Question: Is there a known path-dependent Black-Scholes PDE?
To be a little more precise, let $S$ be a stock price under a risk-neutral measure such that $S$ satisfies the SDE with path-dependent volatility: $$dS_t = r S_t dt + \sigma(t, S_.) S_t dB_t.$$ Here $\sigma: [0,\infty)\times C([0, \infty), (0,\infty)) \to \mathbb{R}$ is a previsible path functional, i.e. for any $t\geq 0$, $\sigma(t, S_.) \in \mathscr{F}_t$. In other words $\sigma$ depends on the path of $S$ up to time $t$ and is known at time $t$.
When $\sigma(t, S_.)=\sigma(t, S_t)$ only depends on the current state $S_t$, the ordinary Feynman-Kac formula says that $$u(t, s) = \mathbb{E}(e^{-r(T-t)}h(S_T) | S_t=s),$$ if and only if $u$ solves the PDE $$u_t+r s u_s + \frac12 \sigma(t,s)^2 s^2 u_{ss}-ru=0$$ with terminal condition $u(T, s) = h(s)$.
What can we say about this path-dependent case? $$U_t = \mathbb{E}(e^{-r(T-t)} h(S_T) | \mathscr{F}_t)?$$
My only hunch is that Bruno Dupire's functional Ito calculus might be applied to this particular case but it is taking me some time to read through this paper and check everything. Also his Feynman Kac has a slightly different form, 1) it is more general as it lets $h$ also be a functional, in my notation $h(t, S_.)$, and 2) his condition in the conditional expectation is $S_t$ is a little different. It involves conditioning on the path rather than the filtration.
