Verifying a Series Solution to Dirichlet's Problem via separation of variables In Stein's Fourier Analysis i'm having trouble attempting to verify the series solution given in the problem in $(1.)$
$(1.)$
The Dirchlet problem is the annulus defined by ${{(r, \theta): p < r < 1}}$, where $0 \leq p \leq 1$ in the inner radius. The problem to solve:
$$\frac{\partial^{2}u}{\partial{r}^{2}} + \frac{1}{r} \frac{\partial{u}}{\partial{r}}+ \frac{1}{r^{2}}\frac{\partial^{2}u}{\partial{\theta}^{2}} = 0$$
subject to the boundary conditions:
$${
     \begin{align} 
       u(1,\theta) = f(\theta) \\
       u(p,\theta) = g(\theta). 
     \end{align}
 }$$
Stein's intial argument was to write solutions for the Dirchlet Problem as he did previously in the following form:
$$u(r,\theta)= \sum_{}^{}c_{n}(r)e^{in\theta}$$
with $c_{n}(r) = A_{n}r^{n} + B_{n}r^{-n}, n \neq 0$ Set: 
$$f(\theta) \sim \sum_{}^{} a_{n}e^{in\theta}$$ and 
$$g(\theta) \sim \sum_{}^{}b_{n}e^{in\theta}$$
This leads to the solution
$$u(r,\theta) = \sum_{n \leq 0} (\frac{1}{p^{n}-p^{-n}})((p/r)^{n} - (r/p)^{n})a_{n} + (r^{n} - r^{-n})b_{n}]e^{in\theta} + a_{o} + (b_{o} - a_{o}\frac{logr}{logp}$$
From looking what was done in Chapter 1 as a prior example,  and comparing the problem in $(1.)$the series the solution was obtained via Sepration of Variables. My initial attack can be followed in $(2.)$
$(2.)$
$$\frac{\partial^{2}u}{\partial{r}^{2}} + \frac{1}{r} \frac{\partial{u}}{\partial{r}}+ \frac{1}{r^{2}}\frac{\partial^{2}u}{\partial{\theta}^{2}} = \Delta{u}$$
$$r^{2}\frac{\partial^{2}u}{\partial{r}^{2}} +  r\frac{\partial{u}}{\partial{r}} = -\frac{\partial^{2}u}{\partial{\theta}^{2}}$$
Plugging in our solution product: $(u(r,\theta))=F(r)G(\theta))$
$$r^{2}\frac{\partial^{2}u}{\partial{r}^{2}}F(r)G(\theta) +  r\frac{\partial{u}}{\partial{r}}F(r)G(\theta)= -\frac{\partial^{2}u}{\partial{\theta}^{2}}F(r)G(\theta)$$
Now dividing by our Solution Product:
 $$\frac{r^{2}\frac{\partial^{2}u}{\partial{r}^{2}}F(r)G(\theta) + r\frac{\partial{u}}{\partial{r}}F(r)G(\theta) } {F{(r)}} = \frac{-\frac{\partial^{2}u}{\partial{\theta}^{2}}F(r)G(\theta)}{G{(\theta)}} $$
Finally one can observe in a more convenient form  that we have the following:
$$\frac{r^{2}F(r)G''(\theta)+r(G(\theta))F'(r)}{F(r)} = \frac{F(r)-G''(\theta)}{G(\theta)}$$
$$
  \left\{
     \begin{align} 
        r^2G''(\theta) + r(G(\theta)F'(r)) = 0 \\ 
        F(r) - \dfrac{F(r) - G''(\theta)}{G(\theta)} = 0
     \end{align} 
  \right. 
$$
$$   \left\{
     \begin{align}
        r^{2}G''(\theta)+r(G(\theta))F'(r)\lambda F(r)=0 \\
        F(r) - \lambda \frac{G''(\theta)}{G''(\theta)} = 0
     \end{align} 
  \right. 
$$
From the previous result above I'm stuck on working out a series solution to the above ODE's is their any fundamental observations I'm missing towards the problem ?
 A: $\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,}
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The general solution of the $\ds{2D}$-Laplace Equation is given by
\begin{align}
\mrm{u}\pars{r,\theta} & =
\pars{A\,{\theta \over 2\pi} + B}\bracks{C\ln\pars{r \over p} + D}
\\[5mm] & +
\sum_{n = 1}^{\infty}\left\{\bracks{%
A_{n}\pars{r \over p}^{n} +
B_{n}\pars{p \over r}^{n}}\sin\pars{n\theta}\right.
\\[3mm] & \left.\phantom{\sum_{n = 1}^{\infty}\!\!\braces{}}+
\bracks{%
C_{n}\pars{r \over p}^{n} +
D_{n}\pars{p \over r}^{n}}\cos\pars{n\theta}\right\}\label{1.a}\tag{1.a}
\end{align}

where $\ds{A, B, C, D, \braces{A_{n}}, \braces{B_{n}}, \braces{C_{n}}\ \mbox{and}\ \braces{D_{n}}}$ are constants which are determined by the boundary conditions.

Since the solution must be invariant under $\ds{\theta \mapsto \theta + 2\pi}$, I'll conclude that $\ds{A = 0}$. In such a case, $\ds{B}$ is 'redundant' such that I can set $\ds{B = 1}$. \eqref{1.a} is reduced to:
\begin{align}
\mrm{u}\pars{r,\theta} & =
C\ln\pars{r \over p} + D +
\sum_{n = 1}^{\infty}\left\{\bracks{%
A_{n}\pars{r \over p}^{n} +
B_{n}\pars{p \over r}^{n}}\sin\pars{n\theta}\right.
\\[3mm] & \left.\phantom{\sum_{n = 1}^{\infty}\!\!\braces{}AAAAAAAAAAA\,}+
\bracks{%
C_{n}\pars{r \over p}^{n} +
D_{n}\pars{p \over r}^{n}}\cos\pars{n\theta}\right\}
\label{1.b}\tag{1.b}
\end{align}

Then,
\begin{align}
\mrm{f}\pars{\theta} & =
-C\ln\pars{p} + D +
\sum_{n = 1}^{\infty}\bracks{%
\pars{A_{n}p^{-n} + B_{n}p^{n}}\sin\pars{n\theta} +
\pars{C_{n}p^{-n} + D_{n}p^{n}}\sin\pars{n\theta}}\label{2}\tag{2}
\\[2mm]
\mrm{g}\pars{\theta} & = D +
 \sum_{n = 1}^{\infty}\bracks{%
\pars{A_{n} + B_{n}}\sin\pars{n\theta} +
\pars{C_{n} + D_{n}}\sin\pars{n\theta}}
\label{3}\tag{3}
\end{align}

Integrating both members of \eqref{2} and \eqref{3} over $\ds{\pars{0,2\pi}}$:

\begin{equation}
\left\{\begin{array}{rcrcl}
\ds{-\ln\pars{p}C} & \ds{+} & \ds{D} & \ds{=} & \ds{{1 \over 2\pi}\int_{0}^{2\pi}\mrm{f}\pars{\theta}\,\dd\theta}
\\[2mm]
 && \ds{D} & \ds{=} & \ds{{1 \over 2\pi}\int_{0}^{2\pi}\mrm{g}\pars{\theta}\,\dd\theta}
\end{array}\right.
\end{equation}

Those relations determines $\ds{C\ \mbox{and}\ D}$.

Similarly,
\begin{align}
\int_{0}^{2\pi}\mrm{f}\pars{\theta}\sin\pars{n\theta}\,{\dd\theta \over \pi} & =
A_{n}p^{-n} + B_{n}p_{n}
\\[2mm]
\int_{0}^{2\pi}\mrm{f}\pars{\theta}\cos\pars{n\theta}\,{\dd\theta \over \pi} & =
C_{n}p^{-n} + D_{n}p_{n}
\\[2mm]
\int_{0}^{2\pi}\mrm{g}\pars{\theta}\sin\pars{n\theta}\,{\dd\theta \over \pi} & =
A_{n} + B_{n}
\\[2mm]
\int_{0}^{2\pi}\mrm{g}\pars{\theta}\cos\pars{n\theta}\,{\dd\theta \over \pi} & =
C_{n} + D_{n}
\end{align}

Those equations determine $\ds{\braces{A_{n}},\braces{B_{n}},\braces{C_{n}}\ \mbox{and}\ \braces{D_{n}}}$.

