On the global well-posedness of 3-D axi-symmetric Navier–Stokes system with small swirl component

Abstract

In this paper, we prove the local well-posedness of 3-D axi-symmetric Navier–Stokes system with initial data in the critical Lebesgue spaces. We also obtain the global well-posedness result with small initial data. Furthermore, with the initial swirl component of the velocity being sufficiently small in the almost critical spaces, we can still prove the global well-posedness of the system.

1 Introduction

In this paper, we investigate the well-posedness and long-time behavior of global solutions to 3D axisymmetric Navier–Stokes equations with a small swirl component. In general, 3-D Navier–Stokes system in $\mathop {\mathbb R}\nolimits ^3$ reads

$$\begin{aligned} \left\{ \begin{array}{l} \displaystyle \partial _t u+u\cdot \nabla u-\Delta u+\nabla p=0, \qquad (t,x)\in \mathop {\mathbb R}\nolimits ^+\times \mathop {\mathbb R}\nolimits ^3\\ \displaystyle \mathrm{div}\,u=0,\\ \displaystyle u|_{t=0} =u_0, \end{array} \right. \end{aligned}$$

(1.1)

where $u(t,x)=\left( u^1,u^2,u^3\right)$ stands for the velocity field and p the scalar pressure function of the fluid, which guarantees the divergence free condition of the velocity field. This system describes the motion of viscous incompressible fluid flows.

We recall that except the initial data with special structure, it is not known whether or not the system (1.1) has a unique global smooth solution with large smooth initial data. For instance, the system (1.1) is globally well-posed for data which is axisymmetric and without swirl component (that is the case when $u^\theta =0$ in (1.3) below). In this case, Ladyzhenskaya [7] and independently Ukhovskii and Yudovich [10] proved the existence of weak solutions along with the uniqueness and regularities of such solution for (1.1). Leonardi et al. [8] gave a refined proof of the same result in [7, 10]. And even with a small swirl component, the authors [11] could also establish the global well-posedess of (1.1). In general, even the global wellposedness of (1.1) with axisymmetric initial data is still open.

On the other hand, in the seminal paper [9], Leray proved the global existence of finite energy weak solutions to (1.1). Yet the uniqueness and regularity to this weak solution are big open questions in the field of mathematical fluid mechanics. Furthermore, Leray emphasized two facts about Navier–Stokes system. Firstly, he pointed out that energy estimate method is very important to study Navier–Stokes system. The general energy inequality for (1.1)

$$\begin{aligned} \frac{1}{2} \Vert v(t)\Vert _{L^2}^2 +\int _0^t\Vert \nabla v\left( t'\right) \Vert _{L^2}^2dt' = \frac{1}{2} \Vert v_0\Vert _{L^2}^2, \end{aligned}$$

is the cornerstone of the proof to the existence of global turbulent solution to (1.1) in [9]. The energy estimate relies (formally) on the fact that if v is a divergence free vector field, $(v\cdot \nabla f|f)_{L^2} =0~$ and that $(\nabla p|v)_{L^2}=0$. In the present work, we shall use the more general fact that for any divergence free vector field v and any function a, we have

$$\begin{aligned} \int _{\mathop {\mathbb R}\nolimits ^3} v(x) \cdot \nabla a(x) |a(x)|^{p-2}a(x)\, dx = 0\qquad \text{ for } \text{ any }\ \ p\in ]1,\infty [. \end{aligned}$$

This will lead to the  $L^p$ type energy estimate. Secondly Leray pointed out that the scaling invariance of (1.1), that is,

$$\begin{aligned} v(t,x)\mapsto \lambda v\left( \lambda ^2 t, \lambda x\right) \quad \hbox {and}\quad p(t,x)\mapsto \lambda ^2 p\left( \left( \lambda ^2 t, \lambda x\right) \right) , \end{aligned}$$

(1.2)

if (v, p) is a solution of (1.1) on $[0,T]\times \mathop {\mathbb R}\nolimits ^3$ associated with an initial data $v_0$, then $(v_\lambda ,p_\lambda )$ is also a solution of (1.1) on $[0,\lambda ^{-2}T]\times \mathop {\mathbb R}\nolimits ^3$ associated with the initial data $\lambda v_0(\lambda x),$ is another important fact in the study of Navier–Stokes system. The scaling property is also the foundation of the Kato theory which gives a general method to solve (locally or globally) the incompressible Navier–Stokes equation in critical spaces i.e. spaces with the norms of which are invariant under the scaling. In what follows, we shall use such scaling invariant space as $L^\infty (]0,t[; L^1(\Omega )),$ where the norm $L^1(\Omega )$ is given by (1.5).

In fact, Gally and $\breve{S}$verák [5] recently proved the global well-posedness of 3-D axisymmetric Navier–Stokes system without swirl and with initial data in the scaling invariant function spaces. We remark that the reason why one can prove the global well-posedness of (1.1) in this case is due to the $\theta$ component of the vorticity, $\omega ^\theta ,$ satisfies

$$\begin{aligned} \partial _t \frac{\omega ^\theta }{r}+\left( u^r\partial _r+u^z\partial _z\right) \frac{\omega ^\theta }{r}-\left( \Delta +\frac{2}{r}\partial _r\right) \frac{\omega ^\theta }{r} =0. \end{aligned}$$

The scaling invariant Lebesgue space for $\frac{\omega ^\theta }{r}$ is $L^\infty (]0,t[;L^1(\mathop {\mathbb R}\nolimits ^3))$. Motivated by [5], the purpose of this paper is to improve the norm for the initial data in [11] to be scaling invariant ones. We remark that the other motivation of this paper comes from [3] where the authors proved that one scaling invariant norm to one component of Navier–Stokes system controls the regularity of the solution. Yet we still do not know in general the global well-posedness of Navier–Stokes with one component being small in some scaling invariant space.

Now we restrict ourselves to the axisymmetric solutions of (1.1) with the following form

$$\begin{aligned} u(t,x)=u^r(t,r,z)e_r+u^\theta (t,r,z)e_\theta +u^z(t,r,z)e_z, \end{aligned}$$

where $(r,\theta ,z)$ denotes the usual cylindrical coordinates in $\mathop {\mathbb R}\nolimits ^3$ so that $x=(r\cos \theta ,r\sin \theta ,z)$, and

$$\begin{aligned} e_r=(\cos \theta ,\sin \theta ,0), e_\theta =(-\sin \theta ,\cos \theta ,0), e_z=(0,0,1), r=\sqrt{x_1^2+x_2^2}. \end{aligned}$$

Then in this case, we can reformulate (1.1) as

$$\begin{aligned} \left\{ \begin{array}{l} \displaystyle \partial _t u^r+\left( u^r\partial _r+u^z\partial _z\right) u^r-\left( \partial _r^2+\partial _z^2+\frac{1}{r}\partial _r-\frac{1}{r^2}\right) u^r-\frac{\left( u^\theta \right) ^2}{r}+\partial _r p=0,\\ \displaystyle \partial _t u^\theta +\left( u^r\partial _r+u^z\partial _z\right) u^\theta -\left( \partial _r^2+\partial _z^2+\frac{1}{r}\partial _r-\frac{1}{r^2}\right) u^\theta +\frac{u^r u^\theta }{r}=0,\\ \displaystyle \partial _t u^z+\left( u^r\partial _r+u^z\partial _z\right) u^z-\left( \partial _r^2+\partial _z^2+\frac{1}{r}\partial _r\right) u^z+\partial _z p=0,\\ \displaystyle \partial _r u^r+\frac{1}{r} u^r+\partial _z u^z=0,\\ \displaystyle u|_{t=0} =u_0. \end{array} \right. \end{aligned}$$

(1.3)

Let us denote $\widetilde{u}\buildrel \mathrm{def}\over =u^r e_r+u^z e_z.$ Then it is easy to check that

$$\begin{aligned} \mathrm{div}\,\widetilde{u}=0 \quad \hbox {and}\quad \mathrm{curl}\,\widetilde{u} =\omega ^\theta e_\theta , \end{aligned}$$

so that the Biot–Savart law shows that $u^r$ and $u^z$ can be uniquely determined by $\omega ^\theta$ (see Sect. 2.1). Hence we can write the System (1.3) as

$$\begin{aligned} \left\{ \begin{array}{l} \partial _t\omega ^\theta -\left( \partial _r^2+\partial _z^2+\frac{1}{r} \partial _r-\frac{1}{r^2}\right) \omega ^\theta =-\mathrm{div}\,_{*}\left( \widetilde{u}\omega ^\theta \right) +\frac{2u^\theta \partial _zu^\theta }{r},\\ \partial _tu^\theta -\left( \partial _r^2+\partial _z^2+\frac{1}{r} \partial _r-\frac{1}{r^2}\right) u^\theta =-\mathrm{div}\,_{*}\left( \widetilde{u}u^\theta \right) -\frac{2u^\theta u^r}{r},\\ \left( \omega ^\theta ,u^\theta \right) |_{t=0} =\left( \omega ^\theta _0, u^\theta _0\right) . \end{array} \right. \end{aligned}$$

(1.4)

Here and all in that follows, we always denote $\mathrm{div}\,_* f\buildrel \mathrm{def}\over =\partial _r f^r+\partial _z f^z$ and abuse the notation $\tilde{u}=(u^r,u^z).$

As in [5], we shall equip the half-plane $\Omega =\{(r,z)|r>0,z\in \mathop {\mathbb R}\nolimits \}$ with the 2D measure drdz, instead of the 3D measure rdrdz. For any $p\in [1,\infty [$, we denote by $L^p(\Omega )$ the space of measurable functions $f{:}\,\Omega \rightarrow \mathop {\mathbb R}\nolimits$ which verifies

$$\begin{aligned} \Vert f\Vert _{L^p(\Omega )}\buildrel \mathrm{def}\over =\left( \int _\Omega |f(r,z)|^p drdz\right) ^{\frac{1}{p}}<\infty ,\quad 1\le p<\infty . \end{aligned}$$

(1.5)

The space $L^\infty (\Omega )$ can be defined with the usual modification. Sometimes, we shall also use the 3D Lebesgue measure rdrdz, and the corresponding Lebesgue spaces are then denoted by $L^p(\mathop {\mathbb R}\nolimits ^3)$ or $L^p$ with norm

$$\begin{aligned} \Vert g\Vert _{L^p}\buildrel \mathrm{def}\over =\left( \int _\Omega |f(r,z)|^p rdrdz\right) ^{\frac{1}{p}},\ 1\le p<\infty . \end{aligned}$$

Our main results state as follows.

Theorem 1.1

For any initial data $\omega ^\theta _0\in L^1(\Omega )$ and $u^\theta _0\in L^2(\Omega )$ satisfying $r^{-\frac{3}{10}}u^\theta _0\in L^{\frac{20}{13}}(\Omega )$, there exists some $T(\omega ^\theta _0,u^\theta _0)$ such that the equations (1.4) have a unique mild solution

$$\begin{aligned} \begin{aligned}&\omega ^\theta \in C\left( [0,T];L^1(\Omega )\right) \bigcap C\left( ]0,T];L^\infty (\Omega )\right) ,\ u^\theta \in C\left( [0,T];L^2(\Omega )\right) \\&\quad \bigcap C\left( ]0,T];L^\infty (\Omega )\right) \quad \hbox {with}\quad \quad r^{-\frac{3}{10}}{u^\theta }\in C\left( [0,T];L^{\frac{20}{13}}(\Omega )\right) \bigcap C\left( ]0,T];L^\infty (\Omega )\right) . \end{aligned} \end{aligned}$$

(1.6)

Furthermore, the solution $(\omega ^\theta , u^\theta )$ verifies

• for any $p\in [1,\infty ],~q\in [2,\infty ]$ and $\kappa \in [{20}/{13},\infty ],$ there holds

  $$\begin{aligned} \begin{aligned}&L_p(T)\buildrel \mathrm{def}\over =\sup _{0\le t\le T}t^{1-\frac{1}{p}}\Vert \omega ^\theta (t)\Vert _{L^p(\Omega )}<\infty ,\quad M_q(T)\buildrel \mathrm{def}\over =\sup _{0\le t\le T}t^{\frac{1}{2}-\frac{1}{q}}\Vert u^\theta (t)\Vert _{L^q(\Omega )}<\infty ,\\&N_\kappa (T)\buildrel \mathrm{def}\over =\sup _{0\le t\le T}t^{\frac{13}{20}-\frac{1}{\kappa }}\Vert r^{-\frac{3}{10}}{u^\theta }(t)\Vert _{L^\kappa (\Omega )}<\infty . \end{aligned} \end{aligned}$$

  (1.7)

  Moreover, when $p\in ]1,\infty ],~q\in ]2,\infty ]$ and $\kappa \in ]{20}/{13},\infty ],$ we have

  $$\begin{aligned} \lim _{t\rightarrow 0}\left( L_p(t)+M_q(t)+N_\kappa (t)\right) =0; \end{aligned}$$

  (1.8)

• if

  $$\begin{aligned} \Vert \omega ^\theta _0\Vert _{L^1(\Omega )}+\Vert u^\theta _0\Vert _{L^2(\Omega )} +\Vert r^{-\frac{3}{10}}u^\theta _0\Vert _{L^{\frac{20}{13}}(\Omega )}\le c \end{aligned}$$

  (1.9)

  for some sufficiently small constant c,  then $T=\infty .$ And if $\Vert u^\theta _0\Vert _{L^2(\Omega )}+\Vert r^{-\frac{3}{10}}u^\theta _0\Vert _{L^{\frac{20}{13}}(\Omega )}$ is small enough, then the lifespan $T^\star$ of the solution depends only on $\omega ^\theta _0$.

Remark 1.1

• Let us remark that the norms $\Vert \omega ^\theta _0\Vert _{L^1(\Omega )},~\Vert u^\theta _0\Vert _{L^2(\Omega )}$ and $\Vert r^{-\frac{3}{10}}u^\theta _0\Vert _{L^{\frac{20}{13}}(\Omega )}$ are scaling invariant under the scaling transformation (1.2). Moreover, the method used here might be used to study axi-symmetric vortex ring for 3-D Navier–Stokes system with swirl (see the corresponding result of [4] for the case without swirl).

• The reason for requiring $r^{-\frac{3}{10}}u^\theta _0\in L^{\frac{20}{13}}(\Omega )$ is to handle the term $\frac{\partial _z|u^\theta (s)|^2}{r}$ in $\omega ^\theta$ equation of (3.1), so that the exponent $\frac{3}{2}-\frac{1}{p}+\frac{1}{5}$ appearing in (3.7) is less than 1.

Theorem 1.2

Let $\omega ^\theta _0$ and $u^\theta _0$ satisfy $\eta _0\buildrel \mathrm{def}\over =\frac{\omega ^\theta _0}{r}\in L^1, ~U_0\buildrel \mathrm{def}\over =\frac{u^\theta _0}{r}\in L^{\frac{3}{2}}$,  $ru^\theta _0\in L^{A}\bigcap L^\infty$ for some finite A. We assume that $\Vert ru^\theta _0\Vert _{L^\infty (\Omega )}$ is sufficiently small, then the system (1.4) has a unique global solution which satisfies

$$\begin{aligned} \eta \buildrel \mathrm{def}\over =\frac{\omega ^\theta }{r}\in C\left( [0,+\infty [;L^1\right) \quad \hbox {and}\quad U\buildrel \mathrm{def}\over =\frac{u^\theta }{r}\in C\left( [0,+\infty [;L^{\frac{3}{2}}\right) . \end{aligned}$$

(1.10)

Remark 1.2

• The main difficulty in the proof of the above theorem is when $\omega ^\theta _0\in L^p(\Omega )$ for $p=1,$ the dissipative term, $\frac{4(p-1)}{p^2}\Vert \nabla |\eta |^{\frac{p}{2}}\Vert _{L^2}^2,$ in (4.7) disappears. That is the reason why we divide the proof of Theorem 1.2 in the following two steps: we first get, by applying Theorem 1.1, that the system (1.4) has a unique local solution with $\eta (t_0)\in L^{p_0}(\mathop {\mathbb R}\nolimits ^3)$ for some $t_0>0$ and $p_0>1;$ then in the second step, starting with initial data at $t_0,$ we prove the global well-posedness of the system (1.4).

• One may see (5.6), (5.17), (5.19) and (5.24) for the exact smallness condition for $\Vert r u^\theta _0\Vert _{L^\infty }$. And the exact global estimate of $\Vert \eta (t)\Vert _{L^1}$ and $\Vert U(t)\Vert _{L^{\frac{3}{2}}}$ is given in (5.25).

• It follows from Lemma 2.1 below and Hölder’s inequality

$$\begin{aligned} \Vert r^\kappa u^\theta \Vert _{L^{\frac{3}{1-\kappa }}}\le \Vert U\Vert _{L^{\frac{3}{2}}}^{\frac{1-\kappa }{2}} \Vert ru^\theta \Vert _{L^\infty }^{\frac{1+\kappa }{2}},\quad \forall \ \kappa \in ]-1,1[, \end{aligned}$$

that the solutions constructed in Theorem 1.2 in fact satisfy

$$\begin{aligned} r^\kappa u^\theta \in C\left( [0,+\infty [;L^{\frac{3}{1-\kappa }}\right) ,\quad \forall \ \kappa \in [-\,1,1]. \end{aligned}$$

2 Preliminaries

2.1 Some elementary results

Lemma 2.1

(Proposition 1 of [2]) Let $(u^r,u^\theta , u^z)$ be a smooth enough solution of (1.3) on [0, T]. Then for any $p\in [2,\infty ]$, we have

$$\begin{aligned} \Vert ru^\theta (t)\Vert _{L^p}\leqslant \Vert ru^\theta _0\Vert _{L^p}\quad \forall \ t\in [0,T]. \end{aligned}$$

(2.1)

Lemma 2.2

(See Lemma 5.5 from [1] for instance) Let E be a Banach space, $\mathfrak {B}(\cdot ,\cdot )$ a continuous bilinear map from $E\times E$ to E,  and $\mathfrak {\alpha }$ a positive real number such that

$$\begin{aligned} \mathfrak {\alpha }<\frac{1}{4\Vert \mathfrak {B}\Vert }\quad \hbox {with}\quad \Vert \mathfrak {B}\Vert \buildrel \mathrm{def}\over =\sup _{\Vert f\Vert ,\Vert g\Vert \le 1}\Vert \mathfrak {B}(f,g)\Vert . \end{aligned}$$

Then for any a in the ball $B(0,\mathfrak {\alpha })$ in E,  there exists a unique x in $B(0,2\mathfrak {\alpha })$ such that

$$\begin{aligned} x=a+\mathfrak {B}(x,x). \end{aligned}$$

Let us recall also some facts from Sect. 2 of [5]. We first recall the axisymmetric Biot–Savart law which determines $\widetilde{u}=(u^r,u^z)$ in terms of $\omega ^\theta ,$ namely

$$\begin{aligned} u^r(r,z)=\int _\Omega G_r(r,z,\bar{r},\bar{z})\omega ^\theta (\bar{r},\bar{z})d\bar{r}d\bar{z},\quad u^z(r,z)=\int _\Omega G_z(r,z,\bar{r},\bar{z})\omega ^\theta (\bar{r},\bar{z})d\bar{r}d\bar{z}, \end{aligned}$$

(2.2)

where

$$\begin{aligned} \begin{aligned} G_r(r,z,\bar{r},\bar{z})&=-\frac{1}{\pi }\frac{z-\bar{z}}{r^{3/2}\bar{r}^{1/2}}F'(\xi ^2),\quad \xi ^2=\frac{(r-\bar{r})^2+(z-\bar{z})^2}{r\bar{r}},\\ G_z(r,z,\bar{r},\bar{z})&=\frac{1}{\pi }\frac{r-\bar{r}}{r^{3/2}\bar{r}^{1/2}}F'(\xi ^2) +\frac{1}{4\pi }\frac{\bar{r}^{1/2}}{r^{3/2}}\left( F(\xi ^2)-2\xi ^2 F'(\xi ^2)\right) \quad \hbox {with}\quad \\ F(s)&= \int _0^{\frac{\pi }{2}}\frac{\cos (2\phi )d\phi }{(\sin ^2\phi +s/4)^{1/2}},\quad s>0. \end{aligned} \end{aligned}$$

(2.3)

It follows from the Remark 2.2 of [5] that

Lemma 2.3

$s^\alpha F(s)$ and $s^\beta F'(s)$ are bounded on $]0,\infty [$ for $\alpha \in ]0,3/2]$ and $\beta \in [1,5/2]$.

Lemma 2.4

(Proposition 2.3 of [5]) Let us denote $\widetilde{u}\buildrel \mathrm{def}\over =(u^r,u^z).$ Then one has

• (i) Assume that $1<p<2<q<\infty$ and $\frac{1}{q}=\frac{1}{p}-\frac{1}{2}$. If $\omega ^\theta \in L^p(\Omega )$, then $\widetilde{u}\in L^q(\Omega )$ and

$$\begin{aligned} \Vert \widetilde{u}\Vert _{L^q(\Omega )}\leqslant C\Vert \omega ^\theta \Vert _{L^p(\Omega )}. \end{aligned}$$

(2.4)

• (ii) If $1\leqslant p<2<q\leqslant \infty$ and $\omega ^\theta \in L^p(\Omega )\bigcap L^q(\Omega )$, then $\widetilde{u}\in L^\infty (\Omega )$ and

  $$\begin{aligned} \Vert \widetilde{u}\Vert _{L^\infty (\Omega )}\leqslant C\Vert \omega ^\theta \Vert _{L^p(\Omega )}^\sigma \Vert \omega ^\theta \Vert _{L^q(\Omega )}^{1-\sigma }, \quad \hbox {where}\quad \sigma =\frac{p(q-2)}{2(q-p)}\in ]0,1[. \end{aligned}$$

  (2.5)

Next we investigate the solution operator S(t) to the linearized system of (1.4), namely $\omega ^\theta (t)=S(t)\omega _0$ verifies

$$\begin{aligned} \left\{ \begin{array}{l} \displaystyle \partial _t\omega ^\theta -\left( \partial _r^2+\partial _z^2+\frac{1}{r} \partial _r-\frac{1}{r^2}\right) \omega ^\theta =0, \quad (t,r,z)\in \mathop {\mathbb R}\nolimits ^+\times \Omega \\ \displaystyle \omega ^\theta |_{r=0}=0,\\ \displaystyle \omega ^\theta |_{t=0} =\omega ^\theta _0. \end{array} \right. \end{aligned}$$

(2.6)

Lemma 2.5

(Lemma 3.1, 3.2 of [5]) For any $t>0$, one has

$$\begin{aligned} \left( S(t)\omega _0\right) (r,z)=\frac{1}{4\pi t}\int _{\Omega }\frac{\bar{r}^{1/2}}{r^{1/2}} {H}\left( \frac{t}{r\bar{r}}\right) \exp \left( -\frac{(r-\bar{r})^2+(z-\bar{z})^2}{4t}\right) \omega _0(\bar{r},\bar{z})d\bar{r}d\bar{z}, \end{aligned}$$

(2.7)

where the function ${H}:]0,+\infty [\rightarrow \mathop {\mathbb R}\nolimits$ is defined by

$$\begin{aligned} {H}(t)=\frac{1}{\sqrt{\pi t}}\int _{-\pi /2}^{\pi /2}e^{-\frac{\sin ^2 \phi }{t}}\cos (2\phi ) d\phi ,\qquad t>0, \end{aligned}$$

(2.8)

which is smooth on $]0,\infty [$ and has the asymptotic expansions:

• (i) ${H}(t)=\frac{\pi ^{1/2}}{4t^{3/2}}+\mathcal {O}\left( \frac{1}{t^{5/2}}\right) ,\ {H}'(t)=-\frac{3\pi ^{1/2}}{8t^{5/2}}+\mathcal {O}\left( \frac{1}{t^{7/2}}\right)$, as $t\rightarrow \infty$;

• (ii) ${H}(t)=1-\frac{3t}{4}+\mathcal {O}(t^2),\ {H}'(t)=-\frac{3}{4}+\mathcal {O}(t)$, as $t\rightarrow 0$.

Corollary 2.1

$t^\alpha H(t)$ and $t^\beta H'(t)$ are bounded on $]0,\infty [$ provided $0\le \alpha \le \frac{3}{2}$, $0\le \beta \le \frac{5}{2}$.

2.2 The estimate of $\frac{u^r}{r}$ in terms of $\frac{\omega ^\theta }{r}$

In this subsection, we shall exploit the basic facts recalled in Sect. 2.1 to derive the estimate of $\frac{u^r}{r}$ in terms of $\frac{\omega ^\theta }{r}$, which will be used in Sect. 4 below. The main result states as follows:

Proposition 2.1

Let $p\in ]1,3[$ and $q\in \bigl ]\frac{3p}{3-p},\infty \bigr ].$ We assume that $\eta \buildrel \mathrm{def}\over =\frac{\omega ^\theta }{r}\in L^p(\mathop {\mathbb R}\nolimits ^3)\cap L^{3p}(\mathop {\mathbb R}\nolimits ^3)$. Then we have

$$\begin{aligned} \bigl \Vert \frac{u^r}{r}\bigr \Vert _{L^q}\lesssim \Vert \eta \Vert _{L^p}^{\lambda }\Vert \eta \Vert _{L^{3p}}^{1-\lambda } \quad \hbox {with}\quad \quad \lambda =\frac{p-1}{2}+\frac{3p}{2q}. \end{aligned}$$

(2.9)

Proof

By virtue of (2.2) and (2.3), we write

$$\begin{aligned} r^{\frac{1}{q}}\frac{u^r(r,z)}{r}=-\int _\Omega \frac{z-\bar{z}}{\pi }\frac{\bar{r}^{\frac{1}{2}}}{r^{\frac{5}{2}-\frac{1}{q}}} F'(\xi ^2)\eta (\bar{r},\bar{z})d\bar{r} d\bar{z}\quad \forall \ (r,z)\in \Omega . \end{aligned}$$

(2.10)

We decompose the integral domain $\Omega =I_1\bigcup I_2$ with

$$\begin{aligned} I_1\buildrel \mathrm{def}\over =\bigl \{(\bar{r},\bar{z})\in \Omega \ \big |\ \bar{r}\le 2r,\bigr \} \quad \hbox {and}\quad I_2\buildrel \mathrm{def}\over =\Omega \backslash I_1. \end{aligned}$$

(2.11)

We first consider the case when $q<\infty$. Let s be determined by $\frac{1}{s}=\frac{1}{q}+\frac{1}{3}.$ Then due to $q\in \bigl ]\frac{3p}{3-p},\infty \bigr [,$ we have $s>p.$ Moreover, it follows from Lemma 2.3 that $|F'(s)|\lesssim (\frac{1}{s})^{\frac{7}{6}}=\left( \frac{1}{s}\right) ^{1+\frac{1}{2}\left( \frac{1}{s}-\frac{1}{q}\right) }.$ Note that that $\frac{\bar{r}}{r}\leqslant 2$ in $I_1$ and $\frac{3}{2}-\frac{1}{2}(\frac{1}{s}+\frac{1}{q})>0,$ we thus obtain

$$\begin{aligned}\begin{aligned}&\Biggl |\int _{I_1}\frac{z-\bar{z}}{\pi }\frac{\bar{r}^{\frac{1}{2}}}{r^{\frac{5}{2}-\frac{1}{q}}} F'(\xi ^2)\eta (\bar{r},\bar{z})d\bar{r} d\bar{z}\Biggr |\\&\quad \lesssim \int _{I_1}\frac{|z-\bar{z}|\cdot \bar{r}^{\frac{1}{2}-\frac{1}{s}}}{r^{{\frac{5}{2}}-\frac{1}{q}}} \left( \frac{r\bar{r}}{(r-\bar{r})^2+(z-\bar{z})^2}\right) ^{\frac{7}{6}} \cdot \bar{r}^{\frac{1}{s}}|\eta (\bar{r},\bar{z})|d\bar{r} d\bar{z}\\&\quad \lesssim \int _{I_1}\left( \frac{\bar{r}}{r}\right) ^{\frac{3}{2}-\frac{1}{2}(\frac{1}{s}+\frac{1}{q})} \left( \frac{1}{|(\bar{r},\bar{z})-(r,z)|}\right) ^{\frac{4}{3}}\cdot \bar{r}^{\frac{1}{s}}|\eta (\bar{r},\bar{z})|d\bar{r} d\bar{z}\\&\quad \lesssim \int _{\Omega } \left( \frac{1}{|(\bar{r},\bar{z})-(r,z)|}\right) ^{\frac{4}{3}}\cdot \bar{r}^{\frac{1}{s}}|\eta (\bar{r},\bar{z})|d\bar{r} d\bar{z}, \end{aligned} \end{aligned}$$

from which, and Hardy–Littlewood–Sobolev inequality, we infer

$$\begin{aligned} \begin{aligned}&\bigl \Vert \int _{I_1}\frac{z-\bar{z}}{\pi }\frac{\bar{r}^{\frac{1}{2}}}{r^{\frac{5}{2}-\frac{1}{q}}} F'(\xi ^2)\eta (\bar{r},\bar{z})d\bar{r} d\bar{z}\bigr \Vert _{L^q(\Omega ;drdz)}\\&\quad \lesssim \bigl \Vert |(r,z)|^{-\frac{4}{3}}\bigr \Vert _{L^{\frac{3}{2},\infty }(\Omega )} \Vert r^{\frac{1}{s}}\eta \Vert _{L^{s}(\Omega )}\\&\quad \thicksim \Vert \eta \Vert _{L^{s}}\lesssim \Vert \eta \Vert _{L^p}^{\frac{p-1}{2}+\frac{3p}{2q}}\Vert \eta \Vert _{L^{3p}}^{1-\frac{p-1}{2}-\frac{3p}{2q}}. \end{aligned} \end{aligned}$$

(2.12)

Note that in the region $I_2,$ there holds $\bar{r}\le 2|\bar{r}-r|$. Thus by using Lemma 2.3, we get

$$\begin{aligned} \begin{aligned}&\Bigl |\int _{I_{2}}\frac{z-\bar{z}}{\pi }\frac{\bar{r}^{\frac{1}{2}}}{r^{\frac{5}{2}-\frac{1}{q}}} F'(\xi ^2)\eta (\bar{r},\bar{z})d\bar{r} d\bar{z}\Bigr |\\&\quad \lesssim \int _{I_2}\frac{|z-\bar{z}|\cdot \bar{r}^{\frac{1}{2}}}{r^{{\frac{5}{2}}-\frac{1}{q}}} \left( \frac{r\bar{r}}{(r-\bar{r})^2+(z-\bar{z})^2}\right) ^{\frac{5}{2}-\frac{1}{q}} \cdot |\eta (\bar{r},\bar{z})|d\bar{r} d\bar{z}\\&\quad \lesssim \int _{I_{2}}\left( \frac{1}{|(\bar{r},\bar{z})-(r,z)|}\right) ^{1-\frac{1}{q}} |\eta (\bar{r},\bar{z})|d\bar{r} d\bar{z}. \end{aligned} \end{aligned}$$

To proceed further, for any given $R>0,$ we split $I_2=I_{21}\cup I_{22}$ with

$$\begin{aligned} I_{21}\,{=}\,I_2\cap \bigl \{(\bar{r},\bar{z}){\in }\,\Omega \big ||(\bar{r},\bar{z})\,{-}\,(r,z)|\ge R\bigr \}, \quad I_{22}=I_2\cap \bigl \{(\bar{r},\bar{z})\in \Omega \big ||(\bar{r},\bar{z})-(r,z)|{<} R\bigr \}. \end{aligned}$$

Then we get, by applying Young’s inequality, that

$$\begin{aligned}\begin{aligned}&\bigl \Vert \int _{I_{21}}\frac{z-\bar{z}}{\pi }\frac{\bar{r}^{\frac{1}{2}}}{r^{\frac{5}{2}-\frac{1}{q}}} F'(\xi ^2)\eta (\bar{r},\bar{z})d\bar{r} d\bar{z}\bigr \Vert _{L^q(\Omega )}\\&\quad \le \bigl \Vert \int _{I_{21}}\left( \frac{1}{|(\bar{r},\bar{z})-(r,z)|}\right) ^{1+\frac{1}{p}-\frac{1}{q}} \cdot \bar{r}^{\frac{1}{p}}|\eta (\bar{r},\bar{z})|d\bar{r} d\bar{z}\bigr \Vert _{L^q(\Omega )}\\&\quad \lesssim \bigl \Vert r^{\frac{1}{p}}\eta \bigr \Vert _{L^{p}(\Omega )}\left( \int _R^\infty \rho ^{\frac{2/q-{2}/{p}}{1-{1}/{p}+1/q}} \rho d\rho \right) ^{1-\frac{1}{p}+\frac{1}{q}}\thicksim R^{1-\frac{3}{p}+\frac{3}{q}}\Vert \eta \Vert _{L^{p}}. \end{aligned} \end{aligned}$$

For the integral on $I_{22}$, in the case $q>3p$, by applying Young’s inequality, we get

$$\begin{aligned}\begin{aligned}&\Biggl \Vert \int _{I_{22}}\frac{z-\bar{z}}{\pi }\frac{\bar{r}^{\frac{1}{2}}}{r^{\frac{5}{2}-\frac{1}{q}}} F'(\xi ^2)\eta (\bar{r},\bar{z})d\bar{r} d\bar{z}\Biggr \Vert _{L^q(\Omega )}\\&\quad \le \Biggl \Vert \int _{I_{22}}\left( \frac{1}{|(\bar{r},\bar{z})-(r,z)|}\right) ^{1+\frac{1}{3p}-\frac{1}{q}} \cdot \bar{r}^{\frac{1}{3p}}|\eta (\bar{r},\bar{z})|d\bar{r} d\bar{z}\Biggr \Vert _{L^q(\Omega )}\\&\quad \lesssim \bigl \Vert r^{\frac{1}{3p}}\eta \bigr \Vert _{L^{3p}(\Omega )}\left( \int _0^R\rho ^{\frac{2/q-{2}/{3p}}{1-{1}/{3p}+1/q}} d\rho \right) ^{1-\frac{1}{3p}+\frac{1}{q}}\thicksim R^{1-\frac{1}{p}+\frac{3}{q}}\Vert \eta \Vert _{L^{3p}}. \end{aligned} \end{aligned}$$

While in the case $\frac{3p}{3-p}<q\le 3p$, another use of Young’s inequality gives

$$\begin{aligned}\begin{aligned}&\Biggl \Vert \int _{I_{22}}\frac{z-\bar{z}}{\pi }\frac{\bar{r}^{\frac{1}{2}}}{r^{\frac{5}{2}-\frac{1}{q}}} F'(\xi ^2)\eta (\bar{r},\bar{z})d\bar{r} d\bar{z}\Biggr \Vert _{L^q(\Omega )}\\&\quad \le \Biggl \Vert \int _{I_{22}}\left( \frac{1}{|(\bar{r},\bar{z})-(r,z)|}\right) ^{1+\frac{3-p}{3p}-\frac{1}{q}} \cdot \bar{r}^{\frac{3-p}{3p}}|\eta (\bar{r},\bar{z})|d\bar{r} d\bar{z}\Biggr \Vert _{L^q(\Omega )}\\&\quad \lesssim \bigl \Vert r^{\frac{3-p}{3p}}\eta \bigr \Vert _{L^{\frac{3p}{3-p}}(\Omega )}\left( \int _0^R\rho ^{\frac{2/q-{2(3-p)}/{3p}}{1-{(3-p)}/{3p}+1/q}} d\rho \right) ^{1-\frac{3-p}{3p}+\frac{1}{q}}\\&\quad \thicksim R^{2-\frac{3}{p}+\frac{3}{q}}\Vert \eta \Vert _{L^{\frac{3p}{3-p}}} \lesssim R^{1-\frac{3}{p}+\frac{3}{q}}\Vert \eta \Vert _{L^{p}}+R^{1-\frac{1}{p}+\frac{3}{q}}\Vert \eta \Vert _{L^{3p}}. \end{aligned} \end{aligned}$$

As a result, it comes out

$$\begin{aligned}&\bigl \Vert \int _{I_{2}}\frac{z-\bar{z}}{\pi }\frac{\bar{r}^{\frac{1}{2}}}{r^{\frac{5}{2}-\frac{1}{q}}} F'(\xi ^2)\eta (\bar{r},\bar{z})d\bar{r} d\bar{z}\bigr \Vert _{L^q(\Omega )}\lesssim R^{1-\frac{3}{p}+\frac{3}{q}}\Vert \eta \Vert _{L^{p}}+R^{1-\frac{1}{p}+\frac{3}{q}}\Vert \eta \Vert _{L^{3p}}. \end{aligned}$$

Taking $R=\left( \frac{\Vert \eta \Vert _{L^p}}{\Vert \eta \Vert _{L^{3p}}}\right) ^{\frac{p}{2}}$ in the above inequality gives rise to

$$\begin{aligned} \bigl \Vert \int _{I_{2}}\frac{z-\bar{z}}{\pi }\frac{\bar{r}^{\frac{1}{2}}}{r^{\frac{5}{2}-\frac{1}{q}}} F'(\xi ^2)\eta (\bar{r},\bar{z})d\bar{r} d\bar{z}\bigr \Vert _{L^q(\Omega )} \lesssim \Vert \eta \Vert _{L^p}^{\frac{p-1}{2}+\frac{3p}{2q}}\Vert \eta \Vert _{L^{3p}}^{1-\frac{p-1}{2}-\frac{3p}{2q}}. \end{aligned}$$

(2.13)

Due to $\bigl \Vert r^{\frac{1}{q}}\frac{u^r}{r}\bigr \Vert _{L^q(\Omega )}=\bigl \Vert \frac{u^r}{r}\bigr \Vert _{L^q},$ (2.12) together with (2.13) ensures (2.9) for any $q\in \bigl ]\frac{3p}{3-p},\infty \bigr [$.

The end-point case when $q=\infty$ follows exactly along the same line. This completes the proof of Proposition 2.1. $\square$

2.3 The estimates of the solution operator S(t)

The goal of this subsection is to present the estimates of the solution operator S(t),  which will be used in Sect. 3.

Proposition 2.2

Let S(t) be the solution operator given by (2.7). Then this family $\left( S(t)\right) _{t\ge 0}$ are strongly continuous semigroups of bounded linear operators in $L^m(\Omega )$ for any $m\in [1,\infty [$. Moreover, for $1\le p\le q\le \infty$, there holds

• 1. For any $\alpha ,~\beta$ satisfying $\alpha +\beta \le 0,~a\ge -1$ and $\beta \ge -1$, and any $f=(f^r,f^z)\in L^p(\Omega )^2$, there holds

  $$\begin{aligned} \Vert r^\alpha S(t)\mathrm{div}\,_* (r^{\beta }f)\Vert _{L^q(\Omega )}\le \frac{C}{t^{\frac{1}{2}-\frac{\alpha +\beta }{2}+\frac{1}{p}-\frac{1}{q}}}\Vert f\Vert _{L^p(\Omega )}. \end{aligned}$$

  (2.14)

  In particular, taking $\alpha =\beta =0$, we have

  $$\begin{aligned} \Vert S(t)\mathrm{div}\,_* f\Vert _{L^q(\Omega )}\le \frac{C}{t^{\frac{1}{2}+\frac{1}{p}-\frac{1}{q}}}\Vert f\Vert _{L^p(\Omega )}. \end{aligned}$$

  (2.15)

• 2. For any $\alpha ,~\beta$ satisfying $\alpha +\beta \le 1,~\alpha \ge -1$ and $\beta \ge -1$, and any $g\in L^p(\Omega )$, there holds

  $$\begin{aligned} \Vert r^\alpha S(t) (r^{\beta -1} g)\Vert _{L^q(\Omega )} \le \frac{C}{t^{\frac{1}{2}-\frac{\alpha +\beta }{2}+\frac{1}{p}-\frac{1}{q}}}\Vert g\Vert _{L^p(\Omega )}. \end{aligned}$$

  (2.16)

  In particular, taking $\alpha =0,~\beta =1$, and $\alpha =\beta =0$, we have

  $$\begin{aligned} \Vert S(t) g\Vert _{L^q(\Omega )} \le \frac{C}{t^{\frac{1}{p}-\frac{1}{q}}}\Vert g\Vert _{L^p(\Omega )},\quad \Vert S(t) \left( \frac{g}{r}\right) \Vert _{L^q(\Omega )} \le \frac{C}{t^{\frac{1}{2}+\frac{1}{p}-\frac{1}{q}}}\Vert g\Vert _{L^p(\Omega )}. \end{aligned}$$

  (2.17)

• 3. For any $\delta \in \bigl [-\,1,\frac{1}{2}\bigr ],~m\in [1,\infty [$, and any g satisfying $r^\delta g\in L^m(\Omega )$, we have

  $$\begin{aligned} \Vert r^\delta S(t) g-r^\delta g\Vert _{L^m(\Omega )}\rightarrow 0,\quad \text{ as }\ t\rightarrow 0. \end{aligned}$$

  (2.18)

Proof

The boundedness of the semigroup $\left( S(t)\right) _{t\ge 0}$ is shown in (2.17). Then in order to prove $\left( S(t)\right) _{t\ge 0}$ is strongly continuous in $L^m(\Omega )$ for any $m\in [1,\infty [$, we only need to verify the continuity at the origin, which is a direct consequence of (2.18) (with $\delta =0$). Hence it remains to prove the estimates (2.14–2.17), which we handle term by term below.

• 1. By integration by parts, we write

$$\begin{aligned} \begin{aligned}&r^\alpha \left( S(t)\mathrm{div}\,_* (r^{\beta }f)\right) (r,z)\\&\quad =\frac{1}{4\pi t} \int _\Omega \frac{\bar{r}^{\frac{1}{2}+\beta }}{r^{\frac{1}{2}-\alpha }}\exp \left( -\frac{(r-\bar{r})^2+(z-\bar{z})^2}{4t}\right) \cdot \left( A_r f^r+A_z f^z\right) \left( \bar{r},\bar{z}\right) \,d\bar{r}\,d\bar{z}, \end{aligned} \end{aligned}$$

(2.19)

where

$$\begin{aligned} A_r\left( \bar{r},\bar{z}\right) =\frac{t}{r\bar{r}^2}{H}'\left( \frac{t}{r\bar{r}}\right) -\left( \frac{1}{2\bar{r}}+\frac{r-\bar{r}}{2t}\right) {H} \left( \frac{t}{r\bar{r}}\right) ,\quad A_z\left( \bar{r},\bar{z}\right) =-\frac{z-\bar{z}}{2t}{H}\left( \frac{t}{r\bar{r}}\right) . \end{aligned}$$

• Let us first handle the term $\bigl |A_r+\frac{r-\bar{r}}{2t}{H}\left( \frac{t}{r\bar{r}}\right) \bigr |$.

If $(\alpha ,\beta )\in \Omega _1\buildrel \mathrm{def}\over =\{(\alpha ,\beta )|0\le \frac{1}{2}-\frac{\alpha +\beta }{2}\le \frac{3}{2} ,~0\le \frac{1}{2}-\beta \le \frac{3}{2},~\beta \le \alpha \}$, we can divide the integral area into $\{\bar{r}\ge \frac{r}{2}\}$ and $\{\bar{r}< \frac{r}{2}\}$. When $\bar{r}\ge \frac{r}{2}$, we can deduce from Corollary 2.1 that

$$\begin{aligned}&\frac{\bar{r}^{\frac{1}{2}+\beta }}{r^{\frac{1}{2}-\alpha }} \exp \left( -\frac{\left( r-\bar{r}\right) ^2+\left( z-\bar{z}\right) ^2}{4t}\right) \left| A_r+\frac{r-\bar{r}}{2t}{H}\left( \frac{t}{r\bar{r}}\right) \right| \nonumber \\&\quad \lesssim \left( \frac{t}{r^{\frac{3}{2}-\alpha }\bar{r}^{\frac{3}{2}-\beta }}{H}'\left( \frac{t}{r\bar{r}}\right) +\frac{1}{r^{\frac{1}{2}-\alpha }\bar{r}^{\frac{1}{2}-\beta }}{H}\left( \frac{t}{r\bar{r}}\right) \right) \cdot \exp \left( -\frac{\left( r-\bar{r}\right) ^2+\left( z-\bar{z}\right) ^2}{4t}\right) \nonumber \\&\quad \lesssim \left( \frac{t}{r^{\frac{3}{2}-\alpha }\bar{r}^{\frac{3}{2}-\beta }}\left| \frac{r\bar{r}}{t}\right| ^{\frac{3}{2}-\frac{\alpha +\beta }{2}} +\frac{1}{r^{\frac{1}{2}-\alpha }\bar{r}^{\frac{1}{2}-\beta }}\left| \frac{r\bar{r}}{t}\right| ^{\frac{1}{2}-\frac{\alpha +\beta }{2}}\right) \cdot \exp \left( -\frac{\left( r-\bar{r}\right) ^2+\left( z-\bar{z}\right) ^2}{4t}\right) \nonumber \\&\quad \lesssim \frac{1}{t^{\frac{1}{2}-\frac{\alpha +\beta }{2}}}\cdot \exp \left( -\frac{\left( r-\bar{r}\right) ^2+\left( z-\bar{z}\right) ^2}{5t}\right) . \end{aligned}$$

(2.20)

And when $\bar{r}<\frac{r}{2}$, there then holds ${r}<2|\bar{r}-r|$, another use of Corollary 2.1 gives

$$\begin{aligned} \begin{aligned}&\frac{\bar{r}^{\frac{1}{2}+\beta }}{r^{\frac{1}{2}-\alpha }} \exp \left( -\frac{\left( r-\bar{r}\right) ^2+\left( z-\bar{z}\right) ^2}{4t}\right) \left| A_r+\frac{r-\bar{r}}{2t}{H}\left( \frac{t}{r\bar{r}}\right) \right| \\&\quad \lesssim \left( \frac{t}{r^{\frac{3}{2}-\alpha }\bar{r}^{\frac{3}{2}-\beta }}\left| \frac{r\bar{r}}{t}\right| ^{\frac{3}{2}-\beta } +\frac{1}{r^{\frac{1}{2}-\alpha }\bar{r}^{\frac{1}{2}-\beta }}\left| \frac{r\bar{r}}{t}\right| ^{\frac{1}{2}-\beta }\right) \times \exp \left( -\frac{\left( r-\bar{r}\right) ^2+\left( z-\bar{z}\right) ^2}{4t}\right) \\&\quad \lesssim \frac{{r}^{\alpha -\beta }}{t^{\frac{1}{2}-\beta }}\cdot \left( \frac{5t}{\left( r-\bar{r}\right) ^2+\left( z-\bar{z}\right) ^2}\right) ^{\frac{\alpha -\beta }{2}}\cdot \exp \left( -\frac{\left( r-\bar{r}\right) ^2+\left( z-\bar{z}\right) ^2}{5t}\right) \\&\quad \lesssim \frac{1}{t^{\frac{1}{2}-\frac{\alpha +\beta }{2}}}\cdot \exp \left( -\frac{\left( r-\bar{r}\right) ^2+\left( z-\bar{z}\right) ^2}{5t}\right) . \end{aligned} \end{aligned}$$

(2.21)

If $(\alpha ,\beta )\in \Omega _2\buildrel \mathrm{def}\over =\{(\alpha ,\beta )|0\le \frac{1}{2}-\frac{\alpha +\beta }{2}\le \frac{3}{2} ,~0\le \frac{1}{2}-\alpha \le \frac{3}{2},~\alpha \le \beta \}$, we divide the integral area in a different way as $\{\bar{r}\le 2r\}$ and $\{\bar{r}>2r\}$. Similar to the previous estimates, when $\bar{r}\le 2r$, we have

$$\begin{aligned} \begin{aligned}&\frac{\bar{r}^{\frac{1}{2}+\beta }}{r^{\frac{1}{2}-\alpha }} \exp \left( -\frac{\left( r-\bar{r}\right) ^2+\left( z-\bar{z}\right) ^2}{4t}\right) |A_r+\frac{r-\bar{r}}{2t}{H}\left( \frac{t}{r\bar{r}}\right) |\\&\quad \lesssim \left( \frac{t}{r^{\frac{3}{2}-\alpha }\bar{r}^{\frac{3}{2}-\beta }}|\frac{r\bar{r}}{t}|^{\frac{3}{2}-\frac{\alpha +\beta }{2}} +\frac{1}{r^{\frac{1}{2}-\alpha }\bar{r}^{\frac{1}{2}-\beta }}|\frac{r\bar{r}}{t}|^{\frac{1}{2}-\frac{\alpha +\beta }{2}}\right) \cdot \exp \left( -\frac{\left( r-\bar{r}\right) ^2+\left( z-\bar{z}\right) ^2}{4t}\right) \\&\quad \lesssim \frac{1}{t^{\frac{1}{2}-\frac{\alpha +\beta }{2}}}\cdot \exp \left( -\frac{\left( r-\bar{r}\right) ^2+\left( z-\bar{z}\right) ^2}{5t}\right) . \end{aligned} \end{aligned}$$

(2.22)

And when $\bar{r}> 2r$, there then holds $\bar{r}<2|\bar{r}-r|$, thus we deduce

$$\begin{aligned} \begin{aligned}&\frac{\bar{r}^{\frac{1}{2}+\beta }}{r^{\frac{1}{2}-\alpha }} \exp \left( -\frac{\left( r-\bar{r}\right) ^2+\left( z-\bar{z}\right) ^2}{4t}\right) |A_r+\frac{r-\bar{r}}{2t}{H}\left( \frac{t}{r\bar{r}}\right) |\\&\quad \lesssim \left( \frac{t}{r^{\frac{3}{2}-\alpha }\bar{r}^{\frac{3}{2}-\beta }}|\frac{r\bar{r}}{t}|^{\frac{3}{2}-\alpha } +\frac{1}{r^{\frac{1}{2}-\alpha }\bar{r}^{\frac{1}{2}-\beta }}|\frac{r\bar{r}}{t}|^{\frac{1}{2}-\alpha }\right) \exp \left( -\frac{\left( r-\bar{r}\right) ^2+\left( z-\bar{z}\right) ^2}{4t}\right) \\&\quad \lesssim \frac{\bar{r}^{\beta -\alpha }}{t^{\frac{1}{2}-\alpha }}\cdot \left( \frac{5t}{\left( r-\bar{r}\right) ^2+\left( z-\bar{z}\right) ^2}\right) ^{\frac{\beta -\alpha }{2}}\cdot \exp \left( -\frac{\left( r-\bar{r}\right) ^2+\left( z-\bar{z}\right) ^2}{5t}\right) \\&\quad \lesssim \frac{1}{t^{\frac{1}{2}-\frac{\alpha +\beta }{2}}}\cdot \exp \left( -\frac{\left( r-\bar{r}\right) ^2+\left( z-\bar{z}\right) ^2}{5t}\right) . \end{aligned} \end{aligned}$$

(2.23)

Thus combining the estimates (2.14–2.17), we conclude that whenever $(\alpha ,\beta )\in \Omega _1\bigcup \Omega _2$, i.e. $\alpha ,~\beta$ satisfy $\alpha +\beta \le 1$, $\alpha \ge -1$ and $\beta \ge -1$, there holds

$$\begin{aligned} \begin{aligned}&\frac{\bar{r}^{\frac{1}{2}+\beta }}{r^{\frac{1}{2}-\alpha }} \exp \left( -\frac{(r-\bar{r})^2+(z-\bar{z})^2}{4t}\right) \cdot \left| A_r+\frac{r-\bar{r}}{2t}{H}\left( \frac{t}{r\bar{r}}\right) \right| \\&\quad \lesssim \frac{1}{t^{\frac{1}{2}-\frac{\alpha +\beta }{2}}}\cdot \exp \left( -\frac{(r-\bar{r})^2+(z-\bar{z})^2}{5t}\right) . \end{aligned} \end{aligned}$$

(2.24)

• For $|A_z|$ term in the integrand (2.19). When $\bar{r}>2r$ or $\bar{r}\le \frac{r}{2}$, there then holds $\bar{r}+r<3|\bar{r}-r|$. If in addition $\alpha +\beta \ge -4$, $\alpha \ge -1$ and $\beta \ge -2$, there then exists a positive constant $\gamma$ so that $\max \bigl \{0,\frac{1}{2}-\alpha ,-\frac{1}{2}-\beta ,-\frac{1+\alpha +\beta }{2}\bigr \}\le \gamma \le \frac{3}{2}$. Then we deduce from Corollary 2.1 that

  $$\begin{aligned}&\frac{\bar{r}^{\frac{1}{2}+\beta }}{r^{\frac{1}{2}-\alpha }} \exp \left( -\frac{\left( r-\bar{r}\right) ^2+\left( z-\bar{z}\right) ^2}{4t}\right) \left( |A_z|+|\frac{r-\bar{r}}{2t}{H}\left( \frac{t}{r\bar{r}}\right) |\right) \\&\quad \lesssim \frac{\bar{r}^{\frac{1}{2}+\beta }}{r^{\frac{1}{2}-\alpha }}\frac{|r-\bar{r}|+|z-\bar{z}|}{t}|\frac{r\bar{r}}{t}|^{\gamma } \left( \frac{5t}{\left( r-\bar{r}\right) ^2+\left( z-\bar{z}\right) ^2}\right) ^{\frac{1+2\gamma +\alpha +\beta }{2}}\\&\qquad \times \exp \left( -\frac{\left( r-\bar{r}\right) ^2+\left( z-\bar{z}\right) ^2}{5t}\right) \\&\quad \lesssim \frac{1}{t^{\frac{1}{2}-\frac{\alpha +\beta }{2}}}\exp \left( -\frac{\left( r-\bar{r}\right) ^2+\left( z-\bar{z}\right) ^2}{5t}\right) . \end{aligned}$$

  And when $\frac{r}{2}\le \bar{r}\le 2r$, if in addition, $-3\le \alpha +\beta \le 0$, we have

  $$\begin{aligned}&\frac{\bar{r}^{\frac{1}{2}+\beta }}{r^{\frac{1}{2}-\alpha }} \exp \left( -\frac{\left( r-\bar{r}\right) ^2+\left( z-\bar{z}\right) ^2}{4t}\right) \left( |A_z|+|\frac{r-\bar{r}}{2t}{H}\left( \frac{t}{r\bar{r}}\right) |\right) \\&\quad \lesssim \frac{\bar{r}^{\frac{1}{2}+\beta }}{r^{\frac{1}{2}-\alpha }} \frac{|r-\bar{r}|+|z-\bar{z}|}{t}|\frac{r\bar{r}}{t}|^{-\frac{\alpha +\beta }{2}} \left( \frac{5t}{\left( r-\bar{r}\right) ^2+\left( z-\bar{z}\right) ^2}\right) ^{\frac{1}{2}}\\&\qquad \times \exp \left( -\frac{\left( r-\bar{r}\right) ^2+\left( z-\bar{z}\right) ^2}{5t}\right) \\&\quad \lesssim \frac{1}{t^{\frac{1}{2}-\frac{\alpha +\beta }{2}}}\cdot \exp \left( -\frac{\left( r-\bar{r}\right) ^2+\left( z-\bar{z}\right) ^2}{5t}\right) . \end{aligned}$$

Therefore as long as $\alpha +\beta \le 0$, $\alpha \ge -1$ and $\beta \ge -2$, there holds

$$\begin{aligned} \begin{aligned}&\frac{\bar{r}^{\frac{1}{2}+\beta }}{r^{\frac{1}{2}-\alpha }} \exp \left( -\frac{(r-\bar{r})^2+(z-\bar{z})^2}{4t}\right) \left( |A_z|+\bigl |\frac{r-\bar{r}}{2t}{H} \left( \frac{t}{r\bar{r}}\right) \bigr |\right) \\&\quad \lesssim \frac{1}{t^{\frac{1}{2}-\frac{\alpha +\beta }{2}}}\cdot \exp \left( -\frac{(r-\bar{r})^2+(z-\bar{z})^2}{5t}\right) . \end{aligned} \end{aligned}$$

(2.25)

By combining (2.24) with (2.25), we achieve

$$\begin{aligned} \frac{\bar{r}^{\frac{1}{2}+\beta }}{r^{\frac{1}{2}-\alpha }} \exp \left( -\frac{(r-\bar{r})^2+(z-\bar{z})^2}{4t}\right) \left( |A_r|+|A_z|\right) \lesssim \frac{1}{t^{\frac{1}{2}-\frac{\alpha +\beta }{2}}}\cdot \exp \left( -\frac{(r-\bar{r})^2+(z-\bar{z})^2}{5t}\right) , \end{aligned}$$

provided $\alpha +\beta \le 0,~a\ge -1$ and $\beta \ge -1$. And then (2.14) follows from (2.19) and Young’s inequality in two space dimension.

• 2. It follows from the proof of (2.24) that

  $$\begin{aligned}&\frac{1}{r^{\frac{1}{2}-\alpha }\bar{r}^{\frac{1}{2}-\beta }} \bigl |{H}\left( \frac{t}{r\bar{r}}\right) \bigr | \exp \left( -\frac{(r-\bar{r})^2+(z-\bar{z})^2}{4t}\right) \\&\quad \lesssim \frac{1}{t^{\frac{1}{2}-\frac{\alpha +\beta }{2}}}\exp \left( -\frac{(r-\bar{r})^2+(z-\bar{z})^2}{5t}\right) , \end{aligned}$$

  whenever $\alpha +\beta \le 1,~\alpha \ge -1$ and $\beta \ge -1$. Then by virtue of (2.7), we get, by applying Young’s inequality, that there holds (2.16).

• 3. In view of (2.7), we get, by using changes of variables that

  $$\begin{aligned} \left( r^\delta S(t) g-r^\delta g\right) (r,z)=\frac{1}{4\pi }\int _{\Omega } \Psi (r,z,\rho ,\xi ,t)\cdot \exp \left( -\frac{\rho ^2+\xi ^2}{4}\right) \, d\rho d\xi , \end{aligned}$$

  (2.26)

  for all $(r,z)\in \Omega$, and where

  $$\begin{aligned} \Psi (r,z,\rho ,\xi ,t)\buildrel \mathrm{def}\over =r^\delta \left( \frac{r+\sqrt{t}\rho }{r}\right) ^{\frac{1}{2}} {H}\left( \frac{t}{r\left( r+\sqrt{t}\rho \right) }\right) g\left( r+\sqrt{t}\rho ,z+\sqrt{t}\xi \right) -r^\delta g(r,z). \end{aligned}$$

  (2.27)

Notice that $\frac{1}{2}-\delta \in [0,\frac{3}{2}]$, applying Corollary 2.1 gives

$$\begin{aligned} \begin{aligned}&\left( \frac{\bar{r}}{r}\right) ^{\frac{1}{2}-\delta } {H}\left( \frac{t}{r\bar{r}}\right) \le C\quad \text{ if }\quad \bar{r}\le 2r \quad \hbox {and}\quad \\&\left( \frac{\bar{r}}{r}\right) ^{\frac{1}{2}-\delta } {H}\left( \frac{t}{r\bar{r}}\right) \le C\left( \frac{\bar{r}}{r}\right) ^{\frac{1}{2}-\delta } \left( \frac{r\bar{r}}{t}\right) ^{\frac{1}{2}-\delta }\le C\left( \frac{|\bar{r}-r|^2}{t}\right) ^{\frac{1}{2}-\delta }\quad \text{ if }\quad \bar{r}>2r, \end{aligned} \end{aligned}$$

which implies for any given $(\rho ,\xi , t)$, we have

$$\begin{aligned} \Psi (r,z,\rho ,\xi ,t)\lesssim (1+\rho ^{1-2\delta })\cdot \left( r+\sqrt{t}\rho \right) ^\delta g\left( r+\sqrt{t}\rho ,z+\sqrt{t}\xi \right) -r^\delta g(r,z)\in L^m(\Omega ). \end{aligned}$$

(2.28)

Moreover, noting that $H(t)=1+\mathcal {O}(t),$ as $t\rightarrow 0$, it is easy to observe that $\Psi (r,z,\rho ,\xi ,t)\rightarrow 0$ as $t\rightarrow 0$. Then Lebesgue dominated convergence theorem ensures that

$$\begin{aligned} \Vert \Psi (\cdot ,\cdot ,\rho ,\xi ,t)\Vert _{L^m(\Omega )}\rightarrow 0,\ \text{ as }\ t\rightarrow 0, \end{aligned}$$

from which and (2.26), another use of Lebesgue’s dominated convergence theorem gives

$$\begin{aligned}&\Vert r^\delta S(t) g-r^\delta g\Vert _{L^m(\Omega )}\\&\quad \le \frac{1}{4\pi }\int _{\Omega } \Vert \Psi (\cdot ,\cdot ,\rho ,\xi ,t)\Vert _{L^m(\Omega )}\exp \left( -\frac{\rho ^2+\xi ^2}{4}\right) \, d\rho d\xi \rightarrow 0,\ \text{ as }\ t\rightarrow 0. \end{aligned}$$

This completes the proof of the proposition. $\square$

3 Local existence of solutions to (1.1) in critical spaces

The purpose of this section is to investigate the local existence and uniqueness of the mild solutions to (1.4) in the spirit of [5, 6]. In view of (2.7), we rewrite the systems (1.4) as

$$\begin{aligned} \left\{ \begin{array}{l} \displaystyle \omega ^\theta (t)=S(t)\omega ^\theta _0-\int _0^t S(t-s)\left( \mathrm{div}\,_*\left( \widetilde{u}(s)\omega ^\theta (s)\right) -\frac{\partial _z|u^\theta (s)|^2}{r}\right) \,ds,\quad t>0. \\ \displaystyle u^\theta (t)=S(t)u^\theta _0-\int _0^t S(t-s)\left( \mathrm{div}\,_*\left( \widetilde{u}(s)u^\theta (s)\right) -\frac{2u^\theta (s)u^r(s)}{r}\right) ds,\quad t>0. \end{array} \right. \end{aligned}$$

(3.1)

Now we present the proof of Theorem 1.1.

Proof of Theorem 1.1

The main idea to prove Theorem 1.1 is to apply fixed point argument for the integral formulation (3.1). Toward this, for any $T>0$, we introduce the functional space

$$\begin{aligned} \begin{aligned}&X_T\buildrel \mathrm{def}\over =\left\{ (\omega ^\theta ,u^\theta )\in C\left( ]0,T];L^{\frac{4}{3}}(\Omega )\right) \times C\left( ]0,T];L^{4}(\Omega ) \right) \ \big |\right. \\&\quad \left. r^{-\frac{3}{10}}u^\theta \in C\left( ]0,T];L^{2}(\Omega )\right) \quad \hbox {and}\quad \Vert \left( \omega ^\theta ,u^\theta \right) \Vert _{X_T}<\infty \right\} , \end{aligned} \end{aligned}$$

(3.2)

where

$$\begin{aligned} \Vert (\omega ^\theta ,u^\theta )\Vert _{X_T}\buildrel \mathrm{def}\over =\sup _{0<t\le T}\left( t^{\frac{1}{4}}\Vert \omega ^\theta (t)\Vert _{L^{\frac{4}{3}}(\Omega )} +t^{\frac{1}{4}}\Vert u^\theta (t)\Vert _{L^{4}(\Omega )}+t^{\frac{3}{20}}\Vert r^{-\frac{3}{10}}u^\theta (t)\Vert _{L^{2}(\Omega )}\right) . \end{aligned}$$

(3.3)

For convenience, sometimes we may abuse the notation $\Vert \omega ^\theta \Vert _{X_T}=\sup \limits _{0<t\le T}t^{\frac{1}{4}}\Vert \omega ^\theta \Vert _{L^{\frac{4}{3}}(\Omega )}$,  $\Vert u^\theta \Vert _{X_T}=\sup \limits _{0<t\le T}\left( t^{\frac{1}{4}}\Vert u^\theta (t)\Vert _{L^{4}(\Omega )}+t^{\frac{3}{20}}\Vert r^{-\frac{3}{10}}u^\theta (t)\Vert _{L^{2}(\Omega )}\right)$, and $\omega ^\theta \in X_T$ (resp. $u^\theta \in X_T$) means that $\Vert \omega ^\theta \Vert _{X_T}<\infty$ (resp. $\Vert u^\theta \Vert _{X_T}<\infty$).

• The estimate of $\omega ^\theta$ term

In view of (2.17), $S(t)\omega ^\theta _0\in X_T$ for any $T>0$, and there exists a universal constant $C_1>0$ such that for any $T>0$, we have

$$\begin{aligned} \sup _{0<t\le T}t^{\frac{1}{4}}\Vert S(t)\omega ^\theta _0\Vert _{L^{\frac{4}{3}}(\Omega )} \le C_1\Vert \omega ^\theta _0\Vert _{L^1(\Omega )}. \end{aligned}$$

(3.4)

On the other hand, since $L^1(\Omega )\bigcap L^{\frac{4}{3}}(\Omega )$ is dense in $L^1(\Omega ).$ For any $\varepsilon >0,$ there exists $\widetilde{\omega }^\theta _0\in L^1(\Omega )\bigcap L^{\frac{4}{3}}(\Omega )$ satisfying $\Vert \widetilde{\omega }^\theta _0-\omega ^\theta _0\Vert _{L^1(\Omega )}<\varepsilon .$ Then it follows from (2.17) that

$$\begin{aligned} \begin{aligned} t^{\frac{1}{4}}\Vert S(t)\omega ^\theta _0\Vert _{L^{\frac{4}{3}}(\Omega )}&\le t^{\frac{1}{4}}\Vert S(t)(\omega ^\theta _0-\widetilde{\omega }^\theta _0)\Vert _{L^{\frac{4}{3}}(\Omega )}+t^{\frac{1}{4}}\Vert S(t)\widetilde{\omega }^\theta _0\Vert _{L^{\frac{4}{3}}(\Omega )}\\&\le C\left( \Vert \widetilde{\omega }^\theta _0-\omega ^\theta _0\Vert _{L^1(\Omega )}+t^{\frac{1}{4}}\Vert \widetilde{\omega }^\theta _0\Vert _{L^{\frac{4}{3}}(\Omega )}\right) \\&\le C\left( \varepsilon +t^{\frac{1}{4}}\Vert \widetilde{\omega }^\theta _0\Vert _{L^{\frac{4}{3}}(\Omega )}\right) , \end{aligned} \end{aligned}$$

which implies

$$\begin{aligned} \lim _{t\rightarrow 0}t^{\frac{1}{4}}\Vert S(t)\omega ^\theta _0\Vert _{L^{\frac{4}{3}}(\Omega )}=0. \end{aligned}$$

(3.5)

Let us denote $\widetilde{u}(\omega ^\theta )(t)$ be velocity field determined by the vorticity $\omega ^\theta e_\theta$ via the axisymmetric Biot–Savart law (2.2). Given $\left( \omega ^\theta _1,u^\theta _1\right) ,~\left( \omega ^\theta _2,u^\theta _2\right) \in X_T$, for any $t\in [0,T],$ we define the mapping $\mathcal {F}^\omega$ on $X_T\times X_T$ by

$$\begin{aligned} \mathcal {F}^\omega \left( \left( \omega ^\theta _1,u^\theta _1\right) ,\left( \omega ^\theta _2,u^\theta _2\right) \right) (t)\buildrel \mathrm{def}\over =\int _0^tS(t-s)\cdot \left( \mathrm{div}\,_*(\widetilde{u}\left( \omega ^\theta _1\right) (s)\omega ^\theta _2(s)) -\frac{\partial _z(u^\theta _1(s)u^\theta _2(s))}{r}\right) ds. \end{aligned}$$

(3.6)

Then for any $p\in [1,\frac{10}{7}[,$ we deduce from (2.14) (with $~\alpha =0,~\beta =-\frac{2}{5}$), (2.15) and (2.4) that for any $t\in ]0,T]$,

$$\begin{aligned} \begin{aligned}&t^{1-\frac{1}{p}}\Vert \mathcal {F}^\omega \left( \left( \omega _1,u^\theta _1\right) ,\left( \omega _2,u^\theta _2\right) \right) (t)\Vert _{L^p(\Omega )}\\&\quad \le C t^{1-\frac{1}{p}}\int _0^t\frac{\Vert \widetilde{u}\left( \omega ^\theta _1\right) (s)\cdot \omega ^\theta _2(s)\Vert _{L^1(\Omega )} }{(t-s)^{\frac{3}{2}-\frac{1}{p}}} +\frac{\Vert r^{-\frac{3}{5}}u^\theta _1(s)u^\theta _2(s)\Vert _{L^1(\Omega )} }{(t-s)^{\frac{3}{2}-\frac{1}{p}+\frac{1}{5}}}\,ds\\&\quad \le C\, t^{1-\frac{1}{p}}\int _0^t\frac{\Vert \widetilde{u}\left( \omega ^\theta _1\right) (s)\Vert _{L^4(\Omega )}\Vert \omega ^\theta _2(s)\Vert _{L^{\frac{4}{3}}(\Omega )}}{(t-s)^{\frac{3}{2}-\frac{1}{p}}}\\&\quad \quad +\frac{\Vert r^{-\frac{3}{10}}u^\theta _1(s)\Vert _{L^2(\Omega )}\Vert r^{-\frac{3}{10}}u^\theta _2(s)\Vert _{L^{2}(\Omega )}}{(t-s)^{\frac{3}{2}-\frac{1}{p}+\frac{1}{5}}}\,ds\\&\quad \le C\, t^{1-\frac{1}{p}}\int _0^t \frac{\Vert \omega ^\theta _1\Vert _{X_T}\Vert \omega ^\theta _2\Vert _{X_T}}{(t-s)^{\frac{3}{2}-\frac{1}{p}}\cdot s^{\frac{1}{2}}} +\frac{\Vert u^\theta _1\Vert _{X_T}\Vert u^\theta _2\Vert _{X_T}}{(t-s)^{\frac{17}{10}-\frac{1}{p}}\cdot s^{\frac{3}{10}}}\,ds\\&\quad \le A^\omega _p\left( \Vert \omega ^\theta _1\Vert _{X_T}\Vert \omega ^\theta _2\Vert _{X_T}+\Vert u^\theta _1\Vert _{X_T}\Vert u^\theta _2\Vert _{X_T}\right) . \end{aligned} \end{aligned}$$

(3.7)

• The estimate of $u^\theta$ terms

Thanks to (2.16) (with $\alpha =-\frac{3}{10},~\beta =\frac{13}{10}$), by a similar derivation of (3.4), (3.5), we get

$$\begin{aligned} \begin{aligned}&\sup _{0<t\le T}\left( t^{\frac{1}{4}}\Vert S(t)u^\theta _0\Vert _{L^{4}(\Omega )} +t^{\frac{3}{20}}\Vert r^{-\frac{3}{10}}S(t)u^\theta _0\Vert _{L^{2}(\Omega )}\right) \le C_1\left( \Vert u^\theta _0\Vert _{L^{2}(\Omega )} +\Vert r^{-\frac{3}{10}}u^\theta _0\Vert _{L^{\frac{20}{13}}(\Omega )}\right) ,\\&\quad \hbox {and}\quad \lim _{t\rightarrow 0}\left( t^{\frac{1}{4}}\Vert S(t)u^\theta _0\Vert _{L^{4}(\Omega )}+t^{\frac{3}{20}}\Vert r^{-\frac{3}{10}}S(t)u^\theta _0\Vert _{L^{2}(\Omega )}\right) =0, \end{aligned} \end{aligned}$$

(3.8)

Given $\left( \omega ^\theta _1,u^\theta _1\right) ,~\left( \omega ^\theta _2,u^\theta _2\right) \in X_T,$ for any $t\in [0,T],$ we define the mapping $\mathcal {F}^u$ on $X_T\times X_T$ by

$$\begin{aligned}&\mathcal {F}^u\left( \left( \omega _1,u^\theta _1\right) ,\left( \omega _2,u^\theta _2\right) \right) (t)\buildrel \mathrm{def}\over =\int _0^tS(t-s)\nonumber \\&\quad \left( \mathrm{div}\,_*\left( \widetilde{u}\left( \omega ^\theta _1\right) (s)\cdot u^\theta _2(s)\right) -\frac{2u^r\left( \omega ^\theta _1\right) (s)\cdot u^\theta _2(s)}{r}\right) \,ds. \end{aligned}$$

(3.9)

Then for any $q\in [2,\infty [$, we deduce from (2.15), (2.17) and (2.4) that for any $t\in ]0,T]$

$$\begin{aligned}&t^{\frac{1}{2}-\frac{1}{q}}\Vert \mathcal {F}^u\left( \left( \omega _1,u^\theta _1\right) ,\left( \omega _2,u^\theta _2\right) \right) (t)\Vert _{L^q(\Omega )}\nonumber \\&\quad \le t^{\frac{1}{2}-\frac{1}{q}}\int _0^t\frac{C}{(t-s)^{1-\frac{1}{q}}}\Vert \widetilde{u}\left( \omega ^\theta _1\right) (s)\cdot u^\theta _2(s)\Vert _{L^2(\Omega )}\,ds\nonumber \\&\quad \le t^{\frac{1}{2}-\frac{1}{q}}\int _0^t\frac{C}{(t-s)^{1-\frac{1}{q}}}\Vert \widetilde{u}\left( \omega ^\theta _1\right) (s)\Vert _{L^4(\Omega )}\Vert u^\theta _2(s)\Vert _{L^4(\Omega )}\,ds\\&\quad \le t^{\frac{1}{2}-\frac{1}{q}}\int _0^t\frac{C}{(t-s)^{1-\frac{1}{q}}}\Vert \omega ^\theta _1(s)\Vert _{L^{\frac{4}{3}}(\Omega )}\Vert u^\theta _2(s)\Vert _{L^4(\Omega )}\,ds\nonumber \\&\quad \le t^{\frac{1}{2}-\frac{1}{q}}\int _0^t\frac{C}{(t-s)^{1-\frac{1}{q}}}\frac{\Vert \omega ^\theta _1\Vert _{X_t}\Vert u^\theta _2\Vert _{X_t}}{s^{\frac{1}{2}}}\,ds\le A^u_q\Vert \omega ^\theta _1\Vert _{X_T}\Vert u^\theta _2\Vert _{X_T}.\nonumber \end{aligned}$$

(3.10)

Along the same line, for any $\kappa \in [\frac{20}{13},4[$, we deduce from (2.14), (2.16) (with $\alpha =-\frac{3}{10},~\beta =\frac{3}{10}$) and (2.4) that for any $t\in ]0,T]$

$$\begin{aligned} \begin{aligned}&t^{\frac{13}{20}-\frac{1}{\kappa }}\Vert r^{-\frac{3}{10}}\mathcal {F}^u\left( \left( \omega _1,u^\theta _1\right) ,\left( \omega _2,u^\theta _2\right) \right) (t)\Vert _{L^\kappa (\Omega )}\\&\quad \le t^{\frac{13}{20}-\frac{1}{\kappa }}\int _0^t\frac{C}{(t-s)^{\frac{5}{4}-\frac{1}{\kappa }}}\Vert \widetilde{u}\left( \omega ^\theta _1\right) (s)\cdot r^{-\frac{3}{10}}u^\theta _2(s)\Vert _{L^{\frac{4}{3}}(\Omega )}\,ds\\&\quad \le t^{\frac{13}{20}-\frac{1}{\kappa }}\int _0^t\frac{C}{(t-s)^{\frac{5}{4}-\frac{1}{\kappa }}} \Vert \widetilde{u}\left( \omega ^\theta _1\right) (s)\Vert _{L^4(\Omega )} \Vert r^{-\frac{3}{10}}u^\theta _2(s)\Vert _{L^2(\Omega )}\,ds\\&\quad \le t^{\frac{13}{20}-\frac{1}{\kappa }}\int _0^t\frac{C}{(t-s)^{\frac{5}{4}-\frac{1}{\kappa }}}\frac{\Vert \omega ^\theta _1\Vert _{X_t}\Vert u^\theta _2\Vert _{X_t}}{s^{\frac{2}{5}}}\,ds\le B^u_k\Vert \omega ^\theta _1\Vert _{X_T}\Vert u^\theta _2\Vert _{X_T}. \end{aligned} \end{aligned}$$

(3.11)

• Fixed point argument

For $\left( \omega ^\theta _1,u^\theta _1\right) ,~\left( \omega ^\theta _2,u^\theta _2\right) \in X_T$, we consider the following bilinear map

$$\begin{aligned} \mathcal {F}\left( \left( \omega _1,u^\theta _1\right) ,\left( \omega _2,u^\theta _2\right) \right) \buildrel \mathrm{def}\over =\left( {\mathcal F}^\omega ,{\mathcal F}^u\right) \left( \left( \omega _1,u^\theta _1\right) ,\left( \omega _2,u^\theta _2\right) \right) , \end{aligned}$$

(3.12)

with ${\mathcal F}^\omega ,~{\mathcal F}^u$ given by (3.6) and (3.9) respectively.

By virtue of (3.5) and (3.8), for any $\omega ^\theta _0 \in L^1(\Omega ),~u^\theta _0\in L^2(\Omega )$ with $r^{-\frac{3}{10}}u^\theta _0\in L^{\frac{20}{13}}(\Omega )$, there exists a positive time T such that

$$\begin{aligned} \begin{aligned} EL(T)&\buildrel \mathrm{def}\over =\sup _{t\in [0,T]}\left( t^{\frac{1}{4}}\bigl \Vert S(t)\omega ^\theta _0\bigr \Vert _{L^{\frac{4}{3}}(\Omega )} +t^{\frac{1}{4}}\Vert S(t)u^\theta _0\Vert _{L^4(\Omega )}+t^{\frac{3}{20}}\Vert r^{-\frac{3}{10}}S(t)u^\theta _0\Vert _{L^{2}(\Omega )}\right) \\&\le \frac{1}{4\left( A^\omega _{4/3}+A^u_4+B^u_2\right) }, \end{aligned} \end{aligned}$$

(3.13)

where the constants $A^\omega _{4/3},~A^u_4$ and $B^u_2$ are determined by (3.7), (3.10) and (3.11) respectively. Then we deduce from Lemma 2.2 that (3.1) has a unique solution $(\omega ^\theta , u^\theta )$ in $X_T$. Furthermore, by virtue of (3.5) and (3.8), for any $\varepsilon >0,$ there exists $T_\varepsilon >0$ so that

$$\begin{aligned} EL(T_\varepsilon )\le \varepsilon . \end{aligned}$$

Then Lemma 2.2 ensures that

$$\begin{aligned} \bigl \Vert \left( \omega ^\theta ,u^\theta \right) \bigr \Vert _{X_{T_\varepsilon }}\le 2\varepsilon , \end{aligned}$$

which implies that

$$\begin{aligned} \lim _{T\rightarrow 0} \bigl \Vert \left( \omega ^\theta , u^\theta \right) \bigr \Vert _{X_{T}}=0. \end{aligned}$$

(3.14)

Futhermore, for any $T>0$, it follows from (3.4) and (3.8) that

$$\begin{aligned} EL(T)\le C_1\left( \Vert \omega ^\theta _0\Vert _{L^1(\Omega )}+\Vert u^\theta _0\Vert _{L^2(\Omega )} +\Vert r^{-\frac{3}{10}}u^\theta _0\Vert _{L^{\frac{20}{13}}(\Omega )}\right) \le \frac{1}{4\left( A^\omega _{4/3}+A^u_4+B^u_2\right) }, \end{aligned}$$

provided that c in (1.9) satisfying $c\le \frac{1}{4C_1\left( A^\omega _{4/3}+A^u_4+B^u_2\right) }.$ This together with Lemma 2.2 shows that (3.1) has a unique global solution in $X_\infty .$

• Behavior near $t=0$

Let us now turn to the estimate (1.8). Let $(\omega ^\theta , u^\theta )$ be the unique solution of (3.1) on [0, T] obtained in the previous one step, we denote

$$\begin{aligned}&J_{p,q,\kappa }(T)\buildrel \mathrm{def}\over =\sup _{0\le t\le T}(t^{1-\frac{1}{p}}\bigl \Vert S(t)\omega ^\theta _0\bigr \Vert _{L^p(\Omega )} +t^{\frac{1}{2}-\frac{1}{q}}\Vert S(t)u^\theta _0\Vert _{L^q(\Omega )}\\&\quad +t^{\frac{13}{20}-\frac{1}{\kappa }}\Vert r^{-\frac{3}{10}}S(t)u^\theta _0\Vert _{L^{\kappa }(\Omega )}. \end{aligned}$$

Along the same line to the proof of (3.5), it is easy to observe that

$$\begin{aligned} \lim \limits _{T\rightarrow 0}J_{p,q,\kappa }(T)=0,\quad \forall p\in ]1,\infty ],~q\in ]2,\infty ],~\kappa \in ]{20}/{13},\infty ]. \end{aligned}$$

(3.15)

While it follows from the proof of (3.7), (3.10) and (3.11) that

$$\begin{aligned} \begin{aligned}&\bigl \Vert \omega ^\theta (t)-S(t)\omega ^\theta _0\bigr \Vert _{L^1(\Omega )}\le A^\omega _p\left( \Vert \omega ^\theta \Vert ^2_{X_t} +\Vert u^\theta \Vert ^2_{X_t}\right) ,\\&\Vert u^\theta (t)-S(t)u^\theta _0\Vert _{L^2(\Omega )}\le A^u_2\Vert \omega ^\theta \Vert _{X_t}\Vert u^\theta \Vert _{X_t},\\&\Vert r^{-\frac{3}{10}}u^\theta (t)-r^{-\frac{3}{10}}S(t)u^\theta _0\Vert _{L^{\frac{20}{13}}(\Omega )}\le B^u_{\frac{20}{13}}\Vert \omega ^\theta \Vert _{X_t}\Vert u^\theta \Vert _{X_t}, \end{aligned} \end{aligned}$$

from which, (2.18) and (3.14), we infer

$$\begin{aligned}&\lim _{t\rightarrow 0}\left( \bigl \Vert \omega ^\theta (t)-\omega ^\theta _0\bigr \Vert _{L^1(\Omega )}+ \Vert u^\theta (t)-u^\theta _0\Vert _{L^2(\Omega )}\right. \nonumber \\&\quad +\left. \Vert r^{-\frac{3}{10}}u^\theta (t)-r^{-\frac{3}{10}}u^\theta _0 \Vert _{L^{\frac{20}{13}}(\Omega )}\right) =0. \end{aligned}$$

(3.16)

Whereas for any $t_0>0,$ we get, by using a similar derivation of (3.7), that

$$\begin{aligned} \begin{aligned} \Bigl \Vert&\int _{t_0}^t S(t-s)\left( \mathrm{div}\,_*\left( \widetilde{u}(s)\omega ^\theta (s)\right) +\frac{2u^\theta (s)\omega ^r(s)}{r}\right) \,ds\Bigr \Vert _{L^1(\Omega )}\\&\le C\int _{t_0}^t \frac{\Vert \omega ^\theta _1\Vert _{X_T}\Vert \omega ^\theta _2\Vert _{X_T}}{(t-s)^{\frac{1}{2}}\cdot s^{\frac{1}{2}}} +\frac{\Vert u^\theta _1\Vert _{X_T}\Vert u^\theta _2\Vert _{X_T}}{(t-s)^{\frac{7}{10}}\cdot s^{\frac{3}{10}}}\,ds\\&\le C\left( t_0^{-\frac{1}{2}}(t-t_0)^{\frac{1}{2}}+t_0^{-\frac{3}{10}}(t-t_0)^{\frac{3}{10}}\right) \left( \Vert \omega ^\theta \Vert _{X_t}^2+\Vert u^\theta \Vert _{X_t}\Vert \omega ^r\Vert _{X_t}\right) . \end{aligned} \end{aligned}$$

Hence by virtue of the expression (3.1), we deduce that

$$\begin{aligned} \lim _{t\rightarrow t_0}\Vert \omega ^\theta (t)-\omega ^\theta (t_0)\Vert _{L^1(\Omega )}=0. \end{aligned}$$

(3.17)

Exactly along the same line, we can prove that

$$\begin{aligned} \lim _{t\rightarrow t_0}\left( \Vert u^\theta (t)-u^\theta (t_0)\Vert _{L^2(\Omega )} +\Vert r^{-\frac{3}{10}}u^\theta (t)-r^{-\frac{3}{10}}u^\theta (t_0)\Vert _{L^{\frac{20}{13}}(\Omega )}\right) =0, \end{aligned}$$

This together with (3.16) and (3.17) ensures that

$$\begin{aligned} \omega \in C([0,T];L^1(\Omega )),~u^\theta \in C([0,T];L^2(\Omega )) \text{ and } r^{-\frac{3}{10}}u^\theta \in C([0,T];L^{\frac{20}{13}}(\Omega )). \end{aligned}$$

(3.18)

For any $q\in ]2,\infty ]$, by using (3.10), (3.14) and (3.15), we deduce that $M_q(T)$ are bounded, and $M_q(T)\rightarrow 0$ as $T\rightarrow 0$.

For the estimate of $N_\kappa (T)$ and $L_p(T)$, we shall use a bootstrap argument. Indeed to estimate $N_\kappa (T)$, we get, by a similar derivation of (3.11), that

$$\begin{aligned} \begin{aligned}&t^{\frac{13}{20}-\frac{1}{\kappa }}\Vert r^{-\frac{3}{10}}\mathcal {F}^u \left( \left( \omega ,u^\theta \right) ,\left( \omega ,u^\theta \right) \right) (t)\Vert _{L^\kappa (\Omega )}\\&\le t^{\frac{13}{20}-\frac{1}{\kappa }}\int _0^t\frac{C}{(t-s)^{\frac{3}{4}+\frac{1}{\kappa _1}-\frac{1}{\kappa }}}\Vert \widetilde{u}(\omega ^\theta )(s) \cdot r^{-\frac{3}{10}}u^\theta (s)\Vert _{L^{\frac{4\kappa _1}{4+\kappa _1}}(\Omega )}\,ds\\&\le t^{\frac{13}{20}-\frac{1}{\kappa }}\int _0^t\frac{C}{(t-s)^{\frac{3}{4}+\frac{1}{\kappa _1}-\frac{1}{\kappa }}} \Vert \widetilde{u}(\omega ^\theta )(s)\Vert _{L^4(\Omega )}\Vert r^{-\frac{3}{10}}u^\theta (s)\Vert _{L^{\kappa _1}(\Omega )}\,ds\\&\le t^{\frac{13}{20}-\frac{1}{\kappa }}\int _0^t\frac{C}{(t-s)^{\frac{3}{4}+\frac{1}{\kappa _1}-\frac{1}{\kappa }}} \frac{\Vert \omega ^\theta \Vert _{X_t}N_{\kappa _1}(T)}{s^{\frac{9}{10}-\frac{1}{\kappa _1}}}\,ds \le B^u_{\kappa ,\kappa _1}\Vert \omega ^\theta \Vert _{X_T}N_{\kappa _1}(T), \end{aligned} \end{aligned}$$

(3.19)

where the exponents $\kappa \in ]{20}/{13},\infty ],~\kappa _1\in ]{20}/{13},\infty ]$ satisfying

$$\begin{aligned} \frac{4\kappa _1}{4+\kappa _1}\le \kappa \quad \hbox {and}\quad \frac{3}{4}+\frac{1}{\kappa _1}-\frac{1}{\kappa }<1. \end{aligned}$$

(3.20)

Meanwhile it follows from (3.1) and (3.19) that

$$\begin{aligned} N_\kappa (T)\le J_{p,q,\kappa }(T)+B^u_{\kappa ,\kappa _1}\Vert \omega ^\theta \Vert _{X_T}N_{\kappa _1}(T). \end{aligned}$$

(3.21)

Note that $J_{p,q,\kappa }(T),~\Vert \omega ^\theta \Vert _{X_T},~N_2(T)\rightarrow 0$ as $T\rightarrow 0$, by taking $\kappa _1=2$ in (3.21), it follows from (3.20) that for any $\kappa \in ]{20}/{13},4[$, we have

$$\begin{aligned} N_\kappa (T)\quad \text{ are } \text{ bounded } \quad \hbox {and}\quad N_\kappa (T)\rightarrow 0\quad \text{ as }\ T\rightarrow 0. \end{aligned}$$

(3.22)

Next, taking $\kappa _1=3$ in (3.21), we deduce from (3.20) that (3.22) holds for any $\kappa \in [4,12[$. Along the same line, taking $\kappa _1=10$ in (3.21) ensures (3.22) for any $\kappa \in [12,\infty ]$. Hence we prove that (3.22) holds for any $\kappa \in ]{20}/{13},\infty ]$.

To handle $L_p(T)$, we get, by a similar derivation of (3.7), that but here we need to split the integral area in two parts,

$$\begin{aligned} \begin{aligned}&t^{1-\frac{1}{p}}\Vert \mathcal {F}^\omega \left( \left( \omega ,u^\theta \right) ,\left( \omega ,u^\theta \right) \right) (t)\Vert _{L^p(\Omega )}\\&\quad \le C\, t^{1-\frac{1}{p}}\int _0^{\frac{t}{2}}\frac{\left( s^{\frac{1}{4}}\Vert \omega ^\theta (s)\Vert _{L^{\frac{4}{3}}(\Omega )}\right) ^2}{(t-s)^{\frac{3}{2}-\frac{1}{p}}s^{\frac{1}{2}}} +\frac{\left( s^{\frac{3}{20}}\Vert r^{-\frac{3}{10}}u^\theta (s)\Vert _{L^{2}(\Omega )}\right) ^2}{(t-s)^{\frac{3}{2}-\frac{1}{p}+\frac{1}{5}}s^{\frac{3}{10}}}\,ds\\&\quad \quad +C\, t^{1-\frac{1}{p}}\int _{\frac{t}{2}}^t\left( \frac{\left( s^{1-\frac{1}{p_1}}\Vert \omega ^\theta (s)\Vert _{L^{p_1}(\Omega )}\right) ^2 }{(t-s)^{\frac{2}{p_1}-\frac{1}{p}}s^{2-\frac{2}{p_1}}} +\frac{\left( s^{\frac{13}{20}-\frac{1}{2p}}\Vert r^{-\frac{3}{10}}u^\theta (s)\Vert _{L^{2p}(\Omega )}\right) ^2}{(t-s)^{\frac{7}{10}}s^{\frac{13}{10}-\frac{1}{p}}}\right) ds,\\&\quad \le A^\omega _{p,p_1,p_2}\left( \Vert \omega ^\theta \Vert ^2_{X_T}+\Vert u^\theta \Vert ^2_{X_T}+L_{p_1}^2(T)+N^2_{2p}(T)\right) . \end{aligned} \end{aligned}$$

(3.23)

where the exponents $p,~p_1\in ]1,\infty ]$ satisfying

$$\begin{aligned} \frac{1}{2}\le \frac{2}{p_1}-\frac{1}{p}<1. \end{aligned}$$

(3.24)

Then we deduce from (3.1) and (3.23) that

$$\begin{aligned} L_p(T)\le J_{p,q,\kappa }(T)+A^\omega _{p,p_1,p_2}\left( \Vert \omega ^\theta \Vert ^2_{X_T}+\Vert u^\theta \Vert ^2_{X_T}+L_{p_1}^2(T)+N^2_{2p}(T)\right) . \end{aligned}$$

(3.25)

Note that $J_{p,q,\kappa }(T),~L_{\frac{4}{3}}=\Vert \omega ^\theta \Vert _{X_T},~\Vert u^\theta \Vert _{X_T},~N_{2p}(T)\rightarrow 0$ as $T\rightarrow 0$, by taking $p_1=\frac{4}{3}$ in (3.25), we deduce from (3.24), that

$$\begin{aligned} L_p(T) \quad \text{ are } \text{ bounded } \quad \hbox {and}\quad L_p(T)\rightarrow 0 \quad \text{ as } \ T\rightarrow 0 \end{aligned}$$

(3.26)

for any $p \in ]1,2[$. Next taking $p_1=\frac{5}{3}$ in (3.25) ensures that (3.26) holds for any $p\in [2,5[$. Similarly, taking $p_1=\frac{5}{2}$ in (3.25) implies that (3.26) holds for any $p\in [5,\infty ]$. Thus (3.26) holds for any $p\in ]1,\infty ]$. This completes the proof of Theorem 1.1. $\square$

Let $L_p(T),~M_q(T),~N_\kappa (T)$ be given by (1.7), for any $s\in \mathop {\mathbb N}\nolimits ^+$, we shall denote

$$\begin{aligned} \begin{aligned}&L_{p_1,p_2}(T)\buildrel \mathrm{def}\over =\sup _{p_1\le p\le p_2}L_p(T),\quad M_{q_1,q_2}(T)\buildrel \mathrm{def}\over =\sup _{q_1\le q\le q_2} M_q(T),\\&N_{\kappa _1,\kappa _2}(T)\buildrel \mathrm{def}\over =\sup _{\kappa _1\le \kappa \le \kappa _2} N_\kappa (T) \quad \hbox {and}\quad L^s_{p_1,p_2}(T)\buildrel \mathrm{def}\over =L_{p_1,p_2}(T)+|L_{p_1,p_2}(T)|^s,\\&M^s_{q_1,q_2}(T)\buildrel \mathrm{def}\over =M_{q_1,q_2}(T)+|M_{q_1,q_2}(T)|^s,\quad N^s_{\kappa _1,\kappa _2}(T)\buildrel \mathrm{def}\over =N_{\kappa _1,\kappa _2}(T)+|N_{\kappa _1,\kappa _2}(T)|^s. \end{aligned} \end{aligned}$$

(3.27)

For later use, we state the following result.

Corollary 3.1

For any $0\le \gamma \le \delta \le 1,~1\le p\le \infty ,~1\le q_2\le q_1\le \infty$, under the assumptions of Theorem 1.1, if we assume moreover that $r^{-\gamma }u^\theta _0\in L^{q_2}(\Omega )$, then there holds

$$\begin{aligned}&\Vert r^{-\delta }u^\theta (t)\Vert _{L^{q_1}(\Omega )} \lesssim \frac{L_{\frac{20}{17},2}(T)N_{\frac{20}{13},20}(T)}{t^{\frac{1}{2}+\frac{\delta }{2}-\frac{1}{q_1}}} +\frac{\Vert r^{-\gamma }u^\theta _0\Vert _{L^{q_2}(\Omega )}}{t^{\frac{\delta -\gamma }{2}+\frac{1}{q_2}-\frac{1}{q_1}}},\quad \forall \ t\in [0, T], \end{aligned}$$

(3.28)

$$\begin{aligned}&\Vert r^{-\delta }\omega ^\theta (t)\Vert _{L^p(\Omega )} \lesssim \frac{L_{1,\frac{20}{3}}^5(T)+M_2^3(T) +N^5_{\frac{20}{13},20}(T) }{t^{1+\frac{\delta }{2}-\frac{1}{p}}},\quad \forall \ t\in [0, T]. \end{aligned}$$

(3.29)

Proof

When $q_1\in [1,10]$, in view of (3.1), we get, by applying Proposition 2.2 and then Lemma 2.4, that

$$\begin{aligned}&\Vert r^{-\delta }u^\theta (t)\Vert _{L^{q_1}(\Omega )} \lesssim \frac{\Vert r^{-\gamma }u^\theta _0\Vert _{L^{q_2}(\Omega )}}{t^{\frac{\delta -\gamma }{2}+\frac{1}{q_2}-\frac{1}{q_1}}} +\int _0^{\frac{t}{2}} \frac{\Vert \widetilde{u}(s)\Vert _{L^{\frac{20}{7}}(\Omega )} \Vert r^{-\frac{3}{10}}u^\theta (s)\Vert _{L^{\frac{20}{13}}(\Omega )}}{(t-s)^{\frac{27}{20}+\frac{\delta }{2}-\frac{1}{q_1}}}\, ds\\&\quad \quad +\int _{\frac{t}{2}}^t \frac{\Vert \widetilde{u}(s)\Vert _{L^{\frac{20}{7}q_1}(\Omega )}\Vert r^{-\frac{3}{10}}u^\theta (s)\Vert _{L^{{\frac{20}{13}q_1}}(\Omega )}}{(t-s)^{\frac{7}{20}+\frac{\delta }{2}}}\,ds\\&\quad \lesssim \frac{\Vert r^{-\gamma }u^\theta _0\Vert _{L^{q_2}(\Omega )} }{t^{\frac{\delta -\gamma }{2}+\frac{1}{q_2}-\frac{1}{q_1}}} +\int _0^{\frac{t}{2}} \frac{L_{\frac{20}{17}}(T)N_{\frac{20}{13}}(T)}{(t-s)^{\frac{27}{20}+\frac{\delta }{2}-\frac{1}{q_1}}s^{\frac{3}{20}}}\,ds +\int _{\frac{t}{2}}^t \frac{L_{\frac{20q_1}{10q_1+7}}(T)N_{\frac{20}{13}q_1}(T)}{(t-s)^{\frac{7}{20}+\frac{\delta }{2}}s^{\frac{23}{20}-\frac{1}{q_1}}}\,ds\\&\quad \lesssim \frac{\Vert r^{-\gamma }u^\theta _0\Vert _{L^{q_2}(\Omega )} }{t^{\frac{\delta -\gamma }{2}+\frac{1}{q_2}-\frac{1}{q_1}}} +\frac{L_{\frac{20}{17}}(T)N_{\frac{20}{13}}(T)+L_{\frac{20q_1}{10q_1+7}}(T)N_{\frac{20}{13}q_1}(T)}{t^{\frac{1}{2}+\frac{\delta }{2}-\frac{1}{q_1}}}, \end{aligned}$$

which yields (3.28) for $q_1\in [1,10]$.

When $q_1\in ]10,\infty ]$, we get, by a similar derivation, that

$$\begin{aligned}&\Vert r^{-\delta }u^\theta (t)\Vert _{L^{q_1}(\Omega )} \lesssim \frac{\Vert r^{-\gamma }u^\theta _0\Vert _{L^{q_2}(\Omega )} }{t^{\frac{\delta -\gamma }{2}+\frac{1}{q_2}-\frac{1}{q_1}}} +\int _0^{\frac{t}{2}} \frac{L_{\frac{20}{17}}(T)N_{\frac{20}{13}}(T)}{(t-s)^{\frac{27}{20}+\frac{\delta }{2}-\frac{1}{q_1}}s^{\frac{3}{20}}}\,ds\\&\quad +\int _{\frac{t}{2}}^t \frac{L_{2}(T)N_{20}(T)}{(t-s)^{\frac{2}{5}+\frac{\delta }{2}-\frac{1}{q_1}}s^{\frac{11}{10}}}\,ds \lesssim \frac{\Vert r^{-\gamma }u^\theta _0\Vert _{L^{q_2}(\Omega )} }{t^{\frac{\delta -\gamma }{2}+\frac{1}{q_2}-\frac{1}{q_1}}} +\frac{L_{\frac{20}{17}}(T)N_{\frac{20}{13}}(T)+L_{2}(T)N_{20}(T)}{t^{\frac{1}{2}+\frac{\delta }{2}-\frac{1}{q_1}}}. \end{aligned}$$

This proves the estimate (3.28).

To handle the estimate (3.29), we first consider the case when $\delta \in [0,\frac{1}{2}]$ and $p\in [1,5]$. In this case, applying Proposition 2.2 to (3.1) and then using the estimate (3.28) gives rise to

$$\begin{aligned} \begin{aligned} \Vert r^{-\delta }\omega ^\theta (t)\Vert _{L^p(\Omega )}&\lesssim \frac{\Vert \omega ^\theta _0\Vert _{L^1(\Omega )}}{t^{\frac{\delta }{2}+1-\frac{1}{p}}} +\int _0^{\frac{t}{2}}\left( \frac{\Vert \widetilde{u}(s)\Vert _{L^{4}(\Omega )}\Vert \omega ^\theta (s)\Vert _{L^{\frac{4}{3}(\Omega )}}}{(t-s)^{\frac{3}{2}+\frac{\delta }{2}-\frac{1}{p}}} +\frac{\bigl \Vert {u^\theta }(s)\bigr \Vert _{L^{2}(\Omega )}^2}{(t-s)^{2+\frac{\delta }{2}-\frac{1}{p}}}\right) \,ds\\&\quad +\int _{\frac{t}{2}}^t \left( \frac{\Vert \widetilde{u}(s)\Vert _{L^{4p}(\Omega )}\Vert \omega ^\theta (s)\Vert _{L^{\frac{4p}{3}(\Omega )}}}{(t-s)^{\frac{1}{2}+\frac{\delta }{2}}} +\frac{\bigl \Vert r^{-\frac{1+\delta }{2}}{u^\theta }(s)\bigr \Vert _{L^{2p}(\Omega )}^2}{(t-s)^{\frac{1}{2}}}\right) \,ds\\&\lesssim \frac{\Vert \omega ^\theta _0\Vert _{L^1(\Omega )}}{t^{\frac{\delta }{2}+1-\frac{1}{p}}} +\int _0^{\frac{t}{2}}\left( \frac{L_{\frac{4}{3}}^2(T)}{(t-s)^{\frac{3}{2}+\frac{\delta }{2}-\frac{1}{p}}s^{\frac{1}{2}}} +\frac{M_2^2(T)}{(t-s)^{2+\frac{\delta }{2}-\frac{1}{p}}}\right) \,ds\\&\quad +\int _{\frac{t}{2}}^t \left( \frac{L_{\frac{4p}{1+2p}}(T)L_{\frac{4p}{3}}(T)}{(t-s)^{\frac{1}{2}+\frac{\delta }{2}}s^{\frac{3}{2}-\frac{1}{p}}} +\frac{L^2_{\frac{20}{17},\frac{20}{11}}(T)N^2_{\frac{20}{13},20}(T)+\Vert u^\theta _0\Vert _{L^2(\Omega )}^2}{(t-s)^{\frac{1}{2}}s^{\frac{3}{2}+\frac{\delta }{2}-\frac{1}{p}}}\right) \,ds\\&\lesssim \frac{L_{1,\frac{20}{3}}^4(T)+M_2^2(T) +N^4_{\frac{20}{13},20}(T) }{t^{\frac{\delta }{2}+1-\frac{1}{p}}}, \end{aligned} \end{aligned}$$

(3.30)

where in the last step we use the fact that $\Vert \omega ^\theta _0\Vert _{L^1(\Omega )}\le L_1,~\Vert u^\theta _0\Vert _{L^2(\Omega )}\le M_2.$

On the other side, when $p\in ]5,\infty ]$, we have

$$\begin{aligned} \begin{aligned} \Vert r^{-\delta }\omega ^\theta (t)\Vert _{L^p(\Omega )}&\lesssim \frac{\Vert \omega ^\theta _0\Vert _{L^1(\Omega )}}{t^{\frac{\delta }{2}+1-\frac{1}{p}}} +\int _0^{\frac{t}{2}}\left( \frac{\Vert \widetilde{u}(s)\Vert _{L^{4}(\Omega )}\Vert \omega ^\theta (s)\Vert _{L^{\frac{4}{3}(\Omega )}}}{(t-s)^{\frac{3}{2}+\frac{\delta }{2}-\frac{1}{p}}} +\frac{\bigl \Vert {u^\theta }(s)\bigr \Vert _{L^{2}(\Omega )}^2}{(t-s)^{2+\frac{\delta }{2}-\frac{1}{p}}}\right) \,ds\\&\quad +\int _{\frac{t}{2}}^t \left( \frac{\Vert \widetilde{u}(s)\Vert _{L^{20}(\Omega )}\Vert \omega ^\theta (s)\Vert _{L^{\frac{20}{3}(\Omega )}}}{(t-s)^{\frac{7}{10}+\frac{\delta }{2}-\frac{1}{p}}} +\frac{\bigl \Vert r^{-\frac{1+\delta }{2}}{u^\theta }(s)\bigr \Vert _{L^{2p}(\Omega )}^2}{(t-s)^{\frac{1}{2}}}\right) \,ds\\&\lesssim \frac{L_{1,\frac{20}{3}}^4(T)+M_2^2(T) +N^4_{\frac{20}{13},20}(T) }{t^{\frac{\delta }{2}+1-\frac{1}{p}}}. \end{aligned} \end{aligned}$$

(3.31)

Combining (3.30) with (3.31) leads to the estimate (3.29) for $\delta \in [0,\frac{1}{2}]$.

For the remaining case when $\delta \in ]1/2,1]$, we first get, by a similar derivation of (3.30) and then using (3.28) that for $p\in [1,5]$

$$\begin{aligned} \begin{aligned} \Vert r^{-\delta }\omega ^\theta (t)\Vert _{L^p(\Omega )}&\lesssim \frac{\Vert \omega ^\theta _0\Vert _{L^1(\Omega )}}{t^{\frac{\delta }{2}+1-\frac{1}{p}}} +\int _0^{\frac{t}{2}}\left( \frac{\Vert \widetilde{u}(s)\Vert _{L^{4}(\Omega )}\Vert \omega ^\theta (s)\Vert _{L^{\frac{4}{3}(\Omega )}}}{(t-s)^{\frac{3}{2}+\frac{\delta }{2}-\frac{1}{p}}} +\frac{\bigl \Vert {u^\theta }(s)\bigr \Vert _{L^{2}(\Omega )}^2}{(t-s)^{2+\frac{\delta }{2}-\frac{1}{p}}}\right) \,ds\\&\quad +\int _{\frac{t}{2}}^t \left( \frac{\Vert \widetilde{u}(s)\Vert _{L^{4p}(\Omega )}\Vert r^{-\frac{1}{2}}\omega ^\theta (s)\Vert _{L^{\frac{4p}{3}(\Omega )}}}{(t-s)^{\frac{1}{4}+\frac{\delta }{2}}} +\frac{\bigl \Vert r^{-\frac{1+\delta }{2}}{u^\theta }(s)\bigr \Vert _{L^{2p}(\Omega )}^2}{(t-s)^{\frac{1}{2}}}\right) \,ds\\&\lesssim \frac{L_{1,\frac{20}{3}}^5(T)+M_2^3(T) +N^5_{\frac{20}{13},20}(T) }{t^{\frac{\delta }{2}+1-\frac{1}{p}}}. \end{aligned} \end{aligned}$$

(3.32)

Finally when $p\in ]5,\infty ]$, we deduce, by a similar derivation of (3.31), that

$$\begin{aligned} \begin{aligned} \Vert r^{-\delta }\omega ^\theta (t)\Vert _{L^p(\Omega )}&\lesssim \frac{\Vert \omega ^\theta _0\Vert _{L^1(\Omega )}}{t^{\frac{\delta }{2}+1-\frac{1}{p}}} +\int _0^{\frac{t}{2}}\left( \frac{\Vert \widetilde{u}(s)\Vert _{L^{4}(\Omega )}\Vert \omega ^\theta (s)\Vert _{L^{\frac{4}{3}(\Omega )}}}{(t-s)^{\frac{3}{2}+\frac{\delta }{2}-\frac{1}{p}}} +\frac{\bigl \Vert {u^\theta }(s)\bigr \Vert _{L^{2}(\Omega )}^2}{(t-s)^{2+\frac{\delta }{2}-\frac{1}{p}}}\right) \,ds\\&\quad +\int _{\frac{t}{2}}^t \left( \frac{\Vert \widetilde{u}(s)\Vert _{L^{20}(\Omega )}\Vert r^{-\frac{1}{2}}\omega ^\theta (s)\Vert _{L^{\frac{20}{3}(\Omega )}}}{(t-s)^{\frac{9}{20}+\frac{\delta }{2}-\frac{1}{p}}} +\frac{\bigl \Vert r^{-\frac{1+\delta }{2}}{u^\theta }(s)\bigr \Vert _{L^{2p}(\Omega )}^2}{(t-s)^{\frac{1}{2}}}\right) \,ds\\&\lesssim \frac{L_{1,\frac{20}{3}}^5(T)+M_2^3(T) +N^5_{\frac{20}{13},20}(T) }{t^{\frac{\delta }{2}+1-\frac{1}{p}}}. \end{aligned} \end{aligned}$$

(3.33)

Combining (3.32) with (3.33) implies (3.29) for $\delta \in ]\frac{1}{2},1]$. This completes the proof of the estimate (3.29) and hence the corollary. $\square$

4 Global a priori estimates of (1.4) with nearly critical initial data

The goal of this section is basically to prove that, as long as the initial data belongs to the almost critical spaces, the system has a unique global solution.

Let us introduce another two variables which are of great importance in our work, namely

$$\begin{aligned} \eta \buildrel \mathrm{def}\over =\frac{\omega ^\theta }{r},\quad V^\varepsilon \buildrel \mathrm{def}\over =\frac{u^\theta }{r^{1-\varepsilon }}\quad \text{ for } \text{ any }\ \varepsilon \in ]0,1[. \end{aligned}$$

(4.1)

And it is not difficult to deduce the equations for $\eta$ and $V^\varepsilon$ from (1.4) that

$$\begin{aligned} \left\{ \begin{array}{l} \displaystyle \partial _t \eta +\left( u^r\partial _r+u^z\partial _z\right) \eta -\left( \Delta +\frac{2}{r}\partial _r\right) \eta -\frac{2V \partial _z V}{r^{2\varepsilon }}=0,\\ \displaystyle \partial _t V+\left( u^r\partial _r+u^z\partial _z\right) V+(2-\varepsilon )\frac{u^r V}{r} -\left( \Delta + 2(1-\varepsilon )\frac{\partial _r}{r}\right) V +\left( 2\varepsilon -\varepsilon ^2\right) \frac{V}{r^2}=0,\\ \displaystyle \eta |_{t=0} =\eta _0=\frac{\omega ^\theta _0}{r},\, V|_{t=0} =V_0=\frac{u^\theta _0}{r^{1-\varepsilon }}, \end{array} \right. \end{aligned}$$

(4.2)

here and in all that follows, we always denote $V^\varepsilon$ as V, if there is no ambiguity.

Proposition 4.1

Let $(u^r,u^\theta ,u^z)$ be a smooth enough solution of (1.3) on [0, T]. Let $p\in ]1,\frac{21}{20}]$, $\varepsilon =\frac{-9p^2+21p-4}{24p-2}\in \bigl [\frac{3251}{9280},\frac{4}{11}\bigr [$, and q be given by

$$\begin{aligned} (2-\varepsilon )q=3p. \end{aligned}$$

(4.3)

We assume that the initial data $\eta _0\in L^p$, $V_0^\varepsilon =\frac{u^\theta }{r^{1-\varepsilon }}\in L^q$, $ru^\theta _0\in L^\infty \bigcap L^{\frac{1}{p-1}},$ which satisfy

$$\begin{aligned} (2M_0)^{\frac{3(p+2)}{p(3p+11)}} \Vert ru^\theta _0\Vert _{L^{\frac{10(12p-1)}{9(p-1)(p+2)(p+3)}}}^{\frac{10(12p-1)}{3p(p+3)(3p+11)}} +(2M_0)^{\frac{p-1}{4p}} \Vert ru^\theta _0\Vert _{L^{\frac{2}{3(p-1)^2}}}^{\frac{1}{6p}}\le c_0(p-1), \end{aligned}$$

(4.4)

for some sufficiently small constant $c_0$ which does not depend on the choice of p, and $M_0\buildrel \mathrm{def}\over =\Vert V_0\Vert _{L^q}^q+\Vert \eta _0\Vert _{L^p}^p$. Then for any $t\in [0,T]$, we have

$$\begin{aligned} \begin{aligned}&\Vert V(t)\Vert _{L^q}^q+\Vert \eta (t)\Vert _{L^p}^p +\frac{p-1}{2}\Vert \nabla |\eta |^{\frac{p}{2}}\Vert _{L^2((0,t)\times \mathop {\mathbb R}\nolimits ^3)}^2\\&\quad +\frac{1}{2}\left( \Vert \nabla |V|^{\frac{q}{2}}\Vert _{L^2((0,t)\times \mathop {\mathbb R}\nolimits ^3)}^2+\Bigl \Vert \frac{|V|^{\frac{q}{2}}}{r}\Bigr \Vert _{L^2((0,t)\times \mathop {\mathbb R}\nolimits ^3)}^2\right) \le 2\left( \Vert V_0\Vert _{L^q}^q+\Vert \eta _0\Vert _{L^p}^p\right) . \end{aligned} \end{aligned}$$

(4.5)

Let us remark that both the index $\frac{10(12p-1)}{9(p-1)(p+2)(p+3)}$ and ${\frac{2}{3(p-1)^2}}$ are close enough to $\infty$ as long as p approaches 1,  which corresponds to the case with initial data in the critical spaces.

The proof of the above proposition relies on the following lemmas:

Lemma 4.1

Under the assumptions of Proposition 4.1, for any $t\in ]0,T],$ there holds

$$\begin{aligned} \begin{aligned}&\frac{1}{q}\frac{d}{dt}\Vert V(t)\Vert _{L^{q}}^{q}+ \frac{4(q-1)}{q^2}\bigl \Vert \nabla |V|^{\frac{q}{2}}\bigr \Vert _{L^2}^2+(2\varepsilon -\varepsilon ^2)\int _{\mathop {\mathbb R}\nolimits ^3}\frac{|V|^{q}}{r^2}\,dx\\&\quad \lesssim \Vert ru^\theta \Vert _{L^{\frac{5}{2-\varepsilon }\cdot \frac{p}{(p-1)(p+2)}}}^{\frac{15}{(2-\varepsilon )(3p+11)}} \Vert |V|^{\frac{q}{2}}\Vert _{L^2}^{\frac{23-17p-3p^2}{p(3p+11)}} \Vert \eta \Vert _{L^p}^{\frac{3p^2+23p-11}{2(3p+11)}}\\&\quad \quad \times \left( \Vert \nabla |\eta |^{\frac{p}{2}}\Vert _{L^{2}}^{2}+ \Vert \nabla |V|^{\frac{q}{2}}\Vert _{L^2}^{2}+ \int _{\mathop {\mathbb R}\nolimits ^3}\frac{|V|^{q}}{r^2}\,dx\right) . \end{aligned} \end{aligned}$$

(4.6)

Lemma 4.2

Under the assumptions of Proposition 4.1, for any $t\in ]0,T],$ there holds

$$\begin{aligned} \begin{aligned}&\frac{1}{p}\frac{d}{dt}\Vert \eta (t)\Vert _{L^p}^p +\frac{4(p-1)}{p^2}\Vert \nabla |\eta |^{\frac{p}{2}}\Vert _{L^2}^2\\&\quad \lesssim \Vert ru^\theta \Vert _{L^{\frac{2}{3(p-1)^2}}}^{\frac{1}{6p}} \Vert \eta \Vert ^{\frac{p-1}{4}}_{L^p}\left( \Vert \nabla |\eta |^{\frac{p}{2}}\Vert _{L^2}^2+\Vert \nabla |V|^{\frac{q}{2}}\Vert _{L^2}^2 +\Bigl \Vert \frac{|V|^q}{r^2}\Bigr \Vert _{L^1}\right) . \end{aligned} \end{aligned}$$

(4.7)

Let us admit the above lemmas and continue our proof of Proposition 4.1.

Proof of Proposition 4.1

By virtue of Lemma 2.1, we get, by summarizing (4.6) for $\varepsilon =\frac{-9p^2+21p-4}{24p-2}$ with (4.7), that

$$\begin{aligned} \begin{aligned}&\frac{d}{dt}\left( \Vert V(t)\Vert _{L^q}^q+\Vert \eta (t)\Vert _{L^p}^p\right) +\frac{4(p-1)}{p}\Vert \nabla |\eta |^{\frac{p}{2}}\Vert _{L^2}^2\\&\quad \quad +\frac{4(q-1)}{q}\Vert \nabla |V|^{\frac{q}{2}}\Vert _{L^2}^2 +\frac{3p(-9p^2+21p-4)}{24p-2}\Bigl \Vert \frac{|V|^q}{r^2}\Bigr \Vert _{L^1}\\&\quad \le C \left( \Vert ru^\theta _0\Vert _{L^{\frac{10(12p-1)}{9(p-1)(p+2)(p+3)}}}^{\frac{10(12p-1)}{3p(p+3)(3p+11)}} \Vert |V|^{\frac{q}{2}}\Vert _{L^2}^{\frac{23-17p-3p^2}{p(3p+11)}} \Vert \eta \Vert _{L^p}^{\frac{3p^2+23p-11}{2(3p+11)}} +\Vert ru^\theta _0\Vert _{L^{\frac{2}{3(p-1)^2}}}^{\frac{1}{6p}} \Vert \eta \Vert ^{\frac{p-1}{4}}_{L^p}\right) \\&\quad \quad \times \left( \Vert \nabla |\eta |^{\frac{p}{2}}\Vert _{L^2}^2+\Vert \nabla |V|^{\frac{q}{2}}\Vert _{L^2}^2 +\Bigl \Vert \frac{|V|^q}{r^2}\Bigr \Vert _{L^1}\right) . \end{aligned} \end{aligned}$$

(4.8)

Let $M_0\buildrel \mathrm{def}\over =\Vert V_0\Vert _{L^q}^q+\Vert \eta _0\Vert _{L^p}^p$, and let $T'>0$ be determined by

$$\begin{aligned} T'\buildrel \mathrm{def}\over =\sup \Bigl \{\ t\in [0,T]:\ \sup _{t'\in [0,t]}\left( \Vert V(t)\Vert _{L^q}^q+\Vert \eta (t)\Vert _{L^p}^p\right) \le 2M_0\ \Bigr \}. \end{aligned}$$

(4.9)

If $T'<T,$ then for $t\le T',$ we deduce from the (4.8) that

$$\begin{aligned}\begin{aligned}&\frac{d}{dt}\left( \Vert V(t)\Vert _{L^q}^q+\Vert \eta (t)\Vert _{L^p}^p\right) +(p-1)\Vert \nabla |\eta |^{\frac{p}{2}}\Vert _{L^2}^2+\Vert \nabla |V|^{\frac{q}{2}}\Vert _{L^2}^2 +\Bigl \Vert \frac{|V|^q}{r^2}\Bigr \Vert _{L^1}\\&\quad \le C \left( (2M_0)^{\frac{3(p+2)}{p(3p+11)}} \Vert ru^\theta _0\Vert _{L^{\frac{10(12p-1)}{9(p-1)(p+2)(p+3)}}}^{\frac{10(12p-1)}{3p(p+3)(3p+11)}} +(2M_0)^{\frac{p-1}{4p}} \Vert ru^\theta _0\Vert _{L^{\frac{2}{3(p-1)^2}}}^{\frac{1}{6p}}\right) \\&\quad \quad \times \left( \Vert \nabla |\eta |^{\frac{p}{2}}\Vert _{L^2}^2+\Vert \nabla |V|^{\frac{q}{2}}\Vert _{L^2}^2 +\Bigl \Vert \frac{|V|^q}{r^2}\Bigr \Vert _{L^1}\right) , \end{aligned}\end{aligned}$$

which together with (4.4) ensures that for any t in $[0,T']$,

$$\begin{aligned} \frac{d}{dt}\left( \Vert V(t)\Vert _{L^q}^q+\Vert \eta (t)\Vert _{L^p}^p\right) +\frac{p-1}{2}\Vert \nabla |\eta |^{\frac{p}{2}}\Vert _{L^2}^2+\frac{1}{2}\Vert \nabla |V|^{\frac{q}{2}}\Vert _{L^2}^2 +\frac{1}{2}\Bigl \Vert \frac{|V|^q}{r^2}\Bigr \Vert _{L^1}\le 0. \end{aligned}$$

(4.10)

This in particular gives rise to

$$\begin{aligned} \Vert V(t)\Vert _{L^q}^q+\Vert \eta (t)\Vert _{L^p}^p \le M_0\quad \text{ for } \text{ any }\quad t\le T'. \end{aligned}$$

(4.11)

This contradicts with the definition of $T'$ given by (4.9). As a result, it comes out $T'=T$, and there holds (4.5) for any $t\in [0,T].$ this completes the proof of the proposition. $\square$

Let us now turn to the proof of Lemmas 4.1 and 4.2.

Proof of Lemma 4.1

For any $p\in ]1,\frac{21}{20}]$ and q given by (4.3), we get, by multiplying the second equation of (4.2) by $|V|^{q-2}V$ and then integrating the resulting equality over $\mathop {\mathbb R}\nolimits ^3,$ that

$$\begin{aligned}&\frac{1}{q}\frac{d}{dt}\Vert V(t)\Vert _{L^{q}}^{q}+\frac{1}{q}\int _{\mathop {\mathbb R}\nolimits ^3}\left( u^r\partial _r +u^z\partial _z\right) |V|^{q}\,dx+(2-\varepsilon )\int _{\mathop {\mathbb R}\nolimits ^3}\frac{u^r}{r}|V|^{q}\,dx -\int _{\mathop {\mathbb R}\nolimits ^3}\Delta V \\&\quad |V|^{q-2}V\,dx-\frac{2(1-\varepsilon )}{q}\int _{\mathop {\mathbb R}\nolimits ^3}\partial _r|V|^{q}\frac{1}{r}\,dx +\left( 2\varepsilon -\varepsilon ^2\right) \int _{\mathop {\mathbb R}\nolimits ^3}\frac{|V|^{q}}{r^2}\,dx=0. \end{aligned}$$

Using the fact that $\partial _r(ru^r)+\partial _z(ru^z)=0$, which implies $\int _{\mathop {\mathbb R}\nolimits ^3}(u^r\partial _r+u^z\partial _z)|V|^{q}\,dx=0$, and the homogeneous Dirichlet boundary condition for $u^r$ on $r=0,$ we deduce

$$\begin{aligned} \begin{aligned}&\frac{1}{q}\frac{d}{dt}\Vert V(t)\Vert _{L^{q}}^{q}+\left( 2-\varepsilon \right) \int _{\mathop {\mathbb R}\nolimits ^3}\frac{u^r}{r}|V|^{q}\,dx +\frac{2(1-\varepsilon )}{q}\int _{-\infty }^{+\infty } |V|^{q}\big |_{r=0}\,dz\\&\quad +\frac{4(q-1)}{q^2}\bigl \Vert \nabla |V|^{\frac{q}{2}}\bigr \Vert _{L^2}^2+\left( 2\varepsilon -\varepsilon ^2\right) \int _{\mathop {\mathbb R}\nolimits ^3}\frac{|V|^{q}}{r^2}\,dx=0, \end{aligned} \end{aligned}$$

which implies

$$\begin{aligned} \frac{1}{q}\frac{d}{dt}\Vert V(t)\Vert _{L^{q}}^{q}+ \frac{4(q-1)}{q^2}\bigl \Vert \nabla |V|^{\frac{q}{2}}\bigr \Vert _{L^2}^2+ \left( 2\varepsilon -\varepsilon ^2\right) \int _{\mathop {\mathbb R}\nolimits ^3}\frac{|V|^{q}}{r^2}\,dx \lesssim \int _{\mathop {\mathbb R}\nolimits ^3}\bigl |\frac{u^r}{r}\bigr |\cdot |V|^{q}\,dx \end{aligned}$$

(4.12)

Let us take $q_1,q_2$ satisfying $\frac{1}{q_1}+\frac{1}{q_2}=1,$ and temporarily assume that $q_1\in \bigl ]\frac{3p}{3-p},\infty \bigr [,$ so that it follows from Lemma 2.1 and Sobolev embedding Theorem that

$$\begin{aligned} \begin{aligned} \int _{\mathop {\mathbb R}\nolimits ^3}\bigl |\frac{u^r}{r}\bigr |\cdot |V|^{q} dx&\le \bigl \Vert \frac{u^r}{r}\bigr \Vert _{L^{q_1}}\Vert |V|^{q}\Vert _{L^{q_2}}\\&\lesssim \Vert \eta \Vert _{L^p}^{\frac{p-1}{2}+\frac{3p}{2q_1}}\Vert \eta \Vert _{L^{3p}}^{\frac{3-p}{2}-\frac{3p}{2q_1}}\Vert |V|^{q}\Vert _{L^{q_2}}\\&\lesssim \Vert \eta \Vert _{L^p}^{\frac{p-1}{2}+\frac{3p}{2q_1}}\Vert \nabla |\eta |^{\frac{p}{2}}\Vert _{L^{2}}^{2\left( \frac{3-p}{2p}-\frac{3}{2q_1}\right) }\Vert |V|^{q}\Vert _{L^{q_2}} . \end{aligned} \end{aligned}$$

Take $q_2=\vartheta +\sigma +\frac{2\sigma }{3p},$ with $\vartheta >0$ and $0<\sigma <1$ to be determined later, then we have

$$\begin{aligned} \begin{aligned} \Vert |V|^{q}\Vert _{L^{q_2}}&=\left( \int _{\mathop {\mathbb R}\nolimits ^3}|V|^{q\vartheta }\left( \frac{|V|^{q}}{r^2}\right) ^{\sigma }|r^{2-\varepsilon }V|^{\frac{2\sigma }{2-\varepsilon }}\,dx\right) ^{\frac{1}{q_2}}\\&\le \Vert ru^\theta \Vert _{L^{\frac{2\sigma }{2-\varepsilon }\cdot \frac{p}{p-1}}}^{\frac{2\sigma }{q_2(2-\varepsilon )}} \Vert |V|^{\frac{q}{2}}\Vert _{L^{\frac{2p\vartheta }{1-p\sigma }}}^{\frac{2\vartheta }{q_2}}\left( \int _{\mathop {\mathbb R}\nolimits ^3}\frac{|V|^{q}}{r^2}\,dx\right) ^{\frac{\sigma }{q_2}}\\&\le \Vert ru^\theta \Vert _{L^{\frac{2\sigma }{2-\varepsilon }\cdot \frac{p}{p-1}}}^{\frac{2\sigma }{q_2(2-\varepsilon )}} \Vert |V|^{\frac{q}{2}}\Vert _{L^2}^{\frac{-\vartheta -3\sigma }{q_2}+\frac{3}{pq_2}}\Vert \nabla |V|^{\frac{q}{2}}\Vert _{L^2}^{\frac{3(\vartheta +\sigma )}{q_2}-\frac{3}{pq_2}} \left( \int _{\mathop {\mathbb R}\nolimits ^3}\frac{|V|^{q}}{r^2}\,dx\right) ^{\frac{\sigma }{q_2}}, \end{aligned} \end{aligned}$$

(4.13)

where in the last step, we use the interpolation inequality provided that

$$\begin{aligned} \frac{2p\vartheta }{1-p\sigma }\in [2,6], \end{aligned}$$

(4.14)

which will be verified later. Then we get, by applying Young’s inequality, that

$$\begin{aligned} \begin{aligned} \int _{\mathop {\mathbb R}\nolimits ^3}\bigl |\frac{u^r}{r}\bigr ||V|^q dx&\lesssim \Vert ru^\theta \Vert _{L^{\frac{2\sigma }{2-\varepsilon }\cdot \frac{p}{p-1}}}^{\frac{2\sigma }{q_2(2-\varepsilon )}} \Vert |V|^{\frac{q}{2}}\Vert _{L^2}^{\frac{-\vartheta -3\sigma }{q_2}+\frac{3}{pq_2}} \Vert \eta \Vert _{L^p}^{\frac{p-1}{2}+\frac{3p}{2q_1}}\\&\quad \times \Vert \nabla |\eta |^{\frac{p}{2}}\Vert _{L^{2}}^{2\left( \frac{3-p}{2p}-\frac{3}{2q_1}\right) } \Vert \nabla |V|^{\frac{q}{2}}\Vert _{L^2}^{\frac{3(\vartheta +\sigma )}{q_2}-\frac{3}{pq_2}} \left( \int _{\mathop {\mathbb R}\nolimits ^3}\frac{|V|^{q}}{r^2}\,dx\right) ^{\frac{\sigma }{q_2}}\\&\lesssim \Vert ru^\theta \Vert _{L^{\frac{2\sigma }{2-\varepsilon }\cdot \frac{p}{p-1}}}^{\frac{2\sigma }{q_2(2-\varepsilon )}} \Vert |V|^{\frac{q}{2}}\Vert _{L^2}^{\frac{-\vartheta -3\sigma }{q_2}+\frac{3}{pq_2}} \Vert \eta \Vert _{L^p}^{\frac{p-1}{2}+\frac{3p}{2q_1}}\\&\quad \times \left( \Vert \nabla |\eta |^{\frac{p}{2}}\Vert _{L^{2}}^{2}+ \Vert \nabla |V|^{\frac{q}{2}}\Vert _{L^2}^{2}+ \int _{\mathop {\mathbb R}\nolimits ^3}\frac{|V|^{q}}{r^2}\,dx\right) , \end{aligned} \end{aligned}$$

(4.15)

provided that

$$\begin{aligned} \frac{3-p}{2p}-\frac{3}{2q_1}+\frac{3(\vartheta +\sigma )}{2q_2}-\frac{3}{2pq_2}+\frac{\sigma }{q_2}=1, \end{aligned}$$

and this can be satisfied by choosing

$$\begin{aligned} \begin{aligned}&\sigma =\frac{5}{2(p+2)},\ \vartheta =1-\frac{5}{6p},\quad \text{ and }\quad q_2=\vartheta +\sigma +\frac{2\sigma }{3p}=\frac{3p+11}{3(p+2)}. \end{aligned} \end{aligned}$$

(4.16)

It is easy to verify that for any $p\in \bigl ]1,\frac{21}{20}\bigr ]$, the corresponding $\frac{2p\vartheta }{1-p\sigma }=\frac{2(p+2)(6p-5)}{3(4-3p)}$ is exactly in [2, 6], and $q_2$ is exactly in $\bigl ]1,\frac{3p}{4p-3}\bigr [$, thus the conjugate number $q_1=\left( 1-\frac{1}{q_2}\right) ^{-1}>\frac{3p}{3-p}$. Thus all the above calculations go through.

With the indexes given by (4.16), by inserting the Estimate (4.15) into (4.12), we achieve (4.6). This completes the proof of Lemma 4.1. $\square$

Proof of Lemma 4.2

Analogue to the proof of Lemma 4.1, we get, by first multiplying the $\eta$ equation of (4.2) by $|\eta |^{p-2}\eta$ and then integrating the resulting equality over $\mathop {\mathbb R}\nolimits ^3,$ that

$$\begin{aligned}&\frac{1}{p}\frac{d}{dt}\Vert \eta (t)\Vert _{L^p}^p+\frac{1}{p}\int _{\mathop {\mathbb R}\nolimits ^3}(u^r\partial _r+u^z\partial _z)|\eta |^p\,dx{}-\int _{\mathop {\mathbb R}\nolimits ^3}\Delta \eta |\eta |^{p-2}\eta \,dx\\&\quad -\frac{2}{p}\int _{\mathop {\mathbb R}\nolimits ^3}\frac{\partial _r|\eta |^p}{r}\,dx =\int _{\mathop {\mathbb R}\nolimits ^3}\frac{2V \partial _z V}{r^{2\varepsilon }}|\eta |^{p-2}\eta dx. \end{aligned}$$

Once again due to $\partial _r(ru^r)+\partial _z(ru^z)=0$ and $u^r|_{r=0}=0$, we get

$$\begin{aligned} \frac{1}{p}\frac{d}{dt}\Vert \eta (t)\Vert _{L^p}^p +\frac{4(p-1)}{p^2}\Vert \nabla |\eta |^{\frac{p}{2}}\Vert _{L^2_x}^2 +\frac{2}{p}\int _{-\infty }^{+\infty }|\eta |^p|_{r=0} dz =\int _{\mathop {\mathbb R}\nolimits ^3}\frac{2V \partial _z V}{r^{2\varepsilon }}|\eta |^{p-2}\eta dx. \end{aligned}$$

(4.17)

It is easy to observe that

$$\begin{aligned} \begin{aligned} \bigl |\int _{\mathop {\mathbb R}\nolimits ^3}\frac{2V \partial _z V}{r^{2\varepsilon }}|\eta |^{p-2}\eta dx\bigr |&\le \frac{2}{q}\int _{\mathop {\mathbb R}\nolimits ^3}|\eta |^{p-1}\bigl |\partial _z|V|^{\frac{q}{2}}\bigr | \frac{|V|^{2-\frac{q}{2}}}{r^{2\varepsilon }}\,dx\\&\lesssim \bigl \Vert |\eta |^{p-1}\bigr \Vert _{L^\frac{2p}{p-1}}\bigl \Vert \partial _z|V|^{\frac{q}{2}}\bigr \Vert _{L^2} \bigl \Vert \frac{|V|^{2-\frac{q}{2}}}{r^{2\varepsilon }}\bigr \Vert _{L^{2p}}. \end{aligned} \end{aligned}$$

It follows from Sobolev embedding Theorem that

$$\begin{aligned} \begin{aligned} \bigl \Vert |\eta |^{p-1}\bigr \Vert _{L^\frac{2p}{p-1}}&=\Vert |\eta |^{\frac{p}{2}}\Vert _{L^4}^{\frac{2(p-1)}{p}}\lesssim \Vert |\eta |^{\frac{p}{2}}\Vert _{L^2}^{\frac{p-1}{2p}} \Vert \nabla |\eta |^{\frac{p}{2}}\Vert _{L^2}^{\frac{3(p-1)}{2p}}\\&=\Vert \eta \Vert _{L^p}^{\frac{p-1}{4}}\Vert \nabla |\eta |^{\frac{p}{2}}\Vert _{L^2}^{\frac{3(p-1)}{2p}}. \end{aligned} \end{aligned}$$

As a result, we obtain

$$\begin{aligned} \begin{aligned}&\Biggl |\int _{\mathop {\mathbb R}\nolimits ^3}\frac{2V \partial _z V}{r^{2\varepsilon }}|\eta |^{p-2}\eta dx\Biggr |\lesssim \Vert \eta \Vert ^{\frac{p-1}{4}}_{L^p}\Vert \nabla |\eta |^{\frac{p}{2}}\Vert _{L^2}^\frac{3(p-1)}{2p} \Vert \partial _z|V|^{\frac{q}{2}}\Vert _{L^2} \Bigl \Vert \frac{|V|^{2-\frac{q}{2}}}{r^{2\varepsilon }}\Bigr \Vert _{L^{2p}}. \end{aligned} \end{aligned}$$

(4.18)

To handle the term $\Bigl \Vert \frac{|V|^{2-\frac{q}{2}}}{r^{2\varepsilon }}\Bigr \Vert _{L^{2p}},$ we split $\frac{|V|^{2-\frac{q}{2}}}{r^{2\varepsilon }}$ as

$$\begin{aligned} \frac{|V|^{2-\frac{q}{2}}}{r^{2\varepsilon }}=\left| \frac{|V|^q}{r^2}\right| ^\alpha \bigl |r^{2-\varepsilon }V\bigr |^{\frac{1}{6p}} |V|^{\frac{q}{2} \cdot \beta }, \end{aligned}$$

with $\alpha , \beta$ being determined by

$$\begin{aligned} \alpha =\frac{1}{6p}\left( 1{-}\,\frac{\varepsilon }{2}\right) +\varepsilon \quad \hbox {and}\quad \beta =\frac{2}{q}\left( 2-\frac{q}{2}-q\alpha -\frac{1}{6p}\right) =\frac{1}{3p}\left( 7-\frac{2}{3p}\right) \left( 1-\frac{\varepsilon }{2}\right) -1-2\varepsilon . \end{aligned}$$

It follows from Sobolev embedding Theorem that

$$\begin{aligned} \begin{aligned} \bigl \Vert |V|^{\frac{q}{2} \cdot \beta }\bigr \Vert _{L^{r}}&\lesssim \bigl \Vert \nabla |V|^{\frac{q}{2}}\bigr \Vert _{L^2}^{\frac{3-p}{2p}-2\alpha } \bigl \Vert |V|^{\frac{q}{2}}\bigr \Vert _{L^2}^{2\alpha +\beta -\frac{3-p}{2p}}\quad \hbox {with}\quad \\ \frac{1}{r}&=\frac{\beta }{2}+\frac{2\alpha }{3}-\frac{3-p}{6p}=\frac{1}{18p}\left( 14-\frac{23}{2}\varepsilon -\frac{2-\varepsilon }{p}\right) -\frac{1}{3}(1+\varepsilon ). \end{aligned} \end{aligned}$$

(4.19)

It is easy to verify that $\alpha +\frac{1}{r}=\frac{1}{2p}-\frac{(p-1)^2}{4p}$ provided selecting

$$\begin{aligned} \varepsilon =\frac{-9p^2+21p-4}{24p-2}\, , \end{aligned}$$

(4.20)

which belongs to $\bigl [\frac{3251}{9280},\frac{4}{11}\bigr [$ whenever $p\in \bigl ]1,\frac{21}{20}\bigr ]$.

Moreover, under such choice of indexes, the term $\bigl \Vert |V|^{\frac{q}{2}}\bigr \Vert _{L^2}^{2\alpha +\beta -\frac{3-p}{2p}}$ in (4.19) disappears. Then we get, by applying Hölder’s inequality, that

$$\begin{aligned}\begin{aligned} \Bigl \Vert \frac{|V|^{2-\frac{q}{2}}}{r^{2\varepsilon }}\Bigr \Vert _{L^{2p}}&\le \Bigl \Vert \left( \frac{|V|^q}{r^2}\right) ^{\alpha }\Bigr \Vert _{L^{\frac{1}{\alpha }}} \bigl \Vert |V|^{\frac{q}{2} \cdot \beta }\bigr \Vert _{L^{r}} \bigl \Vert |r^{2-\varepsilon }V|^{\frac{1}{6p}}\bigr \Vert _{L^{\frac{4p}{(p-1)^2}}}\\&\le \Bigl \Vert \frac{|V|^q}{r^2}\Bigr \Vert ^{\alpha }_{L^1} \bigl \Vert \nabla |V|^{\frac{q}{2}}\bigr \Vert _{L^2}^{\frac{3-p}{2p}-2\alpha } \Vert ru^\theta \Vert _{L^{\frac{2}{3(p-1)^2}}}^{\frac{1}{6p}}.\\ \end{aligned}\end{aligned}$$

Inserting the above inequality into (4.18) gives rise to

$$\begin{aligned} \begin{aligned}&\Bigl |\int _{\mathop {\mathbb R}\nolimits ^3}\frac{2V \partial _z V}{r^{2\varepsilon }}|\eta |^{p-2}\eta dx\Bigr |\\&\quad \lesssim \Vert \eta \Vert ^{\frac{p-1}{4}}_{L^p}\Vert \nabla |\eta |^{\frac{p}{2}}\Vert _{L^2}^\frac{3(p-1)}{2p} \Vert \nabla |V|^{\frac{q}{2}}\Vert _{L^2}^{\frac{3-p}{2p}+1-2\alpha } \Bigl \Vert \frac{|V|^q}{r^2}\Bigr \Vert ^{\alpha }_{L^1} \Vert ru^\theta \Vert _{L^{\frac{2}{3(p-1)^2}}}^{\frac{1}{6p}}. \end{aligned} \end{aligned}$$

Note that $\frac{3(p-1)}{2p}+\left( \frac{3-p}{2p}+1-2\alpha \right) +2\alpha =2,$ by substituting the above inequality into (4.17) and using Young’s inequality, we achieve (4.7). This completes the proof of Lemma 4.2. $\square$

5 Global well-posedness with critical initial data

In what follows, we shall always denote $U\buildrel \mathrm{def}\over =\frac{u^\theta }{r}$ and $W\buildrel \mathrm{def}\over =r^{-\frac{7}{11}}u^\theta$.

Lemma 5.1

Under the assumption of Theorem 1.2, for any $p\in [1,\frac{21}{20}]$, $\varepsilon =\frac{-9p^2+21p-4}{24p-2}$, and $q=\frac{3p}{2-\varepsilon }$, the local solutions constructed in Theorem 1.1 satisfy

$$\begin{aligned} \Vert \eta (t)\Vert _{L^p}\le & {} \frac{C}{t^{\frac{3(p-1)}{2p}}}\cdot \left( L_{1,\frac{20}{3}}^5(T)+M_2^3(T) +N^5_{\frac{20}{13},20}(T)\right) \quad \forall \ t\in ]0, T], \end{aligned}$$

(5.1)

$$\begin{aligned} \Vert U(t)\Vert _{L^{\frac{3}{2}}}\le & {} C\left( \Vert U_0\Vert _{L^{\frac{3}{2}}} +L_{\frac{20}{17},2}(T)N_{\frac{20}{13},20}(T)\right) \quad \forall \ t\in ]0, T], \end{aligned}$$

(5.2)

$$\begin{aligned} \Vert V^\varepsilon (t)\Vert _{L^q}\le & {} \frac{C}{t^{\frac{3(p-1)}{2q}}} \left( \Vert U_0\Vert _{L^{\frac{3}{2}}}^{\frac{9}{11}}\Vert ru^\theta _0\Vert _{L^\infty }^{\frac{2}{11}}+ L_{\frac{20}{17},2}^2(T)+N^2_{\frac{20}{13},20}(T)\right) \quad \forall \ t\in ]0, T].\qquad \end{aligned}$$

(5.3)

Proof

Taking $\delta =1-\frac{1}{p}$ in (3.29) yields (5.1). Likewise, taking $\delta =\gamma =\frac{1}{3}$ and $~q_1=q_2=\frac{3}{2}$ in (3.28) leads to (5.2).

On the other hand, for any $p\in [1, {21}/{20}]$, due to the choice of $\varepsilon$ and q, we have

$$\begin{aligned} 1-\varepsilon -\frac{1}{q}=\frac{9p^2-7}{24p-2}\ge \frac{1}{11},\quad q=\frac{24p-2}{3p+9}\ge \frac{11}{6}. \end{aligned}$$

Noting that $\Vert V^\varepsilon \Vert _{L^q}=\Bigl \Vert \frac{u^\theta }{r^{1-\varepsilon -\frac{1}{q}}}\Bigr \Vert _{L^q(\Omega )}$ and $\Vert r^{-\frac{7}{11}}u^\theta _0\Vert _{L^{\frac{11}{6}}}=\bigl \Vert r^{-\frac{1}{11}}u^\theta _0\bigr \Vert _{L^{\frac{11}{6}}(\Omega )}$, then applying (3.28) with $q_1=q,~q_2=\frac{11}{6},$ $\delta =1-\varepsilon -\frac{1}{q}$ and $\gamma =\frac{1}{11}$ gives

$$\begin{aligned} \Vert V^\varepsilon (t)\Vert _{L^q}&\le \frac{C}{t^{\frac{3(p-1)}{2q}}}\cdot \left( \Vert r^{-\frac{7}{11}}u^\theta _0\Vert _{L^{\frac{11}{6}}}+ L_{\frac{20}{17},2}(T)N_{\frac{20}{13},20}(T)\right) \\&\le \frac{C}{t^{\frac{3(p-1)}{2q}}}\cdot \left( \Vert U_0\Vert _{L^{\frac{3}{2}}}^{\frac{9}{11}}\Vert ru^\theta _0\Vert _{L^\infty }^{\frac{2}{11}}+ L_{\frac{20}{17},2}^2(T)+N^2_{\frac{20}{13},20}(T)\right) ,\quad \forall t\in ]0, T], \end{aligned}$$

where we have used Hölder’s inequality in the last step. $\square$

Proof of Theorem 1.2

Due to $U_0=\frac{u^\theta _0}{r}\in L^{\frac{3}{2}},~ru^\theta _0\in L^\infty$, and

$$\begin{aligned} \Vert r^\kappa u^\theta _0\Vert _{L^{\frac{3}{1-\kappa }}}\le \Vert U_0\Vert _{L^{\frac{3}{2}}}^{\frac{1-\kappa }{2}} \Vert ru^\theta _0\Vert _{L^\infty }^{\frac{1+\kappa }{2}},\ \forall \kappa \in ]-1,1[, \end{aligned}$$

we deduce that both $\Vert u^\theta _0\Vert _{L^2(\Omega )}=\Vert r^{-\frac{1}{2}}u^\theta _0\Vert _{L^{2}}$ and $\Vert r^{-\frac{3}{10}}u^\theta _0\Vert _{L^{\frac{20}{13}}(\Omega )}=\Vert r^{-\frac{19}{20}}u^\theta _0\Vert _{L^{\frac{20}{13}}}$ are sufficiently small as long as $\Vert ru^\theta _0\Vert _{L^\infty }$ is small enough. Then by Theorem 1.1, the equation (1.4) has a unique mild solution

$$\begin{aligned}&\omega ^\theta \in C\left( [0,T];L^1(\Omega )\right) \bigcap C\left( ]0,T];L^\infty (\Omega )\right) ,\\&\quad u^\theta \in C\left( [0,T];L^2(\Omega )\right) \bigcap C\left( ]0,T];L^\infty (\Omega )\right) \\&\quad \hbox {with}\quad r^{-\frac{3}{10}}{u^\theta }\in C\left( [0,T];L^{\frac{20}{13}}(\Omega )\right) \bigcap C\left( ]0,T];L^\infty (\Omega )\right) , \end{aligned}$$

and the lifespan $T>0$ depends only on $\omega _0^\theta$. We denote $t_0\buildrel \mathrm{def}\over =\frac{T}{2}$. In the following, we will always abbreviate $L_p(T)$ as $L_p$, similar abbreviations for the remaining ones in (1.7), (3.27).

If $ru^\theta (t_0)\in L^{A}\bigcap L^\infty$ and if there exists some $p_0\in \left]1,\min \left( 1+\frac{1}{10A},\frac{21}{20}\right) \right[$, it follows from Proposition 4.1 that for $\varepsilon _0\buildrel \mathrm{def}\over =\frac{-9p_0^2+21p_0-4}{24p_0-2}$ and $q_0\buildrel \mathrm{def}\over =\frac{2(12p_0-1)}{3(p+3)},$ if there holds

$$\begin{aligned} (2M_1)^{\frac{3(p_0+2)}{p_0(3p_0+11)}} \Vert ru^\theta (t_0)\Vert _{L^{\alpha (p_0)}}^{\frac{10(12p_0-1)}{3p_0(p_0+3)(3p_0+11)}} +(2M_1)^{\frac{p_0-1}{4p_0}} \Vert ru^\theta (t_0)\Vert _{L^{\frac{2}{3(p_0-1)^2}}}^{\frac{1}{6p_0}}\le c_0\cdot (p_0-1), \end{aligned}$$

(5.4)

where

$$\begin{aligned} M_1\buildrel \mathrm{def}\over =\Vert V^{\varepsilon _0}(t_0)\Vert _{L^{q_0}}^{q_0}+\Vert \eta (t_0)\Vert _{L^{p_0}}^{p_0} \quad \hbox {and}\quad \alpha (p_0)\buildrel \mathrm{def}\over =\frac{10(12p_0-1)}{9(p_0-1)(p_0+2)(p_0+3)}. \end{aligned}$$

Then the system (1.4) has a global solution. Moreover, in view of (4.5) and Lemma 5.1, for all $t\in [t_0,\infty )$, there holds

$$\begin{aligned} \Vert \eta (t)\Vert _{L^{p_0}}^{p_0} +\frac{p_0-1}{2}\bigl \Vert \nabla |\eta |^{\frac{p_0}{2}}\bigr \Vert _{L^2([t_0,t)\times \mathop {\mathbb R}\nolimits ^3)}^2 \le \frac{C}{t_0^{\frac{3(p_0-1)}{2}}} \left( \Vert ru^\theta _0\Vert _{L^\infty }^{q_0}+L_{1,\frac{20}{3}}^{5p_0}+M_2^{3p_0} +N^{5p_0}_{\frac{20}{13},20}\right) . \end{aligned}$$

(5.5)

By the choice of $p_0$, $\alpha (p_0)>10A$. Then we deduce from Lemma 2.1 and Lemma 5.1 that the smallness condition (5.4) holds provided that

$$\begin{aligned} \begin{aligned} \Vert&ru^\theta _0\Vert _{L^\infty }\le \frac{c_0}{L_{1,\frac{20}{3}}^6+M_2^4 + N^6_{\frac{20}{13},20}} \min \left\{ \left( (p_0-1)\left( t_0^{\frac{3}{2}}\Vert ru^\theta _0\Vert _{L^A}^{-A}\right) ^{\frac{(p_0-1)^2}{4p_0}}\right) ^{\frac{12p_0}{2-3A(p_0-1)^2}},\right. \\&\qquad \qquad \left. \left( (p_0-1) \left( t_0^{\frac{3}{2}}\Vert ru^\theta _0\Vert _{L^A}^{-A}\right) ^{\frac{3(p_0-1)(p_0+2)}{p_0(3p_0+11)}}\right) ^{\frac{3p_0(p_0+3)(3p_0+11)}{10(12p_0-1)-9A(p_0-1)(p_0+2)(p_0+3)}}\right\} . \end{aligned} \end{aligned}$$

(5.6)

In the following, we consider estimates in critical spaces. By a similar derivation as Lemma 4.1, that for $q_{1,1}, q_{2,1}$ satisfying $\frac{1}{q_{1,1}}+\frac{1}{q_{2,1}}=1$ and $q_{1,1}>\frac{3p_0}{3-p_0}$, there holds

$$\begin{aligned} \begin{aligned}&\frac{6}{11}\frac{d}{dt}\Vert W(t)\Vert _{L^{\frac{11}{6}}}^{\frac{11}{6}}+ \frac{120}{121}\bigl \Vert \nabla |W|^{\frac{11}{12}}\bigr \Vert _{L^2}^2+\frac{72}{121}\int _{\mathop {\mathbb R}\nolimits ^3}\frac{|W|^{\frac{11}{6}}}{r^2}\,dx \lesssim \int _{\mathop {\mathbb R}\nolimits ^3}\bigl |\frac{u^r}{r}\bigr |\cdot |W|^{\frac{11}{6}}\,dx\\&\quad \lesssim \Vert \eta \Vert _{L^{p_0}}^{\frac{p_0-1}{2}+\frac{3p_0}{2q_{1,1}}} \Vert \nabla |\eta |^{\frac{p_0}{2}}\Vert _{L^{2}}^{2\left( \frac{3-p_0}{2p_0}-\frac{3}{2q_{1,1}}\right) }\Vert |W|^{\frac{11}{6}}\Vert _{L^{q_{2,1}}}.\\ \end{aligned} \end{aligned}$$

(5.7)

Take $q_{2,1}=\vartheta _1+\sigma _1+\frac{2\sigma _1}{3},$ with $\vartheta _1>0,~0<\sigma _1<1$ to be determined later, then we have

$$\begin{aligned} \begin{aligned} \Vert |W|^{\frac{11}{6}}\Vert _{L^{q_{2,1}}}&= \left( \int _{\mathop {\mathbb R}\nolimits ^3}|W|^{\frac{11}{6}\vartheta _1}\left( \frac{|W|^{\frac{11}{6}}}{r^2}\right) ^{\sigma _1}|r^{\frac{18}{11}}W|^{\frac{11\sigma _1}{9}}\,dx\right) ^{\frac{1}{q_{2,1}}}\\&\lesssim \Vert ru^\theta \Vert _{L^{\frac{11\sigma _1 p_0}{12(p_0-1)}}}^{\frac{11\sigma _1}{9q_{2,1}}} \Vert |W|^{\frac{11}{12}}\Vert _{L^{\frac{6p_0\vartheta _1}{4-p_0-3p_0\sigma _1}}}^{\frac{2\vartheta _1}{q_{2,1}}} \left( \int _{\mathop {\mathbb R}\nolimits ^3}\frac{|W|^{\frac{11}{6}}}{r^2}\,dx\right) ^{\frac{\sigma _1}{q_{2,1}}},\\&\lesssim \Vert ru^\theta \Vert _{L^{\frac{11\sigma _1 p_0}{12(p_0-1)}}}^{\frac{11\sigma _1}{9q_{2,1}}} \Vert |W|^{\frac{11}{12}}\Vert _{L^2}^{\frac{-\vartheta _1-3\sigma _1}{q_{2,1}}+\frac{4-p_0}{p_0q_{2,1}}}\\&\quad \times \Vert \nabla |W|^{\frac{11}{12}}\Vert _{L^2}^{\frac{3(\vartheta _1+\sigma _1)}{q_{2,1}}-\frac{4-p_0}{p_0q_{2,1}}} \left( \int _{\mathop {\mathbb R}\nolimits ^3}\frac{|W|^{\frac{11}{6}}}{r^2}\,dx\right) ^{\frac{\sigma _1}{q_{2,1}}}, \end{aligned} \end{aligned}$$

where in the last step, we used Galiardo–Nirenberg inequality provided that

$$\begin{aligned} \frac{6p_0\vartheta _1}{4-p_0-3p_0\sigma _1}\in [2,6], \end{aligned}$$

(5.8)

which will be verified later. Then we get, by applying Young’s inequality, that

$$\begin{aligned} \begin{aligned} \int _{\mathop {\mathbb R}\nolimits ^3}\bigl |\frac{u^r}{r}\bigr ||W|^{\frac{11}{6}} dx&\lesssim \Vert ru^\theta \Vert _{L^{\frac{11\sigma _1 p_0}{12(p_0-1)}}}^{\frac{11\sigma _1}{9q_{2,1}}} \Vert |W|^{\frac{11}{12}}\Vert _{L^2}^{\frac{-\vartheta _1-3\sigma _1}{q_{2,1}}+\frac{4-p_0}{p_0q_{2,1}}} \Vert \eta \Vert _{L^{p_0}}^{\frac{p_0-1}{2}+\frac{3p_0}{2q_{1,1}}}\\&\quad \times \left( \Vert \nabla |\eta |^{\frac{p_0}{2}}\Vert _{L^{2}}^{2}+ \Vert \nabla |W|^{\frac{11}{12}}\Vert _{L^2}^{2}+ \int _{\mathop {\mathbb R}\nolimits ^3}\frac{|W|^{\frac{11}{6}}}{r^2}\,dx\right) , \end{aligned} \end{aligned}$$

(5.9)

provided that

$$\begin{aligned} \frac{3-p_0}{2p_0}-\frac{3}{2q_{1,1}}+\frac{3(\vartheta _1+\sigma _1)+1}{2q_{2,1}}-\frac{2}{p_0q_{2,1}}+\frac{\sigma _1}{q_{2,1}}=1, \end{aligned}$$

which in particular holds by taking

$$\begin{aligned} \sigma _1=\frac{2}{5},\quad \vartheta _1=\frac{2}{3},\quad \text{ and }\quad q_{2,1}=\vartheta _1+\sigma _1+\frac{2\sigma _1}{3}=\frac{4}{3}. \end{aligned}$$

(5.10)

So that (5.8) holds and $q_{1,1}=4>\frac{3p_0}{3-p_0}$. Hence all the above calculations go through.

By inserting the Estimate (5.9) into (5.7), with the indices given by (5.10), we obtain

$$\begin{aligned} \begin{aligned}&\frac{6}{11}\frac{d}{dt}\Vert W(t)\Vert _{L^{\frac{11}{6}}}^{\frac{11}{6}}+ \frac{120}{121}\bigl \Vert \nabla |W|^{\frac{11}{12}}\bigr \Vert _{L^2}^2+\frac{72}{121}\int _{\mathop {\mathbb R}\nolimits ^3}\frac{|W|^{\frac{11}{6}}}{r^2}\,dx\\&\quad \lesssim \Vert ru^\theta \Vert _{L^{\frac{11 p_0}{30(p_0-1)}}}^{\frac{11}{30}} \Vert |W|^{\frac{11}{12}}\Vert _{L^2}^{\frac{3}{p_0}-\frac{43}{20}} \Vert \eta \Vert _{L^{p_0}}^{\frac{7p_0}{8}-\frac{1}{2}}\\&\quad \quad \times \left( \Vert \nabla |\eta |^{\frac{p_0}{2}}\Vert _{L^{2}}^{2}+ \Vert \nabla |W|^{\frac{11}{12}}\Vert _{L^2}^{2}+ \int _{\mathop {\mathbb R}\nolimits ^3}\frac{|W|^{\frac{11}{6}}}{r^2}\,dx\right) . \end{aligned} \end{aligned}$$

(5.11)

Next, applying $L^1$ energy estimate for the $\eta$ equation in (4.2) yields

$$\begin{aligned} \frac{d}{dt}\Vert \eta (t)\Vert _{L^1}+\int _{\mathop {\mathbb R}\nolimits ^3}(-\Delta \eta )\cdot \mathrm {sgn}\ \eta \,dx +2\int _{-\infty }^{+\infty }|\eta |\big |_{r=0} dz \lesssim \bigl \Vert \partial _z|W|^{\frac{11}{12}}\bigr \Vert _{L^2} \Bigl \Vert \frac{|W|^{\frac{13}{12}}}{r^{\frac{8}{11}}}\Bigr \Vert _{L^2}. \end{aligned}$$

(5.12)

Noting that $\frac{|W|^{\frac{13}{12}}}{r^{\frac{8}{11}}}=\bigl |\frac{|W|^{\frac{11}{6}}}{r^2}\bigr |^{\frac{1}{2}}\cdot \bigl |r^{\frac{18}{11}}W\bigr |^{\frac{1}{6}}$, so we have

$$\begin{aligned} \Bigl \Vert \frac{|W|^{\frac{13}{12}}}{r^{\frac{8}{11}}}\Bigr \Vert _{L^2} \le \Vert ru^\theta \Vert _{L^\infty }^{\frac{1}{6}}\cdot \Bigl \Vert \frac{|W|^{\frac{11}{6}}}{r^2}\Bigr \Vert _{L^1}^{\frac{1}{2}}. \end{aligned}$$

(5.13)

Substituting (5.13) into (5.12), and using the fact that $\int _{\mathop {\mathbb R}\nolimits ^3}(-\Delta \eta )\cdot \mathrm {sgn}\ \eta \,dx\le 0$, we achieve

$$\begin{aligned} \begin{aligned} \frac{d}{dt}\Vert \eta (t)\Vert _{L^1}&\lesssim \Vert ru^\theta \Vert _{L^\infty }^{\frac{1}{6}} \bigl \Vert \partial _z|W|^{\frac{11}{12}}\bigr \Vert _{L^2} \Bigl \Vert \frac{|W|^{\frac{11}{6}}}{r^2}\Bigr \Vert _{L^1}^{\frac{1}{2}}\\&\lesssim \Vert ru^\theta \Vert _{L^\infty }^{\frac{1}{6}} \left( \bigl \Vert \partial _z|W|^{\frac{11}{12}}\bigr \Vert _{L^2}^2+\Bigl \Vert \frac{|W|^{\frac{11}{6}}}{r^2}\Bigr \Vert _{L^1}\right) . \end{aligned}\end{aligned}$$

(5.14)

Meanwhile, by taking $p=p_0$ in (4.8), we have

$$\begin{aligned} \begin{aligned}&\frac{d}{dt}\left( \Vert V^{\varepsilon _0}(t)\Vert _{L^{q_0}}^{q_0}+\Vert \eta (t)\Vert _{L^{p_0}}^{p_0}\right) +\frac{4(p_0-1)}{p_0}\Vert \nabla |\eta |^{\frac{p_0}{2}}\Vert _{L^2}^2\\&\quad \quad +\frac{4(q_0-1)}{q_0}\Vert \nabla |V^{\varepsilon _0}|^{\frac{q_0}{2}}\Vert _{L^2}^2 +\frac{3p_0(-9p_0^2+21p_0-4)}{24p_0-2}\Bigl \Vert \frac{|V^{\varepsilon _0}|^{q_0}}{r^2}\Bigr \Vert _{L^1}\\&\quad \lesssim \left( \Vert ru^\theta _0\Vert _{L^{\alpha (p_0)}}^{\frac{10(12p_0-1)}{3p_0(p_0+3)(3p_0+11)}} \Vert V^{\varepsilon _0}\Vert _{L^2}^{q_0\cdot \frac{23-17p_0-3p_0^2}{2p_0(3p_0+11)}} \Vert \eta \Vert _{L^p_0}^{\frac{3p_0^2+23p_0-11}{2(3p_0+11)}} +\Vert ru^\theta _0\Vert _{L^{\frac{2}{3(p_0-1)^2}}}^{\frac{1}{6p_0}} \Vert \eta \Vert ^{\frac{p_0-1}{4}}_{L^{p_0}}\right) \\&\quad \quad \times \left( \Vert \nabla |\eta |^{\frac{p_0}{2}}\Vert _{L^2}^2+\Vert \nabla |V^{\varepsilon _0}|^{\frac{q_0}{2}}\Vert _{L^2}^2 +\Bigl \Vert \frac{|V^{\varepsilon _0}|^{q_0}}{r^2}\Bigr \Vert _{L^1}\right) . \end{aligned} \end{aligned}$$

(5.15)

Summarizing the estimates (5.11), (5.14) and (5.15) gives rise to

$$\begin{aligned} \begin{aligned}&\frac{d}{dt}\left( \Vert \eta (t)\Vert _{L^1}+\Vert W(t)\Vert _{L^{\frac{11}{6}}}^{\frac{11}{6}} +\Vert V^{\varepsilon _0}(t)\Vert _{L^{q_0}}^{q_0}+\Vert \eta (t)\Vert _{L^{p_0}}^{p_0}\right) +(p_0-1)\Vert \nabla |\eta |^{\frac{p_0}{2}}\Vert _{L^2}^2\\&\quad \quad +\Vert \nabla |V^{\varepsilon _0}|^{\frac{q_0}{2}}\Vert _{L^2}^2 +\Bigl \Vert \frac{|V^{\varepsilon _0}|^{q_0}}{r^2}\Bigr \Vert _{L^1} +\bigl \Vert \nabla |W|^{\frac{11}{12}}\bigr \Vert _{L^2}^2+\int _{\mathop {\mathbb R}\nolimits ^3}\frac{|W|^{\frac{11}{6}}}{r^2}\,dx\\&\quad \lesssim \left( \Vert ru^\theta _0\Vert _{L^{\alpha (p_0)}}^{\frac{10(12p_0-1)}{3p_0(p_0+3)(3p_0+11)}} \Vert V^{\varepsilon _0}\Vert _{L^2}^{q_0\cdot \frac{23-17p_0-3p_0^2}{2p_0(3p_0+11)}} \Vert \eta \Vert _{L^{p_0}}^{\frac{3p_0^2+23p_0-11}{2(3p_0+11)}} +\Vert ru^\theta _0\Vert _{L^{\frac{2}{3(p_0-1)^2}}}^{\frac{1}{6p_0}} \Vert \eta \Vert ^{\frac{p_0-1}{4}}_{L^{p_0}}\right) \\&\quad \quad \times \left( \Vert \nabla |\eta |^{\frac{p_0}{2}}\Vert _{L^2}^2+\Vert \nabla |V^{\varepsilon _0}|^{\frac{q_0}{2}}\Vert _{L^2}^2 +\Bigl \Vert \frac{|V^{\varepsilon _0}|^{q_0}}{r^2}\Bigr \Vert _{L^1}\right) \\&\quad \quad +\Vert ru^\theta _0\Vert _{L^\infty }^{\frac{1}{6}} \left( \bigl \Vert \partial _z|W|^{\frac{11}{12}}\bigr \Vert _{L^2}^2+\Bigl \Vert \frac{|W|^{\frac{11}{6}}}{r^2}\Bigr \Vert _{L^1}\right) +\Vert ru^\theta _0\Vert _{L^{\frac{11 p_0}{30(p_0-1)}}}^{\frac{11}{30}} \Vert |W|^{\frac{11}{12}}\Vert _{L^2}^{\frac{3}{p_0}-\frac{43}{20}} \Vert \eta \Vert _{L^{p_0}}^{\frac{7p_0}{8}-\frac{1}{2}}\\&\quad \quad \times \left( \Vert \nabla |\eta |^{\frac{p_0}{2}}\Vert _{L^{2}}^{2}+ \Vert \nabla |W|^{\frac{11}{12}}\Vert _{L^2}^{2}+ \int _{\mathop {\mathbb R}\nolimits ^3}\frac{|W|^{\frac{11}{6}}}{r^2}\,dx\right) . \end{aligned} \end{aligned}$$

(5.16)

Then under the smallness conditions (5.6), and

$$\begin{aligned} \Vert ru^\theta _0\Vert _{L^\infty }^{\frac{1}{6}}\le c_0,\quad \Vert ru^\theta _0\Vert _{L^{\frac{11 p_0}{30(p_0-1)}}}^{\frac{11}{30}} \Vert |W|^{\frac{11}{12}}(t_0)\Vert _{L^2}^{\frac{3}{p_0}-\frac{43}{20}} \Vert \eta (t_0)\Vert _{L^{p_0}}^{\frac{7p_0}{8}-\frac{1}{2}}\le c_0(p_0-1), \end{aligned}$$

(5.17)

we get, by a standard continued argument, as we did in the last step of the proof of Proposition 4.1, that

$$\begin{aligned} \begin{aligned}&\Vert \eta (t)\Vert _{L^1}+\Vert W(t)\Vert _{L^{\frac{11}{6}}}^{\frac{11}{6}}+\Vert V^{\varepsilon _0}(t)\Vert _{L^{q_0}}^{q_0}+\Vert \eta (t)\Vert _{L^{p_0}}^{p_0}\\&\quad \quad +(p_0-1)\Vert \nabla |\eta |^{\frac{p_0}{2}}\Vert _{L^2}^2 +\int _{t_0}^t \int _{\mathop {\mathbb R}\nolimits ^3}\frac{|W(t')|^{\frac{11}{6}}}{r^2}\,dxdt'\\&\quad \le 2\left( \Vert \eta (t_0)\Vert _{L^1}+\Vert W(t_0)\Vert _{L^{\frac{11}{6}}}^{\frac{11}{6}} +\Vert V^{\varepsilon _0}(t_0)\Vert _{L^{q_0}}^{q_0}+\Vert \eta (t_0)\Vert _{L^{p_0}}^{p_0}\right) \\&\quad \le \frac{C}{t^{\frac{3(p_0-1)}{2}}} \left( \Vert ru^\theta _0\Vert _{L^\infty }^{\frac{11}{6}}+L_{1,\frac{20}{3}}^{5p_0}+M_2^{3p_0} +N^{5p_0}_{\frac{20}{13},20}\right) , \quad \forall \ t_0\le t<\infty . \end{aligned} \end{aligned}$$

(5.18)

Recalling that $1<p_0<\min \left( 1+\frac{1}{10A},\frac{21}{20}\right)$, we have $\frac{11 p_0}{30(p_0-1)}>\frac{11}{3}A$. Then it follows from Lemmas 2.1 and 5.1 that the condition (5.17) is verified provided that

$$\begin{aligned} \Vert ru^\theta _0\Vert _{L^\infty }\le \frac{c_0}{L_{1,\frac{20}{3}}^6+M_2^2 +N^6_{\frac{20}{13},20}} \left( (p_0-1)\left( t_0^{\frac{21}{16}p_0-\frac{3}{4}}\Vert ru^\theta _0\Vert _{L^A}^{-A}\right) ^{\frac{(p_0-1)}{p_0}}\right) ^{\frac{240p_0}{300-127p_0-240A(p_0-1)}}. \end{aligned}$$

(5.19)

Finally we derive the $L^{\frac{3}{2}}$ estimate for U. Indeed along the same line of the derivation of $\Vert W(t)\Vert _{L^{\frac{11}{6}}},$ and using the indices given by (5.10), we infer

$$\begin{aligned} \begin{aligned}&\frac{d}{dt}\Vert U(t)\Vert _{L^{\frac{3}{2}}}^{\frac{3}{2}}+ \bigl \Vert \nabla |U|^{\frac{3}{4}}\bigr \Vert _{L^2}^2 \lesssim \int _{\mathop {\mathbb R}\nolimits ^3}\bigl |\frac{u^r}{r}\bigr |\cdot |U|^{\frac{3}{2}}\,dx\\&\quad \lesssim \Vert \eta \Vert _{L^{p_0}}^{\frac{7p_0}{8}-\frac{1}{2}} \bigl \Vert \nabla |\eta |^{\frac{p_0}{2}}\bigr \Vert _{L^{2}}^{2\left( \frac{3}{2p_0}-\frac{7}{8}\right) }\bigl \Vert |U|^{\frac{3}{2}}\bigr \Vert _{L^{\frac{4}{3}}}. \end{aligned} \end{aligned}$$

(5.20)

whereas by applying Hölder’s inequality, one has

$$\begin{aligned} \begin{aligned} \bigl \Vert |U|^{\frac{3}{2}}\bigr \Vert _{L^{\frac{4}{3}}}&= \left( \int _{\mathop {\mathbb R}\nolimits ^3}|U|\left( \frac{|W|^{\frac{11}{6}}}{r^2}\right) ^{\frac{2}{5}} |r^2U|^{\frac{4}{15}}\,dx\right) ^{\frac{3}{4}}\\&\lesssim \Vert ru^\theta \Vert _{L^{\frac{p_0}{5(p_0-1)}}}^{\frac{1}{5}} \bigl \Vert |U|^{\frac{3}{4}}\bigr \Vert _{L^{\frac{20p_0}{20-11p_0}}} \left( \int _{\mathop {\mathbb R}\nolimits ^3}\frac{|W|^{\frac{11}{6}}}{r^2}\,dx\right) ^{\frac{3}{10}},\\&\lesssim \Vert ru^\theta \Vert _{L^{\frac{p_0}{5(p_0-1)}}}^{\frac{1}{5}} \bigl \Vert |U|^{\frac{3}{4}}\bigr \Vert _{L^2}^{\frac{3}{p_0}-\frac{43}{20}} \bigl \Vert \nabla |U|^{\frac{3}{4}}\bigr \Vert _{L^2}^{\frac{63}{20}-\frac{3}{p_0}} \left( \int _{\mathop {\mathbb R}\nolimits ^3}\frac{|W|^{\frac{11}{6}}}{r^2}\,dx\right) ^{\frac{3}{10}}. \end{aligned} \end{aligned}$$

(5.21)

Substituting (5.21) into (5.20), then using of Young’s inequality gives rise to

$$\begin{aligned}\begin{aligned} \frac{d}{dt}\Vert U(t)\Vert _{L^{\frac{3}{2}}}^{\frac{3}{2}}+ \bigl \Vert \nabla |U|^{\frac{3}{4}}\bigr \Vert _{L^2}^2&\lesssim \Vert \eta \Vert _{L^{p_0}}^{\frac{7p_0}{8}-\frac{1}{2}} \Vert ru^\theta _0\Vert _{L^{\frac{p_0}{5(p_0-1)}}}^{\frac{1}{5}} \Vert U\Vert _{L^{\frac{3}{2}}}^{\frac{3}{2}\left( \frac{3}{2p_0}-\frac{43}{40}\right) }\\&\quad \times \left( \bigl \Vert \nabla |\eta |^{\frac{p_0}{2}}\bigr \Vert _{L^{2}}^2 +\bigl \Vert \nabla |U|^{\frac{3}{4}}\bigr \Vert _{L^2}^2+\int _{\mathop {\mathbb R}\nolimits ^3}\frac{|W|^{\frac{11}{6}}}{r^2}\,dx\right) , \end{aligned} \end{aligned}$$

from which and (5.18), we deduce by a standard continued argument that

$$\begin{aligned} \begin{aligned}&\Vert U(t)\Vert _{L^{\frac{3}{2}}}^{\frac{3}{2}}+ \int _{t_0}^t \bigl \Vert \nabla |U|^{\frac{3}{4}}(t')\bigr \Vert _{L^2}^2\,dt'\\&\quad \le 2\Vert U_0\Vert _{L^{\frac{3}{2}}}^{\frac{3}{2}} +\frac{C}{t^{\frac{3(p_0-1)}{2}}} \left( \Vert ru^\theta _0\Vert _{L^\infty }^{\frac{11}{6}}+L_{1,\frac{20}{3}}^{5p_0}+M_2^{3p_0} +N^{5p_0}_{\frac{20}{13},20}\right) , \quad \forall \ t_0\le t<\infty , \end{aligned} \end{aligned}$$

(5.22)

provided that the smallness conditions (5.6), (5.17), (5.19) and

$$\begin{aligned} \left\| \eta (t_0)\right\| _{L^{p_0}}^{\frac{7p_0}{8}-\frac{1}{2}} \left\| ru^\theta _0\right\| _{L^{\frac{p_0}{5(p_0-1)}}}^{\frac{1}{5}} \left\| U(t_0)\right\| _{L^{\frac{3}{2}}}^{\frac{3}{2}\left( \frac{3}{2p_0}-\frac{43}{40}\right) } \le c_0(p_0-1), \end{aligned}$$

(5.23)

hold. Yet it follows Lemma 5.1 that (5.23) can be satisfied as long as

$$\begin{aligned} \left\| ru^\theta _0\right\| _{L^\infty }\le \frac{c_0}{L_{1,\frac{20}{3}}^{20}+M_2^{12} +N^{20}_{\frac{20}{13},20}} \left( (p_0-1)\cdot \left( t_0^{\frac{21}{16}p_0-\frac{3}{4}} \left\| ru^\theta _0\right\| _{L^A}^{-A}\right) ^{\frac{(p_0-1)}{p_0}}\right) ^{\frac{5p_0}{p_0-5A(p_0-1)}}. \end{aligned}$$

(5.24)

Therefore under the smallness conditions (5.6), (5.17), (5.19) and (5.24), we get, by summing up (5.18) and (5.22) that for any $0\le t<\infty$,

$$\begin{aligned} \left\| \eta (t)\right\| _{L^1}+\left\| U(t)\right\| _{L^{\frac{3}{2}}}^{\frac{3}{2}} \le 2\left\| U_0\right\| _{L^{\frac{3}{2}}}^{\frac{3}{2}} +\frac{C}{t^{\frac{3(p_0-1)}{2}}} \left( \Vert ru^\theta _0\Vert _{L^\infty }^{\frac{11}{6}}+L_{1,\frac{20}{3}}^{5p_0}+M_2^{3p_0} +N^{5p_0}_{\frac{20}{13},20}\right) . \end{aligned}$$

(5.25)

This completes the proof of Theorem 1.2. $\square$