Questions about the Laplace's equation in polar coordinates The Laplace's equation in polar coordinates at a cyclic disk:
$$u_{rr}+\frac{1}{r}u_r+\frac{1}{r^2}u_{\theta \theta}, \ \ \ 0 \leq r \leq a, \ \ \ 0 \leq \theta \leq 2 \pi$$
$$u(a,\theta)=h(\theta), \ \ \ 0 \leq \theta \leq 2 \pi$$
Using the method  of separation of variables, the solution is of the form $u(r, \theta)=R(r) \Theta(\theta)$.
Substituting this at the problem, we get the following two problems:
$$\left.\begin{matrix}
\Theta''+\lambda \Theta=0\\ 
\Theta(0)=\Theta(2 \pi)
\end{matrix}\right\}(*)$$
and
$$\left.\begin{matrix}
r^2R''+rR'-\lambda R=0
\end{matrix}\right\}(**)$$
According to my notes, the solution of the problem $(*)$ is the following:
So that there is a non-trivial periodic solution, it should be: $\lambda \geq 0$.
For $\lambda=0$ we have the solution $\Theta(\theta)=1$.
$$\lambda_0=0, \ \ \ \Theta_0(\theta)=1$$
The other eigenvalues and periodic, with period $2 \pi$ ,eigenfunctions are given from
$$\lambda_n=n^2, \ \ \ \Theta_n(\theta)=A_n \cos{(n \theta)}+B_n \sin{(n \theta)}, \ \ \ n \in \mathbb{N}$$
So, totally we have the following eigenvalues and eigenfunctions with period $2 \pi$:
$$\lambda_n=n^2, \ \ \ \Theta_n(\theta)=A_n \cos{(n \theta)}+B_n \sin{(n \theta)}, \ \ \ n \in \mathbb{N}_0$$



*

*How did we find these eigenvalues ( $  \ \lambda=n^2  \ $ )??

*Why are there also constants at the eigenfunctions? Aren't the eigenfunctions usually of the form $X_n(x) =\sin{(\frac{n \pi x}{L})}$ and not of the form $X_n(x) =c_n \sin{(\frac{n \pi x}{L})}$ ??

 A: In general the differential equation
\begin{align}
f''+ \alpha f = 0
\end{align}
has the solutions
\begin{align}
f = A \ \cos(\sqrt{\alpha} x) + B \ \sin(\sqrt{\alpha} x)
\end{align}
Since the square root is typically messy and $\alpha$ is suitably chosen then let $\alpha = \beta^{2}$ for which
\begin{align}
f = A \ \cos(\beta x) + B \ \sin(\beta x)
\end{align}
of which the form "looks nicer" and still remains arbitrary to some extent. Now, given the boundary condition $f(x) = f(x+2\pi)$ it is seen that
\begin{align}
f(x+2\pi) &= A \ \cos(\beta x + 2\pi \beta) + B \ \sin(\beta x + 2 \pi \beta) \\
&= A \left[ \cos(2 \pi \beta) \cos(\beta x) - \sin(2 \pi \beta) \sin(\beta x) \right] + B \left[ \sin(\beta x) \cos(2 \pi \beta) + \cos(\beta x) \sin(2 \beta \pi) \right].
\end{align}
This is reduced to the desired form when $\beta$ is an integer. Let $\beta = n$, $n \geq 0$, to obtain
\begin{align}
f(x+ 2 \pi) &= A \ \cos(n x) + B \ \sin(n x) = f(x)
\end{align}
The remainder of the solution will be determined from the second differential equation of the p.d.e . Since there are two differential equations involved and one set of boundary conditions one of the equations cannot be completely solved independently and leads to a Fourier series to determine all possible solutions, ie
\begin{align}
u(r,\theta) = \sum_{n} (A_{n} \ \cos(n\theta) + B_{n} \ \sin(n\theta)) R_{n}(r).
\end{align}
A: $\newcommand{\+}{^{\dagger}}
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The general solution is given by
\begin{align}
{\rm u}\pars{r,\theta}&=
\pars{A\theta + B}\bracks{C\ln\pars{r} + D}
\\[3mm]&+\sum_{n = 1}^{\infty}\braces{\bracks{%
A_{n}\sin\pars{n\theta} + B_{n}\cos\pars{n\theta}}r^{n} + \bracks{%
C_{n}\sin\pars{n\theta} + D_{n}\cos\pars{n\theta}}r^{-n}}
\end{align}

In order to keep a finite behavior at $\ds{r = 0}$, we set
  $\ds{C_{n} = D_{n} = 0\,,\ \forall\ n \geq 1}$ and $\ds{C = 0}$. $\ds{D}$ is 'trivially' set to one. The solution is reduced to:
  \begin{align}
{\rm u}\pars{r,\theta}&=
A\theta + B
+\sum_{n = 1}^{\infty}\bracks{%
A_{n}\sin\pars{n\theta} + B_{n}\cos\pars{n\theta}}r^{n}
\end{align}

Also,
\begin{align}
{\rm h}\pars{\theta}={\rm u}\pars{a,\theta}&=
A\theta + B
+\sum_{n = 1}^{\infty}\bracks{%
A_{n}\sin\pars{n\theta} + B_{n}\cos\pars{n\theta}}a^{n}
\end{align}

From this expression we'll get:
  \begin{align}
\int_{0}^{2\pi}{\rm h}\pars{\theta}&=2\pi^{2} A + 2\pi B
\\[3mm]
\int_{0}^{2\pi}{\rm h}\pars{\theta}
\braces{\sin\pars{n\theta} \atop \cos\pars{n\theta}}\,\dd\theta&=
\pi a^{n}\braces{A_{n} \atop B_{n}}
\end{align}
  which determines $\ds{\braces{A_{n}, B_{n},\quad n = 1,2,3\ldots}}$ and, in principle, $\ds{A}$ and $\ds{B}$.

$\ds{A = 0}$ whenever we impose the $\ds{{\rm u}\pars{r,\theta}}$
${\large\tt\mbox{periodicity condition}}$
$$
{\rm u}\pars{r,\theta + 2\pi} = {\rm u}\pars{r,\theta}\,,\qquad
\forall\ r \in 0 \left[\vphantom{\large A}0,a\right)\,,\quad\forall\ \theta \in
\left[\vphantom{\large A}0,2\pi\right)
$$
In that case the solution is:
\begin{align}
\color{#66f}{\large{\rm u}\pars{r,\theta}}&=
\color{#66f}{\large B +\sum_{n = 1}^{\infty}\bracks{%
A_{n}\sin\pars{n\theta} + B_{n}\cos\pars{n\theta}}r^{n}}
\\[3mm]&\color{#c00000}{%
\left\lbrace\begin{array}{rcl}
\braces{A_{n} \atop B_{n}} & = & {1 \over \pi a^{n}}\int_{0}^{2\pi}{\rm h}\pars{\theta}
\braces{\sin\pars{n\theta} \atop \cos\pars{n\theta}}\,\dd\theta
\\[3mm]
B & = & {1 \over 2\pi}\int_{0}^{2\pi}{\rm h}\pars{\theta}\,\dd\theta
\end{array}\right.}
\end{align}
