# PDE with the Jacobian Determinant

Hello,

Initially I thought that the following problem can be solved through Monge-Ampere equation, but with Monge-Ampere, I have not been able to control the support.

The question seems so natural that I'm sure it has been studied somewhere in some form but I have not been able to dig out the right reference. Few minutes in MathScinet did not come up with what I'm looking for. All references seem to handle the situation when $f$ is positive which is clearly not the case here. Any suggestion or reference in this direction is welcome.

Thank you.

QUESTION:

Let $U\subset\mathbb{R}^n$ be open, connected and let $f\in C_{0}^{\infty}(U;\mathbb{R})$ satisfy $$\int_{U}f(x)dx=0.$$ Is it true that there exists a $u\in C_{0}^{\infty}(U;\mathbb{R}^n)$ satisfying $$\operatorname*{det}(\nabla u)=f \text{ in }U?$$ Note that $$C_{0}^{\infty}(U;\mathbb{R}^n)=\{u\in C^{\infty}(U;\mathbb{R}^n):\operatorname*{supp}(u)\text{ is compact and }\operatorname*{supp}(u)\subset U\}.$$

• If $U$ is a submanifold of $\mathbb R^n$ with boundary, then you can think of $\int_U f(x)dx$ as the integral of an $n$-form over $U$. In particular, $f = d\omega$ where $\omega$ in an $(n-1)$-form, also on $U$. So we're not writing $f$ as the determinant of the derivative of a function, but the trace of the derivative of a function, which is similar to what you're asking for but not quite the same. – Ryan Budney Dec 5 '12 at 5:15
• @Ryan budney: What you are talking about is the problem with prescribed divergence which is essentially the linearized version of the problem with prescribed Jacobian. – Tatin Dec 5 '12 at 5:42
• If $f$ is not identically zero, then it will be negative somewhere, and I think there is not much chance of being able to solve your equation using Monge-Ampere results (i.e. $u=\nabla v$ for some scalar function $v$), because ellipticity is lost and the literature on Monge-Ampere equations with RHS that changes sign is not very developed (although it is nonempty, see e.g. arxiv.org/abs/1204.6704 ) – YangMills Dec 5 '12 at 6:30
• It is certainly false if you don't assume that $U$ is connected. – Robert Bryant Dec 5 '12 at 13:20
• @Robert Bryant: Thank you. Connectedness is an assumption here. Re-edited accordingly. – Tatin Dec 5 '12 at 17:19

This is not really an answer, but I prefer the luxury of the answer box instead of the rather spartan comment box.

You have only one equation for $n$ unknown functions (the components of the vector-valued function $u$), so the equation is underdetermined. This gives you a lot of flexibility on what to do. Roughly speaking, you get to impose $n-1$ additional conditions in order to get a well-posed boundary value problem for a system of PDE's.

There might be a clever choice of the $n-1$ conditions that makes your question very easy to answer, but I don't know and haven't been able to think of any.

The most common way to do this is to set $u$ equal to the gradient of a scalar function $\phi$, which turns your equation into a real Monge-Ampere equation. If the function $f$ were always positive, then you can try to solve for $\phi$ convex. There is a well-developed theory of this situation, because the PDE becomes second order elliptic and many techniques are available.

However, if $f$ changes sign, then much, much less is known. But here are some partial answers:

1) If you restrict to $n=2$, assume $\Omega$ is a nice smooth domain, and assume that $f$ changes sign "cleanly" (i.e., $f$ vanishes along a smooth curve in $\Omega$ and its gradient is everywhere nonzero along this curve), then the equation becomes what is known as a nonlinear Tricomi equation, which has been studied. You can probably set up some appropriate boundary value problem to solve your equation on $\Omega$ this way.

2) The ideas in 1) probably could be generalized to higher dimensions, but I'm unaware of any prior work on this. The idea is to solve for $\phi$ to have positive definite Hessian where $f$ is positive and signature $(n-1,1)$ where $f$ is negative. Then the PDE becomes elliptic where $f$ is positive and hyperbolic where $f$ is negative.

3) All of this might be able to be recast in the form of a nonlinear symmetric positive system as defined and studied by K. O. Friederichs.

4) Maybe the most promising but still rather difficult direction is to not use the Monge-Ampere equation but try to find a different set of $n-1$ additional first order equations to impose on $u$, (so you get an $n$-by-$n$ system of first order PDE's) such that the resulting system is symmetric positive. Then you try to identify the right boundary conditions that both leads to the existence of a solution and implies that $u$ vanishes on the boundary of $\Omega$.