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Feynman–Kac Representation for a PDE with an Initial Condition

Article Quant Q&A · Author: Paulo Rocha
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Summary

The document presents a linear parabolic Cauchy problem with a terminal condition and gives its Feynman–Kac representation in terms of an Itô diffusion, a potential term, and a source term. It then asks how to formulate the corresponding representation for a problem specified by an initial condition, with the time derivative sign reversed. This highlights the connection between PDE time direction and the stochastic process used to represent its solution.

The post provides the equations and model ingredients needed to frame the question, but it does not include an answer, derivation, or worked example for the initial-value case. It therefore serves as a mathematical prompt rather than a complete tutorial. Any application would require checking the time reversal, boundary data, and regularity assumptions needed for a Feynman–Kac result; these details are not discussed.

Key ideas

  • The terminal-value PDE includes diffusion, drift, a multiplicative potential, and a source term.
  • Its solution is expressed through expectations over an Itô process.
  • The post asks how the representation changes when data are specified at the initial time.
  • No derivation or answer to the initial-value question is included.

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# Explicit form for forwards Feynman-Kac formula


# Explicit form for forwards Feynman-Kac formula












This might be a simple question, but I'm having trouble with it.

Consider the Cauchy problem with final condition. \begin{equation} \begin{cases} \frac{\partial u}{\partial t}(t,x) + \mathcal{L}u(t,x) + k(t,x)u(t,x) = g(t,x) &\textit{in}\quad\left[0,T\right]\times\mathbb{R}\\ u(T,x)=\phi(x)&\textit{in}\quad\mathbb{R} \end{cases} \label{CauchyProb} \end{equation} where \begin{equation} \mathcal{L} = \frac{1}{2}\sigma^2(t,x)\frac{\partial^2}{\partial x^2} + \mu(t,x)\frac{\partial }{\partial x}. \end{equation} I am well aware that a solution to this problem can be given in terms of the following Feynman–Kac formula: \begin{equation} u(t,x)=\mathbb{E}\left[\phi(X_T^{t,x})\exp\left\lbrace\int_t^Tk(s,X_s^{t,x})ds\right\rbrace -\int_t^Tg(s,X_s^{t,x})\exp\left\lbrace\int_t^s k(u,X_u^{t,x})du\right\rbrace ds\right] \end{equation} where $X_t$ is an Itô process that is described by: \begin{equation} dX_t = \mu(t,X_t)dt + \sigma(t,X_t)dW_t\,, \end{equation} with $X_0=x$.

I was wandering. What is the Feynman-Kac formula for the Cauchy problem with initial condition: \begin{equation} \begin{cases} -\frac{\partial u}{\partial t}(t,x) + \mathcal{L}u(t,x) + k(t,x)u(t,x) = g(t,x) &\textit{in}\quad\left[0,T\right]\times\mathbb{R}\\ u(0,x)=\phi(x)&\textit{in}\quad\mathbb{R} \end{cases} \label{CauchyProb2} \end{equation}

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This summary was written by Stratmill's research agent from the original; it is not a copy of the source.