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Math Help - Finding Green's Function for ODE

  1. #1
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    Finding Green's Function for ODE

    [Math]u''+u = f(x)[/tex]
    0<x<R
    u(0) = 0 u(R) = 0
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  2. #2
    Newbie mukmar's Avatar
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    1. Solve homogeneous equation

    u'' + u = 0 \Rightarrow u = a\cos x + b\sin x

    2. LG(x,s) = 0 for x \neq s

    x < s,  G_1(x,s) = a_1(s)\cos x + b_1(s) \sin x, G(0, s) = 0 \Rightarrow a_1(s) = 0

    x > s,  G_2(x, s) = a_2(s)\cos x + b_2(s) \sin x, G(R, s) = 0 \Rightarrow  a_2(s)\cos R + b_2(s) \sin R = 0

    We note R \neq n\pi to ensure linear independence of solutions as the Wronskian is \sin R.

    Hence  G_2(x, s) = a_2(s) (\cos x - \sin x \cot R)

    3. G(x, s) continuous at x = s
    G_1(s, s) = G_2(s, s) \Rightarrow b_1(s) \sin s = a_2(s) (\cos s - \sin s \cot R)

    Therefore Eq. 1,

    <br />
b_1(s) = a_2(s) (\cot s - \cot R) <br />

    4. Jump Condition
    G'(x)\Big{|}^{x = s^+}_{x = s^-} = 1 \Rightarrow -a_2(s)(\sin s + \cos s \cot R) - b_1(s)\cos s = 1

    Which gives:

    Eq. 2
    <br />
-a_2(s)(\sin s + \cos s \cot R) - b_1(s)\cos s = 1 <br />

    Now Eq. 1 and Eq. 2 together give:
    -a_2(s)(\sin s + \cos s \cot R + \cos s\cot s  - \cos s\cot R ) = 1

    \therefore a_2(s) = -\sin s

    Now using Eq. 1 again, we know b_1(s) = - \sin s(\cot s - \cot R) = \frac{\sin(s - R)}{\sin R}

    Also we can simplify: G_2(x, s) =  -\sin s(\cos x - \sin x \cot R) = \sin s \frac{\sin(x - R)}{\sin R}

    So putting this all together, we get our Green's function:
    G(x, s) = <br />
\begin{cases}<br />
\frac{\sin(s - R)}{\sin R} \sin x& x < s \\\\<br /> <br />
\sin s \frac{\sin(x - R)}{\sin R} & x > s<br />
\end{cases}


    From here you should be able to represent the solution as a sum of two integrals that utilize the Green's function. Good luck

    P.S. Let me know if I made any mistakes!
    Last edited by mukmar; December 8th 2010 at 11:37 AM.
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  3. #3
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    your answer is correct according to the solution

    many thanks!
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