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Homework Help: Homogeneous Laplace's Equation

  1. Jun 27, 2010 #1
    1. The problem statement, all variables and given/known data

    2. Relevant equations

    3. The attempt at a solution

    Using separable method I get
    Y"-kY= 0 and X"+kX=0

    For Case 1 and Case 2 where k>0 and k=0 there are no eigenvalues

    So Case 3 k<0 gives
    Y=ccos(sqrk x y) + dsin(sqrk x y)
    y=0, then c=0
    y=pie, then dsin(pie x sqrk)=0
    k= -n2, Yn=sin(ny)
    X"-n2X=0, X=ccosh(nx)+dsinh(nx)

    u(x,y)= summation (from n=1 to n=infinity) sin(ny)[cncosh(nx)+dnsinh(nx)]

    u(0,y)= summation (from n=1 to n=infinity) cnsin(ny)

    Using Fourier's Series to expand 0, I got 0 so cn is 0

    ux(x,y)= summation (from n=1 to n=infinity) sin(ny)[dnncosh(nx)]
    ux(x,0)= summation (from n=1 to n=infinity) sin(ny)[dnn]

    Using Fourier's Series to expand 3siny-5sin4y, I also got 0 so dn is 0 as well

    Therefore u(x,y)=0

    However the answer given in the book is

    u(x,y)=(3/cosh5)(siny)(sinhx) - (5/cosh20)(sin4y)(sinh4x)

    I realized that the book didn't used Fourier's Series but compared the coefficients of the two equations instead. But my question is what happened to the n because if we compare the coefficients aren't we supposed to get

    [3/ncosh(5n)] n=1 so we get [3/cosh(5)]
    [-5/nsinh(5n)] n=4 so we get [-5/4sinh(20)] so u(x,y) should be

    [3/cosh(5)](siny)(sinhx) - [5/4sinh(20)](sin4y)(sinh4x)

    Did I do it correctly or is it just a printing error?
  2. jcsd
  3. Jun 27, 2010 #2


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    Gold Member

    They left out a 4 multiplying the cosh(20) in the denominator
    And you meant cosh(20) instead of sinh(20). They just dropped a 4.
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