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Spatial disorder of soliton solutions for discrete

nonlinear Schr¨

odinger equations in a 2D lattice

Shih-Feng Shieh

October 12, 2009

Abstract

In this paper, we employ the construction of topological horseshoes to study the pattern of the soliton solutions to the discrete nonlinear Schr¨odinger (DNLS) equations in a two-dimensional lattice. The spatial disorder of the DNLS equa-tions is the result of the strong amplitudes and stiffness of the nonlinearities. The complexity of this disorder is determined by the oscillations (number of turning points) of the nonlinearities. Nonnegative soliton solutions of the DNLS equations with a cubic nonlinearity is also discussed.

Keywords: discrete nonlinear Schr¨odinger equation, horseshoe, soliton solution, spa-tial disorder

1

Introduction

Our principal focus in this paper is to study the soliton solutions of the time-dependent discrete nonlinear Schr¨odinger (DNLS) equation with the cubic nonlinearity

(

ιdtdφm,n = −φm,n+1− φm,n−1+ 4φm,n− φm+1,n− φm−1,n+ ν|φm,n|2φm,n,

m, n ∈ Z, (1.1) eq1.1

This work is partially supported by The National Center of Theoretical Sciences, of R.O.C. on

Taiwan.

Department of Mathematics, National Taiwan Normal University, Taipei 11677, Taiwan. E-mail:

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where ι = √−1. Equation (1.1) is a discretization of the nonlinear Schr¨odinger (NLS) equation

ι ∂

∂tφ = −∇

2φ + ν|φ|2φ,

where φ = φ(t, x), t ∈ R and x ∈ R2. The connection with the NLS equations is clearer from the alternative form of (1.1):

( ιd dtφm,n = − 1 h2(φm,n+1+ φm,n−1− 4φm,n+ φm+1,n+ φm−1,n) + ν|φm,n|2φm,n, m, n ∈ Z.

Systems of NLS equations arise in many fields of physics, including condensed matter, hydrodynamics, optics, plasmas, and Bose-Einstein condensates (BECs) (see e.g. [1, 4, 12, 23]). In the presence of strong periodic trapped potentials, a NLS equation can be approximated by a DNLS equation by using the “tight binding approximation” [24]. Equation (1.1) describes a large class of discrete nonlinear systems such as optical fibers [5, 6], small molecules such as benzene [7], and, more recently, dilute BECs trapped in a multiwell potential [2, 3, 25, 24].

The interplay between disorder and nonlinearity is a central topic of nonlinear science. This raises a number of mathematical questions related to the behavior of many physical systems. In the context of ultracold atomic gases, disorder may result from the roughness of a magnetic trap [9] or a magnetic microtrap [27]. This motivates us to study the spatial disorder of the soliton solutions of the DNLS equation (1.1). To obtain such soliton solutions, we set φm,n(t) = e−ιλtum,n, λ, um,n ∈ R, and transform

(1.1) into the time-independent discrete nonlinear Schr¨odinger equation

−um,n+1− um,n−1+ 4um,n− um+1,n− um−1,n+ νu3m,n = λum,n. (1.2) eq1.2

Here, (φm,n) is called a “soliton solution” if um,n → 0 exponentially as max{|m|, |n|} →

∞. By nature, discrete solitons represent self-trapped wavepackets in nonlinear peri-odic structures and result from the interplay between lattice diffraction (or dispersion) and material nonlinearity. Discrete solitons in one-dimensional lattices were first ex-perimentally observed in a nonlinear AlGaAs array by the groups of Silberberg and Aitchison [8]. In subsequent investigations, discrete soliton transport dynamics were studied by [18] in such arrays and it was observed in [21] that the nonlinearly induced escape from a waveguide defect. Optical discrete solitons in two-dimensional nonlinear waveguide arrays were first observed in biased photorefractive crystals by Segev’s and Christodoulides’s groups [11, 10]. Please refer to the survey article [17] for more details in the developments in the observation of discrete solitons.

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Arising from the abundance of physical experiments on discrete solitons, three rel-evant mathematical issues are proposed: (i) the existence of soliton solutions to (1.2), (ii) patterns of these soliton solutions and (iii) their complexity. To study the patterns of soliton solutions, the formulation of five-point difference in (1.2) enable us to study a more general form of the second order elliptic partial difference equation (PdE)

−αum,n+1− βum,n−1− um+1,n− γum−1,n+ f (um,n) = 0, (1.3) eq1.3

where f ∈ C1([a, b]) and α, β, γ ∈ R with γ 6= 0. We further assume the nonlinearity

in f in (1.3) satisfies the following:

(A1) Denote c1 < c2 < · · · < cN the turning points of f in the interval [a, b]. Let

c0 = a, cN +1 = b; δ1 and δ2, respectively, be the minimal and maximal value of

{αx + βy + γz| a ≤ x, y, z ≤ b}; and δ3 and δ4, respectively, be the minimal and

maximal value of {(αx + βy + z)/γ| a ≤ x, y, z ≤ b}. Assume there exist closed intervals Ii ⊂ [ci, ci+1], for i = 0, . . . , N , such that

f (Ii) ⊇ [a + δ1, b + δ2] and f (Ii)/γ ⊇ [a + δ3, b + δ4].

By f (Ii)/γ we mean the closed interval {f (u)/γ| u ∈ Ii}.

(A2) Let |γ0| = max{1, |γ|}. Assume

|f0(u)| ≥ |α| + |β| + √ 5 + 3 2 |γ 0|, for all u ∈ Ii, i = 0, . . . , N .

Our first theorem concerns the spatial disorder of PdE (1.3).

hthm1.1iTheorem 1.1. Suppose assumptions (A1) and (A2) hold. For any k = (k

m,n)m,n∈Z ∈

{0, . . . , N }Z2, there exists a unique solution (u

m,n) to PdE (1.3) such that

um,n ∈ Ikm,n, for all m, n ∈ Z.

We see in Theorem 1.1 that the strong amplitudes (Assumption (A1)) and stiffness (Assumption (A2)) of the nonlinearities in f lead the PdE (1.3) to the spatial disorder. The complexity of this disorder is determined by the oscillations (number of turning points) of the nonlinearities. More precisely, the spatial entropy of the PdE (1.3) equals to log(N + 1). By applying Theorem 1.1 to (1.2), we can prove our second theorem involving the spatial disorder and pattern of soliton solutions to the DNLS equation (1.2).

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hthm1.2iTheorem 1.2. Let ωdenote the largest value of real roots of 2ω + ˜∆ 12 = r 3ω ω − ˜∆ and 2ω − ˜∆ 12 = r 3ω ω + ˜∆,

where ˜∆ = (7 +√5)/2. Suppose ν < 0 and 4 − λ > ω∗. Then there exist disjoint closed intervals I0, I1 ⊂ R+, 0 ∈ I0, such that for every (km,n)m,n∈Z∈ {0, 1}Z

2

, there exists a unique nonnegative solution (um,n) to DNLS equation (1.2) satisfying um,n ∈ Ikm,n. In addition, if km,n = 0 for |m|, |n| > N0, some given positive integer, then

um,n = O(µmax{|m|,|n|}) as |m| or |n| are sufficiently large.

Here 0 < µ < (√5 − 1)/2 is a constant independent of the solution (um,n).

The solutions in the second assertion of Theorem 1.2 are referred to as the so-called “bright solitons”. Here both the existence and the variety of solutions to DNLS equation (1.2) are presented. Specifically, the state at the (m, n)-th site, um,n, can

be either dark (um,n ∈ I0) or bright (um,n ∈ I1) that depends on the configuration

km,n = 0 or 1, respectively. Considering only the soliton solutions, the DNLS equation

also exhibits the spatial disorder. For the DNLS equations in the one-dimensional lattice (i.e., the case α = β = 0 and γ = 1 in (1.3)), soliton solutions were studied in [22] by the construction of homoclinic/hetronic orbits. In [26], the spatial disorder in the one-dimensional NLS equations equipped with periodic/quasiperiodic trapped potentials was studied, in which a coherent structure ansatz was applied to reduce the NLS equation to a forced Duffing equation. In [20, 13], the soliton solutions of DNLS equations in a two-dimensional lattice was studied in the case |ν|  1 and λ/ν = O(1) by variational techniques. Our result in Theorem 1.2 is valid for λ and µ = O(1). The chaotic behavior of DNLS equations in one-dimensional lattice as well as its synchronization phenomena were studied by [19]. Bifurcation analysis of DNLS equations for the ground state solutions was studied by [15]. Recently, it was reported by [16] the occurrence of the phase separation for the ground state solutions of the DNLS equation in lattices with a general connection topology.

This paper is organized as follows. In Section 2, we prove Theorem 1.1 by the construction of a horseshoe in l∞ for the map F introduced in (2.1). We follow the

standard process for planner maps and generalize it to an infinite dimensional case. In Section 3, we prove Theorem 1.2 by the using of Theorem 1.1.

Throughout this paper, we denote l∞= {u = (. . . , u−1, u0, u1, . . .)| supn|uu| < ∞}.

For any finite set {0, . . . , N }, we denote {0, . . . , N }Z = {k = (k

n)n∈Z| kn∈ {0, . . . , N }}

and {0, . . . , N }Z2 = {k = (k

m,n)m,n∈Z2| km,n ∈ {0, . . . , N }}. We use the boldface alphabets (or symbols) to denote operators (or vectors). We say u ≤ v if un ≤ vn for

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all n ∈ Z. We use k · k = k · k∞ to denote the infinity norm of an operator or a vector.

Note that for any bounded operator A on l∞, the infinity norm of A can be computed

by kAk = supkuk=1kAuk = supm∈ZP

n∈Z|amn|.

2

Construction of horseshoe and its hyperbolicity

First, we define the map F : l∞× l∞ → l∞× l∞ by

heq2.1i F : ( ¯ u = g(u) − γv, ¯ v = u, (2.1a) eq2.1a where g : l∞ → l∞ is given by

gn(u) = −αun+1+ f (un) − βun−1, n ∈ Z. (2.1b)?eq2.1b?

Considering a bounded solution (um,n)m,n∈Zof (1.3), let u(m)= (. . . , um,−1, um,0, um,1, . . .)

for all m ∈ Z, that is, u(m) is the m-th row of (um,n)m,n∈Z. Hence, we see that

(u(m+1), u(m)) = F(u(m), u(m−1)). This means that (u

m,n)m,n∈Z forms an orbit of F

in the m-direction. In this section, we shall construct a horseshoe for F and prove the hyperbolicity of its invariant Cantor set. To this end, we begin with some ba-sic settings for this construction adopted from [28]. Let B ⊂ l∞ denote the box

B = {u ∈ l∞| a ≤ un≤ b, for all n ∈ Z}.

hdef2.1iDefinition 2.1. Let µ be a real nonnegative number. A µ-horizontal surface in B × B

is the graph of a differentiable function v = r(u), u ∈ B, satisfying kDr(u)k ≤ µ. A µ-horizontal strip in B × B is the set

H = {(u, v)|r1(u) ≤ v ≤ r2(u), u ∈ B},

where r1 < r2 are µ-horizontal surfaces. Similarly, a µ-vertical surface in B × B is

the graph of a differentiable function u = s(v), v ∈ B, satisfying kDs(v)k ≤ µ. A µ-vertical strip in B × B is the set

V = {(u, v)|s1(v) ≤ u ≤ s2(v), v ∈ B},

where s1 < s2 are µ-horizontal surfaces. The widths of the horizontal and the vertical

strips are defined, respectively, as d(H) = sup

u∈B

||r1(u) − r2(u)|| , d(V) = sup v∈B

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Let E = {0, . . . , N }Z. For a given k ∈ E, let

Bk= {u ∈ B| un∈ Ikn, n ∈ Z}.

Here N is the number of turning points of f and Ikn are the closed intervals given in (A1). We define the horizontal and vertical strips

Hk= B × Bk= {(u, v) ∈ (B × B)| v ∈ Bk},

Vk= Bk× B = {(u, v) ∈ (B × B)| u ∈ Bk}.

Now we are ready to construct a horseshoe for F. Before giving any proof, we note from (2.1) that the inverse of F is given by

F−1 : (

u = ¯v,

v = (g(¯v) − ¯u)/γ.

From Theorem 2.2 to Lemma 2.5, each result is associated with a horizontal and a vertical case. Due to the symmetry of F and F−1, we shall only give the proofs of the horizontal cases by using the map F. The proofs for the vertical cases can be similarly verified by using F−1.

hprop2.1iProposition 2.1. Let A be a bounded operator on l

∞. Suppose A is diagonal

dom-inant, i.e., there exists  > 0 such that |amm| > P ∞

n=−∞,n6=m|amn| +  for all m ∈ Z.

Then A is invertible. In addition, if D is an invertible diagonal bounded operator on l∞ such that kD−1(A − D)k < 1, then

kA−1k ≤ kD

−1k

1 − kD−1(A − D)k. (2.2) eq2.2

Proof. Suppose Au = 0 for some u 6= 0. Then we have ammum =

P∞

n=−∞,n6=mamnun

for all m ∈ Z. Taking absolute on both sides of the equation and applying the triangular inequality to the resulting equation, we obtain |amm||um| ≤ P

n=−∞,n6=m|amn|kuk <

(|amm| − )kuk. This is a contradiction since kuk = supn∈Z|un|. The proof of the

first assertion is complete. For the second assertion, note that A−1 = k(I + D−1(A − D))−1D−1k ≤ k(I+D−1(A−D))−1kkD−1k. On the other hand, since kD−1(A−D)k <

1, we have k(I+D−1(A−D))−1k ≤P∞

n=0kD

−1(A−D))−1kn = 1/(1−kD−1(A−D)k).

This gives (2.2).

hthm2.2iTheorem 2.2. Let ∆ = min

1≤i≤N{|f0(u)|| u ∈ Ii} − (|α| + |β| + |γ0|) where |γ0| is

defined in (A2). Suppose µ is a constant satisfying |γ0|

∆ < µ < √

5 − 1

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Let k ∈ E be given. If S is a µ-horizontal surface, then F(S ∩ Vk) ∩ (B × B) is a

µ-horizontal surface contained in Hk. If S is a µ-vertical surface, then F−1(S ∩ Hk) ∩

(B × B) is a µ-vertical surface contained in Vk.

Here we remark from (A2) that ∆ ≥ (√5 + 1)|γ0|/2. Hence the constant µ in Theorem 2.2 is well defined. Moreover, µ is between 0 and 1.

Proof. Suppose S is a µ-horizontal surface. Denote ¯S = F(S ∩ Vk) ∩ (B × B). To show

that ¯S is a µ-horizontal surface in Hk, we give the proof in 3 steps:

1. If (¯u, ¯v) = F(u, v) ∈ ¯S, then we have ¯v ∈ Bk. This means (¯u, ¯v) ∈ Hk.

2. For every (¯u, ¯v) = F(u, v) ∈ ¯S, ¯v is locally a differentiable function of ¯u, say ¯

v = ¯r(¯u). Moreover, kD¯r(¯u)k < µ. 3. The function ¯r is well defined on B. Now, we begin with the proof of Step 1. Step 1.

Since F is injective and (¯u, ¯v) ∈ ¯S, we have (u, v) ∈ S ∩ Vk. Thus, u ∈ Bk. Then

the assertion follows from (2.1a) directly. Step 2.

Since S is a µ-horizontal surface, from Definition 2.1 we may assume S to be the graph of v = r(u) with kDrk ≤ µ. It follows from (2.1) that each (¯u, ¯v) ∈ ¯S satisfies

¯

u = g(¯v) − γr(¯v). Taking the derivative of ¯u with respect to ¯v, we obtain ∂ ¯u

∂ ¯v = Dg(¯v) − γDr(¯v).

Let A = ∂ ¯u/∂ ¯v and D be the diagonal part of Dg(¯v), that is, D = diag(. . . , f0(¯v−1), f0(¯v0), f0(¯v1), . . .).

Since ¯v ∈ Bk, we have |f0(¯vn)| ≥ ∆ + |α| + |β| + |γ0|, and hence,

kD−1k ≤ 1

∆ + |α| + |β| + |γ0|. (2.3) eq2.3

By (A2) and the fact that kDr(¯v)k ≤ µ < 1, we see that A is diagonal dominant. From Proposition 2.1, it implies A is invertible. Hence, by the inverse function theorem, ¯v is locally a differentiable function of ¯u, say ¯v = ¯r(¯u). Now we show that kD¯r(¯u)k < µ.

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Note that kD−1(A − D)k < 1. By Proposition 2.1 and (2.3), we have kA−1k ≤ kD −1k 1 − kD−1k · k(A − D)k ≤ 1 ∆+|α|+|β|+|γ0| 1 −∆+|α|+|β|+|γ1 0|(|α| + |β| + |γ|µ) = 1 ∆ + (|γ0| − µ|γ|) < 1 ∆ < µ. (2.4)?eq2.4? Step 3.

We need to show that for any ¯u ∈ B, there exists a unique (u, v) ∈ S ∩ Vk such that

the u-component of F(u, v) is equal to ¯u. To this end, we define the map T : B → l∞

by

Tn(u) = gn(u) − γrn(u), n ∈ Z. (2.5) eq2.5

If (u, v) ∈ S ∩ Vk, then from (2.1a) and (2.5) we have F(u, v) = (T(u), u). Hence, we

need to show that for any ¯u ∈ B there exists a unique u ∈ Bksuch that ¯u = T(u). We

first show that the map T is injective in Bk. Suppose T(u) = T(u0) for some u and

u0 ∈ Bk. Since Bk is convex, by the mean value theorem, there exist u(n), n ∈ Z, in

the line segment connecting u and u0 such that

∇Tn(u(n)) · (u − u0) = 0, for n ∈ Z. (2.6) eq2.6

We observe that ∇Tn(u(n)) = q(n)− γ∇rn(u(n)), where q(n) = ∇gn(u(n)) ∈ l∞ is the

row vector given by

qm(n)=          −β, m = n − 1, f0(u(n)n ), m = n, −α, m = n + 1, 0, otherwise. (2.7) eq2.7

Let G = (gnm) be the bounded operator on l∞ whose n-th raw is equal to ∇Tn(u(n)).

Hence, (2.6) can be rewritten as the form

G(u − u0) = 0. (2.8) eq2.8

Since u(n) ∈ Bk, from assumption (A2), (2.7) and the fact that k∇rn(u(n))k ≤ kDr(u(n))k ≤

µ < 1, for all n ∈ Z, we have gnn−

X

m∈Z,m6=n

|gnm| ≥ |f0(u(n)n )| − |α| − |β| − |γ|k∇hn(u(n))k

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This shows G is diagonal dominant, and hence, is invertible by Proposition 2.1. From (2.8) we conclude that u = u0, and hence, T is injective in Bk.

Next, we show that T(Bk) ⊃ B. Let Ii = [di, ei], i = 0, . . . , N . The boundary of Bk

consists of the hyperplanes in l∞:

Dkn := {u ∈ Bk|un = dkn} , Ekn := {u ∈ Bk|un= ekn},

for n ∈ Z. Since T is injective in Bk, it maps the boundary of Bk to the boundary of

T(Bk). Let n ∈ Z be given. From assumption (A1), f is monotonic on Ikn. Without loss of generality, let f be increasing. For any u ∈ Dkn, using the result in Step 1 that a < rn(u) < b, we calculate that

Tn(u) = f (dkn) − αun+1− βun−1− γhn(u) ≤ f (dkn) − min

x,y,z∈[a,b]αx + βy + γz

≤ a (2.9) eq2.9

The last inequality in (2.9) follows from assumption (A1). A similar argument yields Tn(u) ≥ b for u ∈ Ekn. Hence B ⊂ T(Vk). This completes the proof.

From Theorem 2.2, we see that F(Vk) ∩ (B × B) ⊂ Hkand F−1(Hk) ∩ (B × B) ⊂ Vk

form a µ-horizontal strip and a µ-vertical strip, respectively. Let H∗k = F(Vk) ∩ (B × B) , Vk∗ = F

−1

(Hk) ∩ (B × B). (2.10) eq2.10

Thus the resulting surfaces in Theorem 2.2, F(S∩Vk)∩(B×B) and F−1(S∩Hk)∩(B×B),

can be accordingly rewritten as F(S) ∩ H∗k and F(S) ∩ Vk∗, respectively. We have the following immediate consequence of Theorem 2.2.

hcor2.3iCorollary 2.3. Let µ be the constant given in Theorem 2.2 and k ∈ E be given. If H

is a µ-horizontal strip, then F(H) ∩ H∗k is also a µ-horizontal strip. If V is a µ-vertical strip, then F−1(V) ∩ Vk∗ is also a µ-vertical strip.

In Corollary 2.3, we see that F (resp., F−1) maps a horizontal strip (resp., µ-vertical strip) to an uncountable number of µ-horizontal strips (resp., µ-µ-vertical strips), in which exactly one strip is included in H∗k (resp., Vk) for each k ∈ E. In the next theorem, we will see that every strip becomes thinner under the mapping by a factor less than 1.

hthm2.4i

Theorem 2.4. Let µ be be the constant given in Theorem 2.2 and k ∈ E be given. Suppose H is a µ-horizontal strip and V is a µ-vertical strip. If ¯H = F(H) ∩ H∗

k and ˜ V = F−1(V) ∩ V∗ k, then d( ¯H) ≤ µ 1 − µ2d(H) , d( ˜V) ≤ µ 1 − µ2d(V).

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Here we remark that the factor µ/(1 − µ2) < 1 by the assumption that µ < (√5 − 1)/2. Before proving Theorem 2.4, we first prove the following lemma.

hlem2.5iLemma 2.5. Let µ be be the constant given in Theorem 2.2 and k ∈ E be given.

Suppose (u, v) is a point in B × B, ζ =  ξ η  ∈ l∞× l∞, ¯ζ =  ¯ ξ ¯ η  := DF(u, v)ζ, and ˜ζ =  ˜ ξ ˜ η  := DF−1(u, v)ζ.

(a) Suppose (u, v) ∈ Vk and kηk ≤ µkξk. If {nj}∞j=0 ⊂ Z is any sequence such that

|ξnj| → kξk as j → ∞, then there exists N0 > 0, independent of the choice of (u, v), such that

µ| ¯ξnj| > kζk for j > N0.

(b) Suppose (u, v) ∈ Hk and kξk ≤ µkηk. If {nj}∞j=0 ⊂ Z is any sequence such that

|ηnj| → kηk as j → ∞, then there exists N0 > 0, independent of the choice of (u, v), such that

µ|˜ηnj| > kζk for j > N0.

Proof. Suppose (u, v) ∈ Vk and kηk ≤ µkξk. Using (2.1), we calculate that

DF(u, v) = Dg(u) −γI

I 0

 . For each nj, we have

| ¯ξnj| = |(Dg(u)ξ)nj − γηnj|

= |f0(unj)ξnj − αξnj+1− βξnj−1− γηnj|. (2.11) eq2.11 Since kξk ≥ |ξm| and kηk ≥ |ηm| for all m ∈ Z, using the assumption that kηk ≤ µkξk

follows |ηnj| ≤ µkξk. Thus, equation (2.11) becomes | ¯ξnj| ≥ |f 0 (unj)||ξnj| − (|α| + |β| + |γ|µ)kξk = ∆kξk +  |f0(unj)||ξnj| − min u∈∪Ii |f0(u)|kξk  + (|γ0| − |γ|µ)kξk. (2.12) eq2.12

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Here ∆ is the constant given in Theorem 2.2. Since |f0(unj)| ≥ minu∈∪Ii|f

0(u)| and

|ξnj| → kξk as j → ∞, there exists N0 > 0, independent of (u, v), such that |f0(unj)||ξnj| − min

u∈∪Ii

|f0(u)|kξk ≥ −(|γ0| − |γ|µ)kξk, (2.13) eq2.13

for j > N0. Substituting (2.13) into (2.12), we see that | ¯ξnj| ≥ ∆kξk for j > N0. Since kζk = kξk and ∆−1 < µ, this completes the proof.

Later we will see that Lemma 2.5 plays an important role in the proof of the hyperbolicity of F. Now we are in position to prove Theorem 2.4.

Proof of Theorem 2.4. Since ¯H is a µ-horizontal strip, we may assume ¯p0 = (¯u∗, ¯v0)

and ¯p1 = (¯u∗, ¯v1) be the points on the boundary of ¯H such that d( ¯H) = k¯p0 − ¯p1k.

We define the sets ¯

Γ = {¯p(t) = (¯u(t), ¯v(t)) := t¯p1+ (1 − t)¯p0| t ∈ [0, 1]},

and

¯

S = {(¯u, ¯v) ∈ B × B| ¯u = ¯u∗}. Here, ¯u(t) and ¯v(t) are given by

¯

u(t) = ¯u∗ and ¯v(t) = t¯v1+ (1 − t)¯v0. (2.14) eq2.14

We note that (i) ¯Γ is the straight line that connects ¯p1 and ¯p0 and (ii) ¯Γ is contained

in ¯S ∩ ¯H. Now, let p(t) = (u(t), v(t)) = F−1(¯p(t)). Hence, we have d dtp(t) = DF −1 (¯p(t))d dtp(t).¯

From (2.14) we see that kd¯u(t)/dtk = 0 ≤ µkd¯v(t)/dtk and d¯v(t)/dt = ¯v1 − ¯v0 ∈

l∞ is independent of t. Hence, we may assume {nj}∞j=1 ⊂ Z be a sequence such

that limj→∞|d¯vnj(t)/dt| = kd¯v(t)/dtk for all 0 ≤ t ≤ 1. Applying assertion (b) of Lemma 2.5 to ζ = d¯p(t)/dt and ˜ζ = dp(t)/dt, we conclude that there exists N0 > 0,

independent of t, such that µ d dtvnj(t) ≥ d¯p(t) dt > 0, (2.15) eq2.15

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for all j > N0 and t ∈ [0, 1]. However, from (2.14) again we see that kd¯p(t)/dtk = d( ¯H).

The first inequality in equation (2.15) thus becomes d( ¯H) ≤ µ d dtvnj(t)

for all t ∈ [0, 1]. (2.16) eq2.16

From the second inequality in (2.15), we see that vnj(t) is monotonic for j sufficiently large. Integrating both sides of (2.16) from t = 0 to t = 1, we obtain

d( ¯H) ≤ µ Z 1 0 d dtvnj(t) dt = µ|vnj(1) − vnj(0)| ≤ µkv(1) − v(0)k. (2.17) eq2.17

Next, we shall estimate an upper bound for (2.17). Since H is a µ-horizontal strip, there exist µ-horizontal surfaces S0 := {(u, v) ∈ B × B|v = r0(u)} and S1 := {(u, v) ∈

B × B|v = r1(u)} contained in H such that p(0) ∈ S0 and p(1) ∈ S1. Therefore,

heq2.18i

kv(1) − v(0)k = kr1(u(1)) − r0(u(0))k

≤ kr1(u(1)) − r1(u(0))k + kr1(u(0)) − r0(u(0))k

≤ µku(1) − u(0)k + d(H). (2.18a) eq2.18a

The last inequality in (2.18a) follows from the mean value theorem. On the other hand, from Theorem 2.2, it follows that S := F−1( ¯S ∩ Hk) ∩ (B × B) is a µ-vertical surface.

Hence, we assume that S is the graph of u = s(v) with kDsk ≤ µ. However, p(0) and p(1) ∈ S, from the mean value theorem again we have

ku(1) − u(0)k = ks(v(1)) − s(v(0))k ≤ µkv(1) − v(0)k. (2.18b) eq2.18b

From (2.18a) and (2.18b), we conclude that kv(1) − v(0)k ≤ 1

1 − µ2d(H). (2.19) eq2.19

Substituting (2.19) into (2.18), we see that d( ¯H) ≤ µ

1 − µ2d(H).

This completes the proof of the horizontal case. Now, we are ready to prove Theorem 1.1.

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Proof of Theorem 1.1. Let µ be the constant defined in Theorem 2.2. Define Λ−1 = [ k−1∈E H∗k−1 , Λ0 = [ k0∈E Vk0, where H∗k−1 and Vk

0 are defined in (2.10). Note that B × B is not only a µ-horizontal strip, but also a µ-vertical strip. This implies that each H∗k−1 and Vk0 are, respectively, a µ-horizontal strip and a µ-vertical strip. Let

Λ−n−1= Λ−1∩ F(Λ−1) ∩ · · · ∩ Fn(Λ−1),

Λn= Λ0∩ F−1(Λ0) ∩ · · · ∩ F−n(Λ0).

Hence, we may set

Λ−n−1 = [ k−j ∈E j=1,...,n+1 H∗k−1,k−2,...,k−n−1 and Λn = [ kj ∈E j=0,...,n Vk∗ 0,k1,...,kn, where H∗k−1,...,k−n−1 = {(u, v) ∈ B × B| F−j(u, v) ∈ Hk−j−1, j = 0, . . . , n} and Vk0,k1,...,kn = {(u, v) ∈ B × B| Fj(u, v) ∈ Vkj, j = 0, . . . , n}. Note that H∗k−1,...,k−n−1 = H∗k−1∩ F (H∗ k−2,...,k−n−1) and V ∗ k0,k1,...,kn = V ∗ k0∩ F −1(V∗ k1,...,kn). By Corollary 2.3, an inductive argument shows that each Hk

−1,...,k−n−1 and V

k0,k1,...,kn are, respectively, a µ-horizontal and a µ-vertical strip. In addition, it follows from Theorem 2.4 that d H∗k−1,k−2,...,k−n−1 ≤  µ 1 − µ2 n d(B) , d Vk0,k1,...,kn ≤  µ 1 − µ2 n d(B). Hence, for any sequences (k−1, k−2. . .) and (k0, k1, . . .) ∈ EN,

∞ \ n=1 H∗k−1,k−2,...,k−n , ∞ \ n=0 Vk0,k1,...,kn (2.20) eq2.20

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are decreasing to two surfaces, say H∗k−1,k−2,... = {v = r(u)} and Vk0,k1,... = {u = s(v)}. Here we note that r and s may not be differentiable. However, the uniform convergency of the upper and lower surfaces in (2.20) implies they satisfy a Lipschitz condition with Lipschitz constant µ; i.e., for any u1, u2, v1, v2 ∈ B,

kr(u1) − r(u2)k ≤ µku1− u2k, ks(v1) − s(v2)k ≤ µkv1− v2k.

Since |µ| < 1, by the contraction mapping theorem, the equation (

v = r(u) u = s(v)

has a unique solution in B × B. This means H∗k−1,k−2,... = {v = r(u)} and Vk

0,k1,... = {u = s(v)} have a unique intersection. Hence, the invariant set Λ = Λ−∞∩ Λ∞ is a

Cantor set. Denote Σ the symbolic space Σ = {(. . . , k−1|k0, k1, . . .)| kn ∈ E, n ∈ Z}.

To see F|Λ is topological conjugate to the full shift σ on Σ, we define the function

φ(p) = (. . . , k−1|k0, k1, . . .),

where p = Hk−1,k−2,...∩ Vk0,k1,.... It is easy to verify that φ is a homeomorphism from Λ to Σ. We only need to show that φ(F(p)) = σ(φ(p)). From the construction of V∗

k0,k1,..., we have

F(Vk0,k1,...) = Vk1,k2,.... (2.21) eq2.21

On the other hand, p ∈ H∗k−1,k−2,...∩ V∗ k0 ⊂ H

k−1,k−2,... ∩ Vk0. From Theorem 2.2, it implies F(p) ∈ H∗k0,k−1,k−2,.... Together with (2.21), this shows

φ(F(p)) = φ(Hk

0,k−1,k−2,...∩ V

k1,k2,...) = (. . . , k−1, k0|k1, . . .) = σ(φ(p)).

Before proving the hyperbolicity of Λ, we shall adopt the following theorem in [14, p. 266].

hthm2.6iTheorem 2.6. A compact F-invariant set Λ is hyperbolic if there exists κ > 1 such

that for every p ∈ Λ there is a decomposition TpM = Sp ⊕ Tp (in general, not DF

invariant), a family of the horizontal cones Hp ⊃ Sp, and a family of vertical cones

Vp⊃ Tp associated with the decomposition such that

DF(p)Hp ⊂ Int HF(p), DF−1(p)Vp ⊂ Int VF(p), (2.22) eq2.22

and

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Proof of Theorem 1.1: The hyperbolicity of Λ. We shall prove the hyperbolicity by ver-ifying the conditions in Theorem 2.6. First, let

Sp= {  0 η  ∈ l∞× l∞| η ∈ l∞}, Tp = {  ξ 0  ∈ l∞× l∞| ξ ∈ l∞}, and Hp= {  ξ η  ∈ l∞× l∞| kηk ≤ µkξk}, Vp = {  ξ η  ∈ l∞× l∞| kξk ≤ µkηk}.

Here we see that Sp ⊂ Hp and Tp ⊂ Vp. Now, let p = (u, v) ∈ Λ and ζ =

 ξ η

 ∈ Sp

be given. Hence, p ∈ Vk for some k ∈ E. Moreover, there exists a µ-horizontal

surface S = {v = r(u)} containing p such that ζ is a tangent vector to S at p, i.e. η = Dr(u)ξ. Since F(p) ∈ Λ ⊂ B × B, it follows from Theorem 2.2 that the connected component of F(S) ∩ (B × B) containing F(p), denoted by ¯S, is also a µ-horizontal surface. Suppose ¯S is the graph of ¯v = ¯r(¯u). Consequently, ¯ζ =

 ¯ ξ ¯ η  := DF(p)ζ is a tangent vector to ¯S at F(p). Hence ¯η = D¯r(¯u)¯ξ. From the result of Step 2 in the proof of Theorem 2.2, we conclude that

k¯ηk = kD¯r(¯u)¯ξk < µk¯ξk.

This proves the first invariance condition in (2.22). The second can be similarly ob-tained. Letting κ = 1/µ, the Contraction and Expansion condition (2.23) follows from Lemma 2.5 directly. This completes the proof.

3

Application to the cubic nonlinearities

In this section, we shall give the proof of Theorem 1.2.

Proof of Theorem 1.2. Let ω = 4 − λ and define f (u) = ωu + νu3. To prove the first

assertion of Theorem 1.2, it suffices to apply Theorem 1.1 by verifying assumptions (A1) and (A2) for f . Now let a = 0 and b = p−ω/ν. Here a and b are zeros of f in R+. Denote u1 =

q

( ˜∆ − ω)/3ν, u2 =

q

−( ˜∆ + ω)/3ν and  > 0 sufficiently small. Hence, the constants δ1 and δ2 are given by

δ1 = 0 and δ2 = 3b.

Let I0 = [a, u1 − ] and I1 = [u2+ , b]. Since α = β = γ = 1, it is easy to verify that

f0(u1) = ˜∆ = (

5 + 3)|γ0|/2 + (|α| + |β|) and f0(u

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u ∈ I0 and f0(u) < − ˜∆ for u ∈ I1 because u1 < c < u2 and f0 is quadratical, where c

denotes the positive turning point of f . Hence, (A2) is satisfied. To prove the validity of (A1), we first show that

f (I0) ⊇ [a + δ1, b + δ2]. (3.1) eq3.1

It is easily seen that f (a) ≤ a + δ1. Since f is monotonic on I0, if we can show that

f (u1) > b + δ2, (3.2) eq3.2

then (3.1) holds true. A calculation leads to that (3.2) is equivalent to the inequality 2ω + ˜∆ > 12

q

3ω/(ω − ˜∆). This is true due to the assumption of this theorem. The conclusion f (I1) ⊇ [a + δ1, b + δ2] can also be shown by a similar argument. Hence, by

Theorem 1.1 the first assertion is proven.

Now, we show the second assertion of Theorem 1.2. Let (km,n) ∈ {0, 1}Z

2

satisfying km,n = 0 for |m|, |n| > N . From the first assertion, there exists a solution (um,n) to

DNLS equation (1.2) such that um,n ∈ Ikm,n for all m, n ∈ Z. Let F be the map given in (2.1) with α = β = γ = 1. Let u(m), u0(m) ∈ l ∞ and k(m), k0(m) ∈ {0, 1}Z be given by u(m) = the m-th row of (um,n), u0(n) = the n-th column of (um,n), k(m) = the m-th row of (km,n),

u0(n) = the u-th column of (um,n).

Using (1.2), we see that (u(m+1), u(m)) = F(u(m), u(m−1)) for all m ∈ Z. Since k(m)

0 as |m| → ∞, the topological conjugacy between F and the full shift σ leads to (u(m), u(m−1)) → (0, 0) as m → ∞. However, F is hyperbolic. We conclude that

ku(m)k = O(µ|m|

) for |m| sufficiently large. (3.3) eq3.3

A similar argument shows that

ku0(n)k = O(µ|n|) for |n| sufficiently large. (3.4) eq3.4

Combination of (3.3) and (3.4) leads to the assertion of this theorem.

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