Skip to main content Skip to navigation


Feynman-Kac for SPDEs with multiplicative noise

One application of the celebrated Feynman-Kac formula in probability theory gives a stochastic representation of solutions to Cauchy problems. Following the basic idea of the proof, the formula is first derived for terminal value problems of type \[ -\frac{\partial}{\partial t}u\,=\, {\cal A} u - k\times u + g,\quad u(x,T)=f(x), \] where $f$ is a function on $\mathbb{R}^d$, $k$ and $g$ are functions on $\mathbb{R}^d\times[0,T]$, and ${\cal A}$ is a second-order partial differential operator. Under the right conditions, the representation would read \[ u(x,t)\,=\, {\bf E}[f(X_T)\exp\{-\int_t^T k(X_s,s)\,ds\}+\int_t^T g(X_s,s)\exp\{-\int_t^s k(X_r,r)\,dr\}\,ds\,|\,X_t=x], \] for $(x,t)\in\mathbb{R}^d\times[0,T]$, where $X_t$ stands for the Markov process generated by ${\cal A}$, and ${\bf E}$ denotes the corresponding expectation operator.

This note is NOT about the conditions on $f,k,g, {\cal A}$ guaranteeing such a representation. This note is about the corresponding initial value problem with $k$ being a random space-time field. To concentrate on this issue, in what follows, $g$ is set to be zero. After guessing a formula for the initial value problem from the above formula by simple time-reversion, it is discussed why this formula needs further treatment when $k$ is a rather IRREGULAR random space-time field.

If $u(x,t)$ satisfies the initial value problem \[\frac{\partial}{\partial t}u\,=\, {\cal A} u - k(x,t)\times u,\quad u(x,0)=f(x),\tag{1}\label{ivp}\] and if ${\cal A}$ only acts on the $x$-variable then, for an arbitrary but fixed $t>0$, $\tilde{u}(x,t')=u(x,t-t')$ satisfies the terminal value problem \[-\frac{\partial}{\partial t'}\tilde{u}\,=\, {\cal A}\tilde{u} - k(x,t-t')\times \tilde{u},\quad \tilde{u}(x,t)=f(x),\] and hence \[ \tilde{u}(x,t')\,=\, {\bf E}[f(X_t)\exp\{-\int_{t'}^t k(X_s,t-s)\,ds\}\,|\,X_{t'}=x], \] for any $(x,t')\in\mathbb{R}^d\times[0,t]$, which gives the stochastic representation \[ u(x,t)\,=\, {\bf E}[f(X_t)\exp\{-\int_{0}^t k(X_s,t-s)\,ds\}\,|\,X_{0}=x]\tag{2}\label{FK}\] for the solution to the initial value problem, when $t'$ is taken to be zero.

Now assume that $k(x,t)$ is a random space-time field, for example, $k(x,t)=h(x)\times\dot{B}_t$, where $h$ denotes a deterministic function on $\mathbb{R}^d$, and $B=(B_t)_{t\ge 0}$ is a one-dimensional standard Brownian motion. Clearly, this makes (\ref{ivp}) a rather FORMAL stochastic partial differential equation (SPDE), because the product $k(x,t)\times u$ needs explanation, because $\dot{B}_t=\partial B_t/\partial t$ is not a function anymore.

However, when constructing both $X_t$ and $B_t$ as two independent processes on the same probability space, formula (\ref{FK}) could be read \[ u(x,t)(\omega)\,=\, {\bf E}[f(X_t)\exp\{-\int_{0}^t h(X_s)\dot{B}_{t-s}\,ds\}\,|\,X_{0}=x,B=\omega],\tag{3}\label{FK1}\] and this might give a solution to the SPDE, in some sense, if the integral $\int_{0}^t h(X_s)\dot{B}_{t-s}\,ds$ can be well-defined.

A first idea could be to treat the integral as a backward stochastic integral against Brownian motion. But this would cause an adaptedness-problem when verifying equation (\ref{ivp}) as the integral form of this equation would rather be obtained by integrating forward in time. So, one wants to write \[ \int_{0}^t h(X_{t-s})\dot{B}_s\,ds\quad\mbox{for}\quad \int_{0}^t h(X_s)\dot{B}_{t-s}\,ds \] which is a formal operation, only, since $\dot{B}$ is not a function. But this operation would have its proper meaning when $\dot{B}$ is approximated by a continuous function $\dot{B}_\varepsilon$ obtained by convoluting $\dot{B}$ against a mollifier. Then, \[ u_\varepsilon(x,t)(\omega)\,=\, {\bf E}[f(X_t)\exp\{-\int_{0}^t h(X_{t-s})\dot{B}_\varepsilon(s)\,ds\}\,|\,X_{0}=x,B=\omega]\tag{4}\label{FKapprox} \] solves \[ \frac{\partial}{\partial t}u_\varepsilon\,=\, {\cal A} u_\varepsilon - h(x)\dot{B}_\varepsilon(t)\times u_\varepsilon,\quad u_\varepsilon(x,0)=f(x), \] in mild sense, that is \[ u_\varepsilon(x,t)\,=\,\int p_t(x,y)f(y)\,dy+\int_0^t\!\!\!\int p_{t-s}(x,y)\,u_\varepsilon(y,s)\,h(y)dy\,\dot{B}_\varepsilon(s)ds, \] where $p_t(x,y)$ stands for the family of transition probability densities of the Markov process $X_t$. Note that $X_t$ is a time-homogeneous Markov process because its generator ${\cal A}$ does not act on time.

Taking $\varepsilon$ to zero in the last equation gives \[ u(x,t)\,=\,\int p_t(x,y)f(y)\,dy+\int_0^t\!\!\!\int p_{t-s}(x,y)\,u(y,s)\,h(y)dy\,\circ dB_s, \] where \[ u(x,t)(\omega)\,=\, {\bf E}[f(X_t)\exp\{-\int_{0}^t h(X_{t-s})\circ dB_s\}\,|\,X_{0}=x,B=\omega]\tag{5}\label{solStrato}\] is the $\varepsilon$-limit of (\ref{FKapprox}). Here $\circ dB_s$ denotes Stratonovich integration as this is what $\dot{B}_\varepsilon(s)ds$ would "produce" in the limit. All in all, (\ref{solStrato}) yields a mild solution to (\ref{ivp}) if the product $k(x,t)\times u$ with $k(x,t)=h(x)\times\dot{B}_t$ is understood in the sense of Stratonovich. If the SPDE (\ref{ivp}) is understood in the sense of Itô, under the exponential in (\ref{solStrato}), the corresponding Itô-correction term $\frac{-1}{2}\int_{0}^t h(X_{s})^2\,ds$ has to be substracted. Note that the Stratonovich integral under the exponential in (\ref{solStrato}) nevertheless coincides with the Itô integral because the processes $X_t$ and $B_t$ are independent.

Now, if the process $X_t$ is reversible, there is a trick to get rid of the backward time of the integrand under the exponential in (\ref{solStrato}). This trick is used by Bertini/Cancrini in The Stochastic Heat Equation: Feynman-Kac Formula and Intermittence, Journal of Statistical Physics, Vol. 78, Nos 5/6, 1995, when ${\cal A}$ is the Laplacian and $k(x,t)$ is space-time white noise. They don't say that this trick is new but they don't give a reference, either. Below, the right-hand side of (\ref{solStrato}) is going to be transformed following Bertini/Cancrini's ideas (I do not know any earlier reference---comments on earlier references would be most welcome).

Since the processes $X_t$ and $B_t$ are independent, in what follows, the conditioning on $B$ is suppressed and the expectation is only taken with respect to the process $X_t$ writing ${\bf E}^X$ for the corresponding expectation operator. Using this notation, the right-hand side of (\ref{solStrato}) equals \begin{align*} &{\bf E}^X[f(X_t)\exp\{-\int_{0}^t h(X_{t-s})\circ dB_s\}\,|\,X_{0}=x]\\ =&\int f(y)\,{\bf E}^X[\exp\{-\int_{0}^t h(X_{t-s})\circ dB_s\}\,|\,X_{0}=x,X_t=y]\,p_t(x,y)\,dy. \end{align*} Obviously, for fixed $t$, the expectation is now taken with respect to an $X$-bridge satisfying $X_0=x$ and $X_t=y$. But the integration is with respect to the time-reversion of this bridge, and the time-reversed bridge has the same law as the corresponding forward-time bridge since the process $X_t$ is reversible in the sense of $p_t(x,y)=p_t(y,x)$. So, if ${\bf E}^b_{y,x;t}$ denotes the expectation with respect to an $X$-bridge satisfying $X_0=y$ and $X_t=x$, then the right-hand side of (\ref{solStrato}) becomes \[ \int dy\,f(y)\,p_t(y,x)\,{\bf E}^b_{y,x;t}[\exp\{-\int_{0}^t h(b_{s})\circ dB_s\}], \] which is the type of Feynman-Kac formula used by Bertini/Cancrini in the paper referenced above. They prove that their formula gives a solution to the corresponding SPDE. This proof is quite short, and I think that the validity of (3.21) on page 1394 of their paper can hardly be understood without mentioning the underlying principle of time-reversion I stressed above.