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In this nice tutorial, it is just stated that generally the Green's Function of the linear homogeneous differential equation (eq1 in the tutorial)

$$ L y_n(x) + \lambda_n w(x)y_n(x) = 0 $$

can be written in terms of the Eigenfunctions $y_n(x)$ and Eigenvalues $\lambda$ as (eq9 in the tutorial)

$$ G_{\lambda}(x,x') = \sum\limits_n \frac{y_n(x)y^{*}(x')}{\lambda - \lambda_n}$$

The author writes that he derived it in class, but this does not help me.

So can somebody explain the proof to me, or give some hints how it works? I would appreciate answers at about the same level as the tutorial.

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1 Answer

up vote 1 down vote accepted

Use the fact that the eigenfunctions $y_n$ form an orthogonal basis of the linear operator $L$ that satisfy

$$\sum_n w(x) y_n(x) y_n^* (x') = \delta(x-x') $$

where the $*$ denotes complex conjugation.

Also, the Green's function may be expressed in terms of the orthogonal basis functions:

$$G(x,x') = \sum_n a_n(x') y_n(x)$$

Consider the defining equation for the Green's function

$$L \, G(x,x') + \lambda w(x) G(x,x') = \delta(x-x')$$

where

$$L \, G(x,x') = \sum_n a_n(x') L \,y_n(x) = - \sum_n a_n(x') \lambda_n w(x) \,y_n(x)$$

and the defining equation may be rewritten as

$$\sum_n a_n(x') (\lambda - \lambda_n) w(x) y_n(x) = \sum_n w(x) y_n(x) y_n^* (x')$$

or

$$\sum_n [a_n(x') (\lambda - \lambda_n) - y_n^* (x')] w(x) y_n(x) = 0$$

Because the $y_n$ form an orthogonal basis set, we can consider each term in the sum individually and conclude that

$$a_n(x') = \frac{y_n^*(x')}{\lambda-\lambda_n}$$

and therefore

$$G(x,x') = \sum_n \frac{y_n(x) y_n^*(x')}{\lambda-\lambda_n}$$

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Thanks for this nice and very clear step by step derivation, it is exactly what I was looking for. –  Dilaton Feb 3 '13 at 13:07
    
@Ron Gordon So what is $\lambda$ here? Who determines that? How does this method match with the method you said here - math.stackexchange.com/questions/284870/… –  user6818 Apr 11 '13 at 20:00
    
@user6818: $\lambda$ defines the particular differential operator that has that particular Green's function. The $\lambda_n$ are eigenvalues of that differential operator, i.e. allowed values of $\lambda$ when the result of applying the differential operator to a function is a scalar times that function. –  Ron Gordon Apr 11 '13 at 20:10
    
And how does this version of the Green's function compare to the one I found for a particular differential operator? Note that here, $\lambda = 1/4$. The sum over eigenfunctions/eigenvalues should agree with the specific results in that problem, although I have not worked that out. –  Ron Gordon Apr 11 '13 at 20:18
    
@RonGordon So what are the $\lambda_n$ in that other question? (..aren't both the $\lambda _n$ also 1/4 for both the eigen functions and then here the denominator becomes 0..) And how will you get a split function from this formalism? In that other question how does $y(0)$ condition become the $G(x<x')$ condition and vice-versa for $y(\pi)$? I don't get this translation. –  user6818 Apr 11 '13 at 20:28
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