Open Access

General decay for weak viscoelastic Kirchhoff plate equations with delay boundary conditions

Boundary Value Problems20172017:96

https://doi.org/10.1186/s13661-017-0820-y

Received: 25 January 2017

Accepted: 30 May 2017

Published: 29 June 2017

Abstract

We consider a weak viscoelastic Kirchhoff plate model with time-varying delay in the boundary. By using a suitable energy and Lyapunov function, we obtain a decay rate for the energy, which depends on the behavior of g and α.

Keywords

Kirchhoff platerelaxation functiongeneral decaymemory termtime-varying delay

1 Introduction

The equation which describes the small vibration of a thin homogeneous, isotropic plate of uniform thickness h is given by
$$ \textstyle\begin{cases} { \rho h u_{tt} -\frac{\rho h^{3}}{12}\Delta u_{tt}+ D(0)\Delta^{2} u - \int_{0}^{t} D'(t-s) \Delta^{2} u (s)\, d s= f} ,\quad \hbox{in } \Omega\times(0 ,\infty), \\ { u = \frac{\partial u }{\partial\nu} = 0 },\quad \hbox{on } \Gamma_{0} \times(0, \infty), \\ {\mathcal {B}}_{1} u - { {\mathcal {B}}_{1} ( \int_{0}^{t} D'(t-s ) u(s )\, d s )} =-\nu\cdot m ,\quad \hbox{on } \Gamma_{1} \times(0 ,\infty), \\ {\mathcal {B}}_{2} u - { \frac{ h^{2}}{12}\frac{\partial u_{tt}}{\partial\nu}- {\mathcal {B}}_{2} ( \int_{0}^{t} D'(t-s ) u(s ) \,d s ) =-\frac{\partial\eta\cdot m}{\partial\eta} } ,\quad\hbox{on } \Gamma_{1} \times(0, \infty), \end{cases} $$
(1.1)
where Ω is an open bounded set of \({\mathbb {R}}^{2}\) with a sufficiently smooth boundary \(\Gamma=\Gamma_{0} \cup\Gamma_{1}\). Here, \(\Gamma_{0}\) and \(\Gamma_{1} \) are closed and disjoint. Let us denote by \(\nu= (\nu_{1} , \nu_{2} )\) the external unit normal vector to Γ, and let us denote by \(\eta=( - \nu_{2} , \nu_{1} )\) the unit tangent vector positively oriented on Γ. The differential operators \({\mathcal {B}}_{1}\) and \({\mathcal {B}}_{2}\) are given by
$${\mathcal {B}}_{1} u = \Delta u + (1-\mu) B_{1} u \quad\hbox{and} \quad{ \mathcal {B}}_{2} u = \frac{ \partial\Delta u}{\partial\nu} + (1-\mu) \frac{\partial B_{2} u}{\partial\eta} $$
and the operators \(B_{1}\) and \(B_{2}\) are defined by
$$\begin{gathered} B_{1} u = 2 \nu_{1} \nu_{2} \frac{\partial^{2} u}{\partial x\, \partial y} - \nu_{1}^{2}\frac{\partial^{2} u}{\partial y^{2}} - \nu_{2}^{2} \frac{\partial^{2} u}{\partial x^{2}} , \\ B_{2} u = \bigl(\nu_{1}^{2} - \nu_{2}^{2} \bigr)\frac{\partial^{2} u}{\partial x\, \partial y} + \nu_{1} \nu_{2} \biggl( \frac{\partial^{2} u}{\partial y^{2}} - \frac{\partial^{2} u}{\partial x^{2} } \biggr). \end{gathered}$$
The constants in the above equations have the following physical meanings: ρ is mass density, D is flexural rigidity, \(\mu\in(0,\frac{1}{2})\) is Poisson’s ratio, m is distribution of external force, \(m\cdot\nu\) is a bending moment about the normal vector, \(m\cdot\eta\) is a bending moment about the tangent vector and f is vertical loading on the faces of the plate. For simplicity, we assume that the bending moments about both the tangent and the normal vectors are zero. To simplify equation (1.1), we make the change of variable \(t\rightarrow t\sqrt{D(0)/\rho h}\) in the time scale and we take \(\gamma=h^{2}/12\), \(g(t)=D'(t)\) for any \(t>0\); with these notations the initial boundary value problem (1.1) is equivalent to
$$ \textstyle\begin{cases} { u_{tt} -\gamma\Delta u_{tt}+ \Delta^{2} u - \int_{0}^{t} g(t-s) \Delta^{2} u (s)\, d s= 0} ,\quad \hbox{in } \Omega\times(0 ,\infty), \\ { u = \frac{\partial u }{\partial\nu} = 0 },\quad \hbox{on } \Gamma_{0} \times(0, \infty), \\ {\mathcal {B}}_{1} u - { {\mathcal {B}}_{1} ( \int_{0}^{t} g(t-s ) u(s ) \,d s )} =0,\quad \hbox{on } \Gamma_{1} \times(0 ,\infty), \\ {\mathcal {B}}_{2} u -\gamma{\frac{\partial u_{tt}}{\partial\nu}- {\mathcal {B}}_{2} ( \int_{0}^{t} g(t-s ) u(s ) \,d s ) =0 } ,\quad \hbox{on } \Gamma_{1} \times(0, \infty) . \end{cases} $$
(1.2)
Rivera et al. [1] showed exponential and polynomial decay of the solutions to viscoelastic plate equation (1.2). They considered a relaxation function satisfying
$$-c_{0} g(t) \leq g'(t) \leq-c_{1} g(t),\qquad 0 \leq g''(t) \leq c_{2} g(t), $$
for some positive constant \(c_{i}\), \(i=0, 1, 2\). The uniform stabilization of Kirchhoff plates with linear or nonlinear boundary feedback was investigated by several authors [26].
It is well known that delay effects often arise in many practical problems because these phenomena depend not only on the present state but also on the past history of the system. In recent years, the behavior of solutions for the PDEs with time delay effects has become an active area of research; see, for instance, [711] and the references therein. Datko et al. [9] proved that a small delay in a boundary control is a source of instability. To stabilize a system involving delay terms, additional control terms will be necessary. Nicaise and Pignotti [11] considered the following wave equation with a linear damping and delay term inside the domain:
$$u_{tt } -\Delta u +\mu_{1} u_{t} + \mu_{2} u_{t}(t-\tau)=0 . $$
They obtained some stability results in the case \(0<\mu_{2} <\mu_{1} \). It is also showed in the case \(\mu_{2} \geq\mu_{1}\) that there exists a sequence of arbitrary small (or large) delays such that instabilities occur. Moreover, the same results were proved when both the damping and the delay acted on the boundary. Kirane and Said-Houari [10] investigated the following linear viscoelastic wave equation with a linear damping and a delay term
$$ u_{tt } -\Delta u + \int_{0}^{t} g(t-s) \Delta u(s)\, ds + \mu_{1} u_{t} + \mu_{2} u_{t} (t-\tau) =0, $$
(1.3)
where \(\mu_{1}\) and \(\mu_{2}\) are positive constants. They showed that its energy was exponentially decaying when \(\mu_{2} \leq\mu_{1}\). Dai and Yang [7] improved the results of [10] under weaker conditions. They also obtained an exponential decay results for the energy of problem (1.3) in the case \(\mu_{1}=0\). Furthermore, Nicaise and Pignotti [12] considered the following wave equation with time-dependent delay term:
$$u_{tt } -\Delta u +\mu_{1} u_{t} + \mu_{2} u_{t}\bigl(t-\tau(t)\bigr)=0, $$
where \(\tau(t)>0\) is the time-varying delay, \(\mu_{1}\) and \(\mu_{2}\) are real numbers with \(\mu_{1}>0\). They analyzed the exponential stability result under the condition
$$ |\mu_{2}| < \sqrt{1-d} \mu_{1}, $$
(1.4)
where d is a constant such that \(\tau'(t) \leq d <1\), \(\forall t>0\). Liu [13] investigated the viscoelastic wave equation (1.3) with time-varying delay term under condition (1.4).
The stability result of viscoelastic wave equations without time delay has been studied by many authors. Cavalcanti et al. [14] established an exponential rate of decay for a viscoelastic wave equation under the condition \(-\xi_{1} g(t) \leq g'(t) \leq-\xi_{2} g(t) \), \(t\geq0\), for some positive constant \(\xi_{i}\), \(i=1, 2\). Later, this assumption was relaxed by several authors. Berrimi and Messaoudi [15] proved exponential and polynomial decay rates under the condition \(g'(t) \leq-\xi g^{p}(t)\), \(t\geq0\), \(1\leq p< \frac{3}{2} \), for a positive constant ξ. Messaoudi [16] considered the following weak viscoelastic equation:
$$ u_{tt} -\Delta u + \alpha(t) \int_{0}^{t} g(t-s) \Delta u (s) \,ds =0 , $$
(1.5)
where α and g are positive nonincreasing functions defined on \({\mathbb {R}}^{+}\). Under some assumptions on the relaxation function g and the potential α, the author obtained a general decay result which depends on the behavior of g and α. For more results on weak viscoelastic equations, we can refer to [1719] and the references therein.
Recently, Yang [20] showed the existence and energy decay of solutions for the following Euler-Bernoulli equation with a delay:
$$ u_{tt } +\Delta^{2} u - \int_{0}^{t} g(t-s) \Delta^{2} u(s) \,ds + \mu_{1} u_{t} + \mu_{2} u_{t} (t-\tau) =0 $$
(1.6)
under some restrictions on \(\mu_{1}\) and \(\mu_{2}\). The author proved an exponential decay results for the energy in two cases (\(\mu_{1} \neq0\) or \(\mu_{1}=0\)). Moreover, the stability of partial differential equations with time delay effects has been discussed by many authors [2129].
Then, a natural problem is what would happen when a delay term occurs in (1.2). Motivated by these results [16, 18, 20, 29], we consider a decay rate of the solutions for the following weak viscoelastic Kirchhoff plate equations (1.2) with time-varying delay in the boundary:
$$ \textstyle\begin{cases} u_{tt}(x, t) -\gamma\Delta u_{tt}(x, t)+ \Delta^{2} u(x, t) -\alpha(t) {\int_{0}^{t} g(t-s) \Delta^{2} u (x,s) \,d s= 0 } ,\quad \hbox{in } \Omega\times(0 ,\infty), \\ u(x, t) = {\frac{\partial u(x, t) }{\partial\nu} = 0 },\quad \hbox{on } \Gamma_{0} \times(0, \infty), \\ {\mathcal {B}}_{1} u(x,t) - {\mathcal {B}}_{1} (\alpha(t) { \int_{0}^{t} g(t-s ) u(x,s ) \,d s} ) =0,\quad \hbox{on } \Gamma_{1} \times(0 ,\infty), \\ {\mathcal {B}}_{2} u(x, t) -\gamma {\frac{\partial u_{tt}(x,t)}{\partial\nu}- {\mathcal {B}}_{2} ( \alpha(t) \int_{0}^{t} g(t-s ) u(x,s ) \,d s} ) \\\quad=\mu_{1} u_{t}(x, t)+\mu_{2} u_{t} (x, t-\tau(t)) , \quad \hbox{on } \Gamma_{1} \times(0, \infty), \\ u(x,0) = u_{0}(x) , \qquad u_{t} (x,0) = u_{1}(x),\quad x\in\Omega, \\ u_{t}(x, t)=f_{0} (x, t) ,\quad (x, t) \in\Gamma_{1} \times[-\tau(0), 0), \end{cases} $$
(1.7)
where \(\mu_{1}\) is a positive constant, \(\mu_{2}\) is a real number, \(\tau(t)>0\) represents the time-varying delay, g and α are real functions satisfying some conditions to be specified later.

When the viscoelastic term is modulated by a time-dependent coefficient \(\alpha(t)\), we prove an energy decay result of the solutions for weak viscoelastic Kirchhoff plate equations (1.7) in the case \(\mu_{1} \neq0\) or \(\mu_{1} = 0\), respectively. In order to achieve this goal, we need a restriction of the size between the parameter \(\mu_{2}\) and the kernel g.

The paper is organized as follows. In Section 2, we give some preparations for our consideration and our main results. In Section 3, we study an energy decay of the solutions for problem (1.7). By introducing suitable energy and Lyapunov functionals, we obtain a decay estimate for the energy, which depends on the behavior of both α and g.

2 Preliminaries

In this section, we introduce some material needed in the proof of our result and state the main result. We denote
$$(u, v) = \int_{\Omega}uv \,d x , \qquad(u, v)_{\Gamma_{1}} = \int_{\Gamma_{1}} u v \,d \Gamma. $$
For simplicity, we denote \(\|\cdot\|_{L^{2}(\Omega)}\) and \(\|\cdot\|_{L^{2} (\Gamma_{1})}\) by \(\|\cdot\|\) and \(\|\cdot\|_{\Gamma_{1}}\), respectively. To study the existence of solution of system (1.7), we introduce the following spaces:
$$V=\bigl\{ u \in H^{1} (\Omega) \mid u =0 \hbox{ on } \Gamma_{0} \bigr\} ,\qquad W= \biggl\{ u \in H^{2} (\Omega) \bigm| u = \frac{\partial u }{\partial \nu} =0 \hbox{ on } \Gamma_{0} \biggr\} . $$
Let us define the following bilinear symmetric form:
$$\begin{aligned} a(u, v)={}& \int_{\Omega}\biggl( \frac{\partial^{2} u}{\partial x^{2}} \frac {\partial^{2} v}{\partial x^{2}} + \frac{\partial^{2} u}{\partial y^{2}} \frac{\partial^{2} v}{\partial y^{2}} +\mu\biggl( \frac{\partial^{2} u}{\partial x^{2}} \frac{\partial^{2} v}{\partial y^{2}} + \frac{\partial^{2} u}{\partial y^{2}} \frac{\partial^{2} v}{\partial x^{2}}\biggr)\\ &+ 2 ( 1 -\mu) \frac{\partial^{2} u}{\partial x \,\partial y} \frac{\partial^{2} v}{\partial x \,\partial y} \biggr) \,d x \,dy . \end{aligned} $$
Based on the integration by parts formula, a simple calculation yields
$$\bigl( \Delta^{2} u , v \bigr)= a(u, v) + ({\mathcal {B}}_{2} u ,v )_{\Gamma_{1}}- \biggl({\mathcal {B}}_{1} u , \frac {\partial v }{\partial\nu} \biggr)_{\Gamma_{1}} . $$
Since \(\Gamma_{0} \neq\emptyset\), we know that \(\sqrt{a(u,u)}\) is equivalent to the \(H^{2} (\Omega)\) norm on W, that is,
$$c_{0} \|u\|^{2}_{H^{2} (\Omega)} \leq a(u, u) \leq\tilde {c}_{0} \|u\|^{2}_{H^{2} (\Omega)}, $$
where \(c_{0}\) and \(\tilde{c}_{0}\) are generic positive constants. The Sobolev imbedding theorem and a trace estimate imply that for some positive constants \(C_{p} \), \(C_{s}\), \(\tilde{C_{p}} \) and \(\tilde{C_{s}}\),
$$ \begin{aligned}[c] &\|u\|^{2} \leq C_{p} a(u, u) , \qquad\|\nabla u\|^{2} \leq C_{s} a(u, u) , \\ &\|u\|^{2}_{\Gamma_{1}} \leq \tilde{C_{p}} a(u, u) \quad\hbox{and} \quad\|u\|_{\Gamma_{1}}^{2} \leq \tilde{C_{s}} \|\nabla u\|^{2}, \quad\forall u \in W. \end{aligned} $$
(2.1)
For the relaxation function g and the potential α, as in [16], we assume that
  1. (H1)
    \(g, \alpha: {\mathbb {R}}^{+} \rightarrow{\mathbb {R}}^{+}\) are nonincreasing differentiable functions satisfying
    $$ \begin{aligned}[c] &g(0)>0,\quad l_{0} := \int_{0}^{\infty}g(s) \,ds < \infty, \\ &\alpha(t)>0, \quad 1-2 \alpha(t) \int_{0}^{t} g(s) \,ds \geq l >0, \quad\hbox{for } t\geq0, \end{aligned}$$
    (2.2)
    and there exists a nonincreasing differentiable function \(\xi: {\mathbb {R}}^{+} \rightarrow{\mathbb {R}}^{+}\) satisfying
    $$ \xi(t)>0,\quad g'(t) \leq- \xi(t) g(t), \quad\hbox{for } t\geq0 $$
    (2.3)
    and
    $$ \lim_{t\rightarrow\infty} \frac{-\alpha'(t)}{\xi(t)\alpha(t)} =0 . $$
    (2.4)
     

Remark 2.1

Note that (H1) implies \({\lim_{t\rightarrow\infty} \frac{-\alpha'(t)}{\alpha(t)} =0.}\)

Since the function g is continuous and positive, we obtain
$$ \int_{0}^{t} g(s) \,ds \geq \int_{0}^{t_{0}} g(s) \,ds:= g_{0} >0 $$
(2.5)
for all \(t\geq t_{0}>0\). This fact will be used subsequently in the proof of our main result.
As in [12], for the time-varying delay, we assume that \(\tau\in W^{2, \infty}([0, T])\) for \(T>0\), and there exist positive constants \(\tau_{0}\), \(\tau_{1}\) and d satisfying
$$ 0< \tau_{0} \leq\tau(t) \leq\tau_{1} \quad\hbox{and}\quad \tau'(t) \leq d < 1 \quad\hbox{for all } t >0, $$
(2.6)
and that \(\mu_{1}\) and \(\mu_{2}\) satisfy
$$ |\mu_{2} | < \sqrt{1-d} \mu_{1} . $$
(2.7)
Let us introduce the function as in [12]
$$ z(x, \rho, t)=u_{t} \bigl(x, t-\tau(t) \rho\bigr),\quad x\in \Gamma_{1} , \rho \in(0,1), t>0 . $$
Then problem (1.7) is equivalent to
$$ \textstyle\begin{cases} { u_{tt}(x, t) -\gamma\Delta u_{tt}(x, t)+ \Delta^{2} u(x, t) -\alpha(t) \int_{0}^{t} g(t-s) \Delta^{2} u (x,s) \,d s= 0 },\quad \hbox{in } \Omega\times(0 ,\infty), \\ \tau(t) z_{t}(x,\rho, t)+(1-\tau'(t)\rho)z_{\rho}(x, \rho, t)=0,\quad \hbox{in } \Gamma_{1} \times(0, 1)\times(0 ,\infty), \\ {u(x, t) = \frac{\partial u(x, t) }{\partial\nu} = 0 , \quad\hbox{on } \Gamma_{0} \times(0, \infty),} \\ {\mathcal {B}}_{1} u(x,t) - { {\mathcal {B}}_{1} (\alpha(t) \int_{0}^{t} g(t-s ) u(x,s ) \,d s ) =0},\quad \hbox{on } \Gamma_{1} \times(0 ,\infty), \\ {\mathcal {B}}_{2} u(x, t) - { \gamma\frac{\partial u_{tt}(x,t)}{\partial\nu}- {\mathcal {B}}_{2} ( \alpha(t) \int_{0}^{t} g(t-s ) u(x,s ) \,d s )}\\\quad =\mu_{1} u_{t}(x, t)+\mu_{2} z (x, 1, t) , \quad \hbox{on } \Gamma_{1} \times(0, \infty), \\ u(x,0) = u_{0}(x) ,\qquad u_{t} (x,0) = u_{1}(x),\quad x\in\Omega, \\ z(x, 0, t)=u_{t}(x, t),\quad \hbox{on } \Gamma_{1} \times(0, \infty), \\ z(x, \rho, 0)=f_{0} (x, -\rho\tau(0)) ,\quad (x, \rho) \in\Gamma_{1} \times(0, 1). \end{cases} $$
(2.8)
We can prove the existence of weak solution by making use of the classical Faedo-Galerkin method. Then, using elliptic regularity and second order estimates, we can show the regularity of the solution. We state a well-posedness result without a proof here (see [1, 10, 12, 20]).

Lemma 2.1

Let (2.6) and (2.7) be satisfied and g, α satisfy (H1). If \((u_{0}, u_{1} )\in W \times V \), \(f_{0}\in L^{2}(\Gamma_{1}\times(0,1))\) and \(T>0\), then there exists a unique weak solution \((u, u_{t})\in C([0, T];W\times V)\) of problem (2.8). Moreover, if \((u_{0}, u_{1}) \in (W\cap H^{4}(\Omega)) \times(V\cap H^{3} (\Omega))\), \(f_{0} \in H^{2}(\Gamma_{1}\times(0,1))\), then the solution of (2.8) has the following regularity:
$$u\in C^{0}\bigl([0,T];W\cap H^{4}(\Omega)\bigr)\cap C^{1} \bigl([0,T];V\cap H^{3}(\Omega)\bigr). $$
Inspired by [12, 16], we define a modified energy functional as
$$ \begin{aligned} E(t):={}&\frac{1}{2} \|u_{t}\|^{2} +\frac{\gamma}{2}\| \nabla u_{t}\|^{2}+\frac {1}{2} \biggl(1-\alpha(t) \int_{0}^{t} g(s) \,ds \biggr) a(u, u)\\ & + \frac{\alpha(t)}{2} g \square\partial^{2} u + \frac{\zeta}{2} \int_{t-\tau(t)}^{t} e^{\lambda(s-t)}\big\| u_{t}(s) \big\| ^{2}_{\Gamma_{1}} \,ds , \end{aligned} $$
where ζ and λ are positive constants satisfying
$$ \frac{|\mu_{2}|}{\sqrt{1-d}} < \zeta< 2\mu_{1} - \frac{|\mu_{2}|}{\sqrt{1-d}}\quad \hbox{and}\quad \lambda< \frac{1}{\tau_{1}}\log\frac{\zeta\sqrt{1-d}}{|\mu_{2}|} . $$
(2.9)
Note that this choice of ζ is possible from assumption (2.7).

The main result of this paper is the following.

Theorem 2.1

Let (2.6) be satisfied and g, α satisfy (H1). Assume that either one of the following two conditions holds:
  1. (i)

    \(0< |\mu_{2}| <\sqrt{1-d} \mu_{1} \),

     
  2. (ii)

    \(\mu_{1} =0\), \(0<|\mu_{2} | < \mu_{0}\) and \(\alpha(t)\xi(t)>\xi_{0}\), \(\forall t \ge t_{1}\).

     
Then there exist positive constants k and K such that, for any solution of problem (1.7), the energy satisfies
$$ E(t) \leq K e^{-k\int_{t_{2}}^{t} \alpha(s) \xi(s)\,ds}, \quad\forall t \geq t_{2}, $$
(2.10)
where \(\mu_{0}\) and \(\xi_{0}\) are positive constants given by (3.29) and (3.33), respectively.

3 General decay of solutions

In this section we show a general decay rate. To simplify calculation, in our analysis we introduce the following notation:
$$ \begin{gathered} (g*u) (t) = \int_{0}^{t} g(t- s) u (s ) \,d s , \\ ( g \square u ) (t):= \int_{0}^{t} g(t-s) \big\| u( t)-u( s) \big\| ^{2} \,ds, \\ \bigl(g \square\partial^{2} u\bigr) (t) := \int_{0}^{t} g(t-s) a\bigl(u(t)-u(s),u( t)-u(s)\bigr) \,ds . \end{gathered}$$
We give some estimates related to the convolution operator. By the symmetry of \(a(\cdot, \cdot)\) and direct calculations, we shall see that
$$ \begin{aligned}[b] \alpha(t) a(g*u, u_{t} ) ={}& {-}\frac{\alpha(t)}{2}g(t) a(u, u) +\frac{\alpha(t)}{2}g' \square\partial^{2} u + \frac{\alpha'(t)}{2}g \square\partial^{2} u \\ &-\frac{\alpha'(t)}{2} \biggl( \int_{0}^{t} g(s) \,ds \biggr)a(u, u)\\ & -\frac{1}{2} \frac{d}{dt} \biggl[ \alpha(t) g \square \partial^{2} u -\alpha(t) \biggl( \int_{0}^{t} g(s) \,ds \biggr) a(u, u) \biggr], \end{aligned}$$
(3.1)
and
$$ a(g*u, u ) \leq2 \biggl( \int_{0}^{t} g(s) \,ds \biggr) a(u, u) + \frac{1}{4} g \square\partial^{2} u . $$
(3.2)
To prove our result, we need to introduce the following auxiliary functionals:
$$\begin{aligned}& \Phi(t) = \int_{\Omega}u_{t} u \,dx + \gamma \int_{\Omega}\nabla u_{t} \nabla u \,d x , \\& \begin{aligned} \Psi(t) ={}& {-} \int_{\Omega}u_{t} \int_{0}^{t} g(t-s) \bigl(u(t)-u(s)\bigr)\,d s \,d x\\ &-\gamma \int_{\Omega}\nabla u_{t} \int_{0}^{t} g(t-s) \bigl(\nabla u (t)-\nabla u(s) \bigr)\,d s \,d x . \end{aligned} \end{aligned}$$

We divide the proof of Theorem 2.1 into two cases as follows.

Case 1: \(0<|\mu_{2} |< \sqrt{1-d} \mu_{1}\).

First, we consider the functional
$$ L(t) := N E(t) + \epsilon_{1} \alpha(t)\Phi(t) + \epsilon_{2} \alpha(t) \Psi(t) , $$
(3.3)
where \(\epsilon_{1}\) and \(\epsilon_{2}\) are positive constants. We easily get the following lemmas.

Lemma 3.1

For \(N>0\) large enough, there exist positive constants \(C_{1}\) and \(C_{2}\) such that
$$ C_{1} {E}(t) \leq L(t) \leq C_{2} { E}(t), \quad\forall t \geq0. $$
(3.4)

Proof

By applying Young’s inequality, the Cauchy-Schwarz inequality, (2.1) and (2.2), we clearly have
$$\begin{aligned}& \big| \Phi(t) \big|\leq \frac{1}{2} \| u_{t}\|^{2} + \frac{\gamma}{2} \|\nabla u_{t}\|^{2}+ \frac{C_{p} +\gamma C_{s} }{2 l} \biggl(1-\alpha(t) \int_{0}^{t} g(s) \, ds \biggr) a (u,u), \\& \big|\Psi(t) \big| \leq\frac{1}{2} \| u_{t}\|^{2}+ \frac{\gamma}{2}\|\nabla u_{t}\|^{2} +\frac{ ( C_{p} + \gamma C_{s} )l_{0} }{2}g \square \partial^{2} u , \end{aligned}$$
which gives us
$$ \begin{aligned} \big|L(t) - N {E}(t) \big| \leq{}& \frac{\alpha(0)}{2}(\epsilon_{1}+ \epsilon_{2}) \| u_{t}\|^{2}+\frac{\gamma \alpha(0)}{2}( \epsilon_{1}+\epsilon_{2}) \|\nabla u_{t} \|^{2}\\ & + \frac{ \epsilon_{2} (C_{p} + \gamma C_{s} )l_{0} }{2} \alpha(t)g \square \partial^{2} u \\ & + \frac{\epsilon_{1} (C_{p}+\gamma C_{s} )\alpha(0)}{2 l} \biggl(1-\alpha(t) \int_{0}^{t} g(s) \,ds \biggr) a (u,u)\\ \leq{}& C_{3} { E}(t), \end{aligned}$$
where \(C_{3}=\max \{ \alpha(0)(\epsilon_{1}+\epsilon_{2}), \epsilon_{2} (C_{p} + \gamma C_{s} )l_{0} , \frac{\epsilon_{1} (C_{p}+\gamma C_{s} )\alpha(0)}{ l} \}\). Choosing \(N>0\) large, we complete the proof of Lemma 3.1. □

Lemma 3.2

Let (2.6) and (2.7) be satisfied and g, α satisfy (H1). Then, for all regular solutions of problem (1.7), there exist positive constants \(\alpha_{0}\) and \(\alpha_{1}\) satisfying
$$ \begin{aligned}[b] E'(t) \leq{}&{-}\alpha_{0} \|u_{t} \|^{2}_{\Gamma_{1}} -\alpha_{1} \big\| u_{t} \bigl(t- \tau(t)\bigr)\big\| ^{2}_{\Gamma_{1}} + \frac{\alpha(t)}{2}g' \square \partial^{2} u - \frac{\alpha'(t)}{2} \biggl( \int_{0}^{t} g(s)\,ds \biggr) a (u, u)\hspace{-12pt} \\ & -\frac{\lambda\zeta}{2} \int_{t-\tau(t)}^{t} e^{\lambda (s-t)} \big\| u_{t}(s) \big\| ^{2}_{\Gamma_{1}} \,ds . \end{aligned}$$
(3.5)

Proof

Multiplying (1.7) by \(u_{t} (t) \), we get the identity
$$ \begin{aligned}[b] &\frac{d}{dt} \biggl\{ \frac{1}{2}\|u_{t}\|^{2}+ \frac{\gamma}{2} \|\nabla u_{t}\|^{2}+\frac{1}{2} a(u, u) \biggr\} \\&\quad = -\mu_{1}\|u_{t}\|^{2}_{\Gamma_{1}} -\mu_{2} \bigl(u_{t} \bigl(t-\tau(t)\bigr), u_{t} \bigr)_{\Gamma_{1}} + \alpha(t) a(g* u , u_{t}) . \end{aligned}$$
(3.6)
Applying (3.1) to (3.6), we have
$$ \begin{aligned}[b] E' (t) ={}& {-}\mu_{1}\|u_{t} \|^{2}_{\Gamma_{1}} -\mu_{2} \bigl(u_{t} \bigl(t- \tau(t)\bigr), u_{t}\bigr)_{\Gamma_{1}} \\ &-\frac{\alpha(t)}{2}g(t) a(u, u) +\frac{\alpha(t)}{2}g' \square\partial ^{2} u + \frac{\alpha'(t)}{2} g \square\partial^{2} u \\ & - \frac{\alpha'(t)}{2} \biggl( \int_{0}^{t} g(s)\,ds \biggr) a (u, u)+ \frac{\zeta}{2}\|u_{t}\|^{2}_{\Gamma_{1}} \\ & - \frac{\zeta}{2}e^{-\lambda \tau(t)}\bigl(1-\tau'(t)\bigr) \big\| u_{t}\bigl(t-\tau(t)\bigr)\big\| ^{2}_{\Gamma_{1}}- \frac{\lambda\zeta}{2} \int_{t-\tau(t)}^{t} e^{\lambda(s-t)} \big\| u_{t}(s) \big\| ^{2}_{\Gamma_{1}} \,ds . \end{aligned}$$
(3.7)
From Young’s inequality, we obtain
$$ -\mu_{2} \bigl(u_{t}\bigl(t-\tau(t)\bigr), u_{t} \bigr)_{\Gamma_{1}} \leq\frac{|\mu_{2}|}{2\sqrt {1-d}}\|u_{t}\|^{2}_{\Gamma_{1}} + \frac{|\mu_{2}| \sqrt{1-d}}{2}\big\| u_{t} \bigl(t-\tau(t)\bigr)\big\| ^{2}_{\Gamma_{1}} . $$
(3.8)
By (2.6), we get
$$ - \frac{\zeta}{2}e^{-\lambda \tau(t)}\bigl(1-\tau'(t)\bigr) \big\| u_{t}\bigl(t-\tau(t)\bigr)\big\| ^{2}_{\Gamma_{1}} \leq- \frac{\zeta(1-d)}{2e^{\lambda\tau_{1}} }\big\| u_{t}\bigl(t-\tau(t)\bigr)\big\| ^{2}_{\Gamma_{1}} . $$
(3.9)
Combining (3.7)-(3.9) and (H1), we have
$$ \begin{aligned} E' (t) \leq{}&{-} \biggl( \mu_{1} - \frac{\zeta}{2}-\frac{|\mu_{2}|}{2\sqrt{1-d}} \biggr) \|u_{t} \|^{2}_{\Gamma_{1}} - \biggl(\frac{\zeta(1-d)}{2e^{\lambda\tau_{1}}}- \frac{|\mu_{2}|\sqrt{1-d}}{2} \biggr) \big\| u_{t} \bigl(t-\tau(t)\bigr)\big\| ^{2}_{\Gamma_{1}} \\ & + \frac{\alpha(t)}{2}g'\square \partial^{2} u - \frac{\alpha'(t)}{2} \biggl( \int_{0}^{t} g(s)\,ds \biggr) a (u, u) - \frac{\lambda\zeta}{2} \int_{t-\tau(t)}^{t} e^{\lambda(s-t)} \big\| u_{t}(s) \big\| ^{2}_{\Gamma_{1}} \,ds . \end{aligned}$$
By using condition (2.9), we obtain
$$\alpha_{0} := \mu_{1}-\frac{\zeta}{2}-\frac{|\mu_{2}|}{2\sqrt{1-d}} >0 \quad\hbox{and} \quad\alpha_{1}:= \frac{\zeta(1-d)}{2e^{\lambda\tau_{1}}}- \frac{|\mu_{2}|\sqrt{1-d}}{2} >0 , $$
which implies the desired inequality (3.5). The proof is now complete. □

Lemma 3.3

Under assumption (2.2), the functional Φ satisfies the estimate
$$ \begin{aligned}[b] \Phi'(t) \leq{}&{-} \frac{l}{2} a(u, u) +\|u_{t} \|^{2} +\gamma\|\nabla u_{t}\|^{2}\\ &+ \frac{\alpha(t)}{4} g \square\partial^{2} u + \frac{\mu_{1}^{2} \tilde{C}_{p} }{l} \|u_{t}\|^{2}_{\Gamma_{1}}+ \frac{\mu_{2}^{2} \tilde{C}_{p}}{l} \big\| u_{t}\bigl(t-\tau(t)\bigr)\big\| ^{2}_{\Gamma_{1}} . \end{aligned}$$
(3.10)

Proof

By using (1.7) and (3.2), we get
$$ \begin{aligned}[b] \Phi'(t) ={}&\|u_{t}\|^{2} +\gamma\| \nabla u_{t}\|^{2}- a(u, u)+ \alpha(t)a(g*u, u)\\ & - \mu_{1}(u_{t}, u )_{\Gamma_{1}} -\mu_{2} \bigl(u_{t} \bigl(t-\tau(t)\bigr), u\bigr)_{\Gamma_{1}} \\ \leq{}&\|u_{t}\|^{2} +\gamma\|\nabla u_{t}\|^{2} - \biggl( 1-2\alpha(t) \int_{0}^{t} g(s) \,ds \biggr) a(u, u)+ \frac{\alpha(t)}{4} g\square \partial^{2} u \\ & - \mu_{1}(u_{t}, u )_{\Gamma_{1}} -\mu_{2} \bigl(u_{t} \bigl(t-\tau(t)\bigr), u\bigr)_{\Gamma_{1}} . \end{aligned}$$
(3.11)
From Young’s inequality and (2.1), we see that, for any \(\eta>0\),
$$\begin{aligned}& - \mu_{1}(u_{t}, u )_{\Gamma_{1}} \leq \frac{\eta\tilde{C}_{p} }{2}a(u,u) +\frac{\mu_{1}^{2}}{2\eta}\|u_{t}\|^{2}_{\Gamma_{1}} , \end{aligned}$$
(3.12)
$$\begin{aligned}& -\mu_{2} \bigl(u_{t} \bigl(t-\tau(t)\bigr), u \bigr)_{\Gamma_{1}} \leq\frac{\eta \tilde{C}_{p} }{2}a(u,u) + \frac{\mu_{2}^{2}}{2\eta} \big\| u_{t}\bigl(t-\tau(t)\bigr)\big\| ^{2}_{\Gamma_{1}} . \end{aligned}$$
(3.13)
Combining (2.2), (3.12) and (3.13) with (3.11) and choosing \(\eta=\frac {l}{2 \tilde{C}_{p}}\), we have (3.10). □

Lemma 3.4

Under assumption (2.2), the functional Ψ satisfies the estimate
$$ \begin{aligned}[b] \Psi'(t) \leq{}& {-} \biggl( \int_{0}^{t} g(s) \,ds -\delta \biggr) \|u_{t}\|^{2} -\gamma \biggl( \int_{0}^{t} g(s) \,ds - \delta \biggr) \|\nabla u_{t}\|^{2} \\ &+ \delta \biggl(1+ \biggl(\frac{1-l}{2} \biggr)^{2} \biggr)a(u,u) +\delta\|u_{t}\|^{2}_{\Gamma_{1}} +\delta \big\| u_{t}\bigl(t-\tau(t)\bigr)\big\| ^{2}_{\Gamma_{1}} \\ & + \biggl( \alpha(t)+\frac{1}{2 \delta} +\frac{ \mu_{1}^{2}\tilde{C}_{p}}{4\delta} +\frac{ \mu_{2}^{2}\tilde{C}_{p}}{4\delta} \biggr) \biggl( \int_{0}^{t} g(s) \,ds \biggr) g\square \partial^{2} u\\ &-\frac{g(0)(C_{p} +\gamma C_{s})}{4\delta} g' \square \partial^{2} u. \end{aligned}$$
(3.14)

Proof

Similarly, we find that
$$ \begin{aligned}[b] \Psi'(t) ={}& \int_{0}^{t} g(t-s)a \bigl( u(t)-u(s), u(t) \bigr) \,ds \\ &- \alpha(t) \int_{0}^{t} g(t-s)a \biggl( u(t)-u(s), \int_{0}^{t} g(t-\tau) u(\tau) \,d \tau\biggr) \,ds \\ & + \mu_{1} \int_{0}^{t} g(t-s) \bigl(u(t)-u(s), u_{t}(t) \bigr)_{\Gamma_{1}} \, d s\\ & +\mu_{2} \int_{0}^{t} g(t-s) \bigl(u(t)-u(s), u_{t} \bigl(t-\tau(t)\bigr) \bigr)_{\Gamma_{1}} \,d s \\ & - \gamma \int_{0}^{t} g'(t-s) \bigl( \nabla u(t)- \nabla u(s), \nabla u_{t}(t)\bigr) \,ds\\ & - \int_{0}^{t} g'(t-s) \bigl( u(t)-u(s), u_{t}(t)\bigr) \,d s \\ & -\gamma \biggl( \int_{0}^{t} g(s) \,d s \biggr) \|\nabla u_{t}\|^{2}- \biggl( \int_{0}^{t} g(s)\,d s \biggr) \|u_{t} \|^{2} \\ :={}& I_{1} + \cdots+I_{6} -\gamma \biggl( \int_{0}^{t} g(s) \,d s \biggr) \|\nabla u_{t}\|^{2} - \biggl( \int_{0}^{t} g(s)\,d s \biggr) \|u_{t} \|^{2} . \end{aligned}$$
(3.15)
Now, we estimate the terms on the right-hand side of (3.15). Young’s inequality, (2.1) and (2.2) give that
$$\begin{aligned}& |I_{1}| \leq \delta a(u, u) + \frac{1}{4\delta} \biggl( \int_{0}^{t} g(s) \,ds \biggr) g \square \partial^{2} u , \\ & \begin{aligned} |I_{2}| &\leq\alpha(t) \biggl( \int_{0}^{t} g(s)\,d s \biggr) g\square \partial^{2} u + \alpha(t) \int_{0}^{t} g(t-s) \int_{0}^{t} g(t-\tau) a\bigl(u(t)-u(s) , u(t)\bigr) \,d\tau\,ds \\ & \leq\delta \biggl(\frac{1-l}{2} \biggr)^{2} a(u, u) + \biggl( \alpha(t) + \frac{1}{4\delta} \biggr) \biggl( \int_{0}^{t} g(s)\,d s \biggr) g \square \partial^{2} u , \end{aligned}\\ & |I_{3}| \leq\delta\| u_{t} \|^{2}_{\Gamma_{1}} +\frac{\mu_{1}^{2}\tilde{C}_{p}}{4\delta} \biggl( \int_{0}^{t} g(s)\,d s \biggr) g\square \partial^{2} u , \\ & |I_{4}| \leq\delta\big\| u_{t}\bigl(t-\tau(t)\bigr) \big\| ^{2}_{\Gamma_{1}} + \frac{\mu_{2}^{2} \tilde {C}_{p}}{4\delta} \biggl( \int_{0}^{t} g(s)\,d s \biggr) g \square \partial^{2} u , \\ & \begin{aligned} |I_{5}| &\leq \gamma\delta\|\nabla u_{t}\|^{2} + \frac{\gamma}{4\delta} \int_{\Omega}\biggl( \int_{0}^{t} g' (t-s)\big|\nabla u(t)-\nabla u(s)\big| \,d s \biggr)^{2} \,dx\\ & \leq\gamma\delta \|\nabla u_{t} \|^{2} -\frac{ g(0)\gamma C_{s} }{4\delta} g' \square\partial^{2} u , \end{aligned}\\ & |I_{6}|\leq\delta\|u_{t}\|^{2}+ \frac{1}{4\delta} \int_{\Omega}\biggl( \int_{0}^{t} g' (t-s)\big| u(t)-u(s)\big| \,d s \biggr)^{2} \,dx \leq\delta\|u_{t}\|^{2}- \frac{g(0)C_{p}}{4 \delta} g' \square\partial^{2} u , \end{aligned}$$
where \(\delta>0\). From the above estimates, we obtain (3.14). □

Lemma 3.5

For \(t_{0} > 0\) and sufficiently large \(N>0\), there exist \(k_{1}>0\), \(k_{2}>0\) and \(t_{1}\geq t_{0}\) such that
$$ L'(t) \leq- k_{1} \alpha(t) { E}(t) + k_{2} \alpha(t) g \square\partial^{2} u ,\quad \forall t \geq t_{1} , $$
(3.16)
where \(k_{1}\) and \(k_{2}\) depend on \(g_{0}\).

Proof

By using (2.1) and Young’s inequality, we get
$$ \begin{aligned}[b] \epsilon_{1} \alpha'(t) \Phi(t)+ \epsilon_{2} \alpha'(t) \Psi(t) \leq{}&{-}{\alpha'(t) }\|u_{t}\|^{2}- \alpha'(t)\gamma\|\nabla u_{t}\|^{2} -\frac {\alpha'(t)\epsilon_{1}^{2} }{2} ( C_{p} + \gamma C_{s}) a(u, u)\hspace{-24pt} \\ & -\frac{\alpha'(t)\epsilon_{2}^{2} }{2} (C_{p} +\gamma C_{s}) \biggl( \int_{0}^{t} g(s)\,ds \biggr) g \square \partial^{2} u . \end{aligned}$$
(3.17)
Then, using (2.5), (3.3), (3.5), (3.10), (3.14) and (3.17), we have
$$\begin{aligned} L' (t) \leq{}&{-} \alpha(t) \biggl( (g_{0}- \delta)\epsilon_{2}-\epsilon_{1} +\frac{\alpha'(t)}{\alpha(t)} \biggr) \|u_{t}\|^{2} -\gamma\alpha(t) \biggl( (g_{0} - \delta) \epsilon_{2} - \epsilon_{1} +\frac{\alpha'(t)}{\alpha(t)} \biggr) \|\nabla u_{t}\|^{2} \\ & -\alpha(t) \biggl\{ \frac{l\epsilon_{1}}{2} - \biggl( 1+\biggl( \frac{1-l}{2}\biggr)^{2} \biggr)\delta\epsilon_{2}\\ & + \frac{N \alpha'(t)}{2\alpha(t)} \biggl( \int_{0}^{t} g(s) \,ds \biggr)+ \frac{\alpha'(t)\epsilon_{1}^{2} }{2\alpha(t)} ( C_{p} + \gamma C_{s}) \biggr\} a(u, u) \\ & - \alpha(t) \biggl( \frac{\alpha_{0} N}{\alpha(0)}- \delta\epsilon_{2} - \frac{\mu_{1}^{2} \tilde{C}_{p} \epsilon_{1}}{l} \biggr) \|u_{t}\|^{2}_{\Gamma_{1}}\\ & - \alpha(t) \biggl( \frac{\alpha_{1} N}{\alpha(0)}- \delta\epsilon_{2} - \frac{\mu_{2}^{2} \tilde{C}_{p}\epsilon_{1}}{l} \biggr) \big\| u_{t} \bigl(t-\tau(t)\bigr) \big\| ^{2}_{\Gamma_{1}} \\ & + \alpha(t) \biggl[ \frac{\epsilon_{1} \alpha(t)}{4}\\ & + \biggl\{ \epsilon_{2} \biggl( \alpha(t) +\frac{2+(\mu_{1}^{2}+\mu_{2}^{2})\tilde{C}_{p}}{4\delta} \biggr)- \frac{\alpha'(t)\epsilon_{2}^{2} }{2\alpha(t)} (C_{p} +\gamma C_{s}) \biggr\} \biggl( \int_{0}^{t} g(s) \,ds \biggr) \biggr] g \square \partial^{2} u \\ & + \alpha(t) \biggl(\frac{N}{2}-\frac{ g(0)\epsilon_{2}}{4\delta} (C_{p} + \gamma C_{s}) \biggr) g' \square \partial^{2} u - \frac{\lambda\zeta N}{2} \int_{t-\tau(t)}^{t} e^{\lambda(s-t)} \big\| u_{t} (s) \big\| ^{2}_{\Gamma_{1}} \,ds. \end{aligned}$$
We first choose \(\delta>0\) so small that
$$\delta< \min \biggl\{ \frac{g_{0}}{2}, \frac{lg_{0}}{2(4+(1-l)^{2})} \biggr\} . $$
Then, we obtain
$$g_{0} -\delta> \frac{1}{2}g_{0} \quad\hbox{and}\quad \frac{2\delta}{l} \biggl( 1+\biggl(\frac{1-l}{2}\biggr)^{2} \biggr) < \frac{1}{4} g_{0} . $$
Hence δ is fixed, the choice of any two positive constants \(\epsilon_{1}\) and \(\epsilon_{2}\) satisfying
$$\frac{g_{0}}{4} \epsilon_{2} < \epsilon_{1} < \frac{g_{0}}{2} \epsilon_{2} $$
will make
$$(g_{0}-\delta)\epsilon_{2}-\epsilon_{1} >0\quad \hbox{and} \quad\frac{l\epsilon_{1}}{2} - \biggl( 1+\biggl(\frac{1-l}{2} \biggr)^{2} \biggr)\delta \epsilon_{2} >0 . $$
As long as δ, \(\epsilon_{1}\) and \(\epsilon_{2}\) are fixed, we take N large enough such that
$$\frac{\alpha_{0} N}{\alpha(0)}- \delta\epsilon_{2} - \frac{\mu_{1}^{2} \tilde{C}_{p} \epsilon_{1}}{l}>0,\qquad \frac{\alpha_{1} N}{\alpha(0)}- \delta \epsilon_{2} - \frac{\mu_{2}^{2} \tilde{C}_{p}\epsilon_{1}}{l} >0 $$
and
$$\frac{N}{2}-\frac{g(0)\epsilon_{2}}{4\delta} (C_{p} + \gamma C_{s}) >0 . $$
Since \({\lim_{t\rightarrow \infty}\frac{-\alpha'(t)}{\alpha(t)} =0,}\) we can take \(t_{1} \geq t_{0}\) sufficiently large so that
$$\begin{gathered} (g_{0}-\delta)\epsilon_{2}-\epsilon_{1} +\frac{\alpha'(t)}{\alpha(t)} >0, \\ \frac{l\epsilon_{1}}{2} - \biggl( 1+\biggl(\frac{1-l}{2} \biggr)^{2} \biggr)\delta \epsilon_{2} + \frac{N \alpha'(t)}{2\alpha(t)} \biggl( \int_{0}^{t} g(s) \,ds \biggr)+ \frac{\alpha'(t)\epsilon_{1}^{2} }{2\alpha(t)} ( C_{p} + \gamma C_{s}) >0 . \end{gathered}$$
Therefore, we get (3.16) for some positive constants \(k_{1}\) and \(k_{2}\) depending on \(g_{0}\). □
Multiplying (3.16) by \(\xi(t)\) and using (2.3) and (3.5), we find that
$$ \begin{aligned} \xi(t) L'(t) &\leq- k_{1}\alpha(t) \xi(t) { E}(t) + k_{2} \alpha(t)\xi(t) g \square\partial^{2} u \\ & \leq- k_{1}\alpha(t) \xi(t) { E}(t) - k_{2} \alpha(t) g' \square \partial^{2} u \\ & \leq- k_{1}\alpha(t) \xi(t) { E}(t) - k_{2} \biggl( 2 E'(t) +\alpha'(t) \biggl( \int_{0}^{t} g(s) \,ds \biggr) a(u,u) \biggr) ,\quad \forall t\geq t_{1} . \end{aligned}$$
From \(\xi'(t) \leq0\), (2.2) and the definition of \(E(t)\), we obtain
$$\begin{aligned} \bigl( \xi(t) L(t) + 2k_{2} E(t) \bigr)' &\leq- k_{1}\alpha(t) \xi(t) E(t) - k_{2} \alpha'(t) \biggl( \int_{0}^{t} g(s) \,ds \biggr) a(u,u) \\ & \leq- \biggl( k_{1} + \frac{2k_{2} \alpha'(t)}{l \alpha(t) \xi(t)} \biggl( \int _{0}^{t} g(s) \,ds \biggr) \biggr)\alpha(t) \xi(t) E(t) , \quad\forall t\geq t_{1} . \end{aligned}$$
By (2.4), we can choose \(t_{2} \geq t_{1}\) such that \(k_{1} + \frac{2k_{2} \alpha'(t)}{l \alpha(t) \xi(t)} (\int_{0}^{t} g(s) \,ds ) >0\) for \(t\geq t_{2}\). Let \({\mathcal {L}}(t) = \xi(t) L(t) + 2k_{2} E(t) \), then from (3.4) we can see that \({\mathcal {L}}(t)\) is equivalent to \(E(t)\). Then we deduce that
$${\mathcal {L}}'(t) \leq- k \alpha(t) \xi(t) {\mathcal {L}}(t) , \quad\forall t\geq t_{2} , $$
for some positive constant k depending on \(g_{0}\), α and ξ. Integrating this over \((t_{2} , t)\), we get
$${\mathcal {L}}(t) \leq{\mathcal {L}}(t_{2}) e^{-k\int_{t_{2}}^{t} \alpha(s) \xi(s) \,d s } ,\quad \forall t\geq t_{2} . $$
Using the equivalence of \({\mathcal {L}}(t)\) and \(E(t)\) again, we have
$$E(t) \leq K e^{-k\int_{t_{2}}^{t} \alpha(s) \xi(s) \,d s }, \quad\forall t \geq t_{2} , $$
for some positive constant K depending on the initial data.

Case 2: \(\mu_{1}=0\), \(|\mu_{2}|>0\).

First, we define the Lyapunov function
$$ F(t):= E(t)+\varepsilon_{1} \alpha(t) \Phi(t) +\varepsilon_{2} \alpha(t) \Psi(t), $$
(3.18)
where \(\varepsilon_{1}\) and \(\varepsilon_{2}\) are positive constants.
Similar to Case 1, from Lemma 3.1, we can obtain, for \(\varepsilon_{1}, \varepsilon_{2}>0\) small enough,
$$ \beta_{1} E(t) \leq F(t) \leq\beta_{2} E(t),\quad \forall t \geq0, $$
(3.19)
where \(\beta_{1}\) and \(\beta_{2}\) are positive constants. From Lemma 3.2, we get
$$ \begin{aligned}[b] E' (t) \leq{}& \biggl( \frac{\zeta}{2}+ \frac{|\mu_{2}|}{2\sqrt{1-d}} \biggr) \|u_{t}\|^{2}_{\Gamma_{1}} + \biggl( \frac{|\mu_{2}|\sqrt{1-d}}{2}-\frac{\zeta(1-d)}{2e^{\lambda\tau_{1}}} \biggr) \big\| u_{t} \bigl(t- \tau(t)\bigr)\big\| ^{2}_{\Gamma_{1}} \\ & + \frac{\alpha(t)}{2}g'\square \partial^{2} u - \frac{\alpha'(t)}{2} \biggl( \int_{0}^{t} g(s)\,ds \biggr) a (u, u)\\ & - \frac{\lambda\zeta}{2} \int_{t-\tau(t)}^{t} e^{\lambda(s-t)} \big\| u_{t}(s) \big\| ^{2}_{\Gamma_{1}} \,ds . \end{aligned}$$
(3.20)
Similar to Lemmas 3.3 and 3.4, we have
$$ \Phi'(t)\leq - \frac{l}{2} a(u, u)+ C_{0} \bigl( \|u_{t}\|^{2} +\gamma\|\nabla u_{t}\|^{2} + \big\| u_{t}\bigl(t-\tau(t)\bigr) \big\| ^{2}_{\Gamma_{1}}\bigr) + \frac{\alpha(t)}{4} g \square \partial^{2} u , $$
(3.21)
where \(C_{0}=\max\{ 1, \frac{\mu_{2}^{2} \tilde{C}_{p}}{2l} \}\) and
$$ \begin{aligned}[b] \Psi'(t) \leq{}& {-}( g_{0} -\delta) \|u_{t}\|^{2} -\gamma(g_{0} - \delta) \|\nabla u_{t}\|^{2}\\ & + \delta \biggl(1+ \biggl(\frac{1-l}{2} \biggr)^{2} \biggr)a(u,u) +\delta\big\| u_{t}\bigl(t-\tau (t)\bigr) \big\| ^{2}_{\Gamma_{1}} \\ & + \biggl( \alpha(t)+\frac{1}{2 \delta}+\frac{ \mu_{2}^{2}\tilde{C}_{p}}{4\delta} \biggr) \biggl( \int_{0}^{t} g(s) \,ds \biggr) g\square \partial^{2} u -\frac{g(0)(C_{p} +\gamma C_{s})}{4\delta} g' \square \partial^{2} u , \end{aligned}$$
(3.22)
respectively. By (3.18), (3.20)-(3.22) and (2.1), we obtain
$$\begin{aligned} F' (t) \leq{}& \alpha(t) \biggl( \varepsilon_{1} C_{0} -(g_{0}-\delta)\varepsilon_{2} - \frac{\alpha'(t)}{\alpha(t)} \biggr)\|u_{t}\|^{2} \\ & +\gamma\alpha(t) \biggl( \varepsilon_{1} C_{0} -(g_{0} -\delta) \varepsilon_{2} + \frac{\tilde{C}_{s}}{\alpha(0)\gamma} \biggl( \frac{\zeta}{2}+\frac{|\mu _{2}|}{2\sqrt{1-d}} \biggr)-\frac{\alpha'(t)}{\alpha(t)} \biggr) \|\nabla u_{t}\| ^{2} \\ & +\alpha(t) \biggl\{ \biggl( 1+\biggl(\frac{1-l}{2} \biggr)^{2} \biggr)\delta\varepsilon_{2} -\frac{l\varepsilon_{1}}{2}\\ &- \frac{ \alpha'(t)}{2\alpha(t)} \biggl( \int_{0}^{t} g(s) \,ds \biggr)- \frac{\alpha'(t)\varepsilon_{1}^{2} }{2\alpha(t)} ( C_{p} + \gamma C_{s}) \biggr\} a(u, u) \\ & + \alpha(t) \biggl( \varepsilon_{1} C_{0} + \delta \varepsilon_{2}+\frac{(1-d)}{\alpha(0)} \biggl( \frac{|\mu_{2}|}{2\sqrt{1-d}}- \frac{\zeta}{2e^{\lambda\tau_{1}}} \biggr) \biggr) \big\| u_{t} \bigl(t-\tau(t)\bigr) \big\| ^{2}_{\Gamma_{1}} \\ & + \alpha(t) \biggl[ \frac{\varepsilon_{1} \alpha(t)}{4}\\ & + \biggl\{ \varepsilon _{2} \biggl( \alpha(t) +\frac{2+\mu_{2}^{2}\tilde{C}_{p}}{4\delta} \biggr)-\frac{\alpha'(t)\varepsilon_{2}^{2} }{2\alpha(t)} (C_{p} +\gamma C_{s}) \biggr\} \biggl( \int_{0}^{t} g(s) \,ds \biggr) \biggr] g \square \partial^{2} u \\ & + \alpha(t) \biggl(\frac{1}{2}-\frac{ g(0)\varepsilon_{2}}{4\delta} (C_{p} + \gamma C_{s}) \biggr) g' \square \partial^{2} u - \frac{\lambda\zeta}{2} \int_{t-\tau(t)}^{t} e^{\lambda(s-t)} \big\| u_{t} (s) \big\| ^{2}_{\Gamma_{1}} \,ds. \end{aligned}$$
Now, we choose \(\delta>0\) small enough such that
$$ \delta< \min \biggl\{ \frac{g_{0}}{4}, \frac{lg_{0}}{2C_{0}(4+(1-l)^{2})} \biggr\} . $$
(3.23)
As long as δ is fixed, we take \(\varepsilon_{2}\) such that
$$0< \varepsilon_{2} < \frac{2\delta}{g(0)(C_{p}+\gamma C_{s} )} . $$
Then we get
$$ \frac{1}{2}-\frac{ g(0)\varepsilon_{2}}{4\delta} (C_{p} + \gamma C_{s}) >0 . $$
(3.24)
From the choice of δ, we have
$$\frac{g_{0}}{4C_{0}} < \frac{g_{0}-2\delta}{2C_{0}} . $$
Then we select \(\varepsilon_{1}\) such that
$$ \frac{g_{0} \varepsilon_{2} }{4C_{0}} < \varepsilon_{1} < \frac{(g_{0}-2\delta)\varepsilon_{2}}{2C_{0}}. $$
(3.25)
By (3.23) and (3.25), we obtain
$$ \biggl( 1+\biggl(\frac{1-l}{2}\biggr)^{2} \biggr)\delta \varepsilon_{2} -\frac{l\varepsilon_{1}}{2} < 0 $$
(3.26)
and
$$ 0< \varepsilon_{1} C_{0} +\delta\varepsilon_{2} < (g_{0} -\delta) \varepsilon_{2} -\varepsilon_{1} C_{0} . $$
(3.27)
Now, we add a restriction condition on γ, that is, we suppose that
$$ \frac{\tilde{C}_{s}}{1-d} < \gamma . $$
(3.28)
Note that \(e^{\lambda\tau_{1} } \rightarrow1\) as \(\lambda\rightarrow0\). Hence, if we take λ small enough, and from (3.27) and (3.28), there exists a positive constant ζ such that
$$ \frac{2\alpha(0) e^{\lambda\tau_{1}} }{1-d} (\varepsilon_{1} C_{0} +\delta \varepsilon_{2}) < \zeta< \frac{2\alpha(0)\gamma}{\tilde{C}_{s}} \bigl((g_{0} - \delta) \varepsilon_{2} -\varepsilon_{1} C_{0} \bigr). $$
And then, we see that
$$\frac{\zeta}{ e^{\lambda\tau_{1}} }-\frac{2\alpha(0) }{1-d} (\varepsilon_{1} C_{0} +\delta\varepsilon_{2}) >0 $$
and
$$\frac{2\alpha(0)\gamma}{\tilde{C}_{s}} \bigl((g_{0} -\delta) \varepsilon_{2} - \varepsilon_{1} C_{0} \bigr) -\zeta>0 . $$
If we choose \(|\mu_{2} | > 0\) such that
$$ \begin{aligned}[b] |\mu_{2}|&< \sqrt{1-d} \biggl(\min \biggl\{ \frac{\zeta}{ e^{\lambda\tau_{1}} }- \frac{2\alpha(0) }{1-d} (\varepsilon_{1} C_{0} +\delta \varepsilon_{2}), \frac{2\alpha(0)\gamma}{\tilde{C}_{s}} \bigl((g_{0} -\delta) \varepsilon_{2} -\varepsilon_{1} C_{0} \bigr) -\zeta \biggr\} \biggr)\hspace{-24pt}\\ &=: \mu_{0} , \end{aligned}$$
(3.29)
where \(\mu_{0}\) depends on \(g_{0}\), we find that
$$ \varepsilon_{1} C_{0} +\delta \varepsilon_{2}+ \frac{(1-d)}{\alpha(0)} \biggl( \frac{|\mu_{2}|}{2\sqrt{1-d}}-\frac{\zeta}{2e^{\lambda\tau_{1}}} \biggr) < 0 $$
(3.30)
and
$$ \varepsilon_{1} C_{0} -(g_{0} -\delta) \varepsilon_{2} + \frac{\tilde{C}_{s}}{\alpha(0)\gamma} \biggl(\frac{\zeta}{2}+ \frac{|\mu _{2}|}{2\sqrt{1-d}} \biggr) < 0 . $$
(3.31)
Consequently, from Remark 2.1, (3.24), (3.26), (3.27), (3.30) and (3.31), there exist two positive constants \(k_{3}\) and \(k_{4}\) such that, for \(t_{1} \geq t_{0}\),
$$ F'(t) \leq- k_{3} \alpha(t) E(t) + k_{4} \alpha(t) g \square\partial^{2} u ,\quad \forall t \geq t_{1} , $$
(3.32)
where \(k_{3}\) and \(k_{4}\) depend on \(g_{0}\). Multiplying (3.32) by \(\xi(t)\) and using (2.3), (3.20), (3.31) and the definition of \(E(t)\), we get, for \(t\geq t_{1}\),
$$\begin{aligned} \xi(t) F'(t) \leq{}&{-} k_{3}\alpha(t) \xi(t) { E}(t) - k_{4} \alpha(t) g' \square\partial^{2} u \\ \leq{}&{-} k_{3}\alpha(t) \xi(t) { E}(t)- 2k_{4} E'(t) + 2k_{4} \gamma\alpha(0) \bigl( (g_{0}-\delta)\varepsilon_{2} -\varepsilon_{1} C_{0} \bigr)\|\nabla u_{t}\|^{2} \\ &-k_{4} \alpha'(t) \biggl( \int_{0}^{t} g(s) \,ds \biggr) a(u,u) \\ \leq{}&{-} k_{3}\alpha(t) \xi(t) { E}(t) - 2k_{4} E'(t) + 4k_{4} \alpha(0) \bigl( (g_{0}-\delta) \varepsilon_{2} -\varepsilon_{1} C_{0} \bigr) E(t)\\ & - \frac{2k_{4}\alpha'(t)}{l} \biggl( \int_{0}^{t} g(s) \,ds \biggr) E(t). \end{aligned}$$
By \(\xi'(t) \leq0\), we have, for \(t\geq t_{1}\),
$$ \begin{gathered} \bigl( \xi(t) F(t) + 2k_{4} E(t) \bigr)'\\ \quad \leq- \biggl( k_{3} -\frac{4k_{4}\alpha(0)}{\alpha(t)\xi(t)} \bigl( (g_{0}-\delta) \varepsilon_{2} -\varepsilon_{1} C_{0} \bigr) + \frac{2k_{4} \alpha'(t)}{l \alpha(t) \xi(t)} \biggl( \int_{0}^{t} g(s) \,ds \biggr) \biggr) \alpha(t) \xi(t)E(t) . \end{gathered}$$
Now, we add a restriction condition on α and ξ, that is, we assume that
$$ \alpha(t)\xi(t) > \frac{4k_{4}\alpha(0)}{k_{3}} \bigl( (g_{0}-\delta) \varepsilon_{2} -\varepsilon_{1} C_{0} \bigr):= \xi_{0}, \quad\forall t\geq t_{1} . $$
(3.33)
From (2.4), we can take \(t_{2} \geq t_{1}\) such that \(k_{3} -\frac{4k_{4}\alpha(0)}{\alpha(t)\xi(t)} ( (g_{0}-\delta)\varepsilon_{2} -\varepsilon_{1} C_{0} ) + \frac{2k_{4} \alpha'(t)}{l \alpha(t) \xi(t)} (\int_{0}^{t} g(s) \,ds ) >0\) for \(t\geq t_{2}\). Hence, there exists a positive constant k such that
$${\mathcal {F}}'(t) \leq- k \alpha(t) \xi(t) {\mathcal {F}}(t) ,\quad \forall t\geq t_{2} , $$
where \({\mathcal {F}}(t) = \xi(t) F(t) + 2k_{4} E(t)\). From (3.19), we can see that \({\mathcal {F}}(t)\) is equivalent to \(E(t)\). Integrating this over \((t_{2} , t)\) and using the equivalence of \({\mathcal {F}}(t)\) and \(E(t)\) again, we obtain (2.10). Then, we complete the proof.

Example

If g decays exponentially, \(\xi(t)=a\) and \(\alpha(t)=\frac{b}{1+t}+c\), then (2.10) gives us
$$E(t) \leq K e^{-k(ab\ln(1+t)+ac t)} , $$
where \(a, b, c>0\).

4 Conclusions

In the present paper, we consider a decay rate of the solutions for weak viscoelastic Kirchhoff plate equations with time-varying delay in the boundary. By introducing suitable energy and Lyapunov functions, we obtain a decay estimate for the energy, which depends on the behavior of both α and g. On the other hand, different from the previous literature, we use the memory term instead of the damping term to control the delay term.

Declarations

Acknowledgements

This work was supported by the Dong-A University research fund.

Open Access This article is distributed under the terms of the Creative Commons Attribution 4.0 International License (http://creativecommons.org/licenses/by/4.0/), which permits unrestricted use, distribution, and reproduction in any medium, provided you give appropriate credit to the original author(s) and the source, provide a link to the Creative Commons license, and indicate if changes were made.

Authors’ Affiliations

(1)
Center for Education Accreditation, Pusan National University
(2)
Department of Mathematics, Dong-A University

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