91Ó°ÊÓ

A quarter has been given.

Laplace’s equation in cylindrical coordinates is

$\begin{array}{c}{\mathbf{∇}}^{\mathbf{2}}\mathbf{u}\mathbf{=}\frac{\mathbf{1}}{\mathbf{r}}\frac{\mathbf{∂}}{\mathbf{∂}\mathbf{r}}\mathbf{\left(}\mathbf{r}\frac{\mathbf{∂}\mathbf{u}}{\mathbf{∂}\mathbf{r}}\mathbf{\right)}\mathbf{+}\frac{\mathbf{1}}{{\mathbf{r}}^{\mathbf{2}}}\frac{{\mathbf{∂}}^{\mathbf{2}}\mathbf{u}}{\mathbf{∂}{\mathbf{θ}}^{\mathbf{2}}}\mathbf{+}\frac{{\mathbf{∂}}^{\mathbf{2}}\mathbf{u}}{\mathbf{∂}{\mathbf{z}}^{\mathbf{2}}}\\ \mathbf{=}\mathbf{0}\end{array}$

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

Solving the following equation.

$\frac{1}{r}\frac{∂}{∂r}\left(r\frac{∂u}{∂r}\right)+\frac{1}{r^2}\frac{{∂}^{2}u}{∂{\mathrm{θ}}^2}=0$

Assume a separable solution.

$u(r,\mathrm{θ})=R\left(r\right)\mathrm{Θ}\left(\mathrm{θ}\right)$

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

$r\frac{1}{R}\frac{∂}{∂r}\left(r\frac{∂R}{∂r}\right)+\frac{1}{\mathrm{Θ}}\frac{{∂}^2\mathrm{Θ}}{∂{\mathrm{θ}}^2}=0$

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

Call that constant $n^2$as below.

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

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

role="math" localid="1664345753448" $\begin{array}{l}\frac{{∂}^2\mathrm{Θ}}{∂{\mathrm{θ}}^2}+n^2\mathrm{Θ}=0\\ \mathrm{Θ}\left(\mathrm{θ}\right)=\left\{\begin{array}{l}\sin \left(n\mathrm{θ}\right)\\ \cos \left(n\mathrm{θ}\right)\end{array}\right.\end{array}$

Here, $\mathrm{Θ}\left(\mathrm{θ}\right)$must be periodic function so it can be concluded that is a natural number.

The first equation.

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

To solve the first equation try solution of the form.

$R\left(r\right)=\underset{m=0}{\overset{\mathrm{∞}}{∑}}{a}_m{r}^m;\text{ â¶Ä‰â¶Ä‰}n≠0$

Put the solution to equation (1).

$r\frac{d}{dr}\left(r\frac{d}{dr}\underset{m=0}{\overset{\mathrm{∞}}{∑}}{a}_m{r}^m\right)=r\frac{d}{dr}\left(\underset{m=0}{\overset{\mathrm{∞}}{∑}}m{a}_m{r}^m\right)=\underset{m=0}{\overset{\mathrm{∞}}{∑}}{m}^2{a}_m{r}^m$

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

$\underset{m=0}{\overset{\mathrm{∞}}{∑}}{m}^2{a}_m{r}^m=n^2\underset{m=0}{\overset{\mathrm{∞}}{∑}}{a}_m{r}^m$

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 $n=\mathrm{θ}$the right side is $\mathrm{θ}$so one of the solutions can definitely be some constant because its derivation.

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

The other way to get $\mathrm{θ}$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.

$u(r,\mathrm{θ})=\underset{n=1}{\overset{\mathrm{∞}}{∑}}{r}^n(A_n\sin \left(n\mathrm{θ}\right)+B_n\cos \left(n\mathrm{θ}\right))+\underset{n=1}{\overset{\mathrm{∞}}{∑}}{r}^{−n}(C_n\sin \left(n\mathrm{θ}\right)+D_n\cos \left(n\mathrm{θ}\right))+E\ln \left(r\right)+F$

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

Write the remaining solution.

$u(r,\mathrm{θ})=\underset{n=1}{\overset{\mathrm{∞}}{∑}}{r}^n(A_n\sin \left(n\mathrm{θ}\right)+B_n\cos \left(n\mathrm{θ}\right))+F$

Use the boundary conditions.

$\begin{array}{l}u(a,\mathrm{θ})=\{100,\text{ â¶Ä‰â¶Ä‰}0≤\mathrm{θ}<\frac{\mathrm{Ï€}}{2}\\ u(r,0)=0\\ u(r,\frac{\mathrm{Ï€}}{2})=0\end{array}$

From the second boundary condition.

$\begin{array}{c}u(r,0)=0\\ =\underset{n=1}{\overset{\mathrm{∞}}{∑}}{r}^n\left(A_n\sin \left(n0\right)+B_n\cos \left(n0\right)\right)+F\end{array}$

Here,

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

From the third boundary condition.

$\begin{array}{c}u(r,\frac{\mathrm{π}}{2})=0\\ =\underset{n=1}{\overset{\mathrm{∞}}{∑}}{r}^n(A_n\sin \left(n\frac{\mathrm{π}}{2}\right)\end{array}$

$\sin \left(\frac{n\mathrm{Ï€}}{2}\right)=m\mathrm{Ï€}$

$n=2m,m=1,2,3…$

Solve the equation further and you obtain,

$\begin{array}{c}u(a,\mathrm{θ})=100\\ =\underset{n=1}{\overset{\mathrm{∞}}{∑}}{A}_{2n}{a}^{2n}\sin \left(2n\mathrm{θ}\right)\\ =\underset{n=1}{\overset{\mathrm{∞}}{∑}}{A}_{2n}^{\text{'}}\sin \left(2n\mathrm{θ}\right)\end{array}$

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

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

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

Hence, the interior-steady temperature is$u(r,\mathrm{θ})=\underset{n=1,3,5…}{\overset{\mathrm{∞}}{∑}}\frac{r^{2n}}{a^{2n}}\frac{400}{\mathrm{π}n}\sin \left(2n\mathrm{θ}\right)$.