HOMOGENIZATION OF HAMILTON-JACOBI EQUATIONS NOTES AND REMARKS SON NGUYEN THAI TU. Department of Mathematics, University of Wisconsin at Madison

Transcription

1 HOMOGENIZATION OF HAMILTON-JACOBI EQUATIONS NOTES AND REMARKS SON NGUYEN THAI TU Department of Mathematics, University of Wisconsin at Madison 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. Tran 1 at University of Science, Ho Chi Minh City, Viet Nam from July 6 th to July 14 th, addresses: thaison@math.wisc.edu, sonv@andrew.cmu.edu. 1 Department of Mathematics, University of Wisconsin at Madison 1

2 Notations: HOMOGENIZATION OF HAMILTON-JACOBI EQUATIONS NOTES AND REMARKS n = n / n is the n-dimensional donut (torus). w.r.t: with respect to LECTURE 1 Clear contest : ɛ > 0 : scale of the equation, ɛ 1 : ɛ is very small. Equation of interest. Following the Hamilton-Jacobi equation x (I) u ɛ (x, t) + H t ɛ, Duɛ (x, t) = 0 in n (0, ) where u ɛ (x, t) : n [0, ), x is the location variable (spatial) and t is the time variable. u ɛ(x, t) = t t uɛ (x, t). Du ɛ (x, t) = x (u ɛ )(x, t) = u ɛ x 1,..., uɛ x n. H is the Hamiltonian (total energy) Example 1 (Classical mechanics). H : n n (y, p) H(y, p). H(y, p) = 1 p + 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 ɛ 0. Is there some u such that u ɛ 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) (H) (periodicity) lim H(y, p) = uniformly in y, p H(y + k, p) = H(y, p) k n.

3 HOMOGENIZATION OF HAMILTON-JACOBI EQUATIONS NOTES AND REMARKS 3 Formal Analysis. (not rigorous) Consider the equation x (I) u ɛ (x, t) + H t ɛ, Duɛ (x, t) = 0 where t is time variable, x is spatial variable (slow variable) and y = x ɛ oscillatory variable. There are several ways to begin our analysis: Numerical analysis: plot u ɛ. Guessing: some asymptotic expansion w.r.t ɛ. We consider some ansatzs (1) u ɛ (x, t) = u(x, t) + ɛu 1 x, x ɛ, t + ɛ u x, x ɛ, t +... is the fast () u ɛ (x, t) = u x, x ɛ, t + ɛu 1 x, x ɛ, t (3) u ɛ (x, t) = u (x, t) + ɛ 5 u 1 x ɛ, t (4) u ɛ (x, t) = u(x, t) + ɛu 1 x ɛ, t + ɛ u x ɛ, t +... Now using (4), we have u ɛ (x, t) = u(x, t) + ɛu 1 x ɛ, t + ɛ u x ɛ, t +... Du ɛ (x, t) = Du(x, t) + Du 1 x ɛ, t + ɛdu x ɛ, t +... Plugging these into (I) we have x u ɛ (x, t) + t ɛu1 t ɛ, t H x x x ɛ, Du(x, t) + Du1 ɛ, t + ɛdu ɛ, t +... = 0. Denote y = x ɛ, we obtain u ɛ (x, t) + t ɛu1 (y, t) H y, Du(x, t) + Du 1 (y, t) + ɛdu (y, t) +... = 0. 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) + Du 1 (y, t) = 0. Now, fix (x, t) n (0, ), since u t (x, t) is a constant w.r.t y, is expected to be a constant w.r.t y, i.e., H(y, Du(x, t) + Du 1 (y, t)) H(y, Du(x, t) + Du 1 (y, t)) = C(p, t). The word Ansatz in German is some thing like the word formulation in English.

4 HOMOGENIZATION OF HAMILTON-JACOBI EQUATIONS NOTES AND REMARKS 4 Let p = Du(x, t) n, then H(y, p + Du 1 (y, t)) = C(p, t). Assume one more reduction u 1 (y, t) = u 1 (y), we then have the following PDE H(y, p + Du 1 (y)) = C(p). This is called the cell (or ergodic) problem. Theorem 1 (Lions-Pappanicolau-Varadhan). Fix p n, there exists a unique constant C n such that the cell problem has a periodic solution u 1. H(y, p + Du 1 (y)) = C(p) in n Define the effective Hamiltonian H(p) = C(p). We are back to our original question: consider u ɛ t (C ɛ ) + H( x ɛ, Duɛ ) = 0 u ɛ. (x, 0) = u 0 (x) Do the solutions u ɛ of (C ɛ ) converges to a solution u of ut + H(Du) = 0 (HJ)? u(x, 0) = u 0 (x) Remark 1. This is a natural question since from the Hamiltonian H( x ɛ, Duɛ ) we derived the effective Hamiltonian H that is independent of ɛ so as we pass ɛ 0, (C ɛ ) should become (HJ). The answer to this problem is yes and will be elaborated in lecture 3.

5 HOMOGENIZATION OF HAMILTON-JACOBI EQUATIONS NOTES AND REMARKS 5 LECTURE Before we prove theorem 1, let s talk about viscosity solutions. Introduction to viscosity solutions. Consider The static problem (S) u(y) + H(y, Du(y)) = 0. (C) The Cauchy problem ut (x, t) + H(y, Du(x, t)) = 0 in n (0, ) u(y, 0) = u 0 (y) on n. We then look into the static problem in n. The idea here is that the problem (S ɛ ) u ɛ + H(Du ɛ (y), y) = ɛ u ɛ in n has a smooth solution u ɛ. Assume further that u ɛ u locally uniformly in n. Take a smooth function φ such that (u φ)(x0 ) = 0 (u φ)(x) < 0 else where. That is, u φ has a strict max at x 0. Exercise 1. For ɛ > 0 small enough, u ɛ φ has a max at x ɛ near by x 0 and there is a subsequence 0 such that x x 0. 3 Play with u ɛ. Since u ɛ φ has max at x ɛ, we then have, of course by second derivative test, Du ɛ (x ɛ ) = Dφ(x ɛ ) = Du ɛ (x ɛ ) = Dφ(x ɛ ) (1) (u ɛ φ)(x ɛ ) 0 u ɛ (x ɛ ) φ(x ɛ ). Reconsider the PDE, From (1), we can see that u ɛ (x ɛ ) + H(x ɛ, Du ɛ (x ɛ )) = ɛ u ɛ (x ɛ ). u ɛ (x ɛ ) + H(x ɛ, Dφ(x ɛ )) ɛ φ(x ɛ ). 3 Morever, if u m is continuous and φ is smooth such that u m u uniformly on a compact set Ω n, and assume u φ has a strict (local) maximum over Ω as x 0, u m φ has (local) maximum over Ω at x m. Then we must have x m x 0 as m. We have the convegence of whole sequence here, while in case u ɛ u uniformly as ɛ 0 we only have the convergence of sub-sequence.

6 HOMOGENIZATION OF HAMILTON-JACOBI EQUATIONS NOTES AND REMARKS 6 Let 0, we then have u(x 0 ) + H(x 0, Dφ(x 0 )) 0. Inspired by this derivation, we have the definition of viscosity solutions. Definition 1 (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 ( n ) such that (u φ)(x 0 ) is a strict max then u(x 0 ) + H(x 0, Dφ(x 0 )) 0. (Supersolution) u is called a viscosity supersolution if for all φ C ( n ) such that (u φ)(x 0 ) is a strict min then u(x 0 ) + H(x 0, Dφ(x 0 )) 0. 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 n. For any x Ω, the sets D u(x) = p n u(y) u(x) p, y x : lim inf 0 y x y x D + u(x) = p n : lim sup y x u(y) u(x) p, y x 0 y x are called, respectively the subdifferential and superdifferential of u at x. We can see that p D + u(x 0 ) if u(x) u(x 0 ) + p (x x 0 ) for all x B(x 0, r). x 0 p u Similarly, p D u(x) if u(x) u(x 0 ) + p (x x 0 ) for all x B(x 0, r).

7 HOMOGENIZATION OF HAMILTON-JACOBI EQUATIONS NOTES AND REMARKS 7 u From this, we can see that, up to a constant, x 0 p u φ has a strict max at x 0 u is touched from above by φ at x 0 Dφ(x 0 ) D + u(x 0 ), u φ has a strict min at x 0 u is touched from below by φ at x 0 Dφ(x 0 ) D u(x 0 ). φ x 0 p u p u φ x 0 In fact, we have the following properties: Proposition 1 (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 1 (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 1 (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.

8 Proof. HOMOGENIZATION OF HAMILTON-JACOBI EQUATIONS NOTES AND REMARKS 8 (a) It s clear since if a n a then lim sup n ( a n ) = lim inf n a n (b) It s also clearl from the definitions. (c) Assume that p D + f (x), by definition, we can find δ > 0 and a continuous increasing function σ : [0, ) with σ(0) = 0 such that () f (y) f (x) + p, y x + y x σ( y x ) for y x < δ. Define then r ρ(r) = σ(t) d t 0 ρ(0) = ρ (0) = 0 and rσ(r) ρ(r) rσ(r) Now for y B(x, δ) we set ϕ(y) = f (x) + p, y x + ρ( y x ). Since f (x) = ϕ(x), clearly ϕ is differentiable. Furthermore, because ρ( y x ) σ(r) σ( y x ) y x and σ(r) 0 as r 0, we conclude that ϕ(x) = p. Now for y B(x, δ), from () we have f (y) f (x) p, y x + y x σ( y x ) p, y x + ρ( 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 1 (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 lim sup y x ϕ(y) ϕ(x) p, y x = 0. y x Therefore, p D + f (x). (d) This completely similar to (c).

9 HOMOGENIZATION OF HAMILTON-JACOBI EQUATIONS NOTES AND REMARKS 9 (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 1 (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 1 (Ω) such that ϕ(x) = p f ϕ has local maximum at x ϕ(x) = ψ(x) = f (x) and ψ(x) = q 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 1 (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 0 Ω and ɛ > 0 be given. We will show that there exists a function ϕ C 1 (Ω) such that f ϕ has local maximum in B(x 0, ɛ) at some point y in B(x 0, ɛ). Consider the smooth function in C 1 (B(x 0, ɛ)) given by 1 ϕ(x) = ɛ x x 0 x B(x 0, ɛ). It s easy to extend ϕ into a function in C 1 (Ω). Also observe that ϕ(x) + as x x 0 ɛ. Since f is continuous, we have f ϕ has a local maximum in B(x 0, ɛ) 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 0 Ω and ɛ > 0 so small enough, the set Ω + contains a point y B(x 0, ɛ). This shows that Ω + is dense in Ω. Similarly, if we consider the C 1 (B(x 0, ɛ)) function given by 1 ϕ(x) = ɛ x x 0 x B(x 0, ɛ). 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 3 (Equivalent definition of viscosity solution for the static equation). Assume u is continuous on its domain

10 HOMOGENIZATION OF HAMILTON-JACOBI EQUATIONS NOTES AND REMARKS 10 (Subsolution) u is called a viscosity subsolution if x 0 n u(x 0 ) + H(x 0, p) 0 p D + u(x 0 ). (Supersolution) u is called a viscosity supersolution if x 0 n u(x 0 ) + H(x 0, p) 0 p D u(x 0 ). 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 4 (General definition for viscosity solutions). Consider the first order partial differential equation (3) F(x, u(x), u(x)) = 0 where F : Ω n is a continuous function from an open set Ω n. A function u C(Ω) is a viscosity subsolution of (3) if for every x Ω F(x, u(x), p) 0 p D + u(x). Similarly, u C(Ω) is viscosity supersolution of (3) if for every x Ω F(x, u(x), p) 0 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 by the method adding small viscosity. H(y, Du ɛ (y)) = ɛ u ɛ (y). Example (Eikonal s equation). Let H denote the Hamiltonian, H(u, p) = p in, consider the Eikonal s equations H(y, Du(y)) = 1 in (0, 1) u (y) = 1 in (0, 1) =. u(0) = u(1) = 0 u(0) = u(1) = 0 We can see that there are infinite many solutions (in weak sense) of this problem, but there is only one correct visosity solution.

11 HOMOGENIZATION OF HAMILTON-JACOBI EQUATIONS NOTES AND REMARKS correct viscosity solution Finding a viscosity solution. We already know this equation has infinitely many solutions. Consider a solution u that its graph contains a shape, for example D + u = ; D u = [ 1, 1]. Thus, if u is a viscosity supersolution then 1 p D u : p 1 0. It s contradiction. The only u which is not contain this shape is

12 HOMOGENIZATION OF HAMILTON-JACOBI EQUATIONS NOTES AND REMARKS D + u = [ 1, 1] ; D u =. Thus, if u is a viscosity subsolution then 1 p D + u : p Thus, this is the only viscosity solution to the problem. Remark 3. In general, Hamilton-Jacobi equations have infinitely a.e solutions. Viscosity solutions give -further adittional requirement to select one good solution in form these infinitely many. Now we are ready to prove theorem 1. Recall The ergodic (cell) problem. Fix p n, there exists a unique c such that (E p ) H(y, p + Dv(y)) = c in n. has a periodic solution v. Proof of theorem 1. We prove through some steps. Discounted approximation. For ɛ > 0, consider the following problem ɛv ɛ (y) + H(y, p + Dv ɛ (y)) = 0 in n Claim without proof: There exists a unique viscosity solution v ɛ (y) to the above problem. 1 st observation: if v ɛ (y) is a solution then v ɛ (y + k) for k n is also a solution by periodicity of H. Thus v ɛ (y) is periodic, i.e y ɛ (y) = v ɛ (y + k) by the uniqueness of the solution v ɛ. nd observation: there are apriori estimates. We try to bound v ɛ. First of all, take φ = c 1, and observe H(y, p) is periodic, this is bounded, assume H(y, p) C. If we choose φ = c 1 = C ɛ then ɛφ + H(y, p + Dφ(y)) = ɛc 1 + H(y, p) ɛc 1 + C = 0.

13 HOMOGENIZATION OF HAMILTON-JACOBI EQUATIONS NOTES AND REMARKS 13 If we choose φ = c 1 = C ɛ then ɛφ + H(y, p + Dφ(y)) = ɛc 1 + H(y, p) ɛc 1 C = 0. Thus we obtain φ = C ɛ φ = C ɛ for C > 0 large is a subsolution, for C > 0 large is a supersolution. Therefore since v ɛ is a solution, we get C ɛ vɛ C ɛ = ɛv ɛ C, which implies H(y, p + Dv ɛ (y)) C y. Now by coercivity of H, we must have Dv ɛ (y) C for some constant C > 0. 3 rd observation: {v ɛ } is equi-lipschitz and v ɛ is periodic. Since v ɛ (x) v ɛ (y) Dv ɛ (ζ) x y C x y for some ζ [x, y] = {t x + (1 t)y : t [0, 1]} and clearly C is independent to ɛ. In other words, for any ɛ > 0 v ɛ Lip( n ) and Dv ɛ L ( n ) C. Remark 4. At this point we want to use Arzela-Ascoli theorem. However, v ɛ aren t bounded as ɛ 0 so we can t get the uniform bound for every ɛ thus Arzela-Ascoli cannot be applied here. Luckily, we have something lurking underneath so Arzela-Ascoli theorem could be applied. Define We have Dw ɛ (y) = Dv ɛ (y) C and so w ɛ (y) = v ɛ (y) v ɛ (0). w ɛ (y) = v ɛ (y) v ɛ (0) C y C n since n is bounded. Thus w ɛ is bounded and we can apply the Arzela-Ascoli theorem to {w ɛ }, there exists 0 such that as j, (4) w (y) w(y) uniformly in n. Since ɛv ɛ (0) C, by Bozalno-Weitrass s principle, we can assume (passing to subsequence if necessary) that v (0) c for some constant c. We claim furthermore that v (y) c as j.

14 HOMOGENIZATION OF HAMILTON-JACOBI EQUATIONS NOTES AND REMARKS 14 Indeed, we have v (y) v (0) Dv L ( n ) C 0 as j. Now look at It becomes ɛv ɛ (y) + H(y, p + Dv ɛ (y)) = 0 in n. (5) w (y) + v (0) + H(y, p + Dw (y)) = 0. 4 th observation: Now we claim that (6) H(y, p + Dw(y)) = c in the viscosity sense. (We don t have Dw Dw.). Indeed, Sub-test: Fix x n and let ϕ C ( n ) such that (w ϕ)(x) = 0 w ϕ < 0 else where i.e, w ϕ has a strict max at x. Since w w uniformly on some compact neighborhood of x, there exists a subsequence {j k } such that w k ϕ has a max at x ɛjk near x x ɛjk x as k. w k x k w(x) as k Look at (5), in viscosity sense, we must have k w k x k + k v k (0) + H(x ɛjk, p + Dϕ k (x k )) 0. Now let k we obtain c + H(x, p + Dϕ(x)) 0 H(x, p + Dϕ(x)) c. Thus, w is a viscosity subsolution of (6). Super-test: Fix x n and let ϕ C ( n ) such that (w ϕ)(x) = 0 w ϕ > 0 else where i.e, w ϕ has a strict min at x.

15 HOMOGENIZATION OF HAMILTON-JACOBI EQUATIONS NOTES AND REMARKS 15 Since w w uniformly on some compact neighborhood of x, there exists a subsequence {j k } such that w k ϕ has a min at x ɛjk near x x ɛ jk x as k. w k x ɛjk w(x) as k Look at (5), in viscosity sense, we must have k w k x k + k v k (0) + H(x k, p + Dϕ k (x k )) 0. Now let k we obtain c + H(x, p + Dϕ(x)) 0 H(x, p + Dϕ(x)) c. Thus, w is a viscosity supersolution of (6). Thus, w is a viscosity solution of (6). It s clearly periodic. The uniqueness of c: Assume that there exists (v 1, c 1 ) and (v, c ) where c 1 < c H(y, p + Dv 1 (y)) = c 1 < c = H(y, p + Dv (y)) in n. Since v 1 and v are periodic, these are bounded, and thus we can take ɛ > 0 small enough so that ɛv 1 (y) + H(y, p + Dv 1 (y)) c 1 + c ɛv (y) + H(y, p + Dv (y)). This means that v 1 and v are sub- and super-solutions, respectively, to the problem ɛv(y) + H(y, p + Dv(y)) = c 1 + c. Therefore, by comparison principle, v 1 v. However, we have that both v 1 and v are bounded; and if v 1 is a subsolution then so is v 1 + C. So, for large enough C we have v 1 + C > v, which is a contradiction. Remark 5. One should note that although c is unique, the solution to cell problem is not unique in general. One of the homework is an example this claim. Exercise. Let n = 1, H(x, p) = p W (x) where W is defined by its graph

16 HOMOGENIZATION OF HAMILTON-JACOBI EQUATIONS NOTES AND REMARKS 16 y 1 W x That is W : is periodic. Consider the ergodic cell problem (E 0 ) (E 0 ) H(y, 0 + Dv(y)) = Dv(y) W (y) = c in. (a) Find the unique constant c such that (E 0 ) has solution. (b) Find an infinite number of solutions for (E 0 ) for all y n.

17 HOMOGENIZATION OF HAMILTON-JACOBI EQUATIONS NOTES AND REMARKS 17 LECTURE 3 In this lecture, our task is to prove the following homogenization theorem, answering the question in lecture 1. Theorem (Lions-Pappanicolau-Varadhan, Evans). Assume u 0 is bounded, uniformly continuous on n and for any ɛ > 0, u ɛ solves the 1 st P.D.E u ɛ t (C ɛ ) (x, t) + H x ɛ, Duɛ (x, t) = 0 in n (0, ) u ɛ (x, 0) = u 0 (x) on n. Then, as ɛ 0, u ɛ converges locally uniformly 4 to u in n [0, ), and u solves (C) ut (x, t) + H (Du(x, t)) = 0 in n (0, ) u(x, 0) = u 0 (x) on n. The idea of the proof of this theorem comes from Evans s perturbed test function method. The following is a heuristic proof. Lemma 1. There is a priori estimate u ɛ t + Duɛ C. Proof. Notes from last year summer school. Proof of theorem. Using above priori estimate, we deduce by Arzela-Ascoli theorem there exists a sequence { } 0 such that u u locally uniformly. We are left to show u solves (C). Sub-test. One might be tempted to take the following approach. Failed attempt. Let φ C and (x 0, t 0 ) n (0, ) is a strict maximum of u φ. We want to show that φ t (x 0, t 0 ) + H(Dφ(x 0, t 0 )) 0. By exercise 1 we have u φ has a max at (x j, t j ) and (x j, t j ) (x 0, t 0 ) as j. Since u is a viscosity solution of (C ), we obtain x j φ t (x j, t j ) + H, Dφ(x j, t j ) 0. As j, we have φ t (x j, t j ) φ(x 0, t 0 ). However, the catch is that we re not sure about the term 4 That means the convergence is uniformly on any compact subset of n [0, )

18 HOMOGENIZATION OF HAMILTON-JACOBI EQUATIONS NOTES AND REMARKS 18 converges to H x j, Dφ(x j, t j ) H(Dφ(x 0, t 0 )) in any sense. The following method, called the perturbed test function method was introduced by Evans in deal with this problem. Heuristic proof - Assume everything is smooth. Let p 0 = Dφ(x 0, t 0 ) n. Take v 0 to be a (Lipschitz, periodic) solution of (7) H(y, p 0 + Dv(y)) = H(p 0 ) = H(Dφ(x 0, t 0 )) y n. A very important observation is that v 0 is bounded by its periodicity and Dv 0 is bounded the coercivity assumption of H. From this, we also have the function x (x, t) u ɛ (x, t) ɛv 0 ɛ converges uniformly to u as ɛ 0. Thus by exercise 1 we obtain there exists a sequence { } 0 such that xɛ φ(x, t) + v 0 j (8) u (x, t) has a max at (x j, t j ) near (x 0, t 0 ) and (x j, t j ) (x 0, t 0 ) as j. The function u is called the perturbed test function. So, by the definition of viscosity subsolutons, we have 5 x j x j φ t (x j, t j ) + H, Dφ(x j, t j ) + Dv 0 0. Also observe that φ t (x j, t j ) φ t (x 0, t 0 ), Dφ(x j, t j ) Dφ(x 0, t 0 ). Furthermore, 6 x H j x j, Dφ(x j, t j ) + Dv 0 H x j x j, Dφ(x 0, t 0 ) + Dv 0 C Dφ(x j, t j ) Dφ(x 0, t 0 ) 0 5 Here is the point where the proof goes wrong as v 0 may not be differentiable at x j 6 Also, (I think we need to) assume that H is Lipschitz in the p-coordinate to apply here.

19 HOMOGENIZATION OF HAMILTON-JACOBI EQUATIONS NOTES AND REMARKS 19 as j. So, combine the facts we obtain x j x j x j x j φ t (x j, t j ) + H, Dφ(x j, t j ) + Dv 0 = φ t (x j, t j ) + H, Dφ(x 0, t 0 ) + Dv 0 x j x j x j x j + H, Dφ(x j, t j ) + Dv 0 H, Dφ(x 0, t 0 ) + Dv 0. From (7) we have x j x j H, Dφ(x 0, t 0 ) + Dv 0 = H(Dφ(x 0, t 0 )) = H(p 0 ). Thus let j 0 in the former formula we thus obtain φ t (x 0, t 0 ) + H(p 0 ) 0. So, u is a viscosity sub-solution of (C). Based on the idea above plus another technique called doubling of variables, we have a real proof. Rigorous proof. Recall u u locally uniformly on n (0, ). We perform the subsolution test, the supersolution case is similar and left as an exercise. Suppose u φ has a strict max at (x 0, t 0 ). We want to show that φ t (x 0, t 0 ) + H (Dφ(x 0, t 0 )) 0. Denote p 0 = Dφ(x 0, t 0 ), consider an open neiborhood Ω (0, T) n (0, ) with compact closure which contains (x 0, t 0 ) such that u(x, t) φ(x, t) < u (x 0, t 0 ) φ (x 0, t 0 ) (x, t) Ω [0, T]. We have u u uniformly on Ω. Let v 0 be the periodic Lipschitz viscosity solution of the cell problem H(y, p 0 + Dv 0 (y)) = H(p 0 ). Then v 0 is bounded by its periodicity, now let η > 0 small consider Φ ɛj,η : Ω n [0, T] x (x, y, t) u ɛ y j (x, t) φ(x, t) v 0 (y). η Since v 0 is bounded, the function Φ η is continuous and bounded above.

20 HOMOGENIZATION OF HAMILTON-JACOBI EQUATIONS NOTES AND REMARKS 0 Step 0. There exists x jη, y jη, t jη Ω n [0, T] such that Φ ɛj,η attains its maximum there. Indeed, let α ɛj,η = sup Ω n [0,T] Φ ɛj,η (x, y, t). Then there exists some sequence (x m, y m, t m ) Ω n [0, T] such that Φ,η (x m, y m, t m ) α,η. Since Ω [0, T] is compact, we can assume there exists a sub-sequence {m k } such that xmk, t mk (x, t ) Ω [0, T] as k. We claim that {y nk } is bounded. Assume the contrary, then we can assume further that (passing a sub-sequence) that ymk > k for all k. We have Φ,η x mk, y mk, t x m k ɛ mk = u,η j y mk x mk, t mk φ xmk, t mk ɛj v 0 ymk η ymk u x mk ɛ j,η x mk, t mk φ xmk, t mk ɛj v 0 ymk η u k x m k ɛ j,η x mk, t mk φ xmk, t mk ɛj v 0 ymk. η Let k, we obtain α,η. It s contradiction, so {y mk } must be bounded. Thus, we can also assume that (passing a sub-sequence) y mk y for some y n. In conclusion, Φ,η attains its maximum at (x, y, t ) Ω n [0, T]. We reindex this point into x jη, y jη, t jη Ω n [0, T]. Note, also, that y jη x jη because of the penality effect as η gets small. Step 1. Since Ω [0, T] is compact, by passing to sub-sequence, we can assume that there exists some point x j, t j Ω [0, T] such that x jη, t jη x j, t j as η 0. We claim that as η 0, then (9) x jη, y jη, t jη x j, y j, t j and y j = x j.

21 HOMOGENIZATION OF HAMILTON-JACOBI EQUATIONS NOTES AND REMARKS 1 To see that, we observe that x jη, y jη, t jη Φ η Φ η x jη, y, t jη y n. This is equivalent to v 0 (y) + x jη y η x jη v 0 y jη + y jη η. Thus if we choose y = x jη, then it becomes yjη v 0 v 0 y jη x jη y jη Since v 0 is bounded, the LHS of the above inequality is bounded. Thus, at the limit as η 0, we must have x j = y j. 7 Thus (9) is true. Step. Also, for any (x, t) Ω [0, T] we have Φ η x jη, y jη, t jη Φ η x, x, t = u xɛ (x, t) φ(x, t) v 0. j Since v 0 is bounded, let η 0 and using the fact that x jη, y jη, t jη x j, x j, t j, we obtain x j x j, t j φ x j, t j v 0 u xɛ (x, t) φ(x, t) v 0 j u for all (x, t) Ω [0, T]. Thus the function ψ (x, t) = u (x, t) v 0 x satisfies ψ φ has a maximum over Ω [0, T] at x j, t j. Also, since v0 is bounded, ψ u uniformly on Ω [0, T]. Thus by exercise 1 we must have (maybe some subsequence) x j, t j (x0, t 0 ) as j since u φ has strict maximum over Ω [0, T] at (x 0, t 0 ). η. 7 Note that Φ x 0, x 0, t 0 = u (x 0, t 0 ) φ(x 0, t 0 ) v 0 (y 0 ) c for some cosntant c, but I honestly don t see the usefulness of this fact.

22 HOMOGENIZATION OF HAMILTON-JACOBI EQUATIONS NOTES AND REMARKS Step 3. Fix y = y jη then the function (x, t) Φ η x, y jη, t has a max at x jη, t jη. So, (x, t) u (x, t) φ(x, t) Observe that the mapping x y jη η (x, t) φ(x, t) + has a max at (x jη, t jη ). x y jη is smooth, thus by definition of viscosity solution for the equation u t +H we have x j η y jη (10) φ t x jη, t jη + H x j η, Dφ(x jη, t jη ) + η η 0. x, Du = 0, Step 4. Observe also that as η 0, the following term is bounded, that is there exists C > 0 such that x j η y jη (11) < C. η Assume the converse, this term is tend to + as η 0, then from (10) and by the coercivity of H we obtain a contradiction since (x jη, y jη, t jη ) x j, x j, t j. Thus (11) is true, therefore we can assume (maybe some subsequence) that there exists some q n such that (1) lim η Using (1) in (10), let η 0 we obtain x j η y jη η = q. (13) φ t (x j, t j ) + H y j, Dφ x j, t j + q 0.

23 HOMOGENIZATION OF HAMILTON-JACOBI EQUATIONS NOTES AND REMARKS 3 Step 5. Now, fix (x jη, t jη ), we have the map So, Thus y Φ(x jη, y, t jη ) has a max at y jη. x y y v 0 (y) η 1 x j η y v 0 (y) η y has a max at y jη. has a min at y jη. Therefore, by mean of viscosity solution of H(y, p 0 + Dv 0 (y)) = H(p 0 ) we have 1 x j η y H y jη, p 0 + D H(p 0 ), η y=y j η which is the same as H y jη, p 0 + x j η y jη η H(p 0 ). Pass η 0 and using (1), we obtain from the continuity of H that (14) H y j, p 0 + q H(p 0 ). Using (14) we obtain (15) φ t x j, t j + H(p0 ) φ t x j, t j + H y j, p 0 + q. Observe that from (13) and the continuity of H we have 8 φ t (x j, t j ) + H(y j, Dφ(x j, t j ) + q) 0 j, H y j, p 0 + q H(y j, Dφ(x j, t j ) + q) 0 as j. Using this fact in (15) and letting j, we obtain φ t (x 0, t 0 ) + H(p 0 ) 0. Therefore, u is a viscosity sub-solution of (C). Exercise 3. Give a formal proof for case super-solution. 8 Note that since y j = x j, so that as j and x j x 0, we have y j. And thus I think we need H is Lipschitz in p-coordinate to apply here.

24 HOMOGENIZATION OF HAMILTON-JACOBI EQUATIONS NOTES AND REMARKS 4 Remark 6. The deep part of this proof is that it passes the indifferentiability of v 0 to a test smooth function φ. This part is achieved by introducing the variable y.

25 HOMOGENIZATION OF HAMILTON-JACOBI EQUATIONS NOTES AND REMARKS 5 LECTURE 4 In this lecture, we study the properties of the effective Hamiltonian, H. Recall that there exists a Lipschitz viscosity solution v ɛ to the following equation ɛv ɛ (y) + H(y, p + Dv ɛ (y)) = 0 in n. We also know that H(p) can be defined as the limit of v (0) as j in view of theorem 1. We want to understand H qualitatively, quantitatively and numerically. Proposition. Consider the following Cauchy problem wt (y, t) + H(y, p + Dw(y)) = 0 in n (0, ) w(y, 0) = 0 on n. Then Proof. We try to bound the solution, Let v be the solution to the cell problem Denote We have w(y, t) lim = H(p). t t subsolution w( y, t) supersolution H(y, p + Dv(y) = H(p). φ + (y, t) = v(y) + c H(p)t and φ (y, t) = v(y) c H(p)t. 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)) = 0. For t = 0, φ + (y, 0) = v(y) + c 0 and φ (y, 0) 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 w(y, t) H(p) t t Let t, we obtain the final result. Remark 7. Finding H numerically is extremely hard. v(y) + c t H(p).

26 HOMOGENIZATION OF HAMILTON-JACOBI EQUATIONS NOTES AND REMARKS 6 We also have some further more properties of the effective Hamiltonian Exercise 4. Suppose H Lip( n n ) satisfies (H1)-(H), 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, 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 3 (Representation formula of H(p)). H(p) = inf{c : v C( n ) : H(y, p + Dv) c in n in viscous sense}. Proof. There exists a viscousity solution v Lip( n ) such that H(y, p + Dv(y)) = H(p) 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 ɛ > 0 small such that ɛw + H(y, p + Dw) c + H(p) ɛv + H(y, p + Dv). Therefore, w v in n, a contradiction since we can always add a constant to make w > v. In case p H(., p) is convex, we also have a further presentation of H. Proposition 4 (inf sup formula). Assume p H(y, p) is convex. Then H(p) = inf sup H(y, p + Dφ(y)). φ C 1 ( n ) 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 1 ( n ) y n.

27 HOMOGENIZATION OF HAMILTON-JACOBI EQUATIONS NOTES AND REMARKS 7 Take φ C 1 ( n ), let C = max H(y, p + Dφ(y)). y n 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 For any ɛ > 0 let φ C c ( n ) φ 0 and φ ɛ (x) = 1 ɛ n φ x ɛ = n φ(x) d x = 1. n φ ɛ (x) d x = 1. Recall that supp(φ ɛ )(x.) B(x, Cɛ). Now take v ɛ (x) = v ɛ v (x) = φ ɛ (x y)v(y) d y. n Then we have (in viscous sense) H(p) = H(y, p + Dv(y)) = = H(y, p + Dv(y))φ ɛ (x y) d y n B(x,Cɛ) H(x, p + Dv(y))φ ɛ (x y) d y C ɛ. By Jensen s inequality for convex Hamiltonian H in p, we obtain H(p) H x, B(x,Cɛ) (p + Dv(y))φ ɛ (x y) d y C ɛ = H(x, p + Dv ɛ (y)) C ɛ H(y, p + Dv ɛ (y)) C ɛ. Thus there exists some constant C > 0 such that H(p) + C ɛ H(y, p + Dv ɛ (y)). Since v ɛ is smooth and max y n H(y, p + Dv ɛ (y)) B, we obtain H(p) + C ɛ max y n H(y, p + Dvɛ (y)) inf B.

28 HOMOGENIZATION OF HAMILTON-JACOBI EQUATIONS NOTES AND REMARKS 8 Pass ɛ 0, we get H(p) inf B = RHS. Exercise 5. If p H(x, p) is convex then p H(p) is also convex.

29 HOMOGENIZATION OF HAMILTON-JACOBI EQUATIONS NOTES AND REMARKS 9 APPENDIX. SOLUTIONS TO PROBLEMS Notes. Proof 1 will be of Son Tu and proof will be of Son Van. The proof of exercise 3 is completely similar to case sub-solution in the lecture and so it will be omitted. Exercise 1. Let φ C 1 ( n ) such that (u φ)(x 0 ) = 0 and u φ has a strict max at x 0. Assume u ɛ converges to u locally uniformly on n. Prove that for ɛ > 0 small enough, u ɛ φ has a max at x ɛ near by x 0 and there is a subsequence 0 such that x ɛj x 0. Proof 1. In this exercise, we will understand the notion of maximum (or strict maximum) as local maximum (or strict maximum). Assume u(x 0 ) = φ(x 0 ) and u φ has a strict (local) maximum at x 0, so there exists an open neighborhood of x 0 with compact closure Ω such that u(y) φ(y) < u(x 0 ) φ(x 0 ) = 0 y Ω\{x 0 }. Clearly u is continuous on Ω since u ɛ u uniformly on Ω. Let r > 0 small such that B (x 0, r) Ω. Clearly u(y) φ(y) < 0 y S(x 0, r) = {y n : y x 0 = r}. Claim. There exists δ r > 0 such that for every ɛ < δ r, u ɛ φ only attains its maximum over Ω at some points x ɛ B (x, r). Proof of claim. Assume the converse, then for δ arbitrarily small, there exists ɛ < δ such that u ɛ φ attains its maximum over Ω outside B (x, r). Take δ = 1 n, we can construct a sequence ɛ n 0 such that u ɛ n φ attains its maximum over Ω at x ɛn Ω\B (x 0, r), i.e x ɛn x 0 r for all n and (16) u ɛ n xɛn φ xɛn u ɛ n (x) φ (x) x Ω. Since Ω is compact, there exists some point x Ω such that x ɛn x. Clearly we also have x Ω\B (x, r). Observe that since u ɛ n u uniformly on Ω as n, we have u ɛ n xɛn u(x ) sup u ɛ n (x) u(x) + u xɛn u(x ) 0 as n. x Ω So that if we let n in (16) we obtain u(x ) φ(x ) u(x) φ(x) x Ω.

30 HOMOGENIZATION OF HAMILTON-JACOBI EQUATIONS NOTES AND REMARKS 30 Therefore u φ attain a maximum over Ω at x x 0. It s a contradiction since u φ has strict max over Ω at x 0. Now return to our exercise, denote α r = max{u(y) φ(y) : y S(x 0, r)} < 0, the maximum is archived since u φ is continuous on compact set S(x 0, r). Now since u ɛ u uniformly on Ω, there exists ζ r > 0 such that for any ɛ < ζ r, then u ɛ (y) u(y) < α r Now for any y S(x 0, r) we have y Ω. (17) u ɛ (y) φ(y) = u ɛ (y) u(y) + u(y) φ(y) < α r. Choose λ r = min{ζ r, δ r }, then for any ɛ < λ r, u ɛ φ is continuous on Ω, thus it obtains a maximum over Ω at some point x ɛ B (x 0, r), denote its maximum by α = u ɛ (x ɛ ) φ(x ɛ ), then clearly (18) α = max{u ɛ (y) φ(y)} u ɛ (x 0 ) φ(x 0 ) = u ɛ (x 0 ) u(x 0 ) > α r Ω. Combine (17) and (18) we obtain x ɛ must lie in B(x 0, r). Therefore for any ɛ > 0 small enough, u ɛ φ only has maximum over Ω at some points x ɛ B(x 0, r). The rest is pretty simple, let r = 1 n, then for each n, choose ɛ n = min λ 1, 1 = ɛ n n 0 as n n and u ɛ n φ has a max at xɛn inside B x 0, 1 n, clearly xɛn x 0 and furthermore u ɛ n (x ɛn ) u(x 0 ) u ɛ n (x ɛn ) u(x ɛn ) + u(xɛn ) u(x 0 ) 0 as n. Proof. Observe that this theorem is wrong if u ɛ 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 u ɛ and u are continuous. The problem now is equivalent to the following problem: Suppose f is a continuous function with global max at x 0 and that f ɛ are continuous function that converges uniformly to f as ɛ 0. Then for small enough ɛ, f ɛ has a max x ɛ near x 0. Pick r 1 > r > 0, we have that f {x:r x x 0 r 1 } has a max at x. Furthermore, f (x 0 ) > f (x ). Define h := f (x 0 ) f (x ). For small enough ɛ, we have f ɛ (x) f (x) < h 3 for ɛ < ɛ. Also, for small enough r (we can assume r < r 1 ), we have f (x 0 ) f (x) < h 3

31 HOMOGENIZATION OF HAMILTON-JACOBI EQUATIONS NOTES AND REMARKS 31 for x B(x 0, r). Therefore, pick x B(x 0, r) and ɛ < ɛ, we have f (x 0 ) f ɛ (x) < h 3. Pick y {x : r x x 0 r 1 }. We have f ɛ (x) > f (x ) + h 3 > f (y) + h 3 > f ɛ (y). This means that, max B(x0,r 1 δ) f ɛ (t) f ɛ (x) > f ɛ (y) for arbitrary small δ > 0. This means that max B(x0,r ) f ɛ (t) must be at t 0 B(x 0, r 1 ) B(x 0, r 1+r ). Therefore, f ɛ (t 0 ) f ɛ (t) for all t B(x 0, r 1+r ). Thus, t 0 is our desired x ɛ. (Not the cleanest proof but it works...) Exercise. Let n = 1, H(x, p) = p W (x) where W is defined by its graph y 1 W x i.e, W : is periodic. Consider the ergodic cell problem (E 0 ) (E 0 ) H(y, 0 + Dv(y)) = Dv(y) W (y) = c in. (a) Find the unique constant c such that (E 0 ) has solution. (b) Find infinitely solutions for (E 0 ) for all y n. Proof 1. (a) Assume v is continuous a viscosity solution of (E 0 ), since is compact we have v is bounded and archived its minimum, called m = v(x m ) = min{v(x) : x } v(x) x Super-test. At x m, we can see that q = 0 belong to Dv (x m ). Thus 0 = q W (x m ) + c min W (x) + c = c = 0 c. x Sub-test. Setting + = {x : D + v(x) }, let p D + v(x) then 0 = p W (x) + c x +. But + is dense in, thus there exists a sequence {x n } + such that 9 x n 1 4, combine with the fact that W is continuous we obtain c W (x n ) n = c W (1/4) = 0 = c 0. 9 We write 1/4 mean the element 1/4 in

32 HOMOGENIZATION OF HAMILTON-JACOBI EQUATIONS NOTES AND REMARKS 3 Thus c = 0 is the only possible value such that (E 0 ) has a viscosity solution. (b) For each c 1, 4 3, consider the following function u c (x) = x x χ [ 0, 1] + x + 3 x 1 χ [ 1,c ] [ 3 c,1 ] + x 3 x c + 3c 1 χ [ c, 3 c ]. For example as c = 0.6 and c = 0.7 we have FIGURE 1. The function u c as c = 0.6 and c = 0.7 Now we will show that u c is a viscosity solution of Du c (x) = W (x) for any c 1, 4 3. To do this, we only need to consider points in which uc is not smooth, that is x = c and x = 3 c. Also, we only need to consider D+ u c (x) since the set D u c (x) is always empty at non-smooth points of u c. To begin with, we will give explicit formulas for D + u c (c) and D + u 3 c 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) = 3 x c p = 3 c p. Consider x > c, then it becomes (x c) x + c 3 p(x c) = x + c 3 p = c 3 p. Thus in this case we have p D + u c (c) p 3 c.

33 HOMOGENIZATION OF HAMILTON-JACOBI EQUATIONS NOTES AND REMARKS 33 Now we need to check that p W (c) for all p D + u c (c), we have p D + u c (c) = p 3 c = 3 4c = W (c). Case x = 3 c. Since u c and W are symmetric through the line x = 3 4, this case is completely similar to case x = c. Proof. Consider the following equation (HJ) ɛu ɛ (y) + H(y, Du ɛ ) = 0. Plug the formula of H into (HJ) we have ɛu ɛ (y) + Du ɛ (y) W (y) = 0. Observe that for each ɛ > 0, u ɛ = 0 is a subsolution and u ɛ = 1 ɛ is a supersolution. To see this, for u ɛ = 0, let φ C s.t. (φ u ɛ )(y 0 ) = 0 is a max. We have So Dφ(y 0 ) = Du ɛ (y 0 ) = 0. ɛφ(y 0 ) + H(y 0, Dφ(y 0 )) = 0 W (y 0 ) 0. Similarly, for u ɛ = 1 ɛ, let φ C s.t. (φ u ɛ )(y 0 ) = 0 is a min. We have ɛφ(y 0 ) + H(y 0, Dφ(y 0 )) = 1 W (y 0 ) 0. Therefore, if v ɛ is a viscosity solution to (HJ), we have Thus, we have 1 ɛv ɛ 0. (19) 1 W (y) Dv ɛ (y). From here we conclude that v ɛ is equi-lipschitz with modulus 1. This is not enough to conclude anything about the convergence of ɛv ɛ except of the result obtained in class. Luckily, we have more observe that Dvɛ W = 0, Dvɛ W = 0. 4 So, 1 3 Dv ɛ = Dv ɛ =

34 HOMOGENIZATION OF HAMILTON-JACOBI EQUATIONS NOTES AND REMARKS 34 That means, for all ɛ > 0, by plugging the above to our equation, ɛv ɛ 1 4 = 0, which means v ɛ 1 4 = 0. Similarly, v ɛ 3 4 = 0 for all ɛ > 0. Combine with (19), we deduce that for all ɛ > 0, v ɛ (y) = y 1 4 v ɛ (t)d t y 1 4 v ɛ (t) y d t 1 4 1d t = y So, ɛv ɛ (y) 0 as ɛ 0 and, therefore, c = 0. The solutions to the problem are obtained similarly to proof 1. Exercise 4. Let H Lip( n n ) satisfying (H1) lim H(y, p) = + uniformly in y, p (H) H(y + k, p) = H(y, p) for all k n. Then H(p) : n is a Lipschitz and coercive function. Proof 1. For u C( 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 F(x, u(x), Du(x)) C in the viscous sense F(x, u(x), p) C By assumption, there exists a constant C > 0 such that p D + u(x) u is sub-solution of F(x, u(x), Du(x)) = C, H(x, p) H(x, q) C p q x, p, q n. p D u(x) u is super-solution of F(x, u(x), Du(x)) = C. We will prove the Lipschitz property of H by using the representation (0) H(q) = inf {c : v C( n ) : H(x, q + Dv(x)) c in n in viscous sense}. Step 1. First of all, we prove that (1) H(p) H(q) C p q. Let u Lip( n ) is a periodic viscosity solution of the problem (E q ) H(x, q + Du(x)) = H(q). By definition ζ D + u(x) = H(x, q + ζ) H(q).

35 HOMOGENIZATION OF HAMILTON-JACOBI EQUATIONS NOTES AND REMARKS 35 We will show that () H(x, p + Du(x)) C p q + H(q) 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 () is true, from the representation of H in (0), we obtain (1) 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 (3) H(q) H(p) C p q. From (1) and (3) we obtain the Lipschitz s property of H, H(p) H(q) C p q. Step 3. We now prove the coercivity of H. First restate the coercivity of H as (4) lim min H(x, p) = +. p x n We start with a similar result as representation formula of H in (0), that is (5) H(q) = sup {c : v C( n ) : H(x, q + Dv(x)) c in n in viscous sense}. Indeed, define α = sup {c : v C( n ) : H(x, q + Dv(x)) c in n in viscous sense}. Let u C( n ) 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 (5), 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 ɛ > 0 such that ɛu(x) + H(x, p + Du(x)) < c + H(p) < ɛw(x) + H(x, p + Dw(x)).

36 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 1 ( n ), and define c = min H(x, p + Dϕ(x)) = min H(x, p). x n x n Then, c is admissible in the formula of the representation (5), so that H(p) c, i.e (6) H(p) min H(x, p). x n Now combine (6) with (4) we obtain lim H(p) p lim p min H(x, p) = +. x n 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 u ɛ and v ɛ be solutions to the following problems (7) ɛu ɛ + H(x, Du ɛ + p) = 0 and (8) ɛv ɛ + H(x, Dv ɛ + q) = 0. We have, by Lipschitzivity, H(x, Du ɛ + p) H(x, Du ɛ + q) C p q. Thus, which implies ɛu ɛ H(x, Du ɛ + q) C p q, (9) ɛu ɛ + C p q + H(x, Du ɛ + q) 0. Thus, u ɛ + C p q ɛ meaning Similarly, we obtain is a supersolution to (9) and u ɛ C p q + v ɛ, ɛ ɛv ɛ ɛu ɛ C p q. C p q ɛv ɛ ɛu ɛ. Thus, ɛu ɛ ɛv ɛ C p q. Let { } be a sequence such that u H(p) and v H(q), we then have H(p) H(q) C p q.

37 HOMOGENIZATION OF HAMILTON-JACOBI EQUATIONS NOTES AND REMARKS 37 Coercivity part: Let M > 0. 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 0. So consider the test function φ(x) = u(x 0 ). We have that, by definition of viscosity solution, M < H(x, Dφ(x 0 ) + p) = H(x, p) H(p). Exercise 5. Let H Lip( n n ) satisfying (H1) lim H(y, p) = + uniformly in y, p (H) 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 1. 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 n in viscous sense} = inf sup H(x, q + Dϕ(x)) : ϕ C 1 ( n ). x n Let u Lip( n ) is a viscosity solution of the problem (E p ) H(x, p + Du(x)) = H(p). Let ϕ C 1 ( n ) arbitrary and let For λ (0, 1), consider the function We will show that (in the viscous sense) C = max H(x, q + Dϕ(x)). x n w = λu + (1 λ)ϕ. (30) H (x, λp + (1 λ)q + Dw) λh(p) + (1 λ)c. To do 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.

38 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 1 (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 w(y) w(x) ζ, y x = y x y x ϕ(y) ϕ(x) Dϕ(x), y x (1 λ). y x Taking lim sup of both sides, we have α D + (λu)(x), therefore α λ D+ u(x), then α δ = λ + (1 λ)dϕ(x) λd λ + u(x) + (1 λ)dϕ(x). 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) max H(x, q + Dϕ(x)) x n and since ϕ C 1 ( n ) is arbitrary, we get H (λp + (1 λ)q) λh(p) (1 λ) Thus, and H is convex. inf ϕ C 1 ( n ) H (λp + (1 λ)q) λh(q) + (1 λ)h(q) max H(x, q + Dϕ(x)) = (1 λ)h(q). x n

39 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 (3) H(x, Du + (1 k)p + kq) = H((1 k)p + kq) and v p, v q be solutions of and respectively. (3) Implies that H(x, Dv + p) = H(p) H(x, Dv + q) = H(q) H((1 k)p + kq) > (1 k)h(x, Dv p + p) + kh(x, Dv q + q) H(x, (1 k)(dv p + p) + k(dv q + q)) = H(x, (1 k)dv p + kdv q + (1 k)p + kq). So, (1 k)v p + kv q is a subsolution to (33). Therefore, u (1 k)v p + kv q. But if u is a solution to (33), u + C is also a solution to (33), which makes u + C (1 k)v p + kv q for all C, which is a contradiction since v p and v q 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.) 7 (199) 1-67, July 199. [] 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): , [3] Pierre-Louis Lions, George Papanicolaou, and Srinivasa RS Varadhan. Homogenization of hamiltonjacobi equations. Unpublished preprint, [4] Hiroyoshi Mitake and Hung V. Tran. Dynamical properties of Hamilton-Jacobi equations via the nonlinear adjoint method: Large time behavior and Discounted approximation. Springer Lecture Notes in Mathematics, vol. 183, 017, vii + 8.