Setting: let us consider a vector $v=(x,y)\in\mathbb{R}^2$ and a time horizon $[0,T]$ for $T>0$ such that for any $0\leq t\leq T$ we have the following partial differential equation with a terminal condition $\psi:\mathbb{R}\rightarrow\mathbb{R}$ at time $T$: \begin{align} \begin{aligned} &\partial_tu(t,x)+xy\partial_xu(t,x)+\frac{1}{2}\beta(t,x)\partial_{xx}u(t,x)=yu(t,x)+g(t,x,y)\\ &u(T,x)=\psi(x) \end{aligned}\tag{1} \end{align} for some function $u:[0,T]\times\mathbb{R}\rightarrow\mathbb{R}$.
Question: can we apply Feynman-Kac theorem to PDE $(1)$?
Proof: let $v=(x_1,\dots,x_n)\in\mathbb{R}^n$. The multidimensional Feynman-Kac theorem is valid for a parabolic PDE on a unknown function $U:[0,T]\times\mathbb{R}^n\rightarrow\mathbb{R}$ with form: \begin{align} \begin{aligned} &\partial_tU(t,v) +\sum_i\alpha_i(t,v)\partial_iU(t,v) +\frac{1}{2}\sum_i\sum_j\beta_{ij}(t,v)\partial_{ij}U(t,v) =\gamma(t,v)U(t,v)+g(t,v) \\ &U(T,v)=\psi(v) \end{aligned}\tag{2} \end{align}
Let us fix $v=(x,y)$ and set: \begin{align} \begin{aligned} \alpha_x(t,x,y)&=xy\\ \beta_{xx}(t,x,y)&=\beta(t,x)\\ \gamma(t,x,y)&=y\\ \psi(v)&=\psi(x)\\ \end{aligned}\tag{3} \end{align} as well as: \begin{align} \begin{aligned} \alpha_y(t,x,y)\partial_yU(t,x,y)&=0\\ \beta_{yy}(t,x,y)\partial_{yy}U(t,x,y)&=0\\ \beta_{xy}(t,x,y)\partial_{xy}U(t,x,y)&=0\\ \beta_{yx}(t,x,y)\partial_{yx}U(t,x,y)&=0\\ \end{aligned}\tag{4} \end{align}
PDE $(2)$ combined with $(3)$ and $(4)$ yields PDE $(1)$, up to the point that in $(2)$ the unknown $U$ is a function of both $x$ and $y$, whereas in problem $(1)$ the function $u$ merely depends on $x$. However given:
- the terminal condition function $\psi$ only depends on $x$;
- condition $(4)$, which entails partial derivatives involving $y$ have null contribution,
can we actually conclude that $U(t,x,y)=u(t,x)$ that is $U$ does not depend on $y$? If so, is the Feynman-Kac theorem still applicable?