Finite Time Blowup of 2D Boussinesq and 3D Euler Equations with $C^{1,\alpha }$ Velocity and Boundary

Abstract

Inspired by the numerical evidence of a potential 3D Euler singularity by Luo-Hou [30, 31] and the recent breakthrough by Elgindi [11] on the singularity formation of the 3D Euler equation without swirl with $C^{1,\alpha }$ initial data for the velocity, we prove the finite time singularity for the 2D Boussinesq and the 3D axisymmetric Euler equations in the presence of boundary with $C^{1,\alpha }$ initial data for the velocity (and density in the case of Boussinesq equations). Our finite time blowup solution for the 3D Euler equations and the singular solution considered in [30, 31] share many essential features, including the symmetry properties of the solution, the flow structure, and the sign of the solution in each quadrant, except that we use $C^{1,\alpha }$ initial data for the velocity field. We use a dynamic rescaling formulation and follow the general framework of analysis developed by Elgindi in [11]. We also use some strategy proposed in our recent joint work with Huang in [7] and adopt several methods of analysis in [11] to establish the linear and nonlinear stability of an approximate self-similar profile. The nonlinear stability enables us to prove that the solution of the 3D Euler equations or the 2D Boussinesq equations with $C^{1,\alpha }$ initial data will develop a finite time singularity. Moreover, the velocity field has finite energy before the singularity time.

Appendix A.

In Appendix A.1, we estimate $\Gamma (\beta )$ and the constant c appeared in the approximate profile (4.8). In Appendix A.2, we perform the derivations and establish several inequalities in the linear stability analysis in Sect. 5.6. In Appendix A.3, we derive the singular term (7.5) in the elliptic estimates. In Appendix A.4, we will establish several estimates of $L_{12}(\Omega )$ that are used frequently in the nonlinear stability analysis. Notice that we only have the formula of ${\bar{\eta }} = {\bar{\theta }}_x$ in (4.8). We need to recover ${\bar{\theta }}, {\bar{\xi }} ={\bar{\theta }}_y$ from ${\bar{\eta }}$ via integration. Yet, we do not have a simple formula to perform integration. Alternatively, we derive useful estimates for ${\bar{\xi }}$ in Appendix A.5. Some estimates of ${\bar{\Omega }}, {\bar{\eta }}$ are also obtained there. In Appendix A.6, we show that the truncation of the approximate steady state would contribute only to a small perturbation under the norm we use, and we prove Lemma 9.1. In Appendix A.7, we prove Lemma 9.1. In Appendix A.8, we study the toy model introduced in [11].

1.1 Estimates of $\Gamma (\beta )$ and the constant c

Lemma A.1

For $x \in [0,1]$, the following estimate holds uniformly for $\lambda \ge 1 / 10$,

$$\begin{aligned} (1-x^{\kappa }) x^{\lambda } \le \frac{\kappa }{\lambda }. \end{aligned}$$

(A.1)

Consequently, for $\beta \in [0, \pi /2]$, $2 \ge \lambda \ge 1/10$, we have

$$\begin{aligned} | ( \Gamma (\beta ) - 1) (\sin (2\beta ))^{\lambda } | \lesssim | (\cos ^{\alpha }(\beta ) - 1) (\cos (\beta ))^{\lambda } | \lesssim \alpha , \end{aligned}$$

and

$$\begin{aligned} \Big |c - \frac{2}{\pi } \Big | = \Big |\frac{2}{\pi } \int _0^{\pi /2} (\Gamma (\beta ) - 1) \sin (2\beta ) d\beta \Big | \le 2 \alpha . \end{aligned}$$

Proof

Using change of a variable $t = x^{\kappa }$, it suffices to show that for $t \in [0, 1], (1 - t ) t^{ \lambda / \kappa } \le \frac{\kappa }{\lambda }.$ Notice that $\lambda \ge 1/10$ and $t \le 1$. Using Young’s inequality, we derive

$$\begin{aligned} \begin{aligned}&(1 - t ) t^{ \lambda / \kappa } = \frac{\kappa }{\lambda } \cdot ( \frac{ \lambda }{\kappa }(1 -t) ) t^{\lambda / \kappa } \le \frac{\kappa }{\lambda } \left( \frac{ \frac{ \lambda }{\kappa }(1 -t) + \frac{\lambda }{\kappa } t }{ 1 + \frac{\lambda }{\kappa }} \right) ^{1 + \lambda / \kappa } = \frac{\kappa }{\lambda } \left( \frac{ \lambda }{\lambda +\kappa } \right) ^{1 +\lambda /\kappa } \le \frac{\kappa }{\lambda } , \end{aligned} \end{aligned}$$

which implies (A.1). The remaining inequalities in the Lemma follows directly from (A.1). $\quad \square$

1.2 Computations in the linear stability analysis

We perform the derivations and establish several inequalities in the linear stability analysis in Sect. 5.6.

The calculations and estimates presented below can also be verified using Mathematica⁵ since we have simple and explicit formulas.

1.2.1 Derivations of (5.35)

Recall the formulas of $\psi _0, \varphi _0$ in (5.32). A direct calculation yields

$$\begin{aligned} \begin{aligned} \frac{1}{2}(R\varphi _0)_R - \varphi _0&= \Big (\frac{1}{2} \Big ( R \cdot \frac{ (1+R)^3}{R^3} \Big )_R - \frac{(1+R)^3}{R^3} \Big ) \sin (2\beta )\\&= \Big ( \frac{1}{2} \Big ( -2R^{-3} - 3 R^{-2} + 1 \Big ) - \frac{(1+R)^3}{R^3} \Big )\sin (2\beta ) \\&= - \Big ( 2R^{-3} + \frac{9}{2} R^{-2} + 3 R^{-1} + \frac{1}{2} \Big ) \sin (2\beta ). \end{aligned} \end{aligned}$$

Denote $\psi _0 = A(R) \Gamma (\beta )^{-1}$. For the coefficient in the $\eta$ integral in (5.35), we have

$$\begin{aligned}&\frac{1}{2} (R\psi _0 )_R + (- 2 + \frac{3}{1+R}) \psi _0 \\&\quad = \Big (\frac{1}{2} (R A(R))_R + (- 2 + \frac{3}{1+R}) A(R) \Big )\Gamma (\beta )^{-1} \triangleq ( I + II) \Gamma (\beta )^{-1}. \end{aligned}$$

Note that $A(R) =\frac{3}{16} \left( \frac{(1+R)^3}{R^4} + \frac{3}{2} \frac{(1+R)^4}{R^3} \right)$ (5.32). A direct calculation implies

$$\begin{aligned} \begin{aligned} I&=\frac{3}{32}\left( \frac{(1+R)^3}{R^3} + \frac{3}{2} \frac{(1+R)^4}{R^2} \right) _R\\&= \frac{3}{32}\left( 3 \frac{ (1+R)^2}{R^3} - 3\frac{(1 + R)^3}{R^4} + 6 \frac{(1+R)^3}{R^2} - 3 \frac{(1+R)^4}{R^3} \right) \\&= \frac{3}{32}\left( \frac{(1+R)^2}{R^4} ( 3 R - 3(1+R) + 6(1+R)R^2 - 3(1+R)^2 R ) \right) \\&= \frac{3(1+R)^2}{32R^4} ( -3 - 3 R + 3R^3 ) , \\ II&= \left( -2 + \frac{3}{1+R} \right) \frac{3}{32} \left( 2 \frac{(1+R)^3}{R^4} + 3 \frac{(1+R)^4}{R^3} \right) \\&= \frac{ 3 (1+R)^2}{ 32 R^4} ( -2 -2 R + 3 ) ( 2 + 3R(1+R)) , \\ I + II&= \frac{3 (1+R)^2}{32R^4} ( -3 - 3 R + 3R^3 + (1-2R)( 2 + 3R +3R^2) ) \\&= \frac{3(1+R)^2}{32R^4} ( -1 - 4 R -3R^2 -3 R^3 ). \end{aligned} \end{aligned}$$

The above calculations imply (5.35).

1.2.2 Derivations of (5.40)

From (4.8), we know

$$\begin{aligned} \frac{{\bar{\eta }} - R\partial _R {\bar{\eta }} }{{\bar{\eta }}} = \frac{(1+R)^3}{ 6R} \Big ( \frac{6R}{ (1+R)^3} - R\cdot \frac{6}{(1+R)^3} + R \cdot \frac{18 R}{ (1+R)^4} \Big ) = \frac{3R}{1+R}. \end{aligned}$$

Using the above identity, (5.32) and $c_{\omega } = -\frac{2}{\pi \alpha } L_{12}(\Omega )(0)$ (4.11), we can compute

$$\begin{aligned} \begin{aligned} ({\bar{\eta }} - R\partial _R {\bar{\eta }} ) \psi _0 c_{\omega }&= \frac{{\bar{\eta }} - R\partial _R {\bar{\eta }} }{{\bar{\eta }}} \frac{9}{8}\frac{\alpha }{c} \Big ( R^{-3} + \frac{3}{2} \frac{1+R}{R^2} \Big ) c_{\omega }\\&= \frac{27\alpha }{8c} \frac{R}{1+R} \Big ( R^{-3} + \frac{3}{2} \frac{1+R}{R^2} \Big ) c_{\omega } \\&=\Big ( \frac{27\alpha }{8c} \frac{1}{(1+R)R^2} + \frac{81 \alpha }{ 16 c} \frac{1}{R} \Big ) \cdot \frac{-2}{\pi \alpha } L_{12}(\Omega )(0) \\&= \Big (-\frac{27}{4\pi c} \frac{1}{(1+R)R^2} - \frac{81}{8\pi c} \frac{1}{R}\Big ) L_{12}(\Omega )(0), \end{aligned} \end{aligned}$$

which implies (5.40).

1.2.3 Derivation of the ODE (5.41) for $L_{12}(\Omega )(0)$

Multiplying $\sin (2\beta ) / R$ on both sides of (5.5) and then integrating (5.5), we derive

$$\begin{aligned} \begin{aligned} \frac{d}{dt} L_{12} (\Omega )(0) =&- \Big \langle R\partial _R \Omega , \frac{\sin (2\beta )}{R} \Big \rangle - L_{12}(\Omega )(0) + c_{\omega } \Big \langle {\bar{\Omega }} - R \partial _R {\bar{\Omega }}, \frac{\sin (2\beta )}{R} \Big \rangle \\&+ \Big \langle \eta , \frac{\sin (2\beta )}{R} \Big \rangle - \Big \langle \frac{3}{1+R} D_{\beta } \Omega ,\frac{\sin (2\beta )}{R} \Big \rangle + \Big \langle {{\mathcal {R}}}_{\Omega }, \frac{\sin (2\beta )}{R} \Big \rangle . \end{aligned} \end{aligned}$$

The first term vanishes by an integration by parts argument. Using (4.8) and (4.11), we can compute the third term

$$\begin{aligned} \begin{aligned} c_{\omega }\Big \langle {\bar{\Omega }} - R \partial _R {\bar{\Omega }}, \frac{\sin (2\beta )}{R} \Big \rangle&= \frac{\alpha }{c} c_{\omega } \int _0^{\infty } \int _0^{\pi /2}\Gamma (\beta ) \frac{6R^2}{(1+R)^3}\cdot \frac{\sin (2\beta )}{R} d \beta d R\\&=\frac{\pi \alpha }{2} c_{\omega } \int _0^{\infty } \frac{6R}{(1+R)^3} d R \\&= 3 \pi \alpha c_{\omega }\left( - (1+R)^{-1} + \frac{1}{2} (1+R)^{-2} \right) \Big |_0^{\infty } \\&= \frac{3\pi \alpha }{2} c_{\omega } = - 3 L_{12}(\Omega )(0) . \end{aligned} \end{aligned}$$

It follows that

$$\begin{aligned} \frac{d}{dt} L_{12} (\Omega )(0)= & {} - 4 L_{12} (\Omega )(0) + \Big \langle \eta , \frac{\sin (2\beta )}{R} \Big \rangle - \Big \langle \frac{3 \sin (2\beta )}{(1+R)R} , D_{\beta } \Omega \Big \rangle + \Big \langle {{\mathcal {R}}}_{\Omega }, \frac{\sin (2\beta )}{R} \Big \rangle . \end{aligned}$$

Multiplying $\frac{81}{ 4 \pi c} L_{12}(\Omega )(0)$ to the both sides, we derive (5.41).

1.2.4 Computations of the integrals in (5.43)

A simple calculation implies that for any $k > 2$

$$\begin{aligned} \int _0^{\infty } (1+R)^{-k} d R= & {} \frac{1}{k-1}, \quad \int _0^{\infty } \frac{R}{(1+R)^{k}} d R = \int _0^{\infty } \frac{1}{(1+R)^{k-1}} \nonumber \\&- \frac{1}{(1+R)^k} d R= \frac{1}{(k-1)(k-2)}. \end{aligned}$$

(A.2)

For the integral in $\beta$, we get

$$\begin{aligned} \int _0^{\pi /2} (1 - 2 \sin (2\beta ))^2 d \beta= & {} \frac{\pi }{2} - 4 \int _0^{\pi /2} \sin ( 2\beta ) d \beta + 4 \int _0^{\pi /2} ( \sin (2\beta ) )^2 d \beta = \frac{\pi }{2} \\&- 4 + 4 \cdot \frac{\pi }{4} = \frac{3\pi }{2} - 4. \end{aligned}$$

Using (A.2) with $k=4$ and the above calculation, we can compute

$$\begin{aligned}&\Big | \Big | \frac{R^{3/2}}{(1+R)^2} \frac{1}{R}(1 - 2 \sin (2\beta ))\Big |\Big |_2^2 \\&\quad = \int _0^{\infty } \frac{R}{(1+R)^4} dR \cdot \int _0^{\pi /2} (1 - 2 \sin (2\beta ))^2 d \beta = \frac{1}{6}(\frac{3\pi }{2} - 4 ). \end{aligned}$$

For $A_1$ in (5.40), we apply the Cauchy–Schwarz inequality directly to yield

$$\begin{aligned} A_1= & {} -\frac{27}{4\pi c} L_{12}(\Omega ) (0) \Big \langle \eta , \frac{1 }{(1+R)R^2} \Big \rangle \\\le & {} \frac{27}{ 4\pi c} | L_{12}(\Omega ) (0)| \Big \langle \eta ^2, \frac{(1+R)^3}{R^{4}} \Big \rangle ^{1/2} \Big \langle \frac{R^{4}}{(1+R)^3} , \frac{1}{(1+R)^2 R^4} \Big \rangle ^{1/2} . \end{aligned}$$

Using (A.2), we can calculate

$$\begin{aligned} \Big | \Big | \frac{R^2}{(1+R)^{3/2}} \cdot \frac{1}{(1+R)R^2}\Big |\Big |_2^2 = \int _0^{\pi /2} 1 d\beta \cdot \int _0^{\infty } (1+R)^{-5} dR = \frac{\pi }{8}. \end{aligned}$$

1.2.5 Estimates of $D(\Omega ), D(\eta )$ and the proof of (5.51)

We introduce

$$\begin{aligned} \begin{aligned} D_1(\eta )&\triangleq - \frac{3(1+R)^2}{ 32 R^4} ( 1 + 4 R + 3 R^2 + 3 R^3) , \\ D_2(\eta )&\triangleq \left( \frac{3}{16} R^{-3} + \frac{3}{8}\frac{ (1 + R)^2}{R^2} + \frac{3 R}{4(1+R)} \right) + \frac{3}{16} \left( \frac{1}{6}\frac{(1+R)^4}{R^3} + \frac{3}{8} \frac{(1+R)^3}{R^4} \right) . \end{aligned} \end{aligned}$$

Recall $D(\Omega ), D(\eta )$ in (5.50) and the weights $\varphi _0,\psi _0$ defined in (5.32). By definition, $D(\eta ) = D_1(\eta ) \Gamma (\beta )^{-1}+ D_2(\eta )$. Thus, (5.51) is equivalent to

$$\begin{aligned} \sin (2\beta ) D(\Omega ) \le -\frac{1}{6} \varphi _0, \quad D_1(\eta ) \Gamma (\beta )^{-1}+ D_2(\eta ) \le -\frac{1}{8} \psi _0. \end{aligned}$$

(A.3)

To prove the first inequality, it suffices to prove

$$\begin{aligned} D(\Omega ) = - 2 R^{-3} - \frac{9}{2} R^{-2} - 3 R^{-1} - \frac{1}{2} + \frac{4}{3} R^{-3} + 6 R^{-2} + \frac{1+R}{ 3 R} \le - \frac{(1+R)^3}{ 6 R^3}, \end{aligned}$$

which is equivalent to proving

$$\begin{aligned} (-2 + \frac{4}{3} + \frac{1}{6}) R^{-3} + (-\frac{9}{2} + 6+ \frac{1}{2}) R^{-2} + (-3 + \frac{1}{3} + \frac{1}{2}) R^{-1} + (-\frac{1}{2} + \frac{1}{3} + \frac{1}{6}) \le 0. \end{aligned}$$

It is further equivalent to

$$\begin{aligned} -\frac{1}{2} R^{-3} + 2 R^{-2} - \frac{13}{6} R^{-1} \le 0, \end{aligned}$$

which is valid since $2 \sqrt{ \frac{1}{2} \times \frac{13}{6}} > 2$. Hence, we prove the first inequality in (A.3).

For the second inequality in (A.3), firstly, we use $\Gamma (\beta ) D_2(\eta ) \le D_2(\eta )$ ($\Gamma (\beta ) = \cos ^{\alpha }(\beta )$ (4.8)) to obtain

$$\begin{aligned} \begin{aligned}&D_3(\eta ) \triangleq D_1 (\eta ) + D_2(\eta ) \Gamma (\beta ) \le D_1(\eta ) + D_2(\eta ) \\&= \frac{3}{16} \left\{ - \frac{(1+R)^2}{ 2 R^4} ( 1 + 4 R + 3 R^2 + 3 R^3) + R^{-3} + 2 \frac{ (1 + R)^2}{R^2} \right. \\&\left. + \frac{ 4 R}{1+R} +\frac{1}{6}\frac{(1+R)^4}{R^3} + \frac{3}{8} \frac{(1+R)^3}{R^4} \right\} . \end{aligned} \end{aligned}$$

(A.4)

Recall the definition of $\psi _0$ in (5.32). Multiplying both sides of the second inequality in (A.3) by $\Gamma (\beta )$, we obtain that the inequality is equivalent to

$$\begin{aligned} D_3(\eta ) \le \frac{3}{16} \left( - \frac{1}{8} \frac{(1+R)^3}{R^4} - \frac{3}{16} \frac{(1+R)^4}{R^3} \right) . \end{aligned}$$

(A.5)

We split the negative term in the upper bound of $D_3(\eta )$ in (A.4) as follows

$$\begin{aligned} \begin{aligned}&-\frac{(1+R)^2}{ 2 R^4} ( 1 + 4 R + 3 R^2 + 3 R^3)\\&\quad = - \frac{(1+R)^2}{2 R^4} \left\{ (1 +R) + (3 R^2) + R(1+R)^2 + R( 2 -2 R + 2R^2) \right\} \\&\quad = - \frac{ (1+R)^3}{ 2 R^4} - \frac{3}{2} \frac{(1+R)^2}{R^2} - \frac{(1+R)^4}{ 2 R^3} - \frac{ (1+R)^2(1-R + R^2)}{R^3} . \end{aligned} \end{aligned}$$

It follows that

$$\begin{aligned} \begin{aligned}&D_3(\eta ) \le \frac{3}{16}\left\{ \frac{ (1+R)^3}{ R^4} \left( -\frac{1}{2} + \frac{3}{8}\right) + \frac{(1+R)^4}{R^3} \left( -\frac{1}{2} + \frac{1}{6}\right) + \frac{1}{2} \frac{(1+R)^2}{R^2} - \frac{ (1+R)^2(1-R + R^2)}{R^3} \right. \\&\left. + \frac{1}{R^3} + \frac{4 R}{1+R} \right\} =\frac{3}{16} \left\{ -\frac{1}{8} \frac{(1+R)^3}{R^4} - \frac{1}{3} \frac{(1+R)^4}{R^3} + \frac{1}{2}\frac{(1+R)^2}{R^2}- \frac{(1+R)(1+R^3)}{R^3} + \frac{1}{R^3} + \frac{4R}{1+R} \right\} . \end{aligned} \end{aligned}$$

Observe that

$$\begin{aligned} \begin{aligned}&-\frac{1}{3}\frac{(1+R)^4}{R^3} + \frac{1}{2} \frac{(1+R)^2}{R^2} = -\frac{3}{16}\frac{(1+R)^4}{R^3} \\&\quad + \left( - \frac{7}{48} \frac{(1+R)^4}{R^3} + \frac{1}{2} \frac{(1+R)^2}{R^2} \right) \le -\frac{3}{16}\frac{(1+R)^4}{R^3} , \\&\quad - \frac{(1+R)(1+R^3)}{ R^3} + \frac{1}{R^3} + \frac{ 4 R}{1+R} = -\frac{1}{R^2} -(1+R) \\&\quad + \frac{4 R}{1+R} = -\frac{1}{R^2}-\frac{ (R-1)^2}{ (1+R)} \le 0 , \end{aligned} \end{aligned}$$

where we have used $\frac{7}{48} \frac{(1+R)^2}{R} \ge \frac{7}{48} \times 4 \ge 1/ 2$ to derive the first inequality. Therefore, we prove (A.5), which further implies the second inequality in (A.3).

1.3 Derivation of the singular term (7.5) in the elliptic estimates

Suppose that $\Psi$ is the solution of (7.1). Consider $\tilde{\Psi } = \Psi + G \sin (2\beta )$. Notice that if $\alpha = 0$, $\sin (2\beta )$ is the kernel of the operator ${{\mathcal {L}}}_{\alpha }$ in (7.1) (it is self-adjoint if $\alpha =0$). We have

$$\begin{aligned} {{\mathcal {L}}}_{\alpha }(\tilde{\Psi }) = \Omega + {{\mathcal {L}}}_{\alpha }(G \sin (2\beta )) = \Omega - (\alpha ^2 R^2 \partial _{RR} G + \alpha (\alpha + 4) R \partial _R G ) \sin (2\beta ). \end{aligned}$$

We look for G(R) that satisfies $G(R) \rightarrow 0$ as $R \rightarrow +\infty$ and ${{\mathcal {L}}}_{\alpha }(\tilde{\Psi })$ is orthogonal to $\sin (2\beta )$:

$$\begin{aligned} 0 = \int _0^{\pi /2} \sin (2\beta ) ( \Omega - (\alpha ^2 R^2 \partial _{RR} G + \alpha (\alpha + 4) R \partial _R G ) \sin (2\beta )) d \beta \end{aligned}$$

for every R, which implies

$$\begin{aligned} \alpha ^2 R^2 \partial _{RR} G + \alpha (\alpha +4) R \partial _R G = \frac{4}{\pi } \Omega _*, \end{aligned}$$

(A.6)

where $\Omega _*(R) = \int _0^{\pi /2} \Omega (R, \beta ) \sin (2\beta ) d \beta$ and we have used $\int _0^{\pi /2} \sin ^2(2\beta ) d \beta = \frac{\pi }{4}$. The above ODE is first order with respect to $\partial _R G$ and can be solved explicitly. Multiplying the integrating factor $\frac{1}{\alpha ^2} R^{-2 + \frac{4+\alpha }{\alpha }}$ to both sides and then integrating from 0 to R yield

$$\begin{aligned} R^{\frac{4+\alpha }{\alpha }} \partial _R G = \frac{4}{\alpha ^2 \pi } \int _0^{R} \Omega _*(t) t^{ \frac{4}{\alpha } -1} d t. \end{aligned}$$

Imposing the vanishing condition $G(R) \rightarrow 0$ as $R \rightarrow +\infty$, we yield

$$\begin{aligned} G = -\frac{4}{\alpha ^2 \pi } \int _R^{\infty } s^{ - \frac{4+\alpha }{\alpha }} \int _0^s \Omega _*(t) t^{ \frac{4}{\alpha } -1} d t ds. \end{aligned}$$

Using integration by parts, we further derive

$$\begin{aligned} G= & {} \frac{1}{\alpha \pi } \int _R^{\infty } \partial _s ( s^{ -\frac{4}{\alpha }}) \int _0^s \Omega _*(t) t^{ \frac{4}{\alpha } -1} d t ds = -\frac{1}{ \alpha \pi } \int _R^{\infty } \frac{\Omega _*(s)}{s} ds \\&- \frac{1}{\alpha \pi } R^{ -\frac{4}{\alpha }} \int _0^{R} \Omega _*(s) s^{\frac{4}{\alpha } - 1} ds. \end{aligned}$$

Using the above formula and the notation $L_{12}(\Omega )$ (2.16), we derive (7.5).

1.4 Estimates of $L_{12}(\Omega )$

Recall $\tilde{L}_{12}(\Omega ) = L_{12}(\Omega ) - L_{12}(\Omega )(0)$. We have the following important cancellation between $\tilde{L}_{12}(\Omega )$ and $\Omega$.

Lemma A.2

For $k \in [3/2,4]$ and any $\lambda > 0$, we have

$$\begin{aligned} \begin{aligned} \langle \sin (2\beta ) \Omega \tilde{L}_{12}(\Omega ), R^{-k} \rangle&= - \frac{k-1}{2} \Big | \Big | \tilde{L}_{12}(\Omega ) R^{-k/2} \Big | \Big |^2_{L^2{(R)}} ,\\ \langle (\sin (2\beta ) \Omega + \lambda \tilde{L}_{12}(\Omega ))^2 , R^{-k} \rangle&= \langle R^{-k} (\sin (2\beta ) )^2, \Omega ^2 \rangle - ( (k-1) \lambda \\&\quad - \frac{\pi }{2} \lambda ^2 ) \Big | \Big | \tilde{L}_{12}(\Omega ) R^{-k/2} \Big | \Big |^2_{L^2{(R)}} . \end{aligned} \end{aligned}$$

(A.7)

Proof

From the definition of $\tilde{L}_{12}(\omega )(R)$ in (5.8), we know that it does not depend on $\beta$ and

$$\begin{aligned} \int _0^{\pi / 2} \Omega (s, \beta ) \sin (2\beta ) d\beta = - (\partial _{R} \tilde{L}_{12}(R) ) R. \end{aligned}$$

Using integration by parts, we obtain

$$\begin{aligned} \langle \sin (2\beta ) \Omega \tilde{L}_{12}(\Omega ), R^{-k} \rangle= & {} \int _0^{\infty } (- (\partial _{R} \tilde{L}_{12}(R) ) R ) \tilde{L}_{12}(\Omega ) R^{-k} d R \\= & {} - \frac{k-1}{2} \int _0^{\infty } \tilde{L}_{12}(\Omega )^2 R^{-k} d R, \end{aligned}$$

which is exactly the first identity in (A.7). The second identity in (A.7) is a direct consequence of $\langle \tilde{L}^2_{12}(\Omega ), R^{-k}\rangle = \frac{\pi }{2} || \tilde{L}_{12}(\Omega ) R^{-k/2} ||^2_{L^2{(R)}}$ and the first identity. $\quad \square$

To estimate $\tilde{L}_{12}(\Omega ) g$ in ${{\mathcal {L}}}_i$, we use the following simple Lemma.

Lemma A.3

Let g be some function depending on ${\bar{\Omega }}, {\bar{\eta }}, {\bar{\xi }}$ and $\varphi$ be some weights. We have

$$\begin{aligned} \begin{aligned} \langle \tilde{L}^2_{12} (\Omega ) g^2 , \varphi \rangle&\lesssim ||R^{-1} \tilde{L}_{12} (\Omega ) ||^2_{L^2(R)} \Big | \Big | \int _0^{\pi /2} R^2 g^2(R, \beta ) \varphi (R, \beta ) d \beta \Big |\Big |_{L^{\infty }(R)} , \\ \langle ( D^k_R \tilde{L}_{12} (\Omega ) )^2 g^2 , \varphi \rangle&\lesssim || R^{-1} D^{k-1}_R\Omega ||^2_{L^2} \Big | \Big | \int _0^{\pi /2} R^2 g^2(R, \beta ) \varphi (R, \beta ) d \beta \Big |\Big |_{L^{\infty }(R)} , \\ \end{aligned}\nonumber \\ \end{aligned}$$

(A.8)

for $k\ge 1$, provided that the upper bound is well-defined, where $D_R = R \partial _R$.

Proof

The first inequality follows directly from that $\tilde{L}_{12}(\Omega )$ does not dependent on $\beta$. Recall the definition of $\tilde{L}_{12}(\Omega )$ in (5.8) and $D_R = R \partial _R$. Notice that for $k \ge 1$, we have

$$\begin{aligned} \begin{aligned} D_R^k \tilde{L}_{12}(\Omega )&= - \int _0^{\pi / 2} D^{k-1}_R\Omega (R, \beta ) \sin (2\beta ) d\beta . \end{aligned} \end{aligned}$$

Using the Cauchy–Schwarz inequality, we prove

$$\begin{aligned} \begin{aligned}&\langle ( D^k_R \tilde{L}_{12} (\Omega ) )^2 g^2 , \varphi \rangle = \int _0^{\infty } \left( ( \int _0^{\pi / 2} D_R^{k-1}\Omega (R, \beta ) \sin (2\beta ) d\beta )^2 \int _0^{\pi /2} g^2 \varphi d \beta \right) d R \\&\quad \lesssim \int _0^{\infty } ( \int _0^{\pi /2} (D_R^{k-1}\Omega )^2 d \beta ) ( \int _0^{\pi /2} g^2 \varphi d \beta ) d R \\&\quad \le || R^{-1} D_R^{k-1} \Omega ||^2_{L^2} \Big | \Big | \int _0^{\pi /2} R^2 g^2 \varphi (R, \beta ) d \beta \Big |\Big |_{L^{\infty }(R)}. \end{aligned} \end{aligned}$$

$\square$

Lemma A.4

Let $\chi (\cdot ) : [0, \infty ) \rightarrow [0, 1]$ be a smooth cutoff function, such that $\chi (R) = 1$ for $R \le 1$ and $\chi (R) = 0$ for $R \ge 2$. For $k=1,2$, we have

$$\begin{aligned} \begin{aligned}&|| L_{12}(\Omega ) ||_{L^{\infty }}\lesssim || \frac{1+R}{R} \Omega ||_{L^2}, \quad || \tilde{L}_{12}(\Omega ) (R^{-2} + R^{-3})^{1/2} ||^2_{L^2(R)} \lesssim || \Omega \frac{(1+R)^2}{R^2} ||^2_{L^2} , \\&|| L_{12}(\Omega )||_2 \lesssim || \Omega ||_2, \quad || \frac{(1+R)^k}{R^k} ( L_{12}(\Omega ) - L_{12}(\Omega )(0) \chi ) ||_{L^2(R)} \lesssim || \frac{(1+R)^k}{R^k} \Omega ||_{L^2}. \end{aligned}\nonumber \\ \end{aligned}$$

(A.9)

provided that the right hand side is bounded. Moreover, if $\Omega \in {{\mathcal {H}}}^3$, then for $0 \le k \le 3, 0\le l \le 2$, we have

$$\begin{aligned} \begin{aligned}&|| L_{12}(\Omega ) - L_{12}(\Omega )(0) \chi ||_{{{\mathcal {H}}}^3} + || D_R( L_{12}(\Omega ) - L_{12}(\Omega )(0) \chi ) ||_{{{\mathcal {H}}}^3} \lesssim || \Omega ||_{{{\mathcal {H}}}^3}, \\&|| D^k_R L_{12}(\Omega ) ||_{\infty } + || D^k_R ( L_{12}(\Omega ) -\chi L_{12}(\Omega )(0)) ||_{\infty } \lesssim || \Omega ||_{{{\mathcal {H}}}^3}, \\&|| (1 +R) \partial _R D^l_R L_{12}(\Omega ) ||_{\infty } + || (1+R) \partial _R D^l_R ( L_{12}(\Omega ) -\chi L_{12}(\Omega )(0)) ||_{\infty } \lesssim || \Omega ||_{{{\mathcal {H}}}^3}, \\&|| L_{12}(\Omega ) ||_{X} + || D_R L_{12}(\Omega ) ||_X \lesssim || \Omega ||_{{{\mathcal {H}}}^3}, \end{aligned}\nonumber \\ \end{aligned}$$

(A.10)

where $X \triangleq {{\mathcal {H}}}^3 \oplus {{\mathcal {W}}}^{5,\infty }$ is defined in (7.7).

Remark A.5

We subtract $\chi L_{12}(\Omega )(0)$ near $R=0$ since $L_{12}(\Omega )$ does not vanishes at $R=0$.

Proof

Recall $L_{12}(\Omega )$ in (2.16) and $\tilde{L}_{12}(\Omega )$ in (5.8). Using the Cauchy–Schwarz and the Hardy inequality, we get

$$\begin{aligned} \begin{aligned} || L_{12}(\Omega )||_{L^{\infty }}&\lesssim \langle |\Omega |, \frac{1}{R} \rangle \lesssim || \frac{1+R}{R} \Omega ||_{L^2} || \frac{1}{1+R}||_{L^2(R) } \lesssim || \frac{1+R}{R} \Omega ||_{L^2} , \\ || \frac{1}{R^l} \tilde{L}_{12}(\Omega ) ||_{L^2(R)}&\lesssim \int _0^{\infty } \frac{1}{R^{2l}} \tilde{L}^2_{12}(\Omega ) d R \lesssim \int _0^{\infty } \frac{1}{R^{2l-2}} (\partial _R \tilde{L}_{12}(\Omega ) )^2 d R \lesssim \langle \Omega ^2, R^{-2l} \rangle , \end{aligned}\nonumber \\ \end{aligned}$$

(A.11)

for $l = 1,\frac{3}{2},2$, which implies the first two inequalities in (A.9). For $k=1,2$, observe that

$$\begin{aligned} \begin{aligned}&|| \frac{(1+R)^k}{R^k} ( L_{12}(\Omega ) - L_{12}(\Omega )(0) \chi ) ||_{L^2(R)} \lesssim || \frac{(1+R)^k}{R^k} \tilde{L}_{12}(\Omega ) \chi ||_{L^2(R)}\\&\quad + || \frac{(1+R)^k}{R^k} L_{12}(\Omega ) (1- \chi )||_{L^2(R)} \\&\quad \lesssim || \frac{1}{R^k} \tilde{L}_{12}(\Omega ) ||_{L^2(R)} + || L_{12}(\Omega ) ||_{L^2(R)} \lesssim || \Omega \frac{(1+R)^k}{R^k} ||_{L^2} + || L_{12}(\Omega ) ||_{L^2(R)} , \end{aligned} \end{aligned}$$

where we have used (A.11) in the last inequality. Denote $\Omega _* = \int _0^{\pi /2} \Omega d \beta$. From (2.16), we know

$$\begin{aligned} L_{12}(\Omega )(R) = \int _R^{\infty } \frac{\Omega _*(S)}{S} d S = \int _0^{\infty } K(R, S) \Omega _*(S) d S, \quad K(R, S) = \frac{1}{S} {\mathbf {1}}_{R \le S}. \end{aligned}$$

The $L^2$ boundedness of $L_{12}$ is standard. Notice that K is homogeneous of degree $-1$, i.e. $K(\lambda R, \lambda S) = \lambda ^{-1} K(R, S)$ for $\lambda >0$. Using change of a variable $S = R z$ , we get

$$\begin{aligned} L_{12}(\Omega )(R) = \int _0^{\infty } \frac{1}{R} K(1, z) \Omega _*( R z) R dz = \int _0^{\infty } K(1, z) \Omega _*( R z) dz. \end{aligned}$$

Then, the Minkowski inequality implies

$$\begin{aligned} || L_{12}(\Omega )||_{L^2}\le & {} \int _0^{\infty } K(1, z) || \Omega _*( R z)||_{L^2(R)} dz \lesssim \int _0^{\infty } K(1, z) z^{-1/2} ||\Omega ||_{L^2} dz \\= & {} || \Omega ||_{L^2}\int _{z \ge 1} z^{-3/2} dz \lesssim || \Omega ||_{L^2}. \end{aligned}$$

We complete the proof of (A.9). Notice that $D_R L_{12}(\Omega ) = - \Omega _* , || D_R^k \chi ||_{L^2} \lesssim 1$ for $1 \le k\le 4$ and $D_{\beta } L_{12}(\Omega ) = 0, D_{\beta } \chi = 0$. Using that $\sin (2\beta )^{-\sigma }$ in the weight $\varphi _1 = \sin (2\beta )^{-\sigma } \frac{(1+R)^4}{R^4}$ is integrable in the $\beta$ direction and (A.9), we yield

$$\begin{aligned} \begin{aligned}&|| (L_{12}(\Omega ) - L_{12}(\Omega )(0) \chi ) \varphi _1^{1/2} ||_{L^2} + || D_R^k (L_{12}(\Omega ) - L_{12}(\Omega )(0) \chi ) \varphi _1^{1/2} ||_{L^2} \\&\quad \lesssim || (L_{12}(\Omega ) - L_{12}(\Omega )(0) \chi ) \frac{(1+R)^2}{R^2} ||_{L^2} +|| D_R^k (L_{12}(\Omega ) - L_{12}(\Omega )(0) \chi )\frac{(1+R)^2}{R^2} ||_{L^2} \\&\quad \lesssim || \Omega \frac{(1+R)^2}{R^2}||_{L^2} + || D^{k-1}_R \Omega _* \frac{(1+R)^2}{R^2} ||_{L^2} + |L_{12}(\Omega )(0)| || D_R^k \chi \frac{(1+R)^2}{R^2}||_{L^2} \\&\quad \lesssim || \Omega \frac{(1+R)^2}{R^2}||_{L^2} + || D_R^{k-1} \Omega \frac{(1+R)^2}{R^2}||_{L^2} \lesssim || \Omega ||_{{{\mathcal {H}}}^3}, \end{aligned}\nonumber \\ \end{aligned}$$

(A.12)

which implies the first estimate in (A.10). From the definition of $L_{12}(\Omega )$ in (2.16), we have $D_R L_{12}(\Omega )= L_{12}(D_R \Omega )$. Notice that $| D_R^k \chi (R) | \lesssim 1$. Using (A.9), we prove for $k \le 3$

$$\begin{aligned} || D_R^k L_{12}(\Omega )||_{L^{\infty }} + | L_{12}(\Omega )(0)| \cdot || D_R^k \chi ||_{L^{\infty }} \lesssim || \Omega ||_{{{\mathcal {H}}}^3}, \end{aligned}$$

which implies the second estimate in (A.10). Similarly, since $\partial _R D_R^l L_{12}(\Omega ) = \partial _R L_{12}(D_R^l \Omega ) = - R^{-1} D_R^l \Omega _*(R)$, where $\Omega _*(R) = \int _0^{\pi /2} \Omega (R, \beta ) d \beta$, and that $l \le 2$, we have

$$\begin{aligned} || \partial _R D_R^l L_{12}(\Omega ) ||_{L^{\infty }}= & {} || R^{-1} D_R^l \Omega _*||_{L^{\infty }(R)} \lesssim || R^{-1} D_R^l \Omega _* ||^{1/2}_{L^2(R) } || \partial _R (R^{-1} D_R^l \Omega _* ) ||^{1/2}_{L^2(R) } \\\lesssim & {} || \Omega ||_{{{\mathcal {H}}}^3}, \end{aligned}$$

which along with the second estimate in (A.10) and $| \partial _R D_R^l \chi L_{12}(\Omega )(0) | \lesssim | L_{12}(\Omega )(0) | \lesssim || \Omega ||_{{{\mathcal {H}}}^3}$ completes the proof of the third estimate in (A.10).

Since $\chi L_{12}(\Omega )(0)$ does not depend on $\beta$, we apply the first two estimates in (A.10) to yield

$$\begin{aligned} \begin{aligned} || D_R^i L_{12}(\Omega ) ||_X&\le || D_R^i ( L_{12}(\Omega ) - \chi L_{12}(\Omega )(0)) ||_{{{\mathcal {H}}}^3} + || D_R^i \chi L_{12}(\Omega )(0) ||_{{{\mathcal {W}}}^{5,\infty }} \\&\lesssim || \Omega ||_{{{\mathcal {H}}}^3} + | L_{12}(\Omega )(0) | \lesssim || \Omega ||_{{{\mathcal {H}}}^3} \end{aligned} \end{aligned}$$

for $i= 0, 1$. We complete the proof of (A.10). $\quad \square$

1.5 Estimate of the approximate self-similar solution

In appendix A.5.1, we estimate some norm of ${\bar{\Omega }}, {\bar{\eta }}$ using the explicit formulas. For ${\bar{\xi }}$, it is given by an integration of ${\bar{\eta }}$ that does not have an explicit formula. We estimates ${\bar{\xi }}$, its derivatives and some norm in Sect. A.5.2.

1.5.1 Estimate of ${\bar{\Omega }}, {\bar{\eta }}$

Recall the formula of ${\bar{\Omega }}, {\bar{\eta }}$ in (4.8). A simple calculation yields

$$\begin{aligned} {\bar{\Omega }} = \frac{\alpha }{c} \frac{3 R \Gamma (\beta ) }{(1+R)^2}, \ {\bar{\eta }} = \frac{\alpha }{c} \frac{6 R \Gamma (\beta ) }{(1+R)^3}, \ {\bar{\Omega }} - D_R {\bar{\Omega }} = \frac{\alpha }{c} \frac{6R^2 \Gamma (\beta ) }{(1+R)^3}, \ {\bar{\eta }} - D_R {\bar{\eta }} = \frac{\alpha }{c} \frac{18R^2 \Gamma (\beta ) }{(1+R)^4} .\nonumber \\ \end{aligned}$$

(A.13)

Without specification, in later sections, we assume that $R \ge 0, \beta \in [0,\pi /2]$.

Lemma A.6

The following results apply to any $k \le 3, 0 \le i + j \le 3, j \ne 1$. (a) For $f = {\bar{\Omega }}, {\bar{\eta }}, {\bar{\Omega }} - D_R {\bar{\Omega }}, {\bar{\eta }} - D_R {\bar{\eta }}$, we have

$$\begin{aligned} | D_R^k f | \lesssim f , \quad |D_R^i D^j_{\beta } f | \lesssim \alpha \sin (\beta ) f. \end{aligned}$$

(A.14)

(b) Let $\varphi _i$ be the weights defined in (5.14). For $g = {\bar{\Omega }}, {\bar{\eta }}$, we have

$$\begin{aligned} \int _0^{\pi /2} R^2 (D_R^k g )^2 \varphi _1 d \beta \lesssim \alpha ^2, \quad \int _0^{\pi /2} R^2 (D_R^i D^j_{\beta } g )^2 \varphi _2 d \beta \lesssim \alpha ^3, \end{aligned}$$

(A.15)

uniformly in R and

$$\begin{aligned} \langle (D^k_R ( g - D_R g ) )^2 , \varphi _1 \rangle \lesssim \alpha ^2, \quad \langle (D^i_R D^j_{\beta } ( g - D_R g ) )^2 , \varphi _2 \rangle \lesssim \alpha ^3. \end{aligned}$$

(A.16)

Proof

Recall $D_{\beta }= \sin (2\beta ) \partial _{\beta }, D_R = R \partial _R$. Using $\Gamma (\beta )=\cos (\beta )^{\alpha }$, (5.22) and a direct calculation gives

$$\begin{aligned} | D^j_{\beta } \Gamma (\beta ) | \lesssim \alpha \sin (\beta ) \Gamma (\beta ), \quad | D_R^i \frac{R}{(1+R)^m} | \lesssim \frac{R}{ (1+R)^m}, \quad | D_R^i \frac{R^2}{(1+R)^m} | \lesssim \frac{R^2}{ (1+R)^m}.\nonumber \\ \end{aligned}$$

(A.17)

for $1 \le j \le 5$, $0\le i \le 5$ and $m = 2,3, 4$. Combining these estimates and the formulas in (A.13) implies (A.14). As a result, we have the following pointwise estimates for $g = {\bar{\Omega }}$ or ${\bar{\eta }}$

$$\begin{aligned} \begin{aligned} |D_R^k g|&\lesssim g \lesssim \alpha \Gamma (\beta ) \frac{R}{(1+R)^2}, \quad |D_R^i D^j_{\beta } g | \lesssim \alpha \sin (\beta ) g \lesssim \alpha ^2 \sin (\beta )\Gamma (\beta ) \frac{R}{(1+R)^2}, \\ |D_R^k (g - D_R g) |&\lesssim g - D_R g \lesssim \alpha \frac{R^2 \Gamma (\beta ) }{(1+R)^3} , |D_R^i D^j_{\beta } (g - D_R g) | \\&\lesssim \alpha \sin (\beta ) (g - D_R g) \lesssim \alpha ^2 \sin (\beta )\frac{R^2 \Gamma (\beta ) }{(1+R)^3}, \end{aligned} \end{aligned}$$

for $k \le 3$, $i+j \le 3, j \ne 0$, where we have used $c \approx \frac{2}{\pi }$ in Lemma A.1. Recall $\varphi _i$ in Definition 5.2.

$$\begin{aligned} \varphi _1 \triangleq (1+R)^4 R^{-4} \sin (2\beta )^{ - \sigma }, \quad \varphi _2 \triangleq (1+R)^4R^{-4} \sin (2\beta )^{ - \gamma } . \end{aligned}$$

Notice that for $\sigma = \frac{99}{100}, \gamma = 1 + \frac{\alpha }{10}$, we have

$$\begin{aligned}&\int _0^{\pi /2} \Gamma (\beta )^2 \sin (2\beta )^{-\sigma } d \beta \lesssim 1,\\&\int _0^{\pi /2} \alpha ^2 \sin (\beta )^2\Gamma (\beta )^2 \sin (2\beta )^{-\gamma } d \beta \lesssim \alpha ^2 \int _0^{\pi /2} \cos (\beta )^{2\alpha -1 - \alpha / 10} d \beta \lesssim \alpha . \end{aligned}$$

Combining the pointwise estimates, the estimates of the angular integral and a simple calculation then gives (A.15), (A.16). $\quad \square$

Recall the ${{\mathcal {W}}}^{l,\infty }$ norm in (7.6). We have

Proposition A.7

It holds true that $\Gamma (\beta ) ,{\bar{\Omega }}, {\bar{\eta }} \in {{\mathcal {W}}}^{7, \infty }$ with

$$\begin{aligned} \begin{aligned}&|| \Gamma (\beta )||_{{{\mathcal {W}}}^{7,\infty }} \lesssim 1, \quad ||\frac{(1+R)^2}{R} {\bar{\Omega }}||_{{{\mathcal {W}}}^{7,\infty }} + || \frac{(1+R)^2}{R}{\bar{\eta }} ||_{{{\mathcal {W}}}^{7,\infty }} \lesssim \alpha , \\&|| D_{\beta } {\bar{\Omega }}||_{{{\mathcal {W}}}^{7,\infty }} + || D_{\beta }{\bar{\eta }} ||_{{{\mathcal {W}}}^{7,\infty }} \lesssim \alpha ^2 . \end{aligned} \end{aligned}$$

Proof

The proof follows directly from the calculation A.17 and $\sin (\beta ) \Gamma (\beta ) \sin (2\beta )^{-\alpha /5} \lesssim 1$. $\quad \square$

1.5.2 Estimates of ${\bar{\xi }}$

Recall that the approximate self-similar profile ${\bar{\eta }}$ (4.8) is given by

$$\begin{aligned} \begin{aligned} ({\bar{\theta }}_x)(x, y)&= {\bar{\eta }}(R, \theta ) = \frac{\alpha }{c} \frac{6R}{(1+R)^3} \cos ^{\alpha }(\beta ) = \frac{ 6 \alpha }{c} \frac{ x^{\alpha }}{(1+(x^2 +y^2)^{\alpha /2})^3} . \end{aligned} \end{aligned}$$

(A.18)

We also use ${\bar{\eta }}(x, y)$ to denote the above expression. Throughout this section, we use the following notation

$$\begin{aligned} R = (x^2 + y^2)^{\alpha /2}, \quad \beta = \arctan (y/x), \quad S = (z^2 + y^2)^{\alpha /2} , \quad \tau = \arctan ( y /z),\nonumber \\ \end{aligned}$$

(A.19)

where z will be used in the integral. ${\bar{\theta }}(x, y)$, ${\bar{\xi }}(R, \theta ) = {\bar{\theta }}_y(x, y)$ can be obtained from ${\bar{\eta }}(x, y)$ (or ${\bar{\theta }}_x$) as follows

$$\begin{aligned} {\bar{\theta }} = \int _0^x {\bar{\eta }}(z, y) dz , \quad {\bar{\xi }} = {\bar{\theta }}_y = \int _0^x {\bar{\eta }}_y(z, y) dz, \end{aligned}$$

(A.20)

where we have used ${\bar{\theta }}(0, y) = 0$. Observe that

$$\begin{aligned} \begin{aligned} {\bar{\eta }}_y(z, y)&= - \frac{ 6 \alpha }{c} \cdot \frac{3\alpha y}{y^2 + z^2} \frac{ (z^2 +y^2)^{\alpha /2} z^{\alpha }}{(1+(z^2 +y^2)^{\alpha /2})^4} \\&= - \frac{1}{z} \frac{ 3 \alpha yz }{y^2 + z^2} \frac{ (z^2 +y^2)^{\alpha /2}}{ 1+(z^2 +y^2)^{\alpha /2} }{\bar{\eta }}(z, y) = -\frac{1}{z} \frac{ 3\alpha \sin (2 \tau ) S}{ 2 (1+S)} {\bar{\eta }} , \end{aligned}\nonumber \\ \end{aligned}$$

(A.21)

where we have used the notation $S, \tau$ defined in (A.19). Hence, we get

$$\begin{aligned} {\bar{\xi }}= \int _0^x - \frac{ 6 \alpha }{c} \cdot \frac{3\alpha y}{y^2 + z^2} \frac{ (z^2 +y^2)^{\alpha /2} z^{\alpha }}{(1+(z^2 +y^2)^{\alpha /2})^4} dz = \int _0^{x} \frac{1}{z} \left( - \frac{ 3\alpha \sin (2 \tau ) S}{2 (1+S)} {\bar{\eta }} \right) dz.\nonumber \\ \end{aligned}$$

(A.22)

These integrals cannot be calculated explicitly for general $\alpha$. We have the following estimates for ${\bar{\xi }}$.

Lemma A.8

Assume that $0\le \alpha \le \frac{1}{1000}$. For $R \ge 0, \beta \in [0, \pi /2]$ and $0 \le i + j \le 5$, we have

$$\begin{aligned}&| D^i_R D^j_{\beta } {\bar{\xi }} | \lesssim - {\bar{\xi }}, \quad | D^i_R D^j_{\beta } (3{\bar{\xi }} - R\partial _R {\bar{\xi }}) | \lesssim -{\bar{\xi }}, \end{aligned}$$

(A.23)

$$\begin{aligned}&|{\bar{\xi }} | \lesssim \frac{ \alpha ^2(x^2 + y^2)^{\alpha /2}}{ (1 + (x^2 + y^2)^{\alpha /2}) } \frac{y^{\alpha }}{( 1 + y^{\alpha })^3} \min \left( 1 , \frac{x^{1+\alpha }}{y^{1+\alpha }} \right) \nonumber \\&\quad \lesssim \frac{\alpha ^2R^2}{1+R} \left( {\mathbf {1}}_{\beta < \pi /4} \frac{ \sin ^{\alpha }(\beta )}{ (1+R \sin ^{\alpha }(\beta ) )^3} + {\mathbf {1}}_{\beta \ge \pi /4} \frac{\cos ^{\alpha +1}(\beta )}{(1+R)^3} \right) , \nonumber \\&-{\bar{\xi }} \lesssim \alpha ^2 \cos (\beta ) , \quad || {\bar{\xi }} ||_{{{\mathcal {C}}}^1} \lesssim || \frac{1+R}{R} ( 1 + ( R \sin ( 2 \beta )^{\alpha } )^{-\frac{1}{40} } ) {\bar{\xi }} ||_{L^{\infty }} \lesssim \alpha ^2, \end{aligned}$$

(A.24)

where $||\cdot ||_{{{\mathcal {C}}}^1}$ is defined in (6.5). Let $\psi _1, \psi _2$ be the weights defined in (5.14). We have

$$\begin{aligned} \int _0^{\pi /2} R^2 (D^i_R D^j_{\beta } {\bar{\xi }} )^2 \psi _k d \beta \lesssim \alpha ^4 \end{aligned}$$

(A.25)

uniformly in R, and

$$\begin{aligned} \langle (D^i_R D^j_{\beta } ( 3{\bar{\xi }} - R \partial _R {\bar{\xi }} ) )^2 , \psi _k \rangle \lesssim \alpha ^4, \quad \langle (D^i_R D^j_{\beta } {\bar{\xi }})^2 , \psi _k \rangle \lesssim \langle {\bar{\xi }}^2 , \psi _k \rangle \lesssim \alpha ^4, \end{aligned}$$

(A.26)

where $(D^i_R D^j_{\beta }, \psi _k )$ represents $( D^i_R, \psi _1)$ for $0\le i \le 5$, and $(D^i_R D^j_{\beta }, \psi _2 )$ for $i+j \le 5, j \ge 1$.

Remark A.9

Using (A.22), we have $-{\bar{\xi }} \ge 0$ for $R \ge 0, \beta \in [0, \pi /2]$.

We have several commutator estimates which enable us to exchange the derivative and integration in (A.22) so that we can estimate $D_R^i D^j_{\beta }{\bar{\xi }}$ easily.

Recall the relation between $\partial _x, \partial _y$ and $\partial _R, \partial _{\beta }$ in (2.9). We have the following relation

$$\begin{aligned} D_R = R \partial _R = \frac{1}{\alpha } (x \partial _x + y \partial _y), \quad D_{\beta } = \sin (2\beta ) \partial _{\beta } = 2 y \partial _y - 2 \alpha \sin ^2 (\beta ) D_R .\nonumber \\ \end{aligned}$$

(A.27)

The first relation holds because $R = r^{\alpha }, R\partial _R = \frac{1}{\alpha } r\partial _r$, and the second relation is obtained by multiplying $\partial _y = \frac{\sin (\beta )}{r} \alpha D_R + \frac{\cos (\beta )}{ r} \partial _{\beta }$ by y and then using $y/r = \sin (\beta ) , x / r = \cos (\beta )$.

Lemma A.10

Suppose that $f(0, y) =0$ for any y. Denote

$$\begin{aligned} I(f)(x ,y) = \int _0^x \frac{1}{z} f(z, y) dz. \end{aligned}$$

(A.28)

We have

$$\begin{aligned} D_R I(f)(x, y)&= I( D_S f)(x,y) , \end{aligned}$$

(A.29)

$$\begin{aligned} D_{\beta } I(f)(x, y) - I( D_{\tau } f)(x,y)&= -2 \alpha \sin ^2 (\beta ) \cdot I( D_S f) + 2 \alpha I( \sin ^2(\tau ) D_S f) , \end{aligned}$$

(A.30)

where $R, \beta , S , \tau$ are defined in (A.19), provided that f is sufficiently smooth.

Proof

Notice that $y \partial _y$ commutes with the z integral. From (A.27), it suffices to prove

$$\begin{aligned} x \partial _x I(f)(x, y) = I( z\partial _z f ). \end{aligned}$$

A directly calculation yields

$$\begin{aligned} \begin{aligned} x \partial _x I(f)(x,y)&= x \partial _x ( \int _0^x \frac{1}{z} f(z, y) dz) = f (x, y) , \quad \\ I(z \partial _z f)(x,y)&= \int _0^x \frac{1}{z} \cdot z \partial _z f(z, y) dz = f(x, y). \end{aligned} \end{aligned}$$

It follows (A.29). Using the fact that both $y \partial _y$ and $R\partial _R$ commute with the z integral and the formula of $D_{\beta }$ (A.27) twice, we derive

$$\begin{aligned} \begin{aligned}&D_{\beta } I(f)(x,y) = (2 y \partial _y - 2\alpha \sin ^2(\beta ) D_R) I(f) = I( 2y\partial _y f) - 2 \alpha \sin ^2(\beta ) I(D_S f) \\&\quad = I( D_{\tau } f + 2\alpha \sin ^2(\tau ) D_S f) - 2 \alpha \sin ^2(\beta ) I(D_S f) = I(D_{\tau } f) \\&\qquad + 2\alpha I( \sin ^2(\tau ) D_S f)- 2 \alpha \sin ^2(\beta ) I(D_S f) . \end{aligned} \end{aligned}$$

(A.30) follows by rearranging the above identity. $\quad \square$

Next, we prove Lemma A.8.

Proof of Lemma A.8

Step 1. Recall $D_R = R\partial _R, D_{\beta } = \sin (2\beta ) \partial _{\beta }$. First, we show that

$$\begin{aligned} | D_R^i D^j_{\beta } {\bar{\xi }} | \lesssim \alpha \int _0^{x} \frac{1}{z} \sin (2\tau ) \frac{S}{1+S}{\bar{\eta }}(z, y) dz \asymp -{\bar{\xi }} \end{aligned}$$

(A.31)

for $0\le i+ j \le 5$. Using $\Gamma (\beta ) =\cos (\beta )^{\alpha }$,(5.22) and a direct calculation yields

$$\begin{aligned} \Big | D^i_R \frac{R^2}{(1+R)^4} \Big | \lesssim \frac{R^2}{(1+R)^4} , \quad |D^i_{\beta }\Gamma (\beta )| \lesssim \alpha \sin (\beta ) \Gamma (\beta ), \quad |D^i_{\beta }\sin (2\beta )| \lesssim \sin (2\beta )\nonumber \\ \end{aligned}$$

(A.32)

for $i \le 5$. Denote

$$\begin{aligned} f(S, \tau ) = \frac{3\alpha }{2} \sin (2\tau ) \frac{S}{1+S} {\bar{\eta }} = \frac{ 9 \alpha ^2}{c} \sin (2\tau ) \Gamma (\tau ) \frac{S^2}{(1+S)^4}. \end{aligned}$$

(A.33)

We remark that $f = - z {\bar{\eta }}_y(z, y)$ according to (A.21). Obviously, $f(S, \tau ) \ge 0$. Using the above estimates, we get

$$\begin{aligned} | D^i_S D^j_{\tau } f| \lesssim f \end{aligned}$$

(A.34)

for $i+j \le 5$. Notice that (A.22) implies ${\bar{\xi }} = - I(f)$ and that $I(\cdot )$ (A.28) is a positive linear operator for $x \ge 0$. We further derive

$$\begin{aligned} | I( D^i_S D^j_{\tau }f|) | \le I( |D^i_S D^j_{\tau }f |) \lesssim I(f ) \end{aligned}$$

(A.35)

for $i + j \le 5$. Using (A.29) and the above estimates, we yield

$$\begin{aligned} |D_R^i {\bar{\xi }} | = | D_R^i I(f) | = | I(D_S^i f) | \lesssim I(f). \end{aligned}$$

For other derivatives $D_R^i D^j_{\beta }$ with $j \ge 1, i+ j \le 5$, we estimate $D^2_{\beta } {\bar{\xi }}$, which is representative. Using (A.30), we have

$$\begin{aligned} \begin{aligned} D^2_{\beta } {\bar{\xi }} =&D^2_{\beta } I(f) = D_{\beta } \left( I(D_{\tau } f) -2\alpha \sin ^2(\beta ) \cdot I(D_S f) + 2\alpha I( D_S f \sin ^2(\tau ) ) \right) \\ =&I(D^2_{\tau } f) -2\alpha \sin ^2(\beta ) \cdot I(D_S D_{\tau } (f)) + 2\alpha I( \sin ^2(\tau ) D_S D_{\tau } f ) \\&+ D_{\beta } \left( -2\alpha \sin ^2(\beta ) \cdot I(D_S f) \right) + D_{\beta }\left( 2\alpha I( D_S f \sin ^2(\tau ) ) \right) = J_1 + J_2 + J_3 + J_4 + J_5. \end{aligned} \end{aligned}$$

For $J_1, J_2, J_3$, we simply use $\sin ^2(\beta ), \sin ^2(\tau ) \le 1$ and (A.35) to obtain

$$\begin{aligned} I_1, J_2, J_3 \lesssim I( | D^i_R D^j_{\tau } f |) \lesssim I(f) \end{aligned}$$

(A.36)

for $(i,j) = (0,2), (1,1), (1,1)$ respectively. For $J_4$, if $D_{\beta }$ acts on $\sin ^2(\beta )$, we obtain $\alpha D_{\beta } (\sin ^2(\beta )) \cdot I(D_S f)$, which can be bounded as before using (A.35). For the remaining parts in $J_4$ and $J_5$, $D_{\beta }$ acts on $I(\cdot )$ and we can use (A.30) again to obtain several terms. Each term can be bounded using (A.35) and an argument similar to (A.36). The estimates of other derivatives $D_R^i D^j_{\beta }$ can be done similarly. We omit these estimates. Since the right hand side of (A.31) is $\frac{2}{3} I(f) = -\frac{2}{3}{\bar{\xi }} \asymp -{\bar{\xi }}$, the above estimates imply (A.31).

Step 2. The estimate (A.31) can be generalized to $i+j \le 6$ easily. Hence, we get

$$\begin{aligned} | D_R^i D^j_{\beta } (3 {\bar{\xi }} - R\partial _R {\bar{\xi }} ) | \lesssim |D_R^i D^j_{\beta } {\bar{\xi }}| + |D_R^{i+1} D^j_{\beta } {\bar{\xi }}| \lesssim -{\bar{\xi }}, \end{aligned}$$

for any $i+j\le 5$, which proves (A.23).

Step 3: Pointwise estimate. In this step, we prove (A.24). From (A.22), we know that the first inequality in (A.24) is equivalent to

$$\begin{aligned} \int _0^{x} \frac{y}{y^2 + z^2} \frac{ z^{\alpha } (y^2 + z^2)^{\alpha /2} }{(1 + (y^2 +z^2)^{\alpha /2})^4 } d z \lesssim \frac{ (x^2 + y^2)^{\alpha /2}}{ (1 + (x^2 + y^2)^{\alpha /2}) } \frac{y^{\alpha }}{( 1 + y^{\alpha })^3} \min \left( 1 , \frac{x^{1+\alpha }}{y^{1+\alpha }} \right) . \end{aligned}$$

For $z \in [0, x]$, we have $z^2 + y^2 \le x^2 + y^2$. Since $\frac{t}{1+t}$ is increasing with respect to $t \ge 0$, we yield

$$\begin{aligned} \frac{ (y^2 + z^2)^{\alpha /2} }{ 1 + (y^2 +z^2)^{\alpha /2} } \lesssim \frac{ (y^2 + x^2)^{\alpha /2} }{ 1 + (y^2 +x^2)^{\alpha /2} }. \end{aligned}$$

Therefore, it suffices to prove

$$\begin{aligned} J(x, y) \triangleq \int _0^{x} \frac{y}{y^2 + z^2} \frac{ z^{\alpha } }{(1 + (y^2 +z^2)^{\alpha /2})^3 } d z \lesssim \frac{y^{\alpha }}{( 1 + y^{\alpha })^3} \min \left( 1 , \frac{x^{1+\alpha }}{y^{1+\alpha }} \right) .\nonumber \\ \end{aligned}$$

(A.37)

Case 1 : $x \le 1 + y$ Observe that

$$\begin{aligned} J \le \frac{1}{ (1+y^{\alpha })^3 } \int _0^x \frac{y z^{\alpha }}{y^2 + z^2} dz =\frac{y^{\alpha }}{ (1+y^{\alpha })^3 } \int _0^{ \frac{x}{y}} \frac{ t^{\alpha } }{1+t^2} dt , \end{aligned}$$

where we have used change of a variable $z = y t$ to derive the identity. Since $\alpha \le 1/10$, we get

$$\begin{aligned} \int _0^{ \frac{x}{y}} \frac{ t^{\alpha } }{1+t^2} dt \le \int _0^{\infty } \frac{ t^{\alpha } }{1+t^2} dt \lesssim 1, \quad \int _0^{ \frac{x}{y}} \frac{ t^{\alpha } }{1+t^2} dt \le \int _0^{\frac{x}{y}} t^{\alpha } dt \lesssim \frac{x^{1+\alpha }}{y^{1+\alpha }}. \end{aligned}$$

Combining the above estimates, we prove (A.37) for $x \le 1 + y$.

Case 2 : $x > 1 + y$ Firstly, we have

$$\begin{aligned} J(x, y)= & {} \int _0^{1 + y} \frac{y}{y^2 + z^2} \frac{ z^{\alpha } }{(1 + (y^2 +z^2)^{\alpha /2})^3 } d z \\&+ \int _{1+y}^x \frac{y}{y^2 + z^2} \frac{ z^{\alpha } }{(1 + (y^2 +z^2)^{\alpha /2})^3 } d z \triangleq J_1 + J_2. \end{aligned}$$

We apply the result in Case 1 to estimate $J_1$

$$\begin{aligned} J(1 + y, y) \lesssim \frac{y^{\alpha }}{ (1+y^{\alpha })^3} \min \left( 1, \frac{(1+y)^{1+\alpha }}{y^{1+\alpha }} \right) \lesssim \frac{y^{\alpha }}{ (1+y^{\alpha })^3} . \end{aligned}$$

For $J_2$, we have

$$\begin{aligned} \begin{aligned} J_2&\le \int _{1+y}^x \frac{y}{y^2 + z^2} \frac{z^{\alpha }}{z^{3\alpha }} dz = y^{- 2\alpha } \int _{ \frac{1+y}{y}}^{\frac{x}{y}} \frac{t^{-2\alpha }}{1+t^2} dt \lesssim y^{-2\alpha } \int _{ \frac{1+y}{y} }^{\infty } t^{-2\alpha -2 } dt \\&\lesssim y^{-2\alpha } \left( \frac{ 1+y}{y} \right) ^{ - 1 - 2\alpha } = \frac{y}{(1+y)^{1+2\alpha }} = \frac{y^{\alpha }}{ (1+y)^{3\alpha }} \frac{ y^{1-\alpha }}{ (1+y)^{1-\alpha }} \lesssim \frac{y^{\alpha }}{ (1 +y^{\alpha })^3} , \end{aligned} \end{aligned}$$

where we have used change of a variable $z = y t$ to derive the first identity. Noting that $x \ge y$ in this case. We conclude

$$\begin{aligned} J(x, y) = J_1 + J_2 \lesssim \frac{y^{\alpha }}{ (1 + y^{\alpha })^3} \le \frac{y^{\alpha }}{ (1+y^{\alpha })^3} \min \left( 1, \frac{{x}^{1+\alpha }}{y^{1+\alpha }} \right) . \end{aligned}$$

Combining the above two cases, we prove (A.37), which implies the first inequality in (A.24).

Finally, we prove the second inequality in (A.24). Using the notation (A.19), we have

$$\begin{aligned} R= & {} (x^2 + y^2)^{\alpha /2}, \quad \frac{(x^2 + y^2)^{\alpha /2}}{1 +(x^2 + y^2)^{\alpha /2} } = \frac{R}{1+R}, \quad y^{\alpha } = R\sin ^{\alpha }(\beta ), \\ \frac{y^{\alpha }}{(1+y^{\alpha })^3}= & {} \frac{ R\sin ^{\alpha }(\beta )}{ (1 + R\sin ^{\alpha }(\beta ))^3}. \end{aligned}$$

For $x \le y$, we have $\beta \ge \pi / 4, \ 1 \lesssim \sin (\beta ), \ x^2 + y^2 \lesssim y^2$. Hence,

$$\begin{aligned} \frac{y^{\alpha }}{ (1+y^{\alpha })^3} \frac{x^{1+\alpha }}{y^{1+\alpha }} \lesssim \frac{y^{\alpha }}{ (1 + (x^2 + y^2)^{\alpha /2} )^3 } \frac{x^{1+\alpha }}{y^{1+\alpha }} = \frac{R\sin ^{\alpha }(\beta ) }{ (1 +R)^3} \cdot \frac{ \cos ^{1+\alpha }(\beta )}{ \sin ^{1+\alpha }(\beta ) } \lesssim \frac{R\cos ^{1+\alpha }(\beta ) }{ (1 +R)^3} . \end{aligned}$$

Combining the above identity and the estimate, we prove the second inequality in (A.24). The last inequality in (A.24) follows directly from (A.23) and the first two inequalities in (A.24).

Step 4: Estimates of the integral Now, we are in a position to prove (A.25) and (A.26). We are going to prove

$$\begin{aligned} \int _0^{\pi /2} {\bar{\xi }}^2(R, \beta ) \psi _k d \beta \lesssim \frac{\alpha ^4}{(1+R)^2}. \end{aligned}$$

(A.38)

Clearly, (A.25) and (A.26) follow from the above estimate and (A.23).

Notice that $\psi _i$ defined in (5.14) satisfies

$$\begin{aligned} \psi _1, \psi _2 \lesssim \frac{(1+R)^4}{R^4} \sin (\beta )^{-\sigma } \cos (\beta )^{-\gamma }, \end{aligned}$$

(A.39)

where $\gamma = 1 + \frac{\alpha }{10}, \sigma = \frac{99}{100}$. Using (A.24), $1 + R\sin ^{\alpha }(\beta ) \ge (1+R) \sin ^{\alpha }(\beta )$, we yield

$$\begin{aligned} \begin{aligned}&(1+R)^2 \int _0^{\pi /2} | {\bar{\xi }} |^2 \psi _k d \beta \lesssim (1+R)^2 \frac{ \alpha ^4 R^4}{(1+R)^2}\cdot \left\{ \int _0^{\pi /4} \frac{ \sin ^{ 2\alpha }(\beta )}{ ( ( 1+ R) \sin ^{\alpha }(\beta ) )^6} \psi _k d\beta \right. \\&\qquad \left. + \int _{\pi /4}^{\pi /2} \frac{ \cos ^{2\alpha + 2}}{(1+R)^6} \psi _k d \beta \right\} \\&\quad \lesssim \frac{\alpha ^4 R^4}{(1+R)^6} \frac{(1+R)^4}{R^4} \left\{ \int _0^{\pi /4} \sin (\beta )^{-4\alpha } \sin (\beta )^{-\sigma } \cos (\beta )^{-\gamma } \right. \\&\qquad \left. +\int _{\pi /4}^{\pi /2} \cos (\beta )^{2 + 2\alpha } \sin (\beta )^{-\sigma } \cos (\beta )^{-\gamma } d \beta \right\} \\&\quad \lesssim \alpha ^4 \left( \int _0^{\pi /4} \sin (\beta )^{-\sigma - 4\alpha } d\beta +\int _{\pi /4}^{\pi /2} \cos (\beta )^{2 + 2\alpha - \gamma } d \beta \right) \lesssim \alpha ^4, \end{aligned} \end{aligned}$$

where we have used $\alpha \le \frac{1}{1000}$, $4 \alpha + \sigma < \frac{199}{200}$, $2 + 2 \alpha - \gamma \ge 1$, to derive the last inequality which does not depend on $\alpha$ for $\alpha \le \frac{1}{1000}$. It follows (A.38). $\quad \square$

1.6 Other Lemmas

We use the following Lemma to construct small perturbation.

Lemma A.11

Let $\chi (\cdot ) : [0, \infty ) \rightarrow [0, 1]$ be a smooth cutoff function, such that $\chi (R) = 1$ for $R \le 1$ and $\chi (R) = 0$ for $R \ge 2$. Denote

$$\begin{aligned} \chi _{\lambda }(R) = \chi (R / \lambda ) , \quad {\bar{\Omega }}_{\lambda } = \chi _{\lambda } {\bar{\Omega }}, \quad {\bar{\eta }}_{\lambda } = \partial _x ( \chi _{\lambda } {\bar{\theta }}) , \quad {\bar{\xi }}_{\lambda } =\partial _{y} (\chi _{\lambda } {\bar{\theta }} ), \end{aligned}$$

(A.40)

where ${\bar{\theta }}$ is obtained in (A.20). We have

$$\begin{aligned}&\lim _{\lambda \rightarrow +\infty } || {\bar{\Omega }}_{\lambda } - {\bar{\Omega }}||_{{{\mathcal {H}}}^3} + || (1+R) ( {\bar{\eta }}_{\lambda } -{\bar{\eta }} )||_{{{\mathcal {H}}}^3} + || {\bar{\xi }}_{\lambda } -{\bar{\xi }}||_{{{\mathcal {H}}}^3(\psi )}\nonumber \\&\quad = 0, \quad {\overline{\lim }}_{\lambda \rightarrow +\infty } || {\bar{\xi }}_{\lambda } -{\bar{\xi }} ||_{{{\mathcal {C}}}^1} \le K_{10} \alpha ^{ 2}, \end{aligned}$$

(A.41)

where $K_{10} >0$ is some absolute constant. In particular, we also have

$$\begin{aligned} \lim _{\lambda \rightarrow +\infty } L^2_{12}({\bar{\Omega }}_{\lambda } - {\bar{\Omega }})(0) + \langle ({\bar{\Omega }}_{\lambda } - {\bar{\Omega }})^2, \varphi _0 \rangle + \langle ({\bar{\eta }}_{\lambda } - {\bar{\eta }} )^2, \psi _0 \rangle = 0. \end{aligned}$$

(A.42)

We need a Lemma similar to Lemma A.10.

Lemma A.12

Suppose that $f(0, y) =0$ for any y. Denote $J(f)(x ,y) = \frac{1}{z} \int _0^x f(z, y) dz.$ We have

$$\begin{aligned} \begin{aligned} D_R J(f)(x, y)&= J( D_S f)(x,y) , \\ D_{\beta } J(f)(x, y) - J( D_{\tau } f)(x,y)&= -2 \alpha \sin ^2 (\beta ) \cdot J( D_S f) + 2 \alpha J( \sin ^2(\tau ) D_S f) , \end{aligned} \end{aligned}$$

where $R, \beta , S , \tau$ are defined in (A.19), provided that f is sufficiently smooth.

The first identity follows from a direct calculation and the proof of the second is similar to that in Lemma A.10. We omit the proof.

Proof of Lemma A.11

Step 1: Estimate of ${\bar{\theta }}$. Using (A.20) and the operator J in Lemma A.12, we get $\frac{{\bar{\theta }}}{x} = J({\bar{\eta }})$. We have the following estimate for ${\bar{\theta }}$

$$\begin{aligned} | D^i_R D^j_{\beta } \frac{ {\bar{\theta }}}{x} | = | D^i_R D^j_{\beta } J({\bar{\eta }}) | \lesssim J({\bar{\eta }}) = \frac{1}{x} \int _0^x {\bar{\eta }}(z, y) dz \lesssim {\bar{\eta }}, \end{aligned}$$

(A.43)

for $0 \le i+ j \le 5$. The proof of the first inequality follows from Lemma A.12 and the argument in the proof of (A.31). The proof of the second inequality is similar to that of (A.37) by considering $x \le 1+ y$ and $x > 1+ y$. We omit the proof.

Step 2: Estimate of ${\bar{\eta }}_{\lambda } -{\bar{\eta }}, {\bar{\xi }}_{\lambda }-{\bar{\xi }}$. Recall ${\bar{\eta }}_{\lambda } = \partial _ x( \chi _{\lambda } {\bar{\theta }})$, ${\bar{\xi }}_{\lambda } = \partial _ y( \chi _{\lambda } {\bar{\theta }})$ and the formula of $\partial _x , \partial _y$ (2.9). A direct calculation yields

$$\begin{aligned} \begin{aligned} {\bar{\eta }}_{\lambda }(R, \beta ) -{\bar{\eta }}&= \alpha \frac{\cos (\beta )}{r} D_R \chi _{\lambda } \cdot {\bar{\theta }} + (\chi _{\lambda } - 1) {\bar{\eta }} = \alpha \cos ^2(\beta ) D_R \chi _{\lambda } \cdot J({\bar{\eta }}) + (\chi _{\lambda } - 1) {\bar{\eta }}, \\ {\bar{\xi }}_{\lambda }(R, \beta ) -{\bar{\xi }}&= \alpha \frac{\sin (\beta ) }{r} D_R \chi _{\lambda } \cdot {\bar{\theta }} + (\chi _{\lambda } - 1) {\bar{\xi }} = \alpha \sin (\beta ) \cos (\beta ) D_R \chi _{\lambda } \cdot J({\bar{\eta }}) + (\chi _{\lambda } - 1) {\bar{\xi }}, \end{aligned}\nonumber \\ \end{aligned}$$

(A.44)

where we have used $\partial _x {\bar{\theta }} = {\bar{\eta }}, \partial _y {\bar{\theta }} = {\bar{\xi }} , \ (r\cos (\beta ))^{-1} {\bar{\theta }} = \frac{1}{x} {\bar{\theta }} = J({\bar{\eta }})$. From (A.40), we have

$$\begin{aligned} D_{\beta } \chi _{\lambda } = 0, \quad | D_R \chi _{\lambda } | = (R / \lambda ) | \chi ^{\prime }( R / \lambda ) | \lesssim 1. \end{aligned}$$

Similarly, we have

$$\begin{aligned} | D^k_R \chi _{\lambda }| \lesssim 1 , \end{aligned}$$

(A.45)

for $k =1,2,3,4$. Notice that $\partial _R \chi _{\lambda }, \ (\chi _{\lambda } - 1) = 0$ for $R \le \lambda$. From the formula of ${\bar{\eta }}$ and (A.26) in Lemma A.8, we know $(\chi _1 -1) (1+R){\bar{\eta }}\in {{\mathcal {H}}}^3$ (${\bar{\eta }}$ decays $R^{-2}$ for large R) and ${\bar{\xi }} \in {{\mathcal {H}}}^3(\psi )$. Using the estimates of $J({\bar{\eta }})$ in (A.43), we also have $(\chi _1 - 1) J({\bar{\eta }}) \in {{\mathcal {H}}}^3 \subset {{\mathcal {H}}}^3(\psi )$. Therefore, applying (A.44), (A.45) to $\chi _{\lambda }$ and the Dominated Convergence Theorem yields

$$\begin{aligned} \lim _{\lambda \rightarrow \infty }|| (1+R) ({\bar{\eta }}_{\lambda } - {\bar{\eta }}) ||_{{{\mathcal {H}}}^3} = 0 , \quad \lim _{\lambda \rightarrow \infty }|| {\bar{\xi }}_{\lambda } - {\bar{\xi }} ||_{{{\mathcal {H}}}^3(\psi )} = 0 . \end{aligned}$$

Similarly, we have

$$\begin{aligned} \lim _{\lambda \rightarrow \infty } || {\bar{\Omega }}_{\lambda } - {\bar{\Omega }} ||_{{{\mathcal {H}}}^3} = 0. \end{aligned}$$

Using (A.43), (A.45) and the fact that ${\bar{\eta }}$ decays for large R (see (4.8)), we have

$$\begin{aligned} {\overline{\lim }}_{\lambda \rightarrow \infty } || \sin (\beta ) \cos (\beta ) D_R \chi _{\lambda } \cdot J({\bar{\eta }}) ||_{{{\mathcal {C}}}^1} = 0. \end{aligned}$$

Using (A.23)–(A.24) in Lemma A.8 and (A.45), we conclude

$$\begin{aligned} || (\chi _{\lambda } - 1) {\bar{\xi }}||_{{{\mathcal {C}}}^2} \lesssim \alpha ^2. \end{aligned}$$

We complete the proof of (A.41).

Recall that the ${{\mathcal {H}}}^3$ norm is stronger than $L^2(\varphi _1)$. Using Lemma A.4 for $L_{12}(\Omega )(0)$, the fact that $\varphi _0 \lesssim \varphi _1, \psi _0 \lesssim (1+R) \varphi _1$ (see Definition 5.2, 5.7) and the limit obtained in (A.41), we prove (A.42). $\quad \square$

Let $C^{ \frac{\alpha }{40}}$ be the standard Hölder space. Recall the ${{\mathcal {C}}}^1$ norm defined in (6.5). We have the following embedding.

Lemma A.13

Suppose that $f \in {{\mathcal {C}}}^1(R, \beta )$ and $f(R, \pi /2) = 0$ for $R \ge 0$. We have

$$\begin{aligned} || f||_{C^{ \frac{\alpha }{ 40} }} \le C_{\alpha } || f ||_{{{\mathcal {C}}}^1} \end{aligned}$$

for some constant $C_{\alpha }$ depending on $\alpha$ only.

Proof

Recall the relation between the Cartesian coordinates (x, y) and the polar coordinates $(r, \beta ), (R, \beta )$. Since f vanishes on the axis $\beta = \frac{\pi }{2}$. It suffices to prove that f is Hölder in ${\mathbb {R}}^2_{++}$. Let $(R_1, \beta _1), (R_2, \beta _2 )$ be arbitrary two different points in ${\mathbb {R}}^2_{++}$, i.e. $R_1, R_2 \ge 0, \beta _1, \beta _2 \in [0, \pi /2]$, and $r_1 = R_1^{ 1 /\alpha }, r_2 = R_2^{1/\alpha }$. Without loss of generality, we assume $R_1 \le R_2$, $\beta _1 \le \beta _2$ and $||f ||_{{{\mathcal {C}}}^1} = 1$. From (6.5), we have $| f | \le 1, |\partial _R f| \le \frac{1}{1+R} , \ | \partial _{\beta } f | \le R^{1/40} \sin (2\beta )^{\alpha /40 -1}$. Using

$$\begin{aligned} \sin (2\beta )^{\alpha / 40 -1} \lesssim ( \sin (\beta )^{\alpha /40-1} + \cos (\beta )^{\alpha / 40 -1} ) \lesssim ( \beta ^{\alpha /40-1} + (\pi / 2 -\beta )^{\alpha /40 -1} ) \end{aligned}$$

and the estimates of the derivatives, we obtain

$$\begin{aligned} \begin{aligned} | f(R_1, \beta _1) - f(R_1, \beta _2) |&\le \int _{\beta _1}^{\beta _2} | \partial _{\beta } f(R_1, \beta ) | d \beta \le C R_1^{ \frac{1}{40}} \int _{\beta _1}^{\beta _2} \left( \beta ^{ \frac{\alpha }{40}-1} + ( \frac{\pi }{2} -\beta )^{ \frac{\alpha }{40} -1} \right) d \beta \\&\le C_{\alpha } R_1^{ \frac{1}{40}} ( \beta _2^{ \frac{\alpha }{40}} - \beta _1^{ \frac{\alpha }{40} } + ( \frac{\pi }{2} - \beta _1)^{ \frac{\alpha }{40}} - ( \frac{\pi }{2} - \beta _2)^{ \frac{\alpha }{40}} )\\&\le C_{\alpha } R_1^{ \frac{1}{40}} |\beta _2 - \beta _1|^{\frac{\alpha }{40}} , \\ | f(R_1, \beta _2) - f(R_2, \beta _2) |&\le \int _{R_1}^{R_2} |\partial _R f(R, \beta _2) | d R \le \int _{R_1}^{R_2} \frac{1}{1+R} d R \\&= \log \frac{1+R_2}{1+R_1} \lesssim (R_2 - R_1)^{1/40}, \end{aligned} \end{aligned}$$

where we have used $\log \frac{1+R_2}{1+R_1}\le \log (1 + R_2 - R_1)$ and $\log (1 + x) \lesssim x^{1/40}$ for $x \ge 0$ in the last inequality. The distance d between two points is

$$\begin{aligned} \begin{aligned} d^2&= (r_1 \cos (\beta _1) - r_2 \cos (\beta _2) )^2 + (r_1 \sin (\beta _1) - r_2 \sin (\beta _2) )^2\\&= (r_1 - r_2)^2 + 2r_1 r_2 (1 - \cos (\beta _1 - \beta _2)) \\&= |R_1^{1/\alpha } - R_2^{1/\alpha }|^2 + 4 R_1^{1/\alpha } R^{1/\alpha }_2 \sin ( \frac{1}{2} (\beta _1 - \beta _2))^2 \ge C_{\alpha } ( | R_1 - R_2|^{2/\alpha } + R_1^{2/\alpha } |\beta _1 -\beta _2|^2 ), \end{aligned} \end{aligned}$$

where we have used $R_1 \le R_2$ in the last inequality. Using the triangle inequality and the above estimates, we conclude $| f(R_1, \beta _1) - f(R_2, \beta _2)| \lesssim C_{\alpha } d^{ \frac{\alpha }{40}}$. $\quad \square$

1.7 Proof of Lemma 9.1

Proof of Lemma 9.1

We simplify $\omega ^{\theta }$ as $\omega$ and denote by $\vartheta = \arctan (x_2 /x_1)$ the angular variable. Recall the cylinder $D_1 = \{ (r, z) : r \in [0,1], |z| \le 1\}$. We extend $\omega {\mathbf {1}}_{(r,z)\in D_1}$ to ${\mathbb {R}}^3$ as follows :

$$\begin{aligned} \omega _e( r, z) = \omega (r, z) \,\mathrm { for }\, (r, z) \in D_1, \quad \omega _e(r, z) = 0 \,\mathrm { for }\, (r,z) \notin D_1. \end{aligned}$$

(A.46)

Note that $\omega _e$ is only supported in $D_1$, which is different from $\omega$. Denote

$$\begin{aligned} \begin{aligned} \omega _{\pm }&= \max ( \pm \omega _e, 0), \quad {{\mathcal {L}}}= -\partial _{rr} - \frac{1}{r} \partial _r - \partial _{zz} + \frac{1}{r^2}, \quad \Delta = \partial _{rr} + \frac{1}{r} \partial _r + \partial _{zz} + \frac{1}{r^2} \partial _{\vartheta \vartheta } , \\ \psi _{\pm }(r,z)&= \frac{1}{4\pi } \int _0^{\infty } \int _{-\infty }^{\infty } \int _0^{2\pi } \frac{ \sin (\vartheta ) \omega _{\pm }(r_1, z_1) }{ \left( (z-z_1)^2 + r^2 + r_1^2 - 2 \sin (\vartheta ) r r_1 \right) ^{1/2} } r_1 d r_1 dz_1 d \vartheta , \end{aligned}\nonumber \\ \end{aligned}$$

(A.47)

where $\Delta$ is the Laplace operator in ${\mathbb {R}}^3$ in cylindrical coordinates. Clearly, $\psi _{\pm }$ solve the Poisson equation in ${\mathbb {R}}^3$: $-\Delta (\sin (\vartheta ) \psi _{\pm }(r, z) ) = \omega _{\pm }(r,z) \sin (\vartheta )$, which can be verified easily using the Green function of $-\Delta$. Since $\omega _{\pm } \ge 0$, using the above formula and $\frac{ \sin (\vartheta ) }{ \left( (z-z_1)^2 + r^2 + r_1^2 - 2 \sin (\vartheta ) r r_1 \right) ^{1/2} } - \frac{ \sin (\vartheta ) }{ \left( (z-z_1)^2 + r^2 + r_1^2 + 2 \sin (\vartheta ) r r_1 \right) ^{1/2} } \ge 0$ for $\vartheta \in [0, \pi ]$, we get $\psi _{\pm } \ge 0$.

Let $\tilde{\psi }$ be a solution of (9.3)–(9.4). By definition of ${{\mathcal {L}}}$, we have

$$\begin{aligned} -\Delta ( \tilde{\psi } \sin (\vartheta )) = \sin (\vartheta ) {{\mathcal {L}}}\tilde{\psi } = \omega \sin (\vartheta ). \end{aligned}$$

Consider the domain $D_1^+ = \{ (r, z, \vartheta ) : r \in [0, 1], |z|\le 1 , \vartheta \in [0, \pi ] \}$, which is a half of the cylinder $D_1$. Next, we compare $\tilde{\psi }\sin (\vartheta )$ and $\psi _+ \sin (\vartheta )$ in $D_1^+$ using the maximal principle for the Laplace operator $\Delta$.

Recall from (A.46) that $\omega _e = \omega$ in $D_1^+ \subset D_1$. For $(r, z , \vartheta ) \in D_1^+$, we have $\sin (\vartheta ) \ge 0$ and

$$\begin{aligned} -\Delta ( (\tilde{\psi } - \psi _+ ) \sin (\vartheta ) ) = (\omega - \omega _+) \sin (\vartheta ) \le 0. \end{aligned}$$

(A.48)

On the boundary of $\partial D_1^+$, we have $\vartheta \in \{ 0, \pi \}$, $r=1$ or $z \in \{-1, 1\}$. The boundary related to $\vartheta \in \{0, \pi \}$ is $\{ (r,z, \vartheta ) : r\in [0,1], |z|\le 1, \vartheta =0, \pi \}$, or equivalently $\{ (x,y,z) : |x|\le 1, y=0, |z|\le 1\}$ in the Cartesian coordinates. It contains the symmetry axis $r=0$. Recall that $\tilde{\psi }$ is odd and 2-periodic in z. We obtain (9.5) $\tilde{\psi }( r, \pm 1) =0$. Recall the boundary condition (9.4) $\tilde{\psi }(1, z)=0$ and the fact that $\psi ^+$ is nonnegative. We have

$$\begin{aligned} (\tilde{\psi } - \psi _+ ) \sin (\vartheta ) = 0 \mathrm { \quad for \quad } \vartheta \in \{ 0, \pi \}, \qquad (\tilde{\psi } - \psi _+ ) \sin (\vartheta ) \le 0 \mathrm { \quad for \quad } r = 1 \,\text { or }\, z \in \{ -1, 1\} , \end{aligned}$$

where we have used $\sin (\vartheta ) \ge 0$ in $D_1^+$. Applying the maximal principle to (A.48) in the bounded domain $D_1^+$, we yield $(\tilde{\psi }(r,z) - \psi _+(r,z) ) \sin (\vartheta ) \le 0$ in $D_1^+$, which further implies $\tilde{\psi }(r,z) \le \psi _+(r,z)$ for $r\le 1, |z|\le 1$. Similarly, we have $\tilde{\psi } + \psi _- \ge 0$. Hence $| \tilde{\psi }| \le \psi _+ + \psi _-$.

Recall from (A.46),(A.47) that ${\mathrm {supp}}( \omega _{\pm } ) \subset {\mathrm {supp}}(\omega ) \cap D_1$ and the assumption ${\mathrm {supp}}(\omega ) \cap D_1 \subset \{ (r ,z) : (r-1)^2 + z^2 < 1/4 \}$ in Lemma 9.1. Thus, for $r > \frac{1}{4}$, $(r_1, z_1)$ in the support of $\omega _{\pm }$ and $|\vartheta | \le \pi$, we have $r_1 > \frac{1}{2}$ and

$$\begin{aligned}&(z-z_1)^2 + r^2 + r_1^2 - 2 \cos (\vartheta ) r r_1 \\&\quad = (z-z_1)^2 + (r- r_1)^2 + 4 \sin ^2(\vartheta / 2) r r_1 \asymp ( ( (z-z_1)^2 + (r- r_1)^2)^{1/2} + |\vartheta |)^2. \end{aligned}$$

We have similar estimate with $\cos (\vartheta )$ replaced by $\sin (\vartheta )$. Using this estimate and integrating the $\vartheta$ variable in the integral about $\psi _{\pm }$ in (A.47), we complete the proof. $\quad \square$

Remark A.14

The above proof can also be established in the Cartesian coordinates, which is essentially the same up to change of variables.

1.8 A toy model for 2D Boussinesq

We consider the toy model introduced in [11]

$$\begin{aligned} \begin{aligned}&\omega _t - (x_1 \lambda (t), -x_2 \lambda (t)) \cdot \nabla \omega = \partial _1 \theta , \\&\theta _t - (x_1 \lambda (t), -x_2 \lambda (t)) \cdot \nabla \theta = 0 , \quad \lambda (t) = \int _{{\mathbb {R}}^2} \frac{y_1 y_2}{ |y|^4} \omega (y, t) dy ,\\ \end{aligned} \end{aligned}$$

where $\partial _1 \theta = \partial _{x_1 }\theta$. This model can be derived from the 2D Boussinesq equations by approximating the velocity (u, v) by $u_{x_1}(0,0,t) \cdot (x_1, -x_2)$ and rescale the solution by a constant. We assume that $\omega$ is odd in $x_1$ and $x_2$, and $\theta$ is even in $x_1$ and odd in $x_2$. We show that for initial data $\omega _0 , \nabla \theta _0\in C_c^{\alpha }({\mathbb {R}}_2)$, the solution exists globally. We follow the argument in [11]. Without loss of generality, we assume ${\mathrm {supp}}( \partial _1 \theta _0) \subset [-1,1]^2$. Using the derivation in [11], we get

$$\begin{aligned} \begin{aligned} \omega (x_1, x_2, t)&= (\partial _1 \theta _0) ( \mu (t) x_1, \frac{x_2}{\mu (t)}) \int _0^t \mu (s) ds, \quad \mu (t) \triangleq \exp ( \int _0^t \lambda (s) ds) , \\ \frac{ {\dot{\mu }}}{\mu }&= 4 \int _0^t \mu (s) ds J(t), \quad J(t) \triangleq \int _0^{\infty } \int _0^{\infty } \frac{y_1 y_2}{ |y|^4} ( \partial _1 \theta _{0} ) ( \mu (t) y_1, \frac{y_2}{\mu (t)} ) d y_1 dy_2. \end{aligned}\nonumber \\ \end{aligned}$$

(A.49)

Next, we estimate J(t). Denote $\tilde{\theta }(x_1,x_2) = \theta _0(x_1, x_2) - \theta _0(0, x_2)$. Clearly, we have $\partial _1 \tilde{\theta }= \partial _1 \theta$. We simplify $\mu (t)$ as $\mu$. Since $(\partial _1 \tilde{\theta }_{0} )( \mu y_1, \frac{y_2}{\mu }) =\mu ^{-1} \partial _1( \tilde{\theta }_0( \mu y_1, \frac{y_2}{\mu }) )$, ${\mathrm {supp}}( \partial _1 \tilde{\theta }_{0}) = {\mathrm {supp}}( \partial _1 \theta _{0}) \subset [-1,1]^2$, using integration by parts and $\partial _1 \frac{y_1y_2}{|y|^4} = \frac{y_2(y_2^2 - 3y_1^2)}{|y|^6}$, we yield

$$\begin{aligned} \begin{aligned} J&= \mu ^{-1} \int _0^{ \mu ^{-1}} \int _0^{\infty } \frac{y_1 y_2}{ |y|^4} \partial _1 \Big ( \tilde{\theta }_{0}( \mu (t) y_1, \frac{y_2}{\mu (t)} ) \Big ) d y_1 dy_2 \\&= \mu ^{-1} \int _0^{\infty } \frac{ \mu ^{-1} y_2}{ (\mu ^{-2} + y_2^2)^2} \tilde{\theta }_0( 1, \frac{y_2}{\mu }) dy_2 \\&\quad - \mu ^{-1} \int _0^{\mu ^{-1}} \int _0^{\infty } \frac{y_2(y_2^2 - 3y_1^2)}{|y|^6} \tilde{\theta }_0( \mu y_1, \frac{y_2}{\mu }) dy_1 dy_2 \triangleq J_1 + J_2. \end{aligned} \end{aligned}$$

Since $\tilde{\theta }_0 \in C^{1,\alpha }$, $\tilde{\theta }_0(0, x_2) = 0$ and $\tilde{\theta }_0(x_1, 0) = 0$, we have $| \tilde{\theta }_0( x_1, x_2)| \lesssim |x_1|^{\alpha } |x_2|$. It follows

$$\begin{aligned} \begin{aligned} |J_1|&\lesssim \mu ^{-1} \int _0^{\infty } \frac{\mu ^{-1} y_2}{ (\mu ^{-2} + y_2^2)^2} \frac{y_2}{\mu } dy_2 = \mu ^{-2} \int _0^{\infty } \frac{z^2}{ (1+z^2)^2} dz \lesssim \mu ^{-2} , \\ |J_2|&\lesssim \mu ^{-2}\int _0^{\mu ^{-1}} (\mu y_1)^{\alpha } \int _0^{\infty } \Big |\frac{y_2^2 ( y_2^2 - 3y_1^2 ) }{|y|^6}\Big | dy_2 dy_1 \lesssim \mu ^{-2}\int _0^{\mu ^{-1}} (\mu y_1)^{\alpha } y_1^{-1} dy_1 \lesssim _{\alpha } \mu ^{-2}. \end{aligned} \end{aligned}$$

Plugging the above estimates in (A.49), we obtain

$$\begin{aligned} \Big |\frac{ {\dot{\mu }}}{\mu } \Big | \lesssim _{\alpha } \mu ^{-2} \int _0^t \mu (s) ds . \end{aligned}$$

Thus, $\mu$ remains bounded for all time. Formula (A.49) implies that the solution exists globally.