Archive for June, 2009

Antonio Ache joins!

Antonio Ache, a grad student at Wisconsin will be blogging with us starting this week (after all, this is meant to be a group blog!). He is a good friend and we have known each other since our undergraduate days.

Antonio will begin by telling us about the problem of finding Riemannian metrics with given Ricci curvature. Welcome Antonio!

PS: On an unrelated note, I am writing this post from my phone! the wordpress application for the iPhone works perfectly. Yes, this means I can now blog whenever I find myself lost during a math talk…

Solving the Monge-Ampere equation (continued)

In a previous post I began the proof of the following theorem:

Let $\psi \in C^\alpha$ be a positive function in $B_1$, then there exists a unique function $u \in C^{2,\alpha}(B_1)$ such that

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

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

To prove the theorem, we looked at the function

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

we noted that it is easy to solve the problem explicitely for $\psi=\psi_0$.  Then thanks to the inverse function theorem (the Banach space version) we saw that the set $A$ of $t$‘s for which we can solve the equation $A$ is open (recall that $A$ was defined as the set of those $t$ in $[0,1]$ for which we can solve the equation above with right hand side  equal to $\psi_t$).

Since $A$ is open and non-empty, if we show that its also closed the theorem would be proved.

Part II:  Showing $A$ is closed.

(see previous post of the series for Part I)

That $A$ is closed will be shown to be a consequence of the following a priori estimate:

Theorem (A priori estimate for the Monge-Ampere equation) Let $u$ be a smooth solution of

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

$u= 0 \mbox{ on } \partial B_1$

There is a constant $C>0$ (depending only on $n$ and the $C^3$ norm of $\psi$) such that

$||u||_{C^{2,\alpha}}\leq C$

Proving the a priori estimate is usually the hardest step, so first let’s see that the a priori estimate in fact implies that $A$ is closed: let $t_k$ be a sequence in $A$ converging to a number $t \in [0,1]$, by the a priori estimate above the sequence of solutions $u^{(t_k)}=u^{(k)}$ is uniformly bounded in $C^{2,\alpha}$, thus by the Arzela-Ascoli theorem a subsequence converges to a function $u \in C^{2,\alpha}_0$, since $\psi_{t_k}$ converges to $\psi_t$ we see this $u$ is a solution of Problem $(t)$, i.e. $t \in A$. This proves $A$ is closed and therefore the theorem.

Of course, now we “just” have to prove the a priori estimate. ..

The broad outline of its proof is: first we prove a series of a priori estimates, each stronger than the previous one, and at each given step we  use the estimates already achieved, this will bound the first and second derivatives, the main tool here is the maximum principle used as in the classical technique of Bernstein. Then, the Holder regularity of the second derivatives will be consequence of deep regularity results from the theory of fully non-linear equations, among them the Evans-Krylov theorem. Now lets go through the argument in detail, and to try to make this just a bit readable I will number the steps

Step 1) The easiest estimate is the $L^\infty$ bound

(a priori estimate (1))

$||u||_{L^\infty(B_1)} \leq C$

where $C$ depends only on the $L^\infty$ norm of $\psi$. To prove it, consider the auxiliary function $\phi(x)=\frac{M}{2n}(|x^2|-1)$, this is a smooth function that vanishes on $\partial B_1$ and $D^2\phi = M^n$, picking $M$ large enough (and controlled by the supremum of $|\psi|$) we have

$det(D^2\phi) \geq \psi = det(D^2u)$

then the maximum principle implies that

$u \geq \phi \geq -\frac{M}{2n}$ in $B_1$

plus $u$ is a convex function  vanishing on $\partial B_1$, so we have $u \leq 0$ as well. This proves the a priori estimate (1).

Step 2) Now we go for a Lipschitz bound for $u$, which we will refer to as the

(a priori estimate (2))

$||\nabla u||_{L^\infty(B_1)} \leq C$

Let’s differentiate the equation $det(D^2u)=\psi$ in the direction of the unit vector $\xi$, and we obtain

$L u_\xi= a_{ij} (u_\xi)_{ij}=\psi_\xi$

Here $a_{ij}(x)$ correspond to the entries of the cofactor matrix of $D^2u(x)$, what the left hand side above amounts to is the linearization of $det(D^2u)$ at $u$, which is the elliptic operator $L$ (we did this before, in the previous post). Notice also that I can talk freely about third derivatives of $u$ because any strictly convex, $C^{2,\alpha}$ solution of the Monge-Ampere equation is actually $C^\infty$ (that this is true is another of the many consequences of  the Schauder theory for second order  linear elliptic equations with Holder continuous coefficients, and its not too hard to check), so the equation above holds classically. In particular, by the maximum principle (again) we conclude that for some $C$ depending on the supremum of $|\psi|$:

$\sup\limits_{B_1} |u_e| \leq \sup \limits_{\partial B_1}|\nabla u|+C$

but $u \equiv 0$ on $\partial B_1$ thus there $\nabla u = u_n n$, where $n$ is the outer normal to $\partial B_1$. Here we bring back our old friend, the function $\phi$ defined in step 1), as we know $u\geq \phi$ on $B_1$ and they agree on the boundary, therefore

$u_n \leq \phi_n$ on $\partial B_1$

also, since $u \leq 0$ and vanishes on $\partial B_1$ we also have the inequality $0 \leq u_n$. Therefore the $|u_n|$ is  controlled by $M$ times a dimensional constants, i.e.  the supremum of $|\psi|$ controls $|\nabla u|$. Putting the last three identities/inequalities together we get the a priori estimate (2).

Step 3) Notice that up to now we have only used the ellipticity of the equation and not the particular structure of the Monge-Ampere equation (even the differentiation step of Step 2) is a fact that holds for general elliptic equations). This  will change now that we are going to prove the

(a priori estimate (3))

$||D^2 u||_{L^\infty(B_1)} \leq C$

We start off as in step 2 by differentiating along a fixed, arbitrary direction $\xi$,  except now we do it twice and thus we get something much more complicated than before. We get around this noting that the map $M \to det(M)$ is a convex function on positive definite matrices, and get an inequality that allows us to drop the lower order terms:

$Lu_{\xi \xi} \geq \psi_{\xi\xi} \geq -C$

and this says that  $\sup \limits_{B_1} u_{\xi\xi}$ is controlled by the same supremum on $\partial B_1$ plus a constant $C$ controlled by $\sup |\psi|$.  Then we only have to estimate things on the boundary (just as for the first derivatives).

So lets pick a point $x_0 \in \partial B_1$ and estimate the entries of $D^2u(x_0)$. Since on $\partial B_1$ we have $u \equiv 0$ all the first, and second order angular derivatives of $u$ at $x_0$ are zero, and this says a lot about the Hessian of $u$ on the boundary, its just a matter of relating partial derivatives in polar and Cartesian coordinates to see then (after doing an adequate rotation of our coordinate system) that the entries of the Hessian satisfy the relations:

$u_{ij}(x_0)=u_n(x_0)\delta_{ij}$ for $i,j

here “$n$” stands both for the number $n$ and the exterior normal direction $\partial B_1$ since in our rotated coordinate system the vector $e_n$ is normal to $\partial B_1$ at the point $x_0$.Since we already have an a priori bound for $u_n$ this gives us an a priori bound for the second partial derivatives on the boundary, except for the partial derivatives of the form $u_{in}$, $0\leq i \leq n$.

Here we use the fact that our equation is rotation invariant (that is, the determinant is invariant under rotations of the coordinate system). Since $u_{ij}$ ($i,j) is diagonal, we can compute at once $det(D^2u(x_0))$

$det(D^2u(x_0))=\psi(x_0)=u_n^{n-1}u_{nn}$

I claim that there exists a $\delta_0>0$  such that $u_n(x_0)>\delta_0$ on $\partial B_1$, and $\delta_0$ is bounded away from zero by a quantity controlled by $\inf \psi$. Therefore there exists a  $C$ controlled by $\psi$, such that

$u_{nn}\leq C\psi(x_0)\leq C\sup |\psi|$

Plus, $u_{nn}$ is non-negative (since $u$ is convex). Thus we have given a priori estimates for all the pure second order derivatives of $u$ on $\partial B_1$, and by the remarks made at the beginning of this step we have proved the a priori estimate (3)

Step 4) Putting the 3 previous estimates together, we have:

a priori estimate (4)

$||u||_{C^{1,1}(B_1)}\leq C$

Next, we have to push this to a $C^{2,\alpha}$ estimate in  $B_1$. This will be the content of the next post (I am sure we can agree that this post is long enough). In the next post I  will talk about the Evans-Krylov theorem and the boundary behavior of solutions to elliptic equations. Hopefully that will wrap up this series of posts on the Dirichlet problem for Monge-Ampere,  after this I might go back to the Minkowski problem.

Quick post: A theorem and an open problem about Gauss curvature

While I finish my next post on the Dirichlet problem for the Monge-Ampere equation, I thought I could mention two neat things I have learned from reading Jerry Kazdan’s survey on Prescribing the curvature of a Riemannian manifold, a short book that I recommend strongly (its a bit out of date, but hey!, geometry has advanced quite a bit in the last couple of years!)

First, a cute solution to an easily-stated problem, as  observed by Wallach and Warner in 1970

Theorem:  Given a compact smooth 2-d manifold and $\Omega$ is a 2-form such that $\int_M \Omega = 2\pi \chi(M)$, then there exists a smooth Riemannian metric $g$ on $M$ such that $\Omega = KdA$, where $K$ is the Gauss curvature of $g$ and $A$ is the area form of $g$.

Proof: Pick your favorite metric $g_0$ on $M$, we will  prove that the metric we are looking for is in fact conformal to $g_0$, for any smooth function $u$, define $g_u = e^{2u}g_0$. Using the well known formulas for the Gauss curvature and area form of $g_u$, one arrives at the identity

$K_udA_u=K_odA_0-(\Delta_0 u) dA_0$

where the sub-indexes $0$ and $u$ refer to the object corresponding to the metric $g_0$ and $g_u$, in particular, $\Delta_0$ is the Laplace operator for $(M,g_0)$. Then one wants to pick $u$ such that

$K_odA_0-\Delta_0 u dA_0=\Omega$

or $(\Delta_0 u) dA_0 = K_0dA_0-\Omega$

but, using the metric $g_0$ we can write $\Omega = f(x)dA_0$ for some $f$ (or equivalently, using the Hodge star operator given by $g_0$) thus the $u$ we want is the solution of the equation

$\Delta_0 u = K_0-f (**)$

but since $\int_M \Omega =\int_MfdA_0= 2\pi\chi(M)=\int_MK_0dA_0$ we have that $\int_MK_0-fdA_0=0$ thus by standard elliptic theory (or Fredholm theory, etc) there exists a smooth function $u$ solving equation (**). Thus the metric $g_u$ is the one we were looking for and the theorem is proved.

Now, an easily stated problem whose solution is not likely to be as short, and has led to a lot of research in the last 2 decades. I  read about it for the first time in an interview with Louis Nirenberg, which can be found here.

Open problem:  Given a two dimensional Riemannian manifold, can we embed it isometrically (even just locally) in three dimensional Euclidean space?

Recall that Gauss curvature solely determines the local geometry of a 2d manifold, so its not surprising that this problem is equivalent to the following problem involving the  (…tataaaaa! )  Monge-Ampere equation:

Open problem restated: Given a $C^\infty$ function $K(x,y)$, find (even just locally) a function $u(x,y)$ which solves the equation

$\frac{u_{xx}u_{yy}-u_{xy}^2}{\left (1+u_x^2+u_y^2 \right)^2} = K(x,y)$

but mind you, this is not your grandpa’s Monge Ampere equation, for it is not an elliptic equation unless $K$ is strictly positive. For general $K$, it is an equation that varies  between hyperbolic or elliptic according to the sign of $K$, so you run into real problems in the set of points where $K$ vanishes .  If on the other hand $K$ is strictly negative, Kazdan says that the equation above is solved using tools from non-linear hyperbolic equations, of which sadly I know nothing. The case $K >0$ is then dealt with the techniques from the last two and the next post, which I should finish in a day or two.

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

The Minkowski problem and the Monge-Ampere equation

So as it has been already noticed I haven’t posted much at all in a couple of months,  Sean has been doing a good job with the posting! Now that is summer and I have more free time I can get back to blogging. I won’t continue with my previous posts on minimal surfaces (at least for now), but rather talk for a couple of posts about something less heavy than geometric measure theory.

I would like to talk about the Minkowski problem. It is not only an  interesting and historically important problem in geometry but it also has deep connections with the theory of non linear elliptic equations.

The Minkowski problem goes as follows:

Given a strictly positive real function $f$ defined on $S^2$, find a strictly convex compact surface $\Sigma \subset \mathbb{R}^3$ such that the Gauss curvature of $\Sigma$ at the point $x$ equals $f(n(x))$, where $n(x)$ denotes the normal to $\Sigma$ at $x$.

The simplest example would be when $f(x)$ is identically equal to the constant $\frac{1}{R^2}$, in that case a solution to the Minkowski problem would be the sphere of radius $R$ (it also turns out that the Minkowski problem has a unique solution for a given $f$, so the sphere is THE solution to the Minkowski problem for $f \equiv R^{-2}$).

It is my understanding (although I only know very little about the historical background) that this problem was studied intensively (among many, many other people) by A.D. Alexksandrov and  A. Pogorelov. Aleksandrov developed a theory of weak solutions and showed existence of weak solutions. Years later, via the work of many people (particularly people working in non-linear PDE) it was shown that these weak solutions were actually smooth, classical solutions, thus proving that the problem always has a unique solution.

How do PDE’s get in the picture? I will do a computation that will make this matter clear, but first let me point out the analogy with the theory of minimal surfaces and the prescription of the mean curvature of a surface: Minimal surfaces are surfaces for which the mean curvature vanishes, thus if the graph of a function happens to be a minimal surface this function solves a PDE (“the minimal surface equation”), which happens to be a quasilinear PDE. The problem of showing the existence of classical solutions for the minimal surface equation and other problems from the calculus of variations motivated much of the work in PDE through part of the 20th century (read for instance, about the work of Ennio De Giorgi on Hilbert’s 19th problem). For the Minkowski problem what we are prescribing is not the average of the principal curvatures of the surface (mean curvature) but the product of the principal curvatures (Gauss curvature), the corresponding PDE is not a quasilinear PDE, but a fully non linear PDE closely related to the Monge-Ampere equation.

To explain this further, allow me to recall a classical computation for the Gauss curvature.

Let us consider a portion of a surface given by the graph of a function u(x,y) (any small patch can be written this way if we rotate and translate our coordinate system), it is well known that one can express the Gauss curvature of this surface in terms of partial derivatives of $u$. I am going to assume that if you are reading this you are familiar with geometry of surfaces, if not, there are a couple of good wikipedia articles on this, such as the one on the Weingarten map (aka the Shape operator).

To compute the the Gauss curvature, I am going to use the fact that the Gauss curvature equals the determinant of the differential of the Gauss map. The Gauss map is the map that associates to each point on the surface the unique unit vector parallel to the unit normal at that point), this defines a map from  the surface $\Sigma$ to the sphere $S^2$. Then its differential its just a map from the tangent plane at a point in $\Sigma$ to a tangent plane to $S^2$.

In the case when the surface is the graph of $u$, in the coordinates $(x,y) \to (x,y,u(x,y))$ the Gauss map is given by

$N_u(x)=\left ( \frac{-\nabla u}{\sqrt{1+\nabla u^2}}, \frac{1}{\sqrt{1+\nabla u^2}} \right )$

Then the Weingarten map (or second fundamental form, or shape operator) its the differential of this map, its determinant is the Gauss curvature, and computing the differential (I am cheating and won’t write the differential here) we see that the Gauss curvature equals

$\frac{det(D^2u)}{\left ( 1+|\nabla u|^2\right )^{\frac{n+2}{2}}}$

Then, if the surface $u$ parametrizes is a solution of the Minkowski problem, then

$\frac{det(D^2u)}{\left ( 1+|\nabla u|^2\right )^{\frac{n+2}{2}}}=f(N_u(x)) = g(\nabla u)$

where $g$ is the obvious function. Thus we have arrived at a PDE for $u$

$det (D^2u) = g(\nabla u) \left ( 1+|\nabla u|^2 \right)^{\frac{n+2}{2}}$

this is an equation of Monge-Ampere type.

There are several methods to construct solutions to these equations (weak solutions, approximation via polygonal surfaces, continuity method), some of these were developed independently in works of some people like A. Alexandroff, A. Pogorelov, P.L. Lions and S.Y. Cheng and S.T, Yau. One of these methods consists in building first solutions in a “weak” sense, known as Alexandroff solutions, which a priori might not be twice differentiable, so they are not solutions of the equation in the classical sense, THEN one proves that these weak solutions must actually be smooth (this is is usually called a regularity theory, and its considered the hardest part).  I will talk a bit about the existence of classical solutions on a future post on the continuity method.