On the infinity Laplace operator

Transcription

1 On the infinity Laplace operator Petri Juutinen Köln, July 2008

2 The infinity Laplace equation Gunnar Aronsson (1960 s): variational problems of the form S(u, Ω) = ess sup H (x, u(x), Du(x)). (1) x Ω

3 The infinity Laplace equation Gunnar Aronsson (1960 s): variational problems of the form S(u, Ω) = ess sup H (x, u(x), Du(x)). (1) x Ω Special case H (x, r, p) = 1 2 p 2 corresponds to the minimal Lipschitz extension problem.

4 The infinity Laplace equation Gunnar Aronsson (1960 s): variational problems of the form S(u, Ω) = ess sup H (x, u(x), Du(x)). (1) x Ω Special case H (x, r, p) = 1 2 p 2 corresponds to the minimal Lipschitz extension problem. an Euler-Lagrange equation (the Aronsson equation) d dx ( H (x, u(x), Du(x)) Hp (x, u(x), Du(x)) = 0 (2) to (1). Here H p denotes the derivative of H with respect to the gradient variable, and u is scalar valued.

5 The infinity Laplace equation Gunnar Aronsson (1960 s): variational problems of the form S(u, Ω) = ess sup H (x, u(x), Du(x)). (1) x Ω Special case H (x, r, p) = 1 2 p 2 corresponds to the minimal Lipschitz extension problem. an Euler-Lagrange equation (the Aronsson equation) d dx ( H (x, u(x), Du(x)) Hp (x, u(x), Du(x)) = 0 (2) to (1). Here H p denotes the derivative of H with respect to the gradient variable, and u is scalar valued. If H (x, r, p) = 1 2 p 2, then (2) reduces to u := ( D 2 u Du ) Du = n u xi u xj u xi x j = 0. (3) i,j =1

6 Derivation of the equation We approximate the Lipschitz extension problem by the problems of minimizing Du p dx, 1 < p <, with given boundary values. Ω

7 Derivation of the equation We approximate the Lipschitz extension problem by the problems of minimizing Du p dx, 1 < p <, Ω with given boundary values. The Euler-Lagrange equation of this problem is the p-laplace equation div( Du p 2 Du) = 0, which can be written as ( ) Du (p 2) Du p 4 2 p 2 u + u = 0.

8 Derivation of the equation We approximate the Lipschitz extension problem by the problems of minimizing Du p dx, 1 < p <, Ω with given boundary values. The Euler-Lagrange equation of this problem is the p-laplace equation div( Du p 2 Du) = 0, which can be written as ( ) Du (p 2) Du p 4 2 p 2 u + u = 0. If Du 0, this implies u = Du 2 p 2 u, and thus letting p we recover the infinity Laplace equation u = 0.

9 Motivation is the Laplacian of Calculus of variations of L functionals

10 Motivation is the Laplacian of Calculus of variations of L functionals applications: image processing, shape metamorphism, differential games etc.

11 Motivation is the Laplacian of Calculus of variations of L functionals applications: image processing, shape metamorphism, differential games etc. stochastic version: the random turn Tug-of-War of Peres, Schramm, Sheffield and Wilson (J. Amer. Math. Soc. 2008)

12 Remark Minimizing the functional ess sup Du 2 is equivalent to minimizing ess sup Du.

13 Remark Minimizing the functional ess sup Du 2 is equivalent to minimizing ess sup Du. Thus we could use, instead of (3), the singular and one-homogeneous version S u = 1 Du 2 uxi u xj u xi x j = 0.

14 Remark Minimizing the functional ess sup Du 2 is equivalent to minimizing ess sup Du. Thus we could use, instead of (3), the singular and one-homogeneous version S u = 1 Du 2 uxi u xj u xi x j = 0. However, the non-homogeneous equations such as and are, of course, not equivalent. u(x) = f (x) S u(x) = f (x)

15 Viscosity solutions Definition Let Ω R n be an open set. An upper semicontinuous function u : Ω R is a viscosity subsolution of (3) in Ω if, whenever ˆx Ω and ϕ C 2 (Ω) are such that (i) u(ˆx) = ϕ(ˆx), (ii) u(x) < ϕ(x) for all x ˆx

16 Viscosity solutions Definition Let Ω R n be an open set. An upper semicontinuous function u : Ω R is a viscosity subsolution of (3) in Ω if, whenever ˆx Ω and ϕ C 2 (Ω) are such that (i) u(ˆx) = ϕ(ˆx), (ii) u(x) < ϕ(x) for all x ˆx then ϕ(ˆx) 0.

17 Viscosity solutions Definition Let Ω R n be an open set. An upper semicontinuous function u : Ω R is a viscosity subsolution of (3) in Ω if, whenever ˆx Ω and ϕ C 2 (Ω) are such that (i) u(ˆx) = ϕ(ˆx), (ii) u(x) < ϕ(x) for all x ˆx then ϕ(ˆx) 0. A lower semicontinuous function v : Ω R is a viscosity supersolution of (3) in Ω if v is a viscosity subsolution. Finally, a continuous function h : Ω R is a viscosity solution of (3) in Ω if it is both a viscosity subsolution and a viscosity supersolution.

18 Remark A function u C 2 is is a viscosity solution of (3) in Ω if and only if u(x) = 0 for all x Ω.

19 Remark A function u C 2 is is a viscosity solution of (3) in Ω if and only if u(x) = 0 for all x Ω. u : R 2 R, u(x, y) = x 4/3 y 4/3 is a viscosity solution, and u / C 2.

20 Functionals of the form I (v, Ω) = Absolute minimizers Ω f (x, u(x), Du(x)) dx are set-additive. Thus if u minimizes I (, Ω) (with given boundary data), then it automatically also minimizes I (, V ), subject to its own boundary values, for every open V Ω.

21 Functionals of the form I (v, Ω) = Absolute minimizers Ω f (x, u(x), Du(x)) dx are set-additive. Thus if u minimizes I (, Ω) (with given boundary data), then it automatically also minimizes I (, V ), subject to its own boundary values, for every open V Ω. The same is not true for the supremum functionals S(u, Ω) = ess sup x Ω H (x, u(x), Du(x)).

22 Absolute minimizers Functionals of the form I (v, Ω) = Ω f (x, u(x), Du(x)) dx are set-additive. Thus if u minimizes I (, Ω) (with given boundary data), then it automatically also minimizes I (, V ), subject to its own boundary values, for every open V Ω. The same is not true for the supremum functionals S(u, Ω) = ess sup x Ω H (x, u(x), Du(x)). Definition A locally Lipschitz continuous function u : Ω R m, m 1, is called an absolute minimizer of S(, Ω), if S(u, V ) S(v, V ) for every V Ω and v W 1, (V ) C (V )) such that v V = u V.

23 Comparison principle Theorem (Jensen 1993) Let Ω R n be a bounded domain, and suppose that u and v are a subsolution and a supersolution of (3) in Ω, respectively, such that u v on Ω. Then u v in Ω.

24 Comparison principle Theorem (Jensen 1993) Let Ω R n be a bounded domain, and suppose that u and v are a subsolution and a supersolution of (3) in Ω, respectively, such that u v on Ω. Then u v in Ω. The main point of the proof is to approximate a given subsolution by subsolutions with non-vanishing gradient. The case of unbounded domains has been considered by Crandall, Gunnarsson and Wang (2007).

25 Existence Theorem Let Ω R n be a bounded domain, and suppose that g : Ω R is continuous. Then there is a unique u such that { u = 0 in Ω, u = g on Ω.

26 Existence Theorem Let Ω R n be a bounded domain, and suppose that g : Ω R is continuous. Then there is a unique u such that { u = 0 in Ω, u = g on Ω. The existence can be obtained using 1. Perron s method. 2. the approximation involving p-laplace equation. 3. Tug-of-War.

27 Regularity Let u be a viscosity solution of u = 0 in Ω R n. Then u is locally Lipschitz continuous.

28 Regularity Let u be a viscosity solution of u = 0 in Ω R n. Then u is locally Lipschitz continuous. if n = 2, u C 1,α (Savin 2005, Evans and Savin 2008).

29 Regularity Let u be a viscosity solution of u = 0 in Ω R n. Then u is locally Lipschitz continuous. if n = 2, u C 1,α (Savin 2005, Evans and Savin 2008). C 1,1/3 is the best one can hope for (in any dimension).

30 The cone functions Comparison with cones C (x) = a x x 0 + b are solutions of the infinity Laplace equation in R n \ {x 0 }. These are fundamental solutions.

31 The cone functions Comparison with cones C (x) = a x x 0 + b are solutions of the infinity Laplace equation in R n \ {x 0 }. These are fundamental solutions. Definition (Crandall, Evans, Gariepy 2001) A function u C (Ω) enjoys comparison with cones from above if, whenever V Ω is open, x 0 / V, a, b R, a > 0, are such that u(x) C (x) := a x x 0 + b on V, we have u(x) C (x) in V.

32 The cone functions Comparison with cones C (x) = a x x 0 + b are solutions of the infinity Laplace equation in R n \ {x 0 }. These are fundamental solutions. Definition (Crandall, Evans, Gariepy 2001) A function u C (Ω) enjoys comparison with cones from above if, whenever V Ω is open, x 0 / V, a, b R, a > 0, are such that we have u(x) C (x) := a x x 0 + b on V, u(x) C (x) in V. A function v C (Ω) enjoys comparison with cones from below if v enjoys comparison with cones from above. Finally, u C (Ω) enjoys comparison with cones if it enjoys comparison with cones both from above and below.

33 Equivalence Theorem (Jensen 1993, Crandall et al 2001) Let u : Ω R be a locally Lipschitz continuous function. Then the following are equivalent: (1) u is an absolute minimizer of the functional S(v) = ess sup Du. (2) u is an AMLE (=absolutely minimizing Lipschitz extension): for every open V Ω we have Lip(u, V ) = Lip(u, V ). (3) u is a viscosity solution of the infinity Laplacian. (4) u enjoys comparison with cones.

34 (4) = (2): Since u(z) Lip(u, V ) x z u(x) u(z) + Lip(u, V ) x z holds for x, z V and V Ω

35 (4) = (2): Since u(z) Lip(u, V ) x z u(x) u(z) + Lip(u, V ) x z holds for x, z V and V Ω and u enjoys comparison with cones from above and below, the inequalities hold also if x V.

36 (4) = (2): Since u(z) Lip(u, V ) x z u(x) u(z) + Lip(u, V ) x z holds for x, z V and V Ω and u enjoys comparison with cones from above and below, the inequalities hold also if x V. Thus for any fixed x V. Lip(u, (V \ {x})) = Lip(u, V )

37 (4) = (2): Since u(z) Lip(u, V ) x z u(x) u(z) + Lip(u, V ) x z holds for x, z V and V Ω and u enjoys comparison with cones from above and below, the inequalities hold also if x V. Thus Lip(u, (V \ {x})) = Lip(u, V ) for any fixed x V. Using this twice, we have Lip(u, V ) = Lip(u, (V \ {x, y})) for x, y V.

38 (4) = (2): Since u(z) Lip(u, V ) x z u(x) u(z) + Lip(u, V ) x z holds for x, z V and V Ω and u enjoys comparison with cones from above and below, the inequalities hold also if x V. Thus Lip(u, (V \ {x})) = Lip(u, V ) for any fixed x V. Using this twice, we have Lip(u, V ) = Lip(u, (V \ {x, y})) for x, y V. Hence u(x) u(y) Lip(u, V ) x y, which shows that Lip(u, V ) = Lip(u, V ).

39 (3) = (1): Let u C 2 (Ω) satisfy u(x) = 0 in Ω, and suppose there is V Ω, x 0 V and v W 1, (Ω) C (Ω) such that u = v on V and Du(x 0 ) > ess sup Dv. (4) V

40 (3) = (1): Let u C 2 (Ω) satisfy u(x) = 0 in Ω, and suppose there is V Ω, x 0 V and v W 1, (Ω) C (Ω) such that u = v on V and Du(x 0 ) > ess sup Dv. (4) V Let γ be the unit speed integral curve of the vector field x Du(x) that passes through x 0.

41 (3) = (1): Let u C 2 (Ω) satisfy u(x) = 0 in Ω, and suppose there is V Ω, x 0 V and v W 1, (Ω) C (Ω) such that u = v on V and Du(x 0 ) > ess sup Dv. (4) V Let γ be the unit speed integral curve of the vector field x Du(x) that passes through x 0. Since 0 = u(x) = 1 2 D( Du(x) 2 ) Du(x) in Ω, x Du(x) 2 is constant along γ.

42 (3) = (1): Let u C 2 (Ω) satisfy u(x) = 0 in Ω, and suppose there is V Ω, x 0 V and v W 1, (Ω) C (Ω) such that u = v on V and Du(x 0 ) > ess sup Dv. (4) V Let γ be the unit speed integral curve of the vector field x Du(x) that passes through x 0. Since 0 = u(x) = 1 2 D( Du(x) 2 ) Du(x) in Ω, x Du(x) 2 is constant along γ. Let y, z V γ. Now u(y) u(z) = Du(γ(t)) γ(t) dt = Du(γ(t)) dt = Du(x 0 ) dt, γ γ γ

43 (3) = (1): Let u C 2 (Ω) satisfy u(x) = 0 in Ω, and suppose there is V Ω, x 0 V and v W 1, (Ω) C (Ω) such that u = v on V and Du(x 0 ) > ess sup Dv. (4) V Let γ be the unit speed integral curve of the vector field x Du(x) that passes through x 0. Since 0 = u(x) = 1 2 D( Du(x) 2 ) Du(x) in Ω, x Du(x) 2 is constant along γ. Let y, z V γ. Now u(y) u(z) = Du(γ(t)) γ(t) dt = Du(γ(t)) dt = Du(x 0 ) γ γ γ dt, while v(y) v(z) ess sup Dv dt. V γ

44 (3) = (1): Let u C 2 (Ω) satisfy u(x) = 0 in Ω, and suppose there is V Ω, x 0 V and v W 1, (Ω) C (Ω) such that u = v on V and Du(x 0 ) > ess sup Dv. (4) V Let γ be the unit speed integral curve of the vector field x Du(x) that passes through x 0. Since 0 = u(x) = 1 2 D( Du(x) 2 ) Du(x) in Ω, x Du(x) 2 is constant along γ. Let y, z V γ. Now u(y) u(z) = Du(γ(t)) γ(t) dt = Du(γ(t)) dt = Du(x 0 ) γ γ γ dt, while v(y) v(z) ess sup Dv dt. V γ Combining these two inequalities with (4) yields u(y) u(z) > v(y) v(z), which contradicts u = v on V.

45 Tug-of-war Recently Peres, Schramm, Sheffield and Wilson considered the following zero-sum two player stochastic game called tug-of-war: given: bounded domain Ω R n and g C ( Ω).

46 Tug-of-war Recently Peres, Schramm, Sheffield and Wilson considered the following zero-sum two player stochastic game called tug-of-war: given: bounded domain Ω R n and g C ( Ω). fix x = x 0 Ω and step-size ε > 0.

47 Tug-of-war Recently Peres, Schramm, Sheffield and Wilson considered the following zero-sum two player stochastic game called tug-of-war: given: bounded domain Ω R n and g C ( Ω). fix x = x 0 Ω and step-size ε > 0. At the k th turn, the players toss a coin and winner chooses an x k with x k x k 1 < ε.

48 Tug-of-war Recently Peres, Schramm, Sheffield and Wilson considered the following zero-sum two player stochastic game called tug-of-war: given: bounded domain Ω R n and g C ( Ω). fix x = x 0 Ω and step-size ε > 0. At the k th turn, the players toss a coin and winner chooses an x k with x k x k 1 < ε. The game ends when x k Ω, and player I s payoff is g(x k ).

49 Tug-of-war Recently Peres, Schramm, Sheffield and Wilson considered the following zero-sum two player stochastic game called tug-of-war: given: bounded domain Ω R n and g C ( Ω). fix x = x 0 Ω and step-size ε > 0. At the k th turn, the players toss a coin and winner chooses an x k with x k x k 1 < ε. The game ends when x k Ω, and player I s payoff is g(x k ). The value function u ε (x) of the above game is continuous for each ε > 0 and u ε u uniformly, where u is the unique viscosity solution of the infinity Laplacian so that u = g on Ω.

50 The dynamic programming principle (DPP) reads u ε (x) = 1 max 2( u ε + min u ) ε. B ε(x) B ε(x)

51 The dynamic programming principle (DPP) reads u ε (x) = 1 max 2( u ε + min u ) ε. B ε(x) B ε(x) Suppose that u ϕ, where ϕ C 2 (Ω), has a strict local maximum at x. Then u ε ϕ has a local max at x ε and x ε x.

52 The dynamic programming principle (DPP) reads u ε (x) = 1 max 2( u ε + min u ) ε. B ε(x) B ε(x) Suppose that u ϕ, where ϕ C 2 (Ω), has a strict local maximum at x. Then u ε ϕ has a local max at x ε and x ε x. (DPP) = ϕ(x ε ) 1 2( max B ε(x ε) ϕ + min B ε(x ε) ϕ). (5)

53 The dynamic programming principle (DPP) reads u ε (x) = 1 max 2( u ε + min u ) ε. B ε(x) B ε(x) Suppose that u ϕ, where ϕ C 2 (Ω), has a strict local maximum at x. Then u ε ϕ has a local max at x ε and x ε x. For ε small, we have (DPP) = ϕ(x ε ) 1 2( max B ε(x ε) ϕ + min B ε(x ε) ϕ). (5) max ϕ = ϕ(x ε) + ε Dϕ(x ε ) + ε2 B ε(x ε) 2 D 2 ϕ(x ε ) Dϕ(x ε) Dϕ(x ε ) Dϕ(x ε ) Dϕ(x ε ) + o(ε2 ), min ϕ = ϕ(x ε) ε Dϕ(x ε ) + ε2 B ε(x ε) 2 D 2 ϕ(x ε ) Dϕ(x ε) Dϕ(x ε ) Dϕ(x ε ) Dϕ(x ε ) + o(ε2 ).

54 The dynamic programming principle (DPP) reads u ε (x) = 1 max 2( u ε + min u ) ε. B ε(x) B ε(x) Suppose that u ϕ, where ϕ C 2 (Ω), has a strict local maximum at x. Then u ε ϕ has a local max at x ε and x ε x. For ε small, we have (DPP) = ϕ(x ε ) 1 2( max B ε(x ε) ϕ + min B ε(x ε) ϕ). (5) max ϕ = ϕ(x ε) + ε Dϕ(x ε ) + ε2 B ε(x ε) 2 D 2 ϕ(x ε ) Dϕ(x ε) Dϕ(x ε ) Dϕ(x ε ) Dϕ(x ε ) + o(ε2 ), min ϕ = ϕ(x ε) ε Dϕ(x ε ) + ε2 B ε(x ε) 2 D 2 ϕ(x ε ) Dϕ(x ε) Dϕ(x ε ) Dϕ(x ε ) Dϕ(x ε ) + o(ε2 ). Substituting these to (5), dividing by ε 2 and letting ε 0 yields 0 D 2 ϕ(x) Dϕ(x) Dϕ(x) Dϕ(x) Dϕ(x) = S ϕ(x).

55 G. Aronsson, Extension of functions satisfying Lipschitz conditions, Ark. Mat. 6 (1967), G. Aronsson, M. G. Crandall, and P. Juutinen, A tour of the theory of absolutely minimizing functions, Bull. Amer. Math. Soc. (N.S.) 41 (2004), no. 4, M. G. Crandall, L. C. Evans, and R. Gariepy, Optimal Lipschitz extensions and the infinity Laplacian, Calc. Var. Partial Differential Equations 13 (2001), no. 2, R. Jensen, Uniqueness of Lipschitz extensions: minimizing the sup norm of the gradient, Arch. Rational Mech. Anal. 123 (1993), no. 1, Y. Peres, O. Schramm, S. Sheffield and D. Wilson, Tug-of-war and the infinity Laplacian, J. Amer. Math. Soc., to appear.