1. Introduction
Viscoelastic materials exhibit an instantaneous elasticity effect
and creep characteristics at the same time. The importance of the
viscoelastic properties of materials has been realized because of
the rapid developments in rubber and plastics industry. The modeling
of the dynamics of physical phenomena such as heat flow in
conductors with memory, hereditary polarization in dielectrics,
population dynamics, viscolasticity can be described by an abstract
integro-differential equation of the form
\begin{equation}\label{eqn1:1}
\begin{cases}
u”(t)+Au(t)-\displaystyle\int_{-\infty}^tg(t-s)A^{\alpha}u(s)ds=0, & t>0,\\
u(-t)=u_0(t), & t\geq 0, \;\; u'(0)=u_1,
\end{cases}
\end{equation}
(1)
where \(‘\) represents a derivative with respect to time \(t\), \(A:
\mathcal{D}(A)\subset H \longrightarrow H\) is a positive definite
self-adjoint operator on \(H\), \(g\) is the relaxation function
(convolution kernel), \(\alpha \in [0,1]\), \(u_0,\, u_1\) are given history
function and initial data respectively.
The study of viscoelastic
problems has attracted the attention of many authors and several
decay and blow up results have been established. We start with the
pioneer work [6,7,] where Dafermos
considered a one-dimensional viscoelastic problem and established
various existence results and then proved, for smooth monotone
decreasing relaxation functions, that the solutions go to zero as
\(t\) goes to infinity. After that, many results dealing with the
existence, uniqueness, regularity and asymptotic behavior of many
systems of the form (1) have been studied; see, for
example, [1,8,9,10,11]. In the case of finite memory, that
is, \(u_0(t)=0\) for \(t< 0\), see [12,13,14,15,16,17,18]. In particular, Rivera et al., [15]
considered the interpolating cases \(\alpha \in (0,1)\) and a relaxation
function \(g\) which decays exponentially to zero at infinity, that
is,
\begin{equation}\label{eqn2:1}
-c_0 g(s) \leq g'(s) \leq -c_1 g(s) \qquad \forall\, s \in \mathbb{R}_+.
\end{equation}
(2)
They showed that the energy decays polynomially at the rate of
\(\frac{1}{t}\). Recently, Hassan and Messaoudi
[
19] considered
\begin{equation}\label{eqn1:10}
\begin{cases}
u”(t)+Au(t)-\displaystyle\int_{0}^tg(t-s)A^\alpha u(s)ds=0, & t>0,\\
u(0)=u_0(t), & u'(0)=u_1,
\end{cases}
\end{equation}
(3)
and established a new general decay rate result for which the
relaxation function \(g\) satisfies condition
\begin{equation} \label{g1}
g'(t) \leq – \xi(t) G \left(g(t)\right),~~t \geq 0.
\end{equation}
(4)
For case of infinite memory, see [
20,
21,
22,
23,
24,
25]. In particular, Guesmia [
1] considered
\begin{equation}
u_{tt}+Au-\int_{0}^{+\infty}g(s)Bu(t-s)ds=0 \hspace{0.1in}\text{for}
\hspace{0.1in}
t>0,
\end{equation}
(5)
and introduced a new ingenuous approach for proving a more general
decay result based on the properties of convex functions and the use
of the generalized Young inequality. He used a larger class of
infinite history kernels satisfies the following condition
\begin{equation}\label{gg}
\int_{0}^{+\infty}\frac{g(s)}{G^{-1}(-g^{\prime}(s))}ds+\sup_{s\in\mathbb{R}_+}\frac{g(s)}{G^{-1}(-g^{\prime}(s))}< +\infty,
\end{equation}
(6)
such that
\begin{equation}\label{E:3}
G(0)=G^{\prime}(0)=0\hspace{0.05in}\text{and}\hspace{0.05in}\lim_{t\rightarrow
+\infty}{G^{\prime}(t)}=+\infty,
\end{equation}
(7)
where \(G:\mathbb{R}_+\to \mathbb{R}_+\) is an increasing strictly
convex function. Al-Mahdi and Al-Gharabli [
2]
considered the following viscoelastic problem
\begin{equation}\label{4.4}
\begin{gathered}
\begin{cases}
u_{tt}-\Delta u+\int_{0}^{+\infty}g(s)\Delta u(t-s)ds+{\vert
u_{t}\vert}^{m-2}u_{t}=0, & \text{in}\;\;
\Omega\times(0,+\infty)
\\
u(x,t)=0,&\text{on}\;\;\partial\Omega\times(0,+\infty)
\\
u(x,-t)=u_{0}(x,t),\hspace{0.05in}u_{t}(x,0)=u_{1}(x),&\text{in}\;\;\Omega\times(0,+\infty),
\end{cases}
\end{gathered}
\end{equation}
(8)
and they established decay results with using a relaxation function
\(g\), satisfying the condition
\begin{equation}\label{2017:E3y}
g'(t) \leq – \xi(t) g^{p} (t), ~~1\leq p < \frac{3}{2}.
\end{equation}
(9)
Very recently, Guesmia
[
26] considered two models of wave equations
with infinite memory and established an explicit and general decay
rate results where the relaxation function satisfying the condition
(4).
Motivated by the above works, we
intend to study the following class of viscoelastic equations of the
form
\begin{equation}\label{eqn1}
\begin{cases}
u”(t)+Au(t)-\displaystyle\int_{0}^{\infty} g(s)A^\alpha u(t-s)ds=0, & t>0,\\
u(-t)=u_0, & u'(0)=u_1,
\end{cases}
\end{equation}
(10)
where \(A: \mathcal{D}(A)\subset H \longrightarrow H\) is a positive
definite self-adjoint operator on \(H\) such that the embedding \(\mathcal{D}
(A^\beta) \hookrightarrow \mathcal{D} (A^\sigma)\) is compact for any \(\beta > \sigma
\geq 0\) and \( \alpha \in (0,1) \).
Remark 1.
The assumption \(\mathcal{D} (A^\beta) \hookrightarrow \hookrightarrow \mathcal{D}
(A^\sigma)\) for any \(\beta > \sigma \geq 0\) guarantees the existence of some
constants \(\omega,\, \omega_0,\, \omega_1\) such that
\begin{eqnarray}\label{PO}
\|v\|^2 \leq \omega\left\|A^{1/2}v\right\|^2 \qquad \forall \, v \in
\mathcal{D}\left(A^{1/2}\right),
\end{eqnarray}
(11)
\begin{equation}\label{a1}
\left\|A^{\alpha/2}v\right\|^2 \leq \omega_0 \left\|A^{1/2}v\right\|^2
\qquad \forall\, v \in \mathcal{D}\left(A^{1/2}\right),
\end{equation}
(12)
and
\begin{equation}\label{a2}
\left\|A^{1/2}v\right\|^2 \leq _1 \left\|A^{1-\alpha/2}v\right\|^2
\qquad \forall\, v \in \mathcal{D}\left(A^{1-\alpha/2}\right).
\end{equation}
(13)
2. Our main objectives
We intend to establish a two fold
objective:
- improve many earlier works such as the ones in
[11,15,19] from finite memory to infinite memory;
- prove a general decay estimate for the solution of Problem
(10) with a wider class of relaxation functions than the
ones considered in [1,2,3,4,5] by
getting a better decay rate with imposing a weaker assumption on the
boundedness of initial data than the one considered in the earlier
papers such as the one in [1,2,3,4,5].
The paper is organized as follows: We present some
assumptions and remarks in §3. We state and prove
some technical lemmas in §4. The main result, its
proof and some examples are presented in §5.
3. Assumptions
In this section, we state some assumptions needed in the proof of our main decay result. The strictly decreasing differentiable
relaxation (kernel) function \( g : [0,\infty)\longrightarrow
(0,\infty)\) satisfies the following assumptions:
- ( A.1) \(\qquad\qquad\qquad \displaystyle g(0)>0 \qquad\mathrm{and\qquad} 1- _0\int_0^{+\infty}g(s)ds=l>0. \)
- ( A.2) There exists a non-increasing differentiable function \(\xi\!:\mathbb{R}_+\longrightarrow(0,\infty)\) and
a \(\boldsymbol{C}^1\) function
\(G:[0,+\infty)\longrightarrow[0,+\infty)\) which is linear or it is
strictly increasing and strictly
convex \(C^2\) function on \((0,r]\), with \(G(0)=G^{\prime}(0)=0\), such that
\begin{equation}\label{a4}
g'(t)\leq-\xi(t)G(g(t)),\qquad\forall\,t \geq 0,
\end{equation}
(14)
where \(\xi\) is satisfying \(\int_0^{+ \infty}\xi(s)ds = + \infty.\)
- ( A.3) We assume that \[\int_{0}^{+\infty}g(s)\vert\vert A^{\alpha/2}u_0(s)\vert\vert^{2} ds <
+\infty,\] and
\[\int_{0}^{+\infty}g(s)\vert\vert A^{1/2}u_0(s)\vert\vert^{2} ds <
+\infty.\]
Remark 2.The class of relaxation functions satisfying
( A.1)-( A.2) in the present paper is larger than the
ones satisfying (6) and (7) used in some earlier
papers such as the one in [1]. In fact, the boundedness
of the sup in (6) use in [1], can be interpreted
as the inequality in ( A.2) in the present paper (with
\(\xi=1\)). The conditions (6) and (7) used in
[1] ask also the boundedness of the integral. So, it is
better to consider the relaxation functions satisfy (
A.1)-( A.2) used in the present paper than the one used in
[1].
Remark 3.
Hypothesis ( A.3) is needed for proving the existence and
stability results. For the stability, if ( A.3) holds, then
the functions \(h_0\) and \(h_1\) defined in Lemma 4 well be
defined. Moreover, Hypothesis ( A.3) is weaker than the one
used in [1,2,3,4,5] that is, there exists
a positive constant \(M\) such that
\[\vert\vert \nabla
A^{\alpha/2} u_{0}(s)\vert\vert^{2} \leq M,\] and
\[\vert\vert \nabla
A^{1/2} u_{0}(s)\vert\vert^{2} \leq M.\]
Remark 4.
As is in Mustafa [14], if \(G\) is a strictly
increasing and strictly convex \(\boldsymbol{C}^2\) function on
\((0,r]\), with \(G(0)=G'(0)=0\), then there is a strictly convex and
strictly increasing \(\boldsymbol{C}^2\) function
\(\overline{G}:[0,+\infty)\longrightarrow[0,+\infty)\) which is an
extension of \(G\). For instance, we can define \(\overline{G}\), for
any \(t>r\), by
\[\overline{G}(t):=\frac{G”(r)}{2}t^2+\big(G'(r)-G”(r)r\big)t+\left(G(r)+\frac{G”(r)}{2}r^2-G'(r)r\right).\]
We state the existence, regularity and uniqueness theorem whose
proof is in [
15].
Theorem 1([15]).
Suppose that \((u_0(\cdot, 0),u_1) \in \mathcal{D}(A) \times
\mathcal{D}(A^{1/2})\) and ( A.1– A.3) hold. Then, Problem
(10) has a unique global solution satisfying
\[u \in \boldsymbol{C} \left(\mathbb{R}_+;\mathcal{D}(A)\right) \cap \boldsymbol{C}^1 \left(\mathbb{R}_+;\mathcal{D}\left(A^{1/2}\right)\right) \cap \boldsymbol{C}^2 \left(\mathbb{R}_+;H\right).\]
Moreover, if \((u_0(\cdot, 0),u_1) \in \mathcal{D}\left(A^{\sigma+1/2}\right)
\times \mathcal{D}\left(A^{\sigma}\right)\) for \(\sigma \geq 0\), then the solution
satisfies
\[u \in \boldsymbol{C} \left(\mathbb{R}_+;\mathcal{D}\left(A^{\sigma+1/2}\right)\right) \cap \boldsymbol{C}^1 \left(\mathbb{R}_+;\mathcal{D}\left(A^{\sigma}\right)\right) \cap \boldsymbol{C}^2 \left(\mathbb{R}_+;\mathcal{D}\left(A^{\sigma-1/2}\right)\right).\]
The “modified” energy functionals associated to our problem are
given by
\begin{align}
&\label{enr:1}
E(t):=\frac{1}{2}\left[
\|u'(t)\|^2+\left\|A^{1/2}u(t)\right\|^2-\left(\frac{1-l}{\omega_0}\right)\left\|A^{\alpha/2}u(t)\right\|^2
+\left(g\circ A^{\alpha/2}u\right)(t) \right],\\
\end{align}
(15)
\begin{align}
&\label{enr:2}
\mathcal E(t):=\frac{1}{2}\left[\vphantom{\int}\right. \|A^{(1-\alpha)/2}u'(t)\|^2+\left\|A^{1-\alpha/2}u(t)\right\|^2
\left.-\left(\frac{1-l}{\omega_0}\right)\left\|A^{1/2}u(t)\right\|^2
+\left(g\circ A^{1/2}u\right)(t) \right],
\end{align}
(16)
for any \(t \geq 0\), where for \(v \in L^2_{loc}(\mathbb{R}_+;H)\),
\[(g \circ v)(t):=\int_0^{\infty} g(s)\|v(t)-v(t-s)\|^2 ds.\]
Remark 5.
The positiveness of the energy functionals comes from inequalities
(12) and (13).
Lemma 1([15]).
For any initial data \((u_0,u_1) \in \mathcal{D}(A) \times
\mathcal{D}\left(A^{1/2}\right)\), the energy functionals associated to
Problem (10) satisfy, for any \(t \geq 0\), the identities
\begin{equation}\label{dis:1}
E'(t) = \frac{1}{2}\left(g’ \circ A^{\alpha/2}u\right)(t) \leq 0,
\end{equation}
(17)
\begin{equation}\label{dis:2}
\mathcal E'(t) = \frac{1}{2}\left(g’ \circ A^{1/2}u\right)(t) \leq 0.
\end{equation}
(18)
As in Jin
et al., [
27], we set, for any \(0< \varepsilon< 1\),
\[C_\varepsilon:=\int_0^\infty\frac{g^2(s)}{\varepsilon g(s)-g'(s)}ds \qquad\mathrm{and}\qquad h_\varepsilon(t):=\varepsilon g(t)-g'(t).\]
Lemma 2([27]).
Assume that the condition ( A.1) holds. Then, for any \(v \in
L^2_{loc} \big([0,+\infty);L^2(0,L)\big)\), we have
\begin{equation}\label{e3s2}
\int_0^L\left(\int_0^{\infty} g(s)(v(t)-v(t-s))ds\right)^2dx\leq
C_\varepsilon(h_\varepsilon \circ v)(t), \qquad\forall\,t\geq0.
\end{equation}
(19)
Lemma 3(Jensen’s inequality).
Let \(F:[a,b]\longrightarrow\mathbb{R}\) be a convex function. Assume that the
functions \(f: \longrightarrow[a,b]\) and \(h:\Omega\longrightarrow\mathbb{R}\) are
integrable such that \(h(x)\geq0\), for any \(x\in\Omega\) and
\(\displaystyle\int_\Omega h(x)dx=k>0\). Then,
\[F\left(\frac{1}{k}\int_\Omega f(x)h(x)dx\right) \leq \frac{1}{k}\int_\Omega F(f(x))h(x)dx.\]
4. Technical lemmas
In this section, we state and prove some Lemmas that are useful in
the proof of Theorem 2. Through out this work we use
\(c>1\) to represent a generic constant, which is independent of \(t\)
and the initial data.
Lemma 4.
Assume that ( A.1– A.3) hold. Then, there exist two positive constants \(M_0, M_1\) such that
\begin{equation}\label{h_0}
\begin{aligned}
\int_{t}^{+\infty}g(s)\vert \vert A^{\alpha/2}u(t)-A^{\alpha/2}u(t-s) \vert
\vert^{2} ds \leq M_0 h_0(t),
\end{aligned}
\end{equation}
(20)
and
\begin{equation}\label{h_1}
\begin{aligned}
\int_{t}^{+\infty}g(s)\vert \vert A^{1/2}u(t)-A^{1/2}u(t-s) \vert
\vert^{2} ds \leq M_1 h_1(t),
\end{aligned}
\end{equation}
(21)
where \(h_0(t)=\int_{0}^{+\infty}g(t+s)\left(1+\vert \vert
A^{\alpha/2}u_0(s)\vert\vert^{2} \right) ds,\) and
\(h_1(t)=\int_{0}^{+\infty}g(t+s)\left(1+\vert \vert
A^{1/2}u_0(s)\vert\vert^{2} \right) ds.\)
Proof. Indeed, we have
\begin{align}
\label{hproof}
\notag\int_{t}^{+\infty}g(s) &\vert \vert
A^{\alpha/2}u(t)-A^{1/2}u(t-s)\vert \vert^{2} ds \\&\leq 2 \vert
\vert A^{\alpha/2}u(t) \vert \vert^2 \int_{t}^{+\infty}g(s) ds +2
\int_{t}^{+\infty}g(s) \vert \vert
A^{\alpha/2}u(t-s) \vert \vert^2 ds\notag\\
&\leq 2 \sup_{s\geq 0}\vert \vert A^{\alpha/2}u(s) \vert \vert^2
\int_{0}^{+\infty}g(t+s) ds +2 \int_{0}^{+\infty}g(t+s) \vert \vert
A^{\alpha/2}u(-s) \vert \vert^2 ds\notag\\
&\leq \frac{4 \omega_0 \sup_{s\geq 0} E(s)}{1-l}
\int_{0}^{+\infty}g(t+s) ds +2 \int_{0}^{+\infty}g(t+s)
\vert \vert A^{\alpha/2}u_{0}(s) \vert \vert^2 ds\notag\\
&\leq \frac{4 \omega_0 E(0)}{1-l} \int_{0}^{+\infty}g(t+s) ds +2
\int_{0}^{+\infty}g(t+s) \vert \vert A^{\alpha/2}u_{0}(s) \vert \vert^2 ds\notag\\
&\leq M_0 \int_{0}^{+\infty}g(t+s)\left(1+\vert\vert
A^{\alpha/2}u_{0}(s)\vert\vert^{2} \right) ds.
\end{align}
(22)
where \(M_0=\max\big\{2, \frac{4 \omega_0 E(0)}{1-l} \big\}\).
The proof of (21) can be established similarly to the proof of
(20).
Lemma 5.
Assume that conditions ( A.1– A.3) hold. Then, for any
\(0< \delta< 1\), the functional \(I_1\) defined by
\[I_1(t):=-\left\langle u'(t), \int_0^{\infty} g(s)(u(t)-u(t-s))ds\right\rangle\]
satisfies, along the solution of (10), the estimate
\begin{eqnarray}\label{eqn1:3}
I_1′(t)&\leq&-\left(\frac{1-l}{\omega_0}-\delta\right)\|u'(t)\|^2+ \delta\left\|A^{1/2}u(t)\right\|^2 + \frac{c}{\delta}(C_\varepsilon + 1)\left( h_\varepsilon \circ A^{1/2}u
\right)(t),\qquad\forall\,t\geq0.
\end{eqnarray}
(23)
Proof.
Differentiating \(I_1\) and exploiting the differential equation in
Problem (10), we get
\begin{eqnarray}\label{eqn2:3}
I_1′(t)&=&\left\langle A^{1/2}u(t),\int_0^\infty g(s)A^{1/2}(u(t)-u(t-s))ds\right\rangle\notag\\
&& – \left\langle\int_0^\infty g(s)A^{\alpha/2}u(t-s)ds,\int_0^\infty g(s)A^{\alpha/2}(u(t)-u(t-s))ds\right\rangle\notag\\
&&-\left(\frac{1-l}{\omega_0}\right)\|u'(t)\|^2-\left\langle
u'(t),\int_0^{\infty} g'(s)(u(t)-u(t-s))ds\right\rangle.
\end{eqnarray}
(24)
Next, we estimate the terms in the right-hand side of the above
identity. Using the Cauchy-Schwarz, Young and Hölder inequalities,
Lemma 2 and inequalities (11) and (12), it
follows that, for any \(0< \delta< 1\),
\begin{align*}
&\left\langle A^{1/2}u(t), \int_0^\infty g(s)A^{1/2}(u(t)-u(t-s))ds
\right\rangle \\
&\leq \frac{\delta}{2} \|A^{1/2}u(t)\|^2 +
\frac{c}{\delta}C_\varepsilon \left( h_\varepsilon \circ A^{1/2}u \right)(t),
-\left\langle \int_0^\infty g(s)A^{\alpha/2}u(t-s)ds,\right. \left.\int_0^\infty g(s)A^{\alpha/2}(u(t)-u(t-s))ds \right\rangle \\
& = \left\| \int_0^\infty g(s)A^{\alpha/2}(u(t)-u(t-s))ds \right\|^2
– \left\langle\left(\frac{1-l}{\omega_0}\right)A^{\alpha/2}u(t), \int_0^\infty g(s)A^{\alpha/2}(u(t)-u(t-s))ds\right\rangle \\
& \leq \frac{\delta}{2\omega_0}\left\|A^{\alpha/2}u(t)\right\|^2 + \frac{c}{\delta}C_\varepsilon \left( h_\varepsilon \circ A^{\alpha/2}u\right)(t) \\
& \leq \frac{\delta}{2}\left\|A^{1/2}u(t)\right\|^2 + \frac{c}{\delta}C_\varepsilon
\left( h_\varepsilon \circ A^{1/2}u\right)(t),
\end{align*}
and
\begin{align*}
&\left\langle u'(t),\int_0^\infty g'(s)(u(t)-u(t-s))ds \right\rangle \\
&= \left\langle u'(t),\varepsilon\int_0^\infty g(s)(u(t)-u(t-s))ds \right\rangle
– \left\langle u'(t),\int_0^\infty h_\varepsilon(s)(u(t)-u(t-s))ds \right\rangle \end{align*}\begin{align*}
&\leq \frac{\delta}{2} \|u'(t)\|^2 + \frac{\varepsilon^2}{2\delta}\left\|\int_0^\infty g(s)(u(t) – u(t-s)) ds \right\|^2
+ \frac{\delta}{2} \|u'(t)\|^2 + \frac{1}{2\delta}\left\| \int_0^\infty h_\varepsilon(s)(u(t)- u(t-s)) ds \right\|^2 \\
&\leq \delta \|u'(t)\|^2 + \frac{c}{\delta}(C_\varepsilon + 1)\left( h_\varepsilon \circ
A^{1/2}u \right)(t).
\end{align*}
Plugging the above estimates in (24), we obtain the
desired estimate.
Lemma 6.
Under the conditions ( A.1– A.3), the functional \(I_2\)
defined by
\[I_2(t):=\langle u'(t),u(t)\rangle\]
satisfies, along the solution of (10), the estimate
\begin{equation}\label{eqn3:3}
I_2′(t) \leq \|u'(t)\|^2 – \frac{l}{2}\|A^{1/2}u(t)\|^2 +
cC_\varepsilon\left( h_\varepsilon \circ A^{\alpha/2}u \right)(t), \qquad\forall\, t
\geq 0.
\end{equation}
(25)
Proof.
Differentiating \(I_2\), using the equation in (10), and
repeating the above computations, we get
\begin{align*}
I_2′(t)&=\|u'(t)\|^2 – \left\|A^{1/2}u(t)\right\|^2 + \left( \frac{1-l}{\omega_0} \right)\left\|A^{\alpha/2}\right\|^2
+ \left\langle \int_0^\infty g(s)A^{\alpha/2}(u(s)-u(t-s))ds, A^{\alpha/2}u(t)\right\rangle \\
&\leq \|u'(t)\|^2 -l \left\|A^{1/2}u(t)\right\|^2 + \frac{l}{2\omega_0}\left\| A^{\alpha/2}u(t) \right\|^2 + \frac{\omega_0}{2l}\left\| \int_0^\infty g(s) A^{\alpha/2}((u(t) – u(t-s)) ds \right\|^2 \\
&\leq \|u'(t)\|^2 – \frac{l}{2}\left\|A^{1/2}u(t)\right\|^2 +
cC_\varepsilon\left( h_\varepsilon \circ A^{1/2}u \right)(t), \qquad\forall\, t \geq
0.
\end{align*}
Lemma 7.
Assume that ( A.1– A.3) hold. Then, the functionals \(J_1\)
and \(J_2\) defined by
\[ J_1(t) := \int_0^t p(t-s)\left\| A^{\alpha/2} u(s) \right\|^2 ds \]
and
\[ J_2(t) := \int_0^t p(t-s)\left\| A^{1/2} u(s) \right\|^2 ds \]
with \( \displaystyle p(t) := \int_t^\infty g(s) ds \) satisfy, along
the solution of (10), the estimates
\begin{equation*}\label{e8s3}
J_1′(t) \leq 3(1-l)\|A^{1/2} u(t) \|^2 – \frac{1}{2}(g \circ
A^{\alpha/2}u)(t)+ \frac{1}{2}\int_t^{\infty} g(s) \|A^{\alpha/2}
u(t)-A^{\alpha/2} u(s)\|_2^2ds,
\end{equation*}
and
\begin{equation*}\label{e9s3}
J_2′(t) \leq \frac{3}{ _0}(1-l)\|A^{1/2} u(t)\|^2 – \frac{1}{2}(g
\circ A^{1/2}u)(t)+ \frac{1}{2}\int_t^{\infty} g(s) \|A^{1/2}
u(t)-A^{1/2} u(s)\|_2^2ds,
\end{equation*}
for any \( t \geq 0 \).
Proof.
Exploiting Young’s inequality, ( A.1– A.3), inequality
(12) and the fact that \( p(t) \leq p(0) = \frac{1-l}{ _0} \),
we obtain, for any \( t \geq 0 \),
\begin{align}
J_1′(t) =& p(0)\left\| A^{\alpha/2}u(t)\right\|^2 – \int_0^t g(t-s)\left\| A^{\alpha/2}u(s)\right\|^2 ds \notag\\
=&p(t)\left\| A^{\alpha/2}u(t) \right\|^2 – \int_0^t g(t-s)\left\| A^{\alpha/2}(u(s) – u(t))\right\|^2 ds \notag\\
& – 2\int_0^t g(t-s)\left\langle A^{\alpha/2}u(t), A^{1/2}(u(s) – u(t)) \right\rangle ds \notag\\
\leq& p(0)\left\| A^{\alpha/2}u(t) \right\|^2 -\int_0^t g(t-s)\Vert
A^{\alpha/2}u(t)-A^{\alpha/2}u(t-s)\Vert^2 ds \notag\\
&+ \frac{2}{ _0}(1-l)\left\| A^{\alpha/2}u(t) \right\|^2 +
\frac{1}{2}\int_0^t g(t-s)\Vert A^{\alpha/2}u(t)-A^{\alpha/2}u(t-s)\Vert^2
ds\\
=& \frac{3}{ _0}(1-l)\left\| A^{\alpha/2}u(t) \right\|^2 –
\frac{1}{2}\int_0^t g(t-s)\Vert A^{\alpha/2}u(t)-A^{\alpha/2}u(t-s)\Vert^2
ds\notag\\
\leq& 3(1-l)\left\| A^{1/2}u(t) \right\|^2 – \frac{1}{2}\int_0^t
g(t-s)\Vert A^{\alpha/2}u(t)-A^{\alpha/2}u(t-s)\Vert^2 ds\notag\\
\leq& 3(1-l)\left\| A^{1/2}u(t) \right\|^2 –
\frac{1}{2}\int_0^\infty g(t-s)\Vert
A^{\alpha/2}u(t)-A^{\alpha/2}u(t-s)\Vert^2 ds\notag\\&+\frac{1}{2}\int_t^\infty
g(t-s)\Vert A^{\alpha/2}u(t)-A^{\alpha/2}u(t-s)\Vert^2 ds.
\end{align}
(26)
Similarly, differentiating \(J_2\) and repeating the above
computations, we get, for any \( t \geq 0 \),
\begin{align}
J_2′(t) =& p(0)\left\| A^{1/2}u(t)\right\|^2 – \int_0^t g(t-s)\left\| A^{1/2}u(s)\right\|^2 ds \notag\\
=& p(t)\left\| A^{1/2}u(t) \right\|^2 – \int_0^t g(t-s)\left\| A^{1/2}(u(s) – u(t))\right\|^2 ds\notag \\
& – 2\int_0^t g(t-s)\left\langle A^{1/2}u(t), A^{1/2}(u(s) – u(t)) \right\rangle ds \notag\\
\leq& p(0)\left\| A^{1/2}u(t) \right\|^2 -\int_0^\infty g(t-s)\Vert
A^{1/2}u(t)-A^{1/2}u(t-s)\Vert^2 ds\notag \\
&+ \frac{2}{ _0}(1-l)\left\| A^{1/2}u(t) \right\|^2 +
\frac{1}{2}\int_0^\infty g(t-s)\Vert
A^{1/2}u(t)-A^{1/2}u(t-s)\Vert^2 ds \notag\\
=& \frac{3}{ _0}(1-l)\left\| A^{1/2}u(t) \right\|^2 –
\frac{1}{2}\int_0^\infty g(t-s)\Vert
A^{1/2}u(t)-A^{1/2}u(t-s)\Vert^2 ds\notag\\
=& \frac{3}{ _0}(1-l)\left\| A^{1/2}u(t) \right\|^2-
\frac{1}{2}\int_0^\infty g(t-s)\Vert
A^{1/2}u(t)-A^{1/2}u(t-s)\Vert^2 ds\notag\\&+\frac{1}{2}\int_t^\infty
g(t-s)\Vert A^{1/2}u(t)-A^{1/2}u(t-s)\Vert^2 ds.
\end{align}
(27)
Lemma 8.
Assume ( A.1– A.3) hold. Then, the functional \(\mathcal{L}\) defined by
\[ \mathcal{L} (t):=N(E(t) + \mathcal E (t)) + \varepsilon_1 I_1(t) +
\varepsilon_2 I_2(t)\]
satisfies, for a suitable choice of \(N, \varepsilon_1, \varepsilon_2 > 0\),
\begin{equation}\label{e4:3}
\mathcal{L} \sim E + \cal E,
\end{equation}
(28)
and the estimate
\begin{equation}\label{eqn5:3a}
\begin{split}
\mathcal{L}'(t) \leq & – \frac{2}{l}(1 – l)\left( 4 + \frac{3}{2\omega_0}
\right)\|u'(t)\|^2 – (1 – l)\left( 4 + \frac{3}{2\omega_0}
\right)\|A^{1/2}u(t)\|^2 \\ &+ \frac{1}{4}\left( g \circ A^{\alpha/2}u +
g \circ A^{1/2}u \right)(t), \qquad \forall \, t \geq 0,
\end{split}
\end{equation}
(29)
Proof.
It is straightforward to establish the equivalence (28). To
prove (29), we start by exploiting relations
(17), (18), (23) and (25) to
get
\begin{eqnarray}\label{eqn6:3}
\mathcal{L}'(t)&\leq& -\left[\left( \frac{1-l}{\omega_0} -\delta\right)\varepsilon_1 – \varepsilon_2\right]\|u'(t)\|^2-\left(\frac{l}{2}\varepsilon_2 – \delta \varepsilon_1\right)\left\|A^{1/2}u(t)\right\|^2 \notag\\
&& – \left( \frac{N}{2} – \frac{c}{\delta}(\varepsilon_1 + \varepsilon_2) – \frac{c}{\delta}C_\varepsilon(\varepsilon_1 + \varepsilon_2) \right)\left( h_\varepsilon \circ A^{1/2}u \right)(t) \notag\\
&& – \left( \frac{N}{2} – \frac{c}{\delta}(\varepsilon_1 + \varepsilon_2) – \frac{c}{\delta}C_\varepsilon(\varepsilon_1 + \varepsilon_2) \right)\left( h_\varepsilon \circ A^{\alpha/2}u \right)(t) \notag\\
&& + \frac{N\varepsilon}{2}\left( g \circ A^{\alpha/2}u + g \circ A^{1/2}u
\right)(t).
\end{eqnarray}
(30)
Now, we set \(\beta:=\frac{1-l}{\omega_0}\) and choose \(\delta\) small
enough so that
\[d< \min\left\{\frac{1}{2}\beta, \, \frac{l}{8}\beta \right\}.\]
Consequently, for \( \varepsilon_1 = \frac{16(1-l)}{l\beta}\left( 4 +
\frac{3}{2 _0} \right) \), we pick \( \varepsilon_2 = \frac{3}{8}\beta\varepsilon_1 \)
satisfying
\[\frac{1}{4}\beta\varepsilon_1 < \varepsilon_2 \frac{1}{2}\beta\varepsilon_1-\varepsilon_2 = \frac{1}{8}\beta\varepsilon_1 = \frac{2}{l}(1 – l)\left( 4 + \frac{3}{2 _0} \right) \]
and
\[ \frac{l}{2}\varepsilon_2 – \delta\varepsilon_1 > \frac{l}{2}\left(\varepsilon_2 – \frac{1}{4}\beta\varepsilon_1 \right) = \frac{l}{16}\beta\varepsilon_1 = (1 – l)\left( 4 + \frac{3}{2 _0} \right). \]
From \(\displaystyle \frac{\varepsilon g^2(s)}{\varepsilon g(s)-g'(s)}< g(s)\) and the
Lebesgue Dominated Convergence Theorem, we deduce
\[\lim_{\varepsilon\rightarrow0^+}\varepsilon C_\varepsilon=\lim_{\varepsilon\rightarrow0^+}\int_0^\infty\frac{\varepsilon g^2(s)}{\varepsilon g(s)-g'(s)}ds=0.\]
So there exists \(0 < \varepsilon_0 < 1\) such that if \(\varepsilon < \varepsilon_0\), then
\[\varepsilon C_\varepsilon \max \left\lbrace \frac{4c}{\delta}(\varepsilon_1 + \varepsilon_2),\,\,\frac{1}{2\varepsilon_0} \right\rbrace. \]
For \(\varepsilon=\frac{1}{2N}\), we have \[ \frac{N}{4} – \frac{c}{\delta}(\varepsilon_1
+ \varepsilon_2) > 0 \qquad\mathrm{and}\qquad \varepsilon \frac{N}{2} – \frac{c}{\delta}(\varepsilon_1 + \varepsilon_2) – \frac{1}{8\varepsilon} = \frac{N}{4} – \frac{c}{\delta}(\varepsilon_1 + \varepsilon_2) > 0. \]
Thus estimate (30) becomes
\begin{eqnarray*}
\mathcal{L}'(t) &\leq & – \frac{2}{l}(1 – l)\left( 4 + \frac{3}{2 _0}
\right)\|u'(t)\|^2 – (1 – l)\left( 4 + \frac{3}{2 _0}
\right)\|A^{1/2}u(t)\|^2 \\ &&+ \frac{1}{4}\left( g \circ A^{\alpha/2}u
+ g \circ A^{1/2}u \right)(t), \qquad \forall \, t \geq
0.\end{eqnarray*}
Lemma 9.
Assume that ( A.1– A.3) hold. Then, the energy functional
satisfies, for all \(t\in \mathbb{R}^+\) and for some positive
constant \(\tilde{m}\), the following estimate
\begin{equation}\label{E1:r13:St}
\int_{0}^{t} E(s) ds < \tilde{m} f(t),
\end{equation}
(31)
where \(f(t)=1+\int_{0}^{t} h(s)ds\) and \(h=h_0+h_1\) and \(h_0\), \(h_1\) are defined in (20) and (21).
Proof.
Let \(F(t)=\mathcal{L}(t)+J_1(t)+ \frac{1}{2}J_2(t)\), then we obtain, for
all \(t\in \mathbb{R}_+\),
\begin{align}
\label{E:r13:1:0}
F^{\prime}(t) \le & -\frac{2}{l}(4+\frac{3}{2\omega_0}) \vert \vert
u'(t) \vert \vert^2 -(1-l) \vert \vert A^1/2 u(t) \vert
\vert^2-\frac{1}{4}(g\circ A^1/2
u)(t)\notag\\&+\frac{1}{2}\int_{t}^{+\infty} g(s)\Vert A^{\alpha/2}
u(t)-A^{\alpha/2} u(t-s) \Vert^{2} ds
dx\notag\\&+\frac{1}{2}\int_{t}^{+\infty} g(s)\Vert A^{1/2} u(t)-A^{1/2}
u(t-s) \Vert^{2} ds dx.
\end{align}
(32)
Estimates (18) and (32) yield, for some
positive constant \(\lambda\) and for all \(t\in \mathbb{R}_+\),
\begin{align*}
F^{\prime}(t)\le& – \lambda E(t)+\frac{1}{2}\int_{t}^{+\infty}
g(s)\Vert A^{\alpha/2} u(t)-A^{\alpha/2} u(t-s) \Vert^{2} ds
dx\\&+\frac{1}{2}\int_{t}^{+\infty} g(s)\Vert A^{1/2} u(t)-A^{1/2}
u(t-s) \Vert^{2} ds dx.
\end{align*} Therefore,
using (20) and (21) and integrating both sides of the
last inequality, over \((0,t)\), we arrive at
\begin{equation}\label{H}
\begin{aligned}
\lambda\int_{0}^{t}E(s)ds&\le
F(0)-F(t)+\frac{M_0}{2}\int_{0}^{t}h(s) ds\le
F(0)+\frac{M_0}{2}\int_{0}^{t} h(s)ds.
\end{aligned}
\end{equation}
(33)
Hence, we get
\begin{equation}
\int_{0}^{t}E(s)ds\le
\frac{F(0)}{\lambda}+\frac{M_0}{2\lambda}\int_{0}^{t} h(s)ds \leq
\tilde{m}\bigg(1+\int_{0}^{t} h(s)ds\bigg),
\end{equation}
(34)
where \(\tilde{m}=\max \big\{\frac{F(0)}{\lambda},\frac{M_0}{2\lambda}\big\}\).
Corollary 1.
There exists \(0< q_0 < 1\) such that, for all \(t \ge 0\), we have the
following estimate:
\begin{equation}\label{E1c:L2:St0}
\int_{0}^t g(s)\left(\left\| A^{\alpha/2}(u(t)-u(t-s))\right\|^2+\left\|
A^{1/2}(u(t)-u(t-s))\right\|^2\right)ds \leq
\frac{1}{q(t)}{G}^{-1}\left(\frac{q(t) \mu(t)}{\xi(t)}\right)
\end{equation}
(35)
where
\begin{equation}\label{E3d:r13:St}
\mu(t):=-\int_{0}^tg'(s)\left(\left\|
A^{\alpha/2}(u(t)-u(t-s))\right\|^2+\left\|
A^{1/2}(u(t)-u(t-s))\right\|^2\right)ds,
\end{equation}
(36)
\begin{equation}\label{q}
q(t):=\frac{q_0}{f(t)},
\end{equation}
(37)
\(G\) is defined in Remark 4 and
\(f(t)\) is defined in (31).
Proof.
We introduce a functional \(\eta\) defined by
\[\eta(t) := q(t)\int_{0}^t\left(\left\| A^{\alpha/2}(u(t)-u(t-s))\right\|^2+\left\| A^{1/2}(u(t)-u(t-s))\right\|^2\right)ds,\quad\forall\, t \geq 0,\]
and observe, from inequality
(12), that
\begin{equation}\label{qq}
E(t) \geq \frac{l}{2}\left\| A^{1/2}u(t) \right\|^2
\qquad\mathrm{and}\qquad E(t) \geq \frac{l}{2 _0}\left\|
A^{\alpha/2}u(t) \right\|^2, \qquad\forall\, t \geq 0.
\end{equation}
(38)
Use of (15), (17) and (38) yields
\begin{align*}
\eta(t) & \leq 2q(t)\int_{0}^t\left(\left\| A^{\alpha/2}u(t)\right\|^2 + \left\|A^{\alpha/2}u(t-s)\right\|^2+\left\| A^{1/2}u(t)\right\|^2 +\left\|A^{1/2}u(t-s)\right\|^2\right)ds \\
& \leq \frac{4q(t)}{l}(1+ _0) \int_{0}^t \big( E(t) + E(t-s) \big)ds \\
& \leq \frac{8q(t)}{l}(1+ _0) \int_{0}^t E(s) ds, \qquad\forall\, t
\geq 0.
\end{align*}Thanks to (31), we can pick \(0< q_0 < \min\big\{1,
\frac{l}{8\tilde{m}(1+\omega_0)}\big\} \) so that
\begin{equation}\label{e8s5}
\eta(t)< 1,\qquad\forall\, t \geq 0.
\end{equation}
(39)
To prove (35), we define another functional \(\mu\) by
\begin{equation}\label{e9s5}
\begin{aligned}
\mu(t)&:=-\int_{0}^tg'(s)\left(\left\|
A^{\alpha/2}(u(t)-u(t-s))\right\|^2+\left\|
A^{1/2}(u(t)-u(t-s))\right\|^2\right)ds\\&\leq -c\big(E'(t) +
\mathcal E'(t)\big). \end{aligned}
\end{equation}
(40)
Also, the strict convexity of \(G\) and the fact that \(G(0)=0\) entail
that \[G(s\tau) \leq sG(\tau),\qquad \mathrm{for} \quad 0 \leq s
\leq 1 \quad \mathrm{and} \quad \tau \in (0,r].\]
Combining this with the hypothesis (
A.2), Jensen’s inequality and (39), we obtain, for any \(t\geq0\),
\begin{align*}
\mu(t)&=-\frac{1}{\eta(t)}\int_{0}^t\eta(t)g'(s)\left(\left\| A^{\alpha/2}(u(t)-u(t-s))\right\|^2+\left\| A^{1/2}(u(t)-u(t-s))\right\|^2\right)ds\\
&\geq\frac{1}{\eta(t)}\int_{0}^t\eta(t)\xi(s)G(g(s))\left(\left\| A^{\alpha/2}(u(t)-u(t-s))\right\|^2+\left\| A^{1/2}(u(t)-u(t-s))\right\|^2\right)ds\\
&\geq\frac{\xi(t)}{\eta(t)}\int_{0}^t G(\eta(t)g(s))\left(\left\| A^{\alpha/2}(u(t)-u(t-s))\right\|^2+\left\| A^{1/2}(u(t)-u(t-s))\right\|^2\right)ds\\
&\geq\frac{\xi(t)}{q(t)}\overline{G}\left( q(t)\int_{0}^t
g(s)\left(\left\| A^{\alpha/2}(u(t)-u(t-s))\right\|^2+\left\|
A^{1/2}(u(t)-u(t-s))\right\|^2\right)ds\right),
\end{align*}
where \(\overline{G}\) is a \(\boldsymbol{C}^2\) extension of \(G\) which
is strictly increasing and strictly convex on \((0,\infty)\). For
simplicity, in the rest of this paper, we use \(G\) instead of
\(\overline{G}\). Then we have for any \(t \geq 0\),
\[ \int_{0}^t g(s)\left(\left\| A^{\alpha/2}(u(t)-u(t-s))\right\|^2+\left\|
A^{1/2}(u(t)-u(t-s))\right\|^2\right)ds \leq
\frac{1}{q(t)}{G}^{-1}\left(\frac{q(t) \mu(t)}{\xi(t)}\right).\]
5. The main result
In this section, we state and prove our decay result. We introduce
the following functions:
\begin{equation}\label{G234}
G_{2}(t)=t G^{\prime}(\varepsilon_{0}t),\quad G_3(t)=t
G’^{-1}(t), \quad G_4(t)=\overline{G}_3^{*}(t).
\end{equation}
(41)
It is not difficult to show that the above functions are convex and increasing on
\((0,r]\). Now we state our main result.
Theorem 2.
Assume that hypotheses ( A.1)-( A.3) hold and the initial
data satisfy
\[(u_0,u_1)\in \left[\mathcal{D}\left(A^{1-\alpha/2}\right) \times \mathcal{D}\left(A^{(1-\alpha)/2}\right)\right] \cap \left[\mathcal{D}\left(A^{1/2}\right) \times H\right].\]
Then, for all \(0 \leq s \leq t\) and for strictly positive constant
\(C\), we have the following decay results
\begin{equation}\label{decay3a}
E(t) \le C
\left(\frac{E(0)}{q(t)}\right)G_{2}^{-1}\bigg[\frac{C+\int_{0}^{t}
\xi(s) G_4\left[\frac{c}{d }q(s) h(s)\right]
ds,}{\int_{0}^{t}\xi(s)ds}\bigg],
\end{equation}
(42)
where \(q\) is defined in (37), \(h=h_0+h_1\) where \(h_0\), \(h_1\)
are defined in (20) and (21) and the functions
\(G_2(s)\) and \(G_{4}(s)\) are defined in (41).
Proof.
We start by combining (15), (20), (21),
(29) and (35); then, for some \(m>0\) and for
any \(t\geq 0\), we have
\begin{equation}\label{eqn5:3}
\mathcal{L}'(t) \leq -m E(t) + \frac{c}{q(t)} {G}^{-1} \left( \frac{q(t)
\mu(t)}{\xi(t)}\right)+c h(t), \qquad \forall\, t \geq 0.
\end{equation}
(43)
Let \( 0 < \varepsilon_0 0\) and \(G”>0\), we get for any
\(t \geq 0\),
\begin{align}
\mathcal{F}'(t) &= \frac{\varepsilon_0 q(t) E'(t)}{E(0)}
G”\left(\frac{\varepsilon_0 q(t)E(t)}{E(0)}\right)\mathcal{L}(t) +
G’\left(\frac{\varepsilon_0 q(t)E(t)}{E(0)}\right)\mathcal{L}'(t)\notag\end{align}
\begin{align}
\label{e11s5}
&\leq – mE(t) G’\left(\frac{\varepsilon_0 q(t)E(t)}{E(0)}\right) +
\frac{c}{q(t)} G’\left(\frac{\varepsilon_0
q(t)E(t)}{E(0)}\right){G}^{-1}\left(\frac{q(t)
\mu(t)}{\xi(t)}\right)+c h(t) G’\left(\frac{\varepsilon_0
q(t)E(t)}{E(0)}\right).
\end{align}
(44)
Let \(G^{*}\) be the convex conjugate of \(G\) in the sense of Young
(see [
28]), then
\begin{equation}\label{p6:conj0b}
G^{*}(s)=s(G^{\prime})^{-1}(s)-G\left[(G^{\prime})^{-1}(s)\right],\hspace{0.1in}\text{if}\hspace{0.05in}s\in
(0,G^{\prime}(r)]
\end{equation}
(45)
and it satisfies the following generalized Young inequality
\begin{equation}\label{p6:young0b}
A B\le G^{*}(A)+G(B),\hspace{0.15in}\text{if}\hspace{0.05in}A\in
(0,G^{\prime}(r)],\hspace{0.05in}B\in(0,r].
\end{equation}
(46)
So, with
\(A=G^{\prime}\left(\varepsilon_{0}\frac{E(t)q(t)}{E(0)}\right)\),
\(B=G^{-1}\left(\frac{q(t) \mu(t)}{\xi(t)}\right)\), and using
(17), (18), and (44)-(46),
we arrive at
\begin{align}
\label{E:m:xi10b}
\mathcal{F}^{\prime}(t)\le& -m
E(t)G^{\prime}\left(\varepsilon_{0}\frac{E(t)q(t)}{E(0)}\right)+\frac{c}{q(t)}
G^{*}\left(G^{\prime}\left(\varepsilon_{0}\frac{E(t)q(t)}{E(0)}\right)\right)+c
\left(\frac{ \mu(t)q(t)}{\xi(t)}\right)+c h(t) G^{\prime}\left(\varepsilon_{0}\frac{E(t)q(t)}{E(0)}\right)\notag\\
\le& -m
E(t)G^{\prime}\left(\varepsilon_{0}\frac{E(t)q(t)}{E(0)}\right)+c\varepsilon_{0}\frac{E(t)}{E(0)}G^{\prime}\left(\varepsilon_{0}\frac{E(t)q(t)}{E(0)}\right)+c
\left(\frac{ \mu(t)q(t)}{\xi(t)}\right)+c h(t)
G^{\prime}\left(\varepsilon_{0}\frac{E(t)q(t)}{E(0)}\right).
\end{align}
(47)
So, multiplying (47) by \(\xi(t)\) and using (40)
and the fact that \(\varepsilon_{0}\frac{E(t)q(t)}{E(0)}0\), we obtain, for all \(t \in \mathbb{R}_+\),
\begin{align}
\label{p6:main40b}
\mathcal{F}_{1}^{\prime}(t)&\le -k
\xi(t)\left(\frac{E(t)}{E(0)}\right)G^{\prime}\left(\varepsilon_{0}\frac{E(t)q(t)}{E(0)}\right)+c
\xi(t)
h(t)G^{\prime}\left(\varepsilon_{0}\frac{E(t)q(t)}{E(0)}\right)\notag\\&\leq
-k\frac{\xi(t)}{q(t)} G_{2}\left(\frac{E(t)q(t)}{E(0)}\right)+c
\xi(t)
h(t)G^{\prime}\left(\varepsilon_{0}\frac{E(t)q(t)}{E(0)}\right),
\end{align}
(48)
where \(\mathcal{F}_{1}=\xi \mathcal{F}+c (E +\mathcal E)\). Since
\(G^{\prime}_{2}(t)=G^{\prime}(t)+t G^{\prime\prime}(t),\) then, using
the strict convexity of \(G\) on \((0,r],\) we find that
\(G_{2}^{\prime}(t),
G_{2}(t)>0\) on \((0,r].\)
Using the general Young inequality (46) for the last
term in (48) with
\(A=G^{\prime}\left(\varepsilon_{0}\frac{E(t)q(t)}{E(0)}\right)\) and
\(B=[\frac{c}{d}h (t)]\), we have for any \(d>0\),
\begin{align}
\label{essa1a}
\notag c h
(t)G^{\prime}\left(\varepsilon_{0}\frac{E(t)q(t)}{E(0)}\right)&=\frac{d}{q(t)}
\left[\frac{c}{d }q(t) h(t)\right] \bigg(G’
\left(\varepsilon_{0}\frac{E(t)q(t)}{E(0)}\right)\bigg)\\ \notag &\leq
\frac{d}{q(t)}
G_3\bigg(G’\left(\varepsilon_{0}\frac{E(t)q(t)}{E(0)}\right)\bigg)+\frac{d}{q(t)}G_3^*\left[\frac{c}{d
}q(t) h(t)\right]\\ \notag&\leq \frac{d}{q(t)}
\left(\varepsilon_{0}\frac{E(t)q(t)}{E(0)}\right)\left(G’\left(\varepsilon_{0}\frac{E(t)q(t)}{E(0)}\right)\right)+\frac{d}{q(t)}
G_4\left[\frac{c}{d }q(t) h(t)\right]\\&\leq \frac{d}{q(t)}
G_2\left(\varepsilon_{0}\frac{E(t)q(t)}{E(0)}\right)+\frac{d}{q(t)}
G_4\left[\frac{c}{d }q(t) h(t)\right],
\end{align}
(49)
where \(G_2\), \(G_3\) and \(G_4\) are given in
(41). Now, combining (48) and (49)
and choosing \(d\) small enough so that \(k_1=(k-d)>0\), we arrive at
\begin{equation}\label{R’}
\begin{aligned}
& \mathcal{F}_{1}^{\prime}(t) \leq – k_1 \frac{\xi (t)}{q(t)} G_2
\left(\varepsilon_{0}\frac{E(t)q(t)}{E(0)}\right)+\frac{d
\xi(t)}{q(t)}G_4\left[\frac{c}{d }q(t) h(t)\right].
\end{aligned}
\end{equation}
(50)
Since \(E'< 0\) and \(q'< 0\), then \((qE)(t)\) is decreasing function.
Using this fact and since \(G_2\) is increasing, we have, for \(0 \leq
t\leq T\),
\begin{equation}\label{star}
G_{2}\left(\varepsilon_0 \frac{E(T)q(T)}{E(0)}\right) \leq
G_{2}\left(\varepsilon_0 \frac{E(t)q(t)}{E(0)}\right)
\end{equation}
(51)
Combining (50) with (51) and multiplying by \(q(t)\), we
get
\begin{equation}\label{p6:main40bc}
\begin{aligned}
q(t)\mathcal{F}_{1}^{\prime}(t)+k_1\xi(t)
G_{2}\left(\varepsilon_0\frac{E(T)q(T)}{E(0)}\right)&\le d
\xi(t)G_4\left(\frac{c}{d }q(t) h(t)\right).
\end{aligned}
\end{equation}
(52)
Since \(q'< 0\), then for all \(0 \leq t\leq T\),
\begin{equation}\label{p6:main40bcc}
\begin{aligned}
\bigg(q(t)\mathcal{F}_{1}\bigg)^{\prime}(t)+k_1\xi(t)
G_{2}\left(\varepsilon_0\frac{E(T)q(T)}{E(0)}\right)&\le d
\xi(t)G_4\left(\frac{c}{d }q(t) h(t)\right).
\end{aligned}
\end{equation}
(53)
Integrating (53) over \([0, T]\) and using the fact
\(q(0)=q_0\), we have
\begin{equation}\label{p6:main40bde}
\begin{aligned}
G_{2}\left(\varepsilon_0\frac{E(T)q(T)}{E(0)}\right)\int_{0}^{T}\xi(t)dt
\le \frac{q_0 \mathcal{F}_{1}(0)}{k_1}+d \int_{0}^{T} \xi(t)
G_4\left(\frac{c}{d }q(t) h(t)\right) dt.
\end{aligned}
\end{equation}
(54)
Hence,
\begin{equation}
\begin{aligned}
G_{2}\left(\varepsilon_0\frac{E(T)q(T)}{E(0)}\right) \le
\bigg[\frac{\frac{\mathcal{F}_{1}(0)}{c}+d \int_{0}^{T} \xi(t)
G_4\left(\frac{c}{d }q(t) h(t)\right)
dt,}{\int_{0}^{T}\xi(t)dt}\bigg].
\end{aligned}
\end{equation}
(55)
Thus
\begin{equation}
\begin{aligned}
\left(\varepsilon_0\frac{E(T)q(T)}{E(0)}\right) \le
G_{2}^{-1}\bigg[\frac{\frac{\mathcal{F}_{1}(0)}{c}+d \int_{0}^{T}
\xi(t) G_4\left(\frac{c}{d }q(t) h(t)\right)
dt,}{\int_{0}^{T}\xi(t)dt}\bigg],
\end{aligned}
\end{equation}
(56)
which yields
\begin{equation}
\begin{aligned}
E(T) \le C
\left(\frac{E(0)}{q(T)}\right)G_{2}^{-1}\bigg[\frac{C+\int_{0}^{T}
\xi(t) G_4\left(\frac{c}{d }q(t) h(t)\right)
dt,}{\int_{0}^{T}\xi(t)dt}\bigg],
\end{aligned}
\end{equation}
(57)
where \(C=\max\big\{1, \frac{\mathcal{F}_{1}(0)}{c},
\frac{c}{d}, \frac{1}{\varepsilon_0} \big\}\).
Example 1. Let
\(g(t)=\frac{a}{(1+t)^\nu}\), where \(\nu >1\) and \(0< a< \nu-1\). In this
case \(\xi(t)=\nu a^{\frac{-1}{\nu}}\) and
\(G(t)=t^{\frac{\nu+1}{\nu}}\). Then \(G'(t)=a_0
t^{\frac{1}{\nu}}\). We will discuss two cases:
Case 1: if \(m_0 \leq 2+\vert \vert A^{\alpha/2}u_{0}+A^{1/2}u_{0}\vert
\vert2 \leq m_1\). Then we have the following:
\begin{equation}\label{gs}
\begin{cases}
&G_4(t)=a_1 t^{\frac{\nu+1}{\nu}},~~~G_2(t)=a_2
t^{\frac{\nu+1}{\nu}},\\ \\
& a_3 (1+t)^{-\nu+1}\leq h(t) \leq a_4 (1+t)^{-\nu+1},\\ \\
&\int_{0}^{T} \xi(t) G_4\left(\frac{c_1}{d }q(t) h(t)\right)
dt< +\infty,\\ \\
&G_{2}^{-1}\bigg[\frac{C+\int_{0}^{T} \xi(t) G_4\left(\frac{c_1}{d
}q(t) h(t)\right) dt,}{\int_{0}^{T}\xi(t)dt}\bigg] \leq a_5
T^{-(\frac{\nu}{\nu+1})},
\end{cases}
\end{equation}
(58)
\begin{equation}\begin{aligned}
&\frac{q_0}{q(T)} \leq a_6\left\{%
\begin{array}{ll}
1+\ln(1+T), & \hbox{\(\nu=2\);} \\
2, & \hbox{\(\nu > 2\);} \\
(1+T)^{-\nu+2+r}, & \hbox{\(1< \nu < 2\) .}
\end{array}%
\right.
\end{aligned}
\end{equation}
(59)
Then
\begin{equation}\begin{aligned}
&E(T) \leq a_7\left\{%
\begin{array}{ll}
\bigg(1+\ln(1+T)\bigg)t^{-(\frac{\nu}{\nu+1})}, & \hbox{\(\nu=2\);} \\
T^{-(\frac{\nu}{\nu+1})}, & \hbox{\(\nu > 2\);} \\
(1+T)^{-(\nu-2+\frac{\nu}{\nu+1})}, & \hbox{\(1< \nu < 2\) .} \\
\end{array}%
\right.
\end{aligned}
\end{equation}
(60)
Thus for \(\nu \geq 2\) or \(\sqrt{2} < \nu < 2\) we have
\(\lim_{T\rightarrow +\infty} E(T)=0\).
Case 2: if \(m_0 (1+t)^{r} \leq 2+\vert \vert
A^{\alpha/2}u_{0}+A^{1/2}u_{0}\vert \vert^2 \leq m_1 (1+t)^{r}\), where
\(0 < r < \nu-1\), then we have the following:
\begin{equation}\begin{cases}
& a_3 (1+t)^{-\nu+1+r}\leq h(t) \leq a_4 (1+t)^{-\nu+1+r},\\\\
&\int_{0}^{T} \xi(t) G_4\left(\frac{c_1}{d }q(t) h(t)\right) dt<
+\infty,
\end{cases}
\end{equation}
(61)
\begin{equation}\begin{aligned}
&\frac{q_0}{q(T)} \leq a_6\left\{%
\begin{array}{ll}
1+\ln(1+T), & \hbox{\(\nu-r=2\);} \\
2, & \hbox{\(\nu -r> 2\);} \\
(1+T)^{-\nu+2+r}, & \hbox{\(1< \nu -r < 2\) .} \\
\end{array}%
\right.
\end{aligned}
\end{equation}
(62)
Then,
\begin{equation}\begin{aligned}
&E(T) \leq a_7\left\{%
\begin{array}{ll}
\bigg(1+\ln(1+T)\bigg)t^{-(\frac{\nu}{\nu+1})}, & \hbox{\(\nu-r=2\);} \\
T^{-(\frac{\nu}{\nu+1})}, & \hbox{\(\nu -r> 2\);} \\
(1+T)^{-(\nu-2-r+\frac{\nu}{\nu+1})}, & \hbox{\(1< \nu-r < 2\) .} \\
\end{array}%
\right.
\end{aligned}
\end{equation}
(63)
Thus for \(\nu -r\geq 2\) or \(\frac{1}{2}\big(r+\sqrt{r^2+4r+8}\big) <
\nu < r+2\) we have \(\lim_{T\rightarrow +\infty} E(T)=0\).
Acknowledgments :
The authors thank University of King Fahd University of Petroleum
and Minerals (KFUPM) and the referee for his/her very careful
reading and valuable comments. This work is sponsored by KFUPM under
Project No. SB201012.
Author Contributions:
All authors contributed equally to the writing of this paper. All authors read and approved the final manuscript.
Conflicts of Interest:
”The authors declare no conflict of interest.”
Data Availability:
All data required for this research is included within this paper.
Funding Information:
No funding is available for this research.
