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Math Help - Laplace's Equation in 3-Dimension (Orthogonality and Eigen function)

  1. #1
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    Laplace's Equation in 3-Dimension (Orthogonality and Eigen function)

    Any help will be appreciated!
    I am solving Laplace's Equation in 3-Dimensions using superposition.
    The problem is that I don't know how to use orthogonality to solve for the constant when you have two Eigen functions.

    \frac{\partial ^{2}u}{\partial x^{2}}+\frac{\partial ^{2}u}{\partial y^{2}}+\frac{\partial ^{2}u}{\partial z^{2}}=0

    u(x,y,z)=v(x,y,z)+w(x,y,z)

    This is my attempt to get a solution for v(x,y,z).

    v(x,y,z)=A(x)B(y)C(z)

    The solution using the given homogeneous boundary conditions comes out to be:
    v(x,y,z)=\sum Ksinh(\lambda z)sin(n\pi x)sin(m\pi y)

    where n:1\rightarrow \infty and m:1\rightarrow \infty

    The remaining boundary condition is:
    v(x,y,1)=1

    Which gives:
    v(x,y,1)=\sum Ksinh(\lambda)sin(n\pi x)sin(m\pi y)=1

    How do I use orthogonality to find K when I have two eigen functions in the solution?
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    Re: Laplace's Equation in 3-Dimension (Orthogonality and Eigen function)

    Hey bilalsaeedkhan.

    For the orthogonality condition, do you use the L^2 inner product and if so what are the limits of the integral?
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    Re: Laplace's Equation in 3-Dimension (Orthogonality and Eigen function)

    Quote Originally Posted by chiro View Post
    Hey bilalsaeedkhan.

    For the orthogonality condition, do you use the L^2 inner product and if so what are the limits of the integral?
    I don't know what the L^2 inner product is so cant say anything about that. But I was able to solve it though. Thanks for taking a look at it.
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    Re: Laplace's Equation in 3-Dimension (Orthogonality and Eigen function)

    Just for your future reference, we can establish orthogonality between functions if we get <f,g> = 0 where <f,g> = Integral fg over the right interval.

    Different inner products can have different intervals of integration and it depends on what your full space is (for example on space would be all functions on the interval [-1,1] while another might be all of R and subsequently every space has its own issues with how to construct a basis).

    So with this if you have an inner product defined in terms of an L^2 integral inner product, you can check whether they are orthogonal and this is the kind of thing that happens in Fourier Analysis.
    Thanks from bilalsaeedkhan
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