\pars{z} = -\sum_{n = 1}^{\infty}\pars{-z}^{n - 1} = -\,{1 \over 1 - \pars{-z}}\,,\qquad\varphi\pars{0} = 0$$ $$\varphi\pars{z} = -\ln\pars{1 + z} = -\ln\pars{1 + {r \over 3}\,\cos\pars{\theta} + {r \over 3}\,\sin\pars{\theta}\ic}$$ $$\Im\varphi\pars{z} = -\arctan\pars{r\sin\pars{\theta}/3 \over 1 + r\cos\pars{\theta}/3}$$ $$\color{#0000ff}{\large{\rm u}\pars{r,\theta} = 2 + 2\arctan\pars{r\sin\pars{\theta}/3 \over 1 + r\cos\pars{\theta}/3}}$$
Notice that $\pars{~\mbox{when}\ r \to 3^{-}~}$ we'll have: \begin{align} \arctan\pars{r\sin\pars{\theta}/3 \over 1 + r\cos\pars{\theta}/3} &\to\arctan\pars{\sin\pars{\theta} \over 1 + \cos\pars{\theta}} = \arctan\pars{1 - \cos\pars{\theta} \over \sin\pars{\theta}} \\[3mm]&= \arctan\pars{2\sin^{2}\pars{\theta/2} \over 2\sin\pars{\theta/2}\cos\pars{\theta/2}} = \arctan\pars{\sin\pars{\theta/2} \over \cos\pars{\theta/2}} \\[3mm]&= \arctan\pars{\tan\pars{\theta \over 2}} = {\theta \over 2} \end{align} such that $\ds{\lim_{r \to 3^{-}}{\rm u}\pars{r,\theta} = 2 + \theta}$.