91影视

An equation has been given.

$u=50+\frac{100}{蟺}\arctan \left(\frac{2ar\sin 胃}{a^2鈭�r^2}\right)$

Laplace鈥檚 equation in cylindrical coordinates is,

$\begin{array}{c}{\mathbf{鈭�}}^{\mathbf{2}}\mathbf{u}\mathbf{=}\frac{\mathbf{1}}{\mathbf{r}}\frac{\mathbf{鈭�}}{\mathbf{鈭�}\mathbf{r}}\mathbf{\left(}\mathbf{r}\frac{\mathbf{鈭�}\mathbf{u}}{\mathbf{鈭�}\mathbf{r}}\mathbf{\right)}\mathbf{+}\frac{\mathbf{1}}{{\mathbf{r}}^{\mathbf{2}}}\frac{{\mathbf{鈭�}}^{\mathbf{2}}\mathbf{u}}{\mathbf{鈭�}{\mathbf{胃}}^{\mathbf{2}}}\mathbf{+}\frac{{\mathbf{鈭�}}^{\mathbf{2}}\mathbf{u}}{\mathbf{鈭�}{\mathbf{z}}^{\mathbf{2}}}\\ \mathbf{=}\mathbf{0}\end{array}$

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

Start with the equation mentioned below.

$u(r,胃)=50+\underset{n=1,3,5鈥�}{\overset{\mathrm{鈭�}}{鈭�}}\frac{r^n}{a^n}\frac{200}{蟺n}\sin \left(n胃\right)$

Sum the series as below.

$\begin{array}{c}S=\underset{n=1,3,5鈥�}{\overset{\mathrm{鈭�}}{鈭�}}\frac{r^n}{a^n}\frac{200}{蟺n}\sin \left(n胃\right)\\ =\underset{n=1,3,5鈥�}{\overset{\mathrm{鈭�}}{鈭�}}\frac{r^n}{a^n}\frac{200}{蟺n}\mathrm{Im}\left(e^{in胃}\right)\\ =\mathrm{Im}\left(\underset{n=1,3,5鈥�}{\overset{\mathrm{鈭�}}{鈭�}}{\left(e^{i胃}\frac{r}{a}\right)}^n\frac{200}{蟺n}\right)\end{array}$

Recognize the series (chapter 1, problem 13.7)

$\begin{array}{l}\underset{1,3,5鈥�}{鈭�}\frac{z^n}{n}=\frac{1}{2}\ln \left(\frac{1+z}{1鈭�z}\right)\\ S=\frac{200}{2蟺}\mathrm{Im}\left(\ln \left(\frac{1+e^{i胃}\frac{r}{a}}{1鈭�e^{i胃\frac{r}{a}}}\right)\right)\end{array}$

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

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.

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

Find the sum and you have,

$S=\frac{100}{蟺}\arctan \left(\frac{2ar\sin \left(胃\right)}{a^2鈭�r^2}\right)$

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