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Green's Function - modified operator

  1. Feb 28, 2015 #1
    Hi, I'm stuck with a question from one of my examples sheets from uni.

    The question is as follows:

    If G(x,x') is a greens function for the linear operator L, then what is the corresponding greens function for the linear operator L'=f(x)L, where f(x) =/=0?

    So I've started by writing L'G'(x,x')= delta(x-x') =f(x)LG'(x,x') from the definition of the greens function.

    Also, LG(x,x')=delta(x-x').

    Therefore, LG=f(x)LG'.

    I'm now stuck! I don't know where to go from here. I've tried to integrate over a certain range x=[a,b] and use the delta function properties but I've got nowhere!

    Many thanks in advance :)
     
  2. jcsd
  3. Feb 28, 2015 #2

    RUber

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    What do you get if you apply G to L'? Since f is not zero, you can divide by it.
     
  4. Feb 28, 2015 #3
    I get L'G=f(x)LG=f(x) delta(x-x')

    I also have LG'=delta(x-x')/f(x)

    Does any of that seem right?
     
  5. Feb 28, 2015 #4

    RUber

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    I think the goal is to write G' in terms of G and f. I think you should have to tools to do that now.
     
  6. Feb 28, 2015 #5
    I know that, but I can't seem to remove the dependence on either an L or L' operator. Do you have a hint on how to proceed?
     
  7. Feb 28, 2015 #6
    So I've tried integrating these over an arbitrary range to remove the delta function, is there anything I can do with the integral of LG' or L'G ??

    Thanks
     
  8. Feb 28, 2015 #7

    RUber

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    Is there any reason it wouldn't just be G/f?
     
  9. Feb 28, 2015 #8
    I did initially think that, but when you do L'(G/f), won't L' also need to act on f ?
     
  10. Mar 1, 2015 #9

    RUber

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    What if we were to multiply by a function of x' which, assuming L acts on x, will commute right through, like ##G'(x,x') = G(x,x') \frac 1 {f(x')} ##?
     
  11. Mar 1, 2015 #10
    Oh! I see. Thanks a lot for your help!
     
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