Home General decay rate for a viscoelastic wave equation with distributed delay and Balakrishnan-Taylor damping
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General decay rate for a viscoelastic wave equation with distributed delay and Balakrishnan-Taylor damping

  • Abdelbaki Choucha , Salah Boulaaras EMAIL logo and Djamel Ouchenane
Published/Copyright: October 18, 2021

Abstract

A nonlinear viscoelastic wave equation with Balakrishnan-Taylor damping and distributed delay is studied. By the energy method we establish the general decay rate under suitable hypothesis.

MSC 2010: 35B40; 35L70; 76Exx; 93D20

1 Introduction

Let = Ω × ( τ 1 , τ 2 ) × ( 0 , ) , in the present work, we consider the following wave equation:

(1.1) u t t ( ζ 0 + ζ 1 u 2 2 + σ ( u , u t ) L 2 ( Ω ) ) Δ u ( t ) + α ( t ) 0 t h ( t ϱ ) Δ u ( ϱ ) d ϱ + β 1 u t ( t ) m 2 u t ( t ) + τ 1 τ 2 β 2 ( s ) u t ( t s ) m 2 u t ( t s ) d s = 0 , u ( x , 0 ) = u 0 ( x ) , u t ( x , 0 ) = u 1 ( x ) , in Ω , u t ( x , t ) = f 0 ( x , t ) , in Ω × ( 0 , τ 2 ) , u ( x , t ) = 0 , in Ω × ( 0 , ) ,

where Ω R N is a bounded domain with sufficiently smooth boundary Ω . ζ 0 , ζ 1 , σ , β 1 are positive constants, m 1 for N = 1 , 2 , and 1 < m N + 2 N 2 for N 3 .

τ 1 < τ 2 are non-negative constants such that β 2 : [ τ 1 , τ 2 ] R represents distributive time delay, h , α are positive functions.

The importance of the viscoelastic properties of materials has been realized because of the accelerated development of the rubber and plastics industry. Physically, the viscoelastic damping term is the relation-ship between the stress and strain history in the beam inspired by the Boltzmann theory, where the kernel of the term of memory is the function h , see [1,2, 3,4,5, 6,7].

In [8], Balakrishnan and Taylor proposed a new model of damping and called it the Balakrishnan-Taylor damping, as it relates to the span problem and the plate equation. For more details, the readers can refer to some papers that focused on the study of this damping [9,10,11, 12,13].

On the other hand, the stability issue of systems with delay is of theoretical and practical great importance, whereas, the dynamic systems with delay terms have become a major research subject in differential equation since the last five decades. Recently, the stability and the asymptotic behavior of evolution systems with time delay especially the distributed delay effect have been studied by many authors, see [14,15,16, 17,18].

Very recently, in [19] the authors considered our problem (1.1) but in the presence of the delay, they proved the general decay result of solutions by the energy method under suitable assumptions.

Based on all of the above, the combination of these terms of damping (memory term, Balakrishnan-Taylor damping, and the distributed delay ) in one particular problem with the addition of α ( t ) to the term of memory and the distributed delay term τ 1 τ 2 β 2 ( s ) u t ( t s ) m 2 u t ( t s ) d s we believe that it constitutes a new problem worthy of study and research, different from the above that we will try to shed light on.

Our paper is divided into several sections: in Section 2 we lay down the hypotheses, concepts, and lemmas we need, and in Section 3 we prove our main result. Finally, we give a conclusion in Section 4.

2 Preliminaries

For studying our problem, in this section we need some materials.

First, introducing the following hypothesis for β 2 , h , and α :

  1. h , α : R + R + are non-increasing C 1 functions satisfying

    (2.1) h ( t ) > 0 , α ( t ) > 0 , l 0 = 0 h ( ϱ ) d ϱ < , ζ 0 2 α ( t ) 0 t h ( ϱ ) d ϱ l > 0 ;

  2. ϑ : R + R + is a non-increasing C 1 function satisfying

    (2.2) ϑ ( t ) h ( t ) + h ( t ) 0 , t 0 and lim t α ( t ) ϑ ( t ) α ( t ) = 0 ;

  3. β 2 : [ τ 1 , τ 2 ] R is a bounded function satisfying

    (2.3) τ 1 τ 2 β 2 ( s ) d s < β 1 .

Let us introduce

( h ψ ) ( t ) Ω 0 t h ( t ϱ ) ψ ( t ) ψ ( ϱ ) 2 d ϱ d x

and

M ( t ) ( ζ 0 + ζ 1 u 2 2 + σ ( u ( t ) , u t ( t ) ) L 2 ( Ω ) ) .

Lemma 2.1

(Sobolev-Poincare inequality [20]). Let 2 q < ( n = 1 , 2 ) or 2 q < 2 n n 2 ( n 3 ) . Then, c = c ( Ω , q ) > 0 such that

u q c u 2 , u H 0 1 ( Ω ) .

As in [18], taking the following new variables:

y ( x , ρ , s , t ) = u t ( x , t s ρ ) ,

which satisfy

(2.4) s y t ( x , ρ , s , t ) + y ρ ( x , ρ , s , t ) = 0 , y ( x , 0 , s , t ) = u t ( x , t ) .

So, problem (1.1) can be written as

(2.5) u t t ( ζ 0 + ζ 1 u 2 2 + σ ( u , u t ) L 2 ( Ω ) ) Δ u ( t ) + α ( t ) 0 t h ( t ϱ ) Δ u ( ϱ ) d ϱ + β 1 u t ( t ) m 2 u t ( t ) + τ 1 τ 2 β 2 ( s ) y ( x , 1 , s , t ) m 2 y ( x , 1 , s , t ) d s = 0 , s y t ( x , ρ , s , t ) + y ρ ( x , ρ , s , t ) = 0 , u ( x , 0 ) = u 0 ( x ) , u t ( x , 0 ) = u 1 ( x ) , in Ω , y ( x , ρ , s , 0 ) = f 0 ( x , ρ s ) , in Ω × ( 0 , 1 ) × ( 0 , τ 2 ) , u ( x , t ) = 0 , in Ω × ( 0 , ) ,

where

( x , ρ , s , t ) Ω × ( 0 , 1 ) × ( τ 1 , τ 2 ) × ( 0 , ) .

Now, we give the energy functional.

Lemma 2.2

The energy functional E , defined by

(2.6) E ( t ) = 1 2 u t 2 2 + 1 2 ζ 0 α ( t ) 0 t h ( ϱ ) d ϱ u ( t ) 2 2 + ζ 1 4 u ( t ) 2 4 + α ( t ) 2 ( h u ) ( t ) + m 1 m 0 1 τ 1 τ 2 s β 2 ( s ) y ( x , ρ , s , t ) m m d s d ρ ,

satisfies

(2.7) E ( t ) β 1 τ 1 τ 2 β 2 ( s ) d s u t ( t ) m m + α ( t ) 2 ( h u ) ( t ) α ( t ) 2 0 t h ( ϱ ) d ϱ u ( t ) 2 2 σ 4 d d t u ( t ) 2 2 2 .

Proof

Taking the inner product of ( 2.5 ) 1 with u t , then integrating over Ω , we find

(2.8) ( u t t ( t ) , u t ( t ) ) L 2 ( Ω ) ( M ( t ) Δ u ( t ) , u t ( t ) ) L 2 ( Ω ) + ( α ( t ) 0 t h ( t ϱ ) Δ u ( ϱ ) d ϱ , u t ( t ) ) L 2 ( Ω ) + β 1 ( u t m 2 u t , u t ) L 2 ( Ω ) + τ 1 τ 2 β 2 ( s ) ( y ( x , 1 , s , t ) m 2 y ( x , 1 , s , t ) , u t ( t ) ) L 2 ( Ω ) d s = 0 .

By computation, integration by parts and the last condition in (2.5), we get

(2.9) ( u t t ( t ) , u t ( t ) ) L 2 ( Ω ) = 1 2 d d t u t ( t ) 2 2 ,

by integration by parts, we find

(2.10) ( M ( t ) Δ u ( t ) , u t ( t ) ) L 2 ( Ω ) = ( ( ζ 0 + ζ 1 u 2 2 + σ ( u ( t ) , u t ( t ) ) L 2 ( Ω ) ) Δ u ( t ) , u t ( t ) ) L 2 ( Ω ) = ( ζ 0 + ζ 1 u 2 2 + σ ( u ( t ) , u t ( t ) ) L 2 ( Ω ) ) Ω u ( t ) u t ( t ) d x = ( ζ 0 + ζ 1 u 2 2 + σ ( u ( t ) , u t ( t ) ) L 2 ( Ω ) ) d d t Ω u ( t ) 2 d x = d d t 1 2 ζ 0 + ζ 1 2 u 2 2 u ( t ) 2 2 + σ 4 d d t { u ( t ) 2 2 } 2 ,

and we have

(2.11) 0 t h ( t ϱ ) Δ u ( ϱ ) d ϱ , u t ( t ) L 2 ( Ω ) = 0 t h ( t ϱ ) ( Δ u ( ϱ ) , u t ( t ) ) L 2 ( Ω ) d ϱ = 0 t h ( t ϱ ) Ω u ( x , ϱ ) u ( x , t ) d x d ϱ

and

(2.12) u ( x , ϱ ) u ( x , t ) = 1 2 d d t { u ( x , ϱ ) u ( x , t ) ( t ) 2 } 1 2 d d t u ( x , t ) 2 ,

then

(2.13) 0 t h ( t ϱ ) ( u ( ϱ ) , u t ( t ) ) L 2 ( Ω ) d ϱ = 0 t h ( t ϱ ) Ω 1 2 d d t { u ( x , ϱ ) u ( x , t ) 2 } d x d ϱ , 0 t h ( t ϱ ) Ω 1 2 d d t { u ( x , t ) 2 } d x d ϱ = 1 2 0 t h ( t ϱ ) d d t Ω u ( x , t ) u ( x , ϱ ) 2 d x d ϱ 1 2 0 t h ( t ϱ ) d d t u ( x , t ) 2 2 d x d ϱ ,

by (2.1), we find

(2.14) 1 2 0 t h ( t ϱ ) d d t Ω u ( x , t ) u ( x , ϱ ) 2 d x d ϱ = 1 2 d d t 0 t h ( t ϱ ) Ω u ( x , t ) u ( x , ϱ ) 2 d x d ϱ 1 2 0 t h ( t ϱ ) Ω u ( x , t ) u ( x , ϱ ) 2 d x d ϱ = 1 2 d d t ( h u ) ( t ) 1 2 ( h u ) ( t )

and

(2.15) 1 2 0 t h ( t ϱ ) d d t { u ( t ) 2 2 } d x d ϱ = 1 2 0 t h ( t ϱ ) d ϱ d d t u ( t ) 2 2 d x = 1 2 0 t h ( ϱ ) d ϱ d d t u ( t ) 2 2 d x = 1 2 d d t 0 t h ( ϱ ) d ϱ u ( t ) 2 2 + 1 2 h ( t ) u ( t ) 2 2 ,

by inserting (2.14) and (2.15) into (2.13), we find

(2.16) α ( t ) 0 t h ( t ϱ ) Δ u ( ϱ ) d ϱ , u t ( t ) L 2 ( Ω ) = d d t α ( t ) 2 ( h u ) ( t ) α ( t ) 2 0 t h ( ϱ ) d ϱ u ( t ) 2 2 α ( t ) 2 ( h u ) ( t ) + α ( t ) 2 h ( t ) u ( t ) 2 2 α ( t ) 2 ( h u ) ( t ) + α ( t ) 2 0 t h ( ϱ ) d ϱ u ( t ) 2 2 .

Now, multiplying the equation ( 2.5 ) 2 by y β 2 ( s ) , integrating over Ω × ( 0 , 1 ) × ( τ 1 , τ 2 ) , and using ( 2.4 ) 2 , we get

(2.17) d d t m 1 m Ω 0 1 τ 1 τ 2 s β 2 ( s ) y ( x , ρ , s , t ) m d s d ρ d x = ( m 1 ) Ω 0 1 τ 1 τ 2 β 2 ( s ) y m 1 y ρ d s d ρ d x = m 1 m Ω 0 1 τ 1 τ 2 β 2 ( s ) d d ρ y ( x , ρ , s , t ) m d s d ρ d x = m 1 m Ω τ 1 τ 2 β 2 ( s ) ( y ( x , 0 , s , t ) m y ( x , 1 , s , t ) m ) d s d x = m 1 m τ 1 τ 2 β 2 ( s ) d s Ω u t ( t ) m d x m 1 m Ω τ 1 τ 2 β 2 ( s ) y ( x , 1 , s , t ) m d s d x = m 1 m τ 1 τ 2 β 2 ( s ) d s u t ( t ) m m m 1 m τ 1 τ 2 β 2 ( s ) y ( x , 1 , s , t ) m m d s ,

and by Young’s inequality, we have

(2.18) τ 1 τ 2 β 2 ( s ) ( y ( x , 1 , s , t ) m 2 y ( x , 1 , s , t ) , u t ( t ) ) L 2 ( Ω ) d s 1 m τ 1 τ 2 β 2 ( s ) d s u t ( t ) m m + m 1 m τ 1 τ 2 β 2 ( s ) y ( x , 1 , s , t ) m m d s .

By inserting (2.9)–(2.10) and (2.16)–(2.18) into (2.8), we obtain (2.6) and (2.7).

Hence, by (2.2), we get the function E is a non-increasing t t 1 . This completes the proof.□

Now we state the local existence of problem (2.5).

Theorem 2.3

Suppose that (2.1)–(2.3) are satisfied. Then, for any u 0 , u 1 H 0 1 ( Ω ) L 2 ( Ω ) , and f 0 L 2 ( Ω , ( 0 , 1 ) , ( τ 1 , τ 2 ) ) , there exists a weak solution u of problem (2.5) such that

u C ( ] 0 , T [ , H 0 1 ( Ω ) ) C 1 ( ] 0 , T [ , L 2 ( Ω ) ) , u t C ( ] 0 , T [ , H 0 1 ( Ω ) ) L 2 ( ] 0 , T [ , L 2 ( Ω , ( 0 , 1 ) , ( τ 1 , τ 2 ) ) ) .

3 General decay

In this section, we state and prove the asymptotic behavior of system (2.5). For this goal, we set

(3.1) Ψ ( t ) Ω u ( t ) u t ( t ) d x + σ 4 u ( t ) 2 4 ,

and

(3.2) Φ ( t ) Ω u t 0 t h ( t ϱ ) ( u ( t ) u ( ϱ ) ) d ϱ d x ,

and

(3.3) Θ ( t ) 0 1 τ 1 τ 2 s e ρ s β 2 ( s ) y ( x , ρ , s , t ) m m d s d ρ .

Lemma 3.1

The functional Ψ ( t ) defined in (3.1) satisfies, for any ε > 0

(3.4) Ψ ( t ) u t 2 2 ( l ε ( c 1 + c 2 ) ) u 2 2 ζ 1 u 2 4 + α ( t ) 4 ( h u ) ( t ) + c ( ε ) u t m m + τ 1 τ 2 β 2 ( s ) y ( x , 1 , s , t ) m m d s .

Proof

A differentiation of (3.1) and using ( 2.5 ) 1 gives

(3.5) Ψ ( t ) = u t 2 2 + Ω u t t u d x + σ u 2 2 Ω u t u d x = u t 2 2 ζ 0 u 2 2 ζ 1 u 2 4 β 1 Ω u t m 2 u t u d x I 1 + α ( t ) Ω u ( t ) 0 t h ( t ϱ ) u ( ϱ ) d ϱ d x I 2 Ω τ 1 τ 2 β 2 ( s ) y ( x , 1 , s , t ) m 2 y ( x , 1 , s , t ) u d s d x I 3 .

We estimate the last three terms of the right-hand side (RHS) of (3.5). Applying Hölder’s, Sobolev-Poincare’s, and Young’s inequalities, (2.1) and (2.6), we find

(3.6) I 1 ε β 1 m u m m + c ( ε ) u t m m ε β 1 m c p m u 2 m + c ( ε ) u t m m ε β 1 m c p m E ( 0 ) l ( m 2 ) / 2 u 2 2 + c ( ε ) u t m m ε c 1 u 2 2 + c ( ε ) u t m m

and

(3.7) I 2 2 α ( t ) 0 t h ( ϱ ) d ϱ u 2 2 + α ( t ) 4 ( h u ) ( t ) ( ζ 0 l ) u 2 2 + α ( t ) 4 ( h u ) ( t ) .

Similar to I 1 , we have

(3.8) I 3 ε c 2 u 2 2 + c ( ε ) τ 1 τ 2 β 2 ( s ) y ( x , 1 , s , t ) m m d s .

Combining (3.6)–(3.8) and (3.5), we get

Ψ ( t ) u t 2 2 ( l ε ( c 1 + c 2 ) ) u 2 2 ζ 1 u 2 4 + α ( t ) 4 ( h u ) ( t ) + c ( ε ) u t m m + τ 1 τ 2 β 2 ( s ) y ( x , 1 , s , t ) m m d s .

Lemma 3.2

The functional Φ ( t ) defined in (3.2) satisfies, for any δ > 0

(3.9) Φ ( t ) 0 t h ( ϱ ) d ϱ δ u t 2 2 + δ ( ζ 0 + 2 ( 1 l ) 2 α ( t ) ) u 2 2 + ζ 1 δ u 2 4 + δ σ E ( 0 ) l 1 2 d d t u 2 2 2 + c ( δ ) + 2 δ + 1 4 δ ( ζ 0 l ) α ( t ) ( h u ) ( t ) + c ( δ ) u t m m + τ 1 τ 2 β 2 ( s ) y ( x , 1 , s , t ) m m d s g ( 0 ) c p 2 4 δ ( h u ) ( t ) .

Proof

A differentiation of (3.2) and using ( 2.5 ) 1 gives

(3.10) Φ ( t ) = Ω u t t 0 t h ( t ϱ ) ( u ( t ) u ( ϱ ) ) d ϱ d x Ω u t 0 t h ( t ϱ ) ( u ( t ) u ( ϱ ) ) d ϱ d x 0 t h ( ϱ ) d ϱ u t 2 2 = ( ζ 0 + ζ 1 u 2 2 ) Ω u 0 t h ( t ϱ ) ( u ( t ) u ( ϱ ) ) d ϱ d x J 1 + σ Ω u u t d x Ω u 0 t h ( t ϱ ) ( u ( t ) u ( ϱ ) ) d ϱ d x J 2

α ( t ) Ω 0 t h ( t ϱ ) u ( ϱ ) d ϱ 0 t h ( t ϱ ) ( u ( t ) u ( ϱ ) ) d ϱ d x J 3 β 1 Ω u t m 2 u t 0 t h ( t ϱ ) ( u ( t ) u ( ϱ ) ) d ϱ d x J 4 Ω τ 1 τ 2 β 2 ( s ) y ( x , 1 , s , t ) m 2 y ( x , 1 , s , t ) 0 t h ( t ϱ ) ( u ( t ) u ( ϱ ) ) d ϱ d ϱ d x J 5 Ω u t 0 t h ( t ϱ ) ( u ( t ) u ( ϱ ) ) d ϱ d x J 6 0 t h ( ϱ ) d ϱ u t 2 2 .

We estimate the terms J i , i = 1 , , 6 of the RHS of (3.10). Applying Hölder’s, Sobolev-Poincare’s, and Young’s inequalities, (2.1) and (2.6), we find

(3.11) J 1 ( ζ 0 + ζ 1 u 2 2 ) δ u 2 2 + ( ζ 0 l ) 4 δ ( h u ) ( t ) δ ζ 0 u 2 2 + δ ζ 1 u 2 4 + ζ 0 ( ζ 0 l ) 4 δ + ζ 1 ( ζ 0 l ) E ( 0 ) 4 l δ ( h u ) ( t )

and

(3.12) J 2 δ σ Ω u u t d x 2 u 2 2 + σ ( ζ 0 l ) 4 δ ( h u ) ( t ) δ σ E ( 0 ) l 1 2 d d t u 2 2 2 + σ ( ζ 0 l ) 4 δ ( h u ) ( t ) ,

(3.13) J 3 δ α ( t ) Ω 0 t h ( t ϱ ) ( u ( t ) u ( ϱ ) u ( t ) ) d ϱ 2 d x + 1 4 δ α ( t ) Ω 0 t h ( t ϱ ) ( u ( t ) u ( ϱ ) ) d ϱ 2 d x 2 δ c ( ζ 0 l ) 2 α ( t ) u 2 2 + c 2 δ + 1 4 δ ( ζ 0 l ) α ( t ) ( h u ) ( t ) ,

(3.14) J 4 c ( δ ) u t m m + δ β 1 m Ω 0 t h ( t ϱ ) ( u ( t ) u ( ϱ ) ) d ϱ m d x c ( δ ) u t m m + δ β 1 m ( ζ 0 l ) m 1 c p m 0 t h ( t ϱ ) u ( t ) u ( ϱ ) 2 m d ϱ c ( δ ) u t m m + δ β 1 m ( ζ 0 l ) m 1 c p m E ( 0 ) l ( m 2 ) / 2 ( h u ) ( t ) c ( δ ) u t m m + δ c 3 ( h u ) ( t ) .

Similarly, we have

(3.15) J 5 c ( δ ) y ( x , 1 , s , t ) m m + δ c 4 ( h u ) ( t ) ,

(3.16) J 6 δ u t 2 2 h ( 0 ) c p 2 4 δ ( h u ) ( t ) .

By substituting of (3.11)–(3.16) into (3.10), we get

Φ ( t ) 0 t h ( ϱ ) d ϱ δ u t 2 2 + δ ( ζ 0 + 2 c ( ζ 0 l ) 2 α ( t ) ) u 2 2 + ζ 1 δ u 2 4 + δ σ E ( 0 ) l 1 2 d d t u 2 2 2 + c ( δ ) + 2 δ + 1 4 δ ( ζ 0 l ) α ( t ) ( h u ) ( t ) + c ( δ ) u t m m + τ 1 τ 2 β 2 ( s ) y ( x , 1 , s , t ) m m d s h ( 0 ) c p 2 4 δ ( h u ) ( t ) .

Lemma 3.3

The functional Θ ( t ) defined in (3.3) satisfies

(3.17) Θ ( t ) η 1 0 1 τ 1 τ 2 s β 2 ( s ) y ( x , ρ , s , t ) m m d s d ρ η 1 τ 1 τ 2 β 2 ( s ) y ( x , 1 , s , t ) m m d s + β 1 u t ( t ) m m .

Proof

By differentiating Θ ( t ) , and using ( 2.5 ) 2 , we have

Θ ( t ) = m Ω 0 1 τ 1 τ 2 e s ρ β 2 ( s ) y m 1 y ρ ( x , ρ , s , t ) d s d ρ d x = Ω 0 1 τ 1 τ 2 s e s ρ β 2 ( s ) y ( x , ρ , s , t ) m d s d ρ d x Ω τ 1 τ 2 β 2 ( s ) [ e s y ( x , 1 , s , t ) m y ( x , 0 , s , t ) m ] d s d x .

Applying y ( x , 0 , s , t ) = u t ( x , t ) , e s e s ρ 1 , for any 0 < ρ < 1 , and η 1 = e τ 2 , we obtain

Θ ( t ) η 1 Ω 0 1 τ 1 τ 2 s β 2 ( s ) y ( x , ρ , s , t ) m d s d ρ d x η 1 Ω τ 1 τ 2 β 2 ( s ) y ( x , 1 , s , t ) m d s d x + τ 1 τ 2 β 2 ( s ) d s Ω u t m ( t ) d x .

Using (2.3), we find (3.17).□

Now, we introduce the functional

(3.18) G ( t ) E ( t ) + ε 1 α ( t ) Ψ ( t ) + ε 2 α ( t ) Φ ( t ) + ε 3 α ( t ) Θ ( t ) ,

for some positive constants ε i , i = 1 , 2 , 3 to be determined.

Lemma 3.4

There exist μ 1 , μ 2 > 0 , such that

(3.19) μ 1 E ( t ) G ( t ) μ 2 E ( t ) .

Proof

From (3.1)–(3.3), by using Hölder’s, Young’s, and Poincare’s inequalities, we get

(3.20) G ( t ) E ( t ) ε 1 α ( t ) 2 ( u t ( t ) 2 2 + c p u ( t ) 2 2 ) + ε 1 σ α ( t ) 4 u ( t ) 2 4 + ε 2 α ( t ) 2 u t ( t ) 2 2 + ε 2 α ( t ) 2 c p ( ζ 0 l ) ( h u ) ( t ) + ε 3 α ( t ) 0 1 τ 1 τ 2 s e ρ s β 2 ( s ) y ( x , ρ , s , t ) m m d s d ρ .

Using the fact that 0 < α ( t ) α ( 0 ) and e ρ s < 1 , we find

(3.21) G ( t ) E ( t ) ε 1 α ( 0 ) 2 ( c p u ( t ) 2 2 + u t ( t ) 2 2 ) + ε 1 σ α ( 0 ) 4 u ( t ) 2 4 + ε 2 α ( 0 ) 2 u t ( t ) 2 2 + ε 2 α ( 0 ) 2 c p ( ζ 0 l ) ( h u ) ( t ) + ε 3 α ( 0 ) 0 1 τ 1 τ 2 s e ρ s β 2 ( s ) y ( x , ρ , s , t ) m m d s d ρ C ( ε 1 , ε 2 , ε 3 ) E ( t ) .

Choosing ε 1 , ε 2 , and ε 3 sufficiently small, then (3.19) follows from (3.21).□

Lemma 3.5

There exist d 5 , d 6 , t 0 > 0 satisfying

(3.22) G ( t ) d 5 α ( t ) E ( t ) + d 6 α ( t ) ( h u ) ( t ) , t > t 0 .

Proof

Since the function h is a positive and continuous, for all t 0 > 0 , we have

0 t h ( ϱ ) d ϱ 0 t 0 h ( ϱ ) d ϱ h 0 , t t 0 .

By differentiating (3.18), using (2.7) and Lemmas 3.13.3, we get

(3.23) G ( t ) E ( t ) + ε 1 α ( t ) Ψ ( t ) + ε 2 α ( t ) Φ ( t ) + ε 3 α ( t ) Θ ( t ) + ε 1 α ( t ) Ψ ( t ) + ε 2 α ( t ) Φ ( t ) + ε 3 α ( t ) Θ ( t ) α ( t ) ( ε 1 ε 2 ( h 0 δ ) ) u t 2 2 + α ( t ) ( ε 2 δ ( ζ 0 + 2 c ( ζ 0 l ) 2 α ( t ) ) ε 1 ( l ε ( c 1 + c 2 ) ) ) u 2 2 + α ( t ) ( ε 2 ζ 1 δ ε 1 ζ 1 ) u 2 4 + α ( t ) ε 2 δ σ E ( 0 ) l σ α ( 0 ) 1 2 d d t u 2 2 2 + α ( t ) ε 1 α ( t ) 4 + ε 2 ( c ( δ ) + 2 δ + 1 4 δ ( ζ 0 l ) α ( t ) ) ( h u ) ( t ) + α ( t ) 1 2 ε 2 h ( 0 ) c p 2 4 δ ( h u ) ( t ) + α ( t ) ε 1 c ( ε ) + ε 2 c ( δ ) + ε 3 β 1 η 0 α ( 0 ) u t m m + α ( t ) ( ε 1 c ( ε ) + ε 2 c ( δ ) η 1 ε 3 ) τ 1 τ 2 β 2 ( s ) y ( x , 1 , s , t ) m m d s α ( t ) ε 3 η 1 0 1 τ 1 τ 2 s β 2 ( s ) y ( x , ρ , s , t ) m m d s d ρ α ( t ) 2 0 t h ( ϱ ) d ϱ u ( t ) 2 2 + ε 1 α ( t ) Ω u u t d x + ε 2 α ( t ) Ω u t 0 t h ( t ϱ ) ( u ( t ) u ( ϱ ) ) d ϱ d x + ε 3 α ( t ) 0 1 τ 1 τ 2 s e ρ s β 2 ( s ) y ( x , ρ , s , t ) m m d s d ρ .

By using the fact that e ρ s < 1 , Young’s and Sobolev-Poincare’s inequalities, we find

α ( t ) Ω u u t d x + α ( t ) Ω u t 0 t h ( t ϱ ) ( u ( t ) u ( ϱ ) ) d ϱ d x + α ( t ) 0 1 τ 1 τ 2 s e ρ s β 2 ( s ) y ( x , ρ , s , t ) m m d s d ρ α ( t ) c p 2 2 u 2 2 α ( t ) u t 2 2 α ( t ) c p 2 2 0 t h ( ϱ ) d ϱ h u ( t ) α ( t ) 0 1 τ 1 τ 2 s β 2 ( s ) y ( x , ρ , s , t ) m m d s d ρ .

Hence,

(3.24) G ( t ) α ( t ) ε 1 ε 2 ( h 0 δ ) α ( t ) α ( t ) u t 2 2 + α ( t ) ( ε 2 δ ( ζ 0 + 2 c ( ζ 0 l ) 2 α ( t ) ) ε 1 ( l ε ( c 1 + c 2 ) ) α ( t ) 2 α ( t ) 0 t h ( ϱ ) d ϱ ε 1 α ( t ) c p 2 2 α ( t ) u 2 2 + α ( t ) ( ε 2 ζ 1 δ ε 1 ζ 1 ) u 2 4 + α ( t ) ε 2 δ σ E ( 0 ) l σ α ( 0 ) 1 2 d d t u 2 2 2 + α ( t ) ε 1 α ( t ) 4 + ε 2 ( c ( δ ) + 2 δ + 1 4 δ ( ζ 0 l ) α ( t ) ) ε 2 α ( t ) c p 2 2 α ( t ) 0 t h ( ϱ ) d ϱ ( h u ) ( t ) + α ( t ) 1 2 ε 2 h ( 0 ) c p 2 4 δ ( h u ) ( t ) + α ( t ) ε 1 c ( ε ) + ε 2 c ( δ ) + ε 3 β 1 η 0 α ( 0 ) u t m m + α ( t ) ( ε 1 c ( ε ) + ε 2 c ( δ ) η 1 ε 3 ) τ 1 τ 2 β 2 ( s ) y ( x , 1 , s , t ) m m d s + α ( t ) ε 3 η 1 α ( t ) α ( t ) 0 1 τ 1 τ 2 s β 2 ( s ) y ( x , ρ , s , t ) m m d s d ρ .

Choosing δ , ε so small that

h 0 δ > 1 2 h 0 , δ ( l ε ( c 1 + c 2 ) ) ( ζ 0 + 2 ( 1 l ) 2 ) < 1 4 h 0 .

For any fixed δ > 0 , we select ε 1 , ε 2 so small satisfying

h 0 4 ε 2 < ε 1 < h 0 2 ε 2 ,

and

ε 2 ( h 0 δ ) ε 1 > 0 , ε 1 ( l ε ( c 1 + c 2 ) ) ε 2 δ ( ζ 0 + 2 c ( ζ 0 l ) 2 α ( t ) ) > 0 .

Then, we select ε 1 , ε 2 , and ε 3 so small that (3.19) and (3.24) remain valid, and further,

ζ 1 ( ε 1 ε 2 δ ) > 0 , σ α ( 0 ) ε 2 δ σ E ( 0 ) l > 0 , 1 2 ε 2 h ( 0 ) c p 2 4 δ > 0 , η 0 α ( 0 ) ε 1 c ( ε ) ε 2 c ( δ ) ε 3 β 1 > 0 , η 1 ε 3 ε 1 c ( ε ) ε 2 c ( δ ) > 0 ,

where η 0 = β 1 τ 1 τ 2 β 2 ( s ) d s > 0 .

Therefore, (3.24) becomes, for positive constants d 1 , d 2 , d 3 , d 6 ,

(3.25) G ( t ) α ( t ) d 1 + α ( t ) α ( t ) u t 2 2 α ( t ) d 3 u 2 4 α ( t ) d 2 + α ( t ) 2 α ( t ) 0 t h ( ϱ ) d ϱ + ε 1 α ( t ) c p 2 2 α ( t ) u 2 2 + α ( t ) d 6 ε 2 h 0 α ( t ) c p 2 2 α ( t ) ( h u ) ( t ) α ( t ) ε 3 η 1 α ( t ) α ( t ) 0 1 τ 1 τ 2 ϱ β 2 ( s ) y ( x , ρ , s , t ) m m d s d ρ .

According to (2.2), lim t α ( t ) α ( t ) = 0 , we can choose t 1 > t 0 so that (3.25) can be written as

(3.26) G ( t ) α ( t ) d 1 u t 2 2 + d 3 u 2 4 + d 2 u 2 2 d 6 ( h u ) ( t ) + d 4 0 1 τ 1 τ 2 s β 2 ( s ) y ( x , ρ , s , t ) m m d s d ρ α ( t ) d 5 E ( t ) + α ( t ) d 6 ( h u ) ( t ) , t t 1 .

Theorem 3.6

Suppose that (2.1)–(2.3) are satisfied and E ( 0 ) > 0 for any ( u 0 , u 1 , f 0 ) . Then, the energy E ( t ) of (2.5) decays to zero exponentially. That is, λ 1 , λ 2 > 0 such that

(3.27) E ( t ) λ 1 e λ 2 t 1 t α ( ϱ ) ϑ ( ϱ ) d ϱ , t t 1 .

Proof

Multiplying (3.22) by ϑ ( t ) , using (2.1) and (2.7), we find

(3.28) ϑ ( t ) G ( t ) d 5 ϑ ( t ) α ( t ) E ( t ) + d 6 α ( t ) ϑ ( t ) ( h u ) ( t ) d 5 ϑ ( t ) α ( t ) E ( t ) d 6 α ( t ) ( h u ) ( t ) d 5 ϑ ( t ) α ( t ) E ( t ) d 6 2 E ( t ) α ( t ) 0 t h ( ϱ ) d ϱ u ( t ) 2 2 .

Since ϑ ( t ) is non-increasing function, we have

(3.29) d d t ( ϑ ( t ) G ( t ) + 2 d 6 E ( t ) ) d 5 ϑ ( t ) α ( t ) E ( t ) d 6 α ( t ) 0 t h ( ϱ ) d ϱ u ( t ) 2 2 .

From (2.6) and (2.2) that l u ( t ) 2 2 E ( t ) , we find

(3.30) d d t ( ϑ ( t ) G ( t ) + 2 d 6 E ( t ) ) d 5 α ( t ) ϑ ( t ) E ( t ) d 6 α ( t ) 0 t h ( ϱ ) d ϱ u ( t ) 2 2 d 5 α ( t ) ϑ ( t ) E ( t ) 2 d 6 α ( t ) l E ( t ) α ( t ) ϑ ( t ) d 5 + 2 d 6 l 0 α ( t ) l ϑ ( t ) α ( t ) E ( t ) .

Since lim t α ( t ) ϑ ( t ) α ( t ) = 0 , we can choose t 1 t 0 such that d 5 + 2 d 6 l 0 α ( t ) l α ( t ) ϑ ( t ) > 0 for t t 1 .

Finally, let

(3.31) K ( t ) ϑ ( t ) G ( t ) + 2 d 6 E ( t ) E ( t ) .

Hence, for some λ 2 > 0 , we obtain

(3.32) K ( t ) λ 2 ϑ ( t ) α ( t ) K ( t ) , t t 1 .

By integrating (3.32) over ( t 1 , t ) yields

(3.33) K ( t ) K ( t 1 ) e λ 2 t 1 t α ( ϱ ) ϑ ( ϱ ) d ϱ , t t 1 .

Hence, (3.27) is established by virtue of (3.31) and (3.33). The proof is complete.□

4 Conclusion

The objective of this work is the study of the general decay of solutions for a viscoelastic wave equation with distributed delay and Balakrishnan-Taylor damping. This type of problem is frequently found in some mathematical models in applied sciences. Particularly in the theory of viscoelasticity. What interests us in this current work is the combination of these terms of damping (memory term, Balakrishnan-Taylor damping, and the distributed delay terms), which dictates the emergence of these terms in the problem.

In the next work, we will try to use the same method with the same problem, but with the addition of other damping (Dispersion, Source, and Logarithmic terms).


In memory of the mother of the second author: Ms. Fatma bent Tayeb Zeghdoud (1940–2021).


Acknowledgements

The authors express their appreciation for the referee’s careful reading and comments.

  1. Funding information: Author states no funding involved.

  2. Author contributions: The authors contributed equally to this article. They have all read and approved the final manuscript.

  3. Conflict of interest: The authors declare that there are no conflicts of interest regarding the publication of this paper.

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Received: 2021-07-01
Revised: 2021-09-22
Accepted: 2021-09-27
Published Online: 2021-10-18

© 2021 Abdelbaki Choucha et al., published by De Gruyter

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

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