### Update: Blog restarted, new name+new blog address!

After an almost year long hiatus I am getting back into math blogging.  The new address is a bit long but at least easy to remember! (turns out “pdeblog.wordpress.com” was already taken!), please update your bookmarks!

The old site  (the one you are looking at now) will be up at least for a while, but all the old posts have been copied to the new blog, To reflect the fact that nearly all posts here have dealt with Partial Differential Equations the new site will go by the name  “PDE Blog”.

### Reviewing the regularity theory of elliptics PDEs via the Laplace equation. Part III. (representation formulas)

This is the third of a series of posts dealing with the regularity theory of elliptic equations. My motivation in writing these is outlined in the first post. The previous post is here.

Let us recall Green’s identity, if ${u,v}$ are any functions smooth in ${\bar{\Omega}}$ and ${\Omega}$ is a bounded domain with smooth boundary we have

$\displaystyle \int_\Omega u\Delta v - v \Delta u dx = \int_{\partial \Omega}u\frac{\partial v}{\partial \nu}-v\frac{\partial u}{\partial \nu}dS$

this identity can be obtained with a couple of integration by parts involving the vector fields ${u \nabla v}$ and ${v \nabla u}$.

Lets rewrite the identity as

$\displaystyle \int_\Omega u\Delta vdx = -\int_\Omega v \Delta u dx + \int_{\partial \Omega}u\frac{\partial v}{\partial \nu}-v\frac{\partial u}{\partial \nu}dS$

thus, at least formally, if somehow we could find for every ${x \in \Omega}$ a function ${v_x(y)}$ such that

$\displaystyle \Delta v_x(y)=\delta_x(y)$

$\displaystyle v_x(y) \equiv 0 \mbox{ on } \partial \Omega$

then Green’s identity applied to both ${u}$ and ${v_x}$ in ${\Omega}$ would give us an integral representation formula for harmonic functions

### Reviewing the regularity theory of elliptics PDEs via the Laplace equation. Part II.

This is the second of a series of posts dealing with the regularity theory of elliptic equations. My motivation in writing these is outlined in the first post.

Some consequences of Harnack’s inequality the Mean value property

The mean value property is characteristic of harmonic functions, but the fact that harmonic functions control their pointwise values by their local average is a general fact that is characteristic of elliptic equations (as we will see later, less sharp but more general theorems for nonlinear elliptic equations still have this flavor and are at the very heart of the regularity theory of fully nonlinear elliptic PDEs). Let me mention a few of its consequences, I already talked last time about Harnack’s inequality, as it follows from the mean value theorem, the mean value theorem (at least for harmonic functions) is more fundamental.

### Joint Mathematics Meetings in San Francisco -Blog!

My friend and former UT graduate student Adriana Salerno (currently at Bates) will be running the 2010 AMS Joint Math meetings blog. She was also in charge of the blog in previous years (you can check them out here and here). I recommend you check it out in the next few days to see what has been going on at the meetings (specially if, just like me, you don’t happen to be in San Francisco this week).

### Reviewing the regularity theory of elliptics PDEs via the Laplace equation. Part I.

There is a tedious, simple but hopefully fruitful exercise I always wanted to do. It is to review all the different proofs of the Harnack inequality and regularity of solutions to elliptic equations that I know, but only for the Laplace equation. First, because it is a good way to really get your hands on some of the ideas of several deep theorems (like those of De Giorgi-Nash-Moser and Krylov-Safonov) in the simplest possible setting. Second, because looking at all the different proofs it is possible to trace the evolution of analysis and PDEs through the last century (and a bit before that) and appreciate the level maturity reached in several fields: potential theory, singular integrals, calculus of variations, fully non linear elliptic PDE and free boundary problems. The simple’ and elementary’ Laplace equation lies at the intersection of all these fields, so every new breakthrough reflected on our understanding of this equation, each new proof emphasizing a different approach or point of view. Each of the proofs that I will discuss are based on one of the following:

• The mean value property (the proof you learn in your typical complex variables or introductory PDE course).
• The Poisson Kernel for the ball (the proof from potential theory).
• The Calderón-Zygmund theorem (ok not exactly a Harnack inequality’, but it should be on this list anyway) which uses the machinery of singular integrals.
• The De Giorgi-Nash-Moser theorem, which follows the variational point of view and it is best suited for quasilinear equations or equations in divergence form.
• The Aleksandrov-Bakelman-Pucci estimate and the Krylov-Safonov’s Harnack’s inequality’, which follows the comparison principle point of view and it is best suited for fully non linear equations or equations in non-divergence form.

So I am going to review each theorem and its proof but only for Laplace’s equation: ${\Delta u = 0}$. To start off easy, I am going to do first the proof via the mean value property.

First proof: mean value property

The mean value property says basically this

Let ${u}$ be a ${C^2}$ function in the unit ball ${B_1}$ of ${\mathbb{R}^n}$. If ${\Delta u = 0}$ and ${S}$ is a sphere contained in ${B_1}$ and centered at ${x_0}$, then ${u(x_0)}$ equals the average of ${u}$ on ${S}$

It is not hard to prove with some calculus, one basically looks at the function `Average of ${u}$ on the sphere of radius ${r}$ centered at ${x_0}$‘=${f(r)}$ and shows that ${f'(r)=0}$, and since by continuity ${f(0)=u(x_0)}$, the theorem follows. To show ${f'(r)=0}$ one sees (by say, a change of variables) that ${\frac{d}{dr} \frac{1}{|S|}\int_{S_r}u(x)d\sigma =\frac{1}{|S}\int_{S_r}u_nd\sigma}$ and this last integral is zero thanks to Stokes’ theorem and the fact that ${\Delta u = 0}$. Moreove, integrating the result with respect to the radius of the sphere one gets the same statement where instead of average over a sphere we have an average over a ball.

With this, one may prove easily Harnack’s inequality for harmonic functions, which I will state formally for the first time

Theorem 1 For any nonnegative harmonic function ${u}$ in ${B_1}$ we have the inequality

$\displaystyle u(x) \leq 2^nu(0) \;\;\; \mbox{ for all } x \in B_{1/2}$

Proof. Let ${x \in B_{1/2}}$, then the ball of radius ${1/4}$ centered at ${x}$ (call it ${B}$) is completely contained in ${B_1}$, thus by the mean value property

$u(x)=\int_B u(y)dy$

but $B$ is also contained in $B_{1/2}$ and since $u$ is nonnegative we have $\frac{1}{|B|}\int_B u(y)dy \leq \frac{2^n}{|B_{1/2}|} \int_{B_{1/2}}u(y)dy=2^nu(0)$, again by the mean value property. This finishes the proof.

That is for today, in the next post I will explain some of the consequences of this theorem and maybe move on to the proof with potential theory methods.

(Note: this post was made using Luca Trevisan’s Latex to WordPress program, which is very useful although I am still getting used to using it. It allows you to prepare your post in a latex editor and then translate it into HTML code which WordPress can read, I strongly recommend it)

### New year, new posts

After 4 months of inactivity, I am taking up again the task of updating the blog, which has suffered of neglect due to my terrible time management skills.

I am not going to take off from where I left last time (namely, the posts about the Minkowski problem, which I will finish, someday), but instead will start the year with some shorter, lighter posts. I plant to start with a few posts about varifolds vs currents vs BV sets, and also about the Harnack inequality, maybe later I will write a bit about topics from phase transitions such as the Stefan problem or the Cahn-Hilliard equation

### If you want to kill your productivity, move to a new house!

I know things have been extremely slow lately, but I was moving last week (and well, that means packing everything, moving things to the new house, cleaning the old house, unpacking things at the new one… well, you get the picture).

My goal for this week is presenting Aleksandrov’s solution to  the Minkowski problem (see an earlier post I did introducing this problem). So I am going to leave you a problem as a preview, it is a sort of discrete version of the Minkowski problem:

Let $n_1,...,n_k$  be a family of non-coplanar unit vectors in $\mathbb{R}^3$ and let $\alpha_1,...,\alpha_k$ be positive numbers such that

$\sum \limits_{i=1}^k \alpha_in_i=0$

Then, show that there exists a convex closed polyhedron with exactly $k$ faces with normal vectors given by $n_1,...,n_k$ and corresponding areas $\alpha_1,...,\alpha_k$. Plus, this polyhedron is unique up to translation.

This is fact is not surprising (since it is not hard to check that any polyhedron has this property), but the proof is far from trivial. As you may guess, the proof cannot be constructive, it will use a continuity argument to show that there must be at least one such polyhedron. I will present this in my next post.

Don’t be scared, this has not been turned into a photo blog!  Today I received my copy of Singular Integrals I ordered on-line recently, which is funny given that I have been reading that book for a long time now (at least now I won’t have to borrow a copy from the library or from a friend).

In any case, this reminded me of something I heard once: basically, that throughout the early years of your career as a mathematician, there will be a  list of books (that will depend strongly on your research interests) that you must read completely and in full detail to the point where you are be able to reproduce their contents on command. So, wondering what that list should be for me, I piled up some books from my book shelf and took a picture.

…perhaps I should be reading some of those books instead of writing this blog post

### Solving the Monge-Ampere equation (continued… and finished).

I have been postponing this post for over a week due to lack of time, but finally here it is. This post ought to finish a series of past posts (here and here) where I have been describing the proof of existence of classical solutions to the Dirichlet problem for the Monge-Ampere equationvia the continuity method. Via the method of continuity  we reduced the question of existence of classical solutions to the problem of proving good a priori estimates for classical solutions, namely, we were trying to prove

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

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

Last time we got almost there, using the maximum principle and the right barriers we proved the estimate

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

Which is still not strong enough for our needs, so in order to finish the proof of the a priori estimate, we are going to use a powerful interior estimate for concave elliptic equations, proved independently by L.C. Evans and N. Krylov in the 80’s:
Theorem (Evans-Krylov) Let $u$ be a $C^{1,1}$ solution of the elliptic equation

$F(D^2u)=f(x)$ for $x \in B_1$

if $F$ is concave (or convex), then we have the following interior estimate

$||u||_{C^{2,\alpha_0}(B_{\frac{1}{2}})}\leq C||u||_{C^{1,1}(B_1)}$

where $C$ depends only on $F$, and $\alpha_0$ is a universal constant.

This a well known theorem, a couple of places where one can read it are the book of Gilbarg and Trudinger (last edition), or the book of Caffarelli and Cabre. More recently, Caffarelli and Silvestre have come up with a shorter proof, still based on the original ideas of Evans and Krylov, this proof is available in arxiv. Maybe I will talk about the proof in some other post, but for now I am just going to quote the result.

The Evans-Krylov theorem is an interior result, we need also control at the boundary, that is provided by a result of Krylov

Theorem (Krylov) Let $u$ be a solution to our equation, there is a universal $\beta$ and constant $C$ controlled by $|u|_\infty$ such that  for $x \in B_1$ and $y \in \partial B_1$ we have

$|D^2u(x)-D^2u(y)|\leq C|x-y|^\beta$

This actually a corollary of Krylov’s theorem, which is a more general and remarkable result about equations in non-divergence form with measurable coefficients, but again I want to focus on the Monge-Ampere equation, I will talk about Krylov’s theorem some other time, a good place to read about it is the last chapter of Kazdan’s book. With these two tools its a standard argument to show that for a constant $C$ controlled by the previous two, and for $\alpha=min(\alpha_0,\beta)$, we have

$|D^2u(x)-D^2u(y)|\leq C|x-y|^\alpha$

I won’t do it in detail, but the proof is not too hard: basically, if the two points are closer to each other than to the boundary, then the Evans-Krylov estimate (properly scaled) gives us the inequality above, otherwise, the two points are closer to the boundary than to each other, so by the estimate of Krylov we get the same inequality in this case, and thats it!

…and that finishes the proof!. I did not present the most general result to simplify the presentation(at least for the weaker a priori estimates, which is where I did most of the details), but one can work in a more general domain (as long as it is convex) and have arbitrary boundary conditions. A much more general result, which includes not only the Monge-Ampere equation but also the $k-$Hessian equation, was proven  in a  paper by Caffarelli, Nirenberg and Spruck.

### Prescribing the Ricci Curvature of Riemannian metrics

I would like to spend the next few posts talking about a problem I read about in chapter 5 of the book Einstein Manifolds by Arthur Besse (it turns out that the name Arthur Besse is made up as you can read in the preface of the book). The question that we want to address is the following

Given a compact smooth manifold without boundary $M$, when is it possible to find a Riemannian metric $g$ in $M$ satisfying $Ric(g)=r$ for a given Ricci candidate $r$?

It is of course too ambitious to try to answer this question in full generality but we can start by showing some examples of Ricci candidates for which this equation does not have a solution.

Trying to solve $Ric(g)=$ for $g$ amounts to solving a second order, quasilinear PDE on $g$, however, the main difficulty here is that the operator $g\mapsto Ric(g)$ is not elliptic.

A motivation for considering this problem comes from the question of existence of metrics with constant sectional curvature on $3$– manifolds (compact and without boundary). This of course has to do with the celebrated theorem of Richard Hamilton on the description of $3$– manifolds with positive Ricci curvature:

Theorem (Hamilton, 1982): Let $M$ be a connected, compact smooth $3$ dimensional manifold without boundary and suppose that $M$ admits a metric $g$ such that $Ric(g)$ is positive definite everywhere. Then $M$ also admits a metric with constant sectional curvature.

We will discuss some of the ideas involved in the proof of this theorem in future posts. A consequence of this result is that $M$ is diffeomorphic to the quotient of the $3$-sphere $S^{3}$ by a discrete group $G$.

Back to our original problem, recall that given a Riemannian metric $(M,g)$, the full curvature tensor is defined by

$Rm(X,Y,Z,W)=g(R(X,Y)Z,W)=\langle R(X,Y)Z,W\rangle$

Where

$R(X,Y)Z=\nabla_{X}\nabla_{Y}Z-\nabla_{Y}\nabla_{X}Z-\nabla_{[X,Y]}Z$

Here $\nabla$ is the Levi-Civita connection of $g$.

The Ricci tensor is then defined as

$Ric(X,Y)=\mathrm{tr}\left(Z\mapsto R(Z,X)Y\right)$.

Here are two basic properties of $Ric$:

1) $Ric$ is symmetric in $X$ and $Y$, i.e. $Ric(X,Y)=Ric(Y,X)$

2) In local coordinates $Ric$ looks as follows

$R_{ij}=\partial_{l}\Gamma_{ij}^{l}-\partial_{i}\Gamma_{lj}^{l}+\Gamma_{lm}^{l}\Gamma_{ij}^{m}-\Gamma_{im}^{l}\Gamma_{lj}^{m}$

We are using the summation convention (i.e we sum over repeated indices). The Christoffel symbols $\Gamma_{\alpha\beta}^{\gamma}$ are defined by

$\Gamma_{\alpha\beta}^{\gamma}=\frac{1}{2}g^{\gamma\nu}\left(\partial_{\alpha}g_{\beta\nu}+\partial_{\beta}g_{\alpha\nu}-\partial_{\nu}g_{\alpha\beta}\right)$

Where $\left(g^{\gamma\nu}\right)$ are entries of the matrix $g^{-1}$. This says that in local coordinates we can write schematically $Ric(g)=F(g,\partial g,\partial^{2}g)$ where $F$ is a $C^{\infty}$ function that depends linearly on the entries of $\partial^{2}g$. If we then want to solve $Ric(g)=r$ locally, property 2) tells us that we have to look at a system of the form

$F(g,\partial g,\partial^{2}g)=r$

Property 1) tells us that an admissible Ricci candidate $r$ has to be symmetric.

One encounters obstructions for solving this system right away. One of the main difficulties has to do with the fact that the Ricci tensor satisfies the differential Bianchi identity

$\delta Ric(g)=-\frac{1}{2}dR$

Where $\delta$ is the divergence operator respect to $g$ and $R$ is the scalar curvature of $g$ (the trace of the Ricci tensor). This says that if we define a 1-form $Bian(r,g)$ by $Bian(r,g)=\delta r+\frac{1}{2}d\mathrm{tr}(r)$, then in order for $g$ to satisfy $Ric(g)=r$ we must have

$Bian(r,g)=0$

To write $Bian(r,g)$ in coordinates, we start with

$Bian(r,g)_{k}=g^{ij}\left(\nabla_{i}r_{jk}-\frac{1}{2}\nabla_{k}r_{ij}\right)$

From

$\nabla_{a}r_{bc}=\partial_{a}r_{bc}-\Gamma_{ab}^{l}r_{lc}-\Gamma_{ac}^{l}r_{bl}$

(which is just the definition of $\nabla r$ in coordinates) and from the expression in local coordinates of the symbols $\Gamma$, we easily see that

$Bian(r,g)_{k}=g^{ij}\left[\partial_{i}r_{jk}-\frac{1}{2}\partial_{k}r_{ij}-g^{ls}r_{ks}\left(\partial_{i}g_{js}-\frac{1}{2}\partial_{s}g_{ij}\right)\right]$

As discussed in Besse’s book, Dennis DeTurck came up with examples of symmetric tensors that cannot satisfy the Bianchi identity respect to any metric. One of his examples is the following

Consider in $\mathbb{R}^{n}$ a symmetric tensor of the form

$r=x_{1}dx_{1}\otimes dx_{1}+\displaystyle{\sum_{1

The existence of a metric $g$ such that $Ric(g)=r$ implies as we saw before that $Bian(r,g)=0$. In particular, from our expression for $Bian$ we must have

$0= Bian(r,g)_{1}=g^{ij}\left[\partial_{i}r_{j1}-\frac{1}{2}\partial_{1}r_{ij}-g^{ls}r_{1s}\left(\partial_{i}g_{js}-\frac{1}{2}\partial_{s}g_{ij}\right)\right]=$

$=\frac{1}{2}g^{11}+x_{1}g^{l1}g^{ij}\left(\partial_{i}g_{js}-\frac{1}{2}\partial_{s}g_{ij}\right)$

This implies that on the hyperplane $x_{1}=0$, the metric $g$ must satisfy $g^{11}=0$ which is impossible for a Riemannian metric. It follows that for any point $p$ in the hyperplane $x_{1}=0$ the equation $Ric(g)=r$ has no solution near $p$. Notice that at these points the tensor $r$ is singular.

In the next post we will interpret the existence of examples like the one we have just discussed as a consequence of the non-ellipticity of the system $g\mapsto Ric(g)$.