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

Definition 1 Definition of viscosity solution for the static equation.. Using the notions of sub-differential and super-differential, we will give a geometric descriptions for viscosity

NOTES AND REMARKS

SON NGUYEN THAI TU

Department of Mathematics, University of Sciences at Ho Chi Minh City

SON PHUNG TRUONG VAN

Department of Mathematical Sciences, Carnegie Mellon University

A BSTRACT These notes contain the essential materials and some further remarks of the summer course "Homogenization of Hamilton-Jacobi equations" taught by Prof Hung V Tran1at University of Science, Ho Chi Minh City, Viet Nam from July 6 th to July 14 th , 2015.

E-mail addresses:thaison.tn93@gmail.com, sonv@andrew.cmu.edu.

1 Department of Mathematics, University of Wisconsin at Madison

1

• Tn = Rn /Z n is the n-dimensional donut (torus).

• w.r.t: with respect to

L ECTURE 1

Clear contest : ǫ > 0 : scale of the equation, ǫ ≪ 1 : ǫ is very small.

Equation of interest Following the Hamilton-Jacobi equation

• H is the Hamiltonian (total energy)

Question We want to understand (I) as ǫ −→ 0 Is there some u such that u ǫ −→ u and is u solve something simpler? This is one of the key questions in homogenization

Formal Analysis (not rigorous) Consider the equation

• Numerical analysis: plot u ǫ

• Guessing: some asymptotic expansion w.r.t ǫ.

We consider some ansatzs2

Plugging these into (I) we have

‹

= 0

Denote y = x ǫ, we obtain

”

u ǫ t (x, t) + ǫu1t y, t
+ —+ H€y, Du(x, t) + Du1 y, t
+ ǫDu2 y, t
+ Š= 0

We will assume a “big lie": x and y are independent Matching asymptotic

expan-sion zero-order term O(1), we get

Let p = Du(x, t)∈ Rn, then

H( y, p + Du1( y, t)) = C(p, t).

Assume one more reduction u1( y, t) = u1( y), we then have the following PDE

H( y, p + Du1( y)) = C(p).

This is called the cell (or ergodic) problem

Theorem 1 (Lions-Pappanicolau-Varadhan) Fix p∈ Rn , there exists a unique constant

C ∈ Rn such that the cell problem

H( y, p + Du1( y)) = C(p) in R n has a periodic solution u1.

Define the effective Hamiltonian H(p) = C(p) We are back to our original question:

Remark 1 This is a natural question since from the Hamiltonian H( x ǫ , Du ǫ ) we derived the effective Hamiltonian H that is independent of ǫ so as we pass ǫ → 0, (C ǫ ) should become (HJ) The answer to this problem is “yes” and will be elaborated in lecture 3.

L ECTURE 2

Before we prove theorem 1, let’s talk about viscosity solutions

Introduction to viscosity solutions Consider

• The static problem

We then look into the static problem in Rn The idea here is that the problem

(S ǫ) u ǫ + H(Du ǫ ( y), y) = ǫ∆u ǫin Rn

has a smooth solution u ǫ Assume further that u ǫ

→ u locally uniformly in R n Take asmooth functionφ such that

¨

(u − φ)(x0) = 0

(u − φ)(x) < 0 else where.

That is, u − φ has a strict max at x0

Exercise 1 For ǫ > 0 small enough, u ǫ − φ has a max at x ǫ near by x0 and there is a subsequence ǫ j −→ 0 such that x ǫ j −→ x0. 3

Play with u ǫ Since u ǫ

−φ has max at x ǫ, we then have, of course by second derivativetest,

u ǫ (x ǫ ) + H(x ǫ , Du ǫ (x ǫ )) = ǫ∆u ǫ (x ǫ)

From (1), we can see that

u ǫ (x ǫ ) + H(x ǫ , Dφ(x ǫ )) ≤ ǫ∆φ(x ǫ)

3Morever, if u mis continuous andφ is smooth such that u m −→ u uniformly on a compact set Ω ⊂ R n,

and assume u − φ has a strict (local) maximum over Ω as x0, u m − φ has (local) maximum over Ω at

x m Then we must have x m −→ x0as m−→ ∞ We have the convegence of whole sequence here, while

in case u ǫ −→ u uniformly as ǫ −→ 0 we only have the convergence of sub-sequence.

Letǫ j → 0, we then have

u(x0) + H(x0, Dφ(x0)) ≤ 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

con-tinuous on its domain

• (Subsolution) u is called a viscosity subsolution if for all φ ∈ C∞(Rn ) such that (u − φ)(x0) is a strict max then

u(x0) + H(x0, Dφ(x0)) ≤ 0

• (Supersolution) u is called a viscosity supersolution if for all φ ∈ C∞(Rn ) such that (u − φ)(x0) is a strict min then

u(x0) + H(x0, Dφ(x0)) ≥ 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 2 Let u be a real valued function defined on the open set Ω ⊂ Rn For any

are called, respectively the subdifferential and superdifferential of u at x.

We can see that p ∈ D+u(x0) if u(x) ≤ u(x0) + p · (x − x0) for all x ∈ B(x0, r).

x0p

u

Similarly, p ∈ D−u(x) if u(x) ≥ u(x0) + p · (x − x0) for all x ∈ B(x0, r).

u

From this, we can see that, up to a constant,

u − φ has a strict max at x0⇐⇒ u is touched from above by φ at x0

φ

In fact, we have the following properties:

Proposition 1 (Properties of sub-differentials and super-differentials).

Let f : Ω⊂ Rn

−→ R 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 ∈ C1(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 ∈ C1(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)6= ;} Ω−= {x ∈ Ω : D−f (x)6= ;}

are both non-empty Indeed, they are dense in Ω.

(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 functionσ : [0, ∞) −→ R with σ(0) = 0 such that

For the converse, if ϕ ∈ C1

(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

(e) If f is differentiable at x, then clearly ∇ f (x) ∈ D+f (x) ∩ D−f (x) more, if p ∈ D+f (x), then there exists ϕ ∈ C1(Ω) such that

Further-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 sume a ∈ D+f (x) and b ∈ D−f (x), then there exists ϕ, ψ ∈ C1(Ω) such that

As-ϕ(x) = ψ(x) = f (x) and

¨

∇ϕ(x) = p f − ϕ has local maximum at x

∇ψ(x) = q f − ψ has local minimum at x .Therefore, in neighborhood B(x, δ) we have

Furthermore, for every x0∈ Ω and ǫ > 0 so small enough, the set Ω+ contains

a point y ∈ B(x0,ǫ) This shows that Ω+is dense in Ω

Similarly, if we consider the C1(B(x0,ǫ)) function given by

ǫ2− kx − x0k2 x ∈ B(x0,ǫ).

The case Ω− is dense in Ω by a similar argument

From this, we have the equivalent definition for viscosity solution of static equation

as following

Definition 3 (Equivalent definition of viscosity solution for the static equation)

As-sume u is continuous on its domain

• (Subsolution) u is called a viscosity subsolution if

∀ x0∈ Rn u(x0) + H(x0, p)≤ 0 ∀ p ∈ D+u(x0)

• (Supersolution) u is called a viscosity supersolution if

∀ x0∈ Rn u(x0) + H(x0, p)≥ 0 ∀ p ∈ D−u(x0)

• 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

where F : Ω× R × Rn

−→ R is a continuous function from an open set Ω ⊂ R 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 2 The basic questions about well-posedness of PDE in sense of viscosity solution,

the existence, uniqueness and stability were solved by Evans, Crandall-Lions in 1970 for the equation

H( y, Du ǫ ( y)) = ǫ∆u ǫ ( y).

by the method adding small viscosity.

Example 2 (Eikonal’s equation) Let H denote the Hamiltonian, H(u, p) = |p| in R, consider the Eikonal’s equations

We can see that there are infinite many solutions (in weak sense) of this problem, but there is only one correct visosity solution.

−0.2 0.2 0.4 0.6 0.8 1.

−0.4

−0.2

0.2 0.4

0

correct viscosity solution

1 2

Finding a viscosity solution We already know this equation has infinitely many

solu-tions Consider a solution u that its graph contains aW shape, for example

It’s contradiction

The only u which is not contain this shape is

0.5 1.

0.5

0

D+u 12

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 2-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∈ Rn , there exists a unique c∈ R such that

(E p) H( y, p + Dv( y)) = c in Rn

has a periodic solution v.

Proof of theorem 1 We prove through some steps.

Discounted approximation Forǫ > 0, consider the following problem

ǫv ǫ ( y) + H( y, p + Dv ǫ ( y)) = 0 in R n

Claim without proof: There exists a unique viscosity solution v ǫ ( y) to the above

prob-lem

1st observation: if v ǫ ( y) is a solution then v ǫ ( y + k) for k ∈ Z n is also a solution by

periodicity of H Thus v ǫ ( y) is periodic, i.e y ǫ ( y) = v ǫ ( y + k) by the uniqueness of the solution v ǫ

2nd observation: there are apriori estimates We try to bound v ǫ First of all, take

φ = c1, and observe H( y, p) is periodic, this is bounded, assume |H( y, p)| ≤ C If we

chooseφ = c1= −C

ǫφ + H( y, p + Dφ( y)) = ǫc1+ H( y, p) ≤ ǫc1+ C = 0.

ǫ for C > 0 large is a supersolution.

Therefore since v ǫis a solution, we get

Now by coercivity of H, we must have |Dv ǫ

( y)| ≤ C′for some constant C′> 0.

3rd observation:{v ǫ } is equi-Lipschitz and v ǫis periodic Since

|v ǫ (x) − v ǫ ( y)| ≤ |Dv ǫ (ζ)| · |x − y| ≤ C′|x − y|

for someζ ∈ [x, y] = {t x + (1 − t) y : t ∈ [0, 1]} and clearly C′is independent toǫ.

In other words, for anyǫ > 0

v ǫ∈ Lip(Tn) and kDv ǫkL∞ (Tn)≤ C′

Remark 4 At this point we want to use Arzela-Ascoli theorem However, v ǫ aren’t bounded as ǫ → 0 so we can’t get the uniform bound for every ǫ thus Arzela-Ascoli cannot be applied here Luckily, we have something lurking underneath so Arzela-Ascoli theorem could be applied.

Define

w ǫ ( y) = v ǫ ( y) − v ǫ(0)

We have|Dw ǫ ( y)| = |Dv ǫ ( y)| ≤ C′and so

|w ǫ ( y)| = |v ǫ ( y) − v ǫ (0)| ≤ C′| y| ≤ C′pn

since Tn is bounded Thus w ǫis bounded and we can apply the Arzela-Ascoli theorem

to{w ǫ }, there exists ǫ j −→ 0 such that as j −→ ∞,

(4) w ǫ j ( y) −→ w( y) uniformly in Tn

Since|ǫv ǫ

(0)| ≤ C, by Bozalno-Weitrass’s principle, we can assume (passing to

subse-quence if necessary) thatǫ j v ǫ j (0) −→ −c for some constant c ∈ R We claim

further-more that

ǫ j v ǫ j ( y) → −c as j−→ ∞

in the viscosity sense (We don’t have Dw ǫ j → Dw.) Indeed,

• Sub-test: Fix x ∈ R n and letϕ ∈ C∞(Rn) such that

Thus, w is a viscosity subsolution of (6).

• Super-test: Fix x ∈ R n and letϕ ∈ C∞(Rn) such that

¨

(w − ϕ)(x) = 0

w − ϕ > 0 else where i.e, w − ϕ has a strict min at x .

Since w ǫ j −→ w uniformly on some compact neighborhood of x, there exists

a subsequence{ j k} such that

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 (v1, c1) and (v2, c2) where c1 < c2

Therefore, by comparison principle, v1≤ v2 However, we have that both v1and v2 are

bounded; and if v1 is a subsolution then so is v1+ C So, for large enough C we have

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 2 Let n = 1, H(x, p) = |p| − W (x) where W is defined by its graph

x y

0

1 2

1 4

3 4

W

That is W : R −→ R is periodic Consider the ergodic cell problem (E0)

(E0) H( y, 0 + Dv( y)) = |Dv( y)| − W ( y) = c in T.

(a) Find the unique constant c such that (E0) has solution.

(b) Find an infinite number of solutions for (E0) for all y ∈ Tn

Proof of theorem 2 Using above priori estimate, we deduce by Arzela-Ascoli theorem

there exists a sequence {ǫ j } ց 0 such that u ǫ j

→ u locally uniformly We are left to show u solves (C).

Sub-test One might be tempted to take the following approach.

Failed attempt Letφ ∈ C∞ and (x0, t0) ∈ Rn

Heuristic proof - Assume everything is smooth. Let p0= Dφ(x0, t0) ∈ Rn Take v0

to be a (Lipschitz, periodic) solution of

(7) H( y, p0+ Dv( y)) = H(p0) = H(Dφ(x0, t0)) ∀ y ∈ R n

A very important observation is that v0is bounded by its periodicity and Dv0is bounded

the coercivity assumption of H From this, we also have the function

(x, t) 7−→ u ǫ (x, t) − ǫv0

x ǫ

‹

converges uniformly to u as ǫ −→ 0 Thus by exercise 1 we obtain there exists a

sequence{ǫ j} ց 0 such that

has a max at (x j , t j ) near (x0, t0)

and (x j , t j ) −→ (x0, t0) as j −→ ∞ The function u ǫ j is called the perturbed testfunction So, by the definition of viscosity subsolutons, we have5

as j−→ ∞ So, combine the 2 facts we obtain

Based on the idea above plus another technique called doubling of variables, we

have a real proof

Rigorous proof Recall u ǫ j −→ u locally uniformly on R 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 (x0, t0) We want to show that

We have u ǫ j −→ u uniformly on Ω Let v0 be the periodic Lipschitz viscosity solution

of the cell problem

x

ǫ j − y

2

Since v0 is bounded, the function Φǫ jη is continuous and bounded above

Step 0 There exists€x jη , y jη , t jηŠ∈ Ω×Rn ×[0, T ] such that Φ ǫ j,ηattains its maximumthere Indeed, let

x mk

ǫ j − y m k

y m

k

−

η ≥ ǫ j v0



y j η

+

x jη

ǫ j − y j η

x jη

ǫ j − y j η

Since v0 is bounded, letη −→ 0 and using the fact that

−→ u uniformly on Ω × [0, T ] Thus by exercise 1 we must have (maybe some

subsequence)€x j , t jŠ −→ x0, t0
as j −→ ∞ since u − φ has strict maximum over

Step 3 Fix y = y j η then the function

(x, t) 7−→ u ǫ j (x, t) − φ(x, t) −

x

ǫ j − y j η

...

To see that, we observe that

ǫ j v0 y
+

x... jηŠ∈ Ω×Rn ×[0, T ] such that Φ ǫ j,ηattains its maximumthere Indeed, let

x... j

Note, also, that y jη≈ x jη

ǫ j because of the penality effect asη gets small.