In this paper we study the chemotaxis–Stokes system with slow $p$-Laplacian diffusion and rotation: $n_t+u\cdot \nabla n=\nabla \cdot ({|\nabla n|}^{p-2}\nabla n)-\nabla \cdot (nS(x,n,c)\cdot \nabla c)$, $c_t+u\cdot \nabla c=\Delta c-nc$, $u_t+\nabla P=\Delta u+n\nabla \varphi +f(x,t)$ and $\nabla \cdot u=0$ in a bounded domain $\Omega \subset {\mathbb{R}}^3$ with $p>2$, subject to the Neumann–Neumann–Dirichlet boundary conditions, where $\varphi :\bar{\Omega }\to \mathbb{R}$, $f:\bar{\Omega }\times [0,\infty )\to {\mathbb{R}}^3$ and $S:\bar{\Omega }\times {[0,\infty )}^2\to {\mathbb{R}}^{3\times 3}$ are given sufficiently smooth functions with $f$ bounded in $\Omega \times (0,\infty )$, $\left|S(x,n,c)\right|\le S_0\left(c\right){(1+n)}^{-\alpha }$ for $(x,n,c)\in \bar{\Omega }\times {[0,\infty )}^2$ with $\alpha \ge 0$, and nondecreasing function $S_0:[0,\infty )\to [0,\infty )$. It is proved that the problem possesses a globally bounded weak solution provided $\alpha +\frac{4}{3}p>\frac{25}{9}$ and $11p+6\alpha +2\alpha p>23$. This extends the current global boundedness result by Tao and Li (2020), where the case of $\alpha =0$ was well solved. It is mentioned that, without constructing coupled energy functionals, the technique used in the present paper is somewhat different.

在本文中，我们研究了慢速趋化-斯托克斯系统 $p$拉普拉斯的扩散和旋转： ${ñ}_{Ť}+ü\cdot \nabla ñ=\nabla \cdot （{|\nabla ñ|}^{p-2}\nabla ñ）-\nabla \cdot （ñ\mathrm{小号}（X，ñ，C）\cdot \nabla C）$， $C_{Ť}+ü\cdot \nabla C=\Delta C-ñC$， ${ü}_{Ť}+\nabla P=\Delta ü+ñ\nabla \varphi +F（X，Ť）$ 和 $\nabla \cdot ü=0$ 在有界域中 $\Omega \subset {\mathbb{[R}}^3$ 与 $p>2$，受Neumann–Neumann–Dirichlet边界条件的约束，其中 $\varphi ：\bar{\Omega }\to \mathbb{[R}$， $F：\bar{\Omega }\times [0，\infty ）\to {\mathbb{[R}}^3$ 和 $\mathrm{小号}：\bar{\Omega }\times {[0，\infty ）}^2\to {\mathbb{[R}}^{3\times 3}$ 被赋予足够平滑的功能 $F$ 界于 $\Omega \times （0，\infty ）$， $\left|\mathrm{小号}（X，ñ，C）\right|\le {\mathrm{小号}}_0（C）{（\mathrm{1个}+ñ）}^{-\alpha }$ 对于 $（X，ñ，C）\in \bar{\Omega }\times {[0，\infty ）}^2$ 与 $\alpha \ge 0$和非递减功能 ${\mathrm{小号}}_0：[0，\infty ）\to [0，\infty ）$。证明该问题具有提供的全局有界弱解$\alpha +\frac{4}{3}p>\frac{25}{9}$ 和 $11p+6\alpha +2\alpha p>23$。这扩展了Tao和Li（2020）当前的全球有界性结果，在这种情况下$\alpha =0$解决得很好。值得一提的是，在不构建耦合能量泛函的情况下，本文使用的技术有所不同。