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I A somewhat conceptual question about Green's functions

  1. Dec 17, 2016 #1

    BiGyElLoWhAt

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    I just did a problem for a final that required us to use a green's function to solve a diff eq. y'' +y/4 = sin(2x)

    I went through and solved it and got a really nasty looking thing, but I checked it in wolfram and it works out. Now, my question is this:

    After I got the solution from my greens functions, I went through and tried to add in what would be the homogeneous solution, Asin(x/2) + Bcos(x/2) and apply the boundary conditions again to solve for A and B. Both came out to be zero. Is this always the case? Is my integrand of G*f the general solution and not just the particular solution? Was this just a coincidence? I have a few more problems I need to do, most of which are green's functions, and this would be handy to know. If it is the general solution, does it have to do with the fact that I used the homogeneous solution to obtain my Green's solution? (I'm not sure what you want to call it)
    Thanks
     
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  3. Dec 17, 2016 #2

    Orodruin

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    If you have a homogeneous linear differential equation with homogeneous boundary conditions, then the solution is trivial. The point of the Green's function method is to take care of the inhomogeneities.
     
  4. Dec 17, 2016 #3

    BiGyElLoWhAt

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    The diff eq wasn't homogeneous. It was equal to sin (2x). It was linear, and I had homogeneous boundary conditions, though. Y (0)=y (pi)=0
     
  5. Dec 17, 2016 #4

    Orodruin

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    Yes, but you took care of the inhomogeneity with the Green's function. You cannot expect to add a homogeneous solution to this if you have homogeneous boundary conditions because your Green's function is set up to handle the homogeneous boundary conditions.
     
  6. Dec 17, 2016 #5

    BiGyElLoWhAt

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    So is it reasonable to say that I turned a homogeneous solution into a general one? As in the homogeneous is buried within the resultant solution? I've been struggling to get the grand scheme of the green's function and it's solution, so to speak.
     
  7. Dec 17, 2016 #6

    Orodruin

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    Not really. Rather, you found the solution satisfying the boundary conditions without ever having to worry about splitting it into a homogeneous and a particular part. That distinction is arbitrary anyway as there is not one particular solution, but all particular solutions are related in that their difference is a homogeneous solution.

    Instead, you design your Green's function in such a way that it will let you help the particular solution that already satisfies homogeneous boundary conditions. If your problem has homogeneous boundary conditions, then you therefore never need to worry about finding a homogeneous solution.

    I think it is a common thing among students to struggle with Green's function, yet you have likely been using them since high school for computing the electric/gravitational potential of a collection of point particles - just that you never called them Green's functions.
     
  8. Dec 17, 2016 #7

    BiGyElLoWhAt

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    Hmmm. Ok, I think I understand now. Thank you.
     
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