# Blow-Up of Solutions for a Class Quasilinear Wave Equation with Nonlinearity Variable Exponents

Authors: Zakia Tebba; Hakima Degaichia; Mohamed Abdalla; Bahri Belkacem Cherif; Ibrahim Mekawy
Journal: Journal of Function Spaces
 (2021)

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## Abstract

This work deals with the blow-up of solutions for a new class of quasilinear wave equation with variable exponent nonlinearities. To clarify more, we prove in the presence of dispersion term−Δutt a finite-time blow-up result for the solutions with negative initial energy and also for certain solutions with positive energy. Our results are extension of the recent work (Appl Anal. 2017; 96(9): 1509-1515).


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## Body

## 1. Introduction

We study in this paper the following nonlinear wave equation:(1)utt−div∇us.−2∇u−Δutt+ηututq.−2=μuup.−2,inΩ×0,T,ux,t=0,onΩ×0,T,ux,0=u0x,utx,0=u1x,inΩ.Here, Ω⊂ℝnn≥1, be a bounded domain with a smooth boundary ∂Ω, andη,μ>0 are constants, and the exponents q., p., and s. are given measurable functions on Ω satisfying
(2)2≤maxq2,s2<p1≤px≤p2≤s∗x,with
(3)p1≔essinfx∈Ωpx,p2≔esssupx∈Ωpx,s1≔essinfx∈Ωsx,s2≔esssupx∈Ωsx,q1≔essinfx∈Ωqx,q2≔esspsupx∈Ωqx,s∗x=Nsxesssupx∈ΩN−qxifs2<n,+∞ifs2≥n..Also, we suppose thatq., p., and s. satisfy the log-Holder continuity condition:
(4)mx−my≤−Alogx−yfora.e.x,y∈Ω,withx−y<δ,A>0,0<δ<1. In (4), if x=y, the inequality is undefined because log0 is undefined. The inequality is defined for x not equal to y, but the condition that δ is completely greater than zero always makes x not equal to y because x−y<δ. The term Δsu=div∇us.−2∇u is called s.-Laplacian.There are many studies that have studied the problemP in the case of constant and variable-exponent nonlinearities.In the case of constant exponent nonlinearity whenη=0, s=2, and Δutt=0, the explosion term μuup−2 forces the negative-energy solutions to explode in finite time ([1, 2]), whereas when μ=0, s=2, and Δutt=0, the dissipation term ηutq−2ut guarantees the existence of global solutions for any initial data [3].The problem was first treated by (Levine [2] and Vitillaro [4]) in the case when the both terms are present (the dissipation and source). He debated the case when q=2,s=2, and determined the result of blow-up for solutions with negative initial energy. To extend Levine’s results in [5] considered a different method when q>2 and discussed the cases when q≥p and p>q.Chen et al. in [6] looked into the nonlinear p−Laplacian wave equation:
(5)∂ttu−∇.∇up−2∇u−Δut+qx,u=fx,when 2≤p<n and f,q are given functions. Under suitable conditions on the initial data and the functions f,q, they realized global existence and uniqueness and also discussed the long-time behavior of the solution.In [7], Benaissa and Mokeddem considered
(6)utt−div∇up−2∇u−σtdiv∇utm−2∇ut=0,and they achieved an energy decay estimate for the solutions where p,m≥2, σ, is a positive function and expanded Yang [8] and Messaoudi [9] results. Recently, Mokeddem and Mansour [10] added some modification in the problem of Benaissa and Mokeddem [10] and established the same decay result.Messaoudi and Houari [11] studied the nonlinear wave equation:
(7)utt−Δut−cdiv∇uα−2∇u−div∇utβ−2∇ut+autm−2ut=bup−2u,where Ω is a bounded domain in ℝnn≥1,a,b,c>0, and α,β,m,p>2. They investigated with appropriate conditions imposed on α,β,m,p>2, a global nonexistence result for solutions associated with negative initial energy. In [12], Kafini and Messaoudi treated a nonlinear wave equation with delay term and proved, under appropriate hypotheses on the initial data, that the energy of solutions explodes in a finite time. For more results, see the previous studies ([13–35]). In the case of variable-exponent nonlinearity, Antontsev [36] looked into the problem:
(8)∂ttu−divax,t∇upx,t−2∇u−αΔut=bx,tuuσx,t−2,when α is a nonnegative constant, and a,b,p,σ are given functions. He discussed the case when α=0 and α>0 and demonstrated a blow-up result under particular hypothesis on a,b,p,σ.Thereafter, Antontsev in [36] considered the same equation and established a local, global existence of weak solutions for specific conditions on a,b,p,σ and realized a blow-up results for solutions with nonpositive initial energy. In [5], Guo and Gao considered the same problem of Antontsev [13], they picked the constant σx,t=r>2 and realized a blow-up result in finite time, and also, they alleged without any proof the same blow-up result for σx,t=rx.Sun et al. in [37] studied the blow-up result for solutions with positive initial energy for the following equation:
(9)utt−divax,t∇u+cx,tututqx,t−1=bx,tuupx,t−1.They also gave lower and upper bounds for the blow-up time and provided a numerical illustrations for their result.Lately, Messaoudi and Talahmeh [38] looked into
(10)utt−div∇umx−2∇u+μut=upx−2u,where μ≥0. They proved a blow-up result for certain solutions with arbitrary positive initial energy. Korpusov [39] generalized this result and established (10), with m and p are constants.In this paper, we care to find sufficient conditions onq,p,s and the initial data for which the blowup happens.In addition to the introduction, our paper is divided into three sections. The second section deals with variable-exponent Lebesgue and Sobolev spaces and some of their characteristics. We also mention the result of existence, but without demonstration, and the second one deals with the result of blow-up for solutions with negative initial energy.In the fourth one, we present and demonstrate the theorem of blow-up for certain solutions with positive initial energy.


## 2. Background and Preliminaries

This section contains some essential concepts and definitions about the Lebesgue and Sobolev spaces with variable exponents which will be useful to us later (see Fan and Zhao [40] and Lars et al. [41], Mezouar and Boulaaras [42]).LetΩ that is a domain of ℝn,p:Ω⟶1,∞ be a measurable function. We introduce the Lebesgue space with a variable exponent p. by
(11)Lp.Ω≔w:Ω⟶ℝ;measurableinΩ:ρp.λw<+∞,forsomeλ>0,where
(12)ρp.w=∫Ωwxpxdx.Equipped with the Luxembourg-type norm,(13)wp.≔infλ>0:∫Ωwxλpxdx≤1.Lp.Ω is a Banach space (see Lars et al. [41]).Now, we introduce the following variable-exponent Sobolev spaceW1,p.Ω:
(14)W1,p.Ω=w∈Lp.Ωsuchthat∇wexistsand∇w∈Lp.Ω.This space is a Banach space with respect to the normwW1,p.Ω=wp.+∇wp.. Otherwise, we put W01,p.Ω to be the closure of C0∞Ω in W1,p.Ω.Remark 1.
The spaceW1,p.Ω is generally defined in a different way for the variable exponent case.
However, the two definitions are equivalent under condition (4) (see Lars et al. [4]). We define W−1,p′.Ω as the dual of W1,p.Ω, in the same way as the classical Sobolev spaces, where 1/p.+1/p′.=1.Lemma 2.
(Poincare inequality [41]).
LetΩ be a bounded domain and assume that p. satisfies (4), then
(15)wp.≤C∇wp.,forallw∈W1,p.Ω,for C>0 that is a constant depends only on Ω and p1,p2. Particularly, ∇wp determines an equivalent norm on W01,p.Ω.Lemma 3.
(Lars et al. [41]).
Ifm:Ω⟶1,∞ is a measurable function and s.∈CΩ¯ such that
(16)essinfx∈Ωs∗x−mx>0withs∗x=nsxesssupx∈Ωn−sxifs2<n,∞ifs2≥n.Then, the embeddingW01,s∙Ω°LmΩ is continuous and compact.Lemma 4.
(Hölder’s inequality [41]).
Assume thatp,m,r≥1 are measurable functions defined on Ω such that
(17)1ry=1py+1my,fora.e.y∈Ω.
Ifu∈Lp.Ω and w∈Lm.Ω, then uw∈Lr.Ω, with
uwr.≤2up.wm..Lemma 5.
(Unit ball property [41]).
Assume thatp is a measurable function on Ω. Then,
(18)fp.≤1ifandonlyifρp.f≤1.Lemma 6.
(Lars etal [41]).
Ifp is a measurable function on Ω satisfying (4), then for a.e. x∈Ω, we get
(19)minup.p1,up.p2≤ρp.u≤maxup.p1,up.p2,for any u∈Lp.Ω.Proposition 7.
Letu0,u1∈W01,sΩ×L2Ω and suppose that the exponents p,q,s satisfy (1) and (2). Then, problem P admits a unique weak solution such that
(20)u∈L∞0,T,W01,s∙Ω,ut∈L∞0,T,H01Ω,utt∈L∞0,T,W0−1,s′∙Ω,where 1/s+1/s′=1.Remark 8.
We can achieve the proof of the previous proposition by using the Galerkin method as in [13].


## 3. Blowing Up for Negative Initial Energy

To introduce and demonstrate our results, we first define our energy as follows:(21)Et≔12∫Ωut2dx+∫Ω1sx∇usxdx+12∫Ω∇ut2dx−μ∫Ω1pxupxdx.Theorem 9.
Assume that the assumptions ofProposition1 hold true and suppose that
(22)E0<0.Then, the solution of problemP blows up in finite time.To state and demonstrate our result, In order to prove our result, we give the following LemmaLemma 10.
Suppose the conditions of Lemma3 hold. Then, we have
(23)ρp.r/p1u≤C∇us.s1+ρp.u,s1≤r≤p1,for any u∈W01,sΩ, where C>1 is a positive constant which depends on Ω only.Corollary 11.
Let the assumptions of Lemma10 hold. Then, for any u∈W01,s, we get
(24)up1r≤C∇us.s1+up1p1,where s1≤r≤p1 and C are a positive constant.Now, we set(25)Ht≔−Etand use, throughout this paper, C to denote a generic positive constant depending on Ω only. From the result of (21) and (23), we give the following CorollaryCorollary 12.
Let the assumptions of Lemma10 hold. Then, we have
(26)ρp.r/p1u≤CHt+ut22+∇ut22+ρp.u,for any u∈u∈W01,s and s1≤r≤p1.Corollary 13.
Let the assumptions of Lemma10 hold. Then, we have
(27)up1r≤CHt+ut22+∇ut22+up1p1,for any u∈W01,s and s1≤r≤p1.Lemma 14.
Assume that (2) and (4) hold and E0<0. Then, the solution of P satisfies, for some c>0,
(28)ρp.u≥cup1p1.Lemma 15.
Letu be the solution of problem P and assume that (2) holds. Then,
(29)∫Ωuqxdx≤Cρp.uq1/p1+ρp.uq2/p1.Remark 16.
We can achieve the proof of the previous Lemmas and Corollaries as in the paper of Messaoudi and Talahmeh [39].Lemma 17.
Letu be the solution of P. Then, there exists a constant c1>0 such that
(30)∇u.,ts.≥c1,∀t≥0.Proof.
Assume, by contradiction, there exists a sequencetj such that
(31)∇u.,tjs.⟶0asj⟶∞.
Then, Lemmas3 and 6 give us
(32)ρp.u.,tj⟶0asj⟶∞.
This yields(33)limj⟶∞Etj≥0,that contrasts with the fact that Et<0,∀t≥0.Proof.
Proof of Theorem9.
As usual, multiplying byut and integrating over Ω in P to get
(34)E′t=−η∫Ωutx,tqxdx≤0,for almost every t in 0,T since Et is absolutely continuous; hence, H′t≥0 and
(35)0<H0≤Ht≤μp1ρp.u,for every t in 0,T, by remembering the condition that E0<0. We then introduce
(36)Lt≔H1−αt+ε∫Ωuutx,tdx,for ε small to be chosen later and
(37)0<α≤minp1−22p1,p1−q2p1q2−1.By taking the derivative of (35) and using (1), we obtain
(38)L′t=1−αH−αtH′t+ε∫Ωut2x,tdx+ε∫Ωuutt2x,tdx,so
(39)L′t+ddtε∫Ω∇ut∇u=1−αH−αtH′t+ε∫Ωut2−∇usx+∇ut2+εμ∫Ωupx−ηε∫Ωuututqx−2.Adding and subtracting the termε1−ξp1Ht, for 0<ξ<1, from the right side of (37), we get
(40)L′t+ddtε∫Ω∇ut∇u≥1−αH−αtH′t+ε1−ξp1Ht+εμξ∫Ωupx+ε1−ξp12+1ut22+ε1−ξp1s2−1∫Ω∇usxε1−ξp12+1∫Ω∇ut2−ηε∫Ωuututqx−2dx.So, forξ small enough, we obtain
(41)L′t+ddtε∫Ω∇ut∇u≥εβHt+ut22+∇ut22+ρs.∇u+ρp.u+1−αH−αtH′t−ηε∫Ωuututqx−2dx.where
(42)β=min1−ξp1,μξ,1−ξp12+1,1−ξp1s2−1>0.By using Young’s inequality, the last term in (40) yields
(43)∫Ωutqx−1udx≤1q1∫Ωδqxuqx+q2−1q2∫Ωδ−qx/qx−1utqxdx∀δ>0.Thus, by pickingδ such that
(44)δ−qx/qx−1=kH−αt,for a large constant k to be given later, and replacing in (41), we reach to
(45)∫Ωutqx−1udx≤1q1∫Ωk1−qxuqxHαqx−1t+q2−1q2kH−αtH′t∀δ>0.Combining (40) and (43) yields
(46)L′t+ddtε∫Ω∇ut∇u≥εβHt+ut22+∇ut22+ρs.∇u+ρp.u+1−α−εq2−1q2kH−αtH′t−ηεk1−q1q1C1Hαq2−1t∫Ωuqxdx.Exploiting Lemma15 and (34) to get
(47)Hαq2−1t∫Ωuqxdx≤Cρuq1/p1+αq2−1+ρuq2/p1+αq2−1.Now, we employ Lemma10 and (36), and we get
(48)r=q2+αp1q2−1≤p1andr=q1+αp1q2−1≤p1,And it is easy to see from (46) that
(49)Hαq2−1t∫Ωuqxdx≤C∇us.s1+ρp.u.So, by using Lemmas 6 and 17, we obtain
(50)ρs.∇u≥c2∇us.s1.Collecting of (45), (47), and (49), we get
(51)L′t+ddtε∫Ω∇ut∇u≥1−α−q2−1q2εkH−αtH′t+εβ−ηk1−q1q1CHt+ut22+∇ut22+∇us.s1+ρp.u.In this step, we choosek so large that the coefficient
(52)γ=β−ηk1−q1q1C>0.Oncek is fixed (thus γ), we put sufficiently small ε so that
(53)1−α−q2−1q2εk≥0andL0=H1−α0+ε∫Ωu0u1xdx>0.Subsequently, (50) becomes
(54)L′t+ddtε∫Ω∇ut∇u≥εγHt+ut22+∇ut22+∇us.s1+ρp.u≥εγHt+ut22+up1p1,by virtue of (28). Therefore, we get
(55)Lt≥L0>0,forallt≥0.Next, we are in the position to obtain an inequality of the form(56)L′t+ddtε∫Ω∇ut∇u≥ΓL1/1−αt,forallt≥0.Here, Γ is a positive constant that depends on εγ, C (the constant of Corollary1).To achieve (54), we estimate the term∫Ωuutx,tdx≤u2ut2≤Cup1ut2.(32)Hence,
(57)∫Ωuutx,tdx1/1−α≤Cup11/1−αut21/1−α.From Young’s inequality that yields the following estimate(58)∫Ωuutx,tdx1/1−α≤Cup1ω/1−α+ut2χ/1−α,where 1/ω+1/χ=1. Putting χ=2/1−α, we find ω/1−α=2/1−2α≤p1 by (37). Thus, (56) becomes
(59)∫Ωuutx,tdx1/1−α≤Cup1r+ut22,with r=2/1−2α≤p1. We obtain after using Corollary3(60)∫Ωuutx,tdx1/1−α≤CHt+ut22+∇ut22+up1p1,forallt≥0.In the end, by noting that(61)L1/1−αt=H1−αt+ε∫Ωuutx,tdx1/1−α≤21/1−αHt+∫Ωuutx,tdx1/1−α,and combining it with (51) and (58)), the inequality (54) is achieved.The proof is completed.


## 4. Blowing Up for Positive Initial Energy

Now, we are in the position to present and prove one of the main results of this section which is the blow-up for certain solutions with positive energy. For this goal, letA be the best constant of the Sobolev embedding W01,sΩ°LpΩ and set
(62)A1=max1,A,1μ1/s2,α1=1μA1p1s2/p1−s2,(63)α0=∇u0s.s2,E1=1s2−1p1α1,(64)Ht=E1−Et,(65)Kt=H1−λt+ε∫Ωuutx,tdx,for 0<λ<1,ε>0 that are to be specified later.We state here the following theorem which will be our main result.Theorem 18.
Assume that the conditions of Proposition7 hold true and suppose that
(66)E0<E1,α1<α0≤A1−s2.Then, the solution ofP blows up in a finite time.To demonstrate our theorem, we refer the following twolemmas.Lemma 19.
Let the assumptions in Theorem18 be fulfilled, and then there exists a constant α2>α1 such that
(67)∇u.,ts.s2≥α2,∀t≥0.Proof.
. Exploiting (21), we get
(68)Et≥1s2ρs.∇u−μp1ρp.u≥1s2min∇us.s1,∇us.s2−μp1maxup.p1,up.p2≥1s2min∇us.s1,∇us.s2−μp1maxA1∇us.p1,A1∇us.p2=1s2minαs1/s2,α−μp1maxA1s2αp1/s2,A1s2αp2/s2≔hα,∀α∈0,∞,where α=∇us.s2. Let
(69)gα=1s2α−μp1A1s2αp1/r2.By noting thatgα=hα, for 0<α≤A1s2, we can easily verify that the function gα is increasing for 0<α<α1 and decreasing for α1<α≤+∞.BecauseE0<E1=gα1, there exists a positive constant α2∈α1,∞ such that gα2=E0. So, we get gα0=hα0≤E0=gα2. This means that α0≥α2.To demonstrate (67), we suppose that ∇ut0s.s2<α2, for some t0>0. Then, there exists t1>0 such that α1<∇ut1s.s2<α2. Exploiting the monotonicity of gα to find
(70)Et1≥g∇ut1s.s2>gα2=E0,which contradicts Et<E0, for all t∈0,T. Consequently, (67) is determined.Lemma 20.
Let the assumptions in Theorem18 be fulfilled, and so we have
(71)0<H0≤Ht≤μp1ρp.u.Proof.
Exploiting (21), (30), and (64) to get
(72)0<H0≤Ht≤E1−12∫Ωut2dx+∫Ω1sx∇usxdx+12∫Ω∇ut2dx+μ∫Ω1pxupxdx,then from (67), we find
(73)E1−12∫Ωut2dx+∫Ω1sx∇usxdx+12∫Ω∇ut2dx≤E1−∫Ω1s2∇usxdx≤E1−1s2min∇us.s1,∇us.s2≤E1−1s2minα2s1/s2,α2≤E1−1s2minα1s1/s2,α1=E1−1s2α1=−α1p1<0,∀t≥0.Therefore,(74)0<H0≤Ht≤μp1ρp.u.∀t≥0.Proof of Theorem 2. It is not hard to determine the proof precisely by repeating the same steps (35) to (58) of the proof of Theorem 1, with the use of Lemma 20.



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