In this section we prove the decay rates in Theorem 2.1. The following result shows the dissipative property of system (1.1)-(1.7). Multiplying (1.1) by \(u' (t) \), we have the identity
$$ E'(t) = a\bigl(g* u , u'\bigr). $$
(3.1)
We define the modified energy by
$$ {\mathcal{E}}(t) = \frac{1}{2} \bigl\Vert u'\bigr\Vert ^{2}+\frac{h}{2} \bigl\Vert \nabla u'\bigr\Vert ^{2}+\frac{1}{2} \biggl(1-\int_{0}^{t} g(s) \,ds \biggr) a(u, u) + \frac{1}{2} g \,\square\,\partial^{2} u +\frac{1}{4}\Vert \Delta v\Vert ^{2}. $$
From (2.5) and (3.1), we get
$$ {\mathcal{E}}'(t) =-\frac{1}{2}g(t) a(u, u) + \frac{1}{2}g' \,\square\, \partial^{2} u . $$
(3.2)
This implies that \({\mathcal{E}}(t)\) is nonincreasing, and from (2.3) one sees that
$$ E(t) \leq\frac{1}{1-l} {\mathcal{E}}(t), \quad \forall t \geq0. $$
(3.3)
First, let us define the perturbed modified energy by
$$ L(t) = N {\mathcal{E}}(t) +\epsilon\Phi(t) + \Psi(t), $$
(3.4)
where
$$\Phi(t) = \int_{\Omega}u' u \,d\Omega+ h \int _{\Omega}\nabla u' \nabla u \,d\Omega $$
and
$$\Psi(t) =\int_{\Omega}\bigl(h \Delta u' - u' \bigr)\int_{0}^{t} g(t-s) \bigl(u(t)-u(s)\bigr)\,ds \,d\Omega . $$
Using the ideas presented in [9], we easily obtain the following lemmas.
Lemma 3.1
For
\(N>0\)
large enough, there exist
\(\alpha_{1}>0\)
and
\(\alpha_{2}>0\)
such that
$$ \alpha_{1} {\mathcal{E}}(t) \leq L(t) \leq \alpha_{2} {\mathcal{E}}(t),\quad \forall t \geq0. $$
(3.5)
Proof
By Young’s inequality, (2.2) and (2.3), we have
$$ \bigl\vert \Phi(t)\bigr\vert \leq \frac{1}{2} \bigl\Vert u'\bigr\Vert ^{2} + \frac{h}{2} \bigl\Vert \nabla u'\bigr\Vert ^{2} + \frac{C_{p} +C_{s} h}{2} a (u,u) $$
(3.6)
and
$$ \bigl\vert \Psi(t) \bigr\vert \leq \frac{1}{2} \bigl\Vert u'\bigr\Vert ^{2} +\frac{h}{2}\bigl\Vert \nabla u'\bigr\Vert ^{2} + \frac{(C_{p} + C_{s} h) l}{2} g \, \square\, \partial^{2} u . $$
(3.7)
From (3.6) and (3.7) we obtain
$$\begin{aligned} \bigl\vert L(t) - N {\mathcal{E}}(t) \bigr\vert \leq& \frac{1}{2}( \epsilon+1) \bigl\Vert u'\bigr\Vert ^{2}+ \frac{h}{2}(\epsilon+1) \bigl\Vert \nabla u'\bigr\Vert ^{2} \\ &{} + \frac{(C_{p} +C_{s} h)\epsilon}{2} a (u,u) + \frac{(C_{p} + C_{s} h) l}{2} g \,\square\, \partial^{2} u \\ \leq& C_{0} {\mathcal{E}}(t), \end{aligned}$$
where \(C_{0}\) is a positive constant depending on ϵ, h, \(C_{p}\), \(C_{s}\) and l. Choosing \(N>0\) large, we complete the proof of Lemma 3.1. □
Lemma 3.2
For each
\(t_{0} > 0\)
and sufficiently large
\(N>0\), there exist positive constants
\(\beta_{1}\)
and
\(\beta_{2}\)
such that
$$ L'(t) \leq- \beta_{1} {\mathcal{E}}(t) + \beta_{2} g \,\square\,\partial^{2} u ,\quad \forall t \geq t_{0} . $$
(3.8)
Proof
Direct computations, using (1.1), yield
$$ \Phi'(t)= -a(u, u) +a (g*u, u) + \bigl([u, v], u \bigr) +\bigl\Vert u'\bigr\Vert ^{2} +h \bigl\Vert \nabla u'\bigr\Vert ^{2} . $$
(3.9)
By Young’s inequality, we have
$$ a(g*u, u) \leq( \delta+1) \biggl(\int_{0}^{t} g(s) \,ds \biggr) a(u, u) +\frac{1}{4 \delta} g \,\square\,\partial^{2} u . $$
(3.10)
From (3.9) and (3.10), we get
$$\begin{aligned} \Phi'(t) \leq&\bigl\Vert u'\bigr\Vert ^{2} +h \bigl\Vert \nabla u'\bigr\Vert ^{2} +\frac{1}{4\delta} g \,\square\,\partial^{2} u \\ &{}- \biggl(1-(\delta+1) \int_{0}^{t} g(s) \,ds \biggr)a(u, u) -\Vert \Delta v\Vert ^{2}, \end{aligned}$$
(3.11)
where \(\delta>0\). Similarly we deduce
$$\begin{aligned} \Psi'(t) =& \int_{0}^{t} g(t-s)a \bigl( u(t)-u(s), u(t) \bigr)\,ds - \int_{0}^{t} g(t-s) \bigl( u(t)-u(s), [u, v]\bigr) \,ds \\ &{} - \int_{0}^{t} g(t-s)a \biggl( u(t)-u(s), \int_{0}^{t} g(t-\tau) u(\tau)\, d \tau\biggr) \,ds \\ &{} - h \int_{0}^{t} g'(t-s) \bigl( \nabla u(t)-\nabla u(s), \nabla u'(t)\bigr) \,ds -h \biggl( \int _{0}^{t} g(s) \,ds \biggr)\bigl\Vert \nabla u'\bigr\Vert ^{2} \\ &{} - \int_{0}^{t} g'(t-s) \bigl( u(t)-u(s), u'(t)\bigr) \,ds - \biggl(\int_{0}^{t} g(s)\,ds \biggr) \bigl\Vert u'\bigr\Vert ^{2} \\ :=& I_{1}+I_{2}+\cdots+I_{5} -h \biggl(\int _{0}^{t} g(s) \,ds \biggr) \bigl\Vert \nabla u'\bigr\Vert ^{2} - \biggl(\int_{0}^{t} g(s)\,ds \biggr)\bigl\Vert u'\bigr\Vert ^{2}. \end{aligned}$$
(3.12)
Now, we estimate the terms on the right-hand side of (3.12). Since \(E(t)\) is bounded, we have that \(\|v\|_{\infty}\leq \int_{\Omega}|\Delta v|^{2} \,d\Omega\) is also bounded, and then Young’s and Hölder’s inequalities, (2.2), (2.3), (2.6) and (2.7) give that
$$\begin{aligned}& |I_{1}| \leq \eta \biggl(\int_{0}^{t} g(s) \,ds \biggr) a(u, u) + \frac{1}{4\eta}g \,\square\,\partial^{2} u , \\& |I_{2}| = \int_{\Omega}\biggl[u, \int _{0}^{t} g(t-s) \bigl( u(t)-u(s)\bigr) \,ds \biggr] v \,d\Omega\leq \eta a(u,u) +c_{\eta}l\|v\|^{2}_{\infty} g \,\square\, \partial^{2} u, \\& |I_{3} | \leq l \int_{0}^{t} g(t-s) a \bigl(u(t)-u(s) , u(t)-u(s)\bigr) \,ds + l \int_{0}^{t} g(t-s) a\bigl(u(t)-u(s) , u(t)\bigr) \,ds \\& \hphantom{|I_{3} |}\leq \biggl(l + \frac{l}{4\eta} \biggr)g \,\square\, \partial^{2} u + l^{2} \eta a(u, u) , \\& |I_{4}| \leq h\eta\bigl\Vert \nabla u'\bigr\Vert ^{2} + \frac{h}{4\eta} \int_{\Omega}\biggl( \int _{0}^{t} g' (t-s)\bigl|\nabla u(t)-\nabla u(s)\bigr| \,ds \biggr)^{2} \,d\Omega \\& \hphantom{|I_{4}|}\leq h\eta \bigl\Vert \nabla u'\bigr\Vert ^{2} -\frac{ g(0)C_{s} h }{4\eta} g' \, \square\,\partial^{2} u , \\& |I_{5}| \leq \eta \bigl\Vert u'\bigr\Vert ^{2}+\frac{1}{4\eta}\int_{\Omega}\biggl( \int _{0}^{t} g' (t-s)\bigl| u(t)-u(s)\bigr| \,ds \biggr)^{2} \,d\Omega \\& \hphantom{|I_{5}|}\leq \eta\bigl\Vert u'\bigr\Vert ^{2}- \frac{g(0)C_{p}}{4 \eta} g' \,\square\, \partial^{2} u . \end{aligned}$$
From the above estimates, we see that
$$\begin{aligned} \Psi'(t) \leq& \biggl(h \eta-h\int _{0}^{t} g(s) \,ds \biggr) \bigl\Vert \nabla u'\bigr\Vert ^{2} + \biggl( \eta- \int _{0}^{t} g(s) \,ds \biggr) \bigl\Vert u'\bigr\Vert ^{2} + \eta \bigl(1+l+l^{2} \bigr)a(u, u) \\ &{} + \biggl( l+\frac{l}{4 \eta} +\frac{1}{4\eta} +c_{\eta}l \Vert v\Vert _{\infty}^{2} \biggr)g\,\square\, \partial^{2} u -\frac{g(0)(C_{s} h +C_{p})}{4\eta} g' \,\square\, \partial^{2} u . \end{aligned}$$
(3.13)
Let \(\int_{0}^{t_{0}} g(s) \,ds:=g_{0}\), where \(t_{0}\) was introduced in (2.13). Since g is positive, we have \(\int_{0}^{t} g(s) \,ds \geq g_{0}\) for all \(t\geq t_{0}\). Thus, making use of this and combining (3.2), (3.4), (3.11) and (3.13), we obtain
$$\begin{aligned} L' (t) \leq& - (g_{0}-\eta-\epsilon)\bigl\Vert u'\bigr\Vert ^{2} -h( g_{0} -\eta-\epsilon) \bigl\Vert \nabla u'\bigr\Vert ^{2} \\ &{}- \biggl[ \frac{N}{2} g(t) +\epsilon\bigl( 1-(1+\delta)l\bigr)-\eta \bigl(1+l+l^{2}\bigr) \biggr] a(u, u) \\ &{} -\epsilon \Vert \Delta v\Vert ^{2} + \biggl( \frac{N}{2}-\frac{ g(0)(C_{s} h +C_{p})}{4\eta} \biggr) g' \,\square\, \partial^{2} u \\ &{}+ \biggl( l+\frac{\epsilon}{4\delta} +\frac{1}{4\eta}+ \frac{l}{4\eta}+ c_{\eta}l \Vert v\Vert _{\infty}^{2} \biggr)g \,\square\, \partial^{2} u . \end{aligned}$$
We first take \(\epsilon>0\) and \(\delta>0\) so small that \(g_{0} -\epsilon>0\) and \(1-(1+\delta)l >0\), respectively. And then, we choose \(\eta>0\) sufficiently small so that \(g_{0} -\eta-\epsilon>0\) and \(\epsilon( 1-(1+\delta)l )-\eta(1+l+ l^{2})>0\). Finally, taking \(N>0\) large enough, we deduce that (3.8). □
Proof of Theorem 2.1
From (2.17), (3.2) and (3.8), we have
$$\begin{aligned} L'(t) \leq& - \beta_{1} {\mathcal{E}}(t) - \frac{\beta_{2}}{c_{3}} \int_{0}^{t_{0}} g'(t-s) a\bigl(u(t)-u(s), u(t)-u(s)\bigr) \,ds \\ &{} + \beta_{2} \int_{t_{0}}^{t} g(t-s) a \bigl(u(t)-u(s), u(t)-u(s)\bigr) \,ds \\ \leq& - \beta_{1} {\mathcal{E}}(t) -\frac{2\beta_{2}}{c_{3}} { \mathcal{E}}'(t) + \beta_{2} \int_{t_{0}}^{t} g(t-s) a\bigl(u(t)-u(s), u(t)-u(s)\bigr) \,ds . \end{aligned}$$
(3.14)
We take \({\mathcal{L}}(t)=L(t)+\frac{2\beta_{2}}{c_{3}} {\mathcal{E}}(t)\), which is clearly equivalent to \({\mathcal{E}}(t)\). By (3.14), we get, for all \(t\geq t_{0}\),
$$ {\mathcal{L}}'(t) \leq- \beta_{1} { \mathcal{E}}(t)+ \beta_{2} \int_{t_{0}}^{t} g(s) a\bigl(u(t)-u(t-s), u(t)-u(t-s)\bigr) \,ds . $$
(3.15)
(A) The special case \(H(t) =ct^{p} \) and \(1\leq p< \frac{3}{2}\).
Case 1. \(p=1\). Using (2.4) and (3.2), estimate (3.15) yields
$$\begin{aligned} {\mathcal{L}}'(t) \leq& - \beta_{1} { \mathcal{E}}(t) - \frac{\beta_{2}}{c} \int_{t_{0}}^{t} g'(s) a\bigl(u(t)-u(t-s), u(t)-u(t-s)\bigr) \,ds \\ \leq& - \beta_{1} {\mathcal{E}}(t) - \frac{2\beta_{2}}{c} {\mathcal{E}}'(t) , \end{aligned}$$
(3.16)
which gives
$$ \biggl({\mathcal{L}}+ \frac{2\beta_{2}}{c} {\mathcal{E}}\biggr)'(t) \leq- \beta_{1} {\mathcal{E}}(t), \quad \forall t \geq t_{0} . $$
From (3.3) and (3.5), we see that \({\mathcal{L}}+ \frac{2\beta_{2}}{c} {\mathcal{E}} \sim{\mathcal{E}} \sim E\). Then we have
$$E(t) \leq c'e^{-c t }=c' G^{-1} (t), $$
where
$$G(t) = \int_{t}^{1} \frac{1}{s H'(\epsilon_{0} s)} \,ds = \int _{t}^{1} \frac{1}{s c} \,ds =- \frac{\ln t}{c} . $$
Case 2. \(1< p<\frac{3}{2}\). By (2.4) we obtain
$$ g'(t) \leq-c g^{p}(t),\quad 1< p< \frac{3}{2} . $$
(3.17)
Using (2.3) and (3.17) we see that
$$ \int_{0}^{\infty} g^{1-\theta}(s) \,ds < \infty $$
(3.18)
for any \(\theta<2-p\). By (3.2) and (3.18) and taking \(t_{0}\) even larger if needed, we deduce that, for all \(t \geq t_{0}\),
$$\begin{aligned} k(t) :=&\int_{t_{0}}^{t} g^{1-\theta}(s) a\bigl(u(t)-u(t-s), u(t)-u(t-s)\bigr) \,ds \\ \leq& 2 \int_{t_{0}}^{t} g^{1-\theta}(s) \bigl[ a\bigl(u(t),u(t)\bigr)+ a\bigl(u(t-s),u(t-s)\bigr) \bigr] \,ds \\ \leq& c { \mathcal{E}}(0) \int_{t_{0}}^{t} g^{1-\theta}(s) \,ds < 1 . \end{aligned}$$
(3.19)
From Hölder’s inequality, Jensen’s inequality (2.13), (3.2), (3.17) and (3.19), we have
$$\begin{aligned}& \int_{t_{0}}^{t} g(s) a\bigl(u(t)-u(t-s), u(t)-u(t-s) \bigr) \,ds \\& \quad = \int_{t_{0}}^{t} g^{(p-1+\theta)(\frac{\theta}{p-1+\theta})}(s) g^{1-\theta}(s) a\bigl(u(t)-u(t-s), u(t)-u(t-s)\bigr) \,ds \\& \quad \leq \biggl( \int_{t_{0}}^{t} g^{p-1+\theta}(s) g^{1-\theta}(s)a\bigl(u(t)-u(t-s), u(t)-u(t-s)\bigr) \,ds \biggr)^{\frac{\theta}{p-1+\theta}} \\& \qquad {} \times \biggl( \int_{t_{0}}^{t} g^{1-\theta}(s) a\bigl(u(t)-u(t-s), u(t)-u(t-s)\bigr) \,ds \biggr)^{\frac{p-1}{p-1+\theta}} \\& \quad = k(t) \biggl(\frac{1}{k(t)} \int_{t_{0}}^{t} g^{p-1+\theta}(s) g^{1-\theta}(s)a\bigl(u(t)-u(t-s), u(t)-u(t-s)\bigr) \,ds \biggr)^{\frac{\theta}{p-1+\theta}} \\& \quad \leq \biggl( \int_{t_{0}}^{t} g^{p}(s) a\bigl(u(t)-u(t-s), u(t)-u(t-s)\bigr) \,ds \biggr)^{\frac{\theta}{p-1+\theta}} \\& \quad \leq \biggl(\frac{1}{c} \biggr)^{\frac{\theta}{p-1+\theta}} \biggl( \int_{t_{0}}^{t} - g'(s) a \bigl(u(t)-u(t-s), u(t)-u(t-s)\bigr) \,ds \biggr)^{\frac{\theta}{p-1+\theta}} \\& \quad \leq \biggl(\frac{1}{c} \biggr)^{\frac{\theta}{p-1+\theta}} \bigl(-{\mathcal{E}}'(t) \bigr)^{\frac{\theta}{p-1+\theta}}. \end{aligned}$$
(3.20)
Then, using (3.20), we show that (3.15) yields, for \(\theta =\frac{1}{2}\),
$$ {\mathcal{L}}'(t) \leq- \beta_{1} { \mathcal{E}}(t)+ \frac{\beta_{2}}{ c^{\frac{1}{2p-1}}} \bigl(-{\mathcal{E}}'(t) \bigr)^{\frac{1}{2p-1}}. $$
(3.21)
Multiplying (3.21) by \({\mathcal{E}}^{\gamma}(t)\), with \(\gamma=2p-2\), and using (3.2) and Young’s inequality, we obtain
$$\begin{aligned} \bigl({\mathcal{L} } {\mathcal{E}}^{\gamma}\bigr)'(t) =& { \mathcal{L} }'(t) {\mathcal{E} }^{\gamma}(t) + \gamma{ \mathcal{L}}(t){\mathcal{E} }^{\gamma-1}(t) {\mathcal{E} }'(t) \leq- \beta_{1} {\mathcal{E}}^{\gamma+1}(t)+ \frac{\beta_{2}}{ c^{\frac{1}{\gamma+1}}}{ \mathcal{E}}^{\gamma}(t) \bigl(-{\mathcal{E}}'(t) \bigr)^{\frac{1}{\gamma+1}} \\ \leq& - \beta_{1} {\mathcal{E}}^{\gamma+1}(t)+ \varepsilon{ \mathcal{E}}^{\gamma+1}(t) +C_{\varepsilon}\bigl(-{\mathcal{E}}'(t) \bigr). \end{aligned}$$
Taking \(\varepsilon<\beta_{1}\), we have, for some \(C_{1}>0\),
$$ L_{0}'(t)\leq-C_{1} L_{0}^{\gamma+1}(t), $$
where \(L_{0} ={\mathcal{L} }{\mathcal{E}}^{\gamma}+C_{\varepsilon}{\mathcal{E}} \sim{\mathcal{E}}\sim E \). Therefore we deduce that
$$ E(t) \leq\frac{c}{(c'+ c'' t)^{\frac{1}{\gamma}}}. $$
(3.22)
Since \(p<\frac{3}{2}\) and by (3.22), we find that
$$ \int_{0}^{\infty}E(t) \, d t \leq\int _{0}^{\infty}\frac{c}{(c'+ c'' t)^{\frac{1}{2p-2}}}\, d t < +\infty. $$
Using this fact, we have
$$ \int_{0}^{t} a\bigl(u(t)-u(t-s), u(t)-u(t-s)\bigr) \,ds \leq c\int_{0}^{t} E(s) \,ds < +\infty. $$
(3.23)
Hence, from (3.2), (3.17), (3.23) and Hölder’s inequality, estimate (3.15) gives
$$\begin{aligned} {\mathcal{L}}'(t) \leq& - \beta_{1} { \mathcal{E}}(t)+ \beta_{2} \biggl(\int_{0}^{t} a\bigl(u(t)-u(t-s), u(t)-u(t-s)\bigr) \,ds \biggr)^{\frac{p-1}{p}} \bigl(g^{p} \,\square\,\partial^{2} u \bigr)^{\frac{1}{p}} \\ \leq& - \beta_{1} {\mathcal{E}}(t)+ c \bigl(- g' \, \square\, \partial^{2} u \bigr)^{\frac{1}{p}} \leq - \beta_{1} {\mathcal{E}}(t)+ c \bigl( -{\mathcal{E}}'(t) \bigr)^{\frac{1}{p}} . \end{aligned}$$
(3.24)
Now, we multiply (3.24) by \({\mathcal{E}}^{\gamma}(t)\), with \(\gamma=p-1\). Then, repeating the above steps, we see that
$$ E(t) \leq\frac{c}{(c'+ c'' t)^{\frac{1}{\gamma}}} =cG^{-1} \bigl(a'+a''t\bigr), $$
(3.25)
where
$$G(t)=\frac{1}{cp\epsilon_{0}^{p-1}}\int_{t}^{1} \frac{1}{s^{p}} \,ds = \frac {1}{cp(p-1)\epsilon_{0}^{p-1}} \biggl(\frac{1}{t^{p-1}}-1 \biggr). $$
(B) The general case. This case is obtained on account of the ideas presented in [24, 25] as follows. Let \(H_{0}^{*}\) be the convex conjugate of \(H_{0}\) in the sense of Young (see [29]); then
$$ H_{0}^{*} (s) = s\bigl(H_{0}' \bigr)^{-1} (s) -H_{0}\bigl[\bigl(H_{0}' \bigr)^{-1}(s)\bigr], \quad \mbox{if } s \in \bigl(0, H_{0}' (r) \bigr] $$
(3.26)
and \(H_{0}^{*}\) satisfies the following Young’s inequality:
$$ AB \leq H_{0}^{*} (A) +H_{0}(B), \quad \mbox{if } A \in \bigl(0, H_{0}'(r) \bigr], B\in(0, r]. $$
(3.27)
We define \(\eta(t)\) by
$$ \eta(t):=\int_{t_{0}}^{t} \frac{g(s)}{H_{0}^{-1} (-g'(s)) }a \bigl(u(t)-u(t-s), u(t)-u(t-s)\bigr) \,ds , $$
where \(H_{0}\) is such that (2.9) is satisfied. As in (3.19), we find that \(\eta(t)\) satisfies
$$ \eta(t) < 1 , \quad \forall t\geq t_{0}. $$
(3.28)
Furthermore, we define \(I(t)\) by
$$ I(t):=- \int_{t_{0}}^{t} g'(s) \frac{g(s)}{H_{0}^{-1} (-g'(s)) }a\bigl(u(t)-u(t-s), u(t)-u(t-s)\bigr) \,ds . $$
Since \(H_{0}\) is strictly convex on \((0, r]\) and \(H_{0} (0)=0\), then
$$ H_{0} (\lambda x )\leq\lambda H_{0} (x) $$
(3.29)
provided \(0\leq \lambda\leq1\) and \(x\in(0, r]\). From (3.28), (3.29) and Jensen’s inequality (2.13), we obtain
$$\begin{aligned} I(t) =&\frac{1}{\eta(t)}\int_{t_{0}}^{t} \eta(t) H_{0} \bigl[H_{0}^{-1}\bigl(-g'(s) \bigr)\bigr]\frac{g(s)}{H_{0}^{-1} (-g'(s)) }a\bigl(u(t)-u(t-s), u(t)-u(t-s)\bigr) \,ds \\ \geq & \frac{1}{\eta(t)}\int_{t_{0}}^{t} H_{0} \bigl[\eta(t) H_{0}^{-1} \bigl(-g'(s)\bigr)\bigr]\frac{g(s)}{H_{0}^{-1} (-g'(s)) }a\bigl(u(t)-u(t-s), u(t)-u(t-s)\bigr) \,ds \\ \geq& H_{0} \biggl( \int_{t_{0}}^{t} g(s) a\bigl(u(t)-u(t-s), u(t)-u(t-s)\bigr) \,ds \biggr) . \end{aligned}$$
This implies that
$$ \int_{t_{0}}^{t} g(s) a \bigl(u(t)-u(t-s), u(t)-u(t-s)\bigr) \,ds \leq H_{0}^{-1} \bigl(I(t)\bigr) . $$
(3.30)
Using (3.15) and (3.30) we see that
$$ {\mathcal{L}}'(t) \leq-\beta_{1} { \mathcal{E}}(t) +\beta_{2} H_{0}^{-1} \bigl(I(t) \bigr) , \quad \forall t\geq t_{0} . $$
(3.31)
By (2.3), (2.4) and the properties of \(H_{0}\) and D, we have
$$ \frac{g(s)}{H_{0}^{-1} (-g'(s)) } \leq\frac{g(s)}{H_{0}^{-1} (H(g(s)))} = \frac{g(s)}{D^{-1} (g(s)) } \leq\delta_{0} $$
(3.32)
for some positive constant \(\delta_{0}\). Thus, using (2.13), (3.2) and (3.32) and choosing \(t_{0}\) even larger, we can find that \(I(t)\) satisfies, for all \(t\geq t_{0}\),
$$\begin{aligned} I(t) \leq& - \delta_{0} \int_{t_{0}}^{t} g'(s) a\bigl(u(t)-u(t-s), u(t)-u(t-s)\bigr) \,ds \leq -c { \mathcal{E}}(0) \int_{t_{0}}^{t} g'(s) \,ds \\ \leq& cg(t_{0}) {\mathcal{E}}(0) \leq \min\bigl\{ r, H(r), H_{0} (r)\bigr\} . \end{aligned}$$
(3.33)
Now, for \(\epsilon_{0} < r\) and \(d_{0} > 0\), we define the functional
$$F_{1} (t) := {\mathcal{L}}(t) H_{0}' \biggl( \epsilon_{0} \frac{{\mathcal{E}}(t)}{{\mathcal{E}}(0)} \biggr) + d_{0} { \mathcal{E}}(t), $$
which satisfies
$$ d_{1} F_{1}(t) \leq{\mathcal{E}}(t) \leq d_{2} F_{1} (t) $$
(3.34)
for some \(d_{1}, d_{2}>0\). From (3.33), we have \(H_{0}^{-1}(I (t)) \leq r\). Also, by \(\epsilon_{0} < r\), \({\mathcal{E}}' \leq0\), we get \(\epsilon_{0} \frac{{\mathcal{E}}(t)}{{\mathcal{E}}(0)}< r\). Using the fact that \({\mathcal{E}}' \leq0\), \(H_{0} > 0\), \(H_{0}'> 0\) and \(H_{0}'' >0\) on \((0, r]\) and (3.2), (3.26), (3.27), (3.31) and (3.33), we obtain
$$\begin{aligned} F_{1}' (t) \leq& -\beta_{1} { \mathcal{E}}(t)H_{0}' \biggl(\epsilon_{0} \frac{{\mathcal{E}}(t)}{{\mathcal{E}}(0)} \biggr) + \beta_{2} H_{0}' \biggl(\epsilon_{0} \frac{{\mathcal{E}}(t)}{{\mathcal{E}}(0)} \biggr) H_{0}^{-1} \bigl(I(t)\bigr)+d_{0} {\mathcal{E}}'(t) \\ \leq& -\beta_{1} {\mathcal{E}}(t)H_{0}' \biggl(\epsilon_{0} \frac{{\mathcal{E}}(t)}{{\mathcal{E}}(0)} \biggr) +\beta_{2} H_{0}^{*} \biggl(H_{0}' \biggl( \epsilon_{0} \frac{{\mathcal{E}}(t)}{{\mathcal{E}}(0)} \biggr) \biggr)+ \beta_{2} I(t)+ d_{0} {\mathcal{E}}'(t) \\ \leq & -\beta_{1} {\mathcal{E}}(t)H_{0}' \biggl(\epsilon_{0} \frac{{\mathcal{E}}(t)}{{\mathcal{E}}(0)} \biggr) + \beta_{2} \epsilon_{0} \frac{{\mathcal{E}}(t)}{{\mathcal{E}}(0)}H_{0}' \biggl( \epsilon_{0} \frac{{\mathcal{E}}(t)}{{\mathcal{E}}(0)} \biggr) \\ &{}-\beta_{2} H_{0} \biggl(\epsilon_{0} \frac{{\mathcal{E}}(t)}{{\mathcal{E}}(0)} \biggr) + \beta_{2} I(t)+ d_{0} {\mathcal{E}}'(t) \\ \leq& - \bigl(\beta_{1} {\mathcal{E}}(0)- \beta_{2} \epsilon_{0} \bigr) \frac{{\mathcal{E}}(t)}{{\mathcal{E}}(0)}H_{0}' \biggl(\epsilon_{0} \frac{{\mathcal{E}}(t)}{{\mathcal{E}}(0)} \biggr)-2\beta_{2} \delta_{0} {\mathcal{E}}'(t)+d_{0} { \mathcal{E}}'(t) . \end{aligned}$$
Therefore, with a suitable choice of \(\epsilon_{0}\) and \(d_{0}\), we see that
$$ F_{1}' (t) \leq-k \biggl(\frac{{\mathcal{E}}(t)}{{\mathcal{E}}(0)} \biggr)H_{0}' \biggl(\epsilon_{0} \frac{{\mathcal{E}}(t)}{{\mathcal{E}}(0)} \biggr)=-k H_{2} \biggl(\frac{{\mathcal{E}}(t)}{{\mathcal{E}}(0)} \biggr), \quad \forall t \geq t_{0}, $$
(3.35)
where \(k>0\) and \(H_{2} (t) = t H_{0}' (\epsilon_{0} t)\). From the strict convexity of \(H_{0}\) on \((0, r]\), we find that \(H_{2} (t) >0\) and \(H_{2} ' (t)= H_{0}'(\epsilon_{0} t)+\epsilon_{0} t H_{0}'' (\epsilon_{0} t)>0 \) on \((0, 1]\). We take
$$R(t) =\frac{d_{1} F_{1}(t)}{{\mathcal{E}}(0)}, $$
which is clearly equivalent to \({\mathcal{E}}(t)\). By (3.34), (3.35) and \(H_{2}'>0\), we have
$$R'(t) \leq-\frac{k d_{1}}{{\mathcal{E}}(0)}H_{2} \biggl( \frac{{\mathcal{E}}(t)}{{\mathcal{E}}(0)} \biggr) \leq- k_{0} H_{2} \bigl(R(t) \bigr), \quad \forall t \geq t_{0} , $$
where \(k_{0} = \frac{k d_{1}}{{\mathcal{E}}(0)} >0 \). Hence, a simple integration gives, for some \(k_{1}, k_{2}>0\),
$$ R(t) \leq H_{1}^{-1} (k_{1} t +k_{2} ), \quad \forall t\geq t_{0}, $$
(3.36)
where \(H_{1} (t) =\int_{t}^{1} \frac{1}{H_{2} (s)} \,ds\). Here, we have used, on the basis of the properties of \(H_{2}\), the fact that \(H_{1}\) is a strictly decreasing function on \((0, 1]\) and \(\lim_{t\rightarrow0} H_{1} (t)= +\infty\). From (3.3), (3.34) and (3.36), estimate (2.10) is established.
Moreover, if \(\int_{0}^{t} H_{1} (t)\, dt < +\infty\), then
$$\int_{0}^{t} a\bigl(u(t)-u(t-s), u(t)-u(t-s) \bigr) \,ds \leq c \int_{0}^{t} E(s) \,ds < + \infty. $$
Similarly, we define, for large \(t_{0}\),
$$ \eta(t):= \int_{t_{0}}^{t} a\bigl(u(t)-u(t-s), u(t)-u(t-s)\bigr) \,ds < 1 $$
and
$$ I(t):=- \int_{t_{0}}^{t} g' (s) a \bigl(u(t)-u(t-s), u(t)-u(t-s)\bigr) \,ds . $$
Using (2.4), the strict convexity of H and Jensen’s inequality (2.13), we have
$$\begin{aligned} I(t) \geq& \frac{1}{\eta(t)} \int_{t_{0}}^{t} \eta(t) H\bigl(g (s)\bigr) a\bigl(u(t)-u(t-s), u(t)-u(t-s)\bigr) \,ds \\ \geq& \frac{1}{\eta(t)} \int_{t_{0}}^{t} H\bigl( \eta(t) g (s)\bigr) a\bigl(u(t)-u(t-s), u(t)-u(t-s)\bigr) \,ds \\ \geq & H \biggl( \frac{1}{\eta(t)} \int_{t_{0}}^{t} \eta(t) g (s) a\bigl(u(t)-u(t-s), u(t)-u(t-s)\bigr) \,ds \biggr) \\ =& H \biggl( \int_{t_{0}}^{t} g (s) a \bigl(u(t)-u(t-s), u(t)-u(t-s)\bigr) \,ds \biggr) . \end{aligned}$$
Thus, we deduce that
$$\int_{t_{0}}^{t} g (s) a\bigl(u(t)-u(t-s), u(t)-u(t-s)\bigr) \,ds \leq H^{-1} \bigl( I(t)\bigr), $$
and (3.15) becomes
$${\mathcal{L}}'(t) \leq-\beta_{1} {\mathcal{E}} (t) + \beta_{2} H^{-1}\bigl(I(t)\bigr) , \quad \forall t\geq t_{0} . $$
Therefore, repeating the same procedures, we find that for some \(k_{1}\), \(k_{2}\) and \(k_{3} >0\),
$$E(t) \leq k_{3} G^{-1} (k_{1} t +k_{2} ), $$
where \(G(t) = \int_{t}^{1} \frac{1}{sH'(\epsilon_{0} s)} \,ds \). □
Example
We give an example to illustrate the energy decay rates given by Theorem 2.1. If
$$g(t) = \frac{1}{a +t^{q}} $$
for \(q>3\) and \(a>1\) chosen so that g satisfies (2.3), then \(g'(t) =-H(g(t))\), where
$$ H(t) = {q t^{2}} { \biggl(\frac{1}{t} -a \biggr)^{1-\frac{1}{q}}}. $$
Since
$$ H'(t) = \frac{q(1+\frac{1}{q} -2at)}{(\frac{1}{t}-a)^{\frac{1}{q}}}, \qquad H''(t) = \frac{\frac{2a^{2}q}{t^{2}}(t-\frac{1+q-\sqrt{q^{2} -1}}{2a q})(t-\frac{1+q+\sqrt{q^{2} -1}}{2a q})}{(\frac{1}{t}-a)^{1+\frac{1}{q}}}, $$
then the function H satisfies hypothesis (H2) on the interval \((0, r]\) for any \(0< r<\frac{1+q-\sqrt{q^{2}-1}}{2aq}\). By choosing \(D(t)=t^{\alpha}\), (2.9) is satisfied for any \(\alpha>\frac{q}{q-1}\). Then an explicit rate of decay can be obtained by Theorem 2.1. The function \(H_{0} (t) = H(t^{\alpha})\) has derivative
$$ H_{0}'(t) = \frac{q \alpha t^{\alpha-1}[1+\frac{1}{q}- 2 a t^{\alpha}]}{(\frac{1}{t^{\alpha}}-a)^{\frac{1}{q}}} . $$
Hence,
$$ H_{1} (t) = \int_{t}^{1} \frac{ [\frac{1}{( \epsilon_{0} s)^{\alpha}} -a ]^{\frac{1}{q}}}{q \alpha s (\epsilon_{0} s)^{\alpha-1}[1+\frac{1}{q}- 2 a (\epsilon_{0} s)^{\alpha}]} \,ds . $$
Let \(\frac{1}{(\epsilon_{0} s)^{\alpha}} =u\), then we have
$$ H_{1} (t) = \int_{\frac{1}{\epsilon_{0} ^{\alpha}}}^{\frac{1}{(\epsilon_{0} t)^{\alpha}}} \frac{ (u -a )^{\frac{1}{q}} u^{-\frac{1}{\alpha}}}{q \alpha^{2}[1+\frac{1}{q}- \frac{2 a}{u}] }\, d u \leq \frac{1}{q \alpha^{2}[1+\frac{1}{q}- 2 a \epsilon_{0}^{\alpha}] } \int_{\frac{1}{\epsilon_{0} ^{\alpha}}}^{\frac{1}{(\epsilon_{0} t)^{\alpha}}} (u -a )^{\frac{1}{q}} u^{-\frac{1}{\alpha}} \, d u . $$
Using the fact that the function \(f(u) = (u-a)^{\frac{1}{q}}\) is increasing on \((a, +\infty)\) and \((u-a)^{\frac{1}{q}} < u^{\frac{1}{q}}\) and taking \(\epsilon_{0} < a^{-\frac{1}{\alpha}}\), then
$$ H_{1} (t) \leq\frac{ 1 }{q \alpha^{2}[1+\frac{1}{q}- 2 a \epsilon_{0}^{\alpha}] } \int_{\frac{1}{\epsilon_{0} ^{\alpha}}}^{\frac{1}{(\epsilon_{0} t)^{\alpha}}}u^{\frac{1}{q}-\frac{1}{\alpha}} \, d u = \frac{ \epsilon_{0}^{\frac{q-\alpha-\alpha q}{q}}}{\alpha(\alpha-q+\alpha q)[1+\frac{1}{q}- 2 a \epsilon_{0}^{\alpha}] } \bigl[ t^{\frac{q-\alpha-\alpha q}{q}} -1 \bigr]. $$
Now, we find that if \(\alpha< \frac{2q}{1+q}\),
$$\begin{aligned} \int_{0}^{1} H_{1} (t)\, d t \leq& \frac{ \epsilon_{0}^{\frac{q-\alpha-\alpha q}{q}}}{\alpha(\alpha-q+\alpha q)[1+\frac{1}{q}- 2 a \epsilon_{0}^{\alpha}] } \int_{0}^{1} \bigl[ t^{\frac{q-\alpha-\alpha q}{q}}-1 \bigr]\, d t \\ =& \frac{ \epsilon_{0}^{\frac{q-\alpha-\alpha q}{q}}}{ \alpha(2q-\alpha-\alpha q)[1+\frac{1}{q}- 2 a \epsilon_{0}^{\alpha}]} < + \infty. \end{aligned}$$
Choosing \(\frac{1}{\epsilon_{0} s} =v \) and \(\epsilon_{0} < a^{-1}\), we obtain
$$\begin{aligned} G(t) =& \int_{t}^{1} \frac{1}{s H' (\epsilon_{0} s)} \,ds = \int_{t}^{1} \frac{(\frac{1}{\epsilon_{0} s}-a)^{\frac{1}{q}}}{s q (1+\frac{1}{q}-2a\epsilon_{0} s )} \,ds = \int _{\frac{1}{\epsilon_{0}}}^{\frac{1}{\epsilon_{0} t} } \frac{(v-a)^{\frac{1}{q}} v^{-1}}{q(1+\frac{1}{q}-\frac{2a}{v})}\, d v \\ \leq& \frac{1}{q(1+\frac{1}{q}-{2a \epsilon_{0}})} \int_{\frac{1}{\epsilon_{0}}}^{\frac{1}{\epsilon_{0} t} } v^{\frac{1}{q}-1}\, d v = \frac{1}{1+\frac{1}{q}-{2a \epsilon_{0}}} \biggl[ \biggl(\frac{1}{\epsilon_{0} t} \biggr)^{\frac{1}{q}} - \biggl(\frac{1}{\epsilon_{0}} \biggr)^{\frac{1}{q}} \biggr]. \end{aligned}$$
Therefore,
$$ G^{-1} (t) \leq\frac{1}{\epsilon_{0} [ (\frac{1}{\epsilon_{0}} )^{\frac{1}{q}} + (1+\frac{1}{q}-{2a \epsilon_{0}} )t ]^{q}}. $$
Then we can use (2.11) to deduce that the energy decays at the same rate of g, that is,
$$ E(t) \leq\frac{\tilde{c}_{1}}{ \tilde{c}_{2}+ \tilde {c}_{3}t^{q}}, $$
where \(\tilde{c}_{i}\) (\(i=1, 2 , 3\)) are constants.