A plate in the shape of a quarter circle has boundary temperatures as shown. Find the interior steady-state temperature $\mathbf{u}(\mathbf{r},\mathbf{\theta })$. (See Problem 5.12.)

A quarter has been given.

Laplace’s equation in cylindrical coordinates is

$\begin{array}{c}{\mathbf{\nabla }}^{\mathbf{2}}\mathbf{u}\mathbf{=}\frac{\mathbf{1}}{\mathbf{r}}\frac{\mathbf{\partial }}{\mathbf{\partial }\mathbf{r}}\mathbf{\left(}\mathbf{r}\frac{\mathbf{\partial }\mathbf{u}}{\mathbf{\partial }\mathbf{r}}\mathbf{\right)}\mathbf{+}\frac{\mathbf{1}}{{\mathbf{r}}^{\mathbf{2}}}\frac{{\mathbf{\partial }}^{\mathbf{2}}\mathbf{u}}{\mathbf{\partial }{\mathbf{\theta }}^{\mathbf{2}}}\mathbf{+}\frac{{\mathbf{\partial }}^{\mathbf{2}}\mathbf{u}}{\mathbf{\partial }{\mathbf{z}}^{\mathbf{2}}}\\ \mathbf{=}\mathbf{0}\end{array}$

And to separate the variable the solution assumed is of the form $\mathit{u}\mathbf{=}\mathit{R}\mathbf{\left(}\mathbf{r}\mathbf{\right)}\mathit{\Theta }\mathbf{\left(}\mathbf{\theta }\mathbf{\right)}\mathit{Z}\mathbf{\left(}\mathbf{z}\mathbf{\right)}$.

Solving the following equation.

$\frac{1}{r}\frac{\partial }{\partial r}\left(r\frac{\partial u}{\partial r}\right)+\frac{1}{r^2}\frac{{\partial }^{2}u}{\partial {\theta }^2}=0$

Assume a separable solution.

$u(r,\theta )=R\left(r\right)\Theta \left(\theta \right)$

Put the solution and multiplying by $r^2$.

$r\frac{1}{R}\frac{\partial }{\partial r}\left(r\frac{\partial R}{\partial r}\right)+\frac{1}{\Theta }\frac{{\partial }^2\Theta }{\partial {\theta }^2}=0$

Both sides must be equal to some constant (differing in a sign).

Call that constant $n^2$as below.

$r\frac{1}{R}\frac{\partial }{\partial r}\left(r\frac{\partial R}{\partial r}\right)=n^2$

The second equation is an easy differential equation with trigonometric solutions.

role="math" localid="1664345753448" $\begin{array}{l}\frac{{\partial }^2\Theta }{\partial {\theta }^2}+n^2\Theta =0\\ \Theta \left(\theta \right)=\left\{\begin{array}{l}\sin \left(n\theta \right)\\ \cos \left(n\theta \right)\end{array}\right.\end{array}$

Here, $\Theta \left(\theta \right)$must be periodic function so it can be concluded that is a natural number.

The first equation.

$r\frac{\partial }{\partial r}\left(r\frac{\partial R}{\partial r}\right)=n^{2}R$

To solve the first equation try solution of the form.

$R\left(r\right)=\sum _{m=0}^{\infty }{a}_m{r}^m;n\ne 0$

Put the solution to equation (1).

$r\frac{d}{dr}\left(r\frac{d}{dr}\sum _{m=0}^{\infty }{a}_m{r}^m\right)=r\frac{d}{dr}\left(\sum _{m=0}^{\infty }m{a}_m{r}^m\right)=\sum _{m=0}^{\infty }{m}^2{a}_m{r}^m$

As the left-hand side is equal to the right side, you get

$\sum _{m=0}^{\infty }{m}^2{a}_m{r}^m=n^2\sum _{m=0}^{\infty }{a}_m{r}^m$

The above equation can only hold if the condition below is satisfied.

$m=n$or $m=-n$

Write the solutions.

$R\left(r\right)=\left\{\begin{array}{l}{r}^n\\ r^{-n}\end{array}\right.$

If $n=\theta$the right side is $\theta$so one of the solutions can definitely be some constant because its derivation.

$\begin{array}{l}\frac{dR}{dr}=0\\ \Rightarrow R=\text{constant}\end{array}$

The other way to get $\theta$on the left side is if,

$\frac{d}{dr}\left(r\frac{dR}{dr}\right)=0$

$\begin{array}{c}r\frac{dR}{dr}=\text{constant}\\ =A\end{array}$

$R\left(r\right)=A\ln \left(r\right)+B$

So in the case $n=0$the solutions are as given below.

$R\left(r\right)=\left\{\begin{array}{l}\text{constant}\\ \ln \left(r\right)\end{array}\right.$

The complete solution is as follow.

$u(r,\theta )=\sum _{n=1}^{\infty }{r}^n(A_n\sin \left(n\theta \right)+B_n\cos \left(n\theta \right))+\sum _{n=1}^{\infty }{r}^{-n}(C_n\sin \left(n\theta \right)+D_n\cos \left(n\theta \right))+E\ln \left(r\right)+F$

Discard $r^{-n}$and $\ln \left(r\right)$solutions because they diverge at $r=0$.

Write the remaining solution.

$u(r,\theta )=\sum _{n=1}^{\infty }{r}^n(A_n\sin \left(n\theta \right)+B_n\cos \left(n\theta \right))+F$

Use the boundary conditions.

$\begin{array}{l}u(a,\theta )=\{100,0\le \theta <\frac{\pi }{2}\\ u(r,0)=0\\ u(r,\frac{\pi }{2})=0\end{array}$

From the second boundary condition.

$\begin{array}{c}u(r,0)=0\\ =\sum _{n=1}^{\infty }{r}^n\left(A_n\sin \left(n0\right)+B_n\cos \left(n0\right)\right)+F\end{array}$

Here,

$\begin{array}{l}{B}_n=0\\ F=0\end{array}$

From the third boundary condition.

$\begin{array}{c}u(r,\frac{\pi }{2})=0\\ =\sum _{n=1}^{\infty }{r}^n(A_n\sin \left(n\frac{\pi }{2}\right)\end{array}$

$\sin \left(\frac{n\pi }{2}\right)=m\pi$

$n=2m,m=1,2,3\dots$

Solve the equation further and you obtain,

$\begin{array}{c}u(a,\theta )=100\\ =\sum _{n=1}^{\infty }{A}_{2n}{a}^{2n}\sin \left(2n\theta \right)\\ =\sum _{n=1}^{\infty }{A}_{2n}^{\text{'}}\sin \left(2n\theta \right)\end{array}$

There is a Fourier series at the boundary so use the usual relations for the Fourier coefficients. At first, calculate$A_n$.

$\begin{array}{c}{A}_{2n}^{\text{'}}=\frac{2}{\raisebox{1ex}{$\pi $}\!\left/ \!\raisebox{-1ex}{$2$}\right.}\underset{0}{\overset{\raisebox{1ex}{$\pi $}\!\left/ \!\raisebox{-1ex}{$2$}\right.}{\int }}100\sin \left(2n\theta \right)d\theta \\ =-{\frac{400}{2\pi n}\cos \left(2n\theta \right)|}_0^{\raisebox{1ex}{$\pi $}\!\left/ \!\raisebox{-1ex}{$2$}\right.}\\ =-\frac{200}{\pi n}(\cos \left(\pi n\right)-1)\\ =\frac{400}{\pi n};n=1,3,5\dots \end{array}$

$\begin{array}{c}{A}_{2n}^{\text{'}}=a^{2n}{A}_n\\ A_{2n}=\frac{400}{\pi n{a}^{2n}}\end{array}$

Hence, the interior-steady temperature is$u(r,\theta )=\sum _{n=1,3,5\dots }^{\infty }\frac{r^{2n}}{a^{2n}}\frac{400}{\pi n}\sin \left(2n\theta \right)$.