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Given the Green's function $LG(x,c)=\delta(x-c)$:

$$G(x,c)=\begin{cases}{c \sinh x \over x(\cosh c -\sinh c)} &x\in[0,c]\\ {c \exp(c-x)\sinh c \over x(\cosh c -\sinh c)} & x\geq c\end{cases}$$

We want to find the solution $y(x)$ to the forced equation $Ly=F(x)$

where $$F(x) = \begin{cases}1 & x\in[0,x_0]\\0 &x>x_0\end{cases}$$

How is this done?

The solution should look like:

$$y(x) =\begin{cases} {\alpha\sinh x\over x}+1 &x\in [0, x_0]\\ {\beta e^{-x}\over x} &x>x_0 \end{cases}$$

Added: $Ly=-x^{-2}(x^2y')'+y$, where $" ' "$ denotes $d\over dx$

Thank you.

share|improve this question
As you're probably aware, the solution is the convolution of the Green's function with the inhomogeneity $F$. That doesn't seem to work out; the integrals seem to yield more complicated expressions, e.g. this. Are you sure there are no typos? It might help to tell us the operator $L$. Since $G$ isn't of the form $G(x-c)$, apparently $L$ isn't translation-invariant? –  joriki Nov 14 '11 at 6:11
@joriki: Thank you, $Ly=-x^{-2}(x^2y')'+y$. –  Alfonso Catoia Nov 14 '11 at 7:35
Question resolved! :) –  Alfonso Catoia Nov 14 '11 at 19:16
If you've solved your own question, you can answer it yourself and accept your answer, so that others can profit from it and it doesn't keep hanging around as an unanswered question. If you don't want to spend the effort to do that, it would be best to delete it. –  joriki Nov 14 '11 at 21:08

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