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Homework Help: New Problem about solving a Laplace Equation in cylinder coordinates

  1. Sep 8, 2009 #1
    1. The problem statement, all variables and given/known data

    The capacitor is assumed to consist of two circular disc electrodes of radius [tex]\alpha [/tex]. The electrodes are of infinitesimal thickness, placed a distance [tex]2L[/tex]
    apart, and are equally and oppositely charged to potentials +V and -V. To solve the potential distribution in and out the capacitor, the Laplace Equation is:

    [tex]\[\frac{{{\partial ^2}\phi }}{{\partial {r^2}}} + \frac{1}{r}\frac{{\partial \phi }}{{\partial r}} + \frac{{{\partial ^2}\phi }}{{\partial {z^2}}} = 0{\rm{ }}........{\rm{equ}}{\rm{.(1)}}\][/tex]

    A solution in a paper to (1) is

    [tex]\[\phi (r,z) = \int_0^\infty {A(\lambda )({e^{ - \lambda |z - L|}} - {e^{ - \lambda |z + L|}}){J_0}(\lambda r)} d\lambda {\rm{ }}........{\rm{equ}}{\rm{.(2)}}\][/tex]

    I do not really understand why we can derive (2) from (1) directly. And my solution is a little different with (2). Does anyone tell me what is wrong with my method?
    I solve (1) using separation of variables,

    2. Relevant equations

    Assume [tex]\[\phi (r,z) = R(r)Z(z)\][/tex] ,and divide through by [tex]\[\phi (r,z)\][/tex] ,

    [tex]\[\frac{{Z(z)}}{r}\frac{{\partial R(r)}}{{\partial r}} + Z(z)\frac{{{\partial ^2}R(r)}}{{\partial {r^2}}} + R(r)\frac{{{\partial ^2}Z(z)}}{{\partial {z^2}}} = 0\][/tex]

    So, we can get:

    [tex]\[\frac{1}{Z}\frac{{{\partial ^2}Z}}{{\partial {z^2}}} = {\lambda ^2}{\rm{ }}........{\rm{equ}}{\rm{.(3)}}\][/tex]

    [tex]\[\frac{1}{{Rr}}\frac{{\partial R}}{{\partial r}} + \frac{{{\partial ^2}R}}{{R\partial {r^2}}} = - {\lambda ^2}{\rm{ }}........{\rm{equ}}{\rm{.(4)}}\][/tex]

    3. The attempt at a solution

    The solution for (3) and (4) is

    [tex]\[Z(z) = C(1){e^{z\lambda }} + C(2){e^{ - z\lambda }}\][/tex]

    [tex]\[R(r) = C(3){J_0}(r\lambda ) + C(4){Y_0}{\rm{(}}r\lambda )\][/tex]

    Using boundary conditions, we have

    [tex]\[Z(z) = C(1)({e^{z\lambda }} - {e^{ - z\lambda }}){\rm{ }}........{\rm{equ}}{\rm{.(5)}}\][/tex]

    [tex]\[R(r) = C(3){J_0}(r\lambda ){\rm{ }}........{\rm{equ}}{\rm{.(6)}}\][/tex]

    Now we get

    [tex]\[\phi (r,z) = R(r)Z(z) = \int_0^\infty {A(\lambda )({e^{z\lambda }} - {e^{ - z\lambda }}{\rm{)}}{J_0}(r\lambda )d\lambda } {\rm{ }}........{\rm{equ}}{\rm{.(7)}}\][/tex]

    Equation (7) is my solution and is really different with (2). I do not know what is wrong with my method.

    Besides, I slightly do not understand the [tex]$\int_0^\infty {} $[/tex], in (7), why not[tex]\[\sum\limits_{\lambda = - \infty }^\infty {} \]
    [/tex] or [tex]\[\sum\limits_{\lambda = 0}^\infty {} \][/tex]?

    Please give me some advice. Thanks in advance


  2. jcsd
  3. Sep 9, 2009 #2
    ur not that far off.

    the circular discs are at z=L and z=-L respectively and so that should give the [itex]e^{|z-L|}[/itex] bit with the boundary conditions.

    not to sure about the integral myself...sorry.
  4. Sep 9, 2009 #3


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    This looks more like a solution to Poisson's equation than Laplace's equation :wink:

    Your solution looks perfectly fine, but differs from the solution in the paper you mentioned in that the authors of that paper look to have solved Poisson's equation in order to find an expression for the potential that is valid everywhere, including at points on either plate. Your solution is valid everywhere except at the plates themselves, and I'm sure that if you calculate your [itex]A(\lambda)[/itex] and compare it to the [itex]A(\lambda)[/itex] in the paper, you will find different results, in order that the two potentials are equal outside the plates.

    As for your question about the integral; did imposing your boundary conditions result in any restrictions on [itex]\lambda[/itex] (i.e. Is it real, complex, integer etc...)?

    If [itex]\lambda[/itex] is restricted to integer values, then certainly your solution will be a discrete summation. But if the only restriction is that [itex]\lambda[/itex] be a real number, then there will be a continuous distribution of possible values, and so you replace the sum by an integral.
  5. Sep 10, 2009 #4
    Hi,Thanks for your reply.
    Your help is really important for me!

    Actually, I calculated the [itex]A(\lambda)[/itex] and compared it to the [itex]A(\lambda)[/itex] in the paper.

    Because [tex]$\phi (\rho ,H) = V,{\rm{ 0}} \le \rho \le a[/tex] and the special integral

    [tex]\[\int_0^\infty {\frac{{\sin (\lambda a)}}{\lambda }} {\rm{BesselJ}}(0,r\lambda )d\lambda = \left\{ {\begin{array}{*{20}{c}}
    {\frac{1}{2}\pi ,{\rm{ }}0 \le \rho \le a} \\
    {{{\sin }^{ - 1}}(a/\rho ),{\rm{ }}\rho \ge a} \\
    \end{array}} \right.\]

    A proper function for [tex]A(\lambda )[/tex] is

    [tex]\[A(\lambda ) = \frac{{2V}}{\pi }\frac{1}{{\sinh (H\lambda )}}\frac{{\sin (\lambda a)}}{\lambda }\]


    [tex]\[\phi (r,z) = \frac{{2V}}{\pi }\int_0^\infty {\frac{1}{{\sinh (H\lambda )}}\sinh (z\lambda )\frac{{\sin (\lambda a)}}{\lambda }{\rm{BesselJ}}(0,r\lambda )d\lambda } ......equ.(8)\]


    \phi (r,H) = \frac{{2V}}{\pi }\int_0^\infty {\frac{{\sin (\lambda a)}}{\lambda }{\rm{BesselJ}}(0,r\lambda )d\lambda } = V,{\rm{ }}0 \le \rho \le a \\
    \phi (r, - H) = \frac{{2V}}{\pi }\int_0^\infty {\frac{{\sin (\lambda a)}}{\lambda }{\rm{BesselJ}}(0,r\lambda )d\lambda } = - V, \\


    [tex]\[\frac{1}{{\sinh (H\lambda )}} = \frac{1}{{\left( {1 - {e^{ - 2H\lambda }}} \right)}}2{e^{ - {\rm{H}}\lambda }} = 2\sum\limits_{n = 0}^\infty {{e^{ - (2n + 1){\rm{H}}\lambda }}} \]

    And with this, equation (8) becomes

    [tex]\[\phi (r,z) = \frac{{2V}}{\pi }\sum\limits_{n = 0}^\infty {\int_0^\infty {({e^{ - \lambda {b_{n - }}}} - {e^{ - \lambda {b_{n + }}}})\frac{{\sin (\lambda a)}}{\lambda }{\rm{BesselJ}}(0,r\lambda )d\lambda } } ......equ.(9)\]


    [tex]{b_{n \pm }} = H(1 + 2n) \pm z......equ.(10)[/tex]

    According to the tables of integrals

    [tex]\[\int_0^\infty {{e^{ - \lambda {b_{n \pm }}}}\frac{{\sin (\lambda a)}}{\lambda }{\rm{BesselJ}}(0,r\lambda )d\lambda } = \arcsin (\frac{{2a}}{{\sqrt {b_{n \pm }^2 + {{(a + r)}^2}} + \sqrt {b_{n \pm }^2 + {{(a - r)}^2}} }})\]

    and if

    [tex]${x_{n \pm }} = \frac{1}{2}[\sqrt {b_{n \pm }^2 + {{(a + r)}^2}} + \sqrt {b_{n \pm }^2 + {{(a - r)}^2}} ]$


    [tex]\[\int_0^\infty {{e^{ - \lambda {b_{n \pm }}}}\frac{{\sin (\lambda a)}}{\lambda }{\rm{BesselJ}}(0,r\lambda )d\lambda } = \arcsin (\frac{a}{{{x_{n \pm }}}})\]

    Equation (9) becomes

    [tex]\[\phi (r,z) = \frac{{2V}}{\pi }\sum\limits_{n = 0}^\infty {[\arcsin (\frac{a}{{{x_{n - }}}})} - \arcsin (\frac{a}{{{x_{n + }}}})]......equ.(11)\]

    My solution equation (11) looks same with the one in that paper. However, there is an important different in equation (10) which is a part of the final solution. So, the potential plot of my solution is different with the one of that paper.

    The plot of my solution can be seen in the following URL:

    I also draw the plot of the solution in the paper and it can be seen in the following URL:

    Obviously, the solution of the paper is right….

    In addition, actually, I do not really understand the solving method from equation (7) to equation (8). I just imitate from the paper and I attached behind (I knew use an attachment here is not a good idea, so you can also send me a message with your email). It seems like more or less kinds of “cut and try method”. I am not sure for my opinion.
    Could you be kind enough to tell me what wrong with my solving?
    Thanks very much.

    Best Regards


    Attached Files:

  6. Sep 15, 2009 #5


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    The first thing that jumps out at me is that you have an extra factor of 2 in your [itex]A(\lambda)[/itex], since [itex]\sinh(\lambda H)=\frac{1}{2}(e^{\lambda H}-e^{-\lambda H})[/itex], so

    [tex]\phi(r,H)=\frac{2V}{\pi}\int_0^{\infty}\frac{\sin(\lambda r)}{\sinh(\lambda H)}\left(e^{\lambda H}-e^{-\lambda H}\right)J_0\left(\lambda r\right)d\lambda=\frac{4V}{\pi}\int_0^{\infty}\sin(\lambda r)J_0\left(\lambda r\right)d\lambda=\left\{\begin{array}{lcr}2V & , & 0\leq r\leq a \\ \frac{4V}{\pi}\arcsin\left(\frac{a}{r}\right) & , & r\geq a{\end{array}\right.[/tex]
    Last edited: Sep 15, 2009
  7. Sep 15, 2009 #6
    Thanks for your reply!

    I am sorry for the mistake. However, there is nothing wrong with my equation (8).

    As you said, the [itex]\[A(\lambda )\][/itex] should be

    [tex]\[A(\lambda ) = \frac{V}{\pi }\frac{1}{{\sinh (H\lambda )}}\frac{{\sin (\lambda a)}}{\lambda }\]

    so, equation (8) can be

    [tex]\[\phi (r,z) = \frac{V}{\pi }\int_0^\infty {\frac{1}{{\sinh (H\lambda )}}({e^{z\lambda }} - {e^{ - z\lambda }})\frac{{\sin (\lambda a)}}{\lambda }{\rm{BesselJ}}(0,r\lambda )d\lambda } ......equ.(8)\]

    and there is no difference between it and the old one:

    [tex]\[\phi (r,z) = \frac{{2V}}{\pi }\int_0^\infty {\frac{1}{{\sinh (H\lambda )}}\sinh (z\lambda )\frac{{\sin (\lambda a)}}{\lambda }{\rm{BesselJ}}(0,r\lambda )d\lambda } ......equ.(8)\][/tex]

    because [itex]\sinh(\lambda H)=\frac{1}{2}(e^{\lambda H}-e^{-\lambda H})[/itex].

    so the derivation after the equation (8) in my old post is still right.



  8. Sep 18, 2009 #7


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    I think one of the restrictions for this to be true is that [itex]b_{\pm}\geq 0[/itex], a condition which your [itex]b_{\pm}[/itex] doesn't meet for all [itex]z[/itex].
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