### Solving the Monge-Ampere equation

In my previous post, while talking about the Minkowski problem, I introduced the Monge-Ampere equation. I would like to talk about it for a post or two before continuing with the Minkowski problem, in part because the Dirichlet problem for Monge-Ampere is arguably an easier problem than the Minkowski problem.

To make the presentation cleaner, allow me to work only in the unit ball in Euclidean space, this situation contains all of the important issues one deals with to solve the equation, and it will let us avoid some of the technicalities that give little insight. In any case, all of the arguments can be tweaked without too much effort (but perhaps losing some clarity of exposition) to make them work in any smooth, strictly convex domain $\Omega$, the same remark goes for dealing with non-zero boundary data,  I will make more detailed remarks about this in the next post. Here we go then.

The Dirichlet problem for the Monge-Ampere equation: Consider a strictly convex domain $\Omega \subset \mathbb{R}^n$, and a smooth, positive function $\psi \bar{\Omega} \to \mathbb{R}$. Then find a convex function $u$ twice differentiable in $\Omega$ such that:

$det (D^2u(x)) = \psi(x) \mbox{ for all } x \in B_1$

$u(x)= 0 \mbox{ on } \partial B_1$

That a solution exists can be seen via the method of continuity (although, as I pointed out, there are other approaches). The principle behind this method is that if one can show that the set of those positive $\psi \in C^\infty(\bar{\Omega})$ for which there is a solution is  open, closed and not empty then by connectivity there is a solution for any such positive $\psi \in C^\infty$ . This is a very simple and universal method that has proven to be succesful in many situations in Mathematics (e.g. in many applications of PDEs to differential geometry), but in these post I will talk about its application to the Monge-Ampere equation.

From now on we fix $\psi$, and define

$\psi_t(x)=(1-t)+t\psi(x)$

this is a smooth function for every $t$, and strictly positive for every $t \in [0,1]$. Thus we can consider the Dirichlet problem above with $\psi_t$ as the right hand side, lets call it “Problem $(t)$“. Problem $(1)$ is the actual problem we want to solve, and Problem $(0)$ $t=0$  we can solve explicitely, the solution is given by $u_0(x)=\frac{1}{2}(|x|^2-1)$ which satisfies:

$det(D^2u_0(x))=1 = \psi_0(x)\mbox{ in } B_1$

$u_0(x)= 0 \mbox{ on } \partial B_1$

Consider the set

$A=\{t \in [0,1] | \mbox{ Problem } (t) \mbox{ can be soved } \}$

the example above shows that $0 \in A$. Then the strategy is to show that $A$ is at the same time open and closed in $[0,1]$, which would imply the Dirichlet problem has a solution for any $\psi$. Here is where the real work begins, and we divide it in two parts.

Part I: Showing $A$ is open.

This will be dealt with machinery from classical analysis, namely the inverse and implicit function theorems for Banach spaces. Let us consider the map

$T : C^{2,\alpha}_0 \to C^\alpha$

given by $T(u)=det(D^2u)$, this is a continuous but non-linear map between these two Banach spaces, we will apply the inverse function theorem to this map. Suppose $t \in A$ and let $u^{(t)}$ be the solution to the correspoding problem, then we can assume without loss of generality that $u^{(t)} \in C^{2,\alpha}_0$ and that it is a strictly convex function. Then the differential to $T$ at $u^{(t)}$ can be computed by the chain rule, and it turns out that it is just the second order differential operator

$L(v)= \sum \limits_{ij}a_{ij}(x)v_{ij}(x)$

where $a_{ij}(x)$ are the entries of the cofactor matrix of $D^2u^{(t)}(x)$.  The fact that $u^{(t)}$ is strictly convex means that $L$ is an elliptic operator, moreover it’s coefficients are $C^{\alpha}$.Now I claim that $L$ is a one to one bounded linear map* from $C^{2,\alpha}_0$ to $C^\alpha$.

This means that the differential of $T$ at  $u^{(t)}$ is regular, and by the inverse function theorem $T$ is a diffeomorphism between a neighborhood of $u^{(t)}$ in $C^{2,\alpha}_0$ and a neighborhood of $T(u)$ in $C^{\alpha}$. But then we can find $\delta>0$ such that any $s$ such that $|t-s|<\delta$ the function $\psi_s$ lies in this neighborhood, and thus one can find a $u^{(s)} \in C^{2,\alpha}_0$ such that $det(D^2u^{(s)})=\psi_s$,  i.e. Problem $(s)$ has a solution and thus $s \in A$, which shows $A$ is open.

*Remark: The fact that $L$ is one to one is a consequence of the Schauder theory for linear 2nd order equations, but that is something that could be discussed by itself in several posts! The theory says explicitely that given a differential operator as $L$, strictly elliptic and with $C^\alpha$ coefficients, then for any $f \in C^\alpha(\Omega)$ there exists a unique $u \in C^{2,\alpha}(\Omega)$ that vanishes on $\partial \Omega$ and such that

$Lu=f \mbox{ in } \Omega$

That takes care of Part I. Showing that $A$ is closed is usually the more interesting part and it changes from problem to problem (showing $A$ is open is always more or less the same). I will do the second part in the next post