AAA Abstract and Applied Analysis 1687-0409 1085-3375 Hindawi Publishing Corporation 615345 10.1155/2012/615345 615345 Research Article On Existence, Uniform Decay Rates, and Blow-Up for Solutions of a Nonlinear Wave Equation with Dissipative and Source Liu Xiaopan Apreutesei Narcisa C. Department of Mathematics Southeast University Nanjing 210018 China seu.edu.cn 2012 19 9 2012 2012 24 05 2012 17 07 2012 17 07 2012 2012 Copyright © 2012 Xiaopan Liu. This is an open access article distributed under the Creative Commons Attribution License, which permits unrestricted use, distribution, and reproduction in any medium, provided the original work is properly cited.

This paper studies the blow-up and existence, and asymptotic behaviors of the solution of a nonlinear hyperbolic equation with dissipative and source terms. By using Galerkin procedure and the perturbed energy method, the local and global existence of solution is established. In addition, by the concave method, the blow-up of solutions can be obtained.

1. Introduction

In this paper, we investigate the following nonlinear wave equation: (1.1)|ut|ρutt+Δ2u+|ut|mut+γΔ2ut=i=1Nxiσi(uxi)+|u|p-1u,(x,t)Ω×(0,T),u=un=0,(x,t)Ω×(0,T],u(x,0)=u0(x),ut(x,0)=u1(x),xΩ, where Ω is a bounded domain in n with smooth boundary Ω, Δ is a Laplace operator, and u/n|Ω indicates derivative of u in outward normal direction of Ω. In addition to, if n>3,  1<p<(n+2)/(n-2) and if n=1,2, then p>1. γ0 is a constant, σi(s)(i=1,,N) are given in (A1) later.

In 1968, Greenberg et al.  first suggested and studied the following equation: (1.2)utt-uxxt=σ(ux)x. Under the condition σ(s)>0 and higher smooth conditions on σ(s) and initial data, they claimed the global existence of classical solutions for the initial boundary value problem of (1.2).

The multidimensional form of the following: (1.3)utt-Δut-i=1Nxiai(x,t,uxi)=f(x,t) was first studied by Clement [2, 3]. Exploiting the monotone operator method, he obtained the global existence of weak solutions for the initial boundary value problem of (1.2).

Our model comes from . In , Yang has studied the global existence, asymptotic behavior, and blow-up of solutions for a nonlinear wave equations with dissipative term: (1.4)utt+Δ2u+λut=i=1Nxiσi(uxi), with the same initial and boundary conditions as that of (1.1). In our model, we add damping and source terms which enhance the difficulty of proving the existence and decay of solution of (1.2).

More related studies of the damped hyperbolic equation with dissipative term or damping term can be found in papers .

The paper is organized as follows. In Section 2, we present some notations, and results needed later and main results. Section 3 contains the statement and the proofs of the decay of solutions. Section 4 gives the statement and the proofs of the blow-up of solutions.

2. Preliminaries

We first introduce the following abbreviations: (2.1)QT=Ω×(0,T),Lp=Lp(Ω),Wm,p=Wm,p(Ω),Hm=Wm,2,H0m=W0m,2,·2=·L2,·1,2=·H01. Let (·,·) denote the L2-inner product. We denote the dual of W01,p by W-1,p, with p=p/(p-1), and ·-1,p=·W-1,p. Now we make the following assumptions:

σiC1(R),σi(s)s>0 and (2.2)C1|s|r|σi(s)|C2|s|r,

for some r1, if n=1,2, else if n3, then rn/(n-2), so that u2rB1Δu2, i=1,2,,N, where B1 is the optimal embedding constant.

The initial data (2.3)u0H02,u1H01.

m<ρ+1; If n>3, then 0<ρ<m<4/(n-2) and if n=1,2, then 0<ρ<m.

If n>3, 1<p<(n+2)/(n-2) and if n=1,2, then p>1. Without loss of generality, we assume that r<p.

And

σiC1(R),  σi(s)s<0, and (2.4)C1|s|r|σi(s)|C2|s|r,

for some r1, if n=1,2, else if n3, then rn/(n-2), so that u2rB1Δu2, i=1,2,,N. where B1 is the optimal embedding constant, and C2<(r+1)/(p+1).

If n>3, then m<r-1<p-1<4/(n-2) and if n=1,2, then 0<m<r-1<p-1.

ρ+1<r, and if n>3, then ρ<4/(n-2) and if n=1,2, then 0<ρ.

Throughout this paper, we use the embedding H02(Ω)Lq(Ω) which implies uqB2Δu2 when (2.5)2q2nn-2if  n3,q1if  n=1,2, where B2 is an optimal embedding constant. We introduce the following functionals: (2.6)Ai(s)=0sσi(η)dη,(2.7)E(t)=1ρ+2utρ+2ρ+2+12Δu22+i=1NΩAi(uxi)dx-up+1p+1p+1,(2.8)J(t)=12Δu22-C2r+1ur+1r+1-up+1p+1p+1,(2.9)I(t)=Δu22-C2ur+1r+1-up+1p+1. Because r<p, we have (2.10)J(t)=I(t)r+1+r-12(r+1)Δu22+p-r(p+1)(r+1)up+1p+1.

Theorem 2.1.

Assume that (A1)–(A4) hold, then problem (1.1) has a unique solution u satisfying (2.11)uL([0,T];H02)W1,([0,T];Lρ+2);utL2([0,T];H02)L([0,T];Lρ+2), where T<1.

Theorem 2.2.

Assume that (A1)–(A4) hold, u is the local solution of the problem (1.1). And (2.12)C0((2(r+1)r-1E(0))(r-1)/2+(2(r+1)r-1E(0))(p-1)/2)  <1,I(0)>0, where C0=max{B1C2,B2}, then u is a global bounded solution, moreover, (2.13)E(t)M1+t(1+t1/(m+2)+t(m-ρ)/(m+2)+t1/2),t[0,),(2.14)limtΔu22=limtutρ+2ρ+2=0, where M>0 is a constant.

Theorem 2.3.

Assume that (A1)–(A4) and (2.15)pB2p+1(2(r+1)r-1)(p+1)/2(E(0))(p-1)/2<1 hold, u is the local solution of the problem (2.7). Consider (2.12) are satisfied, then (2.16)E(t)Ke-κt,t[0,), where K,κ>0 are constants.

Remark 2.4.

When γ>0, we will use perturbed energy method, which is different to the method of the proof of Theorem 2.2, to prove Theorem 2.3.

Theorem 2.5.

Assume that (B1)–(B3), (A4) hold, and there exist some (u0,u1)H2(Ω)×Lρ+2(Ω), then the solution of the problem (1.1) blows-up at the limited time T*>0.

3. Decay of Solutions

In this section, we prove Theorems 2.12.3. First, we give the following Lemma.

Lemma 3.1 (see [<xref ref-type="bibr" rid="B3">2</xref>]).

Let Ω be any bounded domain in RN, {ωk}k=1 be an orthogonal basis in L2(Ω). Then for any ε>0, there exists a positive number Nε such that (3.1)u(k=1Nε(u,ωk)2)1/2+εu1,p, for all uW01,p  (2p<).

Proof of Theorem <xref ref-type="statement" rid="thm1">2.1</xref>.

We look for approximate solutions un(t) of problem (1.1) of the form (3.2)un(t):=j=1nTjn(t)ωj, where {ωj}j=1n is an orthogonal basis in H02, and also in L2, and the coefficients {Tjn}j=1n satisfy Tjn(t)=(un(t),ωj) with (3.3)(|utn(t)|ρuttn(t),ωj)+(Δ2un(t),ωj)+(|utn|mutn(t),ωj)+γ(Δ2utn(t),ωj)=i(xiσi(uxin(t)),ωj)+(|un(t)|p-1un(t),ωj),t>0,j=1,,n,un(0)=u0n,utn(0)=u1n. Since C0 is dense in H02 and L2, we choose u0n,u1nC0 such that (3.4)u0nu0in  H02,u1nu1in  H01as  n. The above system of o.d.e. has a local solution un(t) defined in some interval [0,Tn). The following will prove that the Tn can be substituted by some T>0.

Multiply (3.3) by Tjn'(t) and summing up about j, we get (3.5)ddt[1ρ+2utnρ+2ρ+2+12Δun22]+utnm+2m+2+γΔutn22=i=1N(σin(uxi),(uxin)t)+(|un|p-1un,utn). A simple integration of (3.5) over (0,t) leads to (3.6)1ρ+2utnρ+2ρ+2+12Δun22+0t(usnm+2m+2+γusn22)ds=C3+i=1N0t(σin(uxi),(uxin)s)ds+0t(|un|p-1un,usn)ds, where C3=1/(ρ+2)u1nρ+2ρ+2+(1/2)Δu0n22>0.

We now estimate the last two terms at the right-hand side of (3.6). Using Hӧlder inequality, Young's inequality, and the embedding theorem, we know there exist q,w>0, satisfying that (3.7)1q+1w+n-22n=1, where (3.8)(r-1)q=2nn-2,1w=n+22n-(r-1)(n-2)2nn-22n,(p-1)q=2nn-2,1w=n+22n-(p-1)(n-2)2nn-22n. Consider the following: (3.9)|i=1N0t(σi(uxin),(uxin)s)ds|C20tΩ|un|r-1|un||usn|dxdsC20tun(r-1)qr-1unwusn2n/(n-2)dsC2C0tΔun2r-1Δun2Δusn2ds=C2C0tΔun2rΔusn2dsC2C(12ϵ0tΔun22rds+ϵ20tΔusn22ds)C2C(12ϵ(0tΔun22pds)r/p+12ϵ(0t1ds)(p-r)/p+ϵ20tΔusn22ds)C2C12ϵpr(0tΔun22pds)+C2C12ϵp-rp+ϵ2C2C0tΔusn22ds,ϵ>0, assuming that t<1.

Similarly, (3.10)0t(|un|p-1un,usn)ds0tun(p-1)qp-1unwusn2n/(n-2)dsC0tΔun2p-1Δun2Δusn2dsC12ϵ0tΔun22pds+ϵ2C0tΔusn22ds,ϵ>0. Using (3.6)–(3.10), we have (3.11)0tusnm+2m+2ds+γ0tΔusn22ds+12Δun22+1ρ+2utnρ+2ρ+2C4+C(ε)0tΔun22pds+ϵ2C0tΔusn22. Choosing ϵ=(1/C)γ in (3.11), we have (3.12)0tusnm+2m+2ds+12Δun22+1ρ+2utnρ+2ρ+2+γ20tΔusn22dsC4+C0tΔun22pds, where C4=C3+((C2C)/2ϵ)((p-r)/p). Assuming Yn(t)=2(C4+C0tΔun22p), we have (3.13)Yn(t)2CYnp. A simple integration of (3.13) over (0,t) leads to (3.14)Yn(t)[Yn1-p(0)-2C(p-1)t]-1/(p-1); this implies that (3.15)Δun22+2ρ+2utnρ+2ρ+2[Yn1-p(0)-2C(p-1)t]-1/(p-1). Though Yn(t) may blow up, there exists 0<T<min{1,Tn} satisfying (3.16)Δun22+utnρ+2ρ+2C,t[0,T], where C is independent of n.

Moreover, (3.17)0tusnm+2m+2ds+0tΔusn22dsC. By (3.17), (3.18)0t|usn(s)|musn(s)(m+2)/(m+1)(m+2)/(m+1)ds0tusn(s)m+2m+2dsC,t[0,T]. By (3.16), (3.19)|un(s)|p-1un(s)(p+1)/p(p+1)/pun(s)p+1p+1Δun(s)2p+1C. From (3.16)–(3.19), we have (3.20)unL([0,T];H02);utnL2([0,T];H02)L([0,T];Lρ+2). So the solution un(t) of problem (3.3) exists on [0,T] for each n. On the other hand, we can extract a subsequence from un, still denoted by un, such that (3.21)unuweak*  in  L([0,T];H02),(3.22)utnutweak*  in  L([0,T];Lρ+2)L2([0,T];H02), as n. By (3.21), the Sobolev embedding theorem and the continuity of σi(s), for t[0,T], (3.23)un(t)u(t)strongly  in  L2,a.e.  on  Ω,σi(uxin(t))σi(uxi(t)),a.e.  on  Ω,i=1,N, as n. By Lemma 3.1, (3.21)-(3.22), for any ε>0, there exist positive constant N1ε and N2ε independent of un and utn, respectively, such that, as n, (3.24)un(t)-u(t)  [k=1N1ε(un-u,ωk)2]1/2+εun(t)-u(t)1,2Mε,0Tutn(s)-ut(s)22ds2(k=1N2ε0T(utn(s)-ut(s),ωk)2ds0Tutn(s)-ut(s)22ds2+ε20Tutn(s)-ut(s)1,22dsk=1N2ϵ)Mε2.

By the arbitrariness of ε we get (3.25)unustrongly  in  L([0,T];L2),a.e.  on  QT,utnutstrongly  in  L2(QT),a.e.  on  QT. From the continuity of |ut|mut and (3.25) we know that |utn|mutn|ut|mut a.e. on QT. With the same methods used above we easily get |un|p-1un|u|p-1u a.e. on QT. Integrating (3.3) over (0,t),  t<T gets (3.26)1ρ+1(|utn(t)|ρ+1,ωj)+0t(Δun(s),Δωj)ds+0t(|usn|musn(s),ωj)ds+0tγ(Δusn(s),Δωj)ds=i0t(xiσi(uxin(s)),ωj)ds+0t(|un(s)|p-1un(s),ωj)ds,t>0,j=1,,n. Exploiting (3.16)-(3.17), we have (3.27)0t(Δun(s),Δωj)dsC0tΔun22dsCT,(|utn(t)|mutn(t),ωj)utnρ+2m+1ωj((ρ+2)/(m+1))utnρ+2ρ+2ωj((ρ+2)/(m+1))((ρ+2)/(m+1))C,0t(|usn(s)|musn(s),ωj)ds0t(usnm+2m+2+ωjm+2m+2)dsCT,0t(|un(s)|p-1un(s),ωj)dsCT.

Let n in (3.26) and we deduce from (3.23), (3.27) and the Lebesgue-dominated convergence theorem that (3.28)1ρ+1(|ut(t)|ρ+1,ωj)+0t(Δu(s),Δωj)ds+0t(|us(t)|mus(s),ωj)ds+0tγ(Δus(s),Δωj)ds=i0t(xiσi(uxi(s)),ωj)ds+0t(|u(s)|p-1u(s),ωj)ds,t>0,j=1,,n. This implies uL([0,T];H02) is a local weak solution of problem (1.1). The proof of Theorem 2.1 is completed.

Secondly, we prove Theorem 2.2. First we give two lemmas. It is easy to prove what follows.

Lemma 3.2.

The modified energy functional satisfies, along solutions of (1.1), (3.29)E(t)=-utm+2m+2-γΔut220.

Lemma 3.3.

Assume that (A1)–(A3) hold, satisfying (3.30)C0((2(r+1)r-1E(0))(r-1)/2+(2(r+1)r-1E(0))(p-1)/2)<1,I(0)>0. Then I(t)>0.

Proof.

Since I(0)>0, then there exists (by continuity) T*T such that I(t)>0,for  all  t[0,T*), this gives (3.31)J(t)=I(t)r+1+r-12(r+1)Δu22+p-r(p+1)(r+1)up+1p+1r-12(r+1)Δu22+p-r(p+1)(r+1)up+1p+1. By using (2.8), (2.9), (3.31), and Lemma 3.2, we easily have (3.32)Δu222(r+1)r-1J(t)2(r+1)r-1E(t)2(r+1)r-1E(0),t[0,T*). We then exploit (2.12), and (3.32) to obtain (3.33)C2ur+1r+1B1C2Δu2r+1B1C2Δu2r-1Δu22B1C2(2(r+1)r-1E(0))(r-1)/2Δu22,up+1p+1B2Δu2p+1B2(2(r+1)r-1E(0))(p-1)/2Δu22,t[0,T*). Using (3.33), we have (3.34)C2ur+1r+1+up+1p+1B1C2(2(r+1)r-1E(0))(r-1)/2Δu22+B2(2(r+1)r-1E(0))(p-1)/2Δu22C0((2(r+1)r-1E(0))(r-1)/2+(2(r+1)r-1E(0))(p-1)/2)Δu22<Δu22,t[0,T*), where C0=max{B2,B1C2}.

Therefore, (3.35)I(t)=Δu22-C2ur+1r+1-up+1p+1>0, for all t[0,T*]. By repeating this procedure, and using the fact that (3.36)limtT*C0((2(r+1)r-1E(t))(r-1)/2+(2(r+1)r-1E(t))(p-1)/2)limtT*C0(2(r+1)r-1E(0))(r-1)/2+(2(r+1)r-1E(0))(p-1)/2<1, the proof is completed.

Lemma 3.4.

Assume that (A1)–(A4) hold, (2.12) satisfy. Then the solution is global existence. More, exist positive constant M>0 has (3.37)Δu22+utp+2p+2M;0tusm+2m+2dsM;0tΔus22dsM,t[0,).

Proof.

It suffices to show that (3.38)Δu22+utρ+2ρ+2 is bounded independently of t. To achieve this, we use (2.9), (3.31), and Lemma 3.2 to get (3.39)E(0)=E(t)+0tusm+2m+2ds+γ0tΔus22dsJ(t)+1ρ+2utρ+2ρ+2+0tusm+2m+2ds+γ0tΔus22dsr-12(r+1)Δu22+p-r(p+1)(r+1)up+1p+1+1ρ+2utρ+2ρ+2+0tusm+2m+2ds+γ0tΔus22dsr-12(r+1)Δu22+1ρ+2utρ+2ρ+2+0tusm+2m+2ds+γ0tΔus22ds. Since I(t),J(t) are positive. Therefore, (3.40)Δu22+utρ+2ρ+2CE(0). Moreover, (3.41)0tusm+2m+2dsE(0);γ0tΔus22dsE(0), where C is a positive constant, which depends only on r.

Lemma 3.5.

Assume that (A1)–(A4) hold, (2.12) satisfy. Then exist C>0 has (3.42)0tusρ+2ρ+2dsCt(m-ρ)/(m+2);(3.43)0tΔu22dsC0tI(s)ds.

Proof.

Using Lemma 3.4, we have (3.44)0tusρ+2ρ+2ds(0tusρ+2m+2ds)(ρ+2)/(m+2)(0t1ds)(m-ρ)/(m+2)C(0tusm+2m+2ds)(ρ+2)/(m+2)t(m-ρ)/(m+2)Ct(m-ρ)/(m+2). Using (3.34), we have (3.45)C2ur+1r+1+up+1p+1BC0((2(r+1)r-1E(0))(r-1)/2(2(r+1)r-1E(0))(p-1)/2)Δu22:=ηΔu22. So (3.46)(1-η)Δu22I(t); the proof is complete.

Lemma 3.6.

Assume that (A1)–(A4) hold, (2.12) satisfy. Then there exists a C>0, having (3.47)0tI(s)dsC(t1/2+t(m-ρ)/(m+2)+t1/(m+2)+1).

Proof.

By multiplying the differential equation in (1.1) by u and integrating over Ω, using integration by parts (2.9) and assumption (A1), we obtain (3.48)0=ddt(|ut|ρutρ+1,u)-utρ+2ρ+2ρ+1+Δu22+(|ut|mut,u)+γ(Δ2ut,u)+i=1N(σi(uxi),uxi)-up+1p+1ddt(|ut|ρutρ+1,u)-utρ+2ρ+2ρ+1+Δu22+(|ut|mut,u)+γ(Δut,Δu)-C2ur+1r+1-up+1p+1=ddt(|ut|ρutρ+1,u)-utρ+2ρ+2ρ+1+I(t)+(utmut,u)+γ(Δut,Δu). So (3.49)0tI(s)dsutρ+2ρ+1uρ+2+u1ρ+2ρ+2u0ρ+2+0tusρ+2ρ+2ρ+1ds+0t(usmus,u)ds+γ0t(Δus,Δu)ds. Using (A3), (3.42)-(3.43), we have (3.50)utρ+2ρ+1uρ+2C(utρ+2ρ+2)(ρ+1)/(ρ+2)Δu2C,0tΩ|us|musudxds0tusm+2m+1um+2dsC0tusm+2m+1Δu2dsC0tusm+2m+1dsC(0tusm+2m+2)(m+1)/(m+2)t1/(m+2)Ct1/(m+2),γ0t(Δus,Δu)dsγ0tΔus2Δu2dsC(0tΔus22ds)1/2t1/2Ct1/2. Therefore, (3.51)0tI(s)dsM(1+t1/(m+2)+t(m-ρ)/(m+2)+t1/2).

Proof of Theorem <xref ref-type="statement" rid="thm2">2.2</xref>.

First, t(1+t)E(t) is also absolutely continuous, and we have (3.52)ddt((1+t)E(t))E(t). A simple integration of (3.52) over (0,t) leads to (3.53)(1+t)E(t)E(0)+0tE(s)dsE(0)+1ρ+20tusρ+2ρ+2ds+120tΔu22ds+0N0tΩAi(uxi)dxds-1p+10tup+1p+1dsE(0)+1ρ+20tusρ+2ρ+2ds+120tΔu22ds+0N0tΩAi(uxi)dxds+1p+10tup+1p+1dsE(0)+1ρ+20tusρ+2ρ+2ds+120tΔu22ds+C2r+10tur+1r+1ds+1p+10tup+1p+1ds. Using the upper inequality, (3.45), (3.46), and (3.52), we have (3.54)(1+t)E(t)E(0)+1ρ+20tusρ+2ρ+2ds+120tΔu22ds+C0tΔu22dsE(0)+1ρ+20tusρ+2ρ+2ds+C0tI(s)ds. Apply (3.42), (3.47), we can get (2.13).

Using (3.31) and Lemma 3.3, we get J(t)0,I(t)>0. (2.13) implies limtE(t)=0. So when t, we have utρ+2ρ+20 and J(t)0. It is that (2.14) is satisfied. Theorem 2.2 is complete.

Following we will prove Theorem 2.3. For this purpose we set (3.55)L(t):=E(t)+εΨ(t), where ε is a positive constant and (3.56)Ψ(t)=1ρ+1Ω|ut|ρutudx.

Lemma 3.7.

Let ε be small enough. Then there exist two positive constants α1 and α2 such that (3.57)α1L(t)E(t)α2L(t).

Proof.

By Lemma 3.4 and Young's inequality, a direct computation gives (3.58)L(t)E(t)+ερ+2Ω|ut|ρ+2dx+ερ+2Ω|u|ρ+2dxE(t)+ερ+2utρ+2ρ+2+εB2ρ+2(2(r+1)E(0)r-1)ρ/2Δu22E(t)+εE(t)+εB2ρ+2(2(r+1)r-1)(ρ+2)/2E(0)ρ/2E(t)1α1E(t). Similarly, we have (3.59)L(t)E(t)-ερ+2Ω|ut|ρ+2dx-ερ+2Ω|u|ρ+2dxE(t)-ερ+2utρ+2ρ+2-εB2ρ+2(2(r+1)E(0)r-1)ρ/2Δu22E(t)-εE(t)-εB2ρ+2(2(r+1)r-1)(ρ+2)/2E(0)ρ/2E(t)1α2E(t), provided that ε is small enough.

Lemma 3.8.

Assume that the conditions of Theorem 2.3 hold, then the function (3.60)Ψ(t):=1ρ+1Ω|ut|ρutudx satisfies, along the solution of (1.1), (3.61)Ψ(t)-14E(t)+μutm+2m+2+ωΔu22.

Proof.

Applying equations of (1.1), we see (3.62)Ψ(t)=1ρ+1utρ+2ρ+2+Ω|ut|ρuttudx=1ρ+1utρ+2ρ+2+Ω(-Δ2u-|ut|mut+i=1Nxiσi(uxi)-γΔ2ut+|u|p-1u)udx=1ρ+1utρ+2ρ+2-Δu22-(|ut|mut,u)-i=1N(σi(uxi),uxi)-γ(Δut,Δu)+up+1p+1=-1ρ+2utρ+2ρ+2-Δu22-i=1NΩAi(uxi)dx+i=1NΩ(Ai(uxi)-(σi(uxi),uxi))dx-(|ut|mut,u)-γ(Δut,Δu)+up+1p+1+2ρ+3(ρ+2)(ρ+1)utρ+2ρ+2-E(t)-(|ut|mut,u)-γ(Δut,Δu)+2ρ+3(ρ+2)(ρ+1)utρ+2ρ+2+pp+1up+1p+1+i=1NΩ(Ai(uxi)-σi(uxi)uxi)dx. Exploiting the assumption (A1) and Young's inequality, we have (3.63)i=1NΩ(Ai(uxi)-σi(uxi)uxi)dx0.pp+1up+1p+1pB2p+1p+1Δu2p+1pB2p+1p+1(2(r+1)r-1)(p+1)/2(E(0))(p-1)/2E(t)|(|ut|mut,u)|1(m+2)δ(m+2)/(m+1)utm+2m+2+δm+2m+2utm+2m+21(m+2)δ(m+2)/(m+1)utm+2m+2+(δB2)m+2m+2(2(r+1)E(0)r-1)m/2Δu221(m+2)δ(m+2)/(m+1)utm+2m+2+(δB2)m+2m+2(2(r+1)r-1)(m+2)/2(E(0))m/2E(t),δ>0|γ(Δut,Δu)|γη-12Δut22+γη2Δu22,η>0. Exploiting (3.63) and (3.62), we get (3.64)Ψ'(t)-[1-pB2p+2p+1(2(r+1)r-1)(p+1)/2(E(0))(p-1)/2-(δB2)m+2m+2(2(r+1)r-1)(m+2)/2(E(0))m/2-γη2(r+1)r-1]E(t)+1(m+2)δ(m+2)/(m+1)utm+2m+2+(γη-1+2ρ+3(ρ+1)(ρ+2)((ρ+2)E(0))ρ/(ρ+2)B22)Δu22. Choosing δ satisfies (3.65)(δB2)m+2m+2(2(r+1)E(0)r-1)m/2=14[1-pB2p+2p+1(2(r+1)r-1)(p+1)/2(E(0))(p-1)/2] and η satisfies (3.66)γη2(r+1)r-1=14[1-pB2p+2p+1(2(r+1)r-1)(p+1)/2(E(0))(p-1)/2], at the above; the proof of (3.61) is completed.

Proof of Theorem <xref ref-type="statement" rid="thm3">2.3</xref>.

Using (3.61) and Lemma 3.7, we have (3.67)L(t)=E(t)+εΨ(t)(-utm+2m+2-γΔut)-ε4E(t)+εμutm+2m+2+εωΔut=-ε4E(t)-(γ-εω)Δut-(1-εμ)utm+2m+2-εα14L(t)-(γ-εω)Δut-(1-εμ)utm+2m+2. Choosing ε satisfies εmin{γ/ω,1/μ}. So we have (3.68)L(t)-εα14L(t),t0. A simple integration of (3.68) over (0,t) leads to (3.69)L(t)L(0)e-(εα1/4)t,t0. Exploiting Lemma 3.7 again, we have (3.70)E(t)α2L(0)e-(εα1/4)t:=Ke-κt,t0, where K,κ>0 are constants. The proof of Theorem 2.3 is complete.

4. Blow-Up of Solutions Proof of Theorem <xref ref-type="statement" rid="thm4">2.5</xref>.

Assuming that the solution of (1.1) is global, we have (4.1)E(t)=-utm+2m+2-γΔut220. So (4.2)Δu22C3; we set (4.3)Q(t):=-0tE(s)ds+(ot+ω)Ωu02dx, where o,ω are constants and will be given later.

Consider (4.4)Q(t)=-E(t)+oΩu02dxoΩu02dx-E(0). Choosing o satisfies the following condition in (4.4): (4.5)oΩu02dx-E(0)=Q(0)>0, so we have (4.6)Q(t)Q(0)>0,t[0,T]. Moreover, we have (4.7)Q(t)-Q(0)=E(0)-E(t)=-0tE(s)ds=0t(usm+2m+2+γΔus22)ds. Define (4.8)K(t):=Q1-γ(t)+ερ+10tΩ|us|ρusudxds, where ε>0 will be given later, and (4.9)0<γmin{r-ρ-1(ρ+2)(r+1),p-m-1(p+1)(m+1),r-m-1(m+1)(r+1)}. Multiplying (1.1) by u and a direct computation yield (4.10)K(t)=(1-γ)Q-γQ(t)+ερ+1Ω|u1|ρu1u0dxds+ερ+10tΩ(|us|ρusu)sdxds=(1-γ)Q-γQ(t)+ερ+1Ω|u1|ρu1u0dxds+ερ+10tΩ|us|ρ+2dxds+ε0tΩ|us|ρussudxds=(1-γ)Q-γQ(t)+ερ+1Ω|u1|ρu1u0dxds+ερ+10tΩ|us|ρ+2dxds-ε0tΩΔ2uudxds-ε0tΩ|us|musudxds-εγ0tΩΔ2usudxds+εi=1N0tΩxiσi(uxi)udxds+ε0tΩ|u|p-1uudxds=(1-γ)Q-γQ(t)+ερ+1Ω|u1|ρu1u0dxds+ερ+10tΩ|us|ρ+2dxds-ε0tΩ|Δu|2dxds-εγ0tΩΔusΔudxds-ε0tΩ|us|musudxds-εi=1N0tΩσi(uxi)uxidxds+ε0tup+1p+1ds. Exploiting (4.7) and (B1), we have (4.11)γ0tΩΔusΔudxdsγη20tΔus22ds+η2γ0tΔu22dsγη2(Q(t)-Q(0))+η2γC3T,(4.12)0tΩ|us|musudxdsςm+2m+20tΩ|u|m+2dxds+m+1m+2ς-((m+1)/(m+2))0tΩ|us|m+2m+2dxdsςm+2m+20tΩ|u|m+2dxds+m+1m+2ς-((m+1)/(m+2))(Q(t)-Q(0))(4.13)-0tΩσi(uxi)uxidxdsC40tur+1r+1ds. Using (4.10)–(4.13), we have (4.14)K(t)(1-γ)Q-γQ(t)+ερ+1Ω|u1|ρu1u0dx+ερ+10tΩ|us|ρ+2dxds+εC40tur+1r+1ds-ε0tΔu22ds+ε0tup+1p+1ds-εη2γTC3-ε(m+1m+2ς-(m+1)/(m+2)+γη2)Q(t)-εςm+2m+20tΩ|u|m+2dxds+ε(m+1m+2ς-(m+1)/(m+2)+γη2)Q(0). Choosing ς,η satisfies (4.15)ς-((m+2)/(m+1))=M1Q-γ(t),1η2=M2Q-γ(t). Then we have (4.16)K(t)[(1-γ)-εm+1m+2M1-εγM2]Q-γQ(t)+ερ+1Ω|u1|ρu1u0dxds+ερ+10tΩ|us|ρ+2dxds+ε(m+1m+2M1+εγM2)Q-γQ(0)-εM1-m-1m+2Qγ(m+1)0tΩ|u|m+2dxds-εM2-1QγγTC3-ε0tΔu22ds+ε0tup+1p+1ds+εC40tur+1r+1ds. A simple computing implies (4.17)Qγ=[-0tE(s)ds+(ot+ω)Ωu02dx]γ[0t(up+1p+1p+1+C2ur+1r+1)ds+(oT+ω)u022]γ2γ-1C5[(oT+ω)γu022γ+(0t(up+1p+1+ur+1r+1)ds)γ]. Using (4.17), (B2), embedding theorem, and Hӧlder inequality, we can get (4.18)Qγ(m+1)[2γ-1C5[(oT+ω)γu022γ+(0t(up+1p+1+ur+1r+1)ds)γ]](m+1)2(γ-1)(m+1)-1C6[(0t(up+1p+1+ur+1r+1)ds)γ(m+1)+(oT+ω)γ(m+1)u022γ(m+1){(0t(up+1p+1+ur+1r+1)ds)γ(m+1)}],(4.19)α0tΩ|u|m+2dxdsα0t(Ω|u|p+1dx)(m+2)/(p+1)ds|Ω|(p-m-1)/(p+1)α|Ω|(p-m-1)/(p+1)T(p-m-1)/(p+1)(0tup+1p+1ds)(m+2)/(p+1)=:C8(0tup+1p+1ds)(m+2)/(p+1), where C8=α|Ω|(p-m-1)/(p+1)T(p-m-1)/(p+1).

Consider (4.20)(1-α)0tΩ|u|m+2dxds(1-α)B0tu2m+2ds(1-α)B(0tu2r+1ds)(m+2)/(r+1)t(r-m-1)/(r+1)C7(0tur+1r+1ds)(m+2)/(r+1). Using (4.18)–(4.20), (B1), we get (4.21)Qγ(m+1)0tΩ|u|m+2dxds2(γ-1)(m+1)-1C6[(0t(up+1p+1+ur+1r+1)ds)γ(m+1)+(oT+ω)γ(m+1)u022γ(m+1)(0t(up+1p+1+ur+1r+1)ds)γ(m+1)]×[C8(0tup+1p+1ds)(m+2)/(p+1)+C7(0tur+1r+1ds)(m+2)/(r+1)]2(γ-1)(m+1)-1C6[C8(0t(up+1p+1+ur+1r+1)ds)γ(m+1)+((m+2)/(p+1))+(oT+ω)γ(m+1)u022γ(m+1)×C8(0t(up+1p+1+ur+1r+1)ds)γ(m+1)+((m+2)/(p+1))]+2(γ-1)(m+1)-1C6[C7(0t(up+1p+1+ur+1r+1)ds)γ(m+1)+((m+2)/(r+1))+(oT+ω)γ(m+1)u022γ(m+1)×C7(0t(up+1p+1+ur+1r+1)ds)γ(m+1)+((m+2)/(r+1))], Using (4.9), (4.17), and Young inequality, we have (4.22)Qγ(m+1)0tΩ|u|m+2dxdsC9(1+0t(up+1p+1+ur+1r+1)ds). Similarly, we have (4.23)TC3QγC10(1+0t(up+1p+1+ur+1r+1)ds). Additionally, we choose ε satisfying (4.24)ε1-γ((m+1)/(m+2))M1+γM2. Using (4.16), (4.21)–(4.24), we get (4.25)K(t)ερ+1Ω|u1|ρu1u0dxds+ερ+10tΩ|us|ρ+2dxds-ε0tΔu22ds-ε[M1(-m-1)m+2C9+C10γM2]×0t(up+1p+1+ur+1r+1)-ε[M1(-m-1)m+2C9+C10γM2]+εmin{1,C4}0t(ur+1r+1+up+1p+1)+ξQ(t)+ξi=1N0tΩ0uxiσi(τ)dτdxds-ξ0tup+1p+1p+1ds+ξρ+10tusρ+2ρ+2ds+ξ20tΔu22ds-ξ(ot+ω)u022,0tΩ0uxiσi(τ)dτdxds-0tup+1p+1p+1ds-C2r+10tur+1r+1ds-0tup+1p+1p+1ds-1p+10t(ur+1r+1+up+1p+1)ds. Using (4.25), we get (4.26)K(t)ξQ(t)+(ξρ+2+ερ+1)0tusρ+2ρ+2ds+ε(ξ2ε-1)0tΔu22ds+ε(min{1,C4}-1p+1ξε-M1-m-1m+2C9-C10γM2)0t(up+1p+1+ur+1r+1)+ε[1ρ+1Ω|u1|ρu1u0dx-ξε(oT+ω)0tu02dx-M1-m-1m+2C9-C10γM2], choosing 2ε<ξ<(p+1)ε, u0,u1 satisfying (4.27)min{1,C4}-1p+1ξε>0,1ρ+1Ω|u1|ρu1u0dx-ξε(oT+ω)0tu02dx>0. Then we choose M1,M2 big enough, satisfying (4.28)min{1,C4}-1p+1ξε-M1-m-1m+2C9-C10γM2>01ρ+1Ω|u1|ρu1u0dx-ξε(oT+ω)0tu02dx-M1-m-1m+2C9-C10γM2>0. Using the two above inequalities, we have (4.29)K(t)Cε[Q(t)+0tusρ+2ρ+2ds+0tΔu22ds+0t(ur+1r+1+up+1p+1)ds+ν], where ν>0. So (4.30)K(t)>K(0)(ωΩu02dx)1-γ>0.

Using (B1)–(B3), (A4), and Hӧlder and Young inequalities, we have (4.31)|1ρ+10tΩ|us|ρusudxds|1ρ+10tusρ+2ρ+1ur+1dsCρ+10t(usρ+2(ρ+1)μ+ur+1θ)ds, where (1/μ)+(1/θ)=1.

So we have (4.32)|1ρ+10tΩ|us|ρusudxds|1/(1-γ)C(0tusρ+2(ρ+1)μds)1/(1-γ)+(0tur+1θds)1/(1-γ)C(0tusρ+2(ρ+1)μ·((ρ+2)/((ρ+1)μ))ds)μ(ρ+1)/(1-γ)(ρ+2)+C(0tur+1θ·((1+r)/θ)ds)θ/(1+r)(1-γ)=C(0tusρ+2ρ+2ds)μ(ρ+1)/(1-γ)(ρ+2)+C(0tur+11+rds)θ/(1+r)(1-γ). Using (4.9), (4.32) and choosing μ satisfy μ(ρ+1)/((ρ+2)(1-γ))=1, then θ/(1+r)(1-γ)<1.

So (4.33)|1ρ+10tΩ|us|ρusudxds|1/(1-γ)C(0tusρ+2ρ+2ds+0tur+11+rds)+βC(0tusρ+2ρ+2ds+0tur+11+rds+0tΔu22ds+0tup+1p+1)+β, where β>0 is a constant. Finally, we can easily get (4.34)K1/(1-γ)21/(1-γ)(Q(t)+ε1/(1-γ)|1ρ+10tΩ|us|ρusudxds|1/(1-γ)). Combining (4.29) and (4.33)-(4.34), we have (4.35)K(t)CK1/(1-γ)(t),tT, for some constant C>0. Integrating the above inequality in (0,t), we get (4.36)K1/(1-γ)1(K-γ/(1-γ)(0)-ct)1/γ,tT. The above inequality implies K(t) blows-up on some time T*. Since u exists globally, so we have (4.37)ερ+10tΩ|us|ρusudxds<. And we know that K(t),as  tT*, so (4.38)Q1-γ(t); this implies Q(t), that is to say -0tE(s)ds. Because (4.39)-0tE(s)ds-tE(t), we know (4.40)E(t)-,as  tT*. This contradicts with the assumption that u is a global solution. So the solution of (1.1) blows-up on time T*.

Acknowledgment

This work was supported by the National Natural Science Foundation of China 10771032.

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