HOMOGENIZATION OF HAMILTON JACOBI EQUATIONS NOTES AND REMARKS SON NGUYEN THAI TU department of mathematics, university of sciences at ho chi minh city

HOMOGENIZATION OF HAMILTON-JACOBI EQUATIONS NOTES AND REMARKS SON NGUYEN THAI TU Department of Mathematics, University of Sciences at Ho Chi Minh City SON PHUNG TRUONG VAN Department of Mathematical Sciences, Carnegie Mellon University ABSTRACT These notes contain the essential materials and some further remarks of the summer course "Homogenization of Hamilton-Jacobi equations" taught by Prof Hung V Tran1at University of Science, Ho Chi Minh City, Viet Nam from July 6th to July 14th , 2015 E-mail addresses: thaison.tn93@gmail.com, sonv@andrew.cmu.edu Department of Mathematics, University of Wisconsin at Madison HOMOGENIZATION OF HAMILTON-JACOBI EQUATIONS NOTES AND REMARKS Notations: • n = n / n is the n-dimensional donut (torus) • w.r.t: with respect to LECTURE Clear contest : ǫ > : scale of the equation, ǫ ≪ : ǫ is very small Equation of interest Following the Hamilton-Jacobi equation x , Duǫ (x, t) = in n × (0, ∞) (I) uǫt (x, t) + H where ã u (x, t) : n ì [0, ∞) −→ , x is the location variable (spatial) and t is the time variable • uǫt (x, t) = ∂∂t uǫ (x, t) ǫ ǫ • Duǫ (x, t) = ∇ x (uǫ )(x, t) = ∂∂ ux , , ∂∂ ux n • H is the Hamiltonian (total energy) H: n × n −→ ( y, p) −→ H( y, p) ∈ Example (Classical mechanics) H( y, p) = |p|2 + V ( y) where the first term is the kinetic energy and the second one is the potential energy Question We want to understand (I) as ǫ −→ Is there some u such that uǫ −→ u and is u solve something simpler? This is one of the key questions in homogenization theory Assumptions The following assumptions are extremely important: (H1) (coercivity) lim H( y, p) = ∞ |p|−→∞ uniformly in y, (H2) (periodicity) H( y + k, p) = H( y, p) ∀k∈ n HOMOGENIZATION OF HAMILTON-JACOBI EQUATIONS NOTES AND REMARKS Formal Analysis (not rigorous) Consider the equation x , Duǫ (x, t) = (I) uǫt (x, t) + H ǫ where t is time variable, x is spatial variable (slow variable) and y = oscillatory variable There are several ways to begin our analysis: • Numerical analysis: plot uǫ • Guessing: some asymptotic expansion w.r.t ǫ We consider some ansatzs2 (1) uǫ (x, t) = u(x, t) + ǫu1 x, ǫx , t + ǫ u2 x, ǫx , t + x ǫ is the fast (2) uǫ (x, t) = u x, ǫx , t + ǫu1 x, ǫx , t + x ,t ǫ (3) uǫ (x, t) = u (x, t) + ǫ u1 (4) uǫ (x, t) = u(x, t) + ǫu1 Now using (4), we have x ,t ǫ + uǫ (x, t) = u(x, t) + ǫu1 Duǫ (x, t) = Du(x, t) + Du1 Plugging these into (I) we have x uǫt (x, t) + ǫu1t , t + +H ǫ Denote y = ǫx , we obtain x ǫ x ,t ǫ + ǫ u2 x ǫ + x , t + ǫ u2 x ǫ ǫ ,t , t + ǫDu2 , Du(x, t) + Du1 x ǫ x ǫ ,t + , t + + ǫDu2 x ǫ , t + = uǫt (x, t) + ǫu1t y, t + + H y, Du(x, t) + Du1 y, t + ǫDu2 y, t + = We will assume a “big lie": x and y are independent Matching asymptotic expansion zero-order term O(1), we get u t (x, t) + H y, Du(x, t) + Du1 ( y, t) = Now, fix (x, t) ∈ n × (0, ∞), since u t (x, t) is a constant w.r.t y, H( y, Du(x, t) + Du1 ( y, t)) is expected to be a constant w.r.t y, i.e., H( y, Du(x, t) + Du1 ( y, t)) = C(p, t) The word Ansatz in German is some thing like the word formulation in English HOMOGENIZATION OF HAMILTON-JACOBI EQUATIONS Let p = Du(x, t) ∈ n NOTES AND REMARKS , then H( y, p + Du1 ( y, t)) = C(p, t) Assume one more reduction u1 ( y, t) = u1 ( y), we then have the following PDE H( y, p + Du1 ( y)) = C(p) This is called the cell (or ergodic) problem n Theorem (Lions-Pappanicolau-Varadhan) Fix p ∈ C ∈ n such that the cell problem H( y, p + Du1 ( y)) = C(p) , there exists a unique constant in n has a periodic solution u1 Define the effective Hamiltonian H(p) = C(p) We are back to our original question: consider (Cǫ ) uǫt + H( ǫx , Duǫ ) = uǫ (x, 0) = u0 (x) Do the solutions uǫ of (Cǫ ) converges to a solution u of (HJ) u t + H(Du) = u(x, 0) = u0 (x) ? Remark This is a natural question since from the Hamiltonian H( ǫx , Duǫ ) we derived the effective Hamiltonian H that is independent of ǫ so as we pass ǫ → 0, (Cǫ ) should become (HJ) The answer to this problem is “yes” and will be elaborated in lecture HOMOGENIZATION OF HAMILTON-JACOBI EQUATIONS NOTES AND REMARKS LECTURE Before we prove theorem 1, let’s talk about viscosity solutions Introduction to viscosity solutions Consider • The static problem u( y) + H( y, Du( y)) = (S) • The Cauchy problem (C) u t (x, t) + H( y, Du(x, t)) = u( y, 0) = u0 ( y) We then look into the static problem in n in on × (0, ∞) n The idea here is that the problem uǫ + H(Duǫ ( y), y) = ǫ∆uǫ in (Sǫ ) n n has a smooth solution uǫ Assume further that uǫ → u locally uniformly in smooth function φ such that (u − φ)(x ) (u − φ)(x) =0 small enough, uǫ − φ has a max at x ǫ near by x and there is a subsequence ǫ j −→ such that x ǫ j −→ x Play with uǫ Since uǫ −φ has max at x ǫ , we then have, of course by second derivative test, (1) Duǫ (x ǫ ) = Dφ(x ǫ ) =⇒ Duǫ (x ǫ ) = Dφ(x ǫ ) ∆(uǫ − φ)(x ǫ ) ≤ ⇐⇒ ∆uǫ (x ǫ ) ≤ ∆φ(x ǫ ) Reconsider the PDE, uǫ (x ǫ ) + H(x ǫ , Duǫ (x ǫ )) = ǫ∆uǫ (x ǫ ) From (1), we can see that uǫ (x ǫ ) + H(x ǫ , Dφ(x ǫ )) ≤ ǫ∆φ(x ǫ ) Morever, if um is continuous and φ is smooth such that um −→ u uniformly on a compact set Ω ⊂ n , and assume u − φ has a strict (local) maximum over Ω as x , um − φ has (local) maximum over Ω at x m Then we must have x m −→ x as m −→ ∞ We have the convegence of whole sequence here, while in case uǫ −→ u uniformly as ǫ −→ we only have the convergence of sub-sequence HOMOGENIZATION OF HAMILTON-JACOBI EQUATIONS NOTES AND REMARKS Let ǫ j → 0, we then have u(x ) + H(x , Dφ(x )) ≤ Inspired by this derivation, we have the definition of viscosity solutions Definition (Definition of viscosity solution for the static equation) Assume u is continuous on its domain • (Subsolution) u is called a viscosity subsolution if for all φ ∈ C ∞ ( (u − φ)(x ) is a strict max then n ) such that u(x ) + H(x , Dφ(x )) ≤ • (Supersolution) u is called a viscosity supersolution if for all φ ∈ C ∞ ( that (u − φ)(x ) is a strict then n ) such u(x ) + H(x , Dφ(x )) ≥ • u is called a viscosity solution if it is both a subsolution and a supersolution Geometric descriptions Using the notions of sub-differential and super-differential, we will give a geometric descriptions for viscosity solution as the following Definition Let u be a real valued function defined on the open set Ω ⊂ x ∈ Ω, the sets D− u(x) = p ∈ n D+ u(x) = n p∈ : lim inf u( y) − u(x) − 〈p, y − x〉 y−x y−→x : lim sup y−x For any ≥0 u( y) − u(x) − 〈p, y − x〉 y−→x n ≤0 are called, respectively the subdifferential and superdifferential of u at x We can see that p ∈ D+ u(x ) if u(x) ≤ u(x ) + p · (x − x ) for all x ∈ B(x , r) x0 p u Similarly, p ∈ D− u(x) if u(x) ≥ u(x ) + p · (x − x ) for all x ∈ B(x , r) HOMOGENIZATION OF HAMILTON-JACOBI EQUATIONS NOTES AND REMARKS u x0 p From this, we can see that, up to a constant, u − φ has a strict max at x ⇐⇒ u is touched from above by φ at x ⇐⇒ Dφ(x ) ∈ D+ u(x ), u − φ has a strict at x ⇐⇒ u is touched from below by φ at x ⇐⇒ Dφ(x ) ∈ D− u(x ) φ x0 u p p φ x0 u In fact, we have the following properties: Proposition (Properties of sub-differentials and super-differentials) Let f : Ω ⊂ n −→ and x ∈ Ω where Ω is open, then the following properties hold (a) D+ f (x) = −D− (− f )(x) (b) D+ f (x) and D− f (x) are convex (possibly empty) (c) If f ∈ C(Ω), then p ∈ D+ f (x) if and only if there is a function ϕ ∈ C (Ω) such that ∇ϕ(x) = p and f − ϕ has a local maximum at x (d) If f ∈ C(Ω), then p ∈ D− f (x) if and only if there is a function ϕ ∈ C (Ω) such that ∇ϕ(x) = p and f − ϕ has a local minimum at x (e) D+ f (x) and D− f (x) are both nonempty if and only if f is differentiable at x In this case we have that D+ f (x) = D− f (x) = {∇ f (x)} (f) If f ∈ C(Ω), the sets of points where a one-sided differential exists Ω+ = {x ∈ Ω : D+ f (x) = } Ω− = {x ∈ Ω : D− f (x) = } are both non-empty Indeed, they are dense in Ω HOMOGENIZATION OF HAMILTON-JACOBI EQUATIONS NOTES AND REMARKS Proof (a) It’s clear since if an −→ a then lim sup (−an ) = − lim inf an n−→∞ n−→∞ (b) It’s also clearl from the definitions (c) Assume that p ∈ D+ f (x), by definition, we can find δ > and a continuous increasing function σ : [0, ∞) −→ with σ(0) = such that f ( y) ≤ f (x) + 〈p, y − x〉 + y − x σ( y − x ) (2) for y − x < δ Define r ρ(r) = σ(t) d t then ρ(0) = ρ ′ (0) = and rσ(r) ≤ ρ(2r) ≤ rσ(2r) Now for y ∈ B(x, δ) we set ϕ( y) = f (x) + 〈p, y − x〉 + ρ(2 y − x ) Since f (x) = ϕ(x), clearly ϕ is differentiable Furthermore, because σ(r) ≤ ρ(2 y − x ) y−x ≤ σ(2 y − x ) and σ(r) −→ as r −→ 0, we conclude that ∇ϕ(x) = p Now for y ∈ B(x, δ), from (2) we have f ( y) − f (x) ≤ 〈p, y − x〉 + y − x σ( y − x ) ≤ 〈p, y − x〉 + ρ(2 y − x ) = ϕ( y) − ϕ(x) Therefore, ( f − ϕ)( y) ≤ ( f − ϕ)(x) for y ∈ B(x, δ), i.e, f − ϕ has a local maximum at x For the converse, if ϕ ∈ C (Ω) such that f − ϕ has a local maximum at x and f (x) = ϕ(x), ∇ϕ(x) = p Then, since f ( y) − f (x) ≤ ϕ( y) − ϕ(x) in a neighborhood of x, we have lim sup y−→x f ( y) − f (x) − 〈p, y − x〉 y−x Therefore, p ∈ D+ f (x) (d) This completely similar to (c) ≤ lim sup y−→x ϕ( y) − ϕ(x) − 〈p, y − x〉 y−x = HOMOGENIZATION OF HAMILTON-JACOBI EQUATIONS NOTES AND REMARKS (e) If f is differentiable at x, then clearly ∇ f (x) ∈ D+ f (x) ∩ D− f (x) Furthermore, if p ∈ D+ f (x), then there exists ϕ ∈ C (Ω) such that ϕ(x) = f (x) ∇ϕ(x) = p and f − ϕ has a local maximum at x, clearly p = ∇ϕ(x) = ∇ f (x) Doing similarly for D− f (x) we have D+ f (x) = D− f (x) = {∇ f (x)} For the converse, assume that D+ f (x) and D− f (x) are both nonempty Assume a ∈ D+ f (x) and b ∈ D− f (x), then there exists ϕ, ψ ∈ C (Ω) such that ϕ(x) = ψ(x) = f (x) and ∇ϕ(x) ∇ψ(x) =p =q f − ϕ has local maximum at x f − ψ has local minimum at x Therefore, in neighborhood B(x, δ) we have ψ( y) ≤ f ( y) ≤ ϕ( y) ∀y s.t y − x < δ Since ψ, ϕ ∈ C (Ω), it’s easy to see that f is also differentiable at x, and so we also have the formula D+ f (x) = D− f (x) = {∇ f (x)} (f) Let x ∈ Ω and ǫ > be given We will show that there exists a function ϕ ∈ C (Ω) such that f − ϕ has local maximum in B(x , ǫ) at some point y in B(x , ǫ) Consider the smooth function in C (B(x , ǫ)) given by ϕ(x) = ǫ − x − x0 x ∈ B(x , ǫ) It’s easy to extend ϕ into a function in C (Ω) Also observe that ϕ(x) −→ +∞ x − x −→ ǫ as Since f is continuous, we have f − ϕ has a local maximum in B(x , ǫ) at some point y By (c), p = ∇ϕ( y) ∈ D+ f (x) and thus D+ f (x) = , i.e y ∈ Ω+ Furthermore, for every x ∈ Ω and ǫ > so small enough, the set Ω+ contains a point y ∈ B(x , ǫ) This shows that Ω+ is dense in Ω Similarly, if we consider the C (B(x , ǫ)) function given by ϕ(x) = −1 ǫ − x − x0 x ∈ B(x , ǫ) The case Ω− is dense in Ω by a similar argument From this, we have the equivalent definition for viscosity solution of static equation as following Definition (Equivalent definition of viscosity solution for the static equation) Assume u is continuous on its domain HOMOGENIZATION OF HAMILTON-JACOBI EQUATIONS NOTES AND REMARKS 10 • (Subsolution) u is called a viscosity subsolution if ∀ x0 ∈ n ∀ p ∈ D+ u(x ) u(x ) + H(x , p) ≤ • (Supersolution) u is called a viscosity supersolution if ∀ x0 ∈ n ∀ p ∈ D− u(x ) u(x ) + H(x , p) ≥ • u is called a viscosity solution if it is both a subsolution and a supersolution Thus, in general we have the following definition for viscosity solution: Definition (General definition for viscosity solutions) Consider the first order partial differential equation F (x, u(x), ∇u(x)) = (3) where F : Ω × × n −→ is a continuous function from an open set Ω ⊂ A function u ∈ C(Ω) is a viscosity subsolution of (3) if for every x ∈ Ω F (x, u(x), p) ≤ n ∀ p ∈ D+ u(x) Similarly, u ∈ C(Ω) is viscosity supersolution of (3) if for every x ∈ Ω F (x, u(x), p) ≥ ∀ p ∈ D− u(x) We say u is a viscosity solution of (3) if it is both a supersolution and a subsolution in the viscosity sense Remark The basic questions about well-posedness of PDE in sense of viscosity solution, the existence, uniqueness and stability were solved by Evans, Crandall-Lions in 1970 for the equation H( y, Duǫ ( y)) = ǫ∆uǫ ( y) by the method adding small viscosity Example (Eikonal’s equation) Let H denote the Hamiltonian, H(u, p) = |p| in consider the Eikonal’s equations H( y, Du( y)) = in (0, 1) u(0) = u(1) = =⇒ |u′ ( y)| = in (0, 1) u(0) = u(1) = , We can see that there are infinite many solutions (in weak sense) of this problem, but there is only one correct visosity solution HOMOGENIZATION OF HAMILTON-JACOBI EQUATIONS NOTES AND REMARKS 25 LECTURE In this lecture, we study the properties of the effective Hamiltonian, H Recall that there exists a Lipschitz viscosity solution v ǫ to the following equation ǫv ǫ ( y) + H( y, p + Dv ǫ ( y)) = in n We also know that H(p) can be defined as the limit of −ǫ j v ǫ j (0) as j −→ ∞ in view of theorem We want to understand H qualitatively, quantitatively and numerically Proposition Consider the following Cauchy problem w t ( y, t) + H( y, p + Dw( y)) = w( y, 0) = in on n n × (0, ∞) Then lim w( y, t) t→∞ t = −H(p) Proof We try to bound the solution, subsolution ≤ w( y, t) ≤ supersolution Let v be the solution to the cell problem H( y, p + Dv( y) = H(p) Denote φ + ( y, t) = v( y) + c − H(p)t φ − ( y, t) = v( y) − c − H(p)t and We have φ t+ ( y, t) + H( y, p + Dφ + ( y, t)) = −H(p) + H( y, p + Dv( y)) = 0, φ t− ( y, t) + H( y, p + Dφ + ( y, t)) = −H(p) + H( y, p + Dv( y)) = For t = 0, φ + ( y, 0) = v( y) + c ≥ and φ − ( y, 0) ≤ for c large enough So, they are super-solution and sub-solutions Thus v( y) − c − H(p)t = φ − ( y, t) ≤ w( y, t) ≤ φ + ( y, t) = v( y) + c − H(p)t Therefore, v( y) − c − H(p) ≤ t Let t −→ ∞, we obtain the final result w( y, t) t ≤ v( y) + c Remark Finding H numerically is extremely hard t − H(p) HOMOGENIZATION OF HAMILTON-JACOBI EQUATIONS NOTES AND REMARKS 26 ————————– We also have some further more properties of the effective Hamiltonian Exercise Suppose H ∈ Lip( n × n ) satisfies (H1)-(H2), i.e coercivity and periodicity in y-coordinate Then H is Lipschitz and coercive We now switch gear to find some representations for H We already knew 2, one is from the definition, another one is right above as the negative of large time average of the solution of proposition The third one is the following, Proposition (Representation formula of H(p)) H(p) = inf{c ∈ n : ∃ v ∈ C( n ) : H( y, p + Dv) ≤ c in Proof There exists a viscousity solution v ∈ Lip( n H( y, p + Dv( y)) = H(p) in viscous sense} ) such that in n Therefore, H(p) ≥ RHS We then want to show that H(p) ≤ RHS By way of contradiction, assume that RHS < H(p) That implies there exists c ≤ H(p) such that there exists w ∈ C( n ) such that H( y, p + Dw) ≤ c < H(p) = H( y, p + Dv) in n Since both v and w are bounded, we can take ǫ > small such that ǫw + H( y, p + Dw) ≤ Therefore, w ≤ v in w > v n c + H(p) ≤ ǫv + H( y, p + Dv) , a contradiction since we can always add a constant to make In case p −→ H(., p) is convex, we also have a further presentation of H Proposition (inf − sup formula) Assume p → H( y, p) is convex Then H(p) = inf φ∈C ( n) sup H( y, p + Dφ( y)) y∈ n Proof Denote the formula in the previous proposition to be A, i.e A = inf c ∈ : ∃ v ∈ C( n ) : H( y, p + Dv) ≤ c in n in viscous sense and B= sup H( y, p + Dφ( y)) : φ ∈ C ( y∈ n n ) HOMOGENIZATION OF HAMILTON-JACOBI EQUATIONS • Take φ ∈ C ( n NOTES AND REMARKS 27 ), let C = maxn H( y, p + Dφ( y)) y∈ So, H( y, p + Dφ( y)) ≤ C in n and C is admisible in the formula of A Therefore, B ⊆ A and thus RHS = inf B ≥ inf A = H(p) • We show that RHS = inf B ≤ inf A = H(p) = H( y, p + Dv( y)) in n Note that v is only Lipschitz so to make it smooth we convolute it Take the standard mollifier φ ∈ Cc∞ ( n ) φ≥0 φ(x) d x = and n For any ǫ > let φ ǫ (x) = ǫ n φ x φ ǫ (x) d x = =⇒ ǫ n ǫ Recall that supp(φ )(x − ) ⊆ B(x, Cǫ) Now take v ǫ (x) = v ǫ ⋆ v (x) = n φ ǫ (x − y)v( y) d y Then we have (in viscous sense) H(p) = H( y, p + Dv( y)) = n H( y, p + Dv( y))φ ǫ (x − y) d y = B(x,Cǫ) H(x, p + Dv( y))φ ǫ (x − y) d y − C ′ ǫ By Jensen’s inequality for convex Hamiltonian H in p, we obtain H(p) ≥ H x, B(x,Cǫ) ǫ (p + Dv( y))φ ǫ (x − y) d y − C ′ǫ = H(x, p + Dv ( y)) − C ′ ǫ ≥ H( y, p + Dv ǫ ( y)) − 2C ′ ǫ Thus there exists some constant C ′ > such that H(p) + 2C ′ ǫ ≥ H( y, p + Dv ǫ ( y)) Since v ǫ is smooth and max y∈ n H( y, p + Dv ǫ ( y)) ∈ B, we obtain H(p) + 2C ′ ǫ ≥ maxn H( y, p + Dv ǫ ( y)) ≥ inf B y∈ HOMOGENIZATION OF HAMILTON-JACOBI EQUATIONS Pass ǫ −→ 0, we get NOTES AND REMARKS H(p) ≥ inf B = RHS Exercise If p −→ H(x, p) is convex then p → H(p) is also convex 28 HOMOGENIZATION OF HAMILTON-JACOBI EQUATIONS APPENDIX SOLUTIONS NOTES AND REMARKS 29 TO PROBLEMS Notes • Proof will be of Son Tu and proof will be of Son Van • The proof of exercise is completely similar to case sub-solution in the lecture and so it will be omitted Exercise Let φ ∈ C ( n ) such that (u − φ)(x ) = and u − φ has a strict max at x Assume uǫ converges to u locally uniformly on n Prove that for ǫ > small enough, uǫ − φ has a max at x ǫ near by x and there is a subsequence ǫ j −→ such that x ǫ j −→ x Proof In this exercise, we will understand the notion of maximum (or strict maximum) as local maximum (or strict maximum) Assume u(x ) = φ(x ) and u − φ has a strict (local) maximum at x , so there exists an open neighborhood of x with compact closure Ω such that u( y) − φ( y) < u(x ) − φ(x ) = ∀ y ∈ Ω\{x } Clearly u is continuous on Ω since uǫ −→ u uniformly on Ω Let r > small such that B ′ (x , 2r) ⊂ Ω Clearly u( y) − φ( y) < ∀ y ∈ S(x , r) = { y ∈ n : y − x = r} Claim There exists δ r > such that for every ǫ < δ r , uǫ − φ only attains its maximum over Ω at some points x ǫ ∈ B ′ (x, r) Proof of claim Assume the converse, then for δ arbitrarily small, there exists ǫ < δ such that uǫ − φ attains its maximum over Ω outside B ′ (x, r) Take δ = 1n , we can construct a sequence ǫn −→ such that uǫn − φ attains its maximum over Ω at x ǫn ∈ Ω\B ′ (x , r), i.e |x ǫn − x | ≥ r for all n ∈ (16) uǫn x ǫn − φ x ǫn ≥ uǫn (x) − φ (x) and ∀ x ∈ Ω Since Ω is compact, there exists some point x ′ ∈ Ω such that x ǫn −→ x ′ Clearly we also have x ′ ∈ Ω\B ′ (x, r) Observe that since uǫn −→ u uniformly on Ω as n −→ ∞, we have uǫn x ǫn − u(x ′ ) ≤ sup |uǫn (x) − u(x)| + u x ǫn − u(x ′ ) −→ x∈Ω So that if we let n −→ ∞ in (16) we obtain u(x ′ ) − φ(x ′ ) ≥ u(x) − φ(x) ∀ x ∈ Ω as n −→ ∞ HOMOGENIZATION OF HAMILTON-JACOBI EQUATIONS NOTES AND REMARKS 30 Therefore u − φ attain a maximum over Ω at x ′ = x It’s a contradiction since u − φ has strict max over Ω at x ————————– Now return to our exercise, denote α r = max{u( y) − φ( y) : y ∈ S(x , r)} < 0, the maximum is archived since u − φ is continuous on compact set S(x , r) Now since uǫ −→ u uniformly on Ω, there exists ζ r > such that for any ǫ < ζ r , then −α r |uǫ ( y) − u( y)| < ∀ y ∈ Ω Now for any y ∈ S(x , r) we have αr (17) uǫ ( y) − φ( y) = uǫ ( y) − u( y) + u( y) − φ( y) < Choose λ r = min{ζ r , δ r }, then for any ǫ < λ r , uǫ − φ is continuous on Ω, thus it obtains a maximum over Ω at some point x ǫ ∈ B ′ (x , r), denote its maximum by α = uǫ (x ǫ ) − φ(x ǫ ), then clearly αr (18) α = max{uǫ ( y) − φ( y)} ≥ uǫ (x ) − φ(x ) = uǫ (x ) − u(x ) > Ω Combine (17) and (18) we obtain x ǫ must lie in B(x , r) Therefore for any ǫ > small enough, uǫ − φ only has maximum over Ω at some points x ǫ ∈ B(x , r) The rest is pretty simple, let r = 1n , then for each n, choose ǫn = λ , n n =⇒ ǫn −→ as n −→ ∞ and uǫn − φ has a max at x ǫn inside B x , 1n , clearly x ǫn −→ x and furthermore uǫn (x ǫn ) − u(x ) ≤ uǫn (x ǫn ) − u(x ǫn ) + u(x ǫn ) − u(x ) −→ as n −→ ∞ Proof Observe that this theorem is wrong if uǫ − φ and u − φ are not continous One can construct a couterexample using a modified version of the topologist’s sine curve (remove the largest points and give them different smaller values) So assume that uǫ and u are continuous The problem now is equivalent to the following problem: “Suppose f is a continuous function with global max at x and that f ǫ are continuous function that converges uniformly to f as ǫ → Then for small enough ǫ, f ǫ has a max x ǫ near x ” Pick r1 > r2 > 0, we have that f |{x:r2 ≤|x−x |≤r1 } has a max at x ′ Furthermore, f (x ) > f (x ′ ) Define h := f (x ) − f (x ′ ) For small enough ǫ ′ , we have | f ǫ (x) − f (x)| < 3h for ǫ < ǫ ′ Also, for small enough r (we can assume r < r1 ), we have HOMOGENIZATION OF HAMILTON-JACOBI EQUATIONS NOTES AND REMARKS 31 | f (x ) − f (x)| < 3h for x ∈ B(x , r) Therefore, pick x ∈ B(x , r) and ǫ < ǫ ′ , we have | f (x ) − f ǫ (x)| < 2h Pick y ∈ {x : r2 ≤ |x − x | ≤ r1 } We have f ǫ (x) > f (x ′ ) + 3h > f ( y) + 3h > f ǫ ( y) This means that, maxB(x ,r1 −δ) f ǫ (t) ≥ f ǫ (x) > f ǫ ( y) for arbitrary small δ > This means that maxB(x ,r2 ) f ǫ (t) must be at t ∈ B(x , r1 ) ⊆ B(x , Therefore, f ǫ (t ) ≥ f ǫ (t) for all t ∈ B(x , (Not the cleanest proof but it works ) r1 +r2 ) r1 +r2 ) Thus, t is our desired x ǫ Exercise Let n = 1, H(x, p) = |p| − W (x) where W is defined by its graph y W x i.e, W : (E0 ) −→ 4 is periodic Consider the ergodic cell problem (E0 ) H( y, + Dv( y)) = |Dv( y)| − W ( y) = c in (a) Find the unique constant c such that (E0 ) has solution (b) Find infinitely solutions for (E0 ) for all y ∈ n Proof (a) Assume v is continuous a viscosity solution of (E0 ), since v is bounded and archived its minimum, called m = v(x m ) = min{v(x) : x ∈ } ≤ v(x) is compact we have ∀x∈ – Super-test At x m , we can see that q = belong to Dv − (x m ) Thus = |q| ≥ W (x m ) + c ≥ W (x) + c = c =⇒ ≥ c x∈ – Sub-test Setting + = {x ∈ : D+ v(x) = }, let p ∈ D+ v(x) then = |p| ≤ W (x) + c ∀x∈ + But is dense in , thus there exists a sequence {x n } ∈ + such that9 x n −→ 14 , combine with the fact that W is continuous we obtain + c ≥ −W (x n ) ∀n∈ We write 1/4 mean the element 1/4 in =⇒ c ≥ −W (1/4) = =⇒ c ≥ HOMOGENIZATION OF HAMILTON-JACOBI EQUATIONS NOTES AND REMARKS 32 Thus c = is the only possible value such that (E0 ) has a viscosity solution (b) For each c ∈ uc (x) = , x2 − , consider the following function x χ[0, ] + −x + + x2 − 3 x− χ[ ,c ]∪[ −c,1] x − 2c + 3c − For example as c = 0.6 and c = 0.7 we have 2 χ[c, −c ] FIGURE The function uc as c = 0.6 and c = 0.7 Now we will show that uc is a viscosity solution of |Duc (x)| = W (x) for any c ∈ , To this, we only need to consider points in which uc is not smooth, that is x = c and x = 23 − c Also, we only need to consider D+ uc (x) since the set D− uc (x) is always empty at non-smooth points of uc To begin with, we will give explicit formulas for D+ uc (c) and D+ uc 32 − c Case x = c Recall that p ∈ D+ u(c) iff u(x) − u(c) ≤ p(x − c) – Consider x < c, then it becomes 3 (x − c) − x − c ≤ p(x − c) =⇒ − x − c ≥ p =⇒ − 2c ≥ p 2 – Consider x > c, then it becomes 3 (x − c) x + c − ≤ p(x − c) =⇒ x + c − ≤ p =⇒ 2c − ≤ p 2 Thus in this case we have p ∈ D+ uc (c) ⇐⇒ |p| ≤ − 2c HOMOGENIZATION OF HAMILTON-JACOBI EQUATIONS NOTES AND REMARKS 33 Now we need to check that |p| ≤ W (c) for all p ∈ D+ uc (c), we have − 4c = W (c) p ∈ D+ uc (c) =⇒ |p| ≤ − 2c = 2 Case x = 23 − c Since uc and W are symmetric through the line x = 43 , this case is completely similar to case x = c Proof Consider the following equation ǫuǫ ( y) + H( y, Duǫ ) = (HJ) Plug the formula of H into (HJ) we have ǫuǫ ( y) + |Duǫ ( y)| − W ( y) = Observe that for each ǫ > 0, uǫ = is a subsolution and uǫ = 1ǫ is a supersolution To see this, for uǫ = 0, let φ ∈ C ∞ s.t (φ − uǫ )( y0 ) = is a max We have Dφ( y0 ) = Duǫ ( y0 ) = So ǫ Similarly, for u = , ǫ ǫφ( y0 ) + H( y0 , Dφ( y0 )) = − W ( y0 ) ≤ let φ ∈ C ∞ s.t (φ − uǫ )( y0 ) = is a We have ǫφ( y0 ) + H( y0 , Dφ( y0 )) = − W ( y0 ) ≥ Therefore, if v ǫ is a viscosity solution to (HJ), we have ≥ ǫv ǫ ≥ Thus, we have ≥ W ( y) ≥ |Dv ǫ ( y)| (19) From here we conclude that v ǫ is equi-Lipschitz with modulus This is not enough to conclude anything about the convergence of ǫv ǫ except of the result obtained in class Luckily, we have more– observe that Dv ǫ Dv ǫ 4 ≤W ≤W = 0, = So, Dv ǫ = Dv ǫ = HOMOGENIZATION OF HAMILTON-JACOBI EQUATIONS NOTES AND REMARKS That means, for all ǫ > 0, by plugging the above to our equation, ǫv ǫ means v ǫ 41 = Similarly, v ǫ that for all ǫ > 0, 4 y y ǫ′ |v ( y)| = = 0, which = for all ǫ > Combine with (19), we deduce y ǫ 34 v (t)d t ≤ ǫ′ v (t) d t ≤ 4 1d t = y − ≤ So, ǫv ǫ ( y) → as ǫ → and, therefore, c = The solutions to the problem are obtained similarly to proof Exercise Let H ∈ Lip( (H1) n n × ) satisfying lim H( y, p) = +∞ uniformly in y, |p|−→∞ (H2) H( y + k, p) = H( y, p) for all k ∈ n Then H(p) : −→ Proof For u ∈ C( n is a Lipschitz and coercive function n ) and F : n × × n −→ is continous, we define F (x, u(x), Du(x)) ≤ C in the viscous sense ⇐⇒ F (x, u(x), p) ≤ C ∀ p ∈ D+ u(x) F (x, u(x), Du(x)) ≥ C in the viscous sense ⇐⇒ F (x, u(x), p) ≥ C ∀ p ∈ D− u(x) ⇐⇒ u is sub-solution of F (x, u(x), Du(x)) = C, ⇐⇒ u is super-solution of F (x, u(x), Du(x)) = C By assumption, there exists a constant C > such that |H(x, p) − H(x, q)| ≤ C p − q n ∀ x, p, q ∈ We will prove the Lipschitz property of H by using the representation (20) H(q) = inf c ∈ : ∃ v ∈ C( n ) : H(x, q + Dv(x)) ≤ c in Step First of all, we prove that (21) Let u ∈ Lip( (Eq ) n H(p) − H(q) ≤ C p − q ) is a periodic viscosity solution of the problem H(x, q + Du(x)) = H(q) By definition ζ ∈ D+ u(x) =⇒ H(x, q + ζ) ≤ H(q) n in viscous sense HOMOGENIZATION OF HAMILTON-JACOBI EQUATIONS NOTES AND REMARKS 35 We will show that H(x, p + Du(x)) ≤ C p − q + H(q) (22) in the viscous sense Indeed, for ζ ∈ D+ u(x), by the Lipschitz’s property of H, we have |H(x, p + ζ) − H(x, q + ζ)| ≤ C p − q =⇒ H(x, p + ζ) ≤ C p − q + H(x, q + ζ) =⇒ H(x, p + ζ) ≤ C p − q + H(q) ∀ ζ ∈ D+ u(x) =⇒ H(x, p + Du(x)) ≤ C p − q + H(q) (vis-sense) Thus (22) is true, from the representation of H in (20), we obtain (21) is true since H(p) ≤ C p − q + H(q) ⇐⇒ H(p) − H(q) ≤ C p − q Step Similarly, doing quite similar to step 1, we obtain H(q) − H(p) ≤ C p − q (23) From (21) and (23) we obtain the Lipschitz’s property of H, |H(p) − H(q)| ≤ C p − q Step We now prove the coercivity of H First restate the coercivity of H as (24) lim |p|−→∞ minn H(x, p) = +∞ x∈ We start with a similar result as representation formula of H in (20), that is (25) H(q) = sup c ∈ : ∃ v ∈ C( n ) : H(x, q + Dv(x)) ≥ c in n in viscous sense Indeed, define α = sup c ∈ Let u ∈ C( n : ∃ v ∈ C( n n ) : H(x, q + Dv(x)) ≥ c in in viscous sense ) be the vicosity solution of the cell problem H(x, p + Du(x)) = H(p) in n Clearly, H(p) is admissible in the right hand of formula (25), thus H(p) ≤ α Now we want to prove α ≤ H(p), assume the contradiction, that is H(p) < α Then there exists (c, w) ∈ × C( n ), H(p) < c ≤ α such that H(x, p + Dw(x)) ≥ c > H(p) = H(x, p + Du(x)) Since u, w are bounded, we can take ǫ > such that ǫu(x) + H(x, p + Du(x)) < c + H(p) < ǫw(x) + H(x, p + Dw(x)) HOMOGENIZATION OF HAMILTON-JACOBI EQUATIONS NOTES AND REMARKS 36 By comparison principle, we have u ≤ w in n , it’s a contradiction since we can always add a constant to make w < u Now using thus representation, let ϕ = C ∈ C ( n ), and define c = minn H(x, p + Dϕ(x)) = minn H(x, p) x∈ x∈ Then, c is admissible in the formula of the representation (25), so that H(p) ≥ c, i.e H(p) ≥ minn H(x, p) (26) x∈ Now combine (26) with (24) we obtain minn H(x, p) = +∞ lim H(p) ≥ lim |p|−→∞ x∈ |p|−→∞ Thus p −→ H(p) is coercive and the proof is complete Proof Lipschitz part ¯ ¯ Let p, q ∈ n , we want to show that |H(p)− H(q)| ≤ C|p−q| Let uǫ and v ǫ be solutions to the following problems ǫuǫ + H(x, Duǫ + p) = (27) and ǫv ǫ + H(x, Dv ǫ + q) = (28) We have, by “Lipschitzivity”, |H(x, Duǫ + p) − H(x, Duǫ + q)| ≤ C|p − q| Thus, | − ǫuǫ − H(x, Duǫ + q)| ≤ C|p − q|, which implies ǫuǫ + C|p − q| + H(x, Duǫ + q) ≥ (29) Thus, uǫ + C|p−q| ǫ is a supersolution to (29) and uǫ + C|p − q| ǫ ≥ vǫ , meaning Similarly, we obtain ǫv ǫ − ǫuǫ ≤ C|p − q| −C|p − q| ≤ ǫv ǫ − ǫuǫ ¯ Thus, |ǫuǫ − ǫv ǫ | ≤ C|p − q| Let {ǫ j } be a sequence such that ǫ j uǫ j → H(p) and ǫj ¯ ǫ j v → H(q), we then have ¯ ¯ |H(p) − H(q)| ≤ C|p − q| HOMOGENIZATION OF HAMILTON-JACOBI EQUATIONS NOTES AND REMARKS 37 Coercivity part: Let M > Pick |p| big enough so that H(x, p) > M Let u be the solution to the cell problem ¯ H(x, Du + p) = H(p) Since u is continuous and periodic, u has a max at x So consider the test function φ(x) = u(x ) We have that, by definition of viscosity solution, ¯ M < H(x, Dφ(x ) + p) = H(x, p) ≤ H(p) Exercise Let H ∈ Lip( (H1) n × n ) satisfying lim H( y, p) = +∞ uniformly in y, |p|−→∞ (H2) H( y + k, p) = H( y, p) for all k ∈ n Assume p −→ H( y, p) is convex for any y ∈ n , then p −→ H(p) is also convex Proof We will prove the convexity of p −→ H(p) by the representation formula of H, that is H(q) = inf c ∈ : ∃ v ∈ C( n ) : H(x, q + Dv(x)) ≤ c in = inf sup H(x, q + Dϕ(x)) : ϕ ∈ C ( x∈ Let u ∈ Lip( n in viscous sense ) n ) is a viscosity solution of the problem H(x, p + Du(x)) = H(p) (E p ) Let ϕ ∈ C ( n n n ) arbitrary and let C = maxn H(x, q + Dϕ(x)) x∈ For λ ∈ (0, 1), consider the function w = λu + (1 − λ)ϕ We will show that (in the viscous sense) (30) H x, λp + (1 − λ)q + Dw ≤ λH(p) + (1 − λ)C To this, we need to show that for any ζ ∈ D+ w(x), then H x, λp + (1 − λ)q + ζ ≤ λH(p) + (1 − λ)C HOMOGENIZATION OF HAMILTON-JACOBI EQUATIONS NOTES AND REMARKS 38 Lemma Let u ∈ C(Ω), then D+ w(x) = λD+ u(x) + (1 − λ)Dϕ(x) for any ϕ ∈ C (Ω), w = λu + (1 − λ)ϕ and λ ∈ [0, 1] Proof of lemma It’s easy to see that λD+ u(x) + (1 − λ)Dϕ(x) ⊆ D+ w(x) For the converse, let ζ ∈ D+ u(x), setting α = ζ − (1 − λ)Dϕ(x), then λu( y) − λu(x) − 〈α, y − x〉 y−x = w( y) − w(x) − 〈ζ, y − x〉 −(1 − λ) y−x ϕ( y) − ϕ(x) − 〈Dϕ(x), y − x〉 y−x Taking lim sup of both sides, we have α ∈ D+ (λu)(x), therefore δ=λ α α λ ∈ D+ u(x), then + (1 − λ)Dϕ(x) ⊆ λD+ u(x) + (1 − λ)Dϕ(x) λ The proof of lemma is complete By above lemma, there exists α ∈ D+ u(x) such that ζ = λα + (1 − λ)Dϕ(x), using the convexity in p of H we obtain H x, λp + (1 − λ)q + ζ = H x, λp + (1 − λ)q + λα + (1 − λ)Dϕ(x) ≤ λH(x, p + α) + (1 − λ)H(x, q + Dϕ) ≤ λH(p) + (1 − λ)C Thus in sense of viscosity, we obtain (30) is true By representation formula for H λp + (1 − λ)q we have H λp + (1 − λ)q ≤ λH(p) + (1 − λ)C In other words, we have H λp + (1 − λ)q − λH(p) ≤ (1 − λ)C = (1 − λ) maxn H(x, q + Dϕ(x)) x∈ and since ϕ ∈ C ( n ) is arbitrary, we get H λp + (1 − λ)q − λH(p) ≤ (1 − λ) inf ϕ∈C ( n) maxn H(x, q + Dϕ(x)) = (1 − λ)H(q) x∈ Thus, H λp + (1 − λ)q ≤ λH(q) + (1 − λ)H(q) and H is convex HOMOGENIZATION OF HAMILTON-JACOBI EQUATIONS NOTES AND REMARKS 39 ¯ Proof Suppose H(p) is not convex There are p, q such that ¯ ¯ ¯ (31) H((1 − k)p + kq) > (1 − k)H(p) + k H(q) for some k ∈ (0, 1) Let u be a solution of (32) ¯ H(x, Du + (1 − k)p + kq) = H((1 − k)p + kq) and vp , vq be solutions of ¯ H(x, Dv + p) = H(p) and ¯ H(x, Dv + q) = H(q) respectively (32) Implies that ¯ H((1 − k)p + kq) > (1 − k)H(x, Dvp + p) + kH(x, Dvq + q) ≥ H(x, (1 − k)(Dvp + p) + k(Dvq + q)) = H(x, (1 − k)Dvp + kDvq + (1 − k)p + kq) So, (1 − k)vp + kvq is a subsolution to (33) Therefore, u ≥ (1 − k)vp + kvq But if u is a solution to (33), u + C is also a solution to (33), which makes u + C ≥ (1 − k)vp + kvq for all C, which is a contradiction since vp and vq are real-valued functions REFERENCES [1] Michael G Crandall, Hitoshi Ishii, and Pierre-Louis Lions user’s guide to viscosity solutions of second order partial differential equations Bull Amer Math Soc (N.S.) 27 (1992) 1-67, July 1992 [2] Lawrence C Evans The perturbed test function method for viscosity solutions of nonlinear pde Proceedings of the Royal Society of Edinburgh: Section A Mathematics, 111(3-4):359–375, 1989 [3] Pierre-Louis Lions, George Papanicolaou, and Srinivasa RS Varadhan Homogenization of hamiltonjacobi equations Unpublished preprint, 1986 [4] Hiroyoshi Mitake and Hung V Tran Dynamical properties of Hamilton-Jacobi equations via the nonlinear adjoint method: Large time behavior and Discounted approximation ... This part is achieved by introducing the variable y HOMOGENIZATION OF HAMILTON- JACOBI EQUATIONS NOTES AND REMARKS 25 LECTURE In this lecture, we study the properties of the effective Hamiltonian,... that is independent of ǫ so as we pass ǫ → 0, (Cǫ ) should become (HJ) The answer to this problem is “yes” and will be elaborated in lecture HOMOGENIZATION OF HAMILTON- JACOBI EQUATIONS NOTES AND. .. SOLUTIONS NOTES AND REMARKS 29 TO PROBLEMS Notes • Proof will be of Son Tu and proof will be of Son Van • The proof of exercise is completely similar to case sub-solution in the lecture and so it