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Laplace's equation & Fourier series - I can use cos or sin?

  1. Nov 11, 2012 #1
    1. The problem statement, all variables and given/known data

    This is a question related to finding the velocity field of an incompressible fluid in a square pipe with sides at y = ±(a/2) and x = ± (a/2).

    It comes down to solving a homogenous equation which is also Laplace's equation

    [itex] \frac {δ^2 w(x,y)^H}{δ x^2} + \frac {δ^2 w(x,y)^H}{δy^2} = 0 [/itex]

    using separation of variables, and combining this with the particular solution (which is only a function of y!) and using the boundary conditions;

    w(a/2,y) = w(-a/2, y) = 0
    w(x, a/2) = w(x, -a/2) = 0

    And

    [itex] w(x,y) = w(y)^P + w(x,y)^H = w(y)^P + X(x)Y(y)[/itex]

    representing the homogenous and particular solutions.

    I let Y(y) be the trigonemtric function (this is mandatory) from Laplace's equation, and the boundary conditions impose that Y(a/2) = Y(-a/2) = 0.

    [itex] Y(y) = Csin(Kx) + Dcos(Kx) [/itex]

    Where C, D, K are constants (K arises from the constant used in Laplace's equation).

    This implies.


    [itex] Csin(Ka/2) + Dcos(Ka/2) = 0 [/itex]
    [itex] -Csin(Ka/2) + Dcos(Ka/2) = 0 [/itex]


    My issue is that K can be resolved to having two values, either;

    [itex] K = \frac{(2m+1)\pi}{a} [/itex]

    m runs from 0 to infinity

    or

    [itex] K = \frac{2m\pi}{a} [/itex]

    m runs from 1 to infinity.

    And in the first case, C = 0. In the second case D = 0.

    My confusion is the simple question: Which is the right one? I go on to add to w(x,y) a linear superposition of solutions to Laplace's equation which uses the value of K throughout, however does this produce the same answer? How can I know?

    If I follow through with C = 0 or D = 0, then I get two very different answers for each - and I can't tell if they're the same or not. (The mathematics is long and complicated that I would rather not post it here).


    However, if I were to change my co-ordinate system (by shifting the y-axis down by a/2), the boundary conditions become w(x,0) = w(x,a). This is exactly the same situation, except that I find that D = 0 is a mandatory requirement, and that

    [itex] K = \frac {n\pi}{a} [/itex]

    So I get a definite solution. This is caused by changing the co-ordinate system, so surely either of the ways above (D = 0, or C = 0) produce the same answer?
     
    Last edited: Nov 11, 2012
  2. jcsd
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