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Solving the 2D laplace equation. Quick question about the final stage

  1. May 10, 2012 #1
    1. The problem statement, all variables and given/known data

    ## \frac{\partial^2 V}{\partial x^2} +\frac{\partial^2 V}{\partial y^2} = 0 ##

    it's defined for ##V=0 ## for

    ## x=0, 0<y<b##,
    ## x=a, 0<y<b##,
    ## y=0, 0<x<a##

    ##V=V_0 x/a ## for ## y=b, 0<x<a##

    2. Relevant equations

    None.

    3. The attempt at a solution

    I've got

    ##V(x,b) = \displaystyle V_0 x/a = \sum_{n=1}^{\infty} K_n \sin ({n \pi x \over a}) \sinh ({n \pi b \over a}) ##

    At first I thought I could set K to 0 for n>1 and do a simple division, but I don't know what K1 is. Do I always need to find Fourier coefficients. I have seen a problem www.robots.ox.ac.uk/~jmb/lectures/pdelecture5.pdf here where in the final stage they do something like this. I'm guessing I need to do the former, but I am interested to know if the other method is possible.

    As for the coefficients, the function is odd so the only integral I need is the sine integral.

    ## K_n = \frac{1}{a} \int_0^{2a} \frac{V x}{a} \sin ({n \pi x \over a}) dx =-\frac{2V}{n\pi} ##

    I think this is right:

    ##V(x,y) = \sum_{n=1}^{\infty} -\frac{2V}{n\pi} \sin ({n \pi x \over a}) \sinh ({n \pi y \over a})##
     
    Last edited: May 10, 2012
  2. jcsd
  3. May 10, 2012 #2

    LCKurtz

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    Presumably those are second order partials.

    Presumably you got for your proposed solution$$
    V(x,y) = \sum_{n=1}^\infty K_n\sin\frac{n\pi x}{a}\sinh\frac{n\pi y}{a}$$It is when you put in ##y=b## you get the next equation, and the left side of that equation should be ##V(x,b)##, not ##V(x,y)## since you are trying to make the last boundary condition work.

    No, you've lost your way here. Of course you need a Fourier Series. How else can the sum of sine terms equal ##V_0x/a##? First, it doesn't matter whether that function is odd because whatever it might be, you would use the odd half-range expansion since all you have to work with is sine terms. What you need is for$$K_n\sinh\frac {n\pi b} a$$ to be the half range Fourier coefficient usually called ##b_n##. So set$$
    K_n\sinh\frac {n\pi b} a = \frac 2 a\int_0^a \frac {V_0x} a \sin\frac{n\pi x}a dx$$
    Solve that for ##K_n## and put in in the formula for ##V(x,y)##.
     
  4. May 10, 2012 #3
    I get ##K_n = -\frac{2V(-1)^n}{n \pi \sinh ({n \pi b \over a } )}##
     
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