Master Thesis Nguyen Tien Thinh - This problem was introduced in [5] by P.L. Lions, G. Papanicolaou and S.R.S. Varadhan in 1986. ... Lions in 1983 regarding the Hamilton-Jacobi equation

Master Thesis

Nguyen Tien Thinh

Homogenization and Viscosity solution

Advisor: Guy Barles

Defense: Friday June 21th, 2013

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 21th, 2013

Nguyen Tien Thinh

iii

Introduction

This paper is devoted to the homogenization of the Hamilton-Jacobi equations

∂uε

∂t+H(x,

x

ε,Duε) = 0 in Rn × (0,+∞). (1)

uε|t=0 = u0(x). (2)

where the initial data u0(x) is a bounded, uniformly continuous on Rn.

We consider the behavior of the viscosity solution uε of (1)-(2) as ε tends to 0 under

assumptions

(H1) H is uniformly continuous on Rn × Rn ×B(0, R) for any R < +∞.

(H2) H(x, y, p) = H(x, y + z, p) for all x, y, p ∈ Rn.

(H3) ∀M > 0,∃rM > 0 such that H(x, y, p) > M for all p ∈ Rn, |p| > rM uniformly for all

x, y ∈ Rn.

This problem was introduced in [5] by P.L. Lions, G. Papanicolaou and S.R.S. Varadhan in

1986. The key to solve was used is an assymptotic expansions introduced by A.Bensoussan,

J.L Lions and G. Papanicolaou (see in [7])

uε(x, t) = u0(x, t) + εw1(x,x

ε, t) + ε2w2(x,

x

ε, t) + . . . (3)

where wi(x, y, t) are Zn-periodic in y, for all i ∈ N. Then inserting (3) in (1) and identifying

the terms in front of powers of ε, we have

∂u0

∂t(x, t) +H(x, y,Dxu0(x, t) +Dyw1(y, t)) = 0 in Rn × Rn × (0,+∞). (4)

It leads to the cell problem: for x, p ∈ Rn, is there H(x, p) ∈ R such that the equation

H(x, y, p+Dyv) = H(x, p) in Rn (5)

has a Zn-periodic viscosity solution v ? If the answer is yes, then using an other argument,

one concludes that uε tends to u, the unique viscosity solution of the equation

∂u

∂t+H(x,Du) = 0 in Rn × (0,+∞), (6)

u(x, 0) = u0(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;

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.

Contents

Preface ii

Introduction iii

1 Viscosity Solution Theory 2

1.1 Introduction. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2

1.2 Properties. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3

2 Ergodic Problem Theory 10

2.1 Introduction. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 10

2.2 Main results. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 11

3 Homogenization Theory 25

3.1 Introduction. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 25

3.2 Main results. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 25

References 33

1

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 1980. 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 1984.

Let us give an example to know where this idea comes from.

Consider the C2-solution u of the equation −u′′ = 0 and φ is an arbitrary C2-function,

touching u from above at some point x0. Hence, u ≤ φ locally, φ(x0) = u(x0). We conclude

that

u(x)− φ(x) ≤ 0 = u(x0)− φ(x0) (1.1)

in a neighborhood of x0. It means x0 is a local maximum point of u− φ.

Therefore, one has

−φ′′(x0) = (u− φ)′′(x0) ≤ 0. (1.2)

The same idea, when we touch u from below at x0, one concludes that x0 is a local minimum

of u− φ and −φ′′(x0) ≥ 0.

In the case u doesn’t belong to C2 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

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

∂u

∂t+H(x,Du) = 0 in Rn × (0,+∞), (1.3)

where H satisfies

(H1) H is uniformly continuous on Rn ×B(0, R) for any R < +∞.

(H2) H(x, p) = H(x+ z, p) for all x, p ∈ Rn.

(H3) ∀M > 0, ∃rM > 0 such that H(x, p) > M for all p ∈ Rn, |p| > rM uniformly for all

x ∈ Rn.

Definition 1.1. u ∈ C(Rn × (0,+∞)) is a subsolution of (1.3) if for any φ ∈ C1(Rn ×

(0,+∞)) and (x0, t0) is a local maximum point of u− φ, one has

∂φ

∂t(x0, t0) +H(x0, Dφ(x0, t0)) ≤ 0. (1.4)

and u ∈ C(Rn × (0,+∞)) is a supersolution of (1.3) if for any φ ∈ C1(Rn × (0,+∞)) and

(x0, t0) is a local minimum point of u− φ, one has

∂φ

∂t(x0, t0) +H(x0, Dφ(x0, t0)) ≥ 0. (1.5)

If u ∈ C(Rn × (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 ε > 0, uε ∈ C(Rn× (0,+∞)) is a viscosity solution of the

equation∂uε∂t

+Hε(x,Duε) = 0 in Rn × (0,+∞), (1.6)

1.2 Properties. 4

where (Hε)ε>0 is a sequence of continuous functions. If uε → u in C(Rn × (0,+∞)) and if

Hε → H in C(Rn × (0,+∞)× R× Rn) 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 ∈ Rn, (1.8)

then

u(x, t) ≤ v(x, t), ∀(x, t) ∈ Rn × (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

∂w

∂t+H(x,Dw) = −η in Rn × (0,+∞), (1.10)

and w(x, 0) = u(x, 0).

Suppose M = supRn×(0,+∞)(u− v) > 0 and introduce the test-function

Ψε,α,β(x, y, t, s) = u(x, t)− v(y, s)− |x− y|2

ε2− |t− s|

2

α2− β(|x|2 + |y|2). (1.11)

Since Ψε,α,β → −∞ as |x|, |y|, |t|, |s| → +∞, there exists a maximum point (x, y, t, s)

depending on ε, α, β > 0 such that

Ψε,α,β(x, y, t, s) ≤ Ψε,α,β(x, y, t, s), ∀x, y ∈ Rn,∀t, s ∈ (0,+∞). (1.12)

Choose y = x, t = s, we have

u(x, t)− v(x, t)− 2β|x|2 ≤ Ψε,α,β(x, y, t, s), ∀(x, t) ∈ Rn × (0,+∞). (1.13)

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

M ′ = maxRn×(0,+∞)

(u(x, t)− v(x, t)− 2β|x|2). (1.14)

Hence,

M ′ ≤ Ψε,α,β(x, y, t, s)

= u(x, t)− v(y, s)− |x− y|2

ε2− |t− s|

2

α2− β(|x|2 + |y|2). (1.15)

Moreover, since M ′ → M as β → 0, one choose β small enough such that M ′ > 0. One

concludes that|x− y|2

ε2+|t− s|2

α2+ β(|x|2 + |y|2) ≤ R, (1.16)

where R = max(‖u‖∞, ‖v‖∞).

Therefore,

|x− y| ≤ R1/2ε, (1.17)

|t− s| ≤ R1/2α, (1.18)

|2βx| ≤ 2β1/2(β|x|2)1/2 ≤ 2β1/2R1/2, (1.19)

|2βy| ≤ 2β1/2(β|y|2)1/2 ≤ 2β1/2R1/2. (1.20)

Notice that t, s 6= 0. Indeed, if t = 0, s 6= 0, since M ′ > 0 and (1.15), we have

M ′ = u(x, 0)− v(y, s)− |x− y|2

ε2− |s|

2

α2− β(|x|2 + |y|2)

≤ u(x, 0)− v(y, s)

≤ u(x, 0)− v(x, 0) + v(x, 0)− v(y, s)

≤ v(x, 0)− v(y, s). (1.21)

On the other hand, v ∈ BUC(Rn × [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 ε, α <δ(M ′)

2R1/2, we conclude that

M ′ ≤ v(x, 0)− v(y, s) <M ′

2. (1.23)

1.2 Properties. 6

Hence, M ′ < 0 is the contradiction and then x, y ∈ Rn, t, s ∈ (0,+∞) for ε, α, β small

enough.

On the other hand, since (x, y, t, s) is the maximum point of Ψε α,β, we conclude that (x, t)

is the maximum point of

u(x, t)−[v(y, s) +

|x− y|2

ε2+|t− s|2

α2+ β(|x|2 + |y|2)

]. (1.24)

Besides, u is a subsolution of (1.3), one has

pε,α,β +H(x, qε,α,β + 2βx) ≤ −η, (1.25)

where pε,α,β =2(t− s)α2

and qε,α,β =2(x− y)

ε2.

In the same way, (y, s) is the minimum point of

v(y, s)−[u(x, t)− |x− y|

2

ε2− |t− s|

2

α2− β(|x|2 + |y|2)

]. (1.26)

But, v is a supersolution of (1.3), one has

pε,α,β +H(y, qε,α,β − 2βy) ≥ 0. (1.27)

On the other hand, since (1.18), one gets

|pε,α,β| = |2(t− s)α2

| ≤ 2R1/2

α. (1.28)

And by the coercivity of H, from (1.25),(1.28) we conclude that there exists Cα > 0 de-

pending only on α and H such that |qε,α,β + 2βx| < Cα.

Thus, since H is uniformly continuous on Rn ×B(0, Cα), one gets

|H(x, p)−H(y, q)| ≤ mH(|x− y|+ |p− q|), ∀x, y ∈ Rn, p, q ∈ B(0, Cα), (1.29)

where mH(r)→ 0 as r → 0.

Subtracting two sides of (1.25),(1.27), we have

η ≤ H(y, qε,α,β − 2βy)−H(x, qε,α,β + 2βx)

≤ mH(|x− y|+ |2βx+ 2βy|). (1.30)

Let (ε, 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(Rn × [0,+∞)) are

respectively subsolution and supersolution of (1.3) then

supRn×(0,+∞)

(u− v) ≤ supRn

(u(x, 0)− v(x, 0)). (1.31)

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

∂u

∂t+H(x,Du) = 0 in Rn × (0,+∞) (1.32)

and the comparison result holds.

Let consider the set

S = {v ∈ C(Rn × (0,+∞));u ≤ v ≤ u, v is a subsolution of (1.32)}

then u = supS 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) = u0(x), (1.34)

where u0(x) ∈ C1(Rn) ∩W 1,∞(Rn), then there exists a unique Lipschitz viscosity solution

of (1.33)-(1.34) in W 1,∞(Rn × (0,+∞)).

Proof. Set M = u0(x) +Ct and N = u0(x)−Ct, where C = sup|q|≤‖Du0‖∞ ‖H(·, q)‖L∞×L∞

one has

C +H(x,Du0) ≥ 0, ∀x ∈ Rn, (1.35)

−C +H(x,Du0) ≤ 0, ∀x ∈ Rn. (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) ∈ Rn × (0,+∞). (1.37)

For all h > 0, one has

u(x, t+ h)− u(x, t) ≤ maxRn×(0,+∞)

(u(x, t+ h)− u(x, t))

≤ maxRn

(u(x, h)− u(x, 0))

≤ maxRn

(u0(x) + Ch− u0(x))

≤ Ch. (1.38)

1.2 Properties. 8

Hence,u(x, t+ h)− u(x, t)

h≤ C. (1.39)

On the other hand,

u(x, t)− u(x, t+ h) ≤ maxRn×(0,+∞)

(u(x, t)− u(x, t+ h))

≤ maxRn

u(x, 0)− u(x, h)

≤ maxRn

(u0(x)− u0(x) + Ch)

≤ Ch. (1.40)

Hence,u(x, t)− u(x, t+ h)

h≤ C. (1.41)

Let h→ 0, one gets

|∂u∂t| ≤ C. (1.42)

Therefore, (1.33) becomes

H(x,Du) = −∂u∂t≤ |∂u

∂t| ≤ C. (1.43)

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,∞(Rn) and ‖Du‖∞ ≤ K.

Proof. Let x ∈ Rn. Since u(y)−K|y − x| → −∞ as |y| → +∞ where K > 0, max(u(y)−

K|y − x|) exists over Rn and we denote this maximum point by y0.

If y0 = x, one has

u(y)−K|y − x| ≤ max(u(y)−K|y − x|) = u(x), ∀y ∈ Rn. (1.44)

If y0 6= x, one has

u(y)− φ(y) ≤ u(y0)− φ(y0), ∀y ∈ Rn, (1.45)

where φ(y) = K|y − x|.

Since φ is C1 in a neighborhood of y0 and u is a subsolution of (1.33), one gets

H(y0,K(y0 − x)

|y0 − x|) ≤ C. (1.46)

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(y0 − x)

|y0 − x|

∣∣∣∣ ≤ R. (1.47)

Choose K1 = R+ 1. Hence, max(u(y)−K1|y − x|) exists only at y0 = x, it means

u(y)−K1|y − x| ≤ u(x), ∀y ∈ Rn. (1.48)

By the same way, let y ∈ Rn, there is K2 > 0 depending only on C and H such that

max(u(x)−K2|x− y|) exists only at y, one gets

u(x)−K2|y − x| ≤ u(y), ∀x ∈ Rn. (1.49)

Therefore, from (1.48),(1.49), with K = max(K1,K2),

|u(y)− u(x)| ≤ K|y − x|, ∀x, y ∈ Rn. (1.50)

Applying the Rademacher’s Theorem, one has u is differentiable almost everywhere in Rn.

Finally, by the subsolution inequality,

H(x,Du) ≤ C, a.e in Rn, (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,∞(Rn × (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.

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 Rn. (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,Dup + p) = H(x, p), in Rn, (2.3)

for any (x, p) ∈ Rn × Rn, where H satisfies

(H1) H is uniformly continuous on Rn × Rn ×B(0, R) for any R < +∞.

(H2) H(x, y, p) = H(x, y + z, p) for all x, y, p ∈ Rn.

(H3) ∀M > 0,∃rM > 0 such that H(x, y, p) > M for all p ∈ Rn, |p| > rM uniformly for all

x, y ∈ Rn.

10

2.2 Main results. 11

2.2 Main results.

Theorem 2.1. Assume that H satisfies (H1)-(H2)-(H3). For (x, p) ∈ Rn×Rn, there exists

a unique H(x, p) such that there exists a Zn-periodic viscosity solution up satisfy (2.3).

Proof.

Lemma 2.1. For γ > 0 and (x, p) ∈ Rn × Rn, the equation

H(x, y,Duγ,p(y) + p) + γuγ,p(y) = 0 in Rn, (2.4)

where the Hamiltonian H satisfy (H1), (H2) and (H3), has a unique, Lipschitz continuous,

Zn-periodic viscosity solution uγ,p(y).

Proof. Firstly, for γ > 0, (x, p) ∈ Rn × Rn, put Mγ,p =‖H(x, ·, p)‖∞

γ, one has

H(x, y, p)− γMγ,p ≤ 0, ∀y ∈ Rn, (2.5)

H(x, y, p) + γMγ,p ≥ 0, ∀y ∈ Rn. (2.6)

Thus, −Mγ,p, Mγ,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 Rn.

Proof. Suppose M = supRn(u− v) is positive and introduce the test-function

Ψε,α(y, z) = u(y)− v(z)− |y − z|2

ε2− α(|y|2 + |z|2). (2.7)

One has

|yε,α − zε,α| ≤ R1/2ε, (2.8)

2α|yε,α| = 2α1/2(α|yε,α|2)1/2 ≤ 2α1/2R1/2, (2.9)

2α|zε,α| = 2α1/2(α|zε,α|2)1/2 ≤ 2α1/2R1/2. (2.10)

where (yε,α, zε,α) ∈ Rn × Rn is the maximum point of Ψε,α and R = max(‖u‖∞, ‖v‖∞).

By the definition of subsolution and supersolution,

H(x, yε,α, qε,α + 2αyε,α + p) + γu(yε,α) ≤ 0, (2.11)

H(x, zε,α, qε,α − 2αzε,α + p) + γv(zε,α) ≥ 0, (2.12)


where qε,α =2(yε,α − zε,α)

ε2.

On the other hand, from (2.11) and (H3), there exists a constant C > 0 such that |qε,α +

2αyε,α + p| ≤ C.

Hence, since H is uniformly continuous on Rn × Rn ×B(0, C), one deduces

|H(x, y, p)−H(x, z, q)| ≤ mCH(|y − z|+ |p− q|), ∀x, y, z ∈ Rn, p, q ∈ B(0, C), (2.13)

where mCH(t)→ 0 as t→ 0.

Subtracting two sides of (2.11), (2.12), then apply (2.13), one gets

γ(u(yε,α)− v(zε,α)) ≤ H(x, zε,α, qε,α − 2αzε,α + p)−H(x, yε,α, qε,α + 2αyε,α + p)

≤ mCH(|yε,α − zε,α|+ |2αyε,α + 2αzε,α|). (2.14)

Therefore, from (2.8),(2.9),(2.10), letting (ε, α)→ 0, we have the contradiction γM ≤ 0.

Hence, if uγ,p is a viscosity solution of (2.4), uγ,p is also a subsolution and a supersolution.

Thus, we have the a priori estimate

−Mγ,p ≤ uγ,p(y) ≤Mγ,p, ∀y ∈ Rn. (2.15)

Thus, by the subsolution inequality and (2.15), we have

H(x, y,Du+ p) ≤ −γu(y) ≤ γMγ,p ≤ ‖H(x, ·, p)‖∞, a.e in Rn. (2.16)

Therefore, by Lemma 1.1, one concludes that there is a constant Kp > 0 depending only

on ‖H(x, ·, p)‖∞ and H such that ‖Du‖∞ ≤ Kp.

Using these a priori estimates, we introduce the sets

W = {v ∈W 1,∞(Rn);−Mγ,p ≤ v ≤Mγ,p, ‖Dv‖∞ ≤ Kp},

S = {v ∈ W; v is a subsolution of (2.4)}.

Following Perron’s method, we conclude that there is a viscosity solution uγ,p of (2.4).

Moreover, by Lemma 2.2, this viscosity solution is unique.

Finally, for all z ∈ Zn, uγ,p(y + z) is a viscosity solution of

H(x, y + z,Duγ,p(y + z) + p) + γuγ,p(y + z) = 0 in Rn. (2.17)

On the other hand, since H satisfy (H2), one gets

H(x, y,Duγ,p(y+ z) + p) + γuγ,p(y+ z) = H(x, y+ z,Duγ,p(y+ z) + p) + γuγ,p(y+ z) = 0.


Therefore, uγ,p(y + z) is also a viscosity solution of (2.4), but by the uniqueness, one has

uγ,p(y + z) = uγ,p(y). Therefore, uγ,p is Zn-periodic.

For (x, p) ∈ Rn × Rn, we consider uγ,p, where γ > 0 is small enough, viscosity solution

of (2.4).

For y ∈ Rn, we set vγ,p(y) = uγ,p(y)− uγ,p(0), then by the result in Theorem 2.1, one has

|vγ,p(y)| = |uγ,p(y)− uγ,p(0)| = |uγ,p(y − [y])− uγ,p(0)| ≤ Kp|y − [y]|, (2.18)

where y = (yi)1≤i≤n and [y] = ([yi])1≤i≤n ∈ Zn, 0 ≤ yi − [yi] < 1 for 1 ≤ i ≤ n.

Hence, since y − [y] ∈ [0, 1]n, we have vγ,p is equibounded.

Besides, for all y, z ∈ Rn,

|vγ,p(z)− vγ,p(y)| = |uγ,p(z)− uγ,p(y)| ≤ Kp|z − y|. (2.19)

Thus, vγ,p is equi-Lipschitz-continuous.

Applying Ascoli’s Theorem, we deduce that there exists a subsequence {vγk,p} ⊂ {vγ,p}

which converges to up ∈ C(Rn).

On the other hand,

| − γkuγk,p(0)| = γk|uγk,p(0)| ≤ γkMγk,p = ‖H(x, ·, p)‖∞. (2.20)

Thus, there exists a constant H(x, p) such that there is a subsequence −γkmuγkm ,p(0) →

H(x, p) as γkm → 0. Furthermore, since (2.20), we have

|H(x, p)| ≤ ‖H(x, ·, p)‖∞. (2.21)

Finally, by the definition of vγ,p and the equation (2.4), one gets

H(x, y,Dvγkm ,p + p) + γkmvγkm ,p = −γkmuγkm ,p(0) a.e in Rn. (2.22)

Then, let m→ +∞, the stability result says

H(x, y,Dup + p) = H(x, p) a.e in Rn.

Finally, we show that ∀(x, p) ∈ Rn × Rn, H(x, p) is unique.

Let (x, p) ∈ Rn × Rn, suppose (up, H(x, p)) and (vp, H ′(x, p)) respectively satisfy

H(x, y,Dup + p) = H(x, p) in Rn (2.23)

H(x, z,Dvp + p) = H ′(x, p) in Rn, (2.24)


where up, vp are continuous, Zn-periodic viscosity solution of (2.23), (2.24).

One can consider the test-function

Ψε(y, z) = up(y)− vp(z)−|y − z|2

ε2, (2.25)

and we can assume without loss of generality that y ∈ [0, 1]n and z ∈ [0, 2]n since the

Zn-periodicity.

Therefore, Ψε has a maximum point (yε, zε) and we have

|yε − zε| ≤ R1/2p ε, (2.26)

where Rp = max(‖up‖∞, ‖vp‖∞).

On the other hand, since up is subsolution of (2.23) and vp is supersolution of (2.24), we

also have

H(x, yε, qε + p) ≤ H(x, p), (2.27)

H(x, zε, qε + p) ≥ H ′(x, p), (2.28)

where qε =2(yε − zε)

ε2.

Besides of that, since (2.27) and the coercivity of H, there is a constant Cp > 0 such that

|qε + p| ≤ Cp.

Hence,

|H(x, y, q)−H(x, z, q)| ≤ mCp

H (|y − z|), ∀q ∈ B(0, Cp), (y, z) ∈ [0, 1]n × [0, 2]n, (2.29)

where mCp

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, zε, qε + p)−H(x, yε, qε + p),

≤ mCp

H (|yε − zε|). (2.30)

Let ε→ 0, we conclude

H ′(x, p) ≤ H(x, p). (2.31)

Exchanging the roles of up, vp, one has

H ′(x, p) ≥ H(x, p). (2.32)

Therefore, for all (x, p) ∈ Rn × Rn, H(x, p) is unique.


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 minRn f(x) = 0.

Firstly, one can assume that u(0) = u(1) = 0 and notice that, at x0 is the minimum point

of f , one has λ(p) = |u′(x0) + p| ≥ 0.

On the other hand, let θ ∈ [0, 1], p, q ∈ Rn, 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 ∈ Rn 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 C1([0, 1]) viscosity solution u such that

u(x)− u(0) = −px+

∫ x

0[f(t) + λ(p)]dt. (2.36)

Hence, by the assumption u(0) = u(1) = 0, one has

p =

∫ 1

0f(t)dt+ λ(p). (2.37)

Besides, since λ(p) ≥ 0, it implies that

p ≥∫ 1

0f(t)dt. (2.38)

On the other hand, assume that

u′(x) + p = −[f(x) + λ(p)], ∀x ∈ [0, 1]. (2.39)

Thus, there is a C1([0, 1]) viscosity solution u such that

u(x)− u(0) = −px−∫ x

0[f(t) + λ(p)]dt. (2.40)


Hence,

p = −∫ 1

0f(t)dt− λ(p). (2.41)

And since λ(p) ≥ 0, it implies that

p ≤ −∫ 1

0f(t)dt. (2.42)

Therefore, for all p ∈ Rn, p ≥∫ 1

0 f(t)dt and λ(p) = p−∫ 1

0 f(t)dt, one obtains

u(x) = −px+

∫ x

0[f(t) + λ(p)]dt, (2.43)

is a viscosity solutions of (2.33).

And, for all p ∈ Rn, p ≤ −∫ 1

0 f(t)dt and λ(p) = −p−∫ 1

0 f(t)dt, one has

u(x) = −px−∫ x

0[f(t) + λ(p)]dt, (2.44)

is a viscosity solutions of (2.33).

On the other hand, since λ(p) ≥ 0 and convex for all p ∈ Rn. We have necessarily λ(p) = 0

for

−∫ 1

0f(t)dt < p <

∫ 1

0f(t)dt. (2.45)

We show that there is a 1-periodic viscosity solution for the equation (2.33). To do that,

starting at x0 ∈ [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 ∈ [x0, x0 + 1]. Hence, we choose a

viscosity solution satisfies

u′(x) = −p+ f(x) (2.46)

on [x0, x] , denoting by u1.

This viscosity solution satisfies also

u′(x) = −p− f(x), (2.47)

on [x, x0 + 1], denoting by u2.

And such that u1(x) = u2(x) for the continuity.

One has

u1(x) =

∫ x

x0

[−p+ f(t)] dt, (2.48)

u2(x) =

∫ 1+x0

x[p+ f(t)] dt. (2.49)


Remark 2.1. One can choose

u1(x) =

∫ x

x0

[p+ f(t)] dt, (2.50)

u2(x) =

∫ 1+x0

x[−p+ f(t)] dt, (2.51)

to find a viscosity solution.

On the other hand,

u1(x0)− u2(x0) = −∫ 1+x0

x0

[p+ f(t)] dt = −[p+

∫ 1+x0

x0

f(t)dt

], (2.52)

u1(1 + x0)− u2(1− x0) =

∫ 1+x0

x0

[−p+ f(t)] dt = −p+

∫ 1+x0

x0

f(t)dt. (2.53)

Besides of that, since f is 1-periodic,∫ 1+x0

x0

f(t)dt =

∫ 1

0f(t)dt, (2.54)

Then apply (2.52), (2.53), (2.54), one has

[u1(x0)− u2(x0)][u1(1 + x0)− u2(1 + x0)] = −[−|p|2 + (

∫ 1

0f(t)dt)2

]< 0. (2.55)

Hence, there exists x ∈ [x0, 1 + x0] such that u1(x)− u2(x) = 0.

Therefore, we can define a 1-periodic, continuous function on [x0, 1 + x0] as below

u(x) =

∫ x

x0

[−p+ f(t)] dt, x ∈ [x0, x],∫ 1+x0

x[p+ f(t)] dt, x ∈ [x, 1 + x0],

(2.56)

Finally, we choose x0 such that for p ≥ 0, u is a viscosity solution for the equation (2.33).

Indeed, we have

limx→x−0

u(x)− u(x0)

x− x0= −p+ f(x0), (2.57)

limx→x+0

u(x)− u(x0)

x− x0= −p− f(x0). (2.58)

One takes x0 ∈ [0, 1] such that f(x0) = 0, u is differentiable at x0.

On the other hand, since f is 1-periodic, one concludes that f(x0 + 1) = f(x0) = 0.

Therefore u is also differentiable at x0 + 1.

Thus, we obtain that u is differentiable in [x0, 1+x0]\x and |u′+p| = f(x), ∀x ∈ [x0, 1+x0]\x.

Hence, u is a 1-periodic viscosity solution of the equation

|u′ + p| = f(x), in [x0, 1 + x0] \ x. (2.59)


On the other hand, we consider if the local maximum point and the local minimum point

at x.

Let φ ∈ C1(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,

u(x)− u(x) ≤ φ(x)− φ(x), ∀x ∈ B(x, r). (2.61)

For x < x, one gets

φ(x)− φ(x) ≥∫ x

x0

[−p+ f(t)]dt−∫ x

x0

[−p+ f(t)]dt

= −∫ x

x[−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

φ(x)− φ(x) ≥∫ 1+x0

x[p+ f(t)]dt−

∫ 1+x0

x[p+ f(t)]dt

=

∫ x

x[p+ f(t)]dt. (2.65)

Thus,

φ(x)− φ(x)

x− x≥ 1

x− x

∫ x

x[p+ f(t)]dt. (2.66)

Hence, since p ≥ 0, one has

φ(x)− φ(x)

x− x+ p ≥ 1

x− x

∫ x

xf(t)dt+ 2p

≥ 1

x− x

∫ x

xf(t)dt. (2.67)

Sending x→ x, one has

φ′(x) + p ≥ −f(x). (2.68)


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 ∈ C1(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,

u(x)− u(x) ≥ φ(x)− φ(x), ∀x ∈ B(x, r). (2.71)

For x < x, one hasφ(x)− φ(x)

x− x≥ 1

x− x

∫ x

x[−p+ f(t)]dt. (2.72)

Thus,

limx→x−

φ(x)− φ(x)

x− x≥ f(x)− p. (2.73)

For x > x, one hasφ(x)− φ(x)

x− x≤ 1

x− x

∫ x

x[p+ f(t)]dt. (2.74)

Thus,

limx→x+

φ(x)− φ(x)

x− x≤ −[f(x)− p]. (2.75)

From (2.73), (2.75), one concludes that

limx→x−

φ(x)− φ(x)

x− x6= lim

x→x+φ(x)− φ(x)

x− x. (2.76)

It is in contradiction with φ ∈ C1(Rn). Hence, u− φ has no local minimum point in x.

Generally, u is a viscosity solution of (2.33).

Proposition 2.1. (x, p) 7→ H(x, p) is continuous on Rn × Rn.

Proof. Let (x, p) ∈ Rn × Rn and (xk, pk)k∈N ⊂ Rn, (xk, pk)→ (x, p) as k → +∞.

We can assume that |xk − x| < 1 and |pk − p| < 1.

For each k ∈ N, there is a continuous, Zn-periodic viscosity solution upk satisfy

H(xk, y,Dupk + pk) = H(xk, pk), in Rn. (2.77)


One can suppose upk(0) = 0 since wpk(y) = upk(y) − upk(0) is also a Zn-periodic viscosity

solution of (2.77).

On the other hand, since (2.21), we have

|H(xk, pk)| ≤ ‖H(xk, ·, pk)‖∞. (2.78)

Therefore,

|H(xk, pk)| ≤ Kx,p, (2.79)

where Kx,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 Cp such that

‖Dupk‖∞ ≤ Cx,p. (2.80)

It implies

|upk(z)− upk(y)| ≤ Cx,p|z − y|, ∀y, z ∈ Rn. (2.81)

Therefore, upk is equi-Lipschitz continuous.

On the other hand,

|upk(y)| = |upk(y)− upk(0)| = |upk(y − [y])− upk(0)| ≤ Cx,p|y − [y]|, ∀y ∈ Rn, (2.82)

where y = (yi)1≤i≤n and [y] = ([yi])1≤i≤n ∈ Zn, 0 ≤ yi − [yi] < 1 for 1 ≤ i ≤ n.

Thus, upk is equi-bounded because y − [y] ∈ [0, 1]n.

Hence, by Ascoli’s Theorem, there exists a subsequence (upm) of (upk) and v ∈ C(Rn) such

that upm → v as m→∞.

On the other hand, since (2.79), one has

|H(xm, pm)| ≤ Kx,p. (2.83)

Then, there exists a λ and a subsequence (H(xmj , pmj )) of (H(xm, pm)) which we also

denote by such that H(xmj , pmj )→ λ as j →∞.

Therefore, by (2.77), one has

H(xmj , y,Dupmj+ pmj ) = H(xmj , pmj ) in Rn. (2.84)

Sending j →∞ and using the stability of H, one has

H(x, y,Dv + p) = λ a.e in Rn. (2.85)


Finally, since H(x, p) is unique, we have λ = H(x, p). It implies that H(xk, pk) → H(x, p)

as k →∞.

Remark 2.2. (x, p) 7→ H(x, p) is uniformly continuous on Rn×B(0, R) for any R < +∞.

Proposition 2.2. If H(x, y, p) is convex in p for all x, y ∈ Rn, H(x, p) is convex in p for

all x ∈ Rn.

Proof. Let θ ∈ [0, 1], x ∈ Rn, p and q belong to Rn.

There exists a Zn-periodic viscosity solution up and unique H(x, p) satisfy

H(x, y,Dup + p) = H(x, p) in Rn. (2.86)

There exists a Zn-periodic viscosity solution uq and unique H(x, q) satisfy

H(x, y,Duq + q) = H(x, q) in Rn. (2.87)

Firstly, put v = θup + (1− θ)uq, we show that v is a subsolution of the equation

H(x, y,Dv + r) = θH(x, p) + (1− θ)H(x, q) in Rn, (2.88)

where r = θp+ (1− θ)q.

Let φ ∈ C1(Rn) and y0 is a strict local maximum point of v − φ, there exists k > 0 such

that

v(y)− φ(y) < v(y0)− φ(y0), ∀y ∈ B(x0, k). (2.89)

Introduce on the set B(y0, k)×B(y0, k), the test function

Ψε(y, z) = θup(y) + (1− θ)uq(z)− φ(y)− |y − z|2

ε2. (2.90)

Then there exists a maximum point (yε, zε) of Ψε in B(y0, k)×B(y0, k).

One has

|yε − zε| ≤ R1/2p,q ε. (2.91)

On the other hand, B(y0, k)×B(y0, k) is a compact set, there exists a subsequence (yεm , zεm)

of (yε, zε) and (y1, z1) ∈ B(y0, k) × B(y0, k) such that (yεm , zεm) → (y1, z1). We can note

this subsequence as (yε, zε) and by (2.91), one has y1 = z1.

Besides of that,

Ψε(y, z) ≤ Ψε(yε, zε), ∀(y, z) ∈ B(y0, k)×B(y0, k). (2.92)


Choose y = z, we obtain for all y ∈ B(y0, k),

θup(y) + (1− θ)uq(y)− φ(y) ≤ θup(yε) + (1− θ)uq(zε)− φ(yε)−|yε − zε|2

ε2. (2.93)

Let ε→ 0, (2.93) becomes

θup(y) + (1− θ)uq(y)− φ(y) ≤ θup(y1) + (1− θ)uq(y1)− φ(y1), (2.94)

for all y ∈ B(y0, k).

Hence, y1 is a maximum point of Ψε on B(y0, k). But y0 is a strict maximum point of Ψε

on this set, so y1 = y0.

One can see that yε is a local maximum point of

up(y)− (1/θ)[−(1− θ)uq(zε) + φ(y) +|y − zε|2

ε2]. (2.95)

And zε is a local maximum point of

uq(z)− 1/(1− θ)[−θup(yε) + φ(yε) +|yε − z|2

ε2]. (2.96)

Because up is a subsolution of (2.86) and uq is a supersolution of (2.87), then

H(x, yε,Dφ(yε)

θ+qεθ

+ p) ≤ H(x, p), (2.97)

H(x, zε,−qε

1− θ+ q) ≤ H(x, q), (2.98)

where qε =2(yε − zε)

ε2.

On the other hand, by (2.97) and the coercivity of H, there is a constant Cx,p > 0 such

that |Dφ(yε)

θ+qεθ

+ p| ≤ Cx,p.

Therefore, since H is uniformly continuous on Rn ×B(0, Cx,p), we have

|H(x, y, r)−H(x, z, q)| ≤ mCx,p

H (|y − z|+ |r − q|), ∀x, y, z ∈ Rn, r, q ∈ B(0, Cx,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, yε,Dφ(yε)

θ+qεθ

+ p) + (1− θ)H(x, zε,−qε

1− θ+ q)

≥ θ[H(x, yε,Dφ(yε)

θ+qεθ

+ p)−H(x, zε,Dφ(yε)

θ+qεθ

+ p)]

+θH(x, zε,Dφ(yε)

θ+qεθ

+ p) + (1− θ)H(x, zε,−qε

1− θ+ q)

≥ θmCx,p

H (|yε − zε|) + θH(x, zε,Dφ(yε)

θ+qεθ

+ p)

+(1− θ)H(x, zε,−qε

1− θ+ q). (2.100)


On the other hand,

H(x, zε, Dφ(yε) + r) = H

[x, zε, θ(

Dφ(yε)

θ+qεθ

+ p) + (1− θ)(− qε1− θ

+ q)

]. (2.101)

Hence, by the convexity of H(x, y, p) in p for all x, y ∈ Rn, we conclude

H(x, zε, Dφ(yε) + r) ≤ θH(x, zε,Dφ(yε)

θ+qεθ

+ p) + (1− θ)H(x, zε,−qε

1− θ+ q). (2.102)

Applying (2.102) in (2.100), one gets

θmCx,p

H (|yε − zε|) +H(x, zε, r) ≤ θH(x, p) + (1− θ)H(x, q). (2.103)

Finally, by (2.91), let ε→ 0, we have

H(x, y0, Dφ(y0) + r) ≤ θH(x, p) + (1− θ)H(x, q). (2.104)

Therefore, v is a subsolution of (2.88).

Secondly, with H(x, r), there exists a Zn-periodic viscosity solution wr of the equation

H(x, y,Dwr + r) = H(x, r), in Rn. (2.105)

Since up, uq are Zn-periodic, v is also Zn-periodic. Hence, one can consider the behavior of

v and wr in [0, 1]n.

Introduce the test-function on [0, 1]n × [0, 1]n,

Ψε(y, z) = v(y)− wr(z)−|y − z|2

ε2. (2.106)

Reproducing the technic as above, we obtain

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 ∈ Rn.

Proposition 2.3. H(x, p) is coercive in p uniformly for all x ∈ Rn.

Proof. Let M > 0, there exists rM > 0 such that

H(x, y, p) > M, ∀p ∈ Rn, |p| > rM , ∀x, y ∈ Rn, (2.108)

since H(x, y, p)→ +∞ as |p| → +∞ uniformly for all x, y ∈ Rn.

On the other hand, for p ∈ Rn, |p| > rM , there exists a Zn-periodic viscosity solution up

and a unique H(x, p) satisfy

H(x, y,Dup + p) = H(x, p), in Rn. (2.109)


Besides of that, since the Zn-periodic, one can see in [0, 1]n, there is a maximum point yε

of up − 0. And thus,

H(x, yε, 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 ∈ Rn.

Chapter 3

Homogenization Theory

3.1 Introduction.

Homogenization theory is concerned with the derivation of equations for averages of solu-

tions of equations with rapidly varying coefficients. This problem arises in obtaining micro-

scopic 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 ε tends to 0

of the viscosity solution uε of the Hamilton-Jacobi equation

∂uε

∂t+H(x,

x

ε,Duε) = 0 in Rn × (0,+∞). (3.1)

uε|t=0 = u0(x). (3.2)

where u0(x) ∈ BUC(Rn).

This problem was introduced in [5] by P.L. Lions, G. Papanicolaou and S.R.S. Varadhan in

1986. 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 ε > 0, there exist a unique

viscosity solution uε 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) = u0(x) ∈ BUC(Rn), (3.4)

where H(x, p) = H(x, y, p+Dyv) such that uε → u in BUC(Rn × [0, T ]) as ε→ 0.

25


Proof. Firstly, since u0(x) ∈ BUC(Rn), there exists a sequence (uk0)k∈N ⊂ C1(Rn) ∩

W 1,∞(Rn) such that uk0 → u0 in BUC(Rn) as k →∞ and ‖uk0‖∞ ≤ ‖u0‖∞, ∀k ∈ N.

On the other hand, since Theorem 1.4, for each k ∈ N, the equation

∂w

∂t+H(x,

x

ε,Dw) = 0, in Rn × (0,+∞), (3.5)

w(x, 0) = uk0(x), (3.6)

has a unique viscosity solution uεk in W 1,∞(Rn × (0,+∞)).

Besides, since Theorem 1.4, there exist Ck, RCk > 0 such that uεk(x, t) is Ck-Lipschitz in t

and RCk -Lipschitz in x.

Hence, for all (x, t), (y, s) ∈ Rn × [0, T ], one has

|uεk(x, t)− uεk(y, s)| ≤ Ck|t− s|+RCk |x− y|, (3.7)

where Ck, RCk are independent on ε.

Thus, uεk is equi-continuous.

On the other hand, since (1.37), for all (x, t) ∈ Rn × [0, T ], one has

|uεk(x, t)| ≤ ‖u0‖∞ + CkT, ∀(x, t) ∈ Rn × [0, T ]. (3.8)

We obtain that uεk is equi-bounded in Rn × [0, T ].

Applying Ascoli’s Theorem, there exists a subsequence (uεjk ) of (uεk) such that u

εjk →

uk in C(Rn × [0, T ]).

Omit the notation of k, j. We consider the cell problem

H(x, p) = H(x, y, p+Dyv(y)) in Rn, (3.9)

where (x, p) ∈ Rn × Rn 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) = u0(x). (3.11)

Proof.

Lemma 3.1. u is a subsolution of (3.10).


Proof. Let φ ∈ C1(Rn × [0, T ]) and (x0, t0) is a strict local maximum point of u− φ.

There exists a Zn-periodic, Lipschitz continuous viscosity solution w(y) satisfying

H(x0, y,Dxφ(x0, t0) +Dyw(y)) = H(x0, Dxφ(x0, t0)). (3.12)

We introduce the test-function

Ψε,β(x, t, y) = uε(x, t)− φ(x, t)− εw(y)− 1

β2|y − x

ε|2. (3.13)

Since (x0, t0) 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(x0, t0)− φ(x0, t0), (3.14)

for all x ∈ B(x0, r) and t ∈ B(t0, r′).

On the other hand, the set B(x0, r) × B(t0, r′) × B(x0

ε,r

ε) is a compact set and Ψε,β is

continuous. There exists a maximum point (xε,β, tε,β, yε,β) of Ψε,β such that

uε(x, t)− φ(x, t)− εw(y)− 1

β2|y − x

ε|2 ≤ uε(xε,β, tε,β)− φ(xε,β, tε,β)

−εw(yε,β)− 1

β2|yε,β −

xε,βε|2. (3.15)

for all x ∈ B(x0, r), t ∈ B(t0, r′) and y ∈ B(x0

ε,r

ε).

Choose y =x

ε, then for all (x, t) ∈ B(x0, r)×B(t0, r′), we have

uε(x, t)−φ(x, t)− εw(x

ε) ≤ uε(xε,β, tε,β)−φ(xε,β, tε,β)− εw(yε,β)− 1

β2|yε,β −

xε,βε|2. (3.16)

Thus, there exists a constant C > 0 such that

|yε,β −xε,βε| ≤ C1/2β. (3.17)

Besides, one has B(x0, r) × B(t0, r′) is a compact set, we can suppose that xε,β → x1 and

tε,β → t1 as (ε, β)→ 0, where (x1, t1) ∈ B(x0, r)×B(t0, r′).

On the other hand, since w(y) is bounded and uε → u in C(Rn × [0, T ]) as ε → 0, let

(ε, β)→ 0 in (3.16), one gets

u(x, t)− φ(x, t) ≤ u(x1, t1)− φ(x1, t1), ∀(x, t) ∈ B(x0, r)×B(t0, r′). (3.18)

It implies that (x1, t1) is a maximum point of u−φ on B(x0, r)×B(t0, r′). But, since (x0, t0)

is a strict maximum point of u−φ on B(x0, r)×B(t0, r′), we conclude that (x1, t1) = (x0, t0).


It implies that (xε,β, tε,β) ∈ B(x0, r)×B(t0, r′). Unless, by the convergence of (xε,β, tε,β) to

(x0, t0) as (ε, β)→ 0, for all α > 0, there exist δ(0, α) and δ′(0, α) such that

r = |xε,β − x0| < α, ∀ε, β < δ(0, α), (3.19)

r′ = |tε,β − t0| < α, ∀ε, β < δ′(0, α). (3.20)

Thus, r = r′ = 0 is the contradiction.

Besides, since uε is a subsolution of (3.1)-(3.2), we have

∂φ

∂t(xε,β, tε,β) +H(xε,β,

xε,βε,Dxφ(xε,β, tε,β)− pε,β) ≤ 0, (3.21)

where pε,β =2

εβ2(yε,β −

xε,βε

).

On the other hand, w is a supersolution of (3.12), one gets

H(x0, yε,β, Dxφ(x0, t0)− pε,β)−H(x0, Dxφ(x0, t0)) ≥ 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

|Dxφ(xε,β, tε,β)− pε,β| < C. (3.23)

Hence, since H is uniformly continuous on Rn × Rn ×B(0, C), we deduce

|H(x, y, p)−H(x′, y′, p′)| ≤ mCH(|x− x′|+ |y − y′|+ |p− p′|), (3.24)

for all (x, y, p), (x′, y′, p′) ∈ Rn × Rn ×B(0, C) where mCH(t)→ 0 as t→ 0.

Subtracting two sides of (3.21), (3.22) and applying (3.24), one has

∂φ

∂t(xε,β, tε,β) +H(x0, q

′) ≤ H(x0, yε,β, q′ − pε,β)−H(xε,β,

xε,βε, q − pε,β), (3.25)

where q = Dxφ(xε,β, tε,β) and q′ = Dxφ(x0, t0).

On the other hand,

H(x0, yε,β, q′ − pε,β)−H(xε,β,

xε,βε, q − pε,β) ≤ mC

H(|xε,β − x0|+ |yε,β −xε,βε|+ |q − q′|).

And since φ ∈ C1(Rn × [0, T ]) and (3.17), let (ε, β)→ 0, we obtain

∂φ

∂t(x0, t0) +H(x0, Dxφ(x0, t0)) ≤ 0. (3.26)

Therefore, u is a subsolution of (3.10).


Lemma 3.2. u is a supersolution of (3.10).

Proof. Let φ ∈ C1(Rn × [0, T ]) and (x0, t0) is a strict local minimum point of u− φ.

There exists a Zn-periodic Lipschitz continuous, viscosity solution w(y) statisfies

H(x0, y,Dxφ(x0, t0) +Dyw(y)) = H(x0, Dxφ(x0, t0)). (3.27)

We introduce the test-function

Ψε,β(x, t, y) = uε(x, t)− φ(x, t)− εw(y) +1

β2|y − x

ε|2. (3.28)

Since (x0, t0) is a strict local minimum point of u− φ, there is r > 0 and r′ > 0 such that

u(x, t)− φ(x, t) > u(x0, t0)− φ(x0, t0), (3.29)

for all x ∈ B(x0, r) and t ∈ B(t0, r′).

On the other hand, the set B(x0, r) × B(t0, r′) × B(x0

ε,r

ε) is a compact set and Ψε,β is

continuous. There exists a minimum point (xε,β, tε,β, yε,β) of Ψε,β such that

uε(x, t)− φ(x, t)− εw(y) +1

β2|y − x

ε|2 ≥ uε(xε,β, tε,β)− φ(xε,β, tε,β)

−εw(yε,β) +1

β2|yε,β −

xε,βε|2. (3.30)

for all x ∈ B(x0, r), t ∈ B(t0, r′) and y ∈ B(x0

ε,r

ε).

Choose y =x

ε, then for all (x, t) ∈ B(x0, r)×B(t0, r′), we have

uε(x, t)−φ(x, t)− εw(x

ε) ≥ uε(xε,β, tε,β)−φ(xε,β, tε,β)− εw(yε,β) +

1

β2|yε,β −

xε,βε|2. (3.31)

Thus, there exists a constant C > 0 such that

|yε,β −xε,βε| ≤ C1/2β. (3.32)

Besides, one has B(x0, r) × B(t0, r′) is a compact set, we can suppose that xε,β → x1 and

tε,β → t1 as (ε, β)→ 0, where (x1, t1) ∈ B(x0, r)×B(t0, r′).

On the other hand, since w(y) is bounded and uε → u in C(Rn × [0, T ]) as ε → 0, let

(ε, β)→ 0 in (3.31), one gets

u(x, t)− φ(x, t) ≥ u(x1, t1)− φ(x1, t1), ∀(x, t) ∈ B(x0, r)×B(t0, r′). (3.33)

It implies that (x1, t1) is a minimum point of u−φ on B(x0, r)×B(t0, r′). But, since (x0, t0)

is a strict minimum point of u−φ on B(x0, r)×B(t0, r′), we conclude that (x1, t1) = (x0, t0).


The same argument in the proof of subsolution, one has (xε,β, tε,β) ∈ B(x0, r) × B(t0, r′).

Besides, since uε is a supersolution of (3.1)-(3.2), we have

∂φ

∂t(xε,β, tε,β) +H(xε,β,

xε,βε,Dxφ(xε,β, tε,β) + pε,β) ≥ 0, (3.34)

where pε,β =2

εβ2(yε,β −

xε,βε

).

On the other hand, w is a subsolution of (3.12), one gets

H(x0, yε,β, Dxφ(x0, t0) + pε,β)−H(x0, Dxφ(x0, t0)) ≤ 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

|Dxφ(x0, t0) + pε,β| < C. (3.36)

Hence, since H is uniformly continuous on Rn × Rn ×B(0, C), we deduce

|H(x, y, p)−H(x′, y′, p′)| ≤ mCH(|x− x′|+ |y − y′|+ |p− p′|), (3.37)

for all (x, y, p), (x′, y′, p′) ∈ Rn × Rn ×B(0, C) where mCH(t)→ 0 as t→ 0.

Subtracting two sides of (3.34), (3.35) and applying (3.37), one has

∂φ

∂t(xε,β, tε,β) +H(x0, q

′) ≥ H(x0, yε,β, q′ + pε,β)−H(xε,β,

xε,βε, q + pε,β), (3.38)

where q = Dxφ(xε,β, tε,β) and q′ = Dxφ(x0, t0).

On the other hand,

−mCH(|xε,β − x0|+ |yε,β −

xε,βε|+ |q − q′|) ≤ H(x0, yε,β, q

′ + pε,β)−H(xε,β,xε,βε, q + pε,β).

Since φ ∈ C1(Rn × [0, T ]) and (3.32), let (ε, β)→ 0, we obtain

∂φ

∂t(x0, t0) +H(x0, Dxφ(x0, t0)) ≥ 0. (3.39)

Therefore, u is a supersolution of (3.10).

Hence, u is a viscosity solution of (3.10) and by Theorem 1.2, u is unique.

Remark 3.1. For k ∈ N, let (uεjk ) be an arbitrary subsequence of (uεk) such that u

εjk → vk

in C(Rn × [0, T ]).

Hence, since Theorem 3.10, vk is a viscosity solution of

∂v

∂t+H(x,Dv) = 0 in Rn × (0,+∞), (3.40)

v(x, 0) = uk0(x). (3.41)


On the other hand, since the uniqueness of uk, we have that uεjk → uk in C(Rn × [0, T ]) for

all (εj) ⊂ (ε) as εj → 0.

Moreover, by Ascoli’s Theorem, uεk is bounded on a compact set of Rn× [0, T ]. We conclude

that uεk → uk as ε→ 0.

On the other hand, since uk is the unique viscosity solution of the equation

∂u

∂t+H(x,Du) = 0 in Rn × (0,+∞), (3.42)

u(x, 0) = uk0(x). (3.43)

Set Nk(x, t) = uk0(x)−Ckt and Mk(x, t) = uk0(x)+Ct, where C = sup|q|≤‖Duk0‖∞‖H(·, q)‖∞.

One has

C +H(x,Duk0) ≥ 0, ∀x ∈ Rn (3.44)

−C +H(x,Duk0) ≤ 0, ∀x ∈ Rn. (3.45)

Hence, Nk(x, t) and Mk(x, t) are respectively subsolution and supersolution of (3.42). And

then, by the comparison result, we have

Nk(x, t) ≤ uk(x, t) ≤Mk(x, t), ∀(x, t) ∈ Rn × [0, T ]. (3.46)

By Corollary 1.1, for m, k ∈ N,m ≤ k, since um, uk are respectively supersolution and

subsolution of (3.42), one has

supRn×[0,T ]

(uk(x, t)− um(x, t)) ≤ supRn

(uk(x, 0)− um(x, 0))

≤ supRn

(uk0(x)− um0 (x))

≤ ‖uk0 − um0 ‖∞

≤ ‖uk0 − u0‖∞ + ‖um0 − u0‖∞. (3.47)

Hence,

‖uk − um‖∞ ≤ ‖uk0 − u0‖∞ + ‖um0 − u0‖∞. (3.48)

Therefore, (uk)k∈N is a Cauchy sequence in BUC(Rn×[0, T ]). Thus, uk uniformly converges

to u ∈ BUC(Rn × [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, uεk, uεm are respectively subsolution

and supersolution of (3.5). Then, by the comparison result, one has

‖uεk − uεm‖∞ ≤ ‖uk0 − u0‖∞ + ‖um0 − u0‖∞. (3.49)


Hence, uεk → uε in BUC(Rn × [0, T ]) as k → ∞. Therefore, since Theorem 1.1, uε is a

viscosity solution of (3.1)-(3.2).

Finally, for all ε > 0, since uk0 → u0 and uk → u in BUC(Rn × [0, T ]), one has

∃K1(ε) ∈ N, ‖uk0 − u0‖∞ < ε,∀k ≥ K1(ε), (3.50)

∃K2(ε) ∈ N, ‖uk − u‖∞ < ε,∀k ≥ K2(ε). (3.51)

Let k ≥ max(K1(ε),K2(ε)), we have

‖uε − u‖∞ ≤ ‖uε − uεk‖∞ + ‖uεk − uk‖∞ + ‖uk − u‖∞

≤ ‖u0 − uk0‖∞ + ‖uεk − uk‖∞ + ‖uk − u‖∞

≤ ε+ ‖uεk − uk‖∞ + ε. (3.52)

Sending ε→ 0, we obtain uε → u in BUC(Rn × [0, T ]).

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. 1-42.

[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. 487-504.

[4] M.G. Crandall, H. Ishii and P.L. Lions, User’s guide to viscosity solutions of second-

order partial differential equations, Bull. AMS 27 (1992), 1-67.

[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), 359-375.

[7] A.Bensoussan, J.L Lions and G. Papanicolaou, Assymptotic analysis for periodic struc-

tures, North-Holland, Amsterdam, 1978.

[8] R.A. Adams, Sobolev Spaces, Academic Press, NewYork, 1975.

[9] H. Brezis, Analyse fonctionnelle. Theorie et Applications, Masson Paris, 1983.

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Documents

Master Thesis Nguyen Tien Thinh - This problem was introduced in [5] by P.L. Lions, G. Papanicolaou and S.R.S. Varadhan in 1986. ... Lions in 1983 regarding the Hamilton-Jacobi equation