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Greens function

  1. Feb 19, 2005 #1


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    I'm wondering if the general method I'm using for getting greens function solutions is wrong, because it's not giving me the right answer.

    Here's what I do. Starting with a differential equation:

    [tex] a(x) \frac{d^2 y(x)}{dx^2} + b(x) \frac{dy(x)}{dx} +c(x) y(x) = d(x) [/tex]

    the green's function solution must satisfy:

    [tex] a(x) \frac{d^2 g(x|\xi)}{dx^2} + b(x) \frac{dg(x|\xi)}{dx} +c(x) g(x|\xi) = \delta(x - \xi) [/tex]

    Now say we're working in the range [itex]0<x<a[/itex], and the boundary conditions specify either the function or it's first derivative is 0 at each of the endpoints. This means the green's function will satisfy the homogenous DE in the regions [itex]0<x<\xi[/itex] and [itex]\xi<x<a[/itex]. If the homogenous solutions are y1(x) and y2(x), then it will have the form:

    [tex] g(x|\xi) = \left\{\begin{array}{cc} A_1 y_1(x) + A_2 y_2(x)&0<x<\xi\\B_1 y_1(x) + B_2 y_2(x)&\xi<x<a\end{array} [/tex]

    To determine these four coefficients, we get two equations from the boundary conditions, another from requiring it to be continuous, and another from the following equation:

    [tex] \lim_{\epsilon \rightarrow 0 } \int_{\xi-\epsilon}^{\xi+\epsilon} \left[a(x) \frac{d^2 g(x|\xi)}{dx^2} + b(x) \frac{dg(x|\xi)}{dx} +c(x) g(x|\xi) \right] dx= \lim_{\epsilon \rightarrow 0 } \int_{\xi-\epsilon}^{\xi+\epsilon} \delta(x - \xi) dx = 1[/tex]

    Here's where I'm a little unsure. Do we always get to assume it will be continuous? And when it is continuous, does the above limit always reduce to:

    [tex] a(\xi) \left(\frac{d g(x|\xi)}{dx}|_{\xi+} - \frac{d g(x|\xi)}{dx}|_{\xi-} \right) = 1 [/tex]

    assuming a(x), b(x), and c(x) are continuous? If this is all right, I have some more questions because some of the solutions I'm getting using it aren't working.
  2. jcsd
  3. Feb 19, 2005 #2


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    One reason I'm confused is because that limit equation doesn't ensure that the left hand side of the DE actually is the delta function, it's just any function whose integral is 1 if you include [itex]\xi[/itex] in the bounds and 0 if you don't. For example, it might be something like:

    [tex] -\frac{x^2}{2 \xi} \delta'(x - \xi) [/tex]

    since that would also satisfy those conditions.

    Edit: actually I think I get this now. Since the only discontinuity on the left hand side comes from a kink in [itex]g(x|\xi)[/itex], which means we're effectively putting a ramp function in [itex]g(x|\xi)[/itex], then taking two derivatives can only give the delta function, no other possible functions that could integrate to 1 over a point and be 0 everywhere else. Is this the right reason?
    Last edited: Feb 19, 2005
  4. Feb 21, 2005 #3


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    You are basically correct. Yes, the Green's function (think potential, or waves...) is continuous at the source, while the gradient is not continuous -- think of E from a charged plate.

    With four parameters, the discontinuity in dG/dx and the continuity condition for G give two equations.Boundary and asymptotic conditions will give two more. E&M texts, potential theory texts, and boundary value problem texts all typically give considerable attention to Green's functions.
    Reilly Atkinson
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