The series in Problem 5.12 can be summed (see Problem 2.6). Show that$\mathbf{u}\mathbf{=}\mathbf{50}\mathbf{+}\frac{\mathbf{100}}{\mathbf{\pi }}\mathbf{arctan}\frac{\mathbf{2}\mathbf{arsin\theta }}{{\mathbf{a}}^{\mathbf{2}}\mathbf{-}{\mathbf{r}}^{\mathbf{2}}}$.

An equation has been given.

$u=50+\frac{100}{\pi }\arctan \left(\frac{2ar\sin \theta }{a^2-r^2}\right)$

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.

$u(r,\theta )=50+\sum _{n=1,3,5\dots }^{\infty }\frac{r^n}{a^n}\frac{200}{\pi n}\sin \left(n\theta \right)$

Sum the series as below.

$\begin{array}{c}S=\sum _{n=1,3,5\dots }^{\infty }\frac{r^n}{a^n}\frac{200}{\pi n}\sin \left(n\theta \right)\\ =\sum _{n=1,3,5\dots }^{\infty }\frac{r^n}{a^n}\frac{200}{\pi n}\mathrm{Im}\left(e^{in\theta }\right)\\ =\mathrm{Im}\left(\sum _{n=1,3,5\dots }^{\infty }{\left(e^{i\theta }\frac{r}{a}\right)}^n\frac{200}{\pi n}\right)\end{array}$

Recognize the series (chapter 1, problem 13.7)

$\begin{array}{l}\sum _{1,3,5\dots }\frac{z^n}{n}=\frac{1}{2}\ln \left(\frac{1+z}{1-z}\right)\\ S=\frac{200}{2\pi }\mathrm{Im}\left(\ln \left(\frac{1+e^{i\theta }\frac{r}{a}}{1-e^{i\theta \frac{r}{a}}}\right)\right)\end{array}$

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

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

Compute the real and imaginary arguments of In.

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

Find the sum and you have,

$S=\frac{100}{\pi }\arctan \left(\frac{2ar\sin \left(\theta \right)}{a^2-r^2}\right)$

Hence, the final solution is$u(r,\theta )=50+\frac{100}{\pi }\arctan \left(\frac{2ar\sin \left(\theta \right)}{a^2-r^2}\right)$.