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A Laplace equation- variable domain

  1. Sep 21, 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 variables 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 general solution is:
    ##\Phi_{tot}= \sum_m (A_m e^{-\lambda_m z} + B_m e^{+\lambda_m z})(C_m J_0(\lambda_m r) + D_m Y_0(\lambda_m r)) ##
    and all the constants che can be calculated exploiting the boundary conditions.

    Now assume ##L## is not fixed, but it can vary in a certain range ##0<L_1<L<L_2##. What I am thinking about is: is it possible to compute ##L## such that [tex] \frac{\partial \Phi_{tot}}{\partial z} \rvert_{z=L} =0[/tex]?

    Thank you!
     
  2. jcsd
  3. Sep 26, 2017 #2
    Thanks for the thread! This is an automated courtesy bump. Sorry you aren't generating responses at the moment. Do you have any further information, come to any new conclusions or is it possible to reword the post? The more details the better.
     
  4. Sep 30, 2017 #3

    Dr Transport

    User Avatar
    Science Advisor
    Gold Member

    yes, you have the solution written, just take the derivative and find the solution
     
  5. Oct 1, 2017 #4
    Form the boundary conditions I can write
    [tex]
    \Phi_m=(\exp(\lambda_m z)+B_m \exp(-\lambda_m z) )C_m F_m(r)
    [/tex]
    where ##F_m## is a function independent from ##L## and
    [tex]
    \begin{cases}
    B_m=\frac{(K_{Bm}/K_{Pm}) \exp(\lambda_m L)-1}{1- (K_{Bm}/K_{Pm}) \exp(-\lambda_m L)} \\
    C_m=\frac{K_{Pm}-K_{Bm} \exp(-\lambda_m L)}{2 K_{1m} \sinh{\lambda_m L}}
    \end{cases}
    [/tex]
    ##K_{Pm}, K_{Bm}## and ##K_{1m}## being constants.
    By computing ## \frac{\partial \Phi_{tot}}{\partial z} \rvert_{z=L} =0 ## I get
    [tex]
    \sum_m \lambda_m (\exp{\lambda_m L} - B_m \exp{-\lambda_m L}) C_m F_m(r) = 0
    [/tex]
    How can I solve this equation? Note that ##L## must not be ##m-##dependent.

    Thank you for your interest.
     
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