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Kyungpook Mathematical Journal 2021; 61(1): 1-10

Published online March 31, 2021

Copyright © Kyungpook Mathematical Journal.

Blow-up of Solutions for Higher-order Nonlinear Kirchhofftype Equation with Degenerate Damping and Source

Yong Han Kang∗, Jong-Yeoul Park

Francisco College, Daegu Catholic University, Gyeongsan 712-702, Republic of Korea
e-mail : yonghann@cu.ac.kr

Department of Mathematics, Pusan National University, Busan 609-735, Republic of Korea
e-mail : jyepark@pusan.ac.kr

Received: November 1, 2020; Revised: November 29, 2020; Accepted: December 14, 2020

This paper is concerned the finite time blow-up of solution for higher-order nonlinear Kirchhoff-type equation with a degenerate term and a source term. By an appropriate Lyapunov inequality, we prove the finite time blow-up of solution for equation (1.1) as a suitable conditions and the initial data satisfying Dmu0>B(p+2)/(p2q),E(0)<E1.

Keywords: Kirchoff-type equation, blow up, higher-order nonlinear, degenerate damping and source

In this paper, we consider the higher-order nonlinear Kirchhoff-type equation with degenerate damping and source:

utt+ΩDmu2dxq(Δ)mu+ukj(ut)=upu,xΩ,t>0u(x,t)=0,iuνi=0,i=1,2,,m1,xΩ,t>0u(x,0)=u0(x),ut(x,0)=u1(x),xΩ,

where m > 1 is an integer constant, p, q, k > 0, Ω is a bounded domain of Rn, with a smooth boundary Ω and a unit outer normal ν.j(·):RR is a given function to be specified later and we use j to denote its sub-differential for example(see [1, 4, 9]). Moreover, in [14], Messaodui and Houari consider the higher-order nonlinear Kirchhoff-type hyperbolic equation

utt+ΩDmu2dxq(Δ)mu+utrut=upu,xΩ,t>0ux,t=0,iuvi=0,i=1,2,,m1,xΩ,t>0ux,0=u0(x),ut(x,0)=u1(x),xΩ,

where m ≥ 1, p, q, r ≥ 0, Ω is a bounded domain of Rn, with a smooth boundary ∂Ω and a unit outer normal ν. They established a blow-up result for certain solutions with positive initial energy. In [4], Han and Wang investigated the global existence and blow-up of solutions for nonlinear viscoelastic wave equation with degenerate damping and source term

uttΔu+0tg(ts)Δu(s)ds+ukj(ut)=up1u,(x,t)Ω×(0,T),u(x,t)=0,(x,t)Ω×(0,T)u(x,0)=u0(x),ut(x,0)=u1(x),xΩ,

where Ω is a bounded domain of Rn, with a smooth boundary ∂Ω in Rn, k, p ≥ 0, and g():R+R+,j():RR are given functions to be specified. Here, we use ∂j to denote its sub-differential(see [1, 4, 9]). Furthermore, in [9], Kim et al. consider a stochastic quasilinear viscoelastic wave equation with degenerate damping and sources

utputt(t)Δu(t)Δutt(t)+0tg(ts)Δu(s)ds+u(t)kj(ut(t))=u(t)p1u(t)+εσ(x,t)tW(x,t),(x,t)D×0,T,u(x,t)=0,(x,t)D×0,Tu(x,0)=u0(x),ut(x,0)=u1(x),xD¯,

where D¯ is a bounded domain in Rn(n1) with a smooth boundary D,k,p0 and g():R+R+,j():RR is given functions to be specified. Here, we use ∂j to denote its sub-differential, W (x, t) is an infinite dimensional Winner process, and σ(x, t) is L2(D) valued progressively measurable. Authors proved the blow-up of solution for stochastic/no stochastic quasilinear viscoelastic wave equation with positive probability or explosive in energy sense [2, 3, 5, 6, 7, 8, 10, 11, 13, 15, 16, 17].

Motivated by the previous works, we studied the blow-up of solutions for higher-order nonlinear Kirchhoff-type equation with degenerate damping and source. To the best of our knowledge. there are no results of a higher-order nonlinear Kirchhoff-type equation with degenerate damping and source. The main result was proved in section 2.

For j():RR, similarly to [1, 4, 9], we have the following assumptions :

(H1) Let j():RR be continuous, convex real valued function, there exist constants c>0,c0>0,c1>0,c20 such that for all s,vR,j() satisfies

  • (1) coercivity: j(s)csr+1,

  • (2) strict monotonicity: (j(s)j(v))(sv)c1svr+1,

  • (3) continuity: j(x) is single-valued and j(s)c0sr+c2.

(H2)pmaxp*2,p*r+kr+1;p*=2nn2.

(H3)k,r0,p>0. In addition, knn2,p+1<2nn2,ifn3.

Here =2. Let B be the best constant of the embedding inequality up+2BDmu. We set

α1=B(p+2)/(p2q),E1=12(q+1)1p+2α12(q+1),

and

E(t)=12ut2+12(q+1)Dmu2(q+1)1p+2up+2p+2.

Lemma 2.1. ([14]) Let u(t) be solution of Eq.(1.1). Assume thatE(0)<E1andDmu0>α1. Then there exists a constantα2>α1such that

Dmu(,t)α2,t0,up+2Bα2,t0.

Under the assumptions (H1)-(H3), we get the two theorems.

Theorem 2.1.Suppose thatpk+r1and

0<p<2n2m,ifn>2m,andp>0,ifn2m.

Then for any initial data(u0,u1)H2m(Ω)H0m(Ω)×H0m(Ω),the solution of Eq.(1.1) is global.

Proof. By using Theorem 1.1 of [12] and Theorem 2.3 of [4], we derive to the solution.

Our main results is the Theorem 2.2. Here c2 = 0.

Theorem 2.2.Suppose thatp>maxk+r1,2qand

0<p<2n2m,ifn>2m,andp>0,ifn2m.

Then for any initial data(u0,u1)H2m(Ω)H0m(Ω)×H0m(Ω),any solution of Eq.(1.1) with initial data satisfying

Dmu0>B(p+2)/(p2q),E(0)<E1,

blows up in finite time T.

Proof. A multiplication of Eq.(1.1) by ut and integration over Ω give

E'(t)=Ωukj(ut)utdx0.

Let

G(t)=E1E(t).

In view of (2.6) and (H1), we easily see that

G'(t)=Ωukj(ut)utdxc1Ωukutr+1dx0.

By using (2.2), (2.6) and (2.7), we have

0<G(0)G(t)=E112ut212(q+1)Dmu2(q+1)+1p+2up+2p+2

and exploiting (2.1) and (2.3) we obtain

E112(q+1)Dmu2(q+1)<E112(q+1)α12(q+1)=1p+2α12(q+1)<0,t0.

Thus using (2.8) and (2.9), we get

0<G(0)G(t)1p+2up+2p+2,t0.

As in [11], we construct a Lyapunov's function

L(t)=G1α(t)+εΩuutdx,

where

α=minprk+1r(p+2),p2(p+2),

and ε being a positive constant to be determined later. By taking a derivative of L(t) and using equation of (1.1), we have

L'(t)=(1α)Gα(t)G'(t)+εut2+εutp+2p+2+εDmu2(q+1)εΩukj(ut)udx+2ε(q+1)G(t)2ε(q+1)E1+2ε(q+1)12ut2+12(q+1)Dmu2(q+1)1p+2up+2p+2.

On exploiting (2.2) and (2.4), estimate (2.13) takes the form

L'(t)(1α)Gα(t)G'(t)+εut2+εup+2p+2+εDmu2(q+1)εΩukj(ut)udx+2ε(q+1)G(t)2ε(q+1)E1(Bα2)(p+2)up+2p+2+ε(q+1)ut2+εDmu2(q+1)2ε(q+1)p+2up+2p+2=(1α)Hα(t)G'(t)+ε(q+2)ut2+εc*up+2p+2+2εDmu2(q+1)+2ε(q+1)H(t)εΩukj(ut)udx,

where

c*=12(q+1)p+22(q+1)E1(Bα2)(p+2)>0,

since α2>B(p+2)/(p2q). Thanks to p+1>k+r, then the continuity of j() in (H1) and using Hölder's inequality, we infer that

Ωukj(ut)udxc0Ωuk+1rkr+1urkr+1utrdxc0Ωukutr+1dxrr+1Ωur+k+1dx1r+1.

From (2.7),(2.15) and Young's inequality it yields

Ωukj(ut)udxKG'(t)rr+1up+2r+k+1r+1Kδup+2r+k+1+δ1rG'(t),

where K=c0c1rr+1Ωprk+1(p+1)(r+1) and δ is a positive constant to be determined later. From (2.14) and (2.16), we obtain

L'(t)(1α)Gα(t)εKδ1rG'(t)+ε(q+2)ut2+2εDmu2(q+1)+2ε(q+1)G(t)εKδup+2r+k+1+εc*up+2p+2.

Putting

δ=c*2Kup+2prk+1>0

in (2.17), then

εKδup+2r+k+1=εc*2up+2p+2.

Thus, we get

L'(t)(1α)Gα(t)εKδ1rG'(t)+ε(q+2)ut2+2εDmu2(q+1)+2ε(q+1)G(t)+εc*2up+2p+2.

In view of (2.10), we deduce

(1α)Gα(t)εKδ1r=Gα(t)1αεKδ1rGα(t)Gα(t)1αεK1+1rc*21r(up+2prk+1)1r(p+2)αup+2α(p+2)=Gα(t)1αεK1+1rc*21r(p+2)αup+2r+kp1+αr(p+2)r.

Since up+2((p+2)G(0))1p+2>0 and α was chosen such that r+kp1+αr(p+2)0, it follows from (2.19) that

(1α)Gα(t)εKδ1rGα(t)1αεK1+1r(c*2)1r(p+2)r+kp1r(p+2)G(0)r+kp1+αr(p+2)r(p+2)=Gα(t)1αεK1+1rF>0,

where

F=(c*2)1r(p+2)r+kp1r(p+2)G(0)r+kp1+αr(p+2)r(p+2).

Now, we choose 0<ε<(1α)(FK1+1/r)1 small enough such that

1αεK1+1rF>0.

In the sequel, we may adjust ε again. From (2.21), it follows

G(0)cεθ,

where θ=r(p+2)prk+1αr(p+2),c:=(c*/2)1/r(p+2)(r+kp1)/r(p+2)K1+1/r(1α)1θ. Here and in what follows we use c to denote a generic positive constant which is independent of ε and initial data. Therefore, we conclude that from (2.18),(2.20) and (2.21) that

L'(t)εcDmu2(q+1)+ut2+G(t)+up+2p+2,

for t0,T. Obviously, (2.23) indicates that L(t) is increasing on 0,T and

L(t)=G1α(t)+εΩuutdxG1α(0)+εΩu0u1dx=L(0).

If Ωu0u1dx0, then no further restriction on ε is needed. However if Ωu0u1dx<0, we further require ε satisfies

0<ε<G1α(0)2Ωu0u1dx.

In both cases, we have L(t)L(0)>0,t0,T. Now we prove that L(t) satisfies the following differential inequality

L'(t)cε1+σL11α(t),

where c is a positive constant and σ=θ1212α(p+2)0. To this end, we consider two cases.

Case 1.Ωuutdx0 for some t0,T. Then we have for such t

L11α(t)=G1α(t)+εΩuutdx11αG(t).

Thus (2.24) is valid for all t0,T for which Ωuutdx0.

Case 2.Ωuutdx0 for some t0,T. By (2.12), it easily follows 0<α<12. Noting 0<ε<1, then by convexity we have

L11α(t)211α1G(t)+εΩuutdx11α.

Using Höolder's inequality, we obtain the following estimate

Ωuutdxuutcup+2ut,

which implies

Ωuutdx11αcup+211αut11α.

Noting 1<11α<2, then by Young's inequality we have

Ωuutdx11αup+2212α+ut2.

From (2.12), it easy to see that 212αp+2. Then it follows from (2.10) that

up+2212α=up+2212α(p+2)((p+2)G(0))2(12α)(p+2)1up+2p+2.

Thanks to 2(12α)(p+2)10 and (2.22), it yields

up+2212αcεσup+2p+2,

where

σ=θ12(12α)(p+2)=r(p+2)prk+1αr(p+2)12(12α)(p+2)0.

Then, from (2.26) and (2.27), we infer that

Ωuutdx11αcεσut2+up+2p+2.

Combing (2.25) and (2.28), we get

L11α(t)cεσG(t)+ut2+up+2p+2cεσDmu2(q+1)+G(t)+ut2+up+2p+2.

Therefore (2.24) follows from (2.23)and (2.29).

From (2.24), we obtain

L(t)1L(0)α/(1α)αc1αε1+σt.

From (2.30), we deduce that limtTL(t)=+, where

T=1ααcε1αL(0)α1α.

We obtain from(2.8) and (2.29)L1/(1α)(t)Cup+2p+2 for some constant C>0 and also up+2BDmu. Hence we have limtTup+2=+ and limtTDmu=+ i.e., the solution blow up at finite time T in Lp+2(Ω) and Hm(Ω) norm.

Thus the proof of Theorem 2.2. is complete.

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