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](https://en.wikipedia.org/wiki/Integral_equation#Power_series_solution_for_integral_equations) 


Thank you.


PS: I also posted on [Math Stack Exchange Post](https://math.stackexchange.com/q/1807693/23874) 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 ?



Is there a way to solve this integral equation?
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
$$\int\_0^\infty f(\kappa\,)d\kappa=1$$


