# Given $g$ find an $f$ which is solution for $L f = g$. How do I do this?

I am learning about Stochastic processes. To characterize uniqueness of solutions to a given Stochastic differential equation, I need to find for each continuous function $g :\Bbb{R}^2_+ \to \Bbb{R}$ a function $f:\Bbb{R}^2_+ \to \Bbb{R}$ such that

$$-\gamma f + (\partial_x f) (y-x) + (\partial_y f) (x-y) + \frac{1}{2} \bigg((\partial_{xx} f) x + (\partial_{yy} f) y\bigg) = g$$

This is I believe an inverse problem which has to do with spectral techniques such as proving that $\gamma \in \rho(L)$ where

$$L f = (\partial_x f) (y-x) + (\partial_y f) (x-y) + \frac{1}{2} \bigg((\partial_{xx} f) x + (\partial_{yy} f) y\bigg)$$

Maybe this can be solved more straightforwardly by finding a Green's function $G(x,y,\tilde{x}, \tilde{y})$ that satisfies $$-\gamma G - L G = \delta_{x -\tilde{x}} \delta_{y - \tilde{y}}$$

or something similar and then defining

$$f = G*g.$$

How can we find such an $f$? Does it exist? Is there any good reference on the literature for this subject?

Note: Here $(\partial_x f) (y-x)$ means $(\partial_x f(x,y)) \cdot (y-x)$

• What do you mean by $fx$ and $fy$ in your notation? And is $\gamma$ fixed or not? – demitau Oct 12 '15 at 12:44
• @demitau, I edited the question, is it clearer? we are multipling the functions $f$ and $x$. $\gamma$ is positive and fixed. – Conrado Costa Oct 12 '15 at 13:06
• Not completely. You are multyplying $f$ by $x$ after the differentiation (so you have $x\cdot \partial_{xx} f$) or before ($\partial_{xx}(f\cdot x)$)? – demitau Oct 12 '15 at 13:10
• Ohh yes, you are right, it was quite confusing. we differentiate first and then multiply the functions. so we have $x \cdot \partial_{xx} f$ I edited it again. Maybe now it's better. – Conrado Costa Oct 12 '15 at 13:19
• It looks a bit like a trial-and-error exercise. Did you try to guess $f$ directly? – demitau Oct 12 '15 at 13:23