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Math Help - Darn those Green's Functions

  1. #1
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    Darn those Green's Functions

    Consider the differential operator Lu = u_{xx} + u_x on (-1,0) with boundary conditions u'(-1)-3u(-1)=u'(0)=0

    Find the Green's function for L with the boundary conditions

    My attempt:
    G_{xx}(x,\xi) + G_x(x,\xi) = \delta(x - \xi)
    G_x(-1,\xi)-3G(-1,\xi)=G(0,\xi) = 0

    For x\ne\xi
    G_{xx}(x,\xi) + G_x(x,\xi) = \0
    G(x,xi) = A(\xi) e^{-x} + B(\xi)

    For x < \xi
    G_x(-1,\xi) - 3G(-1,\xi) = -A(\xi)e - 3A(\xi)e - 3B(\xi) = 0
    A(\xi) = -\frac{3}{4e} B(\xi)
    \Rightarrow G(x,\xi) = \left(-\frac{3}{4e} e^{-x} + 1\right) B(\xi)\; \forall x < \xi

    For x > \xi
    G_x(0,\xi) = -A(\xi) = 0
    \Rightarrow G(x,\xi) = B(\xi)\; \forall x>\xi

    Green's functions are continuous and so for x = \xi
    \left(-\frac{3}{4e} e^{-\xi} + 1\right) B(\xi) - B(\xi) = 0
    B(\xi) = 0

    This can't be right.
    Last edited by mr fantastic; August 3rd 2010 at 07:52 PM.
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  2. #2
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    My attempt #2:
    G_{xx}(x,\xi) + G_x(x,\xi) = \delta(x - \xi)
    G_x(-1,\xi)-3G(-1,\xi)=G(0,\xi) = 0

    For x\ne\xi
    G_{xx}(x,\xi) + G_x(x,\xi) = 0
    G(x,xi) = A(\xi) e^{-x} + B(\xi)

    For x < \xi
    G_x(-1,\xi) - 3G(-1,\xi) = -A_1(\xi)e - 3A_1(\xi)e - 3B_(\xi) = 0
    A_1(\xi) = -\frac{3}{4e} B_1(\xi)
    \Rightarrow G(x,\xi) = \left(-\frac{3}{4e} e^{-x} + 1\right) B_1(\xi)\; \forall x < \xi

    For x > \xi
    G_x(0,\xi) = -A_2(\xi) = 0
    \Rightarrow G(x,\xi) = B_2(\xi)\; \forall x>\xi

    Green's functions are continuous and so for x = \xi
    \left(-\frac{3}{4e} e^{-\xi} + 1\right) B_1(\xi) = B_2(\xi)

    So now I also want to ensure that the first derivative has the proper jump, i.e.
    \lim_{\epsilon \to 0} \int_{\xi - \epsilon}^{\xi + \epsilon}\! G_{xx}(x,\xi)\, dx = 1
    0 - B_1(\xi) \frac{3}{4} e^{-\xi - 1} = 1 \Rightarrow B_1(\xi) = -\frac{4}{3} e^{\xi + 1}

    \Rightarrow 1 -\frac{4}{3} e^{\xi + 1} = B_2(\xi)

    Therefore,
    G(x,\xi) = \begin{cases} e^{\xi - x} - \frac{4}{3}e^{\xi + 1} & x<\xi \\<br />
1 -\frac{4}{3} e^{\xi + 1} & x > \xi<br />
\end{cases}

    Forgot that I was dealing with different constants.
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