91Ó°ÊÓ

An equation has been given.

$u=\frac{200}{\mathrm{π}}\arctan \left(\frac{2a^2{r}^2\sin 2\mathrm{θ}}{a^4−r^4}\right)$

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

Start with the equation mentioned below.

$u(r,\mathrm{θ})=\underset{n=1,3,5…}{\overset{\mathrm{∞}}{∑}}\frac{r^{2n}}{a^{2n}}\frac{400}{\mathrm{π}n}\sin \left(2n\mathrm{θ}\right)$

Sum the series.

$\begin{array}{c}S=\underset{n=1,3,5…}{\overset{\mathrm{∞}}{∑}}\frac{r^{2n}}{a^{2n}}\frac{400}{\mathrm{π}n}\sin \left(2n\mathrm{θ}\right)\\ =\underset{n=1,3,5…}{\overset{\mathrm{∞}}{∑}}\frac{{\left(r^2\right)}^n}{{\left(a^2\right)}^n}\frac{400}{\mathrm{π}n}\mathrm{Im}\left(e^{i2n\mathrm{θ}}\right)\\ =\mathrm{Im}\left(\underset{n=1,3,5…}{\overset{\mathrm{∞}}{∑}}{\left(e^{2i\mathrm{θ}}\frac{r^2}{a^2}\right)}^n\frac{400}{\mathrm{π}n}\right)\end{array}$

Recognize the series (chapter 1, problem 13.7)

$\begin{array}{l}\underset{1,3,5…}{∑}\frac{z^n}{n}=\frac{1}{2}\ln \left(\frac{1+z}{1−z}\right)\\ S=\frac{400}{2\mathrm{π}}\mathrm{Im}\left(\ln \left(\frac{1+e^{2i\mathrm{θ}}\frac{r^2}{a^2}}{1−e^{2i\mathrm{θ}}\frac{r^2}{a^2}}\right)\right)\end{array}$

Solve further.

$\begin{array}{c}\ln \left(\frac{1+e^{2i\mathrm{θ}}\frac{r^2}{a^2}}{1−e^{2i\mathrm{θ}}\frac{r^2}{a^2}}\right)=\ln \left(\frac{a^2+r^2{e}^{2i\mathrm{θ}}}{a^2−r^2{e}^{2i\mathrm{θ}}}\right)\\ =\ln \left(\frac{a^2+r^2{e}^{2i\mathrm{θ}}}{a^2−r^2{e}^{2i\mathrm{θ}}}\left(\frac{a^2−r^2{e}^{−2i\mathrm{θ}}}{a^2−r^2{e}^{−2i\mathrm{θ}}}\right)\right)\\ =\ln \left(\frac{a^4−r^4+a^2{r}^2(e^{2i\mathrm{θ}}−e^{−2i\mathrm{θ}})}{a^4+r^4−a^2{r}^2(e^{2i\mathrm{θ}}+e^{−2i\mathrm{θ}})}\right)\\ =\ln \left(\frac{a^4−r^4+2i{a}^2{r}^2\sin \left(2\mathrm{θ}\right)}{a^4+r^4−2a^2{r}^2\cos \left(2\mathrm{θ}\right)}\right)\end{array}$

It is known that$\mathrm{Im}\left(\ln \left(z\right)\right)=\arctan \left(\frac{\mathrm{Im}\left(z\right)}{Re\left(z\right)}\right)$

Computethe real and imaginary arguments of ln.

$\begin{array}{c}\mathrm{Im}\left(\ln \left(\frac{1+e^{2i\mathrm{θ}}\frac{r^2}{a^2}}{1−e^{2i\mathrm{θ}}\frac{r^2}{a^2}}\right)\right)=\arctan \left(\frac{\frac{2a^2{r}^2\sin \left(2\mathrm{θ}\right)}{a^4+r^4−2a^2{r}^2\cos \left(2\mathrm{θ}\right)}}{\frac{a^4−r^4}{a^4+r^4−2a^2{r}^2\cos \left(2\mathrm{θ}\right)}}\right)\\ =\arctan \left(\frac{2a^2{r}^2\sin \left(2\mathrm{θ}\right)}{a^4−r^4}\right)\end{array}$

Find the sum.

$S=\frac{200}{\mathrm{π}}\arctan \left(\frac{2a^2{r}^2\sin \left(2\mathrm{θ}\right)}{a^4−r^4}\right)$

Hence, the required final solution is$u(r,\mathrm{θ})=\frac{200}{\mathrm{π}}\arctan \left(\frac{2a^2{r}^2\sin \left(2\mathrm{θ}\right)}{a^4−r^4}\right)$.