In this section, we derive uniform estimates on solutions of problem (1) when \(\tau\rightarrow-\infty\). These estimates are necessary to prove the existence of pullback \(\mathcal{D}\)-absorbing sets and the pullback asymptotic compactness of process \(\{U(t,\tau)\}_{t\geq\tau}\) associated with the system.
Lemma 3.1
Under the assumptions (4)-(5) and
\(f\in L_{\mathrm {loc}}^{2}(\mathbb {R};H)\). Let
\(3<\beta\leq5\), \(\tau\in \mathbb {R}\), and
\(u(t)\)
be the solution of problem (1). Then for any
\(\hat{D}\in\mathcal {D}\)
and
\(t\in \mathbb {R}\), there exists
\(\tau_{0}=\tau_{0}(t,\hat{D})< t\), such that
$$\begin{aligned}& \bigl\vert u(t)\bigr\vert _{2}^{2}\leq Ce^{-\sigma t} \int_{-\infty}^{t} e^{\sigma\xi} \bigl\vert f(\xi) \bigr\vert _{2}^{2}\,\mathrm{d}\xi; \\& \int_{\tau}^{t} e^{\sigma\xi}\bigl\| u(\xi) \bigr\| ^{2}\,\mathrm{d}\xi \leq C \int_{-\infty}^{t} e^{\sigma\xi}\bigl\vert f(\xi) \bigr\vert _{2}^{2}\,\mathrm{d}\xi; \\& \int_{\tau}^{t} e^{\sigma\xi}\bigl\vert u(\xi) \bigr\vert _{\beta+1}^{\beta+1} \,\mathrm{d}\xi\leq C \int_{-\infty}^{t} e^{\sigma\xi} \bigl\vert f(\xi) \bigr\vert _{2}^{2}\,\mathrm{d}\xi, \end{aligned}$$
for any
\(u_{\tau}\in D(\tau)\)
and
\(\tau\leq\tau_{0}(t,\hat{D})\).
Proof
Taking the inner product of (1)1 with u, we obtain
$$\begin{aligned} \frac{\mathrm{d}}{\mathrm{d}t}|u|_{2}^{2}+2\mu\| u\|^{2}+2 \alpha|u|_{\beta+1}^{\beta+1}&=2(f,u) \leq\mu\lambda_{1}|u|_{2}^{2}+ \frac{1}{\mu\lambda_{1}}|f|_{2}^{2} \\ &\leq\mu\| u\|^{2}+\frac{1}{\mu\lambda_{1}}|f|_{2}^{2}, \end{aligned}$$
where \(\lambda_{1}\) is the first eigenvalue of the Stokes operator. Thus,
$$ \frac{\mathrm{d}}{\mathrm{d}t}|u|_{2}^{2}+\frac{\mu\lambda _{1}}{2}|u|_{2}^{2}+ \frac{\mu}{2}\| u\|^{2}+2\alpha|u|_{\beta+1}^{\beta+1}\leq \frac{1}{\mu\lambda_{1}}|f|_{2}^{2}. $$
(6)
Multiplying (6) by \(e^{\sigma t}\) and then integrating over \((\tau, t)\), we derive that
$$\begin{aligned}& \bigl\vert u(t)\bigr\vert _{2}^{2} + \frac{\mu}{2}e^{-\sigma t} \int_{\tau}^{t} e^{\sigma\xi}\bigl\Vert u(\xi) \bigr\Vert ^{2}\,\mathrm{d}\xi+2\alpha e^{-\sigma t} \int_{\tau}^{t} e^{\sigma\xi}\bigl\vert u(\xi) \bigr\vert _{\beta+1}^{\beta+1}\,\mathrm{d}\xi \\& \quad \leq \biggl(\sigma-\frac{\mu}{2}\lambda_{1} \biggr)e^{-\sigma t} \int_{\tau}^{t} e^{\sigma\xi}\bigl\vert u(\xi) \bigr\vert _{2}^{2}\,\mathrm{d}\xi+\frac{1}{\mu\lambda_{1}}e^{-\sigma t} \int_{\tau}^{t} e^{\sigma\xi}\bigl\vert f(\xi) \bigr\vert _{2}^{2}\,\mathrm{d}\xi+e^{-\sigma t}e^{\sigma\tau } \bigl\vert u(\tau)\bigr\vert _{2}^{2}. \end{aligned}$$
Since \(0<\sigma<\frac{\mu\lambda_{1}}{2}\), we have
$$\begin{aligned}& \bigl\vert u(t)\bigr\vert _{2}^{2} + \frac{\mu}{2}e^{-\sigma t} \int_{\tau}^{t} e^{\sigma\xi}\bigl\Vert u(\xi) \bigr\Vert ^{2}\,\mathrm{d}\xi+2\alpha e^{-\sigma t} \int_{\tau}^{t} e^{\sigma\xi}\bigl\vert u(\xi) \bigr\vert _{\beta+1}^{\beta+1}\,\mathrm{d}\xi \\& \quad \leq\frac{1}{\mu\lambda_{1}}e^{-\sigma t} \int_{-\infty}^{t} e^{\sigma\xi}\bigl\vert f(\xi) \bigr\vert _{2}^{2}\,\mathrm{d}\xi+e^{-\sigma t}e^{\sigma\tau } \bigl\vert u(\tau)\bigr\vert _{2}^{2}. \end{aligned}$$
(7)
Since \(u(\tau)\in D(\tau)\), for every \(t\in \mathbb {R}\), there exists \(\tau_{0}=\tau_{0}(t,\hat{D})< t\) such that, for all \(\tau\leq\tau_{0}\),
$$ e^{\sigma\tau}\bigl\vert u(\tau)\bigr\vert _{2}^{2}\leq\frac{1}{\mu\lambda_{1}} \int_{-\infty}^{t} e^{\sigma\xi}\bigl\vert f(\xi) \bigr\vert _{2}^{2}\,\mathrm{d}\xi. $$
(8)
By (7) and (8), we find that
$$\begin{aligned}& \bigl\vert u(t)\bigr\vert _{2}^{2} + \frac{\mu}{2}e^{-\sigma t} \int_{\tau}^{t} e^{\sigma\xi}\bigl\Vert u(\xi) \bigr\Vert ^{2}\,\mathrm{d}\xi+2\alpha e^{-\sigma t} \int_{\tau}^{t} e^{\sigma\xi}\bigl\vert u(\xi) \bigr\vert _{\beta+1}^{\beta+1}\,\mathrm{d}\xi \\& \quad \leq\frac{2}{\mu\lambda_{1}}e^{-\sigma t} \int_{-\infty}^{t} e^{\sigma\xi}\bigl\vert f(\xi) \bigr\vert _{2}^{2}\,\mathrm{d}\xi. \end{aligned}$$
□
Lemma 3.2
Under the assumptions (4)-(5) and
\(f\in L_{\mathrm{loc}}^{2}(\mathbb {R};H)\). Let
\(3<\beta\leq5\), \(\tau\in \mathbb {R}\), and
\(u(t)\)
be the solution of problem (1). Then for any
\(\hat{D}\in\mathcal{D}\)
and
\(t\in \mathbb {R}\), there exists
\(\tau_{1}=\tau_{1}(t,\hat{D})< t-2\), such that
$$\begin{aligned}& \int_{t-2}^{t} e^{\sigma\xi}\bigl\vert u(\xi) \bigr\vert _{2}^{2}\,\mathrm{d}\xi\leq C \int_{-\infty}^{t} e^{\sigma\xi}\bigl\vert f(\xi) \bigr\vert _{2}^{2}\,\mathrm{d}\xi; \\& \int_{t-2}^{t} e^{\sigma\xi}\bigl\Vert u(\xi) \bigr\Vert ^{2}\,\mathrm{d}\xi \leq C \int_{-\infty}^{t} e^{\sigma\xi}\bigl\vert f(\xi) \bigr\vert _{2}^{2}\,\mathrm{d}\xi; \\& \int_{t-2}^{t} e^{\sigma\xi}\bigl\vert u(\xi) \bigr\vert _{\beta+1}^{\beta+1}\,\mathrm{d}\xi \leq C \int_{-\infty}^{t} e^{\sigma\xi}\bigl\vert f(\xi) \bigr\vert _{2}^{2}\,\mathrm{d}\xi, \end{aligned}$$
for any
\(u_{\tau}\in D(\tau)\)
and
\(\tau\leq\tau_{1}(t,\hat{D})\).
Proof
It follows from (6) that
$$ \frac{\mathrm{d}}{\mathrm{d}t}\bigl\vert u(t)\bigr\vert _{2}^{2}+\frac{\mu\lambda _{1}}{2}\bigl\vert u(t)\bigr\vert _{2}^{2}\leq\frac{1}{\mu\lambda_{1}}\bigl\vert f(t)\bigr\vert _{2}^{2}. $$
(9)
Let \(s\in[t-2,t]\). Multiplying (9) by \(e^{\sigma t}\), then relabeling t as ξ and integrating with respect to ξ over \((\tau,s)\), we get
$$\begin{aligned} e^{\sigma s}\bigl\vert u(s)\bigr\vert _{2}^{2}& \leq e^{\sigma\tau}\bigl\vert u(\tau)\bigr\vert _{2}^{2}+ \biggl(\sigma-\frac{\mu}{2}\lambda_{1}\biggr) \int_{\tau}^{s} e^{\sigma\xi}\bigl\vert u(\xi) \bigr\vert _{2}^{2}\,\mathrm{d}\xi+\frac{1}{\mu\lambda_{1}} \int_{\tau}^{s} e^{\sigma\xi}\bigl\vert f(\xi) \bigr\vert _{2}^{2}\,\mathrm{d}\xi \\ &\leq e^{\sigma\tau}\bigl\vert u(\tau)\bigr\vert _{2}^{2}+ \frac{1}{\mu\lambda_{1}} \int_{-\infty}^{t} e^{\sigma\xi}\bigl\vert f(\xi) \bigr\vert _{2}^{2}\,\mathrm{d}\xi. \end{aligned}$$
(10)
Since \(u(\tau)\in D(\tau)\), for every \(t\in \mathbb {R}\), there exists \(\tau_{1}=\tau_{1}(t,\hat{D})< t-2\), such that, for all \(\tau\leq\tau_{1}\),
$$ e^{\sigma\tau}\bigl\vert u(\tau)\bigr\vert _{2}^{2}\leq \frac{1}{\mu\lambda_{1}} \int_{-\infty}^{t} e^{\sigma\xi}\bigl\vert f(\xi) \bigr\vert _{2}^{2}\,\mathrm{d}\xi. $$
(11)
By (10) and (11), we have, for \(s\in[t-2,t]\),
$$ e^{\sigma s}\bigl\vert u(s)\bigr\vert _{2}^{2} \leq\frac{2}{\mu\lambda_{1}} \int_{-\infty}^{t} e^{\sigma\xi}\bigl\vert f(\xi) \bigr\vert _{2}^{2}\,\mathrm{d}\xi. $$
(12)
Integrating (12) with respect to s over the interval \((t-2,t)\) produces
$$ \int_{t-2}^{t} e^{\sigma s}\bigl\vert u(s)\bigr\vert _{2}^{2}\,\mathrm{d}s\leq\frac{4}{\mu\lambda_{1}} \int_{-\infty} ^{t} e^{\sigma\xi}\bigl\vert f(\xi) \bigr\vert _{2}^{2}\,\mathrm{d}\xi. $$
(13)
Multiplying (6) by \(e^{\sigma t}\), then relabeling t as ξ and integrating with respect to ξ over \((t-2,t)\), by (12) we obtain, for all \(\tau\leq\tau_{1}\),
$$\begin{aligned}& e^{\sigma t} \bigl\vert u(t)\bigr\vert _{2}^{2}+ \frac{\mu}{2} \int_{t-2}^{t} e^{\sigma\xi}\bigl\Vert u(\xi) \bigr\Vert ^{2}\,\mathrm{d}\xi+2\alpha \int_{t-2}^{t} e^{\sigma\xi}\bigl\vert u(\xi) \bigr\vert _{\beta+1}^{\beta+1}\,\mathrm{d}\xi \\& \quad \leq e^{\sigma(t-2)}\bigl\vert u(t-2)\bigr\vert _{2}^{2}+ \biggl(\sigma-\frac{\mu}{2}\lambda_{1}\biggr) \int_{t-2}^{t} e^{\sigma\xi}\bigl\vert u(\xi) \bigr\vert _{2}^{2}\,\mathrm{d}\xi+\frac{1}{\mu\lambda_{1}} \int _{-\infty}^{t} e^{\sigma\xi}\bigl\vert f(\xi) \bigr\vert _{2}^{2}\,\mathrm{d}\xi \\& \quad \leq\frac{3}{\mu\lambda_{1}} \int_{-\infty}^{t} e^{\sigma\xi}\bigl\vert f(\xi) \bigr\vert _{2}^{2}\,\mathrm{d}\xi, \end{aligned}$$
which along with (13) completes the proof. □
Corollary 3.1
Under the assumptions (4)-(5) and
\(f\in L_{\mathrm {loc}}^{2}(\mathbb {R};H)\). Let
\(3<\beta\leq5\), \(\tau\in \mathbb {R}\), and
\(u(t)\)
be the solution of problem (1). Then for any
\(\hat{D}\in\mathcal {D}\)
and
\(t\in \mathbb {R}\),
$$\begin{aligned}& \int_{t-2}^{t}\bigl\vert u(\xi)\bigr\vert _{2}^{2}\,\mathrm{d}\xi\leq Ce^{-\sigma t} \int_{-\infty}^{t} e^{\sigma\xi}\bigl\vert f(\xi) \bigr\vert _{2}^{2}\,\mathrm{d}\xi; \\& \int_{t-2}^{t} \bigl\Vert u(\xi)\bigr\Vert ^{2}\,\mathrm{d}\xi \leq Ce^{-\sigma t} \int_{-\infty}^{t} e^{\sigma\xi}\bigl\vert f(\xi) \bigr\vert _{2}^{2}\,\mathrm{d}\xi; \\& \int_{t-2}^{t} \bigl\vert u(\xi)\bigr\vert _{\beta+1}^{\beta+1}\,\mathrm{d}\xi \leq Ce^{-\sigma t} \int_{-\infty}^{t} e^{\sigma\xi}\bigl\vert f(\xi) \bigr\vert _{2}^{2}\,\mathrm{d}\xi, \end{aligned}$$
for any
\(u_{\tau}\in D(\tau)\)
and
\(\tau\leq\tau_{1}(t,\hat{D})\).
Proof
It is straightforward from Lemma 3.2. □
Lemma 3.3
Under the assumptions (4)-(5) and
\(f\in L_{\mathrm {loc}}^{2}(\mathbb {R};H)\). Let
\(3<\beta\leq5\), \(\tau\in \mathbb {R}\), and
\(u(t)\)
be the solution of problem (1). Then, for any
\(\hat{D}\in\mathcal {D}\)
and
\(t\in \mathbb {R}\),
$$ \bigl\Vert u(t)\bigr\Vert ^{2}+\bigl\vert u(t) \bigr\vert _{\beta+1}^{\beta+1}\leq Ce^{-\sigma t} \int _{-\infty}^{t} e^{\sigma\xi}\bigl\vert f(\xi) \bigr\vert _{2}^{2}\,\mathrm{d}\xi, $$
(14)
for any
\(u_{\tau}\in D(\tau)\)
and
\(\tau\leq\tau_{1}(t,\hat{D})\).
Proof
Multiplying the first equation of (1) by \(u_{t}\), \(-\Delta u\), respectively, and then integrating the resulting equation on Ω, we obtain
$$\begin{aligned}& |u_{t}|_{2}^{2}+\frac{\mu}{2}\frac{\mathrm{d}}{\mathrm{d}t} \| u\|^{2}+\frac{\alpha}{\beta+1}\frac{\mathrm{d}}{\mathrm {d}t}|u|_{\beta+1}^{\beta+1} \\& \quad =- \int_{\Omega}(u\cdot\nabla)uu_{t}\,\mathrm{d}x+(f,u_{t}), \end{aligned}$$
(15)
$$\begin{aligned}& \frac{1}{2}\frac{\mathrm{d}}{\mathrm{d}t}\| u\|^{2} +\mu|\Delta u|_{2}^{2}+\alpha \int_{\Omega}|u|^{\beta-1}|\nabla u|^{2}\, \mathrm{d}x+\frac{\alpha(\beta-1)}{4} \int_{\Omega}|u|^{\beta-3}\bigl\vert \nabla \vert u\vert ^{2}\bigr\vert ^{2}\,\mathrm{d}x \\& \quad = \int_{\Omega}(u\cdot\nabla)u\Delta u\,\mathrm{d}x-(f,\Delta u). \end{aligned}$$
(16)
From (15) we have
$$ \mu\frac{\mathrm{d}}{\mathrm{d}t}\Vert u\Vert ^{2}+ \frac{2\alpha}{\beta+1}\frac{\mathrm{d}}{\mathrm {d}t}|u|_{\beta+1}^{\beta+1}\leq \bigl\vert (u\cdot\nabla)u\bigr\vert _{2}^{2}+\bigl\vert f(t) \bigr\vert _{2}^{2}. $$
(17)
From (16) we get
$$\begin{aligned}& \frac{\mathrm{d}}{\mathrm{d}t}\| u\|^{2}+\mu|\Delta u|_{2}^{2}+2\alpha \int_{\Omega}|u|^{\beta-1}|\nabla u|^{2}\, \mathrm{d}x +\frac{\alpha(\beta-1)}{2} \int_{\Omega}|u|^{\beta-3}\bigl\vert \nabla \vert u\vert ^{2}\bigr\vert ^{2}\,\mathrm{d}x \\& \quad \leq \frac{2|(u\cdot\nabla)u|_{2}^{2}}{\mu}+\frac{2|f(t)|_{2}^{2}}{\mu}. \end{aligned}$$
(18)
Taking (17), (18) together, it follows that
$$\begin{aligned}& (\mu+1)\frac{\mathrm{d}}{\mathrm{d}t}\| u\|^{2} +\frac {2\alpha}{\beta+1} \frac{\mathrm{d}}{\mathrm{d}t}|u|_{\beta+1}^{\beta +1}+\mu|\Delta u|_{2}^{2}+2 \alpha \int_{\Omega}|u|^{\beta-1}|\nabla u|^{2}\, \mathrm{d}x \\& \qquad {}+\frac{\alpha(\beta-1)}{2} \int_{\Omega}|u|^{\beta-3}\bigl\vert \nabla \vert u\vert ^{2}\bigr\vert ^{2}\,\mathrm{d}x \\& \quad \leq\biggl(\frac{2}{\mu}+1\biggr) \bigl\vert (u\cdot\nabla)u\bigr\vert _{2}^{2}+\biggl(\frac{2}{\mu}+1\biggr)\bigl\vert f(t)\bigr\vert _{2}^{2}. \end{aligned}$$
(19)
From the proof of Theorem 3.1 in [13], we can find that, when \(3<\beta\leq5\),
$$\begin{aligned} J =&C\bigl\vert (u\cdot\nabla)u\bigr\vert _{2}^{2} \\ \leq& C \int_{\Omega}|u|^{2}|\nabla u|^{2}\, \mathrm {d}x \\ \leq&\alpha \int_{\Omega}|u|^{\beta-1}|\nabla u|^{2}\, \mathrm{d}x+\frac{\mu }{4} \int_{\Omega}|\Delta u|^{2}\,\mathrm{d}x+C \int_{\Omega}|u|^{\beta+1}\,\mathrm{d}x. \end{aligned}$$
(20)
Substituting (20) into (19), we find that
$$\begin{aligned} \begin{aligned}[b] & \frac{\mathrm{d}}{\mathrm{d}t}\biggl[(\mu+1)\bigl\Vert u(s)\bigr\Vert ^{2}+\frac {2\alpha}{\beta+1}\bigl\vert u(s)\bigr\vert _{\beta+1}^{\beta+1}\biggr] \\ & \quad \leq C\bigl\vert u(s)\bigr\vert _{\beta +1}^{\beta+1}+\biggl(\frac{2}{\mu}+1\biggr)\bigl\vert f(s)\bigr\vert _{2}^{2} \\ & \quad \leq C\biggl[(\mu+1)\bigl\Vert u(s)\bigr\Vert ^{2}+ \frac{2\alpha}{\beta +1}\bigl\vert u(s)\bigr\vert _{\beta+1}^{\beta+1} \biggr] +\biggl(\frac{2}{\mu}+1\biggr)\bigl\vert f(s)\bigr\vert _{2}^{2}. \end{aligned} \end{aligned}$$
(21)
Applying the uniform Gronwall lemma to (21) on interval \([t-1,t]\), we have
$$\begin{aligned}& (\mu+1)\bigl\Vert u( t)\bigr\Vert ^{2}+\frac{2\alpha}{\beta+1}\bigl\vert u(t)\bigr\vert _{\beta+1}^{\beta+1} \\& \quad \leq C \biggl( \int_{t-1}^{t} \biggl[(\mu+1)\bigl\Vert u(\xi)\bigr\Vert ^{2}+\frac{2\alpha}{\beta+1}\bigl\vert u(\xi)\bigr\vert _{\beta+1}^{\beta +1} \biggr]\,\mathrm{d}\xi+ \biggl( \frac{2}{\mu}+1\biggr) \int_{t-1}^{t} \bigl\vert f(\xi)\bigr\vert _{2}^{2}\,\mathrm{d}\xi \biggr). \end{aligned}$$
Noticing
$$\begin{aligned} \int_{t-1}^{t} \bigl\vert f(\xi)\bigr\vert _{2}^{2}\,\mathrm{d}\xi&=e^{-\sigma(t-1)} \int_{t-1}^{t} e^{\sigma(t-1)}\bigl\vert f(\xi) \bigr\vert _{2}^{2}\,\mathrm{d}\xi \\ &\leq e^{-\sigma(t-1)} \int_{t-1}^{t} e^{\sigma\xi}\bigl\vert f(\xi) \bigr\vert _{2}^{2}\,\mathrm{d}\xi \\ &\leq Ce^{-\sigma t} \int_{-\infty}^{t} e^{\sigma\xi}\bigl\vert f(\xi) \bigr\vert _{2}^{2}\,\mathrm{d}\xi, \end{aligned}$$
according to Corollary 3.1, we have
$$ (\mu+1)\bigl\Vert u(t)\bigr\Vert ^{2}+\frac{2\alpha}{\beta+1}\bigl\vert u(t)\bigr\vert _{\beta+1}^{\beta+1}\leq Ce^{-\sigma t} \int_{-\infty}^{t} e^{\sigma\xi}\bigl\vert f(\xi) \bigr\vert _{2}^{2}\,\mathrm{d}\xi. $$
□
Lemma 3.4
Under the assumptions (4)-(5) and
\(f\in L_{\mathrm {loc}}^{2}(\mathbb {R};H)\). Let
\(3<\beta\leq5\), \(\tau\in \mathbb {R}\), and
\(u(t)\)
be the solution of problem (1). Then for any
\(\hat{D}\in\mathcal {D}\)
and
\(t\in \mathbb {R}\),
$$ \int_{t-1}^{t}\bigl\vert \Delta u(\xi)\bigr\vert _{2}^{2}\,\mathrm{d}\xi\leq Ce^{-\sigma t} \int_{-\infty}^{t} e^{\sigma\xi}\bigl\vert f(\xi) \bigr\vert _{2}^{2}\,\mathrm{d}\xi,\quad \forall t\in \mathbb {R}, $$
for any
\(u_{\tau}\in D(\tau)\)
and
\(\tau\leq\tau_{1}(t,\hat{D})\).
Proof
Similar to the proof of Lemma 3.3, applying the uniform Gronwall lemma to (21) on interval \([t-2,t-1]\), we can also prove
$$ (\mu+1)\bigl\Vert u(t-1)\bigr\Vert ^{2}+ \frac{2\alpha}{\beta +1}\bigl\vert u(t-1)\bigr\vert _{\beta+1}^{\beta+1} \leq Ce^{-\sigma t} \int_{-\infty}^{t} e^{\sigma\xi}\bigl\vert f(\xi) \bigr\vert _{2}^{2}\,\mathrm{d}\xi. $$
(22)
Thanks to (18), (20), we have
$$ \frac{\mathrm{d}}{\mathrm{d}t}\| u\|^{2}+\frac{3}{4}\mu|\Delta u|_{2}^{2}+\alpha \int_{\Omega}|u|^{\beta -1}|\nabla u|^{2}\, \mathrm{d}x\leq C|u|_{\beta+1}^{\beta+1}+\frac{2}{\mu}\bigl\vert f(t)\bigr\vert _{2}^{2}. $$
(23)
Integrating (23) from \(t-1\) to t, according to Corollary 3.1 and (22), we can obtain
$$\begin{aligned}& \bigl\Vert u(t)\bigr\Vert ^{2}+\frac{3}{4}\mu \int_{t-1}^{t} \bigl\vert \Delta u(\xi)\bigr\vert _{2}^{2}\,\mathrm{d}\xi+\alpha \int_{t-1}^{t} \int_{\Omega}\bigl\vert u(\xi )\bigr\vert ^{\beta-1}\bigl\vert \nabla u(\xi)\bigr\vert ^{2}\,\mathrm{d}x\, \mathrm{d}\xi \\& \quad \leq\bigl\Vert u(t-1)\bigr\Vert ^{2}+C \int_{t-1}^{t} \bigl\vert u(\xi)\bigr\vert _{\beta +1}^{\beta+1}\,\mathrm{d}\xi+\frac{2}{\mu} \int_{t-1}^{t} \bigl\vert f(\xi)\bigr\vert _{2}^{2}\, \mathrm {d}\xi \\& \quad \leq Ce^{-\sigma t} \int_{-\infty}^{t} e^{\sigma\xi}\bigl\vert f(\xi) \bigr\vert _{2}^{2}\, \mathrm {d}\xi, \end{aligned}$$
(24)
which completes the proof. □
Lemma 3.5
Under the assumptions (4)-(5) and
\(f\in L_{\mathrm {loc}}^{2}(\mathbb {R};H)\), \(\frac{\mathrm{d}f}{\mathrm{d}t}\in L_{b}^{2}(\mathbb {R};H)\). Let
\(3<\beta\leq5\), \(\tau\in \mathbb {R}\), and
\(u(t)\)
be the solution of problem (1). Then for any
\(\hat{D}\in\mathcal {D}\)
and
\(t\in \mathbb {R}\), there exists a family of positive constants
\(\{r_{1}(t):t\in \mathbb {R}\}\)
such that
$$ \bigl\vert u_{t}(t)\bigr\vert _{2}^{2}\leq r_{1}(t), $$
for any
\(u_{\tau}\in D(\tau)\)
and
\(\tau\leq\tau_{1}(t,\hat{D})\), where
\(r_{1}(t)\)
is a positive constant which is independent of the initial data.
Proof
From (15) and (20) we deduce that
$$\begin{aligned}& |u_{t}|_{2}^{2}+\frac{\mu}{2} \frac{\mathrm{d}}{\mathrm{d}t}\| u\|^{2}+\frac{\alpha}{\beta+1}\frac{\mathrm{d}}{\mathrm {d}t}|u|_{\beta+1}^{\beta+1} \\& \quad \leq \frac{|u_{t}|_{2}^{2}}{2}+\bigl\vert (u\cdot\nabla)u\bigr\vert _{2}^{2}+\bigl\vert f(t)\bigr\vert _{2}^{2} \\& \quad \leq \frac{|u_{t}|_{2}^{2}}{2}+\alpha \int_{\Omega}|u|^{\beta-1}|\nabla u|^{2}\, \mathrm{d}x+\frac{\mu}{4}|\Delta u|_{2}^{2} +C|u|_{\beta+1}^{\beta+1}+\bigl\vert f(t)\bigr\vert _{2}^{2}. \end{aligned}$$
(25)
Thus
$$\begin{aligned}& |u_{t}|_{2}^{2}+\mu\frac{\mathrm{d}}{\mathrm{d}t} \| u\|^{2}+\frac{2\alpha}{\beta+1}\frac{\mathrm{d}}{\mathrm {d}t}|u|_{\beta+1}^{\beta+1} \\& \quad \leq\frac{\mu}{2}|\Delta u|_{2}^{2}+2\alpha \int_{\Omega}|u|^{\beta-1}|\nabla u|^{2}\, \mathrm {d}x+C|u|_{\beta+1}^{\beta+1}+2\bigl\vert f(t)\bigr\vert _{2}^{2}. \end{aligned}$$
(26)
Integrating (26) from \(t-1\) to t, according to Corollary 3.1, Lemma 3.4, (22), and (24), we can obtain
$$\begin{aligned} \int_{t-1}^{t} \bigl\vert u_{\xi}(\xi)\bigr\vert _{2}^{2}\,\mathrm{d}\xi \leq&\mu\bigl\Vert u(t-1) \bigr\Vert ^{2}+\frac{2\alpha}{\beta+1}\bigl\vert u(t-1)\bigr\vert _{\beta+1}^{\beta+1} +\frac{\mu}{2} \int_{t-1}^{t} \bigl\vert \Delta u(\xi)\bigr\vert _{2}^{2}\,\mathrm{d}\xi \\ &{}+2\alpha \int_{t-1}^{t} \int_{\Omega}\bigl\vert u(\xi)\bigr\vert ^{\beta-1}\bigl\vert \nabla u(\xi )\bigr\vert ^{2}\,\mathrm{d}x\,\mathrm{d}\xi \\ &{}+C \int_{t-1}^{t}\bigl\vert u(\xi)\bigr\vert _{\beta+1}^{\beta +1}\,\mathrm{d}\xi +2 \int_{t-1}^{t} \bigl\vert f(\xi)\bigr\vert _{2}^{2}\,\mathrm{d}\xi \\ \leq& Ce^{-\sigma t} \int_{-\infty}^{t} e^{\sigma\xi}\bigl\vert f(\xi) \bigr\vert _{2}^{2}\, \mathrm {d}\xi. \end{aligned}$$
(27)
We now differentiate (3)1 with respect to t and then take the inner product with \(u_{t}\) in H to obtain
$$ \frac{1}{2}\frac{\mathrm{d}}{\mathrm{d}t}|u_{t}|_{2}^{2}+ \mu\| u_{t}\|^{2}\leq\bigl\vert b(u_{t},u,u_{t}) \bigr\vert - \int_{\Omega}\bigl(F'(u)u_{t}\bigr)\cdot u_{t}\,\mathrm{d}x+(f_{t},u_{t}). $$
According to Lemma 2.4 in [16], \((F'(u)u_{t})\cdot u_{t}\) is positive definite, hence we have
$$\begin{aligned} \frac{1}{2}\frac{\mathrm{d}}{\mathrm{d}t}|u_{t}|_{2}^{2}+ \mu\| u_{t}\|^{2}&\leq\bigl\vert b(u_{t},u,u_{t}) \bigr\vert +|f_{t}|_{2}\cdot|u_{t}|_{2} \\ &\leq C|u_{t}|_{2}^{1/2}\| u_{t} \|^{3/2}\| u\|+\frac{1}{2}|u_{t}|_{2}^{2}+ \frac{1}{2}|f_{t}|_{2}^{2} \\ &\leq\frac{\mu}{2}\| u_{t}\|^{2}+C\bigl(1+\| u \|^{4}\bigr)|u_{t}|_{2}^{2}+ \frac{1}{2}|f_{t}|_{2}^{2}. \end{aligned}$$
(28)
Thus,
$$ \frac{\mathrm{d}}{\mathrm{d}t}|u_{t}|_{2}^{2}\leq C\bigl(1+\| u\|^{4}\bigr)|u_{t}|_{2}^{2}+|f_{t}|_{2}^{2}. $$
(29)
Thanks to (14),we have
$$\begin{aligned} C \int_{t-1}^{t} \bigl(1+\bigl\Vert u(\xi)\bigr\Vert ^{4}\bigr)\,\mathrm{d}\xi&\leq C \biggl(1+ \int_{t-1}^{t} \biggl(Ce^{-\sigma s} \int_{-\infty}^{s} e^{\sigma\xi}\bigl\vert f(\xi) \bigr\vert _{2}^{2}\, \mathrm {d}\xi \biggr)^{2}\, \mathrm{d}s \biggr) \\ &=C+C \int_{t-1}^{t} r_{0}^{2}(s)\, \mathrm{d}s, \end{aligned}$$
where \(r_{0}(s)=Ce^{-\sigma s}\int_{-\infty}^{s} e^{\sigma\xi}|f(\xi )|_{2}^{2}\,\mathrm{d}\xi\).
Applying the uniform Gronwall lemma to (29) on interval \([t-1,t]\), we can get
$$\begin{aligned} \bigl\vert u_{t}(t)\bigr\vert _{2}^{2}&\leq \biggl\{ Ce^{-\sigma t} \int_{-\infty}^{t} e^{\sigma\xi }\bigl\vert f(\xi) \bigr\vert _{2}^{2}\,\mathrm{d}\xi+ \int_{t-1}^{t} \bigl\vert f_{\xi}(\xi)\bigr\vert _{2}^{2}\,\mathrm{d}\xi \biggr\} \cdot\exp\biggl\{ C+C \int _{t-1}^{t} r_{0}^{2}(s)\, \mathrm{d}s\biggr\} \\ &\equiv r_{1}(t). \end{aligned}$$
□
Lemma 3.6
Under the assumptions (4)-(5) and
\(f\in L_{\mathrm {loc}}^{2}(\mathbb {R};H)\), \(\frac{\mathrm{d}f}{\mathrm{d}t}\in L_{b}^{2}(\mathbb {R};H)\). Let
\(3<\beta\leq5\), \(\tau\in \mathbb {R}\), and
\(u(t)\)
be the solution of problem (1). Then for any
\(\hat{D}\in\mathcal{D}\)
and
\(t\in \mathbb {R}\), there exists a family of positive constants
\(\{r_{2}(t):t\in \mathbb {R}\}\)
such that
$$ \bigl\vert Au(t)\bigr\vert _{2}\leq r_{2}(t), $$
(30)
for any
\(u_{\tau}\in D(\tau)\)
and
\(\tau\leq\tau_{1}(t,\hat{D})\), where
\(r_{2}(t)\)
is a positive constant which is independent of the initial data.
Proof
Like the proof of Proposition 5 in [16], we can obtain
$$\begin{aligned} \frac{\mu}{2}\bigl\vert Au(t)\bigr\vert _{2} \leq& \bigl\vert u_{t}(t)\bigr\vert _{2}+C\bigl\| u(t)\bigr\| ^{3}+C \bigl\vert u(t)\bigr\vert _{\beta+1}^{\frac{\beta^{2}+4\beta+3}{10-2\beta}} +\bigl\vert f(t) \bigr\vert _{2} \\ \leq& \bigl(r_{1}(t)\bigr)^{1/2}+ \biggl(Ce^{-\sigma t} \int_{-\infty}^{t} e^{\sigma\xi}\bigl\vert f(\xi ) \bigr\vert _{2}^{2}\,\mathrm{d}\xi \biggr)^{3/2} \\ &{}+ \biggl(Ce^{-\sigma t} \int_{-\infty}^{t} e^{\sigma\xi}\bigl\vert f(\xi) \bigr\vert _{2}^{2}\,\mathrm{d}\xi \biggr)^{\frac{\beta+3}{10-2\beta }}+ \bigl\vert f(t)\bigr\vert _{2} \\ \leq& r_{1}(t)^{1/2}+C\bigl(r_{0}(t) \bigr)^{3/2}+C\bigl(r_{0}(t)\bigr)^{\frac{\beta+3}{10-2\beta }}+\bigl\vert f(t)\bigr\vert _{2}\equiv r_{2}(t). \end{aligned}$$
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Lemma 3.7
Under the assumptions (4)-(5) and
\(f\in L_{\mathrm{loc}}^{2}(\mathbb {R};H)\), \(\frac{\mathrm{d}f}{\mathrm{d}t}\in L_{b}^{2}(\mathbb {R};H)\). Let
\(3<\beta\leq5\), \(\tau\in \mathbb {R}\), and
\(u(t)\)
be the solution of problem (1). Then for any
\(\hat{D}\in\mathcal{D}\)
and
\(t\in \mathbb {R}\), there exists a family of positive constants
\(\{r_{3}(t):t\in \mathbb {R}\}\)
such that
$$ \bigl\Vert u_{t}(t+1)\bigr\Vert ^{2} \leq r_{3}(t), $$
for any
\(u_{\tau}\in D(\tau)\)
and
\(\tau\leq\tau_{1}(t,\hat{D})\), where
\(r_{3}(t)\)
is a positive constant which is independent of the initial data.
Proof
From inequality (28) we have
$$ \frac{\mathrm{d}}{\mathrm{d}t}|u_{t}|_{2}^{2}+\mu \| u_{t}\|^{2}\leq C\bigl(1+\| u\|^{4} \bigr)|u_{t}|_{2}^{2}+|f_{t}|_{2}^{2}. $$
(31)
Integrating the above inequality from t to \(t+1\), then we have
$$\begin{aligned} \mu \int_{t}^{t+1}\bigl\Vert u_{t}(s)\bigr\Vert ^{2}\,\mathrm{d}s \leq& \bigl\vert u_{t}(t)\bigr\vert _{2}^{2}+C \int_{t}^{t+1}\bigl(1+\bigl\Vert u(s)\bigr\Vert ^{4}\bigr)\bigl\vert u_{t}(s)\bigr\vert _{2}^{2}\,\mathrm{d}s+ \int_{t}^{t+1}\bigl\vert f_{t}(s)\bigr\vert _{2}^{2}\,\mathrm{d}s \\ \leq& r_{1}(t)+C \int_{t}^{t+1}\bigl(1+\bigl(r_{0}(s) \bigr)^{2}\bigr)r_{1}(s)\,\mathrm{d}s+ \int _{t}^{t+1}\bigl\vert f_{t}(s)\bigr\vert _{2}^{2}\,\mathrm{d}s \\ \equiv&\rho_{1}(t). \end{aligned}$$
(32)
By Lemma 3.6, we know that
$$ \bigl\Vert u(t) \bigr\Vert _{D(A)}\leq r_{2}(t), $$
so using the Agmon inequality we obtain
$$ \bigl\vert u(t)\bigr\vert _{\infty}\leq C\bigl\Vert u(t)\bigr\Vert ^{1/2}\bigl\Vert u(t)\bigr\Vert _{D(A)}^{1/2}\leq C \bigl(r_{0}(t)\bigr)^{1/4}\bigl(r_{2}(t) \bigr)^{1/2}\equiv\rho_{2}(t). $$
We now differentiate (3)1 with respect to t, then taking the inner product with \(Au_{t}\) in H to obtain
$$\begin{aligned} \frac{1}{2}\frac{\mathrm{d}}{\mathrm{d}t}\| u_{t} \|^{2}+\mu |Au_{t}|_{2}^{2} \leq&\bigl\vert b(u_{t}, u,Au_{t})\bigr\vert +\bigl\vert b(u,u_{t},Au_{t})\bigr\vert \\ &{}+ \int_{\Omega}\bigl(F'(u)u_{t}\bigr)\cdot Au_{t}\,\mathrm{d}x +(f_{t},Au_{t}). \end{aligned}$$
(33)
According to Lemma 2.4 in [16], for any \(u,v,w\in \mathbb {R}^{3}\), \(|(F'(u)v)\cdot w|\leq C|u|^{\beta-1}|v||w|\), so
$$\begin{aligned} \int_{\Omega}\bigl(F'(u)u_{t}\bigr)\cdot Au_{t}\,\mathrm{d}x&\leq C \int_{\Omega}|u|^{\beta -1}|u_{t}||Au_{t}| \,\mathrm{d}x\leq C|u|_{\infty}^{\beta -1}|u_{t}|_{2}|Au_{t}|_{2} \\ &\leq C\bigl(\rho_{2}(t)\bigr)^{\beta-1}\| u_{t} \||Au_{t}|_{2}\leq\frac{\mu }{8}|Au_{t}|_{2}^{2}+C \bigl(\rho_{2}(t)\bigr)^{2(\beta-1)}\| u_{t} \|^{2}. \end{aligned}$$
(34)
Because
$$\begin{aligned}& \bigl\vert b(u_{t},u,Au_{t})\bigr\vert \leq C\| u_{t}\|\| u\| ^{1/2}|Au|_{2}^{1/2}|Au_{t}|_{2} \leq\frac{\mu}{8}|Au_{t}|_{2}^{2}+C\| u \||Au|_{2}\| u_{t}\|^{2} \\& \hphantom{\bigl\vert b(u_{t},u,Au_{t})\bigr\vert }\leq\frac{\mu}{8}|Au_{t}|_{2}^{2}+C \bigl(r_{0}(t)\bigr)^{1/2}r_{2}(t)\| u_{t} \|^{2}, \end{aligned}$$
(35)
$$\begin{aligned}& \bigl\vert b(u,u_{t},Au_{t})\bigr\vert \leq C\| u\|\| u_{t}\| ^{1/2}|Au_{t}|_{2}^{1/2}|Au_{t}|_{2} \leq\frac{\mu}{8}|Au_{t}|_{2}^{2}+C\| u \|^{4}\| u_{t}\|^{2} \\& \hphantom{\bigl\vert b(u,u_{t},Au_{t})\bigr\vert }\leq\frac{\mu}{8}|Au_{t}|_{2}^{2}+C \bigl(r_{0}(t)\bigr)^{2}\| u_{t} \|^{2}, \end{aligned}$$
(36)
$$\begin{aligned}& (f_{t},Au_{t})\leq\frac{\mu}{8}|Au_{t}|_{2}^{2}+ \frac{2}{\mu}|f_{t}|_{2}^{2}, \end{aligned}$$
(37)
combining (34)-(37) with (33), we get
$$ \frac{\mathrm{d}}{\mathrm{d}t}\| u_{t}\|^{2}\leq C\bigl[ \bigl(\rho _{2}(t)\bigr)^{2(\beta-1)}+\bigl(r_{0}(t) \bigr)^{1/2}r_{2}(t)+\bigl(r_{0}(t) \bigr)^{2}\bigr]\| u_{t}\|^{2}+\frac{4}{\mu}|f_{t}|_{2}^{2}. $$
(38)
Thanks to (32), applying the uniform Gronwall lemma to (38) on interval \([t,t+1]\), we get
$$\begin{aligned} \bigl\Vert u_{t}(t+1)\bigr\Vert ^{2} \leq& \biggl(\frac{\rho_{1}(t)}{\mu}+\frac {4}{\mu} \int_{t}^{t+1}\bigl\vert f_{t}(s)\bigr\vert _{2}^{2}\,\mathrm{d}s \biggr) \\ &{} \cdot\exp \biggl\{ C \int_{t}^{t+1}\bigl[\bigl(\rho_{2}(s) \bigr)^{2(\beta -1)}+\bigl(r_{0}(s)\bigr)^{1/2}r_{2}(s)+ \bigl(r_{0}(s)\bigr)^{2}\bigr]\,\mathrm{d}s \biggr\} \\ =&r_{3}(t). \end{aligned}$$
(39)
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