Global classical small-data solutions for a three-dimensional chemotaxis Navier–Stokes system involving matrix-valued sensitivities

Abstract

The coupled chemotaxis fluid system

$$\begin{aligned} \left\{ \begin{array}{lll} n_t=\Delta n-\nabla \cdot (n S(x,n,c)\cdot \nabla c)-u\cdot \nabla n, &{}\quad (x,t)\in \Omega \times (0,T),\\ c_t=\Delta c-nc-u\cdot \nabla c, &{}\quad (x,t)\in \Omega \times (0,T),\\ u_t=\Delta u-(u\cdot \nabla )u+\nabla P+n\nabla \Phi ,\quad \nabla \cdot u=0, &{}\quad (x,t)\in \Omega \times (0,T),\\ \nabla c\cdot \nu =(\nabla n-nS(x,n,c)\cdot \nabla c)\cdot \nu =0, \;\; u=0,&{}\quad (x,t)\in \partial \Omega \times (0,T),\\ n(x,0)=n_{0}(x),\quad c(x,0)=c_{0}(x),\quad u(x,0)=u_0(x), &{}\quad x\in \Omega , \end{array} \right. \end{aligned}$$

where $S\in (C^2({\overline{\Omega }}\times [0,\infty )^2))^{N\times N}$, is considered in a bounded domain $\Omega \subset \mathbb {R}^N$, $N\in \{2,3\}$, with smooth boundary. We show that it has global classical solutions if the initial data satisfy certain smallness conditions and give decay properties of these solutions.

Appendix A

We have postponed the proof of Lemma 2.4, which mainly consists in elementary calculus, but is too central to the reasoning of the present work to be left unproven. We begin the Appendix by giving this proof. After that, we will take care of a result on the Helmholtz projection, which was used as tool in the proof of Lemma 5.7. Finally, this appendix contains a variant of Lemma 3.1 adapted to the needs of the proof of Theorem 1.2.

Proof of Lemma 2.4

The assertion can be proven similarly as in [59, Lemma 1.2]. A simple observation shows that for any $t\in [0,\infty )$

$$\begin{aligned}&\int _0^t(1+s^{-\alpha })(1+(t-s)^{-\beta })e^{-\delta (t-s)}e^{-\gamma s} ds\nonumber \\&\quad \le e^{-\delta t}\int _0^te^{-(\gamma -\delta )s}ds +e^{-\delta t}\int _0^ts^{-\alpha } e^{-(\gamma -\delta )s}ds\nonumber \\&\qquad +\,e^{-\delta t}\int _0^t(t-s)^{-\beta } e^{-(\gamma -\delta )s}ds +e^{-\delta t}\int _0^ts^{-\alpha }(t-s)^{-\beta }e^{-(\gamma -\delta )s} ds. \end{aligned}$$

(A.1)

In order to obtain estimates for the summands, independently of the values of $\alpha , \beta , \gamma , \delta$, we can start with

$$\begin{aligned} \int _0^te^{-(\gamma -\delta )s}ds = \frac{1}{\gamma -\delta }[1-e^{-(\gamma -\delta )t}]\le \frac{1}{\eta }, \quad t\in [0,\infty ), \end{aligned}$$

and continue by estimating

$$\begin{aligned} \int _0^ts^{-\alpha } e^{-(\gamma -\delta )s}ds\!\le \!\int _0^1s^{-\alpha }ds\!+\!\int _1^\infty e^{-(\gamma -\delta )s} ds \!\le \! \frac{1}{1-\alpha }\!+\!\frac{1}{\gamma -\delta }\!\le \!\frac{2}{\eta }\quad \text {for } t\in [0,\infty ). \end{aligned}$$

Also in the third term on the right hand side of (A.1) we can split the integral and use the obvious estimates $(t-s)^{-\beta }\le 1$ for $s<t-1$ and $e^{-(\gamma \!-\!\delta )(t-\sigma )}\!\le \! e^{-(\gamma \!-\!\delta )(-\sigma )}{\le }e^{\gamma -\delta }$ for $\sigma \in (0,1)$ to obtain

$$\begin{aligned} \int _0^t(t\!-\!s)^{-\beta } e^{-(\gamma -\delta )s}ds\le & {} \int _0^te^{-(\gamma -\delta )s} ds \!+\!\int _0^1 \sigma ^{-\beta }e^{-(\gamma -\delta )(t-\sigma )}d\sigma \!\le \! \frac{1}{\gamma -\delta }+\frac{1}{1-\beta }e^{\gamma -\delta }\\\le & {} \frac{1}{\eta }+\frac{1}{\eta }e^{\frac{1}{\eta }} \end{aligned}$$

for any $t\in [0,\infty )$. The last integral can be rewritten as

$$\begin{aligned} \int _0^ts^{-\alpha }(t-s)^{-\beta }e^{-(\gamma -\delta )s} ds = t^{1-\alpha -\beta }\int _0^1 \sigma ^{-\alpha }(1-\sigma )^{-\beta }e^{-(\gamma -\delta )\sigma t} d\sigma , \quad t\in [0,\infty ),\nonumber \\ \end{aligned}$$

(A.2)

where we have

$$\begin{aligned} \int _0^1 \sigma ^{-\alpha }(1-\sigma )^{-\beta }e^{-(\gamma -\delta )\sigma t} d\sigma\le & {} \int _0^1 \sigma ^{-\alpha }(1-\sigma )^{-\beta }\le 2^\beta \int _0^{\frac{1}{2}} \sigma ^{-\alpha } d\sigma + 2^\alpha \int _0^{\frac{1}{2}}\sigma ^{-\beta }d\sigma \\\le & {} \frac{2}{1-\alpha }+\frac{2}{1-\beta }\le \frac{4}{\eta }, \end{aligned}$$

so that (A.2) yields the estimate we are aiming for if $1-\alpha -\beta \le 0$ or if $t<1$ and $1-\alpha -\beta >0$. As to $1-\alpha -\beta >0$ and $t\ge 1$, we estimate

$$\begin{aligned}&\int _0^1\sigma ^{-\alpha }(1-\sigma )^{-\beta }e^{-(\gamma -\delta )\sigma t} d\sigma \\&\quad \le \int _0^{\frac{1}{2} t^{-\frac{1-\alpha -\beta }{1-\alpha }}}\sigma ^{-\alpha }(1-\sigma )^{-\beta }e^{-(\gamma -\delta )\sigma t} d\sigma +\int _{\frac{1}{2} t^{-\frac{1-\alpha -\beta }{1-\alpha }}}^1 \sigma ^{-\alpha }(1-\sigma )^{-\beta }e^{-(\gamma -\delta )\sigma t}d\sigma \\&\quad \le \left( \frac{1}{2}\right) ^{-\beta }\int _0^{\frac{1}{2} t^{-\frac{1-\alpha -\beta }{1-\alpha }}} \sigma ^{-\alpha } d\sigma + \left( \frac{1}{2} t^{-\frac{1-\alpha -\beta }{1-\alpha }}\right) ^{-\alpha }e^{-(\gamma -\delta ){\frac{1}{2} t^{-\frac{1-\alpha -\beta }{1-\alpha }}}}\\&\qquad \times \int _{\frac{1}{2} t^{-\frac{1-\alpha -\beta }{1-\alpha }}}^1(1-\sigma )^{-\beta }d\sigma \\&\quad \le \frac{2^{\beta +\alpha -1}}{1-\alpha } t^{-(1-\alpha -\beta )} + \frac{2^\alpha }{1-\beta } t^{-(1-\alpha -\beta )} t^{1-\frac{\beta }{1-\alpha }}e^{-\frac{\gamma -\delta }{2} t^{\frac{\beta }{1-\alpha }}}. \end{aligned}$$

Here,

$$\begin{aligned} t^{1-\frac{\beta }{1-\alpha }}e^{-\frac{\gamma -\delta }{2} t^{\frac{\beta }{1-\alpha }}}\le 1+te^{-\frac{\gamma -\delta }{2} t^{\frac{\beta }{1-\alpha }}},\quad t\in [1,\infty ), \end{aligned}$$

where we have

$$\begin{aligned} \frac{\beta }{1-\alpha }\ge \beta , \quad t^{\frac{\beta }{1-\alpha }}\ge t^\beta \ge t^\eta , \end{aligned}$$

because $t\ge 1$, and hence

$$\begin{aligned} te^{-\frac{\gamma -\delta }{2} t^{\frac{\beta }{1-\alpha }}}\le te^{-\frac{\gamma -\delta }{2} t^\beta } \le te^{-\frac{\eta }{2} t^\eta },\quad t\in [1,\infty ), \end{aligned}$$

which in combination with the finiteness of $\sup _{t>0} te^{-\frac{\eta }{2} t^\eta }$ implies the assertion. $\square$

In order to obtain regularity of u, we have employed the following result in the proof of Lemma 5.7. Other than in [15], we are concerned with the impact of the Helmholtz projection on Hölder-continuous functions (instead of on functions belonging to some $L^p$-space only.)

Lemma A.1

Let $\Omega \subset \mathbb {R}^N$ be a bounded domain with $\partial \Omega \in C^{1+\alpha }$ for some $\alpha >0,$ and let $T>0$. Moreover let $u\in C^{\alpha ,\frac{\alpha }{2}}(\overline{\Omega }\times [0,T])$ and $u=v+w,$ where $\nabla \cdot v=0$ in $\Omega$ and $v\cdot \nu =0$ on $\partial \Omega$ and $w=\nabla \Phi$ for some function $\Phi$. Then $v\in C^{\alpha ,\frac{\alpha }{2}}(\overline{\Omega }\times [0,T])$.

Proof

We have to find a decomposition $u=v+w$ with $\nabla \cdot v=0$ in $\Omega$ and $v\cdot \nu =0$ on $\partial \Omega$ and $w=\nabla \Phi$ for some function $\Phi$. We will construct w and conclude from its smoothness that $\mathscr {P}u=v=u-w\in C^{\alpha ,\frac{\alpha }{2}}(\overline{\Omega }\times [0,T];\mathbb {R}^{N})$. As preparation let us consider the elliptic problem

$$\begin{aligned} \Delta \Phi = \nabla \cdot f,\quad \partial _\nu \Phi \big |_{\partial \Omega }=f\cdot \nu \big |_{\partial \Omega }, \quad \int _\Omega \Phi =0. \end{aligned}$$

(A.3)

Only assuming $f\in C^\alpha (\overline{\Omega };\mathbb {R}^N)$, we fix $p>N$ and let q be such that $\frac{1}{p}+\frac{1}{q}=1$. Then [41, Thm. 4.1], which mirrors the usual Lax–Milgram type result in the context of $L^p$-spaces also for $p\ne 2$, asserts the existence of a unique weak solution $\Phi \!\in \! \{\Phi \!\in \! W^{1,p}(\Omega ), \int _\Omega \Phi =0\}$ such that

$$\begin{aligned} \int _\Omega \nabla \Phi \cdot \nabla \varphi = \int _\Omega f\cdot \nabla \varphi \quad \hbox {for all } \varphi \in W^{1,q}(\Omega ). \end{aligned}$$

Moreover, this solution satisfies

$$\begin{aligned} c_1\Vert \Phi \Vert _{L^\infty (\Omega )}\le & {} c_2\Vert \Phi \Vert _{W^{1,p}(\Omega )}\le \Vert \nabla \Phi \Vert _{L^p(\Omega )}\nonumber \\\le & {} c_3\sup \left\{ \frac{\left|\int _\Omega f\cdot \nabla \varphi \right|}{\Vert \nabla \varphi \Vert _{L^q(\Omega )}};\, \varphi \in W^{1,q}(\Omega ), \nabla \varphi \not \equiv 0\right\} \nonumber \\\le & {} c_3 \Vert f\Vert _{L^p(\Omega )} \le c_4\Vert f\Vert _{C^\alpha (\overline{\Omega })} \end{aligned}$$

(A.4)

with positive constants $c_1$, $c_2$, $c_3$ and $c_4$ that are guaranteed to exist by the continuity of the embedding $W^{1,p}(\Omega )\hookrightarrow L^\infty (\Omega )$, Poincaré’s inequality, [41, Thm. 4.1] and continuity of the embedding $C^{\alpha }(\Omega )\hookrightarrow L^p(\Omega )$, respectively. A standard elliptic regularity result (see [24, Thm. 2.8]) moreover asserts the existence of $c_5>0$ such that $C^{1+\alpha }$-solutions $\Phi$ of (A.3) satisfy

$$\begin{aligned} \Vert \Phi \Vert _{C^{1+\alpha }(\Omega )}\le c_5(\Vert f\Vert _{C^\alpha (\overline{\Omega })}+\Vert \Phi \Vert _{L^\infty (\Omega )}) \end{aligned}$$

and thus, taking into account (A.4),

$$\begin{aligned} \Vert \Phi \Vert _{C^{1+\alpha }(\overline{\Omega })} \le c_6\Vert f\Vert _{C^\alpha (\overline{\Omega })} \end{aligned}$$

with $c_6:=c_5(1 + \frac{c_4}{c_1})$.

Approximating $f\in C^\alpha (\overline{\Omega })$ by a sequence of functions $\{f_n\}_{n\in \mathbb {N}}\subset C^\infty (\overline{\Omega })$ for which the existence of classical solutions $\Phi _n\in C^{2+\alpha }(\overline{\Omega })$ is asserted by well-known results [31, Thm. 3.3.2], we see that for $f\in C^\alpha (\overline{\Omega })$ problem (A.3) has a unique solution $\Phi \in C^{1+\alpha }(\overline{\Omega })$, which moreover satisfies

$$\begin{aligned} \Vert \Phi \Vert _{C^{1+\alpha }(\overline{\Omega })}\le c_6\Vert f\Vert _{C^\alpha (\overline{\Omega })}. \end{aligned}$$

(A.5)

For each t let $\Phi (\cdot ,t)$ denote the solution of

$$\begin{aligned} \Delta \Phi (\cdot ,t)=\nabla \cdot u(\cdot ,t), \quad \partial _\nu \Phi (\cdot ,t)\big |_{\partial \Omega }= u(\cdot ,t)\cdot \nu \big |_{\partial \Omega }, \quad \int _\Omega \Phi =0, \end{aligned}$$

and define $w(\cdot ,t):=\nabla \Phi (\cdot ,t)$ and $v(\cdot ,t):=u(\cdot ,t)-w(\cdot ,t)$, so that clearly $\nabla \cdot v=\nabla \cdot u-\nabla \cdot w=\nabla \cdot u-\Delta \Phi =0$ in $\Omega$ and $v\cdot \nu =u\cdot \nu -w\cdot \nu =u\cdot \nu -\partial _\nu \Phi =0$ on $\partial \Omega$. Concerning smoothness, we see that $\Phi (\cdot ,t)\in C^{1+\alpha }(\overline{\Omega })$ entails $w(\cdot ,t)\in C^{\alpha }(\overline{\Omega })$ and for $t_1,t_2\in [0,T]$ we have that $\Phi (\cdot ,t_2)-\Phi (\cdot ,t_1)={:}\Psi$ solves

$$\begin{aligned} \Delta \Psi = \nabla \cdot (u(\cdot ,t_2)-u(\cdot ,t_1)),\quad \partial _\nu \Psi \big |_{\partial \Omega }= (u(\cdot ,t_2)-u(\cdot ,t_1)) \cdot \nu , \quad \int _\Omega \Psi =0 \end{aligned}$$

so that by (A.5)

$$\begin{aligned} \Vert w(\cdot ,t_2)-w(\cdot ,t_1)\Vert _{C^{\alpha }(\overline{\Omega })}\le \Vert \Psi \Vert _{C^{1+\alpha }(\overline{\Omega })} \le {c_6} \Vert u(\cdot ,t_2)-u(\cdot ,t_1)\Vert _{C^{\alpha }(\overline{\Omega })}. \end{aligned}$$

By the known regularity of u, in conclusion we have $w\in C^{\alpha ,\frac{\alpha }{2}}(\overline{\Omega }\times [0,T])$ and thus $\mathscr {P}u=v=u-w\in C^{\alpha ,\frac{\alpha }{2}}(\overline{\Omega }\times [0,T];\mathbb {R}^N)$. $\square$

The last statement we have postponed to this appendix is concerned with the adaptations necessary for proving Theorem 1.2 instead of Theorem 1.1.

Lemma A.2

Given $M, N, p_0, q_0, \beta , C_S$ as in Theorem 1.2 and some $\delta >0,$ it is possible to choose $M_1,$ $M_2,$ $M_3,$ $M_4,$ $\varepsilon >0,$ $m_0<\varepsilon |\Omega |^{-\frac{1}{p_0}}$ such that for all $m>m_0,$ for all $\alpha _1\in (\frac{m}{2},\min \{m,\lambda _1-\delta \})$ and $\alpha _2\in (0,\min \{\alpha _1,\lambda _1'-\delta \})$ the inequalities

$$\begin{aligned}&k_7(N,q_0)+k_5(q_0)k_7(q_0,q_0)(M_1+2k_1)\Vert \nabla \Phi \Vert _{L^\infty (\Omega )} C_1\\&\qquad +\,3k_7\left( {\scriptstyle \frac{N}{1+\frac{N}{q_0}}},q_0\right) k_5\left( {\scriptstyle \frac{N}{1+\frac{N}{q_0}}}\right) M_3M_4C_2\varepsilon \le \frac{M_3}{2}\\&k_8(N,N)+k_8(N,N)k_5(N)|\Omega |^{\frac{q_0-N}{Nq_0}}(M_1+2k_1)\Vert \nabla \Phi \Vert _{L^\infty (\Omega )}C_3\\&\qquad +\,3 M_3M_4k_8({\scriptstyle \frac{1}{\frac{1}{q_0}+\frac{1}{N}}},N) k_5\left( {\scriptstyle \frac{1}{\frac{1}{q_0}+\frac{1}{N}}}\right) C_4{\varepsilon }\le \frac{M_4}{2}\\&k_3+{C_5k_2\left( |\Omega |^{-\frac{1}{p_0}}+M_1+2k_1\right) Me^{(M_1+2k_1)\sigma \varepsilon }\varepsilon }+3k_2M_2M_3 C_6\varepsilon \le \frac{M_2}{2}\\&3C_SC_7k_4M_2\varepsilon |\Omega |^{-\frac{1}{p_0}}+ 3C_SC_7k_4M_2(M_1+2k_1)\varepsilon +3(M_1+{2k_1})C_7k_4M_3\varepsilon \le \frac{M_1}{2}. \end{aligned}$$

hold, where $k_1,$ $k_2,$ $k_3,$ $k_4,$ $k_5(\cdot ),$ $k_7(\cdot ,\cdot ),$ $k_8(\cdot ,\cdot )$ are taken from Lemmata 2.1, 2.2 and 2.3, and $C_1,$ $C_2,$ $C_3,$ $C_4,$ $C_5,$ $C_6,$ $C_7$ are the constants defined in Sect. 3.

Proof

The condition $m_0<\varepsilon |\Omega |^{-\frac{1}{p_0}}$ that is used to ensure the existence of initial data satisfying (1.8) compels us to choose $m_0$ at the end of this proof, quite in contrast to the situation in Lemma 3.1. Furthermore this makes it necessary to have the estimates during the proof hold regardless of the values of $\alpha _1$, $\alpha _2$, which depend on m. Fortunately, $C_1, \ldots , C_7$ indeed do not depend on $\alpha _1$, $\alpha _2$ (and thus not on m), but—thanks to Lemma 2.4—rather on (a lower bound for) the differences between $\mu$ and $\alpha _1$, $\mu$ and $\alpha _2$ or $\lambda _1$ and $\alpha _1$. (This is the purpose $\delta$ has been introduced for.) The only remaining parameter is $\sigma =\sigma (\alpha _1)=\int _0^\infty (1+s^{-\frac{N}{2p_0}}) e^{-\alpha _1 s} ds$, which is decreasing with respect to $\alpha _1$. If we decide to concentrate on relatively “large” values of $\alpha _1$ only, namely $\alpha _1>\frac{m}{2}$, (which is of no effect to the generality of Theorem 1.2), given $m>0$, for any $\alpha _1\in (\frac{m}{2},\min \{m,\lambda _1-\delta \})$, we may rely on

$$\begin{aligned} \sigma (\alpha _1)\le \int _0^\infty \left( 1+s^{-\frac{N}{2p_0}}\right) e^{-\frac{m}{2} s} ds \le 2\int _0^\infty e^{-\frac{m}{2} s}ds +\int _0^1 s^{-\frac{N}{2p_0}} ds \le \frac{4}{m}+\frac{2p_0}{2p_0-N}. \end{aligned}$$

We pick arbitrary $M_1>0$ and

$$\begin{aligned} A>(M_1+2k_1)\left( 8|\Omega |^{\frac{1}{p_0}}+\frac{1}{1-\frac{N}{2p_0}}\right) . \end{aligned}$$

(A.6)

Moreover, we can choose $M_2$ such that $k_3+C_5k_2(|\Omega |^{-\frac{1}{p_0}}+M_1+2k_1)Me^{A}A \le \frac{M_2}{4}$ and $M_3$ such that $k_7(N,q_0)+k_5(q_0)k_7(q_0,q_0)(M_1+2k_1)\Vert \nabla \Phi \Vert _{L^\infty (\Omega )} C_1\le \frac{M_3}{4}$, and we choose $M_4$ such that $k_8(N,N)+k_8(N,N)k_5(N)|\Omega |^{\frac{q_0-N}{Nq_0}}(M_1+2k_1)\Vert \nabla \Phi \Vert _{L^\infty (\Omega )}C_3\le \frac{M_4}{4}$. Then we let

$$\begin{aligned} 0<\varepsilon< & {} \min \left\{ A, \frac{1}{12k_2M_3C_6}, \frac{1}{12M_3k_8\left( {\scriptstyle \frac{1}{\frac{1}{q_0}+ \frac{1}{N}}},N\right) k_5\left( {\scriptstyle \frac{1}{\frac{1}{q_0}+\frac{1}{N}}}\right) C_4},\right. \\&\left. \frac{1}{12k_7\left( {\scriptstyle \frac{N}{1+\frac{N}{q_0}}},q_0\right) k_5\left( {\scriptstyle \frac{N}{1+\frac{N}{q_0}}}\right) C_2M_4}, \frac{M_1}{2\left( 3C_SC_7k_4M_2{\left( |\Omega |^{-\frac{1}{p_0}}+M_1+2k_1\right) }+3(M_1+{2k_1})C_7k_4M_3\right) }, 1 \right\} \end{aligned}$$

Finally, we want to choose $m_0<\varepsilon |\Omega |^{-\frac{1}{p_0}}$ such that $(M_1+2k_1)\sigma (\alpha _1)\varepsilon <A$ for all $\alpha _1\in ( \frac{m}{2},\min \{m,\lambda _1-\delta \})$, for all $m>m_0$. This is indeed feasible, since $\sigma (\frac{\varepsilon }{2}|\Omega |^{-\frac{1}{p_0}})<\frac{A}{(M_1+2k_1)\varepsilon }$ due to

$$\begin{aligned} \varepsilon \sigma \left( \frac{\varepsilon }{2}|\Omega |^{-\frac{1}{p_0}}\right)<\varepsilon \left( \frac{8}{\varepsilon |\Omega |^{-\frac{1}{p_0}}}+\frac{2p_0}{2p_0-N}\right) \le 8|\Omega |^{\frac{1}{p_0}}+\frac{2p_0}{2p_0-N}<\frac{A}{M_1+2k_1} \end{aligned}$$

and by continuity we can find $m_0<\varepsilon |\Omega |^{-\frac{1}{p_0}}$ so that $\sigma (\frac{m_0}{2})<\frac{A}{(M_1+2k_1)\varepsilon }$. With this choice, for all $\alpha _1\in ( \frac{m}{2},\min \{m,\lambda _1-\delta \})$, for all $m>m_0$, we have $\sigma (\alpha _1)<\sigma (\frac{m}{2})<\sigma (\frac{m_0}{2})<\frac{A}{(M_1+2k_1)\varepsilon }$. $\square$