91Ó°ÊÓ

It has been asked to use problem 5.7 and solve it as instructed.

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 equation 5.4 from the text but in this case the constant is negative.

$\frac{1}{Z}\frac{d^{2}Z\left(z\right)}{d{z}^2}=−K^2$

The solutions in this case are trigonometric functions.

$Z\left(z\right)=\left\{\begin{array}{l}\cos \left(Kz\right)\\ \sin \left(Kz\right)\end{array}\right.$

Meanwhile the radial and angular parts have the same equation as in relation 5.5.

$\frac{r^2}{R\left(r\right)}\frac{∂}{∂r}\left(r\frac{∂R\left(r\right)}{∂r}\right)+\frac{1}{\mathrm{Θ}\left(\mathrm{θ}\right)}\left(\frac{{∂}^2\mathrm{Θ}\left(\mathrm{θ}\right)}{∂{\mathrm{θ}}^2}\right)−K^2{r}^2=0$

Separate this equation in the same way as in the text so for the angular part solution.

$\mathrm{Θ}\left(\mathrm{θ}\right)=\left\{\begin{array}{l}\sin \left(n\mathrm{θ}\right)\\ \cos \left(n\mathrm{θ}\right)\end{array}\right.$

Put $−n^2$instead of the angular part in the radial equation.

$\begin{array}{l}\frac{r^2}{R\left(r\right)}\frac{∂}{∂r}\left(r\frac{∂R\left(r\right)}{∂r}\right)−n^2−K^2{r}^2=0\\ \frac{r^2}{R\left(r\right)}\frac{∂}{∂r}\left(r\frac{∂R\left(r\right)}{∂r}\right)−(n^2+K^2{r}^2)=0\end{array}$

Now recognize equation 17.2 from the chapter 12.

For the solutions there is hyperbolic Bessel functions.

$R\left(r\right)=\left\{\begin{array}{l}{I}_n\left(Kr\right)\\ K_n\left(Kr\right)\end{array}\right.$

With the boundary condition (the solution must be finite at z = 0) eliminate so R(r) becomes the equation mentioned below.

$R\left(r\right)=I_n\left(Kr\right)$

The solution is given below.

$u(r,\mathrm{θ},z)=I_n\left(Kr\right)(A\cos \left(Kz\right)+B\sin \left(Kz\right))(C\cos \left(n\mathrm{θ}\right)+D\sin \left(n\mathrm{θ}\right))$

The boundary conditions of our problem.

$\begin{array}{l}u(r,\mathrm{θ},z=0)=0\\ u(r,\mathrm{θ},z=20)=0\\ u(3,0≤\mathrm{θ}<\mathrm{π},z)=100\\ u(3,\mathrm{π}≤\mathrm{θ}<2\mathrm{π},z)=−100\end{array}$

The first boundary condition

$\begin{array}{c}u(r,\mathrm{θ},z=0)=0\\ =I_n\left(Kr\right)\left(A\right)(C\cos \left(n\mathrm{θ}\right)+D\sin \left(n\mathrm{θ}\right))\end{array}$

See that A = 0.

From the second boundary condition.

$\begin{array}{l}u(r,\mathrm{θ},z=20)=0\\ B\sin \left(20K\right)=0\\ 20K=m\mathrm{Ï€}\\ K=\frac{m\mathrm{Ï€}}{20};\text{ â¶Ä‰â¶Ä‰}m=1,2,3…\end{array}$

The equation becomes as mentioned below.

$u(r,\mathrm{θ},z)=\underset{n=1}{\overset{\mathrm{∞}}{∑}}\underset{m=1}{\overset{\mathrm{∞}}{∑}}{I}_n\left(\frac{m\mathrm{π}}{20}r\right)\sin \left(\frac{m\mathrm{π}}{20}z\right)(A_{mn}\cos \left(n\mathrm{θ}\right)+B_{mn}\sin \left(n\mathrm{θ}\right))$

use boundary condition 3 and 4 to calculate the coefficients.

$\begin{array}{c}u(3,\mathrm{θ},z)=\{\begin{array}{c}100,0≤\mathrm{θ}<\mathrm{π}\\ −100,\mathrm{π}≤\mathrm{θ}<2\mathrm{π}\end{array}\\ =\underset{n=1}{\overset{\mathrm{∞}}{∑}}\underset{m=1}{\overset{\mathrm{∞}}{∑}}{I}_n\left(\frac{3m\mathrm{π}}{20}\right)\sin \left(\frac{m\mathrm{π}}{20}z\right)(A_{mn}\cos \left(n\mathrm{θ}\right)+B_{mn}\sin \left(n\mathrm{θ}\right))\end{array}$

Use orthogonally relations for the trigonometric functions.

$\begin{array}{l}100{∫}_0^{\mathrm{π}}\cos \left(l\mathrm{π}\right)d\mathrm{θ}−100{∫}_{−2\mathrm{π}}^{2\mathrm{π}}\cos \left(l\mathrm{π}\right)d\mathrm{θ}=0\\ A_{m\text{l}}=0\end{array}$

For the sine terms multiply $u(3,\mathrm{θ},z)$with $\sin \left(\mathrm{θ}l\right)$ and integrate from 0 to $2\mathrm{π}$and those terms are not 0. On the left side, the equation mentioned below.

role="math" localid="1664351408346" $\begin{array}{c}{∫}^{2\mathrm{π}}u(3,\mathrm{θ},z)\sin \left(l\mathrm{θ}\right)d\mathrm{θ}=100{∫}^{\mathrm{π}}\sin \left(l\mathrm{θ}\right)d\mathrm{θ}−100{∫}^{2\mathrm{π}}\sin \left(l\mathrm{θ}\right)d\mathrm{θ}\\ =\frac{100}{l}(\cos \left(l\mathrm{π}\right)−1)−\frac{100}{l}(\cos \left(2l\mathrm{π}\right)−\cos \left(l\mathrm{π}\right))\\ =−\frac{200}{l}+\frac{200}{l}{\left(-1\right)}^{\mathrm{I}}\\ =−\frac{400}{l},l=1,3,5…\end{array}$

For the right side.

$\begin{array}{c}\underset{n=1}{\overset{\mathrm{∞}}{∑}}\underset{m=1}{\overset{\mathrm{∞}}{∑}}{I}_n\left(\frac{3m\mathrm{π}}{20}\right)\sin \left(\frac{m\mathrm{π}}{20}z\right){∫}_0^{2\mathrm{π}}{B}_{mn}\sin \left(n\mathrm{θ}\right)\sin \left(l\mathrm{θ}\right)d\mathrm{θ}\\ −\frac{400}{l}=\underset{m=1}{\overset{\mathrm{∞}}{∑}}{I}_l\left(\frac{3m\mathrm{π}}{20}\right)\sin \left(\frac{m\mathrm{π}}{20}z\right)B_{ml}\mathrm{π}\end{array}$

$−400=\underset{m=1}{\overset{\mathrm{∞}}{∑}}{C}_{ml}\sin \left(\frac{m\mathrm{π}}{20}z\right)$

This is a trivial Fourier series so it is easy to calculate coefficients.

$\begin{array}{c}{C}_{ml}=\frac{2}{20}{∫}_0^{l}f\left(z\right)\sin \left(\frac{m\mathrm{Ï€}}{20}z\right)dz\\ =\frac{2}{20}{∫}_0^{20}−400\sin \left(\frac{m\mathrm{Ï€}}{20}z\right)dz\\ ={\frac{−40â‹\dots 20}{m\mathrm{Ï€}}\cos \left(\frac{m\mathrm{Ï€}}{20}z\right)|}_0^{2\mathrm{Ï€}}\\ =−\frac{800}{m\mathrm{Ï€}}(\cos \left(m\mathrm{Ï€}\right)−1)\end{array}$

$C_{ml}=\frac{1600}{m\mathrm{Ï€}},\text{ â¶Ä‰â¶Ä‰}m=1,3,5…$

$\begin{array}{c}{C}_{ml}=B_{ml}{I}_l\left(\frac{3m\mathrm{Ï€}}{20}\right)\mathrm{Ï€}l\\ B_{ml}=\frac{1600}{I_l\left(\frac{3m\mathrm{Ï€}}{20}\right)ml{\mathrm{Ï€}}^2}\end{array}$

Hence, the required final solution is found to be,

$u(r,\mathrm{θ},z)=\underset{n=1,3,5…}{\overset{\mathrm{∞}}{∑}}\underset{m=1,3,5…}{\overset{\mathrm{∞}}{∑}}\frac{1600}{I_n\left(\frac{3m\mathrm{π}}{20}\right)mn{\mathrm{π}}^2}{I}_n\left(\frac{m\mathrm{π}}{20}r\right)\sin \left(\frac{m\mathrm{π}}{20}z\right)\sin \left(n\mathrm{θ}\right)$.