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Nonlinear stability of the two-jet Kolmogorov type flow on the unit sphere under a perturbation with nondissipative part

Tatsu-Hiko Miura Graduate School of Science and Technology, Hirosaki University, 3, Bunkyo-cho, Hirosaki-shi, Aomori, 036-8561, Japan thmiura623@hirosaki-u.ac.jp
Abstract.

We consider the vorticity form of the Navier–Stokes equations on the two-dimensional unit sphere and study the nonlinear stability of the two-jet Kolmogorov type flow which is a stationary solution given by the zonal spherical harmonic function of degree two. In particular, we assume that a perturbation contains a nondissipative part given by a linear combination of the spherical harmonics of degree one and investigate the effect of the nondissipative part on the long-time behavior of the perturbation through the convection term. We show that the nondissipative part of a weak solution to the nonlinear stability problem is preserved in time for all initial data. Moreover, we prove that the dissipative part of the weak solution converges exponentially in time towards an equilibrium which is expressed explicitly in terms of the nondissipative part of the initial data and does not vanish in general. In particular, it turns out that the asymptotic behavior of the weak solution is finally determined by a system of linear ordinary differential equations. To prove these results, we make use of properties of Killing vector fields on a manifold. We also consider the case of a rotating sphere.

Key words and phrases:
vorticity equation, Kolmogorov type flow, nonlinear stability, Killing vector field
2010 Mathematics Subject Classification:
35Q30, 76D05, 35B35, 35R01

1. Introduction

Let S2S^{2} be the two-dimensional (2D) unit sphere in 3\mathbb{R}^{3}. We consider the Navier–Stokes equations

(1.1) t𝐮+𝐮𝐮ν(ΔH𝐮+2𝐮)+p=𝐟,div𝐮=0onS2×(0,).\displaystyle\partial_{t}\mathbf{u}+\nabla_{\mathbf{u}}\mathbf{u}-\nu(\Delta_{H}\mathbf{u}+2\mathbf{u})+\nabla p=\mathbf{f},\quad\mathrm{div}\,\mathbf{u}=0\quad\text{on}\quad S^{2}\times(0,\infty).

Here 𝐮\mathbf{u} is the tangential velocity field of a fluid, pp is the pressure, and 𝐟\mathbf{f} is a given external force. Also, ν>0\nu>0 is the viscosity coefficient, 𝐮𝐮\nabla_{\mathbf{u}}\mathbf{u} is the covariant derivative of 𝐮\mathbf{u} along itself, ΔH\Delta_{H} is the Hodge Laplacian via identification of vector fields and one-forms, and \nabla and div\mathrm{div} are the gradient and the divergence on S2S^{2}. Note that the viscous term in (1.1) contains the zeroth order term since it is twice of the deformation tensor Def𝐮\mathrm{Def}\,\mathbf{u}:

(1.2) 2divDef𝐮=ΔH𝐮+(div𝐮)+2Ric(𝐮)=ΔH𝐮+(div𝐮)+2𝐮.\displaystyle 2\,\mathrm{div}\,\mathrm{Def}\,\mathbf{u}=\Delta_{H}\mathbf{u}+\nabla(\mathrm{div}\,\mathbf{u})+2\,\mathrm{Ric}(\mathbf{u})=\Delta_{H}\mathbf{u}+\nabla(\mathrm{div}\,\mathbf{u})+2\mathbf{u}.

Here Ric1\mathrm{Ric}\equiv 1 is the Ricci curvature of S2S^{2}. There are many works on the Navier–Stokes equations on spheres and manifolds with this kind of viscous term (see e.g. [42, 35, 32, 30, 10, 19, 6, 8, 38, 39, 34, 20, 36, 37]). Also, several authors studied the Navier–Stokes equations on manifolds with viscous term replaced by νΔH𝐮\nu\Delta_{H}\mathbf{u} in analogy of the flat domain case (see e.g. [16, 15, 45, 5, 17, 51, 24, 40]). We refer to [13, 1, 12, 43, 7] for the identity (1.2) and the choice of the viscous term in the Navier–Stokes equations on manifolds. A crucial difference due to the choice of the viscous term is the presence of nondissipative vector fields. The viscosity always works if one chooses νΔH𝐮\nu\Delta_{H}\mathbf{u} since S2S^{2} does not admit nontrivial harmonic forms. On the other hand, if one takes ν(ΔH𝐮+2𝐮)\nu(\Delta_{H}\mathbf{u}+2\mathbf{u}), then the viscosity does not work for tangential vector fields of the form X(x)=𝐚×xX(x)=\mathbf{a}\times x, xS2x\in S^{2} with any 𝐚3\mathbf{a}\in\mathbb{R}^{3}. Note that this XX is a Killing vector field on S2S^{2}, and in general Killing vector fields are nondissipative stationary solutions to the Navier–Stokes equations on a manifold with viscous term given by (1.2). In fact, the problem of the existence and uniqueness of solutions is not so affected by the choice of the viscous term, since the difference of the above two viscous terms is linear and of lower order. However, as we will see below, the long-time behavior of a solution can be different by the choice of the viscous term due to the effect of a nondissipative part of a solution coming from the convection term.

Since S2S^{2} is 2D and simply connected, the Navier–Stokes equations (1.1) are equivalent to the following equation for the vorticity ω=rot𝐮\omega=\mathrm{rot}\,\mathbf{u}:

(1.3) tω+𝐮ων(Δω+2ω)=f,𝐮=𝐧S2×Δ1ωonS2×(0,).\displaystyle\partial_{t}\omega+\mathbf{u}\cdot\nabla\omega-\nu(\Delta\omega+2\omega)=f,\quad\mathbf{u}=\mathbf{n}_{S^{2}}\times\nabla\Delta^{-1}\omega\quad\text{on}\quad S^{2}\times(0,\infty).

Here f=rot𝐟f=\mathrm{rot}\,\mathbf{f} is an external force, 𝐧S2\mathbf{n}_{S^{2}} is the unit outward normal vector field of S2S^{2}, and Δ\Delta is Laplace–Beltrami operator on S2S^{2} with inverse Δ1\Delta^{-1} in L02(S2)L_{0}^{2}(S^{2}), the space of L2L^{2} functions on S2S^{2} with vanishing mean. Also, 𝐚𝐛\mathbf{a}\cdot\mathbf{b} and 𝐚×𝐛\mathbf{a}\times\mathbf{b} stand for the inner and vector products of 𝐚,𝐛3\mathbf{a},\mathbf{b}\in\mathbb{R}^{3}. We refer to [31] for the derivation of (1.3) from (1.1).

For n0n\in\mathbb{Z}_{\geq 0} and |m|n|m|\leq n, let YnmY_{n}^{m} be the spherical harmonics and λn=n(n+1)\lambda_{n}=n(n+1) be the corresponding eigenvalue of Δ-\Delta (see Section 2 for details). Then, for nn\in\mathbb{N} and aa\in\mathbb{R}, the vorticity equation (1.3) with external force f=aν(λn2)Yn0f=a\nu(\lambda_{n}-2)Y_{n}^{0} has a stationary solution with corresponding velocity field

(1.4) ωna(θ,φ)=aYn0(θ),𝐮na(θ,φ)=aλnsinθdYn0dθ(θ)φx(θ,φ),\displaystyle\omega_{n}^{a}(\theta,\varphi)=aY_{n}^{0}(\theta),\quad\mathbf{u}_{n}^{a}(\theta,\varphi)=-\frac{a}{\lambda_{n}\sin\theta}\frac{dY_{n}^{0}}{d\theta}(\theta)\partial_{\varphi}x(\theta,\varphi),

where x(θ,φ)x(\theta,\varphi) is the parametrization of S2S^{2} by the colatitude θ\theta and the longitude φ\varphi. The flow (1.4) can be seen as a spherical version of the well-known plane Kolmogorov flow, which is a stationary solution to the Navier–Stokes equations in a 2D flat torus (see e.g. [29, 18, 27, 33, 28] for the study of the stability of the plane Kolmogorov flow). In [17], the flow (1.4) is called the generalized Kolmogorov flow. Also, it is called an nn-jet zonal flow in [38, 39]. We call (1.4) the nn-jet Kolmogorov type flow in order to emphasize both the similarity to the plane Kolmogorov flow and the number of jets.

In this paper we focus on the case n=2n=2 and consider the stability of the two-jet Kolmogorov type flow for the vorticity equation (1.3) (see Remark 1.3 for the one-jet case). Our particular interest is in studying the effect of a nondissipative part of a perturbation in the stability problem. Since ΔY1m=2Y1m-\Delta Y_{1}^{m}=2Y_{1}^{m} for |m|=0,1|m|=0,1, the viscosity does not work for functions in span{Y10,Y1±1}\mathrm{span}\{Y_{1}^{0},Y_{1}^{\pm 1}\}, for which the corresponding velocity fields are Killing vector fields on S2S^{2}. If a perturbation contains such a nondissipative part, then it seems to be natural to ask how the nondissipative part affects the long-time behavior of the perturbation through the convection term. Our aim is to give an explicit answer to this problem.

The stability of the Kolmogorov type flows was studied by Ilyin [17] and Sasaki, Takehiro, and Yamada [38, 39] for the Navier–Stokes equations on S2S^{2} and by Taylor [41] for the Euler equations on S2S^{2} (and these authors dealt with the case of a rotating sphere). For the viscous case, Ilyin [17] investigated the linear stability and showed that the nn-jet Kolmogorov type flow is globally asymptotically stable for all ν>0\nu>0 when n=1,2n=1,2 but it becomes unstable for small ν>0\nu>0 when n3n\geq 3. In that paper, however, the viscous term in the Navier–Stokes equations was taken as νΔH𝐮\nu\Delta_{H}\mathbf{u}, which becomes νΔω\nu\Delta\omega in the vorticity form. Hence a perturbation does not contain a nondissipative part in the setting of [17]. For the case of the viscous term ν(Δω+2ω)\nu(\Delta\omega+2\omega), Sasaki, Takehiro, and Yamada studied the linear and nonlinear stability of the Kolmogorov type flows in [38] and [39], respectively, and obtained the same results as in [17]. However, they considered a perturbation in the orthogonal complement of span{Y10,Y1±1}\mathrm{span}\{Y_{1}^{0},Y_{1}^{\pm 1}\}, so the effect of a nondissipative part was not taken into account.

In our previous work [31], we studied the linear stability of the two-jet Kolmogorov type flow under a perturbation which contains a nondissipative part. The two-jet Kolmogorov type flow is of the form

(1.5) ωna(θ,φ)=aYn0(θ)=a45π(3cos2θ1).\displaystyle\omega_{n}^{a}(\theta,\varphi)=aY_{n}^{0}(\theta)=\frac{a}{4}\sqrt{\frac{5}{\pi}}(3\cos^{2}\theta-1).

Then the linearized equation for the vorticity equation (1.3) around (1.5) is

(1.6) tω=ν(Δω+2ω)a2cosθφ(I+6Δ1)ω,a2=a45πonS2×(0,),\displaystyle\partial_{t}\omega=\nu(\Delta\omega+2\omega)-a_{2}\cos\theta\,\partial_{\varphi}(I+6\Delta^{-1})\omega,\quad a_{2}=\frac{a}{4}\sqrt{\frac{5}{\pi}}\quad\text{on}\quad S^{2}\times(0,\infty),

where II is the identity operator. We refer to [31] for the derivation of (1.6) from (1.3). In [31] we proved that a solution ω(t)\omega(t) to (1.6) with initial data ω0L02(S2)\omega_{0}\in L_{0}^{2}(S^{2}) satisfies

ω(t)ω,linL2(S2)C(1+|a2|ν)e4νtω0L2(S2),t0,\displaystyle\|\omega(t)-\omega_{\infty,\mathrm{lin}}\|_{L^{2}(S^{2})}\leq C\left(1+\frac{|a_{2}|}{\nu}\right)e^{-4\nu t}\|\omega_{0}\|_{L^{2}(S^{2})},\quad t\geq 0,

where the equilibrium ω,lin\omega_{\infty,\mathrm{lin}} is given by

(1.7) ω,lin=ω0,10Y10+m=±1ω0,1m(Y1m+a2νim25Y2m),ω0,1m=(ω0,Y1m)L2(S2)\displaystyle\omega_{\infty,\mathrm{lin}}=\omega_{0,1}^{0}Y_{1}^{0}+\sum_{m=\pm 1}\omega_{0,1}^{m}\left(Y_{1}^{m}+\frac{a_{2}}{\nu}\frac{im}{2\sqrt{5}}Y_{2}^{m}\right),\quad\omega_{0,1}^{m}=(\omega_{0},Y_{1}^{m})_{L^{2}(S^{2})}

and C>0C>0 is a constant independent of tt, ν\nu, a2a_{2}, and ω0\omega_{0}. Note that ω,lin\omega_{\infty,\mathrm{lin}} contains the nonzero Y2±1Y_{2}^{\pm 1}-components when ω0,1±10\omega_{0,1}^{\pm 1}\neq 0, although the functions Y2±1Y_{2}^{\pm 1} themselves are dissipative in (1.6). This result shows that the nondissipative part indeed affects the long-time behavior of the perturbation through the interaction between the viscosity and convection even in the linear stability case. In [31] and the companion paper [26] we also observed that a part of a solution to (1.6) decays at a rate faster than the usual viscous rate O(eνt)O(e^{-\nu t}) when ν>0\nu>0 is small. Such a phenomenon is called the enhanced dissipation, and it has been attracting interest of many researchers in recent years (see e.g. [9, 52, 48] and [3, 25, 14, 49, 50] for the study of the enhanced dissipation for advection-diffusion equations and for the plane Kolmogorov flow, respectively).

In this paper we study the nonlinear stability of the two-jet Kolmogorov type flow under a perturbation with nondissipative part. The nonlinear stability problem for the vorticity equation (1.3) around the two-jet Kolmogorov type flow (1.5) is

(1.8) {tω=ν(Δω+2ω)a45πcosθφ(I+6Δ1)ω𝐮ωonS2×(0,),𝐮=𝐧S2×Δ1ωonS2×(0,),ω|t=0=ω0onS2,\displaystyle\left\{\begin{aligned} \partial_{t}\omega&=\nu(\Delta\omega+2\omega)-\frac{a}{4}\sqrt{\frac{5}{\pi}}\cos\theta\,\partial_{\varphi}(I+6\Delta^{-1})\omega-\mathbf{u}\cdot\nabla\omega\quad\text{on}\quad S^{2}\times(0,\infty),\\ \mathbf{u}&=\mathbf{n}_{S^{2}}\times\nabla\Delta^{-1}\omega\quad\text{on}\quad S^{2}\times(0,\infty),\\ \omega|_{t=0}&=\omega_{0}\quad\text{on}\quad S^{2},\end{aligned}\right.

where ω0L02(S2)\omega_{0}\in L_{0}^{2}(S^{2}) is an initial perturbation (see [31] for the derivation of the perturbation operator). To state our main result, we fix some notations. For uL02(S2)u\in L_{0}^{2}(S^{2}), we write

unm=(u,Ynm)L2(S2),u=n=m=nnunmYnm,uN=nNu=n.\displaystyle u_{n}^{m}=(u,Y_{n}^{m})_{L^{2}(S^{2})},\quad u_{=n}=\sum_{m=-n}^{n}u_{n}^{m}Y_{n}^{m},\quad u_{\geq N}=\sum_{n\geq N}u_{=n}.

Note that unm=(1)munm¯u_{n}^{-m}=(-1)^{m}\overline{u_{n}^{m}} if uu is real-valued, since Ynm=(1)mYnm¯Y_{n}^{-m}=(-1)^{m}\overline{Y_{n}^{m}} (see Section 2). In what follows, we consider real-valued initial data and solutions to (1.8), although we take the inner product of (1.8) with the complex spherical harmonics YnmY_{n}^{m} in order to make some expressions simple. For a real-valued initial data ω0L02(S2)\omega_{0}\in L_{0}^{2}(S^{2}), we set

(1.9) α=126πω0,11,b=123πω0,10.\displaystyle\alpha=\frac{1}{2\sqrt{6\pi}}\,\omega_{0,1}^{1}\in\mathbb{C},\quad b=\frac{1}{2\sqrt{3\pi}}\,\omega_{0,1}^{0}\in\mathbb{R}.

Then we define ω=m=22ω2,mY2mL02(S2)\omega_{\infty}=\sum_{m=-2}^{2}\omega_{2,\infty}^{m}Y_{2}^{m}\in L_{0}^{2}(S^{2}) by

(1.10) ω2,0=12a|α|2(4ν2+|α|2+b2)(4ν2+4|α|2+b2)(16ν2+4|α|2+b2),ω2,1=6iα(4ν+2ib)4|α|2+(4ν+ib)(4ν+2ib)(ω2,0+a),ω2,1=ω2,1¯,ω2,2=2iα4ν+2ibω2,1,ω2,2=ω2,2¯.\displaystyle\begin{aligned} \omega_{2,\infty}^{0}&=-12a\cdot\frac{|\alpha|^{2}(4\nu^{2}+|\alpha|^{2}+b^{2})}{(4\nu^{2}+4|\alpha|^{2}+b^{2})(16\nu^{2}+4|\alpha|^{2}+b^{2})},\\ \omega_{2,\infty}^{1}&=\frac{\sqrt{6}\,i\alpha(4\nu+2ib)}{4|\alpha|^{2}+(4\nu+ib)(4\nu+2ib)}(\omega_{2,\infty}^{0}+a),\quad\omega_{2,\infty}^{-1}=-\overline{\omega_{2,\infty}^{1}},\\ \omega_{2,\infty}^{2}&=\frac{2i\alpha}{4\nu+2ib}\,\omega_{2,\infty}^{1},\quad\omega_{2,\infty}^{-2}=\overline{\omega_{2,\infty}^{2}}.\end{aligned}

Also, we set H01(S2)=L02(S2)H1(S2)H_{0}^{1}(S^{2})=L_{0}^{2}(S^{2})\cap H^{1}(S^{2}) and write H01(S2)H_{0}^{-1}(S^{2}) for the dual space of H01(S2)H_{0}^{1}(S^{2}) with duality product ,H01\langle\cdot,\cdot\rangle_{H_{0}^{1}}. The main result of this paper is as follows.

Theorem 1.1.

For all real-valued initial data ω0L02(S2)\omega_{0}\in L_{0}^{2}(S^{2}), there exists a real-valued unique global weak solution

(1.11) ωC([0,);L02(S2))Lloc2([0,);H01(S2))withtωLloc2([0,);H01(S2))\displaystyle\begin{aligned} &\omega\in C([0,\infty);L_{0}^{2}(S^{2}))\cap L_{loc}^{2}([0,\infty);H_{0}^{1}(S^{2}))\\ &\text{with}\quad\partial_{t}\omega\in L_{loc}^{2}([0,\infty);H_{0}^{-1}(S^{2}))\end{aligned}

to (1.8) in the sense that ω(0)=ω0\omega(0)=\omega_{0} in L02(S2)L_{0}^{2}(S^{2}) and

(1.12) tω(t),ψH01=ν(ω(t),ψ)L2(S2)+2ν(ω(t),ψ)L2(S2)a45π(cosθφ(I+6Δ1)ω(t),ψ)L2(S2)(𝐮(t)ω(t),ψ)L2(S2)\displaystyle\begin{aligned} \langle\partial_{t}\omega(t),\psi\rangle_{H_{0}^{1}}&=-\nu(\nabla\omega(t),\nabla\psi)_{L^{2}(S^{2})}+2\nu(\omega(t),\psi)_{L^{2}(S^{2})}\\ &\qquad-\frac{a}{4}\sqrt{\frac{5}{\pi}}(\cos\theta\,\partial_{\varphi}(I+6\Delta^{-1})\omega(t),\psi)_{L^{2}(S^{2})}\\ &\qquad-(\mathbf{u}(t)\cdot\nabla\omega(t),\psi)_{L^{2}(S^{2})}\end{aligned}

for all real-valued ψH01(S2)\psi\in H_{0}^{1}(S^{2}) and a.e. t>0t>0. Moreover,

(1.13) ω=1(t)=ω0,=1,ω3(t)L2(S2)e10νtω0,3L2(S2)\displaystyle\omega_{=1}(t)=\omega_{0,=1},\quad\|\omega_{\geq 3}(t)\|_{L^{2}(S^{2})}\leq e^{-10\nu t}\|\omega_{0,\geq 3}\|_{L^{2}(S^{2})}

for all t0t\geq 0 and

(1.14) ω=2(t)ωL2(S2)σ1e2νt(ω0,=2ωL2(S2)+σ2ν)\displaystyle\|\omega_{=2}(t)-\omega_{\infty}\|_{L^{2}(S^{2})}\leq\sigma_{1}e^{-2\nu t}\left(\|\omega_{0,=2}-\omega_{\infty}\|_{L^{2}(S^{2})}+\frac{\sigma_{2}}{\nu}\right)

for all t0t\geq 0. Here ω=m=22ω2,mY2m\omega_{\infty}=\sum_{m=-2}^{2}\omega_{2,\infty}^{m}Y_{2}^{m} is given by (1.9)–(1.10). Also,

(1.15) σ1=exp(C1νω0,3L2(S2)),σ2=C2(|a|+ω0,3L2(S2)+ωL2(S2))ω0,3L2(S2),\displaystyle\begin{aligned} \sigma_{1}&=\exp\left(\frac{C_{1}}{\nu}\|\omega_{0,\geq 3}\|_{L^{2}(S^{2})}\right),\\ \sigma_{2}&=C_{2}\Bigl{(}|a|+\|\omega_{0,\geq 3}\|_{L^{2}(S^{2})}+\|\omega_{\infty}\|_{L^{2}(S^{2})}\Bigr{)}\|\omega_{0,\geq 3}\|_{L^{2}(S^{2})},\end{aligned}

and C1,C2>0C_{1},C_{2}>0 are constants independent of tt, ν\nu, aa, and ω0\omega_{0}.

Note that the last term of (1.12) is well-defined since

𝐮=𝐧S2×Δ1ωH2(S2)L(S2)\displaystyle\mathbf{u}=\mathbf{n}_{S^{2}}\times\nabla\Delta^{-1}\omega\in H^{2}(S^{2})\subset L^{\infty}(S^{2})

for ωH01(S2)\omega\in H_{0}^{1}(S^{2}) by the Sobolev embedding (see e.g. [2]). Also, a test function ψ\psi in (1.12) is assumed to be real-valued, but in fact we may take a complex-valued ψ\psi by considering its real and imaginary parts separately.

Remark 1.2.

Compared to the linear stability case (see (1.7)), the equilibrium ω0,=1+ω\omega_{0,=1}+\omega_{\infty} is complicated but still determined explicitly. Moreover, the Y20Y_{2}^{0}- and Y2±2Y_{2}^{\pm 2}-components of the equilibrium does not vanish in general for the nonlinear stability problem. We also note that the behavior (1.13) of ω=1(t)\omega_{=1}(t) and ω3(t)\omega_{\geq 3}(t) is the same as in the linear stability case (see [31, Theorem 3.1]) and only the behavior (1.14) of ω=2(t)\omega_{=2}(t) is different.

Remark 1.3.

For the one-jet case, we can show that the Y1mY_{1}^{m}-components of a solution to the nonlinear stability problem is preserved in time as in (1.13) and the YnmY_{n}^{m}-components with n2n\geq 2 decay exponentially in time. The proof is the same as that of Theorem 1.1 given in Section 3 and much easier, so we just give the outline in Section 5.

It is expected that we can show that the enhanced dissipation occurs for a part of a solution to (1.8) as in the linear stability case [31, 26], but we need to analyze carefully the interaction between various components of a solution with different longitudinal wave numbers through the convection term. The study of the enhanced dissipation for the nonlinear stability problem will be done in another paper.

The result of Theorem 1.1 can be extended to the case of a rotating sphere. In that case, we consider the vorticity equation with Coriolis force (see e.g. [17, 38, 39])

(1.16) {tζ+𝐯ζ+2ΩφΔ1ζν(Δζ+2ζ)=fonS2×(0,),𝐯=𝐧S2×Δ1ζonS2×(0,),\displaystyle\left\{\begin{aligned} &\partial_{t}\zeta+\mathbf{v}\cdot\nabla\zeta+2\Omega\,\partial_{\varphi}\Delta^{-1}\zeta-\nu(\Delta\zeta+2\zeta)=f\quad\text{on}\quad S^{2}\times(0,\infty),\\ &\mathbf{v}=\mathbf{n}_{S^{2}}\times\nabla\Delta^{-1}\zeta\quad\text{on}\quad S^{2}\times(0,\infty),\end{aligned}\right.

where Ω\Omega\in\mathbb{R} is the rotation speed of a sphere. For each nn\in\mathbb{N} and aa\in\mathbb{R}, the Kolmogorov type flow (1.4) is still a stationary solution to (1.16) with f=aν(λn2)Yn0f=a\nu(\lambda_{n}-2)Y_{n}^{0} since it is independent of the longitude φ\varphi. When n=2n=2, the nonlinear stability problem for (1.16) around the two-jet Kolmogorov type flow (1.5) is

(1.17) {tζ=ν(Δζ+2ζ)2ΩφΔ1ζa45πcosθφ(I+6Δ1)ζ𝐯ζonS2×(0,),𝐯=𝐧S2×Δ1ζonS2×(0,),ζ|t=0=ζ0onS2.\displaystyle\left\{\begin{aligned} \partial_{t}\zeta&=\nu(\Delta\zeta+2\zeta)-2\Omega\,\partial_{\varphi}\Delta^{-1}\zeta\\ &\qquad-\frac{a}{4}\sqrt{\frac{5}{\pi}}\cos\theta\,\partial_{\varphi}(I+6\Delta^{-1})\zeta-\mathbf{v}\cdot\nabla\zeta\quad\text{on}\quad S^{2}\times(0,\infty),\\ \mathbf{v}&=\mathbf{n}_{S^{2}}\times\nabla\Delta^{-1}\zeta\quad\text{on}\quad S^{2}\times(0,\infty),\\ \zeta|_{t=0}&=\zeta_{0}\quad\text{on}\quad S^{2}.\end{aligned}\right.

Then we easily find by direct calculations (see Section 6 for the outline) that solutions ω\omega to (1.8) and ζ\zeta to (1.17) are related by

(1.18) ω(θ,φ,t)=ζ(θ,φΩt,t)+2Ωcosθ=ζ(θ,φΩt,t)+4π3ΩY10(θ)\displaystyle\omega(\theta,\varphi,t)=\zeta(\theta,\varphi-\Omega t,t)+2\Omega\cos\theta=\zeta(\theta,\varphi-\Omega t,t)+4\sqrt{\frac{\pi}{3}}\,\Omega Y_{1}^{0}(\theta)

in spherical coordinates. Thus, for a given ζ0L02(S2)\zeta_{0}\in L_{0}^{2}(S^{2}), we apply Theorem 1.1 to

ω0(θ,φ)=ζ0(θ,φ)+2Ωcosθ=ζ0(θ,φ)+4π3ΩY10(θ),\displaystyle\omega_{0}(\theta,\varphi)=\zeta_{0}(\theta,\varphi)+2\Omega\cos\theta=\zeta_{0}(\theta,\varphi)+4\sqrt{\frac{\pi}{3}}\,\Omega Y_{1}^{0}(\theta),

substitute (1.18) for (1.13) and (1.14), make the change of variable φφ+Ωt\varphi\mapsto\varphi+\Omega t which does not change the L2(S2)L^{2}(S^{2})-norm, and use Ynm(θ,φ+Ωt)=eimΩtYnm(θ,φ)Y_{n}^{m}(\theta,\varphi+\Omega t)=e^{im\Omega t}Y_{n}^{m}(\theta,\varphi) to obtain the following result.

Theorem 1.4.

For all real-valued initial data ζ0L02(S2)\zeta_{0}\in L_{0}^{2}(S^{2}), there exists a unique global weak solution ζ\zeta to (1.17) in the class (1.11) and

ζ=1(t)=m=0,±1eimΩtζ0,1mY1m,ζ3(t)e10νtζ0,3L2(S2)\displaystyle\zeta_{=1}(t)=\sum_{m=0,\pm 1}e^{im\Omega t}\zeta_{0,1}^{m}Y_{1}^{m},\quad\|\zeta_{\geq 3}(t)\|\leq e^{-10\nu t}\|\zeta_{0,\geq 3}\|_{L^{2}(S^{2})}

for all t0t\geq 0. Moreover, if we set

α=126πζ0,11,bΩ=123πζ0,10+23Ω\displaystyle\alpha=\frac{1}{2\sqrt{6\pi}}\,\zeta_{0,1}^{1}\in\mathbb{C},\quad b_{\Omega}=\frac{1}{2\sqrt{3\pi}}\,\zeta_{0,1}^{0}+\frac{2}{3}\,\Omega\in\mathbb{R}

and define ζ(t)=m=22eimΩtζ2,mY2m\zeta_{\infty}(t)=\sum_{m=-2}^{2}e^{im\Omega t}\zeta_{2,\infty}^{m}Y_{2}^{m} by

ζ2,0\displaystyle\zeta_{2,\infty}^{0} =12a|α|2(4ν2+|α|2+bΩ2)(4ν2+4|α|2+bΩ2)(16ν2+4|α|2+bΩ2),\displaystyle=-12a\cdot\frac{|\alpha|^{2}(4\nu^{2}+|\alpha|^{2}+b_{\Omega}^{2})}{(4\nu^{2}+4|\alpha|^{2}+b_{\Omega}^{2})(16\nu^{2}+4|\alpha|^{2}+b_{\Omega}^{2})},
ζ2,1\displaystyle\zeta_{2,\infty}^{1} =6iα(4ν+2ibΩ)4|α|2+(4ν+ibΩ)(4ν+2ibΩ)(ζ2,0+a),ζ2,1=ζ2,1¯,\displaystyle=\frac{\sqrt{6}\,i\alpha(4\nu+2ib_{\Omega})}{4|\alpha|^{2}+(4\nu+ib_{\Omega})(4\nu+2ib_{\Omega})}(\zeta_{2,\infty}^{0}+a),\quad\zeta_{2,\infty}^{-1}=-\overline{\zeta_{2,\infty}^{1}},
ζ2,2\displaystyle\zeta_{2,\infty}^{2} =2iα4ν+2ibΩζ2,1,ζ2,2=ζ2,2¯,\displaystyle=\frac{2i\alpha}{4\nu+2ib_{\Omega}}\,\zeta_{2,\infty}^{1},\quad\zeta_{2,\infty}^{-2}=\overline{\zeta_{2,\infty}^{2}},

then we have

ζ=2(t)ζ(t)L2(S2)σ1(ζ0,=2ζ(0)L2(S2)+σ2ν)\displaystyle\|\zeta_{=2}(t)-\zeta_{\infty}(t)\|_{L^{2}(S^{2})}\leq\sigma_{1}\left(\|\zeta_{0,=2}-\zeta_{\infty}(0)\|_{L^{2}(S^{2})}+\frac{\sigma_{2}}{\nu}\right)

for all t0t\geq 0, where σ1\sigma_{1} and σ2\sigma_{2} are given by (1.15) with ω0,3\omega_{0,\geq 3} and ω\omega_{\infty} replaced by ζ0,3\zeta_{0,\geq 3} and ζ(0)\zeta_{\infty}(0), respectively, and with constants C1,C2>0C_{1},C_{2}>0 independent of tt, ν\nu, aa, Ω\Omega, and ζ0\zeta_{0}.

Let us explain the outline of the proof of Theorem 1.1. Since S2S^{2} is 2D, we can show the global existence and uniqueness of a weak solution to (1.8) by the Galerkin method with basis functions YnmY_{n}^{m} and the energy method as in the case of the Navier–Stokes equations in 2D flat domains (see e.g. [44, 4]). Hence we omit details in this paper. To prove (1.13) and (1.14), we first take the inner product of (1.8) with Y1mY_{1}^{m}, |m|=0,1|m|=0,1. Then we find that ddtω1m(t)=0\frac{d}{dt}\omega_{1}^{m}(t)=0 and thus ω=1(t)=ω0,=1\omega_{=1}(t)=\omega_{0,=1} by using ΔYnm=λnYnm-\Delta Y_{n}^{m}=\lambda_{n}Y_{n}^{m} with λ1=2\lambda_{1}=2 and λ2=6\lambda_{2}=6, a recurrence relation for cosθYnm\cos\theta\,Y_{n}^{m} (see (2.7)), and the fact that a vector field of the form

𝐧S2×χ,χ=cosθ,sinθcosφ,sinθsinφ\displaystyle\mathbf{n}_{S^{2}}\times\nabla\chi,\quad\chi=\cos\theta,\sin\theta\cos\varphi,\sin\theta\sin\varphi

is a Killing vector field on S2S^{2}. Here the last fact is essential to show that the inner product of the convection term with Y1mY_{1}^{m} vanishes (see Lemma 2.2). To prove that result, we apply an identity for a Killing vector field on a general manifold given in Lemma 2.1, which seems to have its own interest.

Next we derive an equation for ω2(t)=ω(t)ω0,=1\omega_{\geq 2}(t)=\omega(t)-\omega_{0,=1} from (1.8) and take the inner product of that equation with (I+6Δ1)ω2(t)(I+6\Delta^{-1})\omega_{\geq 2}(t). Then we obtain a differential inequality which yields the estimate (1.13) for ω3(t)\omega_{\geq 3}(t) by applying λ2=6\lambda_{2}=6 and λn12\lambda_{n}\geq 12 for n3n\geq 3, integration by parts, and the identity

(1.19) (XΔ1ω2(t),ω2(t))L2(S2)=(Xω2(t),Δ1ω2(t))L2(S2)=0\displaystyle\bigl{(}X\cdot\nabla\Delta^{-1}\omega_{\geq 2}(t),\omega_{\geq 2}(t)\bigr{)}_{L^{2}(S^{2})}=\bigl{(}X\cdot\nabla\omega_{\geq 2}(t),\Delta^{-1}\omega_{\geq 2}(t)\bigr{)}_{L^{2}(S^{2})}=0

given in Lemma 2.3, where XX is given by

(1.20) X=𝐧S2×Δ1ω0,=1=12𝐧S2×ω0,=1.\displaystyle X=\mathbf{n}_{S^{2}}\times\nabla\Delta^{-1}\omega_{0,=1}=-\frac{1}{2}\mathbf{n}_{S^{2}}\times\nabla\omega_{0,=1}.

Here we do not have an estimate for ω2(t)=ω=2(t)+ω3(t)\omega_{\geq 2}(t)=\omega_{=2}(t)+\omega_{\geq 3}(t) since (I+6Δ1)ω=2(t)=0(I+6\Delta^{-1})\omega_{=2}(t)=0 by λ2=6\lambda_{2}=6. Moreover, the identity (1.19) is used to show

(X(I+2Δ1)ω2(t),(I+6Δ1)ω2(t))L2(S2)=0,\displaystyle\bigl{(}X\cdot\nabla(I+2\Delta^{-1})\omega_{\geq 2}(t),(I+6\Delta^{-1})\omega_{\geq 2}(t)\bigr{)}_{L^{2}(S^{2})}=0,

where the term X(I+2Δ1)ω2(t)X\cdot\nabla(I+2\Delta^{-1})\omega_{\geq 2}(t) comes from the interaction between the nondissipative part ω0,=1\omega_{0,=1} and the dissipative part ω2(t)\omega_{\geq 2}(t) through the convection term. We further note that, since the above XX is of the form X(x)=𝐚×xX(x)=\mathbf{a}\times x, xS2x\in S^{2} with some 𝐚3\mathbf{a}\in\mathbb{R}^{3} expressed in terms of ω0,10\omega_{0,1}^{0} and ω0,11\omega_{0,1}^{1}, it is a Killing vector field on S2S^{2} and thus we can apply again the identity for a Killing vector field on a general manifold given in Lemma 2.1 to get the identity (1.19).

Lastly, we derive an equation for ω=2(t)=ω2(t)ω3(t)\omega_{=2}(t)=\omega_{\geq 2}(t)-\omega_{\geq 3}(t), prove the estimate (1.14), and determine ω=m=22ω2,mY2m\omega_{\infty}=\sum_{m=-2}^{2}\omega_{2,\infty}^{m}Y_{2}^{m}. It turns out that the evolution of ω=2(t)\omega_{=2}(t) is described by a system of linear ordinary differential equations (ODEs) of the form

(1.21) ddt𝝎(t)={4ν𝑰5+i𝑨+𝑴(t)}𝝎(t)+𝒇3(t)+𝒄,t>0,𝝎(t)=(ω22(t),,ω22(t))T5,\displaystyle\begin{aligned} \frac{d}{dt}\bm{\omega}(t)&=-\{4\nu\bm{I}_{5}+i\bm{A}+\bm{M}(t)\}\bm{\omega}(t)+\bm{f}_{\geq 3}(t)+\bm{c},\quad t>0,\\ \bm{\omega}(t)&=\bigl{(}\omega_{2}^{2}(t),\dots,\omega_{2}^{-2}(t)\bigr{)}^{T}\in\mathbb{C}^{5},\end{aligned}

where 𝑰5\bm{I}_{5} is the 5×55\times 5 identity matrix, 𝑨\bm{A} is a constant self-adjoint matrix, 𝑴(t)\bm{M}(t) and 𝒇3(t)\bm{f}_{\geq 3}(t) are a matrix-valued function and a vector field decaying exponentially in time, and 𝒄\bm{c} is a constant vector coming from the nondissipative part ω0,=1\omega_{0,=1} through the perturbation operator (see (3.15) and (3.17) for the precise definitions). Then, noting that 4ν𝑰5+i𝑨4\nu\bm{I}_{5}+i\bm{A} is invertible since 𝑨\bm{A} is self-adjoint, we set

𝝎=(ω2,2,ω2,2)T=(4ν𝑰5+i𝑨)1𝐜\displaystyle\bm{\omega}_{\infty}=(\omega_{2,\infty}^{2},\dots\omega_{2,\infty}^{-2})^{T}=(4\nu\bm{I}_{5}+i\bm{A})^{-1}\mathbf{c}

and use (1.21) to derive an estimate for 𝝎(t)𝝎\bm{\omega}(t)-\bm{\omega}_{\infty} which corresponds to (1.14). Moreover, using the explicit forms of 𝑨\bm{A} and 𝒄\bm{c}, we can solve (4ν𝑰5+i𝑨)𝝎=𝒄(4\nu\bm{I}_{5}+i\bm{A})\bm{\omega}_{\infty}=\bm{c} to determine 𝝎\bm{\omega}_{\infty}. Here the matrix 𝑨\bm{A} comes from the interaction between the nondissipative part ω0,=1\omega_{0,=1} and the dissipative part ω=2(t)\omega_{=2}(t) through the convection term. Indeed, each entry of 𝑨\bm{A} is given by the Y2mY_{2}^{m}-component of Xω=2X\cdot\nabla\omega_{=2} with XX given by (1.20). Also, we can express XY2mX\cdot\nabla Y_{2}^{m} as a linear combination of Y20Y_{2}^{0}, Y2±1Y_{2}^{\pm 1}, and Y2±2Y_{2}^{\pm 2} for each |m|=0,1,2|m|=0,1,2 (see Lemma 2.4), so we can write 𝑨\bm{A} explicitly in terms of ω0,10\omega_{0,1}^{0} and ω0,11\omega_{0,1}^{1} and find that 𝑨\bm{A} is self-adjoint.

The rest of this paper is organized as follows. In Section 2 we fix notations and give auxiliary results on calculus on a manifold and S2S^{2}. The main part of this paper is Section 3, which is devoted to the proof of Theorem 1.1. In Section 4 we give the proof of Lemma 2.4 which consists of elementary but slightly long calculations. As appendices, we briefly explain the behavior of a perturbation for the one-jet Kolmogorov type flow in Section 5 and observe that solutions to (1.8) and (1.17) are related by (1.18) in Section 6.

2. Preliminaries

We fix notations and give auxiliary results on calculus on a manifold and S2S^{2}.

2.1. Calculus on a manifold

For n2n\geq 2 let MM be an nn-dimensional Riemannian manifold without boundary. Note that MM is a real manifold, and in this subsection we only consider real-valued functions and vector fields on MM. Let ,\langle\cdot,\cdot\rangle, \nabla, and dnd\mathcal{H}^{n} be the Riemannian metric, the Levi-Civita connection, and the volume form on MM. For a function ff and a vector field XX on MM, we write f\nabla f, divX\mathrm{div}\,X, and Xf=X,f\nabla_{X}f=\langle X,\nabla f\rangle for the gradient of ff, the divergence of XX, and the directional derivative of ff along XX. We also denote by Δ=div\Delta=\mathrm{div}\,\nabla the Laplace–Beltrami operator on MM. Note that

(2.1) XY,Z=XY,Z+Y,XZ\displaystyle\nabla_{X}\langle Y,Z\rangle=\langle\nabla_{X}Y,Z\rangle+\langle Y,\nabla_{X}Z\rangle

for vector fields X,Y,ZX,Y,Z on MM, since \nabla is compatible with ,\langle\cdot,\cdot\rangle. For a function ff on MM, let 2f\nabla^{2}f be the covariant Hessian of ff given by

(2.2) (2f)(Y,X)=X(Yf)XYf=XY,fXY,f\displaystyle(\nabla^{2}f)(Y,X)=\nabla_{X}(\nabla_{Y}f)-\nabla_{\nabla_{X}Y}f=\nabla_{X}\langle Y,\nabla f\rangle-\langle\nabla_{X}Y,\nabla f\rangle

for vector fields X,YX,Y on MM (see [23]). As in the flat space case, 2f\nabla^{2}f is symmetric in the sense that (2f)(Y,X)=(2f)(X,Y)(\nabla^{2}f)(Y,X)=(\nabla^{2}f)(X,Y). A (smooth) vector field XX on MM is called Killing if YX,Z+Y,ZX=0\langle\nabla_{Y}X,Z\rangle+\langle Y,\nabla_{Z}X\rangle=0 for all vector fields YY and ZZ on MM. Note that divX=0\mathrm{div}\,X=0 on MM if XX is Killing, since divX=i=1nτiX,τi\mathrm{div}\,X=\sum_{i=1}^{n}\langle\nabla_{\tau_{i}}X,\tau_{i}\rangle at each xMx\in M, where {τ1,,τn}\{\tau_{1},\dots,\tau_{n}\} is an orthonormal basis of the tangent plane of MM at xx. For k0k\geq 0 we denote by Hk(M)H^{k}(M) the Sobolev spaces of L2L^{2} functions on MM (see e.g. [2]).

Lemma 2.1.

Let XX be a Killing vector field on MM. Then

(2.3) M{(Δf)g,X+(Δg)f,X}𝑑n=0\displaystyle\int_{M}\{(\Delta f)\langle\nabla g,X\rangle+(\Delta g)\langle\nabla f,X\rangle\}\,d\mathcal{H}^{n}=0

for all real-valued functions f,gH2(M)f,g\in H^{2}(M).

Proof.

Let F=fg,X+gf,XF=\nabla_{\nabla f}\langle\nabla g,X\rangle+\nabla_{\nabla g}\langle\nabla f,X\rangle on MM. Then

M{(Δf)g,X+(Δg)f,X}𝑑n=MF𝑑n\displaystyle\int_{M}\{(\Delta f)\langle\nabla g,X\rangle+(\Delta g)\langle\nabla f,X\rangle\}\,d\mathcal{H}^{n}=-\int_{M}F\,d\mathcal{H}^{n}

by integration by parts. Moreover, by (2.2) and the symmetry of ,\langle\cdot,\cdot\rangle, we have

(2f)(X,g)+(2g)(X,f)\displaystyle(\nabla^{2}f)(X,\nabla g)+(\nabla^{2}g)(X,\nabla f)
=gX,fgX,f+fX,gfX,g\displaystyle\qquad=\nabla_{\nabla g}\langle X,\nabla f\rangle-\langle\nabla_{\nabla g}X,\nabla f\rangle+\nabla_{\nabla f}\langle X,\nabla g\rangle-\langle\nabla_{\nabla f}X,\nabla g\rangle
=F(fX,g+f,gX)\displaystyle\qquad=F-(\langle\nabla_{\nabla f}X,\nabla g\rangle+\langle\nabla f,\nabla_{\nabla g}X\rangle)

and

(2f)(g,X)+(2g)(f,X)\displaystyle(\nabla^{2}f)(\nabla g,X)+(\nabla^{2}g)(\nabla f,X)
=Xg,fXg,f+Xf,gXf,g\displaystyle\qquad=\nabla_{X}\langle\nabla g,\nabla f\rangle-\langle\nabla_{X}\nabla g,\nabla f\rangle+\nabla_{X}\langle\nabla f,\nabla g\rangle-\langle\nabla_{X}\nabla f,\nabla g\rangle
=Xf,g\displaystyle\qquad=\nabla_{X}\langle\nabla f,\nabla g\rangle

on MM. In the last equality, we also used (2.1) with Y=fY=\nabla f and Z=gZ=\nabla g. Since 2f\nabla^{2}f and 2g\nabla^{2}g are symmetric, it follows from the above equalities that

F(fX,g+f,gX)=Xf,g\displaystyle F-(\langle\nabla_{\nabla f}X,\nabla g\rangle+\langle\nabla f,\nabla_{\nabla g}X\rangle)=\nabla_{X}\langle\nabla f,\nabla g\rangle

on MM. Moreover, fX,g+f,gX=0\langle\nabla_{\nabla f}X,\nabla g\rangle+\langle\nabla f,\nabla_{\nabla g}X\rangle=0 on MM since XX is Killing. Hence

M{(Δf)g,X+(Δg)f,X}𝑑n\displaystyle\int_{M}\{(\Delta f)\langle\nabla g,X\rangle+(\Delta g)\langle\nabla f,X\rangle\}\,d\mathcal{H}^{n} =MF𝑑n=MXf,g𝑑n\displaystyle=-\int_{M}F\,d\mathcal{H}^{n}=-\int_{M}\nabla_{X}\langle\nabla f,\nabla g\rangle\,d\mathcal{H}^{n}
=M(divX)f,g𝑑n=0\displaystyle=\int_{M}(\mathrm{div}\,X)\langle\nabla f,\nabla g\rangle\,d\mathcal{H}^{n}=0

by integration by parts and divX=0\mathrm{div}\,X=0 on MM since XX is Killing. ∎

2.2. Calculus on the unit sphere

Now let M=S2M=S^{2} be the 2D unit sphere in 3\mathbb{R}^{3} equipped with the Riemannian metric induced by the Euclidean metric of 3\mathbb{R}^{3}. We denote by 𝐚𝐛\mathbf{a}\cdot\mathbf{b} and 𝐚×𝐛\mathbf{a}\times\mathbf{b} the inner and vector products of 𝐚,𝐛3\mathbf{a},\mathbf{b}\in\mathbb{R}^{3}. Let 𝐧S2\mathbf{n}_{S^{2}} be the unit outward normal vector field of S2S^{2}. For real tangential vector fields XX and YY on S2S^{2}, the covariant derivative of XX along YY is given by

YX=(Y3)X~([(Y3)X~]𝐧S2)𝐧S2onS2,\displaystyle\nabla_{Y}X=(Y\cdot\nabla^{\mathbb{R}^{3}})\widetilde{X}-\left(\left[(Y\cdot\nabla^{\mathbb{R}^{3}})\widetilde{X}\right]\cdot\mathbf{n}_{S^{2}}\right)\mathbf{n}_{S^{2}}\quad\text{on}\quad S^{2},

where 3\nabla^{\mathbb{R}^{3}} is the standard gradient in 3\mathbb{R}^{3} and X~\tilde{X} is an extension of XX to an open neighborhood of S2S^{2}. Note that the value of YX\nabla_{Y}X is independent of the choice of X~\widetilde{X}. Then we easily see that for any 𝐚3\mathbf{a}\in\mathbb{R}^{3} a real vector field X(x)=𝐚×xX(x)=\mathbf{a}\times x, xS2x\in S^{2} is tangential and satisfies YXZ+YZX=0\nabla_{Y}X\cdot Z+Y\cdot\nabla_{Z}X=0 on S2S^{2} for all real tangential vector fields YY and ZZ on S2S^{2}, i.e. X(x)=𝐚×xX(x)=\mathbf{a}\times x is Killing on S2S^{2}. We use this fact without mention in the sequel.

In what follows, we mainly consider real-valued functions on S2S^{2}, but we take the L2(S2)L^{2}(S^{2})-inner product of functions with the complex spherical harmonics. Thus we write

(u,v)L2(S2)=S2uv¯𝑑2,uL2(S2)=(u,u)L2(S2)1/2\displaystyle(u,v)_{L^{2}(S^{2})}=\int_{S^{2}}u\bar{v}\,d\mathcal{H}^{2},\quad\|u\|_{L^{2}(S^{2})}=(u,u)_{L^{2}(S^{2})}^{1/2}

for complex-valued functions u,vL2(S2)u,v\in L^{2}(S^{2}), where v¯\bar{v} is the complex conjugate of vv. Also, we sometimes abuse the notations of the inner and vector products to write

(2.4) 𝐚(𝐛1+i𝐛2)=𝐚𝐛1+i𝐚𝐛2,𝐚×(𝐛1+i𝐛2)=𝐚×𝐛1+i𝐚×𝐛2\displaystyle\mathbf{a}\cdot(\mathbf{b}_{1}+i\mathbf{b}_{2})=\mathbf{a}\cdot\mathbf{b}_{1}+i\mathbf{a}\cdot\mathbf{b}_{2},\quad\mathbf{a}\times(\mathbf{b}_{1}+i\mathbf{b}_{2})=\mathbf{a}\times\mathbf{b}_{1}+i\mathbf{a}\times\mathbf{b}_{2}

for 𝐚,𝐛1,𝐛23\mathbf{a},\mathbf{b}_{1},\mathbf{b}_{2}\in\mathbb{R}^{3}. We do not encounter the case where 𝐚\mathbf{a} is complex in the sequel.

Let θ\theta and φ\varphi be the colatitude and longitude so that S2S^{2} is parametrized by

(2.5) x(θ,φ)=(sinθcosφ,sinθsinφ,cosθ),θ[0,π],φ[0,2π).\displaystyle x(\theta,\varphi)=(\sin\theta\cos\varphi,\sin\theta\sin\varphi,\cos\theta),\quad\theta\in[0,\pi],\,\varphi\in[0,2\pi).

For a function uu on S2S^{2}, we abuse the notation u(θ,φ)=(ux)(θ,φ)u(\theta,\varphi)=(u\circ x)(\theta,\varphi) so that

(2.6) u(θ,φ)=θu(θ,φ)θx(θ,φ)+φu(θ,φ)sin2θφx(θ,φ)\displaystyle\nabla u(\theta,\varphi)=\partial_{\theta}u(\theta,\varphi)\partial_{\theta}x(\theta,\varphi)+\frac{\partial_{\varphi}u(\theta,\varphi)}{\sin^{2}\theta}\,\partial_{\varphi}x(\theta,\varphi)

for the gradient of uu. Let YnmY_{n}^{m} be the spherical harmonics of the form

Ynm=Ynm(θ,φ)=2n+14π(nm)!(n+m)!Pnm(cosθ)eimφ,n0,|m|n.\displaystyle Y_{n}^{m}=Y_{n}^{m}(\theta,\varphi)=\sqrt{\frac{2n+1}{4\pi}\frac{(n-m)!}{(n+m)!}}\,P_{n}^{m}(\cos\theta)e^{im\varphi},\quad n\in\mathbb{Z}_{\geq 0},\,|m|\leq n.

Here PnmP_{n}^{m} are the associated Legendre functions given by

Pn0(s)\displaystyle P_{n}^{0}(s) =12nn!dndsn(s21)n,\displaystyle=\frac{1}{2^{n}n!}\frac{d^{n}}{ds^{n}}(s^{2}-1)^{n},
Pnm(s)\displaystyle P_{n}^{m}(s) ={(1)m(1s2)m/2dmdsmPn0(s),m0,(1)|m|(n|m|)!(n+|m|)!Pn|m|(s),m=|m|<0\displaystyle=\begin{cases}(-1)^{m}(1-s^{2})^{m/2}\displaystyle\frac{d^{m}}{ds^{m}}P_{n}^{0}(s),&m\geq 0,\\ (-1)^{|m|}\displaystyle\frac{(n-|m|)!}{(n+|m|)!}P_{n}^{|m|}(s),&m=-|m|<0\end{cases}

for s(1,1)s\in(-1,1) (see [22, 11]). Note that Ynm=(1)mYnm¯Y_{n}^{-m}=(-1)^{m}\overline{Y_{n}^{m}} by the above definitions. It is known (see e.g. [47, 46]) that YnmY_{n}^{m} are the eigenfunctions of Δ-\Delta associated with the eigenvalue λn=n(n+1)\lambda_{n}=n(n+1) for each n0n\geq 0, and the set of all YnmY_{n}^{m} forms an orthonormal basis of L2(S2)L^{2}(S^{2}). Hence each uL2(S2)u\in L^{2}(S^{2}) can be expanded as

u=n=0m=nnunmYnm,unm=(u,Ynm)L2(S2).\displaystyle u=\sum_{n=0}^{\infty}\sum_{m=-n}^{n}u_{n}^{m}Y_{n}^{m},\quad u_{n}^{m}=(u,Y_{n}^{m})_{L^{2}(S^{2})}.

Moreover, if uu is real-valued, then unm=(1)munm¯u_{n}^{-m}=(-1)^{m}\overline{u_{n}^{m}} since Ynm=(1)mYnm¯Y_{n}^{-m}=(-1)^{m}\overline{Y_{n}^{m}}. It is also known that the recurrence relation

(nm+1)Pn+1m(s)(2n+1)sPnm(s)+(n+m)Pn1m(s)=0\displaystyle(n-m+1)P_{n+1}^{m}(s)-(2n+1)sP_{n}^{m}(s)+(n+m)P_{n-1}^{m}(s)=0

holds (see [22, (7.12.12)]) and thus (see also [47, Section 5.7])

(2.7) cosθYnm=anmYn1m+an+1mYn+1m,anm=(nm)(n+m)(2n1)(2n+1)\displaystyle\cos\theta\,Y_{n}^{m}=a_{n}^{m}Y_{n-1}^{m}+a_{n+1}^{m}Y_{n+1}^{m},\quad a_{n}^{m}=\sqrt{\frac{(n-m)(n+m)}{(2n-1)(2n+1)}}

for n0n\in\mathbb{Z}_{\geq 0} and |m|n|m|\leq n, where we consider Y|m|1m0Y_{|m|-1}^{m}\equiv 0.

Let L02(S2)L_{0}^{2}(S^{2}) be the space of L2L^{2} functions on S2S^{2} with vanishing mean, i.e.

L02(S2)={uL2(S2)|S2u𝑑2=0}={uL2(S2)(u,Y00)L2(S2)=0}.\displaystyle L_{0}^{2}(S^{2})=\left\{u\in L^{2}(S^{2})~{}\middle|~{}\int_{S^{2}}u\,d\mathcal{H}^{2}=0\right\}=\{u\in L^{2}(S^{2})\mid(u,Y_{0}^{0})_{L^{2}(S^{2})}=0\}.

Then Δ\Delta is invertible and self-adjoint as a linear operator

Δ:DL02(S2)(Δ)L02(S2)L02(S2),DL02(S2)(Δ)=L02(S2)H2(S2).\displaystyle\Delta\colon D_{L_{0}^{2}(S^{2})}(\Delta)\subset L_{0}^{2}(S^{2})\to L_{0}^{2}(S^{2}),\quad D_{L_{0}^{2}(S^{2})}(\Delta)=L_{0}^{2}(S^{2})\cap H^{2}(S^{2}).

Also, the fractional Laplace–Beltrami operator (Δ)s(-\Delta)^{s} with ss\in\mathbb{R} is defined by

(2.8) (Δ)su=n=1m=nnλns(u,Ynm)L2(S2)Ynm,uL02(S2).\displaystyle(-\Delta)^{s}u=\sum_{n=1}^{\infty}\sum_{m=-n}^{n}\lambda_{n}^{s}(u,Y_{n}^{m})_{L^{2}(S^{2})}Y_{n}^{m},\quad u\in L_{0}^{2}(S^{2}).

Note that (Δ)su(-\Delta)^{s}u is real-valued if uu is so. Moreover,

(2.9) cosθφ(I+6Δ1)Ynm=im(16λn)(anmYn1m+an+1mYn+1m)\displaystyle\cos\theta\,\partial_{\varphi}(I+6\Delta^{-1})Y_{n}^{m}=im\left(1-\frac{6}{\lambda_{n}}\right)(a_{n}^{m}Y_{n-1}^{m}+a_{n+1}^{m}Y_{n+1}^{m})

by (2.7), (2.8), and φYnm=imYnm\partial_{\varphi}Y_{n}^{m}=imY_{n}^{m}. We also have

(2.10) (Δ)1/2uL2(S2)=uL2(S2),uL02(S2)H1(S2)\displaystyle\|(-\Delta)^{1/2}u\|_{L^{2}(S^{2})}=\|\nabla u\|_{L^{2}(S^{2})},\quad u\in L_{0}^{2}(S^{2})\cap H^{1}(S^{2})

by a density argument and integration by parts. Let us give auxiliary results.

Lemma 2.2.

Let ψH2(S2)\psi\in H^{2}(S^{2}) be a real-valued function and 𝐮=𝐧S2×ψ\mathbf{u}=\mathbf{n}_{S^{2}}\times\nabla\psi on S2S^{2}. Then

(2.11) (𝐮Δψ,Y1m)L2(S2)=0,|m|=0,1.\displaystyle(\mathbf{u}\cdot\nabla\Delta\psi,Y_{1}^{m})_{L^{2}(S^{2})}=0,\quad|m|=0,1.
Proof.

Since

Y10=C10cosθ,Y1±1=C1±1e±iφsinθ=C1±1(sinθcosφ±isinθsinφ)\displaystyle Y_{1}^{0}=C_{1}^{0}\cos\theta,\quad Y_{1}^{\pm 1}=C_{1}^{\pm 1}e^{\pm i\varphi}\sin\theta=C_{1}^{\pm 1}(\sin\theta\cos\varphi\pm i\sin\theta\sin\varphi)

with constants C10,C1±1C_{1}^{0},C_{1}^{\pm 1}\in\mathbb{R}, it suffices to show that

(𝐮Δψ,χ)L2(S2)=S2(𝐮Δψ)χ𝑑2=0,χ=cosθ,sinθcosφ,sinθsinφ.\displaystyle(\mathbf{u}\cdot\nabla\Delta\psi,\chi)_{L^{2}(S^{2})}=\int_{S^{2}}(\mathbf{u}\cdot\nabla\Delta\psi)\chi\,d\mathcal{H}^{2}=0,\quad\chi=\cos\theta,\sin\theta\cos\varphi,\sin\theta\sin\varphi.

Noting that 𝐮=𝐧S2×ψ\mathbf{u}=\mathbf{n}_{S^{2}}\times\nabla\psi is divergence free on S2S^{2}, we have

(𝐮Δψ,χ)L2(S2)=S2(Δψ)(𝐮χ)𝑑2=S2(Δψ){ψ(𝐧S2×χ)}𝑑2\displaystyle(\mathbf{u}\cdot\nabla\Delta\psi,\chi)_{L^{2}(S^{2})}=-\int_{S^{2}}(\Delta\psi)(\mathbf{u}\cdot\nabla\chi)\,d\mathcal{H}^{2}=\int_{S^{2}}(\Delta\psi)\{\nabla\psi\cdot(\mathbf{n}_{S^{2}}\times\nabla\chi)\}\,d\mathcal{H}^{2}

by integration by parts and (𝐚×𝐛)𝐜=𝐛(𝐚×𝐜)(\mathbf{a}\times\mathbf{b})\cdot\mathbf{c}=-\mathbf{b}\cdot(\mathbf{a}\times\mathbf{c}) for 𝐚,𝐛,𝐜3\mathbf{a},\mathbf{b},\mathbf{c}\in\mathbb{R}^{3}. Moreover, we observe by 𝐧S2(x)=x\mathbf{n}_{S^{2}}(x)=x for xS2x\in S^{2}, (2.5), and (2.6) that

(2.12) X(x)=𝐧S2(x)×χ(x)={𝐞3×xifχ=cosθ,𝐞1×xifχ=sinθcosφ,𝐞2×xifχ=sinθsinφ\displaystyle X(x)=\mathbf{n}_{S^{2}}(x)\times\nabla\chi(x)=\begin{cases}-\mathbf{e}_{3}\times x&\text{if}\quad\chi=\cos\theta,\\ -\mathbf{e}_{1}\times x&\text{if}\quad\chi=\sin\theta\cos\varphi,\\ -\mathbf{e}_{2}\times x&\text{if}\quad\chi=\sin\theta\sin\varphi\end{cases}

for xS2x\in S^{2}, where {𝐞1,𝐞2,𝐞3}\{\mathbf{e}_{1},\mathbf{e}_{2},\mathbf{e}_{3}\} is the standard basis of 3\mathbb{R}^{3}. Since this XX is Killing and ψ\psi is real-valued, we can apply (2.3) with f=g=ψf=g=\psi to find that

(𝐮Δψ,χ)L2(S2)=S2(Δψ)(ψX)𝑑2=0\displaystyle(\mathbf{u}\cdot\nabla\Delta\psi,\chi)_{L^{2}(S^{2})}=\int_{S^{2}}(\Delta\psi)(\nabla\psi\cdot X)\,d\mathcal{H}^{2}=0

and thus (2.11) follows. ∎

Lemma 2.3.

For 𝐚3\mathbf{a}\in\mathbb{R}^{3} let X(x)=𝐚×xX(x)=\mathbf{a}\times x, xS2x\in S^{2}. Then

(2.13) (XΔ1ω,ω)L2(S2)=(Xω,Δ1ω)L2(S2)=0\displaystyle(X\cdot\nabla\Delta^{-1}\omega,\omega)_{L^{2}(S^{2})}=(X\cdot\nabla\omega,\Delta^{-1}\omega)_{L^{2}(S^{2})}=0

for every real-valued function ωL02(S2)\omega\in L_{0}^{2}(S^{2}).

Proof.

Let ψ=Δ1ωH2(S2)\psi=\Delta^{-1}\omega\in H^{2}(S^{2}). Then since ψ\psi is real-valued and XX is a Killing vector field on S2S^{2}, it follows from (2.3) with f=g=ψf=g=\psi that

(XΔ1ω,ω)L2(S2)=(Xψ,Δψ)L2(S2)=0.\displaystyle(X\cdot\nabla\Delta^{-1}\omega,\omega)_{L^{2}(S^{2})}=(X\cdot\nabla\psi,\Delta\psi)_{L^{2}(S^{2})}=0.

Also, noting that divX=0\mathrm{div}\,X=0 on S2S^{2} since XX is Killing, we have

(Xω,Δ1ω)L2(S2)=(XΔ1ω,ω)L2(S2)=0\displaystyle(X\cdot\nabla\omega,\Delta^{-1}\omega)_{L^{2}(S^{2})}=-(X\cdot\nabla\Delta^{-1}\omega,\omega)_{L^{2}(S^{2})}=0

by integration by parts and the above equality, Hence (2.13) is valid. ∎

Lemma 2.4.

For 𝐚=(a1,a2,a3)T3\mathbf{a}=(a_{1},a_{2},a_{3})^{T}\in\mathbb{R}^{3} let X(x)=𝐚×xX(x)=\mathbf{a}\times x, xS2x\in S^{2}. Then

(2.14) XY20=62(ia1+a2)Y21+62(ia1a2)Y21,XY21=(ia1+a2)Y22+ia3Y21+62(ia1a2)Y20,XY22=2ia3Y22+(ia1a2)Y21.\displaystyle\begin{aligned} X\cdot\nabla Y_{2}^{0}&=\frac{\sqrt{6}}{2}(ia_{1}+a_{2})Y_{2}^{1}+\frac{\sqrt{6}}{2}(ia_{1}-a_{2})Y_{2}^{-1},\\ X\cdot\nabla Y_{2}^{1}&=(ia_{1}+a_{2})Y_{2}^{2}+ia_{3}Y_{2}^{1}+\frac{\sqrt{6}}{2}(ia_{1}-a_{2})Y_{2}^{0},\\ X\cdot\nabla Y_{2}^{2}&=2ia_{3}Y_{2}^{2}+(ia_{1}-a_{2})Y_{2}^{1}.\end{aligned}

Moreover, since Y2m=(1)mY2m¯Y_{2}^{-m}=(-1)^{m}\overline{Y_{2}^{m}} and XX is real,

(2.15) XY21=(ia1a2)Y22ia3Y21+62(ia1+a2)Y20,XY22=2ia3Y22+(ia1+a2)Y21.\displaystyle\begin{aligned} X\cdot\nabla Y_{2}^{-1}&=(ia_{1}-a_{2})Y_{2}^{-2}-ia_{3}Y_{2}^{-1}+\frac{\sqrt{6}}{2}(ia_{1}+a_{2})Y_{2}^{0},\\ X\cdot\nabla Y_{2}^{-2}&=-2ia_{3}Y_{2}^{-2}+(ia_{1}+a_{2})Y_{2}^{-1}.\end{aligned}

The equalities (2.14) are shown by elementary calculations under spherical coordinates. To avoid making this section too long, we give the proof of (2.14) in Section 4.

3. Proof of Theorem 1.1

The purpose of this section is to establish Theorem 1.1.

First we note that the global existence and uniqueness of a weak solution to (1.8) are proved in the same way as in the case of the Navier–Stokes equations in flat 2D domains (see e.g. [44, 4]). We can show the global existence of a weak solution by a standard Galerkin method with basis functions YnmY_{n}^{m}. Also, since S2S^{2} is 2D, we can get the uniqueness of a weak solution by estimating the difference of two weak solutions with the aid of the weak form (1.12) and Ladyzhenskaya’s inequality

ωL4(S2)ωL2(S2)1/2ωH1(S2)1/2,ωH1(S2),\displaystyle\|\omega\|_{L^{4}(S^{2})}\leq\|\omega\|_{L^{2}(S^{2})}^{1/2}\|\omega\|_{H^{1}(S^{2})}^{1/2},\quad\omega\in H^{1}(S^{2}),

which follows from the same inequality on 2\mathbb{R}^{2} (see [21]) and a localization argument with a partition of unity, and then by using Gronwall’s inequality. Here we omit details and just give a remark: approximate solutions constructed by the Galerkin method can grow exponentially in time due to the perturbation operator in (1.8), but it does not matter since we take a limit of the approximate solutions on finite time intervals, e.g. on [0,n][0,n], nn\in\mathbb{N} to get weak solutions ωn\omega_{n} on [0,n][0,n] and then we use the uniqueness of a weak solution to define a global weak solution ω\omega by ω=ωn\omega=\omega_{n} on [0,n][0,n] for each nn\in\mathbb{N}. We also refer to [40] for the proof in the case of a modified vorticity equation on S2S^{2}.

Now let ω\omega be the unique global weak solution to (1.8) with initial data ω0L02(S2)\omega_{0}\in L_{0}^{2}(S^{2}). Here we assume that ω0\omega_{0} is real-valued and thus ω(t)\omega(t) is so. We write

ωnm(t)=(ω(t),Ynm)L2(S2),ω=n(t)=m=nnωnm(t)Ynm,ωN(t)=nNω=n(t)\displaystyle\omega_{n}^{m}(t)=(\omega(t),Y_{n}^{m})_{L^{2}(S^{2})},\quad\omega_{=n}(t)=\sum_{m=-n}^{n}\omega_{n}^{m}(t)Y_{n}^{m},\quad\omega_{\geq N}(t)=\sum_{n\geq N}\omega_{=n}(t)

and similarly for ω0\omega_{0} so that

ω(t)=n=1m=nnωnm(t)Ynm,ω0=n=1m=nnω0,nmYnm.\displaystyle\omega(t)=\sum_{n=1}^{\infty}\sum_{m=-n}^{n}\omega_{n}^{m}(t)Y_{n}^{m},\quad\omega_{0}=\sum_{n=1}^{\infty}\sum_{m=-n}^{n}\omega_{0,n}^{m}Y_{n}^{m}.

Then, since ω(t)\omega(t) and ω0\omega_{0} are real-valued and Ynm=(1)mYnm¯Y_{n}^{-m}=(-1)^{m}\overline{Y_{n}^{m}},

(3.1) ωnm(t)=(1)mωnm(t)¯,ω0,nm=(1)mω0,nm¯\displaystyle\omega_{n}^{-m}(t)=(-1)^{m}\overline{\omega_{n}^{m}(t)},\quad\omega_{0,n}^{-m}=(-1)^{m}\overline{\omega_{0,n}^{m}}

and thus ω=n(t)\omega_{=n}(t), ωN(t)\omega_{\geq N}(t), ω0,=n\omega_{0,=n}, and ω0,N\omega_{0,\geq N} are real-valued.

Let us show (1.13) and (1.14). The proof consists of six steps. In what follows, for the sake of simplicity, we say that we take the inner product of an equation like (1.8) with a test function instead of saying that we use a corresponding weak form like (1.12). Also, we frequently use the following fact without mention: a vector field of the form 𝐯=𝐧S2×f\mathbf{v}=\mathbf{n}_{S^{2}}\times\nabla f with a real-valued function ff on S2S^{2} satisfies div𝐯=0\mathrm{div}\,\mathbf{v}=0 on S2S^{2}.

Step 1: ω=1(t)=ω0,=1\omega_{=1}(t)=\omega_{0,=1}. For |m|=0,1|m|=0,1, we have

(cosθφ(I+6Δ1)ω(t),Y1m)L2(S2)=0\displaystyle(\cos\theta\,\partial_{\varphi}(I+6\Delta^{-1})\omega(t),Y_{1}^{m})_{L^{2}(S^{2})}=0

by (2.9) and λ2=6\lambda_{2}=6. Moreover, since λ1=2\lambda_{1}=2,

(Δω(t)+2ω(t),Y1m)L2(S2)=(λ1+2)ω1m(t)=0.\displaystyle(\Delta\omega(t)+2\omega(t),Y_{1}^{m})_{L^{2}(S^{2})}=(-\lambda_{1}+2)\omega_{1}^{m}(t)=0.

Also, for ψ(t)=Δ1ω(t)\psi(t)=\Delta^{-1}\omega(t) we have 𝐮(t)=𝐧S2×Δ1ω(t)=𝐧S2×ψ(t)\mathbf{u}(t)=\mathbf{n}_{S^{2}}\times\nabla\Delta^{-1}\omega(t)=\mathbf{n}_{S^{2}}\times\nabla\psi(t) and thus

(𝐮(t)ω(t),Y1m)L2(S2)=(𝐮(t)Δψ(t),Y1m)L2(S2)=0\displaystyle(\mathbf{u}(t)\cdot\nabla\omega(t),Y_{1}^{m})_{L^{2}(S^{2})}=(\mathbf{u}(t)\cdot\nabla\Delta\psi(t),Y_{1}^{m})_{L^{2}(S^{2})}=0

by (2.11). Hence we take the inner product of (1.8) with Y1mY_{1}^{m} to get

ddtω1m(t)=tω(t),Y1mH01=0\displaystyle\frac{d}{dt}\omega_{1}^{m}(t)=\langle\partial_{t}\omega(t),Y_{1}^{m}\rangle_{H_{0}^{1}}=0

and thus ω1m(t)=ω0,1m\omega_{1}^{m}(t)=\omega_{0,1}^{m} for |m|=0,1|m|=0,1, i.e. ω=1(t)=ω0,=1\omega_{=1}(t)=\omega_{0,=1}.

Step 2: derivation of an equation for ω2(t)=ω(t)ω=1(t)\omega_{\geq 2}(t)=\omega(t)-\omega_{=1}(t). Let

(3.2) X=𝐧S2×Δ1ω0,=1=12𝐧S2×ω0,=1,𝐮2(t)=𝐧S2×Δ1ω2(t).\displaystyle\begin{aligned} X&=\mathbf{n}_{S^{2}}\times\nabla\Delta^{-1}\omega_{0,=1}=-\frac{1}{2}\mathbf{n}_{S^{2}}\times\nabla\omega_{0,=1},\\ \mathbf{u}_{\geq 2}(t)&=\mathbf{n}_{S^{2}}\times\nabla\Delta^{-1}\omega_{\geq 2}(t).\end{aligned}

Here we used Δω0,=1=λ1ω0,=1=2ω0,=1-\Delta\omega_{0,=1}=\lambda_{1}\omega_{0,=1}=2\omega_{0,=1}. Then since ω=1(t)=ω0,=1\omega_{=1}(t)=\omega_{0,=1},

ω(t)=ω0,=1+ω2(t),𝐮(t)=X+𝐮2(t).\displaystyle\omega(t)=\omega_{0,=1}+\omega_{\geq 2}(t),\quad\mathbf{u}(t)=X+\mathbf{u}_{\geq 2}(t).

We substitute these expressions for (1.8). Then tω0,=1=0\partial_{t}\omega_{0,=1}=0 and

Δω0,=1+2ω0,=1\displaystyle\Delta\omega_{0,=1}+2\omega_{0,=1} =(λ1+2)ω0,=1=0,\displaystyle=(-\lambda_{1}+2)\omega_{0,=1}=0,
cosθφ(I+6Δ1)ω0,=1\displaystyle\cos\theta\,\partial_{\varphi}(I+6\Delta^{-1})\omega_{0,=1} =2i5ω0,11Y21+2i5ω0,11Y21\displaystyle=-\frac{2i}{\sqrt{5}}\,\omega_{0,1}^{1}Y_{2}^{1}+\frac{2i}{\sqrt{5}}\,\omega_{0,1}^{-1}Y_{2}^{-1}

by λ1=2\lambda_{1}=2 and (2.9). Also, since Xω0,=1=0X\cdot\nabla\omega_{0,=1}=0 by the expression (3.2) of XX,

𝐮(t)ω(t)\displaystyle\mathbf{u}(t)\cdot\nabla\omega(t) =Xω2(t)+𝐮2(t)ω0,=1+𝐮2(t)ω2(t)\displaystyle=X\cdot\nabla\omega_{\geq 2}(t)+\mathbf{u}_{\geq 2}(t)\cdot\nabla\omega_{0,=1}+\mathbf{u}_{\geq 2}(t)\cdot\nabla\omega_{\geq 2}(t)
=X(I+2Δ1)ω2(t)+𝐮2(t)ω2(t),\displaystyle=X\cdot\nabla(I+2\Delta^{-1})\omega_{\geq 2}(t)+\mathbf{u}_{\geq 2}(t)\cdot\nabla\omega_{\geq 2}(t),

where we also used

𝐮2(t)ω0,=1\displaystyle\mathbf{u}_{\geq 2}(t)\cdot\nabla\omega_{0,=1} ={𝐧S2×Δ1ω2(t)}ω0,=1\displaystyle=\{\mathbf{n}_{S^{2}}\times\nabla\Delta^{-1}\omega_{\geq 2}(t)\}\cdot\nabla\omega_{0,=1}
=Δ1ω2(t)(𝐧S2×ω0,=1)\displaystyle=-\nabla\Delta^{-1}\omega_{\geq 2}(t)\cdot(\mathbf{n}_{S^{2}}\times\nabla\omega_{0,=1})
=2Δ1ω2(t)X.\displaystyle=2\nabla\Delta^{-1}\omega_{\geq 2}(t)\cdot X.

Hence we obtain

(3.3) tω2(t)=ν{Δω2(t)+2ω2(t)}a45πcosθφ(I+6Δ1)ω2(t)X(I+2Δ1)ω2(t)𝐮2(t)ω2(t)+ia2πω0,11Y21ia2πω0,11Y21,\displaystyle\begin{aligned} \partial_{t}\omega_{\geq 2}(t)&=\nu\{\Delta\omega_{\geq 2}(t)+2\omega_{\geq 2}(t)\}-\frac{a}{4}\sqrt{\frac{5}{\pi}}\cos\theta\,\partial_{\varphi}(I+6\Delta^{-1})\omega_{\geq 2}(t)\\ &\qquad-X\cdot\nabla(I+2\Delta^{-1})\omega_{\geq 2}(t)-\mathbf{u}_{\geq 2}(t)\cdot\nabla\omega_{\geq 2}(t)\\ &\qquad+\frac{ia}{2\sqrt{\pi}}\,\omega_{0,1}^{1}Y_{2}^{1}-\frac{ia}{2\sqrt{\pi}}\,\omega_{0,1}^{-1}Y_{2}^{-1},\end{aligned}

where the last two terms are stationary source terms due to the effect of the nondissipative part ω0,=1\omega_{0,=1} of the solution ω(t)\omega(t) through the perturbation operator.

Step 3: expression of XX. By (3.1), we have

ω0,10,ω0,11=Re[ω0,11]+iIm[ω0,11],ω0,11=Re[ω0,11]+iIm[ω0,11].\displaystyle\omega_{0,1}^{0}\in\mathbb{R},\quad\omega_{0,1}^{1}=\mathrm{Re}[\omega_{0,1}^{1}]+i\,\mathrm{Im}[\omega_{0,1}^{1}],\quad\omega_{0,1}^{-1}=-\mathrm{Re}[\omega_{0,1}^{1}]+i\,\mathrm{Im}[\omega_{0,1}^{1}].

Then since

Y10=123πcosθ,Y1±1=1232πsinθ(cosφ±isinφ),\displaystyle Y_{1}^{0}=\frac{1}{2}\sqrt{\frac{3}{\pi}}\cos\theta,\quad Y_{1}^{\pm 1}=\mp\frac{1}{2}\sqrt{\frac{3}{2\pi}}\sin\theta(\cos\varphi\pm i\sin\varphi),

we can write ω0,=1=m=11ω0,1mY1m\omega_{0,=1}=\sum_{m=-1}^{1}\omega_{0,1}^{m}Y_{1}^{m} as

ω0,=1=ω0,1023πcosθRe[ω0,11]32πsinθcosφ+Im[ω0,11]32πsinθsinφ.\displaystyle\omega_{0,=1}=\frac{\omega_{0,1}^{0}}{2}\sqrt{\frac{3}{\pi}}\cos\theta-\mathrm{Re}[\omega_{0,1}^{1}]\sqrt{\frac{3}{2\pi}}\sin\theta\cos\varphi+\mathrm{Im}[\omega_{0,1}^{1}]\sqrt{\frac{3}{2\pi}}\sin\theta\sin\varphi.

By this equality and (2.12), we find that

(3.4) X(x)=12𝐧S2(x)×ω0,=1(x)=𝐚×x,xS2,\displaystyle X(x)=-\frac{1}{2}\mathbf{n}_{S^{2}}(x)\times\nabla\omega_{0,=1}(x)=\mathbf{a}\times x,\quad x\in S^{2},

where 𝐚=(a1,a2,a3)3\mathbf{a}=(a_{1},a_{2},a_{3})\in\mathbb{R}^{3} is given by

(3.5) a1=1232πRe[ω0,11],a2=1232πIm[ω0,11],a3=143πω0,10.\displaystyle a_{1}=-\frac{1}{2}\sqrt{\frac{3}{2\pi}}\,\mathrm{Re}[\omega_{0,1}^{1}],\quad a_{2}=\frac{1}{2}\sqrt{\frac{3}{2\pi}}\,\mathrm{Im}[\omega_{0,1}^{1}],\quad a_{3}=\frac{1}{4}\sqrt{\frac{3}{\pi}}\,\omega_{0,1}^{0}.

In particular, XX is a Killing vector field on S2S^{2}.

Step 4: estimate for ω3(t)\omega_{\geq 3}(t). We take the inner product of (3.3) with

(I+6Δ1)ω2(t)=n2m=nm(16λn)ωnm(t)Ynm.\displaystyle(I+6\Delta^{-1})\omega_{\geq 2}(t)=\sum_{n\geq 2}\sum_{m=-n}^{m}\left(1-\frac{6}{\lambda_{n}}\right)\omega_{n}^{m}(t)Y_{n}^{m}.

Then since the right-hand side is in fact the summation for n3n\geq 3 by λ2=6\lambda_{2}=6,

(Y2m,(I+6Δ1)ω2(t))L2(S2)=0,m=±1.\displaystyle\bigl{(}Y_{2}^{m},(I+6\Delta^{-1})\omega_{\geq 2}(t)\bigr{)}_{L^{2}(S^{2})}=0,\quad m=\pm 1.

Also, it follows from λ2=6\lambda_{2}=6 and λn12\lambda_{n}\geq 12 for n3n\geq 3 that

tω2(t),(I+6Δ1)ω2(t)H01\displaystyle\langle\partial_{t}\omega_{\geq 2}(t),(I+6\Delta^{-1})\omega_{\geq 2}(t)\rangle_{H_{0}^{1}} 14ddtω3(t)L2(S2)2,\displaystyle\geq\frac{1}{4}\frac{d}{dt}\|\omega_{\geq 3}(t)\|_{L^{2}(S^{2})}^{2},
(Δω2(t)+2ω2(t),(I+6Δ1)ω2(t))L2(S2)\displaystyle\bigl{(}\Delta\omega_{\geq 2}(t)+2\omega_{\geq 2}(t),(I+6\Delta^{-1})\omega_{\geq 2}(t)\bigr{)}_{L^{2}(S^{2})} 5ω3(t)L2(S2)2.\displaystyle\leq-5\|\omega_{\geq 3}(t)\|_{L^{2}(S^{2})}^{2}.

Noting that (I+6Δ1)ω2(t)(I+6\Delta^{-1})\omega_{\geq 2}(t) is real-valued and cosθ\cos\theta is independent of φ\varphi, we carry out integration by parts with respect to φ\varphi to find that

(cosθφ(I+6Δ1)ω2(t),(I+6Δ1)ω2(t))L2(S2)=0.\displaystyle\bigl{(}\cos\theta\,\partial_{\varphi}(I+6\Delta^{-1})\omega_{\geq 2}(t),(I+6\Delta^{-1})\omega_{\geq 2}(t)\bigr{)}_{L^{2}(S^{2})}=0.

Since 𝐮2(t)Δ1ω2(t)=0\mathbf{u}_{\geq 2}(t)\cdot\nabla\Delta^{-1}\omega_{\geq 2}(t)=0 by the definition (3.2) of 𝐮2(t)\mathbf{u}_{\geq 2}(t), we have

(𝐮2(t)ω2(t),(I+6Δ1)ω2(t))L2(S2)\displaystyle\bigl{(}\mathbf{u}_{\geq 2}(t)\cdot\nabla\omega_{\geq 2}(t),(I+6\Delta^{-1})\omega_{\geq 2}(t)\bigr{)}_{L^{2}(S^{2})}
=(𝐮2(I+6Δ1)ω2(t),(I+6Δ1)ω2(t))L2(S2)=0\displaystyle\qquad=\bigl{(}\mathbf{u}_{\geq 2}\cdot\nabla(I+6\Delta^{-1})\omega_{\geq 2}(t),(I+6\Delta^{-1})\omega_{\geq 2}(t)\bigr{)}_{L^{2}(S^{2})}=0

by integration by parts and div𝐮2(t)=0\mathrm{div}\,\mathbf{u}_{\geq 2}(t)=0. Also, since ω2(t)\omega_{\geq 2}(t) is real-valued and XX is of the form (3.4), we can apply (2.13) to get

(XΔ1ω2(t),ω2(t))L2(S2)=(Xω2(t),Δ1ω2(t))L2(S2)=0.\displaystyle\bigl{(}X\cdot\nabla\Delta^{-1}\omega_{\geq 2}(t),\omega_{\geq 2}(t)\bigr{)}_{L^{2}(S^{2})}=\bigl{(}X\cdot\nabla\omega_{\geq 2}(t),\Delta^{-1}\omega_{\geq 2}(t)\bigr{)}_{L^{2}(S^{2})}=0.

We further observe that

(Xω2(t),ω2(t))L2(S2)=(XΔ1ω2(t),Δ1ω2(t))L2(S2)=0\displaystyle\bigl{(}X\cdot\nabla\omega_{\geq 2}(t),\omega_{\geq 2}(t)\bigr{)}_{L^{2}(S^{2})}=\bigl{(}X\cdot\nabla\Delta^{-1}\omega_{\geq 2}(t),\Delta^{-1}\omega_{\geq 2}(t)\bigr{)}_{L^{2}(S^{2})}=0

by integration by parts and divX=0\mathrm{div}\,X=0. Hence

(X(I+2Δ1)ω2(t),(I+6Δ1)ω2(t))L2(S2)=0.\displaystyle\bigl{(}X\cdot\nabla(I+2\Delta^{-1})\omega_{\geq 2}(t),(I+6\Delta^{-1})\omega_{\geq 2}(t)\bigr{)}_{L^{2}(S^{2})}=0.

Now we apply the above relations to the inner product of (3.3) with (I+6Δ1)ω2(t)(I+6\Delta^{-1})\omega_{\geq 2}(t). Then we find that

14ddtω3(t)L2(S2)25νω3(t)L2(S2)2\displaystyle\frac{1}{4}\frac{d}{dt}\|\omega_{\geq 3}(t)\|_{L^{2}(S^{2})}^{2}\leq-5\nu\|\omega_{\geq 3}(t)\|_{L^{2}(S^{2})}^{2}

and thus the estimate (1.13) for ω3(t)\omega_{\geq 3}(t) follows (note that ω3(0)=ω0,3\omega_{\geq 3}(0)=\omega_{0,\geq 3}).

Step 5: derivation of ODEs for ω2m(t)\omega_{2}^{m}(t), |m|=0,1,2|m|=0,1,2. We set

(3.6) 𝐮=2(t)=𝐧S2×Δ1ω=2(t)=16𝐧S2×ω=2(t),𝐮3(t)=𝐧S2×Δ1ω3(t).\displaystyle\begin{aligned} \mathbf{u}_{=2}(t)&=\mathbf{n}_{S^{2}}\times\nabla\Delta^{-1}\omega_{=2}(t)=-\frac{1}{6}\mathbf{n}_{S^{2}}\times\nabla\omega_{=2}(t),\\ \mathbf{u}_{\geq 3}(t)&=\mathbf{n}_{S^{2}}\times\nabla\Delta^{-1}\omega_{\geq 3}(t).\end{aligned}

Here we used Δω=2(t)=λ2ω=2(t)=6ω=2(t)-\Delta\omega_{=2}(t)=\lambda_{2}\omega_{=2}(t)=6\omega_{=2}(t). We substitute

ω2(t)=ω=2(t)+ω3(t),𝐮2(t)=𝐮=2(t)+𝐮3(t)\displaystyle\omega_{\geq 2}(t)=\omega_{=2}(t)+\omega_{\geq 3}(t),\quad\mathbf{u}_{\geq 2}(t)=\mathbf{u}_{=2}(t)+\mathbf{u}_{\geq 3}(t)

for the right-hand side of (3.3). Then since Δω=2(t)=6ω=2(t)-\Delta\omega_{=2}(t)=6\omega_{=2}(t),

Δω2(t)+2ω2(t)\displaystyle\Delta\omega_{\geq 2}(t)+2\omega_{\geq 2}(t) =4ω=2(t)+{Δω3(t)+2ω3(t)},\displaystyle=-4\omega_{=2}(t)+\{\Delta\omega_{\geq 3}(t)+2\omega_{\geq 3}(t)\},
(I+6Δ1)ω2(t)\displaystyle(I+6\Delta^{-1})\omega_{\geq 2}(t) =(I+6Δ1)ω3(t),\displaystyle=(I+6\Delta^{-1})\omega_{\geq 3}(t),
X(I+2Δ1)ω2(t)\displaystyle X\cdot\nabla(I+2\Delta^{-1})\omega_{\geq 2}(t) =23Xω=2(t)+X(I+2Δ1)ω3(t).\displaystyle=\frac{2}{3}X\cdot\nabla\omega_{=2}(t)+X\cdot\nabla(I+2\Delta^{-1})\omega_{\geq 3}(t).

Also, since 𝐮=2(t)ω=2(t)=0\mathbf{u}_{=2}(t)\cdot\nabla\omega_{=2}(t)=0 by the expression (3.6) of 𝐮=2(t)\mathbf{u}_{=2}(t),

𝐮2(t)ω2(t)\displaystyle\mathbf{u}_{\geq 2}(t)\cdot\nabla\omega_{\geq 2}(t) =𝐮=2(t)ω3(t)+𝐮3(t)ω=2(t)+𝐮3(t)ω3(t)\displaystyle=\mathbf{u}_{=2}(t)\cdot\nabla\omega_{\geq 3}(t)+\mathbf{u}_{\geq 3}(t)\cdot\nabla\omega_{=2}(t)+\mathbf{u}_{\geq 3}(t)\cdot\nabla\omega_{\geq 3}(t)
=𝐮=2(t)(I+6Δ1)ω3(t)+𝐮3(t)ω3(t),\displaystyle=\mathbf{u}_{=2}(t)\cdot\nabla(I+6\Delta^{-1})\omega_{\geq 3}(t)+\mathbf{u}_{\geq 3}(t)\cdot\nabla\omega_{\geq 3}(t),

where we also used

𝐮3(t)ω=2(t)\displaystyle\mathbf{u}_{\geq 3}(t)\cdot\nabla\omega_{=2}(t) ={𝐧S2×Δ1ω3(t)}ω=2(t)\displaystyle=\{\mathbf{n}_{S^{2}}\times\nabla\Delta^{-1}\omega_{\geq 3}(t)\}\cdot\nabla\omega_{=2}(t)
=Δ1ω3(t){𝐧S2×ω=2(t)}\displaystyle=-\nabla\Delta^{-1}\omega_{\geq 3}(t)\cdot\{\mathbf{n}_{S^{2}}\times\nabla\omega_{=2}(t)\}
=6Δ1ω3(t)𝐮=2(t).\displaystyle=6\nabla\Delta^{-1}\omega_{\geq 3}(t)\cdot\mathbf{u}_{=2}(t).

Hence we have

(3.7) tω2(t)=4νω=2(t)23Xω=2(t)𝐮=2(t)(I+6Δ1)ω3(t)+ia2πω0,11Y21ia2πω0,11Y21+f3(t),\displaystyle\begin{aligned} \partial_{t}\omega_{\geq 2}(t)&=-4\nu\omega_{=2}(t)-\frac{2}{3}X\cdot\nabla\omega_{=2}(t)-\mathbf{u}_{=2}(t)\cdot\nabla(I+6\Delta^{-1})\omega_{\geq 3}(t)\\ &\qquad+\frac{ia}{2\sqrt{\pi}}\,\omega_{0,1}^{1}Y_{2}^{1}-\frac{ia}{2\sqrt{\pi}}\,\omega_{0,1}^{-1}Y_{2}^{-1}+f_{\geq 3}(t),\end{aligned}

where

(3.8) f3(t)=ν{Δω3(t)+2ω3(t)}a45πcosθφ(I+6Δ1)ω3(t)X(I+2Δ1)ω3(t)𝐮3(t)ω3(t).\displaystyle\begin{aligned} f_{\geq 3}(t)&=\nu\{\Delta\omega_{\geq 3}(t)+2\omega_{\geq 3}(t)\}-\frac{a}{4}\sqrt{\frac{5}{\pi}}\cos\theta\,\partial_{\varphi}(I+6\Delta^{-1})\omega_{\geq 3}(t)\\ &\qquad-X\cdot\nabla(I+2\Delta^{-1})\omega_{\geq 3}(t)-\mathbf{u}_{\geq 3}(t)\cdot\nabla\omega_{\geq 3}(t).\end{aligned}

Noe we take the inner product of (3.7) with Y2mY_{2}^{m}, |m|=0,1,2|m|=0,1,2. Then

(3.9) ddtω2m(t)=4νω2m(t)23(Xω=2(t),Y2m)L2(S2)(𝐮=2(t)(I+6Δ1)ω3(t),Y2m)L2(S2)+ia2πω0,11δ1,mia2πω0,11δ1,m+(f3(t),Y2m)L2(S2),\displaystyle\begin{aligned} \frac{d}{dt}\omega_{2}^{m}(t)&=-4\nu\omega_{2}^{m}(t)-\frac{2}{3}(X\cdot\nabla\omega_{=2}(t),Y_{2}^{m})_{L^{2}(S^{2})}\\ &\qquad-(\mathbf{u}_{=2}(t)\cdot\nabla(I+6\Delta^{-1})\omega_{\geq 3}(t),Y_{2}^{m})_{L^{2}(S^{2})}\\ &\qquad+\frac{ia}{2\sqrt{\pi}}\,\omega_{0,1}^{1}\delta_{1,m}-\frac{ia}{2\sqrt{\pi}}\,\omega_{0,1}^{-1}\delta_{-1,m}+(f_{\geq 3}(t),Y_{2}^{m})_{L^{2}(S^{2})},\end{aligned}

where δj,k\delta_{j,k} is the Kronecker delta. Let us calculate the above inner products. We observe by integration by parts and div𝐮=2(t)=0\mathrm{div}\,\mathbf{u}_{=2}(t)=0 that

(𝐮=2(t)(I+6Δ1)ω3(t),Y2m)L2(S2)\displaystyle(\mathbf{u}_{=2}(t)\cdot\nabla(I+6\Delta^{-1})\omega_{\geq 3}(t),Y_{2}^{m})_{L^{2}(S^{2})} =S2{𝐮=2(t)(I+6Δ1)ω3(t)}Y2m¯𝑑2\displaystyle=\int_{S^{2}}\{\mathbf{u}_{=2}(t)\cdot\nabla(I+6\Delta^{-1})\omega_{\geq 3}(t)\}\overline{Y_{2}^{m}}\,d\mathcal{H}^{2}
=S2(I+6Δ1)ω3(t)(𝐮=2(t)Y2m¯)𝑑2.\displaystyle=-\int_{S^{2}}(I+6\Delta^{-1})\omega_{\geq 3}(t)\Bigl{(}\mathbf{u}_{=2}(t)\cdot\nabla\overline{Y_{2}^{m}}\Bigr{)}\,d\mathcal{H}^{2}.

Recall that here we use the complex spherical harmonics and abuse the notations of the inner and vector products in 3\mathbb{R}^{3} (see (2.4)). We write this expression as

(3.12) (𝐮=2(t)(I+6Δ1)ω3(t),Y2m)L2(S2)=k=22Mm,k(t)ω2k(t),Mm,k(t)=16S2(I+6Δ1)ω3(t){(𝐧S2×Y2k)Y2m¯}𝑑2\displaystyle\begin{gathered}(\mathbf{u}_{=2}(t)\cdot\nabla(I+6\Delta^{-1})\omega_{\geq 3}(t),Y_{2}^{m})_{L^{2}(S^{2})}=\sum_{k=-2}^{2}M_{m,k}(t)\omega_{2}^{k}(t),\\ M_{m,k}(t)=\frac{1}{6}\int_{S^{2}}(I+6\Delta^{-1})\omega_{\geq 3}(t)\Bigl{\{}(\mathbf{n}_{S^{2}}\times\nabla Y_{2}^{k})\cdot\nabla\overline{Y_{2}^{m}}\Bigr{\}}\,d\mathcal{H}^{2}\end{gathered}

by using (3.6) and ω=2(t)=m=22ω2m(t)Y2m\omega_{=2}(t)=\sum_{m=-2}^{2}\omega_{2}^{m}(t)Y_{2}^{m}. For the inner product of f3(t)f_{\geq 3}(t) given by (3.8) with Y2mY_{2}^{m}, we see that

(Δω3(t)+2ω3(t),Y2m)L2(S2)\displaystyle(\Delta\omega_{\geq 3}(t)+2\omega_{\geq 3}(t),Y_{2}^{m})_{L^{2}(S^{2})} =0,\displaystyle=0,
(cosθφ(I+6Δ1)ω3(t),Y2m)L2(S2)\displaystyle(\cos\theta\,\partial_{\varphi}(I+6\Delta^{-1})\omega_{\geq 3}(t),Y_{2}^{m})_{L^{2}(S^{2})} =12ima3mω3m(t)\displaystyle=\frac{1}{2}ima_{3}^{m}\omega_{3}^{m}(t)

by ω3(t)=n3m=nnωnm(t)Ynm\omega_{\geq 3}(t)=\sum_{n\geq 3}\sum_{m=-n}^{n}\omega_{n}^{m}(t)Y_{n}^{m}, (2.9), and λ3=12\lambda_{3}=12. Also,

(𝐮3(t)ω3(t),Y2m)L2(S2)\displaystyle(\mathbf{u}_{\geq 3}(t)\cdot\nabla\omega_{\geq 3}(t),Y_{2}^{m})_{L^{2}(S^{2})} =S2{𝐮3(t)ω3(t)}Y2m¯𝑑2\displaystyle=\int_{S^{2}}\{\mathbf{u}_{\geq 3}(t)\cdot\nabla\omega_{\geq 3}(t)\}\overline{Y_{2}^{m}}\,d\mathcal{H}^{2}
=S2ω3(t)(𝐮3(t)Y2m¯)𝑑2\displaystyle=-\int_{S^{2}}\omega_{\geq 3}(t)\Bigl{(}\mathbf{u}_{\geq 3}(t)\cdot\nabla\overline{Y_{2}^{m}}\Bigr{)}\,d\mathcal{H}^{2}

by integration by parts and div𝐮3(t)=0\mathrm{div}\,\mathbf{u}_{\geq 3}(t)=0. We also have

(X(I+2Δ1)ω3(t),Y2m)L2(S2)\displaystyle(X\cdot\nabla(I+2\Delta^{-1})\omega_{\geq 3}(t),Y_{2}^{m})_{L^{2}(S^{2})} =S2{X(I+2Δ1)ω3(t)}Y2m¯𝑑2\displaystyle=\int_{S^{2}}\{X\cdot\nabla(I+2\Delta^{-1})\omega_{\geq 3}(t)\}\overline{Y_{2}^{m}}\,d\mathcal{H^{2}}
=S2(I+2Δ1)ω3(t)(XY2m¯)𝑑2\displaystyle=-\int_{S^{2}}(I+2\Delta^{-1})\omega_{\geq 3}(t)\Bigl{(}X\cdot\nabla\overline{Y_{2}^{m}}\Bigr{)}\,d\mathcal{H}^{2}
=((I+2Δ1)ω3(t),XY2m)L2(S2)\displaystyle=-\bigl{(}(I+2\Delta^{-1})\omega_{\geq 3}(t),X\cdot\nabla Y_{2}^{m}\bigr{)}_{L^{2}(S^{2})}

by integration by parts, divX=0\mathrm{div}\,X=0, and XY2m¯=XY2m¯X\cdot\nabla\overline{Y_{2}^{m}}=\overline{X\cdot\nabla Y_{2}^{m}} since XX is real. Then since XX is of the form (3.4), we see by (2.14) and (2.15) that

XY2mspan{Y20,Y2±1,Y2±2},|m|=0,1,2.\displaystyle X\cdot\nabla Y_{2}^{m}\in\mathrm{span}\{Y_{2}^{0},Y_{2}^{\pm 1},Y_{2}^{\pm 2}\},\quad|m|=0,1,2.

By this fact, ω3(t)=n3m=nnωnm(t)Ynm\omega_{\geq 3}(t)=\sum_{n\geq 3}\sum_{m=-n}^{n}\omega_{n}^{m}(t)Y_{n}^{m}, and (2.8), we get

(X(I+2Δ1)ω3(t),Y2m)L2(S2)=((I+2Δ1)ω3(t),XY2m)L2(S2)=0.\displaystyle(X\cdot\nabla(I+2\Delta^{-1})\omega_{\geq 3}(t),Y_{2}^{m})_{L^{2}(S^{2})}=-\bigl{(}(I+2\Delta^{-1})\omega_{\geq 3}(t),X\cdot\nabla Y_{2}^{m}\bigr{)}_{L^{2}(S^{2})}=0.

Thus, noting that f3(t)f_{\geq 3}(t) is given by (3.8), we see by the above equalities that

(3.13) (f3(t),Y2m)L2(S2)=a85πima3mω3m(t)+S2ω3(t)(𝐮3(t)Y2m¯)𝑑2.\displaystyle(f_{\geq 3}(t),Y_{2}^{m})_{L^{2}(S^{2})}=-\frac{a}{8}\sqrt{\frac{5}{\pi}}\,ima_{3}^{m}\omega_{3}^{m}(t)+\int_{S^{2}}\omega_{\geq 3}(t)\Bigl{(}\mathbf{u}_{\geq 3}(t)\cdot\nabla\overline{Y_{2}^{m}}\Bigr{)}\,d\mathcal{H}^{2}.

To compute the inner product of Xω=2(t)X\cdot\nabla\omega_{=2}(t) with Y2mY_{2}^{m}, we see that we can use (2.14) and (2.15) since XX is of the form (3.4). Hence

Xω=2(t)=m=2mω2m(t)(XY2m)=m=22η2m(t)Y2m,\displaystyle X\cdot\nabla\omega_{=2}(t)=\sum_{m=-2}^{m}\omega_{2}^{m}(t)(X\cdot\nabla Y_{2}^{m})=\sum_{m=-2}^{2}\eta_{2}^{m}(t)Y_{2}^{m},

where η2m(t)=(Xω=2(t),Y2m)L2(S2)\eta_{2}^{m}(t)=(X\cdot\nabla\omega_{=2}(t),Y_{2}^{m})_{L^{2}(S^{2})} is given by

η22(t)\displaystyle\eta_{2}^{2}(t) =2ia3ω22(t)+(ia1+a2)ω21(t),\displaystyle=2ia_{3}\omega_{2}^{2}(t)+(ia_{1}+a_{2})\omega_{2}^{1}(t),
η21(t)\displaystyle\eta_{2}^{1}(t) =(ia1a2)ω22(t)+ia3ω21(t)+62(ia1+a2)ω20(t),\displaystyle=(ia_{1}-a_{2})\omega_{2}^{2}(t)+ia_{3}\omega_{2}^{1}(t)+\frac{\sqrt{6}}{2}(ia_{1}+a_{2})\omega_{2}^{0}(t),
η20(t)\displaystyle\eta_{2}^{0}(t) =62(ia1a2)ω21(t)+62(ia1+a2)ω21(t),\displaystyle=\frac{\sqrt{6}}{2}(ia_{1}-a_{2})\omega_{2}^{1}(t)+\frac{\sqrt{6}}{2}(ia_{1}+a_{2})\omega_{2}^{-1}(t),
η21(t)\displaystyle\eta_{2}^{-1}(t) =62(ia1a2)ω20(t)ia3ω21(t)+(ia1+a2)ω22(t),\displaystyle=\frac{\sqrt{6}}{2}(ia_{1}-a_{2})\omega_{2}^{0}(t)-ia_{3}\omega_{2}^{-1}(t)+(ia_{1}+a_{2})\omega_{2}^{-2}(t),
η22(t)\displaystyle\eta_{2}^{-2}(t) =(ia1a2)ω21(t)2ia3ω22(t).\displaystyle=(ia_{1}-a_{2})\omega_{2}^{-1}(t)-2ia_{3}\omega_{2}^{-2}(t).

Moreover, since

ia1+a2=i232πω0,11,ia1a2=i232πω0,11¯\displaystyle ia_{1}+a_{2}=-\frac{i}{2}\sqrt{\frac{3}{2\pi}}\,\omega_{0,1}^{1},\quad ia_{1}-a_{2}=-\frac{i}{2}\sqrt{\frac{3}{2\pi}}\,\overline{\omega_{0,1}^{1}}

by (3.5), we have

η22(t)\displaystyle\eta_{2}^{2}(t) =i23πω0,10ω22(t)i232πω0,11ω21(t),\displaystyle=\frac{i}{2}\sqrt{\frac{3}{\pi}}\,\omega_{0,1}^{0}\,\omega_{2}^{2}(t)-\frac{i}{2}\sqrt{\frac{3}{2\pi}}\,\omega_{0,1}^{1}\,\omega_{2}^{1}(t),
η21(t)\displaystyle\eta_{2}^{1}(t) =i232πω0,11¯ω22(t)+i43πω0,10ω21(t)3i4πω0,11ω20(t),\displaystyle=-\frac{i}{2}\sqrt{\frac{3}{2\pi}}\,\overline{\omega_{0,1}^{1}}\,\omega_{2}^{2}(t)+\frac{i}{4}\sqrt{\frac{3}{\pi}}\,\omega_{0,1}^{0}\,\omega_{2}^{1}(t)-\frac{3i}{4\sqrt{\pi}}\,\omega_{0,1}^{1}\,\omega_{2}^{0}(t),
η20(t)\displaystyle\eta_{2}^{0}(t) =3i4πω0,11¯ω21(t)3i4πω0,11ω21(t),\displaystyle=-\frac{3i}{4\sqrt{\pi}}\,\overline{\omega_{0,1}^{1}}\,\omega_{2}^{1}(t)-\frac{3i}{4\sqrt{\pi}}\,\omega_{0,1}^{1}\,\omega_{2}^{-1}(t),
η21(t)\displaystyle\eta_{2}^{-1}(t) =3i4πω0,11¯ω20(t)i43πω0,10ω21(t)i232πω0,11ω22(t),\displaystyle=-\frac{3i}{4\sqrt{\pi}}\,\overline{\omega_{0,1}^{1}}\,\omega_{2}^{0}(t)-\frac{i}{4}\sqrt{\frac{3}{\pi}}\,\omega_{0,1}^{0}\,\omega_{2}^{-1}(t)-\frac{i}{2}\sqrt{\frac{3}{2\pi}}\,\omega_{0,1}^{1}\,\omega_{2}^{-2}(t),
η22(t)\displaystyle\eta_{2}^{-2}(t) =i232πω0,11¯ω21(t)i23πω0,10ω22(t).\displaystyle=-\frac{i}{2}\sqrt{\frac{3}{2\pi}}\,\overline{\omega_{0,1}^{1}}\,\omega_{2}^{-1}(t)-\frac{i}{2}\sqrt{\frac{3}{\pi}}\,\omega_{0,1}^{0}\,\omega_{2}^{-2}(t).

Hence we can write

(3.14) 23(Xω=2(t),Y2m)L2(S2)=23η2m(t)=k=22iAm,kω2k(t),\displaystyle\frac{2}{3}(X\cdot\nabla\omega_{=2}(t),Y_{2}^{m})_{L^{2}(S^{2})}=\frac{2}{3}\eta_{2}^{m}(t)=\sum_{k=-2}^{2}iA_{m,k}\omega_{2}^{k}(t),

where we set α=ω0,11/26π\alpha=\omega_{0,1}^{1}/2\sqrt{6\pi}\in\mathbb{C} and b=ω0,10/23πb=\omega_{0,1}^{0}/2\sqrt{3\pi}\in\mathbb{R} as in (1.9) and define

(3.15) 𝑨=(A2,2A2,2A2,2A2,2)=(2b2α0002α¯b6α0006α¯06α0006α¯b2α0002α¯2b).\displaystyle\bm{A}=\begin{pmatrix}A_{2,2}&\cdots&A_{2,-2}\\ \vdots&\ddots&\vdots\\ A_{-2,2}&\cdots&A_{-2,-2}\end{pmatrix}=\begin{pmatrix}2b&-2\alpha&0&0&0\\ -2\bar{\alpha}&b&-\sqrt{6}\,\alpha&0&0\\ 0&-\sqrt{6}\,\bar{\alpha}&0&-\sqrt{6}\,\alpha&0\\ 0&0&-\sqrt{6}\,\bar{\alpha}&-b&-2\alpha\\ 0&0&0&-2\bar{\alpha}&-2b\end{pmatrix}.

Therefore, we deduce from (3.9)–(3.14) and ω=2(0)=ω0,=2\omega_{=2}(0)=\omega_{0,=2} that

{ddtω2m(t)=4νω2m(t)k=22{iAm,k+Mm,k(t)}ω2k(t)+ia2πω0,11δ1,mia2πω0,11δ1,m+(f3(t),Y2m)L2(S2),t>0,ω2m(0)=ω0,2m\displaystyle\left\{\begin{aligned} \frac{d}{dt}\omega_{2}^{m}(t)&=-4\nu\omega_{2}^{m}(t)-\sum_{k=-2}^{2}\{iA_{m,k}+M_{m,k}(t)\}\omega_{2}^{k}(t)\\ &\qquad+\frac{ia}{2\sqrt{\pi}}\,\omega_{0,1}^{1}\delta_{1,m}-\frac{ia}{2\sqrt{\pi}}\,\omega_{0,1}^{-1}\delta_{-1,m}+(f_{\geq 3}(t),Y_{2}^{m})_{L^{2}(S^{2})},\quad t>0,\\ \omega_{2}^{m}(0)&=\omega_{0,2}^{m}\end{aligned}\right.

for |m|=0,1,2|m|=0,1,2. We write this system as

(3.16) {ddt𝝎(t)={4ν𝑰5+i𝑨+𝑴(t)}𝝎(t)+𝒇3(t)+𝒄,t>0,𝝎(0)=𝝎0,\displaystyle\left\{\begin{aligned} \frac{d}{dt}\bm{\omega}(t)&=-\{4\nu\bm{I}_{5}+i\bm{A}+\bm{M}(t)\}\bm{\omega}(t)+\bm{f}_{\geq 3}(t)+\bm{c},\quad t>0,\\ \bm{\omega}(0)&=\bm{\omega}_{0},\end{aligned}\right.

where 𝑰5\bm{I}_{5} is the 5×55\times 5 identity matrix and

(3.17) 𝝎(t)=(ω22(t)ω22(t)),𝝎0=(ω0,22ω0,22),𝑴(t)=(M2,2(t)M2,2(t)M2,2(t)M2,2(t)),𝒇3(t)=((f3(t),Y22)L2(S2)(f3(t),Y22)L2(S2)),𝒄=ia2π(0ω0,110ω0,110)=6ia(0α0α¯0).\displaystyle\begin{aligned} \bm{\omega}(t)&=\begin{pmatrix}\omega_{2}^{2}(t)\\ \vdots\\ \omega_{2}^{-2}(t)\end{pmatrix},\quad\bm{\omega}_{0}=\begin{pmatrix}\omega_{0,2}^{2}\\ \vdots\\ \omega_{0,2}^{-2}\end{pmatrix},\quad\bm{M}(t)=\begin{pmatrix}M_{2,2}(t)&\cdots&M_{2,-2}(t)\\ \vdots&\ddots&\vdots\\ M_{-2,2}(t)&\cdots&M_{-2,-2}(t)\end{pmatrix},\\ \bm{f}_{\geq 3}(t)&=\begin{pmatrix}(f_{\geq 3}(t),Y_{2}^{2})_{L^{2}(S^{2})}\\ \vdots\\ (f_{\geq 3}(t),Y_{2}^{-2})_{L^{2}(S^{2})}\end{pmatrix},\quad\bm{c}=\frac{ia}{2\sqrt{\pi}}\begin{pmatrix}0\\ \omega_{0,1}^{1}\\ 0\\ -\omega_{0,1}^{-1}\\ 0\end{pmatrix}=\sqrt{6}\,ia\begin{pmatrix}0\\ \alpha\\ 0\\ \bar{\alpha}\\ 0\end{pmatrix}.\end{aligned}

In the last equality we also used (3.1) and α=ω0,11/26π\alpha=\omega_{0,1}^{1}/2\sqrt{6\pi}.

Step 5: asymptotic behavior of ω=2(t)\omega_{=2}(t). Since the matrix 𝑨\bm{A} given by (3.15) is self-adjoint, all eigenvalues of 𝑨\bm{A} are real. Then all eigenvalues of 4ν𝑰5+i𝑨4\nu\bm{I}_{5}+i\bm{A} have the real part 4ν04\nu\neq 0 and thus they are nonzero. Hence 4ν𝑰5+i𝑨4\nu\bm{I}_{5}+i\bm{A} is invertible and we can define

(3.18) 𝝎=(ω2,2,,ω2,2)T=(4ν𝑰5+i𝑨)1𝒄.\displaystyle\bm{\omega}_{\infty}=(\omega_{2,\infty}^{2},\dots,\omega_{2,\infty}^{-2})^{T}=(4\nu\bm{I}_{5}+i\bm{A})^{-1}\bm{c}.

Let 𝝃(t)=𝝎(t)𝝎\bm{\xi}(t)=\bm{\omega}(t)-\bm{\omega}_{\infty}. Then we see by (3.16) and (3.18) that

(3.19) {ddt𝝃(t)={4ν𝑰5+i𝑨+𝑴(t)}𝝃(t)+𝒇3(t)𝑴(t)𝝎,t>0,𝝃(0)=𝝎0𝝎.\displaystyle\left\{\begin{aligned} \frac{d}{dt}\bm{\xi}(t)&=-\{4\nu\bm{I}_{5}+i\bm{A}+\bm{M}(t)\}\bm{\xi}(t)+\bm{f}_{\geq 3}(t)-\bm{M}(t)\bm{\omega}_{\infty},\quad t>0,\\ \bm{\xi}(0)&=\bm{\omega}_{0}-\bm{\omega}_{\infty}.\end{aligned}\right.

Let us estimate 𝝃(t)\bm{\xi}(t). In what follows, we write CC for a general positive constant independent of tt, ν\nu, aa, and the initial data ω0L02(S2)\omega_{0}\in L_{0}^{2}(S^{2}) for (1.8). Also, we denote by ,\langle\!\langle\cdot,\cdot\rangle\!\rangle and \|\cdot\| the inner product and norm of 5\mathbb{C}^{5}, and use the same notation \|\cdot\| for the Frobenius norm of a matrix. First we estimate 𝑴(t)\bm{M}(t) and 𝒇3(t)\bm{f}_{\geq 3}(t) given by (3.12), (3.13), and (3.17). Since 𝒏S2\bm{n}_{S^{2}} and Y2mY_{2}^{m} are smooth on S2S^{2} and a3ma_{3}^{m} is a constant given by (2.7), we have

𝑴(t)\displaystyle\|\bm{M}(t)\| C(I+6Δ1)ω3(t)L2(S2),\displaystyle\leq C\|(I+6\Delta^{-1})\omega_{\geq 3}(t)\|_{L^{2}(S^{2})},
𝒇3(t)\displaystyle\|\bm{f}_{\geq 3}(t)\| C(|a||ω3m(t)|+ω3(t)L2(S2)𝒖3(t)L2(S2))\displaystyle\leq C\Bigl{(}|a||\omega_{3}^{m}(t)|+\|\omega_{\geq 3}(t)\|_{L^{2}(S^{2})}\|\bm{u}_{\geq 3}(t)\|_{L^{2}(S^{2})}\Bigr{)}

by Hölder’s inequality. Moreover, noting that

ω3(t)=n3m=nnωnm(t)Yn2,𝐮3(t)=𝐧S2×Δ1ω3(t),\displaystyle\omega_{\geq 3}(t)=\sum_{n\geq 3}\sum_{m=-n}^{n}\omega_{n}^{m}(t)Y_{n}^{2},\quad\mathbf{u}_{\geq 3}(t)=\mathbf{n}_{S^{2}}\times\nabla\Delta^{-1}\omega_{\geq 3}(t),

we use (2.8), (2.10), and λn12\lambda_{n}\geq 12 for n3n\geq 3, and then apply (1.13) to get

(3.20) 𝑴(t)Cω3(t)L2(S2)Ce10νtω0,3L2(S2),𝒇3(t)C(|a|ω3(t)L2(S2)+ω3(t)L2(S2)2)Ce10νt(|a|+ω0,3L2(S2))ω0,3L2(S2).\displaystyle\begin{aligned} \|\bm{M}(t)\|&\leq C\|\omega_{\geq 3}(t)\|_{L^{2}(S^{2})}\leq Ce^{-10\nu t}\|\omega_{0,\geq 3}\|_{L^{2}(S^{2})},\\ \|\bm{f}_{\geq 3}(t)\|&\leq C\Bigl{(}|a|\|\omega_{\geq 3}(t)\|_{L^{2}(S^{2})}+\|\omega_{\geq 3}(t)\|_{L^{2}(S^{2})}^{2}\Bigr{)}\\ &\leq Ce^{-10\nu t}\Bigl{(}|a|+\|\omega_{0,\geq 3}\|_{L^{2}(S^{2})}\Bigr{)}\|\omega_{0,\geq 3}\|_{L^{2}(S^{2})}.\end{aligned}

In the last inequality we also used e10νt1e^{-10\nu t}\leq 1. Now we take the real part of the inner product of (3.19) with 𝝃(t)\bm{\xi}(t). Then, noting that 𝑨𝝃(t),𝝃(t)\langle\!\langle\bm{A}\bm{\xi}(t),\bm{\xi}(t)\rangle\!\rangle is real since 𝑨\bm{A} is self-adjoint, we observe by Cauchy–Schwarz’s and Young’s inequalities that

12ddt𝝃(t)2\displaystyle\frac{1}{2}\frac{d}{dt}\|\bm{\xi}(t)\|^{2} =4ν𝝃(t)2Re𝑴(t)𝝃(t),𝝃(t)\displaystyle=-4\nu\|\bm{\xi}(t)\|^{2}-\mathrm{Re}\langle\!\langle\bm{M}(t)\bm{\xi}(t),\bm{\xi}(t)\rangle\!\rangle
+Re𝒇3(t),𝝃(t)Re𝑴(t)𝝎,𝝃(t)\displaystyle\qquad+\mathrm{Re}\langle\!\langle\bm{f}_{\geq 3}(t),\bm{\xi}(t)\rangle\!\rangle-\mathrm{Re}\langle\!\langle\bm{M}(t)\bm{\omega}_{\infty},\bm{\xi}(t)\rangle\!\rangle
2ν𝝃(t)2+𝑴(t)𝝃(t)2+14ν(𝒇3(t)2+𝑴(t)2𝝎2).\displaystyle\leq-2\nu\|\bm{\xi}(t)\|^{2}+\|\bm{M}(t)\|\|\bm{\xi}(t)\|^{2}+\frac{1}{4\nu}\Bigl{(}\|\bm{f}_{\geq 3}(t)\|^{2}+\|\bm{M}(t)\|^{2}\|\bm{\omega}_{\infty}\|^{2}\Bigr{)}.

Applying (3.20) to this inequality, we get

ddt𝝃(t)2(4ν+Ce10νtω0,3L2(S2))𝝃(t)2+Cσνe20νt,\displaystyle\frac{d}{dt}\|\bm{\xi}(t)\|^{2}\leq\Bigl{(}-4\nu+Ce^{-10\nu t}\|\omega_{0,\geq 3}\|_{L^{2}(S^{2})}\Bigr{)}\|\bm{\xi}(t)\|^{2}+\frac{C\sigma}{\nu}e^{-20\nu t},

where

(3.21) σ=(|a|2+ω0,3L2(S2)2+𝝎2)ω0,3L2(S2)2.\displaystyle\sigma=\Bigl{(}|a|^{2}+\|\omega_{0,\geq 3}\|_{L^{2}(S^{2})}^{2}+\|\bm{\omega}_{\infty}\|^{2}\Bigr{)}\|\omega_{0,\geq 3}\|_{L^{2}(S^{2})}^{2}.

Hence, setting

F(t)=0t(4ν+Ce10ντω0,3L2(S2))𝑑τ=4νt+C10ν(1e10νt)ω0,3L2(S2),\displaystyle F(t)=\int_{0}^{t}\Bigl{(}-4\nu+Ce^{-10\nu\tau}\|\omega_{0,\geq 3}\|_{L^{2}(S^{2})}\Bigr{)}\,d\tau=-4\nu t+\frac{C}{10\nu}(1-e^{-10\nu t})\|\omega_{0,\geq 3}\|_{L^{2}(S^{2})},

we observe by the above inequality and F(t)4νtF(t)\geq-4\nu t that

ddt(eF(t)𝝃(t)2)Cσνe20νtF(t)Cσνe16νt\displaystyle\frac{d}{dt}\Bigl{(}e^{-F(t)}\|\bm{\xi}(t)\|^{2}\Bigr{)}\leq\frac{C\sigma}{\nu}e^{-20\nu t-F(t)}\leq\frac{C\sigma}{\nu}e^{-16\nu t}

and thus

eF(t)𝝃(t)2𝝃(0)2+Cσ16ν2(1e16νt)𝝃(0)2+Cσν2.\displaystyle e^{-F(t)}\|\bm{\xi}(t)\|^{2}\leq\|\bm{\xi}(0)\|^{2}+\frac{C\sigma}{16\nu^{2}}(1-e^{-16\nu t})\leq\|\bm{\xi}(0)\|^{2}+\frac{C\sigma}{\nu^{2}}.

Therefore, noting that

𝝃(t)=𝝎(t)𝝎,𝝃(0)=𝝎0𝝎,F(t)4νt+Cνω0,3L2(S2),\displaystyle\bm{\xi}(t)=\bm{\omega}(t)-\bm{\omega}_{\infty},\quad\bm{\xi}(0)=\bm{\omega}_{0}-\bm{\omega}_{\infty},\quad F(t)\leq-4\nu t+\frac{C}{\nu}\|\omega_{0,\geq 3}\|_{L^{2}(S^{2})},

we obtain

(3.22) 𝝎(t)𝝎2(𝝎0𝝎2+Cσν2)exp(4νt+Cνω0,3L2(S2)).\displaystyle\|\bm{\omega}(t)-\bm{\omega}_{\infty}\|^{2}\leq\left(\|\bm{\omega}_{0}-\bm{\omega}_{\infty}\|^{2}+\frac{C\sigma}{\nu^{2}}\right)\exp\left(-4\nu t+\frac{C}{\nu}\|\omega_{0,\geq 3}\|_{L^{2}(S^{2})}\right).

Now we define ω=n=22ω2,mY2mL02(S2)\omega_{\infty}=\sum_{n=-2}^{2}\omega_{2,\infty}^{m}Y_{2}^{m}\in L_{0}^{2}(S^{2}) with ω2,m\omega_{2,\infty}^{m} given by (3.18). Then, noting that 𝝎(t)\bm{\omega}(t) and 𝝎0\bm{\omega}_{0} are given by (3.17) and

𝒘=wL2(S2),𝒘=(w22,,w22)T5,w=m=22w2mY2m,\displaystyle\|\bm{w}\|=\|w\|_{L^{2}(S^{2})},\quad\bm{w}=(w_{2}^{2},\dots,w_{2}^{-2})^{T}\in\mathbb{C}^{5},\quad w=\sum_{m=-2}^{2}w_{2}^{m}Y_{2}^{m},

we take the square root of (3.22) and use (3.21) to deduce that

ω=2(t)ωL2(S2)σ1e2νt(ω0,=2ωL2(S2)+σ2ν)\displaystyle\|\omega_{=2}(t)-\omega_{\infty}\|_{L^{2}(S^{2})}\leq\sigma_{1}e^{-2\nu t}\left(\|\omega_{0,=2}-\omega_{\infty}\|_{L^{2}(S^{2})}+\frac{\sigma_{2}}{\nu}\right)

for all t0t\geq 0, where σ1\sigma_{1} and σ2\sigma_{2} are given by (1.15) with constants C1,C2>0C_{1},C_{2}>0 independent of tt, ν\nu, aa, and ω0\omega_{0}. Hence we conclude that the estimate (1.14) for ω=2(t)\omega_{=2}(t) is valid.

Step 6: expression of ω2,m\omega_{2,\infty}^{m}, |m|=0,1,2|m|=0,1,2. First note that

(3.23) ω2,m=limtω2m(t)=limt(1)mω2m(t)¯=(1)mω2,m¯,|m|=0,1,2\displaystyle\omega_{2,\infty}^{-m}=\lim_{t\to\infty}\omega_{2}^{-m}(t)=\lim_{t\to\infty}(-1)^{m}\overline{\omega_{2}^{m}(t)}=(-1)^{m}\overline{\omega_{2,\infty}^{m}},\quad|m|=0,1,2

by (3.1) and (3.22). Since (4ν𝑰5+i𝑨)𝝎=𝒄(4\nu\bm{I}_{5}+i\bm{A})\bm{\omega}_{\infty}=\bm{c} by (3.18), we have

(3.24) (4ν+2ib)ω2,22iαω2,1\displaystyle(4\nu+2ib)\omega_{2,\infty}^{2}-2i\alpha\omega_{2,\infty}^{1} =0,\displaystyle=0,
(3.25) 2iα¯ω2,2+(4ν+ib)ω2,16iαω2,0\displaystyle-2i\bar{\alpha}\omega_{2,\infty}^{2}+(4\nu+ib)\omega_{2,\infty}^{1}-\sqrt{6}\,i\alpha\omega_{2,\infty}^{0} =6iaα,\displaystyle=\sqrt{6}\,ia\alpha,
(3.26) 6iα¯ω2,1+4νω2,06iαω2,1\displaystyle-\sqrt{6}\,i\bar{\alpha}\omega_{2,\infty}^{1}+4\nu\omega_{2,\infty}^{0}-\sqrt{6}\,i\alpha\omega_{2,\infty}^{-1} =0\displaystyle=0

by (3.15) and (3.17). Let zk=4ν+ikbz_{k}=4\nu+ikb for k=1,2k=1,2. Then

(3.27) ω2,2=2iαz2ω2,1,ω2,1=6iαz24|α|2+z1z2(ω2,0+a)\displaystyle\omega_{2,\infty}^{2}=\frac{2i\alpha}{z_{2}}\,\omega_{2,\infty}^{1},\quad\omega_{2,\infty}^{1}=\frac{\sqrt{6}\,i\alpha z_{2}}{4|\alpha|^{2}+z_{1}z_{2}}(\omega_{2,\infty}^{0}+a)

by (3.24) and (3.25). Noting that ω2,1=ω2,1¯\omega_{2,\infty}^{-1}=-\overline{\omega_{2,\infty}^{1}} and ω2,0\omega_{2,\infty}^{0}\in\mathbb{R} by (3.23) and that a,ba,b\in\mathbb{R}, we multiply (3.26) by (4|α|2+z1z2)(4|α|2+z1z2¯)(4|\alpha|^{2}+z_{1}z_{2})(4|\alpha|^{2}+\overline{z_{1}z_{2}}) and use (3.27) to find that

{6|α|2z2(4|α|2+z1z2¯)+4ν(4|α|2+z1z2)(4|α|2+z1z2¯)+6|α|2z2¯(4|α|2+z1z2)}ω2,0=6a|α|2{z2(4|α|2+z1z2¯)+z2¯(4|α|2+z1z2)}.\{6|\alpha|^{2}z_{2}(4|\alpha|^{2}+\overline{z_{1}z_{2}})+4\nu(4|\alpha|^{2}+z_{1}z_{2})(4|\alpha|^{2}+\overline{z_{1}z_{2}})+6|\alpha|^{2}\overline{z_{2}}(4|\alpha|^{2}+z_{1}z_{2})\}\omega_{2,\infty}^{0}\\ =-6a|\alpha|^{2}\{z_{2}(4|\alpha|^{2}+\overline{z_{1}z_{2}})+\overline{z_{2}}(4|\alpha|^{2}+z_{1}z_{2})\}.

Moreover, by zk=4ν+ikbz_{k}=4\nu+ikb for k=1,2k=1,2 and direct calculations, we have

6|α|2z2(4|α|2+z1z2¯)+4ν(4|α|2+z1z2)(4|α|2+z1z2¯)+6|α|2z2¯(4|α|2+z1z2)\displaystyle 6|\alpha|^{2}z_{2}(4|\alpha|^{2}+\overline{z_{1}z_{2}})+4\nu(4|\alpha|^{2}+z_{1}z_{2})(4|\alpha|^{2}+\overline{z_{1}z_{2}})+6|\alpha|^{2}\overline{z_{2}}(4|\alpha|^{2}+z_{1}z_{2})
=16ν{16|α|4+8|α|2(10ν2+b2)+64ν4+20ν2b2+b4}\displaystyle\qquad=16\nu\{16|\alpha|^{4}+8|\alpha|^{2}(10\nu^{2}+b^{2})+64\nu^{4}+20\nu^{2}b^{2}+b^{4}\}
=16ν(4|α|2+4ν2+b2)(4|α|2+16ν2+b2)\displaystyle\qquad=16\nu(4|\alpha|^{2}+4\nu^{2}+b^{2})(4|\alpha|^{2}+16\nu^{2}+b^{2})

and

z2(4|α|2+z1z2¯)+z2¯(4|α|2+z1z2)=32ν(|α|2+4ν2+b2).\displaystyle z_{2}(4|\alpha|^{2}+\overline{z_{1}z_{2}})+\overline{z_{2}}(4|\alpha|^{2}+z_{1}z_{2})=32\nu(|\alpha|^{2}+4\nu^{2}+b^{2}).

Hence

(3.28) ω2,0=12a|α|2(4ν2+|α|2+b2)(4ν2+4|α|2+b2)(16ν2+4|α|2+b2)\displaystyle\omega_{2,\infty}^{0}=-12a\cdot\frac{|\alpha|^{2}(4\nu^{2}+|\alpha|^{2}+b^{2})}{(4\nu^{2}+4|\alpha|^{2}+b^{2})(16\nu^{2}+4|\alpha|^{2}+b^{2})}

and we conclude by (3.23), (3.27), (3.28), and zk=4ν+ikbz_{k}=4\nu+ikb for k=1,2k=1,2 that ω=m=22ω2,mY2m\omega_{\infty}=\sum_{m=-2}^{2}\omega_{2,\infty}^{m}Y_{2}^{m} is given by (1.9)–(1.10). The proof of Theorem 1.1 is complete.

4. Proof of Lemma 2.4

Let us give the proof of (2.14) in Lemma 2.4. First we recall that

(4.1) Y20(θ)=C0(3cos2θ1),C0=145π,Y2±1(θ,φ)=C1sinθcosθe±iφ,C1=12152π,Y2±2(θ,φ)=C2sin2θe±2iφ,C2=14152π.\displaystyle\begin{aligned} Y_{2}^{0}(\theta)&=C_{0}(3\cos^{2}\theta-1),&\quad C_{0}&=\frac{1}{4}\sqrt{\frac{5}{\pi}},\\ Y_{2}^{\pm 1}(\theta,\varphi)&=\mp C_{1}\sin\theta\cos\theta\,e^{\pm i\varphi},&\quad C_{1}&=\frac{1}{2}\sqrt{\frac{15}{2\pi}},\\ Y_{2}^{\pm 2}(\theta,\varphi)&=C_{2}\sin^{2}\theta\,e^{\pm 2i\varphi},&\quad C_{2}&=\frac{1}{4}\sqrt{\frac{15}{2\pi}}.\end{aligned}

We use the spherical coordinates x(θ,φ)x(\theta,\varphi) of S2S^{2} given by (2.5) so that

(4.2) θx(θ,φ)=(cosθcosφcosθsinφsinθ),φx(θ,φ)=(sinθsinφsinθcosφ0).\displaystyle\partial_{\theta}x(\theta,\varphi)=\begin{pmatrix}\cos\theta\cos\varphi\\ \cos\theta\sin\varphi\\ -\sin\theta\end{pmatrix},\quad\partial_{\varphi}x(\theta,\varphi)=\begin{pmatrix}-\sin\theta\sin\varphi\\ \sin\theta\cos\varphi\\ 0\end{pmatrix}.

In what follows, we suppress the arguments θ\theta and φ\varphi. Then

𝐞1×x=(0cosθsinθsinφ),𝐞2×x=(cosθ0sinθcosφ),𝐞3×x=(sinθsinφsinθcosφ0)\displaystyle\mathbf{e}_{1}\times x=\begin{pmatrix}0\\ -\cos\theta\\ \sin\theta\sin\varphi\end{pmatrix},\quad\mathbf{e}_{2}\times x=\begin{pmatrix}\cos\theta\\ 0\\ -\sin\theta\cos\varphi\end{pmatrix},\quad\mathbf{e}_{3}\times x=\begin{pmatrix}-\sin\theta\sin\varphi\\ \sin\theta\cos\varphi\\ 0\end{pmatrix}

for the standard basis {𝐞1,𝐞2,𝐞3}\{\mathbf{e}_{1},\mathbf{e}_{2},\mathbf{e}_{3}\} of 3\mathbb{R}^{3}. Hence

(4.3) (𝐞1×x)θx=sinφ,(𝐞1×x)φx=sinθcosθcosφ,(𝐞2×x)θx=cosφ,(𝐞2×x)φx=sinθcosθsinφ,(𝐞3×x)θx=0,(𝐞3×x)φx=sin2θ.\displaystyle\begin{aligned} (\mathbf{e}_{1}\times x)\cdot\partial_{\theta}x&=-\sin\varphi,&\quad(\mathbf{e}_{1}\times x)\cdot\partial_{\varphi}x&=-\sin\theta\cos\theta\cos\varphi,\\ (\mathbf{e}_{2}\times x)\cdot\partial_{\theta}x&=\cos\varphi,&\quad(\mathbf{e}_{2}\times x)\cdot\partial_{\varphi}x&=-\sin\theta\cos\theta\sin\varphi,\\ (\mathbf{e}_{3}\times x)\cdot\partial_{\theta}x&=0,&\quad(\mathbf{e}_{3}\times x)\cdot\partial_{\varphi}x&=\sin^{2}\theta.\end{aligned}

Let us compute (𝐞k×x)Y20(\mathbf{e}_{k}\times x)\cdot\nabla Y_{2}^{0} wth k=1,2,3k=1,2,3. By (2.6), we have

Y20=6C0sinθcosθθx.\displaystyle\nabla Y_{2}^{0}=-6C_{0}\sin\theta\cos\theta\,\partial_{\theta}x.

Hence it follows from (4.1) and (4.3) that

(𝐞1×x)Y20\displaystyle(\mathbf{e}_{1}\times x)\cdot\nabla Y_{2}^{0} =6C0sinθcosθsinφ=6C0sinθcosθeiφeiφ2i\displaystyle=6C_{0}\sin\theta\cos\theta\sin\varphi=6C_{0}\sin\theta\cos\theta\cdot\frac{e^{i\varphi}-e^{-i\varphi}}{2i}
=3C0iC1(Y21Y21)=6i2(Y21+Y21).\displaystyle=\frac{3C_{0}}{iC_{1}}(-Y_{2}^{1}-Y_{2}^{-1})=\frac{\sqrt{6}\,i}{2}(Y_{2}^{1}+Y_{2}^{-1}).

Similarly, using (4.1), (4.3), and cosφ=(eiφ+eiφ)/2\cos\varphi=(e^{i\varphi}+e^{-i\varphi})/2, we obtain

(𝐞2×x)Y20=62(Y21Y21),(𝐞3×x)Y20=0.\displaystyle(\mathbf{e}_{2}\times x)\cdot\nabla Y_{2}^{0}=\frac{\sqrt{6}}{2}(Y_{2}^{1}-Y_{2}^{-1}),\quad(\mathbf{e}_{3}\times x)\cdot\nabla Y_{2}^{0}=0.

Next we consider (𝐞k×x)Y21(\mathbf{e}_{k}\times x)\cdot\nabla Y_{2}^{1}, k=1,2,3k=1,2,3. Since

Y21=C1eiφ{(cos2θsin2θ)θx+icosθsinθφx}\displaystyle\nabla Y_{2}^{1}=-C_{1}e^{i\varphi}\left\{(\cos^{2}\theta-\sin^{2}\theta)\partial_{\theta}x+\frac{i\cos\theta}{\sin\theta}\partial_{\varphi}x\right\}

by (2.6), we see by (4.3) that (see also our notation (2.4))

(𝐞1×x)Y21=C1eiφ{(cos2θsin2θ)sinφ+icos2θcosφ}.\displaystyle(\mathbf{e}_{1}\times x)\cdot\nabla Y_{2}^{1}=C_{1}e^{i\varphi}\{(\cos^{2}\theta-\sin^{2}\theta)\sin\varphi+i\cos^{2}\theta\cos\varphi\}.

Moreover, we use

eiφsinφ=i2(e2iφ1),eiφcosφ=12(e2iφ+1),sin2θ=1cos2θ\displaystyle e^{i\varphi}\sin\varphi=-\frac{i}{2}(e^{2i\varphi}-1),\quad e^{i\varphi}\cos\varphi=\frac{1}{2}(e^{2i\varphi}+1),\quad\sin^{2}\theta=1-\cos^{2}\theta

and then apply (4.1) to find that

(𝐞1×x)Y21\displaystyle(\mathbf{e}_{1}\times x)\cdot\nabla Y_{2}^{1} =iC12(sin2θe2iφ+3cos2θ1)\displaystyle=\frac{iC_{1}}{2}(\sin^{2}\theta\,e^{2i\varphi}+3\cos^{2}\theta-1)
=i2(C1C2Y22+C1C0Y20)=i(Y22+62Y20).\displaystyle=\frac{i}{2}\left(\frac{C_{1}}{C_{2}}Y_{2}^{2}+\frac{C_{1}}{C_{0}}Y_{2}^{0}\right)=i\left(Y_{2}^{2}+\frac{\sqrt{6}}{2}Y_{2}^{0}\right).

In the same way, we obtain

(𝐞2×x)Y21=Y2262Y20,(𝐞3×x)Y21=iY21.\displaystyle(\mathbf{e}_{2}\times x)\cdot\nabla Y_{2}^{1}=Y_{2}^{2}-\frac{\sqrt{6}}{2}Y_{2}^{0},\quad(\mathbf{e}_{3}\times x)\cdot\nabla Y_{2}^{1}=iY_{2}^{1}.

Now let us calculate (𝐞k×x)Y22(\mathbf{e}_{k}\times x)\cdot\nabla Y_{2}^{2}, k=1,2,3k=1,2,3. By (2.6), we have

Y22=2C2e2iφ(sinθcosθθx+iφx).\displaystyle\nabla Y_{2}^{2}=2C_{2}e^{2i\varphi}(\sin\theta\cos\theta\,\partial_{\theta}x+i\partial_{\varphi}x).

Hence we observe by (4.3) that

(𝐞1×x)Y22=2C2e2iφsinθcosθ(sinφ+icosφ).\displaystyle(\mathbf{e}_{1}\times x)\cdot\nabla Y_{2}^{2}=-2C_{2}e^{2i\varphi}\sin\theta\cos\theta(\sin\varphi+i\cos\varphi).

Moreover, since sinφ+icosφ=i(cosθisinθ)=ieiφ\sin\varphi+i\cos\varphi=i(\cos\theta-i\sin\theta)=ie^{-i\varphi}, we get

(𝐞1×x)Y22=2iC2sinθcosθeiφ=2iC2C1Y21=iY21\displaystyle(\mathbf{e}_{1}\times x)\cdot\nabla Y_{2}^{2}=-2iC_{2}\sin\theta\cos\theta\,e^{i\varphi}=2i\frac{C_{2}}{C_{1}}Y_{2}^{1}=iY_{2}^{1}

by (4.1). Calculating similarly, we find that

(𝐞2×x)Y22=Y21,(𝐞3×x)Y22=2iY22.\displaystyle(\mathbf{e}_{2}\times x)\cdot\nabla Y_{2}^{2}=-Y_{2}^{1},\quad(\mathbf{e}_{3}\times x)\cdot\nabla Y_{2}^{2}=2iY_{2}^{2}.

Applying the above results to 𝐚=(a1,a2,a3)T=k=13ak𝐞k\mathbf{a}=(a_{1},a_{2},a_{3})^{T}=\sum_{k=1}^{3}a_{k}\mathbf{e}_{k}, we obtain (2.14).

5. Appendix A: One-jet case

We briefly explain the behavior of a perturbation for the one-jet Kolmogorov type flow

ω1a(θ,φ)=aY10(θ)=a23πcosθ.\displaystyle\omega_{1}^{a}(\theta,\varphi)=aY_{1}^{0}(\theta)=\frac{a}{2}\sqrt{\frac{3}{\pi}}\cos\theta.

The nonlinear stability problem for the vorticity equation (1.3) around the one-jet Kolmogorov type flow is (see [31] for the derivation of the perturbation operator)

(5.1) {tω=ν(Δω+2ω)a43πφ(I+2Δ1)ω𝐮ωonS2×(0,),𝐮=𝐧S2×Δ1ωonS2×(0,),ω|t=0=ω0onS2.\displaystyle\left\{\begin{aligned} \partial_{t}\omega&=\nu(\Delta\omega+2\omega)-\frac{a}{4}\sqrt{\frac{3}{\pi}}\,\partial_{\varphi}(I+2\Delta^{-1})\omega-\mathbf{u}\cdot\nabla\omega\quad\text{on}\quad S^{2}\times(0,\infty),\\ \mathbf{u}&=\mathbf{n}_{S^{2}}\times\nabla\Delta^{-1}\omega\quad\text{on}\quad S^{2}\times(0,\infty),\\ \omega|_{t=0}&=\omega_{0}\quad\text{on}\quad S^{2}.\end{aligned}\right.

For a real-valued ω0L02(S2)\omega_{0}\in L_{0}^{2}(S^{2}), we can prove the global existence and uniqueness of a weak solution ω\omega to (5.1) in the class (1.11) by the Galerkin and energy methods. Also, taking the inner product of (5.1) with Y1mY_{1}^{m}, |m|=0,1|m|=0,1 and using ΔY1m=2Y1m-\Delta Y_{1}^{m}=2Y_{1}^{m} and (2.11), we can show ω=1(t)=ω0,=1\omega_{=1}(t)=\omega_{0,=1} for all t0t\geq 0 as in Step 1 of the proof of Theorem 1.1. Moreover, we take the inner product of (5.1) with (I+2Δ1)ω(t)(I+2\Delta^{-1})\omega(t). Then

tω(t),(I+2Δ1)ω(t)H01\displaystyle\langle\partial_{t}\omega(t),(I+2\Delta^{-1})\omega(t)\rangle_{H_{0}^{1}} 13ddtω2(t)L2(S2)2,\displaystyle\geq\frac{1}{3}\frac{d}{dt}\|\omega_{\geq 2}(t)\|_{L^{2}(S^{2})}^{2},
(Δω(t)+2ω(t),(I+2Δ1)ω(t))L2(S2)\displaystyle\bigl{(}\Delta\omega(t)+2\omega(t),(I+2\Delta^{-1})\omega(t)\bigr{)}_{L^{2}(S^{2})} 83ω2(t)L2(S2)2\displaystyle\leq-\frac{8}{3}\|\omega_{\geq 2}(t)\|_{L^{2}(S^{2})}^{2}

by λ1=2\lambda_{1}=2 and λn6\lambda_{n}\geq 6 for n2n\geq 2. Also, noting that (I+2Δ1)ω(t)(I+2\Delta^{-1})\omega(t) is real-valued, we have

(φ(I+2Δ1)ω(t),(I+2Δ1)ω(t))L2(S2)=0\displaystyle\bigl{(}\partial_{\varphi}(I+2\Delta^{-1})\omega(t),(I+2\Delta^{-1})\omega(t)\bigr{)}_{L^{2}(S^{2})}=0

by integration by parts and, since 𝐮(t)Δ1ω(t)=0\mathbf{u}(t)\cdot\nabla\Delta^{-1}\omega(t)=0 and div𝐮(t)=0\mathrm{div}\,\mathbf{u}(t)=0,

(𝐮(t)ω(t),(I+2Δ1)ω(t))L2(S2)=(𝐮(I+2Δ1)ω(t),(I+2Δ1)ω(t))L2(S2)=0.\displaystyle\bigl{(}\mathbf{u}(t)\cdot\nabla\omega(t),(I+2\Delta^{-1})\omega(t)\bigr{)}_{L^{2}(S^{2})}=\bigl{(}\mathbf{u}\cdot\nabla(I+2\Delta^{-1})\omega(t),(I+2\Delta^{-1})\omega(t)\bigr{)}_{L^{2}(S^{2})}=0.

Therefore, we obtain

13ddtω2(t)L2(S2)283ω2(t)L2(S2)2,\displaystyle\frac{1}{3}\frac{d}{dt}\|\omega_{\geq 2}(t)\|_{L^{2}(S^{2})}^{2}\leq-\frac{8}{3}\|\omega_{\geq 2}(t)\|_{L^{2}(S^{2})}^{2},

which implies that

ω2(t)L2(S2)2e8νtω2(0)L2(S2)2=e8νtω0,2L2(S2)2,\displaystyle\|\omega_{\geq 2}(t)\|_{L^{2}(S^{2})}^{2}\leq e^{-8\nu t}\|\omega_{\geq 2}(0)\|_{L^{2}(S^{2})}^{2}=e^{-8\nu t}\|\omega_{0,\geq 2}\|_{L^{2}(S^{2})}^{2},

i.e. ω2(t)L2(S2)e4νtω0,2L2(S2)\|\omega_{\geq 2}(t)\|_{L^{2}(S^{2})}\leq e^{-4\nu t}\|\omega_{0,\geq 2}\|_{L^{2}(S^{2})} for all t0t\geq 0.

6. Appendix B: Transformation into an equation with Coriolis force

In this section we see that solutions ω\omega to (1.8) and ζ\zeta to (1.17) are related by (1.18).

We use the spherical coordinates x(θ,φ)x(\theta,\varphi) of S2S^{2} given by (2.5) and write u(θ,φ)u(\theta,\varphi) instead of (ux)(θ,φ)(u\circ x)(\theta,\varphi) for a function uu on S2S^{2}. Hence θx\partial_{\theta}x and φx\partial_{\varphi}x are of the form (4.2) and

(6.1) u(θ,φ)=θu(θ,φ)θx(θ,φ)+φu(θ,φ)sin2θφx(θ,φ),Δu(θ,φ)=1sinθθ(sinθθu(θ,φ))+1sin2θφ2u(θ,φ),𝐧S2(θ,φ)=(sinθcosφ,sinθsinφ,cosθ)T.\displaystyle\begin{aligned} \nabla u(\theta,\varphi)&=\partial_{\theta}u(\theta,\varphi)\partial_{\theta}x(\theta,\varphi)+\frac{\partial_{\varphi}u(\theta,\varphi)}{\sin^{2}\theta}\,\partial_{\varphi}x(\theta,\varphi),\\ \Delta u(\theta,\varphi)&=\frac{1}{\sin\theta}\partial_{\theta}\bigl{(}\sin\theta\,\partial_{\theta}u(\theta,\varphi)\bigr{)}+\frac{1}{\sin^{2}\theta}\,\partial_{\varphi}^{2}u(\theta,\varphi),\\ \mathbf{n}_{S^{2}}(\theta,\varphi)&=(\sin\theta\cos\varphi,\sin\theta\sin\varphi,\cos\theta)^{T}.\end{aligned}

Also, we easily find that

(6.2) 𝐧S2(θ,φ)×θx(θ,φ)=1sinθφx(θ,φ).\displaystyle\mathbf{n}_{S^{2}}(\theta,\varphi)\times\partial_{\theta}x(\theta,\varphi)=\frac{1}{\sin\theta}\,\partial_{\varphi}x(\theta,\varphi).

For Φ\Phi\in\mathbb{R} let R(Φ)R(\Phi) be the rotation matrix around the x3x_{3}-axis by the angle Φ\Phi, i.e.

R(Φ)=(cosΦsinΦ0sinΦcosΦ0001).\displaystyle R(\Phi)=\begin{pmatrix}\cos\Phi&-\sin\Phi&0\\ \sin\Phi&\cos\Phi&0\\ 0&0&1\end{pmatrix}.

Then, for 𝐟=θx,φx,𝐧S2\mathbf{f}=\partial_{\theta}x,\partial_{\varphi}x,\mathbf{n}_{S^{2}} and 𝐚,𝐛3\mathbf{a},\mathbf{b}\in\mathbb{R}^{3}, we have

(6.3) R(Φ)𝐟(θ,φ)=𝐟(θ,φ+Φ),(R(Φ)𝐚)(R(Φ)𝐛)=𝐚𝐛,(R(Φ)𝐚)×(R(Φ)𝐛)=R(Φ)(𝐚×𝐛).\displaystyle\begin{aligned} R(\Phi)\mathbf{f}(\theta,\varphi)&=\mathbf{f}(\theta,\varphi+\Phi),\\ \bigl{(}R(\Phi)\mathbf{a}\bigr{)}\cdot\bigl{(}R(\Phi)\mathbf{b}\bigr{)}&=\mathbf{a}\cdot\mathbf{b},\\ \bigl{(}R(\Phi)\mathbf{a}\bigr{)}\times\bigl{(}R(\Phi)\mathbf{b}\bigr{)}&=R(\Phi)(\mathbf{a}\times\mathbf{b}).\end{aligned}

Let uu and vv be functions in L02(S2)L_{0}^{2}(S^{2}) such that u(θ,φ)=v(θ,φΦ)u(\theta,\varphi)=v(\theta,\varphi-\Phi). Then

(6.4) u(θ,φ)=θu(θ,φ)θx(θ,φ)+φu(θ,φ)sin2θφx(θ,φ)=θv(θ,φΦ)R(Φ)θx(θ,φΦ)+φv(θ,φΦ)sin2θR(Φ)φx(θ,φΦ)=R(Φ)v(θ,φΦ)\displaystyle\begin{aligned} \nabla u(\theta,\varphi)&=\partial_{\theta}u(\theta,\varphi)\partial_{\theta}x(\theta,\varphi)+\frac{\partial_{\varphi}u(\theta,\varphi)}{\sin^{2}\theta}\,\partial_{\varphi}x(\theta,\varphi)\\ &=\partial_{\theta}v(\theta,\varphi-\Phi)R(\Phi)\partial_{\theta}x(\theta,\varphi-\Phi)+\frac{\partial_{\varphi}v(\theta,\varphi-\Phi)}{\sin^{2}\theta}\,R(\Phi)\partial_{\varphi}x(\theta,\varphi-\Phi)\\ &=R(\Phi)\nabla v(\theta,\varphi-\Phi)\end{aligned}

by (6.1) and (6.3). Moreover, we see by (6.1) that

(6.5) Δu(θ,φ)=Δv(θ,φΦ),Δ1u(θ,φ)=Δ1v(θ,φΦ).\displaystyle\Delta u(\theta,\varphi)=\Delta v(\theta,\varphi-\Phi),\quad\Delta^{-1}u(\theta,\varphi)=\Delta^{-1}v(\theta,\varphi-\Phi).

Now let ω\omega and ζ\zeta be functions on S2×(0,)S^{2}\times(0,\infty) related by (1.18), i.e.

ω(θ,φ,t)=ζ(θ,φΩt,t)+2Ωcosθ.\displaystyle\omega(\theta,\varphi,t)=\zeta(\theta,\varphi-\Omega t,t)+2\Omega\cos\theta.

Then we observe by (6.5) and Δcosθ=2cosθ\Delta\cos\theta=-2\cos\theta that

(6.6) tω(θ,φ,t)=tζ(θ,φΩt,t)Ωφζ(θ,φΩt,t),(Δω+2ω)(θ,φ,t)=(Δζ+2ζ)(θ,φΩt,t),φ(I+6Δ1)ω(θ,φ,t)=φ(I+6Δ1)ζ(θ,φΩt,t).\displaystyle\begin{aligned} \partial_{t}\omega(\theta,\varphi,t)&=\partial_{t}\zeta(\theta,\varphi-\Omega t,t)-\Omega\,\partial_{\varphi}\zeta(\theta,\varphi-\Omega t,t),\\ (\Delta\omega+2\omega)(\theta,\varphi,t)&=(\Delta\zeta+2\zeta)(\theta,\varphi-\Omega t,t),\\ \partial_{\varphi}(I+6\Delta^{-1})\omega(\theta,\varphi,t)&=\partial_{\varphi}(I+6\Delta^{-1})\zeta(\theta,\varphi-\Omega t,t).\end{aligned}

Moreover, we deduce from (6.3)–(6.5) and Δ1cosθ=21cosθ\Delta^{-1}\cos\theta=-2^{-1}\cos\theta that

ω(θ,φ,t)\displaystyle\nabla\omega(\theta,\varphi,t) =R(Ωt)ζ(θ,φΩt,t)2Ωsinθθx(θ,φ)\displaystyle=R(\Omega t)\nabla\zeta(\theta,\varphi-\Omega t,t)-2\Omega\sin\theta\,\partial_{\theta}x(\theta,\varphi)
=R(Ωt){ζ(θ,φΩt,t)2Ωsinθθx(θ,φΩt)},\displaystyle=R(\Omega t)\{\nabla\zeta(\theta,\varphi-\Omega t,t)-2\Omega\sin\theta\,\partial_{\theta}x(\theta,\varphi-\Omega t)\},
Δ1ω(θ,φ,t)\displaystyle\nabla\Delta^{-1}\omega(\theta,\varphi,t) =R(Ωt){Δ1ζ(θ,φΩt,t)+Ωsinθθx(θ,φΩt)}.\displaystyle=R(\Omega t)\{\nabla\Delta^{-1}\zeta(\theta,\varphi-\Omega t,t)+\Omega\sin\theta\,\partial_{\theta}x(\theta,\varphi-\Omega t)\}.

Thus, setting 𝐮=𝐧S2×Δ1ω\mathbf{u}=\mathbf{n}_{S^{2}}\times\nabla\Delta^{-1}\omega and 𝐯=𝐧S2×Δ1ζ\mathbf{v}=\mathbf{n}_{S^{2}}\times\nabla\Delta^{-1}\zeta, we have

𝐮(θ,φ,t)\displaystyle\mathbf{u}(\theta,\varphi,t) =𝐧S2(θ,φ)×Δ1ω(θ,φ,t)\displaystyle=\mathbf{n}_{S^{2}}(\theta,\varphi)\times\nabla\Delta^{-1}\omega(\theta,\varphi,t)
=𝐧S2(θ,φ)×(R(Ωt){Δ1ζ(θ,φΩt,t)+Ωsinθθx(θ,φΩt)})\displaystyle=\mathbf{n}_{S^{2}}(\theta,\varphi)\times\bigl{(}R(\Omega t)\{\nabla\Delta^{-1}\zeta(\theta,\varphi-\Omega t,t)+\Omega\sin\theta\,\partial_{\theta}x(\theta,\varphi-\Omega t)\}\bigr{)}
=R(Ωt){𝐯(θ,φΩt,t)+Ωφx(θ,φΩt)}\displaystyle=R(\Omega t)\{\mathbf{v}(\theta,\varphi-\Omega t,t)+\Omega\,\partial_{\varphi}x(\theta,\varphi-\Omega t)\}

by using (6.3) and then applying (6.2). Hence

(𝐮ω)(θ,φ,t)\displaystyle(\mathbf{u}\cdot\nabla\omega)(\theta,\varphi,t) =[(𝐯+Ωφx)(ζ2Ωsinθθx)](θ,φΩt,t)\displaystyle=[(\mathbf{v}+\Omega\,\partial_{\varphi}x)\cdot(\nabla\zeta-2\Omega\sin\theta\,\partial_{\theta}x)](\theta,\varphi-\Omega t,t)

by (6.3). Moreover, since φxθx=0\partial_{\varphi}x\cdot\partial_{\theta}x=0, φxζ=φζ\partial_{\varphi}x\cdot\nabla\zeta=\partial_{\varphi}\zeta, and

𝐯θx\displaystyle\mathbf{v}\cdot\partial_{\theta}x =(𝐧S2×Δ1ζ)θx=Δ1ζ(𝐧S2×θx)\displaystyle=(\mathbf{n}_{S^{2}}\times\nabla\Delta^{-1}\zeta)\cdot\partial_{\theta}x=-\nabla\Delta^{-1}\zeta\cdot(\mathbf{n}_{S^{2}}\times\partial_{\theta}x)
=1sinθΔ1ζφx=1sinθφΔ1ζ\displaystyle=-\frac{1}{\sin\theta}\,\nabla\Delta^{-1}\zeta\cdot\partial_{\varphi}x=-\frac{1}{\sin\theta}\,\partial_{\varphi}\Delta^{-1}\zeta

by (𝐚×𝐛)𝐜=𝐛(𝐚×𝐜)(\mathbf{a}\times\mathbf{b})\cdot\mathbf{c}=-\mathbf{b}\cdot(\mathbf{a}\times\mathbf{c}) for 𝐚,𝐛,𝐜3\mathbf{a},\mathbf{b},\mathbf{c}\in\mathbb{R}^{3} and (6.2), we get

(6.7) (𝐮ω)(θ,φ,t)=(𝐯ζ+Ωφζ+2ΩφΔ1ζ)(θ,φΩt,t).\displaystyle(\mathbf{u}\cdot\nabla\omega)(\theta,\varphi,t)=(\mathbf{v}\cdot\nabla\zeta+\Omega\,\partial_{\varphi}\zeta+2\Omega\,\partial_{\varphi}\Delta^{-1}\zeta)(\theta,\varphi-\Omega t,t).

Therefore, for the functions ω\omega and ζ\zeta related by (1.18), we conclude by (6.6) and (6.7) that ω\omega is a solution to (1.8) if and only if ζ\zeta is a solution to (1.17).

Acknowledgments

The work of the author was supported by Grant-in-Aid for JSPS Fellows No. 19J00693.

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