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Question about separation of variables

  1. Jun 29, 2017 #1
    I've solving some separation of variables exercises, and I have a doubt when it comes to the Laplacian

    $$
    u_{xx} +u_{yy} =0
    $$
    I usually have a rectangle as boundary conditions, so I use the principe of superposition and arrive to
    $$
    \dfrac{X''(x)}{X(x)} = - \dfrac{Y''(y)}{Y(y)} = - \lambda
    $$
    Clearly, lambda can be both negative and positive as there's no difference between ##X(x)## and ##Y(y)## in the sense that the problem is symmetric. Then, depending in the boundary conditions I usually chose ##\lambda## so I can use a fourier series with real scalars in the variable in which I have my boundary conditions different from zero (is it's in both then I solve search the solution adding solutions to different problems where only one side is not homogeneous).

    My question is...
    I might chose ## \lambda ## of the opposite sign and I'll also arrive to a solution, but this will have terms more difficult to compute, I'm right?

    I mean

    If my solution is, for example

    $$
    \sum_{n=1}^{\infty} \left( a_{n} \sin (nx) + b_{n} \cos(nx) \right) \left( c_{n} \sinh (ny) + d_{n} \cosh (ny) \right)
    $$
    Should I be able to express it as
    $$
    \sum_{n=1}^{\infty} \left( a'_{n} \sinh (nx) + b'_{n} \cosh (nx) \right) \left( c'_{n} \sin (ny) + d'_{n} \cos (ny) \right)
    $$
    and the choice of ##\lambda## sign is only for convenience?

    A more concrete example. Given the boundary conditions
    $$ u_{x} (0,y) = u_{x} ( \pi , y) =0 \qquad 0 < y < \pi $$
    $$ u_{x} (x,0) =0 \qquad u_{x} ( x, \pi ) = f(x) \qquad 0 < x < \pi $$
    I chose ##\lambda## sign in order to be able to make a fourier series in ##x##, but I guess I should be able to chose it in the other sense and make a fourier series in ##y## and one with hyperbolic sine and hyperbolic cosine in ##x##. I've done a complex analysis course, so I don't worry if the proof uses things like
    $$
    \dfrac{ \sinh (iz) }{ i } = sin(z) \qquad \cosh (iz) = cos(z)
    $$
     
  2. jcsd
  3. Jun 29, 2017 #2

    Orodruin

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    No. The point is that you need to adapt to the boundary conditions. You need homogeneous boundary conditions in order to be able to apply the Sturm-Liouville theorem. If you have inhomogeneous boundary conditions, you can split the problem in two separate problems which have homogeneous boundary conditions in one direction each and write the solution as a superposition of the solutions to these new problems.

    The sinh and cosh functions do not make a complete basis.
     
  4. Jun 29, 2017 #3
    Oh, thank you very much. I supposed they could do it but wasn't really sure.
    They are not dense of they're only not orthonormal?
    I will check that myself.
     
  5. Jun 29, 2017 #4

    Orodruin

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    These are two separate properties. The SL theorem tells you that a the solutions to a regular SL problem are both dense and orthogonal, but in general you can easily find a basis that is dense but not orthogonal or vice versa.
     
  6. Jul 8, 2017 #5
    Yes, I may have phrases it wrongly. I will have to study Sturm Liouville Theorem. I know about basis from functional analysis course, what I tried to ask is why they failed to be a complete orthonormal set.
     
  7. Jul 8, 2017 #6

    Orodruin

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    So the point is that if you have two eigenvectors with the same eigenvalue, then any linear combination of those is also an eigenvector with the same eigenvalue. If the original two are orthogonal, the new is generally not orthogonal to the old ones.
     
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