Exponential stability for a Timoshenko thermoelastic system with second sound and a time-varying delay term in the internal feedback

Abstract

The main goal of this paper is to investigate the exponential stability of the Timoshenko system in thermoelasticity of second sound with a time-varying delay term in the internal feedback. The well-posedness of the problem is assured by using the variable norm technique of Kato. Furthermore the stability of the system is shown by applying the energy method.

Keywords

Timoshenko system thermoelasticity second sound linear damping exponential decay

1. Introduction and related results

Let us consider the Timoshenko system $\begin{matrix} (1) & \{\begin{matrix} ρ_{1} φ_{t t} (x, t) - K {(φ_{x} + ψ)}_{x} (x, t) + μ_{1} φ_{t} (x, t) + μ_{2} φ_{t} (x, t - τ (t)) = 0, \\ ρ_{2} ψ_{t t} (x, t) - b ψ_{x x} (x, t) + K (φ_{x} + ψ) (x, t) + γ θ_{x} (x, t) = 0, \\ ρ_{3} θ_{t} (x, t) + κ q_{x} (x, t) + γ ψ_{t x} (x, t) = 0, \\ τ_{0} q_{t} (x, t) + δ q (x, t) + κ θ_{x} (x, t) = 0, \\ φ (x, 0) = φ_{0} (x), φ_{t} (x, 0) = φ_{1} (x), ψ (x, 0) = ψ_{0} (x), ψ_{t} (x, 0) = ψ_{1} (x), \\ θ (x, 0) = θ_{0} (x), q (x, 0) = q_{0} (x), φ_{t} (x, t - τ (0)) = f_{0} (x, t - τ (0)), \\ φ (0, t) = φ (1, t) = ψ (0, t) = ψ (1, t) = q (0, t) = q (1, t) = 0, \end{matrix} \end{matrix}$ where $(x, t) \in (0, 1) \times (0, \infty)$ , $ρ_{1}$ , $ρ_{2}$ , $ρ_{3}$ , γ, $τ_{0}$ , δ, κ, $μ_{1,} μ_{2}$ , K are positive constants, $φ_{0}$ , $φ_{1}$ , $ψ_{0}$ , $ψ_{1}$ , $θ_{0}$ , $q_{0}$ , $f_{0}$ are the initial functions and $τ (t) > 0$ is the time-varying delay.

Let us dwell on some previous literateur related to our problem.

A century ago, in [8] Timoshenko studied the system $\begin{matrix} (2) & \{\begin{matrix} ρ u_{t t} = {(k (u_{x} - φ))}_{x}, & in (0, L) \times (0, \infty), \\ I_{ρ} K φ_{t t} = {(E I φ_{x})}_{x} + K (u_{x} - φ), & in (0, L) \times (0, \infty), \end{matrix} \end{matrix}$ subjected to the boundary conditions $\begin{matrix} E I φ_{x} |_{x = 0}^{x = L} = 0, (u_{x} - φ) |_{x = 0}^{x = L} = 0, \end{matrix}$ as a simple example illustrating the transverse vibrations of a beam. In its equilibrium configuration, u is the transverse displacement of the beam and φ is the angle of rotation. The coefficients ρ, $I_{ρ}$ are respectively the density and the polar moment of inertia of a cross section.

In [6], Rivera and Racke considered $\begin{matrix} (3) & \{\begin{matrix} ρ_{1} φ_{t t} - k {(φ_{x} + ψ)}_{x} = 0, \\ ρ_{2} ψ_{t t} - α ψ_{x x} + k (φ_{x} + ψ) = 0, \\ ρ_{3} θ_{t} - k θ_{x x} + γ ψ_{t x} = 0, \end{matrix} \end{matrix}$ where φ, ψ, θ are the transversal displacement, the angle of rotation, and the difference of temperature, respectively. They proved that the system is exponentially stable, under different boundary conditions, if and only if $\begin{matrix} (4) & \frac{k}{ρ_{1}} = \frac{α}{ρ_{2}} . \end{matrix}$

Said-Houari et al. in [7] examined the Timoshenko system with delays, where they considered the following problem $\begin{matrix} (5) & \{\begin{matrix} ρ_{1} φ_{t t} - k {(φ_{x} + ψ)}_{x} = 0, \\ ρ_{2} ψ_{t t} - b ψ_{x x} + k (φ_{x} + ψ) + μ_{1} ψ_{t} + μ_{2} ψ_{t} (x, t - τ) = 0, \end{matrix} \end{matrix}$ and under the assumption $μ_{1} ⩾ μ_{2}$ , they proved the well posedness and exponential decay. Moreover, Kirane et al. in [2] extended the work, they introduced the case of time-varying delay and several estimates of general decay have been established.

Recently, Nicaise et al. in [4], introduced the case of time-varying delay in the wave equation and proved the exponential stability under the condition $\begin{matrix} (6) & μ_{2} < \sqrt{1 - d} μ_{1}, \end{matrix}$ where the constant d satisfies $\begin{matrix} (7) & τ^{'} (t) ⩽ d < 1, for all t > 0, \end{matrix}$ and $\begin{matrix} (8) & τ \in W^{2, \infty} ([0, T]), for all T > 0 . \end{matrix}$

The system (1) with $τ (t)$ constant, i.e, when $τ (t) = τ$ , has been studied by Ouchenane [5] where he proved the existence and regularity of the solutions under the assumption $μ_{1} > μ_{2}$ .

This paper is organized as follows, in Section 2, we state some preliminaries and assumptions needed in our work, in Section 3 we give some technical lemma needed to prove our main result.

2. Well-posedness of the system

In this section, we state some preliminaries and assumptions, then we prove the existence and uniqueness of the solutions of our problem. Indeed, in the same spirit of [2], we introduce the auxiliary variable $W (x, ρ, t)$ as follow $\begin{matrix} W (x, ρ, t) = φ_{t} (x, t - τ (t) ρ), \end{matrix}$ hence, we obtain $\begin{matrix} τ (t) W_{t} (x, ρ, t) + (1 - τ^{'} (t) ρ) W_{ρ} (x, ρ, t) = 0 . \end{matrix}$

Then, the problem (1) can be reduced to $\begin{matrix} (9) & \{\begin{matrix} ρ_{1} φ_{t t} (x, t) - K {(φ_{x} + ψ)}_{x} (x, t) + μ_{1} φ_{t} (x, t) + μ_{2} W (x, 1, t) = 0, \\ ρ_{2} ψ_{t t} (x, t) - b ψ_{x x} (x, t) + K (φ_{x} + ψ) (x, t) + γ θ_{x} (x, t) = 0, \\ ρ_{3} θ_{t} (x, t) + κ q_{x} (x, t) + γ ψ_{t x} (x, t) = 0, \\ τ_{0} q_{t} (x, t) + δ q (x, t) + κ θ_{x} (x, t) = 0, \\ τ (t) W_{t} (x, ρ, t) + (1 - τ^{'} (t) ρ) W_{ρ} (x, ρ, t) = 0, \\ φ (x, 0) = φ_{0} (x), φ_{t} (x, 0) = φ_{1} (x), ψ (x, 0) = ψ_{0} (x), ψ_{t} (x, 0) = ψ_{1} (x), \\ θ (x, 0) = θ_{0} (x), q (x, 0) = q_{0} (x), φ_{t} (x, t - τ (0)) = f_{0} (x, t - τ (0)), \\ W (x, 0, t) = φ_{t} (x, t), W (x, 1, t) = f_{0} (x, t - τ (t)), \\ q (0, t) = q (1, t) = ψ (0, t) = ψ (1, t) = φ (0, t) = φ (1, t) = 0, \end{matrix} \end{matrix}$ where $(x, ρ, t) \in (0, 1) \times (0, 1) \times (0, \infty)$ and the function $τ (t)$ satisfies (7), (8) and the condition $\begin{matrix} (10) & 0 < τ_{0} ⩽ τ (t) ⩽ \bar{τ}, for any t > 0 . \end{matrix}$

Let $X = {(φ, φ_{t}, ψ, ψ_{t}, θ, q, W)}^{T}$ , we rewrite (9) as $\begin{matrix} (11) & \{\begin{matrix} X^{'} = AX, \\ X (0) = X_{0} = {(φ_{0}, φ_{1}, ψ_{0}, ψ_{1}, θ, q, f_{0} (\cdot, - ρ τ (0)))}^{T}, \end{matrix} \end{matrix}$ with $\begin{matrix} A (t) (\begin{matrix} φ \\ u \\ ψ \\ v \\ θ \\ q \\ W \end{matrix}) = (\begin{matrix} u \\ \frac{k}{ρ_{1}} (φ_{x x} + ψ_{x}) - \frac{μ_{1}}{ρ_{1}} u - \frac{μ_{2}}{ρ_{1}} W (\cdot, 1) \\ v \\ \frac{b}{ρ_{2}} ψ_{x x} - \frac{k}{ρ_{2}} (φ_{x} + ψ) - \frac{γ}{ρ_{2}} θ_{x} \\ - \frac{k}{ρ_{3}} q_{x} - \frac{γ}{ρ_{3}} v_{x} \\ - \frac{δ}{τ_{0}} q - \frac{k}{τ_{0}} θ_{x} \\ (τ^{'} (t) ρ - 1) W_{ρ} / τ (t) \end{matrix}), \end{matrix}$ where $\begin{array}{rcl} D (A (t)) & = & {{(φ, φ_{t}, ψ, ψ_{t}, θ, q, W)}^{T} \in H : W_{ρ} \in L^{2} ((0, 1) \times L^{2} (0, 1)), \\ (12) & u = W (\cdot, 0), in (0, 1)}, \end{array}$ with $\begin{matrix} H = V \times H_{0}^{1} (0, 1) \times V \times H_{0}^{1} (0, 1) \times H_{0}^{1} (0, 1) \times H_{0}^{1} (0, 1) \times L^{2} ((0, 1) \times H^{1} (0, 1)), \end{matrix}$ for $t > 0$ and $V = H^{2} (0, 1) \cap H_{0}^{1} (0, 1)$ .

Notice that, $D (A (t))$ is independent of t.

Now, defing $E$ as, $\begin{array}{rcl} E & = & H_{0}^{1} (0, 1) \times L^{2} (0, 1) \times H_{0}^{1} (0, 1) \times {(L^{2} (0, 1))}^{2} \\ \times L^{2} (0, 1) \times L^{2} ((0, 1) \times (0, 1)), \end{array}$ with the inner product $\begin{array}{rcl} {⟨ X, \bar{X} ⟩}_{E} & = & \int_{0}^{1} {ρ_{1} u \tilde{u} + ρ_{2} v \tilde{v} + k (φ_{x} + ψ) ({\tilde{φ}}_{x} + \tilde{ψ}) + b ψ_{x} {\tilde{ψ}}_{x} + ρ_{3} θ \tilde{θ}} d x \\ + \int_{0}^{1} τ_{0} q \tilde{q} d x + \int_{0}^{1} \int_{0}^{1} W (x, ρ) \tilde{W} (x, ρ) d ρ d x, \end{array}$ for $X = {(φ, u, ψ, v, θ, q, W)}^{T}$ , $\tilde{X} = {(\tilde{φ}, \tilde{u}, \tilde{ψ}, \tilde{v}, \tilde{θ}, \tilde{q}, \tilde{W})}^{T}$ .

Our first result is

Theorem 1.
We assume that ( 6 ) holds and ( 7 ), ( 8 ), ( 10 ) are satisfied. Then, for any $X_{0} \in D (A (0))$ , there exists a unique solution $X$ of ( 9 ) satisfying $\begin{matrix} X \in C ([0, + \infty), D (A (0))) \cap C^{1} ([0, + \infty), E) . \end{matrix}$

The proof of the above theorem is based on the theorem below (for more detail see [1]). Theorem 2.
The following hypothesis:
$D (A (0))$ is a dense subset of $E$ ;

for any $t > 0$ we have $D (A (t)) = D (A (0))$ ;

$A (t)$ generates a strongly continuous semigroup on $E$ for all $t \in [0; T]$ , and the family $A = {A (t) : t \in [0; T]}$ is stable with stability constants C and m independent of t, i.e, the semigroup ${(S_{t} (s))}_{s ⩾ 0}$ generated by $A (t)$ satisfies $\begin{matrix} {‖ (S_{t} (s)) (u) ‖}_{E} ⩽ C e^{m s} ‖ u ‖_{E}, for all u \in E, s ⩾ 0, C > 0, m > 0; \end{matrix}$

$\partial_{t} \tilde{A} (t) \in L_{}^{\infty} ([0, T], B (D (A (0)), E))$ , where* $L_{}^{\infty} ([0, T], B (D (A (0)), E))$ is the space of equivalent classes of essentially bounded, strongly measurable functions from* $[0; T]$ into the set $B (D (A (0)); E)$ of bounded operators from $D (A (0))$ into $E$ , are hold.

Then given an initial data in $D (A (0))$ , problem ( 11 ) has a unique solution $\begin{matrix} X \in C ([0, T), D (A (0))) \cap C^{1} ([0, T), E) . \end{matrix}$
Proof.
Our proof follows the method used in [4].

(i) Density of $D (A (0))$ in $E$ . Let $G = {(g_{1}, g_{2}, g_{3}, g_{4}, g_{5}, g_{6}, g_{7})}^{T} \in E$ be orthogonal to all elements of $D (A (0))$ with the inner product ${⟨ \cdot, \cdot ⟩}_{E}$ : $\begin{array}{rcl} 0 & = & {⟨ X, G ⟩}_{E} \\ = & \int_{0}^{1} {ρ_{1} u g_{2} + ρ_{2} v g_{4} + k (φ_{x} + ψ) (g_{1 x} + g_{3}) + b ψ_{x} g_{3 x} + ρ_{3} θ g_{5}} d x \\ (13) & + \int_{0}^{1} τ_{0} q g_{6} d x + \int_{0}^{1} \int_{0}^{1} W (x, ρ) g_{7} d ρ d x, \end{array}$ for all $X = {(φ, u, ψ, v, θ, q, W)}^{T} \in D (A (0))$ . To show that $g_{j} = 0$ , for all $j = 1, \dots, 7$ , we take $W \in D ((0, 1) \times (0, 1))$ and $φ = u = ψ = v = θ = q = 0$ , so $X = {(0, 0, 0, 0, 0, 0, W)}^{T} \in D (A (0))$ therfore, from (13), we deduce that $\begin{matrix} \int_{0}^{1} W (x, ρ) g_{7} d ρ d x = 0, \end{matrix}$ since $D ((0, 1) \times (0, 1))$ is dense in $L^{2} ((0, 1) \times (0, 1))$ , it follows then that $g_{7} = 0$ . Similarly, let $u \in D (0, 1)$ , then $X = {(0, u, 0, 0, 0, 0, 0)}^{T} \in D (A (0))$ , which implies from (13) that $\begin{matrix} \int_{0}^{1} u g_{2} d x = 0 . \end{matrix}$

So, as above, $g_{2} = 0$ . Moreover let $X = {(φ, 0, 0, 0, 0, 0, 0)}^{T} \in D (A (0))$ , then we obtain from (13) that $\begin{matrix} \int_{0}^{1} φ g_{1 x} d x = 0 . \end{matrix}$

It is immediate that $X = {(φ, 0, 0, 0, 0, 0, 0)}^{T} \in D (A (0))$ for $φ \in V$ which is dense in $H_{0}^{1} (0, 1)$ , with the inner product $\begin{matrix} {⟨ g, h ⟩}_{H_{0}^{1} (0, 1)} = \int_{0}^{1} g_{x} h_{x} d x . \end{matrix}$

We get $g_{1} = 0$ . By the same ideas as above, we can also show that $g_{3} = 0$ , for $v \in D (0, 1)$ , we get from (13) $\begin{matrix} \int_{0}^{1} v g_{4} d x = 0, \end{matrix}$ and by density of $D (0, 1)$ in $L^{2} (0, 1)$ , we obtain $g_{4} = 0$ .

For $q \in D (0, 1)$ , we get from (13) $\begin{matrix} \int_{0}^{1} q g_{6} d x = 0, \end{matrix}$ and by density of $D (0, 1)$ in $L^{2} (0, 1)$ , we obtain $g_{6} = 0$ .

Next, let $X = {(0, 0, 0, 0, θ, 0, 0)}^{T} \in D (A (0))$ , then we obtain from (13) that $\begin{matrix} \int_{0}^{1} θ g_{5} d x = 0, \end{matrix}$ consequently $g_{5} = 0$ . This completes the proof of (i).

(ii) for any $t > 0$ we have $D (A (t)) = D (A (0))$ .

(iii) $A (t)$ generates a $C_{0}$ -semigroup in $E$ for a fixed t. We define the time-dependent inner-product on $E$ , (which is equivalent to the classical inner product) $\begin{array}{rcl} {⟨ X, \bar{X} ⟩}_{t} & = & \int_{0}^{1} {ρ_{1} u \bar{u} + ρ_{2} v \bar{v} + k (φ_{x} + ψ) ({\bar{φ}}_{x} + \bar{ψ}) + b ψ_{x} {\bar{ψ}}_{x} + ρ_{3} θ \bar{θ}} d x \\ (14) & + \int_{0}^{1} τ_{0} q \bar{q} d x + ξ τ (t) \int_{0}^{1} \int_{0}^{1} W (x, ρ) \bar{W} (x, ρ) d ρ d x, \end{array}$ where ξ satisfies $\begin{matrix} (15) & \frac{μ_{2}}{\sqrt{1 - d}} ⩽ ξ ⩽ 2 μ_{1} - \frac{μ_{2}}{\sqrt{1 - d}}, \end{matrix}$ thanks to hypothesis (6).

Let us set $\begin{matrix} ϰ (t) = \frac{{(τ^{'} {(t)}^{2} + 1)}^{1 / 2}}{2 τ (t)} . \end{matrix}$

At this stage we show that the dissipativity of the operator $\tilde{A} (t) = A (t) - ϰ (t) I$ .

For a fixed t and $X = {(φ, u, ψ, v, θ, q, W)}^{T} \in D (A (t))$ , we have $\begin{array}{rcl} {⟨ A (t) X, X ⟩}_{t} & = & - δ \int_{0}^{1} q^{2} d x - μ_{1} \int_{0}^{1} u^{2} (x) d x - μ_{2} \int_{0}^{1} W (x, 1) u (x) d x \\ (16) & - ξ \int_{0}^{1} \int_{0}^{1} (1 - τ^{'} (t) ρ) W (x, ρ) W_{ρ} (x, ρ) d x d ρ . \end{array}$

Observing that $\begin{array}{l} \int_{0}^{1} \int_{0}^{1} (1 - τ^{'} (t) ρ) W (x, ρ) W_{ρ} (x, ρ) d x d ρ \\ = \int_{0}^{1} \int_{0}^{1} \frac{1}{2} \frac{\partial}{\partial ρ} W^{2} (1 - τ^{'} (t) ρ) d x d ρ \\ (17) & = \frac{τ^{'} (t)}{2} \int_{0}^{1} \int_{0}^{1} W^{2} (x, ρ) d x d ρ + \frac{1}{2} \int_{0}^{1} {W^{2} (x, 1) (1 - τ^{'} (t)) - W^{2} (x, 0)} d x . \end{array}$

Whereupon, $\begin{array}{rcl} {⟨ A (t) X, X ⟩}_{t} & = & - δ \int_{0}^{1} q^{2} d x - μ_{1} \int_{0}^{1} u^{2} (x) d x - μ_{2} \int_{0}^{1} W (x, 1) u (x) d x \\ - \frac{τ^{'} (t) ξ}{2} \int_{0}^{1} \int_{0}^{1} W^{2} (x, ρ) d x d ρ \\ (18) & - \frac{ξ}{2} \int_{0}^{1} W^{2} (x, 1) (1 - τ^{'} (t)) d x + \frac{ξ}{2} \int_{0}^{1} u^{2} (x) d x . \end{array}$

By applying the Cauchy–Schwarz inequality and (7), we get $\begin{array}{rcl} {⟨ A (t) X, X ⟩}_{t} & ⩽ & (- μ_{1} + \frac{μ_{2}}{2 \sqrt{1 - d}} + \frac{ξ}{2}) \int_{0}^{1} u^{2} (x) d x - δ \int_{0}^{1} q^{2} d x \\ + (\frac{μ_{2} \sqrt{1 - d}}{2} - ξ \frac{(1 - d)}{2}) \int_{0}^{1} W^{2} (x, 1) d x + ϰ (t) {⟨ U, U ⟩}_{t} . \end{array}$

Condition (15) allows to write $\begin{matrix} - μ_{1} + \frac{μ_{2}}{2 \sqrt{1 - d}} + \frac{ξ}{2} ⩽ 0, \frac{μ_{2} \sqrt{1 - d}}{2} - ξ \frac{(1 - d)}{2} ⩽ 0 . \end{matrix}$

Therefore, the operator $A (t)$ is dissipative, then $\tilde{A} (t)$ is also dissipative.

We are now in a position to show that the operator $λ I - A (t)$ is surjective. Let ${(f_{1}, f_{2}, f_{3}, f_{4}, f_{5}, f_{6}, f_{7})}^{T} \in E$ and $X = {(φ, u, ψ, v, θ, q, W)}^{T} \in D (A (t))$ be solution of the following system $\begin{matrix} (19) & \{\begin{matrix} λ φ - u = f_{1}, \\ λ u - \frac{k}{ρ_{1}} (φ_{x x} + ψ_{x}) + \frac{μ_{1}}{ρ_{1}} u + \frac{μ_{2}}{ρ_{1}} W (\cdot, 1) = f_{2}, \\ λ ψ - v = f_{3}, \\ λ v - \frac{b}{ρ_{2}} ψ_{x x} + \frac{k}{ρ_{2}} (φ_{x} + ψ) + \frac{γ}{ρ_{2}} θ_{x} = f_{4}, \\ λ θ + \frac{k}{ρ_{3}} q_{x} + \frac{γ}{ρ_{3}} v_{x} = f_{5}, \\ λ q + \frac{δ}{τ_{0}} q + \frac{k}{τ_{0}} θ_{x} = f_{6}, \\ λ W + \frac{(1 - τ^{'} (t) ρ)}{τ (t)} W_{ρ} = f_{7} . \end{matrix} \end{matrix}$

The first and the third equations in (19) give $\begin{matrix} (20) & \{\begin{matrix} u = λ φ - f_{1}, \\ v = λ ψ - f_{3} . \end{matrix} \end{matrix}$

Moreover, by (19) we find W as $\begin{matrix} (21) & W (x, 0) = u (x), for x \in (0, 1) . \end{matrix}$

We obtain, by using the last equation in (19) and the same approach as in [4] $\begin{matrix} W (x, ρ) = u (x) e^{- λ ρ τ (t)} + τ (t) e^{- λ ρ τ (t)} \int_{0}^{1} f_{7} (x, σ) e^{- λ ρ τ (t)} d σ, if τ^{'} (t) = 0, \end{matrix}$ and $\begin{matrix} W (x, ρ) = u (x) e^{ϑ_{ρ} (t)} + e^{ϑ_{ρ} (t)} \int_{0}^{1} \frac{f_{7} (x, σ) τ (t)}{1 - τ^{'} (t) σ} e^{- ϑ_{ρ} (t)} d σ, if τ^{'} (t) \neq 0, \end{matrix}$ where $ϑ_{ρ} (t) = λ \frac{τ (t)}{τ^{'} (t)} ln (1 - τ^{'} (t) ρ)$ . Whereupon, from (20), we have $\begin{array}{rcl} W (x, ρ) & = & λ φ (x) e^{- λ ρ τ (t)} - f_{1} e^{- λ ρ τ (t)} \\ (22) & + τ (t) e^{- λ ρ τ (t)} \int_{0}^{1} f_{7} (x, σ) e^{- λ ρ τ (t)} d σ, if τ^{'} (t) = 0, \end{array}$ and $\begin{matrix} (23) & W (x, ρ) = λ φ (x) e^{ϑ_{ρ} (t)} - f_{1} e^{ϑ_{ρ} (t)} + e^{ϑ_{ρ} (t)} \int_{0}^{1} \frac{f_{7} (x, σ) τ (t)}{1 - τ^{'} (t) σ} e^{- ϑ_{ρ} (t)} d σ, if τ^{'} (t) \neq 0 . \end{matrix}$

By using (19) and (20), we get $\begin{matrix} (24) & \{\begin{matrix} (λ^{2} + \frac{μ_{1}}{ρ_{1}} λ + λ e^{- λ τ} \frac{μ_{2}}{ρ_{1}}) φ - \frac{k}{ρ_{1}} (φ_{x x} + ψ_{x}) = f_{2} + (λ + \frac{μ_{1}}{ρ_{1}}) f_{1} - \frac{μ_{2}}{ρ_{1}} W_{0} (x), \\ λ^{2} ψ - \frac{b}{ρ_{2}} ψ_{x x} + \frac{k}{ρ_{2}} (φ_{x} + ψ) + \frac{γ}{ρ_{2}} θ_{x} = f_{4} + λ f_{3}, \\ λ θ + \frac{k}{ρ_{3}} q_{x} + \frac{γ λ}{ρ_{3}} ψ_{x} = f_{5} + \frac{γ}{ρ_{3}} f_{3 x}, \\ λ q + \frac{δ}{τ_{0}} q + \frac{k}{τ_{0}} θ_{x} = f_{6} . \end{matrix} \end{matrix}$

We obtain, by solving the system (24) $\begin{matrix} (φ, ψ, θ, q) \in V \times V \times H^{2} (0, 1) \times H_{0}^{1} (0, 1), \end{matrix}$ such that $\begin{matrix} (25) & \{\begin{matrix} \int_{0}^{1} ((λ^{2} ρ_{1} + μ_{1} λ + λ e^{- λ τ (t)} μ_{2}) φ w + k (φ_{x} + ψ) w_{x}) d x \\ = \int_{0}^{1} (ρ_{1} f_{2} + (λ ρ_{1} + μ_{1}) f_{1} - μ_{2} W_{0} (x)) w d x, \\ \int_{0}^{1} (ρ_{2} λ^{2} ψ X + b ψ_{x} X_{x} + k (φ_{x} + ψ) X + γ θ_{x} X) d x = \int_{0}^{1} ρ_{2} (f_{4} + λ f_{3}) X d x, \\ \int_{0}^{1} (ρ_{3} λ θ w_{1} + k q_{x} w_{1} + γ λ ψ_{x} w_{1}) d x = \int_{0}^{1} (ρ_{3} f_{5} + γ f_{3 x}) w_{1} d x, \\ \int_{0}^{1} ((τ_{0} λ + δ) q X_{1} + k θ_{x} X_{1}) d x = \int_{0}^{1} τ_{0} f_{6} X_{1} d x, \end{matrix} \end{matrix}$ for all $(w, X, w_{1}, X_{1}) \in H_{0}^{1} (0, 1) \times H_{0}^{1} (0, 1) \times H_{0}^{1} (0, 1) \times H_{0}^{1} (0, 1)$ . From (22) and (23), we obtain $\begin{matrix} W (x, 1) = \{\begin{matrix} λ φ (x) e^{- λ τ (t)} + W_{0} (x), & if τ^{'} (t) = 0, \\ λ φ (x) e^{ϑ (t)} + W_{0} (x), & if τ^{'} (t) \neq 0, \end{matrix} \end{matrix}$ where $x \in (0, 1)$ and $\begin{matrix} (26) & W_{0} (x) = \{\begin{matrix} - f_{1} e^{- λ ρ τ (t)} + τ (t) e^{- λ ρ τ (t)} \int_{0}^{1} f_{7} (x, σ) e^{- λ ρ τ (t)} d σ, & if τ^{'} (t) = 0, \\ - f_{1} e^{ϑ_{ρ} (t)} + e^{ϑ_{ρ} (t)} \int_{0}^{1} \frac{f_{7} (x, σ) τ (t)}{1 - τ^{'} (t) σ} e^{- ϑ_{ρ} (t)} d σ, & if τ^{'} (t) \neq 0 . \end{matrix} \end{matrix}$

Thus, problem (25) is equivalent to $\begin{matrix} (27) & Y ((φ, ψ, θ, q), (w, X, w_{1}, X_{1})) = Q (w, X, w_{1}, X_{1}), \end{matrix}$ where the bilinear from $Y : {[H_{0}^{1} (0, 1) \times H_{0}^{1} (0, 1) \times H_{0}^{1} (0, 1) \times H_{0}^{1} (0, 1)]}^{2} \to R$ , and the linear form $Q : (H_{0}^{1} (0, 1) \times H_{0}^{1} (0, 1) \times H_{0}^{1} (0, 1) \times H_{0}^{1} (0, 1)) \to R$ are defined by $\begin{array}{l} Y ((φ, ψ, θ, q), (w, X, w_{1}, X_{1})) \\ = \int_{0}^{1} ((λ^{2} ρ_{1} + μ_{1} λ + λ e^{- λ τ (t)} μ_{2}) φ w + k (φ_{x} + ψ) (w_{x} + X)) d x \\ + \int_{0}^{1} (ρ_{2} λ^{2} ψ X + b ψ_{x} X_{x} + γ θ_{x} w_{1 x}) d x \\ + \int_{0}^{1} (ρ_{3} λ θ w_{1} + k q_{x} w_{1} + γ λ ψ_{x} w_{1}) d x \\ + \int_{0}^{1} ((τ_{0} λ + δ) q X_{1} + k θ_{x} X_{1 x}) d x, \end{array}$ and $\begin{array}{rcl} Q (w, X, w_{1}, X_{1}) & = & \int_{0}^{1} (ρ_{1} f_{2} + (λ ρ_{1} + μ_{1}) f_{1} - μ_{2} W_{0} (x)) w d x \\ + \int_{0}^{1} ρ_{2} (f_{4} + λ f_{3}) X d x + \int_{0}^{1} (ρ_{3} f_{5} + γ f_{3 x}) w_{1} d x \\ (28) & + \int_{0}^{1} τ_{0} f_{6} X_{1} d x, \end{array}$ if $τ^{'} (t) = 0$ , where $z_{0} (x)$ satisfies the first equation in (26).

If $τ^{'} (t) \neq 0$ , we define $\begin{array}{l} Y ((φ, ψ, θ, q), (w, X, w_{1}, X_{1})) \\ = \int_{0}^{1} ((λ^{2} ρ_{1} + μ_{1} λ + λ e^{ϑ_{ρ} (t)} μ_{2}) φ w + k (φ_{x} + ψ) (w_{x} + X)) d x \\ + \int_{0}^{1} (ρ_{2} λ^{2} ψ X + b ψ_{x} X_{x} + γ θ_{x} w_{1 x}) d x \\ + \int_{0}^{1} (ρ_{3} λ θ w_{1} + k q_{x} w_{1} + γ λ ψ_{x} w_{1}) d x \\ + \int_{0}^{1} ((τ_{0} λ + δ) q X_{1} + k θ_{x} X_{1 x}) d x, \end{array}$ and the operator $Q$ is defined by the above formula (28). So by applying the Lax-Milgram theorem, problem (27) has a unique solution $(φ, ψ, θ, q)$ for all $(w, X, w_{1}, X_{1}) \in H_{0}^{1} (0, 1) \times H_{0}^{1} (0, 1) \times H_{0}^{1} (0, 1) \times H_{0}^{1} (0, 1)$ . By using the classical elliptic regularity, it follows from (25) that $(φ, ψ, θ, q) \in H^{2} (0, 1) \times H^{2} (0, 1) \times H^{1} (0, 1) \times H_{0}^{1} (0, 1)$ .

Consequently, the operator $λ I - A (t)$ is surjective $\begin{matrix} λ I - \tilde{A} (t) = (λ + ϰ (t)) I - A (t), for any fixed t > 0 and λ > 0 . \end{matrix}$

Since $ϰ (t) > 0$ , then, we conclude the surjectivity of the operator $λ I - \tilde{A} (t)$ .

In this step, it is sufficient to prove $\begin{matrix} (29) & \frac{‖ ϕ ‖_{t}}{‖ ϕ ‖_{s}} ⩽ e^{\frac{c}{2 τ_{0}} | t - s |}, for any t, s \in [0, T], \end{matrix}$ where $ϕ = {(φ, u, ψ, v, θ, q, W)}^{T}$ . By (14), we have $\begin{array}{l} ‖ ϕ ‖_{t}^{2} = {⟨ ϕ, ϕ ⟩}_{t} \\ = \int_{0}^{1} {ρ_{1} u^{2} + ρ_{2} v^{2} + k {(φ_{x} + ψ)}^{2} + b ψ_{x}^{2} + ρ_{3} θ^{2} + τ_{0} q^{2}} d x \\ + ξ τ (t) \int_{0}^{1} \int_{0}^{1} W^{2} (x, ρ) d ρ d x, \\ ‖ ϕ ‖_{s}^{2} = {⟨ ϕ, ϕ ⟩}_{s} \\ = \int_{0}^{1} {ρ_{1} u^{2} + ρ_{2} v^{2} + k {(φ_{x} + ψ)}^{2} + b ψ_{x}^{2} + ρ_{3} θ^{2} + τ_{0} q^{2}} d x \\ + ξ τ (s) \int_{0}^{1} \int_{0}^{1} W^{2} (x, ρ) d ρ d x \end{array}$ then, $\begin{array}{l} ‖ ϕ ‖_{t}^{2} - ‖ ϕ ‖_{s}^{2} e^{\frac{c}{2 τ_{0}} | t - s |} \\ = (1 - e^{\frac{c}{2 τ_{0}} | t - s |}) \int_{0}^{1} {ρ_{1} u^{2} + ρ_{2} v^{2} + k {(φ_{x} + ψ)}^{2} + b ψ_{x}^{2} + ρ_{3} θ^{2} + τ_{0} q^{2}} d x \\ (30) & + ξ (τ (t) - τ (s) e^{\frac{c}{2 τ_{0}} | t - s |}) \int_{0}^{1} \int_{0}^{1} W^{2} (x, ρ) d ρ d x . \end{array}$

Here, we shwo that $τ (t) - τ (s) e^{\frac{r}{2 τ_{0}} | t - s |} ⩽ 0$ , for $r > 0$ , we have $\begin{matrix} τ (t) = τ (s) + τ^{'} (a) (t - s), \end{matrix}$ with $a \in (s, t)$ ,we obtain $\begin{matrix} \frac{τ (t)}{τ (s)} ⩽ 1 + \frac{| τ^{'} (a) |}{τ (s)} | t - s | . \end{matrix}$

By using $τ \in W^{2, \infty} ([0, T])$ and $τ^{'}$ is bounded, we deduce that $\begin{matrix} \frac{τ (t)}{τ (s)} ⩽ 1 + \frac{r}{τ_{0}} | t - s | ⩽ e^{\frac{c}{2 τ_{0}} | t - s |}, \end{matrix}$ which proves (29) and therefore (iii).

(iv) We have the operator $\tilde{A} (t)$ is dissipative, which means that $\begin{matrix} {⟨ A (t) U, U ⟩}_{t} - χ (t) {⟨ U, U ⟩}_{t} ⩽ 0 \end{matrix}$ where $\begin{matrix} \tilde{A} (t) = A (t) - χ (t) I \end{matrix}$ and $\begin{matrix} χ (t) = \frac{{(τ^{'} {(t)}^{2} + 1)}^{\frac{1}{2}}}{2 τ (t)} \end{matrix}$ Moreover $\begin{matrix} χ^{'} (t) = \frac{τ^{'} (t) τ^{″} (t)}{2 τ (t) {(τ^{'} {(t)}^{2} + 1)}^{\frac{1}{2}}} - \frac{τ^{'} (t) {(τ^{'} {(t)}^{2} + 1)}^{\frac{1}{2}}}{2 τ {(t)}^{2}} \end{matrix}$ is bounded on $[0, T], \forall T > 0$ then it is easy to check that $\begin{matrix} \frac{d}{d t} A (t) X = (\begin{matrix} 0 \\ 0 \\ 0 \\ 0 \\ 0 \\ 0 \\ \frac{(τ^{″} (t) τ (t) ρ - τ^{'} (t) (τ^{'} (t) ρ - 1))}{τ^{2} (t)} W_{ρ} \end{matrix}) . \end{matrix}$

Then, by using (8) and (10), thus $\begin{matrix} \frac{d}{d t} \tilde{A} (t) \in L_{*}^{\infty} ([0, T], B (D (A (0)), H)) . \end{matrix}$ where (iv) holds in [4]. Then, we conclude that the problem $\begin{matrix} (31) & \{\begin{matrix} {\tilde{X}}_{t} = \tilde{A} (t) \tilde{X}, \\ \tilde{X} (0) = X_{0}, \end{matrix} \end{matrix}$ admits a unique solution $\tilde{X} \in C ([0, + \infty), E)$ and if $X_{0} \in D (A (0))$ , then $\begin{matrix} \tilde{X} \in C ([0, + \infty), D (A (0))) \cap C^{1} ([0, + \infty), E) . \end{matrix}$

Now, let $\begin{matrix} X (t) = e^{β (t)} \tilde{X} (t), \end{matrix}$ where $β (t) = \int_{0}^{t} k (s) d s$ , then we have by using (31) $\begin{array}{rcl} X_{t} (t) & = & k (t) e^{β (t)} \tilde{X} (t) + e^{β (t)} {\tilde{X}}_{t} (t) \\ = & k (t) e^{β (t)} \tilde{X} (t) + e^{β (t)} \tilde{A} (t) \tilde{X} (t) \\ = & e^{β (t)} (k (t) \tilde{X} (t) + \tilde{A} (t) \tilde{X} (t)) \\ = & e^{β (t)} A (t) \tilde{X} (t) \\ = & A (t) e^{β (t)} \tilde{X} (t) \\ = & A (t) X (t) . \end{array}$

Consequently, $X (t)$ is the unique solution of (11). □

3. Exponential stability

In this section, we give some technical lemma and we prove the stability results, following [5]; we introduce the new variable $\begin{matrix} \bar{θ} (x, t) = θ (x, t) - \int_{0}^{1} θ_{0} (x) d x . \end{matrix}$

Using (1), we get $\begin{matrix} \int_{0}^{1} \bar{θ} (x, t) d x = 0, \end{matrix}$ then, $(φ, ψ, θ, q, W)$ satisfies the system (9).

For a positive constant ξ satisfying $\begin{matrix} (32) & \frac{μ_{2}}{\sqrt{1 - d}} ⩽ ξ ⩽ 2 μ_{1} - \frac{μ_{2}}{\sqrt{1 - d}}, \end{matrix}$ we define the energy functional as $\begin{array}{rcl} E (t) & = & E (t, φ, ψ, θ, q, W) \\ = & \frac{1}{2} \int_{0}^{1} (ρ_{1} φ_{t}^{2} + ρ_{2} ψ_{t}^{2}) d x + \frac{1}{2} \int_{0}^{1} {k {(φ_{x} + ψ)}^{2} + b ψ_{x}^{2} + ρ_{3} θ^{2}} d x \\ (33) & + \frac{1}{2} \int_{0}^{1} τ_{0} q^{2} d x + \frac{ξ}{2} τ (t) \int_{0}^{1} \int_{0}^{1} W^{2} (x, ρ, t) d ρ d x . \end{array}$

By multiplying the system (9) by $φ_{t}$ , $ψ_{t}$ , θ, q, respectively, integrating over $(0, 1)$ , and summing them up, we obtain $\begin{array}{l} \frac{1}{2} \frac{d}{d t} \int_{0}^{1} (ρ_{1} φ_{t}^{2} + ρ_{2} ψ_{t}^{2}) d x + \frac{1}{2} \frac{d}{d t} \int_{0}^{1} {k {(φ_{x} + ψ)}^{2} + b ψ_{x}^{2} + ρ_{3} θ^{2} + τ_{0} q^{2}} d x \\ (34) & = - δ \int_{0}^{1} q^{2} d x - μ_{1} \int_{0}^{1} φ_{t}^{2} (x, t) d x - μ_{2} \int_{0}^{1} φ_{t} (x, t) W (x, 1, t) d x . \end{array}$

We multiply (9)₃ by W $\begin{array}{l} \frac{ξ}{2} \frac{d}{d t} \int_{0}^{1} \int_{0}^{1} τ (t) W^{2} (x, ρ, t) d ρ d x \\ = - ξ \int_{0}^{1} \int_{0}^{1} (1 - τ^{'} (t) ρ) W (x, ρ, t) W_{ρ} (x, ρ, t) d ρ d x \\ + \frac{ξ}{2} τ^{'} (t) \int_{0}^{1} \int_{0}^{1} W^{2} (x, ρ, t) d ρ d x \\ = - \frac{ξ}{2} \int_{0}^{1} \int_{0}^{1} \frac{\partial}{\partial ρ} ((1 - τ^{'} (t) ρ) W^{2} (x, ρ, t)) d ρ d x \\ (35) & = \frac{ξ}{2} \int_{0}^{1} (W^{2} (x, 0, t) - W^{2} (x, 1, t)) d x + \frac{ξ τ^{'} (t)}{2} \int_{0}^{1} W^{2} (x, 1, t) d x . \end{array}$

From (33), (34) and (35), we obtain $\begin{array}{rcl} \frac{d E (t)}{d t} & = & - (μ_{1} - \frac{ξ}{2}) \int_{0}^{1} φ_{t}^{2} (x, t) d x + (\frac{ξ τ^{'} (t)}{2} - \frac{ξ}{2}) \int_{0}^{1} W^{2} (x, 1, t) d x \\ (36) & - μ_{2} \int_{0}^{1} φ_{t} (x, t) W (x, 1, t) d x - δ \int_{0}^{1} q^{2} d x . \end{array}$

Thanks to Young’s inequality, the last term in (36) can be estimated as follows $\begin{array}{l} μ_{2} \int_{0}^{1} φ_{t} (x, t) W (x, 1, t) d x \\ (37) & ⩽ \frac{μ_{2}}{2 \sqrt{1 - d}} \int_{0}^{1} φ_{t}^{2} (x, t) d x + \frac{μ_{2} \sqrt{1 - d}}{2} \int_{0}^{1} W^{2} (x, 1, t) d x . \end{array}$

Inserting (37) into (36), we obtain $\begin{array}{rcl} (38) & \frac{d E (t)}{d t} & ⩽ & - (μ_{1} - \frac{ξ}{2} - \frac{μ_{2}}{2 \sqrt{1 - d}}) \int_{0}^{1} φ_{t}^{2} (x, t) d x - δ \int_{0}^{1} q^{2} d x \\ + (\frac{ξ}{2} (τ^{'} (t) - 1) + \frac{μ_{2} \sqrt{1 - d}}{2}) \int_{0}^{1} W^{2} (x, 1, t) d x . \end{array}$

Then, we deduce that there exists $C > 0$ such that $\begin{matrix} (39) & \frac{d E (t)}{d t} ⩽ - δ \int_{0}^{1} q^{2} d x - C {\int_{0}^{1} φ_{t}^{2} (x, t) d x + \int_{0}^{1} W^{2} (x, 1, t) d x} . \end{matrix}$

Then, E is a non-increasing function.

Our second result is

Theorem 3.

Let $U_{0} \in D (A (0))$ . Assume that ( 6 ) holds. Then under the hypotheses ( 7 ), ( 8 ) and ( 10 ), any solution of problem ( 9 ) satisfies $\begin{matrix} (40) & E (t) ⩽ C e^{- γ t}, \forall t ⩾ 0, \end{matrix}$ for some positive constants C and γ independent of t.

To derive the exponential decay of the solution, we build a functional $L (t)$ which is equivalent to the energy $E (t)$ and satisfies $\begin{matrix} \frac{d L (t)}{d t} ⩽ - Λ L (t), \forall t ⩾ 0, \end{matrix}$ for some constant $Λ > 0$ .

First, let us consider the functional $I_{1}$ given by $\begin{matrix} (41) & I_{1} (t) : = \int_{0}^{1} ρ_{1} φ_{t} φ d x + \frac{μ_{1}}{2} \int_{0}^{1} φ^{2} d x . \end{matrix}$

Hence, we obtain

Lemma 1.

Let $(φ, ψ, θ, q, W)$ be the solution of ( 9 ). Hence, we have $\begin{array}{rcl} (42) & \frac{d I_{1}}{d t} & ⩽ & (- k + ε_{1} (\frac{k}{2} + \frac{μ_{2} c}{2})) \int_{0}^{1} φ_{t}^{2} d x + \frac{k}{2 ε_{1}} \int_{0}^{1} ψ_{x}^{2} d x \\ + \frac{μ_{2}}{2 ε_{1}} \int_{0}^{1} W^{2} (x, 1, t) d x + ρ_{1} \int_{0}^{1} φ_{t}^{2} d x, \end{array}$ where $c = 1 / π^{2}$ and $ε_{1} > 0$ .

Proof.

Differentiating $I_{1}$ , we obtain $\begin{matrix} (43) & \frac{d I_{1}}{d t} = ρ_{1} \int_{0}^{1} φ_{t t} φ d x + ρ_{1} \int_{0}^{1} φ_{t}^{2} d x + μ_{1} \int_{0}^{1} φ φ_{t} d x, \end{matrix}$ thanks to (9)₁, we find $\begin{matrix} \frac{d I_{1}}{d t} = k \int_{0}^{1} {(φ_{x} + ψ)}_{x} φ d x - μ_{2} \int_{0}^{1} φ φ_{t} (x, t - τ (t)) d x + ρ_{1} \int_{0}^{1} φ_{t}^{2} d x . \end{matrix}$

Thus, $\begin{matrix} \frac{d I_{1}}{d t} = - k \int_{0}^{1} (φ_{x} + ψ) φ_{x} d x - μ_{2} \int_{0}^{1} φ W (x, 1, t) d x + ρ_{1} \int_{0}^{1} φ_{t}^{2} d x . \end{matrix}$ The use of Young’s and Poincaré’s inequalities leads to (42). □

Now, let w be the solution of $\begin{matrix} (44) & - w_{x x} = ψ_{x}, w (0) = w (1) = 0 . \end{matrix}$

Then we get $\begin{matrix} w (x, t) = - \int_{0}^{x} ψ (y, t) d y + x (\int_{0}^{1} ψ (y, t) d y) . \end{matrix}$

Lemma 2.

The solution of ( 44 ) satisfies $\begin{matrix} \int_{0}^{1} w_{x}^{2} d x ⩽ \int_{0}^{1} ψ^{2} d x, \end{matrix}$ and $\begin{matrix} \int_{0}^{1} w_{t}^{2} d x ⩽ \int_{0}^{1} ψ_{t}^{2} d x . \end{matrix}$

Proof.

Multiplying (44) by w, integrating by parts and applying the Cauchy–Schwarz inequality to get $\begin{matrix} \int_{0}^{1} w_{x}^{2} d x ⩽ \int_{0}^{1} ψ^{2} d x . \end{matrix}$

We differentiate (44), we get $\begin{matrix} \int_{0}^{1} w_{t}^{2} d x ⩽ \int_{0}^{1} ψ_{t}^{2} d x . \end{matrix}$

This complete the proof of the lemma. □

We introduce the following functional $\begin{matrix} (45) & I_{2} (t) : = \int_{0}^{1} (ρ_{2} ψ_{t} ψ + ρ_{1} φ_{t} w - \frac{γ τ_{0}}{k} ψ q) d x . \end{matrix}$

Lemma 3.

Let $(φ, ψ, θ, q, W)$ be the solution of ( 9 ). Hence, we get, $\begin{array}{rcl} \frac{d I_{2} (t)}{d t} & ⩽ & (- b + \frac{c μ_{1} ε_{2}}{2} + \frac{c μ_{2} ε_{2}}{2} + \frac{δ γ ε_{2} c}{2 k}) \int_{0}^{1} ψ_{x}^{2} d x + \frac{μ_{2}}{2 ε_{2}} \int_{0}^{1} W^{2} (x, 1, t) d x \\ + (ρ_{2} + \frac{γ τ_{0} ε_{2}}{2 k} + \frac{ρ_{1} ε_{2}}{2}) \int_{0}^{1} ψ_{t}^{2} d x + (\frac{μ_{1}}{2 ε_{2}} + \frac{ρ_{1}}{2 ε_{2}}) \int_{0}^{1} φ_{t}^{2} d x \\ (46) & + (\frac{γ τ_{0}}{2 k ε_{2}} + \frac{δ γ}{2 k ε_{2}}) \int_{0}^{1} q^{2} d x, \end{array}$ where $ε_{2} > 0$ .

Proof.

By derivating (45), we deduce that $\begin{array}{rcl} \frac{d I_{2} (t)}{d t} & = & - b \int_{0}^{1} ψ_{x}^{2} d x + k \int_{0}^{1} φ ψ_{x} d x - k \int_{0}^{1} ψ^{2} d x + ρ_{2} \int_{0}^{1} ψ_{t}^{2} d x - k \int_{0}^{1} φ_{x} w_{x} d x \\ - k \int_{0}^{1} ψ w_{x} d x - μ_{1} \int_{0}^{1} φ_{t} w d x - μ_{2} \int_{0}^{1} W (x, 1, t) w d x + ρ_{1} \int_{0}^{1} φ_{t} w_{t} d x \\ - \frac{γ τ_{0}}{k} \int_{0}^{1} ψ_{t} q d x + \frac{δ γ}{k} \int_{0}^{1} ψ q d x . \end{array}$

By using (44) and Lemma 2, we have $\begin{array}{rcl} \frac{d I_{2} (t)}{d t} & ⩽ & - b \int_{0}^{1} ψ_{x}^{2} d x + k \int_{0}^{1} φ ψ_{x} d x - k \int_{0}^{1} ψ^{2} d x + ρ_{2} \int_{0}^{1} ψ_{t}^{2} d x + k \int_{0}^{1} φ_{x} ψ d x \\ + k \int_{0}^{1} ψ^{2} d x - μ_{1} \int_{0}^{1} φ_{t} w d x - μ_{2} \int_{0}^{1} W (x, 1, t) w d x + ρ_{1} \int_{0}^{1} φ_{t} w_{t} d x \\ - \frac{γ τ_{0}}{k} \int_{0}^{1} ψ_{t} q d x + \frac{δ γ}{k} \int_{0}^{1} ψ q d x . \end{array}$

Applying Young’s and Poincaré’s inequalities and using Lemma 2, we obtain (46). □

Now, we defined the functional $\begin{matrix} (47) & I_{3} (t) : = ξ τ (t) \int_{0}^{1} \int_{0}^{1} e^{- 2 τ (t) ρ} W^{2} (x, ρ, t) d x d ρ . \end{matrix}$

It satisfies the estimate stated in the following lemma

Lemma 4.

Let $(φ, ψ, θ, q, W)$ be the solution of ( 9 ). Then, we have $\begin{matrix} (48) & \frac{d I_{3} (t)}{d t} ⩽ - 2 I_{3} (t) + ξ \int_{0}^{1} φ_{t}^{2} (x, t) d x . \end{matrix}$

Proof.

Differentiating (48), we obtain $\begin{array}{rcl} (49) & \frac{d I_{3} (t)}{d t} & = & ξ τ^{'} (t) \int_{0}^{1} \int_{0}^{1} e^{- 2 τ (t) ρ} W^{2} (x, ρ, t) d x d ρ \\ - 2 ξ τ (t) τ^{'} (t) \int_{0}^{1} \int_{0}^{1} e^{- 2 τ (t) ρ} ρ W^{2} (x, ρ, t) d x d ρ \\ + 2 ξ τ (t) \int_{0}^{1} \int_{0}^{1} e^{- 2 τ (t) ρ} W_{t} (x, ρ, t) W (x, ρ, t) d x d ρ . \end{array}$

Using (9)₃ and the last term in (49) we find $\begin{array}{l} τ (t) \int_{0}^{1} \int_{0}^{1} e^{- 2 τ (t) ρ} W_{t} (x, ρ, t) W (x, ρ, t) d x d ρ \\ (50) & = \int_{0}^{1} \int_{0}^{1} e^{- 2 τ (t) ρ} (τ^{'} (t) ρ - 1) W_{ρ} (x, ρ, t) W (x, ρ, t) d x d ρ . \end{array}$

Also, one can see that $\begin{array}{l} \int_{0}^{1} \int_{0}^{1} e^{- 2 τ (t) ρ} (τ^{'} (t) ρ - 1) W_{ρ} (x, ρ, t) W (x, ρ, t) d x d ρ \\ = \frac{1}{2} \int_{0}^{1} \int_{0}^{1} \frac{\partial}{\partial ρ} (e^{- 2 τ (t) ρ} (τ^{'} (t) ρ - 1) W^{2} (x, ρ, t)) d x d ρ \\ + τ (t) \int_{0}^{1} \int_{0}^{1} e^{- 2 τ (t) ρ} (τ^{'} (t) ρ - 1) W^{2} (x, ρ, t) d x d ρ \\ (51) & - \frac{τ^{'} (t)}{2} \int_{0}^{1} \int_{0}^{1} e^{- 2 τ (t) ρ} W^{2} (x, ρ, t) d x d ρ . \end{array}$

Using (51) and (50), the equation (49) takes the form $\begin{array}{rcl} (52) & \frac{d I_{3} (t)}{d t} & = & - 2 ξ τ (t) \int_{0}^{1} \int_{0}^{1} e^{- 2 τ (t) ρ} W^{2} (x, ρ, t) d x d ρ + ξ \int_{0}^{1} φ_{t}^{2} x (, t) d x \\ - ξ e^{- 2 τ (t)} (1 - τ^{'} (t)) \int_{0}^{1} W^{2} (x, 1, t) d x . \end{array}$

Then, the proof of the lemma is complete. □

Now, as in [3], we introduce the functional $\begin{matrix} (53) & I_{4} (t) : = ρ_{2} ρ_{3} \int_{0}^{1} (\int_{0}^{x} θ (t, y) d y) ψ_{t} (t, x) d x . \end{matrix}$

Lemma 5.

Let $(φ, ψ, θ, q, W)$ be the solution of ( 9 ). Then, we get $\begin{array}{rcl} \frac{d I_{4} (t)}{d t} & ⩽ & (- γ ρ_{2} + \frac{ε_{4} ρ_{2} k}{2}) \int_{0}^{1} ψ_{t}^{2} d x + (\frac{ε_{4} ρ_{3}}{2} (b + k c)) \int_{0}^{1} ψ_{x}^{2} d x \\ + \frac{ε_{4}^{'} k ρ_{3} c}{2} \int_{0}^{1} φ_{x}^{2} d x + (γ ρ_{3} + \frac{ρ_{3}}{2 ε_{4}^{'}} (b + 2 k)) \int_{0}^{1} θ^{2} d x \\ (54) & + \frac{ρ_{2} k}{2 ε_{4}} \int_{0}^{1} q^{2} d x, \end{array}$ where $ε_{4}, ε_{4}^{'} > 0$ .

Proof.

Differentiating (53) and using (9)₃, we obtain $\begin{array}{rcl} \frac{d I_{4} (t)}{d t} & = & \int_{0}^{1} (\int_{0}^{x} ρ_{3} θ_{t} d y) ρ_{2} ψ_{t} d x + \int_{0}^{1} (\int_{0}^{x} ρ_{3} θ d y) ρ_{2} ψ_{t t} d x \\ = & - \int_{0}^{1} (\int_{0}^{x} (k q_{x} + γ ψ_{t x}) d y) ρ_{2} ψ_{t} d x \\ + \int_{0}^{1} (\int_{0}^{x} ρ_{3} θ d y) (b ψ_{x x} - k (φ_{x} + ψ) - γ θ_{x}) d x \\ = & - γ ρ_{2} \int_{0}^{1} ψ_{t}^{2} d x - ρ_{2} k \int_{0}^{1} q ψ_{t} d x - b ρ_{3} \int_{0}^{1} θ ψ_{x} d x \\ + k ρ_{3} \int_{0}^{1} θ φ d x - k ρ_{3} \int_{0}^{1} (\int_{0}^{x} θ d y) ψ d x + γ ρ_{3} \int_{0}^{1} θ^{2} d x . \end{array}$

Applying Young’s and Poincaré’s inequalities, we find (54). □

Here, we introduce the functional $\begin{matrix} (55) & I_{5} (t) : = - τ_{0} ρ_{3} \int_{0}^{1} q (t, x) (\int_{0}^{x} θ (t, y) d y) d x . \end{matrix}$

Lemma 6.

Let $(φ, ψ, θ, q, W)$ be the solution of ( 9 ). Then, we have $\begin{array}{rcl} (56) & \frac{d I_{5} (t)}{d t} & ⩽ & (- ρ_{3} k + \frac{ε_{5} ρ_{3} δ c}{2}) \int_{0}^{1} θ^{2} d x + \frac{ε_{5}^{'} τ_{0} γ}{2} \int_{0}^{1} ψ_{t}^{2} d x \\ (τ_{0} k + \frac{ρ_{3} δ}{2 ε_{5}} + \frac{τ_{0} γ}{2 ε_{5}^{'}}) \int_{0}^{1} q^{2} d x, \end{array}$ for any $ε_{5}, ε_{5}^{'} > 0$ .

We refer the reader to the proof in [3].

Proof of Theorem 3.

we define the Lyapunov functional $L$ as follows: $\begin{matrix} (57) & L (t) : = N E (t) + I_{1} (t) + N_{2} I_{2} (t) + I_{3} (t) + N_{4} I_{4} (t) + N_{5} I_{5} (t) . \end{matrix}$

Where $N, N_{2}, N_{4}, N_{5} > 0$ .

Combining (39), (42), (46), (48), (54) and (56), we get $\begin{array}{rcl} \frac{d L (t)}{d t} & ⩽ & [\frac{k}{2 ε_{1}} + N_{2} (- b + \frac{c μ_{1} ε_{2}}{2} + \frac{c μ_{2} ε_{2}}{2} + \frac{δ γ ε_{2} c}{2 k}) + N_{4} (\frac{ε_{4}^{'} ρ_{3}}{2} (b + k c))] \int_{0}^{1} ψ_{x}^{2} d x \\ + [- k + ε_{1} (\frac{k}{2} + \frac{μ_{2} c}{2}) + N_{4} \frac{ε_{4}^{'} k ρ_{3} c}{2}] \int_{0}^{1} φ_{x}^{2} d x - 2 I_{3} (t) \\ + [- N C + \frac{μ_{2}}{2 ε_{1}} + N_{2} \frac{μ_{2}}{2 ε_{1}}] \int_{0}^{1} W^{2} (x, 1, t) d x \\ + [- N C + N_{2} (\frac{μ_{1}}{2 ε_{2}} + \frac{ρ_{1}}{2 ε_{2}}) + ρ_{1} + ξ] \int_{0}^{1} φ_{t}^{2} d x \\ + [N_{2} (ρ_{2} + \frac{γ τ_{0} ε_{2}}{2 k} + \frac{ρ_{1} ε_{2}}{2}) + \frac{1}{2 τ} + N_{4} (- γ ρ_{2} + \frac{ε_{4} ρ_{2} k}{2}) + N_{5} \frac{ε_{5}^{'} τ_{0} γ}{2}] \int_{0}^{1} ψ_{t}^{2} d x \\ + [- N δ + N_{2} (\frac{γ τ_{0}}{2 k ε_{2}} + \frac{δ γ}{2 k ε_{2}}) + N_{4} \frac{ρ_{2} k}{2 ε_{4}} + N_{5} (τ_{0} k + \frac{ρ_{3} δ}{2 ε_{5}} + \frac{τ_{0} γ}{2 ε_{5}^{'}})] \int_{0}^{1} q^{2} d x \\ (58) & + [N_{4} (γ ρ_{3} + \frac{ρ_{3}}{2 ε_{4}^{'}} (b + 2 k)) + N_{5} (- ρ_{3} k + \frac{ε_{5} ρ_{3} δ c}{2})] \int_{0}^{1} θ^{2} d x . \end{array}$

Choose $ε_{1}$ , $ε_{2}$ , $ε_{4}$ and $ε_{5}$ small enough, such that $\begin{array}{l} ε_{2} (\frac{c μ_{1}}{2} + \frac{c μ_{2}}{2} + \frac{δ γ c}{2 k}) ⩽ \frac{b}{2}, ε_{1} (\frac{k}{2} + \frac{μ_{2} c}{2}) ⩽ \frac{k}{2}, \\ ε_{4} ⩽ \frac{γ}{k}, ε_{5} ⩽ \frac{k}{δ c} . \end{array}$

We can choose $N_{2}$ large enough, so that $\begin{matrix} N_{2} ⩾ \frac{2 K}{b ε_{1}} . \end{matrix}$

And also $N_{4}$ large enough so that $\begin{matrix} N_{4} \frac{γ ρ_{2}}{4} ⩾ N_{2} (ρ_{2} + \frac{γ τ_{0} ε_{2}}{2 k} + \frac{ρ_{1} ε_{2}}{2}) + \frac{1}{2 τ} . \end{matrix}$ Fixed $N_{2}$ and $N_{4}$ we have $\begin{matrix} ε_{4}^{'} ⩽ min {\frac{N_{2} b}{4 N_{4} ρ_{3} (b + k c)}, \frac{k}{2 N_{4} k ρ_{3} c}} . \end{matrix}$ Let $N_{5}$ be large enough such that $\begin{matrix} \frac{N_{5} ρ_{3} k}{4} ⩾ N_{4} (γ ρ_{3} + \frac{ρ_{3}}{2 ε_{4}^{'}} (b + 2 k)) . \end{matrix}$ We fix $ε_{5}^{'}$ small enough, we have $\begin{matrix} ε_{5}^{'} ⩽ \frac{N_{4} γ ρ_{2}}{4 N_{5} τ_{0} γ} . \end{matrix}$ Now, we have $\begin{matrix} \{\begin{matrix} \frac{CN}{2} ⩾ max {\frac{μ_{2}}{2 ε_{1}} + N_{2} \frac{μ_{2}}{2 ε_{1}}, N_{2} (\frac{μ_{1}}{2 ε_{2}} + \frac{ρ_{1}}{2 ε_{2}}) + ρ_{1}}, \\ \frac{N δ}{2} ⩾ N_{2} (\frac{γ τ_{0}}{2 k ε_{2}} + \frac{δ γ}{2 k ε_{2}}) + N_{4} \frac{ρ_{2} k}{2 ε_{4}} + N_{5} (τ_{0} k + \frac{ρ_{3} δ}{2 ε_{5}} + \frac{τ_{0} γ}{2 ε_{5}^{'}}) . \end{matrix} \end{matrix}$ We see that (58) is equivalent to $\begin{array}{rcl} \frac{d}{d t} L (t) & ⩽ & - η_{1} \int_{0}^{1} (ψ_{t}^{2} + ψ_{x}^{2} + φ_{t}^{2} + {(φ_{x} + ψ)}^{2} + θ^{2} + q^{2}) d x \\ (59) & - η_{1} \int_{0}^{1} \int_{0}^{1} W^{2} (x, ρ, t) d ρ d x, \end{array}$ then, we have $\begin{matrix} (60) & \frac{d}{d t} L (t) ⩽ η_{2} E (t), for all t ⩾ 0, \end{matrix}$ where $- η_{1}$ and $η_{2}$ as a positive constant. □

Lemma 7.

There exist two positive constants $β_{1}$ , $β_{2}$ such that $\begin{matrix} (61) & β_{1} E (t) ⩽ L (t) ⩽ β_{2} E (t), for all t ⩾ 0 . \end{matrix}$

Proof.

We defined the functional $\begin{matrix} H (t) = I_{1} (t) + N_{2} I_{2} (t) + I_{3} (t) + N_{4} I_{4} (t) + N_{5} I_{5} (t), \end{matrix}$ and we prove that $\begin{matrix} | H (t) | ⩽ C E (t), C > 0 . \end{matrix}$

From (41), (45), (47), (53) and (55) we find $\begin{array}{rcl} | H (t) | & ⩽ & | \int_{0}^{1} ρ_{1} φ_{t} φ d x + \frac{μ_{1}}{2} \int_{0}^{1} φ^{2} d x | + N_{2} | \int_{0}^{1} (ρ_{2} ψ_{t} ψ d x + ρ_{1} φ_{t} w - \frac{γ τ_{0}}{k} ψ q) d x | \\ + | \int_{0}^{1} \int_{0}^{1} e^{- 2 τ ρ} W^{2} (x, ρ, t) d ρ d x | + N_{4} | ρ_{2} ρ_{3} \int_{0}^{1} (\int_{0}^{x} θ (t, y) d y) ψ_{t} (t, x) d x | \\ + N_{5} | - τ_{0} ρ_{3} \int_{0}^{1} q (t, x) (\int_{0}^{x} θ (t, y) d y) d x | . \end{array}$

By using the relation $\begin{matrix} \int_{0}^{1} φ^{2} d x ⩽ 2 c \int_{0}^{1} {(φ_{x} + ψ)}^{2} d x + 2 c \int_{0}^{1} ψ_{x}^{2} d x, \end{matrix}$ moreover, by applying Young’s and Poincaré’s inequalities, we get $\begin{array}{rcl} | H (t) | & ⩽ & α_{1} \int_{0}^{1} φ_{t}^{2} d x + α_{2} \int_{0}^{1} ψ_{t}^{2} d x + α_{3} \int_{0}^{1} {(φ_{x} + ψ)}^{2} d x + α_{4} \int_{0}^{1} ψ_{x}^{2} d x + α_{5} \int_{0}^{1} θ^{2} d x \\ (62) & + α_{6} \int_{0}^{1} q^{2} d x + \int_{0}^{1} \int_{0}^{1} W^{2} (x, ρ, t) d ρ d x, \end{array}$ where the positive constants $α_{1}, \dots, α_{6}$ are: $\begin{matrix} \{\begin{matrix} α_{1} : = \frac{1}{2} (ρ_{1} + N_{2} ρ_{1}), \\ α_{2} : = \frac{1}{2} (N_{2} ρ_{2} + N_{4} ρ_{2} ρ_{3}), \\ α_{3} : = ρ_{1} c, \\ α_{4} : = \frac{1}{2} (\frac{N_{2} γ τ_{0} c}{k} + N_{2} ρ_{1} c^{2} + N_{2} ρ_{2} c), \\ α_{5} : = \frac{1}{2} (N_{4} ρ_{2} ρ_{3} c + N_{5} τ_{0} ρ_{3} c), \\ α_{6} : = \frac{1}{2} (N_{2} \frac{γ τ_{0}}{k} + N_{5} τ_{0} ρ_{3}) . \end{matrix} \end{matrix}$

By (62), we obtain $\begin{matrix} | H (t) | ⩽ \tilde{C} E (t), \end{matrix}$ for $\begin{matrix} \tilde{C} = \frac{max {α_{1}, α_{2}, α_{3}, α_{4}, α_{5}, α_{6}}}{min {ρ_{1}, ρ_{2}, ρ_{3}, K, b, k, γ, δ, τ_{0}}}, \end{matrix}$ we get $\begin{matrix} | L (t) - N E (t) | ⩽ \tilde{C} E (t) . \end{matrix}$ □

Combining (60) and (61), we deduce that $\begin{matrix} (63) & \frac{d}{d t} L (t) ⩽ - Λ L (t), for all t, Λ ⩾ 0 . \end{matrix}$

Then $\begin{matrix} (64) & L (t) ⩽ L (0) e^{- Λ t}, for all t ⩾ 0 . \end{matrix}$

Thus, the proof of the theorem is complete.

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