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I Laplace equation boundary conditions

  1. Jun 12, 2017 #1
    Hi,
    I need to solve Laplace equation ##\nabla ^2 \Phi(z,r)=0## in cylindrical coordinates in the domain ##r_1<r<r_2##, ##0<z<L##.
    The boundary conditions are:
    ##
    \left\{
    \begin{aligned}
    &\Phi(0,r)=V_B \\
    &\Phi(L,r)=V_P \\
    & -{C^{'}}_{ox} \Phi(x,r_2)=C_0 \frac{\partial \Phi(x,r)}{\partial r}\rvert_{r=r_2} \\
    &\frac{\partial \Phi(x,r)}{\partial r}\rvert_{r=r_1}=0 \\
    \end{aligned}
    \right.
    ##
    By separation of variable I obtain:
    ##
    \Phi_(z,r)=(A e^{-\lambda z} + B e^{+\lambda z})(C J_0(\lambda r) + D Y_0(\lambda r))
    ##
    ##J_0## and ##Y_0## being zero order first type and second type Bessel functions.
    The constants I have to determine are : ##A, B, C, D## and ##\lambda##; they are ##5## constants, but only ##4## equations are available coming from the behaviour of ##\Phi## at the boundaries of the domain, so I do not know how to proceed.
    Is there any mistake in my reasoning?
     
  2. jcsd
  3. Jun 12, 2017 #2

    Orodruin

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    You cannot find a product solution. The reason to do the variable separation is so that you can expand the solution in the eigenfunctions of Bessel's differential operator. The normalisation of your eigenfunctions is arbitrary.

    Note that you really only have 4 constants as you can take an overall normalisation in one of the factors and absorb it in the other.
     
  4. Jun 12, 2017 #3
    Is it the same like saying that I can fit my boundary conditions whatever the value of one constant among A, B, C and D?
     
  5. Jun 12, 2017 #4

    Orodruin

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    Note what I said about the variable separation. In order to find the solution you will need a superposition of separated solutions. Your base functions in the radial direction will be composed of the Bessel functions and their normalisation constant will not matter (it will just change the normalisation of the expansion coefficients).
     
  6. Jun 12, 2017 #5
    I have understood that. The first boundary condition gives:
    ##
    \sum_m (A_m + B_m )(C_m J_0(\lambda_m r) + D_m Y_0(\lambda_m r))= V_B
    ##
    and I can decide, for example, to choose ##A_m=1##; is this correct?
     
  7. Jun 12, 2017 #6

    Orodruin

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    It would be more common to fix one of the constants in the radial (SL) part, but in essence yes.
     
  8. Jun 12, 2017 #7
    Could you tell me why fixing ##C_m## and ##D_m## is more common?
    Anyway, thank you for you help!
     
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