# PDE’s whose solutions can be presented using path integrals

It is well known that solutions of the Schroedinger equation and of the heat equation can be presented using path integrals: $$\psi(x,t)=\int K(x,t;y,0)\psi(y,0)dy,$$ where the kernel $K(x,t;y,0)$ is given by an integral of some expressions over all paths connecting $y$ and $x$.

Remark. In the former case this presentation exists only at the physical level of rigor as far as I know, while in the latter case it is mathematically rigorous.

I am wondering if there exist any other classes of PDE with the similar property (even at the physical level of rigor). In particular, does there exist such a presentation for solutions of the Dirac equation?

• Certainly various "variants" of the heat equations are OK: you can substitute any generator of a (sufficiently nice) Markov process for the Laplacian, and add a killing term (sometimes called a "Schrödinger potential"). This is, however, rather far from the Dirac equation. Sep 12 '18 at 19:54
• are you familiar with the Feynman-Kac formulas for elliptic and parabolic (see Oksendal)? Feynman-Kac formulas are an attempt to make sense of the path integrals. Sep 12 '18 at 20:42
• @ThomasKojar: No, I am not familiar with that. Actually I am not aware of any other examples of PDE with the relevant property. Can you elaborate on those examples you mentioned?
– makt
Sep 12 '18 at 20:53
• en.wikipedia.org/wiki/…
– user69208
Sep 13 '18 at 0:32

I believe the physicist's heuristic derivation the path integral formula for the Schroedinger equation can be generalized to arbitrary linear partial differential equations, although I don't have a clear idea of how to obtain the expression which the path integral is over. Seeing how vague this response is I hope I'm not just repeating things you already know.

Consider a time-dependent linear partial differential equation

$$\partial_t u (t) = L (t) u (t)$$

Since the equation is linear, $u (t_1)$ can be determined from $u (t_0)$ with a kernel $K_{t_0 \to t_1}$,

$$u (t_1, x_1) = \int K_{t_0 \to t_1} (x_0 \to x_1) u (t_0, x_0) dx_0$$

Iterating this, we can determine $u$ at a later time $t_2$:

$$u (t_2, x_2) = \iint K_{t_1 \to t_2} (x_1 \to x_2) K_{t_0 \to t_1} (x_0 \to x_1) u (t_0, x_0) dx_1 dx_0$$

And so on...

$$u (t_n, x_n) = \idotsint K_{t_{n-1} \to t_n} (x_{n-1} \to x_n) K_{t_{n-2} \to t_{n-1}} (x_{n-2} \to x_{n-1}) \dots K_{t_0 \to t_1} (x_0 \to x_1) u (t_0, x_0) dx_{n-1} \dots dx_0$$

Keeping $t_0, x_0, t_n, x_n$ fixed while taking the limit as the partition $(t_0, t_1, \dots, t_{n-1}, t_n)$ gets more and more fine should produce a path integral expression for $u (t_n, x_n)$, say

$$u (t, y) = \int_{\gamma (t) = y} F_{s,t} (\gamma) u (s, \gamma (0)) \mathcal {D} \gamma$$

where

$$F_{s,t} (\gamma) = \lim _{\substack {t_0 = s < t_1 < \dots < t_{n-1} < t_n = t \\ \max (t_{i+1} - t_i) \to 0}} \prod_{i=0}^{n-1} K_{t_i \to t_{i+1}} (\gamma (t_i) \to \gamma (t_{i+1}))$$

It may be necessary to add a term to this expression corresponding to the volume of the path spaces. Presumably, the next step is to use the fact that $t_{i+1}-t_i$ goes to zero to find a expression for the limit above only in terms of what appears directly in $L (t)$, without making reference to the kernel $K$. Unfortunately, I don't know what the general way of doing that is.