Question

Repeat Problems 12 and 13 for a plate in the shape of a circular sector of angle $\mathbf{30}\mathbf{°}$and radius 10 if the boundary temperatures are $\mathbf{0}\mathbf{°}$on the straight sides and $\mathbf{100}\mathbf{°}$on the circular arc. Can you then state and solve a problem like 14?

Short answer

The solution obtained is$u(r,\theta )=\sum _{\text{modd}}^{\infty }\frac{400}{m\pi }{\left(\frac{r}{10}\right)}^{6m}\sin \left(6m\theta \right)$.

Step-by-step solution

Given Information:

It has been given to repeat problem 12 and 13.

Definition of Laplace’s equation:

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$\mathit{u}\mathbf{=}\mathit{R}\mathbf{\left(}\mathbf{r}\mathbf{\right)}\mathit{\Theta }\mathbf{\left(}\mathbf{\theta }\mathbf{\right)}\mathit{Z}\mathbf{\left(}\mathbf{z}\mathbf{\right)}$.

General solution and boundary condition:

The general solution is mentioned below.

$u(r,\theta )=r^n(A\cos \left(n\theta \right)+B\sin \left(n\theta \right))$

Here, A and B are constants. The boundary conditions.

$\begin{array}{l}\text{u}\left(\text{r,0}\right)\text{=0}\\ u(r,\frac{\pi }{6})=0\\ u(10,\theta )=100°\end{array}$

Apply them into the general solution it is from the first $A=0$and from the second one$\sin \left(n\theta \right)=0$, which gives$n=6\text{m}$. Thus the solution becomes as mentioned below.

$u(r,\theta )=\sum _{m=1}^{\infty }{r}^{6m}\left(B_m\sin \left(6m\theta \right)\right)$

Fourier coefficient:

Now, to determine the Fourier coefficient $B_m$use the third boundary condition

$\begin{array}{c}u(10,\theta )=100°\\ =\sum _{m=1}^{\infty }\left({10}^{6m}{B}_m\sin \left(6m\theta \right)\right)\end{array}$

Fourier Coefficient:

Now, to determine the Fourier coefficient $B_m$use the third boundary condition as given below.

$\begin{array}{c}u(10,\theta )=100°\\ =\sum _{m=1}^{\infty }\left({10}^{6m}{B}_m\sin \left(6m\theta \right)\right)\end{array}$

Solve further and you have,

$\begin{array}{c}{10}^{6m}{B}_m=\frac{12}{\pi }\underset{0}{\overset{\raisebox{1ex}{$\pi $}\!\left/ \!\raisebox{-1ex}{$6$}\right.}{\int }}100\sin \left(6m\theta \right)d\theta \\ =\frac{12}{\pi }\times 100{\left[\frac{\cos \left(6m\theta \right)}{6m}\right]}_0^{\raisebox{1ex}{$\pi $}\!\left/ \!\raisebox{-1ex}{$6$}\right.}\\ =\frac{1200}{\pi }[\frac{\cos (6m\times \frac{\pi }{6})}{6m}-\frac{\cos \left(0\right)}{6m}]\end{array}$

$\begin{array}{c}{10}^{6m}{B}_m=\frac{1200}{6m\pi }[\cos \left(m\pi \right)-1]\\ =\frac{1200}{6m\pi }(1-\cos m\pi )\\ =\frac{400}{m\pi },\text{fornodd}\end{array}$

Hence the required final solution is$u(r,\theta )=\sum _{\text{modd}}^{\infty }\frac{400}{m\pi }{\left(\frac{r}{10}\right)}^{6m}\sin \left(6m\theta \right)$.