Global Strong Solutions to a Coupled Chemotaxis-Fluid Model with Subcritical Sensitivity

Abstract

In this paper we prove the global existence of strong solutions to a coupled chemotaxis-fluid model with subcritical sensitivity in a bounded domain $\varOmega \subset \mathbb{R}^{2}$ without small assumptions on initial data.

Appendices

Appendix A: Local Well-Posedness of a Regularized System

This section is devoted to the proof of the local well-posedness of strong solutions to the problem (1.1)–(1.7).

Theorem A.1

Let $n_{0}\in H^{2}, p_{0}, q_{0}\in H^{3}, u_{0}\in H_{0}^{1}\cap H^{2}$with $\frac{\partial n_{0}}{\partial \nu }= \frac{\partial p_{0}}{\partial \nu }= \frac{\partial q_{0}}{\partial \nu }=0$on $\partial \varOmega$and $n_{0},q_{0},q_{0}\geq 0, \operatorname{div}u_{0}=0$in $\varOmega$. Suppose that $\phi :=\phi (x)$is a smooth function satisfying $\frac{\partial \phi }{\partial \nu }=0$on $\partial \varOmega$. Then the problem (1.1)–(1.7) has a unique strong solution $(n,p,q,u)$satisfying (1.9) for some $0< T\leq \infty$.

We will prove Theorem A.1 by the Banach fixed point theorem. We denote the nonempty set

$$\begin{aligned} \mathcal{A}:=\bigl\{ (\tilde{n},\tilde{u})\in \mathcal{A};( \tilde{n},\tilde{u}) ( \cdot ,0)=(n_{0},u_{0}),\, \tilde{n}\geq 0,\,\operatorname{div}\tilde{u}=0, \,\|(\tilde{n},\tilde{u}) \|_{\mathcal{A}}\leq A\bigr\} \end{aligned}$$

(A.1)

with the norm

$$\begin{aligned} \|(\tilde{n},\tilde{u})\|_{\mathcal{A}}:={}&\|(\tilde{n},\tilde{u}) \|_{L^{\infty }(0,T;H^{2})}+\|(\tilde{n},\tilde{u})\|_{L^{2}(0,T;W^{2,6})} \\ &+\|\partial _{t}(\tilde{n},\tilde{u})\|_{L^{\infty }(0,T;L^{2})}+\| \partial _{t}(\tilde{n},\tilde{u})\|_{L^{2}(0,T;H^{1})}. \end{aligned}$$

Let $(\tilde{n},\tilde{u})\in \mathcal{A}$ be given, we consider the following linear problems:

Let $(n,u)$ be the unique strong solution to the above problem, we define the fixed point map $F:(\tilde{n},\tilde{u})\in \mathcal{A}\to (n,u)\in \mathcal{A}$. We will prove that the map $F$ maps $\mathcal{A}$ into $\mathcal{A}$ for suitable constant $A$ and small $T$ and $F$ is a contraction mapping on $\mathcal{A}$ and thus $F$ has a unique fixed pint in $\mathcal{A}$. This proves Theorem A.1.

Lemma A.1

Let $(\tilde{n},\tilde{u})\in \mathcal{A}$be given. Then the problem $(I)$has a unique solution $p$satisfying $p\geq 0$, and

$$\begin{aligned} &\|p\|_{L^{\infty }(0,T;H^{3})}\leq C, \end{aligned}$$

(A.15)

$$\begin{aligned} &\|p_{t}\|_{L^{\infty }(0,T;H^{1})}\leq C \end{aligned}$$

(A.16)

for some small $0< T\leq 1$.

Here and later on, the letter $C$ will denote a constant independent of the constant $A$ defined in (A.1).

Proof

Since Eq. (A.2) is linear with regular $(\tilde{n},\tilde{u})$, the existence and uniqueness are well-known, we only need to show the a priori estimates.

First, it is clear that

$$ 0\leq p\leq C\quad\mbox{in } \varOmega \times (0,T). $$

(A.17)

Since

$$\begin{aligned} \begin{aligned} \nabla \tilde{u}(x,t)={}&\nabla u_{0}(x)+ \int _{0}^{t}\nabla \partial _{t} \tilde{u}\mathrm{d}s, \\ \|\nabla \tilde{u}\|_{L^{\infty }(0,T;L^{2})}\leq {}&C+ \int _{0}^{T}\| \nabla \partial _{t}\tilde{u}\|_{L^{2}}\mathrm{d}s\leq C+C\sqrt{T}A \leq C \end{aligned} \end{aligned}$$

(A.18)

if $\sqrt{T}A\leq 1$.

Similarly, we have

$$ \|\tilde{n}\|_{L^{\infty }(0,T;H^{1})}\leq C. $$

(A.19)

Testing (A.2) by $p_{t}-\Delta p$, using (A.17), (A.18) and (A.19), we derive that

$$\begin{aligned} &\frac{\mathrm{d}}{\mathrm{d}t} \int |\nabla p|^{2}\mathrm{d}x+ \int \bigl(p_{t}^{2}+( \Delta p)^{2} \bigr)\mathrm{d}x \\ &\quad=- \int (\tilde{u}\cdot \nabla p+\tilde{n}p) (p_{t}-\Delta p) \mathrm{d}x \\ &\quad\leq(\|\tilde{u}\|_{L^{6}}\|\nabla p\|_{L^{3}}+\|\tilde{n} \|_{L^{2}} \|p\|_{L^{\infty }})\|p_{t}-\Delta p \|_{L^{2}} \\ &\quad\leq C(\|\nabla p\|_{L^{3}}+1)\|p_{t}-\Delta p \|_{L^{2}} \\ &\quad\leq C\bigl(\|\nabla p\|_{L^{2}}^{\frac{1}{2}}\|\Delta p \|_{L^{2}}^{\frac{1}{2}}+1\bigr)\|p_{t}- \Delta p \|_{L^{2}} \\ &\quad\leq \frac{1}{2} \int \bigl(p_{t}^{2}+(\Delta p)^{2} \bigr)\mathrm{d}x+C\|\nabla p\|_{L^{2}}^{2}+C, \end{aligned}$$

which gives

$$ \int |\nabla p|^{2}\mathrm{d}x+ \iint _{0}^{T}\bigl((p_{t})^{2}+( \Delta p)^{2}\bigr)\mathrm{d}x\mathrm{d}t\leq C. $$

(A.20)

Here we have used the Gagliardo-Nirenberg inequality

$$ \|\nabla p\|_{L^{3}}^{2}\leq C\|\nabla p\|_{L^{2}} \|\nabla ^{2}p\|_{L^{2}}, $$

(A.21)

and

$$ \|\nabla ^{2}p\|_{L^{2}}\leq C\|\Delta p \|_{L^{2}}. $$

(A.22)

Applying $\partial _{t}$ to (A.2), testing the result by $p_{t}$, using (A.17), we get

$$\begin{aligned} &\frac{1}{2}\frac{\mathrm{d}}{\mathrm{d}t} \int p_{t}^{2}\mathrm{d}x+ \int | \nabla p_{t}|^{2}\mathrm{d}x+ \int \tilde{n}p_{t}^{2}\mathrm{d}x \\ &\quad=- \int \tilde{u}_{t}\cdot \nabla p\cdot p_{t} \mathrm{d}x- \int \tilde{n}_{t}pp_{t}\mathrm{d}x \\ &\quad\leq \int \tilde{u}_{t}p\nabla p_{t}\mathrm{d}x- \int \tilde{n}_{t}pp_{t} \mathrm{d}x \\ &\quad\leq \|\tilde{u}_{t}\|_{L^{2}}\|p\|_{L^{\infty }}\| \nabla p_{t}\|_{L^{2}}+ \|p\|_{L^{\infty }}\| \tilde{n}_{t}\|_{L^{2}}\|p_{t} \|_{L^{2}} \\ &\quad\leq \frac{1}{2}\|\nabla p_{t}\|_{L^{2}}^{2}+CA^{2}+ \|p_{t}\|_{L^{2}}^{2}, \end{aligned}$$

which leads to

$$ \|p_{t}\|_{L^{\infty }(0,T;L^{2})}+\|p_{t}\|_{L^{2}(0,T;H^{1})} \leq C $$

(A.23)

if $A^{2}T\leq 1$ and $T\leq 1$.

Equation (A.2) can be written as

$$ -\Delta p=f:=-\partial _{t}p-\tilde{u}\cdot \nabla p-\tilde{n}p. $$

(A.24)

Using the $H^{2}$ theory of Poisson equation, we have

$$\begin{aligned} \|p\|_{H^{2}}\leq {}&C\|f\|_{L^{2}} \\ \leq {}&C\|p_{t}\|_{L^{2}}+C\|\tilde{u}\|_{L^{6}}\| \nabla p\|_{L^{3}}+C \|\tilde{n}\|_{L^{2}}\|p\|_{L^{\infty }} \\ \leq {}&C+C\|\nabla p\|_{L^{3}} \\ \leq {}&C+C\|\nabla p\|_{L^{2}}^{\frac{1}{2}}\|p\|_{H^{2}}^{\frac{1}{2}}, \end{aligned}$$

whence

$$ \|p\|_{L^{\infty }(0,T;H^{2})}\leq C. $$

(A.25)

Similarly, we have

$$\begin{aligned} \|p\|_{H^{3}}\leq {}&C\|f\|_{H^{1}} \\ \leq {}&C\|f\|_{L^{2}}+C\|\nabla f\|_{L^{2}} \\ \leq {}&C+C\|\nabla p_{t}\|_{L^{2}}+C\|\nabla \tilde{u} \|_{L^{2}}\| \nabla p\|_{L^{\infty }}+C\|\tilde{u}\|_{L^{6}} \|\nabla ^{2}p\|_{L^{3}} \\ &+C\|\tilde{n}\|_{L^{6}}\|\nabla p\|_{L^{3}}+C\|\nabla \tilde{n}\|_{L^{2}} \|p\|_{L^{\infty }} \\ \leq {}&C+C\|\nabla p_{t}\|_{L^{2}}+C\|\nabla p \|_{L^{6}}^{\frac{1}{2}}\|p\|_{H^{3}}^{\frac{1}{2}}, \end{aligned}$$

whence

$$ \|p\|_{H^{3}}\leq C+C\|\nabla p_{t}\|_{L^{2}}. $$

(A.26)

Applying $\partial _{t}$ to (A.2), testing the result by $-\Delta p_{t}$, and using (A.25), (A.18) and (A.19), we have

$$\begin{aligned} &\frac{1}{2}\frac{\mathrm{d}}{\mathrm{d}t} \int |\nabla p_{t}|^{2} \mathrm{d}x+ \int (\Delta p_{t})^{2}\mathrm{d}x \\ &\quad= \int (\tilde{u}\cdot \nabla p_{t}+\tilde{u}_{t} \cdot \nabla p+ \tilde{n}_{t}p+\tilde{n}p_{t})\Delta p_{t}\mathrm{d}x \\ &\quad\leq (\|\tilde{u}\|_{L^{6}}\|\nabla p\|_{L^{3}}+\| \tilde{u}_{t}\|_{L^{3}} \|\nabla p\|_{L^{6}}+\| \tilde{n}_{t}\|_{L^{2}}\|p\|_{L^{\infty }}+\| \tilde{n} \|_{L^{6}}\|p_{t}\|_{L^{3}})\|\Delta p_{t}\|_{L^{2}} \\ &\quad\leq C(1+\|\tilde{u}_{t}\|_{L^{3}}+A+\|p_{t} \|_{L^{3}})\|\Delta p_{t} \|_{L^{2}} \\ &\quad\leq \frac{1}{2}\|\Delta p_{t}\|_{L^{2}}^{2}+C+C \|\tilde{u}_{t}\|_{L^{3}}^{2}+CA^{2}+C \|p_{t}\|_{L^{3}}^{2}, \end{aligned}$$

which proves

$$ \int |\nabla p_{t}|^{2}\mathrm{d}x\leq C $$

(A.27)

if $A^{2}T\leq 1$ and $T\leq 1$.

Using (A.26) and (A.27), we have

$$ \|p\|_{L^{\infty }(0,T;H^{3})}\leq C. $$

(A.28)

This completes the proof of Lemma A.1. □

Lemma A.2

Let $(\tilde{n},\tilde{u})\in \mathcal{A}$be given. Then the problem $(\mathit{II})$has a unique solution $q$satisfying $q\geq 0$, and

$$ 0\leq q,\|q\|_{L^{\infty }(0,T;H^{3})}+\|q_{t}\|_{L^{\infty }(0,T;H^{1})} \leq C, $$

(A.29)

for some small $0< T\leq 1$.

Proof

Since the proof is similar to that of Lemma A.1, we omit the details hare. □

Lemma A.3

Let $(\tilde{n},\tilde{u})\in \mathcal{A}$be given. Then the problem $(\mathit{III})$has a unique solution satisfying $n\geq 0$, and

$$\begin{aligned} &\|n\|_{L^{\infty }(0,T;H^{2})}+\|n\|_{L^{2}(0,T;W^{2,6})}\leq C_{1}, \end{aligned}$$

(A.30)

$$\begin{aligned} &\|n_{t}\|_{L^{\infty }(0,T;L^{2})}+\|n_{t}\|_{L^{2}(0,T;H^{1})} \leq C_{1} \end{aligned}$$

(A.31)

for some small $0\leq T\leq 1$.

Proof

Since Eq. (A.8) is linear with regular functions $(\tilde{n},\tilde{u},p,q)$, the existence and uniqueness are well-known, we only need to show the a priori estimates.

First, it is obvious that $n \geq$, and

$$ \int n\mathrm{d}x= \int n_{0}\mathrm{d}x. $$

(A.32)

Testing (A.8) by $n_{t}-\Delta n$, and using Lemmas A.1–A.2, we compute

$$\begin{aligned} &\frac{\mathrm{d}}{\mathrm{d}t} \int |\nabla n|^{2}\mathrm{d}x+ \int \bigl(n_{t}^{2}+( \Delta n)^{2}\bigr) \mathrm{d}x \\ &\quad=- \int \bigl\{ \bigl(\tilde{u}\cdot \nabla n+nS\nabla ^{2}p+ \nabla p \operatorname{div}(nS)+nS\nabla ^{2}q+\nabla q\operatorname{div}(nS) \\ & \qquad -\nabla \cdot (n\nabla \phi )\bigr) (n_{t}-\Delta n) \bigr\} \mathrm{d}x \\ &\quad\leq C\bigl(\|\tilde{u}\|_{L^{6}}\|\nabla n\|_{L^{3}}+\|n \|_{L^{6}}\|S\|_{L^{\infty }}\|\nabla ^{2}p \|_{L^{3}}+\|\nabla p\|_{L^{\infty }}\|\nabla n\|_{L^{2}} \|S \|_{L^{\infty }} \\ &\qquad+\|\nabla p\|_{L^{\infty }}\|n\|_{L^{\infty }}\|\nabla S \|_{L^{2}}+\|n\|_{L^{6}} \|S\|_{L^{\infty }}\|\nabla ^{2}q\|_{L^{3}} \\ &\qquad+\|\nabla q\|_{L^{\infty }}\|\nabla n\|_{L^{2}}\|S \|_{L^{\infty }}+\| \nabla q\|_{L^{\infty }}\|n\|_{L^{\infty }}\|\nabla S \|_{L^{2}} \\ &\qquad+\|n\|_{H^{1}}\bigr)\|n_{t}-\Delta n\|_{L^{2}} \\ &\quad\leq C(\|\nabla n\|_{L^{3}}+1+\|\nabla S\|_{L^{2}}+\|n \|_{L^{\infty }}+ \|n\|_{H^{1}})\|n_{t}-\Delta n \|_{L^{2}} \\ &\quad\leq \frac{1}{2} \int \bigl(n_{t}^{2}+(\Delta n)^{2} \bigr)\mathrm{d}x+C\|n\|_{H^{1}}^{2}+C+C \|\nabla S \|_{L^{2}}^{2} \end{aligned}$$

which implies

$$ \int |\nabla n|^{2}\mathrm{d}x+ \iint _{0}^{T}\bigl(n_{t}^{2}+| \Delta n|^{2}\bigr) \mathrm{d}x\mathrm{d}t\leq C $$

(A.33)

if $T\leq 1$.

Here we have used the fact:

$$\begin{aligned} \|\nabla S\|_{L^{2}}\leq {}&C \biggl( \biggl\Vert \frac{\partial S}{\partial x_{i}} \biggr\Vert _{L^{2}}+ \biggl\Vert \frac{\partial S}{\partial \tilde{n}} \biggr\Vert _{L^{\infty }}\|\nabla \tilde{n}\|_{L^{2}}+ \biggl\Vert \frac{\partial S}{\partial p}\nabla p \biggr\Vert _{L^{2}}+ \biggl\Vert \frac{\partial S}{\partial q}\nabla q \biggr\Vert _{L^{2}} \biggr) \\ \leq {}&C. \end{aligned}$$

(A.34)

Applying $\partial _{t}$ to (A.8), testing the result by $n_{t}$, and using Lemmas A.1–A.2, we obtain that

$$\begin{aligned} &\frac{1}{2}\frac{\mathrm{d}}{\mathrm{d}t} \int n_{t}^{2}\mathrm{d}x+ \int | \nabla n_{t}|^{2}\mathrm{d}x \\ &\quad=- \int \tilde{u}_{t}\cdot \nabla n\cdot n_{t} \mathrm{d}x+ \int \partial _{t} \biggl(nS \biggl(x,\frac{\tilde{n}}{1+\epsilon \tilde{n}},p,q \biggr)\nabla p \biggr)\cdot \nabla n_{t}\mathrm{d}x \\ &\qquad+ \int \partial _{t} \biggl(nS \biggl(x, \frac{\tilde{n}}{1+\epsilon \tilde{n}},p,q \biggr)\nabla q \biggr) \cdot \nabla n_{t}\mathrm{d}x- \int n_{t}\nabla \phi \nabla n_{t} \mathrm{d}x \\ &\quad\leq \|\tilde{u}_{t}\|_{L^{3}}\|\nabla n \|_{L^{2}}\|n_{t}\|_{L^{6}}+ \|\partial _{t}n\|_{L^{2}}\|S\|_{L^{\infty }}\|\nabla p \|_{L^{\infty }}\| \nabla n_{t}\|_{L^{2}} \\ &\qquad+\|n\|_{L^{\infty }}\|\partial _{t}S\|_{L^{2}}\|\nabla p\|_{L^{\infty }} \|\nabla n_{t}\|_{L^{2}} \\ &\qquad+\|n\|_{L^{\infty }}\|S\|_{L^{\infty }}\|\nabla p_{t} \|_{L^{2}}\|\nabla n_{t} \|_{L^{2}}+\|\partial _{t}n\|_{L^{2}}\|S\|_{L^{\infty }}\|\nabla q \|_{L^{\infty }}\|\nabla n_{t}\|_{L^{2}} \\ &\qquad+\|n\|_{L^{\infty }}\|\partial _{t}S\|_{L^{2}}\|\nabla q\|_{L^{\infty }} \|\nabla n_{t}\|_{L^{2}}+\|n \|_{L^{\infty }}\|S\|_{L^{\infty }}\|\nabla q_{t} \|_{L^{2}}\|\nabla n_{t}\|_{L^{2}} \\ &\qquad+C\|\Delta \phi \|_{L^{\infty }}\|n_{t}\|_{L^{2}}^{2} \\ &\quad\leq C\|\tilde{u}_{t}\|_{L^{3}}\|\nabla n_{t}\|_{L^{2}}+C\|n_{t}\|_{L^{2}} \| \nabla n_{t}\|_{L^{2}}+C\|n\|_{L^{\infty }}\|\partial _{t}S\|_{L^{2}} \|\nabla n_{t} \|_{L^{2}} \\ &\qquad+C\|n\|_{L^{\infty }}\|\nabla n_{t}\|_{L^{2}}+C \|n_{t}\|_{L^{2}}^{2} \\ &\quad\leq \frac{1}{2}\|\nabla n_{t}\|_{L^{2}}^{2}+C \|\tilde{u}_{t}\|_{L^{3}}^{2}+C \|n_{t}\|_{L^{2}}^{2}+C\|n\|_{L^{\infty }}^{2} \|\partial _{t}S\|_{L^{2}}^{2}+C \|n \|_{L^{\infty }}^{2} \\ &\quad\leq \frac{1}{2}\|\nabla n_{t}\|_{L^{2}}^{2}+C \|\tilde{u}_{t}\|_{L^{3}}^{2}+C \|n_{t}\|_{L^{2}}^{2}+C\|n\|_{L^{6}}\|n \|_{H^{2}}A^{2}+C\|n\|_{L^{\infty }}^{2}, \end{aligned}$$

which implies

$$ \|n_{t}\|_{L^{\infty }(0,T;L^{2})}+\|n_{t}\|_{L^{2}(0,T;H^{1})} \leq C_{1} $$

(A.35)

if $A^{2}\sqrt{T}\leq 1$ and $T\leq 1$.

Here we have used the fact:

$$\begin{aligned} \biggl\Vert \frac{\partial S}{\partial t} \biggr\Vert _{L^{2}}\leq{} &C(\| \tilde{n}_{t}\|_{L^{2}}+\|p_{t} \|_{L^{2}}+\|q_{t}\|_{L^{2}}) \\ \leq {}&C(A+1), \end{aligned}$$

(A.36)

and the Gagliardo-Nirenberg inequality

$$ \|n\|_{L^{\infty }}^{2}\leq C\|n\|_{L^{6}}\|n \|_{H^{2}}. $$

(A.37)

Equation (A.8) can be rewritten as

$$\begin{aligned} -\Delta n=g:={}&-\partial _{t}n-\tilde{u}\cdot \nabla n-\nabla \biggl(nS \biggl(x,\frac{\tilde{n}}{1+\epsilon \tilde{n}},p,q \biggr)\cdot \nabla p \biggr) \\ &-\nabla \cdot \biggl(nS \biggl(x,\frac{\tilde{n}}{1+\epsilon \tilde{n}},p,q \biggr)\cdot \nabla q \biggr)+\nabla \cdot (n\nabla \phi ). \end{aligned}$$

(A.38)

Using the $H^{2}$-theory of Poisson equation, we have

$$\begin{aligned} \|n\|_{H^{2}}\leq {}&C\|g\|_{L^{2}} \\ \leq {}&C\|\partial _{t}n\|_{L^{2}}+C\|\tilde{u} \|_{L^{6}}\|\nabla n\|_{L^{3}} \\ &+C\|n\|_{L^{6}}\|S\|_{L^{\infty }}\|\nabla ^{2}p \|_{L^{3}}+C\|n\|_{L^{\infty }}\|\nabla S\|_{L^{2}}\|\nabla p \|_{L^{\infty }} \\ &+C\|\nabla n\|_{L^{2}}\|S\|_{L^{\infty }}\|\nabla p \|_{L^{\infty }}+C\|n \|_{L^{6}}\|S\|_{L^{\infty }}\|\nabla ^{2}q\|_{L^{3}} \\ &+C\|n\|_{L^{\infty }}\|\nabla S\|_{L^{2}}\|\nabla q \|_{L^{\infty }}+C\| \nabla n\|_{L^{2}}\|S\|_{L^{\infty }}\|\nabla q\|_{L^{\infty }}+C\|n\|_{H^{1}} \\ \leq {}&C+C\|\nabla n\|_{L^{3}}+C\|\nabla S\|_{L^{2}}\|n \|_{L^{\infty }} \leq C+C\|\nabla n\|_{L^{2}}^{\frac{1}{2}}\|n \|_{H^{2}}^{\frac{1}{2}}, \end{aligned}$$

which implies

$$ \|n\|_{L^{\infty }(0,T;H^{2})}\leq C_{1}. $$

(A.39)

Similarly, we have

$$\begin{aligned} \|n\|_{W^{2,6}}\leq {}&C\|g\|_{L^{6}} \\ \leq {}&C\|\partial _{t}n\|_{L^{6}}+C\|\tilde{u} \|_{L^{6}}\|\nabla n\|_{L^{\infty }}+C\|n\|_{L^{\infty }}\|S \|_{L^{\infty }}\|\nabla ^{2}p\|_{L^{6}} \\ &+C\|n\|_{L^{\infty }}\|\nabla S\|_{L^{6}}\|\nabla p \|_{L^{\infty }}+C\| \nabla n\|_{L^{6}}\|S\|_{L^{\infty }}\|\nabla p\|_{L^{\infty }} \\ &+C\|n\|_{L^{\infty }}\|S\|_{L^{\infty }}\|\nabla ^{2}q \|_{L^{6}}+C\|n\|_{L^{\infty }}\|\nabla S\|_{L^{6}}\|\nabla q \|_{L^{\infty }} \\ &+C\|\nabla n\|_{L^{6}}\|S\|_{L^{\infty }}\|\nabla q \|_{L^{\infty }}+C\|n \|_{W^{1,6}} \\ \leq {}&C\|\partial _{t}n\|_{L^{6}}+C\|\nabla n \|_{L^{\infty }}+C+C\| \nabla S\|_{L^{6}} \\ \leq {}&C\|\partial _{t}n\|_{L^{6}}+C\|\nabla n \|_{L^{6}}^{\frac{1}{2}}\|n\|_{W^{2,6}}^{\frac{1}{2}}+C+C(\| \nabla \tilde{n}\|_{L^{6}}+1) \\ \leq {}&\frac{1}{2}\|n\|_{W^{2,6}}+C\|\partial _{t}n \|_{L^{6}}+C+CA, \end{aligned}$$

which gives

$$ \|n\|_{L^{2}(0,T;W^{2,6})}\leq C_{1} $$

(A.40)

if $A\sqrt{T}\leq 1$ and $T\leq 1$. This completes the proof of Lemma A.3. □

Lemma A.4

Let $(\tilde{n},\tilde{u})\in \mathcal{A}$be given. Then the problem $(\mathit{IV})$has a unique solution $u$satisfying

$$ \|u\|_{L^{\infty }(0,T;H^{2})}+\|u\|_{L^{2}(0,T;W^{2,6})}+\|u_{t} \|_{L^{\infty }(0,T;L^{2})}+\|u_{t}\|_{L^{2}(0,T;H^{1})}\leq C_{1} $$

(A.41)

for some small $0< T\leq 1$.

Proof

Since Eqs. (A.11) and (A.12) are linear with regular functions $(\tilde{u},n,p,q)$, the existence and uniqueness are well-known, we only need to show the a priori estimates.

Testing (A.11) by $\nabla \pi -\Delta u$, and using (A.12) and Lemmas A.1–A.3, we deduce that

$$\begin{aligned} &\frac{1}{2}\frac{\mathrm{d}}{\mathrm{d}t} \int |\nabla u|^{2}\mathrm{d}x+ \|\nabla \pi -\Delta u \|_{L^{2}}^{2} \\ &\quad\leq (\|n\|_{L^{\infty }}\|\nabla \phi \|_{L^{2}}+\|n \|_{L^{\infty }}\|S \|_{L^{\infty }}\|\nabla p\|_{L^{2}}+\|n \|_{L^{\infty }}\|S\|_{L^{\infty }} \|\nabla q\|_{L^{2}} \\ &\qquad +\|\tilde{u}\|_{L^{6}}\|\nabla u\|_{L^{3}})\|\nabla \pi - \Delta u\|_{L^{2}} \\ &\quad\leq C(1+\|\nabla u\|_{L^{3}})\|\nabla \pi -\Delta u \|_{L^{2}} \\ &\quad\leq \frac{1}{2}\|\nabla \pi -\Delta u\|_{L^{2}}^{2}+C \|\nabla u\|^{2}_{L^{2}}+C, \end{aligned}$$

which yields

$$ \int |\nabla u|^{2}\mathrm{d}x+ \iint _{0}^{T}|\nabla \pi -\Delta u|^{2} \mathrm{d}x\mathrm{d}t\leq C. $$

(A.42)

Using the $H^{2}$-theory of Stokes system:

$$ \|\nabla \pi \|_{L^{2}}+\|u\|_{H^{2}}\leq C\|\nabla \pi - \Delta u\|_{L^{2}}, $$

(A.43)

we have

$$ \|u\|_{L^{2}(0,T;H^{2})}\leq C. $$

(A.44)

Similarly, we get

$$ \|u_{t}\|_{L^{2}(0,T;L^{2})}\leq C. $$

(A.45)

Applying $\partial _{t}$ to (A.11), testing the result by $u_{t}$, and using (A.12), Lemmas A.1–A.3, we observe that

$$\begin{aligned} &\frac{1}{2}\frac{\mathrm{d}}{\mathrm{d}t} \int |u_{t}|^{2}\mathrm{d}x+ \int |\nabla u_{t}|^{2}\mathrm{d}x \\ &\quad=- \int \tilde{u}_{t}\cdot \nabla u\cdot u_{t} \mathrm{d}x+ \int n_{t} \nabla \phi u_{t}\mathrm{d}x+ \int \partial _{t}(nS\nabla p+nS\nabla q)u_{t} \mathrm{d}x \\ &\quad\leq \|\tilde{u}_{t}\|_{L^{3}}\|\nabla u \|_{L^{2}}\|u_{t}\|_{L^{6}}+ \|n_{t} \|_{L^{2}}\|\nabla \phi \|_{L^{\infty }}\|u_{t} \|_{L^{2}} \\ &\qquad+\|n_{t}\|_{L^{2}}\|S\|_{L^{\infty }}(\|\nabla p \|_{L^{\infty }}+\| \nabla q\|_{L^{\infty }})\|u_{t} \|_{L^{2}} \\ &\qquad+\|n\|_{L^{\infty }}\|S_{t}\|_{L^{2}}(\|\nabla p \|_{L^{\infty }}+\| \nabla q\|_{L^{\infty }})\|u_{t} \|_{L^{2}} \\ &\qquad+\|n\|_{L^{\infty }}\|S\|_{L^{\infty }}(\|\nabla p_{t} \|_{L^{2}}+\| \nabla q_{t}\|_{L^{2}}) \|u_{t}\|_{L^{2}} \\ &\quad\leq C\|\tilde{u}_{t}\|_{L^{3}}\|u_{t} \|_{L^{6}}+C\|u_{t}\|_{L^{2}} \\ &\quad\leq \frac{1}{2}\|\nabla u_{t}\|_{L^{2}}^{2}+C \|\tilde{u}_{t}\|_{L^{3}}^{2}+C \|u_{t}\|_{L^{2}}^{2}+C \\ &\quad\leq \frac{1}{2}\|\nabla u_{t}\|_{L^{2}}^{2}+C \|\tilde{u}_{t}\|_{L^{2}}\| \tilde{u}_{t} \|_{H^{1}}+C\|u_{t}\|_{L^{2}}^{2}+C, \end{aligned}$$

which gives

$$ \|u_{t}\|_{L^{\infty }(0,T;L^{2})}+\|u_{t}\|_{L^{2}(0,T;H^{1})} \leq C_{1} $$

(A.46)

if $\sqrt{T}A\leq 1$ and $T\leq 1$.

Equation (A.11) can be written as

$$ -\Delta u+\nabla \pi =h:=-\partial _{t}u-\tilde{u}\cdot \nabla u+n \nabla \phi +nS(\nabla p+\nabla q). $$

(A.47)

Using the $H^{2}$-theory of Stokes system, we have

$$\begin{aligned} \|u\|_{H^{2}}+\|\nabla \pi \|_{L^{2}}\leq {}&C\|h \|_{L^{2}} \\ \leq {}&C\|\partial _{t}u\|_{L^{2}}+C\|\tilde{u} \|_{L^{6}}\|\nabla u\|_{L^{3}}+C \|n\|_{L^{2}}\|\nabla \phi \|_{L^{\infty }} \\ &+C\|n\|_{L^{\infty }}\|S\|_{L^{\infty }}(\|\nabla p\|_{L^{2}}+\| \nabla q \|_{L^{2}}) \\ \leq {}&C+C\|\nabla u\|_{L^{3}}\leq C+C\|\nabla u\|_{L^{2}}^{\frac{1}{2}} \|u \|_{H^{2}}^{\frac{1}{2}}, \end{aligned}$$

which gives

$$ \|u\|_{L^{\infty }(0,T;H^{2})}\leq C. $$

(A.48)

Similarly, using the $W^{2,6}$-theory of Stokes system, we have

$$\begin{aligned} \|u\|_{W^{2,6}}+\|\nabla \pi \|_{L^{6}}\leq {}&C\|h \|_{L^{6}} \\ \leq {}&C\|\partial _{t}u\|_{L^{6}}+C\|\tilde{u} \|_{L^{6}}\|\nabla u\|_{L^{\infty }}+C\|n\|_{L^{6}}\|\nabla \phi \|_{L^{\infty }} \\ &+C\|n\|_{L^{\infty }}\|S\|_{L^{\infty }}(\|\nabla p\|_{L^{6}}+\| \nabla q \|_{L^{6}}) \\ \leq {}&C\|\partial _{t}u\|_{L^{6}}+C\|\nabla u \|_{L^{\infty }}+C \\ \leq {}&C\|\partial _{t}u\|_{L^{6}}+C\|u \|_{W^{2,6}}^{\frac{1}{2}}+C \end{aligned}$$

which implies

$$ \|u\|_{W^{2,6}}\leq C+C\|\nabla \partial _{t}u \|_{L^{2}}, $$

and therefore

$$ \|u\|_{L^{2}(0,T;W^{2,6})}\leq C_{1}. $$

(A.49)

This completes the proof of Lemma A.4. □

Due to the above Lemmas A.1–A.4, we can take $A:=C+(C+1)C_{1}$ and thus $F$ maps $\mathcal{A}$ into $\mathcal{A}$. The following lemma tells us that $F$ is a contraction mapping in the sense of weaker norm.

Lemma A.5

There is a constant $0<\delta <1$such that for any $(\tilde{n}_{i},\tilde{u}_{i})$$(i=1,2)$,

$$ \|F(\tilde{n},\tilde{u}_{1})-F(\tilde{n}_{2}, \tilde{u}_{2})\|_{L^{2}(0,T;H^{1})} \leq \delta \|(\tilde{n}_{1}- \tilde{n}_{2}, \tilde{u}_{1}-\tilde{u}_{2}) \|_{L^{2}(0,T;H^{1})} $$

(A.50)

for some small $0< T\leq 1$.

Proof

Suppose $(p_{i},q_{i},n_{i},u_{i},\pi _{i})$$(i=1,2)$ are the solutions to the problems $(I)$–$(\mathit{IV})$ corresponding to $(\tilde{n}_{i},\tilde{u}_{i})$$(i=1,2)$. Denote

$$\begin{aligned} & p:=p_{1}-p_{2},\qquad q:=q_{1}-q_{2}, \qquad n:=n_{1}-n_{2},\qquad u:=u_{1}-u_{2}, \\ & \pi :=\pi _{1}-\pi _{2},\qquad \tilde{n}:= \tilde{n}_{1}-\tilde{n}_{2},\qquad \tilde{u}:= \tilde{u}_{1}-\tilde{u}_{2}. \end{aligned}$$

Then we have

$$\begin{aligned} &\partial _{t}p+\tilde{u}_{1}\cdot \nabla p-\Delta p+ \tilde{n}_{1}p=- \tilde{n}p_{2}-\tilde{u}\cdot \nabla p_{2}, \end{aligned}$$

(A.51)

$$\begin{aligned} &\partial _{t}q+\tilde{u}_{1}\cdot \nabla q-\Delta q+q= \tilde{n}- \tilde{u}\cdot \nabla q_{2}, \end{aligned}$$

(A.52)

$$\begin{aligned} &\partial _{t}n+\tilde{u}_{1}\cdot \nabla n-\Delta n= \nabla \cdot (n \nabla \phi )-\tilde{u}\cdot \nabla n_{2}-\nabla \cdot \biggl(n_{1}S \biggl(x,\frac{\tilde{n}_{1}}{1+\epsilon \tilde{n}_{1}},p_{1},q_{1} \biggr)\nabla p_{1} \\ &\phantom{\partial _{t}n+\tilde{u}_{1}\cdot \nabla n-\Delta n=} -n_{2}S \biggl(x,\frac{\tilde{n}_{2}}{1+\epsilon \tilde{n}_{2}},p_{2},q_{2} \biggr)\cdot \nabla p_{2} \biggr) \\ &\phantom{\partial _{t}n+\tilde{u}_{1}\cdot \nabla n-\Delta n=}-\nabla \cdot \left (n_{1}S \biggl(x, \frac{\tilde{n}_{1}}{1+\epsilon \tilde{n}_{1}},p_{1},q_{1} \biggr) \cdot \nabla q_{1}\right . \\ &\phantom{\partial _{t}n+\tilde{u}_{1}\cdot \nabla n-\Delta n=} -n_{2}S\left . \biggl(x, \frac{\tilde{n}_{2}}{1+\epsilon \tilde{n}_{2}},p_{2},q_{2} \biggr) \cdot \nabla q_{2}\right ) \end{aligned}$$

(A.53)

$$\begin{aligned} &\partial _{t}u+\tilde{u}_{1}\cdot \nabla u-\Delta u+ \nabla \pi =n \nabla \phi -\tilde{u}\cdot \nabla u_{2}+n_{1}S_{1} \nabla p_{1}-n_{2}S_{2} \nabla p_{2} \\ &\phantom{\partial _{t}u+\tilde{u}_{1}\cdot \nabla u-\Delta u+ \nabla \pi =} +n_{1}S_{1}\nabla q_{1}-n_{2}S_{2} \nabla q_{2}. \end{aligned}$$

(A.54)

Testing (A.53) by $n$ and using $\operatorname{div}u_{1}=0$, we have

$$\begin{aligned} &\frac{1}{2}\frac{\mathrm{d}}{\mathrm{d}t} \int n^{2}\mathrm{d}x+ \int | \nabla n|^{2}\mathrm{d}x \\ &\quad=- \int n\nabla \phi \nabla n\mathrm{d}x- \int \tilde{u}\cdot \nabla n_{2}n\mathrm{d}x \\ &\qquad+ \int nS \biggl(x,\frac{\tilde{n}_{1}}{1+\epsilon \tilde{n}_{1}},p_{1},q_{1} \biggr) (\nabla p_{1}+\nabla q_{1})\cdot \nabla n \mathrm{d}x \\ &\qquad- \int \nabla \biggl[n_{2} \biggl(S \biggl(x, \frac{\tilde{n}_{1}}{1+\epsilon \tilde{n}_{1}},p_{1},q_{1} \biggr)-S \biggl(x, \frac{\tilde{n}_{2}}{1+\epsilon \tilde{n}_{2}},p_{2},q_{2} \biggr) \biggr) (\nabla p_{1}+\nabla q_{1}) \biggr] \\ &\qquad\cdot n\mathrm{d}x + \int n_{2}S \biggl(x, \frac{\tilde{n}_{2}}{1+\epsilon \tilde{n}_{2}},p_{2},q_{2} \biggr) ( \nabla p+\nabla q)\cdot \nabla n\mathrm{d}x \\ &\quad\leq \|\nabla \phi \|_{L^{\infty }}\|n\|_{L^{2}}\|\nabla n \|_{L^{2}}+ \|\nabla n_{2}\|_{L^{\infty }}\|\tilde{u} \|_{L^{2}}\|n\|_{L^{2}} \\ &\qquad+C\|n\|_{L^{2}}\|\nabla (p_{1}+q_{1}) \|_{L^{\infty }}\|\nabla n\|_{L^{2}} \\ &\qquad+C\|\nabla n_{2}\|_{L^{\infty }}\|\nabla S\|_{L^{\infty }}(\| \tilde{n}_{1} \|_{L^{2}}+\|p\|_{L^{2}}+\|q \|_{L^{2}})\|\nabla (p_{1}+q_{1}) \|_{L^{\infty }}\|n\|_{L^{2}} \\ &\qquad+C\|n_{2}\|_{L^{\infty }}\|\nabla ^{2}S \|_{L^{\infty }}\sum_{i=1}^{2}( \| \nabla \tilde{n}_{i}\|_{L^{6}}+\|\nabla p_{i} \|_{L^{6}}+\|\nabla q_{i} \|_{L^{6}}) \\ &(\|\tilde{n}\|_{L^{3}}+\|p\|_{L^{3}}+\|q\|_{L^{3}}) \|\nabla (p_{1}+q_{1}) \|_{L^{\infty }}\|n \|_{L^{2}} \\ &\qquad+C\|n_{2}\|_{L^{\infty }}\|\nabla S\|_{L^{\infty }}(\|\nabla \tilde{n}\|_{L^{2}}+ \|\nabla p\|_{L^{2}}+\|\nabla q \|_{L^{2}})\|\nabla (p_{1}+q_{1}) \|_{L^{\infty }}\|n\|_{L^{2}} \\ &\qquad+C\|n_{2}\|_{L^{\infty }}(\|\tilde{n}\|_{L^{3}}+\|p \|_{L^{3}}+\|q\|_{L^{3}}) \|\nabla ^{2}(p_{1}+q_{1}) \|_{L^{6}}\|n\|_{L^{2}} \\ &\qquad+C\|n_{2}\|_{L^{\infty }}(\|\nabla p\|_{L^{2}}+\|\nabla q\|_{L^{2}})\| \nabla n\|_{L^{2}} \\ &\quad\leq \frac{1}{16}\|\nabla n\|_{L^{2}}^{2}+C\|n \|_{L^{2}}^{2}+C\| \nabla n_{2} \|_{L^{\infty }}^{2}\|n\|_{L^{2}}^{2}+\epsilon _{1}\| \tilde{u}\|_{L^{2}}^{2} \\ &\qquad+C\|p\|_{H^{1}}^{2}+C\|q\|_{H^{1}}^{2}+ \epsilon _{1}\|\tilde{n}\|_{H^{1}}^{2} \end{aligned}$$

(A.55)

for any positive constant $0<\epsilon _{1}<1$. Here we have used the elementary calculus: $S(x)-S(y)=\int _{0}^{1}\frac{\mathrm{d}}{\mathrm{d}t}S(x+t(y-x)) \mathrm{d}t$.

Testing (A.54) by $u$ and using $\operatorname{div}\tilde{u}_{1}=0$, we have

$$\begin{aligned} &\frac{1}{2}\frac{\mathrm{d}}{\mathrm{d}t} \int |u|^{2}\mathrm{d}x+ \int | \nabla u|^{2}\mathrm{d}x \\ &\quad= \int \bigl[n\nabla \phi -\tilde{u}\cdot \nabla u_{2}+n_{1}S_{1} \nabla (p_{1}+q_{1})-n_{2}S_{2} \nabla (p_{2}+q_{2})\bigr]u\mathrm{d}x \\ &\quad\leq \|\nabla \phi \|_{L^{\infty }}\|n\|_{L^{2}}\|u \|_{L^{2}}+\| \nabla u_{2}\|_{L^{\infty }}\|\tilde{u} \|_{L^{2}}\|u\|_{L^{2}} \\ &\qquad+\|n\|_{L^{2}}\|S_{1}\|_{L^{\infty }}\|\nabla (p_{1}+q_{1})\|_{L^{\infty }}\|u\|_{L^{2}} \\ &\qquad+\|n_{2}\|_{L^{\infty }}\|S_{1}-S_{2} \|_{L^{2}}\|\nabla (p_{1}+q_{1}) \|_{L^{\infty }}\|u\|_{L^{2}} \\ &\qquad+\|n_{2}\|_{L^{\infty }}\|S_{2}\|_{L^{\infty }}\| \nabla (p,q)\|_{L^{2}} \|u\|_{L^{2}} \\ &\quad\leq C\|n\|_{L^{2}}\|u\|_{L^{2}}+\|\nabla u_{2} \|_{L^{\infty }}\| \tilde{u}\|_{L^{2}}\|u\|_{L^{2}} \\ &\qquad+C(\|n\|_{L^{2}}+\|p\|_{L^{2}}+\|q\|_{L^{2}})\|u \|_{L^{2}}+C\| \nabla (p+q)\|_{L^{2}}\|u\|_{L^{2}} \\ &\quad\leq C\|n\|_{L^{2}}^{2}+C\|u\|_{L^{2}}^{2}+C \|\nabla u_{2}\|_{L^{\infty }}^{2}\|u \|_{L^{2}}^{2}+\epsilon _{2}\|\tilde{u} \|_{L^{2}}^{2} \\ &\qquad+C\|p\|_{H^{1}}^{2}+C\|q\|_{H^{1}}^{2} \end{aligned}$$

(A.56)

for any positive constant $0<\epsilon _{2}<1$.

Testing (A.51) by $p$ and using $\operatorname{div}\tilde{u}_{1}=0$, we have

$$\begin{aligned} &\frac{1}{2}\frac{\mathrm{d}}{\mathrm{d}t} \int p^{2}\mathrm{d}x+ \int | \nabla p|^{2}\mathrm{d}x+ \int \tilde{n}_{1}p^{2}\mathrm{d}x=- \int ( \tilde{n}p_{2}+\tilde{u}\cdot \nabla p_{2})p\mathrm{d}x \\ &\quad\leq (\|\tilde{n}\|_{L^{2}}\|p_{2}\|_{L^{\infty }}+\| \tilde{u}\|_{L^{2}} \|\nabla p_{2}\|_{L^{\infty }})\|p \|_{L^{2}} \\ &\quad\leq \epsilon _{3}\bigl(\|\tilde{n}\|_{L^{2}}^{2}+ \|\tilde{u}\|_{L^{2}}^{2}\bigr)+C \|p\|_{L^{2}}^{2} \end{aligned}$$

(A.57)

for any $0<\epsilon _{3}<1$.

Testing (A.51) by $-\Delta p$, we have

$$\begin{aligned} &\frac{1}{2}\frac{\mathrm{d}}{\mathrm{d}t} \int |\nabla p|^{2}\mathrm{d}x+ \int |\Delta p|^{2}\mathrm{d}x=- \int \nabla (\tilde{u}_{1}\cdot \nabla p+\tilde{u}\cdot \nabla p_{2}+\tilde{n}_{1}p+\tilde{n}p_{2}) \nabla p\mathrm{d}x \\ &\quad\leq \bigl(\|\tilde{u}_{1}\|_{L^{\infty }}\|\nabla ^{2}p\|_{L^{2}}+\|\nabla \tilde{u}_{1} \|_{L^{\infty }}\|\nabla p\|_{L^{2}}+\|\tilde{u}\|_{L^{3}}\| \nabla ^{2}p_{2}\|_{L^{6}}+\|\nabla \tilde{u} \|_{L^{2}}\|\nabla p_{2} \|_{L^{\infty }} \\ &\qquad+\|\tilde{n}_{1}\|_{L^{\infty }}\|\nabla p\|_{L^{2}}+\| \nabla \tilde{n}_{1} \|_{L^{6}}\|p\|_{L^{3}}+\| \tilde{n}\|_{L^{2}}\|\nabla p_{2}\|_{L^{\infty }}+\|\nabla \tilde{n}\|_{L^{2}}\|p_{2}\|_{L^{\infty }}\bigr)\|\nabla p \|_{L^{2}} \\ &\quad\leq C\bigl(\|\nabla ^{2}p\|_{L^{2}}+\|\nabla \tilde{u}_{1}\|_{L^{\infty }} \|\nabla p\|_{L^{2}}+\| \tilde{u}\|_{L^{3}}+\|\nabla \tilde{u}\|_{L^{2}} \\ &\qquad +\|\nabla p\|_{L^{2}}+\|p\|_{L^{3}}+\|\tilde{n} \|_{H^{1}}\bigr)\| \nabla p\|_{L^{2}} \\ &\quad\leq \frac{1}{16}\|\Delta p\|_{L^{2}}^{2}+C\|p \|_{H^{1}}^{2}+C\| \nabla \tilde{u}_{1} \|_{L^{\infty }}\|p\|_{H^{1}}^{2}+\epsilon _{4}\| \tilde{u}\|_{H^{1}}^{2}+\epsilon _{4}\|\tilde{n}\|_{H^{1}}^{2} \end{aligned}$$

(A.58)

for any $0<\epsilon _{4}<1$.

Similarly to (A.57) and (A.58), we have

$$\begin{aligned} &\frac{1}{2}\frac{\mathrm{d}}{\mathrm{d}t} \int q^{2}\mathrm{d}x+ \int | \nabla q|^{2}\mathrm{d}x+ \int q^{2}\mathrm{d}x= \int (\tilde{n}- \tilde{u}\cdot \nabla q_{2})q\mathrm{d}x \\ &\quad\leq \epsilon _{5}\|\tilde{n}\|_{L^{2}}^{2}+ \epsilon _{5}\|\tilde{u}\|_{L^{2}}^{2}+C \| \nabla q_{2}\|_{L^{\infty }}\|q\|_{L^{2}}^{2} \\ &\quad\leq \epsilon _{5}\|\tilde{n}\|_{L^{2}}^{2}+ \epsilon _{5}\|\tilde{u}\|_{L^{2}}^{2}+C \|q \|_{L^{2}}^{2}, \end{aligned}$$

(A.59)

for any $0<\epsilon _{5}<1$.

$$\begin{aligned} &\frac{1}{2}\frac{\mathrm{d}}{\mathrm{d}t} \int |\nabla q|^{2}\mathrm{d}x+ \int (\Delta q)^{2}\mathrm{d}x+ \int |\nabla q|^{2}\mathrm{d}x \\ &\quad= \int \nabla \tilde{n}\cdot \nabla q\mathrm{d}x- \int \nabla (\tilde{u}_{1} \cdot \nabla q+\tilde{u}\cdot \nabla q_{2})\nabla u\mathrm{d}x \\ &\quad\leq \|\nabla \tilde{n}\|_{L^{2}}\|\nabla q\|_{L^{2}}+\bigl( \|\nabla \tilde{u}_{1}\|_{L^{\infty }}\|\nabla q \|_{L^{2}}+\|\tilde{u}_{1}\|_{L^{\infty }}\|\nabla ^{2}q\|_{L^{2}} \\ &\qquad+\|\tilde{u}\|_{L^{3}}\|\nabla ^{2}q_{2} \|_{L^{6}}+\|\nabla \tilde{u} \|_{L^{2}}\|\nabla q_{2}\|_{L^{\infty }}\bigr)\|\nabla u\|_{L^{2}} \\ &\quad\leq \epsilon _{6}\|\tilde{n}\|_{H^{1}}^{2}+ \epsilon _{6}\|\tilde{u}\|_{H^{1}}^{2}+C \| \nabla q\|_{L^{2}}^{2}+C\|\nabla \tilde{u}_{1} \|_{L^{\infty }}\bigl(\| \nabla q\|_{L^{2}}^{2}+\|\nabla u \|_{L^{2}}^{2}\bigr) \\ &\qquad+\frac{1}{16}\|\Delta q\|_{L^{2}}^{2}+C\|u \|_{H^{1}}^{2} \end{aligned}$$

(A.60)

for any $0<\epsilon _{6}<1$.

Summing up (A.55), (A.56), (A.57), (A.58), (A.59) and (A.60), and taking $\epsilon _{i}$$(i=1,\ldots ,6)$ small enough, and using the Gronwall inequality, we arrive at (A.50) for small $0< T\leq 1$. This completes the proof of Lemma A.5. □

Proof of Theorem A.1

By Lemmas A.1–A.5 and the Banach’s fixed point theorem, we can easily finish the proof of Theorem A.1. □

Appendix B: Proofs of (2.14), (2.17), (2.18), (2.22) and (2.23)

In this appendix, we give the detailed proofs of (2.14), (2.17), (2.18), (2.22) and (2.23).

1. Proof of (2.14).

We will use the $L^{\infty }$-estimate of weak solutions to the parabolic equations:

Lemma B.1

([15])

Let $q\in L^{\infty }(0,T;L^{2})\cap L^{2}(0,T;H^{1})$be a weak solution to the following problem:

$$\begin{aligned} &\partial _{t}q+u\cdot \nabla q-\Delta q+q=f\quad \textit{in } \varOmega \times (0,T), \end{aligned}$$

(B.1)

$$\begin{aligned} &\frac{\partial q}{\partial \nu }=0\quad\textit{on } \partial \varOmega \times (0,T), \end{aligned}$$

(B.2)

$$\begin{aligned} &q(\cdot ,0)=q_{0}(\cdot )\quad\textit{in } \varOmega . \end{aligned}$$

(B.3)

If $u$and $f$satisfy

$$\begin{aligned} &u\in L^{p_{1}}\bigl(\varOmega \times (0,T)\bigr)\quad\textit{with } \operatorname{div}u=0 \\ &\quad\textit{and}\quad f\in L^{p_{2}}\bigl(\varOmega \times (0,T)\bigr)\quad\textit{with } p_{1}>d+2\quad \textit{and}\quad p_{2}>\frac{d+2}{2}, \end{aligned}$$

(B.4)

(here $d$denotes the spatial dimensional number) then it holds

$$ \|q\|_{L^{\infty }(\varOmega \times (0,T))}\leq C\|q_{0}\|_{L^{\infty }( \varOmega )}+C\|f \|_{L^{p_{2}}(\varOmega \times (0,T))}. $$

(B.5)

Using Lemma B.1, (2.12) and (2.13), we easily get (2.14).

2. Proof of (2.17).

We will use the following lemma:

Lemma B.2

([1])

Let $u$be the unique solution of the following problem

$$\begin{aligned} &\partial _{t}u-\Delta u=\operatorname{div}f\quad \textit{in } \varOmega \times (0,T), \end{aligned}$$

(B.6)

$$\begin{aligned} &\frac{\partial u}{\partial \nu }=0\quad \textit{on } \partial \varOmega \times (0,T), \end{aligned}$$

(B.7)

$$\begin{aligned} &u(\cdot ,0)=0\quad\textit{in } \varOmega . \end{aligned}$$

(B.8)

Then there holds

$$ \|\nabla u\|_{L^{s_{1}}(0,T;L^{s_{2}}(\varOmega ))}\leq C\|f\|_{L^{s_{1}}(0,T;L^{s_{2}}( \varOmega ))} $$

(B.9)

for any $1< s_{1},s_{2}<\infty$.

Using Lemma B.2, (2.12) and (2.13), we have (2.17).

3. Proof of (2.18).

We will use the following lemma:

Lemma B.3

([15]). Let $u$be the unique solution of the following problem:

$$\begin{aligned} &\partial _{t}u-\Delta u=f\quad\textit{in } \varOmega \times (0,T), \end{aligned}$$

(B.10)

$$\begin{aligned} &\frac{\partial u}{\partial \nu }=0\quad \textit{on } \partial \varOmega \times (0,T), \end{aligned}$$

(B.11)

$$\begin{aligned} &u(\cdot ,0)=u_{0}(\cdot )\quad \textit{in } \varOmega . \end{aligned}$$

(B.12)

Then it holds

$$\begin{aligned} &\|u\|_{C([0,T];W^{2-\frac{2}{m},m})}+\|u_{t}\|_{L^{m}(0,T;L^{m})}+\|u \|_{L^{m}(0,T;W^{2,m})} \\ &\quad\leq C(\|u_{0}\|_{W^{2-\frac{2}{m},m}}+\|f\|_{L^{m}(0,T;L^{m})})\quad \textit{for } 2\leq m< \infty . \end{aligned}$$

(B.13)

Using Lemma B.3, (2.12) and (2.13), we prove (2.18).

4. Proof of (2.22).

We will use the following lemma:

Lemma B.4

([15])

Let $n\in L^{\infty }(0,T;L^{2})\cap L^{2}(0,T;H^{1})$be the weak solution of the heat equation:

$$\begin{aligned} &\partial _{t}n-\Delta n=\operatorname{div}g\quad \textit{in } \varOmega \times (0,T), \end{aligned}$$

(B.14)

$$\begin{aligned} &\frac{\partial n}{\partial \nu }=0\quad \textit{on } \partial \varOmega \times (0,T), \end{aligned}$$

(B.15)

$$\begin{aligned} &n(\cdot ,0)=n_{0}(\cdot )\quad \textit{in } \varOmega . \end{aligned}$$

(B.16)

If

$$ g\in L^{r}\bigl(\varOmega \times (0,T)\bigr)\quad \textit{with } r>d+2, $$

(B.17)

then there holds

$$ \|n\|_{L^{\infty }(\varOmega \times (0,T))}\leq C\|n_{0}\|_{L^{\infty }}+C\|g \|_{L^{r}(\varOmega \times (0,T))}. $$

(B.18)

Using Lemma B.4, (2.21), (2.12), (2.13), (2.19) and (2.20), we have (2.22).

5. Proof of (2.23).

Using Lemma B.2, (2.12), (2.13), (2.19), (2.20) and (2.21), we easily arrive at (2.23). This completes the proof.

Remark B.1

Here we give another proof of (2.14) as follows.

We will use the following lemma:

Lemma B.5

([15])

Let $q\in L^{\infty }(0,T;L^{2})\cap L^{2}(0,T;H^{1})$be a weak solution to the following problem:

$$\begin{aligned} &\partial _{t}q-\Delta q+q=f+\operatorname{div}g\quad \textit{in } \varOmega \times (0,T), \end{aligned}$$

(B.19)

$$\begin{aligned} &\frac{\partial q}{\partial \nu }=0\quad \textit{on } \partial \varOmega \times (0,T), \end{aligned}$$

(B.20)

$$\begin{aligned} &q(\cdot ,0)=q_{0}(\cdot )\quad \textit{in } \varOmega . \end{aligned}$$

(B.21)

If

$$ f\in L^{r_{1}}\bigl(\varOmega \times (0,T)\bigr)\quad\textit{and}\quad g\in L^{r_{2}}\bigl(\varOmega \times (0,T)\bigr)\quad\textit{with } r_{1}> \frac{d+2}{2}\textit{ and } r_{2}>d+2, $$

(B.22)

then it holds

$$ \|q\|_{L^{\infty }(\varOmega \times (0,T))}\leq C\|q_{0}\|_{L^{\infty }}+C\|f \|_{L^{r_{1}}(\varOmega \times (0,T))}+C\|g\|_{L^{r_{2}}(\varOmega \times (0,T))}. $$

(B.23)

Testing (1.1) by $q^{\ell -1}$$(\ell \geq 2)$, and using (1.3) and (2.12), we have

$$\begin{aligned} &\frac{1}{\ell }\frac{\mathrm{d}}{\mathrm{d}t} \int q^{\ell }\mathrm{d}x+( \ell -1) \int q^{\ell -2}|\nabla q|^{2}\mathrm{d}x+ \int q^{\ell }\mathrm{d}x\\ &\quad= \int nq^{\ell -1}\mathrm{d}x \leq \|n\|_{L^{\ell }}\|q\|_{L^{\ell }}^{\ell -1}\leq C\|q \|_{L^{\ell }}^{ \ell -1}\leq \frac{1}{2} \int q^{\ell }\mathrm{d}x+C, \end{aligned}$$

which gives

$$ \|q\|_{L^{\infty }(0,T;L^{\ell })}\leq C\quad \mbox{for any } 2\leq \ell < \infty . $$

(B.24)

Using Lemma B.5 for $f:=n$ and $g:=-uq$, and using (2.12), (2.13) and (B.24), we conclude that (2.14) holds true. □