On Periodic Motions of a Two Dimensional Toda Type Chain by G. Mancini1 and P.N. Srikanth2
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. (T C)
i i i+1 − ϕi ) − exp (ϕi − ϕi−1 ) ϕtt − ϕxx = exp (ϕ
0 < x < π,
t ∈ IR, i ∈ ZZ
ϕi (0, t) = ϕi (π, t) = 0 ∀ t, i
We consider the case of “closed chains” i.e. ϕi+N = ϕi
∀ i ∈ ZZ and some N ∈ IN and we
look for solutions which are periodic in time. If N = 2, it reduces to a sinh-Gordon equation, which is known to be completely integrable (see [Mi]) and to possess periodic solutions (see [BCN]). Henceforth, we will be interested to the case N ≥ 3. The study is motivated by the work [Mi]. In general such studies take importance due to the following. In the early fifties Fermi, Pasta and Ulam (FPU) numerically studied the problem of energy partion. They observed, in the case of a linear lattice, normal modes of oscillations are mutually independent and no energy exchanges take place among these modes (see [T]). It was thought that if nonlinear interaction was introduced, energy flow would take place. FPU tested this expectation through computer experiments, which led, however, to the discovery of the so called ”FPU recurrence phenomenon”. Hence oscillation of nonlinear lattices became of serious study. Looking for a system with rigorous periodic solutions, Toda considered one-dimensional lattices of N particles with exponential interaction. If Vi denotes the potential of interaction between the ith and (i + 1)th particle, then the equation for the ith particle reads 0 ϕ¨i = Vi−1 (ϕi−1 − ϕi ) − Vi0 (ϕi − ϕi+1 ). 1
(T )
Dipartimento di Mathematica, Universita di Roma Tre, Via S.Leonardo Murialdo 1, Roma,Italy. e-mail:
[email protected]. The author’s research is supported by M.U.R.S.T. under the national project Variational methods and nonlinear differential equations 2 TIFR Centre, IISc Campus, P.B. 1234, Bangalore - 560 012, India. e-mail:
[email protected] 1
In case
Vi (t) = V (t) = ab ebt + at ∀i = 1, . . . , N and ab > 0
one has the classical Toda
lattice, which was shown to be completely integrable, with explicit periodic and soliton solutions ([T]). Furthermore, it is known by methods of Kolmogorov-Arnold-Moser (KAM) theory ([Ar], [M]) 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 [RS] Toda type lattices under large perturbations have been studied and existence of periodic solutions has been established. 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 [Mi] where integrability is shown using inverse-problem method. There is no information about the nature of solutions. In general such questions seem hard and infact Toda says (see [T], page 3) in the context of comments on ergodic behaviour that two-dimensional and three-dimensional systems belong to future problems. Returning to (TC), note that it possesses (linear) normal modes ϕi = ϕ,
ϕtt −ϕxx = 0,
and infact the motion is determined up to normal modes, i.e if Φ = (ϕ1 , . . . ϕN ) solves (TC), then it does Φ + Γ, where Γ = (γ 1 , . . . γ N ) with γ i = γ (i = 1, . . . , N ) and γtt − γxx = 0 (Notice that
N P i=1
ϕitt − ϕixx = 0). To get rid of normal modes, we will ask Φ to satisfy the
normalization condition
N P
ϕi ≡ 0 and prove the following Theorem.
i=1
Theorem. (TC) has infinitely many (normalized) periodic solutions.
2
A dual Problem in Orlicz Spaces
Problem (TC) can be written in vector form 2Φ + ∇f (Φ) = 0 where Φ = (ϕ1 , . . . , ϕN ), f (η) =
2Φ = (φitt − ϕixx )i=1,...N and N X
[exp(η i − η i−1 ) − 1]
η ∈ IRN , η 0 = η N
i=1
∇f (η) = (exp(η − η N ) − exp(η 2 − η 1 ), . . . , exp(η N − η N −1 ) − exp(η 1 − η N )) 1
Clearly, ∇f (t11) = 0 ∀t, where 11 = (1, . . . , 1) and ∇f (η) ∈ 11⊥ = {ξ ∈ IRN :
N P i=1
2
ξ i = 0}.
Basic qualitative properties of the interaction potential f , proven in the Appendix, are as follows. Let
cN :=
√
∀ α > 0,
(2.1)
2 N (N −1)
; then √ α + Nα ∀ η ∈ 11 with |η| ≥ cN ⊥
h∇f (η), η i ≥ αf (η)
h∇f (η), η i ≥ cN |η| [ecN |η| − 1]
(2.2)
h D2 f (η) ξ, ξ i ≥ c2N e−
(2.3)
√
2 |η|
|ξ|2
∀ η ∈ 11⊥
∀ η ∈ IRN ,
∀ ξ ∈ 11⊥
A(s) = es − s − 1, s ≥ 0, then
Furthermore, if
√ A(cN |η| ) ≤ f (η) ≤ N A( 2 |η| ) ∀η ∈ 11⊥
(2.4)
The function A so defined is an Orlicz function (or N-function; see Appendix for basic facts on Orlicz functions and Orlicz spaces used throughout the paper). We denote by LA = LA (Ωn ) the Orlicz space associated to A, consisting of measurable functions Φ : Ωn → IR
N
such that
Z
A(λ|Φ|) < ∞
for some λ = λ(Φ) > 0
Ωn
h
i
Ωn = [0, π] × 0, 2π (n to be chosen, later, suitably large), i.e. LA = hKA i where n
where
)
(
KA = Φ :
R
A(|Φ|) < ∞ . LA , endowed with the Orlicz-Luxemburg norm
Ωn
Z
Ωn
kΦkA = inf k > 0 :
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→
Z Ωn
1Z f (Φ) − |Φt |2 − |Φx |2 2 Ωn
in some Orlicz-Sobolev space associated to A (in the spirit of the celebrated paper [R]). Instead, because of the strong indefiniteness of the quadratic part, we prefer, following [BCN], to work with a dual functional, via a Legendre transformation. To this end, we first derive some properties of the dual potential f ∗ : f ∗ (ξ) := sup [hξ, ηi − f (η)] ∀ ξ ∈ 11⊥ . η∈11⊥
3
Since f |11⊥ has superlinear growth and is strictly convex by (2.1) - (2.4), f ∗ turns out to be a C ∞ -strictly convex function on 11⊥ . Also, ∇f (η) = ξ ⇔ η = ∇f ∗ (ξ) and f ∗ (ξ) = hξ, ηi−f (η) for ξ = ∇f (η) and we have hξ, ηi ≤ f ∗ (ξ) + f (η) ∀ ξ, η ∈ 11⊥ .
(Young’s inequality)
We denote by A∗ the N -function complementary to A, i.e A∗ (s) = sup(st − A(t)),
s ≥ 0.
t≥0
Easily,
A∗ (s) = (1 + s) log(1 + s) − s,
N A∗
Lemma 1.
|ξ| √ N 2
and, by (2.4) and direct computation, we have
≤ f ∗ (ξ) ≤ A∗
|ξ| cN
∀ ξ ∈ 11⊥ .
A crucial feature of A∗ is that it satisfies a (global) (∆2 ) property: A∗ (rt) ≤ k(r) A∗ (t) ∀t ≥ 0
∀ r > 0 ∃k(r) > 0 :
(∆2 )
with k(r) = max{r, r2 }. From (∆2 ) and Lemma 1 it follows 1 1 A∗ (|ξ|) ≤ f ∗ (ξ) ≤ 2 A∗ (|ξ|) ∀ ξ ∈ 11⊥ . 2N cN
(2.5)
Hence, the natural space to work in is the Orlicz space LA∗ . Note that, thanks to (∆2 ), LA∗ = KA∗ . In particular, LA∗ is a separable Banach space (see Prop. A1). Also, from H¨older’s inequality Z
hΦ, Ψi ≤ N kΦkA kΨkA∗
∀ Φ ∈ LA , Ψ ∈ LA∗ ,
it follows that every Φ ∈ LA defines an element of L0A∗ and infact, since A∗ satisfies (∆2 ) , L0A∗ turns out to be isomorphic and homeomorphic to LA . We now proceed to establish some properties of the functional F (Ψ) :=
R Ωn
its “gradient operator” Ψ → ∇f ∗ (Ψ), on LA∗ := {Ψ ∈ LA∗ :
N P i=1
Lemma 2. (i) k∇f ∗ (Ψ)kA ≤
1 cN
1 +
kΨk2A∗ c2N
∀ Ψ ∈ LA∗ 4
ψ j ≡ 0}.
f ∗ (Ψ), and
(ii) kΨj − ΨkA∗ → 0 ⇒
R
h∇f ∗ (Ψj ), h ¯i →
Ωn
(iii) F (Ψ) =
R
h∇f ∗ (Ψ), h ¯i ∀ h ¯ ∈ LA∗ .
Ωn
f ∗ (Ψ) is a locally Lipschitz functional on LA∗ .
R Ωn
(iv) Ψ → ∇f ∗ (Ψ) is the Gateaux derivative of F. i.e. Z d Z ∗ [ f (Ψ + t¯ h)]|t=0 = h∇f ∗ (Ψ), h ¯ i ∀ Ψ, h ¯ ∈ LA∗ . dt Ωn
Ωn
Proof. (i) Let Φ = ∇f ∗ (Ψ), and hence Ψ = ∇f (Φ). From (2.2) we see that 1 |Ψ| ≥ ecN |Φ| − 1 cN and hence, by monotonicity, A∗ ( R
Since
Ωn
|Ψ| ) ≥ A∗ (ecN |Φ| − 1) = cN |Φ| ecN |Φ| − ecN |Φ| + 1 ≥ A(cN |Φ|) cN
A∗ (|Ψ|) ≤ supt>0
A∗ (kΨkA∗ t) A∗ (t)
≤ max{kΨkA∗ , kΨk2A∗ } ∀Ψ ∈ LA∗ , (see Prop.
A2-(ii) and (∆2 ) for A∗ ), then Z
A(cN |∇f ∗ (Ψ)|) ≤ 1 +
Ωn
kΨk2A∗ c2N
∀ Ψ ∈ LA∗ .
kΨk2A∗ c2N
∀ Ψ ∈ LA∗
Finally, using Prop. A2-(i), we obtain kcN ∇f ∗ (Ψ)kA ≤ 1 +
(ii)
From kΨj − ΨkA∗ → 0, we have
R
A∗ (|Ψj − Ψ|) → 0 (see Prop. A2-(i)) and hence
Ψj → Ψ in measure. In addition, we know by (i) that k∇f ∗ (Ψj )kA is bounded and hence (ii) follows by Prop. A3. (iii)
Let Ψ1 , Ψ2 ∈ Br (0). Then R R1
|F (Ψ1 ) − F (Ψ2 )| = |
Ωn
h∇f ∗ (tΨ1 + (1 − t)Ψ2 ), Ψ1 − Ψ2 i|
0
≤ N kΨ1 − Ψ2 kA∗ sup k∇f ∗ (Ψ)kA Ψ∈Br (0)
with
sup k∇f ∗ (Ψ)kA < ∞ by (i). Ψ∈Br (0)
5
R
(iv) follows from
Ωn
R
f ∗ (Ψ+t¯ h)−f ∗ (Ψ) t
h∇f ∗ (Ψ) + τ t¯ h), h ¯i →
Ωn
R
=
R1 R
( h∇f ∗ (Ψ + τ t¯ h), h ¯ i)dτ
and (see (ii))
0 Ωn
h∇f ∗ (Ψ), h ¯ i ∀ τ ∈ [0, 1] , because, by (i),
Ωn
|
Z
h∇f ∗ (Ψ + τ t¯ h), h ¯ i| ≤ k¯ hk
k∇f ∗ (Ψ)k ≤ c
sup h k A∗ kΨ−ΨkA∗ ≤k¯
Ωn
Let us now recall basic properties of the operator 2 =
∂2 ∂t2
2
∂ − ∂x acting on summable 2
functions which are 2π periodic and satisfy Dirichlet boundary conditions. It is known (see [L], [BCN]) that 2 γ = 0 (in the weak sense), if and only if γ(x, t) = p(x + t) − p(t − x) for some p ∈ L1loc (IR), 2π−periodic and such that we have that
2π R
p = 0. Also, for a given summable ψ,
0
ψ γ = 0 ∀ γ ∈ L∞ ∩ Ker 2 if and only if
R [0,π]×[0,2π]
Zπ
(2.6)
[ψ(x, t − x) − ψ(x, t + x)]dx = 0 for a.e. t
0
and, for such a ψ, there is a unique 2π periodic continuous function ϕ = Lψ satisfying the boundary condition ϕ(0, t) = ϕ(π, t) = 0 ∀t, such that 2ϕ = ψ
Z
and
ψ γ = 0 ∀ γ ∈ Ker 2 ∩ L∞ .
[0,π]×[0,2π]
Actually, Lψ is explicitely known (see [L]): Lψ(x, t) = ϕ˜ + p(t + x) − p(t − x) where Rπ
t−x+ξ R
x
t+x−ξ
ϕ(x, ˜ t) = − 21 dξ and
ψ(ξ, τ )dτ +
π−x 2π
Rπ 0
dξ
t+ξ R
ψ(ξ, τ )dτ
t−ξ
π 1 Z [ϕ(s, ˜ y − s) − ϕ(s, ˜ y + s)]ds p(y) = 2π 0
It can be easily checked that (2.7)
∃C >0:
kLψkL∞ ≤ C kψkL1 , 6
kLψkC 0,1 ≤ C kψkL∞
ψ ∈ C k,1 ⇒ Lψ ∈ C k+1,1
(2.8)
IK := ( Ker 2 )N ,
Let us now write
E = Ψ ∈ LA∗ :
N X
LΨ = (Lψ 1 , . . . , Lψ N ) , and set
ψ j ≡ 0,
j=1
Z Ωn
hΨ, Γi = 0 ∀ Γ ∈ IK ∩ LA
Note that E, a closed subspace of LA∗ , is a separable Banach space. Lemma 3. L induces a compact linear operator from E into LA . Proof. Since LA∗ ⊂ L1 , LΨ is infact in L∞ . Let Ψk be a bounded sequence in E. Since sup kΨk kA∗ < ∞ implies sup kΨk kL1 < ∞, LΨk is bounded in L∞ . We want to prove LΨk are k
k
equicontinuous and Ascoli-Arzela will imply compactness. Using the notations introduced above, ˜ k + Pk (t + x) − Pk (t − x), LΨk = Φ
P = (pi )
˜ k are equicontinuous. We have (dropping superscripts) and it is clearly enough to prove that Φ 0
0
|ϕ˜k (x, t) − ϕ˜k (x , t )| ≤ where
Rπ
t+ξ R
0
t−ξ
ck := dξ
t−x+ξ π R 1 R | dξ ψk (ξ, τ )dτ 2 x t+x−ξ
ψk (ξ, τ )dτ
−
Rπ
dξ
x0
0 t0 −x R +ξ
t0 +x0 −ξ
are constants bounded by
ψk (ξ, τ )dτ | + supk kψk kL1
ck |x0 2π
− x|
(recall (2.6)).
Hence it is enough to estimate ∆=|
ψk (ξ, τ )dξdτ −
R Tx,t
R
ψk (ξ, τ )dξdτ | ≤
Tx0 ,t0
R
|ψk |
(Tx,t \Tx0 ,t0 )∪(Tx0 t0 \Tx,t )
where Tx,t = {(ξ, τ ); x ≤ π,
t + x − ξ ≤ τ ≤ t − x + ξ} ⊂ [0, π] × IR
and ψk are extended as periodic functions on [0, π]×IR. Clearly the measure of the symmetric difference is small if (x, t) and (x0 , t0 ) are close. On the other hand by Vall´ee-Poussin theorem ([KR] page 103) sup k
Z
A∗ (|ψk |) < +∞ ⇒ sup
Ωn
Since A∗ (|Ψk |) ≤ 1+kΨk k2A∗ we have R
k
Z
|ψk | ≤ ε
if |Σ| ≤ δε
P
(see Prop. A2-(ii) and (∆2 ) for A? ), and supk kΨk kA? < +∞,
|ϕ˜k (x, t) − ϕ˜k (x0 , t0 )| ≤ ε if |x − x0 | + |t − t0 | ≤ δε 7
uniformly with respect to k. Hence the result. We now state the dual variational principle. Proposition 4. Let Ψ be a critical point of the (Gateaux differentiable) functional (2.9)
S(Ψ) =
1Z f (Ψ) + h Ψ, LΨ i, 2
Z
∗
Ωn
Then Φ = ∇f ∗ (Ψ) is a (weak)
2π n
Ωn
periodic solution of (TC). IK ∩ LA)is closed in LA , as well as LA ∩ IK ⊥ :=
Proof. Since IK is closed in L1 , ( Φ ∈ LA :
Ψ∈E
hΦ, Γ i = 0 ∀ Γ ∈ IK ∩ LA∗ . Thus we can write
R Ωn
∇f ∗ (Ψ) + LΨ = W + Γ, Notice that
P
j
W ∈ IK ⊥ ∩ LA ,
wj + γ j ≡ 0 and hence we can assume Z
∗
h∇f (Ψ), h ¯i +
Z
P
j
Γ ∈ IK ∩ LA .
wj ≡ 0. Since, by assumption,
h ¯ LΨ = 0 ∀ h ¯∈E
Ωn
Ωn
we see that W = 0. Hence ∇f ∗ (Ψ) + LΨ = Γ. Let Φ = ∇f ∗ (Ψ) = Γ − LΨ. We have −2Φ = 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 of S. For a given θ < 1, let mθ = 2(α + N α) where α = (1 − θ)−1 . Then Lemma 5. |ξ| ) cN
≥ |∇f ∗ (ξ)| ≥
(i)
1 cN
(ii)
∀ θ < 1, f ∗ (ξ) ≥ θh∇f ∗ (ξ), ξi
log(1 +
N ∗ |ξ| A ( N √2 ) |ξ|
if
∀ ξ ∈ 11⊥ ,
ξ 6= 0 mθ
|ξ| ≥ c(N, θ) := 4N e cN
Proof. (i) From Lemma 1 and h∇f ∗ (ξ), ξi ≥ f ∗ (ξ), we obtain the RHS inequality. Next, if η = ∇f ∗ (ξ), from (2.2) we derive |ξ| = |∇f (η)| ≥ cN [ecN |η| − 1] = cN [ecN |∇f gives the LHS inequality. (ii) From the RHS inequality in (i), one easily obtains |ξ| ≥ 4N e
mθ cN
⇒
√ α + Nα |∇f ∗ (ξ)| ≥ cN 8
∗ (ξ)|
− 1]. This
Hence (ii) follows from (2.1) and f ∗ (ξ) = h∇f ∗ (ξ), ξi − 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 [BCN]. Let E be a separable Banach space and let S : E → IR be a locally Lipschitz Gateaux differentiable functional. Suppose further (i) Ψj → Ψ
S 0 (Ψj )¯ h → S 0 (Ψ)¯ h ∀¯ h ∈ E.
⇒
(ii) ∃ U open, 0 ∈ U such that
inf ∂U S > S(0)
(iii) ∃Ψ0 6∈ U such that S(Ψ0 ) ≤ S(0) (iv) S(Ψj ) → c, S 0 (Ψj ) → 0
⇒
∃ Ψ ∈ S −1 (c) such that S 0 (Ψ) = 0.
Then c:=
inf
p∈IP
max S(p(t))
t∈[0,1]
is a critical value , where IP = {p ∈ C([0, 1], E) : p(0) = 0, p(1) = Ψ0 } We have already seen that S defined by (2.9) is locally Lipschitz and Gateaux differentiable, with derivative given by S 0 (Ψ)¯ h = h∇f ∗ (Ψ), h ¯ i + hLΨ, h ¯ i, R
∀¯ h ∈ E. We now prove
S satisfies (i)-(iv). Proof of (i). It follows by Lemma 2-iii and continuity properties of L. Proof of (ii). It is exactly here where we need to take n suitably large. Let U = {Ψ ∈ E : R
f ∗ (Ψ) < 2N |Ωn |}. From Young’s inequality and max f < 2N, we get |ξ|=1
Ωn
|ξ| ≤ f ∗ (ξ) + f and hence
R
|ψ j | ≤ 2N |Ωn | +
Ωn
R Ωn
hΨ, LΨi ≥ −2
ξ |ξ| R
!
≤ 2N + f ∗ (ξ) ∀ ξ ∈ 11⊥ f ∗ (Ψ) = 4N |Ωn | ∀ j and Ψ ∈ ∂U. Since
Ωn N R P
(
j
2
|ψ |) (see [BCN ]) we have, for Ψ ∈ ∂U ,
j=1 Ωn
9
S(Ψ) ≥ 2N |Ωn | −
P
|ψ j |)2 ≥ 2N |Ωn | − 16N 3 |Ωn |2 > 0 = S(0) if n > 16π 2 N 2 .
R
(
Ωn
Proof of (iii). It follows from the subquadratic behaviour of f ∗ at infinity. Proof of (iv). Let Ψj be such that Z Ωn
1Z hΨj , LΨj i = c+◦(1) and f (Ψj )+ 2 ∗
sup | k¯ hkA∗ ≤1
Ωn
h∈E ¯
Z
h∇f ∗ (Ψj )+LΨj , h ¯ i| = ◦(1) as j → ∞
Ωn
and hence 2
Z
∗
f (Ψj ) −
Ωn
Z
h∇f ∗ (Ψj ), Ψj i = 2c + ◦(1) + ◦(1)kΨj kA∗
Ωn
By Lemma 5- (ii) and f ∗ ≥ 0, we have Z
∗
Ωn
where c˜(N, θ) =
h
1 cN
f (Ψj ) ≥ θ
h∇f ∗ (Ψj ), Ψj i − θ˜ c(N, θ)
Ωn
i
c(θ,N ) ) cN
log(1 +
|Ωn |. Hence
1 Z ∗ ≥ 2− f (Ψj ) − [˜ c(N, θ) + 2c + ◦(1)] θ
◦ (1)kΨj kA∗
(3.1)
Z
Ωn
From (2.5) and Prop. A2-(i) , we have Z
(3.2)
1 [ kΨj kA∗ − 1 ] 2N
f ∗ (Ψj ) ≥
Ωn
Taking θ > 34 , we obtain from (3.1) and (3.2) 1 1 kΨj kA∗ ≤ + c˜(N, θ) + 2c 3N 2N
(3.3)
Hence sup kΨj kA∗ < ∞. Then (see [KR] 14.4) ∃ Ψ ∈ E such that, for a subsequence, j
Z Ωn
hΨj , Φ i →
Z
hΨ, Φi
∀ Φ ∈ EA
Ωn
where EA is the closure of the set of bounded functions in LA . We want to show that Z
h∇f ∗ (Ψ) + LΨ, h ¯i = 0 ∀ h ¯ ∈ E.
Ωn
10
i.e Ψ is a critical point at level c. We will use here, following [BCN], the so called Minty-trick, exploiting monotonicity of the operator Ψ → ∇f ∗ (Ψ) : Z
(3.4)
h∇f ∗ (Ψj ) − ∇f ∗ (¯ h),
Ψj − h ¯i ≥ 0 ∀ h ¯ ∈ LA∗ .
Writing Wj ∈ LA ∩ IK ⊥ , Γj ∈ LA ∩ IK
∇f ∗ (Ψj ) + LΨj = Wj + Γj , with
P i
wji = 0 and kWj kA → 0 (see Prop. 4) and inserting in (3.4), we get
(3.5)
Z
hWj − LΨj − ∇f ∗ (¯ h), Ψj − h ¯i ≥ 0 ∀ h ¯ ∈ LA∗
If h ¯ ∈ L∞ , we can pass to the limit in (3.5), to get Z
(3.6)
hLΨ + ∇f ∗ (¯ h), Ψ − h ¯ i ≤ 0,
Now, choose as test function
nΨ
− t¯ h, h ¯ ∈ L∞ where (
nΨ
∀ h ¯ ∈ L∞
j
= (n ψ ),
nψ
=
ψ if |ψ| ≤ n 0 if |ψ| > n.
Then we have Z
hLΨ + ∇f ∗ (n Ψ − t¯ h), (Ψ −n Ψ) + t¯ hi ≤ 0
Since A∗ is ∆2 , kn Ψ − ΨkA∗ → 0 (see [KR] 10.1) and hence Z
(3.7)
hLΨ + ∇f ∗ (Ψ − t¯ h) , t¯ hi≤0 ∀h ¯ ∈ L∞ , .
by Lemma 2 - (ii) and because ∇f ∗ (n Ψ − t¯ h) is bounded in LA by Lemma 2 - (i). Now, since L∞ is dense in LA∗ , arguments as above give Z
hLΨ + ∇f ∗ (Ψ − t¯ h), h ¯i ≤ 0 ∀ h ¯ ∈ E and t > 0.
Letting t → 0, we get Z
hLΨ + ∇f ∗ (Ψ), h ¯i = 0 ∀ h ¯
The theorem now follows from Lemma [BCN]. We remark that the existence of infinitely many solutions follows from the simple device (see [N]) of working in a space of functions whose period is smaller than the minimal period of the solution previously obtained.
11
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 [BN], 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 2 Ψ ∈ L , 1
(4.1)
Z
ΨΠ = 0 ∀Π ∈ L∞ ∩ Ker 2
Ψ = ∇f (Γ − LΨ) which will lead to C ∞ regularity for Ψ. Clearly, Φ := Γ − LΨ is a weak solution of (TC) and, if
P
ϕj ≡ 0 , then Φ = ∇f ∗ (Ψ) ∈ C ∞ . We start with a simple summability Lemma.
∇f (Φ) ∈ L1
Lemma 6.
⇒
exp (ϕj − ϕj−1 ) < ∞ ∀ j.
R Ωn
Proof . Arguing by contradiction, we find j such that
R
exp [ϕj − ϕj−1 ] = +∞ and hence
Ωn
A0 := {ϕj −ϕj−1 ≥ 1} has positive measure. Since we have
exp (ϕj+1 − ϕj ) = +∞
R
exp[ϕj −ϕj−1 ]−exp[ϕj+1 −ϕj ] = ψ j ∈ L1
and hence
A1 := A0 ∩ {ϕj+1 − ϕj ≥ 1 } has
A0
positive measure . After N iterations, we find ϕj+i ≥ ϕj+i−1 + 1 for i = 0, . . . , N − 1 on a set of positive measure AN −1 . Adding up these inequalities we find a contradiction. Remark . If Φ is a normalized weak solution with Lemma and Lemma A.1, we have
R
∇f (Φ) ∈ L1
, then, by the previous
ecN |Φ| < ∞ . Even this simple summability property
fails for arbitrary weak solutions. To start the bootstrap argument, we first need a couple of Lemmata. Ψ, Γ satisfy (4.1). Then γ j − γ j−1 ∈ L∞ ∀j.
Lemma 7. Let
˜ := −LΨ belongs to L∞ by (2.7). Let M := 2kΦk ˜ ∞ . Denoted Φ := Γ + Φ, ˜ Proof . First, Φ it results (4.2)
γj −
M 2
≤ ϕj ≤ γ j +
M 2
. 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 ]
12
satisfy (2.6). We have exp (γ j − γ j−1 − M ) − exp (γ j+1 − γ j + M ) ≤
(4.3)
≤ exp (ϕj − ϕj−1 ) − exp (ϕj+1 − ϕj ) ≤ ≤ exp (γ j − γ j−1 + M ) − exp (γ j+1 − γ j − M ) Let γ j = pj (t + x) − pj (t − x), γ j − γ j−1 = p˜j (t + x) − p˜j (t − x), where p˜j := pj − pj−1 , and rewrite (4.3) as exp [˜ pj (t + x) − p˜j (t − x) − M ] − exp [˜ pj+1 (t + x) − p˜j+1 (t − x) + M ] ≤ ≤ ψ j (x, t) ≤ ≤ exp [˜ pj (t + x) − p˜j (t − x) + M ] − exp [˜ pj+1 (t + x) − p˜j+1 (t − x) − M ] Computing at (x, t − x), (x, t + x) respectively and subctracting, we get exp [˜ pj (t) − p˜j (t − 2x) − M ] − exp [˜ pj+1 (t) − p˜j+1 (t − 2x) + M ] +
(4.4)
+ exp [˜ pj+1 (t + 2x) − p˜j+1 (t) − M ] − exp [˜ pj (t + 2x) − p˜j (t) + M ] ≤ ≤ ψ j (x, t − x) − ψ j (x, t + x) ≤ ≤ exp [˜ pj (t) − p˜j (t − 2x) + M ] − exp [˜ pj+1 (t) − p˜j+1 (t − 2x) − M ] − − exp [˜ pj (t + 2x) − p˜j (t) − M ] + exp [˜ pj+1 (t + 2x) − p˜j+1 (t) + M ] Notice that, due to Lemma 6, sinh p˜j ∈ L1 ([0, 2π]) ∀j. In fact, by Fubini and 2π-periodicity, ∞>
2π R Rπ 0 0
p˜j (t+x)−˜ pj (t−x)
e
dxdt =
2π R 0
p˜j (s)
e
Rπ 0
! −˜ pj (s−2x)
e
dx ds =
1 2
2π R 0
j
ep˜ (s) ds
2π R
j
e−˜p (τ ) dτ
0
Integrating w.r. to x ∈ [0, π] the inqualities (4.4) and changing variables s = t − 2x and s = t + 2x, we get, using 2π-periodicity of p˜j and property (2.6) of ψ j , we obtain (4.5)
Z2π
[sinh (˜ pj (t) − p˜j (s) − M ) + sinh (˜ pj+1 (s) − p˜j+1 (t) − M )]ds ≤ 0 for a.e t
0
(4.6)
Z2π
[sinh (˜ pj (t) − p˜j (s) + M ) + sinh(˜ pj+1 (s) − p˜j+1 (t) + M )]ds ≥ 0, for a.e t
0
We now derive from (4.5) that ess sup p˜j < ∞ ∀j. We argue by contradiction. Assume that Ajk = {t :
p˜j (t) ≥ k} has positive measure for some j ∈ {1, . . . N } and every k ∈ IN . 13
We claim that this implies Aj+1 = {t ∈ Ajk : p˜j+1 (t) ≥ k} has positive measure ∀ k. If not, k ¯ and tk ∈ Aj ⊂ Aj¯ ∀k ≥ k¯ such that p˜j+1 (tk ) ≤ k. ¯ From (4.5) at tk we have ∃ k, k k Z2π
sinh(k − M − p˜j (s)) + sinh(˜ pj+1 (s) − k¯ − M ) ≤ 0 ∀ k ≥ k¯
0
Taking into account the sommability of sinh p˜j by Levi’s Theorem. Thus Aj+1 k j+N −1 j+N −2 j Ak ⊂ Ak . . . ⊂ Ak , k
, we get, sending k to infinity, a contradiction
has positive measure ∀ k. Repeating the argument, we find ∈ IN sets of positive measure, such that
−1 p˜i (t) ≥ k for i = 1 . . . , N and k ∈ IN , t ∈ Aj+N k N P
Since
p˜i (t) ≡ 0, we arrive at a contradiction. Similarly, using (4.6), we can derive
i=1 j
ess inf p˜ > −∞ ∀ 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 hence LΨ ∈ C 0,1 (see (2.7)). We first derive from (4.2) and (2.6) Zπ
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 =
0
Zπ
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
0
for every j and a.e. t. Thus, after writing g j = exp[−L(ψ j − ψ j−1 )] ∈ C 0,1 , we have, using previous notations, (4.7)
Zπ
g j (x, t − x) exp [˜ pj (t) − p˜j (t − 2x)] − g j (x, t + x) exp [˜ pj (t + 2x) − p˜j (t)] dx =
0
Zπ
g j+1 (x, t − x) exp [˜ pj+1 (t) − p˜j+1 (t − 2x)] − g j+1 (x, t + x) exp [˜ pj+1 (t + 2x) − p˜j+1 (t)] dx
0
for every j and a.e. t. Computing at t + h, t, and taking the difference, we see that I j (t, h) := Zπ
g j (x, t + h − x) exp [˜ pj (t + h) − p˜j (t + h − 2x)] − g j (x, t − x) exp [˜ pj (t) − p˜j (t − 2x)] dx
0
14
−
Zπ
g j (x, t + h + x) exp [˜ pj (t + h + 2x) − p˜j (t + h)] − g j (x, t + x) exp [˜ pj (t + 2x) − p˜j (t)] dx
0
= k1j (t, h) − k2j (t, h) is independent on j for every fixed h and a.e. t. Now, k1j
h
p˜j (t+h)
= e
p˜j (t)
−e
i Zπ
j
g j (x, t − x) e−˜p (t−2x) dx
0
p˜j (t+h)
+e
π Z Zπ j j g j (x, t + h − x) e−˜p (t+h−2x) dx − g j (x, t − x) e−˜p (t−2x) dx 0
Rπ
Since that
0 π− h 2
j
g j (x, t + h − x) e−˜p (t+h−2x) dx =
0 Rπ
j
g j (x, t + h − x) e−˜p (t+h−2x) dx −
R
Rπ
−h 2
j
g j (x + h2 , t − x + h2 ) e−˜p (s−2x) dx , we see j
g j (x, t − x) e−˜p (t−2x) dx = O(h) , because g j
0
0
is Lipschitz and p˜j ∈ L∞ , and hence
pj (t + h) − p˜j (t)] + O(h) k1j = cj1 (t, h) [˜ where
cj1 (t, h) =
R1
j
j
es˜p (t+h)+(1−s)˜p (t) ds
0
Rπ
j
g j (x, t − x) e−˜p (t−2x) dx is positive, (essentially)
0
bounded and bounded away from zero uniformly w.r. to h. Similarly h
i
k2j = −cj2 p˜j (t + h) − p˜j (t) + O(h). Hence
I j (t, h) = cj p˜j (t + h) − p˜j (t) + Oj (h) where cj = cj1 + cj2 . Since I 1 (t, h) = I 2 (t, h) = . . . = I N (t, h) for every h and a.e. t, then (4.8)
p˜j+1 (t + h) − p˜j+1 (t) = rj [˜ pj (s + h) − p˜j (s)] + Oj (h) a.e.
where rj = rj (t, h) = (4.9)
cj cj+1
are positive, bounded and bounded away from zero. Now, let
A(M, h) := {t : p˜1 (t + h) − p˜1 (t) ≥ M |h| }
It follows from (4.8) that, for a.e. t ∈ A(M, h), p˜j+1 (t + h) − p˜j+1 (t) ≥ r1 r2 . . . rj M |h| + Oj (h) for j = 1, . . . , N − 1
15
Taking into account that
N P
p˜j ≡ 0, we obtain, for a.e. t ∈ A(M, h),
j=1
0 ≥ (1 + r1 + r1 r2 + . . . + r1 r2 . . . rN −1 ) M |h| + O(|h|) and hence ess supt [˜ p1 (t + h) − p˜1 (t)] = O(h). Similarly, ess inft [˜ p1 (t + h) − p˜1 (t)] = O(h). By (4.8) it follows that ess supt |˜ pj (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 hence LΨ ∈ C 1,1 by (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 (4.10)
LΨ ∈ C k,1 ,
Ψ, Γ satisfy (4.1). Then, for k = 1, 2, . . . γ j − γ j−1 ∈ C k−1,1 ⇒ γ j − γ j−1 ∈ C k,1 and Lψ ∈ C k+1,1
Proof. Let us discuss first in detail the case k = 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 the t derivative in (4.7), to obtain that j
J (t) :=
Zπ h
j
j
j
j
gtj (x, t − x) ep˜ (t)−˜p (t−2x) − gtj (x, t + x) ep˜ (t+2x)−˜p (t)
i
dx +
0
Zπ
j
j
j
j
j
j
j
j
g j (x, t − x) ep˜ (t)−˜p (t−2x) [p˜˙ (t) − p˜˙ (t − 2x)] − g j (x, t + x) ep˜ (t+2x)−˜p (t) [p˜˙ (t + 2x) − p˜˙ (t)] dx
0 j is independent on j, for a. e. t. Since p˜˙ is bounded and gtj is Lipschitz, we can write j J j = Aj p˜˙ + B j
with Aj , B j Lipschitz continuos , Aj positive , bounded and bounded away from zero. Hence j+1 p˜˙ =
Aj ˙ j p˜ + C j Aj+1
with C j Lipschitz. Thus, for any given h ∈ IR, j+1 j+1 p˜˙ (t + h) − p˜˙ (t) =
Aj ˙ j j [p˜ (t + h) − p˜˙ (t) ] + O(h) for a.e. t Aj+1 16
N P
Since
j=1 1,1
Ψ ∈C
j j p˜˙ ≡ 0 , we conclude as in Lemma 8 that p˜˙ ∈ C 0,1 . As above, p˜j , LΨ ∈ C 1,1 ⇒
⇒ LΨ ∈ C 2,1 .
In the same way we can deal with k ≥ 2 in (4.10). First observe that g j ∈ C k,1 because LΨ ∈ C k,1 . Since p˜j ∈ C k−1,1 and hence (˜ pj )(k) exists as an L∞ function, we can take the k th derivative in (4.7) to obtain that Ikj (t)
:=
Zπ 0
i dk h j p˜j (t)−˜ pj (t−2x) j p˜j (t+2x)−˜ pj (t) g (x, t − x) e − g (x, t + x) e dtk
is, for a.e. t, independent on j. As above, we can write pj )(k) + Bkj (t) Ikj (t) = Ajk (t) (˜ with Ajk , Bkj Lipschtiz, Ajk positive, bounded and bounded away from zero. Again, this implies (˜ pj )(k) ∈ C 0,1 . From (4.2) we then derive that Ψ ∈ C k,1 and hence LΨ ∈ C k+1,1 by (2.8).
5
Appendix
Given ξ ∈ IRN , let [ξ] := max{ξ 1 − ξ N , ξ 2 − ξ 1 , . . . ξ N − ξ N −1 }, maxi |ξ i |
and denote
Lemma A1.
[ξ] ≥
11 = (1, 1, . . . 1), 2 N −1
cN =
√ 2 . N (N −1)
1
|ξ j |2 ) 2 ,
|ξ|∞ =
j = 1, . . . N, k ∈ ZZ ( i.e ξ ∈ IRN is identified with an N
ξ j ≤ ξ j−1 + [ξ] ξ j ≤ ξ j−2 + 2[ξ] .. . ξ j ≤ ξ j−(N −1) + (N − 1)[ξ]
(5.2)
j=1
We have
periodic sequence), we have for any given j ∈ {1, . . . , N }
and
PN
|ξ|∞ ≥ cN |ξ| ∀ ξ ∈ 11⊥ .
Proof. After writing ξ j+kN = ξ j ,
(5.1)
|ξ| = (
ξ j+1 ≤ ξ j + [ξ] ξ j+2 ≤ ξ j + 2[ξ] .. . ξ j+(N −1) ≤ ξ j + (N − 1)[ξ] 17
Adding up the N − 1 inequalities in (5.1) we get (N − 1)ξ j ≤
X
ξ i + (1 + 2 + (N − 1))[ξ].
i6=j
Since
ξ i = −ξ j , we have
P i6=j
N ξj ≤
N (N − 1) [ξ]. 2
Similarly, from (5.2) we get −ξ j ≤
N −1 [ξ]. 2
Lemma A2. Let f (ξ) = exp(ξ 1 − ξ N ) + . . . + exp(ξ N − ξ N −1 ) − N. Then ∀ ξ ∈ 11⊥ ,
(i)
∀ α > 0,
h∇f (ξ), ξ i ≥ αf (ξ)
(ii)
h∇ f (ξ), ξ i ≥ cN |ξ| [ecN |ξ| − 1] ∀ ξ ∈ 11⊥
(iii)
(ecN |ξ| − cN |ξ| − 1) ≤ f (ξ) ≤ N e
(iv)
h D2 f (η) ξ, ξ i ≥ c2N e
√
2 |η|
|ξ|2
√
2|ξ|
−
√
with |ξ| ≥
2|ξ| − 1
√ α+ N α cN
∀ ξ ∈ 11⊥
∀ η ∈ IRN , ∀ ξ ∈ 11⊥ .
Proof. (i)-(ii) : Let [ξ] = ξ j − ξ j−1 . Then h∇f (ξ), ξi − αf (ξ) =
P
i
((ξ i − ξ i−1 ) − α)eξ −ξ
i−i
+ ([ξ] − α)e[ξ] + αN
i6=j
≥
P
((ξ i − ξ i−1 ) − α)eα + ([ξ] − α)e[ξ]
i6=j
Since
N P
(ξ i − ξ i−1 ) = 0 , we have
i=1
h∇f (ξ), ξi − αf (ξ) ≥ −[[ξ] + (N − 1)α]eα + ([ξ] − α)e[ξ]
(5.3) From
(s − α)es − (s + (N − 1)α)eα ≥ eα [(s − α)2 − N α] we see that h∇f (ξ), ξi ≥ αf (ξ) if [ξ] ≥ α + 18
√
≥ 0 if s ≥ α +
√ Nα
N α, and hence (i) in view of Lemma A1.
Taking α = 0, in (5.3), we get (ii).
(iii) From (ii), f (ξ) =
R1
h∇f (t ξ), ξ i dt ≥
0
R1
cN |ξ| [ecN t|ξ| − 1] dt = ecN |ξ| − cN |ξ| − 1.
0
As for the RHS inequality, we have
f (ξ) =
N P
i
eξ −ξ
i−1
−1 =
i=1
h
P
=
i
eξ −ξ
i−1
i
− (ξ i − ξ i−1 ) − 1 +
i
eξ −ξ
i
i−1
− (ξ i − ξ i−1 ) − 1 .
ξ i −ξ i−1 >0
ξ i −ξ i−1 0 :
A∗ (rs) ≤ k(r)A∗ (s) ∀s ≥ 0.
The set KA (Ωn ) = KA = {Φ ∈ L1 (Ωn , IRn ) :
A(|Φ|) < ∞} is called the A−Orlicz
R Ωn
class and its linear hull, LA = {ϕ : λϕ ∈ KA for some λ > 0} is the Orlicz space. Endowed with the Orlicz-Luxemurg norm, kΦkA = inf{k > 0 :
Z
|Φ| A k
!
≤ 1}
the space LA becomes a Banach space. Proposition A1. ([A] pg, 232-240) Let A∗ satisfy (∆2 ), Then (i)
KA∗ = LA∗ .
(ii) C0∞ (Ωn ) is dense in LA∗ . (iii)
The dual sapce L0A∗ is isomorphic and homeomorphic to LA .
Remark. The duality pairing between LA∗ and LA is given by LA∗ × LA 3 (Ψ, Φ) →
Z
ΦΨ.
Further more, we have H¨older, (see [A] 237) |
Z
ΦΨ| ≤ kΦkLA∗ kΦkLA
Proposition A2. ([KR], Theorem 9.5) (i)
kΦkA ≤ 1 ⇒
R
A(|Φ|) ≤ kΦkA ,
kΦkA > 1 ⇒
Ωn
R
A(|Φ|) ≥ kΦkA
Ωn
(ii) If A satisfies (∆2 ) , then Z Ωn
A(|Φ|) ≤ sup t>0
A(kΨkA t) = k(kΦkA ) A(t)
Proposition A3. ([KR], Theorem (4.6)). Let Ωn have finite measure and A∗ satisfy (∆2 ). R Let Φn ∈ LA , Φn → Φ in measure. If sup kΦn kA < ∞, then Φ ∈ LA and Φn Ψ → n R ΦΨ ∀ Ψ ∈ LA∗ . 20
Acknowledgement. Part of this work was done when the second author was visiting the Department of Mathematics, Rome 3, Italy. Part of this work was done when the first author was visiting TIFR Centre, Bangalore, India. They thank their respective hosts for the hospitality.
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[RS] Ruf, B. & Srikanth, P.N., On periodic Motions of Lattices of Toda Type via Critical Point Theory, Arch. Rational Mech. Anal. 126, 1994, pp. 369-385. [T] Toda, M., Theory of Nonlinear Lattices, Springer-Verlag, 1989.
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