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Need to find Green Function to solve ODE

  1. Oct 8, 2006 #1
    Hi,

    I have a basic ODE:

    [tex]y''(x)+\frac{1}{4}y'(x)=f(x)[/tex]

    on 0<x<L

    With Boundary conditions:

    [tex]y(0)=y(L)=0[/tex]

    For which I would like to construct a Green Function.

    Rather than just plain ask for help, I'll show you what I've been thinking and maybe someone wiser can help/correct me:

    We must solve:

    [tex]G''(x,x')+\frac{1}{4}G'(x,x'')=\delta(x-x')[/tex]

    Since G vanishes at the boundaries we can expand as a fourier sine series:

    [tex]G(x,x')=\sum_{n=1}^\infty \gamma_n sin\frac{n\pi x}{L}[/tex]

    and:

    [tex]\delta(x-x')=\sum_{n=1}^\infty A_n(x')sin\frac{n\pi x}{L}[/tex]

    Integrating the delta function I get:

    [tex]A_n(x')=\frac{2}{L}sin\frac{n\pi x}{L}[/tex]

    I take the first and second derivative of the equation for G

    and plug everything into the differential equation for G:

    [tex]\sum_{n=1}^\infty \gamma_n (\frac{-n^2 \pi^2}{L^2})sin\frac{n\pi x}{L}+\frac{1}{4} \sum_{n=1}^\infty \gamma_n (\frac{n \pi}{L})cos\frac{n\pi x}{L}=\sum_{n=1}^\infty \frac{2}{L}sin\frac{n\pi x'}{L}sin\frac{n\pi x}{L}[/tex]

    Now I am not sure what to do from here to get the required Green function...can someone guide me please?

    :confused:
     
  2. jcsd
  3. Oct 9, 2006 #2

    HallsofIvy

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    It's not clear to me why you would use Fourier series. For all x except x', the Green's function satisfies [itex]G"+ \frac{1}{4}G'= 0[/itex] which has general solution
    [tex]G(x)= A+ Be^{\frac{x}{4}}[/tex]
    The Green's function must be of the form
    [tex]G(x, x')= \left{\begin{array}{c}A+ Be^{\frac{x}{4}} if 0\le x\le x' \\C+ De^{\frac{x}{4}} if x'\le x\le L\end{array}[/tex]

    A, B, C, D depend on x' and are chosen so that the boundary values are satisfied, the two "pieces" have the same value when x= x', and the difference in the derivative when x= x' (left side minus right) is 1: 4 conditions to determine 4 numbers.
     
  4. Oct 9, 2006 #3
    Ok, let me try and from some equations from the conditions:

    Condition 1.

    [tex]G(0,x')=0=A+Be^0[/tex]
    so A=-B

    Conditon 2.

    [tex]G(L,x')=0=C+De^{L/4}[/tex]

    so[tex]C=-De^{L/4}[/tex]

    Conditon 3:

    [tex]Be^{x'/4}=De^{1/4(x'-L)}[/tex]

    Condition 4.

    [tex]\frac{Bx'(x-L)}{x-L}+B=1[/tex]

    Does this look about right??

    Thanks!
     
    Last edited: Oct 9, 2006
  5. Oct 11, 2006 #4

    HallsofIvy

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    If, by "condition 1", "condition 2", etc. you mean
    "A, B, C, D depend on x' and are chosen so that the boundary values are satisfied, the two "pieces" have the same value when x= x', and the difference in the derivative when x= x' (left side minus right) is 1." that I mentioned before, then your equations for "condition 1" and "condition 2" are correct.
    For the third one, having the same value when x= x',
    [tex]A+ Be^{\frac{x'}{4}}= C+ De^{\frac{x'}{4}}[/tex]

    For the fourth, that the difference in left and right derivatives at x= x' be 1,
    [tex]\frac{B}{4}e^{{x'}{4}}- \frac{D}{4}e^{{x'}{4}}= 1[/tex]
     
  6. Oct 11, 2006 #5
    Thanks!

    Yes, thats exactly what I meant by 4 conditions, but I was just borrowing your phrase '4 conditions to determine 4 numbers', so if I have used the phrase incorrectly I apologize.

    I'll keep working on this.

    In the last condition is that supposed to be 4e or is it a typo and supposed to be e/4?
     
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