WEAK SOLUTIONS OF THE HAMILTON-JACOBI EQUATION FOR TIME PERIODIC LAGRANGIANS

GONZALO CONTRERAS, RENATO ITURRIAGA, AND HÉCTOR SÁNCHEZ-MORGADO

Abstract. In this work we prove the existence of Fathi’s weak KAM solutions for periodic Lagrangians and give a construction of all of them.

0. Introduction and statement of results

Let M be a closed connected manifold, T M its tangent bundle. Let L : T M × ℝ → ℝ be a C^{∞} Lagrangian. We will assume for the Lagrangian the hypothesis of Mather’s seminal paper [9]. The Lagrangian L should be:

(1) Convex. The Lagrangian L restricted to T_{x}M, in linear coordinates should have positive definite Hessian.

(2) Superlinear. For some Riemannian metric we have

lim_{|v|→∞} \frac{L(x, v, t)}{|v|} = ∞,

uniformly on x and t.

(3) Periodic. The Lagrangian should be periodic in time, i.e.

L(x, v, t + 1) = L(x, v, t),

for all x, v, t.

(4) Complete. The Euler Lagrange flow associated to the Lagrangian should be complete.

Let M(L) be the set of probabilities on the Borel σ-algebra of T M ×𝕊^{1} that have compact support and are invariant under the Euler-Lagrange flow ϕ_{t}.

The action of μ ∈ M(L) is defined by

A_{L}(μ) = ∫ L dμ.

Mather defined the function α : H^{1}(M, ℝ) → ℝ as

(1)  α([ω]) = − min {∫ (L − ω) dμ : μ ∈ M(L)} .

All authors were partially supported by CONACYT-México grant # 28489-E.

Keywords: Hamilton-Jacobi equation, weak KAM theory, action potential, periodic Lagrangians, Aubry-Mather theory, Mañé’s critical level, lagrangian graphs.

AMS 2000 Math. Subject Classification: 70H20, 37J50, 70H08, 49L99, 37J05.

For any k in ℝ define the (L + k)-action of an absolutely continuous curve γ : [a, b] → M as

A_{L+k}(γ) = ∫_{a}^{b}(L + k)(γ(τ), γ̇(τ), τ)dτ

For t in ℝ we denote by [t] the corresponding point in 𝕊^{1}. For any pair of points (x, [s]), (y, [t]) on M × 𝕊^{1} and n a non negative integer, define C((x, [s]), (y, [t]); n) as the set of absolutely continuous curves γ : [a, b] → M with γ(a) = x and γ(b) = y such that [a] = [s] and [b] = [t], and the integer part of b − a is n.

Let Φ_{k}^{n} be the real function defined on M × 𝕊^{1} × M × 𝕊^{1} as

Φ_{k}^{n}((x, [s]), (y, [t])) = min_{γ∈C((x,[s]),(y,[t]);n)}{A_{L+k}(γ)}.

so that Φ_{k}^{n} = Φ_{0}^{n} + kn.

Then the action functional is defined by

Φ_{k} = inf_{n} Φ_{k}^{n},

and the Extended Peierls barrier by

h_{k} = lim inf_{n→∞} Φ_{k}^{n}.

Thus Φ_{k} ≤ h_{k}.

A curve γ : [a, b] → M will be called closed if γ(a) = γ(b) and b − a is an integer. In analogy to the autonomous case [8], [2], there is a critical value c(L) given by the following proposition:

1. Proposition.

(1) If k < c(L), then Φ_{k}((x, [s]), (y, [t])) = −∞, for all (x, [s]), (y, [t]) on M × 𝕊^{1} (2)

c(L) = min{k : ∫_{γ} L + k ≥ 0 for all closed curves γ}

(3) If k ≥ c(L), then Φ_{k}((x, [s]), (y, [t])) > −∞ for all (x, [s]), (y, [t]) on M × 𝕊^{1}.

(4) In terms of Mather’s α function we have

(2)  c(L) = − min{∫ Ldμ : μ is an invariant probability}
(3)  = α(0)

Invariant probabilities realizing the infimum above are called minimizing measures.

From now on, set c = c(L).

In contrast with the autonomous case, the action potential Φ_{c} may fail to be continuous and to satisfy the triangle inequality. However, for the extended Peierls barrier we shall prove the following:

2. Proposition.

(1) If k < c, h_{k} ≡ −∞.

(2) If k > c, h_{k} ≡ ∞.

(3) h_{c} is finite.

(4) h_{c}((x, [s]), (z, [τ])) ≤ h_{c}((x, [s]), (y, [t])) + Φ_{c}((y, [t]), (z, [τ])).

(5) h_{c} is Lipschitz.

Let H(x, p, t) be the Hamiltonian associated to the Lagrangian;

H : T^{∗}M × ℝ → ℝ
(4)  H(x, p, t) = max_{v∈T_{x}M} pv − L(x, v, t)

In [4] the critical value or α(0) for the autonomous case is characterized as

c(L) = inf_{f∈C^{∞}(M,ℝ)} sup_{x∈M} H(x, d_{x}f)
= inf{k ∈ ℝ : there exists f ∈ C^{∞}(M, ℝ) such that H(df) < k},

This can be restated in physical terms, by saying that c(L) is either the infimum of the values of k ∈ ℝ for which there is an exact Lagrangian graph with energy less than k, or the infimum of the values of k ∈ ℝ for which there exist smooth solutions of the Hamilton-Jacobi inequality H(df) < k.

The second interpretation has a natural generalization. We will prove in section 2 the following

3. Theorem. The critical value, c(L) or α(0) is characterized as the infimum of k such that there exists a subsolution f : M ×𝕊^{1} → ℝ of the Hamilton Jacobi equation,

d_{t}f + H(x, d_{x}f, t) ≤ k.

We can recover the previous interpretation by using the autonomous Hamiltonian ℍ(x, p, t, e) = H(x, p, t) + e on T^{∗}(M × 𝕊^{1}). Then df = (d_{x}f, d_{t}f) is an exact Lagrangian graph and c(L) = inf_{u} sup_{(x,t)} ℍ(d_{(x,t)}u). The results in [4] can not be directly applied to this case because the Hamiltonian ℍ does not come from a Lagrangian.

The other values of Mather’s alpha function can be similarly characterized by recalling that α([ω]) = c(L − ω) and that the Hamiltonian of L − ω is (x, p, t) ↦ H(x, p + ω, t).

In corollary 14 we observe that differentiable solutions may only exist when k = c(L).

4. Definition. Following Fathi [6] we say that u : M × 𝕊^{1} → ℝ is a forward weak KAM solution if

(1) u is L + c dominated, i.e.

u(y, [t]) − u(x, [s]) ≤ Φ_{c}((x, [s]), (y, [t])).

We use the notation u ≺ L + c.

(2) For every (x, [s]) ∈ M × ℝ there exists a curve γ : (s, ∞) → M such that u(γ(t), [t]) − u(x, [s]) = A_{L+c}(γ|_{[s,t]}), in that case we say that γ realizes u.

Similarly u : M × 𝕊^{1} → ℝ is a backward weak KAM solution if it is dominated and for every (x, [s]) ∈ M × 𝕊^{1} there exists a curve γ : (−∞, s) → M such that u(x, [s]) − u(γ(t), [t]) = A_{L+c}(γ|_{[t,s]}))

Let S^{−} (resp. S^{+}) be the set of backward (resp. forward) weak KAM solutions. A point (x, v, [s]) ∈ T M × 𝕊^{1} is a positive (resp. negative) semistatic point if the

solution γ = γ_{(x,v,s)} of the Euler-Lagrange equation with initial conditions (x, v, [s]), satisfies for all t

A_{L+c}(γ|_{[s,t]}) = Φ_{c}((x, [s]), (γ(t), [t]));

(resp. A_{L+c}(γ|_{[t,s]}) = Φ_{c}((γ(t), [t]), (x, [s])) for all t).

A point (x, v, [s]) ∈ T M × 𝕊^{1} is a static point if it is positive semistatic and

A_{L+c}(γ|_{[s,t]}) = −Φ_{c}((γ(t), [t]), (x, [s])).

It turns out that if a point is static then its whole orbit under the Euler-Lagrange flow is static.

We denote by Σ^{+} the set of positive semistatic points.

For a forward weak KAM solution u we define its forward basin as

Γ_{0}^{+}(u) = {(x, v, [s]) ∈ Σ^{+} : u(γ_{(x,v,s)}(t), [t]) − u(x, [s]) = Φ_{c}((x, [s]), (γ_{(x,v,s)}(t), [t])) ∀t > s};

and define its cut locus by π(Γ_{0}^{+}(u) \ Γ^{+}(u)), where π : T M × 𝕊^{1} → M × 𝕊^{1} is the proyection,

Γ^{+}(u) = ⋃_{t>0} ϕ_{t}(Γ_{0}^{+}(u)),

and ϕ_{t} is the Euler-Lagrange flow. It is easy to see that the sets Σ^{+} and Γ_{0}^{+}(u) are positively invariant and so Γ^{+}(u) ⊂ Γ_{0}^{+}(u). Similarly, define the backward basins Γ_{0}^{−}(u), Γ^{−}(u) for u ∈ S^{−}.

The relevance of weak KAM solutions is that they have several properties, including those given by the following theorem.

5. Theorem. If u : M × 𝕊^{1} → ℝ is a weak KAM solution then (1) u is Lipschitz and satisfies the Hamilton Jacobi equation

H(x, d_{x}u, t) + d_{t}u = c

at any point of differentiability. Moreover, d_{x}u and γ̇ are Legendre conjugate.

(2) Graph property: π : Γ^{+}(u) → M ×𝕊^{1} is injective and its inverse is Lipschitz.

(3) u is differentiable on π(Γ^{+}(u)).

Observe that since a weak KAM solution u is Lipschitz, by Rademacher’s theorem it is differentiable Lebesgue almost everywhere.

Define the Aubry set A as

A := { (x, [t]) ∈ M × 𝕊^{1} | h_{c}((x, [t]), (x, [t])) = 0 }.

We define an equivalence relation on A by (x, [s]) ∼ (y, [t]) if and only if

Φ_{c}((x, [s]), (y, [t])) + Φ_{c}((y, [t]), (x, [s])) = 0.

The equivalence classes of this relation are called static classes.

Let � be the set of static classes. For each static class Γ ∈ � choose a point (p, [s]) ∈ Γ and let 𝔸 be the set of such points.

6. Remark. Observe that by item 4 of proposition 2, if (p, [τ]) ∈ A then

h_{c}((p, [τ]), (x, [t])) = Φ_{c}((p, [τ]), (x, [t])).

7. Theorem. The map {f : 𝔸 → ℝ | f dominated } ⟶ S^{−}

f ⟼ u_{f}(x, [t]) = min_{(p,[s])∈𝔸}f(p, [s]) + h_{c}((p, [s]), (x, [t])),

and the map {f : 𝔸 → ℝ | f dominated } ⟶ S^{+}

f ⟼ v_{f}(x, [t]) = max_{(p,[s])∈𝔸}f(p, [s]) − h_{c}((x, [t]), (p, [s])),

are bijections.

1. The Peierls barrier

We will be using the following lemma due to Mather [9]. We say that an absolutely continuous curve γ : [a, b] → M is a minimizer if A_{L}(γ) ≤ A_{L}(η) for any absolutely continuous curve η : [a, b] → M with η(a) = γ(a) and η(b) = γ(b). It turns out that a minimizer is a solution of the Euler-Lagrange equation \frac{d}{dt}L_{v} = L_{x}.

8. Lemma. There is A > 0 such that if b−a ≥ 1 and γ : [a, b] → M is a minimizer, then |γ̇(t)| ≤ A for t ∈ [a, b].

The proof of most of Propositions 1 and 2 follow standard arguments. We only give the proof of the Lipschitz continuity of h_{c}.

9. Lemma. Given (z, [σ]) ∈ M × 𝕊^{1} define

u(x, [t]) := h_{c}((z, [σ]; x, [t]), v(x, [t]) := −h_{c}((x, [t]; z, [σ]).

Then u ∈ S^{−} and v ∈ S^{+}.

Proof: By item 4 of proposition 2, h(z, [σ]), (x, [t])) ≤ h_{c}((z, [σ]), (y, [s]))+Φ_{c}((y, [s]), (x, [t])) for all (y, [s]), (x, [t]) ∈ M × 𝕊^{1}. Thus u ≺ L + c.

Given (x, [t]) ∈ M × 𝕊^{1} choose sequences n_{k} → +∞, n_{k} ∈ ℤ and (x, v_{k}) ∈ T_{x}M such that

h_{c}((z, [σ]), (x, [t])) = lim_{k} A_{L+c}(γ_{k}|_{[σ−n_{k},t]}),

where γ_{k}(s) = π φ_{s−t}(x, v_{k}, t) is the solution of the Euler-Lagrange equation such that (γ_{k}(t), γ̇_{k}(t)) = (x, v_{k}). By lemma 8, the norm ∥v_{k}∥ is uniformly bounded. Choose a convergent subsequence v_{k} → w. Let η(s) := π φ_{s−t}(x, w, t), then for any fixed s < 0,

h_{c}((z, [σ]), (x, [t])) ≤ h_{c}((z, [σ]), (η(s), [s])) + A_{L+c}(η|_{[s,t]})

= lim_{k} h_{c}((z, [σ]), (γ_{k}(s), [s])) + A_{L+c}(γ_{k}|_{[s,t]})

≤ lim_{k} A_{L+c}(γ_{k}|_{[σ−n_{k},s]}) + A_{L+c}(γ_{k}|_{[s,t]})

= h_{c}((z, [σ]), (x, [t])).

So that u(x, [t]) − u(η(s), [s]) = A_{L+c}(η|_{[s,t]}) for all s < 0.

□

For autonomous lagrangians, dominated functions are Lipschitz. In contrast, for time periodic lagrangians the action potential is dominated but it is not continuous at ((x, s), (x, s)) when (x, s) is not in the Aubry set. Nevertheless, we have the following:

10. Lemma. If u : M × 𝕊^{1} → ℝ is a weak KAM solution (i.e. u ∈ S^{+} ∪ S^{−}) then it is Lipschitz. Moreover the Lipschitz constant does not depend on u.

Proof: Assume that u ∈ S^{−}, the case u ∈ S^{+} is similar. Let (x, [t_{0}]), (y, [s_{0}]) ∈ M × 𝕊^{1} be nearby points with |s_{0} − t_{0}| < \frac{1}{4}. Let γ : [0, 1] → M be a length minimizing geodesic joining x to y and let τ(r) = t_{0} + r (s_{0} − t_{0}), r ∈ [0, 1]. Fix δ > 1 and let z : [t_{0} − δ, t_{0}] → M be such that

(5)  u(x, [t_{0}]) = u(z(t), [t]) + ∫_{t}^{t_{0}} L(z, ż) + c dt for all t_{0} − δ < t ≤ t_{0}.

For r ∈ [0, 1], let η(r, t), t ∈ [t_{0} − δ, τ(r)], be a locally minimizing solution of (E-L) such that η(r, t_{0} − δ) = z(t_{0} − δ) and η(r, τ(r)) = γ(r).

Then

u(γ(r), [τ(r)]) ≤ u(z(t_{0} − δ), [t_{0} − δ]) + ∫_{t_{0}−δ}^{τ(r)} L(η, \frac{∂η}{∂t}, t) + c dt.

with equality for r = 0. Substracting the equality (5) at r = 0, we get that

(6)  u(γ(r), [τ(r)]) − u(x, [t_{0}]) ≤ ∫_{t_{0}−δ}^{τ(r)} (L + c) dt − A_{L+c}(z|_{[t_{0}−δ,t_{0}]}).

Observe that this formula holds either for s_{0} ≤ t_{0} or t_{0} ≤ s_{0}. As we shall see below, formula (6) implies that u(y, s)−u(x, t) ≤ K ( |s−t|+d(x, y) ) for some fixed K > 0. Then changing the roles of s and t we get that u is Lipschitz.

Indeed, differentiating the right hand side and integrating by parts, we have

\frac{d}{dr} ∫_{t_{0}−δ}^{t_{0}+r (s_{0}−t_{0})} L(η, \frac{∂η}{∂t}, t) + c dt =

= [L(η, \frac{∂η}{∂t}, t)|_{(r,τ(r))} + c] (s_{0} − t_{0}) + ∫_{t_{0}−δ}^{τ(r)} L_{x} \frac{∂η}{∂r} + L_{v} \frac{∂^{2}η}{∂t∂r}

= [L(η, \frac{∂η}{∂t}, t)|_{(r,τ(r))} + c] (s_{0} − t_{0}) + \frac{∂L}{∂v}(η, \frac{∂η}{∂t}, t))|_{(r,τ(r))} · \frac{∂η}{∂r} |_{(r,τ(r))}

Observe that since u is dominated the realizing curve z must be a minimizer. By lemma 8, \left∥{ \atop ż}\right∥ is uniformly bounded. By the continuity of the solutions of (E-L) with respect to initial values, ∥\frac{∂η}{∂t} ∥ is uniformly bounded. Hence there is a uniform constant K > 0 (independent of z(t), x, y, [s_{0}], [t_{0}], u) such that

|L(η, \frac{∂η}{∂t}, t) + c| ≤ K  and  ∥\frac{∂L}{∂v}(η, \frac{∂η}{∂t}, t)∥ < K.

Since \frac{∂η}{∂r} |_{(r,τ(r))} = γ̇(r), we get that

\frac{d}{dr} [∫_{t_{0}−δ}^{τ(r)}[L + c] − A_{L+c}(z)] ≤ K |s_{0} − t_{0}| + K \left∥{ \atop γ̇}\right∥ .

The value of the right hand side of (6) is 0 at r = 0. Integrating this inequality,

u(y, [s_{0}]) − u(x, [t_{0}]) ≤ K [|s_{0} − t_{0}| + d(x, y)].

Interchanging the roles of (x, [t_{0}]) and (y, [s_{0}]) we obtain that the function u is Lipschitz.

□

Combining lemmas 9 and 10 we get that the functions f, g : M × 𝕊^{1} → ℝ, f(y, [t]) = h_{c}((x, [s]), (y, [t])) and g(x, [s]) = h_{c}((x, [s]), (y, [t])) are Lipschitz. This implies that h_{c} is Lipschitz.

2. Subsolutions of the Hamilton Jacobi equation

Following the same ideas as in [4], one obtains

11. Lemma. If k is a real number such that there exists a function f in C^{1}(M ×𝕊^{1}) subsolution of the Hamilton Jacobi equation

H(x, d_{x}f) + d_{t}f ≤ k

Then k ≥ c(L).

12. Lemma. Let k ≥ c(L). If f : M × 𝕊^{1} → ℝ is differentiable at (x, [t]) ∈ M × 𝕊^{1} and satisfies

f(y, [t_{2}]) − f(x, [t_{1}]) ≤ Φ_{k}(x, [t_{1}], y, [t_{2}])

for all y in a neighbourhood of x, then H(x, d_{x}f) + d_{t}f ≤ k.

13. Proposition. For any k > c(L) there exists f ∈ C^{∞}(M × 𝕊^{1}, ℝ) such that H(x, d_{x}f, t) + d_{t}f < k.

We give a proof of the following fact

14. Corollary. If u is a C^{1+Lip} global solution of the Hamilton-Jacobi equation u_{t} + H(x, u_{x}, t) = k, then k = c(L) and u is a weak KAM solution in S^{−} ∩ S^{+}.

Proof: By lemma 11, k ≥ c(L). Let L_{t}(x, v) = L_{v}(x, v, t) be the conjugate moment associated to L and let ξ(x, t) be the vector field defined by

ξ(x, t) = L_{t}^{−1}(u_{x}) ∈ T_{x}M. Then the vector field (ξ, 1) in M × 𝕊^{1} is Lipschitz. Let ρ_{t} be the flow of (ξ, 1) in M × 𝕊^{1}. From the Hamilton-Jacobi equation we get that

(7)  d_{(x,[t])}u · (v, 1) = u_{x}(x, t) · v + u_{t} · 1 ≤ L(x, v, [t]) + k.

and that

(8)  d_{(x,[t])}u · (ξ(x, t), 1) = L(x, ξ(x, t), t) + k for all (x, [t]) ∈ M × 𝕊^{1}.

Integrating equation (7) along absolutely continuous curves (γ(t), [t]) in M ×𝕊^{1} from (x, [s]) to (y, [t]), we get that

u(y, [t]) − u(x, [s]) ≤ inf_{γ} ∮_{γ}(L + k) = Φ_{k}((x, [s]), (y, [t])).

So that u ≺ L + k.

Also, integrating equation (8), we get that the orbits of ρ_{t} realize u in the sense of the definition of a weak KAM solution. In particular, the orbits of ρ are global minimizers of the (L + k)-action, and thus they are solutions of the Euler-Lagrange equation.

It remains to prove that k = c(L). Let ν be an invariant Borel probability for ρ_{t} and let μ be its lift to T M × 𝕊^{1} using the vectorfield ξ. Then μ is an invariant probability of the Lagrangian flow and, by equation (8),

∫ (L + k) dμ = ∫ du dμ = 0.

This implies that k ≤ c(L). Thus k = c(L) and also μ is a minimizing measure. □

3. Weak KAM solutions

Proof of theorem 5:

We first prove item 1. By lemma 10 we have that u is Lipschitz and hence it is differentiable almost everywhere. Let (x, [t]) be a point of differentiability, then by lemma 12 we have

H(x, d_{x}u, t) + u_{t} ≤ c.

Moreover let γ : [t, ∞) → M be such that

u(γ(s), [s]) − u(x, [t]) = A_{L+c}(γ|_{[t,s]}),

lim_{s→t} \frac{u(γ(s), [s]) − u(x, [t])}{s − t} = lim_{s→t} \frac{1}{s − t} ∫_{t}^{s}(L + c)(γ(s), γ̇(s), [s])ds,

so

d_{x}u(x, [t])γ̇ + d_{t}u(x, [t]) = L(x, γ̇, t) + c.

Therefore

c = d_{x}u γ̇ − L + d_{t}u ≤ H(x, d_{x}u, t) + d_{t}u ≤ c.

So u is a solution of the Hamilton Jacobi Equation and d_{x}u and γ̇ are related by the Legendre transformation of L.

□

Proof of the Graph Property:

We need the following lemma due to Mather, a proof of which can be found in [9].

15. Lemma. Given A > 0 there exists K > 0 ε_{1} > 0 and δ > 0 with the following property: if |v_{i}| < A, (p_{i}, v_{i}, [t_{i}]) ∈ T M ×𝕊^{1}, i = 1, 2 satisfy d((p_{1}, [t_{1}]), (p_{2}, [t_{2}])) < δ and d((p_{1}, v_{1}, [t_{1}]), (p_{2}, v_{2}, [t_{2}])) ≥ K^{−1}d((p_{1}, [t_{1}]), (p_{2}, [t_{2}])) then, if a ∈ ℝ and x_{i} : ℝ → M, i = 1, 2, are the solutions of L with x_{i}(t_{i}) = p_{i}, ẋ_{i}(p_{i}) = v_{i}, there exist solutions γ_{i} : [t_{i} − ε, t_{i} + ε] → M of L with 0 < ε < ε_{1}, satisfying

γ_{1}(t_{1} − ε) =x_{1}(t_{1} − ε) , γ_{1}(t_{2} + ε) = x_{2}(t_{2} + ε) ,
γ_{2}(t_{2} − ε) =x_{2}(t_{2} − ε) , γ_{2}(t_{1} + ε) = x_{1}(t_{1} + ε) ,
S_{L}(x_{1}|_{[t_{1}−ε,t_{2}+ε]}) + S_{L}(x_{2}|_{[t_{2}−ε,t_{1}+ε]}) > S_{L}(γ_{1}) + S_{L}(γ_{2})

We now prove the graph property. Let (p_{1}, v_{1}, [t_{1}]), (p_{2}, v_{2}, [t_{2}]) ∈ Γ^{+}(u) and suppose that K d((v_{1}, [t_{1}]), (v_{2}, [t_{2}])) > d((p_{1}, [t_{1}]), (p_{2}, [t_{2}])), where K is from lemma 15 and the A that we input on lemma 15 is from lemma 8. Let y_{i}^{+} = x_{i}(t_{i} + ε), i = 1, 2, and y_{i}^{−} = x_{i}(t_{i} − ε) for ε small, then

(9)  u(y_{1}^{+}, [t_{1} + ε]) − u(y_{1}^{−}, [t_{1} − ε]) = Φ_{c}((y_{1}^{−}, [t_{1} − ε]), (y_{1}^{+}, [t_{1} + ε]))
(10)  u(y_{2}^{+}, [t_{2} + ε]) − u(y_{2}^{−}, [t_{2} − ε]) = Φ_{c}((y_{2}^{−}, [t_{2} − ε]), (y_{2}^{+}, [t_{2} + ε]))

Then using that u ≺ L + c and lemma 15 , we get that

(11) u(y_{2}^{+}, [t_{2} + ε]) − u(y_{1}^{−}, [t_{1} − ε]) + u(y_{1}^{+}, [t_{1} + ε]) − u(y_{2}^{−}, [t_{2} − ε])

≤ S_{L+c}(γ_{1}) + S_{L+c}(γ_{2})

< S_{L+c}(x_{1}|_{[t_{1}−ε,t_{2}+ε]} + S_{L+c}(x_{2}|_{[t_{2}−ε,t_{1}+ε]})

= Φ_{c}((y_{1}^{−}, [t_{1} − ε]), (y_{1}^{+}, [t_{1} + ε]) + Φ_{c}((y_{2}^{−}, [t_{2} − ε]), (y_{2}^{+}, [t_{2} + ε]).

Which is a contradiction with the sum of (9) and (10). □

Proof of item 3:

Let (x, [s]) in πΓ^{+}(u), let (σ(τ), [α(τ)]) be a curve on M ×𝕊^{1} with σ(0) = x, α(0) = s. Let γ_{s} be the curve such that

u(γ_{s}(t), [t]) − u(γ_{s}(s − δ), [s − δ]) = A_{L+c}(γ_{s}|_{[s−δ,t]}).

Since (x, [s]) is in πΓ^{+}(u) we can make a backwards variation (γ_{τ}) of the solution γ_{s}. That is, γ_{τ} : [s − δ, α(τ)] → M is a solutions of the Euler -Lagrange equation joining the points p = γ_{s}(s − δ) and σ(τ).

Since u is dominated we have

u(σ(τ), [α(τ)]) − u(x, [s]) = u(σ(τ), [α(τ)]) − u(p, [s − δ]) − (u(x, [s]) − u(p, [s − δ]))
≤ A_{L+c}(γ_{τ}|_{[s−δ,α(τ)]}) − A_{L+c}(γ_{s}|_{[s−δ,s]})
= A_{L+c}(γ_{τ}|_{[s−δ,s]}) − A_{L+c}(γ_{s}|_{[s−δ,s]}) + A_{L+c}(γ_{τ}|_{[s,α(τ)]})

Dividing by τ − s and taking limits as τ tends to s and using the fact that γ_{τ} is a solution of the Euler-Lagrange equation, we obtain

lim sup_{τ→s} \frac{u(σ(τ), [α(τ)]) − u(x, [s])}{τ − s} ≤ L_{v}(γ̇_{s}, s) · σ′(0) + L + c(γ̇_{s}, s)α′(0)

Similarly we can make a forward variation to get

lim inf_{τ→s} \frac{u(σ(τ), [α(τ)]) − u(x, [s])}{τ − s} ≥ L_{v}(γ̇_{s}, s) · σ′(0) + L + c(γ̇_{s}, s)α′(0)

□

Proof of theorem 7:

Let u ∈ S^{−}, since u is dominated, then

(12)  u(x, [t]) ≤ min_{(y,[r])} u(y, [r]) + Φ_{c}((y, [r]), (x, [t])).

Let γ :] − ∞, t] → M be such that for all s ≤ t,

u(x, [t]) − u(γ(s), [s]) = A_{L+c}(γ|_{[s,t]}).

Then γ(s) is semistatic and the minimum in (12) is realized at every point (γ(s), [s]) with s < t. Choose a convergent sequence (γ(s_{n}), [s_{n}]) → (p, [τ]) ∈ M × 𝕊^{1}, with s_{n} → −∞. Then by lemma 16 below, (p, [τ]) is in the Pierls set. Therefore, using the continuity of Φ_{c} at (p, [τ]) (see lemma 17 below) and (12), we have that

u(x, [t]) = u(p, [τ]) + Φ_{c}((p, [τ]), (x, [t]))

= min_{(q,[σ])∈A} u(q, [σ]) + Φ_{c}((q, [σ]), (x, [t])). (13)

We show now that it is enough to choose one point on each static class to achieve the minimum on (13). Suppose that (p, [τ]) and (q, [σ]) are in the same static class. Then

Φ_{c}((q, [σ]), (x, [t])) ≤ Φ_{c}((q, [σ]), (p, [τ])) + Φ_{c}((p, [τ]), (x, [t]))
≤ Φ_{c}((q, [σ]), (p, [τ])) + Φ_{c}((p, [τ]), (q, [σ])) + Φ_{c}((q, [σ]), (x, [t]))
= Φ_{c}((q, [σ]), (x, [t])).

So that Φ_{c}((q, [σ]), (x, [t])) = Φ_{c}((q, [σ]), (p, [τ])) + Φ_{c}((p, [τ]), (x, [t])). Moreover,

u(p, [τ]) ≤ u(q, [σ])) + Φ_{c}((q, [σ], p, [τ]))
≤ u(p, [τ]) + Φ_{c}((p, [τ]), (q, [σ])) + Φ_{c}((q, [σ]), (p, [τ]))
= u(p, [τ]).

So that u(q, [σ]) + Φ_{c}((q, [σ]), (p, [τ])) = u(p, [τ]). Thus

u(q, [σ]) + Φ_{c}((q, [σ]), (x, [t])) = u(q, [σ]) + Φ_{c}((q, [σ]), (p, [τ])) + Φ_{c}((p, [τ]), (x, [t]))
= u(p, [τ]) + Φ_{c}((p, [τ]), (x, [t])).

So that u = u_{f}, with f = u|_{𝔸}.

Observe that by definition, if f : 𝔸 → ℝ is dominated, then u_{f}|_{𝔸} ≡ f. This implies that the map {f dominated} ↦ u_{f} is injective.

Finally, it remains to prove that if f : 𝔸 → ℝ then u_{f} ∈ S^{−}. This follows from lemma 9 and lemma 18 below. □

16. Lemma. If γ :] − ∞, t_{0}] → M is semistatic and s_{n} → −∞ is such that lim_{n} (γ(s_{n}), [s_{n}]) = (p, [τ]) exists. Then (p, [τ]) is in the Aubry set.

Proof: Let ε > 0 be small. Chose n_{0} > 0 such that for n > n_{0}, we have

|s_{n} − τ mod 1| < \frac{ε}{2}  ,  d(γ(s_{n}), p) < \frac{ε}{2}.

Let λ_{n}^{−} : [τ, s_{n}+ε mod 1] → M be a minimizer with λ_{n}^{−}(τ) = p, λ_{n}^{−}(s_{n}+ε mod 1) = γ(s_{n} + ε). By lemma 8, \left∥{ \atop γ̇}\right∥ is uniformly bounded. By the same argument, using the first variation formula, as in proposition 2.c,

A_{L+c}(λ_{n}^{−}) ≤ K_{1} [ d(γ(s_{n}), p) + |s_{n} + ε − τ mod 1| ] ≤ 3 ε K_{1}.

Let λ_{n}^{+} : [s_{n} − ε mod 1, τ] → M be a minimizer with λ_{n}^{+}(s_{n} − ε) = γ(s_{n} − ε), λ_{n}^{+}(τ) = p. Similarly,

A_{L+c}(λ_{n}^{+}) ≤ 3 ε K_{1}.

We have that

h_{c}((p, [τ]), (p, [τ])) ≤ lim inf_{N→∞} A_{L+c}(λ_{N}^{−}) + A_{L+c}(γ|_{[s_{N} +ε,s_{n}−ε]}) + A_{L+c}(λ_{n}^{+})

≤ 6 ε K_{1} + lim inf_{N} A_{L+c}(γ|_{[s_{N} +ε,s_{n}−ε]}). (14)

Adding the action of γ on the intervals with endpoints s_{N} −ε < s_{N} +ε < s_{n} −ε < s_{n} + ε and using that γ is semistatic on [s_{N} − ε, s_{n} + ε], we have that

A_{L+c}(γ|_{[s_{N} +ε,s_{n}−ε]}) = Φ_{c}((γ(s_{N} − ε), s_{N} − ε) , (γ(s_{n} + ε), s_{n} + ε))

− A_{L+c}(γ|_{[s_{N} −ε,s_{N} +ε]}) − A_{L+c}(γ|_{[s_{n}−ε,s_{n}+ε]}). (15)

Comparing Φ_{c} with the action of a minimal length geodesic, parameterized by the small interval I = [s_{N} − ε mod 1, s_{n} + ε mod 1] of length ε ≤ ℓ(I) ≤ 3ε, with

speed ≤ \frac{1}{ε} d(γ(s_{N} − ε), γ(s_{n} + ε)) ≤ \frac{1}{ε} [ ε \left∥{ \atop γ̇}\right∥ + d(γ(s_{N}), γ(s_{n})) + ε \left∥{ \atop γ̇}\right∥ ]

≤ 2 \left∥{ \atop γ̇}\right∥ + 1;

we have that

Φ_{c}((γ(s_{N} − ε), s_{N} − ε) , (γ(s_{n} + ε), s_{n} + ε)) ≤ ℓ(I) [ max_{|v|≤2 \left∥{ \atop γ̇}\right∥+1} L + c ] ≤ 3 ε K_{2}.

The two actions in (15) are bounded by 2 (2ε · K_{2}). Thus, from (15),

A_{L+c}(γ|_{[s_{N} +ε,s_{n}−ε]}) ≤ 7 ε K_{2}.

From (14),

0 ≤ h_{c}((p, [τ]), (p, [τ])) ≤ 6 ε K_{1} + 7 ε K_{2}.

Now let ε → 0. □

17. Lemma. If lim_{n}(y_{n}, [s_{n}]) = (p, [τ]) ∈ A then for all (x, [t]) ∈ M × 𝕊^{1},

lim_{n} Φ_{c}((y_{n}, [s_{n}]), (x, [t])) = Φ_{c}((p, [τ]), (x, [t])) = h_{c}((p, [τ]), (x, [t])).

Proof: Recall that by remark 6, h_{c}((p, [τ]), (x, [t])) = Φ_{c}((p, [τ]), (x, [t])). By item 4 of proposition 2,

Φ_{c}((p, [τ]), (x, [t])) = h_{c}((p, [τ]), (x, [t]))

≤ h_{c}((p, [τ]), (y_{n}, [s_{n}])) + Φ_{c}((y_{n}, [s_{n}]), (x, [t])) (16)

≤ h_{c}((p, [τ]), (y_{n}, [s_{n}])) + h_{c}((y_{n}, [s_{n}]), (x, [t]))

≤ h_{c}((p, [τ]), (y_{n}, [s_{n}])) + h_{c}((y_{n}, [s_{n}]), (p, [τ])) + Φ_{c}((p, [τ]), (x, [t])) (17)

Using that h_{c} is continuous, taking lim_{n} on inequalities (16) and (17), we get that lim_{n} Φ_{c}((y_{n}, [s_{n}]), (x, [t])) = Φ_{c}((p, [τ]), (x, [t])).

□

18. Lemma.

If U ⊂ S^{−}, let 𝔲(x, [t]) := inf_{u∈U} u(x, [t]) then either 𝔲 ≡ −∞ or 𝔲 ∈ S^{−}. If V ⊂ S^{+}, let 𝔳(x, [t]) := sup_{v∈V} v(x, [t]) then either 𝔳 ≡ +∞ or 𝔳 ∈ S^{+}.

Proof: Since u ≺ L + c for all u ∈ U, for all (x, [s]), (y, [t]) ∈ M × 𝕊^{1},

u(y, [t]) ≤ u(x, [s]) + Φ_{c}((x, [s]), (y, [t])),  for all u ∈ U,
min_{u∈U} u(y, [t]) = 𝔲(y, [t]) ≤ u(x, [s]) + Φ_{c}((x, [s]), (y, [t])),  for all u ∈ U,
(18)  𝔲(y, [t]) ≤ 𝔲(x, [s]) + Φ_{c}((x, [s]), (y, [t])).

Now fix (x, [t]) ∈ M × 𝕊^{1} and fix a sequence u_{k} ∈ U such that 𝔲(x, [t]) = lim_{k} u_{k}(x, [t]). Let (x, v_{k}, [t]) ∈ Γ^{−}(u_{k}). By lemma 8, ∥v_{k}∥ is uniformly bounded. We can assume that v_{k} → w. Let γ_{v_{k}}(s) := π φ_{s−t}(x, v_{k}, t) and γ_{w}(s) := π φ_{s−t}(x, w, t). Then

u_{k}(x, t) = u_{k}(γ_{v_{k}}(s), [s]) + A_{L+c}(γ_{v_{k}}|_{[s,t]}),  for all s < t,

Since γ_{v_{k}} ⟶^{C^{1}} γ_{w} uniformly on bounded intervals, using that by lemma 10 all the u_{k}’s have the same Lipschitz constant, taking the lim inf on k we get that

(19)  𝔲(x, t) ≥ 𝔲(γ_{w}(s), [s]) + A_{L+c}(γ_{w}|_{[s,t]}),  for all s < t,

The domination condition (18) implies that (19) is an equality. □

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CIMAT, A.P. 402, 3600, Guanajuato. Gto, México

E-mail address: gonzalo@cimat.mx

CIMAT, A.P. 402, 3600, Guanajuato. Gto, México

E-mail address: renato@cimat.mx

Instituto de Matemáticas, UNAM. Ciudad Universitaria C. P. 04510, Cd. de México, México.

E-mail address: hector@math.unam.mx