In this section, we focus our attention on the global existence and decay rate of the solution to problem (1.1). We first define
$$\begin{aligned}& I(t):= \|\Delta u\|_{2}^{2}+\|\Delta v \|_{2}^{2}+\|\nabla u_{t}\| _{2}^{2}+ \|\nabla v_{t}\|_{2}^{2} +\biggl(\gamma- \int_{0}^{t}g_{1}(s)\,ds\biggr)\|\nabla u \|_{2}^{2} \\& \hphantom{I(t):={}}{}+\biggl(\delta- \int_{0}^{t}g_{2}(s)\,ds\biggr)\|\nabla v \|_{2}^{2}+g_{1}\circ \nabla u+g_{2}\circ \nabla v-(m+1) \int_{\Omega}F(u,v)\, dx, \end{aligned}$$
(3.1)
$$\begin{aligned}& J(t):= \frac{1}{2} \biggl(\|\Delta u\|_{2}^{2}+\| \Delta v\|_{2}^{2}+\| \nabla u_{t} \|_{2}^{2} +\|\nabla v_{t}\|_{2}^{2}+ \biggl(\gamma- \int_{0}^{t}g_{1}(s)\,ds\biggr)\|\nabla u \| _{2}^{2} \\& \hphantom{J(t):={}}{}+\biggl(\delta- \int_{0}^{t}g_{2}(s)\,ds\biggr)\|\nabla v \|_{2}^{2}+g_{1}\circ \nabla u+g_{2}\circ \nabla v \biggr)- \int_{\Omega}F(u,v)\, dx, \end{aligned}$$
(3.2)
and define the energy function as
$$ E(t):=\frac{1}{2}\bigl(\|u_{t}\|_{2}^{2}+ \|v_{t}\|_{2}^{2}\bigr)+J(t). $$
(3.3)
Lemma 3.1
Suppose that (A1) and (2.4) hold. Let
\((u,v)\)
be the solution of problem (1.1). Then
\(E(t)\)
is a nonincreasing function, that is, for
\(t\geq0\),
$$\begin{aligned} {{d\over {dt}}E(t)} =& {-\|\nabla u_{t}\|_{2}^{2}- \|\nabla v_{t}\| _{2}^{2}+\frac{1}{2} \bigl(g_{1}'\circ\nabla u\bigr)+\frac{1}{2} \bigl(g_{2}'\circ\nabla v\bigr)-\frac{1}{2}g_{1}(t) \|\nabla u\|_{2}^{2}} \\ &{} -\frac{1}{2}g_{2}(t)\|\nabla v\|_{2}^{2}- \|u_{t}\|_{p+1}^{p+1}-\| v_{t} \|_{q+1}^{q+1}. \end{aligned}$$
(3.4)
Proof
Multiplying (1.1)1 by \(u_{t}\) and (1.2)2 by \(v_{t}\), integrating over Ω, summing up, and then using integration by parts, we obtain
$$\begin{aligned}& {d\over {dt}} \biggl[\frac{1}{2}\bigl(\|u_{t} \|_{2}^{2}+\|v_{t}\| _{2}^{2}+ \gamma\|\nabla u\|_{2}^{2}+\delta\|\nabla v \|_{2}^{2}+\|\Delta u\|_{2}^{2}+\|\Delta v \|_{2}^{2}+\|\nabla u_{t}\|_{2}^{2}+ \|\nabla v_{t}\|_{2}^{2}\bigr) \\& \qquad {}- \int_{\Omega}F(u,v)\,dx \biggr] \\& \quad = -\|\nabla u_{t}\|_{2}^{2}-\|\nabla v_{t}\|_{2}^{2}+ \int_{0}^{t} \int _{\Omega}g_{1}(t-s)\nabla u(s)\cdot\nabla u_{t}\,dx\,ds \\& \qquad {}+ \int_{0}^{t} \int _{\Omega}g_{2}(t-s)\nabla v(s)\cdot\nabla v_{t}\,dx\,ds \\& \qquad {}-\|u_{t}\|_{p+1}^{p+1}-\|v_{t} \|_{q+1}^{q+1}. \end{aligned}$$
Applying Lemma 2.4 to the third and fourth terms on the right-hand side of this equality, we get (3.4). □
Lemma 3.2
Suppose that (A1) and (2.4) hold. Let
\((u,v)\)
be the solution of problem (1.1). Assume further that
\(I(0)>0\)
and
$$ \alpha_{1}:=(m+1)c_{1}\tilde{c}^{m+1} \biggl( \frac {2(m+1)}{m-1}E(0) \biggr)^{\frac{m-1}{2}}< 1. $$
(3.5)
Then
$$ I(t)>0, \quad t\geq0. $$
(3.6)
Proof
Since \(I(0)>0\), by continuity there exists a maximal time \(t_{\max}>0\) (possibly \(t_{\max}=T\)) such that
$$I(t)\geq0, \quad t\in[0,t_{\max}], $$
which implies that, for \(t\in[0,t_{\max}]\),
$$\begin{aligned} J(t) =& \frac{m-1}{2(m+1)} \biggl(\|\Delta u\|_{2}^{2}+\| \Delta v\| _{2}^{2}+\|\nabla u_{t} \|_{2}^{2} +\|\nabla v_{t}\|_{2}^{2}+ \biggl(\gamma- \int_{0}^{t}g_{1}(s)\,ds\biggr)\|\nabla u \| _{2}^{2} \\ &{} +\biggl(\delta- \int_{0}^{t}g_{2}(s)\,ds\biggr)\|\nabla v \|_{2}^{2}+g_{1}\circ \nabla u+g_{2}\circ \nabla v \biggr)+\frac{1}{m+1}I(t) \\ \geq&\frac{m-1}{2(m+1)} \bigl(\|\Delta u\|_{2}^{2}+\|\Delta v\| _{2}^{2}+l\|\nabla u\|_{2}^{2} +k\| \nabla v\|_{2}^{2} \bigr). \end{aligned}$$
(3.7)
Since \(E(t)\) is nonincreasing by (3.4), using (3.7) and (3.3), we get, for \(t\in[0,t_{\max}]\),
$$\begin{aligned}& \|\Delta u\|_{2}^{2}+\|\Delta v\|_{2}^{2}+l \|\nabla u\|_{2}^{2} +k\|\nabla v\|_{2}^{2} \\& \quad \leq{\frac{2(m+1)}{m-1}J(t)}\leq{\frac {2(m+1)}{m-1}E(t)}\leq{\frac{2(m+1)}{m-1}E(0)}. \end{aligned}$$
(3.8)
Using Lemma 2.2, (2.6), (3.8), and (3.5), we obtain, for \(t\in [0,t_{\max}]\),
$$\begin{aligned} (m+1) \int_{\Omega}F(u,v)\, dx&\leq(m+1)c_{1}\bigl(\|u \|_{m+1}^{m+1}+\|v\| _{m+1}^{m+1}\bigr) \\ &\leq(m+1)c_{1}\tilde{c}^{m+1}\bigl(\|\Delta u \|_{2}^{m+1}+\|\Delta v\| _{2}^{m+1}\bigr) \\ &\leq\alpha_{1}\bigl(\|\Delta u\|_{2}^{2}+\| \Delta v\|_{2}^{2}\bigr) \\ &< \|\Delta u\|_{2}^{2}+\|\Delta v\|_{2}^{2}. \end{aligned}$$
(3.9)
Thus,
$$I(t)>0,\quad t\in[0,t_{\max}]. $$
By repeating these steps and using the fact that
$${\lim_{t\rightarrow t_{\max}}}(m+1)c_{1}\tilde{c}^{m+1} \biggl(\frac {2(m+1)}{m-1}E(t) \biggr)^{\frac{m-1}{2}}\leq\alpha_{1}< 1, $$
this implies that we can take \(t_{\max}=T\). □
Lemma 3.3
Let the assumptions of Lemma
3.2
hold. Then there exists
\(\eta_{1}>1\)
such that
$$ \|\Delta u\|_{2}^{2}+\|\Delta v\|_{2}^{2} \leq\eta_{1}I(t), \quad t\in [0,T), $$
(3.10)
where
\(\eta_{1}= {\frac{1}{1-\alpha_{1}}}\).
Proof
From (3.9) we have
$$(m+1) \int_{\Omega}F(u,v)\,dx\leq\alpha_{1}\bigl(\|\Delta u \|_{2}^{2}+\|\Delta v\|_{2}^{2}\bigr). $$
Letting \(\eta_{1}= {\frac{1}{1-\alpha_{1}}}\) and using (3.1), we obtain (3.10). □
Theorem 3.4
Suppose that (A1) and (A2) hold. Let
\(u_{0},v_{0}\in H_{0}^{2}(\Omega)\)
and
\(u_{1},v_{1}\in H_{0}^{1}(\Omega )\)
satisfy
\(I(0)>0\)
and (3.5). Then the solution
\((u, v)\)
of problem (1.1) is global and bounded. Furthermore, if
$$ \gamma+\delta>\frac{7+5\eta_{1}}{2}\max \biggl\{ \int_{0}^{\infty }g_{1}(s)\,ds, \int_{0}^{\infty}g_{2}(s)\,ds \biggr\} , $$
(3.11)
then we have the following decay estimates:
-
(i)
if
\(p=q=1\), then, for
\(t\geq0\),
$$E(t)\leq E(0)e^{-\tau_{1}t}, $$
-
(ii)
if
\(\max\{p,q\}>1\), then, for
\(t\geq0\),
$$E(t)\leq \biggl(E^{-\max\{\frac{p-1}{2},\frac{q-1}{2}\}}(0)+\tau_{2}\max \biggl\{ \frac{p-1}{2},\frac{q-1}{2} \biggr\} [t-1]^{+} \biggr)^{-\frac{2}{\max\{ p,q\}-1}}, $$
where
\(\tau_{1}\)
and
\(\tau_{2}\)
are some positive constants.
Proof
First, to prove that \(T=\infty\), it suffices to show that \(\|\Delta u\|_{2}^{2}+\|\Delta v\|_{2}^{2}+\|\nabla u_{t}\|_{2}^{2}+\| \nabla v_{t}\|_{2}^{2}\) is bounded independently of t. Thanks to (3.3), (3.4), and (3.6), we have
$$\begin{aligned} E(0) \geq& E(t)\geq J(t) \\ =& \frac{m-1}{2(m+1)} \biggl(\|\Delta u\|_{2}^{2}+\| \Delta v\|_{2}^{2}+\|\nabla u_{t}\|_{2}^{2} +\|\nabla v_{t}\|_{2}^{2} \\ &{}+\biggl(\gamma- \int_{0}^{t}g_{1}(s)\,ds\biggr)\|\nabla u \| _{2}^{2} \\ &{} +\biggl(\delta- \int_{0}^{t}g_{2}(s)\,ds\biggr)\|\nabla v \|_{2}^{2}+g_{1}\circ \nabla u+g_{2}\circ \nabla v \biggr)+\frac{1}{m+1}I(t) \\ >& \frac{m-1}{2(m+1)} \bigl(\|\Delta u\|_{2}^{2}+\|\Delta v \|_{2}^{2}+\| \nabla u_{t}\|_{2}^{2} +\|\nabla v_{t}\|_{2}^{2} \bigr). \end{aligned}$$
Therefore,
$$\|\Delta u\|_{2}^{2}+\|\Delta v\|_{2}^{2}+ \|\nabla u_{t}\|_{2}^{2}+\| \nabla v_{t} \|_{2}^{2}\leq\alpha_{2}E(0), $$
where \(\alpha_{2}\) is a positive constant depending only on m. Thus, we obtain the global existence.
We further derive the decay rate of the energy function for problem (1.1) by Nakao’s method [18]. For this purpose, we have to show that the energy function defined by (3.3) satisfies the hypothesis of Lemma 2.6. Integrating (3.4) over \([t,t+1]\), we have
$$ E(t)-E(t+1)=D_{1}^{p+1}(t)+D_{2}^{q+1}(t), $$
(3.12)
where
$$\begin{aligned}& D_{1}^{p+1}(t):= \int_{t}^{t+1}\|\nabla u_{t} \|_{2}^{2}\,ds-\frac {1}{2} \int_{t}^{t+1}g_{1}'\circ \nabla u\,ds \\& \hphantom{D_{1}^{p+1}(t):={}}{}+\frac{1}{2} \int _{t}^{t+1}g_{1}\|\nabla u \|_{2}^{2}\,ds+ \int_{t}^{t+1}\|u_{t}\| _{p+1}^{p+1} \,ds, \end{aligned}$$
(3.13)
$$\begin{aligned}& D_{2}^{q+1}(t):= \int_{t}^{t+1}\|\nabla v_{t} \|_{2}^{2}\,ds-\frac {1}{2} \int_{t}^{t+1}g_{2}'\circ\nabla v\,ds \\& \hphantom{D_{2}^{q+1}(t):={}}{}+\frac{1}{2} \int _{t}^{t+1}g_{2}\|\nabla v \|_{2}^{2}\,ds+ \int_{t}^{t+1}\|v_{t}\| _{q+1}^{q+1} \,ds. \end{aligned}$$
(3.14)
By (3.13), (3.14), and the Hölder inequality, we observe that
$$ \int_{t}^{t+1} \int_{\Omega}|u_{t}|^{2}\,dx\,ds+ \int_{t}^{t+1} \int_{\Omega }|v_{t}|^{2}\,dx\,ds\leq c_{1}(\Omega)D_{1}^{2}(t)+ c_{2}(\Omega )D_{2}^{2}(t), $$
(3.15)
where \(c_{1}(\Omega)=\operatorname{vol}(\Omega)^{\frac{p-1}{p+1}}\) and \(c_{2}(\Omega )=\operatorname{vol}(\Omega)^{\frac{q-1}{q+1}}\).
By the mean value theorem there exist \(t_{1}\in[t,t+\frac{1}{4}]\) and \(t_{2}\in[t+\frac{3}{4},t+1]\) such that
$$ \bigl\Vert u_{t}(t_{i})\bigr\Vert _{2}^{2}+ \bigl\Vert v_{t}(t_{i})\bigr\Vert _{2}^{2} \leq4c_{1}(\Omega )D_{1}^{2}(t)+ 4c_{2}( \Omega)D_{2}^{2}(t),\quad i=1,2. $$
(3.16)
Next, multiplying Eq. (1.1)1 by u and Eq. (1.1)2 by v, integrating over \(\Omega\times[t_{1},t_{2}]\), and using integration by parts, we obtain
$$\begin{aligned} \int_{t_{1}}^{t_{2}}I(t)\,dt =& - \int_{t_{1}}^{t_{2}} \int_{\Omega }uu_{tt}\,dx\,dt- \int_{t_{1}}^{t_{2}} \int_{\Omega}vv_{tt}\,dx\,dt- \int _{t_{1}}^{t_{2}} \int_{\Omega}\nabla u\cdot\nabla u_{t}\,dx\,dt \\ &{} - \int_{t_{1}}^{t_{2}} \int_{\Omega}\nabla v\cdot\nabla v_{t}\,dx\,dt- \int _{t_{1}}^{t_{2}} \int_{\Omega}\nabla u\cdot\nabla u_{tt}\,dx\,dt \\ &{}- \int _{t_{1}}^{t_{2}} \int_{\Omega}\nabla v\cdot\nabla v_{tt}\,dx\,dt+ \int_{t_{1}}^{t_{2}}\|\nabla u_{t} \|_{2}^{2}\,dt \\ &{} + \int _{t_{1}}^{t_{2}}\|\nabla v_{t} \|_{2}^{2}\,dt+ \int _{t_{1}}^{t_{2}}g_{1}\circ\nabla u\,dt+ \int_{t_{1}}^{t_{2}}g_{2}\circ \nabla v\, dt \\ &{} + \int_{t_{1}}^{t_{2}} \int_{\Omega} \int_{0}^{t}g_{1}(t-s)\nabla u(t)\cdot \bigl(\nabla u(s)-\nabla u(t)\bigr)\,ds\,dx\,dt \\ &{} + \int_{t_{1}}^{t_{2}} \int_{\Omega} \int_{0}^{t}g_{2}(t-s)\nabla v(t)\cdot \bigl(\nabla v(s)-\nabla v(t)\bigr)\,ds\,dx\,dt \\ &{} - \int_{t_{1}}^{t_{2}} \int_{\Omega}|u_{t}|^{p-1}u_{t}u\,dx \,dt- \int _{t_{1}}^{t_{2}} \int_{\Omega}|v_{t}|^{q-1}v_{t}v\,dx \,dt. \end{aligned}$$
(3.17)
Integrating by parts and applying the Cauchy-Schwarz inequality in the first term of the right-hand side of (3.17), we obtain
$$\begin{aligned}& \biggl\vert \int_{t_{1}}^{t_{2}} \int_{\Omega}uu_{tt}\,dx\,dt\biggr\vert \leq\sum _{i=1}^{2}\bigl\Vert u_{t}(t_{i}) \bigr\Vert _{2}\bigl\Vert u(t_{i})\bigr\Vert _{2}+ \int_{t_{1}}^{t_{2}}\bigl\Vert u_{t}(t)\bigr\Vert _{2}^{2}\,dt, \end{aligned}$$
(3.18)
$$\begin{aligned}& \biggl\vert \int_{t_{1}}^{t_{2}} \int_{\Omega}vv_{tt}\,dx\,dt\biggr\vert \leq\sum _{i=1}^{2}\bigl\Vert v_{t}(t_{i}) \bigr\Vert _{2}\bigl\Vert v(t_{i})\bigr\Vert _{2}+ \int_{t_{1}}^{t_{2}}\bigl\Vert v_{t}(t)\bigr\Vert _{2}^{2}\,dt, \end{aligned}$$
(3.19)
$$\begin{aligned}& \biggl\vert \int_{t_{1}}^{t_{2}} \int_{\Omega}\nabla u\cdot\nabla u_{tt}\,dx\,dt\biggr\vert \leq\sum_{i=1}^{2}\bigl\Vert \nabla u_{t}(t_{i})\bigr\Vert _{2}\bigl\Vert \nabla u(t_{i})\bigr\Vert _{2}+ \int_{t_{1}}^{t_{2}}\bigl\Vert \nabla u_{t}(t) \bigr\Vert _{2}^{2}\,dt, \end{aligned}$$
(3.20)
and
$$ \biggl\vert \int_{t_{1}}^{t_{2}} \int_{\Omega}\nabla v\cdot\nabla v_{tt}\,dx\,dt\biggr\vert \leq\sum_{i=1}^{2}\bigl\Vert \nabla v_{t}(t_{i})\bigr\Vert _{2}\bigl\Vert \nabla v(t_{i})\bigr\Vert _{2}+ \int_{t_{1}}^{t_{2}}\bigl\Vert \nabla v_{t}(t) \bigr\Vert _{2}^{2}\,dt. $$
(3.21)
Now, we estimate the third term of the right-hand side of inequality (3.17). By the Cauchy-Schwarz inequality we have
$$ \biggl\vert \int_{t_{1}}^{t_{2}} \int_{\Omega}\nabla u\cdot\nabla u_{t}\,dx\,dt \biggr\vert \leq \int_{t_{1}}^{t_{2}}\|\nabla u\|_{2}\Vert \nabla u_{t}\Vert _{2}\,dt $$
(3.22)
and
$$ \biggl\vert \int_{t_{1}}^{t_{2}} \int_{\Omega}\nabla v\cdot\nabla v_{t}\,dx\,dt \biggr\vert \leq \int_{t_{1}}^{t_{2}}\|\nabla v\|_{2}\|\nabla v_{t}\|_{2}\,dt. $$
(3.23)
Since
$$\begin{aligned}& \int_{\Omega} \int_{0}^{t}g_{1}(t-s)\nabla u(t)\cdot \bigl(\nabla u(s)-\nabla u(t)\bigr)\,ds\,dx \\& \quad = \frac{1}{2} \biggl[ \int_{0}^{t}g_{1}(t-s) \bigl(\bigl\Vert \nabla u(t)\bigr\Vert _{2}^{2}+\bigl\Vert \nabla u(s) \bigr\Vert _{2}^{2}\bigr)\,ds- \int_{0}^{t}g_{1}(t-s) \bigl(\bigl\Vert \nabla u(t)-\nabla u(s)\bigr\Vert _{2}^{2}\bigr)\,ds \biggr] \\& \qquad {} - \int_{\Omega} \int_{0}^{t}g_{1}(s)\bigl\vert \nabla u(t)\bigr\vert ^{2}\,ds\,dx \\& \quad = -\frac{1}{2} \int_{\Omega} \int_{0}^{t}g_{1}(s)\bigl\vert \nabla u(t)\bigr\vert ^{2}\,ds\,dx+\frac{1}{2} \int_{0}^{t}g_{1}(t-s)\bigl\Vert \nabla u(s)\bigr\Vert _{2}^{2}\,ds-\frac{1}{2}(g_{1} \circ\nabla u) \\& \quad \leq\frac{1}{2} \int_{0}^{t}g_{1}(t-s)\bigl\Vert \nabla u(s)\bigr\Vert _{2}^{2}\,ds-\frac {1}{2}(g_{1} \circ\nabla u) \end{aligned}$$
(3.24)
and
$$\begin{aligned}& \int_{\Omega} \int_{0}^{t}g_{2}(t-s)\nabla v(t)\cdot \bigl(\nabla v(s)-\nabla v(t)\bigr)\,ds\,dx \\& \quad \leq\frac{1}{2} \int_{0}^{t}g_{2}(t-s)\bigl\Vert \nabla v(s)\bigr\Vert _{2}^{2}\,ds-\frac {1}{2}(g_{2} \circ\nabla v), \end{aligned}$$
(3.25)
by (3.18)-(3.25) we have
$$\begin{aligned} \int_{t_{1}}^{t_{2}}I(t)\,dt \leq& \sum _{i=1}^{2}\bigl\Vert u_{t}(t_{i}) \bigr\Vert _{2}\bigl\Vert u(t_{i})\bigr\Vert _{2}+\sum_{i=1}^{2}\bigl\Vert v_{t}(t_{i})\bigr\Vert _{2}\bigl\Vert v(t_{i})\bigr\Vert _{2} \\ &{} + \int_{t_{1}}^{t_{2}}\bigl(\bigl\Vert u_{t}(t) \bigr\Vert _{2}^{2}+\bigl\Vert v_{t}(t)\bigr\Vert _{2}^{2}\bigr)\,dt+ \int_{t_{1}}^{t_{2}}\Vert \nabla u\Vert _{2} \Vert \nabla u_{t}\Vert _{2}\,dt \\ &{} + \int_{t_{1}}^{t_{2}}\Vert \nabla v\Vert _{2} \Vert \nabla v_{t}\Vert _{2}\,dt+\sum _{i=1}^{2}\bigl\Vert \nabla u_{t}(t_{i}) \bigr\Vert _{2}\bigl\Vert \nabla u(t_{i})\bigr\Vert _{2} \\ &{} +\sum_{i=1}^{2}\bigl\Vert \nabla v_{t}(t_{i})\bigr\Vert _{2}\bigl\Vert \nabla v(t_{i})\bigr\Vert _{2}+2 \int_{t_{1}}^{t_{2}}\bigl(\Vert \nabla u_{t} \Vert _{2}^{2}+\Vert \nabla v_{t}\Vert _{2}^{2}\bigr)\,dt \\ &{} +\frac{1}{2} \int_{t_{1}}^{t_{2}} \int_{0}^{t}g_{1}(t-s)\bigl\Vert \nabla u(s)\bigr\Vert _{2}^{2}\,ds\,dt+\frac{1}{2} \int_{t_{1}}^{t_{2}} \int_{0}^{t}g_{2}(t-s)\bigl\Vert \nabla v(s)\bigr\Vert _{2}^{2}\,ds\,dt \\ &{} +\frac{1}{2} \int_{t_{1}}^{t_{2}}g_{1}\circ\nabla u\,dt+ \frac{1}{2} \int _{t_{1}}^{t_{2}}g_{2}\circ\nabla v\,dt- \int_{t_{1}}^{t_{2}} \int_{\Omega }|u_{t}|^{p-1}u_{t}u\,dx \,dt \\ &{} - \int_{t_{1}}^{t_{2}} \int_{\Omega}|v_{t}|^{q-1}v_{t}v\,dx \,dt. \end{aligned}$$
(3.26)
Now, we will estimate the right-hand side of (3.26). First, by (3.16), (2.1), and (3.8), letting \(\beta=\min\{l,k\}\), we have
$$\begin{aligned} \bigl\Vert u_{t}(t_{i})\bigr\Vert _{2}\bigl\Vert u(t_{i})\bigr\Vert _{2} \leq&c_{\ast}\sqrt { 4c_{1}(\Omega )D_{1}^{2}(t)+ 4c_{2}(\Omega)D_{2}^{2}(t)}\sup _{t_{1}\leq s\leq t_{2}}\bigl\Vert \nabla u(s)\bigr\Vert _{2} \\ \leq&c_{\ast} \biggl(\frac{2(m+1)}{\beta(m-1)} \biggr)^{\frac{1}{2}}\sqrt { 4c_{1}(\Omega)D_{1}^{2}(t)+ 4c_{2}(\Omega)D_{2}^{2}(t)}\sup _{t_{1}\leq s\leq t_{2}}E^{\frac{1}{2}}(s) \end{aligned}$$
(3.27)
and
$$ \bigl\Vert v_{t}(t_{i})\bigr\Vert _{2}\bigl\Vert v(t_{i})\bigr\Vert _{2}\leq{c_{\ast} \biggl(\frac {2(m+1)}{\beta(m-1)} \biggr)^{\frac{1}{2}}\sqrt{ 4c_{1}( \Omega )D_{1}^{2}(t)+ 4c_{2}(\Omega)D_{2}^{2}(t)} \sup_{t_{1}\leq s\leq t_{2}}E^{\frac{1}{2}}(s)}. $$
(3.28)
By the Hölder inequality and (3.13) we find
$$ \int_{t_{1}}^{t_{2}}\|\nabla u_{t} \|_{2}\,dt\leq \biggl( \int _{t_{1}}^{t_{2}}1^{2}\,dt \biggr)^{\frac{1}{2}} \biggl( \int_{t_{1}}^{t_{2}}\| \nabla u_{t} \|_{2}^{2}\,dt \biggr)^{\frac{1}{2}}\leq{D^{\frac{p+1}{2}}(t)}. $$
(3.29)
Then we have
$$ \int_{t_{1}}^{t_{2}}\|\nabla u\|_{2}\|\nabla u_{t}\|_{2}\,dt\leq \biggl(\frac {2(m+1)}{\beta(m-1)} \biggr)^{\frac{1}{2}}D_{1}^{\frac{p+1}{2}}(t)\sup_{t_{1}\leq s\leq t_{2}}E^{\frac{1}{2}}(s), $$
(3.30)
and similarly we obtain
$$ \int_{t_{1}}^{t_{2}}\|\nabla v\|_{2}\|\nabla v_{t}\|_{2}\,dt\leq \biggl(\frac {2(m+1)}{\beta(m-1)} \biggr)^{\frac{1}{2}}D_{2}^{\frac{q+1}{2}}(t)\sup_{t_{1}\leq s\leq t_{2}}E^{\frac{1}{2}}(s). $$
(3.31)
By (3.13) and (3.14) we observe that
$$\int_{t}^{t+1}\Vert \nabla u_{t}\Vert _{2}^{2}\,ds+ \int_{t}^{t+1}\Vert \nabla v_{t}\Vert _{2}^{2}\,ds\leq D_{1}^{p+1}(t)+D_{2}^{q+1}(t). $$
By the mean value theorem there exist \(t_{1}\in[t,t+\frac{1}{4}]\) and \(t_{2}\in[t+\frac{3}{4},t+1]\) such that
$$ \bigl\Vert \nabla u_{t}(t_{i})\bigr\Vert _{2}^{2}+\bigl\Vert \nabla v_{t}(t_{i}) \bigr\Vert _{2}^{2}\leq 4D_{1}^{p+1}(t)+4D_{2}^{q+1}(t). $$
(3.32)
From (3.8) and (3.32) we have
$$ \bigl\Vert \nabla u_{t}(t_{i})\bigr\Vert _{2} \bigl\Vert \nabla u(t_{i})\bigr\Vert _{2}\leq \biggl( \frac {2(m+1)}{\beta(m-1)} \biggr)^{\frac{1}{2}}\sqrt {4D_{1}^{p+1}(t)+4D_{2}^{q+1}(t)} \sup_{t_{1}\leq s\leq t_{2}}E^{\frac {1}{2}}(s) $$
(3.33)
and
$$ \bigl\Vert \nabla v_{t}(t_{i})\bigr\Vert _{2} \bigl\Vert \nabla v(t_{i})\bigr\Vert _{2}\leq \biggl( \frac {2(m+1)}{\beta(m-1)} \biggr)^{\frac{1}{2}}\sqrt {4D_{1}^{p+1}(t)+4D_{2}^{q+1}(t)} \sup_{t_{1}\leq s\leq t_{2}}E^{\frac {1}{2}}(s). $$
(3.34)
Applying Young’s inequality to convolution \(\|\phi\ast\psi\|_{q}\leq\| \phi\|_{r}\|\psi\|_{s}\) with
$${\frac{1}{q}=\frac{1}{r}+\frac{1}{s}-1},\quad {1\leq q,r,s \leq\infty,} $$
and noting that if \(q=1\), then \(r=1\) and \(s=1\), we get
$$\begin{aligned} \int_{t_{1}}^{t_{2}} \int_{0}^{t}g_{1}(t-s)\bigl\Vert \nabla u(s)\bigr\Vert _{2}^{2}\,ds\,dt&\leq \int_{t_{1}}^{t_{2}}g_{1}(t)\,dt \int_{t_{1}}^{t_{2}}\bigl\Vert \nabla u(t)\bigr\Vert _{2}^{2}\,dt \\ &\leq(\gamma-\beta) \int_{t_{1}}^{t_{2}}\bigl\Vert \nabla u(t)\bigr\Vert _{2}^{2}\,dt \end{aligned}$$
(3.35)
and
$$ \int_{t_{1}}^{t_{2}} \int_{0}^{t}g_{2}(t-s)\bigl\Vert \nabla v(s)\bigr\Vert _{2}^{2}\,ds\,dt \leq(\delta-\beta) \int_{t_{1}}^{t_{2}}\bigl\Vert \nabla v(t)\bigr\Vert _{2}^{2}\,dt. $$
(3.36)
From (3.1), (3.9), (3.10), (3.35), (3.36), and (A1) we have
$$\begin{aligned}& \frac{1}{2} \biggl( \int_{t_{1}}^{t_{2}} \int_{0}^{t}g_{1}(t-s)\bigl\Vert \nabla u(s)\bigr\Vert _{2}^{2}\,ds\,dt+ \int_{t_{1}}^{t_{2}} \int_{0}^{t}g_{2}(t-s)\bigl\Vert \nabla v(s)\bigr\Vert _{2}^{2}\,ds\,dt \biggr) \\& \quad \leq\frac{\gamma+\delta-\beta}{2\beta} \int_{t_{1}}^{t_{2}}\bigl(l\Vert \nabla u\Vert _{2}^{2} +k\Vert \nabla v\Vert _{2}^{2} \bigr)\,dt \\& \quad \leq\frac{\gamma+\delta-\beta}{2\beta} \int_{t_{1}}^{t_{2}} \biggl(I(t)+(m+1) \int_{\Omega}F(u,v)\, dx \biggr)\,dt \\& \quad \leq\frac{\gamma+\delta-\beta}{2\beta}(1+\eta_{1}) \int _{t_{1}}^{t_{2}}I(t)\,dt. \end{aligned}$$
(3.37)
To estimate the eleventh and twelfth terms on the right-hand side of (3.26), we use (3.37) to obtain
$$\begin{aligned} \frac{1}{2} \int_{t_{1}}^{t_{2}} (g_{1}\circ\nabla u+g_{2}\circ \nabla v )\,dt =& \frac{1}{2} \int_{t_{1}}^{t_{2}} \int _{0}^{t}g_{1}(t-s) \bigl(\bigl\Vert \nabla u(s)-\nabla u(t)\bigr\Vert _{2}^{2}\bigr)\,ds \,dt \\ &{} +\frac{1}{2} \int_{t_{1}}^{t_{2}} \int_{0}^{t}g_{2}(t-s) \bigl(\bigl\Vert \nabla v(s)-\nabla v(t)\bigr\Vert _{2}^{2}\bigr)\,ds\,dt \\ \leq& \int_{t_{1}}^{t_{2}} \int_{0}^{t}g_{1}(t-s) \bigl(\bigl\Vert \nabla u(s)\bigr\Vert _{2}^{2}+\bigl\Vert \nabla u(t) \bigr\Vert _{2}^{2}\bigr)\,ds\,dt \\ &{} + \int_{t_{1}}^{t_{2}} \int_{0}^{t}g_{2}(t-s) \bigl(\bigl\Vert \nabla v(s)\bigr\Vert _{2}^{2}+\bigl\Vert \nabla v(t)\bigr\Vert _{2}^{2}\bigr)\,ds\,dt \\ \leq&\frac{2(\gamma+\delta-\beta)}{\beta}(1+\eta_{1}) \int _{t_{1}}^{t_{2}}I(t)\,dt. \end{aligned}$$
(3.38)
Using Hölder inequality, (2.1), (3.8), and (3.13), we have
$$\begin{aligned} \biggl\vert \int_{t_{1}}^{t_{2}} \int_{\Omega}|u_{t}|^{p-1}u_{t}u\,dx \,dt\biggr\vert &\leq \int_{t_{1}}^{t_{2}}\|u_{t}\|^{p}_{p+1} \|u\|_{p+1}\,dt \\ &\leq c_{\ast} \int_{t_{1}}^{t_{2}}\|u_{t}\|^{p}_{p+1} \|\nabla u\| _{2}\,dt \\ &\leq c_{\ast} \biggl(\frac{2(m+1)}{\beta(m-1)} \biggr)^{\frac{1}{2}} \int _{t_{1}}^{t_{2}}\|u_{t} \|^{p}_{p+1}\,dt\sup_{t_{1}\leq s\leq t_{2}}E^{\frac{1}{2}}(s) \\ &\leq c_{\ast} \biggl(\frac{2(m+1)}{\beta(m-1)} \biggr)^{\frac {1}{2}}D_{1}^{p}(t) \sup_{t_{1}\leq s\leq t_{2}}E^{\frac{1}{2}}(s) \end{aligned}$$
(3.39)
and
$$ \biggl\vert \int_{t_{1}}^{t_{2}} \int_{\Omega}|v_{t}|^{q-1}v_{t}v\,dx \,dt\biggr\vert \leq{c_{\ast} \biggl(\frac{2(m+1)}{\beta(m-1)} \biggr)^{\frac {1}{2}}D_{2}^{q}(t)\sup_{t_{1}\leq s\leq t_{2}}E^{\frac{1}{2}}(s)}. $$
(3.40)
Therefore, applying (3.13)-(3.15), (3.27)-(3.31), (3.33)-(3.34), and (3.37)-(3.40) to (3.26) we obtain
$$\begin{aligned} \beta_{2} \int_{t_{1}}^{t_{2}}I(t)\,dt \leq& 4c_{\ast} \beta_{1}\sqrt{ 4c_{1}(\Omega)D_{1}^{2}(t)+ 4c_{2}(\Omega)D_{2}^{2}(t)}\sup _{t_{1}\leq s\leq t_{2}}E^{\frac{1}{2}}(s) \\ &{} +c_{1}(\Omega)D_{1}^{2}(t)+ c_{2}(\Omega)D_{2}^{2}(t)+\beta_{1} \bigl(D_{1}^{\frac{p+1}{2}}(t)+D_{2}^{\frac{q+1}{2}}(t) \bigr) \sup_{t_{1}\leq s\leq t_{2}}E^{\frac{1}{2}}(s) \\ &{} +2 \bigl(D_{1}^{p+1}(t)+D_{2}^{q+1}(t) \bigr)+4\beta_{1}\sqrt {4D_{1}^{p+1}(t)+4D_{2}^{q+1}(t)} \sup_{t_{1}\leq s\leq t_{2}}E^{\frac {1}{2}}(s) \\ &{}+ c_{\ast}\beta_{1} \bigl(D_{1}^{p}(t)+D_{2}^{q}(t) \bigr)\sup_{t_{1}\leq s\leq t_{2}}E^{\frac{1}{2}}(s), \end{aligned}$$
(3.41)
where
$$\beta_{1}:= { \biggl(\frac{2(m+1)}{\beta(m-1)} \biggr)^{\frac{1}{2}}}, \qquad \beta _{2}:= {1-\frac{5(\gamma+\delta-\beta)}{2\beta}(1+\eta_{1})}. $$
Note that the assumption
$${\gamma+\delta>\frac{7+5\eta_{1}}{2}\max \biggl\{ \int_{0}^{\infty }g_{1}(s)\,ds, \int_{0}^{\infty}g_{2}(s)\,ds \biggr\} } $$
gives \(\beta_{2}>0\). Thus,
$$\begin{aligned} \int_{t_{1}}^{t_{2}}I(t)\,dt \leq& c_{3} \Bigl[\sqrt{ 4c_{1}(\Omega )D_{1}^{2}(t)+ 4c_{2}(\Omega)D_{2}^{2}(t)}\sup _{t_{1}\leq s\leq t_{2}}E^{\frac{1}{2}}(s)+D_{1}^{2}(t)+D_{2}^{2}(t) \\ &{}+ \bigl(D_{1}^{\frac{p+1}{2}}(t)+D_{2}^{\frac{q+1}{2}}(t) \bigr)\sup_{t_{1}\leq s\leq t_{2}}E^{\frac {1}{2}}(s)+D_{1}^{p+1}(t)+D_{2}^{q+1}(t) \\ &{} +\sqrt{4D_{1}^{p+1}(t)+4D_{2}^{q+1}(t)} \sup_{t_{1}\leq s\leq t_{2}}E^{\frac{1}{2}}(s) \\ &{}+ \bigl(D_{1}^{p}(t)+D_{2}^{q}(t) \bigr)\sup_{t_{1}\leq s\leq t_{2}}E^{\frac{1}{2}}(s) \Bigr], \end{aligned}$$
(3.42)
where
$$c_{3}:=\frac{\max\{4c_{\ast}\beta_{1},c_{1}(\Omega),c_{2}(\Omega ),4\beta_{1}\}}{\beta_{2}}. $$
On the other hand, from the definition of \(J(t)\) and \(I(t)\), (3.9), and (3.10) we have
$$\begin{aligned} \begin{aligned}[b] J(t)={}& \frac{m-1}{2(m+1)} \biggl(\|\Delta u\|_{2}^{2}+\| \Delta v\| _{2}^{2}+\|\nabla u_{t} \|_{2}^{2} +\|\nabla v_{t}\|_{2}^{2}+ \biggl(\gamma- \int_{0}^{t}g_{1}(s)\,ds\biggr)\|\nabla u \| _{2}^{2} \\ &{} +\biggl(\delta- \int_{0}^{t}g_{2}(s)\,ds\biggr)\|\nabla v \|_{2}^{2}+g_{1}\circ \nabla u+g_{2}\circ \nabla v \biggr)+\frac{1}{m+1}I(t) \\ ={}& \frac{m-1}{2(m+1)} \biggl(I(t)+(m+1) \int_{\Omega}F(u,v)\,dx \biggr)+\frac {1}{m+1}I(t) \\ \leq{}& \biggl(1+\frac{m-1}{2(m+1)}\eta_{1} \biggr)I(t) \\ :={}& c_{4}I(t). \end{aligned} \end{aligned}$$
(3.43)
Hence, integrating (3.3) over \((t_{1},t_{2})\) and then using (3.43), (3.42), and (3.15), we deduce that
$$\begin{aligned} \int_{t_{1}}^{t_{2}}E(t)\,dt =& \frac{1}{2} \int_{t_{1}}^{t_{2}}\bigl(\| u_{t} \|_{2}^{2}+\|v_{t}\|_{2}^{2} \bigr)\,dt+ \int_{t_{1}}^{t_{2}}J(t)\,dt \\ \leq&\frac{1}{2} \int_{t_{1}}^{t_{2}}\bigl(\|u_{t} \|_{2}^{2}+\|v_{t}\| _{2}^{2} \bigr)\,dt+c_{4} \int_{t_{1}}^{t_{2}}I(t)\,dt \\ \leq&c_{5} \Bigl[D_{1}^{2}(t)+ D_{2}^{2}(t)+\sqrt{ 4c_{1}(\Omega )D_{1}^{2}(t)+ 4c_{2}(\Omega)D_{2}^{2}(t)} \sup_{t_{1}\leq s\leq t_{2}}E^{\frac{1}{2}}(s) \\ &{} + \bigl(D_{1}^{\frac{p+1}{2}}(t)+D_{2}^{\frac{q+1}{2}}(t) \bigr)\sup_{t_{1}\leq s\leq t_{2}}E^{\frac {1}{2}}(s)+D_{1}^{p+1}(t)+D_{2}^{q+1}(t) \\ &{} +\sqrt{4D_{1}^{p+1}(t)+4D_{2}^{q+1}(t)} \sup_{t_{1}\leq s\leq t_{2}}E^{\frac{1}{2}}(s) \\ &{}+ \bigl(D_{1}^{p}(t)+D_{2}^{q}(t) \bigr)\sup_{t_{1}\leq s\leq t_{2}}E^{\frac{1}{2}}(s) \Bigr], \end{aligned}$$
(3.44)
where \(c_{5}:=c_{3}c_{4}\). By integrating (3.4) over \([t,t_{2}]\) we obtain
$$\begin{aligned} E(t) =& E(t_{2})+ \int_{t}^{t_{2}} \biggl(\|\nabla u_{t} \|_{2}^{2}+\|\nabla v_{t}\|_{2}^{2}- \frac{1}{2}\bigl(g_{1}'\circ\nabla u\bigr)- \frac{1}{2}\bigl(g_{2}'\circ \nabla v\bigr)+ \frac{1}{2}g_{1}(s)\|\nabla u\|_{2}^{2} \\ &{} +\frac{1}{2}g_{2}(s)\|\nabla v\|_{2}^{2}+ \|u_{t}\|_{p+1}^{p+1}+\| v_{t} \|_{q+1}^{q+1} \biggr)\,ds. \end{aligned}$$
(3.45)
Since \(t_{2}-t_{1}\geq\frac{1}{2}\) and \(E(t)\) is nonincreasing in t, it follows that
$$\int_{t_{1}}^{t_{2}}E(t)\,dt\geq \int_{t_{1}}^{t_{2}}E(t_{2})\,dt\geq \frac{1}{2}E(t_{2}). $$
Then, we have
$$\begin{aligned} E(t) \leq&2 \int_{t_{1}}^{t_{2}}E(t)\,dt+ \int_{t}^{t_{2}} \biggl(\|\nabla u_{t} \|_{2}^{2}+\|\nabla v_{t}\|_{2}^{2} \\ &{}- \frac{1}{2}\bigl(g_{1}'\circ\nabla u\bigr)- \frac{1}{2}\bigl(g_{2}'\circ\nabla v\bigr)+ \frac{1}{2}g_{1}(s)\|\nabla u\| _{2}^{2} \\ &{} +\frac{1}{2}g_{2}(s)\|\nabla v\|_{2}^{2}+ \|u_{t}\|_{p+1}^{p+1}+\| v_{t} \|_{q+1}^{q+1} \biggr)\,ds \\ \leq&2 \int_{t_{1}}^{t_{2}}E(t)\,dt+D_{1}^{p+1}(t)+D_{2}^{q+1}(t). \end{aligned}$$
(3.46)
Consequently, since \(E(t)\) is nonincreasing, combining (3.44) with (3.46), we obtain
$$\begin{aligned} E(t) \leq& c_{6} \Bigl(D_{1}^{2}(t)+ D_{2}^{2}(t)+\sqrt{ 4c_{1}(\Omega )D_{1}^{2}(t)+ 4c_{2}(\Omega)D_{2}^{2}(t)}E^{\frac{1}{2}}(t) \\ &{} +\bigl(D_{1}^{\frac{p+1}{2}}(t)+D_{2}^{\frac{q+1}{2}}(t) \bigr)E^{\frac {1}{2}}(t)+D_{1}^{p+1}(t)+D_{2}^{q+1}(t) \\ &{} +\sqrt{4D_{1}^{p+1}(t)+4D_{2}^{q+1}(t)}E^{\frac {1}{2}}(t)+ \bigl(D_{1}^{p}(t)+D_{2}^{q}(t) \bigr)E^{\frac{1}{2}}(t) \Bigr). \end{aligned}$$
Then a simple application of Young’s inequality gives, for \(t\geq0\),
$$ E(t)\leq{c_{7} \bigl[D_{1}^{2}(t)+ D_{2}^{2}(t)+D_{1}^{p+1}(t)+D_{2}^{q+1}(t)+D_{1}^{2p}(t)+D_{2}^{2q}(t) \bigr]}, $$
(3.47)
where \(c_{6}\) and \(c_{7}\) are positive constants.
Therefore, we have the following decay estimate.
(i) For \(p=q=1\), from (3.47) and (3.12) we get
$$E(t)\leq c_{8}\bigl[E(t)-E(t+1)\bigr], $$
where we choose \(c_{8}>1\). Thus, by Lemma 2.6 we obtain
$$E(t)\leq E(0)e^{-\tau_{1}t},\quad t\geq0, $$
where \(\tau_{1}:=\ln\frac{c_{8}}{c_{8}-1}\).
(ii) For \(\max\{p, q\}>1\), it follows from (3.47) that, for \(t\geq0\),
$$E(t)\leq c_{7} \bigl[ \bigl(1+D_{1}^{p-1}(t)+D_{1}^{2p-2}(t) \bigr)D_{1}^{2}(t)+ \bigl(1+D_{2}^{q-1}(t)+D_{2}^{2q-2}(t) \bigr)D_{2}^{2}(t) \bigr]. $$
Since \(D_{1}(t)\leq E^{\frac{1}{p+1}}(t)\leq E^{\frac{1}{p+1}}(0)\) and \(D_{2}(t)\leq E^{\frac{1}{q+1}}(t)\leq E^{\frac{1}{q+1}}(0)\) by (3.12) and (3.4), we have, for \(t\geq0\),
$$\begin{aligned} E(t)&\leq c_{7} \bigl[ \bigl(1+E^{\frac{p-1}{p+1}}(0)+E^{\frac {2p-2}{p+1}}(0) \bigr)D_{1}^{2}(t)+ \bigl(1+E^{\frac{q-1}{q+1}}(0)+E^{\frac {2q-2}{q+1}}(0) \bigr)D_{2}^{2}(t) \bigr] \\ &\leq c_{9} \bigl(D_{1}^{2}(t)+D_{2}^{2}(t) \bigr). \end{aligned}$$
Then, setting \(\rho:=\max \{\frac{p-1}{2},\frac{q-1}{2} \}\), we obtain
$$\begin{aligned} E^{1+\rho}(t)&\leq \bigl[c_{9} \bigl(D_{1}^{2}(t)+D_{2}^{2}(t) \bigr) \bigr]^{1+\rho} \\ &\leq c_{10} \bigl(D_{1}^{2\rho+2}(t)+D_{2}^{2\rho+2}(t) \bigr) \\ &= c_{10} \bigl(D_{1}^{2\rho-p+1}(t)D_{1}^{p+1}(t)+D_{2}^{2\rho -q+1}(t)D_{2}^{q+1}(t) \bigr) \\ &\leq c_{10} \bigl(E^{\frac{2\rho-p+1}{p+1}}(0)D_{1}^{p+1}(t)+E^{\frac {2\rho-q+1}{q+1}}(0)D_{2}^{q+1}(t) \bigr) \\ &\leq c_{11} \bigl(D_{1}^{p+1}(t)+D_{2}^{q+1}(t) \bigr) \\ &= c_{11}\bigl(E(t)-E(t+1)\bigr), \end{aligned}$$
(3.48)
where \(c_{10}:=2^{\rho}\cdot c_{9}^{1+\rho}\) and \(c_{11}:=c_{10}\max (E^{\frac{2\rho-p+1}{p+1}}(0),E^{\frac{2\rho-q+1}{q+1}}(0) )\). Application of Lemma 2.6 to (3.48) yields
$$E(t)\leq\bigl(E^{-\rho}(0)+\tau_{2}\rho[t-1]^{+} \bigr)^{-\frac{1}{\rho}},\quad t\geq0, $$
with \(\tau_{2}:=c_{11}^{-1}\).
The proof of Theorem 3.4 is completed. □