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.
$\newcommand{\+}{^{\dagger}}% \newcommand{\angles}[1]{\left\langle #1 \right\rangle}% \newcommand{\braces}[1]{\left\lbrace #1 \right\rbrace}% \newcommand{\bracks}[1]{\left\lbrack #1 \right\rbrack}% \newcommand{\ceil}[1]{\,\left\lceil #1 \right\rceil\,}% \newcommand{\dd}{{\rm d}}% \newcommand{\ds}[1]{\displaystyle{#1}}% \newcommand{\equalby}[1]{{#1 \atop {= \atop \vphantom{\huge A}}}}% \newcommand{\expo}[1]{\,{\rm e}^{#1}\,}% \newcommand{\fermi}{\,{\rm f}}% \newcommand{\floor}[1]{\,\left\lfloor #1 \right\rfloor\,}% \newcommand{\half}{{1 \over 2}}% \newcommand{\ic}{{\rm i}}% \newcommand{\iff}{\Longleftrightarrow} \newcommand{\imp}{\Longrightarrow}% \newcommand{\isdiv}{\,\left.\right\vert\,}% \newcommand{\ket}[1]{\left\vert #1\right\rangle}% \newcommand{\ol}[1]{\overline{#1}}% \newcommand{\pars}[1]{\left( #1 \right)}% \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\pp}{{\cal P}}% \newcommand{\root}[2][]{\,\sqrt[#1]{\,#2\,}\,}% \newcommand{\sech}{\,{\rm sech}}% \newcommand{\sgn}{\,{\rm sgn}}% \newcommand{\totald}[3][]{\frac{{\rm d}^{#1} #2}{{\rm d} #3^{#1}}} \newcommand{\ul}[1]{\underline{#1}}% \newcommand{\verts}[1]{\left\vert\, #1 \,\right\vert}$ The general solution, which is finite when $r \to 0^{+}$, has the form: $$ {\rm u}\pars{r,\theta} = A_{0}\theta + B + \sum_{n = 1}^{\infty} \bracks{A_{n}\sin\pars{n\theta} + B_{n}\cos\pars{n\theta}}\pars{r \over 3}^{n}\tag{1} $$ When $r \to 3^{-}$, we'll have: $$ 2 + \theta = A_{0}\theta + B + \sum_{n = 1}^{\infty} \bracks{A_{n}\sin\pars{n\theta} + B_{n}\cos\pars{n\theta}}\tag{2} $$ $$ \int_{-\pi}^{\pi}\pars{2 + \theta}\,\dd\theta = \int_{-\pi}^{\pi}\braces{A_{0}\theta + B + \sum_{n = 1}^{\infty} \bracks{A_{n}\sin\pars{n\theta} + B_{n}\cos\pars{n\theta}}}\,\dd\theta $$ That yields $4\pi = 2\pi B\quad\imp\quad B = 2$ such that $\pars{2}$ is reduced to $$ \pars{1 - A_{0}}\theta =\sum_{n = 1}^{\infty} \bracks{A_{n}\sin\pars{n\theta} +B_{n}\cos\pars{n\theta}}\tag{3} $$ This expression leads to: \begin{align} \int_{-\pi}^{\pi}\bracks{\pars{1 - A_{0}}\theta}\sin\pars{m\theta}\,\dd\theta &=\sum_{n = 1}^{\infty} \int_{-\pi}^{\pi}\!\!\!\!\!\sin\pars{m\theta}\braces{% \bracks{A_{n}\sin\pars{n\theta} +B_{n}\cos\pars{n\theta}}}\,\dd\theta\tag{4} \\[3mm] \int_{-\pi}^{\pi}\bracks{\pars{1 - A_{0}}\theta}\cos\pars{m\theta}\,\dd\theta &=\sum_{n = 1}^{\infty} \int_{-\pi}^{\pi}\!\!\!\!\!\cos\pars{m\theta}\braces{% \bracks{A_{n}\sin\pars{n\theta} +B_{n}\cos\pars{n\theta}}}\,\dd\theta\tag{5} \end{align} $\pars{4}$ and $\pars{5}$ yield: \begin{align} 2\pars{1 - A_{0}}\ \overbrace{\int_{0}^{\pi}\theta\sin\pars{m\theta}\,\dd\theta} ^{\ds{{\pars{-1}^{m + 1} \over m}\,\pi}} &= 2A_{m}\int_{0}^{\pi}\sin^{2}\pars{m\theta}\,\dd\theta = \pi A_{m} \\[3mm] \pars{1 - A_{0}}\ \underbrace{\int_{-\pi}^{\pi}\theta\cos\pars{m\theta}\,\dd\theta}_{\ds{0}} &= 2B_{m}\int_{0}^{\pi}\cos^{2}\pars{m\theta}\,\dd\theta = \pi B_{m} \end{align} $\ds{\imp\quad A_{m} = 2\pars{1 - A_{0}}\,{\pars{-1}^{m + 1} \over m} \quad\mbox{and}\quad B_{m} = 0}$. $\pars{1}$ is reduced to: $$ {\rm u}\pars{r,\theta} = A_{0}\theta + 2 + 2\pars{1 - A_{0}}\sum_{n = 1}^{\infty}{\pars{-1}^{n + 1} \over n}\, \pars{r \over 3}^{n}\sin\pars{n\theta} $$ where $A_{0}$ satisfies $$ A_{0}\pars{-\pi} + 2 = A_{0}\pi + 2\quad\imp\quad A_{0} = 0 $$ $$ \color{#0000ff}{\large{\rm u}\pars{r,\theta} = 2 - 2\sum_{n = 1}^{\infty}{\pars{-1}^{n} \over n}\, \pars{r \over 3}^{n}\sin\pars{n\theta}} $$
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 $\ds{\lim_{r \to 3^{-}}{\rm u}\pars{r,\theta} = 2 + \theta}$.