Lie derivative of a scalar and PDE I posted this on the physics stackexchange, but they told me to post here, as it may be more relevant. 
I am reading about differential geometry, and in particular the Lie derivative and its relation to (relativistic) hydrodynamics. In particular, I was wondering if, given two scalar functions $f(\mathbf{x}),~g(\mathbf{x})$, it is possible to solve the following equation:
$$\mathcal{L}_Z[f]=Z^\mu\partial_\mu f=g,$$
where $Z=Z^\mu\partial_\mu$ is an arbitrary vector field. Since I am thinking of hydrodynamics here, I am considering the manifold associated to $Z$ to be Euclidean space, but I would be interested to know how it works for arbitrary manifolds. 
Is their a theorem that guaranties the existence of a solution? I am okay with existence, I don't necessarily need a way to find the solution. If there is no such proof, what are the restriction one needs to impose on $Z$? If there is a proof, is it also valid for tensor quantities, ie is there a solution to
$$\mathcal{L}_Z[f^{\mu_1...\mu_n}_{\nu_1...\nu_m}]=g^{\mu_1...\mu_n}_{\nu_1...\nu_m}.$$

The zeros of $Z$ form an obvious obstruction, as $g$ must vanish there also, to order no less then $Z$. Consider for example, in one variable, 
$Z=x^2\partial_x,$ $g(x)=x.$
Away from the zeros of $Z$, your equation can be solved, locally, for any $g$, as $Z$ can be ''rectified'', i.e. one can find coordinates $x_1, \ldots, x_n$ so that $Z$ is the partial wrt $x_1$, and the equation is solved by integrating $g$ wrt $x_1$. 
Globally, even away from zeros of $Z$, there might be obstructions, if $Z$ has closed integral curves. Consider for example in the plane, $Z=-y\partial_x+x\partial_y$. 
