Relaxation Approximation and Asymptotic Stability of Stratified Solutions to the IPM Equation

Abstract

We prove the nonlinear asymptotic stability of stably stratified solutions to the Incompressible Porous Media equation (IPM) for initial perturbations in $\dot{H}^{1-\tau }(\mathbb {R}^2) \cap \dot{H}^s(\mathbb {R}^2)$ with $s > 3$ and for any $0< \tau <1$. Such a result improves upon the existing literature, where the asymptotic stability is proved for initial perturbations belonging at least to $H^{20}(\mathbb {R}^2)$. More precisely, the aim of the article is threefold. First, we provide a simplified and improved proof of global-in-time well-posedness of the Boussinesq equations with strongly damped vorticity in $H^{1-\tau }(\mathbb {R}^2) \cap \dot{H}^s(\mathbb {R}^2)$ with $s > 3$ and $0< \tau <1$. Next, we prove the strong convergence of the Boussinesq system with damped vorticity towards (IPM) under a suitable scaling. Lastly, the asymptotic stability of stratified solutions to (IPM) follows as a byproduct. A symmetrization of the approximating system and a careful study of the anisotropic properties of the equations via anisotropic Littlewood-Paley decomposition play key roles to obtain uniform energy estimates. Finally, one of the main new and crucial points is the integrable time decay of the vertical velocity $\Vert u_2(t)\Vert _{L^\infty (\mathbb {R}^2)}$ for initial data only in $\dot{H}^{1-\tau }(\mathbb {R}^2) \cap \dot{H}^s(\mathbb {R}^2)$ with $s >3$.

Appendices

Appendix A

Lemma A.1

(Anisotropic product laws) Let $-\frac{1}{2}\leqq s_1 \leqq \frac{5}{2}$ and $-\frac{1}{2} \leqq s_2 \leqq \frac{1}{2}$. Let $\delta _1,\delta _2,\delta _3,\delta _4\geqq 0$ and f, g be two smooth functions, one has

$$\begin{aligned} \Vert fg\Vert _{B^{s_1,s_2}}&\lesssim \Vert f\Vert _{L^\infty }\Vert g\Vert _{B^{s_1,s_2}}+\Vert f\Vert _{B^{s_1,s_2}}\Vert g\Vert _{L^\infty } \\ {}&\quad +\Vert f\Vert _{B^{s_1+\frac{1}{2}+\delta _1,-\delta _2}}\Vert g\Vert _{B^{-\delta _1,s_2+\frac{1}{2}+\delta _2}}+ \Vert f\Vert _{B^{-\delta _3,s_2+\frac{1}{2}+\delta _4}} \Vert g\Vert _{B^{s_1+\frac{1}{2}+\delta _3,-\delta _4}}. \end{aligned}$$

Moreover,

$$\begin{aligned} \Vert fg\Vert _{B^{-\frac{1}{2},\frac{1}{2}}}&\lesssim \Vert f\Vert _{L^\infty }\Vert g\Vert _{B^{-\frac{1}{2},\frac{1}{2}}}+\Vert f\Vert _{B^{\frac{1}{2},\frac{1}{2}}}\Vert \Lambda ^{-1} g\Vert _{L^\infty } \\&\quad +\Vert f\Vert _{B^{\delta _1,-\delta _2}}\Vert g\Vert _{B^{-\delta _1,1+\delta _2}}+ \Vert f\Vert _{B^{-\delta _3,1+\delta _4}} \Vert g\Vert _{B^{\delta _3,-\delta _4}}. \end{aligned}$$

1.1 A.1 Proof of Lemma A.1

Proof of Lemma A.1

Recalling the definitions of the Littlewood-Paley blocks $\dot{\Delta }_j$ and ${\dot{\Delta }}_q^h$ in Section 3.1, we introduce the isotropic and anisotropic paradifferential decomposition due to Bony in [5]. Let $f,g\in \mathcal {S}'(\mathbb {R}^d)$,

$$\begin{aligned} fg=T(f,g)+R(f,g) \quad \!\text{ and }\!\quad fg=T^h(f,g)+R^h(f,g) \end{aligned}$$

where

$$\begin{aligned} T(f,g)= \sum _{j\in \mathbb {Z}}S_{j-1}f{\dot{\Delta }}_jg \quad \!\text{ and }\!\quad R(f,g)=\sum _{j\in \mathbb {Z}}{\dot{\Delta }}_jf S_{j+2}g, \end{aligned}$$

and in the horizontal direction

$$\begin{aligned} T^h(f,g)= \sum _{q\in \mathbb {Z}}S_{q-1}f{\dot{\Delta }}_q^h g \quad \!\text{ and }\!\quad R^h(f,g)=\sum _{q\in \mathbb {Z}}{\dot{\Delta }}_q^h f S_{q+2}g. \end{aligned}$$

Applying the double paraproduct decomposition to fg, one has

$$\begin{aligned} fg=TT^h(f,g)+TR^h(f,g)+RT^h(f,g)+RR^h(f,g). \end{aligned}$$

(50)

Let us estimate each of these terms separately. For the first one, we have

$$\begin{aligned} \Vert {\dot{\Delta }}_j{\dot{\Delta }}_q^h TT^1(f,g)\Vert _{L^2}&\lesssim \sum _{|j'-j|\leqq 4, |q'-q|\leqq 4} \Vert S_{j'-1}S_{q'-1}f\dot{\Delta }_{j'}\dot{\Delta }_{q'}\partial _yg\Vert _{L^2}\\&\lesssim \sum _{|j'-j|\leqq 4, |q'-q|\leqq 4}\Vert S_{j'-1}S_{q'-1}f\Vert _{L^\infty }\Vert \dot{\Delta }_{j'}\dot{\Delta }_{q'}\partial _yg\Vert _{L^2}\\&\lesssim c_{j,q}2^{-js_1}2^{-qs_2}\Vert f\Vert _{L^\infty }\Vert g\Vert _{B^{s_1,s_2}}. \end{aligned}$$

Concerning the fourth term, one has

$$\begin{aligned} \Vert {\dot{\Delta }}_j{\dot{\Delta }}_q^h RR^1(f,g)\Vert _{L^2}&\lesssim \sum _{|j'-j|\leqq 4, |q'-q|\leqq 4} \Vert \dot{\Delta }_{j'}\dot{\Delta }_{q'}fS_{j'+2}S_{q'+2}g\Vert _{L^2}\\ {}&\lesssim \sum _{|j'-j|\leqq 4, |q'-q|\leqq 4} \Vert \dot{\Delta }_{j'}\dot{\Delta }_{q'}f\Vert _{L^2}\Vert S_{j'+2}S_{q'+2}g\Vert _{L^\infty }\\&\lesssim c_{j,q}2^{-js_1}2^{-qs_2}\Vert f\Vert _{B^{s_1,s_2}}\Vert g\Vert _{L^\infty }. \end{aligned}$$

The following computations in the special case $s_1=-\frac{1}{2}$ and $s_2=\frac{1}{2}$ will be needed in the proof of Lemma A.1:

$$\begin{aligned} \Vert {\dot{\Delta }}_j{\dot{\Delta }}_q^h RR^1(f,g)\Vert _{L^2}&\lesssim \sum _{|j'-j|\leqq 4, |q'-q|\leqq 4} \Vert \dot{\Delta }_{j'}\dot{\Delta }_{q'}fS_{j'+2}S_{q'+2}g\Vert _{L^2}\\&\lesssim \sum _{|j'-j|\leqq 4, |q'-q|\leqq 4} \Vert \dot{\Delta }_{j'}\dot{\Delta }_{q'}f\Vert _{2}\Vert S_{j'+2}S_{q'+2}g\Vert _{L^\infty }\\&\lesssim \sum _{|j'-j|\leqq 4, |q'-q|\leqq 4}2^{j}\Vert \dot{\Delta }_{j'}\dot{\Delta }_{q'}f\Vert _{L^2}2^{-j}\Vert S_{j'+2}S_{q'+2}g\Vert _{L^2}\\&\lesssim c_{j,q}2^{\frac{j}{2}}2^{-\frac{q}{2}}\Vert f\Vert _{B^{\frac{1}{2},\frac{1}{2}}}\Vert \Lambda ^{-1} g\Vert _{L^\infty }. \end{aligned}$$

For the third term, using the anisotropic Bernstein inequality, one has

$$\begin{aligned}&\Vert {\dot{\Delta }}_j{\dot{\Delta }}_q^h TR^1(f,g)\Vert _{L^2} \sum _{|j'-j|\leqq 4, |q'-q|\leqq 4} \Vert \dot{\Delta }_{j'}S_{q'+2}fS_{j'+2}\dot{\Delta }_{q'}g\Vert _{L^2}\\&\quad \lesssim \sum _{|j'-j|\leqq 4, |q'-q|\leqq 4} \Vert \dot{\Delta }_{j'}S_{q'+2}f\Vert _{L^2_x L^\infty _y}\Vert S_{j'+2}\dot{\Delta }_{q'}g\Vert _{L^\infty _x L^2_y }\\&\quad \lesssim \sum _{|j'-j|\leqq 4, |q'-q|\leqq 4} 2^{\frac{j'}{2}} \Vert \dot{\Delta }_{j'}S_{q'+2}f\Vert _{L^2}2^{\frac{q'}{2}}\Vert S_{j'+2}\dot{\Delta }_{q'}g\Vert _{L^2}\\&\quad \lesssim 2^{-js_1}2^{-qs_2}\sum _{|j'-j|\leqq 4, |q'-q|\leqq 4} 2^{j'(s_1+\frac{1}{2})} \Vert \dot{\Delta }_{j'}S_{q'+2}f\Vert _{L^2}2^{q'(s_2+\frac{1}{2})}\Vert S_{j'+2}\dot{\Delta }_{q'}g\Vert _{L^2}\\&\quad \lesssim 2^{-js_1}2^{-qs_2}\sum _{|j'-j|\leqq 4, |q'-q|\leqq 4} 2^{j'(s_1+\frac{1}{2}+\delta _1)}2^{-q\delta _2} \Vert \dot{\Delta }_{j'}S_{q'+2}f\Vert _{L^2}2^{q'(s_2+\frac{1}{2}+\delta _2)} 2^{-j'\delta _1}\Vert S_{j'+2}\dot{\Delta }_{q'}g\Vert _{L^2}\\&\quad \lesssim 2^{-js_1}2^{-qs_2} c_{j,q} \left\{ \begin{aligned}&\Vert f\Vert _{B^{s_1+\frac{1}{2}+\delta _1,-\delta _2}}\Vert g\Vert _{B^{-\delta _1,s_2+\frac{1}{2}+\delta _2}} \quad&\text {if } \delta _1,\delta _2>0 \\&\Vert f\Vert _{B^{s_1+\frac{1}{2}}}\Vert g\Vert _{B^{s_2+\frac{1}{2}}_h} \quad&\text {if } \delta _1,\delta _2=0, \end{aligned} \right. \end{aligned}$$

where $B_h^s$ refers to the Besov spaces with horizontal localisation ${\dot{\Delta }}_q$. Since

$$\begin{aligned} \Vert f\Vert _{B^{s_1+\frac{1}{2}}}\Vert g\Vert _{B^{s_2+\frac{1}{2}}_h}\lesssim \Vert f\Vert _{B^{s_1+\frac{1}{2},0}}\Vert g\Vert _{B^{0,s_2+\frac{1}{2}}}, \end{aligned}$$

we obtain the desired estimate. Similar computations lead to

$$\begin{aligned} \Vert {\dot{\Delta }}_j{\dot{\Delta }}_q^h&RT^1(f,g)\Vert _{L^2} \lesssim 2^{-js_1}2^{-qs_2} c_{j,q} \Vert f\Vert _{B^{-\delta _3,s_2+\frac{1}{2}+\delta _4}} \Vert g\Vert _{B^{s_1+\frac{1}{2}+\delta _3,-\delta _4}}. \end{aligned}$$

Multiplying the above estimates by $2^{js_1}2^{qs_2}$ and summing on $j,q\in \mathbb {Z}$, the desired result follows. $\quad \square$

1.2 A.2 Proof of Lemma 5.1

Proof of Lemma 5.1

I) Estimates for $h=(\mathcal {R}_2 \Omega , - \mathcal {R}_1 \Omega ) \cdot \nabla b$. i) Applying Lemma A.1 with $f=\mathcal {R}_1 \Omega$, $g=\partial _yb$, $s_1=\frac{3}{2}$, $s_2=\frac{1}{2}$, $\delta _1=\delta _2=\frac{1}{2}$ and $\delta _3=\delta _4=0$, one has

$$\begin{aligned} \Vert \mathcal {R}_1\Omega \partial _yb\Vert _{L^1_T(B^{\frac{3}{2},\frac{1}{2}})}&\lesssim \Vert \mathcal {R}_1\Omega \Vert _{L^1_T(L^\infty )}\Vert \partial _yb\Vert _{L^\infty _T(B^{\frac{3}{2},\frac{1}{2}})}\nonumber \\&\quad +\Vert \mathcal {R}_1\Omega \Vert _{L^1_T(B^{\frac{3}{2},\frac{1}{2}})}\Vert \partial _yb\Vert _{L^\infty _T(L^\infty )}\nonumber \\&\quad + \Vert \mathcal {R}_1\Omega \Vert _{L^2_T(B^{\frac{5}{2},-\frac{1}{2}})}\Vert \partial _yb\Vert _{L^2_T(B^{-\frac{1}{2},\frac{3}{2}})}\nonumber \\&\quad +\Vert \mathcal {R}_1\Omega \Vert _{L^1_T(B^{0,1})}\Vert \partial _yb\Vert _{L^\infty _T(B^{2,0})}. \end{aligned}$$

(51)

Now, using Lemma 3.2 and B.5 we have

• $\Vert \mathcal {R}_1\Omega \Vert _{L^1_T(L^\infty )}\Vert \partial _yb\Vert _{L^\infty _T(B^{\frac{3}{2},\frac{1}{2}})} \lesssim X(t)\Vert b\Vert _{L^\infty ({\dot{H}}^s\cap H^{3-\varepsilon })} \lesssim X(t)^2$,

• $\Vert \mathcal {R}_1\Omega \Vert _{L^1_T(B^{\frac{3}{2},\frac{1}{2}})}\Vert \partial _yb\Vert _{L^\infty _T(L^\infty )} \lesssim Y(t)\Vert b\Vert _{L^\infty (H^{2+\varepsilon })} \lesssim Y(t)X(t)$,

• $\Vert \mathcal {R}_1\Omega \Vert _{L^2_T(B^{\frac{5}{2},-\frac{1}{2}})}\Vert \partial _yb\Vert _{L^2_T(B^{-\frac{1}{2},\frac{3}{2}})}\lesssim \Vert \Omega \Vert _{L^2_T(B^{\frac{3}{2},\frac{1}{2}})}\Vert b\Vert _{L^2_T(B^{\frac{1}{2},\frac{3}{2}})} \lesssim Y(t)^2$,

• $\Vert \mathcal {R}_1\Omega \Vert _{L^1_T(B^{0,1})}\Vert \partial _yb\Vert _{L^\infty _T(B^{2,0})}\lesssim \Vert \Omega \Vert _{L^1_T(B^{-1,2})}\Vert b\Vert _{L^\infty _T(B^{3,0})}\lesssim \Vert \Omega \Vert _{L^1_T(B^{-\frac{1}{2},\frac{3}{2}})} \Vert b\Vert _{L^\infty _T(B^{3,0})} \lesssim X(t)^2$.

Gathering the above estimates, one obtains

$$\begin{aligned} \Vert {\dot{\Delta }}_j{\dot{\Delta }}_q^h \mathcal {R}_1\Omega \partial _yb\Vert _{L^1_T(B^{\frac{3}{2},\frac{1}{2})}}\lesssim X(t)^2. \end{aligned}$$

ii) Applying Lemma A.1 with $f=\mathcal {R}_1 \Omega$, $g=\partial _yb$, $s_1=\frac{1}{2}$, $s_2=\frac{1}{2}$, $\delta _1=\delta _2=\frac{1}{2}$ and $\delta _3=\delta _4=0$, we get

$$\begin{aligned} \Vert \mathcal {R}_1\Omega \partial _yb\Vert _{L^1_T(B^{\frac{1}{2},\frac{1}{2}})}&\lesssim \Vert \mathcal {R}_1\Omega \Vert _{L^1_T(L^\infty )}\Vert \partial _y b\Vert _{L^\infty _T(B^{\frac{1}{2},\frac{1}{2}})}+\Vert \mathcal {R}_1\Omega \Vert _{L^1_T(B^{\frac{1}{2},\frac{1}{2}})}\Vert \partial _yb\Vert _{L^\infty _T(L^\infty )}\\ {}&\quad +\Vert \mathcal {R}_1\Omega \Vert _{L^2_T(B^{\frac{3}{2},-\frac{1}{2}})}\Vert \partial _yb\Vert _{L^2_T(B^{-\frac{1}{2},\frac{3}{2}})}\\&\quad +\Vert \mathcal {R}_1\Omega \Vert _{L^1_T(B^{0,1})}\Vert \partial _yb\Vert _{L^\infty _T(B^{1,0})}. \end{aligned}$$

Using Lemma 3.2 and B.5, we have

• $\Vert \mathcal {R}_1\Omega \Vert _{L^1_T(L^\infty )}\Vert \partial _yb\Vert _{L^\infty _T(B^{\frac{1}{2},\frac{1}{2}})} \lesssim Y(t)\Vert b\Vert _{L^\infty _T({\dot{H}}^s\cap H^{2-\tau })} \lesssim X(t)^2$,

• $\Vert \mathcal {R}_1\Omega \Vert _{L^1_T(B^{\frac{1}{2},\frac{1}{2}})}\Vert \partial _yb\Vert _{L^\infty _T(L^\infty )} \lesssim Y(t)\Vert b\Vert _{L^\infty (H^{2+\tau })} \lesssim Y(t)X(t)$,

• $\Vert \mathcal {R}_1\Omega \Vert _{L^2_T(B^{\frac{3}{2},-\frac{1}{2}})}\Vert \partial _yb\Vert _{L^2_T(B^{-\frac{1}{2},\frac{3}{2}})}\lesssim \Vert \Omega \Vert _{L^2_T(B^{\frac{1}{2},\frac{1}{2}})}\Vert b\Vert _{L^2_T(B^{\frac{1}{2},\frac{3}{2}})} \lesssim Y(t)^2$,

• $\Vert \mathcal {R}_1\Omega \Vert _{L^1_T(B^{0,1})}\Vert \partial _yb\Vert _{L^\infty _T(B^{1,0})}\lesssim \Vert \Omega \Vert _{L^1_T(B^{-1,2})}\Vert b\Vert _{L^\infty _T(B^{2,0})}\lesssim \Vert \Omega \Vert _{L^1_T(B^{-\frac{1}{2},\frac{3}{2}})}\Vert b\Vert _{L^\infty _T(B^{2,0})} \lesssim X(t)^2$.

iii) For the second addend of h, applying Lemma A.1 with $f=\mathcal {R}_2 \Omega$, $g=\partial _xb$, $s_1=\frac{3}{2}$, $s_2=\frac{1}{2}$, $\delta _1=\delta _2=\frac{1}{2}$ and $\delta _3=\delta _4=0$, one obtains

$$\begin{aligned} \Vert \mathcal {R}_2\Omega \partial _yb\Vert _{L^1_T(B^{\frac{3}{2},\frac{1}{2}})}&\lesssim \Vert \mathcal {R}_2\Omega \Vert _{L^2_T(L^\infty )}\Vert \partial _xb\Vert _{L^2_T(B^{\frac{3}{2},\frac{1}{2}})}\\&\quad +\Vert \mathcal {R}_2\Omega \Vert _{L^2_T(B^{\frac{3}{2},\frac{1}{2}})}\Vert \partial _xb\Vert _{L^2_T(L^\infty )}\\&\quad + \Vert \mathcal {R}_2\Omega \Vert _{L^\infty _T(B^{\frac{5}{2},-\frac{1}{2}})}\Vert \partial _xb\Vert _{L^1_T(B^{-\frac{1}{2},\frac{3}{2}})}\\&\quad +\Vert \mathcal {R}_2\Omega \Vert _{L^2_T(B^{0,1})}\Vert \partial _xb\Vert _{L^2_T(B^{2,0})}. \end{aligned}$$

Let us deal with each r.h.s. term:

• $\Vert \mathcal {R}_2\Omega \Vert _{L^2_T(L^\infty )}\Vert \partial _xb\Vert _{L^2_T(B^{\frac{3}{2},\frac{1}{2}})}\lesssim \Vert \mathcal {R}_2\Omega \Vert _{L^2_T(B^{\frac{1}{2},\frac{1}{2})}}\Vert \mathcal {R}_1b\Vert _{L^2_T({\dot{H}}^s\cap H^{3-\tau )}} \lesssim Y(t)X(t),$

• $\Vert \mathcal {R}_2\Omega \Vert _{L^2_T(B^{\frac{3}{2},\frac{1}{2}})}\Vert \partial _xb\Vert _{L^2_T(L^\infty )}\lesssim Y(t)X(t),$

• $\Vert \mathcal {R}_2\Omega \Vert _{L^\infty _T(B^{\frac{5}{2},-\frac{1}{2}})}\Vert \partial _xb\Vert _{L^1_T(B^{-\frac{1}{2},\frac{3}{2}})}\lesssim \Vert \mathcal {R}_2\Omega \Vert _{L^\infty _T(B^{\frac{5}{2},-\frac{1}{2}})}\Vert b\Vert _{L^1_T(B^{-\frac{1}{2},\frac{5}{2}})}\lesssim X(t)Y(t),$

• $\Vert \mathcal {R}_2\Omega \Vert _{L^2_T(B^{0,1})}\Vert \partial _xb\Vert _{L^2_T(B^{2,0})}\lesssim \Vert \mathcal {R}_2\Omega \Vert _{L^2_T(B^{\frac{1}{2},\frac{1}{2}})}\Vert \partial _xb\Vert _{L^2_T(B^{2,0})}\lesssim Y(t)X(t)$.

iv) Applying Lemma A.1 with $f=\mathcal {R}_2 \Omega$, $g=\partial _xb$, $s_1=\frac{1}{2}$, $s_2=\frac{1}{2}$, $\delta _1=\delta _2=0$ and $\delta _3=\delta _4=0$,

$$\begin{aligned} \Vert \mathcal {R}_2\Omega \partial _xb\Vert _{L^1_T(B^{\frac{1}{2},\frac{1}{2})}}&\lesssim \Vert \mathcal {R}_2\Omega \Vert _{L^2_T(L^\infty )}\Vert \partial _xb\Vert _{L^2_T(B^{\frac{1}{2},\frac{1}{2}})}\\&\quad +\Vert \mathcal {R}_2\Omega \Vert _{L^2_T(B^{\frac{1}{2},\frac{1}{2}})}\Vert \partial _xb\Vert _{L^2_T(L^\infty )}\\&\quad +\Vert \mathcal {R}_2\Omega \Vert _{L^2_T(B^{1,0})}\Vert \partial _xb\Vert _{L^2_T(B^{0,1})}\\&\quad +\Vert \mathcal {R}_2\Omega \Vert _{L^2_T(B^{0,1})}\Vert \partial _xb\Vert _{L^2_T(B^{1,0})}. \end{aligned}$$

Again, we deal with each term separately:

• $\Vert \mathcal {R}_2\Omega \Vert _{L^2_T(L^\infty )}\Vert \partial _xb\Vert _{L^2_T(B^{\frac{1}{2},\frac{1}{2}})}\lesssim \Vert \Omega \Vert _{L^2_T(B^{\frac{1}{2},\frac{1}{2}})}\Vert b\Vert _{L^2_T(B^{\frac{1}{2},\frac{3}{2}})}\leqq Y(t)^2$,

• $\Vert \mathcal {R}_2\Omega \Vert _{L^2_T(B^{\frac{1}{2},\frac{1}{2}})}\Vert \partial _xb\Vert _{L^2_T(L^\infty )} \lesssim \Vert \Omega \Vert _{L^2_T(B^{\frac{1}{2},\frac{1}{2}})}\Vert b\Vert _{L^2_T(B^{\frac{1}{2},\frac{3}{2}})}\lesssim Y(t)^2$,

• $\Vert \mathcal {R}_2\Omega \Vert _{L^2_T(B^{1,0})}\Vert \partial _xb\Vert _{L^2_T(B^{0,1})}\lesssim \Vert \Omega \Vert _{L^2_T(B^{1,0})}\Vert \mathcal {R}_1b\Vert _{L^2_T(B^{2,0})}\lesssim X(t)^2$,

• $\Vert \mathcal {R}_2\Omega \Vert _{L^2_T(B^{0,1})}\Vert \partial _xb\Vert _{L^2_T(B^{1,0})}\lesssim \Vert \Omega \Vert _{L^2_T(B^{1,0})}\Vert \mathcal {R}_1b\Vert _{L^2_T(B^{2,0})}\lesssim X(t)^2$.

Gathering the above estimates yields

$$\begin{aligned} \Vert {\dot{\Delta }}_j{\dot{\Delta }}_q^h \mathcal {R}_1\Omega \partial _yb\Vert _{L^1_T(B^{\frac{1}{2},\frac{1}{2}})}&\lesssim X(t)^2. \end{aligned}$$

Adding the estimates from $i)-iv)$ and using that $Y(t)\lesssim X(t)$ thanks to Lemma 3.2, one obtains

$$\begin{aligned} \Vert h\Vert _{L^1_T(B^{\frac{3}{2},\frac{1}{2}}\cap B^{\frac{1}{2},\frac{1}{2}})}&\lesssim X(t)^2. \end{aligned}$$

(52)

II) Estimates for $g=\varepsilon \mathcal {R}_1 (\mathcal {R}_2 \Omega , -\mathcal {R}_1\Omega ) \cdot \nabla b + \Lambda ^{-1} ((\mathcal {R}_2 \Omega , -\mathcal {R}_1\Omega ) \cdot (\nabla \Lambda \Omega )).$

The estimates of the first term $\varepsilon \mathcal {R}_1 (\mathcal {R}_2 \Omega , -\mathcal {R}_1\Omega ) \cdot \nabla b$ can be obtained in the same way as the previous terms. Indeed, notice that by Lemma B.5 and thanks to the boundedness of the Riesz transform $\mathcal {R}_1: B^{1,0}\rightarrow B^{1,0}$, one has

$$\begin{aligned} \Vert \mathcal {R}_1 (\mathcal {R}_2 \Omega \partial _x b)\Vert _{L_T^1(B^{\frac{1}{2}, \frac{1}{2}})}&\lesssim \Vert \mathcal {R}_1 (\mathcal {R}_2 \Omega \partial _x b)\Vert _{L_T^1(B^{1, 0})}\lesssim \Vert \mathcal {R}_2 \Omega \partial _x b\Vert _{L_T^1(B^{1, 0})}, \end{aligned}$$

which is exactly the same term as the one we treated in $\varvec{I})$. A similar argument can be applied for the bound in $B^{\frac{3}{2},\frac{1}{2}}$. We then turn to the second addend of g.

i) First, observing that

$$\begin{aligned} \Vert \Lambda ^{-1} ((\mathcal {R}_2 \Omega , -\mathcal {R}_1\Omega ) \cdot (\nabla \Lambda \Omega ))\Vert _{B^{\frac{3}{2},\frac{1}{2}}}\leqq \Vert (\mathcal {R}_2 \Omega , -\mathcal {R}_1\Omega ) \cdot (\nabla \Lambda \Omega )\Vert _{B^{\frac{1}{2},\frac{1}{2}}} \end{aligned}$$

and since $\Omega$ has better decay properties than b and we control $\Omega$ in $L^2_T(H^s)$ for $s> 3$, we can directly deduce from our previous computations that

$$\begin{aligned} \Vert \Lambda ^{-1} ((\mathcal {R}_2 \Omega , -\mathcal {R}_1\Omega ) \cdot (\nabla \Lambda \Omega ))\Vert _{B^{\frac{3}{2},\frac{1}{2}}}\leqq X(t)^2. \end{aligned}$$

ii) For the second regularity setting it is a bit trickier as one has

$$\begin{aligned} \Vert \Lambda ^{-1} ((\mathcal {R}_2 \Omega , -\mathcal {R}_1\Omega ) \cdot (\nabla \Lambda \Omega ))\Vert _{B^{\frac{1}{2},\frac{1}{2}}}&\lesssim \Vert (\mathcal {R}_2 \Omega , -\mathcal {R}_1\Omega ) \cdot (\nabla \Lambda \Omega )\Vert _{B^{-\frac{1}{2},\frac{1}{2}}}. \end{aligned}$$

Applying the second inequality of Lemma A.1 with $f=\mathcal {R}_2 \Omega$, $g=\partial _x\Lambda \Omega$, $s_1=-\frac{1}{2}$, $s_2=\frac{1}{2}$, $\delta _1=1$, $\delta _2=0$ and $\delta _3=\delta _4=0$, we obtain

$$\begin{aligned} \Vert \mathcal {R}_2\Omega \partial _x\Lambda \Omega \Vert _{L^1_T(B^{-\frac{1}{2},\frac{1}{2}})}&\lesssim \Vert \mathcal {R}_2\Omega \Vert _{L^\infty _T(L^\infty )}\Vert \partial _x\Lambda \Omega \Vert _{L^1_T(B^{-\frac{1}{2},\frac{1}{2}})}\\&\quad +\Vert \mathcal {R}_2\Omega \Vert _{L^\infty _T(B^{\frac{1}{2},\frac{1}{2}})}\Vert \Lambda ^{-1} \partial _x\Lambda \Omega \Vert _{L^1_T(L^\infty )}\\&\quad + \Vert \mathcal {R}_2\Omega \Vert _{L^2_T(B^{1,0})}\Vert \partial _x\Lambda \Omega \Vert _{L^2_T(B^{-1,1})}\\&\quad +\Vert \mathcal {R}_2\Omega \Vert _{L^2_T(B^{0,1})}\Vert \partial _x\Lambda \Omega \Vert _{L^2_T(B^{0,0})}. \end{aligned}$$

$\square$

Remark A.1

Alternatively, one could think to exploit the first inequality of Lemma A.1, which would require a control of $\Vert \mathcal {R}_2\Omega \Vert _{L^2_T(B^{-\frac{1}{2},\frac{1}{2}})}\Vert \partial _x\Lambda \Omega \Vert _{L^2_T(L^\infty )}$. However, such bounds hold under additional low-regularity assumptions on the initial data, for instance $B^{0,0}$, that we want to avoid here.

We estimate the above terms as follows.

• $\Vert \mathcal {R}_2\Omega \Vert _{L^\infty _T(L^\infty )}\Vert \partial _x\Lambda \Omega \Vert _{L^1_T(B^{-\frac{1}{2},\frac{1}{2}})}\lesssim \Vert \Omega \Vert _{L^\infty _T(B^{\frac{1}{2},\frac{1}{2}})}\Vert \Omega \Vert _{L^1_T(B^{\frac{1}{2},\frac{3}{2}})}\lesssim Y(t)^2$,

• $\Vert \mathcal {R}_2\Omega \Vert _{L^\infty _T(B^{\frac{1}{2},\frac{1}{2}})}\Vert \Lambda ^{-1} \partial _x\Lambda \Omega \Vert _{L^1_T(L^\infty )} \lesssim \Vert \Omega \Vert _{L^\infty _T(B^{\frac{1}{2},\frac{1}{2}})}\Vert \Omega \Vert _{L^1_T(B^{\frac{1}{2},\frac{3}{2}})}\lesssim X(t)Y(t)$,

• $\Vert \mathcal {R}_2\Omega \Vert _{L^2_T(B^{1,0})}\Vert \partial _x\Lambda \Omega \Vert _{L^2_T(B^{-1,1})}\lesssim \Vert \Omega \Vert _{L^2_T(B^{1,0})}\Vert \Omega \Vert _{L^2_T(B^{0,2})}\lesssim \Vert \Omega \Vert _{L^2_T(B^{1,0})}\Vert \Omega \Vert _{L^2_T(B^{2,0})}\lesssim X(t)^2$,

• $\Vert \mathcal {R}_2\Omega \Vert _{L^2_T(B^{0,1})}\Vert \partial _x\Lambda \Omega \Vert _{L^2_T(B^{0,0})}\lesssim \Vert \Omega \Vert _{L^2_T(B^{1,0})}\Vert \Omega \Vert _{L^2_T(B^{1,1})}\lesssim \Vert \Omega \Vert _{L^2_T(B^{1,0})}\Vert \Omega \Vert _{L^2_T(B^{\frac{3}{2},\frac{1}{2}})}\lesssim X(t)Y(t)$.

The estimates for the last term $\mathcal {R}_1\Omega \partial _y\Lambda \Omega$ follow the exact same lines, then we omit them. We have therefore

$$\begin{aligned} \Vert g\Vert _{L^1_T(B^{\frac{3}{2},\frac{1}{2}}\cap B^{\frac{1}{2},\frac{1}{2}})}&\lesssim X(t)^2. \end{aligned}$$

(53)

Adding (52) and (53) concludes the proof of Lemma 5.1. $\quad \square$

Appendix B. Toolbox

We collect some technical lemmas that are used in the course of the proof the results of this article. In some case, we provide (short) proofs, while in other cases, appealing to the existing literature to which we refer explicitly, we omit the proofs. When no explicit reference is provided being the results classical, the reader can look for instance at [20].

Lemma B.1

Let $X: [0,T]\rightarrow \mathbb {R}_+$ be a continuous function such that $X^2$ is differentiable. Assume that there exists a constant $B\geqq 0$ and a measurable function $A: [0,T]\rightarrow \mathbb {R}_+$ such that

$$\begin{aligned} \frac{1}{2}\frac{d}{dt}X^2+BX^2\leqq AX\quad \hbox {a.e. on }\ [0,T]. \end{aligned}$$

Then, for all $t\in [0,T],$ we have

$$\begin{aligned} X(t)+B\int _0^tX\leqq X_0+\int _0^tA. \end{aligned}$$

Lemma B.2

(Product estimates, [35, Lemma 2.1]) Let $s>0, 1 \leqq p,r \leqq \infty$, then

$$\begin{aligned} \Vert \Lambda ^s(fg)\Vert _{L^p(\mathbb {R}^2)} \lesssim \Vert f\Vert _{L^{p_1}(\mathbb {R}^2)}\Vert \Lambda ^{s}g\Vert _{L^{p_2}(\mathbb {R}^2)}+\Vert g\Vert _{L^{r_1}(\mathbb {R}^2)}\Vert \Lambda ^s f\Vert _{L^{r_2}(\mathbb {R}^2)}, \end{aligned}$$

where $1 \leqq p_1, r_1 \leqq \infty$ such that $\frac{1}{p}=\frac{1}{p_1}+\frac{1}{p_2}=\frac{1}{r_1}+\frac{1}{r_2}$.

Lemma B.3

(Commutator estimates, [35, Lemma 2.1]) Let $s>0$, $1 \leqq p_1, r_1 \leqq \infty$ and $1<p, p_1, r_1 <1$ such that $\frac{1}{p}=\frac{1}{p_1} + \frac{1}{p_2} = \frac{1}{r_1} + \frac{1}{r_2}$. Then

$$\begin{aligned} \Vert [\Lambda ^s, f]g\Vert _{L^p} \lesssim (\Vert \nabla f\Vert _{L^{p_1}}\Vert \Lambda ^{s-1}g\Vert _{L^{p_2}}+\Vert \Lambda ^s f\Vert _{L^{r_1}}\Vert g\Vert _{L^{r_2}}). \end{aligned}$$

Lemma B.4

(Interpolation) Let $s_0 \leqq s \leqq s_1$. Then, for $\theta \in (0,1)$, such that $s=\theta s_0+(1-\theta )s_1$, it holds

$$\begin{aligned} \Vert f\Vert _{\dot{H}^s} \lesssim \Vert f\Vert _{\dot{H}^{s_0}}^\theta \Vert f\Vert _{\dot{H}^{s_1}}^{1-\theta }. \end{aligned}$$

Lemma B.5

(Embedding in Besov spaces) For $s>0$, $s_1,s_2\in \mathbb {R}$, one has

$$\begin{aligned} B^{s_1+s,s_2-s} \subset B^{s_1,s_2} \end{aligned}$$

(54)

Proof

Let $f\in B^{s_1+s,s_2-s}\cap B^{s_1,s_2}$. By definition of the localisation ${\dot{\Delta }}_j$ and ${\dot{\Delta }}_q^h$, when applying ${\dot{\Delta }}_j{\dot{\Delta }}_q^h$ to a function, one can use that there exists $N_0$ such that $j\geqq q- N_0$. This implies that

$$\begin{aligned} \sum _{j,q\in \mathbb {Z}} 2^{js_1}2^{qs_2}\Vert {\dot{\Delta }}_j{\dot{\Delta }}_q^hf\Vert _{L^2}&\lesssim \sum _{j,q\in \mathbb {Z}} 2^{js_1}2^{js}2^{-js}2^{qs_2}\Vert {\dot{\Delta }}_j{\dot{\Delta }}_q^hf\Vert _{L^2}\\&\lesssim \sum _{j,q\in \mathbb {Z},j\geqq q-N_0} 2^{js_1}2^{js}2^{-qs}2^{qs_2}\Vert {\dot{\Delta }}_j{\dot{\Delta }}_q^hf\Vert _{L^2}\\&\lesssim \sum _{j,q\in \mathbb {Z}} 2^{j(s_1+s)}2^{q(s_2-s)}\Vert {\dot{\Delta }}_j{\dot{\Delta }}_q^hf\Vert _{L^2}. \end{aligned}$$

$\square$

The next lemma provides a generalized version of the Kenig–Ponce–Vega inequality (the fractional version of the Leibniz rule) for all $s>0$, see [27] and [14].

Recall the notation $\alpha , \beta \in \mathbb {N}^2$ (multi-index) and $\nabla ^\alpha =(\partial _x^{\alpha _1}, \partial _y^{\alpha _2})$, while the operator $\Lambda ^{s, \alpha }$ is defined via Fourier transform as

$$\begin{aligned} \widehat{\Lambda ^{s, \alpha } f}(\xi ) = \widehat{\Lambda ^{s, \alpha }}(\xi ) {\widehat{f}}(\xi ), \qquad \widehat{\Lambda ^{s, \alpha }}(\xi )=i^{-|\alpha |}\partial _\xi ^\alpha (|\xi |^s). \end{aligned}$$

Lemma B.6

(Generalized Kenig–Ponce–Vega inequality [27, Theorem 5.1]) Let $s>0$ and $1<p<\infty$. Then, for any $s_1, s_2 \geqq 0$ such that $s_1+s_2=s$, and any $f, g \in \mathcal {S}(\mathbb {R}^d)$,

$$\begin{aligned}&\left\| \Lambda ^s (fg)- \sum _{|\alpha | \leqq s_1} \frac{1}{\alpha !} (\nabla ^\alpha f) (\Lambda ^{s,\alpha }g) - \sum _{|\beta | \leqq s_2-1} \frac{1}{\beta !} (\nabla ^\beta g) (\Lambda ^{s,\beta }f) \right\| _{L^p}\\&\quad \lesssim \Vert \Lambda ^{s_1} f\Vert _{\text {BMO}} \Vert \Lambda ^{s_2} g\Vert _{L^p}. \end{aligned}$$