• Non ci sono risultati.

Applicable AnalysisPublication details, including instructions for authors and subscription information:http://www.informaworld.com/smpp/title~content=t713454076

N/A
N/A
Protected

Academic year: 2021

Condividi "Applicable AnalysisPublication details, including instructions for authors and subscription information:http://www.informaworld.com/smpp/title~content=t713454076"

Copied!
23
0
0

Testo completo

(1)

PLEASE SCROLL DOWN FOR ARTICLE This article was downloaded by: [Autuori, Giuseppina]

On: 15 March 2011

Access details: Access Details: [subscription number 934988671]

Publisher Taylor & Francis

Informa Ltd Registered in England and Wales Registered Number: 1072954 Registered office: Mortimer House, 37- 41 Mortimer Street, London W1T 3JH, UK

Applicable Analysis

Publication details, including instructions for authors and subscription information:

http://www.informaworld.com/smpp/title~content=t713454076

Local asymptotic stability for polyharmonic Kirchhoff systems†

G. Autuori

a

; P. Pucci

a

a

Dipartimento di Matematica e Informatica, Università degli Studi di Perugia, 06123 Perugia, Italy First published on: 16 September 2010

To cite this Article Autuori, G. and Pucci, P.(2011) 'Local asymptotic stability for polyharmonic Kirchhoff systems†', Applicable Analysis, 90: 3, 493 — 514, First published on: 16 September 2010 (iFirst)

To link to this Article: DOI: 10.1080/00036811.2010.483433 URL: http://dx.doi.org/10.1080/00036811.2010.483433

Full terms and conditions of use: http://www.informaworld.com/terms-and-conditions-of-access.pdf This article may be used for research, teaching and private study purposes. Any substantial or systematic reproduction, re-distribution, re-selling, loan or sub-licensing, systematic supply or distribution in any form to anyone is expressly forbidden.

The publisher does not give any warranty express or implied or make any representation that the contents

will be complete or accurate or up to date. The accuracy of any instructions, formulae and drug doses

should be independently verified with primary sources. The publisher shall not be liable for any loss,

actions, claims, proceedings, demand or costs or damages whatsoever or howsoever caused arising directly

or indirectly in connection with or arising out of the use of this material.

(2)

Vol. 90, Nos. 3–4, March–April 2011, 493–514

Local asymptotic stability for polyharmonic Kirchhoff systemsy

G. Autuori and P. Pucci*

Dipartimento di Matematica e Informatica, Universita` degli Studi di Perugia, Via Vanvitelli 1, 06123 Perugia, Italy

Communicated by R.P. Gilbert

(Received 8 March 2010; final version received 1 April 2010)

In this article we treat the question of local asymptotic stability for polyharmonic Kirchhoff systems, governed by time-dependent source forces and nonlinear damping terms. Even the simplest case (P

3

) studied here includes several systems of great interest, as the physical models for vibrating beams of the Woinowsky–Krieger type. This article extends and generalizes some local stability results of Autuori et al. [Asymptotic stability for anisotropic Kirchhoff systems, J. Math. Anal. Appl. 352 (2009), pp. 149–165]

to higher order systems, and also of Nakao and Zhu [Decay rate of solutions of a wave equation with damping and external force, Nonlinear Anal. 46 (2001), pp. 335–345; An attractor for a nonlinear dissipative wave equation of Kirchho type, J. Math. Anal. Appl. 353 (2009), pp. 652–659].

Keywords: polyharmonic operator; Kirchhoff systems; local asymptotic stability; qualitative properties of solutions

AMS Subject Classifications: Primary 35B40; 35L75; 35B35;

Secondary 35L55; 35Q99

1. Introduction

In this article we deal with the question of the local asymptotic stability of solutions of polyharmonic Kirchhoff systems, governed by time-dependent source forces and nonlinear external dampings terms for three model cases. We first consider in R þ

0  

u tt þ MðkD L uk 2 2 ÞðDÞ L u þ gðtÞðDÞ L u t þ u þ Qðt, x, u, u t Þ þ f ðt, x, uÞ ¼ 0, ðP 1 Þ

with L  1. Here   R

n

is a bounded domain and u ¼ (u

1

, . . . , u

N

) ¼ u(t, x) represents the vectorial displacement, where N  1 and   0. Throughout this article, the function g  0 is in L 1 loc ðR þ 0 Þ and the dissipative Kirchhoff function M is taken of the standard form

MðÞ ¼ a þ b 1 , a, b  0, a þ b 4 0,  4 1 if b 4 0:

*Corresponding author. Email: [email protected]

yDedicated to Professor Paul Leo Butzer on the occasion of his 80th birthday.

ISSN 0003–6811 print/ISSN 1563–504X online ß 2011 Taylor & Francis

DOI: 10.1080/00036811.2010.483433 http://www.informaworld.com

Downloaded By: [Autuori, Giuseppina] At: 19:18 15 March 2011

(3)

When a40 and b  0, system (P

1

) is said non-degenerate, otherwise, when a ¼ 0 and b40, it is called degenerate. The case   1 was first considered in [1] for special cases of (P

1

). For further comments on asymptotic behaviour of solutions, as well as for previous references, we refer to [1]. Finally,

D L u ¼ DD j u, if L ¼ 2j þ 1 D j u, if L ¼ 2j (

, L ¼ 1, 2, . . .

On the dissipation Q acting on the body and on the conservative source force f, we assume throughout this article the structural assumptions

f 2 CðR þ 0    R N ! R N Þ, Fðt, x, 0Þ ¼ 0, f ðt, x, uÞ ¼ F u ðt, x, uÞ, Q 2 CðR þ 0    R N  R N ! R N Þ, ðQðt, x, u, vÞ, vÞ  0

for all arguments t 2 R þ 0 , x 2  and u, v 2 R

N

, where (, ) is the inner product in R

N

. Moreover, when L  2, we study the system in R þ 0  

u tt þ ðDÞ L ðu þ gðtÞu t Þ þ MðkD L1 uk 2 2 ÞðDÞ L1 u þ u þ Qðt, x, u, u t Þ þ f ðt, x, uÞ ¼ 0:

ðP 2 Þ

Both (P

1

) and (P

2

) are considered under the homogeneous Dirichlet boundary conditions

uðt, xÞ ¼ 0, Duðt, xÞ ¼ 0, . . . , D 2ðL1Þ uðt, xÞ ¼ 0 on R þ 0  @: ð1:1Þ Throughout this article, we assume that n42L and denote by X the Sobolev space [W

L,2

()]

N

and by 2  L ¼ 2n=ðn  2LÞ its critical exponent, while we leave to the reader the case 15n  2L, for which the modifications are standard, being 2  L ¼ 1.

In the biharmonic case L ¼ 2, problems (P

1

) and (P

2

) are largely studied in the literature in several simplified subcases. Indeed, the original physical models, governed by (P

1

) and (P

2

), are vibrating beams of the Woinowsky–Krieger type, with internal material higher order damping term g(t)(D)

L

u

t

of Kelvin–Voigt type and a nonlinear damping Q effective in .

The last model considered in this article is

u tt þ ðDÞ L u  MðkDuk 2 2 ÞDu þ u þ Qðt, x, u, u t Þ þ f ðt, x, uÞ ¼ 0, L  1, ðP 3 Þ

with Dirichlet boundary conditions (1.1) on R þ 0  @. In the relevant physical case L ¼ 2, system (P

3

) coincides with (P

2

) when g 0, and is a model for vibrating beams of the Woinowsky–Krieger type, see [2] for a special relevant subcase of (P

3

).

Another case of (P

3

) is treated in [3], when Q 0 and N ¼ 1.

In [4] the question of global asymptotic stability was considered when the potential energy satisfies a global condition of the Hale type. Here, we assume the usual Hale condition, that is only for u sufficiently small, so that global asymptotic stability cannot be longer expected; for general comments and previous references we refer to [5] for wave systems, to [6] for anisotropic Kirchhoff systems of second order, to [7] for general Kirchhoff systems, and finally to the recent article [8] and references therein for the local asymptotic stability of a special case of (P

3

).

Downloaded By: [Autuori, Giuseppina] At: 19:18 15 March 2011

(4)

More precisely we shall prove that if u is a global solution with initial data ku

t

(0, )k

2

and kD

L

u(0, )k

2

sufficiently small, then

t!1 lim ðku t k 2 þ kD L uk 2 Þ ¼ 0, ð1:2Þ that is, the rest state (0, 0) is a locally attracting set in X  L

2

(), where for simplicity of notation L

2

() ¼ [L

2

()]

nN

if L is odd, while L

2

() ¼ [L

2

()]

N

if L is even. From now on we drop the exponents N and nN in all the functional spaces involved in the treatment, as made before.

The main results of this article are obtained through qualitative estimates on the energy functional Eu associated to any solution u of (P

1

)–(P

3

) and we proceed by assuming the existence of a suitable auxiliary function k, following an argument introduced by Pucci and Serrin for damped wave systems [5,9,10].

Theorems 3.8–3.11 for (P

1

) and Theorems 4.6–4.9 for (P

2

) investigate the effects on the stability when a relation between k and g occurs. In particular, these results are extremely simple in the relevant model case in which Q does not depend on t, usually treated in the literature.

Indeed, in this situation, if either g 2 W 1,1 loc ðR þ 0 Þ \ L 1 ðR þ 0 Þ, g ò 0 and jg 0 (t)j ¼ o(g(t)) as t ! 1, or g 2 CBVðR þ 0 Þ n L 1 ðR þ 0 Þ, then it is enough to take k ¼ g and conclude that the rest state (0, 0) is a locally attracting set. Last but not least, as a consequence of the main results, the validity of (1.2) follows at once, when Q does not depend on t, g(t)  g

0

40 for t large enough and lim inf t!1 t 2 R t

0 gðÞd 5 1, by simply taking k ¼ 1. All these situations cover the trivial widely studied case in which g Constant4 0.

2. Preliminaries and structural framework

By the Sobolev embedding theorems (see e.g. [11]), the embedding X ,! L

h

() is continuous, provided that 1  h  2  L ¼ 2n=ðn  2LÞ; that is, there exists a positive constant s h such that

kuk h  s h kD L uk 2 for all u 2 X: ð2:1Þ In this article we denote by K the main solution and test function space for (P

1

)–(P

3

), that is

K ¼ f 2 K 0 : E is locally bounded on R þ 0 g,

where K 0 ¼ C 1 ðR þ 0 ! X Þ for (P

1

)–(P

2

), while K 0 ¼ CðR þ 0 ! X Þ \ C 1 ðR þ 0 ! L 2 ðÞÞ for (P

3

). For all  2 K, the Energy functional along  associated to (P

1

)–(P

3

) is

EðtÞ ¼ 1

2 k t ðt, Þk 2 2 þ aðtÞ þ FðtÞ, ðE Þ

where the elliptic functional a changes according to the problem, and will be specified in the corresponding sections below. On the other hand, the potential energy F of the field  2 K is defined by

FðtÞ ¼ Z



Fðt, x, Þdx,

Downloaded By: [Autuori, Giuseppina] At: 19:18 15 March 2011

(5)

where Fðt, x, uÞ ¼ R 1

0 ð f ðt, x, uÞ, uÞd for all ðt, x, uÞ 2 R þ 0    R N , being f ¼ F

u

. We denote by  0 ¼ inffkD L uk 2 2 =kuk 2 2 : u 2 X, u 6¼ 0g the first eigenvalue of (D)

L

in , under the boundary conditions (1.1). Hence 

0

40, since  is a bounded domain, cf. [12, Theorem 1.1].

On the source term f we also assume the following condition:

(F ) Suppose that there exists  2 ½0,  0 Þ such that for (P

1

) lim inf

u!0

ð f ðt, x, uÞ, uÞ

juj 2  a , with a 4 0, ðaÞ

while for (P

2

) and (P

3

)

lim inf

u!0

ð f ðt, x, uÞ, uÞ juj 2  :

ðaÞ

In addition, suppose that there exist 2 5 q  2  L , and a positive constant  such that for all ðt, x, uÞ 2 R þ 0    R N

j f ðt, x, uÞj  ð1 þ juj q1 Þ:

ðbÞ

The function

f ðt, x, uÞ ¼ V 1 ðt, xÞjuj q2 u  V 2 ðt, xÞu, 2 5 q  2  L , with V

1

, V

2

 0, V

1

, V 2 2 CðR þ 0  Þ,  ¼ sup R

þ

0

 ½V 1 ðt, xÞ þ V 2 ðt, xÞ 5 1, and for which there exists  2 ½0,  0 Þ such that either

sup

R

þ

0



V 2 ðt, xÞ  a for ðP 1 Þ, or sup

R

þ

0



V 2 ðt, xÞ   for ðP 2 Þ and ðP 3 Þ

verifies condition (F ).

Concerning the dissipation term Q we also require the following:

(Q) There exist constant exponents m, p, with 2  m 5 p  2  L ,

and non-negative continuous functions d

1

¼ d

1

(t, x), d

2

¼ d

2

(t, x), such that jQðt, x, u, vÞj  d 1 ðt, xÞ 1=m ðQðt, x, u, vÞ, vÞ 1=m

0

þ d 2 ðt, xÞ 1=p ðQðt, x, u, vÞ, vÞ 1=p

0

, ðaÞ

for all arguments t, x, u, v, and the following functions 

1

and 

2

are well-defined

 1 ðtÞ ¼ kd 1 ðt, Þk 2



L

=ð2

L

mÞ ,  2 ðtÞ ¼ kd 2 ðt, Þk 2



L

=ð2

L

pÞ , if p 5 2  L , kd 2 ðt, Þk 1 , if p ¼ 2  L :



Moreover, there are functions ¼ (t), ! ¼ !() such that

ðQðt, x, u, vÞ, vÞ  ðtÞ!ðjvjÞ for all arguments t, x, u, v, ðbÞ

where ! 2 CðR þ 0 Þ is such that

!ð0Þ ¼ 0, !ðÞ 4 0 for 0 5  5 1, !ðÞ ¼  2 for   1, while  0 and 1} 2 L 1 loc ðR þ 0 Þ for some exponent +41.

Downloaded By: [Autuori, Giuseppina] At: 19:18 15 March 2011

(6)

The prototype

Qðt, x, u, vÞ ¼ jvj m2 A 1 ðt, x, uÞv þ jvj p2 A 2 ðt, x, uÞv,

with A

1

, A

2

semidefinite positive continuous matrices defined in R þ 0    R N , and 2  m 5 p  2  L , as in (Q), can be taken as an example of damping term for the models (P

1

)–(P

3

). Indeed, denote by h

i

(t, x) the number inf u 2 R

N

h ~ i ðt, x, uÞ, where h ~ i ðt, x, uÞ is the least eigenvalue of the symmetric part of the coefficient matrix A

i

(t, x, u) and assume that H i ðt, xÞ ¼ sup u 2 R

N

H ~ i ðt, x, uÞ 5 1, where ~ H i ðt, x, uÞ is the Euclidean norm of A

i

(t, x, u). By assumption, h

i

(t, x)  0. Suppose furthermore that there exist 

1

 1 and 

2

 1 such that

H i   i h i , i ¼ 1, 2, d 1 ¼  1 m1 H 1 , d 2 ¼  2 p1 H 2 in R þ 0  ,

with H 1 ðt, Þ 2 L 2

L

=ð2

L

mÞ ðÞ and either H 2 ðt, Þ 2 L 2

L

=ð2

L

pÞ ðÞ if p 5 2  L , or H

2

(t, ) 2 L

1

() if p ¼ 2  L .

Condition H

i

 

i

h

i

is the weak uniform definiteness of A

i

. Of course, it is automatic in the scalar case N ¼ 1 or when Q ¼ Q(v).

By the definition of h

i

we have for all ðt, xÞ 2 R þ 0  

h i ðt, xÞjvj 2  ðA i ðt, x, uÞv, vÞ, jA i ðt, x, uÞj  H i ðt, xÞ, i ¼ 1, 2, so that

jQðt, x, u, vÞj  H 1 ðt, xÞ 1=m ð H 1 ðt, xÞjvj m Þ 1=m

0

þH 2 ðt, xÞ 1=p ð H 2 ðt, xÞjvj p Þ 1=p

0

 H 1 ðt, xÞ 1=m ð  1 h 1 jvj m Þ 1=m

0

þH 2 ðt, xÞ 1=p ð  2 h 2 jvj p Þ 1=p

0

: ð2:2Þ On the other hand,

ðQðt, x, u, vÞ, vÞ  h 1 ðt, xÞjvj m þ h 2 ðt, xÞjvj p , ð2:3Þ and combining (2.2) with (2.3), we get (Q)-(a). Finally, (Q

1

)-(b) immediately follows from (2.3), just taking

ðtÞ ¼ inf

x2 fh 1 ðt, xÞ þ h 2 ðt, xÞg, !ðÞ ¼ minf p ,  2 g,

provided that 1} 2 L 1 loc ðR þ 0 Þ for some +41. For other special examples of Q, we refer to [5–7,10,13].

The main results are proved under the additional assumption that F t u  0 in R þ 0 ,

along a global solution u 2 K. This request is trivially automatic, whenever either f does not depend on t, or F

t

 0 in R þ 0    R N , as well as in many other cases, and for simplicity, we assume it in the definition of solution.

3. Local asymptotic stability for (P

1

)

From here on, we also assume (F ) and (Q)-(a), without further mentioning. For all

 2 K, the total energy E associated to (P

1

) is given in (E), with a ðtÞ ¼ 1

2



MðkD L ðt, Þk 2 2 Þ þ kðt, Þk 2 2



, MðÞ ¼ a þ b  : ð3:1Þ

Downloaded By: [Autuori, Giuseppina] At: 19:18 15 March 2011

(7)

We say that a function u 2 K is a solution of (P

1

), if u satisfies the following two conditions:

(A) Distribution identity hu t , i  t

0 ¼ Z t

0

fhu t ,  t i  MðkD L uk 2 2 ÞhD L u, D L i  gðÞhD L u t , D L i  hu, i

 hQð,  , u, u t Þ, i  h f ð,  , uÞ, igd, for all t 2 R þ 0 and  2 K.

(B) Conservation law

ðiÞ Du :¼ hQðt,  , u, u t Þ, u t i þ gðÞkD L u t k 2 2  F t u 2 L 1 loc ðR þ 0 Þ, ðiiÞ F t u  0, t ° EuðtÞ þ

Z t 0

DuðÞd is non-increasing in R þ 0 : Throughout this article, the bracket pairing h, i is h’, i ¼ R



(’, )dx and is well-defined provided that (’, ) 2 L

1

().

Assumptions (F ) and (Q)-(a) make the definition of solution meaningful using (B)(i). Moreover, (B)(ii) and the structural assumptions imply that Eu is non-increasing in R þ 0 .

Before proving our main theorem on the local stability of solutions of (P

1

), we give some preliminary results. Note that in the proof of the next three lemmas we do not use the assumption F

t

u  0 in R þ 0 , required in (B)(ii).

L EMMA 3.1 There exist two numbers  2 ð,  0 Þ and c40 such that

ð f ðt, x, uÞ, uÞ  ajuj 2  cjuj q in R þ 0    R N : ð3:2Þ Furthermore, if u is a solution of (P

1

) then for all t 2 R þ 0

FuðtÞ   a

2 kuðt, Þk 2 2  c

q kuðt, Þk q q EuðtÞ  1

2 ku t ðt, Þk 2 2 þ a 4 1  

 0

 

kD L uðt, Þk 2 2 þ akuðt, Þk ~ 2 q  ~ckuðt, Þk q q ,

~c ¼ c

q and a ¼ ~ a

4s 2 q 1  

 0

 

4 0,

ð3:3Þ

where s q is given in (2.1) with h ¼ q.

Proof Fix  2 ð,  0 Þ. By (F )-(a) we have ( f (t, x, u), u)  ajuj

2

for all ðt, x, uÞ 2 R þ 0    R N , with juj5, provided  2 (0, 1] is sufficiently small. On the other hand, (F )-(b) and juj   imply that ( f (t, x, u), u)  (juj

1q

þ 1)juj

q



(

1q

þ 1)juj

q

, so that ( f (t, x, u), u)  cjuj

q

with c ¼ (

1q

þ 1) for all ðt, x, uÞ 2 R þ 0    R N such that juj  . Hence (3.2) holds, with c as large as we wish. Integrating (3.2), we obtain the first part of (3.3) along the solution u.

By (3.2) and the definition of E, we have in R þ 0 EuðtÞ  1

2 ku t ðt, Þk 2 2 þ a 4 1  

 0

 

kD L uðt, Þk 2 2 þ a 4 1  

 0

 

kD L uðt, Þk 2 2

 c

q kuðt, Þk q q ,

and so the second part of (3.3) follows by the application of (2.1) with h ¼ q. g

Downloaded By: [Autuori, Giuseppina] At: 19:18 15 March 2011

(8)

Consider the set

 ~ 0 ¼ fð, E Þ 2 R 2 : 0  5 z 0 , ðÞ  E 5 ~ E 0 g, where ðÞ ¼ 2 ~ a 2  ~c q

z 0 ¼ a ~

~c

  1=ðq2Þ

, E ~ 0 ¼ a ~ 2  1 q

 

z 2 0 ,

and the numbers ~ a and ~c are given in (3.3). Of course ~  0 is well defined, since q42 by (F )-(b). Without loss of generality, we also assume that ~ a= ~c  1, by taking c sufficiently large, if necessary. Hereafter, u will be a fixed solution of (P

1

) and (t) ¼ ku(t, )k

q

.

L EMMA 3.2 If ðð0Þ, Euð0ÞÞ 2 ~  0 , then ððtÞ, EuðtÞÞ 2 ~  0 for all t 2 R þ 0 . Moreover, in R þ 0

2Eu  ku t k 2 2 þ a 2 1  

 0

 

kD L uk 2 2  0: ð3:4Þ

Proof By (3.3) and (2.1), with h ¼ q, we immediately obtain for all t 2 R þ 0

EuðtÞ  2 ~ aðtÞ 2  ~cðtÞ q : ð3:5Þ Of course, (t)  0 for all t 2 R þ 0 . We next show that (t)5z

0

for all t 2 R þ 0 . To do this, since (0)5z

0

by assumption, by a continuity argument it is sufficient to prove that there cannot exist any t 2 R þ 0 such that (t) ¼ z

0

. Indeed, if such a t exists, then, by (3.5), using the monotonicity of Eu and the assumption Euð0Þ 5 ~ E 0 , we have

2 ~ az 2 0  ~cz q 0 ¼ ~ E 0 4 Euð0Þ  EuðtÞ  2 ~ aðtÞ 2  ~cðtÞ q ,

which is a contradiction. Hence 0  (t)5z

0

for all t 2 R þ 0 . This moreover implies that

~

aðtÞ 2  ~cðtÞ q  0 for all t 2 R þ 0 , ð3:6Þ and in turn, by (3.5) we immediately have Eu  0 in R þ 0 . Finally, relation (3.4) is an

immediate consequence of (3.3) and (3.6). g

Remark By (3.5) and (B)(ii), there exists an l  0 such that

t!1 lim EuðtÞ ¼ l: ð3:7Þ

L EMMA 3.3 If ðð0Þ, Euð0ÞÞ 2 ~  0 , then

a kD L uk 2 2 þ b kD L uk 2 2 þ kuk 2 2 þ h f ðt,  , uÞ, ui

 a 2 1  

 0

 

kD L uk 2 2 þ b kD L uk 2 2 þ kuk 2 2 : ð3:8Þ Proof By (3.2) and the Sobolev’s inequality we have

h f ðt,  , uÞ, ui  akuðt, Þk 2 2  ckuðt, Þk q q  a 

 0

kD L uðt, Þk 2 2  ckuðt, Þk q q :

Downloaded By: [Autuori, Giuseppina] At: 19:18 15 March 2011

(9)

Therefore

a kD L uk 2 2 þ b kD L uk 2 2 þ kuk 2 2 þ h f ðt,  , uÞ, ui

 a 1  

 0

 

kD L uk 2 2 þ b kD L uk 2 2 þ kuk 2 2  ckuðt, Þk q q : ð3:9Þ Now, using (2.1), with h ¼ q, from (3.9) we get

a kD L uk 2 2 þ b kD L uk 2 2 þ kuk 2 2 þ h f ðt,  , uÞ, ui

 a 2 1  

 0

 

kD L uk 2 2 þ b kD L uk 2 2 þ kuk 2 2 þ a

2s 2 q 1  

 0

 

ðtÞ 2  cðtÞ q

 a 2 1  

 0

 

kD L uk 2 2 þ b kD L uk 2 2 þ kuk 2 2 þ aðtÞ ~ 2  cðtÞ q , ð3:10Þ where ~ a is defined in (3.3). In [0, z

0

), the quantity ~ a(t)

2

 c(t)

q

 0, so that,

from (3.10) we get (3.8). g

L EMMA 3.4 If ðð0Þ, Euð0ÞÞ 2 ~  0 , then

ku t k 2 , kD L uk 2 , kD L1 uk 2 , kuk 2



L

2 L 1 ðR þ 0 Þ,

gkD L u t k 2 2 , hQðt,  , u, u t Þ, u t i, F t u 2 L 1 ðR þ 0 Þ: ð3:11Þ Proof The fact that ku

t

k

2

and kD

L

uk

2

are in L 1 ðR þ 0 Þ follows at once by (3.4), and so also kD L1 uk 2 2 L 1 ðR þ 0 Þ by Theorem 4.4.1 of [14], while the fact that kuk 2



L

2 L 1 ðR þ 0 Þ follows from the Sobolev’s inequality. Finally, condition (3.11)

2

follows at once.

Indeed, Du  0 in R þ 0 by (B)(ii) and the fact that (Q, u

t

)  0, and Du consists of three non-negative terms, and 0  R t

0 DuðÞd  Euð0Þ  EuðtÞ  Euð0Þ, being Eu  0 in R þ 0

by (3.4). g

L EMMA 3.5 For all t  T  0 Z t

T

gðÞkðÞhD L u, D L u t id  " 0 ðT Þ Z t

T

gðÞk 2 ðÞd

  1=2

, ð3:12Þ

where " 0 ðT Þ ¼ ðsup R

þ

0

kD L uðt, Þk 2 Þ  ð R 1

T gðÞkD L u t k 2 2 dÞ 1=2 ! 0 as T ! 1.

Proof By Ho¨lder’s inequality and (3.11) in R þ 0 jhD L uðt, Þ, D L u t ðt, Þij  sup t 2 R

þ

0

kD L uðt, Þk 2

 

kD L u t ðt, Þk 2 :

Then, by integration from T and t and another use of Ho¨lder’s inequality, we obtain Z t

T

gðÞkðÞhD L u, D L u t id  " 0 ðT Þ Z t

T

gðÞk 2 ðÞd

  1=2

,

where "

0

(T ) ! 0 as T ! 1 once more by (3.11). g Note that in the proof of the next lemma we do not use the assumption F

t

u  0 in R þ 0 , required in (B)(ii).

Downloaded By: [Autuori, Giuseppina] At: 19:18 15 March 2011

(10)

L EMMA 3.6 If l40 in (3.7), then there exists ¼ (l )40 such that

ku t k 2 2 þ a kD L uk 2 2 þ b kD L uk 2 2 þ kuk 2 2 þ h f ðt,  , uÞ, ui  : ð3:13Þ Proof Since Eu(t)  l for all t 2 R þ 0 it follows that

ku t k 2 2 þ a kD L uk 2 2 þ b kD L uk 2 2 þ kuk 2 2  2ðl  FuÞ in R þ 0 : ð3:14Þ Consider the sets

J 1 ¼ ft 2 R þ 0 : FuðtÞ  l=2g, J 2 ¼ ft 2 R þ 0 : FuðtÞ 4 l=2g: ð3:15Þ In J

1

we have

ku t k 2 2 þ a kD L uk 2 2 þ b kD L uk 2 2 þ kuk 2 2  l: ð3:16Þ Consequently, denoting by Lu the left-hand side of (3.13), and using Lemma 3.3, we get in J

1

Lu  ku t k 2 2 þ a 2 1  

 0

 

kD L uk 2 2 þ bkD L uk 2 2 þ kuk 2 2

 1 2 1  

 0

 

ku t k 2 2 þ akD L uk 2 2 þ bkD L uk 2 2 þ kuk 2 2

n o

 1  

 0

 

l

2 : ð3:17Þ

Let us consider J

2

. By (F )-(b) we have in R þ 0 jFuj   ½kuk 1 þ kuk q q 

: ð3:18Þ

and in turn we get

jFuðtÞj  C ½kD L uðt, Þk 2 þ kD L uðt, Þk q 2 , ð3:19Þ where C40 is an appropriate constant depending on  and on the Sobolev constant.

Hence, for t 2 J

2

l

2 5 FuðtÞ  2C kD L uðt, Þk 2 , if kD L uðt, Þk 2  1, kD L uðt, Þk q 2 , if kD L uðt, Þk 2 4 1,



that is,

kD L uðt, Þk 2  min l 4C , l

4C

  1=q

( )

¼ C 2 ¼ C 2 ðl Þ 4 0: ð3:20Þ

Therefore, by (3.8), for all t 2 J

2

it results that

LuðtÞ  a kD L uðt, Þk 2 2 þ b kD L uðt, Þk 2 2 þ kuðt, Þk 2 2 þ h f ðt,  , uðt, ÞÞ, uðt, Þi

 a 2 1  

 0

 

kD L uðt, Þk 2 2 þ b kD L uðt, Þk 2 2 þ kuðt, Þk 2 2

 a 2 1  

 0

 

C 2 2 þ bC 2 2 : ð3:21Þ

Downloaded By: [Autuori, Giuseppina] At: 19:18 15 March 2011

(11)

Thus, combining (3.17) with (3.21) we obtain the assertion, with

¼ ðl Þ ¼ 1 2 1  

 0

 

min l, aC  2 2

þ bC 2 2 , if J 2 6¼ ;,

l, if J 2 ¼ ;:

(

This completes the proof. g

We now possess all the tools to prove the main local stability results for (P

1

).

T HEOREM 3.7 Assume also (Q)-(b) and that there exists an auxiliary function k, satisfying either

k 2 CBVðR þ 0 ! R þ 0 Þ and k62 L 1 ðR þ 0 Þ or ðK 1 Þ

k 2 W 1,1 loc ðR þ 0 ! R þ 0 Þ, k 6 0 and lim

t!1

R t

0 jk 0 ðÞjd

R t

0 kðÞd ¼ 0, ðK 2 Þ

and assume that

lim inf

t!1

Z t 0

g k 2 d

  1=2

þAðkðtÞÞ

" # Z t

0

k d 5 1

, ð3:22Þ

where

AðkðtÞÞ ¼ BðkðtÞÞ þ Z t

0

1} k } d

  1=}

,

BðkðtÞÞ ¼ Z t

0

 1 k m d

  1=m

þ Z t

0

 2 k p d

  1=p

:

ð3:23Þ

If the initial data ku

t

(0, )k

2

and kD

L

u(0, )k

2

are sufficiently small, then (1.2) holds.

Proof First, note that if the data ku

t

(0, )k

2

and kD

L

u(0, )k

2

are sufficiently small then ðð0Þ, Euð0ÞÞ 2 ~  0 . Indeed, (0) ¼ ku(0, )k

q

5z

0

 1 if kD

L

u(0, )k

2

is sufficiently small by the continuity of the embedding X ,! L

q()

(), while the definitions of Eu and (3.19) give

Euð0Þ  1

2 ku t ð0, Þk 2 2 þ a þ b þ s 2

2 þ 2C

 

kD L uð0, Þk 2 ,

where s 2 is defined in (2.1), when h ¼ 2. This shows that Euð0Þ 5 ~ E 0 for sufficiently small data, and moreover Eu(0)  0 by (3.4).

Now, in order to prove that lim

t!1

Eu(t) ¼ 0, we proceed by contradiction assuming that l40 in (3.7) and we distinguish the cases (K

1

) and (K

2

).

Case (K

1

) First, assume that k 2 CBVðR þ 0 Þ \ C 1 ðR þ 0 Þ. Define a Lyapunov function by

VðtÞ ¼ kðtÞhu, u t i ¼ hu t , i,  ¼ kðtÞu 2 K:

Downloaded By: [Autuori, Giuseppina] At: 19:18 15 March 2011

(12)

From the distribution identity (A) we obtain for all t  T  0 VðÞ  t

T ¼ Z t

T

k 0 hu, u t i

 þ 2kku t k 2 2  k½ku t k 2 2 þ MðkD L uk 2 2 ÞkD L uk 2 2

þkuk 2 2 þ h f ð,  , uÞ, ui d

 Z t

T

gkhD L u, D L u t i d  Z t

T

khQð,  , u, u t Þ, uid: ð3:24Þ We now estimate the right–hand side of (3.24). First, note that

sup

R

þ

0

jhuðt, Þ, u t ðt, Þij  sup

R

þ

0

kuðt, Þk 2  ku t ðt, Þk 2 ¼ U 5 1 ð3:25Þ

by (3.11) of Lemma 3.4, that is kuk

2

, ku t k 2 2 L 1 ðR þ 0 Þ. Now, using Lemma 3.6 Z t

T

k ku  t k 2 2 þ MðkD L uk 2 2 ÞkD L uk 2 2 þ kuk 2 2 þ h f ð,  , uÞ, ui d 

Z t T

k d: ð3:26Þ Moreover, by Lemma 3.2 of [5], being certainly k 2 L 1 loc ðR þ 0 Þ, we have that

Z t T

kjhQð,  , u, u t Þ, uijd  " 1 ðT ÞBðkðtÞÞ, ð3:27Þ where

" 1 ðT Þ ¼ sup

R

þ

0

kuðt, Þk 2



L

 Z 1

T

DuðtÞdt

  1=m

0

þ Z 1

T

DuðtÞdt

  1=p

0

" #

¼ oð1Þ ð3:28Þ

as T ! 1, since kuðt, Þk 2



L

2 L 1 ðR þ 0 Þ by (3.11). Similarly, by (Q)-(b) and Lemma 3.3 of [5],

Z t T

kku t k 2 2 d  Z t

T

k d þ " 2 ðT ÞCð Þ Z t

0

1} k } d

  1=}

, ð3:29Þ

where Cð Þ ¼ ! 1=}

0

, ! ¼ supf 2 =!ðÞ :   ffiffiffiffiffiffiffiffiffiffiffi =jj p g, and

" 2 ðT Þ ¼ sup

R

þ

0

ku t ðt, Þk 2=} 2  Z 1

T

DuðtÞdt

  1=}

0

¼ oð1Þ ð3:30Þ

as T ! 1. Thus, using Lemma 3.5, from (3.24) it follows that VðÞ  t

T  U Z t

T

jk 0 jd þ 2 Z t

T

k d þ 2"ðT ÞCð Þ Z t

0

1} k } d

  1=}

 Z t

T

k d þ "ðT Þ Z t

0

gk 2 d

  1=2

þ"ðT ÞBðkðtÞÞ, ð3:31Þ where "(T ) ¼ max{"

0

(T ), "

1

(T ), "

2

(T )}. By (3.22) there exist a sequence t

i

% 1 and a number ‘40 such that

Z t

i

0

g k 2 d

  1=2

þAðkðt i ÞÞ  ‘ Z t

i

0

k d: ð3:32Þ

Downloaded By: [Autuori, Giuseppina] At: 19:18 15 March 2011

(13)

Choose ¼ (l ) ¼ /4 and fix T40 sufficiently large so that

"ðT Þ½2Cð Þ þ 1‘  =4, ð3:33Þ

being "(T ) ¼ o(1) as T ! 1. Consequently, for t

i

 T, Vðt i Þ  U

Z t

i

T

jk 0 jd þ SðT Þ  4

Z t

i

T

k d, ð3:34Þ

where U is given in (3.25) and SðT Þ ¼ VðT Þ þ "ðT Þ½2Cð Þ þ 1‘ R T

0 k d. Thus by (K

1

) we get

i!1 lim Vðt i Þ ¼ 1, ð3:35Þ

since k 0 2 L 1 ðR þ 0 Þ being k 2 CBVðR þ 0 Þ. On the other hand, by (3.25) and recalling that k is bounded, we have for all t 2 R þ 0

jVðtÞj  sup R

þ

0

k

 

kuðt, Þk 2  ku t ðt, Þk 2  sup R

þ

0

k

 

U: ð3:36Þ

This contradiction completes the first part of the proof.

We pass to the generale case k 2 CBVðR þ 0 Þ n C 1 ðR þ 0 Þ. Let k 2 C 1 ðR þ 0 Þ and G  R þ 0 be an open subset such that

(i) 2k  k  k in R þ 0 n G

0 in G ;



(ii) Var k  2Var k;

(iii) Z

G

k ds  1:

Clearly, k 2 CBVðR þ 0 Þ by (ii). We next prove that k satisfies (K

1

) and (3.22). Note that since k62 L 1 ðR þ 0 Þ, it is possible to find a value T

1

such that

Z T

1

0

k d  2: ð3:37Þ

Considering t  T

1

, by (i), (ii) and (3.37) we obtain Z t

0

k d  Z

½0,t n G

k d  Z t

0

k d  Z

G

k d  Z t

0

k d  1  1 2

Z t 0

k d: ð3:38Þ Hence k satisfies (K

1

). Moreover, by (i) and (3.38) for all t  T

1

Z t 0

g k 2 d

  1=2

þAðkðtÞÞ

( ) Z t

0

k d  4 Z t

0

g k 2 d

  1=2

þAðkðtÞÞ

( )Z t

0

k d,

where k ° A(k) is defined in (3.23). This shows that k also satisfies (3.22).

The general case is therefore reduced to the situation when k is smooth, and this completes the proof of case (K

1

).

Case (K

2

) We still obtain 

t

¼ k 0 u þ ku

t

, so that  2 K. By Lemmas 3.1–3.5 and Lemmas 3.2 and 3.3 of [5], that is (3.27) and (3.29) are available, as in case (K

1

), we obtain (3.34). The definition of V yields

jVðt i Þj  Ukðt i Þ  U kð0Þ þ Z t

i

0

jk 0 j d



,

Downloaded By: [Autuori, Giuseppina] At: 19:18 15 March 2011

(14)

so that by (3.34) 4

Z t

i

0

k d  2U Z t

i

0

jk 0 jd þ SðT Þ þ Ukð0Þ: ð3:39Þ First note that k62 L 1 ðR þ 0 Þ by (K

2

). Now, dividing (3.39) by R t

i

0 k d, we contradict (K

2

) as i ! 1.

Hence, in both cases, we have proved that Eu(t) approaches zero as t ! 1.

Finally, (3.4) and the facts that a40 and  2 ð,  0 Þ imply (1.2) at once. g We now provide some local stability results for the model (P

1

), under the requirement that k and g are related in a certain way. Note that in the next result we do not require (Q)-(b), and also (3.22) is relaxed.

T HEOREM 3.8 Suppose that there exists an auxiliary function k, satisfying either (K

1

) or (K

2

), and in addition

kðtÞ  Const: gðtÞ for all t sufficiently large, ð3:40Þ

lim inf

t!1

Z t 0

g k 2 d

  1=2

þ BðkðtÞÞ

" # Z t

0

k d 5 1,

ð3:41Þ

with k ° B(k) defined in (3.23). If the initial data ku

t

(0, )k

2

and kD

L

u(0, )k

2

are sufficiently small, then (1.2) holds.

Proof We can proceed as in Theorem 3.7 to prove that ðð0Þ, Euð0ÞÞ 2 ~  0 , when the initial data ku t ð0, Þk 2 2 and kD

L

u(0, )k

2

are sufficiently small. Hence let ðð0Þ, Euð0ÞÞ 2 ~  0 .

As before, we carry on with assuming l40 in (3.7) by contradiction and with distinguishing the two cases (K

1

) and (K

2

).

Case (K

1

) We start with k 2 CBVðR þ 0 Þ \ C 1 ðR þ 0 Þ and define the auxiliary function V given in proof of Theorem 3.7, obtaining once more (3.24)–(3.28), while Lemma 3.3 of [5] cannot be used any longer, so that (3.29) is now replaced by

Z t T

kku t k 2 2 d  Const:

Z t T

gkD L u t k 2 2  " 3 ðT Þ, ð3:42Þ where " 3 ðT Þ ¼ Const: R 1

T gkD L u t k 2 2 dt ¼ oð1Þ as T ! 1 by (3.11). Hence, putting (3.42) into (3.24) and using (3.12), (3.13), (3.25) and (3.27), we obtain

VðÞ  t T  U

Z t T

jk 0 jd þ 2" 3 ðT Þ  Z t

T

k d þ " 0 ðT Þ Z t

0

gk 2 d

  1=2

þ " 1 ðT ÞBðkðtÞÞ, ð3:43Þ

where "

0

(T ) is given in Lemma 3.5, "

1

(T ) in (3.28), and ¼ (l )40 in Lemma 3.6.

By (3.41) there exists a sequence t

i

% 1 and a number ‘40 such that Z t

i

0

g k 2 d

  1=2

þ Bðkðt i ÞÞ  ‘ Z t

i

0

k d: ð3:44Þ

Hence, (3.43) implies (3.34), where now SðT Þ ¼ VðT Þ þ "ðT Þ½2 þ ‘ R T o k d

and "(T ) ¼ max{"

0

(T ), "

1

(T ), "

3

(T )} is chosen so small that "(T )  3 /4‘.

Downloaded By: [Autuori, Giuseppina] At: 19:18 15 March 2011

(15)

Therefore, (K

1

) implies (3.35), being k 0 2 L 1 ðR þ 0 Þ. From now on the proof of the case (K

1

) can be continued as in Theorem 3.7.

Case (K

2

) It is possible to repeat the proof of Case (K

2

) in Theorem 3.7.

Thus, in both cases, l ¼ 0 and (1.2) follows at once, as shown in the proof of

Theorem 3.7. g

In the next corollary we present an application of Theorem 3.8, giving a concrete example of the auxiliary function k.

C OROLLARY 3.9 Let 

1

,  2 2 L 1 ðR þ 0 Þ. Suppose that g is either of class CBVðR þ 0 Þ n L 1 ðR þ 0 Þ or that g 2 W 1,1 loc ðR þ 0 Þ \ L 1 ðR þ 0 Þ, g ò 0 and jg 0 (t)j ¼ o(g(t)) as t ! 1. Then (1.2) holds, provided that the initial data ku

t

(0, )k

2

and kD

L

u(0, )k

2

are sufficiently small.

Proof It is sufficient to apply Theorem 3.8, with k ¼ g. Indeed, in the case g 2 CBVðR þ 0 Þ n L 1 ðR þ 0 Þ, conditions (K

1

) and (3.40) hold at once. On the other hand, if g 2 W 1,1 loc ðR þ 0 Þ \ L 1 ðR þ 0 Þ, g ò 0 and jg 0 (t)j ¼ o(g(t)) as t ! 1, then (K

2

), (3.40) are satisfied and g62 L 1 ðR þ 0 Þ, as in the previous case. Otherwise, g 2 L 1 ðR þ 0 Þ and by (K

2

) it follows that R 1

0 j g 0 ðÞjd ¼ 0, and so g Const. and in turn g 0 since g 2 L 1 ðR þ 0 Þ by assumption of contradiction, which is impossible, being g ò 0. Therefore

0  Z t

0

g k 2 d

  1=2

þ BðkðtÞÞ

" # Z t

0

k d

 Const:

Z t 0

gd

  1=2

! 0, as t ! 1, being g62 L 1 ðR þ 0 Þ in both cases. Hence, also condition (3.41) holds. g T HEOREM 3.10 Suppose that also (Q)-(b) holds and that there exists a function k 2 ½W 1,1 loc ðR þ 0 Þ \ L 1 ðR þ 0 Þ n L 1 ðR þ 0 Þ, satisfying (3.22),

jk 0 j  Const: ffiffiffiffiffi p gk

for t sufficiently large: ð3:45Þ If the initial data ku

t

(0, )k

2

and kD

L

u(0, )k

2

are sufficiently small, then (1.2) holds.

Proof We follow the technique used for the regular case of (K

1

) in Theorem 3.7, so that, without loss of generality, we assume that ðð0Þ, Euð0ÞÞ 2 ~  0 , since the initial data ku

t

(0, )k

2

and kD

L

u(0, )k

2

are sufficiently small, and get as before (3.24).

By (3.45) and the boundedness of ku

t

k

2

, it results jk 0 hu, u t ij  Const: ffiffiffiffiffi

p gk

kuk 2  ku t k 2  Const: ffiffiffiffiffi p gk

kD L u t k 2 : Consequently, by Ho¨lder’s inequality

Z t T

jk 0 hu, u t ijd  " 4 ðT Þ 1 þ Z t

T

k d

 

, ð3:46Þ

where " 4 ðT Þ ¼ R 1

T gkD L u t k 2 2 d

 1=2

¼ oð1Þ as T ! 1 by Lemma 3.4. Therefore, (3.31) becomes

VðÞ  t

T  " 4 ðT Þ 1 þ Z t

T

k d

 

þ 2 Z t

T

k d  Z t

T

k d

þ 2" 2 ðT ÞCð Þ Z t

0

1} k } d

  1=}

þ " 0 ðT Þ Z t

0

gk 2 d

  1=2

þ" 1 ðT ÞBðkðtÞÞ, ð3:47Þ

Downloaded By: [Autuori, Giuseppina] At: 19:18 15 March 2011

(16)

where ¼ (l )40 is given in Lemma 3.6, "

0

(T ) in Lemma 3.5, "

1

(T ) in (3.28) and "

2

(T ) in (3.30). By (3.22) there exist a sequence t

i

% 1 and a number ‘40 such that (3.32) holds for some ‘40. Put "(T ) ¼ max{"

0

(T ), "

1

(T ), "

2

(T ), "

4

(T )} ¼ o(1) as T ! 1 and choose ¼ /4 and T so large that "(T )  /4(1 þ ‘ þ 2C( )‘). Hence, from (3.47) it follows that

Vðt i Þ  SðT Þ  4

Z t

i

T

k d, ð3:48Þ

where SðT Þ ¼ VðT Þ þ "ðT Þ



1 þ ‘ þ 2Cð Þ‘  R T

0 k d. The proof can be completed as

in Theorem 3.7. g

In the next result we do not require (Q)-(b), and also (3.22) is relaxed.

T HEOREM 3.11 Let k 2 ½W 1,1 loc ðR þ 0 Þ \ L 1 ðR þ 0 Þ n L 1 ðR þ 0 Þ satisfy (3.40), (3.41) and (3.45). If the initial data ku

t

(0, )k

2

and kD

L

u(0, )k

2

are sufficiently small, then (1.2) holds.

Proof We start the proof recalling that ðð0Þ, Euð0ÞÞ 2 ~  0 , when the initial data ku

t

(0, )k

2

and kD

L

u(0, )k

2

are sufficiently small, as shown in the proof of Theorem 3.7. Let ðð0Þ, Euð0ÞÞ 2 ~  0 . We proceed as in last two theorems getting (3.42) and (3.46), so that (3.31) becomes

VðÞ  t

T  " 4 ðT Þ 1 þ Z t

T

k d

 

þ 2" 3 ðT Þ  Z t

T

k d

þ " 0 ðT Þ Z t

0

gk 2 d

  1=2

þ " 1 ðT ÞBðkðtÞÞ, ð3:49Þ where ¼ (l ) is given in Lemma 3.6, "

0

(T ) in Lemma 3.5, "

1

(T ) in (3.28), "

3

(T ) in (3.42) and "

4

(T ) in (3.46). By (3.41) we obtain (3.44). Taking "(T ) ¼ max{"

1

(T ),

"

0

(T ), "

3

(T ), "

4

(T )} ¼ o(1) as T ! 1 and taking T so large that "(T ) 3 /4(1 þ ‘), from (3.49) we get once again (3.48), where SðT Þ ¼ VðT Þ þ "ðT Þ ½3 þ ‘ R T

0 k d  . The

proof can be completed as in Theorem 3.7. g

C OROLLARY 3.12 Let 

1

,  2 2 L 1 ðR þ 0 Þ. Suppose that g 2 ½W 1,1 loc ðR þ 0 Þ \ L 1 ðR þ 0 Þn L 1 ðR þ 0 Þ, jg 0 (t)j ¼ O(g(t)) as t ! 1. Then (1.2) holds, provided that the initial data ku

t

(0, )k

2

and kD

L

u(0, )k

2

are sufficiently small.

Proof It is sufficient to apply Theorem 3.11, with k ¼ g, since g trivially verifies all the structural assumptions by the same main arguments used in the proof of

Corollary 3.9. g

4. Local asymptotic stability for (P

2

)

Throughout the section we assume (F ) in the version for (P

2

) and (Q)-(a), without further mentioning. Again X, K and K 0 are as always. For any  2 K, the total energy associated to (P

2

) is formally given in (E), but now

a ðtÞ ¼ 1 2



kD L ðt, Þk 2 2 þ MðkD L1 ðt, Þk 2 2 Þ þ kðt, Þk 2 2



: ð4:1Þ

Downloaded By: [Autuori, Giuseppina] At: 19:18 15 March 2011

(17)

We define a solution of (P

2

) as a function u 2 K satisfying (A) Distribution identity

hu t , i  t 0 ¼

Z t 0

fhu t ,  t i  hD L u, D L i  MðkD L1 uk 2 ÞhD L1 u, D L1 i

 gðÞhD L u t , D L i  hu þ Qð,  , u, u t Þ þ f ð,  , uÞ, igd

for all t 2 R þ 0 and  2 K.

(B) Conservation Law

ðiÞ Du :¼ hQðt,  , u, u t Þ, u t i þ gðÞkD L u t k 2 2  F t u 2 L 1 loc ðR þ 0 Þ ðiiÞ F t u  0, t ° EuðtÞ þ

Z t 0

DuðÞd is non-increasing in R þ 0 : In a similar manner as for the previous section, under (F ) and (Q)-(a) the definition of solution is meaningful using (B)(i). Once again condition (B)(ii), together with the structural assumptions given in Section 1, implies that Eu is non-increasing in R þ 0 .

The next result is the analogue of Lemma 3.1 for (P

2

).

L EMMA 4.1 There exist  2 ð,  0 Þ and c40 such that

ð f ðt, x, uÞ, uÞ  juj 2  cjuj q in R þ 0    R N : ð4:2Þ Furthermore, if u is a solution of (P

2

), then for all t 2 R þ 0

FuðtÞ   

2 kuðt, Þk 2 2  c

q kuðt, Þk q q , EuðtÞ  1

2 ku t ðt, Þk 2 2 þ 1 4 1  

 0

 

kD L uðt, Þk 2 2 þ akuðt, Þk ~ 2 q  ~ckuðt, Þk q q ,

~c ¼ c

q and a ¼ ~ 1

4s 2 q 1  

 0

 

4 0: ð4:3Þ

Proof Inequality (4.2) is obtained as the corresponding one in Lemma 3.1, but now using (F ), in the form suitable for (P

2

). The first relation in (4.3) then follows by integration on . Moreover,

1 4 1  

 0

 

kD L uðt, Þk 2 2  c

q kuðt, Þk q q  akuðt, Þk ~ 2 q  ~ckuðt, Þk q q ,

by (2.1), with h ¼ q. Thus, (4.2) and the definition of E imply (4.3) at once. g Consider, the set ~  0 , with z

0

, ~ E 0 and (), given as before, but with ~ a and ~c defined in (4.3). Again, without loss of generality, we also assume that ~ a= ~c  1, by taking c sufficiently large, if necessary. Hereafter in the section, u will be a fixed solution of (P

2

) and (t) ¼ ku(t, )k

q

. The next result is analogous to Lemma 3.2 for the model (P

2

).

L EMMA 4.2 If ðð0Þ, Euð0ÞÞ 2 ~  0 , then ððtÞ, EuðtÞÞ 2 ~  0 for all t 2 R þ 0 . Moreover, in R þ 0

2Eu  ku t k 2 2 þ 1 2 1  

 0

 

kD L uk 2 2  0: ð4:4Þ

Downloaded By: [Autuori, Giuseppina] At: 19:18 15 March 2011

(18)

Proof We follow the proof technique used in Lemma 3.2. By (4.3) and (2.1), with h ¼ q, we once more obtain (3.5), where now ~ a and ~c are defined in (4.3). The rest of the proof is exactly as for Lemma 3.2, with the only exception that (4.3) is used in place of (3.3), obtaining (4.4), since (3.6) continues to hold, with now ~ a and ~c given

in (4.3). g

L EMMA 4.3 If ðð0Þ, Euð0ÞÞ 2 ~  0 , then

kD L uk 2 2 þ a kD L1 uk 2 2 þ b kD L1 uk 2 2 þ kuk 2 2 þ h f ðt,  , uÞ, ui

 1 2 1  

 0

 

kD L uk 2 2 þ a kD L1 uk 2 2 þ b kD L1 uk 2 2 þ kuk 2 2 : ð4:5Þ Proof By (4.2) and Poincare´’s inequality, we immediately get h f ðt,  , uÞ, ui 

kuðt, Þk 2 2  ckuðt, Þk q q  kD L uk 2 2 = 0  ckuðt, Þk q q . Therefore

kD L uk 2 2 þ a kD L1 uk 2 2 þ b kD L1 uk 2 2 þ kuk 2 2 þ h f ðt,  , uÞ, ui

 1  

 0

 

kD L uk 2 2 þ a kD L1 uk 2 2 þ b kD L1 uk 2 2

þ kuk 2 2  ckuðt, Þk q q : ð4:6Þ

Now, using (2.1), with h ¼ q, from (4.6) we get

kD L uk 2 2 þ a kD L1 uk 2 2 þ b kD L1 uk 2 2 þ kuk 2 2 þ h f ðt,  , uÞ, ui

 1 2 1  

 0

 

kD L uk 2 2 þ a kD L1 uk 2 2 þ b kD L1 uk 2 2 þ kuk 2 2

þ 1

2s 2 q 1  

 0

 

ðtÞ 2  cðtÞ q

 1 2 1  

 0

 

kD L uk 2 2 þ a kD L1 uk 2 2 þ b kD L1 uk 2 2 þ kuk 2 2

þ aðtÞ ~ 2  cðtÞ q , ð4:7Þ

where ~ a is defined in (4.3). In [0, z

0

), the quantity ~ a(t)

2

 c(t)

q

 0, so that,

from (4.7) we get (4.5). g

Let us now point out that Lemmas 3.4 and 3.5 continue to hold also for the model (P

2

), as well as (3.7).

L EMMA 4.4 If l40 in (3.7), there exists ¼ (l )40 such that

ku t k 2 2 þ kD L uk 2 2 þ a kD L1 uk 2 2 þ b kD L1 uk 2 2 þ kuk 2 2 þ h f ðt,  , uÞ, ui  : ð4:8Þ Proof The proof is analogous of the one produced in Lemma 3.6. Indeed, denote by J

1

and J

2

the sets introduced in (3.15). In J

1

, we have

ku t k 2 2 þ kD L uk 2 2 þ a kD L1 uk 2 2 þ b kD L1 uk 2 2 þ kuk 2 2  l: ð4:9Þ

Downloaded By: [Autuori, Giuseppina] At: 19:18 15 March 2011

(19)

Consequently, denoting by Lu the left-hand side of (4.8) and using Lemma 4.3, we get

LuðtÞ  ku t k 2 þ 1 2 1  

 0

 

kD L uk 2 2 þ akD L1 uk 2 2 þ bkD L1 uk 2 2 þ kuk 2 2

 1 2 1  

 0

 

ku t k 2 þ kD L uk 2 2 þ akD L1 uk 2 2 þ bkD L1 uk 2 2 þ kuk 2 2

n o

,

that is, for all t 2 J

1

we have

LuðtÞ  1  

 0

 

l

2 : ð4:10Þ

In J

2

condition (3.18) is still valid, so that we get once more (3.19) and (3.20).

Moreover, by (4.5), for all t 2 J

2

it results that

LuðtÞ  kD L uk 2 2 þ a kD L1 uk 2 2 þ b kD L1 uk 2 2 þ kuk 2 2 þ h f ðt,  , uÞ, ui

 1 2 1  

 0

 

kD L uk 2 2 þ a kD L1 uk 2 2 þ b kD L1 uk 2 2 þ kuk 2 2

 1  

 0

 

C 2 2

2 , ð4:11Þ

by (3.20), being a, b,   0. Hence, combining (4.10) with (4.11) we obtain (4.8) with

¼ ðl Þ ¼ 1 2 1  

 0

  min l, C  2 2

, if J 2 6¼ ;,

l, if J 2 ¼ ;:

(

This completes the proof. g

T HEOREM 4.5 Suppose that also (Q)-(b) holds and that there exists an auxiliary function k, satisfying either (K

1

) or (K

2

), and (3.22). If the initial data ku

t

(0, )k

2

and kD

L

u(0, )k

2

are sufficiently small, then (1.2) holds.

Proof First, note that if the data ku

t

(0, )k

2

and kD

L

u(0, )k

2

are sufficiently small then ðð0Þ, Euð0ÞÞ 2 ~  0 . Indeed, (0) ¼ ku(0, )k

q

5z

0

 1 if kD

L

u(0, )k

2

is sufficiently small by the continuity of the embedding X ,! L

q()

(), while the definition of Eu and (3.19) give

Euð0Þ  1

2 ku t ð0, Þk 2 2 þ 1 þ ða þ bÞ þ s 2

2 þ 2C

 

kD L uð0, Þk 2 , ð4:12Þ

where  is derived from kD

L1

u(t, )k

2

  kD

L

u(t, )k

2

, while s 2 is defined in (2.1), with h ¼ 2. This shows that Euð0Þ 5 ~ E 0 for sufficiently small data, and finally Eu(0)  0 by (4.4).

Hence, let us assume that ðð0Þ, Euð0ÞÞ 2 ~  0 and, by contradiction, that l40 in (3.7).

Downloaded By: [Autuori, Giuseppina] At: 19:18 15 March 2011

(20)

Case (K

1

) Suppose k 2 CBVðR þ 0 Þ \ C 1 ðR þ 0 Þ. Define V(t) ¼ k(t)hu, u

t

i, as in Theorem 3.7, so that for all t  T  0

VðÞ  t T ¼

Z t T

k 0 hu, u t i

 þ 2kku t k 2 2  k½ku t k 2 2 þ kD L uk 2 2

þMðkD L1 uk 2 2 ÞkD L1 uk 2 2 þ kuk 2 2 þ h f ð,  , uÞ, ui d

 Z t

T

gkhD L u, D L u t i d  Z t

T

khQð,  , u, u t Þ, uid: ð4:13Þ Condition (3.25) follows from (3.11) given in Lemma 3.4. On the other hand, Lemmas 3.2 and 3.3 of [5] produce again (3.27) and (3.29). Finally, using (4.8) we get once more (3.31). Thus, the proof of the case (K

1

) can be completed exactly as in Theorem 3.7.

Case (K

2

) The proof is the same of Theorem 3.7.

Hence, in both cases, l ¼ 0 and (1.2) is simply a consequence of (4.4). g We give local stability results also for (P

2

), when k and g are related, without assuming (Q)-(b) and relaxing (3.22).

T HEOREM 4.6 Let k satisfy either (K

1

) or (K

2

), and (3.40)–(3.41). If the initial data ku

t

(0, )k

2

and kD

L

u(0, )k

2

are sufficiently small, then (1.2) holds.

Proof Let us suppose, without loss of generality, that ðð0Þ, Euð0ÞÞ 2 ~  0 , being the initial data sufficiently small, and proceed by contradiction assuming that l40 in (3.7).

Case (K

1

) Let k 2 CBVðR þ 0 Þ \ C 1 ðR þ 0 Þ and define the auxiliary function V, as in the proof of Theorem 4.5, obtaining once more (4.13). Condition (3.25) is now a consequence of Lemma 3.4, and Lemma 4.4 is in charge. Finally, (3.27) derives from Lemma 3.2 of [5], while Lemma 3.3 of [5] cannot be used longer, so that (3.29) must be replaced by (3.42). Hence, (3.43) follows at once and the proof can be carried on as in Theorem 3.8.

Case (K

2

) It is possible to repeat the proof of Theorem 3.8.

Hence, in both cases, l ¼ 0 and again (1.2) follows from (4.4). g C OROLLARY 4.7 Let 

1

,  2 2 L 1 ðR þ 0 Þ. Suppose that either g 2 CBVðR þ 0 Þ n L 1 ðR þ 0 Þ, or g 2 W 1,1 loc ðR þ 0 Þ \ L 1 ðR þ 0 Þ, g ò 0 and jg 0 (t)j ¼ o(g(t)) as t ! 1. Then (1.2) holds, provided that the initial data ku

t

(0, )k

2

and kD

L

u(0, )k

2

are sufficiently small.

Proof Following the proof steps of Corollary 3.9, we see that it is sufficient to apply Theorem 4.6, with k ¼ g.

T HEOREM 4.8 Let (Q)-(b) hold and suppose that there exists a function k of class

½W 1,1 loc ðR þ 0 Þ \ L 1 ðR þ 0 Þ n L 1 ðR þ 0 Þ, satisfying (3.22) and (3.45). If the initial data ku

t

(0, )k

2

and kD

L

u(0, )k

2

are sufficiently small, then (1.2) holds.

Proof Let ðð0Þ, Euð0ÞÞ 2 ~  0 , without loss of generality, being the initial data ku

t

(0, )k

2

and kD

L

u(0, )k

2

sufficiently small by assumption. Hence, proceed by contradiction, assuming once more that l40 in (3.7). We follow the technique used in the proof of Case (K

1

) in Theorem 4.5, with the only exception that the term R t

T k 0 hu, u t id in (4.13) is estimated as in the proof of Theorem 3.10. Hence, we get

Downloaded By: [Autuori, Giuseppina] At: 19:18 15 March 2011

Riferimenti

Documenti correlati

The physics to be pursued a t the electron-positron collider D A a N E in Frascati with the detectors DEAR, FZNUDA and KLOE is addressed, with particular emphasis on the test

The pitless extra virgin olive oil is obtained from the greenest, the freshest and the healthiest olives picked up by hand, by removing the pits from the olives first and then

[r]

[r]

 Command ls provides information about a file according to the specified options.  If the pathname is a directory, ls lists the files and subdirectories contained in that

For books be sure to include the chapter title, chapter author(s), editor(s) of the book, title of the book (including volume or edition number), publisher's name and location, date

We see +lines as a tool to prove something involving the Hilbert function of unions of curves and fat points.. For an alternative approach to such disjoint unions, see

Petriccione, “Sulla misurabilit`a delle famiglie dei sistemi di k iperpiani passanti per un punto fisso dello spazio affine A 7 ”, Atti Accad.. Petriccione, “Sulla misurabilit`a