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Blow-up for logarithmic viscoelastic equations with delay and acoustic boundary conditions

  • Sun-Hye Park EMAIL logo
Published/Copyright: April 27, 2023
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Advances in Nonlinear Analysis
From the journal Advances in Nonlinear Analysis Volume 12 Issue 1

Abstract

In the present work, we establish a blow-up criterion for viscoelastic wave equations with nonlinear damping, logarithmic source, delay in the velocity, and acoustic boundary conditions. Due to the nonlinear damping term, we cannot apply the concavity method introduced by Levine. Thus, we use the energy method to show that the solution with negative initial energy blows up after finite time. Furthermore, we investigate the upper and lower bounds of the blow-up time.

Keywords: blow-up; viscoelastic equation; logarithmic nonlinearity; time delay; acoustic boundary condition; nonlinear dissipation
MSC 2010: 35L05; 35L70; 35B44

1 Introduction

We discuss the following viscoelastic equation with nonlinear damping, logarithmic source, delay in the velocity, and acoustic boundary conditions:

(1.1) u t t − Δ u + ∫ 0 t k ( t − s ) Δ u ( s ) d s + c 1 ∣ u t ( t ) ∣ q − 2 u t ( t ) + c 2 ∣ u t ( t − τ ) ∣ q − 2 u t ( t − τ ) = c 3 ∣ u ∣ p − 2 u ln ∣ u ∣ in Ω × ( 0 , T ) ,

(1.2) u = 0 on Γ 0 × ( 0 , T ) ,

(1.3) ∂ u ∂ ν − ∫ 0 t k ( t − s ) ∂ u ( s ) ∂ ν d s = h ( x ) z t on Γ 1 × ( 0 , T ) ,

(1.4) u t + f ( x ) z t + g ( x ) z = 0 on Γ 1 × ( 0 , T ) ,

(1.5) u ( 0 ) = u 0 , u t ( 0 ) = u 1 in Ω , z ( 0 ) = z 0 on Γ 1 ,

(1.6) u t ( x , t − τ ) = j 0 ( x , t − τ ) for ( x , t ) ∈ Ω × ( 0 , τ ) ,

where Ω ⊂ R n is a bounded domain with smooth boundary ∂ Ω = Γ 0 ∪ Γ 1 , ∣ Γ 0 ∣ ≠ 0 , ∣ Γ 1 ∣ ≠ 0 , Γ 0 ∩ Γ 1 = ∅ , τ > 0 is time delay, q ≥ 2 , p > 2 , c 1 > 0 , c 2 ∈ R , c 3 > 0 , and the functions h , f , g : Γ 1 → ( 0 , ∞ ) are essentially bounded functions satisfying

g ( x ) ≥ g 0 > 0 for all x ∈ Γ 1 ,

where g 0 is a constant and the function k : [ 0 , ∞ ) → [ 0 , ∞ ) is a nonincreasing differentiable function.

The study of parabolic or hypberbolic equations with logarithmic nonlinearity ( ∣ u ∣ p − 2 u ln ∣ u ∣ ) has been done by many researchers due to their applications in many branches of physics such as geophysics, nuclear physics, and so on. We refer [3,13,15] and [4,7] for parabolic and hyperbolic equations, respectively. For the physical application of this nonlinearity, we also mention ([1,5]). Di et al. [4] looked into a strongly damped equation:

u t t − Δ u − Δ u t = c 3 ∣ u ∣ p − 2 u ln ∣ u ∣

with the Dirichlet boundary condition. They showed finite time blow-up results for the solution with subcritical and critical initial energy under appropriate conditions. Ha and Park [7] obtained similar results as those of [4] for the equation of memory type. Most existing works on blow-up criteria for logarithmic nonlinear hyperbolic equations are related to linearly damped systems and done by utilizing the potential well method and Levine’s concavity technique [12]. Recently, the authors of [8] discussed a linearly damped wave equation with delay in the velocity

u t t − Δ u + c 1 u t ( t ) + c 2 u t ( t − τ ) = c 3 ∣ u ∣ p − 2 u ln ∣ u ∣

with Dirichlet boundary condition. They showed the solution with negative initial energy blows up after finite time by applying the energy method. We present [10] for nonexistence for problems with boundary condition of memory type. Inspired by work [8], we intend to study on global nonexistence for the viscoelastic equations (1.1)–(1.6) with nonlinear damping and delay terms, logarithmic source effect, and acoustic boundary conditions. For global nonexistence of equations with such boundary conditions, we mention [2,6,18]. Due to the nonlinear damping term, we have some difficulty in applying the concavity method of Levine to estimate the bounds of blow-up time. The simultaneous presence of logarithmic nonlinearity and acoustic boundary conditions arises some delicate estimates. By adapting the energy method, we establish a blow-up criterion for problems (1.1)–(1.6). At this point, it is worth mentioning that there are few works that dealt with equations with nonlinear dissipation and logarithmic nonlinearity. The readers who are interested in asymptotic stability to problem (1.1) with q = 2 and c 3 = 0 can refer [11] and the references therein.

Here is the outline of this article. In Section 2, some materials such as notations, hypotheses, and auxiliary formulas are presented. In Section 3, a blow-up criterion is established.

2 Preliminaries

Let

V = { ϕ ∈ H 1 ( Ω ) : ϕ = 0 on Γ 0 }

and ( ⋅ , ⋅ ) denote the scalar product in the space L 2 ( Ω ) . ‖ ⋅ ‖ r and ‖ ⋅ ‖ Γ 1 , r represent the norms in the spaces L r ( Ω ) and L r ( Γ 1 ) , respectively. Moreover, ‖ ⋅ ‖ Y denotes the norm of normed space Y . C > 0 denotes a generic constant. If there is no ambiguity, we omit the variables t and x . We recall the embedding

V ↪ L r ( Ω ) , where 2 ≤ r < ∞ if n = 1 , 2 ; ≤ 2 n n − 2 if n ≥ 3 ,

in the sequel B r denotes the embedding constant satisfying ‖ ϕ ‖ r ≤ B r ‖ ∇ ϕ ‖ 2 , Young’s inequality

(2.1) a b ≤ 1 μ 1 a μ 1 + 1 μ 2 b μ 2 , a , b ≥ 0 , μ 1 , μ 2 > 1 , 1 μ 1 + 1 μ 2 = 1 ,

the inequality

(2.2) a μ 3 ≤ a + 1 ≤ 1 + 1 μ 4 ( a + μ 4 ) , a ≥ 0 , 0 < μ 3 ≤ 1 , μ 4 > 0 ,

and the relation

(2.3) ∣ a ω ln a ∣ ≤ 1 e ω for 0 < a < 1 and 0 ≤ ln a ≤ a ω e ω for a ≥ 1 for any ω > 0 .

As in [16], we set

y ( x , δ , t ) = u t ( x , t − δ τ ) for ( x , δ , t ) ∈ Ω × ( 0 , 1 ) × ( 0 , T ) .

Then, problems (1.1)–(1.6) read

(2.4) u t t − Δ u + ∫ 0 t k ( t − s ) Δ u ( s ) d s + c 1 ∣ u t ∣ q − 2 u t + c 2 ∣ y ( x , 1 , t ) ∣ q − 2 y ( x , 1 , t ) = c 3 ∣ u ∣ p − 2 u ln ∣ u ∣ in Ω × ( 0 , T ) ,

(2.5) τ y t ( x , δ , t ) + y δ ( x , δ , t ) = 0 for ( x , δ , t ) ∈ Ω × ( 0 , 1 ) × ( 0 , T ) ,

(2.6) u = 0 on Γ 0 × ( 0 , T ) ,

(2.7) ∂ u ∂ ν − ∫ 0 t k ( t − s ) ∂ u ( s ) ∂ ν d s = h ( x ) z t on Γ 1 × ( 0 , T ) ,

(2.8) u t + f ( x ) z t + g ( x ) z = 0 on Γ 1 × ( 0 , T ) ,

(2.9) u ( 0 ) = u 0 , u t ( 0 ) = u 1 in Ω , z ( 0 ) = z 0 on Γ 1 ,

(2.10) y ( x , δ , 0 ) = j 0 ( x , − δ τ ) ≔ y 0 ( x , δ ) for ( x , δ ) ∈ Ω × ( 0 , 1 ) .

From here, we put c 3 = 1 for simplicity.

By the arguments of [7,9,17], we can state the well posedness:

Theorem 2.1

Assume that the exponent p satisfies

2 < p < 2 ( n − 1 ) n − 2 i f n ≥ 3 ; 2 < p < ∞ i f n = 1 , 2 ,

the kernel k verifies

1 − ∫ 0 ∞ k ( s ) d s ≔ k l > 0 ,

and the coefficients c 1 and c 2 fulfill

0 < ∣ c 2 ∣ < c 1 .

Then, for every ( u 0 , u 1 , z 0 , y 0 ) ∈ V × L 2 ( Ω ) × L 2 ( Γ 1 ) × L q ( Ω × ( 0 , 1 ) ) , problems (2.4)–(2.10) admit a unique local solution ( u , z , y ) with u ∈ C ( 0 , T ; V ) ∩ C 1 ( 0 , T ; L 2 ( Ω ) ) , h z ∈ L ∞ ( 0 , T ; L 2 ( Γ 1 ) ) , and y ∈ L ∞ ( 0 , T ; L q ( Ω × ( 0 , 1 ) ) ) .

Our goal is to find a blow-up criterion to problem (1.1)–(1.6). For this aim, we impose the following assumptions:

( A n ) Let 1 ≤ n ≤ 3 .

( A p ) Let p satisfy

2 < p < ∞ if n = 1 , 2 ; 2 < p < min 2 ( n − 1 ) n − 2 , n + 2 n − 2 = 4 if n = 3 .

( A q ) Let q verify

(2.11) max 2 , p 2 + 2 p p 2 − 2 p + 4 , ( p 2 + 4 p ) { 2 p + ( n − 1 ) ( p − 2 ) } − p 2 { n ( p − 2 ) + 4 } ( p 2 + 4 ) { 2 p + ( n − 1 ) ( p − 2 ) } − p 2 { n ( p − 2 ) + 4 } < q < p .

( A k ) The kernel function k fulfills

(2.12) ∫ 0 ∞ k ( s ) d s < p − 2 p − 2 + 1 p .

Remark 2.1

  1. For p > 2 and n ≥ 1 , one can note

    (2.13) p > max p 2 + 2 p p 2 − 2 p + 4 , p 2 + 4 p p 2 + 4 = p 2 + 2 p p 2 − 2 p + 4

    and

    (2.14) p > ( p 2 + 4 p ) { 2 p + ( n − 1 ) ( p − 2 ) } − p 2 { n ( p − 2 ) + 4 } ( p 2 + 4 ) { 2 p + ( n − 1 ) ( p − 2 ) } − p 2 { n ( p − 2 ) + 4 } .

    Thus, condition (2.11) on q is reasonable.

  2. For p > 2 and n ≥ 1 , it holds

    (2.15) min 2 n + 2 n , 4 p + 2 ( n − 1 ) ( p − 2 ) n ( p − 2 ) + 4 = 4 p + 2 ( n − 1 ) ( p − 2 ) n ( p − 2 ) + 4 .

  3. ( A q ) , ( A p ) , and ( A n ) ensure

    (2.16) 2 ( p − q ) p 2 ( q − 1 ) < p − 2 2 p

    and

    (2.17) 2 < 2 p 2 ( q − 1 ) p 2 ( q − 1 ) − 4 ( p − q ) < 4 p + 2 ( n − 1 ) ( p − 2 ) n ( p − 2 ) + 4 < p .

  4. From ( A p ) , there exist ω 0 , ω 1 > 0 such that

    2 < p + ω 0 < ∞ if n = 1 , 2 ; ≤ 2 n n − 2 = 6 if n = 3 ,

    and

    2 < 2 ( p − 1 + ω 1 ) < ∞ if n = 1 , 2 ; ≤ 2 n n − 2 = 6 if n = 3 .

Thus, we note

(2.18) V ↪ L p + ω 0 ( Ω ) and V ↪ L 2 ( p − 1 + ω 1 ) ( Ω ) .

3 Blow-up criteria

In this section, we present a blow-up criterion of the solution to (2.4)–(2.10) inspired by the ideas in [6,8,14].

The energy to problems (2.4)–(2.10) is given as follows:

(3.1) E ( t ) = 1 2 ‖ u t ‖ 2 2 + 1 2 1 − ∫ 0 t k ( s ) d s ‖ ∇ u ‖ 2 2 + 1 2 ( k ♢ ∇ u ) + 1 p 2 ‖ u ‖ p p + 1 2 ∫ Γ 1 h g z 2 d Γ − 1 p ∫ Ω ∣ u ∣ p ln ∣ u ∣ d x + ξ τ ∫ 0 1 ‖ y ( δ , t ) ‖ q q d δ ,

where

( k ♢ ∇ u ) ( t ) = ∫ 0 t k ( t − s ) ‖ ∇ u ( t ) − ∇ u ( s ) ‖ 2 2 d s

and

(3.2) ( q − 1 ) ∣ c 2 ∣ q < ξ < c 1 − ∣ c 2 ∣ q .

Lemma 3.1

Under the conditions of Theorem 2.1, it holds

(3.3) d d t E ( t ) ≤ − γ 1 ( ‖ u t ‖ q q + ‖ y ( 1 , t ) ‖ q q ) + 1 2 ( k ′ ♢ ∇ u ) − ∫ Γ 1 h f z t 2 d Γ − k ( t ) 2 ‖ ∇ u ‖ 2 2 ≤ 0

for some γ 1 > 0 .

Proof

From (2.4)–(2.10), we obtain

(3.4) d d t E ( t ) = − c 1 ‖ u t ‖ q q − ∫ Γ 1 h f z t 2 d Γ + 1 2 ( k ′ ♢ ∇ u ) − k ( t ) 2 ‖ ∇ u ‖ 2 2 − c 2 ∫ Ω ∣ y ( x , 1 , t ) ∣ q − 2 y ( x , 1 , t ) u t d x + ∂ ∂ t ξ τ ∫ 0 1 ‖ y ( δ , t ) ‖ q q d δ .

By using (2.1) with q − 1 q + 1 q = 1 , we have

(3.5) − c 2 ∫ Ω ∣ y ( x , 1 , t ) ∣ q − 2 y ( x , 1 , t ) u t d x ≤ ∣ c 2 ∣ ( q − 1 ) q ‖ y ( 1 , t ) ‖ q q + ∣ c 2 ∣ q ‖ u t ‖ q q .

From (2.5), we find

(3.6) ∂ ∂ t ∫ 0 1 ‖ y ( δ , t ) ‖ q q d δ = ∫ Ω ∫ 0 1 q ∣ y ( x , δ , t ) ∣ q − 2 y ( x , δ , t ) y t ( x , δ , t ) d δ d x = − 1 τ ∫ Ω ∫ 0 1 q ∣ y ( x , δ , t ) ∣ q − 2 y ( x , δ , t ) y δ ( x , δ , t ) d δ d x = − 1 τ ∫ Ω ∫ 0 1 ∂ ∂ δ ∣ y ( x , δ , t ) ∣ q d δ d x = − 1 τ ‖ y ( 1 , t ) ‖ q q + 1 τ ‖ y ( 0 , t ) ‖ q q = − 1 τ ‖ y ( 1 , t ) ‖ q q + 1 τ ‖ u t ( t ) ‖ q q .

From (3.4)–(3.6), one sees

(3.7) d d t E ( t ) ≤ − c 1 − ∣ c 2 ∣ q − ξ ‖ u t ‖ q q − ∫ Γ 1 h f z t 2 d Γ + 1 2 ( k ′ ♢ ∇ u ) − k ( t ) 2 ‖ ∇ u ‖ 2 2 − ξ − ∣ c 2 ∣ ( q − 1 ) q ‖ y ( 1 , t ) ‖ q q .

Letting

γ 1 = min c 1 − ∣ c 2 ∣ q − ξ , ξ − ∣ c 2 ∣ ( q − 1 ) q ,

we obtain (3.3) from (3.2).□

Now, we put

(3.8) F ( t ) = − E ( t ) ,

then

(3.9) F ′ ( t ) ≥ − E ′ ( t ) = γ 1 ( ‖ u t ‖ q q + ‖ y ( 1 , t ) ‖ q q ) + ∫ Γ 1 h f z t 2 d Γ − 1 2 ( k ′ ♢ ∇ u ) + k ( t ) 2 ‖ ∇ u ‖ 2 ≥ 0 .

If E ( 0 ) < 0 , one obtains from (3.9) and (3.1)

(3.10) 0 < F ( 0 ) ≤ F ( t ) ≤ 1 p ∫ Ω ∣ u ∣ p ln ∣ u ∣ d x ,

which gives

(3.11) ∫ Ω ∣ u ∣ p ln ∣ u ∣ d x ≥ 0 .

To obtain our main result, we use the following lemmas, which are introduced by Kafini and Messaoudi [8].

Lemma 3.2

Let ϕ ∈ V and r verify

2 ≤ r < ∞ i f n = 1 , 2 ; ≤ 2 n n − 2 i f n ≥ 3 .

Then, for any 2 ≤ s ≤ r , it holds

‖ ϕ ‖ r s ≤ C ( ‖ ϕ ‖ r r + ‖ ∇ ϕ ‖ 2 2 ) .

Lemma 3.3

Let ϕ ∈ V satisfy (3.11) and r verify

2 ≤ r < ∞ i f n = 1 , 2 ; ≤ n + 2 n − 2 i f n ≥ 3 .

Then, for any 2 ≤ s ≤ r , it holds

∫ Ω ∣ ϕ ∣ r ln ∣ ϕ ∣ d x s r ≤ C ‖ ∇ ϕ ‖ 2 2 + ∫ Ω ∣ ϕ ∣ r ln ∣ ϕ ∣ d x .

Lemma 3.4

Let r ≥ 2 . For ϕ ∈ L r ( Ω ) ∩ V with (3.11), it holds

‖ ϕ ‖ r r ≤ C ‖ ∇ ϕ ‖ 2 2 + ∫ Ω ∣ ϕ ∣ r ln ∣ ϕ ∣ d x .

Our main result is the following theorem:

Theorem 3.1

Let the conditions of Theorem 2.1, ( A n ) , ( A p ) , ( A q ) , and ( A k ) hold. If E ( 0 ) < 0 , the solution ( u , z , y ) to problems (2.4)–(2.10) blows up after a finite time T ∗ . Moreover, the blow-up time T ∗ is bounded as follows:

(3.12) ∫ Λ ( 0 ) + ∞ 1 2 ψ + d 1 ψ p − 1 + ω 1 + d 2 d ψ ≤ T ∗ ≤ 1 − β C β L β 1 − β ( 0 ) ,

where

Λ ( 0 ) = ‖ u 1 ‖ 2 2 + ‖ ∇ u 0 ‖ 2 2 + ∫ Γ 1 h g z 0 2 d Γ + 2 ξ τ ‖ y 0 ‖ L q ( Ω × ( 0 , 1 ) ) q ,

d 1 and d 2 are the constants given in (3.45), β satisfies (3.14), L ( 0 ) is given in (3.25), and C > 0 is a certain constant.

Proof

By following the idea of [14], we put

(3.13) L ( t ) = F 1 − β ( t ) + ε ( u , u t ) − ε 2 ∫ Γ 1 h f z 2 d Γ − ε ∫ Γ 1 h z u d Γ ,

where ε > 0 and

(3.14) 2 ( p − q ) p 2 ( q − 1 ) ≤ β ≤ min p − q p ( q − 1 ) , p − 2 2 p , m − 2 2 m ,

here m verifying

(3.15) 2 p 2 ( q − 1 ) p 2 ( q − 1 ) − 4 ( p − q ) < m < 4 p + 2 ( n − 1 ) ( p − 2 ) ( p − 2 ) n + 4 .

One can note that this choice of m and β is reasonable due to (2.17) and (2.16).

From (2.4)–(2.10) and Young’s inequality, we obtain

(3.16) L ′ ( t ) = ( 1 − β ) F − β ( t ) F ′ ( t ) + ε ‖ u t ‖ 2 2 − ε 1 − ∫ 0 t k ( s ) d s ‖ ∇ u ‖ 2 2 + ε ∫ Γ 1 h g z 2 d Γ + ε ∫ Ω ∣ u ∣ p ln ∣ u ∣ d x + ε ∫ 0 t k ( t − s ) ( ∇ u ( t ) , ∇ u ( s ) − ∇ u ( t ) ) d s − ε c 1 ∫ Ω u ∣ u t ∣ q − 2 u t d x − ε c 2 ∫ Ω u ∣ y ( x , 1 , t ) ∣ q − 2 y ( x , 1 , t ) d x ≥ ( 1 − β ) F − β ( t ) F ′ ( t ) + ε ‖ u t ‖ 2 2 + ε ∫ Γ 1 h g z 2 d Γ + ε ∫ Ω ∣ u ∣ p ln ∣ u ∣ d x − ε 1 − ∫ 0 t k ( s ) d s + 1 4 η ∫ 0 t k ( s ) d s ‖ ∇ u ‖ 2 2 − ε η ( k ♢ ∇ u ) − ε c 1 ∫ Ω u ∣ u t ∣ q − 2 u t d x − ε c 2 ∫ Ω u ∣ y ( x , 1 , t ) ∣ q − 2 y ( x , 1 , t ) d x for any η > 0 .

By subtracting and adding the term ε λ ∫ Ω ∣ u ∣ p ln ∣ u ∣ d x , where 0 < λ < p − 2 p , and using (3.1) and (3.8), we obtain

(3.17) L ′ ( t ) ≥ ( 1 − β ) F − β ( t ) F ′ ( t ) + ε 1 + p ( 1 − λ ) 2 ‖ u t ‖ 2 2 + ε ( 1 − λ ) p ‖ u ‖ p p + ε λ ∫ Ω ∣ u ∣ p ln ∣ u ∣ d x + ε 1 + p ( 1 − λ ) 2 ∫ Γ 1 h g z 2 d Γ + ε p ( 1 − λ ) 2 − η ( k ♢ ∇ u ) + ε p ( 1 − λ ) F ( t ) + ε p ( 1 − λ ) 2 − 1 − p ( 1 − λ ) 2 − 1 + 1 4 η ∫ 0 t k ( s ) d s ‖ ∇ u ‖ 2 2 − ε c 1 ∫ Ω u ∣ u t ∣ q − 2 u t d x − ε c 2 ∫ Ω u ∣ y ( x , 1 , t ) ∣ q − 2 y ( x , 1 , t ) d x .

By using (2.1) with q − 1 q + 1 q = 1 and 0 < ∣ c 2 ∣ < c 1 , taking δ 1 = ( θ F − β ( t ) ) − q − 1 q , θ > 0 , and applying (3.9), we derive

(3.18) c 1 ∫ Ω u ∣ u t ∣ q − 2 u t d x + c 2 ∫ Ω u ∣ y ( x , 1 , t ) ∣ q − 2 y ( x , 1 , t ) d x ≤ c 1 ( q − 1 ) q δ 1 − q q − 1 ‖ u t ‖ q q + ( c 1 + ∣ c 2 ∣ ) δ 1 q q ‖ u ‖ q q + ∣ c 2 ∣ ( q − 1 ) q δ 1 − q q − 1 ‖ y ( 1 , t ) ‖ q q ≤ c 1 ( q − 1 ) q δ 1 − q q − 1 ( ‖ u t ‖ q q + ‖ y ( 1 , t ) ‖ q q ) + ( c 1 + ∣ c 2 ∣ ) δ 1 q q ‖ u ‖ q q = c 1 ( q − 1 ) θ q ( F ( t ) ) − β ( ‖ u t ‖ q q + ‖ y ( 1 , t ) ‖ q q ) + c 1 + ∣ c 2 ∣ q θ q − 1 ( F ( t ) ) β ( q − 1 ) ‖ u ‖ q q ≤ c 1 ( q − 1 ) θ q γ 1 ( F ( t ) ) − β F ′ ( t ) + C q θ q − 1 ( F ( t ) ) β ( q − 1 ) ‖ u ‖ p q .

Thanks to ( A p ) , we note that the solution u to (2.4)–(2.10) belongs to L p + 1 ( Ω ) . By using (3.10), Lemma 3.4, and (2.1) with q p + p − q p = 1 , we find

(3.19) ( F ( t ) ) β ( q − 1 ) ‖ u ‖ p q ≤ C ∫ Ω ∣ u ∣ p ln ∣ u ∣ d x β ( q − 1 ) ∫ Ω ∣ u ∣ p ln ∣ u ∣ d x + ‖ ∇ u ‖ 2 2 q p ≤ C ∫ Ω ∣ u ∣ p ln ∣ u ∣ d x β ( q − 1 ) ∫ Ω ∣ u ∣ p ln ∣ u ∣ d x q p + ‖ ∇ u ‖ 2 2 q p ≤ C ∫ Ω ∣ u ∣ p ln ∣ u ∣ d x β ( q − 1 ) + q p + ∫ Ω ∣ u ∣ p ln ∣ u ∣ d x β ( q − 1 ) ‖ ∇ u ‖ 2 2 q p ≤ C ∫ Ω ∣ u ∣ p ln ∣ u ∣ d x β p ( q − 1 ) + q p + ∫ Ω ∣ u ∣ p ln ∣ u ∣ d x β ( q − 1 ) p p − q + ‖ ∇ u ‖ 2 2 .

From (3.14), we see 0 < β ≤ p − q p ( q − 1 ) , which gives

2 ≤ β p ( q − 1 ) + q ≤ p .

By applying Lemma 3.3, we derive

(3.20) ∫ Ω ∣ u ∣ p ln ∣ u ∣ d x β p ( q − 1 ) + q p ≤ C ‖ ∇ u ‖ 2 2 + ∫ Ω ∣ u ∣ p ln ∣ u ∣ d x .

Similarly, from (3.14), we see 2 ( p − q ) p 2 ( q − 1 ) ≤ β ≤ p − q p ( q − 1 ) , which implies

2 ≤ β ( q − 1 ) p 2 p − q ≤ p .

So, we have

(3.21) ∫ Ω ∣ u ∣ p ln ∣ u ∣ d x β ( q − 1 ) p p − q ≤ C ‖ ∇ u ‖ 2 2 + ∫ Ω ∣ u ∣ p ln ∣ u ∣ d x .

By inserting (3.20) and (3.21) to (3.19), we obtain

(3.22) ( F ( t ) ) β ( q − 1 ) ‖ u ‖ p q ≤ C ‖ ∇ u ‖ 2 2 + ∫ Ω ∣ u ∣ p ln ∣ u ∣ d x .

From this and (3.18),

(3.23) c 1 ∫ Ω u ∣ u t ∣ q − 2 u t d x + c 2 ∫ Ω u ∣ y ( x , 1 , t ) ∣ q − 2 y ( x , 1 , t ) d x ≤ c 1 ( q − 1 ) θ q γ 1 ( F ( t ) ) − β F ′ ( t ) + C q θ q − 1 ∫ Ω ∣ u ∣ p ln ∣ u ∣ d x + ‖ ∇ u ‖ 2 2 .

By applying this to (3.17), we arrive at

L ′ ( t ) ≥ ( 1 − β ) − ε c 1 ( q − 1 ) θ q γ 1 F − β ( t ) F ′ ( t ) + ε 1 + p ( 1 − λ ) 2 ‖ u t ‖ 2 2 + ε 1 + p ( 1 − λ ) 2 ∫ Γ 1 h g z 2 d Γ + ε λ − C q θ q − 1 ∫ Ω ∣ u ∣ p ln ∣ u ∣ d x + ε ( 1 − λ ) p ‖ u ‖ p p + p ( 1 − λ ) 2 − η ( k ♢ ∇ u ) + ε p ( 1 − λ ) F ( t ) + ε p ( 1 − λ ) 2 − 1 − p ( 1 − λ ) 2 − 1 + 1 4 η ∫ 0 t k ( s ) d s − C q θ q − 1 ‖ ∇ u ‖ 2 2 .

By taking η = p ( 1 − λ ) 2 and using the relation

p ( 1 − λ ) 2 − 1 − p ( 1 − λ ) 2 − 1 + 1 4 η ∫ 0 t k ( s ) d s = p ( 1 − λ ) 2 − 1 1 − ∫ 0 t k ( s ) d s − 1 2 p ( 1 − λ ) ∫ 0 t k ( s ) d s ≥ p ( 1 − λ ) 2 − 1 k l − 1 2 p ( 1 − λ ) ( 1 − k l ) ≕ a λ ,

we have

(3.24) L ′ ( t ) ≥ ( 1 − β ) − ε c 1 ( q − 1 ) θ q γ 1 F − β ( t ) F ′ ( t ) + ε 1 + p ( 1 − λ ) 2 ‖ u t ‖ 2 2 + ε 1 + p ( 1 − λ ) 2 ∫ Γ 1 h g z 2 d Γ + ε λ − C q θ q − 1 ∫ Ω ∣ u ∣ p ln ∣ u ∣ d x + ε ( 1 − λ ) p ‖ u ‖ p p + ε p ( 1 − λ ) F ( t ) + ε a λ − C q θ q − 1 ‖ ∇ u ‖ 2 2 .

Let

a = p 2 − 1 k l − 1 2 p ( 1 − k l ) ,

then

a > 0 , a λ < a , lim λ → 0 + a λ = a .

Thus, there exists 0 < λ 0 < p − 2 p such that

a λ > 0 for 0 < λ < λ 0 .

We fix θ > 0 appropriately large to obtain

a λ − C q θ q − 1 > 0 and λ − C q θ q − 1 > 0

and pick ε > 0 suitably small to have

( 1 − β ) − ε c 1 ( q − 1 ) θ q γ 1 > 0

and

(3.25) L ( 0 ) = F 1 − β ( 0 ) + ε ( u 0 , u 1 ) − ε 2 ∫ Γ 1 h f z 0 2 d Γ − ε ∫ Γ 1 h z 0 u 0 d Γ > 0 .

Whereupon

(3.26) L ′ ( t ) ≥ C ‖ u t ‖ 2 2 + ∫ Γ 1 h g z 2 d Γ + ‖ u ‖ p p + F ( t ) + ∫ Ω ∣ u ∣ p ln ∣ u ∣ d x + ‖ ∇ u ‖ 2 2 .

On the one hand, from (3.13), we know that

(3.27) L 1 1 − β ( t ) ≤ C F ( t ) + ∣ ( u , u t ) ∣ 1 1 − β + ∫ Γ 1 ∣ h ( x ) z u ∣ d Γ 1 1 − β .

By using (2.1) with 1 2 ( 1 − β ) + 1 − 2 β 2 ( 1 − β ) = 1 , noting 2 ≤ 2 1 − 2 β ≤ p from (3.14), and applying Lemma 3.2, we deduce

(3.28) ∣ ( u , u t ) ∣ 1 1 − β ≤ ‖ u ‖ 2 1 1 − β ‖ u t ‖ 2 1 1 − β ≤ C ‖ u ‖ p 1 1 − β ‖ u t ‖ 2 1 1 − β ≤ C ( ‖ u t ‖ 2 2 + ‖ u ‖ p 2 1 − 2 β ) ≤ C ( ‖ u t ‖ 2 2 + ‖ u ‖ p p + ‖ ∇ u ‖ 2 2 ) .

By recalling m > 2 and using (2.1) with 1 2 ( 1 − β ) + 1 − 2 β 2 ( 1 − β ) = 1 again, we have

(3.29) ∫ Γ 1 h z u d Γ 1 1 − β = ∫ Γ 1 h g z u g d Γ 1 1 − β ≤ ‖ h ‖ Γ 1 , ∞ ‖ g ‖ Γ 1 , ∞ g 0 2 1 2 ( 1 − β ) ∫ Γ 1 h g z 2 d Γ 1 2 ( 1 − β ) ∫ Γ 1 u 2 d Γ 1 2 ( 1 − β ) ≤ C ∫ Γ 1 h g z 2 d Γ 1 2 ( 1 − β ) ‖ u ‖ Γ 1 , m 1 1 − β ≤ C ∫ Γ 1 h g z 2 d Γ + ‖ u ‖ Γ 1 , m 2 1 − 2 β .

By noting 2 m ( 1 − 2 β ) ≤ 1 from the choice of β given in (3.14) and using (2.2) with μ 3 = 2 m ( 1 − 2 β ) and μ 4 = F ( 0 ) , we infer

(3.30) ‖ u ‖ Γ 1 , m 2 1 − 2 β = ∫ Γ 1 u m d Γ 2 m ( 1 − 2 β ) ≤ F ( 0 ) + 1 F ( 0 ) ∫ Γ 1 u m d Γ + F ( 0 ) ≤ C ( ‖ u ‖ Γ 1 , m m + F ( t ) ) .

Here, we used (3.10). By combining (3.27)–(3.30), we derive

(3.31) L 1 1 − β ( t ) ≤ C F ( t ) + ‖ u t ‖ 2 2 + ‖ u ‖ p p + ‖ ∇ u ‖ 2 2 + ∫ Γ 1 h g z 2 d Γ + ‖ u ‖ Γ 1 , m m .

Now, we estimate the last term in (3.31). From (3.15), (2.17), and (2.15), one knows

2 < m < 2 n + 2 n ,

which guarantees

(3.32) 0 < n 2 − n − 1 m < 2 m < 1 .

We let s ∈ [ n 2 − n − 1 m , 2 m ) , which is valid due to (3.32). By using the embedding H s ( Ω ) ↪ L m ( Γ 1 ) and the interpolation theorem, one sees

(3.33) ‖ u ‖ Γ 1 , m m ≤ C ‖ u ‖ H s ( Ω ) m ≤ C ‖ u ‖ 2 ( 1 − s ) m ‖ ∇ u ‖ 2 s m ≤ C ‖ u ‖ p ( 1 − s ) m ‖ ∇ u ‖ 2 s m .

By using (2.1) with s m 2 + 2 − s m 2 = 1 , we derive

(3.34) ‖ u ‖ Γ 1 , m m ≤ C ( ‖ ∇ u ‖ 2 2 + ( ‖ u ‖ p p ) 2 m ( 1 − s ) p ( 2 − s m ) ) .

From (3.15), it holds

n 2 − n − 1 m ≤ 2 ( p − m ) m ( p − 2 ) .

So, we can take s with n 2 − n − 1 m ≤ s ≤ 2 ( p − m ) m ( p − 2 ) ( < 2 m ) , which gives

0 < 2 m ( 1 − s ) p ( 2 − s m ) ≤ 1 .

Finally, by using (2.2) with μ 3 = 2 m ( 1 − s ) p ( 2 − s m ) and μ 4 = F ( 0 ) , we obtain

(3.35) ( ‖ u ‖ p p ) 2 m ( 1 − s ) p ( 2 − s m ) ≤ F ( 0 ) + 1 F ( 0 ) ( ‖ u ‖ p p + F ( 0 ) ) ≤ C ( ‖ u ‖ p p + F ( t ) ) .

From (3.31), (3.34), and (3.35), we conclude

(3.36) L 1 1 − β ( t ) ≤ C F ( t ) + ‖ u t ‖ 2 2 + ‖ u ‖ p p + ‖ ∇ u ‖ 2 2 + ∫ Γ 1 h g z 2 d Γ .

From (3.26) and (3.36), we conclude

(3.37) L ′ ( t ) ≥ C L 1 1 − β ( t ) , t ≥ 0 ,

which gives that the solution u blows up after finite time

(3.38) T ∗ ≤ 1 − β C β L β 1 − β ( 0 ) .

To show the left inequality of (3.12), we set

(3.39) Λ ( t ) = ‖ u t ‖ 2 2 + 1 − ∫ 0 t k ( s ) d s ‖ ∇ u ‖ 2 2 + ( k ♢ ∇ u ) + ∫ Γ 1 h g z 2 d Γ + 2 ξ τ ∫ 0 1 ‖ y ( δ , t ) ‖ q q d δ .

Thanks to (3.37), (3.13), (3.8), and (3.1), we have

(3.40) lim t → ( T ∗ ) − ∫ Ω ∣ u ∣ p ln ∣ u ∣ d x = + ∞ .

From (3.11), (2.3), and (2.18), we have

(3.41) 0 < ∫ Ω ∣ u ∣ p ln ∣ u ∣ d x = ∫ ∣ u ∣ < 1 ∣ u ∣ p ln ∣ u ∣ d x + ∫ ∣ u ∣ ≥ 1 ∣ u ∣ p ln ∣ u ∣ d x < 1 e ω 0 ∫ ∣ u ∣ ≥ 1 ∣ u ∣ p + ω 0 d x ≤ 1 e ω 0 ‖ u ‖ p + ω 0 p + ω 0 ≤ B p + ω 0 p + ω 0 e ω 0 ‖ ∇ u ‖ 2 p + ω 0 .

Due to (3.40) and (3.41), we obtain

(3.42) lim t → ( T ∗ ) − ‖ ∇ u ( t ) ‖ 2 = + ∞ ,

which gives

(3.43) lim t → ( T ∗ ) − Λ ( t ) ≥ lim t → ( T ∗ ) − k l ‖ ∇ u ( t ) ‖ 2 2 = + ∞ .

From (3.39), (3.1), (3.3), (2.1), (2.3), and (2.18), we obtain

(3.44) Λ ′ ( t ) = 2 E ′ ( t ) + d d t 1 p ∫ Ω ∣ u ∣ p ln ∣ u ∣ d x − 1 p 2 ‖ u ‖ p p ≤ − 2 γ 1 ( ‖ u t ‖ q q + ‖ y ( 1 , t ) ‖ q q ) + ( k ′ ♢ ∇ u ) − 2 ∫ Γ 1 h f z t 2 d Γ − k ( t ) ‖ ∇ u ‖ 2 2 + 2 ∫ Ω u t ∣ u ∣ p − 2 u ln ∣ u ∣ d x ≤ 2 ∫ Ω u t ∣ u ∣ p − 2 u ln ∣ u ∣ d x ≤ 2 ‖ u t ‖ 2 2 + ∫ Ω ∣ ∣ u ∣ p − 1 ln ∣ u ∣ ∣ 2 d x = 2 ‖ u t ‖ 2 2 + ∫ ∣ u ∣ ≥ 1 ∣ ∣ u ∣ p − 1 ln ∣ u ∣ ∣ 2 d x + ∫ ∣ u ∣ < 1 ∣ ∣ u ∣ p − 1 ln ∣ u ∣ ∣ 2 d x ≤ 2 ‖ u t ‖ 2 2 + 1 e 2 ω 1 2 ∫ ∣ u ∣ ≥ 1 ∣ u ∣ 2 ( p − 1 + ω 1 ) d x + ∣ Ω ∣ e 2 ( p − 1 ) 2 ≤ 2 ‖ u t ‖ 2 2 + B 2 ( p − 1 + ω 1 ) 2 ( p − 1 + ω 1 ) e 2 ω 1 2 ‖ ∇ u ‖ 2 2 ( p − 1 + ω 1 ) + ∣ Ω ∣ e 2 ( p − 1 ) 2 ≤ 2 Λ ( t ) + d 1 Λ p − 1 + ω 1 ( t ) + d 2 ,

where

(3.45) d 1 = B 2 ( p − 1 + ω 1 ) 2 ( p − 1 + ω 1 ) e 2 ω 1 2 k l p − 1 + ω 1 and d 2 = ∣ Ω ∣ e 2 ( p − 1 ) 2 .

From (3.44), we have

∫ 0 T ∗ Λ ′ ( t ) 2 Λ ( t ) + d 1 Λ p − 1 + ω 1 ( t ) + d 2 d t ≤ ∫ 0 T ∗ 1 d t ,

which gives

∫ Λ ( 0 ) + ∞ 1 2 ψ + d 1 ψ p − 1 + ω 1 + d 2 d ψ ≤ T ∗ .

Thus, we complete the proof.□

4 Conclusion

In this article, the author considered a viscoelastic equation with nonlinear damping and time delay terms, logarithmic source effect, and acoustic boundary conditions. By utilizing the energy method, he showed that the solution with negative initial energy blows up after a finite time under some suitable conditions. Furthermore, he estimated the upper and lower bounds of the blow-up time.

Acknowledgments

The author is grateful to the anonymous referees for the careful reading and their important comments.

  1. Funding information: This work was supported by Basic Science Research Program through the National Research Foundation of Korea (NRF) funded by the Ministry of Education (2020R1I1A3066250).

  2. Conflict of interest: The authors state no conflict of interest.

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Received: 2021-01-27
Revised: 2022-10-23
Accepted: 2023-03-07
Published Online: 2023-04-27

© 2023 the author(s), published by De Gruyter

This work is licensed under the Creative Commons Attribution 4.0 International License.

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https://doi.org/10.1515/anona-2022-0310
Keywords for this article
blow-up; viscoelastic equation; logarithmic nonlinearity; time delay; acoustic boundary condition; nonlinear dissipation
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Articles in the same Issue

  1. Regular Articles
  2. On sufficient “local” conditions for existence results to generalized p(.)-Laplace equations involving critical growth
  3. On the critical Choquard-Kirchhoff problem on the Heisenberg group
  4. On the local behavior of local weak solutions to some singular anisotropic elliptic equations
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  14. Bautin bifurcation with additive noise
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  17. A global compactness result with applications to a Hardy-Sobolev critical elliptic system involving coupled perturbation terms
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