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# Dynamic bifurcation for the granulation convection in the solar photosphere

Boundary Value Problems20172017:110

https://doi.org/10.1186/s13661-017-0842-5

• Received: 16 March 2017
• Accepted: 7 July 2017
• Published:

## Abstract

The main objective of this paper is to investigate the dynamic bifurcation for the granulation convection in the solar photosphere. Based on the dynamic bifurcation theory, which is established by Ma and Wang (Phase transition dynamics, pp. 380-459, 2014), the critical parameters $$(R,F)$$ condition for the granulation convection is derived. Furthermore, the corresponding R-F-phase diagram is generated. In addition, the bifurcation solution is also obtained with certain assumptions.

## Keywords

• granulation
• eigenvalue
• solar photosphere
• dynamic bifurcation

## 1 Introduction

The atmosphere of the sun is divided into three parts: the bottom layer of the atmosphere is photosphere, the middle layer is chromosphere and the outermost layer is the corona. The grainy appearance of solar photosphere is produced by the tops of the convective cells and is called granulation. It worth noting that Kutner analyzed the structure of granulation in his book . He pointed out that the temperature decreases with the increasing height in the photosphere. And he concluded that the temperature difference between the top and bottom of photosphere causes the granulation convection in photosphere which can be explained by circulation cells of material. In circulation cells, the hotter gas is brought up from below, producing the bright regions. The cooler gas, which produces the darker regions, is carried down to replace the gas that was brought up (see Figure 1). Figure 1 The structure of granulation. This is a lateral view showing the situation under the surface. The hotter gas is brought up from below, producing the bright regions. The cooler gas, which produces the darker regions, is carried down to replace the gas that was brought up.
In addition, Ustyugov introduced the solar activity and gave the magneto-hydrodynamics (MHD) equations in [3, 4], which can explain the convection structure in the sun. There have been many studies on this kind of problem. For instance, Chae , Agélas , and Wu  considered the regularity of MHD equations, and Capone  and Fuchs  studied the stability bifurcation of MHD equations. For more research about the MHD equations refer to . It is noticed that the granulation convection structure (see Figure 1) is similar to the structure of Rayleigh-Bénard convection and the Taylor problem. Inspired by the dynamic theory, which is established by Ma and Wang to study the Rayleigh-Bénard convection and the Taylor problem (see [1, 1820]), we will investigate the dynamic bifurcation for the granulation convection. This paper is organized as follows.
1. 1.

In Section 2, we introduce the equations governing the atmospheric circulation with magnetic field and also give the bifurcation theory.

2. 2.
Section 3 is devoted to getting the RF-phase diagram of granulation under the parametric condition $$K_{1}=0$$ and $$F\neq0$$, where R is the Rayleigh number, which is related to the difference of temperature $$T_{1}-T_{0}$$, F is a dimensionless parameter, which is related to the magnetic field H, and $$K_{1}$$ is as in (2.5), which is related to the boundary condition $$H_{0}$$ and $$H_{1}$$. The transition occurs as the parameter $$(R, F)$$ crosses the critical curve L from region I to region II; see Figure 2. Figure 2 The RF-phase diagram. In region I, there is no granulation convection structure; and in region II, the granulation appears. So the transition occurs as the parameter (R,F) crosses the critical curve L from region I to region II.

3. 3.

In Section 4, the bifurcation solution and the critical Rayleigh number $$R_{C}$$ are obtained under the condition $$F= 0$$.

## 2 Governing equations

### 2.1 The model in spherical coordinates

Let $$(\varphi, \theta,r)$$ be the spherical coordinates, where φ, θ, r represent the longitude, the latitude, and the radial coordinate, respectively. The unknown functions include the velocity field $$u=(u_{\varphi},u_{\theta},u_{r})$$, the temperature function T, the pressure function p, the electromagnetic press function Φ and the magnetic field $$H=(H_{\varphi},H_{\theta},H_{r})$$. Then the equations governing the atmospheric circulation with magnetic field [9, 2123] in the spherical coordinates are given by
$$\left\{ \textstyle\begin{array}{l} \frac{\partial u}{\partial t}+(u\cdot\bar{ \nabla})u =\nu\bar {\triangle}u-\frac{1}{\rho_{0}}\nabla p-g\vec{k} [1- \alpha(T-T_{0}) ] + \frac{1}{\rho_{0}}(\nabla\times H)\times H, \\ \frac{\partial T}{\partial t}+(u \cdot\nabla)T=\kappa\triangle T, \\ \frac{\partial H}{\partial t}=\nabla\times(u\times H)+\eta\bar {\triangle} H+\nabla\Phi, \\ \operatorname{div} u=0, \\ \operatorname{div} H=0, \end{array}\displaystyle \right .$$
(2.1)
where $$0\leq\varphi\leq2\pi$$, $$-\frac{\pi}{2}\leq\theta\leq\frac{\pi }{2}$$, $$r_{0}< r< r_{1}$$, $$r_{1}=r_{0}+d$$, $$r_{0}$$ is the radius of the sun, and d is the height of the troposphere, $$\vec{k}=(0,0,1)$$, g is the gravitative constant, ν is the kinetic viscosity, κ is the thermal diffusivity, and η is the resistivity. The differential operators in the spherical coordinates are given as follows:
\begin{aligned}& \begin{aligned} (u\cdot\bar{\nabla})u={}& \biggl((u\cdot \nabla)u_{\varphi}+\frac {u_{r}u_{\varphi}}{r}+\frac{u_{\varphi}u_{\theta}}{r}\cot\theta, \\ & (u\cdot\nabla)u_{\theta}+ \frac{u_{r}u_{\theta}}{r}- \frac{u_{\varphi }^{2}}{r} \cot\theta,\\ &(u\cdot\nabla)u_{r}- \frac{u_{\theta }^{2}}{r}-\frac{u_{\varphi}^{2}}{r} \biggr)^{T},\end{aligned} \\& \nabla p= \biggl(\frac{1}{r\sin\theta}\frac{\partial p}{\partial\varphi },\frac{1}{r} \frac{\partial P}{\partial\theta}, \frac{\partial p}{\partial r} \biggr)^{T}, \\& \operatorname{div} u= \frac{1}{r\sin\theta} \frac{\partial u_{\varphi }}{\partial\varphi}+ \frac{1}{r\sin\theta} \frac{\partial (\sin\theta u_{\theta})}{\partial\theta}+\frac{1}{r^{2}}\frac{\partial }{\partial r} \bigl(r^{2}u_{r} \bigr), \\& \begin{aligned}\bar{\triangle}u={}& \biggl(\triangle u_{\varphi}+\frac{2}{r^{2}\sin \theta} \frac{\partial u_{r}}{\partial\varphi}+ \frac{2\cos\theta}{r^{2}\sin^{2}\theta}\frac{\partial u_{\theta }}{\partial\varphi}-\frac{u_{\varphi}}{r^{2}\sin^{2}\theta}, \\ &\triangle u_{\theta}-\frac{2\cos\theta}{r^{2}\sin \theta}\frac{\partial u_{\varphi}}{\partial\varphi}- \frac{u_{\theta}}{r^{2}\sin^{2}\theta}+\frac{2}{r^{2}}\frac{\partial u_{r}}{\partial\theta}, \\ &\triangle u_{r}-\frac{2u_{r}}{r^{2}} - \frac{2}{r^{2}\sin\theta}\frac{\partial(\sin\theta u_{\theta })}{\partial\theta}-\frac{2}{r^{2}\sin\theta} \frac{\partial u_{\varphi}}{\partial\varphi} \biggr)^{T},\end{aligned} \\& \begin{aligned}(\nabla\times H)\times H={}& \biggl(\frac{H_{r}}{r} \biggl[\frac{\partial (rH_{\varphi})}{\partial r}- \frac{1}{\sin\theta} \frac{\partial H_{r}}{\partial\theta} \biggr]-\frac{H_{\theta}}{r\sin\theta } \biggl[ \frac{\partial H_{\theta}}{\partial\varphi}- \frac{\partial(H_{\varphi}\sin\theta)}{\partial\theta} \biggr], \\ &\frac{H_{\varphi}}{r\sin\theta} \biggl[\frac {\partial H_{\theta}}{\partial\varphi}- \frac{\partial(H_{\varphi}\sin\theta)}{\partial\theta} \biggr]-\frac {H_{r}}{r} \biggl[\frac{\partial H_{r}}{\partial\theta}- \frac{\partial(rH_{\theta})}{\partial r} \biggr], \\ &\frac{H_{\theta}}{r} \biggl[\frac{\partial H_{r}}{\partial\theta}- \frac{\partial(rH_{\theta})}{\partial r} \biggr] -\frac{H_{\varphi}}{r} \biggl[\frac{\partial(rH_{\varphi})}{\partial r}- \frac {1}{\sin\theta}\frac{\partial H_{r}}{\partial\theta} \biggr] \biggr)^{T},\end{aligned} \\& \begin{aligned}\nabla\times(u\times H)= {}&\biggl(\frac{1}{r}\frac{\partial}{\partial r} \bigl[r(u_{\varphi }H_{r}-u_{r}H_{\varphi}) \bigr]- \frac{1}{r}\frac{\partial}{\partial\theta}(u_{\theta}H_{\varphi }-u_{\varphi}H_{\theta}), \\ & \frac{1}{r\sin\theta}\frac{\partial}{\partial\varphi}(u_{\theta }H_{\varphi}-u_{\varphi}H_{\theta})- \frac{1}{r\sin\theta}\frac{\partial}{\partial r} \bigl[r\sin\theta (u_{r}H_{\theta}-u_{\theta}H_{r}) \bigr], \\ &\frac{1}{r^{2}\sin\theta}\frac{\partial}{\partial\theta} \bigl[r\sin\theta (u_{r}H_{\theta}-u_{\theta}H_{r}) \bigr]- \frac{1}{r^{2}\sin\theta} \bigl[r(u_{\varphi}H_{r}-u_{r}H_{\varphi}) \bigr] \biggr)^{T},\end{aligned} \end{aligned}
̄H, Φ, divH are similar to ̄u, p, divu, and the operators Δ, $$(u\cdot\nabla)$$ are given by
$$\begin{gathered} (u\cdot\nabla)=\frac{u_{\theta}}{r} \frac{\partial}{\partial\theta }+\frac{u_{\varphi}}{r\sin\theta}\frac{\partial}{\partial\varphi} +u_{r} \frac{\partial}{\partial r}, \\ \triangle=\frac{1}{r^{2}\sin^{2}\theta}\frac{\partial^{2}}{\partial \varphi^{2}}+\frac{1}{r^{2}\sin\theta} \frac{\partial}{\partial\theta} \biggl(\sin\theta\frac{\partial}{\partial\theta} \biggr)+\frac{1}{r^{2}} \frac {\partial}{\partial r} \biggl(r^{2}\frac{\partial}{\partial r} \biggr). \end{gathered}$$
In this paper, we mainly focus on the dynamic bifurcation for the granulation. For simplicity, we assume $$\theta=\frac{\pi}{2}$$. Then the velocity component $$u_{\theta}$$ and the magnetic field $$H_{\theta}$$ are zero, and equations (2.1) become
$$\left\{ \textstyle\begin{array}{l} \frac{\partial u_{\varphi}}{\partial t}+(u\cdot \nabla)u_{\varphi }+\frac{u_{r}u_{\varphi}}{r} =\nu (\triangle u_{\varphi}+ \frac{2}{r^{2}}\frac{\partial u_{r}}{\partial\varphi} -\frac{u_{\varphi}}{r^{2}} ) - \frac{1}{\rho_{0}r}\frac{\partial p}{\partial\varphi}+\frac {H_{r}}{\rho_{0}r} [\frac{\partial(rH_{\varphi})}{\partial r} - \frac{\partial H_{r}}{\partial\varphi} ], \\ \frac{\partial u_{r}}{\partial t}+(u\cdot\nabla)u_{r}-\frac {u_{\varphi}^{2}}{r} =\nu ( \triangle u_{r}-\frac{2u_{r}}{r^{2}} -\frac{2}{r^{2}} \frac{\partial u_{\varphi}}{\partial\varphi} ) -\frac{1}{\rho_{0}}\frac{\partial p}{\partial r} -g [1- \alpha(T-T_{0}) ] \\ \hphantom{\frac{\partial u_{r}}{\partial t}+(u\cdot\nabla)u_{r}-\frac {u_{\varphi}^{2}}{r} =}+ \frac{H_{r}}{\rho_{0}r} [\frac{\partial H_{r}}{\partial\varphi}-\frac {\partial(rH_{\varphi})}{\partial r} ], \\ \frac{\partial T}{\partial t}+(u \cdot\nabla)T=\kappa\triangle T, \\ \frac{\partial H_{\varphi}}{\partial t}=\frac{1}{r}\frac{\partial }{\partial r} [r(u_{\varphi}H_{r}-u_{r}H_{\varphi}) ] +\eta (\triangle H_{\varphi}+\frac{2}{r^{2}}\frac{\partial H_{r}}{\partial\varphi} -\frac{H_{\varphi}}{r^{2}} )+\frac{1}{r}\frac{\partial\Phi}{\partial \varphi}, \\ \frac{\partial H_{r}}{\partial t}=\frac{1}{r^{2}}\frac{\partial }{\partial\varphi} [r(u_{r}H_{\varphi}-u_{\varphi}H_{r}) ] +\eta (\triangle H_{r}-\frac{2H_{r}}{r^{2}} - \frac{2}{r^{2}}\frac{\partial H_{\varphi}}{\partial\varphi} )+\frac {\partial\Phi}{\partial r}, \\ \operatorname{div}u=0, \\ \operatorname{div}H=0, \end{array}\displaystyle \right .$$
(2.2)
where
\begin{aligned} &(u\cdot\nabla)= \frac{u_{\varphi}}{r} \frac{\partial}{\partial\varphi }+u_{r}\frac{\partial}{\partial r}, \\ &\triangle=\frac{1}{r^{2}}\frac{\partial^{2}}{\partial\varphi^{2}}+ \frac{1}{r^{2}} \frac{\partial}{\partial r} \biggl(r^{2}\frac{\partial }{\partial r} \biggr), \\ &\operatorname{div}u=\frac{1}{r}\frac{\partial u_{\varphi}}{\partial\varphi }+\frac{1}{r^{2}} \frac{\partial}{\partial r} \bigl(r^{2}u_{r} \bigr). \end{aligned}
Furthermore, equations (2.2) are supplemented with the following boundary condition:
$$\left\{ \textstyle\begin{array}{l} (u,T,H) (\varphi,r)=(u,T,H) ( \varphi+2k\pi, r), \\ u_{r}=0,\qquad H_{r}=H_{0},\qquad T=T_{0},\qquad \frac{\partial u_{\varphi}}{\partial r}=\frac{\partial H_{\varphi}}{\partial r}=0,\qquad r=r_{0}, \\ u_{r}=0,\qquad H_{r}=H_{1},\qquad T=T_{1},\qquad \frac{\partial u_{\varphi }}{\partial r}=\frac{\partial H_{\varphi}}{\partial r}=0,\qquad r=r_{0}+d. \end{array}\displaystyle \right .$$
(2.3)

### 2.2 Perturbed dimensionless equations

We determine the basic flow by following assumptions.
1. 1.

$$U=(u_{\varphi},u_{r},T,H_{\varphi} ,H_{r})=(0,0,\tilde {T}(r),0,\tilde{H}_{r})$$, $$p=\tilde{p}(r)$$, $$\Phi=0$$; that is, the pressure function, the temperature function and the magnetic field function in r-direction are not zero, and which are only depending on r.

2. 2.
The functions $$\tilde{T}(r)$$, $$\tilde{H}(r)$$ and $$\tilde{p}(r)$$ satisfy
$$\left\{ \textstyle\begin{array}{l} -\frac{1}{\rho_{0}}\frac{\partial\tilde{p}}{\partial r}-g [1-\alpha (\tilde{T}-T_{0}) ]=0, \\ \triangle\tilde{T}=0, \\ \triangle\tilde{H_{r}}-\frac{2}{r^{2}}\tilde{H_{r}}=0. \end{array}\displaystyle \right .$$

3. 3.
Based on the boundary condition (2.3), the value of basic flow on the boundary is given by
$$\begin{gathered} \tilde{H_{r}}=H_{0},\qquad \tilde{T }=T_{0},\qquad r=r_{0}, \\ \tilde{H_{r}}=H_{1},\qquad \tilde{T }=T_{1},\qquad r=r_{0}+d. \end{gathered}$$

From the above assumptions, we derive the basic flow as follows:
$$\left \{ \textstyle\begin{array}{l} \tilde{T}=\frac{C_{1}}{r}+C_{0}, \\ \tilde{H_{r}}=K_{1}r+\frac{K_{0}}{r^{2}}, \\ \tilde{p}= \int^{r_{0+d}}_{r_{0}}-\rho_{0}g [1-\alpha( \tilde{T}-T_{0}) ]\,dr, \end{array}\displaystyle \right .$$
(2.4)
where
$$\begin{gathered} C_{0}=\frac{T_{1}r_{1}-T_{0}r_{0}}{r_{1}-r_{0}},\qquad C_{1}=\frac {(T_{0}-T_{1})r_{0}r_{1}}{r_{1}-r_{0}}, \\ H_{0}=K_{1}r_{0}+ \frac{K_{0}}{r_{0}^{2}},\qquad H_{1}=K_{1}r_{1}+ \frac {K_{0}}{r_{1}^{2}}. \end{gathered}$$
(2.5)
It is noticed that $$K_{0}$$ and $$K_{1}$$ are related to the boundary value $$H_{0}$$ and $$H_{1}$$. Furthermore, in order to get the perturbation equations related to the variables r and φ, we make the following translations:
$$p=p'+\tilde{p},\qquad T=T'+\tilde{T},\qquad H_{r}=H_{r}'+\tilde{H_{r}}.$$
Omitting the primes, equations (2.2) can be rewritten as
$$\left \{ \textstyle\begin{array}{l} \frac{\partial u_{\varphi}}{\partial t}+(u\cdot \nabla)u_{\varphi }+\frac{u_{r}u_{\varphi}}{r} =\nu (\triangle u_{\varphi}+ \frac{2}{r^{2}}\frac{\partial u_{r}}{\partial\varphi} -\frac{u_{\varphi}}{r^{2}} ) - \frac{1}{\rho_{0}r}\frac{\partial p}{\partial\varphi}+\frac {H_{r}}{\rho_{0}r} [\frac{\partial(rH_{\varphi})}{\partial r} - \frac{\partial H_{r}}{\partial\varphi} ] \\ \hphantom{\frac{\partial u_{\varphi}}{\partial t}+(u\cdot \nabla)u_{\varphi }+\frac{u_{r}u_{\varphi}}{r} =} +\frac{\tilde{H_{r}}}{\rho_{0}r} [\frac{\partial(rH_{\varphi})}{\partial r}-\frac{\partial H_{r}}{\partial\varphi} ], \\ \frac{\partial u_{r}}{\partial t}+(u\cdot\nabla)u_{r}-\frac {u_{\varphi}^{2}}{r} =\nu ( \triangle u_{r}-\frac{2u_{r}}{r^{2}} -\frac{2}{r^{2}} \frac{\partial u_{\varphi}}{\partial\varphi} ) -\frac{1}{\rho_{0}}\frac{\partial p}{\partial r} -g\alpha T, \\ \frac{\partial T}{\partial t}+(u \cdot\nabla)T=\kappa\triangle T+\frac{C_{1}}{r^{2}}u_{r}, \\ \frac{\partial H_{\varphi}}{\partial t}=\frac{1}{r}\frac{\partial }{\partial r} [r(u_{\varphi}H_{r}-u_{r}H_{\varphi}) ] +\eta (\triangle H_{\varphi}+\frac{2}{r^{2}}\frac{\partial H_{r}}{\partial\varphi} -\frac{H_{\varphi}}{r^{2}} )+\frac{1}{r}\frac{\partial\Phi}{\partial \varphi}+ \frac{1}{r}\frac{\partial}{\partial r}(ru_{\varphi}\tilde{H_{r}}), \\ \frac{\partial H_{r}}{\partial t}=\frac{1}{r^{2}}\frac{\partial }{\partial\varphi} [r(u_{r}H_{\varphi}-u_{\varphi}H_{r}) ] +\eta (\triangle H_{r}-\frac{2H_{r}}{r^{2}} - \frac{2}{r^{2}}\frac{\partial H_{\varphi}}{\partial\varphi} )+\frac {\partial\Phi}{\partial r} -\frac{1}{r} \tilde{H_{r}}\frac{\partial u_{\varphi}}{\partial\varphi}, \\ \operatorname{div}u=0, \\ \operatorname{div}H=0. \end{array}\displaystyle \right .$$
(2.6)
As we know, the radius of the sun is about $$7\times10^{5}\mbox{ km}$$ and the thickness of the photosphere is about 500 km. Then the ratio of the thickness of the photosphere to the radius of the sun is small. Hence, we adopt the approximations that $$1/r \simeq1/r_{0}$$, $$(r_{0}+d)/r_{0}\simeq1$$. For simplicity, we assume $$\nu/\kappa=\eta/\kappa=1$$. Also, we introduce the following dimensionless variables:
$$\begin{gathered} u=\frac{\kappa}{d}u',\qquad r=dr', \qquad T= \frac{T_{0}-T_{1}}{\sqrt {R}}T',\qquad p=\frac{\rho_{0}\kappa^{2}}{d^{2}}p', \\ H=H_{0}H',\qquad\Phi=H_{0} \frac{\kappa}{d} \Phi',\qquad t=\frac {d^{2}}{\kappa}t', \end{gathered}$$
where R, called the Rayleigh number, is a dimensionless parameter and
$$R=\frac{g\alpha(T_{0}-T_{1})}{\kappa\nu}d^{3}.$$
Let $$(\varphi'', r'')=(r_{0}\varphi', r')$$. With the above assumptions, we omit the primes and get the approximate equations, which are given as follows:
$$\left\{ \textstyle\begin{array}{l} \frac{\partial u_{\varphi}}{\partial t}+(u\cdot \nabla)u_{\varphi }+\frac{u_{r}u_{\varphi}}{r_{0}} =\triangle u_{\varphi}+ \frac{2}{r_{0}}\frac{\partial u_{r}}{\partial \varphi} -\frac{u_{\varphi}}{r^{2}_{0}} -\frac{\partial p}{\partial\varphi}+ \frac{1}{\rho_{0}}\frac {d^{2}}{\kappa^{2}}H_{0}^{2}H_{r} (\frac{\partial H_{\varphi}}{\partial r} -\frac{\partial H_{r}}{\partial\varphi} ) \\ \hphantom{\frac{\partial u_{\varphi}}{\partial t}+(u\cdot \nabla)u_{\varphi }+\frac{u_{r}u_{\varphi}}{r_{0}} =} + \frac{1}{\rho_{0}}\frac{d^{2}}{\kappa^{2}}H_{0} (K_{1}dr_{0}+ \frac {K_{0}}{d^{2}r_{0}^{2}} ) (\frac{\partial H_{\varphi}}{\partial r}-\frac{\partial H_{r}}{\partial \varphi} ), \\ \frac{\partial u_{r}}{\partial t}+(u\cdot\nabla)u_{r}-\frac {u_{\varphi}^{2}}{r} =\triangle u_{r}-\frac{2u_{r}}{r_{0}^{2}} -\frac{2}{r_{0}}\frac{\partial u_{\varphi}}{\partial\varphi} - \frac{\partial p}{\partial r}+\sqrt{R}T, \\ \frac{\partial T}{\partial t}+(u \cdot\nabla)T=\triangle T+\sqrt{R}u_{r}, \\ \frac{\partial H_{\varphi}}{\partial t}=\frac{\partial}{\partial r}(u_{\varphi}H_{r}-u_{r}H_{\varphi}) +\triangle H_{\varphi}+\frac{2}{r_{0}}\frac{\partial H_{r}}{\partial \varphi} - \frac{H_{\varphi}}{r_{0}^{2}}+\frac{1}{r}\frac{\partial\Phi}{\partial \varphi}+\frac{K_{1}d}{H_{0}}u_{\varphi} \\ \hphantom{\frac{\partial H_{\varphi}}{\partial t}=} + \frac{1}{H_{0}} (K_{1}dr_{0}+\frac{K_{0}}{d^{2}r_{0}^{2}} )\frac {\partial u_{\varphi}}{\partial r}, \\ \frac{\partial H_{r}}{\partial t}=\frac{\partial}{\partial\varphi }(u_{r}H_{\varphi}-u_{\varphi}H_{r}) +\triangle H_{r}-\frac{2H_{r}}{r_{0}^{2}} -\frac{2}{r_{0}} \frac{\partial H_{\varphi}}{\partial\varphi}+\frac {\partial\Phi}{\partial r} -\frac{1}{H_{0}} (K_{1}dr_{0}+ \frac{K_{0}}{d^{2}r_{0}^{2}} )\frac {\partial u_{\varphi}}{\partial\varphi}, \\ \operatorname{div}u=0, \\ \operatorname{div}H=0, \end{array}\displaystyle \right .$$
(2.7)
where $$(\varphi,r)\in M=(0,L)\times(r_{0},r_{0}+1)$$, $$r_{0}$$ is the radius of the sun with the unit of d, $$L=2\pi r_{0}$$, and $$K_{0}$$, $$K_{1}$$ are given as (2.5). $$(u\cdot\nabla)$$, div and are general differential operators.
$$(u\cdot\nabla)=u_{\varphi}\frac{\partial}{\partial\varphi}+u_{r} \frac {\partial}{\partial r},\qquad \triangle=\frac{\partial^{2}}{\partial\varphi^{2}}+\frac{\partial ^{2}}{\partial r ^{2}}, \qquad \operatorname{div}u=\frac{\partial u_{\varphi}}{\partial\varphi}+\frac {\partial u_{r}}{\partial r}.$$
The boundary conditions (2.3) are rewritten as
$$\left \{ \textstyle\begin{array}{l} (u,T,H) (\varphi,r)=(u,T,H) ( \varphi+L, r), \\ u_{r}=0,\qquad H_{r}=0,\qquad T=0,\qquad \frac{\partial u_{\varphi}}{\partial r}= \frac {\partial H_{\varphi}}{\partial r}=0,\qquad r=r_{0}, \\ u_{r}=0,\qquad H_{r}=0,\qquad T=0,\qquad \frac{\partial u_{\varphi}}{\partial r}= \frac{\partial H_{\varphi}}{\partial r}=0,\qquad r=r_{0}+1. \end{array}\displaystyle \right .$$
(2.8)

### 2.3 Abstract operator equation

Now we will show that equations (2.7) can be written in the operator form. Since the velocity field u and the magnetic field H on M are divergence-free, there exist the following stream functions $$f_{1}$$ and $$f_{2}$$ satisfying the given boundary condition:
$$\begin{gathered} u_{\varphi}=\frac{\partial f_{1}}{\partial r}, \qquad u_{r}=- \frac {\partial f_{1}}{\partial\varphi}, \\ H_{\varphi}=\frac{\partial f_{2}}{\partial r}, \quad\quad H_{r}=- \frac {\partial f_{2}}{\partial\varphi}. \end{gathered}$$
Moreover, the following two vector fields:
$$\begin{gathered} \biggl(\frac{2}{r_{0}}\frac{\partial u_{r}}{\partial\varphi}, -\frac {2}{r_{0}} \frac{\partial u_{\varphi}}{\partial\varphi} \biggr) =-\frac{2}{r_{0}}\nabla\frac{\partial f_{1}}{\partial\varphi}, \\ \biggl(\frac{2}{r_{0}}\frac{\partial H_{r}}{\partial\varphi}, - \frac {2}{r_{0}} \frac{\partial H_{\varphi}}{\partial\varphi} \biggr) =-\frac{2}{r_{0}}\nabla \frac{\partial f_{2}}{\partial\varphi}, \end{gathered}$$
are gradient fields, which can be balanced by p and Φ in (2.7). Hence, (2.7) are equivalent to the following equations:
$$\left\{ \textstyle\begin{array}{l} \frac{\partial u_{\varphi}}{\partial t}+(u\cdot \nabla)u_{\varphi }+\frac{u_{r}u_{\varphi}}{r_{0}} =\triangle u_{\varphi} - \frac{u_{\varphi}}{r^{2}_{0}} -\frac{\partial p}{\partial\varphi}+\frac{1}{\rho_{0}}\frac {d^{2}}{\kappa^{2}}H_{0}^{2}H_{r} (\frac{\partial H_{\varphi}}{\partial r} -\frac{\partial H_{r}}{\partial\varphi} ) \\ \hphantom{\frac{\partial u_{\varphi}}{\partial t}+(u\cdot \nabla)u_{\varphi }+\frac{u_{r}u_{\varphi}}{r_{0}} =} + \frac{1}{\rho_{0}}\frac{d^{2}}{\kappa^{2}}H_{0} (K_{1}dr_{0}+ \frac {K_{0}}{d^{2}r_{0}^{2}} ) (\frac{\partial H_{\varphi}}{\partial r}-\frac{\partial H_{r}}{\partial \varphi} ), \\ \frac{\partial u_{r}}{\partial t}+(u\cdot\nabla)u_{r}-\frac {u_{\varphi}^{2}}{r} =\triangle u_{r}-\frac{2u_{r}}{r_{0}^{2}} -\frac{\partial p}{\partial r}+\sqrt{R}T, \\ \frac{\partial T}{\partial t}+(u \cdot\nabla)T=\triangle T+\sqrt{R}u_{r}, \\ \frac{\partial H_{\varphi}}{\partial t}=\frac{\partial}{\partial r}(u_{\varphi}H_{r}-u_{r}H_{\varphi}) +\triangle H_{\varphi} -\frac{H_{\varphi}}{r_{0}^{2}}+\frac{\partial\Phi}{\partial\varphi }+ \frac{K_{1}d}{H_{0}}u_{\varphi} \\ \hphantom{\frac{\partial H_{\varphi}}{\partial t}=} + \frac{1}{H_{0}} (K_{1}dr_{0}+\frac{K_{0}}{d^{2}r_{0}^{2}} )\frac {\partial u_{\varphi}}{\partial r}, \\ \frac{\partial H_{r}}{\partial t}=\frac{\partial}{\partial\varphi }(u_{r}H_{\varphi}-u_{\varphi}H_{r}) +\triangle H_{r}-\frac{2H_{r}}{r_{0}^{2}} +\frac{\partial\Phi}{\partial r} - \frac{1}{H_{0}} (K_{1}dr_{0}+\frac{K_{0}}{d^{2}r_{0}^{2}} )\frac {\partial u_{\varphi}}{\partial\varphi}, \\ \operatorname{div}u=0, \\ \operatorname{div}H=0. \end{array}\displaystyle \right .$$
(2.9)
To get the abstract form of (2.9), we define the following spaces:
$$\begin{gathered} H= \bigl\{ (u,T,H)\in L^{2} \bigl(M, R^{5} \bigr)\mid\operatorname{div} u= \operatorname{div} H=0,(u,T,H) \text{ are periodic in } \varphi \text{-direction} \bigr\} , \\ H_{1}= \bigl\{ (u,T,H)\in H^{2} \bigl(M,R^{5} \bigr) \cap H\mid(u,T,H) \text{ satisfy (2.8)} \bigr\} . \end{gathered}$$
Now, we define the operators $$L=-A+B: H_{1}\rightarrow H$$ and $$G: H_{1}\rightarrow H$$ by
\begin{aligned}& \begin{aligned} AU={}&{-}P \biggl(\triangle u_{\varphi} - \frac{u_{\varphi}}{r^{2}_{0}}+\widetilde{A}F \biggl(\frac{\partial H_{\varphi }}{\partial r}-\frac{\partial H_{r}}{\partial\varphi} \biggr), \triangle u_{r} -\frac{u_{r}}{r^{2}_{0}}, \\ & \triangle T, \triangle H_{\varphi} - \frac{1}{r^{2}_{0}}H_{\varphi}+\frac {1}{H_{0}}F\frac{\partial u_{\varphi}}{\partial r}, \triangle H_{r} -\frac{1}{r^{2}_{0}}H_{r}- \frac{1}{H_{0}}F\frac {\partial u_{\varphi}}{\partial\varphi} \biggr)^{T},\end{aligned} \\& BU=P \biggl(0, \sqrt{R}T, \sqrt{R}u_{r}, \frac {K_{1}d}{H_{0}}u_{\varphi}, 0 \biggr)^{T}, \\& \begin{aligned}GU={}&P \biggl(-(u\cdot\nabla)u_{\varphi}-\frac{u_{r}u_{\varphi }}{r_{0}}+ \frac{1}{\rho_{0}} \frac{d^{2}}{\kappa^{2}}H^{2}_{0}H_{r} \biggl( \frac{\partial H_{\varphi}}{\partial r}-\frac{\partial H_{r}}{\partial \varphi} \biggr), \\ & {-}(u\cdot\nabla)u_{r}- \frac{u_{\varphi }^{2}}{r_{0}}, -(u\cdot \nabla)T, \frac{\partial}{\partial r}(u_{\varphi}H_{r}-u_{r}H_{\varphi }), \frac{\partial}{\partial\varphi}(u_{r}H_{\varphi}-u_{\varphi}H_{r}) \biggr)^{T}, \end{aligned} \end{aligned}
where $$P: L^{2}(M, R^{5})\rightarrow H$$ is the Leray projection, $$U=(u,T,H)\in H_{1}$$ and
$$F=K_{1}r_{0}d+\frac{K_{0}}{r_{0}^{2}d^{2}},\qquad \widetilde{A}=\frac {H_{0}d^{2}}{\rho_{0}\kappa^{2}}.$$
(2.10)
Therefore, the problem (2.9) with boundary conditions (2.8) are equivalent to the following abstract equation.
$$\left \{ \textstyle\begin{array}{l} \frac{\text{d}U}{\text{d}t}=L_{\lambda}U+GU, \\ U(0)=U_{0}, \end{array}\displaystyle \right .$$
(2.11)
where $$\lambda=(R,F)\in R^{2}$$ is the parameter and $$U_{0}$$ is the initial value of (2.9).

## 3 Under the condition $$K_{1}=0$$

In this section, we study (2.9) the case that $$K_{1}=0$$. Hence, the boundary condition in (2.5) satisfies
$$H_{1}=\frac{r_{0}^{2}}{r_{1}^{2}}H_{0}.$$
We choose the Rayleigh number and the F-number as the control parameters, which are given as follows:
$$R=\frac{g\alpha(T_{0}-T_{1})}{\kappa\nu}d^{3}, \qquad F=\frac{H_{0}}{d^{2}}.$$
(3.1)
Then we consider the following eigenvalue equations of (2.9):
$$\left\{ \textstyle\begin{array}{l} \triangle u_{\varphi} - \frac{u_{\varphi}}{r^{2}_{0}} -\frac{\partial p}{\partial\varphi}+\widetilde{A}F (\frac{\partial H_{\varphi}}{\partial r}- \frac{\partial H_{r}}{\partial\varphi} )=\beta u_{\varphi}, \\ \triangle u_{r}-\frac{2u_{r}}{r_{0}^{2}} -\frac{\partial p}{\partial r}+\sqrt{R}T= \beta u_{r}, \\ \triangle T+\sqrt{R}u_{r}=\beta T, \\ \triangle H_{\varphi} -\frac{H_{\varphi}}{r_{0}^{2}}+\frac{\partial\Phi}{\partial\varphi} + \frac{1}{H_{0}}F\frac{\partial u_{\varphi}}{\partial r}=\beta H_{\varphi}, \\ \triangle H_{r}-\frac{2H_{r}}{r_{0}^{2}} +\frac{\partial\Phi}{\partial r} - \frac{1}{H_{0}}F\frac{\partial u_{\varphi}}{\partial\varphi}=\beta H_{r}, \\ \operatorname{div}u=0, \\ \operatorname{div}H=0, \end{array}\displaystyle \right .$$
(3.2)
with the boundary conditions (2.8). We proceed with the separation of variables. Under the periodic boundary condition (2.8), the problem (3.2) possesses two eigenvectors: Ψ and Ψ̃ in the following forms:
\begin{aligned}& \Psi\text{:}\quad\left \{ \textstyle\begin{array}{l} u_{\varphi}= \cos a_{k}\varphi\cdot h'_{k}(r), \\ u_{r}=a_{k}\sin a_{k}\varphi\cdot h_{k}(r), \\ T=\sin a_{k}\varphi\cdot T_{k}(r), \\ H_{\varphi}=\cos a_{k}\varphi\cdot g'_{k}(r), \\ H_{r}=a_{k}\sin a_{k}\varphi\cdot g_{k}(r), \\ p=p_{k}(r)\sin a_{k}\varphi, \\ \Phi=\Phi_{k}(r)\sin a_{k}\varphi, \end{array}\displaystyle \right . \end{aligned}
(3.3)
\begin{aligned}& \tilde{\Psi}\text{:}\quad \left\{ \textstyle\begin{array}{l} u_{\varphi}= \sin a_{k}\varphi\cdot h'_{k}(r), \\ u_{r}=-a_{k}\cos a_{k}\varphi\cdot h_{k}(r), \\ T=-\cos a_{k}\varphi\cdot T_{k}(r), \\ H_{\varphi}=\sin a_{k}\varphi\cdot g'_{k}(r), \\ H_{r}=-a_{k}\cos a_{k}\varphi\cdot g_{k}(r), \\ \tilde{p}=-p_{k}(r)\cos a_{k}\varphi, \\ \tilde{\Phi}=-\Phi_{k}(r)\cos a_{k}\varphi, \end{array}\displaystyle \right . \end{aligned}
(3.4)
where $$a_{k}= 2k\pi/L$$. Putting $$(\Psi,p,\Phi)$$ and $$(\tilde{\Psi},\tilde{p},\tilde{\Phi})$$ into (3.2), respectively, we deduce from (3.2), (3.3) and (3.4) that $$(h_{k},g_{k},T_{k},p_{k},\Phi_{k})$$ satisfy the eigenvalue problems,
$$\left\{ \textstyle\begin{array}{l} D^{\ast}Dh_{k}- \frac{1}{r_{0}^{2}}Dh_{k}-a_{k}p_{k}+ \widetilde{A}F D^{\ast}g_{k}=\beta Dh_{k}, \\ a_{k}D^{\ast}h_{k}-\frac{2}{r_{0}^{2}}a_{k}h_{k}+ \sqrt {R}T_{k}-Dp_{k}=\beta a_{k}h_{k}, \\ D^{\ast}T_{k}+\sqrt{R}a_{k}h_{k}=\beta T_{k}, \\ D^{\ast}Dg_{k}-\frac{1}{r_{0}^{2}}Dg_{k}+a_{k} \Phi_{k}+\frac {1}{H_{0}}K_{1}dDh_{k}+ \frac{1}{H_{0}}FD^{2}h_{k}=\beta Dg_{k}, \\ a_{k}D^{\ast}g_{k}-\frac{2}{r_{0}^{2}}a_{k}g_{k}+D \Phi_{k}+\frac {1}{H_{0}}Fa_{k}Dh_{k}=\beta a_{k}g_{k}, \end{array}\displaystyle \right .$$
(3.5)
for any $$k\in Z$$, with $$k\neq0$$, where
$$D=\frac{d}{dr},\qquad D^{\ast}=D^{2}-a_{k}^{2}.$$
We infer from (3.5) and the boundary conditions (2.8) that the $$h_{k}$$ satisfy the equations
\begin{aligned}& \begin{aligned}[b] &\biggl[ \biggl(D^{\ast} \bigl( \beta-D^{\ast} \bigr)^{2}-\frac{1}{r_{0}^{2}} \bigl(D^{\ast }-a^{2} \bigr) \bigl(D^{\ast}-\beta \bigr)+Ra^{2} \biggr) \biggl(D^{\ast} \bigl(\beta-D^{\ast} \bigr)+\frac{1}{r_{0}^{2}} \bigl(D^{\ast}-a^{2} \bigr) \biggr) \\ &\quad{}+ \widetilde{A}F\frac {1}{H_{0}}FD^{\ast} D \bigl(D^{\ast}-\beta \bigr)D^{\ast}D \biggr]h_{k}=0, \end{aligned} \end{aligned}
(3.6)
\begin{aligned}& h_{k}=D^{2}h_{k}=D^{4}h_{k}=0\quad \text{at } r=r_{0}, r_{0}+1. \end{aligned}
(3.7)
By (3.7), the $$h_{k}$$ are sine functions, i.e.,
$$h_{k}=\sin l\pi(r-r_{0})\quad (l=1,2,3,\ldots).$$
(3.8)
Putting (3.8) into (3.6), we see that
$$\begin{gathered} \alpha_{kl}^{4} \bigl(\alpha_{kl}^{2}+\beta \bigr)^{3}+ \frac{2}{r_{0}^{2}}\alpha _{kl}^{2} \bigl( \alpha_{kl}^{2}+ \beta \bigr)^{2} \bigl( \alpha_{kl}^{2}+a_{k}^{2} \bigr) + \frac{1}{r_{0}^{4}} \bigl(\alpha_{kl}^{2}+a_{k}^{2} \bigr)^{2} \bigl(\alpha _{kl}^{2}+\beta \bigr) \\ \quad{}-Ra^{2} \biggl[\alpha_{kl}^{2} \bigl( \alpha_{kl}^{2}+\beta \bigr) +\frac{1}{r_{0}^{2}} \bigl( \alpha_{kl}^{2}+a_{k}^{2} \bigr) \biggr]+ \frac{d^{2}}{\rho _{0}\kappa^{2}}F^{2}\alpha_{kl}^{4} l^{2}\pi^{2} \bigl(\alpha_{kl}^{2}+\beta \bigr)=0, \end{gathered}$$
(3.9)
where
$$\alpha_{kl}^{2}=a_{k}^{2}+l^{2} \pi^{2},\qquad a_{k}=2k\pi/L.$$
(3.10)
It is well known that all solutions of the cubic equation (3.9) $$\beta_{kl}$$ are eigenvalues of (3.2). Let $$\beta_{kl}^{i}$$ ($$1\leq i\leq3$$) be three zero points of (3.9). It is easy to see that $$Re\beta_{kl}^{1}\geq Re\beta _{kl}^{2}\geq Re\beta_{kl}^{3}$$. Furthermore, $$\beta=0$$ is a zero point of (3.9) if and only if
$$b_{kl}-c_{kl}R+d_{kl}F^{2}=0,$$
where
$$\left\{ \textstyle\begin{array}{l} b_{kl}= \alpha_{kl}^{10}+\frac{2}{r_{0}^{2}}\alpha_{kl}^{6} (\alpha _{kl}^{2}+a_{k}^{2} ) + \frac{1}{r_{0}^{4}} (\alpha_{kl}^{2}+a_{k}^{2} ), \\ c_{kl}=a_{k}^{2} [\alpha_{kl}^{4}+ \frac{1}{r_{0}^{2}} (\alpha _{kl}^{2}+a_{k}^{2} ) ], \\ d_{kl}=\frac{d^{2}}{\rho_{0}\kappa^{2}}\alpha_{kl}^{6}l^{2} \pi^{2}, \end{array}\displaystyle \right .$$
(3.11)
where $$\alpha_{kl}$$ and $$a_{k}$$ are as defined in (3.10). When $$F=0$$, we can get the critical Rayleigh number $$R_{C}$$
$$R_{C}=\min_{(k,l)}\frac{b_{kl}}{c_{kl}}= \min_{(k,l)}\frac{\alpha _{kl}^{10}+\frac{2}{r_{0}^{2}}\alpha_{kl}^{6}(\alpha_{kl}^{2} +a_{k}^{2})+\frac{1}{r_{0}^{4}}(\alpha_{kl}^{2}+ \alpha_{kl}^{2})}{a_{k}^{2}[\alpha_{kl}^{4}+\frac{1}{r_{0}^{2}}(\alpha _{kl}^{2}+a_{k}^{2})]}=\frac{b_{k_{0}l_{0}}}{c_{k_{0}l_{0}}}.$$
(3.12)
Then the critical parameter curve in Figure 2 is given by
$$L= \bigl\{ (R,F)\in R^{2}_{+}\mid b_{k_{0}l_{0}}-c_{k_{0}l_{0}}R+d_{k_{0}l_{0}}F^{2}=0 \bigr\} ,$$
(3.13)
where $$b_{k_{0}l_{0}}$$, $$c_{k_{0}l_{0}}$$ and $$d_{k_{0}l_{0}}$$ as in (3.11) and (3.12).

### Lemma 3.1

Let $$(k,l)=(k_{0},l_{0})$$ minimize the right hand side of (3.12) and the zero point $$\beta_{k_{0}l_{0}}^{1}$$ of (3.9) is a real single eigenvalue of (3.2) near curve L. Then $$\beta _{k_{0}l_{0}}^{1}$$ satisfies
$$\beta_{k_{0}l_{0}}^{1}\left \{ \textstyle\begin{array}{l@{\quad}l} < 0, &(R,F)\in\textit{I} , \\ =0, &(R,F)\in L, \\ >0, &(R,F)\in\textit{II}, \end{array}\displaystyle \right .$$
$$\beta_{kl}^{i}<0$$, $$\forall(k,l,i)\neq(k_{0},l_{0},1)$$, and $$(R,F)$$ near L, where the region I and II are as Figure  2.

### Proof

We first prove that
$$\beta_{k_{0}l_{0}}^{1}< 0, \quad(R,F)\in\text{I}.$$
(3.14)
In fact, as $$(R,F)=(0,0)$$, the solutions of (3.9) are
$$\beta_{k_{0}l_{0}}^{1}=-\alpha_{k_{0}l_{0}}^{2},\qquad \beta _{k_{0}l_{0}}^{2}=\beta_{k_{0}l_{0}}^{3}=- \alpha_{k_{0}l_{0}}^{2} -\frac{\alpha_{k_{0}l_{0}}^{2}+a_{k}^{2}}{\alpha_{k_{0}l_{0}}^{2}r_{0}^{2}}.$$
Namely, $$\beta_{k_{0}l_{0}}^{1}(0,0)<0$$. Since $$\beta_{k_{0}l_{0}}^{1}(R,F)$$ are continuous on $$(R,F)$$, then (3.14) holds true. Next, we are ready to prove that the first eigenvalue $$\beta _{k_{0}l_{0}}^{1}(R,F)>0$$ in region II near the curve L. Let
\begin{aligned} g_{k_{0}l_{0}}={}& \alpha_{k_{0}l_{0}}^{4} \bigl(\alpha_{k_{0}l_{0}}^{2}+\beta \bigr)^{3}+ \frac {2}{r_{0}^{2}}\alpha_{k_{0}l_{0}}^{2} \bigl(\alpha_{k_{0}l_{0}}^{2}+ \beta \bigr)^{2} \bigl(\alpha_{k_{0}l_{0}}^{2}+a_{k_{0}}^{2} \bigr) \\ &+\frac{1}{r_{0}^{4}} \bigl(\alpha_{k_{0}l_{0}}^{2}+a_{k_{0}}^{2} \bigr)^{2} \bigl(\alpha _{k_{0}l_{0}}^{2}+\beta \bigr) -Ra^{2} \biggl[\alpha_{k_{0}l_{0}}^{2} \bigl( \alpha_{k_{0}l_{0}}^{2}+\beta \bigr) +\frac{1}{r_{0}^{2}} \bigl( \alpha_{k_{0}l_{0}}^{2}+a_{k_{0}}^{2} \bigr) \biggr] \\ &+\frac{d^{2}}{\rho_{0}\kappa^{2}}F^{2}\alpha_{k_{0}l_{0}}^{4} l^{2}\pi^{2} \bigl(\alpha_{k_{0}l_{0}}^{2}+\beta \bigr) \\ ={}&\alpha_{k_{0}l_{0}}^{4}\beta^{3}+ \biggl[3 \alpha_{k_{0}l_{0}}^{6}+\frac {2}{r_{0}^{2}}\alpha_{k_{0}l_{0}}^{2} \bigl(\alpha_{k_{0}l_{0}}^{2}+a_{k}^{2} \bigr) \biggr]\beta^{2} \\ &+ \biggl[3\alpha _{k_{0}l_{0}}^{8}+ \frac{4}{r_{0}^{2}}\alpha_{k_{0}l_{0}}^{4} \bigl( \alpha_{k_{0}l_{0}}^{2}+a_{k}^{2} \bigr)\\ &+\frac{1}{r_{0}^{4}} \bigl(\alpha _{k_{0}l_{0}}^{2}+a_{k}^{2} \bigr)^{2}-Ra^{2}_{k}\alpha_{k_{0}l_{0}}^{2} +\frac{d^{2}}{\rho_{0}^{2}\kappa^{2}}F^{2}\alpha^{4}l_{0}^{2} \pi ^{2} \biggr]\beta\\ &+L_{k_{0}l_{0}}, \end{aligned}
where $$L_{k_{0}l_{0}}=b_{k_{0}l_{0}}-c_{k_{0}l_{0}}R+d_{k_{0}l_{0}}F^{2}$$. When $$L_{k_{0}l_{0}}= 0$$, $$\Gamma_{1}$$ is the graph of $$g_{k_{0}l_{0}}$$; when $$L_{k_{0}l_{0}}< 0$$, $$\Gamma_{2}$$ is the graph of $$g_{k_{0}l_{0}}$$, as shown in Figure 3. Then the intersecting point of $$\Gamma_{i}$$ ($$i=1,2$$) and β-axis is the first eigenvalue $$\beta_{k_{0}l_{0}}^{1}$$. When $$L_{k_{0}l_{0}}=0$$, the intersecting point of $$\Gamma_{1}$$ and β-axis, $$\beta _{k_{0}l_{0}}^{1}= 0$$, and $$(R,F)$$ is in curve L. When $$L_{k_{0}l_{0}}< 0$$, the intersecting point of $$\Gamma_{2}$$ and β-axis, $$\beta _{k_{0}l_{0}}^{1}> 0$$, and $$(R,F)$$ is in region II. At last, we will show that $$\beta_{kl}^{i}<0$$, $$\forall(k,l)\neq(k_{0},l_{0})$$, when $$(R,F)$$ in region I or in region II near the curve L. We only need
$$\beta_{kl}^{i}\neq0,\quad \forall(R,F)\in \text{I}, (k,l)\neq(k_{0},l_{0}).$$
(3.15)
Otherwise, if (3.15) does not hold true, there exist a pair of $$(R_{1},F_{1})$$, such that $$\beta_{kl}^{i}(R_{1},F_{1})=0$$. That is to say, there exists a curl $$L_{kl}$$, which passes through the point $$(R_{1},F_{1})$$ and intersects with R-axis at $$R_{C_{1}}$$, as shown in Figure 4. It is easy to find that $$R_{C}> R_{C_{1}}$$, which makes a contradiction that $$R_{C}$$ is the minimum critical Rayleigh number. So, we prove (3.15). When $$(R,F)=(0,0)$$, we have
$$\beta_{kl}^{1}=-\alpha_{kl}^{2}< 0,\qquad \beta_{kl}^{2}=\beta _{kl}^{3}=- \alpha_{kl}^{2} -\frac{\alpha_{kl}^{2}+a_{k}^{2}}{\alpha_{kl}^{2}r_{0}^{2}}< 0,$$
which implies $$\beta_{kl}^{i}<0$$, $$\forall(k,l)\neq(k_{0},l_{0})$$. Because $$\beta_{kl}^{i}(R,F)$$ are continuous, $$\beta_{kl}^{i}<0$$ in region I or in region II near the curve L. This proves Lemma 3.1. Figure 3 The graph of $$\pmb{g_{k_{0}l_{0}}}$$ . As the figure shows: When $$L_{k_{0}l_{0}}= 0$$, $$\Gamma_{1}$$ is the graph of $$g_{k_{0}l_{0}}$$; when $${L_{k_{0}l_{0}}< 0}$$, $$\Gamma_{2}$$ is the graph of $$g_{k_{0}l_{0}}$$,Then the intersecting point of $$\Gamma_{i}$$ ($$i=1,2$$) and β-axis is the first eigenvalue $$\beta_{k_{0}l_{0}}^{1}$$. When $$L_{k_{0}l_{0}}=0$$, the intersecting point of $$\Gamma_{1}$$ and β-axis, $$\beta _{k_{0}l_{0}}^{1}= 0$$, and $${(R,F)}$$ is in curve L. When $$L_{k_{0}l_{0}}< 0$$, the intersecting point of $$\Gamma_{2}$$ and β-axis, $$\beta _{k_{0}l_{0}}^{1}> 0$$, and $$(R,F)$$ is in region II. Figure 4 The graph of $$\pmb{L_{kl}}$$ . The curl $$L_{kl}$$ passes through the point $$(R_{1},F_{1})$$ and intersects with the R-axis at $$R_{C_{1}}$$, it is easy to find that $$R_{C}> R_{C_{1}}$$.  □

Based on Theorem 2.3.1 in , Theorem 2 in  and Lemma 3.1, we derive the following theorem.

### Theorem 3.2

The critical parameter curve L divides the RF-plane into two regions I and II (see Figure  2), such that the following conclusions hold true.
1. 1.

If $$(R,F)\in\textit{I}$$, the problem (2.8) and (2.9) have no bifurcation, and the basic flow in (2.4) with $$K_{1}=0$$ is stable.

2. 2.

If $$(R,F)\in\textit{II}$$, there exists a bifurcation solution.

3. 3.

The transition takes place as the control parameter $$(R,F)$$ crosses the critical curve L from region I into II.

## 4 For the condition that $$K_{1}\neq0$$ and $$F=0$$

### 4.1 Eigenvalue problem

Consider the case that $$F=0$$, and F is given by (2.10). Then equations (2.9) become
$$\left\{ \textstyle\begin{array}{l} \frac{\partial u_{\varphi}}{\partial t}+(u\cdot \nabla)u_{\varphi }+\frac{u_{r}u_{\varphi}}{r_{0}} =\triangle u_{\varphi} - \frac{u_{\varphi}}{r^{2}_{0}} -\frac{\partial p}{\partial\varphi}+\frac{1}{\rho_{0}}\frac {d^{2}}{\kappa^{2}}H_{0}^{2}H_{r} (\frac{\partial H_{\varphi}}{\partial r} -\frac{\partial H_{r}}{\partial\varphi} ), \\ \frac{\partial u_{r}}{\partial t}+(u\cdot\nabla)u_{r}-\frac {u_{\varphi}^{2}}{r} =\triangle u_{r}-\frac{2u_{r}}{r_{0}^{2}} -\frac{\partial p}{\partial r}+\sqrt{R}T, \\ \frac{\partial T}{\partial t}+(u \cdot\nabla)T=\triangle T+\sqrt{R}u_{r}, \\ \frac{\partial H_{\varphi}}{\partial t}=\frac{\partial}{\partial r}(u_{\varphi}H_{r}-u_{r}H_{\varphi}) +\triangle H_{\varphi} -\frac{H_{\varphi}}{r_{0}^{2}}+\frac{\partial\Phi}{\partial\varphi }+ \frac{K_{1}d}{H_{0}}u_{\varphi}, \\ \frac{\partial H_{r}}{\partial t}=\frac{\partial}{\partial\varphi }(u_{r}H_{\varphi}-u_{\varphi}H_{r}) +\triangle H_{r}-\frac{2H_{r}}{r_{0}^{2}} +\frac{\partial\Phi}{\partial r}, \\ \operatorname{div} u=0, \\ \operatorname{div}H=0. \end{array}\displaystyle \right .$$
(4.1)
Essentially, the control parameter in (4.1) is only the Rayleigh number R in (3.1). Then the eigenvalues and the eigenvectors Ψ, Ψ̃ are similar to those in Section 3, and we can get the corresponding eigenvalue problems
$$\left\{ \textstyle\begin{array}{l} D^{\ast}Dh_{k}- \frac{1}{r_{0}^{2}}Dh_{k}-a_{k}p_{k}=\beta Dh_{k}, \\ a_{k}D^{\ast}h_{k}-\frac{2}{r_{0}^{2}}a_{k}h_{k}+ \sqrt {R}T_{k}-Dp_{k}=\beta a_{k}h_{k}, \\ D^{\ast}T_{k}+\sqrt{R}a_{k}h_{k}=\beta T_{k}, \\ D^{\ast}Dg_{k}-\frac{1}{r_{0}^{2}}Dg_{k}+a_{k} \Phi_{k}+\frac {1}{H_{0}}K_{1}dDh_{k}=\beta Dg_{k}, \\ a_{k}D^{\ast}g_{k}-\frac{2}{r_{0}^{2}}a_{k}g_{k}+D \Phi_{k}=\beta a_{k}g_{k}. \end{array}\displaystyle \right .$$
(4.2)
From (4.2), we see that $$g_{k}$$ ($$k=1,2,\ldots$$) satisfy the equations
\begin{aligned}[b] &\biggl[- \bigl(D^{\ast} \bigr)^{2} \bigl(D^{\ast}-\beta \bigr)^{3}+ \frac{2}{r_{0}^{2}} \bigl(D^{\ast }-a_{k}^{2} \bigr)D^{\ast} \bigl(D^{\ast}-\beta \bigr)^{2} \\ &\quad{}-\frac{1}{r_{0}^{4}} \bigl(D^{\ast}-a_{k}^{2} \bigr)^{2} \bigl(D^{\ast}-\beta \bigr)-Ra_{k}^{2}D^{\ast} \bigl(D^{\ast}-\beta \bigr) \\ &\quad{}+\frac{1}{r_{0}^{2}}Ra_{k}^{2} \bigl(D^{\ast}-a_{k}^{2} \bigr) \biggr]g_{k}=0,\quad k=1,2,\ldots. \end{aligned}
(4.3)
By (4.2) and the boundary conditions (2.8), $$g_{k}$$ are the sine functions
$$g_{kl}=\sin l\pi(r-r_{0}),\quad\text{for } l=1,2,3, \ldots.$$
Moreover, $$h_{k}$$ and $$T_{k}$$ are determined by the following:
\begin{aligned}& D^{\ast} \bigl(\beta-D^{\ast} \bigr)g_{k}+ \frac{1}{r_{0}^{2}} \bigl(D^{\ast }-a_{k}^{2} \bigr)g_{k}=\frac{1}{H_{0}}K_{1}dD^{2}h_{k}, \end{aligned}
(4.4)
\begin{aligned}& \sqrt{R}a_{k}T_{k}= \bigl(D^{\ast} \bigr) \bigl(D^{\ast}-\beta \bigr)h_{k}-\frac {1}{r_{0}^{2}} \bigl(D^{\ast}-a_{k}^{2} \bigr)h_{k}. \end{aligned}
(4.5)
Substituting $$g_{kl}$$ into (4.3), we see that the eigenvalue β of (4.2) satisfies the cubic equation
\begin{aligned}[b] &\alpha_{kl}^{4} \bigl(\alpha_{kl}^{2}+\beta \bigr)^{3}+ \frac{2}{r_{0}^{2}} \bigl(\alpha _{kl}^{2}+a_{k}^{2} \bigr) \bigl(\alpha_{kl}^{2}+\beta \bigr)^{2} \alpha_{kl}^{2} +\frac{1}{r_{0}^{4}} \bigl( \alpha_{kl}^{2}+a_{k}^{2} \bigr)^{2} \bigl(\alpha _{kl}^{2}+\beta \bigr) \\ &\quad{}-Ra_{k}^{2}\alpha_{kl}^{2} \bigl( \alpha_{kl}^{2}+\beta \bigr)- \frac{1}{r_{0}^{2}}Ra_{k}^{2} \bigl(\alpha_{kl}^{2}+a_{k}^{2} \bigr)=0. \end{aligned}
(4.6)
Hence, all eigenvalues and eigenvectors can be derived from the following cases.
1. 1.
For $$(k,l)=(0,l)$$, we have
$$\begin{gathered} \beta_{0l}^{1}=-l^{2}\pi^{2},\qquad \beta_{0l}^{2}=-l^{2}\pi^{2}- \frac {1}{r_{0}^{2}}, \\\Psi_{0l}^{1}= \bigl(0,0,\sin l\pi(r-r_{0}),0,0 \bigr),\qquad\tilde{\Psi} _{0l}^{1}= \bigl(0,0,\cos l \pi(r-r_{0}),0,0 \bigr), \\\Psi_{0l}^{2}= \bigl(0,0,0,\cos l\pi(r-r_{0}),0 \bigr),\qquad\tilde{\Psi }_{0l}^{2}= \bigl(0,0,0,\sin l \pi(r-r_{0}),0 \bigr). \end{gathered}$$

2. 2.
For $$(k,l)=(k,0)$$, the eigenvalues and eigenvectors are given by
$$\begin{gathered} \beta_{k0}=-a_{k}^{2}-\frac{2}{r_{0}^{2}}, \\\Psi_{k0}=(0,0,0,0,\sin a_{k}\varphi),\qquad\tilde{\Psi }_{k0}=(0,0,0,0,\cos a_{k}\varphi). \end{gathered}$$

3. 3.
For $$k\neq0$$ and $$l\neq0$$, it is well known that all solutions of equation (4.6) are eigenvalues. Let $$\beta_{kl}^{1}(R)$$ ($$1\leq i\leq3$$) be three zero points of (4.6) such that
$$Re\beta_{kl}^{1}\geq Re\beta_{kl}^{2} \geq Re\beta_{kl}^{3}.$$

Then, by (3.3), (3.4), (4.4) and (4.5), the eigenvectors $$\Psi_{kl}^{i}$$ and $$\tilde{\Psi}_{kl}^{i}$$ corresponding to $$\beta_{kl}^{1}$$ can be written as
\begin{aligned}& \Psi_{kl}^{i}= \left\{ \textstyle\begin{array}{l} l\pi h_{kl}^{i}\cos a_{k}\varphi\cdot\cos l\pi(r-r_{0}), \\ a_{k}h_{kl}^{i}\sin a_{k}\varphi\cdot \sin l\pi(r-r_{0}) , \\ T_{kl}^{i}\sin a_{k}\varphi\cdot\sin l \pi(r-r_{0}), \\ l\pi\cos a_{k}\varphi\cdot\cos l\pi(r-r_{0}), \\ a_{k}\sin a_{k}\varphi\cdot\sin l\pi(r-r_{0}), \end{array}\displaystyle \right . \end{aligned}
(4.7)
\begin{aligned}& \tilde{ \Psi}_{kl}^{i}= \left\{ \textstyle\begin{array}{l} l\pi h_{kl}^{i}\sin a_{k}\varphi\cdot\cos l \pi(r-r_{0}), \\ -a_{k}h_{kl}^{i}\cos a_{k}\varphi \cdot\sin l\pi(r-r_{0}) , \\ -T_{kl}^{i}\cos a_{k}\varphi\cdot\sin l \pi(r-r_{0}), \\ l\pi\sin a_{k}\varphi\cdot\cos l\pi(r-r_{0}), \\ -a_{k}\cos a_{k}\varphi\cdot\sin l\pi(r-r_{0}), \end{array}\displaystyle \right . \end{aligned}
(4.8)
where
$$\begin{gathered} h_{kl}^{i}=\frac{H_{0}[\alpha_{kl}^{2}(\alpha_{kl}^{2}+\beta _{kj}^{i})-\frac{1}{r_{0}^{2}}(\alpha_{kl}^{2}+a_{k}^{2})]}{ l^{2}\pi^{2}K_{1}d}, \\ T_{kl}^{i}=\frac{H_{0}[\alpha_{kl}^{4}(\alpha_{kl}^{2}+\beta _{kj}^{i})^{2}-\frac{1}{r_{0}^{4}}(\alpha_{kl}^{2}+a_{k}^{2})^{2}]}{ \sqrt{R}a_{k}l^{2}\pi^{2}K_{1}d}. \end{gathered}$$
It is clear that the eigenvalues and eigenvectors have the following properties.
1. 1.

$$\beta_{0l}^{i}$$, $$\beta_{k0}$$ and $$\beta_{kl}^{i}$$ ($$k\neq0$$, $$l\neq 0$$) consist of all eigenvalues of equation (4.2), and all eigenvectors form a basis of H;

2. 2.

$$\beta_{0l}^{i}< 0$$ ($$i=1,2$$) and $$\beta_{k0}< 0$$;

3. 3.

the eigenvalues $$\beta_{kl}^{i}$$ ($$k\neq0$$, $$l\neq0$$) only depend on the Rayleigh number.

Now, we are ready to determine the dual eigenvector. The dual eigenvectors $$\Psi_{kl}^{i\ast}$$ and $$\tilde{\Psi}_{kl}^{i\ast}$$ are given by
\begin{aligned}& \Psi_{kl}^{i\ast}=\left \{ \textstyle\begin{array}{l} l\pi h_{kl}^{i\ast}\cos a_{k}\varphi\cdot\cos l\pi(r-r_{0}), \\ a_{k}h_{kl}^{i\ast}\sin a_{k}\varphi\cdot \sin l\pi(r-r_{0}) , \\ T_{kl}^{i\ast}\sin a_{k}\varphi\cdot\sin l \pi(r-r_{0}), \\ l\pi\cos a_{k}\varphi\cdot\cos l\pi(r-r_{0}), \\ a_{k}\sin a_{k}\varphi\cdot\sin l\pi(r-r_{0}), \end{array}\displaystyle \right . \end{aligned}
(4.9)
\begin{aligned}& \tilde{ \Psi}_{kl}^{i\ast}=\left \{ \textstyle\begin{array}{l} l\pi h_{kl}^{i\ast}\sin a_{k}\varphi\cdot\sin l \pi(r-r_{0}), \\ -a_{k}h_{kl}^{i\ast}\cos a_{k}\varphi \cdot\cos l\pi(r-r_{0}) , \\ -T_{kl}^{i\ast}\cos a_{k}\varphi\cdot\cos l \pi(r-r_{0}), \\ l\pi\sin a_{k}\varphi\cdot\sin l\pi(r-r_{0}), \\ -a_{k}\cos a_{k}\varphi\cdot\cos l\pi(r-r_{0}), \end{array}\displaystyle \right . \end{aligned}
(4.10)
where
$$\begin{gathered} h_{kl}^{i\ast}=\frac{-\frac{K_{1}d}{H_{0}}l^{2}\pi^{2}(\beta _{kl}^{i}+\alpha_{kl}^{2})}{ Ra_{k}^{2}+(\beta_{kl}^{i}+\alpha_{kl}^{2})(\alpha_{kl}^{4}-\beta _{kl}^{i}\alpha_{kl}^{2}-\frac{2}{r_{0}^{2}}a_{k}^{2} -\frac{1}{r_{0}^{2}}l^{2}\pi^{2})}, \\ T_{kl}^{i\ast}=\frac{-\frac{K_{1}d}{H_{0}}l^{2}\pi^{2}a_{k}\sqrt{R}}{ Ra_{k}^{2}+(\beta_{kl}^{i}+\alpha_{kl}^{2})(\alpha_{kl}^{4}-\beta _{kl}^{i}\alpha_{kl}^{2}-\frac{2}{r_{0}^{2}}a_{k}^{2} -\frac{1}{r_{0}^{2}}l^{2}\pi^{2})}. \end{gathered}$$
Thus, all dual eigenvectors consist of $$\Psi_{0l}^{1\ast}=(0,0,\sin l\pi (r-r_{0}),0,0)$$, $$\Psi_{0l}^{2\ast}=(0,0,0, \sin l\pi(r-r_{0}),0)$$, $$\Psi_{k0}^{\ast}=(0,0,0,0,\sin a_{k}\varphi)$$, $$\tilde{\Psi}_{k0}^{\ast }=(0,0,0,0,\cos a_{{k}}\varphi)$$.
In this part, we are ready to study the critical-crossing of the first eigenvalue. Since the eigenvalues $$\beta_{kl}^{i}$$ ($$1\leq i\leq3$$, $$k\neq0$$, $$l\neq0$$) only depend on the Rayleigh number, it suffices to focus on the eigenvalue problem (4.6). If $$\beta=0$$ is a zero point of (4.6), we can get
$$\alpha_{kl}^{10}+\frac{2}{r_{0}^{2}} \bigl( \alpha_{kl}^{2}+a_{k}^{2} \bigr)\alpha _{kl}^{6}+\frac{1}{r_{0}^{4}} \bigl(\alpha_{kl}^{2} +a_{k}^{2} \bigr)^{2}\alpha_{kl}^{2} -Ra_{k}^{2} \biggl(\alpha_{kl}^{4}+ \frac{1}{r_{0}^{2}}\alpha_{kl}^{2}+\frac {1}{r_{0}^{2}}a_{k}^{2} \biggr)=0.$$
In this case, we have
$$R=\frac{\alpha_{kl}^{10}+\frac{2}{r_{0}^{2}}(\alpha _{kl}^{2}+a_{k}^{2})\alpha_{kl}^{6}+\frac{1}{r_{0}^{4}}(\alpha_{kl}^{2} +a_{k}^{2})^{2}\alpha_{kl}^{2}}{ a_{k}^{2}(\alpha_{kl}^{4}+\frac{1}{r_{0}^{2}}\alpha_{kl}^{2}+\frac {1}{r_{0}^{2}}a_{k}^{2})}.$$
(4.11)
Hence the critical Rayleigh number $$R_{C}$$ has a similar form to Section 3, which is given by
$$R_{C}=\min_{(k,l)}\frac{\alpha_{kl}^{10}+\frac{2}{r_{0}^{2}}\alpha _{kl}^{6}(\alpha_{kl}^{2}+a_{k}^{2})+\frac{1}{r_{0}^{4}}(\alpha_{kl}^{2}+ \alpha_{kl}^{2})}{a_{k}^{2}[\alpha_{kl}^{4}+\frac{1}{r_{0}^{2}}(\alpha _{kl}^{2}+a_{k}^{2})]}= \frac{b_{k_{0}l_{0}}}{c_{k_{0}l_{0}}}.$$
Thus, we have the following lemma.

### Lemma 4.1

Let $$(k_{0},l_{0})$$ minimize the right hand side of (4.11). Assume the zero point $$\beta_{k_{0}l_{0}}^{1}$$ of (4.6) is a real single eigenvalue of (4.1) near $$R_{C}$$, then $$\beta_{k_{0}l_{0}}^{1}$$ satisfies
$$\begin{gathered} \beta_{k_{0}l_{0}}^{1} \left\{ \textstyle\begin{array}{l@{\quad}l} < 0,& \textit{if } R< R_{C}, \\ =0, &\textit{if } R=R_{C}, \\ >0, &\textit{if } R>R_{C}, \end{array}\displaystyle \right . \\ Re\beta_{kl}^{i}(R_{C})< 0,\quad\forall(k,l,i) \neq(k_{0},l_{0},1). \end{gathered}$$

The proof is similar to the proof of Lemma 3.1, so we omit it.

### 4.2 Transition theory with $$F=0$$

In this section, we consider the transition theorem of problem (4.1) with boundary conditions (2.8) with $$K_{1}\neq0$$ and $$F=0$$, where $$K_{1}$$, F as in (2.10). Then we have the following theorem.

### Theorem 4.2

Let $$K_{1}\neq0$$ and $$F=0$$, we have the following conclusions for equations (4.1) with boundary conditions (2.8).
1. 1.

Equation (4.1) bifurcates from $$((u,T,H),R)=(0,R_{C})$$ to an attractor $$\Sigma_{R}\in H_{1}$$, only consisting of a steady state solution.

2. 2.
The steady state solution $$(u,T,H)=(u_{\varphi},u_{r},T,H_{\varphi },H_{r})$$ can be expressed as
$$(u_{\varphi},u_{r},T,H_{\varphi},H_{r})=C \bigl( \beta _{k_{0}l_{0}}^{1}(R) \bigr)^{\frac{1}{2}} \bigl(x \Psi_{k_{0}l_{0}}^{1} +y\tilde{\Psi}_{k_{0}l_{0}}^{1} \bigr)+o \bigl(\big|\beta_{k_{0}l_{0}}^{1}\big|^{\frac{1}{2}} \bigr),$$
where $$C>0$$ is a constant, $$\Psi_{k_{0}l_{0}}^{1}$$ and $$\tilde{\Psi }_{k_{0}l_{0}}^{1}$$ are the first eigenvectors given by (4.7) and (4.8), and $$x^{2}+y^{2}=1$$.

### Proof

We first reduce the abstract equation (2.11) to the center manifold. Let $$J_{0}=(k_{0},l_{0},1)$$, then $$\Psi_{k_{0}l_{0}}^{1}=\Psi_{J_{0}}$$ and $$\tilde{\Psi }_{k_{0}l_{0}}^{1}= \tilde{\Psi}_{J_{0}}$$, the reduced equations read
$$\left\{ \textstyle\begin{array}{l} \frac{dx}{dt}= \beta_{J_{0}}x+\frac{1}{\langle\Psi_{J_{0}}, \Psi _{J_{0}}^{\ast}\rangle} \langle G(U,U), \Psi_{J_{0}}^{\ast} \rangle, \\ \frac{dy}{dt}=\beta_{J_{0}}y+\frac{1}{\langle\tilde{\Psi}_{J_{0}}, \tilde{\Psi}_{J_{0}}^{\ast}\rangle} \langle G(U,U),\tilde{\Psi}_{J_{0}}^{\ast} \rangle, \end{array}\displaystyle \right .$$
(4.12)
where $$U=(u_{\varphi},u_{r},T,H_{\varphi},H_{r})\in H_{1}$$ is written as
$$U=x\Psi_{J_{0}}+y\tilde{\Psi}_{J_{0}}+ \sum _{J\neq J_{0}}(x_{J} \Psi_{J}+ y_{J} \tilde{\Psi}_{J})=x\Psi_{J_{0}} +y\tilde{\Psi}_{J_{0}}+ \Phi.$$
(4.13)
Φ is the center manifold function, and G is a bilinear operator, which is
\begin{aligned} G(U_{1},U_{2})={}&P \biggl(-(u_{1}\cdot\nabla)u_{2\varphi}-\frac {u_{1r}u_{2\varphi}}{r_{0}}+ \frac{1}{\rho_{0}}\frac{d^{2}}{\kappa^{2}}H^{2}_{0}H_{1r} \biggl(\frac{\partial H_{2\varphi}}{\partial r}-\frac{\partial H_{2r}}{\partial\varphi} \biggr), \\ &{-}(u_{1}\cdot\nabla)u_{2r}- \frac{u_{1\varphi}u_{2\varphi }}{r_{0}}, -(u_{1}\cdot\nabla)T_{2}, \frac{\partial}{\partial r}(u_{1\varphi}H_{2r}-u_{1r}H_{2\varphi}), \\ &\frac{\partial}{\partial\varphi}(u_{1r}H_{2\varphi}-u_{1\varphi }H_{2r}) \biggr)^{T}. \end{aligned}
It is easy to verify that
\begin{aligned} \bigl\langle G(\Psi_{J_{0}},\Psi_{J_{0}}),\Psi_{J_{0}}^{\ast} \bigr\rangle &= \bigl\langle G(\Psi_{J_{0}},\tilde{\Phi}_{J_{0}}), \Psi_{J_{0}}^{\ast} \bigr\rangle \\&=\bigl\langle G(\tilde{\Phi}_{J_{0}},\Psi_{J_{0}}), \Psi_{J_{0}}^{\ast} \bigr\rangle = \bigl\langle G(\tilde{ \Phi}_{J_{0}},\tilde{\Phi}_{J_{0}}),\Psi_{J_{0}}^{\ast } \bigr\rangle =0. \end{aligned}
Hence, (4.12) can be rewritten as
$$\left \{ \textstyle\begin{array}{l} \frac{dx}{dt}= \beta_{J_{0}}x+\frac{x}{\langle\Psi_{J_{0}}, \Psi _{J_{0}}^{\ast}\rangle} [ \langle G( \Psi_{J_{0}},\Phi),\Psi_{J_{0}}^{\ast} \rangle+ \langle G(\Phi ,\Psi_{J_{0}}),\Psi_{J_{0}}^{\ast} \rangle ] \\ \hphantom{\frac{dx}{dt}=} + \frac{y}{\langle\Psi_{J_{0}}, \Psi_{J_{0}}^{\ast}\rangle} [ \langle G(\tilde{\Psi}_{J_{0}}, \Phi),\Psi_{J_{0}}^{\ast} \rangle + \langle G(\Phi,\tilde{ \Psi}_{J_{0}}),\Psi_{J_{0}}^{\ast} \rangle ] +o (|x|^{3}+|y|^{3} ), \\ \frac{dy}{dt}=\beta_{J_{0}}y+\frac{x}{\langle\tilde{\Psi}_{J_{0}}, \tilde{\Psi}_{J_{0}}^{\ast}\rangle} [ \langle G(\Psi_{J_{0}},\Phi),\tilde{\Psi}_{J_{0}}^{\ast} \rangle + \langle G(\Phi,\Psi_{J_{0}}),\tilde{\Psi}_{J_{0}}^{\ast} \rangle ] \\ \hphantom{\frac{dy}{dt}=} +\frac{y}{\langle\tilde{\Psi}_{J_{0}}, \tilde{\Psi}_{J_{0}}^{\ast }\rangle} [ \langle G(\tilde{\Psi}_{J_{0}}, \Phi),\tilde{\Psi}_{J_{0}}^{\ast} \rangle + \langle G(\Phi, \tilde{\Psi}_{J_{0}}), \tilde{\Psi}_{J_{0}}^{\ast} \rangle ]+o (|x|^{3}+|y|^{3} ). \end{array}\displaystyle \right .$$
(4.14)
Based on the approximation formula Theorem 6.1 in for center manifold functions, we see that the manifold function Φ satisfies
\begin{aligned}[b] -L\Phi={}&{-}\sum _{J\neq J_{0}}( \beta_{J}x_{J} \Psi_{J}+ \tilde { \beta_{J}}y_{J}\tilde{ \Psi}_{J})\\ ={}&x^{2}G(\Psi_{J_{0}},\Psi_{J_{0}}) +y^{2}G(\tilde{\Psi }_{J_{0}},\tilde{\Psi}_{J_{0}}) +xyG( \Psi_{J_{0}},\tilde{\Psi}_{J_{0}})+xyG(\Psi_{J_{0}},\tilde{ \Psi }_{J_{0}})\\ &+o \bigl(|x|^{2}+|y|^{2} \bigr). \end{aligned}
(4.15)
By direct calculation, from (4.15), the center manifold function can be written
$$\Phi=- \bigl(x^{2}+y^{2} \bigr) \frac{a_{k_{0}}T_{k_{0}l_{0}}^{\ast }h_{k_{0}l_{0}}}{2l_{0}\pi}\Psi_{0,2l_{0}}^{1}+o \bigl(|x|^{2}+|y|^{2} \bigr).$$
(4.16)
Inserting (4.16) into (4.14), we have
$$\left\{ \textstyle\begin{array}{l} \frac{dx}{dt}= \beta_{J_{0}}x-\frac{\delta x}{\langle\Psi_{J_{0}}, \Psi _{J_{0}}^{\ast}\rangle} (x^{2}+y^{3} )+o (|x|^{3}+|y|^{3} ), \\ \frac{dy}{dt}=\beta_{J_{0}}y-\frac{\delta y}{\langle\tilde{\Psi }_{J_{0}}, \tilde{\Psi}_{J_{0}}^{\ast}\rangle} (x^{2}+y^{3} )+o (|x|^{3}+|y|^{3} ), \end{array}\displaystyle \right .$$
(4.17)
where
$$\begin{gathered} \delta =\frac{L}{8}a_{k_{0}}^{2} \bigl(T_{k_{0}l_{0}}^{\ast } \bigr)^{2}h_{k_{0}l_{0}}^{2} \pi>0, \\ \bigl\langle \Psi_{J_{0}}, \Psi_{J_{0}}^{\ast} \bigr\rangle = \bigl\langle \tilde{\Psi }_{J_{0}}, \tilde{\Psi}_{J_{0}}^{\ast} \bigr\rangle >0. \end{gathered}$$
(4.18)
Therefore, the conclusions of Theorem 4.2 follows from (4.17), (4.18) and the attractor bifurcation theorem in . The proof of the theorem is completed. □

## 5 Conclusions

In summary, the production of granulation convection is illustrated by Theorem 3.2 and Figure 2, where $$R_{C}$$ is the critical Rayleigh number. As is shown in Figure 2, when the differences of temperature and magnetic are small enough, the points $$(R,F)$$ are below the critical curve L and in region I, the gas is in a static state. When they become greater than the curve L, the gas suddenly breaks into regular circulation cells, which is granulation convection. The circulation cells can be expressed by the bifurcation solution in Theorem 4.2. From both the mathematical and the physical points of view, the results are valuable to understand the phase transition in fluid dynamics.

## Declarations

### Acknowledgements

The author would like to thank the referee for invaluable comments insightful suggestions. This article was supported by the National Natural Science Foundation of China (A0109). 