# Lifespan Estimates of Solutions for a Coupled System of Wave Equations with Damping terms and Negative Mass terms∗

Authors: Jie Yang; Sen Ming; Wei Han; Jiayi Du
Journal: Discrete Dynamics in Nature and Society
 (2022)

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## Abstract

The main purpose of this paper is to study the formation of singularity for a coupled system of wave equations with damping terms, negative mass terms, and divergence form nonlinearities. Upper bound lifespan estimates of solutions to the system are obtained by using the iteration method. The results are the same as the corresponding coupled system of the wave equation with power nonlinearitiesvpand uq. To the best of our knowledge, the results in Theorems 1–5 are new. In addition, the variation trend of the wave is analyzed by using numerical simulation.


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## Body

## 1. Introduction

In this paper, we mainly consider the coupled system of wave equations with damping terms and negative mass terms(1)utt−Δu+b1tut−b2tu=Gv,x∈ℝn,t>0,vtt−Δv+b3tvt−b4tv=Gu,x∈ℝn,t>0,u,ut,v,vtx,0=εu0,εu1,εv0,εv1,x∈ℝn,where(2)Gv=vp+∑i=1nbivp−2vvxi+∑i,j=1nbijvp−2vvxixj,Gu=uq+∑i=1nbiuq−2uuxi+∑i,j=1nbijuq−2uuxixj,Here, Δ=∑i=1n∂2/∂xi2 is the Laplace operator. b1t,b2t,b3t,b4t∈C0,T∩L10,T are non-negative functions. ε is a small positive parameter. Initial values u0,u1,v0,v1 are non-negative functions. It holds that suppu0,u1,v0,v1⊂BR0, where BR0=x|x≤R and R>2. The exponents in the nonlinear terms satisfy p,q>1. It is well known that the classical wave equation(3)utt−Δu=up.is related to the Strauss conjecture. The Strauss exponentpSn is the positive solution of the following quadratic equation:(4)−n−1p2+n+1p+2=0.More precisely, if1<p≤pSn, there is no global solution to the Cauchy problem for (3) with small initial data. The solution exists globally when p>pSn, which is equivalent to(5)1+p−1p−1≥n−12.Let us briefly review the small data Cauchy problem for the wave equation with scattering damping term(6)utt−Δu+μ1+tβut=0,x∈ℝn,t>0,ux,0=εfx,utx,0=εgx,x∈ℝn.which has attracted more attention (see [1–14]). Wirth et al. [13] obtained the energy solution to the Cauchy problem. Wirth et al. [14] proved a structural property of solution to the Cauchy problem.(7)utt−Δu+μ1+tβut=fu,ut,x∈ℝn,t>0,ux,0=εfx,utx,0=εgx,x∈ℝn.Lai and Takamura [4] obtained an upper bound lifespan estimate of the solution to the problem with a nonlinear term fu,ut=up in the subcritical case by introducing the multiplier and using the iteration method. Wakasa and Yordanov [15] proved the lifespan estimate of the solution for the variable coefficient wave equation with scattering damping in the critical case by using the rescaled test function method and iterative technique. Utilizing some multiplier to absorb the damping term, Lai and Takamura [16] successfully established an upper bound lifespan estimate of the solution, which is the same as the case without a damping term. Lai and Takamura [17] studied the problem with combined nonlinearities fu,ut=utp+uq. It is shown that the scattering damping has no effect on the behavior of the solution compared with the solution to wave equation.Let us briefly review the blow-up dynamics of the coupled system for semilinear classical wave equations.(8)utt−Δu=vp,x∈ℝn,t>0,vtt−Δv=uq,x∈ℝn,t>0,u,ut,v,vtx,0=εu0,εu1,εv0,εv1,x∈ℝn.Which has been studied in ([18–29]). Santo et al. [28] proved that there exists a critical curve in the p,q plane, which divides the plane into two pieces when n≥2. One piece is the range which shows the global existence of small amplitude solutions. Another piece is the range of nonexistence of global solutions. We observe that the critical curve is determined by a cubic relation between p and q, which possesses a cusp at p=q. Agemi et al. [18] obtained lifespan estimates for solutions when n=3 and p,q>2. Moreover, lower and upper bounds lifespan estimates for solutions are derived in the subcritical case. Santo and Mitidieri [29] investigated the lifespan estimate of the solution to the problem when p=q. Georgiev et al. [21] studied the formation of singularities solutions to the problem in the case n≥4. Kurokawa and Takamura [25] acquired the upper bound lifespan estimates of solutions to coupled semilinear wave equations in the high dimensional critical case by making use of the iteration method.(9)utt−Δu+b1tut=fv,vt,x∈ℝn,t>0,vtt−Δv+b1tvt=fu,ut,x∈ℝn,t>0,u,ut,v,vtx,0=εu0,εu1,εv0,εv1,x∈ℝn.Recently, the small data Cauchy problem for coupled wave equations with damping terms have been investigated (see [30–41]). Takamura and Palmieri [37] considered lifespan estimates of solutions to the problem with scattering damps term and power nonlinear terms fv,vt=vp, fu,ut=uq in the subcritical and critical cases. The critical curve of exponential pair p,q in nonlinear terms is the same as that of coupled semilinear wave equations without damping terms. Palmieri and Takamura [36] studied the blow-up of solutions to the semilinear damped wave equations with mixed nonlinear terms fv,vt=vp,fu,ut=utq. Lifespan estimates of solutions in the subcritical case are obtained by employing the iterative method. The formation of singularities of solutions in the critical case is established by making use of the auxiliary functions, which are different from the test functions in [32]. Palmieri and Takamura [38] investigated upper bound lifespan estimates of solutions for semilinear damped wave equations with derivative type nonlinearities in the subcritical and critical cases.Inspired by the works in [10, 37, 42–45], we mainly considered blow-up dynamics of solutions to the coupled wave equations with divergence form nonlinearities in the subcritical and critical cases. It is worth noting that Han [44] studied the problem with divergence form nonlinearities when n=4. We extended the problems studied in [44] to the coupled system of the semilinear wave equation in n space dimensions. Palmieri and Takamura [37] studied upper bound lifespan estimates of solutions to the coupled system of wave equations. Blow-up results of solutions in the subcritical and critical cases are obtained. Ming et al. [10] established upper bound lifespan estimates of the solution to the variable coefficient wave equation with scattering damping and divergence nonlinear terms in the case n=4. We extended the problem studied in [10, 37, 44] to problem (1) with damping terms, negative mass terms, and divergence form nonlinear terms. We used the iterative method and slicing method to derive upper bound lifespan estimates of solutions in the subcritical and critical cases, respectively. To the best of our knowledge, the results in Theorems 1–5 are new conclusions. In addition, we describe the variation of waves through numerical simulation.We present the definition of weak solutions as follows:Definition 1.
Letu0,v0∈H1ℝn and u1,v1∈L2ℝn. Then, u,v are called weak solutions to the problem (1) on 0,T if(10)u∈C0,T,H1ℝn∩C10,T,L2ℝn∩Llocq0,T×ℝn,v∈C0,T,H1ℝn∩C10,T,L2ℝn∩Llocp0,T×ℝn.
It holds that(11)∫ℝnutx,tϕx,tdx−∫ℝnεu1xϕx,0dx−∫0t∫ℝnutx,sϕtx,sdxds+∫0t∫ℝn∇ux,s∇ϕx,sdxds+∫0t∫ℝnb1sutx,sϕx,sdxds−∫0t∫ℝnb2sux,sϕx,sdxds=∫0t∫ℝnGvϕx,sdxds,and(12)∫ℝnvtx,tψx,tdx−∫ℝnεv1xψx,0dx−∫0t∫ℝnvtx,sψtx,sdxds+∫0t∫ℝn∇vx,s∇ψx,sdxds+∫0t∫ℝnb3svtx,sψx,sdxds−∫0t∫ℝnb4svx,sψx,sdxds=∫0t∫ℝnGuψx,sdxds.
for allϕ,ψ∈C0∞0,T×ℝn and t∈0,T.
Employing integration by parts in (11)–(12) and letting t⟶T, we get(13)∫ℝnutx,tϕx,t−ux,tϕtx,t+b1tux,tϕx,tdx−∫ℝnεu1xϕx,0−εu0xϕtx,0+b10εu0xϕx,0dx+∫0T∫ℝnux,sϕttx,s−Δϕx,sb1sϕx,ss−b2sϕx,sdxds=∫0T∫ℝnGvϕx,sdxds.and(14)∫ℝnvtx,tψx,t−vx,tψtx,t+b3tvx,tψx,tdx−∫ℝnεv1xψx,0−εv0xψtx,0+b30εv0xψx,0dx+∫0T∫ℝnvx,sψssx,s−Δψx,s−b3sψx,ss−b4sψx,sdxds=∫0T∫ℝnGuψx,sdxds.
Throughout the paper, we denote(15)Fn,p,q=p+2+q−1pq−1−n−12,Gn,p,q=21+p−1pq−1−np+n−2.
The main results of this work are stated as follows:Theorem 1.
Assume that1−1/p∑i=1nbiλ−1/p∑i,j=1nbijλ2>0,1−1/q∑i=1nbiλ−1/q∑i,j=1nbijλ2>0, and(16)maxp+2+q−1pq−1,q+2+p−1pq−1>n−12.
Suppose that problem (1) has solutions u,v∈C0,T,H1ℝn∩C10,T,L2ℝn2, which satisfy suppu,v⊂x,t‖x|≤t+R. Then, the upper bound estimate for the lifespan satisfies(17)Tε≤Cε−maxFn,p,q,Fn,q,p−1,where, C is a positive constant independent of ε.Theorem 2.
Letn=1, p,q>1 or n=2, 1<p,q<2, 1−1/p∑i=1nbiλ−1/p∑i,j=1nbiλ2>0,and 1−1/q∑i=1nbiλ−1/q∑i,j=1nbiλ2>0. Assume(18)∫ℝnu1xdx≠0,∫ℝnv1xdx≠0.
Suppose that problem (1) has solutions u,v∈C0,T,H1ℝn∩C10,T,L2ℝn2, which satisfy suppu,v⊂x,t‖x|≤t+R. Then, the lifespan estimates in (17) can be improved as(19)Tε≤Cε−maxGn,p,q,Gn,q,p−1.Theorem 3.
Letn=2, 1<p<2, q≥2, 1−1/p∑i=1nbiλ−1/p∑i.j=1nbiλ2>0, and 1−1/q∑i=1nbiλ−1/q∑i.j=1nbiλ2>0. Assume(20)∫ℝ2u1xdx≠0.
Suppose that problem (1) has solutions u,v∈C0,T,H1ℝn∩C10,T,L2ℝn2, which satisfy suppu,v⊂x,t‖x|≤t+R. Then, the lifespan estimates in (17) can be improved as(21)Tε≤Cε−maxFn,p,q,Gn,q,p−1.Theorem 4.
Letn=2, p≥2, 1<q<2, 1−1/p∑i=1nbiλ−1/p∑i,j=1nbiλ2>0, and 1−1/q∑i=1nbiλ−1/q∑i,j=1nbiλ2>0. Assume(22)∫ℝ2v1xdx≠0.
Suppose that problem (1) has solutions u,v∈C0,T,H1ℝn∩C10,T,L2ℝn2, which satisfy suppu,v⊂x,t‖x|≤t+R. Then, the lifespan estimates in (17) can be improved as(23)Tε≤Cε−maxFn,q,p,Gn,q,p−1.Theorem 5.
Letn≥2,1−1/p∑i=1nbiλ−1/p∑i,j=1nbiλ2>0,1−1/q∑i=1nbiλ−1/q∑i,j=1nbiλ2>0 and(24)maxp+2+q−1pq−1,q+2+p−1pq−1=n−12.
Suppose that problem (1) has solutions u,v∈C0,T,H1ℝn∩C10,T,L2ℝn2, which satisfy suppu,v⊂x,t|x≤t+R. Then, the upper bound estimates for the lifespan satisfies(25)Tε≤expCε−minqpq−1,ppq−1,p≠q,expCεpp−1,p=q.


## 2. Proof of Theorem1

Let(26)φx=ex+e−x,n=1,∫Sn−1eω⋅xdSω,n≥2,where ψ1=ψ1x,t=e−tφx and φx∼x−n−1/2 as x⟶∞. ∂xiφλx≤λφλx, ∂xi∂xjφλx≤λ2φλx, and ∑i,j=1n∂xi∂xjφλx=λ2φλx.We set(27)U1t=∫ℝnux,tψ1x,tdx,(28)V1t=∫ℝnvx,tψ1x,tdx.We are in the position to establish the lower bound for the spatial integral of nonlinear termsvp,uq. It holds that(29)ddt∫ℝnutx,tϕx,tdx+∫ℝn−utx,tϕtx,t−ux,tΔϕx,tdx+∫ℝnb1tutx,tϕx,tdx−∫ℝnb2tux,tϕx,tdx=∫ℝnGvϕx,tdx.Multiplying both sides of (29) with mt yields(30)ddtmt∫ℝnutx,tϕx,tdx+mt∫ℝn−utx,tϕtx,t−ux,tΔϕx,tdx=mt∫ℝnb2tux,tϕx,tdx+mt∫ℝnGvϕx,tdx.Then, we have(31)mt∫ℝnutx,tϕx,tdx−m0ε∫ℝnu1xϕxdx−∫0tds∫ℝnmsutx,sϕtx,sds−∫0tds∫ℝnmsux,sΔϕx,sdx=∫0tds∫ℝnmsb2sux,sϕx,sdx+∫0tds∫ℝnmsGvϕx,sdx.It follows that(32)mt∫ℝnutx,tϕx,tdx−m0ε∫ℝnu1xϕx,0dx−mt∫ℝnux,tϕtx,tdx+m0ε∫ℝnu0xϕtx,0dx+∫0tds∫ℝnmsb1sux,sϕtx,sdx+∫0tds∫ℝnmsux,sϕttx,sdx−∫0tds∫ℝnmsux,sΔϕx,sdx=∫0tds∫ℝnmsb2sux,sϕx,sdx+∫0tds∫ℝnmsGvϕx,sdx.Setting(33)ϕx,t=ψ1x,t=e−tφx.we have(34)ϕt=−ϕ,ϕtt=Δϕ.Direct calculation gives rise to(35)U1′t+2U1t≥m0Cu0,u1ε+∫0tmsb1s+b2sU1sds,where, Cu0,u1=∫ℝnu0+u1ϕxdx>0.Multiplying (35) with e2t, we obtain(36)e2tU1t≥U10+m0Cu0,u1ε∫0te2sds+∫0te2sds∫0smrb1r+b2rU1rdr.Applying a comparison argument, we haveU1t>0 for all t>0. Similar to the calculation in [46], we arrive at(37)e2tU1t≥m0Cu0,u1ε2e2t−1.Finally, we derive the lower bound ofU1t in the form(38)U1t≥1−e−2t2m0Cu0,u1ε.Similarly, we have(39)V1t≥1−e−2t2m0Cv0,v1ε.By using (27) and (28) and Holder inequality, we have(40)∫ℝnux,tqdx≥U1tq∫x|t+Rψ1q′x,tdx−q−1,(41)∫ℝnvx,tpdx≥V1tp∫x≤t+Rψ1p′x,tdx−p−1.Direct calculation shows(42)∫x≤t+Rψ1q′x,tdx≤Cφ,R1+tn−1−n−1q/2q−1.From (38), (40), and (42), we acquire(43)∫ℝnux,tqdx≥C0εq1+tn−1−n−1q/2.Applying (39), (41), and (42) yields(44)∫ℝnvx,tpdx≥K0εp1+tn−1−n−1p/2.Choosingϕ=ϕx,s=1 and ψ=ψx,s=1 in (11) and (12), we find(45)∫0t∫ℝnGvdxds=∫ℝnutx,tdx−∫ℝnεu1x,0dx+∫0t∫ℝnb1sutx,sdxds−∫0t∫ℝnb2sux,sdxds,∫0t∫ℝnGudxds=∫ℝnvtx,tdx−∫ℝnεv1x,0dx+∫0t∫ℝnb3svtx,sdxds−∫0t∫ℝnb4svx,sdxds.Equivalently, we obtain(46)U′t−U′0+∫0tb1sU′sds−∫0tb2sUsds=∫0t∫ℝnGvdxds,(47)V′t−V′0+∫0tb3sV′sds−∫0tb4sVsds=∫0t∫ℝnGudxds.Differentiating (46) and (47) with respect to t, we arrive at(48)U″t+b1tU′t−b2tUt=∫ℝnGvdx,V″t+b3tV′t−b4tVt=∫ℝnGudx.Multiplying (48) by m1t gives rise to(49)m1tU″t+m1′tU′t=ddtm1tU′t=m1t∫ℝnGvdx+m1tb2tUt.Using the assumptionu1≥0, we obtain(50)U′t=1m1t∫0tm1sds∫ℝnGvdx+1m1t∫0tm1sb2sUsds+m10m1sU′0.Similar to Section3 in [46], we acquire(51)U′t≥m10∫0t∫ℝnGvdxds,where(52)Gv=vp+∑i=1nbivp−2vvxi+∑i,j=1nbijvp−2vvxixj.Employing integration by parts yields(53)∫ℝn∑i=1nbivp−2vvxidx=−∫ℝn1p∑i=1nbi1vpdx=0,∫ℝn∑i,j=1nbijvp−2vvxixjdx=∫ℝn1p∑i,j=1nbij∂xi∂xj1vpdx=0.Substituting (52) into (51), we have(54)U′t≥m10∫0t∫ℝnvx,spdxds.Integrating (54) over the interval 0,t leads to(55)Ut≥m10∫0t∫0τ∫ℝnvx,spdxdsdτ.Utilizing the Holder inequality yields(56)Ut≥C1∫0t∫0τ1+s−np−1Vspdsdτ,where, C1=m10measB11−pR−np−1>0. Similarly, we derive(57)Vt≥m20∫0t∫0τ∫ℝnux,sqdxdsdτ≥K1∫0t∫0τ1+s−nq−1Usqdsdτ,where, K1=m10measB11−qR−nq−1>0.Assume that(58)Ut≥Dj1+t−ajtbjfort≥0,(59)Vt≥Δj1+t−αjtβjfort≥0,where, ajj∈ℕ,bjj∈ℕ,Djj∈ℕ,αjj∈ℕ,βjj∈ℕ,Δjj∈ℕ are suitable sequences of non-negative real numbers to be determined afterwards.Plugging (44) into (55), we obtain(60)Ut≥m10K0εp∫0t∫0τ1+sn−1−n−1p/2dsdτ≥m10K0εp1+t−n−1p/2∫0t∫0τsn−1dsdτ≥m10K0nn+1εp1+t−n−1p/2tn+1,where(61)D1=m101K0nn+1εp,a1=n−1p2,b1=n+1.Substituting (61) into (63), we arrive at(62)Ut≥C1Δjp∫0t∫0τ1+s−np−1−αjpsβjpdsdτ≥C1Δjp1+t−np−1−αjp∫0t∫0τsβjpdsdτ=C1Δjpβjp+1βjp+21+t−np−1−αjptβjp+2,where(63)Dj+1=C1Δjpβjp+1βjp+2,aj+1=np−1+αjp,bj+1=βjp+2.Taking into account (61) and (62), for an odd integer j, we have(64)aj=n+n−12ppqj−1/2−n,αj=n+n−12qpqj−1/2−n,bj=n+1+2p+1pq−1pqj−1/2−2p+1pq−1,βj=n+1+2q+1pq−1pqj−1/2−2q+1pq−1.Ifj is even, from (58), we obtain(65)bj=pn+1+2q+1pq−1pqj−2/2−2pq+1pq−1+2,(66)βj=qn+1+2p+1pq−1pqj−2/2−2qp+1pq−1+2.Thus, from (65) and (66), we recognize that for all j≥1, it holds that(67)bj<B0pqj−1/2,βj<B0˜pqj−1/2,for oddj,(68)bj<B0pqj/2,βj<B0˜pqj/2,for evenj,where, B0=B0p,q,n and B0˜=B0˜p,q,n are positive constants.The next step is to derive lower bounds forDj and Δj. From the definition of Dj and Δj, we obtain(69)Dj≥C1bj2Δj−1p,Δj≥K1βj2Dj−1q.Due to (67)–(69), we acquire(70)Dj≥C1B02Δj−1ppqj−1≥C1K1pB02Dj−2pqpqj−1βj−12p≥C˜Dj−2pqpqp+1j−1,(71)Δj≥K1B˜02Dj−1qpqj−1≥K1C1qB˜02Δj−2pqpqj−1bj−12q≥K˜Δj−2pqpqq+1j−1,where, C˜=C1K1p/B02B˜02pand K˜=K1C1q/B˜02B02p. Making use of (70), if j is odd, we have(72)logDj≥pqj−1/2logD1−∑k=1j−1/2j+1−2kpqk−1p+1logpq+∑k=0j−3/2pqklogC˜.Employing an inductive argument yields(73)∑k=1j−1/2j+1−2kpqk−1=1pq−12pqpqj−1/2−1pq−1−j+1.Similarly, we have(74)logDj≥pqj−1/2logD1−2pqp+1pq−12logpq+logC˜pq−1+2pqp+1pq−12logpq+j−1p+1pq−1logpq−logC˜pq−1.Consequently, for an oddj satisfying j>logC˜/p+1logpq−2pq/pq−1+1, we derive(75)logDj≥pqj−1/2logD1−Sp,q∞,where, Sp,q∞=2pqp+1/pq−12logpq−logC˜/pq−1.We arrive at(76)Ut≥exppqj−1/2logD1−n+n−12plog1+t+n+1+2p+1pq−1logt−Sp,q∞1+tnt−2p+1/pq−1.Fort≥1, applying the inequality log2t≥log1+t, we derive(77)Ut≥exppqj−1/2Jt1+tnt−2p+1/pq−1,where(78)Jt=logD1tpq+2p+1/pq−1−n−1p/2−n+n−12plog2−Sp,q∞.IfFn,q,p>0, there exists ε0=ε0u0,u1,v0,v1,n,p,q,b1,b2,R>0 such that(79)C¯ε0−Fn,q,p−1≥1,where, C¯=nn+1/m10K02n+n−1/pexpSp,q∞1/pFn,q,p. Analogously, in the case Fn,p,q>0, we set(80)K¯ε0−Fn,p,q−1≥1,where, K¯=nn+1/m20C02n+n−1/qexpSp,q∞1/qFn,q,p. For all ε∈0,ε0 and t>K˜ε−Fn,p,q−1, we deduce t≥1 and J˜t>0. The proof of Theorem 1 is finished. ■


## 3. Proofs of Theorems2–4

Bear in mindU′0=ε∫ℝnu1xdx>0 and V′0=ε∫ℝnv1xdx>0. We note that U and V are convex functions. Therefore, we deduce Ut≥U′0t and Vt≥V′0t. Consequently, using the Holder inequality yields(81)∫ℝnut,xqdx≥C1˜1+t−nq−1Utq≥C1˜εq∫ℝnu1xdxq1+t−nq−1tq,∫ℝnvt,xpdx≥K1˜1+t−np−1Vtp≥K1˜εp∫ℝnv1xdxp1+t−np−1tp.Ifn=1,2and 1<p<2, we define(82)Δ1=C1˜∫ℝnu1xdxqεq,α1=q,β1=n−1q.Ifn=1,2and 1<q<2, we set(83)D1=K1˜∫ℝnu1xdxpεp,a1=p,b1=n−1p.We replace (78) by(84)Jt=logD1tpGn,p,q−n+plog2−Sp,q∞.Analogously, we acquire(85)J˜t=logΔ1tqGn,q,p−n+qlog2−Sp,q∞.The proofs of Theorems2–4 are finished. ■


## 4. Proof of Theorem5

We are in the position to recall a pair of auxiliary functions from [47]. Let λ0∈0,α/2 and r>−1. We introduce the functions(86)ξrx,t=∫0λ0e−λt+Rcoshλtφλxλrdλ,ηrx,t,s=∫0λ0e−λt+Rsinhλt−sλt−sφλxλrdλ,where, λ0 is a fixed positive parameter and φλxis defined in (26).Lemma 1.
Letn≥2. There exists a constant λ0>0 such that the following properties hold:(i)
Ifr>−1,x≤R, andt≥0, then(87)ξrt,x≥A0,ηrt,0,x≥B0t−1.(ii)
Ifr>−1,x≤s+R, andt>s≥0, then(88)ηrt,s,x≥B1t−1s−r.(iii)
Ifr>n−3/2,x≤t+R, andt>0, then(89)ηrt,t,x≤B2t−n−1/2t−xn−3/2−r,Here, A0 and Bk k=0,1,2 are positive constants depending only on λ0, r, and R, y=3+y.Lemma 2.
Letλ be a positive real parameter. bt∈C10,T∩L10,T is a non-negative function. Assume(90)Lb=∂t2+bt∂t−λ2,Lb∗=∂s2−∂sbs−λ2.
The solutionsyj=yjt,s;λ (j=1,2) to the fundamental system satisfy(91)Lby1t,s;λ=0,y1s,s;λ=1,∂ty1s,s;λ=0,Lby2t,s;λ=0,y2s,s;λ=0,∂ty1s,s;λ=1.
It holds that(92)iy1t,s;λ≥e−bL1coshλt−s,iiy2t,s;λ≥e−2bL1sinhλt−sλ,iiiLb∗y2t,s;λ=0,ivy1t,0;λ=b0y2t,0;λ−∂sy2t,0;λ,v∂sy2t,t;λ=−1.
We set(93)Ut=∫ℝnux,tηr1t,t,xdx,Vt=∫ℝnvx,tηr2t,t,xdx.
By choosingϕ=ϕs,x=φλxy2t,s;λ and ψ=ψs,x=φλxy2t,s;λ in (13) and (14), we have(94)b10ϕx,0−ϕsx,0=φλy1t,0,λ,b10ψx,0−ψsx,0=φλy1t,0,λ,ϕssx,s−Δϕx,s−b1tϕx,ss−b2ϕx,s=0,ψssx,s−Δψx,s−b3tψx,ss−b4ψx,s=0.
From Lemma2, we acquire(95)∫ℝnux,tφλxdx=εy1t,0;λ∫ℝnu0xφλxdx+εy2t,0;λ∫ℝnu1xφλxdx+∫0ty2t,s;λ∫ℝnGvφλxdx,where(96)∫ℝnGvφλxdx=∫ℝnvp+∑i=1nbivp−2vvxi+∑i,j=1nbijvp−2vvxixjφλxdx.
Taking advantage of (26), we deduce(97)∫ℝnφλx∑i=1nbivp−2vvxidx=∫ℝnφλx∑i=1nbivppxidx=−∫ℝn1p∑i=1nbi∂xiφλxvpdx≤1p∑i=1n|bi‖∂xiφλx|vpdx≤1p∑i=1nbiλφλxvpdx.
It follows from the Gauss formula that:(98)∫ℝnφλx∑i,j=1nbijvp−2vvxixjdx=−∫ℝn∑i,j=1nbij∂xjφλxvppxidx=∫ℝn1p∑i,j=1nbij∂xi∂xjφλxvpdx≤1p∑i,j=1nbijλ2φλxvpdx.
It is deduced from (97) and (98) that(99)∫ℝnGvφλxdx=∫ℝnvpφλx+1p∑i=1nbi∂xiφλxvp+1p∑i,j=1nbij∂xi∂xjφλxvpdx≥∫ℝnφλxvp1−1p∑i=1nbiλ−1p∑i,j=1nbijλ2dx≥C2∫ℝnvpφλxdx,where, C2 is a positive constant. We have(100)∫ℝnux,tφλxdx≥εe−b1L1coshλt∫ℝnu0xφλxdx+εe−2b1L1sinhλtλ∫ℝnu1xφλxdx+C2e−2b1L1∫0tsinhλt−sλds∫ℝnvx,spφλxdx.
Multiplying both sides of (102) by e−λt+Rλr1 and integrating with respect to λ over 0,λ0, we acquire(101)∫ℝnux,tηr1t,t,xdx≥e−2b1L1εt∫ℝnu1xηr1t,0,xdx+C2e−2b1L1∫0tt−s∫ℝnvx,spηr1t,s,xdxds+e−b1L1ε∫ℝnu0xξr1t,xdx.
Similar to (101), we get(102)∫ℝnvx,tηr2t,t,xdx≥e−2b2L1εt∫ℝnv1xηr2t,0,xdx+C2e−2b2L1∫0tt−s∫ℝnux,sqηr2t,s,xdxds+e−b2L1ε∫ℝnv0xξr2t,xdx.
We assume thatr1,r2 are given parameters satisfying r1=n−1/2−1/q and(103)r2>n−12−1p,ifp>q,r2=n−12−1p,ifp=q.
We are in the position to establish estimates forUt and Vt. If p>q, from (95), we have(104)Vs≤∫ℝnvx,spηr1t,s,xdx1/p∫Bs+Rηr2s,s,xp′ηr1t,s,xp′/pdx1/p′.
Marking the use of (ii) and (iii) in Lemma1 leads to(105)∫Bs+Rηr2s,s,xp′ηr1t,s,xp′/pdx≤tp′/ps−n−1p′/2+p′/pr1∫Bs+Rs−xn−3/2−r2p′dx≤tp′/ps−n−1p′+p′/2/pr1+n+n−3/2−r2p′.
Combining (101), (104), and (105), we find(106)Ut≥∫0tt−s∫ℝnvx,spηr1t,s,xdxds≥Ct−1∫0tt−ssn−1/2+1+1/q+r2+1−npVspds.
Similar to (106), we acquire(107)Vt≥∫0tt−s∫ℝnux,sqηr2t,s,xdxds=Kt−1∫0tt−ss−r2−n−1q/2+n−1logs−q−1Usqds.
Ifp=q, due to −n−1/2q−1+1/q=−1, we obtain(108)Ut≥Ct−1∫0tt−ss−1logs−p−1Vspds,and(109)Vt≥Kt−1∫0tt−ss−1logs−q−1Usqds.
We establish the lower bound estimates ofUt and Vt.Lemma 3.
LetFn,p,q=0. Then, for all t≥3/2, the following estimates hold:(110)Ut≥C˜εpqlog2t3,ifp>q,Ut≥C˜εplog2t3,ifp=q.Lemma4.
Ifp=q, utilzing (57) and (101), we derive(111)Ut≥∫0tt−s∫ℝnvx,spηr1t,s,xdxds≥t−1∫0tt−ss−r1∫ℝnvs,xpdxds≥εpt−1∫0tt−ssn−1/2/+1/q−n−1p/2ds.
Direct calculation shows(112)∫0tt−ss−1ds≥∫1tt−ssds=∫2t/3tlogsds≥tlog2t3≥tlog2t3.
Substituting (111) into (112) yields(113)Ut≥C˜εplog2t3.
Similarly, taking advantage of (42) and (104), we get(114)Vt≥εqt−1∫0tt−ss−r2+n−1−n−1q/2ds.
Ifp<q, applying (114), we find(115)Vt≥εqt−1−r2−n−1p/2∫t/2tt−ssn−1ds≥εqt−1−r2−n−1/2qt2n−1∫t/2tt−sds≥εqt−r2−n−1p/2.
Applying (106), we arrive at(116)Ut≥εpqlog2t3.
We finish the proof of Lemma3. ■
Assume that(117)Ut≥Cjlogt−bjlogtl2jaj,fort≥l2j,j∈ℕ,where, lj=2−2−j+1, ajj∈ℕ, and bjj∈ℕ,Cjj∈ℕ are sequences of non-negative real numbers. We have(118)C0=C˜εpq,ifp>q,C˜εp,ifp=q,a0=1,b0=0.
Ifp>q, combining (109) and (117), for all s≥l2j+1, we obtain(119)Vs≥Ks−1∫l2jss−ττ−r2−n−1p/2+n−1logτ−q−1Uτqdτ,≥KCjqlogs−q−1−bjqs−1−r2−n−1p/2∫l2jss−ττn−1logτl2jajqdτ,where(120)∫l2jss−ττn−1logτl2jajqdτ≥∫l2js/l2j+1ss−ττn−1logτl2jajqdτ≥2−4j−3n−8sn+1logsl2j+1ajq.Here, we have employed the condition 1−lk/lk+1≥2−k+3. Substituting (120) into (119) yields(121)Vs≥2−4j−3n−8KCjqlogs−q−1−bjqs−r2−n−1p/2+nlogsl2j+1ajq.
Similarly, we get(122)Ut≥2−2p+12j−3n+8p−8CKpCjpqajpq+1−1×logt−pq−1−bjpqlogtl2j+2ajpq+1.
Therefore, we set(123)Cj+1=2−2p+12j−3n+8p−8CKpCjpqajpq+1−1,aj+1=ajpq+1,bj+1=pq−1+bjpq.
Direct computation shows(124)aj=aj−1pq+1=pqj+1−1pq−1,bj=pq−1+bj−1pq=pq−1pq−1pqj−1.
It follows that:(125)aj−1pq+1=pqj+1−1pq−1≤pqpq−1pqj.
Therefore, we have(126)Cj≥MΘ−jCj−1pq,where, Θ=222p+1pq and M=2−3n+4p−6CKppq−1/pq.
Ifp=q, combining (109) and (117), we obtain(127)Vs≥Ks−1∫l2jss−ττ−1logτ−q−1Uτqdτ≥2−2KCjqlogs−q−1−bjqs−1∫l2jss−ττlogτl2jajqdτ≥2−2j+7KCjqajq+1−1logs−q−1−bjqlogsl2j+1ajq+1.
Applying (108) to (127), we have(128)Ut≥2−2j+7p−2CKpCjpqajq+1−plogt−pq−1−bjpqt−1×∫l2j+1tt−sslogsl2j+1ajpq+pds,where(129)∫l2j+1tt−sslogsl2j+1ajpq+pds≥2−2j+3ajpq+p+1−1tlogtl2j+2ajpq+p+1.
From (128) and (129), we arrive at(130)Ut≥2−p+12j−7p−8CKpCjpqajq+1−pajpq+p+1−1×logt−pq−1−bjpqlogtl2j+2ajpq+p+1.
Setting(131)Cj+1=2−p+12j−7p−8CKpCjpqajq+1−pajpq+p+1−1,aj+1=ajpq+p+1,bj+1=pq−1+bjpq,
we have(132)aj=1+p+1pq−1pqj−p+1pq−1,bj=pqj−1.
Employing (126), we have(133)Cj≥MΘ−jCj−1pq,where, Θ=22p+1pqp+1 and M=2−5p−6CKppq−1p+1/pq+1p+1pqp+1.
Applying (126) gives rise to(134)logCj≥pqjlogC0−SjlogΘ+logMpq−1−logMpq−1,where, Sj=∑k=1jk/pqk. It holds that Cj⟶S as j⟶∞. We obtain(135)Cj≥M−pq−1exppqjlogC0Θ−sMpq−1.
The proof of Theorem5 is finished. ■


### 4.1. Numerical Simulation

In this Section, we study numerical simulation of solutions to the coupled system of wave equations with power nonlinear terms. We consider the special dimensioned casen=2 in problem (1). That is(136)∂2u∂t2=∂2∂x2+∂2∂y2u+v2,∂2v∂t2=∂2∂x2+∂2∂y2u+u2,where, ux,y,t is the displacement of film in position x,y at the time t. The initial values satisfy(137)u|t=0=e−20x2+y+1.32+e−x2+y−0.52,∂u∂t|t=0=0,v|t=0=e−20x2+y+1.32+e−x2+y−0.52,∂v∂t|t=0=0.Discretizingu in the computational region, we obtain(138)uN×N=u11u12…u1Nu21u22…u2N0p3…0⋮⋮…⋮uN1uN2…uNN.We denote(139)∂nu∂yn⟶DNnuN×N,and(140)∂nu∂xn⟶DNnuN×NTT=uN×NDNnT.We obtain the trend of waves in Figure1.Figure 1
The variation trend of waves.Figure1 shows the trend of the two waves in the time 0 ∼4s. We define the wave with the initial position of x=0,y=0.5 as a large wave, and the wave with x=0,y=−1.3 as a small wave. When t=0s, it represents the initial state of the wave, in which the peak of the large wave is 0.9989, and the peak of the small wave is 0.9388. When t=0.25s, the peaks of these two waves both begin to decline. At this time, the peak of the large wave is 0.8792, and the peak of the small wave is −0.1137. Comparing the big wave with the small wave, it is obvious that the small wave has a faster-descending speed. The peak of the big wave is still above the horizontal plane, while the small wave has fallen below the horizontal plane within 0.25s. When t=0.5s, the peaks of the two waves continue to move downward. The peak of the large wave is 0.5753, and the peak of the small wave is −0.08573. When t=1s, the crests of these two waves both drop below the horizontal plane, the crests of the small wave begin to rise upwards as well. At the same time, these two waves begin to meet. The small wave begins to spread outward, and the spread of the big wave is affected. When t=2s, both waves are concave down at the same time. The wave continues to move downward when t=2∼4s. By observing the wave changes from 0∼4s, we can see that when the two waves move at the same time, the changing trend of the wavelet is obvious.


## 4.1. Numerical Simulation

In this Section, we study numerical simulation of solutions to the coupled system of wave equations with power nonlinear terms. We consider the special dimensioned casen=2 in problem (1). That is(136)∂2u∂t2=∂2∂x2+∂2∂y2u+v2,∂2v∂t2=∂2∂x2+∂2∂y2u+u2,where, ux,y,t is the displacement of film in position x,y at the time t. The initial values satisfy(137)u|t=0=e−20x2+y+1.32+e−x2+y−0.52,∂u∂t|t=0=0,v|t=0=e−20x2+y+1.32+e−x2+y−0.52,∂v∂t|t=0=0.Discretizingu in the computational region, we obtain(138)uN×N=u11u12…u1Nu21u22…u2N0p3…0⋮⋮…⋮uN1uN2…uNN.We denote(139)∂nu∂yn⟶DNnuN×N,and(140)∂nu∂xn⟶DNnuN×NTT=uN×NDNnT.We obtain the trend of waves in Figure1.Figure 1
The variation trend of waves.Figure1 shows the trend of the two waves in the time 0 ∼4s. We define the wave with the initial position of x=0,y=0.5 as a large wave, and the wave with x=0,y=−1.3 as a small wave. When t=0s, it represents the initial state of the wave, in which the peak of the large wave is 0.9989, and the peak of the small wave is 0.9388. When t=0.25s, the peaks of these two waves both begin to decline. At this time, the peak of the large wave is 0.8792, and the peak of the small wave is −0.1137. Comparing the big wave with the small wave, it is obvious that the small wave has a faster-descending speed. The peak of the big wave is still above the horizontal plane, while the small wave has fallen below the horizontal plane within 0.25s. When t=0.5s, the peaks of the two waves continue to move downward. The peak of the large wave is 0.5753, and the peak of the small wave is −0.08573. When t=1s, the crests of these two waves both drop below the horizontal plane, the crests of the small wave begin to rise upwards as well. At the same time, these two waves begin to meet. The small wave begins to spread outward, and the spread of the big wave is affected. When t=2s, both waves are concave down at the same time. The wave continues to move downward when t=2∼4s. By observing the wave changes from 0∼4s, we can see that when the two waves move at the same time, the changing trend of the wavelet is obvious.



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