3.1
\((L^{2}(\mathbb{R}^{n}),L^{2}(\mathbb{R}^{n}))\)-\(\mathscr{D}\)-pullback attractors
Firstly, the following lemma ensures a \(\mathscr{D}\)-pullback absorbing set in \(L^{2}( \mathbb{R}^{n})\).
Lemma 3.1
Assume that (1.2)-(1.4) hold and the external force
\(g\in L^{2}_{\mathrm {loc}}(\mathbb{R}, L^{2}(\mathbb{R}^{n}))\)
satisfies (1.5). Then, for any
\(\hat{\mathscr{D}}\in \mathscr{D}\subset B(L^{2}(\mathbb{R}^{n}))\)
and any
\(t\in\mathbb{R}\), there exists
\(\tau_{0}(t, \hat{\mathscr{D}})\leq t\)
such that
$$\begin{aligned} \bigl\Vert U(t, \tau)u_{\tau} \bigr\Vert \leq R_{0}(t) \quad \textit{for all } \tau\leq\tau_{0}(t, \hat{ \mathscr{D}}) \textit{ and all } u_{\tau}\in D(\tau), \end{aligned}$$
(3.1)
where
\(R_{0}(t)= ( \frac{2\beta_{3}\Vert a(x)\Vert _{1}}{\sigma}+\frac{2e ^{-\sigma t}}{\lambda-\beta_{1}}\int_{-\infty}^{t}e^{\sigma r}\Vert g(x, r)\Vert ^{2}\,dr ) ^{\frac{1}{2}}\).
Proof
Taking the inner product of (1.1) with u in \(L^{2}( \mathbb{R}^{n})\), we have
$$\begin{aligned} \frac{1}{2}\frac{d}{dt}\Vert u\Vert ^{2}+\nu \Vert \nabla u\Vert ^{2}+\lambda \Vert u\Vert ^{2}+ \bigl(f_{1}(u), u \bigr)+ \bigl(a(x)f_{2}(u), u \bigr)= \bigl(g(x, t), u \bigr). \end{aligned}$$
Due to (1.2)-(1.4) and Young’s inequality, we get
$$\begin{aligned}& \frac{d}{dt}\Vert u\Vert ^{2}+(\lambda- \beta_{1})\Vert u\Vert ^{2}\leq2\beta_{3} \bigl\Vert a(x) \bigr\Vert _{1}+\frac{\Vert g(x, t)\Vert ^{2}}{\lambda-\beta_{1}}, \end{aligned}$$
(3.2)
$$\begin{aligned}& \begin{aligned}[b] &\frac{d}{dt}\Vert u\Vert ^{2}+(\lambda-\beta_{1})\Vert u\Vert ^{2}+2\nu \Vert \nabla u \Vert ^{2}+2\alpha_{1}\Vert u\Vert _{p}^{p}+2\alpha_{3} \int_{\mathbb{R}^{n}}a(x)\vert u\vert ^{p}\,dx \\ &\quad\leq2\beta_{3} \bigl\Vert a(x) \bigr\Vert _{1}+ \frac{\Vert g(x, t)\Vert ^{2}}{\lambda-\beta _{1}}. \end{aligned} \end{aligned}$$
(3.3)
By (3.2), we obtain
$$\begin{aligned} \frac{d}{dr} \bigl(e^{\sigma r}\Vert u\Vert ^{2} \bigr)+( \lambda-\beta_{1}-\sigma)e^{ \sigma r}\Vert u\Vert ^{2} \leq2\beta_{3} \bigl\Vert a(x) \bigr\Vert _{1}e^{\sigma r}+ \frac{\Vert g(x, r)\Vert ^{2}}{\lambda-\beta_{1}}e^{\sigma r}. \end{aligned}$$
Integrating over the interval \([\tau, t]\) and noting that \(\sigma \in(0, \lambda-\beta_{1})\), we have
$$\begin{aligned} e^{\sigma t} \bigl\Vert u(t) \bigr\Vert ^{2} \leq& \frac{2\beta_{3}\Vert a(x)\Vert _{1}}{\sigma }e^{\sigma t}+\frac{1}{\lambda-\beta_{1}} \int_{\tau}^{t}e^{\sigma r} \bigl\Vert g(x, r) \bigr\Vert ^{2}\,dr+e^{\sigma\tau} \Vert u_{\tau} \Vert ^{2} \\ \leq& \frac{2\beta_{3}\Vert a(x)\Vert _{1}}{\sigma}e^{\sigma t}+\frac{1}{ \lambda-\beta_{1}} \int_{-\infty}^{t}e^{\sigma r} \bigl\Vert g(x, r) \bigr\Vert ^{2}\,dr+e ^{\sigma\tau} \Vert u_{\tau} \Vert ^{2}. \end{aligned}$$
(3.4)
Thus, we get
$$\begin{aligned} \bigl\Vert u(t) \bigr\Vert ^{2}\leq\frac{2\beta_{3}\Vert a(x)\Vert _{1}}{\sigma}+e^{-\sigma t}e ^{\sigma\tau} \Vert u_{\tau} \Vert ^{2}+\frac{e^{-\sigma t}}{\lambda-\beta _{1}} \int_{-\infty}^{t}e^{\sigma r} \bigl\Vert g(x, r) \bigr\Vert ^{2}\,dr, \end{aligned}$$
and this implies (3.1). □
Let \(\hat{\mathscr{B}}_{0}=\{B_{0}(t): t\in\mathbb{R}\}\), where
$$\begin{aligned} B_{0}(t)= \bigl\{ u\in L^{2} \bigl( \mathbb{R}^{n} \bigr): \Vert u\Vert \leq R_{0}(t) \bigr\} . \end{aligned}$$
(3.5)
By Lemma 3.1, it is easy to know that the family \(\hat{\mathscr{B}} _{0}\) is \((L^{2}(\mathbb{R}^{n}),L^{2}(\mathbb{R}^{n}))\)-\(\mathscr{D}\)-pullback absorbing for the process \(\{U(t, \tau)\}_{ \tau\leq t}\) defined by (2.1) and
$$\begin{aligned} e^{\sigma t} \bigl( R_{0}(t) \bigr) ^{2} \rightarrow0 \quad \text{as } t\rightarrow-\infty. \end{aligned}$$
(3.6)
Let \(F_{1}(u)=\int_{0}^{u}f_{1}(s)\,ds\) and \(F_{2}(u)=\int_{0}^{u}f_{2}(s)\,ds\). By (1.2)-(1.3), there exist positive constants \(\tilde{\alpha_{i}}\), \(\tilde{\beta_{i}}\), \(i=1, 2, 3, 4\), such that
$$\begin{aligned}& \tilde{\alpha}_{1}\vert u\vert ^{p}-\tilde{ \beta}_{1}\vert u\vert ^{2}\leq F_{1}(u) \leq \tilde{\alpha}_{2}\vert u\vert ^{p}+\tilde{ \beta}_{2}\vert u\vert ^{2}, \quad \quad \lambda>2 \tilde{ \beta}_{1}, \end{aligned}$$
(3.7)
$$\begin{aligned}& \tilde{\alpha}_{3}\vert u\vert ^{p}-\tilde{ \beta}_{3}\leq F_{2}(u)\leq \tilde{\alpha}_{4} \vert u\vert ^{p}+\tilde{\beta}_{4}. \end{aligned}$$
(3.8)
Lemma 3.2
Assume that (1.2)-(1.4) hold and the external force
\(g\in L^{2}_{\mathrm {loc}}(\mathbb{R}, L^{2}(\mathbb{R}^{n}))\)
satisfies (1.5). Then, for any
\(\hat{\mathscr{D}}\in \mathscr{D}\subset B(L^{2}(\mathbb{R}^{n}))\)
and any
\(t\in\mathbb{R}\), there exists
\(\tau_{1}(t, \hat{\mathscr{D}})\leq t\)
such that
$$\begin{aligned} \bigl\Vert u(t) \bigr\Vert ^{2}+ \bigl\Vert \nabla u(t) \bigr\Vert ^{2}+ \bigl\Vert u(t) \bigr\Vert ^{p}_{p}\leq \bigl( R_{1}(t) \bigr) ^{2} \quad \textit{for all } \tau\leq\tau_{1}(t, \hat{\mathscr{D}}) \textit{ and all } u_{\tau}\in D(\tau), \end{aligned}$$
(3.9)
where
\(R_{1}(t)=C ( \frac{\beta_{3}\Vert a(x)\Vert _{1}}{\sigma}+\frac{e ^{-\sigma t}}{\lambda-\beta_{1}}\int_{-\infty}^{t}e^{\sigma r}\Vert g(x, r)\Vert ^{2}\,dr ) ^{\frac{1}{2}}\)
and the positive constant
C
is independent of
t
and
\(\hat{\mathscr{D}}\).
Proof
Multiplying (3.3) by \(e^{\sigma t}\), we have
$$\begin{aligned}& \frac{d}{dt} \bigl(e^{\sigma t} \bigl\Vert u(t) \bigr\Vert ^{2} \bigr)+(\lambda-\beta_{1}-\sigma)e ^{\sigma t} \bigl\Vert u(t) \bigr\Vert ^{2}+2\nu e^{\sigma t} \bigl\Vert \nabla u(t) \bigr\Vert ^{2} \\ & \quad\quad{} +2\alpha_{1}e^{\sigma t}\Vert u\Vert _{p}^{p}+2\alpha _{3}e^{\sigma t} \int_{\mathbb{R}^{n}}a(x)\vert u\vert ^{p}\,dx \\ & \quad \leq2\beta_{3}e^{\sigma t} \bigl\Vert a(x) \bigr\Vert _{1}+e^{\sigma t}\frac{\Vert g(x, t) \Vert ^{2}}{\lambda-\beta_{1}}. \end{aligned}$$
Let \(\tau< t-1\) and \(r\in[\tau, t-1]\), integrating over the interval \([r, r+1]\), we get
$$\begin{aligned}& e^{\sigma(r+1)} \bigl\Vert u(r+1) \bigr\Vert ^{2}+(\lambda- \beta_{1}-\sigma) \int_{r} ^{r+1}e^{\sigma s} \bigl\Vert u(s) \bigr\Vert ^{2}\,ds+2\nu \int_{r}^{r+1}e^{\sigma s} \bigl\Vert \nabla u(s) \bigr\Vert ^{2}\,ds \\& \quad\quad{} +2\alpha_{1} \int_{r}^{r+1}e^{\sigma s} \bigl\Vert u(s) \bigr\Vert _{p}^{p}\,ds+2\alpha _{3} \int_{r}^{r+1}e^{\sigma s} \int_{\mathbb{R}^{n}}a(x) \bigl\vert u(s) \bigr\vert ^{p}\,dx \,ds \\& \quad \leq2\beta_{3} \bigl\Vert a(x) \bigr\Vert _{1} \int_{r}^{r+1}e^{\sigma s}\,ds+ \int_{r} ^{r+1}e^{\sigma s}\frac{\Vert g(x, s)\Vert ^{2}}{\lambda-\beta_{1}} \,ds+e^{ \sigma r} \bigl\Vert u(r) \bigr\Vert ^{2}. \end{aligned}$$
By (3.4), we find
$$\begin{aligned}& \int_{r}^{r+1}e^{\sigma s} \biggl( \bigl\Vert u(s) \bigr\Vert ^{2}+ \bigl\Vert \nabla u(s) \bigr\Vert ^{2}+ \bigl\Vert u(s) \bigr\Vert _{p}^{p}+ \int_{\mathbb{R}^{n}}a(x) \bigl\vert u(s) \bigr\vert ^{p}\,dx \biggr) \,ds \\& \quad \leq C \biggl( \frac{2\beta_{3}\Vert a(x)\Vert _{1}}{\sigma}e^{\sigma (r+1)}+\frac{1}{ \lambda-\beta_{1}} \int_{\tau}^{r+1}e^{\sigma s} \bigl\Vert g(x, s) \bigr\Vert ^{2}\,ds+e ^{\sigma\tau} \Vert u_{\tau} \Vert ^{2} \biggr) \\& \quad \leq C \biggl( \frac{2\beta_{3}\Vert a(x)\Vert _{1}}{\sigma}e^{\sigma t}+e ^{\sigma\tau} \Vert u_{\tau} \Vert ^{2}+\frac{1}{\lambda-\beta_{1}} \int_{-\infty}^{t}e^{\sigma s} \bigl\Vert g(x, s) \bigr\Vert ^{2}\,ds \biggr) . \end{aligned}$$
Thus, by (3.7) and (3.8), we can obtain
$$\begin{aligned}& \int_{r}^{r+1}e^{\sigma s} \biggl( \frac{\nu}{2}\Vert \nabla u\Vert ^{2}+\frac{ \lambda}{2}\Vert u\Vert ^{2}+ \int_{\mathbb{R}^{n}}F_{1}(u)\,dx+ \int_{\mathbb{R}^{n}}a(x)F_{2}(u)\,dx \biggr) \,ds \\& \quad \leq C \biggl( \frac{2\beta_{3}\Vert a(x)\Vert _{1}}{\sigma}e^{\sigma t}+e ^{\sigma\tau} \Vert u_{\tau} \Vert ^{2}+\frac{1}{\lambda-\beta_{1}} \int_{-\infty}^{t}e^{\sigma s} \bigl\Vert g(x, s) \bigr\Vert ^{2}\,ds \biggr) . \end{aligned}$$
(3.10)
Multiplying (1.1) by \(u_{t}\) and integrating on \(\mathbb{R} ^{n}\), we have
$$\begin{aligned}& \Vert u_{t}\Vert ^{2}+\frac{d}{dt} \biggl( \frac{\nu}{2}\Vert \nabla u\Vert ^{2}+\frac{ \lambda}{2}\Vert u\Vert ^{2}+ \int_{\mathbb{R}^{n}}F_{1}(u)\,dx+ \int_{\mathbb{R}^{n}}a(x)F_{2}(u)\,dx \biggr) \\& \quad \leq\frac{1}{2} \bigl( \bigl\Vert g(x, t) \bigr\Vert ^{2}+\Vert u_{t}\Vert ^{2} \bigr). \end{aligned}$$
And then
$$\begin{aligned} \frac{d}{dt} \biggl( \frac{\nu}{2}\Vert \nabla u \Vert ^{2}+\frac{\lambda}{2}\Vert u \Vert ^{2}+ \int_{\mathbb{R}^{n}}F_{1}(u)\,dx+ \int_{\mathbb {R}^{n}}a(x)F_{2}(u)\,dx \biggr) \leq \frac{1}{2} \bigl\Vert g(x, t) \bigr\Vert ^{2}. \end{aligned}$$
(3.11)
It follows from (3.11) that
$$\begin{aligned}& \frac{d}{dr}e^{\sigma r} \biggl( \frac{\nu}{2}\Vert \nabla u \Vert ^{2}+\frac{ \lambda}{2}\Vert u\Vert ^{2}+ \int_{\mathbb{R}^{n}}F_{1}(u)\,dx+ \int_{\mathbb{R}^{n}}a(x)F_{2}(u)\,dx \biggr) \\& \quad \leq\sigma e^{\sigma r} \biggl( \frac{\nu}{2}\Vert \nabla u \Vert ^{2}+\frac{ \lambda}{2}\Vert u\Vert ^{2}+ \int_{\mathbb{R}^{n}}F_{1}(u)\,dx+ \int_{\mathbb{R}^{n}}a(x)F_{2}(u)\,dx \biggr) +\frac{e^{\sigma r}}{2} \bigl\Vert g(x,t) \bigr\Vert ^{2}. \end{aligned}$$
By (1.5), (3.7), (3.8), (3.10) and the uniform Gronwall inequality, we obtain
$$\begin{aligned}& \frac{\nu}{2}\Vert \nabla u\Vert ^{2}+ \frac{\lambda}{2}\Vert u\Vert ^{2}+ \int_{\mathbb{R}^{n}}F_{1}(u)\,dx+ \int_{\mathbb{R}^{n}}a(x)F_{2}(u)\,dx \\& \quad\leq C \biggl( \frac{2\beta_{3}\Vert a(x)\Vert _{1}}{\sigma}+e^{-\sigma(t- \tau)}\Vert u_{\tau} \Vert ^{2}+\frac{e^{-\sigma t}}{\lambda-\beta_{1}} \int_{-\infty}^{t}e^{\sigma s} \bigl\Vert g(x, s) \bigr\Vert ^{2}\,ds \biggr) . \end{aligned}$$
(3.12)
It follows from (3.7) and (3.8) that
$$\begin{aligned}& \bigl\Vert u(t) \bigr\Vert ^{2}+ \bigl\Vert \nabla u(t) \bigr\Vert ^{2}+ \bigl\Vert u(t) \bigr\Vert ^{p}_{p} \\& \quad\leq C \biggl( \frac{2\beta_{3}\Vert a(x)\Vert _{1}}{\sigma}+e^{-\sigma(t- \tau)}\Vert u_{\tau} \Vert ^{2}+\frac{e^{-\sigma t}}{\lambda-\beta_{1}} \int_{-\infty}^{t}e^{\sigma s} \bigl\Vert g(x, s) \bigr\Vert ^{2}\,ds \biggr) , \end{aligned}$$
and this implies (3.9). □
Lemma 3.3
Assume that (1.2)-(1.4) hold and the external force
\(g\in L^{2}_{\mathrm {loc}}(\mathbb{R}, L^{2}(\mathbb{R}^{n}))\)
satisfies (1.5). Let the family
\(\hat{\mathscr{B}}_{0}=\{B _{0}(t): t\in\mathbb{R}\}\)
be defined by (3.5). Then, for any
\(\varepsilon\geq0\)
and any
\(t\in\mathbb{R}\), there exist
\(\tilde{k}=\tilde{k}(t, \varepsilon)>0\)
and
\(\tau^{\prime}(t, \varepsilon)\)
such that
$$\begin{aligned} \int_{\vert x\vert \geq k} \bigl\vert U(t, \tau)u_{\tau} \bigr\vert ^{2}\,dx\leq \varepsilon \quad \textit{for all } k\geq\tilde{k}, \tau\leq \tau^{\prime}(t, \varepsilon ) \textit{ and } u_{\tau}\in B_{0}(\tau). \end{aligned}$$
(3.13)
Proof
Choose a smooth function θ such that \(0\leq\theta(s)\leq1\) for \(s\in\mathbb{R}^{+}\),
$$\begin{aligned} \theta(s)= \textstyle\begin{cases} 0, &0\leq s\leq1, \\ 1, &s\geq2, \end{cases}\displaystyle \end{aligned}$$
and there exists a constant c such that \(\vert \theta^{\prime}(s)\vert \leq c\).
Multiplying (1.1) by \(\theta^{2}(\frac{\vert x\vert ^{2}}{k^{2}})u\) and integrating on \(\mathbb{R}^{n}\), we have
$$\begin{aligned}& \frac{1}{2}\frac{d}{dt} \int_{\mathbb{R}^{n}}\theta^{2} \biggl(\frac{\vert x\vert ^{2}}{k ^{2}} \biggr) \vert u\vert ^{2}\,dx-\nu \int_{\mathbb{R}^{n}}\theta^{2} \biggl(\frac{\vert x\vert ^{2}}{k ^{2}} \biggr)u \Delta u\,dx+\lambda \int_{\mathbb{R}^{n}}\theta^{2} \biggl(\frac{\vert x\vert ^{2}}{k ^{2}} \biggr) \vert u\vert ^{2}\,dx \\& \quad=- \int_{\mathbb{R}^{n}}\theta^{2} \biggl(\frac{\vert x\vert ^{2}}{k^{2}} \biggr)f_{1}(u)u\,dx- \int_{\mathbb{R}^{n}}\theta^{2} \biggl(\frac{\vert x\vert ^{2}}{k^{2}} \biggr)a(x)f_{2}(u)u\,dx+ \int_{\mathbb{R}^{n}}\theta^{2} \biggl(\frac{\vert x\vert ^{2}}{k^{2}} \biggr)g(x, t)u\,dx \\& \quad\leq\beta_{1} \int_{\mathbb{R}^{n}}\theta^{2} \biggl(\frac {\vert x\vert ^{2}}{k^{2}} \biggr) \vert u\vert ^{2}\,dx- \alpha_{1} \int_{\mathbb{R}^{n}}\theta^{2} \biggl(\frac{\vert x\vert ^{2}}{k^{2}} \biggr) \vert u\vert ^{p}\,dx +\beta_{3} \int_{\mathbb{R}^{n}}\theta^{2} \biggl(\frac{\vert x\vert ^{2}}{k^{2}} \biggr)a(x)\,dx \\& \quad\quad{} -\alpha_{3} \int_{\mathbb{R}^{n}}\theta^{2} \biggl(\frac {\vert x\vert ^{2}}{k^{2}} \biggr)a(x)\vert u\vert ^{p}\,dx+ \frac{\lambda-\beta_{1}}{2} \int_{\mathbb{R}^{n}}\theta^{2} \biggl(\frac{\vert x\vert ^{2}}{k ^{2}} \biggr) \vert u\vert ^{2}\,dx \\& \quad\quad{} +\frac{1}{2(\lambda-\beta_{1})} \int_{\mathbb {R}^{n}}\theta^{2} \biggl(\frac{\vert x\vert ^{2}}{k ^{2}} \biggr) \bigl\vert g(x, t) \bigr\vert ^{2}\,dx. \end{aligned}$$
And so
$$\begin{aligned}& \frac{d}{dt} \int_{\mathbb{R}^{n}}\theta^{2} \biggl(\frac{\vert x\vert ^{2}}{k^{2}} \biggr) \vert u\vert ^{2}\,dx-2 \nu \int_{\mathbb{R}^{n}}\theta^{2} \biggl(\frac{\vert x\vert ^{2}}{k^{2}} \biggr)u \Delta u\,dx+( \lambda-\beta_{1}) \int_{\mathbb{R}^{n}}\theta^{2} \biggl(\frac{\vert x\vert ^{2}}{k ^{2}} \biggr) \vert u\vert ^{2}\,dx \\& \quad\leq2\beta_{3} \int_{\mathbb{R}^{n}}\theta^{2} \biggl( \frac{\vert x\vert ^{2}}{k^{2}} \biggr)a(x)\,dx+\frac{1}{(\lambda-\beta_{1})} \int_{\mathbb{R}^{n}}\theta^{2} \biggl(\frac{\vert x\vert ^{2}}{k^{2}} \biggr) \bigl\vert g(x, t) \bigr\vert ^{2}\,dx. \end{aligned}$$
(3.14)
For the second term on the left-hand side of (3.14), we know
$$\begin{aligned} \begin{aligned}[b] &-\nu \int_{\mathbb{R}^{n}}\theta^{2} \biggl(\frac{\vert x\vert ^{2}}{k^{2}} \biggr)u \Delta u\,dx \\ &\quad = \nu \int_{\mathbb{R}^{n}}\theta^{2} \biggl(\frac{\vert x\vert ^{2}}{k^{2}} \biggr) \vert \nabla u\vert ^{2}\,dx+ \nu \int_{\mathbb{R}^{n}}\theta^{\prime} \biggl(\frac{\vert x\vert ^{2}}{k^{2}} \biggr) \theta \biggl(\frac{\vert x\vert ^{2}}{k^{2}} \biggr)u\frac{4x}{k^{2}} \cdot\nabla u\,dx \end{aligned} \end{aligned}$$
(3.15)
and
$$\begin{aligned} \begin{aligned}[b] & \biggl\vert \int_{\mathbb{R}^{n}}\theta^{\prime} \biggl(\frac{\vert x\vert ^{2}}{k^{2}} \biggr) \theta \biggl(\frac{\vert x\vert ^{2}}{k^{2}} \biggr)u\frac{4x}{k^{2}}\cdot\nabla u\,dx \biggr\vert \\ &\quad \leq\frac{4\sqrt{2}c}{k} \int_{k\leq \vert x\vert \leq\sqrt{2}k}\vert u\vert \vert \nabla u\vert \,dx\leq \frac{C}{k}\Vert u\Vert \Vert \nabla u\Vert . \end{aligned} \end{aligned}$$
(3.16)
It follows from (3.15) and (3.16) that
$$\begin{aligned}& \frac{d}{dr} \biggl( e^{\sigma r} \int_{\mathbb{R}^{n}}\theta^{2} \biggl(\frac{\vert x\vert ^{2}}{k ^{2}} \biggr) \vert u\vert ^{2}\,dx \biggr) \\& \quad\leq2\beta_{3}e^{\sigma r} \int_{\vert x\vert \geq k}a(x)\,dx +\frac{e^{\sigma r}}{(\lambda-\beta_{1})} \int_{\vert x\vert \geq k} \bigl\vert g(x, t) \bigr\vert ^{2}\,dx+ \frac{C}{k}e ^{\sigma r}\Vert u\Vert \Vert \nabla u\Vert . \end{aligned}$$
Integrating over the interval \([\tau, t]\), we get
$$\begin{aligned}& \int_{\mathbb{R}^{n}}\theta^{2} \biggl(\frac{\vert x\vert ^{2}}{k^{2}} \biggr) \bigl\vert u(t) \bigr\vert ^{2}\,dx \\& \quad\leq2\beta_{3}e^{-\sigma t} \int_{\tau}^{t}e^{\sigma r} \int_{\vert x\vert \geq k}a(x)\,dx\,dr +\frac{e^{-\sigma t}}{(\lambda-\beta_{1})} \int_{ \tau}^{t}e^{\sigma r} \int_{\vert x\vert \geq k} \bigl\vert g(x, r) \bigr\vert ^{2}\,dx \,dr \\& \quad\quad{} +\frac{C}{k}e^{-\sigma t} \int_{\tau}^{t}e^{\sigma r}\Vert u\Vert \Vert \nabla u\Vert \,dr+e^{-\sigma t}e^{\sigma\tau} \Vert u_{\tau} \Vert ^{2}, \end{aligned}$$
(3.17)
where \(\tau\leq\tau_{0}(t, \hat{\mathscr{D}})\). We now treat each term on the right-hand side of (3.17). For the first term,
$$\begin{aligned} 2\beta_{3}e^{-\sigma t} \int_{\tau}^{t}e^{\sigma r} \int_{\vert x\vert \geq k}a(x)\,dx\,dr \leq2\beta_{3}e^{-\sigma t}e^{\sigma t} \int_{\vert x\vert \geq k}a(x)\,dx\leq2 \beta_{3} \int_{\vert x\vert \geq k}a(x)\,dx, \end{aligned}$$
by (1.4), for any \(\varepsilon>0\), there exists \(k_{1}( \varepsilon, t)\) such that
$$\begin{aligned} 2\beta_{3} \int_{\vert x\vert \geq k}a(x)\,dx< \frac{\varepsilon}{4} \quad \text{for all } k \geq k_{1}. \end{aligned}$$
(3.18)
For the second term, by (1.5), for any \(\varepsilon>0\), there exists \(k_{2}(\varepsilon, t)\) such that
$$\begin{aligned}& \frac{1}{(\lambda-\beta_{1})} \int_{\tau}^{t}e^{\sigma r} \int_{\vert x\vert \geq k} \bigl\vert g(x, r) \bigr\vert ^{2}\,dx \,dr \\& \quad \leq\frac{1}{(\lambda-\beta_{1})} \int_{-\infty}^{t}e^{\sigma r} \int_{\vert x\vert \geq k} \bigl\vert g(x, r) \bigr\vert ^{2}\,dx \,dr< \frac{\varepsilon}{4} \quad\text{for all } k\geq k_{2}. \end{aligned}$$
(3.19)
For the forth term, since \(u_{\tau}\in B_{0}(\tau)\), by (3.6), for any \(t\in\mathbb{R}\), we get
$$\begin{aligned} e^{-\sigma t}e^{\sigma\tau} \Vert u_{\tau} \Vert ^{2}\rightarrow0 \quad \text{as }\tau\rightarrow-\infty. \end{aligned}$$
(3.20)
We now handle the third term on the right-hand side of (3.17). By Young’s inequality, we know
$$\begin{aligned} \frac{C}{k}e^{-\sigma t} \int_{\tau}^{t}e^{\sigma r}\Vert u\Vert \Vert \nabla u\Vert \,dr \leq\frac{C}{2k}e^{-\sigma t} \int_{\tau}^{t}e^{\sigma r}\Vert u\Vert ^{2}\,dr+ \frac{C}{2k}e^{-\sigma t} \int_{\tau}^{t}e^{\sigma r}\Vert \nabla u\Vert ^{2}\,dr. \end{aligned}$$
We can find \(\delta_{0}>0\) such that
$$\begin{aligned} \int_{\tau}^{t}e^{\sigma r}\Vert u\Vert ^{2}\,dr\leq \int_{\tau}^{t}e^{( \sigma+\delta_{0})r}\Vert u\Vert ^{2}\,dr, \end{aligned}$$
and by (3.4), we have
$$\begin{aligned}& \int_{\tau}^{t}e^{(\sigma+\delta_{0})r} \bigl\Vert u(r) \bigr\Vert ^{2}\,dr \\& \quad\leq \int_{\tau}^{t}e^{(\sigma+\delta_{0})r} \biggl( \frac{2\beta _{3}\Vert a(x)\Vert _{1}}{\sigma}+e^{-\sigma r}e^{\sigma\tau} \Vert u_{\tau} \Vert ^{2}+\frac{e^{-\sigma r}}{\lambda-\beta_{1}} \int_{-\infty}^{r}e ^{\sigma s} \bigl\Vert g(x, s) \bigr\Vert ^{2}\,ds \biggr) \,dr \\& \quad\leq\frac{2\beta_{3}\Vert a(x)\Vert _{1}}{\sigma}e^{(\sigma+\delta_{0})t}+e ^{\sigma\tau} \Vert u_{\tau} \Vert ^{2} \int_{\tau}^{t}e^{\delta_{0}r}\,dr+\frac{1}{ \lambda-\beta_{1}} \int_{\tau}^{t} e^{\delta_{0}s}\,ds \int_{-\infty} ^{t}e^{\sigma s} \bigl\Vert g(x, s) \bigr\Vert ^{2}\,ds \\& \quad\leq\frac{2\beta_{3}\Vert a(x)\Vert _{1}}{\sigma}e^{(\sigma+\delta _{0})t}+\frac{1}{ \delta_{0}}e^{\delta_{0}t}e^{\sigma\tau} \Vert u_{\tau} \Vert ^{2}+\frac{1}{ \delta_{0}(\lambda-\beta_{1})} e^{\delta_{0}t} \int_{-\infty}^{t}e ^{\sigma s} \bigl\Vert g(x, s) \bigr\Vert ^{2}\,ds \\& \quad< \infty. \end{aligned}$$
Analogously, we can obtain
$$\begin{aligned} \int_{\tau}^{t}e^{\sigma r}\Vert \nabla u\Vert ^{2}\,dr< \infty. \end{aligned}$$
Thus, for any \(\varepsilon>0\), there exists \(k_{3}(\varepsilon, t)\) such that
$$\begin{aligned} \frac{C}{k}e^{-\sigma t} \int_{\tau}^{t}e^{\sigma r}\Vert u\Vert \Vert \nabla u\Vert \,dr< \frac{ \varepsilon}{4} \quad \text{for all } k\geq k_{2}. \end{aligned}$$
(3.21)
It follows from (3.18)-(3.21) that
$$\begin{aligned} \int_{\vert x\vert \geq2k} \bigl\vert U(t, \tau)u_{\tau} \bigr\vert ^{2}\,dx\leq \int_{\mathbb{R}^{n}}\theta^{2} \biggl(\frac{\vert x\vert ^{2}}{k^{2}} \biggr) \bigl\vert u(t) \bigr\vert ^{2}\,dx< \varepsilon. \end{aligned}$$
So, the proof is complete. □
Next, we utilise Definition 2.6 to prove that the process \(\{U(t, \tau)\}_{\tau\leq t}\) associated with the initial value problem (1.1) is \(\mathscr{D}\)-pullback ω-limit compact.
Lemma 3.4
Assume that (1.2)-(1.4) hold and the external force
\(g\in L^{2}_{\mathrm {loc}}(\mathbb{R}, L^{2}(\mathbb{R}^{n}))\)
satisfies (1.5). Then the process
\(\{U(t, \tau)\}_{\tau \leq t}\)
associated with the initial value problem (1.1) is
\(\mathscr{D}\)-pullback
ω-limit compact in
\(L^{2}(\mathbb{R}^{n})\).
Proof
Denote \(B_{r}=B(0, r)\cap\mathbb{R}^{n}\), we can split \(u(t)\) as
$$\begin{aligned} u(t)=\chi(x)u(t)+ \bigl(1-\chi(x) \bigr)u(t), \end{aligned}$$
where \(\chi(x)\) is a smooth function satisfying \(0\leq\chi(x) \leq1\), \(\vert \chi^{\prime}(x)\vert \leq c_{0}\), and it is defined by
$$\begin{aligned} \chi(x)= \textstyle\begin{cases} 0, &x\in B_{r}, \\ 1, &x\notin B_{r+1}. \end{cases}\displaystyle \end{aligned}$$
And so, we have
$$\begin{aligned}& u_{1}(t)= \textstyle\begin{cases} u(t), &x\in B_{r}, \\ 0, &x\notin B_{r+1}, \\ \chi(x)u(t), &\text{others}, \end{cases}\displaystyle \quad\quad u_{2}(t)= \textstyle\begin{cases} 0, &x\in B_{r}, \\ u(t), &x\notin B_{r+1}, \\ (1-\chi(x))u(t), &\text{others}. \end{cases}\displaystyle \end{aligned}$$
For any \(\hat{\mathscr{D}}\in\mathscr{D}\subset B(L^{2}(\mathbb{R} ^{n}))\), \(\{U(t,\tau)D(\tau)\}=\{U(t, \tau)u_{\tau}\mid u_{\tau} \in D(\tau)\}\) can be split as
$$\begin{aligned} U(t,\tau)D(\tau)=\chi(x)U(t,\tau)D(\tau)+ \bigl(1-\chi(x) \bigr)U(t,\tau)D( \tau). \end{aligned}$$
By Lemma 2.1, we have
$$\begin{aligned} \alpha \bigl(U(t,\tau)D(\tau) \bigr)\leq\alpha \bigl(\chi(x)U(t, \tau)D(\tau) \bigr)+ \alpha \bigl( \bigl(1-\chi(x) \bigr)U(t,\tau)D(\tau) \bigr). \end{aligned}$$
(3.22)
By Lemma 3.1, we get \(u_{1}(t)\in L^{2}(B_{r})\) as \(\tau\leq\tau _{0}(t, \hat{\mathscr{D}})\) and
$$\begin{aligned} \chi(x)U(t,\tau)D(\tau)= \bigl\{ \chi(x)U(t,\tau)u_{\tau}=u_{1}(t) \mid u _{\tau}\in D(\tau) \bigr\} . \end{aligned}$$
By Lemma 3.2, we have
$$\begin{aligned} \bigl\Vert u_{1}(t) \bigr\Vert _{H_{0}^{1}(B_{r+1})}= \bigl\Vert \nabla u_{1}(t) \bigr\Vert _{L^{2}(B_{r+1})} \leq R_{1}(t) \quad \text{for all } \tau\leq\tau_{1}(t, \hat{\mathscr{D}}). \end{aligned}$$
Since \(H_{0}^{1}(B_{r+1})\hookrightarrow L^{2}(B_{r+1})\) is compact, \(\chi(x)U(t,\tau)D(\tau)\) is compact in \(L^{2}(B_{r+1})\). By Lemma 2.1, we obtain
$$\begin{aligned} \alpha \bigl(\chi(x)U(t,\tau)D(\tau) \bigr)=0. \end{aligned}$$
(3.23)
By Lemma 3.3, for any \(\varepsilon>0\), we can choose r large enough such that
$$\begin{aligned} \int_{\vert x\vert \geq r}\vert u\vert ^{2}\leq\varepsilon. \end{aligned}$$
And then
$$\begin{aligned} \Vert u_{2}\Vert \leq\varepsilon \quad \text{for all }\tau\leq\tau^{\prime}(t, \varepsilon). \end{aligned}$$
(3.24)
We know
$$\begin{aligned} \bigl(1-\chi(x) \bigr)U(t,\tau)D(\tau)= \bigl\{ \bigl(1-\chi(x) \bigr)U(t, \tau)u_{\tau}=u _{2}(t)\mid u_{\tau}\in D(\tau) \bigr\} . \end{aligned}$$
By (3.24), we obtain
$$\begin{aligned} \alpha \bigl( \bigl(1-\chi(x) \bigr)U(t,\tau)D(\tau) \bigr)\leq \varepsilon \quad \text{for all }\tau\leq\tau^{\prime}(t, \varepsilon). \end{aligned}$$
(3.25)
It follows from (3.22), (3.23) and (3.25) that
$$\begin{aligned} \alpha \bigl(U(t,\tau)D(\tau) \bigr)\leq\varepsilon \quad \text{for all } \tau \leq \min \bigl\{ \tau_{0}(t, \hat{\mathscr{D}}), \tau _{1}(t, \hat{ \mathscr{D}}), \tau^{\prime}(t, \varepsilon) \bigr\} . \end{aligned}$$
By Definition 2.6, we obtain \(\{U(t, \tau)\}_{\tau\leq t}\) is \(\mathscr{D}\)-pullback ω-limit compact in \(L^{2}(\mathbb{R} ^{n})\). □
Using Theorem 2.2 or Theorem 2.4, it is easy to prove the following theorem by Lemma 3.1 and Lemma 3.4.
Theorem 3.1
Assume that (1.2)-(1.4) hold and the external force
\(g\in L^{2}_{\mathrm {loc}}(\mathbb{R}, L^{2}(\mathbb{R}^{n}))\)
satisfies (1.5). Then the process
\(\{U(t, \tau)\}_{\tau \leq t}\)
associated with the initial value problem (1.1) has a
\(\mathscr{D}\)-pullback attractor
\(\mathcal{A}=\{A(t):t \in\mathbb{R}\}\)
in
\(L^{2}(\mathbb{R}^{n})\).
3.2
\((L^{2}(\mathbb{R}^{n}),L^{p}(\mathbb{R}^{n}))\)-\(\mathscr{D}\)-pullback attractors
In this subsection, we prove the existence of \(\mathscr{D}\)-pullback attractors in \(L^{p}(\mathbb{R}^{n})\). We set \(\hat{\mathscr{B}}_{1}=\{B_{1}(t): t\in\mathbb{R}\}\), where
$$\begin{aligned} B_{1}(t)= \bigl\{ u\in L^{2} \bigl( \mathbb{R}^{n} \bigr)\cap L^{p} \bigl(\mathbb{R}^{n} \bigr): \Vert u \Vert +\Vert u\Vert _{p}\leq R_{1}(t) \bigr\} \quad \text{for all } t\in\mathbb{R}, \end{aligned}$$
(3.26)
and \(R_{1}(t)\) is defined in Lemma 3.2. So by Lemma 3.2, we obtain the family \(\hat{\mathscr{B}}_{1}=\{B_{1}(t): t\in\mathbb{R}\}\) is \((L ^{2}(\mathbb{R}^{n}), L^{p}(\mathbb{R}^{n}))\)-\(\mathscr{D}\)-pullback absorbing for the process \(\{U(t, \tau)\}_{ \tau\leq t}\), i.e. for any \(\hat{\mathscr{D}}\in\mathscr{D}\subset B(L^{2}(\mathbb{R}^{n}))\), there exists \(\tau_{1}(t, \hat{\mathscr{D}})\leq t\) such that \(U(t, \tau)D(\tau)\subset B_{1}(t)\) for all \(\tau\leq\tau_{1}(t, \hat{\mathscr{D}})\). We also know
$$\begin{aligned} e^{\sigma t} \bigl( R_{1}(t) \bigr) ^{2} \rightarrow0 \quad \text{as } t\rightarrow-\infty. \end{aligned}$$
(3.27)
Based on Theorem 2.4, we only prove that the process \(\{U(t, \tau)\} _{\tau\leq t}\) associated with the initial value problem (1.1) is \(\mathscr{D}\)-pullback ω-limit compact in \(L^{p}(\mathbb{R}^{n})\). Firstly, we prove the following lemma.
Lemma 3.5
Assume that (1.2)-(1.4) hold and the external force
\(g\in L^{2}_{\mathrm {loc}}(\mathbb{R}, L^{2}(\mathbb{R}^{n}))\)
satisfies (1.5). Let the family
\(\hat{\mathscr{B}}_{1}=\{B _{1}(t): t\in\mathbb{R}\}\)
be defined by (3.26). Then, for any
\(\varepsilon\geq0\), any
\(\hat{\mathscr{D}}\in\mathscr{D}\subset B(L ^{2}(\mathbb{R}^{n}))\)
and any
\(t\in\mathbb{R}\), there exist
\(M=M(t, \varepsilon)>0\)
and
\(\tau^{\prime\prime}(t, \varepsilon)\)
such that
$$\begin{aligned} \begin{aligned}[b] & \int_{\mathbb{R}^{n}(\vert U(t, \tau)u_{\tau} \vert \geq M)} \bigl\vert U(t, \tau)u _{\tau} \bigr\vert ^{p}\,dx\leq\varepsilon \\ &\quad \textit{for all } u_{\tau}\in D( \tau), \tau\leq\tau^{\prime\prime }(t, \varepsilon)\textit{ and } M \geq2M_{0}. \end{aligned} \end{aligned}$$
(3.28)
Proof
For any \(\varepsilon>0\) be given, by (1.5), there exists \(\delta_{1}>0\) such that
$$\begin{aligned} \int_{-\infty}^{t}e^{\sigma s} \int_{e_{1}} \bigl\vert g(x, s) \bigr\vert ^{2}\,dx \,ds< \varepsilon , \end{aligned}$$
(3.29)
where \(e_{1}\subset\mathbb{R}^{n}\) and \(m(e_{1})\leq\delta_{1}\). By Lemma 2.2 and Lemma 3.2, we know that there exist \(M_{1}=M_{1}(t, \varepsilon)\) and \(\tau_{2}=\tau_{2}(t, \varepsilon)\) such that
$$\begin{aligned} m \bigl( \mathbb{R}^{n} \bigl( \bigl\vert U(t, \tau)u_{\tau} \bigr\vert \geq M_{1} \bigr) \bigr) \leq \delta_{1}\quad \text{for all } u_{\tau}\in D(\tau) \text{ and } \tau \leq\tau_{2}. \end{aligned}$$
(3.30)
By (1.2) and (1.3), we can choose \(M_{2}\) large enough such that
$$\begin{aligned}& \alpha_{1}\vert u\vert ^{p-1}-\beta_{1}\vert u\vert \leq f_{1}(u)\leq\alpha_{2}\vert u\vert ^{p-1}+ \beta_{2}\vert u\vert \quad \text{in } \mathbb{R}^{n}\ \bigl(U(t, \tau)u_{\tau}\geq M_{2} \bigr), \end{aligned}$$
(3.31)
$$\begin{aligned}& \alpha_{3}\vert u\vert ^{p-1}\leq f_{2}(u)\leq \alpha_{4}\vert u\vert ^{p-1} \quad \text{in } \mathbb{R}^{n}\ \bigl(U(t, \tau)u_{\tau}\geq M_{2} \bigr). \end{aligned}$$
(3.32)
Let \(M_{0}=\max\{M_{1}, M_{2}\}\) and \(\tau\leq\tau_{2}\). Multiplying Eq. (1.1) by \((u-M_{0})_{+}^{p-1}\) and integrating on \(\mathbb{R}^{n}\), we have
$$\begin{aligned}& \frac{1}{p}\frac{d}{dt} \int_{\mathbb{R}^{n}} \bigl\vert (u-M_{0})_{+} \bigr\vert ^{p}\,dx- \nu \int_{\mathbb{R}^{n}}\Delta u(u-M_{0})_{+}^{p-1} \,dx+\lambda \int_{\mathbb{R}^{n}}u(u-M_{0})_{+}^{p-1} \,dx \\& \quad{} + \int_{\mathbb{R}^{n}}f_{1}(u) (u-M_{0})_{+}^{p-1} \,dx+ \int_{\mathbb{R}^{n}}a(x)f_{2}(u) (u-M_{0})_{+}^{p-1} \,dx= \int_{\mathbb{R}^{n}}g(x, t) (u-M_{0})_{+}^{p-1} \,dx, \end{aligned}$$
where \((u-M_{0})_{+}\) denotes the positive part of \(u-M_{0}\), that is
$$\begin{aligned} (u-M_{0})_{+}= \textstyle\begin{cases} u-M_{0}, &u\geq M_{0}, \\ 0, &u< M_{0}. \end{cases}\displaystyle \end{aligned}$$
Let \(\Omega_{1}=\mathbb{R}^{n}(U(t, \tau)u_{\tau}\geq M_{0})\), we get
$$\begin{aligned} (u-M_{0})_{+}^{2p-2}\leq \vert u\vert ^{p-1}(u-M_{0})_{+}^{p-1} \quad \text{and} \quad (u-M_{0})_{+}^{p}\leq u(u-M_{0})_{+}^{p-1} \quad \text{in } \Omega_{1}. \end{aligned}$$
It follows from (3.31), (3.32), Young’s inequality and Hölder’s inequality that
$$\begin{aligned}& -\nu \int_{\mathbb{R}^{n}}\Delta u(u-M_{0})_{+}^{p-1} \,dx=\nu(p-1) \int_{\Omega_{1}}\vert \nabla u\vert ^{2}(u-M_{0})_{+}^{p-2} \,dx\geq0, \end{aligned}$$
(3.33)
$$\begin{aligned}& \begin{aligned}[b] \int_{\mathbb{R}^{n}}f_{1}(u) (u-M_{0})_{+}^{p-1} \,dx&\geq \int_{\Omega _{1}}\alpha_{1}\vert u\vert ^{p-1}(u-M_{0})_{+}^{p-1}\,dx \\ &\quad{} - \int_{\Omega_{1}}\beta _{1}\vert u\vert (u-M_{0})_{+}^{p-1}\,dx, \end{aligned} \end{aligned}$$
(3.34)
$$\begin{aligned}& \int_{\mathbb{R}^{n}}a(x)f_{2}(u) (u-M_{0})_{+}^{p-1} \,dx\geq\alpha_{3} \int_{\Omega_{1}}a(x)\vert u\vert ^{p-1}(u-M_{0})_{+}^{p-1} \,dx\geq0, \end{aligned}$$
(3.35)
$$\begin{aligned}& \begin{aligned}[b] \int_{\mathbb{R}^{n}}g(x, t) (u-M_{0})_{+}^{p-1} \,dx &\leq\frac{1}{2 \alpha_{1}} \int_{\Omega_{1}} \bigl\vert g(x, t) \bigr\vert ^{2}\,dx+ \frac{\alpha_{1}}{2} \int_{\Omega_{1}}(u-M_{0})_{+}^{2p-2}\,dx \\ &\leq\frac{1}{2\alpha_{1}} \int_{\Omega_{1}} \bigl\vert g(x, t) \bigr\vert ^{2}\,dx+ \frac{ \alpha_{1}}{2} \int_{\Omega_{1}}\vert u\vert ^{p-1}(u-M_{0})_{+}^{p-1} \,dx . \end{aligned} \end{aligned}$$
(3.36)
By (3.33)-(3.36), we get
$$\begin{aligned} \frac{1}{p}\frac{d}{dt} \int_{\Omega_{1}} \bigl\vert (u-M_{0})_{+} \bigr\vert ^{p}\,dx+(\lambda -\beta_{1}) \int_{\Omega_{1}} \bigl\vert (u-M_{0})_{+} \bigr\vert ^{p}\,dx\leq\frac{1}{2\alpha _{1}} \int_{\Omega_{1}} \bigl\vert g(x, t) \bigr\vert ^{2}\,dx, \end{aligned}$$
which implies that
$$\begin{aligned}& \frac{d}{dt}(t-\tau)e^{\sigma t} \int_{\Omega_{1}} \bigl\vert (u-M_{0})_{+} \bigr\vert ^{p}\,dx+c _{0}e^{\sigma t} \int_{\Omega_{1}} \bigl\vert (u-M_{0})_{+} \bigr\vert ^{p}\,dx \\& \quad \leq\frac{p(t-\tau)}{2\alpha_{1}}e^{\sigma t} \int_{\Omega_{1}}\bigl\vert g(x, t)\bigr\vert ^{2}\,dx, \end{aligned}$$
(3.37)
where \(u>0\) in \(\Omega_{1}\) and \(c_{0}= ( p(\lambda-\beta_{1})- \sigma ) (t-\tau)-1\). Since \(\sigma\in(0, \lambda-\beta_{1})\) and \(p>2\), there exists \(\tau_{3}=\tau_{3}(t, \varepsilon)<0\) such that
$$\begin{aligned} \bigl( p(\lambda-\beta_{1})-\sigma \bigr) (t-\tau)\geq1 \quad \text{for all } \tau\leq\tau_{3}. \end{aligned}$$
So integrating (3.37) over the interval \([\tau, t]\), we have
$$\begin{aligned} \int_{\Omega_{1}} \bigl\vert (u-M_{0})_{+} \bigr\vert ^{p}\,dx\leq\frac{p}{2\alpha_{1}}e^{- \sigma t} \int_{-\infty}^{t}e^{\sigma s} \int_{\Omega_{1}}\bigl\vert g(x, s)\bigr\vert ^{2}\,dx \,ds. \end{aligned}$$
By (3.29), we can obtain
$$\begin{aligned} \int_{\Omega_{1}} \bigl\vert (u-M_{0})_{+} \bigr\vert ^{p}\,dx\leq C\varepsilon \quad\text{for all } \tau\leq \tau_{3} \text{ and } u_{\tau}\in D(\tau), \end{aligned}$$
(3.38)
where \(C>0\) is a constant independent of \(M_{0}\). Set \(\Omega_{2}= \mathbb{R}^{n}(U(t, \tau)u_{\tau}\leq-M_{0})\). Likewise, replacing \((u-M_{0})_{+}\) with \((u+M_{0})_{-}\), we can also obtain that there exists \(\tau_{4}=\tau_{4}(t, \varepsilon)\) such that
$$\begin{aligned} \int_{\Omega_{2}} \bigl\vert (u+M_{0})_{-} \bigr\vert ^{p}\,dx\leq C\varepsilon \quad\text{for all } \tau \leq \tau_{4} \text{ and } u_{\tau}\in D(\tau), \end{aligned}$$
(3.39)
where \((u+M_{0})_{-}\) is the negative part of \(u+M_{0}\), that is
$$\begin{aligned} (u+M_{0})_{-}= \textstyle\begin{cases} u+M_{0}, &u\leq-M_{0}, \\ 0, &u> -M_{0}. \end{cases}\displaystyle \end{aligned}$$
Then it follows from (3.38) and (3.39) that
$$\begin{aligned} \int_{\mathbb{R}^{n}(\vert U(t, \tau)u_{\tau} \vert \geq M_{0})} \bigl\vert \bigl(\vert u\vert -M_{0} \bigr)\bigr\vert ^{p}\,dx \leq\varepsilon \quad\text{for all } \tau \leq \tau^{\prime\prime}(t, \varepsilon) \text{ and } u_{\tau}\in D(\tau), \end{aligned}$$
where \(\tau^{\prime\prime}(t, \varepsilon)=\min\{\tau_{3}, \tau _{4}\}\). Hence, we get
$$\begin{aligned}& \int_{\mathbb{R}^{n}(\vert U(t, \tau)u_{\tau} \vert \geq2M_{0})} \bigl\vert U(t, \tau)u_{\tau}\bigr\vert ^{p} \,dx \\& \quad= \int_{\mathbb{R}^{n}(\vert U(t, \tau)u_{\tau} \vert \geq2M_{0})} \bigl(\vert u\vert -M_{0}+M _{0} \bigr)^{p}\,dx \\& \quad\leq2^{p-1} \biggl( \int_{\mathbb{R}^{n}(\vert U(t, \tau)u_{\tau }\vert \geq2M _{0})} \bigl(\vert u\vert -M_{0} \bigr)^{p}\,dx+ \int_{\mathbb{R}^{n}(\vert U(t, \tau)u_{\tau} \vert \geq2M_{0})}(M_{0})^{p}\,dx \biggr) \\& \quad\leq2^{p-1} \biggl( \int_{\mathbb{R}^{n}(\vert U(t, \tau)u_{\tau }\vert \geq M_{0})}\bigl(\vert u\vert -M _{0} \bigr)^{p}\,dx+ \int_{\mathbb{R}^{n}(\vert U(t, \tau)u_{\tau} \vert \geq M_{0})} \bigl(\vert u\vert -M _{0} \bigr)^{p}\,dx \biggr) \leq2^{p}\varepsilon. \end{aligned}$$
Finally, we obtain (3.28) and the proof is complete. □
By Theorem 2.3, Lemma 3.4 and Lemma 3.5, we can obtain that the process \(\{U(t, \tau)\}_{\tau\leq t}\) associated with the initial value problem (1.1) is \(\mathscr{D}\)-pullback ω-limit compact in \(L^{p}(\mathbb{R}^{n})\). So it is easy to prove the following theorem.
Theorem 3.2
Assume that (1.2)-(1.4) hold and the external force
\(g\in L^{2}_{\mathrm {loc}}(\mathbb{R}, L^{2}(\mathbb{R}^{n}))\)
satisfies (1.5). Then the family of sets
\(\mathcal{A}^{\prime }=\{A^{\prime}(t):t\in\mathbb{R}\}\)
is
\((L^{2}(\mathbb{R}^{n}), L^{p}(\mathbb{R}^{n}))\)-\(\mathscr{D}\)-pullback attractors for
\(\{U(t, \tau)\}_{\tau\leq t}\).
Proof
We know that the family \(\hat{\mathscr{B}}_{1}=\{B_{1}(t): t\in \mathbb{R}\}\) is \((L^{2}(\mathbb{R}^{n}), L^{p}(\mathbb{R}^{n}))\)-\(\mathscr{D}\)-pullback absorbing for the process \(\{U(t, \tau)\}_{ \tau\leq t}\), where \(B_{1}(t)\) is defined by (3.26). Thus, by Theorem 2.4, we can deduce that the theorem is true. □