Definition 4.10.1 (Reversible Lagrangian). The Lagrangian L is said to be reversible if it satisfies L(x, −v) = L(x, v), for each (x, v) ∈ T M .

Example 4.10.2. Let g be a Riemannian metric on M , we denote by k · kx the norm deduced from g on TxM . If V : M → R is C2, the Lagrangian L defined by L(x, v) = 12kvk2x− V (x) is reversible.

Proposition 4.10.3. For a reversible Lagrangian L, we have

−c[0] = inf

x∈ML(x, 0) = inf

(x,v)∈T ML(x, v).

Moreover ˜M0 = {(x, 0) | L(x, 0) = −c[0]}.

Proof. By the strict convexity and the superlinearity of L in the fibers of the tangent bundle T M , we have L(x, 0) = infv∈TxML(x, v), for all x ∈ M . Let us set k = infx∈ML(x, 0) = inf(x,v)∈T ML(x, v).

Since −c[0] = infR

L dµ, where the infimum is taken over all Borel probability measures on T M invariant under the flow φt, we ob-tain k ≤ −c[0]. Let then x0 ∈ M be such that L(x0, 0) = k, the constant curve ] − ∞, +∞[→ M, t 7→ x0 is a minimizing ex-tremal curve. Consequently φt(x0, 0) = (x0, 0) and the Dirac mass δ(x0,0) is invariant by φt, butR

L dδ(x0,0)= k. Therefore −c[0] = k and (x0, 0) ∈ ˜M0. Let µ be a Borel probability measure on T M such thatR

T ML dµ = −c[0]. Since −c[0] = infT ML, we necessar-ily have L(x, v) = infT ML on the support of µ. It follows that supp(µ) ⊂ {(x, 0) | L(x, 0) = −c[0]}.

We then consider the case where M is the circle T = R/Z.

We identify the tangent bundle T T with T × R. As a Lagran-gian L we take one defined by L(x, v) = 12v2 − V (x), where V : T → R is C2. We thus have −c[0] = infT×RL = − sup V , hence c[0] = sup V . Let us identify TT with T × R. The Hamil-tonian H is given by H(x, p) = 12p2+ V (x). The differential equa-tion on TT which defines the flow φt is given by ˙x = p and

˙p = −V(x). If u ∈ S, the compact subset Graph(du) is con-tained in the set H−1(c[0]) = {(x, p) | p = ±p

sup V − V (x)}.

We strongly encourage the reader to do some drawings FAIRE

DES DESSINS of the situation in R × R, the universal cover of T × R. To describe u completely let us consider the case where V reaches its maximum only at 0. In this case the set H−1(c[0]) consists of three orbits of φt, namely the fixed point (0, 0), the orbit O+ = {(x,p

sup V − V (x)) | x 6= 0} and the orbit O = {(x, −p

sup V − V (x)) | x 6= 0}. On O+the direction of the increasing t is that of the increasing x (we identify in a natural way T\ 0 with ]0, 1[). On Othe direction of the increasing t is that of

It follows that there is a point x0 such that Graph(du) is the union of (0, 0) and the two sets {(y,p

sup V − V (y)) | y ∈]0, x0]}

and {(y, −p

sup V − V (y)) | y ∈ [x0, 1[}. Moreover, since the function u is defined on T, we have limx→1u(x) = u(0) and thus the integral on ]0, 1[ of the derivative of u must be 0. This gives the relation

This equality determines completely a unique point x0, since sup V − V (x) > 0 for x ∈]0, 1[. In this case, we see that u is unique up to an additive constant and that

u(x) =

Exercise 4.10.4. 1) If V : T → R reaches its maximum exactly n times, show that the solutions u depend on n real parameters, one of these parameters being an additive constant.

2) Describe the Mather function α : H1(T, R) → R, Ω 7→ c[Ω].

3) If ω is a closed differential 1-form on T, describe the func-tion uω for the Lagrangian Lω defined by L(x, v) = 12v2− V (x) − ωx(v).

Chapter 5

Conjugate Weak KAM Solutions

In this chapter, as in the previous ones, we denote by M a com-pact and connected manifold. The projection of T M on M is denoted by π : T M → M . We suppose given a Cr Lagrangian L : T M → R, with r ≥ 2, such that, for each (x, v) ∈ T M , the second vertical derivative ∂v2L2(x, v) is definite > 0 as a quadratic form, and that L is superlinear in each fiber of the tangent bundle π : T M → M . We will also endow M with a fixed Riemannian metric. We denote by d the distance on M associated with this Riemannian metric. If x ∈ M , the norm k · kx on TxM is the one induced by this same Riemannian metric.

5.1 Conjugate Weak KAM Solutions

We start with the following lemma

Lemma 5.1.1. If u ≺ L + c[0], then we have

∀x ∈ M0, ∀t ≥ 0, u(x) = Ttu(x) + c[0]t = Tt+u(x) − c[0]t.

Proof. Since u ≺ L + c[0], we have u ≤ Ttu + c[0]t and u ≥ Tt+u − c[0]t. We consider the point (x, v) ∈ ˜M0 above x. Let us note by γ : ] − ∞, +∞[→ M the extremal curve s 7→ π(φs(x, v)).

167

By lemma 4.8.2, for each t ≥ 0, we have u(γ(0)) − u(γ(−t)) ≤

Z 0

−t

L(γ(s), ˙γ(s)) ds + c[0]t, u(γ(t)) − u(γ(0)) ≤

Z t 0

L(γ(s), ˙γ(s)) ds + c[0]t.

Since γ(0) = x, we obtain the inequalities u(x) ≥ Ttu(x) + c[0]t and u(x) ≤ Tt+u(x) − c[0]t.

Theorem 5.1.2 (Existence of Conjugate Pairs). If u : M → R is a function such that u ≺ L + c[0], then, there exists a unique function u∈ S(resp. u+∈ S+) with u = u(resp. u = u+) on the projected Mather set M0. These functions verify the following properties

(1) we have u+ ≤ u ≤ u;

(2) if u1∈ S (resp. u1+∈ S+) verifies u ≤ u1 (resp. u1+≤ u), then u≤ u1 (resp. u1+ ≤ u+);

(3) We have u= limt→+∞Ttu+c[0]t and u+= limt→+∞Tt+u−

c[0]t, the convergence being uniform on M .

Proof. It will be simpler to consider the modified semigroup ˆTtv = Ttv + c[0]t. The elements of S are precisely the fixed points of the semigroup ˆTt. The condition u ≺ L + c[0] is equivalent to u ≤ ˆTtu. As ˆTt preserves the order, we see that ˆTtu ≤ u1 for each u1 ∈ S satisfying u ≤ u1. As ˆTtu = u on the projected Mather set M0, it then remains to show that ˆTtu is uniformly convergent for t → ∞. However, we have ˆTtu ≤ ˆTt+s u, if s ≥ 0, because this is true for t = 0 and the semigroup ˆTt preserves the order. Since for t ≥ 1 the family of maps ˆTtu is equi-Lipschitzian, it is enough to see that this family of maps is uniformly bounded.

To show this uniform boundedness, we fix u0 ∈ S, by compact-ness of M , there exists k ∈ R such that u ≤ u0+ k. By what was already shown, we have ˆTtu ≤ u0+ k.

Corollary 5.1.3. For any function u ∈ S (resp. u+ ∈ S+), there exists one and only one function of u+∈ S+(resp. u∈ S) satisfying u+= uon M0. Moreover, we have u+≤ uon all M .

169 Definition 5.1.4 (Conjugate Functions). A pair of functions (u, u+) is said to be conjugate if u∈ S, u+∈ S+ and u= u+ on M0. We will denote by D the set formed by the differences u− u+ of pairs (u, u+) of conjugate functions.

The following lemma will be useful in the sequel.

Lemma 5.1.5 (Compactness of the Differences). All the functions in D are ≥ 0. Moreover, the subset D is compact in C0(M, R) for the topology of uniform convergence.

Proof. If u and u+ are conjugate, we then know that u+ ≤ u

and thus u − u+ ≥ 0. If we fix x0 ∈ M , the set Sx0 = {u | u(x0) = 0} (resp. S+x0 = {u+ | u+(x0) = 0}) is compact, since it is a family of equi-Lipschitzian functions on the compact manifold M which all vanishes at the point x0. However, for c ∈ R, it is obvious that the pair (u, u+) is conjugate if and only if the pair (u+ c, u++ c) is conjugate. We conclude that D is the subset of the compact subset Sx0 − S+x0 formed by the functions which vanish on M0.

Corollary 5.1.6. Let us suppose that all the functions u ∈ S

are C1 (what is equivalent to S = S+). Then, two arbitrary functions in S differ by a constant.

Proof. Conjugate functions are then equal, because the C1 func-tions contained in S or S+ are also in S∩ S+ by 4.7.8. Sup-pose then that u1 and u2 are two functions in S. We of course do have u = (u1+ u2)/2 ≺ L + c[0]. By the Theorem of Ex-istence of Conjugate Pairs 5.1.2, we can find a pair of conju-gate functions (u, u+) with u+ ≤ u ≤ u. As conjugate func-tions are equal, we have u = (u1 + u2)/2 ∈ S. The three functions u, u1 and u2 are C1 and in S, we must then have H(x, dx(u1+ u2)/2) = H(x, dxu1) = H(x, dxu2) = c[0], for each x ∈ M . This is compatible with the strict convexity of H in fibers of TM only if dxu1= dxu2, for each x ∈ M .

Nel documento Weak KAM Theorem in Lagrangian Dynamics Seventh Preliminary Version (pagine 179-184)