Regularity for Partial Differential Equations

Feature Article Regularity for Partial Differential Equations: from De Giorgi-Nash-Moser Theory to Intrinsic Scaling by JoséMiguel Urbano Departamento de Matemática - CMUC Universidade de Coimbra 1 A beautiful problem smoothness of L, the minimizer (assuming it exists) is also smooth. In the academic year 1956-1957, John Nash had a visit- Problems of this type are related to elliptic equations in ing position at the Institute for Advanced Study (IAS) that a minimizer u is a weak solution of the associated in Princeton, on a sabbatical leave from MIT, but he Euler-Lagrange equation actually lived in New York City. The IAS at the time “was known to be about the dullest place you could n 1 X ∂ find” and Nash used to hang around the Courant In- L (∇u(x)) = 0 in Ω . ∂x ξi stitute which was close to home and full of activity. i=1 i That’s how he came across a problem that mathematicians had been trying to solve for quite a while. The This equation can be differentiated with respect to xk, story goes that Louis Nirenberg, at the time a young to give that, for any k = 1, 2, . , n, the partial deriva- ∂u professor at Courant, was the person responsible for the tive := vk satisfies a linear PDE of the form ∂xk unveiling: “...it was a problem that I was interested in and tried to solve. I knew lots of people interested in n X ∂ ∂vk this problem, so I might have suggested it to him, but a (x) = 0 , (1) ∂x ij ∂x I’m not absolutely sure”, said Nirenberg recently in an i,j=1 i j interview to the Notices of the AMS (cf. [19]). with coefficients a (x) := L (∇u(x)). The PDE is As so many other great questions of 20th century math- ij ξiξj elliptic provided L is assumed to be convex. ematics, it all started with one of Hilbert’s problems presented on the occasion of the 1900 International In the 1950’s, regularity theory for elliptic equations Congress of Mathematicians in Paris, namely the 19th was essentially based on Schauder’s estimates which, problem: Are the solutions of regular problems in the k,α roughly speaking, guarantee that if aij ∈ C then the calculus of variations always necessarily analytic? A solutions of (1) are of class Ck+1,α, for k = 0, 1,... simple example of such a problem is, in modern termi- 1,α So if it could be shown that u ∈ C then aij(x) := nology, the problem of minimizing a functional 0,α 1,α Lξiξj (∇u(x)) would belong to C , v to C and u to Z C2,α; a bootstrap argument would then solve Hilbert’s min L(∇w(x)) dx 19th problem. w∈A Ω where Ω ⊂ Rn is a bounded and smooth domain, the Meanwhile, the existence theory had been developed Lagrangian L(ξ) is a smooth (possibly nonlinear) scalar through the use of direct methods: the minimization function defined on Rn and A is a set of admissible problem has a unique solution provided L, apart from functions (typically the elements of a certain function satisfying natural growth conditions like space satisfying a boundary condition like w = g on ∂Ω, for a given g). The question is to prove that, given the |L(ξ)| ≤ C |ξ|p , 1Cathleen Morawetz, quoted in [15]. 8 is also coercive and uniformly convex. The notion of and to satisfy the uniform ellipticity condition (for solution had to be conveniently extended and the ad- λ > 0) missible set A taken to be the set of functions that, n together with their first weak derivatives, belong to Lp, X a ξ ξ ≥ λ|ξ|2 , ∀x ∈ Ω, ∀ξ ∈ Rn . i.e., that belong to the Sobolev space W 1,p. ij i j i,j=1 So the existence theory gave a minimizer u ∈ W 1,p and A weak solution of equation (2) is a function u ∈ the missing step for the regularity problem to be solved W 1,2(Ω) which satisfies the integral identity was 1,p 1,α n u ∈ W =⇒ u ∈ C X Z ∂u ∂ϕ aij = 0 (3) p ∂x ∂x i.e., from first derivatives in L to Höldercontinuous i,j=1 Ω i j first derivatives. In terms of the elliptic PDE (1), reg- 1,2 1,2 ularity theory worked if the leading coefficients were for all test functions ϕ ∈ W0 (Ω) (the elements of W already somewhat regular (at least continuous) since which vanish on the boundary ∂Ω in a suitable weak it was based on perturbation arguments and compari- sense). son of the solutions with harmonic functions. Assum- To simplify the writing we assume from now on that ing only the measurability and the boundedness of the coefficients (together with the essential structural as- n n o Ω = B1 := x ∈ R : |x| < 1 . sumption of ellipticity) was insufficient, and nothing was known about the regularity of the solutions in this case. Theorem 1 Every weak solution of (2) is locally bounded. The problem was solved by C.B. Morrey in 1938 for the special case n = 2 but the techniques he employed Proof. Let k ≥ 0 and η be a smooth function with were typically two dimensional, involving complex anal- compact support in B . Put v = (u − k)+ and take ysis and quasi-conformal mappings. The n-dimensional 1 ϕ = vη2 as test function in (3). The use of the assump- problem remained open until the late 50’s and that’s tions and Young’s inequality give exactly what Nirenberg told Nash about. Z 2 Z 2 2 4Λ 2 2 |∇v| η ≤ 2 |∇η| v . (4) B1 λ B1 2 De Giorgi’s breakthrough These Cacciopoli inequalities on level sets of u will be the building blocks of the whole theory and once they The problem wouldn’t resist the genius of John Nash are obtained the PDE can be forgotten: the problem and Ennio De Giorgi. The two men worked totally un- becomes purely analytic. aware of each other’s progress and solved the problem Next, by Hölderand Sobolev’s inequalities (with 2∗ = using entirely different methods. 2n/(n − 2) being the Sobolev exponent), It was De Giorgi who did it first (actually for p = 2; 2 ∗ Z Z ∗ 2 2 the result would later be extended to any p ∈ (1, ∞)) 1− ∗ (vη)2 ≤ (vη)2 |{vη 6= 0}| 2 and it is his proof that will now be analyzed. To re- B1 B1 ally understand in full depth De Giorgi’s ideas there 2 Z 2 is no way around the technicalities. In what follows I ≤ c(n) |{vη 6= 0}| n |∇(vη)| B did my best to explain things in a clear way but the 1 reader should not expect everything to be trivial or im- and since, due to (4), mediately understandable; so please grab a pencil and Z 2 Z a piece of paper and be prepared to struggle a bit with 2 4Λ 2 2 |∇(vη)| ≤ 2 + 1 |∇η| v inequalities and iterations. B1 λ B1 Consider the equation we arrive at Z Z n 2 2 2 2 X ∂ ∂u (vη) ≤ c(n, λ, Λ) |{vη 6= 0}| n |∇η| v . a (x) = 0 in Ω (2) ∂x ij ∂x B1 B1 i,j=1 i j Now for fixed 0 < r < R < 1, choose the cut-off func- n ∞ where Ω ⊂ R is a smooth bounded domain and the tion η ∈ C0 (BR) such that 0 ≤ η ≤ 1, η ≡ 1 in Br and 2 coefficients aij are only assumed to be measurable and |∇η| ≤ R−r . Putting, for ρ > 0, bounded, with n o kaijkL∞ ≤ Λ , A(k, ρ) = x ∈ Bρ : u(x) > k , 9 we obtain (with C ≡ c(n, λ, Λ)) From (6) we obtain Z C 2 Z 2 m(1+ 2 ) 2 n 2 1+ 2 ψ(k0, r0) n 2 n ψ(k0, r0) (u − k) ≤ |A(k, R)| (u − k) . ψ(k , r ) ≤ 2Cγ n (R − r)2 m m 2 2m m A(k,r) A(k,R) k n γ n γ For h > k and 0 < ρ < 1, 2 1+ 2 and choose first γ > 1 such that γ n = 2 n and then Z Z k large enough so that (u − h)2 ≤ (u − k)2 A(h,ρ) A(k,ρ) 2 2 ψ(k0, r0) n 1+ n ∗ 2Cγ 2 ≤ 1 ⇐ k = C ψ(k0, r0) and k n Z 2 2 (h − k) |A(h, ρ)| ≤ (u − k) where C∗ ≡ C∗(n, λ, Λ). A(k,ρ) 1 so we have Finally let m → ∞ in (7) to get ψ(k, 2 ) ≤ 0, i.e., + 2 k(u − k) kL (B 1 ) = 0 . Z 2 (u − h)2 A(h,r) Hence + ∗ + 2 sup u ≤ C ku kL (B1) . C 2 Z B 1 ≤ |A(h, R)| n (u − h)2 2 (R − r)2 A(h,R) Using a dilation argument, this estimate can be refined; 1+ 2 Z ! n indeed, for any θ ∈ (0, 1) and p > 1, it holds C 1 2 ≤ 2 4 (u − k) (R − r) (h − k) n A(k,R) + C(n, λ, Λ) + sup u ≤ ku k p . n/p L (B1) + Bθ (1 − θ) 2 or, equivalently, with ψ(s, ρ) = k(u − s) kL (Bρ) The same type of reasoning gives similar conclusions C 1 2 − 1+ n concerning u and the result follows.

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