This question is something on the lines for what I am looking for, Estimates for parabolic vs elliptic PDE.
For certain elliptic PDE's, we can show the stochastic representation of a solution to the PDE looks somewhat similar to the Feynman-Kac formula as on wikipedia https://en.wikipedia.org/wiki/Feynman%E2%80%93Kac_formula, with the time variable replaced with a stopping time, call it $\tau$. Given the solution of an elliptic PDE, I can go ahead and take $\tau=t$ in the formula to recover the solution for a parabolic PDE as stated on the wiki page. Is there a method for me to go the opposite way: Starting from the solution to the parabolic PDE, can I convert this solution to that of an elliptic PDE.
For simplicity, consider that we have a parabolic PDE of the form: \begin{aligned} &\mathcal{A}u(t,x)+\frac{\partial u}{\partial t}(t,x)-q(t,x)u(t,x)=0,\\ &u(0,x)=f(x), \end{aligned} and the elliptic PDE \begin{aligned} &\mathcal{A}v(x)-r(x)v(x)=0,\\ &v(x)=f(x). \end{aligned}
Now I believe by recalculating the generator $\mathcal{A}$ that the derivative vanishes when we 'stop' the solution from the parabolic PDE at time $\tau$ to give the elliptic PDE equation. I haven't seen anything treating the two problems as one, although it seems as if they are very closely linked so I thought this should be possible. My question is: if I were very lazy and derived one of these PDE's, could I get the other one for free?
I would like to turn something like $$u(t,x)=\mathbb{E}[e^{-\int^T_t q(s,X_s)ds}f(X_T)|X_t=x]$$ into $$v(x)=\mathbb{E}[e^{-\int^\tau_0 r(s,X_s)ds}f(X_\tau)|X_0=x]$$