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Question on Dirichlet problem in cylinder with radial symmetry.

  1. May 12, 2010 #1
    This is not homework.

    I have problem deriving the solution for cylinder with radial symmetry given:

    [tex] \nabla^2U(\rho,z)=R''+\frac{1}{\rho}R'+\frac{Z''}{Z}=0[/tex]

    Which give [itex] \rho^2 R''+ \rho R' -k\rho^2 R=0 \hbox { and } Z''+kZ=0[/itex]

    With given boundary conditions [itex] U(\rho,0) = U(\rho,h) =0 \hbox { for } \rho<a \hbox{ and } U(a,z) = f(z)[/itex]

    The book claimed [itex]k=-\lambda^2 = -ve[/itex] can only produce trivial solution of R. I cannot verify this.

    k=-ve give [itex]\rho^2 R''+ \rho R' +\lambda^2 R=0[/itex]

    This is parametric Bessel's equation of order zero which give [itex]R(\rho)=c_1 J_0(\lambda_n\rho) + c_2 Y_0(\lambda_n\rho)[/itex]

    R(0) is bounded [itex] \Rightarrow c_2 = 0 \Rightarrow R(\rho)= c_1 J_0(\lambda_n\rho)[/itex]

    That's where I get stuck. I cannot rule this out [itex] R(\rho)= c_1 J_0(\lambda_n\rho)[/itex] as a solution with the given boundary condition.

    The book claimed only k=+ve would give solution. Please give me some suggestion.
    Last edited: May 13, 2010
  2. jcsd
  3. May 12, 2010 #2


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    It is the Z boundary value problem which restricts the values of k. Your Z bvp is:

    Z'' + k Z = 0
    Z(0) = Z(h) = 0

    If you solve this you will find that only k = μ2 > 0 gives non-trivial solutions. So these are the values that must be used in the other equation too.
  4. May 12, 2010 #3
    I am embarrassed!!!! Thank you very much.

    One more verification. [itex] Z''+kZ=0 \Rightarrow Z=d_1cos(\sqrt{k} z) + d_2 sin (\sqrt{k} z)[/itex]

    for k=0 [itex] \Rightarrow Z=d_1.\;\;U(\rho,0)=0 \Rightarrow d_1=0 \[/itex]
    I don't know why I cannot edit the [itex]d_2[/itex] portion.

    Therefore k=0 give trivial solution also. Am I correct?

    Thanks a million
    Last edited: May 12, 2010
  5. May 13, 2010 #4


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    When k = 0 the solution to the equation does not have a {sine,cosine} form. It is just:

    Z'' = 0.

    Apply the boundary conditions to the general solution of that and see what happens.
  6. May 13, 2010 #5
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