Applying Feynman–Kac to a PDE with a Cross Derivative and Source Term

Summary

The document poses a terminal-value PDE in two variables with second derivatives in each variable, a mixed derivative, and a source term proportional to one state variable. It asks how to solve the equation when reducing the two-dimensional dependence through a change of variables appears to leave the source term dependent on an extra coordinate.

The material provides the PDE and terminal condition but no answer, derivation, probabilistic representation, or solution. Feynman–Kac is named as the intended approach, but the document does not explain how to construct the associated diffusion, handle the mixed derivative, or evaluate the source term. It is therefore useful mainly as a narrowly specified example of a PDE question, rather than as a worked method or evidence for a solution.

Key ideas

• The PDE includes a mixed second derivative as well as a source term depending on one state variable.
• The terminal condition is exponential in the sum of the two state variables.
• A change of variables aimed at combining the terminal-condition variables does not directly remove dependence introduced by the source term.
• The document asks about Feynman–Kac but contains no solution or supporting derivation.

Tags

# How to solve this particular PDE using Feynman-Kac formula?


# How to solve this particular PDE using Feynman-Kac formula?












I have to solve the PDE

$$ \begin{align} \frac{\partial F}{\partial t} + \frac{1}{2}\frac{\partial^2 F}{\partial x^2} + \frac{1}{2}\frac{\partial^2 F}{\partial y^2} + \frac{1}{2}\frac{\partial^2 F}{\partial x \partial y} + y &= 0 \\ F(T,x,y) &= e^{x+y} \end{align} $$

but I don't know how to do. Usually, when I have PDEs with cross derivatives, I use a translation $F(t,x,y) = G(t,z)$ (in this case it would be $e^{x+y} = e^z$). But now I have the term "$+y$" that makes this method useless since I get a term "$z-x$", so my PDE is a function both of $z$ and $x$. Do you know how to solve it? Thanks in advance.

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