KYUNGPOOK Math. J. 2019; 59(4): 631-649
Global Nonexistence of Solutions for a Quasilinear Wave Equation with Time Delay and Acoustic Boundary Conditions
Yong Han Kang∗, Jong-Yeoul Park
Francisco College, Daegu Catholic University, Gyeongsan-si 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
* Corresponding Author.
Received: October 8, 2019; Revised: December 4, 2019; Accepted: December 5, 2019; Published online: December 23, 2019.

This is an Open Access article distributed under the terms of the Creative Commons Attribution Non-Commercial License (http://creativecommons.org/licenses/by-nc/3.0/) which permits unrestricted non-commercial use, distribution, and reproduction in any medium, provided the original work is properly cited.
Abstract

In this paper, we prove the global nonexistence of solutions for a quasilinear wave equation with time delay and acoustic boundary conditions. Further, we establish the blow up result under suitable conditions.

Keywords: global nonexistence of solutions, quasilinear wave equation, blow up, time delay, acoustic boundary.
1. Introduction

In this paper, we consider the following quasilinear wave equation with time delay and acoustic boundary conditions:

$(∣ut(x,t)∣l-2ut(x,t))t-Δut(x,t)-div(a(x)∣∇u(x,t)∣α-2∇u(x,t))-div(∣∇ut(x,t)∣β-2∇ut(x,t))+Q(x,t,ut)+μ1ut(x,t)+μ2ut(x,t-τ)=f(x,u(x,t)) in Ω×[0,T),$$u=0 on Γ0×[0,T),$$∂ut(x,t)∂ν+a(x)∣∇u(x,t)∣α-2∂ut(x,t)∂ν+∣∇ut(x,t)∣β-2∂ut(x,t)∂ν=h(x)yt(x,t) on Γ1×[0,T),$$ut(x,t)+k(x)yt(x,t)+q(x)y(x,t)=0 on Γ1×[0,T),$$u(x,0)=u0(x), ut(x,0)=u1(x) in Ω,$$ut(x,t-τ)=f0(x,t-τ) in Ω×(0,τ),$$y(x,0)=y0(x) on Γ1.$

Here, J = [0, T), 0 < T ≤ ∞, a: Ω → R+ is a positive function, l, α, β ≥ 2, μ1 > 0, μ2 is a real number, and τ > 0 represents the time delay. Further, Ω is a regular and bounded domain of Rn(n ≥ 1) and ∂Ω(:= Γ) = Γ0 ∪ Γ1, where Γ0 and Γ1 are closed and disjoint and $∂∂ν$ denotes the outer normal derivative. The functions k, q, h: Γ1R+(:= [0,∞]) are essentially bounded and 0 < q0q(x) on Γ1.

The acoustic boundary conditions were introduced by Morse and Ingard [16] and developed by Beale and Rosencrans in [1], where the authors proved the global existence and regularity of the linear problem. Other authors have studied the existence and decay of solutions for a viscoelastic wave equation with acoustic boundary conditions (see [3, 4, 6, 7, 12, 13, 15, 19, 20, 23] and the references therein).

The time delay arises in many physical, chemical, biological and economical phenomena because these phenomena depend not only on the present state but also on the past history of the system in a more complicated way. In particular, the effects of time delay strikes on our system have a significant effect on the range of existence and the stability of the system. The differential equations with time delay effects have become an active area of research, see for example [9, 11, 17, 18]. In [14], without the delay term and the acoustic boundary condition, Liu and Wang considered the global nonexistence of solutions with the positive initial energy for a class of wave equations:

$(∣ut(x,t)∣l-2ut(x,t))t-Δut(x,t)-div(a(x)∣∇u(x,t)∣α-2∇u(x,t))-div(∣∇u(x,t)∣β-2∇ut(x,t))+Q(x,t,ut)=f(x,u(x,t)) in J×Ω,u(x,t)=0 on J×∂Ω,u(x,0)=u0(x), ut(x,0)=u1(x) in Ω,$

where J = [0, T), 0 < T ≤ ∞, Ω is a bounded regular open subset of Rn(n ≥ 1), l, α, β ≥ 2 and a, Q, f satisfy some conditions. Recently, for l = 2, a(x) = 1, Q(ut) = a|ut|m−2ut, μ1 = μ2 = 0, f(u) = b|u|p−2u, and without the time delay term in our system, Jeong at al [8] investigated the global nonexistence of solutions for a quasilinear wave equation with acoustic boundary conditions

$utt-Δut-div(∣∇u∣α-2∇u)-div(∣∇ut∣β-2∇ut)+α∣ut∣m-2ut=b∣u∣p-2u in Ω×(0,∞),u=0 on Γ0×(0,∞),∂ut∂ν+∣∇u∣α-2∂u∂ν+∣∇ut∣β-2∂ut∂ν=h(x)yt on Γ1×(0,∞),ut+f(x)yt+q(x)y=0 on Γ1×(0,∞),u(x,0)=u0(u), ut(x,0)=u1(x) in Ω,y(x,0)=y0(x) on Γ1,$

where a, b > 0, α, β, m, p > 2 are constants and Ω is a regular and bounded domain of Rn(n ≥ 1) and ∂Ω(= Γ) = Γ0 ∪ Γ1. Here Γ0 and Γ1 are closed and disjoint. The functions h, f, q: Γ1R+ are essentially bounded. Moreover, for a(x) = 1, l = 2, div(|∇ut|β−2ut) = 0, Q = 0, and without boundary conditions, Kafini and Messaoudi [10] studied the following nonlinear damped wave equation

$utt(x,t)-div(∣∇u(x,t)∣m-2∇u(x,t))+μ1ut(x,t)+μ2ut(x,t-τ)=b∣u(x,t)∣p-2u(x,t) in Ω×(0,∞),ut(x,t-τ)=f0(x,t-τ) on (0,τ),u(x,t)=0 on ∂Ω×(0,∞),u(x,0)=u0(x), ut(x,0)=u1(x) in Ω,$

where p > m ≥ 2, b, μ1 are positive constants, μ2 is a real number, and τ > 0 represents the time delay. They proved the blow-up result in a nonlinear wave equation with time delay and without acoustic boundary conditions.

Motivated by the previous works, we consider an equation in a broader and more generalized form than the system discussed above. So we study the global nonexistence of solutions for a quasilinear wave equation with the time delay and acoustic boundary conditions. To the best of our knowledge. there are no results of a quasilinear wave equations with the time delay and acoustic boundary conditions. Thus the result in this work is very meaningful. The main result will be proved in Section 3.

2. Preliminaries

In this section, we shall give some notations, assumptions and a theorem which will be used throughout this paper. We denote by m′ the Hölder conjugate of m, ||u||p = ||u||Lp(Ω), ||u||p = ||u||Lp(Γ), ||u||1,s = ||u||W1,s(Ω), where Lp(Ω) and W1,s(Ω) stand for the Lebesgue spaces and the classical Sobolev spaces, respectively. Specially we introduce the set

$WΓ01,s(Ω)={u∈W1,s∣u=0 on Γ0}, W01,s(Ω)={u∈W1,s∣u=0 on Γ}.$

We make the following same assumptions on a, Q, f as section 4.2 of [22].

(H1)a(x) ∈ L(Ω) such that a(x) ≥ a0 a.e. in Ω for some a0 > 0.

(H2)f(x, u) ∈ C(Ω×ℝn,ℝn) and f(x, u) = ∇uΦ(x, u), with normalizing condition Φ(x, 0) = 0.

There are constants d1 > 0, p > α and μ < μ0a0 such that

$∣f(x,u)∣≤μ∣u∣α-1+d1∣u∣p-1$

for all x ∈ Ω and u ∈ ℝn. Moreover, there is ε1 > 0 such that for all ε ∈ (0, ε1] there exists d2 = d2(ε) > (pα)d1/p such that

$f(x,u)u-(p-ɛ)Φ(x,u)≥d2∣u∣p$

for all x ∈ Ω.

(H3) There are m > 1 and a measurable function d = d(x, t) defined on Ω×J such that d(·, t) ∈ Lp/(pm)(Ω) for a.e. tJ and

$Q(x,t,v)v≥0$$∣Q(x,t,v)∣≤[d(x,t)]1/m[Q(x,t,v)v]1/m′$

for all values of the arguments x, t, v, where

$d(x,t)≥0,‖d(·,t)‖p/(p-m)∈Lloc∞(J).$

### Remark 2.1

We note that when Q(x, t, ut) = b(1+t)ρ|ut|m−2ut, −∞ < ρm−1, condition (H3) holds.

Now, we transform the equation (1.1)(1.7) to the system, using the idea of [21] and introduce the associated energy. So, we introduce the new variable:

$z(x,ρ,t)=ut(x,t-τρ), x∈Ω, ρ∈(0,1), t>0.$

Thus, we have

$τzt(x,ρ,t)+zρ(x,ρ,t)=0, x∈Ω, ρ∈(0,1), t>0.$

Then problem (1.1)–(1.7) takes the following form:

$(∣ut(x,t)∣l-2ut(x,t))t-Δut(x,t)-div(a(x)∣∇u(x,t)∣α-2∇u(x,t))-div(∣∇ut(x,t)∣β-2∇ut(x,t))+Q(x,t,ut)+μ1ut(x,t)+μ2z(x,1,t)=f(x,u(x,t)) in Ω×J,$$τzt(x,ρ,t)+zρ(x,ρ,t)=0 in Ω×(0,1)×J,$$u=0 on Γ0×J,$$∂ut(x,t)∂ν+a(x)∣∇u(x,t)∣α-2∂u(x,t)∂ν+∣∇ut(x,t)∣β-2∂ut(x,t)∂ν=h(x)yt(x,t) on Γ1×J,$$ut(x,t)+k(x)yt(x,t)+q(x)y(x,t)=0 on Γ1×J,$$u(x,0)=u0(x), ut(x,0)=u1(x) in Ω,$$z(x,ρ,0)=f0(x,-ρτ) in Ω×(0,1),$$y(x,0)=y0(x) on Γ1.$

We introduce the following space

$Z=L∞([0,T];WΓ01,α(Ω))∩W1,∞([0,T];L2(Ω))∩W1,β([0,T];WΓ01,β(Ω))∩W1,m([0,T];Lm(Ω)),$

for some T > 0.

We state, without a proof, a local existence which can be established by combining arguments of [2, 5, 24].

### Theorem 2.1

Let$u0∈WΓ01,α(Ω)$, u1L2(Ω), f0L2(Ω×(0, 1)) and y0L21) be given. Suppose that l, α, β, m, p > 2, max{l, β, m} < α < p < nα/(nα), μ1 > |μ2| and (H1)(H3) hold. Then problem(2.6)–(2.13) has a unique local solution (u, z, y) ∈ Z×L2([0, T);L2(Ω×(0, 1)))×L2([0, T);L21)) for some T > 0.

In order to state and prove our result, we introduce the energy functional

$E(t)=l-1l∫Ω∣ut(x,t)∣ldx+1α∫Ωa(x)∣∇u(x,t)∣αdx-∫ΩΦ(x,u(x,t))dx+ξ2∫Ω∫01z2(x,ρ,t)dρdx+12∫Γ1h(x)q(x)y2(x,t)dΓ,$

where

$τ∣μ2∣ <ξ<τ(2μ1-∣μ2∣), μ1> ∣μ2∣.$

We set

$λ1=(A0-μμ0)1/(p-α)(d1B1p)-1/(p-α),$$E1=(1α-1p)(a0-μμ0)p/(p-α)(d1B1p)-α/(p-α),$

where B1 is the best constant of the Sobolev embedding $W01,α(Ω)↪Lp(Ω)$ given by

$B1-1=inf{‖∇u‖α:u∈W01,α(Ω),‖u‖p=1}.$

We also set

$Σ={(λ,E)∈ℝ2∣λ>λ1, E
3. Proof of Main Result

In this section, we state and prove our main result. Our main result as follows.

### Theorem 3.1

Let$u0∈WΓ01,α(Ω)$, u1L2(Ω), f0L2(Ω×(0, 1)) and y0L21) be given. Suppose that l, α, β, m, p > 2, max{l, β, m} < α < p < nα/(nα), μ1 > |μ2| and (H1)(H3) hold. Assume further that

$(‖∇u‖α,E(0))∈Σ.$

Then the solution (u, z, y) ∈ Z × L2(R+);L2(Ω × (0, 1))) × L2(R+);L21)) of problem (2.6)–(2.13) can not exist for all time.

In this section, we shall prove Theorem 3.1. We start with a series of lemmas. We denote

$λ0=‖∇u0‖α, E0=E(0).$

Theorem 3.1 will be proved by contradiction, so we shall suppose that the solution of (2.6)–(2.13) exists on the whole interval [0,∞), i.e. T = ∞.

Proof of Theorem 3.1

We use the idea of Vitillaro [22].

### Lemma 3.1

Let (u, z, y) be the solution of (2.6)–(2.13). Then the energy functional defined by (2.15) satisfies, for some constant c0 > 0,

$ddtE(t)≤(l-1)∫Ω∣ut(x,t)∣l-2ut(t)utt(t)dx+∫Ωa(x)∣∇u(t)∣α-2∇u(t)∇ut(t)dx-∫ΩQ(x,t,ut(t))ut(t)dx-c0∫Ω(ut2(t)+z(x,1,t))dx-∫Γ1h(x)k(x)yt2(t)dΓ≤0.$
Proof

Multiplying the equation (2.6) by ut(t), integrating over Ω, using Green’s formula and exploiting the equation (2.9), we obtain

$ddt{l-1l∫Ω∣ut(t)∣ldx+1α∫Ωa(x)∣∇u(t)∣αdx-∫ΩΦ(x,u(t))dx}-∫Γ1h(x)yt(t)ut(t)dΓ=(l-1)∫Ω∣ut(t)∣l-2ut(t)utt(t)dx+∫Ωa(x)∣∇u(t)∣α-2∇u(t)∇ut(t)dx-∫Ω∣∇ut(t)∣dx-∫Ω∇uΦ(x,u)ut(t)dx-μ1∫Ωut2(t)dx-μ2∫Ωz(x,1,t)ut(t)dx.$

On the other hand, we have from the equation in (2.10) that

$-∫Γ1h(x)yt(t)ut(t)dΓ=∫Γ1h(x)k(x)yt2(t)dΓ+∫Γ1h(x)q(x)y(t)yt(t)dΓ.$

Also, multiplying the equation (2.7) by $ξ2z(x,ρ,t)$ and integrating over Ω × (0, 1), we deduce

$ddt{ξ2∫Ω∫01z2(x,ρ,t)dρdx}=-ξτ∫Ω∫01z(x,ρ,t)zρ(x,ρ,t)dρdx =-ξ2τ∫Ω∫01∂∂ρz2(x,ρ,t)dρdx =ξ2τ∫Ω[z2(x,0,t)-z2(x,1,t)]dx =ξ2τ[∫Ωut2(t)dx-∫Ωz2(x,1,t)dx],$

and

$-μ2∫Ωz(x,1,t)ut(t)dx≤∣μ2∣2[∫Ωut2(t)dx+∫Ωz2(x,1,t)dx].$

Hence, from (2.15) and (3.3)–(3.6), we arrive at

$ddtE(t) ≤ -∫Ω∣∇ut(t)∣2dx-∫Ω∣∇ut(t)∣βdx -∫ΩQ(x,t,ut(t))ut(t)dx-(μ1-ξ2τ-∣μ2∣2)∫Ωut2(t)dx -(ξ2τ-∣μ2∣2)∫Ωz2(x,1,t)dx-∫Γ1h(x)k(x)yt2(t)dΓ.$

By using (2.16), we get, for some c0 > 0,

$ddtE(t) ≤ -∫Ω∣∇ut(t)∣2dx-∫Ω∣∇ut(t)∣βdx -∫ΩQ(x,t,ut(t))ut(t)dx-c0∫Ω[ut2(t)+z2(x,1,t)]dx -∫Γ1h(x)k(x)yt2(t)dΓ≤0.$

Hence we get E(t) ≤ E(0) for all tJ.

### Lemma 3.2

If (λ0, E(0)) ∈ ∑, then we have

$(i) ‖∇u(t)‖α ≥λ2 for all t∈J,for some λ2>λ1,$$(ii) ‖u(t)‖p ≥B1λ2 for all t∈J,for the some λ2 in (i).$
Proof

First, we will prove the (i). From (2.15), we see that

$E(t)≥1α∫Ωa(x)∣∇u(t)∣αdx-∫ΩΦ(x,u(t))dx.$

Using (2.1), since f(x, u(t)) = ∇uΦ(x, u(t)), it follows that

$Φ(x,u(t))=∫01f(x,τu(t))u(t)dτ≤μα∣u(t)∣α+d1p∣u(t)∣p,$

and then

$∫ΩΦ(x,u(t))dx=∫01f(x,τu(t))u(t)dτ≤μα‖u(t)‖αα+d1p‖u(t)‖pp,$

Therefore

$E(t)≥a0α‖∇u(t)‖αα-μα‖u(t)‖αα-d1p‖u(t)‖pp ≥(a0-μμ0)1α‖∇u(t)‖αα-d1p‖u(t)‖pp ≥(a0-μμ0)1α‖∇u(t)‖αα-d1B1p1p‖u(t)‖αp =(a0-μμ0)1αλα-d1B1p1pλp:=ϕ(λ),$

where λ = ||∇u(t)||α. It is easy to verify that ϕ is increasing for 0 < λ < λ1, decreasing for λ > λ1, ϕ(λ) → −∞ as λ → +∞ and ϕ(λ) = E1, where λ1 is given in (2.18). Therefore, since E0< E1, there exists λ2> λ1 such that ϕ(λ2) = E(0). From (3.12) we have ϕ(λ0) ≤ E(0) = ϕ(λ2), which implies that λ0λ2 since λ0 > λ1. To proof the result, we suppose by contradiction that ||∇u0||α< λ2, for some t0 > 0 and by the continuity of ||∇u(t)||α we can choose such that ||∇u(t0)||α> λ1. Again the use of (3.12) leads to

$E(t0)≥ϕ(‖∇u(t0)‖α)>ϕ(λ2)=E(0).$

This is impossible since E(t) ≤ E(0), for all t ≥ 0. Thus (i) is established. Next, we will prove the (ii). From (3.12), we get

$d1p‖u(t)‖pp≥(a0-μμ0)1α‖∇u(t)‖αα-E(t) ≥(a0-μμ0)1α‖∇u(t)‖αα-E0 ≥(a0-μμ0)1αλ2α-ϕ(λ2)=d1B1p1pλ2p.$

Thus, the proof is complete.

In the remainder of this section, we consider initial values (λ0, E0) ∈ ∑. We set

$H(t)=E1-E(t), t≥0.$

Then we have the following Lemma.

### Lemma 3.3

For all tJ, we have

$0
Proof

From Lemma 3.1, we see that H′(t) ≥ 0. Thus, we deduce

$H(t)≥H(0)=E1-E(0)>0, ∀t≥0.$

From (3.12), we obtain

$H(t)=E1-E(t) ≥ϕ(λ1)-(a0-μμ0)1α‖∇u(t)‖αα+d1p‖u(t)‖pp =(a0-μμ0)1α(λ1α-‖∇u(t)‖αα)-d1B1p1pλ1p+d1p‖u(t)‖pp.$

From (3.8), ||∇u(t)||α> λ1, we get

$H(t)≤d1p‖∇u(t)‖pp.$

Thus, combing (3.15) and (3.16) we obtain (3.14).

Now, we define

$L(t) =H1-σ(t)+ɛ∫Ωu(t)∣ut(t)∣l-2ut(t)dx +μ1ɛ2∫Ωut2(t)dx-ɛ2∫Γ1h(x)k(x)y2(t)dΓ -ɛ∫Γ1h(x)u(t)y(t)dΓ,$

for ɛ small to be chosen later and

$0<σ≤min{α-2p,α-βp(β-1),α-mp(m-1),α-lαl,kɛα-1}.$

By taking a derivative of (3.17) we have

$L′(t)=(1-σ)H-σ(t)H′(t)+ɛ‖ut(t)‖ll+ ɛ∫Ωu(t)(∣ut(t)∣l-2ut(t))tdx+μ1ɛ∫Ωu(t)ut(t)dx-ɛ∫Γ1h(x)k(x)y(t)yt(t)dΓ-ɛ∫Γ1h(t)ut(t)y(t)dΓ-ɛ∫Γ1h(x)u(t)yt(t)dΓ.$

By using (2.6)–(2.10) and estimate (3.19), we find

$L′(t)=(1-σ)H-σ(t)H′(t)+ɛ‖ut(t)‖ll + ɛ∫Ω(Δut(t)+div(a(x)∣∇u(t)∣α-2∇u(t)) +div(∣∇ut(t)∣β-2∇ut(t))-Q(x,t,ut) -μ1ut(t)-μ2z(x,1,t)+f(x,u(t)))u(t)dx +μ1ɛ∫Ωu(t)ut(t)dx-ɛ∫Γ1h(x)k(x)y(t)yt(t)dΓ -ɛ∫Γ1h(t)ut(t)y(t)dΓ-ɛ∫Γ1h(x)u(t)yt(t)dΓ =(1-σ)H-σ(t)H′(t)+ɛ‖ut(t)‖ll -ɛ∫Ω∇ut(t)∇u(t)dx-ɛ∫Ωa(x)∣∇u(x,t)∣αdx -ɛ∫Ω∣∇ut(t)∣β-2∇ut(t)∇u(t)dx +ɛ∫Γ1(∂ut(t)∂ν+∣∇u(t)∣α-2∂u(t)∂ν+∣∇ut(x,t)∣β-2∂ut(x,t)∂ν)u(t)dΓ -ɛ∫ΩQ(x,t,ut)u(t)dx+ɛ∫Ωf(x,u(x,t))u(t)dx -μ1ɛ∫Ωu(t)ut(t)dx-μ2ɛ∫Ωz(x,1,t)u(t)dx +μ1ɛ∫Ωu(t)ut(t)dx-ɛ∫Γ1h(x)k(x)y(t)yt(t)dΓ -ɛ∫Γ1h(x)ut(t)y(t)dΓ-ɛ∫Γ1h(x)u(t)yt(t)dΓ =(1-σ)H-σ(t)H′(t)+ɛ‖ut(t)‖ll -ɛ∫Ω∇ut(t)∇u(t)dx-ɛ∫Ωa(x)∣∇u(x,t)∣αdx -ɛ∫Ω∣∇ut(t)∣β-2∇ut(t)∇ut(t)dx -ɛ∫ΩQ(x,t,ut)u(t)dx+ɛ∫Ωf(x,u(x,t))u(t)dx -μ2ɛ∫Ωz(x,1,t)u(t)dx+ɛ∫Γ1h(x)q(x)y2(t)dΓ$

Exploiting Hölder’s and Young’s inequality and (H3), for any δ, μ, η, ρ > 0, we deduce

$∫ΩQ(x,t,u)u(t)dx≤∫Ω∣u(t)∣[d(x,t)]1/m[Q(x,t,ut(t))ut(t)]1/m′dx≤δmm∫Ω∣u(t)∣md(x,t)dx+m-1mδ-mm-1∫ΩQ(x,t,ut(t))ut(t)dx≤δmm‖u(t)‖pm‖d(t)‖p/(p-m)+m-1mδ-mm-1∫ΩQ(x,t,ut(t))ut(t)dx≤δmCm‖u(t)‖pm+m-1mδ-mm-1∫ΩQ(x,t,ut(t))ut(t)dx.$

By Young’s inequality, we get

$∫Ω∇ut(t)∇ut(t)dx≤14μ∫Ω∣∇u(t)∣2dx+μ∫Ω∣∇ut(t)∣2dx,$$∫Ω∣∇ut(t)∣β-2∇ut(t)∇u(t)dx≤ηββ∫Ω∣∇u(t)∣βdx+β-1βη-ββ-1∫Ω∣∇ut(t)∣βdx,$$μ2∫Ωu(t)z(x,1,t)dx≤∣μ2∣4ρ∫Ωu2(t)dx+∣μ2∣ρ∫Ωz2(x,1,t)dx.$

A substitution of (3.21) – (3.24) into (3.20) yields

$L′(t)≥ (1-σ)H-σ(t)H′(t)+ɛ‖∇ut(t)‖ll -ɛ4μ∫Ω∣∇u(t)∣2dx-ɛμ∫Ω∣∇ut(t)∣2dx -ɛ∫Ωa(x)∣∇u(t)∣αdx -ɛηββ∫Ω∣∇u(t)∣βdx-ɛ(β-1)βη-ββ-1∫Ω∣∇ut(t)∣βdx -ɛδmCm‖u(t)‖pm-ɛ(m-1)mδ-mm-1∫ΩQ(x,t,ut(t))ut(t)dx -ɛ∣μ2∣4ρ∫Ωu2(t)dx-ɛ∣μ2∣ρ∫Ωz2(x,1,t)dx +ɛ∫Ωf(x,u(x,t))u(t)dx+ɛ∫Γ1h(x)q(x)y2(t)dΓ$

Therefore, we choose δ, μ, η, and ρ so that

$δ-mm-1=M1H-σ(t),μ=M2H-σ(t)η-ββ-1=M3H-σ(t),ρ=M4H-σ(t),$

for M1, M2, M3, M4 to be specified later. Using (2.10), (3.25) and (3.26), we arrive at

$L′(t) ≥(1-σ)H-σ(t)H′(t)+ɛ‖ut(t)‖ll-ɛ4M2Hσ(t)∫Ω∣∇u(t)∣2dx -ɛ∫Ωa(x)∣∇u(t)∣αdx-ɛM3-(β-1)βHσ(β-1)(t)∫Ω∣∇u(t)∣βdx -ɛM1(m-1)CmHσ/(m-1)(t)‖u(t)‖pm-ɛ∣μ2∣4M4Hσ(t)∫Ωu2(t)dx -ɛ[M2∫Ω∣∇ut(t)∣2dx+(β-1)βM3∫Ω∣∇ut(t)∣βdx +(m-1)mM1∫ΩQ(x,t,ut(t))ut(t)dx+∣μ2∣M4∫Ωz2(x,1,t)dx]H-σ(t) +ɛ∫Ωf(x,u(x,t))u(t)dx+ɛ∫Γ1h(x)q(x)y2(t)dΓ.$

If $M=M2+(β-1)M3β+(m-1)M1m+∣μ2∣M4$, then (3.27) takes the form

$L′(t) ≥(1-σ-ɛM)H-σ(t)H′(t)+ɛ‖ut(t)‖ll-ɛ4M2Hσ(t)∫Ω∣∇u(t)∣2dx -ɛ∫Ωa(x)∣∇u(t)∣αdx-ɛM3-(β-1)βHσ(β-1)(t)∫Ω∣∇u(t)∣βdx -ɛCmM1(m-1)Hσ/(m-1)(t)‖u(t)‖pm-ɛ∣μ2∣4M4Hσ(t)∫Ωu2(t)dx +ɛMH-σ(t)∫Γ1h(x)k(x)yt2(t)dΓ +ɛ∫Ωf(x,u(t))u(t)dx+ɛ∫Γ1h(x)q(x)y2(t)dΓ.$

From (3.14),(3.18), the embedding W1,α(Ω) ↪ Lp(Ω) and

$zδ≤(1+1/a)(z+a),∀z>0, 0<δ≤1, a>0,$

we have (see[15])

$Hσ(t)∫Ω∣∇ut(t)∣2dx ≤c(Ω)(B1pd1p)σ(∫Ω∣∇u(t)∣αdx)(pσ+2)/α ≤d(B1pd1p)σ(∫Ω∣∇u(t)∣αdx+H(t)),$$Hσ(β-1)(t)∫Ω∣∇ut(t)∣βdx ≤c(Ω)(B1pd1p)σ(β-1)(∫Ω∣∇u(t)∣αdx)(pσ(β-1)+β)/α ≤d(B1pd1p)σ(β-1)(∫Ω∣∇u(t)∣αdx+H(t)),$$Hσ(m-1)‖u(t)‖pm ≤c(Ω)(B1pd1p)σ(m-1)B1m(∫Ω∣∇u(t)∣αdx)(pσ(m-1)+m)/α ≤d(B1pd1p)σ(m-1)B1m(∫Ω∣∇u(t)∣αdx+H(t)),$

and

$Hσ(t)∫Ω∣∇ut(t)∣2dx≤c(Ω)(B1pd1p)σB12(∫Ω∣∇u(t)∣αdx)(σp)/α ≤d(B1pd1p)σB12 (∫Ω∣∇u(t)∣αdx+H(t)),$

for all t ≥ 0, where d = c(Ω)[1 + 1/H(0)]. Inserting estimates (3.29)–(3.32) into (3.28), we obtain

$L′(t) ≥ (1-σ)-ɛM)H-σ(t)H′(t)+kH(t)+(ɛ+k(l-1)l)‖ut(t)‖ll -ɛc2M2(∫Ω∣∇u(t)∣αdx+H(t))-ɛ∫Ωa(x)∣∇u(t)∣αdx -ɛc3M3β-1(∫Ω∣∇u(t)∣αdx+H(t))+kα∫Ωa(x)∣∇u(t)∣αdx -ɛc1M1m-1(∫Ω∣∇u(t)∣αdx+H(t)) -ɛc4M4(∫Ω∣∇u(t)∣αdx+H(t)) +ɛ∫Ωf(x,u(t))u(t)dx+ɛ∫Γ1h(x)q(x)y2(t)dΓ -k∫ΩΦ(x,u(t))dx+kξ2∫Ω∫01z2(x,ρ,t)dρdx +k2∫Γ1h(x)q(x)y2(t)dΓ-kE1+ɛMH-σ(t)∫Γ1h(x)k(x)yt2(t)dΓ,$

for some constant k and

$c1=cdm(B1pd1p)σ/(m-1)B1m, c2=d4(B1pd1p)σ,c3=dβ(B1pd1p)σ(β-1), c4=d(B1pd1p)σB12.$

From (2.17),(2.18) and Lemma 3.2, we have

$-kE1≥-kE1B1-pλ1-p‖u(t)‖pp=-kd1(1α-1p)‖u(t)‖pp.$

From (2.2), we can choose k satisfying

$αɛ≤k

and

$ɛ∫Ωf(x,u(t))u(t)dx-k∫ΩΦ(x,u(t))dx-kE1≥ɛd2‖u(t)‖pp-kd1(1α-1p)‖u(t)‖pp≥0.$

Thus, it follows that

$L′(t) ≥ (1-σ)-ɛM)H-σ(t)H′(t)+(ɛ+k(l-1)l)‖ut(t)‖ll ɛ(kɛ-c2M2-c3M3β-1-c1M1m-1-c4M4)H(t) +ɛ((kɛα-1)a0-c2M2-c3M3β-1-c1M1m-1-c4M4)∫Ω∣∇u(t)∣αdx +ɛ∫Γ1h(x)q(x)y2(t)dΓ+kξ2∫Ω∫01z(x,ρ,t)dρdx +k2∫Γ1h(x)q(x)y2(t)dΓ+ɛMH-σ(t)∫Γ1h(x)k(x)yt2(t)dΓ.$

At this point, choosing M1, M2, M3, M4 large enough and ɛ sufficiently small and using

$ɛMH-σ(t)∫Γ1h(x)k(x)yt2(t)dΓ≥0,$

we deduce

$L′(t) ≥ (1-σ)-ɛM)H-σ(t)H′(t)+γɛ(H(t)+‖ut(t)‖ll +∫Ω∣∇ut(t)∣αdx+∫Γ1h(x)q(x)y2(t)dΓ+∫Ω∫01z2(x,ρ,t)dρdx),$

where γ is a positive constant (it is possible since k > ɛα). We choose ɛ sufficiently small and 0 < ɛ < (1 – σ)/M so that

$L(0)=H1-σ(0)+ɛ∫Ωu0∣u1∣l-2u1dx+μ1ɛ2∫Ωu02dx-ɛ2∫Γ1h(x)k(x)y02dΓ-ɛ∫Γ1h(x)u0y0dΓ>0.$

Then from(3.33) we get

$L(t)≥L(0)≥0, ∀t≥0,$

and

$L′(t) ≥ γɛ(H(t)+‖ut(t)‖ll +∫Ω∣∇u(t)∣αdx+∫Γ1h(x)q(x)y2(t)dΓ+∫Ω∫01z2(x,ρ,t)dρdx).$

On the other hand, from(3.17) and h(x), q(x) > 0, we have

$L(t)≤H1-σ(t)+ɛ∫Ωu(t)∣ut(t)∣l-2ut(t)dx+μ1ɛ2∫Ωu2(t)dx-ɛ∫Γ1h(x)u(t)y(t)dΓ.$

Then the above inequality leads to

$L11-σ(t)≤ [H1-σ(t)+ɛ∫Ωu(t)∣ut(t)∣l-2ut(t)dx +μ1ɛ2∫Ωu2(t)dx-ɛ∫Γ1h(x)u(t)y(t)dΓ]1/(1-σ)≤ C(ɛ,μ1,σ)[H(t)+∣∫Ωu(t)∣ut(t)∣l-2ut(t)dx∣11-σ +(∫Ωu2(t)dx)11-σ+∣∫Γ1h(x)u(t)y(t)dΓ∣11-σ].$

Next, using Hölder’s inequality, the embedding W1,α(Ω) ↪ Ll(Ω), α > l and Young’s inequality, we derive

$|∫Ωu(t)∣ut(t)|l-2ut(t)dx∣≤(∫Ω∣u(t)∣ldx)1/l(∫Ω∣ut(t)∣ldx)(l-1)/l≤(∫Ω∣∇u(t)∣σdx)1/α(∫Ω∣ut(t)∣ldx)(l-1)/l≤c[(∫Ω∣∇u(t)∣αdx)l(1-σ)/[l(1-σ)-(l-1)]α+(∫Ω∣ut(t)∣ldx)(l-σ)].$

From (3.18) and (3.29), we obtain

$|∫Ωu(t)|ut(t)∣l-2ut(t)dx∣1/(1-σ)≤c[(∫Ω∣∇u(t)∣αdx)l/[l(1-σ)-(l-1)]α+∫Ω∣ut(t)∣ldx]≤c[(1+1H(0))(∫Ω∣∇u(t)∣αdx+H(t))+∫Ω∣ut(t)∣ldx].$

Therefore, there exists a positive constant C′ such that for all t ≥ 0,

$|∫Ωu(t)∣ut(t)∣l-2ut(t)dx|1/(1-σ)≤C′[H(t)) +‖∇u(t)‖αα+‖ut(t)‖ll].$

Furthermore, by the same method, we deduce

$∫Γ1h(x)u(t)y(t)dΓ=|∫Γ1h(x)q(x)q(x)u(t)y(t)dΓ| ≤‖h‖∞12‖q‖∞12q0(∫Γ1h(x)q(x)y2(t)dΓ)12 (∫Γ1u2(t)dΓ)12.$

Similarly, we find

$∫Γ1h(x)u(t)y(t)dΓ=|∫Γ1h(x)q(x)q(x)u(t)y(t)dΓ| ≤‖h‖∞12‖q‖∞12q0(∫Γ1h(x)q(x)y2(t)dΓ)12 (∫Γ1u2(t)dΓ)12.$

Using the embedding $W01,α(Ω)↪L2(Γ1)$ and Hölder’s inequality, we get

$∫Γ1h(x)u(x)y(t)dΓ≤c5‖h‖∞12‖q‖∞12q0(∫Γ1h(x)q(x)y2(t)dΓ)12 (∫Ω∣∇u(t)∣αdx)1α.$

where c5 is a embedding constant. Consequently, there exists a positive constant c6 = c(||h||, ||q||, q0, σ, α) such that

$(∫Γ1h(x)u(t)y(t)dΓ)11-σ≤c6(∫Γ1h(x)q(x)y2(t)dΓ)12(1-σ) (∫Ω∣∇u(t)∣αdx)1α(1-σ).$

Using Young’s inequality, we write

$(∫Γ1h(x)u(t)y(t)dΓ)11-σ≤c7[ (∫Ω∣∇u(t)∣αdx)2α(1-2σ)+∫Γ1h(x)q(x)y2(t)dΓ],$

where c7 is a positive constant depending on c6 and α. Applying once again the algebraic inequality (3.29) with $z= ‖∇u(t)‖αα$, ν = 2/[α(1 – 2σ)] and making use of (3.18), we see that by the same method as above

$(∫Γ1h(x)u(t)y(t)dΓ)11-σ≤c8[H(t)+‖∇u(t)‖αα+∫Γ1h(x)q(x)y2(t)dΓ],$

where c8 is a positive constant. Hence combining (3.35) – (3.37) and using α > 2, we arrive at

$L11-σ(t)≤C*[H(t)+‖ut(t)‖ll+‖∇u(t)‖αα +∫Γ1h(x)q(x)y2(t)dΓ++∫Ω∫01z2(x,ρ,t)dρdx],∀ t≥0,$

where C* is a positive constant. Consequently a combining of (3.34) and (3.38), for some ξ > 0, we obtain

$L′(t)≥ξL11-σ(t), ∀ t≥0.$

Integration of (3.9) over (0, t) yield

$Lσ1-σ(t)≥1L-σ1-σ(0)-ξσ1-σt,∀ t≥0.$

Therefore L(t) blow up in finite time

$T≤T*=1-σξσLσ1-σ(0).$

Thus the proof of Theorem 2.1 is complete.

Acknowledgements

We would like to thank the references for their valuable comments and suggestions to improve our paper.

### Funding

Y. H. Kang was supported by research grants from the Daegu Catholic University in 2017(Number 20171288).

### Availability of data and materials

Data sharing not applicable to this article as no datasets were generated or analyzed during the current study.

### Competing interests

The authors declare that they have no competing interests.

Not applicable.

### Author’s contributions

The authors declare that the study was realized in collaboration with the same responsibility. All authors read and approved the final manuscript.

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