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Hello,

I am wondering whether it is possible to convert the following integral equation to a partial differential equation.

Integral Equation here http://ima.epfl.ch/~lechen/images/integralEq.jpg

where $J_0(t,x)$ is some given nonnegative function and $\nu>0$ is a constant. It is clear $t\ge 0$.

The aim is to solve this equation. To convert it to PDE is just one possible way to solve it, since latter we can use the hopefully the fundamental solutions.

My current solution is

PDE http://ima.epfl.ch/~lechen/images/PDE.jpg

But I am not sure whether it is right or not.

Thanks for any comments or hints!

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1 Answer 1

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Of course. When $\nu=1$, if you apply the operator $\partial_t-\partial^2_{xx}$ to the last integral you obtain precisely $f(t,x)$ so the equation is $$ f_t - f_{xx} = (\partial_t-\partial^2_{xx}) J_0^2 + f.$$

EDIT: you seem to know already the answer, so I stop here :) You edited your question when I was writing my answer...

By the way, if you want to solve the PDE just set $ f(t,x) = e^{t} g(t,x) $ and the equation in $g$ is a homogeneous heat equation. This sounds like some textbook exercise, I musr say

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  • $\begingroup$ Thanks Peiro, I met a contradiction in my research. I am now debugging. I took my solution above for granted before. Now I am suspicious for that. That's why I am asking the question. :-) $\endgroup$
    – Anand
    Commented Sep 15, 2010 at 11:38
  • $\begingroup$ As for solving this problem, we can also use Fourier transform. :-) I want to make sure the integral equation and PDE are equivalent, with the above initial conditions. Thanks Prof. D'Ancona for your answer. :-) $\endgroup$
    – Anand
    Commented Sep 15, 2010 at 12:31
  • $\begingroup$ Dear Prof. D'Ancona, what kind of initial condition should we pose on this problem? I think it might be $f(0,x)=J^2_0(0,x)$. However, in some cases, it is reasonable to ask when $J_0(0,x)=\delta_0(x)$. Then in this case, what does it mean for $\delta^2_0$? Thank you for your help! $\endgroup$
    – Anand
    Commented Sep 16, 2010 at 13:12

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