On Aleksandrov-Bakelman-Pucci Type Estimates For Integro-Differential Equations

Transcription

1 On Aleksandrov-Bakelman-Pucci Type Estimates For Integro-Differential Equations Russell Schwab Carnegie Mellon University 2 March 2012 (Nonlocal PDEs, Variational Problems and their Applications, IPAM) *partially supported by the National Science Foundation

2 Outline A Little Background A Very Basic Question Other Uses For The ABP Ideas of The Proof

3 A Little Background

4 The Operators α (0, 2), the fractional order of the equation δu(x, y) = u(x + y) + u(x y) 2u(x) The α/2 fractional Laplace ( ) α/2 u(x) = C n,α R n δu(x, y) y n α dy A linear operator comparable to the fractional Laplace with, FOR EXAMPLE, L(u, x) = C n,α R n δu(x, y)k(x, y)dy λ n+α K(x, y) Λ y y n+α A fully nonlinear operator comparable to the fractional Laplace F (u, x) = inf sup { Cn,α δu(x, y)k αβ (x, y)dy }. α β R n

5 Simplest Possible Nonlocal Operator The simplest possible nonlocal operator the difference operator Its limits... δu(x, y) = u(x + y) + u(x y) 2u(x) y T D 2 u(x)y δu(x, he) lim h 0 h 2 = D ee u(x). Measure of local quadratic deviation from being affine

6 Spherical Averages Infinitesimal weighted spherical average deviation from affine ( ) 1 1 u(x) = n lim r 0 r 2 δu(x, y)ds r (y) B r B r

7 Spherical Averages Average deviation from affine at all scales ( ) 1 ( ) α/2 u(x) = C(n, α) r 1 α δu(x, y)ds r (y) dr 0 B r B r Rewritten as ( ) α/2 u(x) = C(n, α) δu(x, y) y n α dy R n

8 Non-Spherical Averages Anisotropic averages in direction and scale, non-constant coefficients weight K(x, y) L K (u, x) = C(n, α) δu(x, y)k(x, y)dy R n note these L K do necessarily not satisfy L K (u(r ), x) = r α L K (u, rx)!

9 Limits α 2 Fractional Laplace Laplace... C(n, α) (2 α) as α 2 u(x) = C(n) lim (2 α) δu(x, y) y n α dy α 2 R n Tr(AD 2 u(x)) = ( u( A ))(( A) 1 x) Tr(AD 2 u(x)) = lim α 2 ( ( ) α/2 u( A ))(( A) 1 x)

10 Fully Nonlinear Equations Ellipticity An example of general fully nonlinear operator F (u, x) F (u, x) = inf sup { Cn,α δu(x, y)k αβ (x, y)dy }. α β R n Choice of family in inf and sup = choice of ellipticity class, K a A, maximal and minimal operators with respect to A (Caffarelli, Caffarelli-Silvestre) M A (u, x) = inf (L K a K a(u, x)) and A M+ A (u, x) = sup (L K a(u, x)) K a A F is elliptic with respect to A if M A (u v, x) F (u, x) F (v, x) M+ A (u v, x)

11 Fully Nonlinear Equations Ellipticity What does it mean to be uniformly elliptic, anyway? It s worth some discussion! A = {K : (2 α)λ y n+α K(x, y) (2 α)λ y n+α } Caffarelli-Silvestre A = {K = y T A(x)y y n+α+2 : Tr(A) λ, 0 A(x) ΛId} Guillen-Schwab g( y y A = {K : ) K(x, y) Λ } Kassmann-Mimica y n+α y n+α (g : S n 1 [0, ) satisfies for some (symmetric) open subset I S n 1 a uniform positivity assumption g δ > 0 on I )

12 A Very Rich Family 2nd Order F (D 2 u) functions of Hessian = A 2nd order S(n) (symmetric matrices) α Order F (u, x) functions of global behavior of u = A α order L 1 (min( y 2, 1))

13 A Very Basic Question

14 A Very Basic Question Suppose L is a uniformly elliptic (2nd or α order) and { Lu k f k in B 1 u k = 0 in R n \ B 1, where for simplicity we assume u k 0 and 0 f k 1. A Very Basic Question When does {f k > 0} 0 also imply inf(u k ) 0?

15 The ABP Result, 2nd Order If Lu = a ij (x)u xi x j then Aleksandrov (circa 1960) says Theorem (Aleksandrov-Bakelman-Pucci) inf(u k ) C(n) λ f + k L n (C u), where λid (a ij ) ΛId, C u = {u k = Γ}, and Γ = sup{v : v u and λ 1 (D 2 v) 0 in B 3 } (convex envelope).

16 What About Integro-Differential Equations? For L integro-differential (order α), until recently there was no answer.

17 The New ABP Result, α order A restricted family: (L A )u(x) = C(n, α) δu(x, y)(y T A(x)y) y n α 2 dy R n The minimal operator M (u, x) = inf{(l A )u(x) : 0 A ΛId and Tr(A) λ} These K(x, y) = y T A(x)y y n α 2 do not satisfy λ n+α K(x, y) Λ y y n+α

18 Assume that u is a supersolution The New ABP Result, α order Theorem (Guillen-Schwab 2011) { M (u, x) f in B 1 u 0 in R n \ B 1. C(n, α) inf(u) ( f + λ L (C u)) (2 α)/2 ( f + L n (C u)) α/2, where C u = {u = Γ α } and Γ α is an appropriate replacement for the convex envelope, and C(n, α) is uniformly bounded as α 2.

19 Other Uses For The ABP

20 ABP in Homogenization Linear Equation simply for presentation φ fixed test function. Must find a unique λ such that the solution v ε (in x variable) { δφ(x0, y)k( x ε, y, ω)dy + δv ε (x, y)k( x ε, y, ω)dy = λ in B 1(x 0 ) v ε = 0 in R n \ B 1 (x 0 ) also satisfies as ε 0 max B 1 v ε 0 If so, then F (φ, x 0 ) = λ

21 ABP in Homogenization ( ) λ << 0 λ =????? ( ) λ >> 0 { δφ(x0, y)k( x ε, y, ω)dy + δv ε (x, y)k( x ε, y, ω)dy = λ in B 1(x 0 ) v ε = 0 in R n \ B 1 (x 0 )

22 ABP in Homogenization Obstacle Problem The key idea to find this λ is in the contact set of an obstacle problem (Caffarelli-Souganidis-Wang for 2nd order; also Schwab for α order) w ε solves the largest subsolution problem with obstacle = 0 v ε solves F (v ε, x ε ) = λ w ε = sup{u : F (u, x ) λ and u 0} ε (note F (w ε, x ε ) = λ when w ε 0) M (w ε v ε, x) F (w ε, x ε ) F (v ε, x ε ) 1 {w ε =0}(x)F (0, x ε ),

23 ABP in Homogenization Obstacle Problem M (w ε v ε, x) F (w ε, x ε ) F (v ε, x ε ) 1 {w ε =0}(x)F (0, x ε ), ABP = (w ε v ε ) C {w ε = 0} 1/n So if {w ε = 0} 0 as ε 0, then we know lim sup v ε 0 ε 0

24 ABP in Regularity Point To Measure Estimate Lemma (Point To Measure Estimate) There are positive constants M > 1, µ < 1 and δ 0 such that if u satisfies: 1. u 0 in R n 2. inf Q 3 u 1 3. M (u, x) f in Q 4 n and f L n (Q 4 n ) δ 0, f L (Q 4 n ) 1. Then we have the bound {x Q 1 : u(x) M} µ Q 1.

25 ABP in Regularity Point To Measure Estimate Proposition (A Special Bump Function) Given 0 < λ Λ and σ 0 (0, 2) there exist constants C 0, M > 0 and a C 1,1 function η(x) : R n R such that 1. supp η B 2 n (0) 2. η 2 in Q 3 and η M 3. For every σ > σ 0 we have M + (η, x) C 0 ξ everywhere where ξ is a continuous function with support inside B 1/4 (0) and such that 0 ξ 1.

26 ABP in Regularity Point To Measure Estimate M (u + Φ) f + ξ 1 inf(u + Φ) C( f α/2 L n + {u + Φ = Γ} 1/n Q 1 )

27 Caffarelli-Silvestre Replacement For ABP Assume that u is a supersolution { M (u, x) f in B 1 u 0 in R n \ B 1. Theorem (Caffarelli-Silvestre 2009) inf(u) C(n, α) λ ( ) 1/n (max Qi f + ) n where {Q i } is a finite cube covering of the contact set with the convex envelope, Γ and C(n, α) is uniformly bounded as α 2. {y nq i : u(y) < Γ(y) + C(max f )d 2 i } Q i µ Q i i

28 Other Uses For 2nd Order ABP A vital role for the 2nd order ABP Comparison of strong solutions for linear non-divergence equations Krylov-Safonov Harnack Inequality Fully Nonlinear Regularity Theory (Harnack and Hölder)... Krylov-Safonov, Caffarelli, etc... W 2,ε estimates for Fully Nonlinear Equations... Fang-Hua Lin, Caffarelli, etc... (and things using the W 2,ε : Rates of convergence in numerical approximations and homogenization, Caffarelli-Souganidis) L p theory for viscosity solutions of fully nonlinear equations, Caffarelli-Crandall-Kocan-Swiech Stochastic Homogenization, Papanicolaou-Varadhan, Caffarelli-Souganidis-Wang many more...

29 Ideas of The Proof of The New ABP

30 A 120 Second Proof of The Original ABP (2nd Order) C is a cone chosen so that C(0) Γ(B 1 ) and inf C = 2M c(n)m n Γ(B 1 ) det(d 2 Γ)dx (change vars) B 1 det(d 2 Γ)dx (LΓ(x)) n dx (λ 1 (D 2 Γ)=0) (comparison) {u=γ} {u=γ} (Lu(x)) n dx (det(d 2 Γ) 1/n =inf(tr(ad 2 Γ))) (equation) {u=γ} (f + (x)) n dx {u=γ}

31 New Difficulties (Headaches) convex envelope not compatible with integro-differential operators. det(d 2 Γ(x)) (????) (M (Γ, x)) n u(x) = 1 x β with 0 < α < β < 1, {u = Γ} = {0}, so estimate will never hold reduction to {u = Γ} will not imply M (Γ, x) 0... hence can t get back to M u and f

32 The Main Idea Replace Γ by a new envelope, Γ α, TBD Raise the order of the equation in principle Γ α solves α-order equation only know geometric arguments for 2nd-order equation use Riesz Potential of Γ α P = Γ α K where K(y) = A(n, 2 α) y n+(2 α) 2nd order operations on P should give α-order operations on Γ α ( )P = ( ) 1 ( ) (α 2)/2 Γ α = ( ) α/2 Γ α 2nd order proof on P should become compatible with M (Γ, )

33 The New Tools All new tools come from the following formula which is the correct version of the incorrect (but useful) idea that we should have P xi x j = Γ α (K yi y j ). Lemma (formula for D 2 P) D 2 P(x) = (n + α 2)(n + α) A(n, 2 α) 2 R n δγ(x, y) [ ] y y y n+σ+2 Id (n + σ) y n+σ dy

34 The New Tools Replacement for D 2 Γ h α (Γ α ) := A(n, 2 α) δγ(x, y) y y R n y n+σ+2 dy Replacement for Convexity (λ 1 (D 2 Γ) 0) λ 1 (h α (Γ α )) 0 New Envelope Γ α = sup(v : v u in R n and λ 1 (h α (v)) 0 in B 3 ) degenerate elliptic obstacle problem with operator λ 1 (h α ( )) λ 1 (h α (Γ α )) = 0 in the set {u Γ α } M (u, x) = inf(tr(ah α (u, x)))

35 The Picture M Γ α (x) M (u, x) in the set {u = Γ α } by comparison λ 1 (h α (Γ α )) = 0 in the set {u Γ α } ( )P = ( ) α/2 Γ α

36 Key Points of The Proof An important feature of Γ α is that D 2 P can be computed classically! Proposition (Γ α is Regularizing) If u is C 1,1 from above in B 1, then Γ α is H α/2 (Ω) for each Ω B 3.

37 Key Points of The Proof Lemma (Relationship Between det(d 2 P) and M (Γ)) If D 2 P(x) 0, then det(d 2 P(x)) 1/n 1 n M (Γ, x) Proof. n det(d 2 P) 1/n = inf{tr(ad 2 P) : det(a) = 1, A 0} inf{tr(bd 2 P) : det(b) = 1, B = A + Tr(A) Id, A 0} α 1 inf{ det(b) 1/n Tr(BD2 P) : B = A + Tr(A) Id, A ΛId, Tr(A) αλ} α inf{ 1 αλ Tr(Ah C(n, α) α(γ)) : A ΛId, Tr(A) αλ}= M (Γ) λ

38 Key Points of The Proof P CE is convex envelope of P in B R, R depends only on α Lemma (Reduction to The Good Contact Set) {D 2 P CE > 0} {u = Γ α } Proof. We note that {D 2 P CE > 0} {D 2 P > 0} {P = P CE }. case 1, x B 3. P(x) = ( ) α/2 Γ α (x) < 0 because Γ α (x) = 0 which is global max of Γ α = P(x) P CE (x) case 2, x B 3 {u Γα }. unit vector τ with τ T h α (Γ α, x)τ = 0 from λ 1 (h α (Γ α, x)) = 0. = P ττ (x) 0.

39 Key Points of The Proof now we use the geometric argument on P ( inf(p)) n C(n) det(d 2 P CE )dx B R {D 2 P CE >0} C(n) det(d 2 P)dx Lem & comparison {u=γ α} C(n) ( M det min op ordering λ n (Γ α ) ) n dx {u=γ α} C(n) ( f + ) n dx. equation λ n {u=γ α}

40 Key Points of The Proof But we did not get what we want!!!! Need to relate inf(p) back to inf(γ α ). This relationship is not to be expected from the Riesz Potential alone... Proposition (inf(p) & inf(γ α ) relationship) Let x 0 be such that Γ(x 0 ) = inf R n{γ}. Then ( ) 1 (2 α)/α inf{p} = inf{p} C(n, α)( Γ(x Rn 0 )) 2/α. B 3 2f (x 0 ) Key idea is to estimate the number of bad rings, Rk, with radius 2 k where {y : Γ α (x 0 + y) Γ α (x 0 ) > f (x 0 ) y α } 1 R k 2 R k

41 Additional Formulas Determinant If B 0 then det(b) 1/n = (1/n) inf{tr(ab) : det(a) = 1 & A 0} Minimal Operator M u, x = inf{l A u(x) : 0 A ΛId & Tr(A) λ} = inf{tr(ah α (u, x)) : 0 A ΛId & Tr(A) λ}

42 Open Problems More general kernels (cf. Kassmann-Mimica 2011) g( y y ) n+α K(x, y) Λ y y n+α, where g : S n 1 R satisfies for some open subset I S n 1 a uniform positivity assumption g δ > 0 on I. Remove the interpolation form... find some p depending on α and n B 1/2 inf(u) C(n, α) f + L p λ W s,ε estimates ( ) ε 1/ε u(x + y) + u(x y) 2u(x) R n x y n+sε C( u L + f Lq)

43 The End Thanks!