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I have ran into the following integral equation as part of my phd research project, trying to enforce a boundary condition of a parabolic pde problem.

For $\xi = (\alpha\theta)^{1/\alpha}$ and for all $\theta>0$.

I have the following equality $$\int\limits_0^\infty g(\kappa, x_0) \exp{\left[-\frac{\left(\xi + \kappa \right)^2}{2\alpha\theta}\right]} d\kappa % = \exp{\left[-\frac{\left(\xi - x_0\right)^2}{2\alpha\theta}\right]}%. $$ If we assume $g(\kappa, x_0)$ to be a smooth function dependent on $\kappa$ and $x_0$. Is there a way to find $g(\kappa, x_0)$ ? I do not have a broad enough background in integral equations, so will really appreciate some guidance.

I was also wondering if the following technique can be applied here Power series solution for integral equations wikipedia

Thank you.

PS: I also posted on Math Stack Exchange Post but my post did not generate much interest from others.

Edit 1: This condition can also be presented as follows $$ \int\limits_0^\infty g(\kappa)\exp{\left(-\kappa^2/(2\alpha \theta) -\xi\kappa/(\alpha\theta)\right)} d\kappa = \exp{\left(-\frac{x_0^2 - 2\xi x_0}{2\alpha\theta}\right)}$$ Can it be interpreted as some form of convolution ?

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    $\begingroup$ Eight hours? Incredible. $\endgroup$
    – Asaf Karagila
    Commented Jun 1, 2016 at 8:48
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    $\begingroup$ I currently do not have a lead to pursue, so I am wondering if someone here can give me a clue, instead of booing my post like you. After all the purpose of this forum is for mathematicians to seek each others help, is it not? Did have anything relevant to my problem to add. I am currently an amateur probabilistic and I do not have any experience in integral equations, which is a rather wide field in its own, why is it wrong to seek someone 's help in getting some directions where to look? $\endgroup$ Commented Jun 1, 2016 at 9:09
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    $\begingroup$ What Asaf means is that you should not crosspost here without waiting at least several days for answers on MSE. (I do not have an opinion about whether your question is on topic here in principle.) $\endgroup$ Commented Jun 1, 2016 at 12:06
  • $\begingroup$ @ComicBookGuy What is assumed about $x_0$? $\endgroup$
    – Andrew
    Commented Jun 4, 2016 at 4:57
  • $\begingroup$ @Andrew, $x_0$ is a constant, which came into the picture as my original PDE, had an initial condition in form of a Dirac delta function. $\endgroup$ Commented Jun 4, 2016 at 6:08

1 Answer 1

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There is no unique solution to the integral equation $$\int\limits_0^\infty g(\kappa, x_0) \exp{\left[-\frac{\left(\xi + \kappa \right)^2}{2\alpha\theta}\right]} d\kappa % = \exp{\left[-\frac{\left(\xi - x_0\right)^2}{2\alpha\theta}\right]},\;\;\xi=(\alpha\theta)^{1/\alpha}.$$

Any function of the form $$g(\kappa,x_0)=f(\kappa)\exp{\left[\frac{\left(\xi + \kappa \right)^2}{2\alpha\theta}\right]}\exp{\left[-\frac{\left(\xi - x_0\right)^2}{2\alpha\theta}\right]}$$ will do, provided that $$\int_0^\infty f(\kappa\,)d\kappa=1$$

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  • $\begingroup$ Of course that makes so much sense thank you. $\endgroup$ Commented Jun 4, 2016 at 9:40
  • $\begingroup$ In your answer g is also dependent on theta, what if g is only allowed to depend on kappa? Can you please shed some light on that case as well. $\endgroup$ Commented Jun 5, 2016 at 2:08
  • $\begingroup$ if you insist that $g$ should be independent of $\alpha,\theta$ (and thereby also independent of $\xi$) then for $x_0<0$ you could take $g=\delta(\kappa+x_0)$, a Dirac delta function, but then this would violate your earlier requirement that $g$ depends smoothly on $\kappa$ and $x_0$. $\endgroup$ Commented Jun 5, 2016 at 10:52
  • $\begingroup$ I know $x_0$ is strictly positive in my problem, if that helps your analysis. $\endgroup$ Commented Jun 5, 2016 at 10:54

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