Global classical solution to the chemotaxis-Navier-Stokes system with some realistic boundary conditions

Chunhua Jin

doi:10.1017/prm.2023.19

Global classical solution to the chemotaxis-Navier-Stokes system with some realistic boundary conditions

Part of: Equations and systems of special type Physiological, cellular and medical topics Partial differential equations

Published online by Cambridge University Press: 02 March 2023

Chunhua Jin

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Chunhua Jin*: Affiliation:
School of Mathematical Sciences, South China Normal University, Guangzhou 510631, China (jinchhua@126.com)

Article contents

Abstract
Introduction
Preliminaries
Global classical solution: Neumann-Dirichlet-Navier slip boundary value condition
Global classical solution: Zero Flux-Dirichlet(Robin)-Navier slip boundary conditions
References

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Abstract

In this paper, we consider the chemotaxis-Navier-Stokes model with realistic boundary conditions matching the experiments of Hillesdon, Kessler et al. in a two-dimensional periodic strip domain. For the lower boundary, we impose the usual homogeneous Neumann-Neumann-Dirichlet boundary condition. While, for the upper boundary, since it is open to the atmosphere, we consider three kinds of different mixed non-homogeneous boundary conditions, that is, (i) Neumann-Dirichlet-Navier slip boundary condition; (ii) Zero flux-Dirichlet-Navier slip boundary condition; (iii) Zero flux-Robin-Navier slip boundary condition. For boundary conditions (i) and (iii), the existence and uniqueness of global classical solutions for any initial data and any large chemotactic sensitivity coefficient is established, and for boundary condition (ii), the existence and uniqueness of global classical solutions for any initial data and small chemotactic sensitivity coefficient is proved.

Keywords

Chemotaxis-Navier-Stokes system mixed nonhomogeneous boundary conditions global classical solution uniqueness

MSC classification

Primary: 35M33: Initial-boundary value problems for systems of mixed type

Secondary: 35A09: Classical solutions 92C17: Cell movement (chemotaxis, etc.)

Information

Type: Research Article
Information: Proceedings of the Royal Society of Edinburgh Section A: Mathematics , Volume 154 , Issue 2 , April 2024 , pp. 445 - 482

DOI: https://doi.org/10.1017/prm.2023.19 [Opens in a new window]
Copyright: Copyright © The Author(s), 2023. Published by Cambridge University Press on behalf of The Royal Society of Edinburgh

1. Introduction

When the well mixed suspension of Bacillus subtilis cells is placed in a chamber with the upper surface open to the atmosphere, this kind of aerobic bacteria consume oxygen, swim toward the direction of sufficient oxygen, that is, the surface of the water layer. Then they form a thin boundary layer with dense cells upstream. Below this layer, the cells in the suspension were severely depleted. Because that bacteria are about 10% denser than water, thus, the density of the mixed suspension becomes larger near the water surface than at the bottom. When the density of the upper boundary layer is too high, it becomes unstable and forms a descending bacterial plume. And finally evolved into various patterns [Reference Hillesdon, Pedley and Kessler10, Reference Hillesdon and Pedley11, Reference Kessler14]. Based on these experimental observations, Goldstein et al. [Reference Tuval, Cisneros, Dombrowski, Wolgemuth, Kessler and Goldstein19] proposed the following chemotaxis-fluid model

(1.1)\begin{equation} \left\{\begin{array}{ll} n_t+{\bf u}\cdot\nabla n=\Delta n-\chi\nabla\cdot(n\nabla c), & \text{in}\ Q, \\ c_t+{\bf u}\cdot\nabla c=\Delta c-cn, & \text{in} \ Q, \\ {\bf u}_t+{\bf u}\cdot\nabla {\bf u}=\Delta {\bf u}-\nabla\pi+n\nabla\varphi, & \text{in}\ Q, \\ \nabla\cdot {\bf u}=0, & \text{in}\ Q, \end{array}\right. \end{equation}

where $Q=\Omega \times \mathbb {R}^+$, $n$, $c$ represent the bacteria cell density, the oxygen concentration respectively, $\chi >0$ is the sensitivity coefficient of aggregation induced by the concentration changes of oxygen, $J=n\nabla c$ is the chemotactic flux, $-cn$ is the consumption term of oxygen, that is more bacteria, more oxygen are consumed, ${\bf u}$, $\pi$ are the fluid velocity and the associated pressure, the fluid couples to $n$ and $c$ through transports ${\bf u}\cdot \nabla n$, ${\bf u}\cdot \nabla c$ and the gravitational potential $n\nabla \varphi$.

In recent years, this kind of models has been widely studied by many authors. For the studies of Cauchy problem in $\mathbb {R}^N$, we refer to [Reference Duan, Lorz and Markowich9, Reference Kozono, Miura and Sugiyama15, Reference Zhang and Zheng22], and for the bounded domain with zero flux boundary value conditions for $n, c$, and no-slip boundary value condition for $u$, please refer to [Reference Di Francesco, Lorz and Markowich8, Reference Winkler20, Reference Winkler21] or the references therein. From these results, one see that the solutions will convergence to the constant steady states, and there is no pattern formation. Matching the experiment descriptions, the following mixed boundary conditions is proposed [Reference Chertock, Fellner, Kurganov, Lorz and Markowich6, Reference Lee and Kim16]: the boundary condition at the top $\Gamma _{top}$ describes the fluid-air surface, where there is no cell flux, the oxygen will be saturated with the air oxygen concentration $c_{air}$ and the vertical fluid velocity and the tangential fluid stress are supposed to be zero

\[ (\nabla n-\chi n\nabla c) \cdot \nu =0,\quad c=c_{air}, \quad {\bf u}\cdot\nu =0, \quad [D({\bf u})\nu]_{\tau}=0, \ \text{on}\ \Gamma_{T}, \]

where $\nu$ denotes the outward unit normal vector of the boundary,$\tau$ denotes unit tangent vector of the boundary, and

\[ [D({\bf u})]=\frac{1}{2}(\nabla {\bf u}+\nabla{\bf u}^T). \]

At the bottom of the domain $\Gamma _{B}$, the cell and oxygen fluxes and the fluid velocity are assumed to be zero:

\[ (\nabla n-\chi n\nabla c) \cdot \nu=0,\quad \nabla c\cdot \nu=0, \quad {\bf u}=0,\ \text{on}\ \Gamma_{B}. \]

Finally, periodic boundary conditions at the sides of the domain are imposed in order to avoid any impact of these boundaries. And some numerical results are given in [Reference Chertock, Fellner, Kurganov, Lorz and Markowich6, Reference Lee and Kim16]. However, very little theoretical research in this regard has been carried out. Peng, Xiang [Reference Peng and Xiang18] considered mixed boundary value problem in an unbounded strip domain of $\mathbb {R}^3$ with the consumption term $cn$ being replaced by $cn^{\gamma }$ ($\gamma \ge 2$) , and established the global existence and convergence of small strong solutions around an equilibrium state. Besides the above boundary value problem, such kind of mixed non-homogeneous boundary value problem

\[ (\nabla n-\chi n\nabla c) \cdot \nu=0,\quad \nabla c\cdot\nu={-}a(x,t)c+b(x,t), \quad {\bf u}=0,\ \text{on}\ \partial\Omega \]

also be considered [Reference Braukhoff4, Reference Braukhoff and Tang5, Reference Jin13], in which, the global classical solution in two dimensional space [Reference Braukhoff4, Reference Braukhoff and Tang5] and time periodic solution in two and three dimensional space [Reference Jin13] are established respectively.

Adopting the realistic boundary conditions mentioned above, in the present paper, we consider the model (1.1) in a two-dimensional strip periodic domain

\[ \Omega=\{(x,y)\in \mathbb{R}^2; x\in \mathbb{R}, 0\le y\le 1\}, \quad \Omega_l=\{(x,y)\in \mathbb{R}^2; 0\le x\le l, 0\le y\le 1\}, \]

with period $l$ and the following boundary conditions, that is

(1.2)\begin{equation} \left\{\begin{array}{@{}l} n_y=0,\ c=c_{air},\ v=0, \ u_y=0,\ (x,y)\in \Gamma_{T}, \\ n_y=c_y=0, \ u=v=0,\ (x,y)\in \Gamma_{B}, \end{array}\right. \end{equation}

(1.3)\begin{equation} \left\{\begin{array}{@{}l} n_y-\chi nc_y=0, \ \tau c_y={-}c+c_{air},\ v=0, \ u_y=0,\ (x,y)\in \Gamma_{T}, \\ n_y=c_y=0, \ u=v=0,\ (x,y)\in \Gamma_{B}, \end{array}\right. \end{equation}

and periodic boundary conditions at the left and right sides of the domain $\Omega _l$ are imposed, where ${\bf u}=(u, v)$, $\Gamma _{T}$ is the upper boundary of the rectangular area, $\Gamma _{B}$ denotes the lower boundary, $c_{air}$ is a positive constant, $\tau =0$ or $1$. It is easy to see that when $\tau =0$, it is corresponding to the Dirichlet boundary condition, when $\tau =1$, it is corresponding to the Robin boundary condition.

We also give the initial value as follows

(1.4)\begin{align} n(x,y,0)& =n_0(x,y)\ge 0, \quad c(x,y,0)=c_0(x,y)\ge 0, \quad {\bf u}(x,y,0)\nonumber\\ & \quad = {\bf u}_0(x,y), \quad (x, y)\in\Omega. \end{align}

Throughout this paper, we assume that

(1.5)\begin{equation} \left\{\begin{array}{@{}l} n_0, c_0, {\bf u}_0\in C^{2+\alpha}(\overline\Omega), \\ n_0, c_0\ge 0, \\ n_0, c_0, {\bf u}_0 \ \text{are periodic functions with respect to variable}\ x\ \text{with period}\ l. \end{array}\right. \end{equation}

We give the global existence theorems as follows.

Theorem 1.1 Assume (1.5) holds. Then the problem (1.1), (1.2), and (1.4) admits a unique global bounded classical solution $(n, c, {\bf u}, \pi )$ ($\pi$ is unique up to a constant) with $(n, c, {\bf u})\in C^{2+\alpha, 1+\frac {\alpha }2}(\overline \Omega _l\times (0,+\infty ))\cap C^{\alpha, \frac {\alpha }2}(\overline \Omega _l\times [0,+\infty ))$, $\nabla \pi \in C^{\alpha, \frac {\alpha }2}(\overline \Omega _l\times (0,+\infty ))$, $n, c\ge 0$,

(1.6)\begin{equation} \|(n, c, {\bf u})\|_{C^{2+\alpha, 1+\frac{\alpha}2}(Q^l)}+\|\nabla\pi\|_{C^{\alpha, \frac{\alpha}2}(Q^l)}\le C, \end{equation}

where the constant $C$ depends only on $n_0$, $c_0$, ${\bf u}_0$, $\alpha$, $l$, $\chi$.

Theorem 1.2 Assume (1.5) holds. When (i) $\tau =0$ with appropriately small $\chi$; or (ii)$\tau =1$ with any large $\chi$, the problem (1.1), (1.3), and (1.4) admits a unique global bounded classical solution $(n, c, {\bf u}, \pi )$ ($\pi$ is unique up to a constant) with $(n, c, {\bf u})\in C^{2+\alpha, 1+\frac {\alpha }2}(\overline \Omega _l\times (0,+\infty ))\cap C^{\alpha, \frac {\alpha }2}(\overline \Omega _l\times [0,+\infty ))$, $\nabla \pi \in C^{\alpha, \frac {\alpha }2}(\overline \Omega _l\times (0,+\infty ))$, ${n, c\ge 0}$,

(1.7)\begin{equation} \|(n, c, {\bf u})\|_{C^{2+\alpha, 1+\frac{\alpha}2}(Q^l)}+\|\nabla\pi\|_{C^{\alpha, \frac{\alpha}2}(Q^l)}\le C, \end{equation}

where the constant $C$ depends only on $n_0$, $c_0$, ${\bf u}_0$, $\alpha$, $l$, $\chi$.

2. Preliminaries

Based on Gagliardo-Nirenberg interpolation inequality and Sobolev trace embedding inequality, we infer the following trace interpolation inequality.

Lemma 2.1 For functions $u: \Omega \to \mathbb {R}$ defined on a bounded Lipschitz domain ${\Omega \in \mathbb {R}^N}$, we have

(2.1)\begin{equation} \|u\|_{L^{q}(\partial\Omega)}\le C\|Du\|_{L^p(\Omega)}^{1-\beta}\|u\|_{L^{r}(\Omega)}^{\beta}+C\|u\|_{L^r(\Omega)}, \end{equation}

where $0< r\le p\le q\le \frac {(N-1)p}{(N-p)_+}$, $p\ge 1$, $\beta =\frac {r(N(p-q)+p(q-1))}{q(N(p-r)+pr)}$.

Proof.

(i) We first consider the case $p< N$. By Sobolev trace embedding inequality [Reference Adams1, Reference Biezuner3], we have
\[ \|u\|_{L^{p^*}(\partial\Omega)}\le C_1\|Du\|_{L^p(\Omega)}+ C_2\|u\|_{L^p(\partial\Omega)}, \]
where $p^*=\frac {(N-1)p}{N-p}$. On the other hand, for any $p\le q\le p^*$,
\[ \|u\|_{L^{q}(\partial\Omega)}\le \|u\|_{L^{p^*}(\partial\Omega)}^{1-\alpha}\|u\|_{L^{p}(\partial\Omega)}^{\alpha} \]
with $\alpha =\frac {p(p^*-q)}{q(p^*-p)}=\frac {(N-1)p-(N-p)q}{q(p-1)}$ for $p< N$. Combining the above two inequalities, and noticing that $(a+b)^{\alpha }\le a^\alpha +b^{\alpha }$ for any $a, b>0$, $0<\alpha \le 1$, then we have
(2.2)\begin{align} \|u\|_{L^{q}(\partial\Omega)}& \le \left( C_1\|Du\|_{L^p(\Omega)}+ C_2\|u\|_{L^p(\partial\Omega)}\right)^{1-\alpha}\|u\|_{L^{p}(\partial\Omega)}^{\alpha} \nonumber\\ & \le\left( C_1^{1-\alpha}\|Du\|_{L^p(\Omega)}^{1-\alpha}+ C_2^{1-\alpha}\|u\|_{L^p(\partial\Omega)}^{1-\alpha}\right)\|u\|_{L^{p}(\partial\Omega)}^{\alpha} \nonumber\\ & \le C_1^{1-\alpha}\|Du\|_{L^p(\Omega)}^{1-\alpha}\|u\|_{L^{p}(\partial\Omega)}^{\alpha} +C_2^{1-\alpha}\|u\|_{L^p(\partial\Omega)}. \end{align}
By [Reference Diaz and Veron7], for any $1\le r\le p<+\infty$,
(2.3)\begin{align} \|u\|_{L^{p}(\partial\Omega)}& \le C_3\left(\|Du\|_{L^p(\Omega)}+ \|u\|_{L^r(\Omega)}\right)^{\theta}\|u\|_{L^{r}(\Omega)}^{1-\theta} \nonumber\\ & \le C_3\|Du\|_{L^p(\Omega)}^{\theta}\|u\|_{L^{r}(\Omega)}^{1-\theta}+\|u\|_{L^r(\Omega)} \end{align}
with $\theta =\frac {N(p-r)+r}{N(p-r)+pr}$. Combining (2.2) and (2.3), we get that
\begin{align*} \|u\|_{L^{q}(\partial\Omega)}& \le C_1^{1-\alpha}\|Du\|_{L^p(\Omega)}^{1-\alpha}\left(C_3\|Du\|_{L^p(\Omega)}^{\theta}\|u\|_{L^{r}(\Omega)}^{1-\theta}+\|u\|_{L^r(\Omega)}\right)^{\alpha} \\ & \quad +C_2^{1-\alpha}\left(C_3\|Du\|_{L^p(\Omega)}^{\theta}\|u\|_{L^{r}(\Omega)}^{1-\theta}+\|u\|_{L^r(\Omega)}\right) \\ & \le C_1^{1-\alpha}C_3^{\alpha}\|Du\|_{L^p(\Omega)}^{1-\alpha+\theta\alpha}\|u\|_{L^{r}(\Omega)}^{\alpha(1-\theta)} +C_1^{1-\alpha}\|Du\|_{L^p(\Omega)}^{1-\alpha}\|u\|_{L^r(\Omega)}^{\alpha} \\ & \quad +C_2^{1-\alpha}C_3\|Du\|_{L^p(\Omega)}^{\theta}\|u\|_{L^{r}(\Omega)}^{1-\theta} +C_2^{1-\alpha}\|u\|_{L^r(\Omega)} \\ & = C_1^{1-\alpha}C_3^{\alpha}\|Du\|_{L^p(\Omega)}^{1-\alpha+\theta\alpha}\|u\|_{L^{r}(\Omega)}^{\alpha(1-\theta)} \\ & \quad +C_1^{1-\alpha}\|Du\|_{L^p(\Omega)}^{1-\alpha}\|u\|_{L^r(\Omega)}^{\frac{\alpha(1-\alpha)(1-\theta)}{1-\alpha+\theta\alpha}} \|u\|_{L^r(\Omega)}^{\frac{\alpha\theta}{1-\alpha+\theta\alpha}} \\ & \quad+C_2^{1-\alpha}C_3\|Du\|_{L^p(\Omega)}^{\theta}\|u\|_{L^{r}(\Omega)}^{\alpha(1-\theta)\frac{\theta}{1-\alpha+\theta\alpha}} \|u\|_{L^{r}(\Omega)}^{\frac{(1-\theta)(1-\alpha)}{1-\alpha+\theta\alpha}} \\ & \quad +C_2^{1-\alpha}\|u\|_{L^r(\Omega)} \\ & \le C_4\|Du\|_{L^p(\Omega)}^{1-\alpha+\theta\alpha}\|u\|_{L^{r}(\Omega)}^{\alpha(1-\theta)}+C_5\|u\|_{L^r(\Omega)}. \end{align*}
Let $\beta =\alpha (1-\theta )$, by a direct calculation, we see that
\[ \beta=\frac{r(N(p-q)+p(q-1))}{q(N(p-r)+pr)},\quad \text{for}\ p< N. \]
Then (2.1) is proved for $p< N$, $r\ge 1$.
Next, we show that (2.1) also holds for any $r>0$. By the prove above, we see that
(2.4)\begin{equation} \|u\|_{L^{q}(\partial\Omega)}\le C\|Du\|_{L^p(\Omega)}^{1-\tilde\beta}\|u\|_{L^{\tilde r}(\Omega)}^{\tilde \beta}+C\|u\|_{L^{\tilde r}(\Omega)}, \end{equation}
for $1\le \tilde r\le p\le q$, $\tilde \beta =\frac {\tilde r(N(p-q)+p(q-1))}{q(N(p-\tilde r)+p\tilde r)}$ for $p< N$. While by Gagliardo-Nirenberg inequality [Reference Li and Lankeit17], for any $0< r<1$,
\[ \|u\|_{L^{\tilde r}(\Omega)}\le C_6\|D u\|_{L^p(\Omega)}^{\alpha_1}\|u\|^{1-\alpha_1}_{L^r(\Omega)}+C_7\|u\|_{L^r(\Omega)}, \]
with $\frac 1{\tilde r}=(\frac 1p-\frac 1N)\alpha _1+\frac {1-\alpha _1}{r}$. Substituting it into (2.4), we arrive at
(2.5)\begin{align} \|u\|_{L^{q}(\partial\Omega)}& \le C\|Du\|_{L^p(\Omega)}^{1-\tilde\beta}\left(C_6\|D u\|_{L^p(\Omega)}^{\alpha_1}\|u\|^{1-\alpha_1}_{L^r(\Omega)}+C_7\|u\|_{L^r(\Omega)}\right)^{\tilde\beta}\nonumber\\ & \quad +C\left(C_6\|D u\|_{L^p(\Omega)}^{\alpha_1}\|u\|^{1-\alpha_1}_{L^r(\Omega)}+C_7\|u\|_{L^r(\Omega)}\right) \nonumber\\ & \le CC_6^{\tilde\beta}\|Du\|_{L^p(\Omega)}^{1-\tilde\beta+\alpha_1\tilde\beta}\|u\|^{\tilde\beta(1-\alpha_1)}_{L^r(\Omega)} +CC_7^{\tilde\beta}\|Du\|_{L^p(\Omega)}^{1-\tilde\beta}\|u\|_{L^r(\Omega)}^{\tilde\beta}\nonumber\\ & \quad +CC_6\|D u\|_{L^p(\Omega)}^{\alpha_1}\|u\|^{1-\alpha_1}_{L^r(\Omega)}+CC_7\|u\|_{L^r(\Omega)}\nonumber\\ & \le C_8\|Du\|_{L^p(\Omega)}^{1-\tilde\beta+\alpha_1\tilde\beta}\|u\|^{\tilde\beta(1-\alpha_1)}_{L^r(\Omega)}+C_9\|u\|_{L^r(\Omega)}. \end{align}
By a direct calculation, we see that $\tilde \beta (1-\alpha _1)=\frac {r(N(p-q)+p(q-1))}{q(N(p-r)+pr)}$. Combining with (2.4), we prove (2.1) for any $r>0$.
(ii) Next, we prove the case $p\ge N$. For any $q\ge p$, there exists $\tilde p< N$ such that $q<\frac {(N-1)p}{N-\tilde p}$, namely $\frac {q\tilde p}{p}<\frac {(N-1)\tilde p}{N-\tilde p}$, then by the result of (i), we have
\begin{align*} \|u\|_{L^q(\partial\Omega)}& =\|u^{\frac{p}{\tilde p}}\|_{L^{\frac{q\tilde p}{p}}(\partial\Omega)}^{\frac{\tilde p}{p}}\le C_{10}\|Du^{\frac{p}{\tilde p}}\|_{L^{\tilde p}(\Omega)}^{\frac{\tilde p}{p}(1-\gamma)}\|u^{\frac{p}{\tilde p}}\|_{L^{\tilde p}(\Omega)}^{\frac{\tilde p}{p}\gamma}+C_{11}\|u^{\frac{p}{\tilde p}}\|_{L^{\tilde p}(\Omega)}^{\frac{\tilde p}{p}} \\ & \le C_{10}\left\|\frac{p}{\tilde p}u^{\frac{p-\tilde p}{\tilde p}}Du\right\|_{L^{\tilde p}(\Omega)}^{\frac{\tilde p}{p}(1-\gamma)}\|u\|_{L^{p}(\Omega)}^{\gamma}+C_{11}\|u\|_{L^{p}(\Omega)} \\ & \le C_{12}\|u\|_{L^p(\Omega)}^{\frac{p-\tilde p}{p}(1-\gamma)}\|Du\|_{L^{p}(\Omega)}^{\frac{\tilde p}{p}(1-\gamma)}\|u\|_{L^{p}(\Omega)}^{\gamma}+C_{11}\|u\|_{L^{p}(\Omega)} \\ & = C_{12}\|u\|_{L^p(\Omega)}^{\frac{p-\tilde p}{p}(1-\gamma)+\gamma}\|Du\|_{L^{p}(\Omega)}^{\frac{\tilde p}{p}(1-\gamma)}+C_{11}\|u\|_{L^{p}(\Omega)} \end{align*}
where $\gamma =\frac {(N-1)p-q(N-\tilde p)}{q\tilde p}$, it is easy to see that $\frac {p-\tilde p}{p}(1-\gamma )+\gamma =\frac {p(q-1)+N(p-q)}{pq}$. We denote $\tilde \beta =\frac {p(q-1)+N(p-q)}{pq}$. The above inequality is equivalent to
(2.6)\begin{equation} \|u\|_{L^q(\partial\Omega)}\le C_{12}\|u\|_{L^p(\Omega)}^{\tilde\beta}\|Du\|_{L^{p}(\Omega)}^{1-\tilde\beta}+C_{11}\|u\|_{L^{p}(\Omega)} \end{equation}
Similar to the proof of (2.5), using Gagliardo-Nirenberg inequality [Reference Li and Lankeit17], for any $0< r< p$,
\[ \|u\|_{L^{p}(\Omega)}\le C_{13}\|D u\|_{L^p(\Omega)}^{\beta_1}\|u\|^{1-\beta_1}_{L^r(\Omega)}+C_{14}\|u\|_{L^r(\Omega)}, \]
with $\frac {1}{p}=\left (\frac {1}{p}-\frac {1}{N}\right )\beta _1+\frac {1-\beta _1}r$. Substituting it into (2.6), we arrive at
\begin{align*} \|u\|_{L^q(\partial\Omega)}& \le C_{12}\left(C_{13}\|D u\|_{L^p(\Omega)}^{\beta_1}\|u\|^{1-\beta_1}_{L^r(\Omega)}+C_{14}\|u\|_{L^r(\Omega)}\right)^{\tilde\beta}\|Du\|_{L^{p}(\Omega)}^{1-\tilde\beta}\\ & \quad +C_{11}\left(C_{13}\|D u\|_{L^p(\Omega)}^{\beta_1}\|u\|^{1-\beta_1}_{L^r(\Omega)}+C_{14}\|u\|_{L^r(\Omega)}\right) \\ & \le C_{12}C_{13}^{\tilde\beta}\|D u\|_{L^p(\Omega)}^{\beta_1\tilde\beta+1-\tilde\beta}\|u\|^{\tilde\beta(1-\beta_1)}_{L^r(\Omega)} +C_{12}C_{14}^{\tilde\beta}\|Du\|_{L^{p}(\Omega)}^{1-\tilde\beta}\|u\|_{L^r(\Omega)}^{\tilde\beta}\\ & \quad +C_{11}C_{13}\|D u\|_{L^p(\Omega)}^{\beta_1}\|u\|^{1-\beta_1}_{L^r(\Omega)}+C_{11}C_{14}\|u\|_{L^r(\Omega)} \\ & \le C_{15}\|D u\|_{L^p(\Omega)}^{\beta_1\tilde\beta+1-\tilde\beta}\|u\|^{\tilde\beta(1-\beta_1)}_{L^r(\Omega)}+C_{16}\|u\|_{L^r(\Omega)}. \end{align*}
By a direct calculation, we get that $\tilde \beta (1-\beta _1)=\frac {r(N(p-q)+p(q-1))}{q(N(p-r)+pr)}$. We complete the proof.

By [Reference Jin12], we have the following lemma.

Lemma 2.2 Let $T>0$, $\tau \in (0, T)$, $\alpha >0$, $\beta >0$, and suppose that $f: [0, T)\to [0, \infty )$ is absolutely continuous, and satisfies

\[ f'(t)-g(t)f(t)+f^{1+\sigma}(t)\le h(t), t\in \mathbb{R}^+, \]

where $\sigma >0$ is a constant, $g(t), h(t)\ge 0$ with $g(t), h(t)\in L^1_{loc}([0, T))$ and

\[ \sup_{t\in [\tau, T)}\int_{t-\tau}^t g(s)\,{\rm d}s\le \alpha,\quad \sup_{t\in [\tau, T)}\int_{t-\tau}^t h(s)\,{\rm d}s\le \beta. \]

Then for any $t>t_0$, we have

\[ f(t)\le f(t_0)\,{\rm e}^{\int_{t_0}^tg(s)\,{\rm d}s}+\int_{t_0}^th(\tau)\,{\rm e}^{\int_{\tau}^tg(s)\,{\rm d}s}\,{\rm d}\tau, \]

and

\begin{align*} \sup_{t\in(0, T)} f(t)& \le \sigma\left(\frac{2A}{1+\sigma}\right)^{\frac{1+\sigma}\sigma}+2B,\quad \sup_{t\in [\tau, T)}\int_{t-\tau}^t f^{1+\sigma}(s)\,{\rm d}s\\ & \le (1+\alpha)\sup_{t\in(0, T)}\{f(t)\}+\beta, \end{align*}

where

\[ A=\tau^{-\frac{1}{1+\sigma}}(1+\alpha)^{\frac{1}{1+\sigma}}\,{\rm e}^{2\alpha},\quad B=\tau^{-\frac{1}{1+\sigma}}\beta^{\frac{1}{1+\sigma}}\,{\rm e}^{2\alpha}+2\beta\,{\rm e}^{2\alpha}+f(0)\,{\rm e}^\alpha. \]

3. Global classical solution: Neumann-Dirichlet-Navier slip boundary value condition

We first give some notations, which will be used throughout this paper.

Notations: $Q^l=\Omega _l\times \mathbb {R}^+$, $Q_T^l=\Omega _l\times (0, T)$, $\|f\|_{L^q}:=\|f\|_{L^q(\Omega _l)}$.

In this section, we pay our attention to the global existence of classical solutions to the problem (1.1), (1.2) and (1.4). We use Leray-Schauder's fixed point framework to show the local existence of classical solutions. For this purpose, let's consider the following linear problem

(3.1)\begin{equation} \left\{\begin{array}{@{}ll} n_t+{\bf u}\cdot\nabla n=\Delta n-\chi\nabla\cdot(n_+\nabla c), & \text{in}\ Q, \\ c_t-\Delta c+{\bf u}\cdot\nabla c={-}\tilde n_+c, & \text{in}\ Q, \\ {\bf u}_t+\tilde{\bf u}\cdot\nabla {\bf u}=\Delta{\bf u}-\nabla\pi+\sigma\tilde n\nabla\varphi, & \text{in}\ Q, \\ \nabla\cdot{\bf u}=0, & \text{in}\ Q, \\ n_y=0,\ c=\sigma c_{air},\ v=0, \ u_y=0, & (x,y)\in \Gamma_{T}, \\ n_y=c_y=0, \ u=v=0, & (x,y)\in \Gamma_{B}, \\ n(x, y, 0)=\sigma n_0(x,y)\ge 0,\ c(x,y, 0)\\ \quad =\sigma c_0(x, y)\ge 0, \ {\bf u}(x,y,0) =\sigma {\bf u}_0(x,y), & x\in\Omega, \end{array}\right. \end{equation}

for any $T>1$ and for any given $\tilde {\bf u}\in C^{\alpha,\frac {\alpha }2}(\overline Q_T^l)\cap W_2^{1,0}(Q_T^l), \tilde n\in C^{\alpha,\frac {\alpha }2}(\overline Q_T^l)$ with $\nabla \cdot \tilde {\bf u}=0$ and $\tilde {\bf u}\cdot \nu |_{\Gamma _{T}}=0$. By classical theory of linear parabolic equations, we have ${\bf u}\in C^{2+\alpha, 1+\frac {\alpha }2}(\overline \Omega ^l\times (0, T])\cap C^{\alpha, \frac {\alpha }2}(\overline Q_T^l), \nabla \pi \in C^{\alpha, \frac {\alpha }2}(\overline \Omega ^l\times (0, T])$, and we further have $c\in C^{2+\alpha, 1+\frac {\alpha }2}(\overline \Omega ^l\times (0, T])\cap C^{\alpha, \frac {\alpha }2}(\overline Q_T^l)$, and $n\in C^{2+\alpha, 1+\frac {\alpha }2}(\overline \Omega ^l\times (0, T])\cap C^{\alpha, \frac {\alpha }2}(\overline Q_T^l)$.

On the other hands, by comparison lemma, it is easy to obtain $c\ge 0$. Next, we show that $n\ge 0$. Let $n_-=\min \{0, n\}$. Multiplying the first equation of (3.1) by $n_1$, and integration it over $\Omega _l\times (0,t)$ yields

\begin{align*} \int_{\Omega_l}|n_-({\cdot},t)|^2\,{\rm d}x\,{\rm d}y& ={-}\int_0^t\int_{\Omega_l}{\bf u}\cdot\nabla n n_-\,{\rm d}x\,{\rm d}y\,{\rm d}\tau+\int_0^t\int_{\Omega_l}n_-\Delta n\,{\rm d}x\,{\rm d}y\,{\rm d}\tau \\ & \quad - \chi\int_0^t\int_{\Omega_l}n_-\nabla\cdot(n_+\nabla c)\,{\rm d}x\,{\rm d}y\,{\rm d}\tau \\ & ={-}\frac{1}{2}\int_0^t\int_{\partial J(\tau)}| n_-|^2{\bf u}\cdot\nu\,{\rm d}S\,{\rm d}\tau+\int_0^t\int_{\partial J(\tau)}n_-\frac{\partial n}{\partial\nu}\,{\rm d}S\,{\rm d}\tau\\ & \quad -\int_0^t\int_{\Omega_l}|\nabla n_-|^2\,{\rm d}x\,{\rm d}y\,{\rm d}\tau \\ & ={-}\int_0^t\int_{\Omega_l}|\nabla n_-|^2\,{\rm d}x\,{\rm d}y\,{\rm d}\tau\le 0, \end{align*}

where $J(t)=\{x\in \Omega _l; n(x,t)\le 0\}$, which implies that $\int _{\Omega _l}|n_-(x,t)|^2\,{\rm d}x\,{\rm d}y=0$, that is $n\ge 0$.

We define the mapping

\begin{align*} & \mathcal{T}: C^{\alpha,\frac{\alpha}2}(\overline Q_T^l)\cap W_2^{1,0}(Q_T^l)\times C^{\alpha,\frac{\alpha}2}(\overline Q_T^l) \times [0, 1]\\ & \quad \rightarrow C^{\alpha,\frac{\alpha}2}(\overline Q_T^l)\cap W_2^{1,0}(Q_T^l)\times C^{\alpha,\frac{\alpha}2}(\overline Q_T^l), \end{align*}

\begin{align*} & \mathcal{T}: (\tilde{\bf u}, \tilde n, \sigma)\rightarrow ({\bf u}, n). \end{align*}

From the above analysis, we see that ${\bf u}, n\in C^{2+\alpha, 1+\frac {\alpha }2}(\overline \Omega ^l\times (0, T])\cap C^{\alpha, \frac {\alpha }2}(\overline Q_T^l)$, and noticing that $C^{2+\alpha, 1+\frac {\alpha }2}(\overline Q_T^l)\cap C^{\alpha, \frac {\alpha }2}(\overline Q_T^l)\hookrightarrow C^{\alpha,\frac {\alpha }2}(\overline Q_T^l)\cap W_2^{1,0}(Q_T^l)$, then the operator $\mathcal {T}$ is completely continuous. It is easy to verify that

\[ \mathcal{T}: (\tilde{\bf u}, \tilde n, 0)\equiv {\bf 0}. \]

In fact, it is easy to see that when $\sigma =0$, it follows $c=0$, ${\bf u}=0$, then one can further derive that $n=0$.

By Leray-Schauder's fixed point theorem, to show the local existence of classical solutions, we only need to show that $\|{\bf u}, n\|_{C^{\alpha, \frac {\alpha }2}}+\|{\bf u}\|_{W_2^{1,0}(Q_T^l} \le C$ if $({\bf u}, n, \sigma )$ is a classical solution of $\mathcal {T}: ({\bf u}, n, \sigma )= ({\bf u}, n)$. For this purpose, in what follows, we pay our attention to the energy estimates.

Lemma 3.1 Let $\mathcal {T}: ({\bf u}, n, \sigma )=({\bf u}, n)$. Then we have

\[ n\ge0, \quad 0\le c\le \sigma c_{air}, \]

(3.2)\begin{equation} \|n({\cdot},t)\|_{L^1}+\chi\int_0^t\int_{\Gamma_T}n|c_y|\,{\rm d}S\,{\rm d}\tau=\|n_0\|_{L^1}, \end{equation}

and

(3.3)\begin{align} & \sup_{0< t< T}\int_{\Omega_l}\left(|\nabla\sqrt c|^2 +n\ln n+|{\bf u}|^2\right)\,{\rm d}x\,{\rm d}y\nonumber\\ & \quad +\sup_{0< t< T-1}\int_{t}^{t+1}\,{\rm d}\tau\int_{\Omega_l}\left(|\nabla{\bf u}|^2+ \frac{|\nabla n|^2}{n}+c|D^2\ln c|^2+n|\nabla\sqrt c|^2 \right)\,{\rm d}x\,{\rm d}y \nonumber\\ & \quad+\sup_{0< t< T-1}\int_{t}^{t+1}\,{\rm d}\tau\int_{\Gamma_T}\left( |c_y|^3+| c_y n \ln n |\right)\,{\rm d}S\le C, \end{align}

where $C$ depends only on $n_0, c_0, {\bf u}_0$, $\chi$, and $c_{air}$, and it is independent of $T$.

Proof. By comparison lemma, it is easy to obtain that

\[ 0\le c\le \sigma c_{air}. \]

The positivity of $n$ has been established in the above analysis. By a direct integration, we see that

(3.4)\begin{equation} \frac{{\rm d}}{{\rm d}t}\int_{\Omega_l}n(x,y,t)\,{\rm d}x\,{\rm d}y+\chi\int_{\Gamma_T}n\partial_y c\,{\rm d}S=0, \end{equation}

which implies (3.2) since $\partial _y c\ge 0$ on $\Gamma _T$. By the second equation, and using a direct calculation, we see that

\begin{align*} & \partial_t|\nabla\sqrt c|^2+\frac{1}{2} c|D^2\ln c|^2=\Delta|\nabla\sqrt c|^2-\nabla\cdot({\bf u}|\nabla\sqrt c|^2)- n|\nabla\sqrt c|^2\\ & \quad -\frac{1}2\nabla n\nabla c- 2\nabla\sqrt c \nabla{\bf u} \nabla\sqrt c. \end{align*}

Then

(3.5)\begin{align} & \frac{{\rm d}}{{\rm d}t}\int_{\Omega_l}|\nabla\sqrt c|^2\,{\rm d}x\,{\rm d}y+\frac12 \int_{\Omega_l} c|D^2\ln c|^2\,{\rm d}x\,{\rm d}y+\int_{\Omega_l} n|\nabla\sqrt c|^2\,{\rm d}x\,{\rm d}y\nonumber\\ & = \int_{\Omega_l}\Delta|\nabla\sqrt c|^2dxdy -\int_{\Omega_l}\nabla\cdot({\bf u}|\nabla\sqrt c|^2)\,{\rm d}x\,{\rm d}y\nonumber\\ & \quad -\frac{1}2\int_{\Omega_l} \nabla n\nabla c\,{\rm d}x\,{\rm d}y- 2\int_{\Omega_l}\nabla\sqrt c\nabla{\bf u}\nabla\sqrt c\,{\rm d}x\,{\rm d}y \nonumber\\ & =\int_{\Gamma_T}\partial_y |\nabla\sqrt c|^2\,{\rm d}S\!-\!\int_{\Gamma_B}\partial_y |\nabla\sqrt c|^2\,{\rm d}S\!-\!\frac{1}2\int_{\Omega_l} \nabla n\nabla c\,{\rm d}x\,{\rm d}y\!-\! \frac{1}{2}\int_{\Omega_l}\frac{\nabla c\nabla{\bf u}\nabla c}{c}\,{\rm d}x\,{\rm d}y. \end{align}

Noticing that $c=\sigma c_{air}$ on $\Gamma _T$, then $c_t\Big |_{\Gamma _T}=0$, $\partial _x c\Big |_{\Gamma _T}=0$, $\partial _{xx} c\Big |_{\Gamma _T}=0$, $\partial _y c\Big |_{\Gamma _T}\ge 0$. Recalling the equation $c$ satisfied, it gives

\[ -\partial_{yy}c+u\partial_x c+v\partial_y c={-}nc,\ \text{on} \ \Gamma_T, \]

which implies that

\[ \partial_{yy}c =\sigma nc_{air},\ \text{on} \ \Gamma_T. \]

Therefore, we have

(3.6)\begin{align} \int_{\Gamma_T}\partial_y |\nabla\sqrt c|^2\,{\rm d}S& ={-}\frac{1}{4}\int_{\Gamma_T}\frac{|\nabla c|^2\partial_y c}{c^2}\,{\rm d}S+\frac{1}{2}\int_{\Gamma_T}\frac{\nabla c\cdot\nabla \partial_y c }{c}\,{\rm d}S \nonumber\\ & ={-}\frac14\int_{\Gamma_T}\frac{|\partial_y c|^2\partial_y c}{c^2}\,{\rm d}S+\frac{1}{2}\int_{\Gamma_T}\frac{\partial_y c\partial_{yy} c}{c}\,{\rm d}S \nonumber\\ & ={-}\frac{1}{4}\int_{\Gamma_T}\frac{|\partial_y c|^3}{c^2}\,{\rm d}S+\frac{1}2\int_{\Gamma_T} n\partial_y c\,{\rm d}S. \end{align}

Noticing that $\partial _y c=0$ on $\Gamma _B$, it implies that $\partial _{xy} c=0$ on $\Gamma _B$. Thus, we have

(3.7)\begin{equation} \int_{\Gamma_B}\partial_y |\nabla\sqrt c|^2\,{\rm d}S={-}\frac14\int_{\Gamma_B}\frac{|\nabla c|^2\partial_y c}{c^2}\,{\rm d}S+\frac12\int_{\Gamma_B}\frac{\partial_xc \partial_{xy} c +\partial_yc \partial_{yy} c }{c}\,{\rm d}S=0. \end{equation}

Noting that $c=\sigma c_{air}$ on $\Gamma _T$, substituting (3.6), (3.7) into (3.5) yields

(3.8)\begin{align} & \frac{{\rm d}}{{\rm d}t}\int_{\Omega_l}|\nabla\sqrt c|^2\,{\rm d}x\,{\rm d}y+\frac12 \int_{\Omega_l} c|D^2\ln c|^2\,{\rm d}x\,{\rm d}y+\int_{\Omega_l} n|\nabla\sqrt c|^2\,{\rm d}x\,{\rm d}y\nonumber\\ & \quad +\frac1{4\sigma^2c_{air}^2}\int_{\Gamma_T}|\partial_y c|^3\,{\rm d}S\nonumber\\ & = \frac{1}{2}\int_{\Gamma_T} n\partial_y c\,{\rm d}S-\frac{1}2\int_{\Omega_l} \nabla n\nabla c\,{\rm d}x\,{\rm d}y- \frac{1}{2}\int_{\Omega_l}\frac{\nabla c\nabla{\bf u}\nabla c}{c}\,{\rm d}x\,{\rm d}y\nonumber\\ & \le \frac{1}{2}\int_{\Gamma_T} n\partial_y c\,{\rm d}S-\frac{1}2\int_{\Omega_l} \nabla n\nabla c\,{\rm d}x\,{\rm d}y+\eta\int_{\Omega_l}\frac{|\nabla c|^4}{c^3}\,{\rm d}x\,{\rm d}y+\frac {c_{air}}{16\eta}\int_{\Omega_l}|\nabla{\bf u}|^2\,{\rm d}x\,{\rm d}y \end{align}

for any small $\eta >0$. By a direct calculation, we see that

\begin{align*} & \int_{\Omega_l}\frac{|\nabla c|^4}{c^3}\,{\rm d}x\,{\rm d}y=\int_{\Omega_l} |\nabla\ln c|^2\nabla\ln c\nabla c\,{\rm d}x\,{\rm d}y \\ & =\int_{\Gamma_T} |\nabla\ln c|^2\nabla c\cdot\nu\,{\rm d}S-\int_{\Omega_l} c^{{-}1}\left( |\nabla c|^2\Delta\ln c+2\nabla cD^2\ln c\nabla c \right)\,{\rm d}x\,{\rm d}y \\ & \le \int_{\Gamma_T}\frac{|\nabla c|^2\nabla c\cdot\nu}{c^2}\,{\rm d}S+\left(\int_{\Omega_l}\frac{|\nabla c|^4}{c^3}\,{\rm d}x\,{\rm d}y\right)^{\frac12}\left(\int_{\Omega_l} c|\Delta\ln c|^2\,{\rm d}x\,{\rm d}y\right)^{\frac12}\\ & \quad +2\left(\int_{\Omega_l}\frac{|\nabla c|^4}{c^3}\,{\rm d}x\,{\rm d}y\right)^{\frac12}\left(\int_{\Omega_l} c|D^2\ln c|^2\,{\rm d}x\,{\rm d}y\right)^{\frac12} \\ & \le \int_{\Gamma_T}\frac{|\partial_y c|^3}{c^2}\,{\rm d}S+\sqrt 2\left(\int_{\Omega_l}\frac{|\nabla c|^4}{c^3}\,{\rm d}x\,{\rm d}y\right)^{\frac12} \left(\int_{\Omega_l} c|D^2\ln c|^2\,{\rm d}x\,{\rm d}y\right)^{\frac12}\\ & \quad +2\left(\int_{\Omega_l}\frac{|\nabla c|^4}{c^3}\,{\rm d}x\,{\rm d}y\right)^{\frac12}\left(\int_{\Omega_l} c|D^2\ln c|^2\,{\rm d}x\,{\rm d}y\right)^{\frac12} \\ & \le\frac 1{\sigma^2 c_{air}^2}\int_{\Gamma_T} |\partial_y c|^3\,{\rm d}S+\frac12 \int_{\Omega_l}\frac{|\nabla c|^4}{c^3}\,{\rm d}x\,{\rm d}y+\frac{(2+\sqrt 2)^2}2\int_{\Omega_l} c|D^2\ln c|^2\,{\rm d}x\,{\rm d}y, \end{align*}

that is

(3.9)\begin{equation} \int_{\Omega_l}\frac{|\nabla c|^4}{c^3}\,{\rm d}x\,{\rm d}y\le \frac 2{\sigma^2c_{air}^2}\int_{\Gamma_T} |\partial_y c|^3\,{\rm d}S+ (6+4\sqrt 2)\int_{\Omega_l} c|D^2\ln c|^2\,{\rm d}x\,{\rm d}y. \end{equation}

Taking $\eta =\frac {1}{4(6+4\sqrt 2)}$ in (3.8), and combining with (3.9), we arrive at

(3.10)\begin{align} & \frac{{\rm d}}{{\rm d}t}\int_{\Omega_l}|\nabla\sqrt c|^2\,{\rm d}x\,{\rm d}y+\frac14 \int_{\Omega_l} c|D^2\ln c|^2\,{\rm d}x\,{\rm d}y+ \int_{\Omega_l} n|\nabla\sqrt c|^2\,{\rm d}x\,{\rm d}y\nonumber\\ & \quad +\frac{\sqrt 2-1}{2\sigma^2c_{air}^2}\int_{\Gamma_T} |\partial_y c|^3\,{\rm d}S\nonumber\\ & \le \frac{1}2\int_{\Gamma_T} n\partial_y c\,{\rm d}S-\frac{1}2\int_{\Omega_l} \nabla n\nabla c\,{\rm d}x\,{\rm d}y+\frac {(3+2\sqrt 2)c_{air}}{2}\int_{\Omega_l}|\nabla{\bf u}|^2\,{\rm d}x\,{\rm d}y. \end{align}

Multiplying the first equation of (3.1) by $1+\ln n$, and integrating the resulting equation over $\Omega _l$ gives,

(3.11)\begin{align} & \frac{{\rm d}}{{\rm d}t}\int_{\Omega_l} n\ln n\,{\rm d}x\,{\rm d}y+\int_{\Omega_l} \frac{|\nabla n|^2}{n}\,{\rm d}x\,{\rm d}y={-}\chi\int_{\Gamma_T}n(1+\ln n)\partial_y c\,{\rm d}S\nonumber\\ & \quad + \chi\int_{\Omega_l}\nabla c\nabla n\,{\rm d}x\,{\rm d}y. \end{align}

Multiplying the third equation of (3.1) by $u$, and integrating the resulting equation over $\Omega _l$, using the boundary conditions and Poincaré inequality, we see that for any $1< p<2$,

\begin{align*} \frac{1}{2}\frac{{\rm d}}{{\rm d}t}\int_{\Omega_l}|{\bf u}|^2\,{\rm d}x\,{\rm d}y+\int_{\Omega_l}|\nabla{\bf u}|^2\,{\rm d}x\,{\rm d}y& =\sigma\int_{\Omega_l}n{\bf u}\nabla\varphi\,{\rm d}x\,{\rm d}y \\ & \le \sigma\|\nabla\varphi\|_{L^\infty}\|{\bf u}\|_{L^{\frac{p}{p-1}}}\|n\|_{L^p}\\ & \le C\sigma\|\nabla\varphi\|_{L^\infty}\|\nabla{\bf u}\|_{L^2}\|n\|_{L^p} \\ & \le \frac12\int_{\Omega_l}|\nabla{\bf u}|^2\,{\rm d}x\,{\rm d}y+C\|n\|_{L^p}^2, \end{align*}

since ${\bf u}\Big |_{\Gamma _B}=0$, which implies that

(3.12)\begin{equation} \frac{{\rm d}}{{\rm d}t}\int_{\Omega_l}|{\bf u}|^2\,{\rm d}x\,{\rm d}y+\int_{\Omega_l}|\nabla{\bf u}|^2\,{\rm d}x\,{\rm d}y\le C_p\|n\|_{L^p}^2,\ \text{for any}\ 1< p<2. \end{equation}

Combining (3.10), (3.11) and (3.12), we arrive at

(3.13)\begin{align} & \frac{{\rm d}}{{\rm d}t}\left(\int_{\Omega_l}|\nabla\sqrt c|^2\,{\rm d}x\,{\rm d}y+\frac{1}{2\chi}\int_{\Omega_l} n\ln n\,{\rm d}x\,{\rm d}y+(3+2\sqrt 2)c_{air}\int_{\Omega_l}|{\bf u}|^2\,{\rm d}x\,{\rm d}y\right)\nonumber\\ & \quad +\frac {(3+2\sqrt 2)c_{air}}{2}\int_{\Omega_l}|\nabla{\bf u}|^2\,{\rm d}x\,{\rm d}y\nonumber\\ & \quad+\frac{1}{2\chi}\int_{\Omega_l} \frac{|\nabla n|^2}{n}\,{\rm d}x\,{\rm d}y+\frac{1}{4}\int_{\Omega_l} c|D^2\ln c|^2\,{\rm d}x\,{\rm d}y\nonumber\\ & \quad + \int_{\Omega_l} n|\nabla\sqrt c|^2\,{\rm d}x\,{\rm d}y+\frac{\sqrt 2-1}{2\sigma^2c_{air}^2}\int_{\Gamma_T} |\partial_y c|^3\,{\rm d}S\nonumber\\ & \le -\frac{1}{2}\int_{\Gamma_T}n\ln n \partial_y c\,{\rm d}S+ C_p(3+2\sqrt 2)c_{air}\|n\|_{L^p}^2. \end{align}

Noticing that $\partial _y c\ge 0$ on $\Gamma _T$, then

(3.14)\begin{align} -\frac{1}2\int_{\Gamma_T}n \ln n \partial_y c\,{\rm d}S& ={-}\frac{1}2\int_{\Gamma_T}n |\ln n| \partial_y c\,{\rm d}S +\int_{\Gamma_T\cap\{n< 1\}}n|\ln n| \partial_y c\,{\rm d}S\nonumber\\ & \le -\frac{1}2\int_{\Gamma_T}n |\ln n| \partial_y c\,{\rm d}S +C_1\int_{\Gamma_T}\partial_y c\,{\rm d}S \nonumber\\ & \le -\frac{1}2\int_{\Gamma_T}n |\ln n| \partial_y c\,{\rm d}S+\frac{\sqrt 2-1}{4c_{air}^2}\int_{\Gamma_T} |\partial_y c|^3\,{\rm d}S +C_2. \end{align}

From Gagliardo-Nirenberg interpolation inequality, and noticing that $4-\frac 4p<2$ we infer that

(3.15)\begin{align} C_p(3+2\sqrt 2)c_{air}\|n\|_{L^p}^2& =C_p(3+2\sqrt 2)c_{air}\|\sqrt n\|_{L^{2p}}^4 \nonumber\\ & \le C_3\|\nabla\sqrt n\|_{L^{2}}^{4-\frac 4p}\|\sqrt n\|_{L^{2}}^{\frac 4p} +C_4\|\sqrt n\|_{L^{2}}^4 \nonumber\\ & \le \frac{1}{2\chi}\|\nabla\sqrt n\|_{L^{2}}^2+C_5. \end{align}

Substituting (3.14) and (3.15) into (3.13) yields

(3.16)\begin{align} & \frac{{\rm d}}{{\rm d}t}\left(\int_{\Omega_l}|\nabla\sqrt c|^2\,{\rm d}x\,{\rm d}y+\frac{1}{2\chi}\int_{\Omega_l} n\ln n\,{\rm d}x\,{\rm d}y+(3+2\sqrt 2)c_{air}\int_{\Omega_l}|{\bf u}|^2\,{\rm d}x\,{\rm d}y\right)\nonumber\\ & \quad +\frac {(3+2\sqrt 2)c_{air}}{2}\int_{\Omega_l}|\nabla{\bf u}|^2\,{\rm d}x\,{\rm d}y \nonumber\\ & \quad+\frac{3}{8\chi}\int_{\Omega_l} \frac{|\nabla n|^2}{n}\,{\rm d}x\,{\rm d}y+\frac{1}{4} \int_{\Omega_l} c|D^2\ln c|^2\,{\rm d}x\,{\rm d}y+ \int_{\Omega_l} n|\nabla\sqrt c|^2\,{\rm d}x\,{\rm d}y\nonumber\\ & \quad +\frac{\sqrt 2-1}{4\sigma^2c_{air}^2}\int_{\Gamma_T} |\partial_y c|^3\,{\rm d}S+\frac{1}2\int_{\Gamma_T}n |\ln n| \partial_y c\,{\rm d}S \nonumber\\ & \le C_6, \end{align}

which implies (3.3).

Lemma 3.2 Assume that $\mathcal {T}: ({\bf u}, n, \sigma )=({\bf u}, n)$. Then

(3.17)\begin{align} & \sup_{0< t< T}\|\nabla c({\cdot},t)\|_{L^2}^2+\sup_{0< t< T-1}\int_t^{t+1}(\|D^2 c\|_{L^2}^2+\|c_t\|_{L^2}^2)\,{\rm d}s\le C, \end{align}

(3.18)\begin{align} & \sup_{0< t< T}\|\nabla {\bf u}({\cdot},t)\|_{L^2}^2+ \sup_{0< t< T-1}\int_t^{t+1}(\|D^2{\bf u}\|_{L^2}^2+\|{\bf u}_t\|_{L^2}^2)\,{\rm d}s\le C, \end{align}

where the constants $C$ depend only on $n_0, c_0, {\bf u}_0$, $\chi$, and $c_{air}$, and they are independent of $T$.

Proof. By Gagliardo-Nirenberg interpolation inequality, we get that

(3.19)\begin{align} & \|\nabla c\|_{L^4}^4\le C_1\|D^2 c\|_{L^2}^2\|c\|_{L^\infty}^2+C_2\|c\|_{L^\infty}^4\le C_3\|D^2 c\|_{L^2}^2+C_4, \end{align}

(3.20)\begin{align} & \|n\|_{L^2}^2=\|\sqrt n\|_{L^4}^4\le C_5\|\nabla\sqrt n\|_{L^2}^2\|\sqrt n\|_{L^2}^2+C_6\|\sqrt n\|_{L^2}^2\le C_7\|\nabla\sqrt n\|_{L^2}^2+C_8, \end{align}

(3.21)\begin{align} & \|u\|_{L^4}^4\le C_9\|\nabla u\|_{L^2}^2\|u\|_{L^2}^2+C_{10}\|u\|_{L^2}^4\le C_{11}\|\nabla u\|_{L^2}^2+C_{12}. \end{align}

Applying $\nabla$ to the second equation of (3.1), and multiplying the resulting equation by $\nabla c$, noticing that $c_y=c_{xy}=0$ on $\Gamma _B$, and $c_x=c_{xx}=0$, $c_{yy}=nc$ on $\Gamma _T$, and using (3.19)–(3.21), we conclude that

\begin{align*} & \frac{1}{2}\frac{{\rm d}}{{\rm d}t}\int_{\Omega_l}|\nabla c|^2\,{\rm d}x\,{\rm d}y\!=\!\int_{\Omega_l} \nabla\Delta c \nabla c\,{\rm d}x\,{\rm d}y \!-\!\int_{\Omega_l} \nabla(nc) \nabla c\,{\rm d}x\,{\rm d}y\!-\!\int_{\Omega_l} \nabla({\bf u}\nabla c) \nabla c\,{\rm d}x\,{\rm d}y \\ & =\frac12\int_{\Omega_l} \Delta|\nabla c|^2\,{\rm d}x\,{\rm d}y-\int_{\Omega_l} |D^2 c|^2\,{\rm d}x\,{\rm d}y-\int_{\Gamma_T} ncc_y\,{\rm d}S+\int_{\Omega_l}(nc+{\bf u}\nabla c)\Delta c\,{\rm d}x\,{\rm d}y\\ & \le\frac12\int_{\Gamma_T}\partial_y|\nabla c|^2\,{\rm d}S -\frac12\int_{\Gamma_B}\partial_y|\nabla c|^2\,{\rm d}S-\int_{\Omega_l}|D^2 c|^2\,{\rm d}x\,{\rm d}y- \int_{\Gamma_T} ncc_y\,{\rm d}S\\ & \quad+\frac{1}4\int_{\Omega_l} |\Delta c|^2\,{\rm d}x\,{\rm d}y+ 2\int_{\Omega_l}|\nabla c|^{2}|{\bf u}|^2\,{\rm d}x\,{\rm d}y+ 2\int_{\Omega_l} |nc|^2\,{\rm d}x\,{\rm d}y \\ & \le \int_{\Gamma_T}(c_xc_{xy}+c_yc_{yy})\,{\rm d}S-\int_{\Omega_l}|D^2 c|^2\,{\rm d}x\,{\rm d}y- \int_{\Gamma_T} ncc_y\,{\rm d}S \\ & \quad+\frac{1}2\int_{\Omega_l} |D^2 c|^2\,{\rm d}x\,{\rm d}y\!+\! 2\left(\int_{\Omega_l}|\nabla c|^4\,{\rm d}x\,{\rm d}y\right)^{\frac12}\left(\int_{\Omega_l}|u|^4\,{\rm d}x\,{\rm d}y\right)^{\frac12}\!+\! 2\int_{\Omega_l} |nc|^2\,{\rm d}x\,{\rm d}y \\ & \le -\frac{1}{4}\int_{\Omega_l} |D^2 c|^2\,{\rm d}x\,{\rm d}y+C_{13}\int_{\Omega_l}|\nabla u|^2\,{\rm d}x\,{\rm d}y+C_{14}\int_{\Omega_l}\frac{|\nabla n|^2}n\,{\rm d}x\,{\rm d}y+C_{15}, \end{align*}

which implies

\[ \sup_{0< t< T}\int_{\Omega_l}|\nabla c|^2\,{\rm d}x\,{\rm d}y+\sup_{0< t< T-1}\int_t^{t+1}\,{\rm d}s\int_{\Omega_l}|D^2 c|^2\,{\rm d}x\,{\rm d}y\le C. \]

Multiplying the second equation of (3.1) by $c_t$, then we obtain

\[ \sup_{0< t< T-1}\int_t^{t+1}\,{\rm d}s\int_{\Omega_l}|c_t|^2\,{\rm d}x\,{\rm d}y\le C. \]

Then (3.17) is proved. Multiplying the third equation of (3.1) by ${\bf u}_t$, and integrating it over $\Omega _l$ yields

\begin{align*} & \frac12\frac{{\rm d}}{{\rm d}t}\int_{\Omega_l}|\nabla {\bf u}|^2\,{\rm d}x\,{\rm d}y+\int_{\Omega_l}|{\bf u}_t|^2\,{\rm d}x\,{\rm d}y={-}\int_{\Omega_l}{\bf u}\cdot\nabla{\bf u}\ {\bf u}_t\,{\rm d}x\,{\rm d}y+\sigma\int_{\Omega_l}n\nabla\varphi {\bf u}_t\,{\rm d}x\,{\rm d}y\\ & \le \frac12\int_{\Omega_l}|{\bf u}_t|^2\,{\rm d}x\,{\rm d}y+\int_{\Omega_l}(|{\bf u}|^2|\nabla {\bf u}|^2+n^2|\nabla\varphi|^2)\,{\rm d}x\,{\rm d}y. \end{align*}

Noticing that

\[ -\Delta{\bf u}+\nabla\pi={-}{\bf u}_t -{\bf u}\cdot\nabla{\bf u}+\sigma n\nabla\varphi, \]

then by $L^2$ theory of Stokes operator [Reference Acevedo, Amrouche, Conca and Ghosh2], we have

\[ \|D^2{\bf u}\|_{L^2}^2+\|\nabla\pi\|_{L^2}^2\le C_{17}(\|{\bf u}_t\|_{L^2}^2+\|{\bf u}\nabla {\bf u}\|_{L^2}^2+\|n\nabla\varphi\|_{L^2}^2). \]

Combining the above two inequalities, and using (3.3) we arrive at

\begin{align*} & 2C_{17}\frac{{\rm d}}{{\rm d}t}\int_{\Omega_l}|\nabla {\bf u}|^2\,{\rm d}x\,{\rm d}y+C_{17}\int_{\Omega_l}|{\bf u}_t|^2\,{\rm d}x\,{\rm d}y +\int_{\Omega_l}|\nabla ^2{\bf u}|^2\,{\rm d}x\,{\rm d}y+\int_{\Omega_l}|\nabla \pi|^2\,{\rm d}x\,{\rm d}y\\ & \le C_{18}\int_{\Omega_l}(|{\bf u}|^2|\nabla {\bf u}|^2+n^2)\,{\rm d}x\,{\rm d}y \\ & \le C_{18}\|{\bf u}\|_{L^4}^2\|\nabla{\bf u}\|_{L^4}^2+C_{18}\|n\|_{L^2}^2 \\ & \le C_{19}\left(\|{\bf u}\|_{L^2}\|\nabla{\bf u}\|_{L^2}+\|{\bf u}\|_{L^2}^2\right)\left(\|\nabla{\bf u}\|_{L^2}\|\Delta{\bf u}\|_{L^2}+\|{\bf u}\|_{L^2}^2\right)+C_{20}\|\nabla\sqrt n\|_{L^2}^2+C_{21} \\ & \le \frac12 \int_{\Omega_l}|\nabla ^2{\bf u}|^2\,{\rm d}x\,{\rm d}y+C_{22}\|\nabla{\bf u}\|_{L^2}^4+ C_{20}\int_{\Omega_l}|\nabla\sqrt n|^2\,{\rm d}x\,{\rm d}y+C_{23}, \end{align*}

that is

(3.22)\begin{align} & 2C_{17}\frac{{\rm d}}{{\rm d}t}\int_{\Omega_l}|\nabla {\bf u}|^2\,{\rm d}x\,{\rm d}y+\frac12\int_{\Omega_l}|D^2{\bf u}|^2\,{\rm d}x\,{\rm d}y+C_{17}\int_{\Omega_l}|{\bf u}_t|^2\,{\rm d}x\,{\rm d}y\nonumber\\ & \quad \le C_{22}\|\nabla{\bf u}\|_{L^2}^4+ C_{20}\int_{\Omega_l}|\nabla\sqrt n|^2\,{\rm d}x\,{\rm d}y+C_{23}. \end{align}

Recalling (3.3), we see that

\[ \sup_{0< t< T-1}\int_t^{t+1}\int_{\Omega_l}|\nabla{\bf u}|^2\,{\rm d}x\,{\rm d}y\,{\rm d}s\le \hat C, \]

by mean value theorem of integrals, for any $t\in (0, T-1)$ there exists $t_0\in (t, t+1)$ such that

\[ \int_{\Omega_l}|\nabla {\bf u}(x,t_0)|^2\,{\rm d}x\,{\rm d}y\le \hat C. \]

From (3.22), and by a direct calculation, we derive that for any $t_0$,

\begin{align*} & \|\nabla {\bf u}({\cdot},t)\|_{L^2}^2 \le \left(\|\nabla {\bf u}({\cdot},t_0)\|_{L^2}^2 +\frac{1}{2C_{17}}\int_{t_0}^t(C_{20}\|\nabla\sqrt n({\cdot}, s)\|_{L^2}^2+C_{23})\,{\rm d}s\right)\\ & \quad \times\exp\left\{\frac{C_{22}}{2C_{17}}\int_{t_0}^t\|\nabla {\bf u}({\cdot},s)\|_{L^2}^2\,{\rm d}s \right\}. \end{align*}

Combining the above two inequalities, we get that

\[ \sup_{0< t< T}\|\nabla {\bf u}({\cdot},t)\|_{L^2}^2\le C_{24}. \]

Using this inequality, and integrating (3.22) from $t$ to $t+1$ yields

\[ \sup_{0< t< T-1}\int_t^{t+1}(\|D^2{\bf u}\|_{L^2}^2+\|{\bf u}_t\|_{L^2}^2)\,{\rm d}s\le C_{25}. \]

Then (3.18) is proved.

Lemma 3.3 Assume that $\mathcal {T} ({\bf u}, n, \sigma )=({\bf u}, n)$. Then for any $r>0$

(3.23)\begin{align} & \sup_{0< t< T}\int_{\Omega_l}|n({\cdot},t)|^{r+1}\,{\rm d}x\,{\rm d}y+\sup_{0< t< T-1}\int_t^{t+1}\int_{\Omega_l} |\nabla n^{\frac{r+1}2}|^2\,{\rm d}x\,{\rm d}y\,{\rm d}\tau\nonumber\\ & \quad +\sup_{0< t< T-1}\int_t^{t+1}\int_{\Gamma_T}n^{r+1}c_y\,{\rm d}S\,{\rm d}\tau\le C, \end{align}

where $C$ depends only on $n_0, c_0, {\bf u}_0$, $\chi$, $r$ and $c_{air}$, and it is independent of $T$.

Proof. Multiplying the first equation of (3.1) by $n^r$ for $r>0$, and integrating the resulting equation over $\Omega$ yields

\begin{align*} & \frac{1}{r+1}\frac{{\rm d}}{{\rm d}t}\int_{\Omega_l}n^{r+1}\,{\rm d}x\,{\rm d}y+\frac{4r}{(r+1)^2}\int_{\Omega_l} |\nabla n^{\frac{r+1}2}|^2\,{\rm d}x\,{\rm d}y+\chi\int_{\Gamma_T}n^{r+1}c_y\,{\rm d}S \\ & =\frac{2r\chi}{r+1}\int_{\Omega_l}n^{\frac{r+1}2}\nabla c\nabla n^{\frac{r+1}2}\,{\rm d}x\,{\rm d}y \\ & \le \frac{2r\chi}{r+1}\|\nabla n^{\frac{r+1}2}\|_{L^2}\|n^{\frac{r+1}2}\|_{L^4}\|\nabla c\|_{L^4} \\ & \le \frac{2Cr\chi}{r+1}\|\nabla n^{\frac{r+1}2}\|_{L^2}\left(\|n^{\frac{r+1}2}\|_{L^2}^{\frac12}\|\nabla n^{\frac{r+1}2}\|_{L^2}^{\frac12}+ \|n\|_{L^1}^{\frac{r+1}2}\right)\left(\|\nabla c\|_{L^2}^{\frac12}\|\Delta c\|_{L^2}^{\frac12}+\|\nabla c\|_{L^2}\right) \\ & \le\frac{r}{(r+1)^2}\|\nabla n^{\frac{r+1}2}\|_{L^2}^2 +C\|n\|_{L^{r+1}}^{r+1}(\|\Delta c\|_{L^2}^2+1)+C\|\Delta c\|_{L^2}^2+C. \end{align*}

By Gagliardo-Nirenberg interpolation inequality, we get that

(3.24)\begin{align} \|n\|_{L^{r+2}}^{r+2}& =\|n^{\frac{r+1}2}\|_{L^{\frac{2(r+2)}{r+1}}}^{\frac{2(r+2)}{r+1}}\le C_1\|\nabla n^{\frac{r+1}2}\|_{L^2}^2\|n^{\frac{r+1}2}\|_{L^{\frac 2{r+1}}}^{\frac 2{r+1}}\nonumber\\ & \quad +C_2\|n\|_{L^1}^{r+2}\le C_3\|\nabla n^{\frac{r+1}2}\|_{L^2}^2+C_4. \end{align}

Combining the above two inequalities, there exists a small constant $\eta >0$ such that

\begin{align*} & \frac{1}{r+1}\frac{{\rm d}}{{\rm d}t}\int_{\Omega_l}n^{r+1}\,{\rm d}x\,{\rm d}y+\frac{2r}{(r+1)^2}\int_{\Omega_l} |\nabla n^{\frac{r+1}2}|^2\,{\rm d}x\,{\rm d}y\\ & \quad +\chi\int_{\Gamma_T}n^{r+1}c_y\,{\rm d}S +\eta\int_{\Omega_l}n^{r+2}\,{\rm d}x\,{\rm d}y \\ & \le C\|n\|_{L^{r+1}}^{r+1}(\|\Delta c\|_{L^2}^2+1)+C\|\Delta c\|_{L^2}^2+C. \end{align*}

Using lemma 2.2 and (3.17), we complete the proof.

Lemma 3.4 Assume that $\mathcal {T} ({\bf u}, n, \sigma )=({\bf u}, n)$. Then for any $r>0$

(3.25)\begin{equation} \sup_{0< t< T}\int_{\Omega_l}|\nabla c|^r\,{\rm d}x\,{\rm d}y+\sup_{0< t< T-1}\int_t^{t+1}\int_{\Omega_l}|\nabla c|^{r-2}|D^2 c|^2\,{\rm d}x\,{\rm d}y\,{\rm d}s \le C_r, \end{equation}

where $C_r$ depends only on $n_0, c_0, {\bf u}_0$, $\chi$, $r$ and $c_{air}$, and it is independent of $T$.

Proof. Using the boundary value conditions, we see that

\[ {\bf u}\nabla c=uc_x+vc_y=0,\ \text{ on}\ \Gamma_T, \]

\[ \partial_y|\nabla c|^r=r(c_x^2+c_y^2)^{\frac r2-1}(c_xc_{xy}+c_yc_{yy})=rc_y^{r-1}c_{yy}=rncc_y^{r-1},\ \text{ on}\ \Gamma_T, \]

and

\[ \partial_y|\nabla c|^r=0,\ \text{ on}\ \Gamma_B. \]

Applying $\nabla$ to the second equation of (3.1), and multiplying the resulting equation by $|\nabla c|^{r-2}\nabla c$ yields

(3.26)\begin{align} \frac{1}{r}\frac{{\rm d}}{{\rm d}t}\int_{\Omega_l}|\nabla c|^r\,{\rm d}x\,{\rm d}y& =\int_{\Omega_l} \nabla\Delta c |\nabla c|^{r-2}\nabla c\,{\rm d}x\,{\rm d}y -\int_{\Omega_l} \nabla(nc)|\nabla c|^{r-2}\nabla c\,{\rm d}x\,{\rm d}y\nonumber\\ & \quad -\int_{\Omega_l} \nabla({\bf u}\nabla c)|\nabla c|^{r-2}\nabla c\,{\rm d}x\,{\rm d}y\nonumber\\ & =\frac{1}{r}\int_{\Omega_l} \Delta|\nabla c|^r\,{\rm d}x\,{\rm d}y-(r-1)\int_{\Omega_l}|\nabla c|^{r-2}|D^2 c|^2\,{\rm d}x\,{\rm d}y\nonumber\\ & \quad -\int_{\partial\Omega_l} nc|\nabla c|^{r-2}\nabla c\cdot\nu\,{\rm d}S\nonumber\\ & \quad+\int_{\Omega_l}({\bf u}\nabla c+nc)\left(|\nabla c|^{r-2}\Delta c+(r-2)|\nabla c|^{r-4}\nabla cD^2 c\nabla c\right)\,{\rm d}x\,{\rm d}y\nonumber\\ & \le\frac1r\int_{\Gamma_T}\partial_y|\nabla c|^r\,{\rm d}S -\frac1r\int_{\Gamma_B}\partial_y|\nabla c|^r\,{\rm d}S-(r-1)\nonumber\\ & \quad \times\int_{\Omega_l}|\nabla c|^{r-2}|D^2 c|^2\,{\rm d}x\,{\rm d}y-\int_{\Gamma_T} nc|c_y|^{r-2}c_y\,{\rm d}S\nonumber\\ & \quad+\frac{r-1}2\int_{\Omega_l}|\nabla c|^{r-2}|D^2 c|^2\,{\rm d}x\,{\rm d}y+(r-1)\int_{\Omega_l}|\nabla c|^{r}|{\bf u}|^2\,{\rm d}x\,{\rm d}y\nonumber\\ & \quad + (r-1)\|c\|_{L^\infty}\int_{\Omega_l}|\nabla c|^{r-2}|n|^2\,{\rm d}x\,{\rm d}y \nonumber\\ & ={-}\frac{r-1}2\int_{\Omega_l}|\nabla c|^{r-2}|D^2 c|^2\,{\rm d}x\,{\rm d}y\nonumber\\ & \quad +(r-1)\int_{\Omega_l}|\nabla c|^{r}|{\bf u}|^2\,{\rm d}x\,{\rm d}y+ \|c\|_{L^\infty}(r-1)\int_{\Omega_l}|\nabla c|^{r-2}|n|^2\,{\rm d}x\,{\rm d}y, \end{align}

and noticing that

(3.27)\begin{align} & \|\nabla c\|_{L^{r+2}}^{r+2}=\||\nabla c|^{\frac r2}\|_{L^{\frac{2(r+2)}{r}}}^{\frac{2(r+2)}{r}}\le C_1\|\nabla|\nabla c|^{\frac r2}\|_{L^2}^2\||\nabla c|^{\frac r2}\|_{L^{\frac{4}{r}}}^{\frac{4}{r}}\nonumber\\ & \quad +C_2\|\nabla c\|_{L^2}^{r+2}\le C_3\||\nabla c|^{\frac{r-2}2}D^2c\|_{L^2}^2+C_4, \end{align}

then there exists a constant $\eta >0$ such that

\begin{align*} & \frac{1}{r}\frac{{\rm d}}{{\rm d}t}\int_{\Omega_l}|\nabla c|^r\,{\rm d}x\,{\rm d}y+\frac{r-1}4\int_{\Omega_l}|\nabla c|^{r-2}|D^2 c|^2\,{\rm d}x\,{\rm d}y +\eta\int_{\Omega_l}|\nabla c|^{r+2}\,{\rm d}x\,{\rm d}y \\ & \le \frac\eta 2\int_{\Omega_l}|\nabla c|^{r+2}\,{\rm d}x\,{\rm d}y+C\int_{\Omega_l} |{\bf u}|^{r+2}\,{\rm d}x\,{\rm d}y+ C\int_{\Omega_l}|n|^{\frac{r+2}2}\,{\rm d}x\,{\rm d}y+C. \end{align*}

Recalling (3.18) and (3.23), we infer that

\[ \frac{1}{r}\frac{{\rm d}}{{\rm d}t}\int_{\Omega_l}|\nabla c|^r\,{\rm d}x\,{\rm d}y+\frac{r-1}4\int_{\Omega_l}|\nabla c|^{r-2}|D^2 c|^2\,{\rm d}x\,{\rm d}y +\frac{\eta}{2}\int_{\Omega_l}|\nabla c|^{r+2}\,{\rm d}x\,{\rm d}y\le C_r. \]

Thus (3.25) is proved.

Lemma 3.5 Assume that $\mathcal {T} ({\bf u}, n, \sigma )=({\bf u}, n)$. Then for any $r>0$

(3.28)\begin{equation} \sup_{0< t< T}(\|n\|_{L^\infty}+\|\nabla c\|_{L^\infty}+\|{\bf u}\|_{L^\infty})\le C, \end{equation}

and for any $\beta \in \left (\frac 12, 1\right )$,

(3.29)\begin{equation} \sup_{0< t< T}\|A^{\beta}{\bf u}\|_{L^\infty}\le \tilde C, \end{equation}

where $A=P\Delta$, $P$ is Helmholtz projection, $C$, $\tilde C$ depend only on $n_0, c_0, {\bf u}_0$, $\chi$, $c_{air}$ and $\beta$, and they are independent of $T$.

Proof. Multiplying the first equation of (3.1) by $u^r$ for $r>1$, and integrating the resulting equation over $\Omega$ yields

\begin{align*} & \frac{{\rm d}}{{\rm d}t}\int_{\Omega_l}n^{r+1}\,{\rm d}x\,{\rm d}y+\frac{4r}{r+1}\int_{\Omega_l} |\nabla n^{\frac{r+1}2}|^2\,{\rm d}x\,{\rm d}y\nonumber\\ & \quad +\chi(r+1)\int_{\Gamma_T}n^{r+1}c_y\,{\rm d}S +\int_{\Omega_l}n^{r+1}\,{\rm d}x\,{\rm d}y \\ & =2r\chi\int_{\Omega_l}n^{\frac{r+1}2}\nabla c\nabla n^{\frac{r+1}2}\,{\rm d}x\,{\rm d}y+\int_{\Omega_l}n^{r+1}\,{\rm d}x\,{\rm d}y \\ & \le 2r\chi\|\nabla n^{\frac{r+1}2}\|_{L^2}\|n^{\frac{r+1}2}\|_{L^4}\|\nabla c\|_{L^4}+\|n^{\frac{r+1}2}\|_{L^2}^2 \\ & \le 2Cr\chi\|\nabla n^{\frac{r+1}2}\|_{L^2}\left(\|n^{\frac{r+1}2}\|_{L^1}^{\frac14}\|\nabla n^{\frac{r+1}2}\|_{L^2}^{\frac34}+ \|n\|_{L^{\frac{r+1}2}}^{\frac{r+1}2}\right)\nonumber\\ & \quad +C\left(\|\nabla n^{\frac{r+1}2}\|_{L^2}^{\frac12}\|n^{\frac{r+1}2}\|_{L^1}^{\frac12}+\|n^{\frac{r+1}2}\|_{L^1}^2\right) \\ & \le\frac{2r}{r+1}\|\nabla n^{\frac{r+1}2}\|_{L^2}^2 +Cr(r+1)^7\|n\|_{L^{\frac{r+1}2}}^{r+1}, \end{align*}

which implies that

\[ \frac{{\rm d}}{{\rm d}t}\int_{\Omega_l}n^{r+1}\,{\rm d}x\,{\rm d}y +\int_{\Omega_l}n^{r+1}\,{\rm d}x\,{\rm d}y\le C(r+1)^8\|n\|_{L^{\frac{r+1}2}}^{r+1}. \]

Taking $r_j=2r_{j-1}=2^j r_0$, $r_0=1$, $M_j=\max \left \{1, \|n_0\|_{L^\infty }, \sup _{t\in (0,T)}\|n\|_{L^{r_j}}\right \}$, then we have

\begin{align*} M_j& \le C^{\frac 1{r_j}}r_j^{\frac 8{r_j}}M_{j-1}= C^{\frac 1{r_02^j}} r_0^{\frac {8}{r_02^j}}2^{\frac {8j}{r_02^j}}M_{j-1} \\ & \le C^{\sum_{k=1}^j\frac 1{r_02^k}} r_0^{\sum_{k=1}^j\frac {8}{r_02^k}}2^{\sum_{k=1}^j\frac {8k}{r_02^k}}M_0\le \tilde C, \end{align*}

where $\tilde C$ is independent of $j$, letting $j\to \infty$, we obtain the $L^\infty$ estimate of $n$. Recalling (3.26), using (3.18) and (3.23),for any $r\ge 3$, we see that

\begin{align*} & \frac{{\rm d}}{{\rm d}t}\int_{\Omega_l}|\nabla c|^r\,{\rm d}x\,{\rm d}y+\frac{r(r-1)}2\int_{\Omega_l}|\nabla c|^{r-2}|D^2 c|^2\,{\rm d}x\,{\rm d}y+\int_{\Omega_l}|\nabla c|^r\,{\rm d}x\,{\rm d}y \\ & \le r(r\!-\!1)\int_{\Omega_l}|\nabla c|^{r}|{\bf u}|^2\,{\rm d}x\,{\rm d}y\!+\! Cr(r\!-\!1)\int_{\Omega_l}|\nabla c|^{r-2}|n|^2\,{\rm d}x\,{\rm d}y\!+\!{+}\int_{\Omega_l}|\nabla c|^r\,{\rm d}x\,{\rm d}y \\ & = r(r-1)\||\nabla c|^{\frac r2}\|_{L^4}^{2}\|{\bf u}\|_{L^4}^{2}+ Cr(r-1)\||\nabla c|^{\frac r2}\|_{L^{\frac{4(r-2)}{r}}}^{\frac{2(r-2)}{r}}\|n\|_{L^4}^{2}+\||\nabla c|^{\frac r2}\|_{L^2}^{2} \\ & \le C_1r(r-1)\left(\||\nabla c|^{\frac r2}\|_{L^1}^{\frac12}\|\nabla|\nabla c|^{\frac r2}\|_{L^2}^{\frac32}+\||\nabla c|^{\frac r2}\|_{L^1}^2\right)\\ & \quad+ C_2r(r-1)\left(\||\nabla c|^{\frac r2}\|_{L^1}^{\frac12}\|\nabla|\nabla c|^{\frac r2}\|_{L^2}^{\frac{3r-8}{2r}}+ \||\nabla c|^{\frac r2}\|_{L^1}^{\frac{2(r-2)}{r}}\right) \\ & \quad+C_3\left(\||\nabla c|^{\frac r2}\|_{L^1}\|\nabla |\nabla c|^{\frac r2}\|_{L^2}+\||\nabla c|^{\frac r2}\|_{L^1}^2\right) \\ & \le \frac{r(r-1)}4\int_{\Omega_l}|\nabla c|^{r-2}|D^2 c|^2\,{\rm d}x\,{\rm d}y+C_4r(r-1)\|\nabla c\|_{L^{\frac r2}}^r+C_5r(r-1)\|\nabla c\|_{L^{\frac r2}}^{\frac{r^2}{r+8}}\\ & \quad + C_6r(r-1)\|\nabla c\|_{L^{\frac r2}}^{r-2}, \end{align*}

that is

\begin{align*} & \frac{{\rm d}}{{\rm d}t}\int_{\Omega_l}|\nabla c|^r\,{\rm d}x\,{\rm d}y+\int_{\Omega_l}|\nabla c|^r\,{\rm d}x\,{\rm d}y\le C_4r(r-1)\|\nabla c\|_{L^{\frac r2}}^r\\ & \quad +C_5r(r-1)\|\nabla c\|_{L^{\frac r2}}^{\frac{r^2}{r+8}}+ C_6r(r-1)\|\nabla c\|_{L^{\frac r2}}^{r-2}. \end{align*}

Then, similar to above, we obtain the $L^\infty$ estimates of $\nabla c$, and we complete the proof.

By semigroup theory of Stokes operator, we see that

\begin{align*} & \|{\bf u}\|_{L^\infty}\le\,{\rm e}^{-\delta t}\|{\bf u_0}\|_{L^\infty}+C_7\int_0^t\,{\rm e}^{-\delta (t-s)}(t-s)^{-\frac{3}4}\|{\bf u}\cdot\nabla{\bf u}\|_{L^{\frac 43}}\,{\rm d}s\nonumber\\ & \quad +C_8\int_0^t\,{\rm e}^{-\delta (t-s)}(t-s)^{-\frac{1}2}\|n\|_{L^2} \\ & \le\,{\rm e}^{-\delta t}\|{\bf u_0}\|_{L^\infty}+C_7\sup_{t< T}\|{\bf u}({\cdot}, t)\|_{L^4}\|\nabla{\bf u}({\cdot}, t)\|_{L^2}\int_0^t\,{\rm e}^{-\delta s}s^{-\frac{3}4}\,{\rm d}s\nonumber\\ & \quad +C_8\sup_{t< T}\|n({\cdot}, t)\|_{L^2}\int_0^t\,{\rm e}^{-\delta s}s^{-\frac{1}2}\,{\rm d}s \\ & \le C_9. \end{align*}

Similarly, for any $\beta \in \left (\frac 12, 1\right )$, we also have

\begin{align*} & \|A^\beta{\bf u}\|_{L^\infty}\le\,{\rm e}^{-\delta t}\|A^\beta{\bf u_0}\|_{L^\infty}+C_{10}\int_0^t\,{\rm e}^{-\delta (t-s)}(t-s)^{-\beta}\|{\bf u}\cdot\nabla{\bf u}\|_{L^\infty}\,{\rm d}s\\ & \quad +C_{11}\int_0^t\,{\rm e}^{-\delta (t-s)}(t-s)^{-\beta}\|n\|_{L^\infty} \\ & \le\,{\rm e}^{-\delta t}\|A^\beta{\bf u_0}\|_{L^\infty}+C_{10}\sup_{t< T}\|{\bf u}({\cdot}, t)\|_{L^\infty}\|\nabla{\bf u}({\cdot}, t)\|_{L^\infty}\int_0^t\,{\rm e}^{-\delta s}s^{-\beta}\,{\rm d}s\\ & \quad ++\,C_{11}\sup_{t< T}\|n({\cdot}, t)\|_{L^\infty}\int_0^t\,{\rm e}^{-\delta s}s^{-\beta}\,{\rm d}s \\ & \le {\rm e}^{-\delta t}\|A^\beta{\bf u_0}\|_{L^\infty}+C_{12}\sup_{t< T}\|{\bf u}({\cdot}, t)\|_{L^\infty}(\|{\bf u}({\cdot}, t)\|_{L^\infty}^{\frac{2\beta-1}{2\beta}}\|A^{\beta}{\bf u}({\cdot}, t)\|_{L^\infty}^{\frac1{2\beta}}\\ & \quad +\|{\bf u}({\cdot}, t)\|_{L^\infty}) +C_{13} \\ & \le C_{14}+C_{15}\sup_{t< T}\|A^{\beta}{\bf u}({\cdot}, t)\|_{L^\infty}^{\frac1{2\beta}}, \end{align*}

which implies that

\[ \sup_{t< T}\|A^{\beta}{\bf u}({\cdot}, t)\|_{L^\infty}\le C_{14}+C_{15}\sup_{t< T}\|A^{\beta}{\bf u}({\cdot}, t)\|_{L^\infty}^{\frac1{2\beta}}. \]

Noticing that $\frac 1{2\beta }<1$, then

\[ \sup_{t< T}\|A^{\beta}{\bf u}({\cdot}, t)\|_{L^\infty}\le C.\]

□

Lemma 3.6 Let $\mathcal {T}: ({\bf u}, n, \sigma )=({\bf u}, n)$. Then for any $p>1$,

(3.30)\begin{equation} \sup_{0< t< T-1}\left(\|({\bf u}, n, c)\|_{W_p^{2,1}(Q_1^l(t))}+\|\pi\|_{W_p^{1,0}(Q_1^l(t))} \right)\le C_p, \end{equation}

where $Q_1^l(t)=\Omega _l\times (t, t+1)$, $C_p$ is independent of $T$, it depends only on $n_0, c_0, {\bf u}_0$, $\chi$, $c_{air}$ and $p$.

Proof. Noticing that

\[ c_t-\Delta c+{\bf u}\cdot\nabla c+c=c-nc , \]

and (3.28), using the $L^p$ theory of linear parabolic equations, it is easy to obtain that

(3.31)\begin{equation} \sup_{0< t< T-1}\|c\|_{W_p^{2,1}(Q_1^l(t))}^p \le C\sup_{0< t< T-1}\|c-nc\|_{L^p(Q_1^l(t))}^p+C\|c_0\|_{W^{2,p}}^p\le C_1. \end{equation}

For ${\bf u}$, we see that

\[ {\bf u}_t+{\bf u}\cdot\nabla {\bf u}-\Delta{\bf u}+\nabla\pi+{\bf u}={\bf u}+\sigma n\nabla\varphi, \]

then

(3.32)\begin{align} & \sup_{0< t< T-1}\left(\|{\bf u}\|_{W_p^{2,1}(Q_1^l(t))}^p+\|\pi\|_{W_p^{1,0}(Q_1^l(t))}^p \right)\nonumber\\ & \quad \le C\sup_{0< t< T-1}\|{\bf u}+\sigma n\nabla\varphi\|_{L^p(Q_1^l(t))}^p+C\|{\bf u}_0\|_{W^{2,p}}^p\le C_2. \end{align}

Recalling that

\[ n_t+{\bf u}\cdot\nabla n-\Delta n+n=n-\chi\nabla\cdot(n\nabla c). \]

Then for any $q>1$,

\begin{align*} & \int_t^{t+1}(\|n\|_{W^{2,q}}^q+\|n_t\|_{L^q}^q)\,{\rm d}s\le C \|n_0\|_{W^{2,q}}^q+C\int_t^{t+1}\|\nabla\cdot(n\nabla c)\|_{L^q}^q\,{\rm d}s \\ & \quad +C\int_t^{t+1}\|n\|_{L^q}^q\,{\rm d}s \\ & \le C\|n_0\|_{W^{2,q}}^q+C_2\|\nabla c\|_{L^\infty(Q_T^l)}^q\int_t^{t+1}\|\nabla n\|_{L^q}^q\,{\rm d}s\\ & \quad + C\|n\|_{L^\infty(Q_T^l)}^q\int_t^{t+1}\|\Delta c\|_{L^q}^q\,{\rm d}s+{+}C\int_t^{t+1}\|n\|_{L^q}^q\,{\rm d}s \\ & \le C\|n_0\|_{W^{2,q}}^q+\frac12\int_t^{t+1}\|\nabla n\|_{W^{1,q}}^q\,{\rm d}s+\tilde C, \end{align*}

which implies that

\[ \sup_{0< t< T-1}\int_t^{t+1}(\|n\|_{W^{2,q}}^q+\|n_t\|_{L^q}^q)\,{\rm d}s\le C_3. \]

Thus, (3.30) is proved.

Proof Proof of theorem 1.1

Noticing that $W_p^{2,1}(Q_T^l)\hookrightarrow C^{\beta, \frac {\beta }2}(\overline Q_T^l)$, for any $\beta < 2-\frac {4}p$, then by lemma 3.6, we derive that

\[ \|(n, c, {\bf u})\|_{C^{\beta, \frac{\beta}2}(Q_T^l)}\le C. \]

Using Leray-Schauder fixed point theorem, the problem (1.1), (1.2), and (1.4) admits a classical solution $(n, c, {\bf u}, \pi )$ in $Q_T^l$. From the above lemmas, we see that all these estimates are independent of $T$, which implies that the solution $(n, c, {\bf u}, \pi )$ is a global classical solution, and we conclude that

(3.33)\begin{equation} \|(n, c, {\bf u})\|_{C^{\beta, \frac{\beta}2}(Q^l)}\le C. \end{equation}

Combining (3.33) and lemma 3.6, and using the classical theory of linear parabolic equations, we have

(3.34)\begin{equation} \|(c, {\bf u})\|_{C^{2+\alpha, 1+\frac{\alpha}2}(Q^l)}+\|\nabla\pi\|_{C^{\alpha, \frac{\alpha}2}(Q^l)}\le C. \end{equation}

Using (3.34), we further have

(3.35)\begin{equation} \|n\|_{C^{2+\alpha, 1+\frac{\alpha}2}(Q^l)}\le C. \end{equation}

The global existence in theorem 1.1 is proved.

The proof of uniqueness is standard, for the completeness of the paper, in what follows, we still give the proof. Suppose the contrary. Let $(n_1, c_1, {\bf u_1}, \pi _1)$, $(n_2, c_2, {\bf u_2}, \pi _2)$ be two solutions of (1.1), (1.2), and (1.4). Denote $\tilde n=n_1-n_2$, $\tilde c=c_1-c_2$, $\tilde {\bf u}={\bf u_1}-{\bf u_2}=(\tilde u, \tilde v)$, $\tilde \pi =\pi _1-\pi _2$. Then

(3.36)\begin{equation} \left\{\begin{array}{@{}ll} \tilde n_t+\tilde{\bf u}\cdot\nabla n_1+{\bf u}_2\cdot\nabla\tilde n=\Delta\tilde n-\chi\nabla\cdot(\tilde n\nabla c_1+n_2\nabla\tilde c), & \text{in}\ Q, \\ \tilde c_t-\Delta\tilde c+\tilde{\bf u}\cdot\nabla c_1+{\bf u}_2\cdot\nabla\tilde c={-}\tilde n c_1-n_2\tilde c, & \text{in}\ Q, \\ \tilde{\bf u}_t+\tilde{\bf u}\cdot\nabla {\bf u}_1+{\bf u}_2\cdot\nabla\tilde {\bf u}=\Delta\tilde{\bf u}-\nabla\tilde\pi+\tilde n\nabla\varphi, & \text{in}\ Q, \\ \nabla\cdot\tilde{\bf u}=0, & \text{in}\ Q, \\ \tilde n_y=0,\ \tilde c=0,\ \tilde v=0, \ \tilde u_y=0, & (x,y)\in \Gamma_{T}, \\ \tilde n_y=\tilde c_y=0, \ \tilde u=\tilde v=0, & (x,y)\in \Gamma_{B}, \\ \tilde n(x, y, 0)=0, \ \tilde c(x,y, 0)=0,\ \tilde {\bf u}(x,y,0)=0, & x\in\Omega, \end{array}\right. \end{equation}

Multiplying the first equation of (3.36) by $\tilde n$, and integrating it over $\Omega$ yields

\begin{align*} & \frac12\frac{{\rm d}}{{\rm d}t}\int_\Omega |\tilde n|^2\,{\rm d}x\,{\rm d}y+\int_\Omega |\nabla\tilde n|^2\,{\rm d}x\,{\rm d}y \\ & ={-}\int_\Omega\tilde n\tilde{\bf u}\cdot\nabla n_1\,{\rm d}x\,{\rm d}y \!-\!\chi \int_{\Gamma_T}(\tilde n\partial_y c_1\!+\!n_2\partial_y\tilde c)\tilde n\,{\rm d}S\!+\!\chi\int_\Omega (\tilde n\nabla c_1+n_2\nabla\tilde c)\nabla\tilde n\,{\rm d}x\,{\rm d}y \\ & \le \frac14\int_\Omega |\nabla\tilde n|^2\,{\rm d}x\,{\rm d}y+C\int_\Omega(|\tilde{\bf u}|^2+|\tilde n|^2+|\nabla\tilde c|^2)\,{\rm d}x\,{\rm d}y\\ & \quad -\chi \int_{\Gamma_T}|\tilde n|^2\partial_y c_1\,{\rm d}S-\chi \int_{\Gamma_T} n_2\partial_y\tilde c\tilde n\,{\rm d}S \\ & \le \frac14\int_\Omega |\nabla\tilde n|^2\,{\rm d}x\,{\rm d}y+C\int_\Omega(|\tilde{\bf u}|^2+|\tilde n|^2+|\nabla\tilde c|^2)\,{\rm d}x\,{\rm d}y+ C\int_{\Gamma_T}(|\tilde n|^2+|\partial_y\tilde c|^2)\,{\rm d}S, \end{align*}

that is

(3.37)\begin{align} & \frac12\frac{{\rm d}}{{\rm d}t}\int_\Omega |\tilde n|^2\,{\rm d}x\,{\rm d}y+\frac34\int_\Omega |\nabla\tilde n|^2\,{\rm d}x\,{\rm d}y\le C\int_\Omega(|\tilde{\bf u}|^2+|\tilde n|^2+|\nabla\tilde c|^2)\,{\rm d}x\,{\rm d}y\nonumber\\ & \quad + C\int_{\Gamma_T}(|\tilde n|^2+|\partial_y\tilde c|^2)\,{\rm d}S. \end{align}

Multiplying the second equation of (3.36) by $\tilde c$, and integrating it over $\Omega$ yields

\begin{align*} & \frac12\frac{{\rm d}}{{\rm d}t}\int_\Omega |\tilde c|^2\,{\rm d}x\,{\rm d}y+\int_\Omega |\nabla\tilde c|^2\,{\rm d}x\,{\rm d}y \\ & ={-}\int_\Omega\tilde c\tilde{\bf u}\cdot\nabla c_1\,{\rm d}x\,{\rm d}y-\int_\Omega\tilde c{\bf u}_2\cdot\nabla\tilde c\,{\rm d}x\,{\rm d}y -\int_\Omega\tilde c(\tilde n c_1+n_2\tilde c)\,{\rm d}x\,{\rm d}y \\ & =\int_\Omega\nabla\tilde c\cdot\tilde{\bf u} c_1\,{\rm d}x\,{\rm d}y -\int_\Omega\tilde c(\tilde n c_1+n_2\tilde c)\,{\rm d}x\,{\rm d}y \\ & \le \frac14\int_\Omega |\nabla\tilde c|^2\,{\rm d}x\,{\rm d}y+C\int_\Omega(|\tilde{\bf u}|^2+|\tilde n|^2+|\tilde c|^2)\,{\rm d}x\,{\rm d}y, \end{align*}

namely,

(3.38)\begin{equation} \frac12\frac{{\rm d}}{{\rm d}t}\int_\Omega |\tilde c|^2\,{\rm d}x\,{\rm d}y+\frac34\int_\Omega |\nabla\tilde c|^2\,{\rm d}x\,{\rm d}y\le C\int_\Omega(|\tilde{\bf u}|^2+|\tilde n|^2+|\tilde c|^2)\,{\rm d}x\,{\rm d}y. \end{equation}

Applying $\nabla$ to the second equation of (3.36), and multiplying the resulting equation by $\nabla \tilde c$, and integrating it over $\Omega$ yields

\begin{align*} & \frac12\frac{{\rm d}}{{\rm d}t}\int_\Omega |\nabla\tilde c|^2\,{\rm d}x\,{\rm d}y=\int_\Omega\nabla\Delta\tilde c \nabla\tilde c\,{\rm d}x\,{\rm d}y\\ & \quad - \int_\Omega \nabla(\tilde{\bf u}\cdot\nabla c_1+{\bf u}_2\cdot\nabla\tilde c)\nabla\tilde c\,{\rm d}x\,{\rm d}y -\int_\Omega\nabla\tilde c\nabla(\tilde n c_1+n_2\tilde c)\,{\rm d}x\,{\rm d}y \\ & =\frac12\int_\Omega\Delta|\nabla\tilde c|^2\,{\rm d}x\,{\rm d}y-\int_\Omega |D^2\tilde c|^2\,{\rm d}x\,{\rm d}y+ \int_\Omega (\tilde{\bf u}\cdot\nabla c_1+{\bf u}_2\cdot\nabla\tilde c)\Delta\tilde c\,{\rm d}x\,{\rm d}y \\ & \quad-\int_{\Gamma_T}(\tilde n c_1+n_2\tilde c)\partial_y\tilde c\,{\rm d}S+\int_\Omega(\tilde n c_1+n_2\tilde c)\Delta\tilde c\,{\rm d}x\,{\rm d}y \\ & =\frac12\int_{\Gamma_T}\partial_y|\nabla\tilde c|^2\,{\rm d}S-\frac12\int_{\Gamma_B}\partial_y|\nabla\tilde c|^2\,{\rm d}S-\int_\Omega |D^2\tilde c|^2\,{\rm d}x\,{\rm d}y \\ & \quad+ \int_\Omega (\tilde{\bf u}\cdot\nabla c_1+{\bf u}_2\cdot\nabla\tilde c)\Delta\tilde c\,{\rm d}x\,{\rm d}y \\ & \quad-\int_{\Gamma_T}(\tilde n c_1+n_2\tilde c)\partial_y\tilde c\,{\rm d}S+\int_\Omega(\tilde n c_1+n_2\tilde c)\Delta\tilde c\,{\rm d}x\,{\rm d}y, \end{align*}

noticing that ${\tilde c}_y={\tilde c}_{xy}=0$ on $\Gamma _B$, $\tilde c={\tilde c}_x={\tilde c}_{xx}=0$ on $\Gamma _T$, and ${\tilde c}_{yy}=\tilde n c_1+n_2\tilde c$ on $\Gamma _T$, it implies that

\[ \frac12\partial_y|\nabla\tilde c|^2={\tilde c}_x{\tilde c}_{xy}+{\tilde c}_y{\tilde c}_{yy}=0, \quad \text{on}\ \Gamma_B, \]

\[ \frac12\partial_y|\nabla\tilde c|^2={\tilde c}_x{\tilde c}_{xy}+{\tilde c}_y{\tilde c}_{yy}={\tilde c}_y(\tilde n c_1+n_2\tilde c), \quad \text{on}\ \Gamma_T. \]

Then

\begin{align*} & \frac12\frac{{\rm d}}{{\rm d}t}\int_\Omega |\nabla\tilde c|^2\,{\rm d}x\,{\rm d}y+\int_\Omega |D^2\tilde c|^2\,{\rm d}x\,{\rm d}y \\ & =\int_\Omega (\tilde{\bf u}\cdot\nabla c_1+{\bf u}_2\cdot\nabla\tilde c)\Delta\tilde c\,{\rm d}x\,{\rm d}y +\int_\Omega(\tilde n c_1+n_2\tilde c)\Delta\tilde c\,{\rm d}x\,{\rm d}y \\ & \le \frac14\int_\Omega |D^2\tilde c|^2\,{\rm d}x\,{\rm d}y+C\int_\Omega\left( |\tilde{\bf u}|^2+|\nabla\tilde c|^2+|\tilde c|^2+|\tilde n|^2\right){\rm d}x\,{\rm d}y. \end{align*}

That is

(3.39)\begin{equation} \frac12\frac{{\rm d}}{{\rm d}t}\int_\Omega |\nabla\tilde c|^2\,{\rm d}x\,{\rm d}y+\frac34\int_\Omega |D^2\tilde c|^2\,{\rm d}x\,{\rm d}y \le C\int_\Omega\left( |\tilde{\bf u}|^2+|\nabla\tilde c|^2+|\tilde c|^2+|\tilde n|^2\right){\rm d}x\,{\rm d}y. \end{equation}

Similarly, we also have

(3.40)\begin{align} & \frac12\frac{{\rm d}}{{\rm d}t}\int_\Omega |\tilde{\bf u}|^2\,{\rm d}x\,{\rm d}y+\int_\Omega |\nabla\tilde{\bf u}|^2\,{\rm d}x\,{\rm d}y ={-}\int_\Omega\tilde{\bf u}\cdot\nabla {\bf u}_1\tilde{\bf u}\,{\rm d}x\,{\rm d}y+\int_\Omega\tilde n\nabla\varphi\tilde{\bf u}\,{\rm d}x\,{\rm d}y \nonumber\\ & \le C\int_\Omega |\tilde{\bf u}|^2\,{\rm d}x\,{\rm d}y+C\int_\Omega |\tilde n|^2\,{\rm d}x\,{\rm d}y. \end{align}

Combining (3.37)–(3.40) yields

(3.41)\begin{align} & \frac12\frac{{\rm d}}{{\rm d}t}\int_\Omega (|\tilde n|^2+|\tilde{\bf u}|^2+|\tilde c|^2+ |\nabla\tilde c|^2)\,{\rm d}x\,{\rm d}y\nonumber\\ & \quad +\frac34\int_\Omega\left(|\nabla\tilde n|^2 + |\nabla\tilde c|^2+|D^2\tilde c|^2 + |\nabla\tilde{\bf u}|^2\right) {\rm d}x\,{\rm d}y \nonumber\\ & \le C_1\int_\Omega\left( |\tilde{\bf u}|^2+|\nabla\tilde c|^2+|\tilde c|^2+|\tilde n|^2\right){\rm d}x\,{\rm d}y +C_2\int_{\Gamma_T}(|\tilde n|^2+|\partial_y\tilde c|^2)\,{\rm d}S. \end{align}

Using Sobolev trace embedding inequality, we see that

\[ C_2\int_{\Gamma_T}(|\tilde n|^2+|\partial_y\tilde c|^2)\,{\rm d}S\!\le\! \frac14\int_\Omega\left( |\nabla\tilde n|^2+|D^2\tilde c|^2\right) {\rm d}x\,{\rm d}y \!+\!C_3\int_\Omega\left(\tilde n|^2\!+\!|\nabla\tilde c|^2\right) {\rm d}x\,{\rm d}y. \]

Substituting it into (3.42) gives

(3.42)\begin{align} & \frac12\frac{{\rm d}}{{\rm d}t}\int_\Omega (|\tilde n|^2+|\tilde{\bf u}|^2+|\tilde c|^2+ |\nabla\tilde c|^2)\,{\rm d}x\,{\rm d}y+\frac12\int_\Omega\left(|\nabla\tilde n|^2\right.\nonumber\\ & \left.\quad +\, |\nabla\tilde c|^2+|D^2\tilde c|^2 + |\nabla\tilde{\bf u}|^2\right) {\rm d}x\,{\rm d}y \nonumber\\ & \le C_4\int_\Omega\left( |\tilde{\bf u}|^2+|\nabla\tilde c|^2+|\tilde c|^2+|\tilde n|^2\right){\rm d}x\,{\rm d}y, \end{align}

which implies that

\[ \int_\Omega (|\tilde n|^2+|\tilde{\bf u}|^2+|\tilde c|^2+ |\nabla\tilde c|^2)\,{\rm d}x\,{\rm d}y\equiv 0, \ \text{for any } t>0, \]

and the uniqueness is proved.

4. Global classical solution: Zero Flux-Dirichlet(Robin)-Navier slip boundary conditions

In this section, we consider the global classical solution of the problem (1.1), (1.3) and (1.4).

Consider the following linear problem

(4.1)\begin{equation} \left\{\begin{array}{@{}ll} n_t+{\bf u}\cdot\nabla n=\Delta n-\chi\nabla\cdot(n_+\nabla c), & \text{in}\ Q, \\ c_t-\Delta c+{\bf u}\cdot\nabla c={-}\tilde n_+c, & \text{in}\ Q, \\ {\bf u}_t+\tilde{\bf u}\cdot\nabla {\bf u}=\Delta{\bf u}-\nabla\pi+\sigma\tilde n\nabla\varphi, & \text{in}\ Q, \\ \nabla\cdot{\bf u}=0, & \text{in}\ Q, \\ n_y-\chi n_+c_y=0, \ \tau c_y={-}c+\sigma c_{air}, \ v=0, \ u_y=0, & (x,y)\in \Gamma_{T}, \\ n_y=c_y=0,\ u=v=0, & (x,y)\in \Gamma_{B}, \\ n(x, y, 0)=\sigma n_0(x,y)\ge 0,\ c(x,y, 0)\\ \quad =\sigma c_0(x, y)\ge 0, \ {\bf u}(x,y,0)=\sigma {\bf u}_0(x,y), & x\in\Omega. \end{array}\right. \end{equation}

Similar to §2, for any $T>1$ and for any given $\tilde {\bf u}\in C^{\alpha,\frac {\alpha }2}(\overline Q_T^l)\cap W_2^{1,0}(Q_T^l), \tilde n\in C^{\alpha,\frac {\alpha }2}(\overline Q_T^l)$ with $\nabla \cdot \tilde {\bf u}=0$ and $\tilde {\bf u}\cdot \nu |_{\Gamma _{T}}=0$. The above problem admits a classical solution in $Q_T^l$. By comparison lemma, we also have $0\le c\le \sigma c_{air}$. Next, we show that $n\ge 0$. Let $n_-=\min \{0, n\}$. Multiplying the first equation of (4.1) by $n_1$, and integrating it over $\Omega _l\times (0,t)$ yields

\begin{align*} & \int_{\Omega_l}|n_-(x,y,t)|^2\,{\rm d}x\,{\rm d}y \\ & ={-}\int_0^t\int_{\Omega_l}{\bf u}\cdot\nabla n n_-\,{\rm d}x\,{\rm d}y\,{\rm d}\tau+\int_0^t\int_{\Omega_l}n_-\Delta n\,{\rm d}x\,{\rm d}y\,{\rm d}\tau \\ & \quad - \chi\int_0^t\,{\rm d}\tau\int_{\Omega_l}n_-\nabla\cdot(n_+\nabla c)\,{\rm d}x\,{\rm d}y\\ & ={-}\frac12\int_0^t\int_{\partial J(\tau)}| n_-|^2{\bf u}\cdot\nu\,{\rm d}S\,{\rm d}\tau+\int_0^t\int_{\partial J(\tau)}n_-\left(\frac{\partial n}{\partial\nu}-\chi n_+\frac{\partial c}{\partial\nu} \right)\,{\rm d}S\,{\rm d}\tau\\ & \quad -\int_0^t\,{\rm d}\tau\int_{\Omega_l}|\nabla n_-|^2\,{\rm d}x\,{\rm d}y\\ & ={-}\int_0^t\int_{\Omega_l}|\nabla n_-|^2\,{\rm d}x\,{\rm d}y\,{\rm d}\tau\le 0, \end{align*}

where $J(t)=\{x\in \Omega _l; n(x,t)\le 0\}$, which implies that $\int _{\Omega _l}|n_-(x,t)|^2\,{\rm d}x\,{\rm d}y=0$, that is $n\ge 0$.

We define the mapping

\begin{align*} & \mathcal{T}: C^{\alpha,\frac{\alpha}2}(\overline Q_T^l)\cap W_2^{1,0}(Q_T^l)\times C^{\alpha,\frac{\alpha}2}(\overline Q_T^l) \times [0, 1] \\ & \quad \rightarrow C^{\alpha,\frac{\alpha}2}(\overline Q_T^l)\cap W_2^{1,0}(Q_T^l)\times C^{\alpha,\frac{\alpha}2}(\overline Q_T^l), \end{align*}

\[ \mathcal{T}: (\tilde{\bf u}, \tilde n, \sigma)\rightarrow ({\bf u}, n). \]

\[ \mathcal{T}: (\tilde{\bf u}, \tilde n, 0)\equiv {\bf 0}. \]

Next, we use Leray-Schauder's fixed point theorem, to show the existence of classical solutions. For this purpose, some a prior energy estimates are necessary.

Lemma 4.1 Let $\mathcal {T}: ({\bf u}, n, \sigma )=({\bf u}, n)$. Then we have

(4.2)\begin{equation} n\ge0, \quad 0\le c\le \sigma c_{air}, \end{equation}

and

(4.3)\begin{equation} \|n({\cdot},t)\|_{L^1}=\sigma\|n_0\|_{L^1}. \end{equation}

Moreover, when $\tau =0$, for appropriately small $\chi >0$, we have

(4.4)\begin{align} & \sup_{t\le T}\int_{\Omega_l}\left(|\nabla\sqrt c|^2+n\ln n+|{\bf u}|^2\right) \,{\rm d}x\,{\rm d}y\nonumber\\ & \quad +\sup_{t\le T-1}\int_t^{t+1}\,{\rm d}s\int_{\Omega_l}\left(|\nabla{\bf u}|^2+\frac{|\nabla n|^2}{n}+c|D^2\ln c|^2+ n|\nabla\sqrt c|^2\right)\,{\rm d}x\,{\rm d}y \nonumber\\ & \quad+\sup_{t\le T-1}\int_t^{t+1}\,{\rm d}s\int_{\Gamma_T} |\partial_y c|^3\,{\rm d}S\le C_1. \end{align}

When $\tau =1$,

(4.5)\begin{align} & \sup_{t\le T}\left(\int_{\Omega_l}\left(|\nabla\sqrt c|^2+n\ln n+|{\bf u}|^2+c \ln c\right) {\rm d}x\,{\rm d}y +\int_{\Gamma_T}(c-\sigma c_{air}\ln c)\,{\rm d}S\right) \nonumber\\ & \quad+\sup_{t\le T-1}\int_t^{t+1}\,{\rm d}s\int_{\Omega_l}\left(\frac{|\nabla n|^2}{n}+c|D^2\ln c|^2+n|\nabla\sqrt c|^2+|\nabla{\bf u}|^2\right){\rm d}x\,{\rm d}y \nonumber\\ & \quad+\sup_{t\le T-1}\int_t^{t+1}\,{\rm d}s\int_{\Gamma_T}\left(\frac{|\nabla c|^2c_y }{c^2}+\frac{|c_x|^2}{c}+\sigma c_{air}\frac{|c_x|^2}{c^2}+(c-\sigma c_{air}\ln c)\right){\rm d}S\le C_2. \end{align}

Here $C_1, C_2$ depend only on $n_0, c_0, {\bf u}_0$, $\chi$, and $c_{air}$, and they are independent of $T$.

Proof. From the above analysis, it is easy to obtain (4.2). And (4.3) is derived from a direct integration as follows

\[ \frac{{\rm d}}{{\rm d}t}\int_{\Omega_l}n(x,y,t)\,{\rm d}x\,{\rm d}y=0. \]

Multiplying the first equation of (4.1) by $1+\ln n$, and integrating the resulting equation over $\Omega _l$ gives,

(4.6)\begin{equation} \frac{{\rm d}}{{\rm d}t}\int_{\Omega_l} n\ln n\,{\rm d}x\,{\rm d}y+\int_{\Omega_l} \frac{|\nabla n|^2}{n}\,{\rm d}x\,{\rm d}y=\chi\int_{\Omega_l}\nabla c\nabla n\,{\rm d}x\,{\rm d}y. \end{equation}

For ${\bf } u$ it is completely similar to the proof of (3.12), we conclude that

(4.7)\begin{equation} \frac{{\rm d}}{{\rm d}t}\int_{\Omega_l}|{\bf u}|^2\,{\rm d}x\,{\rm d}y+\int_{\Omega_l}|\nabla{\bf u}|^2\,{\rm d}x\,{\rm d}y\le C_p\|n\|_{L^p}^2, \ \text{for any}\ 1< p<2. \end{equation}

From (3.5) and (3.7), we infer that

(4.8)\begin{align} & \frac{{\rm d}}{{\rm d}t}\int_{\Omega_l}|\nabla\sqrt c|^2\,{\rm d}x\,{\rm d}y+\frac12 \int_{\Omega_l} c|D^2\ln c|^2\,{\rm d}x\,{\rm d}y+\int_{\Omega_l} n|\nabla\sqrt c|^2\,{\rm d}x\,{\rm d}y \nonumber\\ & = \int_{\Gamma_T}\partial_y |\nabla\sqrt c|^2\,{\rm d}S-\frac{1}2\int_{\Omega_l} \nabla n\nabla c\,{\rm d}x\,{\rm d}y- \frac12\int_{\Omega_l}\frac{\nabla c\nabla{\bf u}\nabla c}{c}\,{\rm d}x\,{\rm d}y. \end{align}

(i) When $\tau =0$, completely similar to the proof of (3.10), we have
\begin{align*} & \frac{{\rm d}}{{\rm d}t}\int_{\Omega_l}|\nabla\sqrt c|^2\,{\rm d}x\,{\rm d}y+\frac14 \int_{\Omega_l} c|D^2\ln c|^2\,{\rm d}x\,{\rm d}y+ \int_{\Omega_l} n|\nabla\sqrt c|^2\,{\rm d}x\,{\rm d}y\\ & \quad +\frac{\sqrt 2-1}{2\sigma^2c_{air}^2}\int_{\Gamma_T} |\partial_y c|^3\,{\rm d}S \\ & \le \frac{1}2\int_{\Gamma_T} nc_y\,{\rm d}S-\frac{1}2\int_{\Omega_l} \nabla n\nabla c\,{\rm d}x\,{\rm d}y+\frac {(3+2\sqrt 2)c_{air}}{2}\int_{\Omega_l}|\nabla{\bf u}|^2\,{\rm d}x\,{\rm d}y. \end{align*}
Combining with (4.6), (4.7), and using (3.15), we arrive at
\begin{align*} & \frac{{\rm d}}{{\rm d}t}\left(\int_{\Omega_l}|\nabla\sqrt c|^2\,{\rm d}x\,{\rm d}y+\frac1{2\chi}\int_{\Omega_l} n\ln n\,{\rm d}x\,{\rm d}y+(3+2\sqrt 2)c_{air}\int_{\Omega_l}|{\bf u}|^2\,{\rm d}x\,{\rm d}y\right)\\ & \quad +\,\frac {(3+2\sqrt 2)c_{air}}{2}\int_{\Omega_l}|\nabla{\bf u}|^2\,{\rm d}x\,{\rm d}y \\ & \quad+\frac1{2\chi}\int_{\Omega_l} \frac{|\nabla n|^2}{n}\,{\rm d}x\,{\rm d}y+\frac14 \int_{\Omega_l} c|D^2\ln c|^2\,{\rm d}x\,{\rm d}y+ \int_{\Omega_l} n|\nabla\sqrt c|^2\,{\rm d}x\,{\rm d}y\\ & \quad +\frac{\sqrt 2-1}{2\sigma^2c_{air}^2}\int_{\Gamma_T} |\partial_y c|^3\,{\rm d}S \\ & \le \frac{1}2\int_{\Gamma_T}nc_y\,{\rm d}S+ C_p(3+2\sqrt 2)c_{air}\|n\|_{L^p}^2 \\ & \le \frac{\sqrt 2-1}{4\sigma^2c_{air}^2}\int_{\Gamma_T} |c_y|^3\,{\rm d}S+\frac{\sigma c_{air}}{\sqrt{2\sqrt2-2}}\int_{\Gamma_T}n^{\frac32}\,{\rm d}S+ \frac1{8\chi}\int_{\Omega_l} \frac{|\nabla n|^2}{n}\,{\rm d}x\,{\rm d}y+C. \end{align*}
By lemma 2.1, we have
\begin{align*} & \|n\|_{L^{\frac32}(\Gamma_T)}^{\frac32}=\|\sqrt n\|_{L^3(\Gamma_T)}^3\le C_1\|\nabla\sqrt n\|_{L^2}^2\|\sqrt n\|_{L^2}+C_2\|n\|_{L^1}\\ & \quad \le C_3\int_{\Omega_l} \frac{|\nabla n|^2}{n}\,{\rm d}x\,{\rm d}y+C_4. \end{align*}
Combining the above two inequalities, then when $\chi$ is appropriately small, such that $\frac {C_3\sigma c_{air}}{\sqrt {2\sqrt 2-2}}\le \frac 1{8\chi }$, we arrive at
(4.9)\begin{align} & \frac{{\rm d}}{{\rm d}t}\left(\int_{\Omega_l}|\nabla\sqrt c|^2\,{\rm d}x\,{\rm d}y+\frac1{2\chi}\int_{\Omega_l} n\ln n\,{\rm d}x\,{\rm d}y+(3+2\sqrt 2)c_{air}\int_{\Omega_l}|{\bf u}|^2\,{\rm d}x\,{\rm d}y\right)\nonumber\\ & \quad +\frac {(3+2\sqrt 2)c_{air}}{2}\int_{\Omega_l}|\nabla{\bf u}|^2\,{\rm d}x\,{\rm d}y \nonumber\\ & \quad+\frac1{4\chi}\int_{\Omega_l} \frac{|\nabla n|^2}{n}\,{\rm d}x\,{\rm d}y+\frac14 \int_{\Omega_l} c|D^2\ln c|^2\,{\rm d}x\,{\rm d}y+ \int_{\Omega_l} n|\nabla\sqrt c|^2\,{\rm d}x\,{\rm d}y\nonumber\\ & \quad +\frac{\sqrt 2-1}{4\sigma^2c_{air}^2}\int_{\Gamma_T} |\partial_y c|^3\,{\rm d}S\le C. \end{align}
Noticing that
\[ \|n\ln n\|_{L^1}\le \|n\|_{L^2}^2+\|n\|_{L^1}\le C\|\nabla\sqrt n\|_{L^2}^2+C, \]
and
\[ |\nabla\sqrt c|^2\le \frac{|\nabla c|^4}{c^3}+c, \]
combining (3.9) and Poincaré inequality, we see that
(4.10)\begin{align} & \int_{\Omega_l}\left(|\nabla\sqrt c|^2+n\ln n+|{\bf u}|^2+c \ln c\right) {\rm d}x\,{\rm d}y \nonumber\\ & \le C\int_{\Omega_l}\left(\frac{|\nabla n|^2}{n} +c|D^2\ln c|^2+|\nabla{\bf u}|^2\right){\rm d}x\,{\rm d}y+C\int_{\Gamma_T} |c_y|^3\,{\rm d}S. \end{align}
Combining (4.9), (4.11), and (4.4) is derived.
(ii) When $\tau =1$, noticing that $c_y=-c+\sigma c_{air}$ on $\Gamma _T$, we take the derivative in the tangential direction and get that $c_{xy}=-c_x$ on $\Gamma _T$, then

(4.11)\begin{align} \int_{\Gamma_T}\partial_y |\nabla\sqrt c|^2\,{\rm d}S& ={-}\frac14\int_{\Gamma_T}\frac{|\nabla c|^2\partial_y c}{c^2}\,{\rm d}S+\frac12\int_{\Gamma_T}\frac{\nabla c\cdot\nabla \partial_y c}{c}\,{\rm d}S \nonumber\\ & ={-}\frac14\int_{\Gamma_T}\frac{|\nabla c|^2c_y }{c^2}\,{\rm d}S+\frac12\int_{\Gamma_T}\frac{c_x c_{xy}}{c}\,{\rm d}S +\frac12\int_{\Gamma_T}\frac{c_y c_{yy}}{c}\,{\rm d}S \nonumber\\ & ={-}\frac14\int_{\Gamma_T}\frac{|\nabla c|^2c_y }{c^2}\,{\rm d}S\!-\!\frac12\int_{\Gamma_T}\frac{|c_x|^2}{c}\,{\rm d}S \!+\!\frac12\int_{\Gamma_T}\frac{(\sigma c_{air}-c) c_{yy}}{c}\,{\rm d}S. \end{align}

Noticing that $c_t+uc_x=c_{xx}+c_{yy}-nc$ on $\Gamma _T$, multiplying this equation by $\frac 12\frac {(\sigma c_{air}-c)}{c}$, and integrating it on $\Gamma _T$ yields

(4.12)\begin{align} & \frac12\int_{\Gamma_T}\frac{(\sigma c_{air}-c) c_{yy}}{c}\,{\rm d}S- \frac12\frac{{\rm d}}{{\rm d}t}\int_{\Gamma_T}(\sigma c_{air}\ln c-c)\,{\rm d}S \nonumber\\ & =\frac12\int_{\Gamma_T} \frac{(\sigma c_{air}-c)}{c}uc_x\,{\rm d}S+ \frac12\int_{\Gamma_T}(\sigma c_{air}-c)n\,{\rm d}S-\frac12\int_{\Gamma_T}\frac{(\sigma c_{air}-c)}{c}c_{xx}\,{\rm d}S \nonumber\\ & =\frac12\int_{\Gamma_T} \frac{(\sigma c_{air}-c)}{c}uc_x\,{\rm d}S+ \frac12\int_{\Gamma_T}(\sigma c_{air}-c)n\,{\rm d}S\nonumber\\ & \quad -\frac12\int_0^l\frac{(\sigma c_{air}-c(x,1, t))}{c(x,1, t)}c_{xx}(x,1, t)\,{\rm d}x\,{\rm d}y \nonumber\\ & =\frac12\int_{\Gamma_T} \frac{(\sigma c_{air}-c)}{c}uc_x\,{\rm d}S+ \frac12\int_{\Gamma_T}(\sigma c_{air}-c)n\,{\rm d}S- \frac{\sigma c_{air}}2\int_{\Gamma_T}\frac{|c_x|^2}{c^2}\,{\rm d}S. \end{align}

Combining (4.8), (4.4), (4.11) and (4.12), we arrive at

\begin{align*} & \frac{{\rm d}}{{\rm d}t}\int_{\Omega_l}|\nabla\sqrt c|^2\,{\rm d}x\,{\rm d}y +\frac12\frac{{\rm d}}{{\rm d}t}\int_{\Gamma_T}(c-\sigma c_{air}\ln c)\,{\rm d}S\\ & \quad +\frac12 \int_{\Omega_l} c|D^2\ln c|^2\,{\rm d}x\,{\rm d}y+\int_{\Omega_l} n|\nabla\sqrt c|^2\,{\rm d}x\,{\rm d}y \\ & \quad+\frac14\int_{\Gamma_T}\frac{|\nabla c|^2c_y }{c^2}\,{\rm d}S+\frac12\int_{\Gamma_T}\frac{|c_x|^2}{c}\,{\rm d}S+ \frac{\sigma c_{air}}2\int_{\Gamma_T}\frac{|c_x|^2}{c^2}\,{\rm d}S\\ & = \frac12\int_{\Gamma_T} \frac{(\sigma c_{air}-c)}{c}uc_x\,{\rm d}S+ \frac12\int_{\Gamma_T}(\sigma c_{air}-c)n\,{\rm d}S\\ & \quad-\frac{1}2\int_{\Omega_l} \nabla n\nabla c\,{\rm d}x\,{\rm d}y- \frac12\int_{\Omega_l}\frac{\nabla c\nabla{\bf u}\nabla c}{c}\,{\rm d}x\,{\rm d}y \\ & \le \frac{\sigma c_{air}}4\int_{\Gamma_T}\frac{|c_x|^2}{c^2}\,{\rm d}S +\sigma c_{air}\int_{\Gamma_T}u^2\,{\rm d}S+\frac14\int_{\Gamma_T}\frac{|c_x|^2}{c}\,{\rm d}S\\ & \quad+\int_{\Gamma_T}cu^2\,{\rm d}S +\frac12\int_{\Gamma_T}(\sigma c_{air}-c)n\,{\rm d}S \\ & \quad-\frac{1}2\int_{\Omega_l} \nabla n\nabla c\,{\rm d}x\,{\rm d}y+\eta\int_{\Omega_l}\frac{|\nabla c|^4}{c^3}\,{\rm d}x\,{\rm d}y+\frac{1}{16\eta}\int_{\Omega_l}c|\nabla{\bf u}|^2\,{\rm d}x\,{\rm d}y \\ & \le \frac{\sigma c_{air}}4\int_{\Gamma_T}\frac{|c_x|^2}{c^2}\,{\rm d}S +2\sigma c_{air}\int_{\Gamma_T}u^2\,{\rm d}S+\frac14\int_{\Gamma_T}\frac{|c_x|^2}{c}\,{\rm d}S \\ & \quad +\frac{\sigma c_{air}}2\int_{\Gamma_T}n\,{\rm d}S-\frac{1}2\int_{\Omega_l} \nabla n\nabla c\,{\rm d}x\,{\rm d}y \\ & \quad+\eta\int_{\Omega_l}\frac{|\nabla c|^4}{c^3}\,{\rm d}x\,{\rm d}y+\frac{1}{16\eta}\int_{\Omega_l}c|\nabla{\bf u}|^2\,{\rm d}x\,{\rm d}y, \end{align*}

which implies that

(4.13)\begin{align} & \frac{{\rm d}}{{\rm d}t}\int_{\Omega_l}|\nabla\sqrt c|^2\,{\rm d}x\,{\rm d}y +\frac12\frac{{\rm d}}{{\rm d}t}\int_{\Gamma_T}(c-\sigma c_{air}\ln c)\,{\rm d}S +\frac12 \int_{\Omega_l} c|D^2\ln c|^2\,{\rm d}x\,{\rm d}y\nonumber\\ & \quad +\int_{\Omega_l} n|\nabla\sqrt c|^2\,{\rm d}x\,{\rm d}y \nonumber\\ & \quad+\frac14\int_{\Gamma_T}\frac{|\nabla c|^2c_y }{c^2}\,{\rm d}S+\frac14\int_{\Gamma_T}\frac{|c_x|^2}{c}\,{\rm d}S + \frac{\sigma c_{air}}4\int_{\Gamma_T}\frac{|c_x|^2}{c^2}\,{\rm d}S \nonumber\\ & \le 2\sigma c_{air}\int_{\Gamma_T}u^2\,{\rm d}S +\frac{\sigma c_{air}}2\int_{\Gamma_T}n\,{\rm d}S-\frac{1}2\int_{\Omega_l} \nabla n\nabla c\,{\rm d}x\,{\rm d}y +\eta\int_{\Omega_l}\frac{|\nabla c|^4}{c^3}\,{\rm d}x\,{\rm d}y\nonumber\\ & \quad +\frac{1}{16\eta}\int_{\Omega_l}c|\nabla{\bf u}|^2\,{\rm d}x\,{\rm d}y. \end{align}

Noticing that

\[ u(x,1, t)=u(x,0, t)+\int_0^1u_y(x,y,t)\,{\rm d}y=\int_0^1u_y(x,y,t)\,{\rm d}y, \]

from Hölder's inequality, we infer that

\[ |u(x,1, t)|^2\le\int_0^1|u_y(x,y,t)|^2\,{\rm d}y, \]

then

(4.14)\begin{equation} 2\sigma c_{air}\int_{\Gamma_T}u^2\,{\rm d}S\le 2\sigma c_{air}\int_0^l\,{\rm d}x\int_0^1|u_y(x,y,t)|^2\,{\rm d}y\le 2\sigma c_{air}\int_{\Omega_l}|\nabla{\bf u}|^2\,{\rm d}x\,{\rm d}y. \end{equation}

By a direct calculation, we see that

\begin{align*} & \int_{\Omega_l}\frac{|\nabla c|^4}{c^3}\,{\rm d}x\,{\rm d}y=\int_{\Omega_l} |\nabla\ln c|^2\nabla\ln c\nabla c\,{\rm d}x\,{\rm d}y \\ & =\int_{\Gamma_T} |\nabla\ln c|^2\nabla c\cdot\nu\,{\rm d}S-\int_{\Omega_l} c^{{-}1}\left( |\nabla c|^2\Delta\ln c+2\nabla cD^2\ln c\nabla c \right){\rm d}x\,{\rm d}y \\ & \le \int_{\Gamma_T}\frac{|\nabla c|^2\nabla c\cdot\nu}{c^2}\,{\rm d}S+\left(\int_{\Omega_l}\frac{|\nabla c|^4}{c^3}\,{\rm d}x\,{\rm d}y\right)^{\frac12}\left(\int_{\Omega_l} c|\Delta\ln c|^2\,{\rm d}x\,{\rm d}y\right)^{\frac12}\\ & \quad +2\left(\int_{\Omega_l}\frac{|\nabla c|^4}{c^3}\,{\rm d}x\,{\rm d}y\right)^{\frac12}\left(\int_{\Omega_l} c|D^2\ln c|^2\,{\rm d}x\,{\rm d}y\right)^{\frac12} \\ & \le \int_{\Gamma_T}\frac{|\nabla c|^2 c_y}{c^2}\,{\rm d}S\\ & \quad +\sqrt 2\left(\int_{\Omega_l}\frac{|\nabla c|^4}{c^3}\,{\rm d}x\,{\rm d}y\right)^{\frac12} \left(\int_{\Omega_l} c|D^2\ln c|^2\,{\rm d}x\,{\rm d}y\right)^{\frac12}\\ & \quad +2\left(\int_{\Omega_l}\frac{|\nabla c|^4}{c^3}\,{\rm d}x\,{\rm d}y\right)^{\frac12} \left(\int_{\Omega_l} c|D^2\ln c|^2\,{\rm d}x\,{\rm d}y\right)^{\frac12} \\ & \le \int_{\Gamma_T}\frac{|\nabla c|^2 c_y}{c^2}\,{\rm d}S+\frac12 \int_{\Omega_l}\frac{|\nabla c|^4}{c^3}\,{\rm d}x\,{\rm d}y+\frac{(2+\sqrt 2)^2}2\int_{\Omega_l} c|D^2\ln c|^2\,{\rm d}x\,{\rm d}y, \end{align*}

that is

(4.15)\begin{equation} \int_{\Omega_l}\frac{|\nabla c|^4}{c^3}\,{\rm d}x\,{\rm d}y\le 2 \int_{\Gamma_T}\frac{|\nabla c|^2 c_y}{c^2}\,{\rm d}S+ (6+4\sqrt 2)\int_{\Omega_l} c|D^2\ln c|^2\,{\rm d}x\,{\rm d}y. \end{equation}

Taking $\eta =\frac {1}{4(6+4\sqrt 2)}$ in (4.13), and combining with (4.2), (4.14) and (4.15), we arrive at

(4.16)\begin{align} & \frac{{\rm d}}{{\rm d}t}\int_{\Omega_l}|\nabla\sqrt c|^2\,{\rm d}x\,{\rm d}y +\frac12\frac{{\rm d}}{{\rm d}t}\int_{\Gamma_T}(c-\sigma c_{air}\ln c)\,{\rm d}S +\frac14 \int_{\Omega_l} c|D^2\ln c|^2\,{\rm d}x\,{\rm d}y\nonumber\\ & \quad +\int_{\Omega_l} n|\nabla\sqrt c|^2\,{\rm d}x\,{\rm d}y \nonumber\\ & \quad+\frac18\int_{\Gamma_T}\frac{|\nabla c|^2c_y }{c^2}\,{\rm d}S+\frac14\int_{\Gamma_T}\frac{|c_x|^2}{c}\,{\rm d}S+\frac{\sigma c_{air}}4\int_{\Gamma_T}\frac{|c_x|^2}{c^2}\,{\rm d}S \nonumber\\ & \le \frac{\sigma c_{air}}2\int_{\Gamma_T}n\,{\rm d}S-\frac{1}2\int_{\Omega_l} \nabla n\nabla c\,{\rm d}x\,{\rm d}y +\frac {(7+2\sqrt 2)\sigma c_{air}}{2}\int_{\Omega_l}|\nabla{\bf u}|^2\,{\rm d}x\,{\rm d}y. \end{align}

Similar to (3.15), we have

(4.17)\begin{align} C_p(7+2\sqrt 2)c_{air}\|n\|_{L^p}^2& =C_p(7+2\sqrt 2)c_{air}\|\sqrt n\|_{L^{2p}}^4 \nonumber\\ & \le C_1\|\nabla\sqrt n\|_{L^{2}}^{4-\frac 4p}\|\sqrt n\|_{L^{2}}^{\frac 4p} +C_2\|\sqrt n\|_{L^{2}}^4 \nonumber\\ & \le \frac1{2\chi}\|\nabla\sqrt n\|_{L^{2}}^2+C_3. \end{align}

Combining (4.6), (4.7) , (4.16), (4.17), and using Sobolev interpolation inequality, we arrive at

\begin{align*} & \frac{{\rm d}}{{\rm d}t}\left(\int_{\Omega_l}|\nabla\sqrt c|^2\,{\rm d}x\,{\rm d}y+\frac1{2\chi}\int_{\Omega_l} n\ln n\,{\rm d}x\,{\rm d}y+(7+2\sqrt 2)c_{air}\int_{\Omega_l}|{\bf u}|^2\,{\rm d}x\,{\rm d}y\right.\\ & \left.\quad +\,\frac12\int_{\Gamma_T}(c-\sigma c_{air}\ln c)\,{\rm d}S\right) \\ & \quad+\frac1{2\chi}\int_{\Omega_l} \frac{|\nabla n|^2}{n}\,{\rm d}x\,{\rm d}y+\frac14 \int_{\Omega_l} c|D^2\ln c|^2\,{\rm d}x\,{\rm d}y+ \int_{\Omega_l} n|\nabla\sqrt c|^2\,{\rm d}x\,{\rm d}y\\ & \quad +\frac18\int_{\Gamma_T}\frac{|\nabla c|^2c_y }{c^2}\,{\rm d}S +\frac14\int_{\Gamma_T}\frac{|c_x|^2}{c}\,{\rm d}S \\ & \quad+\frac{\sigma c_{air}}4\int_{\Gamma_T}\frac{|c_x|^2}{c^2}\,{\rm d}S+\frac {(7+2\sqrt 2)c_{air}}{2}\int_{\Omega_l}|\nabla{\bf u}|^2\,{\rm d}x\,{\rm d}y \\ & \le \frac{\sigma c_{air}}2\int_{\Gamma_T}n\,{\rm d}S+ C_p(7+2\sqrt 2)c_{air}\|n\|_{L^p}^2 \\ & \le \frac1{8\chi}\int_{\Omega_l} \frac{|\nabla n|^2}{n}\,{\rm d}x\,{\rm d}y+\int_{\Omega_l} n\,{\rm d}x\,{\rm d}y+ \frac1{8\chi}\int_{\Omega_l} \frac{|\nabla n|^2}{n}\,{\rm d}x\,{\rm d}y+C, \end{align*}

that is

(4.18)\begin{align} & \frac{{\rm d}}{{\rm d}t}\left(\int_{\Omega_l}|\nabla\sqrt c|^2\,{\rm d}x\,{\rm d}y+\frac1{2\chi}\int_{\Omega_l} n\ln n\,{\rm d}x\,{\rm d}y+(7+2\sqrt 2)c_{air}\int_{\Omega_l}|{\bf u}|^2\,{\rm d}x\,{\rm d}y \right.\nonumber\\ & \left.\quad +\,\frac12\int_{\Gamma_T}(c-\sigma c_{air}\ln c)\,{\rm d}S\right) \nonumber\\ & \quad+\frac1{4\chi}\int_{\Omega_l} \frac{|\nabla n|^2}{n}\,{\rm d}x\,{\rm d}y+\frac14 \int_{\Omega_l} c|D^2\ln c|^2\,{\rm d}x\,{\rm d}y+ \int_{\Omega_l} n|\nabla\sqrt c|^2\,{\rm d}x\,{\rm d}y\nonumber\\ & \quad +\frac18\int_{\Gamma_T}\frac{|\nabla c|^2c_y }{c^2}\,{\rm d}S+\frac14\int_{\Gamma_T}\frac{|c_x|^2}{c}\,{\rm d}S \nonumber\\ & \quad+\frac{\sigma c_{air}}4\int_{\Gamma_T}\frac{|c_x|^2}{c^2}\,{\rm d}S+\frac {(7+2\sqrt 2)c_{air}}{2}\int_{\Omega_l}|\nabla{\bf u}|^2\,{\rm d}x\,{\rm d}y\le C. \end{align}

Multiplying the second equation of (4.1) by $1+\ln c$, and integrating it over $\Omega _l$ yields

(4.19)\begin{align} & \frac{{\rm d}}{{\rm d}t}\int_{\Omega_l}c \ln c\,{\rm d}x\,{\rm d}y+\int_{\Omega_l}\frac{|\nabla c|^2}{c}\,{\rm d}x\,{\rm d}y+\int_{\Omega_l}nc(1+\ln c)\,{\rm d}x\,{\rm d}y\nonumber\\ & = \int_{\Gamma_T}c_y(1+\ln c)\,{\rm d}S=\int_{\Gamma_T}(\sigma c_{air}-c)(1+\ln c)\,{\rm d}S \nonumber\\ & ={-}\int_{\Gamma_T}(c\ln c+c-\sigma c_{air}\ln c)\,{\rm d}S+\sigma c_{air}l \nonumber\\ & ={-}\int_{\Gamma_T}(c-\sigma c_{air}\ln c)\,{\rm d}S+C. \end{align}

Combining (4.18) and (4.19), we arrive that

(4.20)\begin{align} & \frac{{\rm d}}{{\rm d}t}\left(\int_{\Omega_l}\left(|\nabla\sqrt c|^2+\frac1{2\chi}n\ln n+(7+2\sqrt 2)c_{air}|{\bf u}|^2+c \ln c\right) {\rm d}x\,{\rm d}y\right.\nonumber\\ & \left.\quad +\,\frac12\int_{\Gamma_T}(c-\sigma c_{air}\ln c)\,{\rm d}S\right) \nonumber\\ & \quad+\frac1{4\chi}\int_{\Omega_l} \frac{|\nabla n|^2}{n}\,{\rm d}x\,{\rm d}y+\frac14 \int_{\Omega_l} c|D^2\ln c|^2\,{\rm d}x\,{\rm d}y+ \int_{\Omega_l} n|\nabla\sqrt c|^2\,{\rm d}x\,{\rm d}y\nonumber\\ & \quad +\frac18\int_{\Gamma_T}\frac{|\nabla c|^2c_y }{c^2}\,{\rm d}S+\frac14\int_{\Gamma_T}\frac{|c_x|^2}{c}\,{\rm d}S \nonumber\\ & \quad+\frac{\sigma c_{air}}4\int_{\Gamma_T}\frac{|c_x|^2}{c^2}\,{\rm d}S+\frac {(7+2\sqrt 2)c_{air}}{2}\int_{\Omega_l}|\nabla{\bf u}|^2\,{\rm d}x\,{\rm d}y\nonumber\\ & \quad +\int_{\Gamma_T}(c-\sigma c_{air}\ln c)\,{\rm d}S\le C. \end{align}

Noticing that

\[ \|n\ln n\|_{L^1}\le \|n\|_{L^2}^2+\|n\|_{L^1}\le C\|\nabla\sqrt n\|_{L^2}^2+C, \]

and

\[ |\nabla\sqrt c|^2\le \frac{|\nabla c|^4}{c^3}+c, \]

combining (4.15) and Poincaré inequality, we see that

(4.21)\begin{align} & \int_{\Omega_l}\left(|\nabla\sqrt c|^2+n\ln n+|{\bf u}|^2+c \ln c\right) {\rm d}x\,{\rm d}y \nonumber\\ & \le C\int_{\Omega_l}\left(\frac{|\nabla n|^2}{n} +c|D^2\ln c|^2+|\nabla{\bf u}|^2\right)\,{\rm d}x\,{\rm d}y+C\int_{\Gamma_T}\frac{|\nabla c|^2c_y }{c^2}\,{\rm d}S. \end{align}

Combining (4.20) and (4.21), we finally conclude (4.5).

Lemma 4.2 Assume that $\mathcal {T}: ({\bf u}, n, \sigma )=({\bf u}, n)$. Then for $\tau =0$ with small $\chi >0$, or $\tau =1$, we have

(4.22)\begin{align} & \sup_{0< t< T}\|\nabla c({\cdot},t)\|_{L^2}^2+\sup_{0< t< T-1}\int_t^{t+1}(\|D^2 c\|_{L^2}^2+\|c_t\|_{L^2}^2)\,{\rm d}s\le C, \end{align}

(4.23)\begin{align} & \sup_{0< t< T}\|\nabla {\bf u}({\cdot},t)\|_{L^2}^2+ \sup_{0< t< T-1}\int_t^{t+1}(\|D^2{\bf u}\|_{L^2}^2+\|{\bf u}_t\|_{L^2}^2)\,{\rm d}s\le C, \end{align}

where the constants $C$ depend only on $n_0, c_0, {\bf u}_0$, $\chi$, and $c_{air}$, and they are independent of $T$.

Proof. When $\tau =0$, the proof is completely similar to lemma 3.2. We omit it. In what follows, we consider the case $\tau =1$. Applying $\nabla$ to the second equation of (3.1), and multiplying the resulting equation by $\nabla c$, noticing that $c_{xy}=-c_x$ on $\Gamma _T$, and using (3.19)–(3.21) yields

(4.24)\begin{align} & \frac12\frac{{\rm d}}{{\rm d}t}\int_{\Omega_l}|\nabla c|^2\,{\rm d}x\,{\rm d}y+\int_{\Omega_l}|D^2 c|^2\,{\rm d}x\,{\rm d}y ={-}\int_{\Omega_l}\nabla({\bf u}\nabla c)\nabla c\,{\rm d}x\,{\rm d}y\nonumber\\ & \quad -\int_{\Omega_l}\nabla(cn)\nabla c \,{\rm d}x\,{\rm d}y +\int_{\Gamma_T}(c_xc_{xy}+c_yc_{yy})\,{\rm d}S \nonumber\\ & \le \|\nabla{\bf u}\|_{L^2}\|\nabla c\|_{L^4}^2+\|{\bf u}\|_{L^4}\|D^2 c\|_{L^2}\|\nabla c\|_{L^4}\nonumber\\ & \quad +2\|c\|_{L^\infty}\|\nabla\sqrt n\|_{L^2}\|\sqrt n\nabla c\|_{L^2}-\|\sqrt n\nabla c\|_{L^2}^2 \nonumber\\ & \quad-\int_{\Gamma_T}|c_x|^2\,{\rm d}S+\int_{\Gamma_T}(\sigma c_{air}-c)c_{yy}\,{\rm d}S \nonumber\\ & \le \frac12\int_{\Omega_l}|D^2 c|^2\,{\rm d}x\,{\rm d}y+\tilde C_1\int_{\Omega_l}|\nabla{\bf u}|^2\,{\rm d}x\,{\rm d}y\nonumber\\ & \quad +\tilde C_2\int_{\Omega_l}\frac{|\nabla n|^2}n\,{\rm d}x\,{\rm d}y -\int_{\Gamma_T}|c_x|^2\,{\rm d}S+\int_{\Gamma_T}(\sigma c_{air}-c)c_{yy}\,{\rm d}S. \end{align}

Noticing that $c_t+uc_x=c_{xx}+c_{yy}-nc$ on $\Gamma _T$, multiplying this equation by $(\sigma c_{air}-c)$, integrating it on $\Gamma _T$, and using (4.14) yields

(4.25)\begin{align} & \int_{\Gamma_T}(\sigma c_{air}-c) c_{yy}\,{\rm d}S- \frac{{\rm d}}{{\rm d}t}\int_{\Gamma_T}(\sigma c_{air} c-\frac{|c|^2}2)\,{\rm d}S \nonumber\\ & =\int_{\Gamma_T}(\sigma c_{air}-c)uc_x\,{\rm d}S+ \int_{\Gamma_T}(\sigma c_{air}-c)cn\,{\rm d}S-\int_{\Gamma_T}(\sigma c_{air}-c)c_{xx}\,{\rm d}S \nonumber\\ & \le \frac{1}2\int_{\Gamma_T}|c_x|^2\,{\rm d}S+\frac{|\sigma c_{air}|^2}2\int_{\Gamma_T}|u|^2\,{\rm d}S +\frac{\sigma c_{air}}2\int_{\Gamma_T}n\,{\rm d}S\nonumber\\ & \quad-\frac12\int_0^l(\sigma c_{air}-c(x,1, t))c_{xx}(x,1, t)\,{\rm d}x \nonumber\\ & \le\frac{1}2\int_{\Gamma_T}|c_x|^2\,{\rm d}S+\frac{|\sigma c_{air}|^2}2\int_{\Gamma_T}|\nabla{\bf u}|^2\,{\rm d}S +C\int_{\Gamma_T}\frac{|\nabla n|^2}{n}\,{\rm d}x\,{\rm d}y\nonumber\\ & \quad + \frac{1}2\int_{\Gamma_T}|c_x|^2\,{\rm d}S+\tilde C_5. \end{align}

Combining (4.24) with (4.25) gives

(4.26)\begin{align} & \frac12\frac{{\rm d}}{{\rm d}t}\int_{\Omega_l}|\nabla c|^2\,{\rm d}x\,{\rm d}y-\frac{{\rm d}}{{\rm d}t}\int_{\Gamma_T}(\sigma c_{air} c-\frac{|c|^2}2)\,{\rm d}S +\frac12\int_{\Omega_l}|D^2 c|^2\,{\rm d}x\,{\rm d}y+\int_{\Gamma_T}|c_x|^2\,{\rm d}S \nonumber\\ & \le \tilde C_3\int_{\Omega_l}|\nabla{\bf u}|^2\,{\rm d}x\,{\rm d}y+\tilde C_4\int_{\Omega_l}\frac{|\nabla n|^2}n\,{\rm d}x\,{\rm d}y+\tilde C_5. \end{align}

Noticing that $c$ is bounded uniformly, using (4.5), we arrive at

\begin{align*} & \sup_{0< t< T}\int_{\Omega_l}|\nabla c|^2\,{\rm d}x\,{\rm d}y+\sup_{0< t< T-1}\int_t^{t+1}\int_{\Omega_l}|D^2 c|^2\,{\rm d}x\,{\rm d}y\,{\rm d}s\\ & \quad + \sup_{0< t< T-1}\int_t^{t+1}\int_{\Gamma_T}|c_x|^2\,{\rm d}S\,{\rm d}s\le C. \end{align*}

Multiplying the second equation of (4.8) by $c_t$, then we obtain

\[ \sup_{0< t< T-1}\int_t^{t+1}\int_{\Omega_l}|c_t|^2\,{\rm d}x\,{\rm d}y\,{\rm d}s\le C. \]

Then (4.22) is proved. The proof of (4.23) is also completely similar to (3.18), we omit it.

Then similar to the proof of lemma 3.3 by deleting the term $\chi \int _{\Gamma _T}n^{r+1}c_y\,{\rm d}S$, and lemma 3.5, we also have

Lemma 4.3 Assume that $\mathcal {T} ({\bf u}, n, \sigma )=({\bf u}, n)$. For $\tau =0$ with small $\chi >0$, or $\tau =1$, we have

(4.27)\begin{equation} \sup_{0< t< T}\int_{\Omega_l}|n({\cdot},t)|^{r+1}\,{\rm d}x\,{\rm d}y+\sup_{0< t< T-1}\int_t^{t+1}\int_{\Omega_l} |\nabla n^{\frac{r+1}2}|^2\,{\rm d}x\,{\rm d}y\,{\rm d}\tau \!\le\! C_r,\quad \text{for any}\ r\!>\!0, \end{equation}

and

(4.28)\begin{equation} \sup_{0< t< T}\|{\bf u}\|_{L^\infty} \le C, \end{equation}

where $C_r$, $C$ depend on $n_0, c_0, {\bf u}_0$, $\chi$, $r$ and $c_{air}$, and $C_r$ also depends on $r$, and all of them are independent of $T$.

Lemma 4.4 Assume that $\mathcal {T} ({\bf u}, n, \sigma )=({\bf u}, n)$. When $\tau =0$ with small $\chi >0$, or $\tau =1$, for any $r>0$,

(4.29)\begin{equation} \sup_{0< t< T}\int_{\Omega_l}|\nabla c|^r\,{\rm d}x\,{\rm d}y\le C_r, \end{equation}

where $C_r$ depends only on $n_0, c_0, {\bf u}_0$, $\chi$, $r$ and $c_{air}$, and it is independent of $T$.

Proof. When $\tau =0$, the proof is completely similar to (3.25). In what follows, we only consider the case $\tau =1$. Let

\[ \tilde c={\rm e}^{\frac{y^2}2}(c-\sigma c_{air}). \]

It is easy to see that $\tilde c(x,y,t)$ is periodic on $x$ with period $l$, $-\sigma c_{air}\,{\rm e}^{\frac {l^2}2}\le \tilde c\le 0$, and

(4.30)\begin{equation} \left\{\begin{array}{@{}l} \dfrac{\partial\tilde c}{\partial t}+{\bf u}\nabla\tilde c-yv\tilde c=\Delta\tilde c+(y^2-1)\tilde c-y\tilde c_y-n\tilde c-\sigma c_{air}n\,{\rm e}^{\frac{y^2}{2}}, \\ \left.\dfrac{\partial\tilde c}{\partial\nu}\right|_{\Gamma_T\cup\Gamma_B}=0. \end{array}\right. \end{equation}

From a direct calculation, we derive that

\[ \partial_y|\nabla\tilde c|^r\left|_{\Gamma_T\cup\Gamma_B} =r|\nabla\tilde c|^{r-2}(\tilde c_{x}\tilde c_{xy}+\tilde c_y\tilde c_{yy})\right|_{\Gamma_T\cup\Gamma_B}=0. \]

Applying $\nabla$ to the first equation of (4.30), multiplying the resulting equation by $|\nabla \tilde c|^{r-2}\nabla \tilde c$, and using (3.27), (4.22), (4.27), (4.28), it yields

\begin{align*} & \frac{1}{r}\frac{{\rm d}}{{\rm d}t}\int_{\Omega_l}|\nabla\tilde c|^r\,{\rm d}x\,{\rm d}y\\ & = \int_{\Omega_l} \nabla\Delta \tilde c |\nabla\tilde c|^{r-2}\nabla\tilde c\,{\rm d}x\,{\rm d}y +\int_{\Omega_l} \nabla(yv\tilde c-{\bf u}\nabla\tilde c \\ & \quad +(y^2-1)\tilde c-y\tilde c_y-n\tilde c-\sigma c_{air}n\,{\rm e}^{\frac{y^2}2})|\nabla\tilde c|^{r-2}\nabla\tilde c\,{\rm d}x\,{\rm d}y \\ & = \frac1r\int_{\Omega_l} \Delta|\nabla\tilde c|^r\,{\rm d}x\,{\rm d}y-(r-1)\int_{\Omega_l}|\nabla\tilde c|^{r-2}|D^2\tilde c|^2\,{\rm d}x\,{\rm d}y \\ & \quad-\int_{\Omega_l}(yv\tilde c-{\bf u}\nabla\tilde c +(y^2-1)\tilde c-y\tilde c_y-n\tilde c-\sigma c_{air}n\,{\rm e}^{\frac{y^2}2})\left(|\nabla\tilde c|^{r-2}\Delta\tilde c\right.\\ & \left.\quad+(r-2)|\nabla\tilde c|^{r-4}\nabla\tilde cD^2 \tilde c\nabla\tilde c\right) {\rm d}x\,{\rm d}y \\ & \le -\frac{r-1}4\int_{\Omega_l}|\nabla\tilde c|^{r-2}|D^2\tilde c|^2\,{\rm d}x\,{\rm d}y +C\int_{\Omega_l}(yv\tilde c-{\bf u}\nabla\tilde c \\ & \quad +(y^2-1)\tilde c-y\tilde c_y-n\tilde c-\sigma c_{air}n\,{\rm e}^{\frac{y^2}2})^2|\nabla\tilde c|^{r-2}\,{\rm d}x\,{\rm d}y \\ & \le -\frac{r-1}4\int_{\Omega_l}|\nabla\tilde c|^{r-2}|D^2\tilde c|^2\,{\rm d}x\,{\rm d}y+\eta \int_{\Omega_l}|\nabla\tilde c|^{r+2}\,{\rm d}x\,{\rm d}y+C_\eta\int_{\Omega_l}(n^{\frac{r+2}2}+1)\,{\rm d}x\,{\rm d}y \\ & \le -\frac{r-1}2\int_{\Omega_l}|\nabla\tilde c|^{r-2}|D^2\tilde c|^2\,{\rm d}x\,{\rm d}y+C_r, \end{align*}

then

(4.31)\begin{equation} \sup_{0< t< T}\|\nabla\tilde c\|_{L^r}^r+\sup_{0< t< T-1}\int_t^{t+1}\int_{\Omega_l}|\nabla\tilde c|^{r-2}|D^2\tilde c|^2\,{\rm d}xmy\,{\rm d}s\le \tilde C_r,\quad \text{for any }\ r>2, \end{equation}

which implies (4.29).

Then similar to the proof of lemmas 3.5 and 3.6, we conclude that

Lemma 4.5 Assume that $\mathcal {T} ({\bf u}, n, \sigma )=({\bf u}, n)$. For $\tau =0$ with small $\chi >0$, or $\tau =1$,

(4.32)\begin{align} & \sup_{0< t< T}(\|n\|_{L^\infty}+\|\nabla c\|_{L^\infty}+\|A^{\beta}{\bf u}\|_{L^\infty})\le C, \quad\text{for any}\ \beta\in\left(\frac12, 1\right), \end{align}

(4.33)\begin{align} & \sup_{0< t< T-1}\left(\|({\bf u}, n, c)\|_{W_p^{2,1}(Q_1^l(t))}+\|\pi\|_{W_p^{1,0}(Q_1^l(t))} \right) \le C_p,\quad \text{for any}\ p>1, \end{align}

where $C$, $C_p$ depend only on $n_0, c_0, {\bf u}_0$, $\chi$, $c_{air}$, $\beta$ and $p$, and they are independent of $T$.

Next, we show theorem 1.2.

Proof Proof of theorem 1.2

Using lemma 4.5, completely similar to the proof theorem 1.1, we complete the proof of global existence in theorem 1.2.

Next, we show the uniqueness. Suppose the contrary. Let $(n_1, c_1, {\bf u_1}, \pi _1)$, $(n_2, c_2, {\bf u_2}, \pi _2)$ be two solutions of (1.1), (1.3), and (1.4). Denote $\tilde n=n_1-n_2$, $\tilde c=c_1-c_2$, $\tilde {\bf u}={\bf u_1}-{\bf u_2}=(\tilde u, \tilde v)$, $\tilde \pi =\pi _1-\pi _2$. Then

(4.34)\begin{equation} \left\{\begin{array}{@{}ll} \tilde n_t+\tilde{\bf u}\cdot\nabla n_1+{\bf u}_2\cdot\nabla\tilde n=\Delta\tilde n-\chi\nabla\cdot(\tilde n\nabla c_1+n_2\nabla\tilde c), & \text{in}\ Q, \\ \tilde c_t-\Delta\tilde c+\tilde{\bf u}\cdot\nabla c_1+{\bf u}_2\cdot\nabla\tilde c={-}\tilde n c_1-n_2\tilde c, & \text{in}\ Q, \\ \tilde{\bf u}_t+\tilde{\bf u}\cdot\nabla {\bf u}_1+{\bf u}_2\cdot\nabla\tilde {\bf u}=\Delta\tilde{\bf u}-\nabla\tilde\pi+\tilde n\nabla\varphi, & \text{in}\ Q, \\ \nabla\cdot\tilde{\bf u}=0, & \text{in}\ Q, \\ \tilde n_y-\chi(\tilde n\partial_y c_1+n_2\partial_y\tilde c)=0, \ \tau\tilde c_y={-}\tilde c,\ \tilde v=0, \ \tilde u_y=0, & (x,y)\in \Gamma_{T}, \\ \tilde n_y=\tilde c_y=0, \ \tilde u=\tilde v=0, & (x,y)\in \Gamma_{B}, \\ \tilde n(x, y, 0)=0, \ \tilde c(x,y, 0)=0, \ \tilde {\bf u}(x,y,0)=0, & x\in\Omega, \end{array}\right. \end{equation}

Multiplying the first equation of (4.34) by $\tilde n$, and integrating it over $\Omega$ yields

that is

(4.35)\begin{equation} \frac12\frac{{\rm d}}{{\rm d}t}\int_\Omega |\tilde n|^2\,{\rm d}x\,{\rm d}y+\frac34\int_\Omega |\nabla\tilde n|^2\,{\rm d}x\,{\rm d}y \le C_1\int_\Omega(|\tilde{\bf u}|^2+|\tilde n|^2+|\nabla\tilde c|^2)\,{\rm d}x\,{\rm d}y. \end{equation}

Multiplying the second equation of (3.36) by $\tilde c$, and integrating it over $\Omega$ yields

\begin{align*} & \frac12\frac{{\rm d}}{{\rm d}t}\int_\Omega |\tilde c|^2\,{\rm d}x\,{\rm d}y+\int_\Omega |\nabla\tilde c|^2\,{\rm d}x\,{\rm d}y+\tau\int_{\Gamma_T}|\tilde c_y|^2\,{\rm d}S \\ & ={-}\int_\Omega\tilde c\tilde{\bf u}\cdot\nabla c_1\,{\rm d}x\,{\rm d}y-\int_\Omega\tilde c{\bf u}_2\cdot\nabla\tilde c\,{\rm d}x\,{\rm d}y -\int_\Omega\tilde c(\tilde n c_1+n_2\tilde c)\,{\rm d}x\,{\rm d}y \\ & =\int_\Omega\nabla\tilde c\cdot\tilde{\bf u} c_1\,{\rm d}x\,{\rm d}y -\int_\Omega\tilde c(\tilde n c_1+n_2\tilde c)\,{\rm d}x\,{\rm d}y \\ & \le \frac14\int_\Omega |\nabla\tilde c|^2\,{\rm d}x\,{\rm d}y+C_2\int_\Omega(|\tilde{\bf u}|^2+|\tilde n|^2+|\tilde c|^2)\,{\rm d}x\,{\rm d}y, \end{align*}

namely,

(4.36)\begin{equation} \frac12\frac{{\rm d}}{{\rm d}t}\int_\Omega |\tilde c|^2\,{\rm d}x\,{\rm d}y\!+\!\frac34\int_\Omega |\nabla\tilde c|^2\,{\rm d}x\,{\rm d}y\!+\!\tau\int_{\Gamma_T}|\tilde c_y|^2\,{\rm d}S\le C_2\int_\Omega(|\tilde{\bf u}|^2\!+\!|\tilde n|^2\!+\!|\tilde c|^2)\,{\rm d}x\,{\rm d}y. \end{equation}

Similarly, we also have

(4.37)\begin{equation} \frac12\frac{{\rm d}}{{\rm d}t}\int_\Omega |\tilde{\bf u}|^2\,{\rm d}x\,{\rm d}y+\int_\Omega |\nabla\tilde{\bf u}|^2\,{\rm d}x\,{\rm d}y \le C_3\int_\Omega (|\tilde{\bf u}|^2+|\tilde n|^2)\,{\rm d}x\,{\rm d}y. \end{equation}

Combining (4.35)–(4.37) yields

\begin{align*} & \frac{{\rm d}}{{\rm d}t}\int_\Omega \left(\frac12|\tilde n|^2+\frac12|\tilde{\bf u}|^2+C_1|\tilde c|^2\right){\rm d}x\,{\rm d}y+\frac34\int_\Omega\left(|\nabla\tilde n|^2 +\frac{2C_1}3 |\nabla\tilde c|^2 + |\nabla\tilde{\bf u}|^2\right) {\rm d}x\,{\rm d}y\nonumber \\ & \le C_4\int_\Omega\left( |\tilde{\bf u}|^2+|\tilde c|^2+|\tilde n|^2\right){\rm d}x\,{\rm d}y, \end{align*}

which implies that

\[ \int_\Omega (|\tilde n|^2+|\tilde{\bf u}|^2+|\tilde c|^2+ |\nabla\tilde c|^2)\,{\rm d}x\,{\rm d}y\equiv 0,\ \text{for any}\ t>0, \]

and the uniqueness is proved.

Acknowledgements

This work is supported by Guangdong Basic and Applied Basic Research Foundation (2021A1515010336), NSFC (12271186, 12171166).

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Article contents

Global classical solution to the chemotaxis-Navier-Stokes system with some realistic boundary conditions

Abstract

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MSC classification

Information

1. Introduction

2. Preliminaries

3. Global classical solution: Neumann-Dirichlet-Navier slip boundary value condition

4. Global classical solution: Zero Flux-Dirichlet(Robin)-Navier slip boundary conditions

Acknowledgements

References

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