# On Existence, Uniform Decay Rates, and Blow-Up for Solutions of a Nonlinear Wave Equation with Dissipative and Source

Authors: Xiaopan Liu
Journal: Abstract and Applied Analysis
 (2012)

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## Abstract

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.


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## Body

## 1. Introduction

In this paper, we investigate the following nonlinear wave equation:(1.1)|ut|ρutt+Δ2u+|ut|mut+γΔ2ut=∑i=1N‍∂∂xiσi(uxi)+|u|p-1u,(x,t)∈Ω×(0,T),u=∂u∂n=0,(x,t)∈∂Ω×(0,T],u(x,0)=u0(x),ut(x,0)=u1(x),x∈Ω,
where Ω is a bounded domain in ℝn 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. γ≥0 is a constant, σi(s)(i=1,…,N) are given in (A1) later.In 1968, Greenberg et al. [1] 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=1N‍∂∂xiai(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 [4]. In [4], 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=1N‍∂∂xiσi(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 [5–15].The paper is organized as follows. In Section2, 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:(A1)
σi∈C1(R),σi(s)s>0 and
(2.2)C1|s|r≤|σi(s)|≤C2|s|r,for some r≥1, if n=1,2, else if n≥3, then r≤n/(n-2), so that ∥∇u∥2r≤B1∥Δu∥2, i=1,2,…,N, where B1 is the optimal embedding constant.(A2)
The initial data(2.3)u0∈H02,u1∈H01.(A3)
m<ρ+1; If n>3, then 0<ρ<m<4/(n-2) and if n=1,2, then 0<ρ<m.(A4)
Ifn>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(B1)
σi∈C1(R),σi(s)s<0, and
(2.4)C1|s|r≤|σi(s)|≤C2|s|r,for some r≥1, if n=1,2, else if n≥3, then r≤n/(n-2), so that ∥∇u∥2r≤B1∥Δu∥2, i=1,2,…,N. where B1 is the optimal embedding constant, and C2<(r+1)/(p+1).(B2)
Ifn>3, then m<r-1<p-1<4/(n-2) and if n=1,2, then 0<m<r-1<p-1.(B3)
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 embeddingH02(Ω)↪Lq(Ω) which implies ∥u∥q≤B2∥Δu∥2 when
(2.5)2≤q≤2nn-2ifn≥3,q≥1ifn=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+2∥ut∥ρ+2ρ+2+12∥Δu∥22+∑i=1N‍∫Ω‍Ai(uxi)dx-∥u∥p+1p+1p+1,(2.8)J(t)=12∥Δu∥22-C2r+1∥∇u∥r+1r+1-∥u∥p+1p+1p+1,(2.9)I(t)=∥Δu∥22-C2∥∇u∥r+1r+1-∥u∥p+1p+1.
Because r<p, we have
(2.10)J(t)=I(t)r+1+r-12(r+1)∥Δu∥22+p-r(p+1)(r+1)∥u∥p+1p+1.Theorem 2.1.
Assume that (A1)–(A4) hold, then problem (1.1) has a unique solution u satisfying
(2.11)u∈L∞([0,T];H02)∩W1,∞([0,T];Lρ+2);ut∈L2([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)limt→∞∥Δu∥22=limt→∞∥ut∥ρ+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.
Whenγ>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 Theorems2.1–2.3. First, we give the following Lemma.Lemma 3.1 (see [2]).
LetΩ 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+ε∥u∥1,p,
for all u∈W01,p(2≤p<∞).Proof of Theorem2.1.
We look for approximate solutionsun(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,u1n∈C0∞ such that
(3.4)u0n→u0inH02,u1n→u1inH01asn→∞.
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+2∥utn∥ρ+2ρ+2+12∥Δun∥22]+∥utn∥m+2m+2+γ∥Δutn∥22=∑i=1N‍(σin(uxi),(uxin)t)+(|un|p-1un,utn).
A simple integration of (3.5) over (0,t) leads to
(3.6)1ρ+2∥utn∥ρ+2ρ+2+12∥Δun∥22+∫0t‍(∥usn∥m+2m+2+γ∥usn∥22)ds=C3+∑i=1N‍∫0t‍(σin(uxi),(uxin)s)ds+∫0t‍(|un|p-1un,usn)ds,
where C3=1/(ρ+2)∥u1n∥ρ+2ρ+2+(1/2)∥Δu0n∥22>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)2n≥n-22n,(p-1)q=2nn-2,1w=n+22n-(p-1)(n-2)2n≥n-22n.
Consider the following:
(3.9)|∑i=1N‍∫0t‍(σi(uxin),(uxin)s)ds|≤C2∫0t‍∫Ω‍|∇un|r-1|∇un||∇usn|dxds≤C2∫0t‍∥∇un∥(r-1)qr-1∥∇un∥w∥∇usn∥2n/(n-2)ds≤C2C∫0t‍∥Δun∥2r-1∥Δun∥2∥Δusn∥2ds=C2C∫0t‍∥Δun∥2r∥Δusn∥2ds≤C2C(12ϵ∫0t‍∥Δun∥22rds+ϵ2∫0t‍∥Δusn∥22ds)≤C2C(12ϵ(∫0t‍∥Δun∥22pds)r/p+12ϵ(∫0t‍1ds)(p-r)/p+ϵ2∫0t‍∥Δusn∥22ds)≤C2C12ϵpr(∫0t‍∥Δun∥22pds)+C2C12ϵp-rp+ϵ2C2C∫0t‍∥Δusn∥22ds,∀ϵ>0,
assuming that t<1.
Similarly,(3.10)∫0t‍(|un|p-1un,usn)ds≤∫0t‍∥un∥(p-1)qp-1∥un∥w∥usn∥2n/(n-2)ds≤C∫0t‍∥Δun∥2p-1∥Δun∥2∥Δusn∥2ds≤C12ϵ∫0t‍∥Δun∥22pds+ϵ2C∫0t‍∥Δusn∥22ds,∀ϵ>0.
Using (3.6)–(3.10), we have
(3.11)∫0t‍∥usn∥m+2m+2ds+γ∫0t‍∥Δusn∥22ds+12∥Δun∥22+1ρ+2∥utn∥ρ+2ρ+2≤C4+C(ε)∫0t‍∥Δun∥22pds+ϵ2C∫0t‍∥Δusn∥22.
Choosing ϵ=(1/C)γ in (3.11), we have
(3.12)∫0t‍∥usn∥m+2m+2ds+12∥Δun∥22+1ρ+2∥utn∥ρ+2ρ+2+γ2∫0t‍∥Δusn∥22ds≤C4+C∫0t‍∥Δun∥22pds,
where C4=C3+((C2C)/2ϵ)((p-r)/p). Assuming Yn(t)=2(C4+C∫0t‍∥Δun∥22p), 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)∥Δun∥22+2ρ+2∥utn∥ρ+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)∥Δun∥22+∥utn∥ρ+2ρ+2≤C,∀t∈[0,T],
where C is independent of n.
Moreover,(3.17)∫0t‍∥usn∥m+2m+2ds+∫0t‍∥Δusn∥22ds≤C.
By (3.17),
(3.18)∫0t‍∥|usn(s)|musn(s)∥(m+2)/(m+1)(m+2)/(m+1)ds≤∫0t‍∥usn(s)∥m+2m+2ds≤C,t∈[0,T].
By (3.16),
(3.19)∥|un(s)|p-1un(s)∥(p+1)/p(p+1)/p≤∥un(s)∥p+1p+1≤∥Δun(s)∥2p+1≤C.
From (3.16)–(3.19), we have
(3.20)un∈L∞([0,T];H02);utn∈L2([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)un→uweak*inL∞([0,T];H02),(3.22)utn→utweak*inL∞([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)stronglyinL2,a.e.onΩ,σi(uxin(t))→σi(uxi(t)),a.e.onΩ,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+ε∥un(t)-u(t)∥1,2≤Mε,∫0T‍∥utn(s)-ut(s)∥22ds≤2(∑k=1N2ε‍∫0T‍(utn(s)-ut(s),ωk)2ds∫0T‍∥utn(s)-ut(s)∥22ds≤2+ε2∫0T‍∥utn(s)-ut(s)∥1,22ds∑k=1N2ϵ‍)≤Mε2.
By the arbitrariness ofε we get
(3.25)un→ustronglyinL∞([0,T];L2),a.e.onQT,utn→utstronglyinL2(QT),a.e.onQT.
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=∑i‍∫0t‍(∂∂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)ds≤C∫0t‍∥Δun∥22ds≤CT,(|utn(t)|mutn(t),ωj)≤∥utn∥ρ+2m+1∥ωj∥((ρ+2)/(m+1))′≤∥utn∥ρ+2ρ+2∥ωj∥((ρ+2)/(m+1))′((ρ+2)/(m+1))′≤C,∫0t‍(|usn(s)|musn(s),ωj)ds≤∫0t‍(∥usn∥m+2m+2+∥ωj∥m+2m+2)ds≤CT,∫0t‍(|un(s)|p-1un(s),ωj)ds≤CT.
Letn→∞ 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=∑i‍∫0t‍(∂∂xiσi(uxi(s)),ωj)ds+∫0t‍(|u(s)|p-1u(s),ωj)ds,t>0,j=1,…,n.
This implies u∈L∞([0,T];H02) is a local weak solution of problem (1.1). The proof of Theorem 2.1 is completed.Secondly, we prove Theorem2.2. 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)=-∥ut∥m+2m+2-γ∥Δut∥22≤0.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.
SinceI(0)>0, then there exists (by continuity) T*≤T such that I(t)>0,forallt∈[0,T*), this gives
(3.31)J(t)=I(t)r+1+r-12(r+1)∥Δu∥22+p-r(p+1)(r+1)∥u∥p+1p+1≥r-12(r+1)∥Δu∥22+p-r(p+1)(r+1)∥u∥p+1p+1.
By using (2.8), (2.9), (3.31), and Lemma 3.2, we easily have
(3.32)∥Δu∥22≤2(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)C2∥∇u∥r+1r+1≤B1C2∥Δu∥2r+1≤B1C2∥Δu∥2r-1∥Δu∥22≤B1C2(2(r+1)r-1E(0))(r-1)/2∥Δu∥22,∥u∥p+1p+1≤B2∥Δu∥2p+1≤B2(2(r+1)r-1E(0))(p-1)/2∥Δu∥22,∀t∈[0,T*).
Using (3.33), we have
(3.34)C2∥∇u∥r+1r+1+∥u∥p+1p+1≤B1C2(2(r+1)r-1E(0))(r-1)/2∥Δu∥22+B2(2(r+1)r-1E(0))(p-1)/2∥Δu∥22≤C0((2(r+1)r-1E(0))(r-1)/2+(2(r+1)r-1E(0))(p-1)/2)∥Δu∥22<∥Δu∥22,∀t∈[0,T*),
where C0=max{B2,B1C2}.
Therefore,(3.35)I(t)=∥Δu∥22-C2∥∇u∥r+1r+1-∥u∥p+1p+1>0,
for all t∈[0,T*]. By repeating this procedure, and using the fact that
(3.36)limt→T*C0((2(r+1)r-1E(t))(r-1)/2+(2(r+1)r-1E(t))(p-1)/2)≤limt→T*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)∥Δu∥22+∥ut∥p+2p+2≤M;∫0t‍∥us∥m+2m+2ds≤M;∫0t‍∥Δus∥22ds≤M,∀t∈[0,∞).Proof.
It suffices to show that(3.38)∥Δu∥22+∥ut∥ρ+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)+∫0t‍∥us∥m+2m+2ds+γ∫0t‍∥Δus∥22ds≥J(t)+1ρ+2∥ut∥ρ+2ρ+2+∫0t‍∥us∥m+2m+2ds+γ∫0t‍∥Δus∥22ds≥r-12(r+1)∥Δu∥22+p-r(p+1)(r+1)∥u∥p+1p+1+1ρ+2∥ut∥ρ+2ρ+2+∫0t‍∥us∥m+2m+2ds+γ∫0t‍∥Δus∥22ds≥r-12(r+1)∥Δu∥22+1ρ+2∥ut∥ρ+2ρ+2+∫0t‍∥us∥m+2m+2ds+γ∫0t‍∥Δus∥22ds.
Since I(t),J(t) are positive. Therefore,
(3.40)∥Δu∥22+∥ut∥ρ+2ρ+2≤CE(0).
Moreover,
(3.41)∫0t‍∥us∥m+2m+2ds≤E(0);γ∫0t‍∥Δus∥22ds≤E(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)∫0t‍∥us∥ρ+2ρ+2ds≤Ct(m-ρ)/(m+2);(3.43)∫0t‍∥Δu∥22ds≤C∫0t‍I(s)ds.Proof.
Using Lemma3.4, we have
(3.44)∫0t‍∥us∥ρ+2ρ+2ds≤(∫0t‍∥us∥ρ+2m+2ds)(ρ+2)/(m+2)(∫0t‍1ds)(m-ρ)/(m+2)≤C(∫0t‍∥us∥m+2m+2ds)(ρ+2)/(m+2)t(m-ρ)/(m+2)≤Ct(m-ρ)/(m+2).
Using (3.34), we have
(3.45)C2∥∇u∥r+1r+1+∥u∥p+1p+1≤BC0((2(r+1)r-1E(0))(r-1)/2(2(r+1)r-1E(0))(p-1)/2)∥Δu∥22:=η∥Δu∥22.
So
(3.46)(1-η)∥Δu∥22≤I(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)ds≤C(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)-∥ut∥ρ+2ρ+2ρ+1+∥Δu∥22+(|ut|mut,u)+γ(Δ2ut,u)+∑i=1N‍(σi(uxi),uxi)-∥u∥p+1p+1≥ddt(|ut|ρutρ+1,u)-∥ut∥ρ+2ρ+2ρ+1+∥Δu∥22+(|ut|mut,u)+γ(Δut,Δu)-C2∥∇u∥r+1r+1-∥u∥p+1p+1=ddt(|ut|ρutρ+1,u)-∥ut∥ρ+2ρ+2ρ+1+I(t)+(∥ut∥mut,u)+γ(Δut,Δu).
So
(3.49)∫0t‍I(s)ds≤∥ut∥ρ+2ρ+1∥u∥ρ+2+∥u1∥ρ+2ρ+2∥u0∥ρ+2+∫0t‍∥us∥ρ+2ρ+2ρ+1ds+∫0t‍(∥us∥mus,u)ds+γ∫0t‍(Δus,Δu)ds.
Using (A3), (3.42)-(3.43), we have
(3.50)∥ut∥ρ+2ρ+1∥u∥ρ+2≤C(∥ut∥ρ+2ρ+2)(ρ+1)/(ρ+2)∥Δu∥2≤C,∫0t‍∫Ω‍|us|musudxds≤∫0t‍∥us∥m+2m+1∥u∥m+2ds≤C∫0t‍∥us∥m+2m+1∥Δu∥2ds≤C∫0t‍∥us∥m+2m+1ds≤C(∫0t‍∥us∥m+2m+2)(m+1)/(m+2)t1/(m+2)≤Ct1/(m+2),γ∫0t‍(Δus,Δu)ds≤γ∫0t‍∥Δus∥2∥Δu∥2ds≤C(∫0t‍∥Δus∥22ds)1/2t1/2≤Ct1/2.
Therefore,
(3.51)∫0t‍I(s)ds≤M(1+t1/(m+2)+t(m-ρ)/(m+2)+t1/2).Proof of Theorem2.2.
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)ds≤E(0)+1ρ+2∫0t‍∥us∥ρ+2ρ+2ds+12∫0t‍∥Δu∥22ds+∑0N‍∫0t‍∫Ω‍Ai(uxi)dxds-1p+1∫0t‍∥u∥p+1p+1ds≤E(0)+1ρ+2∫0t‍∥us∥ρ+2ρ+2ds+12∫0t‍∥Δu∥22ds+∑0N‍∫0t‍∫Ω‍Ai(uxi)dxds+1p+1∫0t‍∥u∥p+1p+1ds≤E(0)+1ρ+2∫0t‍∥us∥ρ+2ρ+2ds+12∫0t‍∥Δu∥22ds+C2r+1∫0t‍∥∇u∥r+1r+1ds+1p+1∫0t‍∥u∥p+1p+1ds.
Using the upper inequality, (3.45), (3.46), and (3.52), we have
(3.54)(1+t)E(t)≤E(0)+1ρ+2∫0t‍∥us∥ρ+2ρ+2ds+12∫0t‍∥Δu∥22ds+C∫0t‍∥Δu∥22ds≤E(0)+1ρ+2∫0t‍∥us∥ρ+2ρ+2ds+C∫0t‍I(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 limt→∞E(t)=0. So when t→∞, we have ∥ut∥ρ+2ρ+2→0 and J(t)→0. It is that (2.14) is satisfied. Theorem 2.2 is complete.
Following we will prove Theorem2.3. 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+1∫Ω‍|ut|ρutudx.Lemma 3.7.
Letε be small enough. Then there exist two positive constants α1 and α2 such that
(3.57)α1L(t)≤E(t)≤α2L(t).Proof.
By Lemma3.4 and Young's inequality, a direct computation gives
(3.58)L(t)≤E(t)+ερ+2∫Ω‍|ut|ρ+2dx+ερ+2∫Ω‍|u|ρ+2dx≤E(t)+ερ+2∥ut∥ρ+2ρ+2+εB2ρ+2(2(r+1)E(0)r-1)ρ/2∥Δu∥22≤E(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+2∫Ω‍|ut|ρ+2dx-ερ+2∫Ω‍|u|ρ+2dx≥E(t)-ερ+2∥ut∥ρ+2ρ+2-εB2ρ+2(2(r+1)E(0)r-1)ρ/2∥Δu∥22≥E(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 Theorem2.3 hold, then the function
(3.60)Ψ(t):=1ρ+1∫Ω‍|ut|ρutudx
satisfies, along the solution of (1.1),
(3.61)Ψ′(t)≤-14E(t)+μ∥ut∥m+2m+2+ω∥Δu∥22.Proof.
Applying equations of (1.1), we see
(3.62)Ψ′(t)=1ρ+1∥ut∥ρ+2ρ+2+∫Ω‍|ut|ρuttudx=1ρ+1∥ut∥ρ+2ρ+2+∫Ω‍(-Δ2u-|ut|mut+∑i=1N‍∂∂xiσi(uxi)-γΔ2ut+|u|p-1u)udx=1ρ+1∥ut∥ρ+2ρ+2-∥Δu∥22-(|ut|mut,u)-∑i=1N‍(σi(uxi),uxi)-γ(Δut,Δu)+∥u∥p+1p+1=-1ρ+2∥ut∥ρ+2ρ+2-∥Δu∥22-∑i=1N‍∫Ω‍Ai(uxi)dx+∑i=1N‍∫Ω‍(Ai(uxi)-(σi(uxi),uxi))dx-(|ut|mut,u)-γ(Δut,Δu)+∥u∥p+1p+1+2ρ+3(ρ+2)(ρ+1)∥ut∥ρ+2ρ+2≤-E(t)-(|ut|mut,u)-γ(Δut,Δu)+2ρ+3(ρ+2)(ρ+1)∥ut∥ρ+2ρ+2+pp+1∥u∥p+1p+1+∑i=1N‍∫Ω‍(Ai(uxi)-σi(uxi)uxi)dx.
Exploiting the assumption (A1) and Young's inequality, we have
(3.63)∑i=1N‍∫Ω‍(Ai(uxi)-σi(uxi)uxi)dx≤0.pp+1∥u∥p+1p+1≤pB2p+1p+1∥Δu∥2p+1≤pB2p+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)∥ut∥m+2m+2+δm+2m+2∥ut∥m+2m+2≤1(m+2)δ(m+2)/(m+1)∥ut∥m+2m+2+(δB2)m+2m+2(2(r+1)E(0)r-1)m/2∥Δu∥22≤1(m+2)δ(m+2)/(m+1)∥ut∥m+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-12∥Δut∥22+γη2∥Δu∥22,∀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)∥ut∥m+2m+2+(γη-1+2ρ+3(ρ+1)(ρ+2)((ρ+2)E(0))ρ/(ρ+2)B22)∥Δu∥22.
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 Theorem2.3.
Using (3.61) and Lemma 3.7, we have
(3.67)L′(t)=E′(t)+εΨ′(t)≤(-∥ut∥m+2m+2-γ∥Δut∥)-ε4E(t)+εμ∥ut∥m+2m+2+εω∥Δut∥=-ε4E(t)-(γ-εω)∥Δut∥-(1-εμ)∥ut∥m+2m+2≤-εα14L(t)-(γ-εω)∥Δut∥-(1-εμ)∥ut∥m+2m+2.
Choosing ε satisfies ε≤min{γ/ω,1/μ}. So we have
(3.68)L′(t)≤-εα14L(t),∀t≥0.
A simple integration of (3.68) over (0,t) leads to
(3.69)L(t)≤L(0)e-(εα1/4)t,∀t≥0.
Exploiting Lemma 3.7 again, we have
(3.70)E(t)≤α2L(0)e-(εα1/4)t:=Ke-κt,∀t≥0,
where K,κ>0 are constants. The proof of Theorem 2.3 is complete.


## 4. Blow-Up of Solutions

Proof of Theorem2.5.
Assuming that the solution of (1.1) is global, we have
(4.1)E′(t)=-∥ut∥m+2m+2-γ∥Δut∥22≤0.
So
(4.2)∥Δu∥22≤C3;
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)+o∫Ω‍u02dx≥o∫Ω‍u02dx-E(0).
Choosing o satisfies the following condition in (4.4):
(4.5)o∫Ω‍u02dx-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(∥us∥m+2m+2+γ∥Δus∥22)ds.
Define
(4.8)K(t):=Q1-γ(t)+ερ+1∫0t‍∫Ω‍|us|ρusudxds,
where ε>0 will be given later, and
(4.9)0<γ≤min{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+1∫Ω‍|u1|ρu1u0dxds+ερ+1∫0t‍∫Ω‍(|us|ρusu)sdxds=(1-γ)Q-γQ′(t)+ερ+1∫Ω‍|u1|ρu1u0dxds+ερ+1∫0t‍∫Ω‍|us|ρ+2dxds+ε∫0t‍∫Ω‍|us|ρussudxds=(1-γ)Q-γQ′(t)+ερ+1∫Ω‍|u1|ρu1u0dxds+ερ+1∫0t‍∫Ω‍|us|ρ+2dxds-ε∫0t‍∫Ω‍Δ2uudxds-ε∫0t‍∫Ω‍|us|musudxds-εγ∫0t‍∫Ω‍Δ2usudxds+ε∑i=1N‍∫0t‍∫Ω‍∂∂xiσi(uxi)udxds+ε∫0t‍∫Ω‍|u|p-1uudxds=(1-γ)Q-γQ′(t)+ερ+1∫Ω‍|u1|ρu1u0dxds+ερ+1∫0t‍∫Ω‍|us|ρ+2dxds-ε∫0t‍∫Ω‍|Δu|2dxds-εγ∫0t‍∫Ω‍ΔusΔudxds-ε∫0t‍∫Ω‍|us|musudxds-ε∑i=1N‍∫0t‍∫Ω‍σi(uxi)uxidxds+ε∫0t‍∥u∥p+1p+1ds.
Exploiting (4.7) and (B1), we have
(4.11)γ∫0t‍∫Ω‍ΔusΔudxds≤γη2∫0t‍∥Δus∥22ds+η2γ∫0t‍∥Δu∥22ds≤γη2(Q′(t)-Q′(0))+η2γC3T,(4.12)∫0t‍∫Ω‍|us|musudxds≤ςm+2m+2∫0t‍∫Ω‍|u|m+2dxds+m+1m+2ς-((m+1)/(m+2))∫0t‍∫Ω‍|us|m+2m+2dxds≤ςm+2m+2∫0t‍∫Ω‍|u|m+2dxds+m+1m+2ς-((m+1)/(m+2))(Q′(t)-Q′(0))(4.13)-∫0t‍∫Ω‍σi(uxi)uxidxds≥C4∫0t‍∥∇u∥r+1r+1ds.
Using (4.10)–(4.13), we have
(4.14)K′(t)≥(1-γ)Q-γQ′(t)+ερ+1∫Ω‍|u1|ρu1u0dx+ερ+1∫0t‍∫Ω‍|us|ρ+2dxds+εC4∫0t‍∥∇u∥r+1r+1ds-ε∫0t‍∥Δu∥22ds+ε∫0t‍∥u∥p+1p+1ds-εη2γTC3-ε(m+1m+2ς-(m+1)/(m+2)+γη2)Q′(t)-εςm+2m+2∫0t‍∫Ω‍|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+1∫Ω‍|u1|ρu1u0dxds+ερ+1∫0t‍∫Ω‍|us|ρ+2dxds+ε(m+1m+2M1+εγM2)Q-γQ′(0)-εM1-m-1m+2Qγ(m+1)∫0t‍∫Ω‍|u|m+2dxds-εM2-1QγγTC3-ε∫0t‍∥Δu∥22ds+ε∫0t‍∥u∥p+1p+1ds+εC4∫0t‍∥∇u∥r+1r+1ds.
A simple computing implies
(4.17)Qγ=[-∫0t‍E(s)ds+(ot+ω)∫Ω‍u02dx]γ≤[∫0t‍(∥u∥p+1p+1p+1+C2∥∇u∥r+1r+1)ds+(oT+ω)∥u0∥22]γ≤2γ-1C5[(oT+ω)γ∥u0∥22γ+(∫0t(∥u∥p+1p+1+∥∇u∥r+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)γ∥u0∥22γ+(∫0t‍(∥u∥p+1p+1+∥∇u∥r+1r+1)ds)γ]](m+1)≤2(γ-1)(m+1)-1C6[(∫0t‍(∥u∥p+1p+1+∥∇u∥r+1r+1)ds)γ(m+1)+(oT+ω)γ(m+1)∥u0∥22γ(m+1){(∫0t‍(∥u∥p+1p+1+∥∇u∥r+1r+1)ds)γ(m+1)}],(4.19)α∫0t‍∫Ω‍|u|m+2dxds≤α∫0t‍(∫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)(∫0t‍∥u∥p+1p+1ds)(m+2)/(p+1)=:C8(∫0t‍∥u∥p+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)∫0t‍∫Ω‍|u|m+2dxds≤(1-α)B∫0t‍∥∇u∥2m+2ds≤(1-α)B(∫0t‍∥∇u∥2r+1ds)(m+2)/(r+1)t(r-m-1)/(r+1)≤C7(∫0t‍∥∇u∥r+1r+1ds)(m+2)/(r+1).
Using (4.18)–(4.20), (B1), we get
(4.21)Qγ(m+1)∫0t‍∫Ω‍|u|m+2dxds≤2(γ-1)(m+1)-1C6[(∫0t‍(∥u∥p+1p+1+∥∇u∥r+1r+1)ds)γ(m+1)+(oT+ω)γ(m+1)∥u0∥22γ(m+1)(∫0t‍(∥u∥p+1p+1+∥∇u∥r+1r+1)ds)γ(m+1)]×[C8(∫0t‍∥u∥p+1p+1ds)(m+2)/(p+1)+C7(∫0t‍∥∇u∥r+1r+1ds)(m+2)/(r+1)]≤2(γ-1)(m+1)-1C6[C8(∫0t‍(∥u∥p+1p+1+∥∇u∥r+1r+1)ds)γ(m+1)+((m+2)/(p+1))+(oT+ω)γ(m+1)∥u0∥22γ(m+1)×C8(∫0t‍(∥u∥p+1p+1+∥∇u∥r+1r+1)ds)γ(m+1)+((m+2)/(p+1))]+2(γ-1)(m+1)-1C6[C7(∫0t‍(∥u∥p+1p+1+∥∇u∥r+1r+1)ds)γ(m+1)+((m+2)/(r+1))+(oT+ω)γ(m+1)∥u0∥22γ(m+1)×C7(∫0t‍(∥u∥p+1p+1+∥∇u∥r+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)∫0t‍∫Ω‍|u|m+2dxds≤C9(1+∫0t‍(∥u∥p+1p+1+∥∇u∥r+1r+1)ds).
Similarly, we have
(4.23)TC3Qγ≤C10(1+∫0t‍(∥u∥p+1p+1+∥∇u∥r+1r+1)ds).
Additionally, we choose ε satisfying
(4.24)ε≤1-γ((m+1)/(m+2))M1+γM2.
Using (4.16), (4.21)–(4.24), we get
(4.25)K′(t)≥ερ+1∫Ω‍|u1|ρu1u0dxds+ερ+1∫0t‍∫Ω‍|us|ρ+2dxds-ε∫0t‍∥Δu∥22ds-ε[M1(-m-1)m+2C9+C10γM2]×∫0t‍(∥u∥p+1p+1+∥∇u∥r+1r+1)-ε[M1(-m-1)m+2C9+C10γM2]+εmin{1,C4}∫0t‍(∥∇u∥r+1r+1+∥u∥p+1p+1)+ξQ(t)+ξ∑i=1N‍∫0t‍∫Ω‍∫0uxi‍σi(τ)dτdxds-ξ∫0t‍∥u∥p+1p+1p+1ds+ξρ+1∫0t‍∥us∥ρ+2ρ+2ds+ξ2∫0t‍∥Δu∥22ds-ξ(ot+ω)∥u0∥22,∫0t‍∫Ω‍∫0uxi‍σi(τ)dτdxds-∫0t‍∥u∥p+1p+1p+1ds≥-C2r+1∫0t‍∥∇u∥r+1r+1ds-∫0t‍∥u∥p+1p+1p+1ds≥-1p+1∫0t(∥∇u∥r+1r+1+∥u∥p+1p+1)ds.
Using (4.25), we get
(4.26)K′(t)≥ξQ(t)+(ξρ+2+ερ+1)∫0t‍∥us∥ρ+2ρ+2ds+ε(ξ2ε-1)∫0t‍∥Δu∥22ds+ε(min{1,C4}-1p+1ξε-M1-m-1m+2C9-C10γM2)∫0t‍(∥u∥p+1p+1+∥∇u∥r+1r+1)+ε[1ρ+1∫Ω‍|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+1∫Ω‍|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+1∫Ω‍|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)+∫0t‍∥us∥ρ+2ρ+2ds+∫0t‍∥Δu∥22ds+∫0t‍(∥∇u∥r+1r+1+∥u∥p+1p+1)ds+ν],
where ν>0. So
(4.30)K(t)>K(0)≥(ω∫Ω‍u02dx)1-γ>0.
Using (B1)–(B3), (A4), and Hӧlder and Young inequalities, we have
(4.31)|1ρ+1∫0t‍∫Ω‍|us|ρusudxds|≤1ρ+1∫0t‍∥us∥ρ+2ρ+1∥∇u∥r+1ds≤Cρ+1∫0t‍(∥us∥ρ+2(ρ+1)μ+∥∇u∥r+1θ)ds,
where (1/μ)+(1/θ)=1.
So we have(4.32)|1ρ+1∫0t‍∫Ω‍|us|ρusudxds|1/(1-γ)≤C(∫0t‍∥us∥ρ+2(ρ+1)μds)1/(1-γ)+(∫0t‍∥∇u∥r+1θds)1/(1-γ)≤C(∫0t‍∥us∥ρ+2(ρ+1)μ·((ρ+2)/((ρ+1)μ))ds)μ(ρ+1)/(1-γ)(ρ+2)+C(∫0t‍∥∇u∥r+1θ·((1+r)/θ)ds)θ/(1+r)(1-γ)=C(∫0t‍∥us∥ρ+2ρ+2ds)μ(ρ+1)/(1-γ)(ρ+2)+C(∫0t‍∥∇u∥r+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+1∫0t‍∫Ω‍|us|ρusudxds|1/(1-γ)≤C(∫0t‍∥us∥ρ+2ρ+2ds+∫0t‍∥∇u∥r+11+rds)+β≤C(∫0t‍∥us∥ρ+2ρ+2ds+∫0t‍∥∇u∥r+11+rds+∫0t‍∥Δu∥22ds+∫0t‍∥u∥p+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+1∫0t‍∫Ω‍|us|ρusudxds|1/(1-γ)).
Combining (4.29) and (4.33)-(4.34), we have
(4.35)K′(t)≥CK1/(1-γ)(t),t≤T,
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,∀t≤T.
The above inequality implies K(t) blows-up on some time T*. Since u exists globally, so we have
(4.37)ερ+1∫0t‍∫Ω‍|us|ρusudxds<∞.
And we know that K(t)→∞,ast→T*, 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)→-∞,ast→T*.
This contradicts with the assumption that u is a global solution. So the solution of (1.1) blows-up on time T*.



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*Source: 615345-2012-09-19.xml*