Question

Solve Problem 5.7 if half the curved surface of the cylinder is held at $\mathbf{100}\mathbf{°}$and the other half at $\mathbf{-}\mathbf{100}\mathbf{°}$with the ends at $\mathbf{0}\mathbf{°}$.

Short answer

The solution is found to be,

$u(r,\theta ,z)=\sum _{n=1,3,5\dots }^{\infty }\sum _{m=1,3,5\dots }^{\infty }\frac{1600}{I_n\left(\frac{3m\pi }{20}\right)mn{\pi }^2}{I}_n\left(\frac{m\pi }{20}r\right)\sin \left(\frac{m\pi }{20}z\right)\sin \left(n\theta \right)$

Step-by-step solution

Given Information:

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

Definition of Laplace’ equation:

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

Solve the equation:

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.

$\frac{r^2}{R\left(r\right)}\frac{\partial }{\partial r}\left(r\frac{\partial R\left(r\right)}{\partial r}\right)+\frac{1}{\Theta \left(\theta \right)}\left(\frac{{\partial }^2\Theta \left(\theta \right)}{\partial {\theta }^2}\right)-K^2{r}^2=0$

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

$\Theta \left(\theta \right)=\left\{\begin{array}{l}\sin \left(n\theta \right)\\ \cos \left(n\theta \right)\end{array}\right.$

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

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

Now recognize equation 17.2 from the chapter 12.

For the solutions there is hyperbolic Bessel functions.

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

Use boundary condition.

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.

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

The boundary conditions of our problem.

$\begin{array}{l}u(r,\theta ,z=0)=0\\ u(r,\theta ,z=20)=0\\ u(3,0\le \theta <\pi ,z)=100\\ u(3,\pi \le \theta <2\pi ,z)=-100\end{array}$

The first boundary condition

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

See that A = 0.

From the second boundary condition.

$\begin{array}{l}u(r,\theta ,z=20)=0\\ B\sin \left(20K\right)=0\\ 20K=m\pi \\ K=\frac{m\pi }{20};m=1,2,3\dots \end{array}$

The equation becomes as mentioned below.

$u(r,\theta ,z)=\sum _{n=1}^{\infty }\sum _{m=1}^{\infty }{I}_n\left(\frac{m\pi }{20}r\right)\sin \left(\frac{m\pi }{20}z\right)(A_{mn}\cos \left(n\theta \right)+B_{mn}\sin \left(n\theta \right))$

Calculate the coefficient:

use boundary condition 3 and 4 to calculate the coefficients.

$\begin{array}{c}u(3,\theta ,z)=\{\begin{array}{c}100,0\le \theta <\pi \\ -100,\pi \le \theta <2\pi \end{array}\\ =\sum _{n=1}^{\infty }\sum _{m=1}^{\infty }{I}_n\left(\frac{3m\pi }{20}\right)\sin \left(\frac{m\pi }{20}z\right)(A_{mn}\cos \left(n\theta \right)+B_{mn}\sin \left(n\theta \right))\end{array}$

Use orthogonally relations for the trigonometric functions.

$\begin{array}{l}100{\int }_0^{\pi }\cos \left(l\pi \right)d\theta -100{\int }_{-2\pi }^{2\pi }\cos \left(l\pi \right)d\theta =0\\ A_{m\text{l}}=0\end{array}$

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

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

For the right side.

$\begin{array}{c}\sum _{n=1}^{\infty }\sum _{m=1}^{\infty }{I}_n\left(\frac{3m\pi }{20}\right)\sin \left(\frac{m\pi }{20}z\right){\int }_0^{2\pi }{B}_{mn}\sin \left(n\theta \right)\sin \left(l\theta \right)d\theta \\ -\frac{400}{l}=\sum _{m=1}^{\infty }{I}_l\left(\frac{3m\pi }{20}\right)\sin \left(\frac{m\pi }{20}z\right)B_{ml}\pi \end{array}$

$-400=\sum _{m=1}^{\infty }{C}_{ml}\sin \left(\frac{m\pi }{20}z\right)$

Find the final solution:

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

$\begin{array}{c}{C}_{ml}=\frac{2}{20}{\int }_0^{l}f\left(z\right)\sin \left(\frac{m\pi }{20}z\right)dz\\ =\frac{2}{20}{\int }_0^{20}-400\sin \left(\frac{m\pi }{20}z\right)dz\\ ={\frac{-40\cdot 20}{m\pi }\cos \left(\frac{m\pi }{20}z\right)|}_0^{2\pi }\\ =-\frac{800}{m\pi }(\cos \left(m\pi \right)-1)\end{array}$

$C_{ml}=\frac{1600}{m\pi },m=1,3,5\dots$

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

Hence, the required final solution is found to be,

$u(r,\theta ,z)=\sum _{n=1,3,5\dots }^{\infty }\sum _{m=1,3,5\dots }^{\infty }\frac{1600}{I_n\left(\frac{3m\pi }{20}\right)mn{\pi }^2}{I}_n\left(\frac{m\pi }{20}r\right)\sin \left(\frac{m\pi }{20}z\right)\sin \left(n\theta \right)$.