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Green's Function Solution to ODE. Boundary Conditions Problem.

  1. Sep 21, 2009 #1
    Use Green's Functions to solve:

    [tex]\frac{d^{2}y}{dx^{2}} + y = cosec x[/tex]

    Subject to the boundary conditions:

    [tex]y\left(0\right) = y\left(\frac{\pi}{2}\right) = 0[/tex]


    [tex]\frac{d^{2}G\left(x,z\right)}{dx^{2}} + G\left(x,z\right) = \delta\left(x-z\right)[/tex]

    For [tex]x\neq z [/tex] the RHS is zero so the complementary solution consists of sinx and cosx terms but with different superpositions on either side of x = z since the first derivative is required to have a discontinuity there.

    Assume a form [tex] G\left(x,z\right) = A\left(z\right)sinx + B\left(z\right)cosx [/tex] for x < z

    and [tex]G\left(x,z\right) = C\left(z\right)sinx + D\left(z\right)cosx [/tex] for x > z

    I'm just following an example from a book that then continues to state that according to the boundary conditions B and C are equal to zero, seems simple but I'm unclear as to how they achieve this. I don't see how to apply this and discern B and C being zero, I seem to just find A,B,C and D all zero. Also how would one proceed in the case of the boundary conditions stating G and its first derivative equal to zero and achieve A and B zero?

  2. jcsd
  3. Sep 21, 2009 #2
    Okay, for the first one is it the case that x is valid over the interval [tex]\left[0<x<\frac{\pi}{2}\right][/tex], therefore z is valid over the same interval? That seems to solve my problem with the maths.

    For the second boundary case would it be that it is equally possible to say A = B = 0 as it is to say C = D = 0.

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