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Math Help - A pide

  1. #1
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    A pide

    Find a non-trivial solution in the first quadrant for:

    f+\frac{\partial f}{\partial x}+\frac{\partial f}{\partial y}=\int_{x}^{\infty}\int_{y}^{\infty} f(u,v)dvdu

    Hint: Can you think of a trial function of two parameters f(x,y;a,b) that would make the right side converge?
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  2. #2
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    Spoiler:
    I tried f(x,y) = \exp(-\alpha x-\beta y). This seems to work provided that \alpha-\tfrac1\alpha + \beta-\tfrac1\beta = 1. That has lots of solutions, for example \alpha=\beta=\tfrac14(1+\sqrt{17}).
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  3. #3
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    Quote Originally Posted by Opalg View Post
    Spoiler:
    I tried f(x,y) = \exp(-\alpha x-\beta y). This seems to work provided that \alpha-\tfrac1\alpha + \beta-\tfrac1\beta = 1. That has lots of solutions, for example \alpha=\beta=\tfrac14(1+\sqrt{17}).
    You have the right idea but your solution is not correct: if I back-substitute your particular solution into the DE, I do not get equality. Also, if you guys are interested, explain what the plot below has to do with this PIDE:
    Attached Thumbnails Attached Thumbnails A pide-solutionspace.jpg  
    Last edited by shawsend; October 19th 2009 at 02:07 PM.
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  4. #4
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    Quote Originally Posted by shawsend View Post
    You have the right idea but your solution is not correct: if I back-substitute your particular solution into the DE, I do not get equality.
    That was me being careless, trying to do something in a hurry late at night (excuses, excuses...). The equation \alpha-\tfrac1\alpha + \beta-\tfrac1\beta = 1 should have been 1-\alpha - \beta = \tfrac1{\alpha\beta}. That has no real solutions for positive \alpha and \beta. But it does have complex solutions with positive real parts (which are necessary for the integral to converge), for example \alpha = \tfrac12,\ \beta = \tfrac14(1+i\sqrt{31}). Unless I'm still making silly mistakes, that should give a complex solution to the problem. I don't have time to check it now, but I suppose that the real part of that ought to give a real solution.

    Spoiler:
    Complex solution: f(x,y) = \exp\bigl(-\tfrac12x-\tfrac14(1+i\sqrt{31})y\bigr).

    Real solution (?): f(x,y) = \exp\bigl(-\tfrac12x-\tfrac14y\bigr)\cos\bigl(\tfrac{\sqrt{31}}4y\bigr).
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  5. #5
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    That's the right relationship between a and b. So there are only complex solutions. So if the operator is linear:

    \left(1+\frac{\partial}{\partial x}+\frac{\partial}{\partial y}-I_2\right)f=0

    with f=u+iv a solution, then won't u and v be independent solutions as well?

    Anyone care to explain the plot? Where are the solutions in complex a-b space?
    Last edited by shawsend; October 20th 2009 at 05:54 AM.
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