In this section, we establish three related lemmas before proving our result.
Lemma 5.1
If (G1)–(G3) and (3.15) hold, the functional
\(\phi (t)\)
defined by
$$ \phi (t):= \int _{\varOmega }u_{t}u\,dx+ \int _{\varOmega }v_{t}v\,dx $$
satisfies, along solutions of (1.1),
$$\begin{aligned} \phi '(t) &\leq -\frac{l}{4} \Vert \nabla u \Vert _{2}^{2}+\frac{1-l}{2l}(g_{1} \circ \nabla u) (t)+\biggl(1+\frac{1}{4\beta _{1}}\biggr) \Vert u_{t} \Vert _{2}^{2} \\ &\quad {} -\frac{k}{4} \Vert \nabla v \Vert _{2}^{2}+ \frac{1-k}{2k}(g_{2}\circ \nabla v) (t)+\biggl(1+ \frac{1}{4\beta _{2}}\biggr) \Vert v_{t} \Vert _{2}^{2} \\ &\quad {} +2(p+2) \int _{\varOmega }F(u,v)\,dx. \end{aligned}$$
(5.1)
Proof
A differentiation of \(\phi (t)\) with respect to t, it follows from system (1.1) that
$$\begin{aligned} \phi '(t) &= \int _{\varOmega }u_{tt}u\,dx+ \Vert u_{t} \Vert _{2}^{2}+ \int _{\varOmega }v _{tt}v\,dx+ \Vert v_{t} \Vert _{2}^{2} \\ &= \int _{\varOmega }\nabla u(t) \int _{0}^{t}g_{1}(t-\tau )\nabla u(\tau )d \tau \,dx+ \Vert u_{t} \Vert _{2}^{2}- \Vert \nabla u \Vert _{2}^{2}- \int _{\varOmega }uu_{t}\,dx \\ &\quad {} + \int _{\varOmega }\nabla v(t) \int _{0}^{t}g_{2}(t-\tau )\nabla v( \tau )\,d\tau \,dx+ \Vert v_{t} \Vert _{2}^{2}- \Vert \nabla v \Vert _{2}^{2}- \int _{\varOmega }vv _{t}\,dx \\ &\quad {} + \int _{\varOmega }uf_{1}(u,v)\,dx+ \int _{\varOmega }vf_{2}(u,v)\,dx. \end{aligned}$$
(5.2)
For the first term on the right-hand side of (5.2), by using Young’s inequality and the fact that \(\int _{0}^{t}g_{1}(s)\,ds\leq \int _{0}^{\infty }g_{1}(s)\,ds =1-l\), for \(\eta =\frac{l}{1-l}>0\), we get
$$\begin{aligned}& \int _{\varOmega }\nabla u(t) \int _{0}^{t}g_{1}(t-\tau )\nabla u(\tau )d \tau \,dx \\& \quad \leq \frac{1}{2} \Vert \nabla u \Vert _{2}^{2}+ \frac{1}{2} \int _{\varOmega } \biggl( \int _{0}^{t}g_{1}(t-\tau ) \bigl( \bigl\vert \nabla u(\tau )-\nabla u(t) \bigr\vert + \bigl\vert \nabla u(t) \bigr\vert \bigr)\,d\tau \biggr)^{2}\,dx \\& \quad \leq \frac{1}{2} \Vert \nabla u \Vert _{2}^{2}+ \frac{1}{2}(1+\eta ) \int _{\varOmega } \biggl( \int _{0}^{t}g_{1}(t-\tau ) \bigl\vert \nabla u(t) \bigr\vert \,d\tau \biggr)^{2}\,dx \\& \qquad {} +\frac{1}{2}\biggl(1+\frac{1}{\eta }\biggr) \int _{\varOmega } \biggl( \int _{0}^{t}g_{1}(t- \tau ) \bigl\vert \nabla u(\tau )-\nabla u(t) \bigr\vert \,d\tau \biggr)^{2} \,dx \\& \quad \leq \frac{1}{2} \Vert \nabla u \Vert _{2}^{2}+ \frac{1+\eta }{2}(1-l)^{2} \Vert \nabla u \Vert _{2}^{2}+\frac{1}{2l} \int _{\varOmega } \biggl( \int _{0}^{t}g_{1}(t- \tau ) \bigl\vert \nabla u(\tau )-\nabla u(t) \bigr\vert \,d\tau \biggr)^{2} \,dx \\& \quad \leq \frac{2-l}{2} \Vert \nabla u \Vert _{2}^{2}+ \frac{1-l}{2l}(g_{1} \circ \nabla u) (t). \end{aligned}$$
(5.3)
Similar calculations also yield, for \(\eta _{1}=\frac{k}{1-k}>0\),
$$\begin{aligned} \int _{\varOmega }\nabla v(t) \int _{0}^{t}g_{2}(t-\tau )\nabla v(\tau )d \tau \,dx\leq \frac{2-k}{2} \Vert \nabla v \Vert _{2}^{2}+\frac{1-k}{2k}(g_{2} \circ \nabla v) (t). \end{aligned}$$
(5.4)
Applying Young’s inequality and Poincare’s inequality, for some \(\beta _{1}>0\), we obtain
$$\begin{aligned} \int _{\varOmega }uu_{t}\,dx &\leq \beta _{1} \Vert u \Vert _{2}^{2}+ \frac{1}{4\beta _{1}} \Vert u_{t} \Vert _{2}^{2} \\ &\leq \beta _{1} C_{*}^{2} \Vert \nabla u \Vert _{2}^{2}+\frac{1}{4\beta _{1}} \Vert u _{t} \Vert _{2}^{2}. \end{aligned}$$
(5.5)
Likewise, for some \(\beta _{2}>0\), we have
$$\begin{aligned} \int _{\varOmega }vv_{t}\,dx &\leq \beta _{2} \Vert v \Vert _{2}^{2}+ \frac{1}{4\beta _{2}} \Vert v_{t} \Vert _{2}^{2} \\ &\leq \beta _{2} C_{*}^{2} \Vert \nabla v \Vert _{2}^{2}+\frac{1}{4\beta _{2}} \Vert v _{t} \Vert _{2}^{2}. \end{aligned}$$
(5.6)
Inserting (5.3)–(5.6) into (5.2) yields
$$\begin{aligned} \phi '(t) &\leq - \biggl(\frac{l}{2}-\beta _{1}C_{*}^{2} \biggr) \Vert \nabla u \Vert _{2} ^{2}+\frac{1-l}{2l}(g_{1} \circ \nabla u) (t) \\ &\quad {} +\biggl(1+\frac{1}{4\beta _{1}}\biggr) \Vert u_{t} \Vert _{2}^{2}- \biggl(\frac{k}{2}- \beta _{2}C_{*}^{2} \biggr) \Vert \nabla v \Vert _{2}^{2} \\ &\quad {} +\frac{1-k}{2k}(g_{2}\circ \nabla v) (t)+\biggl(1+ \frac{1}{4\beta _{2}}\biggr) \Vert v_{t} \Vert _{2}^{2}+2(p+2) \int _{\varOmega }F(u,v)\,dx. \end{aligned}$$
(5.7)
Now, we pick \(\beta _{1}, \beta _{2}>0\) small enough such that
$$\begin{aligned} \frac{l}{2}-\beta _{1} C_{*}^{2} \geq \frac{l}{4}, \quad\quad \frac{k}{2}-\beta _{2} C_{*}^{2}\geq \frac{k}{4}. \end{aligned}$$
(5.8)
Finally, a combination of (5.7) and (5.8) gives (5.1). □
Lemma 5.2
Suppose that (G1)–(G3) and (3.15) hold. The functional
\(\psi _{1}(t)\)
defined by
$$ \psi _{1}(t):=- \int _{\varOmega }u_{t} \int _{0}^{t}g_{1}(t-\tau ) \bigl(u(t)-u( \tau )\bigr)\,d\tau \,dx $$
satisfies, along solutions of (1.1),
$$\begin{aligned} \psi _{1}'(t) &\leq \bigl[\delta +2(1-l)^{2}\delta +\alpha _{2}\delta l \bigr] \Vert \nabla u \Vert _{2}^{2}-\frac{g_{1}(0)C_{*}^{2}}{4\delta } \bigl(g_{1}' \circ \nabla u\bigr) (t) \\ &\quad {} + \biggl(2\delta +\frac{1}{2\delta }+\frac{C_{*}^{2}}{2\delta } \biggr) (1-l) (g_{1}\circ \nabla u) (t) \\ &\quad {} +\alpha _{2}\delta k \Vert \nabla v \Vert _{2}^{2}+ \biggl(2\delta - \int _{0}^{t}g_{1}(\tau )\,d\tau \biggr) \Vert u_{t} \Vert _{2}^{2}, \end{aligned}$$
(5.9)
where
\(\alpha _{2}=\lambda _{1} (\frac{2(p+2)}{p+1}E(0) )^{2p+2}\), \(\lambda _{1}\)
is the constant in Lemma 4.2.
Proof
Taking the derivative of \(\psi _{1}(t)\) with respect to t and using system (1.1) gives
$$\begin{aligned} \psi _{1}'(t) &=- \int _{\varOmega }u_{tt} \int _{0}^{t}g_{1}(t-\tau ) \bigl(u(t)-u( \tau )\bigr)\,d\tau \,dx \\ &\quad {} - \int _{\varOmega }u_{t} \int _{0}^{t}g_{1}'(t- \tau ) \bigl(u(t)-u(\tau )\bigr)d \tau \,dx- \int _{0}^{t}g_{1}(\tau )\,d\tau \Vert u_{t} \Vert _{2}^{2} \\ &= \int _{\varOmega }\nabla u(t) \int _{0}^{t}g_{1}(t-\tau ) \bigl( \nabla u(t)- \nabla u(\tau )\bigr)\,d\tau \,dx \\ &\quad {} - \int _{\varOmega } \biggl( \int _{0}^{t}g_{1}(t-\tau )\nabla u(\tau )d \tau \biggr) \biggl( \int _{0}^{t}g_{1}(t-\tau ) \bigl( \nabla u(t)-\nabla u(\tau )\bigr)d \tau \biggr)\,dx \\ &\quad {} + \int _{\varOmega }u_{t} \int _{0}^{t}g_{1}(t-\tau ) \bigl(u(t)-u(\tau )\bigr)d \tau \,dx \\ &\quad {} - \int _{\varOmega }f_{1}(u,v) \int _{0}^{t}g_{1}(t-\tau ) \bigl(u(t)-u( \tau )\bigr)\,d\tau \,dx \\ &\quad {} - \int _{\varOmega }u_{t} \int _{0}^{t}g'_{1}(t- \tau ) \bigl(\nabla u(t)- \nabla u(\tau )\bigr)\,d\tau \,dx- \int _{0}^{t}g_{1}(\tau )\,d\tau \Vert u_{t} \Vert _{2} ^{2}. \end{aligned}$$
(5.10)
For the first term on the right-hand side of (5.10), by exploiting (G1), Young’s inequality, and Cauchy–Schwarz inequality, for any \(\delta >0\), we get
$$\begin{aligned}& \int _{\varOmega }\nabla u(t) \int _{0}^{t}g_{1}(t-\tau ) \bigl( \nabla u(t)- \nabla u(\tau )\bigr)\,d\tau \,dx \\& \quad \leq \delta \Vert \nabla u \Vert _{2}^{2}+ \frac{1}{4\delta } \int _{ \varOmega } \biggl( \int _{0}^{t}g_{1}(t-\tau ) \bigl\vert \nabla u(\tau )-\nabla u(t) \bigr\vert d \tau \biggr)^{2} \,dx \\& \quad \leq \delta \Vert \nabla u \Vert _{2}^{2}+ \frac{1}{4\delta } \int _{0} ^{t}g_{1}(\tau )\,d\tau \int _{0}^{t}g_{1}(t-\tau ) \int _{\varOmega } \bigl\vert \nabla u(\tau )-\nabla u(t) \bigr\vert ^{2}\,dx\,d\tau \\& \quad \leq \delta \Vert \nabla u \Vert _{2}^{2}+ \frac{1-l}{4\delta }(g_{1} \circ \nabla u) (t). \end{aligned}$$
(5.11)
As for the second term in (5.10), recall that \((a+b)^{2}\leq 2(a ^{2}+b^{2})\), for \(\eta _{2}=1\), we obtain
$$\begin{aligned}& \int _{\varOmega } \biggl( \int _{0}^{t}g_{1}(t-\tau )\nabla u(\tau )\,d\tau \biggr) \biggl( \int _{0}^{t}g_{1}(t-\tau ) \bigl( \nabla u(t)-\nabla u(\tau )\bigr)d \tau \biggr)\,dx \\& \quad \leq \delta \int _{\varOmega } \biggl\vert \int _{0}^{t}g_{1}(t-\tau ) \nabla u(\tau )\,d\tau \biggr\vert ^{2}\,dx \\& \qquad {} +\frac{1}{4\delta } \int _{\varOmega } \biggl\vert \int _{0}^{t}g_{1}(t-\tau ) \bigl( \nabla u(t)-\nabla u(\tau )\bigr)\,d\tau \biggr\vert ^{2}\,dx \\& \quad \leq \delta \int _{\varOmega } \biggl( \int _{0}^{t}g_{1}(t-\tau ) \bigl( \bigl\vert \nabla u(\tau )-\nabla u(t) \bigr\vert + \bigl\vert \nabla u(t) \bigr\vert \bigr)\,d\tau \biggr)^{2}\,dx \\& \qquad {} +\frac{1}{4\delta } \int _{\varOmega } \biggl\vert \int _{0}^{t}g_{1}(t-\tau ) \bigl( \nabla u(t)-\nabla u(\tau )\bigr)\,d\tau \biggr\vert ^{2}\,dx \\& \quad \leq \delta (1+\eta _{2}) \int _{\varOmega } \biggl( \int _{0}^{t}g_{1}(t- \tau ) \bigl\vert \nabla u(t) \bigr\vert \,d\tau \biggr)^{2}\,dx \\& \qquad {} +\delta \biggl(1+\frac{1}{\eta _{2}}\biggr) \int _{\varOmega } \biggl( \int _{0}^{t}g_{1}(t- \tau ) \bigl\vert \nabla u(\tau )-\nabla u(t) \bigr\vert \,d\tau \biggr)^{2} \,dx \\& \qquad {} +\frac{1}{4\delta } \int _{\varOmega } \biggl\vert \int _{0}^{t}g_{1}(t-\tau ) \bigl( \nabla u(t)-\nabla u(\tau )\bigr)\,d\tau \biggr\vert ^{2}\,dx \\& \quad \leq 2\delta \int _{\varOmega } \biggl\vert \int _{0}^{t}g_{1}(t-\tau ) \bigl( \nabla u(t)-\nabla u(\tau )\bigr)\,d\tau \biggr\vert ^{2}\,dx \\& \qquad {} +2\delta (1-l)^{2} \Vert \nabla u \Vert _{2}^{2}+\frac{1}{4\delta } \biggl( \int _{0}^{t}g_{1}(\tau )\,d\tau \biggr) (g_{1}\circ \nabla u) (t) \\& \quad \leq 2\delta (1-l)^{2} \Vert \nabla u \Vert _{2}^{2}+\biggl(2\delta +\frac{1}{4 \delta }\biggr) (1-l) (g_{1}\circ \nabla u) (t). \end{aligned}$$
(5.12)
The third term can be handled by
$$\begin{aligned}& \int _{\varOmega }u_{t} \int _{0}^{t}g_{1}(t-\tau ) \bigl(u(t)-u(\tau )\bigr)\,d\tau \,dx \\& \quad \leq \delta \Vert u_{t} \Vert _{2}^{2}+ \frac{1}{4\delta } \int _{\varOmega } \biggl( \int _{0}^{t}g_{1}(t-\tau ) \bigl\vert u(\tau )-u(t) \bigr\vert \,d\tau \biggr)^{2}\,dx \\& \quad \leq \delta \Vert u_{t} \Vert _{2}^{2}+ \frac{C_{*}^{2}}{4\delta } \int _{0}^{t}g_{1}(\tau )\,d\tau (g_{1}\circ \nabla u) (t) \\& \quad \leq \delta \Vert u_{t} \Vert _{2}^{2}+ \frac{C_{*}^{2}(1-l)}{4\delta }(g _{1}\circ \nabla u) (t). \end{aligned}$$
(5.13)
For the forth term, it follows from (4.9) that
$$\begin{aligned}& \int _{\varOmega }f_{1}(u,v) \int _{0}^{t}g_{1}(t-\tau ) \bigl(u(t)-u(\tau )\bigr)d \tau \,dx \\& \quad \leq \lambda _{1}\delta \bigl(l \Vert \nabla u \Vert _{2}^{2}+k \Vert \nabla v \Vert _{2} ^{2} \bigr)^{2p+3}+\frac{1}{4\delta } \int _{\varOmega } \biggl( \int _{0}^{t}g _{1}(t-\tau ) \bigl\vert u(\tau )-u(t) \bigr\vert \,d\tau \biggr)^{2}\,dx \\& \quad \leq \lambda _{1}\delta \biggl(\frac{2(p+2)}{p+1}E(0) \biggr)^{2p+2} \bigl(l \Vert \nabla u \Vert _{2}^{2}+k \Vert \nabla v \Vert _{2}^{2} \bigr)+ \frac{C_{*}^{2}(1-l)}{4 \delta }(g_{1}\circ \nabla u) (t) \\& \quad =\alpha _{2}\delta \bigl(l \Vert \nabla u \Vert _{2}^{2}+k \Vert \nabla v \Vert _{2} ^{2} \bigr)+\frac{C_{*}^{2}(1-l)}{4\delta }(g_{1}\circ \nabla u) (t), \end{aligned}$$
(5.14)
where \(\alpha _{2}=\lambda _{1} (\frac{2(p+2)}{p+1}E(0) )^{2p+2}\).
The fifth term on the right-hand side of (5.10) can be estimated as
$$\begin{aligned}& \int _{\varOmega }u_{t} \int _{0}^{t}g_{1}'(t- \tau ) \bigl(u(t)-u(\tau )\bigr)d \tau \,dx \\& \quad \leq \delta \Vert u_{t} \Vert _{2}^{2}+ \frac{1}{4\delta } \int _{\varOmega } \biggl( \int _{0}^{t}g_{1}'(t- \tau ) \bigl\vert u(t)-u(\tau ) \bigr\vert \,d\tau \biggr)^{2} \,dx \\& \quad \leq \delta \Vert u_{t} \Vert _{2}^{2}- \frac{g_{1}(0)C_{*}^{2}}{4 \delta }\bigl(g_{1}'\circ \nabla u\bigr) (t). \end{aligned}$$
(5.15)
Taking into account estimates (5.11)–(5.13), estimate (5.9) is established. □
Similar computations also yield the following.
Lemma 5.3
Suppose that (G1)–(G3) and (3.15) hold. The functional
\(\psi _{2}(t)\)
defined by
$$ \psi _{2}(t):=-v_{t} \int _{0}^{t}g_{2}(t-\tau ) \bigl(v(t)-v(\tau )\bigr)\,d\tau \,dx $$
satisfies, along solutions of system (1.1),
$$\begin{aligned} \psi _{2}'(t) &\leq \bigl[\delta +2(1-k)^{2}\delta +\alpha _{3}\delta k \bigr] \Vert \nabla v \Vert _{2}^{2}-\frac{g_{2}(0)C_{*}^{2}}{4\delta } \bigl(g_{2}' \circ \nabla v\bigr) (t) \\ &\quad {} + \biggl(2\delta +\frac{1}{2\delta }+\frac{C_{*}^{2}}{2\delta } \biggr) (1-k) (g_{2}\circ \nabla v) (t) \\ &\quad {} +\alpha _{3}\delta l \Vert \nabla u \Vert _{2}^{2}+ \biggl(2\delta - \int _{0}^{t}g_{2}(\tau )\,d\tau \biggr) \Vert v_{t} \Vert _{2}^{2}, \end{aligned}$$
(5.16)
where
\(\alpha _{3}=\lambda _{2} (\frac{2(p+2)}{p+1}E(0) )^{2p+2}\), \(\lambda _{2}\)
is the constant in Lemma 4.2.
Now, we define the functional
$$\begin{aligned} W(t)=E(t)+\epsilon _{1}\phi (t)+\epsilon _{2}\psi (t), \end{aligned}$$
(5.17)
where \(\epsilon _{1}\) and \(\epsilon _{2}\) are positive constants, \(\phi (t)\) is given in Lemma 5.1 and \(\psi (t):=\psi _{1}(t)+\psi _{2}(t)\).
Lemma 5.4
([5])
Let
\((u,v)\)
be the solution of system (1.1) and assume that (3.15) holds. Then there exist constant
\(\epsilon >0\)
small enough and
\(M>0\)
large enough such that the following relation
$$\begin{aligned} \beta _{1}W(t)\leq E(t)\leq \beta _{2}W(t), \quad \forall t\geq 0, \end{aligned}$$
(5.18)
holds for two positive constants
\(\beta _{1}\)
and
\(\beta _{2}\).
Theorem 5.5
Assume (G1)–(G3) and (3.15) hold. Let
\((u_{0},v_{0}) \in H_{0}^{1}(\varOmega )\times H_{0}^{1}(\varOmega )\)
and
\((u_{1},v_{1}) \in L^{2}(\varOmega )\times L^{2}(\varOmega )\). Then, for each
\(t_{0}>0\), there exist positive constants
K, k, \(k_{1}\), \(k_{2}\)
such that the solution of system (1.1) satisfies, for all
\(t\geq t_{0}\),
$$\begin{aligned}& E(t)\leq Ke^{-k\int _{t_{0}}^{t}\zeta (\tau )\,d\tau }, \quad \gamma =1, \end{aligned}$$
(5.19)
$$\begin{aligned}& E(t)\leq k_{1} \biggl[\frac{1}{1+\int _{t_{0}}^{t}\zeta ^{2\gamma -1}( \tau )\,d\tau } \biggr]^{\frac{1}{2\gamma -2}}, \quad 1< \gamma < \frac{3}{2}. \end{aligned}$$
(5.20)
Furthermore, if
$$ \int _{0}^{+\infty } \biggl[\frac{1}{1+t\zeta ^{2\gamma -1}(t)} \biggr]^{\frac{1}{2 \gamma -2}}\,dt< +\infty , \quad 1< \gamma < \frac{3}{2}, $$
(5.21)
then
$$ E(t)\leq k_{2} \biggl[\frac{1}{1+\int _{t_{0}}^{t}\zeta ^{\gamma }(\tau )d \tau } \biggr]^{\frac{1}{\gamma -1}}, \quad 1< \gamma < \frac{3}{2}, $$
(5.22)
where
\(\zeta (t)=\min \{\zeta _{1}(t),\zeta _{2}(t) \}\).
Proof
From (G1), we know \(g_{1}\) and \(g_{2}\) are positive, then for any \(t\geq t_{0}>0\), we have
$$\begin{aligned}& \int _{0}^{t}g_{1}(\tau )\,d\tau \geq \int _{0}^{t_{0}}g_{1}(\tau )d \tau =g_{0}, \\& \int _{0}^{t}g_{2}(\tau )\,d\tau \geq \int _{0}^{t_{0}}g_{2}(\tau )d \tau =h_{0}. \end{aligned}$$
Taking derivative of (5.17) with respect to t and using (3.6), (5.1), and (5.16) yields
$$\begin{aligned} W'(t) &=E'(t)+\epsilon _{1}\phi '(t)+\epsilon _{2}\psi '(t) \\ &\leq \biggl(\frac{1}{2}-\frac{g_{1}(0)C_{*}^{2}}{4\delta }\epsilon _{2} \biggr) \bigl(g_{1}'\circ \nabla u \bigr) (t)+ \biggl(\frac{1}{2}-\frac{g_{2}(0)C _{*}^{2}}{4\delta }\epsilon _{2} \biggr) \bigl(g_{2}'\circ \nabla v \bigr) (t) \\ &\quad {} - \biggl[\epsilon _{2}(g_{0}-2\delta )- \epsilon _{1}\biggl(1+\frac{1}{4 \beta _{1}}\biggr) \biggr] \Vert u_{t} \Vert _{2}^{2} \\ &\quad {} - \biggl[\epsilon _{2}(h_{0}-2\delta )- \epsilon _{1}\biggl(1+\frac{1}{4 \beta _{2}}\biggr) \biggr] \Vert v_{t} \Vert _{2}^{2}+2(p+2)\epsilon _{1} \int _{\varOmega }F(u,v)\,dx \\ &\quad {} - \biggl[\frac{l\epsilon _{1}}{4}-\epsilon _{2} \bigl(1+2(1-l)^{2}+ \alpha _{2}l+\alpha _{3}l \bigr)\delta \biggr] \Vert \nabla u \Vert _{2}^{2} \\ &\quad {} - \biggl[\frac{k\epsilon _{2}}{4}-\epsilon _{2} \bigl(1+2(1-k)^{2}+ \alpha _{2}k+\alpha _{3}k \bigr)\delta \biggr] \Vert \nabla v \Vert _{2}^{2} \\ &\quad {} + \biggl[\frac{\epsilon _{1}}{2l}+ \biggl(2\delta +\frac{1}{2\delta }+ \frac{C_{*}^{2}}{2\delta } \biggr)\epsilon _{2} \biggr](1-l) (g_{1}\circ \nabla u) (t) \\ &\quad {} + \biggl[\frac{\epsilon _{2}}{2k}+ \biggl(2\delta +\frac{1}{2\delta }+ \frac{C_{*}^{2}}{2\delta } \biggr)\epsilon _{2} \biggr](1-k) (g_{2}\circ \nabla v) (t). \end{aligned}$$
(5.23)
At this point, we pick \(\delta >0\) small enough such that
$$\begin{aligned} \delta \leq \min \biggl\{ \frac{g_{0}}{2}, \frac{h_{0}}{2} \biggr\} \end{aligned}$$
and
$$\begin{aligned}& \frac{4}{l}\delta \bigl(1+2(1-l)^{2}+\alpha _{2}l+\alpha _{3}l \bigr)< \frac{g _{0}}{4(1+\frac{1}{4\beta _{1}})}, \\& \frac{4}{k}\delta \bigl(1+2(1-k)^{2}+\alpha _{2}k+\alpha _{3}k \bigr)< \frac{g _{0}}{4(1+\frac{1}{4\beta _{2}})}. \end{aligned}$$
As long as δ is fixed, the choice of any two positive constants \(\epsilon _{1}\) and \(\epsilon _{2}\) satisfying
$$\begin{aligned} \frac{g_{0}}{4(1+\frac{1}{4\beta _{1}})}\epsilon _{2}< \epsilon _{1}< \frac{g _{0}}{2(1+\frac{1}{4\beta _{1}})}\epsilon _{2} \end{aligned}$$
and
$$\begin{aligned} \frac{h_{0}}{4(1+\frac{1}{4\beta _{2}})}\epsilon _{2}< \epsilon _{1}< \frac{h _{0}}{2(1+\frac{1}{4\beta _{2}})}\epsilon _{2} \end{aligned}$$
will make
$$\begin{aligned}& k_{1}=\epsilon _{2}(g_{0}-2\delta )- \epsilon _{1}\biggl(1+ \frac{1}{4\beta _{1}}\biggr)>0, \\& k_{2}=\epsilon _{2}(h_{0}-2\delta )- \epsilon _{1}\biggl(1+ \frac{1}{4\beta _{2}}\biggr)>0, \\& k_{3}=\frac{l\epsilon _{1}}{4}-\epsilon _{2} \bigl(1+2(1-l)^{2}+\alpha _{2}l+\alpha _{3}l \bigr)\delta >0, \\& k_{4}=\frac{k\epsilon _{2}}{4}-\epsilon _{2} \bigl(1+2(1-k)^{2}+\alpha _{2}k+\alpha _{3}k \bigr)\delta >0. \end{aligned}$$
Hence, there exist two positive constants m and C such that
$$\begin{aligned} W'(t)\leq -mE(t)+C \bigl[(g_{1}\circ \nabla u) (t)+(g_{2}\circ \nabla v) (t) \bigr],\quad t\geq t_{0}. \end{aligned}$$
(5.24)
Case 1. \(\gamma =1\):
Let \(\zeta (t)=\min \{\zeta _{1}(t),\zeta _{2}(t) \}\), since \(\zeta _{1}(t)\) and \(\zeta _{2}(t)\) are nonincreasing differentiable functions, then we get that \(\zeta (t)\) is nonincreasing.
When \(\gamma =1\), from (2.2), we easily get
$$\begin{aligned} g_{i}'(t)\leq -\zeta _{i}(t)g_{i}^{\gamma }(t)=- \zeta _{i}(t)g_{i}(t) \quad \text{for }i=1,2. \end{aligned}$$
(5.25)
Multiplying both sides of (5.24) by \(\zeta (t)\) and applying (3.6), (4.27), (4.29), and (5.25) yields
$$\begin{aligned} \zeta (t)W'(t) &\leq -m\zeta (t)E(t)+C\zeta (t) (g_{1}\circ \nabla u) (t)+C \zeta (t) (g_{2}\circ \nabla v) (t) \\ &\leq -m\zeta (t)E(t)+C\zeta _{1}(t) (g_{1}\circ \nabla u) (t)+C\zeta _{2}(t) (g_{2}\circ \nabla v) (t) \\ &\leq -m\zeta (t)E(t)+C(\zeta _{1}g_{1}\circ \nabla u) (t)+C(\zeta _{2}g _{2}\circ \nabla v) (t) \\ &\leq -m\zeta (t)E(t)-C\bigl(g_{1}'\circ \nabla u \bigr) (t)-C\bigl(g_{2}'\circ \nabla v\bigr) (t) \\ &\leq -m\zeta (t)E(t)-CE'(t). \end{aligned}$$
(5.26)
Setting \(F(t)=\zeta (t)W(t)+CE(t)\), then clearly \(F\sim E\). Recalling that \(W\sim E\geq 0\) by (3.21) and (5.18), \(\zeta '(t) \leq 0\) by (G2), we get that \(\zeta '(t)W(t)\leq 0\), then together with (5.26), we have, for some \(k>0\),
$$\begin{aligned} F'(t) &=\zeta '(t)W(t)+\zeta (t)W'(t)+CE'(t) \\ &\leq \zeta (t)W'(t)+CE'(t) \\ &\leq -m\zeta (t)E(t)-CE'(t)+CE'(t) \\ &\leq -k\zeta (t)F(t). \end{aligned}$$
(5.27)
Integrating (5.27) over \([t_{0},t]\) gives
$$\begin{aligned} F(t)\leq F(t_{0})e^{-k\int _{t_{0}}^{t}\zeta (\tau )\,d\tau }, \quad \forall t \geq t_{0}. \end{aligned}$$
(5.28)
Therefore, by using the fact that \(F(t)\sim E(t)\), we derive
$$ E(t)\leq Ke^{-k\int _{t_{0}}^{t}\zeta (\tau )\,d\tau }, \quad \forall t\geq t _{0}. $$
(5.29)
Case 2. \(1<\gamma <\frac{3}{2}\):
Multiplying both sides of (5.24) by \(\zeta (t)\), using (4.27) and (4.29) leads to
$$\begin{aligned} \zeta (t)W'(t) &\leq -m\zeta (t)E(t)+C\zeta (t) (g_{1}\circ \nabla u) (t)+C \zeta (t) (g_{2}\circ \nabla v) (t) \\ &\leq -m\zeta (t)E(t)+C\zeta _{1}(t) (g_{1}\circ \nabla u) (t)+C\zeta _{2}(t) (g_{2}\circ \nabla v) (t) \\ &\leq -m\zeta (t)E(t)+C\bigl[-E'(t)\bigr]^{\frac{1}{2\gamma -1}}. \end{aligned}$$
(5.30)
Multiplying (5.30) by \(\zeta ^{\alpha }(t)E^{\alpha }(t)\), with \(\alpha =2\gamma -2\), we get
$$\begin{aligned} \zeta ^{\alpha +1}(t)E^{\alpha }(t)W'(t)\leq -m\zeta ^{\alpha +1}(t)E ^{\alpha +1}(t)+C(\zeta E)^{\alpha }(t) \bigl[-E'(t)\bigr]^{\frac{1}{\alpha +1}}. \end{aligned}$$
(5.31)
Exploiting Young’s inequality with \(q=1+\alpha \) and \(q'=\frac{1+ \alpha }{\alpha }\) gives
$$\begin{aligned} \zeta ^{\alpha +1}(t)E^{\alpha }(t)W'(t) &\leq -m \zeta ^{\alpha +1}(t)E ^{\alpha +1}(t)+C \bigl[\epsilon \zeta ^{\alpha +1}(t)E^{\alpha +1}(t)-C _{\epsilon }E'(t) \bigr] \\ &\leq -(m-\epsilon C)\zeta ^{\alpha +1}(t)E^{\alpha +1}(t)-CE'(t). \end{aligned}$$
(5.32)
We pick \(\epsilon <\frac{m}{C}\) and recall that \(\zeta '(t)\leq 0\), \(\zeta (t)>0\) by (G2), \(E'(t)\leq 0\) by (3.6), and \(W\thicksim E\geq 0\) by (3.21) and (5.18), then together with (5.32) we have, for some \(c_{1}>0\),
$$\begin{aligned} \bigl(\zeta ^{\alpha +1}E^{\alpha }W \bigr)'(t) &=( \alpha +1)\zeta ^{\alpha }(t)\zeta '(t)E^{\alpha }(t)W(t) \\ &\quad {} +\alpha \zeta ^{\alpha +1}(t)E^{\alpha -1}(t)E'(t)W(t) \\ &\quad {} +\zeta ^{\alpha +1}(t)E^{\alpha }(t)W'(t) \\ &\leq \zeta ^{\alpha +1}(t)E^{\alpha }(t)W'(t) \\ &\leq -c_{1}\zeta ^{\alpha +1}(t)E^{\alpha +1}(t)-CE'(t). \end{aligned}$$
(5.33)
Next, we take \(F(t)=\zeta ^{\alpha +1}WE^{\alpha }+CE\), which is clearly equivalent to \(E(t)\), then there exists \(a_{0}>0\) such that
$$\begin{aligned} F'(t)\leq -c_{1}\zeta ^{\alpha +1}(t)E^{\alpha +1}(t) \leq -a_{0} \zeta ^{\alpha +1}(t)F^{\alpha +1}(t). \end{aligned}$$
(5.34)
Integrating (5.34) over \((t_{0},t)\) and recalling that \(F(t)\thicksim E(t)\) and \(\alpha =2\gamma -2\), we obtain
$$\begin{aligned} E(t)\leq k_{1} \biggl[\frac{1}{1+\int _{t_{0}}^{t}\zeta ^{2\gamma -1}( \tau )\,d\tau } \biggr]^{\frac{1}{2\gamma -2}}, \quad \forall t\geq t_{0}. \end{aligned}$$
(5.35)
From (5.21) and (5.35), we infer that
$$\begin{aligned} \int _{t_{0}}^{+\infty }E(t)\,dt< +\infty . \end{aligned}$$
(5.36)
Setting \(\lambda _{1}(t)=\int _{0}^{t} \Vert \nabla u(t)-\nabla u(t-\tau ) \Vert _{2} ^{2}\,d\tau \), by using (3.3), we deduce
$$\begin{aligned} \lambda _{1}(t)&= \int _{0}^{t} \bigl\Vert \nabla u(t)-\nabla u(t-\tau ) \bigr\Vert _{2}^{2}d \tau \leq C \int _{0}^{t} \bigl( \bigl\Vert \nabla u(t) \bigr\Vert _{2}^{2}+ \bigl\Vert \nabla u(t- \tau ) \bigr\Vert _{2}^{2} \bigr)\,d\tau \\ &\leq C \int _{0}^{t} \bigl[E(t)+E(t-\tau ) \bigr]\,d \tau \\ &\leq 2C \int _{0}^{t}E(t-\tau )\,d\tau \\ &=2C \int _{0}^{t}E(\tau )\,d\tau < 2C \int _{0}^{+\infty }E(\tau )\,d\tau < + \infty . \end{aligned}$$
(5.37)
Similarly, let \(\lambda _{2}(t)=\int _{0}^{t} \Vert \nabla v(t)-\nabla v(t- \tau ) \Vert _{2}^{2}\,d\tau \), we have \(\lambda _{2}(t)<+\infty \).
From (5.24) and recalling that \(\zeta (t)=\min \{\zeta _{1}(t), \zeta _{2}(t) \}\), \(\zeta _{1}(t)\) and \(\zeta _{2}(t)\) are nonincreasing differentiable functions, we arrive at
$$\begin{aligned} \zeta (t)W'(t) &\leq -m\zeta (t)E(t)+C\zeta (t) \bigl[(g_{1}\circ \nabla u) (t)+(g_{2}\circ \nabla v) (t) \bigr] \\ &\leq -m\zeta (t)E(t)+C\zeta _{1}(t) (g_{1}\circ \nabla u) (t)+C\zeta _{2}(t) (g_{2}\circ \nabla v) (t) \\ &\leq -m\zeta (t)E(t)+C\frac{\lambda _{1}(t)}{\lambda _{1}(t)} \int _{0} ^{t} \bigl(\zeta _{1}^{\gamma }( \tau )g_{1}^{\gamma }(\tau ) \bigr)^{\frac{1}{ \gamma }} \bigl\Vert \nabla u(t)-\nabla u(t-\tau ) \bigr\Vert _{2}^{2} \,d\tau \\ &\quad {} +C\frac{\lambda _{2}(t)}{\lambda _{2}(t)} \int _{0}^{t} \bigl(\zeta _{2}^{\gamma }( \tau )g_{2}^{\gamma }(\tau ) \bigr)^{\frac{1}{\gamma }} \bigl\Vert \nabla v(t)-\nabla v(t-\tau ) \bigr\Vert _{2}^{2} \,d\tau . \end{aligned}$$
(5.38)
Exploiting Jensen’s inequality, for the second term on the right-hand side of (5.38), with \(G(y)=y^{\frac{1}{\gamma }}\), \(y>0\), \(f(\tau )= \zeta ^{\gamma }(\tau )g^{\gamma }(\tau )\) and \(h(\tau )= \Vert \nabla u(t)- \nabla u(t-\tau ) \Vert _{2}^{2}\), where we assume that \(\lambda _{1}(t), \lambda _{2}(t)>0\), otherwise we get \(\Vert \nabla u(t)-\nabla u(t-\tau ) \Vert = \Vert \nabla v(t)-\nabla v(t-\tau ) \Vert =0\), then by using (5.24) we deduce
$$\begin{aligned} E(t)\leq Ce^{-mt}. \end{aligned}$$
Since \(\zeta _{1}(t)\) and \(\zeta _{2}(t)\) are nonincreasing, then for some \(C_{1}>0\), estimate (5.38) becomes
$$\begin{aligned} \zeta (t)W'(t) &\leq -m\zeta (t)E(t)+C\lambda _{1}(t) \biggl[\frac{1}{ \lambda _{1}(t)} \int _{0}^{t}\zeta _{1}^{\gamma }( \tau )g_{1}^{\gamma }( \tau ) \bigl\Vert \nabla u(t)-\nabla u(t-\tau ) \bigr\Vert _{2}^{2}\,d\tau \biggr]^{\frac{1}{ \gamma }} \\ &\quad {} +C\lambda _{2}(t) \biggl[\frac{1}{\lambda _{2}(t)} \int _{0}^{t}\zeta _{2}^{\gamma }( \tau )g_{2}^{\gamma }(\tau ) \bigl\Vert \nabla v(t)-\nabla v(t- \tau ) \bigr\Vert _{2}^{2}\,d\tau \biggr]^{\frac{1}{\gamma }} \\ &\leq -m\zeta (t)E(t)+C\lambda _{1}^{\frac{\gamma -1}{\gamma }}(t) \biggl[ \zeta _{1}^{\gamma -1}(0) \int _{0}^{t}\zeta _{1}(\tau )g_{1}^{ \gamma }(\tau ) \bigl\Vert \nabla u(t)-\nabla u(t- \tau ) \bigr\Vert _{2}^{2}\,d\tau \biggr]^{\frac{1}{ \gamma }} \\ &\quad {} +C\lambda _{2}^{\frac{\gamma -1}{\gamma }}(t) \biggl[\zeta _{2} ^{\gamma -1}(0) \int _{0}^{t}\zeta _{2}(\tau )g_{2}^{\gamma }(\tau ) \bigl\Vert \nabla v(t)-\nabla v(t- \tau ) \bigr\Vert _{2}^{2}\,d\tau \biggr]^{\frac{1}{\gamma }} \\ &=-m\zeta (t)E(t)+C \bigl(\lambda _{1}\zeta _{1}(0) \bigr)^{\frac{\gamma -1}{ \gamma }}\bigl(-g_{1}'\circ \nabla u \bigr)^{\frac{1}{\gamma }}+C \bigl(\lambda _{2}\zeta _{2}(0) \bigr)^{\frac{\gamma -1}{\gamma }}\bigl(-g_{2}' \circ \nabla v\bigr)^{\frac{1}{\gamma }} \\ &\leq -m\zeta (t)E(t)+C_{1}\bigl[-E'(t) \bigr]^{\frac{1}{\gamma }}. \end{aligned}$$
(5.39)
Multiplying both sides of (5.39) by \(\zeta ^{\alpha }(t)E^{\alpha }(t)\), with \(\alpha =\gamma -1\), we deduce
$$\begin{aligned} \zeta ^{\alpha +1}(t)E^{\alpha }(t)W'(t)\leq -m\zeta ^{\alpha +1}(t)E ^{\alpha +1}(t)+C_{1}\zeta ^{\alpha }(t)E^{\alpha }(t)\bigl[-E'(t) \bigr]^{\frac{1}{ \alpha +1}}. \end{aligned}$$
(5.40)
Applying Young’s inequality, with \(q=1+\alpha \), and \(q'=\frac{1+ \alpha }{\alpha }\) leads to
$$\begin{aligned} \zeta ^{\alpha +1}(t)E^{\alpha }(t)W'(t) &\leq -m \zeta ^{\alpha +1}(t)E ^{\alpha +1}(t)+C_{1} \bigl(\sigma \zeta ^{\alpha +1}(t)E^{\alpha +1}(t)-C _{\sigma }E'(t) \bigr) \\ &=-(m-C_{1}\sigma )\zeta ^{\alpha +1}(t)E^{\alpha +1}(t)-CE'(t). \end{aligned}$$
(5.41)
Then, by taking \(\sigma <\frac{m}{C_{1}}\) and recalling that \(\zeta '(t)\leq 0\), \(\zeta (t)>0\) by (G2), \(E'(t)\leq 0\) by (3.6), and \(W(t)\thicksim E(t)\geq 0\) by (3.21) and (5.18), together with (5.41), we get, for some \(C_{2}>0\),
$$\begin{aligned} \bigl(\zeta ^{\alpha +1}E^{\alpha }W \bigr)'(t) &=( \alpha +1)\zeta ^{\alpha }(t)\zeta '(t)E^{\alpha }(t)W(t) \\ &\quad {} +\alpha \zeta ^{\alpha +1}(t)E^{\alpha -1}(t)E'(t)W(t) \\ &\quad {} +\zeta ^{\alpha +1}(t)E^{\alpha }(t)W'(t) \\ &\leq \zeta ^{\alpha +1}(t)E^{\alpha }(t)W'(t) \\ &\leq -C_{2}\zeta ^{\alpha +1}(t)E^{\alpha +1}(t)-CE'(t), \end{aligned}$$
(5.42)
which implies
$$\begin{aligned} \bigl(\zeta ^{\alpha +1}E^{\alpha }W+CE \bigr)'(t) \leq -C_{2}\zeta ^{ \alpha +1}(t)E^{\alpha +1}(t). \end{aligned}$$
(5.43)
Let \(L=\zeta ^{\alpha +1}E^{\alpha }W+CE\), then clearly \(L\thicksim E\), we obtain, for some \(C_{3}>0\),
$$\begin{aligned} L'(t)\leq -C_{3}\zeta ^{\alpha +1}(t)L^{\alpha +1}(t), \quad \forall t \geq t_{0}. \end{aligned}$$
(5.44)
Integrating (5.44) over \((t_{0},t)\) and recalling that \(L\thicksim E\) and \(\alpha =\gamma -1\) yields
$$\begin{aligned} E(t)\leq k_{2} \biggl[\frac{1}{1+\int _{t_{0}}^{t}\zeta ^{\gamma }(\tau )d \tau } \biggr]^{\frac{1}{\gamma -1}}, \quad \forall t\geq t_{0}. \end{aligned}$$
(5.45)
This completes the proof. □