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A Green's functions issue

  1. Oct 8, 2018 #1

    joshmccraney

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    Gold Member

    Hi PF!

    The ODE $$g''(x) + (1-k^2)g(x) = f(x)\\ g(0) = y(\pi/3) = 0$$
    where ##f(x)## is a forcing function and ##k \in \mathbb N## is a constant has a Green's function via variation of parameters as
    $$
    G_L = \frac{L(y)R(x)}{W} : 0<x<y<\pi/3\\
    G_R = \frac{L(x)R(y)}{W} : 0<y<x<\pi/3
    $$

    with solutions $$L(x) = \frac{\sin(\beta x)}{\beta}\\
    R(x) = \sin\beta(x-\pi/3)\\
    W = \sin (\beta \pi/3)\\
    \beta = \sqrt{1-k^2}$$

    where ##L(x)## is the left sided solution and ##R(x)## is the right sided solution (obviously). I can verify this Green's function is correct for the given ODE. However, despite being correct, the inequalities look wrong. Why isn't the Green's function instead defined as
    $$
    G_L = \frac{L(y)R(x)}{W} : 0<y<x<\pi/3\\
    G_R = \frac{L(x)R(y)}{W} : 0<x<y<\pi/3
    $$
    since ##L## is the solution at ##x=0## and ##R## is the solution at ##x=\pi/3##?

    If We look at a similar problem, $$d_x^2 g = f(x)\\g(0)=g(1) = 0$$ we see its Green's function is $$G_L = l(x) r(y):0<x<y<\pi/3\\G_R=l(y)r(x):0<y<x<\pi/3$$
    where $$l(x) = -x\\ r(x) = (1-x)$$ which makes sense, since ##l(0)=0## and ##r(1)=0##.

    So why is there a difference (or am I not seeing something) in the two approaches?
     
    Last edited: Oct 8, 2018
  2. jcsd
  3. Oct 13, 2018 at 7:00 PM #2
    Thanks for the thread! This is an automated courtesy bump. Sorry you aren't generating responses at the moment. Do you have any further information, come to any new conclusions or is it possible to reword the post? The more details the better.
     
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