Showing the Monge-Ampere equation is elliptic

Question

I have a question about my books definition of ellipticy on how it relates to the Monge-Ampere equation in $\mathbb R^2$.

The Monge-Ampere equation is as follows. Let $\Omega$ be an open subset of $ \mathbb R^2=\{x=(x_1,x_2)\}$ and let $u\in C^2(\Omega)$ satisfy $u_{x_1x_1}(x)u_{x_2x_2}(x)-u^2_{x_1x_2}(x)-f(x)=0$ where $f>0$ in $\Omega$ and the Hessian of $u$ is positive definite ($u$ is convex).

The definition of ellipticy given in my book is as follows:

Consider a general differential equation $F[u]=F(x,u(x),Du(x),D^2u(x))=0$ where $F:S:=\Omega \times \mathbb R \times \mathbb R^d \times S(d,\mathbb R) \to \mathbb R$ where $S(d,\mathbb R)$ is the space of real symmetric $d\times d$ matricies. Elements of $ S$ are written as $(x,z,p,r)$ where $p=(p_1,...,p_d)$ and $r=(r_{ij})_{1\le i,j \le d}$ The differential equation $F[u]$ is said to be elliptic at $u$ if $\left( \dfrac{\partial F}{\partial r_{i,j}}(x,u(x),Du(x),D^2u(x))\right)_{1 \le i,j \le d} $ is positive definite.

So using this definition how do we compute $ \dfrac{\partial F}{\partial r_{i,j}}(x,u(x),Du(x),D^2u(x)) $ for each pair of $(i,j)$? We can see that $r_{i,j}=u_{x_ixj}$ so for example does $$ \dfrac{\partial F}{\partial r_{1,1}}(x,u(x),Du(x),D^2u(x)) = \dfrac{\partial}{u_{x_1x_1}}\left(u_{x_1x_1}(x)u_{x_2x_2}(x)-u^2_{x_1x_2}(x)-f(x)\right)=u_{x_2x_2}(x) $$ Then by this logic $\left( \dfrac{\partial F}{\partial r_{i,j}}(x,u(x),Du(x),D^2u(x))\right)_{1 \le i,j \le d}= \begin{bmatrix} u_{x_2x_2} & -2u_{x_1x_2} \\ -2u_{x_1x_2} & u_{x_1x_1} \\ \end{bmatrix}$ but if this is correct why is this matrix positive definite? Also where does the condition of $f>0$ come into play.

Any help is appreciated!

Answer 1

The Monge-Ampere equation is not elliptic in general. You need to assume that the solution $u$ is convex. Take a look at wikipedia wikipedia or at the review paper De-Phillipis-Figalli

Answer 2

@user10354138's last comment is right. I think it is not helpful to write $r_{ij}=u_{x_ix_j}$ at first. Write $F[A] = a_{11}a_{22}-a_{12}a_{21}$, then the matrix: $$ F^{ij} = \frac{\partial F}{\partial a_{ij}} = \begin{pmatrix}a_{22} & -a_{21}\\ -a_{12} & a_{,11}\end{pmatrix}$$.

This is positive definite when its evaluated at $D^2u$, provided $u$ is smooth and convex.

Alternatively, regardless of the dimension you can use the derivative of the determinant: $$\frac{d}{dt} [detA(t)]=\det A(t) \cdot \mbox{trace}\left[A^{-1}(t)\cdot \frac{d}{dt} A(t)\right]$$.