# Nonexistence of Global Weak Solutions of a System of Nonlinear Wave Equations with Nonlinear Fractional Damping

Authors: Mohamed Jleli; Mokhtar Kirane; Bessem Samet
Journal: Journal of Function Spaces
 (2020)

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## Abstract

We consider the system of nonlinear wave equations with nonlinear time fractional dampingutt+−Δmu+CD0,tαtσuq=vp,t>0,x∈ℝN,vtt+−Δmv+CD0,tβtδvr=vs,t>0,x∈ℝN,u0,x,ut0,x=u0x,u1x,x∈ℝN,u0,x,ut0,x=u0x,u1x,x∈ℝN,where u,v=ut,x,vt,x, m and N are positive natural numbers, p,q,r,s>1, σ,δ≥0, 0<α,β<1, and CD0,tκ, 0<κ<1, is the Caputo fractional derivative of order κ. Namely, sufficient criteria are derived so that the system admits no global weak solution. To the best of our knowledge, the considered system was not previously studied in the literature.


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## Body

## 1. Introduction

In this paper, we investigate the system of nonlinear wave equations with nonlinear time fractional damping:(1)utt+−Δmu+CD0,tαtσuq=vp,t>0,x∈ℝN,vtt+−Δmv+CD0,tβtδvr=vs,t>0,x∈ℝN,u0,x,ut0,x=u0x,u1x,x∈ℝN,u0,x,ut0,x=u0x,u1x,x∈ℝN,where u,v=ut,x,vt,x, m and N are positive natural numbers, p,q,r,s>1, σ and δ are nonnegative numbers that will be specified later, 0<α,β<1, and CD0,tκ, 0<κ<1, is the Caputo fractional derivative (with respect to t) of order κ. Namely, we are interested in obtaining sufficient conditions for which the considered system admits no global weak solution. Our approach is based on the nonlinear capacity method (see, e.g., [1]).Before we state and prove our result, let us dwell on existence literature. Single wave equations or systems of wave equations have been studied in large; we may mention the books of Lions [2], Reed [3], Georgiev [4], and Strauss [5] and the papers of Aliev et al. [6], Said-Houari [7], Takeda [8], Goergiev and Todorova [9], Todorova and Yordanov [10], Zhang [11], and Kirane and Qafsaoui [12] for equations and systems with classical linear or nonlinear damping and Tatar [13], Kirane and Tatar [14], and Kirane and Laskri [15] for wave equations with fractional damping. In particular, in [13], the following problem was considered:
(2)utt+CD0,tαu−Δu=aup−1u,t>0,x∈Ω,ut,x=0,t>0,x∈∂Ω,u0,x,ut0,x=u0x,u1x,x∈Ω,where Ω is a bounded domain of ℝN with smooth boundary ∂Ω, a>0, p>1, and 0<α<1. It was shown that, if u is a solution to (2), then there exist T∗≤∞ and sufficiently large initial data so that u blows up at T∗. Problem (2) was also considered in [14]. Namely, it was shown that the solution of (2) is unbounded and grows up exponentially in the Lp+1-norm for sufficiently large initial data. In [15], the following problem was studied:
(3)utt−Δu+−Δβ/2D0,tαu=up,t>0,x∈ℝN,u0,x,ut0,x=u0x,u1x,x∈ℝN,where p>1, 1≤β≤2, −Δβ/2 is the fractional Laplacian of order β/2, 0<α<1, and D0,tα is the Riemann-Liouville fractional derivative of order α. It was shown that for all p>1, if liminfx→+∞u1x=+∞,, then (3) does not admit a local weak solution for any T>0.In the next section, we recall some notions on fractional calculus. In Section3, we define global weak solutions of system (1) and state our main result. Moreover, as a consequence, we deduce a nonexistence result in the case of a single equation. Finally, the proof of the main result is given in Section 4.


## 2. Preliminaries

Some preliminaries on fractional calculus (see, e.g., [16–21]) are provided in this section. Given ς>0, the fractional integrals I0ςf and ITςf of f∈L10,T are defined by
(4)I0ςft=1Γς∫0tt−ξζ−1fξdξ,t∈0,Ta.e.,ITςft=1Γς∫tTξ−tζ−1fξdξ,t∈0,Ta.e.,where Γ denotes the gamma function.Lemma 1.
Letς>0, f∈L10,T, and g∈L∞0,T. Then,
(5)∫0TI0ςftgtdt=∫0TftITςgtdt.
Given0<ς<1, the Caputo fractional derivative CD0ςf of f∈AC0,T is defined by
(6)DC0ςft=I01−ςf′t,t∈0,Ta.e.
The following lemma will be used later in the proof of our main result.Lemma 2.
Let0<ς<1 and A:0,T→ℝ be the function defined by
(7)At=T−γT−tγ,where γ>0. Then,
(8)IT1−ςAt=Γγ+1Γ2−ς+γT−γT−tγ+1−ς,(9)IT1−ςAt=−Γγ+1Γ1−ς+γT−γT−tγ−ς,for all t∈0,T.Proof.
For0<t<T, one has
(10)IT1−ςAt=1Γ1−ς∫tTξ−t−ςAξdξ=T−γΓ1−ς∫tTξ−t−ςT−ξγdξ=T−γT−tγΓ1−ς∫tTξ−t−ς1−ξ−tT−tγdξ.
Takingρ=ξ−t/T−t, one obtains (8). Next, (9) follows immediately from (8).


## 3. Main Result

We begin with the definition of the intended solutions of system (1). Given 0<T<∞, let QT=0,T×ℝN and ΦT be the set of functions φ=φt,x∈Ct,x2,2mQT satisfying the following conditions:
(a)
φT,·=φtT,·≡0(b)
suppxφ⊂⊂ℝN, i.e., suppxφt,x⊂K⊂ℝN uniformly in t∈0,T, where K is a compact(c)
IT1−κφt∈L∞QT, for all 0<κ<1Definition 3.
Letui,vi∈Lloc1ℝN, i=0,1. A pair of functions
(11)u,v∈Llocmaxq,s0,∞×ℝN×Llocmaxp,r0,∞×ℝNis said to be a global weak solution of system (1), if for all 0<T<∞ and φ∈ΦT,
(12)∫QTvpφdxdt+∫ℝNφ0,xu1dx−∫ℝNφt0,xu0dx=∫QTuφttdxdt+∫QTu−Δmφdxdt−∫QTtσuqIT1−αφtdxdt,(13)∫QTusφdxdt+∫ℝNφ0,xv1dx−∫ℝNφt0,xv0dx=∫QTvφttdxdt+∫QTv−Δmφdxdt−∫QTtδvrIT1−βφtdxdt.
Next, we introduce the parameters(14)ρ1=Nm+1ps−1p−2s+1,ρ2=Nm+1ps−rp−2s+δ−β,ρ3=Nm+1ps−qp+sσ−α−2q,ρ4=Nm+1ps−rqp+sσ−α+qδ−β(15)ρ¯1=Nm+1ps−1s−2p+1,ρ¯2=Nm+1ps−qs−2p+σ−α,ρ¯3=Nm+1ps−rs+pδ−β−2r,ρ¯4=Nm+1ps−rqs+pδ−β+rσ−α.
Here,m and N are positive natural numbers, p,q,r,s>1, σ and δ are positive numbers that will be specified later, 0<α,β<1, and DC0,tκ,0<κ<1.Theorem 4.
Let0<α,β<1, σ,δ≥0, s>q>1, and p>r>1. Suppose that ui,vi∈L1ℝN, i=0,1, and
(16)min∫ℝNu1xdx,∫ℝNv1xdx>0.
If(17)minmaxρi:i=1,2,3,4,maxρ¯i:i=1,2,3,4<0,then there exists no global weak solution of system (1).
Consider now the case of a single equation. Namely,(18)utt+−Δmu+CD0,tαtσuq=up,t>0,x∈ℝN,u0,x,ut0,x=u0x,u1x,x∈ℝN,where 0<α<1, σ≥0, and p,q>1. A global weak solution of (18) can be defined in a similar way as in Definition 3. Taking in Theorem 4u=v, α=β, σ=δ, q=r, p=s, and u0,u1=v0,v1, one deduces the following corollary.Corollary 5.
Letp>q>1 and ui∈L1ℝN, i=0,1. Suppose that
(19)∫ℝNu1xdx>0.
If(20)Nm<minp+1p−1,p2+α−σp+qp2−q,α−σ−1p2+2pq+qp2−q,α−σ−1p+qp−q,then there exists no global weak solution of problem (18).Example 6.
Consider the equation(21)utt+−Δ4u+DC0,t2/3u2=u3,t>0,x∈ℝ2,u0,x,ut0,x=e−x,x2e−x,x∈ℝ2.
Observe that (21) is a special case of (18) with m=4, N=2, α=2/3, σ=0, q=2, p=3, u0x=e−x, and u1x=x2e−x. One has p>q>1, u0,u1∈L1ℝN, and
(22)∫ℝNu1xdx=∫ℝ2x2e−xdx>0.
On the other hand, one can check easily that(23)minp+1p−1,p2+α−σp+qp2−q,α−σ−1p2+2pq+qp2−q,α−σ−1p+qp−q=1.
Since(24)Nm=24<1,it follows from Corollary 5 where (21) admits no global weak solution.


## 4. Proof of Theorem4

Before proving Theorem4, we need some preliminary results.Given0<T<∞, let
(25)φt,x=AtBx,t,x∈QT,where A is defined by (7) with γ≫1 and
(26)Bx=Ξx2mT2ℓmμ,x∈ℝN,where μ≫1, ℓ>0 is a certain parameter that will be specified later, and Ξ∈C0∞0,∞ is a decreasing function satisfying
(27)Ξρ=1if0≤ρ≤1,0ifρ≥2.Lemma 7.
For all0<T<∞, the function φ defined by (25) belongs to ΦT.Proof.
One can check easily thatφ∈Ct,x2,2mQT and satisfies conditions (a) and (b). On the other hand, by (9), for all 0<κ<1, one obtains
(28)IT1−κφtt,x=BxIT1−κAtt=−Γγ+1Γ1−κ+γT−γT−tγ−κBx≔−Cγ,κT−γT−tγ−κBx,t,x∈QT,which yields
(29)IT1−κφtt,x≤Cγ,κT−κ,t,x∈QT.
This shows thatφ satisfies condition (c).The following estimate follows from elementary calculations.Lemma 8.
There exists a constantC>0 such that
(30)ΔmΞy2mμ≤CΞy2mμ−2m,y∈ℝN,for any positive natural number m.Proof of Theorem 4.
Suppose that(31)u,v∈Llocmaxq,s0,∞×ℝN×Llocmaxp,r0,∞×ℝNis a global weak solution of system (1). For T≫1, using (12) and Lemma 7, one obtains
(32)∫QTvpφdxdt+∫ℝNφ0,xu1dx−∫ℝNφt0,xu0dx≤∫QTuφttdxdt+∫QTuΔmφdxdt+∫QTtσuqIT1−αφtdxdt,where φ is defined by (25). Similarly, using (13), one obtains
(33)∫QTusφdxdt+∫ℝNφ0,xv1dx−∫ℝNφt0,xv0dx≤∫QTvφttdxdt+∫QTvΔmφdxdt+∫QTtδvrIT1−βφtdxdt.Using Hölder’s inequality, the following holds:
(34)∫QTuφttdxdt≤∫QTusφdxdt1/s∫QTφ−1/s−1φtts/s−1dxdts−1/s.
Similarly, one has(35)∫QTuΔmφdxdt≤∫QTusφdxdt1/s∫QTφ−1/s−1Δmφs/s−1dxdts−1/s,(36)∫QTtσuqIT1−αφtdxdt≤∫QTusφdxdtq/s∫QTtσs/s−qφ−q/s−qIT1−αφts/s−qdxdts−q/s.
It follows from (32), (34), (35), and (36) that
(37)∫QTvpφdxdt+∫ℝNφ0,xu1dx−∫ℝNφt0,xu0dx≤∫QTusφdxdt1/s∫QTφ−1/s−1φtts/s−1dxdts−1/s+∫QTusφdxdt1/s∫QTφ−1/s−1Δmφs/s−1dxdts−1/s+∫QTusφdxdtq/s∫QTtσs/s−qφ−q/s−qIT1−αφts/s−qdxdts−q/s.
Similarly, using (33) and Hölder’s inequality, one obtains
(38)∫QTusφdxdt+∫ℝNφ0,xv1dx−∫ℝNφt0,xv0dx≤∫QTvpφdxdt1/p∫QTφ−1/p−1φttp/p−1dxdtp−1/p+∫QTvpφdxdt1/p∫QTφ−1/p−1Δmφp/p−1dxdtp−1/p+∫QTvpφdxdtr/p∫QTtδp/p−rφ−r/p−rIT1−βφtp/p−rdxdtp−r/p.
Now, set(39)I=∫QTusφdxdt1/s,J=∫QTvpφdxdt1/p.
Forι>1, let
(40)Kι=∫QTφ−1/ι−1φttι/ι−1dxdtι−1/ι,Lι=∫QTφ−ι/ι−1Δmφι/ι−1dxdtι−1/ι.
Forζ≥0, 0<κ<1, and a>b>1, let
(41)Mζ,κ,a,b=∫QTtζa/a−bφ−b/a−bIT1−κφta/a−bdxdta−b/a.
From (37) and (38), we may write
(42)Jp+∫ℝNφ0,xu1dx−∫ℝNφt0,xu0dx≤IKs+Ls+IqMσ,α,s,q,(43)Is+∫ℝNφ0,xv1dx−∫ℝNφt0,xv0dx≤JKp+Lp+JrMδ,β,p,r.
On the other hand, from (25), for i=0,1, one has
(44)∫ℝNφ0,xuidx=∫ℝNΞx2mT2ℓmμuidx.
Sinceui∈L1ℝN, i=0,1, by the dominated convergence theorem, one obtains
(45)limT→∞∫ℝNφ0,xuidx=∫ℝNuidx<∞.
Similarly, sincevi∈L1ℝN, i=0,1, the following holds:
(46)limT→∞∫ℝNφ0,xvidx=∫ℝNvidx<∞.
Again, from (25), one has
(47)−∫ℝNφt0,xu0dx=T−1∫ℝNφ0,xu0dx,(48)−∫ℝNφt0,xv0dx=T−1∫ℝNφ0,xv0dx.
Using (16), (45), and (47), one obtains
(49)limT→∞∫ℝNφ0,xu1dx−∫ℝNφt0,xu0dx=∫ℝNu1dx>0.
Similarly, using (16), (46), and (48), one obtains
(50)limT→∞∫ℝNφ0,xv1dx−∫ℝNφt0,xv0dx=∫ℝNv1dx>0.
Consequently, one has(51)∫ℝNφ0,xu1dx−∫ℝNφt0,xu0dx>0,T≫1,(52)∫ℝNφ0,xv1dx−∫ℝNφt0,xv0dx>0,T≫1.
Next, it follows from (42) and (51) that
(53)Jp≤IKs+Ls+IqMσ,α,s,q.
Similarly, (43) and (52) yield
(54)Is≤JKp+Lp+JrMδ,β,p,r.
Using the inequality(55)w+zι≤2ι−1wι+zι,w,z≥0,ι>1,it follows from (53) that
(56)Jps≤2s−1IsKs+Lss+IqsMσ,α,s,qs≤2s−12s−1IsKss+Lss+IqsMσ,α,s,qs,where upon
(57)Jps≤22s−2Kss+LssIs+IqsMσ,α,s,qs.
Similarly, it follows from (54) that
(58)Iqs≤22q−2Kpq+LpqJq+JrqMδ,β,p,rq.
Using (54), (57), and (58), the following holds:
(59)Jps≤C×Kp+LpKss+LssJ+Kss+LssMδ,β,p,rJr+Mσ,α,s,qsKpq+LpqJq+Mσ,α,s,qsMδ,β,p,rqJrq,where C>0 is a constant that depends only on s and q. Here and below, any positive constant independent of T is denoted by C. Next, using theε-Young inequality withε>0which is small enough, one deduces from (59) that
(60)Jps≤C×Kp+Lpps/ps−1Kss+Lssps/ps−1+Kss+Lssps/ps−rMδ,β,p,rps/ps−r+Mσ,α,s,qsKpq+Lpqps/ps−q+Mσ,α,s,qsMδ,β,p,rqps/ps−rq.
Similarly, one obtains(61)Ips≤C×Ks+Lsps/ps−1Kpp+Lppps/ps−1+Kpp+Lppps/ps−qMσ,α,s,qps/ps−q+Mδ,β,p,rpKsr+Lsrps/ps−r+Mδ,β,p,rpMσ,α,s,qrps/ps−rq.
Further, we shall estimateKι and Lι for ι>1 and Mζ,κ,a,b for ζ>0, 0<κ<1, and a>b>1. From (25), one has
(62)Kι=∫QTφ−1/ι−1φttι/ι−1dxdtι−1/ι=∫0TAt−1/ι−1Atttι/ι−1dtι−1/ι∫ℝNBxdxι−1/ι=CT−γ∫0TT−tγ−2ι/ι−1dtι−1/ι∫ℝNΞx2mT2ℓmμdxι−1/ι=CT−i+1/iTℓN∫0<y2m<2Ξy2mdyι−1/ι,which yields
(63)Kι=CTℓN−1−ℓN+1/i,ι>1.
Similarly, using (25) and Lemma 8, one obtains
(64)Lι=∫QTφ−1/ι−1Δmφι/ι−1dxdtι−1/ι=∫0TAtdtι−1/ι∫ℝNBx−1/ι−1ΔmBxι/ι−1dxι−1/ι=∫0TT−γT−tγdtι−1/ι∫ℝNΞx2mT2ℓm−μ/ι−1ΔmΞx2mT2ℓmμι/ι−1dxι−1/ι=CTι−1/ιTℓN∫ℝNEy2m−μ/ι−1T−2ℓmΔmEy2mμι/ι−1dyι−1/ι=CTι−1/ιTι−1/ιℓN−2ℓmι/ι−1∫1<y2m<2Ey2m−μ/ι−1ΔmEy2mμι/ι−1dyι−1/ι≤CTℓN−2ℓm+1−ℓN+1/i∫1<y2m<2Ey2mμ−2mι/ι−1dyι−1/ι,so
(65)Lι≤CTℓN−2ℓm+1−ℓN+1/i,ι>1.
Next, using (9) and (25), for ζ≥0, 0<κ<1, and a>b>1, one obtains
(66)Mζ,κ,a,b=∫QTtζa/a−bφ−b/a−bIT1−κφta/a−bdxdta−b/a=∫0Ttζa/a−bAt−b/a−bIT1−κAta/a−bdta−b/a∫ℝNΞx2mT2ℓmμdxa−b/a=CT−γ∫0Ttζa/a−bT−tγ−κa/a−bdta−b/aTℓN∫0<y2m<2Ξy2mμdya−b/a,whereupon
(67)Mζ,κ,a,b=CTa−b/aℓN+1+ζ−κ,ζ≥0,0<κ<1,a>b>1.
Next, takingℓ=1/m, it follows from (63), (65), and (67) that
(68)Kι=Lι=OTN/m+1ι−1/ι−2,ι>1,T≫1,(69)Mζ,κ,a,b=CTa−b/aN/m+1+ζ−κ,ζ≥0,0<κ<1,a>b>1.
Using (60), (68), and (69), the following holds:
(70)Jp≤CTρ1ps/ps−1+Tρ2ps/ps−r+Tρ3¯ps/ps−q+Tρ4ps/ps−rq1/s,where ρi, i=1,2,3,4, are defined by (14). Similarly, using (61), (68), and (69), one obtains
(71)Is≤CTρ¯1ps/ps−1+Tρ¯2ps/ps−q+Tρ¯3ps/ps−r+Tρ¯4ps/ps−rq1/p,where ρ¯i, i=1,2,3,4, are defined by (15).Now, from condition (17), we have two cases.Case 1.
Ifmaxρi:i=1,2,3,4<0, in this case, taking the infimum limit as T→∞ in (70) and using Fatou’s lemma, the following holds:
(72)vt,x=0,t,x∈0,∞×ℝNa.e.
Hence, by (33), the following holds:
(73)∫ℝNφ0,xv1dx−∫ℝNφt0,xv0dx≤0,T≫1,which contradicts (52).Case 2.
Ifmaxρ¯i:i=1,2,3,4<0, as in the previous case, taking the infimum limit as T→∞ in (71), one obtains
(74)ut,x=0,t,x∈0,∞×ℝNa.e.
Hence, by (32), the following holds:
(75)∫ℝNφ0,xu1dx−∫ℝNφt0,xu0dx≤0,T≫1,which contradicts (51).



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