Existence, general decay and blow-up of solutions for a viscoelastic Kirchhoff equation with Balakrishnan-Taylor damping and dynamic boundary conditions
aa r X i v : . [ m a t h . A P ] J un Existence, general decay and blow-up of solutions for aviscoelastic Kirchhoff equation with Balakrishnan-Taylordamping and dynamic boundary conditions
Gang Li, Biqing Zhu, Danhua WangCollege of Mathematics and Statistics, Nanjing University of Information Science andTechnology, Nanjing 210044, China.E-mail: [email protected] . Abstract
Our aim in this article is to study a nonlinear viscoelastic Kirchhoff equation withstrong damping, Balakrishnan-Taylor damping, nonlinear source and dynamical boundarycondition. Firstly, we prove the local existence of solutions by using the Faedo-Galerkinapproximation method combined with a contraction mapping theorem. We then prove thatif the initial data enter into the stable set, the solution globally exists, and if the initial dataenter into the unstable set, the solution blows up in a finite time. Moreover, we obtain ageneral decay result of the energy, from which the usual exponential and polynomial decayrates are only special cases.
Keywords:
Blow-up, stable and unstable set, global solutions, viscoelastic equation,strong damping, Balakrishnan-Taylor damping, dynamic boundary conditions.
AMS Subject Classification (2000):
In this paper, we are concerned with the following problem: u tt − M ( t )∆ u + Z t g ( t − s )∆ u ( x, s )d s + αu t − β ∆ u t = | u | p − u, x ∈ Ω , t > ,u ( x, t ) = 0 , x ∈ Γ , t > ,u tt ( x, t ) = − M ( t ) ∂u∂ν ( x, t ) + Z t g ( t − s ) ∂u∂ν ( x, s )d s − β ∂u t ∂ν ( x, t ) − γ | u t | m − u t ( x, t ) , x ∈ Γ , t > ,u ( x,
0) = u ( x ) , u t ( x,
0) = u ( x ) , x ∈ Ω , (1.1)where M ( t ) = a + b k∇ u ( t ) k + σ Z Ω ∇ u ( t ) ∇ u t ( t )d x , Ω is a regular and bounded domain of R N , ( N ≥ ∂ Ω = Γ ∪ Γ , mes (Γ ) >
0, Γ ∩ Γ = ∅ , ∂∂ν denotes the unit outer normalderivative, m ≥ p > a , b , σ , α , β , γ are positive constants. The relaxation function g is a positive and uniformly decaying function. The function u , u are given initial data.From the mathematical point of view, these problems like (1.1) take into account acceler-ation terms on the boundary. Such type of boundary conditions are usually called dynamicboundary conditions. They come from several physical applications and play an importantrole in different dimension space (see [3], [6], [10], [27] for more details).1rom the theoretical point of view, such as in the one-dimensional case, many existence,uniqueness and decay results have been established (see [17], [18], [19], [20], [41], [42], [55] formore details). For example, in [19] the author considered the following problem u tt − u xx − u txx = 0 , x ∈ (0 , L ) , t > ,u (0 , t ) = 0 , t > ,u tt ( L, t ) = − [ u x + u tx ]( L, t ) , t > ,u ( x,
0) = u ( x ) , u t ( x,
0) = v ( x ) , x ∈ (0 , L ) ,u ( L,
0) = η, u t ( L,
0) = µ t > . (1.2)By using the theory of B-evolutions and theory of fractional powers, the author proved thatproblem (1.2) gives rise to an analytic semigroup in an appropriate functional space and ob-tained the existence and the uniqueness of solutions. For a problem related to (1.2), anexponential decay result was obtained in [18], which describes the weakly damped vibrationsof an extensible beam. Later, Zang and Hu [55], considered the problem u tt − p ( u x ) xt − q ( u x ) x = 0 , x ∈ (0 , l ) , t > ,u (0 , t ) = 0 , t ≥ , ( p ( u x ) t + q ( u x )( l, t ) + ku tt ( l, t )) = 0 , t ≥ ,u ( x,
0) = u ( x ) , u t ( x,
0) = u ( x ) , x ∈ (0 , l ) . By using the Nakao inequality, and under appropriate conditions on p and q , they establishedboth exponential and polynomial decay rates for the energy depending on the form of theterms p and q . In [42] the authors considered the following linear wave equation with strongdamping and dynamical boundary conditions u tt − u xx − αu txx = 0 , x ∈ (0 , l ) , t > ,u (0 , t ) = 0 , t > ,u tt ( l, t ) = − ε [ u x + αu tx + ru t ]( l, t ) , t > , (1.3)as an alternative model for the classical ODE, namely m u ′′ ( t ) + d u ′ ( t ) + k ( t ) = 0 . (1.4)Based on the semigroup theory, spectral perturbation analysis and dominant eigenvalues, theycompared analytically these two approaches to the same physical system. Later, Pellicer in [41]considered (1.3) with a control acceleration εf ( u ( l, t ) , u t ( l, t ) / √ ε ) as a model for a controlledspring-mass-damper system and established some results concerning its large time behavior.By using the invariant manifold theory, the author proved that the infinite dimensional systemadmits a two dimensional invariant manifold where the equation is well represented by aclassical nonlinear oscillations ODE (1.4), which can be exhibited explicitly.2n the multi-dimensional cases, we can cite ([22], [26], [48]) for problems with the Dirichletboundary conditions and ([45], [46], [47]) for the Cauchy problems. Recently, Gerbi and Said-Houair in ([23], [24]) studied problem (1.1) with M ≡ g , linear damping αu t . They showed in [23] that if the initial data are large enough then theenergy and the L p norm of the solution of the problem is unbounded, grows up exponentiallyas time goes to infinity. Later in [24], they established the global existence and asymptoticstability of solutions starting in a stable set by combining the potential well method and theenergy method. A blow-up result for the case m = 2 with initial data in the unstable set wasalso obtained. For the other works, we refer the readers to ([6], [27], [28], [43], [50], [52]) andthe references therein.Many authors have investigated the existence, decay and blow-up results for nonlinearviscoelastic wave equation with boundary dissipation (see [31], [33], [35], [36]). For example,Cavalcanti et al [15] considered the following equation u tt − ∆ u + Z t g ( t − τ )∆ u ( τ )d τ + a ( x ) u t + | u | γ u = 0 , ( x, t ) ∈ Ω × (0 , ∞ )where Ω is a bounded domain of R n ( n ≥
1) with smooth boundary ∂ Ω, r > a : Ω → R + is a bounded function, which may vanish on a part of the domain. g is a positive nonincreasingfunction defined on R + . Under the condition that a ( x ) ≥ a > ω ⊂ Ω, with ω satisfyingsome geometry restrictions, when − ξ g ( t ) ≤ g ′ ( t ) ≤ − ξ g ( t ), t ≥
0, for some positive constants ξ and ξ , they proved an exponential rate of decay. Later, S. Berrimi and S. A. Messaoudi[7] improved Cavalcanti’s result by introducing a different functional, which allowed to weakenthe conditions on both a and g . Recently, in [25], Gerbi and Said-Houair studied problem(1.1) with M ≡
1, without the linear damping αu t and got the existence, exponential growthresults. Jeong et al [29] concerded the following problem and proved the general energy decayfor nonlinear viscoelastic wave equation with boundary damping. u ′′ − ∆ u − α ∆ u ′ + Z t h ( t − τ ) div [ a ( x ) ∇ u ( τ )]d τ = 0 , in Ω × (0 , ∞ ) ,u = 0 , on Γ × (0 , ∞ ) ,u ′′ + ∂u∂ν + α ∂u ′ ∂ν − Z t h ( t − τ )[ a ( x ) ∇ u ( τ )] · ν d τ + | u ′ | m − u ′ = | u | p − u, on Γ × (0 , ∞ ) ,u ′ ( x,
0) = u ( x ) , u ( x,
0) = u ( x ) , in Ω . When M ( t ) = a + b k∇ u ( t ) k + σ Z Ω ∇ u ( t ) ∇ u t ( t )d x , with Balakrishnan-Taylor damping( σ = 0), equation (1.1) describes the motion of a deformable solid with a hereditary effect.The more details about the stability and the well-posedness of the system, we refer the readersto ([2], [13], [14], [40]). On the contrary, with Balakrishnan-Taylor damping ( σ = 0) and g = 0,was initially proposed by Balakrishnan and Taylor in 1989 [4] and Bass and Zes [5]. It is used3o solve the spillover problem. The related problems also concerned by Tatar and Zarai in([44], [53], [54]), Mu in [39] and Wu in [51].Motivated by the above works, we intend to study the local existence, global existence,general decay and blow-up of solutions to problem (1.1). The main difficulties we encounterhere arise from the simultaneous appearance of the viscoelastic term, the strong dampingterm, the Balakrishnan-Taylor damping, the nonlinear source term and the nonlinear boundarydamping term. We will show that if the initial data is in the stable set, the solution is global.Otherwise, if the initial data is in the unstable set, by the concavity technique, we get thesolution will blow up in a finite time (we can see [34] for more details). Then, for the relaxationfunction g satisfies ( G g , which are not necessarily decaying like polynomial orexponential functions.The paper is organized as follows. In Section 2, we present some notations and materialneeded for our work. In Section 3, we establish the local existence. In section 4, the globalexistence for solutions is given. In Section 5, we prove the general decay result. A finite timeblows-up result for initial data in the unstable set is given in Section 6. In this section we present some material needed in the proof of our result. For the relaxationfunction g , we assume( G g : R + −→ R + is a nonincreasing differentiable function satisfying g (0) > , a − Z ∞ g ( s )d s = l > . ( G
2) There exists a nonincreasing differentiable function ξ : R + −→ R + such that g ′ ( s ) ≤ − ξ ( s ) g ( s ) , ∀ s ∈ R + . We introduce the set H = { u ∈ H (Ω) | u | Γ0 = 0 } , V = H (Ω) ∩ L m (Γ ). By ( · , · ) wedenote the scalar product in L (Ω); i.e.,( u, v )( t ) = Z Ω u ( x, t ) v ( x, t )d x. Also, by k · k q we mean the L q (Ω) norm for 1 ≤ q ≤ ∞ , by k · k q, Γ the L q (Γ ) norm. We willalso use the embedding (see [1]): H ֒ → L q (Γ ), 2 ≤ q ≤ q , where q satifies q = ( N − N − > , if N ≥ , + ∞ , if N = 1 , . We introduce the following functionals as in ([8], [37], [38]) I ( t ) := I ( u ( t )) = (cid:18) a − Z t g ( s )d s (cid:19) k∇ u ( t ) k + ( g ◦ ∇ u )( t ) − k u ( t ) k pp + b k∇ u ( t ) k , (2.1)4 ( t ) := J ( u ( t )) = 12 (cid:20)(cid:18) a − Z t g ( s )d s (cid:19) k∇ u ( t ) k + b k∇ u ( t ) k + ( g ◦ ∇ u )( t ) (cid:21) − p k u ( t ) k pp , (2.2) E ( t ) := E ( u ( t ) , u t ( t )) = J ( t ) + 12 k u t ( t ) k + 12 k u t ( t ) k , Γ , (2.3)where ( g ◦ ∇ u )( t ) = Z t g ( t − s ) k∇ u ( t ) − ∇ u ( s ) k ds ≥ . We define the potential well depth as d ( t ) = inf u ∈ H (Ω) \{ } sup λ ≥ J ( λu ) . (2.4) In this section we will prove the local existence and the uniqueness of the solution of problem(1.1). The process is closely related to ([23] , Theorem2 .
1) that Gerbi and Said-Houair haveproved, where no memory term and Balakrishnan-Taylor damping were present.
Theorem 3.1
Assume ( G and ( G hold, ≤ p ≤ q and max n , qq +1 − p o ≤ m ≤ q . Thengiven u ∈ H (Ω) and u ∈ L (Ω) , there exist T > and a unique solution u of problem (1.1) on (0 , T ) such that u ∈ C (cid:0) [0 , T ] , H (Ω) (cid:1) ∩ C (cid:0) [0 , T ] , L (Ω) (cid:1) ,u t ∈ L (cid:0) , T ; H (Ω) (cid:1) ∩ L m ((0 , T ) × Γ ) . By using the Fadeo-Galerkin approxiations and the well-known contraction mapping theo-rem we can prove this theorem. In order to define the function for which a fixed point exists,for u ∈ C (cid:0) [0 , T ] , H (Ω) (cid:1) ∩ C (cid:0) [0 , T ] , L (Ω) (cid:1) , we consider the following problem v tt − N ( t )∆ v + Z t g ( t − s )∆ v ( x, s )d s + αv t − β ∆ v t = | u | p − u, x ∈ Ω , t > ,v ( x, t ) = 0 , x ∈ Γ , t > ,v tt ( x, t ) = − N ( t ) ∂v∂ν ( x, t ) + Z t g ( t − s ) ∂v∂ν ( x, s )d s − β ∂v t ∂ν ( x, t ) − γ | v t | m − v t ( x, t ) , x ∈ Γ , t > ,v ( x,
0) = u ( x ) , v t ( x,
0) = u ( x ) , x ∈ Ω , (3.1)where N ( t ) = a + b k∇ v ( t ) k + σ Z Ω ∇ v ( t ) ∇ v t ( t )d x . Definition 3.2
A function v ( x, t ) such that v ∈ L ∞ (cid:0) , T ; H (Ω) (cid:1) , t ∈ L (cid:0) , T ; H (Ω) (cid:1) ∩ L m ((0 , T ) × Γ ) ,v t ∈ L ∞ (cid:0) , T ; H (Ω) (cid:1) ∩ L ∞ (cid:0) (0 , T ) × L (Γ ) (cid:1) ,v tt ∈ L ∞ (cid:0) , T ; L (Ω) (cid:1) ∩ L ∞ (cid:0) (0 , T ) × L (Γ ) (cid:1) ,v ( x,
0) = u ( x ) , v t ( x,
0) = u ( x ) , is a generalized solution to problem (3.1) if for any function ω ∈ H (Ω) ∩ L m (Γ ) and ϕ ∈ C (0 , T ) with ϕ ( T ) = 0 , we get Z T ( | u | p − u, w )( t ) ϕ ( t )d t = Z T (cid:20) ( v tt , ω )( t ) + a ( ∇ v, ∇ ω )( t ) + b Z Ω k∇ v ( t ) k ∆ v ( t ) ω ( t )d x + σ Z Ω dd t k∇ v ( t ) k ∆ v ( t ) ω ( t )d x + α ( v t , ω )( t ) + β ( ∇ v t , ∇ ω )( t ) − Z t g ( t − s ) ∇ v ( s ) · ∇ ω ( t )d s (cid:21) ϕ ( t )d t + Z T ϕ ( t ) Z Γ (cid:2) v tt ( t ) + γ | v t ( t ) | m − v t ( t ) (cid:3) ω d σ d t. Lemma 3.3
Assume ( G , ( G , ≤ p ≤ q and ≤ m ≤ q hold. Let u ∈ H (Ω) ∩ V ,u ∈ H (Ω) and f ∈ H (0 , T ; L (Ω)) , then for any T > , there exists a unique generalizedsolution v ( t, x ) of problem (3.1) . Proof.
We use Faedo-Galerkin method, some difficulties appear deriving a second-order esti-mate of v tt (0), inspired by the ideas of Doronin and Larkin in [20] and Cavalcanti et al [11],we set e v = v ( t, x ) − ψ ( t, x ) with ψ ( t, x ) = u ( x ) + tu ( x ). Therefore, we have e v tt − e N ( t )∆ e v + Z t g ( t − s )∆ e v ( x, s )d s + α e v t − β ∆ e v t = f ( x, t )+ e N ( t )∆ ψ − Z t g ( t − s )∆ ψ ( s )d s − αψ t + β ∆ ψ t , x ∈ Ω , t > , e v ( x, t ) = 0 , x ∈ Γ , t > , e v tt ( x, t ) = − e N ( t ) ∂ ( e v + ψ ) ∂ν ( x, t ) + Z t g ( t − s ) ∂ ( e v + ψ ) ∂ν ( x, s )d s − β ∂ ( e v t + ψ t ) ∂ν ( x, t ) − γ | e v t + ψ t | m − ( e v t + ψ t )( x, t ) , x ∈ Γ , t > , e v ( x,
0) = 0 , e v t ( x,
0) = 0 , x ∈ Ω , (3.2)where e N ( t ) = a + b k∇ ( e v + ψ ) k + σ t k∇ ( e v + ψ ) k . We obtain that if e v is a solution ofproblem (3.2) on [0 , T ], then v is a solution of problem (3.1) on [0 , T ]. By using the Faedo-Galerkin method, for every h ≥
1, let W h = Span { ω , · · · , ω h } , where { ω j } is the orthogonalcomplete system of eigenfunctions of − ∆ in the space V such that k w j k = 1 for all j . Then { ω j } is orthogonal and complete in L (Ω) ∩ L (Γ ). Let e v h ( t ) = n X j =1 ϑ jh ( t ) ω j , (3.3)6here e v h ( t ) are solutions to the finite-dimensional Cauchy problem, we have Z Ω e v tth ( t ) ω j d x + e N ( t ) Z Ω ∇ ( e v h + ψ ) · ∇ ω j d x + α Z Ω ( e v h + ψ ) t ω j d x + β Z Ω ∇ ( e v h + ψ ) t · ∇ ω j d x − Z t g ( t − s ) Z Ω ∇ ( e v h + ψ ) · ∇ ω j d x + Z Γ ( e v tth + γ | ( e v h + ψ ) t | m − ( e v h + ψ ) t ) ω j d σ = Z Ω f ω j d x, (3.4) ϑ jh (0) = ϑ ′ jh (0) = 0 , j = 1 , · · · , h. According to [16], by the Caratheodory theorem, problem (3.4) has a unique solution ( ϑ jh ( t )) ∈ H (0 , t h ) defined on [0 , t h ). To do this, we need two priori estimates to show that for all h ∈ N , t h = T and those approximations converge to a solution of problem (3.2), for the proof of theestimates can be done as in ([23] , Lemma2 . Lemma 3.4
Assume ( G , ( G , ≤ p ≤ q and max n , qq +1 − p o ≤ m ≤ q hold. Then given u ∈ H (Ω) and u ∈ L (Ω) , there exist T > and a unique solution v of problem (3.1) on (0 , T ) such that v ∈ C (cid:0) [0 , T ] , H (Ω) (cid:1) ∩ C (cid:0) [0 , T ] , L (Ω) (cid:1) ,v t ∈ L (cid:0) , T ; H (Ω) (cid:1) ∩ L m ((0 , T ) × Γ ) and satisfies the energy identity (cid:20)(cid:18) a − Z τ g ( z )d z (cid:19) k∇ v k + b k∇ v k + k v t k + k v t k , Γ + ( g ◦ ∇ v )( τ ) (cid:21) ts + α Z ts k v t ( τ ) k d τ + σ Z ts (cid:18) d k∇ v ( z ) k d z (cid:19) d z + β Z ts k∇ v t ( τ ) k d s + γ Z ts k v t ( τ ) k mm, Γ d τ − Z ts ( g ′ ◦ ∇ v )( τ )d τ + 12 Z ts g ( τ ) k∇ v ( τ ) k d τ = Z ts Z Ω | u ( τ ) | p − u ( τ ) v t ( τ )d τ d x (3.5) for ≤ s ≤ t ≤ T . Proof.
First, by a sequence ( u ι ) ι ∈ N ⊂ C ∞ ([0 , T ] × Ω) by standard convolution arguments (see[9]), we approximate u ∈ C (cid:0) [0 , T ] , H (Ω) (cid:1) ∩ C (cid:0) [0 , T ] , L (Ω) (cid:1) endowed with the standardnorm k u k = max t ∈ [0 ,T ] (cid:0) k u t ( t ) k + k u ( t ) k H (Ω) (cid:1) .
7e obtain f ( u ι ) = | u ι | p − u ι ∈ H (0 , T ; L (Ω)) . Next, we approximate the initial data u ∈ L (Ω) by a sequence { u ι } ∈ C ∞ (Ω). At last, because the space H (Ω) ∩ V ∩ H (Ω) is densein H (Ω) for the H norm, we approximate u ∈ H (Ω) by a sequence { u ι } ⊂ H (Ω) ∩ V ∩ H (Ω).We now consider the following problem v ιtt − N ι ( t )∆ v ι + Z t g ( t − s )∆ v ι ( x, s )d s + αv ιt − β ∆ v ιt = | u ι | p − u ι , x ∈ Ω , t > ,v ι ( x, t ) = 0 , x ∈ Γ , t > ,v ιtt ( x, t ) = − N ι ( t ) ∂v ι ∂ν ( x, t ) + Z t g ( t − s ) ∂v ι ∂ν ( x, s )d s − β ∂v ιt ∂ν ( x, t ) − γ | v ιt | m − v ιt ( x, t ) , x ∈ Γ , t > ,v ι ( x,
0) = u ι ( x ) , v ιt ( x,
0) = u ι ( x ) , x ∈ Ω , (3.6)where N ι ( t ) = a + b k∇ v ι ( t ) k + σ Z Ω ∇ v ι ( t ) ∇ v ιt ( t )d x . As every hypothesis of Lemma 3.3 issatisfied, we can find a sequence of unique solutions { v k } of problem (3.6). Next, we shallprove that { v ι , v ιt } is a Cauchy sequence in the space Y T = (cid:8) ( v, v t ) : v ∈ C (cid:0) [0 , T ] , H (Ω) (cid:1) ∩ C (cid:0) [0 , T ] , L (Ω) (cid:1) ,v t ∈ L (cid:0) , T ; H (Ω) (cid:1) ∩ L m ((0 , T ) × Γ ) (cid:9) endowed with the norm k ( v, v t ) k Y T = max t ∈ [0 ,T ] (cid:2) k v t k + k∇ v k (cid:3) + k v t k L m ((0 ,T ) × Γ ) + Z t k∇ v t ( s ) k d s. To achieve this, we set U = u ι − u ι ′ , V = v ι − v ι ′ . It’s easy to see that V satisfies V tt − N ( t )∆ V + Z t g ( t − s )∆ V ( x, s )d s + αV t − β ∆ V t = | u ι | p − u ι − | u ι ′ | p − u ι ′ , x ∈ Ω , t > ,V ( x, t ) = 0 , x ∈ Γ , t > ,V tt ( x, t ) = − N ( t ) ∂V∂ν ( x, t ) + Z t g ( t − s ) ∂V∂ν ( x, s )d s − β ∂V t ∂ν ( x, t ) − γ (cid:16) | v ιt | m − v ιt ( x, t ) − | v ι ′ t | m − v ι ′ t ( x, t ) (cid:17) , x ∈ Γ , t > ,V ( x,
0) = u ι − u ι ′ , V t ( x,
0) = u ι ( x ) − u ι ′ , x ∈ Ω , where N ( t ) = a + b k∇ V ( t ) k + σ Z Ω ∇ V ( t ) ∇ V t ( t )d x . By multiplying the above differentialequation by V t and integrating over Ω × (0 , t ), we arrive at12 (cid:20)(cid:18) a − Z t g ( s )d s (cid:19) k∇ V ( t ) k + b k∇ V ( t ) k + k V t ( t ) k + k V t ( t ) k , Γ (cid:21) + α Z t k V t ( s ) k d s β Z t k∇ V t ( s ) k d s − Z t ( g ′ ◦ ∇ V )( s )d s + ( g ◦ ∇ V )( t ) + Z t Z Ω g ( s ) |∇ V ( s ) | d x d s + σ k∇ V ( t ) k + γ Z t Z Γ (cid:16) | v ιt | m − v ιt − | v ι ′ t | m − v ι ′ t (cid:17) (cid:16) v ιt − v ι ′ t (cid:17) d σ d s = 12 (cid:20)(cid:18) a − Z t g ( s )d s (cid:19) k∇ V (0) k + b k∇ V (0) k + k V t (0) k + k V t (0) k , Γ (cid:21) + Z t Z Ω (cid:16) | u ι | p − u ι − | u ι ′ | p − u ι ′ (cid:17) (cid:16) v ιt − v ι ′ t (cid:17) d x d s, ∀ t ∈ (0 , T ) . By the algebraic inequality ∀ m ≥ , ∃ c > , ∀ a, b ∈ R , (cid:0) | a | m − a − | b | m − b (cid:1) ( a − b ) ≥ c | a − b | m , (3.7)we get12 (cid:20)(cid:18) a − Z t g ( s )d s (cid:19) k∇ V ( t ) k + b k∇ V ( t ) k + k V t ( t ) k + k V t ( t ) k , Γ (cid:21) + α Z t k V t ( s ) k d s + c k V t k mm, Γ + σ k∇ V ( t ) k + β Z t k∇ V t ( s ) k d s − Z t ( g ′ ◦ ∇ V )( s )d s + ( g ◦ ∇ V )( t )+ Z t Z Ω g ( s ) |∇ V ( s ) | d x d s ≤ (cid:20)(cid:18) a − Z t g ( s )d s (cid:19) k∇ V (0) k + b k∇ V (0) k + k V t (0) k + k V t (0) k , Γ (cid:21) + Z t Z Ω (cid:16) | u ι | p − u ι − | u ι ′ | p − u ι ′ (cid:17) (cid:16) v ιt − v ι ′ t (cid:17) d x d s, ∀ t ∈ (0 , T ) . Then, by using Young’s inequality, Gronwall inequality and the result of Georgiev and Todorova[26], there exists a constant C depending only on Ω and p such that k V k Y T ≤ C (cid:0) k∇ V (0) k + k V t (0) k + k V t (0) k , Γ (cid:1) + CT k U k Y T . By the above notation, we can obtain V (0) = u ι − u ι ′ , V t (0) = u ι − u ι ′ . Thus, we conclude that { ( v ι , v ιt ) } is a Cauchy sequence in Y T , because { u ι } is a converging sequence in H (Ω), { u ι } is a converging sequence in L (Ω) and { u ι } is a converging sequence in C (cid:0) [0 , T ] , H (Ω) (cid:1) ∩ C (cid:0) [0 , T ] , L (Ω) (cid:1) . So, we conclude that { ( v ι , v ιt ) } converges to a limit ( v, v t ) ∈ Y T . As doneby [26], we can prove that this limit is a weak solution of problem (3.1). This completes theproof of Lemma 3.4.Now we are ready to prove the local existence result. Proof of Theorem 3.1.
For
R >
T >
0, we define the convex closed subsetof Y T , we define a class of functions as X T = { ( v, v t ) ∈ Y T : v (0) = u , v t (0) = u and k v k Y T ≤ R } . Then, Lemma 3.4 implies that, for any u ∈ X T , we may define v = Φ( u ) to be a uniquesolution of (3.1) corresponding to u . We would like to show that Φ is a contraction map9atisfying Φ( X T ) ⊂ X T for a suitable T >
0. Firstly, we show Φ( X T ) ⊂ X T . For this, by usingthe energy identity, we have l k∇ v ( t ) k + k v t ( t ) k + k v t ( t ) k , Γ + 2 α Z t k v t ( s ) k d s + 2 β Z t k∇ v t ( s ) k d s + 2 γ Z t k v t ( s ) k mm, Γ d s ≤k∇ u k + k u k + k u k , Γ + 2 Z t Z Ω | u ( s ) | p − u ( s ) v t ( s )d x d s. (3.8)By the Holder’s inequality, we estimate the last term in the right-hand side of the inequality(3.8) as follows Z t Z Ω | u ( s ) | p − u ( s ) v t ( s )d x d s ≤ Z t k u ( s ) k p − N/ ( N − k v t ( s ) k N/ (3 N − Np +2( p − d s. As p ≤ NN − , we get 2 N (3 N − N p + 2( p − ≤ NN − . Thus, by Young’s and Sobolev’s inequalities, for all δ > , t ∈ (0 , T ), there exists a positiveconstant C ( δ ) such that Z t Z Ω | u ( s ) | p − u ( s ) v t ( s )d x d s ≤ C ( δ ) tR p − + δ Z t k∇ v t ( s ) k d s. Inserting the last estimate in the inequality (3.8) and choosing δ small enough, we have k v k Y T ≤ R + CT R p − . Then, choose T small enough, we get k v k Y T ≤ R , which shows that Φ maps X T into itself.Next, we verify that Φ is a contraction. To this end, we set U = u − u and V = v − v ,where v = Φ( u ) and v = Φ( u ). It’s straightforward to verify that V satisfies V tt − N ( t )∆ V + Z t g ( t − s )∆ V ( x, s )d s + αV t − β ∆ V t = | u | p − u − | u | p − u, x ∈ Ω , t > ,V ( x, t ) = 0 , x ∈ Γ , t > ,V tt ( x, t ) = − N ( t ) ∂V∂ν ( x, t ) + Z t g ( t − s ) ∂V∂ν ( x, s )d s − β ∂V t ∂ν ( x, t ) − γ (cid:0) | v t | m − v t ( x, t ) − | v t | m − v t ( x, t ) (cid:1) , x ∈ Γ , t > ,V ( x,
0) = 0 , V t ( x,
0) = 0 , x ∈ Ω , (3.9)10here N ( t ) = a + b k∇ V ( t ) k + σ Z Ω ∇ V ( t ) ∇ V t ( t )d x . By multiplying the differential equationin (3.9) by V t and integrating over Ω × (0 , t ), we arrive at12 (cid:20)(cid:18) a − Z t g ( s )d s (cid:19) k∇ V ( t ) k + b k∇ V ( t ) k + k V t ( t ) k + k V t ( t ) k , Γ (cid:21) + α Z t k V t ( s ) k d s + β Z t k∇ V t ( s ) k d s − Z t ( g ′ ◦ ∇ V )( s )d s + 12 ( g ◦ ∇ V )( t ) + 12 Z t Z Ω g ( s ) |∇ V ( s ) | d x d s + σ k∇ V ( t ) k + γ Z t Z Γ (cid:0) | v t | m − v t − | v t | m − v t (cid:1) ( v t − v t ) d σ d s = Z t Z Ω (cid:0) | u | p − u − | u | p − u (cid:1) ( v t − v t ) d x d s, ∀ t ∈ (0 , T ) . Similar to the discussion in [23], we can get k V k Y T ≤ CR p − T / k U k Y T . (3.10)By choosing T so small that CR p − T / <
1, estimate (3.10) shows that Φ is a contraction. Thecontraction mapping theorem then guarantees the existence of a unique v satisfying v = Φ( v ).The proof of Theorem 3.1 is now completed. In this section, for the initial data in the stable set, we show the solution is global. We needthe following definition and lemmas.
Definition 4.1
Let ≤ p ≤ q and max n , qq +1 − p o ≤ m ≤ q . We denote u as the solution of (1.1) . We define T max = sup { T > , u = u ( t ) exists on [0 , T ] } . Lemma 4.2
Suppose that ( G , ( G , ≤ p ≤ q and max n , qq +1 − p o ≤ m ≤ q hold. If u isthe solution of (1.1) , then E ′ ( t ) = 12 ( g ′ ◦ ∇ u )( t ) − g ( t ) k∇ u ( t ) k − σ (cid:18)
12 ddt k∇ u ( t ) k (cid:19) − α k u t ( t ) k − β k∇ u t ( t ) k − γ k u t ( t ) k mm, Γ ≤ , (4.1) for almost every t ∈ [0 , T max ) . Proof.
In view of ( G u t and integrating byparts over Ω, we obtain the result. Lemma 4.3
Let ( u , u ) ∈ H × L (Ω) be given, suppose that ( G and ≤ p ≤ q hold, suchthat C p ∗ l (cid:18) pl ( p − E (0) (cid:19) p − < ,I ( u ) > , (4.2)11 here C ∗ is the best poincare constant, then I ( u ( t )) > , ∀ t ∈ [0 , T max ) . Proof.
Since I (0) >
0, there exists a T ∗ < T max such that I ( t ) > t ∈ [0 , T ∗ ). Thisimplies that J ( t ) = p − p (cid:20)(cid:18) a − Z t g ( s )d s (cid:19) k∇ u ( t ) k + ( g ◦ ∇ u )( t ) + b k∇ u ( t ) k (cid:21) + 1 p I ( t ) ≥ p − p (cid:20)(cid:18) a − Z t g ( s )d s (cid:19) k∇ u ( t ) k + ( g ◦ ∇ u )( t ) + b k∇ u ( t ) k (cid:21) . (4.3)Thus, by ( G l k∇ u k ≤ (cid:18) a − Z t g ( s )d s (cid:19) k∇ u ( t ) k ≤ pp − J ( t ) ≤ pp − E ( t ) ≤ pp − E (0) , (4.4)for all t ∈ [0 , T ∗ ). This combines with the Soblev imbedding, ( G
1) and (4.2), implies that k u k pp ≤ C p ∗ k∇ u k p ≤ C p ∗ l k∇ u k p − l k∇ u k ≤ C p ∗ l (cid:18) pl ( p − E (0) (cid:19) p − l k∇ u k < (cid:18) a − Z t g ( s )d s (cid:19) k∇ u k , for all t ∈ [0 , T ∗ ). Therefore, I ( t ) >
0, for all t ∈ [0 , T ∗ ). By repeating this procedure and usingthe fact that lim t → T ∗ C p ∗ l (cid:18) pl ( p − E (0) (cid:19) p − < , where T ∗ is extended to T max . Theorem 4.4
Let ( u , u ) ∈ H × L (Ω) be given and satisfying (4.2) , suppose that ( G , ( G , ≤ p ≤ q and max n , qq +1 − p o ≤ m ≤ q hold. Then the solution is global and bounded. Proof.
We have just to check that k∇ u ( t ) k + k u t ( t ) k is bounded and independently of t . To achieve this, by (2.1), (4.1) and (4.3), we have E (0) ≥ E ( t ) = J ( t ) + 12 k u t ( t ) k + 12 k u t ( t ) k , Γ ≥ p − p (cid:20) l k∇ u ( x, t ) k + ( g ◦ ∇ u )( x, t ) + b k∇ u ( t ) k (cid:21) + 12 k u t ( t ) k + 12 k u t ( t ) k , Γ + 1 p I ( t ) ≥ p − p l k∇ u ( t ) k + 12 k u t ( t ) k + 12 k u t ( t ) k , Γ , (4.5)since I ( t ) and ( g ◦ ∇ u )( t ) are positive. Therefore k∇ u ( t ) k + k u t ( t ) k ≤ CE (0) , ∀ t ∈ [0 , T max ) , where C is a positive constant, which depends only on p and l . Then by the definition of T max ,the solution is global, that is T max = ∞ . 12 Decay of solutions
In this section, we state and prove the general decay result. For positive constants ε and ε ,we use the following modified functional L ( t ) = E ( t ) + ε G ( t ) + ε H ( t ) , (5.1)where G ( t ) = Z Ω u t u d x + Z Γ u t u d σ + α k u k + β k∇ u k + σ k∇ u k (5.2)and H ( t ) = − Z Ω u t Z t g ( t − s ) ( u ( t ) − u ( s )) d s d x − Z Γ u t Z t g ( t − s ) ( u ( t ) − u ( s )) d s d σ. (5.3)It is easy to check that, by using Poincare’s inequality, trace inequality, (2.2), (2.3) and for ε , ε small enough, α L ( t ) ≤ E ( t ) ≤ α L ( t ) (5.4)holds for two positive constants α and α . Theorem 5.1
Let ( u , u ) ∈ H × L (Ω) be given. Assume that g and ξ satisfy ( G and ( G . Then, for each t > , there exist positive constants K and k such that the solution (1.1) satisfies, for all t ≥ t , E ( t ) ≤ Ke − k R tt ξ ( s )d s . (5.5) Lemma 5.2
Under the conditions of Theorem 5.1, the functional G ( t ) defined by (5.2) satisfies G ′ ( t ) ≤k u t ( t ) k + k u t ( t ) k , Γ − l − Cγδ − m m (cid:18) pE (0) l ( p − (cid:19) m − ! k∇ u ( t ) k + k u ( t ) k pp + γ ( m − m δ mm − k u t ( t ) k mm, Γ + a − l l ( g ◦ ∇ u ) ( t ) , (5.6) for some δ > . Proof.
By using the differential equation in (1.1), we get G ′ ( t ) = k u t ( t ) k + Z Ω u tt ( t ) u ( t )d x + Z Γ u tt ( t ) u ( t )d σ + k u t ( t ) k , Γ + α Z Ω u t ( t ) u ( t )d x + β Z Ω ∇ u t ( t ) · ∇ u ( t )d x + σ (cid:18) ddt k∇ u ( t ) k (cid:19) k∇ u ( t ) k = k u t ( t ) k + k u t ( t ) k , Γ + Z Γ ∂u ( t ) ∂ν u ( t )d σ − (cid:0) a + b k∇ u ( t ) k (cid:1) k∇ u ( t ) k − Z t g ( t − s ) Z Γ ∂u ( s ) ∂ν u ( s )d σ d s + Z t g ( t − s ) Z Ω ∇ u ( s ) · ∇ u ( t )d x d s β Z Γ ∂u t ( t ) ∂ν u ( t )d σ + Z Γ u tt ( t ) u ( t )d σ + k u ( t ) k pp = k u t ( t ) k + k u t ( t ) k , Γ − (cid:0) a + b k∇ u ( t ) k (cid:1) k∇ u ( t ) k + Z Ω ∇ u ( t ) · Z t g ( t − s ) ∇ u ( s )d s d x − γ Z Γ | u t ( t ) | m − u t ( t ) u ( t )d σ + k u ( t ) k pp . (5.7)We now estimate the right hand side of (5.7). For two positive constants δ and η , we have theestimates as follows Z Ω ∇ u ( t ) · Z t g ( t − s ) ∇ u ( s )d s d x ≤ ( η + a − l ) k∇ u ( t ) k + a − l η ( g ◦ ∇ u )( t ) . (5.8)By Young’s inequality, trace inequality and combining (4.4), we have (cid:12)(cid:12)(cid:12)(cid:12)Z Γ | u t ( t ) | m − u t ( t ) u ( t )d σ (cid:12)(cid:12)(cid:12)(cid:12) ≤ δ − m m k u ( t ) k mm, Γ + m − m δ mm − k u t ( t ) k mm, Γ , (5.9) k u ( t ) k mm, Γ ≤ C k∇ u ( t ) k m ≤ C (cid:18) pE (0) l ( p − (cid:19) m − k∇ u ( t ) k . (5.10)Combining (5.7)-(5.10), we obtain G ′ ( t ) ≤k u t ( t ) k + k u t ( t ) k , Γ − l − η − Cγδ − m m (cid:18) pE (0) l ( p − (cid:19) m − ! k∇ u ( t ) k + k u ( t ) k pp + γ ( m − m δ mm − k u t ( t ) k mm, Γ + a − l η ( g ◦ ∇ u ) ( t ) , Letting η = l > Lemma 5.3
Under the conditions of Theorem 5.1, the functional H ( t ) defined by (5.3) satis-fies H ′ ( t ) ≤ (cid:18) α + δ − Z t g ( s )d s (cid:19) k u t ( t ) k + " δ + 2 δ ( a − l ) + δC p − ∗ (cid:18) pE (0) l ( p − (cid:19) p − k∇ u ( t ) k + " a − l δ (cid:18) a + 2 pbE (0)( p − l (cid:19) + a − l l + a − l δλ + β ( a − l )2 + (cid:18) δ + 14 δ (cid:19) ( a − l ) + γC m ∗ c δm (cid:18) pE (0) l ( p − (cid:19) m − ( g ◦ ∇ u )( t ) + β k∇ u t ( t ) k + ( δ + a − l ) k u t ( t ) k , Γ + γδ ( m − m ( a − l ) k u t ( t ) k mm, Γ + a − l δλ ( − g ′ ◦ ∇ u )( t ) + σ ( p − p E (0) E ′ ( t ) , (5.11) for some δ > . Proof.
By using the differential equation in (1.1), we get H ′ ( t ) = − Z Ω u tt ( t ) Z t g ( t − s ) ( u ( t ) − u ( s )) d s d x − Z Ω u t ( t ) Z t g ′ ( t − s ) ( u ( t ) − u ( s )) d s d x (cid:18)Z t g ( s )d s (cid:19) Z Ω | u t ( t ) | d x − Z Γ u tt ( t ) Z t g ( t − s ) ( u ( t ) − u ( s )) d s d σ − Z Γ u t ( t ) Z t g ′ ( t − s ) ( u ( t ) − u ( s )) d s d σ − (cid:18)Z t g ( s )d s (cid:19) Z Γ | u t ( t ) | d σ = − Z Ω (cid:18) M ( t )∆ u ( t ) − Z t g ( t − s )∆ u ( s )d s − αu t ( t ) + β ∆ u t ( t ) + | u ( t ) | p − u ( t ) (cid:19) × Z t g ( t − s ) ( u ( t ) − u ( s )) d s d x − Z Ω u t ( t ) Z t g ′ ( t − s ) ( u ( t ) − u ( s )) d s d x − (cid:18)Z t g ( s )d s (cid:19) Z Ω | u t ( t ) | d x − Z Γ u t ( t ) Z t g ′ ( t − s ) ( u ( t ) − u ( s )) d s d σ − Z Γ (cid:18) − M ( t ) ∂u∂ν ( x, t ) + Z t g ( t − s ) ∂u∂ν ( x, s )d s − β ∂u t ∂ν ( x, t ) − γ | u t | m − u t ( x, t ) (cid:19) × Z t g ( t − s ) ( u ( t ) − u ( s )) d s d σ − (cid:18)Z t g ( s )d s (cid:19) Z Γ | u t ( t ) | d σ = Z Ω (cid:0) a + b k∇ u ( t ) k (cid:1) ∇ u ( t ) · Z t g ( t − s ) ( ∇ u ( t ) − ∇ u ( s )) d s d x + Z Ω σ ( ∇ u ( t ) ∇ u t ( t )d x ) ∇ u ( t ) · Z t g ( t − s ) ( ∇ u ( t ) − ∇ u ( s )) d s d x + β Z Ω ∇ u t ( t ) · Z t g ( t − s ) ( ∇ u ( t ) − ∇ u ( s )) d s d x − Z Ω Z t g ( t − s ) ∇ u ( s )d s Z t g ( t − s ) ( ∇ u ( t ) − ∇ u ( s )) d s d x + α Z Ω u t ( t ) Z t g ( t − s ) ( u ( t ) − u ( s )) d s d x − Z Ω | u ( t ) | p − u ( t ) Z t g ( t − s ) ( u ( t ) − u ( s )) d s d x − Z Ω u t ( t ) Z t g ′ ( t − s ) ( u ( t ) − u ( s )) d s d x − (cid:18)Z t g ( s )d s (cid:19) k u t ( t ) k + γ Z Γ | u t ( t ) | m − u t ( t ) Z t g ( t − s ) ( u ( t ) − u ( s )) d s d σ − Z Γ u t ( t ) Z t g ′ ( t − s ) ( u ( t ) − u ( s )) d s d σ − (cid:18)Z t g ( s )d s (cid:19) Z Γ | u t ( t ) | d σ = − (cid:18)Z t g ( s )d s (cid:19) k u t ( t ) k + X i =1 I i . (5.12)We now estimate I i , i = 1 , · · · ,
10 in the right side of (5.12). For δ >
0, similar as in [8],exploiting E ′ ( t ) ≤ − σ (cid:18)
12 dd t k∇ u ( t ) k (cid:19) by (4.1), (4.4), Young’s inequality, Hoider’s inequalityand Cauchy-Schwarzs’s inequality, we get I = Z Ω (cid:0) a + b k∇ u ( t ) k (cid:1) ∇ u ( t ) · Z t g ( t − s ) ( ∇ u ( t ) − ∇ u ( s )) d s d σ δ k∇ u ( t ) k + a − l δ (cid:18) a + 2 pbE (0)( p − l (cid:19) ( g ◦ ∇ u )( t ) , (5.13) I = Z Ω σ ( ∇ u ( t ) ∇ u t ( t )d x ) ∇ u ( t ) · Z t g ( t − s ) ( ∇ u ( t ) − ∇ u ( s )) d s d x ≤ δ ( ∇ u ( t ) ∇ u t ( t )d x ) l k∇ u ( t ) k + 14 l Z Ω (cid:18)Z t g ( t − s ) ( ∇ u ( t ) − ∇ u ( s )) d s (cid:19) d x ≤ − σ ( p − p E (0) E ′ ( t ) + a − l l ( g ◦ ∇ u )( t ) , (5.14) I = β Z Ω ∇ u t ( t ) · Z t g ( t − s ) ( ∇ u ( t ) − ∇ u ( s )) d s d x ≤ β k∇ u t ( t ) k + β ( a − l )2 ( g ◦ ∇ u )( t ) , (5.15) I = − Z Ω Z t g ( t − s ) ∇ u ( s )d s Z t g ( t − s ) ( ∇ u ( t ) − ∇ u ( s )) d s d x ≤ δ ( a − l ) k∇ u ( t ) k + (cid:18) δ + 14 δ (cid:19) ( a − l ) ( g ◦ ∇ u )( t ) , (5.16) I = α Z Ω u t ( t ) Z t g ( t − s ) ( u ( t ) − u ( s )) d s d x ≤ α k u t ( t ) k + a − l δλ ( g ◦ ∇ u )( t ) , (5.17) I = − Z Ω | u ( t ) | p − u ( t ) Z t g ( t − s ) ( u ( t ) − u ( s )) d s d x ≤ δC p − ∗ (cid:18) pE (0) l ( p − (cid:19) p − k∇ u ( t ) k + a − l δλ ( g ◦ ∇ u )( t ) , (5.18) I = − Z Γ u t ( t ) Z t g ′ ( t − s ) ( u ( t ) − u ( s )) d s d σ ≤ δ k u t ( t ) k + a − l δλ ( − g ′ ◦ ∇ u )( t ) , (5.19) I = γ Z Γ | u t ( t ) | m − u t ( t ) Z t g ( t − s ) ( u ( t ) − u ( s )) d s d σ ≤ γδ ( m − m ( a − l ) k u t ( t ) k mm, Γ + γC m ∗ c δm (cid:18) pE (0) l ( p − (cid:19) m − ( g ◦ ∇ u )( t ) , (5.20) I = − Z Γ u t ( t ) Z t g ′ ( t − s ) ( u ( t ) − u ( s )) d s d σ k u t ( t ) k , Γ + a − l δλ ( − g ′ ◦ ∇ u )( t ) , (5.21) I = − (cid:18)Z t g ( s )d s (cid:19) Z Γ | u t ( t ) | d σ ≤ ( a − l ) k u t ( t ) k , Γ . (5.22)A combination of (5.13)-(5.22) yields (5.11). Proof of Theorem 5.1.
Since g is continuous and g (0) >
0, then for any t >
0, we have Z t g ( s )d s ≥ Z t g ( s )d s := g > , for all t ≥ t . From (4.1), (5.6), (5.11), then from (5.1), we get L ′ ( t ) = E ′ ( t ) + ε G ′ ( t ) + ε H ′ ( t ) ≤ − (cid:20) α − ε − ε (cid:18) α + δ − Z t g ( s )d s (cid:19)(cid:21) k u t ( t ) k − " g ( t ) − ε − l Cγδ − m m (cid:18) pE (0) l ( p − (cid:19) m − ! − ε K δ k∇ u ( t ) k − (cid:18) β − βε (cid:19) k∇ u t ( t ) k + (cid:20) ε ( a − l )2 l + ε K µ (cid:21) ( g ◦ ∇ u ) ( t )+ [ ε + ε ( δ + a − l )] k u t ( t ) k , Γ − " γ − ε ( m − δ mm − m − ε γδ ( m − a − l ) m k u t ( t ) k mm, Γ + (cid:20) − ε ( a − l )2 δλ (cid:21) ( g ′ ◦ ∇ u )( t ) + ε k u ( t ) k pp − σ ( p − p E (0) E ′ ( t ) , (5.23)where K δ = δ + 2 δ ( a − l ) + δC p − ∗ (cid:18) pE (0) l ( p − (cid:19) p − > ,K µ = a − l δ (cid:18) a + 2 pbE (0)( p − l (cid:19) + β ( a − l )2 + (cid:18) δ + 14 δ (cid:19) (1 − l ) + γC m ∗ c δm (cid:18) pE (0) l ( p − (cid:19) m − + a − l l + a − l δλ > . By the trace inequality k u t k , Γ ≤ C k u t k W , (Ω) ≤ C k u t k , then we have L ′ ( t ) ≤ − (cid:20) α − ε − ε (cid:18) α + δ − Z t g ( s )d s (cid:19) − C [ ε + ε ( δ + a − l )] (cid:21) k u t ( t ) k " g ( t ) − ε − l Cγδ − m m (cid:18) pE (0) l ( p − (cid:19) m − ! − ε K δ k∇ u ( t ) k − (cid:20) β − βε (cid:21) k∇ u t ( t ) k + (cid:20) ε a − l l + ε K µ (cid:21) ( g ◦ ∇ u ) ( t ) − " γ − ε ( m − δ mm − m − ε γδ ( m − a − l ) m k u t ( t ) k mm, Γ + (cid:20) − ε ( a − l )2 δλ (cid:21) ( g ′ ◦ ∇ u )( t ) + ε k u ( t ) k pp − σ ( p − p E (0) E ′ ( t ) . (5.24)At this point, we choose δ > < δ < α − ε − ε (cid:16) α − R t g ( s )d s (cid:17) − C [ ε + ε ( a − l )](1 + C ) ε , so we get k = α − ε − ε (cid:18) α + δ − Z t g ( s )d s (cid:19) − C [ ε + ε ( δ + a − l )] > . (5.25)We then choose ε and ε so small that (5.4) and (5.25) remain valid and k = 12 g ( t ) − ε − l Cγδ − m m (cid:18) pE (0) l ( p − (cid:19) m − ! − ε K δ > ,k = β − βε > ,k = γ − ε ( m − δ mm − m − ε γδ ( m − a − l ) m > ,k = 12 − ε ( a − l )2 δλ > . Therefore, we have L ′ ( t ) ≤ − k k u t ( t ) k − k k∇ u ( t ) k − k k∇ u t ( t ) k + (cid:20) ε a − l l + ε K µ (cid:21) ( g ◦ ∇ u ) ( t ) − k k u t ( t ) k mm, Γ + k ( g ′ ◦ ∇ u )( t ) + ε k u ( t ) k pp − σ ( p − p E (0) E ′ ( t ) . (5.26)Then, by ( G
1) and ( G M and α we obtain L ′ ( t ) ≤ − M E ( t ) + α ( g ◦ ∇ u )( t ) − σ ( p − p E (0) E ′ ( t ) , ∀ t ≥ t . (5.27)Assume that ( G
1) and ( G
2) are satisfied, multiplying (5.27) by ξ ( t ), we have ξ ( t ) L ′ ( t ) ≤ − M ξ ( t ) E ( t ) + α ξ ( t )( g ◦ ∇ u )( t ) − σ ( p − p E (0) ξ ( t ) E ′ ( t ) , ∀ t ≥ t , (5.28)18sing ( G ξ and g are nonincreasing, we get ξ ( t ) Z t g ( t − s ) k∇ u ( t ) − ∇ u ( s ) k ds ≤ − Z t g ′ ( t − s ) k∇ u ( t ) − ∇ u ( s ) k ds ≤ − E ′ ( t )and σ ( p − p E (0) ξ ( t ) E ′ ( t ) ≤ σ ( p − p E (0) ξ (0) E ′ ( t ) . (5.29)Inserting the last two inequalities in (5.28), we can obtain ξ ( t ) L ′ ( t ) + 2 α E ′ ( t ) ≤ − M ξ ( t ) E ( t ) , ∀ t ≥ t , (5.30)where α = α + σ ( p − p E (0) ξ (0) is a positive constant.Now we consider the functional Φ = ξL + 2 α E. Using the facts that Φ ∼ E and ξ is nonincreasing, (5.30) givesΦ ′ ( t ) ≤ − kξ ( t )Φ( t ) , ∀ t ≥ t , for some k >
0. Therefore, direct integration leads toΦ( t ) ≤ Φ( t ) e − k R tt ξ ( s )d s , ∀ t ≥ t and the fact that Φ ∼ E yields E ( t ) ≤ α Φ( t ) e − k R tt ξ ( s )d s = Ke − k R tt ξ ( s )d s , ∀ t ≥ t , where α is a positive constant and K = α Φ( t ). This completes the proof. In this section, we prove a finite time blow-up result for initial data in the unstable set. Fromthe definition (2.4) of the potential depth d , for u ∈ H (Ω) \{ } , we have Theorem 6.1
Suppose ≤ p ≤ q , m = 2 , ( G and ( G hold. Let u be the unique localsolution to problem (1.1) , for any fixed δ < , assume that u , u satisfy E (0) = δd , (6.1) I (0) = 0 . (6.2)19 uppose that Z ∞ g ( s )d s ≤ p − a ( p −
2) + 1 / [(1 − e δ ) p + 2 δ (1 − e δ )] , (6.3) where e δ = max { , δ } and suppose further that R Ω u u d x + R Γ u u d σ > for ≤ E (0) < d ,then T < ∞ . For t ≥ , by (2.4) , we have d is the lower bound of d ( t ) . Lemma 6.2
Under the same assumption as in Theorem 6.1, one has I ( t ) < and d < p − p (cid:20)(cid:18) a − Z t g ( s ) ds (cid:19) k∇ u ( t ) k + ( g ◦ ∇ u )( t ) + b k∇ u ( t ) k (cid:21) < p − p k u ( t ) k pp , (6.4) for all t ∈ [0 , T max ) . Proof.
By Lemma 4.2 and (6.1), we have E ( t ) ≤ δd , for all t ∈ [0 , T max ). Moreover, wecan obtain I ( t ) < t ∈ [0 , T max ). However, if it is not true, then there exists some t ∗ ∈ [0 , T max ) such that I ( t ∗ ) = 0. Thus I ( t ) < ≤ t < t ∗ , i.e. (cid:18) a − Z t g ( s ) ds (cid:19) k∇ u ( t ) k + ( g ◦ ∇ u )( t ) + b k∇ u ( t ) k < k u ( t ) k pp , ≤ t < t ∗ . (6.5)By the definition of d , we have d < p − p (cid:20)(cid:18) a − Z t g ( s ) ds (cid:19) k∇ u ( t ) k + ( g ◦ ∇ u )( t ) + b k∇ u ( t ) k (cid:21) , ≤ t < t ∗ . (6.6)Combined with (6.5) and (6.6), we get k u ( t ) k pp > pp − d > , ≤ t < t ∗ . By the continuity of t
7→ k u ( t ) k pp , we get u ( t ∗ ) = 0. Because d is the lower bound of d and(2.2), we obtain d ≤ p − p k u ( t ∗ ) k pp = J ( u ( t ∗ )) , which contradicts to J ( u ( t ∗ )) ≤ E ( t ∗ ) < d . By repeating the previous step, we obtain (6.4). Lemma 6.3
Let F ( t ) be a positive, twice differentiable function, for t > , we can obtain F ( t ) F ′′ ( t ) − (1 + κ ) F ′ ( t ) ≥ , with some κ > . If F (0) > and F ′ (0) > , then there exists a time T ∗ ≤ F (0) / [ κF ′ (0)] suchthat lim t → T ∗− F ( t ) = ∞ . We can see [30] for more details. roof of Theorem 6.1. Assume by contradiction that the solution u is global. Then forany T >
0, let us define the functional F as follows F ( t ) = k u ( t ) k + k u ( t ) k , Γ + σ Z t k∇ u ( s ) k d s + α Z t k u ( s ) k d s + β Z t k∇ u ( s ) k d s + γ Z t k u ( s ) k , Γ d s + ( T − t ) h α k u k + β k∇ u k + γ k u k , Γ + σ k∇ u k i + b ( t + T ) , where T and T are positive constants to be chosen later, b > E (0) < b = 0 if E (0) >
0. Then F ( t ) > t ∈ [0 , T ]. Furthermore, F ′ ( t ) =2 Z Ω u t u d x + 2 Z Γ u t u d σ + σ Z t dd t k∇ u ( s ) k d s + 2 α Z t Z Ω u t u d x d s + 2 β Z t Z Ω ∇ u t · ∇ u d x d s + 2 γ Z t Z Γ u t u d σ d s + 2 b ( t + T ) , (6.7)and F ′′ ( t ) =2 (cid:20) k u t ( t ) k − (cid:18) a − Z t g ( s )d s (cid:19) k∇ u ( t ) k + k u t ( t ) k , Γ − b k∇ u ( t ) k − Z Ω ∇ u ( t ) · (cid:18)Z t g ( t − s )( ∇ u ( t ) − ∇ u ( s ))d s (cid:19) d x + k u k pp + b (cid:21) . Therefore, using the definition of F ( t ), we get F ( t ) F ′′ ( t ) − p + 24 F ′ ( t ) = F ( t ) F ′′ ( t ) + ( p + 2) (cid:20) η ( t ) − (cid:0) F ( t ) − ( T − t ) (cid:0) α k u ( t ) k + β k∇ u ( t ) k + γ k u ( t ) k , Γ + σ k∇ u k (cid:17)(cid:17) (cid:18) k u t ( t ) k + k u t ( t ) k , Γ + σ Z t k∇ u ( s ) k d s + α Z t k u t ( s ) k d s + β Z t k∇ u t ( s ) k d s + γ Z t k u t ( s ) k , Γ d s + b (cid:19)(cid:21) where the function η is defined by η ( t ) = (cid:18) k u ( t ) k + k u ( t ) k , Γ + σ Z t k∇ u ( s ) k d s + α Z t k u ( s ) k d s + β Z t k∇ u ( s ) k d s + γ Z t k u ( s ) k , Γ d s + b ( t + T ) (cid:19) (cid:18) k u t ( t ) k + k u t ( t ) k , Γ + σ Z t k∇ u t ( s ) k d s + α Z t k u t ( s ) k d s + β Z t k∇ u t ( s ) k d s + γ Z t k u t ( s ) k , Γ d s + b (cid:19) − (cid:20)Z Ω u t ( t ) u ( t )d x + Z Γ u t ( t ) u ( t )d σ + σ Z t dd t k∇ u ( s ) k d s + α Z t Z Ω u t ( t ) u ( t )d x d s + β Z t Z Ω ∇ u t ( t ) ∇ u ( t )d x d s + γ Z t Z Γ u t ( t ) u ( t )d σ d s + b ( t + T ) (cid:21) .
21y Cauchy-Schwarz inequality, for all t ∈ [0 , T ], we have η ( t ) ≥
0. As a consequence, we readthe following differential inequality F ( t ) F ′′ ( t ) − p + 24 F ′ ( t ) ≥ F ( t ) ζ ( t ) , ∀ t ∈ [ t, T ] . (6.8)where ζ ( t ) = − p k u t ( t ) k − (cid:18) a − Z t g ( s )d s (cid:19) k∇ u ( t ) k − b k∇ u ( t ) k − p k u t ( t ) k , Γ + 2 k u ( t ) k pp − ( p + 2) (cid:18) α Z t k u t ( s ) k d s + β Z t k∇ u t ( s ) k d s + γ Z t k u t ( s ) k , Γ d s + σ Z t k∇ u t ( s ) k d s (cid:19) − Z Ω ∇ u ( t ) · (cid:18)Z t g ( t − s )( ∇ u ( t ) − ∇ u ( s )) ds (cid:19) d x − pb = − pE ( t ) + ( p − (cid:18) a − Z t g ( s )d s (cid:19) k∇ u ( t ) k + p ( g ◦ ∇ u )( t ) + ( p − b k∇ u ( t ) k − ( p + 2) (cid:18) α Z t k u t ( s ) k d s + β Z t k∇ u t ( s ) k d s + γ Z t k u t ( s ) k , Γ d s + σ Z t k∇ u t ( s ) k d s (cid:19) − Z Ω ∇ u ( t ) · (cid:18)Z t g ( t − s )( ∇ u ( t ) − ∇ u ( s ))d s (cid:19) d x − pb. From the equality (4.1), ( G G
2) and m = 2 we have E ′ ( t ) = 12 ( g ′ ◦ ∇ u )( t ) − g ( t ) k∇ u ( t ) k − α k u t ( t ) k − β k∇ u t ( t ) k − γ k u t ( t ) k , Γ − σ (cid:18)
12 ddt k∇ u ( t ) k (cid:19) ≤ − α k u t ( t ) k − β k∇ u t ( t ) k − γ k u t ( t ) k , Γ − σ (cid:18)
12 ddt k∇ u ( t ) k (cid:19) . Thus, we can write ζ ( t ) ≥ − pE (0) + ( p − (cid:18) a − Z t g ( s )d s (cid:19) k∇ u ( t ) k + p ( g ◦ ∇ u )( t ) + ( p − b k∇ u ( t ) k + ( p − (cid:18) α Z t k u t ( s ) k d s + β Z t k∇ u t ( s ) k d s + γ Z t k u t ( s ) k , Γ d s + σ Z t k∇ u t ( s ) k d s (cid:19) − Z Ω ∇ u ( t ) · (cid:18)Z t g ( t − s )( ∇ u ( t ) − ∇ u ( s ))d s (cid:19) d x − pb. (6.9)By Young’s inequality, we have2 Z Ω ∇ u ( t ) · (cid:18)Z t g ( t − s )( ∇ u ( t ) − ∇ u ( s ))d s (cid:19) ≤ ε Z t g ( s ) k∇ u ( t ) k d s + ε ( g ◦ ∇ u )( t ) , (6.10)for any ε >
0. Substitute (6.10) for the fifth term of the right hand side of (6.9), we have ζ ( t ) ≥ − pE (0) + (cid:20) a ( p − − (cid:18) p − ε (cid:19) Z t g ( s )d s (cid:21) k∇ u ( t ) k + ( p − ε )( g ◦ ∇ u )( t )22 ( p − (cid:18) α Z t k u t ( s ) k d s + β Z t k∇ u t ( s ) k d s + γ Z t k u t ( s ) k , Γ d s + σ Z t k∇ u t ( s ) k d s (cid:19) + ( p − b k∇ u ( t ) k − pb. (6.11)On the one hand, if δ <
0, then E (0) <
0, we choose ε = p in (6.11) and b small enoughsuch that b ≤ − E (0). Then, by (6.3), we get ζ ( t ) ≥ (cid:20) a ( p − − (cid:18) p − p (cid:19) Z t g ( s )d s (cid:21) k∇ u ( t ) k + p ( − E (0) − b ) + ( p − b k∇ u ( t ) k + ( p − (cid:18) α Z t k u t ( s ) k d s + β Z t k∇ u t ( s ) k d s + γ Z t k u t ( s ) k , Γ d s + σ Z t k∇ u t ( s ) k d s (cid:19) ≥ (cid:20) a ( p − − (cid:18) p − p (cid:19) Z t g ( s )d s (cid:21) k∇ u ( t ) k + ( p − b k∇ u ( t ) k + ( p − (cid:18) α Z t k u t ( s ) k d s + β Z t k∇ u t ( s ) k d s + γ Z t k u t ( s ) k , Γ d s + σ Z t k∇ u t ( s ) k d s (cid:19) ≥ . (6.12)On the other hand, if 0 ≤ δ <
1, then 0 ≤ E (0) = δd < d , we choose ε = (1 − δ ) p + 2 δ in(6.11). Then, we have ζ ( t ) ≥ − pE (0) + δ ( p − g ◦ ∇ u )( t ) + ( p − b k∇ u ( t ) k + (cid:20) a ( p − − (cid:18) p − − δ ) p + 2 δ (cid:19) Z t g ( s )d s (cid:21) k∇ u ( t ) k + ( p − (cid:18) α Z t k u t ( s ) k d s + β Z t k∇ u t ( s ) k d s + γ Z t k u t ( s ) k , Γ d s + σ Z t k∇ u t ( s ) k d s (cid:19) . By (6.3), we obtain a ( p − − (cid:18) p − − δ ) p + 2 δ (cid:19) Z t g ( s )d s ≥ δ ( p − (cid:18) a − Z t g ( s )d s (cid:19) , and therefore, by (6.4), we have ζ ( t ) ≥ − pE (0) + δ ( p − (cid:20)(cid:18) a − Z t g ( s )d s (cid:19) k∇ u ( t ) k + ( g ◦ ∇ u )( t ) (cid:21) + ( p − b k∇ u ( t ) k + ( p − (cid:18) α Z t k u t ( s ) k d s + β Z t k∇ u t ( s ) k d s + γ Z t k u t ( s ) k , Γ d s + σ Z t k∇ u t ( s ) k d s (cid:19) ( p − (cid:18) α Z t k u t ( s ) k d s + β Z t k∇ u t ( s ) k d s + γ Z t k u t ( s ) k , Γ d s + σ Z t k∇ u t ( s ) k d s (cid:19) + ( p − b k∇ u ( t ) k + 2 p ( δd − E (0)) ≥ . (6.13)From (6.8), (6.12) and (6.13), we get F ( t ) F ′′ ( t ) − p + 24 F ′ ( t ) ≥ , ∀ t ∈ [ t, T ] . By (6.7), We can choose T sufficiently large such that F ′ (0) = 2 R Ω u u d x + 2 R Γ u u d σ +2 bT >
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