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 T∗T with T × R. The Hamil-tonian H is given by H(x, p) = 12p2+ V (x). The differential equa-tion on T∗T 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 φ∗t, 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 O−the 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) = Tt−u(x) + c[0]t = Tt+u(x) − c[0]t.
Proof. Since u ≺ L + c[0], we have u ≤ Tt−u + 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)).
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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) ≥ Tt−u(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→+∞Tt−u+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 ˆTt−v = Tt−v + 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 ≤ ˆTt−u. As ˆTt− preserves the order, we see that ˆTt−u ≤ u1− for each u1− ∈ S− satisfying u ≤ u1−. As ˆTt−u = u on the projected Mather set M0, it then remains to show that ˆTt−u is uniformly convergent for t → ∞. However, we have ˆTt−u ≤ 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 ˆTt−u 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 ˆTt−u ≤ 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+= u−on M0. Moreover, we have u+≤ u−on 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 S−x0 = {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 S−x0 − 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 T∗M only if dxu1−= dxu2−, for each x ∈ M .