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Green Function Question

  1. Jan 23, 2007 #1
    Hey,

    I am trying to find a GF for the function:

    [tex] y''+\frac{1}{24}y=f(x)[/tex]

    The function is bounded by:

    [tex]y(0)=y(\pi)=0[/tex]

    I have followed a math textbook that goes through the exact process for the function:

    [tex] y''+k^2y=f(x)[/tex]

    and have found a nice looking general solution:

    [tex]G(x,x')=-\frac{sin(x/4)sin(\frac{1}{4}[\pi-x'])}{\frac{1}{4}sin(\pi/4)}[/tex]

    for x<x'

    and

    [tex]G(x,x')=-\frac{sin(x'/4)sin(\frac{1}{4}[\pi-x])}{\frac{1}{4}sin(\pi/4)}[/tex]

    for x>x'


    Now, here is my problem:

    I need to find y(x) for f(x)=sin(x)

    This sounds easy right, just use the Green function, multiply by f(x) and integrate over the boundary 0 to pi.

    The problem is that the integral only converges for k^2 is some integer.
    For non integer k^2 it seems to diverge...not good as I have k^2=1/24

    I was just wandering if anyone has any suggestions????

    Thanks!
     
  2. jcsd
  3. Jan 23, 2007 #2

    Clausius2

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    I...I don't see how can diverge. It's a multiplication of trigonometric functions isn't it?. How can that integral diverge?.
     
  4. Jan 23, 2007 #3
    hmmmmm....maybe its just my formula table....but when you integrate there is a factor of nx on the denominator and when you plug in the limits the zero gives you infinty.
     
  5. Jan 23, 2007 #4

    HallsofIvy

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    What formula is that? If you are integrating sin(x/4)f(x), you certainly won't get any "x" in the denominator from the sine!
     
  6. Jan 23, 2007 #5
    Sorry, I figured it now...you are right. I carried an x' through into the denominator, but it should only have been the coefficient of x. That was the origin of the infinity. Thanks, for your assistance!
     
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