1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

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]

    Attempt:

    [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?

    Cheers
     
  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.

    Thanks.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook




Similar Discussions: Green's Function Solution to ODE. Boundary Conditions Problem.
Loading...