A Paradifferential Approach for Well-Posedness of the Muskat Problem

Abstract

We study the Muskat problem for one fluid or two fluids, with or without viscosity jump, with or without rigid boundaries, and in arbitrary space dimension d of the interface. The Muskat problem is scaling invariant in the Sobolev space $H^{s_c}({\mathbb {R}}^d)$ where $s_c=1+\frac{d}{2}$. Employing a paradifferential approach, we prove local well-posedness for large data in any subcritical Sobolev spaces $H^s({\mathbb {R}}^d)$, $s>s_c$. Moreover, the rigid boundaries are only required to be Lipschitz and can have arbitrarily large variation. The Rayleigh–Taylor stability condition is assumed for the case of two fluids with viscosity jump but is proved to be automatically satisfied for the case of one fluid. The starting point of this work is a reformulation solely in terms of the Drichlet–Neumann operator. The key elements of proofs are new paralinearization and contraction results for the Drichlet–Neumann operator in rough domains.

Appendices

Appendix A. Traces for Homogeneous Sobolev Spaces

1.1 A.1. Infinite Strip-Like Domains

Let $\eta _1$ and $\eta _2$ be two Lipchitz functions on ${\mathbb {R}}^d$, $\eta _j\in \dot{W}^{1, \infty }({\mathbb {R}}^d)$, such that $\eta _1>\eta _2$. Set

$$\begin{aligned} L=\Vert \nabla \eta _1\Vert _{L^\infty }+\Vert \nabla \eta _2\Vert _{L^\infty }, \quad \Theta (x)=\frac{\eta _1(x)-\eta _2(x)}{2L}. \end{aligned}$$

Consider the infinite strip-like domain

$$\begin{aligned} U=\{(x, y)\in {\mathbb {R}}^{d+1}: \eta _2(x)<y<\eta _1(x)\}. \end{aligned}$$

(A.1)

We record in this Appendix the trace theory in [49] (see also [60]) for $\dot{H}^1(U)$ where

$$\begin{aligned} \dot{H}^1(U)=\{u\in L^2_{loc}(U): \nabla u\in L^2(U)\}/~{\mathbb {R}}. \end{aligned}$$

Theorem A.1

([49, Theorem 5.1]) There exists a unique linear operator

$$\begin{aligned} \text {Tr}: \dot{H}^1(U)\rightarrow L^2_{loc}({\mathbb {R}}^d) \end{aligned}$$

such that the following hold:

• (1) $\text {Tr}(u)=u\vert _{\partial U}$ for all $u\in \dot{H}^1(U)\cap C({{\overline{U}}})$.

• (2) There exists a positive constant $C=C(d)$ such that for all $u\in \dot{H}^1(U)$, the functions $g_j=\text {Tr}(u)(\cdot , \eta _j(\cdot ))$ are in $\widetilde{H}^\frac{1}{2}_\Theta ({\mathbb {R}}^d)$ and satisfy

  $$\begin{aligned}&\Vert g_j\Vert _{ \widetilde{H}^\frac{1}{2}_\Theta ({\mathbb {R}}^d)}\leqq C(1+L) \Vert u\Vert _{\dot{H}^1(U)}, \end{aligned}$$

  (A.2)
  $$\begin{aligned}&\int _{{\mathbb {R}}^d}\frac{|g_1(x)-g_2(x)|^2}{\eta _1(x)-\eta _2(x)}\,\hbox {d}x\leqq C\int _U |\partial _y u(x, y)|^2\,\hbox {d}x\,\hbox {d}y. \end{aligned}$$

  (A.3)

Recall that the space $\widetilde{H}^\frac{1}{2}_\Theta ({\mathbb {R}}^d)$ is defined by (3.5).

Theorem A.2

([49, Theorem 5.4]) Suppose that $g_1$ and $g_2$ are in $\widetilde{H}^\frac{1}{2}_{a(\eta _1-\eta _2)}({\mathbb {R}}^d)$ for some $a\in (0, 1)$ such that

$$\begin{aligned} \int _{{\mathbb {R}}^d}\frac{|g_1(x)-g_2(x)|^2}{\eta _1(x)-\eta _2(x)}\,\hbox {d}x<\infty . \end{aligned}$$

Then there exists $u\in \dot{H}^1(U)$ such that $\text {Tr}(u)(\cdot , \eta _j(\cdot ))=g_j$ and

$$\begin{aligned} \Vert u\Vert ^2_{\dot{H}^1(U)}\leqq & {} C(1+L)^2\int _{{\mathbb {R}}^d}\frac{|g_1(x)-g_2(x)|^2}{\eta _1(x)-\eta _2(x)}\,\hbox {d}x\nonumber \\&+\,C(1+L)^2\left( \Vert g_1\Vert ^2_{\widetilde{H}^\frac{1}{2}_{a(\eta _1-\eta _2)}({\mathbb {R}}^d)}+\Vert g_2\Vert ^2_{\widetilde{H}^\frac{1}{2}_{a(\eta _1-\eta _2)}({\mathbb {R}}^d)}\right) \end{aligned}$$

(A.4)

where $C=C(d)$.

1.2 A.2. Lipschitz Half Spaces

For a Lipschitz function $\eta$ on ${\mathbb {R}}^d$ we consider the associated half-space

$$\begin{aligned} U=\{(x, y)\in {\mathbb {R}}^{n+1}: y<\eta (x)\}. \end{aligned}$$

Theorem A.3

There exists a unique linear operator

$$\begin{aligned} \text {Tr}: \dot{H}^1(U)\rightarrow L^2_{loc}({\mathbb {R}}^d) \end{aligned}$$

such that the following hold:

• (1) $\text {Tr}(u)=u\vert _{\partial U}$ for all $u\in \dot{H}^1(U)\cap C({{\overline{U}}})$.

• (2) There exists a positive constant $C=C(d)$ such that for all $u\in \dot{H}^1(U)$, the function $g=\text {Tr}(u)(\cdot , \eta (\cdot ))$ is in $\widetilde{H}^\frac{1}{2}_\infty ({\mathbb {R}}^d)$ and satisfies

  $$\begin{aligned} \Vert g\Vert _{ \widetilde{H}^\frac{1}{2}_\infty ({\mathbb {R}}^d)}\leqq C(1+\Vert \nabla \eta \Vert _{L^\infty ({\mathbb {R}}^d)}) \Vert u\Vert _{\dot{H}^1(U)}. \end{aligned}$$

  (A.5)

Theorem A.4

For each $g\in \widetilde{H}^\frac{1}{2}_\infty ({\mathbb {R}}^d)$, there exists $u\in \dot{H}^1(U)$ such that $\text {Tr}(u)(\cdot , \eta (\cdot ))=g$ and

$$\begin{aligned} \Vert u\Vert _{\dot{H}^1(U)}\leqq C(1+\Vert \nabla \eta \Vert _{L^\infty ({\mathbb {R}}^d)})\Vert g_1\Vert _{\widetilde{H}^\frac{1}{2}_\infty ({\mathbb {R}}^d)} \end{aligned}$$

(A.6)

where $C=C(d)$.

1.3 A.3. Trace of Normalized Normal Derivative

We consider the infinite strip-like domain

$$\begin{aligned} U_h=\{(x, y)\in {\mathbb {R}}^d\times {\mathbb {R}}:~\eta (x)-h<y<\eta (x)\} \end{aligned}$$

of width h underneath the graph of $\eta$

$$\begin{aligned} \Sigma =\{ (x, \eta (x)):~x\in {\mathbb {R}}^d\}. \end{aligned}$$

Note that $n=\frac{1}{\sqrt{1+|\nabla \eta |}}(-\nabla \eta , 1)$ is the upward pointing unit normal to $\Sigma$.

Our goal is to show that for any function u in the homogeneous maximal domain of the Laplace operator $\Delta$

$$\begin{aligned} E(U_h)=\{u\in \dot{H}^1(U_h):~\Delta _{x, y} u\in L^2(U_h)\}, \end{aligned}$$

the trace

$$\begin{aligned} \sqrt{1+|\nabla \eta |^2}\frac{\partial u}{\partial n}\vert _\Sigma =\partial _y u(x, \eta (x))-(\nabla _xu)(x ,\eta (x))\cdot \nabla \eta (x) \end{aligned}$$

makes sense in $H^{-\frac{1}{2}}({\mathbb {R}}^d)$.

Theorem A.5

Assume that $\eta \in \dot{W}^{1, \infty }({\mathbb {R}}^d)$. There exists a unique linear operator ${\mathcal {N}}: E(U_h)\rightarrow H^{-\frac{1}{2}}({\mathbb {R}}^d)$ such that the following hold:

• (1) ${\mathcal {N}}(u)=\sqrt{1+|\nabla \eta |^2}\frac{\partial u}{\partial n}\vert _\Sigma$ if $u\in \dot{H}^1(U_h)\cap C^1({{\overline{U}}}_h)$.

• (2) There exists an absolute constant $C>0$ such that

  $$\begin{aligned} \Vert {\mathcal {N}}(u)\Vert _{H^{-\frac{1}{2}}({\mathbb {R}}^d)}\leqq & {} Ch^{-\frac{1}{2}}(1+\Vert \nabla \eta \Vert _{L^\infty })\Vert \nabla _{x, y}u\Vert _{L^2(U_h)}\\&\quad +\,Ch^{-\frac{1}{2}}\Vert \Delta _{x, y}u\Vert _{L^2(U_h)} \end{aligned}$$

  for all $u\in E(U_h)$.

Proof

Set $S={\mathbb {R}}^d\times (-1, 0)$ and introduce $\theta (x, z)=\eta (x)+zh$ for $(x, z)\in S$. It is clear that $(x, z)\mapsto (x, \theta (x, z))$ is a diffeomorphism from S onto $U_h$. If $f:U_h\rightarrow {\mathbb {R}}$ we denote $\widetilde{f}(x, z)=f(x, \theta (x, z))$. It follows that

$$\begin{aligned} \begin{aligned}&(\partial _yf)(x, \theta (x, z))=\frac{1}{h}\partial _z\widetilde{f}(x, z),\\&(\nabla _xf)(x, \theta (x, z))=\left( \nabla _x-\frac{\nabla \eta }{h}\partial _z\right) \widetilde{f}(x, z):=\Lambda \widetilde{f}. \end{aligned} \end{aligned}$$

For $u\in E(U)$ we have that $f=\Delta u$ satisfies

$$\begin{aligned} \widetilde{f} ={{\,\mathrm{div}\,}}_{x,z}(A \nabla _{x,z}\widetilde{u}) \end{aligned}$$

(A.7)

with

$$\begin{aligned} A= \begin{bmatrix} Id &{} \quad -\frac{\nabla \eta }{h}\\ -\frac{(\nabla \eta )^T}{h} &{}\quad \frac{1+|\nabla \eta |^2}{h^2}. \end{bmatrix} \end{aligned}$$

(A.8)

Equivalently, we have

$$\begin{aligned} {{\,\mathrm{div}\,}}_x\Lambda \widetilde{u}+\partial _z\left( -\frac{\nabla \eta }{h}\nabla _x\widetilde{u}+\frac{1+|\nabla \eta |^2}{h^2}\partial _z\widetilde{u}\right) =\widetilde{f}. \end{aligned}$$

(A.9)

Let us consider the quantity

$$\begin{aligned} \begin{aligned}&(\partial _y u)(x, \theta (x, z)) -(\nabla _xu)(x, \theta (x, z))\cdot \nabla \eta (x)\\&\quad =\frac{1}{h}\partial _z\widetilde{u}(x, z)-\nabla \eta (x, z)\cdot \left( \nabla _x-\frac{\nabla \eta }{h}\partial _z\right) \widetilde{u}(x, z)\\&\quad =-\frac{\nabla \eta (x)}{h}\cdot \nabla _x\widetilde{u}(x, z)+\frac{1+|\nabla \eta (x, z)|^2}{h^2}\partial _z\widetilde{u}(x, z)\\&\quad := \Xi (x, z). \end{aligned} \end{aligned}$$

Using the first expression we deduce easily that

$$\begin{aligned} \Vert \Xi \Vert _{L^2((-1, 0); L^2({\mathbb {R}}^d))}\leqq h^{-\frac{1}{2}}(1+\Vert \nabla \eta \Vert _{L^\infty })\Vert \nabla _{x, y}u\Vert _{L^2(U_h)}. \end{aligned}$$

(A.10)

Moreover, (A.9) implies

$$\begin{aligned} \partial _z \Xi =\widetilde{f}-{{\,\mathrm{div}\,}}_x\Lambda \widetilde{u}. \end{aligned}$$

(A.11)

Note that if $u\in C^1({{\overline{U}}})$ then

$$\begin{aligned} \partial _y u(x, \eta (x)) -(\nabla _xu)(x ,\eta (x))\cdot \nabla \eta (x)=\Xi (x, 0). \end{aligned}$$

We shall appeal to Theorem A.6 below to prove that the trace $\Xi \vert _{z=0}$ is well-defined in $H^{-\frac{1}{2}}({\mathbb {R}}^d)$ for $u\in \dot{H}^1(U_h)$. To this end, we use (A.11) to have

$$\begin{aligned} \Vert \partial _z\Xi \Vert _{L^2((-1, 0); H^{-1}({\mathbb {R}}^d))}\leqq \Vert \widetilde{f}\Vert _{L^2((-1, 0); H^{-1}({\mathbb {R}}^d))}+\Vert \Lambda \widetilde{u}\Vert _{L^2((-1, 0); L^2({\mathbb {R}}^d))} \end{aligned}$$

where

$$\begin{aligned} \Vert \Lambda \widetilde{u}\Vert _{L^2((-1, 0); L^2({\mathbb {R}}^d))}\leqq h^{-\frac{1}{2}}\Vert \nabla _xu\Vert _{L^2(U_h)}. \end{aligned}$$

On the other hand,

$$\begin{aligned} \Vert \widetilde{f}\Vert _{L^2((-1, 0); H^{-1}({\mathbb {R}}^d))}\leqq \Vert \widetilde{f}\Vert _{L^2(S)}\leqq h^{-\frac{1}{2}}\Vert f\Vert _{L^2(U_h)}, \end{aligned}$$

hence

$$\begin{aligned} \Vert \partial _z\Xi \Vert _{L^2((-1, 0); H^{-1}({\mathbb {R}}^d))}\leqq h^{-\frac{1}{2}}\big (\Vert f\Vert _{L^2(U_h)}+\Vert \nabla _xu\Vert _{L^2(U_h)}\big ). \end{aligned}$$

(A.12)

Combining (A.10) and (A.12) we conclude by virtue of Theorem A.6 that $\Xi \in C([-1, 0]; L^2({\mathbb {R}}^d))$ and

$$\begin{aligned} \Vert \Xi \Vert _{ C([-1, 0]; L^2({\mathbb {R}}^d))}\leqq Ch^{-\frac{1}{2}}(1+\Vert \nabla \eta \Vert _{L^\infty })\Vert \nabla _{x, y}u\Vert _{L^2(U_h)}+Ch^{-\frac{1}{2}}\Vert f\Vert _{L^2(U_h)} \end{aligned}$$

(A.13)

for some absolute constant $C>0$. $\quad \square$

Theorem A.6

([48, Theorem 3.1]) Let $s \in {\mathbb {R}}$ and I be a closed (bounded or unbounded) interval in ${\mathbb {R}}$. Let $u\in L^2_z(I, H^{s+ \frac{1}{2}}({\mathbb {R}}^d))$ such that $\partial _z u \in L_z^2(I, H^{s-\frac{1}{2}}({\mathbb {R}}^d)).$ Then $u \in BC(I, H^{s}({\mathbb {R}}^d))$ and there exists an absolute constant $C>0$ such that

$$\begin{aligned} \sup _{z\in I}\Vert u(z, \cdot ) \Vert _{H^{s}({\mathbb {R}}^d)} \leqq C \left( \Vert u\Vert _{L^2(I,H^{s+ \frac{1}{2}}({\mathbb {R}}^d))}+ \Vert \partial _z u\Vert _{L^2(I,H^{s- \frac{1}{2}}({\mathbb {R}}^d))} \right) . \end{aligned}$$

1.4 A.4. Proof of Proposition 3.2

Assuming (3.14), let us prove (3.15). By comparing $\sigma _1$ with $\min \{\sigma _1, \sigma _2\}+M$ it suffices to prove the following claim. If $\sigma \geqq 2h>0$ for some $h>0$, then for any $M>0$, there exists $C=C(d, h, M)$ such that

$$\begin{aligned} \Vert f\Vert _{\widetilde{H}^\frac{1}{2}_{\sigma +M}}\leqq C\Vert f\Vert _{\widetilde{H}^\frac{1}{2}_{\sigma }}. \end{aligned}$$

(A.14)

By iteration, (A.14) will follow from the same estimate with M replaced by $\delta =\frac{h}{10}$. To prove this, we first note that

$$\begin{aligned} \begin{aligned} \Vert f\Vert _{\widetilde{H}^\frac{1}{2}_{\sigma +\delta }}^2&=\Vert f\Vert _{\widetilde{H}^\frac{1}{2}_{\sigma }}^2+ \int _{x\in {\mathbb {R}}^d}\int _{\{\sigma (x)\leqq \vert k\vert \leqq \sigma (x)+\delta \}}\\&\qquad \frac{\vert f(x+k)-f(x)\vert ^2}{\vert k\vert ^{d+1}}\,\hbox {d}k\,\hbox {d}x:=\Vert f\Vert _{\widetilde{H}^\frac{1}{2}_{\sigma }}^2+ J. \end{aligned} \end{aligned}$$

Letting $\theta (x)=1-\frac{h}{2\sigma (x)}$, we have $\theta \in [\frac{1}{2}, 1]$ and

$$\begin{aligned} \vert y\theta (x)\vert \leqq \sigma (x)-\frac{h}{4}\quad \text {when}\quad \vert y\vert \leqq \sigma (x)+\delta . \end{aligned}$$

(A.15)

Then, for $\vert u\vert \leqq h/4$, we decompose $J\leqq 2J_1+2J_2$ where

$$\begin{aligned} \begin{aligned} J_1&= \int _{x\in {\mathbb {R}}^d}\int _{\{\sigma (x)\leqq \vert k\vert \leqq \sigma (x)+\delta \}}\frac{\vert f(x+k)-f(x+\theta k+u)\vert ^2}{\vert k\vert ^{d+1}}\,\hbox {d}k\,\hbox {d}x\\ J_2&=\int _{x\in {\mathbb {R}}^d}\int _{\{\sigma (x)\leqq \vert k\vert \leqq \sigma (x)+\delta \}}\frac{\vert f(x+\theta k+u)-f(x)\vert ^2}{\vert k\vert ^{d+1}}\,\hbox {d}k\,\hbox {d}x. \end{aligned} \end{aligned}$$

We can easily dispense with $J_2$ by changing variable $k\mapsto z=\theta k+u$ and using (A.15):

$$\begin{aligned} \begin{aligned} J_2&\lesssim \int _{x\in {\mathbb {R}}^d}\int _{\{\vert z\vert \leqq \sigma (x)\}}\frac{\vert f(x+z)-f(x)\vert ^2}{\vert z\vert ^{d+1}}\,\hbox {d}x\,\hbox {d}z=\Vert f\Vert _{H^s_{\sigma }}^2\\ \end{aligned} \end{aligned}$$

uniformly in $\vert u\vert \leqq h/4$. To estimate $J_1$, we average over $\vert u\vert \leqq h/4$:

$$\begin{aligned}&J_1\lesssim h^{-d} \int _{u\in B(0,h/4)}\int _{x\in {\mathbb {R}}^d}\int _{\{\sigma (x)\leqq \vert k\vert \leqq \sigma (x)+\delta \}}\\&\quad \frac{\vert f(x+k)-f(x+ k+(\theta -1)k+u)\vert ^2}{\vert k\vert ^{d+1}}\,\hbox {d}k\,\hbox {d}x\,\hbox {d}u. \end{aligned}$$

Since $|(\theta -1)k+u|\leqq h$, by the changes of variables $u\mapsto (\theta -1)k+u$ and then $k\mapsto k+x$, we get

$$\begin{aligned} \begin{aligned} J_1&\lesssim h^{-d} \int _{u\in B(0,h)}\int _{x,k\in {\mathbb {R}}^d}\\&\quad \frac{\vert f(x+k)-f(x+k+u)\vert ^2}{\vert k\vert ^{d+1}}{{1}}_{\{\sigma (x)\leqq \vert k\vert \leqq \sigma (x)+\delta \}}\,\hbox {d}k\,\hbox {d}x\,\hbox {d}u\\&\lesssim h^{-d}\int _{z\in {\mathbb {R}}^d} \int _{u\in B(0,h)}\vert f(z)-f(z+u)\vert ^2 \\&\quad \left\{ \int _{x\in {\mathbb {R}}^d}\frac{1}{\vert z-x\vert ^{d+1}}{{1}}_{\{\sigma (x)\leqq \vert z-x\vert \leqq \sigma (x)+\delta \}}\,\hbox {d}x\right\} \,\hbox {d}u\,\hbox {d}z\\&\lesssim \int _{z\in {\mathbb {R}}^d} \int _{u\in B(0,h)}\frac{\vert f(z)-f(z+u)\vert ^2}{\vert u\vert ^{d+1}} \,\hbox {d}u\,\hbox {d}z\\&\lesssim \Vert f\Vert _{H^\frac{1}{2}_{\sigma }}^2, \end{aligned} \end{aligned}$$

which finishes the proof.

Appendix B. Proof of (2.8) and (2.9)

Since $p^+(x, \eta (x))=p^-(x, \eta (x))$ we have

$$\begin{aligned} \nabla _xp^+-\nabla _xp^-=(\partial _yp^--\partial _yp^+)\nabla \eta \quad \text {on}~\Sigma =\{y=\eta (x)\}. \end{aligned}$$

Then

$$\begin{aligned}&\sqrt{1+|\nabla \eta |^2}RT=-(\nabla _x p^+-\nabla _xp^-)\vert _\Sigma \cdot \nabla \eta +(\partial _y p^+-\partial _yp^-)\vert _\Sigma \\&\quad =(\partial _y p^+-\partial _yp^-)\vert _\Sigma (1+|\nabla \eta |^2). \end{aligned}$$

Finally, using the fact that

$$\begin{aligned} \partial _yp^\pm \vert _\Sigma =\partial _yq^\pm \vert _\Sigma -\rho ^\pm =B^\pm -\rho ^\pm , \end{aligned}$$

we obtain

$$\begin{aligned} RT=\sqrt{1+|\nabla \eta |^2}[(\rho ^--\rho ^+)-(B^--B^+)], \end{aligned}$$

which proves (2.8). As for (2.9), we use the Darcy law (1.5) and the continuity (1.7) of $u\cdot n$ to have

$$\begin{aligned} \mu ^\pm u\cdot n+\nabla p^\pm =-(0, \rho ^\pm )\cdot n=-\rho ^\pm (1+|\nabla \eta |^2)^{-\frac{1}{2}}, \end{aligned}$$

yielding

$$\begin{aligned} RT=(\nabla p^+-\nabla p^-)\cdot n=(\mu ^--\mu ^+)u\cdot n+(\rho ^--\rho ^+)\rho ^\pm (1+|\nabla \eta |^2)^{-\frac{1}{2}}. \end{aligned}$$

Appendix C. Paradifferential Calculus

This section is devoted to a review of basic features of Bony’s paradifferential calculus (see for example [3, 12, 45, 52]).

Definition C.1

1. (Symbols) Given $\rho \in [0, \infty )$ and $m\in {\mathbb {R}}$, $\Gamma _{\rho }^{m}({\mathbb {R}}^d)$ denotes the space of locally bounded functions $a(x,\xi )$ on ${\mathbb {R}}^d\times ({\mathbb {R}}^d{\setminus } 0)$, which are $C^\infty$ with respect to $\xi$ for $\xi \ne 0$ and such that, for all $\alpha \in {\mathbb {N}}^d$ and all $\xi \ne 0$, the function $x\mapsto \partial _\xi ^\alpha a(x,\xi )$ belongs to $W^{\rho ,\infty }({\mathbb {R}}^d)$ and there exists a constant $C_\alpha$ such that

$$\begin{aligned} \forall |\xi |\geqq \frac{1}{2},\quad \Vert \partial _\xi ^\alpha a(\cdot ,\xi )\Vert _{W^{\rho ,\infty }({\mathbb {R}}^d)}\leqq C_\alpha (1+|\xi |)^{m-|\alpha |}. \end{aligned}$$

Let $a\in \Gamma _{\rho }^{m}({\mathbb {R}}^d)$, we define the semi-norm

$$\begin{aligned} M_{\rho }^{m}(a)= \sup _{|\alpha |\leqq 2(d+2) +\rho ~}\sup _{|\xi | \geqq \frac{1}{2}~} \Vert (1+|\xi |)^{|\alpha |-m}\partial _\xi ^\alpha a(\cdot ,\xi )\Vert _{W^{\rho ,\infty }({\mathbb {R}}^d)}. \end{aligned}$$

(C.1)

2. (Paradifferential operators) Given a symbol a, we define the paradifferential operator $T_a$ by

$$\begin{aligned} \widehat{T_a u}(\xi )=(2\pi )^{-d}\int \chi (\xi -\eta ,\eta ){\widehat{a}}(\xi -\eta ,\eta )\Psi (\eta ){\widehat{u}}(\eta ) \, d\eta , \end{aligned}$$

(C.2)

where ${\widehat{a}}(\theta ,\xi )=\int e^{-ix\cdot \theta }a(x,\xi )\, \,\hbox {d}x$ is the Fourier transform of a with respect to the first variable; $\chi$ and $\Psi$ are two fixed $C^\infty$ functions such that

$$\begin{aligned} \Psi (\eta )=0\quad \text {for } |\eta |\leqq \frac{1}{5},\quad \Psi (\eta )=1\quad \text {for }|\eta |\geqq \frac{1}{4}, \end{aligned}$$

(C.3)

and $\chi (\theta ,\eta )$ satisfies, for $0<\varepsilon _1<\varepsilon _2$ small enough,

$$\begin{aligned} \chi (\theta ,\eta )=1 \quad \text {if}\quad |\theta |\leqq \varepsilon _1| \eta |,\quad \chi (\theta ,\eta )=0 \quad \text {if}\quad |\theta |\geqq \varepsilon _2|\eta |, \end{aligned}$$

and such that

$$\begin{aligned} \forall (\theta ,\eta ), \quad | \partial _\theta ^\alpha \partial _\eta ^\beta \chi (\theta ,\eta )|\leqq C_{\alpha ,\beta }(1+| \eta |)^{-|\alpha |-|\beta |}. \end{aligned}$$

Remark C.2

The cut-off $\chi$ can be appropriately chosen so that when $a=a(x)$, the paradifferential operator $T_au$ becomes the usual paraproduct.

Definition C.3

Let $m\in {\mathbb {R}}$. An operator T is said to be of order m if, for all $\mu \in {\mathbb {R}}$, it is bounded from $H^{\mu }$ to $H^{\mu -m}$.

Symbolic calculus for paradifferential operators is summarized in the following theorem:

Theorem C.4

(Symbolic calculus) Let $m\in {\mathbb {R}}$ and $\rho \in [0,1)$.

• (i) If $a \in \Gamma ^m_0({\mathbb {R}}^d)$, then $T_a$ is of order m. Moreover, for all $\mu \in {\mathbb {R}}$ there exists a constant K such that

  $$\begin{aligned} \Vert T_a \Vert _{H^{\mu }\rightarrow H^{\mu -m}}\leqq K M_{0}^{m}(a). \end{aligned}$$

  (C.4)

• (ii) If $a\in \Gamma ^{m}_{\rho }({\mathbb {R}}^d), b\in \Gamma ^{m'}_{\rho }({\mathbb {R}}^d)$ then $T_a T_b -T_{ab}$ is of order $m+m'-\rho$. Moreover, for all $\mu \in {\mathbb {R}}$ there exists a constant K such that

  $$\begin{aligned} \begin{aligned} \Vert T_a T_b - T_{a b} \Vert _{H^{\mu }\rightarrow H^{\mu -m-m'+\rho }}&\leqq K (M_{\rho }^{m}(a)M_{0}^{m'}(b)+M_{0}^{m}(a)M_{\rho }^{m'}(b)). \end{aligned} \end{aligned}$$

  (C.5)

• (iii) Let $a\in \Gamma ^{m}_{\rho }({\mathbb {R}}^d)$. Denote by $(T_a)^*$ the adjoint operator of $T_a$ and by ${\overline{a}}$ the complex conjugate of a. Then $(T_a)^* -T_{{\overline{a}}}$ is of order $m-\rho$ where Moreover, for all $\mu$ there exists a constant K such that

  $$\begin{aligned} \Vert (T_a)^* - T_{{\overline{a}}} \Vert _{H^{\mu }\rightarrow H^{\mu -m+\rho }} \leqq K M_{\rho }^{m}(a). \end{aligned}$$

  (C.6)

Remark C.5

In the definition (C.2) of paradifferential operators, the cut-off $\Psi$ removes the low frequency part of u. In particular, when $a\in \Gamma ^m_0$ we have

$$\begin{aligned} \Vert T_a u\Vert _{H^\sigma }\leqq CM_0^m(a)\Vert \nabla u\Vert _{H^{\sigma +m-1}}. \end{aligned}$$

To handle symbols of negative Zygmund regularity, we shall appeal to the following.

Proposition C.6

([3, Proposition 2.12]) Let $m\in {\mathbb {R}}$ and $\rho <0$. We denote by $\dot{\Gamma }^m_\rho ({\mathbb {R}}^d)$ the class of symbols $a(x, \xi )$ that are homogeneous of order m in $\xi$, smooth in $\xi \in {\mathbb {R}}^d{\setminus } \{0\}$ and such that

$$\begin{aligned} M_{\rho }^{m}(a)= \sup _{|\alpha |\leqq 2(d+2) +\rho ~}\sup _{|\xi | \geqq \frac{1}{2}~} \Vert |\xi |^{|\alpha |-m}\partial _\xi ^\alpha a(\cdot ,\xi )\Vert _{C^\rho _*({\mathbb {R}}^d)}<\infty . \end{aligned}$$

If $a\in {{\dot{\Gamma }}}^m_\rho$ then the operator $T_a$ defined by (C.2) is of order $m-\rho$.

Notation C.7

If a and u depend on a parameter $z\in J\subset {\mathbb {R}}$ we denote

$$\begin{aligned} (T_au)(z)=T_{a(z)}u(z),\quad M^m_\rho (a; J)=\sup _{z\in J}M^m_\rho (a(z)). \end{aligned}$$

If $M^m_\rho (a; J)$ is finite we write $a\in \Gamma ^m_\rho ({\mathbb {R}}^d\times J)$.

Definition C.8

Given two functions a,  u defined on ${\mathbb {R}}^d$ the Bony’s remainder is defined by

$$\begin{aligned} R(a,u)=au-T_a u-T_u a. \end{aligned}$$

We gather here several useful product and paraproduct rules.

Theorem C.9

Let $s_0$, $s_1$ and $s_2$ be real numbers.

• (1) For any $s\in {\mathbb {R}}$,

  $$\begin{aligned} \Vert T_a u\Vert _{H^s}\leqq C\Vert a\Vert _{L^\infty }\Vert u\Vert _{H^s}. \end{aligned}$$

  (C.7)

• (2) If $s_0\leqq s_2$ and $s_0 < s_1 +s_2 -\frac{d}{2}$, then

  $$\begin{aligned} \Vert T_a u\Vert _{H^{s_0}}\leqq C \Vert a\Vert _{H^{s_1}}\Vert u\Vert _{H^{s_2}}. \end{aligned}$$

  (C.8)

• (3) If $s_1+s_2>0$ then

  $$\begin{aligned}&\Vert R(a,u) \Vert _{H^{s_1 + s_2-\frac{d}{2}}({\mathbb {R}}^d)} \leqq C \Vert a \Vert _{H^{s_1}({\mathbb {R}}^d)}\Vert u\Vert _{H^{s_2}({\mathbb {R}}^d)}, \end{aligned}$$

  (C.9)
  $$\begin{aligned}&\Vert R(a,u) \Vert _{H^{s_1+s_2}({\mathbb {R}}^d)} \leqq C \Vert a \Vert _{C^{s_1}_*({\mathbb {R}}^d)}\Vert u\Vert _{H^{s_2}({\mathbb {R}}^d)}. \end{aligned}$$

  (C.10)

• (4) If $s_1+s_2> 0$, $s_0\leqq s_1$ and $s_0< s_1+s_2-\frac{d}{2}$ then

  $$\begin{aligned} \Vert au - T_a u\Vert _{H^{s_0}}\leqq C \Vert a\Vert _{H^{s_1}}\Vert u\Vert _{H^{s_2}}. \end{aligned}$$

  (C.11)

• (5) If $s_1+s_2> 0$, $s_0\leqq s_1$, $s_0\leqq s_2$ and $s_0< s_1+s_2-\frac{d}{2}$ then

  $$\begin{aligned} \Vert u_1 u_2 \Vert _{H^{s_0}}\leqq C \Vert u_1\Vert _{H^{s_1}}\Vert u_2\Vert _{H^{s_2}}. \end{aligned}$$

  (C.12)

Theorem C.10

Consider $F\in C^\infty ({\mathbb {C}}^N)$ such that $F(0)=0$.

• (i) For $s>\frac{d}{2}$, there exists a non-decreasing function ${\mathcal {F}}:{\mathbb {R}}_+\rightarrow {\mathbb {R}}_+$ such that, for any $U\in H^s({\mathbb {R}}^d)^N$,

  $$\begin{aligned} \Vert F(U)\Vert _{H^s}\leqq {\mathcal {F}}\bigl (\Vert U\Vert _{L^\infty }\bigr )\Vert U\Vert _{H^s}. \end{aligned}$$

  (C.13)

• (ii) For $s>0$, there exists an increasing function ${\mathcal {F}}:{\mathbb {R}}_+\rightarrow {\mathbb {R}}_+$ such that, for any $U\in C_*^s({\mathbb {R}}^d)^N$,

  $$\begin{aligned} \Vert F(U)\Vert _{C_*^s}\leqq {\mathcal {F}}\bigl (\Vert U\Vert _{L^\infty }\bigr )\Vert U\Vert _{C_*^s}. \end{aligned}$$

  (C.14)

Theorem C.11

([10, Theorem 2.92] and [52, Theorem 5.2.4]) (Paralinearization for nonlinear functions) Let $\mu ,~\tau$ be positive real numbers and let $F\in C^{\infty }({\mathbb {C}}^N)$ be a scalar function satisfying $F(0)=0$. If $U=(u_j)_{j=1}^N$ with $u_j\in H^{\mu }({\mathbb {R}}^d)\cap C_*^{\tau }({\mathbb {R}}^d)$ then we have

$$\begin{aligned} F(U)=\Sigma _{j=1}^NT_{\partial _jF(U)}u_j+R_F(U) \end{aligned}$$

(C.15)

with

$$\begin{aligned} \Vert R_F(U)\Vert _{ H^{\mu +\tau }}\leqq {\mathcal {F}}(\Vert U\Vert _{L^{\infty }})\Vert U\Vert _{C_*^{\tau }}\Vert U\Vert _{H^{\mu }}. \end{aligned}$$

Recall the definitions (3.32). The next proposition provides parabolic estimates for elliptic paradifferential operators.

Proposition C.12

([3, Proposition 2.18]) Let $r\in {\mathbb {R}}$, $\varrho \in (0,1)$, $J=[z_0,z_1]\subset {\mathbb {R}}$ and let $p\in \Gamma ^{1}_{\varrho }({\mathbb {R}}^d\times J)$ satisfying

$$\begin{aligned} {{\,\mathrm{Re}\,}}p(z;x,\xi ) \geqq c |\xi |, \end{aligned}$$

for some positive constant c. Then for any  $f\in Y^r(J)$ and $w_0\in H^{r}({\mathbb {R}}^d)$, there exists $w\in X^{r}(J)$ solution of the parabolic evolution equation

$$\begin{aligned} \partial _z w + T_p w =f,\quad w\arrowvert _{z=z_0}=w_0, \end{aligned}$$

(C.16)

satisfying

$$\begin{aligned} \Vert w \Vert _{X^{r}(J)}\leqq {\mathcal {F}}\left( {\mathcal {M}}^1_\varrho (p), \frac{1}{c}\right) \left( \Vert w_0\Vert _{H^{r}}+ \Vert f\Vert _{Y^{r}(J)}\right) \end{aligned}$$

for some increasing function ${\mathcal {F}}$ depending only on r and $\varrho$. Furthermore, this solution is unique in  $X^s(J)$ for any $s\in {\mathbb {R}}$.

Appendix D. Proof of Lemma 3.10

(1) Assuming (3.33), we prove (3.34). We decompose $u_1u_2=T_{u_1}u_2+T_{u_2}u_1+R(u_1, u_2)$. Since $s_0\leqq s_2+1$ and $s_1+s_2>s_0+\frac{d}{2}-1$, (C.8) gives

$$\begin{aligned} \Vert T_{u_1}u_2\Vert _{L^2 H^{s_0-\frac{1}{2}}}\lesssim \Vert u_1\Vert _{L^\infty H^{s_1}}\Vert u_2\Vert _{L^2 H^{s_2+\frac{1}{2}}}\lesssim \Vert u_1\Vert _{X^{s_1}}\Vert u_2\Vert _{X^{s_2}}. \end{aligned}$$

The paraproduct $T_{u_2}u_1$ can be estimated similarly. As for $R(u_1, u_2)$ we use (C.9) and the conditions $s_1+s_2+1>0$ and $s_1+s_2>s_0+\frac{d}{2}-1$:

$$\begin{aligned} \Vert R(u_1, u_2)\Vert _{L^1 H^{s_0}}\lesssim \Vert u_1\Vert _{L^2H^{s_1+\frac{1}{2}}}\Vert u_2\Vert _{L^2 H^{s_2+\frac{1}{2}}}\lesssim \Vert u_1\Vert _{X^{s_1}}\Vert u_2\Vert _{X^{s_2}}. \end{aligned}$$

(2) Assuming (3.35), we prove (3.36). Again, we decompose $u_1u_2=T_{u_1}u_2+T_{u_2}u_1+R(u_1, u_2)$. If $u_1=a+b$ with $a\in L^1 H^{s_1}$ and $b\in L^2 H^{s_1-\frac{1}{2}}$ then

$$\begin{aligned} u_1u_2=T_{a}u_2+T_bu_2+T_{u_2}a+T_{u_2}b+R(a, u_2)+R(b, u_2). \end{aligned}$$

Using the conditions $s_0\leqq s_2$ and $s_1+s_2>s_0+\frac{d}{2}$, we can apply (C.8) to get

$$\begin{aligned}&\Vert T_{a}u_2\Vert _{L^1 H^{s_0}}\lesssim \Vert a\Vert _{L^1 H^{s_1}}\Vert u_2\Vert _{L^\infty H^{s_2}},\\&\Vert T_{b}u_2\Vert _{L^2 H^{s_0-\frac{1}{2}}}\lesssim \Vert b\Vert _{L^2 H^{s_1-\frac{1}{2}}}\Vert u_2\Vert _{L^\infty H^{s_2}}. \end{aligned}$$

Next from the conditions $s_0\leqq s_1$ and $s_1+s_2>s_0+\frac{d}{2}$, (C.8) yields

$$\begin{aligned}&\Vert T_{u_2}a\Vert _{L^1 H^{s_0}}\lesssim \Vert u_2\Vert _{L^\infty H^{s_2}} \Vert a\Vert _{L^1 H^{s_1}},\\&\Vert T_{u_2}b\Vert _{L^2 H^{s_0-\frac{1}{2}}}\lesssim \Vert u_2\Vert _{L^\infty H^{s_2}}\Vert b \Vert _{L^2 H^{s_1-\frac{1}{2}}}. \end{aligned}$$

Finally, under the conditions $s_1+s_2>0$ and $s_1+s_2>s_0+\frac{d}{2}$, using (C.9) we obtain

$$\begin{aligned}&\Vert R(a, u_2)\Vert _{L^1H^{s_0}}\lesssim \Vert a\Vert _{L^1 H^{s_1}}\Vert u_2\Vert _{L^\infty H^{s_2}},\\&\Vert R(b, u_2)\Vert _{L^1H^{s_0}}\lesssim \Vert b\Vert _{L^2 H^{s_1-\frac{1}{2}}}\Vert u_2\Vert _{L^2 H^{s_2+\frac{1}{2}}}. \end{aligned}$$

We have proved that

$$\begin{aligned} \Vert u_1u_2\Vert _{Y^{s_0}}\lesssim (\Vert a\Vert _{L^1 H^{s_1}}+\Vert b\Vert _{L^2 H^{s_1-\frac{1}{2}}})\Vert u_2\Vert _{X^{s_2}} \end{aligned}$$

for any decomposition $u_1=a+b$. Therefore, (3.36) follows. We have also proved (3.37), (3.38) and (3.39).