91影视

It has been given to repeat problem 12 and 13.

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

The general solution is mentioned below.

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

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

$\begin{array}{l}\text{u}\left(\text{r,0}\right)\text{=0}\\ u(r,\frac{蟺}{6})=0\\ u(10,胃)=100掳\end{array}$

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

$u(r,胃)=\underset{m=1}{\overset{\mathrm{鈭�}}{鈭�}}{r}^{6m}\left(B_m\sin \left(6m胃\right)\right)$

Fourier coefficient:

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

$\begin{array}{c}u(10,胃)=100掳\\ =\underset{m=1}{\overset{\mathrm{鈭�}}{鈭�}}\left({10}^{6m}{B}_m\sin \left(6m胃\right)\right)\end{array}$

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

$\begin{array}{c}u(10,胃)=100掳\\ =\underset{m=1}{\overset{\mathrm{鈭�}}{鈭�}}\left({10}^{6m}{B}_m\sin \left(6m胃\right)\right)\end{array}$

Solve further and you have,

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

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

Hence the required final solution is$u(r,胃)=\underset{\text{modd}}{\overset{\mathrm{鈭�}}{鈭�}}\frac{400}{m蟺}{\left(\frac{r}{10}\right)}^{6m}\sin \left(6m胃\right)$.