Going through Duffie and Kan (1996) I came up with a question.
After Eq. $(3.4)$ the claim the following:
By $(2.4)$, we also know that $\mathcal{D} F (X_t, t) - R(X_t) F(X_t, t) = 0$. Since $F$ is strictly positive valued, from $(3.4)$ we have [...].
and then they equate what they claim together with equation $(3.4)$ to get equation $(3.5)$.
I don't see how to get that $\mathcal{D} F (X_t, t) - R(X_t) F(X_t, t) = 0$ from equation $(2.4)$. Eq. $(2.4)$ is just the definition of a zero-coupon bond as the expectation of the exponential compounding of the instantaneous rate (please see the paper for the exact formula). Does someone know how to proceed to prove it? From how it is stated in the article I guess that it must be obvious somehow, however, I can't see a clear path to prove it.
Edit: As Kurt G. mentioned, my question can be easily solved using the Feynman-Kac formula. The FK formula tells us that an unknown function $u:\mathbb{R}\times[0,T]\to\mathbb{R}$ whose evolution is described by the PDE
$$\frac{\partial u}{\partial t}(x,t) + \mu(x,t) \frac{\partial u}{\partial x}(x,t) + \tfrac{1}{2} \sigma^2(x,t) \frac{\partial^2 u}{\partial x^2}(x,t) -V(x,t) u(x,t) + f(x,t) = 0, $$
defined for all $x \in \mathbb{R}$ and $t \in [0, T]$ and subject to the terminal condition $u(x,T)=\psi(x)$, has a solution that can be written as a conditional expectation as follows:
$$u(x,t) = E^Q\left[ \int_t^T e^{- \int_t^r V(X_\tau,\tau)\, d\tau}f(X_r,r)dr + e^{-\int_t^T V(X_\tau,\tau)\, d\tau}\psi(X_T) \,\Bigg|\, X_t=x \right] .$$
Taking the inverse path, i.e. starting from the conditional expectation, where in the case of the formula appearing in Duffie and Kan we have $f(x,t) = 0 $ and $\psi(X_T) = 1$, we can see that the PDE that $u(x,t)$ must satisfy corresponds to:
$$\frac{\partial u}{\partial t}(x,t) + \mu(x,t) \frac{\partial u}{\partial x}(x,t) + \tfrac{1}{2} \sigma^2(x,t) \frac{\partial^2 u}{\partial x^2}(x,t) -V(x,t) u(x,t) = 0, $$
where $\mathcal{D} = \frac{\partial}{\partial t}(x,t) + \mu(x,t) \frac{\partial}{\partial x}(x,t) + \tfrac{1}{2} \sigma^2(x,t) \frac{\partial^2}{\partial x^2}(x,t) $.
