# Laplace's Equation in Cylindrical Coordinates (Potential)

1. Oct 14, 2014

### V0ODO0CH1LD

1. The problem statement, all variables and given/known data

A hollow cylinder with radius $a$ and height $L$ has its base and sides kept at a null potential and the lid on top kept at a potential $u_0$. Find $u(r,\phi,z)$.

2. Relevant equations

Laplace's equation in cylindrical coordinates:
$$\nabla^2u=\frac{1}{r}\frac{\partial}{\partial{}r}\left(r\frac{\partial{}u}{\partial{}r}\right)+\frac{1}{r^2}\frac{\partial^2u}{\partial\phi^2}+\frac{\partial^2u}{\partial{}z^2}=0$$
Boundary conditions (specific to this problem):
$$\begin{eqnarray} u(r,\phi,0)=0\text{ for }0≤r≤a \\ u(a,\phi,z)=0\text{ for }0≤z≤L \\ u(r,\phi,L)=u_0\text{ for }0≤r≤a \end{eqnarray}$$
3. The attempt at a solution

After substituting $u=R(r)\Phi(\phi)Z(z)$ and dividing the equation by $R(r)\Phi(\phi)Z(z)$ we get
$$\frac{1}{rR(r)}\frac{d}{dr}(rR'(r))+\frac{1}{r^2\Phi(\phi)}\Phi''(\phi)+\frac{1}{Z(z)}Z''(z)=0.$$
Since the $z$ term is isolated we can set it to a constant $k^2$ to get $Z(z)=Ae^{kz}+Be^{−kz}$. Then we multiply through by $r^2$ and isolate the $\phi$ term and set it to a constant $-m^2$ to get $\Phi(\phi)=C\cos(m\phi)+D\sin(m\phi)$. The original equation now looks like
$$\frac{r}{R(r)}\frac{d}{dr}(rR'(r))-m^2+(rk)^2=r\frac{d}{dr}(rR'(r))+[(rk)^2-m^2]R(r)=0,$$
where we can substitute $y=rk$ to get
$$y^2R''(y)+yR'(y)+[y^2-m^2]R(r)=0,$$
the Bessel equation.

How do I use the boundary conditions to move forward from here? What does the format of the answer I am looking for looks like?

Last edited: Oct 14, 2014
2. Oct 14, 2014

### vela

Staff Emeritus
The last one should be equal to $u_0$, not 0.

You should also have $u(r,\phi,z) = u(r,\phi+2\pi,z)$ because you want u to be single-valued. This imposes a constraint on the allowed values of $m$.

Take the condition that the base is held at 0. That requires Z(0)=0 since it has to hold for all values of $r$ and $\phi$. What does this tell you about A and B?

Take the condition that the side is held at 0. The requires that R(a)=0. What can you infer from this about the allowed values of $k$?

3. Oct 14, 2014

### V0ODO0CH1LD

Could you expand on that?
The condition $Z(0)=0$ implies that $A=-B$, right?. Do I use $R(a)=0$ before $Z(L)=u_0$? And does that mean that I would have to solve Bessel's equation to completely solve this problem?

4. Oct 14, 2014

### vela

Staff Emeritus
If you increase $\phi$ by $2\pi$, you've gone around the cylinder once and end up back in the same spot. You can't have two different values of $u$ for the same point.

Yes, so $Z(z) = A(e^{kr}-e^{-kr}) = A'\sinh kr$.

You can't say that $Z(L)=u_0$. You can only say $u(r,\phi,L)=u_0$.

I'd expect you can just say that each R is a Bessel function without having to derive that result over again.

5. Oct 14, 2014

### V0ODO0CH1LD

I don't see how I am supposed to use $R(a)=0$ to move forward.. What am I supposed to be doing at this point?

6. Oct 14, 2014

### vela

Staff Emeritus
What are the solutions for $R$ in terms of the constants $m$ and $k$?

7. Oct 14, 2014

### V0ODO0CH1LD

I still don't see what the next step is.. How do I keep solving the problem from here?
I'm not sure.. Are you talking about
$$R(r)=\sum_{n=0}^\infty{}b_n(rk)^{n+m},$$
the solution for the Bessel equation? Which would mean that
$$\sum_{n=0}^\infty{}b_n(ak)^{n+m}=0.$$
Is that what I am supposed to do? Where do I go from here?

8. Oct 14, 2014

### vela

Staff Emeritus
I don't understand where you got that expression for R from.

9. Oct 14, 2014

### V0ODO0CH1LD

I used the Frobenius method on $R$, which gives the answer as a power series of that form. But I honestly don't know what the next step is, after I found $Z(z)=A(e^{kz}-e^{-kz})$ I don't know what to do.

10. Oct 14, 2014

### vela

Staff Emeritus
At this point, I think you need to go over a similar example in your textbook and get an understanding of the logic behind each step.