# Jackson 3.9 - my coefficients are vanishing!

1. Oct 17, 2009

### Old Guy

1. The problem statement, all variables and given/known data
A hollow right circular cylinder of radius b has its axis coincident with the z axis and its ends at z=0 and z=L. the potential on the end faces is zero, while the potential on the cylindrical surface is given as V($$\varphi,z$$ ). Using the appropriate separation of variables in cylindrical coordinates, find a series solution for the potential anywhere inside the cylinder.

2. Relevant equations I've come up with
$$\Phi \left( {\rho ,\varphi ,z} \right) = \sum\limits_{\nu = 0}^\infty {\sum\limits_{m = 1,3,5,...}^\infty {\left[ {A_\nu \sin \left( {\nu \varphi } \right) + B_\nu \cos \left( {\nu \varphi } \right)} \right]\left[ {\sin \left( {\frac{{m\pi }}{L}z} \right)} \right]\left[ {I_\nu \left( {\frac{{m\pi }}{L}\rho } \right)} \right]} }$$

3. The attempt at a solution Solving the above at the surface of the cylinder and using Fourier's Trick, I got the following for A
$$A_\nu = \frac{V}{{I_\nu \left( {\frac{{m\pi b}}{L}} \right)}}\int\limits_0^{2\pi } {\int\limits_0^L {\sin \left( {\nu '\varphi } \right)\sin \left( {\frac{{m'\pi }}{L}z} \right)dzd\varphi } }$$

The problem is that when I do the integration, the coefficient vanishes (same thing happens with B) although I know that I need both A and B.

2. Oct 17, 2009

### gabbagabbahey

You seem to be assuming $V(\phi,z)$ is a constant $V$, but my copy of Jackson gives:

$$V(\phi,z)=\left\{\begin{array}{lr}V,& -\frac{\pi}{2}<\phi<\frac{\pi}{2}\\-V,&\frac{\pi}{2}<\phi<\frac{3\pi}{2}\end{array}$$

Also, your coefficient depend on both $\nu$ and $m$, so they should really be labeled $A_{\nu m}$ and $B_{\nu m}$. You also seem to be missing a factor of $\frac{2}{L\pi}$ from your expression for $A_{\nu m}$, and I have no idea why you are only summing over odd values of $m$ or why you have primed indices in your expression.

$$A_\nu = \frac{2}{{L\pi I_\nu \left( \frac{{m\pi b}}{L} \right)}}\int\limits_0^{2\pi } \int\limits_0^L V(\phi,z)\sin \left( \nu \varphi \right)\sin \left( \frac{{m\pi }}{L}z \right)dzd\varphi$$

Last edited: Oct 17, 2009
3. Oct 17, 2009

### Old Guy

First, the 2-part potential you show is Problem 3.10 in my version of Jackson; I'm working on 3.9 where the potential is given as V on the sides of the cylinder.

Next, I realize the coefficients depend on both nu and m; I'll confess to a bit of sloppiness on that one.

Also, I made an error on the integration wrt z; that gives the $\frac{2}{L\pi}$ factor, but shouldn't there also be an m in the denominiator? When you do this, odd values of m come out of the integration wrt z. The primes can be ignored; they are an artifact of theway I originally learned Fourier's Trick.

But, I"m still left with the integration of sine of $$\nu\varphi$$ between 0 and 2$$\pi$$; won't that be zero since nu is an integer (according to Jackson)?

Thanks very much for your help; I've got a nagging feeling I'm missing something simple.

4. Oct 17, 2009

### gabbagabbahey

Okay yes, the potential I showed is from 3.10. For 3.9, you aren't actually told that $V(\phi,z)$ is a constant are you?

I don't see where the $1/m$ would come from....can you post your calculation?

Again, I think the problem is that you are assuming that $V(\phi,z)$ is a constant $V$....if you aren't specifically told this, you need to simply leave it as $V(\phi,z)$ and keep it inside the integral. You won't be able to explicitly evaluate the integrals of course, but I suspect you aren't meant to for prob 3.9.

5. Oct 17, 2009

### Old Guy

After all the time I spend telling my students to read the problems carefully and literally, I make the same mistake. I'm humbled ut appreciative.$$\varpi$$

Here is my integral of the z term:
$$\begin{array}{l} A_\nu = \frac{V}{{I_\nu \left( {\frac{{m\pi b}}{L}} \right)}}\int\limits_0^{2\pi } {d\varphi \int\limits_0^L {\sin \left( {\nu \varphi } \right)\sin \left( {\frac{{m\pi }}{L}z} \right)dz} } \\ A_\nu = \frac{V}{{I_\nu \left( {\frac{{m\pi b}}{L}} \right)}}\int\limits_0^{2\pi } {d\varphi \sin \left( {\nu \varphi } \right)\left[ { - \frac{L}{{m\pi }}\cos \left( {\frac{{m\pi }}{L}z} \right)} \right]} _0^L \\ A_\nu = \frac{V}{{I_\nu \left( {\frac{{m\pi b}}{L}} \right)}}\int\limits_0^{2\pi } {d\varphi \sin \left( {\nu \varphi } \right)\left[ { - \frac{L}{{m\pi }}\left( {\cos m\pi - 1} \right)} \right]} \\ A_\nu = \frac{V}{{I_\nu \left( {\frac{{m\pi b}}{L}} \right)}}\int\limits_0^{2\pi } {d\varphi \sin \left( {\nu \varphi } \right)\left[ { - \frac{L}{{m\pi }}\left( { - 1 - 1} \right)} \right]} \\ A_\nu = \frac{V}{{I_\nu \left( {\frac{{m\pi b}}{L}} \right)}}\frac{{2L}}{{m\pi }}\int\limits_0^{2\pi } {d\varphi \sin \left( {\nu \varphi } \right)} \\ \end{array}$$
Seems like that m should be there.

6. Oct 17, 2009

### gabbagabbahey

But how did you get to that first integral from

$$\Phi \left( {\rho ,\varphi ,z} \right) = \sum\limits_{\nu = 0}^\infty {\sum\limits_{m = 1,3,5,...}^\infty {\left[ {A_\nu \sin \left( {\nu \varphi } \right) + B_\nu \cos \left( {\nu \varphi } \right)} \right]\left[ {\sin \left( {\frac{{m\pi }}{L}z} \right)} \right]\left[ {I_\nu \left( {\frac{{m\pi }}{L}\rho } \right)} \right]} }$$

And why are you only summing over odd $m$ in the above equation....surely

$$\Phi_{\nu m}(\rho,\phi,z)=\left[A_{\nu m} \sin \left( {\nu \varphi } \right) + B_{\nu m} \cos \left( {\nu \varphi } \right)\right]\sin \left( {\frac{{m\pi }}{L}z} \right) I_\nu \left( {\frac{{m\pi }}{L}\rho } \right)[/itex] satisfies Laplace's equation and your boundary conditions $\Phi(\rho,\phi,0)=\Phi(\rho,\phi,L)=0$ for even $m$ aswell? 7. Oct 18, 2009 ### Old Guy I see the point about m odd, so let's change the starting equation back to: [tex]\Phi \left( {\rho ,\varphi ,z} \right) = \sum\limits_{\nu = 0}^\infty {\sum\limits_{m = 0}^\infty {\left[ {A_{\nu m}\sin \left( {\nu \varphi } \right) + B_\nu \cos \left( {\nu \varphi } \right)} \right]\left[ {\sin \left( {\frac{{m\pi }}{L}z} \right)} \right]\left[ {I_\nu \left( {\frac{{m\pi }}{L}\rho } \right)} \right]} }$$
I've assumed the double sum can be split (like we routinely do with single sums), so on the left side I would have $$\Phi _A \left( {\rho ,\varphi ,z} \right) + \Phi _B \left( {\rho ,\varphi ,z} \right)$$
corresponding to the two parts. Then, for A:
$$\Phi _A \left( {\rho ,\varphi ,z} \right) = \sum\limits_{\nu = 0}^\infty {\sum\limits_{m = 0}^\infty {\left[ {A_{\nu m}\sin \left( {\nu \varphi } \right)} \right]\left[ {\sin \left( {\frac{{m\pi }}{L}z} \right)} \right]\left[ {I_\nu \left( {\frac{{m\pi }}{L}\rho } \right)} \right]} }$$
Then I substitute b (the radius) for $$\rho$$. The modified Bessel function value is now constant, leaving only the $$\varphi$$ and z variables. Fourier's Trick gets me to
$$\\A_{\nu m} = \frac{\Phi _A \left( {b,\varphi ,z} \right) }{{I_\nu \left( {\frac{{m\pi b}}{L}} \right)}}\int\limits_0^{2\pi } {\int\limits_0^L {\sin \left( {\nu \varphi } \right)\sin \left( {\frac{{m\pi }}{L}z} \right)dzd\varphi } }$$
And, of course, a similar approach with the B coefficient.

8. Oct 18, 2009

### gabbagabbahey

You are still skipping important steps, and making errors in the process. How do you go from

$$\Phi _A \left( {b,\varphi ,z} \right) = \sum\limits_{\nu = 0}^\infty {\sum\limits_{m = 0}^\infty {\left[ {A_{\nu m}\sin \left( {\nu \varphi } \right)} \right]\left[ {\sin \left( {\frac{{m\pi }}{L}z} \right)} \right]\left[ {I_\nu \left( {\frac{{m\pi }}{L}b } \right)} \right]} }$$

to

$$\\A_{\nu m} = \frac{\Phi _A \left( {b,\varphi ,z} \right) }{{I_\nu \left( {\frac{{m\pi b}}{L}} \right)}}\int\limits_0^{2\pi } {\int\limits_0^L {\sin \left( {\nu \varphi } \right)\sin \left( {\frac{{m\pi }}{L}z} \right)dzd\varphi } }$$

??? (Don't just say "using orthogonality", show your work)

9. Oct 18, 2009

### Old Guy

Starting with

$$\Phi _A \left( {b,\varphi ,z} \right) = \sum\limits_{\nu = 0}^\infty {\sum\limits_{m = 0}^\infty {\left[ {A_{\nu m}\sin \left( {\nu \varphi } \right)} \right]\left[ {\sin \left( {\frac{{m\pi }}{L}z} \right)} \right]\left[ {I_\nu \left( {\frac{{m\pi }}{L}b } \right)} \right]} }$$

multiply both sides by $$\sin \left( {\nu '\varphi } \right)\sin \left( {\frac{{m'\pi }}{L}z} \right)}$$

to get
$$\Phi _A \left( {b,\varphi ,z} \right)\sin \left( {\nu '\varphi } \right)\sin \left( {\frac{{m'\pi }}{L}z} \right) = \sum\limits_{\nu = 0}^\infty {\sum\limits_{m = 0}^\infty {\left[ {A_{\nu m} \sin \left( {\nu \varphi } \right)} \right]\left[ {\sin \left( {\frac{{m\pi }}{L}z} \right)} \right]\left[ {I_\nu \left( {\frac{{m\pi b}}{L}} \right)} \right]} } \sin \left( {\nu '\varphi } \right)\sin \left( {\frac{{m'\pi }}{L}z} \right)$$

Then integrate both sides.
$$\begin{array}{l} \int\limits_0^{2\pi } {d\varphi \left\{ {\int\limits_0^L {dz} \left[ {\Phi _A \left( {b,\varphi ,z} \right)\sin \left( {\nu '\varphi } \right)\sin \left( {\frac{{m'\pi }}{L}z} \right)} \right]} \right\}} = \\ \int\limits_0^{2\pi } {d\varphi \left\{ {\int\limits_0^L {dz} \left[ {\sum\limits_{\nu = 0}^\infty {\sum\limits_{m = 0}^\infty {\left[ {A_{\nu m} \sin \left( {\nu \varphi } \right)} \right]\left[ {\sin \left( {\frac{{m\pi }}{L}z} \right)} \right]\left[ {I_\nu \left( {\frac{{m\pi b}}{L}} \right)} \right]} } \sin \left( {\nu '\varphi } \right)\sin \left( {\frac{{m'\pi }}{L}z} \right)} \right]} \right\}} \\ \end{array}$$

On the right hand side, the summations come out of the integral:
$$\sum\limits_{\nu = 0}^\infty {\sum\limits_{m = 0}^\infty {\int\limits_0^{2\pi } {d\varphi \left\{ {\int\limits_0^L {dz} \left[ {A_{\nu m} \sin \left( {\nu \varphi } \right)} \right]\left[ {\sin \left( {\frac{{m\pi }}{L}z} \right)} \right]\left[ {I_\nu \left( {\frac{{m\pi b}}{L}} \right)} \right]\sin \left( {\nu '\varphi } \right)\sin \left( {\frac{{m'\pi }}{L}z} \right)} \right\}} } }$$
Orthogonality says that the integration of the products will be $$\delta_{\nu\nu'}$$ and $$\delta_{mm'}$$and the summations are dropped. This results in
$$\int\limits_0^{2\pi } {d\varphi \left\{ {\int\limits_0^L {dz} \left[ {\Phi _A \left( {b,\varphi ,z} \right)\sin \left( {\nu '\varphi } \right)\sin \left( {\frac{{m'\pi }}{L}z} \right)} \right]} \right\}} = A_{\nu m} I_\nu \left( {\frac{{m\pi b}}{L}} \right)$$

The primes clearly aer no longer necessary and can be dropped. The modified Bessel function term is a constant, so
$$A_{\nu m} = \frac{1}{{I_\nu \left( {\frac{{m\pi b}}{L}} \right)}}\int\limits_0^{2\pi } {\int\limits_0^L {\Phi _A \left( {b,\varphi ,z} \right)\sin \left( {\nu \varphi } \right)\sin \left( {\frac{{m\pi }}{L}z} \right)dzd\varphi } }$$
Also note that I put $$\Phi_{A}$$ back in the integral since I'm no longer assuming it's constant. Does that help? Thanks for hanging in there with me on this.

10. Oct 18, 2009

### gabbagabbahey

The only remaining errors you have are that

$$\int_0^L \sin\left(\frac{m\pi z}{L}\right)\sin\left(\frac{m'\pi z}{L}\right)dz\neq \delta_{mm'}$$

and

$$\int_0^{2\pi} \sin(\nu \phi)\sin(\nu' \phi)d\phi\neq \delta_{\nu\nu'}$$

11. Oct 18, 2009

### Old Guy

Well duh! It always gets back to the basics. Thanks for you help on this!

12. Oct 19, 2009

### gabbagabbahey

Also, do you really need to break up $V(\phi,z)$ into two unknown functions $\Phi_A(b,\phi,z)$ and $\Phi_B(b,\phi,z)$? After all, you know

$$V \left( {\varphi ,z} \right) = \sum\limits_{\nu = 0}^\infty {\sum\limits_{m = 0}^\infty {\left[ {A_{\nu m}\sin \left( {\nu \varphi } \right)} \right]\left[ {\sin \left( {\frac{{m\pi }}{L}z} \right)} \right]\left[ {I_\nu \left( {\frac{{m\pi }}{L}b } \right)} \right]} }+\sum\limits_{\nu = 0}^\infty {\sum\limits_{m = 0}^\infty {\left[ {B_{\nu m}\cos \left( {\nu \varphi } \right)} \right]\left[ {\sin \left( {\frac{{m\pi }}{L}z} \right)} \right]\left[ {I_\nu \left( {\frac{{m\pi }}{L}b } \right)} \right]} }$$

What happens when you multiply the second term by $\sin(\nu\varphi)d\varphi$ and integrate from $0$ to $2\pi$?