Taking the inner product with
\(\Delta^{2} v\) on both sides of (
3.2) in
D, we have
$$\begin{aligned} &\frac{1}{2}\frac{d}{dt}\|\Delta v\|^{2}+ \bigl(\alpha-z( \theta_{t}\omega) \bigr) \bigl\| \Delta v(t) \bigr\| ^{2}+\bigl\langle 2 \Delta v, \Delta^{2} v\bigr\rangle +\bigl\| \Delta^{2} v \bigr\| ^{2} \\ &\quad{}+\bigl\langle e^{2z(\theta _{t}\omega)}vG*v^{2}, \Delta^{2} v\bigr\rangle =0. \end{aligned}$$
(3.22)
Notice
$$\bigl|\bigl\langle 2\Delta v,\Delta^{2} v\bigr\rangle \bigr|\leq \frac{1}{2}\bigl\| \Delta^{2} v\bigr\| ^{2}+2\|\Delta v\| ^{2}. $$
In addition, we have
$$\begin{aligned} &\bigl\langle e^{2z(\theta_{t}\omega)}vG*v^{2}, \Delta^{2} v\bigr\rangle \\ &\quad=e^{2z(\theta_{t}\omega)} \int_{D} \bigl[\Delta vG*v^{2} +2\nabla v \nabla G*v^{2}+v\Delta G*v^{2} \bigr]\Delta v \,dx \\ &\quad\leq ae^{2z(\theta{_{t}}\omega)\|v\|^{2}\|\Delta v\|^{2}} +2\lambda _{1}^{-\frac{1}{2}}\|\nabla G \|_{L^{\infty}}e^{2z(\theta _{t}\omega)}\|v\|^{2}\|\Delta v\|^{2} \\ &\qquad{}+\lambda_{1}^{-1}\|\Delta G\|_{L^{\infty}}e^{2z(\theta _{t}\omega)} \|v\| ^{2}\|\Delta v\|^{2} \\ &\quad\leq ae^{2z(\theta{_{t}}\omega)\|v\|^{2}\|\Delta v\|^{2}} +2\lambda _{1}^{-\frac{1}{2}}K_{1} e^{2z(\theta_{t}\omega)}\|v\|^{2}\|\Delta v\|^{2} + \lambda_{1}^{-1}K_{2} e^{2z(\theta_{t}\omega)}\|v \|^{2}\|\Delta v\|^{2}. \end{aligned}$$
(3.23)
Then from (
3.22), we obtain
$$\begin{aligned} &\frac{d}{dt}\|\Delta v\|^{2}+2 \bigl( \varepsilon-z( \theta_{t}\omega) \bigr) \bigl\| \Delta v(t) \bigr\| ^{2} \\ &\quad\leq2 \bigl(a+2K_{1} \lambda_{1}^{-\frac{1}{2}}+K_{2} \lambda_{1}^{-1} \bigr)e^{2z(\theta_{t}\omega)}\|v\|^{2} \|\Delta v\|^{2}+2(2-\alpha+\varepsilon)\| \Delta v\|^{2} \\ &\quad\doteq2Ke^{2z(\theta_{t}\omega)}\|v\|^{2}\|\Delta v\|^{2}+2(2- \alpha+\varepsilon)\|\Delta v\|^{2}. \end{aligned}$$
(3.24)
Using Gronwall’s inequality, we have
$$\begin{aligned} & \bigl\| \Delta v \bigl(t+1,\omega,v_{0}(\omega) \bigr) \bigr\| ^{2} \\ &\quad\leq e^{-2\varepsilon(t+1-s)+\int_{s}^{t+1}2z(\theta_{\tau}\omega )\,d\tau }\bigl\| \Delta v \bigl(s,\omega,v_{0}( \omega) \bigr) \bigr\| ^{2} \\ &\qquad{}+2\int_{s}^{t+1} \bigl[Ke^{2z(\theta_{\tau}\omega)} \bigl\| v \bigl(\tau,\omega,v_{0}(\omega) \bigr) \bigr\| ^{2}+(2-\alpha+ \varepsilon) \bigr] \\ &\qquad{} \cdot e^{-2\varepsilon(t+1-\tau)+\int_{\tau}^{t+1}2z(\theta _{r}\omega )\,dr} \bigl\| \Delta v \bigl(\tau,\omega,v_{0}( \omega) \bigr) \bigr\| ^{2}\,d\tau, \end{aligned}$$
(3.25)
where
\(t\leq s\leq t+1\). In the above inequality (
3.25), taking the integral from
t to
\(t+1\) with respect to
s, we obtain
$$\begin{aligned} & \bigl\| \Delta v \bigl(t+1,\omega,v_{0}(\omega) \bigr) \bigr\| ^{2} \\ &\quad\leq\int_{t}^{t+1}e^{-2\varepsilon(t+1-s)+\int_{s}^{t+1}2z(\theta _{\tau }\omega)\,d\tau} \bigl\| \Delta v \bigl(s, \omega,v_{0}(\omega) \bigr) \bigr\| ^{2}\,ds \\ &\qquad{}+2\int_{t}^{t+1}\int_{s}^{t+1} \bigl[Ke^{2z(\theta_{\tau}\omega)} \bigl\| v \bigl(\tau,\omega ,v_{0}(\omega) \bigr) \bigr\| ^{2}+(2-\alpha+\varepsilon) \bigr] \\ &\qquad{} \cdot e^{-2\varepsilon(t+1-\tau)+\int_{\tau}^{t+1}2z(\theta _{r}\omega )\,dr} \bigl\| \Delta v \bigl(\tau,\omega,v_{0}( \omega) \bigr) \bigr\| ^{2}\,d\tau\,ds. \end{aligned}$$
(3.26)
In (
3.13), replacing
t by
τ and substituting it into the above inequality (
3.26), we obtain
$$\begin{aligned} & \bigl\| \Delta v \bigl(t+1,\omega,v_{0}(\omega) \bigr) \bigr\| ^{2} \\ &\quad\leq\int_{t}^{t+1}e^{-2\varepsilon(t+1-s)+\int _{s}^{t+1}2z(\theta_{\tau }\omega)\,d\tau} \bigl\| \Delta v \bigl(s, \omega,v_{0}(\omega) \bigr)\bigr\| ^{2}\,ds \\ &\qquad{}+2K \bigl\| v_{0}(\omega) \bigr\| ^{2}\int_{t}^{t+1} \int_{s}^{t+1}e^{2z(\theta_{\tau }\omega)-2\varepsilon(t+1)+\int_{0}^{t+1} 2z(\theta_{r}\omega)\,dr} \bigl\| \Delta v \bigl( \tau, \omega,v_{0}({\omega}) \bigr) \bigr\| ^{2}\,d\tau\,ds \\ &\qquad{} +\frac{2K(\alpha-2-\varepsilon)^{2}}{2b}\int_{t}^{t+1}\int _{s}^{t+1} \int_{0}^{\tau}e^{2z(\theta_{\tau}\omega)-2z(\theta_{r}\omega )-2\varepsilon(t+1-r)+\int_{r}^{t+1} 2z(\theta_{\xi}\omega)\,d\xi } \\ &\qquad{}\cdot\bigl\| \Delta v \bigl(\tau,\omega,v_{0}(\omega) \bigr) \bigr\| ^{2} \,dr \,d\tau\,ds \\ &\qquad{}+\int_{t}^{t+1}\int_{s}^{t+1} 2(2-\alpha+\varepsilon)e^{-2\varepsilon (t+1-\tau)+\int_{\tau}^{t+1}2z(\theta_{r}\omega)\,dr} \bigl\| \Delta v \bigl(\tau, \omega,v_{0}(\omega) \bigr) \bigr\| ^{2}\,d\tau\,ds. \end{aligned}$$
(3.27)
It is easy to obtain from the above inequality (
3.27) that
$$\begin{aligned} & \bigl\| \Delta v \bigl(t+1,\omega,v_{0}(\omega) \bigr) \bigr\| ^{2} \\ &\quad\leq(5-2\alpha+2\varepsilon)\int_{t}^{t+1}e^{-2\varepsilon (t+1-s)+\int _{s}^{t+1}2z(\theta_{\tau}\omega)\,d\tau} \bigl\| \Delta v \bigl(s,\omega,v_{0}(\omega) \bigr) \bigr\| ^{2}\,ds \\ &\qquad{}+2K \bigl\| v_{0}(\omega) \bigr\| ^{2}\int_{t}^{t+1} \int_{s}^{t+1}e^{2z(\theta_{\tau }\omega)-2\varepsilon(t+1)+\int_{0}^{t+1} 2z(\theta_{r}\omega)\,dr} \bigl\| \Delta v \bigl( \tau, \omega,v_{0}({\omega}) \bigr) \bigr\| ^{2}\,d\tau\,ds \\ &\qquad{} +\frac{2K(\alpha-2-\varepsilon)^{2}}{2b}\int_{t}^{t+1}\int _{s}^{t+1} \int_{0}^{\tau}e^{2z(\theta_{\tau}\omega)-2z(\theta_{r}\omega )-2\varepsilon(t+1-r)+\int_{r}^{t+1} 2z(\theta_{\xi}\omega)\,d\xi } \\ &\qquad{}\cdot\bigl\| \Delta v \bigl(\tau,\omega,v_{0}({\omega}) \bigr)\bigr\| ^{2} \,dr\,d \tau\,ds. \end{aligned}$$
(3.28)
In the above inequality (
3.28), replacing
ω by
\(\theta _{-t-1}\omega\), we obtain
$$\begin{aligned} & \bigl\| \Delta v \bigl(t+1,\theta_{-t-1} \omega,v_{0}( \theta_{-t-1}\omega) \bigr) \bigr\| ^{2} \\ & \quad\leq(5-2\alpha+2\varepsilon)\int_{t}^{t+1}e^{-2\varepsilon (t+1-s)+\int _{s}^{t+1}2z(\theta_{\tau-t-1}\omega)\,d\tau} \bigl\| \Delta v \bigl(s,\theta_{-t-1}\omega,v_{0}( \theta_{-t-1}\omega) \bigr) \bigr\| ^{2}\,ds \\ &\quad{}+2K\bigl\| v_{0}(\theta_{-t-1}\omega)\bigr\| ^{2}\int _{t}^{t+1}\int_{s}^{t+1}e^{2z(\theta_{\tau-t-1}\omega)-2\varepsilon (t+1)+\int_{0}^{t+1} 2z(\theta_{r-t-1}\omega)\,dr} \\ &\qquad{}\cdot\bigl\| \Delta v \bigl(\tau,\theta_{-t-1}\omega,v_{0}( \theta_{-t-1}\omega) \bigr) \bigr\| ^{2}\,d\tau\,ds \\ &\quad{} +\frac{2K(\alpha-2-\varepsilon)^{2}}{2b}\int_{t}^{t+1}\int _{s}^{t+1} \int_{0}^{\tau}e^{2z(\theta_{\tau-t-1}\omega)-2z(\theta _{r-t-1}\omega )-2\varepsilon(t+1-r)+\int_{r}^{t+1} 2z(\theta_{\xi-t-1}(\omega))\,d\xi} \\ & \qquad{} \cdot\bigl\| \Delta v \bigl(\tau,\theta_{-t-1}\omega,v_{0}( \theta_{-t-1}\omega) \bigr)\bigr\| ^{2} \,dr\,d\tau\,ds. \end{aligned}$$
(3.29)
Now, we estimate each term on the right side of the above inequality. For the first term, we have
$$\begin{aligned} &(5-2\alpha+2\varepsilon)\int_{t}^{t+1}e^{-2\varepsilon (t+1-s)+\int _{s}^{t+1}2z(\theta_{\tau-t-1}\omega)\,d\tau} \bigl\| \Delta v \bigl(s,\theta_{-t-1}\omega,v_{0}( \theta_{-t-1}\omega) \bigr) \bigr\| ^{2}\,ds \\ &\quad\leq(5-2\alpha+2\varepsilon)e^{2\max_{-1\leq\tau\leq0}|z(\theta _{\tau}\omega)} \int_{t}^{t+1} \bigl\| \Delta v \bigl(s,\theta_{-t-1}\omega,v_{0}( \theta_{-t-1}\omega) \bigr) \bigr\| ^{2}\,ds . \end{aligned}$$
(3.30)
For the second term, we estimate it as follows:
$$\begin{aligned} &2K \bigl\| v_{0}(\theta_{-t-1}\omega) \bigr\| ^{2}\int_{t}^{t+1}\int _{s}^{t+1}e^{2z(\theta_{\tau-t-1}\omega)-2\varepsilon(t+1)+\int_{0}^{t+1} 2z(\theta_{r-t-1}\omega)\,dr} \\ &\qquad{}\cdot\bigl\| \Delta v \bigl(\tau, \theta_{-t-1}\omega,v_{0}( \theta_{-t-1}\omega) \bigr) \bigr\| ^{2}\,d\tau\,ds \\ &\quad=2K \bigl\| v_{0}(\theta_{-t-1}\omega) \bigr\| ^{2}e^{-2(t+1)(\varepsilon-\frac{\int _{-t-1}^{0}2z(\theta_{\tau}\omega)\,d\tau}{t+1})} \int_{t}^{t+1}\int_{s}^{t+1}e^{2z(\theta_{\tau-t-1}\omega)} \\ &\qquad{}\cdot\bigl\| \Delta v \bigl(\tau,\theta_{-t-1}\omega,v_{0}( \theta_{-t-1}\omega) \bigr) \bigr\| ^{2}\,d\tau\,ds \\ &\quad\leq2K \bigl\| v_{0}(\theta_{-t-1}\omega) \bigr\| ^{2}e^{-2(t+1)(\varepsilon-\frac{\int _{-t-1}^{0}2z(\theta_{\tau}\omega)\,d\tau}{t+1})} e^{2\max_{-1\leq\tau \leq0} |z(\theta_{\tau}\omega)|} \\ &\qquad{}\cdot\int_{t}^{t+1} \bigl\| \Delta v \bigl(s, \theta_{-t-1}\omega,v_{0}( \theta_{-t-1}\omega) \bigr) \bigr\| ^{2}\,ds. \end{aligned}$$
(3.31)
Then, for the third term, we have the estimate
$$\begin{aligned} &\frac{2K(\alpha-2-\varepsilon)^{2}}{2b}\int_{t}^{t+1} \int_{s}^{t+1} \int_{0}^{\tau}e^{2z(\theta_{\tau-t-1}\omega)-2z(\theta _{r-t-1}\omega )-2\varepsilon(t+1-r)+\int_{r}^{t+1} 2z(\theta_{\xi-t-1}\omega)\,d\xi} \\ & \qquad{} \cdot\bigl\| \Delta v \bigl(\tau,\theta_{-t-1} \omega,v_{0}( \theta_{-t-1}\omega) \bigr)\bigr\| ^{2} \,dr \,d\tau\,ds \\ &\quad\leq\frac{2K(\alpha-2-\varepsilon)^{2}}{2b}e^{2\max_{-1\leq\tau\leq 0}|z(\theta_{\tau}\omega)|} \int_{t}^{t+1} \int_{-t-1}^{\tau-t-1} e^{-2z(\theta_{r}\omega)+2\varepsilon r+\int _{r}^{0} 2z(\theta_{\xi }\omega)\,d\xi} \\ & \qquad{} \cdot\bigl\| \Delta v \bigl(\tau,\theta_{-t-1} \omega,v_{0}( \theta_{-t-1}\omega) \bigr)\bigr\| ^{2} \,dr \,d\tau \\ &\quad\leq\frac{2K(\alpha-2-\varepsilon)^{2}}{2b}e^{2\max_{-1\leq\tau\leq 0}|z(\theta_{\tau}\omega)|} \int_{t}^{t+1} \int_{-t-1}^{0} e^{2r(\varepsilon-\frac{z(\theta_{r}\omega)}{r}-\frac {\int_{r}^{0} 2z(\theta_{\xi}\omega)\,d\xi}{-r})}\,dr \\ & \qquad{} \cdot\bigl\| \Delta v \bigl(\tau,\theta_{-t-1} \omega,v_{0}( \theta_{-t-1}\omega) \bigr)\bigr\| ^{2}\,d \tau \\ & \quad\leq\frac{2KK_{3}(\alpha-2-\varepsilon)^{2}}{2b}e^{2\max_{-1\leq\tau \leq0}|z(\theta_{\tau}\omega)|} \int_{t}^{t+1} \bigl\| \Delta v \bigl(\tau,\theta_{-t-1}\omega,v_{0}(\theta _{-t-1}\omega) \bigr)\bigr\| ^{2}\,d\tau. \end{aligned}$$
(3.32)
From the above estimates (
3.30), (
3.31), and (
3.32), we deduce that
$$\begin{aligned} \bigl\| \Delta v \bigl(t+1,\theta_{-t-1}\omega,v_{0}( \theta_{-t-1}\omega) \bigr) \bigr\| ^{2}\leq\rho_{3}( \omega), \quad\mbox{as } t\rightarrow+\infty. \end{aligned}$$
(3.33)
Then the lemma is easy to obtain. The proof is complete. □