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Math Help - [SOLVED] Solve initial value problem with Green's function

  1. #1
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    [SOLVED] Solve initial value problem with Green's function

    I need to solve the following intial value problem using Green's Function:

    (Ay)(x) = h(x), Ay:=d^2y/dx^2 + dy/dx, y(0) = y'(0) = 0

    where h is defined as the following:

    h(x) := { 0, when 0<=x<1
    { 1, when x >= 1

    Can someone help, im not really sure where to start

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  2. #2
    Super Member Rebesques's Avatar
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    Firstly, I must admit I cannot see why on earth someone would want to solve anything (...other than a two-point BVP...) by using the Green's function method.


    Ok, so we are looking for a symmetric function G(x,s) to satisfy

    \int_0^2G(x,s)h(s)ds=\begin{cases} 0, \ x<1 \\<br />
y(x),  \ x\geq 1\end{cases}

    where I took the liberty of setting [0,2] as the domain, but this is inessential as we can modify the Green's function and get the domain as large as we like.

    For x<1, we have h(x)=0, so y(x)=0 solves the IVP. and we have the liberty to define G any way we like.
    For x>1, the IVP y''+y'=1, \ y(0)=y'(0)=0 has the solution y(x)={\rm e}^{-x}+x-1. So we demand

    {\rm e}^{-x}+x-1=\int_1^2G(x,s)ds (1)

    Since G has to be symmetric, write {\rm e}^{-x}+x-1=\int_1^2{\rm e}^{-x}\phi(s)+(x-1)\psi(s)ds

    and observe that (1) is satisfied for \phi(s)=\frac{{\rm e}^{-s}}{{\rm e}^{-1}-{\rm e}^{-2}}, \ \psi(s)=2(s-1). Define G this way on 0\leq x\leq 2, \ 0\leq s\leq 2. So we get the solution of the IVP as

    y(x)=\int_0^2\left[\frac{{\rm e}^{-s-x}}{{\rm e}^{-1}-{\rm e}^{-2}} + 2(x-1)(s-1)\right]h(s) ds.
    Last edited by Rebesques; August 28th 2007 at 10:46 AM. Reason: bad eyesight
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