2.1. Notations

Throughout this paper, $C\gt 0$ denotes some constant independent of the time, and $C_{T}\gt 0$ is a constant which may depend on the time $T$ . $\mathcal{F}(f)$ and $\mathcal{F}^{-1}(f)$ stand for the Fourier transform and the inverse transform of the function $f$ . Sometimes we write

\begin{equation*} \Lambda ^{\sigma }= (\!-\Delta )^\frac {\sigma }{2}=\mathcal{F}^{-1} \Big ( |\xi |^{\sigma } \mathcal{F}(\cdot ) \Big ),\quad \sigma \in \mathbb{R}. \end{equation*}

Let $W^{s, p}(\mathbb{R}^3)$ with $1< p< \infty$ , $s\in \mathbb{R}$ be the standard fractional exponent Sobolev space on $\mathbb{R}^3$ with equivalent norms

\begin{equation*} \|f\|_{W^{s, p}} \sim \|(\textrm {Id}-\Delta )^\frac {s}{2} f\|_{L^p} \sim \|f\|_{L^p}+\|\Lambda ^s f\|_{L^p}. \end{equation*}

Especially, when $p=2,$ $H^s(\mathbb{R}^3)$ is the Sobolev space of exponent $s$ on $\mathbb{R}^3$ with the standard norm

\begin{equation*} \|f\|_{H^s}\;:\!=\;\Big (\int _{\mathbb{R}^3} (1+|\xi |^2)^s|\widehat f (\xi )|^2 \,d\xi \Big )^{\frac {1}{2}}\lt \infty \end{equation*}

and the inner product

\begin{equation*}\langle f, g\rangle =\int _{\mathbb{R}^3} (1+|\xi |^2)^s \widehat f (\xi ) \overline {\widehat g (\xi )} \,d\xi .\end{equation*}

Furthermore, we denote by $\dot {H}^s(\mathbb{R}^3)$ the homogeneous Sobolev space endowed with the norm

\begin{equation*} \|f\|_{\dot {H}^s}\;:\!=\;\Big (\int _{\mathbb{R}^3} |\xi |^{2s}|\widehat f (\xi )|^2 \,d\xi \Big )^{\frac {1}{2}}\sim \|\Lambda ^{s}f\|_{L^2},\, s\in \mathbb{R}. \end{equation*}

First, we have the following generalised Sobolev-type estimate (see [Reference Stein46, Chapter 5, Theorem 1] and [Reference Schonbek and Schonbek43, Lemma 2.2]).

Lemma 2.1. Let $0\lt \alpha \lt 3$ , $1\leq q \lt p\lt \infty$ and $\frac {1}{p}+\frac {\alpha }{3}=\frac {1}{q}$ . There exists a constant $C \geq 0$ such that if $u \in \mathcal{S}'$ is such that $\widehat u$ is a function, then

(2.1) \begin{equation} \|u\|_{L^p}\leq C \|\Lambda ^\alpha u\|_{L^q}. \end{equation}

Next lemma is devoted to the Gagliardo–Nirenberg inequality.

Lemma 2.2 (Gagliardo–Nirenberg inequality). For $1\leq r \leq \infty, $ $1\leq m \lt \infty, $ $0\lt \alpha \lt 3,$ $0\leq \theta \leq 1$ and $\sigma \lt s$ , assume $\Lambda ^\alpha u \in L^m(\mathbb{R}^3)$ and $\Lambda ^s u \in L^r(\mathbb{R}^3).$ If

\begin{equation*}\frac {1}{p}-\frac {\sigma }{3}= \theta \left (\frac {1}{m}-\frac {\alpha }{3} \right ) +(1-\theta )\left (\frac {1}{r}-\frac {s}{3}\right ), \end{equation*}

then

(2.2) \begin{equation} \|\Lambda ^\sigma u\|_{L^p} \leq C \|\Lambda ^\alpha u\|_{L^m}^\theta \|\Lambda ^{s}u\|_{L^{r}}^{1-\theta }. \end{equation}

Proof. By [Reference Nirenberg37] or [Reference Bahouri, Chemin and Danchin3, Chapter 1], one has for any $1\leq q,r\leq \infty$ , $\sigma \lt s$ , and $\theta \in [0,1]$ and $\frac {1}{p}-\frac {\sigma }{3}=\frac {\theta }{q}+(1-\theta )(\frac {1}{r}-\frac {s}{3})$ , if $u\in L^q (\mathbb{R}^3)$ and $\Lambda ^s u\in L^r (\mathbb{R}^3)$ . Then $\Lambda ^{\sigma }u\in L^q(\mathbb{R}^3)$ and

(2.3) \begin{equation} \begin{aligned} &\|\Lambda ^{\sigma }u\|_{L^p}\leq C\|u\|_{L^q}^{\theta }\|\Lambda ^{s}u\|_{L^{r}}^{1-\theta }. \end{aligned} \end{equation}

If we let

(2.4) \begin{equation} \frac {1}{q}+\frac {\alpha }{3}=\frac {1}{m} \end{equation}

and

(2.5) \begin{equation} \frac {1}{p}=\frac {\theta }{q}+\frac {1-\theta }{r}, \end{equation}

then by Lemma2.1 and (2.3), we obtain (2.2). It can be easily checked that $\frac {1}{p}-\frac {\sigma }{3}= \theta \left (\frac {1}{m}-\frac {\alpha }{3} \right ) +$ $(1-\theta )\left (\frac {1}{r}-\frac {s}{3}\right )$ holds due to (2.4) and (2.5).

Remark 2.1. We would like to point out that Lemma 2.2 is valid in all dimensions, meaning that for $1\leq r \leq \infty, $ $1\leq m \lt \infty, $ $0\lt \alpha \lt n,$ $0\leq \theta \leq 1$ and $\sigma \lt s$ , if we assume $\Lambda ^\alpha u \in L^m(\mathbb{R}^n)$ , $\Lambda ^s u \in L^r(\mathbb{R}^n)$ and

\begin{equation*}\frac {1}{p}-\frac {\sigma }{n}= \theta \left (\frac {1}{m}-\frac {\alpha }{n} \right ) +(1-\theta )\left (\frac {1}{r}-\frac {s}{n}\right ), \end{equation*}

then

\begin{equation*} \|\Lambda ^\sigma u\|_{L^p(\mathbb{R}^n)} \leq C \|\Lambda ^\alpha u\|_{L^m(\mathbb{R}^n)}^\theta \|\Lambda ^{s}u\|_{L^{r}(\mathbb{R}^n)}^{1-\theta }. \end{equation*}

We also need the Moser-type product laws for fractional Laplacian operators. The first estimate (2.6) is due to Kato-Ponce [Reference Kato and Ponce21] and Kenig-Ponce-Vega [Reference Kenig, Carlos and Vega25], and the second estimate (2.7) can be found in [Reference Bahouri, Chemin and Danchin3, Corollary 2.55].

Finally, we recall Hardy’s inequality for fractional norms (see [Reference Bahouri, Chemin and Danchin3, Page 91]).

Lemma 2.4. Let $0\leq s\lt \frac {3}{2}$ . For any $v\in \dot {H}^{s}(\mathbb{R}^{3})$ , it holds that

(2.8) \begin{equation} \begin{aligned} &\Big (\int _{\mathbb{R}^{3}}\frac {|v(x)|^2}{|x|^{2s}}\,dx\Big )^{\frac {1}{2}}\leq C\|\Lambda ^{s}v\|_{L^2}. \end{aligned} \end{equation}

3.1. Proof of Theorem 1.2

We will give the proof of Theorem1.2. This can be proved by a contradictory argument. Assume that the maximal time $T_*$ is finite, and (1.6) is not true, i.e.

(3.1) \begin{equation} B_{1}\;:\!=\;\lim _{T\rightarrow T_*}\int _0^{T}\left (\|n\|_{L^{q_1}}^{p_1}+\|u\|_{L^{q_2}}^{p_2} \right )\,dt \lt +\infty \end{equation}

for any $(p_i, q_i) (i=1, 2)$ satisfying (1.7). Then, once the $H^2$ -norm of the solution $(n, c, u)$ to the Cauchy problem (1.1)–(1.2) is bounded when $t$ is close to $T_*$ , one can show that the assumption $T_*\lt +\infty$ will lead to a contradiction with the maximality of $T_*$ in accordance with the local well-posedness result (Theorem1.1).

In the following, we establish the key estimates of solutions. The proof is divided into three steps.

• • Step 1: $L^2$ -estimates of $n$ and $H^1$ -estimates of $c,u$ .

In this step, our goal is to get

(3.2) \begin{equation} \sup _{t\in (0,T]}(\|n\|_{L^2}^2+\|(c,u)\|_{H^1}^2)+\int _{0}^{T}(\|\nabla n\|_{L^2}^2+\|\nabla c\|_{H^1}^2+\|\Lambda ^{\alpha }u\|_{H^1}^2)\,dt\leq C(B_1,T_*) \end{equation}

for any $0\lt T\lt T_*$ and some constant $C(B_1,T_*)\gt 0$ .

To begin with, we give some uniform estimates. It follows from the classical maximum principle that

(3.3) \begin{equation} \begin{aligned} &n(t,x)\geq 0,\quad \quad 0\leq c(t,x)\leq \|c_0\|_{L^{\infty }},\quad (t,x)\in [0,T]\times \mathbb{R}^3. \end{aligned} \end{equation}

In addition, by taking the inner product of (1.1) $_2$ by $c$ and using $f(c)\geq 0$ , we arrive at

(3.4) \begin{equation} \begin{aligned} \sup _{t\in (0,T]}\|c\|_{L^2}^2+\int _{0}^{T}\|\nabla c\|_{L^2}^2\,dt\leq \|c_0\|_{L^2}^2. \end{aligned} \end{equation}

In the following, we divide the proof of (3.2) into the cases $\alpha \geq 1$ and $\frac {3}{4}\lt \alpha \lt 1$ .

• • Case 1: $\alpha \geq 1$

Now, we perform the $L^2$ -estimate of $n$ and the $H^1$ -estimates of $c,u$ . First, we multiply (1.1) $_1$ by $n$ , integrate the resulting equation by parts, and then derive

\begin{equation*} \begin{aligned} \frac {1}{2}\frac {d}{dt}\|n\|_{L^2 }^2+\|\nabla n\|_{L^2 }^2 &= \int _{\mathbb{R}^3}\chi (c) n\nabla c\cdot \nabla n\,dx\\[5pt] &\leq \sup _{0\leq c\leq \|c_0\|_{L^\infty }} |\chi (c)| \|\nabla n\|_{L^2 } \|n\|_{L^{q_1} }\|\nabla c\|_{L^\frac {2q_1}{q_1-2}}\\[5pt] &\leq C \|\nabla n\|_{L^2 } \|n\|_{L^{q_1} } \|\nabla c\|_{L^{2} }^\frac {q_1-3}{q_1} \|\nabla ^2c\|_{L^2 }^\frac {3}{q_1}\\[5pt] & \leq \frac {1}{2} \|\nabla n\|_{L^2 }^2+\frac {1}{4}\|\Delta c\|_{L^2}^2+C\|n\|_{L^{q_1} }^{\frac {2q_1}{q_1-3}}\|\nabla c\|_{L^2 }^2, \end{aligned} \end{equation*}

where we have used (3.3), $3\lt q_1\lt \infty$ , the Gagliardo–Nirenberg inequality (2.2) and the fact that

(3.5) \begin{align} \|\Delta g\|_{L^2}\leq \|\nabla ^2g\|_{L^2}\leq C\|\Delta g\|_{L^2},\quad \forall g\in \dot {H}^2. \end{align}

Due to $\frac {2q_1}{q_1-3}\leq p_1$ , it thus follows that

(3.6) \begin{equation} \frac {d}{dt}\|n\| _{L^2 }^2+\|\nabla n\|_{L^2 }^2 \leq \frac {1}{2}\|\Delta c\|_{L^2}^2+C(1+\|n\|_{L^{q_1}}^{p_1})\|\nabla c\| _{L^2 }^2. \end{equation}

Furthermore, we multiply both sides of (1.1) $_2$ by $-\Delta c$ and integrate the resulting equation to obtain

(3.7) \begin{equation} \frac {1}{2} \frac {d}{dt}\|\nabla c\| _{L^2 }^2+\|\Delta c\| _{L^2 }^2 =\int _{\mathbb{R}^3} (u\cdot \nabla c)\cdot \Delta c\, dx+\int _{\mathbb{R}^3} n \Delta c f(c)\,dx. \end{equation}

Owing to $2\alpha -1\geq 1$ , we have $\frac {2q_2}{q_2-3}\leq p_2$ , so it follows from (2.2) and (3.5) that

\begin{equation*} \begin{aligned} \left |\int _{\mathbb{R}^3} (u\cdot \nabla c)\cdot \Delta c\, dx\right |&\leq \|u\|_{L^{q_2}}\|\nabla c\|_{L^\frac {2q_2}{q_2-2}} \|\Delta c\|_{L^2} \\[5pt] &\leq C \|u\|_{L^{q_2}} \|\nabla c\|_{L^2}^{\frac {q_2-3}{q_2}}\|\nabla ^2c\|_{L^2}^{\frac {3}{q_2}}\|\Delta c\|_{L^2}\\[5pt] &\leq \frac {1}{4}\|\Delta c\|_{L^2}^2+C\|u\|_{L^{q_2}}^{\frac {2q_2}{q_2-3}} \|\nabla c\|_{L^2}^2\\[5pt] &\leq \frac {1}{4}\|\Delta c\|_{L^2}^2+C(1+\|u\|_{L^{q_2}}^{p_2}) \|\nabla c\|_{L^2}^2. \end{aligned} \end{equation*}

And by (3.3), it is easy to verify that

\begin{eqnarray*} \begin{aligned} \left |\int _{\mathbb{R}^3} n \Delta c f(c)\,dx\right |&\leq \frac {1}{4}\|\Delta c\|_{L^2}^2+\sup _{0\leq s\leq \|c_0\|_{L^\infty }}|f(s)|^2 \|n\| _{L^2 }^2. \end{aligned} \end{eqnarray*}

Hence, we have

(3.8) \begin{equation} \begin{aligned} & \frac {d}{dt}\|\nabla c\| _{L^2 }^2+\|\Delta c\| _{L^2 }^2\leq C(1+\|u\|_{L^{q_2}}^{p_2}) \|\nabla c\|_{L^2}^2+C\|n\| _{L^2 }^2. \end{aligned} \end{equation}

We are going to analyse the velocity $u$ . For the third equation in (1.1), one has

(3.9) \begin{equation} \begin{aligned} \frac {1}{2}\frac {d}{dt}\|u\| _{L^2 }^2+\|\Lambda ^{\alpha } u\| _{L^2 }^2 =-\int _{\mathbb{R}^3} u\cdot n\nabla \phi \leq \|u\| _{L^2 }^2+C \|\nabla \phi \|_{L^\infty }^2\|n\| _{L^2 }^2. \end{aligned} \end{equation}

To deal with the $\dot {H}^1$ -estimates of $u$ , we multiply both sides of the third equation of (1.1) by $-\Delta u$ and integrate by parts, and then have

(3.10) \begin{equation} \begin{aligned} \frac {1}{2} \frac {d}{dt} \|\nabla u\| _{L^2 }^2 +\|\Lambda ^{\alpha +1} u\| _{L^2 }^2 = \int _{\mathbb{R}^3} \Delta u \cdot n \nabla \phi \,dx+\int _{\mathbb{R}^3} \Delta u \cdot (u \cdot \nabla ) u\,dx. \end{aligned} \end{equation}

For the terms on the right-hand side of (3.10), we first deduce from the Gagliardo–Nirenberg inequality (2.2) and $\frac {2q_1}{q_1-3}\leq p_1$ that

(3.11) \begin{equation} \begin{aligned} \left |\int _{\mathbb{R}^3} \Delta u \cdot n \nabla \phi \,dx \right |&\leq \|\nabla \phi \|_{L^\infty }\|\Delta u\| _{L^2 }\|n\| _{L^2 }\\[5pt] &\leq C \|\nabla u\| _{L^2 }^{1-\frac {1}{\alpha }} \|\Lambda ^{\alpha +1} u\| _{L^2 }^\frac {1}{\alpha }\|n\| _{L^2 }\\[5pt] &\leq \frac {1}{4} \|\Lambda ^{\alpha +1} u\| _{L^2 }^2 +C\|\nabla u\| _{L^2 }^2+C\|n\| _{L^2 }^2. \end{aligned} \end{equation}

Similarly, it holds by (2.2) that

(3.12) \begin{equation} \begin{aligned} \left |\int _{\mathbb{R}^3} \Delta u \cdot (u \cdot \nabla ) u \,dx\right |&=\left | \sum _{i,j,k=1}^{3}\int _{\mathbb{R}^3} u_k(\partial _{ik}u_j \partial _i u_j +\partial _k u_j \partial _{ii} u_j)\,dx \right |\\[5pt] &\leq C\|u\|_{L^{q_2}}\|\nabla u\|_{L^{r_1}}\|\Delta u\|_{L^{r_2}}\\[5pt] &\leq C\|u\|_{L^{q_2}} \|\nabla u\| _{L^2 }^\theta \|\Lambda ^{\alpha +1}u\| _{L^2 }^{1-\theta } \|\nabla u\| _{L^2 }^\theta \|\Lambda ^{\alpha +1}u\| _{L^2 }^{1-\theta }\\[5pt] & \leq \frac {1}{4} \|\Lambda ^{\alpha +1}u\| _{L^2 }^2+ C\|u\|_{L^{q_2}}^\frac {1}{\theta }\|\nabla u\| _{L^2 }^2, \end{aligned} \end{equation}

where we used $\alpha \geq 1$ , and the constants $r_1, \, r_2$ and $\theta$ satisfy

\begin{eqnarray*} \begin{cases} \dfrac {1}{r_1}+\dfrac {1}{r_2}+\dfrac {1}{q_2}=1,&\\[9pt] \dfrac {1}{r_1}-\dfrac {1}{3}=\theta \left (\dfrac {1}{2}-\dfrac {1}{3}\right )+ (1-\theta )\left (\dfrac {1}{2}-\dfrac {\alpha +1}{3}\right )&\\[9pt] \dfrac {1}{r_2}-\dfrac {2}{3}=\theta \left (\dfrac {1}{2}-\dfrac {1}{3}\right )+ (1-\theta )\left (\dfrac {1}{2}-\dfrac {\alpha +1}{3}\right ),& \end{cases} \end{eqnarray*}

which can be rewritten as

\begin{equation*} r_1=\frac {6q_2}{2q_2-3},\,\,\quad r_2=\frac {6q_2}{4q_2-3},\,\, \quad \frac {1}{p_2}\leq \theta =1-\frac {1}{2\alpha }\left (1+\frac {3}{q_2}\right ). \end{equation*}

In order to promise $1\lt r_1, \, r_2\lt \infty$ and $0\lt \theta \lt 1$ , one requires $\min \{\frac {3}{2},\frac {3}{2\alpha -1}\}\lt q_2\lt \infty$ . Substituting (3.11) and (3.12) into (3.10), we conclude that

(3.13) \begin{equation} \begin{aligned} \frac {d}{dt}\|\nabla u\| _{L^2 }^2 +\|\Lambda ^{\alpha +1} u\| _{L^2 }^2 \leq C \|n\|_{L^2}^2+\left (1+\|u\|_{L^{q_2}}^{p_2}\right )\|\nabla u\| _{L^2 }^2. \end{aligned} \end{equation}

The combination of (3.6), (3.8), (3.9) and (3.13) yields

\begin{equation*} \begin{aligned} &\frac {d}{dt}(\|(n,\nabla c)\|_{L^2}^2+\| u\| _{H^1 }^2) +\frac {1}{2}\|(\nabla n,\Delta c)\|_{L^2}^2+ \frac {1}{2}\|\Lambda ^{\alpha } u\| _{H^1}^2\\[5pt] &\leq C(1+\|n\|_{L^{q_1}}^{p_1}+\|u\|_{L^{q_2}}^{p_2})(\|(n,\nabla c)\|_{L^2}^2+\| u\| _{H^1 }^2), \end{aligned} \end{equation*}

which, together with (3.1), (3.5) and Grönwall’s lemma, yields

\begin{equation*} \begin{aligned} &\sup _{t\in (0,T]}(\|(n,\nabla c)\|_{L^2}^2+\| u\| _{H^1 }^2)+\int _{0}^{T}(\|(\nabla n,\nabla ^2 c)\|_{L^2}^2+ \|\Lambda ^{\alpha } u\| _{H^1}^2)\,dt\\[5pt] &\quad \leq Ce^{CB_1+C T}(\|(n_0,\nabla c_0)\|_{L^2}^2+\| u_0\|_{H^1}^2) \end{aligned} \end{equation*}

for any $0\lt T\lt T_*$ . This together with (3.4) yields (3.2) in the case $\alpha \geq 1$ .

• • Case 2: $\frac {3}{4}\lt \alpha \lt 1$

In this case, we are able to obtain the inequality (3.6) for $n$ as in Case 1. Regarding the estimates of $c$ , the second term on the right-hand side of (3.7) can be handled in the same way in Case 1. To deal with the remainder on the right-hand side of (3.7), one has

\begin{equation*} \begin{aligned} \int _{\mathbb{R}^3} (u\cdot \nabla c)\cdot \Delta c\, dx&=-\int _{\mathbb{R}^3} (\nabla c \cdot \nabla u)\cdot \nabla c\,dx-\int _{\mathbb{R}^3}(u\cdot \nabla ^2 c)\cdot \nabla c\,dx \\[5pt] &=\int _{\mathbb{R}^3} \nabla ^2 c\;:\; \nabla u c\,dx, \end{aligned} \end{equation*}

derived from $\nabla \cdot u=0$ and integration by parts, so it holds by (2.2), (3.3) and (3.5) that

\begin{equation*} \begin{aligned} \left |\int _{\mathbb{R}^3} (u\cdot \nabla c)\cdot \Delta c\, dx\right |&\leq \|c\|_{L^{\infty }} \|\nabla u\|_{L^2}\|\nabla ^2c\|_{L^2}\\[5pt] &\leq C\|c_0\|_{L^{\infty }} \|u\|_{L^2}^{\frac {\alpha }{\alpha +1}}\|\nabla \Lambda ^\alpha u\|_{L^2}^{\frac {1}{\alpha +1}} \|\Delta c\|_{L^2}\\[5pt] &\leq \frac {1}{4} \|\Delta c\|_{L^2}^2+\frac {1}{4}\|\nabla \Lambda ^{\alpha }u\|_{L^2}^2+C \|u\|_{L^2}^2. \end{aligned} \end{equation*}

Therefore, we gain

(3.14) \begin{equation} \begin{aligned} & \frac {d}{dt}\|\nabla c\| _{L^2 }^2+\|\Delta c\| _{L^2 }^2\leq \frac {1}{2}\|\nabla \Lambda ^{\alpha }u\|_{L^2}^2+C\|(n,u)\| _{L^2 }^2. \end{aligned} \end{equation}

Furthermore, we are going to establish the $\dot {H}^1$ -estimates of $u$ . In fact, multiplying both sides of the third equation of (1.1) by $-\Delta u$ and integrating by parts, one has

(3.15) \begin{equation} \begin{aligned} \frac {1}{2} \frac {d}{dt} \|\nabla u\| _{L^2 }^2 +\|\nabla \Lambda ^{\alpha } u\| _{L^2 }^2 = \int _{\mathbb{R}^3} \Delta u \cdot n \nabla \phi \,dx+\int _{\mathbb{R}^3} \Delta u \cdot (u \cdot \nabla ) u \,dx. \end{aligned} \end{equation}

The first term on the right-hand side of (3.15) is analysed by

(3.16) \begin{equation} \begin{aligned} \left |\int _{\mathbb{R}^3} \Delta u \cdot n \nabla \phi \,dx \right |&=\left |\int _{\mathbb{R}^3} \nabla u \cdot \nabla (n \nabla \phi )\,dx \right |\\[5pt] &\leq \|\nabla u\|_{L^2}\|n\| _{L^2 }\|\nabla ^2 \phi \|_{L^\infty }+\|\nabla u\|_{L^2}\|\nabla n\| _{L^2 }\|\nabla \phi \|_{L^\infty }\\[5pt] &\leq C\|\nabla u\| _{L^2 }^2+C\|n\| _{L^2 }^2+\frac {1}{4}\|\nabla n\| _{L^2 }^2. \end{aligned} \end{equation}

As $\Lambda ^{\sigma }$ is self-adjoint for any $\sigma \gt 0$ , one deduces from $\nabla \cdot u=0$ and (2.2) that

\begin{equation*} \begin{aligned} \Big | \int _{\mathbb{R}^3} \Delta u \cdot (u \cdot \nabla ) u \,dx\Big |&=\Big | \int _{\mathbb{R}^3} \Lambda ^{\alpha -1}\Delta u \cdot \Lambda ^{1-\alpha } \nabla \cdot (u \otimes u) \,dx\Big |\\[5pt] &\leq C\|\nabla \Lambda ^{\alpha }u\|_{L^2} \|\Lambda ^{2-\alpha }(u\otimes u)\|_{L^2}. \end{aligned} \end{equation*}

Note that the product law (2.6) and the Gagliardo–Nirenberg inequality (2.2) imply that

\begin{equation*} \begin{aligned} \|\Lambda ^{2-\alpha }(u\otimes u)\|_{L^2}\leq C\|u\|_{L^{q_2}} \|\Lambda ^{2-\alpha }u\|_{L^{\frac {2q_2}{q_2-2}}}\leq C\|u\|_{L^{q_2}} \|\nabla u\|_{L^2}^{1-\theta } \|\nabla \Lambda ^{\alpha }u\|_{L^2}^{\theta }, \end{aligned} \end{equation*}

where $\theta =\frac {1}{\alpha }(1+\frac {3}{q_2})-1$ fulfills

\begin{equation*} \frac {q_2-2}{2q_2}-\frac {2-\alpha }{3}=\left (\frac {1}{2}-\frac {1}{3}\right )(1-\theta )+\left (\frac {1}{2}-\frac {1+\alpha }{3}\right )\theta . \end{equation*}

Since $\frac {2}{1-\theta }\leq p_2$ due to (1.7), we obtain

\begin{equation*} \begin{aligned} \Big | \int _{\mathbb{R}^3} \Delta u \cdot (u \cdot \nabla ) u \,dx\Big | &\leq \frac {1}{2}\|\nabla \Lambda ^{\alpha }u\|_{L^2}^2+C\|u\|_{L^{q_2}}^{\frac {2}{1-\theta }}\|\nabla u\|_{L^2}^2\\[5pt] &\leq \frac {1}{2}\|\nabla \Lambda ^{\alpha }u\|_{L^2}^2+C(1+\|u\|_{L^{q_2}}^{p_2})\|\nabla u\|_{L^2}^2. \end{aligned} \end{equation*}

It thus follows that

(3.17) \begin{equation} \begin{aligned} \frac {d}{dt} \|\nabla u\| _{L^2 }^2 +\|\nabla \Lambda ^{\alpha } u\| _{L^2 }^2&\leq C(1+\|u\|_{L^{q_2}}^{p_2})\|\nabla u\|_{L^2}^2+C\|n\|_{L^2}^2+\frac {1}{2}\|\nabla n\| _{L^2 }^2. \end{aligned} \end{equation}

Collecting the estimates (3.6), (3.11) and (3.14) and (3.17) and thence applying (3.1) and Grönwall’s lemma, we arrive at

\begin{equation*} \begin{aligned} &\sup _{t\in (0,T]}(\|(n,\nabla c)\|_{L^2}^2+\| u\| _{H^1 }^2)+\int _{0}^{T}(\|(\nabla n,\nabla ^2 c)\|_{L^2}^2+ \|\Lambda ^{\alpha } u\| _{H^1}^2)\,dt\\[5pt] &\quad \leq e^{CT+CB_1}(\|(n_0,\nabla c_0)\|_{L^2}^2+\|u_0\|_{H^1}^2). \end{aligned} \end{equation*}

Together with (3.4), we thus establish (3.2) for all $\alpha \gt \frac {3}{4}$ .

• • Step 2: $H^1$ -estimates of $n$ and $H^2$ -estimates of $c,u$ .

In this step, we claim that for any $0\lt T\lt T_*$ , there exists some constant $C(B_1,T_*)$ such that

(3.18) \begin{equation} \sup _{t\in (0,T]}\|(\nabla n,\nabla ^2 c,\nabla ^2 u)\|_{L^2}^2+\int _{0}^{T}\|(\nabla ^2 n,\nabla ^3 c,\Lambda ^{\alpha +2}u)\|_{L^2}^2\,dt\leq C(B_1,T_*). \end{equation}

Despite the $H^1$ -estimate of $c$ having been obtained in Step 1, it is not enough to control the chemotaxis term $\nabla \cdot (\chi (c) n \nabla c)$ in (1.1) $_1$ . To this end, we first establish the $H^2$ -estimate of $c$ . Applying $\Delta$ to (1.1) $_2$ yields

(3.19) \begin{equation} \partial _t \Delta c-\Delta \Delta c=-\Delta (u \cdot \nabla c)-\Delta (nf(c)). \end{equation}

Multiplying (3.19) by $\Delta c$ , we obtain

(3.20) \begin{equation} \begin{aligned} \frac {1}{2}\frac {d}{dt}\|\Delta c\| _{L^2 }^2 +\|\nabla \Delta c\| _{L^2 }^2 =\int _{\mathbb{R}^3} \nabla \Delta c \cdot \nabla (u \cdot \nabla c)\,dx +\int _{\mathbb{R}^3}\nabla \Delta c \cdot \nabla (n f(c))\,dx. \end{aligned} \end{equation}

In view of (2.2) and Sobolev’s inequality, the terms on the right-hand side of (3.20) can be analysed as

(3.21) \begin{equation} \begin{aligned} &\left |\int _{\mathbb{R}^3} \nabla \Delta c \cdot \nabla (u \cdot \nabla c)\,dx \right |\\[5pt] & \quad \leq \|\nabla \Delta c\|_{L^2 }\left ( \|\nabla u\|_{L^3} \|\nabla c\|_{L^6}+ \|u\|_{L^\infty }\|\nabla ^2 c\| _{L^2 } \right )\\[5pt] &\quad \leq C \|\nabla \Delta c\| _{L^2 } \left ( \|\Lambda ^{\alpha +1} u\| _{L^2 }^\frac {3}{2(1+\alpha )} \|u\| _{L^2 }^{1-\frac {3}{2(1+\alpha )}} \|\nabla ^2 c\| _{L^2 }+\|u\|_{H^{\alpha +1}}\|\Delta c\| _{L^2 }\right )\\[5pt] &\quad \leq \frac {1}{4}\|\nabla \Delta c\| _{L^2 }^2+C \|u\|_{H^{\alpha +1}}^2 \|\Delta c\| _{L^2 }^2, \end{aligned} \end{equation}

and

(3.22) \begin{equation} \begin{aligned} &\left |\int _{\mathbb{R}^3}\nabla \Delta c \cdot \nabla (n f(c))\,dx\right |\\[5pt] &\quad \leq \|\nabla \Delta c\|_{L^2}\Big (\sup _{0\leq s\leq \|c_0\|_{L^\infty }}|f(s)| \|\nabla n\|_{L^2}+\sup _{0\leq s\leq \|c_0\|_{L^\infty }}|f'(s)| \|n\|_{L^3} \|\nabla c\|_{L^6}\Big )\\[5pt] &\quad \leq C\|\nabla \Delta c\|_{L^2}(\|\nabla n\|_{L^2}+\|n\|_{L^2}^{\frac {1}{2}}\|\nabla n\|_{L^2}^{\frac {1}{2}} \|\Delta c\|_{L^2})\\[5pt] &\quad \leq \frac {1}{4}\|\nabla \Delta c\| _{L^2 }^2+C \|\nabla n \| _{L^2 }^2+C(\|n\|_{L^2}^2+\|\nabla n\|_{L^2}^2)\|\Delta c\|_{L^2}^2. \end{aligned} \end{equation}

Substituting (3.21) and (3.22) into (3.20), one has

\begin{equation*} \begin{aligned} \frac {d}{dt}\|\Delta c \| _{L^2 }^2+\|\nabla \Delta c \| _{L^2 }^2 \leq C(\|n\|_{L^2}^2+\|\nabla n\|_{L^2}^2+\|u\|_{H^{\alpha +1}}^2)\|\Delta c\|_{L^2}^2+C\|\nabla n\| _{L^2 }^2, \end{aligned} \end{equation*}

which, together with (3.2) and Grönwall’s inequality, implies that, for any $t\in (0,T)$ with $0\lt T\lt T_*$ ,

(3.23) \begin{equation} \begin{aligned} &\sup _{t\in (0,T]} \|\nabla ^2 c \| _{L^2 }^2+ \int _0^{T}\|\nabla ^3 c \| _{L^2 }^2 \,dt\\[5pt] &\,\,\leq C \textrm {exp} \left \{\int _0^{T}(\|\nabla n\|_{L^2}^2+\|u\|_{H^{\alpha +1}}^2)\,dt+T\sup _{t\in (0,T)}\|n\|_{L^2}^2\right \}\\[5pt] &\,\,\quad \quad \quad \cdot \left ( \|\nabla ^2 c_0\| _{L^2 }^2+\int _0^{T} \|\nabla n\| _{L^2 }^2 \,dt \right )\leq C(B_1,T). \end{aligned} \end{equation}

Next, we are ready to bound $\nabla n$ . Taking the $L^2$ -inner product of (1.1) $_1$ with $\Delta n$ yields

(3.24) \begin{equation} \begin{aligned} &\frac {1}{2}\frac {d}{dt}\|\nabla n\|_{L^2}^2+\|\Delta n\|_{L^2}^2\leq \frac {1}{2}\|\Delta n\|_{L^2}^2+\|u\cdot \nabla n\|_{L^2}^2+\|\nabla \cdot (\chi (c)n\nabla c)\|_{L^2}^2. \end{aligned} \end{equation}

Since $\alpha \gt \frac {1}{2}$ , we can use (2.2) and the Young inequality to obtain

\begin{equation*} \begin{aligned} \|u\cdot \nabla n\|_{L^2}^2\leq \|u\|_{L^{\infty }}^2 \|\nabla n\|_{L^2}^2\leq C(\|u\|_{L^2}^2+\|\Lambda ^{1+\alpha }u\|_{L^2}^2)\|\nabla n\|_{L^2}^2, \end{aligned} \end{equation*}

and the last term in (3.24) can be controlled by

\begin{equation*} \begin{aligned} \|\nabla \cdot (\chi (c)n\nabla c)\|_{L^2}^2&\leq C\sup _{0\leq s\leq \|c_0\|_{L^{\infty }}}|\chi '(s)| \|\nabla c\|_{L^6}^4\|n\|_{L^6}^2\\[5pt] &\quad +C\sup _{0\leq s\leq \|c_0\|_{L^{\infty }}}|\chi (s)|(\|\nabla n\|_{L^2}^2\|\nabla c\|_{L^\infty }^2+\|n\|_{L^{6}}^2\|\nabla ^2 c\|_{L^3}^2)\\[5pt] &\leq C(\|\nabla ^2 c\|_{L^2}^4+\|\nabla ^2c\|_{H^1}^2)\|\nabla n\|_{L^2}^2. \end{aligned} \end{equation*}

Hence, using Grönwall’s lemma and the estimates (3.2) and (3.23) at hand, for any $0\lt T\lt T_*$ , we obtain

(3.25) \begin{equation} \begin{aligned} &\sup _{t\in (0,T]}\|\nabla n\| _{L^2 }^2+ \int _0^{T}\|\nabla ^2 n\| _{L^2 }^2 \,dt\\[5pt] &\leq C\|\nabla n_0\| _{L^2 }^2 \cdot \textrm {exp} \left \{ \int _0^{T} \left (\|\nabla ^2 c\| _{L^2 }^4 +\|\nabla c\|_{H^2}^2+\|u\|_{L^2}^2+\|\Lambda ^{1+\alpha }u\|_{L^2}^2\right )\,dt \right \}\leq C(B_1,T). \end{aligned} \end{equation}

Regarding the estimates of $u$ , applying $\Delta$ to both sides of (1.1) $_3$ and then taking the inner product of the resulting equation by $\Delta u,$ we have

(3.26) \begin{equation} \begin{aligned} \frac {1}{2}\frac {d}{dt}&\|\Delta u\| _{L^2 }^2 +\|\Lambda ^{\alpha }\Delta u\| _{L^2 }^2\\[5pt] &= \int _{\mathbb{R}^3}\nabla \Delta u \cdot \nabla ((u \cdot \nabla u)u)\,dx+ \int _{\mathbb{R}^3}\nabla \Delta u \cdot \nabla (n\nabla \phi )\,dx. \end{aligned} \end{equation}

To proceed with the estimate, we deal with the two terms at the right-hand side of (3.26), respectively.

For any $\alpha \gt \frac {3}{4}$ , we note that

\begin{equation*} \begin{aligned} &\int _{\mathbb{R}^3}\nabla \Delta u \cdot \nabla ((u \cdot \nabla ) u)\,dx\\[5pt] &\quad =\sum _{i,j,k,m=1}^3\int _{\mathbb{R}^3}\partial _{iij}u^m \partial _{j}(u^k\partial _{k}u^m)\,dx\\[5pt] &\quad =-\sum _{i,j,k,m=1}^3\int _{\mathbb{R}^3}\partial _{ii}u^m \partial _{jj}u^k \partial _{k}u^m\,dx-2\sum _{i,j,k,m=1}^3 \int _{\mathbb{R}^3}\partial _{ii}u^m \partial _{j}u^k\partial _{jk} u^m\,dx, \end{aligned} \end{equation*}

so it follows from the Gagliardo–Nirenberg inequality (2.2) that

(3.27) \begin{equation} \begin{aligned} &\left |\int _{\mathbb{R}^3}\nabla \Delta u \cdot \nabla ((u \cdot \nabla ) u)\,dx\right |\\[5pt] &\quad \leq 3 \|\nabla u\| _{L^2 } \|\nabla ^2 u\|_{L^4}^2\\[5pt] &\quad \leq C\|\nabla u\| _{L^2 } \|\nabla ^2 u\| _{L^2 }^{2-\frac {3}{2\alpha }} \|\Lambda ^{2+\alpha } u\| _{L^2 }^{\frac {3}{2\alpha }}\\[5pt] &\quad \leq \frac {1}{4} \|\Lambda ^{2+\alpha } u\| _{L^2 }^2 + C \|\nabla u\| _{L^2 }^{\frac {4\alpha }{4\alpha -3}}\|\nabla ^2 u\| _{L^2 }^2 \\[5pt] &\quad \leq \frac {1}{4} \|\Lambda ^{2+\alpha } u\| _{L^2 }^2 + C \|\nabla ^2 u\| _{L^2 }^2 \|\Lambda ^{\alpha +1} u\| _{L^2 }^{\frac {4\alpha }{(4\alpha -3)(\alpha +1)}}\|u\| _{L^2 }^\frac {4\alpha ^2}{(4\alpha -3)(\alpha +1)}\\[5pt] &\quad \leq \frac {1}{4} \|\Lambda ^{2+\alpha } u\| _{L^2 }^2 + C\left (1+\|\Lambda ^{\alpha +1} u\| _{L^2 }^2+\| u\| _{L^2 }^2\right ) \|\Delta u\| _{L^2 }^2. \end{aligned} \end{equation}

For the second term at the right-hand side of (3.26), in the case $\alpha \geq 1$ , we get from the Gagliardo–Nirenberg inequality (2.2) that

(3.28) \begin{equation} \begin{aligned} \left | \int _{\mathbb{R}^3}\nabla \Delta u \cdot \nabla (n\nabla \phi )\,dx\right | &\leq \|\nabla ^3 u\| _{L^2 }\|\nabla ( n \nabla \phi )\| _{L^2 }^2\\[5pt] &\leq C\|\Lambda ^{\alpha +2} u\| _{L^2 }^\frac {6}{2+\alpha } \|u\| _{L^2 }^{2-\frac {6}{2+\alpha }} + C\|\nabla \phi \|_{W^{1,\infty }}^2\|n\|_{H^1}^2 \\[5pt] &\leq \frac {1}{4} \|\Lambda ^{\alpha +2} u\| _{L^2 }^2+ C\left (\|u\| _{L^2 }^2 +\| n\|_{H^1}^2\right ). \end{aligned} \end{equation}

As for the case $\alpha \lt 1$ , one also has

(3.29) \begin{equation} \begin{aligned} \left | \int _{\mathbb{R}^3}\nabla \Delta u \cdot \nabla (n\nabla \phi )\,dx\right | &\leq \|\Lambda ^{\alpha -1}\nabla \Delta u\|_{L^2}\|\Lambda ^{1-\alpha }\nabla (n\nabla \phi )\|_{L^2}\\[5pt] &\leq C\|\Lambda ^{\alpha }\Delta u\|_{L^2} \| \Lambda ^{2-\alpha }(n\nabla \phi )\|_{L^2}\\[5pt] &\leq \frac {1}{4} \|\Lambda ^{\alpha +2} u\| _{L^2 }^2+ C\| n\|_{H^2}^2, \end{aligned} \end{equation}

where we have used

\begin{equation*} \| \Lambda ^{2-\alpha }(n\nabla \phi )\|_{L^2}\leq C\|n\nabla \phi \|_{L^2}^{\frac {\alpha }{2}}\|\nabla ^2(n\nabla \phi )\|_{L^2}^{1-\frac {\alpha }{2}} \leq C\|\nabla \phi \|_{W^{2,\infty }}\|n\|_{H^2}, \end{equation*}

derived from (2.2) and $1\lt 2-\alpha \lt 2$ . The combination of (3.26), (3.27), (3.28) and (3.29) yields

\begin{eqnarray*} \begin{aligned} \frac {d}{dt}\|\Delta u\| _{L^2 }^2 +\|\Lambda ^{\alpha } \Delta u\| _{L^2 }^2 \leq C \left (1+\| u\|_{H^{\alpha +1}}^2\right )\|\Delta u\| _{L^2 }^2 +C\left (\|u\| _{L^2 }^2 +\|n\|_{H^2}^2 \right ). \end{aligned} \end{eqnarray*}

Hence, thanks to Grönwall’s inequality, (3.2), (3.5) (3.6) and (3.25), there holds that, for any $0\lt T\lt T_*$ ,

\begin{equation*} \begin{aligned} &\sup _{t\in (0,T]}\|\nabla ^2 u\| _{L^2 }+ \int _0^{T} \|\Lambda ^{\alpha +2} u\| _{L^2 }^2\, dt\\[5pt] &\,\,\leq C \textrm {exp} \left \{\int _0^{T} \left (1+\|u\|_{L^2}^2+\|u\|_{H^{\alpha +1}}^2\right )\,dt \right \}\\[5pt] &\,\,\quad \cdot \left ( \|\nabla ^2 u_0\| _{L^2 }^2+ T \sup _{t\in (0,T]} \left (\|u\| _{L^2 }^2+\|n\|_{H^1}^2 \right )+\int _{0}^{T} \|\nabla ^2 n\|_{L^2}^2\,dt\right )\leq C(B_1,T_*), \end{aligned} \end{equation*}

which, as well as (3.23) and (3.25), yields (3.18).

• • Step 3: $H^2$ -estimates of $n$ .

Finally, it suffices to show that, for any $0\lt T\lt T_*$ ,

(3.30) \begin{equation} \sup _{t\in (0,T]}\|\nabla ^2n\|_{L^2}^2+\int _{0}^{T}\|\nabla ^3n\|_{L^2}^2\,dt\leq C(B_1,T_*). \end{equation}

To achieve it, we first apply $\Delta$ to (1.1) $_1$ and then perform the $L^2$ -energy estimate to have

(3.31) \begin{equation} \begin{aligned} \frac {1}{2}\frac {d}{dt} \|\Delta n\| _{L^2 }^2+ \|\nabla \Delta n\| _{L^2 }^2 &= \int _{\mathbb{R}^3} \nabla \Delta n \cdot \nabla ( u \cdot \nabla n+ \nabla \cdot (\chi (c) n \nabla c))\,dx\\[5pt] &\leq \frac {1}{4}\|\nabla \Delta n\| _{L^2 }^2+C\|\nabla ( u \cdot \nabla n)\|_{L^2}^2+ \|\nabla \nabla \cdot (\chi (c) n \nabla c))\|_{L^2}^2, \end{aligned} \end{equation}

where the second term on the right-hand side of (3.31) can be estimated as

\begin{equation*} \begin{aligned} \|\nabla ( u \cdot \nabla n)\|_{L^2}^2&\leq \|\nabla u\|_{L^6}^2\|\nabla n\|_{L^{3}}^2+\|u\|_{L^{\infty }}^2\|\nabla ^2n\|_{L^2}^2\\[5pt] &\leq C\|\nabla ^2 u\|_{L^2}^2\|\nabla n\|_{H^1}^2+\|u\|_{H^2}^2\|\nabla ^2n\|_{L^2}^2. \end{aligned} \end{equation*}

Note that

(3.32) \begin{equation} \begin{aligned} |\nabla \nabla \cdot (\chi (c) n \nabla c))|&\leq |\chi ''(c)| |\nabla c|^3 n+2|\chi '(c)| |\nabla ^2 c| |\nabla c| n+2|\chi '(c)| |\nabla c|^2 |\nabla n|\\[5pt] &\quad +|\chi (c)| |\nabla ^2n| |\nabla c|+|\chi (c)| |\nabla n| |\nabla ^2c|. \end{aligned} \end{equation}

By (3.32) and the Gagliardo–Nirenberg inequality, we handle the last term on (3.31) as follows

\begin{equation*} \begin{aligned} \|\nabla \nabla \cdot (\chi (c) n \nabla c))\|_{L^2}^2& \leq C(1+\|c\|_{H^2}^3)\|n\|_{H^2}^2\|\nabla c\|_{H^2}^2. \end{aligned} \end{equation*}

Thus, Grönwall’s inequality, together with (3.2) and (3.18), leads to (3.30).

The combination of (3.2), (3.18) and (3.30) gives rise to the required bounds to extend the solution $(n,c,u)$ beyond $T_*$ . Thus, we finish the proof of Theorem1.2.

3.2. Proof of Theorem 1.3

Similarly, we show Theorem1.3 towards a contradiction. Assume that the maximal time $T_*$ is finite, and (1.8) fails, i.e.

(3.33) \begin{equation} B_{2}\;:\!=\;\lim _{T\rightarrow T_*}\int _0^{T}\left (\|\nabla c\|_{L^{q_1}}^{p_1}+\|\nabla u\|_{L^{q_2}}^{p_2} \right )\,dt \lt +\infty \end{equation}

for any $(p_i, q_i) (i=1, 2)$ satisfying (1.9). Our goal is to show that for any $0\lt T\lt T_*$ , the $H^2$ -norms of $(n, c, u)$ can be bounded by some constant depending on $T_*$ and $B_2$ , and therefore the solution can be extended beyond $T_*$ , which leads to the contradiction.

First, one deduces from (1.1) $_1$ , (3.3) and the Gagliardo–Nirenberg inequality that

\begin{equation*} \begin{aligned} \frac {1}{2}\frac {d}{dt}\|n\|_{L^2 }^2+\|\nabla n\|_{L^2 }^2 &= \int _{\mathbb{R}^3}\chi (c) n\nabla c\cdot \nabla n\,dx\\[5pt] &\leq \sup _{0\leq c\leq \|c_0\|_{L^\infty }} |\chi (c)| \|\nabla n\|_{L^2 } \|n\|_{L^\frac {2q_1}{q_1-2}}\|\nabla c\|_{L^{q_1} }\\[5pt] &\leq C \|\nabla n\|_{L^2 } \|n\|_{L^{2} }^\frac {q_1-3}{q_1} \|\nabla n\|_{L^2 }^\frac {3}{q_1} \|\nabla c\|_{L^{q_1} }\\[5pt] & \leq \frac {1}{2} \|\nabla n\|_{L^2 }^2+C\|\nabla c\|_{L^{q_1} }^{\frac {2q_1}{q_1-3}}\|n\|_{L^2 }^2, \end{aligned} \end{equation*}

which, together with $\frac {2q_1}{q_1-3}\leq p_1$ (see (1.9)), leads to

\begin{equation*} \frac {d}{dt}\|n\| _{L^2 }^2+\|\nabla n\|_{L^2 }^2 \leq C(1+\|\nabla c\|_{L^{q_1}}^{p_1})\|n\| _{L^2 }^2. \end{equation*}

Hence, it follows from (3.33) that

(3.34) \begin{equation} \sup _{t\in (0,T]}\|n\|_{L^2}^2+\int _{0}^{T} \|\nabla n\|_{L^2}^2\,d\tau \leq e^{CT+CB_2}\|n_0\|_{L^2}^2 \end{equation}

for any $0\lt T\lt T_*$ .

Then, recall that (3.9) holds. Arguing similarly as in (3.14) in all the range $\alpha \gt \frac {1}{2}$ , one has

(3.35) \begin{equation} \begin{aligned} & \frac {d}{dt}\|\nabla c\| _{L^2 }^2+\|\Delta c\| _{L^2 }^2\leq C\|(n, u)\| _{L^2 }^2+\frac {1}{8} \|\Lambda ^{1+\alpha }u\|_{L^2}^2. \end{aligned} \end{equation}

We now deal with the $\dot {H}^1$ -estimates of $u$ . As in (3.10), we have

\begin{equation*} \begin{aligned} \frac {1}{2} \frac {d}{dt} \|\nabla u\| _{L^2 }^2 +\|\Lambda ^{\alpha +1} u\| _{L^2 }^2 = \int _{\mathbb{R}^3} \Delta u \cdot n \nabla \phi \,dx+\int _{\mathbb{R}^3} \Delta u \cdot (u \cdot \nabla ) u\,dx. \end{aligned} \end{equation*}

We estimate every term on the right-hand side of the above equality as follows. Arguing similarly as in (3.16), one has

\begin{equation*} \begin{aligned} \left | \int _{\mathbb{R}^3} \Delta u \cdot n \nabla \phi \,dx\right |&\leq C\|\nabla u\| _{L^2 }^2+C\|n\| _{L^2 }^2+C\|\nabla n\| _{L^2 }^2. \end{aligned} \end{equation*}

After integrating by parts as in (3.12) and using $\nabla \cdot u=0$ and the Gagliardo–Nirenberg inequality (2.2), we deduce that

\begin{equation*} \begin{aligned} \left |\int _{\mathbb{R}^3} \Delta u \cdot (u \cdot \nabla ) u\,dx \right |&\leq \|\nabla u\|_{L^3}^3\\[5pt] &\leq C \|\nabla u\|_{L^{q_2}} \|\nabla u\|_{L^{\frac {2q_2}{q_2-1}}}^2\\[5pt] &\leq C \|\nabla u\|_{L^{q_2}} \|\nabla u\|_{L^{2}}^{2\theta } \|\Lambda ^{\alpha }\nabla u\|_{L^2}^{2(1-\theta )}\\[5pt] &\leq \frac {1}{8} \|\Lambda ^{\alpha +1}u\|_{L^2}^2+C\|\nabla u\|_{L^{q_2}}^{\frac {1}{\theta }} \|\nabla u\|_{L^2}^2, \end{aligned} \end{equation*}

where $\theta =1-\frac {3}{2q_2\alpha }$ satisfies $\frac {1}{\theta }\leq p_2$ due to (1.9). It thus holds that

(3.36) \begin{equation} \begin{aligned} \frac {d}{dt} \|\nabla u\| _{L^2 }^2 +\|\Lambda ^{\alpha +1} u\| _{L^2 }^2\leq C(1+\|\nabla u\|_{L^{q_2}}^{p_2}) \|\nabla u\|_{L^2}^2+C\|n\| _{L^2 }^2+C\|\nabla n\| _{L^2 }^2. \end{aligned} \end{equation}

Adding (3.9), (3.35) and (3.36) together and then using Grönwall’s inequality, (3.33) and (3.34), one has

\begin{eqnarray*} \begin{aligned} &\sup _{t\in (0,T]}(\|\nabla c\|_{L^2}^2+\| u\| _{H^1}^2) +\int _{0}^{T}(\|\nabla ^2 c\|_{L^2}^2+\|\Lambda ^{\alpha } u\| _{H^1}^2)\,dt\leq C(T_*,B_2). \end{aligned} \end{eqnarray*}

Finally, repeating the same arguments as Steps 2--3 in Subsection 3.1, we are able to establish the higher-order estimates

\begin{eqnarray*} \begin{aligned} &\sup _{t\in (0,T]}\|(n, c,u)\|_{H^2}^2 +\int _{0}^{T}(\|(\nabla n,\nabla c)\|_{H^2}^2+\|\Lambda ^{\alpha } u\| _{H^2}^2)\,dt\leq C(T_*,B_2),\quad \quad 0\lt T\lt T_*. \end{aligned} \end{eqnarray*}

For the sake of simplicity, we omitted the details here.

Thus, based on the above estimates and Theorem1.1, one can extend the solution $(n,c,u)$ beyond $T_*$ . This contradicts the fact that $T_*$ is the maximal time for existence. The proof of Theorem1.3 is thus complete.

4.1. Proof of Theorem 1.4

In order to prove (1.4), we justify the blow-up criterion stated in Theorem1.3 as follows.

Lemma 4.1. Let $(n,c,u)$ be a strong solution to the Cauchy problem ( 1.1 )–( 1.2 ) on $[0,T_*)$ obtained in Theorem 1.1 . Then there exists a constant $\delta _0=\delta _0(p)$ such that if $T_*$ is finite and ( 1.10 ) follows, then it holds for any $0\lt T\lt T_*$ that

(4.1) \begin{equation} \begin{aligned} \sup _{t\in (0,T]}\|n\|_{L^r}\leq N_{r} \end{aligned} \end{equation}

for any $r\in (1,\infty )$ and some uniform constant $C_r$ .

Furthermore, we have

(4.2) \begin{equation} \begin{aligned} \int _0^{T}\|u\|_{L^{q}}^{p}\,dt\leq C_{T_*}, \end{aligned} \end{equation}

where $C_{T_*}$ is a constant depending on $T_*$ , and $p$ , $q$ satisfy

(4.3) \begin{align} &\frac {2\alpha }{p}+\frac {3}{q}=\frac {3}{2},\quad 2\leq q\leq \frac {6}{3-2\alpha }\quad \text{for}\quad \frac {1}{2}\lt \alpha \lt \frac {3}{2},\quad 2\leq q\leq \infty \quad \text{for}\quad \alpha \gt \frac {3}{2}. \end{align}

Proof. If $\|c_{0}\|_{L^{\infty }}\leq \delta ^*$ with some suitably small $\delta ^*\gt 0$ , then we are able to obtain (4.1) according to the coupling structure of (1.1) $_1$ –(1.1) $_2$ . The proof can be done by the following similar arguments as in the works [Reference Chae, Kang and Lee8, Proposition 2] or [Reference Tao47, Lemma 3.1]. For the sake of simplicity, the details are omitted here.

We then deal with the estimates of $u$ . Given the $L^2$ -estimate of $n$ due to (4.1), it follows from (3.9) that

(4.4) \begin{equation} \begin{aligned} &\sup _{t\in [0,T]}\|u\|_{L^2}^2+2\int _{0}^{T}\|\Lambda ^{\alpha }u\|_{L^2}^2\,dt\leq e^{T}( \|u_{0}\|_{L^2}^2+CN_{2}T). \end{aligned} \end{equation}

Hence, for $\theta =1-\frac {3}{\alpha }(\frac {1}{2}-\frac {1}{q})$ such that

\begin{equation*} \frac {1}{q}=\frac {\theta }{2}+(1-\theta )(\frac {1}{2}-\frac {\alpha }{3}),\quad \quad p(1-\theta )=2, \end{equation*}

one has

\begin{equation*} \begin{aligned} \int _0^{T}\|u\|_{L^{q}}^{p} \,dt&\leq C\int _0^{T}\|u\|_{L^{2}}^{p\theta } \|\Lambda ^{\alpha }u\|_{L^2}^{p(1-\theta )} \,dt\leq C\sup _{t\in (0,T]} \|u\|_{L^{2}}^{p\theta }\int _{0}^{T}\|\Lambda ^{\alpha }u\|_{L^2}^{2}\,dt\leq C_{T}, \end{aligned} \end{equation*}

derived from (4.4) and the Gagliardo–Nirenberg inequality (2.2). This completes the proof of Lemma4.1.

4.2. Proof of Theorem 1.5: Global existence

In this subsection, we shall prove the global existence part in Theorem1.5 if the initial $L^2$ energy of $(n_0,c_0, u_0)$ is small but the highest-order norm of initial data can be arbitrarily large. Our proof is based on elaborate energy estimates, the bootstrap argument as well as interpolation inequalities.

To achieve global existence, the key ingredient is to establish the following uniform a priori estimates.

Proposition 4.1. Let the assumptions of Theorem 1.5 hold. Define

\begin{align*} \mathcal{E}_{0}\;:\!=\; \|(n_0, c_0,u_0)\|_{L^2}^2,\quad \quad \mathcal{E}_{2}\;:\!=\; \|(\nabla ^2 n_0,\nabla ^2 c_0,\nabla ^2 u_0)\|_{L^2}^2, \end{align*}

and

(4.5) \begin{align} S_{f,\chi }\;:\!=\;\sup _{0\leq s\leq \|c_0\|_{L^{\infty }}}(|f(s)|+|f'(s)|+|\chi (s)|+|\chi '(s)|+|\chi ''(s)|). \end{align}

For any $\frac {3}{4}\lt \alpha \lt \frac {5}{4}$ , suppose that $(n,c,u)$ is a strong solution to the Cauchy problem ( 1.1 )–( 1.2 ) defined on $[0,T)$ satisfying

(4.6) \begin{equation} S_{f,\chi }^2 \|\nabla c\|_{L^3}^2+\|(c,u)\|_{L^3}^2+\int _{0}^{t}\|\nabla u\|_{L^2}^{\frac {4\alpha }{4\alpha -3}}\, d\tau \leq M_{0} \end{equation}

for any $ 0\lt t\lt T$ and some generic constant $M_{0}\gt 0$ given by ( 4.9 ), ( 4.13 ), ( 4.16 ), ( 4.22 ) and ( 4.26 ) below. Then, there exists a constant $\delta _{1}$ depending only on $S_{f,\chi }$ , $\|(1+|x|)\nabla \phi \|_{W^{2,\infty }}$ , $\||x|^{1+\alpha }\nabla \phi \|_{L^{\infty }}$ and $\|(\nabla ^2n_0,\nabla ^2c_0,\nabla ^2 u_0)\|_{L^2}$ such that if ( 1.12 ) holds, then for any $0\lt t\lt T$ we have

(4.7) \begin{equation} \begin{aligned} &S_{f,\chi }^2 \|\nabla c\|_{L^3}^2+\|(c,u)\|_{L^3}^2+\int _{0}^{t}\|\nabla u\|_{L^2}^{\frac {4\alpha }{4\alpha -3}} \,d\tau \leq \frac {1}{2}M_{0}, \end{aligned} \end{equation}

The proof of Proposition4.1 relies on the uniform estimates obtained in Lemmas4.2-4.4 below.

Lemma 4.2. Under the assumptions of Proposition 4.1 , if ( 4.6 ) holds with some constant $M_0\gt 0$ , then for all $t\in (0,T)$ , we have

(4.8) \begin{equation} \begin{aligned} &\sup _{\tau \in (0,t]}\|(n, c,u)\|_{L^2}^2+\int _{0}^{t}\|(\nabla n,\nabla c,\Lambda ^{\alpha }u)\|_{L^2}^2\,d\tau \leq C_{1,\phi } \mathcal{E}_0 \end{aligned} \end{equation}

for

\begin{equation*} C_{1,\phi }\;:\!=\;C(1+\||x|^{1+\alpha }\nabla \phi \|_{L^{\infty }}^2). \end{equation*}

Proof. We recall that $c$ satisfies (3.4) which is uniform in time. Taking the inner product of (1.1) $_{1}$ by $n$ , we derive

\begin{equation*} \begin{aligned} &\frac {1}{2}\frac {d}{dt}\|n\|_{L^2}^2+\|\nabla n\|_{L^2}^2=\int _{\mathbb{R}^3} n\nabla n\cdot \nabla c \chi (c)\,dx. \end{aligned} \end{equation*}

According to the Sobolev embedding and $0\leq c\leq \|c_{0}\|_{L^{\infty }}$ , one has

\begin{equation*} \begin{aligned} \left |\int _{\mathbb{R}^3} n\nabla n\cdot \nabla c \chi (c) \,dx\right |&\leq \sup _{0\leq s\leq \|c_{0}\|_{L^{\infty }}} \chi (s) \|\nabla n\|_{L^2} \|\nabla c\|_{L^3}\|n\|_{L^6} \leq C_1 S_{f,\chi } \|\nabla c\|_{L^3} \|\nabla n\|_{L^2}^2, \end{aligned} \end{equation*}

where the constant $S_{f,\chi }$ is given by (4.5). Therefore, we arrive at

\begin{equation*} \begin{aligned} &\frac {1}{2}\frac {d}{dt}\|n\|_{L^2}^2+\left (1-C_1 S_{f,\chi }\|\nabla c\|_{L^3}\right )\|\nabla n\|_{L^2}^2\leq 0, \end{aligned} \end{equation*}

where $C_1\gt 0$ is a generic constant. As long as (4.6) holds with

(4.9) \begin{align} M_0\leq \frac {1}{(2C_1)^2}, \end{align}

we deduce that

(4.10) \begin{equation} \begin{aligned} &\sup _{\tau \in (0,t]}\|n\|_{L^2}^2+\int _0^t\|\nabla n\|_{L^2}^2\,d\tau \leq \|n_0\|_{L^2}^2. \end{aligned} \end{equation}

On the other hand, from (1.1) $_{3}$ and Hardy’s inequality in Lemma2.4, one has

\begin{equation*} \begin{aligned} \frac {1}{2}\frac {d}{dt}\|u\|_{L^2}^2+\|\Lambda ^\alpha u\|_{L^2}^2&=-\int _{\mathbb{R}^3} n\nabla \phi \cdot u\,dx\\[5pt] &\leq \||x|^{1+\alpha }\nabla \phi \|_{L^{\infty }}\Big \|\frac {n}{|x|}\Big \|_{L^2}\Big \|\frac {u}{|x|^{\alpha }}\Big \|_{L^{2}}\\[5pt] &\leq \frac {1}{2}\|\Lambda ^{\alpha }u\|_{L^2}^2+C\||x|^{1+\alpha }\nabla \phi \|_{L^{\infty }}^2\|\nabla n\|_{L^{2}}^2, \end{aligned} \end{equation*}

which, as well as (4.10), yields

\begin{equation*} \begin{aligned} &\|u\|_{L^2}^2+\int _{0}^{t}\|\Lambda ^{\alpha }u\|_{L^2}^2\,d\tau \leq \|u_{0}\|_{L^2}+C\||x|^{1+\alpha }\nabla \phi \|_{L^{\infty }}^2\|n_0\|_{L^2}^2. \end{aligned} \end{equation*}

Hence, we end up with (4.8).

Lemma 4.3. Under the assumptions of Proposition 4.1 , if we have ( 4.6 ) with some constant $M_0\gt 0$ , then for all $t\in (0,T)$ , it holds that

(4.11) \begin{equation} \begin{aligned} \sup _{\tau \in (0,t]}&\|(\nabla n,\nabla c,\nabla u)\|_{L^2}^2+\int _{0}^{t}(\|(\nabla ^2n,\nabla ^2 c,\nabla \Lambda ^{\alpha }u)\|_{L^2}^2)\,d\tau \\[5pt] &\leq \mathcal{C}_{2,\phi, f,\chi } (\mathcal{E}_0^{\frac {1}{2}}\mathcal{E}_2^{\frac {1}{2}}+\mathcal{E}_0^3+\mathcal{E}_0+\mathcal{E}_0^2\mathcal{E}_2), \end{aligned} \end{equation}

where $\mathcal{C}_{2,\phi, f,\chi }$ is given by

\begin{equation*} \begin{aligned} \mathcal{C}_{2,\phi, f,\chi }&\;:\!=\;C(1+S_{f,\chi }^4)(1+\|(1+|x|^{1+\alpha })\nabla \phi \|_{L^{\infty }}^6) \end{aligned} \end{equation*}

for some universal constant $C\gt 0$ .

Proof. From (1.1) $_{2}$ , it holds that

\begin{equation*} \begin{aligned} \frac {1}{2}\frac {d}{dt}\|\nabla c\|_{L^2}^2+\|\Delta c\|_{L^2}^2 \leq \|u\cdot \nabla c\|_{L^2}^2+\|n f(c)\|_{L^2}^2+\frac {1}{2}\|\Delta c\|_{L^2}^2. \end{aligned} \end{equation*}

Regarding the first nonlinear term, one has

(4.12) \begin{equation} \begin{aligned} \|u\cdot \nabla c\|_{L^2}^2\leq \|u\|_{L^{3}}^2\|\nabla c\|_{L^6}^2\leq C_2\|u\|_{L^{3}}^2\|\Delta c\|_{L^2}^2 \end{aligned} \end{equation}

for some constant $C_2\gt 0$ . Due to $f(0)=0$ , we can write $f(c)=\tilde {f}(c)c$ with $\tilde {f}(c)=\int _{0}^{1}f'(\theta c)\,d\theta$ . Therefore, the second nonlinear term is handled by

\begin{equation*} \begin{aligned} \|n f(c)\|_{L^2}^2&\leq S_{f,\chi }^2 \|n\|_{L^6}^2\|c\|_{L^3}^2\leq C S_{f,\chi }^2\|c\|_{L^3}^2\|\nabla n\|_{L^2}^2. \end{aligned} \end{equation*}

Thus, together with (3.3)–(3.4), we arrive at

\begin{equation*} \begin{aligned} \frac {1}{2}\frac {d}{dt}\|\nabla c\|_{L^2}^2+(1-C_2\|u\|_{L^3}^2)\|\Delta c\|_{L^2}^2\leq C_2S_{f,\chi }^2\|c\|_{L^3}^2\|\nabla n\|_{L^2}^2. \end{aligned} \end{equation*}

Let $M_0$ satisfy

(4.13) \begin{equation} \begin{aligned} &M_0\leq \frac {1}{2C_2}. \end{aligned} \end{equation}

This leads to

(4.14) \begin{equation} \begin{aligned} \sup _{\tau \in (0,t]}\|\nabla c\|_{L^2}^2+\int _{0}^{t}\|\nabla ^2 c\|_{L^2}^2\,d\tau &\leq \|\nabla c_0\|_{L^2}^2+ S_{f,\chi }^2 \int _{0}^{t}\|\nabla n\|_{L^2}^2\,d\tau \\[5pt] &\leq \|\nabla c_0\|_{L^2}^2+ C_{1,\phi }S_{f,\chi }^2\mathcal{E}_{0}, \end{aligned} \end{equation}

where (4.6) and (4.8) have been used.

To establish the estimate of $n$ , one takes the inner product of (1.1) $_{1}$ with $\Delta n$ to obtain

(4.15) \begin{equation} \begin{aligned} &\frac {1}{2}\frac {d}{dt}\|\nabla n\|_{L^2}^2+\left (\frac {3}{4}-C_3 (S_{f,\chi }^2\|\nabla c\|_{L^3}^2+\|u\|_{L^3}^2) \right )\|\Delta n\|_{L^2}^2\\[5pt] &\,\leq C_3 (S_{f,\chi }^2+S_{f,\chi }^4) \|\nabla ^2c\|_{L^2}^4\|\nabla n\|_{L^2}^2, \end{aligned} \end{equation}

where $C_3\gt 0$ is a generic constant, and we have used Young’s inequality,

\begin{equation*} \begin{aligned} C\|u\cdot \nabla n\|_{L^2}^2\leq C\|u\|_{L^{3}}^2\|\nabla n\|_{L^6}^2\leq C_3\|u\|_{L^3}^2\|\Delta n\|_{L^2}^2, \end{aligned} \end{equation*}

and

\begin{equation*} \begin{aligned} &C\|\nabla (\chi (c) n \nabla c)\|_{L^2}^2\\[5pt] &\quad \leq CS_{f,\chi }^2(\|n\|_{L^\infty }^2\|\nabla ^2 c\|_{L^2}^2+\|\nabla n\|_{L^6}^2\|\nabla c\|_{L^3}^2+\|\nabla c\|_{L^6}^4\|n\|_{L^6}^2)\\[5pt] &\quad \leq CS_{f,\chi }^2(\|\nabla n\|_{L^2}\|\nabla ^2n\|_{L^{2}}\|\nabla ^2 c\|_{L^2}^2+C\|\nabla ^2n\|_{L^2}^2\|\nabla c\|_{L^3}^2+ \|\nabla ^2 c\|_{L^2}^4\|\nabla n\|_{L^2}^2)\\[5pt] &\quad \leq \left (\frac {1}{4}+C_3 S_{f,\chi }^2\|\nabla c\|_{L^3}^2\right )\|\Delta n\|_{L^2}^2+C_3 (S_{f,\chi }^2+S_{f,\chi }^4) \|\nabla ^2 c\|_{L^2}^4\|\nabla n\|_{L^2}^2. \end{aligned} \end{equation*}

We recall (4.6) and let $M_0$ be such that

(4.16) \begin{equation} \begin{aligned} &M_0\leq \frac {1}{4C_3}. \end{aligned} \end{equation}

Thus, it follows from (4.8), (4.14) and (4.15) that

(4.17) \begin{equation} \begin{aligned} &\sup _{\tau \in (0,t]}\|\nabla n\|_{L^2}^2+\int _{0}^{t}\|\nabla ^2 n\|_{L^2}^2\,d\tau \\[5pt] &\quad \leq \|\nabla n_0\|_{L^2}^2 +2C_3 (S_{f,\chi }^2+S_{f,\chi }^4 ) \sup _{\tau \in (0,t]}\|\nabla c\|_{L^2}^4 \int _0^t \|\nabla n\|_{L^2}^2\,d\tau \\[5pt] &\quad \leq \|\nabla n_0\|_{L^2}^2+2C_3(S_{f,\chi }^2+S_{f,\chi }^4 )(\|\nabla c_0\|_{L^2}^2+ C_{1,\phi }S_{f,\chi }^2\mathcal{E}_{0})^2C_{1,\phi }\mathcal{E}_0. \end{aligned} \end{equation}

Finally, we are ready to establish the $L^2$ -estimates of $\nabla u$ . Taking the inner product of (1.1) $_{3}$ with $\Delta u$ , we have

(4.18) \begin{equation} \begin{aligned} \frac {1}{2}\frac {d}{dt}\|\nabla u\|_{L^2}^2+\|\Lambda ^{\alpha } \nabla u\|_{L^2}^2=\int _{\mathbb{R}^3} \nabla (n\nabla \phi )\;:\;\nabla u\,dx-\int _{\mathbb{R}^3} \nabla (u\cdot \nabla u)\;:\; \nabla u\,dx. \end{aligned} \end{equation}

We handle the first term on the right-hand side of (4.18) in two cases. In the case $1\leq \alpha \lt \frac {5}{4}$ , one deduces from Hardy’s inequality (2.8) and the Gagliardo–Nirenberg inequality (2.2) that

\begin{equation*} \begin{aligned} \left |\int _{\mathbb{R}^3} \nabla (n\nabla \phi )\;:\; \nabla u\,dx\right |&=-\int _{\mathbb{R}^3} n\nabla \phi \cdot \Delta u\,dx\\[5pt] &\leq \||x|\nabla \phi \|_{L^{\infty }}\Big {\|}\frac {n}{|x|}\Big {\|}_{L^2}\|\Delta u\|_{L^2}\\[5pt] &\leq C\||x|\nabla \phi \|_{L^{\infty }}\|\nabla n\|_{L^2}\|u\|_{\dot {H}^{\alpha }}^{\alpha -1}\|u\|_{\dot {H}^{1+\alpha }}^{2-\alpha }. \end{aligned} \end{equation*}

In the case $\frac {3}{4}\lt \alpha \lt 1$ , one also has

\begin{equation*} \begin{aligned} &\left |\int _{\mathbb{R}^3} \nabla (n\nabla \phi )\;:\; \nabla u\,dx\right |\\[5pt] &\quad \leq \Big (\|\nabla n\|_{L^2}\|\nabla \phi \|_{L^{\infty }}+\Big \|\frac {n}{|x|}\Big \|_{L^2}\||x|\nabla \phi \|_{L^{\infty }}\Big )\|\nabla u\|_{L^2}\\[5pt] &\quad \leq C\|(1+|x|)\nabla \phi \|_{L^{\infty }}\|\nabla n\|_{L^2}\|u\|_{\dot {H}^{\alpha }}^{\alpha }\|u\|_{\dot {H}^{1+\alpha }}^{1-\alpha }. \end{aligned} \end{equation*}

Hence, for all $\frac {3}{4}\lt \alpha \lt \frac {5}{4}$ , we gain

(4.19) \begin{equation} \begin{aligned} \left |\int _{\mathbb{R}^3} \nabla (n\nabla \phi )\;:\; \nabla u\,dx\right |\leq \frac {1}{4}\|\Lambda ^{\alpha } \nabla u\|_{L^2}^2+C\|(1+|x|)\nabla \phi \|_{L^{\infty }}^2\|\nabla n\|_{L^2}^2+C\|u\|_{\dot {H}^{\alpha }}^2. \end{aligned} \end{equation}

Now the second term on the right-hand side of (4.18) is analysed as follows. Using the duality of $\dot {H}^{\sigma }$ and $\dot {H}^{-\sigma }$ for any $\sigma \in \mathbb{R}$ , we have

\begin{equation*} \begin{aligned} \Big | \int _{\mathbb{R}^3} \Delta u \cdot (u \cdot \nabla ) u \,dx\Big |&\leq \|\Delta u\|_{\dot {H}^{\alpha -1}}\|u\cdot \nabla u\|_{\dot {H}^{1-\alpha }}\leq C\|\Lambda ^{1+\alpha }u\|_{L^2}\|u\cdot \nabla u\|_{\dot {H}^{1-\alpha }}. \end{aligned} \end{equation*}

One deduces from $\nabla \cdot u=0$ , the product law (2.7) and the interpolation inequality (2.2) that

\begin{equation*} \begin{aligned} \|u\cdot \nabla u\|_{\dot {H}^{1-\alpha }}=\|\nabla \cdot (u\otimes u)\|_{\dot {H}^{1-\alpha }}\leq \|u\otimes u\|_{\dot {H}^{2-\alpha }}\leq C\|u\|_{\dot {H}^{\frac {7}{4}-\frac {\alpha }{2}}}^2\leq C\|u\|_{\dot {H}^1}^{3-\frac {3}{2\alpha }} \|u\|_{\dot {H}^{1+\alpha }}^{\frac {3}{2\alpha }-1}. \end{aligned} \end{equation*}

Here we note that $\frac {3}{2\alpha }\lt 2$ due to $\alpha \gt \frac {3}{4}$ . It thus follows that

(4.20) \begin{equation} \begin{aligned} \left |\int _{\mathbb{R}^3} \nabla (u\cdot \nabla u)\;:\; \nabla u\,dx\right |&\leq \|\Delta u\|_{\dot {H}^{\alpha -1}}\|u\cdot \nabla u\|_{\dot {H}^{1-\alpha }}\\[5pt] &\leq C\|u\|_{\dot {H}^1}^{3-\frac {3}{2\alpha }} \|u\|_{\dot {H}^{1+\alpha }}^{\frac {3}{2\alpha }}\\[5pt] &\leq \frac {1}{4} \|\Lambda ^{\alpha } \nabla u\|_{L^2}^2+C\|\nabla u\|_{L^2}^{\frac {4\alpha }{4\alpha -3}}\|\nabla u\|_{L^2}^2. \end{aligned} \end{equation}

Putting (4.19) and (4.20) into (4.18) gives rise to

\begin{equation*} \begin{aligned} &\frac {d}{dt}\|\nabla u\|_{L^2}^2+\|\Lambda ^{\alpha } \nabla u\|_{L^2}^2\leq C\|(1+|x|)\nabla \phi \|_{L^{\infty }}^2\|\nabla n\|_{L^2}^2+C\|u\|_{\dot {H}^{\alpha }}^2+C\|\nabla u\|_{L^2}^{\frac {4\alpha }{4\alpha -3}+2}. \end{aligned} \end{equation*}

Integrating the above inequality over $[0,t]$ , we infer that there exists a universal constant $C_4\gt 0$ such that

\begin{equation*} \begin{aligned} &\sup _{\tau \in (0,t]}\|\nabla u\|_{L^2}^2+\int _{0}^{t}\|\Lambda ^{\alpha } \nabla u\|_{L^2}^2\,d\tau \\[5pt] &\quad \leq \|\nabla u_{0}\|_{L^2}^2+C_4\|(1+|x|)\nabla \phi \|_{L^{\infty }}^2\int _{0}^{t}\|\nabla n\|_{L^2}^2\,d\tau +\int _{0}^{t}\|u\|_{\dot {H}^{\alpha }}^2\,d\tau \\[5pt] &\quad +C_4\int _{0}^{t}\|\nabla u\|_{L^2}^{\frac {4\alpha }{4\alpha -3}} \,d\tau \sup _{\tau \in (0,t]}\|\nabla u\|_{L^2}^2. \end{aligned} \end{equation*}

This, as well as (4.6) and (4.8), implies that

(4.21) \begin{equation} \begin{aligned} &\frac {1}{2}\sup _{t\in (0,\tau ]}\|\nabla u\|_{L^2}^2+2\int _{0}^{t}\|\Lambda ^{\alpha } \nabla u\|_{L^2}^2\,d\tau \\[5pt] &\,\leq \|\nabla u_0\|_{L^2}^2+C_4(1+\|(1+|x|)\nabla \phi \|_{L^{\infty }}^2)C_{1,\phi }\mathcal{E}_0, \end{aligned} \end{equation}

when $M_0$ in (4.6) satisfies

(4.22) \begin{align} M_0\leq \frac {1}{2C_4}. \end{align}

Combining (4.14), (4.17), (4.21) and the fact that $\|(\nabla n_0,\nabla c_0,\nabla u_0)\|_{L^2}^2\leq C\mathcal{E}_0^{\frac {1}{2}}\mathcal{E}_2^{\frac {1}{2}}$ due to (2.2), we conclude (4.11) and thus complete the proof of Lemma4.3.

Lemma 4.4. Under the assumptions of Proposition 4.1 , if ( 4.6 ) holds with some constant $M_0\gt 0$ and let $\mathcal{E}_0\leq 1$ , then for all $t\in (0,T)$ , we have

(4.23) \begin{equation} \begin{aligned} \sup _{\tau \in (0,t]}&\|(\nabla ^2 c,\nabla ^2 u)\|_{L^2}^2+\int _{0}^{t}\|(\nabla ^3 c,\nabla ^2\Lambda ^{\alpha }u)\|_{L^2}^2\,d\tau \leq \mathcal{C}_{3,\phi, f,\chi }(\mathcal{E}_2+\mathcal{E}_0+\mathcal{E}_2^{3}), \end{aligned} \end{equation}

and

(4.24) \begin{equation} \begin{aligned} \sup _{\tau \in (0,t]}&\|\nabla ^2 n\|_{L^2}^2+\int _{0}^{t}\|\nabla ^3n\|_{L^2}^2\,d\tau \leq e^{\mathcal{C}_{3,\phi, f,\chi }(1+\mathcal{E}_2^{2})}\mathcal{C}_{3,\phi, f,\chi }(\mathcal{E}_2+\mathcal{E}_0+\mathcal{E}_2^{10}), \end{aligned} \end{equation}

where $C$ is a generic constant, and $\mathcal{C}_{3,\phi, f,\chi }$ denotes some constant depending only on $S_{\chi, f}$ , $\|(1+|x|^{1+\alpha })\nabla \phi \|_{L^{\infty }}$ , $\|\nabla ^2\phi \|_{L^{\infty }}$ and $\| |x|\nabla ^3\phi \|_{L^{\infty }}$ .

Proof. We first perform the $L^2$ -estimates of $\nabla ^2 c$ . One deduces from (1.1) $_{2}$ that

\begin{equation*} \begin{aligned} \frac {1}{2} \frac {d}{dt}&\|\Delta c\| _{L^2 }^2+\|\nabla \Delta c\| _{L^2 }^2\\[5pt] &= \int _{\mathbb{R}^3}\nabla \Delta c \cdot \nabla (n f(c))\,dx+\int _{\mathbb{R}^3}\nabla \Delta c \cdot \nabla (u\cdot \nabla c)\,dx\\[5pt] &\leq \frac {1}{16}\|\nabla \Delta c\| _{L^2 }^2+C\| \nabla \cdot (n f(c))\|_{L^2}^2+C\|\nabla \cdot (u\cdot \nabla c)\|_{L^2}^2. \end{aligned} \end{equation*}

Here by $f(0)=0$ and (2.2) one can verify that

\begin{equation*} \begin{aligned} \| \nabla \cdot (n f(c))\|_{L^2}^2&\leq C S_{f,\chi }^2\Big (\|c\|_{L^6}^2\|\nabla n\|_{L^3}^2+\|n\|_{L^{\infty }}^2\|\nabla c\|_{L^2}^2\Big )\\[5pt] &\leq C S_{f,\chi }^2 \Big ( \|c\|_{L^6}^2\|\nabla n\|_{L^2}\|\Delta n\|_{L^2}+ \|\nabla n\|_{L^2}\|\nabla ^2 n\|_{L^2} \|\nabla c\|_{L^2}^2\Big ) \\[5pt] &\leq C S_{f,\chi }^2\|\Delta n\|_{L^2}^2+C S_{f,\chi }^2\|\nabla c\|_{L^2}^4 \|\nabla n\|_{L^2}^2, \end{aligned} \end{equation*}

and

\begin{equation*} \begin{aligned} \|\nabla \cdot (u\cdot \nabla c)\|_{L^2}^2&\leq \|\nabla u\|_{L^{2}}^2\|\nabla c\|_{L^{\infty }}^2+\|u\|_{L^{6}}^2\|\nabla ^2c\|_{L^3}^2\\[5pt] &\leq C\|\nabla u\|_{L^2}^2\|\nabla ^2 c\|_{L^2}\|\nabla ^3c\|_{L^2} \\[5pt] &\leq \frac {1}{2}\|\nabla \Delta c\|_{L^2}^2+C\|\nabla u\|_{L^2}^4\|\nabla ^2 c\|_{L^2}^2. \end{aligned} \end{equation*}

Here $S_{f,\chi }$ is given by (4.5). Thus, it follows that

\begin{equation*} \begin{aligned} &\frac {d}{dt}\|\Delta c\| _{L^2 }^2+ \|\nabla \Delta c\| _{L^2 }^2\\[5pt] &\leq CS_{f,\chi }^2\|\Delta n\|_{L^2}^2+C S_{f,\chi }^2\|\nabla c\|_{L^2}^4 \|\nabla n\|_{L^2}^2+C\|\nabla u\|_{L^2}^4\|\nabla ^2 c\|_{L^2}^2, \end{aligned} \end{equation*}

which, together with (4.8) and (4.11), leads to

(4.25) \begin{equation} \begin{aligned} \sup _{\tau \in (0,t]}\|\nabla ^2c\|_{L^2}^2+\int _0^t \|\nabla ^3c\|_{L^2}^2\, d\tau &\leq \|\nabla ^2c_0\|_{L^2}^2+C S_{f,\chi }^2\int _{0}^{t}\|\nabla ^2 n\|_{L^2}^2\,d\tau \\[5pt] &\quad +C S_{f,\chi }^2\sup _{\tau \in (0,t]}\|\nabla c\|_{L^2}^4\int _{0}^{t}\|\nabla n\|_{L^2}^2\,d\tau \\[5pt] &\quad +C\sup _{\tau \in (0,t]}\|\nabla u\|_{L^2}^4\int _{0}^{t}\|\nabla ^2 c\|_{L^2}^2\,d\tau \\[5pt] &\leq \|\nabla ^2c_0\|_{L^2}^2+CS_{f,\chi }^2\mathcal{C}_{2,\phi, f,\chi }(\mathcal{E}_0^{\frac {1}{2}}\mathcal{E}_2^{\frac {1}{2}}+\mathcal{E}_0^3+\mathcal{E}_0+\mathcal{E}_0^2\mathcal{E}_2)\\[5pt] &\quad +CS_{f,\chi }^2 \mathcal{C}_{2,\phi, f,\chi }^2 C_{1,\phi }(\mathcal{E}_0^{\frac {1}{2}}\mathcal{E}_2^{\frac {1}{2}}+\mathcal{E}_0^3+\mathcal{E}_0+\mathcal{E}_0^2\mathcal{E}_2)^2\mathcal{E}_{0}\\[5pt] &\quad +C\mathcal{C}_{2,\phi, f,\chi }^3 (\mathcal{E}_0^{\frac {1}{2}}\mathcal{E}_2^{\frac {1}{2}}+\mathcal{E}_0^3+\mathcal{E}_0+\mathcal{E}_0^2\mathcal{E}_2)^3. \end{aligned} \end{equation}

Then, applying $\Delta u$ to (1.1) $_3$ and then taking the inner product of the resulting equation by $\Delta u$ , we have

\begin{equation*} \begin{aligned} \frac {1}{2}\frac {d}{dt}&\|\Delta u\|_{L^2}^2+\|\Lambda ^{\alpha } \Delta u\|_{L^2}^2\\[5pt] &= \int _{\mathbb{R}^3}\nabla \Delta u \cdot \nabla ((u \cdot \nabla ) u)\,dx-\int _{\mathbb{R}^3}\Delta u \cdot \Delta (n\nabla \phi )\,dx. \end{aligned} \end{equation*}

Similarly to (3.27), one can show

\begin{equation*} \begin{aligned} &\left |\int _{\mathbb{R}^3}\nabla \Delta u \cdot \nabla ((u \cdot \nabla ) u)\,dx\right |\\[5pt] &\quad \leq 3 \|\nabla u\| _{L^2 } \|\nabla ^2 u\| _{L^2 }^{2-\frac {3}{2\alpha }} \|\Lambda ^{2+\alpha } u\| _{L^2 }^{\frac {3}{2\alpha }}\\[5pt] &\quad \leq \frac {1}{8}\|\nabla ^{2}\Lambda ^{\alpha } u\|_{L^2}^2+ C\|\nabla u\|_{L^2}^{\frac {4\alpha }{4\alpha -3}} \|\nabla ^2 u\|_{L^2}^2. \end{aligned} \end{equation*}

Moreover, due to $\alpha \lt 2\lt 2+\alpha$ , one deduces from (2.2) and (2.8) that

\begin{equation*} \begin{aligned} &\left |\int _{\mathbb{R}^3} \Delta u \cdot \Delta (n\nabla \phi )\,dx\right |\\[5pt] &\quad \leq \|\Delta u\|_{L^2}\left (\|\nabla ^2 n\|_{L^2}\|\nabla \phi \|_{L^{\infty }}+2\|\nabla n\|_{L^2}\|\nabla ^2\phi \|_{L^{\infty }}+\Big \|\frac {n}{|x|}\Big \|_{L^{2}}||x|\nabla ^3 \phi \|_{L^{\infty }}\right ) \\[5pt] &\quad \leq \frac {1}{8}\|\Lambda ^{\alpha } \Delta u\|_{L^2}^2+C\|\Lambda ^{\alpha } u\|_{L^2}^2\\[5pt] &\quad \quad +C\|\nabla \phi \|_{L^{\infty }}^2\|\nabla ^2 n\|_{L^2}^2+C(\|\nabla ^2\phi \|_{L^{\infty }}^2+\||x|\nabla ^3 \phi \|_{L^{\infty }}^2)\|\nabla n\|_{L^2}^2. \end{aligned} \end{equation*}

Therefore, we get

\begin{equation*} \begin{aligned} \frac {d}{dt}&\|\Delta u\|_{L^2}^2+\|\Lambda ^{\alpha }\Delta u\|_{L^2}^2\\[5pt] &\leq C \|\nabla u\|_{L^2}^{\frac {4\alpha }{4\alpha -3}} \|\nabla ^2 u\|_{L^2}^2+C\|\Lambda ^{\alpha } u\|_{L^2}^2\\[5pt] &\quad +C(\|\nabla ^2\phi \|_{L^{\infty }}^2+\||x|\nabla ^3 \phi \|_{L^{\infty }}^2)\|\nabla n\|_{L^2}^2+C\|\nabla \phi \|_{L^{\infty }}^2\|\nabla ^2 n\|_{L^2}^2, \end{aligned} \end{equation*}

which, together with Grönwall’s inequality, (4.8), (4.11) and (4.6) with $M_0$ satisfying

(4.26) \begin{align} M_0\leq 1, \end{align}

gives rise to

(4.27) \begin{equation} \begin{aligned} &\sup _{\tau \in (0,t]}\|\nabla ^2 u\|_{L^2}^2+\int _{0}^{t}\|\nabla ^2\Lambda ^{\alpha }u\|_{L^2}^2\,d\tau \\[5pt] &\quad \leq Ce^{\int _{0}^{t}\|\nabla u\|_{L^2}^{\frac {4\alpha }{4\alpha -3}} \,d\tau }\Big ( \|\nabla ^2 u_0\|_{L^2}^2+\int _{0}^{t}\|\Lambda ^{\alpha } u\|_{L^2}^2\,d\tau \\[5pt] &\quad \quad +(\|\nabla ^2\phi \|_{L^{\infty }}^2+\||x|\nabla ^3 \phi \|_{L^{\infty }}^2)\int _{0}^{t}\|\nabla n\|_{L^2}^2\,d\tau \\[5pt] &\quad \quad +\|\nabla \phi \|_{L^{\infty }}^2\int _{0}^{t}\|\nabla ^2 n\|_{L^2}^2\,d\tau \Big )\\[5pt] &\quad \leq C\bigg (\|\nabla ^2u_0\|_{L^2}^2+(1+\|\nabla ^2\phi \|_{L^{\infty }}^2+\| |x|\nabla ^3\phi \|_{L^{\infty }}^2)C_{1,\phi }\mathcal{E}_0\\[5pt] &\quad \quad +\|\nabla \phi \|_{L^{\infty }}^2 \mathcal{C}_{2,\phi, f,\chi } (\mathcal{E}_0^{\frac {1}{2}}\mathcal{E}_2^{\frac {1}{2}}+\mathcal{E}_0^3+\mathcal{E}_0+\mathcal{E}_0^2\mathcal{E}_2) \bigg ). \end{aligned} \end{equation}

Due to (4.25) and (4.27), (4.23) follows.

Finally, it suffices to have the higher estimates of $n$ . By (1.1) $_1$ , (3.32) and Gagliardo–Nirenberg–Sobolev inequalities, one has

\begin{equation*} \begin{aligned} &\frac {1}{2}\frac {d}{dt} \|\Delta n\| _{L^2 }^2+ \frac {1}{2}\|\nabla \Delta n\| _{L^2 }^2\\[5pt] &\leq \|\nabla ( u \cdot \nabla n+ \nabla \cdot (\chi (c) n \nabla c))\|_{L^2}^2\\[5pt] &\leq C\|\nabla u\|_{L^3}^2\|\nabla n\|_{L^6}^2+C\|u\|_{L^{\infty }}^2\|\nabla ^2 n\|_{L^2}^2\\[5pt] &\quad +C S_{f,\chi }^2\Big ( \|n\|_{L^{\infty }}^2\|\nabla c\|_{L^6}^6+\|n\|_{L^6}^2\|\nabla ^2 c\|_{L^6}^2\|\nabla c\|_{L^6}^2+\|\nabla c\|_{L^6}^4\|\nabla n\|_{L^6}^2\Big )\\[5pt] &\quad +C S_{f,\chi }^2\Big ( \|\nabla ^2 n\|_{L^3}^2\|\nabla c\|_{L^{6}}^2+\|\nabla n\|_{L^3}^2\|\nabla ^2 c\|_{L^6}^2+\|n\|_{L^{\infty }}^2\|\nabla ^3c\|_{L^2}^2\Big )\\[5pt] &\leq \frac {1}{4} \|\nabla \Delta n\|_{L^2}^2+CS_{f,\chi }^4\| \nabla ^2 c\|_{L^2}^4 \|\nabla ^2 n\|_{L^2}^2+C\|\nabla u\|_{L^2}\|\nabla ^2 u\|_{L^2} \|\nabla ^2 n\|_{L^2}^2\\[5pt] &\quad +C S_{f,\chi }^2\Big ( \|\nabla n\|_{L^2} \|\nabla ^2 n\|_{L^2}\|\nabla ^2 c\|_{L^2}^6+\|\nabla n\|_{L^2}^2\|\nabla ^2 c\|_{L^2}^2\|\nabla ^3 c\|_{L^2}^2+\|\nabla ^2 c\|_{L^2}^4\|\nabla ^2 n\|_{L^2}^2\Big )\\[5pt] &\quad +C S_{f,\chi }^2\|\nabla n\|_{L^2} \|\nabla ^2 n\|_{L^2} \|\nabla ^3 c\|_{L^2}^2. \end{aligned} \end{equation*}

Using Grönwall’s inequality, we end up with

\begin{equation*} \begin{aligned} &\sup _{\tau \in (0,t]}\|\nabla ^2 n\| _{L^2 }^2+\frac {1}{2}\int _{0}^{t}\|\nabla ^3 n\| _{L^2 }^2\,d\tau \\[5pt] &\quad \leq e^{CS_{f,\chi }^4\sup \limits _{\tau \in (0,t]}\|\nabla ^2c\|_{L^2}^2\int _0^t\|\nabla ^2c\|_{L^2}^2\,d\tau }\bigg (\|\nabla ^2n_0\|_{L^2}^2+C\sup _{\tau \in (0,t]}\|\nabla u\|_{L^2}\sup _{\tau \in (0,t]}\|\nabla ^2u\|_{L^2}\int _{0}^t\|\nabla ^2n\|_{L^2}^2\,d\tau \\[5pt] &\quad +C S_{f,\chi }^2 \sup _{\tau \in (0,t]}\|\nabla ^2 c\|_{L^2}^6 \left (\int _0^t \|\nabla n\|_{L^2}^2\,d\tau \right )^{\frac {1}{2}}\left (\int _0^t \|\nabla ^2 n\|_{L^2}^2\,d\tau \right )^{\frac {1}{2}}\\[5pt] &\quad +C S_{f,\chi }^2 \sup _{\tau \in (0,t]}\|\nabla n\|_{L^2}^2 \sup _{\tau \in (0,t]}\|\nabla ^2 c\|_{L^2}^2 \int _0^t \|\nabla ^3 c\|_{L^2}^2\,d\tau \\[5pt] &\quad +CS_{f,\chi }^2\sup _{\tau \in (0,t]} \|\nabla ^2c\|_{L^2}^4 \int _0^t \|\nabla ^2 n\|_{L^2}^2\,d\tau \\[5pt] &\quad +CS_{f,\chi }^2 \sup _{\tau \in (0,t]} \|\nabla n\|_{L^2}\left (\int _{0}^{t}\|\nabla ^2 n\|_{L^2}^2\,dt\right )^{\frac {1}{2}}\int _{0}^{t}\|\nabla ^3 c\|_{L^2}^2\,dt\bigg ). \end{aligned} \end{equation*}

Recalling $\mathcal{E}_{0}\leq 1$ and combining the above estimate with (4.11), (4.25) and (4.27), we obtain the desired estimates of $n$ in (4.24) and thus complete the proof of Lemma4.4.

Proof of Proposition 4.1: According to Theorem1.2, the Cauchy problem (1.1)–(1.2) has a unique strong solution $(n,c,u)$ on $[0,T_*)$ , where $T_*\gt 0$ is a maximal existence time. Let $M_{0}$ be a generic constant satisfying (4.9), (4.13), (4.16), (4.22) and (4.26).

By virtue of Lemmas4.2-4.3 and the Gagliardo–Nirenberg inequality in Lemma2.2, there holds that

\begin{equation*} \begin{aligned} \|u\|_{L^3}^2&\leq C\|u\|_{L^2}\|\nabla u\|_{L^2}\leq CC_{1,\phi }^{\frac {1}{2}}C_{2,\phi, f,\chi }^{\frac {1}{2}} (1+\mathcal{E}_2)\mathcal{E}_0^{\frac {3}{4}}\leq \frac {1}{6}M_0, \end{aligned} \end{equation*}

provided that we let

(4.28) \begin{align} \mathcal{E}_0\leq \delta _1\;:\!=\;\min \Big \{1, \Big (6 C C_{1,\phi }^{\frac {1}{2}}C_{2,\phi }^{\frac {1}{2}}(1+\mathcal{E}_2)\Big )^{-\frac {4}{3}}M_0\Big \}. \end{align}

In accordance with Lemmas4.2 and 4.4 as well as the Gagliardo–Nirenberg inequality, we also have

\begin{equation*} \begin{aligned} S_{f,\chi }^2\|\nabla c\|_{L^3}^2&\leq CS_{f,\chi }^2\|c\|_{L^2}^{\frac {1}{4}}\|\nabla ^2 c\|_{L^2}^{\frac {3}{4}}\\[5pt] &\leq C S_{f,\chi }^2 C_{1,\phi }^{\frac {1}{4}} \Big (\mathcal{C}_{3,\phi, f,\chi }(1+\mathcal{E}_2^{3})\Big )^{\frac {3}{4}}\mathcal{E}_0^{\frac {1}{4}}\leq \frac {1}{6}M_0, \end{aligned} \end{equation*}

as long as

(4.29) \begin{align} \mathcal{E}_0\leq \delta _2\;:\!=\;\min \left \{1,\Big (6C S_{f,\chi }^2 C_{1,\phi }^{\frac {1}{4}}\Big (\mathcal{C}_{3,\phi, f,\chi }(1+\mathcal{E}_2^{3}))\Big )^{\frac {3}{4}}\Big )^{-4}M_0^{4}\right \}. \end{align}

To enclose the condition (4.6), it suffices to justify

\begin{equation*} \int _{0}^{t}\|\nabla u\|_{L^2}^{\frac {4\alpha }{4\alpha -3}}\, d\tau \leq \frac {1}{6}M_0. \end{equation*}

In the case $1\leq \alpha \lt \frac {5}{4}$ , we observe that $\frac {4\alpha }{4\alpha -3}-2\alpha =\frac {2\alpha (5-4\alpha )}{4\alpha -3}\gt 0$ . We thence take advantage of the interpolation inequality (2.2) and combine (4.8) and (4.11) together to obtain

\begin{equation*} \begin{aligned} \int _{0}^{t}\|\nabla u\|_{L^2}^{\frac {4\alpha }{4\alpha -3}}\, d\tau &\leq C\sup _{\tau \in (0, t)}\|\nabla u\|_{L^2}^{\frac {2\alpha (5-4\alpha )}{4\alpha -3}}\int _{0}^{t} \|\nabla u\|_{L^2}^{2\alpha }\,d\tau \\[5pt] &\leq \sup _{\tau \in (0,t)}\|\nabla u\|_{L^2}^{\frac {2\alpha (5-4\alpha )}{4\alpha -3} } \sup _{\tau \in (0,t)}\|u\|_{L^2}^{2(\alpha -1)}\int _{0}^{t} \|\Lambda ^{\alpha }u\|_{L^2}^{2}\,d\tau \\[5pt] &\leq C\Big ( \mathcal{C}_{2,\phi, f,\chi } ( 1+\mathcal{E}_2) \Big )^{\frac {\alpha (5-4\alpha )}{4\alpha -3} } \Big (\mathcal{C}_{1,\phi }\mathcal{E}_0\Big )^{\alpha }\\[5pt] &\leq C\mathcal{C}_{1,\phi }^{\alpha }\mathcal{C}_{2,\phi, f,\chi }^{\frac {\alpha (5-4\alpha )}{4\alpha -3} } (1+\mathcal{E}_2)^{\frac {\alpha (5-4\alpha )}{4\alpha -3} } \mathcal{E}_0^{\alpha }\\[5pt] &\leq \frac {1}{6}M_0, \end{aligned} \end{equation*}

if for $1\leq \alpha \lt \frac {5}{4}$ we take

(4.30) \begin{equation} \begin{aligned} &\mathcal{E}_0\leq \delta _3=\min \Big \{ 1, \Big ( 6C\mathcal{C}_{1,\phi }^{\alpha }\mathcal{C}_{2,\phi, f,\chi }^{\frac {\alpha (5-4\alpha )}{4\alpha -3} } (1+\mathcal{E}_2)^{\frac {\alpha (5-4\alpha )}{4\alpha -3} } \Big )^{-\frac {1}{\alpha }}M_0 \Big \}. \end{aligned} \end{equation}

As for the case $\frac {3}{4}\lt \alpha \lt 1$ , one has $\frac {4\alpha }{4\alpha -3}-2(1+\alpha )=\frac {2(3+\alpha -4\alpha ^2)}{4\alpha -3}\gt 0$ . We thus conclude that

\begin{equation*} \begin{aligned} \int _{0}^{t}\|\nabla u\|_{L^2}^{\frac {4\alpha }{4\alpha -3}}\,d\tau &\leq C\sup _{\tau \in (0,t)}\|\nabla u\|_{L^2}^{\frac {2(3+\alpha -4\alpha ^2)}{4\alpha -3}}\int _{0}^{t} \|\nabla u\|_{L^2}^{2(1+\alpha )}\,d\tau \\[5pt] &\leq C\sup _{\tau \in (0,t)}\|\nabla u\|_{L^2}^{\frac {2(3+\alpha -4\alpha ^2)}{4\alpha -3}}\sup _{\tau \in (0,t)}\|u\|_{L^2}^{2\alpha }\int _{0}^{t} \|\Lambda ^{1+\alpha } u\|_{L^2}^{2}\,d\tau \\[5pt] &\leq C \Big ( \mathcal{C}_{2,\phi, f,\chi } (1+\mathcal{E}_2) \Big )^{\frac {(3+\alpha -4\alpha ^2)}{4\alpha -3}+1}\Big (\mathcal{C}_{1,\phi }\mathcal{E}_0\Big )^{\alpha }\\[5pt] &\leq C \mathcal{C}_{1,\phi }^{\alpha }\mathcal{C}_{2,\phi, f,\chi }^{\frac {\alpha (5-4\alpha )}{4\alpha -3} } (1+\mathcal{E}_2)^{\frac {\alpha (5-4\alpha )}{4\alpha -3} } \mathcal{E}_0^{\alpha }\\[5pt] &\leq \frac {1}{6}M_0, \end{aligned} \end{equation*}

when

(4.31) \begin{equation} \begin{aligned} &\mathcal{E}_0\leq \delta _3\;:\!=\;\min \Big \{ 1, \Big ( 6 C \mathcal{C}_{1,\phi }^{\alpha }\mathcal{C}_{2,\phi, f,\chi }^{\frac {\alpha (5-4\alpha )}{4\alpha -3} } (1+\mathcal{E}_2)^{\frac {\alpha (5-4\alpha )}{4\alpha -3} } \Big )^{-\frac {1}{\alpha }}M_0 \Big \}. \end{aligned} \end{equation}

Based on the above estimates, we choose

\begin{equation*} \mathcal{E}_0\leq \delta _0\;:\!=\;\min \{\delta _1,\delta _2,\delta _3\}. \end{equation*}

Thus, the estimate (4.7) is true for all $[0,T_*)$ , and therefore the bounds in Lemma4.2–4.3 are valid. A standard continuity argument implies $T_*=\infty$ and that $(n,c,u)$ is indeed a global strong solution to the Cauchy problem (1.1)–(1.2) satisfying the uniform estimates (1.13). To complete the proof of Theorem1.5, we will establish the decay estimates (1.14) in the next subsection.

4.3. Proof of Theorem1.5: Time-decay estimates

This subsection is devoted to the time-decay part in Theorem1.5. We consider the global solution $(n,c,u)$ to the Cauchy problem (1.1)–(1.2) given by Subsection 4.2.

Under the conditions $n_0,u_0\in L^1$ , integrating (1.1) $_1$ and (1.1) $_2$ over $[0,t]\times \mathbb{R}^3$ and using (3.3), we obtain

(4.32) \begin{align} \|n\|_{L^1}=\|n_0\|_{L^1},\quad \quad \|c\|_{L^1}=\|c_0\|_{L^1},\quad \quad t\geq 0. \end{align}

Now we show the $L^p$ -decay estimates of $n$ and $c$ . For all $2\leq p\lt \infty$ , we perform the $L^p$ -energy estimates for (1.1) $_1$ as follows

\begin{equation*} \begin{aligned} \frac {d}{dt}&\| n\|_{L^p}^p+\frac {4(p-1)}{p}\|\nabla n^{\frac {p}{2}}\|_{L^2}^2\\[5pt] &\leq p(p-1)\int _{\mathbb{R}^3} \chi (c)\nabla c n^{p-1}|\nabla n| \, dx\\[5pt] &\leq 2(p-1) \sup _{0\leq c\leq \|c_0\|_{L^\infty }} |\chi (c)| \|\nabla c\|_{L^3}\|n^{\frac {p}{2}}\|_{L^6}\|\nabla n^{\frac {p}{2}}\|_{L^2}\\[5pt] &\leq C \|\nabla c\|_{L^3}\|\nabla n^{\frac {p}{2}}\|_{L^2}^2. \end{aligned} \end{equation*}

Since $\|\nabla c\|_{L^3}$ is suitably small due to (4.8), (4.23) and

(4.33) \begin{align} \|\nabla c\|_{L^3}&\leq C\|c\|_{L^2}^{\frac {1}{2}}\|\nabla ^2 c\|_{L^2}^{\frac {1}{2}}\lt \lt 1, \end{align}

it follows that

(4.34) \begin{equation} \begin{aligned} &\frac {d}{dt}\| n\|_{L^p}^p+\frac {2(p-1)}{p}\|\nabla n^{\frac {p}{2}}\|_{L^2}^2\leq 0. \end{aligned} \end{equation}

In view of (4.32) and the Gagliardo–Nirenberg inequality (2.2), we have

(4.35) \begin{equation} \begin{aligned} \| n\|_{L^p}^{p}= \|n^{\frac {p}{2}}\|_{L^{2}}^2\leq C\|n^{\frac {p}{2}}\|_{L^{\frac {2}{p}}}^{\frac {4}{3p-1}} \|\nabla n^{\frac {p}{2}}\|_{L^2}^{\frac {6(p-1)}{3p-1}}\leq C\|n_{0}\|_{L^1}^{\frac {8}{p(3p-1)}}\|\nabla n^{\frac {p}{2}}\|_{L^2}^{\frac {6(p-1)}{3p-1}}. \end{aligned} \end{equation}

Putting (4.35) into (4.34) leads to the differential inequality

\begin{equation*} \begin{aligned} &\frac {d}{dt}\|n\|_{L^p}^p+C\Big (\|n\|_{L^p}^{p}\Big )^{1+\frac {2}{3(p-1)}} \leq 0, \end{aligned} \end{equation*}

from which we infer

(4.36) \begin{equation} \begin{aligned} \| n\|_{L^p}\leq C(1+ t)^{-\frac {3}{2}(1-\frac {1}{p})}\|n_{0}\|_{L^1},\quad \quad 2\leq p\lt \infty . \end{aligned} \end{equation}

In addition, using Hölder’s inequality gives

\begin{equation*} \begin{aligned} \| n\|_{L^p}\leq \|n\|_{L^1}^{\frac {2}{p}-1} \|n\|_{L^2}^{2-\frac {2}{p}}\leq C(1+t)^{-\frac {3}{2}(1-\frac {1}{p})},\quad \quad 1\lt p\lt 2. \end{aligned} \end{equation*}

A similar argument yields

(4.37) \begin{equation} \begin{aligned} \| c\|_{L^p}\leq C(1+t)^{-\frac {3}{2}(1-\frac {1}{p})},\quad \quad 1\lt p\lt \infty . \end{aligned} \end{equation}

The details are omitted.

Next, we are going to establish the higher-order decay estimates of $n$ and $c$ . Multiplying (1.1) $_1$ by $\Delta n$ and integrating it by parts yields

\begin{equation*} \begin{aligned} \frac {1}{2}\frac {d}{dt}\|\nabla n\|_{L^2}^2+\|\Delta n\|_{L^2}^2 \leq \|u\cdot \nabla n\|_{L^2}^2+\|\nabla \cdot (\chi (c)n\nabla c)\|_{L^2}^2+\frac {1}{2}\|\Delta n\|_{L^2}^2, \end{aligned} \end{equation*}

where the nonlinear terms are analysed as

\begin{equation*} \begin{aligned} \|u\cdot \nabla n\|_{L^2}^2\leq \|u\|_{L^3}^2\|\nabla n\|_{L^6}^2\leq C\|u\|_{L^3}^2\|\Delta n\|_{L^2}^2, \end{aligned} \end{equation*}

and

\begin{equation*} \begin{aligned} \|\nabla \cdot (\chi (c)n\nabla c)\|_{L^2}^2&\leq C\|n\|_{L^6}^2\|\nabla c\|_{L^6}^4+C\|\nabla c\|_{L^3}^2\|\nabla n\|_{L^6}^2+C\|n\|_{L^{\infty }}^2\|\nabla ^2c\|_{L^2}^2\\[5pt] &\leq C(\|n\|_{L^6}^2\|\nabla ^2 c\|_{L^2}^2+\|n\|_{L^6}\|\nabla ^2n\|_{L^2})\|\Delta c\|_{L^2}^2+C\|\nabla c\|_{L^3}^2 \|\Delta n\|_{L^2}^2. \end{aligned} \end{equation*}

Hence, it follows that

(4.38) \begin{equation} \begin{aligned} &\frac {d}{dt}\|\nabla n\|_{L^2}^2+(1-C\|(\nabla c,u)\|_{L^3}^2)\|\Delta n\|_{L^2}^2\\[5pt] &\leq C(\|n\|_{L^6}^2\|\nabla ^2 c\|_{L^2}^2+\|n\|_{L^6}\|\nabla ^2n\|_{L^2}) \|\Delta c\|_{L^2}^2. \end{aligned} \end{equation}

This requires an estimate of $c$ at the $\dot {H}^1$ -regularity level. To achieve it, one deduces from (1.1) $_2$ that

\begin{equation*} \begin{aligned} \frac {1}{2}\frac {d}{dt}\|\nabla c\|_{L^2}^2+\|\Delta c\|_{L^2}^2 \leq \|u\cdot \nabla c\|_{L^2}^2-\int _{\mathbb{R}^3} n f(c) \Delta c \, dx+\frac {1}{2}\|\Delta c\|_{L^2}^2. \end{aligned} \end{equation*}

The first nonlinear term has been estimated in (4.12). We recall that $f(c)=\tilde {f}(c)c$ with $\tilde {f}(c)=\int _{0}^{1}f'(\theta c)\,d \theta$ due to $f(0)=0$ . Therefore, the second nonlinear term can be handled by

\begin{equation*} \begin{aligned} \left |\int n f(c) \Delta c\, dx\right |&\leq \left | \int _{\mathbb{R}^3} \tilde {f}(c)c\nabla n \cdot \nabla c\, dx\right |+\left |\int _{\mathbb{R}^3} n f'(c)\nabla c\cdot \nabla c \, dx\right |\\[5pt] &\leq C\|\nabla n\|_{L^6}\|c\|_{L^{\frac {3}{2}}}\|\nabla c\|_{L^6}+C\|n\|_{L^{\frac {3}{2}}}\|\nabla c\|_{L^{6}}^2\\[5pt] &\leq C\|(n,c)\|_{L^{\frac {3}{2}}}\|\Delta c\|_{L^2}^2+C\|c\|_{L^{\frac {3}{2}}}\|\Delta n\|_{L^2}^2. \end{aligned} \end{equation*}

It thus holds that

(4.39) \begin{equation} \begin{aligned} \frac {d}{dt}\|\nabla c\|_{L^2}^2+(1-C\|(n,c)\|_{L^{\frac {3}{2}}}-C\|u\|_{L^3}^2)\|\Delta c\|_{L^2}^2\leq C\|c\|_{L^{\frac {3}{2}}}\|\Delta n\|_{L^2}^2. \end{aligned} \end{equation}

Adding (4.38) and (4.39) together, we get

\begin{equation*} \begin{aligned} &\frac {d}{dt}\|(\nabla n,\nabla c)\|_{L^2}^2\\[5pt] &\quad +\left (1-C\|(n,c)\|_{L^{\frac {3}{2}}}-C\|(\nabla c,u)\|_{L^3}^2-C\|n\|_{L^6}^2\|\nabla ^2 c\|_{L^2}^2-C\|n\|_{L^6}\|\nabla ^2n\|_{L^2}\right )\|(\Delta n,\Delta c)\|_{L^2}^2\leq 0. \end{aligned} \end{equation*}

Recall that in Lemma4.2, the $L^2$ -norm of $\|(n, c, u)\|_{L^2}^2$ can be bounded by $C_{1, \phi }\mathcal{E}_0$ which can be suitably small, while Lemma4.4 implies that the norms $\|\nabla ^2n\|_{L^2}$ and $\|\nabla ^2c\|_{L^2}$ are uniformly bounded in time. Hence, we have (4.33) and

\begin{equation*} \begin{aligned} \|(n,c)\|_{L^{\frac {3}{2}}}&\leq C\|(n,c)\|_{L^1}^{\frac {1}{3}}\|(n,c)\|_{L^2}^{\frac {2}{3}}\lt \lt 1,\\[5pt] \|n\|_{L^6}&\leq C\|n\|_{L^2}^{\frac {1}{2}} \|\nabla ^2n\|_{L^2}^{\frac {1}{2}}\lt \lt 1,\\[5pt] \|u\|_{L^3}&\leq C\|u\|_{L^2}^{\frac {3}{4}}\|\nabla ^2 u\|_{L^2}^{\frac {1}{4}}\lt \lt 1. \end{aligned} \end{equation*}

Therefore, the following energy inequality holds:

\begin{equation*} \begin{aligned} &\frac {d}{dt}\|(\nabla c,\nabla n)\|_{L^2}^2+\|(\Delta c,\Delta n)\|_{L^2}^2\leq 0. \end{aligned} \end{equation*}

The Gagliardo–Nirenberg inequality (2.2) and (4.32) ensure that

\begin{equation*} \begin{aligned} \|(\nabla c,\nabla n)\|_{L^2}\leq C \|(c,n)\|_{L^1}^{\frac {2}{7}} \|(\Delta c,\Delta n)\|_{L^2}^{\frac {5}{7}}\leq C \|(\Delta c,\Delta n)\|_{L^2}^{\frac {5}{7}}. \end{aligned} \end{equation*}

Thus, we get

\begin{equation*} \begin{aligned} &\frac {d}{dt}\|(\nabla c,\nabla n)\|_{L^2}^2+C\Big (\|(\nabla c,\nabla n)\|_{L^2}^2 \Big )^{1+\frac {2}{5}}\leq 0. \end{aligned} \end{equation*}

Solving this differential inequality, we have the decay of $(\nabla c,\nabla n)$ as follows

\begin{equation*} \begin{aligned} \|(\nabla c,\nabla n)\|_{L^2}\leq C(1+t)^{-\frac {5}{4}}. \end{aligned} \end{equation*}

We finally deal with the estimates of $u$ . Taking the inner product of (1.1) $_3$ with $\Lambda ^{2\alpha }u$ leads to

(4.40) \begin{equation} \begin{aligned} \frac {1}{2}\frac {d}{dt}\|\Lambda ^{\alpha }u\|_{L^2}^2+\|\Lambda ^{2\alpha }u\|_{L^2}^2 \leq \frac {1}{2}\|\Lambda ^{2\alpha }u\|_{L^2}^2+\|u\cdot \nabla u\|_{L^2}^2+\|n\nabla \phi \|_{L^2}^2. \end{aligned} \end{equation}

Multiplying (4.40) by $t$ , we arrive at

\begin{equation*} \begin{aligned} \frac {d}{dt}\Big (t\|\Lambda ^{\alpha }u\|_{L^2}^2\Big )+t\|\Lambda ^{2\alpha }u\|_{L^2}^2\leq 2t \|u\cdot \nabla u\|_{L^2}^2+2t\|n\nabla \phi \|_{L^2}^2. \end{aligned} \end{equation*}

By virtue of the product law (2.7) with $s_1=\frac {3}{2}-\alpha$ , $s_{2}=\alpha$ and $\alpha \lt \frac {5}{2}-\alpha \lt 1+\alpha$ , it holds that

\begin{equation*} \begin{aligned} 2 t \|u\cdot \nabla u\|_{L^2}^2\leq C\|u\|_{\dot {H}^{\frac {5}{2}-\alpha }}^2 t\|\Lambda ^{\alpha }u\|_{L^2}^2\leq C\|u\|_{\dot {H}^{\alpha }\cap \dot {H}^{1+\alpha }}^2 t\|\Lambda ^{\alpha }u\|_{L^2}^2. \end{aligned} \end{equation*}

And Hardy’s inequality in Lemma2.4 guarantees that

\begin{equation*} \begin{aligned} 2 t\|n\nabla \phi \|_{L^2}^2\leq 2\||x|\nabla \phi \|_{L^{\infty }}^2 t\Big \|\frac {n}{|x|}\Big \|_{L^2}^2\leq C t\|\nabla n\|_{L^2}^2\leq C(1+t)^{-\frac {3}{2}}. \end{aligned} \end{equation*}

Hence, it follows that

\begin{equation*} \begin{aligned} \frac {d}{dt}\Big (t\|\Lambda ^{\alpha }u\|_{L^2}^2\Big )+t\|\Lambda ^{2\alpha }u\|_{L^2}^2\leq C\|u\|_{\dot {H}^{\alpha }\cap \dot {H}^{1+\alpha }}^2 t\|\Lambda ^{\alpha }u\|_{L^2}^2+C(1+t)^{-\frac {3}{2}}. \end{aligned} \end{equation*}

Employing Grönwall’s inequality yields

(4.41) \begin{equation} \begin{aligned} &t\|\Lambda ^{\alpha }u\|_{L^2}^2+\int _{0}^{t}\tau \|\Lambda ^{2\alpha }u\|_{L^2}^2\,d\tau \leq Ce^{\int _{0}^{t}\|u\|_{\dot {H}^{\alpha }\cap \dot {H}^{1+\alpha }}^2 \,d\tau }\int _{0}^{t}(1+\tau )^{-\frac {3}{2}}d\tau \leq C. \end{aligned} \end{equation}

which, together with Sobolev’s embedding theorem, leads to

\begin{equation*} \begin{aligned} &\|u\|_{L^{\frac {6}{3-2\alpha }}}\leq \|\Lambda ^{\alpha }u\|_{L^2}\leq C(1+t)^{-\frac {1}{2}}. \end{aligned} \end{equation*}

Furthermore, one also has

(4.42) \begin{equation} \begin{aligned} &\|u\|_{L^p}\leq \|u\|_{L^2}^{1-\frac {3}{\alpha }(\frac {1}{2}-\frac {1}{p})} \|u\|_{L^{\frac {6}{3-2\alpha }}}^{\frac {3}{\alpha }(\frac {1}{2}-\frac {1}{p})}\leq C(1+t)^{-\frac {3}{2\alpha }(\frac {1}{2}-\frac {1}{p})},\quad 2\lt p\lt \frac {6}{3-2\alpha }. \end{aligned} \end{equation}

By (4.36)–(4.37) and (4.41)–(4.42), we conclude (1.14) and complete the proof of Theorem1.5.