Eliminating 1st order terms in elliptic partial differential equation Under what conditions is it possible, using a suitable change of variables, to eliminate 1st order terms in an elliptic partial differential equation, so that the equation involves the 2nd derivatives, the dependent variable, and independent terms only?
To be concrete, consider the elliptic equation $-\Delta u + \sum_i \frac{d u}{dx^i} a^i + f(x)=0$. 
If the $a^i$ are constant, define $u(x) = v(x) e^{\frac{1}{2}\sum_j a^j x^j}$ and obtain 
$-\Delta v -  \frac{1}{4} v \sum_i a^i a^i + f(x)e^{-\frac{1}{2}\sum_j a^j x^j}=0$, an elliptic equation without 1st order terms.
If the $a^i$ are not constant or if the equation is quasilinear, the problem is harder. It can be approached using contact transformations and Cartan's method of equivalence, but I am not aware of results.
 A: Dear Pait,
note that the kind of transformation you are using for constant $a^i$ 's also
works if the vector $a=(a^i)_i$ is a gradient, say $a(x)=\nabla g (x)$ for some smooth $g$.
In this case define 
$u(x)= v(x)e^{\frac{1}{2}g(x)}$ 
and you will obtain again
an elliptic equation without first order term.
This transformation is sometimes called ground state transformation and it is frequently used to 
go from reversible diffusion generators to Schroedinger type operators  (or the other way around).
A: The necessary and sufficient conditions for transforming one second-order differential operator of the type you are interested in into another are given by the so-called Cotton theorem, see Theorem 1 in this paper by Finkel and Kamran. In your case, you want the linear part of the transformed operator to vanish ($\tilde{\mathcal{A}}=0$ in the notation of the above paper), whence you can readily extract the conditions you ask for. Roughly speaking, in order to have no first-order terms in your transformed operator, the linear part of your original operator should be "pure gauge up to a change of independent variables".
A: Another classical transform, sometimes called Hopf-Cole: the equation $\Delta u+|\nabla u|^2=f$ becomes $\Delta v=fv$ after $u=\log v$.
A: I'm confused about what constitutes "eliminating the first-order terms". For an equation with a non-constant matrix of coefficients for the second derivatives, I think it would be useful to have a more precise specification of the problem. 
For many equations, it is more useful to write the equation in divergence form $\partial_i (A^{ij} \partial_j u)=0$ than to write it in the form $B^{ij} \partial_i \partial_j u=0$. An equation in divergence form can be rewritten in the second form plus first-order terms, i.e. $\partial_i (A^{ij}\partial_j u) + (\partial_i A^{ij})(\partial_j u)$, and vice versa. Which of these do you consider to be the one without first-order terms. This is particularly important for quasilinear equations. 
Along the same lines, in spherical coordinates, the Laplacian has first-order terms (with divergent coefficients). More generally, on a manifold with Riemannian metric $g$ and associated connection $\nabla$, some people write Laplace's equation as $\nabla_i\nabla^i u=0$, where others might write $(det g)^{-1} \partial_j ((det g) g^{ji} \partial_i u)=0$, which appears to have first-order terms (depending on how you view divergence form). 
