Home Exponential stability of Timoshenko system in thermoelasticity of second sound with a memory and distributed delay term
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Exponential stability of Timoshenko system in thermoelasticity of second sound with a memory and distributed delay term

  • Abdelkader Moumen , Djamel Ouchenane , Abdelbaki Choucha , Khaled Zennir EMAIL logo and Sulima A. Zubair
Published/Copyright: December 31, 2021
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Open Mathematics
From the journal Open Mathematics Volume 19 Issue 1

Abstract

This article concerns linear one-dimensional thermoelastic Timoshenko system with memory and distributed delay terms where the Cattaneo law governs the heat flux q ( x , t ) . We proved an exponential stability result by using the energy method combined with Lyapunov functional.

Keywords: Timoshenko system; general decay; exponential decay; distributed delay term
MSC 2010: 35B40; 35L70; 93D15; 93D20

1 Introduction

Thermoelastic is a material in which the reactions depend only on the values of the determining parameters at the current time t . This means that the entire thermomechanical process is not included in the governing equations. Thermoelasticity has aroused great interest since the publication of the pioneering article of Dafermos (1968). Dafermos proved that the solutions of the thermoelastic equations with homogeneous boundary conditions are asymptotically stable; however, he did not establish any stabilization rate. Since then, many works have shown asymptotic stability [1,2, 3,4].

Timoshenko systems were the subject of research by such outstanding scientists. However, as a unified science, the theory of damped Timoshenko system developed in the late years, which became the years of its rapid development. Undoubtedly, the theory of thermoelastic Timoshenko system is still a young science, although research in this direction was carried out even very recently [5,6,7].

We also include some references to Timoshenko’s system with memory [8] and with delay [9,10].

To begin with, let G = ( 0 , 1 ) × ( 0 , ∞ ) , ( x , t ) ∈ G and we are concerned with the problem

(1.1) ρ 1 φ t t − K ( φ x + ϕ ) x + γ θ x = 0 , ρ 2 ϕ t t − b ϕ x x + K ( φ x + ϕ ) + ∫ 0 t μ ( t − p ) ϕ x x ( s ) d p + μ 1 ϕ t + ∫ τ 1 τ 2 ∣ μ 2 ( p ) ∣ ϕ t ( x , t − p ) d p = 0 , ρ 3 θ t + k q x + γ φ t x = 0 , τ 0 q t + δ q + k θ x = 0 ,

with the following initial data

(1.2) φ ( x , 0 ) = φ 0 ( x ) , φ t ( x , 0 ) = φ 1 ( x ) , ϕ ( x , 0 ) = ϕ 0 ( x ) , ϕ t ( x , 0 ) = ϕ 1 ( x ) , θ ( x , 0 ) = θ 0 ( x ) , q ( x , 0 ) = q 0 ( x ) , ϕ t ( x , − τ ) = f ( x , − τ ) ,

where

( x , τ ) ∈ ( 0 , 1 ) × ( τ 1 , τ 2 ) ,

and Dirichlet boundary conditions for t ≥ 0

(1.3) φ ( 0 , t ) = φ ( 1 , t ) = 0 , ϕ ( 0 , t ) = ϕ ( 1 , t ) = 0 , q ( 0 , t ) = q ( 1 , t ) = 0 .

Here, φ is the transverse displacement of the beam, ϕ is the angle of rotation, the function θ is the difference temperature, ρ 1 , ρ 2 , ρ 3 , b , δ , k , K , τ 0 , γ > 0 and τ 1 , τ 2 > 0 are a time delay, μ 1 is the positive constant, μ 2 is an L ∞ function and g is a C 1 satisfying

  1. The kernel μ ∈ C 1 ( R + , R + ) satisfies

    (1.4) μ ( 0 ) > 0 , b − ∫ 0 ∞ μ ( p ) d p = l > 0 ,

    where R + = { α ∣ α ≥ 0 } .

  2. There exists a decreasing function ϑ ∈ C 1 ( R + , R + ) satisfying

    (1.5) μ ′ ( t ) ≤ − ϑ ( t ) μ ( t ) , ∀ t ≥ 0 .

  3. The L ∞ function μ 2 ∈ C ( [ τ 1 , τ 2 ] , R ) satisfying

    (1.6) ∫ τ 1 τ 2 ∣ μ 2 ( p ) ∣ d p ≤ μ 1 .

In the last few decades, under similar assumptions, many authors touched upon the existence and stability of problems with delays and memory terms (please see [11,12,13, 14,15,16]). For instance, in [16], the authors proved stability and the well-posedness of the following Timoshenko system with a distributed delay term and past history

(1.7) ρ 1 φ t t = k ( φ x − ψ ) x − β θ t x , ρ 2 ψ t t = b ψ x x − k ( φ x − ψ ) − ∫ τ 1 τ 2 g ( s ) ψ x x ( x , t − s ) d s + β θ t − f ( ψ ) , ρ 3 θ t t = δ θ x x + l θ t x x − γ φ t x − γ φ t + ∫ τ 1 τ 2 μ ( ζ ) θ t x x ( x , t − ζ ) d ζ .

Apalara in [17] considered the following system of porous-elastic with memory

(1.8) ρ u t t = μ u x + b ϕ x , J ϕ t t = δ ϕ x x − b u x − ξ ϕ − ∫ 0 t g ( t − s ) ϕ x x ( x , s ) d s .

They proved a stability result of system (1.8) under suitable conditions. Based on all this and complement them, we are concerning to prove the exponential decay of system (1.1), by using the energy method, where the interesting term is the distributed delay. In all the following, consider c as a positive constant.

It follows that to use Poincare’s inequality, for θ ,

(1.9) d d t ∫ 0 1 θ ( x , t ) d x = 0 .

So, after solving equation (1.9) and using the initial system, we get

∫ 0 1 θ ( x , t ) d x = ∫ 0 1 θ 0 ( x ) d x .

Then, if we set

(1.10) θ ¯ ( x , t ) = θ ( x , t ) − ∫ 0 1 θ 0 ( x ) d x ,

we get

∫ 0 1 θ ¯ ( x , t ) d x = 0 , ∀ t ≥ 0 .

From now, Poincare’s inequality is applicable for θ ¯ and ( φ , ϕ , θ ¯ , q ) satisfies system (1.1) with initial conditions θ ¯ given by

θ ¯ 0 ( x ) = θ 0 ( x ) − ∫ 0 1 θ 0 ( x ) d x ,

and we use θ ¯ instead of θ in the rest of the paper, but write θ for simplification.

2 Exponential stability result

This section is devoted to prove exponential stability of system (1.1)–(1.3), for this end as in [18], let us introduce a new dependent variable

z ( x , ϱ , τ , t ) = ϕ t ( x , t − τ ϱ ) ,

then, we obtain

τ z t ( x , ϱ , τ , t ) = − z ϱ ( x , ϱ , τ , t ) , z ( x , 0 , τ , t ) = ϕ t ( x , t ) .

Thus, system (1.1)–(1.3) can be rewritten as

(2.1) ρ 1 φ t t − K ( φ x − ϕ ) x + γ θ x = 0 , ρ 2 ϕ t t − b ϕ x x + K ( φ x + ϕ ) + ∫ 0 t μ ( t − p ) ϕ x x ( p ) d p + μ 1 ϕ t + ∫ τ 1 τ 2 ∣ μ 2 ( p ) ∣ ϕ t ( x , t − p ) d p = 0 , ρ 3 θ t + k q x + γ φ t x = 0 , τ 0 q t + δ q + k θ x = 0 , τ z t ( x , ϱ , τ , t ) + z ϱ ( x , ϱ , τ , t ) = 0 ,

with the following initial data and boundary conditions

(2.2) φ ( x , 0 ) = φ 0 ( x ) , φ t ( x , 0 ) = φ 1 ( x ) , ϕ ( x , 0 ) = ϕ 0 ( x ) , ϕ t ( x , 0 ) = ϕ 1 ( x ) , θ ( x , 0 ) = θ 0 ( x ) , q ( x , 0 ) = q 0 ( x ) , z ( x , ϱ , τ , 0 ) = f ( x , ϱ τ ) , φ ( 0 , t ) = φ ( 1 , t ) = 0 , ϕ ( 0 , t ) = ϕ ( 1 , t ) = 0 , q ( 0 , t ) = q ( 1 , t ) = 0 ,

where

( x , ϱ , τ , t ) ∈ ( 0 , 1 ) × ( 0 , 1 ) × ( τ 1 , τ 2 ) × ( 0 , ∞ ) .

We state, without proof, the existence result. The proof can be established by the classical Galerkin method, it is similar in [15]. Let

ℱ = ( H 0 1 ( 0 , 1 ) × L 2 ( 0 , 1 ) ) 2 × ( L 2 ( 0 , 1 ) ) 2 × L 2 ( ( 0 , 1 ) × ( 0 , 1 ) × ( τ 1 , τ 2 ) ) .

Theorem 2.1

Let Ξ 0 = ( φ 0 , φ 1 , ϕ 0 , ϕ 1 , θ 0 , q 0 , f ) T ∈ ℱ be given. Assume that (A1)–(A3) are satisfied, then there exists a unique global (weak) solution of (2.1)–(2.2) satisfying

Ξ = ( φ , φ t , ϕ , ϕ t , θ , q , z ) ∈ C ( R + , ℱ ) .

We need several lemmas.

Lemma 2.2

The energy of solution is defined as

(2.3) ℰ ( t ) = 1 2 ∫ 0 1 ρ 1 φ t 2 + ρ 2 ϕ t 2 + b − ∫ 0 t μ ( p ) d p ϕ x 2 + K ( φ x + ϕ ) 2 + ρ 3 θ 2 + τ 0 q 2 d x + 1 2 ( μ ∘ ϕ x ) + 1 2 ∫ 0 1 ∫ 0 1 ∫ τ 1 τ 2 τ ∣ μ 2 ( τ ) ∣ z 2 ( x , ϱ , τ , t ) d τ d ϱ d x ,

which satisfies

(2.4) ℰ ′ ( t ) = − δ ∫ 0 1 q 2 d x + 1 2 ( μ ′ ∘ ϕ x ) − 1 2 μ ( t ) ∫ 0 1 ϕ x 2 d x − μ 1 − ∫ τ 1 τ 2 ∣ μ 2 ( p ) ∣ d p ∫ 0 1 ϕ t 2 d x

and

(2.5) ℰ ′ ( t ) ≤ − δ ∫ 0 1 q 2 d x + 1 2 ( μ ′ ∘ ϕ x ) − η 0 ∫ 0 1 ϕ t 2 d x ≤ 0 ,

where η 0 = μ 1 − ∫ τ 1 τ 2 ∣ μ 2 ( p ) ∣ d p ≥ 0 . Here

μ ∘ ϕ x = ∫ 0 t μ ( t − s ) ∥ ϕ ( t ) − ϕ ( s ) ∥ 2 2 d s ,

for any ϕ ∈ L 2 ( 0 , 1 ) .

Proof

Multiplying equations in (2.1) by φ t , ϕ t , θ , q , respectively, using integration by parts, and (1.3), we get

(2.6) 1 2 d d t ∫ 0 1 ρ 1 φ t 2 + ρ 2 ϕ t 2 + b − ∫ 0 t μ ( p ) d p ϕ x 2 + K ( φ x + ϕ ) 2 + ρ 3 θ 2 + τ 0 q 2 d x + 1 2 ( μ ∘ ϕ x ) + δ ∫ 0 1 q 2 d x + μ 1 ∫ 0 1 ϕ t 2 d x + ∫ 0 1 ϕ t ∫ τ 1 τ 2 ∣ μ 2 ( τ ) ∣ z ( x , 1 , τ , t ) d τ d x = 0 .

Multiplying now ( 2.1 ) 5 by z ∣ μ 2 ( τ ) ∣ and integrating we get

(2.7) d d t 1 2 ∫ 0 1 ∫ 0 1 ∫ τ 1 τ 2 τ ∣ μ 2 ( τ ) ∣ z 2 ( x , ϱ , τ , t ) d τ d ϱ d x = − ∫ 0 1 ∫ 0 1 ∫ τ 1 τ 2 ∣ μ 2 ( τ ) ∣ z z ϱ ( x , ϱ , τ , t ) d τ d ϱ d x = − 1 2 ∫ 0 1 ∫ 0 1 ∫ τ 1 τ 2 ∣ μ 2 ( τ ) ∣ d d ϱ z 2 ( x , ϱ , τ , t ) d τ d ϱ d x = 1 2 ∫ 0 1 ∫ τ 1 τ 2 ∣ μ 2 ( τ ) ∣ ( z 2 ( x , 0 , τ , t ) − z 2 ( x , 1 , τ , t ) ) d τ d x = 1 2 ∫ τ 1 τ 2 ∣ μ 2 ( p ) ∣ d p ∫ 0 1 ϕ t 2 d x − 1 2 ∫ 0 1 ∫ τ 1 τ 2 ∣ μ 2 ( τ ) ∣ z 2 ( x , 1 , τ , t ) d τ d x .

From (2.4), (2.6) and (2.7), we get

(2.8) ℰ ′ ( t ) ≤ − δ ∫ 0 1 q 2 d x + 1 2 ( μ ′ ∘ ϕ x ) − μ 1 − ∫ τ 1 τ 2 ∣ μ 2 ( p ) ∣ d p ∫ 0 1 ϕ t 2 d x ,

then, by (A3), ∃ η 0 > 0 such as

(2.9) ℰ ′ ( t ) ≤ − δ ∫ 0 1 q 2 d x + 1 2 ( μ ′ ∘ ϕ x ) − η 0 ∫ 0 1 ϕ t 2 d x ,

then we obtain ℰ is a decreasing function.□

We prepare now many results to introduce a Lyapunov functional.

Lemma 2.3

The functional

(2.10) G 1 ( t ) = ρ 1 ∫ 0 1 φ t φ d x

satisfies

(2.11) G 1 ′ ( t ) ≤ − K 2 ∫ 0 1 φ x 2 d x + c ∫ 0 1 ϕ x 2 d x + c ∫ 0 1 θ 2 d x + ρ 1 ∫ 0 1 φ t 2 d x .

Proof

By direct computation and using integration by parts, we obtain

(2.12) G 1 ′ ( t ) = − K ∫ 0 1 φ x 2 d x − K ∫ 0 1 ϕ φ x d x + γ ∫ 0 1 θ φ x d x + ρ 1 ∫ 0 1 φ t 2 d x .

Applying Young’s and Poincare’s inequalities, we find (2.11).□

Now, let v be a solution of

(2.13) − v x x = ϕ x , v ( 0 ) = v ( 1 ) = 0 .

Then we get

v ( x , t ) = − ∫ 0 x ϕ ( p , t ) d p + x ∫ 0 1 ϕ ( p , t ) d p ,

and we have then the next lemma.

Lemma 2.4

The solution of (2.13) satisfies

(2.14) ∫ 0 1 v x 2 d x ≤ ∫ 0 1 ϕ 2 d x ,

(2.15) ∫ 0 1 v t 2 d x ≤ ∫ 0 1 ϕ t 2 d x .

Proof

Multiplying equation (2.13) by v and using the Cauchy-Schwartz inequality after the integration by parts, we obtain

∫ 0 1 v x 2 d x ≤ ∫ 0 1 ϕ 2 d x .

Next differentiating (2.13) by the same method, we get

∫ 0 1 v t 2 d x ≤ ∫ 0 1 ϕ t 2 d x ,

which completes the proof.□

Then we have the following lemma.

Lemma 2.5

Assume that (A3) and (2.13) hold. Then, we have

G 2 ( t ) = ρ 2 ∫ 0 1 ϕ t ϕ d x + ∫ 0 1 ρ 1 φ t v d x + μ 1 2 ∫ 0 1 ϕ 2 d x ,

which satisfies,

(2.16) G 2 ′ ( t ) ≤ − l 2 ∫ 0 1 ϕ x 2 d x + ε 1 ∫ 0 1 φ t 2 d x + c 1 + 1 ε 1 ∫ 0 1 ϕ t 2 d x + c ∫ 0 1 φ t 2 d x + c ∫ 0 1 θ 2 d x + c ( μ ∘ ϕ x ) + c ∫ 0 1 ∫ τ 1 τ 2 ∣ μ 2 ( τ ) ∣ z 2 ( x , 1 , τ , t ) d τ d x .

Proof

By differentiating G 2 , then using (2.1), integration by parts, and by (1.3) we get

(2.17) G 2 ′ ( t ) = − b ∫ 0 1 ϕ x 2 d x − K ∫ 0 1 ( φ x + ϕ ) ϕ d x − μ 1 ∫ 0 1 ϕ t ϕ d x + ρ 2 ∫ 0 1 ϕ t 2 d x + K ∫ 0 1 ( φ x + ϕ ) x v d x − γ ∫ 0 1 θ x v d x + ∫ 0 1 ϕ x ∫ 0 t μ ( t − p ) ϕ x d p d x + ρ 1 ∫ 0 1 φ t v t d x − ∫ 0 1 ϕ ∫ τ 1 τ 2 ∣ μ 2 ( τ ) ∣ z ( x , 1 , τ , t ) d τ d x + μ 1 ∫ 0 1 ϕ t ϕ d x .

By using (2.13) and (2.14), we get

G 2 ′ ( t ) = − b ∫ 0 1 ϕ x 2 d x + K ∫ 0 1 φ ϕ x d x − K ∫ 0 1 ϕ 2 d x + ρ 2 ∫ 0 1 ϕ t 2 d x + K ∫ 0 1 φ x ϕ d x + K ∫ 0 1 ϕ 2 d x + γ ∫ 0 1 θ v x d x + ρ 1 ∫ 0 1 φ t v t d x + ∫ 0 1 ϕ x ∫ 0 t μ ( t − p ) ϕ x d p d x − ∫ 0 1 ϕ ∫ τ 1 τ 2 ∣ μ 2 ( τ ) ∣ z ( x , 1 , τ , t ) d τ d x .

We apply Young’s and Poincare’s inequalities and using (2.14) and (2.15), we find (2.16).□

Lemma 2.6

The functional

G 3 ( t ) = ρ 1 ρ 3 ∫ 0 1 φ t ∫ 0 x θ ( p ) d p d x ,

satisfies

(2.18) G 3 ′ ( t ) ≤ − ρ 1 γ 2 ∫ 0 1 φ t 2 d x + ε 2 ∫ 0 1 φ x 2 d x + ε 3 ∫ 0 1 ϕ 2 d x + c 1 + 1 ε 2 + 1 ε 3 ∫ 0 1 θ 2 d x + c ∫ 0 1 q 2 d x .

Proof

A classical computation gives

G 3 ′ ( t ) = − ρ 1 γ ∫ 0 1 φ t 2 d x − ρ 3 K ∫ 0 1 φ x θ d x + ρ 3 K ∫ 0 1 θ ϕ d x + ρ 3 γ ∫ 0 1 θ 2 d x − ρ 1 k ∫ 0 1 φ t q d x ,

and by using Young’s inequality, we get (2.18).□

Lemma 2.7

The functional

G 4 ( t ) = ρ 2 ∫ 0 1 ϕ ϕ t d x + μ 1 2 ∫ 0 1 ϕ 2 d x

satisfies

(2.19) G 4 ′ ( t ) ≤ − K 2 ∫ 0 1 ϕ 2 d x − l 2 ∫ 0 1 ϕ x 2 d x + c ∫ 0 1 φ x 2 d x + ρ 2 ∫ 0 1 ϕ t 2 d x + c ( μ ∘ ϕ x ) + c ∫ 0 1 ∫ τ 1 τ 2 ∣ μ 2 ( τ ) ∣ z 2 ( x , 1 , τ , t ) d τ d x .

Proof

By differentiating G 4 , we have

G 4 ′ ( t ) = − b ∫ 0 1 ϕ x 2 d x − K ∫ 0 1 φ x ϕ d x − K ∫ 0 1 ϕ 2 d x + ρ 2 ∫ 0 1 ϕ t 2 d x − ∫ 0 1 ϕ ∫ τ 1 τ 2 ∣ μ 2 ( τ ) ∣ z ( x , 1 , τ , t ) d τ d x .

Estimating (2.19) is easy by applying Young’s and Poincare’s inequalities.□

Lemma 2.8

The functional

G 5 ( t ) = − τ 0 ρ 3 ∫ 0 1 q ∫ 0 x θ ( p ) d p d x

satisfies, for any ε 4 > 0 ,

(2.20) G 5 ′ ( t ) ≤ − τ 0 k 2 ∫ 0 1 θ 2 d x + ε 4 ∫ 0 1 φ t 2 d x + c 1 + 1 ε 4 ∫ 0 1 q 2 d x .

Proof

By differentiating G 5 , we have

G 5 ′ ( t ) = − ρ 3 k ∫ 0 1 θ 2 d x + ρ 3 δ ∫ 0 1 q ∫ 0 x θ ( p ) d p d x + τ 0 k ∫ 0 1 q 2 d x + τ 0 γ ∫ 0 1 q φ t d x .

Estimating (2.20) is easy by applying Young’s and Cauchy-Schwartz’s inequalities.□

Lemma 2.9

The functional

G 6 ( t ) ≔ ∫ 0 1 ∫ 0 1 ∫ τ 1 τ 2 τ e − τ ϱ ∣ μ 2 ( s ) ∣ z 2 ( x , ϱ , τ , t ) d τ d ϱ d x

satisfies

(2.21) G 6 ′ ( t ) ≤ − η 1 ∫ 0 1 ∫ 0 1 ∫ τ 1 τ 2 τ ∣ μ 2 ( τ ) ∣ z 2 ( x , ϱ , τ , t ) d τ d ϱ d x + μ 1 ∫ 0 1 ϕ t 2 d x − η 1 ∫ 0 1 ∫ τ 1 τ 2 ∣ μ 2 ( τ ) ∣ z 2 ( x , 1 , τ , t ) d τ d x ,

where η 1 > 0 .

Proof

By differentiating G 6 and using ( 2.1 ) 5 , we get

G 6 ′ ( t ) = − 2 ∫ 0 1 ∫ 0 1 ∫ τ 1 τ 2 e − τ ϱ ∣ μ 2 ( τ ) ∣ z z ϱ ( x , ϱ , τ , t ) d τ d ϱ d x = − ∫ 0 1 ∫ 0 1 ∫ τ 1 τ 2 τ e − τ ϱ ∣ μ 2 ( τ ) ∣ z 2 ( x , ϱ , τ , t ) d τ d ϱ d x − ∫ 0 1 ∫ τ 1 τ 2 ∣ μ 2 ( τ ) ∣ [ e − τ z 2 ( x , 1 , τ , t ) − z 2 ( x , 0 , τ , t ) ] d τ d x .

Using the equality z ( x , 0 , τ , t ) = ϕ t ( x , t ) and the fact that e − τ ≤ e − τ ϱ ≤ 1 , ∀ 0 < ϱ < 1 , we get

G 6 ′ ( t ) = − η 1 ∫ 0 1 ∫ 0 1 ∫ τ 1 τ 2 τ ∣ μ 2 ( τ ) ∣ z 2 ( x , ϱ , τ , t ) d τ d ϱ d x − ∫ 0 1 ∫ τ 1 τ 2 e − τ ∣ μ 2 ( τ ) ∣ z 2 ( x , 1 , τ , t ) d τ d x + ∫ τ 1 τ 2 ∣ μ 2 ( p ) ∣ d p ∫ 0 1 ϕ t 2 d x .

As − e − τ is an increasing function, we have − e − τ ≤ − e − τ 2 , for any τ ∈ [ τ 1 , τ 2 ] . Then, setting η 1 = e − τ 2 and using ( A 3 ) , we obtain (2.21).□

Now, we prove the main result

Theorem 2.10

Under Assumptions (A1)–(A3), there exist λ 1 , λ 2 > 0 such that the energy functional (2.3) satisfies

(2.22) E ( t ) ≤ λ 2 e − λ 1 t , ∀ t ≥ 0 .

Proof

We introduce the Lyapunov functional as

(2.23) ℒ ( t ) ≔ N E ( t ) + N 1 G 1 ( t ) + N 2 G 2 ( t ) + N 3 G 3 + G 4 ( t ) + N 5 G 5 ( t ) + N 6 G 6 ( t ) ,

where N , N 1 , N 2 , N 3 , N 5 , N 6 > 0 .

By differentiating (2.23) and using (2.3), (2.5), (2.11), (2.16), (2.18), (2.19), (2.21), we have

ℒ ′ ( t ) ≤ − l N 2 2 + l 2 − c N 1 ∫ 0 1 ϕ x 2 d x − ρ 1 γ 2 N 3 − ε 1 N 2 − ε 4 N 5 − ρ 1 N 1 ∫ 0 1 φ t 2 d x − K N 1 2 − ε 2 N 3 − c ∫ 0 1 φ x 2 d x − N δ − c N 3 − c 1 + 1 ε 4 N 5 ∫ 0 1 q 2 d x − K 2 − ε 3 N 3 ∫ 0 1 ϕ 2 d x − N η 0 − c N 2 1 + 1 ε 1 − μ 1 N 6 − ρ 2 ∫ 0 1 ϕ t 2 d x − N 5 k τ 0 2 − c N 1 − c N 2 − c 1 + 1 ε 2 + 1 ε 3 N 3 ∫ 0 1 θ 2 d x + [ c N 2 + c ] ( μ ∘ ϕ x ) + N 2 ( μ ′ ∘ ϕ x ) − [ N 6 η 1 − c N 2 − c ] ∫ 0 1 ∫ τ 1 τ 2 ∣ μ 2 ( τ ) ∣ z 2 ( x , 1 , τ , t ) d τ d x − N 6 η 1 ∫ 0 1 ∫ 0 1 ∫ τ 1 τ 2 τ ∣ μ 2 ( τ ) ∣ z 2 ( x , ϱ , τ , t ) d τ d ϱ d x .

By setting

ε 1 = ρ 1 γ N 3 8 N 2 , ε 2 = K N 1 4 N 3 , ε 3 = K 4 N 3 , ε 4 = ρ 1 γ N 3 8 N 5 .

We obtain

ℒ ′ ( t ) ≤ − l N 2 2 + l 2 − c N 1 ∫ 0 1 ϕ x 2 d x − ρ 1 γ 4 N 3 − ρ 1 N 1 ∫ 0 1 φ t 2 d x − K N 1 4 − c ∫ 0 1 φ x 2 d x − N δ − c N 3 − c 1 + N 5 N 3 N 5 ∫ 0 1 q 2 d x − K 4 ∫ 0 1 ϕ 2 d x − N η 0 − c N 2 1 + N 2 N 3 − μ 1 N 6 − ρ 2 ∫ 0 1 ϕ t 2 d x − N 5 k τ 0 2 − c N 1 − c N 2 − c 1 + N 3 N 1 + N 3 N 3 ∫ 0 1 θ 2 d x + [ c N 2 + c ] ( μ ∘ ϕ x ) + N 2 ( μ ′ ∘ ϕ x ) − [ N 6 η 1 − c N 2 − c ] ∫ 0 1 ∫ τ 1 τ 2 ∣ μ 2 ( τ ) ∣ z 2 ( x , 1 , τ , t ) d τ d x − [ N 6 η 1 ] ∫ 0 1 ∫ 0 1 ∫ τ 1 τ 2 τ ∣ μ 2 ( τ ) ∣ z 2 ( x , ϱ , τ , t ) d τ d ϱ d x .

Now, we choose carefully our constants. Beginning by choosing N 1 large enough such that

α 1 = K N 1 4 − c > 0 .

Next, we choose N 3 , N 2 large enough such that

α 2 = ρ 1 γ N 3 4 − ρ 1 N 1 > 0 , α 3 = l N 2 2 + l 2 − c N 1 > 0 ,

then we choose N 5 , N 6 large enough such that

α 4 = k τ 0 N 5 2 − c N 1 − c N 2 − c N 3 1 + N 3 N 1 + N 3 > 0 , α 5 = η 1 N 6 − c N 2 − c > 0 ,

thus, we arrive at

(2.24) ℒ ′ ( t ) ≤ − α 1 ∫ 0 1 ϕ x 2 d x − α 0 ∫ 0 1 ϕ 2 d x − α 2 ∫ 0 1 φ t 2 d x − α 3 ∫ 0 1 φ x 2 d x − α 4 ∫ 0 1 θ d x − α 5 ∫ 0 1 ∫ τ 1 τ 2 τ ∣ μ 2 ( τ ) ∣ z 2 ( x , 1 , τ , t ) d x − [ N η 0 − c ] ∫ 0 1 ϕ t 2 d x − [ N δ − c ] ∫ 0 1 q 2 d x − α 6 ∫ 0 1 ∫ 0 1 ∫ τ 1 τ 2 τ ∣ μ 2 ( τ ) ∣ z 2 ( x , ϱ , τ , t ) d τ d ϱ d x + N 2 ( μ ′ ∘ ϕ x ) + c ( μ ∘ ϕ x ) ,

where α 0 = K 2 , α 6 = η 0 N 6 .

On the other hand, if we let

T ( t ) = N 1 G 1 ( t ) + N 2 G 2 ( t ) + N 3 G 3 + G 4 ( t ) + N 5 G 5 ( t ) + N 6 G 6 ( t ) ,

then

∣ T ( t ) ∣ ≤ ρ 1 N 1 ∫ 0 1 ∣ φ φ t ∣ d x + ρ 2 N 2 ∫ 0 1 ∣ ϕ ϕ t ∣ d x + ρ 1 N 2 ∫ 0 1 ∣ ϕ t v ∣ d x + μ 1 2 ( N 2 + 1 ) ∫ 0 1 ϕ 2 d x + ρ 1 ρ 3 N 3 ∫ 0 1 φ t ∫ 0 x θ ( p ) d p d x + ρ 2 ∫ 0 1 ∣ ϕ ϕ t ∣ d x + τ 0 ρ 3 N 5 ∫ 0 1 q ∫ 0 x θ ( p ) d p d x + N 6 ∣ G 6 ( t ) ∣ .

By using

∫ 0 1 φ 2 d x ≤ 2 c ∫ 0 1 ( φ x + ϕ ) 2 + 2 c ∫ 0 1 ϕ x 2 d x

and Young’s, Cauchy-Schwartz’s and Poincare’s inequalities, we get

∣ T ( t ) ∣ ≤ c ∫ 0 1 ( φ t 2 + ϕ t 2 + ϕ x 2 + ( φ x + ϕ ) 2 + θ 2 + q 2 ) d x + c ( μ ∘ ϕ x ) + c ∫ 0 1 ∫ 0 1 ∫ τ 1 τ 2 τ ∣ μ 2 ( τ ) ∣ z 2 ( x , ϱ , τ , t ) d τ d ϱ d x . ≤ c ℰ ( t ) .

Then, we obtain

∣ T ( t ) ∣ = ∣ ℒ ( t ) − N ℰ ( t ) ∣ ≤ c ℰ ( t ) ,

which means

(2.25) ( N − c ) ℰ ( t ) ≤ ℒ ( t ) ≤ ( N + c ) ℰ ( t ) ,

and that

ℒ ( t ) ≡ ℰ ( t ) .

Finally, we choose N so that

N − c > 0 , N δ − c > 0 , N η 0 − c > 0 .

By using (2.3), (2.24) and (2.25), we get

(2.26) ℒ ′ ( t ) ≤ − k 1 ℰ ( t ) + k 2 ( μ ∘ ϕ x ) , ∀ t ≥ t 0 ,

and

(2.27) h 1 ℰ ( t ) ≤ ℒ ( t ) ≤ h 2 ℰ ( t ) , ∀ t ≥ 0 ,

for some k 1 , k 2 , h 1 , h 2 > 0 .

Multiplying (2.26) by ϑ ( t ) , we get

(2.28) ϑ ( t ) ℒ ′ ( t ) ≤ − k 1 ϑ ( t ) ℰ ( t ) + k 2 ϑ ( t ) ( μ ∘ ϕ x ) , ∀ t ≥ t 0 .

By using (1.5), we have

ϑ ( t ) ( μ ∘ ϕ x ) = ϑ ( t ) ∫ 0 1 ∫ 0 t μ ( t − p ) ( ϕ x ( t ) − ϕ x ( p ) ) 2 d p d x ≤ ∫ 0 1 ∫ 0 t ϑ ( t − p ) μ ( t − p ) ( ϕ x ( t ) − ϕ x ( p ) ) 2 d p d x ≤ − ∫ 0 1 ∫ 0 t μ ′ ( t − p ) ( ϕ x ( t ) − ϕ x ( p ) ) 2 d p d x = − ( μ ′ ∘ ϕ x ) ≤ − 2 ℰ ′ ( t ) .

Thus, (2.28) becomes

ϑ ( t ) ℒ ′ ( t ) ≤ − k 1 ϑ ( t ) ℰ ( t ) − 2 k 2 ℰ ′ ( t ) , ∀ t ≥ t 0 ,

which can be formulated as

( ϑ ( t ) ℒ ( t ) + 2 k 2 ℰ ( t ) ) ′ − ϑ ′ ( t ) ℒ ( t ) ≤ − k 1 ϑ ( t ) ℰ ( t ) , ∀ t ≥ t 0 ,

and using ϑ ′ ( t ) ≤ 0 , ∀ t ≥ 0 , we get

( ϑ ( t ) ℒ ( t ) + 2 k 2 ℰ ( t ) ) ′ ≤ − k 1 ϑ ( t ) ℰ ( t ) , ∀ t ≥ t 0 .

By using (2.27), we note that

(2.29) ℋ ( t ) = ϑ ( t ) ℒ ( t ) + 2 k 2 ℰ ( t ) ∼ ℰ ( t ) .

Consequently, for some λ > 0 , we get

(2.30) ℋ ′ ( t ) ≤ − λ ℋ ( t ) ϑ ( t ) , ∀ t ≥ t 0 .

Integrating (2.30) over ( t 0 , t ) we obtain

(2.31) ℋ ( t ) ≤ ℋ ( t 0 ) e − λ ∫ t 0 t ϑ ( s ) d s , ∀ t ≥ t 0 .

Finally, (2.22) is established by (2.27) and (2.31).□

Acknowledgements

The authors would like to thank the anonymous referees and the handling editor for their careful reading and for relevant remarks/suggestions to improve the paper.

  1. Funding information: Authors state no funding involved.

  2. Author contributions: All authors have accepted responsibility for the entire content of this manuscript and approved its submission.

  3. Conflict of interest: Authors state no conflict of interest.

  4. Data availability statement: No data were used in this study.

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Received: 2021-08-12
Revised: 2021-10-28
Accepted: 2021-10-30
Published Online: 2021-12-31

© 2021 Abdelkader Moumen et al., 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/math-2021-0117
Keywords for this article
Timoshenko system; general decay; exponential decay; distributed delay term
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