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The Feynman-Kac formula (or Kolmogorov backward equation) describes a link between SDEs and PDEs. This answer introduces Ricci flows on Riemann manifolds which also seems to provide a mapping between PDEs to SDEs.

Are there other non-trivial examples besides the above mentioned where a mapping between SDEs and PDEs is possible?

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  • $\begingroup$ For example, the book Brownian Motion and Stochastic Calculus by Karatzas & Shreve contains a whole chapter devoted to this subject. $\endgroup$
    – Kurt G.
    Dec 10, 2022 at 7:42
  • $\begingroup$ Thank you, Kurt, for the reference! I had a quick look at the book: The connections presented are related to the Feynman-Kac formula (as far as I can tell) and consider general cases of elliptical and parabolic PDEs. I wonder if a connection had been found beyond this, e.g. hyperbolic PDEs. $\endgroup$
    – Jurek
    Dec 11, 2022 at 17:46
  • $\begingroup$ Hyperbolic PDEs can be ruled out as they typically describe wave phenomena while elliptic/parabolic are very closely related to convection of heat and diffusion of many particles. If not K&S, then maybe the good book by Olver might explain this a bit more. $\endgroup$
    – Kurt G.
    Dec 11, 2022 at 17:50
  • $\begingroup$ That's interesting and I am not sure I understand this: From a physics perspective I can follow the reasoning, but is there a mathematical reason why a mapping for hyperbolic PDEs to SDEs can be ruled out? Do you mean the Introduction to Partial Differential Equations by P. Olver? Thanks again, Kurt, for your replies and the book recommendations! $\endgroup$
    – Jurek
    Dec 11, 2022 at 20:40
  • $\begingroup$ Yes. That book. I am sure you will find a better non-physical explanation that I can produce here in those comments. Just a hint: look at the wave equation $\partial_t^2f-\partial_x^2f=0$ and flip the sign of time $t$. Symmetry! Processes from SDEs don't have that symmetry. Look up what the Generator $A$ of a Markov process is and how the transition probabilities satisfy the parabolic (never hyperbolic) equation $\frac{d}{dt}f=Af$. $\endgroup$
    – Kurt G.
    Dec 12, 2022 at 8:45

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