Global well-posedness of the 3D axially symmetric magnetohydrodynamic boundary layer equations in an analytic space
- Open Access
- 11.12.2025
- Research
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Abstract
1 Introduction
In this paper, we consider the following 3D incompressible MHD boundary layer equations with homogeneous Dirichlet boundary conditions [23]
$$ \textstyle\begin{cases} \partial _{t} \mathbf{v} + (\mathbf{v} \cdot \nabla ) \mathbf{v} - ( \mathbf{H} \cdot \nabla ) \mathbf{H} + \nabla p - \frac{1}{Re} \Delta \mathbf{v} = \mathbf{0}, \\ \partial _{t} \mathbf{H} + (\mathbf{v} \cdot \nabla ) \mathbf{H} - ( \mathbf{H} \cdot \nabla ) \mathbf{v}- \frac{1}{Rm} \Delta \mathbf{H} = \mathbf{0}, \\ \nabla \cdot \mathbf{v} = \nabla \cdot \mathbf{H} = 0. \end{cases} $$
(1.1)
Here, the parameters Re and Rm denote the hydrodynamic Reynolds number and magnetic Reynolds number respectively. v and \(\mathbf{H} = (b_{h}, b_{x_{3}})\) stand for the velocity field and magnetic field in turn, \(p = \tilde{p} + |\mathbf{H}|^{2} / 2\) is the total pressure, where p̃ refers to the fluid pressure.
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One important problem in MHD equations is understanding the high Reynolds number limits at domain boundaries. In this paper, we consider the case where \(1/Re=1/Rm=\epsilon \) with \(\epsilon \ll 1\). The derivation of the MHD boundary layer equations can be found in [10] (For the perfectly conducting boundary condition, the derivation of the corresponding boundary layer can be found in [27]):
$$ \textstyle\begin{cases} \partial _{t} \mathbf{v}_{h} + (\mathbf{v} \cdot \nabla ) \mathbf{v}_{h} - (\mathbf{H} \cdot \nabla ) \mathbf{H}_{h} + \nabla _{h} P - \partial _{z}^{2} \mathbf{v}_{h} = \mathbf{0}, \\ \partial _{t} \mathbf{H}_{h} + (\mathbf{v} \cdot \nabla ) \mathbf{H}_{h} - (\mathbf{H} \cdot \nabla ) \mathbf{v}_{h} - \partial _{z}^{2} \mathbf{H}_{h} = \mathbf{0}, \\ \partial _{t} b_{z} + (\mathbf{v} \cdot \nabla ) b_{z} - (\mathbf{H} \cdot \nabla ) v_{z} - \partial _{z}^{2} b_{z} = \mathbf{0}, \\ \nabla \cdot \mathbf{v} = \nabla \cdot \mathbf{H} = 0, \\ \left . (\mathbf{v}_{h}, v_{z}) \right |_{z = 0} = \left . ( \mathbf{H}_{h}, b_{z}) \right |_{z = 0} = \mathbf{0}, \quad \left . ( \mathbf{v}_{h}, \mathbf{H}_{h}) \right |_{z \to +\infty} = ( \mathbf{V}, \mathbf{B}). \end{cases} $$
(1.2)
Here, we define domain as \(\mathbb{R}_{+}^{3} = \left \{ (x, y, z) \bigm| (x, y) \in \mathbb{R}^{2}, \, z > 0 \right \}\), the symbol ∇ is defined as \((\nabla _{h}, \partial _{z})\), where \(\nabla _{h} = (\partial _{x}, \partial _{y})\), we denote the velocity field and magnetic field by \(\mathbf{v} = (\mathbf{v}_{h}, v_{z})\) and \(\mathbf{H} = (\mathbf{H}_{h}, b_{z})\) respectively. Additionally, the given functions \((\mathbf{V}, \mathbf{B}, P)\) are the boundary traces of the tangential outflow field and outflow pressure, which satisfy Bernoulli’s law:
$$ \textstyle\begin{cases} \partial _{t} \mathbf{V} + (\mathbf{V} \cdot \nabla _{h}) \mathbf{V} - (\mathbf{B} \cdot \nabla _{h}) \mathbf{B} + \nabla _{h} P = \mathbf{0}, \\ \partial _{t} \mathbf{B} + (\mathbf{V} \cdot \nabla _{h}) \mathbf{B} - (\mathbf{B} \cdot \nabla _{h}) \mathbf{V} = \mathbf{0}. \end{cases} $$
(1.3)
In the absence of the magnetic field H, problem (1.1) simplifies to the classical Prandtl equations, first formally introduced by Prandtl [39] in 1904. These equations describe fluid flow near a solid boundary as a thin viscous boundary layer and an outer inviscid region. In the 1960s, Oleinik [34] established the first rigorous mathematical framework, proving local-in-time well-posedness for the 2D Prandtl equations via the Crocco transformation under a tangential velocity monotonicity condition (detailed in [33]). Beyond this, Xin et al. [43, 44] obtained global weak solution well-posedness for the classical Prandtl equation under favorable pressure \((\partial _{x}p\leq 0)\). Recent advances in the well-posedness theory of Prandtl equations are well summarized in [1, 8, 9, 15, 19, 32, 45] and related references.
For the 3D Prandtl equations, the well-posedness theory remains underdeveloped due to secondary flows and complex boundary layer structures arising from multi-dimensional velocity fields, with most studies focusing on specific settings. Under special outer Euler flow velocity structures, Liu, Wang and Yang [25, 26] established local well-posedness of classical solutions and global existence of weak solutions via the Crocco transformation. Lin and Zhang [22] investigated almost global existence of unique classical solutions for the 3D Prandtl system with initial data near a stable shear flow (extending [11]). Pan and Xu [36] proved global existence of solutions to the 3D axially symmetric Prandtl equations with small initial data (in \(H^{1}\) for normal variable, analytic in tangential variables), improving upon [11] and extending the results of [35]. Recently, the analyticity condition has been weakened to Gevrey function spaces [14, 18, 37, 38] and related references.
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When the velocity field equations couple with the magnetic field equations, boundary layer phenomena differ due to the presence of more complex magnetic field boundary layers [10], with the added challenge of further tangential derivative loss in the magnetic field. By using the non-degenerate initial tangential magnetic field assumption (instead of Oleinik’s velocity monotonicity condition), well-posedness theory for MHD boundary layer equations in Sobolev spaces has been established in [2‐5, 7, 16, 24, 27, 28]. Liu et al. [27] proved local well-posedness for the 2D nonlinear MHD boundary layer equations without monotonicity in weighted Sobolev spaces via energy methods, and extended this in [28] to validate Prandtl expansion with \(L^{\infty}\) error estimates. Chen et al. [3] complemented this with long-time well-posedness for small initial data in lower-order weighted spaces, showing that solution lifespan depends on initial data. For large magnetic Reynolds numbers (neglecting resistivity), Liu et al. [24] studied local well-posedness of the 2D MHD boundary layer equations without resistivity and found linear instability when tangential magnetic field degenerates at a point. Chen and Li [2] extended this to long-time well-posedness in lower-order spaces. Li and Xu [16] addressed the velocity diffusion absence, proving well-posedness via pseudo-differential calculations, which complementing [24, 27]. Gao et al. [7] investigated local well-posedness in weighted conormal Sobolev spaces for the 2D incompressible MHD boundary layer equations.
To date, besides well-posedness results in the Sobolev framework, there have also been findings in the analytic framework for the 2D MHD boundary layer equations. Xie and Yang [41] proved the global existence of solutions to the 2D MHD boundary layer equations in the mixed Prandtl-Hartmann regime, where the initial data is a small perturbation of the Hartmann profile; these solutions decay exponentially over time in the analytic norm. Building on [11], Xie and Yang [42] analyzed the lifespan of solutions to the 2D MHD boundary layer system using the cancellation mechanism, deriving a lower bound for the lifespan. Inspired by [12], Lin and Zhang [21] used the energy method to study the local existence of solutions for the 2D incompressible magneto-micropolar boundary layer equations when the initial data are analytic in the x-variable. Liu and Zhang [29] established the global existence of solutions to the 2D MHD boundary layer equations with small initial data, along with asymptotic decay estimates. Li and Xie [13] obtained results on the global well-posedness of solutions to the 2D MHD boundary layer equations in analytic function spaces. More recently, the analyticity condition has been further relaxed to the Gevrey function space [17, 20, 23, 40] and related references.
For the 3D axially symmetric MHD equations, Maafa et al. [30, 31] obtained the global well-posedness of the equation in Sobolev space. Besides, there have been a few results regarding the well-posedness to the 3D axially symmetric MHD boundary layer equations. Lin and Zou [23] established the well-posedness of the equation in Gevrey function spaces without any structural assumptions. To our knowledge, the well-posedness of the 3D axially symmetric MHD boundary layer equations in an analytic spaces remains an open question. Building on the well-posedness results established in [23] for the 3D axially symmetric MHD boundary layer equations and in [11, 36] for the 2D/3D Prandtl equations, the main goal of this paper is to investigate the well-posedness of solutions in an analytic spaces for the 3D axially symmetric MHD boundary layer equations. Below, we outline the strategy for proving the main results of this paper. Firstly, we reformulate the original problem (1.1) into the symmetric MHD boundary layer equations (2.5), the detailed derivation can be found in [23]. Secondly, leveraging the specific structure of the symmetric MHD boundary layer equations, we construct an energy functional featuring a polynomial weight on tangential variables. This functional resolves the order mismatch between the tangential radial velocity \(v^{r} \) and the normal velocity \(v^{z} \) with respect to the distance r from the symmetry axis. Thirdly, the unknown quantity in the energy functional is a specially designed combination of \(v^{r} \) and its primitive in the normal variable. It exhibits a sufficiently rapid time decay rate for our weighted analytical energy, ensuring that the analytical radius maintains a positive lower bound at all times.
The rest of this paper is structured as follows. In Sect. 2, we present the reformulation of the axially symmetric MHD boundary layer equations and state the main result. Section 3 establishes bounds for \(v^{r}\), \(v^{z}\), ϕ as well as \(b^{r}\), \(b^{z}\), ψ. In Sect. 4, we provide the proof for the existence component of Theorem 2.2. Section 5 develops estimates for the nonlinear terms, and Sect. 6 verifies the uniqueness of solutions to complete the proof of Theorem 2.2.
2 Reformation of the symmetric MHD boundary layer equations and the main result
2.1 Reformation of the symmetric MHD boundary layer equations
To ensure readers can follow the full context, we derive the 3D axially symmetric MHD boundary layer equations in cylindrical coordinates \((r,\theta ,z)\), following the approach in [23]. Let \(\mathbf{x} = (x,y,z) \in \mathbb{R}^{3}\) satisfy a solution to (1.2) and (1.3) is said to be axisymmetric if and only if where the basis vectors \(e_{r}\), \(e_{\theta}\), \(e_{z}\) are
$$\begin{aligned} r = \sqrt{x^{2} + y^{2}}, \quad \theta = \arctan \frac{y}{x}, \end{aligned}$$
$$\begin{aligned} \mathbf{v} =&\tilde{v}^{r}(t,r,z)e_{r}+\tilde{v}^{\theta}(t,r,z)e_{ \theta}+\tilde{v}^{z}(t,r,z)e_{z}, \\ \mathbf{H} =&\tilde{b}^{r}(t,r,z)e_{r}+\tilde{b}^{\theta}(t,r,z)e_{ \theta}+\tilde{b}^{z}(t,r,z)e_{z}, \\ \mathbf{V} =&V^{r}(t,r)e_{r}+V^{\theta}(t,r)e_{\theta}, \\ \mathbf{B} =&B^{r}(t,r)e_{r}+B^{\theta}(t,r)e_{\theta}, \\ P =& P(t,r), \end{aligned}$$
$$\begin{aligned} e_{r} = \left ( \frac{x}{r}, \frac{y}{r}, 0 \right ), \quad e_{ \theta }= \left ( -\frac{y}{r}, \frac{x}{r}, 0 \right ), \quad e_{z} = (0, 0, 1). \end{aligned}$$
After transforming them into cylindrical coordinates, equations (1.2) and (1.3) take the form of and
$$\begin{aligned} \left \{ \textstyle\begin{array}{l} \partial _{t} \begin{pmatrix} \tilde{v}^{r} \\ \tilde{v}^{\theta } \\ \tilde{b}^{r} \\ \tilde{b}^{\theta }\end{pmatrix} + \left (\tilde{v}^{r} \partial _{r} + \tilde{v}^{z} \partial _{z} \right ) \begin{pmatrix} \tilde{v}^{r} \\ \tilde{v}^{\theta } \\ \tilde{b}^{r} \\ \tilde{b}^{\theta }\end{pmatrix} - \left ( \tilde{b}^{r} \partial _{r} + \tilde{b}^{z} \partial _{z} \right ) \begin{pmatrix} \tilde{b}^{r} \\ \tilde{b}^{\theta } \\ \tilde{v}^{r} \\ \tilde{v}^{\theta }\end{pmatrix} + \frac{1}{r} \begin{pmatrix} (\tilde{b}^{\theta})^{2} - (\tilde{v}^{\theta})^{2} \\ \tilde{v}^{r} \tilde{v}^{\theta }- \tilde{b}^{r} \tilde{b}^{\theta } \\ 0 \\ \tilde{v}^{r} \tilde{b}^{\theta }- \tilde{b}^{r} \tilde{v}^{\theta }\end{pmatrix} \\ \quad + \begin{pmatrix} \partial _{r} P \\ 0 \\ 0 \\ 0 \end{pmatrix} = \partial _{z}^{2} \begin{pmatrix} \tilde{v}^{r} \\ \tilde{v}^{\theta } \\ \tilde{b}^{r} \\ \tilde{b}^{\theta }\end{pmatrix} , \\ \partial _{t} \tilde{b}^{z} + (\tilde{v}^{r} \partial _{r} + \tilde{v}^{z} \partial _{z}) \tilde{b}^{z} - (\tilde{b}^{r} \partial _{r} + \tilde{b}^{z} \partial _{z}) \tilde{v}^{z} - \partial _{z}^{2} \tilde{b}^{z}=0, \\ \frac{\partial _{r} (r \tilde{v}^{r})}{r} + \partial _{z} \tilde{v}^{z} = \frac{\partial _{r} (r \tilde{b}^{r})}{r} + \partial _{z} \tilde{b}^{z} = 0, \\ \left . (\tilde{v}^{r}, \tilde{v}^{\theta}, \tilde{v}^{z}, \tilde{b}^{r}, \tilde{b}^{\theta}, \tilde{b}^{z}) \right |_{z=0} = \mathbf{0}, \quad \lim _{z \to +\infty} (\tilde{v}^{r}, \tilde{v}^{\theta}, \tilde{b}^{r}, \tilde{b}^{\theta}) = (V^{r}, V^{\theta}, B^{r}, B^{ \theta}), \\ \left . (\tilde{v}^{r}, \tilde{v}^{\theta}, \tilde{b}^{r}, \tilde{b}^{ \theta}) \right |_{t=0} = (\tilde{u}_{0}^{r}, \tilde{u}_{0}^{\theta}, \tilde{b}_{0}^{r}, \tilde{b}_{0}^{\theta}), \end{array}\displaystyle \right . \end{aligned}$$
(2.1)
$$\begin{aligned} \left \{ \textstyle\begin{array}{l} \partial _{t} \begin{pmatrix} V^{r} \\ V^{\theta } \\ B^{r} \\ B^{\theta }\end{pmatrix} + V^{r} \partial _{r} \begin{pmatrix} V^{r} \\ V^{\theta } \\ B^{r} \\ B^{\theta }\end{pmatrix} - B^{r} \partial _{r} \begin{pmatrix} B^{r} \\ B^{\theta } \\ V^{r} \\ V^{\theta }\end{pmatrix} + \frac{1}{r} \begin{pmatrix} (B^{\theta})^{2} - (V^{\theta})^{2} \\ V^{r} V^{\theta }- B^{r} B^{\theta } \\ 0 \\ V^{r} B^{\theta }- B^{r} V^{\theta }\end{pmatrix} + \begin{pmatrix} \partial _{r} P \\ 0 \\ 0 \\ 0 \end{pmatrix} = \mathbf{0}. \end{array}\displaystyle \right . \end{aligned}$$
Here, we assume the flow is swirl free, i.e., \(\tilde{v}^{\theta }= \tilde{b}^{\theta }= V^{\theta }= B^{\theta } \equiv 0\). We also assume the outflow satisfies \(V^{r} = B^{r} \equiv 0\), which implies that \(\partial _{r} P \equiv 0\).
In reality, for a general \((\mathbf{V}, \mathbf{B})\), the equations governing \((\tilde{b}^{r},\tilde{b}^{z})\) may be overdetermined. However, in the case where \((\mathbf{V}, \mathbf{B}) = (0,0)\), equation (2.1)5 can be derived from (2.1)3 and the divergence free condition (2.1)6, meaning the equations for \((\tilde{b}^{r},\tilde{b}^{z})\) are not overdetermined. By applying \(r\partial _{r}\) to (2.1)3 and utilizing the divergence-free condition, we obtain the following result:
$$\begin{aligned} \partial _{t}\partial _{r}\big(r\tilde{b}^{r}\big)+\partial _{z} \partial _{r}\big(\tilde{v}^{z}(r\tilde{b}^{r})-(r\tilde{v}^{r}) \tilde{b}^{z}\big)-\partial _{z}^{2}\partial _{r}\big(r\tilde{b}^{r} \big)=0. \end{aligned}$$
(2.2)
Thanks to (2.1)6, we can define two stream functions \(\big(\tilde{\phi}(r,z),\tilde{\psi}(r,z)\big)\) such that \(\partial _{z}\big(\tilde{\phi},\tilde{\psi}\big)=\big(r\tilde{v}^{r},r \tilde{b}^{r}\big)\) and \(\partial _{r}\big(\tilde{\phi},\tilde{\psi}\big)=-\big(r\tilde{v}^{z},r \tilde{b}^{z}\big)\) hold and have boundary values
$$ \big(\tilde{\phi},\tilde{\psi}\big)|_{z=0}=\big(\tilde{\phi}, \tilde{\psi}\big)|_{z=+\infty}=0. $$
As a result, we have \(\left .(\tilde{v}^{z}, \tilde{b}^{z})\right |_{z\to +\infty} = (0,0)\), which means \(\tilde{b}^{z} = -\int _{z}^{+\infty} \partial _{\tilde{z}} \tilde{b}^{z} d\tilde{z}\). By integrating (2.2) over the interval \([z, +\infty )\) and using the divergence-free condition, we can derive the following:
$$\begin{aligned} \partial _{t}\big(r\tilde{b}^{z}\big)-\partial _{r}\big(\tilde{v}^{z}(r \tilde{b}^{r})-(r\tilde{v}^{r})\tilde{b}^{z}\big)-\partial _{z}^{2} \big(r\tilde{b}^{z}\big) -\partial _{z}\partial _{r}(r\tilde{b}^{r})|_{z \rightarrow +\infty}=0. \end{aligned}$$
(2.3)
Since we’re looking at the solution \((\tilde{v}^{r},\tilde{b}^{r})\) that decays rapidly to \((V^{r}, B^{r})\) as z approaches +∞, this implies \(\left . \partial _{z} \partial _{r}(r\tilde{b}^{r})\right |_{z \to + \infty} = 0\). This let us derive (2.1)5 from (2.1)3 for any \(r>0\), so in the case studied here, the equations governing \((\tilde{b}^{r},\tilde{b}^{z})\) aren’t overdetermined.”
The equations (2.1) can then be expressed as
$$\begin{aligned} \left \{ \textstyle\begin{array}{l} \partial _{t} \tilde{v}^{r} + (\tilde{v}^{r} \partial _{r} + \tilde{v}^{z} \partial _{z}) \tilde{v}^{r} - (\tilde{b}^{r} \partial _{r} + \tilde{b}^{z} \partial _{z}) \tilde{b}^{r} - \partial _{z}^{2} \tilde{v}^{r} = 0, \\ \partial _{t} \tilde{b}^{r} + (\tilde{v}^{r} \partial _{r} + \tilde{v}^{z} \partial _{z}) \tilde{b}^{r} - (\tilde{b}^{r} \partial _{r} + \tilde{b}^{z} \partial _{z}) \tilde{v}^{r} - \partial _{z}^{2} \tilde{b}^{r} = 0, \\ \frac{\partial _{r} (r \tilde{v}^{r})}{r} + \partial _{z} \tilde{v}^{z} = \frac{\partial _{r} (r \tilde{b}^{r})}{r} + \partial _{z} \tilde{b}^{z} = 0, \\ \left . (\tilde{v}^{r}, \tilde{v}^{z}, \tilde{b}^{r}, \tilde{b}^{z}) \right |_{z=0} = \mathbf{0}, \quad \lim _{z \to +\infty} (\tilde{v}^{r}, \tilde{b}^{r}) = \mathbf{0}, \\ \left . (\tilde{v}^{r}, \tilde{b}^{r}) \right |_{t=0} = (\tilde{u}_{0}^{r}, \tilde{b}_{0}^{r}). \end{array}\displaystyle \right . \end{aligned}$$
(2.4)
Assuming the axially symmetric velocity field \(\mathbf{\tilde{v}} = \tilde{v}^{r}(t,r,z)e_{r}+\tilde{v}^{\theta}(t,r,z)e_{ \theta}+\tilde{v}^{z}(t,r,z)e_{z}\) and the axially symmetric magnetic field \(\mathbf{\tilde{H}}=\tilde{b}^{r}(t,r,z)e_{r}+\tilde{b}^{\theta}(t,r,z)e_{ \theta}+\tilde{b}^{z}(t,r,z)e_{z}\) are smooth and divergence-free, we draw the conclusion that this means that no singularities exist in these equalities \(\frac{\tilde{v}^{r}}{r}\) and \(\frac{\tilde{b}^{r}}{r}\) at \(r = 0\).
$$\begin{aligned} \left . \tilde{v}^{r} \right |_{r=0} = \left . \tilde{v}^{\theta } \right |_{r=0} = \left . \tilde{b}^{r} \right |_{r=0} = \left . \tilde{b}^{\theta }\right |_{r=0} \equiv 0, \end{aligned}$$
Introduce the new unknowns as
$$\begin{aligned} (v^{r}, v^{z}, b^{r}, b^{z}) := \left ( \frac{\tilde{v}^{r}}{r}, \tilde{v}^{z}, \frac{\tilde{b}^{r}}{r}, \tilde{b}^{z} \right ). \end{aligned}$$
Subsequently, the newly formulated axially symmetric MHD boundary layer equations take the form:
$$\begin{aligned} \left \{ \textstyle\begin{array}{l} \partial _{t}v^{r}+(rv^{r}\partial _{r}+v^{z}\partial _{z})v^{r}-(rb^{r} \partial _{r}+b^{z}\partial _{z})b^{r}-\partial _{z}^{2}v^{r}+(v^{r})^{2}-(b^{r})^{2}=0, \\ \partial _{t}b^{r}+(rv^{r}\partial _{r}+v^{z}\partial _{z})b^{r}-(rb^{r} \partial _{r}+b^{z}\partial _{z})v^{r}-\partial _{z}^{2}b^{r}=0, \\ r\partial _{r}v^{r}+2v^{r}+\partial _{z}v^{z}=r\partial _{r}b^{r}+2b^{r}+ \partial _{z}b^{z}=0, \\ (v^{r}, v^{z}, b^{r}, b^{z})|_{z=0} = \mathbf{0}, \quad \lim \limits _{z \to +\infty} (v^{r}, b^{r}) = \mathbf{0}, \\ (v^{r}, b^{r})|_{t=0} = (u_{0}^{r}, b_{0}^{r}). \end{array}\displaystyle \right . \end{aligned}$$
(2.5)
2.2 The linearly good unknown
We assume that \(v^{r}\), \(v^{z}\), \(b^{r}\), \(b^{z} \) decay sufficiently rapidly as \(z \to \infty \), and we’ll introduce the following definition:
$$ \phi (t, r, z) := -\int _{z}^{+\infty} v^{r}(t, r, \bar{z}) \mathrm{d}\bar{z},\ \psi (t, r, z) := -\int _{z}^{+\infty} b^{r}(t, r, \bar{z}) \mathrm{d}\bar{z}. $$
(2.6)
These quantities also decay sufficiently rapidly as \(z \to \infty \). By integrating (2.5) over the interval \([z, +\infty ] \) with respect to the z variable, we get
$$\begin{aligned} \left \{ \textstyle\begin{array}{l} \partial _{t} \phi -\partial _{z}^{2}\phi -v^{r}v^{z}+\int _{z}^{+ \infty}(v^{r})^{2} \mathrm{d}\bar{z}-2 \int _{z}^{+\infty}\partial _{z}v^{r}v^{z} \mathrm{d}\bar{z} \\ +b^{r}b^{z}-\int _{z}^{+\infty}(b^{r})^{2} \mathrm{d}\bar{z}+2 \int _{z}^{+ \infty}\partial _{z}b^{r}b^{z}\mathrm{d}\bar{z}= 0, \\ \partial _{t} \psi -\partial _{z}^{2}\psi +v^{z}b^{r}-v^{r}b^{z}=0, \\ \partial _{z} (\phi ,\psi )|_{z=0} = 0, \quad \lim \limits _{z \to + \infty} (\phi ,\psi ) = 0, \\ (\phi ,\psi )|_{t=0} = (\phi _{0},\psi _{0}) = -\int _{z}^{+\infty} (v^{z},b^{z})(0, r, \bar{z}) \mathrm{d}\bar{z}. \end{array}\displaystyle \right . \end{aligned}$$
(2.7)
And \((v^{r}, v^{z},b^{r}, b^{z}) \) is obtained from \((\phi ,\psi ) \) as
$$ (v^{r},b^{r}) = \partial _{z} (\phi ,\psi ), \quad (v^{z},b^{z})=(-r \partial _{r}\phi -2\phi ,-r\partial _{r}\psi -2\psi ). $$
Inspired by the good unknown in [35, 36], we introduce the following two linearly good functions which satisfies
$$\begin{aligned} \left \{ \textstyle\begin{array}{l} g := \partial _{z} \phi + \frac{z}{2\langle t \rangle} \phi = v^{r} + \frac{z}{2\langle t \rangle} \phi , \\ \tilde{g} := \partial _{z} \psi + \frac{z}{2\langle t \rangle} \psi = b^{r} + \frac{z}{2\langle t \rangle} \psi \end{array}\displaystyle \right . \end{aligned}$$
(2.8)
$$\begin{aligned} \left \{ \textstyle\begin{array}{l} \partial _{t}g-\partial _{z}^{2}g+\frac{1}{\langle t\rangle}g+(rv^{r} \partial _{r}+v^{z}\partial _{z})g-(rb^{r}\partial _{r}+b^{z} \partial _{z})\tilde{g}-\frac{1}{2\langle t\rangle} v^{z}\partial _{z}(z \phi )+\frac{1}{2\langle t\rangle} b^{z}\partial _{z}(z\psi ) \\ +\frac{z}{\langle t \rangle}v^{r}\phi -\frac{z}{\langle t \rangle}b^{r} \psi +\frac{z}{2\langle t\rangle}\int _{z}^{+\infty}(v^{r})^{2} \mathrm{d}\bar{z}-\frac{z}{\langle t\rangle}\int _{z}^{+\infty} \partial _{z} v^{r} v^{z}\mathrm{d}\bar{z} \\ +(v^{r})^{2}-(b^{r})^{2}-\frac{z}{2\langle t\rangle}\int _{z}^{+ \infty}(b^{r})^{2}\mathrm{d}\bar{z}+\frac{z}{\langle t\rangle}\int _{z}^{+ \infty}\partial _{z} b^{r} b^{z}\mathrm{d}\bar{z}= 0, \\ \partial _{t}\tilde{g}-\partial _{z}^{2}\tilde{g}+ \frac{1}{\langle t\rangle}\tilde{g}+(rv^{r}\partial _{r}+v^{z} \partial _{z})\tilde{g}-(rb^{r}\partial _{r}+b^{z}\partial _{z})g- \frac{1}{2\langle t\rangle} v^{z}\partial _{z}(z\psi )+ \frac{1}{2\langle t\rangle} b^{z}\partial _{z}(z\phi ) \\ +\frac{z}{\langle t \rangle}v^{r}\psi -\frac{z}{\langle t \rangle}b^{r} \phi =0, \\ \left . (g,\tilde{g}) \right |_{z=0} = 0, \quad \lim \limits _{z \to + \infty} (g,\tilde{g})= 0, \\ \left . (g,\tilde{g}) \right |_{t=0} = (\tilde{g}_{0},g_{0})=(v^{r}(0, r, z) + \frac{z}{2}\phi _{0}(r,z),b^{r}(0,r,z) + \frac{z}{2} \psi _{0}(r,z)). \end{array}\displaystyle \right . \end{aligned}$$
(2.9)
The radial-axial velocity components \((v^{r}, v^{z})\) and magnetic field components \((b^{r},b^{z})\) can be governed by the introduced terms \(\mathbf{g}=:(g,\tilde{g})\), which have a lower-order time weight. This structural property enables the closure of our energy functional (defined below) for any \(t > 0\).
2.3 Energy functional spaces and the main result
As in [36], set
$$\begin{aligned} \theta (t,z) := \exp \left ( \frac{z^{2}}{8\langle t \rangle} \right ). \end{aligned}$$
For \(\lambda \in \mathbb{R}\), set
$$\begin{aligned} \theta _{\lambda}(t,z) = \exp \left ( \frac{\lambda z^{2}}{8\langle t \rangle} \right ). \end{aligned}$$
Then, by a simple calculation, we obtain, for any \(\lambda , \mu \in \mathbb{R}\), \(\theta _{\lambda + \mu} = \theta _{\lambda }\cdot \theta _{\mu}\).
Denote and
$$\begin{aligned} M_{n} = \frac{(n + 1)^{4}}{n!}, \quad \partial _{h}^{\alpha}= \partial _{x}^{\alpha _{1}}\partial _{y}^{\alpha _{2}}, \quad \alpha = (\alpha _{1}, \alpha _{2}) \in \mathbb{N}^{2} \end{aligned}$$
$$\begin{aligned} \langle r \rangle = (r^{2} + 1)^{\frac{1}{2}}=\sqrt{x^{2} + y^{2} + 1}, \quad \langle t \rangle = (t + 1), \quad (x,y) \in \mathbb{R}^{2}, \, t \geq 0. \end{aligned}$$
For a positive time-dependent function \(\tau := \tau (t)\), we define the Sobolev weighted semi-norms
$$\begin{aligned} \left \{ \textstyle\begin{array}{l} X_{n}(f)=X_{n}(f, \tau )=\sum \limits _{|\alpha |=n} \|\theta \langle r\rangle ^{n}\partial _{h}^{\alpha }f \|_{L^{2}} \tau ^{n} M_{n}, \, n \in \mathbb{N}, \\ D_{n}(f)=D_{n}(f, \tau )=\sum \limits _{|\alpha |=n} \|\theta \langle r\rangle ^{n}\partial _{h}^{\alpha}\partial _{z} f\|_{L^{2}} \tau ^{n} M_{n}=X_{n}(\partial _{z} f,\tau ),\, n \in \mathbb{N}, \\ Y_{n}(f)=Y_{n}(f, \tau )=\sum \limits _{|\alpha |=n} \|\theta \langle r\rangle ^{n}\partial _{h}^{\alpha }f\|_{L^{2}} \tau ^{n-1}nM_{n}, \, n \in (\mathbb{N}\setminus \{0\}). \end{array}\displaystyle \right . \end{aligned}$$
(2.10)
We’re working with the following functional space, where functions are real-analytic in the horizontal variable \(\boldsymbol{x}_{h} = (x, y)\) and lie in a weighted \(L^{2}\) space with respect to the vertical coordinate z. where
$$\begin{aligned} \mathcal{X}_{\tau }= \left \{ \forall \alpha \in \mathbb{N}^{2}, \langle r \rangle ^{|\alpha |} \partial _{h}^{\alpha }f(t, r, z) \in L^{2} \left ( \mathbb{R}_{+}^{3}; \theta ^{2} \mathrm{d}x \mathrm{d}y \mathrm{d}z \right ):\|f\|_{\mathcal{X}_{\tau}} < \infty \right \} \end{aligned}$$
$$\begin{aligned} \|f\|_{\mathcal{X}_{\tau}}=\sum _{n\geq 0}X_{n}(f,\tau ). \end{aligned}$$
We also define the semi-norm
$$\begin{aligned} \|f\|_{\mathcal{Y}_{\tau}} = \sum _{n \geq 1} Y_{n}(f, \tau ). \end{aligned}$$
This captures the gain of one derivative in analytic estimates, especially when the summation over n is taken in \(\ell ^{1}\) instead of \(\ell ^{2}\) a standard trick in Fourier analysis. It’s worth noting that for \(\beta > 1\), the result from [11] establishes that
$$\begin{aligned} \|f\|_{\mathcal{Y}_{\tau}} \leq \tau ^{-1} \|f\|_{\mathcal{X}_{\beta \tau}} \sup _{n \geq 1} (n \beta ^{-n}) \leq C_{\beta }\tau ^{-1}\|f \|_{\mathcal{X}_{\tau}}. \end{aligned}$$
(2.11)
In particular, if \(f \in \mathcal{X}_{2\tau}\), it follows that \(\|f\|_{\mathcal{Y}_{\tau}} \leqq \tau ^{-1}\|g\|_{\mathcal{X}_{2\tau}}\). The gain of a z-derivative will be encoded in the dissipative semi-norm
$$\begin{aligned} \|f\|_{\mathcal{D}_{\tau}}=\sum _{n\geq 0}D_{n}(f,\tau )=\|\partial _{z} f\|_{\mathcal{X}_{\tau}}. \end{aligned}$$
Following the introduction of the functional spaces in this paper and prior to presenting the main results, we provide a definition of solutions to the reformulated axially symmetric MHD boundary layer equations (2.9).
Definition 2.1
[11, 36] (Classical in tangential variables and weak in normal variable) For a fixed time \(t>0\), let \(\mathcal{H}\) be the closure of the set of functions under the space norm
$$\begin{aligned} \left \{f(t,x,y,z)\in C_{c}^{\infty}\left ( \mathbb{R}^{2} \times [0, + \infty )\right ):f|_{z=0}=0 \right \} \end{aligned}$$
$$\begin{aligned} \|f(t)\|_{\mathcal{H}}^{2}:=\sum _{|\alpha |\leq 3}\int _{\mathbb{R}_{+}^{3}} \left |\partial _{h}^{\alpha }f(t,x,y,z)\right |^{2}\exp \left ( \frac{z^{2}}{4\langle t\rangle}\right )\mathrm{d}x\mathrm{d}y \mathrm{d}z. \end{aligned}$$
For any \(T>0\), we say that a function g is classical in x, y variables and weak in z variable solution of (2.9) if and (2.9) holds when tested by \(C_{c}^{\infty}\left ( [0, T)\times \mathbb{R}^{2}\times [0,+\infty ) \right )\).
$$\begin{aligned} \|f(t)\|_{\mathcal{H}}\in L^{\infty}((0, T))\quad \text{and}\quad \| \partial _{z}f(t)\|_{\mathcal{H}}\in L^{2}((0, T)), \end{aligned}$$
Theorem 2.2
For small enough \(\varepsilon \le \varepsilon _{1} \), assume the initial data \(\mathbf{g}_{0}:=(g_{0}(r,z),\tilde{g}_{0}(r,z))\) satisfies where the parameter \(\varepsilon _{1}=\frac{\delta \tau _{0}^{2}}{2^{8}\times 3C_{0}}\), \(\delta \in (0,\displaystyle \frac{1}{4}]\) and \(\tau _{0}>2^{4}\). Then the axially symmetric MHD boundary layer equations (2.9) have a unique globally-in-time solution \(\mathbf{g}:=(g,\tilde{g})\) satisfies where analyticity radius \(\tau (t)\ge \frac{1}{2}\tau _{0}\) for any \(t>0 \) and \(\|\mathbf{g}\|_{X}=\|g\|_{X}+\|\tilde{g}\|_{X}\).
$$\begin{aligned} \|\mathbf{g}_{0}\|_{\mathcal{X}_{\tau _{0}}} \le \varepsilon , \end{aligned}$$
$$\begin{aligned} \mathbf{g}(t)\in \mathcal{X}_{\tau (t)} \end{aligned}$$
Remark 2.3
For the MHD boundary layer equations, Liu, Xie and Yang [27] established their well-posedness in Sobolev spaces under the magnetic field boundary condition corresponding to a conducting boundary. In contrast, focusing on the same MHD boundary layer equations, we replaced the conducting boundary with an insulating one as the magnetic field boundary condition, and proved the global well-posedness of solutions in analytic spaces for small initial data.
Remark 2.4
It is noteworthy that the magnetic field boundary condition \(\textbf{H}_{h} = \textbf{0}\) (vanishing tangential component of magnetic field intensity at the boundary) adopted in this work inherently corresponds to an insulating boundary. For insulators with extremely low electrical conductivity (\(\sigma \approx 0\)), conduction current is negligible, leading to an irrotational magnetic field (\(curl\ \textbf{H} = \textbf{0}\)) from Maxwell’s equations, and \(\textbf{H}_{h} = \textbf{0}\) ensures magnetic field continuity without penetration, consistent with insulating media properties. In contrast, \(\partial _{z}\textbf{H}_{h} = \textbf{0}\) (zero normal derivative of the tangential magnetic field component) characterizes a conducting boundary: ideal conductors (\(\sigma \to \infty \)) have zero internal electric field and time-invariant magnetic field, and this condition guarantees no magnetic flux variation, matching the electromagnetic response of highly conductive media. These two conditions, strictly derived from medium electrical properties rather than arbitrary settings, directly govern the formulation and solution characteristics of the MHD boundary layer equations.
Remark 2.5
This work contributes to establishing the global well-posedness of the 3D axially symmetric magnetohydrodynamic (MHD) boundary layer equations within an analytic space, yet it has certain non-negligible limitations. Firstly, the initial data are subject to strict constraints: they must be of small amplitude, belong to the \(H^{1}\) Sobolev space with respect to the normal variable, and be analytic in the tangential variables. Consequently, any initial data, non-analytic data, and those with lower regularity are excluded from our framework. Secondly, the insulating boundary condition (\(H_{h} = 0\)) is adopted herein to simplify the analytical derivation of well-posedness. Nevertheless, compared with the conducting boundary condition (\(\partial _{z} H_{h} = 0\)), the insulating boundary condition is less commonly encountered in practical engineering scenarios. For the conducting boundary condition, the well-posedness of the 3D MHD boundary layer equations in an analytic space will be addressed in a subsequent independent work [6].
3 The bounds of \(v^{r}\), \(v^{z}\), ϕ and \(b^{r}\), \(b^{z}\), ψ
In this section, we establish bounds for \(v^{r}\), \(v^{z}\), ϕ in terms of g and bounds for \(b^{r}\), \(b^{z}\), ψ in terms of g̃ using three lemmas that were proven in [36]. As such, we omit their proofs here.
Lemma 3.1
Let \((v^{r},v^{z},b^{r},b^{z})\) be the solution of (2.9), \((\phi ,\psi )\) and g be respectively the functions defined in (2.6) and (2.8). For any \(n \in \mathbb{N}\), \(|\alpha |=n\) and \(0\leq \lambda <1\), it holds where the notation \(|f|\) denotes the absolute value of the function \(|f|\).
$$\begin{aligned} \left \{ \textstyle\begin{array}{l} |\theta _{\lambda }\langle r \rangle ^{n} \partial _{h}^{\alpha } \phi | \lesssim _{\lambda }\theta _{\lambda - 1} \langle t \rangle ^{ \frac{1}{4}} \|\theta \langle r \rangle ^{n} \partial _{h}^{\alpha }g \|_{L_{z}^{2}}, \\ |\theta _{\lambda }\langle r \rangle ^{n} \partial _{h}^{\alpha } \psi | \lesssim _{\lambda }\theta _{\lambda - 1} \langle t \rangle ^{ \frac{1}{4}} \|\theta \langle r \rangle ^{n} \partial _{h}^{\alpha } \tilde{g}\|_{L_{z}^{2}}, \\ |\theta _{\lambda }\langle r \rangle ^{n} \partial _{h}^{\alpha }v^{r}| \lesssim _{\lambda }|\theta _{\lambda }\langle r \rangle ^{n} \partial _{h}^{\alpha }g| + \frac{z}{\langle t \rangle ^{\frac{3}{4}}} \theta _{\lambda - 1} \| \theta \langle r \rangle ^{n} \partial _{h}^{\alpha }g\|_{L_{z}^{2}}, \\ |\theta _{\lambda }\langle r \rangle ^{n} \partial _{h}^{\alpha }b^{r}| \lesssim _{\lambda }|\theta _{\lambda }\langle r \rangle ^{n} \partial _{h}^{\alpha }\tilde{g}| + \frac{z}{\langle t \rangle ^{\frac{3}{4}}} \theta _{\lambda - 1} \| \theta \langle r \rangle ^{n} \partial _{h}^{\alpha }\tilde{g}\|_{L_{z}^{2}}, \\ |\theta _{\lambda }\langle r \rangle ^{n} \partial _{h}^{\alpha } \partial _{z} v^{r}| \lesssim _{\lambda }\frac{z}{\langle t \rangle} | \theta _{\lambda }\langle r \rangle ^{n} \partial _{h}^{\alpha }g(z)| + |\theta _{\lambda }\langle r \rangle ^{n} \partial _{h}^{\alpha } \partial _{z} g| \\ \hphantom{|\theta _{\lambda }\langle r \rangle ^{n} \partial _{h}^{\alpha } \partial _{z} v^{r}|}\quad +\Big(\frac{1}{\langle t \rangle} + \frac{z^{2}}{\langle t \rangle ^{2}}\Big)\theta _{\lambda -1}\langle t \rangle ^{\frac{1}{4}}\|\theta \langle r \rangle ^{n} \partial _{h}^{ \alpha }g\|_{L_{z}^{2}}, \\ |\theta _{\lambda }\langle r \rangle ^{n} \partial _{h}^{\alpha } \partial _{z} b^{r}| \lesssim _{\lambda }\frac{z}{\langle t \rangle} | \theta _{\lambda }\langle r \rangle ^{n} \partial _{h}^{\alpha } \tilde{g}(z)| + |\theta _{\lambda }\langle r \rangle ^{n} \partial _{h}^{ \alpha }\partial _{z} \tilde{g}| \\ \hphantom{|\theta _{\lambda }\langle r \rangle ^{n} \partial _{h}^{\alpha } \partial _{z} b^{r}|}\quad +\Big(\frac{1}{\langle t \rangle} + \frac{z^{2}}{\langle t \rangle ^{2}}\Big)\theta _{\lambda -1}\langle t \rangle ^{\frac{1}{4}}\|\theta \langle r \rangle ^{n} \partial _{h}^{ \alpha }\tilde{g}\|_{L_{z}^{2}}, \end{array}\displaystyle \right . \end{aligned}$$
(3.1)
Lemma 3.2
(Bounds of \(v^{r}\), \(v^{z}\), ϕ in terms of g) For any \(n\in \mathbb{N}\), \(|\alpha |= n\) and \(0\leq \lambda <1\), it holds
$$\begin{aligned} \left \{ \textstyle\begin{array}{l} \|\theta _{\lambda }\langle r \rangle ^{n} \partial _{h}^{\alpha } \phi \|_{L_{z}^{2}} \lesssim _{\lambda }\langle t \rangle ^{ \frac{1}{2}} \|\theta \langle r \rangle ^{n} \partial _{h}^{\alpha }g \|_{L_{z}^{2}}, \\ \|\theta _{\lambda }\langle r \rangle ^{n} \partial _{h}^{\alpha }v^{r} \|_{L^{2}} \lesssim _{\lambda }\|\theta \langle r \rangle ^{n} \partial _{h}^{\alpha }g\|_{L^{2}}, \\ \sum \limits _{|\alpha | = n} \|\theta _{\lambda }\langle r \rangle ^{n} \partial _{h}^{\alpha }v^{r}\|_{L_{h}^{\infty }L_{z}^{2}} \lesssim _{ \lambda }(n + 1)^{2} \sum \limits _{|\alpha | = n}^{n + 2} \|\theta \langle r \rangle ^{|\alpha |} \partial _{h}^{\alpha }g\|_{L^{2}}, \\ \|\theta _{\lambda }\langle r \rangle ^{n} \partial _{h}^{\alpha }v^{r} \|_{L_{h}^{2} L_{z}^{\infty}} \lesssim _{\lambda }\|\theta \langle r \rangle ^{n} \partial _{h}^{\alpha }(g, \partial _{z} g)\|_{L^{2}}, \\ \sum \limits _{|\alpha | = n} \|\theta _{\lambda }\langle r \rangle ^{n} \partial _{h}^{\alpha }v^{r}\|_{L_{h}^{\infty }L_{z}^{\infty}} \lesssim _{\lambda }(n + 1)^{2} \sum \limits _{|\alpha | = n}^{n + 2} \|\theta \langle r \rangle ^{|\alpha |} \partial _{h}^{\alpha }(g, \partial _{z} g)\|_{L^{2}}, \\ \|\theta _{\lambda }\langle r \rangle ^{n} \partial _{h}^{\alpha }v^{z} \|_{L_{h}^{2} L_{z}^{\infty}} \lesssim _{\lambda }\langle t \rangle ^{ \frac{1}{4}} \|\theta \langle r \rangle ^{n} \partial _{h}^{\alpha }(r \partial _{r} g, g)\|_{L^{2}}, \\ \sum \limits _{|\alpha | = n} \|\theta _{\lambda }\langle r \rangle ^{n} \partial _{h}^{\alpha }v^{z}\|_{L_{h}^{\infty }L_{z}^{\infty}} \lesssim _{\lambda }(n + 1)^{2} \langle t \rangle ^{\frac{1}{4}} \sum \limits _{|\alpha | = n}^{n + 2} \|\theta \langle r \rangle ^{| \alpha |} \partial _{h}^{\alpha }(r \partial _{r} g, g)\|_{L^{2}}, \\ \|\theta _{\lambda }\langle r \rangle ^{n} \partial _{h}^{\alpha } \partial _{z} v^{r}\|_{L^{2}} \lesssim _{\lambda }\langle t \rangle ^{- \frac{1}{2}} \|\theta \langle r \rangle ^{n} \partial _{h}^{\alpha }g \|_{L^{2}} + \|\theta \langle r \rangle ^{n} \partial _{h}^{\alpha } \partial _{z} g\|_{L^{2}}, \\ \sum \limits _{|\alpha |=n} \|\theta _{\lambda }\langle r \rangle ^{n} \partial _{h}^{\alpha }\partial _{z} v^{r}\|_{L_{h}^{\infty }L_{z}^{2}} \lesssim _{\lambda}(n+1)^{2} \sum \limits _{|\alpha | = n}^{n + 2} \Big(\langle t \rangle ^{-\frac{1}{2}}\|\theta \langle r \rangle ^{| \alpha |}\partial _{h}^{\alpha }g\|_{L^{2}}+\|\theta \langle r \rangle ^{|\alpha |}\partial _{h}^{\alpha }\partial _{z} g\|_{L^{2}} \Big). \end{array}\displaystyle \right . \end{aligned}$$
(3.2)
Lemma 3.3
(Bounds of \(b^{r}\), \(b^{z}\), ψ in terms of g̃) For any \(n\in \mathbb{N}\), \(|\alpha |= n\) and \(0\leq \lambda <1\), it holds
$$\begin{aligned} \left \{ \textstyle\begin{array}{l} \|\theta _{\lambda }\langle r \rangle ^{n} \partial _{h}^{\alpha } \psi \|_{L_{z}^{2}} \lesssim _{\lambda }\langle t \rangle ^{ \frac{1}{2}} \|\theta \langle r \rangle ^{n} \partial _{h}^{\alpha } \tilde{g}\|_{L_{z}^{2}}, \\ \|\theta _{\lambda }\langle r \rangle ^{n} \partial _{h}^{\alpha }b^{r} \|_{L^{2}} \lesssim _{\lambda }\|\theta \langle r \rangle ^{n} \partial _{h}^{\alpha }\tilde{g}\|_{L^{2}}, \\ \sum \limits _{|\alpha | = n} \|\theta _{\lambda }\langle r \rangle ^{n} \partial _{h}^{\alpha }b^{r}\|_{L_{h}^{\infty }L_{z}^{2}} \lesssim _{ \lambda }(n+1)^{2} \sum \limits _{|\alpha | = n}^{n + 2} \|\theta \langle r \rangle ^{|\alpha |} \partial _{h}^{\alpha }\tilde{g}\|_{L^{2}}, \\ \|\theta _{\lambda }\langle r \rangle ^{n} \partial _{h}^{\alpha }b^{r} \|_{L_{h}^{2} L_{z}^{\infty}} \lesssim _{\lambda }\|\theta \langle r \rangle ^{n} \partial _{h}^{\alpha }(\tilde{g}, \partial _{z} \tilde{g})\|_{L^{2}}, \\ \sum \limits _{|\alpha | = n} \|\theta _{\lambda }\langle r \rangle ^{n} \partial _{h}^{\alpha }b^{r}\|_{L_{h}^{\infty }L_{z}^{\infty}} \lesssim _{\lambda }(n + 1)^{2} \sum \limits _{|\alpha | = n}^{n + 2} \|\theta \langle r \rangle ^{|\alpha |} \partial _{h}^{\alpha }( \tilde{g}, \partial _{z} \tilde{g})\|_{L^{2}}, \\ \|\theta _{\lambda }\langle r \rangle ^{n} \partial _{h}^{\alpha }b^{z} \|_{L_{h}^{2} L_{z}^{\infty}} \lesssim _{\lambda }\langle t \rangle ^{ \frac{1}{4}} \|\theta \langle r \rangle ^{n} \partial _{h}^{\alpha }(r \partial _{r} \tilde{g}, \tilde{g})\|_{L^{2}}, \\ \sum \limits _{|\alpha | = n} \|\theta _{\lambda }\langle r \rangle ^{n} \partial _{h}^{\alpha }b^{z}\|_{L_{h}^{\infty }L_{z}^{\infty}} \lesssim _{\lambda }(n + 1)^{2} \langle t \rangle ^{\frac{1}{4}} \sum \limits _{|\alpha | = n}^{n + 2} \|\theta \langle r \rangle ^{| \alpha |} \partial _{h}^{\alpha }(r \partial _{r} \tilde{g}, \tilde{g})\|_{L^{2}}, \\ \|\theta _{\lambda }\langle r \rangle ^{n} \partial _{h}^{\alpha } \partial _{z} b^{r}\|_{L^{2}} \lesssim _{\lambda }\langle t \rangle ^{- \frac{1}{2}} \|\theta \langle r \rangle ^{n} \partial _{h}^{\alpha } \tilde{g}\|_{L^{2}} + \|\theta \langle r \rangle ^{n} \partial _{h}^{ \alpha }\partial _{z} \tilde{g}\|_{L^{2}}, \\ \sum \limits _{|\alpha |=n} \|\theta _{\lambda }\langle r \rangle ^{n} \partial _{h}^{\alpha }\partial _{z} b^{r}\|_{L_{h}^{\infty }L_{z}^{2}} \lesssim _{\lambda}(n+1)^{2} \sum \limits _{|\alpha | = n}^{n + 2} \Big(\langle t \rangle ^{-\frac{1}{2}}\|\theta \langle r \rangle ^{| \alpha |}\partial _{h}^{\alpha }\tilde{g}\|_{L^{2}}+\|\theta \langle r \rangle ^{|\alpha |}\partial _{h}^{\alpha }\partial _{z} \tilde{g}\|_{L^{2}} \Big). \end{array}\displaystyle \right . \end{aligned}$$
(3.3)
4 The proof of the existence part of Theorem 2.2
In this section, we establish the a priori estimates of solutions needed to prove the existence part of Theorem 2.2. Proposition 4.1 provides weighted energy estimates for the good unknown g.
Proposition 4.1
For small enough \(\varepsilon \le \varepsilon _{1} \), assume initial data \(\mathbf{g}_{0}(r,z)\) satisfy Then the axially symmetric MHD boundary layer equations (2.9) have a globally in-time solution g satisfies with the analyticity radius \(\tau (t)\ge \frac{1}{2}\tau _{0}\) for any \(t>0\).
$$\begin{aligned} \|\mathbf{g}_{0}\|_{\mathcal{X}_{\tau _{0}}} \le \varepsilon . \end{aligned}$$
$$\begin{aligned} &\langle t \rangle ^{\frac{5}{4}-\delta} \|\mathbf{g}\|_{\mathcal{X}_{ \tau}} + \frac{\delta}{12} \int _{0}^{t} \big( \langle s \rangle ^{ \frac{1}{4} - \delta} \|\mathbf{g}\|_{\mathcal{X}_{\tau}} + \langle s \rangle ^{\frac{3}{4}- \delta} \|\mathbf{g}\|_{\mathcal{D}_{\tau}} \big) ds \\ &+ C_{0} \int _{0}^{t} \frac{\langle s \rangle ^{\frac{5}{4} - \delta}}{\tau ^{2}(s)} \big( \|\mathbf{g}\|_{\mathcal{X}_{\tau}} + \langle s \rangle ^{\frac{1}{4}} \|\mathbf{g}\|_{\mathcal{D}_{\tau}} \big) \|\mathbf{g}\|_{\mathcal{Y}_{ \tau}} ds \leq \|\mathbf{g}_{0}\|_{\mathcal{X}_{\tau _{0}}} \leq \varepsilon \end{aligned}$$
(4.1)
First, we rewrite the equations (2.9) as follows
$$\begin{aligned} \left \{ \textstyle\begin{array}{l} \partial _{t}g-\partial _{z}^{2}g+\frac{1}{\langle t\rangle}g=-(rv^{r} \partial _{r}+v^{z}\partial _{z})g+(rb^{r}\partial _{r}+b^{z} \partial _{z})\tilde{g}+\frac{1}{2\langle t\rangle} v^{z}\partial _{z}(z \phi )-\frac{1}{2\langle t\rangle} b^{z}\partial _{z}(z\psi ) \\ -\frac{z}{\langle t \rangle}v^{r}\phi +\frac{z}{\langle t \rangle}b^{r} \psi -\frac{z}{2\langle t\rangle}\int _{z}^{+\infty}(v^{r})^{2} \mathrm{d}\bar{z}+\frac{z}{\langle t\rangle}\int _{z}^{+\infty} \partial _{z} v^{r} v^{z}\mathrm{d}\bar{z} \\ -(v^{r})^{2}+(b^{r})^{2}+\frac{z}{2\langle t\rangle}\int _{z}^{+ \infty}(b^{r})^{2}\mathrm{d}\bar{z}-\frac{z}{\langle t\rangle}\int _{z}^{+ \infty}\partial _{z}b^{r}b^{z}\mathrm{d}\bar{z}= 0, \\ \partial _{t}\tilde{g}-\partial _{z}^{2}\tilde{g}+ \frac{1}{\langle t\rangle}\tilde{g}=-(rv^{r}\partial _{r}+v^{z} \partial _{z})\tilde{g}+(rb^{r}\partial _{r}+b^{z}\partial _{z})g \\ +\frac{1}{2\langle t\rangle} v^{z}\partial _{z}(z\psi )- \frac{1}{2\langle t\rangle} b^{z}\partial _{z}(z\phi )- \frac{z}{\langle t \rangle}v^{r}\psi +\frac{z}{\langle t \rangle}b^{r} \phi . \end{array}\displaystyle \right . \end{aligned}$$
(4.2)
Let \(n\geq 0\) and \(|\alpha |=n\). Applying \(\langle r\rangle ^{n}\partial _{h}^{\alpha}\) to (4.2)1,2 and multiplying the resulted equations with \(\theta ^{2}\langle r\rangle ^{n} \partial _{h}^{\alpha }g\) and \(\theta ^{2}\langle r\rangle ^{n} \partial _{h}^{\alpha}\tilde{g}\), respectively, and then integrating over \(\mathbb{R}_{+}^{3}\), we give
$$\begin{aligned} \left \{ \textstyle\begin{array}{l@{\quad}l} &\frac{1}{2} \frac{\mathrm{d}}{\mathrm{dt}} \|\theta \langle r \rangle ^{n} \partial _{h}^{\alpha }g\|_{L^{2}}^{2} + \|\theta \langle r \rangle ^{n} \partial _{h}^{\alpha }\partial _{z} g\|_{L^{2}}^{2} + \frac{3}{4\langle t \rangle} \|\theta \langle r \rangle ^{n} \partial _{h}^{\alpha }g\|_{L^{2}}^{2} \\ &= - \int \theta \langle r \rangle ^{n} \partial _{h}^{\alpha }(v^{r} r \partial _{r} g) \theta \langle r \rangle ^{n} \partial _{h}^{\alpha }g+ \int \theta \langle r \rangle ^{n} \partial _{h}^{\alpha }(v^{r} r \partial _{r} \tilde{g})\theta \langle r \rangle ^{n} \partial _{h}^{ \alpha }g \\ &- \int \theta \langle r \rangle ^{n} \partial _{h}^{\alpha }(v^{z} \partial _{z} g) \theta \langle r \rangle ^{n} \partial _{h}^{\alpha }g+ \theta \langle r \rangle ^{n} \partial _{h}^{\alpha }g+\int \theta \langle r \rangle ^{n} \partial _{h}^{\alpha }(v^{z} \partial _{z} \tilde{g})\theta \langle r\rangle ^{n} \partial _{h}^{\alpha }g \\ &- \int \theta \langle r \rangle ^{n} \partial _{h}^{\alpha }(v^{r})^{2} \theta \langle r \rangle ^{n} \partial _{h}^{\alpha }g +\int \theta \langle r\rangle ^{n}\partial _{h}^{\alpha }(b^{r})^{2} \theta \langle r\rangle ^{n}\partial _{h}^{\alpha }g \\ &+ \frac{1}{2\langle t \rangle} \int \theta \langle r \rangle ^{n} \partial _{h}^{\alpha }(v^{z} \partial _{z} (z \phi )) \theta \langle r \rangle ^{n} \partial _{h}^{\alpha }g - \frac{1}{2\langle t\rangle}\int \theta \langle r\rangle ^{n}\partial _{h}^{ \alpha}(b^{z} \partial _{z}(z \psi ) \theta \langle r\rangle ^{n} \partial _{h}^{\alpha }g \\ &- \frac{1}{\langle t \rangle} \int z \theta \langle r \rangle ^{n} \partial _{h}^{\alpha }(v^{r} \phi ) \theta \langle r \rangle ^{n} \partial _{h}^{\alpha }g +\frac{1}{\langle t\rangle}\int z\theta \langle r\rangle ^{n}\partial _{h}^{\alpha }(b^{r}\psi )\theta \langle r\rangle ^{n}\partial _{h}^{\alpha }g \\ &- \frac{1}{2\langle t \rangle} \int z \theta \int _{z}^{\infty } \langle r \rangle ^{n} \partial _{h}^{\alpha }(v^{r})^{2} \mathrm{d} \bar{z} \theta \langle r \rangle ^{n} \partial _{h}^{\alpha }g+ \frac{1}{2\langle t\rangle}\int z\theta \int _{z}^{\infty }\langle r \rangle ^{n}\partial _{h}^{\alpha }(b^{r})^{2}\mathrm{d}\bar{z} \theta \langle r\rangle ^{n} \partial _{h}^{\alpha }g \\ &+ \frac{1}{\langle t \rangle} \int z \theta \int _{z}^{\infty } \langle r \rangle ^{n} \partial _{h}^{\alpha }(\partial _{z} v^{r} v^{z}) \mathrm{d}\bar{z} \theta \langle r \rangle ^{n} \partial _{h}^{ \alpha }g -\frac{1}{\langle t \rangle}\int z\theta \int _{z}^{\infty} \langle r\rangle ^{n}\partial _{h}^{\alpha}(\partial _{z}b^{r}b^{z}) \mathrm{d}\bar{z}\theta \langle r\rangle ^{n}\partial _{h}^{\alpha }g \\ &:= \sum \limits _{j=1}^{7} I_{j}, \\ &\frac{1}{2} \frac{\mathrm{d}}{\mathrm{dt}} \|\theta \langle r \rangle ^{n} \partial _{h}^{\alpha }\tilde{g}\|_{L^{2}}^{2} + \| \theta \langle r \rangle ^{n} \partial _{h}^{\alpha }\partial _{z} \tilde{g}\|_{L^{2}}^{2} + \frac{3}{4\langle t \rangle} \|\theta \langle r \rangle ^{n} \partial _{h}^{\alpha }\tilde{g}\|_{L^{2}}^{2} \\ &= - \int \theta \langle r \rangle ^{n} \partial _{h}^{\alpha }(v^{r} r \partial _{r} \tilde{g}) \theta \langle r \rangle ^{n} \partial _{h}^{ \alpha }\tilde{g}+\int \theta \langle r \rangle ^{n} \partial _{h}^{ \alpha }(b^{r} r \partial _{r} g) \theta \langle r \rangle ^{n} \partial _{h}^{\alpha }\tilde{g} \\ & - \int \theta \langle r \rangle ^{n} \partial _{h}^{\alpha }(v^{z} \partial _{z} \tilde{g}) \theta \langle r \rangle ^{n} \partial _{h}^{ \alpha }\tilde{g} +\int \theta \langle r \rangle ^{n} \partial _{h}^{ \alpha }(b^{z} \partial _{z} g)\theta \langle r\rangle ^{n} \partial _{h}^{ \alpha }\tilde{g} \\ &+ \frac{1}{2\langle t \rangle} \int \theta \langle r \rangle ^{n} \partial _{h}^{\alpha }(v^{z} \partial _{z} (z \psi )) \theta \langle r \rangle ^{n} \partial _{h}^{\alpha }\tilde{g} - \frac{1}{2\langle t\rangle}\int \theta \langle r\rangle ^{n}\partial _{h}^{ \alpha}(b^{z} \partial _{z}(z \phi ) \theta \langle r\rangle ^{n} \partial _{h}^{\alpha }\tilde{g} \\ &- \frac{1}{\langle t \rangle} \int z \theta \langle r \rangle ^{n} \partial _{h}^{\alpha }(v^{r} \psi ) \theta \langle r \rangle ^{n} \partial _{h}^{\alpha }\tilde{g} +\frac{1}{\langle t\rangle}\int z \theta \langle r\rangle ^{n}\partial _{h}^{\alpha }(b^{r}\phi ) \theta \langle r\rangle ^{n}\partial _{h}^{\alpha }\tilde{g} \\ &:= \sum \limits _{j=1}^{4} K_{j}. \end{array}\displaystyle \right . \end{aligned}$$
(4.3)
For a function \(f(t, x, y, z) \), we use the simplified notation ∫f to represent \(\int _{\mathbb{R}_{+}^{3}} f(t, x, y, z) \mathrm{d}x \mathrm{d}y \mathrm{d}z \) in cases where no ambiguity arises.
If we divide the first of the above equalities by \(\|\theta \langle r \rangle ^{n} \partial _{h}^{\alpha }g\|_{L^{2}}\) and the second by \(\|\theta \langle r \rangle ^{n} \partial _{h}^{\alpha }\tilde{g}\|_{L^{2}}\), then multiply each resulting equations by \(\tau ^{n}(t) M_{n}\) and sum over all \(|\alpha | = n\), we can deduce that for \(n \geq 0\), when \(n = 0 \), we set \(Y_{0} = 0 \).
$$\begin{aligned} \left \{ \textstyle\begin{array}{l} \displaystyle \frac{\mathrm{d}}{\mathrm{dt}} X_{n}(g)+ \sum \limits _{| \alpha | = n} \frac{\|\theta \langle r \rangle ^{n} \partial _{h}^{\alpha }\partial _{z} g\|_{L^{2}}^{2}}{\|\theta \langle r \rangle ^{n} \partial _{h}^{\alpha }g\|_{L^{2}}} + \frac{3}{4\langle t \rangle} X_{n}(g)\\ \displaystyle\quad = \dot{\tau}(t) Y_{n}(g)+ \sum \limits _{|\alpha | = n} \frac{\tau ^{n}(t) M_{n}}{\|\theta \langle r \rangle ^{n} \partial _{h}^{\alpha }g\|_{L^{2}}} \sum \limits _{j = 1}^{7} I_{j}, \\ \displaystyle \frac{\mathrm{d}}{\mathrm{dt}} X_{n}(\tilde{g})+ \sum \limits _{|\alpha | = n} \frac{\|\theta \langle r \rangle ^{n} \partial _{h}^{\alpha }\partial _{z} \tilde{g}\|_{L^{2}}^{2}}{\|\theta \langle r \rangle ^{n} \partial _{h}^{\alpha }\tilde{g}\|_{L^{2}}} + \frac{3}{4\langle t \rangle} X_{n}(\tilde{g})\\ \displaystyle\quad = \dot{\tau}(t) Y_{n}( \tilde{g}) + \sum \limits _{|\alpha | = n} \frac{\tau ^{n}(t) M_{n}}{\|\theta \langle r \rangle ^{n} \partial _{h}^{\alpha }\tilde{g}\|_{L^{2}}} \sum \limits _{j = 1}^{4} K_{j}, \end{array}\displaystyle \right . \end{aligned}$$
(4.4)
The following lemma introduces the quantitative relation between \(\|\theta \langle r\rangle ^{n}\partial _{h}^{n}\partial _{z}g\|_{L_{z}^{2}}^{2}\) and
\(\|\theta \langle r\rangle ^{n}\partial _{h}^{n} g \|_{L_{z}^{2}}^{2}\), which proofs can be found in [36], we omit the details.
Lemma 4.2
(\(Poincar\acute{e}\) inequality with gaussian weights) Let g be such that \(\partial _{z} g\big|_{z=0}=0\) and \(g\big|_{z=\infty}=0 \). For \(n\in \mathbb{N}\), and \(t\geqq 0\), it holds that
$$\begin{aligned} \frac{1}{2\langle t \rangle} \| \theta \langle r\rangle ^{n}\partial _{h}^{n} g \|_{L_{z}^{2}}^{2} \leqq \| \theta \langle r\rangle ^{n}\partial _{z} \partial _{h}^{n} g \|_{L_{z}^{2}}^{2}. \end{aligned}$$
(4.5)
Now, employing (4.5), it follows from (4.4) that
$$\begin{aligned} \left \{ \textstyle\begin{array}{l} \displaystyle \frac{\mathrm{d}}{\mathrm{dt}} X_{n}(g)+ \displaystyle \frac{1}{\sqrt{2\langle t\rangle}}D(g)+ \frac{3}{4\langle t \rangle} X_{n}(g) \leq \dot{\tau}(t) Y_{n}(g)+ \sum \limits _{|\alpha | = n} \frac{\tau ^{n}(t) M_{n}}{\|\theta \langle r \rangle ^{n} \partial _{h}^{\alpha }g\|_{L^{2}}} \sum \limits _{j = 1}^{7} I_{j}, \\ \displaystyle \frac{\mathrm{d}}{\mathrm{dt}} X_{n}(\tilde{g})+ \displaystyle \frac{1}{\sqrt{2\langle t\rangle}}D(\tilde{g})+ \frac{3}{4\langle t \rangle} X_{n}(\tilde{g})\leq \dot{\tau}(t) Y_{n}( \tilde{g}) + \sum \limits _{|\alpha | = n} \frac{\tau ^{n}(t) M_{n}}{\|\theta \langle r \rangle ^{n} \partial _{h}^{\alpha }\tilde{g}\|_{L^{2}}} \sum \limits _{j = 1}^{4} K_{j}. \end{array}\displaystyle \right . \end{aligned}$$
(4.6)
Proposition 4.3
(Estimates of the nonlinear terms) For the nonlinear terms in (4.6), we have the following estimate
$$\begin{aligned} \left \{ \textstyle\begin{array}{l} \sum \limits _{n\ge 0} \sum \limits _{|\alpha |=n} \frac{\tau ^{n}(t) M_{n}}{\|\theta \langle r \rangle ^{n} \partial _{h}^{\alpha }g\|_{L^{2}}} \sum \limits _{j=1}^{7} I_{j} \\ \le C \tau ^{-2}(t) \bigl( \|\mathbf{g}\|_{\mathcal{X}_{\tau}} + \langle t \rangle ^{1/4} \|\mathbf{g}\|_{\mathcal{D}_{\tau}} \bigr) \|\mathbf{g}\|_{\mathcal{Y}_{\tau}} \\ + C \tau ^{-2}(t) \bigl( \|\mathbf{g}\|_{\mathcal{X}_{\tau}} + \langle t \rangle ^{1/4} \|\mathbf{g}\|_{\mathcal{D}_{\tau}} \bigr) \|\mathbf{g}\|_{\mathcal{X}_{\tau}}, \\ \sum \limits _{n\ge 0} \sum \limits _{|\alpha |=n} \frac{\tau ^{n}(t) M_{n}}{\|\theta \langle r \rangle ^{n} \partial _{h}^{\alpha }\tilde{g}\|_{L^{2}}} \sum \limits _{j=1}^{4} K_{j} \\ \le C \tau ^{-2}(t) \bigl( \|\mathbf{g}\|_{\mathcal{X}_{\tau}} + \langle t \rangle ^{1/4} \|\mathbf{g}\|_{\mathcal{D}_{\tau}} \bigr) \|\mathbf{g}\|_{\mathcal{Y}_{\tau}} \\ + C \tau ^{-2}(t) \bigl( \|\mathbf{g}\|_{\mathcal{X}_{\tau}} + \langle t \rangle ^{1/4} \|\mathbf{g}\|_{\mathcal{D}_{\tau}} \bigr) \|\mathbf{g}\|_{\mathcal{X}_{\tau}}. \end{array}\displaystyle \right . \end{aligned}$$
(4.7)
We defer the proof of Proposition 4.3 to Sect. 5 and proceed with proving the a priori estimate in Proposition 4.1.
Proof of Proposition 4.1
From (4.6), by summing on \(n\geq 0\) and inserting (4.7) into (4.6), we have for a uniform constant \(C_{0}\geq 1\)
$$\begin{aligned} &\frac{d}{dt} \|\mathbf{g}\|_{\mathcal{X}_{\tau}} + \frac{1}{\sqrt{2\langle t \rangle}} \|\mathbf{g}\|_{\mathcal{D}_{\tau}} + \frac{3}{4\langle t \rangle} \|\mathbf{g} \|_{\mathcal{X}_{\tau}} \\ &\leq (\dot{\tau} + C_{0} \tau ^{-2}(t) \big( \|\mathbf{g}\|_{ \mathcal{X}_{\tau}} + \langle t \rangle ^{\frac{1}{4}} \|\mathbf{g}\|_{ \mathcal{D}_{\tau}} \big)) \|\mathbf{g} \|_{\mathcal{Y}_{\tau}} \\ &+ C_{0} \tau ^{-2}(t) \big( \|\mathbf{g}\|_{\mathcal{X}_{\tau}} + \langle t \rangle ^{\frac{1}{4}} \|\mathbf{g}\|_{\mathcal{D}_{\tau}} \big) \|\mathbf{g}\|_{\mathcal{X}_{\tau}}. \end{aligned}$$
(4.8)
By applying (4.5), for any small \(\delta _{1} > 0 \), we arrive at
$$\begin{aligned} &\frac{1}{\sqrt{2\langle t \rangle}} \| g \|_{\mathcal{D}_{\tau}} = \frac{\delta _{1}}{\sqrt{2\langle t \rangle}} \| g \|_{\mathcal{D}_{ \tau}} + \frac{(1 - \delta _{1})}{\sqrt{2\langle t \rangle}} \| g \|_{ \mathcal{D}_{\tau}} \\ &\geq \frac{1\delta _{1}}{\sqrt{2\langle t \rangle}} \| g \|_{ \mathcal{D}_{\tau}} + \frac{(1 - \delta _{1})}{2\langle t \rangle} \| g \|_{\mathcal{X}_{\tau}} \\ &\geq \frac{\delta _{1}}{\sqrt{2\langle t \rangle}} \| g \|_{ \mathcal{D}_{\tau}} + \frac{\delta _{1}}{\langle t \rangle} \| g \|_{ \mathcal{X}_{\tau}} + \frac{1 - 3\delta _{1}}{2\langle t \rangle} \| g \|_{\mathcal{X}_{\tau}}. \end{aligned}$$
(4.9)
Similarly, we also obtain
$$\begin{aligned} \frac{1}{\sqrt{2\langle t \rangle}} \| \tilde{g} \|_{\mathcal{D}_{ \tau}} \geq \frac{\delta _{1}}{\sqrt{2\langle t \rangle}} \| \tilde{g} \|_{\mathcal{D}_{\tau}} + \frac{\delta _{1}}{\langle t \rangle} \| \tilde{g} \|_{\mathcal{X}_{ \tau}} + \frac{1 - 3\delta _{1}}{2\langle t \rangle} \| \tilde{g} \|_{ \mathcal{X}_{\tau}}. \end{aligned}$$
(4.10)
Plugging (4.9)-(4.10) into (4.8), we conclude that
$$\begin{aligned} &\frac{d}{dt} \|\mathbf{g}\|_{\mathcal{X}_{\tau}} + \frac{\frac{5}{4} - \frac{3}{2}\delta _{1}}{\langle t \rangle} \| \mathbf{g}\|_{\mathcal{X}_{\tau}}+\big( \frac{\delta _{1}}{\langle t \rangle} \|\mathbf{g}\|_{\mathcal{X}_{ \tau}} + \frac{\delta _{1}}{\sqrt{2\langle t \rangle}} \|\mathbf{g}\|_{ \mathcal{D}_{\tau}} \big) \\ &\leq (\dot{\tau} + C_{0} \tau ^{-2}(t) \big( \|\mathbf{g}\|_{ \mathcal{X}_{\tau}} + \langle t \rangle ^{\frac{1}{4}} \|\mathbf{g}\|_{ \mathcal{D}_{\tau}} \big)) \|\mathbf{g} \|_{\mathcal{Y}_{\tau}} \\ & + C_{0} \tau ^{-2}(t) \big( \|\mathbf{g}\|_{\mathcal{X}_{\tau}}+ \langle t \rangle ^{\frac{1}{4}} \|\mathbf{g}\|_{\mathcal{D}_{\tau}} \big) \|\mathbf{g} \|_{\mathcal{X}_{\tau}}. \end{aligned}$$
(4.11)
For \(\delta \in (0, \frac{1}{4}] \), by choosing \(\delta _{1} = \frac{\delta}{3} \), we have
$$\begin{aligned} &\frac{d}{dt} \|\mathbf{g} \|_{\mathcal{X}_{\tau}} + \frac{\frac{5}{4} - \frac{1}{2}\delta}{\langle t \rangle} \| \mathbf{g}\|_{\mathcal{X}_{\tau}} + \frac{\delta}{6} \big( \frac{1}{\langle t \rangle} \|\mathbf{g}\|_{\mathcal{X}_{\tau}} + \frac{1}{\sqrt{\langle t \rangle}} \|\mathbf{g}\|_{\mathcal{D}_{\tau}} \big) \\ &\leq (\dot{\tau} + C_{0} \tau ^{-2}(t) \big( \|\mathbf{g}\|_{ \mathcal{X}_{\tau}} + \langle t \rangle ^{\frac{1}{4}} \|\mathbf{g} \|_{\mathcal{D}_{\tau}} \big)) \|\mathbf{g} \|_{\mathcal{Y}_{\tau}} \\ &+ C_{0} \tau ^{-2}(t) \big( \|\mathbf{g} \|_{\mathcal{X}_{\tau}} + \langle t \rangle ^{\frac{1}{4}} \|\mathbf{g}\|_{\mathcal{D}_{\tau}} \big) \|\mathbf{g}\|_{\mathcal{X}_{\tau}}. \end{aligned}$$
(4.12)
Now, we make the a priori assumption that for any \(t > 0 \),
$$\begin{aligned} \langle t \rangle ^{\frac{5}{4} - \delta} \|\mathbf{g}\|_{\mathcal{X}_{ \tau}} \leq 2\varepsilon _{0}, \quad \tau (t) \geq \frac{1}{4}\tau _{0}. \end{aligned}$$
(4.13)
Applying the a priori assumption (4.13), and through selecting an appropriate \(\tau (t) \) and a sufficiently small \(\varepsilon _{0} \) (which depends on \(\tau _{0} \) and δ), we will demonstrate that
$$\begin{aligned} \langle t \rangle ^{\frac{5}{4} - \delta} \|\mathbf{g}\|_{\mathcal{X}_{ \tau}} \leq \varepsilon _{0}, \quad \tau (t) \geq \frac{1}{2}\tau _{0}. \end{aligned}$$
(4.14)
Then continuity argument ensures that (4.14) holds for any \(t > 0 \).
First, inserting (4.13) into (4.12), we obtain
$$\begin{aligned} &\frac{d}{dt} \|\mathbf{g}\|_{\mathcal{X}_{\tau}} + \frac{\frac{5}{4} - \frac{1}{2}\delta}{\langle t \rangle} \| \mathbf{g}\|_{\mathcal{X}_{\tau}} + \frac{\delta}{6} \big( \frac{1}{\langle t \rangle} \|\mathbf{g}\|_{\mathcal{X}_{\tau}} + \frac{1}{\sqrt{\langle t \rangle}} \|\mathbf{g}\|_{\mathcal{D}_{\tau}} \big) \\ &\leq (\dot{\tau} + C_{0} \tau ^{-2}(t) \big( \|\mathbf{g}\|_{ \mathcal{X}_{\tau}} + \langle t \rangle ^{\frac{1}{4}} \|\mathbf{g}\|_{ \mathcal{D}_{\tau}} \big)) \|\mathbf{g} \|_{\mathcal{Y}_{\tau}} \\ &+ \frac{32C_{0}\varepsilon _{0}}{\tau _{0}^{2} \langle t \rangle ^{\frac{5}{4} - \delta}} \big( \|\mathbf{g}\|_{\mathcal{X}_{\tau}} + \langle t \rangle ^{ \frac{1}{4}} \|\mathbf{g}\|_{\mathcal{D}_{\tau}} \big). \end{aligned}$$
(4.15)
By choosing \(\varepsilon _{0} \) such that \(\frac{32C_{0}\varepsilon _{0}}{\tau _{0}^{2}} < \frac{\delta}{12} \), we derive
$$\begin{aligned} &\frac{d}{dt} \|\mathbf{g}\|_{\mathcal{X}_{\tau}} + \frac{\frac{5}{4} - \frac{1}{2}\delta}{\langle t \rangle} \| \mathbf{g}\|_{\mathcal{X}_{\tau}} + \frac{\delta}{12} \big( \frac{1}{\langle t \rangle} \|\mathbf{g}\|_{\mathcal{X}_{\tau}} + \frac{1}{\sqrt{\langle t \rangle}} \|\mathbf{g}\|_{\mathcal{D}_{\tau}} \big) \\ &\leq (\dot{\tau} + C_{0} \tau ^{-2}(t) \big( \|\mathbf{g}\|_{ \mathcal{X}_{\tau}} + \langle t \rangle ^{\frac{1}{4}} \|\mathbf{g}\|_{ \mathcal{D}_{\tau}} \big)) \|\mathbf{g} \|_{\mathcal{Y}_{\tau}}. \end{aligned}$$
(4.16)
We next choose the function \(\tau (t) \) such that
$$\begin{aligned} \dot{\tau} + \frac{2C_{0}}{\tau ^{2}(t)} \big( \|\mathbf{g}\|_{ \mathcal{X}_{\tau}} + \langle t \rangle ^{\frac{1}{4}} \|\mathbf{g}\|_{ \mathcal{D}_{\tau}} \big) = 0. \end{aligned}$$
(4.17)
Then (4.16) implies that
$$\begin{aligned} &\frac{d}{dt} \big( \langle t \rangle ^{\frac{5}{4} - \delta} \| \mathbf{g}\|_{\mathcal{X}_{\tau}} \big) + \frac{\delta}{12} \big( \langle t \rangle ^{\frac{1}{4} - \delta} \|\mathbf{g} \|_{ \mathcal{X}_{\tau}} + \langle t \rangle ^{\frac{3}{4}-\delta} \| \mathbf{g}\|_{\mathcal{D}_{\tau}} \big) \\ &+ \frac{C_{0}}{\tau ^{2}(t)} \langle t \rangle ^{\frac{5}{4} - \delta} \big( \|\mathbf{g}\|_{\mathcal{X}_{\tau}} + \langle t \rangle ^{\frac{1}{4}} \|\mathbf{g}\|_{\mathcal{D}_{\tau}} \big) \| \mathbf{g}\|_{\mathcal{Y}_{\tau}} \leq 0. \end{aligned}$$
(4.18)
Integrating (4.18) in time t, we have which implies that
$$\begin{aligned} &\langle t \rangle ^{\frac{5}{4} - \delta} \|\mathbf{g}\|_{ \mathcal{X}_{\tau}} + \frac{\delta}{12} \int _{0}^{t} \big( \langle s \rangle ^{\frac{1}{4} - \delta} \|\mathbf{g}\|_{\mathcal{X}_{\tau}} + \langle s \rangle ^{\frac{3}{4}- \delta} \|\mathbf{g}\|_{\mathcal{D}_{ \tau}} \big) ds \\ &+ C_{0} \int _{0}^{t} \frac{\langle s \rangle ^{\frac{5}{4} - \delta}}{\tau ^{2}(s)} \big( \|\mathbf{g}\|_{\mathcal{X}_{\tau}} + \langle s \rangle ^{\frac{1}{4}} \|\mathbf{g}\|_{\mathcal{D}_{\tau}} \big) \|\mathbf{g}\|_{\mathcal{Y}_{ \tau}} ds \\ & \leq \|\mathbf{g}_{0}\|_{\mathcal{X}_{\tau _{0}}} \leq \varepsilon _{0}, \end{aligned}$$
(4.19)
$$\begin{aligned} \int _{0}^{t} \big( \langle s \rangle ^{\frac{1}{4} - \delta} \| \mathbf{g}\|_{\mathcal{X}_{\tau}} + \langle s \rangle ^{\frac{3}{4}- \delta} \|\mathbf{g}\|_{\mathcal{D}_{\tau}} \big) ds \leq \frac{12}{\delta}\varepsilon _{0}. \end{aligned}$$
Then from (4.17), we see that by choosing small \(\varepsilon _{0} \). Then by choosing small \(\varepsilon _{0}\in (0,\frac{7\delta \tau _{0}^{3}}{576C_{0}}] \), we obtain (4.14) and (4.19), which finishes the proof of Proposition 4.1.
$$\begin{aligned} &\tau ^{3}(t) = \tau _{0}^{3} - 6C_{0} \int _{0}^{t} \big( \| \mathbf{g}\|_{\mathcal{X}_{\tau}} + \langle s \rangle ^{\frac{1}{4}} \|\mathbf{g}\|_{\mathcal{D}_{\tau}} \big) ds \\ &\geq \tau _{0}^{3} - \frac{72C_{0}}{\delta}\varepsilon _{0} \geq \big( \frac{1}{2}\tau _{0} \big)^{3}, \end{aligned}$$
(4.20)
5 The estimates of the nonlinear terms
In this section, we derive estimates for the nonlinear terms on the right-hand side of (4.6). Upon summing over \(n \geq 0 \), we obtain the following tangentially analytic estimates for these nonlinear terms.
Lemma 5.1
(The estimates of the nonlinear terms) The following estimates for the nonlinear terms on the righthand of \(\textit{(4.3)}_{1}\) hold:
$$\begin{aligned} \left \{ \textstyle\begin{array}{l} \sum _{n \geq 0} \limits \sum _{|\alpha | = n}\limits \frac{|I_{1}| \tau ^{n}(t) M_{n}}{\|\theta \langle r\rangle ^{n} \partial _{h}^{\alpha }g\|_{L^{2}}} \lesssim \tau ^{-2} \big( \|g\|_{\mathcal{X}_{\tau}} + \|g\|_{ \mathcal{D}_{\tau}} \big) \|\mathbf{g}\|_{\mathcal{Y}_{\tau}}, \\ \sum _{n \geq 0} \limits \sum _{|\alpha | = n}\limits \frac{|I_{2}| \tau ^{n}(t) M_{n}}{\|\theta \langle r\rangle ^{n} \partial _{h}^{\alpha }g\|_{L^{2}}} \lesssim \tau ^{-2} \langle t \rangle ^{\frac{1}{4}} \big( \|g\|_{ \mathcal{X}_{\tau}} + \|g\|_{\mathcal{Y}_{\tau}} \big) \|\mathbf{g}\|_{ \mathcal{D}_{\tau}}, \\ \sum _{n \geq 0} \limits \sum _{|\alpha | = n}\limits \frac{|I_{3}| \tau ^{n}(t) M_{n}}{\|\theta \langle r\rangle ^{n} \partial _{h}^{\alpha }g\|_{L^{2}}} + \sum _{n \geq 0} \limits \sum _{|\alpha | = n} \limits \frac{|I_{6}| \tau ^{n}(t) M_{n}}{\|\theta (r)^{n} \partial _{h}^{\alpha }g\|_{L^{2}}} \lesssim \tau ^{-2} \big( \|\mathbf{g}\|_{\mathcal{X}_{\tau}} + \| \mathbf{g}\|_{\mathcal{D}_{\tau}} \big) \|\mathbf{g}\|_{\mathcal{X}_{ \tau}}, \\ \sum _{n \geq 0} \limits \sum _{|\alpha | = n}\limits \frac{|I_{4}| \tau ^{n}(t) M_{n}}{\|\theta \langle r\rangle ^{n} \partial _{h}^{\alpha }g\|_{L^{2}}} \lesssim \tau ^{-2} \langle t \rangle ^{-\frac{1}{4}} \big( \| \mathbf{g}\|_{\mathcal{X}_{\tau}} + \|\mathbf{g}\|_{\mathcal{Y}_{\tau}} \big) \|\mathbf{g}\|_{\mathcal{X}_{\tau}}, \\ \sum _{n \geq 0} \limits \sum _{|\alpha | = n}\limits \frac{|I_{5}| \tau ^{n}(t) M_{n}}{\|\theta \langle r\rangle ^{n} \partial _{h}^{\alpha }g\|_{L^{2}}} \lesssim \tau ^{-2} \big( \|\mathbf{g}\|_{\mathcal{X}_{\tau}}+\| \mathbf{g}\|_{\mathcal{D}_{\tau}} \big) \|\mathbf{g}\|_{\mathcal{X}_{ \tau}}, \\ \sum _{n \geq 0} \limits \sum _{|\alpha | = n}\limits \frac{|I_{7}| \tau ^{n}(t) M_{n}}{\|\theta \langle r\rangle ^{n} \partial _{h}^{\alpha }g\|_{L^{2}}} \\ \lesssim \tau ^{-2} \big( \langle t \rangle ^{-\frac{1}{4}} \| \mathbf{g}\|_{\mathcal{X}_{\tau}} + \langle t \rangle ^{\frac{1}{4}} \|\mathbf{g}\|_{\mathcal{D}_{\tau}} \big) \big( \|\mathbf{g}\|_{ \mathcal{X}_{\tau}} + \|\mathbf{g}\|_{\mathcal{Y}_{\tau}} \big). \end{array}\displaystyle \right . \end{aligned}$$
(5.1)
Proof
Building on the nonlinear term estimates from [36], we’ve made minor modifications in this paper. For the sake of completeness, we present the proof of this lemma below.
Before the proof, we show the following two inequalities, which have already been proved in [36], and it will be frequently used throughout the following proofs. For any \(k \in \mathbb{N} \), \(1 \leq p\), \(q \leq +\infty \),
$$\begin{aligned} \sum _{|\alpha | = k} \| \theta \langle r \rangle ^{k} \partial _{h}^{ \alpha }(r\partial _{r} g) \|_{L_{h}^{p} L_{z}^{q}} \lesssim \sum _{| \alpha | = k + 1} \| \theta \langle r \rangle ^{k + 1} \partial _{h}^{ \alpha }g \|_{L_{h}^{p} L_{z}^{q}} + k \sum _{|\alpha | = k} \| \theta \langle r \rangle ^{k} \partial _{h}^{\alpha }g \|_{L_{h}^{p} L_{z}^{q}}. \end{aligned}$$
(5.2)
For multi-indices α, β with \(\beta \leq \alpha \), we will frequently use for all sequences \(\{a_{\beta}\}\) and \(\{b_{\gamma}\}\).
$$\begin{aligned} \binom{\alpha}{\beta} \leq \frac{|\alpha |!}{|\beta |!}, \quad \sum _{| \alpha | = n} \sum _{\substack{|\beta | = k, \\ \beta \leq \alpha}} a_{ \beta }b_{\alpha - \beta} = \Big( \sum _{|\beta | = k} a_{\beta } \Big) \Big( \sum _{|\gamma | = n - k}b_{\gamma }\Big) \end{aligned}$$
(5.3)
Next, we are ready to prove Lemma 5.1. For the estimate of the term \(I_{1}\), by using (5.3), one has
$$\begin{aligned} &\sum _{|\alpha | = n} \frac{|I_{1}| \tau ^{n}(t) M_{n}}{\| \theta \langle r \rangle ^{n} \partial _{h}^{\alpha }g \|_{L^{2}}} \\ &\leq \tau ^{n}(t) M_{n} \sum _{k = 0}^{[ \frac{n}{2} ]} \binom{n}{k} \Big( \sum _{|\gamma | = n - k} \| \langle r \rangle ^{n-k} \partial _{h}^{\gamma }v^{r}\|_{L_{h}^{\infty }L_{z}^{2}}\Big) \Big( \sum _{|\beta | = k}\|\theta \langle r\rangle ^{k}\partial _{h}^{ \beta}(r\partial _{r} \mathbf{g})\|_{L_{h}^{\infty }L_{z}^{2}}\Big) \\ &+\tau ^{n}(t) M_{n} \sum _{k =[ \frac{n}{2} ]+1}^{n} \binom{n}{k} \Big(\sum _{|\gamma |= n-k}\|\langle r\rangle ^{n-k}\partial _{h}^{ \gamma }v^{r}\|_{L^{\infty}}\Big)\Big(\sum _{|\beta |=k} \| \theta \langle r\rangle ^{k}\partial _{h}^{\beta}(r\partial _{r} \mathbf{g}) \|_{L^{2}}\Big). \end{aligned}$$
Applying (3.2)4,5, and noting that \(M_{n}\binom{n}{k}=\frac{(n+1)^{4}}{(n-k)!k!}\), we get
$$\begin{aligned} &\sum _{|\alpha |=n} \frac{|I_{1}|\tau ^{n}(t) M_{n}}{\| \theta \langle r \rangle ^{n} \partial _{h}^{\alpha }g \|_{L^{2}}} \\ &\lesssim \tau ^{-2} \sum _{k = 0}^{[\frac{n}{2} ]} (X_{n-k}(g)+D_{n-k}(g)) \frac{\tau ^{k}}{k!} \sum _{|\beta |=k} \| \theta \langle r \rangle ^{k} \partial _{h}^{\beta }(r\partial _{r}\mathbf{g})\|_{L_{h}^{\infty }L_{z}^{2}} \\ &+ \tau ^{-2} \sum _{k=[\frac{n}{2}]+1}^{n}\sum _{i=0}^{2}(X_{n-k+i}(g)+D_{n-k+i}(g)) \frac{\tau ^{k}(k+1)^{4}}{k!}\sum _{|\beta |=k}\|\theta \langle r \rangle ^{k}\partial _{h}^{\beta}(r\partial _{r} \mathbf{g})\|_{L^{2}}. \end{aligned}$$
(5.4)
For the estimate of the term \(\sum \limits _{|\beta |=k} \| \theta \langle r \rangle ^{k} \partial _{h}^{\beta }(r\partial _{r}\mathbf{g})\|_{L_{h}^{\infty }L_{z}^{2}}\), employing the two dimensional Sobolev embedding inequality we obtain
$$\begin{aligned} \| f \|_{L_{h}^{\infty}} \lesssim \| f \|_{L_{h}^{2}} + \| \partial _{h}^{2} f \|_{L_{h}^{2}}, \end{aligned}$$
(5.5)
$$\begin{aligned} \bigl\| \theta _{\lambda }\langle r \rangle ^{k} \partial _{h}^{ \alpha }g\bigr\| _{L_{h}^{\infty }L_{z}^{2}} \lesssim \bigl\| \theta _{ \lambda }\langle r \rangle ^{k} \partial _{h}^{\alpha }g \bigr\| _{L_{h}^{2} L_{z}^{2}} +\bigl\| \theta _{\lambda}\partial _{h}^{2}\bigl[\langle r \rangle ^{k} \partial _{h}^{\alpha }g \bigr] \bigr\| _{L_{h}^{2} L_{z}^{2}}. \end{aligned}$$
(5.6)
It is easy to check that for \(k\in \mathbb{N}\setminus \{0\}\),
$$\begin{aligned} \bigl| \partial _{h}^{2} \bigl[\langle r \rangle ^{k} \partial _{h}^{ \alpha }g \bigr] \bigr| \lesssim \frac{(k+1)^{2}}{\langle r \rangle ^{2}}\sum \limits _{|\gamma |=0}^{2} \bigl| \langle r \rangle ^{k+ |\gamma |}\partial _{h}^{\alpha +\gamma}g \bigr|. \end{aligned}$$
(5.7)
Inserting (5.7) into (5.6) and summing over \(|\alpha |=k\), we deduce
$$\begin{aligned} \sum _{|\alpha |=k} \bigl\| \theta _{\lambda }\langle r \rangle ^{k} \partial _{h}^{\alpha }g\bigr\| _{L_{h}^{\infty }L_{z}^{2}} \lesssim (k+1)^{2} \sum \limits _{|\alpha |=k}^{k+2} \bigl\| \theta _{\lambda }\langle r \rangle ^{|\alpha |} \partial _{h}^{\alpha }g\bigr\| _{L^{2}}. \end{aligned}$$
(5.8)
Combining (5.2) with (5.8), one implies
$$\begin{aligned} &\sum _{|\beta |=k}\|\theta \langle r\rangle ^{k}\partial _{h}^{ \beta }(r\partial _{r}\mathbf{g}) \|_{L_{h}^{\infty }L_{z}^{2}} \\ &\lesssim (k+1)^{2}\sum _{|\beta |=k+2}\|\theta \langle r\rangle ^{| \beta |}\partial _{h}^{\beta}(r\partial _{r}\mathbf{g})\|_{L^{2}} \\ &\lesssim (k+1)^{2}\sum _{|\beta |=k+1}\|\theta \langle r\rangle ^{| \beta |}\partial _{h}^{\beta}\mathbf{g}\|_{L^{2}}+(k + 1)^{2}\sum _{| \beta |=k}\|\theta \langle r\rangle ^{|\beta |} \partial _{h}^{\beta} \mathbf{g}\|_{L^{2}}. \end{aligned}$$
Then it is easy to show that where, when \(k=i=0\), we have set \(Y_{0}=0\).
$$\begin{aligned} \frac{\tau ^{k}}{k!}\sum _{|\beta |=k}\|\theta \langle r\rangle ^{k} \partial _{h}^{\beta}(r\partial _{r}\mathbf{g})\|_{L_{h}^{\infty }L_{z}^{2}} \lesssim \tau ^{-2}\sum _{i=0}^{3}Y_{k + i}(\mathbf{g}), \end{aligned}$$
(5.9)
Using (5.2) again, it leads to where we used the fact that \(\tau \leq \tau _{0}\) since later we will choose \(\tau (t)\) to be a decreased function of t.
$$\begin{aligned} \frac{\tau ^{k}(k+1)^{4}}{k!} \sum _{|\beta |=k} \|\theta \langle r \rangle ^{k}\partial _{h}^{\beta}(r\partial _{r}\mathbf{g})\|_{L^{2}} \lesssim Y_{k}(\mathbf{g})+Y_{k+1}(\mathbf{g}), \end{aligned}$$
(5.10)
Inserting (5.9)-(5.10) into (5.4), one yields
$$\begin{aligned} \sum _{|\alpha |=n} \frac{|I_{1}| \tau ^{n}(t)M_{n}}{\| \theta \langle r \rangle ^{n} \partial _{h}^{\alpha }g\|_{L^{2}}} \lesssim \tau ^{-2}\sum _{k=0}^{[\frac{n}{2}]}\sum _{i=0}^{2} (X_{n-k+i}(g)+D_{n-k+i}(g)) \sum _{j=0}^{3}Y_{k+j}(\mathbf{g}). \end{aligned}$$
(5.11)
Then by using the following inequality
$$\begin{aligned} \sum _{n\geq 0}\sum _{k=0}^{n}a_{n-k}b_{k}\leq \sum _{k\geq 0}a_{k} \sum _{j\geq 0}b_{j}. \end{aligned}$$
(5.12)
It infers from (5.11) that which is (5.1)1.
$$\begin{aligned} &\sum _{n \geq 0}\sum _{|\alpha |=n} \frac{|I_{1}^{\alpha}|\tau ^{n}(t)M_{n}}{\|\theta \langle r\rangle ^{n}\partial _{h}^{\alpha }g\|_{L^{2}}} \lesssim \tau ^{-2}\sum _{k\geq 0}(X_{k}(g)+D_{k}(g))\sum _{k\geq 0}Y_{k} (g) \\ &=\tau ^{-2}\big(\|g\|_{\mathcal{X}_{\tau}}+\|g\|_{\mathcal{D}_{\tau}} \big)\|\mathbf{g}\|_{\mathcal{Y}_{\tau}}, \end{aligned}$$
(5.13)
Now we establish the estimate of term \(I_{2} \). In view of (5.3), we get
$$\begin{aligned} &\sum _{|\alpha |=n} \frac{|I_{2}| \tau ^{n}(t) M_{n}}{\| \theta \langle r \rangle ^{n} \partial _{h}^{\alpha }g \|_{L^{2}}} \\ &\leq \tau ^{n}(t) M_{n} \sum _{k = 0}^{[ \frac{n}{2} ]} \binom{n}{k} \sum _{|\gamma | = n - k} \| \langle r \rangle ^{n - k} \partial _{h}^{\gamma }v^{z} \|_{L_{h}^{2} L_{z}^{\infty}} \sum _{| \beta | = k} \| \theta \langle r \rangle ^{k} \partial _{h}^{\beta} \partial _{z}\mathbf{g}\|_{L_{h}^{\infty }L_{z}^{2}} \\ &+\tau ^{n}(t)M_{n} \sum _{k = [ \frac{n}{2} ] + 1}^{n} \binom{n}{k} \sum _{|\gamma | = n - k} \| \langle r \rangle ^{n - k} \partial _{h}^{ \gamma }v^{z} \|_{L^{\infty}} \sum _{|\beta | = k} \|\theta \langle r \rangle ^{k} \partial _{h}^{\beta}\partial _{z}\mathbf{g}\|_{L^{2}}. \end{aligned}$$
(5.14)
Owing to (3.2)6,7, and noting that \(M_{n}\binom{n}{k}=\frac{(n+1)^{4}}{(n-k)!k!}\), we have
$$\begin{aligned} &\sum _{|\alpha |=n} \frac{|I_{2}| \tau ^{n}(t) M_{n}}{\| \theta \langle r \rangle ^{n} \partial _{h}^{\alpha }g\|_{L^{2}}} \\ &\lesssim \langle t\rangle ^{\frac{1}{4}} \tau ^{-2}\Big\{ \sum _{k=0}^{[ \frac{n}{2} ]} \frac{(n-k+1)^{4}}{(n - k)! k!} \sum _{|\gamma |=n-k} \|\langle r \rangle ^{n-k} \partial _{h}^{\gamma }(r\partial _{r} g, g) \|_{L^{2}} \sum _{|\beta | = k} \| \theta \langle r \rangle ^{k} \partial _{h}^{\beta }\partial _{z}\mathbf{g} \|_{L_{h}^{\infty }L_{z}^{2}} \\ &+\sum _{k=[ \frac{n}{2} ]+1}^{n} \frac{(k+1)^{4} (n-k+1)^{2}}{(n-k)! k!}\sum _{|\gamma |=n-k} \| \langle r \rangle ^{|\gamma |} \partial _{h}^{\gamma }(r\partial _{r} g, g)\|_{L^{2}} \sum _{|\beta |=k}\|\theta \langle r\rangle ^{k} \partial _{h}^{\beta}\partial _{z}\mathbf{g}\|_{L^{2}}\Big\} . \end{aligned}$$
(5.15)
Similar to (5.8), we conclude
$$\begin{aligned} \frac{1}{k!} \|\theta \langle r \rangle ^{k} \partial _{h}^{\beta} \partial _{z}g\|_{L_{h}^{\infty }L_{z}^{2}} \lesssim \frac{(k+1)^{2}}{k!} \sum _{|\beta |=k}\|\theta \langle r \rangle ^{| \beta |} \partial _{h}^{\beta}\partial _{z}g\|_{L^{2}} \lesssim \tau ^{-2} \sum _{i=0}^{2} D_{k+i}(g). \end{aligned}$$
(5.16)
Plugging (5.15)-(5.16) into (5.14), we derive
$$\begin{aligned} &\sum _{|\alpha |=n} \frac{|I_{2}|\tau ^{n}(t) M_{n}}{\|\theta \langle r \rangle ^{n}\partial _{h}^{\alpha }g\|_{L^{2}}} \\ &\lesssim \langle t \rangle ^{\frac{1}{4}} \tau ^{-2}\sum _{k=0}^{[ \frac{n}{2} ]}\frac{\tau ^{n-k}(n-k+1)^{4}}{(n-k)!}\sum _{|\gamma |=n-k} \|\langle r \rangle ^{n-k} \partial _{h}^{\gamma }(r\partial _{r} g, g) \|_{L^{2}}\sum _{i=0}^{2} D_{k+i}(\mathbf{g}) \\ &+\langle t \rangle ^{\frac{1}{4}}\tau ^{-2} \sum _{k=[ \frac{n}{2}]+1}^{n} \frac{\tau ^{n-k} (n-k+1)^{2}}{(n-k)!} \sum _{|\gamma |=n-k}^{n-k+2} \| \langle r \rangle ^{|\gamma |} \partial _{h}^{\gamma }(r\partial _{r} g, g) \|_{L^{2}} D_{k}(\mathbf{g}). \end{aligned}$$
(5.17)
It holds that
$$\begin{aligned} \left \{ \textstyle\begin{array}{l} \frac{\tau ^{n-k}(n-k+1)^{4}}{(n-k)!}\sum \limits _{|\gamma |=n-k}\|( \tau )^{n-k}\partial _{h}^{\gamma}(r\partial _{r} g, g)\|_{L^{2}} \leq X_{n-k}(g)+Y_{n-k+1}(g)+Y_{n-k}(g), \\ \frac{\tau ^{n-k}(n-k+1)^{2}}{(n-k)!}\sum \limits _{|\gamma |=n-k}^{n-k+2} \|\langle r\rangle ^{|\gamma |}\partial _{h}^{\gamma}(r\partial _{r} g, g)\|_{L^{2}}\leq X_{n-k}(g)+\tau ^{-2} \sum _{i=0}^{3} Y_{n-k+i}(g). \end{array}\displaystyle \right . \end{aligned}$$
(5.18)
Substituting (5.18) into (5.17), we get
$$ \sum _{|\alpha |=n} \frac{|I_{2}| \tau ^{n}(t) M_{n}}{\|\theta \langle r\rangle ^{n}\partial _{h}^{\alpha }g\|_{L^{2}}} \lesssim \langle t \rangle ^{\frac{1}{4}} \tau ^{-2}\sum _{k=0}^{n} \Big(X_{n-k} (g)+\sum _{i = 0}^{3}Y_{n-k+i}(g)\Big)\sum _{i = 0}^{2}D_{k+i}( \mathbf{g}). $$
(5.19)
Now we derive the estimate of term \(I_{3} \). In view of (5.3), we get
$$\begin{aligned} &\sum _{|\alpha |=n} \frac{|I_{3}| \tau ^{n}(t) M_{n}}{\|\theta \langle r\rangle ^{n}\partial _{h}^{\alpha }g\|_{L^{2}}} \\ &\lesssim \tau ^{n}(t) M_{n} \sum _{k = 0}^{[\frac{n}{2}]} \binom{n}{k} \sum _{|\gamma | = n - k} \|\theta _{\frac{1}{2}} \langle r\rangle ^{n - k}\partial _{h}^{\gamma }(v^{r},b^{r})\|_{L^{2}_{h} L^{\infty}_{z}} \sum _{|\beta |=k} \|\theta _{\frac{1}{2}} (\tau )^{k} \partial _{h}^{\beta }(v^{r},b^{r})\|_{L^{\infty}_{h} L^{2}_{z}} \\ &+ \tau ^{n}(t) M_{n} \sum _{k=[ \frac{n}{2} ] + 1}^{n} \binom{n}{k} \sum _{|\gamma | = n - k} \|\theta _{\frac{1}{2}}\langle r\rangle ^{n - k}\partial _{h}^{\gamma}(v^{r},b^{r})\|_{L^{\infty}} \sum _{|\beta | = k} \|\theta _{\frac{1}{2}}\langle r\rangle ^{k} \partial _{h}^{\beta}(v^{r},b^{r}) \|_{L^{2}}. \end{aligned}$$
(5.20)
Using (3.2)2-(3.2)5 and (3.3)2-(3.3)5, and noting that \(M_{n} \binom{n}{k}=\frac{(n+1)!}{(n-k)! k!}\), we deduce
$$\begin{aligned} &\sum _{|\alpha | = n} \frac{|I_{3}| \tau ^{n}(t) M_{n}}{\|\theta \langle r\rangle ^{n}\partial _{h}^{\alpha }g\|_{L^{2}}} \\ &\lesssim \tau ^{-2} \sum _{k = 0}^{[ \frac{n}{2} ]} (X_{n-k}( \mathbf{g})+D_{n-k}(\mathbf{g})) \sum _{i = 0}^{2} X_{k+i}(\mathbf{g}) \\ &+\tau ^{-2} \sum _{i= [ \frac{n}{2} ] + 1}^{n} \sum _{i = 0}^{2} (X_{n-k+i}( \mathbf{g})+D_{n-k+i}(\mathbf{g}))X_{k}(\mathbf{g}) \\ &\lesssim \tau ^{-2} \sum _{k=0}^{n}\sum _{i=0}^{2}(X_{n-k+i}( \mathbf{g})+D_{n-k+i}(\mathbf{g}))\sum _{i=0}^{2}X_{k+i}(\mathbf{g}). \end{aligned}$$
(5.21)
Now we establish the estimate of the terms \(I_{4}\), from (2.8), we get
$$\begin{aligned} \partial _{z}(z\phi )=\Big(1-\frac{z^{2}}{2\langle t\rangle}\Big) \phi +zg,\ \partial _{z}(z\psi )=\Big(1- \frac{z^{2}}{2\langle t\rangle}\Big)\psi +z\tilde{g}. \end{aligned}$$
For \(|\alpha |=k\), it follows from (3.2)1 that
$$\begin{aligned} &\|\theta _{\lambda }\langle r\rangle ^{k} \partial _{h}^{\alpha} \partial _{z} (z\phi )\|_{L^{2}_{z}} \\ &\lesssim \|\theta _{\lambda }\langle r\rangle ^{k}\partial _{h}^{ \alpha }\phi \|_{L^{2}_{z}}+\Big\| \theta _{\lambda } \frac{z^{2}}{\langle t \rangle} \langle r\rangle ^{k} \partial _{h}^{ \alpha }\phi \Big\| _{L^{2}_{z}} + \|\theta _{\lambda }z \langle r \rangle ^{k} \partial _{h}^{\alpha }g\|_{L^{2}_{z}} \\ &\lesssim \sqrt{\langle t \rangle} \|\theta _{\lambda }\langle r \rangle ^{k} \partial _{h}^{\alpha }\phi \|_{L^{2}_{z}}+\|\theta _{ \frac{1+\lambda}{2}}\langle r\rangle ^{k}\partial _{h}^{\alpha }g\|_{L^{2}_{z}}+ \sqrt{\langle t \rangle} \|\theta _{\frac{1+\lambda}{2}}\langle r \rangle ^{k}\partial _{h}^{\alpha }g\|_{L^{2}} \\ &\lesssim \sqrt{\langle t \rangle} \|\theta \langle r\rangle ^{k} \partial _{h}^{\alpha }g\|_{L^{2}_{z}}. \end{aligned}$$
(5.22)
Similarly, we also get
$$\begin{aligned} \|\theta _{\lambda }\langle r\rangle ^{k} \partial _{h}^{\alpha} \partial _{z} (z\psi )\|_{L^{2}_{z}} \lesssim \sqrt{\langle t \rangle} \|\theta \langle r\rangle ^{k} \partial _{h}^{\alpha }\tilde{g}\|_{L^{2}_{z}}. \end{aligned}$$
(5.23)
Now we come to the estimate of term \(I_{4}\). Thanks to (5.3) and (5.22)-(5.23), we obtain
$$\begin{aligned} &\sum _{|\alpha | = n} \frac{|I_{4}| \tau ^{n}(t) M_{n}}{\|\theta \langle r\rangle ^{n}\partial _{h}^{\alpha }g\|_{L^{2}}} \\ &\lesssim \langle t \rangle ^{-\frac{1}{2}} \tau ^{n}(t) M_{n} \sum _{| \alpha | = n} \sum _{\beta \leq \alpha ,\ |\beta | \leq [ \frac{n}{2} ]} \binom{\alpha}{\beta} \|\theta _{\frac{1}{2}} \langle r\rangle ^{n - | \beta |}\partial _{h}^{n - \beta} (v^{z},b^{z})\|_{L^{2}_{h}L^{\infty}_{z}} \|\theta \langle r\rangle ^{|\beta |}\partial _{h}^{\beta }\mathbf{g} \|_{L^{\infty}_{h} L^{2}_{z}} \\ &+ \langle t \rangle ^{-\frac{1}{2}} \tau ^{n}(t) M_{n} \sum _{| \alpha |=n} \sum _{\beta \leq \alpha ,\ |\beta | \geq [ \frac{n}{2} ] + 1} \binom{\alpha}{\beta} \|\theta _{\frac{1}{2}} \langle r\rangle ^{n - |\beta |}\partial _{h}^{n - \beta} (v^{z},b^{z})\|_{L^{\infty}} \| \theta \langle r\rangle ^{|\beta |}\partial _{h}^{\beta}\mathbf{g}\|_{L^{2}}. \end{aligned}$$
Then, by replacing \(\partial _{h} \mathbf{g}\) with g, we can derive an estimate almost identical to the one in (5.14), which leads to a corresponding estimate analogous to (5.19) as follows:
$$ \sum _{|\alpha | = n} \frac{|I_{4}| \tau ^{n}(t) M_{n}}{\|\theta \langle r\rangle ^{n}\partial _{h}^{\alpha }g\|_{L^{2}}} \lesssim \langle t \rangle ^{-\frac{1}{4}} \tau ^{-2} \sum _{k = 0}^{n} \Big(X_{n - k} (\mathbf{g})+ \sum _{i = 0}^{3} Y_{n - k + i}( \mathbf{g})\Big) \sum _{i = 0}^{2} X_{k + i}(\mathbf{g}). $$
(5.24)
Now we come up with the estimate of term \(I_{5}\). It is not hard to show that, from (3.2)1 and (3.3)1,
$$\begin{aligned} \left \{ \textstyle\begin{array}{l} \|\theta _{\lambda }\langle r\rangle ^{k} \partial _{h}^{\alpha}( \partial _{z}\phi )\|_{L^{2}_{z}} \lesssim \sqrt{\langle t \rangle} \|\theta _{\frac{1+\lambda}{2}}\langle r\rangle ^{k} \partial _{h}^{ \alpha }\phi \|_{L^{2}_{z}} \lesssim \langle t \rangle \|\theta \langle r\rangle ^{k} \partial _{h}^{\alpha }g\|_{L^{2}}, \\ \|\theta _{\lambda }\langle r\rangle ^{k} \partial _{h}^{\alpha}( \partial _{z}\psi )\|_{L^{2}_{z}} \lesssim \sqrt{\langle t \rangle} \|\theta _{\frac{1+\lambda}{2}}\langle r\rangle ^{k} \partial _{h}^{ \alpha }\psi \|_{L^{2}_{z}} \lesssim \langle t \rangle \|\theta \langle r\rangle ^{k} \partial _{h}^{\alpha }\tilde{g}\|_{L^{2}}. \end{array}\displaystyle \right . \end{aligned}$$
(5.25)
It follows from (5.3) and (5.25)
$$\begin{aligned} &\sum _{|\alpha | = n} \frac{|I_{5}| \tau ^{n}(t) M_{n}}{\|\theta \langle r\rangle ^{n}\partial _{h}^{\alpha }g\|_{L^{2}}} \\ &\lesssim \tau ^{n}(t) M_{n} \sum _{k = 0}^{[ \frac{n}{2} ]} \binom{n}{k} \sum _{|\gamma | = n - k} \|\langle r\rangle ^{n - k} \partial _{h}^{\gamma }(v^{r},b^{r})\|_{L^{2}_{h} L^{\infty}_{z}} \sum _{|\beta | = k} \|\theta \langle r\rangle ^{k} \partial _{h}^{ \beta }\mathbf{g}\|_{L^{\infty}_{h} L^{2}_{z}} \\ &+ \tau ^{n}(t) M_{n} \sum _{k = [ \frac{n}{2} ] + 1}^{n} \binom{n}{k} \sum _{|\gamma | = n - k} \|\langle r\rangle ^{n - k} \partial _{h}^{\gamma }(v^{r},b^{r})\|_{L^{\infty}} \sum _{|\beta | = k} \|\theta \langle r\rangle ^{k} \partial _{h}^{\beta }\mathbf{g}\|_{L^{2}}. \end{aligned}$$
Because of (3.2)4,5 and (3.3)4,5, and noting that \(M_{n} \binom{n}{k}=\frac{(n+1)!}{(n-k)!k!}\), we get
$$\begin{aligned} &\sum _{|\alpha | = n} \frac{|I_{5}| \tau ^{n}(t) M_{n}}{\|\theta \langle r\rangle ^{n}\partial _{h}^{\alpha }g\|_{L^{2}}} \\ &\lesssim \sum _{k = 0}^{[ \frac{n}{2} ]} (X_{n - k}(\mathbf{g})+D_{n-k}( \mathbf{g})) \frac{\tau ^{k}}{k!} \sum _{|\beta | = k} \|\theta \langle r\rangle ^{k} \partial _{h}^{\beta }\mathbf{g}\|_{L^{\infty}_{h} L^{2}_{z}} \\ &+\tau ^{-2} \sum _{k = [ \frac{n}{2} ] + 1}^{n} \sum _{i = 0}^{2} (X_{n - k + i}(\mathbf{g})+ D_{n - k + i}(\mathbf{g})) X_{k}(\mathbf{g}). \end{aligned}$$
(5.26)
By applying Sobolev embedding theorem, it is easy to see that
$$\begin{aligned} \left \{ \textstyle\begin{array}{l} \frac{\tau ^{k}}{k!} \sum \limits _{|\beta | = k} \|\theta \langle r \rangle ^{k} \partial _{h}^{\beta }g\|_{L^{\infty}_{h} L^{2}_{z}} \lesssim \tau ^{-2} \sum \limits _{i = 0}^{2} X_{k + i}(g), \\ \frac{\tau ^{k}}{k!} \sum \limits _{|\beta | = k} \|\theta \langle r \rangle ^{k} \partial _{h}^{\beta }\tilde{g}\|_{L^{\infty}_{h} L^{2}_{z}} \lesssim \tau ^{-2} \sum \limits _{i = 0}^{2} X_{k + i}(\tilde{g}). \end{array}\displaystyle \right . \end{aligned}$$
(5.27)
Plugging (5.27) into (5.26), we conclude
$$ \sum _{|\alpha |=n} \frac{|I_{5}| \tau ^{n}(t) M_{n}}{\|\theta \langle r\rangle ^{n}\partial _{h}^{\alpha }g\|_{L^{2}}} \lesssim \langle t \rangle ^{-\frac{1}{2}} \tau ^{-2} \sum _{k=0}^{n} \sum _{i=0}^{2} (X_{n - k + i}(\mathbf{g})+ D_{n - k + i}(\mathbf{g})) \sum _{i=0}^{2} X_{k + i}(\mathbf{g}). $$
(5.28)
Now we achieve the estimate of term \(I_{6}\). First, by a simple calculation, we get
$$\begin{aligned} & \frac{|I_{6}|}{\|\theta \langle r\rangle ^{n}\partial _{h}^{\alpha }g\|_{L^{2}}} \leq \frac{1}{\langle t \rangle} \Big\| z\theta (z) \int _{z}^{ \infty }\langle r\rangle ^{n} \partial _{h}^{\alpha }(v^{r},b^{r})^{2} (\bar{z}) \mathrm{d}\bar{z} \Big\| _{L^{2}} \\ &= \frac{1}{\langle t \rangle} \Big\| z\theta _{-1/2}(z)\theta _{3/2}(z) \int _{z}^{\infty}\langle r\rangle ^{n} \partial _{h}^{\alpha }(v^{r},b^{r})^{2}( \bar{z}) \mathrm{d}\bar{z} \Big\| _{L^{2}} \\ &\leq \frac{1}{\langle t \rangle} \Big\| z\theta _{-1/2}(z)\|_{L^{ \infty}_{h}L^{2}_{z}}\|\theta _{3/2}(z)\int _{z}^{\infty}\langle r \rangle ^{n} \partial _{h}^{\alpha }(v^{r},b^{r})^{2}(\bar{z}) \mathrm{d}\bar{z} \Big\| _{L_{h}^{2} L^{\infty}_{z}} \\ &\lesssim \langle t \rangle ^{-\frac{1}{4}}\|\theta _{3/2}(z)\int _{z}^{ \infty}\langle r\rangle ^{n} \partial _{h}^{\alpha }(v^{r},b^{r})^{2}( \bar{z}) \mathrm{d}\bar{z} \Big\| _{L_{h}^{2} L^{\infty}_{z}}. \end{aligned}$$
While
$$\begin{aligned} &\Big\| \theta _{3/2}(z)\int _{z}^{\infty}\langle r\rangle ^{n} \partial _{h}^{\alpha }(v^{r},b^{r})^{2}(\bar{z}) \mathrm{d}\bar{z} \Big\| _{L_{h}^{2} L^{\infty}_{z}} \\ &\leq \sup _{z \geq 0} \Big\{ \theta _{\frac{3}{2}} (z) \Big( \int _{z}^{ \infty }\theta _{-\frac{7}{2}} (\bar{z}) \mathrm{d}\bar{z} \Big)^{ \frac{1}{2}} \Big\} \Big\| \theta _{\frac{7}{4}} \langle r\rangle ^{n} \partial _{h}^{\alpha }(v^{r},b^{r})^{2} \Big\| _{L^{2}_{z}} \\ &\leq \langle t \rangle ^{\frac{1}{4}}\Big\| \theta _{\frac{7}{4}} \langle r\rangle ^{n} \partial _{h}^{\alpha }(v^{r},b^{r})^{2} \Big\| _{L^{2}_{z}}. \end{aligned}$$
Then
$$ \sum _{|\alpha |=n} \frac{|I_{6}| \tau ^{n}(t) M_{n}}{\|\theta \langle r\rangle ^{n}\partial _{h}^{\alpha }g\|_{L^{2}}} \lesssim \tau (t) M_{n} \sum _{|\alpha |=n} \Big\| \theta _{ \frac{7}{4}} \langle r\rangle ^{n} \partial _{h}^{\alpha }(v^{r},b^{r})^{2} \Big\| _{L^{2}_{z}}. $$
(5.29)
The remaining part is analogous to \(I_{3} \) in (5.20), with \(\displaystyle \frac{1}{2} \) replaced by \(\displaystyle \frac{7}{8} \); this corresponds to (5.2)3 for the term \(I_{6} \).
Finally, we establish the estimate of term \(I_{7}\). Repeating the proof for (5.29), we can derive
$$ \sum _{|\alpha |=n} \frac{|I_{7}| \tau (t) M_{n}}{\|\theta \langle r\rangle ^{n}\partial _{h}^{\alpha }g\|_{L^{2}}} \lesssim \tau (t) M_{n} \sum _{|\alpha |=n} \Big\| \theta _{ \frac{7}{4}} \langle r\rangle ^{n} \partial _{h}^{\alpha }(v^{z} \partial _{z}v^{r},b^{z}\partial _{z}b^{r})\Big\| _{L^{2}}. $$
Owing to (5.3), we get
$$\begin{aligned} &\sum _{|\alpha |=n} \frac{|I_{7}| \tau (t) M_{n}}{\|\theta \langle r\rangle ^{n}\partial _{h}^{\alpha }g\|_{L^{2}}} \\ &\lesssim \tau ^{n}(t) M_{n} \sum _{k=0}^{[ \frac{n}{2} ]} \binom{n}{k} \sum _{|\gamma |=n - k} \|\theta _{7/8}\langle r\rangle ^{n - k}\partial _{h}^{\gamma }(v^{z},b^{z})\|_{L_{h}^{2} L^{\infty}_{z}} \sum _{|\beta |=k} \|\theta _{7/8}\langle r\rangle ^{k} \partial _{h}^{ \beta }\partial _{z} (v^{r},b^{r})\|_{L^{\infty}_{h}L_{z}^{2}} \\ &+ \tau ^{n}(t) M_{n} \sum _{k=0}^{[ \frac{n}{2} ]} \binom{n}{k} \sum _{|\gamma |=n - k} \|\theta _{7/8}\langle r\rangle ^{n - k} \partial _{h}^{\gamma }(v^{z},b^{z})\|_{L^{\infty}} \sum _{|\beta |=k} \|\theta _{7/8}\langle r\rangle ^{k} \partial _{h}^{\beta }\partial _{z} (v^{r},b^{r})\|_{L^{2}}. \end{aligned}$$
Then by using (3.2)6-(3.2)9 and (3.3)6-(3.3)9, and noting that \(M_{n}\binom{n}{k}=\frac{(n+1)^{4}}{(n-k)!k!}\), we obtain
$$\begin{aligned} &\sum _{|\alpha |=n} \frac{|I_{7}| \tau ^{n}(t) M_{n}}{\|\theta \langle r\rangle ^{n}\partial _{h}^{\alpha }g\|_{L^{2}}} \\ &\lesssim \langle t \rangle ^{\frac{1}{4}} \tau ^{-2} \sum _{k=0}^{[ \frac{n}{2} ]} \frac{(n - k + 1)^{4} \tau ^{n - k}}{(n - k)!}\sum _{| \gamma |=n-k} \|\langle r\rangle ^{n - k}\partial _{h}^{\gamma }(r \partial _{r} g,r\partial _{r}\tilde{g},g,\tilde{g})\|_{L^{2}} \\ &\quad \times \sum _{i=0}^{2} (\langle t \rangle ^{-\frac{1}{2}} X_{k + i}(\mathbf{g})+ D_{k + i}(\mathbf{g})) \\ &+ \langle t \rangle ^{\frac{1}{2}} \sum _{k=[\frac{n}{2} ] + 1}^{n} \frac{(n - k + 1)^{2} \tau ^{n-k}}{(n - k)!} \sum _{|\gamma |=n-k}^{n - k + 2}\|\langle r\rangle ^{|\gamma |}\partial _{h}^{\gamma }(r \partial _{r} g,r\partial _{r}\tilde{g},g,\tilde{g})\|_{L^{2}} ( \langle t \rangle ^{-\frac{1}{2}} X_{k}(\mathbf{g})+ D_{k}(\mathbf{g})). \end{aligned}$$
In addition, similar to (5.18), we also have
$$\begin{aligned} \left \{ \textstyle\begin{array}{l} \frac{\tau ^{n-k}(n-k+1)^{4}}{(n-k)!}\sum \limits _{|\gamma |=n-k}\|( \tau )^{n-k}\partial _{h}^{\gamma}(r\partial _{r} \tilde{g}, \tilde{g})\|_{L^{2}} \leq X_{n-k}(\tilde{g})+Y_{n-k+1}(\tilde{g})+Y_{n-k}( \tilde{g}), \\ \frac{\tau ^{n-k}(n-k+1)^{2}}{(n-k)!}\sum \limits _{|\gamma |=n-k}^{n-k+2} \|\langle r\rangle ^{|\gamma |}\partial _{h}^{\gamma}(r\partial _{r} \tilde{g}, \tilde{g})\|_{L^{2}}\leq X_{n-k}(\tilde{g})+\tau ^{-2} \sum _{i=0}^{3} Y_{n-k+i}(\tilde{g}). \end{array}\displaystyle \right . \end{aligned}$$
(5.30)
Then using (5.18) and (5.30), we deduce
$$\begin{aligned} \sum _{|\alpha |=n} \frac{|I_{7}^{\alpha}| \tau ^{n}(t) M_{n}}{\|\theta \langle r\rangle ^{n}\partial _{h}^{\alpha }g\|_{L^{2}}} &\lesssim \langle t \rangle ^{\frac{1}{4}} \tau ^{-2} \sum _{k=0}^{n} \Big(X_{n-k}(\mathbf{g})+\sum _{i=0}^{3} Y_{n-k+i}(\mathbf{g})\Big) \\ &\quad \times\sum _{i=0}^{2} (\langle t \rangle ^{-\frac{1}{2}} X_{k+i}(\mathbf{g})+ D_{k + i}(\mathbf{g})). \end{aligned}$$
(5.31)
Lemma 5.2
(The estimates of the nonlinear terms) The following estimates for the nonlinear terms on the right-hand of \(\textit{(4.3)}_{2}\) hold:
$$\begin{aligned} \left \{ \textstyle\begin{array}{l} \sum _{n \geq 0} \limits \sum _{|\alpha | = n}\limits \frac{|J_{1}| \tau ^{n}(t) M_{n}}{\|\theta \langle r\rangle ^{n} \partial _{h}^{\alpha }\tilde{g}\|_{L^{2}}} \lesssim \tau ^{-2} \big( \|\mathbf{g}\|_{\mathcal{X}_{\tau}} + \| \mathbf{g}\|_{\mathcal{D}_{\tau}} \big) \|\mathbf{g}\|_{\mathcal{Y}_{ \tau}}, \\ \sum _{n \geq 0} \limits \sum _{|\alpha | = n}\limits \frac{|J_{2}| \tau ^{n}(t) M_{n}}{\|\theta \langle r\rangle ^{n} \partial _{h}^{\alpha }\tilde{g}\|_{L^{2}}} \lesssim \tau ^{-2} \langle t \rangle ^{\frac{1}{4}} \big( \| \mathbf{g}\|_{\mathcal{X}_{\tau}} + \|\mathbf{g}\|_{\mathcal{Y}_{\tau}} \big) \|\mathbf{g}\|_{\mathcal{D}_{\tau}}, \\ \sum _{n \geq 0} \limits \sum _{|\alpha | = n}\limits \frac{|J_{3}| \tau ^{n}(t) M_{n}}{\|\theta \langle r\rangle ^{n} \partial _{h}^{\alpha }\tilde{g}\|_{L^{2}}} \lesssim \tau ^{-2} \langle t \rangle ^{-\frac{1}{4}} \big( \| \mathbf{g}\|_{\mathcal{X}_{\tau}} + \|\mathbf{g}\|_{\mathcal{Y}_{\tau}} \big) \|\mathbf{g}\|_{\mathcal{X}_{\tau}}, \\ \sum _{n \geq 0} \limits \sum _{|\alpha | = n}\limits \frac{|J_{4}| \tau ^{n}(t) M_{n}}{\|\theta \langle r\rangle ^{n} \partial _{h}^{\alpha }\tilde{g}\|_{L^{2}}} \lesssim \tau ^{-2} \big( \|\mathbf{g}\|_{\mathcal{X}_{\tau}}+\| \mathbf{g}\|_{\mathcal{D}_{\tau}} \big) \|\mathbf{g}\|_{\mathcal{X}_{ \tau}}. \end{array}\displaystyle \right . \end{aligned}$$
(5.32)
Proof
The proof of this lemma is analogous to the estimates of terms \(I_{1,2,4,5}\) in Lemma 5.1, respectively, and we omit them here. □
6 Uniqueness
Next, we aim to prove the uniqueness of solutions to equations (4.2). Let \(\mathbf{g}_{1}:=(g_{1}(t,x,y), \tilde{g}_{1}(t,x,y))\) and \(\mathbf{g}_{2}:=(g_{2}(t,x,y),\tilde{g}_{2}(t,x,y))\) be two solutions to equations (4.2) with the same initial data \(\mathbf{g}_{0}:=(g_{0},\tilde{g}_{0}) \in X_{2\tau _{0}}\). Denote the tangential radii of analytic regularity of \(\mathbf{g}_{1}\) and \(\mathbf{g}_{2}\) by \(\tau _{1}(t)\) and \(\tau _{2}(t)\) respectively, both of which satisfy the bounds given in Theorem 2.2.
Define \(\tau (t)\) to solve
$$\begin{aligned} \dot{\tau} + \frac{2C_{0}}{\tau ^{2}(t)} \big( \|\mathbf{g}_{1}\|_{ \mathcal{X}_{\tau}} + \langle t \rangle ^{\frac{1}{4}} \|\mathbf{g}_{1} \|_{\mathcal{D}_{\tau}} \big)=0,\ \tau (0)=\frac{\tau _{0}}{4}. \end{aligned}$$
(6.1)
From the equation above, we are able to infer that
$$\begin{aligned} \dot{\tau} + \frac{2C_{0}}{\tau ^{2}(t)} \big(\|\mathbf{g}_{1}\|_{ \mathcal{X}_{\tau}} + \langle t \rangle ^{\frac{1}{4}} \|\mathbf{g}_{1} \|_{\mathcal{D}_{\tau}} \big)\leq \dot{\tau} + \frac{2C_{0}}{\tau ^{2}(t)} \big( \|\mathbf{g}_{1}\|_{\mathcal{X}_{ \tau _{1}}} + \langle t \rangle ^{\frac{1}{4}} \|\mathbf{g}_{1}\|_{ \mathcal{D}_{\tau _{1}}} \big)=0. \end{aligned}$$
(6.2)
Here, we have utilized the fact that \(\tau (t) \leq \tau _{1}(t)\) and that the norms \(\mathcal{X}_{\tau}\) and \(\mathcal{D}_{\tau}\) are increasing with respect to τ.
In view of the estimate (4.19) for \((g_{1}, \tilde{g}_{1})\) and the lower bounds (4.20) for \(\tau _{1}(t)\) and \(\tau _{2}(t)\), we arrive at the conclusion that for all \(t>0\).
$$\begin{aligned} \frac{\tau _{0}}{8}\leq \tau (t)\leq \frac{\tau _{0}}{4}\leq \frac{\min \{\tau _{1}(t),\tau _{2}(t)\}}{2}, \end{aligned}$$
(6.3)
Using (2.11), (4.1) and (6.3), we also have
$$\begin{aligned} \left \{ \textstyle\begin{array}{l} \|g_{2}\|_{\mathcal{Y}_{\tau}}\leq \frac{1}{\tau}\|g_{2}\|_{ \mathcal{X}_{2\tau}}\leq \frac{1}{\tau}\|g_{2}\|_{\mathcal{X}_{\tau _{2}}} \leq \frac{\varepsilon \langle t\rangle ^{-5/4+\delta}}{\tau}, \\ \|\tilde{g}_{2}\|_{\mathcal{Y}_{\tau}}\leq \frac{1}{\tau}\|\tilde{g}_{2} \|_{\mathcal{X}_{2\tau}}\leq \frac{1}{\tau}\|\tilde{g}_{2}\|_{ \mathcal{X}_{\tau _{2}}}\leq \frac{\varepsilon \langle t\rangle ^{-5/4+\delta}}{\tau}.\end{array}\displaystyle \right . \end{aligned}$$
(6.4)
We consider the difference of solutions \(\mathbf{G}=(G,\tilde{G}):=(g_{1}-g_{2},\tilde{g}_{1}-\tilde{g}_{2})\) which obeys
$$\begin{aligned} \left \{ \textstyle\begin{array}{l} \partial _{t}G-\partial _{z}^{2}G+\frac{1}{\langle t\rangle}G=-(rv^{r}_{1} \partial _{r}G+V^{z}\partial _{z}g_{1})+(rb_{1}^{r}\partial _{r} \tilde{G}+B^{z}\partial _{z}\tilde{g}_{1})+ \frac{1}{2\langle t\rangle} V^{z}\partial _{z}(z\phi ) \\ -\frac{1}{2\langle t\rangle} B^{z}\partial _{z}(z\psi )- \frac{z}{\langle t \rangle}V^{r}\phi +\frac{z}{\langle t \rangle}B^{r} \psi -\frac{z}{2\langle t\rangle}\int _{z}^{+\infty}V^{r}(v^{r}_{1}+u_{2}^{r}) \mathrm{d}\bar{z}+\frac{z}{\langle t\rangle}\int _{z}^{+\infty} \partial _{z} u_{1}^{r} V^{z}\mathrm{d}\bar{z} \\ -V^{r}(v^{r}_{1}+u_{2}^{r})+B^{r}(b^{r}_{1}+b_{2}^{r})+ \frac{z}{2\langle t\rangle}\int _{z}^{+\infty}B^{r}(b^{r}_{1}+b_{2}^{r}) \mathrm{d}\bar{z}-\frac{z}{\langle t\rangle}\int _{z}^{+\infty} \partial _{z}b_{1}^{r}B^{z}\mathrm{d}\bar{z} \\ -(rV^{r}\partial _{r}g_{2}+u_{2}^{z}\partial _{z}G)+(rB^{r}\partial _{r} \tilde{g}_{2}+b^{z}_{2}\partial _{z}\tilde{G}) \\ +\frac{z}{\langle t\rangle}\int _{z}^{+\infty}\partial _{z}V^{r}u_{2}^{z} \mathrm{d}\bar{z}-\frac{z}{\langle t\rangle}\int _{z}^{+\infty} \partial _{z}B^{r}b_{2}^{z}\mathrm{d}\bar{z}= 0, \\ \partial _{t}\tilde{G}-\partial _{z}^{2}\tilde{G}+ \frac{1}{\langle t\rangle}\tilde{G}=-(ru_{1}^{r}\partial _{r} \tilde{G}+V^{z}\partial _{z}\tilde{g}_{1})+(rb_{1}^{r}\partial _{r}G+B^{z} \partial _{z}g_{1}) -(rV^{r}\partial _{r}\tilde{g}_{2}+u_{2}^{z} \partial _{z}\tilde{G}) \\ +(rB^{r}\partial _{r}g_{2}+b_{2}^{z}\partial _{z}G)+ \frac{1}{2\langle t\rangle} V^{z}\partial _{z}(z\psi )- \frac{1}{2\langle t\rangle} B^{z}\partial _{z}(z\phi )- \frac{z}{\langle t \rangle}V^{r}\psi +\frac{z}{\langle t \rangle}B^{r} \phi . \end{array}\displaystyle \right . \end{aligned}$$
(6.5)
Here we also denote \(V^{r}=v^{r}_{1}-v^{r}_{2}\), \(V^{z}=v^{z}_{1}-v^{z}_{2}\), \(B^{r}=b^{r}_{1}-b^{r}_{2}\) and \(B^{z}=b^{z}_{1}-b^{z}_{2}\).
Note that the initial data and the boundary conditions are and
$$\begin{aligned} G(t,x,y)|_{t=0} = 0, \quad \tilde{G}(t,x,y)|_{t=0} = 0 \end{aligned}$$
$$ \textstyle\begin{cases} G|_{y=0} = 0, \\ G |_{y=\infty} = 0, \end{cases}\displaystyle \quad \text{and} \quad \textstyle\begin{cases} \tilde{G} |_{y=0} = 0, \\ \tilde{G} |_{y=\infty} = 0. \end{cases} $$
Similar to (4.12), we attain
$$\begin{aligned} &\frac{d}{dt} \|\mathbf{G} \|_{\mathcal{X}_{\tau}} + \frac{\frac{5}{4}-\delta}{\langle t \rangle} \|\mathbf{G}\|_{ \mathcal{X}_{\tau}} + \frac{\delta}{6} \big( \frac{1}{\langle t \rangle} \|\mathbf{G}\|_{\mathcal{X}_{\tau}} + \frac{1}{\sqrt{\langle t \rangle}} \|\mathbf{G}\|_{\mathcal{D}_{\tau}} \big) \\ &\leq \Big(\dot{\tau}+2C_{0} \tau ^{-2}(t) \big(\|\mathbf{g}_{1}\|_{ \mathcal{X}_{\tau}} + \langle t \rangle ^{\frac{1}{4}} \|\mathbf{g}_{1} \|_{\mathcal{D}_{\tau}} \big)\Big)\|\mathbf{G} \|_{\mathcal{Y}_{\tau}} \\ &+\Big(2C_{0} \tau ^{-2}(t) \big(\|\mathbf{G}\|_{\mathcal{X}_{\tau}} + \langle t \rangle ^{\frac{1}{4}} \|\mathbf{G} \|_{\mathcal{D}_{\tau}} \big)\Big)\|\mathbf{g}_{2} \|_{\mathcal{Y}_{\tau}} \\ &+ C_{0} \tau ^{-2}(t) \big( \|\mathbf{G} \|_{\mathcal{X}_{\tau}} + \langle t \rangle ^{\frac{1}{4}} \|\mathbf{G}\|_{\mathcal{D}_{\tau}} \big) \|\mathbf{g}_{2}\|_{\mathcal{X}_{\tau}} \\ &+ C_{0} \tau ^{-2}(t) \big( \|\mathbf{g}_{1} \|_{\mathcal{X}_{\tau}} + \langle t \rangle ^{\frac{1}{4}} \|\mathbf{g}_{1}\|_{\mathcal{D}_{ \tau}} \big) \|\mathbf{g}\|_{\mathcal{X}_{\tau}}. \end{aligned}$$
(6.6)
Using (4.19), and inserting (6.2) and (6.4) into (6.6), we can derive that
$$\begin{aligned} &\frac{d}{dt} \big( \langle t \rangle ^{\frac{5}{4} - \delta} \| \mathbf{G}\|_{\mathcal{X}_{\tau}} \big) + \frac{\delta}{6} \big( \langle t \rangle ^{\frac{1}{4} - \delta} \|\mathbf{G} \|_{ \mathcal{X}_{\tau}} + \langle t \rangle ^{\frac{3}{4}-\delta} \| \mathbf{G}\|_{\mathcal{D}_{\tau}} \big) \\ &\leq \Big(2C_{0}\varepsilon \tau ^{-3}(t) \big(\|\mathbf{G}\|_{ \mathcal{X}_{\tau}} + \langle t \rangle ^{\frac{1}{4}} \|\mathbf{G} \|_{\mathcal{D}_{\tau}} \big)\Big) \\ &+ C_{0}\varepsilon \tau ^{-2}(t) \big( \|\mathbf{G} \|_{\mathcal{X}_{ \tau}} + \langle t \rangle ^{\frac{1}{4}} \|\mathbf{G}\|_{\mathcal{D}_{ \tau}} \big) \\ &+ C_{0} \tau ^{-2}(t)\langle t \rangle ^{\frac{5}{4} - \delta}\big( \|\mathbf{g}_{1}\|_{\mathcal{X}_{\tau}} + \langle t \rangle ^{ \frac{1}{4}} \|\mathbf{g}_{1}\|_{\mathcal{D}_{\tau}} \big) \| \mathbf{G}\|_{\mathcal{X}_{\tau}}. \end{aligned}$$
(6.7)
If we take the sufficiently small \(\varepsilon \in (0,\frac{\delta \tau _{0}^{2}}{2^{8}\times 3C_{0}}]\) and \(\tau _{0}\geq 2^{4}\). Thus, (6.7) implies
$$\begin{aligned} \frac{d}{dt} \big( \langle t \rangle ^{\frac{5}{4} - \delta} \| \mathbf{G}\|_{\mathcal{X}_{\tau}} \big) \leq C_{0} \tau ^{-2}(t) \langle t \rangle ^{\frac{5}{4} - \delta}\big( \|\mathbf{g}_{1}\|_{ \mathcal{X}_{\tau}} + \langle t \rangle ^{\frac{1}{4}} \|\mathbf{g}_{1} \|_{\mathcal{D}_{\tau}} \big) \|\mathbf{G}\|_{\mathcal{X}_{\tau}}. \end{aligned}$$
(6.8)
Applying Gronwall inequality for (6.9) and using (4.1) yields which implies the uniqueness since \(\mathbf{G}_{0}=\mathbf{0}\).
$$\begin{aligned} \big( \langle t \rangle ^{\frac{5}{4} - \delta} \|\mathbf{G}\|_{ \mathcal{X}_{\tau}} \big) \leq & 64C_{0} \tau _{0}^{-2}\int _{0}^{t} \big( \|\mathbf{g}_{1}\|_{\mathcal{X}_{\tau}} + \langle t \rangle ^{ \frac{1}{4}} \|\mathbf{g}_{1}\|_{\mathcal{D}_{\tau}}\big)ds \| \mathbf{G}_{0}\|_{\mathcal{X}_{\tau}} \\ \leq &64C_{0} \tau _{0}^{-2}\varepsilon \|\mathbf{G}_{0}\|_{ \mathcal{X}_{\tau}}, \end{aligned}$$
(6.9)
The proof of uniqueness is thus complete. □
Acknowledgements
The authors are grateful to the anonymous referees and editors for helpful comments and suggestions that greatly improved the presentation of a previous version of the manuscript.
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