In this section, we shall study the general decay of energy to problem (1.1)-(1.3) to prove Theorem 2.2. For this purpose, we need the following technical lemmas.
Lemma 3.1
Under the assumptions of Theorem
2.2, the energy functional
\(E(t)\)
satisfies, for any
\(t\geq0\),
$$\begin{aligned} E'(t) \leq&\frac{1}{2}\bigl(g' \circ\nabla u\bigr) (t)-\frac{1}{2}g(t) \bigl\Vert \nabla u(t) \bigr\Vert ^{2}+ \biggl(\frac{\xi}{2}-\mu_{1}+ \frac{|\mu_{2}|}{2\sqrt {1-d}} \biggr) \bigl\Vert \nabla u_{t}(t) \bigr\Vert ^{2} \\ &{}+ \biggl[\frac{|\mu_{2}|}{2}\sqrt{1-d}-\frac{\xi }{2}(1-d)e^{-\lambda\tau_{1}} \biggr] \bigl\Vert \nabla u_{t}\bigl(t-\tau(t)\bigr) \bigr\Vert ^{2} \\ &{}-\frac{\lambda\xi}{2} \int^{t}_{t-\tau(t)}e^{-\lambda(t-s)} \bigl\Vert \nabla u_{t}(s) \bigr\Vert ^{2}\,ds. \end{aligned}$$
(3.1)
Proof
Differentiating (2.5) and using (1.1), (2.3)-(2.4) and integration by parts, we have
$$\begin{aligned} E(t) =& \int_{\Omega}u_{t}u_{tt}\,dx- \frac{1}{2}g(t) \Vert \nabla u \Vert ^{2}+ \biggl(1- \int^{t}_{0}g(s)\,ds \biggr) \int_{\Omega}\nabla u\cdot\nabla u_{t}\,dx+ \frac{1}{2}\bigl(g'\circ\nabla u\bigr) \\ &{}+ \int^{t}_{0}g(t-s)\,ds\cdot \int_{\Omega}\nabla u\cdot\nabla u_{t}\,dx- \int_{\Omega}\nabla u_{t}(t)\cdot \int^{t}_{0}g(t-s)\nabla u(s)\,ds\,dx \\ &{}+\frac{\xi}{2} \Vert \nabla u_{t} \Vert ^{2}- \frac{\xi}{2}e^{-\lambda\tau(t)}\bigl(1-\tau'(t)\bigr) \bigl\Vert \nabla u_{t}\bigl(t-\tau(t)\bigr) \bigr\Vert ^{2} \\ &{}-\frac{\lambda\xi}{2} \int^{t}_{t-\tau(t)}e^{-\lambda(t-s)} \bigl\Vert \nabla u_{t}(s) \bigr\Vert ^{2}\,ds \\ \leq&-\mu_{1} \Vert \nabla u_{t} \Vert ^{2}- \frac{1}{2}g(t) \Vert \nabla u \Vert ^{2}+\frac{1}{2} \bigl(g'\circ\nabla u\bigr)+\frac{\xi}{2} \Vert \nabla u_{t} \Vert ^{2} \\ &{}-\mu_{2} \int_{\Omega}\nabla u_{t}\cdot\nabla u_{t} \bigl(t-\tau(t)\bigr)\,dx-\frac{\xi}{2}(1-d)e^{-\lambda\tau_{1}} \bigl\Vert \nabla u_{t}\bigl(t-\tau(t)\bigr) \bigr\Vert ^{2} \\ &{}-\frac{\lambda\xi}{2} \int^{t}_{t-\tau(t)}e^{-\lambda(t-s)} \bigl\Vert \nabla u_{t}(s) \bigr\Vert ^{2}\,ds. \end{aligned}$$
(3.2)
It follows from Young’s inequality that
$$ -\mu_{2} \int_{\Omega}\nabla u_{t}\cdot\nabla u_{t} \bigl(t-\tau(t)\bigr)\,dx\leq \frac{|\mu_{2}|}{2\sqrt{1-d}} \bigl\Vert \nabla u_{t}(t) \bigr\Vert ^{2}+\frac{|\mu_{2}|}{2}\sqrt{1-d} \bigl\Vert \nabla u_{t}\bigl(t-\tau(t)\bigr) \bigr\Vert ^{2}, $$
which, together with (3.2) gives us (3.1). The proof is done. □
Lemma 3.2
Under the assumptions of Theorem
2.2, then the functional
\(\phi(t)\)
defined as
$$ \phi(t)= \int_{\Omega}u(t)u_{t}(t)\,dx, $$
(3.3)
satisfies the requirement that there exist positive constants
\(c_{1}\), \(c_{2}\)
and
\(c_{3}\)
such that, for any
\(t\geq0\),
$$ \phi'(t)\leq-\frac{l}{2} \bigl\Vert \nabla u(t) \bigr\Vert ^{2}+c_{1} \bigl\Vert \nabla u_{t}(t) \bigr\Vert ^{2}+c_{2} \bigl\Vert \nabla u_{t}\bigl(t-\tau(t)\bigr) \bigr\Vert ^{2}+c_{3}(g \circ\nabla u)(t). $$
(3.4)
Proof
By using (1.1), we can get
$$\begin{aligned} \phi'(t) =& \int_{\Omega}u_{tt}u\,dx+\|u_{t}\|^{2} \\ =&\|u_{t}\|^{2}+ \int_{\Omega} u(t)\cdot \biggl(\Delta u- \int^{t}_{0}g(t-s)\Delta u(s)\,ds+\mu_{1} \Delta u_{t}+\mu_{2}\Delta u_{t}\bigl(t-\tau(t) \bigr) \biggr)\,dx \\ =&\|u_{t}\|^{2}-\|\nabla u\|^{2}+ \int_{\Omega}\nabla u(t)\cdot \int ^{t}_{0}g(t-s) \bigl(\nabla u(s)-\nabla u(t) \bigr)\,ds\,dx \\ &{}+ \int^{t}_{0}g(s)\,ds\cdot\|\nabla u\|^{2}+ \mu_{1} \int_{\Omega}u\cdot\Delta u_{t}\,dx-\mu_{2} \int_{\Omega}\nabla u\cdot\nabla u_{t}\bigl(t-\tau(t) \bigr)\,dx \\ \leq&\|u_{t}\|^{2}-l\|\nabla u\|^{2}+\underbrace{ \int_{\Omega}\nabla u(t)\cdot \int^{t}_{0}g(t-s) \bigl(\nabla u(s)-\nabla u(t) \bigr)\,ds\,dx}_{:=I_{1}} \\ &{}+\underbrace{\mu_{1} \int_{\Omega}u\cdot\Delta u_{t}\,dx}_{:=I_{2}}\, -\, \underbrace{\mu_{2} \int_{\Omega}\nabla u\cdot\nabla u_{t}\bigl(t-\tau(t) \bigr)\,dx}_{:=I_{3}}. \end{aligned}$$
(3.5)
By using Young’s inequality and Hölder’s inequality, we shall see below, for any \(\delta>0\),
$$\begin{aligned}& I_{1}\leq\delta\|\nabla u\|^{2}+ \frac{1}{4\delta} \int_{\Omega}\biggl( \int^{t}_{0}g(t-s) \bigl(\nabla u(s)-\nabla u(t) \bigr)\,ds \biggr)^{2}\,dx \\& \hphantom{I_{1}}\leq\delta\|\nabla u\|^{2}+\frac{1}{4\delta} \int^{t}_{0}g(s)\,ds(g\circ\nabla u)(t) \\& \hphantom{I_{1}}\leq\delta\|\nabla u\|^{2}+\frac{1-l}{4\delta}(g\circ \nabla u)(t), \end{aligned}$$
(3.6)
$$\begin{aligned}& I_{2}\leq \delta\|\nabla u\|^{2}+ \frac{\mu_{1}^{2}}{4\delta}\|\nabla u_{t}\|^{2}, \\& I_{3}\leq \delta\|\nabla u\|^{2}+\frac{\mu_{2}^{2}}{4\delta} \bigl\Vert \nabla u_{t}\bigl(t-\tau(t)\bigr) \bigr\Vert ^{2}, \end{aligned}$$
(3.7)
which, together with (3.5)-(3.7), implies, for any \(\delta>0\),
$$ \phi'(t)\leq\|u_{t}\|^{2}-(l-3 \delta)\|\nabla u\|^{2}+\frac{\mu_{1}^{2}}{4\delta}\|\nabla u_{t} \|^{2}+\frac{\mu_{2}^{2}}{4\delta} \bigl\Vert \nabla u_{t}\bigl(t- \tau(t)\bigr) \bigr\Vert ^{2} +\frac{1-l}{4\delta}(g\circ\nabla u)(t). $$
(3.8)
Now taking \(\delta>0\) small enough such that
$$l-3\delta>\frac{l}{2}, $$
we can get the desired estimate (3.4) with
$$c_{1}=\frac{1}{\lambda_{1}}+\frac{\mu_{1}^{2}}{4\delta}, \qquad c_{2}= \frac{\mu_{2}^{2}}{4\delta}, \qquad c_{3}=\frac{1-l}{4\delta}, $$
hereafter the positive constant \(\lambda_{1}\) represents the Poincaré’s constant, i.e., \(\lambda_{1}\|u\|^{2}\leq\|\nabla u\|\) for \(u\in H^{1}_{0}(\Omega)\). The proof is hence complete. □
Lemma 3.3
We define the functional
\(\psi(t)\)
as
$$ \psi(t)=- \int_{\Omega}u_{t}(t)\cdot \int^{t}_{0}g(t-s) \bigl(u(t)-u(s)\bigr)\,ds\,dx. $$
(3.9)
Under the assumptions of Theorem
2.2, the functional
\(\psi(t)\)
satisfies, for any
\(\delta>0\),
$$\begin{aligned} \psi'(t) \leq&- \biggl( \int^{t}_{0}g(s)\,ds-\delta \biggr)\|u_{t} \|^{2}+\delta \bigl\Vert \nabla u_{t}(t) \bigr\Vert ^{2}+\delta \bigl\Vert \nabla u_{t}\bigl(t-\tau(t)\bigr) \bigr\Vert ^{2} \\ &{}+c_{4}(g\circ\Delta u)(t)-c_{5}\bigl(g'\circ \Delta u\bigr)(t), \end{aligned}$$
(3.10)
where
\(c_{4}\)
and
\(c_{5}\)
are positive constants.
Proof
It follows from (1.1) and integration by parts that
$$\begin{aligned} \psi'(t) =&- \int_{\Omega}u_{tt}\cdot \int^{t}_{0}g(t-s) \bigl(u(t)-u(s)\bigr)\,ds\,dx \\ &{}- \int_{\Omega}u_{t} \biggl[u_{t} \int^{t}_{0}g(t-s)\,ds+ \int^{t}_{0}g'(t-s) \bigl(u(t)-u(s) \bigr)\,ds \biggr]\,dx \\ =& \int_{\Omega}\biggl(-\Delta u+ \int^{t}_{0}g(t-s)\Delta u(s)\,ds-\nabla u_{t}-\mu_{2}\Delta u_{t}\bigl(t-\tau(t)\bigr) \biggr) \\ &{}\times \int^{t}_{0}g(t-s) \bigl(u(t)-u(s)\bigr)\,ds\,dx- \int^{t}_{0}g(s)\,ds\|u_{t}\| ^{2} \\ &{}- \int_{\Omega}u_{t} \int^{t}_{0}g'(t-s) \bigl(u(t)-u(s) \bigr)\,ds\,dx \\ =& \biggl(1- \int^{t}_{0}g(s)\,ds \biggr) \int_{\Omega}\nabla u(t)\cdot \int^{t}_{0}g(t-s) \bigl(\nabla u(t)-\nabla u(s) \bigr)\,ds\,dx \\ &{}+ \int_{\Omega}\biggl( \int^{t}_{0}g(t-s) \bigl(\nabla u(s)-\nabla u(t) \bigr)\,ds \biggr)^{2}\,dx- \int^{t}_{0}g(s)\,ds\|u_{t} \|^{2} \\ &{}- \int_{\Omega}u_{t} \int^{t}_{0}g'(t-s) \bigl(u(t)-u(s) \bigr)\,ds\,dx \\ &{}+\mu_{1} \int_{\Omega}\nabla u_{t}(t)\cdot \int^{t}_{0}g(t-s) \bigl(\nabla u(t)-\nabla u(s) \bigr)\,ds\,dx \\ &{}+\mu_{2} \int_{\Omega}\nabla u_{t}\bigl(t-\tau(t)\bigr)\cdot \int ^{t}_{0}g(t-s) \bigl(\nabla u(t)-\nabla u(s) \bigr)\,ds\,dx. \end{aligned}$$
(3.11)
By using Hölder’s inequality, Young’s inequality and the Poincaré inequality, we can obtain, for any \(\delta>0\),
$$\begin{aligned}& \mu_{2} \int_{\Omega}\nabla u \int^{t}_{0}g(t-s) \bigl(\nabla u(t)-\nabla u(s) \bigr)\,ds\,dx\leq\delta\|\nabla u_{t}\|^{2}+ \frac{\mu_{1}^{2}(1-l)}{4\delta}(g\circ\nabla u)(t), \end{aligned}$$
(3.12)
$$\begin{aligned}& \mu_{2} \int_{\Omega}\nabla u_{t}\bigl(t-\tau(t)\bigr)\cdot \int^{t}_{0}g(t-s) \bigl(\nabla u(t)-\nabla u(s) \bigr)\,ds\,dx \\& \quad \leq\delta \bigl\Vert \nabla u_{t}\bigl(t-\tau(t)\bigr) \bigr\Vert ^{2}+\frac{\mu_{2}^{2}(1-l)}{4\delta}(g\circ\nabla u)(t), \end{aligned}$$
(3.13)
$$\begin{aligned}& - \int_{\Omega}u_{t} \int^{t}_{0}g'(t-s) \bigl(u(t)-u(s) \bigr)\,ds\,dx \\& \quad \leq\delta\|u_{t}\|^{2}+\frac{1}{4\delta} \biggl( \int^{t}_{0}\bigl(-g'(t-s)\bigr) \bigl\Vert u(t)-u(s) \bigr\Vert \,ds \biggr)^{2} \\& \quad \leq\delta\|u_{t}\|^{2}-\frac{Cg(0)}{4\delta\lambda_{1}} \bigl(g'\circ \Delta u\bigr)(t), \end{aligned}$$
(3.14)
and
$$ \int_{\Omega}\biggl( \int^{t}_{0}g(t-s) \bigl(\nabla u(s)-\nabla u(t) \bigr)\,ds \biggr)^{2}\,dx\leq\frac{1-l}{\lambda_{1}}(g\circ\nabla u)(t), $$
which, combined (3.12)-(3.14) with (3.11), gives us (3.10). The proof is therefore complete. □
Now we define the Lyapunov functional \(F(t)\) by
$$ F(t):=E(t)+\varepsilon_{1}\phi (t)+\varepsilon_{2} \psi(t), $$
(3.15)
where \(\varepsilon_{1}\) and \(\varepsilon_{2}\) are positive constants to be taken later. First we know that, for \(\varepsilon_{1}>0\) and \(\varepsilon_{2}>0\) small enough, there exist two positive constants \(\beta_{1}\) and \(\beta_{2}\) such that, for any \(t>0\),
$$ \beta_{1} E(t)\leq F(t)\leq\beta_{2} E(t). $$
(3.16)
Proof of Theorem 2.2
For any \(t_{0}>0\), we get, for any \(t\geq t_{0}\),
$$ \int^{t}_{0}g(s)\,ds\geq \int^{t_{0}}_{0}g(s)\,ds:=g_{0}. $$
It follows from (3.1), (3.4) and (3.10) that, for any \(t\geq t_{0}\),
$$\begin{aligned} F'(t) \leq&- \biggl( \int^{t}_{0}g(s)\,ds-\delta \biggr) \varepsilon_{2} \bigl\Vert u_{t}(t) \bigr\Vert ^{2}- \biggl(\frac{l}{2}\varepsilon_{1}-\delta \varepsilon _{2} \biggr) \bigl\Vert \nabla u(t) \bigr\Vert ^{2} \\ &{}+ \biggl(\frac{\xi}{2}-\mu_{1}+\frac{|\mu_{2}|}{2\sqrt {1-d}}+c_{1} \varepsilon_{1}+\delta\varepsilon_{2} \biggr) \bigl\Vert \nabla u_{t}(t) \bigr\Vert ^{2} \\ &{}+ \biggl[\frac{|\mu_{2}|}{2}\sqrt{1-d}-\frac{\xi }{2}(1-d)e^{-\lambda\tau_{1}}+c_{2} \varepsilon_{1}+\delta\varepsilon _{2} \biggr] \bigl\Vert \nabla u_{t}\bigl(t-\tau(t)\bigr) \bigr\Vert ^{2} \\ &{}+ \biggl(\frac{1}{2}-c_{5}\varepsilon_{2} \biggr) \bigl(g'\circ\nabla u\bigr) (t)+(c_{3} \varepsilon_{1}+c_{4}\varepsilon_{2}) (g\circ\nabla u)(t) \\ &{}-\frac{\lambda\xi}{2} \int^{t}_{t-\tau(t)}e^{-\lambda(t-s)} \bigl\Vert \nabla u_{t}(s) \bigr\Vert ^{2}\,ds. \end{aligned}$$
(3.17)
Obviously, \(e^{\lambda\tau_{1}}\rightarrow1\) as \(\lambda\rightarrow0\). Because of the continuity of the set of real numbers, we pick \(\lambda>0\) small enough such that there exists a positive constant ξ such that
$$ \frac{e^{\lambda\tau_{1}}|\mu_{2}|}{\sqrt{1-d}}< \xi< \mu_{1}, $$
(3.18)
which implies
$$ \frac{|\mu_{2}|}{2\sqrt{1-d}}-\mu_{1}+\frac{\xi}{2}< 0 \quad \mbox{and} \quad \frac{|\mu_{2}|}{2}\sqrt{1-d}-\frac{\xi}{2e^{\lambda\tau_{1}}}(1-d)< 0. $$
(3.19)
It follows from (3.18) and (3.19) that the energy functional (2.5) is nonincreasing.
Now we take \(\delta>0\) small enough such that, for \(t\geq t_{0}\),
$$\int^{t}_{0}g(s)\,ds-\delta\geq\frac{1}{2}g_{0}. $$
At this point, for any fixed \(\delta>0\), we choose \(\varepsilon_{2}>0\) so small that (3.16) holds, and further
$$\varepsilon_{2}< \min \biggl\{ \frac{1}{4c_{5}}, \frac{\mu_{1}}{2\delta}- \frac{\xi}{4\delta}-\frac{|\mu _{2}|}{4\delta\sqrt{1-d}}, \frac{\xi}{4\delta}(1-d)e^{-\lambda\tau_{1}}- \frac{|\mu _{2}|}{4\delta}\sqrt{1-d} \biggr\} , $$
which yields
$$\frac{1}{2}-c_{5}\varepsilon_{2}>\frac{1}{4}, \qquad \frac{\xi}{2}-\mu_{1}+\frac{|\mu_{2}|}{2\sqrt{1-d}}+\delta \varepsilon_{2}< \frac{\xi}{4}-\frac{\mu_{1}}{2}+\frac{|\mu _{2}|}{4\sqrt{1-d}} $$
and
$$\frac{|\mu_{2}|}{2}\sqrt{1-d}-\frac{\xi}{2}(1-d)e^{-\lambda\tau _{1}}+\delta \varepsilon_{2} < \frac{|\mu_{2}|}{4}\sqrt{1-d}-\frac{\xi}{4}(1-d)e^{-\lambda\tau_{1}}. $$
In the sequel, for any fixed \(\delta>0\) and \(\varepsilon_{2}>0\), we take \(\varepsilon_{1}>0\) small so that (3.16) holds, and further
$$\frac{4\delta\varepsilon_{2}}{l}< \varepsilon_{1}< \min \biggl\{ \frac {\mu_{1}}{4c_{1}}- \frac{\xi}{8c_{1}}-\frac{|\mu_{2}|}{8c_{1}\sqrt{1-d}}, \frac{\xi}{8c_{2}}(1-d)e^{-\lambda\tau_{1}}- \frac{|\mu_{2}|}{8c_{2}}\sqrt {1-d} \biggr\} , $$
which gives us
$$\frac{l}{2}\varepsilon_{1}-\delta\varepsilon_{2}> \frac {l}{4}\varepsilon_{2}, \qquad \frac{\xi}{4}- \frac{\mu_{1}}{2}+\frac{|\mu_{2}|}{4\sqrt {1-d}}+c_{1}\varepsilon_{1}< \frac{\xi}{8}-\frac{\mu_{1}}{4}+\frac{|\mu _{2}|}{8\sqrt{1-d}} $$
and
$$\frac{|\mu_{2}|}{4}\sqrt{1-d}-\frac{\xi}{4}(1-d)e^{-\lambda\tau _{1}}+c_{2} \varepsilon_{1} < \frac{|\mu_{2}|}{8}\sqrt{1-d}-\frac{\xi}{8}(1-d)e^{-\lambda\tau_{1}}. $$
From the above we know that, for positive constants \(\alpha_{1}\) and \(\alpha_{2}\),
$$ F'(t)\leq-\alpha_{1}E(t)+ \alpha_{2}(g\circ\nabla u) (t),\quad \forall t\geq t_{0}. $$
(3.20)
Multiplying (3.20) by \(\zeta(t)\) and using (2.2), we can get, for any \(t\geq t_{0}\),
$$\begin{aligned} \zeta(t)F'(t) \leq &-\alpha_{1} \zeta(t)E(t)+\alpha_{2}\zeta(t) (g\circ\Delta u) (t) \\ \leq&-\alpha_{1}\zeta(t)E(t)-\alpha_{2} \bigl(g'\circ\Delta u\bigr) (t) \\ \leq&-\alpha_{1}\zeta(t)E(t)-\alpha_{3}E'(t), \end{aligned}$$
(3.21)
where \(\alpha_{3}>0\). Denote \(\mathcal{E}(t)=\zeta(t)F(t)+\alpha_{3} E(t)\), then it is easy to see that \(\mathcal{E}(t)\) is equivalent to the energy \(E(t)\), i.e., there exist two positive constants \(\beta_{3}\) and \(\beta_{4}\) such that
$$ \beta_{3}E(t)\leq\mathcal{E}(t)\leq \beta_{4}E(t). $$
(3.22)
Thus we can infer that, for any \(t\geq t_{0}\),
$$ \mathcal{E}'(t)\leq-\frac{\alpha_{1}}{\beta_{4}}\zeta(t)\mathcal {E}(t), $$
which, integrating over \((t_{0},t)\) with respect to t, yields, for any \(t\geq t_{0}\),
$$ \mathcal {E}(t)\leq\mathcal {E}(t_{0})\exp \biggl(- \frac{\alpha_{1}}{\beta_{4}} \int^{t}_{t_{0}}\zeta (s)\,ds \biggr). $$
(3.23)
Therefore (2.6) follows from (3.23) by renaming the constants, and by the continuity and boundedness of \(E(t)\) and \(\zeta(t)\). The proof is hence complete. □
Remark 3.1
If taking \(\zeta(t)=\gamma\) and \(\zeta(t)=\gamma(1+t)^{-1}\), and γ a positive constant, we can obtain the exponential decay and polynomial decay of problem (1.1)-(1.3), respectively. Thus the exponential decay and polynomial decay is a particular case of (2.6). We also find some other examples to illustrate several rates of energy decay; see, for example, [25, 32].