ABP

ABP 2016 319 1

ABP A. D. Aleksandrov,I. Y. Bakelman,C. Pucci 1 2 ABP 3 ABP 4 5 2

Ω R n : bounded C 0 = C 0 (n) > 0 such that u f in Ω (classical subsolution) max Ω u max u + C 0diam(Ω) 2 f + L Ω (Ω) 3

Assume diam(ω) = 1 and 0 Ω. Consider v(x) := u(x) + µ x 2. v f 2µn < 0 provided µ > f + 2n. max Ω Otherwise, v(x 0 ) = max v = max v Ω v for x 0 Ω, 0 v(x 0 ) < 0 Ω 4

v(x) = u(x) + µ x 2 Hence, max max Ω Ω u max Ω v = max Ω v v = max Ω v max u + µ max Ω Ω x 2 max u max u + diam(ω)2 f + Ω Ω 2n 5

2ABP 6

max Ω ABP C 1 = C 1 (n) > 0 such that u f in Ω u max u + C 1diam(Ω) f + L Ω n (Γ[u]) 7

Γ r [u] := {x Ω p B r s. t. u(y) u(x) + p, y x ( y Ω)} B r := {p R n p < r} Γ[u] := Γ r [u] upper contact set () r>0 8

Sketch of proof Change of variables max Ω u max u =: r 0 Ω If r = 0, then the proof is done with any C 1 > 0. Assume r > 0. Claim: B r = Du(Γ r [u]) 9

B r Du(Γ r [u]) x Γ r [u] p B r such that u(y) u(x) + p, y x ( y Ω) p = Du(x) i.e. Du(x) B r 10

B r Du(Γ r [u]) p B r fixed x Ω (Not x Ω!) such that max y Ω {u(y) p, y } = u(x) p, x i.e. u(y) u(x) + p, y x ( y Ω) 11

If x Ω, max u < max u + rdiam(ω). Ω Ω Recalldiam(Ω)=1, r = max a contradiction Hence, x Ω. p = Du(x) Ω i.e. p Du(Γ r [u]) u max Ω u 12

T (A) Change of variables formula g(ξ)#(t 1 ξ)dξ = A g(t (x)) det(dt (x)) dx #(T 1 ξ) 1, T := Du, A := Γ r [u], g 1 dξ = dξ detd 2 u dx B r Du(Γ r [u]) Γ r [u] 13

ω n r n = dξ detd 2 u dx B r Γ r [u] x Γ r [u] u(y) u(x) + Du(x), y x i.e. D 2 u(x) O λ k : eigen-value of D 2 u(x) 0 detd 2 u(x) = ( λ 1 ) ( λ n ) 14

( ) u n detd 2 u dx dx Γ r [u] Γ r [u] n Arithmetic mean-geometric mean inequality () max Ω ω n r n n 1 n Γ r [u] (f + ) n dx u max u = r 1 Ω nωn 1/n f + L n (Γ r [u]) 15

max Ω ABP C 2 = C 2 (n, b n ) > 0 such that u + b, Du f in Ω u max u + C 2diam(Ω) f + L Ω n (Γ[u]) 16

Apply Change of variables formula to g(ξ) := 1 κ+ ξ n (κ 0) B r 1 κ + ξ ndξ 2n 1 n n Γ r [u] detd 2 u κ + Du ndx 1 n n Γ r [u] Here, κ := f + n L n (Γ[u]) (a + b) n 2 n 1 (a n + b n ) ( a, b 0) ( b n Du n Γ r [u] κ + Du ndx + Γ r [u] (f + ) n κ + Du n ( u) n κ + Du ndx ) dx 2n 1 n n ( b n n + 1) 17

] ( r LHS = nω n [log(κ + s n ) = nω n log 1 + rn 0 κ ( log 1 + rn ) C( b n n κ + 1) 1 + rn κ exp { C( b n n + 1)} r n κexp { C( b n n + 1)} ) κ := f + n L n (Γ r [u]) 18

ABP Krylov (1976) Tso (1985) 19

Parabolic upper contact set() u : Q T := Ω (0, T ) R Π r [u] := (x, t) Q p B r such that T u(y, s) u(x, t) + p, y x (y, s) Ω [0, t] Π[u] := Π r [u] r>0 20

ABP C 2 = C 2 (n, b n+1, diam(ω), T ) > 0 such that u t max Q T u + b, Du f in Ω (0, T ] u max u + C 2 diam(ω) f + p Q L n+1 (Π[u]) T p Q T := ( Ω [0, T ]) (Ω {0}) 21

D := ( x 1,..., x n ), := (D, t )T Fix a R n. Define Φ : R n+1 R n+1 by Φ(x, t) = Φ := (Du, u Du, x a ). = ( D 2 u Du t D T (u Du, x a ) t (u Du, x a ) ( D 2 u D 2 u(x a) T Du t u t Du t, x a ) ) 22

Lemma det Φ = u t detd 2 u ( D Φ := 2 u D 2 u(x a) T ) Du t u t Du t, x a jth column (x j a j ) be added to n + 1th column ( D 2 ) u 0 u t 23

Assume diam(ω) = 1. m 0 := max u, r 0 := max u m 0 > 0 p Q T Q T D := {(ξ, h) R n R ξ < r 0, ξ < h m 0 < r 0 } (a, t 0 ) Ω (0, T ] such that r 0 = u(a, t 0 ) m 0 Claim : D Φ(Π r0 [u]) 24

Fix (ξ, h) D P (y) := h + ξ, y a affine function P (y) h ξ > m 0 i.e. P u on p Q T ξ + m 0 < h < r 0 + m 0 P (a) = h < r 0 + m 0 = u(a, t 0 ) (ˆx, t 1 ) Ω (0, t 0 ) such that u(ˆx, t 1 ) = P (ˆx), u(y, s) P (y) (y Ω, 0 s t 1 ) 25

26

D D := {(ξ, h) ξ r 0, ξ < h m 0 < r 0 } κ := f + n+1 L n+1 (Π[u]). 1 r0 κ + ξ n+1dhdξ = C r n 1 0 κ + r n+1 0 ( r0 +m 0 r0 r 0 r n 1 r n = C 0 0 κ + r n+1 dr (u t u) n+1 C 1 dxdt κ + Du n+1 Π r0 [u] r+m 0 ) dh dr 27

D 1 κ + ξ n+1dhdξ C 2 C 1 Π r0 [u] Π r0 [u] Π r0 [u] det Φ κ + Du n+1dxdt (u t u) n+1 dxdt κ + Du n+1 ( b Du ) n+1 + (f + ) n+1 κ + Du n+1 dxdt 28

RHS C 2 Π r0 [u] b n+1 dxdt + C 2 r0 r 0 r n 1 r n LHD = C 0 0 κ + r n+1 dr r0 r 0 r n 1 r n C 3 0 κ + r 0 n r n 1 {( = C 3 log κ + r ) 0 n r0 n n+1 1 rn+1 0 ( = C 3 log 1 + rn+1 0 n(n+1)κ n+1 rn+1dr ) } /κ 29

sup Q T ( ) log 1 + rn+1 0 n(n+1)κ C 4 u sup u = r 0 C 5 κn+1 1 = C 5 f + L n+1 (Π[u]) p Q T 30

31

4.1 Weak Harnack inequality i.e. u 0: supersolution in Ω u + b, Du f B 2 Ω in Ω ( ) u L p 0 (B 1 ) C 0 inf u + f L B n (Ω) 1 for p 0, C 0 > 0 constants. 32

Decay estimate of distribution function v := u inf u + ε 1 B 0 f, ε 0 > 0 fixed later n 1 Note inf v 1, f n ε 0 B 1 C 0 > 0 such that v L p 0 (B 1 ) C 0 ( ) u L p 0 (B 1 ) C 0 inf u + f L B n (Ω) 1 33

u L p 0 (B 1 ) C 0 α > 0 such that {x B 1 u(x) > t} Ct α M > 1 and θ (0, 1) such that {x B 1 u(x) > M m } θ m Let us show the case when m = 1 34

u f in B 2 (u B2 = 0 for simplicity!) φ(x) := A( x 2 n 2 2 n ): fundamental solution (A > 0) φ B2 = 0, φ 2 in B 1 ˆφ C 2 such that φ = ˆφ except near 0 ˆφ = 0 except near 0, i.e. supp ˆφ B 1 35

w := u ˆφ, Recall inf B 1 u 1 w f + φ in B 2 sup w sup w + C f + ˆφ L n (Γ[w;B 2 ]) B 2 B 2 LHS sup( u + 2) inf u + 2 1 B 1 B 1 RHS Cε 0 + C ( B 1 {w 0} 1dx ) 1 n 36

σ > 0 such that σ {x B 1 w(x) 0} {x B 1 u(x) M := sup( ˆφ)} B 2 θ > 0 such that {x B 1 u(x) > M} θ we had to work with a unit cube instaed of B 1 to have θ < 1 37

4.2 Local maximum principle Unbounded coefficient case is excluded by a classical argument. Use weak Harnack inequality (by Caffarelli) 38

4.3 Calderón-Zygmund estimatecaffarelli 1989 D 2 u Θ := max{θ +, Θ } { Θ ± (x) := inf M > 0 u(y) ±M y 2 + affine = holds at x uφ xi x j dx C Θ L p φ L p ( φ C 0 ) and D 2 u L p C Θ L p Decay estimate on Θ } 39

Enough to show the case of i = j δ h f(x) := 1 h 2{f(x + he i) + f(x he i ) 2f(x)} u(x)φ xi x i dx uδ h φ(x)dx = φ(x)δ h u(x)dx φ(x)θ(x)dx u(x + y) u(x) M y 2 + ξ, y ( ξ R n ) 40

Asuume diam(ω) = 1 u 0 in Ω u = 0 on Ω 41

5.1 p-laplacian cf. Imbert 2011 p u f in Ω n ( ) p u = Du p 2 u x k x k k=1 = Du p 2 u + (p 2) Du 2 n k,j=1 u xk u xj u xk x j { } is uniformly elliptic if p > 1 and Du 0 max Ω u max u + C f + Ω 1 p 1 L n (Γ[u]) 42

r 0 := max u max u > 0 Ω Ω g( ξ )dξ g( Du ) detd 2 u dx B r0 Γ r0 [u] ( ) u n C p g( Du ) Du p 2 dx C p Γ[u] Γ[u] g( Du ) (f + ) n Du n(p 2)dx 43

g(r) := r n(p 2) = nω n [ r n(p 1) max Ω ] r0 n(p 1) 0 r n(p 1) 0 C p f + n L n u max u + C f + Ω = ω n p 1 rn(p 1) 0 1 p 1 L n 44

p u = 5.2 -Laplacian n k=1 x k ( ) Du p 2 u x k n = Du p 2 u (p 2) Du p 4 u xk u xj u xk x j k,j=1 p 2 1 Du p 4 p with Du = 0 n u := u xk u xj u xk x j k,j=1 Charro-De Phillips-Di Castro-Máximo 2013 45

Consider normalized -Laplacian: p u p 2 1 Du p 2 (p ) n N u := u xk u xj Du k,j=1 2 u x k x j N u f in Ω ( 2 sup u sup u) C Ω Ω sup u Ω sup u f + 1 Γ[u] L ({u=r}) dr Ω 46

5.3 Fractional Laplacian σ (0, 2) L[u] f in Ω u(x + y) u(x) Du(x), y χ B1 (y) L[u](x) := R n y n+σ dy = 1 u(x + y) + u(x y) 2u(x) 2 R n y n+σ dy 47

max Ω L[u] f(x) in Ω u 0 in R n \ Ω 2 σ 2 u C f + L (K u ) f + σ 2L n (K u ) N. Guillen and R. W. Schwab, 2012 48

5.4 Monge-Ampère equation detd 2 u = f in Ω F (X) := detx should be elliptic; Restrict solutions to be convex. If u is convex in Ω (Ω must be convex), then max u = max u, and Ω Ω min u min u C f Ω Ω L 1 (Ω) 49

50

σ k (D 2 u) := 5.5 k-hessian 1< k <n 1 j 1 < <j k n λ jl (D 2 u) λ j (D 2 u(x)): eigen-values of D 2 u(x) Restrict solutions to be convex again! Assume r := min u min u > 0. Ω Ω 51

n i=1 i 1 < <i k N i=1 a i λ i = i 1 < <i k (λ i1 λ ik ) ( ) 1 λi1 λ n 1 C k 1 ik nc k α 1 σ k (D 2 u) n k nc k N α 1 N i=1 nc k 1 n 1 C k 1 a α i (α > 1, a i 0) 52

min Ω u: convex solution of σ k (D 2 u) f in Ω dξ 1 f n k B r nc k L n k (Ω) u min u C f Ω 1 k L n k (Ω) 53

54