Solution to the equation $\Delta f=\nabla f\cdot \nabla g$, where $g$ is radial. I am looking for a non constant solution to the following equation 
$$
\Delta f-\nabla f\cdot\nabla g=0\qquad\text{on }\mathbb{R}^2,
$$
where $g=g(r)$ is a positive, non constant function and $r=\sqrt{x_1^2+x_2^2}$. 
At first I thought to proceed in the following way: I looked for a solution of the first order equation 
$$
\mathrm{div}X-X\cdot\nabla g=0
$$
for a general vector field on $\mathbb{R}^2$. In analogy to the one-dimensional case I found the following solution 
$$
X=(X_1,X_2)=\left(c_1e^{g(r)},c_2e^{g(r)}\right)
$$
On the other hand, this solution is not a gradient, indeed 
$$
\partial_2X_1\ne\partial_1X_2
$$
for any choice of the constants $c_1,c_2$, so this procedure doesn't help me to solve the first equation. Is there any other way to look for an explicit solution? I thought that maybe the solution I found to $\mathrm{div}X=X\cdot\nabla g$ is the simplest one but it's not unique. 
Any help is very appreciated. 
 A: Let $g = -\log h$ which makes $0 < h < 1$. Then by chain rule we have that
$$\Delta f - \nabla f \cdot \nabla(-\log h) = \Delta f + \frac{1}{h}\nabla f \cdot \nabla h = 0$$
$$\implies \nabla \cdot ( h \nabla f) = 0$$
from here there doesn't seem to be any general characterization of the potentials of divergenceless vector fields in two dimensions. The radial condition might be of some use if a practical characterization can be found.
A problem I encountered while trying to tackle the equation head on with the Laplacian in polar coordinates is that the angular dependence ends up being a linear combination of sines and cosines. This may explain why your first attempt failed, because sines and cosines would have had complex values for a first order differential equation.
A: The equation is
$$
\Delta f=\nabla f\cdot  \nabla g,\tag 1
$$
where $g=g\left(\sqrt{x_1^2+x_2^2}\right)=g(r)$, $g$ be any function in $\textbf{R}$. We will find all solutions $f$ of (1), that are of the form $f=f\left(\sqrt{x_1^2+x_2^2}\right)=f(r)$. We have
$$
\Delta f=\frac{f'(r)}{r}+f''(r)\tag 2
$$
and
$$
\nabla f\cdot\nabla g=f'(r)g'(r).
$$
Hence (1) then becomes
$$
f''(r)=(g'(r)-r^{-1})f'(r),\tag 3
$$
which have solution
$$
f=f(x_1,x_2)=C_1\int^{\sqrt{x_1^2+x_2^2}}\frac{e^{g(t)}}{t}dt+C_2.\tag 4
$$
