Global Well-Posedness of 3-D Anisotropic Navier–Stokes System with Small Unidirectional Derivative

Abstract

In Liu and Zhang (Arch Ration Mech Anal 235:1405–1444, 2020), the authors proved that as long as the one-directional derivative of the initial velocity is sufficiently small in some scaling invariant spaces, then the classical Navier–Stokes system has a global unique solution. The goal of this paper is to extend this type of result to the 3-D anisotropic Navier–Stokes system (ANS) with only horizontal dissipation. More precisely, given initial data $u_0=(u_0^\mathrm{h},u_0^3)\in \mathcal {B}^{0,\frac{1}{2}},$ (ANS) has a unique global solution provided that $|D_\mathrm{h}|^{-1}\partial _3u_0$ is sufficiently small in the scaling invariant space $\mathcal {B}^{0,\frac{1}{2}}$.

Appendix A. The Proof of Lemmas 4.2 and 6.1

In this section, we present the proof of Lemmas 4.2 and 6.1.

Proof of Lemma 4.2

By applying Bony’s decomposition in the vertical variable (2.7) to $a\otimes b$, we write

$$\begin{aligned} \begin{aligned}&\int \nolimits _0^T\bigl (\Delta _{\ell }^{\mathrm{v}}A(D)(a\otimes b) \big |\Delta _{\ell }^{\mathrm{v}}c\bigr )_{L^2}\,\mathrm{d}t=Q_1+Q_2\quad \text{ with }\\&\quad Q_1\buildrel {\mathrm{def}}\over =\int \nolimits _0^T\bigl (\Delta _{\ell }^{\mathrm{v}}A(D)(T^\mathrm{v}_{a} b) \big |\Delta _{\ell }^{\mathrm{v}}c\bigr )_{L^2}\,\mathrm{d}t =\int \nolimits _0^T\bigl (\Delta _{\ell }^{\mathrm{v}}(T^\mathrm{v}_{a} b) \big | A(D)\Delta _{\ell }^{\mathrm{v}}c\bigr )_{L^2}\,\mathrm{d}t\quad \text{ and }\\&\quad Q_2\buildrel {\mathrm{def}}\over =\int \nolimits _0^T\bigl (\Delta _{\ell }^{\mathrm{v}}A(D) R^\mathrm{v}(a, b) \big |\Delta _{\ell }^{\mathrm{v}}c\bigr )_{L^2}\,\mathrm{d}t =\int \nolimits _0^T\bigl (\Delta _{\ell }^{\mathrm{v}}R^\mathrm{v}(a, b) \big | A(D)\Delta _{\ell }^{\mathrm{v}}c\bigr )_{L^2}\,\mathrm{d}t. \end{aligned} \end{aligned}$$

(A.1)

Considering the support properties to the Fourier transform of the terms in $T^\mathrm{v}_{a} b$, and noting that A(D) is a smooth homogeneous Fourier multiplier of degree zero, we find

$$\begin{aligned} \begin{aligned} |Q_1|&\leqq \int \nolimits _0^T\Vert \Delta _{\ell }^{\mathrm{v}}(T^\mathrm{v}_{a} b)\Vert _{L_\mathrm{h}^{\frac{4}{3}}(L_\mathrm{v}^2)} \Vert A(D) \Delta _{\ell }^{\mathrm{v}}c\Vert _{L_\mathrm{h}^4(L_\mathrm{v}^2)}\,\mathrm{d}t\\&\lesssim \sum _{|\ell '-\ell |\leqq 5} \int \nolimits _0^T\Vert S^\mathrm{v}_{\ell '-1}a\Vert _{L_\mathrm{h}^4(L_\mathrm{v}^\infty )} \Vert \Delta _{\ell '}^{\mathrm{v}}b\Vert _{L^2}\Vert A(D) \Delta _{\ell }^{\mathrm{v}}c\Vert _{L^2}^{\frac{1}{2}} \Vert \nabla _{\mathrm{h}}A(D)\Delta _{\ell }^{\mathrm{v}}c\Vert _{L^2}^{\frac{1}{2}}\,\mathrm{d}t\\&\lesssim \sum _{|\ell '-\ell |\leqq 5} \Bigl (\int \nolimits _0^T\Vert S^\mathrm{v}_{\ell '-1}a(t)\Vert _{L_\mathrm{h}^4(L_\mathrm{v}^\infty )}^4 \Vert \Delta _{\ell }^{\mathrm{v}}c(t)\Vert _{L^2}^2\,\mathrm{d}t\Bigr )^{\frac{1}{4}}\Vert \Delta _{\ell '}^{\mathrm{v}}b\Vert _{L^2_T(L^2)}\Vert \nabla _{\mathrm{h}}\Delta _{\ell }^{\mathrm{v}}c\Vert _{L^2_T(L^2)}^{\frac{1}{2}}. \end{aligned} \end{aligned}$$

It follows from Lemma 2.1 and Definition 2.4 that

$$\begin{aligned}&\Vert S^\mathrm{v}_{\ell '-1}a(t)\Vert _{L_\mathrm{h}^4(L_\mathrm{v}^\infty )} \leqq \sum _{j\leqq \ell '-2}\Vert \Delta ^\mathrm{v}_{j}a(t)\Vert _{L_\mathrm{h}^4(L_\mathrm{v}^\infty )}\\&\quad \lesssim \sum _{j\leqq \ell '-2}2^{\frac{j}{2}} \Vert \Delta ^\mathrm{v}_{j}a(t)\Vert _{L_\mathrm{h}^4(L_\mathrm{v}^2)}\lesssim \Vert a(t)\Vert _{\mathcal {B}_4^{0,\frac{1}{2}}}. \end{aligned}$$

This together with Definition 2.2 ensures that

$$\begin{aligned} |Q_1| \lesssim d_{\ell }^2 2^{-\ell }\Vert c\Vert _{\widetilde{L}^2_{T,\mathfrak {f}}(\mathcal {B}^{0,\frac{1}{2}})}^{\frac{1}{2}} \Vert b\Vert _{\widetilde{L}^2_T\big (\mathcal {B}^{0,\frac{1}{2}}\big )}\Vert \nabla _{\mathrm{h}}c\Vert _{\widetilde{L}^2_T\big (\mathcal {B}^{0,\frac{1}{2}}\big )}^{\frac{1}{2}}. \end{aligned}$$

(A.2)

Along the same lines, we get, by applying (2.5), that

$$\begin{aligned} \begin{aligned} |Q_{1,\mathfrak {g}}|&\buildrel {\mathrm{def}}\over =\int \nolimits _0^T\bigl |\bigl (\Delta _{\ell }^{\mathrm{v}}(T^\mathrm{v}_{a} b) \big | A(D)\Delta _{\ell }^{\mathrm{v}}c\bigr )_{L^2}\bigr |\mathfrak {g}^2\,\mathrm{d}t\\&\lesssim \sum _{|\ell '-\ell |\leqq 5} \Vert \sqrt{\mathfrak {g}} S^\mathrm{v}_{\ell '-1}a\Vert _{L_T^4(L_\mathrm{h}^4(L_\mathrm{v}^\infty ))}\Vert \mathfrak {g}\Delta _{\ell '}^{\mathrm{v}}b\Vert _{L^2_T(L^2)} \Vert \Delta _{\ell }^{\mathrm{v}}c\Vert _{L^\infty _T(L^2)}^{\frac{1}{2}}\Vert \mathfrak {g}\nabla _{\mathrm{h}}\Delta _{\ell }^{\mathrm{v}}c\Vert _{L^2_T(L^2)}^{\frac{1}{2}}\\&\lesssim d_{\ell }^2 2^{-\ell }\Vert a\Vert _{\widetilde{L}^\infty _T\big (\mathcal {B}^{0,\frac{1}{2}}\big )}^{\frac{1}{2}}\Vert \mathfrak {g}\nabla _{\mathrm{h}}a\Vert _{\widetilde{L}^2_T\big (\mathcal {B}^{0,\frac{1}{2}}\big )}^{\frac{1}{2}} \Vert \mathfrak {g}b\Vert _{\widetilde{L}^2_T\big (\mathcal {B}^{0,\frac{1}{2}}\big )}\Vert c\Vert _{\widetilde{L}^\infty _t\big (\mathcal {B}^{0,\frac{1}{2}}\big )}^{\frac{1}{2}}\Vert \mathfrak {g}\nabla _{\mathrm{h}}c\Vert _{\widetilde{L}^2_T\big (\mathcal {B}^{0,\frac{1}{2}}\big )}^{\frac{1}{2}}. \end{aligned} \end{aligned}$$

(A.3)

On the other hand, once again considering the support properties to the Fourier transform of the terms in $R^\mathrm{v}({a}, b),$ we find

$$\begin{aligned} \begin{aligned} |Q_2|&\leqq \int \nolimits _0^T\Vert \Delta _{\ell }^{\mathrm{v}}R^\mathrm{v}(a,b)\Vert _{L_\mathrm{h}^{\frac{4}{3}}(L_\mathrm{v}^2)} \Vert A(D)\Delta _{\ell }^{\mathrm{v}}c\Vert _{L_\mathrm{h}^4(L_\mathrm{v}^2)}\,\mathrm{d}t\\&\lesssim \sum _{\ell '\geqq \ell -N_0}\int \nolimits _0^T\Vert \Delta ^\mathrm{v}_{\ell '}a\Vert _{L_\mathrm{h}^4(L_\mathrm{v}^2)} \Vert S^\mathrm{v}_{\ell '+2}b\Vert _{L^2_\mathrm{h}(L^\infty _\mathrm{v})} \Vert A(D)\Delta _{\ell }^{\mathrm{v}}c\Vert _{L^2}^{\frac{1}{2}} \Vert \nabla _{\mathrm{h}}A(D)\Delta _{\ell }^{\mathrm{v}}c\Vert _{L^2}^{\frac{1}{2}}\,\mathrm{d}t\\&\lesssim \sum _{\ell '\geqq \ell -N_0} 2^{-\frac{\ell '}{2}}\int \nolimits _0^Td_{\ell '}(t)\Vert a(t)\Vert _{\mathcal {B}^{0,\frac{1}{2}}_4}\Vert b(t)\Vert _{L^2_\mathrm{h}(L^\infty _\mathrm{v})} \Vert \Delta _{\ell }^{\mathrm{v}}c(t)\Vert _{L^2}^{\frac{1}{2}}\Vert \nabla _{\mathrm{h}}\Delta _{\ell }^{\mathrm{v}}c(t)\Vert _{L^2}^{\frac{1}{2}}\,\mathrm{d}t\\&\lesssim \sum _{\ell '\geqq \ell -N_0} d_{\ell '}2^{-\frac{\ell '}{2}}\int \nolimits _0^T\Vert a(t)\Vert _{\mathcal {B}^{0,\frac{1}{2}}_4}\Vert b(t)\Vert _{L^2_\mathrm{h}(L^\infty _\mathrm{v})} \Vert \Delta _{\ell }^{\mathrm{v}}c(t)\Vert _{L^2}^{\frac{1}{2}}\Vert \nabla _{\mathrm{h}}\Delta _{\ell }^{\mathrm{v}}c(t)\Vert _{L^2}^{\frac{1}{2}}\,\mathrm{d}t. \end{aligned} \end{aligned}$$

It follows from Lemma 2.1 however that

$$\begin{aligned} \Vert b\Vert _{L^2_T(L^2_\mathrm{h}(L^\infty _\mathrm{v}))} \lesssim \sum _{\ell \in {\mathbb {Z}}}2^{\frac{\ell }{2}} \Vert \Delta ^\mathrm{v}_{\ell }b\Vert _{L_T^2(L^2)}\leqq \Vert b\Vert _{\widetilde{L}^2_T\big (\mathcal {B}^{0,\frac{1}{2}}\big )}. \end{aligned}$$

As a result, by virtue of Definition 2.2, we obtain

$$\begin{aligned} \begin{aligned} |Q_2|&\lesssim \sum _{\ell '\geqq \ell -N_0}d_{\ell '} 2^{-\frac{\ell '}{2}} \Bigl (\int \nolimits _0^T\Vert a(t)\Vert _{\mathcal {B}_4^{0,\frac{1}{2}}}^4 \Vert \Delta _{\ell }^{\mathrm{v}}c(t)\Vert _{L^2}^{2}\,\mathrm{d}t\Bigr )^{\frac{1}{4}} \Vert \nabla _{\mathrm{h}}\Delta _{\ell }^{\mathrm{v}}c\Vert _{L_T^2(L^2)}^{\frac{1}{2}}\Vert b\Vert _{\widetilde{L}^2_T\big (\mathcal {B}^{0,\frac{1}{2}}\big )}\\&\lesssim \sum _{\ell '\geqq \ell -N_0}d_{\ell '} 2^{-\frac{\ell '}{2}} \Bigl (d_{\ell } 2^{-\frac{\ell }{2}}\Vert c\Vert _{\widetilde{L}^2_{T,\mathfrak {f}}\big (\mathcal {B}^{0,\frac{1}{2}}\big )}\Bigr )^{\frac{1}{2}} \bigl (d_{\ell } 2^{-\frac{\ell }{2}}\Vert \nabla _{\mathrm{h}}c\Vert _{\widetilde{L}^2_T\big (\mathcal {B}^{0,\frac{1}{2}}\big )}\bigr )^{\frac{1}{2}} \Vert b\Vert _{\widetilde{L}^2_T\big (\mathcal {B}^{0,\frac{1}{2}}\big )}\\&\lesssim d_{\ell }^2 2^{-\ell } \Vert c\Vert _{\widetilde{L}^2_{T,\mathfrak {f}}\big (\mathcal {B}^{0,\frac{1}{2}}\big )}^{\frac{1}{2}}\Vert \nabla _{\mathrm{h}}c\Vert _{\widetilde{L}^2_T\big (\mathcal {B}^{0,\frac{1}{2}}\big )}^{\frac{1}{2}} \Vert b\Vert _{\widetilde{L}^2_T\big (\mathcal {B}^{0,\frac{1}{2}}\big )}. \end{aligned} \end{aligned}$$

(A.4)

Similarly, thanks to (2.5), one has

$$\begin{aligned} \begin{aligned}&|Q_{2,\mathfrak {g}}|\buildrel {\mathrm{def}}\over =\int \nolimits _0^T\bigl |\bigl (\Delta _{\ell }^{\mathrm{v}}R^\mathrm{v}(a, b) \big | A(D)\Delta _{\ell }^{\mathrm{v}}c\bigr )_{L^2}\bigr |\mathfrak {g}^2\,\mathrm{d}t\\&\quad \lesssim \sum _{\ell '\geqq \ell -N_0}\Vert \sqrt{\mathfrak {g}}\Delta ^\mathrm{v}_{\ell '}a\Vert _{L^4_T(L_\mathrm{h}^4(L_\mathrm{v}^2))} \Vert \mathfrak {g}S^\mathrm{v}_{\ell '+2}b\Vert _{L^2_T(L^2_\mathrm{h}(L^\infty _\mathrm{v}))} \bigl (\Vert \Delta _{\ell }^{\mathrm{v}}c\Vert _{L^2_T(L^2)}\Vert \mathfrak {g}\nabla _{\mathrm{h}}\Delta _{\ell }^{\mathrm{v}}c\Vert _{L^2_T(L^2)}\bigr )^{\frac{1}{2}}\\&\quad \lesssim d_{\ell }^2 2^{-\ell }\Vert a\Vert _{\widetilde{L}^\infty _T\big (\mathcal {B}^{0,\frac{1}{2}}\big )}^{\frac{1}{2}}\Vert \mathfrak {g}\nabla _{\mathrm{h}}a\Vert _{\widetilde{L}^2_T\big (\mathcal {B}^{0,\frac{1}{2}}\big )}^{\frac{1}{2}} \Vert \mathfrak {g}b\Vert _{\widetilde{L}^2_T\big (\mathcal {B}^{0,\frac{1}{2}}\big )}\Vert c\Vert _{\widetilde{L}^\infty _T\big (\mathcal {B}^{0,\frac{1}{2}}\big )}^{\frac{1}{2}}\Vert \mathfrak {g}\nabla _{\mathrm{h}}c\Vert _{\widetilde{L}^2_T\big (\mathcal {B}^{0,\frac{1}{2}}\big )}^{\frac{1}{2}}. \end{aligned} \end{aligned}$$

(A.5)

Combining (A.2) with (A.4) gives (4.11), and (4.13) follows from (A.3) and (A.5).

It remains to prove (4.12). Similarly to the proof of (A.2), we write

$$\begin{aligned} |Q_1|&\lesssim \sum _{|\ell '-\ell |\leqq 5} \int \nolimits _0^T \Vert S^\mathrm{v}_{\ell '-1}a\Vert _{L_\mathrm{h}^4(L_\mathrm{v}^\infty )} \Vert \Delta _{\ell '}^{\mathrm{v}}b\Vert _{L_\mathrm{h}^4(L_\mathrm{v}^2)}\Vert A(D)\Delta _{\ell }^{\mathrm{v}}c\Vert _{L^2}\,\mathrm{d}t\\&\lesssim \sum _{|\ell '-\ell |\leqq 5} \int \nolimits _0^T \Vert a(t)\Vert _{\mathcal {B}^{0,\frac{1}{2}}_4} \Vert \Delta _{\ell '}^{\mathrm{v}}b(t)\Vert _{L^2}^{\frac{1}{2}}\Vert \Delta _{\ell '}^{\mathrm{v}}\nabla _{\mathrm{h}}b(t)\Vert _{L^2}^{\frac{1}{2}}\Vert \Delta _{\ell }^{\mathrm{v}}c(t)\Vert _{L^2}\,\mathrm{d}t\\&\lesssim \sum _{|\ell '-\ell |\leqq 5} \Bigl (\int \nolimits _0^T \Vert a(t)\Vert _{\mathcal {B}^{0,\frac{1}{2}}_4}^4 \Vert \Delta _{\ell '}^{\mathrm{v}}b(t)\Vert _{L^2}^2\,\mathrm{d}t\Bigr )^{\frac{1}{4}}\Vert \Delta _{\ell '}^{\mathrm{v}}\nabla _{\mathrm{h}}b\Vert _{L^2_T(L^2)}^{\frac{1}{2}}\Vert \Delta _{\ell }^{\mathrm{v}}c\Vert _{L^2_T(L^2)}, \end{aligned}$$

from which, along with Definition 2.2, we infer

$$\begin{aligned} \begin{aligned} |Q_1|&\lesssim d_\ell 2^{-\frac{\ell }{2}} \sum _{|\ell '-\ell |\leqq 5} d_{\ell '} 2^{-\frac{\ell '}{2}}\Vert b\Vert _{\widetilde{L}^2_{T,\mathfrak {f}}\big (\mathcal {B}^{0,\frac{1}{2}}\big )}^{\frac{1}{2}} \Vert \nabla _{\mathrm{h}}b\Vert _{\widetilde{L}^2_T\big (\mathcal {B}^{0,\frac{1}{2}}\big )}^{\frac{1}{2}}\Vert c\Vert _{\widetilde{L}^2_T\big (\mathcal {B}^{0,\frac{1}{2}}\big )}\\&\lesssim d_{\ell }^2 2^{-\ell } \Vert b\Vert _{\widetilde{L}^2_{T,\mathfrak {f}}\big (\mathcal {B}^{0,\frac{1}{2}}\big )}^{\frac{1}{2}}\Vert \nabla _{\mathrm{h}}b\Vert _{\widetilde{L}^2_T\big (\mathcal {B}^{0,\frac{1}{2}}\big )}^{\frac{1}{2}}\Vert c\Vert _{\widetilde{L}^2_T\big (\mathcal {B}^{0,\frac{1}{2}}\big )}. \end{aligned} \end{aligned}$$

(A.6)

We deduce from Definition 2.4 that

$$\begin{aligned} |Q_2|&\lesssim \sum _{\ell '\geqq \ell -N_0}\int \nolimits _0^T \Vert \Delta ^\mathrm{v}_{\ell '}a\Vert _{L_\mathrm{h}^4(L_\mathrm{v}^2)} \Vert S^\mathrm{v}_{\ell '+2}b\Vert _{L^4_\mathrm{h}(L^\infty _\mathrm{v})} \Vert A(D)\Delta _{\ell }^{\mathrm{v}}c\Vert _{L^2}\,\mathrm{d}t\\&\lesssim \sum _{\ell '\geqq \ell -N_0}d_{\ell '}2^{-\frac{\ell '}{2}} \int \nolimits _0^T\Vert a(t)\Vert _{\mathcal {B}^{0,\frac{1}{2}}_4} \Vert b(t)\Vert _{L^4_\mathrm{h}(L^\infty _\mathrm{v})} \Vert \Delta _{\ell }^{\mathrm{v}}c(t)\Vert _{L^2}\,\mathrm{d}t\\&\lesssim d_\ell 2^{-\frac{\ell }{2}}\Vert c\Vert _{\widetilde{L}^2_T\big (\mathcal {B}^{0,\frac{1}{2}}\big )}\sum _{\ell '\geqq \ell -4} d_{\ell '} 2^{-\frac{\ell '}{2}} \Bigl (\int \nolimits _0^T \Vert a(t)\Vert _{\mathcal {B}_4^{0,\frac{1}{2}}}^2 \Vert b(t)\Vert _{L^4_\mathrm{h}(L^\infty _\mathrm{v})}^{2}\,\mathrm{d}t\Bigr )^{\frac{1}{2}}, \end{aligned}$$

whereas we get, by applying the triangle inequality and Lemma 2.1, that

$$\begin{aligned}&\Bigl (\int \nolimits _0^T\Vert a(t)\Vert _{\mathcal {B}_4^{0,\frac{1}{2}}}^2 \Vert b(t)\Vert _{L^4_\mathrm{h}(L^\infty _\mathrm{v})}^{2}\,\mathrm{d}t\Bigr )^{\frac{1}{2}}\\&\quad \lesssim \sum _{\ell \in {\mathbb {Z}}}2^{\frac{\ell }{2}} \Bigl (\int \nolimits _0^T \Vert a(t)\Vert _{\mathcal {B}_4^{0,\frac{1}{2}}}^2 \Vert \Delta _\ell b(t)\Vert _{L^2}\Vert \nabla _\mathrm{h}\Delta _\ell b(t)\Vert _{L^2}\,\mathrm{d}t\Bigr )^{\frac{1}{2}}\\&\quad \lesssim \sum _{\ell \in {\mathbb {Z}}}2^{\frac{\ell }{2}}\Bigl (\int \nolimits _0^T \Vert a(t)\Vert _{\mathcal {B}_4^{0,\frac{1}{2}}}^4 \Vert \Delta _\ell b(t)\Vert _{L^2}^2\,\mathrm{d}t\Bigr )^{\frac{1}{4}}\Vert \nabla _\mathrm{h}\Delta _\ell b\Vert _{L^2_T(L^2)}^{\frac{1}{2}}\\&\quad \lesssim \Vert b\Vert _{\widetilde{L}^2_{T,\mathfrak {f}}\big (\mathcal {B}^{0,\frac{1}{2}}\big )}^{\frac{1}{2}} \Vert \nabla _{\mathrm{h}}b\Vert _{\widetilde{L}^2_T\big (\mathcal {B}^{0,\frac{1}{2}}\big )}^{\frac{1}{2}}. \end{aligned}$$

This in turn shows that

$$\begin{aligned} \begin{aligned} |Q_2|\lesssim d_{\ell }^2 2^{-\ell }\Vert c\Vert _{\widetilde{L}^2_T\big (\mathcal {B}^{0,\frac{1}{2}}\big )}\Vert b\Vert _{\widetilde{L}^2_{T,\mathfrak {f}}\big (\mathcal {B}^{0,\frac{1}{2}}\big )}^{\frac{1}{2}}\Vert \nabla _{\mathrm{h}}b\Vert _{\widetilde{L}^2_T\big (\mathcal {B}^{0,\frac{1}{2}}\big )}^{\frac{1}{2}}, \end{aligned} \end{aligned}$$

which, together with (A.6), ensures (4.12). This completes the proof of Lemma 4.2.

$\square$

Proof of Lemma 6.1

Let $Q_1$ be given by (A.1). We first get, by a similar derivation of (A.2), that

$$\begin{aligned} \begin{aligned} |Q_1|&\lesssim \sum _{|\ell '-\ell |\leqq 5}\Vert S^\mathrm{v}_{\ell '-1}a\Vert _{L^4_T(L_\mathrm{h}^4(L_\mathrm{v}^\infty ))} \Vert \Delta _{\ell '}^{\mathrm{v}}b\Vert _{L^4_T(L^4_\mathrm{h}(L^2_\mathrm{v}))} \Vert A(D)\Delta _{\ell }^{\mathrm{v}}c\Vert _{L^2_t(L^2)}\\&\lesssim d_\ell 2^{-\frac{\ell }{2}}\sum _{|\ell '-\ell |\leqq 5}d_{\ell '}2^{-\frac{\ell '}{2}}\Vert a\Vert _{\widetilde{L}^4_T(\mathcal {B}_4^{0,\frac{1}{2}})} \Vert b\Vert _{\widetilde{L}^4_T(\mathcal {B}_4^{0,\frac{1}{2}})} \Vert c\Vert _{\widetilde{L}^2_T(\mathcal {B}^{0,\frac{1}{2}})}, \end{aligned} \end{aligned}$$

which, together with Proposition 2.1, implies that

$$\begin{aligned} |Q_1|\lesssim d_{\ell }^2 2^{-\ell }\Vert a\Vert _{\widetilde{L}^4_T(\mathcal {B}_4^{0,\frac{1}{2}})} \Vert b\Vert _{{\mathcal {B}^{-\frac{1}{2},\frac{1}{2}}_4}(T)}\Vert c\Vert _{\widetilde{L}^2_T\big (\mathcal {B}^{0,\frac{1}{2}}\big )}. \end{aligned}$$

(A.7)

For $Q_2$ given by (A.1), we get, by a similar derivation of (A.4), that

$$\begin{aligned}\begin{aligned} |Q_2|&\lesssim \sum _{\ell '\geqq \ell -N_0}\Vert \Delta _{\ell '}^{\mathrm{v}}a\Vert _{L^4_T(L_\mathrm{h}^4(L_\mathrm{v}^2))} \Vert S^\mathrm{v}_{\ell '+2}b\Vert _{L^4_T(L^4_\mathrm{h}(L^\infty _\mathrm{v}))} \Vert A(D)\Delta _{\ell }^{\mathrm{v}}c\Vert _{L^2_T(L^2)}\\&\lesssim d_\ell 2^{-\frac{\ell }{2}}\sum _{\ell '\geqq \ell -N_0} d_{\ell '} 2^{-\frac{\ell '}{2}}\Vert a\Vert _{\widetilde{L}^4_T(\mathcal {B}_4^{0,\frac{1}{2}})} \Vert b\Vert _{\widetilde{L}^4_T(\mathcal {B}_4^{0,\frac{1}{2}})} \Vert c\Vert _{\widetilde{L}^2_T(\mathcal {B}^{0,\frac{1}{2}})}, \end{aligned} \end{aligned}$$

from which, with Proposition 2.1, we infer

$$\begin{aligned} |Q_2| \lesssim d_{\ell }^2 2^{-\ell }\Vert a\Vert _{\widetilde{L}^4_T(\mathcal {B}_4^{0,\frac{1}{2}})} \Vert b\Vert _{{\mathcal {B}^{-\frac{1}{2},\frac{1}{2}}_4}(T)}\Vert c\Vert _{\widetilde{L}^2_T\big (\mathcal {B}^{0,\frac{1}{2}}\big )}. \end{aligned}$$

This, together with (A.1) and (A.7), ensures (6.5).

The inequality (6.6) can be proved similarly. As a matter of fact, we observe that

$$\begin{aligned} |Q_1|&\lesssim \sum _{|\ell '-\ell |\leqq 5}\Vert S^\mathrm{v}_{\ell '-1}a\Vert _{L^2_T(L_\mathrm{h}^2(L_\mathrm{v}^\infty ))} \Vert \Delta _{\ell '}^{\mathrm{v}}b\Vert _{L^4_T(L^4_\mathrm{h}(L^2_\mathrm{v}))} \Vert A(D)\Delta _{\ell }^{\mathrm{v}}c\Vert _{L^4_T(L^4_\mathrm{h}(L^2_\mathrm{v}))}\\&\lesssim \sum _{|\ell '-\ell |\leqq 5}\Vert S^\mathrm{v}_{\ell '-1}a\Vert _{L^2_T(L_\mathrm{h}^2(L_\mathrm{v}^\infty ))} \Vert \Delta _{\ell '}^{\mathrm{v}}b\Vert _{L^4_T(L^4_\mathrm{h}(L^2_\mathrm{v}))} \Vert \Delta _{\ell }^{\mathrm{v}}c\Vert _{L^\infty _T(L^2)}^{\frac{1}{2}} \Vert \Delta _{\ell }^{\mathrm{v}}\nabla _{\mathrm{h}}c\Vert _{L^2_T(L^2)}^{\frac{1}{2}}\\&\lesssim d_\ell 2^{-\frac{\ell }{2}}\sum _{|\ell '-\ell |\leqq 5}d_{\ell '}2^{-\frac{\ell '}{2}}\Vert a\Vert _{\widetilde{L}^2_T(\mathcal {B}^{0,\frac{1}{2}})} \Vert b\Vert _{\widetilde{L}^4_T(\mathcal {B}_4^{0,\frac{1}{2}})} \Vert c\Vert _{\mathcal {B}^{0,\frac{1}{2}}(T)}, \end{aligned}$$

and

$$\begin{aligned} |Q_2|&\lesssim \sum _{\ell '\geqq \ell -N_0}\Vert \Delta _{\ell '}^{\mathrm{v}}a\Vert _{L^2_T(L^2)} \Vert S^\mathrm{v}_{\ell '+2}b\Vert _{L^4_T(L^4_\mathrm{h}(L^\infty _\mathrm{v}))} \Vert A(D) \Delta _{\ell }^{\mathrm{v}}c\Vert _{L^4_T(L_\mathrm{h}^4(L_\mathrm{v}^2))}\\&\lesssim d_\ell 2^{-\frac{\ell }{2}}\sum _{\ell '\geqq \ell -N_0} d_{\ell '} 2^{-\frac{\ell '}{2}}\Vert a\Vert _{\widetilde{L}^2_T(\mathcal {B}^{0,\frac{1}{2}})} \Vert b\Vert _{\widetilde{L}^4_T(\mathcal {B}_4^{0,\frac{1}{2}})} \Vert c\Vert _{\mathcal {B}^{0,\frac{1}{2}}(T)}. \end{aligned}$$

Then (6.6) follows from Proposition 2.1. This completes the proof of this lemma.

$\square$