# 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 $X_n(x) =c_n \sin{(\frac{n \pi x}{L})}$ ??

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}

$\newcommand{\+}{^{\dagger}} \newcommand{\angles}[1]{\left\langle\, #1 \,\right\rangle} \newcommand{\braces}[1]{\left\lbrace\, #1 \,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\, #1 \,\right\rbrack} \newcommand{\ceil}[1]{\,\left\lceil\, #1 \,\right\rceil\,} \newcommand{\dd}{{\rm d}} \newcommand{\down}{\downarrow} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,{\rm e}^{#1}\,} \newcommand{\fermi}{\,{\rm f}} \newcommand{\floor}[1]{\,\left\lfloor #1 \right\rfloor\,} \newcommand{\half}{{1 \over 2}} \newcommand{\ic}{{\rm i}} \newcommand{\iff}{\Longleftrightarrow} \newcommand{\imp}{\Longrightarrow} \newcommand{\isdiv}{\,\left.\right\vert\,} \newcommand{\ket}[1]{\left\vert #1\right\rangle} \newcommand{\ol}[1]{\overline{#1}} \newcommand{\pars}[1]{\left(\, #1 \,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\pp}{{\cal P}} \newcommand{\root}[2][]{\,\sqrt[#1]{\vphantom{\large A}\,#2\,}\,} \newcommand{\sech}{\,{\rm sech}} \newcommand{\sgn}{\,{\rm sgn}} \newcommand{\totald}[3][]{\frac{{\rm d}^{#1} #2}{{\rm d} #3^{#1}}} \newcommand{\ul}[1]{\underline{#1}} \newcommand{\verts}[1]{\left\vert\, #1 \,\right\vert} \newcommand{\wt}[1]{\widetilde{#1}}$ 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}