Global solutions and blow-up for a class of strongly damped wave equations systems

Yaojun YE , Lanlan LI

Front. Math. China ›› 2022, Vol. 17 ›› Issue (5) : 767 -782.

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Front. Math. China ›› 2022, Vol. 17 ›› Issue (5) : 767 -782. DOI: 10.1007/s11464-022-1025-y
RESEARCH ARTICLE
RESEARCH ARTICLE

Global solutions and blow-up for a class of strongly damped wave equations systems

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Abstract

The initial-boundary value problem for semilinear wave equation systems with a strong dissipative term in bounded domain is studied. The existence of global solutions for this problem is proved by using potential well method, and the exponential decay of global solutions is given through introducing an appropriate Lyapunov function. Meanwhile, blow-up of solutions in the unstable set is also obtained.

Keywords

Nonlinear wave equations systems / global solutions / exponential decay / blow-up

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Yaojun YE, Lanlan LI. Global solutions and blow-up for a class of strongly damped wave equations systems. Front. Math. China, 2022, 17(5): 767-782 DOI:10.1007/s11464-022-1025-y

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1 Introduction

In this paper, we are concerned with the following damped wave equations systems

utt Δu+μ1 ut ω1Δut=f1 (u,v ), (x,t) Ω× R+,

v ttΔv+ μ2vtω2Δ vt=f 2(u ,v), ( x,t) Ω× R+

with the initial-boundary value conditions

u(x,0) =u0(x) ,ut(x, 0)= u1( x),xΩ,

v(x,0) =v0(x) ,vt(x, 0)= v1( x),xΩ,

u(x,t) =v(x ,t)= 0,(x,t)Ω×R+,

where μi, ωi> 0(i=1,2) are real numbers, ΩRn is a bounded domain with smooth boundary Ω so that divergence theorem can be applied. fi( ,) :R2R(i=1 ,2) are given functions to be determined later.

Systems of nonlinear wave equations (1.1)−(1.5) go back to Reed [15] in 1976 who proposed a similar model in three space-dimensional space, but without the presence of the damping terms. When the strongly damped terms in (1.1) and (1.2) are replaced by nonlinear dissipative terms, problem (1.1)−(1.5) becomes the following initial-boundary value problem:

u ttΔu+a|ut|r1 2ut=f 1(u ,v), ( x,t) Ω× R+,

v ttΔv+a|vt|r2 2vt=f 2(u ,v), ( x,t) Ω× R+,

u(x,0) =u0(x) ,ut(x, 0)= u1( x),xΩ,

v(x,0) =v0(x) ,vt(x,0) =v1(x) ,xΩ,

u(x,t) =v(x ,t)= 0,(x,t)Ω×R+.

Agre and Rammaha [1] showed the existence of global weak solutions and gave the blow-up of solutions under the condition of the negative initial energy. Alves et al. [2] studied the global existence, uniform decay rate and blow-up of solutions in finite time under some conditions on the parameters in the systems and with the nonnegative initial energy and careful analysis involving the Nehari Manifold. When the damping terms are degenerate, Rammaha and Sakuntasathien [14] obtained the existence and uniqueness of global and local solutions to (1.6)−(1.10). In addition, they proved that weak solutions for the systems blow up in finite time whenever the initial energy is negative and the exponent of the source terms is more dominant than the exponent of damping terms. Quite recently, Li, Sun and Liu [9] dealt with global existence, uniform decay and blow-up of solutions of problem (1.6)−(1.10). Said-Houari [16] obtained the global existence and decay of solutions for problem (1.6)−(1.10) under some restrictions on the nonlinearity of the damping and source terms.

Consider the initial-boundary value problem of a single damped wave equation

u ttΔu+μut ωΔ ut=a | u|p2u,x Ω, t>0,

u(x,0) =u0(x) ,ut(x, 0)= u1( x),xΩ,

u(x,t) =0,xΩ, t0.

Gazzola and Squassina [4] studied the existence of local and global solutions of problem (1.11)−(1.13), and not only finite time blow-up for solutions starting in the unstable set was proved, but also high energy initial data for which the solution blows up were constructed. Later, Gerbi and Said-Houari [5-7] considered problem (1.11)−(1.13) or equation (1.11) with dynamic boundary conditions. They obtained the existence of local and global solutions and gave the asymptotic stability and blow-up result of solutions.

Introducing a strong damping term makes problem (1.1)−(1.5) different from the one considered in [1]. For this reason fewer results are, at the present time, known for the wave equation with the strong damping and many problems remain unsolved (see[4]).

Motivated by the above researches, in this paper, we prove the global existence for problem (1.1)−(1.5) by applying the potential well theory introduced by Sattinger [17] and Payne and Sattinger [13]. Moreover, we obtain the exponential decay and the blow-up result of solution for this problem.

We adopt the usual notations and convention. Let H1(Ω) denote the Sobolev space with the usual scalar products and norm. H0 1(Ω) denotes the closure in H1(Ω) of C0(Ω). For simplicity of notations, hereafter we denote by r the Lebesgue space Lr( Ω) norm and denotes L2(Ω) norm, and we write equivalent norm instead of H0 1 (Ω ) norm H0 1(Ω). In addition, Ci(i =0,1, 2,) denote various positive constants which depend on the known constants and may be different at each appearance.

This paper is organized as follows. In the next section, we are going to give some preliminaries. In Section 3, we will study the existence and exponential decay of global solutions to problem (1.1)−(1.5). Then in Section 4, we are devoted to the proof of blow-up result of solutions in finite time.

2 Preliminaries

We make the following assumptions on the functions f1(u, v),f2(u,v ) and parameter p:

(A1):1pnn2,n>2;1p< +, n2.

(A2): f 1(u ,v) and f2( u,v) are nonnegative real value functions from R2 to R and satisfy

f1(u, v)= b1 |u+ v|2(p 1)(u+v )+b 2|u|p2u|v|p, f2( u,v) =b1|u+v|2( p1)(u +v)+ b2|v|p2v|u|p,

where b1,b 2>0 and p>1 are constants.

It is easy to see from (A2) that

uf1(u, v)+v f2(u,v)=2pF(u,v ),( u,v) R2,

where

F(u,v) =b1 2p|u+v|2p +b2 p|uv | p.

Moreover, a quick computation will show that there exist two positive constants C0 and C1 such that the following inequality holds (see [11]):

C0 2p(|u|2p +|v | 2p)F( u,v) C1 2p(|u|2p +|v | 2p).

Now, we define the following functionals:

J(t)=J ([u(t),v(t) ])=12( u2 +v2 ) ΩF(u,v )dx,

K(t)=K ([u(t),v(t) ])=( u 2+v 2) 2pΩF(u,v )dx,

for [u,v ]H 01(Ω)× H0 1(Ω), and denote the total energy related to equations (1.1) and (1.2) by

E(t)=12(ut| 2+vt 2+ u2 +v2 ) ΩF(u,v )dx = 12( ut2+ vt2)+J([u ,v])

for [u,v ]H 01(Ω)× H0 1(Ω),t0 and

E(0)=12(u 12+v1 2+u02+ v0 2) ΩF(u0,v 0)dx

is the initial total energy.

As in [13], the mountain pass value of J(t) (also known as potential well depth) is defined as:

d=inf{supλ0J[λu,λ v]:[u, v] H01(Ω)×H 01( Ω) /{0}} .

We introduce the so-called Nehari manifold (see [12, 18]) as follows:

N={ [u,v ]H 01( Ω) ×H01(Ω)/{0};K (t)=0}.

N separates the two unbounded sets:

N+={[u ,v] H01(Ω)×H 01( Ω) /{0};K (t)>0}{0}.

and

N={[u ,v] H01(Ω)×H 01( Ω) /{0};K (t)<0}.

Then, the stable set W and unstable set U can be defined respectively by

W={ [u,v ]H 01( Ω) ×H01(Ω):J(t)d} N+

and

U={ [u,v ]H 01( Ω) ×H01(Ω):J(t)d} N.

It is readily seen that the potential well depth d defined in (2.10) may also be characterized as

d=inf[u,v ]NJ ([u,v] ).

As it was remarked by [7], this alternative characterization of d shows that

β=dist( 0,N)=inf[u ,v] N(u2+ v 2)12=( 2pdp 1)1 2>0.

For the applications through this paper, we introduce the definition of solutions to problem (1.1)−(1.5) given by Gazzola and Squassina in [4] and list up some known lemmas.

Definition 2.1 A pair of functions [u,v ] is said to be a weak solution of (1.1)−(1.5) on [0,T] if

[u,v ]C([0 ,T], H0 1(Ω)×H01(Ω)) C2([0, T], H1×H 1(Ω)),

and

[ut,vt ]C([0 ,T], L2(Ω)×L2(Ω))L2 ([0,T] ,H01(Ω)× H0 1(Ω))

satisfy

Ωuttφdx+Ωuφdx+μ1Ω utφdx+ω1 Ω utφ dx=Ωf1(u,v )φdx,

Ωvttψdx+Ωvψdx+μ2Ω vtψdx+ω2 Ω vtψ dx=Ωf2(u,v )ψdx

for all test functions [φ ,ψ]H01(Ω)×H 01(Ω) and almost all t[0, T].

Lemma 2.1  Let r be a real number with 2r<+ if n2 and 2r2n n2 if n>2. Then there is a constant C depending on Ω and r such that

u rC u, uH0 1(Ω).

Lemma 2.2 (Young inequality)  Let X,Y and ε be positive constants and 1<ξ,ζ<+, 1ξ +1ζ=1. Then one has the inequality

XY εξ Xξξ+Yζζεζ.

Lemma 2.3  Let [u(t) ,v(t )] be a pair of solutions for problem (1.1)−(1.5). Then E(t) is a nonincreasing function for t>0 and

E (t)=(μ1u t2 2 +μ2v t2 2 +ω1ut 2+ω2vt 2) 0.

Proof Multiplying equation (1.1) by u t and (1.2) by vt, and integrating over Ω ×[0,t]. Then, adding them together, and integrating by parts, we get

E(t)=E (0)0 t[μ1ut(s) 22+ μ2vt(s)2 2 +ω1ut(s) 2+ω2 vt(s)2] ds

for t0. Being the primitive of an integrable function, E(t) is absolutely continuous and equality (2.13) is fulfilled.

We conclude this section by stating a local existence result of problem (1.1)−(1.5), which can be established by using a similar way as done in combination of the arguments in [1,3,4,10,19].

Theorem 2.1 (Local existence)  Suppose that (A1) holds. If [u 0,v0]H01(Ω)×H01(Ω) and [u 1,v1]L2(Ω)×L2(Ω), then there exists T>0 such that problem (1.1)−(1.5) has a unique local solution [u(t ),v(t) ], which satisfies

[u,v] C([0,T);H 01(Ω)× H0 1(Ω)),

utC([0,T ); L2( Ω) )L 2([0 ,T); H01(Ω)),

vtC([0,T ); L2( Ω) )L 2([0 ,T); H01(Ω)).

Moreover, at least one of the following statements holds true: (1)u t2+vt 2+u 2+v 2+ as t T; (2) T=+.

3 Global existence and exponential decay

We can now proceed in study of the existence of global solution for problem (1.1)−(1.5). For this purpose, we need the following lemmas.

Lemma 3.1  Suppose that (A1) and (A2) hold. If [u,v] H0 1(Ω)×H01(Ω), then

d=p12p ( 1C2 B ) 2pp1>0,

where B is the optimal Sobolev's constant from H01(Ω) to L2p(Ω) and the positive constant C2 is determined later.

Proof Since

J(λ[u,v])=λ 22(u2 +v2 )λ 2pΩF(u ,v)dx,

we get

ddλJ(λ[u,v])=λ( u2 +v2 )2pλ 2p1ΩF(u ,v)dx,

d2dλ 2J(λ[u,v])=( u2 +v2 )2p(2p 1) λ2(p 1) Ω F(u, v)dx.

Let ddλJ(λ[u,v])=0. Then we obtain

λ1 =( u 2+v 22p Ω F(u, v)dx) 12(p1).

As λ=λ1, an elementary calculation shows that

d2dλ 2J(λ[u,v])|λ =λ1=2(p1 )(u2+ v 2)< 0.

Therefore, we have

supλ0J(λ[u,v] )=J(λ1[u, v])=p 12p[( u2 +v2 )p 2pΩF(u,v )dx]1 p1.

By Minkowski's inequality and Lemma 2.1, we attain

u+v2p2 2(u2p2+v2p2)2B 2( u2+ v2).

Also, we gain from Hölder inequality and Lemma 2.2 that

uvp u 2p v 2p12(u2p2+ v 2p2) B22(u2+ v2).

From (2.4), (3.3) and (3.4), we deduce that

ΩF( u,v) dx( C2B) 2 p2 p(u2+ v2)p,

where C2=(2pb1 +b2 2p 1 ) 12p.

It follows from (3.2) and (3.5) that

d=p12p ( 1C2 B ) 2pp1>0.

Thus, we complete the proof of Lemma 3.1.

Lemma 3.2  Assume that (A1) and (A2) hold. If [u 0,v0]N+,[u1, v1]L2 (Ω )×L 2(Ω) and E(0)< d, then [u(t ),v(t) ]N+ for each t[0 ,T).

Proof By [ u0, v0] N+, we see that K([u0,v 0])>0. Then, in virtue of the continuity, there exists t [0, T) such that K(t) 0 for t[0, t). We get from (2.6) and (2.7) that

J(t)= p12p( u2+ v2 )+1 2pK (t),

which implies that

0<p12p( u2+ v2 )J(t) , t[0 ,t).

Therefore, we have

u2+ v2 2p p1J(t ).

From (2.8), (3.6), (3.7) and Lemma 2.3, we obtain

u2+ v2 2p p1E(t )2pp 1E(0), t[0 ,t).

It yields from (3.5) and (3.8) that

2pΩF(u ,v)dx ( C2B )2p( u 2+v2)p =(C2B) 2p( u 2+v2)p 1( u2+ v2) θ(u2+ v2),

where

θ=(C2 B)2p( 2pp 1E(0))p1.

By E(0)<d and (3.1), we find

θ=(C2 B)2p( 2pp 1E(0))p1<1.

We conclude from (3.9) and (3.10) that

2p ΩF(u,v )dx< u 2+v2.

Hence K(t)>0, t[0,t), which shows that [u(t) ,v(t )]N+, t[0 ,t).

Noting

limt t (C2B)2p( 2pp 1E(t))p1<(C2 B)2p( 2pp 1E(0))p1<1,

we repeat steps (3.6)−(3.11) to extend t to 2t. We continue this procedure until [u(t ),v(t) ]N+, t[0 ,T).

The following theorem shows that the solution obtained in Theorem 2.1 is a global solution.

Theorem 3.1  Assume that (A1) and (A2) hold. If [u 0,v0]W,[ u1, v1] L2(Ω)×L 2(Ω) and E(0)< d, then the local solution furnished in Theorem 2.1 is a global solution and T may be taken arbitrarily large.

Proof It suffices to show that ut 2+vt 2+ u2 +v2 is bounded independently of t. Under the hypotheses in Theorem 3.1, by Lemma 3.2, we have [u,v] W on [0,T). So, the following formula holds on [0,T):

J(t)=12(u2+ v 2) ΩF(u ,v)dxp12p( u 2+v 2).

We get from (3.12) that

12( ut2+ vt2)+p 12p(u2+ v 2) 12( ut2+ vt2)+J(t) =E(t )E(0) <d,

which implies that

ut2+ vt2+ u2 +v2 max (2,2pp 1)d <+ .

The above inequality and the continuation principle lead to the global existence of solutions for problem (1.1)−(1.5).

The following theorem shows that the global solutions of problem (1.1)−(1.5) is exponential decay.

Theorem 3.2  Assume that (A1) and (A2) are valid. If [ u0, v0] N+, [u 1,v1]L2(Ω)×L2(Ω) and E(0)<d, then there exist two positive constants M and η independent of t such that the global solution has the following exponential decay property:

0<E(t) M e kt,t 0.

Proof By Lemma 3.2, we see that [u( t),v (t)]N+ for all t0. Thus, we have 0<E(t) for all t0. In order to prove the exponential decay of global solutions, we define

F(t)=E (t)+ε Ω(uut +vvt) dx+1 2ε Ω(ω1|u|2+ω2|v |2)dx,

where ε>0 will be determined later. It is easy to prove that there exist two positive constants ξ1 and ξ2 depending on ε such that

ξ1E(t)F(t) ξ2E(t)

for all t0.

In fact, from (2.1), Lemma 2.1 and (3.13), we get

F(t)E(t) +ε2( ut2+ vt2+ u 2+v2)+ε 2(ω1 u 2+ω2 v2 ) ( 1+ε)E(t )+ε 2[B 2+ max{ω1,ω 2} ]( u 2+v 2) [(1+ε)+ pp1ε(B2+max {ω 1,ω2})]E(t)=ξ2E(t).

On the other hand, by using Lemma 2.2, we have

Ωuutdx1 4γut 2+γu 2, Ω vvtdx14γ vt2+ γ v 2.

Therefore, we obtain the following inequality:

F(t)E(t) ε4 γ( ut2+ vt2)εγ(u2+ v 2)+ ε2(ω1u2+ω2 v 2) E (t)ε4γ( ut2+ vt2)+ε[12min{ω1,ω 2} B 2γ ]( u2 +v2 ).

By choosing γ small enough such that γ min{ω1,ω 2}2B 2, it follows from (2.8) and (3.18) that

F(t)E (t)ε4γ( ut2+ vt2)J(t) +(12ε4γ)(ut 2+vt 2).

We then pick ε>0 so small (i.e., ε2 γ) that

F(t)min{1,1ε2γ)} (J (t)+12 ut2)=ξ1E(t ).

From (3.16) and (3.19), inequality (3.15) holds.

By differentiating (3.14) and using equation (1.1) and (1.2), and combining Lemma 2.3, we obtain

F (t)= (ω1ut 2+ω2 vt 2)+ (ε μ1) ut2+ (ε μ2) vt2 ε( u 2+v 2)+2pε ΩF(u,v )dxε Ω (μ1uut +μ2vvt)dx.

It yields form (3.9), (3.17) and (3.20) that

F (t)= (ω1ut 2+ω2 vt 2) +(ε( 1+μ14δ) μ1 )u t2+(ε( 1+μ24δ) μ2 )v t2+ε[δB 2max{μ1, μ2 }+θ 1]( u2 +v2 ).

By (3.10), we find that θ1 <0. Choosing 0<δ<1θ B2max{μ1,μ 2} such that δB 2max{μ1, μ2 }+θ 1<0.

Let η=1θδB 2max{μ1, μ2 }. Then η >0. Thus, for any positive constant Λ>0, from (2.8) and (3.21), we have

F (t) ΛεE(t)+ [ε (1+ Λ2+μ1 4δ )μ 1]ut 2 +[ε (1+ Λ2+μ2 4δ )μ 2]vt 2(ω1ut 2+ω2vt 2) +ε (Λ 2η)( u 2+v 2).

Choosing Λ2η and ε so small enough that

[ε(1+Λ2+μi 4δ )μ i]< 0,i=1,2,

inequality (3.22) implies that

F (t)ΛεE(t), t0.

We conclude from (3.15) and (3.23) that

F (t)kF(t),t 0,

where k=Λε/ ξ2>0.

Integrating the differential inequality (3.24) from 0 to t gives the following exponential decay estimate for function F(t):

F(t)F (0)ekt, t 0.

Consequently, we obtain from (3.15) once again that

E(t)M ekt,t 0,

where M=F( 0)/ξ 1. This completes the proof of Theorem 3.2.

4 Blow-up result

In this section, we are concerned with the blow-up property of solutions for problem (1.1)−(1.5) and give the estimate of lifespan of solutions. For this purpose, we give the following lemmas.

Lemma 4.1 [8]  Suppose that P(t) C2, P(t)0 satisfies the inequality

P(t)P(t )(1+ρ)P2(t) 0

for certain real number ρ >0 and P(0)>0 ,P(0)>0. Then there exists a real number T such that 0<T P(0)ρ P(0) and P(t) + as t T.

Lemma 4.2  Let [u(t) ,v(t )] be a pair of solutions of (1.1)−(1.5) which is given by Theorem 2.1. If [ u0, v0] U and E(0)<d, then [u(t),v(t) ]U and E(t) <d for all t0.

Proof It follows from the conditions in Lemma 4.2 and Lemma 2.3 that

E(t)E (0)<d, t [0,T ).

Therefore, by (2.8), we have

J(t)E (t)<d, t [0,T ).

Next, let us assume by contradiction that there exists t[0,T ) such that u(t)U. Then by continuity, we have K(t)=0. This implies that [u( t),v(t)] N. We get from (2.11) that J( t)d, which is contradiction with (4.1). Consequently, Lemma 4.2 holds.

Theorem 4.1  Assume that (A1) and (A2) are valid. If [ u0, v0] U,[u1,v1]L2(Ω)×L2(Ω) and 0<E(0) <d. Then the solutions [u(t),v(t) ] in Theorem 2.1 of problem (1.1)−(1.5) blow up in finite time T <+, which means that

limt T(u(t)2+ v(t)2 )=+.

Proof From (2.10) and (3.2), we have

dsupλ 0J(λ[u,v])= p12p [ ( u 2+v 2)p2pΩF(u, v)dx] 1p1.

By [u0 ,v0] U, E(0) <d and Lemma 4.2, we obtain [u,v] U for all t[0,T]. Thus, we get

K(t)=( u2 +v2 )2pΩF(u ,v)dx<0

for all t[0,T]. Consequently, by using this last inequality, (4.2) becomes

dp12p( u2+ v2 ),t[0,T ],

which implies that

u2+ v2 2dp p1,t[0,T].

Assume by contradiction that solution [u(t) ,v(t )] is global. Then for any T>0, we define Θ(t) :[0, T][0, +) by

Θ(t)=( u(t)2+v(t)2 )+ 0t( μ1u(s)2 +μ2v(s) 2)ds+ 0t (ω1u(s)2 +ω2v(s)2)ds +(T t)(μ1 u02+ μ2v0 2+ω1 u0 2+ω2 v0 2).

Note that Θ(t) >0 for all t[0, T]. By the continuity of the function Θ(t), there exists λ>0 (independent of the choice of T) such that

Θ(t) λ, t [0,T ].

By differentiating both sides of (4.4) on t, we get

Θ(t)=2Ω(uut+vv t)dx+(μ1 u(t)2+μ2v(t)2) +(ω1u(t)2+ ω2 v(t)2)( μ1u0 2+μ2 v02+ ω1 u02+ ω2 v02) =2 Ω(uut +vvt) dx+2 μ1 0tΩu(s) ut(s)dxds +2μ20tΩv(s )vt(s)dxds+2ω 10tΩu(s) ut(s)dxds +2ω20tΩv(s) vt(s)dxds.

Taking the derivative of the function Θ (t) in (4.6), we obtain

Θ(t) =2(ut 2+vt 2)+2 Ω(uutt+vv tt) dx +2μ 1Ωuutdx+2μ 2Ωvvtdx+2ω 1Ωu utdx+2ω2Ωvvt dx .

It deduces from (1.1), (1.2) and (4.7) that

Θ (t)=2 (ut(t) 2+vt(t) 2+2p Ω F(u, v)dx u(t) 2 v(t)2) .

Combining (4.4), (4.6) with (4.8), we receive that

ΘΘ(t) p+3 4Θ (t)2=2Θ(t)[u t(t )2+ vt(t)2 +2p ΩF(u ,v)dx u(t) 2 v(t)2](p+3 )[Θ(t)(Tt )( μ1u0 2+μ2 v02+ ω1 u02+ ω2 v02)] ×[ ut(t)2 + vt(t)2 +μ1 0t ut(s)2ds+μ2 0tv t(s )2ds +ω10t ut(s)2 ds+ω 20t vt(s)2 ds]+ (r+2 )χ(t),

where

χ (t)= [u(t)2+ v(t) 2+0t( μ1u(s)2 +μ2v(s) 2+ω1u(s)2 +ω2v(s)2)ds]× [ ut(t)2 + vt(t)2+ 0t (μ1 ut( s)2+ μ2vt(s)2+ω1 ut(s)2 +ω2vt(s) 2)ds][Ω(uut+vv t)dx+μ1 0tΩu(s)ut (s) dx ds+μ2 0tΩv(s)vt (s) dx ds+ ω1 0tΩu(s)ut(s) dxds+ω2 0tΩv(s)vt(s)dxds]2.

By applying Schwarz inequality, we gain that

(Ωuutdx)2 u(t)2ut(t)2, (Ωvvtdx )2v(t) 2 vt(t)2 ,

(μ10tΩuutdxds)2μ1 20 t u(s)2 ds0tut(s) 2ds,

(μ20tΩvvtdxds)2μ2 20 t v(s)2 ds0tvt(s) 2ds,

(ω10tΩu(s) ut(s)dxds )2 ω12 0tu(s)2ds 0tut(s) 2ds,

(ω20tΩv(s) vt(s)dxds )2 ω22 0tv(s)2ds 0tvt(s) 2ds,

2ω1 0tΩu(s)ut(s)dxdsΩuutdx ω1 ut( t)20t u(s)2 ds+ω 1 u(t)2 0t ut(s)2 ds,

2ω2 0tΩv(s)vt(s)dxdsΩvvtdx ω2 vt( t)20t v(s)2 ds+ω 2 v(t)2 0t vt(s)2 ds,

and

2μ1 0tΩu(s)ut (s) dx ds Ω uutdx μ1ut(t)2 0tu(s)2 ds+μ1u(t) 20t ut(s)2 ds.

2μ2 0tΩv(s)vt (s) dx ds Ω vvtdx μ2vt(t)2 0tv(s)2 ds+μ2v(t) 20t vt(s)2 ds.

These inequalities (4.11)−(4.19) entail χ(t) 0 for all t[0, T]. Therefore, we reach the following differential inequality from (4.9) that

Θ Θ (t)p+34Θ(t)2Θ(t)Γ(t),t[0,T],

where

Γ(t)=2[ ut(t)2 + vt(t)2 +2p ΩF(u,v )dx u(t)2v(t)2] (p+3) [ ut(t)2 + vt(t)2 +μ1 0t ut(s)2ds+μ2 0tv t(s )2ds+ ω1 0t ut(s)2 ds+ω2 0tvt(s) 2ds].

From (2.8), we attain that

Γ(t)= 2pE( t)+(p1 )(u(t)2 v(t)2 )] (p+3 )[μ10tut(s) 2ds+μ20t vt(s)2 ds+ ω1 0t ut(s)2 ds+ω2 0tvt(s) 2ds].

By (2.14), we get

Γ(t)= 2pE( 0)+(p1 )(u(t)2 v(t)2 )]+(p1 )[μ10tut(s) 2ds+μ20t vt(s)2 ds+ ω1 0t ut(s)2 ds+ω2 0tvt(s) 2ds].

We conclude from (4.3) and E(0)<d that

Γ(t) >2p(dE(0) )+(p1)[μ10tut(s) 2ds+μ20t vt(s)2 ds +ω1 0t ut(s)2 ds+ω 20t vt(s)2 ds] >(p1 )[μ10tut(s) 2ds+μ20t vt(s)2 ds +ω1 0t ut(s)2 ds+ω 20t vt(s)2 ds] 0.

Hence, there exists ϑ>0 which is independent of T such that

Γ(t) ϑ, t 0.

It yields from (4.3), (4.20) and (4.25) that

Θ Θ (t)p+34Θ(t)2λ ϑ,t[0,T ].

Let ρ=p 14>0. Then, by Lemma 4.1, there is T such that

0<T <Θ(0)ρ Θ(0)T

and

limt TΘ (t)=+ .

From Lemma 2.1 and (4.4) (the definition of Θ (t)), (4.28) implies that

limt T(u(t)2+ v(t)2 )=+.

Thus, we cannot suppose that solutions of (1.1)−(1.5) is global.

This finishes the proof of Theorem 4.1.

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