The Lax-Oleinik semi-group on graphs

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by  Renato Iturriaga

THE LAX-OLEINIK SEMIGROUP ON GRAPHS
6 RENATO ITURRIAGA AND HE´CTOR SA´NCHEZ-MORGADO
1
0
Abstract. We consider Tonelli Lagrangians on a graph, define weak KAM so-
2
lutions, prove that they are the fixed points of the Lax-Oleinik semi-group and
n
identify their uniqueness set as the Aubry set, giving a representation formula.
a
J Our main result is the long time convergence of the Lax Oleinik semi-group. We
4 also prove that weak KAM solutions are viscosity solutions, and in the case of
1 Hamiltonians of eikonal type, that viscosity solutions are weak KAM solutions

]
P
A
1. Introduction
.
h
t We study the Lax-Oleinik semi-group defined by a Tonelli Lagrangian on a graph
a
m and prove its long time convergence. For Lagrangians on compact manifolds a
[ standard proof uses the Euler-Lagrange flow and conservation of energy. In our

1 case we do not have these tools but we can follow ideas of Roquejoffre [R] and
v Davini-Siconolfi [DS].
7
Camilli and collaborators [ACCT, CM, CS] have studied viscosity solutions of the
7
5 Hamilton-Jacobi equation and given sufficient conditions for a set to be a uniqueness
3
set and a representation formula.
0
. In this paper we consider a graph G without boundary consisting of unoriented
1
edges {I } and vertices {v }. We will make the convention that at a vertix v
0 j a a
6 the zero vectors in T I all coincide. A Lagrangian in G is a Ck, strictly convex
1 and super-linear funcvtaiojn L : TG → R given, according to our convention, by a
:
v collection of Ck functions L : TI → R such that L |T I is convex and super-linear
i j j j x j
X for any x ∈ I , any j, and L (v ,0) = L (v ,0) for v ∈ I ∩I . We will say that a
j i a j a a i j
r Lagrangian is symmetric at the vertices if L (v ,z) = l (|z|) for v ∈ I , z ∈ T I .
a i a a a j va j
As an example we have a mechanical Lagrangian. For x in the interior of I , it
j
makes sense to say that ζ ∈ T I −{0} points towards one of the vertices of I , and
x j j
for v ∈ I it makes sense to say that ζ ∈ T I −{0} is incoming or outgoing. We
a j va j
set
T+I = {z ∈ T I : z is outging or z = 0},
va j va j
T−I = {z ∈ T I : z is incoming or z = 0}.
va j va j
2. Basic properties of the action
The purpose of this section is to obtain the lower semicontinuity of the action
and apriori bounds for the W1,∞ norm of minimizers. Denote by Cac([a,b]) the set
1
2 R. ITURRIAGAAND H.SA´NCHEZ-MORGADO
of of absolutely continuous functions γ : [a,b] → G provided with the topology of
uniform convergence. For γ ∈ Cac([a,b]) we define its action by
b
A(γ) = L(γ(t)γ˙(t))dt
Z
a
A minimizer is a γ ∈ Cac([a,b]) such that for any α ∈ Cac([a,b]) with α(a) = γ(a),
α(b) = γ(b) we have
A(γ) ≤ A(α)
The following two properties of the Lagrangian are important to achieve our goal
and follow from its convexity and super-linearity.
Proposition 1. If C ≥ 0, ε > 0, there is η > 0 such that for x,y ∈ I , d(x,y) < η
j
and v ∈ T G,|v| ≤ C, w ∈ T G we have
x y
L(y,w) ≥ L(x,v)+L (x,v)(w−v)−ε.
v
Proposition 2. If L ≥ θ > 0, C ≥ 0, ε > 0, there is η > 0 such that for x,y ∈ I ,
vv j
d(x,y) < η and v ∈ T G,|v| ≤ C, w ∈ T G we have
x y
3θ
L(y,w) ≥ L(x,v)+L (x,v)(w −v)+ |w−v|2 −ε.
v
4
Lemma 1. Let L be a Lagrangian on G. If a sequence γ ∈ Cac([a,b]) converges
n
uniformly to the curve γ : [a,b] → G and
liminfA(γ ) < ∞
n
n→∞
then the curve γ is absolutely continuous and
liminfA(γ ) ≥ A(γ).
n
n→∞
Proof. Let l = liminfA(γ ). Passing to a subsequence we can assume that
n
n→∞
A(γ ) < l+1, ∀n ∈ N
n
By the superlinearity we may assume that L ≥ 0. Fix ε > 0 and take B > 2(l+1)/ε.
Again by superlinearity there is a positive number C(B) such that
L(x,v) ≥ Bkvk−C(B), x ∈ G,v ∈ T G
x
Since L ≥ 0, for E ⊂ [a,b] measurable we have
−C(B)Leb(E)+B kγ˙ k ≤ L(γ ,γ˙ )+ L(γ ,γ˙ ) ≤ l +1.
Z n Z n n Z n n
E E [a,b]\E
Thus
1 ε C(B)Leb(E)
kγ˙ k ≤ (l+1+C(B)Leb(E)) ≤ + .
Z n B 2 B
E
LAX-OLEINIK ON GRAPHS 3
εB
Choosing 0 < δ < we have that
2C(B)
Leb(E) < δ ⇒ ∀n ∈ N kγ˙ k < ε.
Z n
E
Since the γ˙ are uniformly integrable, we have that they converge to γ˙ in the
n
σ(L1,L∞) weak topology.
We may assume that γ ([a,b]) is contained in a compact neighbourhood K of
n
γ([a,b]).
1
Let ε > 0 and E = t : |γ˙(t)| ≤ C,d(γ(t),{v }) ≥ , then by Proposition 1,
C a
C
(cid:8) (cid:9)
for n large,
(1) L(γ,γ˙)+L (γ,γ˙)(γ˙ −γ˙)−ε ≤ L(γ γ˙ )
Z v n Z n n
EC(cid:2) (cid:3) EC
Letting n → +∞ on (1), we have that
L(γ,γ˙)−ε (b−a) ≤ l.
Z
EC
Since E ↑ [a,b] when C → +∞ and L ≥ 0, we have
C
A(γ) = lim L(γ,γ˙) ≤ l +ε (b−a).
Z
C→+∞
EC
Now let ε → 0. (cid:3)
Lemma 1 imply
Theorem 1. Let L be a Lagrangian on G. The action A : Cac([a,b]) → R∪{∞} is
lower semicontinuous.
Proposition 3. Let L be a Lagrangian on G.
The set {γ ∈ Cab([a,b]) : A(γ) ≤ K} is compact with the topology of uniform
convergence.
Let C = sup{L(x,v) : x ∈ G,|v| ≤ diam(G)}, then for any minimizer γ : [a,b] →
t t
G with b−a ≥ t we have
A(γ) ≤ C(b−a).
Proposition 4. Suppose γ : [a,b] → G converge uniformly to γ : [a,b] → G and
i
A(γ ) converges to A(γ), then γ˙ converges to γ˙ in L1[a,b]
i i
Proof. From Lemma 1 we have that if F ⊂ [a,b] is a finite union of intervals
L(γ,γ˙) ≤ liminf L(γ,γ˙) and L(γ,γ˙) ≤ liminf L(γ,γ˙)
ZF n ZF ZFc n ZFc
4 R. ITURRIAGAAND H.SA´NCHEZ-MORGADO
Since
lim L(γ ,γ˙ )+ L(γ ,γ˙ ) = limA(γ ) = A(γ)
n ZF n n ZFc n n n n
we have
(2) lim L(γ ,γ˙ ) = L(γ,γ˙)
n ZFc n n ZFc
For l > 0 let F be a finite union of intervals such that
1 1
D := {t ∈ [a,b] : |γ˙(t)| > l,d(γ(t),{v }) < } ⊂ F, and Leb(F \D ) < .
l a l
l l
Given ε > 0, from Proposition 2 we have that for n large enough
3θ
(3) |γ˙ −γ˙|2 ≤ L(γ ,γ˙ )− L(γ,γ˙)− L (γ,γ˙)(γ˙ −γ˙)+ε.
4 Z n Z n n Z Z v n
Fc Fc Fc Fc
As in Lemma 1, the γ˙ are uniformly integrable, so they converge to γ˙ in the
n
σ(L ,L ) weak topology and then
1 ∞
(4) lim L (γ,γ˙)(γ˙ −γ˙) = 0
n ZFc v n
From (2), (3), (4) we get that limsup |γ˙ −γ˙|2 ≤ ε. Since ε > 0 is arbitrary
n
n FRc
(5) lim |γ˙ −γ˙|2 = 0.
n ZFc n
Take B > 0 such that for any x ∈ G and v ∈ T M L(x,v) ≥ θ|v|2 −B. Then
x 2
|γ˙ −γ˙|2 ≤ 2|γ˙ |2 +2|γ˙|2
Z n Z n
F F
4 4 8B
≤ L(γ ,γ˙ )+ L(γ,γ˙)+ Leb(F).
θ Z n n θ Z θ
F F
If l is large enough, L(γ,γ˙) > 0 on F. By (2), for n big enough L(γ ,γ˙ ) ≤
n n
FR
2 L(γ,γ˙). Therefore
FR
12 8B
limsup |γ˙ −γ˙|2 ≤ L(γ,γ˙)+ Leb(F).
Z n θ Z θ
n F F
Letting l → ∞
b
lim |γ˙ −γ˙|2 = 0,
n Z n
a
which implies by Cauchy-Schwartz inequality that
b
lim |γ˙ −γ˙| = 0.
n Z n
a
(cid:3)
LAX-OLEINIK ON GRAPHS 5
Lemma 2. Let L be a Lagrangian in G. For ε > 0 there exists K that is a Lipschitz
ε
constant for any minimizer γ : [a,b] → G with b−a ≥ ε.
Proof. If γ is a minimizer and γ((c,d)) ⊂ I then γ| is a solution of the Euler
j (c,d)
Lagrange equation for L .
j
Suppose the Lemma is not true, then there are minimizers γ : [a ,b ] → G and
i i i
c ∈ [a ,b ] such that γ˙(c ) → ∞ when i → ∞. Translating in time, we can assume
i i i i
that c = c for all i and taking a subsequence that there is a ∈ R such that γ is
i i
defined in [a,a+ε],c ∈ [a,a+ε]. As A(γ |[a,a+ε]) is bounded, there is subsequence
2 2 i 2
γ |[a,a + ε] which converges uniformly to γ : [a,a + ε] → G. Since γ is limit of
i 2 2
minimizers, it is a minimizer and A(γ) ≤ liminfA(γ |[a,a+ ε]). We can not have
i 2
that A(γ) < limsupA(γ |[a,a + ε]) because that would contradict that the γ are
i 2 i
minimizers. Thus A(γ) = limA(γ |[a,a+ ε]).
i 2
If γ(c) is within I , there is δ > 0 such that γ([c−δ,c+δ]) ⊂ I .
j j
If γ(c) = v we have 2 possibilities (not mutually exclusive)
a
1. γ (c) 6= v for a infinitely many i’s. We have 2 possibilities (not exclusive)
i a
a) There is an edge I with v ∈ I and infinitely many i’s such that γ (c) ∈ I
j a j i j
and γ˙ (c) points towards v .
i a
b) There is an edge I with v ∈ I and infinitely many i’s such that γ (c) ∈ I
j a j i j
and γ˙ (c) points towards the other vertex.
i
2. γ (c) = v for infinitely many i’s. We have 2 possibilities (not exclusive)
i a
a) There is an edge I with v ∈ I and infinitely many i’s such that γ˙ (c) is
j a j i
outgoing.
b) There is an edge I with v ∈ I and infinitely many i’s such that γ˙ (c) is
j a j i
incoming.
In cases a) there is δ > 0 such that γ([c−δ,c]) ⊂ I .
j
In cases b) there is δ > 0 such that γ([c,c+δ]) ⊂ I .
j
We have that γ is a solution of the Euler-Lagrange equation for L either on
j
[c − δ,c] or on [c,c + δ] and then |γ˙(t)| ≤ K on [c − δ,c] or [c,c + δ]. For some
0 < δ < δ we have that γ are solutions of the Euler-Lagrange equation for L on
1 i j
[c−δ ,c] or on [c,c+δ ]. For i suficiently large, we have that |γ˙ (t)| > 2K either in
1 1 i
[c−δ ,c] or in [c,c+δ ]. This would contradict Proposition 4 (cid:3)
1 1
3. The Peierls barrier
ThecontentofthissectionissimilartothatforthecaseofLagrangiansoncompact
manifolds. Given x,y ∈ G let Cac(x,y,T) be the set curves α ∈ Cac([0,T]) such that
α(0) = x and α(T) = y. For a given real number k define
h (x,y) = min A(α)
T
α∈Cac(x,y,T)
and
hk(x,y) = liminfh (x,y)+kT
T
T→∞
6 R. ITURRIAGAAND H.SA´NCHEZ-MORGADO
Lemma 3. For ε > 0 the function F : [ε,∞)×G×G → R defined by F(t,x,y) =
h (x,y) is Lipschitz
t
Proof. ByProposition2thereisK > 0thatisaLipschitzconstant foranyminimizer
γ : [0,t] → G with t ≥ ε, so that kγ˙k ≤ K almost everywhere.
Let ∆ = max(1,diam(G)) and
B = max{|L(x,v)| : kvk ≤ K +3∆}.
Let t ≥ ε, x,y,x′,y′ ∈ M and δ = min(ε/3,d(x,x′)), δ′ = min(ε/3,d(y,y′)). There
is a minimizer γ ∈ Cac(x,y,t) such that A(γ) = h (x,y). Let β : [0,δ] → G be a
t
minimizing geodesic between x′ and γ(δ) so that
d(x′,γ(δ)) d(x,x′)+d(x,γ(δ)) d(x,x′)
˙
kβk ≤ ≤ ≤ +K ≤ 3∆+K
δ δ δ
and hence A(β) ≤ Bδ. Similarly, for β′ : [t − δ′,t] → G a minimizing geodesic
between γ(t−δ′) and y′ we have A(β′) ≤ Bδ′. Thus
ht(x′,y′) ≤ hδ(x′,γ(δ))+ht−δ+δ′(γ(δ),γ(t−δ)′)+hδ′(γ(t−δ′),y′)
≤ ht−δ−δ′(γ(δ),γ(t−δ)′)+Bδ +Bδ′
≤ ha,b(x,y)−A(γ|[0,δ])−A(γ|[t−δ′,t])+Bδ +Bδ′
≤ h (x,y)+2Bd(x,x′)+2Bd(y,y′)
t
which proves that 2B is a Lipschitz constant for all functions h with t ≥ ε.
t
Let s,t ∈ R such that ε ≤ t ≤ s and γ ∈ Cac(x,y,s) such that A(γ) = h (x,y).
s
Then
h (x,y) = h (x,γ(t))+h (γ(t),y).
s t s−t
Since kγ˙k ≤ K almost everywhere, the definition of B gives
h (γ(t),y) ≤ B(s−t).
s−t
Since 2B is a Lipschitz constant for h
t
h (x,γ(t))−h (x,y) ≤ 2Bd(γ(t),y) ≤ 2BK(s−t).
t t
Then
h (x,y)−h (x,y) ≤ B(2K +1)(s−t).
s t
(cid:3)
Lemma 4. There exists a real c independent of x and y such that
(1) For all k > c we have hk(x,y) = ∞.
(2) For all k < c we have hk(x,y) = −∞
(3) hc(x,y) is finite. The function h := hc is called the Peierls barrier.
LAX-OLEINIK ON GRAPHS 7
Proof. Fix x,y ∈ G. Let A be the set of k such that hk(x,y) = −∞, and B be the
set of k such that hk(x,y) = ∞.
As function of k, hk(x,y) is monotone increasing. So if hk(x,y) = ∞ for some k
then hk′(x,y) = ∞ for all k′ ≥ k, and analogously, if hk(x,y) = −∞ for some k then
hk′(x,y) = −∞ for all k′ ≤ k. The set B is nonempty, since for k large enough we
have L+k ≥ 1 and then h (x,y)+kT ≥ T. Analogously the sets A is nonempty,
T
since taking a closed curve γ : [0,1] → G and k < −A(γ), we only have to go around
γ more and more times .
Let c be the supremum of A, and we next show that it is also the infimum of B.
Let ε > 0 then hc+ε(x,y) > −∞, so it is either finite or +∞. In the first case, for
for each T we have
hc+2ε(x,y) = liminfh (x,y)+(c+ε)T +εT = ∞.
T T
Since ε is arbitrary, it follows that c is the infimum of B.
Given other points x′,y′ ∈ G, for T ≥ 3 we have
h (x′,y′) ≤ h (x′,x)+h (x,y)+h (y,y′)
T 1 T−2 1
Therefore hk(x′,y′) is −∞ if hk(x,y) = −∞. Changing the roles we conclude that
the break point does not depend on the points.
It remains to prove that at the break point the value is finite. We need the
following two Lemmas
Lemma 5. Let f : [0,n] → R be a family of continuous functions such that f (0) =
n n
a
n
0,f (n) = a and lim = 0 then for any ε > 0 and any T there exist an n and
n n
n→∞ n
an interval [a,b] contained in [0,n] such that b−a > T and |f(b)−f(a)| ≤ ε.
Lemma 6. The value c is the infimum of k such that L + k ≥ 0 for all closed
Rγ
curves γ.
We show first that hc(x,y) > −∞. Assume the opposite, then using a curve
α : [0,1] → G with α(0) = y, α(1) = x we get a closed curve γ with negative
L+c, a contradiction with Lemma 6.
Rγ
To prove that hc(x,y) < ∞, let a be the infimum of L+c over all closed curves
n
Rγ
γ with period less or equal to n. The sequence (a ) is non negative by Lemma 6
n
a a
n n
and lim = 0. Indeed, if l := limsup > 0, there is a sequence n → ∞ and
k
n→∞ n n→∞ n
ln
k
closed curves γ with period less or equal to n such that L + c ≥ . Thus
k k
2
γRk
l ln
L+c− ≥ k a contradiction with c = infB.
4 4
γRk
8 R. ITURRIAGAAND H.SA´NCHEZ-MORGADO
t
Let γ be a curve such that | L+c−a | < 1 , and f (t) = L(γ (s),γ˙ (s))ds+
n n n 0 n n
γRn R
ct. The functions f satisfy the hypothesis of the Lemma 5 so given any T we can
n
find a curve γ : [0,T′] → G with T′ > T joining some points x′ and y′ such that
A(γ)+cT′ ≤ ε. Thus
hT′+2(x,y)+c(T′ +2) ≤ h1(x,x′)+ε+h1(y′,y)+2c
is bounded independenly of T. (cid:3)
Proof of Lemma 5. Forsimplicityweassumethatthefunctionsaremonotone. Given
T > 0 and ε > 0 let n be such that n > Ta /ε. This possible due to the sub-linear
n
growth of a . We divide the interval in a /ε pieces of lenght T. One of this pieces
n n
must have an increment less than ε otherwise the total increment would be larger
than a , a contradiction. (cid:3)
n
Proof of Lema 6. If k ≥ c and γ is a closed curve then L+k ≥ 0 otherwise we can,
Rγ
by going several times around the same curve, make the action arbitrarily negative.
On the other hand if k < c then hk(x,x) = −∞, so there are closed curves with
(cid:3)
negative action.
Definition 1. The Man˜´e potencial Φ : G×G → R is defined by
Φ(x,y) = infh (x,y)+ct.
t
t>0
Clearly we have Φ(x,y) ≤ h(x,y) for any x,y ∈ G.
Proposition 5. Functions h and Φ have the following properties.
(1) Φ(x,z) ≤ Φ(x,y)+Φ(y,z).
(2) h(x,z) ≤ h(x,y)+Φ(y,z), h(x,z) ≤ Φ(x,y)+h(y,z).
(3) h and Φ are Lipschitz
(4) If γ : [0,t ] → G is a sequence of absolutely continuous curves with t → ∞
n n n
and γ (0) → x, γ (t ) → y, then
n n n
(6) h(x,y) ≤ liminfA(γ )+ct .
n n
n→∞
Proof. For x,y,z ∈ G, s,t > 0
(7) h (x,z)+c(s+t) ≤ h (x,y)+h (y,z)+c(s+t)
s+t s t
(1) Taking inf in (7) we get
s>0
Φ(x,z) ≤ Φ(x,y)+h (y,z)+ct.
t
Taking now inf we have
t>0
Φ(x,z) ≤ Φ(x,y)+Φ(y,z).
LAX-OLEINIK ON GRAPHS 9
(2) Taking liminf in (7) we get
s→∞
h(x,z) ≤ h(x,y)+h (y,z)+ct.
t
Taking now inf we have
t>0
h(x,z) ≤ h(x,y)+Φ(y,z).
(3) Set B = sup{L(y,v) : y ∈ G,|v| ≤ 1}. Let x,y,w,z ∈ G, and let γ :
[0,d(y,z)] → G be a unit speed geodesic connecting y to z.
d(y,z)
Φ(y,z) ≤ (L(γ,γ˙)+c) ≤ (B +c)d(y,z).
Z
0
Thus
h(x,z) ≤ h(x,y)+(B +c)d(y,z),
and exchanging the roles of y and z
h(x,y) ≤ h(x,z)+(B +c)d(y,z).
In the same way
|h(x,z)−h(w,z)| ≤ (B +c)d(x,w).
Writting
h(x,y)−h(w,z) = h(x,y)−h(x,z)+h(x,z)−h(w,z),
we get
|h(x,y)−h(w,z)| ≤ (B +c)(d(y,z)+d(x,w).
Similarly Φ is Lipschitz.
(4) Let K be a Lipschitz constant for all h : t ≥ 1, then
t
h (x,y) ≤ h (γ (0),γ (0))+K(d(x,γ (0))+d(γ (t ),y))
tn tn n n n n n
≤ A(γ )+K(d(x,γ (0)+d(γ (t ),y).
n n n n
Thus
h(x,y) ≤ liminfh (x,y)+ct ≤ liminfA(γ )+ct .
tn n n n
n→∞ n→∞
(cid:3)
Definition 2. A curve γ : J → G is called
• semi-static if
s
Φ(γ(t),γ(s) = L(γ,γ˙)
Z
t
for any t,s ∈ J, t ≤ s.
• static if
s
L(γ,γ˙) = −Φ(γ(s),γ(t))
Z
t
for any t,s ∈ J, t ≤ s.
10 R. ITURRIAGAAND H.SA´NCHEZ-MORGADO
The Aubry set A is the set of points x ∈ G such that h(x,x) = 0.
Notice that by item (2) in Proposition 5, h(x,z) = Φ(x,z) if x ∈ A or z ∈ A.
Proposition 6. If η : R → G is static then η(s) ∈ A for any s ∈ R.
Proof. We define the ω-limit of η as
ω(η) = {x ∈ G : ∃t → ∞ such that x = lim η(t )}
n n
n→∞
Let x ∈ ω(η), then
0 ≤ h(η(s),η(s)) ≤ h(η(s),x)+Φ(x,η(s))
≤ lim A(η| )+ct +Φ(x,η(s))
[s,tn] n
n→∞
≤ lim −Φ(γ(t ),γ(s))+Φ(x,η(s)) = 0
n
n→∞
(cid:3)
4. Weak KAM solutions
Following Fathi, we define weak KAMsolutions andprove some oftheir properties
Definition 3. Let c be given by Lemma 4.
• A function u : G → R is dominated if for any x,y ∈ G, we have
u(y)−u(x) ≤ h (x,y)+ct ∀t > 0,
t
or equivalently
u(y)−u(x) ≤ Φ(x,y).
• γ : I → G calibrates a dominated function u : G → R if
s
u(γ(s))−u(γ(t)) = L(γ,γ˙)+c(s−t) ∀s,t ∈ I
Z
t
• Acontinuousfunctionu : G → Risabackward (forward) weak KAM solution
if it is dominated and for any x ∈ G there is γ : (−∞,0] → G (γ : [0,∞) →
G) that calibrates u and γ(0) = x
Proposition 7. For any x ∈ G, h(x,·) is a backward weak KAM solution and
−h(·,x) is a forward weak KAM solution.
Proof. Let γ : [−t ,0] → G be a sequence of minimizing curves connecting x to y
n n
such that
h(x,y) = lim A(γ )+ct
n n
n→∞
By Lemma 2, {γ } is uniformly Lipschitz and then equicontinuous. For l ∈ N fixed,
n
it followsfromtheArzela Ascoli Theorem that there isa sequence n → ∞ such that
j
γ converges uniformly on [−l,0]. Using a diagonal trick one obtains a sequence
nj