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Jackson 3.9 - my coefficients are vanishing!

  1. Oct 17, 2009 #1
    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([tex]\varphi,z[/tex] ). 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
    [tex]$\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]} } $[/tex]

    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
    [tex]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 } } $[/tex]

    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. jcsd
  3. Oct 17, 2009 #2


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

    [tex]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}[/tex]

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

    [tex]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[/tex]
    Last edited: Oct 17, 2009
  4. Oct 17, 2009 #3
    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 [itex]\frac{2}{L\pi}[/itex] 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 [tex]\nu\varphi[/tex] between 0 and 2[tex]\pi[/tex]; 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.
  5. Oct 17, 2009 #4


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

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

    Again, I think the problem is that you are assuming that [itex]V(\phi,z)[/itex] is a constant [itex]V[/itex]....if you aren't specifically told this, you need to simply leave it as [itex]V(\phi,z)[/itex] 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.
  6. Oct 17, 2009 #5
    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.[tex]\varpi[/tex]

    Here is my integral of the z term:
    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)} \\
    Seems like that m should be there.
  7. Oct 17, 2009 #6


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    But how did you get to that first integral from

    [tex]\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]} }[/tex]

    And why are you only summing over odd [itex]m[/itex] in the above equation....surely

    [tex]\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 [itex]\Phi(\rho,\phi,0)=\Phi(\rho,\phi,L)=0[/itex] for even [itex]m[/itex] aswell?
  8. Oct 18, 2009 #7
    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 [tex]\Phi _A \left( {\rho ,\varphi ,z} \right) + \Phi _B \left( {\rho ,\varphi ,z} \right)$
    corresponding to the two parts. Then, for A:
    [tex]\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 [tex]\rho[/tex]. The modified Bessel function value is now constant, leaving only the [tex]\varphi[/tex] and z variables. Fourier's Trick gets me to
    [tex]\\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.
  9. Oct 18, 2009 #8


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

    [tex]\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]} }[/tex]


    [tex]\\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 } }[/tex]

    ??? (Don't just say "using orthogonality", show your work)
  10. Oct 18, 2009 #9
    Starting with

    [tex]\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]} }[/tex]

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

    to get
    [tex]\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)$[/tex]

    Then integrate both sides.
    \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\}} \\

    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 [tex]\delta_{\nu\nu'}[/tex] and [tex]\delta_{mm'}[/tex]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
    [tex]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 [tex]\Phi_{A}[/tex] 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.
  11. Oct 18, 2009 #10


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    The only remaining errors you have are that

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


    [tex]\int_0^{2\pi} \sin(\nu \phi)\sin(\nu' \phi)d\phi\neq \delta_{\nu\nu'}[/tex]
  12. Oct 18, 2009 #11
    Well duh! It always gets back to the basics. Thanks for you help on this!
  13. Oct 19, 2009 #12


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

    [tex]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]} }[/tex]

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