# Global Existence and Blow-Up for the Euler-Bernoulli Plate Equation with Variable Coefficients

Authors: Jianghao Hao; Jie Lan
Journal: Journal of Function Spaces
 (2014)

---

## Abstract

We prove the local existence, blow-up, global existence, and stability of solutions to the initial boundary value problem for Euler-Bernoulli plate equation with variable coefficients.


---

## Body

## 1. Introduction

LetΩ be a bounded domain in Rn with smooth boundary and Γ=∂Ω. We will consider the following initial-boundary value problem:
(1)utt+A2u-γAut+autm-2ut=bup-2u,x∈Ω,t∈0,T,u=∂u∂νA=0,x∈Γ,t∈(0,T),ux,0=u0x,ut(x,0)=u1(x),x∈Ω-,
where the constants γ≥0, a≥0, b≥0, p≥2, m≥2, and
(2)Au=∑i,j=1n‍∂∂xiaij(x)∂u∂xj,x∈Rn,
where aij(x)=aji(x) are C∞ functions in Rn satisfying
(3)∑i,j=1n‍aij(x)ξiξj≥λ∑i=1n‍ξi2>0,∀x∈Rn,0≠ξ1,ξ2,…,ξnT∈Rn,
where λ is a positive constant. ∂u/∂νA=∑i,j=1n‍aij(x)(∂u/∂xi)νj is the so-called conormal derivative and ν=(ν1,ν2,…,νn) is the unit normal of Γ pointing towards the exterior of Ω and νA=Aν.In physical terms the entriesaij(x) are related to coefficients of elasticity. Let A(x)=(aij(x)) be an n×n matrix for x∈Rn, and let x=(x1,x2,…,xn) be the natural coordinate system. For each x∈Rn, we define the inverse matrix of A(x) by (gij(x))=(aij(x))-1 and the inner product and norm over the tangent space Rxn=Rn by
(4)gX,Y=X,Yg=∑i,j=1n‍gijαiβj,Xg=X,Xg1/2,X=∑i=1n‍αi∂∂xi,Y=∑i=1n‍βi∂∂xi.
Then (Rn,g) is a Riemannian manifold with metric g. Denote the gradient operator in metric g by ∇g. Then we have
(5)∇gu=∑i,j=1n‍aijx∂u∂xi∂∂xj,∇gug2=∇gu,∇gug=∑i,j=1n‍aij∂u∂xi∂u∂xj.It is well known that the equation(6)utt+butm-2ut=Δu+aup-2u(a,b≥0) with Dirichlet boundary condition has been studied by many authors. The interaction between the damping and the source term was first discussed by Levine [1, 2] in the linear damping case m=2. He showed that the solutions with negative initial energy blow up in finite time. This result was improved by Kalantarov and Ladyzhenskaya [3] to study more situations. Georgiev and Todorova [4] extended Levine’s results to the nonlinear damping case m≥2 and 2<p≤2(n-1)/(n-2) if n≥3. In their work, the authors proved that global existence for arbitrary initial data when p≤m, while blow-up result is showed when p>m and the initial energy associated with the solutions is sufficiently negative. The blow-up result is extended by Levine and Serrin [5] to the case of negative initial energy and m>1 (and to abstract evolution problems). Hao et al. [6] studied the nonlinear wave equations with damping and source term. By using the energy compensation method, the authors proved the growth orders of the nonlinear strain term, the damping term, and the source term. They also prove that the solution to the problem exists globally and blows up, respectively, and the estimate for the blow-up time is given. For results of the same nature, we refer the reader to [5, 7–9] and the references therein. Chen and Zhou [10] considered a semilinear Petrovsky equation with damping and source terms. It was proved that the solution blows up in finite time if the positive initial energy satisfies a suitable condition and for the linear damping case the solution blows up in finite time even for vanishing initial energy. Zhou [11] gave an elegant argument to the solution to nonlinear evolution equations with vanishing initial energy. Yao [12] discussed the initial-boundary value problem for Euler-Bernoulli plate with variable coefficients and gave the observability inequality.The Cauchy problem of (6) has been studied in the last decade; see [13–17]. Global existence for the Cauchy problem with arbitrarily chosen data in the energy space was proved by Todorova [15] for p≤m. The case m<p has a richer structure. The negativity of initial energy was used to prove blow-up in many papers. Zhou [16, 17] considered the Cauchy problem for a nonlinear wave equation with linear damping and source terms and showed that the solution blows up in finite time even for vanishing initial energy, and global existence and large time behavior also were discussed.For the dynamics of marine risers with damping, Köhl [18] used the Lyapunov function technique to show that the zero solution of the system is stable. Kalantarov and Kurt [19] proved that the zero solution for the considered problem is globally asymptotically stable, and an estimate of the rate of decay of the solution is obtained.The systems of plate equation have been studied by numerous authors; see [20–27]. Messaoudi [26] considered the semilinear Petrovsky equation with the damping and the source term and showed that the solution blows up in finite time if the growth order of damping term is larger than the growth order of source term and the energy is negative, whereas the solution is global if the growth order of damping term is not larger than the growth order of source term. Hoffmann and Rybka [24] studied the analyticity of the nonlinear term forces convergence of solutions for two equations of continuum mechanics. They showed that any solution with appropriate boundary and initial conditions has a limit as t goes to infinity. Avalos et al. [20] are interested in the case of thermoelastic plate and they established stability of the rest state. Eden and Milani [21] discussed the exponential attractors for extensible beam equation. Note that [24] differs from [20, 21] because [24] included a viscous term Δut which played an important role in their consideration. Guesmia [22] considered the system of plate equation with damping and proved that the solution decays exponentially if the damping term behaves like a linear function, whereas the decay is of a polynomial order otherwise. Li and Wu [25] discussed the plate stabilization problem with infinite damping and showed that the energy of the problem decays exponentially provided that the negative damping is sufficiently small. For the Cauchy problem of multidimensional generalized double dispersion equation, Xu and Liu [27] proved the existence and nonexistence of global weak solution by potential well method.Our purpose in this paper is to give the local existence, blow-up, global existence, and stability of the solution to the initial-boundary value problem (1).We will write· denoting the usual L2(Ω) norm ·L2(Ω) and ·p denoting the usual Lp(Ω) norm ·Lp(Ω) and denoting
(7)∫Ω‍∇gw,∇gwgdx=∇gwg2.
Let
(8)W=u∈H2(Ω)∣u=∂u∂νA=0on⁡Γ,H=u∈H4(Ω)∩W∣Au=∂Au∂νA=0on⁡Γ.By a weak solution of system (1) we mean a function u:[0,T)→W satisfying
(9)ddt∫Ω‍utwdx+∫Ω‍AuAwdx+γ∫Ω‍∇gut,∇gwgdx+a∫Ω‍utm-2utwdx=b∫Ω‍up-2uwdx,u0=u0,ut(0)=u1,
for any w∈W.Our paper is organized as follows. In Section2, we prove the local existence of the solution to the initial-boundary value problem (1). Section 3 contains the statements and the proof of the blow-up of the solution to problem (1) with γ=0. Section 4 is devoted to the blow-up result for problem (1) with m=2. In Section 5, we prove the global existence of the solution for problem (1). The last section is devoted to the asymptotic stability of the solution for problem (1).


## 2. Local Existence Result

In this section we establish a local existence result for the solution to problem (1) under suitable conditions on m and p.First, we give the following local existence result.Theorem 1 (local existence).
Suppose that(10)p≥2,forn≤4,2≤p≤2n-2n-4,forn>4,(11)m≥2,forn≤4,2≤m≤2nn-4,forn>4.
Then for initial data (u0,u1)∈W×L2(Ω), there exists a unique weak solution of problem (1) satisfying
(12)u∈C0,T,W,ut∈C0,T,L2Ω∩LmΩ×0,T
for T>0 small enough.Proof.
Step
1. For v(x,t) given, we consider the local existence of the problem
(13)utt+A2u-γAut+autm-2ut=bvp-2v,x∈Ω,t∈0,T,u=∂u∂νA=0,x∈Γ,t∈(0,T),ux,0=u0x,ut(x,0)=u1(x),x∈Ω-.
We take sequences {u0μ},{u1μ}∈C0∞(Ω) to approximate u0 and u1, respectively, and take a sequence {vμ}∈C([0,T],C0∞(Ω)) to approximate v. Then we consider the problems
(14)uttμ+A2uμ-γAutμ+autμm-2utμ=bvμp-2vμ,x∈Ω,t∈0,T,uμ=∂uμ∂νA=0,x∈Γ,t∈(0,T),uμx,0=u0μ,utμ(x,0)=u1μ,x∈Ω-.
Using the same arguments as in [28], we get the existence of a sequence of unique solutions {(uμ,utμ)} of (14) satisfying
(15)uμ∈L∞0,T,H,utμ∈L∞0,T,W∩LmΩ×0,T,uttμ∈L∞0,T,L2Ω.
ForR>0 large and T>0, we let ZT,R denote the set of all functions w which satisfy
(16)w∈C0,T,W,wt∈C([0,T],L2(Ω))∩Lm(Ω×(0,T)),w(x,0)=u0(x),wt(x,0)=u1(x),x∈Ω-,hwt∶=wtC([0,T],L2(Ω))2+AwC([0,T],L2(Ω))2+∇gwgL2((0,T)×Ω)2+wtLm(Ω×(0,T))2≤R2.
It follows from the trace theorem that ZT,R is nonempty if R is sufficiently large and T is small. For example, if we let
(17)w(x,t)=u0xcos⁡⁡t+u1xsin⁡t,
then w∈ZT,R for R suitable large and T small. In the present section, we always make this assumption.
Step
2. We proceed to show that the sequence {(uμ,utμ)} is Cauchy in ZT,R. For this aim, we set
(18)U∶=uμ-uν,V∶=vμ-vν,
and then U satisfies
(19)Utt+A2U-γAUt+autμm-2utμ-autνm-2utνUtt+=bvμp-2vμ-bvνp-2vν,x∈Ω,t∈(0,T),U=∂U∂νA=0,x∈Γ,t∈(0,T),Ux,0=U0x=u0μ-u0ν,Utx,0=U1x=u1μ-u1ν,x∈Ω-.
We multiply the equation of (19) by Ut and integrate over Ω×(0,t); then it follows that
(20)∫Ω‍Ut2+AU2dx+2γ∫0t‍∫Ω‍∇gUsg2dxds+2a∫0t‍∫Ω‍usμm-2usμ-usνm-2usνusμ-usνdxds=∫Ω‍U12+AU02dx+2b∫0t‍∫Ω‍vμp-2vμ-vνp-2vνUsdxds.
To estimate from the above second term on the right side of (20), we use the inequality
(21)vμp-2vμ-vνp-2vν≤C1vμ-vνvμp-2-vνp-2
for vμ,vν∈R and p≥2. Note that Ci  (i≥1) denote positive constants depending only on p, m, γ, a, b, and Ω rather than the initial data (u0,u1) in this section. To estimate the second term on the left side of (20), we use the inequality
(22)wm-2w-w~m-2w~w-w~≥C2w-w~m
for w,w~∈R and m≥2 with w=usμ, w~=usν. Then (20)–(22) yield
(23)∫Ω‍Ut2+AU2dx+2γ∫0t‍∫Ω‍∇gUsg2dxds+2aC2∫0t‍∫Ω‍usμ-usνmdxds≤∫Ω‍U12+AU02dx+2bC1∫0t‍∫Ω‍vμ-vνUsvμp-2-vνp-2dxds.
From Hölder’s inequality and the Sobolev embedding theorem the last term of (23) takes the form
(24)I∶=2bC1∫0t‍∫Ω‍vμ-vνUsvμp-2-vνp-2dxds≤2bC3∫0t‍UsV2n/(n-4)vμn(p-2)/2p-2+vνn(p-2)/2p-2ds≤2bC4∫0t‍UsAVAvμp-2+Avνp-2ds≤4bC4Rp-2∫0t‍UsAVds.
Using Young’s inequality, we have
(25)I≤2bC4Rp-2∫0t‍Us2+AV2ds≤2bC4Rp-2∫0t‍Us2ds+2bC4Rp-2Tmax⁡t∈0,T⁡AV2.
Combining (23) and (25), we obtain, for any t∈[0,T],
(26)Ut2+AU2+2γ∇gUtgL2((0,T)×Ω)2+2aC2UtLm((0,T)×Ω)m≤U12+AU02+2bC4Rp-2Tmax⁡t∈0,T⁡AV2+2bC4Rp-2∫0tUs2‍ds.
It follows from (26) and Gronwall’s inequality that, for any t∈[0,T],
(27)Ut2+AU2+2γ∇gUtgL2((0,t)×Ω)2+2aC2UtLm((0,t)×Ω)m≤U12+AU02+2bC4Rp-2Tmax⁡t∈[0,T]⁡AV2e2bC4Rp-2t.
Furthermore, we have
(28)max⁡t∈[0,T]⁡Ut2+max⁡t∈0,T⁡AU2+2γ∇gUtgL2((0,T)×Ω)2+2aC2UtLm((0,T)×Ω)m≤U12+AU02+2bC4Rp-2Tmax⁡t∈[0,T]⁡AV2e2bC4Rp-2T.
Since {vμ}, {u0μ}, and {u1μ} are Cauchy in C([0,T];W), W, and L2(Ω), respectively, we conclude that {uμ}, {utμ}, {|∇gutμ|g}, and {utμ} are Cauchy in C([0,T];W), C([0,T];L2(Ω)), L2((0,T)×Ω), and Lm((0,T)×Ω), respectively.
Step
3. We now prove that the limit (u(x,t),ut(x,t)) is a weak solution of (13).
To this end, we multiply equation of (14) by w∈W and integrate over Ω; then we obtain that
(29)ddt∫Ω‍utμwdx+∫Ω‍AuμAwdx+γ∫Ω‍∇gutμ,∇gwgdx+a∫Ω‍utμm-2utμwdx=b∫Ω‍vμp-2vμwdx.
As μ→∞, the following hold:
(30)∫Ω‍AuμAwdx⟶∫Ω‍AuAwdx,inC([0,T]),∫Ω‍vμp-2vμwdx⟶∫Ω‍vp-2vwdx,inC([0,T]),∫Ω‍∇gutμ,∇gwgdx⟶∫Ω‍∇gut,∇gwgdx,inC0,T,∫Ω‍utμm-2utμwdx⟶∫Ω‍utm-2utwdx,inL1((0,T)).
Then it follows that
(31)∫Ω‍utwdx∶=lim⁡μ→∞∫Ω‍utμwdx
is an absolutely continuous function on [0,T]; thus for almost all t∈[0,T], (u(x,t),ut(x,t)) is a weak solution of problem (13). To prove uniqueness, we denote that uμ, uν are the corresponding solutions of problem (13) to vμ, vν, respectively. Then U=uμ-uν satisfies
(32)∫Ω‍Ut2+AU2dx+2γ∫0t‍∫Ω‍∇gUsg2dxds+2aC2∫0t‍∫Ω‍usμ-usνmdxds≤2bC1∫0t‍∫Ω‍vμ-vνUsvμp-2-vνp-2dxds.
This shows that U=0 for vμ=vν. The uniqueness follows.
Step
4. We denote by K the map which carries v∈ZT,R into u; that is,
(33)Kv=u,
where (u,ut) is the solution of problem (13). We establish a priori estimate below to show that K maps ZT,R into itself if R is sufficiently large and T is sufficiently small relative to R. We then equip ZT,R with the complete metric ρ defined by
(34)ρw,w-=sup⁡t∈0,Thwt-w-t
and show that K is strict contraction if T is sufficiently small. The contraction mapping principle thus implies that K has a unique fixed point which is obviously a solution to problem (1). For this purpose, we multiply the equation of (13) by 2ut and integrate over Ω×(0,T) to get that
(35)∫Ω‍ut2+Au2dx+2γ∫0t‍∫Ω∇gusg2‍dxds+2a∫0t‍∫Ω‍usmdxds=∫Ω‍u12+Au02dx+2b∫0t‍∫Ω‍vp-2vusdxds.
From Hölder’s inequality, it follows that, for t∈[0,T],
(36)∫Ω‍ut2+Au2dx+2γ∫0t‍∫Ω‍∇gusg2dxds+2a∫0t‍∫Ω‍usmdxds≤∫Ω‍u12+Au02dx+2b∫0t‍∫Ω‍usvp-1dxds≤∫Ω‍u12+Au02dx+2bC5∫0t‍usAvp-1ds≤∫Ω‍u12+Au02dx+2bC5max⁡t∈[0,T]⁡Avp-1∫0t‍usds≤∫Ω‍u12+Au02dx+2bC5max⁡t∈0,T⁡Avp-1×max⁡t∈[0,T]⁡ut∫0t‍ds≤∫Ω‍u12+Au02dx+2bC5TRp-1max⁡t∈[0,T]⁡ut.
By choosing R large enough and then T sufficiently small, we obtain
(37)u∈ZT,R.
This shows that K maps ZT,R into itself.
Step
5. We verify that K is a contraction if T is sufficiently small. Let v,v-∈ZT,R, and set u=Kv and u-=Kv-. Clearly, U:=u-u- is the solution of the problem
(38)Utt+A2U-γAUt+autm-2ut-au-tm-2u-t=bvp-2v-bv-p-2v-x∈Ω,t∈(0,T),U=∂U∂νA=0,x∈Γ,t∈(0,T),U(x,0)=0,Ut(x,0)=0,x∈Ω-.
Multiplying the equation of  (38) by 2Ut and integrating it over Ω, we have
(39)ddtUt2+AU2+2γ∫0t‍∫Ω‍∇gUsg2dxds+2a∫0t‍∫Ω‍usm-2us-u-sm-2u-sus-u-sdxds=2b∫Ω‍Utvp-2v-v-p-2v-dx.
Noticing U(x,0)=Ut(x,0)=0 and using the same arguments as (20)–(28), we get the estimate
(40)ρ(u,u-)≤2bC6RpTe2bC6Rp-2Tρ(v,v-),
where C6 is a positive constant independent of R and T. By choosing T so small that
(41)2bC6RpTe2bC6Rp-2T<1,
then the map K is a contraction. By the contraction mapping principle, the map K has a unique fixed point which is obviously a solution u=Kv∈ZR,T. It is clear that (u,ut) is the desired solution of problem (1), and the proof of Theorem 1 is completed.In the following we denote the maximal existence time of the solution by(Tmax⁡=)T and the energy of system (1) by
(42)Et=12ut2+12Au2-bpupp.
Multiplying the equation of (1) by ut and integrating over Ω, we obtain, for t∈[0,T),
(43)ddtEt=-autmm-γ∇gutg2≤0.DenoteB>0 is the best embedding constant such that
(44)wp≤BΔw,forw∈W.Next we present two lemmas which will be used in the following sections.Lemma 2.
Suppose thatp satisfies (10); then one has
(45)ups≤B2λ+1Au2+upp
for s∈[2,p].Proof.
From the fact thatup≤BΔu, we get
(46)ups≤up2≤B2Δu2≤B2λAu2,
if up≤1. Otherwise,
(47)ups≤upp.
Lemma 2 follows from (46) and (47).Lemma 3.
Suppose that the conditions of Theorem1 hold, and let u(x,t) be the solution of problem (1) with the initial data satisfying E(0)<E1 and Au0>s1, where
(48)E1∶=p-22ps12,s1∶=λp/2bBp1/(p-2).
Then there exists s2>s1, such that
(49)Au≥s2.Proof.
By Sobolev-Poincaré inequality and the property of the operatorA, we have
(50)upp≤BpΔup≤Bpλp/2Aup.
Thus we get
(51)E1>E(0)≥E(t)≥12Au2-bpupp≥12Au2-bBppλp/2Aup∶=FAu,
where
(52)F(s)=12s2-bBppλp/2sp.
It is easy to verify that the function F(s) has a maximum at s1=(λp/2/bBp)1/(p-2) and the maximum value is E1=((p-2)/2p)s12. From the definition of F(s), we see that F(s) is increasing in (0,s1) and decreasing in (s1,+∞) and F(s)→-∞ as s→+∞. From the assumptions E(0)<E1 and Au0>s1, we know that there exists s2>s1, such that
(53)Au≥s2,
which completes the proof of Lemma 3.


## 3. Blow-Up Result forγ=0

In this section we establish the blow-up result for the following problem by the energy compensation method:(54)utt+A2u+autm-2ut=bup-2u,x∈Ω,t∈0,T,u=∂u∂νA=0,x∈Γ,t∈(0,T),ux,0=u0x,ut(x,0)=u1(x),x∈Ω-.
Our technique of proof follows closely the argument of [7] with the modifications needed for our problem. Denoting the energy of system (54) by
(55)E(t)=12ut2+12Au2-bpupp,
multiplying the equation of (54) by ut, and integrating over Ω, we obtain, for t∈[0,T),
(56)ddtE(t)=-autmm.Next we give the blow-up result.Theorem 4.
Assume that the conditions of Theorem1 hold, m<p, and u(x,t) is the solution of (54) with the initial data (u0,u1)∈W×L2(Ω) satisfying either one of the following conditions:(ii)
E(0)<0;(iii)
0≤E(0)<E1 and Au0>s1, in which
(57)E1∶=p-22ps12,s1∶=λp/2bBp1/(p-2)and B is the embedding constant satisfying
(58)up≤BΔu.
Then the solution u(x,t) blows up in finite time.Proof.
In the following, we discuss two cases.
Case
1  (0≤E(0)<E1). We set
(59)H(t)=M-E(t),
where M>E(0) can be chosen later. Since
(60)H′(t)=autmm≥0
and H(0)=M-E(0)>0, hence
(61)0<H(0)≤H(t).
Next we define
(62)G(t)=H1-α(t)+ε∫Ω‍uutdx,
where α, ε are to be determined positive constants and ε is small enough. Integrating by parts and using the equation in (54), we obtain
(63)G′t=1-αH-αtH′(t)+εut2-εAu2-aε∫Ωutm-2utu‍dx+bεupp.
To estimate the fourth term of (63), we use Young’s inequality,
(64)∫Ω‍utm-2utudx≤∫Ω‍uutm-1dx≤δmmumm+m-1mδ-m/(m-1)utmm,
where δ can be time dependent, since the integral is taken over the x variable. Take δ so that
(65)δ-m/(m-1)=kH-α(t),
for sufficiently large k to be specified later. Hence together with the definitions of E(t) and H(t), it follows that
(66)G′t≥1-α-m-1mεkH-αtH′t+ε1+p2ut2+εp2-1Au2-aεk1-mmHαm-1tumm+εpH(t)-εpM.
Furthermore, by Lemma 3 we have
(67)Au2=s22-s12s22Au2+s12s22Au2≥s22-s12s22Au2+s12.
Thus it follows from (66) and (67) that
(68)G′t≥1-α-m-1mεkH-αtH′t+ε1+p2ut2+εp2-1s22-s12s22Au2-aεk1-mmHαm-1(t)umm+εpH(t)+εp2-1s12-εpM.
By taking M=E1, we obtain
(69)G′t≥1-α-m-1mεkH-αtH′t+ε1+p2ut2+εp2-1s22-s12s22Au2-aεk1-mmHα(m-1)(t)umm+εpH(t).
Note that, from the definition of E(t) and (48), we obtain
(70)0<H0≤Ht=M-E(t)≤M-12Au2+bpupp<M-12s22+bpupp≤bpupp.
Thus by (70) and the inequality
(71)umm≤upm,
we obtain
(72)Hα(m-1)(t)umm≤bpα(m-1)upαp(m-1)+m.
Selecting α∈(0,(p-m)/p(m-1)], then together with Lemma 2, we get that
(73)Hα(m-1)(t)umm≤bpα(m-1)B2λ+1Au2+upp.
Combining (69) and (73), we have
(74)G′t≥1-α-m-1mεkH-αtH′t+ε1+p2ut2+εp2-1s22-s12s22Au2-aεk1-mmbpαm-1B2λ+1Au2+upp+εpH(t).
From (74) and the equality
(75)H(t)=M+bpupp-12ut2-12Au2,
we obtain
(76)G′t≥1-α-m-1mεkH-αtH′t+ε1+p2-p-24s22-s12s22ut2+εp-24s22-s12s22B2λ+1bpαm-1≥+εai-amk1-mbpαm-1B2λ+1Au2+εbp-22ps22-s12s22bpαm-1B2λ+1≥+εai-amk1-mbpαm-1B2λ+1upp+εp-p-22s22-s12s22H(t)+εp-22s22-s12s22M.
Then we can choose k large enough such that the coefficients of Au2 and upp in (76) are strictly positive, and hence we have
(77)G′t≥1-α-m-1mεkH-αtH′t+εβut2+Au2+upp+Ht,
where β>0 is the minimum of these coefficients. For fixed k, we pick ε small enough such that 1-α-((m-1)/m)εk≥0 and
(78)G(0)=H1-α(0)+ε∫Ω‍u0u1dx>0.
Hence
(79)G′t≥εβut2+Au2+upp+H(t).
Thus for t≥0, G(t) is a nondecreasing function, and we have, for all t≥0,
(80)G(t)≥G(0)>0.
Here we estimateG1/(1-α)(t) from above. In the following of this section, we denote by C the general positive constant which depends on Ω, p, α, and H(0).
We use Hölder’s inequality to estimate the term(81)∫Ω‍uutdx≤uut≤Cuput,
which implies
(82)∫Ω‍uutdx1/(1-α)≤Cup1/(1-α)ut1/(1-α).
Using Young’s inequality
(83)XY≤δμμXμ+δ-θθYθ,X,Y≥0,∀δ>0,1μ+1θ=1,
with δ=1, μ=2(1-α)/(1-2α), and θ=2(1-α), we have
(84)∫Ω‍uutdx1/(1-α)≤Cup2/(1-2α)+ut2.
By choosing
(85)0<α≤min⁡⁡p-22p,p-mpm-1,
which implies 2/(1-2α)≤p, then using (70) and the inequality
(86)zν≤z+1≤1+1az+a,∀z≥0,a>0,0<ν≤1,
with z=upp and a=H(0), we deduce
(87)∫Ω‍uutdx1/(1-α)≤C1+1H0upp+ut2+H(t).
Consequently, we have, for all t>0,
(88)G1/(1-α)t=H1-α(t)+ε∫Ω‍uutdx1/(1-α)≤21/(1-α)H(t)+∫Ω‍uutdx1/(1-α)≤CH(t)+upp+ut2.
It follows from (79) and (88) that
(89)G′t≥CG1/(1-α)(t).
A simple integration of (89) over (0,t) then yields
(90)Gα/(1-α)(t)≥1G-α/(1-α)(0)-Cαt/1-α.
Therefore (90) shows that G(t) blows up in a finite time T* given by the estimate
(91)T*≤1-αCαGα/(1-α)(0).
So the solution u(x,t) blows up in a certain finite time.
Case
2  (E(0)<0). We can take M=0 in the definition of H(t); that is,
(92)H(t)=-E(t).
Then we can get our result by the same arguments as in Case 1. This completes the proof of Theorem 4.Remark 5.
The earliest blow-up timeT* can be estimated by
(93)T*≤1-αCαGα/(1-α)(0),
and the larger the G(0) is, the quicker the blow-up takes place.


## 4. Blow-Up Result form=2

In this section we discuss the blow-up result for the following problem:(94)utt+A2u-γAut+aut=bup-2u,x∈Ω,t∈0,T,u=∂u∂νA=0,x∈Γ,t∈(0,T),u(x,0)=u0(x),ut(x,0)=u1(x),x∈Ω-,
with the concavity method; see Levine [1, 2].Denoting the energy of system (94) by
(95)E(t)=12ut2+12Au2-bpupp
and multiplying the equation of (94) by ut and integrating over Ω, we obtain, for t∈[0,T),
(96)ddtEt=-aut2-γ∇gutg2.Our result is as follows.Theorem 6.
Supposep satisfies (10), u(x,t) is the solution of problem (94), and the initial data (u0,u1)∈W×L2(Ω) satisfies either one of the following conditions:(ii)
E(0)<0;(iii)
E(0)=0 and ∫Ω‍u0u1dx>0;(iiii)
0<E(0)<E1 and Au0>s1, where
(97)E1∶=p-22ps12,s1∶=λp/2bBp1/(p-2),and B is the embedding constant satisfying
(98)up≤BΔu.
Then the solution u(x,t) blows up in finite time T, and
(99)lim⁡t→T-ut2+∇gug2=+∞.Proof.
Define(100)Lt=ut2+a∫0t‍us2ds+γ∫0t‍∇gusg2ds+aT0-tu02+γT0-t∇gu0g2+rt+τ2,t∈[0,T0],
where T0, r, and τ are positive constants which are specified latter.
It is not difficult to see thatL(t)>0 for all t∈[0,T0]. Furthermore, we have
(101)L′t=2∫Ω‍uutdx+au2-au02+γ∇gug2-γ∇gu0g2+2r(t+τ)=2∫Ω‍uutdx+2a∫0t‍∫Ω‍uus(x,s)dxds+2γ∫0t‍∫Ω‍∇gux,s,∇gusx,sgdxds+2r(t+τ),L′′t=2ut2-2Au2+2bupp+2r.
Combining (100) and (101), we obtain
(102)LL′′-p+24L′2LL′′=2Lut2-Au2+bupp+rLL′′=+(p+2)u2+a∫0t‍us2ds∫0t‍us22LL′′=+(p+2)pp+γ∫0t‍∇gusg2ds+rt+τ2LL′′=+p=2pp×ut2+a∫0t‍us2dsLL′′=+p=2pp=i+γ∫0t‍∇gusg2ds+rLL′′=+p=2pp-∫Ω‍uutdx+a∫0t‍∫Ω‍uusdxdsLL′′=+p=2pp=i+γ∫0t‍∫Ω‍∇gu,∇gusgdxdsLL′′=+p=2pp=i∫0t‍∫Ω‍uusdx+rt+τ2LL′′=-(p+2)u2+a∫0t‍us2dsLL′′=-p+2w+γ∫0t‍∇gusg2ds+rt+τ2LL′′=×ut2+a∫0t‍us2ds+γ∫0t‍∇gusg2ds+r.
It follows from the fact
(103)∫Ω‍uutdx+a∫0t‍∫Ω‍uusdxds=+γ∫0t‍∫Ω‍∇gu,∇gusgdxds+rt+τ2n+i≤u2+a∫0t‍us2dsn+i=W+γ∫0t∇gug2‍ds+rt+τ2n+i=×ut2+a∫0tus2‍dsn+i=W=+γ∫0t‍∇gusg2ds+r
that
(104)LL′′-p+24L′2LL′′L≥2Lut2-Au2+bupp+rLL′′=≥-p+2u2+a∫0t‍us2dsLL′′=≥-p+2P+γ∫0t‍∇gusg2ds+rt+τ2LL′′=≥×ut2+a∫0tus2‍ds+γ∫0t∇gusg2‍ds+r.
Noticing that
(105)Lt-ut2-a∫0tu‍2ds-γ∫0t‍∇gug2ds-rt+τ2Lt-=aT0-tu02+γT0-t∇gu0g2,
then we get that
(106)LtL′′t-p+24L′t2Lt≥2Ltut2-Au2+bupp+rLt≥i+p+2aT0-tu02+γT0-t∇gu0g2Lt≥i×ut2+a∫0t‍us2ds+γ∫0t‍∇gusg2ds+rLt≥i-p+2Ltut2+a∫0t‍us2dsLt≥i-p+2Ltw+γ∫0t‍∇gusg2ds+rLt≥2Lt-p2ut2-Au2+bupp∫0t‍us2Lt≥2Ltw-ap+22∫0t‍us2dsLt≥2Ltw-γp+22∫0t‍∇gusg2ds-p2rLt∶=2LtQt,
where Q(t) is defined by
(107)Qt∶=-p2ut2-Au2+bupp-ap+22∫0t‍us2ds-γp+22∫0t‍∇gusg2ds-p2r.
By the definition of E(t), we may also write
(108)Qt=-pE(t)+p-22Au2-ap+22∫0tus2‍ds-γp+22∫0t‍∇gusg2ds-pr2.
From (96), we find
(109)E(t)=E0-a∫0t‍us2ds-γ∫0t‍∇gusg2ds.
Then we obtain
(110)Qt=-pE(0)+p-22Au2+ap-22∫0tus2‍ds+γp-22∫0t‍∇gusg2ds-pr2.
In the following, we discuss three cases.
Case
1  (E(0)<0). By taking r=-2E(0)>0, we have
(111)Qt≥-pE0+r2=0,L(0)>0.
Furthermore, we have
(112)LtL′′t-p+24L′t2≥0,t∈0,T0.
Denoting y(t)=L(t)-(p-2)/4, we obtain that
(113)y′′t=-p-24Lt-p+6/4LtL′′t-p+24L′t2≤0,t∈0,T0.
We can select suitable positive constants τ and T such that
(114)L′0>0,0<L0L′0≤p-24T,
and then we have
(115)Lt≥4L(p+2)/404L(0)-p-2L′0t4/(p-2),
for some t>0. Moreover, there exists a T>0 such that T≤T0 and
(116)lim⁡t→T-ut2+a∫0t‍us2ds+γ∫0t‍∇gusg2ds=+∞,
and then we have
(117)lim⁡t→T-ut2+∇gug2=+∞.
Case
2  (E(0)=0and∫Ω‍u0u1dx>0). By taking r=0 we have
(118)Qt≥-pE0+r2=0.
Furthermore, we have
(119)LtL′′t-p+24L′t2≥0,t∈0,T0.
Denoting y(t)=L(t)-(p-2)/4, we obtain that
(120)y′′t=-p-24Lt-(p+6)/4LtL′′t-p+24L′t2≤0,t∈0,T0.
Since ∫Ω‍u0u1dx>0, then we have
(121)u02>0,L(0)>0.
Furthermore, we choose suitable positive constants τ and T such that
(122)0<L0L′0≤p-24T,
and then from the fact that L′(0)>0 we have
(123)Lt≥4L(p+2)/404L(0)-p-2L′0t4/(p-2),
for some t>0. By the similar arguments as we did in the proof of Case 1, we can prove the desired limit.
Case
3  (0<E(0)<E1andAu0>s1). It follows from Lemma 3 that there exists s2>s1, such that
(124)Au≥s2>s1.
Thus together with (110) and (124), it follows that
(125)Qt≥-pE(0)+p-22s22-pr2=-pE(0)+p-22s12+p-22(s22-s12)-pr2.
By taking r=((p-2)/p)(s22-s12)>0, we get
(126)Q(t)≥0,L(0)>0.
Furthermore, we have
(127)LtL′′t-p+24L′t2≥0,t∈0,T0.
Denoting y(t)=L(t)-(p-2)/4, we obtain that
(128)y′′t=-p-24Lt-p+6/4LtL′′t-p+24L′t2≤0,t∈[0,T0].
Furthermore, we choose suitable positive constants τ and T such that
(129)0<L0L′0≤p-24T,
and then we have
(130)L(t)≥4L(p+2)/404L(0)-p-2L′0t4/(p-2),
for some t>0. By the similar arguments as we did in the proof of Case 1, we can prove the desired limit. Theorem 6 is established.


## 5. Global Existence Result

In this section we give the result of the global existence of the solution to problem (1) which is similar to [26].Theorem 7.
Supposep, m satisfy (10)-(11) and p≤m, the initial data (u0,u1)∈W×L2(Ω), and u(x,t) is the solution of problem (1). Then the solution u(x,t) is global.Proof.
Set(131)Pt=12ut2+12Au2+bpupp.
After a simple computation, we get that
(132)P′t=-autmm-γ∇gutg2+2b∫Ω‍up-2uutdx.
From Young’s inequality, it follows that, for any δ>0,
(133)P′t≤-autmm-γ∇gutg2+bδutpp+bCδupp.
Since p≤m, then
(134)P′t≤-autmm-γ∇gutg2+bCδutmp+bCδupp,
where C>0 is a constant depending on the domains Ω and p and C(δ)>0 is a constant depending on δ. Then we get
(135)P′t≤c1+c2Pt;
here and after in this section, ci  (i=1,2,3) denote positive constants. Indeed, if utm>1, then we can choose δ suitable small such that -autmm+bCδutmp≤0, and hence
(136)P′t≤bC(δ)upp≤pCδPt,
and otherwise, if utm≤1, from (134), we obtain
(137)P′t≤bCδ+bCδupp≤bCδ+pCδPt.
By integrating (135) we have, for t≥0,
(138)Pt≤P0+c3ec2t,
where c3=c1/c2. Theorem 7 is established.


## 6. Global Asymptotic Stability Result

In this section we establish the global asymptotic stability result to problem (1) with b>0 small enough and p=2 or b=0 (linear source term or no source term).Theorem 8.
Supposem satisfies (11), the initial data (u0,u1)∈W×L2(Ω), and u(x,t) be the solution of problem (1) with 0<b<λ/B2 and p=2 or b=0 (B is the constant such that u≤BΔu). Then the solution u(x,t) is asymptotically stable, and the following decay estimate holds for t>1 sufficiently large:
(139)ut2+Au2+∇gug2≤Ce-Ct,m=2,ut2+Au2≤Ct(1-m)/m,m∈(2,3),ut2+Au2≤Ct-2/m,m≥3,
where the generic constant C>0 depends on Ω, γ, a,b, m, and the initial data (u0,u1).Proof.
For0<b<λ/B2 and p=2 or b=0, using (48) with p=2 and (3), we have
(140)E0≥Et=12ut2+12Au2-b2u2≥12ut2+12-bB22λAu2=:12ut2+μAu2≥0,
where b<λ/B2 and μ=1/2-bB2/2λ. Then thanks to (43), we have
(141)E0-Et=a∫0t‍∫Ωutm‍dxds+γ∫0t‍∫Ω∇gutg2‍dxds.
Combining (140) and (141), we obtain
(142)E0≥a∫0t‍∫Ω‍usmdxds+γ∫0t‍∫Ω∇gusg2‍dxds.
Therefore, we get the following estimate:
(143)∫0t‍∫Ω‍usmdxds≤E0a,(144)∫0t‍∫Ω‍∇gusg2dxds≤E0γ.
Multiplying the equation of (1) by u and integrating over Ω, we obtain
(145)ddt∫Ω‍uutdx-ut2+Au2ddtW+γddt∇gug2+a∫Ω‍utm-2utudx=bu22.
Multiplying (145) by a small enough positive constant δ to be determined and adding to the energy identity
(146)ddt12ut2+12Au2-bpu22ddtW=-autmm-γ∇gutg2,
we get that
(147)ddt12ut2+12Au2-b2u22∇gug2=w+δ∫Ω‍uutdx+δγ∇gug2ddtW=-autmm-γ∇gutg2+δut2-δAu2ddtW=-aδ∫Ω‍utm-2utudx+δbu22.
Set
(148)Jt∶=12ut2+12Au2-b2u22+δ∫Ω‍uutdx+δγ∇gug2.
Then the following inequality holds:
(149)ddtJt=-autmm-γ∇gutg2+δut2-δAu2-aδ∫Ω‍utm-2utudx+δbu22≤-autmm-γ∇gutg2+δut2-δAu2+aδ∫Ω‍utm-1udx+δbu22=2δut2-autmm-γ∇gutg2-2δEt+aδ∫Ω‍utm-1udx.
Integrating (149) with respect to t over (0,t), we obtain
(150)2δ∫0t‍E(s)ds≤J0-Jt+2δ∫0t‍us2ds-a∫0t‍usmmds-γ∫0t‍∇gusg2ds+aδ∫0t‍∫Ωusm-1‍udxds.
Combining the nondecreasing property of E(t), we get the estimate
(151)2δtEt≤J(0)-J(t)+2δ∫0t‍us2ds-a∫0t‍usmmds-γ∫0t‍∇gusg2ds+aδ∫0t‍∫Ω‍usm-1udxds.
From Young’s inequality and the definition of J(t) and (140), we have
(152)Jt≥12ut2+12Au2-b2u22-δ2u2-δ2ut2+δγ∇gug2≥12-δ2ut2+μ-δB22λAu2+δγ∇gug2≥12-δ2ut2+μ-δB22λAu2+δγ∇gug2≥0,
for δ small enough. Therefore, the following estimate holds:
(153)2δtEt≤J0+2δ∫0t‍us2ds+aδ∫0t‍∫Ω‍usm-1udxds=:J0+I1+I2.
Using Hölder’s inequality, (43), (140), and (143), we obtain
(154)I1=2δ∫0t‍∫Ω‍us2dxds≤2δ∫0t‍∫Ω‍1dxds(m-2)/m∫0t‍∫Ω‍usmdxds2/m≤2δE0a2/mC(Ω)(m-2)/mt(m-2)/m,
where C(Ω) is a positive constant depending on Ω, and
(155)I2=aδ∫0t‍∫Ω‍usm-1udxds≤aδ∫0t‍∫Ω‍usmdxds(m-1)/m∫0t‍∫Ωum‍dxds1/m≤aδC*E0a(m-1)/m∫0t‍Aumds1/m≤aδC*E0a(m-1)/mE0μ1/2t1/m,
where C* is the constant such that um≤C*Au. It follows from (153)–(155) that
(156)2δEt≤J(0)t-1+2δE0a2/mC(Ω)(m-2)/mt-2/m+aδC*E0a(m-1)/mE0μ1/2t(1-m)/m.
Therefore, there are positive constants Ki  (i=1,2,3) such that
(157)E(t)≤K1t-1+K2t-2/m+K3t(1-m)/m.
Hence together with (140), we have, for t>1 sufficiently large, there exist positive constants C>0, such that
(158)ut2+Au2≤Ct(1-m)/m,m∈[2,3),ut2+Au2≤Ct-2/m,m≥3.
In particular, in the casem=2, we obtain from (149) with m=2(159)ddtJ(t)+aut2+γ∇gutg2-δut2+δAu2+aδ∫Ω‍utudx-δbu2=0.
Multiplying J(t) by a small enough positive constant δ* to be determined and adding to (159), we obtain
(160)0=ddtJt+aut2+γ∇gutg2-δut2+δAu2+aδ∫Ω‍utudx-δbu2+δ*Jt-12δ*ut2-12δ*Au2+b2δ*u2-δδ*∫Ω‍uutdx-δδ*γ∇gug2≥ddtJ(t)+δ*Jt+a2-δ-δ*ut2+γ∇gutg2+δ-δ*2Au2+δ*b2u2-δb+δ2δ*2+aδ22u2-δδ*γ∇gug2.
Using (48) with p=2 and (3), we have
(161)u2≤B2λAu2.
It is not difficult to see that
(162)∇gug2≤CAu2,
for some positive constant C. Together with (160)–(162), we can get the following estimate:
(163)0≥ddtJt+δ*J(t)+a2-δ-δ*ut2+γ∇gutg2+δ-δ*2-δb+δ2δ*2+aδ22B2λ-δδ*γCAu2+δbu2.
We can choose δ and δ*  (δ*<δ) sufficiently small with 0<b<λ/B2 such that the following estimate holds:
(164)ddtJt+δ*Jt≤0.
Then we can get that
(165)J(t)≤J(0)e-δ*t.
Thus due to (152), we have
(166)ut2+Au2+∇gug2≤Ce-Ct,
where the generic constants C>0 depend on Ω, γ, a,b, and the initial data (u0,u1). This completes the proof of Theorem 8.



---

*Source: 176429-2014-12-25.xml*