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Math Help - Green's Function

  1. #1
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    Green's Function

     y'' - y = f(x)
    Subject to  y (\pm \infty) = 0

    I started by looking for a Green's Function of the form

     G_{xx} (x, \xi) - G(x, \xi) = \delta(x - \xi)

    For all  x \ne \xi, \delta(x - \xi) = 0

    And thus we have two cases.

    Case 1:  x < \xi

     G_{xx} (x, \xi) - G( x, \xi) = 0
    Implying that  G(x, \xi) = c_1 e^x + c_2 e^{-x}
    Applying our left-hand BC,
     G( -\infty, \xi) = 0 = c_2 e^{\infty}
    So to ensure that our solution is bounded at either extreme, we insist that  c_2 is 0.

    Case 2:  x > \xi

     G_{xx} (x, \xi) - G(x, \xi) = 0
    Which also reduces to the form  G(x, \xi) = c_3 e^x + c_4 e^{-x}
    Applying our right-hand BC,  G( \infty, \xi) = 0 = c_3 e^{\infty}
    So similarly, we insist that  c_3 = 0 .

    Combining our results, we have then that

    G(x, \xi) =
     c_1 e^x;  x < \xi
     c_4 e^{-x}; x > \xi

    However, we want to make sure that there's continuity at  x = \xi and a finite jump of 1 in the first derivative, so we get that

     c_1 e^{\xi} = c_4 e^{-\xi}
    And
     - c_4 e^{-\xi} - c_1e^{\xi} = 1

    Which can be combined to give
     c_4  = \frac{e^{\xi}}{-2} and  c_1 = -\frac{1}{2 e^{\xi}} .

    All I'm really looking to do is make sure that I've correctly found the Green's function up to this point.
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  2. #2
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    Let u := \dot{y}-y. Then your equation is \dot{u}+u=f(x), which has solution u(x) = e^{-x}(\int^x e^s f(s) \,\mathrm{d}s + A). Plugging back in you get y(x) = e^x(\int^x e^{-s}u(s) \,\mathrm{d}s + B).
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  3. #3
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    I'm not entirely sure I understand. I'm specifically trying to solve the ODE using this method, and I was wondering if, up until this point, I have worked it through correctly.
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  4. #4
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    Your Green function is correct, I am pretty sure.
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  5. #5
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    Alright, so, with an application of Green's Theorem and a little reduction, I should end up with

     u(x) = \frac{-1}{2}(\int_{ -\infty}^{x} \frac{f(\xi) e^{\xi}}{e^x} d \xi + \int_{x}^{\infty} \frac{f(\xi) e^x}{e^{\xi}} d \xi)

    Where  u(x) is a solution to the ODE?
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  6. #6
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    I think maybe it should be u(x) = \frac{-1}{2}(\int_{ -\infty}^{x} \frac{f(\xi) e^{\xi}}{e^x} d \xi {\color{red}-} \int_{x}^{\infty} \frac{f(\xi) e^x}{e^{\xi}} d \xi); I'll double-check.

    edit: okies.

    Let u(x) = \frac{-1}{2}(e^{-x}\int_{-\infty}^x e^{\xi}f(\xi)\,\mathrm{d}\xi - e^x \int_{x}^{\infty} e^{-\xi}f(\xi) \,\mathrm{d}\xi)

    then u'(x) = \frac{-1}{2}(-e^{-x}\int_{-\infty}^x e^{\xi}f(\xi)\,\mathrm{d}\xi + e^{-x}e^xf(x) - e^x \int_{x}^{\infty} e^{-\xi}f(\xi) \,\mathrm{d}\xi - e^xe^{-x}f(x)) = \frac{-1}{2}(-e^{-x}\int_{-\infty}^x e^{\xi}f(\xi)\,\mathrm{d}\xi - e^x \int_{x}^{\infty} e^{-\xi}f(\xi) \,\mathrm{d}\xi )

    and u''(x) = \frac{-1}{2}(e^{-x}\int_{-\infty}^x e^{\xi}f(\xi)\,\mathrm{d}\xi - e^{-x}e^xf(x) - e^x \int_{x}^{\infty} e^{-\xi}f(\xi) \,\mathrm{d}\xi - e^xe^{-x}f(x)). This should satisfy the DE.
    Last edited by maddas; April 10th 2010 at 12:08 PM.
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  7. #7
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    I see why that negative sign should be there, but I don't see how it comes about when working through to the solution. Did I solve for the coefficients incorrectly?
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