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Electrostatics Fourier Decomposition (problem setting up boundaries)

  1. Sep 28, 2014 #1
    1. The problem statement, all variables and given/known data
    An #a*b*c box is given in x,y,z (so it's length #a along the x axis, etc.). Every face is kept at #V=0 except for the face at #x=a , which is kept at #V(a,y,z)=V_o*sin(pi*y/b)*sin(pi*z/c). We are to, "solve for all possible configurations of the box's potential"

    2. Relevant equations
    Differential equation solutions (Method of assuming V(x,y,z) = X(x)Y(y)Z(z) and then solving for each.

    #X''/X + Y''/Y + Z''/Z = 0

    --> X'' - (Cx)*X = 0, etc.

    Cx+Cy+Cz=0 (same as the first)
    3. The attempt at a solution
    I first tried to limit how many possible XYZ configurations I had to check. Because V(a,y,z) is in the form of sines of y and z, I thought I could safely choose Y, Z to be of trigonometric form (i.e. the constants they equal are negative). However if they are both of this form, then X is confined to only be of exponential solutions form (because the three constants need to add to zero) which means "all possible configurations" would only be one configuration. I tried looking at other forms but there are so many I feel as though I must be missing some key point to eliminate all but a few (i.e. "Oh, well actually you can still have a f(sin(z)) solution even if Z is linear!". I believe I can successfully solve for the Fourier coefficients and ultimately the solution(s), but I'm stuck on the boundary conditions for now.
  2. jcsd
  3. Sep 28, 2014 #2
    Sorry about the formatting, my post seems unable to be edited now.

    Boundaries:All faces $$V=0$$ except for $$V(a,y,z)=V_o sin(\frac{\pi y}{b}) sin(\frac{\pi z}{c})$$ at x=a

    Dimensions: ##a*b*c##

    Relevant Equations: ##\frac{X''}{X} + \frac{Y''}{Y} + \frac{Z''}{Z }= 0##
    implies ##C_x + C_y + C_z = 0##

    The diffeq solutions for each case are very standard so to save time I won't type them out (I have no issue with how to solve them)
  4. Sep 28, 2014 #3
    First of all, sorry if this post does not help you

    I tried to solve this and I got exponents in the ##X(x)##, the solution looked like this:

    ##A exp(\sqrt{C_x}x)+B exp(-\sqrt{C_x}x)##


    ##A= \frac{V_o cosh(Ka)}{2}## and ##B=-A##

    I think you should have ##C_x = C_y + C_z## right?

    Is this what you need?
  5. Sep 28, 2014 #4
    Well, kind of. I already know how to solve for every form of ##X(x), Y(y), Z(z)## but my trouble is discerning how I can get rid of certain combinations of ##V(x,y,z)=XYZ##. There are 27 total combinations, but the constant equation limits to I think 18. I was hoping I could reason, "well, we need Y,Z to be sines..." or "Well, X can never be a linear function" to rule out more possibilities, but I'm having trouble coming up with sufficient reasoning.
  6. Sep 29, 2014 #5
    I think an easy to see answer (that I worked out explicitly, even using the orthogonality method of finding ##C_n## just to be sure) I got $$V(x,y,z)=V_o \frac{sinh(\frac{- \pi (b+c) x}{bc})}{sinh(\frac{- \pi (b+c) a}{bc})} sin(\frac{\pi y}{b}) sin(\frac{\pi z}{c})$$ which like I said is pretty intuitive, but here's my problem: the wording of the problem implies there are more solutions, but I can't see any. I first thought maybe it would work for any arbitrary constant inside of the ##sinh(C_\Re x)## but I believe that would violate some BC's placed upon the values allowed inside the sinh term. So I'm stuck at trying to figure out other configurations that could solve this problem.
  7. Sep 29, 2014 #6


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    There is a uniqueness theorem about the potential within a volume with given boundary conditions so if you find a solution it is the only one. Your solution is not completely correct however (try differentiating your X(x) to find ##C_x##).
  8. Sep 29, 2014 #7
    I think you need the summation sign because the sines of y and z solutions should have integer term inside them
  9. Sep 29, 2014 #8


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    The boundary condition is only compatible with the first term in the series expansion, where the integers multiplying the arguments are equal to one.
  10. Sep 29, 2014 #9
    I wasn't sure if the uniqueness theorem applied here. Usually you solve for all but one boundary condition and then do a Fourier sum to find one that satisfies all boundary conditions--is that sum the unique solution?

    As for the incorrect constant, I see what you mean: I have ##C_x = k^2## where k is what I solved for later in the form ##X = C_1 \sinh{k x}##. I think what I did incorrectly was ##k + l + m = 0## where instead I should have done ##k^2 + l^2 + m^2 = 0 = C_x + C_y + C_z##. Thanks for pointing that out!
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