Master Thesis. Nguyen Tien Thinh. Homogenization and Viscosity solution

Transcription

1 Master Thesis Nguyen Tien Thinh Homogenization and Viscosity solution Advisor: Guy Barles Defense: Friday June 21 th, 2013

2 ii Preface Firstly, I am grateful to Prof. Guy Barles for helping me studying my research. Additionally, I would like to thank him for helping me developing my experience in PDEs. The second word of thanks goes to all of people in the labotory of Mathematics and Physics Theory, Francois Rabelais University for helping me during the time I was here. Thirdly, I would like to thank Tran Cong Thanh, who is always besides me even I have any difficulties and sadness. Thank for encouraging me. Finally, I would like to thank my friends. I had a wonderful time with them experiencing French life style and surroundings. I want to thank them for the talks about my research, but even more for the fun we had after working time. Tours, June 21 th, 2013 Nguyen Tien Thinh

3 iii Introduction This paper is devoted to the homogenization of the Hamilton-Jacobi equations u ɛ t + H(x, x ɛ, Duɛ ) = 0 in R n (0, + ). (1) u ɛ t=0 = u 0 (x). (2) where the initial data u 0 (x) is a bounded, uniformly continuous on R n. We consider the behavior of the viscosity solution u ɛ of (1)-(2) as ɛ tends to 0 under assumptions (H1) H is uniformly continuous on R n R n B(0, R) for any R < +. (H2) H(x, y, p) = H(x, y + z, p) for all x, y, p R n. (H3) M > 0, r M > 0 such that H(x, y, p) > M for all p R n, p > r M uniformly for all x, y R n. This problem was introduced in [5] by P.L. Lions, G. Papanicolaou and S.R.S. Varadhan in The key to solve was used is an assymptotic expansions introduced by A.Bensoussan, J.L Lions and G. Papanicolaou (see in [7]) u ɛ (x, t) = u 0 (x, t) + ɛw 1 (x, x ɛ, t) + ɛ2 w 2 (x, x, t) +... (3) ɛ where w i (x, y, t) are Z n -periodic in y, for all i N. Then inserting (3) in (1) and identifying the terms in front of powers of ɛ, we have u 0 t (x, t) + H(x, y, D xu 0 (x, t) + D y w 1 (y, t)) = 0 in R n R n (0, + ). (4) It leads to the cell problem: for x, p R n, is there H(x, p) R such that the equation H(x, y, p + D y v) = H(x, p) in R n (5) has a Z n -periodic viscosity solution v? If the answer is yes, then using an other argument, one concludes that u ɛ tends to u, the unique viscosity solution of the equation u t + H(x, Du) = 0 in Rn (0, + ), (6) u(x, 0) = u 0 (x), (7) where H is called the effective-hamiltonian. This is two-scale method which is divided in two steps: first, find the homogenized and cell equations by means of asymptotic expansions;

4 iv second, prove the convergence. Our aim in this paper is also using the two-scale method with the second step following the perturbed test-function method adapted to Hamilton-Jacobi equations introduced by L.C Evans in 1989 (see in [6]). This framework is organized as follows. In chapter 1, we introduce viscosity solution (see in [1],[2],[3]) and related properties. Chapter 2, we give the answer for the existence of solution of cell problem. Finally, results about homogenization is taken in chapter 3.

5 Contents Preface Introduction ii iii 1 Viscosity Solution Theory Introduction Properties Ergodic Problem Theory Introduction Main results Homogenization Theory Introduction Main results References 33 1

6 Chapter 1 Viscosity Solution Theory 1.1 Introduction. The term viscosity solutions first appear in the work of Michael Crandall and Pierre-Louis Lions in 1983 regarding the Hamilton-Jacobi equation. The name is justified by the fact that the existence of solutions was obtained by the vanishing viscosity method. The definition of solution had actually been given earlier by Lawrence Evans in Subsequently the definition and properties of viscosity solutions for the Hamilton-Jacobi equation were refined in a joint work by Crandall, Evans and Lions in Let us give an example to know where this idea comes from. Consider the C 2 -solution u of the equation u = 0 and φ is an arbitrary C 2 -function, touching u from above at some point x 0. Hence, u φ locally, φ(x 0 ) = u(x 0 ). We conclude that u(x) φ(x) 0 = u(x 0 ) φ(x 0 ) (1.1) in a neighborhood of x 0. It means x 0 is a local maximum point of u φ. Therefore, one has φ (x 0 ) = (u φ) (x 0 ) 0. (1.2) The same idea, when we touch u from below at x 0, one concludes that x 0 is a local minimum of u φ and φ (x 0 ) 0. In the case u doesn t belong to C 2 but these above properties hold. We call that u is a viscosity solution. On the other hand, for another example u 1 = 0. This problem has two different almost 2

7 1.2 Properties. 3 everywhere solutions x + 1 and x 1. Hence, it is not unique. We need a different kind of weak solution such that the uniqueness holding. The notion of viscosity solution can solved this problem. In this framework, we consider the Hamilton-Jacobi Equation where H satisfies u t + H(x, Du) = 0 in Rn (0, + ), (1.3) (H1) H is uniformly continuous on R n B(0, R) for any R < +. (H2) H(x, p) = H(x + z, p) for all x, p R n. (H3) M > 0, r M > 0 such that H(x, p) > M for all p R n, p > r M uniformly for all x R n. Definition 1.1. u C(R n (0, + )) is a subsolution of (1.3) if for any φ C 1 (R n (0, + )) and (x 0, t 0 ) is a local maximum point of u φ, one has φ t (x 0, t 0 ) + H(x 0, Dφ(x 0, t 0 )) 0. (1.4) and u C(R n (0, + )) is a supersolution of (1.3) if for any φ C 1 (R n (0, + )) and (x 0, t 0 ) is a local minimum point of u φ, one has φ t (x 0, t 0 ) + H(x 0, Dφ(x 0, t 0 )) 0. (1.5) If u C(R n (0, + )) is a subsolution and also a supersolution, it is called a viscosity solution. 1.2 Properties. In the real world, the data for solving the PDEs are not always exact such as the initial condition, the boundary condition, etc. These errors maybe come from the measurement. Therefore, the stability result of the viscosity solutions plays an important role. It shows that under needed conditions, the limit of a sequence of subsolution and supersolution is still a subsolution and supersolution. Theorem 1.1. Assume that, for ɛ > 0, u ɛ C(R n (0, + )) is a viscosity solution of the equation u ɛ t + H ɛ(x, Du ɛ ) = 0 in R n (0, + ), (1.6)

8 1.2 Properties. 4 where (H ɛ ) ɛ>0 is a sequence of continuous functions. If u ɛ u in C(R n (0, + )) and if H ɛ H in C(R n (0, + ) R R n ) then u is a viscosity solution of the equation u t + H(x, Du) = 0, in Rn (0, + ). (1.7) An other important basic property is Maximum Principle. It comes from a nature question that if u and v are two viscosity solution of (1.3) and u v on the boundary, one can conclude that u v in general or not, because in PDEs, we just know the data on the boundary or initial time. Hence, this property is needed for the a priori estimation of solutions. In the viscosity solution theory, it is called Comparison result. Theorem 1.2. If (H1) (H3) hold and u, v are respectively bounded, uniformly continuous subsolution and supersolution of (1.3) and if u(x, 0) v(x, 0), x R n, (1.8) then u(x, t) v(x, t), (x, t) R n (0, + ). (1.9) Proof. Firstly, assume that u is a strict subsolution of (1.3) since w(x, t) = u(x, t) ηt for some η > 0 is a subsolution of and w(x, 0) = u(x, 0). w t + H(x, Dw) = η in Rn (0, + ), (1.10) Suppose M = sup R n (0,+ )(u v) > 0 and introduce the test-function Since Ψ ɛ,α,β Ψ ɛ,α,β (x, y, t, s) = u(x, t) v(y, s) x y 2 t s 2 ɛ 2 α 2 β( x 2 + y 2 ). (1.11) as x, y, t, s +, there exists a maximum point ( x, ȳ, t, s) depending on ɛ, α, β > 0 such that Ψ ɛ,α,β (x, y, t, s) Ψ ɛ,α,β ( x, ȳ, t, s), x, y R n, t, s (0, + ). (1.12) Choose y = x, t = s, we have u(x, t) v(x, t) 2β x 2 Ψ ɛ,α,β ( x, ȳ, t, s), (x, t) R n (0, + ). (1.13)

9 1.2 Properties. 5 On the other hand, since u(x, t) v(x, t) 2β x 2 as x, t +, there exists a maximum point (x, t ) depending on β of u(x, t) v(x, t) 2β x 2. And we denote Hence, M = max (u(x, t) v(x, t) R n (0,+ ) 2β x 2 ). (1.14) M Ψ ɛ,α,β ( x, ȳ, t, s) = u( x, t) v(ȳ, s) x ȳ 2 ɛ 2 t s 2 α 2 β( x 2 + ȳ 2 ). (1.15) Moreover, since M M as β 0, one choose β small enough such that M > 0. One concludes that where R = max( u, v ). Therefore, x ȳ 2 ɛ 2 + t s 2 α 2 + β( x 2 + ȳ 2 ) R, (1.16) x ȳ R 1/2 ɛ, (1.17) t s R 1/2 α, (1.18) 2β x 2β 1/2 (β x 2 ) 1/2 2β 1/2 R 1/2, (1.19) 2βȳ 2β 1/2 (β ȳ 2 ) 1/2 2β 1/2 R 1/2. (1.20) Notice that t, s 0. Indeed, if t = 0, s 0, since M > 0 and (1.15), we have M = u( x, 0) v(ȳ, s) u( x, 0) v(ȳ, s) x ȳ 2 ɛ 2 s 2 α 2 β( x 2 + ȳ 2 ) u( x, 0) v( x, 0) + v( x, 0) v(ȳ, s) v( x, 0) v(ȳ, s). (1.21) On the other hand, v BUC(R n [0, + )), there exists δ(m ) > 0 such that v(x, t) v(y, s) < M 2, x, y Rn, t, s [0, + ), x y + t s δ(m ). (1.22) Since (1.17),(1.18), choose ɛ, α < δ(m ), we conclude that 2R1/2 M v( x, 0) v(ȳ, s) < M 2. (1.23)

10 1.2 Properties. 6 Hence, M < 0 is the contradiction and then x, ȳ R n, t, s (0, + ) for ɛ, α, β small enough. On the other hand, since ( x, ȳ, t, s) is the maximum point of Ψ ɛ α,β, we conclude that ( x, t) is the maximum point of u(x, t) [v(ȳ, s) + x ȳ 2 ɛ 2 Besides, u is a subsolution of (1.3), one has + ] t s 2 α 2 + β( x 2 + ȳ 2 ). (1.24) p ɛ,α,β + H( x, q ɛ,α,β + 2β x) η, (1.25) where p ɛ,α,β = 2( t s) 2( x ȳ) α 2 and q ɛ,α,β = ɛ 2. In the same way, (ȳ, s) is the minimum point of x y 2 v(y, s) [u( x, t) ɛ 2 t s 2 ] α 2 β( x 2 + y 2 ). (1.26) But, v is a supersolution of (1.3), one has On the other hand, since (1.18), one gets p ɛ,α,β + H(ȳ, q ɛ,α,β 2βȳ) 0. (1.27) p ɛ,α,β = 2( t s) α 2 2R1/2 α. (1.28) And by the coercivity of H, from (1.25),(1.28) we conclude that there exists C α > 0 depending only on α and H such that q ɛ,α,β + 2β x < C α. Thus, since H is uniformly continuous on R n B(0, C α ), one gets H(x, p) H(y, q) m H ( x y + p q ), x, y R n, p, q B(0, C α ), (1.29) where m H (r) 0 as r 0. Subtracting two sides of (1.25),(1.27), we have η H(ȳ, q ɛ,α,β 2βȳ) H( x, q ɛ,α,β + 2β x) m H ( x ȳ + 2β x + 2βȳ ). (1.30) Let (ɛ, β) 0 and then α 0, we have the contradiction η 0. Corollary 1.1. Under the assumption of Theorem 1.2, if u, v BUC(R n [0, + )) are respectively subsolution and supersolution of (1.3) then sup (u v) sup(u(x, 0) v(x, 0)). (1.31) R n (0,+ ) R n

11 1.2 Properties. 7 The next property of viscosity solutions is the existence via Perron s method. This method had been introduced in 1923 by Oskar Perron in order to find solutions for the Laplace equation and consists in building a solution as the supremum of a suitable family of viscosity subsolutions. Theorem 1.3. (Perron s method) If u and u are respectively sub and supersolution of and the comparison result holds. Let consider the set u t + H(x, Du) = 0 in Rn (0, + ) (1.32) S = {v C(R n (0, + )); u v u, v is a subsolution of (1.32)} then u = sup S v(x, t) is a Lipschitz viscosity solution of (1.32). Theorem 1.4. If (H1)-(H3) hold, then the Hamilton-Jacobi Equation u t + H(x, Du) = 0 in Rn (0, + ), (1.33) u(x, 0) = u 0 (x), (1.34) where u 0 (x) C 1 (R n ) W 1, (R n ), then there exists a unique Lipschitz viscosity solution of (1.33)-(1.34) in W 1, (R n (0, + )). Proof. Set M = u 0 (x) + Ct and N = u 0 (x) Ct, where C = sup q Du0 H(, q) L L one has C + H(x, Du 0 ) 0, x R n, (1.35) C + H(x, Du 0 ) 0, x R n. (1.36) Thus, M, N are respectively supersolution and subsolution of (1.33)-(1.34). We have the first a priori estimate N(x, t) u(x, t) M(x, t), (x, t) R n (0, + ). (1.37) For all h > 0, one has u(x, t + h) u(x, t) max (u(x, t + h) u(x, t)) R n (0,+ ) max(u(x, h) u(x, 0)) Rn max (u 0(x) + Ch u 0 (x)) R n Ch. (1.38)

12 1.2 Properties. 8 Hence, On the other hand, u(x, t + h) u(x, t) h C. (1.39) u(x, t) u(x, t + h) max (u(x, t) u(x, t + h)) R n (0,+ ) max u(x, 0) u(x, h) Rn max (u 0(x) u 0 (x) + Ch) R n Ch. (1.40) Hence, u(x, t) u(x, t + h) h C. (1.41) Let h 0, one gets u C. (1.42) t Therefore, (1.33) becomes H(x, Du) = u t u C. (1.43) t Lemma 1.1. If H satisfy (H3) and u is a continuous, bounded viscosity solution of (1.43), then there exists a constant K > 0 depending only on C and H such that u is K-Lipschitz. Furthermore, u W 1, (R n ) and Du K. Proof. Let x R n. Since u(y) K y x as y + where K > 0, max(u(y) K y x ) exists over R n and we denote this maximum point by y 0. If y 0 = x, one has u(y) K y x max(u(y) K y x ) = u(x), y R n. (1.44) If y 0 x, one has where φ(y) = K y x. u(y) φ(y) u(y 0 ) φ(y 0 ), y R n, (1.45) Since φ is C 1 in a neighborhood of y 0 and u is a subsolution of (1.33), one gets H(y 0, K(y 0 x) ) C. (1.46) y 0 x

13 1.2 Properties. 9 Because of the coercivity of H, there is a constant R > 0 depending only on C and H such that K = K(y 0 x) y 0 x R. (1.47) Choose K 1 = R + 1. Hence, max(u(y) K 1 y x ) exists only at y 0 = x, it means u(y) K 1 y x u(x), y R n. (1.48) By the same way, let y R n, there is K 2 > 0 depending only on C and H such that max(u(x) K 2 x y ) exists only at y, one gets u(x) K 2 y x u(y), x R n. (1.49) Therefore, from (1.48),(1.49), with K = max(k 1, K 2 ), u(y) u(x) K y x, x, y R n. (1.50) Applying the Rademacher s Theorem, one has u is differentiable almost everywhere in R n. Finally, by the subsolution inequality, H(x, Du) C, a.e in R n, (1.51) By the coercivity of H, there is a constant K > 0 depending only on C and H such that Du K. Using these a priori estimates, we introduce the sets W = {v W 1, (R n (0, + )); N v M, v t C, Dv K}, S = {v W; v is a subsolution of (1.33)-(1.34)}. Following Perron s method, we conclude that there is a viscosity solution u of (1.33)-(1.34). Moreover, by the comparison result, this viscosity solution is unique.

14 Chapter 2 Ergodic Problem Theory 2.1 Introduction. A central concern of ergodic theory is the behavior of a dynamical system when it is allowed to run for a long time. For example, for large time t, if the viscosity solution u of the equation u t + H(x, Du) = 0 in Rn (0, + ) (2.1) looks like λt + v(x), then λ and v must satisfy the equation H(x, Dv) = λ in R n. (2.2) The question here is if λ and v like that exit or not. Ergodic theory studies about this kind of PDEs problem. In this part, we consider the cell problem H(x, y, Du p + p) = H(x, p), in R n, (2.3) for any (x, p) R n R n, where H satisfies (H1) H is uniformly continuous on R n R n B(0, R) for any R < +. (H2) H(x, y, p) = H(x, y + z, p) for all x, y, p R n. (H3) M > 0, r M > 0 such that H(x, y, p) > M for all p R n, p > r M uniformly for all x, y R n. 10

15 2.2 Main results Main results. Theorem 2.1. Assume that H satisfies (H1)-(H2)-(H3). For (x, p) R n R n, there exists a unique H(x, p) such that there exists a Z n -periodic viscosity solution u p satisfy (2.3). Proof. Lemma 2.1. For γ > 0 and (x, p) R n R n, the equation H(x, y, Du γ,p (y) + p) + γu γ,p (y) = 0 in R n, (2.4) where the Hamiltonian H satisfy (H1), (H2) and (H3), has a unique, Lipschitz continuous, Z n -periodic viscosity solution u γ,p (y). Proof. Firstly, for γ > 0, (x, p) R n R n, put M γ,p = H(x,, p), one has γ H(x, y, p) γm γ,p 0, y R n, (2.5) H(x, y, p) + γm γ,p 0, y R n. (2.6) Thus, M γ,p, M γ,p are subsolution and supersolution of (2.4). Lemma 2.2. If u and v are bounded, uniformly continuous subsolution and supersolution of (2.4), u v on R n. Proof. Suppose M = sup R n(u v) is positive and introduce the test-function One has Ψ ɛ,α (y, z) = u(y) v(z) y z 2 ɛ 2 α( y 2 + z 2 ). (2.7) y ɛ,α z ɛ,α R 1/2 ɛ, (2.8) 2α y ɛ,α = 2α 1/2 (α y ɛ,α 2 ) 1/2 2α 1/2 R 1/2, (2.9) 2α z ɛ,α = 2α 1/2 (α z ɛ,α 2 ) 1/2 2α 1/2 R 1/2. (2.10) where (y ɛ,α, z ɛ,α ) R n R n is the maximum point of Ψ ɛ,α and R = max( u, v ). By the definition of subsolution and supersolution, H(x, y ɛ,α, q ɛ,α + 2αy ɛ,α + p) + γu(y ɛ,α ) 0, (2.11) H(x, z ɛ,α, q ɛ,α 2αz ɛ,α + p) + γv(z ɛ,α ) 0, (2.12)

16 2.2 Main results. 12 where q ɛ,α = 2(y ɛ,α z ɛ,α ) ɛ 2. On the other hand, from (2.11) and (H3), there exists a constant C > 0 such that q ɛ,α + 2αy ɛ,α + p C. Hence, since H is uniformly continuous on R n R n B(0, C), one deduces H(x, y, p) H(x, z, q) m C H( y z + p q ), x, y, z R n, p, q B(0, C), (2.13) where m C H (t) 0 as t 0. Subtracting two sides of (2.11), (2.12), then apply (2.13), one gets γ(u(y ɛ,α ) v(z ɛ,α )) H(x, z ɛ,α, q ɛ,α 2αz ɛ,α + p) H(x, y ɛ,α, q ɛ,α + 2αy ɛ,α + p) m C H( y ɛ,α z ɛ,α + 2αy ɛ,α + 2αz ɛ,α ). (2.14) Therefore, from (2.8),(2.9),(2.10), letting (ɛ, α) 0, we have the contradiction γm 0. Hence, if u γ,p is a viscosity solution of (2.4), u γ,p is also a subsolution and a supersolution. Thus, we have the a priori estimate M γ,p u γ,p (y) M γ,p, y R n. (2.15) Thus, by the subsolution inequality and (2.15), we have H(x, y, Du + p) γu(y) γm γ,p H(x,, p), a.e in R n. (2.16) Therefore, by Lemma 1.1, one concludes that there is a constant K p > 0 depending only on H(x,, p) and H such that Du K p. Using these a priori estimates, we introduce the sets W = {v W 1, (R n ); M γ,p v M γ,p, Dv K p }, S = {v W; v is a subsolution of (2.4)}. Following Perron s method, we conclude that there is a viscosity solution u γ,p of (2.4). Moreover, by Lemma 2.2, this viscosity solution is unique. Finally, for all z Z n, u γ,p (y + z) is a viscosity solution of H(x, y + z, Du γ,p (y + z) + p) + γu γ,p (y + z) = 0 in R n. (2.17) On the other hand, since H satisfy (H2), one gets H(x, y, Du γ,p (y + z) + p) + γu γ,p (y + z) = H(x, y + z, Du γ,p (y + z) + p) + γu γ,p (y + z) = 0.

17 2.2 Main results. 13 Therefore, u γ,p (y + z) is also a viscosity solution of (2.4), but by the uniqueness, one has u γ,p (y + z) = u γ,p (y). Therefore, u γ,p is Z n -periodic. For (x, p) R n R n, we consider u γ,p, where γ > 0 is small enough, viscosity solution of (2.4). For y R n, we set v γ,p (y) = u γ,p (y) u γ,p (0), then by the result in Theorem 2.1, one has v γ,p (y) = u γ,p (y) u γ,p (0) = u γ,p (y [y]) u γ,p (0) K p y [y], (2.18) where y = (y i ) 1 i n and [y] = ([y i ]) 1 i n Z n, 0 y i [y i ] < 1 for 1 i n. Hence, since y [y] [0, 1] n, we have v γ,p is equibounded. Besides, for all y, z R n, v γ,p (z) v γ,p (y) = u γ,p (z) u γ,p (y) K p z y. (2.19) Thus, v γ,p is equi-lipschitz-continuous. Applying Ascoli s Theorem, we deduce that there exists a subsequence {v γk,p} {v γ,p } which converges to u p C(R n ). On the other hand, γ k u γk,p(0) = γ k u γk,p(0) γ k M γk,p = H(x,, p). (2.20) Thus, there exists a constant H(x, p) such that there is a subsequence γ km u γkm,p(0) H(x, p) as γ km 0. Furthermore, since (2.20), we have H(x, p) H(x,, p). (2.21) Finally, by the definition of v γ,p and the equation (2.4), one gets H(x, y, Dv γkm,p + p) + γ km v γkm,p = γ km u γkm,p(0) a.e in R n. (2.22) Then, let m +, the stability result says H(x, y, Du p + p) = H(x, p) a.e in R n. Finally, we show that (x, p) R n R n, H(x, p) is unique. Let (x, p) R n R n, suppose (u p, H(x, p)) and (v p, H (x, p)) respectively satisfy H(x, y, Du p + p) = H(x, p) in R n (2.23) H(x, z, Dv p + p) = H (x, p) in R n, (2.24)

18 2.2 Main results. 14 where u p, v p are continuous, Z n -periodic viscosity solution of (2.23), (2.24). One can consider the test-function Ψ ɛ (y, z) = u p (y) v p (z) y z 2 ɛ 2, (2.25) and we can assume without loss of generality that y [0, 1] n and z [0, 2] n since the Z n -periodicity. Therefore, Ψ ɛ has a maximum point (y ɛ, z ɛ ) and we have where R p = max( u p, v p ). y ɛ z ɛ R 1/2 p ɛ, (2.26) On the other hand, since u p is subsolution of (2.23) and v p is supersolution of (2.24), we also have H(x, y ɛ, q ɛ + p) H(x, p), (2.27) H(x, z ɛ, q ɛ + p) H (x, p), (2.28) where q ɛ = 2(y ɛ z ɛ ) ɛ 2. Besides of that, since (2.27) and the coercivity of H, there is a constant C p > 0 such that q ɛ + p C p. Hence, H(x, y, q) H(x, z, q) m Cp H ( y z ), q B(0, C p), (y, z) [0, 1] n [0, 2] n, (2.29) where m Cp H (t) 0. Subtracting two sides of (2.27), (2.28), then, applying (2.29), one gets H (x, p) H(x, p) H(x, z ɛ, q ɛ + p) H(x, y ɛ, q ɛ + p), m Cp H ( y ɛ z ɛ ). (2.30) Let ɛ 0, we conclude Exchanging the roles of u p, v p, one has H (x, p) H(x, p). (2.31) H (x, p) H(x, p). (2.32) Therefore, for all (x, p) R n R n, H(x, p) is unique.

19 2.2 Main results. 15 Example 2.1. In this example, let p R, we consider the equation u + p = f(x) + λ(p), in R, (2.33) and compute the unique λ(p) such that (2.33) has a 1-periodic viscosity solution if f is 1-periodic, non negative and min R n f(x) = 0. Firstly, one can assume that u(0) = u(1) = 0 and notice that, at x 0 is the minimum point of f, one has λ(p) = u (x 0 ) + p 0. On the other hand, let θ [0, 1], p, q R n, we have λ(θp + (1 θ)q) = u + θp + (1 θ)q f(x) = θ(u + p) + (1 θ)(u + q) θf(x) (1 θ)f(x) θ u + p + (1 θ) u + q θf(x) (1 θ)f(x) = θλ(p) + (1 θ)λ(q). (2.34) Thus, λ is convex. Secondly, we look for p R n and λ(p) such that there are viscosity solutions. Assume that u (x) + p = f(x) + λ(p), x [0, 1]. (2.35) Thus, there is a C 1 ([0, 1]) viscosity solution u such that u(x) u(0) = px + x Hence, by the assumption u(0) = u(1) = 0, one has p = Besides, since λ(p) 0, it implies that 1 p On the other hand, assume that 0 0 [f(t) + λ(p)]dt. (2.36) f(t)dt + λ(p). (2.37) 1 0 f(t)dt. (2.38) u (x) + p = [f(x) + λ(p)], x [0, 1]. (2.39) Thus, there is a C 1 ([0, 1]) viscosity solution u such that u(x) u(0) = px x 0 [f(t) + λ(p)]dt. (2.40)

20 2.2 Main results. 16 Hence, p = And since λ(p) 0, it implies that 1 0 p f(t)dt λ(p). (2.41) 1 0 f(t)dt. (2.42) Therefore, for all p R n, p 1 0 f(t)dt and λ(p) = p 1 0 f(t)dt, one obtains u(x) = px + is a viscosity solutions of (2.33). x 0 [f(t) + λ(p)]dt, (2.43) And, for all p R n, p 1 0 f(t)dt and λ(p) = p 1 0 f(t)dt, one has u(x) = px is a viscosity solutions of (2.33). x 0 [f(t) + λ(p)]dt, (2.44) On the other hand, since λ(p) 0 and convex for all p R n. We have necessarily λ(p) = 0 for 1 0 f(t)dt < p < 1 0 f(t)dt. (2.45) We show that there is a 1-periodic viscosity solution for the equation (2.33). To do that, starting at x 0 [0, 1] which we choose after. Then, if this viscosity solution exists, it is continuous and 1-periodic. Thus, we look for a solution with one change of sign in u + p after some x [x 0, x 0 + 1]. Hence, we choose a viscosity solution satisfies u (x) = p + f(x) (2.46) on [x 0, x], denoting by u 1. This viscosity solution satisfies also u (x) = p f(x), (2.47) on [ x, x 0 + 1], denoting by u 2. And such that u 1 ( x) = u 2 ( x) for the continuity. One has u 1 (x) = u 2 (x) = x x 0 [ p + f(t)] dt, (2.48) 1+x0 x [p + f(t)] dt. (2.49)

21 2.2 Main results. 17 Remark 2.1. One can choose to find a viscosity solution. u 2 (x) = u 1 (x) = x 1+x0 x x 0 [p + f(t)] dt, (2.50) [ p + f(t)] dt, (2.51) On the other hand, u 1 (x 0 ) u 2 (x 0 ) = u 1 (1 + x 0 ) u 2 (1 x 0 ) = 1+x0 x 0 1+x0 Besides of that, since f is 1-periodic, 1+x0 [ 1+x0 [p + f(t)] dt = p + x 0 [ p + f(t)] dt = p + x 0 f(t)dt = Then apply (2.52), (2.53), (2.54), one has 1 [u 1 (x 0 ) u 2 (x 0 )][u 1 (1 + x 0 ) u 2 (1 + x 0 )] = 0 x 0 1+x0 ] f(t)dt, (2.52) x 0 f(t)dt. (2.53) f(t)dt, (2.54) [ p 2 + ( Hence, there exists x [x 0, 1 + x 0 ] such that u 1 ( x) u 2 ( x) = f(t)dt) 2 ] < 0. (2.55) Therefore, we can define a 1-periodic, continuous function on [x 0, 1 + x 0 ] as below x [ p + f(t)] dt, x [x 0, x], x u(x) = 0 1+x0 [p + f(t)] dt, x [ x, 1 + x 0 ], x (2.56) Finally, we choose x 0 such that for p 0, u is a viscosity solution for the equation (2.33). Indeed, we have lim x x 0 lim x x + 0 u(x) u(x 0 ) x x 0 = p + f(x 0 ), (2.57) u(x) u(x 0 ) x x 0 = p f(x 0 ). (2.58) One takes x 0 [0, 1] such that f(x 0 ) = 0, u is differentiable at x 0. On the other hand, since f is 1-periodic, one concludes that f(x 0 + 1) = f(x 0 ) = 0. Therefore u is also differentiable at x Thus, we obtain that u is differentiable in [x 0, 1+x 0 ]\ x and u +p = f(x), x [x 0, 1+x 0 ]\ x. Hence, u is a 1-periodic viscosity solution of the equation u + p = f(x), in [x 0, 1 + x 0 ] \ x. (2.59)

22 2.2 Main results. 18 On the other hand, we consider if the local maximum point and the local minimum point at x. Let φ C 1 (R), suppose that u φ has a local maximum point at x. Then there exists r > 0 such that u(x) φ(x) u( x) φ( x), x B( x, r). (2.60) Hence, For x < x, one gets u(x) u( x) φ(x) φ( x), x B( x, r). (2.61) φ(x) φ( x) x = x 0 [ p + f(t)]dt x x x x 0 [ p + f(t)]dt [ p + f(t)]dt. (2.62) Thus, φ(x) φ( x) x x 1 x x x x [ p + f(t)]dt. (2.63) Sending x x, one has φ ( x) + p f( x). (2.64) For x > x, one gets Thus, φ(x) φ( x) = 1+x0 x x x [p + f(t)]dt 1+x0 x [p + f(t)]dt [p + f(t)]dt. (2.65) Hence, since p 0, one has φ(x) φ( x) x x 1 x x x x [p + f(t)]dt. (2.66) φ(x) φ( x) x x + p 1 x x 1 x x x x x x f(t)dt + 2p f(t)dt. (2.67) Sending x x, one has φ ( x) + p f( x). (2.68)

23 2.2 Main results. 19 From (2.64), (2.68), we conclude φ ( x) + p f( x). (2.69) Therefore, u is a subsolution of (2.33). On the other hand, Let φ C 1 (R), suppose that u φ has a local minimum point at x. Then there exists r > 0 such that u(x) φ(x) u( x) φ( x), x B( x, r). (2.70) Hence, For x < x, one has Thus, For x > x, one has Thus, u(x) u( x) φ(x) φ( x), x B( x, r). (2.71) φ(x) φ( x) x x 1 x x x x [ p + f(t)]dt. (2.72) φ(x) φ( x) lim f( x) p. (2.73) x x x x φ(x) φ( x) x x From (2.73), (2.75), one concludes that 1 x x x x [p + f(t)]dt. (2.74) φ(x) φ( x) lim [f( x) p]. (2.75) x x + x x φ(x) φ( x) φ(x) φ( x) lim lim. (2.76) x x x x x x + x x It is in contradiction with φ C 1 (R n ). Hence, u φ has no local minimum point in x. Generally, u is a viscosity solution of (2.33). Proposition 2.1. (x, p) H(x, p) is continuous on R n R n. Proof. Let (x, p) R n R n and (x k, p k ) k N R n, (x k, p k ) (x, p) as k +. We can assume that x k x < 1 and p k p < 1. For each k N, there is a continuous, Z n -periodic viscosity solution u pk satisfy H(x k, y, Du pk + p k ) = H(x k, p k ), in R n. (2.77)

24 2.2 Main results. 20 One can suppose u pk (0) = 0 since w pk (y) = u pk (y) u pk (0) is also a Z n -periodic viscosity solution of (2.77). On the other hand, since (2.21), we have H(x k, p k ) H(x k,, p k ). (2.78) Therefore, H(x k, p k ) K x,p, (2.79) where K x,p = max z x <1, q p <1 H(z,, q). Besides, by the coercivity of H, one concludes from (2.77), there is a constant C p such that Du pk C x,p. (2.80) It implies u pk (z) u pk (y) C x,p z y, y, z R n. (2.81) Therefore, u pk is equi-lipschitz continuous. On the other hand, u pk (y) = u pk (y) u pk (0) = u pk (y [y]) u pk (0) C x,p y [y], y R n, (2.82) where y = (y i ) 1 i n and [y] = ([y i ]) 1 i n Z n, 0 y i [y i ] < 1 for 1 i n. Thus, u pk is equi-bounded because y [y] [0, 1] n. Hence, by Ascoli s Theorem, there exists a subsequence (u pm ) of (u pk ) and v C(R n ) such that u pm v as m. On the other hand, since (2.79), one has H(x m, p m ) K x,p. (2.83) Then, there exists a λ and a subsequence (H(x mj, p mj )) of (H(x m, p m )) which we also denote by such that H(x mj, p mj ) λ as j. Therefore, by (2.77), one has H(x mj, y, Du pmj + p mj ) = H(x mj, p mj ) in R n. (2.84) Sending j and using the stability of H, one has H(x, y, Dv + p) = λ a.e in R n. (2.85)

25 2.2 Main results. 21 Finally, since H(x, p) is unique, we have λ = H(x, p). It implies that H(x k, p k ) H(x, p) as k. Remark 2.2. (x, p) H(x, p) is uniformly continuous on R n B(0, R) for any R < +. Proposition 2.2. If H(x, y, p) is convex in p for all x, y R n, H(x, p) is convex in p for all x R n. Proof. Let θ [0, 1], x R n, p and q belong to R n. There exists a Z n -periodic viscosity solution u p and unique H(x, p) satisfy H(x, y, Du p + p) = H(x, p) in R n. (2.86) There exists a Z n -periodic viscosity solution u q and unique H(x, q) satisfy H(x, y, Du q + q) = H(x, q) in R n. (2.87) Firstly, put v = θu p + (1 θ)u q, we show that v is a subsolution of the equation H(x, y, Dv + r) = θh(x, p) + (1 θ)h(x, q) in R n, (2.88) where r = θp + (1 θ)q. Let φ C 1 (R n ) and y 0 is a strict local maximum point of v φ, there exists k > 0 such that v(y) φ(y) < v(y 0 ) φ(y 0 ), y B(x 0, k). (2.89) Introduce on the set B(y 0, k) B(y 0, k), the test function Ψ ɛ (y, z) = θu p (y) + (1 θ)u q (z) φ(y) Then there exists a maximum point (y ɛ, z ɛ ) of Ψ ɛ in B(y 0, k) B(y 0, k). y z 2 ɛ 2. (2.90) One has y ɛ z ɛ R 1/2 p,q ɛ. (2.91) On the other hand, B(y 0, k) B(y 0, k) is a compact set, there exists a subsequence (y ɛm, z ɛm ) of (y ɛ, z ɛ ) and (y 1, z 1 ) B(y 0, k) B(y 0, k) such that (y ɛm, z ɛm ) (y 1, z 1 ). We can note this subsequence as (y ɛ, z ɛ ) and by (2.91), one has y 1 = z 1. Besides of that, Ψ ɛ (y, z) Ψ ɛ (y ɛ, z ɛ ), (y, z) B(y 0, k) B(y 0, k). (2.92)

26 2.2 Main results. 22 Choose y = z, we obtain for all y B(y 0, k), θu p (y) + (1 θ)u q (y) φ(y) θu p (y ɛ ) + (1 θ)u q (z ɛ ) φ(y ɛ ) y ɛ z ɛ 2 Let ɛ 0, (2.93) becomes for all y B(y 0, k). ɛ 2. (2.93) θu p (y) + (1 θ)u q (y) φ(y) θu p (y 1 ) + (1 θ)u q (y 1 ) φ(y 1 ), (2.94) Hence, y 1 is a maximum point of Ψ ɛ on B(y 0, k). But y 0 is a strict maximum point of Ψ ɛ on this set, so y 1 = y 0. One can see that y ɛ is a local maximum point of And z ɛ is a local maximum point of u p (y) (1/θ)[ (1 θ)u q (z ɛ ) + φ(y) + y z ɛ 2 u q (z) 1/(1 θ)[ θu p (y ɛ ) + φ(y ɛ ) + y ɛ z 2 ɛ 2 ]. (2.95) ɛ 2 ]. (2.96) Because u p is a subsolution of (2.86) and u q is a supersolution of (2.87), then H(x, y ɛ, Dφ(y ɛ) θ + q ɛ + p) H(x, p), (2.97) θ q ɛ H(x, z ɛ, + q) H(x, q), (2.98) 1 θ where q ɛ = 2(y ɛ z ɛ ) ɛ 2. On the other hand, by (2.97) and the coercivity of H, there is a constant C x,p > 0 such that Dφ(y ɛ) + q ɛ θ θ + p C x,p. Therefore, since H is uniformly continuous on R n B(0, C x,p ), we have H(x, y, r) H(x, z, q) m Cx,p H ( y z + r q ), x, y, z Rn, r, q B(0, C x,p ), (2.99) Multiplying two sides of (2.97) by θ and (2.98) by (1 θ) then adding them side by side, one has θh(x, p) + (1 θ)h(x, q) θh(x, y ɛ, Dφ(y ɛ) θ θ[h(x, y ɛ, Dφ(y ɛ) θ +θh(x, z ɛ, Dφ(y ɛ) θ + q ɛ θ + p) + (1 θ)h(x, z ɛ, 1 θ + q) q ɛ + q ɛ θ + p) H(x, z ɛ, Dφ(y ɛ) + q ɛ θ θ + p)] + q ɛ θ + p) + (1 θ)h(x, z ɛ, 1 θ + q) θm Cx,p H ( y ɛ z ɛ ) + θh(x, z ɛ, Dφ(y ɛ) + q ɛ θ θ + p) +(1 θ)h(x, z ɛ, q ɛ + q). (2.100) 1 θ q ɛ

27 2.2 Main results. 23 On the other hand, H(x, z ɛ, Dφ(y ɛ ) + r) = H [ x, z ɛ, θ( Dφ(y ɛ) + q ɛ θ θ ] + p) + (1 θ)( q ɛ 1 θ + q). (2.101) Hence, by the convexity of H(x, y, p) in p for all x, y R n, we conclude H(x, z ɛ, Dφ(y ɛ ) + r) θh(x, z ɛ, Dφ(y ɛ) θ Applying (2.102) in (2.100), one gets + q ɛ θ + p) + (1 θ)h(x, z ɛ, + q). (2.102) 1 θ θm Cx,p H ( y ɛ z ɛ ) + H(x, z ɛ, r) θh(x, p) + (1 θ)h(x, q). (2.103) Finally, by (2.91), let ɛ 0, we have Therefore, v is a subsolution of (2.88). H(x, y 0, Dφ(y 0 ) + r) θh(x, p) + (1 θ)h(x, q). (2.104) Secondly, with H(x, r), there exists a Z n -periodic viscosity solution w r of the equation H(x, y, Dw r + r) = H(x, r), in R n. (2.105) Since u p, u q are Z n -periodic, v is also Z n -periodic. Hence, one can consider the behavior of v and w r in [0, 1] n. Introduce the test-function on [0, 1] n [0, 1] n, Ψ ɛ (y, z) = v(y) w r (z) Reproducing the technic as above, we obtain y z 2 ɛ 2. (2.106) H(x, r) = H(x, θp + (1 θ)q) θh(x, p) + (1 θ)h(x, q). (2.107) Therefore, H(x, p) is convex in p for all x R n. Proposition 2.3. H(x, p) is coercive in p uniformly for all x R n. Proof. Let M > 0, there exists r M > 0 such that H(x, y, p) > M, p R n, p > r M, x, y R n, (2.108) since H(x, y, p) + as p + uniformly for all x, y R n. On the other hand, for p R n, p > r M, there exists a Z n -periodic viscosity solution u p and a unique H(x, p) satisfy H(x, y, Du p + p) = H(x, p), in R n. (2.109) q ɛ

28 2.2 Main results. 24 Besides of that, since the Z n -periodic, one can see in [0, 1] n, there is a maximum point y ɛ of u p 0. And thus, H(x, y ɛ, p) H(x, p). (2.110) Finally, combine with (2.108), we conclude that H(x, p) > M. Therefore, H(x, p) is coercive in p uniformly for all x R n.

29 Chapter 3 Homogenization Theory 3.1 Introduction. Homogenization theory is concerned with the derivation of equations for averages of solutions of equations with rapidly varying coefficients. This problem arises in obtaining microscopic or homogenized or effective equations for systems with a fine microscopic structure. In this part, under the assumptions in chapter 2, we consider the behavior as ɛ tends to 0 of the viscosity solution u ɛ of the Hamilton-Jacobi equation u ɛ t + H(x, x ɛ, Duɛ ) = 0 in R n (0, + ). (3.1) u ɛ t=0 = u 0 (x). (3.2) where u 0 (x) BUC(R n ). This problem was introduced in [5] by P.L. Lions, G. Papanicolaou and S.R.S. Varadhan in The key to solve was used is base on two-scale method. 3.2 Main results. Theorem 3.1. Assume that H satisfies (H1)-(H3), for all ɛ > 0, there exist a unique viscosity solution u ɛ of (3.1)-(3.2) and a unique viscosity solution u of the equation u t + H(x, Du) = 0 in Rn (0, + ), (3.3) u(x, 0) = u 0 (x) BUC(R n ), (3.4) where H(x, p) = H(x, y, p + D y v) such that u ɛ u in BUC(R n [0, T ]) as ɛ 0. 25

30 3.2 Main results. 26 Proof. Firstly, since u 0 (x) BUC(R n ), there exists a sequence (u k 0 ) k N C 1 (R n ) W 1, (R n ) such that u k 0 u 0 in BUC(R n ) as k and u k 0 u 0, k N. On the other hand, since Theorem 1.4, for each k N, the equation w t + H(x, x ɛ, Dw) = 0, in Rn (0, + ), (3.5) w(x, 0) = u k 0(x), (3.6) has a unique viscosity solution u ɛ k in W 1, (R n (0, + )). Besides, since Theorem 1.4, there exist C k, R C k > 0 such that uɛ k (x, t) is C k-lipschitz in t and Rk C -Lipschitz in x. Hence, for all (x, t), (y, s) R n [0, T ], one has u ɛ k (x, t) uɛ k (y, s) C k t s + Rk C x y, (3.7) where C k, R C k are independent on ɛ. Thus, u ɛ k is equi-continuous. On the other hand, since (1.37), for all (x, t) R n [0, T ], one has u ɛ k (x, t) u 0 + C k T, (x, t) R n [0, T ]. (3.8) We obtain that u ɛ k is equi-bounded in Rn [0, T ]. Applying Ascoli s Theorem, there exists a subsequence (u ɛ j k ) of (uɛ k ) such that uɛ j k u k in C(R n [0, T ]). Omit the notation of k, j. We consider the cell problem H(x, p) = H(x, y, p + D y v(y)) in R n, (3.9) where (x, p) R n R n is fixed. Theorem 3.2. u is the unique viscosity solution of the Hamilton-Jacobi Equation u t + H(x, Du) = 0 in Rn (0, + ), (3.10) u(x, 0) = u 0 (x). (3.11) Proof. Lemma 3.1. u is a subsolution of (3.10).

31 3.2 Main results. 27 Proof. Let φ C 1 (R n [0, T ]) and (x 0, t 0 ) is a strict local maximum point of u φ. There exists a Z n -periodic, Lipschitz continuous viscosity solution w(y) satisfying H(x 0, y, D x φ(x 0, t 0 ) + D y w(y)) = H(x 0, D x φ(x 0, t 0 )). (3.12) We introduce the test-function Ψ ɛ,β (x, t, y) = u ɛ (x, t) φ(x, t) ɛw(y) 1 β 2 y x ɛ 2. (3.13) Since (x 0, t 0 ) is a strict local maximum point of u φ, there is r > 0 and r > 0 such that u(x, t) φ(x, t) < u(x 0, t 0 ) φ(x 0, t 0 ), (3.14) for all x B(x 0, r) and t B(t 0, r ). On the other hand, the set B(x 0, r) B(t 0, r ) B( x 0 ɛ, r ɛ ) is a compact set and Ψ ɛ,β is continuous. There exists a maximum point (x ɛ,β, t ɛ,β, y ɛ,β ) of Ψ ɛ,β such that u ɛ (x, t) φ(x, t) ɛw(y) 1 β 2 y x ɛ 2 u ɛ (x ɛ,β, t ɛ,β ) φ(x ɛ,β, t ɛ,β ) for all x B(x 0, r), t B(t 0, r ) and y B( x 0 ɛ, r ɛ ). Choose y = x ɛ, then for all (x, t) B(x 0, r) B(t 0, r ), we have ɛw(y ɛ,β ) 1 β 2 y ɛ,β x ɛ,β ɛ 2. (3.15) u ɛ (x, t) φ(x, t) ɛw( x ɛ ) uɛ (x ɛ,β, t ɛ,β ) φ(x ɛ,β, t ɛ,β ) ɛw(y ɛ,β ) 1 β 2 y ɛ,β x ɛ,β ɛ 2. (3.16) Thus, there exists a constant C > 0 such that y ɛ,β x ɛ,β ɛ C1/2 β. (3.17) Besides, one has B(x 0, r) B(t 0, r ) is a compact set, we can suppose that x ɛ,β x 1 and t ɛ,β t 1 as (ɛ, β) 0, where (x 1, t 1 ) B(x 0, r) B(t 0, r ). On the other hand, since w(y) is bounded and u ɛ u in C(R n [0, T ]) as ɛ 0, let (ɛ, β) 0 in (3.16), one gets u(x, t) φ(x, t) u(x 1, t 1 ) φ(x 1, t 1 ), (x, t) B(x 0, r) B(t 0, r ). (3.18) It implies that (x 1, t 1 ) is a maximum point of u φ on B(x 0, r) B(t 0, r ). But, since (x 0, t 0 ) is a strict maximum point of u φ on B(x 0, r) B(t 0, r ), we conclude that (x 1, t 1 ) = (x 0, t 0 ).

32 3.2 Main results. 28 It implies that (x ɛ,β, t ɛ,β ) B(x 0, r) B(t 0, r ). Unless, by the convergence of (x ɛ,β, t ɛ,β ) to (x 0, t 0 ) as (ɛ, β) 0, for all α > 0, there exist δ(0, α) and δ (0, α) such that r = x ɛ,β x 0 < α, ɛ, β < δ(0, α), (3.19) r = t ɛ,β t 0 < α, ɛ, β < δ (0, α). (3.20) Thus, r = r = 0 is the contradiction. Besides, since u ɛ is a subsolution of (3.1)-(3.2), we have φ t (x ɛ,β, t ɛ,β ) + H(x ɛ,β, x ɛ,β ɛ, D xφ(x ɛ,β, t ɛ,β ) p ɛ,β ) 0, (3.21) where p ɛ,β = 2 ɛβ 2 (y ɛ,β x ɛ,β ɛ ). On the other hand, w is a supersolution of (3.12), one gets H(x 0, y ɛ,β, D x φ(x 0, t 0 ) p ɛ,β ) H(x 0, D x φ(x 0, t 0 )) 0. (3.22) Besides, since H is coercive, one concludes from (3.21) that there exists a constant C > 0 depending on φ t and H such that D x φ(x ɛ,β, t ɛ,β ) p ɛ,β < C. (3.23) Hence, since H is uniformly continuous on R n R n B(0, C), we deduce H(x, y, p) H(x, y, p ) m C H( x x + y y + p p ), (3.24) for all (x, y, p), (x, y, p ) R n R n B(0, C) where m C H (t) 0 as t 0. Subtracting two sides of (3.21), (3.22) and applying (3.24), one has φ t (x ɛ,β, t ɛ,β ) + H(x 0, q ) H(x 0, y ɛ,β, q p ɛ,β ) H(x ɛ,β, x ɛ,β ɛ, q p ɛ,β), (3.25) where q = D x φ(x ɛ,β, t ɛ,β ) and q = D x φ(x 0, t 0 ). On the other hand, H(x 0, y ɛ,β, q p ɛ,β ) H(x ɛ,β, x ɛ,β ɛ, q p ɛ,β) m C H( x ɛ,β x 0 + y ɛ,β x ɛ,β ɛ + q q ). And since φ C 1 (R n [0, T ]) and (3.17), let (ɛ, β) 0, we obtain Therefore, u is a subsolution of (3.10). φ t (x 0, t 0 ) + H(x 0, D x φ(x 0, t 0 )) 0. (3.26)

33 3.2 Main results. 29 Lemma 3.2. u is a supersolution of (3.10). Proof. Let φ C 1 (R n [0, T ]) and (x 0, t 0 ) is a strict local minimum point of u φ. There exists a Z n -periodic Lipschitz continuous, viscosity solution w(y) statisfies We introduce the test-function H(x 0, y, D x φ(x 0, t 0 ) + D y w(y)) = H(x 0, D x φ(x 0, t 0 )). (3.27) Ψ ɛ,β (x, t, y) = u ɛ (x, t) φ(x, t) ɛw(y) + 1 β 2 y x ɛ 2. (3.28) Since (x 0, t 0 ) is a strict local minimum point of u φ, there is r > 0 and r > 0 such that for all x B(x 0, r) and t B(t 0, r ). u(x, t) φ(x, t) > u(x 0, t 0 ) φ(x 0, t 0 ), (3.29) On the other hand, the set B(x 0, r) B(t 0, r ) B( x 0 ɛ, r ɛ ) is a compact set and Ψ ɛ,β is continuous. There exists a minimum point (x ɛ,β, t ɛ,β, y ɛ,β ) of Ψ ɛ,β such that u ɛ (x, t) φ(x, t) ɛw(y) + 1 β 2 y x ɛ 2 u ɛ (x ɛ,β, t ɛ,β ) φ(x ɛ,β, t ɛ,β ) for all x B(x 0, r), t B(t 0, r ) and y B( x 0 ɛ, r ɛ ). Choose y = x ɛ, then for all (x, t) B(x 0, r) B(t 0, r ), we have ɛw(y ɛ,β ) + 1 β 2 y ɛ,β x ɛ,β ɛ 2. (3.30) u ɛ (x, t) φ(x, t) ɛw( x ɛ ) uɛ (x ɛ,β, t ɛ,β ) φ(x ɛ,β, t ɛ,β ) ɛw(y ɛ,β ) + 1 β 2 y ɛ,β x ɛ,β ɛ 2. (3.31) Thus, there exists a constant C > 0 such that y ɛ,β x ɛ,β ɛ C1/2 β. (3.32) Besides, one has B(x 0, r) B(t 0, r ) is a compact set, we can suppose that x ɛ,β x 1 and t ɛ,β t 1 as (ɛ, β) 0, where (x 1, t 1 ) B(x 0, r) B(t 0, r ). On the other hand, since w(y) is bounded and u ɛ u in C(R n [0, T ]) as ɛ 0, let (ɛ, β) 0 in (3.31), one gets u(x, t) φ(x, t) u(x 1, t 1 ) φ(x 1, t 1 ), (x, t) B(x 0, r) B(t 0, r ). (3.33) It implies that (x 1, t 1 ) is a minimum point of u φ on B(x 0, r) B(t 0, r ). But, since (x 0, t 0 ) is a strict minimum point of u φ on B(x 0, r) B(t 0, r ), we conclude that (x 1, t 1 ) = (x 0, t 0 ).

34 3.2 Main results. 30 The same argument in the proof of subsolution, one has (x ɛ,β, t ɛ,β ) B(x 0, r) B(t 0, r ). Besides, since u ɛ is a supersolution of (3.1)-(3.2), we have φ t (x ɛ,β, t ɛ,β ) + H(x ɛ,β, x ɛ,β ɛ, D xφ(x ɛ,β, t ɛ,β ) + p ɛ,β ) 0, (3.34) where p ɛ,β = 2 ɛβ 2 (y ɛ,β x ɛ,β ɛ ). On the other hand, w is a subsolution of (3.12), one gets H(x 0, y ɛ,β, D x φ(x 0, t 0 ) + p ɛ,β ) H(x 0, D x φ(x 0, t 0 )) 0. (3.35) Besides, since H is coercive, one concludes from (3.35) that there exists a constant C > 0 depending on H such that D x φ(x 0, t 0 ) + p ɛ,β < C. (3.36) Hence, since H is uniformly continuous on R n R n B(0, C), we deduce H(x, y, p) H(x, y, p ) m C H( x x + y y + p p ), (3.37) for all (x, y, p), (x, y, p ) R n R n B(0, C) where m C H (t) 0 as t 0. Subtracting two sides of (3.34), (3.35) and applying (3.37), one has φ t (x ɛ,β, t ɛ,β ) + H(x 0, q ) H(x 0, y ɛ,β, q + p ɛ,β ) H(x ɛ,β, x ɛ,β ɛ, q + p ɛ,β), (3.38) where q = D x φ(x ɛ,β, t ɛ,β ) and q = D x φ(x 0, t 0 ). On the other hand, m C H( x ɛ,β x 0 + y ɛ,β x ɛ,β ɛ + q q ) H(x 0, y ɛ,β, q + p ɛ,β ) H(x ɛ,β, x ɛ,β ɛ, q + p ɛ,β). Since φ C 1 (R n [0, T ]) and (3.32), let (ɛ, β) 0, we obtain Therefore, u is a supersolution of (3.10). φ t (x 0, t 0 ) + H(x 0, D x φ(x 0, t 0 )) 0. (3.39) Hence, u is a viscosity solution of (3.10) and by Theorem 1.2, u is unique. Remark 3.1. For k N, let (u ɛ j k ) be an arbitrary subsequence of (uɛ k ) such that uɛ j k v k in C(R n [0, T ]). Hence, since Theorem 3.10, v k is a viscosity solution of v t + H(x, Dv) = 0 in Rn (0, + ), (3.40) v(x, 0) = u k 0(x). (3.41)

35 3.2 Main results. 31 On the other hand, since the uniqueness of u k, we have that u ɛ j k u k in C(R n [0, T ]) for all (ɛ j ) (ɛ) as ɛ j 0. Moreover, by Ascoli s Theorem, u ɛ k is bounded on a compact set of Rn [0, T ]. We conclude that u ɛ k u k as ɛ 0. On the other hand, since u k is the unique viscosity solution of the equation u t + H(x, Du) = 0 in Rn (0, + ), (3.42) u(x, 0) = u k 0(x). (3.43) Set N k (x, t) = u k 0 (x) C kt and M k (x, t) = u k 0 (x)+ct, where C = sup q Du k 0 H(, q). One has C + H(x, Du k 0) 0, x R n (3.44) C + H(x, Du k 0) 0, x R n. (3.45) Hence, N k (x, t) and M k (x, t) are respectively subsolution and supersolution of (3.42). And then, by the comparison result, we have N k (x, t) u k (x, t) M k (x, t), (x, t) R n [0, T ]. (3.46) By Corollary 1.1, for m, k N, m k, since u m, u k are respectively supersolution and subsolution of (3.42), one has sup (u k (x, t) u m (x, t)) sup(u k (x, 0) u m (x, 0)) R n [0,T ] R n sup(u k 0(x) u m 0 (x)) R n u k 0 u m 0 u k 0 u 0 + u m 0 u 0. (3.47) Hence, u k u m u k 0 u 0 + u m 0 u 0. (3.48) Therefore, (u k ) k N is a Cauchy sequence in BUC(R n [0, T ]). Thus, u k uniformly converges to u BUC(R n [0, T ]) as k. Hence, by Theorem 1.1, u is a viscosity solution of (3.10)-(3.11). On the other hand, for k, m N, since Theorem 1.4, u ɛ k, uɛ m are respectively subsolution and supersolution of (3.5). Then, by the comparison result, one has u ɛ k uɛ m u k 0 u 0 + u m 0 u 0. (3.49)

36 3.2 Main results. 32 Hence, u ɛ k uɛ in BUC(R n [0, T ]) as k. Therefore, since Theorem 1.1, u ɛ is a viscosity solution of (3.1)-(3.2). Finally, for all ɛ > 0, since u k 0 u 0 and u k u in BUC(R n [0, T ]), one has K 1 (ɛ) N, u k 0 u 0 < ɛ, k K 1 (ɛ), (3.50) K 2 (ɛ) N, u k u < ɛ, k K 2 (ɛ). (3.51) Let k max(k 1 (ɛ), K 2 (ɛ)), we have u ɛ u u ɛ u ɛ k + u ɛ k u k + u k u u 0 u k 0 + u ɛ k u k + u k u ɛ + u ɛ k u k + ɛ. (3.52) Sending ɛ 0, we obtain u ɛ u in BUC(R n [0, T ]).

37 References [1] G. Barles, An introduction to the Theory of viscosity solutions for first-order Hamilton- Jacobi equations and applications. [2] M.G. Crandall, L.C. Evans and P.L. Lions, Viscosity solutions of Hamilton-Jacobi equations, Trans. Amer. Math. Soc., 277 (1983), p [3] M.G. Crandall, L.C. Evans and P.L. Lions, Some properties of viscosity solutions of Hamilton-Jacobi equations, Trans. Amer. Math. Soc., 282 (1984), p [4] M.G. Crandall, H. Ishii and P.L. Lions, User s guide to viscosity solutions of secondorder partial differential equations, Bull. AMS 27 (1992), [5] P.L. Lions, G. Papanicolaou and S.R.S. Varadhan, Homogenization of Hamilton-Jacobi equations, unpublished work. [6] L.C. Evans, The perturbed test function technique for viscosity solutions, Proc. Roy. Soc. Edinburgh Sect. A 111 (1989), [7] A.Bensoussan, J.L Lions and G. Papanicolaou, Assymptotic analysis for periodic structures, North-Holland, Amsterdam, [8] R.A. Adams, Sobolev Spaces, Academic Press, NewYork, [9] H. Brezis, Analyse fonctionnelle. Theorie et Applications, Masson Paris,