This is a follow-up to a previous question. Assume I have the following system:
$$\nabla\left(f(x)\frac{1}{2}\nabla u(x)^2\right)=v(x)$$
where $\nabla$ is the Nabla-operator in some $d$-dimensional space, and $u(x)$ and $v(x)$ are some known functions $\mathbb{R}^d\rightarrow\mathbb{R}$. For simplicity's sake (if it is helpful), we can assume that I want to solve this problem on a discretized grid, and that I can evaluate the partial derivatives numerically through finite differences.
Now my question: Is there a way I can find $f(x)$ given this equation, $u(x)$ and $v(x)$?
Here are some candidates for a solution. Every nice function $v$ is a divergence, since you have $$ v(x) = {\rm div}\big(\int_0^1 t^{n-1}xv(tx)dt\big). $$ This is part of the Poincar'e Lemma, for which see Spivak's Calculus on Manifolds or my exposition at bterrell.net.
Suppose your equation has a solution. Then $$ f\tfrac{1}{2}\nabla u^2 $$ and $$ \int_0^1 t^{n-1}xv(tx)dt $$ have the same divergence. Therefore they differ by a function $z$ that has divergence zero, $$ f(x)\tfrac{1}{2}\nabla u(x)^2 = \int_0^1 t^{n-1}xv(tx)dt + z(x). $$ So a first candidate for $f$ is to take the case $z=0$ and solve for $f$ by dividing by $\tfrac{1}{2}\nabla u(x)^2$ wherever that is not zero.
It looks to me that this is a solution. So there are many other solutions by taking various functions $z$.
I don't know what happens at points where $\tfrac{1}{2}\nabla u(x)^2$ is zero, or where the domain of $v$ is not starshaped.