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Sorry for the title, I didn't find a better description (showing that I have no idea for the solution). Feel free to put in a better title and change the tags if you can grasp a view on the problem.

I am stuck with the equation:

$A(x,y,L) = A(x,y,0) \exp\left(-\alpha \int_0^L A(x,y,\xi)+A(\xi,y+x-\xi,x) \text{d}\xi\right)$

where $A:R^2\times R^+\rightarrow R^+$ is a smooth function with a support $(0,L)\times(q,0)\times R^+$ ($q\lt0$).

I differentiated with respect to $L$, getting the DE

$\frac{\partial}{\partial L} A(x,y,L) = -\alpha \left[ A(x,y,L)^2+A(\xi,y+x-L,x) A(x,y,L) \right]$

I know

$E_0(x)=\int_{-\infty}^\infty A(x,y,0)\text{d}y$

$E_L(x)=\int_{-\infty}^\infty A(x,y,L)\text{d}y$

as boundary conditions. Is it possible to calculate at least $A(x,y,0)$?

I tried Fourier transformation but couldn't really do much with the result ...

share|improve this question
    
I forgot: Numerical solutions would be ok, though I a personally interested on how one could do the analytically. –  elcron Jan 9 '12 at 9:13
    
A possible approach is to iterate derivatives with respect to $L$ and set $L=0$. In this way you will get a Taylor series solution as $A(x,y,L)=A(x,y,0)+A'(x,y,0)L+\ldots$ and $A'(x,y,0),\ A''(x,y,0),\ldots$ are given through $A(x,y,0)$. Having this you can do some steps beyond. –  Jon Jan 9 '12 at 13:53
    
Your derivative contains a bound variable (from the integral); by alpha-conversion, this is clearly incorrect. So you must have made a mistake in deriving your DE. –  Jacques Carette Jan 15 '12 at 2:03
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