Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

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.

share|improve this question
1  
Do you have any motivation for this question ? –  Denis Serre Mar 15 '12 at 12:35
    
An application I have in mind is elliptic approximations to 1st order equations - something along the lines of viscosity solutions, but with a finite term −ϵΔ added to the 1st order equation rather than a vanishing viscosity. These will be better behaved from a numerical point of view, among others. The 1st order equations appear, for instance, in control theory, but they are not very well behaved. As I understand, a lot more theoretical understanding would become available if the 1st order terms are absent from the elliptic approximation. –  Pait Mar 16 '12 at 18:51
add comment

4 Answers

up vote 5 down vote accepted

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".

share|improve this answer
    
Finkel and Kamran's is a beautiful paper. I am grateful to mathphysicist, to the authors, and to everyone who read this post, for bringing it to my attention. I suppose one could still ask whether the conditions could be weakened under more general transformations. The subject seems to be under investigation. –  Pait Mar 19 '12 at 17:42
add comment

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).

share|improve this answer
    
Correct. I had not heard of the name, thanks. The $a^i$ being components of a gradient vector is an integrability condition. Still, I wish to find out if a transformation exists for more general equations, possibly even nonlinear ones. –  Pait Mar 13 '12 at 15:52
add comment

Another classical transform, sometimes called Hopf-Cole: the equation $\Delta u+|\nabla u|^2=f$ becomes $\Delta v=fv$ after $u=\log v$.

share|improve this answer
add comment

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).

share|improve this answer
    
In the question and in the answers by Hans and by Denis Serre there are examples of how the 1st order terms can be eliminated. Transforming from divergence to non-divergence form, as you mentioned, or vice-versa, is another technique. It is applicable if the equation is linear, and the 1st order term satisfies an integrability condition that allows it to be expressed as in your formula. I am still interested in, given a more general elliptic equation, finding a form without such terms, if one exists. –  Pait Mar 16 '12 at 18:51
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.