Global smooth axisymmetric solutions of 3-D inhomogeneous incompressible Navier–Stokes system

Abstract

In this paper, we investigate the global regularity to 3-D inhomogeneous incompressible Navier–Stokes system with axisymmetric initial data which does not have swirl component for the initial velocity. We first prove that the $L^\infty$ norm to the quotient of the inhomogeneity by r,  namely $a/r\buildrel \hbox {def}\over =(1/\rho -1)/r,$ controls the regularity of the solutions. Then we prove the global regularity of such solutions provided that the $L^\infty$ norm of $a_0/r$ is sufficiently small. Finally, with additional assumption that the initial velocity belongs to $L^p$ for some $p\in [1,2),$ we prove that the velocity field decays to zero with exactly the same rate as the classical Navier–Stokes system.

1 Introduction

In this paper, we consider the global existence of smooth solutions to the following 3-D inhomogeneous incompressible Navier–Stokes equations with axisymmetric initial data which does not have swirl component for the initial velocity:

$$\begin{aligned} \quad \left\{ \begin{array}{l} \displaystyle \partial _t \rho + \text{ div }( \rho u)=0,\qquad (t,x)\in \mathop {\mathbb R}\nolimits ^+\times \mathop {\mathbb R}\nolimits ^3,\\ \displaystyle \partial _t (\rho u) + \text{ div }(\rho u \otimes u) -\Delta u+ \nabla \Pi =0, \\ \displaystyle \text{ div }u = 0, \\ \displaystyle (\rho , u)|_{t=0}=(\rho _0, u_{0}). \end{array}\right. \end{aligned}$$

(1.1)

where $\rho , u=(u^1,u^2, u^z)$ stand for the density and velocity of the fluid respectively, and $\Pi$ is a scalar pressure function. Such system describes a fluid that is incompressible but has non-constant density. Basic examples are mixture of incompressible and non reactant flows, flows with complex structure (e.g. blood flow or model of rivers), fluids containing a melted substance, etc.

A lot of recent works have been dedicated to the mathematical study of the above system. Global weak solutions with finite energy have been constructed by Simon in [22] (see also the book by Lions [18] for the variable viscosity case). In the case of smooth data with no vacuum, the existence of strong unique solutions goes back to the work of Ladyzhenskaya and Solonnikov in [16]. More precisely, they considered the system (1.1) in a bounded domain $\Omega$ with homogeneous Dirichlet boundary condition for u. Under the assumption that $u_0\in W^{2-\frac{2}{p},p}(\Omega )$ $(p>d)$ is divergence free and vanishes on $\partial \Omega$ and that $\rho _0\in C^1(\Omega )$ is bounded away from zero, then they [16] proved

• Global well-posedness in dimension $d=2;$

• Local well-posedness in dimension $d=3.$ If in addition $u_0$ is small in $W^{2-\frac{2}{p},p}(\Omega ),$ then global well-posedness holds true.

Lately, Danchin and Mucha [9] established the well-posedness of (1.1) in the whole space $\mathop {\mathbb R}\nolimits ^d$ in the so-called critical functional framework for small perturbations of some positive constant density. The basic idea are to use functional spaces (or norms) that is scaling invariant under the following transformation:

$$\begin{aligned} (\rho ,u,\Pi )(t,x)\longmapsto (\rho ,\lambda u,\lambda ^2\Pi ) (\lambda ^2 t,\lambda x),\qquad (\rho _0,u_0)(x)\longmapsto (\rho _0,\lambda u_0)(\lambda x). \end{aligned}$$

(1.2)

One may check [5, 10] and the references therein for the recent progresses along this line.

On the other hand, we recall that except the initial data have some special structure, it is still not known whether or not the System (1.1) has a unique global smooth solution with large smooth initial data, even for the classical Navier–Stokes system (NS),  which corresponds to $\rho =1$ in (1.1). For instance, Ukhovskii and Yudovich [23], and independently Ladyzhenskaya [15] proved the global existence of generalized solution along with its uniqueness and regularity for (NS) with initial data which is axisymmetric and without swirl. Leonardi et al. [17] gave a refined proof of the same result in [15, 23]. The first author [1] improved the regularity of the initial data to be $u_0\in H^{\frac{1}{2}}.$ In general, the global wellposedness of (NS) with axisymmetric initial data is still open (see [7, 24] for instance).

Let $x=(x_1, x_2, z)\in \mathbb {R}^3,$ we denote the cylindrical coordinates of x by $(r, \theta , z),$ i. e., $r(x_1, x_2)\buildrel \hbox {def}\over =\sqrt{x_1^2+x_2^2}, \quad \theta (x_1, x_2) \buildrel \hbox {def}\over =\tan ^{-1} \frac{x_2}{x_1}$ with $r \in [0, \infty ), \, \theta \in [0, 2\pi ]$ and $z \in \mathbb {R},$ and

$$\begin{aligned} e_{r}\buildrel \hbox {def}\over =(\cos \theta , \sin \theta , 0), \quad e_{\theta }\buildrel \hbox {def}\over =(-\sin \theta , \cos \theta , 0),\quad e_{z}\buildrel \hbox {def}\over =(0, 0, 1). \end{aligned}$$

We are concerned here with the global existence of axisymmetric smooth solutions to (1.1) which does not have the swirl component for the velocity field. This means solution of the form:

$$\begin{aligned}&\rho (t,x_1, x_2, z)=\rho (t, r, z), \quad \Pi (t, x_1, x_2, z)=\Pi (t, r, z), \nonumber \\&u(t,x_1,x_2,z)=u^r(t,r, z)e_r+u^{z}(t,r, z)e_{z}. \end{aligned}$$

(1.3)

By virtue of (1.1) and (1.3), we find that $(\rho ,u,\Pi )$ verifies

$$\begin{aligned} {\left\{ \begin{array}{ll} &{}\partial _t \rho +u^r \partial _r \rho +u^{z} \partial _z \rho =0,\\ &{}\rho \partial _t u^r + \rho u^r \partial _r u^r+\rho u^{z} \partial _z u^r+\partial _r \Pi - \left( \frac{1}{r}\partial _r(r\partial _r u^r)+\partial _z^2u^r -\frac{u^r}{r^2}\right) =0,\\ &{}\rho \partial _t u^{z} + \rho u^r \partial _r u^{z}+\rho u^{z} \partial _z u^{z} +\partial _z \Pi - \left( \frac{1}{r}\partial _r(r\partial _r u^z)+\partial _z^2u^z \right) =0,\\ &{}\partial _ru^r+\frac{u^r}{r}+\partial _zu^z=0,\\ &{}\rho |_{t=0}=\rho _0\quad \hbox {and}\quad (u^r,u^z)|_{t=0}=(u_0^r,u_0^z). \end{array}\right. } \end{aligned}$$

(1.4)

Equation of vorticity $\omega \buildrel \hbox {def}\over =\partial _zu^r-\partial _r u^z$: we get, by taking $\partial _z (1.4)_2-\partial _r (1.4)_3,$ that

$$\begin{aligned} \partial _t \omega + u^r \partial _r \omega +u^{z} \partial _z \omega -\frac{1}{r}u^r \omega +\partial _z\left( \frac{\partial _r \Pi }{\rho }\right) -\partial _r\left( \frac{\partial _z \Pi }{\rho }\right) -\partial _z\left( \frac{\partial _z \omega }{\rho }\right) -\partial _r\left( \frac{\partial _r \omega +\omega /r}{\rho }\right) =0. \end{aligned}$$

(1.5)

Equation of $\Gamma \buildrel \hbox {def}\over =\frac{\omega }{r}$: in view of (1.5), one has

$$\begin{aligned} \partial _t \Gamma + u^r \partial _r \Gamma +u^{z} \partial _z \Gamma +\frac{1}{r}\partial _z\left( \frac{\partial _r \Pi }{\rho }\right) -\frac{1}{r}\partial _r\left( \frac{\partial _z \Pi }{\rho }\right) - \partial _z\left( \frac{\partial _z \Gamma }{\rho }\right) -\frac{1}{r}\partial _r\left( \frac{r\partial _r \Gamma +2\Gamma }{\rho }\right) =0. \end{aligned}$$

(1.6)

As for the classical Navier–Stokes system (NS) in [15, 23], the quantity $\Gamma$ will play a crucial role to prove the global well-posedness of (1.4). The main result of this paper states as follows:

Theorem 1.1

Let $a_0\buildrel \hbox {def}\over =\frac{1}{\rho _0}-1\in L^{2}\cap L^\infty$ with $\frac{a_0}{r}\in L^\infty ,$ and there exist positive constants m, M so that

$$\begin{aligned} 0<m\le \rho _0\le M. \end{aligned}$$

(1.7)

Let $u_0=u_0^re_r+u_0^ze_z\in H^1$ be a solenoidal vector filed with $\frac{u_0^r}{r}$ and $\Gamma _0\buildrel \hbox {def}\over =\frac{\omega _0}{r}$ belonging to $L^2.$ Then

• (1) there exists a positive time $T^*$ so that (1.4) has a unique solution $(\rho , u)$ on $[0,T^*)$ which satisfies for any $T<T^*$

  $$\begin{aligned}&\rho \in L^\infty ((0,T)\times \mathop {\mathbb R}\nolimits ^3),\quad u\in {\mathcal C}([0,T];H^1(\mathop {\mathbb R}\nolimits ^3))\quad \hbox {with}\quad \nabla u \in L^2((0,T);H^1(\mathop {\mathbb R}\nolimits ^3))\nonumber \\&\sup _{t\in (0,T]}\left( t\langle {t} \rangle \left( \Vert u_t(t)\Vert _{L^2}^2+\Vert u(t)\Vert _{\dot{H}^2}^2+\Vert \nabla \Pi (t)\Vert _{L^2}^2\right) +\int _0^tt'\langle {t'} \rangle \Vert \nabla u_t(t')\Vert _{L^2}^2\,dt'\right) <\infty . \end{aligned}$$

  (1.8)

  If $T^*<\infty ,$ there holds

  $$\begin{aligned} \lim _{t\rightarrow T^*}\left\| \frac{a(t)}{r}\right\| _{L^\infty }=\infty . \end{aligned}$$

  (1.9)

• (2) If we assume moreover that

  $$\begin{aligned} \left\| \frac{a_0}{r}\right\| _{L^\infty }\le \varepsilon _0 \end{aligned}$$

  (1.10)

  for some sufficiently small positive constant $\varepsilon _0,$ we have $T^*=\infty ,$ and

  $$\begin{aligned} {\begin{matrix} &{}\Vert u\Vert _{L^\infty (\mathop {\mathbb R}\nolimits ^+;H^1)}^2+\left\| \frac{u^r}{r}\right\| _{L^\infty (\mathop {\mathbb R}\nolimits ^+;L^2)}^2+\Vert \nabla u\Vert _{L^2(\mathop {\mathbb R}\nolimits ^+;H^1)}^2+\Vert \partial _tu\Vert _{L^2(\mathop {\mathbb R}\nolimits ^+;L^2)}^2\\ &{}\quad +\Vert \nabla \Pi \Vert _{L^2(\mathop {\mathbb R}\nolimits ^+;L^2)}^2\le C{\mathcal G}_0 +1\quad \hbox {with}\quad \\ &{}\quad {\mathcal G}_0\buildrel \hbox {def}\over =\exp \left( C\Vert u_0\Vert _{L^2}^2\left( 1+\Vert u_0\Vert _{L^2}^6\right) \right) \left( \Vert u_0\Vert _{H^1}^2+\left\| \frac{u_0^r}{r}\right\| _{L^2}^2+2\Vert \Gamma _0\Vert _{L^2}^2\right) ,\end{matrix}} \end{aligned}$$

  (1.11)

  and

  $$\begin{aligned} \left\| \frac{a}{r}\right\| _{L^\infty (\mathop {\mathbb R}\nolimits ^+;L^\infty )}\le C \left\| \frac{a_0}{r}\right\| _{L^\infty }. \end{aligned}$$

  (1.12)

• (3) Besides (1.10), if $u_0\in L^p$ for some $p\in [1,2),$ let $\beta (p)\buildrel \hbox {def}\over =\frac{3}{4}\left( \frac{2}{p}-1\right) ,$ one has

  $$\begin{aligned}&\Vert u(t)\Vert _{L^2}^2\le C\langle {t} \rangle ^{-2\beta (p)}, \quad \Vert \nabla u(t)\Vert _{L^2}^2\le C\langle {t} \rangle ^{-1-2\beta (p)},\\&\Vert u_t(t)\Vert _{L^2}^2+ \Vert u(t)\Vert _{\dot{H}^2}^2+\Vert \nabla \Pi (t)\Vert _{L^2}^2 \le C t^{-1}\langle {t} \rangle ^{-1-2\beta (p)}.\nonumber \end{aligned}$$

  (1.13)

Remark 1.1

• (1) Let us recall that the reason why one can prove the global well-posedness of classical 3-D Navier–Stokes system with axisymmetric data and without swirl is that $\Gamma \buildrel \hbox {def}\over =\frac{\omega }{r}$ satisfies

  $$\begin{aligned} \partial _t\Gamma +u^r\partial _r\Gamma +u^z\partial _z\Gamma -\partial _r^2\Gamma -\partial _z^2\Gamma -\frac{3}{r}\partial _r\Gamma =0, \end{aligned}$$

  which implies for all $p\in [1,\infty ]$ that

  $$\begin{aligned} \Vert \Gamma (t)\Vert _{L^p}\le \Vert \Gamma _0\Vert _{L^p}. \end{aligned}$$

  Nevertheless in the case of inhomogeneous Navier–Stokes system, $\Gamma$ verifies (1.6). Then to get a global in time estimate for $\Vert \Gamma (t)\Vert _{L^2},$ we need the smallness condition (1.10). We remark that in order to prove the global regularity for the axisymmetric Navier–Stokes–Boussinesq system without swirl, the authors [3] require the support of the initial density $\rho _0$ does not intersect the axis (Oz) and the projection of supp$\rho _0$ on the axis is a compact set, which seems stronger than (1.10) near the axis (Oz). Finally since we shall not use the vorticity equation (1.5), here we do not require the initial density to be close enough to some positive constant.

• (2) We remark that the decay estimates (1.13) is in fact proved for general global smooth solutions of (1.1), which does not use the axisymmetric structure of the solutions, whenever $u_0\in L^p$ for some $p\in [1,2).$ In particular, we get rid of the technical assumption in [4] that (1.13) holds for $p\in (1,6/5)$ and moreover the proof here is more concise than that in [4].

Let us complete this section with the notations we are going to use in this context.

Notations: $\dot{H}^s$ (resp. $H^s$) denotes the homogeneous (resp. inhomogeneous) Sobolev space with norm given by $\Vert f\Vert _{\dot{H}^s}\buildrel \hbox {def}\over =\left( \int _{\mathop {\mathbb R}\nolimits ^3}|\xi |^{2s}|\widehat{f}(\xi )|^2\,d\xi \right) ^{\frac{1}{2}}$ (resp. $\Vert f\Vert _{{H}^s}\buildrel \hbox {def}\over =\left( \int _{\mathop {\mathbb R}\nolimits ^3}(1+|\xi |^2)^{s}|\widehat{f}\right. \left. (\xi )|^2\,d\xi \right) ^{\frac{1}{2}}$). For X a Banach space and I an interval of $\mathop {\mathbb R}\nolimits ,$ we denote by ${\mathcal {C}}(I;\,X)$ the set of continuous functions on I with values in X. For $q\in [1,+\infty ],$ the notation $L^q(I;\,X)$ stands for the set of measurable functions on I with values in X,  such that $t\longmapsto \Vert f(t)\Vert _{X}$ belongs to $L^q(I).$ Let $\mathop {\mathbb R}\nolimits ^2_+=(0,\infty )\times \mathop {\mathbb R}\nolimits ,$ we denote $\Vert f\Vert _{\widetilde{L}^q}\buildrel \hbox {def}\over =(\int _{\mathop {\mathbb R}\nolimits ^2_+}|f|^q\,dr\,dz)^{\frac{1}{q}}.$ For $a\lesssim b$, we mean that there is a uniform constant C,  which may be different on different lines, such that $a\le Cb$. We shall denote by (a|b) (or $\int _{\mathop {\mathbb R}\nolimits ^3} a | b\, dx$) the $L^2(\mathop {\mathbb R}\nolimits ^3)$ inner product of a and b,  and finally $\widetilde{\nabla }\buildrel \hbox {def}\over =(\partial _r, \partial _z).$

2 The global $H^1$ estimate

In this section, we shall prove the a priori globally in time $H^1$ estimate for the velocity of (1.1) provided that there holds (1.10). Before proceeding, let us first rewrite the momentum equation of (1.4).

Due to $\partial _ru^r+\frac{u^r}{r}+\partial _zu^z=0$ and $\mathrm{curl}\, u=\omega e_{\theta }$ with $\omega \buildrel \hbox {def}\over =\partial _z u^r-\partial _r u^z,$ we have

$$\begin{aligned} \frac{1}{r}\partial _r(r\partial _r u^r)+\partial _z^2u^r -\frac{u^r}{r^2}&=-\frac{1}{r}\partial _r(r\partial _z u^z+u^r)+\partial _z^2u^r -\frac{u^r}{r^2}\\&=-\frac{1}{r}\left( r\partial _z \partial _r u^z-\frac{u^r}{r} \right) +\partial _z^2u^r -\frac{u^r}{r^2}\\&=\partial _z(\partial _z u^r-\partial _r u^z)=\partial _z \omega . \end{aligned}$$

Similarly, one has

$$\begin{aligned} \frac{1}{r}\partial _r(r\partial _r u^z)+\partial _z^2u^z&=\partial _r^2 u^z+\frac{\partial _r u^z}{r}-\partial _z\left( \partial _r u^r+\frac{u^r}{r}\right) \\&=-\partial _r(\partial _z u^r-\partial _r u^z) -\frac{1}{r}(\partial _z u^r-\partial _r u^z)\\&=-\partial _r\omega -\frac{1}{r}\omega . \end{aligned}$$

So that we can reformulate the momentum equation of (1.4) as

$$\begin{aligned} {\left\{ \begin{array}{ll} &{}\rho \partial _t u^r + \rho u^r \partial _r u^r+\rho u^{z} \partial _z u^r+\partial _r \Pi - \partial _z \omega =0,\\ &{}\rho \partial _t u^{z} + \rho u^r \partial _r u^{z}+\rho u^{z} \partial _z u^{z} +\partial _z \Pi + \partial _r\omega +\frac{1}{r}\omega =0. \end{array}\right. } \end{aligned}$$

(2.1)

2.1 Local in time $H^1$ estimate

The purpose of this subsection is to present the estimate of $\Vert u\Vert _{L^\infty _T(H^1)}$ with T going to $\infty$ when $\varepsilon _0$ in (1.10) tending to zero.

$\bullet$ $\underline{L^2 \hbox { energy estimate}}$

We first deduce from the transport equation of (1.4) and (1.7) that

$$\begin{aligned} m \le \rho (t,r,z)\le M. \end{aligned}$$

(2.2)

While by first multiplying the $u^r$ equation of (1.4) by $u^r$ and then integrating the resulting equation over $\mathop {\mathbb R}\nolimits ^2_+$ with respect to the measure $r\,dr\,dz,$ we write

$$\begin{aligned}&\frac{1}{2}\frac{d}{dt}\int _{\mathop {\mathbb R}\nolimits ^2_+}\rho (u^r)^2\,rdr\,dz-\int _{\mathop {\mathbb R}\nolimits ^2_+}\left( r\partial _t\rho +\partial _r(\rho u^r r)+\partial _z(\rho u^z r)\right) (u^r)^2\,dr\,dz\\&\quad -\int _{\mathop {\mathbb R}\nolimits ^2_+}\Pi \partial _r(u^r r)\,dr\,dz +\int _{\mathop {\mathbb R}\nolimits ^2_+}\left( (\partial _ru^r)^2+(\partial _zu^r)^2+\frac{(u^r)^2}{r^2}\right) r\,dr\,dz=0. \end{aligned}$$

Whereas using the transport equation and $\partial _r(u^rr)+\partial _z(u^zr)=0$ of (1.4), we find

$$\begin{aligned} r\partial _t\rho +\partial _r(\rho u^r r)+\partial _z(\rho u^z r)=r\left( \partial _t\rho +u^r\partial _r\rho +u^z\partial _z u^r\right) +\rho \left( \partial _r(u^rr)+\partial _z(u^zr)\right) =0, \end{aligned}$$

so that we obtain

$$\begin{aligned}&\frac{1}{2}\frac{d}{dt}\int _{\mathop {\mathbb R}\nolimits ^2_+}\rho (u^r)^2\,rdr\,dz+\int _{\mathop {\mathbb R}\nolimits ^2_+}\left( (\partial _ru^r)^2+(\partial _zu^r)^2+\frac{(u^r)^2}{r^2}\right) r\,dr\,dz\\&\quad =\int _{\mathop {\mathbb R}\nolimits ^2_+}\Pi \partial _r(u^r r)\,dr\,dz . \end{aligned}$$

Along the same line, we have

$$\begin{aligned}&\frac{1}{2}\frac{d}{dt}\int _{\mathop {\mathbb R}\nolimits ^2_+}\rho (u^z)^2\,rdr\,dz+\int _{\mathop {\mathbb R}\nolimits ^2_+}\left( (\partial _ru^z)^2+(\partial _zu^z)^2\right) r\,dr\,dz\\&\quad =\int _{\mathop {\mathbb R}\nolimits ^2_+}\Pi \partial _z(u^z r)\,dr\,dz . \end{aligned}$$

Hence due to $\partial _r(ru^r)+\partial _z(ru^z)=0,$ we achieve

$$\begin{aligned} \frac{1}{2}\frac{d}{dt}\int _{\mathop {\mathbb R}\nolimits ^2_+}\rho \left( (u^r)^2+(u^z)^2\right) r\,dr\,dz+\int _{\mathop {\mathbb R}\nolimits ^2_+}\left( |\widetilde{\nabla }u^r|^2+|\widetilde{\nabla }u^z|^2+\frac{(u^r)^2}{r^2}\right) r\,dr\,dz=0. \end{aligned}$$

Integrating the above inequality over [0, t] and using (2.2) gives rise to

$$\begin{aligned} \Vert u\Vert _{L^\infty _t(L^2)}^2+\Vert \widetilde{\nabla }u\Vert _{L^2_t(L^2)}^2+\left\| \frac{u^r}{r}\right\| _{L^2_t(L^2)}^2\le C\Vert u_0\Vert _{L^2}^2. \end{aligned}$$

(2.3)

$\bullet$ $\underline{\dot{H}^1 \hbox { energy estimate}}$

By taking $L^2(\mathop {\mathbb R}\nolimits ^2_+,r\,dr\,dz)$ inner product of the $u^r$ equation of (1.4) with $\partial _tu^r$ and using integration by parts, we have

$$\begin{aligned}&\frac{1}{2}\frac{d}{dt}\int _{\mathop {\mathbb R}\nolimits ^2_+}\left( (\partial _ru^r)^2+(\partial _zu^r)^2+\frac{(u^r)^2}{r^2}\right) r\,dr\,dz+\int _{\mathop {\mathbb R}\nolimits ^2_+}\rho (\partial _tu^r)^2r\,dr\,dz\\&\quad =-\int _{\mathop {\mathbb R}\nolimits ^2_+}\rho \left( u^r\partial _ru^r+u^z\partial _zu^r\right) \partial _tu^rr\,dr\,dz+\int _{\mathop {\mathbb R}\nolimits ^2_+}\Pi \partial _r(\partial _tu^rr)\,dr\,dz. \end{aligned}$$

Similarly we have

$$\begin{aligned}&\frac{1}{2}\frac{d}{dt}\int _{\mathop {\mathbb R}\nolimits ^2_+}\left( (\partial _ru^z)^2+(\partial _zu^z)^2\right) r\,dr\,dz+\int _{\mathop {\mathbb R}\nolimits ^2_+}\rho (\partial _tu^z)^2r\,dr\,dz\\&\quad =-\int _{\mathop {\mathbb R}\nolimits ^2_+}\rho \left( u^r\partial _ru^z+u^z\partial _zu^z\right) \partial _tu^zr\,dr\,dz+\int _{\mathop {\mathbb R}\nolimits ^2_+}\Pi \partial _z(\partial _tu^zr)\,dr\,dz, \end{aligned}$$

which together $\partial _r(ru^r)+\partial _z(ru^z)=0$ gives rise to

$$\begin{aligned}&\frac{1}{2}\frac{d}{dt}\int _{\mathop {\mathbb R}\nolimits ^2_+}\left( |\widetilde{\nabla }u^r|^2+|\widetilde{\nabla }u^z|^2+\frac{(u^r)^2}{r^2}\right) r\,dr\,dz +\int _{\mathop {\mathbb R}\nolimits ^2_+}\rho \left( (\partial _tu^r)^2+(\partial _tu^z)^2\right) r\,dr\,dz\\&\quad =-\int _{\mathop {\mathbb R}\nolimits ^2_+}\rho \left( u^r\partial _ru^r+u^z\partial _zu^r\right) \partial _tu^rr\,dr\,dz-\int _{\mathop {\mathbb R}\nolimits ^2_+}\rho \left( u^r\partial _ru^z+u^z\partial _zu^z\right) \partial _tu^zr\,dr\,dz\\&\quad \le C\left( \Vert \sqrt{\rho }u^r\partial _ru^r\Vert _{L^2}^2+\Vert \sqrt{\rho }u^z\partial _zu^r\Vert _{L^2}^2+\Vert \sqrt{\rho }u^r\partial _ru^z\Vert _{L^2}^2+\Vert \sqrt{\rho }u^z\partial _zu^z)\Vert _{L^2}^2\right) \\&\qquad +\frac{1}{2}\left( \Vert \sqrt{\rho }\partial _tu^r\Vert _{L^2}^2+\Vert \sqrt{\rho }\partial _tu^z\Vert _{L^2}^2\right) , \end{aligned}$$

which along with (2.2) implies

$$\begin{aligned}&\frac{d}{dt}\int _{\mathop {\mathbb R}\nolimits ^2_+}\left( |\widetilde{\nabla }u^r|^2+|\widetilde{\nabla }u^z|^2+\frac{(u^r)^2}{r^2}\right) r\,dr\,dz +\Vert \partial _tu^r\Vert _{L^2}^2+\Vert \partial _tu^z\Vert _{L^2}^2\nonumber \\&\quad \le C\left( \Vert u^r\partial _ru^r\Vert _{L^2}^2+\Vert u^z\partial _zu^r\Vert _{L^2}^2+\Vert u^r\partial _ru^z\Vert _{L^2}^2+\Vert \sqrt{\rho }u^z\partial _zu^z\Vert _{L^2}^2\right) . \end{aligned}$$

(2.4)

$\bullet$ The second derivative estimate of the velocity

By taking $L^2(\mathop {\mathbb R}\nolimits ^2_+;r\,dr\,dz)$ inner product of the $u^r$ equation of (2.1) with $\partial _z\omega$ and using integration by parts, one has

$$\begin{aligned} \int _{\mathop {\mathbb R}\nolimits ^2_+}(\partial _z\omega )^2r\,dr\,dz&=-\int _{\mathop {\mathbb R}\nolimits ^2_+}\partial _z\partial _r\Pi \ |\ \omega r\,dr\,dz\\&\quad -\int _{\mathop {\mathbb R}\nolimits ^2_+}\left( \rho \partial _tu^r+\rho u^r\partial _ru^r+\rho u^z\partial _zu^r\right) \ |\ \partial _z\omega r\,dr\,dz. \end{aligned}$$

Similarly taking $L^2(\mathop {\mathbb R}\nolimits ^2_+;r\,dr\,dz)$ inner product of the $u^z$ equation of (2.1) with $\partial _r(r\omega )r^{-1}$ leads to

$$\begin{aligned} \int _{\mathop {\mathbb R}\nolimits ^2_+}(\partial _r(r\omega ))^2r^{-1}\,dr\,dz= & {} \int _{\mathop {\mathbb R}\nolimits ^2_+}\partial _z\partial _r\Pi \ |\ \omega r\,dr\,dz\\&-\int _{\mathop {\mathbb R}\nolimits ^2_+}\left( \rho \partial _tu^z+\rho u^r\partial _ru^z+\rho u^z\partial _zu^z\right) \ | \partial _r(\omega r)\,dr\,dz. \end{aligned}$$

Yet notice that

$$\begin{aligned} \int _{\mathop {\mathbb R}\nolimits ^2_+}(\partial _r(r\omega ))^2r^{-1}\,dr\,dz= & {} \int _{\mathop {\mathbb R}\nolimits ^2_+}\left( \frac{\omega ^2}{r}+2\omega \partial _r\omega +(\partial _r\omega )^2r\right) \,dr\,dz\\= & {} \int _{\mathop {\mathbb R}\nolimits ^2_+}\left( \frac{\omega ^2}{r^2}+(\partial _r\omega )^2\right) r\,dr\,dz. \end{aligned}$$

As a consequence, for $\Gamma$ given by (1.6), we obtain

$$\begin{aligned} \int _{\mathop {\mathbb R}\nolimits ^2_+}\left( (\partial _r\omega )^2+(\partial _z\omega )^2+\Gamma ^2\right) r\,dr\,dz&\le C\left( \Vert u^r_t\Vert _{L^2}^2+\Vert u^z_t\Vert _{L^2}^2+\Vert u^r\partial _ru^r\Vert _{L^2}^2\right. \nonumber \\&\quad \left. +\Vert u^z\partial _zu^r\Vert _{L^2}^2 +\Vert u^r\partial _ru^z\Vert _{L^2}^2+\Vert u^z\partial _zu^z)\Vert _{L^2}^2\right) . \end{aligned}$$

(2.5)

Along the same line, we have

$$\begin{aligned} \Vert \widetilde{\nabla }\Pi \Vert _{L^2}^2&\le C\left( \Vert u^r_t\Vert _{L^2}^2+\Vert u^z_t\Vert _{L^2}^2+\Vert u^r\partial _ru^r\Vert _{L^2}^2\right. \nonumber \\&\quad \left. +\Vert u^z\partial _zu^r\Vert _{L^2}^2+\Vert u^r\partial _ru^z\Vert _{L^2}^2+\Vert u^z\partial _zu^z\Vert _{L^2}^2\right) . \end{aligned}$$

(2.6)

$\bullet$ The combined estimate

Let $\delta >0$ be a small positive constant, which will be chosen hereafter. By summing up (2.4) with $\delta \times ((2.5)+(2.6))$ leads to

$$\begin{aligned}&\frac{d}{dt}\int _{\mathop {\mathbb R}\nolimits ^2_+}\left( |\widetilde{\nabla }u^r|^2+|\widetilde{\nabla }u^z|^2+\frac{(u^r)^2}{r^2}\right) r\,dr\,dz +(1-2C\delta )\left( \Vert \partial _tu^r\Vert _{L^2}^2+\Vert \partial _tu^z\Vert _{L^2}^2\right) \\&\quad \quad +\delta \left( \int _{\mathop {\mathbb R}\nolimits ^2_+}\left( |\widetilde{\nabla }\omega |^2+\Gamma ^2\right) r\,dr\,dz+\Vert \widetilde{\nabla }\Pi \Vert _{L^2}^2\right) \\&\quad \le C\left( \Vert u^r\partial _ru^r\Vert _{L^2}^2+\Vert u^z\partial _zu^r\Vert _{L^2}^2+\Vert u^r\partial _ru^z\Vert _{L^2}^2+\Vert u^z\partial _zu^z\Vert _{L^2}^2\right) . \end{aligned}$$

Taking $\delta =\frac{1}{4C}$ in the above inequality yields

$$\begin{aligned}&\frac{d}{dt}\int _{\mathop {\mathbb R}\nolimits ^2_+}\left( |\widetilde{\nabla }u^r|^2+|\widetilde{\nabla }u^z|^2+\frac{(u^r)^2}{r^2}\right) r\,dr\,dz +\Vert \partial _tu^r\Vert _{L^2}^2+\Vert \partial _tu^z\Vert _{L^2}^2\nonumber \\&\quad \quad +\int _{\mathop {\mathbb R}\nolimits ^2_+}\left( |\widetilde{\nabla }\omega |^2+\Gamma ^2\right) r\,dr\,dz+\Vert \widetilde{\nabla }\Pi \Vert _{L^2}^2\nonumber \\&\quad \le C\left( \Vert u^r\partial _ru^r\Vert _{L^2}^2+\Vert u^z\partial _zu^r\Vert _{L^2}^2+\Vert u^r\partial _ru^z\Vert _{L^2}^2+\Vert u^z\partial _zu^z\Vert _{L^2}^2\right) . \end{aligned}$$

(2.7)

In order to cope with the right hand side terms in (2.7), we take cut-off functions $\varphi \in C_0^\infty [0,\infty )$ and $\psi \in C^\infty [0,\infty )$ with

$$\begin{aligned} \varphi (r)=\left\{ \begin{array}{l} \displaystyle 1\quad r\in [0,1/2],\\ \displaystyle 0\quad r\in [1,\infty ), \end{array}\right. \quad \hbox {and}\quad \psi (r)=\left\{ \begin{array}{l} \displaystyle 1\quad r\in [1/2, \infty ),\\ \displaystyle 0\quad r\in [0,1/4), \end{array}\right. \end{aligned}$$

(2.8)

and present the lemma as follows:

Lemma 2.1

Let f(r, z) be a smooth enough function which decays sufficiently fast at infinity. Then for $\varphi (r)$ given by (2.8), one has

$$\begin{aligned} \int _{\mathop {\mathbb R}\nolimits ^2_+}f^4\varphi (r)r^3\,dr\,dz\le C\Vert f\Vert _{L^2}^2\left( \Vert f\Vert _{L^2}+\Vert \partial _rf\Vert _{L^2}\right) \Vert \partial _zf\Vert _{L^2}. \end{aligned}$$

(2.9)

Proof

It is easy to observe that

$$\begin{aligned} r^2f^2\varphi (r)\le & {} \int _0^\infty |\partial _r(r^2f^2\varphi (r))|\,dr\\\le & {} C\int _0^\infty |f|(|f|+|\partial _rf|)r\,dr, \end{aligned}$$

and

$$\begin{aligned} rf^2\le \int _{\mathop {\mathbb R}\nolimits }|\partial _zf^2|r\,dz=2\int _{\mathop {\mathbb R}\nolimits }|f||\partial _zf|r\,dz, \end{aligned}$$

from which, we infer

$$\begin{aligned} \int _{\mathop {\mathbb R}\nolimits ^2_+}f^4\varphi (r)r^3\,dr\,dz\le & {} C\int _{\mathop {\mathbb R}\nolimits ^2_+}\int _0^\infty |f|(|f|+|\partial _rf|)r\,dr\int _{\mathop {\mathbb R}\nolimits }|f||\partial _zf|r\,dz\,dr\,dz\\\le & {} C \int _{\mathop {\mathbb R}\nolimits ^2_+}|f|(|f|+|\partial _rf|)r\,dr\,dz\int _{\mathop {\mathbb R}\nolimits ^2_+}|f||\partial _zf|r\,dr\,dz. \end{aligned}$$

Applying Hölder inequality gives rise to (2.9). $\square$

Now let us turn to the estimate of the nonlinear terms in (2.7). We first get, by applying Hölder’s inequality and the 2-D interpolation inequality,

$$\begin{aligned} \Vert f\Vert _{L^4(\mathop {\mathbb R}\nolimits ^2)}\lesssim \Vert f\Vert _{L^2(\mathop {\mathbb R}\nolimits ^2)}^{\frac{1}{2}}\Vert \nabla f\Vert _{L^2(\mathop {\mathbb R}\nolimits ^2)}^{\frac{1}{2}}, \end{aligned}$$

(2.10)

that

$$\begin{aligned} \Vert u^r\partial _ru\Vert _{L^2}^2\le & {} \Vert \partial _ru\Vert _{\widetilde{L}^4}^2\Vert \sqrt{r}u^r\Vert _{\widetilde{L}^4}^2\\\le & {} C\left( \int _{\mathop {\mathbb R}\nolimits ^3}\omega ^4 r^{-1}\,dx\right) ^{\frac{1}{2}}\Vert \sqrt{r}u^r\Vert _{\widetilde{L}^2}\Vert \widetilde{\nabla }(\sqrt{r}u^r)\Vert _{\widetilde{L}^2}, \end{aligned}$$

where we used Biot–Sarvart’s law

$$\begin{aligned} u(t,x)=\frac{1}{4\pi }\int _{\mathop {\mathbb R}\nolimits ^3}\frac{(y-x)\wedge e_\theta \omega (t,y)}{|y-x|^3}\,dy \end{aligned}$$

and the fact that $r^{-1}$ is in $A^p$ class (see [11] for instance) so that

$$\begin{aligned} \Vert \partial _ru\Vert _{\widetilde{L}^4}=\left( \int _{\mathop {\mathbb R}\nolimits ^3}|\partial _r u|^4r^{-1}\,dx\right) ^{\frac{1}{4}}\le C\left( \int _{\mathop {\mathbb R}\nolimits ^3}\omega ^4r^{-1}\,dx\right) ^{\frac{1}{4}}. \end{aligned}$$

Then by virtue of (2.9) and (2.10), we infer

$$\begin{aligned} \left( \int _{\mathop {\mathbb R}\nolimits ^3}\omega ^4r^{-1}\,dx\right) ^{\frac{1}{4}}\le & {} \left( \int _{\mathop {\mathbb R}\nolimits ^2_+}\Gamma ^4r^4\varphi (r)\,dr\,dz\right) ^{\frac{1}{4}}+ \left( \int _{\mathop {\mathbb R}\nolimits ^2_+}\omega ^4(1-\varphi (r))\,dr\,dz\right) ^{\frac{1}{4}}\\\lesssim & {} \left( \int _{\mathop {\mathbb R}\nolimits ^2_+}\Gamma ^4r^3\varphi (r)\,dr\,dz\right) ^{\frac{1}{4}}+\Vert \omega \psi \Vert _{\widetilde{L}^4}\\\lesssim & {} \Vert \Gamma \Vert _{L^2}^{\frac{1}{2}}\left( \Vert \Gamma \Vert _{L^2}^{\frac{1}{2}}+\Vert \widetilde{\nabla }\Gamma \Vert _{L^2}^{\frac{1}{2}}\right) +\Vert \omega \psi \Vert _{\widetilde{L}^2}^{\frac{1}{2}}\Vert \widetilde{\nabla }(\omega \psi ) \Vert _{\widetilde{L}^2}^{\frac{1}{2}}\\\lesssim & {} \Vert \Gamma \Vert _{L^2}^{\frac{1}{2}}\left( \Vert \Gamma \Vert _{L^2}^{\frac{1}{2}}+\Vert \widetilde{\nabla }\Gamma \Vert _{L^2}^{\frac{1}{2}}\right) + \Vert \omega \Vert _{L^2}^{\frac{1}{2}}\left( \Vert \omega \Vert _{L^2}^{\frac{1}{2}}+\Vert \widetilde{\nabla }\omega \Vert _{L^2}^{\frac{1}{2}}\right) . \end{aligned}$$

Moreover, note that

$$\begin{aligned} \Vert \widetilde{\nabla }(\sqrt{r}u^r)\Vert _{\widetilde{L}^2}\le C\left( \Vert \widetilde{\nabla }u^r\Vert _{L^2}+\left\| \frac{u^r}{r}\right\| _{L^2}\right) , \end{aligned}$$

for any $\delta >0,$ we write

$$\begin{aligned} \Vert u^r\partial _ru\Vert _{L^2}^2&\le C\left( \Vert \Gamma \Vert _{L^2}\left( \Vert \Gamma \Vert _{L^2}+\Vert \widetilde{\nabla }\Gamma \Vert _{L^2}\right) + \Vert \omega \Vert _{L^2}\left( \Vert \omega \Vert _{L^2}+\Vert \widetilde{\nabla }\omega \Vert _{L^2}\right) \right) \nonumber \\&\quad \times \Vert u^r\Vert _{L^2}\left( \Vert \widetilde{\nabla }u^r\Vert _{L^2}+\left\| \frac{u^r}{r}\right\| _{L^2}\right) \nonumber \\&\le C_\delta \Vert u^r\Vert _{L^2}^2\left( \Vert \widetilde{\nabla }u^r\Vert _{L^2}^2+\left\| \frac{u^r}{r}\right\| _{L^2}^2\right) \left( \Vert \omega \Vert _{L^2}^2+\Vert \Gamma \Vert _{L^2}^2\right) \nonumber \\&\quad +\delta \left( \Vert \omega \Vert _{L^2}^2+\Vert \widetilde{\nabla }\omega \Vert _{L^2}^2+\Vert \Gamma \Vert _{L^2}^2+\Vert \widetilde{\nabla }\Gamma \Vert _{L^2}^2\right) . \end{aligned}$$

(2.11)

To deal with $\Vert u^z\partial _z u\Vert _{L^2},$ we split $\int _{\mathop {\mathbb R}\nolimits ^2_+}(u^z\partial _zu)^2r\,dr\,dz$ as

$$\begin{aligned} \int _{\mathop {\mathbb R}\nolimits ^2_+}(u^z\partial _zu)^2r\,dr\,dz=\int _{\mathop {\mathbb R}\nolimits ^2_+}(u^z\partial _zu)^2\varphi (r)r\,dr\,dz+ \int _{\mathop {\mathbb R}\nolimits ^2_+}(u^z\partial _zu)^2(1-\varphi (r))r\,dr\,dz.\nonumber \\ \end{aligned}$$

(2.12)

By applying (2.10) and convexity inequality, we get for any $\delta >0$

$$\begin{aligned}&\int _{\mathop {\mathbb R}\nolimits ^2_+}(u^z\partial _zu)^2(1-\varphi (r))r\,dr\,dz\nonumber \\&\quad \lesssim \int _{\mathop {\mathbb R}\nolimits ^2_+}(u^z\partial _zu\psi (r)r^{\frac{1}{2}})^2\,dr\,dz\nonumber \\&\quad \lesssim \left\| u^z\psi (r)r^{\frac{1}{4}}\right\| _{\widetilde{L}^4}^2\left\| \partial _zu\psi (r) r^{\frac{1}{4}}\right\| _{\widetilde{L}^4}^2\nonumber \\&\quad \lesssim \left\| u^z\psi (r)r^{\frac{1}{4}}\right\| _{\widetilde{L}^2}\left\| \widetilde{\nabla }(u^z\psi (r)r^{\frac{1}{4}})\right\| _{\widetilde{L}^2}\left\| \partial _zu\psi r^{\frac{1}{4}}\right\| _{\widetilde{L}^2}\left\| \widetilde{\nabla }(\partial _zu\psi r^{\frac{1}{4}})\right\| _{\widetilde{L}^2}\nonumber \\&\quad \le C_\delta \Vert u^z\Vert _{L^2}^2\left( \Vert u^z\Vert _{L^2}^2+\Vert \widetilde{\nabla }u^z\Vert _{L^2}^2\right) \Vert \partial _zu\Vert _{L^2}^2+\delta \left( \Vert \partial _zu\Vert _{L^2}^2 +\Vert \widetilde{\nabla }\partial _zu\Vert _{L^2}^2\right) . \end{aligned}$$

(2.13)

Before proceeding, let us recall from (2.22) of [19] that

$$\begin{aligned} \frac{u^r}{r}=\partial _z\Delta ^{-1}\Gamma -2\frac{\partial _r}{r}\Delta ^{-1}\partial _z\Delta ^{-1}\Gamma , \end{aligned}$$

(2.14)

and from (21) of [13] that

$$\begin{aligned} \frac{\partial _r}{r}\Delta ^{-1}W=\frac{x_2^2}{r^2}{\mathcal R}_{11}W+\frac{x_1^2}{r^2}{\mathcal R}_{22}W-2\frac{x_1x_2}{r^2}{\mathcal R}_{12}W \end{aligned}$$

(2.15)

for every axisymmetric smooth function W,  and where ${\mathcal R}_{ij}\buildrel \hbox {def}\over =\partial _i\partial _j\Delta ^{-1}.$

By virtue of (2.9), we infer

$$\begin{aligned} \int _{\mathop {\mathbb R}\nolimits ^2_+}(u^z\partial _zu^r)^2\varphi (r)r\,dr\,dz&=\int _{\mathop {\mathbb R}\nolimits ^2_+}(u^z)^2r^{\frac{3}{2}}\varphi ^{\frac{1}{2}}(r)\left( \partial _z\frac{u^r}{r}\right) ^2r^{\frac{3}{2}}\varphi ^{\frac{1}{2}}(r)\,dr\,dz\\&\le \left( \int _{\mathop {\mathbb R}\nolimits ^2_+}(u^z)^4r^3\varphi (r)\,dr\,dz\right) ^{\frac{1}{2}}\left( \int _{\mathop {\mathbb R}\nolimits ^2_+}\left( \partial _z\frac{u^r}{r}\right) ^4r^3\varphi (r)\,dr\,dz\right) ^{\frac{1}{2}}\\&\lesssim \Vert u^z\Vert _{L^2}\left( \Vert u^z\Vert _{L^2}^{\frac{1}{2}}+\Vert \partial _ru^z\Vert _{L^2}^{\frac{1}{2}}\right) \Vert \partial _zu^z\Vert _{L^2}^{\frac{1}{2}} \left\| \partial _z\frac{u^r}{r}\right\| _{L^2}\\&\quad \times \left( \left\| \partial _z\frac{u^r}{r}\right\| _{L^2}^{\frac{1}{2}}+\left\| \partial _z\partial _r\frac{u^r}{r}\right\| _{L^2}^{\frac{1}{2}}\right) \left\| \partial _z^2\frac{u^r}{r}\right\| _{L^2}^{\frac{1}{2}}. \end{aligned}$$

Yet it follows from (2.14) and (2.15) that

$$\begin{aligned} \left\| \partial _z\frac{u^r}{r}\right\| _{L^2}\lesssim \Vert \Gamma \Vert _{L^2},\quad \left\| \partial _z^2\frac{u^r}{r}\right\| _{L^2}\lesssim \Vert \partial _z\Gamma \Vert _{L^2}\quad \hbox {and}\quad \left\| \partial _z\partial _r\frac{u^r}{r}\right\| _{L^2}\lesssim \Vert \widetilde{\nabla }\Gamma \Vert _{L^2}. \end{aligned}$$

Therefore, for any $\delta >0,$ we have

$$\begin{aligned} \int _{\mathop {\mathbb R}\nolimits ^2_+}(u^z\partial _zu^r)^2\varphi (r)r\,dr\,dz\le & {} C\Vert u^z\Vert _{L^2}^2\left( 1+\Vert u^z\Vert _{L^2}^4\right) \Vert \widetilde{\nabla }u^z\Vert _{L^2}^2\Vert \Gamma \Vert _{L^2}^2\nonumber \\&+\,\delta \left( \Vert \Gamma \Vert _{L^2}^2+\Vert \widetilde{\nabla }\Gamma \Vert _{L^2}^2\right) . \end{aligned}$$

(2.16)

While since $\partial _r(ru^r)+\partial _z(ru^z)=0,$ we have

$$\begin{aligned} \int _{\mathop {\mathbb R}\nolimits ^2_+}(u^z\partial _zu^z)^2\varphi (r)r\,dr\,dz=\int _{\mathop {\mathbb R}\nolimits ^2_+}\left( u^z\left( \partial _ru^r+\frac{u^r}{r}\right) \right) ^2\varphi (r)r\,dr\,dz. \end{aligned}$$

(2.17)

Due to (2.14) and (2.15), we have

$$\begin{aligned} \int _{\mathop {\mathbb R}\nolimits ^2_+}\left( \frac{u^z u^r}{r}\right) ^2\varphi (r)r\,dr\,dz&\le \left( \int _{\mathop {\mathbb R}\nolimits ^2_+}(u^z)^3\varphi ^2(r)r\,dr\,dz\right) ^{\frac{2}{3}}\left\| \frac{u^r}{r}\right\| _{L^6}^2\\&\le \left( \int _{\mathop {\mathbb R}\nolimits ^3}\frac{(u^z)^3}{r^{\frac{3}{2}}}\,dx\right) ^{\frac{2}{3}}\left\| \partial _z\Delta ^{-1}\Gamma \right\| _{L^6}^2\\&\le C\Vert \nabla u^z\Vert _{L^2}^2\Vert \Gamma \Vert _{L^2}^2. \end{aligned}$$

where we used Sobolev–Hardy inequality from [6] that

$$\begin{aligned} \int _{\mathop {\mathbb R}\nolimits ^N}\frac{|u|^{q_*(s)}}{|x'|^s}\,dx\le C(s,q,N,k)\left( \int _{\mathop {\mathbb R}\nolimits ^N}|\nabla u|^q\,dx\right) ^{\frac{N-s}{N-q}}, \end{aligned}$$

(2.18)

where $x=(x',z)\in R^N=R^k\times \mathop {\mathbb R}\nolimits ^{N-k}$ with $2\le k\le N,$ $1<q<N,$ $0\le s\le q$ and $s<k,$ $q_*\buildrel \hbox {def}\over =\frac{q(N-s)}{N-q},$ so that there holds

$$\begin{aligned} \left( \int _{\mathop {\mathbb R}\nolimits ^3}\frac{(u^z)^3}{r^{\frac{3}{2}}}\,dx\right) ^{\frac{1}{3}}\le C \Vert \nabla u^z\Vert _{L^2}. \end{aligned}$$

Whereas it follows from (2.14) that

$$\begin{aligned} \partial _ru^r=\partial _z\Delta ^{-1}\Gamma +r\partial _z\partial _r\Delta ^{-1}\Gamma -2\partial _r^2\Delta ^{-1}\partial _z\Delta ^{-1}\Gamma . \end{aligned}$$

Applying Hardy’s inequality (2.18) once again yields

$$\begin{aligned} \int _{\mathop {\mathbb R}\nolimits ^2_+}(u^z)^2\left( \partial _z\Delta ^{-1}\Gamma \right) ^2\varphi (r)r\,dr\,dz&\le \left( \int _{\mathop {\mathbb R}\nolimits ^2_+}|u^z|^3\varphi ^{\frac{3}{2}}(r)r\,dr\,dz\right) ^{\frac{2}{3}}\Vert \partial _z\Delta ^{-1}\Gamma \Vert _{L^6}^2\\&\le \left( \int _{\mathop {\mathbb R}\nolimits ^3}\frac{|u^z|^3}{r^{\frac{3}{2}}}\,dx\right) ^{\frac{2}{3}}\Vert \Gamma \Vert _{L^2}^2\\&\lesssim \Vert \nabla u^z\Vert _{L^2}^2\Vert \Gamma \Vert _{L^2}^2. \end{aligned}$$

Similarly, by applying Lemma 2.2, one has

$$\begin{aligned}&\int _{\mathop {\mathbb R}\nolimits ^2_+}(u^z)^2\left( r\partial _z\partial _r\Delta ^{-1}\Gamma \right) ^2\varphi (r)r\,dr\,dz\nonumber \\&\quad \le \left( \int _{\mathop {\mathbb R}\nolimits ^2_+}(u^z)^4\varphi (r)r^3\,dr\,dz\right) ^{\frac{1}{2}}\left( \int _{\mathop {\mathbb R}\nolimits ^2_+}|\partial _z\partial _r\Delta ^{-1}\Gamma |^4\varphi (r)r^3\,dr\,dz\right) ^{\frac{1}{2}}\nonumber \\&\quad \le C\Vert u^z\Vert _{L^2}\left( \Vert u^z\Vert _{L^2}^{\frac{1}{2}}+\Vert \partial _ru^z\Vert _{L^2}^{\frac{1}{2}}\right) \Vert \partial _zu^z\Vert _{L^2}^{\frac{1}{2}}\Vert \Gamma \Vert _{L^2} \left( \Vert \Gamma \Vert _{L^2}^{\frac{1}{2}}+\Vert \widetilde{\nabla }\Gamma \Vert _{L^2}^{\frac{1}{2}}\right) \Vert \partial _z\Gamma \Vert _{L^2}^{\frac{1}{2}}\nonumber \\&\quad \le C_\delta \Vert u^z\Vert _{L^2}^2\left( 1+\Vert u^z\Vert _{L^2}^4\right) \Vert \widetilde{\nabla } u^z\Vert _{L^2}^2\Vert \Gamma \Vert _{L^2}^2+\delta \left( \Vert \Gamma \Vert _{L^2}^{2}+\Vert \widetilde{\nabla }\Gamma \Vert _{L^2}^{2}\right) . \end{aligned}$$

(2.19)

Let $W\buildrel \hbox {def}\over =\partial _z\Delta ^{-1}\Gamma .$ Then by virtue of (2.14), we find

$$\begin{aligned} \partial _r^2\Delta ^{-1}W&=\partial _r\left( \frac{x_2^2}{r}{\mathcal R}_{11}W+\frac{x_1^2}{r}{\mathcal R}_{22}W-2\frac{x_1x_2}{r}{\mathcal R}_{12}W\right) \\&=\sin ^2\theta {\mathcal R}_{11}W+\cos ^2\theta {\mathcal R}_{22}W-2\sin \theta \cos \theta {\mathcal R}_{12}W\\&\quad +r\left( \sin ^2\theta \partial _r{\mathcal R}_{11}W+\cos ^2\theta \partial _r{\mathcal R}_{22}W-2\sin \theta \cos \theta \partial _r{\mathcal R}_{12}W\right) . \end{aligned}$$

It is easy to observe that

$$\begin{aligned}&\int _{\mathop {\mathbb R}\nolimits ^2_+}(u^z)^2\left( \sin ^2\theta {\mathcal R}_{11}W+\cos ^2\theta {\mathcal R}_{22}W-2\sin \theta \cos \theta {\mathcal R}_{12}W\right) ^2\varphi (r)r\,dr\,dz\\&\quad \lesssim \left( \int _{\mathop {\mathbb R}\nolimits ^3}|u^z|^3r^{-\frac{3}{2}}\,dx\right) ^{\frac{2}{3}}\Vert \partial _z\Delta ^{-1}\Gamma \Vert _{L^6}^2 \lesssim \Vert \nabla u^z\Vert _{L^2}^2\Vert \Gamma \Vert _{L^2}^2, \end{aligned}$$

and it follows from a similar derivation of (2.19) that

$$\begin{aligned}&\int _{\mathop {\mathbb R}\nolimits ^2_+}(u^z)^2\left( \sin ^2\theta \partial _r{\mathcal R}_{11}W+\cos ^2\theta \partial _r{\mathcal R}_{22}W-2\sin \theta \cos \theta \partial _r{\mathcal R}_{12}W\right) ^2\varphi (r)r^3\,dr\,dz\\&\quad \le \left( \int _{\mathop {\mathbb R}\nolimits ^2_+}|u^z|^4\varphi (r)r^{3}\,dr\,dz\right) ^{\frac{1}{2}}\\&\qquad \times \left( \int _{R^2_+} \left( \sin ^2\theta \partial _r{\mathcal R}_{11}W+\cos ^2\theta \partial _r{\mathcal R}_{22}W-2\sin \theta \cos \theta \partial _r{\mathcal R}_{12}W\right) ^4\varphi (r)r^{3}\,dr\,dz\right) ^{\frac{1}{2}}\\&\quad \le C_\delta \Vert u^z\Vert _{L^2}^2\left( 1+\Vert u^z\Vert _{L^2}^4\right) \Vert \widetilde{\nabla } u^z\Vert _{L^2}^2\Vert \Gamma \Vert _{L^2}^2+\delta \left( \Vert \Gamma \Vert _{L^2}^{2}+\Vert \widetilde{\nabla }\Gamma \Vert _{L^2}^{2}\right) . \end{aligned}$$

By resuming the above estimates into (2.17), we obtain

$$\begin{aligned} \int _{\mathop {\mathbb R}\nolimits ^2_+}(u^z\partial _zu^z)^2\varphi (r)r\,dr\,dz\le C_\delta \left( 1+\Vert u^z\Vert _{L^2}^6\right) \Vert \widetilde{\nabla }u^z\Vert _{L^2}^2\Vert \Gamma \Vert _{L^2}^2+ \delta \left( \Vert \Gamma \Vert _{L^2}^2+\Vert \widetilde{\nabla }\Gamma \Vert _{L^2}^2\right) .\nonumber \\ \end{aligned}$$

(2.20)

Therefore, by substituting the Estimates (2.13), (2.16) and (2.20) into (2.12), we obtain

$$\begin{aligned} \Vert u^z\partial _z u\Vert _{L^2}^2&\le C_\delta \left( \left( 1+\Vert u^z\Vert _{L^2}^6\right) \Vert \widetilde{\nabla }u^z\Vert _{L^2}^2\left( \Vert \widetilde{\nabla }u\Vert _{L^2}^2+\Vert \Gamma \Vert _{L^2}^2\right) \right. \nonumber \\&\quad \left. +(1+\Vert u^z\Vert _{L^2}^4)\Vert \partial _zu\Vert _{L^2}^2\right) +\delta \left( \Vert \Gamma \Vert _{L^2}^2+\Vert \widetilde{\nabla }\partial _zu\Vert _{L^2}^2+\Vert \widetilde{\nabla }\Gamma \Vert _{L^2}^2\right) . \end{aligned}$$

(2.21)

Note that for the axisymmetric flow, we have for $1<q<\infty$

$$\begin{aligned}&\mathrm{(i)}\quad \ \Vert \omega \Vert _{L^q}\quad \approx \quad \Vert \nabla u\Vert _{L^q} \quad \hbox {and}\quad \nonumber \\&\mathrm{(ii)}\quad \Vert \nabla \omega \Vert _{L^q} + \left\| \frac{\omega }{r}\right\| _{L^q}\quad \approx \quad \Vert \nabla ^2u\Vert _{L^q}. \end{aligned}$$

(2.22)

Thanks to (2.22), by resuming the Estimates (2.11) and (2.21) into (2.7) and taking $\delta$ to be sufficiently small, we obtain

$$\begin{aligned}&\frac{d}{dt}\left( \Vert \widetilde{\nabla }u(t)\Vert _{L^2}^2+\left\| \frac{u^r(t)}{r}\right\| _{L^2}\right) +\Vert \partial _tu\Vert _{L^2}^2+\Vert u\Vert _{\dot{H}^2}^2+\Vert \Gamma \Vert _{L^2}^2+\Vert \widetilde{\nabla }\Pi \Vert _{L^2}^2\nonumber \\&\quad \le C_\delta \left( (1+\Vert u\Vert _{L^2}^6)\left( \Vert \widetilde{\nabla }u\Vert _{L^2}^2+\left\| \frac{u^r}{r}\right\| _{L^2}^2\right) \left( \Vert \widetilde{\nabla }u\Vert _{L^2}^2+\Vert \Gamma \Vert _{L^2}^2\right) \right. \nonumber \\&\qquad \left. +(1+\Vert u^z\Vert _{L^2}^4)\Vert \widetilde{\nabla }u\Vert _{L^2}^2\right) +\delta \Vert \widetilde{\nabla }\Gamma \Vert _{L^2}^2. \end{aligned}$$

(2.23)

By applying Gronwall’s inequality to (2.23), we write

$$\begin{aligned}&\Vert \nabla u\Vert _{L^\infty _t(L^2)}^2+\left\| \frac{u^r}{r}\right\| _{L^\infty _t(L^2)}^2+\Vert \partial _tu\Vert _{L^2_t(L^2)}^2+\Vert u\Vert _{L^2_t(\dot{H}^2)}^2+\Vert \Gamma \Vert _{L^2_t(L^2)}^2+\Vert \nabla \Pi \Vert _{L^2_t(L^2)}^2\\&\quad \le C\exp \left( C\left( 1+\Vert u\Vert _{L^\infty _t(L^2)}^6\right) \left( \Vert \nabla u\Vert _{L^2_t(L^2)}^2+\left\| \frac{u^r}{r}\right\| _{L^2_t(L^2)}^2\right) \right) \\&\quad \quad \times \left( \Vert \nabla u_0\Vert _{L^2}^2+\left\| \frac{u_0^r}{r}\right\| _{L^2}^2+\left( 1+\Vert u^z\Vert _{L^\infty _t(L^2)}^4\right) \Vert \widetilde{\nabla } u\Vert _{L^2_t(L^2)}^2+\Vert \Gamma \Vert _{L^\infty _t(L^2)}^2+\Vert \widetilde{\nabla }\Gamma \Vert _{L^2_t(L^2)}^2\right) , \end{aligned}$$

from which and (2.3), we infer

$$\begin{aligned}&\Vert \nabla u\Vert _{L^\infty _t(L^2)}^2+\left\| \frac{u^r}{r}\right\| _{L^\infty _t(L^2)}^2+\Vert \partial _tu\Vert _{L^2_t(L^2)}^2+\Vert u\Vert _{L^2_t(\dot{H}^2)}^2+\Vert \Gamma \Vert _{L^2_t(L^2)}^2+\Vert \nabla \Pi \Vert _{L^2_t(L^2)}^2\nonumber \\&\quad \le C \exp \left( C\Vert u_0\Vert _{L^2}^2\left( 1+\Vert u_0\Vert _{L^2}^6\right) \right) \left( \Vert u_0\Vert _{H^1}^2+\left\| \frac{u_0^r}{r}\right\| _{L^2}^2+\Vert \Gamma \Vert _{L^\infty _t(L^2)}^2+\Vert \nabla \Gamma \Vert _{L^2_t(L^2)}^2\right) . \end{aligned}$$

(2.24)

$\bullet$ $\underline{\hbox {The estimate of } \Gamma }$

Let $a\buildrel \hbox {def}\over =1/\rho -1.$ Then we get, by taking $L^2$ inner product of (1.6) with $\Gamma$ and using integrating by parts, that

$$\begin{aligned}&\frac{1}{2}\frac{d}{dt}\Vert \Gamma (t)\Vert _{L^2}^2+\int _{\mathop {\mathbb R}\nolimits ^2_+}\frac{1}{\rho }|\widetilde{\nabla }\Gamma |^2r\,dr\,dz-2\int _{\mathop {\mathbb R}\nolimits ^2_+}\partial _r\left( \frac{\Gamma }{\rho }\right) \Gamma \,dr\,dz\\&\quad =\int _{\mathop {\mathbb R}\nolimits ^2_+}a\left( \partial _r\Pi \partial _z\Gamma -\partial _z\Pi \partial _r\Gamma \right) \,dr\,dz\\&\quad \le \left\| \frac{a}{r}\right\| _{L^\infty }\Vert \widetilde{\nabla }\Pi \Vert _{L^2}\Vert \widetilde{\nabla }\Gamma \Vert _{L^2}. \end{aligned}$$

Note that $a(t,0,z)=0,$ by using integration by parts, one has

$$\begin{aligned} -2\int _{\mathop {\mathbb R}\nolimits ^2_+}\partial _r\left( \frac{\Gamma }{\rho }\right) \Gamma \,dr\,dz&=-2\int _{\mathop {\mathbb R}\nolimits ^2_+}\partial _r\Gamma \Gamma \,dr\,dz-2\int _{\mathop {\mathbb R}\nolimits ^2_+}\partial _r(a\Gamma ) \Gamma \,dr\,dz\\&=\int _{\mathop {\mathbb R}\nolimits }\Gamma ^2(t,0,z)\,dz+2\int _{\mathop {\mathbb R}\nolimits ^2_+}a\Gamma \partial _r\Gamma \,dr\,dz\\&\ge -C\left\| \frac{a}{r}\right\| _{L^\infty }^2\Vert \Gamma \Vert _{L^2}^2-\frac{1}{4}\left\| \frac{\partial _r\Gamma }{\sqrt{\rho }}\right\| _{L^2}^2. \end{aligned}$$

Therefore due to (2.2), we infer

$$\begin{aligned} \frac{d}{dt}\Vert \Gamma (t)\Vert _{L^2}^2+\frac{1}{m}\Vert \widetilde{\nabla }\Gamma \Vert _{L^2}^2\le C\left\| \frac{a}{r}\right\| _{L^\infty }^2\left( \Vert \widetilde{\nabla }\Pi \Vert _{L^2}^2+\Vert \Gamma \Vert _{L^2}^2\right) . \end{aligned}$$

(2.25)

On the other hand, it follows from the transport equation of (1.4) that

$$\begin{aligned}&\partial _ta+u^r\partial _ra+u^z\partial _za=0\quad \hbox {and}\quad \\&\partial _t\frac{a}{r}+u^r\partial _r\frac{a}{r}+u^z\partial _z\frac{a}{r}+\frac{u^r}{r}\frac{a}{r}=0, \end{aligned}$$

which yields

$$\begin{aligned} \left\| \frac{a}{r}(t)\right\| _{L^\infty } \le \left\| \frac{a_0}{r}\right\| _{L^\infty }\exp \left( \left\| \frac{u^r}{r}\right\| _{L^1_t(L^\infty )}\right) . \end{aligned}$$

(2.26)

While note from [2, 8] that

$$\begin{aligned} \left\| \frac{u^r}{r}\right\| _{L^1_t(L^\infty )} \lesssim \Vert \Gamma \Vert _{L^1_t(L^{3,1})} \lesssim t^{\frac{3}{4}} \Vert \Gamma \Vert _{L^\infty _t(L^2)}^{\frac{1}{2}}\Vert \nabla \Gamma \Vert _{L^2_t(L^2)}^{\frac{1}{2}}. \end{aligned}$$

So that by integrating (2.25) over [0, t],  we obtain

$$\begin{aligned}&\Vert \Gamma \Vert _{L^\infty _t(L^2)}^2+\Vert \nabla \Gamma \Vert _{L^2_t(L^2)}^2\le \Vert \Gamma _0\Vert _{L^2}^2\\&\quad +C\left\| \frac{a_0}{r}\right\| _{L^\infty }^2\exp \left( Ct^{\frac{3}{4}}\Vert \Gamma \Vert _{L^\infty _t(L^2)}^{\frac{1}{2}}\Vert \nabla \Gamma \Vert _{L^2_t(L^2)}^{\frac{1}{2}}\right) \left( \Vert \nabla \Pi \Vert _{L^2_t(L^2)}^2+\Vert \Gamma \Vert _{L^2_t(L^2)}^2\right) . \end{aligned}$$

Resuming the Estimate (2.24) into the above inequality leads to

$$\begin{aligned}&\Vert \Gamma \Vert _{L^\infty _t(L^2)}^2+\Vert \nabla \Gamma \Vert _{L^2_t(L^2)}^2\le \Vert \Gamma _0\Vert _{L^2}^2+C\left\| \frac{a_0}{r}\right\| _{L^\infty }^2\exp \left( Ct^{\frac{3}{4}}\Vert \Gamma \Vert _{L^\infty _t(L^2)}^{\frac{1}{2}}\Vert \nabla \Gamma \Vert _{L^2_t(L^2)}^{\frac{1}{2}}\right) \nonumber \\&\quad \times \exp \left( C\Vert u_0\Vert _{L^2}^2\left( 1+\Vert u_0\Vert _{L^2}^6\right) \right) \left( \Vert u_0\Vert _{H^1}^2+\left\| \frac{u_0^r}{r}\right\| _{L^2}^2+\Vert \Gamma \Vert _{L^\infty _t(L^2)}^2+\Vert \nabla \Gamma \Vert _{L^2_t(L^2)}^2\right) . \end{aligned}$$

(2.27)

Proposition 2.1

Let $(\rho , u, \nabla \Pi )$ be a smooth enough solution of (1.4) on $[0,T^*),$ which satisfies (2.2). Let ${\mathcal G}_0$ be given by (1.11) and

$$\begin{aligned} t_1\buildrel \hbox {def}\over =\left( \frac{1}{2C\Vert \Gamma _0\Vert _{L^2}}\ln \left( \frac{\Vert \Gamma _0\Vert _{L^2}^2}{2C\left\| \frac{a_0}{r}\right\| _{L^\infty }^2{\mathcal G}_0}\right) \right) ^{\frac{4}{3}}. \end{aligned}$$

(2.28)

Then under the assumption of (1.10), one has $T^*\ge t_1$ and there holds

$$\begin{aligned}&\Vert \Gamma \Vert _{L^\infty _{t_1}(L^2)}^2+\Vert \nabla \Gamma \Vert _{L^2_{t_1}(L^2)}^2\le 2 \Vert \Gamma _0\Vert _{L^2}^2, \end{aligned}$$

(2.29)

$$\begin{aligned}&\Vert \nabla u\Vert _{L^\infty _{t_1}(L^2)}^2+\left\| \frac{u^r}{r}\right\| _{L^\infty _{t_1}(L^2)}^2+\Vert \partial _tu\Vert _{L^2_{t_1}(L^2)}^2+\Vert u\Vert _{L^2_{t_1}(\dot{H}^2)}^2+\Vert \nabla \Pi \Vert _{L^2_t(L^2)}^2\le C{\mathcal G}_0.\qquad \quad \end{aligned}$$

(2.30)

Proof

Indeed if $\Vert \frac{a_0}{r}\Vert _{L^\infty }$ is sufficiently small, we deduce from (2.27) and (2.28) that

$$\begin{aligned} \Vert \Gamma \Vert _{L^\infty _{t_1}(L^2)}^2+\Vert \nabla \Gamma \Vert _{L^2_{t_1}(L^2)}^2\le \frac{3}{2} \Vert \Gamma _0\Vert _{L^2}^2. \end{aligned}$$

Substituting the above estimate into (2.24) gives rise to (2.30). (2.30) together with the blow-up criteria in [14] implies that $T^*\ge t_1.$ $\square$

2.2 The global in time $H^1$ estimate

The goal of this subsection is to present the global in time $H^1$ estimate for the velocity field. Toward this, we first prove such a estimate for small solutions of (1.1), which does not use the axisymmetric structure of the solutions.

Lemma 2.2

Let $(\rho , u, \nabla \Pi )$ be a smooth enough solution of (1.1) on $[0,T^*),$ which satisfies (2.2). Then there exist positive constants $\eta _1$ and $\eta _2,$ which depend only on $\Vert u_0\Vert _{L^2},$ so that there holds

$$\begin{aligned} \Vert \nabla u(t)\Vert ^2_{L^2}+\int _{t_0}^{t}\left( m\Vert \partial _tu(t')\Vert _{ L^2}^2 +\eta _2\left( \Vert \nabla ^2 u(t')\Vert _{ L^2}^2+\Vert \nabla \Pi (t')\Vert _{L^2}^2\right) \right) \,dt' \le \Vert \nabla u(t_0)\Vert _{L^2}^2 \end{aligned}$$

(2.31)

provided that $\Vert \nabla u(t_0)\Vert _{L^2}\le \eta _1.$

Proof

We first get, by taking the $L^2$ inner product of the momentum equations of (1.1) with $\partial _t u$ and using integration by parts, that

$$\begin{aligned} \Vert \sqrt{\rho } \partial _t u(t)\Vert _{L^2}^2 +\frac{1}{2}\frac{d}{dt}\Vert \nabla u(t)\Vert ^2_{L^2}= & {} -\left( \rho u \cdot \nabla u\ |\ \partial _t u \right) _{L^2}\\\le & {} \Vert \sqrt{\rho }\Vert _{L^{\infty }}\Vert u\Vert _{L^3}\Vert \nabla u\Vert _{L^6}\Vert \sqrt{\rho }\partial _t u\Vert _{L^2}\\\le & {} C\Vert u\Vert _{L^{2}}\Vert \nabla u\Vert _{L^2}\Vert \nabla ^2 u\Vert _{L^2}^2+\frac{1}{4}\Vert \sqrt{\rho }\partial _t u\Vert _{L^2}^2, \end{aligned}$$

which gives

$$\begin{aligned} \frac{1}{2}\frac{d}{dt}\Vert \nabla u(t)\Vert _{L^2}^2+\frac{3}{4}\Vert \sqrt{\rho }\partial _t u(t)\Vert _{L^2}^2 \le C \Vert u\Vert _{L^{2}}\Vert \nabla u\Vert _{L^2}\Vert \nabla ^2 u\Vert _{L^2}^2. \end{aligned}$$

On the other hand, it follows from the classical estimates on linear Stokes operator and

$$\begin{aligned} \left\{ \begin{array}{l} \displaystyle - \Delta u+ \nabla \Pi =\rho \partial _t u - \rho u \cdot \nabla u, \\ \displaystyle \text{ div }\, u = 0, \end{array}\right. \end{aligned}$$

(2.32)

that

$$\begin{aligned} \Vert \nabla ^2 u\Vert _{L^2}^2+\Vert \nabla \Pi \Vert _{L^2}^2&\le C\left( \Vert \rho \partial _tu\Vert _{L^2}^2+\Vert \rho u\cdot \nabla u\Vert _{L^2}^2\right) \\&\le C\left( \Vert \sqrt{\rho }\partial _tu\Vert _{L^2}^2+\Vert \rho \Vert _{L^\infty }\Vert u\Vert _{L^3}^2\Vert \nabla u\Vert _{L^6}^2\right) \\&\le C\left( \Vert \sqrt{\rho }\partial _tu\Vert _{L^2}^2+\Vert u\Vert _{L^2}\Vert \nabla u\Vert _{L^2}\Vert \nabla ^2 u\Vert _{L^2}^2\right) , \end{aligned}$$

so that we obtain for any $\eta _2>0$

$$\begin{aligned}&\frac{1}{2}\frac{d}{dt}\Vert \nabla u(t)\Vert _{L^2}^2+\left( \frac{3m}{4}-C\eta _2\right) \Vert \partial _t u\Vert _{L^2}^2 \nonumber \\&\quad +\left( \eta _2-C\Vert u_0\Vert _{L^2}\Vert \nabla u\Vert _{L^2}\right) \left( \Vert \nabla ^2 u\Vert _{L^2}^2+\Vert \nabla \Pi \Vert _{L^2}^2\right) \le 0. \end{aligned}$$

(2.33)

We denote

$$\begin{aligned} \tau ^*\buildrel \hbox {def}\over =\sup \left\{ \ t\in [t_0, T^*)\, \big |\, \Vert \nabla u(t)\Vert _{L^2}\le 2\eta _1\right\} . \end{aligned}$$

(2.34)

We claim that $\tau ^*=T^*$ provided that $\eta _1$ is sufficiently small. Indeed if $\tau ^*<T^*,$ taking $\eta _2=\frac{m}{4C}$ and $\eta _1\le \frac{\eta _2}{2C\Vert u_0\Vert _{L^2}},$ we deduce from (2.33) that

$$\begin{aligned} \frac{d}{dt}\Vert \nabla u(t)\Vert _{L^2}^2+m\Vert \partial _t u\Vert _{L^2}^2+ \eta _2\left( \Vert \nabla ^2 u\Vert _{L^2}^2+\Vert \nabla \Pi \Vert _{L^2}^2\right) \le 0\qquad \text{ for } \text{ all }\ \, t\in [t_0, \tau ^*), \end{aligned}$$

which implies

$$\begin{aligned} \Vert \nabla u(t)\Vert _{L^2}^2+\int _{t_0}^{\tau ^*}\left( m\Vert \partial _t u(t')\Vert _{L^2}^2+ \eta _2(\Vert \nabla ^2 u(t')\Vert _{L^2}^2+\Vert \nabla \Pi (t')\Vert _{L^2}^2)\right) \,dt'\le \Vert \nabla u(t_0)\Vert _{L^2}^2\le \eta ^2_1. \end{aligned}$$

This contradict with (2.34), and thus $\tau ^*=T^*.$ This concludes the proof of the lemma. $\square$

Proposition 2.2

Let $(\rho , u, \nabla \Pi )$ be the local unique smooth solution of (1.4) on $[0,T^*),$ which satisfies (2.2). Then $T^*=\infty$ and there holds (1.11) provided that $\varepsilon _0$ in (1.10) is sufficiently small.

Proof

It follows from the derivation of (2.3) that

$$\begin{aligned} \frac{1}{2}\Vert \sqrt{\rho }u(t)\Vert _{L^2}^2+\int _0^t\Vert \nabla u(t')\Vert _{L^2}^2\,dt' =\frac{1}{2} \Vert \sqrt{\rho _0}u_0\Vert _{L^2}^2, \end{aligned}$$

(2.35)

which ensures that for any positive integer N,  there holds

$$\begin{aligned} \sum _{k=0}^{N-1}\int _k^{k+1}\Vert \nabla u(t')\Vert _{L^2}^2\,d t' \le \frac{1}{2} \Vert \sqrt{\rho _0}u_0\Vert _{L^2}^2. \end{aligned}$$

Thus there exists $0\le k_0\le N-1$ and some $t_0\in (k_0,k_0+1)$ such that

$$\begin{aligned} \int _{k_0}^{k_0+1}\Vert \nabla u\Vert _{L^2}^2\,d\tau \le \frac{1}{2N}\Vert \sqrt{\rho _0}u_0\Vert _{L^2}^2 \quad \hbox {and}\quad \Vert \nabla u(t_0)\Vert _{L^2}^2\le \frac{1}{2N}\Vert \sqrt{\rho _0}u_0\Vert _{L^2}^2. \end{aligned}$$

For $\eta _1$ given by Lemma 2.2, taking N so large that

$$\begin{aligned} \Vert \nabla u(t_0)\Vert _{L^2}^2\le \frac{1}{2N}\Vert \sqrt{\rho _0}u_0\Vert _{L^2}^2\le \eta _1^2. \end{aligned}$$

Then we deduce from Lemma 2.2 that there holds (2.31).

On the other hand, in view of (2.28), we can take $\Vert \frac{a_0}{r}\Vert _{L^\infty }$ to be so small that $t_1\ge t_0.$ Thus by summing up (2.30) and (2.31), we obtain for any $t<T^*,$

$$\begin{aligned}&\Vert \nabla u\Vert _{L^\infty _t(L^2)}^2+\Vert \partial _tu\Vert _{L^2_t(L^2)}^2+\Vert \nabla ^2 u\Vert _{L^2_t(L^2)}^2+\Vert \nabla \Pi \Vert _{L^2_t(L^2)}^2\nonumber \\&\quad \le \Vert \nabla u\Vert _{L^\infty (0,t_0;L^2)}^2+\Vert \partial _tu\Vert _{L^2(0,t_0;L^2)}^2+\Vert \nabla ^2 u\Vert _{L^2(0,t_0;L^2)}^2+\Vert \nabla \Pi \Vert _{L^2(0,t_0;L^2)}^2\nonumber \\&\quad \quad +\Vert \nabla u\Vert _{L^\infty (t_0,t;L^2)}^2+\Vert \partial _tu\Vert _{L^2(t_0,t;L^2)}^2+\Vert \nabla ^2 u\Vert _{L^2(t_0,t;L^2)}^2+\Vert \nabla \Pi \Vert _{L^2(t_0,t;L^2)}^2\nonumber \\&\quad \le C{\mathcal G}_0 +\eta _1, \end{aligned}$$

(2.36)

for ${\mathcal G}_0$ given by (1.11) and $\eta _1$ being determined by Lemma 2.2. Then thanks to (2.36) and the blow-up criteria in [14], we conclude that $T^*=\infty .$ Moreover, by summing up (2.3) and (2.36), we achieve (1.11). This finishes the proof of Proposition 2.2. $\square$

3 Decay estimates of the global solutions of (1.1)

The purpose of this section is to present the decay estimates (1.13) for any global smooth solutions of (1.1), which does not use the particular axisymmetric structure of the solutions.

Lemma 3.1

Let $(\rho , u, \nabla \Pi )$ be a smooth enough solution of (1.1) on $[0,T^*),$ which satisfies (2.2). Then for $t<T^*,$ one has

$$\begin{aligned} \frac{d}{dt}\Vert \nabla u(t)\Vert _{L^2}^2+\Vert \sqrt{\rho }u_t(t)\Vert _{L^2}^2+\Vert u(t)\Vert _{\dot{H}^2}^2+\Vert \nabla \Pi (t)\Vert _{L^2}^2\le C\Vert \nabla u(t)\Vert _{{H}^1}^2\Vert \nabla u(t)\Vert _{L^2}^2, \end{aligned}$$

(3.1)

and

$$\begin{aligned}&\frac{d}{dt}\Vert \sqrt{\rho }u_t(t)\Vert _{L^2}^2+\Vert \nabla u_t(t)\Vert _{L^2}^2\nonumber \\&\quad \le C\left( \Vert \nabla u(t)\Vert _{{H}^1}^2+\Vert u(t)\Vert _{\dot{H}^1}^4\right) \left( \Vert \sqrt{\rho }u_t(t)\Vert _{L^2}^2+\Vert \nabla u(t)\Vert _{L^2}^4\right) . \end{aligned}$$

(3.2)

Proof

We first get, by a similar derivation of (2.33), that

$$\begin{aligned} \frac{d}{dt}\Vert \nabla u(t)\Vert _{L^2}^2+\left( \Vert \sqrt{\rho }u_t\Vert _{L^2}^2+\Vert u\Vert _{\dot{H}^2}^2+\Vert \nabla \Pi \Vert _{L^2}^2\right)&\le C\Vert \sqrt{\rho }u\cdot \nabla u\Vert _{L^2}^2\\&\le CM\Vert u\Vert _{L^6}^2\Vert \nabla u\Vert _{L^3}^2\\&\le CM \Vert \nabla u\Vert _{\dot{H}^{\frac{1}{2}}}^2\Vert \nabla u\Vert _{L^2}^2, \end{aligned}$$

which gives (3.1).

On the other hand, by taking $\partial _t$ to the momentum equation of (1.1), we write

$$\begin{aligned} \rho \left( \partial _t u_t+u\cdot \nabla u_t\right) -\Delta u_t+\nabla \Pi _t=-\rho _t u_t-(\rho u)_t\cdot \nabla u. \end{aligned}$$

Taking $L^2$ inner product of the above equation with $u_t$ and using the transport equation of (1.1), we obtain

$$\begin{aligned} \frac{1}{2}\frac{d}{dt}\Vert \sqrt{\rho }u_t(t)\Vert _{L^2}^2+\Vert \nabla u_t\Vert _{L^2}^2= & {} -\int _{\mathop {\mathbb R}\nolimits ^3}\rho _t|u_t|^2\,dx-\int _{\mathop {\mathbb R}\nolimits ^3}\rho _t u\cdot \nabla u\ |\ u_t\,dx\nonumber \\&-\int _{\mathop {\mathbb R}\nolimits ^3}\rho u_t\cdot \nabla u\ |\ u_t\,dx. \end{aligned}$$

(3.3)

By using the transport equation of (1.1) and integration by parts, one has

$$\begin{aligned} \left| \int _{\mathop {\mathbb R}\nolimits ^3}\rho _t|u_t|^2\,dx\right|&=\left| \int _{\mathop {\mathbb R}\nolimits ^3}\mathrm{div}(\rho u)|u_t|^2\,dx\right| \\&=2\left| \int _{\mathop {\mathbb R}\nolimits ^3}\rho u\cdot \nabla u_t\ |\ u_t\,dx\right| \\&\le 2\sqrt{M}\Vert u\Vert _{L^\infty }\Vert \sqrt{\rho }u_t\Vert _{L^2}\Vert \nabla u_t\Vert _{L^2}, \end{aligned}$$

which together with the 3-D interpolation inequality that

$$\begin{aligned} \Vert u\Vert _{L^\infty }\le C \Vert u\Vert _{\dot{H}^1}^{\frac{1}{2}}\Vert u\Vert _{\dot{H}^2}^{\frac{1}{2}}, \end{aligned}$$

(3.4)

implies

$$\begin{aligned} \left| \int _{\mathop {\mathbb R}\nolimits ^3}\rho _t|u_t|^2\,dx\right| \le CM\Vert u\Vert _{\dot{H}^1}\Vert u\Vert _{\dot{H}^2}\Vert \sqrt{\rho }u_t\Vert _{L^2}^2+\frac{1}{6}\Vert \nabla u_t\Vert _{L^2}^2. \end{aligned}$$

Along the same line, we have

$$\begin{aligned}&\int _{\mathop {\mathbb R}\nolimits ^3}\rho _tu\cdot \nabla u\ |\ u_t\,dx=-\int _{\mathop {\mathbb R}\nolimits ^3}\mathrm{div}(\rho u)u\cdot \nabla u\ |\ u_t\,dx\\&\quad =\sum _{i,j,k=1}^3\left( \int _{\mathop {\mathbb R}\nolimits ^3}\rho u^i\partial _iu^j\partial _ju^ku_t^k\,dx+\int _{\mathop {\mathbb R}\nolimits ^3}\rho u^iu^j\partial _i\partial _ju^ku_t^k\,dx+\int _{\mathop {\mathbb R}\nolimits ^3}\rho u^iu^j\partial _ju^k\partial _iu_t^k\,dx\right) . \end{aligned}$$

Applying Hölder’s inequality gives

$$\begin{aligned} \sum _{i,j,k=1}^3\left| \int _{\mathop {\mathbb R}\nolimits ^3}\rho u^i\partial _iu^j\partial _ju^ku_t^k\,dx\right|\le & {} \sqrt{M}\Vert u\Vert _{L^\infty }\Vert \nabla u\Vert _{L^3}\Vert \nabla u\Vert _{L^6}\Vert \sqrt{\rho }u_t\Vert _{L^2}\\\le & {} C\left( \Vert u\Vert _{L^\infty }^2\Vert u\Vert _{\dot{H}^2}^2+\Vert \nabla u\Vert _{\dot{H}^{\frac{1}{2}}}^2\Vert \sqrt{\rho }u_t\Vert _{L^2}^2\right) , \end{aligned}$$

and

$$\begin{aligned} \sum _{i,j,k=1}^3\left| \int _{\mathop {\mathbb R}\nolimits ^3}\rho u^iu^j\partial _i\partial _ju^ku_t^k\,dx\right|&\le \sqrt{M}\Vert u\Vert _{L^\infty }^2\Vert \nabla ^2 u\Vert _{L^2}\Vert \sqrt{\rho }u_t\Vert _{L^2}\\&\le C\Vert u\Vert _{L^\infty }^2\left( \Vert u\Vert _{\dot{H}^2}^2+\Vert \sqrt{\rho }u_t\Vert _{L^2}^2\right) , \end{aligned}$$

and

$$\begin{aligned} \sum _{i,j,k=1}^3\left| \int _{\mathop {\mathbb R}\nolimits ^3}\rho u^iu^j\partial _ju^k\partial _iu_t^k\,dx\right|&\le M\Vert u\Vert _{L^6}^2\Vert \nabla u\Vert _{L^6}\Vert \nabla u_t\Vert _{L^2}\\&\le C\Vert \nabla u\Vert _{L^2}^4\Vert u\Vert _{\dot{H}^2}^2+\frac{1}{6}\Vert \nabla u_t\Vert _{L^2}^2. \end{aligned}$$

This yields

$$\begin{aligned} \left| \int _{\mathop {\mathbb R}\nolimits ^3}\rho _tu\cdot \nabla u\ |\ u_t\,dx\right|&\le C\left( \Vert u\Vert _{L^\infty }^2+\Vert \nabla u\Vert _{\dot{H}^{\frac{1}{2}}}^2+\Vert \nabla u\Vert _{L^2}^4\right) \left( \Vert u\Vert _{\dot{H}^2}^2+\Vert \sqrt{\rho }u_t\Vert _{L^2}^2\right) \\&\quad +\frac{1}{6}\Vert \nabla u_t\Vert _{L^2}^2. \end{aligned}$$

Finally it is easy to observe that

$$\begin{aligned} \left| \int _{\mathop {\mathbb R}\nolimits ^3}\rho u_t\cdot \nabla u\ |\ u_t\,dx\right|&\le \sqrt{M}\Vert u_t\Vert _{L^6}\Vert \nabla u\Vert _{L^3}\Vert \sqrt{\rho }u_t\Vert _{L^2}\\&\le C\Vert \nabla u\Vert _{\dot{H}^{\frac{1}{2}}}^2\Vert \sqrt{\rho }u_t\Vert _{L^2}^2+\frac{1}{6}\Vert \nabla u_t\Vert _{L^2}^2. \end{aligned}$$

Resuming the above estimates into (3.3) and using (3.4) results in

$$\begin{aligned}&\frac{d}{dt}\Vert \sqrt{\rho }u_t(t)\Vert _{L^2}^2+\Vert \nabla u_t(t)\Vert _{L^2}^2\nonumber \\&\quad \le C\left( \Vert \nabla u(t)\Vert _{{H}^1}^2+\Vert u(t)\Vert _{\dot{H}^1}^4\right) \left( \Vert u(t)\Vert _{\dot{H}^2}^2+\Vert \sqrt{\rho }u_t(t)\Vert _{L^2}^2\right) . \end{aligned}$$

(3.5)

Whereas it follows from the classical estimates on linear Stokes operator and (2.32) that

$$\begin{aligned} \Vert u\Vert _{\dot{H}^2}+\Vert \nabla \Pi \Vert _{L^2}&\le C\left( \Vert \rho u_t\Vert _{L^2}+\Vert \rho u\cdot \nabla u\Vert _{L^2}\right) \\&\le C\left( \sqrt{M}\Vert \sqrt{\rho }u_t\Vert _{L^2}+M\Vert u\Vert _{L^6}\Vert \nabla u\Vert _{L^3}\right) \\&\le C\left( \Vert \sqrt{\rho }u_t\Vert _{L^2}+\Vert \nabla u\Vert _{L^2}^2\right) +\frac{1}{2}\Vert u\Vert _{\dot{H}^2}, \end{aligned}$$

which yields

$$\begin{aligned} \Vert u\Vert _{\dot{H}^2}+\Vert \nabla \Pi \Vert _{L^2}\le C\left( \Vert \sqrt{\rho }u_t\Vert _{L^2}+\Vert \nabla u\Vert _{L^2}^2\right) . \end{aligned}$$

(3.6)

Substituting (3.6) into (3.5) leads to (3.2). This finishes the proof of the Lemma. $\square$

Corollary 3.1

Under the assumptions of Lemma 3.1 and that

$$\begin{aligned} \Vert u\Vert _{L^\infty (0,T^*;H^1)}^2+\Vert \nabla u\Vert _{L^2(0,T^*;H^1)}^2\le C_0, \end{aligned}$$

(3.7)

one has for any $t<T^*,$

$$\begin{aligned}&\langle {t} \rangle \Vert \nabla u(t)\Vert _{L^2}^2+\int _0^t\langle {t'} \rangle \left( \Vert u_t(t')\Vert _{L^2}^2+\Vert u(t')\Vert _{\dot{H}^2}^2+\Vert \nabla \Pi (t')\Vert _{L^2}^2\right) \,dt'\nonumber \\&\quad \le C\exp (CC_0)\Vert u_0\Vert _{H^1}^2\buildrel \hbox {def}\over =C_1, \end{aligned}$$

(3.8)

and

$$\begin{aligned}&t\langle {t} \rangle \left( \Vert u_t(t)\Vert _{L^2}^2+\Vert u(t)\Vert _{\dot{H}^2}^2+\Vert \nabla \Pi (t)\Vert _{L^2}^2\right) +\int _0^tt'\langle {t'} \rangle \Vert \nabla u_t(t')\Vert _{L^2}^2\,dt'\nonumber \\&\quad \le CC_1(1+C_1)\exp \left( CC_0(1+C_0)\right) \buildrel \hbox {def}\over =C_2. \end{aligned}$$

(3.9)

Proof

We first get, by multiplying (3.1) by $\langle {t} \rangle ,$ that

$$\begin{aligned}&\frac{d}{dt}\left( \langle {t} \rangle \Vert \nabla u(t)\Vert _{L^2}^2\right) +\langle {t} \rangle \left( \Vert \sqrt{\rho }u_t(t)\Vert _{L^2}^2+\Vert u(t)\Vert _{\dot{H}^2}^2+\Vert \nabla \Pi (t)\Vert _{L^2}^2\right) \\&\quad \le \Vert \nabla u(t)\Vert _{L^2}^2+C\Vert \nabla u(t)\Vert _{{H}^1}^2\langle {t} \rangle \Vert \nabla u(t)\Vert _{L^2}^2. \end{aligned}$$

Applying Gronwall’s inequality and using (2.35), (3.7) gives rise to (3.8).

While multiplying (3.2) by $t\langle {t} \rangle$ results in

$$\begin{aligned}&\frac{d}{dt}\left( t\langle {t} \rangle \Vert \sqrt{\rho }u_t(t)\Vert _{L^2}^2\right) +t\langle {t} \rangle \Vert \nabla u_t(t)\Vert _{L^2}^2\le 2\langle {t} \rangle \Vert \sqrt{\rho }u_t(t)\Vert _{L^2}^2\\&\quad +C\left( \Vert \nabla u(t)\Vert _{{H}^1}^2+\Vert u(t)\Vert _{\dot{H}^1}^4\right) t\langle {t} \rangle \left( \Vert \sqrt{\rho }u_t(t)\Vert _{L^2}^2+\Vert \nabla u(t)\Vert _{L^2}^4\right) . \end{aligned}$$

Applying Gronwall’s inequality leads to

$$\begin{aligned}&t\langle {t} \rangle \Vert \sqrt{\rho }u_t(t)\Vert _{L^2}^2+\int _0^tt'\langle {t'} \rangle \Vert \nabla u_t(t')\Vert _{L^2}^2\,dt'\\&\quad \le C\exp \left( C\left( \Vert \nabla u\Vert _{L^2_t({H}^1)}^2+\Vert u\Vert _{L^\infty _t(\dot{H}^1)}^2\Vert u\Vert _{L^2_t(\dot{H}^1)}^2\right) \right) \left( \int _0^t\langle {t'} \rangle \Vert \sqrt{\rho }u_t(t')\Vert _{L^2}^2\,dt'\right. \\&\qquad \left. +\left\| \langle {t'} \rangle \Vert \nabla u(t')\Vert _{L^2}^2\right\| _{L^\infty _t}^2\left( \Vert \nabla u\Vert _{L^2_t({H}^1)}^2 +\Vert u\Vert _{L^\infty _t(\dot{H}^1)}^2\Vert u\Vert _{L^2_t(\dot{H}^1)}^2\right) \right) , \end{aligned}$$

from which, (3.6–3.8), we conclude the proof of (3.9). $\square$

Proposition 3.1

Let $p\in [1,2)$ and $\beta (p)\buildrel \hbox {def}\over =\frac{3}{4}(\frac{2}{p}-1).$ Then under the assumptions of Corollary 3.1, if we assume further that $a_0\buildrel \hbox {def}\over =\frac{1}{\rho _0}-1\in L^2(\mathop {\mathbb R}\nolimits ^3)$ and $u_0\in L^p(\mathop {\mathbb R}\nolimits ^3),$ there holds

$$\begin{aligned}&\Vert u(t)\Vert _{L^2}\le \left\{ \begin{array}{ll} \displaystyle C\langle {t} \rangle ^{-\beta (p)} &{} \text{ if }\quad 1<p< 2, \\ \displaystyle C\langle {t} \rangle ^{-\left( \frac{3}{4}\right) _-}&{} \text{ if }\quad p=1, \end{array}\right. \end{aligned}$$

(3.10)

for any $t<T^*,$ where the constant C depends on $\Vert a_0\Vert _{L^2},$ $C_0, C_1$ and $C_2$ given by Corollary 3.1.

Proof

Motivated by [4], in order to use Schonbek’s strategy in [21], we split the phase-space $\mathop {\mathbb R}\nolimits ^3$ into two time-dependent regions so that

$$\begin{aligned} \Vert \nabla u(t)\Vert _{L^2}^2 = \int _{S(t)}|\xi |^2|\hat{u}(t,\xi )|^2 \, d\xi +\int _{S(t)^{c}}|\xi |^2|\hat{u}(t,\xi )|^2 d\xi , \end{aligned}$$

where $S(t)\buildrel \hbox {def}\over =\{\xi : \ |\xi |\le \sqrt{\frac{M}{2}} \; g(t)\}$ and g(t) satisfies $g(t)\sim \langle { t}\rangle ^{-\frac{1}{2}},$ which will be chosen later on. Then due to the energy law (2.35) of (1.1), one has

$$\begin{aligned} \frac{d}{dt}\Vert \sqrt{\rho }u(t)\Vert ^2_{L^2}+g^2(t)\Vert \sqrt{\rho }\;u(t)\Vert ^2_{L^2} \le Mg^2(t)\int _{S(t)}|\hat{u}(t,\xi )|^2 \, d\xi \end{aligned}$$

(3.11)

To deal with the low frequency part of u on the right-hand side of (3.11), we rewrite the momentum equations of (1.1) as

$$\begin{aligned} u(t)=e^{t \Delta } u_0+\int _{0}^{t}e^{(t-t') \Delta }\mathbb {P}\left( -\nabla \cdot (u\otimes u)+a(\Delta u-\nabla \Pi ) \right) (t')\,d t'. \end{aligned}$$

where $a\buildrel \hbox {def}\over =\frac{1}{\rho }-1$ and $\mathbb {P}\buildrel \hbox {def}\over =Id-\nabla \Delta ^{-1}\mathrm{div}$ denotes the Leray projection operator. Taking Fourier transform with respect to x variables leads to

$$\begin{aligned} |\hat{u}(t,\xi )| \lesssim e^{-t|\xi |^2} |\widehat{u}_0(\xi )| +\int _{0}^{t}e^{-(t-t') |\xi |^2}\left( |\xi ||{\mathcal F}_x(u\otimes u)|+|{\mathcal F}_x(a(\Delta u-\nabla \Pi ))|\right) (t')\, d t', \end{aligned}$$

which implies that

$$\begin{aligned} \int _{S(t)}|\hat{u}(t,\xi )|^2 d\xi\lesssim & {} \int _{S(t)}e^{-2 t |\xi |^2} |\widehat{u}_0(\xi )|^2 d\xi +g^5(t)\left( \int _{0}^{t} \Vert {\mathcal F}_x(u\otimes u)(t')\Vert _{L_{\xi }^{\infty }} \,dt' \right) ^2\nonumber \\&+g^3(t) \left( \int _{0}^{t}\Vert {\mathcal F}_x(a(\Delta u-\nabla \Pi ) )(t')\Vert _{L^\infty _\xi } \,dt' \right) ^2. \end{aligned}$$

(3.12)

Thanks to (3.9), we have

$$\begin{aligned} \left( \int _{0}^{t}\Vert {\mathcal F}_x(a (\Delta u-\nabla \Pi ))(t')\Vert _{L_{\xi }^{\infty }}\, dt'\right) ^2&\le \Vert a\Vert _{L^\infty _t(L^2)}^2\left( \int _{0}^{t}\Vert (\Delta u-\nabla \Pi )(t')\Vert _{L^2} \, d t'\right) ^2\nonumber \\&\lesssim \Vert a_0\Vert _{L^2}^2 \left( \int _0^t(t')^{-\frac{1}{2}}\langle {t'} \rangle ^{-\frac{1}{2}}\,dt'\right) ^{2}\lesssim \ln ^2\langle {t} \rangle . \end{aligned}$$

(3.13)

While it is easy to observe that

$$\begin{aligned} \left( \int _{0}^{t}\Vert {\mathcal F}_x(u\otimes u)(t')\Vert _{L_{\xi }^{\infty }}\, dt'\right) ^2 \le \left( \int _{0}^{t}\Vert u(t')\Vert _{L^{2}}^2 \,dt' \right) ^2 \lesssim t^2\Vert u_0\Vert _{L^2}^2. \end{aligned}$$

Note that for $u_0 \in L^{p}(\mathop {\mathbb R}\nolimits ^3),$ let $\frac{1}{q}\buildrel \hbox {def}\over =\frac{4}{3}\beta (p)=\frac{2}{p}-1$ and $\frac{1}{p}+\frac{1}{p'}=1,$ one has

$$\begin{aligned} \int _{S(t)}e^{-2t |\xi |^2} |\widehat{u}_0(\xi )|^2 \, d\xi&\lesssim \left( \int _{S(t)}e^{-2qt |\xi |^2} \, d\xi \right) ^{\frac{1}{q}} \Vert \widehat{u}_0\Vert _{L^{p'}}^{2}\nonumber \\&\lesssim \Vert u_0\Vert _{L^p}^2\langle { t}\rangle ^{-2\beta (p)}, \end{aligned}$$

(3.14)

where we used the Hausdörff–Young inequality in the last line so that $\Vert \widehat{u}_0\Vert _{L^{p'}}\le C\Vert u_0\Vert _{L^p}.$ Then since $g(t) \lesssim \langle { t}\rangle ^{-\frac{1}{2}},$ we deduce from (3.12) that

$$\begin{aligned} \int _{S(t)}|\hat{u}(t,\xi )|^2 \, d\xi \lesssim \langle { t}\rangle ^{-2\beta (p)}+\langle { t}\rangle ^{-\frac{1}{2}} \lesssim \left\{ \begin{array}{ll} \displaystyle \langle {t} \rangle ^{-\frac{1}{2}}&{} \text{ if }\quad 1\le p<\frac{3}{2}, \\ \displaystyle \langle {t} \rangle ^{-2\beta (p)}&{} \text{ if }\quad \frac{3}{2}\le p<2, \end{array}\right. \end{aligned}$$

(3.15)

In the case when $\frac{3}{2}\le p<2,$ by substituting (3.15) into (3.11), we obtain

$$\begin{aligned} \frac{d}{dt}\Vert \sqrt{\rho }u(t)\Vert ^2_{L^2}+g^2(t)\Vert \sqrt{\rho }\;u(t)\Vert ^2_{L^2} \lesssim g^2(t)\langle { t}\rangle ^{-2\beta (p)}\lesssim \langle { t}\rangle ^{-1-2\beta (p)}, \end{aligned}$$

from which, we infer

$$\begin{aligned} e^{\int _{0}^{t}g^2(t')\, dt'} \Vert \sqrt{\rho }u(t)\Vert ^2_{L^2} \lesssim \Vert \sqrt{\rho _0}u_0\Vert ^2_{L^2} + \int _{0}^{t} e^{\int _{0}^{t'}g^2(t')dt' }\langle { t'}\rangle ^{-1-2\beta (p)}\,dt'. \end{aligned}$$

Taking $\alpha > 2\beta (p)$ and $g^{2}(t)=\alpha \langle {t} \rangle ^{-1}$ in the above inequality leads to

$$\begin{aligned} \Vert \sqrt{\rho }u(t)\Vert ^2_{L^2}\langle { t}\rangle ^{\alpha } \lesssim 1+\int _{0}^{t}\langle { t'}\rangle ^{\alpha -1-2\beta (p)}\,dt' \lesssim 1+\langle { t}\rangle ^{\alpha -2\beta (p)}, \end{aligned}$$

which yields (3.10) for $p\in [3/2,2).$

In the case when $1\le p<\frac{3}{2},$ by substituting the Estimate (3.15) into (3.11), one has

$$\begin{aligned} \frac{d}{dt}\Vert \sqrt{\rho }u(t)\Vert ^2_{L^2}+g^2(t)\Vert \sqrt{\rho }\;u(t)\Vert ^2_{L^2} \lesssim g^2(t)\langle { t}\rangle ^{-\frac{1}{2}}\lesssim \langle { t}\rangle ^{-\frac{3}{2}}, \end{aligned}$$

which implies

$$\begin{aligned} e^{\int _{0}^{t}g^2(t')\, dt'} \Vert \sqrt{\rho }u(t)\Vert ^2_{L^2} \lesssim \Vert \sqrt{\rho _0}u_0\Vert ^2_{L^2} + \int _{0}^{t} e^{\int _{0}^{t'}g^2(t')dt' }\langle { t'}\rangle ^{-\frac{3}{2}}\,dt'. \end{aligned}$$

Taking $\alpha > \frac{1}{2}$ and $g^{2}(t)\buildrel \hbox {def}\over =\alpha \langle {t} \rangle ^{-1}$ in the above inequality results in

$$\begin{aligned} \Vert \sqrt{\rho }u(t)\Vert ^2_{L^2}\langle { t}\rangle ^{\alpha } \lesssim 1+\int _{0}^{t}\langle { t'}\rangle ^{\alpha -\frac{3}{2}}\,dt' \lesssim 1+\langle { t}\rangle ^{\alpha -\frac{1}{2}}, \end{aligned}$$

which gives

$$\begin{aligned} \Vert u(t)\Vert _{L^2}\lesssim \langle { t}\rangle ^{-\frac{1}{4}}. \end{aligned}$$

(3.16)

Then by virtue of (3.16), we write

$$\begin{aligned} \left( \int _{0}^{t}\Vert {\mathcal F}_x(u\otimes u)(t')\Vert _{L_{\xi }^{\infty }}\, dt'\right) ^2 \le \left( \int _{0}^{t}\Vert u(t')\Vert _{L^{2}}^2 \,dt' \right) ^2 \lesssim \left( \int _{0}^{t}\langle { t'}\rangle ^{-\frac{1}{2}} \,dt' \right) ^2 \lesssim \langle { t}\rangle . \end{aligned}$$

(3.17)

Resuming the Estimates (3.13), (3.14) and (3.17) into (3.12) results in

$$\begin{aligned} \int _{\bar{S}(t)}|\hat{u}(t,\xi )|^2 d\xi \lesssim \langle { t}\rangle ^{-2\beta (p)}+\langle { t}\rangle ^{-\left( \frac{3}{2}\right) _-} \lesssim \left\{ \begin{array}{ll} \displaystyle \langle {t} \rangle ^{-2\beta (p)}&{} \text{ if }\quad 1<p<\frac{3}{2}, \\ \displaystyle \langle {t} \rangle ^{-\left( \frac{3}{2}\right) _-}&{} \text{ if }\quad p=1. \end{array}\right. \end{aligned}$$

(3.18)

With (3.18), we can repeat the previous argument to prove (3.10) for the remaining case when $p\in [1,3/2).$ This completes the proof of the proposition. $\square$

Proposition 3.2

Under the assumptions of Proposition 3.1, there holds (1.13) for any $t<T^*.$

Proof

With Proposition 3.1, we shall use a similar argument for the classical Navier–Stokes system to derive the decay estimates for the derivatives of the velocity (see [12] for instance). In fact, for any $s<t<T^*,$ we deduce from the energy equality of (1.1) that

$$\begin{aligned} \frac{1}{2}\Vert \sqrt{\rho }u(t)\Vert _{L^2}^2+\int _s^t\Vert \nabla u(t')\Vert _{L^2}^2\,dt'=\frac{1}{2}\Vert \sqrt{\rho }u(s)\Vert _{L^2}^2. \end{aligned}$$

(3.19)

While multiplying (3.1) by $(t-s)$ leads to

$$\begin{aligned}&\frac{d}{dt}\left( (t-s)\Vert \nabla u(t)\Vert _{L^2}^2\right) +(t-s)\left( \Vert \sqrt{\rho }u_t(t)\Vert _{L^2}^2+\Vert \nabla ^2u(t)\Vert _{L^2}^2+\Vert \nabla \Pi (t)\Vert _{L^2}^2\right) \\&\quad \le \Vert \nabla u(t)\Vert _{L^2}^2+ C\Vert \nabla u(t)\Vert _{{H}^1}^2(t-s)\Vert \nabla u(t)\Vert _{L^2}^2, \end{aligned}$$

Applying Gronwall’s inequality and using (3.19) results in

$$\begin{aligned} (t-s)\Vert \nabla u(t)\Vert _{L^2}^2\le&\exp \left( C\Vert \nabla u\Vert _{L^2_t({H}^1)}^2\right) \int _s^t\Vert \nabla u(t')\Vert _{L^2}^2\,dt'\\ \le&\frac{\exp \left( CC_0\right) }{2}\Vert \sqrt{\rho }u(s)\Vert _{L^2}^2. \end{aligned}$$

In particular, taking $s=\frac{t}{2}$ gives

$$\begin{aligned} \Vert \nabla u(t)\Vert _{L^2}^2\le C\langle {t} \rangle ^{-1}\Vert u(t/2)\Vert _{L^2}^2, \end{aligned}$$

from which and (3.10), we infer for any $t<T^*$

$$\begin{aligned} \Vert \nabla u(t)\Vert _{L^2}^2\le C \left\{ \begin{array}{ll} \displaystyle \langle {t} \rangle ^{-1-2\beta (p)}&{} \text{ if }\quad 1<p< 2, \\ \displaystyle \langle {t} \rangle ^{-\left( \frac{5}{2}\right) _-}&{} \text{ if }\quad p=1, \end{array}\right. \end{aligned}$$

(3.20)

Similarly by applying Gronwall’s lemma to (3.1) over [s, t],  we write

$$\begin{aligned}&\Vert \nabla u(t)\Vert _{L^2}^2+\int _s^t\left( \Vert \sqrt{\rho }u_t(t')\Vert _{L^2}^2+\Vert \nabla ^2u(t')\Vert _{L^2}^2+\Vert \nabla \Pi (t')\Vert _{L^2}^2\right) \,dt'\nonumber \\&\quad \le \exp \left( C\Vert \nabla u\Vert _{L^2_t({H}^1)}^2\right) \Vert \nabla u(s)\Vert _{L^2}^2\nonumber \\&\quad \le \exp \left( CC_0\right) \Vert \nabla u(s)\Vert _{L^2}^2. \end{aligned}$$

(3.21)

Whereas by multiplying (3.2) by $(t-s)$ and applying Gronwall’s lemma to resulting inequality, we get

$$\begin{aligned}&(t-s)\Vert \sqrt{\rho }u_t(t)\Vert _{L^2}^2\le \left( \int _s^t\Vert \sqrt{\rho }u_t(t')\Vert _{L^2}^2\,dt'+\Vert \nabla u\Vert _{L^\infty (s,t;L^2)}^4\right) \\&\qquad \times \exp \left( C\left( \Vert \nabla u\Vert _{L^2_t({H}^1)}^2+\Vert u\Vert _{L^\infty _t(\dot{H}^1)}^2 \Vert u\Vert _{L^2_t(\dot{H}^1)}^2\right) \right) \\&\quad \le \exp \left( CC_0(1+C_0)\right) \left( \Vert \nabla u(s)\Vert _{L^2}^2+\Vert \nabla u\Vert _{L^\infty (s,t;L^2)}^4\right) . \end{aligned}$$

Taking $s=\frac{t}{2}$ in the above inequality and using (3.20), we obtain

$$\begin{aligned} \Vert u_t(t)\Vert _{L^2}^2\le C \left\{ \begin{array}{ll} \displaystyle t^{-1}\langle {t} \rangle ^{-1-2\beta (p)} &{} \text{ if }\quad 1<p< 2, \\ \displaystyle t^{-1}\langle {t} \rangle ^{-\left( \frac{5}{2}\right) _-}&{} \text{ if }\quad p=1. \end{array}\right. \end{aligned}$$

which together with (3.6) and (3.20) ensures that

$$\begin{aligned} \Vert u_t(t)\Vert _{L^2}^2+ \Vert u(t)\Vert _{\dot{H}^2}^2+\Vert \nabla \Pi (t)\Vert _{L^2}^2 \le C \left\{ \begin{array}{ll} \displaystyle Ct^{-1}\langle {t} \rangle ^{-1-2\beta (p)} &{} \text{ if }\quad 1<p< 2, \\ \displaystyle t^{-1}\langle {t} \rangle ^{-\left( \frac{5}{2}\right) _-}&{} \text{ if }\quad p=1, \end{array}\right. \end{aligned}$$

(3.22)

for any $t<T^*.$

With (3.20) and (3.22), it remains to prove (1.13) for $p=1.$ As a matter of fact, we first deduce from (3.22) that

$$\begin{aligned} \left( \int _{0}^{t}\Vert {\mathcal F}_x(a (\Delta u-\nabla \Pi ))(t')\Vert _{L_{\xi }^{\infty }}\, dt'\right) ^2&\le \Vert a\Vert _{L^\infty _t(L^2)}^2\left( \int _{0}^{t}\Vert (\Delta u-\nabla \Pi )(t')\Vert _{L^2} \, d t'\right) ^2\\&\lesssim \Vert a_0\Vert _{L^2}^2 \left( \int _0^t(t')^{-\frac{1}{2}}\langle {t'} \rangle ^{-\left( \frac{5}{4}\right) _-}\,dt'\right) ^{2}\le C. \end{aligned}$$

With (3.13) being replaced by the above inequality, by repeating the proof of Proposition 3.1, we can prove the first inequality of (1.13) for $p=1.$ Then repeating the proof of (3.22), we conclude the proof of the remaining two inequalities in (1.13) for $p=1.$ This finishes the proof of Proposition 3.2. $\square$

4 The proof of Theorem 1.1

The goal of this section is to complete the proof of Theorem 1.1. In order to do so, we first prove the following globally in time Lipschitz estimate for the convection velocity field, which will be used to prove the propagation of the size for $\Vert \frac{a_0}{r}\Vert _{L^\infty }.$

Lemma 4.1

Let $(\rho ,u,\nabla \Pi )$ be a c smooth enough axisymmetric solution of (1.1) on $[0,T^*).$ Then under the assumptions (1.7) and (1.10), we have $T^*=\infty ,$ and there holds

$$\begin{aligned} \Vert \nabla u\Vert _{L^1(\mathop {\mathbb R}\nolimits _+;L^\infty )}\le C, \end{aligned}$$

(4.1)

for some positive constant depending on m, M and $\Vert u_0\Vert _{H^1}.$

Proof

Under the assumptions of (1.7) and (1.10), we deduce from Proposition 2.2 that $T^*=\infty$ and moreover Corollary 3.1 ensures that

$$\begin{aligned}&\sup _{t\in [0,\infty )}\langle {t} \rangle \Vert \nabla u(t)\Vert _{L^2}^2+\int _0^\infty \langle {t} \rangle \left( \Vert u_t(t)\Vert _{L^2}^2+\Vert u(t)\Vert _{\dot{H}^2}^2+\Vert \nabla \Pi (t)\Vert _{L^2}^2\right) \,dt \le C_1,\nonumber \\&\sup _{t\in [0,\infty )} t\langle {t} \rangle \left( \Vert u_t(t)\Vert _{L^2}^2+\Vert u(t)\Vert _{\dot{H}^2}^2+\Vert \nabla \Pi (t)\Vert _{L^2}^2\right) +\int _0^\infty t\langle {t} \rangle \Vert \nabla u_t(t)\Vert _{L^2}^2\,dt\le C_2, \end{aligned}$$

(4.2)

where $C_1$ and $C_2$ given by (3.8) and (3.9) respectively. In particular, by using Sobolev imbedding theorem, we obtain

$$\begin{aligned} \int _0^\infty t\langle {t} \rangle \Vert u_t(t)\Vert _{L^6}^2\,dt\le C_2. \end{aligned}$$

(4.3)

On the other hand, in view of (2.32), we deduce from the classical estimates of linear Stokes operator that

$$\begin{aligned} \Vert \nabla ^2 u(t)\Vert _{L^6}+\Vert \nabla \Pi (t)\Vert _{L^6}\le C\left( \Vert u_t(t)\Vert _{L^6}+\Vert u(t)\Vert _{L^\infty }\Vert \nabla u(t)\Vert _{L^6}\right) , \end{aligned}$$

which together with (3.4) yields

$$\begin{aligned}&\int _0^\infty t\langle {t} \rangle \left( \Vert \nabla ^2 u(t)\Vert _{L^6}^2+\Vert \nabla \Pi (t)\Vert _{L^6}^2\right) \,dt\\&\quad \le C\left( \int _0^\infty t\langle {t} \rangle \Vert u_t(t)\Vert _{L^6}^2\,dt+\int _0^\infty t\langle {t} \rangle \Vert u(t)\Vert _{\dot{H}^1}\Vert u(t)\Vert _{\dot{H}^2}^3\,dt\right) . \end{aligned}$$

Yet it follows from (4.2) that

$$\begin{aligned} \int _0^\infty t\langle {t} \rangle \Vert u(t)\Vert _{\dot{H}^1}\Vert u(t)\Vert _{\dot{H}^2}^3\,dt\le C\sqrt{C_1}C_2^{\frac{3}{2}}\int _0^\infty t^{-\frac{1}{2}}\langle {t} \rangle ^{-1}\,dt\le C\sqrt{C_1}C_2^{\frac{3}{2}}, \end{aligned}$$

which together with (4.3) ensures that

$$\begin{aligned} \int _0^\infty t\langle {t} \rangle \left( \Vert \nabla ^2 u(t)\Vert _{L^6}^2+\Vert \nabla \Pi (t)\Vert _{L^6}^2\right) \,dt\le CC_2\left( 1+\sqrt{C_1C_2}\right) \buildrel \hbox {def}\over =C_3. \end{aligned}$$

(4.4)

By virtue of (4.2) and (4.4), we infer

$$\begin{aligned} \int _0^\infty \Vert \nabla u(t)\Vert _{L^\infty }\,dt&\le C\int _0^\infty \Vert u(t)\Vert _{\dot{H}^2}^{\frac{1}{2}}\Vert \nabla ^2u(t)\Vert _{L^6}^{\frac{1}{2}}\,dt\\&\le CC_2^{\frac{1}{4}}\int _0^\infty t^{-\frac{1}{2}}\langle {t} \rangle ^{-\frac{1}{2}}\left( t\langle {t} \rangle \Vert \nabla ^2 u(t)\Vert _{L^6}^2\right) ^{\frac{1}{4}}\,dt\\&\le CC_2^{\frac{1}{4}}\left( \int _0^\infty t^{-\frac{2}{3}}\langle {t} \rangle ^{-\frac{2}{3}}\,dt\right) ^{\frac{3}{4}}\left( \int _0^\infty t\langle {t} \rangle \Vert \nabla ^2 u(t)\Vert _{L^6}^2\,dt\right) ^{\frac{1}{4}}\\&\le CC_2^{\frac{1}{4}}C_3^{\frac{1}{4}}. \end{aligned}$$

This gives rise to (4.1). $\square$

Now we are in a position to complete the proof of Theorem 1.1.

Proof of Theorem 1.1 The general strategy to prove the existence result to a nonlinear partial differential equation is first to construct an appropriate approximate solutions, and then perform the uniform estimates to these approximate solution sequence, and finally the existence result follows from a compactness argument. For simplicity, here we just present the a priori estimates to smooth enough solutions of (1.4).

Given axisymmetric initial data $(\rho _0,u_0)$ with $\rho _0$ satisfying (1.7) and $a_0\in L^2\cap L^\infty ,$ $\frac{a_0}{r}\in L^\infty ,$ $u_0\in H^1,$ we deduce from (2.23) and (2.25) that there exists a maximal positive time $T^*$ so that (1.4) has a solution on $[0,T^*)$ which satisfies for any $T<T^*,$

$$\begin{aligned} \Vert u\Vert _{L^\infty _T(H^1)}+\Vert \nabla u\Vert _{L^2_T(H^1)}+\Vert \partial _tu\Vert _{L^2_T(L^2)}+\Vert \nabla \Pi \Vert _{L^2_T(L^2)}+ \Vert \Gamma \Vert _{L^\infty _T(L^2)}+\Vert \nabla \Gamma \Vert _{L^2_T(L^2)}\le C, \end{aligned}$$

from which and Corollary 3.1, we deduce that there holds (1.8). And hence the uniqueness part of Theorem 1.1 follows from the uniqueness result in [20].

Now if $T^*<\infty$ and there holds

$$\begin{aligned} \lim _{t\rightarrow T^*}\left\| \frac{a(t)}{r}\right\| _{L^\infty }=C_*<\infty . \end{aligned}$$

Let us take $\delta$ so small that

$$\begin{aligned} 2mCC_*\le \frac{1}{2}. \end{aligned}$$

Then we get, by summing up (2.23) and $2m \delta \times$ (2.25), that

$$\begin{aligned}&\frac{d}{dt}\left( \Vert \widetilde{\nabla } u(t)\Vert _{L^2}^2+\left\| \frac{u^r(t)}{r}\right\| _{L^2}^2+2m\delta \Vert \Gamma (t)\Vert _{L^2}^2\right) \\&\qquad +\Vert \partial _tu\Vert _{L^2}^2+\Vert u\Vert _{\dot{H}^2}^2+\frac{1}{2}\left( \Vert \widetilde{\nabla }\Pi \Vert _{L^2}^2+\Vert \Gamma \Vert _{L^2}^2\right) +\delta \Vert \widetilde{\nabla }\Gamma \Vert _{L^2}^2\\&\quad \le C_\delta \left( (1+\Vert u\Vert _{L^2}^6)\left( \Vert \widetilde{\nabla }u\Vert _{L^2}^2+\left\| \frac{u^r}{r}\right\| _{L^2}^2\right) \left( \Vert \widetilde{\nabla }u\Vert _{L^2}^2+\Vert \Gamma \Vert _{L^2}^2\right) +(1+\Vert u^z\Vert _{L^2}^4)\Vert \partial _zu\Vert _{L^2}^2\right) . \end{aligned}$$

Applying Gronwall’s inequality and using (2.3) leads to

$$\begin{aligned}&\Vert \widetilde{\nabla }u\Vert _{L^\infty _T(L^2)}^2+\left\| \frac{u^r}{r}\right\| _{L^\infty _T(L^2)}^2+\Vert \Gamma \Vert _{L^\infty _T(L^2)}^2+\Vert \partial _tu\Vert _{L^2_T(L^2)}^2+\Vert \widetilde{\nabla }\Pi \Vert _{L^2_T(L^2)}^2\\&\quad +\Vert u\Vert _{L^2_T(\dot{H}^2)}^2+\Vert \Gamma \Vert _{L^2_T(L^2)}^2+\Vert \widetilde{\nabla }\Gamma \Vert _{L^2_T(L^2)}^2\le C, \end{aligned}$$

for any $T<T^*.$ Therefore we can extend the solution beyond the time $T^*,$ which contradicts with the maximality of $T^*.$ Hence there holds (1.9).

Under the assumption of (1.10), we deduce from Proposition 2.2 that $T^*=\infty$ and there holds (1.11). Moreover, Lemma 4.1 ensures that

$$\begin{aligned} \Vert \nabla u\Vert _{L^1(\mathop {\mathbb R}\nolimits ^+;L^\infty )}\le C, \end{aligned}$$

which together with (2.26) and

$$\begin{aligned} \left\| \frac{u^r}{r}\right\| _{L^1(\mathop {\mathbb R}\nolimits ^+;L^\infty )}\le \Vert \nabla u \Vert _{L^1(\mathop {\mathbb R}\nolimits ^+;L^\infty )} \end{aligned}$$

gives rise to (1.12).

Finally with additional assumption that $u_0\in L^p$ for some $p\in [1,2),$ we deduce from Proposition 3.2 that there holds the decay estimate (1.13). This finishes the proof of Theorem 1.1.