Sum the series in Problem 12 to get$\mathbf{u}\mathbf{=}\frac{\mathbf{200}}{\mathbf{\pi }}\mathbf{arctan}\frac{\mathbf{2}{\mathbf{a}}^{\mathbf{2}}{\mathbf{r}}^{\mathbf{2}}\mathbf{sin}\mathbf{2}\mathbf{\theta }}{{\mathbf{a}}^{\mathbf{4}}\mathbf{-}{\mathbf{r}}^{\mathbf{4}}}$.

An equation has been given.

$u=\frac{200}{\pi }\arctan \left(\frac{2a^2{r}^2\sin 2\theta }{a^4-r^4}\right)$

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

Start with the equation mentioned below.

$u(r,\theta )=\sum _{n=1,3,5\dots }^{\infty }\frac{r^{2n}}{a^{2n}}\frac{400}{\pi n}\sin \left(2n\theta \right)$

Sum the series.

$\begin{array}{c}S=\sum _{n=1,3,5\dots }^{\infty }\frac{r^{2n}}{a^{2n}}\frac{400}{\pi n}\sin \left(2n\theta \right)\\ =\sum _{n=1,3,5\dots }^{\infty }\frac{{\left(r^2\right)}^n}{{\left(a^2\right)}^n}\frac{400}{\pi n}\mathrm{Im}\left(e^{i2n\theta }\right)\\ =\mathrm{Im}\left(\sum _{n=1,3,5\dots }^{\infty }{\left(e^{2i\theta }\frac{r^2}{a^2}\right)}^n\frac{400}{\pi n}\right)\end{array}$

Recognize the series (chapter 1, problem 13.7)

$\begin{array}{l}\sum _{1,3,5\dots }\frac{z^n}{n}=\frac{1}{2}\ln \left(\frac{1+z}{1-z}\right)\\ S=\frac{400}{2\pi }\mathrm{Im}\left(\ln \left(\frac{1+e^{2i\theta }\frac{r^2}{a^2}}{1-e^{2i\theta }\frac{r^2}{a^2}}\right)\right)\end{array}$

Solve further.

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

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.

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

Find the sum.

$S=\frac{200}{\pi }\arctan \left(\frac{2a^2{r}^2\sin \left(2\theta \right)}{a^4-r^4}\right)$

Hence, the required final solution is$u(r,\theta )=\frac{200}{\pi }\arctan \left(\frac{2a^2{r}^2\sin \left(2\theta \right)}{a^4-r^4}\right)$.