Global mass-preserving solutions for a two-dimensional chemotaxis system with rotational flux components coupled with a full Navier-Stokes equation

Abstract

We study the chemotaxis-Navier-Stokes system \[\left\{\; \begin{aligned} n_t + u\cdot\nabla n &=\Delta n - \nabla\cdot (nS(x,n,c)\nabla c), &&x\in\Omega, t > 0, \\ c_t + u\cdot\nabla c &=\Delta c - n f(c), && x\in \Omega, t > 0, \\ u_t + (u\cdot\nabla) u &=\Delta u + \nabla P + n \nabla \phi, \;\;\;\; \nabla\cdot u = 0, && x\in\Omega, t > 0 \end{aligned}\right.\tag{$\star$} \] with no-flux boundary conditions for $n$, $c$ in a bounded, convex domain $\Omega\subseteq\mathbb{R}^2$ with a smooth boundary, which is motivated by recent modeling approaches from biology for aerobic bacteria suspended in a sessile water drop. We further do not assume the chemotactic sensitivity $S$ to be scalar as is common, but to be able to attain values in $\mathbb{R}^{2\times 2}$, which allows for more complex modeling of bacterial behavior near the boundary. This is seen as a potential source of the structure formation observed in experiments. While there have been various results for scalar chemotactic sensitivities $S$ due to a convenient energy-type inequality and some for the non-scalar case with only a Stokes fluid equation (or other strong restrictions) simplifying the analysis of the third equation in ($\star$) significantly, we consider the full combined case under little restrictions for the system giving us very little to go on in terms of a priori estimates. We nonetheless manage to still achieve sufficient estimates using Trudinger-Moser type inequalities to extend the existence results seen in a recent work by Winkler for the Stokes case with non-scalar $S$ to the full Navier-Stokes case. Namely, we construct a similar global mass-preserving generalized solution for ($\star$) in planar convex domains for sufficiently smooth parameter functions $S$, $f$ and $\phi$ and only under the fairly weak assumptions that $S$ is appropriately bounded, $f$ is non-negative and $f(0) = 0$.