Well-posedness and stability results for some nonautonomous abstract linear hyperbolic equations with memory

Abstract

In this paper, we study a class of second-order abstract linear hyperbolic equations with infinite memory that involve time-dependent unbounded linear operators. We obtain the well-posedness and stability of solutions to those nonautonomous second-order evolution equations under some appropriate assumptions. Our results generalize a number of previously known results in the autonomous case. Some specific examples are given to illustrate our abstract results, such as the nonautonomous Petrovsky type and wave equations.

1 Introduction

Fix a Hilbert space H once and for all, where $\left\langle \cdot ,\cdot \right\rangle$, and $\Vert \cdot \Vert$ are the inner product and norm associated with it, respectively. We consider the following families of time-dependent linear operators, $A(t): D(A(t))\subset H\rightarrow H$ and $B(t): D(B(t))\subset H\rightarrow H$, where D(A(t)) and D(B(t)) are the domains of the linear operators A(t) and B(t) and let $g:\,\mathbb {R}_{+} \rightarrow \mathbb {R}_{+}$ be a given function.

Consider the following class of nonautonomous second-order hyperbolic equations with infinite memory:

$$\begin{aligned} u_{tt} (t) +A(t)u(t)-\int _0^{\infty }g(s)B(t)u(t-s)ds=0,\quad \forall t\in \mathbb {R}_{+}^* :=(0, \infty ), \end{aligned}$$

(1.1)

equipped with the following initial conditions

$$\begin{aligned} \left\{ \begin{array}{ll} u(-t)=u_0 (t),&{} \forall t \in \mathbb {R}_{+},\\ u_t (0)=u_1,&{} \forall t \in \mathbb {R}_{+}, \end{array} \right. \end{aligned}$$

(1.2)

where $u_{tt}$ and $u_{t}$ denote the second and first derivatives of u with respect to time t, the pair $(u_0 , u_1 )$ is the initial data belonging to a suitable space, and $u:\,\mathbb {R}_{+}\rightarrow H$ is the unknown of the system (1.1)–(1.2).

The main goal of this paper is to investigate the well-posedness and asymptotic stability of the solutions to the system (1.1)–(1.2) as time t approaches infinity, under some appropriate assumptions on the family of time-dependent linear operators A(t) and B(t), as well as the relaxation (kernel) function g.

In recent years, many mathematicians have been drawn to the problem of the well-posedness and stability (respectively, instability) of solutions for evolution equations with delay (respectively, memory), see, for example, [25,26,27,28]. Let us recall some works that are relevant to the issue under consideration in this paper. Indeed, a large literature exists in the case where the operators A(t) and B(t) are not time-dependent (autonomous case), addressing the issues of existence, uniqueness, and asymptotic behavior in time; see, for example, [1, 2, 9,10,11,12,13,14,15, 21, 24, 33, 34]. Depending on the growth of g at infinity, different decay estimates (exponential, polynomial, or others) have been obtained. Furthermore, in the case where the infinite memory is replaced with a finite one and A(t) and B(t) are not time-dependent, numerous papers on this topic are available in the literature, see, for example, [5, 6, 8, 18,19,20, 29,30,31,32, 36, 38,39,46], and the references therein. See, for example, [3, 4, 16, 17, 22, 23, 37], and the references therein in the autonomous case where a discrete or distributed time delay is added to (1.1).

In this paper, it goes back to investigating the nonautonomous case, that is, the case which involves time-dependent linear operators, A(t) and B(t). For more on time-dependent linear operators, evolution families, and evolution equations and their applications, we refer the reader to for instance [11, 25].

The following is the structure of the paper: in Sect. 2, we present our assumptions on A(t), B(t), and g, as well as state and prove the well-posedness of (1.1)–(1.2) (Theorem 2.1). Section 3 contains a statement and proof of the asymptotic stability of solutions to (1.1)–(1.2) under some additional assumptions on A(t), B(t), and g (Theorem 3.1). Finally, in Sects. 4 and 5, we present some examples as well as discuss some general remarks and open problems.

2 Well-posedness

In this section, we state our assumptions on A(t), B(t), and g, as well as establish the well-posedness of the system (1.1)–(1.2). In the sequel, this setting requires the following assumptions:

(A0):

A(t) and B(t) are time-dependent positive definite self-adjoint linear operators that satisfy,

$$\begin{aligned} D(A(t))=D(A(0)),\quad D(B(t))=D(B(0))\,\,\text{ for } \text{ all } \ \ t\in \mathbb {R}_{+}, \end{aligned}$$

(2.1)

and the embeddings

$$D(A(0)) \hookrightarrow D(B(0)) \hookrightarrow H$$

are dense and compact.

(A1):

There exist two functions $a_1 ,\,b_1 :\mathbb {R}_{+}\rightarrow \mathbb {R}_{+}^*$ of class $C^1$ and another continuous function $b_2 :\mathbb {R}_{+}\times \mathbb {R}_{+}\rightarrow \mathbb {R}_{+}^*$ such that

$$\begin{aligned}&b_1 (t)\Vert w\Vert ^2 \le \Vert B^{\frac{1}{2}} (t) w\Vert ^2 ,\quad \forall w\in D(B^{\frac{1}{2}} (0)),\,\,\forall t\in \mathbb {R}_{+} , \end{aligned}$$

(2.2)

$$\begin{aligned}&\Vert B^{\frac{1}{2}} (t) w\Vert ^2 \le a_1 (t)\Vert A^{\frac{1}{2}} (t)w\Vert ^2 ,\quad \forall w\in D(A^{\frac{1}{2}} (0)),\,\,\forall t\in \mathbb {R}_{+} \end{aligned}$$

(2.3)

and

$$\begin{aligned} \Vert B^{\frac{1}{2}} (t_1 ) w\Vert ^2 \le b_2 (t_1 ,t_2 )\Vert B^{\frac{1}{2}} (t_2 )w\Vert ^2 ,\quad \forall w\in D(B^{\frac{1}{2}} (0)),\,\,\forall t_1 ,t_2 \in \mathbb {R}_{+} . \end{aligned}$$

(2.4)

(A2):

For any $t\in \mathbb {R}_{+}$, there exist two time-dependent linear operators on H

$$\begin{aligned} {\tilde{A}}(t):\,D(A(0))\rightarrow H \quad \hbox {and}\quad {\tilde{B}}(t):\,D(B(0))\rightarrow H \end{aligned}$$

(2.5)

satisfying

$$\begin{aligned} \lim _{\tau \rightarrow t} \left[ \left\| \left( \frac{A(\tau )-A(t) }{\tau -t}-{\tilde{A}}(t)\right) w_1\right\| + \left\| \left( \frac{B(\tau ) -B(t)}{\tau -t}-{\tilde{B}}(t)\right) w_2\right\| \right] =0 \end{aligned}$$

(2.6)

for all $(w_1 ,w_2 )\in D(A(0))\times D(B(0))$.

(A3):

The non-increasing class $C^1$ relaxation (kernel) function $g: \mathbb {R}_{+} \rightarrow \mathbb {R}_{+}$ satisfies

$$\begin{aligned} g_0 :=\int _0^{\infty } g(s)ds <\frac{1}{a_1 (t)},\quad \forall t\in \mathbb {R}_{+}, \end{aligned}$$

(2.7)

and there exists a positive constant $\theta _1$ such that

$$\begin{aligned} -g^{\prime } (s) \le \theta _1 g(s),\quad \forall s\in \mathbb {R}_{+}. \end{aligned}$$

(2.8)

Remark 1

Let us recall that the assumptions (A0)–(A3) hold for a wide range of linear operators A(t) and B(t), as well as the relaxation function g. Indeed, consider $\Omega \subset \mathbb {R}^N$ to be an open bounded domain with smooth boundary $\Gamma = \partial \Omega$ where $N\in \mathbb {N}^*$, and consider $H= L^2 (\Omega )$ to be endowed with its standard inner product:

$$\begin{aligned} \left\langle f, h\right\rangle =\displaystyle \int _{\Omega }f(x) h(x) dx \end{aligned}$$

for all $f, h \in L^2(\Omega )$.

Consider the case when A(t) and B(t) and g are given by

$$\begin{aligned} A(t)= & {} -a(t)\Delta ,\,\, B(t)=-b(t)\Delta , \,\, D(A(t))=D(B(t))\\= & {} H^2 (\Omega )\cap H_0^1 (\Omega )\,\, \hbox {and}\,\, g(s)=\theta _0 e^{-\theta _1 s},\end{aligned}$$

where $a,\,b:\mathbb {R}_{+}\rightarrow \mathbb {R}_{+}^*$ are of class $C^1$ and $\theta _0 ,\,\theta _1\in \mathbb {R}_{+}^*$ such that

$$\begin{aligned} \frac{\theta _0 }{\theta _1} < \frac{a(t)}{b(t)}, \quad \forall t \in \mathbb R_{+}^{*}. \end{aligned}$$

(2.9)

To make this work, we carefully choose ${\tilde{A}}(t)$ and ${\tilde{B}}(t)$ as follows:

$$\begin{aligned}\displaystyle {\tilde{A}}(t)=-a^{\prime } (t)\Delta \ \ \text{ and } \ \ {\tilde{B}}(t)= -b^{\prime } (t)\Delta ,\end{aligned}$$

where $a_1 (t)= \frac{b(t)}{a(t)}$, $b_1 (t)=c_0 b(t)$, $b_2 (t_1 ,t_2)= \frac{b(t_1)}{b(t_2)}$, and $c_0$ is the Poincaré constant.

Under the previous assumptions and following a method derived from [10], we consider a new auxiliary variable $\eta$ with its initial data $\eta _0$ defined by

$$\begin{aligned} \left\{ \begin{array}{ll} \eta (t,s)=u(t)-u(t-s), &{} \qquad \forall t,\,s\in \mathbb {R}_{+},\\ \eta _{0}(s)=\eta (0,s)=u_{0}(0)-u_{0}(s), &{}\qquad \forall s\in \mathbb {R}_{+}, \end{array} \right. \end{aligned}$$

(2.10)

and formulate (1.1)–(1.2) as a first-order nonautonomous evolution equation given by

$$\begin{aligned} \left\{ \begin{array}{ll} {\mathcal U}_{t}(t)={\mathcal A} (t){\mathcal U}(t), &{} \quad \forall t \in \mathbb {R}_{+}^*, \\ {\mathcal U}(0)={\mathcal U}_{0}, \end{array} \right. \end{aligned}$$

(2.11)

where ${\mathcal U}=(u,u_{t},\eta )^{T}$, ${\mathcal U}_{0}=(u_{0}(0),u_{1} (0),\eta _{0})^{T}\in {\mathcal H}(t)$,

$$\begin{aligned}&{\mathcal H} (t) =D(A^{\frac{1}{2}} (t))\times H\times L_{g} (t), \\&L_{g} (t)=\left\{ w:\,\mathbb {R}_{+}\rightarrow D(B^{\frac{1}{2}}(t)),\quad \int _{0}^{\infty }g(s)\Vert B^{\frac{1}{2}} (t)w(s)\Vert ^{2}ds<\infty \right\} , \end{aligned}$$

and the time-dependent linear operators ${\mathcal A} (t)$ are given by

$$\begin{aligned} {\mathcal A}(t)\left( \begin{array}{c} w_{1}\\ w_{2}\\ w_{3} \end{array} \right) =\left( \begin{array}{c} w_{2} \\ \left( -A(t)+g_{0} B(t)\right) w_{1} -\displaystyle \int _{0}^{\infty }g(s)B(t)w_{3}(s)ds \\ -{\dfrac{{\partial w_{3}}}{{\partial s}}} +w_{2} \end{array} \right) \end{aligned}$$

(2.12)

for all $\displaystyle \left( \begin{array}{c} w_{1} \\ w_{2} \\ w_{3} \end{array} \right) \in {\mathcal D}({\mathcal A} (t))$ where,

$$\begin{aligned} {\mathcal D}({\mathcal A} (t))=\left\{ \begin{array}{ll} (w_{1},w_{2},w_{3})^{T}\in D(A(t))\times D(A^{\frac{1}{2}} (t))\times L_{g} (t) ,\\ {\frac{{\partial w_{3}}}{{\partial s}}}\in L_{g} (t),\,w_{3}(0)=0,\,w_3 (s)\in D(B(t)),\,\forall s\in \mathbb {R}_{+} \end{array} \right\} . \end{aligned}$$

(2.13)

Based upon (2.1) and (2.4), the spaces ${\mathcal H}(t)$ and $L_{g} (t)$ do not depend on t, that is,

$$\begin{aligned} {\mathcal H}(t)={\mathcal H}(0)\quad \hbox {and}\quad L_{g} (t)=L_{g} (0), \quad \forall t\in \mathbb {R}_{+}. \end{aligned}$$

(2.14)

The space $L_{g} (t)$ is endowed with the classical inner product

$$\begin{aligned} \left\langle w_{1},w_{2}\right\rangle _{L_{g} (t)}=\int _{0}^{\infty }g(s)\left\langle B^{\frac{1}{2}} (t)w_{1}(s),B^{\frac{1}{2}} (t)w_{2}(s)\right\rangle ds. \end{aligned}$$

Based upon (2.1) and (2.14), we have

$$\begin{aligned} {\mathcal D}({\mathcal A} (t))={\mathcal D}({\mathcal A} (0)), \quad \forall t\in \mathbb {R}_{+}. \end{aligned}$$

(2.15)

On the other hand, keeping in mind the definition of $\eta$ in (2.10), we have

$$\begin{aligned} \left\{ \begin{array}{ll} \eta _{t}(t,s)+\eta _{s}(t,s)=u_{t}(t),&{} \forall t,\,s\in \mathbb {R}_{+}, \\ \eta _{s}(t,s)=u_{t}(t-s),&{} \forall t,\,s\in \mathbb {R}_{+}, \\ \eta (t,0)=0,&{} \forall t\in \mathbb {R}_{+} . \end{array} \right. \end{aligned}$$

(2.16)

Therefore, we conclude from (2.12) and (2.16) that the systems (1.1)–(1.2) and (2.11) are equivalent.

Using (2.3) and (2.7), we conclude that ${\mathcal H}(t)$ endowed with the inner product

$$\begin{aligned} \left\langle (w_1 ,w_2 ,w_3 )^T ,({\tilde{w}}_1 ,{\tilde{w}}_2 ,{\tilde{w}}_3 )^T \right\rangle _{{\mathcal H}(t)}= & {} \left\langle A^{1\over 2} (t)w_1 ,A^{1\over 2} (t){\tilde{w}}_1 \right\rangle -g_0 \left\langle B^{1\over 2} (t)w_1 ,B^{1\over 2} (t){\tilde{w}}_1 \right\rangle \\&+ \left\langle w_2 ,{\tilde{w}}_2 \right\rangle +\left\langle w_3 ,{\tilde{w}}_3 \right\rangle _{L_g (t)} \end{aligned}$$

is a Hilbert space with the following embedding ${\mathcal D}({\mathcal A} (t)) \hookrightarrow {\mathcal H} (t)$ being dense (see, for example, [33]).

The following theorem ensures the well-posedness of (2.11):

Theorem 2.1

Under assumptions (A0)-(A3), for any ${\mathcal U}_{0}\in {\mathcal H} (0)$, the system (2.11) has a unique (weak) solution

$$\begin{aligned} {\mathcal U}\in C(\mathbb {R}_{+} ,{\mathcal H} (0)). \end{aligned}$$

(2.17)

Moreover, if ${\mathcal U}_{0}\in {\mathcal D}({\mathcal A} (0))$, then the solution to (2.11) is a classical solution, that is,

$$\begin{aligned} {\mathcal U}\in C^1 (\mathbb {R}_{+} ,{\mathcal H} (0))\cap C (\mathbb {R}_{+} ,{\mathcal D}({\mathcal A} (0))). \end{aligned}$$

(2.18)

Proof

To prove Theorem 2.1, we make use of the semigroup theory approach. The proof is divided into three main steps.

Step 1. The first step consists of showing that the linear operators ${\mathcal A}(t)$ are dissipative for all $t \in \mathbb R_+$. Indeed, as in [22], letting $W=(w_{1},w_{2},w_{3})^{T}\in {\mathcal D}({\mathcal A}(t))$, we obtain,

$$\begin{aligned} \left\langle {\mathcal A}(t)W,W \right\rangle _{{\mathcal H}(t)}= & {} \left\langle A^{1\over 2} (t)w_2 ,A^{1\over 2} (t)w_1 \right\rangle -g_0 \left\langle B^{1\over 2} (t)w_2 ,B^{1\over 2} (t)w_1 \right\rangle -\left\langle \frac{\partial w_3}{\partial s} ,w_3 \right\rangle _{L_g (t)} \nonumber \\&+\left\langle w_2 ,w_3 \right\rangle _{L_g (t)}\nonumber \\&+\left\langle \left( -A(t)+g_{0} B(t)\right) w_{1} -\displaystyle \int _{0}^{\infty }g(s)B(t)w_{3}(s)ds,w_2 \right\rangle . \end{aligned}$$

(2.19)

It is clear from the definitions of $A^{\frac{1}{2}} (t)$ and $B^{\frac{1}{2}} (t)$, and the fact that H is a real Hilbert space, that

$$\begin{aligned} \left\langle \left( -A(t)+g_{0} B(t)\right) w_{1},w_{2}\right\rangle =-\left\langle A^{\frac{1}{2}} (t)w_{2},A^{\frac{1}{2}} (t)w_{1}\right\rangle +g_0 \left\langle B^{\frac{1}{2}} (t)w_{2},B^{\frac{1}{2}} (t)w_{1}\right\rangle \end{aligned}$$

and

$$\begin{aligned} \left\langle -\int _{0}^{\infty }g(s)B(t)w_{3}(s)ds,w_{2}\right\rangle= & {} -\int _{0}^{\infty }g(s)\left\langle B^{\frac{1}{2}}(t)w_{3}(s),B^{\frac{1}{2}}(t)w_{2}\right\rangle ds\\= & {} -\left\langle w_{2},w_{3}\right\rangle _{L_{g} (t)} . \end{aligned}$$

On the other hand, using $\mathbf{(A3)}$, we see that

$$\begin{aligned} \lim _{s\rightarrow \infty }g(s) B^{\frac{1}{2}} (t)w_{3}(s)=0. \end{aligned}$$

Next, integrating by parts with respect to s and using the property $w_{3}(0)=0$ (definition of ${\mathcal D}({\mathcal A} (t))$), we deduce that

$$\begin{aligned} -\left\langle {\frac{{\partial w_{3}}}{{\partial s}}},w_{3}\right\rangle _{L_{g} (t)}= {\frac{1}{2}} \int _{0}^{\infty }g^{\prime } (s)\Vert B^{\frac{1}{2}} (t)w_{3}(s)\Vert ^{2}ds. \end{aligned}$$

Consequently, inserting these three formulas in the previous identity (2.19), we get

$$\begin{aligned} \left\langle {\mathcal A}(t)W,W\right\rangle _{{\mathcal H}(t)} = {\dfrac{1}{2}}\displaystyle \int _{0}^{\infty }g^{\prime } (s)\Vert B^{\frac{1}{2}} (t)w_{3}(s)\Vert ^{2}ds\le 0, \end{aligned}$$

(2.20)

as g is non increasing, which yields ${\mathcal A} (t)$ is dissipative.

Using (2.8) and the fact that g is non-increasing and $w_3 \in L_{g} (t)$, we have

$$\begin{aligned} \begin{array}{lcl} \left| \displaystyle \int _{0}^{\infty }g^{\prime } (s)\Vert B^{\frac{1}{2}} (t)w_{3}(s)\Vert ^{2} ds\right| &{}=&{} -\displaystyle \int _{0}^{\infty }g^{\prime } (s) \Vert B^{\frac{1}{2}} (t)w_{3}(s)\Vert ^{2} ds\\ &{}\le &{} \theta _1 \displaystyle \int _{0}^{\infty }g (s) \Vert B^{\frac{1}{2}} (t)w_{3}(s)\Vert ^{2} ds \\ &{}<&{} \infty \end{array} \end{aligned}$$

and so the integral in the right hand side of (2.20) is well defined.

Step 2. In this step, we prove that $I-{\mathcal A}(t)$ is surjective for all $t\in \mathbb {R}_{+}$, where I stands for the identity operator. Indeed, let $F=(f_{1},f_{2},f_{3})^{T}\in {\mathcal H}(t)$, we show that there exists

$$\begin{aligned} W=(w_{1},w_{2},w_{3})^{T}\in {\mathcal D}({\mathcal A} (t)) \end{aligned}$$

satisfying

$$\begin{aligned} (I-{\mathcal A} (t))W=F, \end{aligned}$$

(2.21)

which is equivalent to

$$\begin{aligned} \left\{ \begin{array}{ll} w_2=w_1-f_{1},\\ \left( A(t)-g_0 B(t)+I\right) w_1+\displaystyle \int _{0}^{\infty }g(s)B (t)w_3 (s)ds= f_{1}+f_{2},\\ w_3+\dfrac{\partial w_{3}}{\partial s}=w_1 +f_{3} -f_{1}. \end{array} \right. \end{aligned}$$

(2.22)

We note that the third Eq. in (2.22) with $w_{3}(0)=0$ has the unique solution

$$\begin{aligned} w_{3} (s) = \left( 1-e^{-s}\right) w_{1}+e^{-s}\displaystyle \int _{0}^{s}e^{y}(f_{3}(y)-f_{1})dy. \end{aligned}$$

(2.23)

Next, plugging (2.23) into the second Eq. in (2.22), we get

$$\begin{aligned} \left( A(t)-g_1 B(t)+ I\right) w_{1}=\tilde{f} (t), \end{aligned}$$

(2.24)

where

$$\begin{aligned} g_1 =\displaystyle \int _{0}^{\infty }g(s)e^{-s}ds \end{aligned}$$

and

$$\begin{aligned} \tilde{f} (t) =f_{1} +f_{2} -\displaystyle \int _{0}^{\infty }g(s)e^{-s}\left( \displaystyle \int _{0}^{s}e^{y}B(t)(f_{3}(y)-f_{1})dy\right) ds. \end{aligned}$$

To complete this step, we need to prove that (2.24) has a solution $w_{1}\in D(A^{\frac{1}{2}} (t))$. Then, substituting $w_1$ in (2.23) and the first Eq. in (2.22), we obtain $W\in {\mathcal D}({\mathcal A} (t))$ satisfying (2.21). Since $g_1 <g_0$, then $A(t)-g_1 B(t)$ is a positive definite operator thanks to (2.3) and (2.7). Therefore, $A(t)-g_1 B(t) + I$ is a self-adjoint linear positive definite operator. Applying the Lax-Milgram Theorem and classical regularity arguments, we conclude that (2.24) has a unique solution $w_{1}\in D(A^{\frac{1}{2}} (t))$ satisfying, using (2.23),

$$\begin{aligned} (A(t)-g_{0}B(t))w_{1}+\displaystyle \int _{0}^{\infty }g(s)B(t)w_{3}(s)ds\in H. \end{aligned}$$

This proves that $I-{\mathcal A}(t)$ is surjective. We note that (2.20) and (2.21) mean that, for any $t\in \mathbb {R}_{+}$, $-{\mathcal A}(t)$ is a maximal monotone operator. Hence, using Lummer-Phillips Theorem (see [35]), we deduce that ${\mathcal A}(t)$ is an infinitesimal generator of a $C_0$-semigroup of contraction on ${\mathcal H}(t)$.

Step 3. Condition (2.6) yields the applications $h_1 ,\,h_2 :\mathbb {R}_{+}\rightarrow H$ given by

$$\begin{aligned} h_1 (t) =A(t)w_1 \quad \hbox {and}\quad h_2 (t) =B(t)w_2 \end{aligned}$$

are differentiable and their derivatives are, respectively,

$$\begin{aligned} {\tilde{h}}_1 (t)= {\tilde{A}}(t)w_1 \quad \hbox {and}\quad {\tilde{h}}_2 (t)= {\tilde{B}}(t)w_2 . \end{aligned}$$

Now, let

$$\begin{aligned} W=(w_1 ,w_2 ,w_3 )^T\in D\left( {\mathcal A}(0) \right) \end{aligned}$$

and $h :\mathbb {R}_{+}\rightarrow {\mathcal H}(0)$ defined by $h (t) ={\mathcal A}(t)W$.

We prove in this step that h is differentiable and that its derivative is the function

$$\begin{aligned} {\tilde{h}} (t) =\left( \begin{array}{c} 0 \\ \left( -{\tilde{A}}(t)+g_{0} {\tilde{B}}(t)\right) w_{1} -\displaystyle \int _{0}^{\infty }g(s){\tilde{B}}(t)w_{3}(s)ds \\ 0 \end{array} \right) . \end{aligned}$$

Notice that, given (2.1), (2.5) and (2.13), we have ${\tilde{h}} (t)\in {\mathcal H}(0)$, for any $t\in \mathbb {R}_{+}$. On the other hand, we have, for any $\tau ,t\in \mathbb {R}_{+}$ with $\tau \ne t$,

$$\begin{aligned} \frac{h (\tau )-h (t)}{\tau -t} =\frac{1}{\tau -t}\left( \begin{array}{c} 0 \\ w \\ 0 \end{array} \right) , \end{aligned}$$

where

$$\begin{aligned} w=-(A(\tau )-A(t))w_1 +g_{0} (B(\tau )-B(t))w_{1} -\displaystyle \int _{0}^{\infty }g(s)(B(\tau )-B(t))w_{3}(s)ds. \end{aligned}$$

Then

$$\begin{aligned} \frac{h (\tau )-h (t)}{\tau -t} - {\tilde{h}} (t)=\left( \begin{array}{c} 0 \\ {\tilde{w}} \\ 0 \end{array} \right) , \end{aligned}$$

where

$$\begin{aligned} {\tilde{w}}= & {} -\left( \frac{A(\tau )-A(t)}{\tau -t}-{\tilde{A}}(t)\right) w_1 +g_{0} \left( \frac{B(\tau )-B(t)}{\tau -t}- {\tilde{B}}(t)\right) w_1 \\&-\displaystyle \int _{0}^{\infty }g(s)\left( \frac{B(\tau )-B(t)}{\tau -t}- {\tilde{B}}(t)\right) w_{3}(s)ds, \end{aligned}$$

which yields

$$\begin{aligned} \left\| \frac{h (\tau )-h (t)}{\tau -t} - {\tilde{h}} (t)\right\| _{{\mathcal H}(t)}= & {} \Vert {\tilde{w}}\Vert \nonumber \\\le & {} \left\| \left( \frac{A(\tau )-A(t)}{\tau -t}-{\tilde{A}}(t)\right) w_1\right\| \nonumber \\&+g_{0} \left\| \left( \frac{B(\tau )-B(t)}{\tau -t}- {\tilde{B}}(t)\right) w_1 \right\| \nonumber \\&+\displaystyle \int _{0}^{\infty }g(s)\left\| \left( \frac{B(\tau )-B(t)}{\tau -t}- {\tilde{B}}(t)\right) w_{3}(s)\right\| ds, \end{aligned}$$

so we get from (2.6) that

$$\begin{aligned} \lim _{\tau \rightarrow t}\left\| \frac{h (\tau )-h (t)}{\tau -t} - {\tilde{h}} (t)\right\| _{{\mathcal H}(t)}=0. \end{aligned}$$

Based upon the properties shown in the previous steps, we conclude that ${\mathcal A}(\cdot )$ generates a unique evolution family on ${\mathcal H}(0)$ (see [35]). Consequently, (2.11) is well-posed in the sense of Theorem 2.1. $\square$

3 Asymptotic stability

In this section, we look at the asymptotic behavior of solutions to (2.11). For that, we assume the following additional conditions are met:

(A4) There exist three continuous functions, $a_2,\,{\tilde{a}},\,{\tilde{b}} :\mathbb {R}_{+}\rightarrow \mathbb {R}_{+}$ that satisfy the following conditions,

$$\begin{aligned}&\Vert A^{\frac{1}{2}} (t)w\Vert ^2 \le a_2 (t)\Vert B^{\frac{1}{2}} (t)w\Vert ^2,\quad \forall w\in D(A^{\frac{1}{2}} (0)),\,\,\forall t\in \mathbb {R}_{+}, \end{aligned}$$

(3.1)

$$\begin{aligned}&\Vert {\tilde{A}}^{\frac{1}{2}} (t)w\Vert ^2 \le {\tilde{a}} (t)\Vert A^{\frac{1}{2}} (t)w\Vert ^2,\quad \forall w\in D(A^{\frac{1}{2}} (0)),\,\,\forall t\in \mathbb {R}_{+}, \end{aligned}$$

(3.2)

and

$$\begin{aligned} \Vert {\tilde{B}}^{\frac{1}{2}} (t)w\Vert ^2 \le {\tilde{b}} (t)\Vert B^{\frac{1}{2}} (t)w\Vert ^2,\quad \forall w\in D(B^{\frac{1}{2}} (0)),\,\,\forall t\in \mathbb {R}_{+} . \end{aligned}$$

(3.3)

(A5) The kernel g satisfies $g_0 >0$ and there exists a positive constant $\theta _2$ such that

$$\begin{aligned}&g^{\prime } (s) \le -\theta _2 g(s),\quad \forall s\in \mathbb {R}_{+}, \end{aligned}$$

(3.4)

$$\begin{aligned}&{\sqrt{{\tilde{b}} (t)}}<\frac{\theta _2}{2}\quad \hbox {and}\quad \left\| g_0 a_1 {\sqrt{{\tilde{b}}}}+{\sqrt{{\tilde{a}}}}\right\| _{L^{\infty } (\mathbb {R}_{+})}\,\,\hbox {is small enough} . \end{aligned}$$

(3.5)

Remark 2

Consider the example given in Remark 1. Observe that, for

$$\begin{aligned} a_2 (t)=\frac{a(t)}{b(t)},\quad {\tilde{a}} (t)= \frac{\vert a^{\prime }(t)\vert }{a(t)}\quad \hbox {and}\quad {\tilde{b}} (t)= \frac{\vert b^{\prime }(t)\vert }{b(t)},\quad \forall t \in \mathbb R_{+}^{*}, \end{aligned}$$

(3.6)

such that (3.5) holds, the assumptions (A4) and (A5) are also fulfilled with $\theta _2 =\theta _1$. In the autonomous case, we have ${\tilde{a}}={\tilde{b}}=0$, and then (3.5) is trivial.

Remark 3

In the sequel, we will make extensive use of Young’s inequality, which is stated as follows: let $\varepsilon : \mathbb {R}_{+} \rightarrow \mathbb {R}_{+}^*$ and $1<p,q<\infty$ be such that $\frac{1}{p}+\frac{1}{q}=1$, then

$$\begin{aligned} \alpha \beta \le \varepsilon (t) \alpha ^{p}+\left( p \varepsilon (t) \right) ^{-\frac{p}{q}}q^{-1}\,\, \beta ^{q} ,\quad \forall t,\alpha ,\beta \in \mathbb {R}_{+} . \end{aligned}$$

(3.7)

When $p=q=2$, we get the special case

$$\begin{aligned} \alpha \beta \le \varepsilon (t) \alpha ^{2}+\frac{1}{4 \varepsilon (t)} \beta ^{2} ,\quad \forall t,\alpha ,\beta \in \mathbb {R}_{+}. \end{aligned}$$

(3.8)

Theorem 3.1

Assume that (A0)-(A5) hold. Then, for any ${\mathcal U}_{0}\in {\mathcal H} (0)$, there exists a positive constant $\lambda$ such that the solution to (2.11) satisfies

$$\begin{aligned} \Vert {\mathcal U}(t)\Vert _{{\mathcal H} (0)}^2 \le \frac{\lambda e^{{\tilde{\xi }}(t)}}{M(t)-M_2 (t)},\quad \forall t\in \mathbb {R}_{+}, \end{aligned}$$

(3.9)

where the functions $M (\cdot ),\,M_2 (\cdot )$ and ${\tilde{\xi }}(\cdot )$ are defined in the proof (see (3.43), (3.44), (3.45), (3.48) and (3.50) below).

Proof

Let us assume that (A0)-(A5) hold and let ${\mathcal U}_{0}\in {\mathcal H} (0)$. The energy functional E associated with the solution of (2.11) corresponding to ${\mathcal U}_{0}$ is given by

$$\begin{aligned} E(t)= & {} \frac{1}{2} \left\| {\mathcal U} (t)\right\| _{{\mathcal H} (0)}^{2} \nonumber \\= & {} \frac{1}{2}\left( \left\| A^{\frac{1}{2}}(t) u(t) \right\| ^{2}+ \left\| u_{t}(t) \right\| ^{2} - g_{0} \left\| B^{\frac{1}{2}}(t) u(t)\right\| ^{2}\right. \nonumber \\&\left. + \int _{0}^{\infty } g(s) \left\| B^{\frac{1}{2}}(t) \eta (t,s) \right\| ^{2} ds\right) . \end{aligned}$$

(3.10)

In order to complete the proof of Theorem 3.1, we need the next lemmas, where throughout the proofs, $c,\,c_1 ,\,c_2 ,\, \cdots ,$ stand for some positive generic constants which do not depend upon t, and c can be different from a given line to another.

Lemma 3.2

The energy functional $E(\cdot )$ satisfies the estimate

$$\begin{aligned} E^{\prime }(t)= & {} \frac{1}{2}\int _{0}^{\infty }g^{\prime }(s) \left\| B^{\frac{1}{2}}(t) \eta (t,s)\right\| ^{2}ds -g_{0}\left\langle {\tilde{B}}^{\frac{1}{2}}(t) u(t), B^{\frac{1}{2}}(t) u(t) \right\rangle \nonumber \\&+\left\langle {\tilde{A}}^{\frac{1}{2}}(t) u(t), A^{\frac{1}{2}}(t) u(t) \right\rangle +\int _{0}^{\infty } g(s)\left\langle {\tilde{B}}^{\frac{1}{2}}(t) \eta (t,s), B^{\frac{1}{2}}(t) \eta (t,s) \right\rangle ds.\nonumber \\ \end{aligned}$$

(3.11)

Proof

Multiplying (1.1) by $u_{t}$ and integrating by parts, one gets,

$$\begin{aligned} \frac{1}{2}\frac{d}{dt}\Vert u_{t}\Vert ^{2} + \left\langle A^{\frac{1}{2}}(t) u(t), A^{\frac{1}{2}}(t) u_{t}(t) \right\rangle - \left\langle \int _{0}^{\infty }g(s) B^{\frac{1}{2}}(t) u(t-s) ds, B^{\frac{1}{2}}(t) u_{t}(t) \right\rangle =0. \end{aligned}$$

(3.12)

Now

$$\begin{aligned} \frac{1}{2}\frac{d}{dt}\left\| A^{\frac{1}{2}}(t)u(t)\right\| ^{2} = \left\langle {\tilde{A}}^{\frac{1}{2}}(t) u(t), A^{\frac{1}{2}}(t) u(t) \right\rangle + \left\langle A^{\frac{1}{2}}(t) u(t), A^{\frac{1}{2}}(t) u_{t}(t) \right\rangle . \end{aligned}$$

(3.13)

A similar result can be obtained for $B(\cdot )$, that is, using the first Eq. in (2.10), one obtains,

$$\begin{aligned}&\left\langle \int _{0}^{\infty }g(s) B^{\frac{1}{2}}(t) u(t-s) ds, B^{\frac{1}{2}}(t) u_{t}(t) \right\rangle \nonumber \\&\quad =\left\langle \int _{0}^{\infty }g(s) B^{\frac{1}{2}}(t)\left[ u(t)- \eta (t,s)\right] ds, B^{\frac{1}{2}}(t) u_{t}(t) \right\rangle \nonumber \\&\quad = -\left\langle \int _{0}^{\infty }g(s) B^{\frac{1}{2}}(t) \eta (t,s) ds, B^{\frac{1}{2}}(t) u_{t}(t) \right\rangle + g_{0} \left\langle B^{\frac{1}{2}}(t)u(t), B^{\frac{1}{2}}(t) u_{t}(t) \right\rangle \nonumber \\&\quad = -\left\langle \int _{0}^{\infty }g(s) B^{\frac{1}{2}}(t) \eta (t,s) ds, B^{\frac{1}{2}}(t) u_{t}(t) \right\rangle +\frac{g_{0}}{2} \frac{d}{dt}\left\| B^{\frac{1}{2}}(t)u(t)\right\| ^{2}\nonumber \\&\qquad -g_{0}\left\langle {\tilde{B}}^{\frac{1}{2}}(t) u(t), B^{\frac{1}{2}}(t) u(t) \right\rangle . \end{aligned}$$

(3.14)

Using the first Eq. in (2.16), we obtain

$$\begin{aligned}&\frac{1}{2} \int _{0}^{\infty } g(s)\frac{\partial }{\partial s} \left\| B^{\frac{1}{2}}(t) \eta (t,s) \right\| ^{2} ds +\frac{1}{2} \int _{0}^{\infty } g(s)\frac{\partial }{\partial t} \left\| B^{\frac{1}{2}}(t) \eta (t,s) \right\| ^{2} ds\nonumber \\&\quad =\int _{0}^{\infty } g(s)\left\langle B^{\frac{1}{2}}(t)\frac{\partial }{\partial s} \eta (t,s) , B^{\frac{1}{2}}(t) \eta (t,s) \right\rangle ds \nonumber \\&\quad \quad +\int _{0}^{\infty } g(s)\left\langle \left[ {\tilde{B}}^{\frac{1}{2}}(t) \eta (t,s) +B^{\frac{1}{2}}(t)\frac{\partial }{\partial t} \eta (t,s) \right] , B^{\frac{1}{2}}(t) \eta (t,s) \right\rangle ds \nonumber \\&\quad = \int _{0}^{\infty } g(s) \left\langle B^{\frac{1}{2}}(t) u_{t}(t), B^{\frac{1}{2}}(t) \eta (t,s) \right\rangle ds\nonumber \\&\qquad +\int _{0}^{\infty } g(s)\left\langle {\tilde{B}}^{\frac{1}{2}}(t) \eta (t,s),B^{\frac{1}{2}}(t) \eta (t,s)\right\rangle ds. \end{aligned}$$

(3.15)

Substituting (3.13)-(3.15) into (3.12) yields

$$\begin{aligned}&\frac{1}{2}\frac{d}{dt}\Vert u_{t}\Vert ^{2} +\frac{1}{2}\frac{d}{dt}\left\| A^{\frac{1}{2}}(t)u(t)\right\| ^{2}-\left\langle {\tilde{A}}^{\frac{1}{2}}(t) u(t), A^{\frac{1}{2}}(t) u(t) \right\rangle - \frac{g_{0}}{2}\frac{d}{dt}\left\| B^{\frac{1}{2}}(t)u(t)\right\| ^{2} \nonumber \\\nonumber&+ \frac{1}{2} \int _{0}^{\infty } g(s)\frac{\partial }{\partial s} \left\| B^{\frac{1}{2}}(t) \eta (t,s) \right\| ^{2} ds +\frac{1}{2} \int _{0}^{\infty } g(s)\frac{\partial }{\partial t} \left\| B^{\frac{1}{2}}(t) \eta (t,s) \right\| ^{2} ds\\&-\int _{0}^{\infty } g(s)\left\langle {\tilde{B}}^{\frac{1}{2}}(t) \eta (t,s) , B^{\frac{1}{2}}(t) \eta (t,s) \right\rangle ds + g_{0} \left\langle {\tilde{B}}^{\frac{1}{2}}(t) u(t), B^{\frac{1}{2}}(t) u(t) \right\rangle =0.\nonumber \\ \end{aligned}$$

(3.16)

Integrating by parts with respect to s and using the properties

$$\begin{aligned} \eta (t,0)=0\quad \hbox {and}\quad \displaystyle \lim _{s\rightarrow \infty }g(s)=0, \end{aligned}$$

the formula in (3.16) becomes,

$$\begin{aligned}&\frac{1}{2}\frac{d}{dt} \left[ \Vert u_{t}\Vert ^{2} +\left\| A^{\frac{1}{2}}(t)u(t)\right\| ^{2} + \int _{0}^{\infty } g(s) \left\| B^{\frac{1}{2}}(t) \eta (t,s) \right\| ^{2} ds - g_{0}\left\| B^{\frac{1}{2}}(t)u(t)\right\| ^{2} \right] \nonumber \\&\quad - \frac{1}{2} \int _{0}^{\infty } g^{\prime }(s) \left\| B^{\frac{1}{2}}(t) \eta (t,s) \right\| ^{2} ds -\int _{0}^{\infty } g(s)\left\langle {\tilde{B}}^{\frac{1}{2}}(t) \eta (t,s),B^{\frac{1}{2}}(t) \eta (t,s) \right\rangle ds \nonumber \\&\quad + g_{0} \left\langle {\tilde{B}}^{\frac{1}{2}}(t) u(t), B^{\frac{1}{2}}(t) u(t) \right\rangle -\left\langle {\tilde{A}}^{\frac{1}{2}}(t) u(t), A^{\frac{1}{2}}(t) u(t) \right\rangle =0, \end{aligned}$$

(3.17)

and the result follows. $\square$

Lemma 3.3

There exists a positive constant $c_1$ such that the functional

$$\begin{aligned} I_{1}(t)= \left\langle u(t) , u_{t}(t) \right\rangle \end{aligned}$$

(3.18)

satisfies, for any continuous function $\varepsilon _{1} :\mathbb {R}_{+}\rightarrow \mathbb {R}_{+}^*$,

$$\begin{aligned}&I_{1}^{\prime }(t) \le \left\| u_{t}(t) \right\| ^2 - \left( 1-g_{0} a_{1}(t) - \varepsilon _{1}(t) a_{1}(t) \right) \left\| A^{\frac{1}{2}}(t) u(t) \right\| ^{2}\nonumber \\&\quad +\frac{c_1}{\varepsilon _{1}(t)} \int _{0}^{\infty } g(s) \left\| B^{\frac{1}{2}}(t) \eta (t,s) \right\| ^{2}ds. \end{aligned}$$

(3.19)

Proof

Differentiating $I_{1}$ with respect to t and using (1.1), we get

$$\begin{aligned} I_{1}^{\prime }(t) = \Vert u_{t}(t) \Vert ^{2} - \left\| A^{\frac{1}{2}}(t) u(t) \right\| ^{2} + \left\langle \int _{0}^{\infty } g(s) B^{\frac{1}{2}}(t) u(t-s) ds, B^{\frac{1}{2}}(t) u(t) \right\rangle .\nonumber \\ \end{aligned}$$

(3.20)

Using the same computations as those in (3.14) and then (2.3), one gets,

$$\begin{aligned} I_{1}^{\prime }(t)= & {} \Vert u_{t}(t) \Vert ^{2} - \left\| A^{\frac{1}{2}}(t) u(t) \right\| ^{2} +g_{0} \left\| B^{\frac{1}{2}}(t) u(t) \right\| ^{2} \nonumber \\&- \left\langle \int _{0}^{\infty } g(s) B^{\frac{1}{2}}(t) \eta (t,s) ds, B^{\frac{1}{2}}(t) u(t) \right\rangle \nonumber \\\le & {} \Vert u_{t}(t) \Vert ^{2} - \left( 1-g_{0} a_{1}(t) \right) \left\| A^{\frac{1}{2}}(t) u(t) \right\| ^{2} \nonumber \\&- \left\langle \int _{0}^{\infty } g(s) B^{\frac{1}{2}}(t) \eta (t,s) ds, B^{\frac{1}{2}}(t) u(t) \right\rangle . \end{aligned}$$

(3.21)

Applying the Cauchy–Schwarz inequality, Young’s inequality on the last term of this inequality and (2.3) yields (3.19). $\square$

Lemma 3.4

There exists a positive constant $c_2$ such that the functional

$$\begin{aligned} I_{2}(t)= \left\langle - u_{t}(t) , \int _{0}^{\infty } g(s) \eta (t,s) ds \right\rangle \end{aligned}$$

(3.22)

satisfies, for any continuous functions $\varepsilon _{2} ,\,\varepsilon _{3} :\mathbb {R}_{+}\rightarrow \mathbb {R}_{+}^*$,

$$\begin{aligned} I_{2}^{\prime }(t)\le & {} -(g_{0}-\varepsilon _{2}(t) ) \left\| u_{t}(t) \right\| ^2 + \varepsilon _{3}(t) (1+ a_1 (t))\left\| A^{\frac{1}{2}}(t) u(t) \right\| ^{2} \nonumber \\&+\left[ g_0 +c_2 \left( \frac{1+a_2 (t)}{\varepsilon _{3}(t)} +\frac{1}{\varepsilon _{2}(t) b_{1}(t)}\right) \right] \int _{0}^{\infty } g(s) \left\| B^{\frac{1}{2}}(t) \eta (t,s) \right\| ^{2}ds.\nonumber \\ \end{aligned}$$

(3.23)

Proof

Differentiating with respect to t and exploiting Eq. (1.1) gives

$$\begin{aligned} I_{2}^{\prime }(t)= & {} \left\langle A^{\frac{1}{2}}(t) u(t) , \int _{0}^{\infty } g(s) A^{\frac{1}{2}}(t) \eta (t,s) ds \right\rangle - \left\langle u_{t}(t) , \int _{0}^{\infty }g(s) \eta _t (t,s) ds\right\rangle \nonumber \\&- \left\langle \int _{0}^{\infty } g(s) B^{\frac{1}{2}}(t) u(t-s) ds, \int _{0}^{+\infty } g(s) B^{\frac{1}{2}} (t) \eta (t,s) ds \right\rangle . \end{aligned}$$

(3.24)

Again from the first Eq. in (2.10) and in (2.16) we have, as for in (3.17),

$$\begin{aligned} - \left\langle u_{t}(t) , \int _{0}^{\infty }g(s) \eta _t (t,s) ds\right\rangle =-\left\langle u_{t}(t) , \int _{0}^{\infty }g^{\prime }(s) \eta (t,s) ds \right\rangle - g_{0} \Vert u_{t}(t) \Vert ^{2}\nonumber \\ \end{aligned}$$

(3.25)

and

$$\begin{aligned}&- \left\langle \int _{0}^{\infty } g(s) B^{\frac{1}{2}}(t) u(t-s) ds, \int _{0}^{\infty } g(s) B^{\frac{1}{2}} (t) \eta (t,s) ds \right\rangle \nonumber \\&= \left\langle \int _{0}^{\infty } g(s) B^{\frac{1}{2}}(t) \eta (t,s) ds , \int _{0}^{\infty } g(s) B^{\frac{1}{2}}(t) \eta (t,s) ds \right\rangle \nonumber \\&-g_{0} \left\langle B^{\frac{1}{2}}(t) u(t) , \int _{0}^{\infty } g(s) B^{\frac{1}{2}}(t) \eta (t,s) ds \right\rangle . \end{aligned}$$

(3.26)

Now, from (3.25) and (3.26), Eq. (3.24) becomes,

$$\begin{aligned} I_{2}^{\prime }(t)= & {} - g_{0} \Vert u_{t}(t) \Vert ^{2}-\left\langle u_{t}(t) , \int _{0}^{\infty }g^{\prime }(s) \eta (t,s) \,ds \right\rangle \nonumber \\&+\left\langle A^{\frac{1}{2}}(t) u(t) , \int _{0}^{\infty } g(s) A^{\frac{1}{2}}(t) \eta (t,s) \,ds \right\rangle \nonumber \\&-g_{0} \left\langle B^{\frac{1}{2}}(t) u(t) , \int _{0}^{\infty } g(s) B^{\frac{1}{2}}(t) \eta (t,s) \,ds\right\rangle \\ \nonumber&+\left\| \int _{0}^{\infty } g(s) B^{\frac{1}{2}}(t) \eta (t,s) \,ds \right\| ^{2}. \end{aligned}$$

(3.27)

Using Cauchy–Schwarz inequality, Young’s inequality, (A1), (A3) and (A4) on the last four terms yields, for the second term,

$$\begin{aligned}&-\left\langle u_{t}(t) , \int _{0}^{\infty }g^{\prime }(s) \eta (t,s) \,ds \right\rangle \nonumber \\&\quad \le \varepsilon _{2}(t) \Vert u_{t}(t) \Vert ^{2} +\frac{c }{\varepsilon _{2}(t) b_{1}(t)} \int _{0}^{\infty } g(s) \left\| B^{\frac{1}{2}}(t) \eta (t,s) \right\| ^{2} \,ds, \end{aligned}$$

(3.28)

the third term

$$\begin{aligned} \left\langle A^{\frac{1}{2}}(t) u(t) , \int _{0}^{\infty } g(s) A^{\frac{1}{2}}(t) \eta (t,s) \,ds \right\rangle\le & {} \varepsilon _{3}(t) \Vert A^{\frac{1}{2}}(t) u(t) \Vert ^{2} \nonumber \\&+ \frac{ca_2 (t)}{\varepsilon _{3}(t)} \int _{0}^{\infty } g(s) \Vert B^{\frac{1}{2}}(t) \eta (t,s) \Vert ^{2} \,ds,\nonumber \\ \end{aligned}$$

(3.29)

the fourth term

$$\begin{aligned} -g_{0} \left\langle B^{\frac{1}{2}}(t) u(t) , \int _{0}^{\infty } g(s) B^{\frac{1}{2}}(t) \eta (t,s) \,ds\right\rangle\le & {} \varepsilon _{3}(t) a_{1}(t) \Vert A^{\frac{1}{2}}(t) u(t) \Vert ^{2} \nonumber \\&+ \frac{c}{\varepsilon _{3}(t)} \int _{0}^{\infty } g(s) \Vert B^{\frac{1}{2}}(t) \eta (t,s) \Vert ^{2} \,ds,\nonumber \\ \end{aligned}$$

(3.30)

and the fifth term

$$\begin{aligned} \left\| \int _{0}^{\infty } g(s) B^{\frac{1}{2}}(t) \eta (t,s) \,ds \right\| ^{2}\le & {} \left( \int _{0}^{\infty } g(s) \left\| B^{\frac{1}{2}}(t) \eta (t,s) \right\| ds \right) ^{2} \nonumber \\\le & {} \left( \int _{0}^{\infty } \sqrt{g(s)} \sqrt{g(s)} \left\| B^{\frac{1}{2}}(t) \eta (t,s) \right\| ds \right) ^{2} \nonumber \\\le & {} \left( \int _{0}^{\infty } g(s) \,ds\right) \left( \int _{0}^{\infty } g(s) \left\| B^{\frac{1}{2}}(t) \eta (t,s) \right\| ^{2} ds \right) \nonumber \\\le & {} g_{0}\int _{0}^{\infty } g(s) \left\| B^{\frac{1}{2}}(t) \eta (t,s) \right\| ^{2} ds. \end{aligned}$$

(3.31)

Combining all the above estimates yields (3.23). $\square$

Lemma 3.5

Let $M_{1}\in \mathbb {R}_+^* ,\, M :\mathbb {R}_{+}\rightarrow \mathbb {R}_{+}^*$ be a differentiable function and let $\varepsilon _{1}, \varepsilon _{2}:\mathbb {R}_{+}\rightarrow \mathbb {R}_{+}^*$ be given continuous functions. Then the functional

$$\begin{aligned} F(t)= M_{1} I_{1}(t)+I_{2}(t)+M (t)E(t), \end{aligned}$$

(3.32)

satisfies

$$\begin{aligned}&F^{\prime }(t) \le M^{\prime }(t)E(t) \nonumber \\&\quad -\min \{A_1 (t) ,A_2 (t), A_3 (t)\}\left( \left\| u_t(t) \right\| ^{2}+\left\| A^{\frac{1}{2}}(t) u(t) \right\| ^{2}\right. \nonumber \\&\left. +\int _{0}^{\infty } g(s) \left\| B^{\frac{1}{2}}(t) \eta (t,s) \right\| ^{2} ds\right) , \end{aligned}$$

(3.33)

where

$$\begin{aligned}&A_1 (t)=g_{0}- \varepsilon _{2}(t)-M_{1} , \\&A_2 (t)=\left( 1-g_{0} a_{1}(t) -\varepsilon _{1}(t) a_{1}(t) \right) M_1 -\varepsilon _{3}(t)(1+a_1 (t))\\&\quad -\left( g_0 {\sqrt{{\tilde{b}}(t)}}a_1 (t)+{\sqrt{{\tilde{a}}(t)}}\right) M (t) \end{aligned}$$

and

$$\begin{aligned} A_3 (t)= \left( \frac{\theta _{2}}{2}-{\sqrt{{\tilde{b}}(t)}}\right) M (t)-\frac{c_1 M_1}{\varepsilon _{1} (t)}- \left[ g_0 +c_2 \left( \frac{1+a_2 (t)}{\varepsilon _{3}(t)} +\frac{1}{\varepsilon _{2}(t) b_{1}(t)}\right) \right] . \end{aligned}$$

Proof

Direct differentiation gives

$$\begin{aligned} F^{\prime }(t) =M_{1} I_{1}^{\prime }(t) +I_{2}^{\prime }(t)+M^{\prime }(t)E(t)+M(t) E^{\prime }(t). \end{aligned}$$

(3.34)

We can also estimate every term of $E^{\prime }(t)$ given in (3.11), using Cauchy–Schwarz inequality and Young’s inequality with the help of (2.3) and (3.2)-(3.4) to get, for the first term of $E^{\prime }(t)$,

$$\begin{aligned} \frac{1}{2}\int _{0}^{\infty }g^{\prime }(s) \left\| B^{\frac{1}{2}}(t) \eta (t,s)\right\| ^{2}ds\le -\frac{\theta _{2}}{2}\int _{0}^{\infty }g(s) \left\| B^{\frac{1}{2}}(t) \eta (t,s)\right\| ^{2}ds, \end{aligned}$$

the second term of $E^{\prime }(t)$

$$\begin{aligned} \left| g_{0}\left\langle {\tilde{B}}^{\frac{1}{2}}(t) u(t), B^{\frac{1}{2}}(t) u(t) \right\rangle \right| \le \,\, g_{0} {\sqrt{{\tilde{b}}(t)}}\left\| B^{\frac{1}{2}}(t) u(t) \right\| ^{2} \le g_{0} {\sqrt{{\tilde{b}}(t)}}a_1 (t)\left\| A^{\frac{1}{2}}(t) u(t) \right\| ^{2}, \end{aligned}$$

the third term of $E^{\prime }(t)$

$$\begin{aligned} \left\langle {\tilde{A}}^{\frac{1}{2}}(t) u(t), A^{\frac{1}{2}}(t) u(t) \right\rangle \le {\sqrt{{\tilde{a}}(t)}}\left\| A^{\frac{1}{2}}(t) u(t) \right\| ^{2}, \end{aligned}$$

and the fourth term of $E^{\prime }(t)$

$$\begin{aligned} \left\langle \int _{0}^{\infty } g(s){\tilde{B}}^{\frac{1}{2}}(t) \eta (t,s) ds, B^{\frac{1}{2}}(t) \eta (t,s) \right\rangle \le {\sqrt{{\tilde{b}}(t)}}\int _{0}^{\infty } g(s) \left\| B^{\frac{1}{2}}(t) \eta (t,s) \right\| ^{2}ds. \end{aligned}$$

Now, $E^{\prime }(t)$ can be estimated as follows:

$$\begin{aligned} E^{\prime }(t)\le & {} -\frac{\theta _{2}}{2}\int _{0}^{\infty }g(s) \left\| B^{\frac{1}{2}}(t) \eta (t,s)\right\| ^{2}ds \end{aligned}$$

(3.35)

$$\begin{aligned}&+\left( g_{0}{\sqrt{{\tilde{b}}(t)}} a_{1}(t) +{\sqrt{{\tilde{a}}(t)}} \right) \left\| A^{\frac{1}{2}}(t) u(t) \right\| ^{2}\nonumber \\&+{\sqrt{{\tilde{b}}(t)}}\int _{0}^{\infty } g(s) \left\| B^{\frac{1}{2}}(t) \eta (t,s) \right\| ^{2}ds \end{aligned}$$

(3.36)

Combining (3.11), (3.19), (3.23) and (3.353.36) leads to

$$\begin{aligned} F^{\prime }(t)\le & {} - A_1 (t)\Vert u_{t}(t)\Vert ^{2} -A_2 (t) \left\| A^{\frac{1}{2}}(t) u(t) \right\| ^{2}\nonumber \\&-A_3 (t)\int _{0}^{\infty } g(s) \left\| B^{\frac{1}{2}}(t) \eta (t,s)\right\| ^{2} ds+M^{\prime }(t)E(t), \end{aligned}$$

(3.37)

so, (3.33) follows. $\square$

Lemma 3.6

Let $\varepsilon _{4}, \varepsilon _{5} :\mathbb {R}_{+}\rightarrow \mathbb {R}_{+}^*$ be continuous functions. Then there exists a positive constant $c_3$ such that the functional F satisfies

$$\begin{aligned} (M(t)-M_2 (t))E(t)\le F(t)\le (M(t)+M_2 (t))E(t), \end{aligned}$$

(3.38)

where

$$\begin{aligned} M_2 (t)=\frac{c_3}{1-g_0 a_1 (t)}\max \left\{ \varepsilon _5 (t) M_1 +\varepsilon _4 (t), \frac{a_1 (t)M_1 }{\varepsilon _5 (t) b_1 (t)}, \frac{1}{\varepsilon _4 (t)b_1 (t)}\right\} . \end{aligned}$$

Proof

We see that

$$\begin{aligned} E(t)\le \frac{1}{2}\left[ \Vert u_{t}(t)\Vert ^{2}+\left\| A^{\frac{1}{2}}(t) u(t) \right\| ^{2} +\int _{0}^{\infty } g(s) \left\| B^{\frac{1}{2}}(t) \eta (t,s)\right\| ^{2} ds\right] \end{aligned}$$

(3.39)

and, using (2.3),

$$\begin{aligned} E(t)\ge \frac{1-g_0 a_1 (t)}{2}\left[ \Vert u_{t}(t)\Vert ^{2}+\left\| A^{\frac{1}{2}}(t) u(t) \right\| ^{2} +\int _{0}^{\infty } g(s) \left\| B^{\frac{1}{2}}(t) \eta (t,s)\right\| ^{2} ds\right] . \end{aligned}$$

(3.40)

On the other hand, using Young’s inequality and assumption (A1), we have, for any continuous functions $\varepsilon _{4} ,\varepsilon _{5} :\mathbb {R}_{+}\rightarrow \mathbb {R}_{+}^*$,

$$\begin{aligned} \vert I_1 (t)\vert \le \varepsilon _{5} (t)\Vert u_{t}(t)\Vert ^{2}+ \frac{ca_1 (t)}{\varepsilon _{5} (t)b_1 (t)}\left\| A^{\frac{1}{2}}(t) u(t) \right\| ^{2} \end{aligned}$$

(3.41)

and

$$\begin{aligned} \vert I_2 (t)\vert \le \varepsilon _{4} (t)\Vert u_{t}(t)\Vert ^{2}+ \frac{c}{\varepsilon _{4} (t)b_1 (t)}\int _{0}^{\infty } g(s) \left\| B^{\frac{1}{2}}(t) \eta (t,s)\right\| ^{2} ds. \end{aligned}$$

(3.42)

Therefore, by combining (3.40)-(3.42), we get

$$\begin{aligned} \vert F(t)-M(t)E(t)\vert \le M_2 (t) E(t), \end{aligned}$$

which gives (3.38). $\square$

We choose the functions $M_j$ and $\varepsilon _j$ carefully. Thanks to the properties of $g_0 ,\,a_1 ,\, b_1 ,\, {\tilde{a}}$ and ${\tilde{b}}$ assumed in (A0)-(A5), one can choose

$$\begin{aligned} M_1 =\frac{g_0}{2} ,\quad \varepsilon _1 (t)=\frac{1-g_0 a_1 (t)}{2a_1 (t)} ,\quad \quad \varepsilon _2 (t)=\frac{g_0^2}{2} a_1 (t),\quad \varepsilon _3 (t)=\frac{g_0 (1-g_0 a_1 (t))}{8(1+a_1 (t))} , \end{aligned}$$

$$\begin{aligned} \varepsilon _4 (t)=\frac{2}{\sqrt{b_1 (t) \left( 4+g_0^2 a_1 (t)\right) }},\quad \varepsilon _5 (t)=\frac{g_0 a_1 (t)}{\sqrt{b_1 (t) \left( 4+g_0^2 a_1 (t)\right) }} \end{aligned}$$

and

$$\begin{aligned} M (t)>\max \left\{ \frac{c_3}{2(1-g_0 a_1 (t))}{\sqrt{\frac{4+g_0^2 a_1 (t)}{b_1 (t)}}} ,\frac{M_3 (t)}{\frac{\theta _2}{2}-\sqrt{{\tilde{b}}(t)}}\right\} , \end{aligned}$$

(3.43)

where

$$\begin{aligned} M_3 (t)= & {} g_0 +\frac{g_0}{8}(1-g_0 a_1 (t))+\frac{c_1 g_0 a_1 (t)}{1-g_0 a_1 (t)}\nonumber \\&+c_2 \left( \frac{8(1+a_1 (t))(1+a_2 (t))}{g_0 (1-g_0 a_1 (t))} +\frac{2}{g_0^2 a_1 (t)b_1 (t)}\right) . \end{aligned}$$

(3.44)

Then

$$\begin{aligned} A_1 (t)= & {} \frac{g_0}{2}(1-g_0 a_1 (t)),\\ A_2(t)= & {} \frac{g_0}{4}(1-g_0 a_1 (t))-\left( g_0 a_1 (t)\sqrt{{\tilde{b}} (t)} +\sqrt{{\tilde{a}} (t)}\right) M(t), \\ A_3 (t)\ge & {} \frac{g_0}{8}(1-g_0 a_1 (t)),\quad M(t)>\max \left\{ M_2 (t), \frac{M_3 (t)}{\frac{\theta _2}{2}-\sqrt{{\tilde{b}}(t)}}\right\} \end{aligned}$$

and

$$\begin{aligned} M_2 (t)=\frac{c_3}{2(1-g_0 a_1 (t))}{\sqrt{\frac{4+g_0^2 a_1 (t)}{b_1 (t)}}}. \end{aligned}$$

(3.45)

On the other hand, we assume that the second assumption in (3.5) holds such that

$$\begin{aligned} \left( g_0 a_1 (t)\sqrt{{\tilde{b}} (t)} +\sqrt{{\tilde{a}} (t)}\right) M (t)\le \frac{g_0}{8}(1-g_0 a_1 (t)) \end{aligned}$$

(3.46)

(notice that (3.46) is possible as $M_2 (t)$ and $M_3 (t)$ depend neither on ${\tilde{a}}$ nor on ${\tilde{b}}$), so we get

$$\begin{aligned} A_2 (t)\ge \frac{g_0}{8}(1-g_0 a_1 (t)), \end{aligned}$$

and then, combining (3.33) and (3.39), we find

$$\begin{aligned} F^{\prime } (t)\le \left[ M^{\prime } (t)-\frac{g_0}{4}(1-g_0 a_1 (t)) \right] E(t), \end{aligned}$$

therefore, according to (3.38),

$$\begin{aligned} F^{\prime } (t)\le \xi (t)F(t), \end{aligned}$$

(3.47)

where

$$\begin{aligned} \xi (t)=\max \left\{ \frac{M^{\prime } (t)-\frac{g_0}{4}(1-g_0 a_1 (t))}{M(t)-M_2 (t)},\frac{M^{\prime } (t)-\frac{g_0}{4}(1-g_0 a_1 (t))}{M(t)+M_2 (t)} \right\} . \end{aligned}$$

(3.48)

By integrating (3.47), we arrive to

$$\begin{aligned} F (t)\le F(0)e^{{\tilde{\xi }}(t)}, \end{aligned}$$

(3.49)

where

$$\begin{aligned} {\tilde{\xi }}(t)=\int _0^t \xi (s)ds . \end{aligned}$$

(3.50)

Consequently, exploiting again (3.38), we conclude (3.9). $\square$

Remark 4

If $\frac{M_3 }{\frac{\theta _2}{2}-\sqrt{{\tilde{b}} }}$ and $M_2$ are bounded, then one can choose M as a constant satisfying

$$\begin{aligned} M>\max \left\{ \Vert M_2\Vert _{L^{\infty } (\mathbb {R}_+)},\left\| \frac{M_3 }{\frac{\theta _2}{2}-\sqrt{{\tilde{b}} }}\right\| _{L^{\infty } (\mathbb {R}_+)}\right\} , \end{aligned}$$

therefore

$$\begin{aligned} \xi (t)=\frac{-\frac{g_0}{4}(1-g_0 a_1 (t))}{M+M_2 (t)} \le \frac{-\frac{g_0}{4}(1-g_0 a_1 (t))}{M+\Vert M_2\Vert _{L^{\infty } (\mathbb {R}_+)}}\quad \hbox {and}\quad \frac{1}{M-M_2 (t)}\le \frac{1}{M-\Vert M_2\Vert _{L^{\infty } (\mathbb {R}_+)}}, \end{aligned}$$

hence (3.9) implies that there exist positive constants $\lambda _0$ and $\lambda _1$ such that

$$\begin{aligned} \Vert {\mathcal U}(t)\Vert _{{\mathcal H} (0)}^2 \le \lambda _0 e^{-\lambda _1 \int _0^t (1-g_0 a_1 (s))ds}. \end{aligned}$$

(3.51)

From (3.44) and (3.45), we observe that $\frac{M_3 }{\frac{\theta _2}{2}-\sqrt{{\tilde{b}} }}$ and $M_2$ are bounded if and only if

$$\begin{aligned} \left\{ \begin{array}{ll} \Vert a_1\Vert _{L^{\infty }(\mathbb {R}_+)}<\infty , \quad g_0<\frac{1}{\Vert a_1\Vert _{L^{\infty }(\mathbb {R}_+)}},\quad \inf _{t\in \mathbb {R}_+} a_1 (t)>0,\quad \inf _{t\in \mathbb {R}_+} b_1 (t)>0, \\ \Vert a_2\Vert _{L^{\infty }(\mathbb {R}_+)}<\infty \quad \hbox {and}\quad \Vert {\tilde{b}}\Vert _{L^{\infty }(\mathbb {R}_+)} <\frac{\theta _2^2}{4}, \end{array} \right. \end{aligned}$$

(3.52)

so (3.51) is reduced to the exponential stability estimate, for ${\tilde{\lambda }}_1 =\lambda _1 (1-g_0 \Vert a_1\Vert _{L^\infty (\mathbb R_+)})$,

$$\begin{aligned} \Vert {\mathcal U}(t)\Vert _{{\mathcal H} (0)}^2 \le \lambda _0 e^{-{\tilde{\lambda }}_1 t}. \end{aligned}$$

(3.53)

Remark 5

Let us construct a solution to (2.11) which converges to 0 as $t \rightarrow \infty$. For that, it is enough to construct a $C_0$-semigroup $({\mathcal U}(t))_{t\ge 0}$ that is exponentially stable. Indeed, let $\Omega \subset \mathbb R^N$, for $N \in \mathbb N^*$, be an open bounded domain with smooth boundary $\Gamma = \partial \Omega$ and let $H = L^2(\Omega )$ equipped with its standard $L^2$-topology. Consider, for $m \in \mathbb N^*$, $L = \Delta ^m$ with $D(L) = H^{2m}(\Omega ) \cap H_0^m(\Omega )$. Obviously, $-L$ is a positive selfadjoint linear operator on $L^2(\Omega )$ with compact resolvent. Further, $D((-L)^{\frac{1}{2}}) = H_0^m(\Omega )$.

Consider the case when $A(t)=-a(t) \Delta ^m$, $B (t)=-b(t) \Delta ^m$, $D(A(t))=D(B(t))=H^{2m}(\Omega ) \cap H_0^m(\Omega )$ with

$$\begin{aligned} a(t) = \alpha + r(t), \ \ b(t) = \beta + k(t) \ \ \text{ for } \text{ all } \ \ t \in \mathbb R_+,\end{aligned}$$

where $\alpha \ge \beta >0$, $\theta _1 = \theta _2 = 1$ (yielding $g_0 = 1$), $r, k: \mathbb R_+ \rightarrow \mathbb R_+^*$ are class $C^1$ bounded functions such that $\Vert r'\Vert _{L^\infty (\mathbb R_+)} < \infty$ and $\Vert k'\Vert _{L^\infty (\mathbb R_+)} < \infty$ and that,

• i) $\displaystyle \inf _{t \in \mathbb R_+} r(t) > \beta + \Vert k\Vert _{L^\infty (\mathbb R_+)}$;

• ii) $\displaystyle \sqrt{\frac{\Vert k'\Vert _{L^\infty (\mathbb R_+)}}{\beta }} < \frac{1}{2}$; and

• iii) $\displaystyle \frac{\beta + \Vert k\Vert _{L^\infty (\mathbb R^+)}}{2\alpha } + \sqrt{\frac{\Vert r'\Vert _{{L^\infty (\mathbb R_+)}}}{\beta }}$ is small enough (to guarantee (3.46)).

In view of the above, it is not hard to see that (3.52) holds. Therefore, the solution to (2.11) converges to 0 as $t \rightarrow \infty$.

4 Applications

In this section, we present two examples that fit into our abstract model, namely (1.1)–(1.2). Let $\Omega \subset \mathbb {R}^{N}$ be an open bounded domain with smooth boundary $\Gamma$, where $N\in \mathbb {N}^{*}$. In both cases, we will assume that $H=L^{2}(\Omega )$ is equipped its standard $L^2$-topology.

4.1 Wave equations

The abstract model (1.1)–(1.2) includes the following nonautonomous wave equation,

$$\begin{aligned} \left\{ \begin{array}{ll} u_{tt} (x,t)-a(t)\Delta u(x,t)+b(t)\displaystyle \int _0^{\infty }g(s)\Delta u(x,t-s)ds=0,&{}\,\forall (x,t)\in \Omega \times \mathbb {R}_{+}^*,\\ u(x,t)=0,&{}\,\forall (x,t)\in \Gamma \times \mathbb {R}_{+}^* ,\\ u(x,-t)=u_0 (x,t),\quad u_t (x,0)=u_1 (x),&{}\,\forall (x,t)\in \Omega \times \mathbb {R}_{+} , \end{array} \right. \end{aligned}$$

(4.1)

where $A(t)=-a(t) \Delta$, $B (t)=-b(t) \Delta$, $D(A(t))=D(B(t))=H^2 (\Omega )\cap H^1_0 (\Omega )$. Theorems 2.1 and 3.1 hold true under the assumptions given in Remarks 1 and 2.

4.2 Petrovsky type systems

The following nonautonomous Petrovsky type system fits into our abstract model (1.1)–(1.2),

$$\begin{aligned} \left\{ \begin{array}{ll} u_{tt} (x,t)+a(t) \Delta ^2 u(x,t)-b(t)\displaystyle \int _0^{\infty }g(s)\Delta ^2 u(x,t-s)ds=0,&{}\,\forall (x,t)\in \Omega \times \mathbb {R}_{+}^*,\\ u(x,t)={{\partial u}\over {\partial \nu }} (x,t)=0,&{}\,\forall (x,t)\in \Gamma \times \mathbb {R}_{+}^* ,\\ u(x,-t)=u_0 (x,t),\quad u_t (x,0)=u_1 (x),&{}\,\forall (x,t)\in \Omega \times \mathbb {R}_{+} . \end{array} \right. \end{aligned}$$

(4.2)

where $A(t)=a(t) \Delta ^2$, $B (t)=b(t) \Delta ^2$, $D(A(t))=D(B(t))=H^4 (\Omega )\cap H^2_0 (\Omega )$, and assumptions of Remarks 1 and 2 yield both Theorems 2.1 and 3.1.

5 General comments and issues

Under some appropriate assumptions on the time-dependent operators A(t) and B(t), as well as the relaxation (kernel) function g, we established the well-posedness and asymptotic stability of the solutions to the system (1.1)–(1.2) as time t goes to infinity. In light of our findings, we would like to propose the following questions, which, to the best of our knowledge, remain unanswered:

• (1) Will we be in the presence of a discrete or distributed delay by adding

  $$\begin{aligned} \int _0^{\infty } f(s)C(t)u_t (t-s)ds\quad \hbox {or}\quad f(t)C(t)u_t (t-\tau ), \end{aligned}$$

  to (1.1), where C(t) is an operator, $f:\mathbb {R}_+\rightarrow \mathbb {R}$ is a function, and $\tau$ is a fixed positive real number?

• (2) Can we apply the previous theory to a larger class of relaxation functions g, that is,

  $$\begin{aligned} g^{\prime } (s)\le -\theta _2 (s)g(s),\quad \forall s\in \mathbb {R}_+ \end{aligned}$$

  instead of (3.4), where $\theta _2 :\mathbb {R}_+\rightarrow \mathbb {R}_+$ is a function?

• (3) Can we establish similar results when D(A(t)) and D(B(t)) are no longer constant in time t?

• (4) What about the damping case?