On asymptotically almost periodic solutions to the Navier–Stokes equations in hyperbolic manifolds

Abstract

In this paper, we study the forward asymptotically almost periodic (AAP-) mild solutions of Navier–Stokes equations on the real hyperbolic manifold $\mathcal {M}=\mathbb {H}^d(\mathbb {R})$ with dimension $d \geqslant 2$. Using the dispersive and smoothing estimates for the Stokes equation, we invoke the Massera-type principle to prove the existence and uniqueness of the AAP- mild solution for the inhomogeneous Stokes equations in $L^p(\Gamma (T\mathcal {M})))$ space with $1<p\leqslant d$. Next, we establish the existence and uniqueness of the small AAP- mild solutions of the Navier–Stokes equations using the fixed point argument, and the results of inhomogeneous Stokes equations. The asymptotic behaviour (exponential decay and stability) of these small solutions is also related. This work, together with our recent work (Xuan et al. in J Math Anal Appl 517(1):1–19, 2023), provides a full existence and asymptotic behaviour of AAP- mild solutions of Navier–Stokes equations in $L^p(\Gamma (T\mathcal {M})))$ spaces for all $p>1$.

Appendix

In this part, we will prove the boundedness of the following integrals which are used in the previous sections

$$\begin{aligned} I_1= & {} \int _0^t [h_d(t-\tau )]^{\frac{1+\delta }{d}}[h_d(\tau )]^{\frac{1-\delta }{d}}e^{-\beta _1(t-\tau )}e^{-\sigma \tau }\textrm{d}\tau<+\infty ,\\ I_2= & {} \int _0^t [h_d(t-\tau )]^{\frac{1+\delta }{d}}[h_d(\tau )]^{\frac{1-\delta }{d}}e^{-\hat{\beta }_1(t-\tau )}e^{-\sigma \tau }\textrm{d}\tau<+\infty ,\\ I_3= & {} \int _0^t [h_d(t-\tau )]^{\frac{1+\delta }{d}}[h_d(\tau )]^{\frac{2(1-\delta )}{d}}e^{-({\beta }-\sigma )(t-\tau )}e^{-\sigma \tau }\textrm{d}\tau<+\infty ,\\ I_4= & {} \int _0^t [h_d(t-\tau )]^{\frac{1+\delta }{d}}[h_d(\tau )]^{\frac{1-\delta }{d}}e^{-({\beta } -\sigma )(t-\tau )} \textrm{d}\tau<+\infty ,\\ I_5= & {} \int _0^t [h_d(t-\tau )]^{\frac{1+\delta }{d}} [h_d(\tau )]^{\frac{1-\delta }{d}} e^{-(\beta _1-\sigma ) (t-\tau )} \textrm{d}\tau<+\infty ,\\ \tilde{I}_5= & {} \int _0^t [h_d(t-\tau )]^{\frac{1+\delta }{d}}e^{-(\beta _1-\sigma )(t-\tau )}\textrm{d}\tau<+\infty ,\\ I_6= & {} \int _0^t [h_d(t-\tau )]^{\frac{1+\delta }{d}}[h_d(\tau )]^{\frac{1-\delta }{d}}e^{-({\beta }-\sigma -\sigma ^*)(t-\tau )} \textrm{d}\tau<+\infty \\ \tilde{I}_6= & {} \int _0^t [h_d(t-\tau )]^{\frac{1+\delta }{d}}[h_d(\tau )]^{\frac{2(1-\delta )}{d}}e^{-({\beta }-\sigma ^*-\sigma )(t-\tau )} \textrm{d}\tau <+\infty , \end{aligned}$$

where t>0, $1<p\leqslant d$ and $0<\sigma <{\beta }$ for $I_3,\, I_4$; $0<\sigma <\beta _1$ for $I_5,\, \tilde{I}_5$; $0<\sigma +\sigma ^*<{\beta }$ for $I_6,\, \tilde{I}_6$.

The boundedness of integrals $I_1,\, I_2,\, I_3,\, I_4$ is useful to the proof of Assertion (i) in Lemma 3.2. The boundedness of integrals $I_5,\, I_6, \, \tilde{I}_6$ serves to the proof of Assertion (ii) in Theorem 4.1. Moreover, if we consider the integral on $(-\infty ,t]$ for all $t\in \mathbb {R}$, then we have the following bounded integrals

$$\begin{aligned} \tilde{I}_1= & {} \int _{-\infty }^t [h_d(t-\tau )]^{\frac{1+\delta }{d}}\lambda ^{-1}(\tau )e^{-\beta _1(t-\tau )}\textrm{d}\tau<+\infty ,\\ \tilde{I}_2= & {} \int _{-\infty }^t [h_d(t-\tau )]^{\frac{1+\delta }{d}}\lambda ^{-1}(\tau )e^{-\hat{\beta }_1(t-\tau )}\textrm{d}\tau<+\infty ,\\ \tilde{I}_3= & {} \lambda (t)\int _{-\infty }^t [h_d(t-\tau )]^{\frac{1+\delta }{d}}\lambda ^{-2}(\tau )e^{-{\beta }(t-\tau )}\textrm{d}\tau<+\infty ,\\ \tilde{I}_4= & {} \lambda (t)\int _{-\infty }^t [h_d(t-\tau )]^{\frac{1+\delta }{d}}\lambda ^{-1}(\tau )e^{-{\beta }(t-\tau )} \textrm{d}\tau <+\infty . \end{aligned}$$

These boundedness are useful to prove the Assertion (ii) in Lemma 3.2.

The boundedness of integral $I_1,\, I_2,\, I_4,\, I_5$ and $I_6$ is similar. We prove only the boundedness of $I_5$. Indeed, we consider the following cases of t:

${{\textrm{Case}}\, 1:\,0<t<1.}$ We have

$$\begin{aligned} I_5= & {} \int _0^t [h_d(t-\tau )]^{\frac{1+\delta }{d}} [h_d(\tau )]^{\frac{1-\delta }{d}} e^{-(\beta _1-\sigma ) (t-\tau )} \textrm{d}\tau \\\leqslant & {} \int _0^t (t-\tau )^{-\frac{1+\delta }{2}} \tau ^{-\frac{1-\delta }{2}} \textrm{d}\tau \\\leqslant & {} \textbf{B}\left( \frac{1-\delta }{2},\,\frac{1+\delta }{2} \right) <+\infty , \end{aligned}$$

where $\textbf{B}(\cdot ,\, \cdot )$ is beta function.

${{\textrm{Case}}\,2:\,1\leqslant t.}$ We have

$$\begin{aligned} I_5= & {} \int _0^t [h_d(t-\tau )]^{\frac{1+\delta }{d}} [h_d(\tau )]^{\frac{1-\delta }{d}} e^{-(\beta _1-\sigma ) (t-\tau )} \textrm{d}\tau \\= & {} \int _0^1 [h_d(t-\tau )]^{\frac{1+\delta }{d}} [h_d(\tau )]^{\frac{1-\delta }{d}} e^{-(\beta _1-\sigma ) (t-\tau )} \textrm{d}\tau \\{} & {} + \int _1^t [h_d(t-\tau )]^{\frac{1+\delta }{d}} [h_d(\tau )]^{\frac{1-\delta }{d}} e^{-(\beta _1-\sigma ) (t-\tau )} \textrm{d}\tau \\\leqslant & {} \int _0^1 \left( (t-\tau )^{-\frac{1+\delta }{2}}+1\right) \tau ^{-\frac{1-\delta }{2}} e^{-(\beta _1 - \sigma ) (t-\tau )} \textrm{d}\tau \\{} & {} + \int _1^t (t-\tau )^{-\frac{1+\delta }{2}} e^{-(\beta _1-\sigma ) (t-\tau )} \textrm{d}\tau \\\leqslant & {} \int _0^t (t-\tau )^{-\frac{1+\delta }{2}}\tau ^{-\frac{1-\delta }{2}} \textrm{d}\tau + \int _0^1 \tau ^{-\frac{1-\delta }{2}} \textrm{d}\tau \\{} & {} + \int _0^t (t-\tau )^{-\frac{1+\delta }{2}} e^{-(\beta _1-\sigma ) (t-\tau )} \textrm{d}\tau \\\leqslant & {} \textbf{B}\left( \frac{1-\delta }{2},\, \frac{1+\delta }{2} \right) + \frac{2}{1+\delta } + (\beta _1-\sigma )^{-\frac{1-\delta }{2}}\mathbf {\Gamma }\left( \frac{1-\delta }{2}\right) <+\infty , \end{aligned}$$

where $\mathbf {\Gamma }(\cdot )$ is gamma function.

The boundedness of integrals $I_3$ and $\tilde{I}_6$ is similar, we prove only for $I_3$ as follows:

${{\textrm{Case}}\, 1:\,0<t<1.}$ We have

$$\begin{aligned} I_3= & {} \int _0^t [h_d(t-\tau )]^{\frac{1+\delta }{d}} [h_d(\tau )]^{\frac{2(1-\delta )}{d}} e^{-({\beta }-\sigma ) (t-\tau )} e^{-\sigma \tau } \textrm{d}\tau \\\leqslant & {} \int _0^t (t-\tau )^{-\frac{1+\delta }{2}} \tau ^{-(1-\delta )} \textrm{d}\tau \\\leqslant & {} t^{\frac{1-\delta }{2}}{} \textbf{B}\left( \frac{1-\delta }{2},\,\delta \right) <+\infty , \end{aligned}$$

because $0<t<1$.

${{\textrm{Case}}\,2:\,1\leqslant t.}$ We have

$$\begin{aligned} I_3= & {} \int _0^t [h_d(t-\tau )]^{\frac{1+\delta }{d}} [h_d(\tau )]^{\frac{2(1-\delta )}{d}} e^{-({\beta }-\sigma ) (t-\tau )} e^{-\sigma \tau } \textrm{d}\tau \\= & {} \int _0^1 [h_d(t-\tau )]^{\frac{1+\delta }{d}} [h_d(\tau )]^{\frac{2(1-\delta )}{d}} e^{-({\beta }-\sigma ) (t-\tau )} \textrm{d}\tau \\{} & {} + \int _1^t [h_d(t-\tau )]^{\frac{1+\delta }{d}} [h_d(\tau )]^{\frac{2(1-\delta )}{d}} e^{-({\beta }-\sigma ) (t-\tau )} \textrm{d}\tau \\\leqslant & {} \int _0^1 \left( (t-\tau )^{-\frac{1+\delta }{2}}+1\right) \tau ^{-(1-\delta )}e^{-({\beta }-\sigma )\tau } \textrm{d}\tau \\{} & {} + \int _1^t (t-\tau )^{-\frac{1+\delta }{2}} e^{-({\beta }-\sigma ) (t-\tau )} \textrm{d}\tau \\\leqslant & {} \left( 2^{\frac{1+\delta }{2}}+1\right) \int _0^{1/2} \tau ^{-(1-\delta )} \textrm{d}\tau + 2^{1-\delta }\int _{1/2}^1 \left( (t-\tau )^{-\frac{1+\delta }{2}}+1\right) e^{-({\beta }-\sigma )\tau }\textrm{d}\tau \\{} & {} + \int _0^t (t-\tau )^{-\frac{1+\delta }{2}} e^{-({\beta }-\sigma ) (t-\tau )} \textrm{d}\tau \\\leqslant & {} \left( 2^{\frac{1+\delta }{2}}+1\right) \frac{1}{\delta 2^\delta } + \frac{2^{1-\delta }}{{\beta }-\sigma }\left( e^{-\frac{{\beta }-\sigma }{2}}- e^{-({\beta }-\sigma )}\right) \\{} & {} + (2^{1-\delta }+1) ({\beta }-\sigma )^{-\frac{1-\delta }{2}}\mathbf {\Gamma }\left( \frac{1-\delta }{2}\right) <+\infty . \end{aligned}$$

The boundedness of integral $\tilde{I}_5$ is established as follows—

$$\begin{aligned} \tilde{I}_5= & {} \int _0^t [h_d(t-\tau )]^{\frac{1+\delta }{d}}e^{-(\beta _1-\sigma )(t-\tau )}\textrm{d}\tau \\\leqslant & {} \int _0^t \left( C^{\frac{1+\delta }{d}}(t-\tau )^{-\frac{1+\delta }{2}}+1\right) e^{-(\beta _1-\sigma )(t-\tau )}\textrm{d}\tau \\\leqslant & {} C^{\frac{1+\delta }{d}} (\beta _1-\sigma )^{-\frac{1-\delta }{2}}{\varvec{\Gamma }}\left( \frac{1-\delta }{2} \right) + \frac{1}{\beta _1-\sigma }\left( 1 - e^{-(\beta _1-\sigma )t}\right) <+\infty . \end{aligned}$$

Now, we consider the boundedness of integrals $\tilde{I}_1,\,\tilde{I}_2,\, \tilde{I}_3,\, \tilde{I}_4$ on $(-\infty ,t]$ for all $t\in \mathbb {R}$. These integrals appear in the proof of Assertion (ii) in Lemma 3.2. The boundedness of $\tilde{I}_1,\, \tilde{I}_2,\, \tilde{I}_4$ are similar, we check for example the boundedness of $\tilde{I}_4$:

$$\begin{aligned} \tilde{I}_4 = \lambda (t)\int _{-\infty }^t [h_d(t-\tau )]^{\frac{1+\delta }{d}}\lambda ^{-1}(\tau )e^{-{\beta }(t-\tau )}\textrm{d}\tau , \end{aligned}$$

where we recall that for $0<\delta<\dfrac{\delta +1}{2}<\delta '<1$:

$$\begin{aligned} \lambda (t) = \left\{ \begin{array}{ll} {[}h_d(t)]^{-\frac{1-\delta }{d}}e^{\sigma t}&{}\quad \hbox { if } t\geqslant 0, \\ {[}h_d(|t|)]^{-\frac{1-\delta '}{d}} &{}\quad \hbox { if } t<0. \end{array}\right. \end{aligned}$$

${{\textrm{Case}}\,1:\,t<0.}$ We have

$$\begin{aligned} \tilde{I}_4\lesssim & {} \int _{-\infty }^t [h_d(t-\tau )]^{\frac{1+\delta }{d}}[h_d(|\tau |)]^{\frac{1-\delta '}{d}}e^{-{\beta }(t-\tau )}\textrm{d}\tau \\{} & {} \hbox { (because } [h_d(|t|)]^{-\frac{1-\delta '}{d}}<C^{\frac{1-\delta '}{2}})\\= & {} \int _{-\infty }^{t-1} [h_d(t-\tau )]^{\frac{1+\delta }{d}}[h_d(|\tau |)]^{\frac{1-\delta '}{d}}e^{-{\beta }(t-\tau )}\textrm{d}\tau \\{} & {} + \int _{t-1}^t [h_d(t-\tau )]^{\frac{1+\delta }{d}}[h_d(|\tau |)]^{\frac{1-\delta '}{d}}e^{-{\beta }(t-\tau )}\textrm{d}\tau \\= & {} \int _{-\infty }^{t-1} e^{-{\beta }(t-\tau )}\textrm{d}\tau + \int _{t-1}^t (t-\tau )^{-\frac{1+\delta }{2}}[h_d(|\tau |)]^{\frac{1-\delta '}{d}}e^{-{\beta }(t-\tau )}\textrm{d}\tau \\= & {} \frac{1}{{\beta }}e^{-{\beta }}+A, \end{aligned}$$

where

$$\begin{aligned} A= \int _{t-1}^t (t-\tau )^{-\frac{1+\delta }{2}}[h_d(|\tau |)]^{\frac{1-\delta '}{d}}e^{-{\beta }(t-\tau )}\textrm{d}\tau . \end{aligned}$$

If $t\leqslant -1$, then

$$\begin{aligned} A= & {} \int _{t-1}^t (t-\tau )^{-\frac{1+\delta }{2}}e^{-{\beta }(t-\tau )}\textrm{d}\tau \leqslant \frac{2}{1-\delta }<+\infty . \end{aligned}$$

If $-1<t<0$, then

$$\begin{aligned} A= & {} \int _{t-1}^{-1} (t-\tau )^{-\frac{1+\delta }{2}}e^{-{\beta }(t-\tau )}\textrm{d}\tau \\{} & {} +\int _{-1}^{t} (t-\tau )^{-\frac{1+\delta }{2}}|\tau |^{-\frac{1-\delta '}{2}}e^{-{\beta }(t-\tau )}\textrm{d}\tau \\\leqslant & {} \frac{2}{1-\delta }\left( 1- (t+1)^{\frac{1-\delta }{2}}\right) + \int _{-1}^{t} (t-\tau )^{-\frac{1+\delta }{2}}|\tau |^{-\frac{1-\delta '}{2}}\textrm{d}\tau \\= & {} \frac{2}{1-\delta }\left( 1- (t+1)^{\frac{1-\delta }{2}}\right) + \int _{0}^{t+1} \tau ^{-\frac{1+\delta }{2}}|t-\tau |^{-\frac{1-\delta '}{2}}\textrm{d}\tau \\= & {} \frac{2}{1-\delta }\left( 1- (t+1)^{\frac{1-\delta }{2}}\right) + \int _{0}^{t+1} \tau ^{-\frac{1+\delta }{2}}(\tau -t)^{-\frac{1-\delta '}{2}}\textrm{d}\tau \\\leqslant & {} \frac{2}{1-\delta }\left( 1- (t+1)^{\frac{1-\delta }{2}}\right) + \int _{0}^{t+1} \tau ^{-\frac{1+\delta }{2}}\tau ^{-\frac{1-\delta '}{2}}\textrm{d}\tau \\{} & {} \hbox { (because }\tau +1>\tau -t>\tau >0)\\= & {} \frac{2}{1-\delta }\left( 1- (t+1)^{\frac{1-\delta }{2}}\right) + \frac{2}{\delta '-\delta }(t+1)^{\frac{\delta '-\delta }{2}}<+\infty . \end{aligned}$$

${{\textrm{Case}}\,2:\,\hbox {t}>0.}$ We have

$$\begin{aligned} \tilde{I}_4= & {} \int _{-\infty }^0 [h_d(t-\tau )]^{\frac{1+\delta }{d}}[h_d(|\tau |)]^{\frac{1-\delta '}{d}}e^{-{\beta }(t-\tau )}\textrm{d}\tau \\{} & {} + \int _0^t [h_d(t-\tau )]^{\frac{1+\delta }{d}}[h_d(\tau )]^{\frac{1-\delta '}{d}}e^{-({\beta }-\sigma )(t-\tau )}\textrm{d}\tau \\= & {} B+C, \end{aligned}$$

where

$$\begin{aligned} B= \int _{-\infty }^0 [h_d(t-\tau )]^{\frac{1+\delta }{d}}[h_d(|\tau |)]^{\frac{1-\delta '}{d}}e^{-{\beta }(t-\tau )}\textrm{d}\tau , \\ C= \int _0^t [h_d(t-\tau )]^{\frac{1+\delta }{d}}[h_d(\tau )]^{\frac{1-\delta '}{d}}e^{-({\beta }-\sigma )(t-\tau )}\textrm{d}\tau . \end{aligned}$$

The boundedness of C is similar the boundedness of integrals $I_1,\,I_2,\,I_3,\,I_4$. Now, we prove the boundedness of B as follows—

$$\begin{aligned} B= & {} \int _{-\infty }^0 [h_d(t-\tau )]^{\frac{1+\delta }{d}}[h_d(|\tau |)]^{\frac{1-\delta '}{d}}e^{-{\beta }(t-\tau )}\textrm{d}\tau \\= & {} \int _{-\infty }^{-1} [h_d(t-\tau )]^{\frac{1+\delta }{d}}e^{-{\beta }(t-\tau )}\textrm{d}\tau \\{} & {} + \int _{-1}^0 [h_d(t-\tau )]^{\frac{1+\delta }{d}}|\tau |^{\frac{1-\delta '}{d}}e^{-{\beta }(t-\tau )}\textrm{d}\tau \\= & {} \int _1^\infty [h_d(\tau )]^{\frac{1+\delta }{d}}e^{-{\beta }\tau }\textrm{d}\tau + \int _{-1}^0 [h_d(t-\tau )]^{-\frac{1+\delta }{d}}|\tau |^{\frac{1-\delta '}{2}}e^{-{\beta }(t-\tau )}\textrm{d}\tau \\\leqslant & {} \int _0^\infty \left( \tau ^{-\frac{1+\delta }{2}}+1\right) e^{-{\beta }\tau }\textrm{d}\tau + \int _{-1}^0 [h_d(t-\tau )]^{-\frac{1+\delta }{d}}|\tau |^{\frac{1-\delta '}{2}}e^{-{\beta }(t-\tau )}\textrm{d}\tau \\\leqslant & {} {\beta }^{-\frac{1-\delta }{2}}{\varvec{\Gamma }}\left( \frac{1-\delta }{2}\right) + \frac{1}{{\beta }} + D, \end{aligned}$$

where

$$\begin{aligned} D= \int _{-1}^0 [h_d(t-\tau )]^{-\frac{1+\delta }{d}}|\tau |^{\frac{1-\delta '}{2}}e^{-{\beta }(t-\tau )}\textrm{d}\tau . \end{aligned}$$

If $t>1$, we have

$$\begin{aligned} D= & {} \int _{-1}^0 |\tau |^{-\frac{1-\delta '}{2}}e^{-{\beta }(t-\tau )}\textrm{d}\tau \leqslant \int _{-1}^0 |\tau |^{-\frac{1-\delta '}{2}}\textrm{d}\tau = \frac{2}{1+\delta }<+\infty . \end{aligned}$$

If $0<t\leqslant 1$, then

$$\begin{aligned} D= & {} \int _{-1}^0 (t-\tau )^{-\frac{1+\delta }{2}}|\tau |^{-\frac{1-\delta '}{2}}e^{-{\beta }(t-\tau )}\textrm{d}\tau \\\leqslant & {} \int _{-1}^0 (-\tau )^{-\frac{1+\delta }{2}}(-\tau )^{-\frac{1-\delta '}{2}} \textrm{d}\tau = \frac{2}{\delta '-\delta }<+\infty . \end{aligned}$$

The boundedness of $\tilde{I}_4$ is completed. The boundedness of $\tilde{I}_3$ is done in the same way as above, but we need the condition that $\dfrac{1+\delta }{2}<\delta '<1$ to guarantee that

$$\begin{aligned} \int _{0}^{t+1}\tau ^{-\frac{1+\delta }{2}}\tau ^{-2\frac{1-\delta '}{2}}\textrm{d}\tau= & {} \int _{0}^{t+1}\tau ^{\delta '-\frac{\delta +3}{2}} \textrm{d}\tau \\= & {} \frac{2}{2\delta '-(1+\delta )}(t+1)^{\frac{2\delta '-(1+\delta )}{2}}<+\infty , \end{aligned}$$

where $-1<t<0$ and

$$\begin{aligned} \int _{-1}^0 (-\tau )^{-\frac{1+\delta }{2}}(-\tau )^{-2\frac{1-\delta '}{2}}\textrm{d}\tau = \int _{-1}^0 (-\tau )^{\delta '-\frac{\delta +3}{2}}\textrm{d}\tau = \frac{2}{2\delta '-(1+\delta )} <+\infty . \end{aligned}$$