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Electrodynamics (Fourier)

  1. Oct 18, 2011 #1
    I don't know how to build up Fourier series for this problem. I tried several times but no results. Can any body help me.

    Consider the 2-d Laplace equation for the potential V(x,y) in the region inside
    an empty square whose sides lie on the lines x = 0, x = a , y = 0, and y = a, respectively.
    The sides at x = a , y = 0, and y = a are grounded, and
    V(0,y) = C[itex]_{1}[/itex]sin[itex]\frac{(\pi)y}{a}[/itex] + C[itex]_{2}[/itex]sin[itex]\frac{2(\pi)y}{a}[/itex]
    where C1 and C2 are constants

    (a) Find V(x,y) everywhere inside the square.
    Hint: use the trivial solution for X(x) with X(x)=0 at x=a

    (b) Calculate the potential for C1=100V, C2=50V in the middle of the box. Do you find
    any dependence on the box size?
  2. jcsd
  3. Oct 18, 2011 #2
    Well I took a stab at it, and I think I have an idea where the solution should go. If you the standard Laplace's Equation: [itex]\nabla^2V=0[/itex]. Where in this case we are dealing with cartesian coordinates.

    I'm hoping you know the general solutions are going to be sines and cosines for the y-direction and sinh and cosh for the x-direction (if you don't know why I can explain).

    So our [itex]V(x,y) = \sum_{i=1}^\infty A_n*sin\left(\frac{n\pi y}{a}\right)\left[cosh\left(\frac{n\pi x}{a}\right)+sinh\left(\frac{n\pi x}{a}\right)\right] [/itex].

    Now if you apply the boundary condition properly and use the orthogonality property, you should get the series in only two terms, when n = 1 and n = 2.

    I hope I'm approaching this correctly, but let me know if it does make sense.
  4. Oct 18, 2011 #3


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    Show us your work. Show/explain how you solved for X(x) and Y(y) before applying the boundary conditions and what applying the three V=0 boundary conditions got you. It doesn't make sense to try apply the last condition if you don't have the solution correct up to that point.
  5. Oct 18, 2011 #4
    Sorry about that, I'm new to posting so I'm unsure how much I should post of a solution.

    Ok so separation of variables (in cartesian coordinates):
    [itex]\frac{\partial^2 V}{\partial x^2} + \frac{\partial^2 V}{\partial y^2} = 0[/itex]
    where V(x,y) = X(x)Y(y).

    [itex]\frac{1}{X}\frac{d^2X}{dx^2}+\frac{1}{Y}\frac{d^2Y}{dy^2} = 0[/itex]

    If you set it equal to a constant say [itex]\alpha^2[/itex] then you're two differential equations will be:
    [itex]\frac{d^2X}{dx^2} - \alpha X = 0[/itex]
    [itex]\frac{d^2Y}{dy^2} + \alpha Y = 0[/itex]
    And solutions:
    [itex]X(x) = Acosh(\alpha x)+Bsinh(\alpha x)[/itex]
    [itex]Y(y) = A_1cos(\alpha y)+B_1sin(\alpha y)[/itex]

    Since [itex]Y(0)=Y(a)=0\implies A_1 = 0[/itex] and [itex]\alpha = \frac{n\pi}{a}[/itex]

    So since we're left with the sine term and with the boundary conditions given for X(x) doesn't allow us to get rid of any of those terms the full solution would be:
    [itex]V(x,y) = \sum_{i=1}^\infty A_n*sin\left(\frac{n\pi y}{a}\right)\left[cosh\left(\frac{n\pi x}{a}\right)+sinh\left(\frac{n\pi x}{a}\right)\right][/itex]

    So then when you apply the last one and use orthogonality that should give you only two terms n=1 and n=2. If you need me to explain how I deduced how I only got two terms I can.
  6. Oct 18, 2011 #5


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    My post was meant for humo90. :smile:

    You can't quite set A=B=1 yet.

    One trick you can use in this type of problem when you have a boundary where the function vanishes is to write the solution in the form
    [tex]X(x) = A\cosh\alpha(x-a)+B\sinh\alpha(x-a)[/tex]You don't really need to do this. It just makes the solution look nicer.
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