Choose an arbitrary nonautonomous force but fixed \(\sigma _{0}(x,t)\in L^{2}_{b}(\mathbb{R},H)\), that is,
$$\sup_{t\in \mathbb{R}} \int^{t+1}_{t}\bigl\| \sigma_{0}(s) \bigr\| ^{2}_{H}\,ds< +\infty. $$
Taking \(\Sigma={\mathcal{H}}(\sigma_{0})\) (defined in Theorem 2.1) as the symbol space of problem (2.5)-(2.8), \(\forall \sigma\in\Sigma\) is called the symbol of system (2.5)-(2.8).
Obviously, \({\mathcal{H}}(\sigma_{0})\) is strictly invariant under the acting of the translation semigroup \(\{S(h)\}_{h\geq0}\), that is, \(S(h){\mathcal{H}}(\sigma_{0})\equiv{\mathcal{H}}(\sigma _{0})\) for all \(h\geq0\).
By Theorem 2.1 the global solution generates the process class \(\{U_{\sigma}(t,\tau), t\geq\tau, \tau\in\mathbb{R}\}\), \(\sigma \in {\mathcal{H}}(\sigma_{0})\), that is, \(U_{\sigma}(t,\tau)u_{\tau}=u(t)\), where \(u(t)\) is the solution of problem (2.5)-(2.8) with symbol \(\sigma\in\Sigma\) and initial data \(u_{\tau }\in H\).
Lemma 5.1
Let the external force
\(\sigma\in\Sigma\)
and
\(u_{\tau }\in H\). Then the process has a bounded uniform (w.r.t. \(\sigma\in{\mathcal{H}}(\sigma _{0})\)) absorbing set
\(B_{0}\)
in
H, where
$$B_{0}= \bigl\{ u\in H:\|u\|_{H}\leq C\|\sigma_{0} \|_{L^{2}_{b}(\mathbb {R};H)}\doteq\rho \bigr\} $$
is a bounded set in
H.
Proof
Multiplying (1.1) with u and integrating on Ω, by the Young inequality we conclude
$$\begin{aligned} &\frac{1}{2}\frac{d}{dt}\|u\|^{2}+\nu\|\nabla u\| ^{2}+\alpha\|u\| ^{2}+\beta\|u\|^{3}_{L^{3}}+ \gamma\|u\|^{4}_{L^{4}} \\ &\quad= \int_{\Omega}\sigma (t,x)u\,dx \leq\frac{\alpha}{2}\|u \|^{2}+\frac{2\|\sigma\| ^{2}}{\alpha}. \end{aligned}$$
(5.1)
Then integrating over \([\tau,t]\), we have
$$\begin{aligned} &\|u\|^{2}+ \int^{t}_{\tau} \bigl(2\nu\|\nabla u\| ^{2}+ \alpha\|u\| ^{2}+2\beta\|u\|^{3}_{L^{3}}+2\gamma\|u \|^{4}_{L^{4}} \bigr)\,ds \\ &\quad\leq\frac{4}{\alpha} \int^{t}_{\tau}\bigl\| \sigma(s)\bigr\| ^{2}\,ds+ \|u_{\tau}\| ^{2}. \end{aligned}$$
(5.2)
Hence,
$$\begin{aligned} \|u\|^{2}+ \int^{t}_{\tau} (2\lambda _{1}\nu +\alpha ) \|u\|^{2}\,ds\leq\frac{4}{\alpha} \int^{t}_{\tau}\bigl\| \sigma (s)\bigr\| ^{2}\,ds+ \|u_{\tau}\|^{2}, \end{aligned}$$
(5.3)
where \(\lambda_{1}\) is the first eigenvalue in the Poincaré inequality.
By Gronwall’s inequality we derive
$$\begin{aligned} \bigl\| u(t)\bigr\| ^{2} \leq& \|u_{\tau}\|^{2}e^{-(2\lambda_{1}\nu+\alpha)(t-\tau)}+ \frac {4}{\alpha} \int^{t}_{\tau}e^{(2\lambda_{1}\nu+\alpha)(s-t)}\bigl\| \sigma (s) \bigr\| ^{2}\,ds \\ \leq& \|u_{\tau}\|^{2}e^{-(2\lambda_{1}\nu+\alpha)(t-\tau)}+\frac {4}{\alpha} \biggl( \int^{t}_{t-1}e^{(2\lambda_{1}\nu+\alpha)(s-t)}\bigl\| \sigma(s) \bigr\| ^{2}\,ds \\ &{}+ \int^{t-1}_{t-2}e^{(2\lambda_{1}\nu+\alpha)(s-t)}\bigl\| \sigma(s)\bigr\| ^{2}\,ds+\cdots \biggr) \\ \leq& \|u_{\tau}\|^{2}e^{-(2\lambda_{1}\nu+\alpha)(t-\tau)}+\frac {4}{\alpha} \biggl( \int^{t}_{t-1}\bigl\| \sigma(s)\bigr\| ^{2}\,ds \\ &{}+e^{-(2\lambda_{1}\nu+\alpha)} \int^{t-1}_{t-2}\bigl\| \sigma(s)\bigr\| ^{2}\,ds+e^{-2(2\lambda_{1}\nu+\alpha)} \int^{t-2}_{t-3}\bigl\| \sigma(s)\bigr\| ^{2}\,ds+\cdots \biggr) \\ \leq& \|u_{\tau}\|^{2}e^{-(2\lambda_{1}\nu+\alpha)(t-\tau)} +\frac{4}{\alpha} \bigl(1+e^{-(2\lambda_{1}\nu+\alpha )}+e^{-2(2\lambda_{1}\nu+\alpha)}+\cdots \bigr) \bigl\| \sigma(s)\bigr\| ^{2}_{L^{2}_{b}(\mathbb{R};H)} \\ \leq& \|u_{\tau}\|^{2}e^{-(2\lambda_{1}\nu+\alpha)(t-\tau)}+\frac {4}{\alpha} \frac{1}{ (1-e^{-(2\lambda_{1}\nu+\alpha)} )} \bigl\| \sigma(s)\bigr\| ^{2}_{L^{2}_{b}(\mathbb{R};H)} \\ \leq& \|u_{\tau}\|^{2}e^{-(2\lambda_{1}\nu+\alpha)(t-\tau)}+\frac {4}{\alpha} \biggl(1+\frac{1}{(2\lambda_{1}\nu+\alpha)} \biggr) \bigl\| \sigma(s)\bigr\| ^{2}_{L^{2}_{b}(\mathbb{R};H)} \\ \leq& \|u_{\tau}\|^{2}e^{-(2\lambda_{1}\nu+\alpha)(t-\tau)}+\frac {4}{\alpha} \biggl(1+\frac{1}{(2\lambda_{1}\nu+\alpha)} \biggr) \|\sigma_{0}\|^{2}_{L^{2}_{b}(\mathbb{R};H)}. \end{aligned}$$
(5.4)
Now choosing \(\|u_{\tau}\|^{2}e^{-(2\lambda_{1}\nu+\alpha)(t-\tau )}\leq\frac{4}{\alpha} (1+\frac{1}{(2\lambda_{1}\nu+\alpha )} ) \|\sigma_{0}\|^{2}_{L^{2}_{b}(\mathbb{R};H)}\), there exists a time \(t_{0}=t_{0}(\alpha,\lambda_{1},\|\sigma_{0}\|^{2}_{L^{2}_{b}(R;H)})>0\) such that \(B_{0}= \{u:\|u\|^{2}\leq\rho^{2} \}\), where \(\rho ^{2}=\frac{8}{\alpha} (1+\frac{1}{(2\lambda_{1}\nu+\alpha)} ) \|\sigma_{0}\|^{2}_{L^{2}_{b}(\mathbb{R};H)}\), that is, \(B_{0}\) is a uniformly (w.r.t. \(\sigma\in\Sigma\)) absorbing ball for the process \(\{ U_{\sigma}(t,\tau)\}\) in H. □
Lemma 5.2
Let
\(\{u^{n}_{\tau}\}\)
be a sequence converging weakly to some
\(u_{\tau}\)
in Banach spaces
V, H, and
\((L^{4}(\Omega))^{3}\), and
\(\{\sigma^{n}\}\in{\mathcal{H}}(\sigma_{0})\)
be a sequence converging weakly to some
\(\sigma\in {\mathcal{H}}(\sigma_{0})\). Then for any fixed
\(t\geq\tau\in\mathbb{R}\), \(U_{\sigma^{n}}(t,\tau)u^{n}_{\tau}\rightharpoonup U_{\sigma}(t,\tau )u_{\tau}\)
weakly in
\(L^{2}(\tau,T;V)\), \(C(\tau,+\infty;H)\), and
\(L^{4}(\tau ,T;(L^{4}(\Omega))^{3})\)
for all
\(t\geq \tau\), respectively.
Proof
Noting that \(\{U_{\sigma^{n}}(t,\tau)u^{n}_{\tau}\}=\{ u^{n}(t)\}\), \(U_{\sigma}(t,\tau)u_{\tau}=u(t)\), by Theorem 2.1 we know that \(\{ u^{n}(t)\}\) and u are bounded in \(L^{2}(\tau,T;V)\), \(C(\tau,+\infty;H)\), and \(L^{4}(\tau,T;(L^{4}(\Omega))^{3})\), and hence the weak convergence holds. □
Lemma 5.3
For any
\(\sigma\in\Sigma={\mathcal{H}}(\sigma_{0})\), the family of processes
\(\{U_{\sigma}(t,\tau)\}\), \(t\geq\tau\in\mathbb{R}\), generated by the global solution of (2.5)-(2.8) are
\((H\times \Sigma,H)\)
continuous.
Proof
Let t and τ be fixed, and for any \(T>0\), \(t\geq\tau\), \(t, \tau\in[0,T]\), let \(\{(u^{n}_{\tau},\sigma^{n})\}\subset H\times{\mathcal{H}}(\sigma _{0})\) be a sequence that converges to some \(\{(u_{\tau},\sigma)\}\subset H\times {\mathcal{H}}(\sigma_{0})\), and let \(u^{n}(t)\) and \(u(t)\) be the solutions of problem (2.5)-(2.8) with symbols \(\sigma^{n}\), σ and initial data \(u^{n}_{\tau}\), \(u_{\tau}\), respectively.
Setting
$$w^{n}(t)=u(t)-u^{n}(t)=U_{\sigma}(t, \tau)u_{\tau}-U_{\sigma^{n}}(t,\tau )u^{n}_{\tau}, \quad n=1,2,\ldots, $$
we can see that \(w^{n}(t)\) is a solution of the following problem for each integer n:
$$\begin{aligned}& \frac{\partial w^{n}}{\partial t}+\nu Aw^{n}+\alpha w^{n}+B(u)-B \bigl(u^{n}\bigr)=\sigma-\sigma^{n}, \end{aligned}$$
(5.5)
$$\begin{aligned}& w^{n}|_{t=\tau}=w^{n}_{\tau}=u_{\tau}-u^{n}_{\tau}, \quad \tau\geq0. \end{aligned}$$
(5.6)
Taking the inner product of (5.6) with \(w^{n}\) in H, we get
$$\begin{aligned} &\frac{1}{2}\frac{d}{dt}\bigl( w^{n},w^{n} \bigr)+\nu\bigl(\nabla w^{n},\nabla w^{n}\bigr)+\alpha \bigl(w^{n},w^{n}\bigr)+\bigl(B(u)-B\bigl(u^{n} \bigr),w^{n}\bigr) \\ &\quad =\bigl(\sigma-\sigma^{n},w^{n}\bigr)\leq \alpha \bigl(w^{n},w^{n}\bigr)+\frac{1}{\alpha}\bigl\| \sigma- \sigma^{n}\bigr\| ^{2}, \end{aligned}$$
(5.7)
where
$$\begin{aligned} &\bigl|\bigl(B(u)-B\bigl(u^{n}\bigr),w^{n}\bigr)\bigr| \\ &\quad\leq C\bigl|\bigl(\beta|u|u+\gamma|u|^{2}u-\beta\bigl|u^{n}\bigr|u^{n}- \gamma \bigl|u^{n}\bigr|^{2}u^{n},w^{n}\bigr)\bigr| \\ &\quad\leq C \bigl|\bigl(\beta|u|u-\beta\bigl|u^{n}\bigr|u+\beta\bigl|u^{n}\bigr|u- \beta \bigl|u^{n}\bigr|u^{n},w^{n}\bigr) \\ &\qquad{} +\bigl(\gamma|u|^{2}u-\gamma\bigl|u^{n}\bigr|^{2}u+ \gamma\bigl|u^{n}\bigr|^{2}u-\gamma \bigl|u^{n}\bigr|^{2}u^{n},w^{n} \bigr)\bigr| \\ &\quad\leq C \bigl[\beta \bigl(\|u\|_{V}+\bigl\| u^{n} \bigr\| _{V} \bigr) \\ &\qquad{} +\gamma \bigl(\|u\|_{V}\|u_{n}\|_{V}+ \|u\|_{(L^{4}(\Omega ))^{3}}+\bigl\| u^{n}\bigr\| _{(L^{4}(\Omega))^{3}} \bigr) \bigr] \bigl\| w^{n}\bigr\| ^{2}. \end{aligned}$$
(5.8)
Hence, from (5.7)-(5.8) it follows that
$$\begin{aligned} \frac{1}{2}\frac{d}{dt}\bigl\| w^{n}(t) \bigr\| ^{2} =&-\bigl(B(u)-B\bigl(u^{n}\bigr),w^{n} \bigr)+\bigl(\sigma-\sigma^{n},w^{n}\bigr) \\ \leq& \frac{1}{\alpha}\bigl\| \sigma-\sigma^{n}\bigr\| ^{2}+C \bigl[ \beta \bigl(\|u\| _{V}+\bigl\| u^{n}\bigr\| _{V} \bigr) \\ &{}+\gamma \bigl(\|u\|_{V}\|u_{n}\|_{V}+\|u \|_{(L^{4}(\Omega))^{3}}+\bigl\| u^{n}\bigr\| _{(L^{4}(\Omega))^{3}} \bigr) \bigr] \bigl\| w^{n}\bigr\| ^{2}. \end{aligned}$$
(5.9)
Applying the Gronwall inequality to (5.9), we obtain
$$\begin{aligned} \bigl\| w^{n}(t)\bigr\| ^{2} =&\bigl\| U_{\sigma}(t, \tau)u_{\tau}-U_{\sigma^{n}}(t,\tau )u^{n}_{\tau} \bigr\| ^{2} \\ \leq& \biggl[\bigl\| u_{\tau}-u^{n}_{\tau}\bigr\| ^{2}+ \frac{1}{\alpha} \int^{t}_{\tau}\bigl\| \sigma(s)-\sigma^{n}(s) \bigr\| ^{2}\,ds \biggr] \exp \biggl\{ C \int^{t}_{\tau} \bigl[\beta \bigl(\|u\|_{V}+ \bigl\| u^{n}\bigr\| _{V} \bigr) \\ &{}+\gamma \bigl(\|u\|_{V}\|u_{n}\|_{V}+\|u \|_{(L^{4}(\Omega))^{3}}+\bigl\| u^{n}\bigr\| _{(L^{4}(\Omega))^{3}} \bigr) \bigr]\,ds \biggr\} \\ \leq& \biggl[\bigl\| u_{\tau}-u^{n}_{\tau}\bigr\| ^{2}+ \frac{1}{\alpha} \int^{T}_{\tau}\bigl\| \sigma(s)-\sigma^{n}(s) \bigr\| ^{2}\,ds \biggr] \\ &{}\times\exp \biggl\{ C \int^{T}_{\tau} \bigl[\beta \bigl(\|u \|_{V}+\bigl\| u^{n}\bigr\| _{V} \bigr) \\ &{}+\gamma \bigl(\|u\|_{V}\|u_{n}\|_{V}+\|u \|_{(L^{4}(\Omega))^{3}}+\bigl\| u^{n}\bigr\| _{(L^{4}(\Omega))^{3}} \bigr) \bigr]\,ds \biggr\} \\ \rightarrow& 0 \end{aligned}$$
(5.10)
since by Theorem 2.1 the following estimates hold for any fixed \(t\geq \tau\in\mathbb{R}\):
$$\begin{aligned} & \exp \biggl\{ C \int ^{T}_{\tau} \bigl[\beta \bigl(\|u \|_{V}+\bigl\| u^{n}\bigr\| _{V} \bigr)+\gamma \bigl(\| u \|_{V}\|u_{n}\|_{V} +\|u\|_{(L^{4}(\Omega))^{3}}+ \bigl\| u^{n}\bigr\| _{(L^{4}(\Omega))^{3}} \bigr) \bigr]\,ds \biggr\} \\ &\quad \leq\exp \biggl[C(T,\gamma,\beta) \int^{T}_{\tau } \bigl(\|u\|^{2}_{V}+ \bigl\| u^{n}\bigr\| ^{2}_{V}+\|u\|^{4}_{(L^{4}(\Omega))^{3}}+ \bigl\| u^{n}\bigr\| ^{4}_{(L^{4}(\Omega))^{3}} \bigr)\,ds \biggr] < +\infty. \end{aligned}$$
(5.11)
Therefore, (5.10) implies the \((H\times{\mathcal{H}}(\sigma_{0}),H)\) continuity of the processes \(\{U_{\sigma}(t,\tau), t\geq\tau\in\mathbb{R}\}\), \(\sigma\in {\mathcal{H}}(\sigma_{0})\), defined on H. □
Lemma 5.4
Let
\(\{u^{n}_{\tau}\}\)
be a sequence in
H
converging strongly to some
\(u_{\tau}\in H\)
in the norm of
H, and let
\(\{\sigma^{n}\}\)
be a sequence of
\({\mathcal{H}}(\sigma_{0})\)
converging strongly to some
\(\sigma\in{\mathcal{H}}(\sigma_{0})\). Then for any fixed
\(\tau\in\mathbb{R}\), \(U_{\sigma^{n}}(\cdot,\tau)u^{n}_{\tau}\rightarrow U_{\sigma}(\cdot ,\tau)u_{\tau}\)
strongly in
\(L^{2}(\tau,T;H)\)
foe all
\(t\geq\tau\).
Proof
From (5.10) we can derive that for any \(t\geq\tau\), \(t,\tau \in [0,T]\), the following inequality holds:
$$\begin{aligned} \bigl\| u(t)-u^{n}(t)\bigr\| ^{2} \leq& \bigl\| w^{n}(t)\bigr\| ^{2} \\ =&\bigl\| U_{\sigma}(t,\tau)u_{\tau}-U_{\sigma^{n}}(t, \tau)u^{n}_{\tau}\bigr\| ^{2} \\ \leq& C(T) \biggl[\bigl\| u_{\tau}-u^{n}_{\tau} \bigr\| ^{2}+C \int^{t}_{\tau}\bigl\| \sigma (s)-\sigma^{n}(s) \bigr\| ^{2}\,ds \biggr]. \end{aligned}$$
(5.12)
Integrating (5.12) over \([\tau,T]\) with respect to t, we have
$$\begin{aligned} & \int^{T}_{\tau}\bigl\| u(t)-u^{n}(t) \bigr\| ^{2}\,dt \\ &\quad\leq \int^{T}_{\tau}\bigl\| U_{\sigma}(t, \tau)u_{\tau}-U_{\sigma ^{n}}(t,\tau)u^{n}_{\tau} \bigr\| ^{2}\,dt \\ &\quad\leq C \int^{T}_{\tau} \biggl[\bigl\| u_{\tau}-u^{n}_{\tau} \bigr\| ^{2}+C \int^{t}_{\tau}\bigl\| \sigma(s)-\sigma^{n}(s) \bigr\| ^{2}\,ds \biggr]\,dt \\ &\quad\leq C(T)\varepsilon. \end{aligned}$$
(5.13)
The proof of the lemma is thus complete. □
Lemma 5.5
For any
\(\sigma\in{\mathcal{H}}(\sigma_{0})\)
and
\(u_{\tau}\in H\), the family of processes
\(\{U_{\sigma}(t,\tau), t\geq\tau\in\mathbb{R}\}\), \(\sigma \in {\mathcal{H}}(\sigma_{0})\), defined on
H, corresponding to equations (2.1)-(2.4) is uniformly (w.r.t. \(\sigma\in\Sigma\)) asymptotically compact in
H.
Proof
Assume that \(\{u^{n}_{\tau}\}\) is a bounded sequence in H, \(\{\sigma^{n}\}\subset{\mathcal{H}}(\sigma_{0})\) and \(\{t_{n}\}\in(\tau ,+\infty)\), \(t_{n}\rightarrow+\infty\) as \(n\rightarrow+\infty\).
From the proof of the existence of uniformly absorbing sets we can see that for any fixed \(\tau\in\mathbb{R}\), there exists a time \(T_{0}=T_{0}(\rho,\tau)\) depending on the radius ρ of the absorbing ball and τ such that for all \(t_{n}\geq T_{0}\), \(\{U_{\sigma^{n}}(t_{n},\tau)u^{n}_{\tau}\}\subseteq B_{0}\), where \(B_{0}\) is defined in Lemma 5.1.
By Lemma 5.2 the sequence \(\{U_{\sigma^{n}}(t_{n},\tau)u^{n}_{\tau}\}\) is weakly precompact in H, and hence we have
$$\begin{aligned} U_{\sigma^{n}}(t_{n},\tau)u^{n}_{\tau } \rightharpoonup u \quad\mbox{weakly in } H \mbox{ as } n\rightarrow +\infty \end{aligned}$$
(5.14)
for some \(u\in H\) and some subsequence (still denoted by) \(U_{\sigma^{n}}(t_{n},\tau)u^{n}_{\tau}\).
Similarly, for each \(T>0\) and \(t_{n}\geq T_{0}+T\),
$$\begin{aligned} u^{n}_{T}\doteq U_{\sigma^{n}}(t_{n}-T, \tau)u^{n}_{\tau}\rightharpoonup u_{T} \quad \mbox{weakly in } H \mbox{ as } n\rightarrow+\infty \end{aligned}$$
(5.15)
for some \(u_{T}\in H\).
Noting that the translation semigroup \(\{S(h):h\geq0\}\) satisfies
$$ U_{S(h)\sigma}(t,\tau)=U_{\sigma}(t+h,\tau+h), \quad \forall h\geq0, t\geq\tau\in\mathbb{R}, \sigma\in{\mathcal{H}}( \sigma_{0}), $$
(5.16)
we see that
$$\begin{aligned} U_{\sigma^{n}}(t_{n},\tau) =&U_{\sigma^{n}}(t_{n},t_{n}-T)U_{\sigma ^{n}}(t_{n}-T, \tau) \\ =&U_{S(t_{n}-T)\sigma^{n}}(T,0)U_{\sigma^{n}}(t_{n}-T,\tau),\quad t_{n}-T\geq\tau. \end{aligned}$$
(5.17)
Letting \(\sigma^{n}_{T}=S(t_{n}-T)\sigma^{n}\), from (5.17) and (5.16) we derive
$$\begin{aligned} U_{\sigma^{n}}(t_{n},\tau)u^{n}_{\tau} =&U_{\sigma^{n}_{T}}(T,0)U_{\sigma ^{n}}(t_{n}-T, \tau)u^{n}_{\tau} \\ =&U_{\sigma^{n}_{T}}(T,0)u^{n}_{T},\quad t_{n}-T \geq\tau. \end{aligned}$$
(5.18)
Since \(\{\sigma^{n}_{T}\}\subset{\mathcal{H}}(\sigma_{0})\) and \(\Sigma={\mathcal{H}}(\sigma_{0})\) is compact in \(L^{2}_{\mathrm{loc}}(\mathbb{R};H)\), there exist a subsequence of \(\{\sigma^{n}_{T}\}\) (also denoted by \(\{\sigma^{n}_{T}\}\)) and some \(\sigma_{T}\in{\mathcal{H}}(\sigma_{0})\) such that
$$\begin{aligned} \sigma^{n}_{T}\rightarrow\sigma_{T} \quad \mbox{strongly in } L^{2}_{\mathrm{loc}}(\mathbb{R};H) \mbox{ as } n \rightarrow+\infty, \forall T>0. \end{aligned}$$
(5.19)
By Lemmas 5.2 and 5.4 and by formulas (5.14)-(5.15) and (5.18) we conclude
$$\begin{aligned} u=U_{\sigma _{T}}(T,0)u_{T},\quad \forall T>0. \end{aligned}$$
(5.20)
Next, we want to prove the asymptotic compactness in H, that is,
$$\begin{aligned} \bigl\| U_{\sigma^{n}_{T}}(t_{n},\tau)u^{n}_{\tau}-u \bigr\| \rightarrow 0 \quad\mbox{as } n\rightarrow+\infty, \end{aligned}$$
(5.21)
that is, it suffices to prove that
$$\begin{aligned} \bigl\| U_{\sigma^{n}_{T}}(t_{n},\tau)u^{n}_{\tau} \bigr\| \rightarrow \|u\| \quad\mbox{as } n\rightarrow+\infty \end{aligned}$$
(5.22)
and
$$\begin{aligned} U_{\sigma^{n}_{T}}(t_{n},\tau)u^{n}_{\tau} \rightharpoonup u\quad \mbox{in } H\mbox{ as } n\rightarrow+\infty. \end{aligned}$$
(5.23)
However, (5.14) clearly ensures (5.23). Next, we only need to prove (5.22).
To this end, we shall prove that
$$\begin{aligned}& \liminf_{n\rightarrow+\infty}\bigl\| U_{\sigma^{n}}(t_{n}, \tau)u^{n}_{\tau }\bigr\| =\liminf_{n\rightarrow+\infty} \bigl\| U_{\sigma^{n}_{T}}(T,0)u^{n}_{T}\bigr\| \geq \|u\|, \end{aligned}$$
(5.24)
$$\begin{aligned}& \limsup_{n\rightarrow+\infty}\bigl\| U_{\sigma^{n}}(t_{n}, \tau)u^{n}_{\tau }\bigr\| =\limsup_{n\rightarrow+\infty} \bigl\| U_{\sigma^{n}_{T}}(T,0)u^{n}_{T}\bigr\| \leq \|u\|. \end{aligned}$$
(5.25)
However, the weak convergence of the corresponding sequences (i.e., (5.14) and (5.15)) and Lemma 5.2 ensure that (5.24) is true, so our aim is only to prove (5.25).
Taking the inner product with the both sides of equation (2.5) by u, we obtain
$$\begin{aligned} \frac{1}{2}\frac {d}{dt}(u,u)+\alpha(u,u) =& \bigl(-B(u)-\nu Au+\sigma,u\bigr) =\bigl(\sigma(t),u\bigr)-G\bigl(u(t)\bigr), \end{aligned}$$
(5.26)
where \(G(u(t))=(B(u)+\nu Au+\alpha u,u)=(PF(u)+\nu Au+\alpha u,u)=(P(\beta |u|u+\gamma|u|^{2}u)+\nu Au+\alpha u,u)\).
Integrating (5.26) over \([\tau,t]\) with respect to the time variable, we derive
$$\begin{aligned} \|u\|^{2} =&\bigl(u(t),u(t)\bigr)=\bigl( u(\tau),u(\tau) \bigr) e^{-2\alpha(t-\tau)}+2 \int^{t}_{\tau}e^{-2\alpha(t-s)}\bigl( \sigma(s),u(s) \bigr)\,ds \\ &{} -2 \int^{t}_{\tau} e^{-2\alpha(t-s)}G\bigl(u(s)\bigr)\,ds. \end{aligned}$$
(5.27)
Substituting \(u(t)\) in (5.27) by \(U_{\sigma^{n}_{T}}(T,0)u^{n}_{T}\) and changing the integration domain to \([0,T]\), we deduce
$$\begin{aligned} \bigl\| U_{\sigma^{n}_{T}}(T,0)u^{n}_{T} \bigr\| ^{2} =&\bigl(U_{\sigma^{n}_{T}}(T,0)u^{n}_{T},U_{\sigma^{n}_{T}}(T,0)u^{n}_{T} \bigr) \\ =&\bigl(u^{n}_{T},u^{n}_{T}\bigr) e^{-2\alpha T}+2 \int^{T}_{0}e^{-2\alpha(T-s)}\bigl( \sigma^{n}_{s}(s),U_{\sigma^{n}_{s}}(s,0)u^{n}_{s} \bigr)\,ds \\ &{}-2 \int^{T}_{0} e^{-2\alpha(T-s)}G\bigl(U_{\sigma^{n}_{s}}(s,0)u^{n}_{s} \bigr)\,ds \\ =&I_{1}+I_{2}+I_{3}. \end{aligned}$$
(5.28)
Similarly, substituting \(u(t)\) in (5.27) by \(U_{\sigma_{T}}(T,0)u_{T}\) and changing the integration domain to \([0,T]\), we have
$$\begin{aligned} \bigl\| U_{\sigma _{T}}(T,0)u_{T}\bigr\| ^{2} =& \bigl(U_{\sigma_{T}}(T,0)u_{T},U_{\sigma_{T}}(T,0)u_{T} \bigr) \\ =&(u_{T},u_{T}) e^{-2\alpha T}+2 \int^{T}_{0}e^{-2\alpha(T-s)}\bigl( \sigma_{s}(s),U_{\sigma_{s}}(s,0)u_{s}\bigr)\,ds \\ &{}-2 \int^{T}_{0} e^{-2\alpha(T-s)}G\bigl(U_{\sigma_{s}}(s,0)u_{s} \bigr)\,ds. \end{aligned}$$
(5.29)
Next, we shall deal with the right-hand side of (5.28) term by term. From Lemma 5.1 we derive
$$\begin{aligned} \limsup_{n}I_{1}=\limsup _{n}\bigl(u^{n}_{T},u^{n}_{T} \bigr) e^{-2\alpha T}\leq \rho^{2}e^{-2\alpha T}. \end{aligned}$$
(5.30)
Since \(L^{2}([\tau,T];V)\hookrightarrow\hookrightarrow L^{2}([\tau ,T];H)\), from (5.15), (5.18), and Lemma 5.2 we deduce that
$$\begin{aligned} U_{\sigma ^{n}_{s}}(s,0)u^{n}_{s}\rightharpoonup U_{\sigma_{s}}(s,0)u_{s} \quad\mbox{weakly in } L^{2} \bigl([\tau,T];H\bigr), \forall T>\tau, \end{aligned}$$
(5.31)
whence we can deduce from (5.31) that
$$ \begin{aligned}[b] \lim_{n\rightarrow\infty}I_{2}&=\lim _{n\rightarrow\infty}2 \int ^{T}_{0}e^{-2\alpha(T-s)}\bigl( \sigma^{n}_{s}(s),U_{\sigma ^{n}_{s}}(s,0)u^{n}_{s} \bigr)\,ds\\ &=2 \int^{T}_{0}e^{-2\alpha^{2}(T-s)}\bigl( \sigma_{s}(s),U_{\sigma _{s}}(s,0)u_{s}\bigr)\,ds. \end{aligned} $$
(5.32)
Next, we will deal with \(I_{3}\) on the right-hand side of (5.26), which we shall prove in Lemma 5.6 (to be proved later on), that is, we want to prove
$$\begin{aligned} \limsup_{n}I_{3} =&\limsup _{n} \biggl(-2 \int^{T}_{0}e^{-2\alpha (t-s)}G\bigl(U_{\sigma^{n}_{s}}(s,0)u^{n}_{s} \bigr)\,ds \biggr) \\ =&-\liminf_{n} \biggl(2 \int^{T}_{0}e^{-2\alpha(t-s)}G\bigl(U_{\sigma ^{n}_{s}}(s,0)u^{n}_{s} \bigr)\,ds \biggr) \\ \leq& -2 \int^{T}_{0}e^{-2\alpha(t-s)}G\bigl(U_{\sigma_{s}}(s,0)u_{s} \bigr)\,ds. \end{aligned}$$
(5.33)
Combining (5.29), (5.30), and (5.32)-(5.33), we obtain
$$\begin{aligned} \limsup_{n} \bigl(\bigl\| U_{\sigma^{n}_{T}}(T,0)u^{n}_{T} \bigr\| ^{2} \bigr) =&\bigl(U_{\sigma ^{n}_{T}}(T,0)u^{n}_{T},U_{\sigma^{n}_{T}}(T,0)u^{n}_{T} \bigr) \\ \leq& \rho^{2}e^{-2\alpha T}+2 \int^{T}_{0}e^{-2\alpha^{2}(T-s)}\bigl(\sigma _{s}(s),U_{\sigma_{s}}(s,0)u_{s}\bigr)\,ds \\ &{}-2 \int^{T}_{0}e^{-2\alpha(t-s)}G\bigl(U_{\sigma_{s}}(s,0)u_{s} \bigr)\,ds \\ =&\bigl\| U_{\sigma_{T}}(T,0)u_{T}\bigr\| ^{2}-(u_{T},u_{T}) e^{-2\alpha T}+\rho^{2}e^{-2\alpha T} \\ \leq& \bigl\| U_{\sigma_{T}}(T,0)u_{T}\bigr\| ^{2}+2 \rho^{2}e^{-2\alpha T}. \end{aligned}$$
(5.34)
Hence, letting T tend to +∞ in (5.34), we derive
$$\begin{aligned} \lim_{n\rightarrow \infty}\bigl\| U_{\sigma^{n}}(t_{n}, \tau)u^{n}_{\tau}\bigr\| ^{2} =&\limsup_{n,T\rightarrow \infty} \bigl\| U_{\sigma^{n}_{T}}(T,0)u^{n}_{T}\bigr\| ^{2} \\ \leq&\limsup_{n,T\rightarrow\infty} \bigl(\bigl\| U_{\sigma_{T}}(T,0)u_{T} \bigr\| ^{2}+2\rho^{2}e^{-2\alpha T} \bigr) \\ =&\bigl\| U_{\sigma}(t,\tau)u_{\tau}\bigr\| ^{2}=\|u \|^{2}, \end{aligned}$$
(5.35)
which, along with (5.24), gives
$$\begin{aligned} \limsup_{n,T\rightarrow \infty}\bigl\| U_{\sigma^{n}_{T}}(T,0)u^{n}_{T} \bigr\| ^{2} =&\lim_{n\rightarrow \infty}\bigl\| U_{\sigma^{n}}(t_{n}, \tau)u^{n}_{\tau}\bigr\| ^{2} =\bigl\| U_{\sigma}(t,\tau)u_{\tau}\bigr\| ^{2}=\|u \|^{2}. \end{aligned}$$
(5.36)
Combining the norm convergence (5.36) and weak convergence (5.23), we obtain
$$\begin{aligned} \lim_{n\rightarrow \infty}\bigl\| U_{\sigma^{n}_{T}}(t_{n}, \tau)u^{n}_{\tau}-u\bigr\| =0. \end{aligned}$$
(5.37)
To complete the proof of Lemma 5.5, we need to prove the following lemma. □
Lemma 5.6
$$ \begin{aligned}[b] \limsup_{n}I_{3}&=\limsup _{n} \biggl(-2 \int^{T}_{0}e^{-2\alpha (t-s)}G\bigl(U_{\sigma^{n}_{s}}(s,0)u^{n}_{s} \bigr)\,ds \biggr) \\ &=-\liminf_{n} \biggl(2 \int^{T}_{0}e^{-2\alpha(t-s)}G\bigl(U_{\sigma ^{n}_{s}}(s,0)u^{n}_{s} \bigr)\,ds \biggr) \\ &\leq -2 \int^{T}_{0}e^{-2\alpha(t-s)}G\bigl(U_{\sigma_{s}}(s,0)u_{s} \bigr)\,ds. \end{aligned} $$
(5.38)
Proof
From (5.15), (5.18), and Lemma 5.2, by the embedding theorem we have:
$$\begin{aligned}& U_{\sigma^{n}_{T}}(s,0)u^{n}_{T}\rightharpoonup U_{\sigma _{T}}(s,0)u_{T}\quad \mbox{weakly in } C(0,T;H), \end{aligned}$$
(5.39)
$$\begin{aligned}& U_{\sigma^{n}_{T}}(s,0)u^{n}_{T}\rightarrow U_{\sigma _{T}}(s,0)u_{T} \quad\mbox{strongly in } L^{2}(0,T;H), \end{aligned}$$
(5.40)
$$\begin{aligned}& U_{\sigma^{n}_{T}}(s,0)u^{n}_{T}\rightharpoonup U_{\sigma _{T}}(s,0)u_{T} \quad\mbox{weakly in } L^{4} \bigl(0,T;\bigl(L^{4}(\Omega)\bigr)^{3}\bigr). \end{aligned}$$
(5.41)
To end the proof of (5.38), we should complete the proof of the following formula:
$$ \liminf_{n} \biggl( \int^{T}_{0}e^{-2\alpha(t-s)}G\bigl(U_{\sigma ^{n}_{s}}(s,0)u^{n}_{s} \bigr)\,ds \biggr)\geq \int^{T}_{0}e^{-2\alpha(t-s)}G\bigl(U_{\sigma_{s}}(s,0)u_{s} \bigr)\,ds, $$
(5.42)
where \(G(u(t))=(B(u)+\nu Au,u)=(P(\beta|u|u+\gamma|u|^{2}u)+\nu Au,u)\).
Since \(A: D(A)\mapsto H\) is a linear positive definite operator, from (5.39)-(5.41) it follows that
$$\begin{aligned} &\liminf_{n} \int^{T}_{0}e^{-2\alpha(t-s)}\bigl(\nu AU_{\sigma^{n}_{s}}(s,0)u^{n}_{s}, U_{\sigma^{n}_{s}}(s,0)u^{n}_{s} \bigr)\,ds \\ &\quad\geq \int^{T}_{0}e^{-2\alpha(t-s)}\bigl(\nu AU_{\sigma_{s}}(s,0)u_{s},U_{\sigma _{s}}(s,0)u_{s} \bigr)\,ds. \end{aligned}$$
(5.43)
Next, we want to prove that
$$\begin{aligned} &\lim_{n\rightarrow+\infty} \int^{T}_{0}e^{-2\alpha(t-s)}\bigl( B \bigl(U_{\sigma^{n}_{s}}(s,0)u^{n}_{s}\bigr), U_{\sigma^{n}_{s}}(s,0)u^{n}_{s}\bigr)\,ds \\ &\quad= \int^{T}_{0} e^{-2\alpha(t-s)}\bigl( B \bigl(U_{\sigma_{s}}(s,0)u_{s}\bigr), U_{\sigma_{s}}(s,0)u_{s} \bigr)\,ds. \end{aligned}$$
(5.44)
Setting
$$\begin{aligned}& K^{n}_{1}(s)=\bigl(B\bigl(U_{\sigma^{n}_{s}}(s,0)u^{n}_{s} \bigr), U_{\sigma^{n}_{s}}(s,0)u^{n}_{s}\bigr), \\& K_{2}(s)=\bigl( B\bigl(U_{\sigma_{s}}(s,0)u_{s}\bigr), U_{\sigma_{s}}(s,0)u_{s}\bigr), \end{aligned}$$
we consider
$$\begin{aligned} I =& \biggl| \int^{T}_{0}e^{-2\alpha(t-s)}\bigl(K^{n}_{2}(s)-K^{n}_{1}(s) \bigr)\,ds \biggr| \\ \leq& \biggl| \int^{T}_{0}e^{-2\alpha(t-s)} \bigl( \beta\bigl|U_{\sigma _{s}}(s,0)u_{s}\bigr|U_{\sigma_{s}}(s,0)u_{s}+ \gamma\bigl|U_{\sigma _{s}}(s,0)u_{s}\bigr|^{2}U_{\sigma_{s}}(s,0)u_{s} \\ &{}-\beta\bigl|U_{\sigma^{n}_{s}}(s,0)u^{n}_{s}\bigr|U_{\sigma^{n}_{s}}(s,0)u^{n}_{s} -\gamma\bigl|U_{\sigma^{n}_{s}}(s,0)u^{n}_{s}\bigr|^{2}U_{\sigma ^{n}_{s}}(s,0)u^{n}_{s}, \\ &{}U_{\sigma_{s}}(s,0)u_{s}-U_{\sigma^{n}_{s}}(s,0)u^{n}_{s} \bigr)\,ds \biggr| \\ \leq& C \biggl| \int^{T}_{0}e^{-2\alpha(t-s)} \bigl( \beta\bigl|U_{\sigma _{s}}(s,0)u_{s}\bigr|U_{\sigma_{s}}(s,0)u_{s}- \beta\bigl|U_{\sigma ^{n}_{s}}(s,0)u^{n}_{s}\bigr|U_{\sigma_{s}}(s,0)u_{s} \\ &{}+\beta\bigl|U_{\sigma^{n}_{s}}(s,0)u^{n}_{s}\bigr|U_{\sigma_{s}}(s,0)u_{s}- \beta \bigl|U_{\sigma^{n}_{s}}(s,0)u^{n}_{s}\bigr|U_{\sigma^{n}_{s}}(s,0)u^{n}_{s}, \\ &{}U_{\sigma_{s}}(s,0)u_{s}-U_{\sigma^{n}_{s}}(s,0)u^{n}_{s} \bigr) \\ &{}+ \bigl(\gamma\bigl|U_{\sigma_{s}}(s,0)u_{s}\bigr|^{2}U_{\sigma_{s}}(s,0)u_{s}- \gamma \bigl|U_{\sigma^{n}_{s}}(s,0)u^{n}_{s}\bigr|^{2}U_{\sigma_{s}}(s,0)u_{s} \\ &{}+\gamma\bigl|U_{\sigma^{n}_{s}}(s,0)u^{n}_{s}\bigr|^{2}U_{\sigma_{s}}(s,0)u_{s}- \gamma \bigl|U_{\sigma^{n}_{s}}(s,0)u^{n}_{s}\bigr|^{2}U_{\sigma^{n}_{s}}(s,0)u^{n}_{s}, \\ &{}U_{\sigma_{s}}(s,0)u_{s}-U_{\sigma^{n}_{s}}(s,0)u^{n}_{s} \bigr)\,ds \biggr| \\ \leq& C \biggl| \int^{T}_{0}e^{-2\alpha(t-s)} \bigl[\beta \bigl( \bigl\| U_{\sigma _{s}}(s,0)u_{s}\bigr\| _{V}+\bigl\| U_{\sigma^{n}_{s}}(s,0)u^{n}_{s} \bigr\| _{V} \bigr) \\ &{}+\gamma \bigl(\bigl\| U_{\sigma_{s}}(s,0)u_{s}\bigr\| _{V} \bigl\| U_{\sigma ^{n}_{s}}(s,0)u^{n}_{s}\bigr\| _{V}+ \bigl\| U_{\sigma_{s}}(s,0)u_{s}\bigr\| _{(L^{4}(\Omega ))^{3}} \\ &{}+\bigl\| U_{\sigma^{n}_{s}}(s,0)u^{n}_{s}\bigr\| _{(L^{4}(\Omega))^{3}} \bigr) \bigr]\bigl\| U_{\sigma_{s}}(s,0)u_{s}-U_{\sigma^{n}_{s}}(s,0)u^{n}_{s} \bigr\| ^{2}\,ds \biggr| \\ \leq& C \biggl| \int^{T}_{0}e^{-4\alpha(t-s)}\beta \bigl( \bigl\| U_{\sigma _{s}}(s,0)u_{s}\bigr\| ^{2}_{V}+ \bigl\| U_{\sigma^{n}_{s}}(s,0)u^{n}_{s}\bigr\| ^{2}_{V} \bigr) \\ &{}+\gamma \bigl(\bigl\| U_{\sigma_{s}}(s,0)u_{s}\bigr\| ^{2}_{V} \bigl\| U_{\sigma ^{n}_{s}}(s,0)u^{n}_{s}\bigr\| ^{2}_{V}+ \bigl\| U_{\sigma_{s}}(s,0)u_{s}\bigr\| ^{4}_{(L^{4}(\Omega ))^{3}} \\ &{}+\bigl\| U_{\sigma^{n}_{s}}(s,0)u^{n}_{s}\bigr\| ^{4}_{(L^{4}(\Omega))^{3}}+C \bigr)\,ds \biggr| \\ &{}\times \Bigl| \sup_{s\in[0,T]}\bigl\| U_{\sigma_{s}}(s,0)u_{s}-U_{\sigma ^{n}_{s}}(s,0)u^{n}_{s} \bigr\| ^{4} \Bigr|. \end{aligned}$$
(5.45)
By Theorem 2.1, Lemma 5.2, and Lemma 5.4, since
$$\begin{aligned} & \biggl| \int^{T}_{0}e^{-4\alpha(t-s)}\beta \bigl( \bigl\| U_{\sigma_{s}}(s,0)u_{s}\bigr\| ^{2}_{V} + \bigl\| U_{\sigma^{n}_{s}}(s,0)u^{n}_{s}\bigr\| ^{2}_{V} \bigr) \\ &\qquad{}+\gamma \bigl(\bigl\| U_{\sigma_{s}}(s,0)u_{s} \bigr\| ^{2}_{V}\bigl\| U_{\sigma ^{n}_{s}}(s,0)u^{n}_{s} \bigr\| ^{2}_{V}+\bigl\| U_{\sigma_{s}}(s,0)u_{s} \bigr\| ^{4}_{(L^{4}(\Omega ))^{3}} \\ &\qquad{}+\bigl\| U_{\sigma^{n}_{s}}(s,0)u^{n}_{s} \bigr\| ^{4}_{(L^{4}(\Omega))^{3}}+C \bigr)\,ds \biggr| \\ &\quad< +\infty, \end{aligned}$$
(5.46)
we conclude
$$\begin{aligned} I\leq C(T) \sup_{s\in[0,T]}\bigl\| U_{\sigma_{s}}(s,0)u_{s}-U_{\sigma ^{n}_{s}}(s,0)u^{n}_{s} \bigr\| ^{4}\rightarrow0 \end{aligned}$$
(5.47)
as \(n\rightarrow\infty\), that is,
$$\begin{aligned} \lim_{n\rightarrow+\infty}I =&\lim_{n\rightarrow+\infty} \int ^{T}_{0}e^{-2\alpha(t-s)}\bigl(K^{n}_{2}(s)-K^{n}_{1}(s) \bigr)\,ds \\ =&\lim_{n\rightarrow+\infty}e^{-2\alpha(t-s)}\bigl(B\bigl(U_{\sigma _{s}}(s,0)u_{s} \bigr), U_{\sigma_{s}}(s,0)u_{s}\bigr) \\ &{}-\bigl(B\bigl(U_{\sigma^{n}_{s}}(s,0)u^{n}_{s}\bigr), U_{\sigma^{n}_{s}}(s,0)u^{n}_{s}\bigr))\,ds \\ =&0. \end{aligned}$$
(5.48)
Combining (5.42)-(5.43) and (5.48), we complete the proof. □
Remark 5.1
For all bounded sets \(B\in H\),
$$\begin{aligned} {\mathcal {A}}_{\Sigma}=\omega_{\tau, {\mathcal{H}}(\sigma_{0})}(B)\doteq \bigcup _{B\in{\mathcal{B}}(H)}\bigcap_{t\geq\tau}\overline { \bigcup_{\sigma\in{\mathcal{H}}(\sigma_{0})}\bigcup_{s\geq t}U_{\sigma}(s, \tau)B}. \end{aligned}$$
(5.49)
Combining the uniformly (w.r.t. \(\sigma\in{\mathcal{H}}(\sigma_{0})\)) absorbing property (i.e., Lemma 5.1), the continuity of the process (i.e., Lemma 5.3), the asymptotic compactness property of the processes \(\{U_{\sigma}(t,\tau)\}\) (\(\sigma\in{\mathcal{H}}(\sigma _{0})\)) in H (i.e., Lemma 5.5), from Theorem 3.1 we deduce the existence of a uniform attractor in the following theorem.
Theorem 5.1
Assume that
\(u_{\tau}\in H\), \(\sigma\in\Sigma \subset L^{2}_{\mathrm{loc}}(\mathbb{R};H)\). Then the family of processes
\(\{U_{\sigma}(t,\tau), t\geq\tau\in\mathbb{R}\}\) (\(\sigma \in {\mathcal{H}}(\sigma_{0})\)) generated by the global solution of problem (1.1) or (2.5)-(2.8) possesses a uniform (w.r.t. \(\sigma\in\Sigma={\mathcal {H}}(\sigma_{0})\)) attractor
\({\mathcal{A}}_{{\mathcal{H}}(\sigma_{0})}={\mathcal{A}}_{\Sigma}\)
in
H.