Imagine we have known (regular enough) functions $a(t,x)$, $b(t,x)$, $c(t,x)$ and $d(t,x)$ and consider the PDE: $$\partial_t u(t,x) = a(t,x) + b(t,x) u(t,x)+c(t,x) \partial_x u(t,x)+d(t,x) \partial_x^2 u(t,x),$$ on $[0,T]\times\mathbb{R}$ with some initial condition $u(0,x)=f(x)$ for some $f$ regular enough.
My question is: is it possible to derive a PDE for, say $v(t,x)=-\partial_x \log(u(t,x))$? Just in terms of $v$. Thank you very much!
I don't think so. $u$ and $v$ are not in one-to-one relationship. If we solve for $u$ as a function of $v$, we find
$$u(x, t) = G(t)\exp \biggl \{-\int v(x,t)dx \biggr \}$$
for an arbitrary function $G(t)$. We might hope the the PDE is invariant of the function $G(t)$, and in some special case it might be. But in this general case, when we substitute $u(x, t)$ into the PDE to find the PDE for $v(x,t)$ by elimination, we find that $G(t)$ does not trivially get eliminated from the equation. One such simplification can be written as
$$\frac{\partial_t G(t)}{G(t)} - \frac{a(x,t)}{G(t)} \exp \biggl \{\int v(x,t)dx \biggr \} = \int \partial_t v(x,t)dx + b(x,t) - c(x,t)v(x,t) +d(x,t)(v(x,t)^2 -\partial_x v(x,t)) $$
So the above non-linear PDE must be satisfied by some fixed function $G(t)$ which is unknown solely based on knowledge of $v(x,t)$. This means that $v(x, t)$ alone does not form a closed dynamical system, but $u(x, t)$ does. A physical intuition for this is that not all related variables form closed systems. For example, it is possible to deduce the cost of a given trip given the position on the map as a function of time, but it is not possible to deduce the cost solely from the knowledge of the velocity as a function of time