transport and mather theory -...

Transport and Mather theory


Joaquın Delgado

Mathematics Department.Universidad Autonoma Metropolitana–Iztapalapa. Mexico, D.F.

[email protected]

Mathematical Congress of the Americas, 2013


Overview

A word of caution

This is survey talk based on several references. Specially on surveypapers by Qinglan Xia (UC Davis), Alfonso Sorrentino (Princeton),Patrick Benard (Institut Fourier), Boris Buffoni (EcolePolytechnique Federale-Lausane), Wilfrid Gangbo (Georgia Tech),Robert McCann (U. Toronto), Luigi De Pascale, Maria Stella Gelli,Luca Granieri (Pisa and Bari) and Lawrence Evans. For lack ofspace, Fathi’s Weak KAM theory ais not mentioned but offers amore natural approach to relate Aubry-Mather and transportation,through viscosity solutionsThe main purpose of the talk is point out a fresh perspectivetowards potential applications to Celestial Mechanics.


Overview

Overview

1 Monge’s problem: optimal transportation.

2 Mathematical formulation of Monge’s problem.

1 Optimality criterion2 The bookshelf problem3 The continuous case

3 Kantorovich relaxation: optimal plan.

4 Montonicity of an optimal plan.

5 The discrete case. Relation with combinatorial optimization.

6 Crash course on Mather theory

7 Mather theory as a special case of transportation.


Monge’s problem

Monge’s problem

Diviser le deblai el le ramblai enelements correspondants de mememasse de telle facon que la somme desproduits obtenus en mulipliant la massd’un element du deblai par sa distance al’element correspondant du remblai soitun minimum

Gaspar Monge. Memoire sur la theorie des deblais et de remblais.Histoire de l’Academie Royale des Sciences de Paris, avec lesMemoires de Mathematiques et de Physique pour la meme annee,1781, 666-704.


Monge’s problem

Monge’s problem

Figure: Decompose a pile of debris particles to fill an embankment of thesame mass, so that the cost of carrying each particle by matching oneparticle to another in straight line, be minimal.


Monge’s problem

Mathematical formulation of Monge’s problem

Mathematical formulation of Monge’s problem-I

The mass: a metric space X.

Debris: a Radon measure µ+.

Embankment: a Radon measure µ−.

Same mass: µ+(X) = µ−(X).

Transportation: a map T : X → X.

To fill the embankment with debris:T#µ

+(A) = µ+(T−1(A)) = µ−(A) for all Borel sets A.

Optimization:

min

∫X|x− T (x)| dµ+(x) | T : X → X,T#µ

+ = µ−

Such a T is called an optimal transportation map.


Monge’s problem

Mathematical formulation of Monge’s problem

Mathematical formulation of Monge’s problem-II

More generally. We consider a Polish space X ( separablecompletely metrizable topological space), two Radon measures µ±,and a cost function c : X ×X → R. Monge’s problem consist in

minimize

∫Xc(x, T (x)) dµ+(x).

Minimization is taken over the transportation maps

T : X → X | T Borel and inyective, T#µ+ = µ−

The problem is highly non–linear, solutions may not exist or maynot be unique. Depend on the cost function; in Monge’s originalformulation, c(x, x′) = |x− x′|.


Monge’s problem

Common cost functions

Common cost functions

Euclidean distance: c(x, x′) = |x− x′|.Quadratic: c(x, x′) = |x− x′|2.

Lp: c(x, x′) = |x− x′|p.

Convex function of Euclidean distance: h(|x− x′|).


Monge’s problem

Lipschitz criterion

Lipschitz criterion

Let u : X → X be Lipschitz with Lip(u) ≤ 1, then∫Xu(x) d(µ− − µ+)(x) ≤

∫X

(u(φ(x))− u(x)) dµ+(x)

≤∫X|φ(x)− x| dµ+(x)

Therefore, if equality is achieved for some u with Lip(u) ≤ 1 thenφ must be an optimal transportation map.


Monge’s problem

The bookshelf

An example: bookshelf

Figure: Two optimal transport maps to fill the right space in thebookshelf


Monge’s problem

The bookshelf

Mathematical formulation of the bookshelf

X = R, µ+ = χ[0,n] dx, µ− = χ[1,n+1] dx

Here

ϕ(x) = x+ 1 and ψ(x) =

x+ n for x ∈ [0, 1)x for x ∈ [1, n],

are optimal transportation maps.Take u(x) = x, then Lip(u) = 1 and∫

Xu(x) d(µ− − µ+) =

∫ n+1

1x dx−

∫ n

0x dx = n;

also ∫X|ϕ(x)− x| dµ+(x) =

∫X|ψ(x)− x| dµ+ = n.

By Lipschitz criterion, ϕ and ψ are (distinct) minimizers.


Monge’s problem

The continuous case

The continuous caseSuppose µ± are absolutely continuous measures on Rn withrespect to Lebesgue measure

µ+ = f(x) dx µ− = g(x) dx.

A map T : Rn → Rn satisfies T#µ+ = µ− if and only if

f(x) = g(T (x)) det [DT (x)];

if f , g have second finite moments, there exists an optimaltransportation of the form T (x) = ∇ψ(x), where the potential ψis a convex function with respect to a quadratic cost. In this casethe Monge–Ampere equation is satisfied

det [D2ψ(x)] =f(x)

g(∇ψ(x)).


Monge’s problem

Monge-Ampere equation

Monge-Ampere equation

Monge-Ampere equation is a second order PDE, linear in theHessian:

L[u] = A(uxxuyy − u2xy) +Buxx + CUxy +Duyy + E = 0

where the coefficients depend on x, y, ux and uy. It is elliptic, ifAC −B2 −DE > 0. In this case, solutions are unique and amaximum principle applies.The problem of prescribed Gaussian curvature is specified by afunction k definen on a domain Ω in Rn. The problem is to find ahypersurface in Rn+1 in the form of a graph z = u(x) over x ∈ Ωso that at each point the Gaussian curvature is K(x). Theresulting differential PDE is

detD2u−K(x)(1 + |Du|2)(n+2)/2 = 0.


Kantorovich relaxation

Kantorovich relaxation:optimal plan

Given a cost function c : X ×X → R, minimize

min

∫X×X

c(x, y) dγ(x, y)

over the set of probability measures P (X ×X) in the productspace with given marginals

Π(µ+, µ−) = γ ∈ P (X ×X) | π1#γ = µ+, π2#γ = µ−.

The set Π(µ+, µ−) is convex and the problem is linear in γ.



Kantorovich by Kantorovich



Relaxed formulation of Monge’s problem

Relaxed formulation of Monge’s problem

Takeγ = (id× T )#µ

+

Thenπ1#γ = µ+, π2#γ = µ−

and ∫X×X

c(x, y) dγ =

∫Xc(x, T (x)) dµ+

Since any transportation map determines a plan with the samecost, it follows:

minγ∈Π(µ+,µ−)

∫X×X

c(x, y) dγ ≤ infψ#µ+=µ−

∫X|x− ψ(x)| dµ+(x)



Monotonicity of the support of an optimal plan


Let the measures µ± be non-atomic1, γ an optimal plan and(x, y), (x′, y′) ∈ support(γ). Then the Monge property holds

c(x, y) + c(x′, y′) ≤ c(x, y′) + c(x′, y).

For a cost function c(x, y) = h(|x− y|) with h convex, thisproperty implies (x− x′)(y − y′) > 0, therefore support(γ) is amonotone set: x > x′ ⇒ y > y′.

1E is an atom for µ, if µ(E) > 0 and ∀ Borel F , either µ(E ∩ F ) > 0 orµ(E − F ) = 0. µ is non–atomic if there are no atoms.




Monotone support of an optimal plan



The discrete case

The discrete case–Debris: xi ∈ R, embankment: yj ∈ R, i = 1, 2, . . . , n– Undistinguishable particles:

µ+ =1

n

n∑i=1

δxi , µ− =1

n

n∑i=1

δyi

– All transportation plan is represented by a stochastic matrix(πi,j) ∈M:

∑i πi,j =

∑j πi,j = 1.

– The problem of minimization is equivalent to:

minπ∈M

∑i,j

πi,jci,j .

– By Birkhoff-Von Neumann theorem: M is the convex hull ofpermutation matrices. Therefore the problem reduces to

minci,σ(i) | σ ∈ Sn.



Monge matrices

Monge matrices

Monge property readsin the discrete case:

ci,j + cr,s ≤ ci,s + cr,j .

Such matrices arecalled Monge matricesand are very importantin combinatorialoptimization.



Linear programming

Linear programming

Assume x ∈ RN , c ∈ RN , b ∈ RM , A ∈ RM×N . Consider theprimal programming problem of finding x∗ ∈ RN , such that

(P )

minimize c · x, subject to theconstraints Ax = b, x ≥ 0.

and the dual problem which amounts to find y∗ ∈ RM such as to

(D)

maximize b · y, subject to theconstraints AT y ≤ c.

Then the saddle property hods: c · x∗ = b · y∗, that is

minc · x | Ax = b, x ≥ 0 = maxb · y | AT y ≤ c.



Primal and dual Kantorovich: discrete case

Primal and dual formulation of Kantorovich: discrete case

Let the nonnegative number ci,j , µ+i , µ−j , i = 1, 2, . . . , n,

j = 1, 2, . . . ,m. Find µ∗i,j such as to

(P )

min

∑ni=1

∑mj=1 ci,jµi,j subject to the constraints∑n

i=1 µ+i =

∑mj=1 µ

−j ,∑m

j=1 µi,j = µ+i ,

∑ni=1 µi,j = µ−j , µi,j ≥ 0.

(D)

max(∑n

i=1 uiµ+i +

∑mj=1 vjµ

−j

)subject to the constraints

ui + vj ≤ ci,j , i = 1, 2, . . . , n, j = 1, 2, . . . ,m.



Dual Kantorovich: continuous case

Dual Kantorovich: continuous case

Introduce the space of dual variables with x, y ∈ Rn:

L = (u, v) | u, v : Rn → R continuous, u(x) + v(y) ≤ c(x, y)

and the functional defined on L:

K(u, v) =

∫Rn

u(x) dµ+(x) +

∫Rn

v(x) dµ−(x)

then the dual Kantorovich problem is to find (u∗, v∗) ∈ L such that

K(u∗, v∗) = max(u,v)∈L

K(u, v)

.


Crash course on Mather theory

Mather theorem I

Consider a monotone twist map f : A→ A, orientation, area andboundary preserving map of the annulus

A = (x, y) ∈ R2 | x ∈ R(mod 1), a ≤ y ≤ b

It will be identified with the infinite strip. From the twist conditionthere exists maps g(x, y), g′(x′, y′) such that f(x, y) = (x′, y′) ifand only if y = g(x, x′) and y′ = g′(x, x′).



Mather theorem IITheorem (Mather)

Let α < β be the rotation number of the inner and outerboundaries. If α ≤ ω ≤ β then there exists a weak order preservingmap φ : R→ R, φ(x+ 1) = φ(x) + 1, such that

f(φ(t), h(t)) = (φ(t+ ω), h(t+ ω)).

where h(t) = g(φ(t), φ(t+ ω)). The map φ is not necessarilycontinuous but if t is a point of continuity, then also t± ω.

The map φ may have at most a countable set of points ofdiscontinuity. For ω = p/q ∈ Q there exists q–periodic points(Birkhoff’s theorem). Let Mφ be the closure of its points ofcontinuity and Σφ = Mφ(mod 1). If φ is continuous then Σφ ishomeomorphic to a circle, otherwise is a Cantor set (Denjoy’stheorem).



Twist maps and periodic Lagrangians

Twist maps and periodic Lagrangians

Figure: Monotone twist maps can be realized as time-one maps of propertime-periodic Lagrangians with convexity properties (Jurgen Moser:1928-1999)



Tonelli Lagrangians

Tonelli Lagrangians

Let M be a compact connected manifold without boundary. A C2

function L : TM → R is a Tonelli Lagrangian if:• It is strictly convex in the fibers, namely D2,2L(x, v) is positivedefinite for al (x, v) ∈ TM .• Has superlinear growth in the fibers:

lim|v|→∞

L(x, v)

|v|=∞

For a Tonelli Lagrangian the Legendre transform L : TM → T ∗M ,(x, v) 7→ (x,D2L(x, v)) is globally defined. The HamiltonianH(x, p) = p · v − L(x, v), defines the Eule-Lagrange flow. The lastcondition is:• The EL flow is globally defined



Aubry-Mather in higher dimensions

Aubry-Mather in higher dimensions IThe proper setting

ML = µ | µ measure on TM , invariant under the EL flow

Mather’s average action

AL(µ) =

∫TM

Ldµ

There exists µ ∈ML s.t. AL(µ) <∞.Rotation vector of an invariant measure: The map

H1(M,R)→ R [c] 3 η 7→∫TM

η dµ

By duality, there exists ρ(µ) ∈ H1(M,R) s.t.∫TM

η dµ = 〈c, ρ(µ)〉.



Minimizing measures

Minimizing measures

Mather’s beta function. For a rotation vector h ∈ H1(M,R)

β(h) = minµ∈ML,ρ(µ)=h

AL(µ)

µ∗ is a minimizing measure if it realizes the minimum.

AL(µ∗) = β(ρ(µ∗)).

By Fenchel’s duality the map α : H1(TM,R)→ R,

α(c) = maxh∈H1(R,R)

(〈c, h〉 − β(h)) = −min

∫TM

Lη dµ

where η is a closed 1-form such that [η] = c; α(c) is called Manecritical value.



Mather, Aubry and Mane sets

Mather, Aubry and Mane sets IDefine

M(L)h = µ ∈ML | ρ(µ) = h, AL(µ) = β(ρ(µ))M(L)c = µ ∈ML | ρ(µ) = h, AL(µ) = β(ρ(µ))

Then (Aubry–Mather)⋃h∈H1(M,R)M(L)h =

⋃c∈H1(M,R)M(L)c.

Mather sets of given cohomology

Mc = closure

⋃µ∈M(L)c

support(µ)

Mather sets of given homology

Mh = closure

⋃µ∈M(L)h

support(µ)




Mather, Aubry and Mane sets II

Mather sets are compact, invariant under the EL flow and theprojection TM →M restricted to the Mather set is a bilipschitzhomeomporphism (Mather graph theorem).




Mane sets I

Fix a cohomology class c and a 1-form η such that [η] = c. Anabsolutely continuous curve (AC) γ : R→M is a c–minimizer iffor any interval [a, b] and any AC curve γ1 : [a, b]→M s.t.γ(a) = γ1(a), γ(b) = γ1(b),∫ b

aLη(γ(t), γ(t)) dt ≤

∫ b

aLη(γ1(t), γ1(t)) dt.

The Mane set with given cohomology c is:

Nc =⋃t∈R(γ(t), γ(t)) | γ is a c–minimizer




Peierls–Nabarro barrier

Define

hη,t = inf

∫ t

0Lη(γ(s), γ(s)) ds

the infimum is takens over the picewise C1 paths γ : [0, t]→Ms.t. γ(0) = x, γ(t) = x. The Peierls–Nabarro barrier is:

hη(x, y) = lim inft→+∞

(hη,t + α(c)t).

It can be shown that this function is finite and continuous.As pseudo distance is defined on M

δc : (x, y) = hη(x, y) + hη(y, x)




Aubry sets

Define the projected Aubry set

Ac = x ∈M | δc(x, x) = 0.

It is a non-empty, compact invariant space that can be lifted to acompact invariant set of TM by means of a Lipschitz map. Byidentifying points in Ac within a zero pseudodistance -(c–staticclasses) the quotient Aubry set (Ac, δc) is obtained.A c–minimizer is regular if for any x∗, y

∗ in the α and ω–limit set,respectively, δc(x∗, y

∗) = 0. Define the Aubry set

Ac =⋃t∈R(γ(t), γ(t)) | γ is a regular c–minimizer




Fundamental relationship

Mc ⊆ Ac ⊆ Nc ⊆ Ec ⊆ TM↓ ↓ ↓ πMc ⊆ Ac ⊆ ⊆ M

where Ec is the level set corresponding to α(c).


Mather theory as a special case of transportation

Cost function

Let L(x, v, t) be a C2, Tonelli Lagrangian defined on TM × [0, T ],where M is a compact Riemannian manifold. For 0 ≤ s < t ≤ T ,define the cost function

cts(x, y) = min

∫ t

sL(γ(τ), γ(τ)) dτ,

the minimum taken over paths γ : [s, t]→M of class C2,satisfying γ(s) = x, γ(t) = y



Dual Kantorovich: continuous case I

Introduce the space of dual variables with x, y ∈ Rn:

L = (u, v) | u, v : Rn → R continuous, u(x) + v(y) ≤ c(x, y)

and the functional defined on L:

K(u, v) =

∫Rn

u(x) dµ+(x) +

∫Rn

v(x) dµ−(x)

then the dual Kantorovich problem is to find (u∗, v∗) ∈ L such that

K(u∗, v∗) = max(u,v)∈L

K(u, v)

.



Dual Kantorovich: continuous case II

A pair of continuous functions u(x), v(x) on M is called anadmissible Kantorovich pair, if

v(x) = miny∈M

(u(y) + c(x, y)) and u(x) = maxx∈M

(v(y)− c(x, y)).

Support of an optimal plan:If γ is an optimal plan and (u, v) is an optimal Kaontorovich pairthen

support(γ) ⊂ (x, y) ∈M2 | v(x)− u(x) = c(x, y).



Interpolation of measures I

Denote µ+ = µ0, µ− = µT . Denote by Kts(µ0, µT ) an optimal

plan with cost function cts. An interpolation between µ0 and µT isa family of measures µt, t ∈ [0, T ], such that for any0 ≤ t1 ≤ t2 ≤ t3 ≤ T , the equality holds

Ct3t1 (µt1 , µt3) = Ct2t1 (µt1 , µt2) + Ct3t2 (µt2 , µt3)

The following result is due to Bernard and Buffoni.

Theorem (Existence of interpolants)

There exists interpolations between µ0 and µT , they are given bythe flow ψts of a bounded locally Lipschitz vector field X(x, t) onM , such that (ψts)#µs = µt



Interpolation of measures II

Recall α : H1(M,R)→ R from Mather theory:

Theorem

α(0) = minµC1

0 (µ, µ)

where the minimum is taken over the set of probability measureson M . The mapping m0 7→ π#m0 is a bijection between the set ofMather measures m0 and the set of probability measures µsatisfying C1

0 (µ, µ) = α(0).



Appendix: Primal and dual Kantorovich (Evans) I

Let the nonnegative number ci,j , µ+i , µ−j , i = 1, 2, . . . , n,

j = 1, 2, . . . ,m be such that

n∑i=1

µ+i =

m∑j=1

µ−j .

Find µ∗i,j , so as to

min

n∑i=1

m∑j=1

ci,jµi,j

subject to the constraints

m∑j=1

µi,j = µ+i ,

n∑i=1

µi,j = µ−j , µi,j ≥ 0.



Appendix: Primal and dual Kantorovich (Evans) II

The so called Hitchcock problem. It is the primal problem with thedimensions N = nm, M = n+m, variables and cost, restrictionparameters,

x = (µ1,1, µ1,2, . . . , µ1,m, µ2,1, . . . , µn,m),c = (c1,1, . . . , c1,m, c2,1, . . . , cn,m),b = (µ+

1 , µ+2 , . . . , µ

+n , µ

−1 , . . . , µ

−m

and the (n+m)× nm coefficient matrix



Appendix: Primal and dual Kantorovich (Evans) III

A =

11 0 · · · 00 11 · · · 0...

... · · ·...

0 0 · · · 11e1 e1 · · · e1

e2 e2 · · · e2...

... · · ·...

em em · · · em

where the first n rows contain 11 = (1, 1, . . . , 1) ∈ Rn, and thefollowing m rows contain the basise1 = (1, 0, . . . , 0), e2, . . . , em ∈ Rm. Write



Appendix: Primal and dual Kantorovich (Evans) IV

y = (u1, u2, . . . , un, v1, . . . , vm) ∈ Rn+m. Using the explicit formof the matrix AT , the dual problem is

max

n∑i=1

uiµ+i +

m∑j=1

vjµ−j

subject to the constraints

ui + vj ≤ ci,j , i = 1, 2, . . . , n, j = 1, 2, . . . ,m.

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