Uniform decay rates of the solutions of a nonlinear lattice

Nonlinear Analysis 54 (2003) 261 – 278

www.elsevier.com/locate/na

Uniform decay rates of the solutions of a nonlinear lattice G. Perla Menzalaa;∗ , V.V. Konotopb a National

Laboratory of Scientic Computation, LNCC/MCT Rua Getulio Vargas 333, Petropolis, RJ, CEP 25651 070, Brazil b Centro de F,-sica de Mat, eria Condensada, Universidade de Lisboa, Complexo Interdisciplinar, Av. Professor Gama Pinto 2, 1649-003 Lisbon, Portugal Received 17 September 2002; accepted 30 October 2002

Abstract We consider a family of 2nite nonlinear Klein–Gordon lattices subject to cyclic boundary conditions under the e4ect of a dissipative mechanism. We show that the model is globally well posed in a natural Banach space and our main result says that the total energy associated with the model decays exponentially fast when t → + ∞. c 2003 Published by Elsevier Science Ltd. 

1. Introduction Nonlinear di4erential-di4erence equations, also called lattices or networks, have received a lot of attention in the last 20 years or so. The main tools used in the theory for obtaining (localized) solutions, which are of the most importance for physical applications are, however, either approximate or numerical (see e.g [4,10]), unless the lattice is integrable (see e.g. [1,9]) or a solution belongs to a special class (say, breather, i.e. localized in space and periodic in time) when proofs of existence are available [2,5]. In a generic situation neither approximate nor numerical methods can provide adequate description of the asymptotic behavior of a lattice since their accuracy is limited to a 2nite temporal interval. On the other hand, analogy between networks and nonlinear partial di4erential equations is also limited. For example, a solution of a nonlinear Klein–Gordon lattice with high nonlinearity is proven to exist globally [8], while ∗

Corresponding author. E-mail addresses: [email protected] (G.P. Menzala), [email protected] (V.V. Konotop).

c 2003 Published by Elsevier Science Ltd. 0362-546X/03/$ - see front matter
doi:10.1016/S0362-546X(03)00061-0

262

G.P. Menzala, V.V. Konotop / Nonlinear Analysis 54 (2003) 261 – 278

a solution of its continuum counterpart, the generalized Korteweg–de Vries equation, displays blow-up [3]. In the present work, we consider asymptotic properties of a nonlinear Klein–Gordon lattice under the presence of an internal damping mechanism. Consequently, the total energy associated with the model is a nonincreasing function of time. We will prove that the energy decays to zero at a uniform rate as time approaches in2nity. The lattice we consider consists of N particles and in the absence of dissipation is described by the Hamiltonian H = Hl + Hnl ;

(1.1)

where Hl is the quadratic part resulting in linear terms of the equation of motion and is given by Hl =

N N K2
M
(u˙ n )2 + (un+1 − un )2 ; 2 2

(1.2)

n=1

n=1

where un = un (t) denotes a displacement of the nth particle, u˙ n = dun =dt, and Hnl is the nonlinear part given by N

Hnl =

Kp+1
(un+1 − un )p+1 : p+1

(1.3)

n=1

In (1.2) and (1.3), M ¿ 0 is the mass of a particle, K2 and Kp+1 are both positive, denote linear and nonlinear force constants, and p is a positive integer which in our main result (global existence and stability) will be odd and ¿ 3. This paper is organized as follows: In Section 2 for the sake of completeness, we brieLy describe the steps to prove global well-posedness of the model we are considering. In Section 3 we prove the result on exponential stability and in Section 4, we comment about some other important lattices and damping mechanisms for which the techniques we describe in Sections 2 and 3 can apply as well with suitable modi2cations. 2. Global well-posedness The evolution equation for lattice (1.1)–(1.3) in the presence of internal dissipation is given by M uM n − K2 (un+1 + un−1 − 2un ) − Kp+1 [(un − un+1 )p + (un − un−1 )p ] = 1 (u˙ n+1 + u˙ n−1 − 2u˙ n ); where 1 is a positive constant. After renormalization motion of the lattice is written as



(2.1) K2 =M t → t, the equation of

uM n − Nun = F(un ; u˙ n );

(2.2)

F(un ; u˙ n ) = p [(un − un−1 )p + (un − un+1 )p ] + Nu˙ n

(2.3)

where

G.P. Menzala, V.V. Konotop / Nonlinear Analysis 54 (2003) 261 – 278

263

and p =

Kp+1 ¿ 0; K2

=

1 ¿0 K2

and

Nfn ≡ fn+1 + fn−1 − 2fn :

Let N be a 2xed (positive) integer. We consider lattice (2.2) satisfying the cyclic boundary conditions (now n can be any integer)  ∀t ∈ R+ ; un (t) = un+N (t); (2.4) un (0) = an ; u˙ n (0) = bn ; n = 1; 2; 3; : : : ; N; where an ; bn are given real numbers. We want to prove the global well-posedness of problem (2.2) satisfying conditions (2.4). From now on, we will assume the following conditions on the an ’s and bn ’s: (H1) initially, i.e. at t = 0, the velocity of the lattice center of mass is zero, that is N


bn = 0

(2.5)

n=1

and (H2) initially the shift of the lattice center of mass due to deformation is zero, i.e. N


an = 0:

(2.6)

n=1

First, we consider the solution of the unperturbed model  vMn − Nvn = 0;    vn (0) = an ; v˙n (0) = bn ;    vn (t) = vn+N (t); ∀t ¿ 0;

(2.7)

where an and bn are given real numbers, satisfying (2.5) and (2.6). It is well known that the solution of (2.7) can be written explicitly as vn (t) =

N


˙ − m; t)am + G(n − m; t)bm }; {G(n

(2.8)

m=1

where G(n; t) is the lattice Green function  N kn 1
sin(!k t) : cos 2 G(n; t) = N !k N

(2.9)

k=1

!k is given by



k !k = 2 sin N



and represents the dispersion relation. Useful properties of the solutions of (2.7) are given in the following lemma.

(2.10)

264

G.P. Menzala, V.V. Konotop / Nonlinear Analysis 54 (2003) 261 – 278

Lemma 2.1. Assume that conditions (2.5) and (2.6) are valid and let vn (t) be the solution of (2.7). Then N
vn (t) = 0 ∀t ¿ 0; (a) I (t) = n=1 N


(b)

vn2

6 4N

N


N


(vn+1 − vn )2

∀t ¿ 0;

n=1

n=1

(c)

2

(v˙n )2 6 4N 2

N


(v˙n+1 − v˙n )2

∀t ¿ 0:

n=1

n=1

N Proof. Let I (t) = n=1 vn (t). We calculate the second-order derivative of I (t) and use Eq. (2.7) to obtain N N

vMn (t) = (vn+1 + vn−1 − 2vn ) IM(t) = n=1

=

N


n=1

(vn+1 − vn ) +

n=1

N


(vn−1 − vn ) = (vN +1 − v1 ) + (v0 − vN ) = 0

n=1

because v1 = v1+N and v0 = vN . Therefore, I (t) = c1 + c2 t ∀t ¿ 0 where c1 = I (0) and c2 = I˙(0). Since (2.5) and (2.6) are valid then c1 = c2 = 0 which proves (a). Now, we will prove (b) for each (arbitrary) time t: Let k; m ∈ {1; 2; : : : ; N }. If k ¡ m, we write m−1
v m − vk = (vn+1 − vn ): n=k

Then, using Cauchy–Schwarz’s inequality we obtain m−1 

1=2 N 
 √
  (vn+1 − vn ) 6 N (vn+1 − vn )2 : vm − vk 6 |vm − vk | =    n=1

n=k

Similarly, if m ¡ k, then vm − vk = −(vk − vm ) 6 |vk − vm | 6

√

N

N


1=2 (vn+1 − vn )2

:

n=1

If at time t, all vn ’s are equal to zero, then, conclusion (b) trivially holds. If at time t not all vn ’s are equal to zero, then we pick the 2rst one of {v1 (t); : : : ; vN (t)} which is di4erent from zero, say vn1 (t). Suppose it is less than zero, vn1 (t) ¡ 0 (otherwise the

N arguments below can be applied to −vn (t)). Since n=1 vn (t) = 0, then there exists at least one element vn2 (t) such that vn2 (t) ¿ 0. Choosing m = n2 and k = n1 in our above discussion we obtain that

1=2 N
√ 2 (vn+1 − vn ) −vn1 (t) 6 vn2 (t) − vn1 (t) 6 N n=1

G.P. Menzala, V.V. Konotop / Nonlinear Analysis 54 (2003) 261 – 278

or vn1 (t) ¿ −

√

N

N


265

1=2 (vn+1 − vn )2

:

(2.11)

n=1

Also, √

vn2 (t) 6 vn2 (t) − vn1 (t) 6 N

N


1=2 (vn+1 − vn )2

n=1

and √

vn1 (t) = vn1 (t) − vn2 (t) + vn2 (t) 6 N

N


1=2 (vn+1 − vn )

2

+ vn2 (t)

n=1

√ 62 N

N


1=2 (vn+1 − vn )

2

:

(2.12)

n=1

From (2.11) we deduce that

1=2 N
√ |vn1 (t)| 6 2 N (vn+1 − vn )2 : n=1

We do the same discussion with the next term vn1 +1 (t). If it is equal to zero, we go to the next one: vn1 +2 (t) and so on until we 2nd the next term di4erent from zero. We continue with the process until vN (t). This implies that N


vn2 (t) 6 4N 2

n=1

N


(vn+1 (t) − vn (t))2 :

n=1

The proof of (c) is similar so we omit it.

N Remark 2.1. The proof of (b) and (c) in Lemma 2.1 requires only that n=1 vn (t) = 0 for all t and does not use any more the equation(s) that vn satis2es. In the following sections, we will use this fact for the solution

N un of the nonlinear problem (2.2)–(2.4) as long as we can guarantee that J (t) = n=1 un (t) = 0 ∀t ¿ 0. Let T ¿ 0. We consider the linear space X (T ) consisting of all functions u(t) of the form u(t) = (u0 (t); u1 (t); u2 (t); : : : ; uN −1 (t)); such that (A) un (t) ∈ C 2 ([0; T ); R); un (t) = un+N (t); un (0) = an ;

∀t ∈ [0; T );

u˙ n (0) = bn

266

G.P. Menzala, V.V. Konotop / Nonlinear Analysis 54 (2003) 261 – 278

with N


an = 0

N


and

n=1

bn = 0

n=1

and (B) sup

 N


06t¡T

 [(u˙ n )2 + (un+1 − un )2 + un2 ] ¡ + ∞:

n=1

Given an element u(t) ∈ X (T ) we de2ne its norm uX , where u2X = sup

N


06t¡T n=1

[(u˙ n )2 + (un+1 − un )2 + un2 ]:

The space (X (T );  · X ) becomes a Banach space. Clearly, the function v(t) = (v0 (t); v1 (t); : : : ; vN −1 (t));

(2.13)

where vn (t) is the solution of (2.7) belongs to the space X (T ) (for any T ¿ 0) provided assumptions (2.5) and (2.6) are satis2ed. In fact, multiplying equation vMn − Nvn = 0 by v˙n and adding from n = 1 up to n = N , we deduce that N d
[(v˙n )2 + (vn+1 − vn )2 ] = 0 dt

∀t ¿ 0:

n=1

Consequently, N


[(v˙n )2 + (vn+1 − vn )2 ] =

n=1

N


[b2n + (an+1 − an )2 ] ¡ + ∞

∀t ¿ 0;

(2.14)

n=1

which together with Lemma 2.1, item (b) proves that v ∈ X (T ) (for any T ¿ 0). Let R ¿ 0 and v(t) as in (2.13) where vn is the solution of (2.7). Let us de2ne the subset of X (T ): YR (t) = {u ∈ X (T ); u − vX 6 R; un (0) = vn (0) = an and u˙ n (0) = v˙n (0) = bn }: Theorem 2.1 (Local existence). There exists T ¿ 0 and a unique function u(t) = (u0 (t); u1 (t); : : : ; uN −1 (t)) dened in [0; T ) such that un (t) satises N  t
G(n − m; t − s)F(um (s); u˙ m (s)) ds; un (t) = vn (t) + m=1

0

(2.15)

which belongs to the Banach space X (T ). Here G(n; t) is given by (2.9) and F(um ; u˙ m ) is given by (2.3). Proof. We de2ne the map P on YR (T ) given by ˜ 1 (t); : : : ; Pu ˜ n (t); : : : ; Pu ˜ N −1 (t)); ˜ 0 ; Pu Pu(t) = (Pu

G.P. Menzala, V.V. Konotop / Nonlinear Analysis 54 (2003) 261 – 278

where

N 


˜ n (t) = vn (t) + Pu

t

0

m=1

267

G(n − m; t − s)F(um (s); u˙ m (s)) ds:

Observe that d ˜ Pun (0) = v˙n (0) = bn ; dt because G(n; 0) = 0. The proof consists of two steps: (i) To show that P maps YR (T ) into itself continuously and (ii) To show that P is a contraction for T ¿ 0 suQciently small, that is, there exists 0 ¡ $ ¡ 1 such that ˜ n (0) = vn (0) = an Pu

and

Pu − PwX 6 $u − wX ;

∀u; w ∈ YR (T ):

Let u ∈ YR (T ). In order to prove that Pu ∈ YR (T ) it is suQcient to prove that Pu − v ∈ YR (T ). Let ˜ n − vn ≡ wn = Pu

N


G(n − m; t − s)F(um (s); u˙ m (s)) ds:

m=1

We need to estimate (a) sup

N


06t¡T n=1

(w˙ n )2 ;

(b) sup

N


06t¡T n=1

(wn+1 − wn )2

and

(c) sup

N


06t¡T n=1

wn2 :

Let us estimate (c): Clearly  N  
t  G(n − m; t − s){p (um − um−1 )p + p (um − um+1 )p } ds |wn | =   0 m=1

+

N 


t

0

m=1

   G1 (n − m; t − s)u˙ m (s) ds ; 

(2.16)

N where G1 (n − t) = − 12 k−1 !k sin(!k t)cos(2kn=N ) and !k is given in (2.10). Now, we use the following properties of G(n; t) and G1 (n; t): |G1 (n; t)| 6 4|t|;

|G(n; t)| 6 |t|;

(2.17)

which together with HMolder’s inequality allow us to obtain from (2.16) the estimate 1=2  t 1=2 N  t
2 |wn | 6 p |t − s| ds |um−1 − um |2p ds 0

m=1

 +

+ 4

0

t

0

|um+1 − um |2p ds

N 
m=1

0

t

2

|t + s| ds

1=2 

1=2  0

t

2

|u˙ m | ds

1=2 :

268

G.P. Menzala, V.V. Konotop / Nonlinear Analysis 54 (2003) 261 – 278

Consequently, |wn (t)| 6 C1 T

2

N


 p

p

sup |um−1 − um | + sup |um+1 − um | +

06t¡T

m=1

06t¡T



2

1=2 

sup |u˙ m |

06t¡T

6 C1 T 2 N (2upX + uX ); where C1 = max{p ; 4 }. Thus, sup

N


06t¡T n=1

wn2 6 CT 4 N 2 (2upX + uX )2 2 6 8CT 4 N 2 (u2p X + uX ):

(2.18)

Now, let us estimate (a): 2  d ˜ 2 Pun − v˙n (w˙ n (t)) = dt  N  2
t ˙ − m; t − s)F(um (s); u˙ m (s)) ds = G(n m=1

=

0

 N 
m=1

−

t

˙ − m; t − s)[p (um − um−1 )p + p (um − um+1 )p ] ds G(n

0

N 
m=1

t

0

2 G˙ 1 (n − m; t − s)u˙ m (s) ds

;

(2.19)

˙ where G(n:t)=(d=dt)G(n; t) and G˙ 1 =(d=dt)G1 . We can use the fact that G˙ and G˙ 1 are ˙ bounded, actually, |G(n; t)| 6 1 and |G˙ 1 (n; t)| 6 4 thus, from (2.19) we deduce that  N  2
t 2 p p (|um−1 − um | + |um+1 − um | + |u˙ m |) ds (w˙ n (t)) 6 C (2.20) m=1

0

for some positive constant C. The right-hand side of (2.20) is bounded by CN 2 T 2 (2upX + uX )2 . Consequently, sup

N


06t¡T n=1

2 (w˙ n )2 6 8CN 3 T 2 (u2p X + uX ):

(2.21)

Similar consideration shows that sup

N


06t¡T n=1

2 (wn+1 − wn )2 6 8CN 3 T 4 (u2p X + uX )

for some positive constant C.

(2.22)

G.P. Menzala, V.V. Konotop / Nonlinear Analysis 54 (2003) 261 – 278

269

Using estimates (2.18), (2.21) and (2.22) we conclude that whenever u ∈ YR (T ) then 2 ˜ 4 + T 2 )(u2p Pu − v2X 6 C(T X + uX );

(2.23)

where v = v(t) given by (2.13) is the solution of (2.7). Using (2.14) and Lemma 2.1 (part (b)) we know that v2X 6

N


[b2n + (an+1 − an )2 ] + 4N 2 sup

N


06t¡T n=1

n=1

6 (1 + 4N 2 )

N


(vn+1 − vn )2

[b2n + (an+1 − an )2 ] ≡ C˜ 21 :

n=1

Since u ∈ YR (T ), then uX 6 R + vX 6 R + C˜ 1 ; which allow us to obtain from (2.23) the estimate ˜ 4 + T 2 ][(R + C˜ 1 )2p + (R + C˜ 1 )2 ]: Pu − v2X 6 C[T

(2.24)

Clearly, we can choose T ¿ 0 suQciently small so that the right-hand side of (2.24) satis2es ˜ 4 + T 2 )[(R + C˜ 1 )2p + (R + C˜ 1 )2 ] 6 R2 : C(T

(2.25)

This shows that P maps YR (T ) into itself. Obviously P is a continuous map; therefore, the proof of step (i) is now complete. Now, we will prove that P is a contraction from YR (T ) into itself if T ¿ 0 is chosen small enough. Let u(t) and w(t) in YR (T ), then N  t
˜ n= ˜ n − Pw G(n − m; t − s){F(um ; u˙ m ) − F(wm ; w˙ m )} ds: Pu 0

m=1

˙ n − Pw ˙ n We want to estimate Pu − PwX that is, we need Let us denote by Qn = Pu to estimate (d) sup

N


06t¡t n=1

(Q˙ n )2 ;

(e) sup

N


06t¡t n=1

(Qn+1 − Qn )2

and

(f ) sup

06t¡t n=1

We begin estimating (f). We have N  t
Qn (t) = p G(n − m; t − s){(um − um−1 )p + (um − um+1 )p m=1

0

− (wm − wm−1 )p − (wm − wm+1 )p } ds +

N 
m=1

0

t

G1 (n − m; t − s)(u˙ m − w˙ m ) ds:

N


Qn2 :

270

G.P. Menzala, V.V. Konotop / Nonlinear Analysis 54 (2003) 261 – 278

Since p is an integer, p ¿ 2 we can use the identity ap − bp = (a − b)(ap−1 + · · · + bp−1 ) together with properties (2.17) of G and G1 to obtain that Qn (t) 6 C

N 
m=1

t

|t − s(um−1 − un ) − (wm−1 − wm )| ×

0

×|(um−1 − um )p−1 + · · · + (wm−1 − wm )p−1 | ds +C

N 
m=1

t

0

|t − s|(um+1 − un ) − (wm+1 − wm )| ×

×|(um+1 − um )p−1 + · · · + (wm+1 − wm )p−1 | ds + +C

N 
m=1

t

0

|t − su˙ m − w˙ m | ds

6 Cu − wX (uX + wX )p−1 T 2 N for some positive constant C. Therefore, we deduce that sup

N


06t¡T n=1

˜ 3 T 4 (uX + wX )2(p−1) u − w2X : Qn2 6 CN

(2.26)

Now, let us estimate (e): Since Qn+1 − Qn =

N 
m=1

0

t

[G(n + 1 − m; t − s) − G(n − m; t − s)] ×

×[F(um ; u˙ m ) − F(wm ; w˙ m )] ds: Then, proceeding as above we get the estimate sup

N


06t¡T n=1

˜ 3 T 4 (uX + wX )2(p−1) u − w2X (Qn+1 − Qn )2 6 CN

(2.27)

˜ for some positive constant C. Similar discussion give us the estimate sup

N


06t¡T n=1

˜ 3 T 2 u − w2X (uX + wX )2(p−1) : (Q˙ n ) 6 CN

(2.28)

Using estimates (2.26)–(2.28) we conclude that for any u; w ∈ YR (t) then Pu − Pw2X 6 C[T 2 + T 4 ]u − w2X (uX + wX )2(p−1)

(2.29)

G.P. Menzala, V.V. Konotop / Nonlinear Analysis 54 (2003) 261 – 278

271

for some positive constant C. Since u and w belong to YR (T ), we already know that uX 6 R+ C˜ 1 and wX 6 R+ C˜ 1 where C˜ 1 is the constant which appeared in (2.24). Thus, from (2.29) we conclude that Pu − Pw2X 6 C(T 2 + T 4 )u − w2X (2R + C˜ 1 )2(p−1) : Given a positive number $ such that 0 ¡ $2 ¡ 1 then we choose T ¿ 0 suQciently small so that C(T 2 + T 4 )(2R + 2C˜ 1 )2(p−1) 6 $2 ¡ 1;

(2.30)

which implies that Pu − PwX 6 $u − wX for any u; w ∈ YR (T ). Therefore, P is a contraction which proves step (ii). Steps (i) and (ii) imply the existence of a unique 2xed point z ∈ YR (T ) of the map P that is, Pz = z where z(t) = (z0 (t); z1 (t); : : : ; zN −1 (t)) satis2es zn = vn +

N 
m=1

0

t

G(n − m; t − s)F(zm ; z˙m ) ds

for 0 6 t ¡ T where T is small so that (2.25) and (2.30) are satis2ed. This completes the proof of Theorem 2.1. Theorem 2.2 (Global existence). Let p= odd ¿ 3 and ¿ 0. Then, the local solution of problem (2.2) – (2.4) we found in Theorem 2.1 can be extended for any T ¡ + ∞. Such global solution is unique. Proof. Using Zorn’s lemma we can extend the solution we found in Theorem 2.1 to the maximal interval of existence which we denote by [0; Tmax ). We want to prove that Tmax = +∞. Suppose this is not the case, then uX (T ) → +∞ as T → Tmax ¡ +∞(T ¡ Tmax ). We will get a contradiction obtaining an a priori estimate. We consider the quantity EN (t) =

N

N

N

n=1

n=1

n=1

p
1
1
(u˙ n )2 + (un+1 − un )2 + (un+1 − un )p+1 : 2 2 p+1

(2.31)

We claim the validity of N


dEN (i) (u˙ n+1 − u˙ n )2 (t) = − dt

(2.32)

(ii) u2X (T ) 6 sup EN (t) 6 EN (0) ¡ + ∞

(2.33)

n=1

and 06t¡T

in any interval [0; T ) with T ¡ Tmax .

272

G.P. Menzala, V.V. Konotop / Nonlinear Analysis 54 (2003) 261 – 278

To prove (i), we calculate the derivative of EN (t) and use Eq. (2.2) to obtain after simpli2cations that N

N

N

n=1

n=1

n=1



dEN
u˙ n uM n + (un+1 − un )(u˙ n+1 − u˙ n ) + p (un+1 − un )p (u˙ n+1 − u˙ n ) = dt N N

u˙ n (un − un−1 ) + p u˙ n (un−1 − un )p =− n=1

+

n=1

N


u˙ n (u˙ n+1 + u˙ n−1 − 2u˙ n ) +

N


u˙ n+1 (un+1 − un )

n=1

n=1

+ p

N


u˙ n+1 (un+1 − un )p :

(2.34)

n=1

Changing variables n = m + 1 we obtain that N


u˙ n (un − un−1 ) =

N


u˙ m+1 (um+1 − um ) − u˙ 1 (u1 − u0 )

m=1

n=1

N
u˙ m+1 (um+1 − um ): + u˙ N +1 (uN +1 − uN ) = − m=1

because uN +1 = u1 and uN = u0 . Since p is odd, we also know that N


N N

u˙ n (un−1 − un )p = − u˙ n (un − un−1 )p = − u˙ m+1 (um+1 − um )p :

n=1

n=1

m=1

Thus, from (2.34) we deduce that N


dEN [u˙ n (u˙ n+1 − u˙ n ) − u˙ n (u˙ n − u˙ n−1 )] = dt n=1



= u˙ N (u˙ N +1 − u˙ N ) +

N
n=1

=−

N


u˙ n (u˙ n+1 − u˙ n ) − u˙ 1 (u˙ 1 − u˙ 0 ) −

N


 u˙ n (u˙ n − u˙ n−1 )

n=1

(u˙ m+1 − u˙ m )2

m=1

because u˙ 1 (u˙ 1 − u˙ 0 ) = u˙ N +1 (u˙ N +1 − u˙ N ). This proves step (i). To prove step (ii), we consider the function JN (t) given by JN (t) =

N
n=1

un (t):

G.P. Menzala, V.V. Konotop / Nonlinear Analysis 54 (2003) 261 – 278

273

We calculate the second-order derivative of JN (t) and use Eq. (2.2) to obtain that JMN (t) =

N


uM n (t) =

[(un+1 − un ) − (un − un−1 )]

n=1

n=1

− p

N


N


[(un+1 − un )p + (un+1 − un )p ] +

N


(u˙ n+1 + u˙ n−1 − 2u˙ n )

n=1

n=1

= [(uN +1 − uN ) − (u1 − u0 )] − p

N


[(un+1 − un )p − (un − un−1 )p ]

n=1

+ [(uN +1 − uN ) − (u1 − u0 )] =0 because u1 = uN +1 and u0 = uN . Consequently, JN (t) = JN (0) + J˙N (0)t. Since conditions (2.5) and (2.6) are valid, then JN (0)=0 and J˙N (0)=0 Thus, JN (t)=0 for all 0 6 t 6 T . As we mentioned in Remark 2.1 (after Lemma 2.1) this fact implies that N


un2 6 4N 2

n=1

N


(un+1 − un )2

for all 0 6 t 6 T

(2.35)

n=1

and N


(u˙ n )2 6 4N 2

N


(u˙ n+1 − u˙ n )2

for all 0 6 t 6 T:

(2.36)

n=1

n=1

From (2.35) and step (i) we deduce the proof of step (ii), which is the a priori estimate we needed. This proves that Tmax = +∞. Remains to prove that the global solution is unique. Suppose that problem (2.2)–(2.4) has two solutions un (t) and wn (t); n=0; 1; : : : ; N −1. Let zn (t)=un (t)−wn (t). We claim that zn (t)=0. In fact, zn (t) satis2es the equation zMn − Nzn = F(un ; u˙ n ) − F(wn ; w˙ n ) = −p [(zn+1 − zn )fn − (zn − zn−1 )gn ] + Nz˙n ;

(2.37)

where fn = (un+1 − un )p−1 + (un+1 − un )p−2 (wn+1 − wn ) + · · · + (wn+1 − wn )p−1 and gn = (un−1 − un )p−1 + (un−1 − un )p−2 (wn−1 − wn ) + · · · + (wn−1 − wn )p−1 : Multiplying (2.3) by z˙n and adding in 1 6 n 6 N . We obtain that N N
1 d
[(z˙n )2 + (zn+1 − zn )2 ] + (z˙n+1 − z˙n )2 = Gn ; 2 dt n=1

n=1

(2.38)

274

G.P. Menzala, V.V. Konotop / Nonlinear Analysis 54 (2003) 261 – 278

where Gn = −p

N


z˙n [(zn+1 − zn )fn − (zn − zn−1 )gn ]:

(2.39)

n=1

Using Cauchy–Schwarz’s inequality and the de2nition of the norm  · X we deduce from (2.3) that N N
1 d
[(z˙n )2 + (zn+1 − zn )2 ] + (z˙n+1 − z˙n )2 2 dt n=1





  6C  sup (|fn | + |gn |)  16n6N 

n=1



06t¡T

 6 C(uX + wX )

p−1

N

N

n=1

n=1

1
1
(z˙n )2 + (zn+1 − zn )2 2 2

N

N

n=1

n=1

1
1
(z˙n )2 + (zn+1 − zn )2 2 2





or d ˜ ’(t) 6 C’(t); dt

N where ’(t) = 12 n=1 [(z˙n )2 + (zn+1 − zn )2 ] and C˜ = C(uX + wX )p−1 . Gronwall’ s inequality implies that ’ = 0 that is u ≡ w for all t ¿ 0. 3. Uniform stabilization as t → +∞ In this section, we will prove the main result of this work, that is, the exponential stabilization of the solution of problem (2.2)–(2.4) as time approaches in2nity. We will use a criteria due to Nakao [7] (see also [6]). Theorem 3.1 (Exponential decay). Let u = (u0 ; u1 ; u2 ; : : : ; uN −1 ) be the global solution of problem (2.2) – (2.4) obtained in Theorem 2 with p = odd ¿ 3 and ¿ 0. Then, there exist positive constants C and - such that EN (t) 6 CEN (0)exp(−-t)

∀t ¿ 0;

where EN (t) is given by (2.31). Proof. Integration of (2.32) in the interval [t; t + 1] gives us  t+1
N EN (t) − EN (t + 1) = (u˙ n+1 − u˙ n )2 ds: t

n=1

(3.1)

N Let us denote by F 2 (t) = EN (t) − EN (t + 1) and by g(s) = n=1 (u˙ n+1 (s) − u˙ n (s))2 . The mean value theorem for integrals implies the existence of t1 and t2 with t1 ∈ [t; t+ 14 ]

G.P. Menzala, V.V. Konotop / Nonlinear Analysis 54 (2003) 261 – 278

and t2 ∈ [t + 34 ; t + 1] such that 

t+1=4

1 g(s) ds = g(t1 ) 4

t

 and

t+1

t+3=4

g(s) ds =

1 g(t2 ): 4

275

(3.2)

Using (3.1) and (3.2) we obtain that g(tj ) 6

4 2 F (t);

j = 1; 2:

(3.3)

Let us multiply Eq. (2.2) by un (t) and add from n = 1 up to n = N to obtain N

N N

d
un (un+1 + un−1 − 2un ) un u˙ n − (u˙ n (t))2 − dt n=1

− p

n=1

n=1

N


un (un−1 − un )p − p

n=1

−

N


N


un (un+1 − un )p

n=1

(u˙ n+1 + u˙ n−1 − 2u˙ n ) = 0:

(3.4)

n=1

Direct calculations show that N N

− un (un+1 + un−1 − 2un ) = (un+1 − un )2 n=1

n=1

and −p

N


un (un−1 − un )p − p

n=1

N


un (un+1 − un )p = −p

N


(un+1 − un )p+1

n=1

because p is odd. Also, N


N
un (u˙ n+1 + u˙ n−1 − 2u˙ n ) = − (un+1 − un )(u˙ n+1 − u˙ n ):

n=1

n=1

From the above identities and (3.4) we deduce that N

N N

d
un u˙ n − (u˙ n )2 + (un+1 − un )2 dt n=1

+ p

n=1

N
n=1

n=1

(un+1 − un )p+1 +

N
n=1

(un+1 − un )(u˙ n+1 − u˙ n ) = 0:

(3.5)

276

G.P. Menzala, V.V. Konotop / Nonlinear Analysis 54 (2003) 261 – 278

Given 0 ¡ / ¡ 1 and using Cauchy–Schwarz’s inequality we obtain   N N N 

1
  (un+1 − un )2 + (u˙ n+1 − u˙ n )2 : (un+1 − un )(u˙ n+1 − u˙ n ) 6 /    / n=1

(3.6)

n=1

n=1

From (3.6) and (3.5) we deduce that (1 − /)

N


(un+1 − un )2 + p

n=1

N


(un+1 − un )p+1

n=1

N

N N
1
d
(u˙ n+1 − u˙ n )2 : un u˙ n + (u˙ n )2 + 6− / dt n=1

n=1

(3.7)

n=1

Integration of both sides of (3.7) in the interval [t1 ; t2 ] and using Cauchy–Schwarz’s inequality and (2.35), (2.36) imply that  (1 − /)

N


t2

t1

6−

(un+1 − un )2 ds + p



t1

n=1

N


N


t2

(un+1 − un )p+1 ds

n=1

{un (t2 )u˙ n (t2 ) − un (t1 )u˙ n (t1 )}

n=1

 +

t2

t1

N


(u˙ n )2 ds +

n=1

 6 4N 2

N


1 /



t2

t1

N


(u˙ n+1 − u˙ n )2 ds

n=1

1=2  (un+1 (t2 ) − un (t2 ))2

N


n=1

 + 4N

2

1=2 (u˙ n+1 (t1 ) − u˙ n (t1 ))2

n=1

N


1=2  (un+1 (t1 ) − un (t1 ))

2

n=1

N


1=2 (u˙ n+1 (t2 ) − u˙ n (t2 ))

2

n=1

 t2
 N 1 (u˙ n+1 + u˙ n )2 ds: + 4N 2 + / t1

(3.8)

n=1

Using (3.1) and (3.3) we can obtain from (3.8) that  (1 − /)

t2

t1

N
n=1

(un+1 − un )2 ds + p



t2

N


t1



(un+1 − un )p+1 ds

n=1

6 16N 2 {EN1=2 (t2 ) + EN1=2 (t1 )}F(t) + 4N 2 +

1 /



F 2 (t):

(3.9)

G.P. Menzala, V.V. Konotop / Nonlinear Analysis 54 (2003) 261 – 278

Combining (2.34), (3.1) and (3.9) we deduce that  t2 EN (s) ds 6 C{EN1=2 (t2 ) + EN1=2 (t1 )}F(t) + CF 2 (t) t1

277

(3.10)

for some positive constant C. The mean value theorem for integrals applied to (3.10) gives us the existence of some t ∗ ∈ [t1 ; t2 ] such that 1 4

EN (t ∗ ) 6 (t2 − t1 )EN (t ∗ ) 6 C{EN1=2 (t2 ) + EN1=2 (t1 )}F(t) + CF 2 (t) 6 2C

sup EN1=2 (s)F(t) + CF 2 (t):

t6s6t+1

(3.11)

Using (2.32) and (3.1) we know that  t∗
N ∗ (u˙ n+1 − u˙ n )2 ds 6 EN (t ∗ ) + F 2 (t): EN (t) = EN (t ) + t1

n=1

Thus, if s ∈ [t; t + 1] and t1 6 s then EN (s) 6 EN (t1 ) 6 EN (t) 6 EN (t ∗ ) + F 2 (t):

(3.12)

If s ∈ [t; t + 1] and s 6 t1 then EN (s) 6 EN (t) 6 EN (t ∗ ) + F 2 (t):

(3.13)

Combining (3.11) with (3.12) and (3.13) we deduce that for any s ∈ [t; t + 1] we have EN (s) 6 C1 6

sup EN1=2 (s)F(t) + C2 F 2 (t)

t6s6t+1

1 sup EN (s) + 2 t6s6t+1



C12 + C2 F 2 (t) 2

for some positive constants C1 and C2 . This implies that sup EN (s) 6 C3 F 2 (t) = C3 (EN (t) − EN (t + 1))

t6s6t+1

for some positive constant C3 and any t ¿ 0. The conclusion of Theorem 3.1 follows using a well-known criteria due to Nakao (see [6] or [7]). 4. Final comments There are several important (for physical application) lattices for which the techniques we used in Sections 2 and 3 can be applied, as well, with suitable modi2cations for example. Among them we mention the sine-Gordon lattice with nonlinear inter-site interactions Hnl =

N
n=1

[1 − cos(un+1 − un )]

278

G.P. Menzala, V.V. Konotop / Nonlinear Analysis 54 (2003) 261 – 278

or on-site interactions Hnl =

N


[1 − cos un ]

n=1

and linear part given by (1.2). Considering suitable damping mechanisms we could prove similar results as Theorems 2.1, 2.2 and 3.1 (the 2rst two theorems in the case of an in2nite lattice are given in [8]). Finally, we remark that in what concerns the type of damping assumed there are several ways or forms that may be considered. One of the most relevant is the external damping (it is of the form u˙ n ) or sometimes a nonlinear damping could be inherent to the problem. In the case of nonlinear internal damping the techniques used in our previous discussion still could be employed, while the asymptotic properties of lattices with external damping or having an in2nite number of atoms are still open problems. Acknowledgements The 2rst author was partially supported by a grant of CNPq and PRONEX (MCT, Brasil). Part of this work was done while he was visiting the University of Chile, Centro de Modelamiento MatemTatico (CMM) as part of the Chair “Marcel Dassault”. He expresses his gratitude for their kind hospitality and support. The second author (V.V.K) was supported by the European grant, COSYC No. HPRN-CT-2000-00158. The main support came from the bilateral program Portugal/Brasil via ICCTI/CAPES being developed at the Federal University of Rio de Janeiro and the University of Lisbon. References [1] M. Ablowitz, P.A. Clarkson, Solitons, Nonlinear Evolution Equations and Inverse Scattering, Cambridge University Press, Cambridge, 1992. [2] D. Bambusi, Exponential stability of breathers in Hamiltonian networks of weakly coupled oscillators, Nonlinearity 9 (1996) 433–457. [3] J.L. Bona, J.L. Saut, Dispersive blowup of solutions of generalized Korteweg de Vries equations, J. Di4erential Equations 103 (1993) 3–57. [4] P.G. Kevrekidis, K.Y. Rasmussen, A.R. Bishop, The discrete nonlinear SchrModinger equation: a survey of recent results, Int. J. Mod. Phys. B 15 (2001) 2833–2900. [5] R.S. Mackay, S. Aubry, Proof. of existence of breathers for time reversible or Hamiltonian networks of weakly coupled oscillators, Nonlinearity 7 (1994) 1623–1643. [6] M. Nakao, Asymptotic stability of the bounded or almost periodic solution of the wave equation with nonlinear dissipative term, J. Math. Anal. Appl. 58 (1977) 336–343. [7] M. Nakao, A di4erence inequality and its application to nonlinear evolution equations, J. Math. Soc. Japan 30 (1978) 747–762. [8] G. Perla Menzala, V.V. Konotop, On global existence of localized solutions of some nonlinear lattices, Appl. Anal. 75 (2000) 157–173. [9] M. Toda, Theory of Nonlinear Lattices, Springer, Berlin, 1981. [10] L. VTazquez, L. Streit, V.M. PTerez-Garcia (Eds.), Nonlinear Klein–Gordon and SchrModinger Systems: Theory and Applications, World Scienti2c, Singapore, 1996.