Jackson 3.9 - my coefficients are vanishing

[Old Guy]

Homework Statement

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.

Homework 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]} } $

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.

 

[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

 

[Old Guy]

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

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.

 

[gabbagabbahey]

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.

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?

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.

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

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.

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.

 

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

 

[gabbagabbahey]

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

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] <br /> <br /> satisfies Laplace&#039;s equation and your boundary conditions \Phi(\rho,\phi,0)=\Phi(\rho,\phi,L)=0 for even m aswell?

 

[Old Guy]

I see the point about m odd, so let's change the starting equation back to:
\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]} } $<br />
I've assumed the double sum can be split (like we routinely do with single sums), so on the left side I would have \Phi _A \left( {\rho ,\varphi ,z} \right) + \Phi _B \left( {\rho ,\varphi ,z} \right)$<br />
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]} } $<br />
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 } } $<br />
And, of course, a similar approach with the B coefficient.

 

[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)

 

[Old Guy]

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)

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 &#039;\varphi } \right)\sin \left( {\frac{{m&#039;\pi }}{L}z} \right)}<br />

to get
\Phi _A \left( {b,\varphi ,z} \right)\sin \left( {\nu &#039;\varphi } \right)\sin \left( {\frac{{m&#039;\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 &#039;\varphi } \right)\sin \left( {\frac{{m&#039;\pi }}{L}z} \right)$

Then integrate both sides.
<br /> \begin{array}{l}<br /> \int\limits_0^{2\pi } {d\varphi \left\{ {\int\limits_0^L {dz} \left[ {\Phi _A \left( {b,\varphi ,z} \right)\sin \left( {\nu &#039;\varphi } \right)\sin \left( {\frac{{m&#039;\pi }}{L}z} \right)} \right]} \right\}} = \\ <br /> \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 &#039;\varphi } \right)\sin \left( {\frac{{m&#039;\pi }}{L}z} \right)} \right]} \right\}} \\ <br /> \end{array}<br />

On the right hand side, the summations come out of the integral:
<br /> \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 &#039;\varphi } \right)\sin \left( {\frac{{m&#039;\pi }}{L}z} \right)} \right\}} } } <br />
Orthogonality says that the integration of the products will be \delta_{\nu\nu&#039;} and \delta_{mm&#039;}and the summations are dropped. This results in
<br /> \int\limits_0^{2\pi } {d\varphi \left\{ {\int\limits_0^L {dz} \left[ {\Phi _A \left( {b,\varphi ,z} \right)\sin \left( {\nu &#039;\varphi } \right)\sin \left( {\frac{{m&#039;\pi }}{L}z} \right)} \right]} \right\}} = A_{\nu m} I_\nu \left( {\frac{{m\pi b}}{L}} \right)<br />

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 } } <br />
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.

 

[gabbagabbahey]

The only remaining errors you have are that

\int_0^L \sin\left(\frac{m\pi z}{L}\right)\sin\left(\frac{m&#039;\pi z}{L}\right)dz\neq \delta_{mm&#039;}

and

\int_0^{2\pi} \sin(\nu \phi)\sin(\nu&#039; \phi)d\phi\neq \delta_{\nu\nu&#039;}

 

[Old Guy]

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

 

[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?