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Rate of the enhanced dissipation for the two-jet Kolmogorov type flow on the unit sphere

Yasunori Maekawa Department of Mathematics, Kyoto University, Kitashirakawa Oiwake-cho, Sakyo-ku, Kyoto 606-8502, Japan [email protected]  and  Tatsu-Hiko Miura Department of Mathematics, Kyoto University, Kitashirakawa Oiwake-cho, Sakyo-ku, Kyoto 606-8502, Japan [email protected]
Abstract.

We study the enhanced dissipation for the two-jet Kolmogorov type flow which is a stationary solution to the Navier–Stokes equations on the two-dimensional unit sphere given by the zonal spherical harmonic function of degree two. Based on the pseudospectral bound method developed by Ibrahim, Maekawa, and Masmoudi [15] and a modified version of the Gearhart–Prüss type theorem shown by Wei [48], we derive an estimate for the resolvent of the linearized operator along the imaginary axis and show that a solution to the linearized equation rapidly decays at the rate O(eνt)O(e^{-\sqrt{\nu}\,t}) when the viscosity coefficient ν\nu is sufficiently small as in the case of the plane Kolmogorov flow.

Key words and phrases:
Navier–Stokes equations, Kolmogorov type flow, enhanced dissipation
2010 Mathematics Subject Classification:
35Q30, 35R01, 47A10, 76D05

1. Introduction

In this paper, as a continuation of [32], we consider the incompressible Navier–Stokes equations on the two-dimensional (2D) unit sphere S2S^{2} in 3\mathbb{R}^{3}:

(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 of the fluid, pp is the scalar-valued pressure, and 𝐟\mathbf{f} is a given tangential 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}. Here the viscous term is taken to be the twice of the divergence of the deformation tensor Def𝐮\mathrm{Def}\,\mathbf{u}:

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},

where Ric1\mathrm{Ric}\equiv 1 is the Ricci curvature of S2S^{2}. The Navier–Stokes equations on spheres and more general manifolds with this kind of viscous term have been studied by many authors (see e.g. [43, 36, 33, 31, 9, 21, 5, 7, 39, 40, 35, 22, 37, 38]). There are also several works on the Navier–Stokes equations on manifolds in which the viscous term is taken to be νΔH𝐮\nu\Delta_{H}\mathbf{u} by analogy of the flat domain case (see e.g. [17, 16, 4, 18, 51, 26, 41]). We refer to [12, 1, 11, 44, 6] for the above identity and the choice of the viscous term in the Navier–Stokes equations on manifolds.

Identifying vector fields with one-forms and taking rot=d\mathrm{rot}=\ast d in (1.1), where \ast and dd are the Hodge star operator and the external derivative, we have the vorticity equation

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

for the vorticity ω=rot𝐮\omega=\mathrm{rot}\,\mathbf{u} (see [32] for derivation). Here 𝐮ω\nabla_{\mathbf{u}}\omega is the directional derivative of ω\omega along 𝐮\mathbf{u} and Δ\Delta is the Laplace–Beltrami operator on S2S^{2} which is invertible on L02(S2)L_{0}^{2}(S^{2}), the space of L2L^{2} functions on S2S^{2} with zero mean, 𝐧S2\mathbf{n}_{S^{2}} is the unit outward normal vector field on S2S^{2}, and ×\times is the vector product in 3\mathbb{R}^{3}. Note that the vorticity equation (1.2) is equivalent to the Navier–Stokes equations (1.1) since any closed one-form on S2S^{2} is exact.

For n0n\in\mathbb{Z}_{\geq 0} and |m|n|m|\leq n let YnmY_{n}^{m} be the spherical harmonics which satisfy ΔYnm=λnYnm-\Delta Y_{n}^{m}=\lambda_{n}Y_{n}^{m} with λn=n(n+1)\lambda_{n}=n(n+1) (see Section 2). The vorticity equation (1.2) with external force rot𝐟na=aν(λn2)Yn0\mathrm{rot}\,\mathbf{f}_{n}^{a}=a\nu(\lambda_{n}-2)Y_{n}^{0} has a stationary solution with velocity field

(1.3) 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}\mathbf{x}(\theta,\varphi)

for nn\in\mathbb{N} and aa\in\mathbb{R}, where 𝐱(θ,φ)\mathbf{x}(\theta,\varphi) is the parametrization of S2S^{2} by the colatitude θ\theta and the longitude φ\varphi (see Section 2). The flow (1.3) can be seen as the spherical version of the Kolmogorov flow which is a stationary solution to the Navier–Stokes equations in a 2D flat torus with shear external force (see e.g. [30, 19, 28, 34, 29] for the study of the stability of the plane Kolmogorov flow). Ilyin [18] called (1.3) the generalized Kolmogorov flow and studied its stability for the Navier–Stokes equations on S2S^{2} with viscous term νΔH𝐮\nu\Delta_{H}\mathbf{u}. Also, Sasaki, Takehiro, and Yamada [39, 40] called (1.3) an nn-jet zonal flow and investigated its stability for the Navier–Stokes equations on S2S^{2} with viscous term ν(ΔH𝐮+2𝐮)\nu(\Delta_{H}\mathbf{u}+2\mathbf{u}). The stability of (1.3) for the Euler equations on S2S^{2} was also studied by Taylor [42]. We call (1.3) 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 consider the linear stability of the two-jet Kolmogorov type flow. We substitute ω=ω2a+ω~2\omega=\omega_{2}^{a}+\tilde{\omega}_{2} for (1.2) and omit the nonlinear term with respect to ω~2\tilde{\omega}_{2} to get

tω~2=ν(Δω~2+2ω~2)a2cosθφ(I+6Δ1)ω~2,a2=a45π,\displaystyle\partial_{t}\tilde{\omega}_{2}=\nu(\Delta\tilde{\omega}_{2}+2\tilde{\omega}_{2})-a_{2}\cos\theta\,\partial_{\varphi}(I+6\Delta^{-1})\tilde{\omega}_{2},\quad a_{2}=\frac{a}{4}\sqrt{\frac{5}{\pi}},

where II is the identity operator (see [32] for derivation of the linearized equation). Replacing ω~2\tilde{\omega}_{2} and a2a_{2} by ω\omega and aa, we rewrite this equation as

(1.4) tω=ν,aω=νAωiaΛω,A=Δ+2,Λ=icosθφ(I+6Δ1),\displaystyle\partial_{t}\omega=\mathcal{L}^{\nu,a}\omega=\nu A\omega-ia\Lambda\omega,\quad A=\Delta+2,\quad\Lambda=-i\cos\theta\,\partial_{\varphi}(I+6\Delta^{-1}),

which is considered in L02(S2)L_{0}^{2}(S^{2}). Then since A-A is nonnegative and self-adjoint in L02(S2)L_{0}^{2}(S^{2}) and Λ\Lambda is AA-compact in L02(S2)L_{0}^{2}(S^{2}), the operator ν,a\mathcal{L}^{\nu,a} generates an analytic semigroup {etν,a}t0\{e^{t\mathcal{L}^{\nu,a}}\}_{t\geq 0} in L02(S2)L_{0}^{2}(S^{2}) by a perturbation theory of semigroups (see [13]) and thus the solution of (1.4) with initial data ω0L02(S2)\omega_{0}\in L_{0}^{2}(S^{2}) is given by ω(t)=etν,aω0\omega(t)=e^{t\mathcal{L}^{\nu,a}}\omega_{0}. In [32] the second author of the present paper proved the linear stability of the two-jet Kolmogorov type flow for all ν>0\nu>0 by getting the exponential decay of etν,aω0e^{t\mathcal{L}^{\nu,a}}\omega_{0} towards a (not orthogonal) projection of ω0\omega_{0} onto the kernel of ν,a\mathcal{L}^{\nu,a}. Moreover, he showed that Λ\Lambda does not have eigenvalues in {0}\mathbb{C}\setminus\{0\} by making use of the mixing structure of Λ\Lambda expressed by a recurrence relation for YnmY_{n}^{m}, and applied it to find that the enhanced dissipation occurs for the rescaled flow etνν,aω0e^{\frac{t}{\nu}\mathcal{L}^{\nu,a}}\omega_{0}, which is a solution to tω=AωiαΛω\partial_{t}\omega=A\omega-i\alpha\Lambda\omega with α=a/ν\alpha=a/\nu, as in the case of an advection-diffusion equation [8, 52, 48]. More precisely, let

𝒳={uL02(S2)(u,Yn0)L2(S2)=(u,Y1m)L2(S2)=0,n1,|m|=0,1},\displaystyle\mathcal{X}=\{u\in L_{0}^{2}(S^{2})\mid(u,Y_{n}^{0})_{L^{2}(S^{2})}=(u,Y_{1}^{m})_{L^{2}(S^{2})}=0,\,n\geq 1,\,|m|=0,1\},

which is a closed subspace of L02(S2)L_{0}^{2}(S^{2}) invariant under the actions of AA and Λ\Lambda (see Section 3.1), and \mathbb{Q} be the orthogonal projection from 𝒳\mathcal{X} onto the orthogonal complement of the kernel of Λ\Lambda restricted on 𝒳\mathcal{X}. Then it was shown in [32, Theorem 1.4] that

lim|a/ν|suptτetνν,a𝒳𝒳=0for eachτ>0.\displaystyle\lim_{|a/\nu|\to\infty}\sup_{t\geq\tau}\|\mathbb{Q}e^{\frac{t}{\nu}\mathcal{L}^{\nu,a}}\|_{\mathcal{X}\to\mathcal{X}}=0\quad\text{for each}\quad\tau>0.

This result in particular gives the convergence of etνν,aω0\mathbb{Q}e^{\frac{t}{\nu}\mathcal{L}^{\nu,a}}\omega_{0} to zero in L2(S2)L^{2}(S^{2}) as ν0\nu\to 0 for each fixed t>0t>0 and aa\in\mathbb{R}, but does not give the actual convergence rate. The purpose of this paper is to give an explicit decay rate of the original flow etν,aω0\mathbb{Q}e^{t\mathcal{L}^{\nu,a}}\omega_{0}.

In the plane case, Beck and Wayne [3] numerically conjectured that a perturbation of the plane Kolmogorov flow decays at the rate O(eνt)O(e^{-\sqrt{\nu}\,t}) when the viscosity coefficient ν\nu is sufficiently small. They also verified this enhanced dissipation for a linearized operator without a nonlocal term based on the hypocoercivity method developed by Villani [47]. Lin and Xu [27] proved the enhanced dissipation for a full linearized operator but without an explicit decay rate by using the Hamiltonian structure of a perturbation operator and the RAGE theorem. The decay rate O(eνt)O(e^{-\sqrt{\nu}\,t}) was confirmed by Ibrahim, Maekawa, and Masmoudi [15] based on the pseudospectral bound method, by Wei and Zhang [49] based on the hypocoercivity method, and by Wei, Zhang, and Zhao [50] based on the wave operator method.

In this paper we show that etν,aω0\mathbb{Q}e^{t\mathcal{L}^{\nu,a}}\omega_{0} decays at the rate O(eνt)O(e^{-\sqrt{\nu}\,t}) as in the plane case. For mm\in\mathbb{Z} let 𝒫m\mathcal{P}_{m} be the orthogonal projection from L2(S2)L^{2}(S^{2}) onto the space of L2L^{2} functions on S2S^{2} of the form u=U(θ)eimφu=U(\theta)e^{im\varphi} (see (2.8) for the precise definition). Note that 𝒳\mathcal{X} is orthogonally decomposed as 𝒳=m{0}𝒫m𝒳\mathcal{X}=\oplus_{m\in\mathbb{Z}\setminus\{0\}}\mathcal{P}_{m}\mathcal{X} and

𝒫mu=𝒫mu,|m|3,(I)𝒫mu=(𝒫mu,Y2m)L2(S2)Y2m,|m|=1,2\displaystyle\mathbb{Q}\mathcal{P}_{m}u=\mathcal{P}_{m}u,\quad|m|\geq 3,\quad(I-\mathbb{Q})\mathcal{P}_{m}u=(\mathcal{P}_{m}u,Y_{2}^{m})_{L^{2}(S^{2})}Y_{2}^{m},\quad|m|=1,2

for u𝒳u\in\mathcal{X} (see Section 3.1). The main result of this paper is as follows.

Theorem 1.1.

There exist constants C1,C2>0C_{1},C_{2}>0 such that

(1.5) 𝒫metν,aω0L2(S2)C1eC2|a|1/2ν1/2|m|2/3t𝒫mω0L2(S2)\displaystyle\|\mathbb{Q}\mathcal{P}_{m}e^{t\mathcal{L}^{\nu,a}}\omega_{0}\|_{L^{2}(S^{2})}\leq C_{1}e^{-C_{2}|a|^{1/2}\nu^{1/2}|m|^{2/3}t}\|\mathbb{Q}\mathcal{P}_{m}\omega_{0}\|_{L^{2}(S^{2})}

for all t0t\geq 0, ω0𝒳\omega_{0}\in\mathcal{X}, ν>0\nu>0, aa\in\mathbb{R}, and m{0}m\in\mathbb{Z}\setminus\{0\}. Also, if |m|=1,2|m|=1,2 and |a/ν||a/\nu| is sufficiently large, then for all t0t\geq 0 and ω0𝒳\omega_{0}\in\mathcal{X} we have

(I)𝒫metν,aω0L2(S2)e4νt(I)𝒫mω0L2(S2)+C3|amν|1/3log(C4|amν|2/3)e2νt𝒫mω0L2(S2),\|(I-\mathbb{Q})\mathcal{P}_{m}e^{t\mathcal{L}^{\nu,a}}\omega_{0}\|_{L^{2}(S^{2})}\leq e^{-4\nu t}\|(I-\mathbb{Q})\mathcal{P}_{m}\omega_{0}\|_{L^{2}(S^{2})}\\ +C_{3}\left|\frac{am}{\nu}\right|^{1/3}\log\left(C_{4}\left|\frac{am}{\nu}\right|^{2/3}\right)e^{-2\nu t}\|\mathbb{Q}\mathcal{P}_{m}\omega_{0}\|_{L^{2}(S^{2})},

where C3,C4>0C_{3},C_{4}>0 are constants independent of tt, ω0\omega_{0}, ν\nu, aa, and mm.

Here we note that the exponent of |m||m| in (1.5) is 2/32/3 in the sphere case, while it is 1/21/2 in the plane case (see [15, Corollary 3.14]). This difference comes from the factor |m|1/2|m|^{-1/2} appearing in the estimate for the L(S2)L^{\infty}(S^{2})-norm of u=U(θ)eimφu=U(\theta)e^{im\varphi} by the L2(S2)L^{2}(S^{2})-norm of u\nabla u (see Lemma 2.3) which yields a different coercive estimate for m(μΛm)\mathbb{Q}_{m}(\mu-\Lambda_{m}) with μ\mu\in\mathbb{R}, m=|𝒫m𝒳\mathbb{Q}_{m}=\mathbb{Q}|_{\mathcal{P}_{m}\mathcal{X}}, and Λm=m1Λ|𝒫m𝒳\Lambda_{m}=m^{-1}\Lambda|_{\mathcal{P}_{m}\mathcal{X}} (see Lemma 3.8).

To prove Theorem 1.1 we follow the argument of the work by Ibrahim, Maekawa, and Masmoudi [15] which verified the enhanced dissipation for the plane Kolmogorov flow by the pseudospectral bound method. By rescaling time as tνtt\mapsto\nu t in (1.4) and taking the Fourier series with respect to the longitude φ\varphi, we consider the equation

tω=Lα,mω=AmωiαmΛmω,Am=A|𝒳m,Λm=Mcosθ(I+6Δ1)|𝒳m\displaystyle\partial_{t}\omega=L_{\alpha,m}\omega=A_{m}\omega-i\alpha m\Lambda_{m}\omega,\quad A_{m}=A|_{\mathcal{X}_{m}},\quad\Lambda_{m}=M_{\cos\theta}(I+6\Delta^{-1})|_{\mathcal{X}_{m}}

in 𝒳m=𝒫m𝒳\mathcal{X}_{m}=\mathcal{P}_{m}\mathcal{X} for each m{0}m\in\mathbb{Z}\setminus\{0\}, where α=a/ν\alpha=a/\nu\in\mathbb{R} and McosθM_{\cos\theta} is the multiplication operator by cosθ\cos\theta. By the definition of 𝒳m\mathcal{X}_{m}, the operator Am-A_{m} is self-adjoint and positive in 𝒳m\mathcal{X}_{m}. Also, the kernel of Λm\Lambda_{m} is {cY2mc}\{cY_{2}^{m}\mid c\in\mathbb{C}\} when |m|=1,2|m|=1,2 and trivial when |m|3|m|\geq 3. We intend to estimate the semigroup etLα,me^{tL_{\alpha,m}} generated by Lα,mL_{\alpha,m}, especially metLα,m\mathbb{Q}_{m}e^{tL_{\alpha,m}} or equivalently etmLα,me^{t\mathbb{Q}_{m}L_{\alpha,m}}, where m\mathbb{Q}_{m} is the orthogonal projection from 𝒳m\mathcal{X}_{m} onto 𝒴m\mathcal{Y}_{m}, the orthogonal complement of the kernel of Λm\Lambda_{m}. To this end, we apply abstract results given in Section 4 to Lα,mL_{\alpha,m} to derive an estimate for the quantity

(supλ(iλmLα,m)1𝒴m𝒴m)1=(supλ(iλ+mLα,m)1𝒴m𝒴m)1\displaystyle\left(\sup_{\lambda\in\mathbb{R}}\|(i\lambda-\mathbb{Q}_{m}L_{\alpha,m})^{-1}\|_{\mathcal{Y}_{m}\to\mathcal{Y}_{m}}\right)^{-1}=\left(\sup_{\lambda\in\mathbb{R}}\|(i\lambda+\mathbb{Q}_{m}L_{\alpha,m})^{-1}\|_{\mathcal{Y}_{m}\to\mathcal{Y}_{m}}\right)^{-1}

which was introduced in [14] and called the pseudospectral bound of mLα,m\mathbb{Q}_{m}L_{\alpha,m} in [15]. Then we get an estimate for the semigroup generated by mLα,m\mathbb{Q}_{m}L_{\alpha,m} by combining the estimate for the pseudospectral bound of mLα,m\mathbb{Q}_{m}L_{\alpha,m} with an abstract theorem for an mm-accretive operator on a weighted Hilbert space which is a version of the Gearhart–Prüss type theorem shown by Wei [48]. In order to use the abstract results, we need to confirm several assumptions for AmA_{m} and Λm\Lambda_{m}. The main effort is to verify a coercive estimate for m(μΛm)\mathbb{Q}_{m}(\mu-\Lambda_{m}) on 𝒴m\mathcal{Y}_{m} with μ\mu\in\mathbb{R} (see Section 3.2). This coercive estimate involves a loss of derivative, but thanks to a small factor in front it can be controlled by the smoothing effect of Am-A_{m} in the estimate for the pseudospectral bound of mLα,m\mathbb{Q}_{m}L_{\alpha,m}. When |μ|1|\mu|\geq 1, we easily get the coercive estimate by taking the L2(S2)L^{2}(S^{2})-inner product of m(μΛm)u\mathbb{Q}_{m}(\mu-\Lambda_{m})u with (I+6Δ1)u(I+6\Delta^{-1})u for u𝒴mu\in\mathcal{Y}_{m} and using the coerciveness of I+6Δ1I+6\Delta^{-1} on 𝒴m\mathcal{Y}_{m} (see Lemmas 3.7 and 3.8). On the other hand, the proof is more involved when |μ|<1|\mu|<1 (see Lemma 3.9). In this case, noting that u𝒴mu\in\mathcal{Y}_{m} is of the form u=U(θ)eimφu=U(\theta)e^{im\varphi}, we analyze an ordinary differential equation (ODE) with variable θ\theta corresponding to m(μΛm)\mathbb{Q}_{m}(\mu-\Lambda_{m}) and prove the coercive estimate by a contradiction argument after giving auxiliary statements. The main difficulty comes from the nonlocalities of the orthogonal projection m\mathbb{Q}_{m} which is nontrivial when |m|=1,2|m|=1,2 and of the inverse operator Δ1\Delta^{-1} appearing in Λm\Lambda_{m}. These nonlocalities affect the analysis for different values of μ\mu. The nonlocality of m\mathbb{Q}_{m} with |m|=1,2|m|=1,2 is relevant to the case where μ\mu is closed to zero, which is the eigenvalue of Λm\Lambda_{m}. On the other hand, the nonlocality of Δ1\Delta^{-1} causes a difficulty when μ\mu is close to ±1\pm 1, which are the critical values of cosθ\cos\theta appearing in Λm\Lambda_{m}, i.e. the derivative (cosθ)=sinθ(\cos\theta)^{\prime}=-\sin\theta vanishes when cosθ=±1\cos\theta=\pm 1. Hence we can deal with these difficulties separately by using a contradiction argument. Moreover, since ImI-\mathbb{Q}_{m} is the orthogonal projection onto the kernel of Λm\Lambda_{m} which is spanned by the smooth function Y2mY_{2}^{m}, we can handle the nonlocality of m\mathbb{Q}_{m} by introducing a suitable auxiliary function which involves Y2mY_{2}^{m}. Also, when μ\mu is close to ±1\pm 1, the use of a contradiction argument enables us to focus on analysis of functions concentrating around the critical points of cosθ\cos\theta, i.e. θ=0,π\theta=0,\pi, for which the nonlocal term consisting of Δ1\Delta^{-1} essentially becomes a small order by the smoothing effect of Δ1\Delta^{-1}. By these facts we can verify the coercive estimate for |μ|<1|\mu|<1, but the actual proof requires very long and careful calculations.

The coercive estimate for m(μΛm)\mathbb{Q}_{m}(\mu-\Lambda_{m}) given in this paper is basically the same as the one in the plane case [15]. In the sphere case, the surface measure on S2S^{2} is of the form sinθdθdφ\sin\theta\,d\theta\,d\varphi under the spherical coordinate system. The weight function sinθ\sin\theta vanishes at the critical points θ=0,π\theta=0,\pi of cosθ\cos\theta appearing in Λm\Lambda_{m}, so one may expect to have a better coercive estimate than the one in the plane case when |μ|<1|\mu|<1 and μ\mu is close to ±1\pm 1, but we cannot get such an estimate. In fact, for θμ=arccosμ\theta_{\mu}=\arccos\mu with μ\mu close to ±1\pm 1, a small parameter ε>0\varepsilon>0, and a function on S2S^{2} of the form u=U(θ)eimφu=U(\theta)e^{im\varphi}, we have a better bound εsinθμ\varepsilon\sin\theta_{\mu} due to the weight function sinθ\sin\theta when we estimate the L2L^{2}-norm of uu on a narrow band on S2S^{2} corresponding to θ(θμε,θμ+ε)\theta\in(\theta_{\mu}-\varepsilon,\theta_{\mu}+\varepsilon) by the LL^{\infty}-norm of UU on (θμε,θμ+ε)(\theta_{\mu}-\varepsilon,\theta_{\mu}+\varepsilon). However, we also have a factor 1/sinθμ1/\sin\theta_{\mu} when we use an interpolation type inequality which bounds the LL^{\infty}-norm UU on (θμε,θμ+ε)(\theta_{\mu}-\varepsilon,\theta_{\mu}+\varepsilon) by the L2L^{2}- and H1H^{1}-norms of uu on S2S^{2} (see Lemma 2.6), so the resulting coercive estimate is the same as in the plane case.

Lastly, let us explain difference from the plane case [15] in the proof of the coercive estimate. The authors of [15] derived coercive estimates for μΛm\mu-\Lambda_{m} on 𝒳m\mathcal{X}_{m} with μ\mu\in\mathbb{R} away from zero and for m(μΛm)\mathbb{Q}_{m}(\mu-\Lambda_{m}) on 𝒴m\mathcal{Y}_{m} with μ\mu\in\mathbb{R} close to zero in order to avoid the nonlocality of m\mathbb{Q}_{m} when μ\mu is away from zero. In this paper, however, we deal with m(μΛm)\mathbb{Q}_{m}(\mu-\Lambda_{m}) for all μ\mu\in\mathbb{R} since the nonlocality of m\mathbb{Q}_{m} can be handled by taking the inner product in 𝒴m\mathcal{Y}_{m} when |μ|1|\mu|\geq 1 and by introducing an auxiliary function when |μ|<1|\mu|<1. Also, we encounter an additional difficulty due to the larger coefficient of the nonlocal operator Δ1\Delta^{-1} in Λm\Lambda_{m}. To prove the coercive estimate when |μ|<1|\mu|<1 and μ\mu is close to 11 (or 1-1), we analyze an auxiliary function ww on S2S^{2} of the form w=W(θ)eimφw=W(\theta)e^{im\varphi} and derive an estimate for the L2L^{2}-norm of ww on a subset of S2S^{2} corresponding to θ(θμ,π)\theta\in(\theta_{\mu},\pi) (or θ(0,θμ)\theta\in(0,\theta_{\mu})) with θμ=arccosμ\theta_{\mu}=\arccos\mu by using an ODE for WW (see Lemma 3.17). In the proof of the estimate, we need to show J+KCθμπ|W(θ)|2sinθdθJ+K\geq C\int_{\theta_{\mu}}^{\pi}|W(\theta)|^{2}\sin\theta\,d\theta with a constant C>0C>0, where

J\displaystyle J =θμπ(μcosθ)(|W(θ)|2+m2sin2θ|W(θ)|2)sinθdθ,\displaystyle=\int_{\theta_{\mu}}^{\pi}(\mu-\cos\theta)\left(|W^{\prime}(\theta)|^{2}+\frac{m^{2}}{\sin^{2}\theta}|W(\theta)|^{2}\right)\sin\theta\,d\theta,
K\displaystyle K =5θμπ|W(θ)|2cosθsinθdθ.\displaystyle=5\int_{\theta_{\mu}}^{\pi}|W(\theta)|^{2}\cos\theta\sin\theta\,d\theta.

Here JJ is always nonnegative but KK may become negative by the presence of cosθ\cos\theta. Moreover, the coefficient 55 of KK comes from the coefficient 66 of Δ1\Delta^{-1} in Λm\Lambda_{m}. In the plane case, one also needs to deal with integrals similar to JJ and KK, but the coefficient of KK becomes 1/21/2 since the coefficient of Δ1\Delta^{-1} in Λm\Lambda_{m} is 11. Thus one can easily get an estimate just by using the zeroth order term of WW in JJ. In our case, however, the use of only the zeroth order term does not work for |m|=1|m|=1 since the coefficient 55 of KK is too large. To overcome this difficulty, we take into account the first order term of WW in JJ. In fact, it is natural to use both the zeroth and first order terms since

|w|2=|W(θ)|2+m2sin2θ|W(θ)|2\displaystyle|\nabla w|^{2}=|W^{\prime}(\theta)|^{2}+\frac{m^{2}}{\sin^{2}\theta}|W(\theta)|^{2}

for the gradient of w=W(θ)eimφw=W(\theta)e^{im\varphi} on S2S^{2}. In the actual proof, we split KK into the integrals K1K_{1} and K2K_{2} over (θμ,Θ)(\theta_{\mu},\Theta) and (Θ,π)(\Theta,\pi) with Θ(θμ,π/2)\Theta\in(\theta_{\mu},\pi/2) and estimate them separately. To K1K_{1} we just use cosθcosΘ\cos\theta\geq\cos\Theta for θ(θμ,Θ)\theta\in(\theta_{\mu},\Theta). Also, we carry out integration by parts for K2K_{2} and apply Young’s inequality to get the integral of a function involving the term |w|2|\nabla w|^{2}. Then, taking an appropriate Θ\Theta and estimating the integrand, we obtain an estimate for K2K_{2} which gives the lower bound of J+KJ+K when combined with the estimate for K1K_{1}.

The rest of this paper is organized as follows. In Section 2 we fix notations and give auxiliary inequalities. Section 3 is devoted to the study of the linearized operator for the two-jet Kolmogorov type flow. We provide settings and basic results in Section 3.1, verify coercive estimates for m(μΛm)\mathbb{Q}_{m}(\mu-\Lambda_{m}) in Section 3.2, and derive estimates for the semigroup generated by Lα,mL_{\alpha,m} and prove Theorem 1.1 in Section 3.3. Section 4 gives abstract results used in Section 3. In Section 5 we show basic formulas of differential geometry on S2S^{2}.

2. Preliminaries

In this section we fix notations and give auxiliary inequalities. We also refer to Section 5 for some notations and basic formulas of differential geometry.

Let S2S^{2} be the 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 θ\theta and φ\varphi the colatitude and the longitude so that S2S^{2} is parametrized by 𝐱(θ,φ)=(sinθcosφ,sinθsinφ,cosθ)\mathbf{x}(\theta,\varphi)=(\sin\theta\cos\varphi,\sin\theta\sin\varphi,\cos\theta) for θ[0,π]\theta\in[0,\pi] and φ[0,2π)\varphi\in[0,2\pi). For a (complex-valued) function uu on S2S^{2}, we sometimes abuse the notation

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

when no confusion may occur. Thus the integral of uu over S2S^{2} is given by

S2u𝑑2=02π(0πu(θ,φ)sinθdθ)𝑑φ,\displaystyle\int_{S^{2}}u\,d\mathcal{H}^{2}=\int_{0}^{2\pi}\left(\int_{0}^{\pi}u(\theta,\varphi)\sin\theta\,d\theta\right)\,d\varphi,

where k\mathcal{H}^{k} is the Hausdorff measure of dimension kk\in\mathbb{N}. As usual, we set

(u,v)L2(S2)=S2uv¯𝑑2,uL2(S2)=(u,u)L2(S2)1/2,u,vL2(S2),\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},\quad u,v\in L^{2}(S^{2}),

where v¯\bar{v} is the complex conjugate of vv, and write Hk(S2)H^{k}(S^{2}), k0k\in\mathbb{Z}_{\geq 0} for the Sobolev spaces of L2L^{2} functions on S2S^{2} with H0(S2)=L2(S2)H^{0}(S^{2})=L^{2}(S^{2}) (see [2]).

Let \nabla and Δ\Delta be the gradient and Laplace–Beltrami operators on S2S^{2}. It is well known (see e.g. [46, 45]) that λn=n(n+1)\lambda_{n}=n(n+1) is an eigenvalue of Δ-\Delta with multiplicity 2n+12n+1 for each n0n\in\mathbb{Z}_{\geq 0} and the corresponding eigenvectors are the spherical harmonics

(2.1) Ynm=Ynm(θ,φ)=2n+14π(nm)!(n+m)!Pnm(cosθ)eimφ,m=0,±1,,±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 m=0,\pm 1,\dots,\pm n.

Here Pn0P_{n}^{0}, n0n\in\mathbb{Z}_{\geq 0} are the Legendre polynomials defined as

Pn0(s)=Pn(s)=12nn!dndsn(s21)n,s(1,1)\displaystyle P_{n}^{0}(s)=P_{n}(s)=\frac{1}{2^{n}n!}\frac{d^{n}}{ds^{n}}(s^{2}-1)^{n},\quad s\in(-1,1)

and the associated Legendre functions PnmP_{n}^{m}, n0n\in\mathbb{Z}_{\geq 0}, |m|n|m|\leq n are given by

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

so that Ynm=(1)mYnm¯Y_{n}^{-m}=(-1)^{m}\overline{Y_{n}^{m}} (see [23, 10]). Moreover, the set of all YnmY_{n}^{m} forms an orthonormal basis of L2(S2)L^{2}(S^{2}), i.e. u=n=0m=nn(u,Ynm)L2(S2)Ynmu=\sum_{n=0}^{\infty}\sum_{m=-n}^{n}(u,Y_{n}^{m})_{L^{2}(S^{2})}Y_{n}^{m} for each uL2(S2)u\in L^{2}(S^{2}). 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 [23, (7.12.12)]) and thus (see also [46, Section 5.7])

(2.2) 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. Note that the superscript mm of anma_{n}^{m} just corresponds to that of YnmY_{n}^{m} and does not mean the mm-th power.

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, for ss\in\mathbb{R}, the operator (Δ)s(-\Delta)^{s} is defined on L02(S2)L_{0}^{2}(S^{2}) by

(2.3) (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}).

We easily observe by a density argument and integration by parts that

(2.4) (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 this relation and Poincaré’s inequality (see e.g. [2, Corollary 4.3])

(2.5) uL2(S2)CuL2(S2),uL02(S2)H1(S2)\displaystyle\|u\|_{L^{2}(S^{2})}\leq C\|\nabla u\|_{L^{2}(S^{2})},\quad u\in L_{0}^{2}(S^{2})\cap H^{1}(S^{2})

we have the norm equivalence

(2.6) (Δ)1/2uL2(S2)uH1(S2)C(Δ)1/2uL2(S2),uL02(S2)H1(S2).\displaystyle\|(-\Delta)^{1/2}u\|_{L^{2}(S^{2})}\leq\|u\|_{H^{1}(S^{2})}\leq C\|(-\Delta)^{1/2}u\|_{L^{2}(S^{2})},\quad u\in L_{0}^{2}(S^{2})\cap H^{1}(S^{2}).

Let uu be a function on S2S^{2}. We write u=U(θ)eimφu=U(\theta)e^{im\varphi} if uu is of the form

u(sinθcosφ,sinθsinφ,cosθ)=U(θ)eimφ,θ[0,π],φ[0,2π)\displaystyle u(\sin\theta\cos\varphi,\sin\theta\sin\varphi,\cos\theta)=U(\theta)e^{im\varphi},\quad\theta\in[0,\pi],\,\varphi\in[0,2\pi)

with some function UU of the colatitude θ\theta and mm\in\mathbb{Z}. In this case,

(2.7) uL2(S2)2=2π0π|U(θ)|2sinθdθ,uL2(S2)2=2π0π(|U(θ)|2+m2sin2θ|U(θ)|2)sinθdθ,2uL2(S2)2=2π0π{|U′′(θ)|2+R(U)}sinθdθ,\displaystyle\begin{aligned} \|u\|_{L^{2}(S^{2})}^{2}&=2\pi\int_{0}^{\pi}|U(\theta)|^{2}\sin\theta\,d\theta,\\ \|\nabla u\|_{L^{2}(S^{2})}^{2}&=2\pi\int_{0}^{\pi}\left(|U^{\prime}(\theta)|^{2}+\frac{m^{2}}{\sin^{2}\theta}|U(\theta)|^{2}\right)\sin\theta\,d\theta,\\ \|\nabla^{2}u\|_{L^{2}(S^{2})}^{2}&=2\pi\int_{0}^{\pi}\{|U^{\prime\prime}(\theta)|^{2}+R(U)\}\sin\theta\,d\theta,\end{aligned}

by (5.1), where U=dUdθU^{\prime}=\frac{dU}{d\theta} and 2u\nabla^{2}u is the covariant Hessian of uu, and

R(U)=2m2sin2θ|U(θ)cosθsinθU(θ)|2+1sin4θ|U(θ)sinθcosθm2U(θ)|20.\displaystyle R(U)=\frac{2m^{2}}{\sin^{2}\theta}\left|U^{\prime}(\theta)-\frac{\cos\theta}{\sin\theta}U(\theta)\right|^{2}+\frac{1}{\sin^{4}\theta}|U^{\prime}(\theta)\sin\theta\cos\theta-m^{2}U(\theta)|^{2}\geq 0.

If u=U(θ)eimφL2(S2)u=U(\theta)e^{im\varphi}\in L^{2}(S^{2}), then (u,Ynm)L2(S2)=0(u,Y_{n}^{m^{\prime}})_{L^{2}(S^{2})}=0 for mmm^{\prime}\neq m. In particular, if m0m\neq 0, then uL02(S2)u\in L_{0}^{2}(S^{2}) and we can apply (2.4)–(2.6) to uu. Also, if u=U(θ)eimφH1(S2)u=U(\theta)e^{im\varphi}\in H^{1}(S^{2}) for mm\in\mathbb{Z}, then UHloc1(0,π)C(0,π)U\in H_{loc}^{1}(0,\pi)\subset C(0,\pi) by (2.7). We use these facts without mention.

For a function uu on S2S^{2} and mm\in\mathbb{Z} we define a function 𝒫mu\mathcal{P}_{m}u on S2S^{2} by

(2.8) 𝒫mu(θ,φ)=eimφ2π02πu(θ,ϕ)eimϕ𝑑ϕ.\displaystyle\mathcal{P}_{m}u(\theta,\varphi)=\frac{e^{im\varphi}}{2\pi}\int_{0}^{2\pi}u(\theta,\phi)e^{-im\phi}\,d\phi.

Note that 𝒫mu=u\mathcal{P}_{m}u=u and 𝒫mu=0\mathcal{P}_{m^{\prime}}u=0 for mmm^{\prime}\neq m if u=U(θ)eimφu=U(\theta)e^{im\varphi}.

The following results are shown in [32, Section 2].

Lemma 2.1.

If u=U(θ)eimφC(S2)u=U(\theta)e^{im\varphi}\in C(S^{2}) with m{0}m\in\mathbb{Z}\setminus\{0\}, then U(0)=U(π)=0U(0)=U(\pi)=0.

Lemma 2.2.

For θ1,θ2[0,π]\theta_{1},\theta_{2}\in[0,\pi] with θ1θ2\theta_{1}\leq\theta_{2} let

(2.9) S2(θ1,θ2)={(x1,x2,x3)S2x3=cosθ,θ(θ1,θ2)}.\displaystyle S^{2}(\theta_{1},\theta_{2})=\{(x_{1},x_{2},x_{3})\in S^{2}\mid x_{3}=\cos\theta,\,\theta\in(\theta_{1},\theta_{2})\}.

Then for each mm\in\mathbb{Z} we have

(2.10) 𝒫muL2(S2(θ1,θ2))uL2(S2(θ1,θ2)),uL2(S2(θ1,θ2)),𝒫muL2(S2(θ1,θ2))uL2(S2(θ1,θ2)),uH1(S2(θ1,θ2)).\displaystyle\begin{aligned} \|\mathcal{P}_{m}u\|_{L^{2}(S^{2}(\theta_{1},\theta_{2}))}&\leq\|u\|_{L^{2}(S^{2}(\theta_{1},\theta_{2}))},&\quad&u\in L^{2}(S^{2}(\theta_{1},\theta_{2})),\\ \|\nabla\mathcal{P}_{m}u\|_{L^{2}(S^{2}(\theta_{1},\theta_{2}))}&\leq\|\nabla u\|_{L^{2}(S^{2}(\theta_{1},\theta_{2}))},&\quad&u\in H^{1}(S^{2}(\theta_{1},\theta_{2})).\end{aligned}
Lemma 2.3.

For m{0}m\in\mathbb{Z}\setminus\{0\} let u=U(θ)eimφH1(S2)u=U(\theta)e^{im\varphi}\in H^{1}(S^{2}). Then

(2.11) uL(S2)2=UL(0,π)21π|m|(Δ)1/2uL2(S2)2.\displaystyle\|u\|_{L^{\infty}(S^{2})}^{2}=\|U\|_{L^{\infty}(0,\pi)}^{2}\leq\frac{1}{\pi|m|}\|(-\Delta)^{1/2}u\|_{L^{2}(S^{2})}^{2}.
Lemma 2.4.

Let μ(1,1)\mu\in(-1,1) and θμ=arccosμ(0,π)\theta_{\mu}=\arccos\mu\in(0,\pi). Then

(2.12) θ1θ2|U(θ)sinθU(θμ)sinθμμcosθ|2𝑑θ16πsin2θμuL2(S2(θ1,θ2))2\displaystyle\int_{\theta_{1}}^{\theta_{2}}\left|\frac{U(\theta)\sqrt{\sin\theta}-U(\theta_{\mu})\sqrt{\sin\theta_{\mu}}}{\mu-\cos\theta}\right|^{2}\,d\theta\leq\frac{16}{\pi\sin^{2}\theta_{\mu}}\|\nabla u\|_{L^{2}(S^{2}(\theta_{1},\theta_{2}))}^{2}

for all θ1[0,θμ]\theta_{1}\in[0,\theta_{\mu}], θ2[θμ,π]\theta_{2}\in[\theta_{\mu},\pi], and u=U(θ)eimφH1(S2(θ1,θ2))u=U(\theta)e^{im\varphi}\in H^{1}(S^{2}(\theta_{1},\theta_{2})) with m{0}m\in\mathbb{Z}\setminus\{0\}, where S2(θ1,θ2)S^{2}(\theta_{1},\theta_{2}) is given by (2.9).

Let us give interpolation type inequalities for u=U(θ)eimφu=U(\theta)e^{im\varphi}.

Lemma 2.5.

Let β(0,π/2)\beta\in(0,\pi/2) and u=U(θ)eimφH1(S2)u=U(\theta)e^{im\varphi}\in H^{1}(S^{2}) with m{0}m\in\mathbb{Z}\setminus\{0\}. Then

(2.13) |U(θ)|2\displaystyle|U(\theta)|^{2} 12πsinβuL2(S2)(2(Δ)1/2uL2(S2)+1π2βuL2(S2)),\displaystyle\leq\frac{1}{2\pi\sin\beta}\|u\|_{L^{2}(S^{2})}\left(2\|(-\Delta)^{1/2}u\|_{L^{2}(S^{2})}+\frac{1}{\pi-2\beta}\|u\|_{L^{2}(S^{2})}\right),
(2.14) |U(θ)|2\displaystyle|U(\theta)|^{2} 12πsin2βMsinθuL2(S2)(2(Δ)1/2uL2(S2)+1π2βuL2(S2))\displaystyle\leq\frac{1}{2\pi\sin^{2}\beta}\|M_{\sin\theta}u\|_{L^{2}(S^{2})}\left(2\|(-\Delta)^{1/2}u\|_{L^{2}(S^{2})}+\frac{1}{\pi-2\beta}\|u\|_{L^{2}(S^{2})}\right)

for all θ(β,πβ)\theta\in(\beta,\pi-\beta), where Msinθu=sinθU(θ)eimφM_{\sin\theta}u=\sin\theta\,U(\theta)e^{im\varphi}.

Proof.

By the mean value theorem for integrals, we have

βπβ|U(ϑ)|2sinϑdϑ=(π2β)|U(θβ)|2sinθβ\displaystyle\int_{\beta}^{\pi-\beta}|U(\vartheta)|^{2}\sin\vartheta\,d\vartheta=(\pi-2\beta)|U(\theta_{\beta})|^{2}\sin\theta_{\beta}

with some θβ(β,πβ)\theta_{\beta}\in(\beta,\pi-\beta). Then we observe that

(2.15) |U(θβ)|2\displaystyle|U(\theta_{\beta})|^{2} =1(π2β)sinθββπβ|U(ϑ)|2sinϑdϑ12π(π2β)sinβuL2(S2)2\displaystyle=\frac{1}{(\pi-2\beta)\sin\theta_{\beta}}\int_{\beta}^{\pi-\beta}|U(\vartheta)|^{2}\sin\vartheta\,d\vartheta\leq\frac{1}{2\pi(\pi-2\beta)\sin\beta}\|u\|_{L^{2}(S^{2})}^{2}

by sinθβsinβ\sin\theta_{\beta}\geq\sin\beta and (2.7). Let θ(β,πβ)\theta\in(\beta,\pi-\beta). Then

(2.16) |U(θ)|2|U(θβ)|2=θβθddϑ|U(ϑ)|2𝑑ϑ2βπβ|U(ϑ)||U(ϑ)|𝑑ϑ2(βπβ|U(ϑ)|2sinϑ𝑑ϑ)1/2(βπβ|U(ϑ)|2sinϑdϑ)1/2.\displaystyle\begin{aligned} |U(\theta)|^{2}-|U(\theta_{\beta})|^{2}&=\int_{\theta_{\beta}}^{\theta}\frac{d}{d\vartheta}|U(\vartheta)|^{2}\,d\vartheta\leq 2\int_{\beta}^{\pi-\beta}|U(\vartheta)||U^{\prime}(\vartheta)|\,d\vartheta\\ &\leq 2\left(\int_{\beta}^{\pi-\beta}\frac{|U(\vartheta)|^{2}}{\sin\vartheta}\,d\vartheta\right)^{1/2}\left(\int_{\beta}^{\pi-\beta}|U^{\prime}(\vartheta)|^{2}\sin\vartheta\,d\vartheta\right)^{1/2}.\end{aligned}

Moreover, by sinϑsinβ\sin\vartheta\geq\sin\beta for ϑ(β,πβ)\vartheta\in(\beta,\pi-\beta), (2.4), and (2.7),

(2.19) βπβ|U(ϑ)|2sinϑ𝑑ϑ1sin2ββπβ|U(ϑ)|2sinϑdϑ12πsin2βuL2(S2)2,βπβ|U(ϑ)|2sinϑdϑ12πuL2(S2)2=12π(Δ)1/2uL2(S2)2.\displaystyle\begin{gathered}\int_{\beta}^{\pi-\beta}\frac{|U(\vartheta)|^{2}}{\sin\vartheta}\,d\vartheta\leq\frac{1}{\sin^{2}\beta}\int_{\beta}^{\pi-\beta}|U(\vartheta)|^{2}\sin\vartheta\,d\vartheta\leq\frac{1}{2\pi\sin^{2}\beta}\|u\|_{L^{2}(S^{2})}^{2},\\ \int_{\beta}^{\pi-\beta}|U^{\prime}(\vartheta)|^{2}\sin\vartheta\,d\vartheta\leq\frac{1}{2\pi}\|\nabla u\|_{L^{2}(S^{2})}^{2}=\frac{1}{2\pi}\|(-\Delta)^{1/2}u\|_{L^{2}(S^{2})}^{2}.\end{gathered}

Hence (2.13) follows from (2.15)–(2.19). Also, for Msinθu=sinθU(θ)eimφM_{\sin\theta}u=\sin\theta\,U(\theta)e^{im\varphi},

βπβ|U(ϑ)|2sinϑdϑ\displaystyle\int_{\beta}^{\pi-\beta}|U(\vartheta)|^{2}\sin\vartheta\,d\vartheta 1sinββπβ|sinϑU(ϑ)||U(ϑ)|sinϑdϑ\displaystyle\leq\frac{1}{\sin\beta}\int_{\beta}^{\pi-\beta}|\sin\vartheta\,U(\vartheta)||U(\vartheta)|\sin\vartheta\,d\vartheta
12πsinβMsinθuL2(S2)uL2(S2)\displaystyle\leq\frac{1}{2\pi\sin\beta}\|M_{\sin\theta}u\|_{L^{2}(S^{2})}\|u\|_{L^{2}(S^{2})}

by sinϑsinβ\sin\vartheta\geq\sin\beta for ϑ(β,πβ)\vartheta\in(\beta,\pi-\beta) and (2.7), and

βπβ|U(ϑ)|2sinϑ𝑑ϑ1sin4ββπβ|sinϑU(ϑ)|2sinϑdϑ12πsin4βMsinθuL2(S2)2.\displaystyle\int_{\beta}^{\pi-\beta}\frac{|U(\vartheta)|^{2}}{\sin\vartheta}\,d\vartheta\leq\frac{1}{\sin^{4}\beta}\int_{\beta}^{\pi-\beta}|\sin\vartheta\,U(\vartheta)|^{2}\sin\vartheta\,d\vartheta\leq\frac{1}{2\pi\sin^{4}\beta}\|M_{\sin\theta}u\|_{L^{2}(S^{2})}^{2}.

We apply these estimates, sinθβsinβ\sin\theta_{\beta}\geq\sin\beta, and the second line of (2.19) to (2.15) and (2.16), and then combine the resulting inequalities to obtain (2.14). ∎

Lemma 2.6.

For m{0}m\in\mathbb{Z}\setminus\{0\} let u=U(θ)eimφH1(S2)u=U(\theta)e^{im\varphi}\in H^{1}(S^{2}). Also, let θμ(0,π)\theta_{\mu}\in(0,\pi) and

0<ε12sinθμ<12min{θμ,πθμ}.\displaystyle 0<\varepsilon\leq\frac{1}{2}\sin\theta_{\mu}<\frac{1}{2}\min\{\theta_{\mu},\pi-\theta_{\mu}\}.

Then there exists a constant C>0C>0 independent of mm, uu, θμ\theta_{\mu}, and ε\varepsilon such that

(2.20) |U(θ)|2\displaystyle|U(\theta)|^{2} CsinθμuL2(S2)(Δ)1/2uL2(S2),\displaystyle\leq\frac{C}{\sin\theta_{\mu}}\|u\|_{L^{2}(S^{2})}\|(-\Delta)^{1/2}u\|_{L^{2}(S^{2})},
(2.21) |U(θ)|2\displaystyle|U(\theta)|^{2} Csin2θμMsinθuL2(S2)(Δ)1/2uL2(S2)\displaystyle\leq\frac{C}{\sin^{2}\theta_{\mu}}\|M_{\sin\theta}u\|_{L^{2}(S^{2})}\|(-\Delta)^{1/2}u\|_{L^{2}(S^{2})}

for all θ(θμε,θμ+ε)\theta\in(\theta_{\mu}-\varepsilon,\theta_{\mu}+\varepsilon), where Msinθu=sinθU(θ)eimφM_{\sin\theta}u=\sin\theta\,U(\theta)e^{im\varphi}.

Proof.

When θμ(0,π/6]\theta_{\mu}\in(0,\pi/6], we set β=θμε\beta=\theta_{\mu}-\varepsilon. Then since

θμ+ε32θμπ4,πβπ2βπ2θμ2π3,\displaystyle\theta_{\mu}+\varepsilon\leq\frac{3}{2}\theta_{\mu}\leq\frac{\pi}{4},\quad\pi-\beta\geq\pi-2\beta\geq\pi-2\theta_{\mu}\geq\frac{2\pi}{3},

we have (θμε,θμ+ε)(β,πβ)(\theta_{\mu}-\varepsilon,\theta_{\mu}+\varepsilon)\subset(\beta,\pi-\beta). Also, by Taylor’s theorem for sinθ\sin\theta at θ=θ0\theta=\theta_{0},

sinβ=sin(θμε)sinθμε12sinθμ.\displaystyle\sin\beta=\sin(\theta_{\mu}-\varepsilon)\geq\sin\theta_{\mu}-\varepsilon\geq\frac{1}{2}\sin\theta_{\mu}.

On the other hand, when θμ[π/6,π/2]\theta_{\mu}\in[\pi/6,\pi/2],

θμεθμ2π12,θμ+ε32θμ34π1112π.\displaystyle\theta_{\mu}-\varepsilon\geq\frac{\theta_{\mu}}{2}\geq\frac{\pi}{12},\quad\theta_{\mu}+\varepsilon\leq\frac{3}{2}\theta_{\mu}\leq\frac{3}{4}\pi\leq\frac{11}{12}\pi.

Hence we set β=π/12\beta=\pi/12 to get (θμε,θμ+ε)(β,πβ)(\theta_{\mu}-\varepsilon,\theta_{\mu}+\varepsilon)\subset(\beta,\pi-\beta) and

π2β=56π,sinβ=sinπ12sinπ12sinθμ.\displaystyle\pi-2\beta=\frac{5}{6}\pi,\quad\sin\beta=\sin\frac{\pi}{12}\geq\sin\frac{\pi}{12}\sin\theta_{\mu}.

Therefore, in both cases θμ(0,π/6]\theta_{\mu}\in(0,\pi/6] and θμ[π/6,π/2]\theta_{\mu}\in[\pi/6,\pi/2], we have (2.20) and (2.21) by applying (2.13) and (2.14) with the above β\beta and using (2.6). Similarly, we can prove (2.20) and (2.21) when θμπ/2\theta_{\mu}\geq\pi/2 by considering the cases θμ[π/2,5π/6]\theta_{\mu}\in[\pi/2,5\pi/6] and θμ[5π/6,π)\theta_{\mu}\in[5\pi/6,\pi) separately. ∎

3. Analysis of the linearized operator

In this section we study the linearized operator for the two-jet Kolmogorov type flow.

3.1. Settings and basic results

Let II be the identity operator and MfM_{f} the multiplication operator by a function ff on S2S^{2}, i.e. Mfu=fuM_{f}u=fu for a function uu on S2S^{2}. We define linear operators AA and Λ\Lambda on L02(S2)L_{0}^{2}(S^{2}) by

A\displaystyle A =Δ+2,\displaystyle=\Delta+2, DL02(S2)(A)\displaystyle\quad D_{L_{0}^{2}(S^{2})}(A) =L02(S2)H2(S2),\displaystyle=L_{0}^{2}(S^{2})\cap H^{2}(S^{2}),
Λ\displaystyle\Lambda =iφMcosθB,\displaystyle=-i\partial_{\varphi}M_{\cos\theta}B, DL02(S2)(Λ)\displaystyle\quad D_{L_{0}^{2}(S^{2})}(\Lambda) ={uL02(S2)φMcosθuL02(S2)},\displaystyle=\{u\in L_{0}^{2}(S^{2})\mid\partial_{\varphi}M_{\cos\theta}u\in L_{0}^{2}(S^{2})\},

where B=I+6Δ1B=I+6\Delta^{-1} on L02(S2)L_{0}^{2}(S^{2}). Then AA is self-adjoint in L02(S2)L_{0}^{2}(S^{2}). Also, Λ\Lambda is densely defined in L02(S2)L_{0}^{2}(S^{2}) since its domain contains the dense subspace L02(S2)H1(S2)L_{0}^{2}(S^{2})\cap H^{1}(S^{2}) of L02(S2)L_{0}^{2}(S^{2}). Since McosθBM_{\cos\theta}B is a bounded operator on L2(S2)L^{2}(S^{2}) (note that it does not map L02(S2)L_{0}^{2}(S^{2}) into itself) and φ\partial_{\varphi} is a closed operator from L2(S2)L^{2}(S^{2}) into L02(S2)L_{0}^{2}(S^{2}), we see that Λ\Lambda is closed in L02(S2)L_{0}^{2}(S^{2}). Moreover, Λ\Lambda is AA-compact in L02(S2)L_{0}^{2}(S^{2}) since H2(S2)H^{2}(S^{2}) is compactly embedded in H1(S2)H^{1}(S^{2}) and

(3.1) uH2(S2)CΔuL2(S2)C(AuL2(S2)+2uL2(S2)),uD(A)\displaystyle\|u\|_{H^{2}(S^{2})}\leq C\|\Delta u\|_{L^{2}(S^{2})}\leq C\left(\|Au\|_{L^{2}(S^{2})}+2\|u\|_{L^{2}(S^{2})}\right),\quad u\in D(A)

by the elliptic regularity theorem. For n1n\geq 1 and |m|n|m|\leq n we have

(3.2) ΔYnm=λnYnm,BYnm=(16λn)Ynm,φYnm=imYnm.\displaystyle\Delta Y_{n}^{m}=-\lambda_{n}Y_{n}^{m},\quad BY_{n}^{m}=\left(1-\frac{6}{\lambda_{n}}\right)Y_{n}^{m},\quad\partial_{\varphi}Y_{n}^{m}=imY_{n}^{m}.

By this equality and (2.2), we see that (here Y|m|1m0Y_{|m|-1}^{m}\equiv 0)

(3.3) AYnm=(λn+2)u,ΛYnm=m(16λn)(anmYn1m+an+1mYn+1m).\displaystyle AY_{n}^{m}=(-\lambda_{n}+2)u,\quad\Lambda Y_{n}^{m}=m\left(1-\frac{6}{\lambda_{n}}\right)(a_{n}^{m}Y_{n-1}^{m}+a_{n+1}^{m}Y_{n+1}^{m}).

In particular, ΛY2m=0\Lambda Y_{2}^{m}=0 for |m|=0,1,2|m|=0,1,2. Let

𝒳={uL02(S2)(u,Yn0)L2(S2)=(u,Y1m)L2(S2)=0,n1,|m|=0,1}.\displaystyle\mathcal{X}=\{u\in L_{0}^{2}(S^{2})\mid(u,Y_{n}^{0})_{L^{2}(S^{2})}=(u,Y_{1}^{m})_{L^{2}(S^{2})}=0,\,n\geq 1,\,|m|=0,1\}.

Then 𝒳\mathcal{X} is a closed subspace of L02(S2)L_{0}^{2}(S^{2}) and each u𝒳u\in\mathcal{X} is expressed as

(3.4) u=m{0}nNm(u,Ynm)L2(S2)Ynm,Nm=max{2,|m|},\displaystyle u=\sum_{m\in\mathbb{Z}\setminus\{0\}}\sum_{n\geq N_{m}}(u,Y_{n}^{m})_{L^{2}(S^{2})}Y_{n}^{m},\quad N_{m}=\max\{2,|m|\},

and we observe by (3.3) (in particular ΛY2m=0\Lambda Y_{2}^{m}=0) and (3.4) that 𝒳\mathcal{X} is invariant under the actions of AA and Λ\Lambda. For the sake of simplicity, we write AA and Λ\Lambda for their restrictions on 𝒳\mathcal{X} with domains D𝒳(A)=𝒳DL02(S2)(A)D_{\mathcal{X}}(A)=\mathcal{X}\cap D_{L_{0}^{2}(S^{2})}(A) and D𝒳(Λ)=𝒳DL02(S2)(Λ)D_{\mathcal{X}}(\Lambda)=\mathcal{X}\cap D_{L_{0}^{2}(S^{2})}(\Lambda).

The follows lemmas are proved in [32, Section 5].

Lemma 3.1.

The operator AA is self-adjoint in 𝒳\mathcal{X} and satisfies

(3.5) (Au,u)L2(S2)4uL2(S2)2,uD𝒳(A).\displaystyle(-Au,u)_{L^{2}(S^{2})}\geq 4\|u\|_{L^{2}(S^{2})}^{2},\quad u\in D_{\mathcal{X}}(A).

Also, Λ\Lambda is densely defined, closed, and AA-compact in 𝒳\mathcal{X}.

Lemma 3.2.

For u𝒳u\in\mathcal{X} let u=u|m|=1,2(u,Y2m)L2(S2)Y2m\mathbb{Q}u=u-\sum_{|m|=1,2}(u,Y_{2}^{m})_{L^{2}(S^{2})}Y_{2}^{m}. Then

(3.6) 12uL2(S2)BuL2(S2)uL2(S2),u𝒳\displaystyle\frac{1}{2}\|\mathbb{Q}u\|_{L^{2}(S^{2})}\leq\|Bu\|_{L^{2}(S^{2})}\leq\|u\|_{L^{2}(S^{2})},\quad u\in\mathcal{X}

and the kernel of Λ\Lambda in 𝒳\mathcal{X} is N𝒳(Λ)=span{Y2m|m|=1,2}N_{\mathcal{X}}(\Lambda)=\mathrm{span}\{Y_{2}^{m}\mid|m|=1,2\}. Thus \mathbb{Q} is the orthogonal projection from 𝒳\mathcal{X} onto 𝒴=N𝒳(Λ)={u𝒳(u,Y2m)L2(S2)=0,|m|=1,2}\mathcal{Y}=N_{\mathcal{X}}(\Lambda)^{\perp}=\{u\in\mathcal{X}\mid(u,Y_{2}^{m})_{L^{2}(S^{2})}=0,\,|m|=1,2\}.

By Lemma 3.1 and a perturbation theory of semigroups (see [13]), the operator Lα=AiαΛL_{\alpha}=A-i\alpha\Lambda with domain D𝒳(Lα)=D𝒳(A)D_{\mathcal{X}}(L_{\alpha})=D_{\mathcal{X}}(A) generates an analytic semigroup {etLα}t0\{e^{tL_{\alpha}}\}_{t\geq 0} in 𝒳\mathcal{X} for each α\alpha\in\mathbb{R}. Our aim is to study the decay rate of etLαe^{tL_{\alpha}} in terms of α\alpha.

For m{0}m\in\mathbb{Z}\setminus\{0\}, let 𝒳m=𝒫m𝒳\mathcal{X}_{m}=\mathcal{P}_{m}\mathcal{X} and 𝒴m=𝒫m𝒴\mathcal{Y}_{m}=\mathcal{P}_{m}\mathcal{Y} with 𝒫m\mathcal{P}_{m} given by (2.8). Then 𝒳m\mathcal{X}_{m} and 𝒴m\mathcal{Y}_{m} are closed subspaces of 𝒳\mathcal{X} and 𝒴\mathcal{Y}, respectively, and the orthogonal decompositions

𝒳=m{0}𝒳m,𝒴=m{0}𝒴m\displaystyle\mathcal{X}=\oplus_{m\in\mathbb{Z}\setminus\{0\}}\mathcal{X}_{m},\quad\mathcal{Y}=\oplus_{m\in\mathbb{Z}\setminus\{0\}}\mathcal{Y}_{m}

hold. Moreover, since u𝒳mu\in\mathcal{X}_{m} is of the form

(3.7) u=nNm(u,Ynm)L2(S2)Ynm,Nm=max{2,|m|},\displaystyle u=\sum_{n\geq N_{m}}(u,Y_{n}^{m})_{L^{2}(S^{2})}Y_{n}^{m},\quad N_{m}=\max\{2,|m|\},

the subspace 𝒳m\mathcal{X}_{m} is invariant under the actions of AA and Λ\Lambda by (3.3) (in particular ΛY2m=0\Lambda Y_{2}^{m}=0). Hence Lα=AiαΛL_{\alpha}=A-i\alpha\Lambda is diagonalized as

Lα=m{0}Lα,m,Lα,m=Lα|𝒳m.\displaystyle L_{\alpha}=\oplus_{m\in\mathbb{Z}\setminus\{0\}}L_{\alpha,m},\quad L_{\alpha,m}=L_{\alpha}|_{\mathcal{X}_{m}}.

Moreover, we see by (2.2), (3.2), and (3.7) that Lα,mL_{\alpha,m} is expressed as

Lα,m=AmiαmΛm,D𝒳m(Lα,m)=𝒳mH2(S2),\displaystyle L_{\alpha,m}=A_{m}-i\alpha m\Lambda_{m},\quad D_{\mathcal{X}_{m}}(L_{\alpha,m})=\mathcal{X}_{m}\cap H^{2}(S^{2}),

where AmA_{m} and Λm\Lambda_{m} are linear operators on 𝒳m\mathcal{X}_{m} defined by

Am\displaystyle A_{m} =A|𝒳m=(Δ+2)|𝒳m,\displaystyle=A|_{\mathcal{X}_{m}}=(\Delta+2)|_{\mathcal{X}_{m}}, D𝒳m(Am)\displaystyle\quad D_{\mathcal{X}_{m}}(A_{m}) =𝒳mH2(S2),\displaystyle=\mathcal{X}_{m}\cap H^{2}(S^{2}),
Λm\displaystyle\Lambda_{m} =McosθB|𝒳m=Mcosθ(I+6Δ1)|𝒳m,\displaystyle=M_{\cos\theta}B|_{\mathcal{X}_{m}}=M_{\cos\theta}(I+6\Delta^{-1})|_{\mathcal{X}_{m}}, D𝒳m(Λm)\displaystyle\quad D_{\mathcal{X}_{m}}(\Lambda_{m}) =𝒳m.\displaystyle=\mathcal{X}_{m}.

Note that, by |cosθ|1|\cos\theta|\leq 1, (3.2), (3.7), and 016/λn10\leq 1-6/\lambda_{n}\leq 1 for n2n\geq 2,

(3.8) ΛmuL2(S2)BuL2(S2)uL2(S2),u𝒳m.\displaystyle\|\Lambda_{m}u\|_{L^{2}(S^{2})}\leq\|Bu\|_{L^{2}(S^{2})}\leq\|u\|_{L^{2}(S^{2})},\quad u\in\mathcal{X}_{m}.

Also, 𝒴m\mathcal{Y}_{m} is a closed subspace of 𝒳m\mathcal{X}_{m} and each u𝒴mu\in\mathcal{Y}_{m} is of the form

(3.9) u=nNm(u,Ynm)L2(S2)Ynm,Nm=max{3,|m|}.\displaystyle u=\sum_{n\geq N^{\prime}_{m}}(u,Y_{n}^{m})_{L^{2}(S^{2})}Y_{n}^{m},\quad N^{\prime}_{m}=\max\{3,|m|\}.

Let m\mathbb{Q}_{m} be the orthogonal projection from 𝒳m\mathcal{X}_{m} onto 𝒴m\mathcal{Y}_{m}. Then

(3.10) mu=u={u(u,Y2m)L2(S2)Y2m,|m|=1,2,u,|m|3\displaystyle\mathbb{Q}_{m}u=\mathbb{Q}u=\begin{cases}u-(u,Y_{2}^{m})_{L^{2}(S^{2})}Y_{2}^{m},&|m|=1,2,\\ u,&|m|\geq 3\end{cases}

for u𝒳mu\in\mathcal{X}_{m} by (3.7) and (3.9). We intend to derive decay estimates for the semigroup generated by Lα,mL_{\alpha,m} in 𝒳m\mathcal{X}_{m} by applying abstract results given in Section 4. To this end, we verify Assumptions 4.14.3 and 4.6 for AmA_{m} and Λm\Lambda_{m}.

Lemma 3.3.

Let m{0}m\in\mathbb{Z}\setminus\{0\}. Then AmA_{m} and Λm\Lambda_{m} satisfy Assumptions 4.14.3 in 𝒳m\mathcal{X}_{m} with B1,m=McosθB_{1,m}=M_{\cos\theta} on m=L2(S2)\mathcal{H}_{m}=L^{2}(S^{2}) and B2,m=B|𝒳m=(I+6Δ1)|𝒳mB_{2,m}=B|_{\mathcal{X}_{m}}=(I+6\Delta^{-1})|_{\mathcal{X}_{m}} on 𝒳m\mathcal{X}_{m}.

Note that B2,mB_{2,m} maps 𝒳m\mathcal{X}_{m} into itself by (3.2) and (3.7). Also, as in (2.3), we set

(3.11) (Am)su=nNm(λn2)s(u,Ynm)L2(S2)Ynm,Nm=max{2,|m|}\displaystyle(-A_{m})^{s}u=\sum_{n\geq N_{m}}(\lambda_{n}-2)^{s}(u,Y_{n}^{m})_{L^{2}(S^{2})}Y_{n}^{m},\quad N_{m}=\max\{2,|m|\}

for ss\in\mathbb{R} and u𝒳mu\in\mathcal{X}_{m} of the form (3.7).

Proof.

Since 𝒳m\mathcal{X}_{m} is a closed subspace of 𝒳\mathcal{X}, Lemma 3.1 implies that AmA_{m} satisfies Assumption 4.1 in 𝒳m\mathcal{X}_{m}. The operator Λm\Lambda_{m} is densely defined, closed, and AmA_{m}-compact in 𝒳m\mathcal{X}_{m} since it is a bounded operator on 𝒳m\mathcal{X}_{m}, the inequality (3.1) holds, and the embedding H2(S2)L2(S2)H^{2}(S^{2})\hookrightarrow L^{2}(S^{2}) is compact. If Λmu=0\Lambda_{m}u=0 for u𝒳mu\in\mathcal{X}_{m}, then B2,mu=0B_{2,m}u=0 since B1,m=McosθB_{1,m}=M_{\cos\theta} is injective on L2(S2)L^{2}(S^{2}). Thus, by (3.6), (3.7), and B2,mY2m=0B_{2,m}Y_{2}^{m}=0 for |m|=1,2|m|=1,2, we see that

(3.12) N𝒳m(Λm)=N𝒳m(B2,m)={{cY2mc},|m|=1,2,{0},|m|3\displaystyle N_{\mathcal{X}_{m}}(\Lambda_{m})=N_{\mathcal{X}_{m}}(B_{2,m})=\begin{cases}\{cY_{2}^{m}\mid c\in\mathbb{C}\},&|m|=1,2,\\ \{0\},&|m|\geq 3\end{cases}

and we get 𝒴m=N𝒳m(Λm)\mathcal{Y}_{m}=N_{\mathcal{X}_{m}}(\Lambda_{m})^{\perp} by (3.9) and (3.12). Also, we have mAmAmm\mathbb{Q}_{m}A_{m}\subset A_{m}\mathbb{Q}_{m} in 𝒳m\mathcal{X}_{m} since Y2mY_{2}^{m} is smooth on S2S^{2}, the relations (3.3) and (3.10) hold, and AmA_{m} is self-adjoint in 𝒳m\mathcal{X}_{m}. Since u𝒴mu\in\mathcal{Y}_{m} is of the form (3.9), we see by (3.2), (3.3), (3.11), and 16/λn1/21-6/\lambda_{n}\geq 1/2 for n3n\geq 3 that the inequalities (4.2) and (4.3) hold for AmA_{m} and B2,mB_{2,m} with constant C=1/2C=1/2. Hence Assumptions 4.2 and 4.3 are valid. ∎

Lemma 3.4.

For m{0}m\in\mathbb{Z}\setminus\{0\} and k=0,1k=0,1 we have

(3.13) (Δ)k/2B2,muL2(S2)2\displaystyle\|(-\Delta)^{k/2}B_{2,m}u\|_{L^{2}(S^{2})}^{2} (Δ)k/2uL2(S2)2,\displaystyle\leq\|(-\Delta)^{k/2}u\|_{L^{2}(S^{2})}^{2}, u\displaystyle\quad u 𝒳mHk(S2),\displaystyle\in\mathcal{X}_{m}\cap H^{k}(S^{2}),
(3.14) 14uL2(S2)2\displaystyle\frac{1}{4}\|u\|_{L^{2}(S^{2})}^{2} B2,muL2(S2)2,\displaystyle\leq\|B_{2,m}u\|_{L^{2}(S^{2})}^{2}, u\displaystyle\quad u 𝒴m.\displaystyle\in\mathcal{Y}_{m}.
Proof.

Since u𝒳mu\in\mathcal{X}_{m} is of the form (3.7), we have (3.13) by (2.3), (3.2), and |16/λn|1|1-6/\lambda_{n}|\leq 1 for n2n\geq 2. Also, if u𝒴mu\in\mathcal{Y}_{m}, then uu is of the form (3.9) and thus (3.14) follows from (3.2) and 16/λn1/21-6/\lambda_{n}\geq 1/2 for n3n\geq 3. ∎

The next result is crucial for the proof of Lemma 3.9 below.

Lemma 3.5.

Let m{0}m\in\mathbb{Z}\setminus\{0\}. Then Λm\Lambda_{m} in 𝒳m\mathcal{X}_{m} has no eigenvalues in {0}\mathbb{C}\setminus\{0\}.

Proof.

The statement follows from [32, Theorem 4.1]. ∎

3.2. Verification of Assumption 4.6

Next we show that AmA_{m} and Λm\Lambda_{m} satisfy Assumption 4.6 in several long steps. In what follows, we frequently use the fact that each function uu in 𝒳m=𝒫m𝒳\mathcal{X}_{m}=\mathcal{P}_{m}\mathcal{X} is of the form u=U(θ)eimφu=U(\theta)e^{im\varphi} (see (2.8)).

Lemma 3.6.

Let m{0}m\in\mathbb{Z}\setminus\{0\}. For u𝒳mH1(S2)u\in\mathcal{X}_{m}\cap H^{1}(S^{2}) we have

(3.15) 23(Δ)1/2uL2(S2)2(Am)1/2uL2(S2)2(Δ)1/2uL2(S2)2.\displaystyle\frac{2}{3}\|(-\Delta)^{1/2}u\|_{L^{2}(S^{2})}^{2}\leq\|(-A_{m})^{1/2}u\|_{L^{2}(S^{2})}^{2}\leq\|(-\Delta)^{1/2}u\|_{L^{2}(S^{2})}^{2}.

Also, let B3,m=MsinθB_{3,m}=M_{\sin\theta} on m=L2(S2)\mathcal{B}_{m}=L^{2}(S^{2}). Then for uD𝒳m(Am)u\in D_{\mathcal{X}_{m}}(A_{m}) we have

(3.16) |Im(Amu,Λmu)L2(S2)|62(Am)1/2uL2(S2)B3,muL2(S2).\displaystyle\bigl{|}\mathrm{Im}(A_{m}u,\Lambda_{m}u)_{L^{2}(S^{2})}\bigr{|}\leq\frac{\sqrt{6}}{2}\|(-A_{m})^{1/2}u\|_{L^{2}(S^{2})}\|B_{3,m}u\|_{L^{2}(S^{2})}.
Proof.

For u𝒳mH1(S2)u\in\mathcal{X}_{m}\cap H^{1}(S^{2}) we have (3.15) by (2.3) and (3.11) since uu is of the form (3.7) and 2λn/3λn2λn2\lambda_{n}/3\leq\lambda_{n}-2\leq\lambda_{n} for n2n\geq 2.

Next let uD𝒳m(Am)u\in D_{\mathcal{X}_{m}}(A_{m}) and v=Δ1uv=\Delta^{-1}u. Then

(Amu,Λmu)L2(S2)=(Δu,x3u)L2(S2)+6(Δu,x3v)L2(S2)+2(u,x3u)L2(S2)+12(u,x3v)L2(S2)(A_{m}u,\Lambda_{m}u)_{L^{2}(S^{2})}=(\Delta u,x_{3}u)_{L^{2}(S^{2})}+6(\Delta u,x_{3}v)_{L^{2}(S^{2})}\\ +2(u,x_{3}u)_{L^{2}(S^{2})}+12(u,x_{3}v)_{L^{2}(S^{2})}

with x3=cosθx_{3}=\cos\theta on S2S^{2}. Moreover, we have

(Δu,x3u)L2(S2)\displaystyle(\Delta u,x_{3}u)_{L^{2}(S^{2})} =(u,x3u)L2(S2)(u,ux3)L2(S2),\displaystyle=-(\nabla u,x_{3}\nabla u)_{L^{2}(S^{2})}-(\nabla u,u\nabla x_{3})_{L^{2}(S^{2})},
(Δu,x3v)L2(S2)\displaystyle(\Delta u,x_{3}v)_{L^{2}(S^{2})} =2(u,x3v)L2(S2)+2(u,x3v)L2(S2)+(u,x3u)L2(S2)\displaystyle=-2(u,x_{3}v)_{L^{2}(S^{2})}+2(u,\nabla x_{3}\cdot\nabla v)_{L^{2}(S^{2})}+(u,x_{3}u)_{L^{2}(S^{2})}

by applying integration by parts once or twice, where we also used Δx3=2x3\Delta x_{3}=-2x_{3} by (5.1) for x3=cosθx_{3}=\cos\theta and Δv=u\Delta v=u in the second line. Hence

(Amu,Λmu)L2(S2)=(u,x3u)L2(S2)+8(u,x3u)L2(S2)(u,ux3)L2(S2)+12(u,x3v)L2(S2).(A_{m}u,\Lambda_{m}u)_{L^{2}(S^{2})}=-(\nabla u,x_{3}\nabla u)_{L^{2}(S^{2})}+8(u,x_{3}u)_{L^{2}(S^{2})}\\ -(\nabla u,u\nabla x_{3})_{L^{2}(S^{2})}+12(u,\nabla x_{3}\cdot\nabla v)_{L^{2}(S^{2})}.

Moreover, since

(u,x3u)L2(S2)+8(u,x3u)L2(S2)=S2x3(|u|2+8|u|2)𝑑2,\displaystyle-(\nabla u,x_{3}\nabla u)_{L^{2}(S^{2})}+8(u,x_{3}u)_{L^{2}(S^{2})}=\int_{S^{2}}x_{3}(-|\nabla u|^{2}+8|u|^{2})\,d\mathcal{H}^{2}\in\mathbb{R},
Im(u,x3v)L2(S2)=ImS2ux3v¯d2=Im(v,ux3)L2(S2),\displaystyle\mathrm{Im}(u,\nabla x_{3}\cdot\nabla v)_{L^{2}(S^{2})}=\mathrm{Im}\int_{S^{2}}u\nabla x_{3}\cdot\nabla\bar{v}\,d\mathcal{H}^{2}=-\mathrm{Im}(\nabla v,u\nabla x_{3})_{L^{2}(S^{2})},

we have

(3.17) |Im(Amu,Λmu)L2(S2)|=|Im((u+12v),ux3)L2(S2)|(u+12v)L2(S2)ux3L2(S2).\displaystyle\begin{aligned} |\mathrm{Im}(A_{m}u,\Lambda_{m}u)_{L^{2}(S^{2})}|&=|\mathrm{Im}(\nabla(u+12v),u\nabla x_{3})_{L^{2}(S^{2})}|\\ &\leq\|\nabla(u+12v)\|_{L^{2}(S^{2})}\|u\nabla x_{3}\|_{L^{2}(S^{2})}.\end{aligned}

Now we recall that v=Δ1uv=\Delta^{-1}u and uu is of the form (3.7) to get

(3.18) (u+12v)L2(S2)2\displaystyle\|\nabla(u+12v)\|_{L^{2}(S^{2})}^{2} =(Δ)1/2(I+12Δ1)uL2(S2)2(Δ)1/2uL2(S2)2\displaystyle=\|(-\Delta)^{1/2}(I+12\Delta^{-1})u\|_{L^{2}(S^{2})}^{2}\leq\|(-\Delta)^{1/2}u\|_{L^{2}(S^{2})}^{2}

by (2.3), (2.4), and |112/λn|1|1-12/\lambda_{n}|\leq 1 for n2n\geq 2. Also, since |x3|=sinθ|\nabla x_{3}|=\sin\theta on S2S^{2} by x3=cosθx_{3}=\cos\theta and (5.1), we have ux3L2(S2)=B3,muL2(S2)\|u\nabla x_{3}\|_{L^{2}(S^{2})}=\|B_{3,m}u\|_{L^{2}(S^{2})} with B3,m=MsinθB_{3,m}=M_{\sin\theta}. Thus we get (3.16) by this equality, (3.15), (3.17), and (3.18). ∎

Let us verify (4.8) and (4.10) in Assumption 4.6. We consider three cases for μ\mu\in\mathbb{R}:

|μ|>1,1κδ2|μ|1+κδ2,|μ|1κδ2.\displaystyle|\mu|>1,\quad 1-\kappa\delta^{2}\leq|\mu|\leq 1+\kappa\delta^{2},\quad|\mu|\leq 1-\kappa\delta^{2}.

Here κ>0\kappa>0 is a small constant given in Lemma 3.8 below and δ(0,1]\delta\in(0,1]. In what follows, we write CC for a general positive constant independent of mm, uu, μ\mu, and δ\delta. Also, we frequently use the expressions cosθ=x3\cos\theta=x_{3} on S2S^{2} and Λmu=x3B2,mu\Lambda_{m}u=x_{3}B_{2,m}u for u𝒳mu\in\mathcal{X}_{m}.

Lemma 3.7.

There exists a constant C>0C>0 such that

(3.19) uL2(S2)2\displaystyle\|u\|_{L^{2}(S^{2})}^{2} C(|μ|1)2m(μΛm)uL2(S2)2,\displaystyle\leq\frac{C}{(|\mu|-1)^{2}}\|\mathbb{Q}_{m}(\mu-\Lambda_{m})u\|_{L^{2}(S^{2})}^{2},
(3.20) MsinθuL2(S2)2\displaystyle\|M_{\sin\theta}u\|_{L^{2}(S^{2})}^{2} C|μ|1m(μΛm)uL2(S2)2\displaystyle\leq\frac{C}{|\mu|-1}\|\mathbb{Q}_{m}(\mu-\Lambda_{m})u\|_{L^{2}(S^{2})}^{2}

for all m{0}m\in\mathbb{Z}\setminus\{0\}, u𝒴mu\in\mathcal{Y}_{m}, and μ\mu\in\mathbb{R} satisfying |μ|>1|\mu|>1.

Proof.

For u𝒴mu\in\mathcal{Y}_{m} let v=Δ1uv=\Delta^{-1}u and f=m(μΛm)uf=\mathbb{Q}_{m}(\mu-\Lambda_{m})u. Then

(f,B2,mu)L2(S2)\displaystyle(f,B_{2,m}u)_{L^{2}(S^{2})} =((μΛm)u,mB2,mu)L2(S2)=((μx3)B2,mu6μv,B2,mu)L2(S2)\displaystyle=\bigl{(}(\mu-\Lambda_{m})u,\mathbb{Q}_{m}B_{2,m}u\bigr{)}_{L^{2}(S^{2})}=\bigl{(}(\mu-x_{3})B_{2,m}u-6\mu v,B_{2,m}u\bigr{)}_{L^{2}(S^{2})}

since B2,mu𝒴mB_{2,m}u\in\mathcal{Y}_{m} by (3.2) and (3.9). Moreover,

(v,B2,mu)L2(S2)=(v,Δv)L2(S2)+6vL2(S2)2=vL2(S2)2+6vL2(S2)2\displaystyle(v,B_{2,m}u)_{L^{2}(S^{2})}=(v,\Delta v)_{L^{2}(S^{2})}+6\|v\|_{L^{2}(S^{2})}^{2}=-\|\nabla v\|_{L^{2}(S^{2})}^{2}+6\|v\|_{L^{2}(S^{2})}^{2}

by B2,mu=Δv+6vB_{2,m}u=\Delta v+6v and integration by parts. Hence

((μx3)B2,mu,B2,mu)L2(S2)+6μ(vL2(S2)26vL2(S2)2)=(f,B2,mu)L2(S2)\displaystyle\bigl{(}(\mu-x_{3})B_{2,m}u,B_{2,m}u\bigr{)}_{L^{2}(S^{2})}+6\mu\Bigl{(}\|\nabla v\|_{L^{2}(S^{2})}^{2}-6\|v\|_{L^{2}(S^{2})}^{2}\Bigr{)}=(f,B_{2,m}u)_{L^{2}(S^{2})}

and we divide both sides by μ\mu and take the real part to get

(3.21) S2(1x3μ)|B2,mu|2𝑑2+6(vL2(S2)26vL2(S2)2)=1μRe(f,B2,mu)L2(S2).\int_{S^{2}}\left(1-\frac{x_{3}}{\mu}\right)|B_{2,m}u|^{2}\,d\mathcal{H}^{2}+6\Bigl{(}\|\nabla v\|_{L^{2}(S^{2})}^{2}-6\|v\|_{L^{2}(S^{2})}^{2}\Bigr{)}\\ =\frac{1}{\mu}\mathrm{Re}(f,B_{2,m}u)_{L^{2}(S^{2})}.

Moreover, since u𝒴mu\in\mathcal{Y}_{m} is of the form (3.9) and 16/λn1/21-6/\lambda_{n}\geq 1/2 for n3n\geq 3, we observe by v=Δ1uv=\Delta^{-1}u, (2.3), and (2.4) that

(3.22) vL2(S2)26vL2(S2)2=(Δ)1/2uL2(S2)26Δ1uL2(S2)212(Δ)1/2uL2(S2)2.\displaystyle\begin{aligned} \|\nabla v\|_{L^{2}(S^{2})}^{2}-6\|v\|_{L^{2}(S^{2})}^{2}&=\|(-\Delta)^{-1/2}u\|_{L^{2}(S^{2})}^{2}-6\|\Delta^{-1}u\|_{L^{2}(S^{2})}^{2}\\ &\geq\frac{1}{2}\|(-\Delta)^{-1/2}u\|_{L^{2}(S^{2})}^{2}.\end{aligned}

Applying (3.22) and 1x3/μ11/|μ|=(|μ|1)/|μ|1-x_{3}/\mu\geq 1-1/|\mu|=(|\mu|-1)/|\mu| on S2S^{2} to the left-hand side of (3.21), and using Hölder’s and Young’s inequalities to the right-hand side, we obtain

|μ|1|μ|B2,muL2(S2)2+3(Δ)1/2uL2(S2)212|μ|1|μ|B2,muL2(S2)2+121|μ|(|μ|1)fL2(S2)2\frac{|\mu|-1}{|\mu|}\|B_{2,m}u\|_{L^{2}(S^{2})}^{2}+3\|(-\Delta)^{-1/2}u\|_{L^{2}(S^{2})}^{2}\\ \leq\frac{1}{2}\cdot\frac{|\mu|-1}{|\mu|}\|B_{2,m}u\|_{L^{2}(S^{2})}^{2}+\frac{1}{2}\cdot\frac{1}{|\mu|(|\mu|-1)}\|f\|_{L^{2}(S^{2})}^{2}

and thus (note that |μ|>1|\mu|>1)

(3.23) B2,muL2(S2)2+6|μ||μ|1(Δ)1/2uL2(S2)21(|μ|1)2fL2(S2)2.\displaystyle\|B_{2,m}u\|_{L^{2}(S^{2})}^{2}+\frac{6|\mu|}{|\mu|-1}\|(-\Delta)^{-1/2}u\|_{L^{2}(S^{2})}^{2}\leq\frac{1}{(|\mu|-1)^{2}}\|f\|_{L^{2}(S^{2})}^{2}.

Hence we get (3.19) by (3.14) and (3.23). To prove (3.20), we see that

(f,u)L2(S2)=((μx3)u6x3v,mu)L2(S2)=((μx3)u6x3v,u)L2(S2)\displaystyle(f,u)_{L^{2}(S^{2})}=\bigl{(}(\mu-x_{3})u-6x_{3}v,\mathbb{Q}_{m}u\bigr{)}_{L^{2}(S^{2})}=\bigl{(}(\mu-x_{3})u-6x_{3}v,u\bigr{)}_{L^{2}(S^{2})}

by u𝒴mu\in\mathcal{Y}_{m}. Taking the real part of this equality, we get

(3.24) S2(μx3)|u|2𝑑2=6Re(x3v,u)L2(S2)+Re(f,u)L2(S2).\displaystyle\int_{S^{2}}(\mu-x_{3})|u|^{2}\,d\mathcal{H}^{2}=6\mathrm{Re}(x_{3}v,u)_{L^{2}(S^{2})}+\mathrm{Re}(f,u)_{L^{2}(S^{2})}.

Moreover, since |x3|=sinθ1|\nabla x_{3}|=\sin\theta\leq 1 on S2S^{2} by x3=cosθx_{3}=\cos\theta and (5.1),

(3.25) |(x3v,u)L2(S2)|=|(x3v,Δv)L2(S2)|=|((x3v),v)L2(S2)|CvH1(S2)2C(Δ)1/2vL2(S2)2=C(Δ)1/2uL2(S2)2\displaystyle\begin{aligned} \bigl{|}(x_{3}v,u)_{L^{2}(S^{2})}\bigr{|}&=\bigl{|}(x_{3}v,\Delta v)_{L^{2}(S^{2})}\bigr{|}=\bigl{|}(\nabla(x_{3}v),\nabla v)_{L^{2}(S^{2})}\bigr{|}\\ &\leq C\|v\|_{H^{1}(S^{2})}^{2}\leq C\|(-\Delta)^{1/2}v\|_{L^{2}(S^{2})}^{2}=C\|(-\Delta)^{-1/2}u\|_{L^{2}(S^{2})}^{2}\end{aligned}

by (2.6) and u=Δvu=\Delta v. Noting that

|μx3|={μx3,μ>1,μ+x3,μ<1,sin2θ=1x322|μx3|onS2,\displaystyle|\mu-x_{3}|=\begin{cases}\mu-x_{3},&\mu>1,\\ -\mu+x_{3},&\mu<-1,\end{cases}\quad\sin^{2}\theta=1-x_{3}^{2}\leq 2|\mu-x_{3}|\quad\text{on}\quad S^{2},

we deduce from (3.24) and (3.25) that

MsinθuL2(S2)2\displaystyle\|M_{\sin\theta}u\|_{L^{2}(S^{2})}^{2} 2S2|μx3||u|2d2C(|(x3v,u)L2(S2)|+|(f,u)L2(S2)|)\displaystyle\leq 2\int_{S^{2}}|\mu-x_{3}||u|^{2}\,d\mathcal{H}^{2}\leq C\bigl{(}\bigl{|}(x_{3}v,u)_{L^{2}(S^{2})}\bigr{|}+\bigl{|}(f,u)_{L^{2}(S^{2})}\bigr{|}\bigl{)}
C((Δ)1/2uL2(S2)2+fL2(S2)uL2(S2))\displaystyle\leq C\Bigl{(}\|(-\Delta)^{-1/2}u\|_{L^{2}(S^{2})}^{2}+\|f\|_{L^{2}(S^{2})}\|u\|_{L^{2}(S^{2})}\Bigr{)}

and applying (3.19) and (3.23) to the last line we obtain (3.20). ∎

Lemma 3.8.

There exist constants κ(0,1/2)\kappa\in(0,1/2) and C>0C>0 such that

(3.26) uL2(S2)2\displaystyle\|u\|_{L^{2}(S^{2})}^{2} C(δ4m(μΛm)uL2(S2)2+δ2|m|(Δ)1/2uL2(S2)2),\displaystyle\leq C\left(\delta^{-4}\|\mathbb{Q}_{m}(\mu-\Lambda_{m})u\|_{L^{2}(S^{2})}^{2}+\frac{\delta^{2}}{|m|}\|(-\Delta)^{1/2}u\|_{L^{2}(S^{2})}^{2}\right),
(3.27) MsinθuL2(S2)2\displaystyle\|M_{\sin\theta}u\|_{L^{2}(S^{2})}^{2} C(δ2m(μΛm)uL2(S2)2+δ4|m|(Δ)1/2uL2(S2)2)\displaystyle\leq C\left(\delta^{-2}\|\mathbb{Q}_{m}(\mu-\Lambda_{m})u\|_{L^{2}(S^{2})}^{2}+\frac{\delta^{4}}{|m|}\|(-\Delta)^{1/2}u\|_{L^{2}(S^{2})}^{2}\right)

for all m{0}m\in\mathbb{Z}\setminus\{0\}, u𝒴mH1(S2)u\in\mathcal{Y}_{m}\cap H^{1}(S^{2}), δ(0,1]\delta\in(0,1], and μ\mu\in\mathbb{R} satisfying

(3.28) 1κδ2|μ|1+κδ2.\displaystyle 1-\kappa\delta^{2}\leq|\mu|\leq 1+\kappa\delta^{2}.
Proof.

Let κ(0,1/2)\kappa\in(0,1/2) be a constant which will be determined later, and let δ(0,1]\delta\in(0,1] and μ\mu\in\mathbb{R} satisfy (3.28). We may assume μ>0\mu>0 since the case μ<0\mu<0 is similarly handled. Let v=Δ1uv=\Delta^{-1}u and f=m(μΛm)uf=\mathbb{Q}_{m}(\mu-\Lambda_{m})u. Since u𝒴mu\in\mathcal{Y}_{m}, we can use (3.21) in the proof of Lemma 3.7 to get

S2(1x3)|B2,mu|2𝑑2+6μ(vL2(S2)26vL2(S2)2)=(1μ)B2,muL2(S2)2+Re(f,B2,mu)L2(S2).\int_{S^{2}}(1-x_{3})|B_{2,m}u|^{2}\,d\mathcal{H}^{2}+6\mu\Bigl{(}\|\nabla v\|_{L^{2}(S^{2})}^{2}-6\|v\|_{L^{2}(S^{2})}^{2}\Bigr{)}\\ =(1-\mu)\|B_{2,m}u\|_{L^{2}(S^{2})}^{2}+\mathrm{Re}(f,B_{2,m}u)_{L^{2}(S^{2})}.

Then, by μ1κδ21/2\mu\geq 1-\kappa\delta^{2}\geq 1/2, 1μκδ21-\mu\leq\kappa\delta^{2}, (3.22), and Young’s inequality,

(3.29) S2(1x3)|B2,mu|2𝑑2+(Δ)1/2uL2(S2)2C(κδ2B2,muL2(S2)2+δ2fL2(S2)2).\int_{S^{2}}(1-x_{3})|B_{2,m}u|^{2}\,d\mathcal{H}^{2}+\|(-\Delta)^{-1/2}u\|_{L^{2}(S^{2})}^{2}\\ \leq C\Bigl{(}\kappa\delta^{2}\|B_{2,m}u\|_{L^{2}(S^{2})}^{2}+\delta^{-2}\|f\|_{L^{2}(S^{2})}^{2}\Bigr{)}.

Since B2,mu𝒳mB_{2,m}u\in\mathcal{X}_{m}, we can write B2,mu=U2,m(θ)eimφB_{2,m}u=U_{2,m}(\theta)e^{im\varphi}. Then

S2(1x3)|B2,mu|2𝑑2\displaystyle\int_{S^{2}}(1-x_{3})|B_{2,m}u|^{2}\,d\mathcal{H}^{2} =2π0π(1cosθ)|U2,m(θ)|2sinθdθ\displaystyle=2\pi\int_{0}^{\pi}(1-\cos\theta)|U_{2,m}(\theta)|^{2}\sin\theta\,d\theta
2πδπ(1cosθ)|U2,m(θ)|2sinθdθ\displaystyle\geq 2\pi\int_{\delta}^{\pi}(1-\cos\theta)|U_{2,m}(\theta)|^{2}\sin\theta\,d\theta

by (2.7). Moreover, since 1cosθ1cosδ1-\cos\theta\geq 1-\cos\delta for θ(δ,π)\theta\in(\delta,\pi) and

δπ|U2,m(θ)|2sinθdθ\displaystyle\int_{\delta}^{\pi}|U_{2,m}(\theta)|^{2}\sin\theta\,d\theta =0π|U2,m(θ)|2sinθdθ0δ|U2,m(θ)|2sinθdθ\displaystyle=\int_{0}^{\pi}|U_{2,m}(\theta)|^{2}\sin\theta\,d\theta-\int_{0}^{\delta}|U_{2,m}(\theta)|^{2}\sin\theta\,d\theta
12πB2,muL2(S2)21π|m|(1cosδ)(Δ)1/2B2,muL2(S2)2\displaystyle\geq\frac{1}{2\pi}\|B_{2,m}u\|_{L^{2}(S^{2})}^{2}-\frac{1}{\pi|m|}(1-\cos\delta)\|(-\Delta)^{1/2}B_{2,m}u\|_{L^{2}(S^{2})}^{2}

by (2.7), (2.11), and 0δsinθdθ=1cosδ\int_{0}^{\delta}\sin\theta\,d\theta=1-\cos\delta, it follows that

S2(1x3)|B2,mu|2𝑑2(1cosδ)B2,muL2(S2)22|m|(1cosδ)2(Δ)1/2B2,muL2(S2)2.\int_{S^{2}}(1-x_{3})|B_{2,m}u|^{2}\,d\mathcal{H}^{2}\\ \geq(1-\cos\delta)\|B_{2,m}u\|_{L^{2}(S^{2})}^{2}-\frac{2}{|m|}(1-\cos\delta)^{2}\|(-\Delta)^{1/2}B_{2,m}u\|_{L^{2}(S^{2})}^{2}.

We further observe that δ2/41cosδδ2/2\delta^{2}/4\leq 1-\cos\delta\leq\delta^{2}/2 by Taylor’s theorem for cosθ\cos\theta at θ=0\theta=0 and δ(0,1](0,π/3)\delta\in(0,1]\subset(0,\pi/3). Therefore,

S2(1x3)|B2,mu|2𝑑214(δ2B2,muL2(S2)22δ4|m|(Δ)1/2B2,muL2(S2)2).\displaystyle\int_{S^{2}}(1-x_{3})|B_{2,m}u|^{2}\,d\mathcal{H}^{2}\geq\frac{1}{4}\left(\delta^{2}\|B_{2,m}u\|_{L^{2}(S^{2})}^{2}-\frac{2\delta^{4}}{|m|}\|(-\Delta)^{1/2}B_{2,m}u\|_{L^{2}(S^{2})}^{2}\right).

We apply this inequality to (3.29) and use (3.13) with k=1k=1 to find that

(3.30) 14δ2B2,muL2(S2)2+(Δ)1/2uL2(S2)2C1κδ2B2,muL2(S2)2+C2(δ2fL2(S2)2+δ4|m|(Δ)1/2uL2(S2)2)\frac{1}{4}\delta^{2}\|B_{2,m}u\|_{L^{2}(S^{2})}^{2}+\|(-\Delta)^{-1/2}u\|_{L^{2}(S^{2})}^{2}\\ \leq C_{1}\kappa\delta^{2}\|B_{2,m}u\|_{L^{2}(S^{2})}^{2}+C_{2}\left(\delta^{-2}\|f\|_{L^{2}(S^{2})}^{2}+\frac{\delta^{4}}{|m|}\|(-\Delta)^{1/2}u\|_{L^{2}(S^{2})}^{2}\right)

with some constants C1,C2>0C_{1},C_{2}>0 independent of mm, uu, δ\delta, μ\mu, and κ\kappa. Now we define

(3.31) κ=18min{1C1,1}(0,12).\displaystyle\kappa=\frac{1}{8}\min\left\{\frac{1}{C_{1}},1\right\}\in\left(0,\frac{1}{2}\right).

Then since 1/4C1κ1/81/4-C_{1}\kappa\geq 1/8, it follows from (3.30) that

(3.32) δ2B2,muL2(S2)2+(Δ)1/2uL2(S2)2C(δ2fL2(S2)2+δ4|m|(Δ)1/2uL2(S2)2)\delta^{2}\|B_{2,m}u\|_{L^{2}(S^{2})}^{2}+\|(-\Delta)^{-1/2}u\|_{L^{2}(S^{2})}^{2}\\ \leq C\left(\delta^{-2}\|f\|_{L^{2}(S^{2})}^{2}+\frac{\delta^{4}}{|m|}\|(-\Delta)^{1/2}u\|_{L^{2}(S^{2})}^{2}\right)

and we get (3.26) by (3.14) and (3.32). Also, since

S2(1x3)|u|2𝑑2\displaystyle\int_{S^{2}}(1-x_{3})|u|^{2}\,d\mathcal{H}^{2} =(1μ)uL2(S2)2+S2(μx3)|u|2𝑑2\displaystyle=(1-\mu)\|u\|_{L^{2}(S^{2})}^{2}+\int_{S^{2}}(\mu-x_{3})|u|^{2}\,d\mathcal{H}^{2}
κδ2uL2(S2)2+C((Δ)1/2uL2(S2)2+fL2(S2)uL2(S2))\displaystyle\leq\kappa\delta^{2}\|u\|_{L^{2}(S^{2})}^{2}+C\Bigl{(}\|(-\Delta)^{-1/2}u\|_{L^{2}(S^{2})}^{2}+\|f\|_{L^{2}(S^{2})}\|u\|_{L^{2}(S^{2})}\Bigr{)}

by 1μκδ21-\mu\leq\kappa\delta^{2}, (3.24), and (3.25), and since sin2θ=1x322(1x3)\sin^{2}\theta=1-x_{3}^{2}\leq 2(1-x_{3}) on S2S^{2},

MsinθuL2(S2)2\displaystyle\|M_{\sin\theta}u\|_{L^{2}(S^{2})}^{2} 2S2(1x3)|u|2𝑑2\displaystyle\leq 2\int_{S^{2}}(1-x_{3})|u|^{2}\,d\mathcal{H}^{2}
C(δ2uL2(S2)2+(Δ)1/2uL2(S2)2+fL2(S2)uL2(S2)).\displaystyle\leq C\Bigl{(}\delta^{2}\|u\|_{L^{2}(S^{2})}^{2}+\|(-\Delta)^{-1/2}u\|_{L^{2}(S^{2})}^{2}+\|f\|_{L^{2}(S^{2})}\|u\|_{L^{2}(S^{2})}\Bigr{)}.

Thus we apply (3.26) and (3.32) to the last line to get (3.27). ∎

Lemma 3.9.

Let κ\kappa be the constant given by (3.31). There exists a constant C>0C>0 such that

(3.33) δ2uL2(S2)2+(Δ)1/2vL2(S2)2+11|μ|wL2(S2)2C(δ2m(μΛm)uL2(S2)2+δ61|μ|(Δ)1/2uL2(S2)2)\delta^{2}\|u\|_{L^{2}(S^{2})}^{2}+\|(-\Delta)^{1/2}v\|_{L^{2}(S^{2})}^{2}+\frac{1}{1-|\mu|}\|w\|_{L^{2}(S^{2})}^{2}\\ \leq C\left(\delta^{-2}\|\mathbb{Q}_{m}(\mu-\Lambda_{m})u\|_{L^{2}(S^{2})}^{2}+\frac{\delta^{6}}{1-|\mu|}\|(-\Delta)^{1/2}u\|_{L^{2}(S^{2})}^{2}\right)

and

(3.34) MsinθuL2(S2)2C(δ2m(μΛm)uL2(S2)2+δ2(1|μ|)(Δ)1/2uL2(S2)2)\displaystyle\|M_{\sin\theta}u\|_{L^{2}(S^{2})}^{2}\leq C\Bigl{(}\delta^{-2}\|\mathbb{Q}_{m}(\mu-\Lambda_{m})u\|_{L^{2}(S^{2})}^{2}+\delta^{2}(1-|\mu|)\|(-\Delta)^{1/2}u\|_{L^{2}(S^{2})}^{2}\Bigr{)}

for all m{0}m\in\mathbb{Z}\setminus\{0\}, u𝒴mH1(S2)u\in\mathcal{Y}_{m}\cap H^{1}(S^{2}), δ(0,1]\delta\in(0,1], and μ\mu\in\mathbb{R} satisfying |μ|1κδ2|\mu|\leq 1-\kappa\delta^{2}. Here

(3.35) v=Δ1u𝒴mH3(S2),w=6μva3m2(u,Y3m)L2(S2)Y2m𝒳mH3(S2),\displaystyle\begin{aligned} v&=\Delta^{-1}u\in\mathcal{Y}_{m}\cap H^{3}(S^{2}),\\ w&=6\mu v-\frac{a_{3}^{m}}{2}(u,Y_{3}^{m})_{L^{2}(S^{2})}Y_{2}^{m}\in\mathcal{X}_{m}\cap H^{3}(S^{2}),\end{aligned}

where a3ma_{3}^{m} is given by (2.2) if |m|=1,2|m|=1,2 and we consider a3m=0a_{3}^{m}=0 if |m|3|m|\geq 3 and Ynm0Y_{n}^{m}\equiv 0 if |m|>n|m|>n.

The proof of (3.33) relies on a contradiction argument. In order to focus on the part of getting a contradiction in that proof, we derive auxiliary estimates in the following lemmas. We assume μ0\mu\geq 0 in the sequel, since the case μ0\mu\leq 0 can be handled similarly.

In Lemmas 3.103.17 below, let κ\kappa be the constant given by (3.31) and δ(0,1]\delta\in(0,1], μ[0,1κδ2]\mu\in[0,1-\kappa\delta^{2}], and θμ=arccosμ(0,π/2]\theta_{\mu}=\arccos\mu\in(0,\pi/2]. Note that

(3.36) (1μ)1/2sinθμ1,κδ2sin2θμ2(1μ)\displaystyle(1-\mu)^{1/2}\leq\sin\theta_{\mu}\leq 1,\quad\kappa\delta^{2}\leq\sin^{2}\theta_{\mu}\leq 2(1-\mu)

by sin2θμ=1μ2=(1μ)(1+μ)\sin^{2}\theta_{\mu}=1-\mu^{2}=(1-\mu)(1+\mu) and κδ21μ\kappa\delta^{2}\leq 1-\mu. For ε>0\varepsilon>0 let

(3.37) Sμ,ε2={(x1,x2,x3)S2x3=cosθ,θ(θμε,θμ+ε)},\displaystyle S_{\mu,\varepsilon}^{2}=\{(x_{1},x_{2},x_{3})\in S^{2}\mid x_{3}=\cos\theta,\,\theta\in(\theta_{\mu}-\varepsilon,\theta_{\mu}+\varepsilon)\},

i.e. Sμ,ε2=S2(θμε,θμ+ε)S_{\mu,\varepsilon}^{2}=S^{2}(\theta_{\mu}-\varepsilon,\theta_{\mu}+\varepsilon) in (2.9). Also, let

(3.40) f=δ1m(μΛm)u𝒴mH1(S2),R(u,δ,μ)=fL2(S2)+δ3(1μ)1/2(Δ)1/2uL2(S2)\displaystyle\begin{gathered}f=\delta^{-1}\mathbb{Q}_{m}(\mu-\Lambda_{m})u\in\mathcal{Y}_{m}\cap H^{1}(S^{2}),\\ R(u,\delta,\mu)=\|f\|_{L^{2}(S^{2})}+\frac{\delta^{3}}{(1-\mu)^{1/2}}\|(-\Delta)^{1/2}u\|_{L^{2}(S^{2})}\end{gathered}

and vv and ww be given by (3.35) for u𝒴mH1(S2)u\in\mathcal{Y}_{m}\cap H^{1}(S^{2}) with m{0}m\in\mathbb{Z}\setminus\{0\}. Hereafter we write CC for a general positive constant independent of δ\delta, μ\mu, mm, and uu.

Lemma 3.10.

Let ε\varepsilon be a positive constant satisfying

(3.41) 0<ε12sinθμ<12θμ.\displaystyle 0<\varepsilon\leq\frac{1}{2}\sin\theta_{\mu}<\frac{1}{2}\theta_{\mu}.

Also, let θ(θμε,θμ+ε)\theta\in(\theta_{\mu}-\varepsilon,\theta_{\mu}+\varepsilon). Then

(3.42) 12sinθμsinθ32sinθμ,|μcosθ|54εsinθμ,\displaystyle\frac{1}{2}\sin\theta_{\mu}\leq\sin\theta\leq\frac{3}{2}\sin\theta_{\mu},\quad|\mu-\cos\theta|\leq\frac{5}{4}\varepsilon\sin\theta_{\mu},
(3.43) 34εsinθμ|μcos(θμ±ε)|54εsinθμ.\displaystyle\frac{3}{4}\varepsilon\sin\theta_{\mu}\leq|\mu-\cos(\theta_{\mu}\pm\varepsilon)|\leq\frac{5}{4}\varepsilon\sin\theta_{\mu}.

Note that cos(θμ+ε)<μ<cos(θμε)\cos(\theta_{\mu}+\varepsilon)<\mu<\cos(\theta_{\mu}-\varepsilon). Also,

(3.44) |μcosθ|12|θθμ|(sinθ+sinθμ),θ[0,π].\displaystyle|\mu-\cos\theta|\geq\frac{1}{2}|\theta-\theta_{\mu}|(\sin\theta+\sin\theta_{\mu}),\quad\theta\in[0,\pi].
Proof.

For θ(θμε,θμ+ε)\theta\in(\theta_{\mu}-\varepsilon,\theta_{\mu}+\varepsilon) we have (3.42) since

|sinθsinθμ|ε12sinθμ,|μcosθ|εsinθ+12ε254εsinθμ\displaystyle|\sin\theta-\sin\theta_{\mu}|\leq\varepsilon\leq\frac{1}{2}\sin\theta_{\mu},\quad|\mu-\cos\theta|\leq\varepsilon\sin\theta+\frac{1}{2}\varepsilon^{2}\leq\frac{5}{4}\varepsilon\sin\theta_{\mu}

by Taylor’s theorem for sinθ\sin\theta and cosθ\cos\theta at θ=θμ\theta=\theta_{\mu} and (3.41). Also,

|cos(θμ±ε)(μεsinθμ)|12ε214εsinθμ\displaystyle|\cos(\theta_{\mu}\pm\varepsilon)-(\mu\mp\varepsilon\sin\theta_{\mu})|\leq\frac{1}{2}\varepsilon^{2}\leq\frac{1}{4}\varepsilon\sin\theta_{\mu}

by Taylors’ theorem and (3.41). Hence (3.43) follows. For θ[0,π]\theta\in[0,\pi] since

|μcosθ|=θminθmaxsinϑdϑ=(θmaxθmin)01sin((1t)θmin+tθmax)𝑑t,\displaystyle|\mu-\cos\theta|=\int_{\theta_{\min}}^{\theta_{\max}}\sin\vartheta\,d\vartheta=(\theta_{\max}-\theta_{\min})\int_{0}^{1}\sin\bigl{(}(1-t)\theta_{\min}+t\theta_{\max}\bigr{)}\,dt,

where θmin=min{θμ,θ}\theta_{\min}=\min\{\theta_{\mu},\theta\} and θmax=max{θμ,θ}\theta_{\max}=\max\{\theta_{\mu},\theta\}, and since sinϑ\sin\vartheta is concave for ϑ[0,π]\vartheta\in[0,\pi], we easily find that (3.44) holds. ∎

Lemma 3.11.

We have

(3.45) (μx3)B2,mu\displaystyle(\mu-x_{3})B_{2,m}u =w+δf\displaystyle=w+\delta f on S2,\displaystyle S^{2},
(3.46) (μx3)(Δw+6w)\displaystyle(\mu-x_{3})(\Delta w+6w) =6μ(w+δf)\displaystyle=6\mu(w+\delta f) on S2.\displaystyle S^{2}.
Proof.

If |m|3|m|\geq 3, then m=I\mathbb{Q}_{m}=I and w=6μvw=6\mu v. Thus

δf=(μΛm)u=(μx3)B2,muw\displaystyle\delta f=(\mu-\Lambda_{m})u=(\mu-x_{3})B_{2,m}u-w

by Λmu=x3B2,mu\Lambda_{m}u=x_{3}B_{2,m}u and B2,mu=(I+6Δ1)u=u+6vB_{2,m}u=(I+6\Delta^{-1})u=u+6v. Hence (3.45) holds.

Let |m|=1,2|m|=1,2 and u𝒴mu\in\mathcal{Y}_{m} be of the form (3.9) with Nm=3N^{\prime}_{m}=3. Then it follows from (2.2), (3.2) and λ3=12\lambda_{3}=12 that

Λmu\displaystyle\Lambda_{m}u =n3(16λn)(u,Ynm)L2(S2)(anmYn1m+an+1mYn+1m)\displaystyle=\sum_{n\geq 3}\left(1-\frac{6}{\lambda_{n}}\right)(u,Y_{n}^{m})_{L^{2}(S^{2})}(a_{n}^{m}Y_{n-1}^{m}+a_{n+1}^{m}Y_{n+1}^{m})
=a3m2(u,Y3m)L2(S2)Y2m+u~3,u~3span¯{Ynmn3}\displaystyle=\frac{a_{3}^{m}}{2}(u,Y_{3}^{m})_{L^{2}(S^{2})}Y_{2}^{m}+\tilde{u}_{\geq 3},\quad\tilde{u}_{\geq 3}\in\overline{\mathrm{span}}\{Y_{n}^{m}\mid n\geq 3\}

and thus mΛmu=Λmu12a3m(u,Y3m)L2(S2)Y2m\mathbb{Q}_{m}\Lambda_{m}u=\Lambda_{m}u-\frac{1}{2}a_{3}^{m}(u,Y_{3}^{m})_{L^{2}(S^{2})}Y_{2}^{m} by (3.10). Hence

δf\displaystyle\delta f =m(μΛm)u=μumΛmu=μuΛmu+a3m2(u,Y3m)L2(S2)Y2m\displaystyle=\mathbb{Q}_{m}(\mu-\Lambda_{m})u=\mu u-\mathbb{Q}_{m}\Lambda_{m}u=\mu u-\Lambda_{m}u+\frac{a_{3}^{m}}{2}(u,Y_{3}^{m})_{L^{2}(S^{2})}Y_{2}^{m}
=(μx3)B2,mu{6μva3m2(u,Y3m)L2(S2)Y2m}=(μx3)B2,muw\displaystyle=(\mu-x_{3})B_{2,m}u-\left\{6\mu v-\frac{a_{3}^{m}}{2}(u,Y_{3}^{m})_{L^{2}(S^{2})}Y_{2}^{m}\right\}=(\mu-x_{3})B_{2,m}u-w

by u𝒴mu\in\mathcal{Y}_{m}, Λmu=x3B2,mu\Lambda_{m}u=x_{3}B_{2,m}u, and B2,mu=u+6vB_{2,m}u=u+6v.

For each m{0}m\in\mathbb{Z}\setminus\{0\} the equality (3.46) follows from (3.35), (3.45), B2,mu=Δv+6vB_{2,m}u=\Delta v+6v, and ΔY2m=λ2Y2m=6Y2m\Delta Y_{2}^{m}=-\lambda_{2}Y_{2}^{m}=-6Y_{2}^{m}, where Y2m0Y_{2}^{m}\equiv 0 if |m|3|m|\geq 3. ∎

Lemma 3.12.

We have

(3.47) wL2(S2)CvL2(S2),wH1(S2)C(Δ)1/2wL2(S2)C(Δ)1/2vL2(S2).\displaystyle\begin{aligned} \|w\|_{L^{2}(S^{2})}&\leq C\|v\|_{L^{2}(S^{2})},\\ \|w\|_{H^{1}(S^{2})}&\leq C\|(-\Delta)^{1/2}w\|_{L^{2}(S^{2})}\leq C\|(-\Delta)^{1/2}v\|_{L^{2}(S^{2})}.\end{aligned}
Proof.

By ΔY3m=λ3Y3m\Delta Y_{3}^{m}=-\lambda_{3}Y_{3}^{m} and Y3mL2(S2)1\|Y_{3}^{m}\|_{L^{2}(S^{2})}\leq 1,

|(u,Y3m)L2(S2)|=|(Δv,Y3m)L2(S2)|=|(v,ΔY3m)L2(S2)|λ3vL2(S2),\displaystyle\bigl{|}(u,Y_{3}^{m})_{L^{2}(S^{2})}\bigr{|}=\bigl{|}(\Delta v,Y_{3}^{m})_{L^{2}(S^{2})}\bigr{|}=\left|\bigl{(}v,\Delta Y_{3}^{m}\bigr{)}_{L^{2}(S^{2})}\right|\leq\lambda_{3}\|v\|_{L^{2}(S^{2})},

where Ynm0Y_{n}^{m}\equiv 0 if |m|>n|m|>n. We apply this inequality, |μ|1|\mu|\leq 1, (Δ)1/2Y2m=λ21/2Y2m(-\Delta)^{1/2}Y_{2}^{m}=\lambda_{2}^{1/2}Y_{2}^{m}, and Y2mL2(S2)1\|Y_{2}^{m}\|_{L^{2}(S^{2})}\leq 1 to ww given by (3.35) to find that

(3.48) (Δ)k/2wL2(S2)C((Δ)k/2vL2(S2)+vL2(S2)),k=0,1.\displaystyle\|(-\Delta)^{k/2}w\|_{L^{2}(S^{2})}\leq C\left(\|(-\Delta)^{k/2}v\|_{L^{2}(S^{2})}+\|v\|_{L^{2}(S^{2})}\right),\quad k=0,1.

Hence we have (3.47) by this inequality and (2.6). ∎

Lemma 3.13.

We have

(3.49) δuL2(S2)C(R(u,δ,μ)+(Δ)1/2vL2(S2)).\displaystyle\delta\|u\|_{L^{2}(S^{2})}\leq C\left(R(u,\delta,\mu)+\|(-\Delta)^{1/2}v\|_{L^{2}(S^{2})}\right).
Proof.

Let ε=κδ2/2sinθμ\varepsilon=\kappa\delta^{2}/2\sin\theta_{\mu}, which satisfies (3.41) and

(3.50) εCδ2(1μ)1/2\displaystyle\varepsilon\leq\frac{C\delta^{2}}{(1-\mu)^{1/2}}

by (3.36). Also, let Sμ,ε2S_{\mu,\varepsilon}^{2} be given by (3.37). Since B2,mu𝒳mB_{2,m}u\in\mathcal{X}_{m}, we can write B2,mu=U2,m(θ)eimφB_{2,m}u=U_{2,m}(\theta)e^{im\varphi}. Then we see by (2.20) and (3.13) that

|U2,m(θ)|2\displaystyle|U_{2,m}(\theta)|^{2} CsinθμB2,muL2(S2)(Δ)1/2B2,muL2(S2)\displaystyle\leq\frac{C}{\sin\theta_{\mu}}\|B_{2,m}u\|_{L^{2}(S^{2})}\|(-\Delta)^{1/2}B_{2,m}u\|_{L^{2}(S^{2})}
CsinθμuL2(S2)(Δ)1/2uL2(S2)\displaystyle\leq\frac{C}{\sin\theta_{\mu}}\|u\|_{L^{2}(S^{2})}\|(-\Delta)^{1/2}u\|_{L^{2}(S^{2})}

for θ(θμε,θμε)\theta\in(\theta_{\mu}-\varepsilon,\theta_{\mu}-\varepsilon). By this inequality and (3.43),

(3.51) B2,muL2(Sμ,ε2)2=2πθμεθμ+ε|U2,m(θ)|2sinθdθCεuL2(S2)(Δ)1/2uL2(S2).\displaystyle\|B_{2,m}u\|_{L^{2}(S_{\mu,\varepsilon}^{2})}^{2}=2\pi\int_{\theta_{\mu}-\varepsilon}^{\theta_{\mu}+\varepsilon}|U_{2,m}(\theta)|^{2}\sin\theta\,d\theta\leq C\varepsilon\|u\|_{L^{2}(S^{2})}\|(-\Delta)^{1/2}u\|_{L^{2}(S^{2})}.

Also, since B2,mu=(w+δf)/(μx3)B_{2,m}u=(w+\delta f)/(\mu-x_{3}) for x3μx_{3}\neq\mu by (3.45),

(3.52) B2,muL2(S2Sμ,ε2)wL(S2)1μx3L2(S2Sμ,ε2)+δfL2(S2)1μx3L(S2Sμ,ε2).\|B_{2,m}u\|_{L^{2}(S^{2}\setminus S_{\mu,\varepsilon}^{2})}\leq\|w\|_{L^{\infty}(S^{2})}\left\|\frac{1}{\mu-x_{3}}\right\|_{L^{2}(S^{2}\setminus S_{\mu,\varepsilon}^{2})}\\ +\delta\|f\|_{L^{2}(S^{2})}\left\|\frac{1}{\mu-x_{3}}\right\|_{L^{\infty}(S^{2}\setminus S_{\mu,\varepsilon}^{2})}.

Moreover, since sinθ(μcosθ)2=ddθ(μcosθ)1\sin\theta(\mu-\cos\theta)^{-2}=-\frac{d}{d\theta}(\mu-\cos\theta)^{-1}, we have

1μx3L2(S2Sμ,ε2)2\displaystyle\left\|\frac{1}{\mu-x_{3}}\right\|_{L^{2}(S^{2}\setminus S_{\mu,\varepsilon}^{2})}^{2} =2π(0θμε+θμ+επ)sinθ(μcosθ)2dθ\displaystyle=2\pi\left(\int_{0}^{\theta_{\mu}-\varepsilon}+\int_{\theta_{\mu}+\varepsilon}^{\pi}\right)\frac{\sin\theta}{(\mu-\cos\theta)^{2}}\,d\theta
2π(1cos(θμε)μ+1μcos(θμ+ε))CεsinθμCδ2\displaystyle\leq 2\pi\left(\frac{1}{\cos(\theta_{\mu}-\varepsilon)-\mu}+\frac{1}{\mu-\cos(\theta_{\mu}+\varepsilon)}\right)\leq\frac{C}{\varepsilon\sin\theta_{\mu}}\leq C\delta^{-2}

by (2.7), (3.43), and εsinθμ=κδ2/2\varepsilon\sin\theta_{\mu}=\kappa\delta^{2}/2. Also, we see by (3.44) that

1μx3L(S2Sμ,ε2)=sup|θθμ|ε1|μcosθ|2εsinθμCδ2.\displaystyle\left\|\frac{1}{\mu-x_{3}}\right\|_{L^{\infty}(S^{2}\setminus S_{\mu,\varepsilon}^{2})}=\sup_{|\theta-\theta_{\mu}|\geq\varepsilon}\frac{1}{|\mu-\cos\theta|}\leq\frac{2}{\varepsilon\sin\theta_{\mu}}\leq C\delta^{-2}.

We apply these inequalities to (3.52) and use (2.11) and (3.47) to ww to get

(3.53) B2,muL2(S2Sμ,ε2)Cδ1(fL2(S2)+(Δ)1/2vL2(S2)).\displaystyle\|B_{2,m}u\|_{L^{2}(S^{2}\setminus S_{\mu,\varepsilon}^{2})}\leq C\delta^{-1}\Bigl{(}\|f\|_{L^{2}(S^{2})}+\|(-\Delta)^{1/2}v\|_{L^{2}(S^{2})}\Bigr{)}.

By (3.14), (3.51), (3.53), and Young’s inequality, we find that

δuL2(S2)12δuL2(S2)+C(δε(Δ)1/2uL2(S2)+fL2(S2)+(Δ)1/2vL2(S2)).\displaystyle\delta\|u\|_{L^{2}(S^{2})}\leq\frac{1}{2}\delta\|u\|_{L^{2}(S^{2})}+C\Bigl{(}\delta\varepsilon\|(-\Delta)^{1/2}u\|_{L^{2}(S^{2})}+\|f\|_{L^{2}(S^{2})}+\|(-\Delta)^{1/2}v\|_{L^{2}(S^{2})}\Bigr{)}.

Hence we have (3.49) by this inequality and (3.50). ∎

Lemma 3.14.

Let ε\varepsilon satisfy (3.41) and Sμ,ε2S_{\mu,\varepsilon}^{2} be given by (3.37). Then

(3.54) δfL2(Sμ,ε2)C(MsinθuL2(Sμ,ε2)..+εsinθμ(Δ)1/2uL2(S2)+(Δ)1/2vL2(S2)).\displaystyle\begin{aligned} \delta\|\nabla f\|_{L^{2}(S_{\mu,\varepsilon}^{2})}&\leq C\Bigl{(}\|M_{\sin\theta}u\|_{L^{2}(S_{\mu,\varepsilon}^{2})}\Bigr{.}\\ &\qquad\qquad\Bigl{.}+\varepsilon\sin\theta_{\mu}\|(-\Delta)^{1/2}u\|_{L^{2}(S^{2})}+\|(-\Delta)^{1/2}v\|_{L^{2}(S^{2})}\Bigr{)}.\end{aligned}
Proof.

By (3.45) and B2,mu=u+6vB_{2,m}u=u+6v,

δf=ux36vx3+(μx3)B2,muwonS2.\displaystyle\delta\nabla f=-u\nabla x_{3}-6v\nabla x_{3}+(\mu-x_{3})\nabla B_{2,m}u-\nabla w\quad\text{on}\quad S^{2}.

We take the L2(Sμ,ε2)L^{2}(S_{\mu,\varepsilon}^{2})-norms of both sides, apply (3.42) to μx3=μcosθ\mu-x_{3}=\mu-\cos\theta on Sμ,ε2S_{\mu,\varepsilon}^{2}, and use |x3|=sinθ1|\nabla x_{3}|=\sin\theta\leq 1 on S2S^{2} by x3=cosθx_{3}=\cos\theta and (5.1). Then, replacing the L2L^{2}-norms of vv, B2,mu\nabla B_{2,m}u, and w\nabla w on Sμ,ε2S_{\mu,\varepsilon}^{2} by those on S2S^{2}, we get

δfL2(Sμ,ε2)\displaystyle\delta\|\nabla f\|_{L^{2}(S_{\mu,\varepsilon}^{2})} C(MsinθuL2(Sμ,ε2)+vL2(S2).\displaystyle\leq C\Bigl{(}\|M_{\sin\theta}u\|_{L^{2}(S_{\mu,\varepsilon}^{2})}+\|v\|_{L^{2}(S^{2})}\Bigr{.}
.+εsinθμB2,muL2(S2)+wL2(S2)).\displaystyle\qquad\qquad\Bigl{.}+\varepsilon\sin\theta_{\mu}\|\nabla B_{2,m}u\|_{L^{2}(S^{2})}+\|\nabla w\|_{L^{2}(S^{2})}\Bigr{)}.

In this inequality, we apply (2.6) and (3.47) to vv and w\nabla w, respectively, and use (2.4) and (3.13) with k=1k=1 to B2,mu\nabla B_{2,m}u. Then we obtain (3.54). ∎

Lemma 3.15.

We have

(3.55) w=W(θ)eimφ,f=F(θ)eimφ,WC1([0,π]),FHloc1(0,π)C(0,π)\displaystyle w=W(\theta)e^{im\varphi},\quad f=F(\theta)e^{im\varphi},\quad W\in C^{1}([0,\pi]),\quad F\in H_{loc}^{1}(0,\pi)\subset C(0,\pi)

and W(θμ)+δF(θμ)=0W(\theta_{\mu})+\delta F(\theta_{\mu})=0. Moreover, for all σ1(0,1/2)\sigma_{1}\in(0,1/\sqrt{2}),

(3.56) |W(θμ)|C(σ11R(u,δ,μ)+σ1(Δ)1/2vL2(S2)).\displaystyle|W(\theta_{\mu})|\leq C\left(\sigma_{1}^{-1}R(u,\delta,\mu)+\sigma_{1}\|(-\Delta)^{1/2}v\|_{L^{2}(S^{2})}\right).
Proof.

Since w𝒳mH3(S2)𝒳mC1(S2)w\in\mathcal{X}_{m}\cap H^{3}(S^{2})\subset\mathcal{X}_{m}\cap C^{1}(S^{2}) by the Sobolev embedding theorem (see [2]), and since f𝒳mH1(S2)f\in\mathcal{X}_{m}\cap H^{1}(S^{2}), we can write (3.55). Also, since ww and ff are continuous near x3=μx_{3}=\mu and B2,mu=(w+δf)/(μx3)B_{2,m}u=(w+\delta f)/(\mu-x_{3}) for x3μx_{3}\neq\mu by (3.45), we have w+δf=0w+\delta f=0 at x3=μx_{3}=\mu, otherwise B2,muB_{2,m}u does not belong to L2(S2)L^{2}(S^{2}). Hence W(θμ)+δF(θμ)=0W(\theta_{\mu})+\delta F(\theta_{\mu})=0.

Let us show (3.56). For σ1(0,1/2)\sigma_{1}\in(0,1/\sqrt{2}) let ε=σ12κδ2/sinθμ\varepsilon=\sigma_{1}^{2}\kappa\delta^{2}/\sin\theta_{\mu}, which satisfies (3.41) and (3.50) by (3.36). By the mean value theorem for integrals, there exists θμ(θμε,θμ+ε)\theta^{\prime}_{\mu}\in(\theta_{\mu}-\varepsilon,\theta_{\mu}+\varepsilon) such that

θμεθμ+ε|F(θ)|2sinθdθ=2ε|F(θμ)|2sinθμ.\displaystyle\int_{\theta_{\mu}-\varepsilon}^{\theta_{\mu}+\varepsilon}|F(\theta)|^{2}\sin\theta\,d\theta=2\varepsilon|F(\theta^{\prime}_{\mu})|^{2}\sin\theta^{\prime}_{\mu}.

Then since 2εsinθμεsinθμ=σ12κδ22\varepsilon\sin\theta^{\prime}_{\mu}\geq\varepsilon\sin\theta_{\mu}=\sigma_{1}^{2}\kappa\delta^{2} by (3.42), we have

(3.57) |F(θμ)|21εsinθμθμεθμ+ε|F(θ)|2sinθdθCσ12δ2fL2(S2)2\displaystyle|F(\theta^{\prime}_{\mu})|^{2}\leq\frac{1}{\varepsilon\sin\theta_{\mu}}\int_{\theta_{\mu}-\varepsilon}^{\theta_{\mu}+\varepsilon}|F(\theta)|^{2}\sin\theta\,d\theta\leq\frac{C}{\sigma_{1}^{2}\delta^{2}}\|f\|_{L^{2}(S^{2})}^{2}

by (2.7). Also, we use Hölder’s inequality, (2.7), and (3.42) to get

(3.58) |F(θμ)F(θμ)|=|θμθμF(θ)𝑑θ|C(εsinθμ)1/2(θμεθμ+ε|F(θ)|2sinθdθ)1/2Cσ1δsinθμfL2(Sμ,ε2),\displaystyle\begin{aligned} |F(\theta_{\mu})-F(\theta^{\prime}_{\mu})|=\left|\int_{\theta^{\prime}_{\mu}}^{\theta_{\mu}}F^{\prime}(\theta)\,d\theta\right|&\leq C\left(\frac{\varepsilon}{\sin\theta_{\mu}}\right)^{1/2}\left(\int_{\theta_{\mu}-\varepsilon}^{\theta_{\mu}+\varepsilon}|F^{\prime}(\theta)|^{2}\sin\theta\,d\theta\right)^{1/2}\\ &\leq\frac{C\sigma_{1}\delta}{\sin\theta_{\mu}}\|\nabla f\|_{L^{2}(S_{\mu,\varepsilon}^{2})},\end{aligned}

where Sμ,ε2S_{\mu,\varepsilon}^{2} is given by (3.37). Moreover, since

MsinθuL2(Sμ,ε2)CsinθμuL2(Sμ,ε2)CsinθμuL2(S2),δsinθμC\displaystyle\|M_{\sin\theta}u\|_{L^{2}(S_{\mu,\varepsilon}^{2})}\leq C\sin\theta_{\mu}\|u\|_{L^{2}(S_{\mu,\varepsilon}^{2})}\leq C\sin\theta_{\mu}\|u\|_{L^{2}(S^{2})},\quad\frac{\delta}{\sin\theta_{\mu}}\leq C

by (3.36) and (3.42), we see by (3.49), (3.50), and (3.54) that

(3.59) δ2sinθμfL2(Sμ,ε2)C(δsinθμMsinθuL2(Sμ,ε2)+δε(Δ)1/2uL2(S2)+δsinθμ(Δ)1/2vL2(S2))C(R(u,δ,μ)+(Δ)1/2vL2(S2)).\displaystyle\begin{aligned} \frac{\delta^{2}}{\sin\theta_{\mu}}\|\nabla f\|_{L^{2}(S_{\mu,\varepsilon}^{2})}&\leq C\left(\frac{\delta}{\sin\theta_{\mu}}\|M_{\sin\theta}u\|_{L^{2}(S_{\mu,\varepsilon}^{2})}\right.\\ &\qquad\left.+\delta\varepsilon\|(-\Delta)^{1/2}u\|_{L^{2}(S^{2})}+\frac{\delta}{\sin\theta_{\mu}}\|(-\Delta)^{1/2}v\|_{L^{2}(S^{2})}\right)\\ &\leq C\Bigl{(}R(u,\delta,\mu)+\|(-\Delta)^{1/2}v\|_{L^{2}(S^{2})}\Bigr{)}.\end{aligned}

By W(θμ)+δF(θμ)=0W(\theta_{\mu})+\delta F(\theta_{\mu})=0, (3.57), and (3.58), we have

|W(θμ)|δ|F(θμ)|+δ|F(θμ)F(θμ)|C(σ11fL2(S2)+σ1δ2sinθμfL2(Sμ,ε2))\displaystyle|W(\theta_{\mu})|\leq\delta|F(\theta^{\prime}_{\mu})|+\delta|F(\theta_{\mu})-F(\theta^{\prime}_{\mu})|\leq C\left(\sigma_{1}^{-1}\|f\|_{L^{2}(S^{2})}+\frac{\sigma_{1}\delta^{2}}{\sin\theta_{\mu}}\|\nabla f\|_{L^{2}(S_{\mu,\varepsilon}^{2})}\right)

and we apply (3.59) and σ1<1\sigma_{1}<1 to the last term to get (3.56). ∎

Lemma 3.16.

We have

(3.60) |(v,ψ)L2(S2)6(v,ψ)L2(S2)|C(σ22R(u,δ,μ)+σ2(Δ)1/2vL2(S2))ψL2(S2)+C(1μ)3/4(σ24R(u,δ,μ)+(Δ)1/2vL2(S2))ψL2(S2)\bigl{|}(\nabla v,\nabla\psi)_{L^{2}(S^{2})}-6(v,\psi)_{L^{2}(S^{2})}\bigr{|}\\ \leq C\left(\sigma_{2}^{-2}R(u,\delta,\mu)+\sigma_{2}\|(-\Delta)^{1/2}v\|_{L^{2}(S^{2})}\right)\|\nabla\psi\|_{L^{2}(S^{2})}\\ +\frac{C}{(1-\mu)^{3/4}}\left(\sigma_{2}^{-4}R(u,\delta,\mu)+\|(-\Delta)^{1/2}v\|_{L^{2}(S^{2})}\right)\|\psi\|_{L^{2}(S^{2})}

for all ψH1(S2)\psi\in H^{1}(S^{2}) and σ2(0,1/2)\sigma_{2}\in(0,1/2). Moreover,

(3.61) 3μ(Δ)1/2vL2(S2)2C(σ22R(u,δ,μ)+σ2(Δ)1/2vL2(S2))(Δ)1/2vL2(S2)+Cσ22(1μ)wL2(S2(0,θμ))2,3\mu\|(-\Delta)^{1/2}v\|_{L^{2}(S^{2})}^{2}\leq C\left(\sigma_{2}^{-2}R(u,\delta,\mu)+\sigma_{2}\|(-\Delta)^{1/2}v\|_{L^{2}(S^{2})}\right)\|(-\Delta)^{1/2}v\|_{L^{2}(S^{2})}\\ +\frac{C}{\sigma_{2}^{2}(1-\mu)}\|w\|_{L^{2}(S^{2}(0,\theta_{\mu}))}^{2},

where S2(0,θμ)S^{2}(0,\theta_{\mu}) is given by (2.9).

Proof.

By Lemma 3.15 we have (3.55) and W(θμ)+δF(θμ)=0W(\theta_{\mu})+\delta F(\theta_{\mu})=0. Since

Δv6v=B2,mu=w+δfμx3\displaystyle-\Delta v-6v=-B_{2,m}u=-\frac{w+\delta f}{\mu-x_{3}}

by (3.45), we take the L2(S2)L^{2}(S^{2})-inner products of both sides with ψH1(S2)\psi\in H^{1}(S^{2}) and carry out integration by parts for (Δv,ψ)L2(S2)(\Delta v,\psi)_{L^{2}(S^{2})} to get

(3.62) (v,ψ)L2(S2)6(v,ψ)L2(S2)=(w+δfμx3,ψ)L2(S2).\displaystyle(\nabla v,\nabla\psi)_{L^{2}(S^{2})}-6(v,\psi)_{L^{2}(S^{2})}=-\left(\frac{w+\delta f}{\mu-x_{3}},\psi\right)_{L^{2}(S^{2})}.

Let us estimate the right-hand side. We may assume ψ=Ψ(θ)eimφ\psi=\Psi(\theta)e^{im\varphi} since

w+δfμx3={W(θ)+δF(θ)μcosθ}eimφ.\displaystyle\frac{w+\delta f}{\mu-x_{3}}=\left\{\frac{W(\theta)+\delta F(\theta)}{\mu-\cos\theta}\right\}e^{im\varphi}.

For σ2(0,1/2)\sigma_{2}\in(0,1/2) let ε=σ22κδ2/sinθμ\varepsilon=\sigma_{2}^{2}\kappa\delta^{2}/\sin\theta_{\mu} and ε=2σ22sinθμ\varepsilon^{\prime}=2\sigma_{2}^{2}\sin\theta_{\mu}. Then ε<ε\varepsilon<\varepsilon^{\prime} by (3.36). Also, ε\varepsilon^{\prime} and thus ε\varepsilon satisfy (3.41) by σ22<1/4\sigma_{2}^{2}<1/4. Let

I1\displaystyle I_{1} =(θμε,θμ+ε),\displaystyle=(\theta_{\mu}-\varepsilon,\theta_{\mu}+\varepsilon),
I2\displaystyle I_{2} =(θμε,θμε][θμ+ε,θμ+ε),\displaystyle=(\theta_{\mu}-\varepsilon^{\prime},\theta_{\mu}-\varepsilon]\cup[\theta_{\mu}+\varepsilon,\theta_{\mu}+\varepsilon^{\prime}),
I3\displaystyle I_{3} =(0,θμε][θμ+ε,π).\displaystyle=(0,\theta_{\mu}-\varepsilon^{\prime}]\cup[\theta_{\mu}+\varepsilon^{\prime},\pi).

Then we can write

(3.63) (w+δfμx3,ψ)L2(S2)=2π0πW(θ)+δF(θ)μcosθΨ(θ)¯sinθdθ=2πk=13Jk[ψ],\displaystyle-\left(\frac{w+\delta f}{\mu-x_{3}},\psi\right)_{L^{2}(S^{2})}=-2\pi\int_{0}^{\pi}\frac{W(\theta)+\delta F(\theta)}{\mu-\cos\theta}\,\overline{\Psi(\theta)}\sin\theta\,d\theta=2\pi\sum_{k=1}^{3}J_{k}[\psi],

where

Jk[ψ]=IkW(θ)+δF(θ)μcosθΨ(θ)¯sinθdθ,k=1,2,3.\displaystyle J_{k}[\psi]=-\int_{I_{k}}\frac{W(\theta)+\delta F(\theta)}{\mu-\cos\theta}\,\overline{\Psi(\theta)}\sin\theta\,d\theta,\quad k=1,2,3.

We estimate J1[ψ]J_{1}[\psi], J2[ψ]J_{2}[\psi], and J3[ψ]J_{3}[\psi] separately. For J1[ψ]J_{1}[\psi], we have

|J1[ψ]|(I1|{W(θ)+δF(θ)}sinθμcosθ|2𝑑θ)1/2(I1|Ψ(θ)|2sinθdθ)1/2.\displaystyle|J_{1}[\psi]|\leq\left(\int_{I_{1}}\left|\frac{\{W(\theta)+\delta F(\theta)\}\sqrt{\sin\theta}}{\mu-\cos\theta}\right|^{2}\,d\theta\right)^{1/2}\left(\int_{I_{1}}|\Psi(\theta)|^{2}\sin\theta\,d\theta\right)^{1/2}.

Since W(θμ)+δF(θμ)=0W(\theta_{\mu})+\delta F(\theta_{\mu})=0, we can use (2.12) to get

I1|{W(θ)+δF(θ)}sinθμcosθ|2𝑑θCsin2θμw+δfL2(Sμ,ε2)2,\displaystyle\int_{I_{1}}\left|\frac{\{W(\theta)+\delta F(\theta)\}\sqrt{\sin\theta}}{\mu-\cos\theta}\right|^{2}\,d\theta\leq\frac{C}{\sin^{2}\theta_{\mu}}\|\nabla w+\delta\nabla f\|_{L^{2}(S_{\mu,\varepsilon}^{2})}^{2},

where Sμ,ε2S_{\mu,\varepsilon}^{2} is given by (3.37). Also, we see by (2.11) and (3.43) that

I1|Ψ(θ)|2sinθdθ\displaystyle\int_{I_{1}}|\Psi(\theta)|^{2}\sin\theta\,d\theta ΨL(0,π)2{cos(θμε)cos(θμ+ε)}\displaystyle\leq\|\Psi\|_{L^{\infty}(0,\pi)}^{2}\{\cos(\theta_{\mu}-\varepsilon)-\cos(\theta_{\mu}+\varepsilon)\}
Cεsinθμ(Δ)1/2ψL2(S2)2=Cσ22κδ2(Δ)1/2ψL2(S2)2.\displaystyle\leq C\varepsilon\sin\theta_{\mu}\|(-\Delta)^{1/2}\psi\|_{L^{2}(S^{2})}^{2}=C\sigma_{2}^{2}\kappa\delta^{2}\|(-\Delta)^{1/2}\psi\|_{L^{2}(S^{2})}^{2}.

We use these inequalities, (3.47), (3.59), and δ/sinθμC\delta/\sin\theta_{\mu}\leq C by (3.36) to deduce that

(3.64) |J1[ψ]|Cσ2(δsinθμwL2(S2)+δ2sinθμfL2(Sμ,ε2))(Δ)1/2ψL2(S2)Cσ2(R(u,δ,μ)+(Δ)1/2vL2(S2))(Δ)1/2ψL2(S2).\displaystyle\begin{aligned} |J_{1}[\psi]|&\leq C\sigma_{2}\Bigl{(}\frac{\delta}{\sin\theta_{\mu}}\|\nabla w\|_{L^{2}(S^{2})}+\frac{\delta^{2}}{\sin\theta_{\mu}}\|\nabla f\|_{L^{2}(S_{\mu,\varepsilon}^{2})}\Bigr{)}\|(-\Delta)^{1/2}\psi\|_{L^{2}(S^{2})}\\ &\leq C\sigma_{2}\Bigl{(}R(u,\delta,\mu)+\|(-\Delta)^{1/2}v\|_{L^{2}(S^{2})}\Bigr{)}\|(-\Delta)^{1/2}\psi\|_{L^{2}(S^{2})}.\end{aligned}

Note that we can use (3.59) since ε\varepsilon satisfies (3.41) and (3.50) by (3.36).

To estimate J2[ψ]J_{2}[\psi], we decompose J2[ψ]=k=15J2,k[ψ]J_{2}[\psi]=\sum_{k=1}^{5}J_{2,k}[\psi] as

J2,1[ψ]\displaystyle J_{2,1}[\psi] =I2W(θ)sinθW(θμ)sinθμμcosθΨ(θ)¯sinθ𝑑θ,\displaystyle=-\int_{I_{2}}\frac{W(\theta)\sqrt{\sin\theta}-W(\theta_{\mu})\sqrt{\sin\theta_{\mu}}}{\mu-\cos\theta}\,\overline{\Psi(\theta)}\sqrt{\sin\theta}\,d\theta,
J2,2[ψ]\displaystyle J_{2,2}[\psi] =W(θμ)sinθμI2Ψ(θ)¯sinθΨ(θμ)¯sinθμμcosθ𝑑θ,\displaystyle=-W(\theta_{\mu})\sqrt{\sin\theta_{\mu}}\int_{I_{2}}\frac{\overline{\Psi(\theta)}\sqrt{\sin\theta}-\overline{\Psi(\theta_{\mu})}\sqrt{\sin\theta_{\mu}}}{\mu-\cos\theta}\,d\theta,
J2,3[ψ]\displaystyle J_{2,3}[\psi] =W(θμ)Ψ(θμ)¯I2sinθμsinθμcosθ𝑑θ,\displaystyle=-W(\theta_{\mu})\overline{\Psi(\theta_{\mu})}\int_{I_{2}}\frac{\sin\theta_{\mu}-\sin\theta}{\mu-\cos\theta}\,d\theta,
J2,4[ψ]\displaystyle J_{2,4}[\psi] =W(θμ)Ψ(θμ)¯I2sinθμcosθ𝑑θ,\displaystyle=-W(\theta_{\mu})\overline{\Psi(\theta_{\mu})}\int_{I_{2}}\frac{\sin\theta}{\mu-\cos\theta}\,d\theta,
J2,5[ψ]\displaystyle J_{2,5}[\psi] =I2δF(θ)μcosθΨ(θ)¯sinθdθ.\displaystyle=-\int_{I_{2}}\frac{\delta F(\theta)}{\mu-\cos\theta}\,\overline{\Psi(\theta)}\sin\theta\,d\theta.

We apply Hölder’s inequality and (2.12) to J2,1[ψ]J_{2,1}[\psi] to get

|J2,1[ψ]|CsinθμwL2(S2)ΨL(0,π)cos(θμε)cos(θμ+ε).\displaystyle|J_{2,1}[\psi]|\leq\frac{C}{\sin\theta_{\mu}}\|\nabla w\|_{L^{2}(S^{2})}\|\Psi\|_{L^{\infty}(0,\pi)}\sqrt{\cos(\theta_{\mu}-\varepsilon^{\prime})-\cos(\theta_{\mu}+\varepsilon^{\prime})}.

Moreover, by (2.11), (3.43), (3.47), and εsinθμ=2σ22sin2θμ\varepsilon^{\prime}\sin\theta_{\mu}=2\sigma_{2}^{2}\sin^{2}\theta_{\mu},

(3.65) |J2,1[ψ]|Cσ2(Δ)1/2vL2(S2)(Δ)1/2ψL2(S2).\displaystyle|J_{2,1}[\psi]|\leq C\sigma_{2}\|(-\Delta)^{1/2}v\|_{L^{2}(S^{2})}\|(-\Delta)^{1/2}\psi\|_{L^{2}(S^{2})}.

Similarly, by Hölder’s inequality, (2.4), (2.12), and |I2|2ε=4σ22sinθμ|I_{2}|\leq 2\varepsilon^{\prime}=4\sigma_{2}^{2}\sin\theta_{\mu},

(3.66) |J2,2[ψ]|Cσ2|W(θμ)|(Δ)1/2ψL2(S2).\displaystyle|J_{2,2}[\psi]|\leq C\sigma_{2}|W(\theta_{\mu})|\|(-\Delta)^{1/2}\psi\|_{L^{2}(S^{2})}.

Here |I2||I_{2}| is the length of I2I_{2}. For J2,3[ψ]J_{2,3}[\psi], since

|sinθμsinθ||θθμ|,|μcosθ|12|θθμ|sinθμ,θ[0,π]\displaystyle|\sin\theta_{\mu}-\sin\theta|\leq|\theta-\theta_{\mu}|,\quad|\mu-\cos\theta|\geq\frac{1}{2}|\theta-\theta_{\mu}|\sin\theta_{\mu},\quad\theta\in[0,\pi]

by (3.44), it follows from (2.11) and |I2|2ε=4σ22sinθμ|I_{2}|\leq 2\varepsilon^{\prime}=4\sigma_{2}^{2}\sin\theta_{\mu} that

(3.67) |J2,3[ψ]|2|I2|sinθμ|W(θμ)|ΨL(0,π)Cσ22|W(θμ)|(Δ)1/2ψL2(S2).\displaystyle|J_{2,3}[\psi]|\leq\frac{2|I_{2}|}{\sin\theta_{\mu}}|W(\theta_{\mu})|\|\Psi\|_{L^{\infty}(0,\pi)}\leq C\sigma_{2}^{2}|W(\theta_{\mu})|\|(-\Delta)^{1/2}\psi\|_{L^{2}(S^{2})}.

To estimate J2,4[ψ]J_{2,4}[\psi], we see that

I2sinθμcosθ𝑑θ\displaystyle\int_{I_{2}}\frac{\sin\theta}{\mu-\cos\theta}\,d\theta =θμεθμεddθ(log(cosθμ))𝑑θ+θμ+εθμ+εddθ(log(μcosθ))𝑑θ\displaystyle=\int_{\theta_{\mu}-\varepsilon^{\prime}}^{\theta_{\mu}-\varepsilon}\frac{d}{d\theta}\bigl{(}\log(\cos\theta-\mu)\bigr{)}\,d\theta+\int_{\theta_{\mu}+\varepsilon}^{\theta_{\mu}+\varepsilon^{\prime}}\frac{d}{d\theta}\bigl{(}\log(\mu-\cos\theta)\bigr{)}\,d\theta
=logcos(θμε)μμcos(θμ+ε)logcos(θμε)μμcos(θμ+ε).\displaystyle=\log\frac{\cos(\theta_{\mu}-\varepsilon)-\mu}{\mu-\cos(\theta_{\mu}+\varepsilon)}-\log\frac{\cos(\theta_{\mu}-\varepsilon^{\prime})-\mu}{\mu-\cos(\theta_{\mu}+\varepsilon^{\prime})}.

Moreover, since ε\varepsilon and ε\varepsilon^{\prime} satisfy (3.41), we can use (3.43) to get

|I2sinθμcosθ𝑑θ|η=ε,ε|logcos(θμη)μμcos(θμ+η)|2log53.\displaystyle\left|\int_{I_{2}}\frac{\sin\theta}{\mu-\cos\theta}\,d\theta\right|\leq\sum_{\eta=\varepsilon,\varepsilon^{\prime}}\left|\log\frac{\cos(\theta_{\mu}-\eta)-\mu}{\mu-\cos(\theta_{\mu}+\eta)}\right|\leq 2\log\frac{5}{3}.

Hence we deduce from this inequality and (2.11) that

(3.68) |J2,4[ψ]|C|W(θμ)|ΨL(0,π)C|W(θμ)|(Δ)1/2ψL2(S2).\displaystyle|J_{2,4}[\psi]|\leq C|W(\theta_{\mu})|\|\Psi\|_{L^{\infty}(0,\pi)}\leq C|W(\theta_{\mu})|\|(-\Delta)^{1/2}\psi\|_{L^{2}(S^{2})}.

For J2,5[ψ]J_{2,5}[\psi], it follows from (2.7) and (2.11) that

|J2,5[ψ]|\displaystyle|J_{2,5}[\psi]| δΨL(0,π)(0π|F(θ)|2sinθdθ)1/2(I2sinθ(μcosθ)2𝑑θ)1/2\displaystyle\leq\delta\|\Psi\|_{L^{\infty}(0,\pi)}\left(\int_{0}^{\pi}|F(\theta)|^{2}\sin\theta\,d\theta\right)^{1/2}\left(\int_{I_{2}}\frac{\sin\theta}{(\mu-\cos\theta)^{2}}\,d\theta\right)^{1/2}
CδfL2(S2)(Δ)1/2ψL2(S2)(I2sinθ(μcosθ)2𝑑θ)1/2.\displaystyle\leq C\delta\|f\|_{L^{2}(S^{2})}\|(-\Delta)^{1/2}\psi\|_{L^{2}(S^{2})}\left(\int_{I_{2}}\frac{\sin\theta}{(\mu-\cos\theta)^{2}}\,d\theta\right)^{1/2}.

Moreover, since sinθ(μcosθ)2=ddθ(μcosθ)1\sin\theta(\mu-\cos\theta)^{-2}=-\frac{d}{d\theta}(\mu-\cos\theta)^{-1},

(3.69) I2sinθ(μcosθ)2𝑑θ1cos(θμε)μ+1μcos(θμ+ε)CεsinθμCσ22δ2\displaystyle\begin{aligned} \int_{I_{2}}\frac{\sin\theta}{(\mu-\cos\theta)^{2}}\,d\theta&\leq\frac{1}{\cos(\theta_{\mu}-\varepsilon)-\mu}+\frac{1}{\mu-\cos(\theta_{\mu}+\varepsilon)}\\ &\leq\frac{C}{\varepsilon\sin\theta_{\mu}}\leq C\sigma_{2}^{-2}\delta^{-2}\end{aligned}

by (3.43) and εsinθμ=σ22κδ2\varepsilon\sin\theta_{\mu}=\sigma_{2}^{2}\kappa\delta^{2}. Hence

(3.70) |J2,5[ψ]|Cσ21fL2(S2)(Δ)1/2ψL2(S2).\displaystyle|J_{2,5}[\psi]|\leq C\sigma_{2}^{-1}\|f\|_{L^{2}(S^{2})}\|(-\Delta)^{1/2}\psi\|_{L^{2}(S^{2})}.

Noting that σ2<1\sigma_{2}<1, we apply (3.65)–(3.70) to J2[ψ]=k=15J2,k[ψ]J_{2}[\psi]=\sum_{k=1}^{5}J_{2,k}[\psi] to get

(3.71) |J2[ψ]|C(|W(θμ)|+σ21fL2(S2)+σ2(Δ)1/2vL2(S2))(Δ)1/2ψL2(S2).\displaystyle|J_{2}[\psi]|\leq C\Bigl{(}|W(\theta_{\mu})|+\sigma_{2}^{-1}\|f\|_{L^{2}(S^{2})}+\sigma_{2}\|(-\Delta)^{1/2}v\|_{L^{2}(S^{2})}\Bigr{)}\|(-\Delta)^{1/2}\psi\|_{L^{2}(S^{2})}.

Let us estimate J3[ψ]J_{3}[\psi]. We write J3[ψ]=J3,1[ψ]+J3,2[ψ]J_{3}[\psi]=J_{3,1}[\psi]+J_{3,2}[\psi] with

(3.72) J3,1[ψ]=I3W(θ)μcosθΨ(θ)¯sinθdθ,J3,2[ψ]=I3δF(θ)μcosθΨ(θ)¯sinθdθ.\displaystyle\begin{aligned} J_{3,1}[\psi]&=-\int_{I_{3}}\frac{W(\theta)}{\mu-\cos\theta}\,\overline{\Psi(\theta)}\sin\theta\,d\theta,\\ J_{3,2}[\psi]&=-\int_{I_{3}}\frac{\delta F(\theta)}{\mu-\cos\theta}\,\overline{\Psi(\theta)}\sin\theta\,d\theta.\end{aligned}

To estimate J3,1[ψ]J_{3,1}[\psi], we split J3,1[ψ]=K1[ψ]+K2[ψ]J_{3,1}[\psi]=K_{1}[\psi]+K_{2}[\psi] into

K1[ψ]\displaystyle K_{1}[\psi] =I3W(θ)sinθW(θμ)sinθμμcosθΨ(θ)¯sinθ𝑑θ,\displaystyle=-\int_{I_{3}}\frac{W(\theta)\sqrt{\sin\theta}-W(\theta_{\mu})\sqrt{\sin\theta_{\mu}}}{\mu-\cos\theta}\,\overline{\Psi(\theta)}\sqrt{\sin\theta}\,d\theta,
K2[ψ]\displaystyle K_{2}[\psi] =W(θμ)sinθμI3Ψ(θ)¯sinθμcosθ𝑑θ.\displaystyle=-W(\theta_{\mu})\sqrt{\sin\theta_{\mu}}\int_{I_{3}}\frac{\overline{\Psi(\theta)}\sqrt{\sin\theta}}{\mu-\cos\theta}\,d\theta.

We use Hölder’s inequality, (2.4), (2.7), and (2.12) to K1[ψ]K_{1}[\psi] to get

|K1[ψ]|Csinθμ(Δ)1/2wL2(S2)ψL2(S2).\displaystyle|K_{1}[\psi]|\leq\frac{C}{\sin\theta_{\mu}}\|(-\Delta)^{1/2}w\|_{L^{2}(S^{2})}\|\psi\|_{L^{2}(S^{2})}.

Also, if θI3\theta\in I_{3}, i.e. |θθμ|ε|\theta-\theta_{\mu}|\geq\varepsilon^{\prime}, then it follows from (3.44) that

|μcosθ|12|θθμ|sinθμ12εsinθμ=σ22sin2θμ.\displaystyle|\mu-\cos\theta|\geq\frac{1}{2}|\theta-\theta_{\mu}|\sin\theta_{\mu}\geq\frac{1}{2}\varepsilon^{\prime}\sin\theta_{\mu}=\sigma_{2}^{2}\sin^{2}\theta_{\mu}.

By this inequality, Hölder’s inequality, and (2.7), we have

|K2[ψ]|C|W(θμ)|ψL2(S2)σ22sin3/2θμ.\displaystyle|K_{2}[\psi]|\leq\frac{C|W(\theta_{\mu})|\|\psi\|_{L^{2}(S^{2})}}{\sigma_{2}^{2}\sin^{3/2}\theta_{\mu}}.

These estimates, (3.36), and (3.47) imply that

(3.73) |J3,1[ψ]|Csin3/2θμ(σ22|W(θμ)|+sinθμ(Δ)1/2wL2(S2))ψL2(S2)C(1μ)3/4(σ22|W(θμ)|+(Δ)1/2vL2(S2))ψL2(S2).\displaystyle\begin{aligned} |J_{3,1}[\psi]|&\leq\frac{C}{\sin^{3/2}\theta_{\mu}}\Bigl{(}\sigma_{2}^{-2}|W(\theta_{\mu})|+\sqrt{\sin\theta_{\mu}}\|(-\Delta)^{1/2}w\|_{L^{2}(S^{2})}\Bigr{)}\|\psi\|_{L^{2}(S^{2})}\\ &\leq\frac{C}{(1-\mu)^{3/4}}\Bigl{(}\sigma_{2}^{-2}|W(\theta_{\mu})|+\|(-\Delta)^{1/2}v\|_{L^{2}(S^{2})}\Bigr{)}\|\psi\|_{L^{2}(S^{2})}.\end{aligned}

Also, as in (3.69), we see by (3.43) and ε=2σ22sinθμ\varepsilon^{\prime}=2\sigma_{2}^{2}\sin\theta_{\mu} that

I3sinθ(μcosθ)2𝑑θ1cos(θμε)μ+1μcos(θμ+ε)CεsinθμCσ22sin2θμ.\displaystyle\int_{I_{3}}\frac{\sin\theta}{(\mu-\cos\theta)^{2}}\,d\theta\leq\frac{1}{\cos(\theta_{\mu}-\varepsilon^{\prime})-\mu}+\frac{1}{\mu-\cos(\theta_{\mu}+\varepsilon^{\prime})}\leq\frac{C}{\varepsilon^{\prime}\sin\theta_{\mu}}\leq\frac{C}{\sigma_{2}^{2}\sin^{2}\theta_{\mu}}.

We deduce from this inequality, (2.7), (2.11), and δ/sinθμC\delta/\sin\theta_{\mu}\leq C by (3.36) that

(3.74) |J3,2[ψ]|δΨL(0,π)(0π|F(θ)|2sinθdθ)1/2(I3sinθ(μcosθ)2𝑑θ)1/2Cσ21fL2(S2)(Δ)1/2ψL2(S2).\displaystyle\begin{aligned} |J_{3,2}[\psi]|&\leq\delta\|\Psi\|_{L^{\infty}(0,\pi)}\left(\int_{0}^{\pi}|F(\theta)|^{2}\sin\theta\,d\theta\right)^{1/2}\left(\int_{I_{3}}\frac{\sin\theta}{(\mu-\cos\theta)^{2}}\,d\theta\right)^{1/2}\\ &\leq C\sigma_{2}^{-1}\|f\|_{L^{2}(S^{2})}\|(-\Delta)^{1/2}\psi\|_{L^{2}(S^{2})}.\end{aligned}

Applying (3.73) and (3.74) to J3[ψ]=J3,1[ψ]+J3,2[ψ]J_{3}[\psi]=J_{3,1}[\psi]+J_{3,2}[\psi], we get

(3.75) |J3[ψ]|C(1μ)3/4(σ22|W(θμ)|+(Δ)1/2vL2(S2))ψL2(S2)+Cσ21fL2(S2)(Δ)1/2ψL2(S2).|J_{3}[\psi]|\leq\frac{C}{(1-\mu)^{3/4}}\Bigl{(}\sigma_{2}^{-2}|W(\theta_{\mu})|+\|(-\Delta)^{1/2}v\|_{L^{2}(S^{2})}\Bigr{)}\|\psi\|_{L^{2}(S^{2})}\\ +C\sigma_{2}^{-1}\|f\|_{L^{2}(S^{2})}\|(-\Delta)^{1/2}\psi\|_{L^{2}(S^{2})}.

Now we deduce from (3.62)–(3.64), (3.71), (3.75), and σ2<1\sigma_{2}<1 that

|(v,ψ)L2(S2)6(v,ψ)L2(S2)|2π(|J1[ψ]|+|J2[ψ]|+|J3[ψ]|)C(|W(θμ)|+σ21R(u,δ,μ)+σ2(Δ)1/2vL2(S2))(Δ)1/2ψL2(S2)+C(1μ)3/4(σ22|W(θμ)|+(Δ)1/2vL2(S2))ψL2(S2).\bigl{|}(\nabla v,\nabla\psi)_{L^{2}(S^{2})}-6(v,\psi)_{L^{2}(S^{2})}\bigr{|}\leq 2\pi(|J_{1}[\psi]|+|J_{2}[\psi]|+|J_{3}[\psi]|)\\ \leq C\Bigl{(}|W(\theta_{\mu})|+\sigma_{2}^{-1}R(u,\delta,\mu)+\sigma_{2}\|(-\Delta)^{1/2}v\|_{L^{2}(S^{2})}\Bigr{)}\|(-\Delta)^{1/2}\psi\|_{L^{2}(S^{2})}\\ +\frac{C}{(1-\mu)^{3/4}}\Bigl{(}\sigma_{2}^{-2}|W(\theta_{\mu})|+\|(-\Delta)^{1/2}v\|_{L^{2}(S^{2})}\Bigr{)}\|\psi\|_{L^{2}(S^{2})}.

In this inequality, we further use (3.56) with σ1=σ22\sigma_{1}=\sigma_{2}^{2} and apply (2.4) to ψ\psi (note that we assume ψ=Ψ(θ)eimφ\psi=\Psi(\theta)e^{im\varphi} with m0m\neq 0). Then we obtain (3.60) since σ2<1\sigma_{2}<1.

Let us show (3.61). We set ψ=w\psi=w for J3,1[ψ]J_{3,1}[\psi] (see (3.72)) to get

J3,1[w]=I3|W(θ)|2μcosθsinθdθ=(0θμε+θμ+επ)|W(θ)|2cosθμsinθdθ.\displaystyle J_{3,1}[w]=-\int_{I_{3}}\frac{|W(\theta)|^{2}}{\mu-\cos\theta}\sin\theta\,d\theta=\left(\int_{0}^{\theta_{\mu}-\varepsilon^{\prime}}+\int_{\theta_{\mu}+\varepsilon^{\prime}}^{\pi}\right)\frac{|W(\theta)|^{2}}{\cos\theta-\mu}\sin\theta\,d\theta\in\mathbb{R}.

Moreover, since cosθμ0\cos\theta-\mu\leq 0 for θ(θμ+ε,π)\theta\in(\theta_{\mu}+\varepsilon^{\prime},\pi) and

cosθμ12(θμθ)sinθμ12εsinθμ=σ22sin2θμσ22(1μ)\displaystyle\cos\theta-\mu\geq\frac{1}{2}(\theta_{\mu}-\theta)\sin\theta_{\mu}\geq\frac{1}{2}\varepsilon^{\prime}\sin\theta_{\mu}=\sigma_{2}^{2}\sin^{2}\theta_{\mu}\geq\sigma_{2}^{2}(1-\mu)

for θ(0,θμε)\theta\in(0,\theta_{\mu}-\varepsilon^{\prime}) by (3.36) and (3.44), we see by (2.7) that

(3.76) J3,1[w]0θμε|W(θ)|2cosθμsinθdθCσ22(1μ)wL2(S2(0,θμ))2,\displaystyle J_{3,1}[w]\leq\int_{0}^{\theta_{\mu}-\varepsilon^{\prime}}\frac{|W(\theta)|^{2}}{\cos\theta-\mu}\sin\theta\,d\theta\leq\frac{C}{\sigma_{2}^{2}(1-\mu)}\|w\|_{L^{2}(S^{2}(0,\theta_{\mu}))}^{2},

where S2(0,θμ)S^{2}(0,\theta_{\mu}) is given by (2.9). Also, since v=Δ1u𝒴mv=\Delta^{-1}u\in\mathcal{Y}_{m},

(v,Y2m)L2(S2)=0,(v,Y2m)L2(S2)=(v,ΔY2m)L2(S2)=λ2(v,Y2m)L2(S2)=0.\displaystyle(v,Y_{2}^{m})_{L^{2}(S^{2})}=0,\quad(\nabla v,\nabla Y_{2}^{m})_{L^{2}(S^{2})}=-(v,\Delta Y_{2}^{m})_{L^{2}(S^{2})}=\lambda_{2}(v,Y_{2}^{m})_{L^{2}(S^{2})}=0.

Note that Y2m0Y_{2}^{m}\equiv 0 when |m|3|m|\geq 3. By this fact and w=6μvβmY2mw=6\mu v-\beta_{m}Y_{2}^{m},

(v,w)L2(S2)6(v,w)L2(S2)=6μ(vL2(S2)26vL2(S2)2).\displaystyle(\nabla v,\nabla w)_{L^{2}(S^{2})}-6(v,w)_{L^{2}(S^{2})}=6\mu\Bigl{(}\|\nabla v\|_{L^{2}(S^{2})}^{2}-6\|v\|_{L^{2}(S^{2})}^{2}\Bigr{)}\in\mathbb{R}.

Thus, setting ψ=w\psi=w in (3.62) and using (3.63), we have

6μ(vL2(S2)26vL2(S2)2)\displaystyle 6\mu\Bigl{(}\|\nabla v\|_{L^{2}(S^{2})}^{2}-6\|v\|_{L^{2}(S^{2})}^{2}\Bigr{)} =2πk=13Jk[w]=2π{Re(J1[w]+J2[w]+J3,2[w])+J3,1[w]}\displaystyle=2\pi\sum_{k=1}^{3}J_{k}[w]=2\pi\{\mathrm{Re}(J_{1}[w]+J_{2}[w]+J_{3,2}[w])+J_{3,1}[w]\}
2π(|J1[w]|+|J2[w]|+|J3,2[w]|+J3,1[w])\displaystyle\leq 2\pi(|J_{1}[w]|+|J_{2}[w]|+|J_{3,2}[w]|+J_{3,1}[w])

and we use (3.64), (3.71), (3.74), (3.76), and σ2<1\sigma_{2}<1 to find that

(3.77) 6μ(vL2(S2)26vL2(S2)2)C(|W(θμ)|+σ21R(u,δ,μ)+σ2(Δ)1/2vL2(S2))(Δ)1/2wL2(S2)+Cσ22(1μ)wL2(S2(0,θμ))2.6\mu\Bigl{(}\|\nabla v\|_{L^{2}(S^{2})}^{2}-6\|v\|_{L^{2}(S^{2})}^{2}\Bigr{)}\\ \leq C\Bigl{(}|W(\theta_{\mu})|+\sigma_{2}^{-1}R(u,\delta,\mu)+\sigma_{2}\|(-\Delta)^{1/2}v\|_{L^{2}(S^{2})}\Bigr{)}\|(-\Delta)^{1/2}w\|_{L^{2}(S^{2})}\\ +\frac{C}{\sigma_{2}^{2}(1-\mu)}\|w\|_{L^{2}(S^{2}(0,\theta_{\mu}))}^{2}.

Moreover, since u𝒴mu\in\mathcal{Y}_{m}, we can use (3.22) in the proof of Lemma 3.7 to get

6μ(vL2(S2)26vL2(S2)2)3μ(Δ)1/2uL2(S2)2=3μ(Δ)1/2vL2(S2)2.\displaystyle 6\mu\Bigl{(}\|\nabla v\|_{L^{2}(S^{2})}^{2}-6\|v\|_{L^{2}(S^{2})}^{2}\Bigr{)}\geq 3\mu\|(-\Delta)^{-1/2}u\|_{L^{2}(S^{2})}^{2}=3\mu\|(-\Delta)^{1/2}v\|_{L^{2}(S^{2})}^{2}.

We apply this inequality, (3.47) for (Δ)1/2w(-\Delta)^{1/2}w, and (3.56) with σ1=σ22\sigma_{1}=\sigma_{2}^{2} to (3.77), and then use σ2<1\sigma_{2}<1 to obtain (3.61). ∎

Lemma 3.17.

There exist constants μ0(0,1)\mu_{0}\in(0,1) and C>0C>0 such that

(3.78) 11μwL2(S2(θμ,π))2C(σ31R(u,δ,μ)2+σ3(Δ)1/2vL2(S2)2)\displaystyle\frac{1}{1-\mu}\|w\|_{L^{2}(S^{2}(\theta_{\mu},\pi))}^{2}\leq C\left(\sigma_{3}^{-1}R(u,\delta,\mu)^{2}+\sigma_{3}\|(-\Delta)^{1/2}v\|_{L^{2}(S^{2})}^{2}\right)

for all μ[μ0,1)\mu\in[\mu_{0},1) and σ3(0,1/2)\sigma_{3}\in(0,1/2), where S2(θμ,π)S^{2}(\theta_{\mu},\pi) is given by (2.9).

Proof.

Since (μx3)Δw=6x3w+6μδf(\mu-x_{3})\Delta w=6x_{3}w+6\mu\delta f by (3.46), we use (3.55), (5.1), and x3=cosθx_{3}=\cos\theta to rewrite this equation as

(μcosθ){1sinθddθ(sinθW(θ))m2sin2θW(θ)}=6cosθW(θ)+6μδF(θ)\displaystyle(\mu-\cos\theta)\left\{\frac{1}{\sin\theta}\frac{d}{d\theta}\bigl{(}\sin\theta\,W^{\prime}(\theta)\bigr{)}-\frac{m^{2}}{\sin^{2}\theta}W(\theta)\right\}=6\cos\theta\,W(\theta)+6\mu\delta F(\theta)

for θ(0,π)\theta\in(0,\pi). We multiply both sides by W(θ)¯sinθ\overline{W(\theta)}\sin\theta and integrate them over (θμ,π)(\theta_{\mu},\pi). Then we carry out integration by parts and use μcosθμ=0\mu-\cos\theta_{\mu}=0 and sinπ=0\sin\pi=0 to get

θμπW(θ)W(θ)¯sin2θdθθμπ(μcosθ)(|W(θ)|2+m2sin2θ|W(θ)|2)sinθdθ=6θμπ|W(θ)|2cosθsinθdθ+6μδθμπF(θ)W(θ)¯sinθdθ.-\int_{\theta_{\mu}}^{\pi}W^{\prime}(\theta)\overline{W(\theta)}\sin^{2}\theta\,d\theta-\int_{\theta_{\mu}}^{\pi}(\mu-\cos\theta)\left(|W^{\prime}(\theta)|^{2}+\frac{m^{2}}{\sin^{2}\theta}|W(\theta)|^{2}\right)\sin\theta\,d\theta\\ =6\int_{\theta_{\mu}}^{\pi}|W(\theta)|^{2}\cos\theta\sin\theta\,d\theta+6\mu\delta\int_{\theta_{\mu}}^{\pi}F(\theta)\overline{W(\theta)}\sin\theta\,d\theta.

We take the real part of this equality and use

ReθμπW(θ)W(θ)¯sin2θdθ=12|W(θμ)|2sin2θμθμπ|W(θ)|2cosθsinθdθ\displaystyle\mathrm{Re}\int_{\theta_{\mu}}^{\pi}W^{\prime}(\theta)\overline{W(\theta)}\sin^{2}\theta\,d\theta=-\frac{1}{2}|W(\theta_{\mu})|^{2}\sin^{2}\theta_{\mu}-\int_{\theta_{\mu}}^{\pi}|W(\theta)|^{2}\cos\theta\sin\theta\,d\theta

by integration by parts and sinπ=0\sin\pi=0 to get

(3.79) J+K=6μδReθμπF(θ)W(θ)¯sinθdθ+12|W(θμ)|2sin2θμ,\displaystyle J+K=-6\mu\delta\,\mathrm{Re}\int_{\theta_{\mu}}^{\pi}F(\theta)\overline{W(\theta)}\sin\theta\,d\theta+\frac{1}{2}|W(\theta_{\mu})|^{2}\sin^{2}\theta_{\mu},

where

(3.80) J=θμπ(μcosθ)(|W(θ)|2+m2sin2θ|W(θ)|2)sinθdθ,K=5θμπ|W(θ)|2cosθsinθdθ.\displaystyle\begin{aligned} J&=\int_{\theta_{\mu}}^{\pi}(\mu-\cos\theta)\left(|W^{\prime}(\theta)|^{2}+\frac{m^{2}}{\sin^{2}\theta}|W(\theta)|^{2}\right)\sin\theta\,d\theta,\\ K&=5\int_{\theta_{\mu}}^{\pi}|W(\theta)|^{2}\cos\theta\sin\theta\,d\theta.\end{aligned}

Now we claim that there exist μ0(0,1)\mu_{0}\in(0,1) and C>0C>0 such that

(3.81) J+KCwL2(S2(θμ,π))2\displaystyle J+K\geq C\|w\|_{L^{2}(S^{2}(\theta_{\mu},\pi))}^{2}

for all μ[μ0,1)\mu\in[\mu_{0},1). If this claim is valid, then we apply Hölder’s and Young’s inequalities, (2.7), (3.81), μ1\mu\leq 1, and fL2(S2(θμ,π))fL2(S2)\|f\|_{L^{2}(S^{2}(\theta_{\mu},\pi))}\leq\|f\|_{L^{2}(S^{2})} to (3.79) to get

wL2(S2(θμ,π))212wL2(S2(θμ,π))2+C(δ2fL2(S2)2+|W(θμ)|2sin2θμ)\displaystyle\|w\|_{L^{2}(S^{2}(\theta_{\mu},\pi))}^{2}\leq\frac{1}{2}\|w\|_{L^{2}(S^{2}(\theta_{\mu},\pi))}^{2}+C\left(\delta^{2}\|f\|_{L^{2}(S^{2})}^{2}+|W(\theta_{\mu})|^{2}\sin^{2}\theta_{\mu}\right)

and we subtract 12wL2(S2(θμ,π))2\frac{1}{2}\|w\|_{L^{2}(S^{2}(\theta_{\mu},\pi))}^{2} from both sides, dividing them by 1μ1-\mu, and use (3.36) and (3.56) with σ1=σ31/2\sigma_{1}=\sigma_{3}^{1/2} to obtain (3.78).

Let us show (3.81). We split K=K1+K2K=K_{1}+K_{2} into

K1=5θμΘ|W(θ)|2cosθsinθdθ,K2=5Θπ|W(θ)|2cosθsinθdθ,\displaystyle K_{1}=5\int_{\theta_{\mu}}^{\Theta}|W(\theta)|^{2}\cos\theta\sin\theta\,d\theta,\quad K_{2}=5\int_{\Theta}^{\pi}|W(\theta)|^{2}\cos\theta\sin\theta\,d\theta,

where Θ(θμ,π/2)\Theta\in(\theta_{\mu},\pi/2) is fixed later. By cosθcosΘ\cos\theta\geq\cos\Theta for θ[θμ,Θ]\theta\in[\theta_{\mu},\Theta],

(3.82) K15cosΘθμΘ|W(θ)|2sinθdθ.\displaystyle K_{1}\geq 5\cos\Theta\int_{\theta_{\mu}}^{\Theta}|W(\theta)|^{2}\sin\theta\,d\theta.

Since w=W(θ)eimφH3(S2)C1(S2)w=W(\theta)e^{im\varphi}\in H^{3}(S^{2})\subset C^{1}(S^{2}) and m0m\neq 0, we have W(π)=0W(\pi)=0 by Lemma 2.1. By this fact, 2cosθsinθ=ddθ(cos2θcos2Θ)2\cos\theta\sin\theta=-\frac{d}{d\theta}(\cos^{2}\theta-\cos^{2}\Theta), and integration by parts, we have

K2\displaystyle K_{2} =52[(cos2θcos2Θ)|W(θ)|2]Θπ+5Θπ(cos2θcos2Θ)Re(W(θ)W(θ)¯)𝑑θ\displaystyle=-\frac{5}{2}\Bigl{[}(\cos^{2}\theta-\cos^{2}\Theta)|W(\theta)|^{2}\Bigr{]}_{\Theta}^{\pi}+5\int_{\Theta}^{\pi}(\cos^{2}\theta-\cos^{2}\Theta)\mathrm{Re}\Bigl{(}W^{\prime}(\theta)\overline{W(\theta)}\Bigr{)}\,d\theta
=5Θπχ(θ)(μcosθ)Re(W(θ)W(θ)¯)𝑑θ,\displaystyle=5\int_{\Theta}^{\pi}\chi(\theta)(\mu-\cos\theta)\mathrm{Re}\Bigl{(}W^{\prime}(\theta)\overline{W(\theta)}\Bigr{)}\,d\theta,

where χ(θ)=(cos2θcos2Θ)(μcosθ)1\chi(\theta)=(\cos^{2}\theta-\cos^{2}\Theta)(\mu-\cos\theta)^{-1}. Hence

(3.83) |K2|5χL(Θ,π)Θπ(μcosθ)|W(θ)||W(θ)|𝑑θ52χL(Θ,π)J\displaystyle|K_{2}|\leq 5\|\chi\|_{L^{\infty}(\Theta,\pi)}\int_{\Theta}^{\pi}(\mu-\cos\theta)|W^{\prime}(\theta)||W(\theta)|\,d\theta\leq\frac{5}{2}\|\chi\|_{L^{\infty}(\Theta,\pi)}\cdot J

by Young’s inequality and m21m^{2}\geq 1 (see (3.80)). To estimate χL(Θ,π)\|\chi\|_{L^{\infty}(\Theta,\pi)}, let

ρ(t)=χ(arccost)=t2T2μt,t[1,T],\displaystyle\rho(t)=\chi(\arccos t)=\frac{t^{2}-T^{2}}{\mu-t},\quad t\in[-1,T],

where T=cosΘ(0,μ)T=\cos\Theta\in(0,\mu). Then

(3.84) dρdt(t)=(tt)(tt+)(μt)2,t±=μ±μ2T2.\displaystyle\frac{d\rho}{dt}(t)=-\frac{(t-t_{-})(t-t_{+})}{(\mu-t)^{2}},\quad t_{\pm}=\mu\pm\sqrt{\mu^{2}-T^{2}}.

Now let μ0=23/5\mu_{0}=\sqrt{23}/5 and T=cosΘ=7/5T=\cos\Theta=\sqrt{7}/5. Then for μ[μ0,1)\mu\in[\mu_{0},1) we see by the mean value theorem for μμ2s\mu-\sqrt{\mu^{2}-s} with s[0,T2]s\in[0,T^{2}] and μ2T216/25\mu^{2}-T^{2}\geq 16/25 that

(3.85) μμ2T2T22μ2T258T2\displaystyle\mu-\sqrt{\mu^{2}-T^{2}}\leq\frac{T^{2}}{2\sqrt{\mu^{2}-T^{2}}}\leq\frac{5}{8}T^{2}

and thus 0<t<T<t+0<t_{-}<T<t_{+} by T<μ<1T<\mu<1. Hence, by (3.84) and ρ(T)=0\rho(T)=0,

χL(Θ,π)\displaystyle\|\chi\|_{L^{\infty}(\Theta,\pi)} =ρL(1,T)=max{|ρ(1)|,|ρ(t)|}\displaystyle=\|\rho\|_{L^{\infty}(-1,T)}=\max\{|\rho(-1)|,|\rho(t_{-})|\}
=max{1T21+μ,2(μμ2T2)}.\displaystyle=\max\left\{\frac{1-T^{2}}{1+\mu},2\left(\mu-\sqrt{\mu^{2}-T^{2}}\right)\right\}.

Moreover, by μμ0=23/5\mu\geq\mu_{0}=\sqrt{23}/5, T2=7/25T^{2}=7/25, (3.85), and 4<23<54<\sqrt{23}<5,

521T21+μ95+23,522(μμ2T2)78,78<910<95+23<1.\displaystyle\frac{5}{2}\cdot\frac{1-T^{2}}{1+\mu}\leq\frac{9}{5+\sqrt{23}},\quad\frac{5}{2}\cdot 2\left(\mu-\sqrt{\mu^{2}-T^{2}}\right)\leq\frac{7}{8},\quad\frac{7}{8}<\frac{9}{10}<\frac{9}{5+\sqrt{23}}<1.

Hence 52χL(Θ,π)9/(5+23)\frac{5}{2}\|\chi\|_{L^{\infty}(\Theta,\pi)}\leq 9/(5+\sqrt{23}) and

(3.86) J+K2J|K2|C0J\displaystyle J+K_{2}\geq J-|K_{2}|\geq C_{0}J

with C0=19/(5+23)>0C_{0}=1-9/(5+\sqrt{23})>0 by (3.83). Moreover,

(3.87) JΘπμcosθsin2θ|W(θ)|2sinθdθ2375Θπ|W(θ)|2sinθdθ\displaystyle J\geq\int_{\Theta}^{\pi}\frac{\mu-\cos\theta}{\sin^{2}\theta}|W(\theta)|^{2}\sin\theta\,d\theta\geq\frac{\sqrt{23}-\sqrt{7}}{5}\int_{\Theta}^{\pi}|W(\theta)|^{2}\sin\theta\,d\theta

by m21m^{2}\geq 1, sin2θ1\sin^{2}\theta\leq 1 and cosθcosΘ=7/5\cos\theta\leq\cos\Theta=\sqrt{7}/5 for θ[Θ,π]\theta\in[\Theta,\pi], and μμ0=23/5\mu\geq\mu_{0}=\sqrt{23}/5. Thus, noting that K=K1+K2K=K_{1}+K_{2}, we combine (3.82), (3.86), and (3.87) and apply (2.7) to w=W(θ)eimφw=W(\theta)e^{im\varphi} to get (3.81). ∎

Now let us prove Lemma 3.9. We give the proofs of (3.33) and (3.34) separately.

Proof of (3.33).

As we mentioned above, we assume μ0\mu\geq 0 and prove (3.33) by contradiction. Assume that for each kk\in\mathbb{N} there exist mk{0}m_{k}\in\mathbb{Z}\setminus\{0\}, uk𝒴mkH1(S2)u_{k}\in\mathcal{Y}_{m_{k}}\cap H^{1}(S^{2}), δk(0,1]\delta_{k}\in(0,1], and μk[0,1κδk2]\mu_{k}\in[0,1-\kappa\delta_{k}^{2}] such that, if we define vkv_{k}, wkw_{k}, fkf_{k}, and Rk=R(uk,δk,μk)R_{k}=R(u_{k},\delta_{k},\mu_{k}) by (3.35) and (3.40) with mm, uu, δ\delta, and μ\mu replaced by mkm_{k}, uku_{k}, δk\delta_{k}, and μk\mu_{k}, then

(3.88) δk2ukL2(S2)2+(Δ)1/2vkL2(S2)2+11μkwkL2(S2)2=1,\displaystyle\delta_{k}^{2}\|u_{k}\|_{L^{2}(S^{2})}^{2}+\|(-\Delta)^{1/2}v_{k}\|_{L^{2}(S^{2})}^{2}+\frac{1}{1-\mu_{k}}\|w_{k}\|_{L^{2}(S^{2})}^{2}=1,
(3.89) limkRk=limk(fkL2(S2)+δk3(1μk)1/2(Δ)1/2ukL2(S2))=0.\displaystyle\lim_{k\to\infty}R_{k}=\lim_{k\to\infty}\left(\|f_{k}\|_{L^{2}(S^{2})}+\frac{\delta_{k}^{3}}{(1-\mu_{k})^{1/2}}\|(-\Delta)^{1/2}u_{k}\|_{L^{2}(S^{2})}\right)=0.

Taking subsequences, we may assume that

limkmk=m{±}{0},\displaystyle\lim_{k\to\infty}m_{k}=m_{\infty}\in\{\pm\infty\}\cup\mathbb{Z}\setminus\{0\},
limkδk=δ[0,1],limkμk=μ[0,1κδ2].\displaystyle\lim_{k\to\infty}\delta_{k}=\delta_{\infty}\in[0,1],\quad\lim_{k\to\infty}\mu_{k}=\mu_{\infty}\in[0,1-\kappa\delta_{\infty}^{2}].

Suppose first that δ>0\delta_{\infty}>0. Then δkδ/2>0\delta_{k}\geq\delta_{\infty}/2>0 for sufficiently large kk. Moreover, we see by (3.47) and the boundedness of Δ1\Delta^{-1} on L02(S2)L_{0}^{2}(S^{2}) that

wkL2(S2)C(Δ)1/2vkL2(S2)=C(Δ)1/2ukL2(S2)C(Δ)1/2ukL2(S2).\displaystyle\|w_{k}\|_{L^{2}(S^{2})}\leq C\|(-\Delta)^{1/2}v_{k}\|_{L^{2}(S^{2})}=C\|(-\Delta)^{-1/2}u_{k}\|_{L^{2}(S^{2})}\leq C\|(-\Delta)^{1/2}u_{k}\|_{L^{2}(S^{2})}.

By these facts, (1μk)11(1-\mu_{k})^{-1}\geq 1, (2.6), and (3.89), we find that

δk2ukL2(S2)2+(Δ)1/2vkL2(S2)2+11μkwkL2(S2)2Cδk61μk(Δ)1/2ukL2(S2)20\delta_{k}^{2}\|u_{k}\|_{L^{2}(S^{2})}^{2}+\|(-\Delta)^{1/2}v_{k}\|_{L^{2}(S^{2})}^{2}+\frac{1}{1-\mu_{k}}\|w_{k}\|_{L^{2}(S^{2})}^{2}\leq\frac{C\delta_{k}^{6}}{1-\mu_{k}}\|(-\Delta)^{1/2}u_{k}\|_{L^{2}(S^{2})}^{2}\to 0

as kk\to\infty, which contradicts (3.88).

Now let δ=0\delta_{\infty}=0. We see by (3.49) for uku_{k}, (3.88), and (3.89) that

(3.90) lim infk((Δ)1/2vkL2(S2)2+11μkwkL2(S2)2)C>0.\displaystyle\liminf_{k\to\infty}\left(\|(-\Delta)^{1/2}v_{k}\|_{L^{2}(S^{2})}^{2}+\frac{1}{1-\mu_{k}}\|w_{k}\|_{L^{2}(S^{2})}^{2}\right)\geq C>0.

Let θk=arccosμk(0,π/2]\theta_{k}=\arccos\mu_{k}\in(0,\pi/2]. Then we see by Lemma 3.15 that

(3.91) wk=Wk(θ)eimkφ,fk=Fk(θ)eimkφ,Wk(θk)+δkFk(θk)=0\displaystyle w_{k}=W_{k}(\theta)e^{im_{k}\varphi},\quad f_{k}=F_{k}(\theta)e^{im_{k}\varphi},\quad W_{k}(\theta_{k})+\delta_{k}F_{k}(\theta_{k})=0

Also, by (3.56) for Wk(θk)W_{k}(\theta_{k}), (3.88), and (3.89),

lim supk|Wk(θk)|Clim supk(σ1Rk+σ(Δ)1/2vkL2(S2))Cσ\displaystyle\limsup_{k\to\infty}|W_{k}(\theta_{k})|\leq C\limsup_{k\to\infty}\Bigl{(}\sigma^{-1}R_{k}+\sigma\|(-\Delta)^{1/2}v_{k}\|_{L^{2}(S^{2})}\Bigr{)}\leq C\sigma

for all σ(0,1/2)\sigma\in(0,1/\sqrt{2}) and thus

(3.92) limkWk(θk)=0.\displaystyle\lim_{k\to\infty}W_{k}(\theta_{k})=0.

In what follows, we consider two cases.

Case 1: 0μ<10\leq\mu_{\infty}<1. If m=±m_{\infty}=\pm\infty, then |mk|3|m_{k}|\geq 3 for sufficiently large kk\in\mathbb{N}. Then since vk𝒴mkv_{k}\in\mathcal{Y}_{m_{k}} is of the form (3.9) with Nmk=|mk|N_{m_{k}}=|m_{k}|, it follows from (2.3), λn=n(n+1)|mk|2\lambda_{n}=n(n+1)\geq|m_{k}|^{2} for n|mk|n\geq|m_{k}|, and (3.88) that

(3.93) vkL2(S2)1|mk|(Δ)1/2vkL2(S2)1|mk|0ask.\displaystyle\|v_{k}\|_{L^{2}(S^{2})}\leq\frac{1}{|m_{k}|}\|(-\Delta)^{1/2}v_{k}\|_{L^{2}(S^{2})}\leq\frac{1}{|m_{k}|}\to 0\quad\text{as}\quad k\to\infty.

We set ψ=vk\psi=v_{k} in (3.60) for vkv_{k} and use (2.4) and (3.88) to find that

(Δ)1/2vkL2(S2)26vkL2(S2)2+C(σ2Rk+σ)+C(1μk)3/4(σ4Rk+1)vkL2(S2)\displaystyle\|(-\Delta)^{1/2}v_{k}\|_{L^{2}(S^{2})}^{2}\leq 6\|v_{k}\|_{L^{2}(S^{2})}^{2}+C(\sigma^{-2}R_{k}+\sigma)+\frac{C}{(1-\mu_{k})^{3/4}}(\sigma^{-4}R_{k}+1)\|v_{k}\|_{L^{2}(S^{2})}

for all σ(0,1/2)\sigma\in(0,1/2). Then we send kk\to\infty and use (3.89), (3.93), and μ<1\mu_{\infty}<1 to get

lim supk(Δ)1/2vkL2(S2)2Cσfor allσ(0,12).\displaystyle\limsup_{k\to\infty}\|(-\Delta)^{1/2}v_{k}\|_{L^{2}(S^{2})}^{2}\leq C\sigma\quad\text{for all}\quad\sigma\in\left(0,\frac{1}{2}\right).

Hence limk(Δ)1/2vkL2(S2)=0\lim_{k\to\infty}\|(-\Delta)^{1/2}v_{k}\|_{L^{2}(S^{2})}=0. Also, limkwkL2(S2)=0\lim_{k\to\infty}\|w_{k}\|_{L^{2}(S^{2})}=0 by (3.47) and (3.93). By these facts and μ<1\mu_{\infty}<1 we get a contradiction with (3.90).

Next suppose that m{0}m_{\infty}\in\mathbb{Z}\setminus\{0\}. Taking a subsequence, we may assume mk=mm_{k}=m_{\infty} and thus vk𝒴mv_{k}\in\mathcal{Y}_{m_{\infty}} for all kk\in\mathbb{N}. Since {vk}k=1\{v_{k}\}_{k=1}^{\infty} is bounded in H1(S2)H^{1}(S^{2}) by (2.6) and (3.88), and since H1(S2)H^{1}(S^{2}) is compactly embedded into L2(S2)L^{2}(S^{2}), we may assume that

(3.94) limkvk=vstrongly in L2(S2) and weakly in H1(S2)\displaystyle\lim_{k\to\infty}v_{k}=v_{\infty}\quad\text{strongly in $L^{2}(S^{2})$ and weakly in $H^{1}(S^{2})$}

with some vH1(S2)v_{\infty}\in H^{1}(S^{2}) by taking a subsequence again. Then v𝒴mv_{\infty}\in\mathcal{Y}_{m_{\infty}} since vk𝒴mv_{k}\in\mathcal{Y}_{m_{\infty}} for all kk\in\mathbb{N} and 𝒴m\mathcal{Y}_{m_{\infty}} is closed in L2(S2)L^{2}(S^{2}). If v=0v_{\infty}=0, i.e. vkL2(S2)0\|v_{k}\|_{L^{2}(S^{2})}\to 0 as kk\to\infty, then we can get a contradiction with (3.90) as in the case of m=±m_{\infty}=\pm\infty. Suppose that v0v_{\infty}\neq 0. Since uk=Δvku_{k}=\Delta v_{k} and ΔY3m=λ3Y3m\Delta Y_{3}^{m_{\infty}}=-\lambda_{3}Y_{3}^{m_{\infty}}, where we consider Ynm0Y_{n}^{m_{\infty}}\equiv 0 if |m|>n|m_{\infty}|>n,

(uk,Y3m)L2(S2)=(Δvk,Y3m)L2(S2)=(vk,ΔY3m)L2(S2)=λ3(vk,Y3m)L2(S2).\displaystyle(u_{k},Y_{3}^{m_{\infty}})_{L^{2}(S^{2})}=(\Delta v_{k},Y_{3}^{m_{\infty}})_{L^{2}(S^{2})}=(v_{k},\Delta Y_{3}^{m_{\infty}})_{L^{2}(S^{2})}=-\lambda_{3}(v_{k},Y_{3}^{m_{\infty}})_{L^{2}(S^{2})}.

Hence wk=6μkvk+12a3mλ3(vk,Y3m)L2(S2)Y2mw_{k}=6\mu_{k}v_{k}+\frac{1}{2}a_{3}^{m_{\infty}}\lambda_{3}(v_{k},Y_{3}^{m_{\infty}})_{L^{2}(S^{2})}Y_{2}^{m_{\infty}} by (3.35) for wkw_{k}, and

(3.95) limkwk=w=6μv+12a3mλ3(v,Y3m)L2(S2)Y2mstrongly in L2(S2)\displaystyle\lim_{k\to\infty}w_{k}=w_{\infty}=6\mu_{\infty}v_{\infty}+\frac{1}{2}a_{3}^{m_{\infty}}\lambda_{3}(v_{\infty},Y_{3}^{m_{\infty}})_{L^{2}(S^{2})}Y_{2}^{m_{\infty}}\quad\text{strongly in $L^{2}(S^{2})$}

by (3.94), where a3ma_{3}^{m_{\infty}} is given by (2.2) if |m|=1,2|m_{\infty}|=1,2 and a3m=0a_{3}^{m_{\infty}}=0 if |m|3|m_{\infty}|\geq 3. For each ψH1(S2)\psi\in H^{1}(S^{2}) and σ(0,1/2)\sigma\in(0,1/2), we see by (3.60) for vkv_{k} and (3.88) that

|(vk,ψ)L2(S2)|6vkL2(S2)ψL2(S2)+C(σ2Rk+σ)ψL2(S2)+C(1μk)3/4(σ4Rk+1)ψL2(S2).\bigl{|}(\nabla v_{k},\nabla\psi)_{L^{2}(S^{2})}\bigr{|}\leq 6\|v_{k}\|_{L^{2}(S^{2})}\|\psi\|_{L^{2}(S^{2})}+C(\sigma^{-2}R_{k}+\sigma)\|\nabla\psi\|_{L^{2}(S^{2})}\\ +\frac{C}{(1-\mu_{k})^{3/4}}(\sigma^{-4}R_{k}+1)\|\psi\|_{L^{2}(S^{2})}.

We send kk\to\infty, use (3.89), (3.94), and μ<1\mu_{\infty}<1, and then let σ0\sigma\to 0 to get

|(v,ψ)L2(S2)|C{1(1μ)3/4+vL2(S2)}ψL2(S2)\displaystyle\bigl{|}(\nabla v_{\infty},\nabla\psi)_{L^{2}(S^{2})}\bigr{|}\leq C\left\{\frac{1}{(1-\mu_{\infty})^{3/4}}+\|v_{\infty}\|_{L^{2}(S^{2})}\right\}\|\psi\|_{L^{2}(S^{2})}

for all ψH1(S2)\psi\in H^{1}(S^{2}), which shows that ΔvL2(S2)\Delta v_{\infty}\in L^{2}(S^{2}) since H1(S2)H^{1}(S^{2}) is dense in L2(S2)L^{2}(S^{2}). Hence vH2(S2)v_{\infty}\in H^{2}(S^{2}) by the elliptic regularity theorem. Now let ψC(S2)\psi\in C^{\infty}(S^{2}) satisfy

(3.96) ψ=0near{(x1,x2,x3)S2x3=μ}.\displaystyle\psi=0\quad\text{near}\quad\{(x_{1},x_{2},x_{3})\in S^{2}\mid x_{3}=\mu_{\infty}\}.

Then, as in the proof of Lemma 3.16, we can get (see (3.62))

(vk,ψ)L2(S2)6(vk,ψ)L2(S2)=(wk+δkfk,ψμkx3)L2(S2).\displaystyle(\nabla v_{k},\nabla\psi)_{L^{2}(S^{2})}-6(v_{k},\psi)_{L^{2}(S^{2})}=-\left(w_{k}+\delta_{k}f_{k},\frac{\psi}{\mu_{k}-x_{3}}\right)_{L^{2}(S^{2})}.

We send kk\to\infty and use (3.89), (3.94), (3.95), and ψ/(μkx3)ψ/(μx3)\psi/(\mu_{k}-x_{3})\to\psi/(\mu_{\infty}-x_{3}) uniformly on S2S^{2} by (3.96). Then we get

(v,ψ)L2(S2)6(v,ψ)L2(S2)=(w,ψμx3)L2(S2)=(wμx3,ψ)L2(S2)\displaystyle(\nabla v_{\infty},\nabla\psi)_{L^{2}(S^{2})}-6(v_{\infty},\psi)_{L^{2}(S^{2})}=-\left(w_{\infty},\frac{\psi}{\mu_{\infty}-x_{3}}\right)_{L^{2}(S^{2})}=-\left(\frac{w_{\infty}}{\mu_{\infty}-x_{3}},\psi\right)_{L^{2}(S^{2})}

for all ψC(S2)\psi\in C^{\infty}(S^{2}) satisfying (3.96). Since vH2(S2)v_{\infty}\in H^{2}(S^{2}), this equality shows that

(3.97) Δv+6v=wμx3on{(x1,x2,x3)S2x3μ}.\displaystyle\Delta v_{\infty}+6v_{\infty}=\frac{w_{\infty}}{\mu_{\infty}-x_{3}}\quad\text{on}\quad\{(x_{1},x_{2},x_{3})\in S^{2}\mid x_{3}\neq\mu_{\infty}\}.

Let u=Δvu_{\infty}=\Delta v_{\infty}. Then u𝒴mu_{\infty}\in\mathcal{Y}_{m_{\infty}} and u0u_{\infty}\neq 0 since v𝒴mH2(S2)v_{\infty}\in\mathcal{Y}_{m_{\infty}}\cap H^{2}(S^{2}), v0v_{\infty}\neq 0, and Δ\Delta maps 𝒴m\mathcal{Y}_{m_{\infty}} into itself and is invertible on 𝒴m\mathcal{Y}_{m_{\infty}} by (3.2) and (3.9). Moreover, we see by (3.95), (3.97), and B2,m=(I+6Δ1)|𝒳mB_{2,m_{\infty}}=(I+6\Delta^{-1})|_{\mathcal{X}_{m_{\infty}}} that uu_{\infty} satisfies

(3.98) Λmu=x3B2,mu=μu+βY2min𝒳m,\displaystyle\Lambda_{m_{\infty}}u_{\infty}=x_{3}B_{2,m_{\infty}}u_{\infty}=\mu_{\infty}u_{\infty}+\beta_{\infty}Y_{2}^{m_{\infty}}\quad\text{in}\quad\mathcal{X}_{m_{\infty}},

where β=12a3mλ3(v,Y3m)L2(S2)\beta_{\infty}=-\frac{1}{2}a_{3}^{m_{\infty}}\lambda_{3}(v_{\infty},Y_{3}^{m_{\infty}})_{L^{2}(S^{2})}.

Now suppose that |m|3|m_{\infty}|\geq 3. Then Λmu=μu\Lambda_{m_{\infty}}u_{\infty}=\mu_{\infty}u_{\infty} in 𝒳m\mathcal{X}_{m_{\infty}} by (3.98) and Y2m0Y_{2}^{m_{\infty}}\equiv 0. Thus, if μ>0\mu_{\infty}>0, then μ\mu_{\infty} is a nonzero eigenvalue of Λm\Lambda_{m_{\infty}} since u0u_{\infty}\neq 0, but this contradicts Lemma 3.5. Also, if μ=0\mu_{\infty}=0, then uu_{\infty} is in the kernel of Λm\Lambda_{m_{\infty}} and thus u=0u_{\infty}=0 by (3.12), which contradicts u0u_{\infty}\neq 0.

Let |m|=1,2|m_{\infty}|=1,2 and μ>0\mu_{\infty}>0. Then, by (3.98) and ΛmY2m=0\Lambda_{m_{\infty}}Y_{2}^{m_{\infty}}=0,

Λm(u+μ1βY2m)=Λmu=μu+βY2m=μ(u+μ1βY2m).\displaystyle\Lambda_{m_{\infty}}(u_{\infty}+\mu_{\infty}^{-1}\beta_{\infty}Y_{2}^{m_{\infty}})=\Lambda_{m_{\infty}}u_{\infty}=\mu_{\infty}u_{\infty}+\beta_{\infty}Y_{2}^{m_{\infty}}=\mu_{\infty}(u_{\infty}+\mu_{\infty}^{-1}\beta_{\infty}Y_{2}^{m_{\infty}}).

Moreover, u+μ1βY2m0u_{\infty}+\mu_{\infty}^{-1}\beta_{\infty}Y_{2}^{m_{\infty}}\neq 0 since u0u_{\infty}\neq 0 and (u,Y2m)L2(S2)=0(u_{\infty},Y_{2}^{m_{\infty}})_{L^{2}(S^{2})}=0 by u𝒴mu_{\infty}\in\mathcal{Y}_{m_{\infty}}. Hence μ\mu_{\infty} is a nonzero eigenvalue of Λm\Lambda_{m_{\infty}}, which contradicts Lemma 3.5.

Suppose that |m|=1|m_{\infty}|=1 and μ=0\mu_{\infty}=0. Since Y2±1=cosθY1±1/a2±1Y_{2}^{\pm 1}=\cos\theta\,Y_{1}^{\pm 1}/a_{2}^{\pm 1} by (2.2) with (n,m)=(1,±1)(n,m)=(1,\pm 1) and Y0±10Y_{0}^{\pm 1}\equiv 0, we apply this equality and x3=cosθx_{3}=\cos\theta to (3.98) to get

x3{B2,±1uβa2±1Y1±1}=0,i.e.B2,±1uβa2±1Y1±1=0onS2.\displaystyle x_{3}\left\{B_{2,\pm 1}u_{\infty}-\frac{\beta_{\infty}}{a_{2}^{\pm 1}}Y_{1}^{\pm 1}\right\}=0,\quad\text{i.e.}\quad B_{2,\pm 1}u_{\infty}-\frac{\beta_{\infty}}{a_{2}^{\pm 1}}Y_{1}^{\pm 1}=0\quad\text{on}\quad S^{2}.

Then since u𝒴±1u\in\mathcal{Y}_{\pm 1} is of the form (3.9), we have

B2,±1uL2(S2)2\displaystyle\|B_{2,\pm 1}u_{\infty}\|_{L^{2}(S^{2})}^{2} =βa2±1(B2,±1u,Y1±1)L2(S2)\displaystyle=\frac{\beta_{\infty}}{a_{2}^{\pm 1}}(B_{2,\pm 1}u_{\infty},Y_{1}^{\pm 1})_{L^{2}(S^{2})}
=βa2±1n3(16λn)(u,Yn±1)L2(S2)(Yn±1,Y1±1)L2(S2)=0\displaystyle=\frac{\beta_{\infty}}{a_{2}^{\pm 1}}\sum_{n\geq 3}\left(1-\frac{6}{\lambda_{n}}\right)(u_{\infty},Y_{n}^{\pm 1})_{L^{2}(S^{2})}(Y_{n}^{\pm 1},Y_{1}^{\pm 1})_{L^{2}(S^{2})}=0

by (3.2) (note that B2,±1=B|𝒳±1B_{2,\pm 1}=B|_{\mathcal{X}_{\pm 1}}). Since u𝒴±1u_{\infty}\in\mathcal{Y}_{\pm 1}, we find by this equality and (3.14) that u=0u_{\infty}=0, which contradicts u0u_{\infty}\neq 0.

Let |m|=2|m_{\infty}|=2 and μ=0\mu_{\infty}=0. Since Y2±2Y_{2}^{\pm 2} is of the form Y2±2=C±2sin2θe±2iφY_{2}^{\pm 2}=C_{\pm 2}\sin^{2}\theta e^{\pm 2i\varphi} with a constant C±2{0}C_{\pm 2}\in\mathbb{R}\setminus\{0\} by (2.1), we see by (3.98) and x3=cosθx_{3}=\cos\theta that

B2,±2u(θ,φ)=βC±2sin2θcosθe±2iφ,(θ,φ)[0,π]×[0,2π),θπ2\displaystyle B_{2,\pm 2}u_{\infty}(\theta,\varphi)=\beta_{\infty}C_{\pm 2}\frac{\sin^{2}\theta}{\cos\theta}e^{\pm 2i\varphi},\quad(\theta,\varphi)\in[0,\pi]\times[0,2\pi),\,\theta\neq\frac{\pi}{2}

and thus β=0\beta_{\infty}=0, otherwise B2,±2uB_{2,\pm 2}u_{\infty} does not belong to L2(S2)L^{2}(S^{2}). Hence B2,±2u=0B_{2,\pm 2}u_{\infty}=0 by the above equality. Since u𝒴±2u_{\infty}\in\mathcal{Y}_{\pm 2}, this fact and (3.14) imply that u=0u_{\infty}=0, which contradicts u0u_{\infty}\neq 0. This completes the proof in Case 1.

Case 2: μ=1\mu_{\infty}=1. In this case, μkμ0\mu_{k}\geq\mu_{0} for sufficiently large kk\in\mathbb{N}, where μ0(0,1)\mu_{0}\in(0,1) is given in Lemma 3.17. Then we see by (3.78) for wkw_{k}, (3.88), and (3.89) that

lim supk11μkwkL2(S2(θk,π))2Clim supk(σ1Rk2+σ(Δ)1/2vkL2(S2)2)Cσ\displaystyle\limsup_{k\to\infty}\frac{1}{1-\mu_{k}}\|w_{k}\|_{L^{2}(S^{2}(\theta_{k},\pi))}^{2}\leq C\limsup_{k\to\infty}\Bigl{(}\sigma^{-1}R_{k}^{2}+\sigma\|(-\Delta)^{1/2}v_{k}\|_{L^{2}(S^{2})}^{2}\Bigr{)}\leq C\sigma

for all σ(0,1/2)\sigma\in(0,1/2), where S2(θk,π)S^{2}(\theta_{k},\pi) is given by (2.9). Hence

(3.99) limk11μkwkL2(S2(θk,π))2=0.\displaystyle\lim_{k\to\infty}\frac{1}{1-\mu_{k}}\|w_{k}\|_{L^{2}(S^{2}(\theta_{k},\pi))}^{2}=0.

Suppose limk(1μk)1wkL2(S2(0,θk))2=0\lim_{k\to\infty}(1-\mu_{k})^{-1}\|w_{k}\|_{L^{2}(S^{2}(0,\theta_{k}))}^{2}=0. Then

(3.100) limk11μkwkL2(S2)2=limk11μk(wkL2(S2(0,θk))2+wkL2(S2(θk,π))2)=0.\displaystyle\lim_{k\to\infty}\frac{1}{1-\mu_{k}}\|w_{k}\|_{L^{2}(S^{2})}^{2}=\lim_{k\to\infty}\frac{1}{1-\mu_{k}}\Bigl{(}\|w_{k}\|_{L^{2}(S^{2}(0,\theta_{k}))}^{2}+\|w_{k}\|_{L^{2}(S^{2}(\theta_{k},\pi))}^{2}\Bigr{)}=0.

Also, for each σ(0,1/2)\sigma\in(0,1/2) we see by (3.61) for vkv_{k} and (3.88) that

3μk(Δ)1/2vkL2(S2)2C(σ2Rk+σ+1σ2(1μk)wkL2(S2(0,θk))2)\displaystyle 3\mu_{k}\|(-\Delta)^{1/2}v_{k}\|_{L^{2}(S^{2})}^{2}\leq C\left(\sigma^{-2}R_{k}+\sigma+\frac{1}{\sigma^{2}(1-\mu_{k})}\|w_{k}\|_{L^{2}(S^{2}(0,\theta_{k}))}^{2}\right)

We send kk\to\infty and use (3.89) and (1μk)1wkL2(S2(0,θk))20(1-\mu_{k})^{-1}\|w_{k}\|_{L^{2}(S^{2}(0,\theta_{k}))}^{2}\to 0 to get

3μlim supk(Δ)1/2vkL2(S2)2=3lim supk(Δ)1/2vkL2(S2)2Cσ\displaystyle 3\mu_{\infty}\limsup_{k\to\infty}\|(-\Delta)^{1/2}v_{k}\|_{L^{2}(S^{2})}^{2}=3\limsup_{k\to\infty}\|(-\Delta)^{1/2}v_{k}\|_{L^{2}(S^{2})}^{2}\leq C\sigma

for all σ(0,1/2)\sigma\in(0,1/2). Thus limk(Δ)1/2vkL2(S2)2=0\lim_{k\to\infty}\|(-\Delta)^{1/2}v_{k}\|_{L^{2}(S^{2})}^{2}=0 and we combine this with (3.100) to get a contradiction with (3.90). Now we assume that

(3.101) infk11μkwkL2(S2(0,θk))2>0.\displaystyle\inf_{k\in\mathbb{N}}\frac{1}{1-\mu_{k}}\|w_{k}\|_{L^{2}(S^{2}(0,\theta_{k}))}^{2}>0.

Since μk[0,1)\mu_{k}\in[0,1) and θk=arccosμk(0,π/2]\theta_{k}=\arccos\mu_{k}\in(0,\pi/2], we have

θkπ2sinθk=π2(1μk2)1/2=π2(1μk)1/2(1+μk)1/222π(1μk)1/2.\displaystyle\theta_{k}\leq\frac{\pi}{2}\sin\theta_{k}=\frac{\pi}{2}(1-\mu_{k}^{2})^{1/2}=\frac{\pi}{2}(1-\mu_{k})^{1/2}(1+\mu_{k})^{1/2}\leq\frac{\sqrt{2}}{2}\pi(1-\mu_{k})^{1/2}.

Moreover, by Taylor’s theorem for cosθ\cos\theta at θ=0\theta=0,

μk=cosθk=112θk2+124θk4cos(τkθk),τk(0,1).\displaystyle\mu_{k}=\cos\theta_{k}=1-\frac{1}{2}\theta_{k}^{2}+\frac{1}{24}\theta_{k}^{4}\cos(\tau_{k}\theta_{k}),\quad\tau_{k}\in(0,1).

By the above relations, |cos(τkθk)|1|\cos(\tau_{k}\theta_{k})|\leq 1, and μ=1\mu_{\infty}=1, we find that

|θk21μk2|=|θk4cos(τkθk)12(1μk)|C(1μk)0ask.\displaystyle\left|\frac{\theta_{k}^{2}}{1-\mu_{k}}-2\right|=\left|\frac{\theta_{k}^{4}\cos(\tau_{k}\theta_{k})}{12(1-\mu_{k})}\right|\leq C(1-\mu_{k})\to 0\quad\text{as}\quad k\to\infty.

Hence θk/(1μk)1/22\theta_{k}/(1-\mu_{k})^{1/2}\to\sqrt{2} as kk\to\infty. Based on this observation, we set

(3.102) γk=42π(1μk)1/2,θ~k=θkγk=π42θk(1μk)1/2.\displaystyle\gamma_{k}=\frac{4\sqrt{2}}{\pi}(1-\mu_{k})^{1/2},\quad\tilde{\theta}_{k}=\frac{\theta_{k}}{\gamma_{k}}=\frac{\pi}{4\sqrt{2}}\cdot\frac{\theta_{k}}{(1-\mu_{k})^{1/2}}.

Then

(3.103) limkγk=0,limkθ~k=θ~=π4\displaystyle\lim_{k\to\infty}\gamma_{k}=0,\quad\lim_{k\to\infty}\tilde{\theta}_{k}=\tilde{\theta}_{\infty}=\frac{\pi}{4}

and thus γk(0,1)\gamma_{k}\in(0,1) and θ~k(π/6,π/3)\tilde{\theta}_{k}\in(\pi/6,\pi/3) for sufficiently large kk\in\mathbb{N}. Also,

(3.104) 2π1γksinθsin(θγk)π21γksinθ,θ(0,π2γk)\displaystyle\frac{2}{\pi}\cdot\frac{1}{\gamma_{k}}\sin\theta\leq\sin\left(\frac{\theta}{\gamma_{k}}\right)\leq\frac{\pi}{2}\cdot\frac{1}{\gamma_{k}}\sin\theta,\quad\theta\in\left(0,\frac{\pi}{2}\gamma_{k}\right)

by 2θ/πsinθθ2\theta/\pi\leq\sin\theta\leq\theta for θ(0,π/2)\theta\in(0,\pi/2). We define

(3.105) W~k(θ~)=Wk(γkθ~),F~k(θ~)=Fk(γkθ~),θ~(0,π2),\displaystyle\widetilde{W}_{k}(\tilde{\theta})=W_{k}(\gamma_{k}\tilde{\theta}),\quad\widetilde{F}_{k}(\tilde{\theta})=F_{k}(\gamma_{k}\tilde{\theta}),\quad\tilde{\theta}\in\left(0,\frac{\pi}{2}\right),
w~k=W~k(θ~)eimkφ,f~k=F~k(θ~)eimkφonS+2={(x1,x2,x3)S2x3>0}.\displaystyle\tilde{w}_{k}=\widetilde{W}_{k}(\tilde{\theta})e^{im_{k}\varphi},\quad\tilde{f}_{k}=\widetilde{F}_{k}(\tilde{\theta})e^{im_{k}\varphi}\quad\text{on}\quad S_{+}^{2}=\{(x_{1},x_{2},x_{3})\in S^{2}\mid x_{3}>0\}.

Then, by W~k(θ~k)=Wk(θk)\widetilde{W}_{k}(\tilde{\theta}_{k})=W_{k}(\theta_{k}), F~k(θ~k)=Fk(θk)\widetilde{F}_{k}(\tilde{\theta}_{k})=F_{k}(\theta_{k}), (3.91), and (3.92),

(3.106) W~k(θ~k)+δkF~k(θ~k)=0,k,limkW~k(θ~k)=0.\displaystyle\widetilde{W}_{k}(\tilde{\theta}_{k})+\delta_{k}\widetilde{F}_{k}(\tilde{\theta}_{k})=0,\quad k\in\mathbb{N},\quad\lim_{k\to\infty}\widetilde{W}_{k}(\tilde{\theta}_{k})=0.

Also, we use (2.7), set θ=γkθ~\theta=\gamma_{k}\tilde{\theta}, and apply (3.102) and (3.104) to get

(3.107) w~kL2(S+2)2=2π0π2|W~k(θ~)|2sinθ~dθ~=2π0π2γk|Wk(θ)|2sin(θγk)dθγkCγk20π2γk|Wk(θ)|2sinθdθC1μkwkL2(S2)2.\displaystyle\begin{aligned} \|\tilde{w}_{k}\|_{L^{2}(S_{+}^{2})}^{2}&=2\pi\int_{0}^{\frac{\pi}{2}}\Bigl{|}\widetilde{W}_{k}(\tilde{\theta})\Bigr{|}^{2}\sin\tilde{\theta}\,d\tilde{\theta}=2\pi\int_{0}^{\frac{\pi}{2}\gamma_{k}}|W_{k}(\theta)|^{2}\sin\left(\frac{\theta}{\gamma_{k}}\right)\frac{d\theta}{\gamma_{k}}\\ &\leq\frac{C}{\gamma_{k}^{2}}\int_{0}^{\frac{\pi}{2}\gamma_{k}}|W_{k}(\theta)|^{2}\sin\theta\,d\theta\leq\frac{C}{1-\mu_{k}}\|w_{k}\|_{L^{2}(S^{2})}^{2}.\end{aligned}

Similarly, since W~k(θ~)=γkWk(γkθ~)\widetilde{W}^{\prime}_{k}(\tilde{\theta})=\gamma_{k}W^{\prime}_{k}(\gamma_{k}\tilde{\theta}), we have

(3.108) w~kL2(S+2)2=2π0π2(|W~k(θ~)|2+mk2sin2θ~|W~k(θ~)|2)sinθ~dθ~=2π0π2γk(γk2|Wk(θ)|2+mk2sin2(θ/γk)|Wk(θ)|2)sin(θγk)dθγkC0π2γk(|Wk(θ)|2+mk2sin2θ|Wk(θ)|2)sinθdθCwkL2(S2)2.\displaystyle\begin{aligned} \|\nabla\tilde{w}_{k}\|_{L^{2}(S_{+}^{2})}^{2}&=2\pi\int_{0}^{\frac{\pi}{2}}\left(\Bigl{|}\widetilde{W}^{\prime}_{k}(\tilde{\theta})\Bigr{|}^{2}+\frac{m_{k}^{2}}{\sin^{2}\tilde{\theta}}\Bigl{|}\widetilde{W}_{k}(\tilde{\theta})\Bigr{|}^{2}\right)\sin\tilde{\theta}\,d\tilde{\theta}\\ &=2\pi\int_{0}^{\frac{\pi}{2}\gamma_{k}}\left(\gamma_{k}^{2}|W^{\prime}_{k}(\theta)|^{2}+\frac{m_{k}^{2}}{\sin^{2}(\theta/\gamma_{k})}|W_{k}(\theta)|^{2}\right)\sin\left(\frac{\theta}{\gamma_{k}}\right)\frac{d\theta}{\gamma_{k}}\\ &\leq C\int_{0}^{\frac{\pi}{2}\gamma_{k}}\left(|W^{\prime}_{k}(\theta)|^{2}+\frac{m_{k}^{2}}{\sin^{2}\theta}|W_{k}(\theta)|^{2}\right)\sin\theta\,d\theta\leq C\|\nabla w_{k}\|_{L^{2}(S^{2})}^{2}.\end{aligned}

By these inequalities, (3.47), and (3.88), we find that

(3.109) w~kH1(S+2)2C(11μkwkL2(S2)2+(Δ)1/2vkL2(S2)2)C.\displaystyle\|\tilde{w}_{k}\|_{H^{1}(S_{+}^{2})}^{2}\leq C\left(\frac{1}{1-\mu_{k}}\|w_{k}\|_{L^{2}(S^{2})}^{2}+\|(-\Delta)^{1/2}v_{k}\|_{L^{2}(S^{2})}^{2}\right)\leq C.

Also, as in (3.107), we use (2.7), (3.102), (3.104), and γkθ~k=θk\gamma_{k}\tilde{\theta}_{k}=\theta_{k} to get

w~kL2(S2(0,θ~k))2C1μkwkL2(S2(0,θk))2.\displaystyle\|\tilde{w}_{k}\|_{L^{2}(S^{2}(0,\tilde{\theta}_{k}))}^{2}\geq\frac{C}{1-\mu_{k}}\|w_{k}\|_{L^{2}(S^{2}(0,\theta_{k}))}^{2}.

By this inequality and (3.101), we obtain

(3.110) infkw~kL2(S+2)2infkw~kL2(S2(0,θ~k))2Cinfk11μkwkL2(S2(0,θk))2>0.\displaystyle\inf_{k\in\mathbb{N}}\|\tilde{w}_{k}\|_{L^{2}(S_{+}^{2})}^{2}\geq\inf_{k\in\mathbb{N}}\|\tilde{w}_{k}\|_{L^{2}(S^{2}(0,\tilde{\theta}_{k}))}^{2}\geq C\inf_{k\in\mathbb{N}}\frac{1}{1-\mu_{k}}\|w_{k}\|_{L^{2}(S^{2}(0,\theta_{k}))}^{2}>0.

Suppose that m=±m_{\infty}=\pm\infty. Then since

w~kL2(S+2)2\displaystyle\|\nabla\tilde{w}_{k}\|_{L^{2}(S_{+}^{2})}^{2} =2π0π2(|W~k(θ~)|2+mk2sin2θ~|W~k(θ~)|2)sinθ~dθ~\displaystyle=2\pi\int_{0}^{\frac{\pi}{2}}\left(\Bigl{|}\widetilde{W}^{\prime}_{k}(\tilde{\theta})\Bigr{|}^{2}+\frac{m_{k}^{2}}{\sin^{2}\tilde{\theta}}\Bigl{|}\widetilde{W}_{k}(\tilde{\theta})\Bigr{|}^{2}\right)\sin\tilde{\theta}\,d\tilde{\theta}
mk22π0π2|W~k(θ~)|2sinθ~dθ~=mk2w~kL2(S+2)2\displaystyle\geq m_{k}^{2}\cdot 2\pi\int_{0}^{\frac{\pi}{2}}\Bigl{|}\widetilde{W}_{k}(\tilde{\theta})\Bigr{|}^{2}\sin\tilde{\theta}\,d\tilde{\theta}=m_{k}^{2}\|\tilde{w}_{k}\|_{L^{2}(S_{+}^{2})}^{2}

by (2.7) and sin2θ~1\sin^{2}\tilde{\theta}\leq 1, it follows from (3.109) that

w~kL2(S+2)21mk2w~kL2(S+2)2Cmk20ask,\displaystyle\|\tilde{w}_{k}\|_{L^{2}(S_{+}^{2})}^{2}\leq\frac{1}{m_{k}^{2}}\|\nabla\tilde{w}_{k}\|_{L^{2}(S_{+}^{2})}^{2}\leq\frac{C}{m_{k}^{2}}\to 0\quad\text{as}\quad k\to\infty,

which contradicts (3.110).

Now let m{0}m_{\infty}\in\mathbb{Z}\setminus\{0\}. Taking a subsequence, we may assume that

(3.111) mk=m,w~k=W~k(θ~)eimφ,f~k=F~k(θ~)eimφfor allk.\displaystyle m_{k}=m_{\infty},\quad\tilde{w}_{k}=\widetilde{W}_{k}(\tilde{\theta})e^{im_{\infty}\varphi},\quad\tilde{f}_{k}=\widetilde{F}_{k}(\tilde{\theta})e^{im_{\infty}\varphi}\quad\text{for all}\quad k\in\mathbb{N}.

Since {w~k}k=1\{\tilde{w}_{k}\}_{k=1}^{\infty} is bounded in H1(S+2)H^{1}(S_{+}^{2}) by (3.109) and H1(S+2)H^{1}(S_{+}^{2}) is compactly embedded into L2(S+2)L^{2}(S_{+}^{2}), we may assume, by taking a subsequence again, that

(3.112) limkw~k=w~strongly in L2(S+2) and weakly in H1(S+2).\displaystyle\lim_{k\to\infty}\tilde{w}_{k}=\tilde{w}_{\infty}\quad\text{strongly in $L^{2}(S_{+}^{2})$ and weakly in $H^{1}(S_{+}^{2})$}.

with some w~H1(S+2)\tilde{w}_{\infty}\in H^{1}(S_{+}^{2}). Then w~0\tilde{w}_{\infty}\neq 0 by (3.110) and (3.112). For each mm\in\mathbb{Z} we see by (2.10) and (3.112) that {𝒫mw~k}k=1\{\mathcal{P}_{m}\tilde{w}_{k}\}_{k=1}^{\infty} converges to 𝒫mw~\mathcal{P}_{m}\tilde{w}_{\infty} strongly in L2(S+2)L^{2}(S_{+}^{2}). Moreover, 𝒫mw~k=w~k\mathcal{P}_{m_{\infty}}\tilde{w}_{k}=\tilde{w}_{k} and 𝒫mw~k=0\mathcal{P}_{m}\tilde{w}_{k}=0 for mmm\neq m_{\infty} by (3.111). Hence

𝒫mw~=0formm,w~=𝒫mw~=W~(θ~)eimφ.\displaystyle\mathcal{P}_{m}\tilde{w}_{\infty}=0\quad\text{for}\quad m\neq m_{\infty},\quad\tilde{w}_{\infty}=\mathcal{P}_{m_{\infty}}\tilde{w}_{\infty}=\widetilde{W}_{\infty}(\tilde{\theta})e^{im_{\infty}\varphi}.

Let us show W~=0\widetilde{W}_{\infty}=0. First we prove

(3.113) W~H2(Θ,π/2)C1([Θ,π/2])for allΘ(0,π2).\displaystyle\widetilde{W}_{\infty}\in H^{2}(\Theta,\pi/2)\subset C^{1}([\Theta,\pi/2])\quad\text{for all}\quad\Theta\in\left(0,\frac{\pi}{2}\right).

We see that W~H1(Θ,π/2)\widetilde{W}_{\infty}\in H^{1}(\Theta,\pi/2) by w~H1(S+2)\tilde{w}_{\infty}\in H^{1}(S_{+}^{2}), (2.7), and

(3.114) sinθ~sinΘ>0,θ~(Θ,π2).\displaystyle\sin\tilde{\theta}\geq\sin\Theta>0,\quad\tilde{\theta}\in\left(\Theta,\frac{\pi}{2}\right).

Moreover, by (2.7), (3.112), and (3.114),

(3.115) limkW~k=W~strongly inL2(Θ,π/2).\displaystyle\lim_{k\to\infty}\widetilde{W}_{k}=\widetilde{W}_{\infty}\quad\text{strongly in}\quad L^{2}(\Theta,\pi/2).

For each ΨL2(Θ,π/2)\Psi\in L^{2}(\Theta,\pi/2), the linear functional

L(w~)=12π02π(Θπ2θ~w~(θ~,φ)Ψ(θ~)¯dθ~)eimφ𝑑φ,w~H1(S+2)\displaystyle L(\tilde{w})=\frac{1}{2\pi}\int_{0}^{2\pi}\left(\int_{\Theta}^{\frac{\pi}{2}}\partial_{\tilde{\theta}}\tilde{w}(\tilde{\theta},\varphi)\,\overline{\Psi(\tilde{\theta})}\,d\tilde{\theta}\right)e^{-im_{\infty}\varphi}\,d\varphi,\quad\tilde{w}\in H^{1}(S_{+}^{2})

is bounded on H1(S+2)H^{1}(S_{+}^{2}). Indeed, by Hölder’s inequality, (3.114), and (5.1),

|L(w~)|12πsinΘw~L2(S+2)ΨL2(Θ,π/2).\displaystyle|L(\tilde{w})|\leq\frac{1}{\sqrt{2\pi\sin\Theta}}\|\nabla\tilde{w}\|_{L^{2}(S_{+}^{2})}\|\Psi\|_{L^{2}(\Theta,\pi/2)}.

Hence limkL(w~k)=L(w~)\lim_{k\to\infty}L(\tilde{w}_{k})=L(\tilde{w}_{\infty}) by (3.112). By this fact and

L(w~k)=Θπ2W~k(θ~)Ψ(θ~)¯𝑑θ~=(W~k,Ψ)L2(Θ,π/2),k{},\displaystyle L(\tilde{w}_{k})=\int_{\Theta}^{\frac{\pi}{2}}\widetilde{W}^{\prime}_{k}(\tilde{\theta})\,\overline{\Psi(\tilde{\theta})}\,d\tilde{\theta}=\Bigl{(}\widetilde{W}^{\prime}_{k},\Psi\Bigr{)}_{L^{2}(\Theta,\pi/2)},\quad k\in\mathbb{N}\cup\{\infty\},

we find that

(3.116) limkW~k=W~weakly inL2(Θ,π/2).\displaystyle\lim_{k\to\infty}\widetilde{W}^{\prime}_{k}=\widetilde{W}^{\prime}_{\infty}\quad\text{weakly in}\quad L^{2}(\Theta,\pi/2).

To get W~′′L2(Θ,π/2)\widetilde{W}^{\prime\prime}_{\infty}\in L^{2}(\Theta,\pi/2), we derive an ODE for W~k\widetilde{W}_{k}, kk\in\mathbb{N}. By (3.105) we have W~k(j)(θ~)=γkjWk(j)(γkθ~)\widetilde{W}_{k}^{(j)}(\tilde{\theta})=\gamma_{k}^{j}W_{k}^{(j)}(\gamma_{k}\tilde{\theta}) for j=0,1,2j=0,1,2, where W~k(j)\widetilde{W}_{k}^{(j)} and Wk(j)W_{k}^{(j)} are the jj-th derivatives of W~k\widetilde{W}_{k} and WkW_{k}. We use this relation and (3.114) and calculate as in (3.107) to get

W~k(j)L2(Θ,π/2)2\displaystyle\Bigl{\|}\widetilde{W}_{k}^{(j)}\Bigr{\|}_{L^{2}(\Theta,\pi/2)}^{2} γk2jsinΘΘπ2|Wk(j)(γkθ~)|2sinθ~dθ~Cγk2j2sinΘjwkL2(S2)2.\displaystyle\leq\frac{\gamma_{k}^{2j}}{\sin\Theta}\int_{\Theta}^{\frac{\pi}{2}}|W_{k}^{(j)}(\gamma_{k}\tilde{\theta})|^{2}\sin\tilde{\theta}\,d\tilde{\theta}\leq\frac{C\gamma_{k}^{2j-2}}{\sin\Theta}\|\nabla^{j}w_{k}\|_{L^{2}(S^{2})}^{2}.

Thus W~kH2(Θ,π/2)\widetilde{W}_{k}\in H^{2}(\Theta,\pi/2) by wkH2(S2)w_{k}\in H^{2}(S^{2}). Also, by (3.46),

Δwk+6wk=6μkwk+δfkμkx3on{(x1,x2,x3)S2x3μk}.\displaystyle\Delta w_{k}+6w_{k}=6\mu_{k}\frac{w_{k}+\delta f_{k}}{\mu_{k}-x_{3}}\quad\text{on}\quad\{(x_{1},x_{2},x_{3})\in S^{2}\mid x_{3}\neq\mu_{k}\}.

Using x3=cosθx_{3}=\cos\theta, (3.91) with mk=mm_{k}=m_{\infty}, and (5.1), we rewrite this equation as

Wk′′(θ)+cosθsinθWk(θ)+(6m2sin2θ)Wk(θ)=6μkWk(θ)+δkFk(θ)μkcosθ,θ(0,π){θk}.\displaystyle W^{\prime\prime}_{k}(\theta)+\frac{\cos\theta}{\sin\theta}W^{\prime}_{k}(\theta)+\left(6-\frac{m_{\infty}^{2}}{\sin^{2}\theta}\right)W_{k}(\theta)=6\mu_{k}\frac{W_{k}(\theta)+\delta_{k}F_{k}(\theta)}{\mu_{k}-\cos\theta},\quad\theta\in(0,\pi)\setminus\{\theta_{k}\}.

We multiply both sides by γk2\gamma_{k}^{2}, set θ=γkθ~\theta=\gamma_{k}\tilde{\theta}, and use (3.105) to get

(3.117) W~k′′(θ~)=ζ1,k(θ~)W~k(θ~)+ζ2,k(θ~)W~k(θ~)+ζ3,k(θ~)W~k(θ~)+δkF~k(θ~)θ~θ~k\displaystyle\widetilde{W}^{\prime\prime}_{k}(\tilde{\theta})=\zeta_{1,k}(\tilde{\theta})\widetilde{W}^{\prime}_{k}(\tilde{\theta})+\zeta_{2,k}(\tilde{\theta})\widetilde{W}_{k}(\tilde{\theta})+\zeta_{3,k}(\tilde{\theta})\frac{\widetilde{W}_{k}(\tilde{\theta})+\delta_{k}\widetilde{F}_{k}(\tilde{\theta})}{\tilde{\theta}-\tilde{\theta}_{k}}

for θ~(0,π/2){θ~k}\tilde{\theta}\in(0,\pi/2)\setminus\{\tilde{\theta}_{k}\}, where

ζ1,k(θ~)=γkcos(γkθ~)sin(γkθ~),ζ2,k(θ~)=m2γk2sin2(γkθ~k)6γk2,ζ3,k(θ~)=6μkγk2(θ~θ~k)μkcos(γkθ~).\displaystyle\zeta_{1,k}(\tilde{\theta})=-\frac{\gamma_{k}\cos(\gamma_{k}\tilde{\theta})}{\sin(\gamma_{k}\tilde{\theta})},\quad\zeta_{2,k}(\tilde{\theta})=\frac{m_{\infty}^{2}\gamma_{k}^{2}}{\sin^{2}(\gamma_{k}\tilde{\theta}_{k})}-6\gamma_{k}^{2},\quad\zeta_{3,k}(\tilde{\theta})=\frac{6\mu_{k}\gamma_{k}^{2}(\tilde{\theta}-\tilde{\theta}_{k})}{\mu_{k}-\cos(\gamma_{k}\tilde{\theta})}.

Since limkγk=0\lim_{k\to\infty}\gamma_{k}=0 and

|cos(γkθ~)1|12γk2θ~2,|γksin(γkθ~)1θ~|=|sin(γkθ~)γkθ~|θ~sin(γkθ~)π12γk2θ~\displaystyle|\cos(\gamma_{k}\tilde{\theta})-1|\leq\frac{1}{2}\gamma_{k}^{2}\tilde{\theta}^{2},\quad\left|\frac{\gamma_{k}}{\sin(\gamma_{k}\tilde{\theta})}-\frac{1}{\tilde{\theta}}\right|=\frac{|\sin(\gamma_{k}\tilde{\theta})-\gamma_{k}\tilde{\theta}|}{\tilde{\theta}\sin(\gamma_{k}\tilde{\theta})}\leq\frac{\pi}{12}\gamma_{k}^{2}\tilde{\theta}

for θ~(0,π/2)\tilde{\theta}\in(0,\pi/2) by Taylor’s theorem and sinθ2θ/π\sin\theta\geq 2\theta/\pi for θ(0,π/2)\theta\in(0,\pi/2),

(3.118) limkζj,k(θ~)=ζj,(θ~)={θ~1,j=1,m2θ~2,j=2,uniformly on(Θ,π2).\displaystyle\lim_{k\to\infty}\zeta_{j,k}(\tilde{\theta})=\zeta_{j,\infty}(\tilde{\theta})=\begin{cases}-\tilde{\theta}^{-1},&j=1,\\ m_{\infty}^{2}\tilde{\theta}^{-2},&j=2,\end{cases}\quad\text{uniformly on}\quad\left(\Theta,\frac{\pi}{2}\right).

Also, since μk=cosθk=cos(γkθ~k)\mu_{k}=\cos\theta_{k}=\cos(\gamma_{k}\tilde{\theta}_{k}) and

cos(γkθ~)=1+N=1(1)N(2N)!γk2Nθ~2N,θ~(0,π2)\displaystyle\cos(\gamma_{k}\tilde{\theta})=1+\sum_{N=1}^{\infty}\frac{(-1)^{N}}{(2N)!}\gamma_{k}^{2N}\tilde{\theta}^{2N},\quad\tilde{\theta}\in\left(0,\frac{\pi}{2}\right)

by the Taylor series for cosθ\cos\theta at θ=0\theta=0, it follows that

(3.119) cos(γkθ~)μk=N=1(1)N(2N)!γk2N(θ~2Nθ~k2N)=12γk2(θ~2θ~k2){1+2γk2ρk(θ~)}\displaystyle\cos(\gamma_{k}\tilde{\theta})-\mu_{k}=\sum_{N=1}^{\infty}\frac{(-1)^{N}}{(2N)!}\gamma_{k}^{2N}(\tilde{\theta}^{2N}-\tilde{\theta}_{k}^{2N})=-\frac{1}{2}\gamma_{k}^{2}(\tilde{\theta}^{2}-\tilde{\theta}_{k}^{2})\{1+2\gamma_{k}^{2}\rho_{k}(\tilde{\theta})\}

for θ~(0,π/2)\tilde{\theta}\in(0,\pi/2). Here

ρk(θ~)=N=2(1)N(2N)!γk2N4(M=0N1θ~2Mθ~k2(N1M))=M=0CM,kθ~2M,\displaystyle\rho_{k}(\tilde{\theta})=-\sum_{N=2}^{\infty}\frac{(-1)^{N}}{(2N)!}\gamma_{k}^{2N-4}\left(\sum_{M=0}^{N-1}\tilde{\theta}^{2M}\tilde{\theta}_{k}^{2(N-1-M)}\right)=\sum_{M=0}^{\infty}C_{M,k}\tilde{\theta}^{2M},

where we rearranged the double summation and defined

CM,k=θ~k2(1M)N=max{2,M+1}(1)N(2N)!(γkθ~k)2(N2),M0.\displaystyle C_{M,k}=-\tilde{\theta}_{k}^{2(1-M)}\sum_{N=\max\{2,M+1\}}^{\infty}\frac{(-1)^{N}}{(2N)!}(\gamma_{k}\tilde{\theta}_{k})^{2(N-2)},\quad M\geq 0.

Since γkθ~k=θk=arccosμk0\gamma_{k}\tilde{\theta}_{k}=\theta_{k}=\arccos\mu_{k}\to 0 as kk\to\infty by μ=1\mu_{\infty}=1, we may assume that γkθ~k(0,1/2)\gamma_{k}\tilde{\theta}_{k}\in(0,1/2) for sufficiently large kk\in\mathbb{N}. By this fact and θ~k(π/6,π/3)\tilde{\theta}_{k}\in(\pi/6,\pi/3), we have

|CM,k|θ~k2(1M)(2M)!N=022N=43θ~k2(1M)(2M)!43(π3)21(2M)!(π6)2M,M0.\displaystyle|C_{M,k}|\leq\frac{\tilde{\theta}_{k}^{2(1-M)}}{(2M)!}\sum_{N=0}^{\infty}2^{-2N}=\frac{4}{3}\cdot\frac{\tilde{\theta}_{k}^{2(1-M)}}{(2M)!}\leq\frac{4}{3}\left(\frac{\pi}{3}\right)^{2}\cdot\frac{1}{(2M)!}\left(\frac{\pi}{6}\right)^{-2M},\quad M\geq 0.

Hence we easily find that ρk(θ~)\rho_{k}(\tilde{\theta}) converges absolutely for all θ~(0,π/2)\tilde{\theta}\in(0,\pi/2) and that ρk\rho_{k} and its derivative ρk\rho^{\prime}_{k} are bounded on (0,π/2)(0,\pi/2) uniformly in kk\in\mathbb{N}. Then since

ζ3,k(θ~)=6μkγk2(θ~θ~k)μkcos(γkθ~)=12μk(θ~θ~k)(θ~2θ~k2){1+2γk2ρk(θ~)}=12μk(θ~+θ~k){1+2γk2ρk(θ~)}\displaystyle\zeta_{3,k}(\tilde{\theta})=\frac{6\mu_{k}\gamma_{k}^{2}(\tilde{\theta}-\tilde{\theta}_{k})}{\mu_{k}-\cos(\gamma_{k}\tilde{\theta})}=\frac{12\mu_{k}(\tilde{\theta}-\tilde{\theta}_{k})}{(\tilde{\theta}^{2}-\tilde{\theta}_{k}^{2})\{1+2\gamma_{k}^{2}\rho_{k}(\tilde{\theta})\}}=\frac{12\mu_{k}}{(\tilde{\theta}+\tilde{\theta}_{k})\{1+2\gamma_{k}^{2}\rho_{k}(\tilde{\theta})\}}

by (3.119), there exists a constant C>0C>0 independent of kk\in\mathbb{N} such that

(3.120) |ζ3,k(θ~)|+|ζ3,k(θ~)|C,θ~(0,π2),\displaystyle|\zeta_{3,k}(\tilde{\theta})|+|\zeta^{\prime}_{3,k}(\tilde{\theta})|\leq C,\quad\tilde{\theta}\in\left(0,\frac{\pi}{2}\right),
(3.121) limkζ3,k(θ~)=ζ3,(θ~)=12θ~+θ~uniformly on(0,π2)\displaystyle\lim_{k\to\infty}\zeta_{3,k}(\tilde{\theta})=\zeta_{3,\infty}(\tilde{\theta})=\frac{12}{\tilde{\theta}+\tilde{\theta}_{\infty}}\quad\text{uniformly on}\quad\left(0,\frac{\pi}{2}\right)

by (3.103), μ=1\mu_{\infty}=1, and the uniform boundedness of ρk\rho_{k} and ρk\rho^{\prime}_{k}.

Now let ΨCc(Θ,π/2)\Psi\in C_{c}^{\infty}(\Theta,\pi/2). We take the L2(Θ,π/2)L^{2}(\Theta,\pi/2)-inner product of (3.117) with Ψ\Psi and carry out integration by parts for the left-hand side to get

(3.122) (W~k,Ψ)L2(Θ,π/2)=(W~k,ζ1,kΨ)L2(Θ,π/2)+(W~k,ζ2,kΨ)L2(Θ,π/2)+(W~k+δkF~kθ~θ~k,ζ3,kΨ)L2(Θ,π/2).-\Bigl{(}\widetilde{W}^{\prime}_{k},\Psi^{\prime}\Bigr{)}_{L^{2}(\Theta,\pi/2)}=\Bigl{(}\widetilde{W}^{\prime}_{k},\zeta_{1,k}\Psi\Bigr{)}_{L^{2}(\Theta,\pi/2)}+\Bigl{(}\widetilde{W}_{k},\zeta_{2,k}\Psi\Bigr{)}_{L^{2}(\Theta,\pi/2)}\\ +\left(\frac{\widetilde{W}_{k}+\delta_{k}\widetilde{F}_{k}}{\tilde{\theta}-\tilde{\theta}_{k}},\zeta_{3,k}\Psi\right)_{L^{2}(\Theta,\pi/2)}.

We send kk\to\infty, use (3.115), (3.116), and (3.118), and then apply Hölder’s inequality and |ζ1,(θ~)|Θ1|\zeta_{1,\infty}(\tilde{\theta})|\leq\Theta^{-1} and |ζ2,(θ~)|m2Θ2|\zeta_{2,\infty}(\tilde{\theta})|\leq m_{\infty}^{2}\Theta^{-2} for θ~(Θ,π/2)\tilde{\theta}\in(\Theta,\pi/2) to get

(3.123) |(W~,Ψ)L2(Θ,π/2)|(1ΘW~L2(Θ,π/2)+m2Θ2W~L2(Θ,π/2))ΨL2(Θ,π/2)+lim supk|(W~k+δkF~kθ~θ~k,ζ3,kΨ)L2(Θ,π/2)|.\left|\Bigl{(}\widetilde{W}^{\prime}_{\infty},\Psi^{\prime}\Bigr{)}_{L^{2}(\Theta,\pi/2)}\right|\leq\left(\frac{1}{\Theta}\Bigl{\|}\widetilde{W}^{\prime}_{\infty}\Bigr{\|}_{L^{2}(\Theta,\pi/2)}+\frac{m_{\infty}^{2}}{\Theta^{2}}\Bigl{\|}\widetilde{W}_{\infty}\Bigr{\|}_{L^{2}(\Theta,\pi/2)}\right)\|\Psi\|_{L^{2}(\Theta,\pi/2)}\\ +\limsup_{k\to\infty}\left|\left(\frac{\widetilde{W}_{k}+\delta_{k}\widetilde{F}_{k}}{\tilde{\theta}-\tilde{\theta}_{k}},\zeta_{3,k}\Psi\right)_{L^{2}(\Theta,\pi/2)}\right|.

Let us estimate the last term as in the proof of Lemma 3.16. We define

Ψk(θ~)=ζ3,k(θ~)Ψ(θ~)sinθ~,θ~(Θ,π2)\displaystyle\Psi_{k}(\tilde{\theta})=\frac{\zeta_{3,k}(\tilde{\theta})\Psi(\tilde{\theta})}{\sin\tilde{\theta}},\quad\tilde{\theta}\in\left(\Theta,\frac{\pi}{2}\right)

and extend it to (0,π/2)(0,\pi/2) by zero outside (Θ,π/2)(\Theta,\pi/2). Then

(3.124) ΨkW1,(0,π/2)CsinΘΨW1,(Θ,π/2),ΨkL2(0,π/2)CsinΘΨL2(Θ,π/2)\displaystyle\|\Psi_{k}\|_{W^{1,\infty}(0,\pi/2)}\leq\frac{C}{\sin\Theta}\|\Psi\|_{W^{1,\infty}(\Theta,\pi/2)},\quad\|\Psi_{k}\|_{L^{2}(0,\pi/2)}\leq\frac{C}{\sin\Theta}\|\Psi\|_{L^{2}(\Theta,\pi/2)}

by (3.114) and (3.120) with a constant C>0C>0 independent of kk\in\mathbb{N}. Let

εk=σ2κδk2sinθk,εk=2σ2sinθk,ε~k=εkγk=σ2κδk2γksinθk,ε~k=εkγk=2σ2sinθkγk\displaystyle\varepsilon_{k}=\frac{\sigma^{2}\kappa\delta_{k}^{2}}{\sin\theta_{k}},\quad\varepsilon^{\prime}_{k}=2\sigma^{2}\sin\theta_{k},\quad\tilde{\varepsilon}_{k}=\frac{\varepsilon_{k}}{\gamma_{k}}=\frac{\sigma^{2}\kappa\delta_{k}^{2}}{\gamma_{k}\sin\theta_{k}},\quad\tilde{\varepsilon}^{\prime}_{k}=\frac{\varepsilon^{\prime}_{k}}{\gamma_{k}}=\frac{2\sigma^{2}\sin\theta_{k}}{\gamma_{k}}

for σ(0,1/2)\sigma\in(0,1/2). Then

0<εk<εk<12sinθk<12θk,γksinθk8π(1μk),π42sinθkγkπ4\displaystyle 0<\varepsilon_{k}<\varepsilon_{k}^{\prime}<\frac{1}{2}\sin\theta_{k}<\frac{1}{2}\theta_{k},\quad\gamma_{k}\sin\theta_{k}\leq\frac{8}{\pi}(1-\mu_{k}),\quad\frac{\pi}{4\sqrt{2}}\leq\frac{\sin\theta_{k}}{\gamma_{k}}\leq\frac{\pi}{4}

by 0μk10\leq\mu_{k}\leq 1, κδk21μk1μk2=sin2θk\kappa\delta_{k}^{2}\leq 1-\mu_{k}\leq 1-\mu_{k}^{2}=\sin^{2}\theta_{k}, and (3.102). Hence

(3.125) πκ8σ2δk21μkε~kπκ4(σδksinθk)2,π22σ2ε~kπ2σ2\displaystyle\frac{\pi\kappa}{8}\cdot\frac{\sigma^{2}\delta_{k}^{2}}{1-\mu_{k}}\leq\tilde{\varepsilon}_{k}\leq\frac{\pi\kappa}{4}\left(\frac{\sigma\delta_{k}}{\sin\theta_{k}}\right)^{2},\quad\frac{\pi}{2\sqrt{2}}\sigma^{2}\leq\tilde{\varepsilon}^{\prime}_{k}\leq\frac{\pi}{2}\sigma^{2}

and 0<ε~k<ε~k<θ~k/20<\tilde{\varepsilon}_{k}<\tilde{\varepsilon}^{\prime}_{k}<\tilde{\theta}_{k}/2 for sufficiently small σ>0\sigma>0 since θ~k(π/6,π/3)\tilde{\theta}_{k}\in(\pi/6,\pi/3). Let

I~1,k\displaystyle\tilde{I}_{1,k} =(θ~kε~k,θ~k+ε~k),\displaystyle=(\tilde{\theta}_{k}-\tilde{\varepsilon}_{k},\tilde{\theta}_{k}+\tilde{\varepsilon}_{k}),
I~2,k\displaystyle\tilde{I}_{2,k} =(θ~kε~k,θ~kε~k][θ~k+ε~k,θ~k+ε~k),\displaystyle=(\tilde{\theta}_{k}-\tilde{\varepsilon}^{\prime}_{k},\tilde{\theta}_{k}-\tilde{\varepsilon}_{k}]\cup[\tilde{\theta}_{k}+\tilde{\varepsilon}_{k},\tilde{\theta}_{k}+\tilde{\varepsilon}^{\prime}_{k}),
I~3,k\displaystyle\tilde{I}_{3,k} =(0,θ~kε~k][θ~k+ε~k,π2).\displaystyle=(0,\tilde{\theta}_{k}-\tilde{\varepsilon}^{\prime}_{k}]\cup\left[\tilde{\theta}_{k}+\tilde{\varepsilon}^{\prime}_{k},\frac{\pi}{2}\right).

Then

(3.126) (W~k+δkF~kθ~θ~k,ζ3,kΨ)L2(Θ,π/2)=0π2W~k(θ~)+δkF~k(θ~)θ~θ~kΨk(θ~)¯sinθ~dθ~=j=13J~j,k,\displaystyle\left(\frac{\widetilde{W}_{k}+\delta_{k}\widetilde{F}_{k}}{\tilde{\theta}-\tilde{\theta}_{k}},\zeta_{3,k}\Psi\right)_{L^{2}(\Theta,\pi/2)}=\int_{0}^{\frac{\pi}{2}}\frac{\widetilde{W}_{k}(\tilde{\theta})+\delta_{k}\widetilde{F}_{k}(\tilde{\theta})}{\tilde{\theta}-\tilde{\theta}_{k}}\,\overline{\Psi_{k}(\tilde{\theta})}\sin\tilde{\theta}\,d\tilde{\theta}=\sum_{j=1}^{3}\tilde{J}_{j,k},

where

J~j,k=I~j,kW~k(θ~)+δkF~k(θ~)θ~θ~kΨk(θ~)¯sinθ~dθ~,j=1,2,3.\displaystyle\tilde{J}_{j,k}=\int_{\tilde{I}_{j,k}}\frac{\widetilde{W}_{k}(\tilde{\theta})+\delta_{k}\widetilde{F}_{k}(\tilde{\theta})}{\tilde{\theta}-\tilde{\theta}_{k}}\,\overline{\Psi_{k}(\tilde{\theta})}\sin\tilde{\theta}\,d\tilde{\theta},\quad j=1,2,3.

Let us estimate J~1,k\tilde{J}_{1,k}. For θ~(0,π/2)\tilde{\theta}\in(0,\pi/2) let G~k(θ~)=W~k(θ~)+δkF~k(θ~)\widetilde{G}_{k}(\tilde{\theta})=\widetilde{W}_{k}(\tilde{\theta})+\delta_{k}\widetilde{F}_{k}(\tilde{\theta}) and

K~1,k=I~1,k|G~k(θ~)sinθ~θ~θ~k|2𝑑θ~.\displaystyle\widetilde{K}_{1,k}=\int_{\tilde{I}_{1,k}}\left|\frac{\widetilde{G}_{k}(\tilde{\theta})\sqrt{\sin\tilde{\theta}}}{\tilde{\theta}-\tilde{\theta}_{k}}\right|^{2}\,d\tilde{\theta}.

Since G~k(θ~k)=0\widetilde{G}_{k}(\tilde{\theta}_{k})=0 by (3.106), we can use Hardy’s inequality to get

(3.127) K~1,k4I~1,k|ddθ~(G~k(θ~)sinθ~)|2𝑑θ~Cθ~kε~kθ~k+ε~k(|G~k(θ~)|2+m2sin2θ~|G~k(θ~)|2)sinθ~dθ~.\displaystyle\begin{aligned} \widetilde{K}_{1,k}&\leq 4\int_{\tilde{I}_{1,k}}\left|\frac{d}{d\tilde{\theta}}\biggl{(}\widetilde{G}_{k}(\tilde{\theta})\sqrt{\sin\tilde{\theta}}\biggr{)}\right|^{2}\,d\tilde{\theta}\\ &\leq C\int_{\tilde{\theta}_{k}-\tilde{\varepsilon}_{k}}^{\tilde{\theta}_{k}+\tilde{\varepsilon}_{k}}\left(\Bigl{|}\widetilde{G}^{\prime}_{k}(\tilde{\theta})\Bigr{|}^{2}+\frac{m_{\infty}^{2}}{\sin^{2}\tilde{\theta}}\Bigl{|}\widetilde{G}_{k}(\tilde{\theta})\Bigr{|}^{2}\right)\sin\tilde{\theta}\,d\tilde{\theta}.\end{aligned}

Noting that G~k(θ~)=Wk(γkθ~)+δkFk(γkθ~)\widetilde{G}_{k}(\tilde{\theta})=W_{k}(\gamma_{k}\tilde{\theta})+\delta_{k}F_{k}(\gamma_{k}\tilde{\theta}), θk=γkθ~k\theta_{k}=\gamma_{k}\tilde{\theta}_{k}, and εk=γkε~k\varepsilon_{k}=\gamma_{k}\tilde{\varepsilon}_{k}, we set θ=γkθ~\theta=\gamma_{k}\tilde{\theta} in the second line of (3.127) and calculate as in (3.108) by using (2.7) and (3.104) to get

(3.128) K~1,kC(wk+δkfk)L2(Sk2)2C(wkL2(S2)+δkfkL2(Sk2)),\displaystyle\begin{aligned} \widetilde{K}_{1,k}\leq C\|\nabla(w_{k}+\delta_{k}f_{k})\|_{L^{2}(S_{k}^{2})}^{2}\leq C\Bigl{(}\|\nabla w_{k}\|_{L^{2}(S^{2})}+\delta_{k}\|\nabla f_{k}\|_{L^{2}(S_{k}^{2})}\Bigr{)},\end{aligned}

where Sk2=S2(θkεk,θk+εk)S_{k}^{2}=S^{2}(\theta_{k}-\varepsilon_{k},\theta_{k}+\varepsilon_{k}) (see (2.9)). Moreover, noting that

MsinθukL2(Sk2)CsinθkukL2(Sk2)CsinθkukL2(S2)\displaystyle\|M_{\sin\theta}u_{k}\|_{L^{2}(S_{k}^{2})}\leq C\sin\theta_{k}\|u_{k}\|_{L^{2}(S_{k}^{2})}\leq C\sin\theta_{k}\|u_{k}\|_{L^{2}(S^{2})}

by (3.42), we apply (3.47) and (3.54) to (3.128) to deduce that

δksinθkK~1,k1/2\displaystyle\frac{\delta_{k}}{\sin\theta_{k}}\widetilde{K}_{1,k}^{1/2} C(δkukL2(S2)+δkεk(Δ)1/2ukL2(S2)+δksinθk(Δ)1/2vkL2(S2)).\displaystyle\leq C\left(\delta_{k}\|u_{k}\|_{L^{2}(S^{2})}+\delta_{k}\varepsilon_{k}\|(-\Delta)^{1/2}u_{k}\|_{L^{2}(S^{2})}+\frac{\delta_{k}}{\sin\theta_{k}}\|(-\Delta)^{1/2}v_{k}\|_{L^{2}(S^{2})}\right).

Also, since κ1/2δk(1μk)1/2sinθk\kappa^{1/2}\delta_{k}\leq(1-\mu_{k})^{1/2}\leq\sin\theta_{k},

δksinθkκ1/2,δkεk=σ2κδk3sinθkCδk3(1μk)1/2.\displaystyle\frac{\delta_{k}}{\sin\theta_{k}}\leq\kappa^{-1/2},\quad\delta_{k}\varepsilon_{k}=\frac{\sigma^{2}\kappa\delta_{k}^{3}}{\sin\theta_{k}}\leq\frac{C\delta_{k}^{3}}{(1-\mu_{k})^{1/2}}.

Hence δkK~1,k1/2/sinθkC(1+Rk)\delta_{k}\widetilde{K}_{1,k}^{1/2}/\sin\theta_{k}\leq C(1+R_{k}) by the above inequalities, (3.49) for uku_{k}, and (3.88) (recall that Rk=R(uk,δk,μk)R_{k}=R(u_{k},\delta_{k},\mu_{k}) is given by (3.40)). By this inequality, |I~1,k|=2ε~k|\tilde{I}_{1,k}|=2\tilde{\varepsilon}_{k}, (3.124), and (3.125), we find that

(3.129) |J~1,k|K~1,k1/2(I~1,k|Ψk(θ~)|2sinθ~dθ~)1/22ε~kK~1,k1/2ΨkL(0,π/2)CσδksinθkK~1,k1/2ΨkL(0,π/2)CσsinΘ(1+Rk)ΨW1,(Θ,π/2).\displaystyle\begin{aligned} |\tilde{J}_{1,k}|&\leq\widetilde{K}_{1,k}^{1/2}\left(\int_{\tilde{I}_{1,k}}|\Psi_{k}(\tilde{\theta})|^{2}\sin\tilde{\theta}\,d\tilde{\theta}\right)^{1/2}\leq\sqrt{2\tilde{\varepsilon}_{k}}\,\widetilde{K}_{1,k}^{1/2}\|\Psi_{k}\|_{L^{\infty}(0,\pi/2)}\\ &\leq\frac{C\sigma\delta_{k}}{\sin\theta_{k}}\widetilde{K}_{1,k}^{1/2}\|\Psi_{k}\|_{L^{\infty}(0,\pi/2)}\leq\frac{C\sigma}{\sin\Theta}(1+R_{k})\|\Psi\|_{W^{1,\infty}(\Theta,\pi/2)}.\end{aligned}

Next we estimate J~2,k\tilde{J}_{2,k}. We split J~2,k=l=14J~2,kl\tilde{J}_{2,k}=\sum_{l=1}^{4}\tilde{J}_{2,k}^{l} into

J~2,k1\displaystyle\tilde{J}_{2,k}^{1} =I~2,kW~k(θ~)sinθ~W~k(θ~)sinθ~kθ~θ~kΨk(θ~)¯sinθ~𝑑θ~,\displaystyle=\int_{\tilde{I}_{2,k}}\frac{\widetilde{W}_{k}(\tilde{\theta})\sqrt{\sin\tilde{\theta}}-\widetilde{W}_{k}(\tilde{\theta})\sqrt{\sin\tilde{\theta}_{k}}}{\tilde{\theta}-\tilde{\theta}_{k}}\,\overline{\Psi_{k}(\tilde{\theta})}\sqrt{\sin\tilde{\theta}}\,d\tilde{\theta},
J~2,k2\displaystyle\tilde{J}_{2,k}^{2} =W~k(θ~k)sinθ~kI~2,kΨk(θ~)¯sinθ~Ψk(θ~k)¯sinθ~kθ~θk~𝑑θ~,\displaystyle=\widetilde{W}_{k}(\tilde{\theta}_{k})\sqrt{\sin\tilde{\theta}_{k}}\int_{\tilde{I}_{2,k}}\frac{\overline{\Psi_{k}(\tilde{\theta})}\sqrt{\sin\tilde{\theta}}-\overline{\Psi_{k}(\tilde{\theta}_{k})}\sqrt{\sin\tilde{\theta}_{k}}}{\tilde{\theta}-\tilde{\theta_{k}}}\,d\tilde{\theta},
J~2,k3\displaystyle\tilde{J}_{2,k}^{3} =W~k(θ~k)Ψk(θ~k)¯sinθkI~2,k1θ~θ~k𝑑θ~,\displaystyle=\widetilde{W}_{k}(\tilde{\theta}_{k})\,\overline{\Psi_{k}(\tilde{\theta}_{k})}\sin\theta_{k}\int_{\tilde{I}_{2,k}}\frac{1}{\tilde{\theta}-\tilde{\theta}_{k}}\,d\tilde{\theta},
J~2,k4\displaystyle\tilde{J}_{2,k}^{4} =I~2,kδkF~k(θ~)θ~θ~kΨk(θ~)¯sinθ~dθ~.\displaystyle=\int_{\tilde{I}_{2,k}}\frac{\delta_{k}\widetilde{F}_{k}(\tilde{\theta})}{\tilde{\theta}-\tilde{\theta}_{k}}\,\overline{\Psi_{k}(\tilde{\theta})}\sin\tilde{\theta}\,d\tilde{\theta}.

Also, let

K~k\displaystyle\widetilde{K}_{k} =0π2|W~k(θ~)sinθ~W~k(θ~)sinθ~kθ~θ~k|2𝑑θ~,\displaystyle=\int_{0}^{\frac{\pi}{2}}\left|\frac{\widetilde{W}_{k}(\tilde{\theta})\sqrt{\sin\tilde{\theta}}-\widetilde{W}_{k}(\tilde{\theta})\sqrt{\sin\tilde{\theta}_{k}}}{\tilde{\theta}-\tilde{\theta}_{k}}\right|^{2}\,d\tilde{\theta},
L~k\displaystyle\tilde{L}_{k} =0π2|Ψk(θ~)sinθ~Ψk(θ~)sinθ~kθ~θ~k|2𝑑θ~.\displaystyle=\int_{0}^{\frac{\pi}{2}}\left|\frac{\Psi_{k}(\tilde{\theta})\sqrt{\sin\tilde{\theta}}-\Psi_{k}(\tilde{\theta})\sqrt{\sin\tilde{\theta}_{k}}}{\tilde{\theta}-\tilde{\theta}_{k}}\right|^{2}\,d\tilde{\theta}.

As in (3.127) and (3.128), we apply Hardy’s inequality to K~k\widetilde{K}_{k}, set θ=γkθ~\theta=\gamma_{k}\tilde{\theta}, and use (2.7) and (3.104). Then we further use (3.47) and (3.88) to get

(3.130) K~kCwkL2(S2)2C(Δ)1/2vkL2(S2)2C.\displaystyle\widetilde{K}_{k}\leq C\|\nabla w_{k}\|_{L^{2}(S^{2})}^{2}\leq C\|(-\Delta)^{1/2}v_{k}\|_{L^{2}(S^{2})}^{2}\leq C.

Also, by Hardy’s inequality, suppΨk(Θ,π/2)\mathrm{supp}\,\Psi_{k}\subset(\Theta,\pi/2), and (3.124),

(3.131) L~kCΨkW1,(0,π/2)2Θπ2(sinθ~+1sinθ~)𝑑θ~Csin3ΘΨW1,(Θ,π/2)2\displaystyle\tilde{L}_{k}\leq C\|\Psi_{k}\|_{W^{1,\infty}(0,\pi/2)}^{2}\int_{\Theta}^{\frac{\pi}{2}}\left(\sin\tilde{\theta}+\frac{1}{\sin\tilde{\theta}}\right)\,d\tilde{\theta}\leq\frac{C}{\sin^{3}\Theta}\|\Psi\|_{W^{1,\infty}(\Theta,\pi/2)}^{2}

We use Hölder’s inequality, |I~2,k|2ε~k|\tilde{I}_{2,k}|\leq 2\tilde{\varepsilon}^{\prime}_{k}, (3.124), (3.125), and (3.130) to get

(3.132) |J~2,k1|K~k1/2|I~2,k|1/2ΨkL(Θ,π/2)CσsinΘΨW1,(Θ,π/2).\displaystyle|\tilde{J}_{2,k}^{1}|\leq\widetilde{K}_{k}^{1/2}|\tilde{I}_{2,k}|^{1/2}\|\Psi_{k}\|_{L^{\infty}(\Theta,\pi/2)}\leq\frac{C\sigma}{\sin\Theta}\|\Psi\|_{W^{1,\infty}(\Theta,\pi/2)}.

Also, by Hölder’s inequality, |I~2,k|2ε~kC|\tilde{I}_{2,k}|\leq 2\tilde{\varepsilon}^{\prime}_{k}\leq C, sinθ~k1\sin\tilde{\theta}_{k}\leq 1, and (3.131),

(3.133) |J~2,k2||W~k(θ~k)|sinθ~kL~k1/2|I~2,k|1/2Csin3/2Θ|W~k(θ~k)|ΨW1,(Θ,π/2).\displaystyle|\tilde{J}_{2,k}^{2}|\leq\Bigl{|}\widetilde{W}_{k}(\tilde{\theta}_{k})\Bigr{|}\sqrt{\sin\tilde{\theta}_{k}}\,\tilde{L}_{k}^{1/2}|\tilde{I}_{2,k}|^{1/2}\leq\frac{C}{\sin^{3/2}\Theta}\Bigl{|}\widetilde{W}_{k}(\tilde{\theta}_{k})\Bigr{|}\|\Psi\|_{W^{1,\infty}(\Theta,\pi/2)}.

We have J~2,k3=0\tilde{J}_{2,k}^{3}=0 since

I~2,k1θ~θ~k𝑑θ~\displaystyle\int_{\tilde{I}_{2,k}}\frac{1}{\tilde{\theta}-\tilde{\theta}_{k}}\,d\tilde{\theta} =θ~kε~kθ~kε~kddθ~(log(θ~kθ~))𝑑θ~+θ~k+ε~kθ~k+ε~kddθ~(log(θ~θ~k))𝑑θ~=0.\displaystyle=\int_{\tilde{\theta}_{k}-\tilde{\varepsilon}^{\prime}_{k}}^{\tilde{\theta}_{k}-\tilde{\varepsilon}_{k}}\frac{d}{d\tilde{\theta}}\bigl{(}\log(\tilde{\theta}_{k}-\tilde{\theta})\bigr{)}\,d\tilde{\theta}+\int_{\tilde{\theta}_{k}+\tilde{\varepsilon}_{k}}^{\tilde{\theta}_{k}+\tilde{\varepsilon}^{\prime}_{k}}\frac{d}{d\tilde{\theta}}\bigl{(}\log(\tilde{\theta}-\tilde{\theta}_{k})\bigr{)}\,d\tilde{\theta}=0.

For J~2,k4\tilde{J}_{2,k}^{4}, we see that

|J~2,k4|δkΨkL(0,π/2)(0π2|F~k(θ~)|2sinθ~dθ~)1/2(I~2,k1|θ~θ~k|2𝑑θ~)1/2.\displaystyle|\tilde{J}_{2,k}^{4}|\leq\delta_{k}\|\Psi_{k}\|_{L^{\infty}(0,\pi/2)}\left(\int_{0}^{\frac{\pi}{2}}\Bigl{|}\widetilde{F}_{k}(\tilde{\theta})\Bigr{|}^{2}\sin\tilde{\theta}\,d\tilde{\theta}\right)^{1/2}\left(\int_{\tilde{I}_{2,k}}\frac{1}{|\tilde{\theta}-\tilde{\theta}_{k}|^{2}}\,d\tilde{\theta}\right)^{1/2}.

As in (3.107), we use F~k(θ~)=Fk(γkθ~)\widetilde{F}_{k}(\tilde{\theta})=F_{k}(\gamma_{k}\tilde{\theta}), (2.7), and (3.104) to get

(3.134) 0π2|F~k(θ~)|2sinθ~dθ~C1μkfkL2(S2)2.\displaystyle\int_{0}^{\frac{\pi}{2}}\Bigl{|}\widetilde{F}_{k}(\tilde{\theta})\Bigr{|}^{2}\sin\tilde{\theta}\,d\tilde{\theta}\leq\frac{C}{1-\mu_{k}}\|f_{k}\|_{L^{2}(S^{2})}^{2}.

Also, we observe by (3.125) that

I~2,k1|θ~θ~k|2𝑑θ~=(θ~kε~kθ~kε~k+θ~k+ε~kθ~k+ε~k)ddθ~(1θ~θ~k)dθ~2ε~kC(1μk)σ2δk2.\displaystyle\int_{\tilde{I}_{2,k}}\frac{1}{|\tilde{\theta}-\tilde{\theta}_{k}|^{2}}\,d\tilde{\theta}=\left(\int_{\tilde{\theta}_{k}-\tilde{\varepsilon}^{\prime}_{k}}^{\tilde{\theta}_{k}-\tilde{\varepsilon}_{k}}+\int_{\tilde{\theta}_{k}+\tilde{\varepsilon}_{k}}^{\tilde{\theta}_{k}+\tilde{\varepsilon}^{\prime}_{k}}\right)\frac{d}{d\tilde{\theta}}\left(-\frac{1}{\tilde{\theta}-\tilde{\theta}_{k}}\right)\,d\tilde{\theta}\leq\frac{2}{\tilde{\varepsilon}_{k}}\leq\frac{C(1-\mu_{k})}{\sigma^{2}\delta_{k}^{2}}.

We deduce from the above inequalities and (3.124) that

|J~2,k4|Cσ1fkL2(S2)ΨkL(0,π/2)Cσ1RksinΘΨW1,(Θ,π/2).\displaystyle|\tilde{J}_{2,k}^{4}|\leq C\sigma^{-1}\|f_{k}\|_{L^{2}(S^{2})}\|\Psi_{k}\|_{L^{\infty}(0,\pi/2)}\leq\frac{C\sigma^{-1}R_{k}}{\sin\Theta}\|\Psi\|_{W^{1,\infty}(\Theta,\pi/2)}.

Thus, by this inequality, J~2,k3=0\tilde{J}_{2,k}^{3}=0, (3.132), (3.133), and sinΘ1\sin\Theta\leq 1,

(3.135) |J~2,k|l=14|J~2,kl|Csin3/2Θ(σ+|W~k(θ~k)|+σ1Rk)ΨW1,(Θ,π/2).\displaystyle|\tilde{J}_{2,k}|\leq\sum_{l=1}^{4}|\tilde{J}_{2,k}^{l}|\leq\frac{C}{\sin^{3/2}\Theta}\left(\sigma+\Bigl{|}\widetilde{W}_{k}(\tilde{\theta}_{k})\Bigr{|}+\sigma^{-1}R_{k}\right)\|\Psi\|_{W^{1,\infty}(\Theta,\pi/2)}.

To estimate J~3,k\tilde{J}_{3,k}, we decompose J~3,k=l=13J~3,kl\tilde{J}_{3,k}=\sum_{l=1}^{3}\tilde{J}_{3,k}^{l} into

J~3,k1\displaystyle\tilde{J}_{3,k}^{1} =I~3,kW~k(θ~)sinθ~W~k(θ~)sinθ~kθ~θ~kΨk(θ~)¯sinθ~𝑑θ~,\displaystyle=\int_{\tilde{I}_{3,k}}\frac{\widetilde{W}_{k}(\tilde{\theta})\sqrt{\sin\tilde{\theta}}-\widetilde{W}_{k}(\tilde{\theta})\sqrt{\sin\tilde{\theta}_{k}}}{\tilde{\theta}-\tilde{\theta}_{k}}\,\overline{\Psi_{k}(\tilde{\theta})}\sqrt{\sin\tilde{\theta}}\,d\tilde{\theta},
J~3,k2\displaystyle\tilde{J}_{3,k}^{2} =W~k(θ~k)sinθ~kI~3,kΨk(θ~)¯sinθ~θ~θ~k𝑑θ~,\displaystyle=\widetilde{W}_{k}(\tilde{\theta}_{k})\sqrt{\sin\tilde{\theta}_{k}}\int_{\tilde{I}_{3,k}}\frac{\overline{\Psi_{k}(\tilde{\theta})}\sqrt{\sin\tilde{\theta}}}{\tilde{\theta}-\tilde{\theta}_{k}}\,d\tilde{\theta},
J~3,k3\displaystyle\tilde{J}_{3,k}^{3} =I~3,kδkF~k(θ~)θ~θ~kΨk(θ~)¯sinθ~dθ~.\displaystyle=\int_{\tilde{I}_{3,k}}\frac{\delta_{k}\widetilde{F}_{k}(\tilde{\theta})}{\tilde{\theta}-\tilde{\theta}_{k}}\,\overline{\Psi_{k}(\tilde{\theta})}\sin\tilde{\theta}\,d\tilde{\theta}.

We apply Hölder’s inequality to J~3,k1\tilde{J}_{3,k}^{1} and use (3.124) and (3.130) to get

|J~3,k1|\displaystyle|\tilde{J}_{3,k}^{1}| K~k1/2ΨkL2(0,π/2)CsinΘΨL2(Θ,π/2).\displaystyle\leq\widetilde{K}_{k}^{1/2}\|\Psi_{k}\|_{L^{2}(0,\pi/2)}\leq\frac{C}{\sin\Theta}\|\Psi\|_{L^{2}(\Theta,\pi/2)}.

When θ~I~3,k\tilde{\theta}\in\tilde{I}_{3,k}, we have |θ~θ~|ε~kπσ2/22|\tilde{\theta}-\tilde{\theta}|\geq\tilde{\varepsilon}^{\prime}_{k}\geq\pi\sigma^{2}/2\sqrt{2} by (3.125). Thus

|J~3,k2|\displaystyle|\tilde{J}_{3,k}^{2}| |W~k(θ~k)|sinθ~k22πσ2I~3,k|Ψk(θ~)|sinθ~dθ~\displaystyle\leq\Bigl{|}\widetilde{W}_{k}(\tilde{\theta}_{k})\Bigr{|}\sqrt{\sin\tilde{\theta}_{k}}\,\frac{2\sqrt{2}}{\pi\sigma^{2}}\int_{\tilde{I}_{3,k}}|\Psi_{k}(\tilde{\theta})|\sqrt{\sin\tilde{\theta}}\,d\tilde{\theta}
Cσ2|W~k(θ~k)|ΨkL2(0,π/2)Cσ2sinΘ|W~k(θ~k)|ΨL2(Θ,π/2).\displaystyle\leq C\sigma^{-2}\Bigl{|}\widetilde{W}_{k}(\tilde{\theta}_{k})\Bigr{|}\|\Psi_{k}\|_{L^{2}(0,\pi/2)}\leq\frac{C\sigma^{-2}}{\sin\Theta}\Bigl{|}\widetilde{W}_{k}(\tilde{\theta}_{k})\Bigr{|}\|\Psi\|_{L^{2}(\Theta,\pi/2)}.

by |I~3,k|C|\tilde{I}_{3,k}|\leq C, sinθ~k1\sin\tilde{\theta}_{k}\leq 1, and (3.124). Similarly,

|J~3,k3|\displaystyle|\tilde{J}_{3,k}^{3}| 22δkπσ2I~3,k|F~k(θ~)||Ψk(θ~)|sinθ~dθ~Cσ2RksinΘΨL2(Θ,π/2)\displaystyle\leq\frac{2\sqrt{2}\,\delta_{k}}{\pi\sigma^{2}}\int_{\tilde{I}_{3,k}}\Bigl{|}\widetilde{F}_{k}(\tilde{\theta})\Bigr{|}|\Psi_{k}(\tilde{\theta})|\sin\tilde{\theta}\,d\tilde{\theta}\leq\frac{C\sigma^{-2}R_{k}}{\sin\Theta}\|\Psi\|_{L^{2}(\Theta,\pi/2)}

by Hölder’s inequality, (3.124), (3.134), and 1μkκδk21-\mu_{k}\geq\kappa\delta_{k}^{2}. Hence

(3.136) |J~3,k|l=13|J~3,kl|CsinΘ(1+σ2|W~k(θ~k)|+σ2Rk)ΨL2(Θ,π/2).\displaystyle|\tilde{J}_{3,k}|\leq\sum_{l=1}^{3}|\tilde{J}_{3,k}^{l}|\leq\frac{C}{\sin\Theta}\left(1+\sigma^{-2}\Bigl{|}\widetilde{W}_{k}(\tilde{\theta}_{k})\Bigr{|}+\sigma^{-2}R_{k}\right)\|\Psi\|_{L^{2}(\Theta,\pi/2)}.

Now we deduce from (3.126), (3.129), (3.135), and (3.136) that

|(W~k+δkF~kθ~θ~k,ζ3,kΨ)L2(Θ,π/2)|CsinΘ(1+σ2|W~k(θ~k)|+σ2Rk)ΨL2(Θ,π/2)+Csin3/2Θ(σ+|W~k(θ~k)|+σ1Rk)ΨW1,(Θ,π/2)\left|\left(\frac{\widetilde{W}_{k}+\delta_{k}\widetilde{F}_{k}}{\tilde{\theta}-\tilde{\theta}_{k}},\zeta_{3,k}\Psi\right)_{L^{2}(\Theta,\pi/2)}\right|\leq\frac{C}{\sin\Theta}\left(1+\sigma^{-2}\Bigl{|}\widetilde{W}_{k}(\tilde{\theta}_{k})\Bigr{|}+\sigma^{-2}R_{k}\right)\|\Psi\|_{L^{2}(\Theta,\pi/2)}\\ +\frac{C}{\sin^{3/2}\Theta}\left(\sigma+\Bigl{|}\widetilde{W}_{k}(\tilde{\theta}_{k})\Bigr{|}+\sigma^{-1}R_{k}\right)\|\Psi\|_{W^{1,\infty}(\Theta,\pi/2)}

for all σ(0,1/2)\sigma\in(0,1/2), where we also used sinΘ1\sin\Theta\leq 1 and σ<1\sigma<1. Then we send kk\to\infty, use (3.89) and (3.106), and let σ0\sigma\to 0 to find that

lim supk|(W~k+δkF~kθ~θ~k,ζ3,kΨ)L2(Θ,π/2)|CsinΘΨL2(Θ,π/2).\displaystyle\limsup_{k\to\infty}\left|\left(\frac{\widetilde{W}_{k}+\delta_{k}\widetilde{F}_{k}}{\tilde{\theta}-\tilde{\theta}_{k}},\zeta_{3,k}\Psi\right)_{L^{2}(\Theta,\pi/2)}\right|\leq\frac{C}{\sin\Theta}\|\Psi\|_{L^{2}(\Theta,\pi/2)}.

Therefore, applying this inequality to (3.123), we obtain

|(W~,Ψ)L2(Θ,π/2)|(1ΘW~L2(Θ,π/2)+m2Θ2W~L2(Θ,π/2)+CsinΘ)ΨL2(Θ,π/2)\displaystyle\left|\Bigl{(}\widetilde{W}^{\prime}_{\infty},\Psi^{\prime}\Bigr{)}_{L^{2}(\Theta,\pi/2)}\right|\leq\left(\frac{1}{\Theta}\Bigl{\|}\widetilde{W}^{\prime}_{\infty}\Bigr{\|}_{L^{2}(\Theta,\pi/2)}+\frac{m_{\infty}^{2}}{\Theta^{2}}\Bigl{\|}\widetilde{W}_{\infty}\Bigr{\|}_{L^{2}(\Theta,\pi/2)}+\frac{C}{\sin\Theta}\right)\|\Psi\|_{L^{2}(\Theta,\pi/2)}

for all ΨCc(Θ,π/2)\Psi\in C_{c}^{\infty}(\Theta,\pi/2), which gives (3.113) since Cc(Θ,π/2)C_{c}^{\infty}(\Theta,\pi/2) is dense in L2(Θ,π/2)L^{2}(\Theta,\pi/2).

Next we derive an ODE for W~\widetilde{W}_{\infty}. For Θ(0,θ~)\Theta\in(0,\tilde{\theta}_{\infty}), let ΨCc(Θ,π/2)\Psi\in C_{c}^{\infty}(\Theta,\pi/2) satisfy Ψ=0\Psi=0 near θ~\tilde{\theta}_{\infty}. Then we observe by (3.122) that

(W~k,Ψ)L2(Θ,π/2)=(W~k,ζ1,kΨ)L2(Θ,π/2)+(W~k,ζ2,kΨ)L2(Θ,π/2)+(W~k+δkF~k,ζ3,kΨθ~θ~k)L2(Θ,π/2).-\Bigl{(}\widetilde{W}^{\prime}_{k},\Psi^{\prime}\Bigr{)}_{L^{2}(\Theta,\pi/2)}=\Bigl{(}\widetilde{W}^{\prime}_{k},\zeta_{1,k}\Psi\Bigr{)}_{L^{2}(\Theta,\pi/2)}+\Bigl{(}\widetilde{W}_{k},\zeta_{2,k}\Psi\Bigr{)}_{L^{2}(\Theta,\pi/2)}\\ +\left(\widetilde{W}_{k}+\delta_{k}\widetilde{F}_{k},\frac{\zeta_{3,k}\Psi}{\tilde{\theta}-\tilde{\theta}_{k}}\right)_{L^{2}(\Theta,\pi/2)}.

We send kk\to\infty in this equality. Then since

δkF~kL2(Θ,π/2)2δk2sinΘΘπ2|F~k(θ~)|2sinθ~dθ~CsinΘδk21μkfkL2(S2)2CRk2sinΘ0\displaystyle\Bigl{\|}\delta_{k}\widetilde{F}_{k}\Bigr{\|}_{L^{2}(\Theta,\pi/2)}^{2}\leq\frac{\delta_{k}^{2}}{\sin\Theta}\int_{\Theta}^{\frac{\pi}{2}}\Bigl{|}\widetilde{F}_{k}(\tilde{\theta})\Bigr{|}^{2}\sin\tilde{\theta}\,d\tilde{\theta}\leq\frac{C}{\sin\Theta}\cdot\frac{\delta_{k}^{2}}{1-\mu_{k}}\|f_{k}\|_{L^{2}(S^{2})}^{2}\leq\frac{CR_{k}^{2}}{\sin\Theta}\to 0

as kk\to\infty by (3.89), (3.114), (3.134), and 1μkκδk21-\mu_{k}\geq\kappa\delta_{k}^{2}, and since

ζ3,kΨθ~θ~kζ3,Ψθ~θ~uniformly on(Θ,π2)\displaystyle\frac{\zeta_{3,k}\Psi}{\tilde{\theta}-\tilde{\theta}_{k}}\to\frac{\zeta_{3,\infty}\Psi}{\tilde{\theta}-\tilde{\theta}_{\infty}}\quad\text{uniformly on}\quad\left(\Theta,\frac{\pi}{2}\right)

by (3.103), (3.121), and Ψ=0\Psi=0 near θ~\tilde{\theta}_{\infty}, we find by (3.115), (3.116), and (3.118) that

(W~,Ψ)L2(Θ,π/2)=(W~,ζ1,Ψ)L2(Θ,π/2)+(W~,ζ2,Ψ)L2(Θ,π/2)+(W~,ζ3,Ψθ~θ~)L2(Θ,π/2).-\Bigl{(}\widetilde{W}^{\prime}_{\infty},\Psi^{\prime}\Bigr{)}_{L^{2}(\Theta,\pi/2)}=\Bigl{(}\widetilde{W}^{\prime}_{\infty},\zeta_{1,\infty}\Psi\Bigr{)}_{L^{2}(\Theta,\pi/2)}+\Bigl{(}\widetilde{W}_{\infty},\zeta_{2,\infty}\Psi\Bigr{)}_{L^{2}(\Theta,\pi/2)}\\ +\left(\widetilde{W}_{\infty},\frac{\zeta_{3,\infty}\Psi}{\tilde{\theta}-\tilde{\theta}_{\infty}}\right)_{L^{2}(\Theta,\pi/2)}.

Since this equality is valid for all Θ(0,θ~)\Theta\in(0,\tilde{\theta}_{\infty}) and ΨCc(Θ,π/2)\Psi\in C_{c}^{\infty}(\Theta,\pi/2) vanishing near θ~\tilde{\theta}_{\infty}, and since W~Hloc2(0,π/2)\widetilde{W}_{\infty}\in H_{loc}^{2}(0,\pi/2) by (3.113), we obtain

(3.137) W~′′(θ~)=ζ1,(θ~)W~(θ~)+{ζ2,(θ~)+ζ3,(θ~)θ~θ~}W~(θ~),θ~(0,π2){θ~}.\displaystyle\widetilde{W}^{\prime\prime}_{\infty}(\tilde{\theta})=\zeta_{1,\infty}(\tilde{\theta})\widetilde{W}^{\prime}_{\infty}(\tilde{\theta})+\left\{\zeta_{2,\infty}(\tilde{\theta})+\frac{\zeta_{3,\infty}(\tilde{\theta})}{\tilde{\theta}-\tilde{\theta}_{\infty}}\right\}\widetilde{W}_{\infty}(\tilde{\theta}),\quad\tilde{\theta}\in\left(0,\frac{\pi}{2}\right)\setminus\{\tilde{\theta}_{\infty}\}.

Now we observe that sinθ~sinθ~k1/2\sin\tilde{\theta}\geq\sin\tilde{\theta}_{k}\geq 1/2 for θ~(θ~k,π/2)\tilde{\theta}\in(\tilde{\theta}_{k},\pi/2) since θ~k(π/6,π/3)\tilde{\theta}_{k}\in(\pi/6,\pi/3). We apply this inequality and calculate as in (3.107) by using (3.104) and θk=γkθ~k\theta_{k}=\gamma_{k}\tilde{\theta}_{k}. Then

W~kL2(θ~k,π/2)22θ~kπ2|W~k(θ~)|2sinθ~dθ~Cγk2θkπ|Wk(θ)|2sinθdθ.\displaystyle\Bigl{\|}\widetilde{W}_{k}\Bigr{\|}_{L^{2}(\tilde{\theta}_{k},\pi/2)}^{2}\leq 2\int_{\tilde{\theta}_{k}}^{\frac{\pi}{2}}\Bigl{|}\widetilde{W}_{k}(\tilde{\theta})\Bigr{|}^{2}\sin\tilde{\theta}\,d\tilde{\theta}\leq\frac{C}{\gamma_{k}^{2}}\int_{\theta_{k}}^{\pi}|W_{k}(\theta)|^{2}\sin\theta\,d\theta.

Moreover, we apply (2.7) and (3.102) to the last term and use (3.99) to get

W~kL2(θ~k,π/2)2C1μkwkL2(S2(θk,π))20ask.\displaystyle\Bigl{\|}\widetilde{W}_{k}\Bigr{\|}_{L^{2}(\tilde{\theta}_{k},\pi/2)}^{2}\leq\frac{C}{1-\mu_{k}}\|w_{k}\|_{L^{2}(S^{2}(\theta_{k},\pi))}^{2}\to 0\quad\text{as}\quad k\to\infty.

By this fact, (3.103), and (3.115) with Θ=π/6\Theta=\pi/6, we find that

W~L2(θ~,π/2)=limkW~kL2(θ~k,π/2)=0.\displaystyle\Bigl{\|}\widetilde{W}_{\infty}\Bigr{\|}_{L^{2}(\tilde{\theta}_{\infty},\pi/2)}=\lim_{k\to\infty}\Bigl{\|}\widetilde{W}_{k}\Bigr{\|}_{L^{2}(\tilde{\theta}_{k},\pi/2)}=0.

Thus W~=0\widetilde{W}_{\infty}=0 on (θ~,π/2)(\tilde{\theta}_{\infty},\pi/2) and, since W~C1(0,π/2)\widetilde{W}_{\infty}\in C^{1}(0,\pi/2) by (3.113), we further get

(3.138) W~(θ~)=W~(θ~)=0.\displaystyle\widetilde{W}_{\infty}(\tilde{\theta}_{\infty})=\widetilde{W}^{\prime}_{\infty}(\tilde{\theta}_{\infty})=0.

Then we easily find that any solution of (3.137)–(3.138) in C1(0,π/2)C^{1}(0,\pi/2) must be trivial by the standard theory of ODEs, since ζ1,\zeta_{1,\infty}, ζ2,\zeta_{2,\infty}, and ζ3,\zeta_{3,\infty} given by (3.118) and (3.121) are smooth on (0,π/2)(0,\pi/2) and the singularity of (θ~θ~)1(\tilde{\theta}-\tilde{\theta}_{\infty})^{-1} at θ~=θ~\tilde{\theta}=\tilde{\theta}_{\infty} is of order one. Hence W~=0\widetilde{W}_{\infty}=0, i.e. w~=0\tilde{w}_{\infty}=0, which contradicts w~0\tilde{w}_{\infty}\neq 0. This completes the proof in Case 2, and we conclude that (3.33) is valid. ∎

Proof of (3.34).

We assume μ0\mu\geq 0 for simplicity, since the case μ0\mu\leq 0 can be handled similarly. Let κ\kappa be the constant given by (3.31) and δ(0,1]\delta\in(0,1], μ[0,1κδ2]\mu\in[0,1-\kappa\delta^{2}], and θμ=arccosμ(0,π/2]\theta_{\mu}=\arccos\mu\in(0,\pi/2]. Also, for u𝒴mH1(S2)u\in\mathcal{Y}_{m}\cap H^{1}(S^{2}) with m{0}m\in\mathbb{Z}\setminus\{0\}, let vv, ww, and ff be given by (3.35) and (3.40). We set ε=κ1/2δ/2\varepsilon=\kappa^{1/2}\delta/2, which satisfies (3.41) by (3.36). Since u,v,w,f𝒳mu,v,w,f\in\mathcal{X}_{m}, we can write

u=U(θ)eimφ,v=V(θ)eimφ,w=W(θ)eimφ,f=F(θ)eimφ.\displaystyle u=U(\theta)e^{im\varphi},\quad v=V(\theta)e^{im\varphi},\quad w=W(\theta)e^{im\varphi},\quad f=F(\theta)e^{im\varphi}.

For θ(θμε,θμ+ε)\theta\in(\theta_{\mu}-\varepsilon,\theta_{\mu}+\varepsilon), we observe by (2.21) and (3.42) that

|sinθU(θ)|2Csin2θμ|U(θ)|2CMsinθuL2(S2)(Δ)1/2uL2(S2).\displaystyle|\sin\theta\,U(\theta)|^{2}\leq C\sin^{2}\theta_{\mu}|U(\theta)|^{2}\leq C\|M_{\sin\theta}u\|_{L^{2}(S^{2})}\|(-\Delta)^{1/2}u\|_{L^{2}(S^{2})}.

By this inequality, (3.43), and Young’s inequality,

(3.139) θμεθμ+ε|sinθU(θ)|2sinθdθCεsinθμMsinθuL2(S2)(Δ)1/2uL2(S2)12MsinθuL2(S2)2+Cε2sin2θμ(Δ)1/2uL2(S2)2.\displaystyle\begin{aligned} \int_{\theta_{\mu}-\varepsilon}^{\theta_{\mu}+\varepsilon}|\sin\theta\,U(\theta)|^{2}\sin\theta\,d\theta&\leq C\varepsilon\sin\theta_{\mu}\|M_{\sin\theta}u\|_{L^{2}(S^{2})}\|(-\Delta)^{1/2}u\|_{L^{2}(S^{2})}\\ &\leq\frac{1}{2}\|M_{\sin\theta}u\|_{L^{2}(S^{2})}^{2}+C\varepsilon^{2}\sin^{2}\theta_{\mu}\|(-\Delta)^{1/2}u\|_{L^{2}(S^{2})}^{2}.\end{aligned}

Since u+6v=B2,mu=(w+δf)/(μx3)u+6v=B_{2,m}u=(w+\delta f)/(\mu-x_{3}) for x3μx_{3}\neq\mu by (3.45),

U(θ)=6V(θ)+W(θ)μcosθ+δF(θ)μcosθ,θ(0,π){θμ}.\displaystyle U(\theta)=-6V(\theta)+\frac{W(\theta)}{\mu-\cos\theta}+\frac{\delta F(\theta)}{\mu-\cos\theta},\quad\theta\in(0,\pi)\setminus\{\theta_{\mu}\}.

By this equality and |μcosθ|12|θθμ|sinθ|\mu-\cos\theta|\geq\frac{1}{2}|\theta-\theta_{\mu}|\sin\theta (see (3.44)),

|U(θ)|C(|V(θ)|+|W(θ)θθμ|+δ|θθμ||F(θ)|),θ(0,π){θμ}.\displaystyle|U(\theta)|\leq C\left(|V(\theta)|+\left|\frac{W(\theta)}{\theta-\theta_{\mu}}\right|+\frac{\delta}{|\theta-\theta_{\mu}|}|F(\theta)|\right),\quad\theta\in(0,\pi)\setminus\{\theta_{\mu}\}.

Moreover, if |θθμ|ε|\theta-\theta_{\mu}|\geq\varepsilon, then δ/|θθμ|δ/ε=2/κ1/2\delta/|\theta-\theta_{\mu}|\leq\delta/\varepsilon=2/\kappa^{1/2}. By these inequalities, |sinθ|1|\sin\theta|\leq 1, and (2.7), we find that

(3.140) |θθμ|ε|sinθU(θ)|2sinθdθC(vL2(S2)2+fL2(S2)2+|θθμ|ε|W(θ)sinθθθμ|2𝑑θ).\int_{|\theta-\theta_{\mu}|\geq\varepsilon}|\sin\theta\,U(\theta)|^{2}\sin\theta\,d\theta\\ \leq C\left(\|v\|_{L^{2}(S^{2})}^{2}+\|f\|_{L^{2}(S^{2})}^{2}+\int_{|\theta-\theta_{\mu}|\geq\varepsilon}\left|\frac{W(\theta)\sqrt{\sin\theta}}{\theta-\theta_{\mu}}\right|^{2}\,d\theta\right).

To estimate the last term, we see that

|θθμ|ε|W(θ)sinθθθμ|2𝑑θC(J+|W(θμ)|2sinθμ|θθμ|ε1|θθμ|2𝑑θ),\displaystyle\int_{|\theta-\theta_{\mu}|\geq\varepsilon}\left|\frac{W(\theta)\sqrt{\sin\theta}}{\theta-\theta_{\mu}}\right|^{2}\,d\theta\leq C\left(J+|W(\theta_{\mu})|^{2}\sin\theta_{\mu}\int_{|\theta-\theta_{\mu}|\geq\varepsilon}\frac{1}{|\theta-\theta_{\mu}|^{2}}\,d\theta\right),

where

J=|θθμ|ε|W(θ)sinθW(θμ)sinθμθθμ|2𝑑θ.\displaystyle J=\int_{|\theta-\theta_{\mu}|\geq\varepsilon}\left|\frac{W(\theta)\sqrt{\sin\theta}-W(\theta_{\mu})\sqrt{\sin\theta_{\mu}}}{\theta-\theta_{\mu}}\right|^{2}\,d\theta.

We use Hardy’s inequality, (2.7), and (3.47) to get

J\displaystyle J 40π|ddθ(W(θ)sinθ)|2𝑑θC0π(|W(θ)|2+m2sin2θ|W(θ)|2)sinθdθ\displaystyle\leq 4\int_{0}^{\pi}\left|\frac{d}{d\theta}\Bigl{(}W(\theta)\sqrt{\sin\theta}\Bigr{)}\right|^{2}\,d\theta\leq C\int_{0}^{\pi}\left(|W^{\prime}(\theta)|^{2}+\frac{m^{2}}{\sin^{2}\theta}|W(\theta)|^{2}\right)\sin\theta\,d\theta
CwL2(S2)2C(Δ)1/2vL2(S2)2.\displaystyle\leq C\|\nabla w\|_{L^{2}(S^{2})}^{2}\leq C\|(-\Delta)^{1/2}v\|_{L^{2}(S^{2})}^{2}.

Also, |θθμ|ε|θθμ|2𝑑θ2/ε\int_{|\theta-\theta_{\mu}|\geq\varepsilon}|\theta-\theta_{\mu}|^{-2}\,d\theta\leq 2/\varepsilon by |θθμ|2=ddθ(θθμ)1|\theta-\theta_{\mu}|^{-2}=-\frac{d}{d\theta}(\theta-\theta_{\mu})^{-1}. Hence

|θθμ|ε|W(θ)sinθθθμ|2𝑑θC((Δ)1/2vL2(S2)2+sinθμε|W(θμ)|2).\displaystyle\int_{|\theta-\theta_{\mu}|\geq\varepsilon}\left|\frac{W(\theta)\sqrt{\sin\theta}}{\theta-\theta_{\mu}}\right|^{2}\,d\theta\leq C\left(\|(-\Delta)^{1/2}v\|_{L^{2}(S^{2})}^{2}+\frac{\sin\theta_{\mu}}{\varepsilon}|W(\theta_{\mu})|^{2}\right).

We combine this inequality, (3.139), and (3.140) and use (2.6) to vv to get

(3.141) MsinθuL2(S2)2Cε2sin2θμ(Δ)1/2uL2(S2)2+C((Δ)1/2vL2(S2)2+fL2(S2)2+sinθμε|W(θμ)|2).\|M_{\sin\theta}u\|_{L^{2}(S^{2})}^{2}\leq C\varepsilon^{2}\sin^{2}\theta_{\mu}\|(-\Delta)^{1/2}u\|_{L^{2}(S^{2})}^{2}\\ +C\left(\|(-\Delta)^{1/2}v\|_{L^{2}(S^{2})}^{2}+\|f\|_{L^{2}(S^{2})}^{2}+\frac{\sin\theta_{\mu}}{\varepsilon}|W(\theta_{\mu})|^{2}\right).

Let us estimate the last term. By Lemma 3.15 we have W(θμ)+δF(θμ)=0W(\theta_{\mu})+\delta F(\theta_{\mu})=0. Fix σ(0,1)\sigma\in(0,1). Since σε\sigma\varepsilon satisfies (3.41), there exists θμ(θμσε,θμ+σε)\theta^{\prime}_{\mu}\in(\theta_{\mu}-\sigma\varepsilon,\theta_{\mu}+\sigma\varepsilon) such that

|F(θμ)|2\displaystyle|F(\theta^{\prime}_{\mu})|^{2} 1σεsinθμθμσεθμ+σε|F(θ)|2sinθdθCσεsinθμfL2(S2)2,\displaystyle\leq\frac{1}{\sigma\varepsilon\sin\theta_{\mu}}\int_{\theta_{\mu}-\sigma\varepsilon}^{\theta_{\mu}+\sigma\varepsilon}|F(\theta)|^{2}\sin\theta\,d\theta\leq\frac{C}{\sigma\varepsilon\sin\theta_{\mu}}\|f\|_{L^{2}(S^{2})}^{2},
|F(θμ)F(θμ)|2\displaystyle|F(\theta_{\mu})-F(\theta^{\prime}_{\mu})|^{2} Cσεsinθμθμσεθμ+σε|F(θ)|2sinθdθCσεsinθμfL2(Sμ,σε2)2\displaystyle\leq\frac{C\sigma\varepsilon}{\sin\theta_{\mu}}\int_{\theta_{\mu}-\sigma\varepsilon}^{\theta_{\mu}+\sigma\varepsilon}|F^{\prime}(\theta)|^{2}\sin\theta\,d\theta\leq\frac{C\sigma\varepsilon}{\sin\theta_{\mu}}\|\nabla f\|_{L^{2}(S_{\mu,\sigma\varepsilon}^{2})}^{2}

as in (3.57) and (3.58), where Sμ,σε2S_{\mu,\sigma\varepsilon}^{2} is given by (3.37). Thus

|W(θμ)|2=δ2|F(θμ)|2\displaystyle|W(\theta_{\mu})|^{2}=\delta^{2}|F(\theta_{\mu})|^{2} Cσεsinθμ(σ2fL2(S2)2+δ2fL2(Sμ,σε2)2),\displaystyle\leq\frac{C\sigma\varepsilon}{\sin\theta_{\mu}}\left(\sigma^{-2}\|f\|_{L^{2}(S^{2})}^{2}+\delta^{2}\|\nabla f\|_{L^{2}(S_{\mu,\sigma\varepsilon}^{2})}^{2}\right),

where we also used δ/ε=2/κ1/2\delta/\varepsilon=2/\kappa^{1/2}. We apply this inequality to (3.141) and then use (3.54), σ<1\sigma<1, and MsinθuL2(Sμ,σε2)MsinθuL2(S2)\|M_{\sin\theta}u\|_{L^{2}(S_{\mu,\sigma\varepsilon}^{2})}\leq\|M_{\sin\theta}u\|_{L^{2}(S^{2})} to get

(1Cσ)MsinθuL2(S2)2C(ε2sin2θμ(Δ)1/2uL2(S2)2+(Δ)1/2vL2(S2)2+σ2fL2(S2)2).(1-C\sigma)\|M_{\sin\theta}u\|_{L^{2}(S^{2})}^{2}\\ \leq C\Bigl{(}\varepsilon^{2}\sin^{2}\theta_{\mu}\|(-\Delta)^{1/2}u\|_{L^{2}(S^{2})}^{2}+\|(-\Delta)^{1/2}v\|_{L^{2}(S^{2})}^{2}+\sigma^{-2}\|f\|_{L^{2}(S^{2})}^{2}\Bigr{)}.

In this inequality, we fix σ(0,1)\sigma\in(0,1) so that Cσ1/2C\sigma\leq 1/2, apply (3.33) to (Δ)1/2v(-\Delta)^{1/2}v, and use ε=κ1/2δ/2\varepsilon=\kappa^{1/2}\delta/2 and (3.36) to obtain (3.34). ∎

Now we are ready to verify Assumption 4.6 for AmA_{m} and Λm\Lambda_{m}. Let κ\kappa be the constant given by (3.31). For m{0}m\in\mathbb{Z}\setminus\{0\}, ξ(0,)\xi\in(0,\infty), and μ\mu\in\mathbb{R}, we define

h1,m(ξ,μ)\displaystyle h_{1,m}(\xi,\mu) ={0if|μ|>1+κξ1,|m|1/2ξ1/2if1κξ1<|μ|1+κξ1,ξ1(1|μ|)1/2if|μ|1κξ1,\displaystyle=\begin{cases}0&\text{if}\quad|\mu|>1+\kappa\xi^{-1},\\ |m|^{-1/2}\xi^{-1/2}&\text{if}\quad 1-\kappa\xi^{-1}<|\mu|\leq 1+\kappa\xi^{-1},\\ \xi^{-1}(1-|\mu|)^{-1/2}&\text{if}\quad|\mu|\leq 1-\kappa\xi^{-1},\end{cases}
h2,m(ξ,μ)\displaystyle h_{2,m}(\xi,\mu) ={0if|μ|>1+κξ2,|m|1/2ξ2if1κξ2<|μ|1+κξ2,ξ1(1|μ|)1/2if|μ|1κξ2.\displaystyle=\begin{cases}0&\text{if}\quad|\mu|>1+\kappa\xi^{-2},\\ |m|^{-1/2}\xi^{-2}&\text{if}\quad 1-\kappa\xi^{-2}<|\mu|\leq 1+\kappa\xi^{-2},\\ \xi^{-1}(1-|\mu|)^{1/2}&\text{if}\quad|\mu|\leq 1-\kappa\xi^{-2}.\end{cases}

Note that hj,m(ξ,μ)0h_{j,m}(\xi,\mu)\geq 0 and limξsupμhj(ξ,μ)=0\lim_{\xi\to\infty}\sup_{\mu\in\mathbb{R}}h_{j}(\xi,\mu)=0 for j=1,2j=1,2.

Lemma 3.18.

Let m{0}m\in\mathbb{Z}\setminus\{0\}. Then AmA_{m} and Λm\Lambda_{m} satisfy Assumption 4.6 with the functions h1,mh_{1,m} and h2,mh_{2,m} given above and the operator B3,m=MsinθB_{3,m}=M_{\sin\theta} on m=L2(S2)\mathcal{B}_{m}=L^{2}(S^{2}).

Proof.

Since B3,mB_{3,m} is a bounded operator on m\mathcal{B}_{m}, it is closed in m\mathcal{B}_{m} and

D𝒳m((Am)1/2)𝒳mL2(S2)=m=Dm(B3,m).\displaystyle D_{\mathcal{X}_{m}}((-A_{m})^{1/2})\subset\mathcal{X}_{m}\subset L^{2}(S^{2})=\mathcal{B}_{m}=D_{\mathcal{B}_{m}}(B_{3,m}).

Also, (4.9) holds by (3.16). Let us verify (4.8) and (4.10). If ξ(0,)\xi\in(0,\infty) and |μ|>1+κξ1|\mu|>1+\kappa\xi^{-1}, then (4.8) is valid by (3.19) and (|μ|1)1<κ1ξ(|\mu|-1)^{-1}<\kappa^{-1}\xi. Also, if ξ[1,)\xi\in[1,\infty) and |μ|1+κξ1|\mu|\leq 1+\kappa\xi^{-1}, then (4.8) follows from (3.15), (3.26), and (3.33) with δ=ξ1/2\delta=\xi^{-1/2}. Let ξ(0,1)\xi\in(0,1) and |μ|1+κξ1|\mu|\leq 1+\kappa\xi^{-1}. For u𝒴mH1(S2)u\in\mathcal{Y}_{m}\cap H^{1}(S^{2}) of the form (3.9), we have

uL2(S2)21|m|2(Δ)1/2uL2(S2)2C|m|2(Am)1/2uL2(S2)2\displaystyle\|u\|_{L^{2}(S^{2})}^{2}\leq\frac{1}{|m|^{2}}\|(-\Delta)^{1/2}u\|_{L^{2}(S^{2})}^{2}\leq\frac{C}{|m|^{2}}\|(-A_{m})^{1/2}u\|_{L^{2}(S^{2})}^{2}

by (2.3), λn=n(n+1)|m|2\lambda_{n}=n(n+1)\geq|m|^{2} for n|m|n\geq|m|, and (3.15). Moreover,

|m|2{|m|1ξ1if1κξ1<|μ|1+κξ1,ξ2(1|μ|)1if|μ|1κξ1\displaystyle|m|^{-2}\leq\begin{cases}|m|^{-1}\xi^{-1}&\text{if}\quad 1-\kappa\xi^{-1}<|\mu|\leq 1+\kappa\xi^{-1},\\ \xi^{-2}(1-|\mu|)^{-1}&\text{if}\quad|\mu|\leq 1-\kappa\xi^{-1}\end{cases}

since |m|1|m|\geq 1, ξ(0,1)\xi\in(0,1), and (1|μ|)11(1-|\mu|)^{-1}\geq 1. Hence (4.8) is valid.

The inequality (4.10) follows from (3.20) and (|μ|1)1<κ1ξ2(|\mu|-1)^{-1}<\kappa^{-1}\xi^{-2} when ξ(0,)\xi\in(0,\infty) and |μ|>1+κξ2|\mu|>1+\kappa\xi^{-2} and from (3.15), (3.27), and (3.34) with δ=ξ1\delta=\xi^{-1} when ξ[1,)\xi\in[1,\infty) and |μ|1+κξ2|\mu|\leq 1+\kappa\xi^{-2}. Also, when ξ(0,1)\xi\in(0,1) and |μ|1+κξ2|\mu|\leq 1+\kappa\xi^{-2}, we have (4.10) by

B3,muL2(S2)2=MsinθuL2(S2)2uL2(S2)2C|m|2(Am)1/2uL2(S2)2\displaystyle\|B_{3,m}u\|_{L^{2}(S^{2})}^{2}=\|M_{\sin\theta}u\|_{L^{2}(S^{2})}^{2}\leq\|u\|_{L^{2}(S^{2})}^{2}\leq\frac{C}{|m|^{2}}\|(-A_{m})^{1/2}u\|_{L^{2}(S^{2})}^{2}

for u𝒴mH1(S2)u\in\mathcal{Y}_{m}\cap H^{1}(S^{2}) and

|m|2{|m|1ξ4if1κξ2<|μ|1+κξ2,κ1ξ2(1|μ|)if|μ|1κξ2\displaystyle|m|^{-2}\leq\begin{cases}|m|^{-1}\xi^{-4}&\text{if}\quad 1-\kappa\xi^{-2}<|\mu|\leq 1+\kappa\xi^{-2},\\ \kappa^{-1}\xi^{-2}(1-|\mu|)&\text{if}\quad|\mu|\leq 1-\kappa\xi^{-2}\end{cases}

due to |m|1|m|\geq 1, ξ(0,1)\xi\in(0,1), and κ1ξ2(1|μ|)1\kappa^{-1}\xi^{2}(1-|\mu|)\geq 1 when |μ|1κξ2|\mu|\leq 1-\kappa\xi^{-2}. ∎

3.3. Estimates for the semigroup

By Lemmas 3.3 and 3.18, AmA_{m} and Λm\Lambda_{m} satisfy Assumptions 4.14.3 and 4.6. Moreover, B2,m𝒳m𝒳m1\|B_{2,m}\|_{\mathcal{X}_{m}\to\mathcal{X}_{m}}\leq 1 by (3.13) and the constants in (4.2), (4.3), and (4.8)–(4.10) can be taken independently of mm. Hence we can apply the abstract results in Section 4 to Lα,m=AmiαmΛmL_{\alpha,m}=A_{m}-i\alpha m\Lambda_{m} to get the following results.

Theorem 3.19.

Let m{0}m\in\mathbb{Z}\setminus\{0\}. Then

(3.142) {ζReζ0}ρ𝒳m(Lα,m)ρ𝒴m(mLα,m)\displaystyle\{\zeta\in\mathbb{C}\mid\mathrm{Re}\,\zeta\geq 0\}\subset\rho_{\mathcal{X}_{m}}(L_{\alpha,m})\subset\rho_{\mathcal{Y}_{m}}(\mathbb{Q}_{m}L_{\alpha,m})

for all α\alpha\in\mathbb{R}. Moreover, for all α{0}\alpha\in\mathbb{R}\setminus\{0\} and λ\lambda\in\mathbb{R} we have

(3.143) (iλ+mLm,α)1𝒴m𝒴mCGm(α,λαm).\displaystyle\|(i\lambda+\mathbb{Q}_{m}L_{m,\alpha})^{-1}\|_{\mathcal{Y}_{m}\to\mathcal{Y}_{m}}\leq CG_{m}\left(\alpha,\frac{\lambda}{\alpha m}\right).

Here C>0C>0 is a constant independent of mm, α\alpha, and λ\lambda. Also,

(3.144) Gm(α,μ)={|αm|1(|μ|1)1if|μ|>1+|α|1/2,|α|1/2|m|1if1|α|1/2<|μ|1+|α|1/2,|αm|2/3(1|μ|)1/3if|μ|1|α|1/2\displaystyle G_{m}(\alpha,\mu)=\begin{cases}|\alpha m|^{-1}(|\mu|-1)^{-1}&\text{if}\quad|\mu|>1+|\alpha|^{-1/2},\\ |\alpha|^{-1/2}|m|^{-1}&\text{if}\quad 1-|\alpha|^{-1/2}<|\mu|\leq 1+|\alpha|^{-1/2},\\ |\alpha m|^{-2/3}(1-|\mu|)^{-1/3}&\text{if}\quad|\mu|\leq 1-|\alpha|^{-1/2}\end{cases}

for α{0}\alpha\in\mathbb{R}\setminus\{0\} and μ\mu\in\mathbb{R}.

Proof.

By Lemma 4.4 and Theorem 4.7 with α\alpha replaced by αm\alpha m we have (3.142) for all α\alpha\in\mathbb{R} and

(3.145) (iλ+mLα,m)1𝒴m𝒴mCFm(αm,λαm)\displaystyle\|(i\lambda+\mathbb{Q}_{m}L_{\alpha,m})^{-1}\|_{\mathcal{Y}_{m}\to\mathcal{Y}_{m}}\leq CF_{m}\left(\alpha m,\frac{\lambda}{\alpha m}\right)

for all α{0}\alpha\in\mathbb{R}\setminus\{0\} and λ\lambda\in\mathbb{R}. Here C>0C>0 is a constant independent of mm, α\alpha, and λ\lambda. Also,

(3.146) Fm(αm,μ)=infξ1,ξ2>0(ξ1|αm|+ξ12ξ22(αm)2+ξ12h2(ξ2,μ)|αm|+h1(ξ1,μ)2)\displaystyle F_{m}(\alpha m,\mu)=\inf_{\xi_{1},\xi_{2}>0}\left(\frac{\xi_{1}}{|\alpha m|}+\frac{\xi_{1}^{2}\xi_{2}^{2}}{(\alpha m)^{2}}+\frac{\xi_{1}^{2}h_{2}(\xi_{2},\mu)}{|\alpha m|}+h_{1}(\xi_{1},\mu)^{2}\right)

for μ=λ/αm\mu=\lambda/\alpha m\in\mathbb{R}. Let us estimate Fm(αm,μ)F_{m}(\alpha m,\mu). In what follows, we write CC for general positive constant depending only on the constant κ\kappa given by (3.31).

When |μ|>1+|α|1/2|\mu|>1+|\alpha|^{-1/2}, let ξ1=ξ22=2κ(|μ|1)1\xi_{1}=\xi_{2}^{2}=2\kappa(|\mu|-1)^{-1}. Then 1+κξjj=(|μ|+1)/2<|μ|1+\kappa\xi_{j}^{-j}=(|\mu|+1)/2<|\mu| and thus hj(ξj,μ)=0h_{j}(\xi_{j},\mu)=0 for j=1,2j=1,2. Also,

ξ1|αm|=2κ|αm|(|μ|1),ξ12ξ22(αm)2=8κ3(αm)2(|μ|1)38κ3|m|1|αm|(|μ|1).\displaystyle\frac{\xi_{1}}{|\alpha m|}=\frac{2\kappa}{|\alpha m|(|\mu|-1)},\quad\frac{\xi_{1}^{2}\xi_{2}^{2}}{(\alpha m)^{2}}=\frac{8\kappa^{3}}{(\alpha m)^{2}(|\mu|-1)^{3}}\leq\frac{8\kappa^{3}}{|m|}\cdot\frac{1}{|\alpha m|(|\mu|-1)}.

By these relations and |m|1|m|\geq 1, we find that

(3.147) Fm(αm,μ)C|αm|(|μ|1).\displaystyle F_{m}(\alpha m,\mu)\leq\frac{C}{|\alpha m|(|\mu|-1)}.

When 1|α|1/2<|μ|1+|α|1/21-|\alpha|^{-1/2}<|\mu|\leq 1+|\alpha|^{-1/2}, we set ξ1=ξ22=κ|α|1/2/2\xi_{1}=\xi_{2}^{2}=\kappa|\alpha|^{1/2}/2. Then

κξjj=2|α|1/2>|α|1/2| 1|μ||,i.e.1κξjj<|μ|<1+κξjj\displaystyle\kappa\xi_{j}^{-j}=2|\alpha|^{-1/2}>|\alpha|^{-1/2}\geq\bigl{|}\,1-|\mu|\,\bigr{|},\quad\text{i.e.}\quad 1-\kappa\xi_{j}^{-j}<|\mu|<1+\kappa\xi_{j}^{-j}

for j=1,2j=1,2. Hence h1(ξ1,μ)=|m|1/2ξ11/2h_{1}(\xi_{1},\mu)=|m|^{-1/2}\xi_{1}^{-1/2}, h2(ξ2,μ)=|m|1/2ξ22h_{2}(\xi_{2},\mu)=|m|^{-1/2}\xi_{2}^{-2}, and

ξ1|αm|=κ2|α|1/2|m|,ξ12ξ22(αm)2=κ38|α|1/2m2,\displaystyle\frac{\xi_{1}}{|\alpha m|}=\frac{\kappa}{2|\alpha|^{1/2}|m|},\quad\frac{\xi_{1}^{2}\xi_{2}^{2}}{(\alpha m)^{2}}=\frac{\kappa^{3}}{8|\alpha|^{1/2}m^{2}},
ξ12h2(ξ2,μ)|αm|=κ2|α|1/2|m|3/2,h1(ξ1,μ)2=2κ|α|1/2|m|.\displaystyle\frac{\xi_{1}^{2}h_{2}(\xi_{2},\mu)}{|\alpha m|}=\frac{\kappa}{2|\alpha|^{1/2}|m|^{3/2}},\quad h_{1}(\xi_{1},\mu)^{2}=\frac{2}{\kappa|\alpha|^{1/2}|m|}.

By these equalities and |m|1|m|\geq 1, we find that

(3.148) Fm(αm,μ)C|α|1/2|m|.\displaystyle F_{m}(\alpha m,\mu)\leq\frac{C}{|\alpha|^{1/2}|m|}.

When |μ|1|α|1/2|\mu|\leq 1-|\alpha|^{-1/2}, we take

ξ1=|αm|1/3(1|μ|)1/3,ξ2=|αm|1/3(1|μ|)1/6.\displaystyle\xi_{1}=|\alpha m|^{1/3}(1-|\mu|)^{-1/3},\quad\xi_{2}=|\alpha m|^{1/3}(1-|\mu|)^{1/6}.

Then for j=1,2j=1,2 we see by (1|μ|)1/3|α|1/6(1-|\mu|)^{-1/3}\leq|\alpha|^{1/6}, κ(0,1)\kappa\in(0,1), and |m|1|m|\geq 1 that

κξjj(1|μ|)=κ|αm|j/3(1|μ|)2j/3κ|m|j/3<1,i.e.|μ|<1κξjj.\displaystyle\frac{\kappa}{\xi_{j}^{j}(1-|\mu|)}=\frac{\kappa}{|\alpha m|^{j/3}(1-|\mu|)^{2j/3}}\leq\frac{\kappa}{|m|^{j/3}}<1,\quad\text{i.e.}\quad|\mu|<1-\kappa\xi_{j}^{-j}.

Hence h1(ξ1,μ)=ξ11(1|μ|)1/2h_{1}(\xi_{1},\mu)=\xi_{1}^{-1}(1-|\mu|)^{-1/2}, h2(ξ2,μ)=ξ21(1|μ|)1/2h_{2}(\xi_{2},\mu)=\xi_{2}^{-1}(1-|\mu|)^{1/2}, and

ξ1|αm|=ξ12ξ22(αm)2=ξ12h2(ξ2,μ)|αm|=h1(ξ1,μ)2=1|αm|2/3(1|μ|)1/3.\displaystyle\frac{\xi_{1}}{|\alpha m|}=\frac{\xi_{1}^{2}\xi_{2}^{2}}{(\alpha m)^{2}}=\frac{\xi_{1}^{2}h_{2}(\xi_{2},\mu)}{|\alpha m|}=h_{1}(\xi_{1},\mu)^{2}=\frac{1}{|\alpha m|^{2/3}(1-|\mu|)^{1/3}}.

Therefore,

(3.149) Fm(αm,μ)4|αm|2/3(1|μ|)1/3.\displaystyle F_{m}(\alpha m,\mu)\leq\frac{4}{|\alpha m|^{2/3}(1-|\mu|)^{1/3}}.

Hence we get (3.143)–(3.144) by (3.145) and (3.147)–(3.149). ∎

Theorem 3.20.

Let m{0}m\in\mathbb{Z}\setminus\{0\}. Then Lα,mL_{\alpha,m} generates an analytic semigroup {etLα,m}t0\{e^{tL_{\alpha,m}}\}_{t\geq 0} in 𝒳m\mathcal{X}_{m} for all α\alpha\in\mathbb{R}. Moreover,

(3.150) metLα,mfL2(S2)C1eC2|α|1/2|m|2/3tmfL2(S2),t0,f𝒳m\displaystyle\|\mathbb{Q}_{m}e^{tL_{\alpha,m}}f\|_{L^{2}(S^{2})}\leq C_{1}e^{-C_{2}|\alpha|^{1/2}|m|^{2/3}t}\|\mathbb{Q}_{m}f\|_{L^{2}(S^{2})},\quad t\geq 0,\,f\in\mathcal{X}_{m}

for all α{0}\alpha\in\mathbb{R}\setminus\{0\}, where C1C_{1} and C2C_{2} are positive constants independent of mm, α\alpha, and tt.

Proof.

Theorem 4.5 implies that Lα,mL_{\alpha,m} generates an analytic semigroup {etLα,m}t0\{e^{tL_{\alpha,m}}\}_{t\geq 0} in 𝒳m\mathcal{X}_{m} for all α\alpha\in\mathbb{R}. Moreover, for all α{0}\alpha\in\mathbb{R}\setminus\{0\} we have

(3.151) metLα,mfL2(S2)C1eC2t/Fm(αm)mfL2(S2),t0,f𝒳m\displaystyle\|\mathbb{Q}_{m}e^{tL_{\alpha,m}}f\|_{L^{2}(S^{2})}\leq C_{1}e^{-C_{2}t/F_{m}(\alpha m)}\|\mathbb{Q}_{m}f\|_{L^{2}(S^{2})},\quad t\geq 0,\,f\in\mathcal{X}_{m}

by (4.16) with α\alpha replaced by αm\alpha m. Here C1,C2>0C_{1},C_{2}>0 are constants independent of mm, α\alpha, and tt. Also, Fm(αm)=supλFm(αm,μ)F_{m}(\alpha m)=\sup_{\lambda\in\mathbb{R}}F_{m}\left(\alpha m,\mu\right) with Fm(αm,μ)F_{m}(\alpha m,\mu) given by (3.146).

Let us estimate Fm(αm)F_{m}(\alpha m). In what follows, CC denotes a general positive constant independent of mm, α\alpha, and μ\mu. If |μ|>1+|α|1/2|\mu|>1+|\alpha|^{-1/2}, then Fm(αm,μ)C|α|1/2|m|1F_{m}(\alpha m,\mu)\leq C|\alpha|^{-1/2}|m|^{-1} by (3.147) and |μ|1>|α|1/2|\mu|-1>|\alpha|^{-1/2}. The same inequality holds when 1|α|1/2<|μ|1+|α|1/21-|\alpha|^{-1/2}<|\mu|\leq 1+|\alpha|^{-1/2} by (3.148). When |μ|1|α|1/2|\mu|\leq 1-|\alpha|^{-1/2}, we have Fm(αm,μ)4|α|1/2|m|2/3F_{m}(\alpha m,\mu)\leq 4|\alpha|^{-1/2}|m|^{-2/3} by (3.149) and 1|μ||α|1/21-|\mu|\geq|\alpha|^{-1/2}. By these facts and |m|1|m|\geq 1, we get Fm(αm)C|α|1/2|m|2/3F_{m}(\alpha m)\leq C|\alpha|^{-1/2}|m|^{-2/3}. Hence we apply this inequality to (3.151) to obtain (3.150). ∎

When |m|3|m|\geq 3, m=I\mathbb{Q}_{m}=I on 𝒳m\mathcal{X}_{m} by (3.10) and thus (3.150) gives a decay estimate for etLα,me^{tL_{\alpha,m}}. On the other hand, when |m|=1,2|m|=1,2, mf=(Im)f=(f,Y2m)L2(S2)Y2m\mathbb{P}_{m}f=(I-\mathbb{Q}_{m})f=(f,Y_{2}^{m})_{L^{2}(S^{2})}Y_{2}^{m} does not vanish for f𝒳mf\in\mathcal{X}_{m}, so we need to estimate the m\mathbb{P}_{m}-part of etLα,me^{tL_{\alpha,m}}.

Theorem 3.21.

Let |m|=1,2|m|=1,2. Then for all t0t\geq 0, f𝒳mf\in\mathcal{X}_{m}, and α\alpha\in\mathbb{R} such that |α|>4|\alpha|>4 is sufficiently large, we have

(3.152) metLα,mfL2(S2)e4tmfL2(S2)+C1|αm|1/3log(C2|αm|2/3)e2tmfL2(S2),\|\mathbb{P}_{m}e^{tL_{\alpha,m}}f\|_{L^{2}(S^{2})}\leq e^{-4t}\|\mathbb{P}_{m}f\|_{L^{2}(S^{2})}\\ +C_{1}|\alpha m|^{1/3}\log\Bigl{(}C_{2}|\alpha m|^{2/3}\Bigr{)}e^{-2t}\|\mathbb{Q}_{m}f\|_{L^{2}(S^{2})},

where C1,C2>0C_{1},C_{2}>0 are constants independent of tt, ff, α\alpha, and mm.

Proof.

Let 𝒩m=N𝒳m(Λm)={cY2mc}\mathcal{N}_{m}=N_{\mathcal{X}_{m}}(\Lambda_{m})=\{cY_{2}^{m}\mid c\in\mathbb{C}\} (see (3.12)). Then

(3.153) (ζmAm)1mf=(ζ+4)1mf,ζρ𝒩m(mAm)={4}\displaystyle(\zeta-\mathbb{P}_{m}A_{m})^{-1}\mathbb{P}_{m}f=(\zeta+4)^{-1}\mathbb{P}_{m}f,\quad\zeta\in\rho_{\mathcal{N}_{m}}(\mathbb{P}_{m}A_{m})=\mathbb{C}\setminus\{-4\}

for f𝒳mf\in\mathcal{X}_{m} since mAmY2m=4Y2m\mathbb{P}_{m}A_{m}Y_{2}^{m}=-4Y_{2}^{m} by (3.3) and λ2=6\lambda_{2}=6. Also, by (4.5),

(3.154) ρ𝒳m(Lα,m)=ρ𝒴m(mLα,m)ρ𝒩m(mAm)=ρ𝒴m(mLα,m)({4}).\displaystyle\rho_{\mathcal{X}_{m}}(L_{\alpha,m})=\rho_{\mathcal{Y}_{m}}(\mathbb{Q}_{m}L_{\alpha,m})\cap\rho_{\mathcal{N}_{m}}(\mathbb{P}_{m}A_{m})=\rho_{\mathcal{Y}_{m}}(\mathbb{Q}_{m}L_{\alpha,m})\cup(\mathbb{C}\setminus\{-4\}).

Since AmA_{m} is self-adjoint and satisfies (3.5) in 𝒳m\mathcal{X}_{m} and Λm\Lambda_{m} is AmA_{m}-compact in 𝒳m\mathcal{X}_{m}, we see by a perturbation theory of semigroups (see [13]) that etLα,me^{tL_{\alpha,m}} is expressed by the Dunford integral

etLα,mf=12πiΓetζ(ζLα,m)1f𝑑ζ,t>0,f𝒳m,\displaystyle e^{tL_{\alpha,m}}f=\frac{1}{2\pi i}\int_{\Gamma}e^{t\zeta}(\zeta-L_{\alpha,m})^{-1}f\,d\zeta,\quad t>0,\,f\in\mathcal{X}_{m},

where Γ\Gamma is any piecewise smooth curve in ρ𝒳m(Lα,m)\rho_{\mathcal{X}_{m}}(L_{\alpha,m}) going from eiδ\infty e^{-i\delta} to eiδ\infty e^{i\delta} with any δ(π/2,π)\delta\in(\pi/2,\pi) and located on the right-hand side of σ𝒳m(Lα,m)=σ𝒴m(mLα,m){4}\sigma_{\mathcal{X}_{m}}(L_{\alpha,m})=\sigma_{\mathcal{Y}_{m}}(\mathbb{Q}_{m}L_{\alpha,m})\cup\{-4\}. We apply m\mathbb{P}_{m} to the above expression to get

(3.155) metLα,mf=12πiΓetζ(ζmAm)1mf𝑑ζαm2πJ\displaystyle\mathbb{P}_{m}e^{tL_{\alpha,m}}f=\frac{1}{2\pi i}\int_{\Gamma}e^{t\zeta}(\zeta-\mathbb{P}_{m}A_{m})^{-1}\mathbb{P}_{m}f\,d\zeta-\frac{\alpha m}{2\pi}J

by (4.6) with α\alpha replaced by αm\alpha m, where

J=Γetζ(ζmAm)1mΛm(ζmLα,m)1mf𝑑ζ.\displaystyle J=\int_{\Gamma}e^{t\zeta}(\zeta-\mathbb{P}_{m}A_{m})^{-1}\mathbb{P}_{m}\Lambda_{m}(\zeta-\mathbb{Q}_{m}L_{\alpha,m})^{-1}\mathbb{Q}_{m}f\,d\zeta.

Noting that ζ=4\zeta=-4 is located on the left-hand side of Γ\Gamma, we use (3.153) and the residue theorem (after replacing Γ\Gamma by a small circle centered at ζ=4\zeta=-4) to deduce that

(3.156) 12πiΓetζ(ζmAm)1mf𝑑ζ=(12πiΓetζζ+4𝑑ζ)mf=e4tmf.\displaystyle\frac{1}{2\pi i}\int_{\Gamma}e^{t\zeta}(\zeta-\mathbb{P}_{m}A_{m})^{-1}\mathbb{P}_{m}f\,d\zeta=\left(\frac{1}{2\pi i}\int_{\Gamma}\frac{e^{t\zeta}}{\zeta+4}\,d\zeta\right)\mathbb{P}_{m}f=e^{-4t}\mathbb{P}_{m}f.

Next we estimate JJ. For α{0}\alpha\in\mathbb{R}\setminus\{0\} and λ\lambda\in\mathbb{R} let

Hm(α,λ)={|m|1/3(|λ||αm|)1if|λ/αm|>1+|α|1/2,|α|1/2|m|2/3if1|α|1/2<|λ/αm|1+|α|1/2,|αm|1/3(|αm||λ|)1/3if|λ/αm|1|α|1/2.\displaystyle H_{m}(\alpha,\lambda)=\begin{cases}|m|^{1/3}(|\lambda|-|\alpha m|)^{-1}&\text{if}\quad|\lambda/\alpha m|>1+|\alpha|^{-1/2},\\ |\alpha|^{-1/2}|m|^{-2/3}&\text{if}\quad 1-|\alpha|^{-1/2}<|\lambda/\alpha m|\leq 1+|\alpha|^{-1/2},\\ |\alpha m|^{-1/3}(|\alpha m|-|\lambda|)^{-1/3}&\text{if}\quad|\lambda/\alpha m|\leq 1-|\alpha|^{-1/2}.\end{cases}

Then since Gm(α,λ/αm)Hm(α,λ)G_{m}(\alpha,\lambda/\alpha m)\leq H_{m}(\alpha,\lambda), where GmG_{m} is the function given by (3.144), we see by (3.143) with λ\lambda replaced by λ-\lambda that

(3.157) (iλmLα,m)1𝒴m𝒴mC1Hm(α,λ)=C1Hm(α,λ)\displaystyle\|(i\lambda-\mathbb{Q}_{m}L_{\alpha,m})^{-1}\|_{\mathcal{Y}_{m}\to\mathcal{Y}_{m}}\leq C_{1}H_{m}(\alpha,-\lambda)=C_{1}H_{m}(\alpha,\lambda)

with a constant C1>0C_{1}>0 independent of mm, α\alpha, and λ\lambda. Also,

(3.160) Ω={ζ|Reζ|1/2C1Hm(α,Imζ)}ρ𝒴m(mLα,m),(ζmLα,m)1𝒴m𝒴m2C1Hm(α,Imζ),ζΩ\displaystyle\begin{gathered}\Omega=\{\zeta\in\mathbb{C}\mid|\mathrm{Re}\,\zeta|\leq 1/2C_{1}H_{m}(\alpha,\mathrm{Im}\,\zeta)\}\subset\rho_{\mathcal{Y}_{m}}(\mathbb{Q}_{m}L_{\alpha,m}),\\ \|(\zeta-\mathbb{Q}_{m}L_{\alpha,m})^{-1}\|_{\mathcal{Y}_{m}\to\mathcal{Y}_{m}}\leq 2C_{1}H_{m}(\alpha,\mathrm{Im}\,\zeta),\quad\zeta\in\Omega\end{gathered}

by (3.157) and a standard Neumann series argument. Let

S1=14C1|αm|2/321/3,S2=|α|1/2|m|2/34C1.\displaystyle S_{1}=\frac{1}{4C_{1}}\frac{|\alpha m|^{2/3}}{2^{1/3}},\quad S_{2}=\frac{|\alpha|^{1/2}|m|^{2/3}}{4C_{1}}.

We take |α|>4|\alpha|>4 sufficiently large so that S1>S2>2S_{1}>S_{2}>2 and set

ζ1,±(s)\displaystyle\zeta_{1,\pm}(s) =s±i|αm|2s2S12,sI1=[2,S1],\displaystyle=-s\pm i\frac{|\alpha m|}{2}\frac{s-2}{S_{1}-2},\quad s\in I_{1}=[2,S_{1}],
ζ2,±(s)\displaystyle\zeta_{2,\pm}(s) =s±i{|αm|1|αm|(4C1s)3},sI2=[S2,S1],\displaystyle=-s\pm i\left\{|\alpha m|-\frac{1}{|\alpha m|}(4C_{1}s)^{3}\right\},\quad s\in I_{2}=[S_{2},S_{1}],
ζ3,±(s)\displaystyle\zeta_{3,\pm}(s) =S2±is,sI3=[|αm||α|1/2|m|,|αm|+|α|1/2|m|],\displaystyle=-S_{2}\pm is,\quad s\in I_{3}=\left[|\alpha m|-|\alpha|^{1/2}|m|,|\alpha m|+|\alpha|^{1/2}|m|\right],
ζ4,±(s)\displaystyle\zeta_{4,\pm}(s) =s±i(|αm|+4C1|m|1/3s),sI4=[S2,).\displaystyle=-s\pm i\left(|\alpha m|+4C_{1}|m|^{1/3}s\right),\quad s\in I_{4}=[S_{2},\infty).

Then we define a piecewise smooth curve

Γ=k=14(Γk,+Γk,){4},Γk,±={ζk,±(s)sIk}.\displaystyle\Gamma=\bigcup_{k=1}^{4}(\Gamma_{k,+}\cup\Gamma_{k,-})\subset\mathbb{C}\setminus\{-4\},\quad\Gamma_{k,\pm}=\{\zeta_{k,\pm}(s)\mid s\in I_{k}\}.

For sI1s\in I_{1} we have |Imζ1,±(s)/αm|1/21|α|1/2|\mathrm{Im}\,\zeta_{1,\pm}(s)/\alpha m|\leq 1/2\leq 1-|\alpha|^{-1/2} by |α|>4|\alpha|>4. Hence

(3.161) Hm(α,Imζ1,±(s))=1|αm|1/3(|αm||Imζ1,±(s)|)1/321/3|αm|2/3\displaystyle H_{m}(\alpha,\mathrm{Im}\,\zeta_{1,\pm}(s))=\frac{1}{|\alpha m|^{1/3}(|\alpha m|-|\mathrm{Im}\,\zeta_{1,\pm}(s)|)^{1/3}}\leq\frac{2^{1/3}}{|\alpha m|^{2/3}}

and

12C1Hm(α,Imζ1,±(s))12C1|αm|2/321/3=2S1>S1|Reζ1,±(s)|,\displaystyle\frac{1}{2C_{1}H_{m}(\alpha,\mathrm{Im}\,\zeta_{1,\pm}(s))}\geq\frac{1}{2C_{1}}\frac{|\alpha m|^{2/3}}{2^{1/3}}=2S_{1}>S_{1}\geq|\mathrm{Re}\,\zeta_{1,\pm}(s)|,

which shows Γ1,±Ω\Gamma_{1,\pm}\subset\Omega. Similarly, for k=2,3,4k=2,3,4 and sIks\in I_{k} we have

(3.162) 2C1Hm(α,Imζk,±(s))={(2s)1,k=2,4,2C1|α|1/2|m|2/3=(2S2)1,k=3\displaystyle 2C_{1}H_{m}(\alpha,\mathrm{Im}\,\zeta_{k,\pm}(s))=\begin{cases}(2s)^{-1},\quad k=2,4,\\ 2C_{1}|\alpha|^{-1/2}|m|^{-2/3}=(2S_{2})^{-1},\quad k=3\end{cases}

and thus Γk,±Ω\Gamma_{k,\pm}\subset\Omega. Hence it follows from (3.154) and (3.160) that

ΓΩ({4})ρ𝒴m(mLα,m)ρ𝒩m(mAm)=ρ𝒳m(Lα,m)\displaystyle\Gamma\subset\Omega\cap(\mathbb{C}\setminus\{-4\})\subset\rho_{\mathcal{Y}_{m}}(\mathbb{Q}_{m}L_{\alpha,m})\cap\rho_{\mathcal{N}_{m}}(\mathbb{P}_{m}A_{m})=\rho_{\mathcal{X}_{m}}(L_{\alpha,m})

and we can take this Γ\Gamma as the path of the integral for JJ to get J=k=14JkJ=\sum_{k=1}^{4}J_{k}, where

Jk=(Γk,++Γk,)etζ(ζmAm)1mΛm(ζmLα,m)1mfdζ.\displaystyle J_{k}=\left(\int_{\Gamma_{k,+}}+\int_{\Gamma_{k,-}}\right)e^{t\zeta}(\zeta-\mathbb{P}_{m}A_{m})^{-1}\mathbb{P}_{m}\Lambda_{m}(\zeta-\mathbb{Q}_{m}L_{\alpha,m})^{-1}\mathbb{Q}_{m}f\,d\zeta.

Let us estimate each JkJ_{k}. In what follows, we write CC and CC^{\prime} for general positive constant depending only on C1C_{1}. For sI1s\in I_{1} we have

|ζ1,±(s)+4|=(4s)2+b2(s2)2=(1+b2){(sc1)2+c22}\displaystyle|\zeta_{1,\pm}(s)+4|=\sqrt{(4-s)^{2}+b^{2}(s-2)^{2}}=\sqrt{(1+b^{2})\{(s-c_{1})^{2}+c_{2}^{2}\}}

and |etζ1,±(s)|=etse2t|e^{t\zeta_{1,\pm}(s)}|=e^{-ts}\leq e^{-2t} and |ddsζ1,±(s)|=1+b2|\frac{d}{ds}\zeta_{1,\pm}(s)|=\sqrt{1+b^{2}}, where

b=|αm|21S12,c1=2(1+11+b2),c2=2b1+b2.\displaystyle b=\frac{|\alpha m|}{2}\frac{1}{S_{1}-2},\quad c_{1}=2\left(1+\frac{1}{1+b^{2}}\right),\quad c_{2}=\frac{2b}{1+b^{2}}.

By these relations, (3.8), (3.153), (3.160), (3.161), we find that

J1L2(S2)\displaystyle\|J_{1}\|_{L^{2}(S^{2})} Ce2t|αm|2/3(2S11(sc1)2+c22𝑑s)mfL2(S2)\displaystyle\leq\frac{Ce^{-2t}}{|\alpha m|^{2/3}}\left(\int_{2}^{S_{1}}\frac{1}{\sqrt{(s-c_{1})^{2}+c_{2}^{2}}}\,ds\right)\|\mathbb{Q}_{m}f\|_{L^{2}(S^{2})}
=Ce2t|αm|2/3[log(sc1+(sc1)2+c22)]s=2S1mfL2(S2).\displaystyle=\frac{Ce^{-2t}}{|\alpha m|^{2/3}}\left[\log\left(s-c_{1}+\sqrt{(s-c_{1})^{2}+c_{2}^{2}}\right)\right]_{s=2}^{S_{1}}\|\mathbb{Q}_{m}f\|_{L^{2}(S^{2})}.

Moreover, since S1=C|αm|2/3S_{1}=C|\alpha m|^{2/3}, we have C|αm|1/3bC|αm|1/3C|\alpha m|^{1/3}\leq b\leq C^{\prime}|\alpha m|^{1/3} and thus b1b\geq 1 when |α|>4|\alpha|>4 is large. Then since c12=2(1+b2)1c_{1}-2=2(1+b^{2})^{-1} and b1c22b1b^{-1}\leq c_{2}\leq 2b^{-1},

1S1c1S1,C|αm|1/3c2S1,(c12)2+c221\displaystyle 1\leq S_{1}-c_{1}\leq S_{1},\quad\frac{C}{|\alpha m|^{1/3}}\leq c_{2}\leq S_{1},\quad(c_{1}-2)^{2}+c_{2}^{2}\leq 1

for sufficiently large |α|>4|\alpha|>4 and thus

1\displaystyle 1 S1c1+(S1c1)2+c22CS1=C|αm|2/3,\displaystyle\leq S_{1}-c_{1}+\sqrt{(S_{1}-c_{1})^{2}+c_{2}^{2}}\leq CS_{1}=C|\alpha m|^{2/3},
1\displaystyle 1 (c12)2+c22(c12)c222C|αm|2/3\displaystyle\geq\sqrt{(c_{1}-2)^{2}+c_{2}^{2}}-(c_{1}-2)\geq\frac{c_{2}^{2}}{2}\geq\frac{C}{|\alpha m|^{2/3}}

by the mean value theorem for (c12)2+s\sqrt{(c_{1}-2)^{2}+s} with s0s\geq 0. Hence

(3.163) J1L2(S2)Ce2t|αm|2/3log(C|αm|2/3)mfL2(S2).\displaystyle\|J_{1}\|_{L^{2}(S^{2})}\leq\frac{Ce^{-2t}}{|\alpha m|^{2/3}}\log\Bigl{(}C|\alpha m|^{2/3}\Bigr{)}\|\mathbb{Q}_{m}f\|_{L^{2}(S^{2})}.

For sI2s\in I_{2} we have |etζ2,±(s)|=etse2t|e^{t\zeta_{2,\pm}(s)}|=e^{-ts}\leq e^{-2t} and

|ζ2,±(s)+4||Imζ2,±(s)||αm|2,|dζ2,±ds(s)|=1+Cs4|αm|2Cs2|αm|\displaystyle|\zeta_{2,\pm}(s)+4|\geq|\mathrm{Im}\,\zeta_{2,\pm}(s)|\geq\frac{|\alpha m|}{2},\quad\left|\frac{d\zeta_{2,\pm}}{ds}(s)\right|=\sqrt{1+\frac{Cs^{4}}{|\alpha m|^{2}}}\leq\frac{Cs^{2}}{|\alpha m|}

by s2/|αm|S22/|αm|=C|m|1/3Cs^{2}/|\alpha m|\geq S_{2}^{2}/|\alpha m|=C|m|^{1/3}\geq C. It follows from these inequalities, (3.8), (3.153), (3.160), (3.162), and S1=C|αm|2/3S_{1}=C|\alpha m|^{2/3} that

(3.164) J2L2(S2)Ce2t|αm|2(S2S1s𝑑s)mfL2(S2)Ce2t|αm|2/3mfL2(S2).\displaystyle\|J_{2}\|_{L^{2}(S^{2})}\leq\frac{Ce^{-2t}}{|\alpha m|^{2}}\left(\int_{S_{2}}^{S_{1}}s\,ds\right)\|\mathbb{Q}_{m}f\|_{L^{2}(S^{2})}\leq\frac{Ce^{-2t}}{|\alpha m|^{2/3}}\|\mathbb{Q}_{m}f\|_{L^{2}(S^{2})}.

Similarly, since |etζ3,±(s)|=eS2te2t|e^{t\zeta_{3,\pm}(s)}|=e^{-S_{2}t}\leq e^{-2t}, |ddsζ3,±(s)|=1|\frac{d}{ds}\zeta_{3,\pm}(s)|=1, and

|ζ3,±(s)+4||Imζ3,±(s)||αm||α|1/2|m||αm|2,sI3\displaystyle|\zeta_{3,\pm}(s)+4|\geq|\mathrm{Im}\,\zeta_{3,\pm}(s)|\geq|\alpha m|-|\alpha|^{1/2}|m|\geq\frac{|\alpha m|}{2},\quad s\in I_{3}

by |α|>4|\alpha|>4, we see by (3.8), (3.153), (3.160), and (3.162) that

(3.165) J3L2(S2)C|I3|e2t|α|3/2|m|5/3mfL2(S2)=Ce2t|α||m|2/3mfL2(S2),\displaystyle\|J_{3}\|_{L^{2}(S^{2})}\leq\frac{C|I_{3}|e^{-2t}}{|\alpha|^{3/2}|m|^{5/3}}\|\mathbb{Q}_{m}f\|_{L^{2}(S^{2})}=\frac{Ce^{-2t}}{|\alpha||m|^{2/3}}\|\mathbb{Q}_{m}f\|_{L^{2}(S^{2})},

where |I3|=2|α|1/2|m||I_{3}|=2|\alpha|^{1/2}|m| is the length of I3I_{3}. For sI4s\in I_{4} we have

|etζ4,±(s)|=este2t,|ζ4,±(s)+4||Imζ4,±(s)|C(|αm|+s),\displaystyle|e^{t\zeta_{4,\pm}(s)}|=e^{-st}\leq e^{-2t},\quad|\zeta_{4,\pm}(s)+4|\geq|\mathrm{Im}\,\zeta_{4,\pm}(s)|\geq C(|\alpha m|+s),
|dζ4,±ds(s)|=1+(4C1|m|1/3)2C.\displaystyle\left|\frac{d\zeta_{4,\pm}}{ds}(s)\right|=\sqrt{1+(4C_{1}|m|^{1/3})^{2}}\leq C.

Note that here |m|=1,2|m|=1,2. By these inequalities, (3.8), (3.153), (3.160), and (3.162),

(3.166) J4L2(S2)Ce2t(S21s(|αm|+s)𝑑s)mfL2(S2)=Ce2t{1|αm|log(|αm|+S2S2)}mfL2(S2)Ce2t|αm|log(C|αm|1/2)mfL2(S2),\displaystyle\begin{aligned} \|J_{4}\|_{L^{2}(S^{2})}&\leq Ce^{-2t}\left(\int_{S_{2}}^{\infty}\frac{1}{s(|\alpha m|+s)}\,ds\right)\|\mathbb{Q}_{m}f\|_{L^{2}(S^{2})}\\ &=Ce^{-2t}\left\{\frac{1}{|\alpha m|}\log\left(\frac{|\alpha m|+S_{2}}{S_{2}}\right)\right\}\|\mathbb{Q}_{m}f\|_{L^{2}(S^{2})}\\ &\leq\frac{Ce^{-2t}}{|\alpha m|}\log\Bigl{(}C|\alpha m|^{1/2}\Bigr{)}\|\mathbb{Q}_{m}f\|_{L^{2}(S^{2})},\end{aligned}

where we also used (|αm|+S2)/S2=C|α|1/2|m|1/3+1C|αm|1/2(|\alpha m|+S_{2})/S_{2}=C|\alpha|^{1/2}|m|^{1/3}+1\leq C|\alpha m|^{1/2} in the last inequality. Now, noting that |α|>4|\alpha|>4 is sufficiently large, we deduce from (3.163)–(3.166) that

JL2(S2)k=14JkL2(S2)Ce2t|αm|2/3log(C|αm|2/3)mfL2(S2).\displaystyle\|J\|_{L^{2}(S^{2})}\leq\sum_{k=1}^{4}\|J_{k}\|_{L^{2}(S^{2})}\leq\frac{Ce^{-2t}}{|\alpha m|^{2/3}}\log\Bigl{(}C|\alpha m|^{2/3}\Bigr{)}\|\mathbb{Q}_{m}f\|_{L^{2}(S^{2})}.

Hence we get (3.152) by applying this inequality and (3.156) to (3.155). ∎

Now we recall that Lα=AiαΛL_{\alpha}=A-i\alpha\Lambda on 𝒳\mathcal{X} is diagonalized as

Lα=m{0}Lα,m,Lα,m=Lα|𝒳m=Lα|𝒫m𝒳,\displaystyle L_{\alpha}=\oplus_{m\in\mathbb{Z}\setminus\{0\}}L_{\alpha,m},\quad L_{\alpha,m}=L_{\alpha}|_{\mathcal{X}_{m}}=L_{\alpha}|_{\mathcal{P}_{m}\mathcal{X}},

where 𝒫m\mathcal{P}_{m} is the operator given by (2.8). Moreover, for f𝒳f\in\mathcal{X} we have

m𝒫mf=𝒫mf,m{0},m𝒫mf=(I)𝒫mf,|m|=1,2,\displaystyle\mathbb{Q}_{m}\mathcal{P}_{m}f=\mathbb{Q}\mathcal{P}_{m}f,\quad m\in\mathbb{Z}\setminus\{0\},\quad\mathbb{P}_{m}\mathcal{P}_{m}f=(I-\mathbb{Q})\mathcal{P}_{m}f,\quad|m|=1,2,

where \mathbb{Q} is the orthogonal projection from 𝒳\mathcal{X} onto 𝒴=N𝒳(Λ)\mathcal{Y}=N_{\mathcal{X}}(\Lambda)^{\perp} (see Lemma 3.2). Hence by Theorems 3.20 and 3.21 we get the next result which implies Theorem 1.1.

Theorem 3.22.

There exist constants C1,C2>0C_{1},C_{2}>0 such that

𝒫metLαfL2(S2)C1eC2|α|1/2|m|2/3t𝒫mfL2(S2)\displaystyle\|\mathbb{Q}\mathcal{P}_{m}e^{tL_{\alpha}}f\|_{L^{2}(S^{2})}\leq C_{1}e^{-C_{2}|\alpha|^{1/2}|m|^{2/3}t}\|\mathbb{Q}\mathcal{P}_{m}f\|_{L^{2}(S^{2})}

for all t0t\geq 0, f𝒳f\in\mathcal{X}, α\alpha\in\mathbb{R}, and m{0}m\in\mathbb{Z}\setminus\{0\} (note that 𝒫m=𝒫m\mathbb{Q}\mathcal{P}_{m}=\mathcal{P}_{m} on 𝒳\mathcal{X} if |m|3|m|\geq 3). Also, if |m|=1,2|m|=1,2 and |α|>4|\alpha|>4 is sufficiently large, then for all t0t\geq 0 and f𝒳f\in\mathcal{X} we have

(I)𝒫metLαfL2(S2)e4t(I)𝒫mfL2(S2)+C3|αm|1/3log(C4|αm|2/3)e2t𝒫mfL2(S2),\|(I-\mathbb{Q})\mathcal{P}_{m}e^{tL_{\alpha}}f\|_{L^{2}(S^{2})}\leq e^{-4t}\|(I-\mathbb{Q})\mathcal{P}_{m}f\|_{L^{2}(S^{2})}\\ +C_{3}|\alpha m|^{1/3}\log\Bigl{(}C_{4}|\alpha m|^{2/3}\Bigr{)}e^{-2t}\|\mathbb{Q}\mathcal{P}_{m}f\|_{L^{2}(S^{2})},

where C3,C4>0C_{3},C_{4}>0 are constants independent of tt, ff, α\alpha, and mm.

Proof of Theorem 1.1.

For ν>0\nu>0 and aa\in\mathbb{R} we have ν,a|𝒳=νLa/ν\mathcal{L}^{\nu,a}|_{\mathcal{X}}=\nu L_{a/\nu} and thus etν,a|𝒳=eνtLa/νe^{t\mathcal{L}^{\nu,a}}|_{\mathcal{X}}=e^{\nu tL_{a/\nu}} in 𝒳\mathcal{X}. Hence Theorem 1.1 follows from Theorem 3.22. ∎

4. Abstract results

In this section we present abstract results for a perturbed operator.

For a linear operator TT on a Banach space \mathcal{B}, we denote by D(T)D_{\mathcal{B}}(T), ρ(T)\rho_{\mathcal{B}}(T), and σ(T)\sigma_{\mathcal{B}}(T) the domain, the resolvent set, and the spectrum of TT in \mathcal{B}. Also, let N(T)N_{\mathcal{B}}(T) and R(T)R_{\mathcal{B}}(T) be the kernel and range of TT in \mathcal{B}.

Let (𝒳,(,)𝒳)(\mathcal{X},(\cdot,\cdot)_{\mathcal{X}}) be a Hilbert space and AA and Λ\Lambda linear operators on 𝒳\mathcal{X}. We make the following assumptions.

Assumption 4.1.

The operator AA is self-adjoint in 𝒳\mathcal{X} and satisfies

(4.1) (Au,u)𝒳CAu𝒳2,uD𝒳(A).\displaystyle(-Au,u)_{\mathcal{X}}\geq C_{A}\|u\|_{\mathcal{X}}^{2},\quad u\in D_{\mathcal{X}}(A).

with some constant CA>0C_{A}>0.

Assumption 4.2.

The following conditions hold:

  1. (i)

    The operator Λ\Lambda is densely defined, closed, and AA-compact in 𝒳\mathcal{X}.

  2. (ii)

    Let 𝒴=N𝒳(Λ)\mathcal{Y}=N_{\mathcal{X}}(\Lambda)^{\perp} be the orthogonal complement of N𝒳(Λ)N_{\mathcal{X}}(\Lambda) in 𝒳\mathcal{X} and \mathbb{Q} the orthogonal projection from 𝒳\mathcal{X} onto 𝒴\mathcal{Y}. Then AA\mathbb{Q}A\subset A\mathbb{Q} in 𝒳\mathcal{X}.

Assumption 4.3.

There exist a Hilbert space (,(,))(\mathcal{H},(\cdot,\cdot)_{\mathcal{H}}), a closed symmetric operator B1B_{1} on \mathcal{H}, and a bounded self-adjoint operator B2B_{2} on 𝒳\mathcal{X} such that the following conditions hold:

  1. (i)

    The inclusion 𝒳\mathcal{X}\subset\mathcal{H} holds and (u,v)𝒳=(u,v)(u,v)_{\mathcal{X}}=(u,v)_{\mathcal{H}} for all u,v𝒳u,v\in\mathcal{X}.

  2. (ii)

    The relation N𝒳(Λ)=N𝒳(B2)N_{\mathcal{X}}(\Lambda)=N_{\mathcal{X}}(B_{2}) holds in 𝒳\mathcal{X} and

    B2uD(B1),B1B2u=Λu𝒳for alluD𝒳(Λ).\displaystyle B_{2}u\in D_{\mathcal{H}}(B_{1}),\quad B_{1}B_{2}u=\Lambda u\in\mathcal{X}\quad\text{for all}\quad u\in D_{\mathcal{X}}(\Lambda).
  3. (iii)

    There exists a constant C>0C>0 such that

    (4.2) (u,B2u)𝒳\displaystyle(u,B_{2}u)_{\mathcal{X}} Cu𝒳2,\displaystyle\geq C\|u\|_{\mathcal{X}}^{2}, u𝒴,\displaystyle u\in\mathcal{Y},
    (4.3) Re(Au,B2u)𝒳\displaystyle\mathrm{Re}(-Au,B_{2}u)_{\mathcal{X}} C(A)1/2u𝒳2,\displaystyle\geq C\|(-A)^{1/2}u\|_{\mathcal{X}}^{2}, uD𝒳(A)𝒴.\displaystyle u\in D_{\mathcal{X}}(A)\cap\mathcal{Y}.

Note that B2B_{2} is a linear operator on the original space 𝒳\mathcal{X}, not on the auxiliary space \mathcal{H}. Also, the operator B1B_{1} on \mathcal{H} does not necessarily map 𝒳\mathcal{X} into itself.

By AA\mathbb{Q}A\subset A\mathbb{Q} in Assumption 4.2 we can consider A\mathbb{Q}A as a linear operator

A:D𝒴(A)𝒴𝒴,D𝒴(A)=D𝒳(A)𝒴.\displaystyle\mathbb{Q}A\colon D_{\mathcal{Y}}(\mathbb{Q}A)\subset\mathcal{Y}\to\mathcal{Y},\quad D_{\mathcal{Y}}(\mathbb{Q}A)=D_{\mathcal{X}}(A)\cap\mathcal{Y}.

In what follows, we use the notation 𝒩=N𝒳(Λ)\mathcal{N}=N_{\mathcal{X}}(\Lambda) for simplicity. Let =I\mathbb{P}=I-\mathbb{Q} be the orthogonal projection from 𝒳\mathcal{X} onto 𝒩\mathcal{N} (note that 𝒩\mathcal{N} is closed in 𝒳\mathcal{X} since Λ\Lambda is closed). Then AA\mathbb{P}A\subset A\mathbb{P} and we can also consider A\mathbb{P}A as a linear operator

A:D𝒩(A)𝒩𝒩,D𝒩(A)=D𝒳(A)𝒩.\displaystyle\mathbb{P}A\colon D_{\mathcal{N}}(\mathbb{P}A)\subset\mathcal{N}\to\mathcal{N},\quad D_{\mathcal{N}}(\mathbb{P}A)=D_{\mathcal{X}}(A)\cap\mathcal{N}.

Note that A\mathbb{Q}A and A\mathbb{P}A are closed in 𝒴\mathcal{Y} and in 𝒩\mathcal{N}, respectively. Also, Λ\mathbb{Q}\Lambda is A\mathbb{Q}A-compact in 𝒴\mathcal{Y}. For α\alpha\in\mathbb{R} we define a linear operator LαL_{\alpha} on 𝒳\mathcal{X} by

Lα=AiαΛ,D𝒳(Lα)=D𝒳(A)\displaystyle L_{\alpha}=A-i\alpha\Lambda,\quad D_{\mathcal{X}}(L_{\alpha})=D_{\mathcal{X}}(A)

and consider Lα=AiαΛ\mathbb{Q}L_{\alpha}=\mathbb{Q}A-i\alpha\mathbb{Q}\Lambda on 𝒴\mathcal{Y} with domain D𝒴(Lα)=D𝒴(A)D_{\mathcal{Y}}(\mathbb{Q}L_{\alpha})=D_{\mathcal{Y}}(\mathbb{Q}A).

Our purpose is to derive a decay estimate for the \mathbb{Q}-part of the semigroup generated by LαL_{\alpha} in terms of α\alpha. The following results were obtained in [32, Lemmas 6.4–6.6].

Lemma 4.4.

Under Assumptions 4.14.3, we have

(4.4) B2=B2on𝒳,Im(Λu,B2u)𝒳=Im(Λu,B2u)𝒳=0,uD𝒳(Λ).\displaystyle\mathbb{Q}B_{2}=B_{2}\quad\text{on}\quad\mathcal{X},\quad\mathrm{Im}(\Lambda u,\mathbb{Q}B_{2}u)_{\mathcal{X}}=\mathrm{Im}(\Lambda u,B_{2}u)_{\mathcal{X}}=0,\quad u\in D_{\mathcal{X}}(\Lambda).

Moreover, LαL_{\alpha} and Lα\mathbb{Q}L_{\alpha} are closed in 𝒳\mathcal{X} and in 𝒴\mathcal{Y}, respectively, and

(4.5) {ζReζ0}ρ𝒳(Lα)=ρ𝒴(Lα)ρ𝒩(A)\displaystyle\{\zeta\in\mathbb{C}\mid\mathrm{Re}\,\zeta\geq 0\}\subset\rho_{\mathcal{X}}(L_{\alpha})=\rho_{\mathcal{Y}}(\mathbb{Q}L_{\alpha})\cap\rho_{\mathcal{N}}(\mathbb{P}A)

for all α\alpha\in\mathbb{R}, and if ζρ𝒳(Lα)\zeta\in\rho_{\mathcal{X}}(L_{\alpha}) and f𝒳f\in\mathcal{X}, then

(4.6) (ζLα)1f=(ζLα)1f,(ζLα)1f=(ζA)1fiα(ζA)1Λ(ζLα)1f.\displaystyle\begin{aligned} \mathbb{Q}(\zeta-L_{\alpha})^{-1}f&=(\zeta-\mathbb{Q}L_{\alpha})^{-1}\mathbb{Q}f,\\ \mathbb{P}(\zeta-L_{\alpha})^{-1}f&=(\zeta-\mathbb{P}A)^{-1}\mathbb{P}f-i\alpha(\zeta-\mathbb{P}A)^{-1}\mathbb{P}\Lambda(\zeta-\mathbb{Q}L_{\alpha})^{-1}\mathbb{Q}f.\end{aligned}

Moreover, the next result was shown in [32, Theorem 6.7] by application of the Gearhart–Prüss type theorem given by Wei [48] to the mm-accretive operator Lα-\mathbb{Q}L_{\alpha} on the Hilbert space 𝒴=𝒴\mathcal{Y}^{\prime}=\mathcal{Y} equipped with the weighted inner product (u,v)𝒴=(u,B2v)𝒳(u,v)_{\mathcal{Y}^{\prime}}=(u,B_{2}v)_{\mathcal{X}}.

Theorem 4.5.

Under Assumptions 4.14.3, the operator LαL_{\alpha} generates an analytic semigroup {etLα}t0\{e^{tL_{\alpha}}\}_{t\geq 0} in 𝒳\mathcal{X} for all α\alpha\in\mathbb{R}. Moreover, there exist positive constants C1C_{1} and C2C_{2} depending only on B2𝒳𝒳\|B_{2}\|_{\mathcal{X}\to\mathcal{X}} and the constants appearing in (4.2) and (4.3) (and in particular independent of the constant CAC_{A} appearing in (4.1)) such that

(4.7) etLαf𝒳C1eC2t/Φ𝒴(Lα)f𝒳,t0,f𝒳\displaystyle\|\mathbb{Q}e^{tL_{\alpha}}f\|_{\mathcal{X}}\leq C_{1}e^{-C_{2}t/\Phi_{\mathcal{Y}}(-\mathbb{Q}L_{\alpha})}\|\mathbb{Q}f\|_{\mathcal{X}},\quad t\geq 0,\,f\in\mathcal{X}

for all α\alpha\in\mathbb{R}, where Φ𝒴(Lα)=supλ(iλLα)1𝒴𝒴\Phi_{\mathcal{Y}}(-\mathbb{Q}L_{\alpha})=\sup_{\lambda\in\mathbb{R}}\|(i\lambda-\mathbb{Q}L_{\alpha})^{-1}\|_{\mathcal{Y}\to\mathcal{Y}}.

In order to get an upper bound of Φ𝒴(Lα)\Phi_{\mathcal{Y}}(-\mathbb{Q}L_{\alpha}) by a function of α\alpha, we make an addition assumption on AA and Λ\Lambda.

Assumption 4.6.

There exist bounded nonnegative functions

h1,h2:(0,)×[0,)satisfyinglimξsupμhj(ξ,μ)=0,j=1,2,\displaystyle h_{1},h_{2}\colon(0,\infty)\times\mathbb{R}\to[0,\infty)\quad\text{satisfying}\quad\lim_{\xi\to\infty}\sup_{\mu\in\mathbb{R}}h_{j}(\xi,\mu)=0,\quad j=1,2,

a constant C>0C>0, a Banach space (,)(\mathcal{B},\|\cdot\|_{\mathcal{B}}), and a closed operator B3B_{3} on \mathcal{B} such that the following conditions hold:

  1. (i)

    For all ξ(0,)\xi\in(0,\infty) and μ\mu\in\mathbb{R} we have

    (4.8) u𝒳2C(ξ2(μΛ)u𝒳2+h1(ξ,μ)2(A)1/2u𝒳2),uD𝒳(A)𝒴.\displaystyle\|u\|_{\mathcal{X}}^{2}\leq C\left(\xi^{2}\|\mathbb{Q}(\mu-\Lambda)u\|_{\mathcal{X}}^{2}+h_{1}(\xi,\mu)^{2}\|(-A)^{1/2}u\|_{\mathcal{X}}^{2}\right),\quad u\in D_{\mathcal{X}}(A)\cap\mathcal{Y}.
  2. (ii)

    The inclusions 𝒳\mathcal{X}\subset\mathcal{B} and D𝒳((A)1/2)D(B3)D_{\mathcal{X}}((-A)^{1/2})\subset D_{\mathcal{B}}(B_{3}) hold and

    (4.9) |Im(Au,Λu)𝒳|C(A)1/2u𝒳B3u,uD𝒳(A)𝒴.\displaystyle|\mathrm{Im}(Au,\Lambda u)_{\mathcal{X}}|\leq C\|(-A)^{1/2}u\|_{\mathcal{X}}\|B_{3}u\|_{\mathcal{B}},\quad u\in D_{\mathcal{X}}(A)\cap\mathcal{Y}.

    Moreover, for all ξ(0,)\xi\in(0,\infty) and μ\mu\in\mathbb{R} we have

    (4.10) B3u2C(ξ2(μΛ)u𝒳2+h2(ξ,μ)2(A)1/2u𝒳2),uD𝒳(A)𝒴.\displaystyle\|B_{3}u\|_{\mathcal{B}}^{2}\leq C\left(\xi^{2}\|\mathbb{Q}(\mu-\Lambda)u\|_{\mathcal{X}}^{2}+h_{2}(\xi,\mu)^{2}\|(-A)^{1/2}u\|_{\mathcal{X}}^{2}\right),\quad u\in D_{\mathcal{X}}(A)\cap\mathcal{Y}.

Let us give an estimate for the resolvent of Lα\mathbb{Q}L_{\alpha} along the imaginary axis which was originally shown in [15, Theorem 2.9] under assumptions on coercive estimates for μΛ\mu-\Lambda on 𝒳\mathcal{X} with μ\mu\in\mathbb{R} away from zero and for (μΛ)\mathbb{Q}(\mu-\Lambda) on 𝒴\mathcal{Y} with μ\mu\in\mathbb{R} close to zero.

Theorem 4.7.

Under Assumptions 4.14.3 and 4.6, we have

(4.11) (iλ+Lα)1𝒳𝒳=(iλ+Lα)1𝒴𝒴CF(α,λα)\displaystyle\|\mathbb{Q}(i\lambda+L_{\alpha})^{-1}\|_{\mathcal{X}\to\mathcal{X}}=\|(i\lambda+\mathbb{Q}L_{\alpha})^{-1}\|_{\mathcal{Y}\to\mathcal{Y}}\leq CF\left(\alpha,\frac{\lambda}{\alpha}\right)

for all α{0}\alpha\in\mathbb{R}\setminus\{0\} and λ\lambda\in\mathbb{R}. Here

(4.12) F(α,μ)=infξ1,ξ2>0(ξ1|α|+ξ12ξ22α2+ξ12h2(ξ2,μ)|α|+h1(ξ1,μ)2)\displaystyle F(\alpha,\mu)=\inf_{\xi_{1},\xi_{2}>0}\left(\frac{\xi_{1}}{|\alpha|}+\frac{\xi_{1}^{2}\xi_{2}^{2}}{\alpha^{2}}+\frac{\xi_{1}^{2}h_{2}(\xi_{2},\mu)}{|\alpha|}+h_{1}(\xi_{1},\mu)^{2}\right)

for α{0}\alpha\in\mathbb{R}\setminus\{0\} and μ\mu\in\mathbb{R}. Also, C>0C>0 is a constant depending only on B2𝒳𝒳\|B_{2}\|_{\mathcal{X}\to\mathcal{X}} and the constants appearing in (4.2), (4.3), and (4.8)–(4.10) (and in particular independent of the constant CAC_{A} appearing in (4.1)).

Proof.

Let α{0}\alpha\in\mathbb{R}\setminus\{0\} and λ\lambda\in\mathbb{R}. Then

iλρ𝒳(Lα)ρ𝒴(Lα),(iλ+Lα)1f=(iλ+Lα)1f,f𝒳\displaystyle-i\lambda\in\rho_{\mathcal{X}}(L_{\alpha})\subset\rho_{\mathcal{Y}}(\mathbb{Q}L_{\alpha}),\quad\mathbb{Q}(i\lambda+L_{\alpha})^{-1}f=(i\lambda+\mathbb{Q}L_{\alpha})^{-1}\mathbb{Q}f,\quad f\in\mathcal{X}

by (4.5) and (4.6). By these relations, we easily find that the first equality of (4.11) holds. Let us prove the second inequality of (4.11). Without loss of generality, we may assume α>0\alpha>0. We set μ=λ/α\mu=\lambda/\alpha and abbreviate hj(ξj,μ)h_{j}(\xi_{j},\mu) to hjh_{j} for ξ1,ξ2>0\xi_{1},\xi_{2}>0 and j=1,2j=1,2. In what follows, we denote by CC a general positive constant depending only on B2𝒳𝒳\|B_{2}\|_{\mathcal{X}\to\mathcal{X}} and the constants appearing in (4.2), (4.3), and (4.8)–(4.10). Let uD𝒴(Lα)u\in D_{\mathcal{Y}}(\mathbb{Q}L_{\alpha}) and

f=(iλ+Lα)u=Au+iα(μΛ)u𝒴𝒳.\displaystyle f=(i\lambda+\mathbb{Q}L_{\alpha})u=\mathbb{Q}Au+i\alpha\mathbb{Q}(\mu-\Lambda)u\in\mathcal{Y}\subset\mathcal{X}.

We take the inner products of both sides with (μΛ)u\mathbb{Q}(\mu-\Lambda)u in 𝒳\mathcal{X}. Then

(4.13) (f,(μΛ)u)𝒳=μ(Au,u)𝒳(Au,Λu)𝒳+iα(μΛ)u𝒳2.\displaystyle(f,\mathbb{Q}(\mu-\Lambda)u)_{\mathcal{X}}=\mu(\mathbb{Q}Au,\mathbb{Q}u)_{\mathcal{X}}-(\mathbb{Q}Au,\mathbb{Q}\Lambda u)_{\mathcal{X}}+i\alpha\|\mathbb{Q}(\mu-\Lambda)u\|_{\mathcal{X}}^{2}.

Since AA\mathbb{Q}A\subset A\mathbb{Q}, u=u\mathbb{Q}u=u, and AA is self-adjoint in 𝒳\mathcal{X}, we have

(Au,Λu)𝒳=(Au,Λu)𝒳,Im(Au,u)𝒳=Im(Au,u)𝒳=0.\displaystyle(\mathbb{Q}Au,\mathbb{Q}\Lambda u)_{\mathcal{X}}=(Au,\Lambda u)_{\mathcal{X}},\quad\mathrm{Im}(\mathbb{Q}Au,\mathbb{Q}u)_{\mathcal{X}}=\mathrm{Im}(Au,u)_{\mathcal{X}}=0.

Thus, taking the imaginary part of (4.13), we obtain

Im(f,(μΛ)u)𝒳=Im(Au,Λu)𝒳+α(μΛ)u𝒳2.\displaystyle\mathrm{Im}(f,\mathbb{Q}(\mu-\Lambda)u)_{\mathcal{X}}=-\mathrm{Im}(Au,\Lambda u)_{\mathcal{X}}+\alpha\|\mathbb{Q}(\mu-\Lambda)u\|_{\mathcal{X}}^{2}.

By this equality and (4.9) we get

α(μΛ)u𝒳2\displaystyle\alpha\|\mathbb{Q}(\mu-\Lambda)u\|_{\mathcal{X}}^{2} f𝒳(μΛ)u𝒳+C(A)1/2u𝒳B3u\displaystyle\leq\|f\|_{\mathcal{X}}\|\mathbb{Q}(\mu-\Lambda)u\|_{\mathcal{X}}+C\|(-A)^{1/2}u\|_{\mathcal{X}}\|B_{3}u\|_{\mathcal{B}}
α2(μΛ)u𝒳2+12αf𝒳2+C(A)1/2u𝒳B3u\displaystyle\leq\frac{\alpha}{2}\|\mathbb{Q}(\mu-\Lambda)u\|_{\mathcal{X}}^{2}+\frac{1}{2\alpha}\|f\|_{\mathcal{X}}^{2}+C\|(-A)^{1/2}u\|_{\mathcal{X}}\|B_{3}u\|_{\mathcal{B}}

and therefore

(4.14) α(μΛ)u𝒳21αf𝒳2+C(A)1/2u𝒳B3u.\displaystyle\alpha\|\mathbb{Q}(\mu-\Lambda)u\|_{\mathcal{X}}^{2}\leq\frac{1}{\alpha}\|f\|_{\mathcal{X}}^{2}+C\|(-A)^{1/2}u\|_{\mathcal{X}}\|B_{3}u\|_{\mathcal{B}}.

Moreover, for each ξ2>0\xi_{2}>0 it follows from (4.10) and (4.14) that

B3u2\displaystyle\|B_{3}u\|_{\mathcal{B}}^{2} C(ξ22(μΛ)u𝒳2+h22(A)1/2u𝒳2)\displaystyle\leq C\left(\xi_{2}^{2}\|\mathbb{Q}(\mu-\Lambda)u\|_{\mathcal{X}}^{2}+h_{2}^{2}\|(-A)^{1/2}u\|_{\mathcal{X}}^{2}\right)
C(ξ22α2f𝒳2+ξ22α(A)1/2u𝒳B3u+h22(A)1/2u𝒳2)\displaystyle\leq C\left(\frac{\xi_{2}^{2}}{\alpha^{2}}\|f\|_{\mathcal{X}}^{2}+\frac{\xi_{2}^{2}}{\alpha}\|(-A)^{1/2}u\|_{\mathcal{X}}\|B_{3}u\|_{\mathcal{B}}+h_{2}^{2}\|(-A)^{1/2}u\|_{\mathcal{X}}^{2}\right)
12B3u2+C{ξ22α2f𝒳2+(ξ24α2+h22)(A)1/2u𝒳2}.\displaystyle\leq\frac{1}{2}\|B_{3}u\|_{\mathcal{B}}^{2}+C\left\{\frac{\xi_{2}^{2}}{\alpha^{2}}\|f\|_{\mathcal{X}}^{2}+\left(\frac{\xi_{2}^{4}}{\alpha^{2}}+h_{2}^{2}\right)\|(-A)^{1/2}u\|_{\mathcal{X}}^{2}\right\}.

Hence, subtracting 12B2u𝒳2\frac{1}{2}\|B_{2}u\|_{\mathcal{X}}^{2} from both sides and taking the square root of the resulting inequality, we obtain

B3uC{ξ2αf𝒳+(ξ22α+h2)(A)1/2u𝒳}.\displaystyle\|B_{3}u\|_{\mathcal{B}}\leq C\left\{\frac{\xi_{2}}{\alpha}\|f\|_{\mathcal{X}}+\left(\frac{\xi_{2}^{2}}{\alpha}+h_{2}\right)\|(-A)^{1/2}u\|_{\mathcal{X}}\right\}.

We apply this inequality to (4.14) to find that

(μΛ)u𝒳2\displaystyle\|\mathbb{Q}(\mu-\Lambda)u\|_{\mathcal{X}}^{2} 1α2f𝒳2+Cα(A)1/2u𝒳{ξ2αf𝒳+(ξ22α+h2)(A)1/2u𝒳}\displaystyle\leq\frac{1}{\alpha^{2}}\|f\|_{\mathcal{X}}^{2}+\frac{C}{\alpha}\|(-A)^{1/2}u\|_{\mathcal{X}}\left\{\frac{\xi_{2}}{\alpha}\|f\|_{\mathcal{X}}+\left(\frac{\xi_{2}^{2}}{\alpha}+h_{2}\right)\|(-A)^{1/2}u\|_{\mathcal{X}}\right\}
C{1α2f𝒳2+(ξ22α2+h2α)(A)1/2u𝒳2}.\displaystyle\leq C\left\{\frac{1}{\alpha^{2}}\|f\|_{\mathcal{X}}^{2}+\left(\frac{\xi_{2}^{2}}{\alpha^{2}}+\frac{h_{2}}{\alpha}\right)\|(-A)^{1/2}u\|_{\mathcal{X}}^{2}\right\}.

For each ξ1>0\xi_{1}>0, we deduce from (4.8) and the above inequality that

(4.15) u𝒳2C(ξ12(μΛ)u𝒳2+h12(A)1/2u𝒳2)C{ξ12α2f𝒳2+(ξ12ξ22α2+ξ12h2α+h12)(A)1/2u𝒳2}.\displaystyle\begin{aligned} \|u\|_{\mathcal{X}}^{2}&\leq C\left(\xi_{1}^{2}\|\mathbb{Q}(\mu-\Lambda)u\|_{\mathcal{X}}^{2}+h_{1}^{2}\|(-A)^{1/2}u\|_{\mathcal{X}}^{2}\right)\\ &\leq C\left\{\frac{\xi_{1}^{2}}{\alpha^{2}}\|f\|_{\mathcal{X}}^{2}+\left(\frac{\xi_{1}^{2}\xi_{2}^{2}}{\alpha^{2}}+\frac{\xi_{1}^{2}h_{2}}{\alpha}+h_{1}^{2}\right)\|(-A)^{1/2}u\|_{\mathcal{X}}^{2}\right\}.\end{aligned}

To estimate (A)1/2u𝒳\|(-A)^{1/2}u\|_{\mathcal{X}}, we observe by f=(iλ+Lα)u=(iλ+Lα)uf=(i\lambda+\mathbb{Q}L_{\alpha})u=\mathbb{Q}(i\lambda+L_{\alpha})u that

(f,B2u)𝒳=((iλ+Lα)u,B2u)𝒳=iλ(u,B2u)𝒳+(Au,B2u)𝒳iα(Λu,B2u)𝒳.\displaystyle(f,B_{2}u)_{\mathcal{X}}=\bigl{(}(i\lambda+L_{\alpha})u,\mathbb{Q}B_{2}u\bigr{)}_{\mathcal{X}}=i\lambda(u,\mathbb{Q}B_{2}u)_{\mathcal{X}}+(Au,\mathbb{Q}B_{2}u)_{\mathcal{X}}-i\alpha(\Lambda u,\mathbb{Q}B_{2}u)_{\mathcal{X}}.

We take the real part of this equality and use (4.4) and Im(u,B2u)𝒳=0\mathrm{Im}(u,B_{2}u)_{\mathcal{X}}=0 by the self-adjointness of B2B_{2} in 𝒳\mathcal{X} to get Re(f,B2u)𝒳=Re(Au,B2u)𝒳\mathrm{Re}(f,B_{2}u)_{\mathcal{X}}=\mathrm{Re}(Au,B_{2}u)_{\mathcal{X}}. Hence

(A)1/2u𝒳2CRe(Au,B2u)𝒳=CRe(f,B2u)𝒳Cf𝒳B2u𝒳Cf𝒳u𝒳\displaystyle\|(-A)^{1/2}u\|_{\mathcal{X}}^{2}\leq C\,\mathrm{Re}(-Au,B_{2}u)_{\mathcal{X}}=-C\,\mathrm{Re}(f,B_{2}u)_{\mathcal{X}}\leq C\|f\|_{\mathcal{X}}\|B_{2}u\|_{\mathcal{X}}\leq C\|f\|_{\mathcal{X}}\|u\|_{\mathcal{X}}

by (4.3) and the boundedness of B2B_{2}. Applying this inequality to (4.15) we obtain

u𝒳2\displaystyle\|u\|_{\mathcal{X}}^{2} C{ξ12α2f𝒳2+(ξ12ξ22α2+ξ12h2α+h12)f𝒳u𝒳}\displaystyle\leq C\left\{\frac{\xi_{1}^{2}}{\alpha^{2}}\|f\|_{\mathcal{X}}^{2}+\left(\frac{\xi_{1}^{2}\xi_{2}^{2}}{\alpha^{2}}+\frac{\xi_{1}^{2}h_{2}}{\alpha}+h_{1}^{2}\right)\|f\|_{\mathcal{X}}\|u\|_{\mathcal{X}}\right\}
12u𝒳2+C{ξ12α2+(ξ12ξ22α2+ξ12h2α+h12)2}f𝒳2.\displaystyle\leq\frac{1}{2}\|u\|_{\mathcal{X}}^{2}+C\left\{\frac{\xi_{1}^{2}}{\alpha^{2}}+\left(\frac{\xi_{1}^{2}\xi_{2}^{2}}{\alpha^{2}}+\frac{\xi_{1}^{2}h_{2}}{\alpha}+h_{1}^{2}\right)^{2}\right\}\|f\|_{\mathcal{X}}^{2}.

We further subtract 12u𝒳2\frac{1}{2}\|u\|_{\mathcal{X}}^{2} from both sides and take the square root of the resulting inequality to find that

u𝒳C(ξ1α+ξ12ξ22α2+ξ12h2α+h12)f𝒳,f=(iλ+Lα)u\displaystyle\|u\|_{\mathcal{X}}\leq C\left(\frac{\xi_{1}}{\alpha}+\frac{\xi_{1}^{2}\xi_{2}^{2}}{\alpha^{2}}+\frac{\xi_{1}^{2}h_{2}}{\alpha}+h_{1}^{2}\right)\|f\|_{\mathcal{X}},\quad f=(i\lambda+\mathbb{Q}L_{\alpha})u

for all uD𝒴(Lα)u\in D_{\mathcal{Y}}(\mathbb{Q}L_{\alpha}). Since iλρ𝒴(Lα)-i\lambda\in\rho_{\mathcal{Y}}(\mathbb{Q}L_{\alpha}), the above inequality shows that

(iλ+Lα)1𝒴𝒴C(ξ1|α|+ξ12ξ22α2+ξ12h2(ξ2,μ)|α|+h1(ξ1,μ)2)\displaystyle\|(i\lambda+\mathbb{Q}L_{\alpha})^{-1}\|_{\mathcal{Y}\to\mathcal{Y}}\leq C\left(\frac{\xi_{1}}{|\alpha|}+\frac{\xi_{1}^{2}\xi_{2}^{2}}{\alpha^{2}}+\frac{\xi_{1}^{2}h_{2}(\xi_{2},\mu)}{|\alpha|}+h_{1}(\xi_{1},\mu)^{2}\right)

for all ξ1,ξ2>0\xi_{1},\xi_{2}>0 with μ=λ/α\mu=\lambda/\alpha. Therefore, the second inequality of (4.11) is valid. ∎

Remark 4.8.

Contrary to [15, Assumption 1], we do not assume that AA has a compact resolvent in Assumption 4.1. Thus the essential spectrum σ~𝒳(A)\tilde{\sigma}_{\mathcal{X}}(A) of AA in 𝒳\mathcal{X} may be not empty, but we must have σ~𝒴(A)=\tilde{\sigma}_{\mathcal{Y}}(\mathbb{Q}A)=\emptyset for the essential spectrum of A\mathbb{Q}A in 𝒴\mathcal{Y} under Assumptions 4.14.3 and 4.6. Indeed, 0ρ𝒴(Lα)0\in\rho_{\mathcal{Y}}(\mathbb{Q}L_{\alpha}) and (Lα)1𝒴𝒴CF(α,0)\|(\mathbb{Q}L_{\alpha})^{-1}\|_{\mathcal{Y}\to\mathcal{Y}}\leq CF(\alpha,0) for all α{0}\alpha\in\mathbb{R}\setminus\{0\} by Theorem 4.7. Thus, by a standard Neumann series argument,

{ζ|ζ|<1/CF(α,0)}ρ𝒴(Lα).\displaystyle\{\zeta\in\mathbb{C}\mid|\zeta|<1/CF(\alpha,0)\}\subset\rho_{\mathcal{Y}}(\mathbb{Q}L_{\alpha}).

On the other hand, since Λ\mathbb{Q}\Lambda is A\mathbb{Q}A-compact in 𝒴\mathcal{Y} and the essential spectrum is invariant under a relatively compact perturbation (see [20, Theorem IV.5.35]),

σ~𝒴(A)=σ~𝒴(AiαΛ)=σ~𝒴(Lα)σ𝒴(Lα).\displaystyle\tilde{\sigma}_{\mathcal{Y}}(\mathbb{Q}A)=\tilde{\sigma}_{\mathcal{Y}}(\mathbb{Q}A-i\alpha\mathbb{Q}\Lambda)=\tilde{\sigma}_{\mathcal{Y}}(\mathbb{Q}L_{\alpha})\subset\sigma_{\mathcal{Y}}(\mathbb{Q}L_{\alpha}).

Hence, if σ~𝒴(A)\tilde{\sigma}_{\mathcal{Y}}(\mathbb{Q}A) contains some ζ\zeta\in\mathbb{C}, then ζσ𝒴(Lα)\zeta\in\sigma_{\mathcal{Y}}(\mathbb{Q}L_{\alpha}) and thus |ζ|1/CF(α,0)|\zeta|\geq 1/CF(\alpha,0) for all α{0}\alpha\in\mathbb{R}\setminus\{0\}, but this is impossible since lim|α|F(α,0)=0\lim_{|\alpha|\to\infty}F(\alpha,0)=0 by limξhj(ξ,0)=0\lim_{\xi\to\infty}h_{j}(\xi,0)=0 for j=1,2j=1,2. Thus σ~𝒴(A)=\tilde{\sigma}_{\mathcal{Y}}(\mathbb{Q}A)=\emptyset. In particular, σ~𝒳(A)=\tilde{\sigma}_{\mathcal{X}}(A)=\emptyset if 𝒩=N𝒳(Λ)={0}\mathcal{N}=N_{\mathcal{X}}(\Lambda)=\{0\}.

Combining Theorems 4.5 and 4.7, we obtain the following result.

Theorem 4.9.

Under Assumptions 4.14.3 and 4.6, there exist positive constants C1C_{1} and C2C_{2} depending only on B2𝒳𝒳\|B_{2}\|_{\mathcal{X}\to\mathcal{X}} and the constants appearing in (4.2), (4.3), and (4.8)–(4.10) (and in particular independent of the constant CAC_{A} appearing in (4.1)) such that

(4.16) etLαf𝒳C1eC2t/F(α)f𝒳t0,f𝒳\displaystyle\|\mathbb{Q}e^{tL_{\alpha}}f\|_{\mathcal{X}}\leq C_{1}e^{-C_{2}t/F(\alpha)}\|\mathbb{Q}f\|_{\mathcal{X}}\quad t\geq 0,\,f\in\mathcal{X}

for all α{0}\alpha\in\mathbb{R}\setminus\{0\}, where F(α)=supμF(α,μ)F(\alpha)=\sup_{\mu\in\mathbb{R}}F(\alpha,\mu) with F(α,μ)F(\alpha,\mu) given by (4.12).

Proof.

By (4.11) we have

Φ𝒴(Lα)\displaystyle\Phi_{\mathcal{Y}}(-\mathbb{Q}L_{\alpha}) =supλ(iλLα)1𝒴𝒴=supλ(iλ+Lα)1𝒴𝒴\displaystyle=\sup_{\lambda\in\mathbb{R}}\|(i\lambda-\mathbb{Q}L_{\alpha})^{-1}\|_{\mathcal{Y}\to\mathcal{Y}}=\sup_{\lambda\in\mathbb{R}}\|(i\lambda+\mathbb{Q}L_{\alpha})^{-1}\|_{\mathcal{Y}\to\mathcal{Y}}
CsupλF(α,λα)=CsupμF(α,μ)=CF(α).\displaystyle\leq C\sup_{\lambda\in\mathbb{R}}F\left(\alpha,\frac{\lambda}{\alpha}\right)=C\sup_{\mu\in\mathbb{R}}F(\alpha,\mu)=CF(\alpha).

Hence (4.16) follows from this inequality and (4.7). ∎

5. Appendix: basic formulas of differential geometry

This section gives some notations and basic formulas of differential geometry. We refer to e.g. [24, 25] for details.

Let (M,g)(M,g) be a two-dimensional Riemannian manifold. For a local coordinate system (x1,x2)(x^{1},x^{2}) of MM, let (1,2)(\partial_{1},\partial_{2}) and (dx1,dx2)(dx^{1},dx^{2}) be the coordinate frame and its dual coframe. We set gjk=g(j,k)g_{jk}=g(\partial_{j},\partial_{k}) for j,k=1,2j,k=1,2 and denote by (gjk)j,k=1,2(g^{jk})_{j,k=1,2} the inverse matrix of (gjk)j,k=1,2(g_{jk})_{j,k=1,2} so that the inner products of one-forms and (0,2)(0,2)-tensor fields on MM are given by g(dxj,dxk)=gjkg(dx^{j},dx^{k})=g^{jk} and

g(dxj1dxk1,dxj2dxk2)=g(dxj1,dxj2)g(dxk1,dxk2)=gj1j2gk1k2.\displaystyle g(dx^{j_{1}}\otimes dx^{k_{1}},dx^{j_{2}}\otimes dx^{k_{2}})=g(dx^{j_{1}},dx^{j_{2}})g(dx^{k_{1}},dx^{k_{2}})=g^{j_{1}j_{2}}g^{k_{1}k_{2}}.

Let Γjkl\Gamma_{jk}^{l} be the Christoffel symbols of gg given by

Γjkl=12m=1,2glm(jgkm+kgjmmgjk),j,k,l=1,2.\displaystyle\Gamma_{jk}^{l}=\frac{1}{2}\sum_{m=1,2}g^{lm}(\partial_{j}g_{km}+\partial_{k}g_{jm}-\partial_{m}g_{jk}),\quad j,k,l=1,2.

For a (complex-valued) function uu on MM, we write u\nabla u and 2u\nabla^{2}u for the gradient and the covariant Hessian of uu, respectively, which are locally expressed as

u=j,k=1,2gjk(ju)k,2u=j,k=1,2u;j;kdxjdxk\displaystyle\nabla u=\sum_{j,k=1,2}g^{jk}(\partial_{j}u)\partial_{k},\quad\nabla^{2}u=\sum_{j,k=1,2}u_{;j;k}dx^{j}\otimes dx^{k}

with u;j;k=kjul=1,2Γkjlluu_{;j;k}=\partial_{k}\partial_{j}u-\sum_{l=1,2}\Gamma_{kj}^{l}\partial_{l}u for j,k=1,2j,k=1,2. Then

|u|2\displaystyle|\nabla u|^{2} =g(u,u¯)=j,k=1,2gjkjuku¯,\displaystyle=g(\nabla u,\nabla\bar{u})=\sum_{j,k=1,2}g^{jk}\partial_{j}u\,\overline{\partial_{k}u},
|2u|2\displaystyle|\nabla^{2}u|^{2} =g(2u,2u¯)=j1,j2,k1,k2=1,2gj1j2gk1k2u;j1;k1u;j2;k2¯.\displaystyle=g(\nabla^{2}u,\nabla^{2}\bar{u})=\sum_{j_{1},j_{2},k_{1},k_{2}=1,2}g^{j_{1}j_{2}}g^{k_{1}k_{2}}u_{;j_{1};k_{1}}\,\overline{u_{;j_{2};k_{2}}}.

Also, the Laplace–Beltrami operator Δ\Delta on MM is locally given by

Δu=1detgj,k=1,2j(gjkkudetg),detg=det((gjk)j,k=1,2).\displaystyle\Delta u=\frac{1}{\det g}\sum_{j,k=1,2}\partial_{j}\Bigl{(}g^{jk}\partial_{k}u\sqrt{\det g}\Bigr{)},\quad\det g=\det\bigl{(}(g_{jk})_{j,k=1,2}\bigr{)}.

We use these expressions under the spherical coordinate system of M=S2M=S^{2}.

Lemma 5.1.

Let (x1,x2)=(θ,φ)(x^{1},x^{2})=(\theta,\varphi) be the spherical coordinate system

[0,π]×[0,2π)(θ,φ)𝐱(θ,φ)=(sinθcosφ,sinθsinφ,cosθ)S2.\displaystyle[0,\pi]\times[0,2\pi)\ni(\theta,\varphi)\mapsto\mathbf{x}(\theta,\varphi)=(\sin\theta\cos\varphi,\sin\theta\sin\varphi,\cos\theta)\in S^{2}.

Then for a function uu on S2S^{2} we have

(5.1) |u|2=|θu|2+1sin2θ|φu|2,|2u|2=|θ2u|2+2sin2θ|φθucosθsinθφu|2+1sin4θ|φ2u+sinθcosθθu|2,Δu=1sinθθ(sinθθu)+1sin2θφ2u.\displaystyle\begin{aligned} |\nabla u|^{2}&=|\partial_{\theta}u|^{2}+\frac{1}{\sin^{2}\theta}|\partial_{\varphi}u|^{2},\\ |\nabla^{2}u|^{2}&=|\partial_{\theta}^{2}u|^{2}+\frac{2}{\sin^{2}\theta}\left|\partial_{\varphi}\partial_{\theta}u-\frac{\cos\theta}{\sin\theta}\partial_{\varphi}u\right|^{2}+\frac{1}{\sin^{4}\theta}|\partial_{\varphi}^{2}u+\sin\theta\cos\theta\,\partial_{\theta}u|^{2},\\ \Delta u&=\frac{1}{\sin\theta}\partial_{\theta}\bigl{(}\sin\theta\,\partial_{\theta}u\bigr{)}+\frac{1}{\sin^{2}\theta}\partial_{\varphi}^{2}u.\end{aligned}
Proof.

We use the index j=θ,φj=\theta,\varphi instead of j=1,2j=1,2. Since

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

we observe by direct calculations that

gθθ\displaystyle g_{\theta\theta} =1,gφφ=sin2θ,detg=sin2θ,\displaystyle=1,\quad g_{\varphi\varphi}=\sin^{2}\theta,\quad\det g=\sin^{2}\theta,
gθθ\displaystyle g^{\theta\theta} =1,gφφ=1sin2θ,Γφφθ=sinθcosθ,Γθφφ=Γφθφ=cosθsinθ\displaystyle=1,\quad g^{\varphi\varphi}=\frac{1}{\sin^{2}\theta},\quad\Gamma_{\varphi\varphi}^{\theta}=-\sin\theta\cos\theta,\quad\Gamma_{\theta\varphi}^{\varphi}=\Gamma_{\varphi\theta}^{\varphi}=\frac{\cos\theta}{\sin\theta}

and the other gjkg_{jk}, gjkg^{jk}, and Γjkl\Gamma_{jk}^{l} vanish identically. Hence

u;θ;θ=θ2u,u;θ;φ=u;φ;θ=φθucosθsinθφu,u;φ;φ=φ2u+sinθcosθθu\displaystyle u_{;\theta;\theta}=\partial_{\theta}^{2}u,\quad u_{;\theta;\varphi}=u_{;\varphi;\theta}=\partial_{\varphi}\partial_{\theta}u-\frac{\cos\theta}{\sin\theta}\partial_{\varphi}u,\quad u_{;\varphi;\varphi}=\partial_{\varphi}^{2}u+\sin\theta\cos\theta\,\partial_{\theta}u

and we obtain (5.1) by the above expressions. ∎

Acknowledgments

The work of the first author was supported by JSPS KAKENHI Grant Numbers 20K03698, 19H05597, 20H00118. Also, the work of the second author was supported by Grant-in-Aid for JSPS Fellows No. 19J00693.

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