(You may see the rest of the lecture notes here).

Part II: De Giorgi-Nash-Moser theory.

Hilbert’s 19th problem

Among Hilbert’s problems from his 1900 address he included the following (Problem #19):

One of the most remarkable facts in the elements of the theory of analytic functions appears to me to be this: That there exist partial differential equations whose integrals are all of necessity analytic functions of the independent variables, that is, in short, equations susceptible of none but analytic solutions…

…and the question naturally arises, whether all solutions of regular variational problems must necessarily be analytic functions. In other words does every Lagrangian partial differential equation of a regular variational problem have the property of admitting analytic integrals exclusively?.

Let us restate the above in equivalent but more modern notation: Consider $u$, a minimizer of the following variational problem (where $\Omega \subset \mathbb{R}^n$ a bounded domain with smooth boundary and $g$ is smooth), \begin{align} \text{Minimize } \;\;\;\; & \int_{\Omega}F(\nabla u(x),u(x),x)dx \\ \
\text{subject to } \;\;\;& u = g \text{ on } \partial \Omega, \end{align} If we suppose $F(p,z,x)$ is uniformly convex in $p$ and analytic in all variables, does it follow that the minimizer $u$ is an analytic function in $\Omega$?

The works by many authors towards the solution of this problem is partly responsible for the development of an important part of the tool kit used by mathematicians working in nonlinear elliptic or parabolic PDE.

For the sake of the ongoing discussion we are going to consider a simpler situation capturing the essence of the problem. An important subclass is when the Lagrangian $F$ only depends on the gradient. Among such examples the two most studied ones are \begin{align} F(\nabla u,u,x) = |\nabla u(x)|^2 \end{align} which corresponds to the Dirichlet energy and harmonic functions, \begin{align} F(\nabla u,u,x) = \sqrt{1+|\nabla u(x)|^2}, \end{align} which corresponds to the problem of finding a function whose graph is a minimal surface.

In both cases $F$ is twice differentiable and convex in the argument (and uniformly convex in the first example), so going forward we are going to look at the minimizers of the above problem when integral to be minimized is of the form \begin{align} \int_{\Omega}F(\nabla u(x))dx, \end{align} where $F:\mathbb{R}^n\to\mathbb{R}$ is a real analytic, globally $C^2$, uniformly convex function. In particular, there are positive constants $\Lambda,\lambda$ such that \begin{align} \lambda \text{I} \leq D^2F(p) \leq \Lambda \text{I} \;\forall\;p\in\mathbb{R}^n. \end{align} By the mid 20th century Hilbert’s 19th problem had been solved for $n=2$ by Morrey, but the general case $n\geq 3$ remained open. Important partial progress had been made by several works, among them works of Schauder and of Hopf.

As a result of Schauder’s and Hopf’s works it was known by the 1950’s that if a minimizer is differentiable with a Holder continuous gradient then $u$ will be analytic. This was possible by showiing first that such a $u$ had to be $C^\infty$ (thanks to Schauder’s work), and then in turn analytic (thanks to Hopf’s work). The last missing piece was establishing minimizers indeed had a Holder continuous gradient.

If $u$ is a minimizer, we expect it to be a critical point of the functional. In fact, the following holds \begin{align} \int_{\Omega} (DF(\nabla u))\cdot \nabla \phi(x)\;dx = 0 \;\forall\; \phi \in C^\infty_c(\Omega). \end{align}

The significance of this identity is that $u$ is a weak solution of the Euler-Lagrange equation for the variational problem, \begin{align} \text{div}\left ( DF(\nabla u) \right ) = 0. \end{align}

This equation is translation invariant, and as a result, one obtains an interesting equation when taking the derivative of both sides in some direction $e \in\mathbb{S}^{n-1}$. Indeed, we have (at least heuristically) that if $w = \partial_e u$, then $w$ must solve

\begin{align} \text{div}\left ( D^2F(\nabla u)\nabla w \right ) = 0 \end{align} For emphasis, $w$ solves the elliptic equation \begin{align} \text{div}(A_u(x)\nabla w) = 0,\; \text{ where } A_u(x) := (D^2F)(\nabla u(x)). \end{align}

In 1957 De Giorgi managed to bridge the gap between the regularity needed to reach analyticity and the natural weak regularity of minimizers. His result says (among other things) that any weak solution to an equation of the type above for $w$ must be Holder continuous. Applying this to any directional derivative of $u$ he concluded $u$ has a Holder continuous gradient, and thus that $u$ must be analytic as stated in Hilbert’s problem.

De Giorgi’s theorem: the big picture

De Giorgi’s first revolutionary idea was to forgo the relationship between $A_u(x) = (D^2F)(\nabla u(x))$ and the function $u$, working with only minimal information about it, namely, $A = A_u$ satisfies \begin{align} \lambda \text{I} \leq A(x) \leq \Lambda \text{I} \text{ for a.e. } x \in \Omega.
\end{align} This is an inequality that holds for any function $u \in H^1(\Omega)$ and it comes from the uniform convexity and second derivative bounds assumed of $F$. Using no other information, De Giorgi showed an $H^1$ solution of $\text{div}(A(x)\nabla u)=0$ must be Holder continuous.

De Giorgi’s second big idea was in the way he proved the Holder regularity. This proves is broken in two partsm today we will discuss only the first. De Giorgi proves what he calls an “in-measure to uniform” estimate, allowing him to control the pointwise values of a weak solution in terms of its $L^2$ norm.

The proof of the “in-measure to uniform” estimate, or $L^2$ to $L^\infty$, involved a powerful new idea that ever since has become a key part of the arsenal of any analyst working in diffusive PDE: he plays two inequalities against each other, and at every scale (these inequalities involve integrals of the solution and its derivatives over balls of smaller and smaller radii), exploiting the fact these inequalities have different “homogeneities”.

The different homogeneities creates a nonlinear effect through an iteration procedure that starts in a ball of radius $r$ and stops before it goes a ball of radius below $r/2$. The iteration procedure consists in him checking that the $L^2$ norm of the solution above some level set must be decreasing as as one increases the level set and shrinks the ball. In this way he shows that the $L^2$ norm of $u$ above a definite finite level set must be zero, which means $u$ is in $L^\infty$.

This nonlinear mechanism can be understood very simply in the continuum setting, with a continuous time variable playing the role of the number of steps. It amounts to a basic exercise in ODEs.

Exercise. Consider a non-negative non-increasing continuous function $\phi(t):[0,T]\to\mathbb{R}$ such that $\phi$ is differentiable in its positivity set $\{t\in[0,T]: \phi(t)>0 \}$ and that there are positive numbers $c,\delta>0$ such that

\begin{align} \phi’(t) \leq -c \phi(t)^{1-\delta} \text{ in } \{t\in[0,T]: \phi(t)>0 \}. \end{align}

Then, $\phi(t) = 0$ for any $t$ larger than $t_*$, where

\begin{align} t_* := (\delta c)^{-1}\phi(0)^\delta. \end{align}

Hint: Argue as one does in separation of variables (for $t$ in the positivity set) to show that for $t$ in such a set we have $\phi(t)^\delta \leq \phi(0)^\delta -\delta c t$.

The power of De Giorgi’s idea can be illustrated from this exercise. Consider a set of finite perimeter $E$ which is area minimizing in some domain $\Omega \subset \mathbb{R}^n$, that is, $E$ is minimal in $\Omega$ in the sense that for any other set of finite perimeter $E’$ we have

\begin{align} E’\setminus \Omega = E \setminus \Omega \Rightarrow \text{Per}(E’) \geq \text{Per}(E). \end{align}

We are going to show an important property of minimal boundaries. Such an $E$ has the property that if $E$ occupies a relatively small proportion of the measure of a ball $B_r(x_0) \subset \Omega$, then it cannot reach the smaller ball $B_{r/2}(x_0)$ at all. Visually, this says $E$ cannot have spikes that are too long.

Lemma. Suppose $E$ is minimal in $\Omega$. There is a dimensional constant $\alpha_n \in (0,1)$ such that if for some ball $B_r(x_0) \subset \Omega$ we have \begin{align} |E\cap B_r(x_0)| \leq \alpha_n|B_r(x_0)|, \end{align} then \begin{align} E \cap B_{r/2}(x_0) = \emptyset, \end{align} In fact, one may choose $\alpha_n = 4^{-n}$.

Proof.

We fix $E$ a set of finite perimeter which has a minimal perimeter in $\Omega$ in the sense described above.

Fix $r$ such that $B_r(x_0) \subset \Omega$ and define for $t\in [0,r]$, \begin{align} \phi(t) = |B_{r-t}(x_0)\cap E|. \end{align} Now, we have \begin{align} \phi’(t) = -\text{Per}(B_r(x_0),E) \end{align} Due to $E$’s minimality, we have for any $\rho \in (0,r)$ \begin{align} \text{Per}(E,B_\rho(x_0)) \leq \text{Per}(B_\rho(x_0),E) \end{align} On the other hand, \begin{align} \text{Per}(E\cap B_\rho(x_0)) = \text{Per}(E,B_\rho(x_0))+\text{Per}(B_\rho(x_0),E). \end{align} Therefore, \begin{align} \text{Per}(B_\rho(x_0),E) \geq \frac{1}{2}(\text{Per}(E,B_\rho(x_0))+\text{Per}(B_\rho(x_0),E)) = \frac{1}{2}\text{Per}(E\cap B_\rho(x_0)). \end{align} From here, we conclude that \begin{align} \phi’(t) \leq -\frac{1}{2}\text{Per}(E\cap B_{r-t}(x_0)). \end{align} This inequality is then a consequence of the minimality of $E$. A different inequality that is true even if $E$ is not minimal is the isoperimetric inequality applied to $E\cap B_{r-t}(x_0)$, \begin{align} \text{Per}(E\cap B_{r-t}(x_0)) \geq n|B_1|^{\frac{1}{n}}|E\cap B_{r-t}(x_0)|^{\frac{n-1}{n}} = n|B_1|^{\frac{1}{n}}\phi(t)^{\frac{n-1}{n}}. \end{align} Putting the two inequalities together we have \begin{align} \phi’(t) \leq -\frac{1}{2}n|B_1|^{\frac{1}{n}}\phi(t)^{\frac{n-1}{n}} = -\frac{1}{2}n|B_1|^{\frac{1}{n}}\phi(t)^{1-\frac{1}{n}} = -c_0 \phi(t)^{1-\delta}. \end{align} Then, applying the previous exercise to this $\phi(t)$ we conclude that $\phi(t)=0$ if \begin{align} t \geq (\delta c)^{-1}\phi(0)^{\delta} = \frac{2}{|B_1|^{\frac{1}{n}}} |E\cap B_r(x_0)|^\frac{1}{n} \end{align}

We would like $\phi(r/2)=0$, so we need at least \begin{align} r/2 \geq \frac{2}{|B_1|^{\frac{1}{n}}} |E\cap B_r(x_0)|^\frac{1}{n}. \end{align}

Given the other assumption, this will hold as long as \begin{align} \frac{1}{2}r \geq \frac{2}{|B_1|^{\frac{1}{n}}} (\alpha_n |B_1|r^n )^\frac{1}{n} = 2\alpha_n^{\frac{1}{n}}r. \end{align} Therefore, all we need of $\alpha_n$ is that \begin{align} \alpha_n \leq \frac{1}{4^n}. \end{align} This proves the lemma.

De Giorgi’s $L^\infty$ estimate for elliptic equations

Consider a domain $\Omega$ bounded and with a smooth boundary. A function $u:\Omega \to \mathbb{R}$ is a weak solution of the Dirichlet problem \begin{align} \text{div}(A\nabla u) & = 0 \text{ in } \Omega \\ \
u & = g \text{ on } \partial \Omega, \end{align} if $u \in H^1(\mathbb{R}^n)$, $u_{\mid \partial \Omega} = g$, and for any $\phi \in C^\infty_c(\Omega)$ we have \begin{align} -\int_{\Omega}A(x)\nabla u\cdot \nabla \phi \;dv = 0 \end{align}

In what follows, $u$ is such a solution with a matrix field $A(x)$ which is measurable in $x$ and such that for some positive constants $\lambda,\Lambda$ we have \begin{align} \lambda \text{I} \leq A(x) \leq \Lambda \text{I} \text{ for } a.e. x\in\Omega. \end{align}

We are going to describe now the two inequalities with different homogeneities that will play the role for us played by the minimiality of $E$ and the isoperimetric inequality for the minimal surfaces lemma above.

The first is known as Caccioppoli’s inequality, and also sometimes called the energy inequality.

Lemma (Caccioppoli's inequality). Suppose $u$ is a weak solution and $A(x)$ satisfies the ellipticity condition above. Then, given $\eta \in C^2_c(\Omega)$, and $\alpha \in \mathbb{R}$, we have \begin{align} \lambda \int_{\Omega} |\nabla(\eta (u-\alpha)_+)|^2\;dx \leq \Lambda \int_{\Omega} |\nabla \eta|^2(u-\alpha)_+^2\;dx, \end{align} as well as \begin{align} \lambda \int_{\Omega} |\nabla(\eta (u-\alpha)_-)|^2\;dx \leq \Lambda \int_{\Omega} |\nabla \eta|^2(u-\alpha)_-^2\;dx. \end{align}

The above inequality (or family of inequalities) is the one point in the proof where we make use of the information that $u$ is a weak solution of the PDE. Accordingly it plays a role analogous to the minimization property in the minimal surface lemma.

The inequality that plays the role of isoperimetric inequality is the Sobolev embedding (so, a property of any function, not just solutions).

Theorem (Sobolev embedding). \begin{align} ||u||_{2^*} \leq C_n ||\nabla u||_2,\; \text{ where } 2^* := \frac{2n}{n-2}. \end{align}

Using Caccioppoli’s inequality and the Sobolev embedding we will prove the $L^2$ to $L^\infty$ estimate – which we state and will prove in the next lecture.

Theorem (De Giorgi's $L^2 \to L^\infty$ estimate). There is a constant $C = C(n,\lambda,\Lambda)$ such that if $B_{2r}(x_0)\subset \Omega$, then any solution $u$ satisfies the estimate \begin{align} \sup \limits_{B_r(x_0)}u(x)^2 \leq \frac{C}{r^n}\int_{B_{2r}(x_0)}u(x)^2\;dx \end{align} (Note: Strictly speaking, we should be using ess-sup on the left hand side)