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A Laplace Eq. in Cylindrical coordinates

  1. May 22, 2017 #1
    Hi,
    I need to solve Laplace equation:## \nabla ^2 \Phi(x,r)=0 ## in cylindrical domain ##0<r<r_0##, ##0<x<L## and ##0<\phi<2\pi##. The boundary conditions are the following ones:
    ##
    \left\{
    \begin{aligned}
    &C_{di}\Phi(x,r_0)=\epsilon \frac{\partial \Phi(x,r)}{\partial r}\rvert_{r=r_0} \\
    &\Phi(0,r)=f_1(r) \\
    &\Phi(L,r)=f_2(r)
    \end{aligned}
    \right.
    ##
    ##\epsilon## being the dielectric constant of the medium and ##C_{di}## a constant capacitance; ##f_1## and ##f_2## are two known function.
    Forcing no ##\phi## dependence, and solving the equation by separation of variable:
    ##
    \Phi(x,r)=(Ae^{\lambda x/r_0}+Be^{-\lambda x/r_0})J_0(\lambda r/r_0)
    ##
    ##\lambda## being the separation constant and ##J_0## the first type Bessel function of 0-order.
    By applying the first boundary condition:
    ##
    \lambda \frac{J_1(\lambda)}{J_0(\lambda)}=C_r
    ##
    ## C_r=r_0 C_{di}/\epsilon##
    which can be solved to compute all the values of ##\lambda##.
    By applying the remaining boundary conditions, I obtain the following set of equations:
    ##
    \left\{
    \begin{aligned}
    &\sum_{m=1}^{\infty} (A_m+B_m)J_0(\lambda_m r/r_0)=f_1(r) \\
    &\sum_{m=1}^{\infty} (A_m e^{\lambda_m L/r_0}+B_m e^{-\lambda_m L/r_0})J_0(\lambda_m r/r_0)=f_2(r) \\
    \end{aligned}
    \right.
    ##
    which allow me to compute the values of all the coefficients, by exploiting the orthogonality between differently scaled Bessel functions:
    ##
    \left\{
    \begin{aligned}
    &A_i=\frac{1}{(2\sinh{\lambda_i L/r_0})(\frac{1}{2}{J_1(\lambda_i)}^2)}(I_2-I_1 e^{-\lambda_i L/r_0}) \\
    &B_i=\frac{1}{(2\sinh{\lambda_i L/r_0})(\frac{1}{2}{J_1(\lambda_i)}^2)}(I_1e^{\lambda_i L/r_0}-I_1) \\
    \end{aligned}
    \right.
    ##
    ##I=\int_0^1 \frac{r}{r_0} f(r) J_0(\lambda_i r/r_0) d(\frac{r}{r_0})##

    Now the point that is driving me crazy is the following: how do you expect ##A_i## and ##B_i## to change as a function of the order ##i##? I expect them to decrease in modulus; indeed, this is the case for ##A_i##, but ##B_i## presents a strange behaviour, oscillating in sign and increasing in modulus. Can you see any mistake? If you want, I can provide you with all the mathematical passages that I omitted for brevity, the trends of ##A_i## and ##B_i## or show you whatever plot you may need. Any suggestion is really appreciated here.

    Thank you all.
     
  2. jcsd
  3. May 23, 2017 #2

    Charles Link

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    I spotted one algebraic mistake I believe: On ## B_i ##, the last ## -I_1 ## should be ## -I_2 ##. I didn't check every other detail, but the rest of it looks like it might be correct. ## \\ ## Editing: A little further reading on the Bessel functions shows that for the functions to be orthogonal, I believe it requires ## J_o(\lambda_m)=0 ##. I'm not sure that your functions meet this condition. ## \\ ## Additional editing: From what I googled on the subject, you may need to also use ## J_n ## of all orders in your series rather than limiting it to ## J_o ##.
     
    Last edited: May 23, 2017
  4. May 24, 2017 #3
    The algebraic mistake s a transcription error; in my implementation ##I_2## correctly takes the place of ##I_1##.

    I am not very strong on Bessel equation and I completely forgot the condition you pointed me out!
    Can you clarify your second comment? Do you mean:
    ##
    \left\{
    \begin{aligned}
    &\sum_{m=1}^{\infty} (A_m+B_m)J_m(\lambda_m r/r_0)=f_1(r) \\
    &\sum_{m=1}^{\infty} (A_m e^{\lambda_m L/r_0}+B_m e^{-\lambda_m L/r_0})J_m(\lambda_m r/r_0)=f_2(r) \\
    \end{aligned}
    \right.
    ##
    ?

    Thank you a lot, your help is really really appreciated.
     
  5. May 24, 2017 #4

    Orodruin

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    This is not correct. Any homogeneous boundary condition, be it of Dirichlet, Neumann, or Robin type, will do. Each of the choices results in a Sturm-Liouville problem.

    This would be true if the problem was not rotationally symmetric. As it stands, the rotational symmetry directly implies that only the ##n = 0## terms are of relevance.
     
  6. May 24, 2017 #5
  7. May 24, 2017 #6

    Orodruin

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  8. May 24, 2017 #7

    Orodruin

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    I did not check your arithmetics, but let me point out that it is significantly simpler to use ##C \cosh(\lambda x/r_0) + D \sinh(\lambda x/r_0)## for the ##x##-dependent part, since the boundary condition at ##x = 0## will give you one of the constants directly. You should find that both ##C## and ##D## are decrease sufficiently fast with ##m## as long as ##I_1## and ##I_2## do (which they should if ##f_1## and ##f_2## are in the correct Hilbert space).
     
  9. May 24, 2017 #8
    I do not get how I can compute both constants if I cannot exploit orthogonality. Could you help me with this too?
    Thank you very much.
     
  10. May 24, 2017 #9

    Orodruin

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    Also note that it is not necessarily a problem that the coefficients increase in modulus. The norm of the basis functions ##J_0(\lambda_m r/r_0)## is not constant.

    What do you mean? I just told you you can exploit orthogonality - it is a Sturm-Liouville problem!
     
  11. May 24, 2017 #10
    As I mentioned before, I am very weak on this topic. Is this correct?

    In general
    ##
    \int_0^1 x J_0(\lambda_m x)J_0(\lambda_i x)=\frac{{J^{'}_0(\lambda_m)}J_0(\lambda_i)-J_0(\lambda_m){J^{'}_0(\lambda_i)}}{{\lambda_i}^2-{\lambda_m}^2}
    ##
    In my case, since
    ##
    \lambda \frac{J_1(\lambda)}{J_0(\lambda)}=C_r
    ##
    I can write
    ##
    J^{'}_0(\lambda_m)=-J_1(\lambda_m)=-J_0(\lambda_m) \frac{C_r}{\lambda_m}
    ##
    and
    ##
    J^{'}_0(\lambda_i)=-J_1(\lambda_i)=-J_0(\lambda_i) \frac{C_r}{\lambda_m}
    ##
    and so
    ##
    \int_0^1 x J_0(\lambda_m x)J_0(\lambda_i x)=\frac{{J^{'}_0(\lambda_m)}J_0(\lambda_i)-J_0(\lambda_m){J^{'}_0(\lambda_i)}}{{\lambda_i}^2-{\lambda_m}^2}=0
    ##
    So I can exploit orthogonality even if ##\lambda## are not zeros of ##J_0##.
    If this is true, my calculations are correct, and I cannot explain the strange behaviour of the constants I depicted before...
     
    Last edited: May 24, 2017
  12. May 24, 2017 #11
    I tried to sum numerically the first 16 values of ##A_i+B_i## and I am very far from fitting the boundary condition in ##x=0## and ##x=L##.
    Also, ##\sum_i^n B_i## seems to increase with ##n##.
     
  13. May 24, 2017 #12

    Orodruin

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    As I said, your basis functions do not have constant norm so there is no a priori reason to expect the coefficients to decrease. Without knowing more of the functions ##f_i##, I don't know what more I can say about this.

    Yes, this is correct. The orthogonality follows from the ##\lambda## being such that the Robin boundary condition is satisfied.
     
  14. May 24, 2017 #13
    These are the remaining functions:

    ##f_1=\Phi_1-\Phi_0(r)## and ##f_2=\Phi_2-\Phi_0(r)##,
    ## \Phi_0(r)=V-k_a \log(\frac{k_b}{{(4k_b-r^2)}^2}k_c) ##
    where ##V## and ##k_{a,b,c}## are some constants.

    Can you guess any ##A_i## and ##B_i## behaviour?
    I am starting to think there is some hidden mistake in my implementation.

    Thank you a lot!
     
  15. May 24, 2017 #14

    Orodruin

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    What are the possible values of ##k_b##?
     
  16. May 24, 2017 #15
    ##k_b## is the solution of another equation.
    Do you mean numerical values ?
     
  17. May 24, 2017 #16

    Orodruin

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    I am just trying to figure out if that function is square integrable or not. Without looking at it in detail I am not sure I can help you. A suggestion would be to try to see what happens with your expressions in some cases (different fs) where you know the solutions to check your implementation.
     
  18. May 24, 2017 #17
    I will take another look tomorrow in the morning. Anyway, my approach should be correct, and the equations for ##A_i## and ##B_i## in the first post should be valid...right?
     
  19. May 28, 2017 #18
    Hi,
    I may have found one tricky point. Can someone tell me the result of ##\int_0^1 x J_0(\lambda_m x)J_0(\lambda_i x)## when ##\lambda_m=\lambda_i## and ##\lambda_i## is not a zero of the Bessel function?
    Edit:
    I have finally found the mistake. I used the following relation:
    ##\int_0^1 x J_0^2(\lambda x) dx=\frac{1}{2}J_1^2(\lambda)##
    but when ##\lambda## is not a zero of ##J_0##, like in my case, this holds:
    ##\int_0^1 x J_0^2(\lambda x) dx=\frac{1}{2} (J_0^2(\lambda)+J_1^2(\lambda)) ##.
    By using the correct relation I find ##B_i## to descrease...
     
    Last edited: May 28, 2017
  20. Jun 5, 2017 #19
    Hi,
    since you helped me so much in my previous doubt, I would like to make another question about Bessel functions.
    Assume that my boudary condition is expressed like this:
    ##
    \sum_{m=1}^{\infty} A_m J_0(\lambda_m r/r_0) + B_m Y_0(\lambda_m r/r_0)=C
    ##
    ##C## being a certain constant and ##Y## second type Bessel function.
    How can I proceed to obtain a relation between ##A_m## and ##B_m## now, following a similar path as before?

    Thank you
     
  21. Jun 5, 2017 #20

    Orodruin

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    The same way as before. First you need to find the appropriate boundary conditions so that you can find what combination of Bessel functions form a complete basis on the domain of interest.
     
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