Open Access

Global existence of solutions and energy decay for a Kirchhoff-type equation with nonlinear dissipation

Journal of Inequalities and Applications20132013:195

https://doi.org/10.1186/1029-242X-2013-195

Received: 25 August 2012

Accepted: 9 April 2013

Published: 22 April 2013

Abstract

This paper deals with the initial boundary value problem for a class of nonlinear Kirchhoff-type equation with dissipative term

u t t φ ( u 2 2 ) Δ u + a | u t | α 2 u t = b | u | β 2 u , x Ω , t > 0

in a bounded domain, where a , b > 0 and α , β > 2 are constants. We obtain the global existence of solutions by constructing a stable set in H 0 1 ( Ω ) and show the energy decay estimate by applying a lemma of Komornik.

MSC:35B40, 35L70.

Keywords

nonlinear Kirchhoff-type equationinitial boundary value problemstable setenergy decay estimate

1 Introduction

In this paper, we investigate the existence and asymptotic stability of global solutions for the initial boundary value problem of the following Kirchhoff-type equation with nonlinear dissipative term in a bounded domain
u t t φ ( u 2 2 ) Δ u + a | u t | α 2 u t = b | u | β 2 u , x Ω , t > 0 ,
(1.1)
u ( x , 0 ) = u 0 ( x ) , u t ( x , 0 ) = u 1 ( x ) , x Ω ,
(1.2)
u ( x , t ) = 0 , x Ω , t 0 ,
(1.3)
where Ω is a bounded domain in R n with a smooth boundary Ω, a , b > 0 and α , β > 2 are constants, φ ( s ) is a C 1 -class function on [ 0 , + ) satisfying
φ ( s ) m 0 , s φ ( s ) 0 s φ ( θ ) d θ , s [ 0 , + )
(1.4)

with m 0 1 is a constant.

If Ω = [ 0 , L ] is an interval of the real line, equation (1.1) describes a small amplitude vibration of an elastic string with fixed endpoints. The original equation is
ρ h u t t + δ u t + f = ( γ 0 + E h 2 L 0 L | u x | 2 d s ) u x x ,

where L is the rest length, E is the Young modulus, ρ is the mass density, h is the cross-section area, γ 0 is the initial axial tension, δ is the resistance modulus and f is a nonlinear perturbation effect.

When a = b = 0 , φ ( s ) = s r , r 1 and u 0 0 (the mildly degenerate case), the local existence of solutions in Sobolev space was investigated by many author [16]. Concerning a global existence of solutions for mildly degenerate Kirchhoff equations, it is natural to add a dissipative term u t or Δ u t .

For a = 1 , b = 0 , α = 2 , φ ( s ) = s r , r 1 , the problem (1.1)-(1.3) was treated by Nishihara and Yamada [7]. They proved the existence and uniqueness of a global solution u ( t ) for small data ( u 0 , u 1 ) ( H 0 1 ( Ω ) H 2 ( Ω ) ) × H 0 1 ( Ω ) with u 0 0 and the polynomial decay of the solution. Aassila and Benaissa [8] extended the global existence part of [7] to the case where φ ( s ) 0 with φ ( u 0 2 ) 0 and the case of nonlinear dissipative term case ( a 0 ).

In the case a = 0 , for large β and φ ( s ) r > 0 , D’Ancona and Spagnolo [9] proved that if u 0 , u 1 C 0 ( R n ) are small, then problem (1.1)-(1.3) has a global solution. The nondegenerate case with α = 2 , a > 0 and b = 0 was considered by De Brito, Yamada and Nishihara [1013], they proved that for small initial data ( u 0 , u 1 ) ( H 0 1 ( Ω ) H 2 ( Ω ) ) × H 0 1 ( Ω ) there exists a unique global solution of (1.1)-(1.3) that decays exponentially as t + .

When φ ( s ) 0 , Ghisi and Gobbino [14] proved the existence and uniqueness of a global solution u ( t ) of the problem (1.1)-(1.3) for small initial data ( u 0 , u 1 ) ( H 0 1 ( Ω ) H 2 ( Ω ) ) × H 0 1 ( Ω ) with m ( u 0 2 ) 0 and the asymptotic behavior ( u ( t ) , u t ( t ) , u t t ( t ) ) ( u , 0 , 0 ) in ( H 0 1 ( Ω ) H 2 ( Ω ) ) × H 0 1 ( Ω ) × L 2 ( Ω ) as t + , where either u = 0 or φ ( u 2 ) = 0 .

The case φ ( s ) r > 0 has been considered by Hosoya and Yamada [15] under the following condition:
0 β < 2 n 4 , n 5 ; 0 β < + , n 4 .

They proved that, if the initial datas are small enough, the problem (1.1)-(1.3) has a global solution which decays exponentially as t + .

In this paper, we prove the global existence for the problem (1.1)-(1.3) by applying the potential well theory introduced by Sattinger [16] and Payne and Sattinger [17]. Meanwhile, we obtain the asymptotic stability of global solutions by use of the lemma of Komornik [18].

We adopt the usual notation and convention. Let H m ( Ω ) denote the Sobolev space with the norm u H m ( Ω ) = ( | α | m D α u L 2 ( Ω ) 2 ) 1 2 , H 0 m ( Ω ) denotes the closure in H m ( Ω ) of C 0 ( Ω ) . For simplicity of notation, hereafter we denote by p the Lebesgue space L p ( Ω ) norm, denotes L 2 ( Ω ) norm and we write equivalent norm instead of H 0 1 ( Ω ) norm H 0 1 ( Ω ) . Moreover, M denotes various positive constants depending on the known constants and it may be difference in each appearance.

This paper is organized as follows: In the next section, we will give some preliminaries. Then in Section 3, we state the main results and give their proof.

2 Preliminaries

In order to state and prove our main results, we first define the following functionals:
K ( u ) = m 0 u 2 b u β β , J ( u ) = m 0 2 u 2 b β u β β ,
for u H 0 1 ( Ω ) . Then we define the stable set S by
S = { u H 0 1 ( Ω ) , K ( u ) > 0 } { 0 } .
We denote the total energy functional associated with (1.1)-(1.3) by
E ( t ) = 1 2 u t 2 + 1 2 0 u 2 φ ( s ) d s b β u β β
(2.1)

for u H 0 1 ( Ω ) , t 0 , and E ( 0 ) = 1 2 u 1 2 + 1 2 0 u 0 2 φ ( s ) d s b β u 0 β β is the total energy of the initial data.

Lemma 2.1 Let q be a number with 2 q < + , n 2 and 2 q 2 n n 2 , n > 2 . Then there is a constant C depending on Ω and q such that
u q C u H 0 1 ( Ω ) , u H 0 1 ( Ω ) .

Lemma 2.2 [18]

Let y ( t ) : R + R + be a nonincreasing function and assume that there are two constants μ 1 and A > 0 such that
s + y ( t ) μ + 1 2 d t A y ( s ) , 0 s < + ,

then y ( t ) C y ( 0 ) ( 1 + t ) 2 μ 1 , t 0 , if μ > 1 , where C is positive constants independent of y ( 0 ) .

Lemma 2.3 Let u ( t , x ) be a solutions to the problem (1.1)-(1.3). Then E ( t ) is a nonincreasing function for t > 0 and
d d t E ( t ) = a u t ( t ) α α .
(2.2)
Proof By multiplying equation (1.1) by u t and integrating over Ω, we get
d d t E ( u ( t ) ) = a u t ( t ) α α 0 .

Therefore, E ( t ) is a nonincreasing function on t. □

We state a local existence result, which is known as a standard one (see [6, 19]).

Theorem 2.1 Suppose that α , β > 2 satisfy
2 < β < + , n 2 ; 2 < β 2 ( n 1 ) n 2 , n > 2 ,
(2.3)
2 < α < + , n 2 ; 2 < α 2 n n 2 , n > 2 ,
(2.4)
and let ( u 0 , u 1 ) H 0 1 ( Ω ) × L 2 ( Ω ) . Then there exists T > 0 such that the problem (1.1)-(1.3) has a unique local solution u ( t ) in the class
u C ( [ 0 , T ) ; H 0 1 ( Ω ) ) , u t C ( [ 0 , T ) ; L 2 ( Ω ) ) L α ( Ω × [ 0 , T ) ) .
(2.5)

In order to prove the existence of global solutions of the problem (1.1)-(1.3), we need the following lemma.

Lemma 2.4 Supposed that (2.3) holds, If u 0 S , u 1 L 2 ( Ω ) such that
δ = b C β ( 2 β ( β 2 ) m 0 E ( 0 ) ) β 2 2 < 1 ,
(2.6)

then u S , for each t [ 0 , T ) .

Proof Assume that there exists a number t [ 0 , T ) such that u ( t ) S on [ 0 , t ) and u ( t ) S . Then we have
K ( u ( t ) ) = 0 , u ( t ) 0 .
(2.7)
Since u ( t ) S on [ 0 , t ) , it holds that
J ( u ( t ) ) = m 0 2 u ( t ) 2 b β u ( t ) β β m 0 2 u ( t ) 2 m 0 β u ( t ) 2 = ( β 2 ) m 0 2 β u ( t ) 2 ,
(2.8)
we have from K ( u ( t ) ) = 0 that
J ( u ( t ) ) = m 0 2 u ( t ) 2 b β u ( t ) β β = m 0 2 u ( t ) 2 m 0 β u ( t ) 2 = ( β 2 ) m 0 2 β u ( t ) 2 ,
(2.9)
we conclude from (1.4) and (2.1) that
E ( t ) 1 2 u t ( t ) 2 + m 0 2 u ( t ) 2 b β u ( t ) β β = 1 2 u t ( t ) 2 + J ( u ( t ) ) .
(2.10)
Therefore, we obtain from (2.8), (2.9) and (2.10) that
u ( t ) 2 2 β ( β 2 ) m 0 J ( u ( t ) ) 2 β ( β 2 ) m 0 E ( t ) 2 β ( β 2 ) m 0 E ( 0 ) ,
(2.11)

for t [ 0 , t ] .

By exploiting Lemma 2.1, (2.6) and (2.11), we easily arrive at
b u ( t ) β β b C β u ( t ) β = b C β u ( t ) β 2 u ( t ) 2 b C β ( 2 β ( β 2 ) m 0 E ( 0 ) ) β 2 2 u ( t ) 2 < u ( t ) 2 ,
(2.12)

for all t [ 0 , t ] .

Therefore, we obtain
K ( u ( t ) ) = m 0 u ( t ) 2 b u ( t ) β β > 0 ,
(2.13)

which contradicts (2.7). Thus, we conclude that u ( t ) S on [ 0 , T ) . □

3 The global existence and nonexistence

Theorem 3.1 Suppose that (2.3) and (2.4) hold, and u ( t ) is a local solution of problem (1.1)-(1.3) on [ 0 , T ) . If u 0 S and u 1 L 2 ( Ω ) satisfy (2.6), then u ( x , t ) is a global solution of the problem (1.1)-(1.3).

Proof It suffices to show that u ( t ) 2 + u t ( t ) 2 is bounded independently of t.

Under the hypotheses in Theorem 3.1, we get from Lemma 2.4 that u ( t ) S on [ 0 , T ) . So the formula (2.8) holds on [ 0 , T ) .

Therefore, we have from (2.8) that
1 2 u t 2 + ( β 2 ) m 0 2 β u ( t ) 2 1 2 u t ( t ) 2 + J ( u ( t ) ) = E ( t ) E ( 0 ) .
(3.1)
Hence, we get
u t ( t ) 2 + u ( t ) 2 max ( 2 , 2 β ( β 2 ) m 0 ) E ( 0 ) < + .

The above inequality and the continuation principle lead to the global existence of the solution, that is, T = + . Thus, the solution u ( t ) is a global solution of the problem (1.1)-(1.3). □

Now we employ the analysis method to discuss the solution of the problem (1.1)-(1.3) occurs blow-up in finite time. Our result reads as follows.

Theorem 3.2 Assume that (i) 2 < β < 2 n n 2 , if n > 2 ; (ii) 0 < β < + , if n 2 . If u 0 S and u 1 L 2 ( Ω ) such that
E ( 0 ) < Q 0 , u 0 β > S 0 ,
where
Q 0 = ( β 2 ) b 2 β ( m 0 b C 2 ) β β 2 , S 0 = ( m 0 b C 2 ) 1 β 2

with C > 0 is a positive Sobolev constant. Then the solution of the problem (1.1)-(1.3) does not exist globally in time.

Proof On the contrary, under the conditions in Theorem 3.2, suppose that u ( x , t ) is a global solution of the problem (1.1)-(1.3); then by Lemma 2.1, it is well known that there exists a constant C > 0 depending only n, β such that u β C u for all u H 0 1 ( Ω ) .

From the above inequality, we conclude that
u 2 C 2 u β 2 .
(3.2)
It follows from (1.4), (2.1) and (3.2) that
E ( t ) = 1 2 u t 2 + 1 2 0 u 2 φ ( s ) d s b β u β β m 0 2 u 2 b β u β β m 0 2 C 2 u β 2 b β u β β .
(3.3)
Setting
s = s ( t ) = u ( t ) β = { Ω | u ( x , t ) | β d x } 1 β .
We denote the right side of (3.3) by Q ( s ) = Q ( u ( t ) β ) , then
Q ( s ) = m 0 2 C 2 s 2 b β s β , s 0 .
(3.4)
By (3.4), we obtain
Q ( s ) = m 0 C 2 s b s β 1 .

Let Q ( s ) = 0 , then we obtain S 0 = ( m 0 b C 2 ) 1 β 2 .

As s = S 0 , we have
Q ( s ) | s = S 0 = ( m 0 C 2 b ( β 1 ) s β 2 ) | s = S 0 = m 0 ( β 2 ) C 2 < 0 .
Consequently, the function Q ( s ) has a single maximum value Q 0 at S 0 , where
Q 0 = Q ( S 0 ) = ( β 2 ) b 2 β ( m 0 b C 2 ) β β 2 .

Since the initial data is such that E ( 0 ) , s ( 0 ) satisfies E ( 0 ) < Q 0 , u 0 β > S 0 .

Therefore, we have from Lemma 2.3 that
E ( t ) E ( 0 ) < Q 0 , t > 0 .
At the same time, by (3.3) and (3.4) it is evident that there can be no time t > 0 for which
E ( t ) < Q 0 , s ( t ) = S 0 .

Hence, we have also s ( t ) > S 0 for all t > 0 from the continuity of E ( t ) and s ( t ) .

According to the above contradiction we know that the global solution of the problem (1.1)-(1.3) does not exist, i.e., the solution blows up in some finite time.

This completes the proof of Theorem 3.2. □

4 Energy decay estimate

The following theorem shows the asymptotic behavior of global solutions of the problem (1.1)-(1.3).

Theorem 4.1 If the hypotheses in Theorem  3.2 are valid, then the global solutions of the problem (1.1)-(1.3) has the following asymptotic property:
E ( t ) M ( 1 + t ) 2 α 2 ,

where M > 0 is a constant depending on initial energy E ( 0 ) .

Proof Multiplying by E ( t ) α 2 2 u on both sides of the equation (1.1) and integrating over Ω × [ S , T ] , we obtain that
0 = S T Ω E ( t ) α 2 2 u [ u t t φ ( u 2 ) Δ u + a | u t | α 2 u t b u | u | β 2 ] d x d t ,
(4.1)

where 0 S < T < + .

Since
S T Ω E ( t ) α 2 2 u u t t d x d t = Ω E ( t ) α 2 2 u u t d x | S T S T Ω E ( t ) α 2 2 | u t | 2 d x d t α 2 2 S T Ω E ( t ) α 4 2 E ( t ) u u t d x d t .
(4.2)
So, substituting the formula (4.2) into the right-hand side of (4.1), we get that
0 = S T E ( t ) α 2 2 ( u t 2 + φ ( u 2 ) u 2 2 b β u β β ) d t S T Ω E ( t ) α 2 2 [ 2 | u t | 2 a | u t | α 2 u t u ] d x d t α 2 2 S T Ω E ( t ) α 4 2 E ( t ) u u t d x d t + Ω E ( t ) α 2 2 u u t d x | S T + ( 2 β 1 ) b S T E ( t ) α 2 2 u β β d t .
(4.3)
We obtain from (2.12) and (2.11) that
b ( 1 2 β ) u β β δ β 2 β u 2 δ β 2 β 2 β ( β 2 ) m 0 E ( t ) = 2 δ m 0 E ( t ) .
(4.4)
We derive from (1.4) that
0 u 2 φ ( s ) d s φ ( u 2 ) u 2 .
(4.5)
It follows from (4.3), (4.4) and (4.5) that
2 ( 1 δ m 0 ) S T E ( t ) α 2 d t S T Ω E ( t ) α 2 2 [ 2 | u t | 2 a | u t | α 2 u t u ] d x d t + α 2 2 S T Ω E ( t ) α 4 2 E ( t ) u u t d x d t Ω E ( t ) α 2 2 u u t d x | S T .
(4.6)
We have from Lemma 2.1 and (3.1) that
| α 2 2 S T Ω E ( t ) α 4 2 E ( t ) u u t d x d t | α 2 2 S T E ( t ) α 4 2 ( E ( t ) ) ( 1 2 u 2 + 1 2 u t 2 ) d t α 2 2 S T E ( t ) α 4 2 E ( t ) ( β C 2 ( β 2 ) m 0 ( β 2 ) m 0 2 β u 2 + 1 2 u t 2 ) d t α 2 2 max ( β C 2 ( β 2 ) m 0 , 1 ) S T E ( t ) α 2 2 E ( t ) d t = α 2 α max ( β C 2 ( β 2 ) m 0 , 1 ) E ( t ) α 2 | S T M E ( S ) α 2 ,
(4.7)
similarly, we have
| Ω E ( t ) α 2 2 u u t d x | S T | max ( β C 2 ( β 2 ) m 0 , 1 ) E ( t ) α 2 | S T M E ( S ) α 2 .
(4.8)
Substituting the estimates (4.7) and (4.8) into (4.6), we conclude that
2 ( 1 δ m 0 ) S T E ( t ) α 2 d t S T Ω E ( t ) α 2 2 [ 2 | u t | 2 a | u t | α 2 u t u ] d x d t + M E ( S ) α 2 .
(4.9)
We get from Young inequality and Lemma 2.3 that
2 S T Ω E ( t ) α 2 2 | u t | 2 d x d t S T Ω ( ε 1 E ( t ) α 2 + M ( ε 1 ) | u t | α ) d x d t M ε 1 S T E ( t ) α 2 d t + M ( ε 1 ) S T u t α α d t = M ε 1 S T E ( t ) α 2 d t M ( ε 1 ) a ( E ( T ) E ( S ) ) M ε 1 S T E ( t ) α 2 d t + M E ( S ) .
(4.10)
From Young inequality, Lemma 2.1, Lemma 2.3 and (2.11), We receive that
a S T Ω E ( t ) α 2 2 u u t | u t | α 2 d x d t a S T E ( t ) α 2 2 ( ε 2 u α α + M ( ε 2 ) u t α α ) d t a C α ε 2 E ( 0 ) α 2 2 S T u α d t + a M ( ε 2 ) E ( S ) α 2 2 S T u t α α d t = a C α ε 2 E ( 0 ) α 2 2 S T ( 2 β ( β 2 ) m 0 E ( t ) ) α 2 d t + M ( ε 2 ) E ( S ) α 2 2 ( E ( S ) E ( T ) ) C α ε 2 E ( 0 ) α 2 2 ( 2 β ( β 2 ) m 0 ) α 2 S T E ( t ) α 2 d t + M E ( S ) α 2 .
(4.11)
Choosing small enough ε 1 and ε 2 , such that
1 2 [ M ε 1 + E ( 0 ) α 2 2 ( 2 β C 2 ( β 2 ) m 0 ) α 2 ε 2 ] + δ m 0 < 1 ,
then, substituting (4.10) and (4.11) into (4.9), we get
S T E ( t ) α 2 d t M E ( S ) + M E ( S ) α 2 M ( 1 + E ( 0 ) ) α 2 2 E ( S ) .
Therefore, we have from Lemma 2.2 that
E ( t ) M ( 1 + t ) α 2 2 , t [ 0 , + ) .

The proof of Theorem 4.1 is thus finished. □

Declarations

Acknowledgements

This work was supported by the Natural Science Foundation of China (No. 61273016), the Natural Science Foundation of Zhejiang Province (No. Y6100016), the Middle-aged Academic Leader of Zhejiang University of Science and Technology (2008-2012), Interdisciplinary Pre-research Project of Zhejiang University of Science and Technology (2010-2012) and Zhejiang province universities scientific research key project (Z201017584).

Authors’ Affiliations

(1)
Department of Mathematics and Information Science, Zhejiang University of Science and Technology

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