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Thread: Finding Green's Function for ODE

  1. #1
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    Finding Green's Function for ODE

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

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

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

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

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

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

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

    3. $\displaystyle G(x, s)$ continuous at $\displaystyle x = s$
    $\displaystyle 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,

    $\displaystyle
    b_1(s) = a_2(s) (\cot s - \cot R)
    $

    4. Jump Condition
    $\displaystyle 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
    $\displaystyle
    -a_2(s)(\sin s + \cos s \cot R) - b_1(s)\cos s = 1
    $

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

    $\displaystyle \therefore a_2(s) = -\sin s$

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

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

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

    \sin s \frac{\sin(x - R)}{\sin R} & x > s
    \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; Dec 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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