On periodic motions of a two dimensional Toda type chain

ESAIM: Control, Optimisation and Calculus of Variations

January 2005, Vol. 11, 72–87 DOI: 10.1051/cocv:2004033

ON PERIODIC MOTIONS OF A TWO DIMENSIONAL TODA TYPE CHAIN

Gianni Mancini

and P.N. Srikanth

Abstract. In this paper we consider a chain of strings with fixed end points coupled with nearest neighbour interaction potential of exponential type,i.e.

ϕⁱ_tt−ϕⁱ_xx= exp(ϕⁱ⁺¹−ϕⁱ)−exp(ϕⁱ−ϕi−1) 0< x < π, t∈IR,i∈ZZ (TC) ϕⁱ(0, t) =ϕⁱ(π, t) = 0 ∀t, i.

We consider the case of “closed chains”i.e. ϕ^i+N =ϕⁱ∀i∈ZZand someN∈INand look for solutions which are peirodic in time. The existence of periodic solutions for the dual problem is proved in Orlicz space setting.

Mathematics Subject Classification. 58E30, 37J45.

Received July 22, 2003.

1. Introduction

In this paper we consider a chain of strings with fixed end points coupled with nearest neighbor interaction potential of exponential type,i.e.





ϕⁱ_tt−ϕⁱ_xx= exp

ϕⁱ⁺¹−ϕⁱ

−exp

ϕⁱ−ϕⁱ⁻¹

0< x < π, t∈IR, i∈ZZ

ϕⁱ(0, t) =ϕⁱ(π, t) = 0 ∀t, i. (TC)

We consider the case of “closed chains”i.e. ϕ^i+N =ϕⁱ∀i∈ZZand someN ∈INand we look for solutions which are periodic in time. IfN = 2, it reduces to a sine-Gordon equation, which is known to be completely integrable (see [10]) and to possess periodic solutions (see [4]). Henceforth, we will be interested to the caseN ≥3.

The study is motivated by the work [10]. However, it was Toda who initiated studies with exponential interaction. More precisely, Toda considered one-dimensional lattices ofNparticles with exponential interaction.

If V_i denotes the potential of interaction between the ith and (i+ 1)th particle, then the equation for the ith particle reads

ϕ¨ⁱ=V_i−1

ϕⁱ⁻¹−ϕⁱ

−V_i

ϕⁱ−ϕⁱ⁺¹

. (T)

Keywords and phrases. Periodic, Toda type chain.

∗ The author’s research is supported by MURST under the national project Variational methods and nonlinear differential equations.

1 Dipartimento di Mathematica, Universita di Roma Tre, Via S. Leonardo Murialdo 1, Roma, Italy;

e-mail:mancini@matrm3.mat.uniroma3.it

2 TIFR Centre, IISc Campus, PB 1234, Bangalore – 560 012, India; e-mail:srikanth@math.tifrbng.res.in

c EDP Sciences, SMAI 2005

In caseV_i(t) =V(t) =^a_be^bt+at∀i= 1, . . . , N andab >0 one has the classical Toda lattice, which was shown to be completely integrable, with explicit periodic and soliton solutions [14]. Furthermore, it is known by methods of Kolmogorov-Arnold-Moser (KAM) theory [2, 9] that periodic and quasi-periodic motions of (T) persist under small perturbations. It is in such a context that periodic motions of lattices has assumed its significance. In [13] Toda type lattices under large perturbations have been studied and existence of periodic solutions has been established. See [5] and [6] for a more complete study of one dimensional lattices.

The problem under consideration in this paper can be thought of as a 2-dimensional generalization of the Toda chain. As far as we know the only work dealing with this kind of problem is [10] where integrability is shown using inverse-problem method. However there is no information about the nature of solutions. Our effort at this stage is to explore if global methods of nonlinear analysis can be employed to study the higher dimensional lattices of Toda type. We plan to takeup the study of more general potentials of interaction in future work.

Returning to (TC), note that it possesses (linear) normal modes ϕⁱ = ϕ, ϕ_tt−ϕ_xx = 0, and infact the motion is determined up to normal modes, i.e. if Φ = (ϕ¹, . . . ϕ^N) solves (TC), then it does Φ + Γ, where Γ = (γ¹, . . . γ^N) with γⁱ =γ(i = 1, . . . , N) and γ_tt−γ_xx = 0 (notice that_N

i=1ϕⁱ_tt−ϕⁱ_xx = 0). To get rid of normal modes, we will ask Φ to satisfy the normalization condition _N

i=1ϕⁱ ≡ 0 and prove the following theorem.

Theorem 1.1. (TC) has infinitely many (normalized) periodic solutions.

2. A dual problem in Orlicz spaces

Problem (TC) can be written in vector form

Φ +∇f(Φ) = 0 where Φ = (ϕ¹, . . . , ϕ^N),Φ = (φⁱ_tt−ϕⁱ_xx)_i=1,...N and

f(η) =^N

i=1

exp

ηⁱ−ηⁱ⁻¹

−1

η∈IR^N, η⁰=η^N

∇f(η) = exp

η¹−η^N

−exp

η²−η¹

, . . . ,exp

η^N −η^N−1

−exp

η¹−η^N .

Clearly,∇f(t11) = 0∀t,where 11 = (1, . . . ,1) and ∇f(η)∈11^⊥={ξ∈IR^N :_N

i=1ξⁱ = 0}·

Basic qualitative properties of the interaction potential f, proven in the appendix, are as follows. Let c_N := ^√_N(N²₋₁₎; then

∀α >0, ∇f(η), η ≥αf(η) ∀η∈11^⊥ with|η| ≥ α+√ Nα

c_N (arabic@¸ section.arabic@¸ equation)

∇f(η), η ≥c_N|η|

e^cN|η|−1

∀η∈11^⊥ (arabic@¸ section.arabic@¸ equation) D²f(η)ξ, ξ

≥c²_N e^−√2|η||ξ|² ∀η∈IR^N, ∀ξ∈11^⊥. (arabic@¸ section.arabic@¸ equation) Furthermore, ifA(s) = e^s−s−1, s≥0,then

A(c_N|η|)≤f(η)≤NA√ 2|η|

∀η∈11^⊥. (arabic@¸ section.arabic@¸ equation) The functionAso defined is an Orlicz function (orN-function; see appendix for basic facts on Orlicz functions and Orlicz spaces used throughout the paper). We denote by L_A =L_A(Ω_n) the Orlicz space associated toA,

consisting of measurable functions

Φ : Ω_n →IR^N such that

Ωn

A(λ|Φ|)<∞ for someλ=λ(Φ)>0 where Ω_n= [0, π]× 0,^2π_n

(nto be chosen, later, suitably large),i.e. L_A=K_AwhereK_A={Φ :

ΩnA(|Φ|)<

∞}·L_A, endowed with the Orlicz-Luxemburg norm ΦA= inf

k >0 :

Ωn

A |Φ|

k

≤1

is a Banach space. Thanks to (2.4), one could seek solutions of (T C) by looking for critical points of Φ→

Ωn

f(Φ)−1 2

Ωn

|Φt|²− |Φx|²

in some Orlicz-Sobolev space associated to A(in the spirit of the celebrated paper [12]). Instead, because of the strong indefiniteness of the quadratic part, we prefer, following [4], to work with a dual functional, via a Legendre transformation. To this end, we first derive some properties of the dual potentialf^∗:

f^∗(ξ) := sup

η∈11^⊥[ξ, η −f(η)] ∀ξ∈11^⊥.

Sincef|₁₁⊥ has superlinear growth and is strictly convex by (2.1)–(2.4),f^∗turns out to be aC^∞-strictly convex function on 11^⊥.Also,∇f(η) =ξ⇔η=∇f^∗(ξ) andf^∗(ξ) =ξ, η −f(η) forξ=∇f(η) and we have

(Young’s inequality) ξ, η ≤f^∗(ξ) +f(η) ∀ξ, η∈11^⊥. We denote byA^∗ theN-function complementary to A,i.e.

A^∗(s) = sup

t≥0(st−A(t)), s≥0.

Easily,A^∗(s) = (1 +s) log(1 +s)−s,and, by (2.4) and direct computation, we have Lemma 1. NA^∗

|ξ|

N√ 2

≤f^∗(ξ)≤A^∗

c|ξ|N

∀ξ∈11^⊥.

A crucial feature ofA^∗ is that it satisfies a (global) (∆₂) property:

(∆₂) ∀r >0 ∃k(r)>0 : A^∗(rt)≤k(r)A^∗(t) ∀t≥0 withk(r) = max{r, r²}.From (∆₂) and Lemma 1 it follows

1

2NA^∗(|ξ|)≤f^∗(ξ)≤ 1

c²_NA^∗(|ξ|) ∀ξ∈11^⊥. (arabic@¸ section.arabic@¸ equation) Hence, the natural space to work in is the Orlicz space L_A∗. Note that, thanks to (∆₂), L_A∗ = K_A∗. In particular,L_A∗ is a separable Banach space (see Prop. A1). Also, from H¨older’s inequality

Φ,Ψ ≤NΦAΨA^∗ ∀Φ∈L_A,Ψ∈L_A∗,

it follows that every Φ∈L_A defines an element ofL_A∗ and infact, sinceA^∗ satisfies (∆₂),L_A∗ turns out to be isomorphic and homeomorphic to L_A.

We now proceed to establish some properties of the functional F(Ψ) :=

Ωnf^∗(Ψ), and its “gradient operator” Ψ→ ∇f^∗(Ψ),on L_A∗ :={Ψ∈L_A∗ :_N

i=1ψ^j ≡0}·

Lemma 2.

(i) ∇f^∗(Ψ)A≤ _c¹_N

1 + ^Ψ_c2^2A∗

N

∀Ψ∈ LA^∗. (ii) Ψ_j−ΨA^∗ →0⇒

Ωn∇f^∗(Ψ_j), →

Ωn∇f^∗(Ψ), ∀∈L_A∗. (iii) F(Ψ) =

Ωnf^∗(Ψ)is a locally Lipschitz functional on LA^∗. (iv) Ψ→ ∇f^∗(Ψ) is the Gateaux derivative ofF.i.e.

d dt

Ωn

f^∗(Ψ +t)

|t=0

=

Ωn

∇f^∗(Ψ), ∀Ψ,∈LA^∗.

Proof. (i) Let Φ =∇f^∗(Ψ),and hence Ψ =∇f(Φ). From (2.2) we see that 1

c_N|Ψ| ≥e^cN|Φ|−1 and hence, by monotonicity,

A^∗ |Ψ|

c_N

≥A^∗

e^cN|Φ|−1

=c_N|Φ|e^cN|Φ|−e^cN|Φ|+ 1≥A(c_N|Φ|).

Since

ΩnA^∗(|Ψ|)≤sup_t>0^A∗(Ψ_A∗(t)^A∗t)≤max{ΨA∗,Ψ²_A∗} ∀Ψ∈L_A∗, (see Prop. A2 (ii) and (∆₂) forA^∗),

then

Ωn

A(c_N|∇f^∗(Ψ)|)≤1 + Ψ²_A∗

c²_N ∀Ψ∈L_A^∗. Finally, using Proposition A2 (i), we obtain

c_N∇f^∗(Ψ)_A≤1 + Ψ²_A∗

c²_N ∀Ψ∈L_A∗. (ii) From Ψ_j−Ψ_A^∗ →0,we have

A^∗(|Ψ_j−Ψ|)→0 (see Prop. A2 (i)) and hence Ψ_j →Ψ in measure. In addition, we know by (i) that∇f^∗(Ψ_j)A is bounded and hence (ii) follows by Proposition A3.

(iii) Let Ψ₁,Ψ₂∈B_r(0).Then

|F(Ψ₁)−F(Ψ₂)|=

Ωn

₁

0 ∇f^∗(tΨ₁+ (1−t)Ψ₂),Ψ₁−Ψ₂

≤NΨ1−Ψ₂A^∗ sup

Ψ∈Br(0)∇f^∗(Ψ)A

with sup_Ψ∈Br(0)∇f^∗(Ψ)A<∞by (i).

(iv) Follows from

Ωn

f^∗(Ψ+t)−f^∗(Ψ)

t =₁

0(

Ωn∇f^∗(Ψ +τt),)dτ and (see (ii))

Ωn

∇f^∗

Ψ) +τt ,

→

Ωn

∇f^∗(Ψ),

∀τ∈[0,1], because, by (i),

Ωn

∇f^∗

Ψ +τt

,≤ sup

Ψ−ΨA∗≤A∗

∇f^∗(Ψ) ≤c.

Let us now recall basic properties of the operator = _∂t^∂22 −_∂x^∂22 acting on summable functions which are 2π periodic and satisfy Dirichlet boundary conditions. It is known (see [4, 8]) that γ= 0 (in the weak sense), if and only if

γ(x, t) =p(x+t)−p(t−x) for some p ∈ L¹_loc(IR),2π-periodic and such that _2π

0 p = 0. Also, for a given summable ψ, we have that

[0,π]×[0,2π]ψγ= 0 ∀γ∈L^∞∩Kerif and only if _π

0

[ψ(x, t−x)−ψ(x, t+x)]dx= 0 for a.e. t (arabic@¸ section.arabic@¸ equation) and, for such aψ, there is a unique 2πperiodic continuous functionϕ=Lψsatisfying the boundary condition ϕ(0, t) =ϕ(π, t) = 0∀t,such that

ϕ=ψ and

[0,π]×[0,2π]

ψγ= 0 ∀γ∈Ker∩L^∞. Actually,Lψis explicitely known (see [8]):

Lψ(x, t) = ˜ϕ+p(t+x)−p(t−x) where

ϕ(x, t) =˜ −1 2

_π

x

dξ

_t−x+ξ

t+x−ξ

ψ(ξ, τ)dτ+π−x 2π

_π

0

dξ _t+ξ

t−ξ

ψ(ξ, τ)dτ and

p(y) = 1 2π

_π

0

[ ˜ϕ(s, y−s)−ϕ(s, y˜ +s)]ds.

It can be easily checked that

∃C >0 : LψL^∞≤Cψ_L¹, Lψ_C^0,1≤CψL^∞ (arabic@¸ section.arabic@¸ equation) ψ∈C^k,1⇒Lψ∈C^k+1,1. (arabic@¸ section.arabic@¸ equation) Let us now write IK:= (Ker)^N, LΨ = (Lψ¹, . . . , Lψ^N), and set

E=



Ψ∈L_A∗ : N j=1

ψ^j≡0,

Ωn

Ψ,Γ= 0 ∀Γ∈IK∩L_A



· Note thatE is a closed subspace of L_A∗ and is a separable Banach space.

Lemma 3. Linduces a compact linear operator fromE intoL_A.

Proof. Since L_A∗ ⊂ L¹, LΨ is infact in L^∞. Let Ψ_k be a bounded sequence in E. Since sup_kΨ_kA^∗ < ∞ implies sup_kΨ_kL¹<∞, LΨ_kis bounded inL^∞.We want to proveLΨ_k are equicontinuous and Ascoli-Arzela will imply compactness. Using the notations introduced above,

LΨ_k = ˜Φ_k+P_k(t+x)−P_k(t−x), P = (pⁱ)

it is clearly enough to prove that ˜Φ_k are equicontinuous. We have (dropping superscripts)

|ϕ˜_k(x, t)−ϕ˜_k(x, t)| ≤ 1 2

_π

x

dξ

_t−x+ξ

t+x−ξ

ψ_k(ξ, τ)dτ− _π

x

dξ

_t−x+ξ

t+x−ξ

ψ_k(ξ, τ)dτ + c_k

2π|x−x|

where c_k :=_π

0 dξ_t+ξ

t−ξ ψ_k(ξ, τ)dτ are constants bounded by sup_kψ_kL¹ (recall (2.6)). Hence it is enough to estimate

∆ =

Tx,t

ψ_k(ξ, τ)dξdτ−

T_x,t

ψ_k(ξ, τ)dξdτ ≤

(Tx,t\T_x,t)∪(T_xt\Tx,t)|ψ_k| where

T_x,t={(ξ, τ);x≤ξ≤π, t+x−ξ≤τ≤t−x+ξ} ⊂[0, π]×IR

and ψ_k are extended as periodic functions on [0, π]×IR. Clearly the measure of the symmetric difference is small if (x, t) and (x, t) are close. On the other hand by Vall´ee-Poussin theorem [7] (p. 103)

supk

Ωn

A^∗(|ψ_k|)<+∞ ⇒sup

k

|ψ_k| ≤ε if|Σ| ≤δ_ε. Since

A^∗(|Ψ_k|)≤1 +Ψ_k²_A∗ (see Prop. A2 (ii) and (∆₂) forA^∗) and sup_kΨ_kA^∗ <+∞, we have

|ϕ˜_k(x, t)−ϕ˜_k(x, t)| ≤ε if |x−x|+|t−t| ≤δ_ε

uniformly with respect tok.Hence the result.

We now state the dual variational principle.

Proposition 4. Let Ψbe a critical point of the (Gateaux differentiable) functional S(Ψ) =

Ωn

f^∗(Ψ) + 1 2

Ωn

Ψ, LΨ, Ψ∈E. (arabic@¸ section.arabic@¸ equation)

Then Φ =∇f^∗(Ψ) is a (weak) ^2π_n periodic solution of (TC).

Proof. SinceIK is closed inL¹,IK∩L_Ais closed inL_A, as well asL_A∩IK^⊥ :={Φ∈L_A:

ΩnΦ,Γ= 0 ∀Γ∈ IK∩L_A∗}.Thus we can write

∇f^∗(Ψ) +LΨ =W+ Γ, W ∈IK^⊥∩L_A, Γ∈IK∩L_A. Notice that

jw^j+γ^j ≡0 and hence we can assume

jw^j ≡0.Since, by assumption,

Ωn

∇f^∗(Ψ),+

Ωn

LΨ = 0 ∀∈E

we see that W = 0.Hence ∇f^∗(Ψ) +LΨ = Γ.Let Φ =∇f^∗(Ψ) = Γ−LΨ.We have

−Φ = Ψ =∇f(Φ)

i.e. Φ is a ^2π_n periodic solution of (TC) in the weak sense.

We end this section stating some growth properties of ∇f^∗ which will be crucial to get critical points ofS.

For a givenθ <1,letm_θ=√

2(α+√

Nα) whereα= (1−θ)⁻¹.Then Lemma 5.

(i) _c¹

N log(1 +_c^|ξ|

N)≥ |∇f^∗(ξ)| ≥ _|ξ|^NA^∗(_N^|ξ|√

2) ∀ξ∈11^⊥, ξ= 0. (ii) ∀θ <1, f^∗(ξ)≥θ∇f^∗(ξ), ξif|ξ| ≥c(N, θ) := 4Ne^mθcN.

Proof.

(i) From Lemma 1 and∇f^∗(ξ), ξ ≥f^∗(ξ),we obtain the RHS inequality. Next, ifη =∇f^∗(ξ), from (2.2) we derive|ξ|=|∇f(η)| ≥c_N[e^cN|η|−1] =c_N[e^{cN|∇f∗(ξ)|}−1].This gives the LHS inequality.

(ii) From the RHS inequality in (i), one easily obtains

|ξ| ≥4Ne^mθcN ⇒ |∇f^∗(ξ)| ≥ α+√ Nα c_N ·

Hence (ii) follows from (2.1) and f^∗(ξ) =∇f^∗(ξ), ξ −f(η), η=∇f^∗(ξ).

3. Existence of weak periodic solutions

In this section we will prove our theorem, by means of a suitable version of the Mountain Pass Lemma.

Lemma([4]). LetEbe a separable Banach space and letS:E→IRbe a locally Lipschitz Gateaux differentiable functional. Suppose further

(i) Ψ_j→Ψ⇒S(Ψ_j)→S(Ψ) ∀∈E;

(ii) ∃U open, 0∈U such that inf_∂US > S(0);

(iii) ∃Ψ₀∈U such that S(Ψ₀)≤S(0);

(iv) S(Ψ_j)→c, S(Ψ_j)→0⇒ ∃Ψ∈S⁻¹(c) such thatS(Ψ) = 0.

Then c:= inf

p∈IP max

t∈[0,1]S(p(t)) is a critical value, where IP ={p∈C([0,1], E) : p(0) = 0, p(1) = Ψ₀}·

We have already seen thatS defined by (2.9) is locally Lipschitz and Gateaux differentiable, with derivative given byS(Ψ)=

∇f^∗(Ψ),+LΨ,, ∀∈E.We now proveS satisfies (i)–(iv).

Proof of (i). It follows by Lemma 2 (iii) and continuity properties ofL.

Proof of (ii). It is exactly here where we need to takensuitably large. LetU ={Ψ∈E:

Ωnf^∗(Ψ)<2N|Ω_n|}.

From Young’s inequality and max_|ξ|=1f <2N,we get

|ξ| ≤f^∗(ξ) +f ξ

|ξ|

≤2N+f^∗(ξ) ∀ξ∈11^⊥

and hence

Ωn|ψ^j| ≤2N|Ωn|+

Ωnf^∗(Ψ) = 4N|Ωn| ∀j and Ψ∈∂U.Since

ΩnΨ, LΨ ≥ −2_N

j=1(

Ωn|ψ^j|)² (see [4]) we have, for Ψ∈∂U,

S(Ψ)≥2N|Ω_n| −

Ωn

ψ^j₂

≥2N|Ω_n| −16N³|Ω_n|²>0 =S(0) ifn >16π²N².

Proof of (iii). It follows from the subquadratic behaviour off^∗ at infinity.

Proof of (iv). Let Ψ_j be such that

Ωn

f^∗(Ψ_j) +1 2

Ωn

Ψ_j, LΨ_j=c+◦(1) and sup

A∗ ≤1

∈E

Ωn

∇f^∗(Ψ_j) +LΨ_j,

=◦(1) asj→ ∞ and hence

2

Ωn

f^∗(Ψ_j)−

Ωn

∇f^∗(Ψ_j),Ψ_j= 2c+◦(1) +◦(1)Ψ_jA^∗.

By Lemma 5 (ii) andf^∗≥0,we have

Ωn

f^∗(Ψ_j)≥θ

Ωn

∇f^∗(Ψ_j),Ψ_j −θ˜c(N, θ)

where ˜c(N, θ) =

c1Nlog(1 +^c(θ,N)_c

N )

|Ω_n|.Hence

◦(1)Ψ_jA^∗ ≥

2−1

θ _Ωnf^∗(Ψ_j)−[˜c(N, θ) + 2c+◦(1)]. (arabic@¸ section.arabic@¸ equation) From (2.5) and Proposition A2 (i) , we have

Ωn

f^∗(Ψ_j)≥ 1

2N [Ψ_jA^∗−1]. (arabic@¸ section.arabic@¸ equation) Takingθ > ³₄,we obtain from (3.1) and (3.2)

1

3NΨjA^∗ ≤ 1

2N + ˜c(N, θ) + 2c. (arabic@¸ section.arabic@¸ equation) Hence sup_jΨjA^∗ <∞.Then (see [7] 14.4)∃Ψ∈E such that, for a subsequence,

Ωn

Ψ_j,Φ →

Ωn

Ψ,Φ ∀Φ ∈ E_A

whereE_A is the closure of the set of bounded functions inL_A.We want to show that

Ωn

∇f^∗(Ψ) +LΨ,= 0 ∀∈E.

i.e. Ψ is a critical point at level c. We will use here, following [4], the so called Minty-trick, exploiting monotonicity of the operator Ψ→ ∇f^∗(Ψ):

∇f^∗(Ψ_j)− ∇f^∗(), Ψ_j− ≥0 ∀∈L_A∗. (arabic@¸ section.arabic@¸ equation) Writing

∇f^∗(Ψ_j) +LΨ_j=W_j+ Γ_j, W_j∈L_A∩IK^⊥,Γ_j∈L_A∩IK with

iwⁱ_j = 0 andW_jA→0 (see Prop. 4) and inserting in (3.4), we get

W_j−LΨ_j− ∇f^∗(),Ψ_j− ≥0 ∀∈L_A∗. (arabic@¸ section.arabic@¸ equation) If∈L^∞,we can pass to the limit in (3.5), to get

LΨ +∇f^∗(),Ψ− ≤0, ∀∈L^∞. (arabic@¸ section.arabic@¸ equation) Now, choose as test function_nΨ−t,∈L^∞where

nΨ = (_nψ^j), _nψ=

ψ if |ψ| ≤n 0 if |ψ|> n.

Then we have

LΨ +∇f^∗(_nΨ−t),(Ψ−_nΨ) +t ≤0.

SinceA^∗ is ∆₂,nΨ−ΨA^∗ →0 (see [7] 10.1) and hence

LΨ +∇f^∗(Ψ−t), t ≤0 ∀∈L^∞, (arabic@¸ section.arabic@¸ equation) by Lemma 2 (ii) and because∇f^∗(_nΨ−t) is bounded inL_Aby Lemma 2 (i). Now, sinceL^∞is dense inL_A∗, arguments as above give

LΨ +∇f^∗(Ψ−t), ≤0 ∀∈E andt >0.

Letting t→0,we get

LΨ +∇f^∗(Ψ),= 0 ∀.

The theorem now follows from Lemma [4]. We remark that the existence of infinitely many solutions follows from the simple device (see [11]) of working in a space of functions whose period is smaller than the minimal period of the solution previously obtained.

4. Regularity

Since solutions to (TC) are determined up to normal modes, no regularity result is available for arbitrary weak solutions. Still, following some ideas in [3], and heavily exploiting the periodic structure of the lattice, we will prove that normalized weak solutions are C^∞. Actually, we will develope a bootstrap argument for Ψ = (ψ^j),Γ = (γ^j) satisfying

Γ∈Ker Ψ∈L¹,

ΨΠ = 0∀Π∈L^∞∩Ker (arabic@¸ section.arabic@¸ equation) Ψ =∇f(Γ−LΨ)

which will lead toC^∞ regularity for Ψ. Clearly, Φ := Γ−LΨ is a weak solution of (TC) and, ifϕ^j ≡0, then Φ =∇f^∗(Ψ)∈C^∞. We start with a simple summability lemma.

Lemma 6. ∇f(Φ)∈L¹⇒

Ωnexp(ϕ^j−ϕ^j−1)<∞ ∀j.

Proof. Arguing by contradiction, we findjsuch that

Ωnexp[ϕ^j−ϕ^j−1] = +∞and henceA₀:={ϕ^j−ϕ^j−1≥1} has positive measure. Since exp[ϕ^j−ϕ^j−1]−exp[ϕ^j+1−ϕ^j] = ψ^j ∈L¹ we have

A0exp(ϕ^j+1−ϕ^j) = +∞

and henceA₁:=A₀∩ {ϕ^j+1−ϕ^j ≥1}has positive measure. AfterN iterations, we findϕ^j+i≥ϕ^j+i−1+ 1 for i= 0, . . . , N−1 on a set of positive measureA_N−1. Adding up these inequalities we find a contradiction.

Remark. If Φ is a normalized weak solution with∇f(Φ)∈L¹, then, by the previous lemma and Lemma A1, we have

e^cN|Φ|<∞. Even this simple summability property fails for arbitrary weak solutions.

To start the bootstrap argument, we first need a couple of lemmata.

Lemma 7. Let Ψ,Γ satisfy (4.1). Then γ^j−γ^j−1∈L^∞∀j.

Proof. First, ˜Φ :=−LΨ belongs toL^∞by (2.7). Let M := 2Φ˜ ∞. Denoted Φ := Γ + ˜Φ, it resultsγ^j−^M₂ ≤ ϕ^j≤γ^j+^M₂. Also, by assumption,

ψ^j= exp γ^j−γ^j−1+ ˜ϕ^j−ϕ˜^j−1

−exp γ^j+1−γ^j+ ˜ϕ^j+1−ϕ˜^j

(arabic@¸ section.arabic@¸ equation)

satisfy (2.6). We have exp

γ^j−γ^j−1−M

−exp

γ^j+1−γ^j+M

≤exp

ϕ^j−ϕ^j−1

−exp

ϕ^j+1−ϕ^j

≤exp

γ^j−γ^j−1+M

−exp

γ^j+1−γ^j−M .

(arabic@¸ section.arabic@¸ equation) Letγ^j=p^j(t+x)−p^j(t−x), γ^j−γ^j−1= ˜p^j(t+x)−p˜^j(t−x), where ˜p^j:=p^j−p^j−1, and rewrite (4.3) as

exp p˜^j(t+x)−p˜^j(t−x)−M

−exp p˜^j+1(t+x)−p˜^j+1(t−x) +M

≤ψ^j(x, t)

≤exp p˜^j(t+x)−p˜^j(t−x) +M

−exp p˜^j+1(t+x)−p˜^j+1(t−x)−M . Computing at (x, t−x),(x, t+x) respectively and subtracting, we get

exp p˜^j(t)−p˜^j(t−2x)−M

−exp p˜^j+1(t)−p˜^j+1(t−2x) +M + exp p˜^j+1(t+ 2x)−p˜^j+1(t)−M

−exp p˜^j(t+ 2x)−p˜^j(t) +M

≤ψ^j(x, t−x)−ψ^j(x, t+x)

≤exp p˜^j(t)−p˜^j(t−2x) +M

−exp p˜^j+1(t)−p˜^j+1(t−2x)−M

−exp p˜^j(t+ 2x)−p˜^j(t)−M

+ exp p˜^j+1(t+ 2x)−p˜^j+1(t) +M

. (arabic@¸ section.arabic@¸ equation) Notice that, due to Lemma 6, sinh ˜p^j∈L¹([0,2π])∀j. In fact, by Fubini and 2π-periodicity,

∞>

_2π

0

_π

0 e^{p˜j(t+x)−p˜j(t−x)}dxdt= _2π

0

e^p˜j(s)

_π

0 e^{−˜pj(s−2x)}dx

ds=1 2

_2π

0 e^p˜j(s)ds _2π

0 e^−p˜j(τ)dτ.

Integrating w.r. to x∈[0, π] the inequalities (4.4) and changing variabless=t−2xands=t+ 2x, we get, using 2π-periodicity of ˜p^j and property (2.6) ofψ^j, we obtain

_2π

0 [sinh (˜p^j(t)−p˜^j(s)−M) + sinh (˜p^j+1(s)−p˜^j+1(t)−M)]ds≤0 for a.e.t

(arabic@¸ section.arabic@¸ equation) _2π

0 [sinh (˜p^j(t)−p˜^j(s) +M) + sinh(˜p^j+1(s)−p˜^j+1(t) +M)]ds≥0, for a.e.t.

(arabic@¸ section.arabic@¸ equation) We now derive from (4.5) that ess sup ˜p^j <∞ ∀j.We argue by contradiction. Assume thatA^j_k =

t: ˜p^j(t)≥k has positive measure for some j ∈ {1, . . . N} and everyk ∈IN .We claim that this implies A^j+1_k ={t ∈A^j_k : p˜^j+1(t)≥k}has positive measure∀k.If not,∃¯k, andt_k ∈A^j_k ⊂A^j_¯k ∀k≥¯ksuch that ˜p^j+1(t_k)≤¯k.From (4.5) att_k we have _2π

0 sinh

k−M −p˜^j(s)

+ sinh

p˜^j+1(s)−k¯−M

≤0 ∀k≥k.¯

Taking into account the summability of sinh ˜p^j, we get, sendingkto infinity, a contradiction by Levi’s theorem.

ThusA^j+1_k has positive measure∀k.Repeating the argument, we findA^j+N_k ⁻¹⊂A^j+N_k ⁻². . .⊂A^j_k,k∈IN sets of positive measure, such that

p˜ⁱ(t)≥kfori= 1. . . , N andk∈IN, t∈A^j+N_k ⁻¹.

Since _N

i=1p˜ⁱ(t) ≡ 0, we arrive at a contradiction. Similarly, using (4.6), we can derive

ess inf ˜p^j >−∞ ∀j.

Lemma 8. Let Ψ,Γ satisfy (4.1). Then γ^j−γ^j−1∈C^0,1∀j.

Proof. By Lemma 7 and (4.2), we see that Ψ∈L^∞and henceLΨ∈C^0,1 (see (2.7)).

We first derive from (4.2) and (2.6) _π

0 exp γ^j−γ^j−1−L

ψ^j−ψ^j−1

(x, t−x)−exp γ^j+1−γ^j−L

ψ^j+1−ψ^j

(x, t−x)dx

= _π

0 exp γ^j−γ^j−1−L

ψ^j−ψ^j−1

(x, t+x)−exp γ^j+1−γ^j−L

ψ^j+1−ψ^j

(x, t+x)dx for everyj and a.e. t. Thus, after writingg^j = exp[−L(ψ^j−ψ^j−1)]∈C^0,1, we have, using previous notations,

_π

0

g^j(x, t−x) exp p˜^j(t)−p˜^j(t−2x)

−g^j(x, t+x) exp p˜^j(t+ 2x)−p˜^j(t) dx

= _π

0

g^j+1(x, t−x) exp p˜^j+1(t)−p˜^j+1(t−2x)

−g^j+1(x, t+x) exp p˜^j+1(t+ 2x)−p˜^j+1(t) dx (arabic@¸ section.arabic@¸ equation) for everyj and a.e. t. Computing att+h, t, and taking the difference, we see that

I^j(t, h) :=

_π

0

g^j(x, t+h−x) exp p˜^j(t+h)−p˜^j(t+h−2x)

−g^j(x, t−x) exp p˜^j(t)−p˜^j(t−2x) dx

− _π

0

g^j(x, t+h+x) exp p˜^j(t+h+ 2x)−p˜^j(t+h)

−g^j(x, t+x) exp p˜^j(t+ 2x)−p˜^j(t) dx

=k₁^j(t, h)−k^j₂(t, h)

is independent onj for every fixedhand a.e. t. Now, k^j₁=

e^p˜j(t+h)−e^p˜j(t)

π

0 g^j(x, t−x)e^{−˜pj(t−2x)}dx + e^p˜j(t+h)

_π

0

g^j(x, t+h−x)e^{−˜pj(t+h−2x)}dx− _π

0

g^j(x, t−x)e^{−˜pj(t−2x)}

dx.

Since _π

0 g^j(x, t+h−x)e^{−p˜j(t+h−2x)}dx = _π−^h₂

−^h₂ g^j(x+ ^h₂, t−x+ ^h₂)e^{−˜pj(s−2x)}dx, we see that _π

0 g^j(x, t+

h−x)e^{−˜pj(t+h−2x)}dx−_π

0 g^j(x, t−x)e^{−p˜j(t−2x)}dx=O(h), becauseg^j is Lipschitz and ˜p^j∈L^∞, and hence k₁^j=c^j₁(t, h) p˜^j(t+h)−p˜^j(t)

+O(h) where c^j₁(t, h) = ₁

0 e^s˜pj(t+h)+(1−s)˜p^j(t)ds_π

0 g^j(x, t−x)e^{−˜pj(t−2x)}dx is positive, (essentially) bounded and bounded away from zero uniformly w.r. to h. Similarly

k^j₂=−c^j₂ p˜^j(t+h)−p˜^j(t)

+O(h).

Hence

I^j(t, h) =c^j

p˜^j(t+h)−p˜^j(t)

+O_j(h)

wherec^j=c^j₁+c^j₂. SinceI¹(t, h) =I²(t, h) =. . .=I^N(t, h) for everyhand a.e. t, then p˜^j+1(t+h)−p˜^j+1(t) =r_j p˜^j(s+h)−p˜^j(s)

+O_j(h) a.e. (arabic@¸ section.arabic@¸ equation) wherer_j =r_j(t, h) =_cj+1^cj are positive, bounded and bounded away from zero. Now, let

A(M, h) :=

t: ˜p¹(t+h)−p˜¹(t)≥M|h|

· (arabic@¸ section.arabic@¸ equation) It follows from (4.8) that, for a.e. t∈A(M, h),

p˜^j+1(t+h)−p˜^j+1(t)≥r₁r₂. . . r_jM|h|+O_j(h) forj= 1, . . . , N−1.

Taking into account that_N

j=1p˜^j ≡0, we obtain, for a.e. t∈A(M, h),

0≥(1 +r₁+r₁r₂+· · ·+r₁r₂. . . r_N−1)M|h|+O(|h|) and hence ess sup_t[˜p¹(t+h)−p˜¹(t)] =O(h). Similarly, ess inf_t p˜¹(t+h)−p˜¹(t)

=O(h). By (4.8) it follows that ess sup_t|p˜^j(t+h)−p˜^j(t)| = O(h) ∀j , and hence ˜p^j, j = 1, . . . , N, are (coincide a.e. with) Lipschitz continuous functions.

Notice that Lemma 8 and (4.2) imply Ψ∈C^0,1 and henceLΨ∈C^1,1by (2.8).

The final step is an iterative scheme that, thanks to Lemma 8 and what we have just observed, will imply

Ψ ∈ C^∞.

Proposition 9. Let Ψ,Γ satisfy (4.1). Then, fork= 1,2, . . .

LΨ∈C^k,1, γ^j−γ^j−1∈C^k−1,1⇒γ^j−γ^j−1∈C^k,1 and Lψ∈C^k+1,1. (arabic@¸ section.arabic@¸ equation) Proof. Let us discuss first in detail the casek= 1. Notice first that LΨ∈C^1,1 ⇒g^j ∈C^1,1. Then, since ˜p^j, being Lipschitz, has a (bounded) derivative almost everywhere, we can take thetderivative in (4.7), to obtain that

J^j(t) :=

_π

0

g_t^j(x, t−x)e^{p˜j(t)−p˜j(t−2x)}−g_t^j(x, t+x)e^{p˜j(t+2x)−p˜j(t)} dx +

_π

0

g^j(x, t−x)e^{p˜j(t)−p˜j(t−2x)} p˙˜^j(t)−p˙˜^j(t−2x)

−g^j(x, t+x)e^{p˜j(t+2x)−˜pj(t)} p˙˜^j(t+ 2x)−p˙˜^j(t) dx is independent onj, for a.e. t. Since ˙˜p^j is bounded andg^j_t is Lipschitz, we can write

J^j =A^jp˙˜^j+B^j

withA^j, B^j Lipschitz continuous,A^j positive, bounded and bounded away from zero. Hence

˙˜

p^j+1= A^j

A^j+1p˙˜^j+C^j withC^j Lipschitz. Thus, for any givenh∈IR,

˙˜

p^j+1(t+h)−p˙˜^j+1(t) = A^j

A^j+1[ ˙˜p^j(t+h)−p˙˜^j(t)] +O(h) for a.e. t.

Since_N

j=1p˙˜^j≡0, we conclude as in Lemma 8 that ˙˜p^j ∈C^0,1. As above, ˜p^j, LΨ∈C^1,1 ⇒Ψ∈C^1,1⇒LΨ∈ C^2,1.