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Let $(M,g)$ be a compact Riemannian manifold. Then the solution $u:M\times [0,\infty)\to \mathbb{R}$ of the heat equation $\partial_t u=\Delta_gu$ starting at $u_0\in C^{\infty}(M)$ is given by the path integral $$ u(x,t)=\int_{\gamma \in H_x} e^{-E(\gamma)/4t}u_0(\gamma(1))d\gamma, $$ where the integral is taken over the classical Wiener space $H_x\subset L^{2,1}([0,1],M)$ of finite energy paths $\gamma:[0,1]\to M$ starting at $x$ (i.e. $\gamma(0)=x$). Also, here $$ E(\gamma)=\int_0^1\Big|\frac{d\gamma}{ds}\Big|^2ds $$ is the Dirichlet energy of a curve in $(M,g)$ and the measure of integration is the Wiener measure. See this article for a survey of the above.

Question: Does the above path integral formula for $u$ hold for more general parabolic PDEs?

More precisely, let $L:C^{\infty}(M)\to C^{\infty}(M)$ be a second order elliptic operator (e.g. above we took $L=\Delta_g$).

Then can we write the solution $u:M\times [0,T)\to \mathbb{R}$ of the parabolic PDE $$ \partial_tu=Lu $$ with initial condition $u_0\in C^{\infty}(M)$ as the path integral $$ u(x,t)=\int_{\gamma \in H_x} e^{-S(\gamma)/4t}u_0(\gamma(1))D\gamma, $$ for some functional $S:H_x\to \mathbb{R}$ on the path space? What is $S$ in this case? Note that $S=E$ when $L=\Delta$.

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    $\begingroup$ I don't know the answer, but it seems worth mentioning that the author of the paper you cite is active on mathoverflow. So if you don't get an answer here, you might want to try it there. $\endgroup$ – MaoWao Nov 27 '18 at 0:11
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I am not sure if this is your point, but the path integral you mentioned is, at least to me, as if it refers to stochastic differential equations (SDE) and the Fokker-Planck equation.

To help explain the mechanism behind the Fokker-Planck equation, let us start from a naïve example. For the heat equation in 1-D Euclidean space $\mathbb{R}$, consider the following Brownian motion $$ {\rm d}X_t=\sigma\,{\rm d}W_t, $$ where the constant $\sigma$ denotes the thermal diffusivity, the stochastic process $W_t$ means the Wiener process. Here $X_t$ is also a stochastic process, tracing the position of a Brownian particle at time $t$. Further, denote by $u(t,x)$ the probability density function (PDF) of $X_t$, i.e., $$ \int_{-\infty}^xu(t,y)\,{\rm d}y=\mathbb{P}(X_t\le x). $$ We hope to find a differential equation that governs $u=u(t,x)$.

Define $$ f(x)=e^{-2\pi i\xi x}. $$ Then $$ \mathbb{E}f(X_t)=\int_{\mathbb{R}}e^{-2\pi i\xi x}u(t,x)\,{\rm d}x=\hat{u}(t,\xi) $$ is the Fourier transform of the PDF. This is known as the characteristic function in probability. To find a governing equation for $u$, it suffices to find that for $\hat{u}$.

By Ito's formula, we have \begin{align} {\rm d}f(X_t)&=f'(X_t)\,{\rm d}X_t+\frac{1}{2}f''(X_t)\,{\rm d}\left<X\right>_t\\ &=-2\pi^2\xi^2\sigma^2f(X_t)\,{\rm d}t-2\pi i\xi\sigma f(X_t)\,{\rm d}W_t, \end{align} where $\left<X\right>_t$ denotes the quadratic variation process of $X_t$, which, for the SDE stated above, reads $$ {\rm d}\left<X\right>_t=\sigma^2\,{\rm d}t. $$ From this result, we obtain $$ f(X_t)=f(X_0)-2\pi^2\xi^2\sigma^2\int_0^tf(X_s)\,{\rm d}s-2\pi i\xi\sigma\int_0^tf(X_s)\,{\rm d}W_s. $$ Note that the last integral is a martingle (see here as well), for which $$ \mathbb{E}\left(\int_0^tf(X_s)\,{\rm d}W_s\right)=0. $$ Thanks to this result, taking the expectation on both sides yields \begin{align} \mathbb{E}f(X_t)&=\mathbb{E}f(X_0)-2\pi^2\xi^2\sigma^2\mathbb{E}\left(\int_0^tf(X_s)\,{\rm d}s\right)\\ &=\mathbb{E}f(X_0)-2\pi^2\xi^2\sigma^2\int_0^t\mathbb{E}f(X_s)\,{\rm d}s, \end{align} or using $\mathbb{E}f(X_t)=\hat{u}(t,\xi)$, $$ \hat{u}(t,\xi)=\hat{u}(0,\xi)-2\pi^2\xi^2\sigma^2\int_0^t\hat{u}(s,\xi)\,{\rm d}s, $$ which is equivalent to the differential form $$ \frac{\rm d}{{\rm d}t}\hat{u}(t,\xi)=-2\pi^2\xi^2\sigma^2\hat{u}(t,\xi), $$ whose inverse Fourier transform gives $$ \frac{\partial}{\partial t}u(t,x)=\frac{\sigma^2}{2}\frac{\partial^2}{\partial x^2}u(t,x). $$

To sum up, the solution of a heat equation can be interpreted as the probability density of some stochastic process.

This concept can be generalized to the general Ito process, a special type of stochastic processes governed by $$ {\rm d}X_t=\mu(t,X_t)\,{\rm d}t+\sigma(t,X_t)\,{\rm d}W_t, $$ where $\mu$ and $\sigma$ are both preloaded functions. Repeat the above derivation with further techniques such as $$ \mathbb{E}\left(f(X_t)\,\mu(t,X_t)\right)=\int_{\mathbb{R}}e^{-2\pi i\xi x}\mu(t,x)u(t,x)\,{\rm d}x=\widehat{\left(\mu\,u\right)}(t,\xi), $$ one may end up with the general 1-D Fokker-Planck equation $$ \frac{\partial}{\partial t}u(t,x)=-\frac{\partial}{\partial x}\left(\mu(t,x)\,u(t,x)\right)+\frac{1}{2}\,\frac{\partial^2}{\partial x^2}\left(\sigma^2(t,x)\,u(t,x)\right). $$ One may also generalize this 1-D equation to $n$-D cases by employing an $n$-dimensional SDE system, as is stated in this page. As far as I see, one may also generalize this Euclidean-space result to differentiable manifolds, but I am afraid it goes beyond my knowledge.

Hope this explanation could be somewhat helpful for your.

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  • $\begingroup$ Thanks for your answer. Do you know if it is possible to obtain a nonlinear Fokker-Plank equation (for example $\partial_tu(x,t)=\partial^2_{xx}u(x,t)+u^3(x,t)$) from some stochastic process? $\endgroup$ – rpf Dec 3 '18 at 22:20
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    $\begingroup$ @rpf: It might be hard if you take $u$ as the probability density of $X_t$. Otherwise, you may construct another stochastic process $Y_t$, such that its conditional expectation with respect to $X_t$ satisfies your nonlinear parabolic equation---but this is still hard and nontrivial. Perhaps you may want to have a look at this reference: Frank, T. D. (2005). Nonlinear Fokker-Planck equations: fundamentals and applications. Springer Science & Business Media. $\endgroup$ – hypernova Dec 5 '18 at 6:06

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