Can Kolmogorov backward and forward equations hold at the same time?

Solution 1:

Part (if not all) of the confusion seems to come from the fact that you got a sign wrong. The Kolmogorov forward equation is $$\tag{KFE} \frac{\partial}{\partial t} p(t,x)+ \frac{\partial}{\partial x} (\mu\,p(t,x))\color{red}{-}\frac12 \frac{\partial^2}{\partial x^2} (\sigma^2 \, p(t,x)) =0 $$ (see [1]). The Kolmogorov backward equation is $$\tag{KBE} \frac{\partial}{\partial t} p(t,x)+ \mu\frac{\partial}{\partial x} p(t,x)\color{red}{+}\sigma^2\frac12 \frac{\partial^2}{\partial x^2} p(t,x) =0\, $$ (this $\color{red}{+}$ sign you had right).

Even if $\mu$ and $\sigma$ are constant the two equations are in general not satisfied by the same function. In [2] we can read that (KFE) (also known as Fokker Planck equation) is satisfied by the probability density of the stochastic process $X$ that satisfies the SDE $$ dX_t=\mu(t,X_t)\,dt+\sigma(t,X_t)\,dW_t,\quad\quad X_0=x\,. $$ In [1] we can read that (KBE) with final condition $u(x)$ at $t=T$ is satisfied by the conditional expectation $$ p(t,x)=\mathbb E[u(X_T)|X_t=x]\,. $$ This $p(t,x)$ is clearly different from the $p(t,x)$ solving (KFE), i.e., from the density of $X_t\,$.