gomez d.a., valdinoci e. - generalized mather problem and homogenization of hamilton-jacobi...

7/25/2019 Gomez D.A., Valdinoci E. - Generalized Mather problem and homogenization of Hamilton-Jacobi equations(23).pdf

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GENERALIZED MATHER PROBLEM

AND HOMOGENIZATION

OF HAMILTON-JACOBI EQUATIONS

DIOGO A. GOMES & ENRICO VALDINOCI

Abstract. We present a general approach to the homogenization of Hamilton-Jacobiequations in the stationary ergodic setting through infinite dimensional linear pro-gramming and generalized Aubry-Mather theory.

Contents

1. Introduction 12. The main result 23. List of assumptions 34. Optimal control and linear programming 95. Moment estimates 146. Duality 167. Stationarity and Effective Lagrangian 24

8. Homogenized problems 309. Uniform estimates 3210. Convergence 3611. Improved Moment Estimates 3912. Viscosity Solutions 4613. Conclusion 50References 50

DG was supported in part by CAMGSD - FCT/POCTI/FEDER,POCI/FEDER/MAT/55745/2004 and

EV was supported in part by MIUR Variational Methods and Nonlinear DifferentialEquations

1. Introduction

The objective of this paper is to present a general approach to the homogenizationof random stationary ergodic Hamilton-Jacobi equations arising in deterministic andstochastic optimal control. In our approach we will use infinite dimensional linearprogramming and the generalized Aubry-Mather theory.

1


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2 DIOGO A. GOMES & ENRICO VALDINOCI

Our setting is the following. Let Mat(nn) be the space of (n n) real matrices and be a probability space. We consider a Hamiltonian

H(M,p,x,X,,) : Mat(n n) Rn Rn Rn R+ R.

We will study the homogenization properties of the Hamilton-Jacobi equation

(1)

tu

+ H

D2xu, Dxu

, x,x

, ,

= 0

u(x , T, ) = (x) ,

as 0. Here above, u(x,t,) : Rn [0, T] R and is a small, positiveparameter and (x,X,) : Rn Rn R is assumed to be a bounded smoothfunction of the variables (x, X), and to satisfy further additional stationarity hypothe-ses in . We will prove that, under suitable assumptions, there exists a non-random

function u(x, t) such that u u. Furthermore, there is a function H(M, p , x), theso-called Effective Hamiltonian, non-random, such that uis a viscosity solution of

(2)

tu + H

D2xu, Dxu, x

= 0

u(x, T) = (x) .

Our strategy to study this homogenization problem is the following: first we establishsome general properties of the original problem by converting it into a linear program-ming problem. Then, we introduce a homogenized problem. Finally, we prove theconvergence of the sequence u to a non-random limit and show that this limit agreeswith the homogenized problem.

To accomplish these goals, we will take a series of assumptions that will be put forwardin section 3.Homogenization problems have been quite popular in recent years, and many steps havebeen made towards the understanding of these problems (see, e.g., [BLP76, LPV86,Eva92, Con96, Sou99, Alv99, Ish99, Ish00, RT00, CDI01, EG01, AB02, LS03, Gom03,LS05, KRV06, Kos06]). Several author, see [KS, CS02, MS98, GR06] have used linearprogramming techniques to study deterministic and stochastic optimal control prob-lems.We should point out that there were some known connections between homogenizationtheory and Aubry-Mather theory (see, for instance, [LS03]), and that convex analysisand duality methods were used in [KRV06] to prove a homogenization result relatedto ours. However, to our knowledge, this is first paper in which this relation is workedout systematically, specially in what concerns second-order equations.

2. The main result

We now state our main result. The assumptions under which it holds are very generaland, for the convenience of the reader, their detailed statement will be postponed tosection 3.

Theorem 1. Let Assumptions 115 of section 3 hold and letu be a viscosity solutionof (1). Then:


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wherevis a bounded progressively measurable control, lying in a closed convex set UR

m, which, without loss of generality, will be assumed to contain the origin 0,

f : Rn Rn U Rn

is the drift coefficient,

: Rn Rn R+ U Mat(n p)

is the diffusion coefficient, and Wt is an p-dimensional Brownian motion defined in aprobability space which is independent of . The probability on will be denotedbyP, andEwill denote the corresponding expected value. We assume enough regular-ity and growth conditions in f and such that the stochastic differential equation (3)is well defined for all bounded progressively measurable controls v.The running cost of this control problem is given by the Lagrangian

L(x,X,v,) : Rn Rn U R,

where U Rm is the control space, which is taken to be closed and convex. Wesuppose also that L is continuous in (x,X,v) and bounded from below. Without lossof generality, we will assume that infL = 0. Furthermore, if U is unbounded wesuppose that

(4) limR+

inf(x,X,)|v|R

L(x,X,v,)

|v| = + .

For simplicity, we assume that the terminal cost is a bounded C function with

bounded derivatives (less regular data may be treated as well).We suppose that, for T >0, equation (1) possesses, for any fixed , one and onlyone viscosity solution u defined for all t [0, T], which is the value function of thestochastic control problem, that is, it can be represented in the form

u(x,t,) =

infE

Tt

L

x(s),

x(s)

, v(s),

ds + (x(T))

,

(5)

where the infimum is taken over all trajectories of the controlled dynamics (3) withinitial condition x(t) =x.

We refer to [FR75, Kry80, FS06, Eva06a] for further discussions about the stochasticversion of Hamilton-Jacobi equations.The Hamiltonian

H(M,p,z,x,X,,) : Mat(n n) Rn Rn Rn R R

is related with the Lagrangian by duality using the generalized Legendre transform

H(M,p,x,X,,)

= supvU

f(x,X,v,) p T(x,X,,v,)

2 :M L(x,X,v,) .

(6)


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GENERALIZED MATHER PROBLEM AND HOMOGENIZATION 5

Here and in the sequel, we use the following notation: given A, BMat(n n), we set

A: B = tr A

T

B,ifA and B are symmetric, and we will use the convention that for general matrices A :B = 1

4tr (AT + A)(B+ BT). Note that (6) implies that His decreasing in M, that is,

ifNMat(n n) is symmetric and non-negatively definite, then

(7) H(M+ N,p,x,X,,) H(M,p,x,X,,) .

Assumption 2. We suppose that

(8) supy

sup>0

supx,

HD2(x), D(x), x y,x

, , 0and p >1.

Assumption 5. We assume that L(x,X, 0, ) is uniformly bounded.

As a notational remark, we point out that, in most of this paper, we think xto be thefrozen initial datum of the controlled dynamics (and so the frozen variable ofu),while x will denote the variable of the integration of the measure we will construct.The next assumption is quite natural and requires that the viscosity solutions can beapproximated by smooth subsolutions. This assumption will be required from section 6onwards.


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Assumption 6. For any >0 there exists v C2(Rn [t, T] ) satisfying

(9) tv

(x,t,) H

D2

xv

(x,t,), Dxv

(x,t,),

x

,

+ 0 ,v(x , T, ) =(x), and

(10) lim0+

v =u

uniformly in (x,t,) Rn [t, T] . We further suppose that D2v is uniformlybounded in for fixed .

In many applications, the smoothing process of section V.7 in [FS06] may be used tobuilt semiconcave/convex sub/supersolutions.To obtain a non-random Effective Hamiltonian, one must impose certain ergodicityconditions on the data of the problem. So far, we have made no assumptions on thebehavior of L, f or with respect to the random parameter . It turns out thata convenient hypothesis is stationarity, as we put forward next. For the purposes ofsection 7, the next assumption is enough, but later we will also require certain ergodicityhypothesis to obtain a non-random homogenized limit, see Assumption 13 below.

Assumption 7. We assume that there is an actionofRn in the probability space ,denoted by X, for X R

n, which preserves the measure in . Furthermore, wesuppose that the L, f and are stationary with respect to this action, that is, forallY Rn

(11) L(x, X+ Y , v , ) =L(x,X,v, Y),

(12) f(x, X+ Y , v , ) =f(x,X,v, Y),

and

(13) (x, X+ Y , , v, ) =(x,X,,v, Y).

As an example, note that the periodic setting can be embedded in this framework in thefollowing way: we let = Tn, and consider in the (normalized) Lebesgue measure.

Then, ifL(x,X,v), f(x,X,v) and (x,X,v) are periodic in X, we can write

L(x,X,v,) =L(x, X+ , v),

with similar definitions for f and , and then we choose Y = + Y.Next, we state the concept of tight sequence of measures, by generalizing a standarddefinition in probability theory (see, e.g., page 336 of [Bil95]):

Definition 1. A net of probability measures on a given spaceZ , withZ Rd

is said to be tight if for any >0 there exists a compact setK in such a way that

lim sup

(Z\ K)

.

The following assumption is simply a compactness property on measures on that isnecessary to extract weak limits.


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Assumption 8. We assume that for each tight net of probability measures (z, )on Z , with Z Rd, there exists a subnet which converges to a probability mea-

sure(z, ), in the sense that: if(X, z, ) : RdZ R is continuous in (X, z) andstationary (that is,(X+Y , z , ) =(X, z, Y) for any (X, Y , z, ) R

dRdZ),we have that

lim

Z

(z, 0, ) d(z, ) =

Z

(z, 0, ) d(z, ) ,

up to subnets.

In section 7 we will use the following scaling hypothesis for the diffusion coefficient

Assumption 9. We will assume that the viscosity coefficient is either

A. = 0B. = 1/20C. = 0,

with 0 =0(x, X , v , ) : Rn Rn U Mat(n m).

We should remark that these are the relevant scalings, as many others could be studiedwith similar methods.From section 9 on, we will assume that our viscosity solutions are smooth, up to asmall uniformly elliptic correction of the Hamiltonian. That is, we suppose:

Assumption 10. For any N, there exist u C2(Rn (0, T)), C

2(Rn) and aHamiltonian H such that u

is a solution of the following Hamilton-Jacobi equation:

tu+ H

D2xu

, Dxu

, x,

x

, ,

= 0

u(x , T, ) = (x) ,

and in such a way that

lim+

u(x, t) =u(x, t)

lim+

(x) =(x) ,

and lim+

supx

H

D2x(x), Dx(x), x,

x

, ,

H

D2x(x), Dx(x), x,x

, , = 0

for anyx Rn, t [0, T] and C2(Rn) with bounded C2-norm. Also, H satisfies

(14) DMijH i j c ||2 ,

for some c > 0.We also assume that

(15) supxRn, t[0,T]

|Du| < + .


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A natural possibility for such Hamiltonian H is of course H tr M/, but otherregularizations are possible (for general regularization results, see, e.g., [Eva82, Eva97,

Wan90, Wan92a, Wan92b, Wan92c, CC95, LS05]).We remark that (15) is not assumed, a-priori, to be uniform in and (in fact, uniformestimates will be proved in Proposition 20 below). In many applications, bounds asin (15) are quite natural and may be obtained by comparing trajectories (see, e.g.,page 555 of [Eva98] and Lemma 8.1 of [FS06]).The next three assumptions will be used in section 9 to prove various uniform estimatesin .

Assumption 11. We assume that Henjoys one of the following properties:

P1) H(M,p,x,X,,) does not depend onMand satisfies the coercivity hypothesis

(16) lim|p|

H(p, x, X, , ) = +,

uniformly in (x,X,,) Rn Rn R+ .P2) There exists a function F(p, ) such that for all (M,p,x,X,,) Mat(n

n) Rn Rn Rn R+ and Rsuch that

+ H(M,p,x,X,,) = 0

implies

1

DMjkH

1

M,p,x,X,,

MijMik+ piDxiH1

M,p,x,X,,

+piDXiH1

M,p,x,X,, F(p, ),and, furthermore, ifKis a fixed compact set ofRn,

(17) inf K

F(p, ) +,

as |p| +.

Examples of Hamiltonians satisfying P2 are easily obtained, for instance, by consideringthe case := Idn, f :=v ,L := |v|

2/2 U(x,X,). In such a case, (6) yields that

(18) H=1

2tr M+

1

2|p|2 + U(x,X,) ,

which is plainly seen to satisfy P2.The next assumption will be used in section 9 to prove that the limit is non-random.

Assumption 12. We suppose that there exists a constant C independent of suchthat for any (x, z) Rn Rn we haveDxHD2u(z, t), Du(z, t), z, x

, ,

C.

Note that (18) provides an example of an Hamiltonian satisfying Assumption 12, dueto Proposition 20 below.


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Assumption 13. We assume the following uniform ergodicity hypothesis: for any >0 there exists M > 0, such that for any 0, 1 , x R

n, t [0, T], there

exists y Rn,|y| M, such thatHD2u(x,t,1), Du(x,t,1), x,x

, , y0

H

D2u(x,t,1), Du

(x,t,1), x,x

, , 1

.

Heuristically, Assumption 13 means we are be able to get close to any point of fromany other by translating, the size of such translation depending only on how close wewant to get (this is the case, for instance, of the geodesics with irrational slope on theflat torus).

From section 11 on, we will also suppose the following:

Assumption 14. We assume that there are constants C0, C1 and C2 such that

|L(x, X, 0, )| C0, |f(x, X, 0, )| C1, |(x

, X, 0, )| C2.

for any (x, X , ) Rn Rn .

Assumption 15. Let p >1 be as in Assumption 4. We suppose that

|f(x, X , v , )|2 C(1 + L(x, X , v , )) , |(x, X, , v, )| C

and

|T

(x

, X, , v, )|p

C(1 + L(x

, X , v , )) ,for any (x, X , v , ) Rn Rn U .

We now deal with the proof of Theorem 1.

4. Optimal control and linear programming

In this section we are going to convert the original optimal control problem (1) intoa linear programming problem. As the random parameter plays no role in most ofthis discussion we omit it until section 7.The controlled stochastic dynamics (3) has an infinitesimal generator (see, e.g., page 105

in [Eva06b]), given by

(19) A(x) =f Dx +T

2 :D2x.

To encode both the behavior at the slow scale x and fast scale X = x

, we need to

introduce the extended generator A, given by

A(x, X) = f Dx+T

2 :D2x

+1

f DX+

1

(T) :D2xX+

1

2T

2 :D2X.


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This extended generator satisfies the following identity

(A)

x,x

= A

x,x

.

Observe further that

(20) A

Xx

= 0 ,

for allC2 functions (x) : Rn R.Before proceeding, we need some further definitions. If U is bounded, set = 1,otherwise, let:U R+ be a function which satisfies

lim|v|

infx,X

L(x,X,v,)

(v) = +,

and

lim|v|

Av(x, X)

(v) = 0,

for all C2(RnRn). Letpbe the exponent for which Assumption 4 holds. Considerthe spaces P and Q of signed1 measures, resp. on Rn Rn U [t, T] and onRn Rn, which satisfy

(v) + |x|p + |X|pd||< ,

|x|p + |X|pd||< .

We remark that Pis the dual of the set C(Rn Rn U[t, T]) of continuous functions: Rn Rn U [t, T] R, which satisfy

lim|x|+|X|+|v|

supt

(x,X,v,t)

(v) + |x|p + |X|p = 0,

and Q is the dual of the set C(Rn Rn) of continuous functions on : Rn Rn R,

which satisfy

lim|x|+|X|

(x, X)

|x|p + |X|p= 0.

We further consider the space C(Rn Rn [t, T]) of continuous functions on :

Rn Rn R, which satisfy

lim|x|+|X|

supt

(x,X,t)

|x|p + |X|p = 0.

In this section, we relax the original stochastic control problem by looking at gen-eralized trajectories represented by certain measures, which will be elements of Pand Q. To do this, we must consider measures that satisfy a generalized holonomy

1All the measures will be tacitly assumed to be Radon. If not differently specified, they are alsoassumed to be positive.


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Proof. Choose (t) such that (t) = (t). Then, from (21), we have that

(s)d=

(T)dT

(t)dt = (T) (t) =

Tt

(s)ds,

yielding (25). Then (26) follows by applying (25) to := 1.

We now prove that both T and are supported on{X= x

}, as long as so is t .

Proposition 3. Suppose t is supported in the set X = x

, and that (, T) sat-

isfy (21). Then bothT and are supported in the setX= x

.

Proof. Let : Rn R+0 be a C2 function such that (0) = 0 and (x)> 0 ifx = 0.

Assume that X x C(Rn Rn). To prove the desired claim, we will show that

if

X

x

dt (x

, X) = 0,

then

X

x

dT(x

, X) = 0,

and

Xx

d(x, X , v , s; ) = 0.

Set (x, X) :=

X x

. Since, by (20),

(27) A= 0 ,

the constraint in (21) implies that

(28)

X

x

dT =

X

x

dt = 0.

Now choose (x, X , t) :=t

X x

. Then, using (27), (28) and (21) once more, we

gather that

X x

d =

t+ Ad

= 0

as we wished.

Proposition 4. Fort (0, T), let

u(x,t,) := inf

RnRnU[t,T]

L(x, X , v , ) d(x, X , v , s; )

+

RnRn

(x) dT(x, X; ) ,(29)


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where the infimum is taken over all measures(, T) which, for every , satisfythe following: is a measure onRn Rn U [t, T] with total massT t, T is a

probability measure onRn Rn, and andTsatisfy the following constraint:RnRnU[t,T]

t(x, X , s) + A(x, X , s) d(x, X , v , s; )

=

RnRn

(x, X , T ) dT(x, X; )

RnRn

(x, X , t) dt (x, X) ,

(30)

for any C2(Rn Rn [t, T]). Then, u u.

Proof. Let u be as in (5) and u given by (29). Fix > 0, and consider a boundedprogressively measurable control v, and the corresponding trajectory x of (3) withthe initial condition x(t) =x, such that

u(x,t,) +

E

Tt

L

x(s),

x(s)

, v(s),

ds + (x(T))

.

(31)

We introduce the probability measuresandon RnRnU[t, T] and on RnRn,

respectively, defined byRnRnU[t,T]

(x, X , v , s) d(x, X , v , s)

:=E T

t

x(s),x(s)

, v(s), sdsand

RnRn(x, X) d(x

, X) :=E

x(T),

x(T)

,

for any C(Rn Rn U [t, T]) and C(Rn Rn).Then, if C2(Rn Rn [t, T]), by Dynkins Formula (see, e.g., formula (3) onpage 105 of [Eva06b]),

RnRnU[t,T]

t+ Ad

=E

x(T),x(T) , T

x(t),x(t) , t

=

RnRn

(x, X , T )d(x, X)

RnRn

(x, X , t)dt (x, X),

where t is taken as (x, X , t) dt (x

, X) :=

x,x

, t

,

which shows that and satisfy the constraint in (30). Accordingly, (29) and (31)give that u + u, and so the claim follows by sending to zero.


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5. Moment estimates

In this section, we discuss some a-priori moment estimates for minimizing measures(, T), as given in (29), that will be essential to establish the tightness necessaryto our scopes. More refined estimates, needed to prove that the homogenized limitis a viscosity solution of a Hamilton-Jacobi equation, will be developed in section 11.From this section on, we will assume Assumptions 4 and 5. By comparison with thecontrol v := 0 in (3), recalling Assumption 5, we see that

(32)

L(x, X , v , )d C ,

with C >0 independent of.

Proposition 5. Let(, T) be measures satisfying (30), witht given by (22). LetL

satisfy(32). Letp be as in Assumption 4. Then, there existsC >0, possibly dependingont andT, but independent of, in such a way that

|x|p d +

|x|p dT C(1 + |x|

p) .

Proof. In what follows, we will denote by Ci suitable positive quantities, which maydepend on t, T, but that are independent of.We observe that

(33) |x|p (1 + |x|2)p/2 2p/2(1 + |x|p) .

We apply the constraint in (30) to the function (x) := (1 + |x|2)p/2 and we obtainthat

p(1 + |x|2)(p2)/2f x

+1

2

ni,j=1

(T)ij

p(p 2)(1 + |x|2)(p4)/2xix

j+ p(1 + |x

|2)(p2)/2ij

d

=

(1 + |x

|2

)p/2

d

T

(1 + |x

|2

)p/2

d

t ,

and so, by (33), |x|pdT

(1 + |x|2)p/2dT

C2

|f| (1 + |x|2)(p1)/2

+|T| (1 + |x|2)(p2)/2d +

(1 + |x|2)p/2dt

.


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Consequently, Youngs inequality, Assumption 4, estimates (32) and (33), and Propo-sition 2 give that

|x|pdT

C3

|f|p +

|T|p

(1 + |x|2)p/2+ (1 + |x|2)p/2 d +

(1 + |x|2)p/2dt

C4

1 + L+ |x|p d +

1 + |x|pdt

C5

1 +

|x|p d +

|x|pdt

.

(34)

To estimate this last term, we apply again (21) to the function (x, s) := (s T)(1 +

|x|2)p/2, to obtain (1 + |x|2)p/2 + (s T)

p(1 + |x|2)(p2)/2f x

+1

2

ni,j=1

(T)ij

p(p 2)(1 + |x|2)(p4)/2xix

j+ p(1 + |x

|2)(p2)/2ij

d

= (T t)

(1 + |x|2)p/2dt .

Thus, using (33) and a scaled Youngs inequality, we obtain that

(1 + |x|2)p/2 d

C6

1 +

|x|p dt +

|f|p +

|T|p

(1 + |x|2)p/2d

+1

2

(1 + |x|2)p/2 d .

Thus, exploiting Assumption 4, estimate (32) and Proposition 2, we conclude that

1

2

|x|pd

1

2

(1 + |x|2)p/2d

C7

1 + Ld+

|x|p dt

C8

1 +

|x|p dt

.

This, (34) and the definition oft , see (22), yield the desired result.

As a consequence of this result, using (22), we have, using Proposition 3, that anyminimizing sequence (n,

T,n) will satisfy the bound

(35)

|x|p + p|X|pdnC,

|x|p + p|X|pdT,n C.


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The above tightness and Assumption 8 imply that there are minimizing measures

and Tattaining the infimum in (29).

6. Duality

In addition to the HamiltonianHas defined in (6), we need also the extended Hamil-tonian operator

(36) H(D2(x, X), D(x, X), x , X, , ) := supvU

A L,

which is defined for C2(Rn Rn). It will be useful later the observation that

H(D2(x, X), D(x, X), x , X, , ),

with

(37) = (x, X) + (Xx

),

is constant in , for any C2 function , thanks to (20).Recalling the framework introduced on page 10, we define the set C2 to be the set offunctions C2(Rn Rn [t, T]) such that(x, X , t), (x, X , T ) C(R

n Rn), and

t+ Av C(R

n Rn U [t, T]).In order to prove the reverse inequality of the one in Proposition 4, we introduce thedual problem of (29):

Proposition 6. In the notation of Proposition 4, we have that

u(x,t,) =(38)

sup(x,X,s)

(T t)

inf

(x,X,s)RnRn(t,T)t H

D2,D,x, X, ,

+ infx,X

[(x, X , ) (x, X , T )] +

RnRn

(x, X , t) dt (x, X).

where the supremum is taken over any C2.

To heuristically motivate the statement in (38), we apply the minimax principle, that is,we exchange formally the infimum with the supremum in the variational problem (29),


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by using the constraint in (30), thus obtaining:

u

(x,t,)= inf

,T

sup(x,X,t)

RnRnU(t,T)

L(x, X , v , ) + A+ t d

+

RnRn

( )dT+

RnRn

dt

(using the minimax principle)

= sup(x,X,s)RnRn d

t

+inf

RnRnU(t,T)

L(x, X , v , ) + A+ t d

+infT

RnRn

( )dT

,

and then, taking into account that both andTare measures, withTa probability

measure, and with mass T t, the definition ofAand Hwe obtain (38).

Proof (of Proposition 6). The rigorous proof of Proposition 6 will be quite long (endingon page 20) and it will make use of the Legendre-Fenchel-Rockafellar Theorem. Let Pand Q be defined as in page 10. Define the following sets:

M1 :=

(, ) P Q:

d= T t,

d= 1, , 0

,

and

M2:=

(, ) P Q:

A+ td=

d

dt , C

2

.

Given C(Rn Rn U [t, T]) and C(R

n Rn [t, T]), we set

h(, ) := (T t) sup(x,X,v,s)

[(x,X,v,s) L(x,X,v,s,)]

+ sup(x,X)[(x

, X , ) (x

, X , T )] .(39)

Note that, for short, we have omitted the dependence ofh on . We consider the set

C= cl

(, ) : C2 , = A+ t C

,

where the closure is taken in the C2 C topology.Let (, ) M1 M2 and define

g(, ) :=

d

(x,X,T)d if (, ) C

otherwise.


18/53


We will now compute the dual functions ofh and g, that is

h(, ) := sup(,)CC

d

d h(, ) ,

g(, ) := inf (,)CC

d

d g(, ) .

(40)

Lemma 7. We have

(41) h(, ) =

Ld +

d if(, ) M1

+ otherwise

and

(42) g(, ) =

0 if(, ) M2 otherwise.

Proof. We first claim that if either or are non-positive then h(, ) = +. To seethis, we choose a sequence of non-negative functions (j, j) such that

jd

jd+.

Since L 0 we have that

sup j L 0,

and, furthermore,sup j sup .

As a consequence, since is bounded,

h(j, j) (T t)infL+ sup

is bounded from above uniformly in j, and so h(, ) = +.Now we claim that the following inequality holds:

h(, )

Ld+

d

+ sup

d (T t) sup

+ sup

d sup

.

(43)

To establish (43), consider a sequenceLj infL= 0 (see Assumption 1) of compactlysupported functions, increasing pointwise to L. For any C and C, define and by= Lj, and = . Note that C and C, so that

h(, ) sup,

Ljd +

d+

d+

d

(T t)sup[Lj+ L] sup

.


19/53


Since Lj L 0 we gather that

h(, )

Ljd +

d

+ sup

d (T t) sup

+ sup

d sup

.

By letting j + we obtain (43), due to the Monotone Convergence Theorem.Now we claim that if the mass ofis notTtor the mass ofis not 1, thenh(, ) =

+. To see this just add constants to or to .Finally, if the mass of is T tand the mass ofis 1, by choosing := 0 and := 0in (43), we obtain

h(, ) Ljd+ d.To obtain the opposite inequality, just use (39) and (40), and observe that, if the massof isT t and the mass ofis 1, then, for each (, ), we have

Ld (T t)sup( L),

as well as dsup( ) ,

so that

d

d h(, )

Ld+

d+ (T t) sup( L) + sup( ) h(, )

=

Ld+

d.

This completes the proof of (41).To compute the Legendre transform ofg, observe that if (, ) M2 then there exist

functions (,) Csuch that

(44) d( ) d( )= 0 ,thanks to the fact that (, ) M2. Possibly changing the signs of and , we may

suppose that the quantity in (44) is positive. Then, take j :=j and j :=j. Weget that

limj+

jd( )

jd( ) = + .

The above considerations show that

g(, ) = inf,

d( ) +

d( ) =

0 if (, ) M2 otherwise,


20/53


which gives (42).

We now complete the proof of Proposition 6 by arguing as follows. We note that thefunction g (resp., h), as defined above, is concave (resp., convex) and upper (resp.,lower) semicontinuous (we remark indeed that Cis a closed convex set).Then, by the Legendre-Fenchel-Rockafellar Theorem (see, e.g., [Vil03] or [Gom06]),

sup(g h) = inf(h g) .

Consequently, (29), (36) and Lemma 7 yield:

u = inf (,)M1M2

L d+

dt

= inf (,)

h g

= sup(,) g h

= sup(,)C

(t T) sup(x,X,v,s)

( L) sup(x,X)

( ) +

dt

= supC2

(t T) sup(x,X,v,s)

(A t L) sup(x,X)

( ) +

dt

= supC2

(t T) sup(x,X,s)

(H t) sup(x,X)

( ) +

dt,

which ends the proof of of Proposition 6.

The result to come shows that (38) is optimized when X= x:

Proposition 8. Suppose thatt is supported in the setX= x

. Then

u(x,t,)

= sup(x,X,s)C2(R

nRn[t,T])

(T t)

inf

(x,X=x

,s)RnRn(t,T)

t H

D2,D,x, X, ,

+ infx

(x) (x,

x

, T)

+

+RnRn (x

, X , t)d

t

= sup(x,s)C2(R

n[t,T])

(T t)

inf

(x,s)Rn(t,T)t

H

D2x, Dx, x

,x

, ,

+ inf

x[(x) (x, T)]

+

RnRn

(x, t)dt


21/53


22/53


From this point on, we will assume Assumption 6. With this, we are now in the positionto improve Proposition 4:

Proposition 9. u =u inRn [t, T] .

Proof. By (22), t is supported in X = x

. Thus, by Proposition 8 and (9), we have

that

u (T t)infx,s

tv H

D2xv

, Dxv

, x

,x

, ,

+inf

x[(x) v(x

, T)] + v(x, t)

+ 0 + v(x, t) .

Sending 0, and using (10), we conclude that u u. The converse inequality isassured by Proposition 4.

Finally, we present a result that converts the constrained minimization problem of (29)and (30) into an unconstrained problem. The idea, quite standard in linear program-ming, consists in using the approximate solution v of the dual problem as a Lagrangemultiplier so to remove the constraints.

Proposition 10 (Unconstrained minimization). Fix > 0, and let v be as in As-sumption 6. Then,u(x,t,)

inf

RnU[t,T]

L(x,x

, v , ) + Av+ tv

d +

vd

t

(T t) + sup

s[t,T]

|u(x,s,) v(x,s,)| ,(47)

where the above infimum is taken over all measures(x, v , s) onRn U [t, T], withtotal massT t.

Proof. By (6),

L(x,x

, v , ) + Av

H(D2

x

v

, Dxv

, x,x

, , ),

and so, integrating with respect to any measure with total mass T t, L(x,

x

, v , ) + Av + tv

d

tv

H(D

2xv

, Dxv

, x

,x

, , ) d (T t) ,

thanks to (9). This and the definition oft , equation (22), imply that

u(x,t,)

inf

RnU[t,T]

L(x,x

, v , ) + Av+ tv

d +

vd

t

(T t) + u(x,t,) v(x,t,) .(48)


23/53


On the other hand, let > 0, and suppose that (x(s), v(s)) is an almost optimaltrajectory for (5), for s [t, T], so that

(49) u

L(x,

x

, v , )do(x

, v , s; ) + E (x(T)) ,

where the measure o on Rn U [t, T] is defined by

(50)

(x, v , s) do:= E

Tt

(x(s), v(s), s) ds ,

for any C(Rn U [t, T]).

Notice that o(Rn U [t, T]) =T t, therefore

inf

L(x

,

x

, v , ) + Av

+ tv

d

L(x,

x

, v , ) + Av+ tv

d

o.

From this, (49) and (22), we get that

+ u(x,t,)

inf

RnU[t,T]

L(x,x

, v , ) + Av+ tv

d+

vd

t

E

x(T),

x(T)

, T

Av+ tv

d

o

vd

t(51)

= E

x(T),x(T)

, T

Av+ tvdo v(x,t,) .

Since, by (50) and Dynkins Formula, Av+ tv

d

o+ v

(x,t,) =Ev

(x(T), T),

we deduce from (51) that

+ u(x,t,)

inf

RnU[t,T]

L(x,x

, v , ) + Av+ tv

d+

vd

t

E[ (x(T)) v

(x(T), T , )] =E[u

(x(T), T , ) v

(x(T), T) ] .From this estimates, sending 0, and (48), we obtain (47).

We remark that Proposition 10 and Assumption 6 imply that

lim0

inf

RnU[t,T]

L

x,

x

, v ,

+ Av+ tv

d+

vd

t

= u(x,t,)(52)

uniformly in (x,t,) Rn [0, T] , where the the infimum is taken over all mea-sures (x, v , s) on Rn U [t, T], with total mass T t.


24/53


25/53


and

(58)

(x

) dt(x

) := lim0

(x

) d

t (x

, X , ) ,

for2 any Cc(Rn).

Proof. The claim in (i) easily follows by comparing with the control v := 0.We now prove the claim in (ii). Consider the control v := 0 and the correspondingtrajectory x0(t; ). By comparing with the measures defined in (23) and (24), and byexploiting (4), we get that there exists a constant Csuch that

C

Tt

L

x0(s; ),

x0(s; )

, 0,

ds + 2 sup ||

d

L(x, 0, v , ) d

(T t) R

n (U\ BR) [t, T]

infL

+R

n (U\ BR) [t, T]

inf|v|R

L.

From the growth conditions in v of L (see equation (4) in Assumption 1) we thusconclude that, as R ,

R

n (U\ BR) [t, T]

0.

Furthermore, we know, from Proposition 5 that

(Rn \ BR) U [t, T]

0,

as R , which then implies that

(Rn \ BR) (U\ BR) [t, T]

0,

as R , and then implies tightness. Therefore, the desired claim follows fromAssumptions 8 and 3.From Proposition 5, it is also clear that the measure in Rn defined by

ERn

(x, )d

T

is tight, and therefore we have (57). The weak convergence in (58) is simply a conse-quence of (22), as then t=x(x

).

From this point on, we will also assume that the scaling hypotheses in Assumption 9hold, that is = 0, with = 1,

12

or 0 (resp. cases A, B or C). In each of thesecases, we define the generalized Mather problem in the following way: given Rn

2A standard observation is that the moment estimates in Proposition 5 imply that (57) and (58)also hold for any continuous (x) growing at infinity less than |x|p once they hold for compactlysupported ones.


26/53


and Mat(n n) we look, for each fixed x, at probability measures (v, ; x,

, )on U , which minimize

(59) L(x,

, ) :=

L(x, 0, v , )d(v, ; x,

, )

under the constraints, which depend on the scaling hypothesis put forward in Assump-tion 9, given by:

A. In this first case, the Effective Lagrangian L(x, ) will not depend on , andwe take as constraints the rotation vector

(60)

f(x, 0, v , )d=

,

and the holonomy constraint

(61)

f(x, 0, v , ) DX(0, )d= 0,

for all stationary functions (X, ), which are C1 in the first variable. Notethat in this scaling the contribution of the diffusion vanishes.

B. In this intermediate scaling, the Effective Lagrangian will also not depend on ,and we require, similarly, the rotation vector

(62)

f(x, 0, v , )d=

,

and the holonomy constraint, which in this case has a contribution from thediffusion

(63)

f(x, 0, v , ) DX(0, ) +

0T0(x, 0, v , )

2 :D2X(0, )d= 0,

for all C2 (in the first variable) stationary functions (X, ).

C. In this scaling, in which diffusion dominates, we require both an average diffu-sion coefficient , a rotation vector

1

2

0

T0d = ,

f(x, 0, v , )d= ,(64)

and the holonomy constraint, which does not have a drift term since the diffu-sion dominates

(65)

0

T0(x, 0, v , )

2 :D2X(0, )d= 0,

for all C2 in Xstationary functions (X, ).

Given the limiting measures and constructed in Proposition 11, we define, foreach (x, s) Rn [t, T],

(66) (x, s) :=

f(x, 0, v , ) d(, v; x, s) .


27/53


Also, in case C, we define

(67) (x, s) := 12

0T0(x, 0, v , ) d(, v; x, s).

It turns out that (x, s) almost everywhere, the measure (v, ; x, s) constructed inProposition 11 satisfies the holonomy constraints corresponding to cases A-C. Moreprecisely:

Proposition 12.For almost any(x, s) Rn[t, T]with respect to the measure(x, s),the measure(v, ; x, s)satisfies conditions(61),(63)or(65), and conditions(60),(62)or (64)with (resp.,) replaced by (x, s)(resp., (x, s)), according to cases A-C.Moreover, in cases A and B, if we define the measure (x,

, s ) onRn Rn [t, T]

by (x

,

, s ) := (x,s)( )(x

, s), we have that

(68)

Dx(x

, s) + t(x, s)d

=

(x, T)dT

(x, t)dt,

for allC1 functions.Analogously, in case C, if we define the measure (x , , , s) onRn Rn Mat(n n) [t, T] by (x, , , s) := (x,s)( )(x,s)()(x

, s), we have that Dx(x

, s) + :D2x(x, s) + t(x

, s)d

(69)

=

(x, T)dT

(x, t)dt,

for allC2 functions.

Proof. As usual, given a, b Rn, we define a b Mat(n n) by (a b)ij :=aibj.Let (X, ) : Rn RbeC2 inXand stationary, let(x, s) C(Rn [t, T]) andconsider the function

(x, X, s, ) (x, s) (X, ) .

By (21), (12) and (13), we see that

1

f(x, 0, v , ) DX(0, ) +

1

22(T)(x, 0, , v , ) :D2X(0, )

(x, s) d

+

t+ f(x

, 0, v , ) Dx+1

2(T)(x, 0, , v , ) :D2x

(0, ) d

+1

(T)(x, 0, , v , ) : (Dx DX(0, )) d

=

dTd

dtd .

(70)


28/53


We first consider cases A and B. We now take a:= 1 in case A or a:= 1/2 in case Band we multiply (70) by , thus obtaining

f(x, 0, v , ) DX(0, ) +2a1

2 (0

T0)(x

0, v , ) :D2X(0, )

(x, s) d

+

t+ f(x

, 0, v , ) Dx+1

2(T)(x, 0, , v , ) :D2x

(0, ) d

+ 2a

(0T0)(x

0, v , ) : (Dx DX(0, )) d

=

dTd

dtd

.

By sending 0 via Proposition 11, it follows thatf(x, 0, v , ) DX(0, ) +

c

2(0

T0)(x

0, v , ) :D2X(0, )

d(v, ; x, s) (x, s) d(x, s) = 0 ,

where c= 0 in case A, and c = 1 in case B.Since is arbitrary, we conclude that

f(x, 0, v , ) DX(0, ) +c

2(0

T0)(x

0, v , ) :D2X(0, )

d(v, ; x, s) = 0

for almost any (x, s) with respect to , thence conditions (61) or (63) are fulfilled,according to cases A and B.Multiplying (70) by 2 and arguing as above, one shows that (65) holds in case C.The fact that satisfies (60), (62) or (64) with replaced by (x, s) follows at onceby (66) and (67).To prove (68) and (69), it is sufficient to plug := 1 in (70) and send 0.

Proposition 13. In the notation of Proposition 12, we have that

(71)

L d

L d d .

Proof. We use the notation of case C, since cases A and B are analogous and easier, tocompute that

L(x, , ) d

=

L(x, 0, v , ) d(v, ; x, , ) d

(x, , , s)

=

L(x, 0, v , ) d(v, ; x, (x, s),(x, s)) d(x, s)

L(x, 0, v , ) d(v, ; x, s) d(x, s) ,

as we wished.


29/53


We remark that, for some choices of and , the constrained minimization prob-lem given in (59) may be ill posed, since there might be no measure fulfilling con-

ditions (61), (63) or (65), and conditions (60), (62) or (64), according to cases A-C.However, one of the consequences of Proposition 12 is that such a problem is alwayswell defined for some particular choice of and , i.e., at least for := (x, s)and :=(x, s). In case, for some (x, , ), there are no measures fulfilling condi-tions (61), (63) or (65), and conditions (60), (62) or (64), according to cases A-C, wedefine L(x , , ) to be +.

Proposition 14. The Effective LagrangianL is lower semicontinuous.

Note that this result, via Yosida regularization (see, e.g., Theorem 2.64 in [Att84]),gives thatL may be approximated monotonically from below by continuous functions.

Proof (of Proposition 14). Let (xn, n, n) (x, , ) and consider the correspond-ing optimal measures n on U .Let

Ln := L(xn, n, n) =

L(xn, 0, v , )dn(v, ; x

n, n, n) .

Without loss of generality, we may suppose that Ln is bounded by above, say LnK(otherwise, either lim infLn= , in which case there is nothing to show, or lim infLn0. Substituting in (81), we obtain that

1

DMjkHD

2ijUD

2ikU+ DiUDxiH+ DiUDXiHCq

at the point yq, for some C>0, due to (8).Then, P2 of Assumption 11 and (80) yield that

1 Cq F(DU(yq), U(yq)) inf||U

F(DU(yq), ) ,

as long as qis small enough.From this and (17), we infer that |DU(yq)|is uniformly bounded for small q. But then,w w(yq) is uniformly bounded, and so isDU, as desired.

As a consequence, we have


34/53


Proposition 21. Fixed any , through an appropriate subsequence, we havethat u(x,t,) converges to u(x,t,) locally uniformly in (x, t) Rn [0, T], for

someu : Rn [0, T] R. Also,

lim0

u(x,t,) d = u0(x, t) ,

for someu0 : Rn [0, T] R.

Proof. This follows from the Ascoli-Arzela Theorem, since

u(x,t,) d is equicon-tinuous, thanks to Propositions 18 and 20, and equibounded, due to the terminalcondition.

Proposition 22. We have that

(82) |u(x,t,y) u(x,t,)| C |y| .

Moreover,

(83) |

u(x y, t , ) d

u(x,t,) d| |y|

D+ C(T t)

.

Proof. Notice that

ut(x,t,y) + H

D2u(x,t,y), Du(x,t,y), x,

x

, , y

= 0.

By (53),

ut(x,t,y) + H(D2u(x,t,y), Du

(x,t,y), x , y+x

, , ) = 0.

Set v(x,y,t; ) = u(x y,t,y). Then, v solves

vt+ H(D2v,Dv,x y,

x

, , ) = 0.

As before, by Assumption 10, we may assume that v is smooth. Then set

w= v

yk.

Thus,wt+ DMH :D

2w+ DpH Dw DxkH= 0.

By Assumption 12, the ellipticity condition in (14) and the maximum principle (see,e.g., Proposition 4.1 and Theorem 4.2 in [DiB95]), we conclude that

w(x, t) D+ C(T t),

which implies

(84) |u(x y,t,y) u(x,t,)| |y| [D+ C(T t)] .

Furthermore, the uniform Lipschitz continuity ofu (recall Proposition 20) implies that

|u(x y,t,y) u(x,t,y)| C |y|.

The latter two estimates yield the claim in (82).Then, (83) follows from (84) by replacing y with y and recalling that y is measurepreserving, according to Assumption 7.


35/53


Proposition 23. We have

lim sup0 supxRnt[0,T]0,1

|u

(x,t,0) u

(x,t,1)| 0.

Proof. To prove the above estimate, observe that

|u(x,t,0) u(x,t,1)| |u

(x,t,0) u(x,t,y0)|

+ |u(x,t,y0) u(x,t,1)|.

Let 1 > 0 be fixed, and suppose that for every there exists y = y(x,t,,1) suchthat|y| M, with Mdepending only on 1 (and so independent of), for which

(85) supxRnt[0,T]0,1

|u(x,t,y0) u(x,t,1)|

1

2.

Then, from Proposition 22, for sufficiently small, we have

supxRnt[0,T]0|y|M

|u(x,t,0) u(x,t,y0)|

12

,

which then, by sending 1 0+ yields the estimate. Therefore our task now consists

in establishing (85).To this extent, fix >0 and choose y as in Assumption 13. Then

u(x, t) :=u(x,t,1) + (T t)

satisfies

0 =ut(x,t,1) + H

D2u(x,t,1), Du(x,t,1), x,

x

, , 1

=ut+ H

D2u, Du,x,

x

, , 1

ut+ H

D2u, Du,x,

x

, , y0

,

that is

ut+ H

D2u, Du, x,x

, , y0

0 .

Then, u u is a subsolution of a linear parabolic equation, due to (14), which thenimplies, by the comparison principle for viscosity solutions (see Proposition 4.1 andTheorem 4.2 in [DiB95]) that

u(x,t,y0) u(x, t).

Accordingly, we obtain (85) by possibly exchanging the roles ofy0and 1, as desired.

By collecting the results in this section, we get the following result:


36/53


Proposition 24. Through an appropriate subsequence,

lim0 u

(x,t,) = u

0

(x, t) ,uniformly inK [0, T] , for any compact setK Rn.

Proof. Fix0. Then, by Proposition 21,uj(x,t,0) converges tou

(x,t,0) locallyuniformly in (x, t). Let u0(x, t) :=u(x,t,0). Then,

limj+

sup(x,t,)K[0,T]

|uj(x,t,) u0(x, t)|

limj+

sup(x,t,)K[0,T]

|uj(x,t,) uj(x,t,0)|

+ limj+

sup(x,t)K[0,T]

|uj(x,t,0) u(x,t,0)|

0 + 0 ,

where Proposition 23 has been exploited.

10. Convergence

In this section, we establish that the limit of the viscosity solutions u agrees with u,which was defined in (72). We divide the proof into two parts, a lower bound and anupper bound.

Proposition 25 (Lower Bound). u0 u.

Proof. Consider the measures and constructed in section 7. Let also

and

T bethe measures optimizing (5).Then, we use Proposition 9, (29) (11), (54), Proposition 11, (71), Proposition 17, (72)and (73) to obtain that

u0(x, t) = lim0

u(x,t,) d

= lim0

L(x, X, v, s, )dd+

(x)dTd

= lim0 L(x, 0, v , s , )d + dT

=

L(x, 0, v , s , )dd +

dT

L(x , , )d

+

dT = u(x, t).

In order to obtain the opposite inequality to Proposition 25, we need the result inProposition 24.

Proposition 26 (Upper bound). u0 = u.


37/53


Proof. Let >0 and v be as in Assumption 6. We denote byo quantities (possiblydepending on (x,t,)) which tend to zero as 0.

Let > 0. We consider almost optimal measures ( (x , , , s), T,(x)) for (72)or (73), and an almost optimal family of probability measure (v, ; x

,s,

, ) for (59),that is, we suppose that (x

, , , s) almost everywhere

L(x, , )

L(x, 0, v , )d(v, ; x,

, ) ,

and that, in case A or B,

u(x, t)

L(x , )d

+

dT, ,

whereas, in case C,

u(x, t)

L(x , , )d

+

dT, .

We construct a trial measure (x, v , s , ) on RnU[t, T] defined in the following

way:

(86)

RnU[t,T]

(x, v , s , )d :=

(x, v , s ,

x

)dd ,

for any Cc(Rn U [t, T] ).

Note that, by taking (s) := (s T) in (68) or (69), we obtain that

d

= T t ,

and so, since is a probability measure, it follows that has total mass T t.In particular, the total mass of is finite and so, recalling Assumption 3, we canslice as

(87) (x, v , s , ) =: (x

, v , s; ) () ,

where () is a probability measure, and is, for -almost any fixed, ameasure of finite total mass. Let m() be the total mass of for a fixed . Byconstruction,

(88) 1 = 1T t

d= 1

T t

d d=

m()

T td() .

Let also

:= T t

m() ,

and note that has total mass T t for any fixed .Then, by Proposition 10,

o+ u

L(x,

x

, v , ) + Av+ tv

d+

vd

t,


38/53


and so, by (87) and (88),

o+

m()T tu(x,t,)d()

L(x,

x

, v , )d+

Av+ tv

d+

m()

T tvd

t d.

Making use of (11) and (59), we get that L(x,

x

, v , )d=

L(x,

x

, v ,

x

)dd

=

L(x, 0, v , )dd

L(x , , )d

+ C,.

for some suitable constant C > 0. Similarly, by (19), (86), (12), (13), (60), (62)and (64),

A d =

f(x,

x

, v , ) Dx +

T

2 (x,

x

, v , ) :D2x d

=

f(x,

x

, v ,

x

) Dx

+T

2

(x,x

, v , x

) :D2x d d

=

f(x, 0, v , ) Dx +

T

2 (x, 0, v , ) : D2x d d

=

Dx +

:D2 d

,

for any(x) C2(Rn), with

(x) := 1

2

T(x, 0, v , ) d .

Thus, exploiting (68) and (69), Av+ tv

d =

Dxv

+

:D2xv+ tv

d

=

vdT

vdt+

( ) :D2xv

d

,

with := 0 in cases A and B and = = in case C.Note that

(89) lim0

= 0

uniformly in (x,v,) in all the cases A-C.


39/53


Also, by (72) and (73),

u(x, t)

L(x , , )d

+

dT, .

By collecting the above estimates, we conclude that

o+

m()

T tu(x,t,)d()

u(x, t) +

(v )dT+

m()

T tvd

td

vdt

+

( ) :D2xv

d

+ C .

That is, by Assumption 6,

o+

m()

T tu d

u+

(u ) dT+

m()

T tudtd

udt+

( ) :D2xv

d

+ C .

We now send 0 and we obtain

o+ u0 u + C ,

thanks to Proposition 24 and formulas (58), (88) and (89).We now sendand to zero, gettingu0 u. This and Proposition 25 yield the desiredclaim.

11. Improved Moment Estimates

In this section we improve the moment estimates of section 5 in order to show that forsmall time optimal trajectories dont go too far. These estimates will be essential inestablishing that the homogenized limit is a viscosity solution of a suitable equation.Here, we will take a small time step h (0, T). We start with an auxiliary result:

Lemma 27. Consider the diffusion

dx0 =f

x0,x0

, 0,

dt +

x0,

x0

, 0,

dWt,

with initial conditionx0(T h) =x. Then, we have

|E[x0(T h) x0(T)]| C h,

and

E|x0(T h) x0(T)|2 C h,

where the constantCcan be chosen independent of, h andx.


40/53


Proof. Fixh >0, let T h t T. By Dynkins Formula we have

(90) d

dtE[x0(T h) x0(t)] =Ef

x0,x0

, 0,

,

which, by Assumption 14, yields the first part of the claim.Similarly, we have

d

dtE|x0(T h) x0(t)|2 =2E

(x0(T h) x0(t))f(x0,

x0

, 0, )

+ tr E[(x0,x0

, 0, )T(x0,

x0

, 0, )].(91)

Thus, if we set A(t) =E|x0(T h) x0(t)|2 we have

d

dtA A + C, A(T h) = 0.A Gronwall estimate yields |A(t)| C h.

Lemma 28. Consider an optimal diffusion x(t), for T h t T, with initialconditionx(T h) =x. Let

M :=E

TTh

L

x,x

, v,

dt.

Then, we have|E[x(T h) x(T)]| M1/2Ch1/2 + Ch,

andE|x(T h) x(T)|2 M Ch+ Ch,

whereCcan be chosen independently of, h andx.

Proof. Using again (90) and Assumption 15 we obtain

E[x(T h) x(T)] =E

TTh

f

x,x

, v,

dt

h1/2

E

TTh

fx,x

, v, 2dt1/2

C h1/2

E TTh L

x,

x

, v,

+ Cdt1/2

C M1/2h1/2 + Ch.

Similarly, by (91), we have

d

dtE|x(T h) x(t)|2 =2E

(x(T h) x(t))f

x,

x

, v,

+ tr E

x,x

, v,

T

x,x

, v,

Ch1E|x(T h) x(t)|2 + ChEL

x,

x

, v,

+ C.


41/53


Thus, if we setA(t) =E|x(Th)x(t)|2 we have, by a Gronwall estimate, that |A(t)| CMh+ Ch.

Now we use the two previous results to prove that the integral of the Lagrangian alongan optimal trajectory is not too big:

Proposition 29. Consider an optimal diffusionx forT h t T, minimizing

E

TTh

L

x,x

, v,

dt + (x(T)),

with initial conditionx(T h) =x. Then there exists a constantC independent ofh, andx such that

E T

Th

Lx,x

, v, dt C h.

Proof. Consider the diffusion

dx0 =f

x0,x0

, 0,

dt +

x0,

x0

, 0,

dWt,

with x(T h) =x. We have, by the optimality ofx, that

E

TTh

L

x,x

, v,

dt + (x(T))

E

TTh

L

x0,

x0

, 0,

dt + (x0(T))

.

Moreover, the fact that x(T h) =x0(T h) =x, via a second order Taylor expansionof , gives that

(x0(T)) (x(T))

D(x) (x0(T) x0(T h)) D(x) (x(T) x(T h))

+C

|x0(T) x0(T h)|2 + |x(T) x(T h)|2

.

By the previous estimates and Lemmata 27 and 28,

M=E

TTh

L

x,

x

, v,

dt

E

TTh

L

x0,

x0

, 0,

dt + (x0(T)) (x(T))

C(1 + M)h+ CM1/2h1/2,

which implies the desired bound.

We will also need the following estimate:

Proposition 30. Fixr >0. Consider the optimal diffusion

dx= f

x,x

, v,

dt +

x,

x

, v,

dWt,

withx(T h) =x. Then, ash 0+, we have

P(|x(T) x|> r) =o(h).


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Proof. Writex= y+ z,

wheredy= f

x,

x

, v,

dt, dz=

x,

x

, v,

dWt,

and y(T h) =x, z(T h) = 0.We have

d

dtE|y x|2 = 2E(y x)f

x,

x

, v,

2

E|y x|2

1/2 E|f|2

1/2,

which impliesd

dtE|y x|21/2 CE|f|21/2 .

Thus, using again Assumption 15,E|y(t) x|2

1/2C

TTh

E|f|2

1/2dt

C h1/2 T

Th

E|f|2dt

1/2

C h1/2

E

TTh

L + Cdt

1/2C h,

due to Proposition 29, and so

E|y(t) x|2 C h2.

Accordingly, by Chebychevs inequality, we have

(92) P

|y(T) x|>r

2

C

h2

r2.

To handle the second term, we use Itos formula (see, e.g., page 105 in [Eva06b]) andwe get

d

dtEe

|z|2

t+2hT

=E |z|2

e

|z|2

t+2hT

(t + 2h T)2+1

2ET :

2I

t + 2h Te

|z|2

t+2hT + 2

z z(t+ 2h T)2

e|z|2

t+2hT

ET :

I

t + 2h Te

|z|2

t+2hT

C

hEe

|z|2

t+2hT ,

if is sufficiently small and t [T h, T].Thus, recalling that z(T h) = 0, we obtain by a Gronwall estimate that

Ee|z|2

t+2hT C,


43/53


for allT h t T. Again by Chebychevs inequality,

(93) P

|z(T)|

r

2

C er

2

2h ,

which is exponentially small as h 0. The desired result then follows from (92)and (93).

We will need a final moment estimate for the optimal measures of the homogenizedproblem. Such moment estimate will only be in use in the proof of Proposition 33 andit requires the following additional condition:

Assumption 16. We assume that for any fixed (x0, 0, 0) Rn Rn Mat(n n)

there exists a map x Sx0,

0,0(x) = (Sx0,

0,0(x), Sx0,

0,0

(x)) Rn Mat(n n)with the properties listed below.

Let 0= 120T0, and Sx0,

0,0(x) = 12Sx0,

0,0(x)

Sx0,

0,0(x)T

. Then(1) Sx0,

0,0(x0) = ( 0, 0);

(2) the mapping x Sx0,

0,0(x) is globally bounded and globally Lipschitz;(3) the mapping xL(Sx0,

0,0(x), S

x0,

0,0(x), x) is globally Lipschitz.

Assumption 16 is obviously fulfilled, by choosing Sx0,

0,0(x) to be identically equalto ( 0, 0), when L is globally Lipschitz in x and, in particular, ifL does not dependon x at all this is the case, for instance, ifL does not depend on x (though it maydepend on X=x/).

Proposition 31. Suppose thatT (0, 1]is conveniently small. Let

andTbe optimalmeasures for (72) under the constraint in (68) in cases A and B, or optimal measuresfor (73) under the constraint in (69) in case C, witht= 0.Let also Assumption 16 hold. Then,

(94)

(x x)dT

C T,(95)

|x x|2d

C T2

and

(96)

|x x|2dT C T.

Furthermore,

(97)

Ld

C T.Proof. First of all, we observe that,

(98) | |2 + || + ||p C(1 + L(x , , ))

thanks to Proposition 15.


44/53


Also, by taking (x, s) =xj xj in the holonomy constraint (68) or (69) (accordingto the case), with j = 1, . . . , n, we obtain the identity

(99)

d

=

x xdT.

By taking = 12

(T s)|x x|2, we obtain

(100)

(T s)( (x x) + tr)

1

2|x x|2d

= 0.

Similarly, take = 12

|x x|2 to get

(101) (x x) + tr d

= |x

x|2dT.

By taking = 1 and = T s we also see that the total masses ofT and are 1and T, respectively. Thus, using (99), the Holder Inequality and (98), we concludethat

(102)

(x x)dT

C T1/2

| |2d

1/2C T+ CT1/2

Ld

1/2.

Also, from (100) and a scaled Holder Inequality, |x x|2d

C T

| ||x x| + ||d

1

2

|x x|2d

+ CT

| |2 + ||d

and so, by (98),

(103)

|x x|2d

C T

| |2 + ||d

C T

T+

Ld

.

We now consider an additional small parameter d >0. From (101), combined with (98), |x x|2dT C

|x x|2

d + d| |2 + d||p + d1/(p1)d

Cd1/(p1)T+C

d

|x x|2d

+ Cd

T+

Ld

.

Hence, exploiting (103),

(104)

|x x|2dT CdT+ C(d+ T)

Ld

,

where Cd> 0 may depend on d.To get the desired estimates, it remains to estimate

Ld

. Fixedx Rn, we take

0=

0(x), 0=0(x) and 0 = 0(x) =0T0 in such a way that

sup,

Dx(x) :D2x(x) L(x,

, )

=Dx(x) 0 0 :D2x(x) L(x, 0, 0) .


45/53


Indeed, note that, since L is coercive, by (98), and lower semicontinuous, by Proposi-tion 14, we have that such

0and 0exist, although they may not be unique. Consider

now the diffusion

(105) dx0=Sx0,

0,0(x0)dt + Sx0,

0,0(x0)dWt ,

with x0(0) =x and let the measures d 0(x, , , s) be defined by

RnRnMat(nn)

(x , , , s)d 0

:=E

T0

(x0(s), Sx0,

0,0(x0(s)), S

x0,

0,0(x0(s)), s)ds

and 0,T(x) by Rn

(x)d0,T(x) :=E(x0(T)).

Then, by Dynkins Formula and the minimality of and T,

(106)

Ld

+

dT

Ld

0+

d0,T.

By Lemma 27, we have

(x x)d0,T C T |(x x)|2 d0,T C T.Thus

(x) (x) d0,T(x

)

|D(x)|

(x x)d0,T

+ C

|(x x)|2

d0,T CT .

(107)

On the other hand, by (102) and (104),

(x) (x) dT(x)

|D(x)|

(x x)dT

+ C

|(x x)|2

dT

C T+ CT1/2

Ld

1/2+ CdT+ C(d+ T)

Ld

CdT+ C(d + T)

Ld

.

(108)


46/53


Therefore, putting together (106), (107) and (108),

Ld

(x) (x)d0,T + (x) (x)dT

+

Ld

0

CdT+ C(d+ T)

Ld

.

This yields the estimate in (97). Then, combining together (97), (102), (103) and (104),we obtain (94), (95) and (96).

12. Viscosity Solutions

This last section is dedicated to the proof of the main result of the paper, that is thatthe homogenized limit is a viscosity solution to an Effective Hamilton-Jacobi equation.

Proposition 32. There exists a function continuous functionH: Mat(n n) Rn Rn R such thatu is a viscosity solution of

(109) ut+ H

D2xu, Dxu, x

= 0.

Proof. To prove that u is a viscosity solution, we are going to use the results from [Bit01].To do so, we first define an operator:

(110) Tt1,t2(x) := inf, t Ld + dt2 ,

where the infimum is taken over all measures on Rn Rn [t1, t2] (or on Rn Rn

Mat(n n) [t1, t2]), and t on Rn that satisfy (68) (or (69), according to case A, B

or C) with Treplaced by t2 and t replaced by t1, andt1 =(x).We also set

(111) Tt := T0,t .

Let Ybe the set of all C2(Rn) such that

(112) sup>0

supx,

H

D2,D,x,x

, ,

0 the evolutionassociated to the Hamilton-Jacobi equation Tt1,t2 is a semigroup, which converges uni-formly, and therefore the limit also satisfies the semigroup property. In further detail:one considers

Tt1,t2:= inf

Ld +

dt2,


47/53


where the infimum is taken on the measures on Rn Rn U[t1, t2] with totalmass t2 t1 and t probability measures on R

n satisfying (30). From Proposition 9

and (5),

Tt1,t2= infE

t2t1

L

x(s),

x(s)

, v(s),

ds + (x(t2)) ,

where the infimum above is taken over all trajectories of the controlled dynamics (3)with x(t1) = x. Then, T

t1,t2

is a semigroup, by the optimality of the latter dynamicprogramming principle, thence so is Tt1,t2 , due to Propositions 24 and 26, as desired.Let now the translation operator be defined byy(x) =(x+y), for any : R

n R.The hypothesis (H2) in [Bit01] requires that for all f Xand all y Rn, yf X.However, this is not really necessary, and we may replace it by requiring for all y Rn

and Y, y Y. This condition is then verified, thanks to (8).

Hypothesis (H3) of [Bit01] is obviously fulfilled here too.Moreover, clearly, Tt1,t2 is monotone, that is, if1 2 then Tt1,t21 Tt1,t22.Therefore, to establish the result it suffices to check that the continuity and regu-larity hypothesis I-IV that we spell next. We will need the following notation: for anysequence d= (dk) of positive reals we define

Qd = { Cc (R

n), D dk, || k}.

I Continuity: for every Xthe function (t, x) Tt[](x) is continuous andfor all b > a 0 there exists C=C(a,b,) such that

|Tt| C,

for any t [a, b].II Locality: for every1, 2 C(Rn) Xand any fixed x Rn, andr >0, such

that 1=2 in the ball B(x, r) then

Tth,t1 Tth,t2 = o(h),

as h 0+.III Regularity: for any sequence of positive numbersd= (dk), any compact setK

Rn and for every C(Rn) Xthere exists a function mK,f,d() : R+ R+

such that mK,f,d(0+) = 0,

|Tt[+ ] Tt[] (x)| mK,f,d()t,

for any (x, ) K Qd and any , t 0.IV Translation: for any compact subset K Rn and every Cc (K), there

exists a function nK, : R+ R+, with nK,(0

+) = 0 such that

|yTt[](x) Tt[y](x)| nK,(|y|)t,

for any x K and t 0.

Property [I] holds by (74). Properties [II] and [III] follow from improved momentestimates of the previous section, as we explain next. Property [IV] follows from (72)and (83).Let us now give the details about how to check Properties [II] and [III]. First, wewill use Proposition 30 to prove that [II] holds. Assume that 1 and 2 satisfy the


48/53


conditions of [II], and let x be an optimal trajectory for 2, that is x optimizes (5)for := 2 and x(t h) =x. Then, by (72) and Propositions 24 and 26,

Tth,t2 = lim0

E

Tt

L

x(s),

x(s)

, v(s),

ds + 2(x(T))

while

Tth,t1lim0

E

Tt

L

x(s),

x(s)

, v(s),

ds + 2(x(T))

,

via a suitable subsequence.Consequently, using that 1 and 2 agree on B(x, r), we gather that

Tth,t1 Tth,t2 (1+ 2) P(|x x|> r),

which is o(h) by Proposition 30.To prove [III], we argue in an analogous way. Namely, we take an optimal trajectoryfor and we take limit for 0, as above. In this case, we also exploit Lemma 28 toconclude that

TTh,T(+ ) TTh,T

lim0

E(x(T)) + (x(T)) (x(T)) (x(T h))

lim0

E((x(T)) (x(T h))) CDh C d1h ,

as required. The desired result then follows from Theorem 3.1 of [Bit01].

We remark that Theorem 3.1 of [Bit01] explicitly gives that

(113) H(D2(x), D(x), x) = limT0

T0,T

T ,

in the notation of (110) and (111).Therefore, under the additional Assumption 16, we can characterize completely H asthe Legendre transform ofL.

Proposition 33. If also Assumption 16 holds, thenH is given by

(114) H(M, p , x) = sup=T,Mat(nn),

Rn

p :M L(x,

, ) .

Proof. We will be using formula (113) to establish the result. We only consider case C,since cases A and B are analogous, with just less variables involved.By taking (x, t) := t and (x, t) := 1 in (69), we see that if and T satisfy (69),then they have total mass Tand 1, respectively. Also, the supremum in (114) is finiteand attained, due to Propositions 14 and 15. Thus, take 0 = 0(x), 0 = 0(x)and 0 = 0(x) =0

T0 that satisfy

sup,

Dx(x) :D2x(x) L(x,

, )

=Dx(x) 0 0 :D2x(x) L(x, 0, 0) .


49/53


We also consider the dynamics of the associated control problem

dx0

=Sx0,

0,0(x

0)dt + S

x0,

0,0(x

0)dW

t,

and corresponding measures 0(x, , , s) and0,T(x

), given byRnRnMat(nn)

(x , , , s)d 0

:=E

T0

(x0(s), Sx0,

0,0(x0(s)), S

x0,

0,0(x0(s)), s)ds

and Rn

(x)d0,T(x) :=E(x0(T)).

We have that (69) holds, by Dynkins Formula, and so

TT

d0,T

Ld

0

=

Dx :D

2x Ld 0

=

Dx(x) 0 0 : D

2x(x) L(x, 0, 0)d 0

+

Dx + :D

2x

+L Dx(x)

0 0 : D2

x

(x) L(x, 0, 0) d 0. Tsup

, Dx(x) :D

2x(x) L(x,

, )

C

T0

E|x0(s) x| ds,

thanks to Assumption 16.That is,

TT

T sup

, Dx(x) :D

2x(x) L(x,

, )

CT T0

E|x0(s) x| ds .

(115)

We now observe that, by Lemma 27, T0

E|x0(s) x| ds

T0

(E|x0(s) x|2)1/2 ds C T3/2,

and so

(116) 1

T

T0

E|x0(s) x| ds 0,

as T 0.


50/53


Consequently, making use of (113), (115) and (116), we thus deduce that

(117) H(D2

(x), D(x), x) sup, Dx(x) :D2

x(x) L(x,

, ).

Now we will establish the reverse inequality and complete the proof of (114). Forthat matter, let (x, , , s) and T(x

) be minimizing measures for (73) under theconstraint in (69), with t= 0. Then,

Tsup,

Dx(x) :D2x(x) L(x,

, )

Dx(x) S

x,, (x) S

x,, (x) :D

2x(x) L(x, S

x,, (x), S

x,, (x))d

Dx(x) :D2x(x) L(x,, ) d C |x x|d ,thanks to Assumption 16.The above estimate, the holonomy constraint and Proposition 31 then give

sup,

Dx(x) :D2x(x) L(x,

, )

dT

Ld

T

C

T

T

|x x|2d

1/2

TT

T CT1/2

as long as T is small enough.Therefore, by sending Tto zero and recalling (113),

sup,

Dx(x) :D2x(x) L(x,

, ) H(D2(x), D(x), x) 0.

This and (117) complete the proof of (114).

13. Conclusion

Proof of Theorem 1. Propositions 4 and 9 imply the first claim of Theorem 1.The second claim follows from Propositions 18 and 20 and Remark 19.

By collecting the results in Propositions 24, 26 and 32, we obtain the last claim ofTheorem 1.

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Diogo A. GomesDepartamento de Matematica, Instituto Superior Tecnico, Lisboa, 1049-001, Portugal

e-mail: [email protected]


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Enrico ValdinociDipartimento di Matematica, Universita di Roma Tor Vergata, Roma, I-00133, Italy

e-mail: [email protected]

gomez d.a., valdinoci e. - generalized mather problem and homogenization of hamilton-jacobi...

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