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##  eISSN 0454-8124
pISSN 1225-6951

### Article

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

Published online March 31, 2021

### 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

### Abstract

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)/(p−2q),E(0)

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

### 1. Introduction

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

$utt+∫ΩDmu2dxq(−Δ)mu+uk∂j(ut)=upu, x∈Ω, t>0$$u(x,t)=0, ∂iu∂νi=0, i=1,2,⋯,m−1, x∈∂Ω, t>0u(x,0)=u0(x),ut(x,0)=u1(x), x∈Ω,$

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 $ν.j(·):R→R$ 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 , Messaodui and Houari consider the higher-order nonlinear Kirchhoff-type hyperbolic equation

$utt+∫ΩDmu2dxq(−Δ)mu+utrut=upu, x∈Ω, t>0ux,t=0, ∂iu∂vi=0, i=1,2,⋯,m−1, x∈∂Ω, t>0ux,0=u0(x), ut(x,0)=u1(x), x∈Ω,$

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 , Han and Wang investigated the global existence and blow-up of solutions for nonlinear viscoelastic wave equation with degenerate damping and source term

$utt−Δu+∫0tg(t−s)Δu(s)ds+uk∂j(ut)=up−1u, (x,t)∈Ω×(0,T),u(x,t)=0, (x,t)∈∂Ω×(0,T)u(x,0)=u0(x), ut(x,0)=u1(x), x∈Ω,$

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

$utputt(t)−Δu(t)−Δutt(t)+∫0tg(t−s)Δu(s)ds +u(t)k∂j(ut(t))=u(t)p−1u(t)+εσ(x,t)∂tW(x,t), (x,t)∈D×0,T,u(x,t)=0, (x,t)∈∂D×0,Tu(x,0)=u0(x), ut(x,0)=u1(x), x∈D¯,$

where $D¯$ is a bounded domain in $Rn(n≥1)$ with a smooth boundary $∂D,k,p≥0$ and $g(⋅) : R+ → R+,j(⋅) : R→R$ 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.

### 2. Blow-up Result

For $j(⋅) : R→R,$ similarly to [1, 4, 9], we have the following assumptions :

(H1) Let $j(⋅) : R→R$ be continuous, convex real valued function, there exist constants $c>0,c0>0,c1>0,c2≥0$ such that for all $s,v∈R,j(⋅)$ satisfies

• (1) coercivity: $j(s)≥csr+1,$

• (2) strict monotonicity: $(∂j(s)−∂j(v))(s−v)≥c1s−vr+1,$

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

(H2)$p≤maxp*2,p*r+kr+1;p*=2nn−2.$

(H3)$k,r≥0,p>0.$ In addition, $k≤nn−2,p+1<2nn−2,if n≥3.$

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

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

and

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

Lemma 2.1. () Let u(t) be solution of Eq.(1.1). Assume that$E(0)and$Dmu0>α1$. Then there exists a constant$α2>α1$such that

$Dmu(⋅,t)≥α2,∀t≥0,$$up+2≥Bα2, ∀t≥0.$

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

Theorem 2.1.Suppose that$p≤k+r−1$and

$02m, and p>0, if n≤2m.$

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  and Theorem 2.3 of , we derive to the solution.

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

Theorem 2.2.Suppose that$p>maxk+r−1,2q$and

$02m, and p>0, if n≤2m.$

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)/(p−2q), E(0)

blows up in finite time T.

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

$E'(t)=−∫Ωuk∂j(ut)utdx≤0.$

Let

$G(t)=E1−E(t).$

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

$G'(t)=∫Ωuk∂j(ut)utdx≥c1∫Ωukutr+1dx≥0.$

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

$0

and exploiting (2.1) and (2.3) we obtain

$E1−12(q+1)Dmu2(q+1)

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

$0

As in , we construct a Lyapunov's function

$L(t)=G1−α(t)+ε∫Ωuutdx,$

where

$α=minp−r−k+1r(p+2),p2(p+2),$

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

$L'(t)=(1−α)G−α(t)G'(t)+εut2+εutp+2p+2+εDmu2(q+1) −ε∫Ωuk∂j(ut)udx+2ε(q+1)G(t)−2ε(q+1)E1 +2ε(q+1)12ut2+12(q+1)Dmu2(q+1)−1p+2up+2p+2.$

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

$L'(t)≥(1−α)G−α(t)G'(t)+εut2+εup+2p+2+εDmu2(q+1) −ε∫Ωuk∂j(ut)udx+2ε(q+1)G(t)−2ε(q+1)E1(Bα2)−(p+2)up+2p+2 +ε(q+1)ut2+εDmu2(q+1)−2ε(q+1)p+2up+2p+2 =(1−α)H−α(t)G'(t)+ε(q+2)ut2+εc*up+2p+2+2εDmu2(q+1) +2ε(q+1)H(t)−ε∫Ωuk∂j(ut)udx,$

where

$c*=1−2(q+1)p+2−2(q+1)E1(Bα2)−(p+2)>0,$

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

$∫Ωuk∂j(ut)udx≤c0∫Ωuk+1−rkr+1urkr+1utrdx ≤c0∫Ωukutr+1dxrr+1∫Ωur+k+1dx1r+1.$

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

$∫Ωuk∂j(ut)udx≤KG'(t)rr+1up+2r+k+1r+1 ≤Kδup+2r+k+1+δ−1rG'(t),$

where $K=c0c1−rr+1Ωp−r−k+1(p+1)(r+1)$ and δ is a positive constant to be determined later. From (2.14) and (2.16), we obtain

$L'(t)≥(1−α)G−α(t)−εKδ−1rG'(t) +ε(q+2)ut2+2εDmu2(q+1) +2ε(q+1)G(t)−εKδup+2r+k+1+εc*up+2p+2.$

Putting

$δ=c*2Kup+2p−r−k+1>0$

in (2.17), then

$εKδup+2r+k+1=εc*2up+2p+2.$

Thus, we get

$L'(t)≥(1−α)G−α(t)−εKδ−1rG'(t) +ε(q+2)ut2+2εDmu2(q+1)+2ε(q+1)G(t) +εc*2up+2p+2.$

In view of (2.10), we deduce

$(1−α)G−α(t)−εKδ−1r=G−α(t)1−α−εKδ−1rGα(t) ≥G−α(t)1−α−εK1+1rc*2−1r(up+2p−r−k+1)−1r(p+2)−αup+2α(p+2) =G−α(t)1−α−εK1+1rc*2−1r(p+2)−αup+2r+k−p−1+αr(p+2)r.$

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

$(1−α)G−α(t)−εKδ−1r≥G−α(t)1−α−εK1+1r(c*2)−1r(p+2)r+k−p−1r(p+2)G(0)r+k−p−1+αr(p+2)r(p+2)=G−α(t)1−α−εK1+1rF>0,$

where

$F=(c*2)−1r(p+2)r+k−p−1r(p+2)G(0)r+k−p−1+αr(p+2)r(p+2).$

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

$1−α−εK1+1rF>0.$

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

$G(0)≥cεθ,$

where $θ=r(p+2)p−r−k+1−αr(p+2),c:=(c*/2)1/r(p+2)(r+k−p−1)/r(p+2)K1+1/r(1−α)−1θ.$ 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 $t∈0,T$. Obviously, (2.23) indicates that L(t) is increasing on $0,T$ and

$L(t)=G1−α(t)+ε∫Ωuutdx ≥G1−α(0)+ε∫Ωu0u1dx=L(0).$

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

$0<ε

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

$L'(t)≥cε1+σL11−α(t),$

where c is a positive constant and $σ=θ1−21−2α(p+2)≥0.$ To this end, we consider two cases.

Case 1.$∫Ωuutdx≤0$ for some $t∈0,T.$ Then we have for such t

$L11−α(t)=G1−α(t)+ε∫Ωuutdx11−α≤G(t).$

Thus (2.24) is valid for all $t∈0,T$ for which $∫Ωuutdx≤0.$

Case 2.$∫Ωuutdx≤0$ for some $t∈0,T$. By (2.12), it easily follows $0<α<12.$ Noting $0<ε<1$, then by convexity we have

$L11−α(t)≤211−α−1G(t)+ε∫Ωuutdx11−α.$

Using Höolder's inequality, we obtain the following estimate

$∫Ωuutdx≤uut≤cup+2ut,$

which implies

$∫Ωuutdx11−α≤cup+211−αut11−α.$

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

$∫Ωuutdx11−α≤up+221−2α+ut2.$

From (2.12), it easy to see that $21−2α≤p+2$. Then it follows from (2.10) that

$up+221−2α=up+221−2α−(p+2)≤((p+2)G(0))2(1−2α)(p+2)−1up+2p+2.$

Thanks to $2(1−2α)(p+2)−1≤0$ and (2.22), it yields

$up+221−2α≤cε−σup+2p+2,$

where

$σ=θ1−2(1−2α)(p+2) =r(p+2)p−r−k+1−αr(p+2)1−2(1−2α)(p+2)≥0.$

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

$∫Ωuutdx11−α≤cε−σut2+up+2p+2.$

Combing (2.25) and (2.28), we get

$L11−α(t)≤cε−σG(t)+ut2+up+2p+2 ≤cε−σDmu2(q+1)+G(t)+ut2+up+2p+2.$

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

From (2.24), we obtain

$L(t)≥1L(0)−α/(1−α)−αc1−αε1+σt.$

From (2.30), we deduce that $limt→T−L(t)=+∞,$ where

$T=1−ααcε−1−αL(0)−α1−α.$

We obtain from(2.8) and (2.29)$L1/(1−α)(t)≤Cup+2p+2$ for some constant $C>0$ and also $up+2≤BDmu.$ Hence we have $limt→T−up+2=+∞$ and $limt→T−Dmu=+∞$ 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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