1 Introduction

The Muskat equation is a fundamental equation for incompressible fluids in porous media. It describes the evolution of a time-dependent free surface \(\Sigma (t)\) separating two fluid domains \(\Omega _1(t)\) and \(\Omega _2(t)\). A common assumption in this theory is that the motion is in two dimensions so that the interface is a curve. In this introduction, for the sake of simplicity, we assume that the interface is a graph (the analysis is done later on for a general interface). On the supposition that the fluids extend indefinitely in horizontal directions, we have that

$$\begin{aligned} \Omega _1(t)&=\left\{ (x,y)\in \mathbb {R}\times \mathbb {R}\,;\, y>h(t,x)\right\} ,\\ \Omega _2(t)&=\left\{ (x,y)\in \mathbb {R}\times \mathbb {R}\,;\, y<h(t,x)\right\} ,\\ \Sigma (t)&=\partial \Omega _1(t)=\partial \Omega _2(t)=\{y=h(t,x)\}. \end{aligned}$$

Introduce the density \(\rho _i\), the velocity \(v_i\) and the pressure \(P_i\) in the domain \(\Omega _i\) (\(i=1,2\)). One assumes that the velocities \(v_1\) and \(v_2\) obey Darcy’s law. Then, the equations by which the motion is to be determined are

$$\begin{aligned} \begin{array}{l@{\quad }l} v_i=\nabla (P_i+\rho _i g y) &{}\text {in }\Omega _i,\\ {{\,\mathrm{div}\,}}v_i=0 &{} \text {in }\Omega _i,\\ P_1=P_2 &{}\text {on }\Sigma ,\\ v_1\cdot n =v_2\cdot n &{}\text {on }\Sigma , \end{array} \end{aligned}$$

where g is the gravity and n is the outward unit normal to \(\Omega _2\) on \(\Sigma \),

$$\begin{aligned} n=\frac{1}{\sqrt{1+(\partial _x h)^2}} \begin{pmatrix} -\partial _x h \\ 1\end{pmatrix}. \end{aligned}$$

The first two equations express the classical Darcy’s law and the last two equations impose the continuity of the pressure and the normal velocities at the interface. This system is supplemented with an equation for the evolution of the free surface:

$$\begin{aligned} \partial _t h=\sqrt{1+(\partial _x h)^2}\, v_2 \cdot n. \end{aligned}$$

The previous system has been introduced by Muskat [41] whose main application was in petroleum engineering (see [42, 43] for many historical comments).

In [21], Córdoba and Gancedo discovered a formulation of the previous system based on contour integral, which applies whether the interface is a graph or not. The latter work opened the door to the solution of many important problems concerning the Cauchy problem or blow-up solutions (see [9,10,11, 19], more references are given below as well as in the survey papers [30, 31]). This formulation is a compact equation where the unknown is the parametrization of the free surface, namely a function \(f=f(t,x)\) depending on time \(t\in \mathbb {R}_+\) and \(x\in \mathbb {R}\), satisfying that

$$\begin{aligned} \partial _{t}f= \frac{\rho }{2\pi }\partial _{x} \int \arctan \left( \Delta _{\alpha } f \right) {{\,\mathrm{d}\,}}\alpha , \end{aligned}$$
(1.1)

where \(\rho ={\rho _2-\rho _1}\) is the difference of the densities, the integral is understood in the principal value sense and \(\Delta _{\alpha } f\) is the slope, namely,

$$\begin{aligned} \Delta _{\alpha } f (t,x)=\frac{f(x,t)-f(x-\alpha ,t)}{\alpha }\cdot \end{aligned}$$

The beauty of Equation (1.1) lies in its apparent simplicity, which should be compared with the complexity of the equations written in Eulerian formulation. This might suggest that (1.1) is the simplest version of the Muskat equation one may hope for. However, since the equation is highly nonlocal (this means that the nonlinearity enters in the nonlocal terms), even with this formulation the study of the Cauchy problem for (1.1) is a very delicate problem. We refer the reader to the above mentioned papers for the description of the main difficulties one has to cope with.

Our goal in this paper is to continue this line of research. We want to simplify further the study of the Muskat problem by transforming the Equation (1.1) into the simplest possible form. We shall prove that one can derive from the formulation (1.1) an explicit parabolic evolution. In particular, we shall see that one can decouple the nonlinear and nonlocal aspects. There are many possible applications that one could work out of this explicit parabolic formulation. Here we shall study the Cauchy problem in homogeneous Sobolev spaces.

The well-posedness of the Cauchy problem was first proved in [21] by Córdoba and Gancedo for initial data in \(H^3(\mathbb {R})\) in the stable regime \(\rho _2>\rho _1\) (they also proved that the problem is ill-posed in Sobolev spaces when \(\rho _2<\rho _1\)). Several extensions of their results have been obtained by different proofs. In [14], Cheng, Granero-Belinchón, Shkoller proved the well-posedness of the Cauchy problem in \(H^2(\mathbb {R})\) (introducing a Lagrangian point of view which can be used in a broad setting, see [34]) and Constantin, Gancedo, Shvydkoy and Vicol ([18]) considered rough initial data which are in \(W^{2,p}(\mathbb {R})\) for some \(p>1\), as well they obtained a regularity criteria for the Muskat problem. We refer also to the recent work [31] where a regularity criteria is obtained in terms of a control of some critical quantities. Many recent results are motivated by the fact that, loosely speaking, the Muskat equation has to do with the slope more than with the curvature of the fluid interface. Indeed, one scale invariant norm is the Lipschitz norm \(\sup _{x\in \mathbb {R}}\left|\partial _x f(t,x)\right|\). We refer the reader to the work [17] of Constantin, Córdoba, Gancedo, Rodríguez-Piazza and Strain for global well-posedness results assuming that the Lipschitz semi-norm is smaller than 1 (see also [15] where time decay of those solutions is proved). In [26], Deng, Lei and Lin proved the existence of global in time solutions with large slopes, assuming some monotonicity assumption on the data. In [8], Cameron was able to prove a global existence result assuming that some critical quantity, namely the product of the maximal and minimal slopes, is smaller than 1. His result allows to consider arbitrary large slopes. By using a new formulation of the Muskat equation involving oscillatory integrals, Córdoba and the second author in [20] proved that the Muskat equation is globally well-posed for sufficiently smooth data provided the critical Sobolev norm \(\dot{H}^{\frac{3}{2}}(\mathbb {R})\) is small enough. The latter is a global existence result of a unique strong solution having arbitrarily large slopes.

These observations suggest to study the local in time well-posedness of the Cauchy problem without assuming that any \(L^p\)-norm of the curvature is finite. The well-posedness of the Cauchy problem in this case was obtained by Matioc [37, 38]. Using tools from functional analysis, Matioc proved that the Cauchy problem is locally in time well-posed for initial data in Sobolev spaces \(H^s(\mathbb {R})\) with \(s>3/2\), without smallness assumption. We shall give a simpler proof which generalizes the latter result to homogeneous Sobolev spaces \(\dot{H}^s(\mathbb {R})\). We also mention that many recent results focus on different rough solutions, which are important for instance in the unstable regime \(\rho _1>\rho _2\) (see for example the existence mixing zones in [12, 13, 44] or the dynamic between the two different regimes [23, 24]). We refer also to [22, 47] where uniqueness issues have been studied using the convex integration scheme.

In this paper we assume that the difference between the densities in the two fluids satisfies \(\rho >0\), so, by rescaling in time, we can assume without loss of generality that \(\rho =2\).

A fundamental difference with the above mentioned results is that we shall determine the full structure of the nonlinearity instead of performing energy estimates. To explain this, we begin by identifying the nonlinear terms. Since \(\rho =2\), one can rewrite Equation (1.1) as

$$\begin{aligned} \partial _tf=\frac{1}{\pi }\int \frac{\partial _x \Delta _\alpha f}{1+(\Delta _\alpha f)^2}{{\,\mathrm{d}\,}}\alpha \end{aligned}$$

(in this introduction some computations are formal, but we shall rigorously justify them later). Consequently, the linearized Muskat equation reads as

$$\begin{aligned} \partial _t u=\frac{1}{\pi }{{\,\mathrm{pv}\,}}\int \partial _x \Delta _\alpha u{{\,\mathrm{d}\,}}\alpha . \end{aligned}$$
(1.2)

Consider the singular integral operators

$$\begin{aligned} \mathcal {H}u=-\frac{1}{\pi }{{\,\mathrm{pv}\,}}\int \Delta _\alpha u {{\,\mathrm{d}\,}}\alpha \quad \text {and}\quad \Lambda =\mathcal {H}\partial _x. \end{aligned}$$
(1.3)

Then \(\mathcal {H}\) is the Hilbert transform (the Fourier multiplier with symbol \(-i{{\,\mathrm{sgn}\,}}(\xi )\)) and \(\Lambda \) is the square root of \(-\partial _{xx}\). With the latter notation, the linearized Muskat Equation (1.2) reads as

$$\begin{aligned} \partial _t u+ \Lambda u=0. \end{aligned}$$

With this notation, the Muskat Equation (1.1) can be written under the form

$$\begin{aligned} \partial _t f+ \Lambda f=\mathcal {T}(f)f, \end{aligned}$$
(1.4)

where \(\mathcal {T}(f)\) is the operator defined by

$$\begin{aligned} \mathcal {T}(f)g=-\frac{1}{\pi }\int \big (\partial _x \Delta _\alpha g\big ) \frac{(\Delta _\alpha f)^2}{1+(\Delta _\alpha f)^2}{{\,\mathrm{d}\,}}\alpha . \end{aligned}$$

Our first main result will provide a thorough study of this nonlinear operator. Before going any further, let us fix some notations.

Definition 1.1

  1. (i)

    Given a real number \(\sigma \), we denote by \(\Lambda ^\sigma \) the Fourier multiplier with symbol \(\left|\xi \right|^\sigma \) and by \(\dot{H}^\sigma (\mathbb {R})\) the homogeneous Sobolev space of tempered distributions whose Fourier transform \(\hat{u}\) belongs to \(L^1_{loc}(\mathbb {R})\) and satisfies

    $$\begin{aligned} \left\Vert u\right\Vert _{\dot{H}^\sigma }^2= \bigl \Vert \Lambda ^\sigma u\bigr \Vert _{L^2}^2=\frac{1}{2\pi }\int \nolimits _\mathbb {R}\left|\xi \right|^{2\sigma }\left|\hat{u}(\xi )\right|^2{{\,\mathrm{d}\,}}\xi <+\infty . \end{aligned}$$
  2. (ii)

    We denote by \(H^\sigma (\mathbb {R})\) the nonhomogeneous Sobolev space \(L^2(\mathbb {R})\cap \dot{H}^\sigma (\mathbb {R})\). We set \(H^\infty (\mathbb {R})\mathrel {:=}\cap _{\sigma \geqq 0}H^\sigma (\mathbb {R})\) and introduce \(X\mathrel {:=}\cap _{\sigma \geqq 1}\dot{H}^\sigma (\mathbb {R})\), the set of tempered distributions whose Fourier transform \(\hat{u}\) belongs to \(L^1_{loc}(\mathbb {R})\) and whose derivative belongs to \(H^\infty (\mathbb {R})\).

  3. (iii)

    Given \(0<s<1\), the homogeneous Besov space \(\dot{B}^{s}_{2,1}(\mathbb R)\) consists of those tempered distributions f whose Fourier transform is integrable near the origin and such that

    $$\begin{aligned} \left\Vert f \right\Vert _{\dot{B}^{s}_{2,1}} = \int \left( \int \frac{\left|f(x)-f(x-\alpha )\right|^2}{\left|\alpha \right|^{2s}}{{\,\mathrm{d}\,}} x\right) ^\frac{1}{2}\frac{{{\,\mathrm{d}\,}}\alpha }{\left|\alpha \right|}<+\infty . \end{aligned}$$
  4. (iv)

    We use the notation \(\left\Vert \cdot \right\Vert _{E\cap F}=\left\Vert \cdot \right\Vert _E+\left\Vert \cdot \right\Vert _F\).

Theorem 1.2

  1. (i)

    (Low frequency estimate) There exists a constant C such that, for all f in \(\dot{H}^{1}(\mathbb {R})\) and all g in \(\dot{H}^\frac{3}{2}(\mathbb {R})\), \(\mathcal {T}(f)g\) belongs to \(L^2(\mathbb {R})\) and

    $$\begin{aligned} \left\Vert \mathcal {T}(f)g\right\Vert _{L^2}\leqq C \left\Vert f\right\Vert _{\dot{H}^{1}} \left\Vert g\right\Vert _{\dot{H}^{\frac{3}{2}}}. \end{aligned}$$
    (1.5)

    Moreover, \(f\mapsto \mathcal {T}(f)f\) is locally Lipschitz from \(\dot{H}^1(\mathbb {R})\cap \dot{H}^{\frac{3}{2}}(\mathbb {R})\) to \(L^2(\mathbb {R})\).

  2. (ii)

    (High frequency estimate) For all \(0<\nu<\varepsilon <1/2\), there exists a positive constant \(C>0\) such that, for all functions fg in \(X=\cap _{\sigma \geqq 1}\dot{H}^\sigma (\mathbb {R})\),

    $$\begin{aligned} \mathcal {T}(f)g=\gamma (f)\Lambda g+V(f)\partial _x g+R(f,g), \end{aligned}$$
    (1.6)

    where

    $$\begin{aligned} \gamma (f)\mathrel {:=}\frac{f_x^2}{1+f_x^2}, \end{aligned}$$

    and R(fg) and V(f) satisfy

    $$\begin{aligned} \left\Vert R(f,g)\right\Vert _{L^2}\leqq C\left\Vert f\right\Vert _{\dot{H}^{\frac{3}{2}+\varepsilon }}\left\Vert g\right\Vert _{\dot{B}^{1-\varepsilon }_{2,1}}, \end{aligned}$$
    (1.7)

    and

    $$\begin{aligned}&\left\Vert V(f)\right\Vert _{C^{0,\nu }}\mathrel {:=}\left\Vert V(f)\right\Vert _{L^\infty }+\sup _{y\in \mathbb {R}}\bigg (\frac{\left|V(f)(x+y)-V(f)(x)\right|}{\left|y\right|^\nu }\bigg ) \nonumber \\&\quad \leqq C\left\Vert f\right\Vert _{\dot{H}^1\cap \dot{H}^{\frac{3}{2}+\varepsilon }}^2. \end{aligned}$$
    (1.8)
  3. (iii)

    Let \(0<\varepsilon <1/2\). There exists a non-decreasing function \(\mathcal {F}:\mathbb {R}_+\rightarrow \mathbb {R}_+\) such that, for all functions fg in X,

    $$\begin{aligned}&\left\Vert \Lambda ^{1+\varepsilon }\mathcal {T}(f)g-\mathcal {T}(f)\Lambda ^{1+\varepsilon }g\right\Vert _{L^2}\nonumber \\&\quad \leqq \mathcal {F}\big (\left\Vert f\right\Vert _{\dot{H}^1\cap \dot{H}^{\frac{3}{2}+\varepsilon }}\big ) \left\Vert f\right\Vert _{\dot{H}^1\cap \dot{H}^{\frac{3}{2}+\varepsilon }}\left\Vert g \right\Vert _{\dot{H}^{\frac{3}{2}+\varepsilon }\cap \dot{H}^2}. \end{aligned}$$
    (1.9)

The proof of the first statement follows directly from the definition of fractional Sobolev spaces in terms of finite differences, see Section 2. The proof of the second statement is the most delicate part of the proof, which requires to uncover some symmetries in the nonlinearity, see Section 4. The last statement is proved in Section 3 by using sharp variants of the usual nonlinear estimates in Sobolev spaces. Namely we used for the later proof a version of the classical Kato–Ponce estimate proved recently by Li and also a refinement of the composition rule in Sobolev spaces proved in Section 2.

We deduce from the previous result a paralinearization formula for the nonlinearity. We do not consider paradifferential operators as introduced by Bony ([6, 39]). Instead, following Shnirelman [46], we consider a simpler version of these operators which is convenient for the analysis of the Muskat equation for rough solutions.

Corollary 1.3

Consider \(0<\varepsilon <1/2\) and, given a bounded function \(a=a(x)\), denote by \(\tilde{T}_a:\dot{H}^{1+\varepsilon }(\mathbb {R})\rightarrow L^2(\mathbb {R})\cap \dot{H}^{1+\varepsilon }(\mathbb {R})\) the paraproduct operator defined by

$$\begin{aligned} \tilde{T}_ag=(I+\Lambda ^{1+\varepsilon })^{-1}(a\Lambda ^{1+\varepsilon }g). \end{aligned}$$

Then, there exists a function \(\mathcal {F}:\mathbb {R}\rightarrow \mathbb {R}\) such that, for all \(f\in X\),

$$\begin{aligned} \mathcal {T}(f)f=\tilde{T}_{\gamma (f)}\Lambda f+\tilde{T}_{V(f)}\partial _x f+ R_\varepsilon (f), \end{aligned}$$
(1.10)

where

$$\begin{aligned} \left\Vert R_\varepsilon (f)\right\Vert _{H^{1+\varepsilon }}\leqq \mathcal {F}\big (\left\Vert f\right\Vert _{\dot{H}^1\cap \dot{H}^{\frac{3}{2}+\varepsilon }}\big ) \left\Vert f\right\Vert _{\dot{H}^1\cap \dot{H}^{\frac{3}{2}+\varepsilon }} \left\Vert f \right\Vert _{\dot{H}^{1}\cap \dot{H}^{2+\frac{\varepsilon }{2}}}. \end{aligned}$$
(1.11)

Proof

Writing

$$\begin{aligned} \mathcal {T}(f)f=(I+\Lambda ^{1+\varepsilon })^{-1}\mathcal {T}(f)(I+\Lambda ^{1+\varepsilon })f+(I+\Lambda ^{1+\varepsilon })^{-1}\big [\Lambda ^{1+\varepsilon },\mathcal {T}(f)\big ]f, \end{aligned}$$

and using the formula (1.6), we find that (1.10) holds with

$$\begin{aligned} R_\varepsilon (f)=(I+\Lambda ^{1+\varepsilon })^{-1}\Big (\mathcal {T}(f)f+ R\big (f,\Lambda ^{1+\varepsilon }f\big )+\big [\Lambda ^{1+\varepsilon },\mathcal {T}(f)\big ]f\Big ). \end{aligned}$$

To prove (1.11), we have to estimate the \(L^2\)-norm of the three terms between parentheses. The \(L^2\)-norm of \(\mathcal {T}(f)f\) is estimated by means of (1.5). We use (1.7) and (1.9) to estimate the two other terms. This yields

$$\begin{aligned} \left\Vert R_\varepsilon (f)\right\Vert _{{H}^{1+\varepsilon }}\leqq \mathcal {F}\big (\left\Vert f\right\Vert _{\dot{H}^1\cap \dot{H}^{\frac{3}{2}+\varepsilon }}\big ) \left\Vert f\right\Vert _{\dot{H}^1\cap \dot{H}^{\frac{3}{2}+\varepsilon }}\Big ( \left\Vert f \right\Vert _{\dot{H}^{\frac{3}{2}}\cap \dot{H}^{2}}+\left\Vert \Lambda ^{1+\varepsilon }f\right\Vert _{\dot{B}^{1-\varepsilon }_{2,1}}\Big ). \end{aligned}$$

Thus, (1.11) follows from \(\dot{H}^{1-\frac{3\varepsilon }{2}}(\mathbb {R})\cap \dot{H}^{1-\frac{\varepsilon }{2}}(\mathbb {R})\hookrightarrow \dot{B}^{1-\varepsilon }_{2,1}(\mathbb {R})\) (see Lemma 2.2). \(\quad \square \)

We now consider the Cauchy problem for the Muskat equation. Substituting the above identity for \(\mathcal {T}(f)\) in the Equation (1.4) and simplifying, we find that

$$\begin{aligned} \Big (\partial _t -V(f)\partial _x +\frac{1}{1+f_x^2}\Lambda \Big )\Lambda ^{1+\varepsilon }f=\Lambda ^{1+\varepsilon }R_\varepsilon (f). \end{aligned}$$

Now, the key point is that the estimates (1.8) and (1.11) mean that the remainder term \(R_\varepsilon (f)\) and the operator \(V\partial _x\) contribute as operators of order stricly less than 1 (namely \(1-\varepsilon /2\) and \(1-\nu \)) to an energy estimate, and so they are sub-principal terms for the analysis of the Cauchy problem. We also observe that the Muskat equation is parabolic as long as one controls the \(L^\infty \)-norm of \(f_x\) only. This observation is related to our second goal, which is to solve the Cauchy problem in homogeneous Sobolev spaces instead of nonhomogeneous spaces. This is a natural result since the Muskat equation is invariant by the transformation \(f\mapsto f+C\). This allows us to make an assumption only on the \(L^\infty \)-norm of the slope of the initial data, allowing initial data which are not bounded or not square integrable.

Theorem 1.4

Consider \(s\in (3/2,2)\) and an initial data \(f_0\) in \(\dot{H}^1(\mathbb {R})\cap \dot{H}^s(\mathbb {R})\). Then, there exists a positive time T such that the Cauchy problem for (1.1) with initial data \(f_0\) has a unique solution f satisfying \(f(t,x)=f_0(x)+u(t,x)\) with \(u(0,x)=0\) and

$$\begin{aligned} u \in C^0\big ([0,T]; H^s(\mathbb {R})\big )\cap C^1\big ([0,T];H^{s-1}(\mathbb {R})\big ) \cap L^{2}\big (0,T; H^{s+\frac{1}{2}}(\mathbb {R})\big ), \end{aligned}$$

where \(H^\sigma (\mathbb {R})\) denotes the nonhomogeneous Sobolev space \(L^2(\mathbb {R})\cap \dot{H}^\sigma (\mathbb {R})\).

The latter result is proved in the last section. We conclude this introduction by fixing some notations.

Notation 1.5

  1. (i)

    We denote by \(f_x\) the spatial derivative of f.

  2. (ii)

    \(A \lesssim B\) means that there is \(C>0\), depending only on fixed quantities, such that \(A \leqq C B\).

  3. (iii)

    Given \(g=g(\alpha ,x)\) and Y a space of functions depending only on x, the notation \(\left\Vert g\right\Vert _Y\) is a compact notation for \(\alpha \mapsto \left\Vert g(\alpha ,\cdot )\right\Vert _{Y}\).

2 Preliminaries

In this section, we recall or prove various results about Besov spaces which we will need throughout the article. We use the definition of these spaces originally given by Besov in [5], using integrability properties of finite differences.

Given a real number \(\alpha \), the finite difference operators \(\delta _\alpha \) and \(s_\alpha \) are defined by

$$\begin{aligned}&\delta _{\alpha } f(x)=f(x)-f(x-\alpha ),\\&s_\alpha f(x)=2f(x)-f(x-\alpha )-f(x+\alpha ). \end{aligned}$$

Definition 2.1

Consider three real numbers (pqs) in \([1,\infty ]^2 \times (0,2)\). The homogeneous Besov space \(\dot{B}^{s}_{p,q}(\mathbb {R})\) consists of those tempered distributions f whose Fourier transform is integrable near the origin and such that the following quantity \(\left\Vert f\right\Vert _{\dot{B}^s_{p,q}}\) is finite:

$$\begin{aligned} \left\Vert f \right\Vert _{\dot{B}^{s}_{p,q}}&= \left\| \frac{\Vert \delta _{\alpha } f\Vert _{L^{p}(\mathbb {R},{{\,\mathrm{d}\,}} x)}}{ \vert {\alpha }\vert ^{s} } \right\| _{L^{q}(\mathbb R, \vert \alpha \vert ^{-1} {{\,\mathrm{d}\,}}\alpha )} \qquad&\text {for }s\in (0,1), \end{aligned}$$
(2.1)
$$\begin{aligned} \left\Vert f \right\Vert _{\dot{B}^{s}_{p,q}}&= \left\| \frac{\Vert s_\alpha f\Vert _{L^{p}(\mathbb {R};{{\,\mathrm{d}\,}} x)}}{ \vert {\alpha }\vert ^{s} } \right\| _{L^{q}(\mathbb R, \vert \alpha \vert ^{-1} {{\,\mathrm{d}\,}}\alpha )} \qquad&\text {for }s\in [1,2). \end{aligned}$$
(2.2)

We refer the reader to the book of Peetre [45, Chapter 8] for the equivalence between these definitions and the one in terms of Littlewood–Paley decomposition (see also [7, Proposition 9] or [4, Theorems 2.36, 2.37] for the case \(s\in (0,1]\)).

In this paper, we use only Besov spaces of the form

$$\begin{aligned} \dot{B}^s_{2,2}(\mathbb {R}),\quad \dot{B}^s_{\infty ,2}(\mathbb {R}),\quad \dot{B}^s_{2,1}(\mathbb {R}). \end{aligned}$$

We will make extensive use of the fact that \(\left\Vert \cdot \right\Vert _{\dot{H}^{s}}\) and \(\left\Vert \cdot \right\Vert _{\dot{B}^s_{2,2}}\) are equivalent for \(s\in (0,2)\). Moreover, for \(s\in (0,1)\),

$$\begin{aligned} \left\Vert u\right\Vert _{\dot{H}^{s}}^2=\frac{1}{4\pi c(s)}\left\Vert u\right\Vert _{\dot{B}^s_{2,2}}^2 \quad \text {with}\quad c(s)=\int \nolimits _\mathbb {R}\frac{1-\cos (t)}{\left|t\right|^{1+2s}}{{\,\mathrm{d}\,}} t. \end{aligned}$$
(2.3)

We will also make extensive use of the fact that, for all s in (0, 2),

$$\begin{aligned} \dot{H}^{s+\frac{1}{2}} (\mathbb {R})\hookrightarrow \dot{B}^{s}_{\infty ,2}(\mathbb R). \end{aligned}$$
(2.4)

We will also use the following:

Lemma 2.2

For any \(s\in (0,1)\) and any \(\delta >0\) such that \([s-\delta ,s+\delta ]\subset (0,1)\),

$$\begin{aligned} \dot{H}^{s-\delta }(\mathbb {R})\cap \dot{H}^{s+\delta }(\mathbb {R})\hookrightarrow \dot{B}^{s}_{2,1}(\mathbb {R}). \end{aligned}$$

Proof

We have

$$\begin{aligned} \int \nolimits _{\left|\alpha \right|\leqq 1} \frac{\Vert \delta _\alpha f\Vert _{L^{2}(\mathbb {R};{{\,\mathrm{d}\,}} x)}}{ \vert {\alpha }\vert ^{s} } \frac{{{\,\mathrm{d}\,}}\alpha }{\vert \alpha \vert }&=\int \nolimits _{\left|\alpha \right|\leqq 1} \left|\alpha \right|^\delta \frac{\Vert \delta _\alpha f\Vert _{L^{2}(\mathbb {R};{{\,\mathrm{d}\,}} x)}}{ \vert {\alpha }\vert ^{s+\delta } } \frac{{{\,\mathrm{d}\,}}\alpha }{\vert \alpha \vert }\\&\leqq \left( \int \nolimits _{\left|\alpha \right|\leqq 1} \left|\alpha \right|^{2\delta }\frac{{{\,\mathrm{d}\,}}\alpha }{\vert \alpha \vert }\right) ^\frac{1}{2}\left( \int \nolimits _{\mathbb {R}}\frac{\Vert \delta _\alpha f\Vert _{L^{2}(\mathbb {R};{{\,\mathrm{d}\,}} x)}^2}{ \vert {\alpha }\vert ^{2(s+\delta )} } \frac{{{\,\mathrm{d}\,}}\alpha }{\vert \alpha \vert }\right) ^\frac{1}{2}\\&\leqq C(\delta )\left\Vert f\right\Vert _{\dot{B}^{s+\delta }_{2,2}}=C(\delta ,s)\left\Vert f\right\Vert _{\dot{H}^{s+\delta }}, \end{aligned}$$

and similarly,

$$\begin{aligned} \int \nolimits _{\left|\alpha \right|\geqq 1} \frac{\Vert \delta _\alpha f\Vert _{L^{2}(\mathbb {R};{{\,\mathrm{d}\,}} x)}}{ \vert {\alpha }\vert ^{s} } \frac{{{\,\mathrm{d}\,}}\alpha }{\vert \alpha \vert } \leqq C'(\delta ,s)\left\Vert f\right\Vert _{\dot{H}^{s-\delta }}, \end{aligned}$$

which gives the result. \(\quad \square \)

As an example of properties which are very simple to prove using the definition of Besov semi-norms in terms of finite differences, let us prove the first point in Theorem 1.2. Recall that, by notation,

$$\begin{aligned} \mathcal {T}(f)g=-\frac{1}{\pi }\int \Delta _\alpha g_x \ \frac{(\Delta _\alpha f)^2}{1+(\Delta _\alpha f)^2}{{\,\mathrm{d}\,}}\alpha , \end{aligned}$$

where \(g_x\mathrel {:=}\partial _x g\).

Proposition 2.3

  1. (i)

    For all f in \(\dot{H}^{1}(\mathbb {R})\) and all g in \(\dot{H}^{\frac{3}{2}}(\mathbb {R})\), the function

    $$\begin{aligned} \alpha \mapsto \Delta _\alpha g_x \ \frac{(\Delta _\alpha f)^2}{1+(\Delta _\alpha f)^2} \end{aligned}$$

    belongs to \(L^1_\alpha (\mathbb {R};L^2_x(\mathbb {R}))\). Consequently, \(\mathcal {T}(f)g\) belongs to \(L^2(\mathbb {R})\). Moreover, there is a constant C such that

    $$\begin{aligned} \left\Vert \mathcal {T}(f)g\right\Vert _{L^2}\leqq C \left\Vert f\right\Vert _{\dot{H}^1}\left\Vert g\right\Vert _{\dot{H}^\frac{3}{2}}. \end{aligned}$$
    (2.5)
  2. (ii)

    For all \(\delta \in [0,1/2)\), there exists a constant \(C>0\) such that, for all functions \(f_1,f_2\) in \(\dot{H}^{1-\delta }(\mathbb {R})\cap \dot{H}^{\frac{3}{2}+\delta }(\mathbb {R})\),

    $$\begin{aligned} \left\Vert (\mathcal {T}(f_1)-\mathcal {T}(f_2))f_2\right\Vert _{L^2} \leqq C \left\Vert f_1-f_2\right\Vert _{\dot{H}^{1-\delta }}\left\Vert f_2\right\Vert _{\dot{H}^{\frac{3}{2}+\delta }}. \end{aligned}$$
  3. (iii)

    The map \(f\mapsto \mathcal {T}(f)f\) is locally Lipschitz from \(\dot{H}^1(\mathbb {R})\cap \dot{H}^{\frac{3}{2}}(\mathbb {R})\) to \(L^2(\mathbb {R})\).

Proof

  1. (i)

    Since

    $$\begin{aligned} \left\Vert \Delta _\alpha g_x \frac{(\Delta _\alpha f)^2}{1+(\Delta _\alpha f)^2} \right\Vert _{L^2}\leqq \left\Vert \Delta _\alpha g_x\right\Vert _{L^2}\left\Vert \Delta _\alpha f\right\Vert _{L^\infty } =\frac{\left\Vert \delta _\alpha g_x\right\Vert _{L^2}}{\left|\alpha \right|}\frac{\left\Vert \delta _\alpha f\right\Vert _{L^\infty }}{\left|\alpha \right|}, \end{aligned}$$

    by using the Cauchy–Schwarz inequality and the definition (2.1) of the Besov semi-norms one finds that

    $$\begin{aligned} \left\Vert \mathcal {T}(f)g\right\Vert _{L^2}&\leqq \frac{1}{\pi }\int \frac{\left\Vert \delta _\alpha g_x\right\Vert _{L^2}}{\left|\alpha \right|} \frac{\left\Vert \delta _\alpha f\right\Vert _{L^\infty }}{\left|\alpha \right|} {{\,\mathrm{d}\,}}\alpha \\&\leqq \frac{1}{\pi }\int \frac{\left\Vert \delta _\alpha g_x\right\Vert _{L^2}}{\left|\alpha \right|^{1/2}} \frac{\left\Vert \delta _\alpha f\right\Vert _{L^\infty }}{\left|\alpha \right|^{1/2}} \frac{{{\,\mathrm{d}\,}}\alpha }{\left|\alpha \right|}\\&\leqq \frac{1}{\pi } \left( \int \frac{\left\Vert \delta _\alpha g_x\right\Vert _{L^2}^2}{\left|\alpha \right|}\frac{{{\,\mathrm{d}\,}}\alpha }{\left|\alpha \right|} \right) ^\frac{1}{2}\left( \int \frac{\left\Vert \delta _\alpha f\right\Vert _{L^\infty }^2}{\left|\alpha \right|} \frac{{{\,\mathrm{d}\,}}\alpha }{\left|\alpha \right|}\right) ^\frac{1}{2}\\&\leqq \frac{1}{\pi }\left\Vert g_x\right\Vert _{\dot{B}^{\frac{1}{2}}_{2,2}}\left\Vert f\right\Vert _{\dot{B}^{\frac{1}{2}}_{\infty ,2}}. \end{aligned}$$

    Recalling that \(\left\Vert \cdot \right\Vert _{\dot{B}^{\frac{1}{2}}_{2,2}}\) and \(\left\Vert \cdot \right\Vert _{\dot{H}^{\frac{1}{2}}}\) are equivalent semi-norms, and using the Sobolev embedding (2.4), we have

    $$\begin{aligned} \left\Vert g_x\right\Vert _{\dot{B}^{\frac{1}{2}}_{2,2}}\lesssim \left\Vert g\right\Vert _{\dot{H}^{\frac{3}{2}}},\quad \left\Vert f\right\Vert _{\dot{B}^{\frac{1}{2}}_{\infty ,2}}\lesssim \left\Vert f\right\Vert _{\dot{H}^1}, \end{aligned}$$

    and hence we obtain the wanted inequality (2.5).

  2. (ii)

    Write that

    $$\begin{aligned} (\mathcal {T}(f_1)-\mathcal {T}(f_2))f_2= -\frac{1}{\pi }\int \Delta _\alpha f_{2x} \Delta _\alpha (f_1-f_2) M(\alpha ,x){{\,\mathrm{d}\,}}\alpha , \end{aligned}$$

    where

    $$\begin{aligned} M(\alpha ,x)=\frac{(\Delta _\alpha f_1)+\Delta _\alpha f_2}{(1+(\Delta _\alpha f_1)^2)(1+(\Delta _\alpha f_2)^2)}. \end{aligned}$$

    Since \(\left|M(\alpha ,x)\right|\leqq 1\), by repeating similar arguments to those used in the first part (balancing the powers of \(\alpha \) in a different way), we get

    $$\begin{aligned} \left\Vert (\mathcal {T}(f_1)-\mathcal {T}(f_2))f_2\right\Vert _{L^2}&\leqq \frac{1}{\pi }\int \frac{\left\Vert \delta _\alpha f_{2x}\right\Vert _{L^2}}{\left|\alpha \right|} \frac{\left\Vert \delta _\alpha (f_1-f_2)\right\Vert _{L^\infty }}{\left|\alpha \right|} {{\,\mathrm{d}\,}}\alpha \\&\leqq \frac{1}{\pi }\int \frac{\left\Vert \delta _\alpha f_{2x}\right\Vert _{L^2}}{\left|\alpha \right|^{1/2+\delta }}\frac{\left\Vert \delta _\alpha (f_1-f_2)\right\Vert _{L^\infty }}{\left|\alpha \right|^{1/2-\delta }} \frac{{{\,\mathrm{d}\,}}\alpha }{\left|\alpha \right|}\\&\leqq \frac{1}{\pi }\left\Vert f_{2x}\right\Vert _{\dot{B}^{\frac{1}{2}+\delta }_{2,2}}\left\Vert f_1-f_2\right\Vert _{\dot{B}^{\frac{1}{2}-\delta }_{\infty ,2}}, \end{aligned}$$

    which implies

    $$\begin{aligned} \left\Vert (\mathcal {T}(f_1)-\mathcal {T}(f_2))f_2\right\Vert _{L^2}\leqq C \left\Vert f_1-f_2\right\Vert _{\dot{H}^{1-\delta }}\left\Vert f_2\right\Vert _{\dot{H}^{\frac{3}{2}+\delta }}. \end{aligned}$$
  3. (iii)

    Consider \(f_1\) and \(f_2\) in \(\dot{H}^1(\mathbb {R})\cap \dot{H}^{\frac{3}{2}}(\mathbb {R})\). Then

    $$\begin{aligned} \mathcal {T}(f_1)f_1-\mathcal {T}(f_2)f_2=\mathcal {T}(f_1)(f_1-f_2)+(\mathcal {T}(f_1)-\mathcal {T}(f_2))f_2. \end{aligned}$$

    Then (2.5) implies that the \(L^2\)-norm of the first term is bounded by

    $$\begin{aligned} C\left\Vert f_1\right\Vert _{\dot{H}^1}\left\Vert f_1-f_2\right\Vert _{\dot{H}^{\frac{3}{2}}}. \end{aligned}$$

    We estimate the second term by using (ii) applied with \(\delta =0\). It follows that

    $$\begin{aligned} \left\Vert \mathcal {T}(f_1)f_1-\mathcal {T}(f_2)f_2\right\Vert _{L^2}\lesssim \big (\left\Vert f_1\right\Vert _{\dot{H}^1\cap \dot{H}^{\frac{3}{2}}} +\left\Vert f_2\right\Vert _{\dot{H}^1\cap \dot{H}^{\frac{3}{2}}}\big ) \left\Vert f_1-f_2\right\Vert _{\dot{H}^1\cap \dot{H}^{\frac{3}{2}}}, \end{aligned}$$

    which completes the proof. \(\quad \square \)

We gather in the following proposition the nonlinear estimates which will be needed:

Proposition 2.4

  1. (i)

    Let \(s\in (0,1)\), then \(L^\infty (\mathbb {R})\cap \dot{H}^{s}(\mathbb {R})\) is an algebra. Moreover, for all uv in \(L^\infty (\mathbb {R})\cap \dot{H}^s(\mathbb {R})\),

    $$\begin{aligned} \left\Vert uv\right\Vert _{\dot{H}^{s}}\leqq 2 \left\Vert u\right\Vert _{L^\infty }\left\Vert v\right\Vert _{\dot{H}^{s}}+2\left\Vert v\right\Vert _{L^\infty }\left\Vert u\right\Vert _{\dot{H}^{s}}. \end{aligned}$$
    (2.6)
  2. (ii)

    Consider a \(C^\infty \) function \(F:\mathbb {R}\rightarrow \mathbb {R}\) satisfying

    $$\begin{aligned} \forall (x,y)\in \mathbb {R}^2,\qquad \left|F(x)-F(y)\right|\leqq K \left|x-y\right|. \end{aligned}$$

    Then, for all \(s\in (0,1)\) and all \(u \in \dot{H}^{s}(\mathbb {R})\), one has \(F(u)\in \dot{H}^{s}(\mathbb {R})\) together with the estimate

    $$\begin{aligned} \left\Vert F(u)\right\Vert _{\dot{H}^s}\leqq K\left\Vert u\right\Vert _{\dot{H}^s}. \end{aligned}$$
    (2.7)
  3. (iii)

    Consider a \(C^\infty \) function \(F:\mathbb {R}\rightarrow \mathbb {R}\) and a real number \(\sigma \) in (1, 2). Then, there exists a non-decreasing function \(\mathcal {F}:\mathbb {R}\rightarrow \mathbb {R}\) such that, for all \(u \in \dot{H}^{\sigma -1}(\mathbb {R})\cap \dot{H}^{\sigma }(\mathbb {R})\) one has \(F(u)\in \dot{H}^{\sigma }(\mathbb {R})\) together with the estimate

    $$\begin{aligned} \left\Vert F(u)\right\Vert _{\dot{H}^\sigma }\leqq \mathcal {F}(\left\Vert u\right\Vert _{L^\infty }) \Big (\left\Vert u\right\Vert _{\dot{H}^{\sigma -1}}+\left\Vert u\right\Vert _{\dot{H}^{\sigma }}\Big ). \end{aligned}$$
    (2.8)

Remark 2.5

  1. (i)

    The inequality (2.6) is the classical Kato–Ponce estimate ([29]). We will use it only when \(0<s<1\), for which one has a straightforward proof (see below).

  2. (ii)

    Statement (ii) is also elementary and classical (see [7]). Notice that (2.7) is a sub-linear estimate, which means that the constant K depends only on F and not on u (which is false in general for \(s>1\)).

  3. (iii)

    The usual estimate for composition implies that

    $$\begin{aligned} \left\Vert F(u)\right\Vert _{\dot{H}^\sigma }\leqq \mathcal {F}(\left\Vert u\right\Vert _{L^\infty })\left\Vert u\right\Vert _{L^2\cap \dot{H}^{\sigma }}. \end{aligned}$$

    The bound (2.8) improves the latter estimate in that one requires less control of the low frequency component. This will play a role in the proof of Lemmas 3.2 and 3.3.

Proof

  1. (i)

    Since \(\delta _{\alpha }(uv)=u \delta _\alpha v +(\tau _\alpha v)\delta _\alpha u\) where \(\tau _\alpha v(x)=v(x-\alpha )\), we have

    $$\begin{aligned} \left\Vert \delta _\alpha (uv)\right\Vert _{L^2}\leqq \left\Vert u\right\Vert _{L^\infty }\left\Vert \delta _\alpha v\right\Vert _{L^2}+\left\Vert v\right\Vert _{L^\infty }\left\Vert \delta _\alpha u\right\Vert _{L^2}. \end{aligned}$$

    Directly from the definition (2.1), we deduce that

    $$\begin{aligned} \left\Vert uv\right\Vert _{\dot{B}^s_{2,2}}\leqq 2\left\Vert u\right\Vert _{L^\infty }\left\Vert v\right\Vert _{\dot{B}^s_{2,2}}+2\left\Vert v\right\Vert _{L^\infty }\left\Vert u\right\Vert _{\dot{B}^s_{2,2}}. \end{aligned}$$

    This implies (2.6) by virtue of the identity (2.3) on the equivalence of \(\left\Vert \cdot \right\Vert _{\dot{H}^s}\) and \(\left\Vert \cdot \right\Vert _{\dot{B}^s_{2,2}}\).

  2. (ii)

    Similarly, the inequality (2.7) follows directly from the fact that

    $$\begin{aligned} \left\Vert \delta _\alpha F(u)\right\Vert _{L^2} \leqq K \left\Vert \delta _\alpha u\right\Vert _{L^2}. \end{aligned}$$
  3. (iii)

    We adapt the classical proof of the composition rule in nonhomogeneous Sobolev spaces, which is based on the Littlewood–Paley decomposition. Namely, choose a function \(\Phi \in C^\infty _0(\{\xi ;\left|\xi \right|< 1\})\) which is equal to 1 when \(|\xi |\leqq 1/2\) and set \(\phi (\xi )=\Phi (\xi /2)-\Phi (\xi )\) which is supported in the annulus \(\{\xi ;1/2\leqq |\xi |\leqq 2\}\). Then, for all \(\xi \in \mathbb {R}\), one has \(\Phi (\xi )+\sum _{j\in \mathbb {N}}\phi (2^{-j}\xi )=1\), which one can use to decompose tempered distribution. For \(u\in \mathcal {S}'(\mathbb {R})\), we set \(\Delta _{-1}u=\mathcal {F}^{-1}(\Phi (\xi )\widehat{u})\) and \(\Delta _j u=\mathcal {F}^{-1}(\phi (2^{-j}\xi )\widehat{u})\) for \(j\in \mathbb {N}\). We also use the notation \(S_j u =\sum _{-1\leqq p\leqq j-1} \Delta _pu\) for \(j\geqq 0\) (so that \(S_0u=\Delta _{-1}u=\Phi (D_x)u\)).

The classical proof (see [3, 4, 35]) of the composition rule consists in splitting F(u) as

$$\begin{aligned} F(u)&=F(S_0 u)+F(S_1 u)-F(S_0u)+\cdots +F(S_{j+1}u)-F(S_ju)+\cdots \\&=F(S_0u)+\sum _{j\in \mathbb {N}} m_j \Delta _j u \qquad \text {with}\quad m_j=\int \nolimits _0^1 F'(S_ju+y\Delta _j u){{\,\mathrm{d}\,}} y\\&=F(S_0u)+\sum _{j\in \mathbb {N}} m_j \Delta _j \tilde{u} \qquad \text {with}\quad \tilde{u}=u-\Phi (2D_x)u, \end{aligned}$$

where we used \(\Delta _j \circ \Phi (2D_x)=0\) for \(j\geqq 0\). Then, the Meyer’s multiplier lemma (see [40, Theorem 2] or [3, Lemma 2.2]) implies that

$$\begin{aligned} \Big \Vert \sum _{j\geqq 0} m_j \Delta _j \tilde{u}\Big \Vert _{H^\sigma } \leqq \mathcal {F}(\left\Vert u\right\Vert _{L^\infty })\left\Vert \tilde{u}\right\Vert _{H^\sigma }, \end{aligned}$$

where, to clarify notations, we insist on the fact that above \(H^\sigma \) is the nonhomogeneous Sobolev space. Since \(\left\Vert \tilde{u}\right\Vert _{H^\sigma }\leqq \left\Vert u\right\Vert _{\dot{H}^\sigma }\), we see that the contribution of \(\sum m_j \Delta _j \tilde{u}\) is bounded by the right-hand side of (2.8). This shows that the only difficulty is to estimate the low frequency component \(F(S_0u)\). We claim that

$$\begin{aligned} \left\Vert F(S_0u)\right\Vert _{\dot{H}^\sigma }\leqq \mathcal {F}(\left\Vert u\right\Vert _{L^\infty })\left\Vert u\right\Vert _{\dot{H}^{\sigma -1}}. \end{aligned}$$
(2.9)

To see this, we start with

$$\begin{aligned} \left\Vert F(S_0u)\right\Vert _{\dot{H}^\sigma } =\left\Vert \partial _x \big (F(S_0u)\big )\right\Vert _{\dot{H}^{\sigma -1}} =\left\Vert F'(S_0u)\partial _x S_0u \right\Vert _{\dot{H}^{\sigma -1}}, \end{aligned}$$

and then use the product rule (2.6) with \(s=\sigma -1\in (0,1)\),

$$\begin{aligned} \left\Vert F'(S_0u)\partial _x S_0u \right\Vert _{\dot{H}^{\sigma -1}}&\leqq 2\left\Vert F'(S_0u)\right\Vert _{L^\infty } \left\Vert \partial _x S_0u \right\Vert _{\dot{H}^{\sigma -1}}\\&\quad +2\left\Vert F'(S_0u)\right\Vert _{\dot{H}^{\sigma -1}}\left\Vert \partial _x S_0u \right\Vert _{L^\infty }. \end{aligned}$$

Since \(\left|\xi \Phi (\xi )\right|\leqq 1\), one has the obvious inequality

$$\begin{aligned} \left\Vert \partial _x S_0u \right\Vert _{\dot{H}^{\sigma -1}}\leqq \left\Vert u \right\Vert _{\dot{H}^{\sigma -1}}. \end{aligned}$$

On the other hand, since the support of the Fourier transform of \(S_0u\) is included in the ball of center 0 and radius 1, it follows from the Bernstein’s inequality that

$$\begin{aligned} \left\Vert S_0u \right\Vert _{L^\infty }\leqq C_1\left\Vert u\right\Vert _{L^\infty },\qquad \left\Vert \partial _x S_0u \right\Vert _{L^\infty }\leqq C_2 \left\Vert u\right\Vert _{L^\infty }. \end{aligned}$$

The first estimate above also implies that

$$\begin{aligned} \left\Vert F'(S_0u)\right\Vert _{L^\infty }\leqq \mathcal {F}_1(\left\Vert S_0 u\right\Vert _{L^\infty })\leqq \mathcal {F}_2(\left\Vert u\right\Vert _{L^\infty }), \end{aligned}$$

where \(\mathcal {F}_1(r)= \sup _{y\in [-r,r]}\left|F'(y)\right|\) and \(\mathcal {F}_2(r)=\mathcal {F}_1(C_1r)\). It thus remains only to estimate \(\left\Vert F'(S_0u)\right\Vert _{\dot{H}^{\sigma -1}}\). Notice that we may apply the composition rule given in statement (ii) since the index \(\sigma -1\) belongs to (0, 1) and since \(F'\) is Lipschitz on an open set containing \(S_0u(\mathbb {R})\). The composition rule (2.7) implies that

$$\begin{aligned} \left\Vert F'(S_0u)\right\Vert _{\dot{H}^{\sigma -1}}\leqq K \left\Vert S_0u\right\Vert _{\dot{H}^{\sigma -1}}\leqq K \left\Vert u\right\Vert _{\dot{H}^{\sigma -1}} \end{aligned}$$

with

$$\begin{aligned} K=\sup _{[-2\left\Vert S_0u\right\Vert _{L^\infty },2\left\Vert S_0u\right\Vert _{L^\infty }]}\left|F''\right|\leqq \mathcal {F}_3(\left\Vert u\right\Vert _{L^\infty }). \end{aligned}$$

This proves that the \(\dot{H}^\sigma \)-norm of \(F(S_0u)\) satisfies (2.9) and hence it is bounded by the right-hand side of (2.8), which completes the proof of statement (iii). \(\quad \square \)

For later purposes, we prove the following commutator estimate with the Hilbert transform:

Lemma 2.6

Let \(0<\theta<\nu <1\). There exists a constant K such that for all \(f\in C^{0,\nu }(\mathbb {R})\), and all u in the nonhomogeneous space \(H^{-\theta }(\mathbb {R})\),

$$\begin{aligned} \left\Vert \mathcal {H}(fu)-f\mathcal {H} u\right\Vert _{L^2} \leqq K\left\Vert f\right\Vert _{C^{0,\nu }} \left\Vert u\right\Vert _{H^{-\theta }}. \end{aligned}$$
(2.10)

Proof

We establish this estimate by using the para-differential calculus of Bony [6]. We use the Littlewood–Paley decomposition (see the proof of Proposition 2.4) and denote by \(T_f\) the operator of para-multiplication by f, so that

$$\begin{aligned} T_f u=\sum _{j\geqq 1}S_{j-1}(f)\Delta _j u. \end{aligned}$$

Denote by \(f^{\flat }\) the multiplication operator \( u\mapsto fu\) and introduce \(\mathcal {H}_0\), the Fourier multiplier with symbol \(-i(1-\Phi (\xi ))\xi /\left|\xi \right|\) where \(\Phi \in C^\infty _0(\mathbb {R})\) is such that \(\Phi (\xi )=1\) on a neighborhood of the origin. With these notations, one can rewrite the commutator \(\left[ \mathcal {H}, f^{\flat } \right] \) as

$$\begin{aligned} \left[ \mathcal {H}, f^{\flat } \right]&=\left[ \mathcal {H},T_{f}\right] +\mathcal {H}(f^{\flat }-T_{f})-(f^{\flat }-T_{f})\mathcal {H}\nonumber \\&=\left[ \mathcal {H}_0,T_{f}\right] +(\mathcal {H}-\mathcal {H}_0)T_{f} -T_f(\mathcal {H}-\mathcal {H}_0)\nonumber \\&\quad +\mathcal {H}(f^{\flat }-T_{f})-(f^{\flat }-T_{f})\mathcal {H}. \end{aligned}$$
(2.11)

Notice that \(\mathcal {H}-\mathcal {H}_0\) is a smoothing operator (that is an operator bounded from \(H^\sigma \) to \(H^{\sigma +t}\) for any real numbers \(\sigma ,t\in \mathbb {R}\)). We then use two classical estimates for paradifferential operators (see [6, 39]). Firstly,

$$\begin{aligned} \forall \sigma \in \mathbb {R},\quad \left\Vert T_f\right\Vert _{{H^{\sigma }}\rightarrow {H^{\sigma }}}\leqq c(\sigma )\left\Vert f\right\Vert _{L^\infty }, \end{aligned}$$

so

$$\begin{aligned} \left\Vert (\mathcal {H}-\mathcal {H}_0)T_{f}\right\Vert _{H^{-\theta }\rightarrow L^2}&\leqq \left\Vert \mathcal {H}-\mathcal {H}_0\right\Vert _{H^{-\theta }\rightarrow L^2}\left\Vert T_{f}\right\Vert _{H^{-\theta }\rightarrow H^{-\theta }}\lesssim \left\Vert f\right\Vert _{L^\infty },\\ \left\Vert T_{f}(\mathcal {H}-\mathcal {H}_0)\right\Vert _{H^{-\theta }\rightarrow L^2}&\leqq \left\Vert T_{f}\right\Vert _{L^2\rightarrow L^{2}} \left\Vert \mathcal {H}-\mathcal {H}_0\right\Vert _{H^{-\theta }\rightarrow L^2}\lesssim \left\Vert f\right\Vert _{L^\infty }. \end{aligned}$$

Secondly, since \(\mathcal {H}_0\) is a Fourier multiplier whose symbol is a smooth function of order 0 (which means that its kth derivative is bounded by \(C_k(1+|\xi |)^{-k}\)), one has

$$\begin{aligned} \forall \sigma \in \mathbb {R},\quad \left\Vert \left[ \mathcal {H}_0,T_f \right] \right\Vert _{{H^{\sigma }}\rightarrow {H^{\sigma +\nu }}}\leqq c(\nu ,\sigma )\left\Vert f\right\Vert _{C^{0,\nu }}. \end{aligned}$$

In particular,

$$\begin{aligned} \left\Vert \left[ \mathcal {H}_0,T_f \right] \right\Vert _{{H^{-\nu }}\rightarrow {L^2}}\lesssim \left\Vert f\right\Vert _{C^{0,\nu }}. \end{aligned}$$

It remains only to estimate the last two terms in the right-hand side of (2.11). We claim that

$$\begin{aligned} \bigl \Vert \mathcal {H}(f^{\flat }-T_{f})\bigr \Vert _{H^{-\theta }\rightarrow L^2}+\bigl \Vert (f^{\flat }-T_{f})\mathcal {H} \bigr \Vert _{H^{-\theta }\rightarrow L^2}\lesssim \left\Vert f\right\Vert _{C^{0,\nu }}. \end{aligned}$$

Since \(\mathcal {H}\) is bounded from \(H^\sigma \) to itself for any \(\sigma \in \mathbb {R}\), it is enough to prove that

$$\begin{aligned} \bigl \Vert f^{\flat }-T_{f}\bigr \Vert _{H^{-\theta }\rightarrow L^2}\lesssim \left\Vert f\right\Vert _{C^{0,\nu }}. \end{aligned}$$

To do this, observe that

$$\begin{aligned} fg-T_{f}g=\sum _{j,p\geqq -1}(\Delta _j f)(\Delta _{p}g)-\sum _{-1\leqq j\leqq p-2}(\Delta _j f)(\Delta _{p}g) =\sum _{j\geqq -1} (S_{j+2}g)\Delta _{j}f. \end{aligned}$$

Then, using the Bernstein’s inequality and the characterization of Hölder spaces in terms of Littlewood–Paley decomposition, it follows from the assumption \(\theta <\nu \) that the series \(\sum 2^{j(\theta -\nu )}\) converges, so

$$\begin{aligned} \left\Vert fg-T_{f}g\right\Vert _{L^2}&\leqq \sum \left\Vert S_{j+2}g\right\Vert _{L^2}\left\Vert \Delta _{j}f\right\Vert _{L^\infty }\\&\lesssim \sum 2^{\theta (j+2)}\left\Vert g\right\Vert _{H^{-\theta }}2^{-j\nu }\left\Vert f\right\Vert _{C^{0,\nu }}\lesssim \left\Vert g\right\Vert _{H^{-\theta }}\left\Vert f\right\Vert _{C^{0,\nu }}. \end{aligned}$$

By combining the previous estimates, we have \(\left\Vert \left[ \mathcal {H}, f^{\flat } \right] \right\Vert _{H^{-\theta }\rightarrow L^2}\lesssim \left\Vert f\right\Vert _{C^{0,\nu }}\), which gives the result. \(\quad \square \)

3 Commutator Estimate

In this section we prove statement (iii) in Theorem 1.2. Namely, we prove the following proposition:

Proposition 3.1

Let \(0<\varepsilon <1/2\). There is a non-decreasing function \(\mathcal {F}:\mathbb {R}_+\rightarrow \mathbb {R}_+\) such that, for all functions fg in \(\cap _{\sigma \geqq 1}\dot{H}^\sigma (\mathbb {R})\),

$$\begin{aligned}&\left\Vert \Lambda ^{1+\varepsilon }\mathcal {T}(f)g-\mathcal {T}(f)\Lambda ^{1+\varepsilon }g\right\Vert _{L^2}\nonumber \\&\quad \leqq \mathcal {F}\big (\left\Vert f\right\Vert _{\dot{H}^1\cap \dot{H}^{\frac{3}{2}+\varepsilon }}\big ) \left\Vert f\right\Vert _{\dot{H}^1\cap \dot{H}^{\frac{3}{2}+\varepsilon }}\left\Vert g \right\Vert _{\dot{H}^{\frac{3}{2}+\varepsilon }\cap \dot{H}^2}. \end{aligned}$$
(3.1)

Proof

Recall that

$$\begin{aligned} \mathcal {T}(f)g=-\frac{1}{\pi }\int \frac{(\Delta _\alpha f)^2}{1+(\Delta _\alpha f)^2}\Delta _\alpha g_x{{\,\mathrm{d}\,}}\alpha . \end{aligned}$$

Since

$$\begin{aligned} \Lambda ^{1+\varepsilon }\Delta _\alpha g_x=\Delta _\alpha \big (\Lambda ^{1+\varepsilon }g_x\big ), \end{aligned}$$

we have

$$\begin{aligned} \Lambda ^{1+\varepsilon }\mathcal {T}(f)g =\mathcal {T}(f)\Lambda ^{1+\varepsilon }g+ R_1(f)g + R_2(f)g, \end{aligned}$$

where

$$\begin{aligned} R_1(f)g=-\frac{1}{\pi }\int \big (\Delta _\alpha g_x\big ) \Lambda ^{1+\varepsilon } \bigg (\frac{(\Delta _\alpha f)^2}{1+(\Delta _\alpha f)^2}\bigg ) {{\,\mathrm{d}\,}}\alpha , \end{aligned}$$
(3.2)

and

$$\begin{aligned} R_2(f)g&= -\frac{1}{\pi }\int \big (\Lambda ^{1+\varepsilon } \big (u_\alpha v_\alpha \big ) -u_\alpha \Lambda ^{1+\varepsilon }v_\alpha - v_\alpha \Lambda ^{1+\varepsilon } u_\alpha \big ) {{\,\mathrm{d}\,}}\alpha \quad \text {with}\\ u_\alpha&=\Delta _\alpha g_x,\qquad v_\alpha =\frac{(\Delta _\alpha f)^2}{1+(\Delta _\alpha f)^2}. \end{aligned}$$

We shall estimate these two terms separately. Classical results from paradifferential calculus (see [6, 16, 39]) would allow us to estimate them provided that we work in nonhomogeneous Sobolev spaces. In the homogeneous spaces we are considering, we shall see that one can derive similar results by using only elementary nonlinear estimates.

We begin with the study of \(R_1(f)g\).

Lemma 3.2

There exists a non-decreasing function \(\mathcal {F}:\mathbb {R}_+\rightarrow \mathbb {R}_+\) such that

$$\begin{aligned} \left\Vert R_1(f)g\right\Vert _{L^2}\leqq \mathcal {F}\big (\left\Vert f\right\Vert _{\dot{H}^1\cap \dot{H}^{\frac{3}{2}+\varepsilon }}\big ) \left\Vert f\right\Vert _{\dot{H}^1\cap \dot{H}^{\frac{3}{2}+\varepsilon }} \left\Vert g \right\Vert _{\dot{H}^{\frac{3}{2}+\varepsilon }\cap \dot{H}^2}. \end{aligned}$$

Proof

By definition

$$\begin{aligned} R_1(f)g=-\frac{1}{\pi }\int \big (\Delta _\alpha g_x\big ) \Lambda ^{1+\varepsilon } \bigg (\frac{(\Delta _\alpha f)^2}{1+(\Delta _\alpha f)^2}\bigg ) {{\,\mathrm{d}\,}}\alpha , \end{aligned}$$

so

$$\begin{aligned} \left\Vert R_1(f)g\right\Vert _{L^2} \lesssim \int \Vert \Delta _{\alpha }g_x\Vert _{L^{\infty }} \left\Vert \frac{(\Delta _\alpha f)^2}{1+(\Delta _\alpha f)^2}\right\Vert _{\dot{H}^{1+\varepsilon }}{{\,\mathrm{d}\,}}\alpha . \end{aligned}$$
(3.3)

The Sobolev embedding \(L^2(\mathbb {R})\cap \dot{H}^{\frac{1}{2}+\varepsilon }(\mathbb {R})\hookrightarrow L^\infty (\mathbb {R})\) implies that, for all \(\alpha \) in \(\mathbb {R}\),

$$\begin{aligned} \sup _{x\in \mathbb {R}}\left|\Delta _\alpha f(x)\right|\leqq \sup _{x\in \mathbb {R}}\left|f_x(x)\right|\lesssim \left\Vert f_x\right\Vert _{L^2\cap \dot{H}^{\frac{1}{2}+\varepsilon }}=\left\Vert f\right\Vert _{\dot{H}^1\cap \dot{H}^{\frac{3}{2}+\varepsilon }}, \end{aligned}$$
(3.4)

so that the composition rule (2.8) implies that

$$\begin{aligned} \left\Vert \frac{(\Delta _\alpha f)^2}{1+(\Delta _\alpha f)^2}\right\Vert _{\dot{H}^{1+\varepsilon }} \leqq \mathcal {F}\big (\left\Vert f\right\Vert _{\dot{H}^1\cap \dot{H}^{\frac{3}{2}+\varepsilon }}\big )\Big ( \left\Vert \Delta _\alpha f\right\Vert _{\dot{H}^\varepsilon }+\left\Vert \Delta _\alpha f\right\Vert _{\dot{H}^{1+\varepsilon }}\Big ). \end{aligned}$$
(3.5)

We claim that we have the two following inequalities:

$$\begin{aligned}&\int \Vert \Delta _{\alpha }g_x\Vert _{L^{\infty }} \left\Vert \Delta _\alpha f\right\Vert _{\dot{H}^{\varepsilon }}{{\,\mathrm{d}\,}}\alpha \lesssim \left\Vert g \right\Vert _{\dot{H}^{\frac{3}{2}+\varepsilon }} \left\Vert f \right\Vert _{\dot{H}^{1}}, \end{aligned}$$
(3.6)
$$\begin{aligned}&\int \Vert \Delta _{\alpha }g_x\Vert _{L^{\infty }} \left\Vert \Delta _\alpha f\right\Vert _{\dot{H}^{1+\varepsilon }}{{\,\mathrm{d}\,}}\alpha \lesssim \left\Vert g\right\Vert _{\dot{H}^2} \left\Vert f \right\Vert _{\dot{H}^{\frac{3}{2}+\varepsilon }}. \end{aligned}$$
(3.7)

Let us prove (3.6). Directly from the definition of \(\Delta _\alpha \), we have

$$\begin{aligned} \int \Vert \Delta _{\alpha }g_x\Vert _{L^{\infty }} \left\Vert \Delta _\alpha f\right\Vert _{\dot{H}^{\varepsilon }}{{\,\mathrm{d}\,}}\alpha&= \int \frac{\Vert \delta _{\alpha }g_{x}\Vert _{L^{\infty }}}{\vert \alpha \vert } \frac{\Vert \delta _{\alpha }\Lambda ^{\varepsilon }f\Vert _{L^2}}{\vert \alpha \vert } {{\,\mathrm{d}\,}}\alpha \\&= \int \frac{\Vert \delta _{\alpha }g_{x}\Vert _{L^{\infty }}}{\vert \alpha \vert ^{\varepsilon }} \frac{\Vert \delta _{\alpha }\Lambda ^{\varepsilon }f\Vert _{L^2}}{\vert \alpha \vert ^{1-\varepsilon }} \frac{{{\,\mathrm{d}\,}}\alpha }{\left|\alpha \right|}, \end{aligned}$$

so, using the Cauchy–Schwarz inequality,

$$\begin{aligned} \int \Vert \Delta _{\alpha }g_x\Vert _{L^{\infty }} \left\Vert \Delta _\alpha f\right\Vert _{\dot{H}^{\varepsilon }}{{\,\mathrm{d}\,}}\alpha \leqq \left( \int \frac{\Vert \delta _{\alpha }g_{x}\Vert _{L^{\infty }}^2}{\vert \alpha \vert ^{2\varepsilon }} \frac{{{\,\mathrm{d}\,}}\alpha }{|\alpha |}\right) ^{\frac{1}{2}} \left( \int \frac{\Vert \delta _{\alpha }\Lambda ^{\varepsilon }f\Vert _{L^2}^2}{\vert \alpha \vert ^{2(1-\varepsilon )}} \frac{{{\,\mathrm{d}\,}}\alpha }{\left|\alpha \right|}\right) ^\frac{1}{2}, \end{aligned}$$

and hence, using the definition of Besov semi-norms (see (2.1)),

$$\begin{aligned} \int \Vert \Delta _{\alpha }g_x\Vert _{L^{\infty }} \left\Vert \Delta _\alpha f\right\Vert _{\dot{H}^{\varepsilon }}{{\,\mathrm{d}\,}}\alpha \lesssim \Vert g_x \Vert _{\dot{B}^{\varepsilon }_{\infty ,2}} \Vert \Lambda ^{\varepsilon } f \Vert _{\dot{B}^{1-\varepsilon }_{2,2}}. \end{aligned}$$

By using (2.3) and (2.4), we obtain that

$$\begin{aligned} \int \Vert \Delta _{\alpha }g_x\Vert _{L^{\infty }} \left\Vert \Delta _\alpha f\right\Vert _{\dot{H}^{\varepsilon }}{{\,\mathrm{d}\,}}\alpha \lesssim \Vert g \Vert _{\dot{H}^{\frac{3}{2}+\varepsilon }} \Vert f \Vert _{\dot{H}^{1}}, \end{aligned}$$

which is the first claim (3.6). To prove the second claim (3.7), we repeat the same arguments except that we balance the powers of \(\alpha \) in a different way:

$$\begin{aligned} \int \Vert \Delta _{\alpha }g_x\Vert _{L^{\infty }} \left\Vert \Delta _\alpha f\right\Vert _{\dot{H}^{1+\varepsilon }}{{\,\mathrm{d}\,}}\alpha&= \int \frac{\Vert \delta _{\alpha }g_{x}\Vert _{L^{\infty }}}{\vert \alpha \vert ^{1/2}} \frac{\Vert \delta _{\alpha }\Lambda ^{1+\varepsilon }f\Vert _{L^2}}{\vert \alpha \vert ^{1/2}} \frac{{{\,\mathrm{d}\,}}\alpha }{\left|\alpha \right|}\\&\leqq \left( \int \frac{\Vert \delta _{\alpha }g_{x}\Vert _{L^{\infty }}^2}{\vert \alpha \vert } \frac{{{\,\mathrm{d}\,}}\alpha }{|\alpha |}\right) ^{\frac{1}{2}} \left( \int \frac{\Vert \delta _{\alpha }\Lambda ^{1+\varepsilon }f\Vert _{L^2}^2}{\vert \alpha \vert } \frac{{{\,\mathrm{d}\,}}\alpha }{\left|\alpha \right|}\right) ^\frac{1}{2}\\&\leqq \left\Vert g_x \right\Vert _{\dot{B}^{\frac{1}{2}}_{\infty ,2}} \left\Vert \Lambda ^{1+\varepsilon } f \right\Vert _{\dot{B}^{\frac{1}{2}}_{2,2}}\\&\lesssim \left\Vert g \right\Vert _{\dot{H}^2} \left\Vert f \right\Vert _{\dot{H}^{\frac{3}{2}+\varepsilon }}, \end{aligned}$$

which proves the claim (3.7). Now, by combining the two claims (3.6), (3.7) with (3.3) and (3.5), we obtain that

$$\begin{aligned} \left\Vert R_1(f)g\right\Vert _{L^2}\leqq \mathcal {F}\big (\left\Vert f\right\Vert _{\dot{H}^1\cap \dot{H}^{\frac{3}{2}+\varepsilon }}\big ) \left\Vert f\right\Vert _{\dot{H}^1\cap \dot{H}^{\frac{3}{2}+\varepsilon }} \left\Vert g \right\Vert _{\dot{H}^{\frac{3}{2}+\varepsilon }\cap \dot{H}^2}, \end{aligned}$$

which is the desired result. \(\quad \square \)

We now move to the second remainder term \(R_2(f)g\).

Lemma 3.3

There exists a non-decreasing function \(\mathcal {F}:\mathbb {R}_+\rightarrow \mathbb {R}_+\) such that

$$\begin{aligned} \left\Vert R_2(f)g\right\Vert _{L^2}\leqq \mathcal {F}\big (\left\Vert f\right\Vert _{\dot{H}^1\cap \dot{H}^{\frac{3}{2}+\varepsilon }}\big ) \left\Vert f\right\Vert _{\dot{H}^1\cap \dot{H}^{\frac{3}{2}+\varepsilon }} \left\Vert g \right\Vert _{\dot{H}^{\frac{3}{2}+\varepsilon }\cap \dot{H}^2}. \end{aligned}$$

Proof

We use the classical Kenig–Ponce–Vega commutator estimate

$$\begin{aligned} \Vert \Lambda ^{s}( uv) -u \Lambda ^s v - v \Lambda ^s u \Vert _{L^{r}} \leqq C \left\Vert \Lambda ^{s_{1}} u \right\Vert _{L^{p_1}} \left\Vert \Lambda ^{s_{2}} v \right\Vert _{L^{p_2}}. \end{aligned}$$
(3.8)

where \(s=s_1+s_2\) and \(1/r=1/p_1+1/p_2\). Kenig, Ponce and Vega considered the case \(s<1\). Since, for our purpose we need \(s>1\), we will use the recent improvement by Li [36] (see also D’Ancona [25]) showing that (3.8) holds under the assumptions

$$\begin{aligned} s=s_1+s_2\in (0,2),\quad s_j\in (0,1), \quad \frac{1}{r}=\frac{1}{p_1}+\frac{1}{p_2},\quad 2\leqq p_j<\infty . \end{aligned}$$

With \(p_1=4\), \(p_2=4\), \(r=2\), \(s=1+\varepsilon \), \(s_1=\frac{3\varepsilon }{2}\), \(s_2=1-\frac{\varepsilon }{2}\), this implies that

$$\begin{aligned} \left\Vert R_2(f)g\right\Vert _{L^2}\lesssim \int \bigl \Vert \Lambda ^{\frac{3\varepsilon }{2}}\Delta _{\alpha } g_{x} \bigr \Vert _{L^{4}} \left\Vert \Lambda ^{1-\frac{\varepsilon }{2}}\frac{(\Delta _\alpha f)^2}{1+(\Delta _\alpha f)^2} \right\Vert _{L^4} {{\,\mathrm{d}\,}}\alpha . \end{aligned}$$

We now use the Sobolev inequality

$$\begin{aligned} \left\Vert u\right\Vert _{L^4}\lesssim \bigl \Vert \Lambda ^{\frac{1}{4}} u\bigr \Vert _{L^2}, \end{aligned}$$

to obtain

$$\begin{aligned} \left\Vert R_2(f)g\right\Vert _{L^2}\lesssim \int \bigl \Vert \Lambda ^{\frac{1}{4}+\frac{3\varepsilon }{2}}\Delta _{\alpha } g_{x} \bigr \Vert _{L^{2}} \left\Vert \Lambda ^{\frac{5}{4}-\frac{\varepsilon }{2}}\frac{(\Delta _\alpha f)^2}{1+(\Delta _\alpha f)^2} \right\Vert _{L^2} {{\,\mathrm{d}\,}}\alpha . \end{aligned}$$

By combining the composition rule (2.8) (applied with \(\sigma =\frac{5}{4}-\frac{\varepsilon }{2}\in (1,2)\)) and (3.4), we obtain that

$$\begin{aligned}&\left\Vert R_2(f)g\right\Vert _{L^2}\\&\quad \lesssim \mathcal {F}(M) \int \bigl \Vert \Lambda ^{\frac{1}{4}+\frac{3\varepsilon }{2}}\Delta _{\alpha } g_{x} \bigr \Vert _{L^{2}} \Big (\bigl \Vert \Lambda ^{\frac{1}{4}-\frac{\varepsilon }{2}}\Delta _\alpha f \bigr \Vert _{L^2} +\bigl \Vert \Lambda ^{\frac{5}{4}-\frac{\varepsilon }{2}}\Delta _\alpha f \bigr \Vert _{L^2}\Big ) {{\,\mathrm{d}\,}}\alpha \end{aligned}$$

where \(M=\left\Vert f\right\Vert _{\dot{H}^1\cap \dot{H}^{\frac{3}{2}+\varepsilon }}\). We now proceed as in the previous proof. More precisely, we balance the powers of \(\alpha \), use the Cauchy–Schwarz inequality, the definition of the Besov semi-norms (2.1) and the Sobolev embedding to obtain that

$$\begin{aligned}&\int \bigl \Vert \Lambda ^{\frac{1}{4}+\frac{3\varepsilon }{2}}\Delta _{\alpha } g_{x} \bigr \Vert _{L^{2}} \bigl \Vert \Lambda ^{\frac{1}{4}-\frac{\varepsilon }{2}}\Delta _\alpha f \bigr \Vert _{L^2} {{\,\mathrm{d}\,}}\alpha \\&\quad =\int \frac{\bigl \Vert \Lambda ^{\frac{1}{4}+\frac{3\varepsilon }{2}}\delta _{\alpha } g_{x} \bigr \Vert _{L^2}}{\left|\alpha \right|} \frac{\bigl \Vert \Lambda ^{\frac{1}{4}-\frac{\varepsilon }{2}}\delta _\alpha f \bigr \Vert _{L^2}}{\left|\alpha \right|}{{\,\mathrm{d}\,}}\alpha \\&\quad =\int \frac{\bigl \Vert \Lambda ^{\frac{1}{4}+\frac{3\varepsilon }{2}}\delta _{\alpha } g_x \bigr \Vert _{L^2}}{\left|\alpha \right|^{1/4-\varepsilon /2}} \frac{\bigl \Vert \Lambda ^{\frac{1}{4}-\frac{\varepsilon }{2}}\delta _\alpha f \bigr \Vert _{L^2}}{\left|\alpha \right|^{3/4+\varepsilon /2}}\frac{{{\,\mathrm{d}\,}}\alpha }{\left|\alpha \right|}\\&\quad \leqq \left\Vert \Lambda ^{\frac{1}{4}+\frac{3\varepsilon }{2}}g_x\right\Vert _{\dot{B}^{\frac{1}{4}-\varepsilon /2}_{2,2}} \left\Vert \Lambda ^{\frac{1}{4}-\frac{\varepsilon }{2}} f\right\Vert _{\dot{B}^{\frac{3}{4}+\varepsilon /2}_{2,2}}\\&\quad \lesssim \left\Vert g\right\Vert _{\dot{H}^{\frac{3}{2}+\varepsilon }} \left\Vert f\right\Vert _{\dot{H}^1}. \end{aligned}$$

One estimates the second term in a similar way. We begin by writing that

$$\begin{aligned}&\int \bigl \Vert \Lambda ^{\frac{1}{4}+\frac{3\varepsilon }{2}}\Delta _{\alpha } g_{x} \bigr \Vert _{L^{2}} \bigl \Vert \Lambda ^{\frac{5}{4}-\frac{\varepsilon }{2}}\Delta _\alpha f \bigr \Vert _{L^2} {{\,\mathrm{d}\,}}\alpha \\&\quad =\int \frac{\bigl \Vert \Lambda ^{\frac{1}{4}+\frac{3\varepsilon }{2}}\delta _{\alpha } g_{x} \bigr \Vert _{L^2}}{\left|\alpha \right|} \frac{\bigl \Vert \Lambda ^{\frac{5}{4}-\frac{\varepsilon }{2}}\delta _\alpha f \bigr \Vert _{L^2}}{\left|\alpha \right|}{{\,\mathrm{d}\,}}\alpha \\&\quad =\int \frac{\bigl \Vert \Lambda ^{\frac{1}{4}+\frac{3\varepsilon }{2}}\delta _{\alpha } g_x \bigr \Vert _{L^2}}{\left|\alpha \right|^{3/4-3\varepsilon /2}} \frac{\bigl \Vert \Lambda ^{\frac{5}{4}-\frac{\varepsilon }{2}}\delta _\alpha f \bigr \Vert _{L^2}}{\left|\alpha \right|^{1/4+3\varepsilon /2}}\frac{{{\,\mathrm{d}\,}}\alpha }{\left|\alpha \right|}. \end{aligned}$$

Since \(\varepsilon \) belongs to (0, 1/2) we have \(3/4-3\varepsilon /2>0\) and \(1/4+3\varepsilon /2<1\). Therefore one can use the definition (2.1) of the Besov semi-norms to deduce that

$$\begin{aligned} \int \bigl \Vert \Lambda ^{\frac{1}{4}+\frac{3\varepsilon }{2}}\Delta _{\alpha } g_{x} \bigr \Vert _{L^{2}} \bigl \Vert \Lambda ^{\frac{5}{4}-\frac{\varepsilon }{2}}\Delta _\alpha f \bigr \Vert _{L^2} {{\,\mathrm{d}\,}}\alpha&\leqq \bigl \Vert \Lambda ^{\frac{1}{4}+\frac{3\varepsilon }{2}}g_x\bigr \Vert _{\dot{B}^{\frac{3}{4}-\frac{3\varepsilon }{2}}_{2,2}} \bigl \Vert \Lambda ^{\frac{5}{4}-\frac{\varepsilon }{2}} f\bigr \Vert _{\dot{B}^{\frac{1}{4}+\frac{3\varepsilon }{2}}_{2,2}}\\&\lesssim \left\Vert g\right\Vert _{\dot{H}^{2}}\left\Vert f\right\Vert _{\dot{H}^{\frac{3}{2}+\varepsilon }}. \end{aligned}$$

By combining the above inequalities, we have proved that

$$\begin{aligned} \left\Vert R_2(f)g\right\Vert _{L^2}\leqq \mathcal {F}\big (\left\Vert f\right\Vert _{\dot{H}^1\cap \dot{H}^{\frac{3}{2}+\varepsilon }}\big ) \left\Vert f\right\Vert _{\dot{H}^1\cap \dot{H}^{\frac{3}{2}+\varepsilon }} \left\Vert g\right\Vert _{\dot{H}^{\frac{3}{2}+\varepsilon }\cap \dot{H}^{2}}, \end{aligned}$$

which concludes the proof. \(\quad \square \)

This completes the proof of the proposition. \(\quad \square \)

4 High Frequency Estimate

We now prove the second point of Theorem 1.2 whose statement is recalled in the next proposition.

Proposition 4.1

For all \(0<\nu<\varepsilon <1/2\), there exists a positive constant \(C>0\) such that, for all functions fg in \(\cap _{\sigma \geqq 1}\dot{H}^\sigma (\mathbb {R})\),

$$\begin{aligned} \mathcal {T}(f)g=\frac{f_x^2}{1+f_x^2}\Lambda g+V(f)\partial _x g+R(f,g) \end{aligned}$$

where

$$\begin{aligned} \left\Vert R(f,g)\right\Vert _{L^2}\leqq C\left\Vert f\right\Vert _{\dot{H}^{\frac{3}{2}+\varepsilon }}\left\Vert g\right\Vert _{\dot{B}^{1-\varepsilon }_{2,1}}, \qquad \left\Vert V(f)\right\Vert _{C^{0,\nu }} \leqq C\left\Vert f\right\Vert _{\dot{H}^1\cap \dot{H}^{\frac{3}{2}+\varepsilon }}^2. \end{aligned}$$

We shall prove this proposition in this section by using a symmetrization argument which consists in replacing the finite differences \(\delta _\alpha f(x)=f(x)-f(x-\alpha )\) by the symmetric finite differences \(2f(x)-f(x-\alpha )-f(x+\alpha )\). To do so, it will be convenient to introduce a few notations.

Notation 4.2

Given a function \(f=f(x)\) and a real number \(\alpha \), we define the functions \(\bar{\delta }_\alpha f\), \(\bar{\Delta }_\alpha f\), \(s_\alpha f\), \(S_\alpha f\) and \(D_\alpha f\) by

$$\begin{aligned}&\bar{\delta }_\alpha f(x)=f(x)-f(x+\alpha ),\\&s_\alpha f(x)=\delta _\alpha f(x)+\bar{\delta }_\alpha f(x)=2f(x)-f(x-\alpha )-f(x+\alpha ), \end{aligned}$$

and

$$\begin{aligned}&\bar{\Delta }_\alpha f(x)=\frac{f(x)-f(x+\alpha )}{\alpha },\\&S_\alpha f(x)=\Delta _\alpha f(x)+\bar{\Delta }_\alpha f(x)=\frac{s_\alpha f(x)}{\alpha }=\frac{2 f(x)-f(x+\alpha )-f(x-\alpha )}{\alpha },\\&D_{\alpha } f(x)=\Delta _\alpha f(x)-\bar{\Delta }_\alpha f(x)=\frac{f(x+\alpha )-f(x-\alpha )}{\alpha }. \end{aligned}$$

Lemma 4.3

One has

$$\begin{aligned} D_{\alpha } f = 2 f_x-\frac{1}{\alpha } \int \nolimits _0^\alpha s_{\eta }f_{x} {{\,\mathrm{d}\,}} \eta , \end{aligned}$$
(4.1)

where \(s_{\eta }f_{x}(x)=2f_x(x)-f_x(x+\eta )-f_x(x-\eta )\). Furthermore,

$$\begin{aligned} \partial _{\alpha }(D_\alpha f)=-\,S_{\alpha }f_{x}+\frac{1}{\alpha ^2}{\int \nolimits _{0}^{\alpha } s_{\eta }f_{x} {{\,\mathrm{d}\,}} \eta }, \end{aligned}$$
(4.2)

and

$$\begin{aligned} \partial _{\alpha }(S_\alpha f)=\bar{\Delta }_{\alpha } f_x-\Delta _{\alpha } f_x-\frac{S_{\alpha }f}{\alpha }. \end{aligned}$$
(4.3)

Proof

The formula (4.1) can be verified by two direct calculations:

$$\begin{aligned} \frac{1}{\alpha } \int \nolimits _0^\alpha 2f_x(x){{\,\mathrm{d}\,}} \eta =2f_x(x), \end{aligned}$$

and

$$\begin{aligned} \frac{1}{\alpha } \int \nolimits _0^\alpha ( f_x(x-\eta )+f_x(x+\eta )){{\,\mathrm{d}\,}} \eta&= \frac{1}{\alpha } \int \nolimits _0^\alpha \partial _\eta (f(x+\eta )-f(x-\eta )){{\,\mathrm{d}\,}} \eta \\&=\frac{1}{\alpha }(f(x+\alpha )-f(x-\alpha )). \end{aligned}$$

Now, the value for \( \partial _{\alpha }(D_\alpha f)\) in (4.2) follows by differentiating (4.1).

The formula for \(\partial _{\alpha }(S_\alpha f)\) follows from the definition of \(S_\alpha f\) and the chain rule. \(\quad \square \)

Recall that

$$\begin{aligned} \mathcal {T}(f)g= -\frac{1}{\pi }\int \frac{(\Delta _\alpha f)^2}{1+(\Delta _\alpha f)^2}\Delta _\alpha g_x {{\,\mathrm{d}\,}}\alpha . \end{aligned}$$

The idea is to decompose the factor

$$\begin{aligned} \frac{(\Delta _\alpha f)^2}{1+(\Delta _\alpha f)^2} \end{aligned}$$

into its even and odd components with respect to the variable \(\alpha \). We define

$$\begin{aligned} \mathcal {E}(\alpha ,\cdot )&=\frac{1}{2}\frac{(\Delta _\alpha f)^2}{1+(\Delta _\alpha f)^2}+\frac{1}{2}\frac{(\bar{\Delta }_\alpha f)^2}{1+(\bar{\Delta }_\alpha f)^2} , \end{aligned}$$
(4.4)
$$\begin{aligned} \mathcal {O}(\alpha ,\cdot )&=\frac{1}{2}\frac{(\Delta _\alpha f)^2}{1+(\Delta _\alpha f)^2} -\frac{1}{2}\frac{(\bar{\Delta }_\alpha f)^2}{1+(\bar{\Delta }_\alpha f)^2} , \end{aligned}$$
(4.5)

where the dots in the notations \(\mathcal {E}(\alpha ,\cdot )\) and \(\mathcal {O}(\alpha ,\cdot )\) are placeholders for the variable x (notice that \((\bar{\Delta }_\alpha f)^2=(\Delta _{-\alpha }f)^2\) and \(\bar{\Delta }_\alpha f=-\,\Delta _{-\alpha }f\)). Then,

$$\begin{aligned} \mathcal {T}(f)g=- \frac{1}{\pi } \int \Delta _{\alpha } g_{x}\ \mathcal {E}(\alpha ,\cdot ) {{\,\mathrm{d}\,}}\alpha -\frac{1}{\pi } \int \Delta _{\alpha } g_{x}\ \mathcal {O}(\alpha ,\cdot ) {{\,\mathrm{d}\,}}\alpha , \end{aligned}$$

and hence, since \(\alpha \mapsto \mathcal {E}(\alpha ,\cdot )\) is even, this yields \(\mathcal {T}(f)g= \mathcal {T}_{e}(f)g+\mathcal {T}_{o}(f)g\) with

$$\begin{aligned} \mathcal {T}_{e}(f)g&=-\frac{1}{2\pi }\int \big (\Delta _{\alpha } g_{x} - \bar{\Delta }_{\alpha } g_{x}\big ) \ \mathcal {E}(\alpha ,\cdot ) {{\,\mathrm{d}\,}}\alpha , \\ \mathcal {T}_{o}(f)g&=-\frac{1}{\pi } \int \Delta _{\alpha } g_{x}\ \mathcal {O}(\alpha ,\cdot ){{\,\mathrm{d}\,}}\alpha . \end{aligned}$$

The following result is the key point of the proof where we identify the main contribution of the nonlinearity:

Proposition 4.4

There exists a constant C such that

$$\begin{aligned}&\left\Vert \mathcal {T}_{e}(f)g - \frac{f_x^2}{1+f_x^2}\Lambda g\right\Vert _{L^2} \leqq C\left\Vert f\right\Vert _{\dot{H}^{\frac{3}{2}+\varepsilon }}\left\Vert g\right\Vert _{\dot{B}^{1-\varepsilon }_{2,1}},\nonumber \\&\left\Vert \mathcal {T}_{o}(f)g- V\partial _x g\right\Vert _{L^2}\leqq C\left\Vert f\right\Vert _{\dot{H}^{\frac{3}{2}+\varepsilon }}\left\Vert g\right\Vert _{\dot{B}^{1-\varepsilon }_{2,1}}, \end{aligned}$$
(4.6)

where

$$\begin{aligned} V(x)=-\frac{1}{\pi }\int \nolimits _\mathbb {R}\frac{\mathcal {O}(\alpha ,x)}{\alpha } {{\,\mathrm{d}\,}}\alpha . \end{aligned}$$
(4.7)

Proof

  1. (i)

    The main difficulty is to extract the elliptic component from \(\mathcal {T}_{e}(f)g\). To uncover it, we shall perform an integration by parts in \(\alpha \). The first key point is that

    $$\begin{aligned} \Delta _{\alpha } g_x - \bar{\Delta }_{\alpha } g_x&=\frac{g_x(\cdot +\alpha )-g_x(\cdot -\alpha )}{\alpha }\\&=\frac{\partial _{\alpha }\left( g(\cdot +\alpha )+g(\cdot -\alpha )-2g(\cdot )\right) }{\alpha }\\&=-\frac{\partial _{\alpha }(s_\alpha g)}{\alpha }. \end{aligned}$$

    Consequently, directly from the definition of \(\mathcal {T}_{e}(f)g\), by integrating by parts in \(\alpha \), we obtain that

    $$\begin{aligned} \mathcal {T}_{e}(f)g&=\frac{1}{2\pi }\int \frac{\partial _{\alpha }( s_\alpha g)}{\alpha } \mathcal {E}(\alpha ,\cdot ) {{\,\mathrm{d}\,}}\alpha \nonumber \\&=\frac{1}{2\pi }\int \frac{s_\alpha g}{\alpha ^2} \mathcal {E}(\alpha ,\cdot ){{\,\mathrm{d}\,}}\alpha -\frac{1}{2\pi }\int \frac{ s_\alpha g}{\alpha } \partial _\alpha \mathcal {E}(\alpha ,\cdot ){{\,\mathrm{d}\,}}\alpha . \end{aligned}$$
    (4.8)

    We now have to estimate the coefficients \(\mathcal {E}(\alpha ,\cdot )\) and \(\partial _\alpha \mathcal {E}(\alpha ,\cdot )\).

Lemma 4.5

  1. (i)

    We have

    $$\begin{aligned} \mathcal {E}(\alpha ,x)=\frac{f_x(x)^2}{1+f_x(x)^2}+Q(\alpha ,x) \end{aligned}$$
    (4.9)

    for some function Q satisfying

    $$\begin{aligned} \left|Q(\alpha ,x)\right|\lesssim \frac{\left|s_\alpha f(x)\right|}{\left|\alpha \right|}+ \left|\frac{1}{\alpha } \int \nolimits _0^\alpha s_{\eta }f_{x}(x) {{\,\mathrm{d}\,}} \eta \right|. \end{aligned}$$
    (4.10)
  2. (ii)

    Furthermore,

    $$\begin{aligned} \left|\partial _\alpha \mathcal {E}(\alpha ,x)\right|\leqq C \left\{ \frac{\left|\bar{\delta }_{\alpha } f_x(x)\right|}{\left|\alpha \right|}+ \frac{\left|\delta _{\alpha } f_x(x)\right|}{\left|\alpha \right|}+\frac{\left|s_{\alpha }f(x)\right|}{\left|\alpha \right|^2}+ \left|\frac{1}{\alpha ^2}{\int \nolimits _{0}^{\alpha } s_{\eta }f_{x} (x){{\,\mathrm{d}\,}} \eta }\right|\right\} \end{aligned}$$
    (4.11)

    for some fixed constant C.

Proof

  1. (i)

    We introduce the function

    $$\begin{aligned} F(a)=\frac{a^2}{1+a^2}. \end{aligned}$$

    Then we have the identity (4.9) with

    $$\begin{aligned} Q\mathrel {:=}\frac{1}{2}\left( F(\Delta _\alpha f)+F(\bar{\Delta }_\alpha f)\right) -F(f_x). \end{aligned}$$
    (4.12)

    Since \(F''\) is bounded, the Taylor formula implies that, for all \((a,b)\in \mathbb {R}^2\),

    $$\begin{aligned} \left|\frac{1}{2}(F(a)+F(b))-F\left( \frac{a+b}{2}\right) \right|\leqq \frac{\left\Vert F''\right\Vert _{L^\infty }}{8} \left|a-b\right|^2. \end{aligned}$$

    On the other hand, since F is bounded, one has the obvious inequality

    $$\begin{aligned} \left|\frac{1}{2}(F(a)+F(b))-F\left( \frac{a+b}{2}\right) \right|\leqq 2 \left\Vert F\right\Vert _{L^\infty }. \end{aligned}$$

    By combining these two inequalities, we find that

    $$\begin{aligned} \left|\frac{1}{2}(F(a)+F(b))-F\left( \frac{a+b}{2}\right) \right|\lesssim \left|a-b\right|. \end{aligned}$$

    Since F is even we have \(F(b)=F(-b)\) and hence

    $$\begin{aligned} \left|\frac{1}{2}(F(a)+F(b)) -F\left( \frac{a-b}{2}\right) \right|\lesssim \left|a+b\right|. \end{aligned}$$

    We now apply this inequality with \(a=\Delta _\alpha f\) and \(b=\bar{\Delta }_\alpha f\). Since, by definition, \(D_\alpha f=\Delta _\alpha f-\bar{\Delta }_\alpha f\), \(S_\alpha f=\Delta _\alpha f+\bar{\Delta }_\alpha f\), we conclude that

    $$\begin{aligned} \left|\frac{1}{2}(F(\Delta _\alpha f)+F(\bar{\Delta }_\alpha f)) -F\left( \frac{1}{2}D_\alpha f\right) \right|\lesssim \left|S_\alpha f\right|. \end{aligned}$$
    (4.13)

    We now use the fact that F is Lipschitz to infer from (4.1) that

    $$\begin{aligned} \left|F\left( \frac{1}{2}D_\alpha f\right) -F(f_x)\right|\lesssim \left|\frac{1}{\alpha } \int \nolimits _0^\alpha s_\eta f_x{{\,\mathrm{d}\,}} \eta \right|. \end{aligned}$$
    (4.14)

    In light of (4.12), by using the triangle inequality, it follows from (4.13) and (4.14) that

    $$\begin{aligned} \left|Q\right|\lesssim \left|S_\alpha f\right|+\left|\frac{1}{\alpha } \int \nolimits _0^\alpha s_\eta f_x{{\,\mathrm{d}\,}} \eta \right|, \end{aligned}$$

    which gives the result (4.10).

  2. (ii)

    Since

    $$\begin{aligned} \mathcal {E}(\alpha ,\cdot )=\frac{1}{2}F(\Delta _\alpha f)+\frac{1}{2}F(\bar{\Delta }_\alpha f), \end{aligned}$$

    and since \(F'\) is bounded, the chain rule implies that

    $$\begin{aligned} \left|\partial _\alpha \mathcal {E}(\alpha ,\cdot )\right|\lesssim \left|\partial _\alpha \Delta _\alpha f\right|+\left|\partial _\alpha \bar{\Delta }_\alpha f\right|. \end{aligned}$$

    By combining this estimate with the identities

    $$\begin{aligned} 2\Delta _\alpha f=S_\alpha f+D_\alpha f, \quad 2\bar{\Delta }_\alpha f=S_\alpha f-D_\alpha f, \end{aligned}$$

    we deduce that \(\left|\partial _\alpha \mathcal {E}(\alpha ,\cdot )\right|\lesssim \left|\partial _\alpha S_\alpha f\right|+\left|\partial _\alpha D_\alpha f\right|\). Then the second estimate (4.11) follows from the values for \(\partial _\alpha S_\alpha f\) and \(\partial _\alpha D_\alpha f\) given by Lemma 4.3. \(\quad \square \)

It follows directly from (4.8) and (4.9) that

$$\begin{aligned} \mathcal {T}_{e}(f)g&=\frac{1}{2\pi }\frac{f_x^2}{1+f_x^2} \int \frac{s_\alpha g}{\alpha ^2} {{\,\mathrm{d}\,}}\alpha \nonumber \\&\quad +\frac{1}{2\pi }\int \frac{s_\alpha g}{\alpha } \left( \frac{Q(\alpha ,\cdot )}{\alpha }-\partial _\alpha \mathcal {E}(\alpha ,\cdot )\right) {{\,\mathrm{d}\,}}\alpha . \end{aligned}$$
(4.15)

Observe that

$$\begin{aligned} \int \frac{s_{\alpha }g}{\alpha ^2} {{\,\mathrm{d}\,}}\alpha&=-\int s_{\alpha }g \partial _\alpha \left( \frac{1}{\alpha }\right) {{\,\mathrm{d}\,}}\alpha \\&= \int \frac{\partial _\alpha s_{\alpha }g}{\alpha }{{\,\mathrm{d}\,}}\alpha =\int \frac{g_x(x-\alpha )-g_x(x+\alpha )}{\alpha }{{\,\mathrm{d}\,}}\alpha \\&=-\,2\int \Delta _\alpha g_x{{\,\mathrm{d}\,}}\alpha = 2\pi \Lambda g, \end{aligned}$$

where we used (1.3). Thus, the first term in the right-hand side of (4.15) is the wanted elliptic component

$$\begin{aligned} \frac{f_x^2}{1+f_x^2}\Lambda g. \end{aligned}$$

To conclude the proof of the first statement in (4.6), it remains only to prove that the second term in the right-hand side of (4.15) is a remainder term. Putting for shortness

$$\begin{aligned} I=\left\Vert \int \frac{s_\alpha g}{\alpha } \left( \frac{Q(\alpha ,\cdot )}{\alpha }-\partial _\alpha \mathcal {E}(\alpha ,\cdot )\right) {{\,\mathrm{d}\,}}\alpha \right\Vert _{L^2}, \end{aligned}$$

we will prove that

$$\begin{aligned} I\lesssim \left\Vert f\right\Vert _{\dot{H}^{\frac{3}{2}+\varepsilon }}\left\Vert g\right\Vert _{\dot{B}^{1-\varepsilon }_{2,1}}. \end{aligned}$$
(4.16)

The \(L^\infty \)-norm of \(\frac{Q(\alpha ,\cdot )}{\alpha }-\partial _\alpha \mathcal {E}(\alpha ,\cdot )\) is controlled from (4.10) and (4.11). We have

$$\begin{aligned} I \lesssim I_1 + I_2 + I_3 + I_4 \end{aligned}$$

with

$$\begin{aligned} I_1&= \int \frac{\Vert s_{\alpha }g \Vert _{L^{2}}}{\vert \alpha \vert ^{1-\varepsilon }}\frac{\Vert \bar{\delta }_{\alpha }f_x \Vert _{L^{\infty }}}{\vert \alpha \vert ^{\varepsilon }} \frac{{{\,\mathrm{d}\,}}\alpha }{\left|\alpha \right|}, \nonumber \\ I_2&= \int \frac{\Vert s_{\alpha }g \Vert _{L^{2}}}{\vert \alpha \vert ^{1-\varepsilon }}\frac{\Vert \delta _{\alpha }f_x \Vert _{L^{\infty }}}{\vert \alpha \vert ^{\varepsilon }} \frac{{{\,\mathrm{d}\,}}\alpha }{\left|\alpha \right|}, \nonumber \\ I_3&= \int \frac{\Vert s_{\alpha }g \Vert _{L^{2}}}{\vert \alpha \vert ^{1-\varepsilon }} \frac{\Vert s_{\alpha }f \Vert _{L^{\infty }}}{\vert \alpha \vert ^{1+\varepsilon }} \frac{{{\,\mathrm{d}\,}}\alpha }{\left|\alpha \right|}, \nonumber \\ I_4&= \int \frac{\Vert s_{\alpha }g \Vert _{L^{2}}}{\left|\alpha \right|^{1-\varepsilon }} \frac{1}{\left|\alpha \right|^{1+\varepsilon }} \left|\int \nolimits _0^\alpha \Vert s_{\eta }f_x \Vert _{L^{\infty }} {{\,\mathrm{d}\,}} \eta \right|\frac{{{\,\mathrm{d}\,}}\alpha }{\left|\alpha \right|}, \end{aligned}$$
(4.17)

where, as above, we have distributed the powers of \(\left|\alpha \right|\) in a balanced way. Using the Cauchy–Schwarz inequality and the definition (2.1) of Besov semi-norms, it follows that

$$\begin{aligned} I_1+I_2\leqq \Vert g \Vert _{\dot{B}^{1-\varepsilon }_{2,2}} \Vert f_{x} \Vert _{\dot{B}^{\varepsilon }_{\infty ,2}}. \end{aligned}$$

and, similarly, it results from (2.2) that

$$\begin{aligned} I_3\leqq \Vert g \Vert _{\dot{B}^{1-\varepsilon }_{2,2}} \Vert f \Vert _{\dot{B}^{1+\varepsilon }_{\infty ,2}}. \end{aligned}$$

Consequently, the Sobolev embeddings

$$\begin{aligned} \dot{B}^{1-\varepsilon }_{2,1}\hookrightarrow \dot{B}^{1-\varepsilon }_{2,2},\qquad \dot{H}^{\frac{3}{2}+\varepsilon }(\mathbb {R})\hookrightarrow \dot{B}^{\varepsilon }_{\infty ,2}(\mathbb {R}), \end{aligned}$$

imply that \(I_1+I_2+I_3\lesssim \left\Vert g\right\Vert _{\dot{B}^{1-\varepsilon }_{2,1}}\left\Vert f\right\Vert _{\dot{H}^{\frac{3}{2}+\varepsilon }}\).

To estimate \(I_4\), the key point consists in using the Cauchy–Schwarz inequality to verify that

$$\begin{aligned} \frac{1}{\left|\alpha \right|^{1+\varepsilon }} \left|\int \nolimits _0^\alpha \Vert s_{\eta }f_x \Vert _{L^{\infty }} {{\,\mathrm{d}\,}} \eta \right|\lesssim \left( \int \nolimits _0^\infty \frac{\Vert s_{\mu }f_x \Vert ^{2}_{L^{\infty }}}{\mu ^{2\varepsilon }} \frac{{{\,\mathrm{d}\,}} \mu }{\mu } \right) ^{\frac{1}{2}} \end{aligned}$$

(notice that the variable \(\eta \) above could be negative, while \(\mu \) here is always positive). It follows from (2.1) that

$$\begin{aligned} \left( \int \nolimits _0^\infty \frac{\Vert s_{\mu }f_x \Vert ^{2}_{L^{\infty }}}{\mu ^{2\varepsilon }} \frac{{{\,\mathrm{d}\,}} \mu }{\mu } \right) ^{\frac{1}{2}} \lesssim \left\Vert f_x\right\Vert _{\dot{B}^{\varepsilon }_{\infty ,2}}\lesssim \left\Vert f\right\Vert _{\dot{H}^{\frac{3}{2}+\varepsilon }}, \end{aligned}$$

So we obtain, again using (2.1) with \((p,q,s)=(2,1,1-\varepsilon )\),

$$\begin{aligned} I_4 \lesssim \left\Vert f\right\Vert _{\dot{H}^{\frac{3}{2}+\varepsilon }} \int \frac{\Vert s_{\alpha }g \Vert _{L^{2}}}{\left|\alpha \right|^{1-\varepsilon }}\frac{{{\,\mathrm{d}\,}}\alpha }{\left|\alpha \right|} \lesssim \left\Vert f\right\Vert _{\dot{H}^{\frac{3}{2}+\varepsilon }}\left\Vert g\right\Vert _{\dot{B}^{1-\varepsilon }_{2,1}}. \end{aligned}$$
(4.18)

This completes the proof of (4.16) and hence the proof of the desired result (4.6).

(ii) It remains to study \(\mathcal {T}_{o}(f)g\). Recall that

$$\begin{aligned} \mathcal {T}_{o}(f)g=-\frac{1}{\pi } \int \big ( \Delta _{\alpha } g_{x}\big ) \mathcal {O}(\alpha ,\cdot ){{\,\mathrm{d}\,}}\alpha , \end{aligned}$$

where \(\mathcal {O}(\alpha ,\cdot )\) is given by (4.5). By splitting the factor \(\Delta _\alpha g_x\) into two parts,

$$\begin{aligned} \Delta _\alpha g_x(x)=\frac{g_x(x)}{\alpha }-\frac{g_x(x-\alpha )}{\alpha }, \end{aligned}$$

and we obtain at once that

$$\begin{aligned} \mathcal {T}_{o}(f)g=V\partial _x g +B, \end{aligned}$$

where V is given by (4.7) and where the remainder B is given by

$$\begin{aligned} B(x)=\frac{1}{\pi } \int \frac{1}{\alpha }g_{x}(x-\alpha )\mathcal {O}(\alpha ,x) {{\,\mathrm{d}\,}}\alpha . \end{aligned}$$

The analysis of B is based on the observation that

$$\begin{aligned} g_{x}(x-\alpha )=\partial _{\alpha }(g(x)-g(x-\alpha ))=\partial _\alpha (\delta _\alpha g), \end{aligned}$$

which allows us to integrate by parts in \(\alpha \) to obtain

$$\begin{aligned} B=\frac{1}{\pi } \int \frac{\delta _\alpha g }{\alpha }\left( \frac{1}{\alpha }\mathcal {O}(\alpha ,\cdot )- \partial _\alpha \mathcal {O}(\alpha ,\cdot ) \right) {{\,\mathrm{d}\,}}\alpha . \end{aligned}$$

Consequently, by writing

$$\begin{aligned} \left\Vert B \right\Vert _{L^2} \lesssim \int \frac{\Vert \delta _\alpha g \Vert _{L^{2}}}{\left|\alpha \right|} \left\Vert \frac{1}{\alpha }\mathcal {O}(\alpha ,\cdot )- \partial _\alpha \mathcal {O}(\alpha ,\cdot ) \right\Vert _{L^\infty }\frac{{{\,\mathrm{d}\,}}\alpha }{\left|\alpha \right|}, \end{aligned}$$

we are back to the situation already treated in the first step. The estimate for \(\partial _\alpha \mathcal {O}\) is proved by repeating the arguments used to prove the estimate (4.11) for \(\partial _\alpha \mathcal {E}\). To bound \(\alpha ^{-1}\mathcal {O}(\alpha ,x)\), remembering the expression of \(\mathcal {O}(\alpha ,x)\) given by (4.5), it is sufficient to notice that

$$\begin{aligned} \left|\frac{\mathcal {O}(\alpha ,\cdot )}{\alpha }\right|&=\frac{1}{2\left|\alpha \right|} \left|\frac{(\Delta _\alpha f)^2}{1+(\Delta _\alpha f)^2} -\frac{(\bar{\Delta }_\alpha f)^2}{1+(\bar{\Delta }_\alpha f)^2}\right|\nonumber \\&\leqq \frac{1}{2\left|\alpha \right|}\left|\frac{\Delta _\alpha f-\bar{\Delta }_\alpha f}{(1+(\Delta _\alpha f)^2)(1+(\bar{\Delta }_\alpha f)^2} \right|\left|\Delta _\alpha f+\bar{\Delta }_\alpha f\right|\nonumber \\&\leqq \frac{\left|S_\alpha f\right|}{\left|\alpha \right|}\cdot \end{aligned}$$
(4.19)

This gives that \(\left|\mathcal {O}(\alpha ,x)\right|\lesssim \left|s_\alpha f(x)\right|/\left|\alpha \right|^2\). Therefore, we obtain that the \(L^\infty \)-norm of \(\frac{\mathcal {O}(\alpha ,\cdot )}{\alpha }-\partial _\alpha \mathcal {O}(\alpha ,\cdot )\) is estimated by the right-hand side of (4.11). Then we may repeat the arguments used in the proof of the first step to estimate I. We call the attention to the fact that, previously, in (4.17), the expressions involved the more favorable symmetric differences \(s_\alpha g\) instead of \(\delta _\alpha g\). However, this is not important for our purpose since, to estimate \(I_1,\ldots ,I_4\), we used only the characterization of Besov norms valid for \(0<s<1\), which involves only the finite differences \(\delta _\alpha f\). This proves that \(\left\Vert B \right\Vert _{L^2}\) is controlled by the right-hand side of (4.16), which implies that \(B\sim 0\). \(\quad \square \)

Lemma 4.6

Let \(0<\nu<\varepsilon <1/2\). There exists a positive constant \(C>0\) such that

$$\begin{aligned} \left\Vert V\right\Vert _{C^{0,\nu }}=\left\Vert V\right\Vert _{L^\infty }+\sup _{y\in \mathbb {R}}\bigg (\frac{\left|V(x+y)-V(x)\right|}{\left|y\right|^\nu }\bigg ) \leqq C\left\Vert f\right\Vert _{\dot{H}^1\cap \dot{H}^{\frac{3}{2}+\varepsilon }}^2. \end{aligned}$$

Proof

As we have already seen, we have

$$\begin{aligned} V(x)=-\frac{1}{\pi }\int \nolimits _\mathbb {R}\frac{\mathcal {O}(\alpha ,x)}{\alpha } {{\,\mathrm{d}\,}}\alpha , \end{aligned}$$

where

$$\begin{aligned} \mathcal {O}(\alpha ,\cdot )=M_\alpha (f)S_\alpha f\quad \text {with}\quad M_\alpha (f)=\frac{1}{2}\frac{\Delta _\alpha f-\bar{\Delta }_\alpha f}{(1+(\bar{\Delta }_\alpha f)^2)(1+(\bar{\Delta }_\alpha f)^2}. \end{aligned}$$

Since \(\left|M_\alpha (f)\right|\leqq \left|\Delta _\alpha f\right|+\left|\bar{\Delta }_\alpha f\right|\), we obtain that

$$\begin{aligned} \left\Vert V\right\Vert _{L^\infty }&\leqq 2 \int \big (\left\Vert \Delta _\alpha f\right\Vert _{L^\infty }+\left\Vert \bar{\Delta }_\alpha f\right\Vert _{L^\infty }\big )\left\Vert S_\alpha f\right\Vert _{L^\infty }\frac{{{\,\mathrm{d}\,}}\alpha }{\left|\alpha \right|}\\&\leqq \int \frac{\left\Vert \delta _\alpha f\right\Vert _{L^\infty }+\left\Vert \bar{\delta }_\alpha f\right\Vert _{L^\infty }}{\left|\alpha \right|^{1-\varepsilon }} \frac{\left\Vert s_\alpha f\right\Vert _{L^\infty }}{\left|\alpha \right|^{1+\varepsilon }}\frac{{{\,\mathrm{d}\,}}\alpha }{\left|\alpha \right|}\\&\leqq \left\Vert f\right\Vert _{\dot{B}^{1-\varepsilon }_{\infty ,2}}\left\Vert f\right\Vert _{\dot{B}^{1+\varepsilon }_{\infty ,2}}\lesssim \left\Vert f\right\Vert _{\dot{H}^{1}\cap \dot{H}^{\frac{3}{2}+\varepsilon }}^2, \end{aligned}$$

where we used the Cauchy–Schwarz inequality, the definitions (2.1) and (2.2) of the Besov semi-norms, and the Sobolev embedding.

We now have to estimate the Hölder-modulus of continuity of V. Given \(y\in \mathbb {R}\) and a function \(u=u(x)\), we introduce the function \([u]_y\) defined by

$$\begin{aligned}{}[u]_y(x)=\frac{u(x+y)-u(x)}{\left|y\right|^{\nu }}\cdot \end{aligned}$$

We want to estimate the \(L^\infty \)-norm of \([V]_y\) uniformly in \(y\in \mathbb {R}\). Notice that

$$\begin{aligned}{}[\mathcal {O}(\alpha ,\cdot )]_y=M_\alpha (f)S_\alpha ([f]_y)+[M_\alpha (f)]_y \tau _y(S_\alpha f), \end{aligned}$$

where \(\tau _y u(x)=u(x+y)\). The contribution of the first term is estimated as above, so by setting \(\delta =\varepsilon -\nu >0\), we have

$$\begin{aligned} \int \left\Vert M_\alpha (f)S_\alpha ([f]_y)\right\Vert _{L^\infty }{{\,\mathrm{d}\,}}\alpha&\leqq \int \big (\left\Vert \Delta _\alpha f\right\Vert _{L^\infty }+\left\Vert \bar{\Delta }_\alpha f\right\Vert _{L^\infty }\big ) \left\Vert S_\alpha [f]_y\right\Vert _{L^\infty }\frac{{{\,\mathrm{d}\,}}\alpha }{\left|\alpha \right|}\\&\leqq \int \frac{\left\Vert \delta _\alpha f\right\Vert _{L^\infty }+\left\Vert \bar{\delta }_\alpha f\right\Vert _{L^\infty }}{\left|\alpha \right|^{1-\delta }} \frac{\left\Vert s_\alpha [f]_y\right\Vert _{L^\infty }}{\left|\alpha \right|^{1+\delta }}\frac{{{\,\mathrm{d}\,}}\alpha }{\left|\alpha \right|}\\&\leqq \left\Vert f\right\Vert _{\dot{B}^{1-\delta }_{\infty ,2}}\left\Vert [f]_y\right\Vert _{\dot{B}^{1+\delta }_{\infty ,2}}. \end{aligned}$$

Now, using the Plancherel theorem and the inequality \(\left|e^{iy\xi }-1\right|\leqq \left|y\xi \right|^\nu \), we have

$$\begin{aligned} \left\Vert [f]_y\right\Vert _{\dot{B}^{1+\delta }_{\infty ,2}}\lesssim \left\Vert [f]_y\right\Vert _{\dot{H}^{\frac{3}{2}+\delta }}\lesssim \left\Vert f\right\Vert _{\dot{H}^{\frac{3}{2}+\delta +\nu }} =\left\Vert f\right\Vert _{\dot{H}^{\frac{3}{2}+\varepsilon }}, \end{aligned}$$

since \(\delta +\nu =\varepsilon \). On the other hand, since

$$\begin{aligned} \left|[M_\alpha (f)]_y\right|\lesssim \left|\Delta _\alpha [f]_y\right|+\left|\bar{\Delta }_\alpha [f]_y\right|, \end{aligned}$$

by repeating the previous arguments, we get

$$\begin{aligned} \int \left\Vert [M_\alpha (f)]_y \tau _y(S_\alpha f)\right\Vert _{L^\infty }{{\,\mathrm{d}\,}}\alpha&\leqq \int \frac{\left\Vert \delta _\alpha [f]_y\right\Vert _{L^\infty }+\left\Vert \bar{\delta }_\alpha [f]_y\right\Vert _{L^\infty }}{\left|\alpha \right|^{1-\nu }} \frac{\left\Vert s_\alpha f\right\Vert _{L^\infty }}{\left|\alpha \right|^{1+\nu }}\frac{{{\,\mathrm{d}\,}}\alpha }{\left|\alpha \right|}\\&\leqq \left\Vert [f]_y\right\Vert _{\dot{B}^{1-\nu }_{\infty ,2}}\left\Vert f\right\Vert _{\dot{B}^{1+\nu }_{\infty ,2}} \lesssim \left\Vert f\right\Vert _{\dot{H}^{\frac{3}{2}}}\left\Vert f\right\Vert _{\dot{H}^{\frac{3}{2}+\nu }}. \end{aligned}$$

This concludes the proof of Lemma 4.6\(\quad \square \)

This completes the proof of Theorem 1.2.

5 Cauchy Problem

In this section we prove Theorem 1.4 about the Cauchy problem.

We prove the uniqueness by estimating the difference of two solutions. With regards to the existence, we construct solutions to the Muskat equation as limits of solutions to a sequence of approximate nonlinear systems, following here [1, 2, 32, 33]. We split the analysis in to three parts.

  1. 1.

    Firstly, we prove that the Cauchy problem for these approximate systems are well-posed locally in time by means of an ODE argument.

  2. 2.

    Secondly, we use Theorem 1.2 and an elementary \(L^2\)-estimate for the paralinearized equation to prove that the solutions of the later approximate systems are bounded in \(C^0([0,T];\dot{H}^1(\mathbb {R})\cap \dot{H}^s(\mathbb {R}))\) on a uniform time interval.

  3. 3.

    The third task consists in showing that these approximate solutions converge to a limit which is a solution of the Muskat equation. To do this, one cannot apply standard compactness results since the equation is non-local. Instead, we prove that the solutions form a Cauchy sequence in an appropriate space, by estimating the difference of two solutions.

5.1 Approximate systems

To define approximate systems, we use a version of Galerkin’s method based on Friedrichs mollifiers. We find convenient to use smoothing operators which are projections and consider, for \(n\in \mathbb {N}{\setminus }\{0\}\), the operators \(J_n\) defined by

$$\begin{aligned} \widehat{J_n u}(\xi )&=\hat{u}(\xi ) \quad&\text {for}&\left|\xi \right|\leqq n,\\ \widehat{J_n u}(\xi )&=0 \quad&\text {for}&\left|\xi \right|> n. \end{aligned}$$

Notice that \(J_n\) is a projection, \(J_n^2=J_n\). This will allow us to simplify some technical arguments.

Now we consider the following approximate Cauchy problems:

$$\begin{aligned} \left\{ \begin{aligned}&\partial _t f+\Lambda f=J_n\big (\mathcal {T}(f)f\big ),\\&f\arrowvert _{t=0}=J_n f_{0}. \end{aligned} \right. \end{aligned}$$
(5.1)

The following lemma states that this system has smooth local in time solutions:

Lemma 5.1

For all \(f_0\in \dot{H}^{1}(\mathbb {R})\), and any \(n\in \mathbb {N}{\setminus }\{0\}\), the initial value problem (5.1) has a unique maximal solution, for some time \(T_n>0\), of the form \(f_n=J_nf_0+u_n\) where \(u_n\in C^{1}([0,T_n[;H^{\infty }(\mathbb {R}))\) is such that \(u_n(0)=0\). Moreover, either

$$\begin{aligned} T_n=+\infty \qquad \text {or}\qquad \limsup _{t\rightarrow T_n} \left\Vert u_n(t)\right\Vert _{L^2}=+\infty . \end{aligned}$$
(5.2)

Proof

We begin by studying an auxiliary system. Consider the following Cauchy problem

$$\begin{aligned} \left\{ \begin{aligned}&\partial _t f+J_n\Lambda f=J_n\big (\mathcal {T}(J_n f)J_n f\big ),\\&f\arrowvert _{t=0}=J_n f_{0}. \end{aligned} \right. \end{aligned}$$
(5.3)

Set \(u=f-J_nf_0\). Then the Cauchy problem (5.3) has the form

$$\begin{aligned} \partial _t u= F_n(u),\quad u\arrowvert _{t=0}=0, \end{aligned}$$
(5.4)

where

$$\begin{aligned} F_n(u)=-\,\Lambda J_n u-\Lambda J_n f_0+J_n\big (\mathcal {T}(J_n(f_0+u))J_n(f_0+u)\big ) \end{aligned}$$

(we have used \(J_n^2=J_n\) to simplify the expression of F). The operator \(J_n\) is a smoothing operator: it is bounded from \(\dot{H}^1(\mathbb {R})\) into \(\dot{H}^\mu (\mathbb {R})\) for any \(\mu \geqq 1\), and from \(L^2(\mathbb {R})\) into \(H^\mu (\mathbb {R})\) for any \(\mu \geqq 0\). Consequently, if u belongs to \(L^{2}(\mathbb {R})\), then \(J_n(f_0+u)\) belongs to \(\dot{H}^\mu (\mathbb {R})\) for any \(\mu \geqq 1\). Thus, it follows from statement (i) in Proposition 2.3 and the assumption \(f_0\in \dot{H}^1(\mathbb {R})\) that \(F_n\) maps \(L^2(\mathbb {R})\) into itself. This shows that (5.4) is in fact an ODE with values in a Banach space for any \(n\in \mathbb {N}{\setminus }\{0\}\). The key point is that statement (ii) in Proposition 2.3 implies that the function \(F_n\) is locally Lipschitz from \(L^{2}(\mathbb {R})\) to itself. Consequently, the Cauchy–Lipschitz theorem gives the existence of a unique maximal solution \(u_n\) in \(C^{1}([0,T_n[;L^{2}(\mathbb {R}))\). Then the function \(f_n=J_nf_0+u_n\) is a solution to (5.3). Since \(J_n^2=J_n\), we check that the function \((I-J_n)f_n\) solves

$$\begin{aligned} \partial _t (I-J_n)f_n=0,\quad (I-J_n)f_n\arrowvert _{t=0}=0. \end{aligned}$$

This shows that \((I-J_n)f_n=0\), so \(J_nf_n=f_n\). Consequently, the fact that \(f_n\) solves (5.3) implies that \(f_n\) is also a solution to (5.1).

The alternative (5.2) is a consequence of the usual continuation principle for ordinary differential equations. Eventually, integrating (5.4) in time and using the fact that \(J_n\) is a smoothing operator, we obtain that \(u_n\) belongs to \(C^0([0,T_n[;H^{\infty }(\mathbb {R}))\). Using again (5.4), we conclude that \(\partial _t u_n\) belong to \(C^0([0,T_n[;H^{\infty }(\mathbb {R}))\). \(\quad \square \)

5.2 A priori estimate for the approximate systems

In this paragraph we prove two a priori estimates which will play a key role to prove uniform estimates for the solutions \((f_n)\) and also to estimate the differences between two such solutions. We begin with the following estimate in \(L^2(\mathbb {R})\cap \dot{H}^s(\mathbb {R})\):

Proposition 5.2

For all real number \(s\in (3/2,2)\), there exists a positive constant \(C>0\) and a non-decreasing function \(\mathcal {F}:\mathbb {R}\rightarrow \mathbb {R}\) such that, any \(n\in \mathbb {N}{\setminus }\{0\}\), for any \(T\in (0,T_n)\), the norm

$$\begin{aligned} M_n(T)=\sup _{t\in [0,T]}\left\Vert f_n(t)-f_0\right\Vert _{L^2\cap \dot{H}^s}^2 \end{aligned}$$

satisfies

$$\begin{aligned}&M_n(T)+\frac{C}{1+K^2}\int \nolimits _0^T\left\Vert f_n(t)\right\Vert _{\dot{H}^{s+\frac{1}{2}}}^2{{\,\mathrm{d}\,}} t\nonumber \\&\quad \leqq (2+T)^2\left\Vert f_0\right\Vert _{\dot{H}^1\cap \dot{H}^s}^2 +T\mathcal {F}\big (\sup _{t\in [0,T]}\left\Vert f_n(t)\right\Vert _{\dot{H}^1\cap \dot{H}^s}^2\big ), \end{aligned}$$
(5.5)

where

$$\begin{aligned} K\mathrel {:=}\sup _{(t,x)\in [0,T]\times \mathbb {R}}\left|\partial _x f_n(t,x)\right|. \end{aligned}$$
(5.6)

Proof

Set \(\mathcal {T}_n(f)=J_n\big (\mathcal {T}(f)f\big )\). We estimate the \(\left\Vert \cdot \right\Vert _{L^2}\)-norm and \(\left\Vert \cdot \right\Vert _{\dot{H}^s}\)-norm by different methods.

First step : low-frequency estimate. Since

$$\begin{aligned} \partial _t f_n+\Lambda f_n=J_n\mathcal {T}(f_n)f_n,\quad f_n\arrowvert _{t=0}=J_nf_0, \end{aligned}$$
(5.7)

we have

$$\begin{aligned} f_n(t)=\exp \left( -t\Lambda \right) J_nf_0+\int \nolimits _0^t \exp \left( -(t-t')\Lambda \right) \mathcal {T}_n(f_n)(t'){{\,\mathrm{d}\,}} t', \end{aligned}$$

so

$$\begin{aligned} f_n(t)-J_nf_0 =\left( \exp \left( -t\Lambda \right) -I\right) J_nf_0+\int \nolimits _0^t \exp \left( -(t-t')\Lambda \right) \mathcal {T}_n(f_n)(t'){{\,\mathrm{d}\,}} t', \end{aligned}$$

where I denotes the identity operator. Using the Fourier transform and Plancherel identity, one obtains immediately that

$$\begin{aligned} \left\Vert \left( \exp \left( -t\Lambda \right) -I\right) J_nf_0\right\Vert _{L^2}\leqq \left\Vert t\Lambda J_nf_0\right\Vert _{L^2} \leqq T\left\Vert f_0\right\Vert _{\dot{H}^1}. \end{aligned}$$

On the other hand,

$$\begin{aligned} \left\Vert \exp \left( -(t-t')\Lambda \right) \mathcal {T}_n(f_n)(t')\right\Vert _{L^2}\leqq \left\Vert \mathcal {T}_n(f_n)(t')\right\Vert _{L^2}. \end{aligned}$$

Consequently,

$$\begin{aligned} \left\Vert f_n(t)-J_nf_0\right\Vert _{L^2}\leqq T\left\Vert f_0\right\Vert _{\dot{H}^1}+T\sup _{t'\in [0,T]} \left\Vert \mathcal {T}_n(f_n)(t')\right\Vert _{L^2}. \end{aligned}$$

Now we want to replace the left hand side of the above inequality by \(\left\Vert f_n(t)-f_0\right\Vert _{L^2}\). To do so, notice that, since the spectrum of \(J_nf_0-f_0\) is contained in \(\{\left|\xi \right|\geqq 1\}\), we have

$$\begin{aligned} \left\Vert J_nf_0-f_0\right\Vert _{L^2}\leqq \left\Vert f_0\right\Vert _{\dot{H}^1}. \end{aligned}$$

By combining the above estimates, we deduce that

$$\begin{aligned} \left\Vert f_n(t)-f_0\right\Vert _{L^2}\leqq (1+T)\left\Vert f_0\right\Vert _{\dot{H}^1}+T\sup _{t'\in [0,T]} \left\Vert \mathcal {T}_n(f_n)(t')\right\Vert _{L^2}. \end{aligned}$$

Now, we estimate the \(L^2\)-norm of the nonlinearity \(\mathcal {T}_n(f_n)\) by means of the first statement in Theorem 1.2. We conclude that, for \(T<1\),

$$\begin{aligned} \left\Vert f_n(t)-f_0\right\Vert _{L^2}^2\leqq 2(1+T)^2\left\Vert f_0\right\Vert _{\dot{H}^1}^2+C T^2\sup _{[0,T]}\left\Vert f_n\right\Vert _{\dot{H}^1}^2 \sup _{[0,T]}\left\Vert f_n\right\Vert _{\dot{H}^{\frac{3}{2}}}^2. \end{aligned}$$

This is in turn estimated by the right side of (5.5). This concludes the first step.

Second step : High frequency estimate. Denote by \((\cdot ,\cdot )\) the scalar product in \(L^2(\mathbb {R})\). To estimate the \(\dot{H}^s\)-norm of \(f_n\), we make act \(\Lambda ^s\) on the equation, and then take its scalar product with \(\Lambda ^s f_n\). We get

$$\begin{aligned} (\partial _t \Lambda ^s f_n,\Lambda ^{s}f_n)+(\Lambda ^{s+1} f_n,\Lambda ^{s} f_n) = \big (\Lambda ^{s}\mathcal {T}_n(f_n),\Lambda ^{s}f_n\big ). \end{aligned}$$

Since the Muskat equation is parabolic of order one, we will be able to gain one half-derivative. We exploit this parabolic regularity by writing that

$$\begin{aligned} (\Lambda ^{s+1} f_n,\Lambda ^{s} f_n)= \bigl \Vert f_n\bigr \Vert _{\dot{H}^{s+\frac{1}{2}}}^2, \end{aligned}$$

and

$$\begin{aligned} \big ( \Lambda ^{s}\mathcal {T}_n(f_n),\Lambda ^{s}f_n\big )&=\big (\Lambda ^{s}J_n\big (\mathcal {T}(f_n)f_n\big ),\Lambda ^{s}f_n\big )\\&=\big (\Lambda ^{s}\big (\mathcal {T}(f_n)f_n\big ),J_n\Lambda ^{s}f_n\big )\\&=\big (\Lambda ^{s}\big (\mathcal {T}(f_n)f_n\big ),\Lambda ^{s}f_n\big ) \quad \text {since }J_nf_n=f_n,\\&=\big (\Lambda ^{s-\frac{1}{2}} \mathcal {T}(f_n)f_n, \Lambda ^{s+\frac{1}{2}}f_n\big ). \end{aligned}$$

Consequently, we find that

$$\begin{aligned} \frac{1}{2}\frac{{{\,\mathrm{d}\,}}}{{{\,\mathrm{d}\,}} t}\Vert f_n \Vert ^{2}_{\dot{H}^{s}} +\bigl \Vert f_n\bigr \Vert _{\dot{H}^{s+\frac{1}{2}}}^2 = \big (\Lambda ^{s-\frac{1}{2}}\mathcal {T}(f_n)f_n, \Lambda ^{s+\frac{1}{2}}f_n\big ). \end{aligned}$$

We next use a variant of the paralinearization formula given by Corollary 1.3. Set

$$\begin{aligned} \varepsilon =s-\frac{3}{2}. \end{aligned}$$

We claim that, for any function g,

$$\begin{aligned} \Lambda ^{1+\varepsilon }(\mathcal {T}(g)g)=V(g)\partial _x \Lambda ^{1+\varepsilon } g+\frac{g_x^2}{1+g_x^2}\Lambda ^{2+\varepsilon }g+\Lambda ^{1+\varepsilon }R_\varepsilon (g), \end{aligned}$$

where V(g) and \(R_\varepsilon (g)\) are two functions satisfying, for any fixed \(\nu <\varepsilon \),

$$\begin{aligned} \left\Vert R_\varepsilon (g)\right\Vert _{\dot{H}^{1+\varepsilon }}&\leqq \mathcal {F}(\left\Vert g\right\Vert _{\dot{H}^1\cap \dot{H}^{s}})\left\Vert g\right\Vert _{\dot{H}^{s+\frac{1}{2}-\frac{\varepsilon }{2}}}, \end{aligned}$$
(5.8)
$$\begin{aligned} \left\Vert V(g)\right\Vert _{C^{0,\nu }}&\leqq \mathcal {F}(\left\Vert g\right\Vert _{\dot{H}^1\cap \dot{H}^{s}})\left\Vert g\right\Vert _{\dot{H}^{s+\frac{1}{2}-\frac{\varepsilon }{2}}}. \end{aligned}$$
(5.9)

where \(\mathcal {F}\) depends only on \(\varepsilon \) (that is s) and \(\nu \) (which will be specified later). The proof of this claim is similar to the one of (1.10).

With notations as above, set

$$\begin{aligned} V_n=V(f_n),\quad R_n=\Lambda ^{1+\varepsilon }R_{\varepsilon }(f_n), \quad \gamma _n=\frac{f_{nx}^2}{1+f_{nx}^2} \quad \text {where }f_{nx}=\partial _xf_n. \end{aligned}$$

Then,

$$\begin{aligned} \frac{1}{2} \frac{{{\,\mathrm{d}\,}}}{{{\,\mathrm{d}\,}} t}\Vert f_n \Vert ^{2}_{\dot{H}^{s}}+\left\Vert f_n\right\Vert _{\dot{H}^{s+\frac{1}{2}}}^2&= \big ( \gamma _n \Lambda ^{s+\frac{1}{2}}f_n,\Lambda ^{s+\frac{1}{2}}f_n , \big )\nonumber \\&\quad +\Big (\big (V_n\partial _x \Lambda ^{s-\frac{1}{2}}f_n+R_n\big ), \Lambda ^{s+\frac{1}{2}}f_n \Big ). \end{aligned}$$
(5.10)

Now the key point is that

$$\begin{aligned} \left\Vert f_n\right\Vert _{\dot{H}^{s+\frac{1}{2}}}^2 - \big (\gamma _n\Lambda ^{s+\frac{1}{2}}f_n, \Lambda ^{s+\frac{1}{2}}f_n \big ) = \int \frac{\big ( \Lambda ^{s+\frac{1}{2}}f_n \big )^2}{1+f_{nx}^2} {{\,\mathrm{d}\,}} x. \end{aligned}$$

On the other hand, the Cauchy–Schwarz inequality and the estimate (5.8) imply that

$$\begin{aligned} \big \vert \big ( R_n,\Lambda ^{s+\frac{1}{2}}f_n\big )\big \vert&\leqq \left\Vert R_n\right\Vert _{L^2}\bigl \Vert \Lambda ^{s+\frac{1}{2}}f_n\bigr \Vert _{L^2}\\&\leqq \mathcal {F}(\left\Vert f_n\right\Vert _{\dot{H}^1\cap \dot{H}^{s}})\left\Vert f_n\right\Vert _{\dot{H}^{s+\frac{1}{2}-\frac{\varepsilon }{2}}}\left\Vert f_n\right\Vert _{\dot{H}^{s+\frac{1}{2}}}. \end{aligned}$$

It remains to estimate the contribution of \(V_n\) to the second term in the right-hand side of (5.10). Here we use the commutator estimate given by Lemma 2.6. To do this, one uses the identity \(\mathcal {H}\Lambda =-\,\partial _x\) where \(\mathcal {H}\) is the Hilbert transform, to write

$$\begin{aligned} V_n\partial _x \Lambda ^{s-\frac{1}{2}}f_n=-\,V_n \Lambda ^{s+\frac{1}{2}}\mathcal {H}f_n. \end{aligned}$$

Since \(\mathcal {H}\) is skew-symmetric, we deduce that

$$\begin{aligned} \Big ( V_n\partial _x \Lambda ^{s-\frac{1}{2}}f_n,\Lambda ^{s+\frac{1}{2}}f_n\Big ) =\frac{1}{2}\Big (\big [ \mathcal {H},V_n\big ]\Lambda ^{s+\frac{1}{2}}f_n,\Lambda ^{s+\frac{1}{2}}f_n \Big ). \end{aligned}$$

Now we exploit the regularity result for \(V_n\) given by (5.9). Fix \(\nu =2\varepsilon /3\) and \(\theta =\varepsilon /2\). By applying the commutator estimate in Lemma 2.6, we obtain

$$\begin{aligned} \bigl \Vert \big [\mathcal {H},V_n\big ]\Lambda ^{s+\frac{1}{2}}f_n\bigr \Vert _{L^2}&\lesssim \left\Vert V_n\right\Vert _{C^{0,\nu }}\bigl \Vert \Lambda ^{s+\frac{1}{2}}f_n\bigr \Vert _{H^{-\theta }}\\&\lesssim \left\Vert V_n\right\Vert _{C^{0,\nu }}\left\Vert f_n\right\Vert _{\dot{H}^{s+\frac{1}{2}-\frac{\varepsilon }{2}}}\\&\leqq \mathcal {F}(\left\Vert f_n\right\Vert _{\dot{H}^1\cap \dot{H}^{s}})\left\Vert f_n\right\Vert _{\dot{H}^{s+\frac{1}{2}-\frac{\varepsilon }{2}}}. \end{aligned}$$

Thus, by combining the above estimates,

$$\begin{aligned} \frac{1}{2} \frac{{{\,\mathrm{d}\,}}}{{{\,\mathrm{d}\,}} t}\Vert f_n \Vert ^{2}_{\dot{H}^{s}} +\int \frac{\big ( \Lambda ^{s+\frac{1}{2}}f_n \big )^2}{1+f_{nx}^2} {{\,\mathrm{d}\,}} x\leqq \mathcal {F}(\left\Vert f_n\right\Vert _{\dot{H}^1\cap \dot{H}^{s}})\left\Vert f_n\right\Vert _{\dot{H}^{s+\frac{1}{2}-\frac{\varepsilon }{2}}}\left\Vert f_n\right\Vert _{\dot{H}^{s+\frac{1}{2}}}. \end{aligned}$$

The end of the proof will consist in exploiting the parabolic regularity and a variant of Gronwall’s lemma to absorb the right-hand side. Set \(K(t) =\sup _{x\in \mathbb {R}}\left|\partial _x f_n(t,x)\right|\). Then

$$\begin{aligned} \frac{1}{1+f_{nx}^2}\geqq \frac{1}{1+K^2}, \end{aligned}$$

so

$$\begin{aligned} \frac{1}{2} \frac{{{\,\mathrm{d}\,}}}{{{\,\mathrm{d}\,}} t}\Vert f_n \Vert ^{2}_{\dot{H}^{s}} +\frac{C}{1+K^2}\left\Vert f_n\right\Vert _{\dot{H}^{s+\frac{1}{2}}}^2 \leqq \mathcal {F}(\left\Vert f_n\right\Vert _{\dot{H}^1\cap \dot{H}^{s}})\left\Vert f_n\right\Vert _{\dot{H}^{s+\frac{1}{2}-\frac{\varepsilon }{2}}}\left\Vert f_n\right\Vert _{\dot{H}^{s+\frac{1}{2}}}. \end{aligned}$$
(5.11)

Then we observe that

$$\begin{aligned}&\mathcal {F}(\left\Vert f_n\right\Vert _{\dot{H}^1\cap \dot{H}^{s}})\left\Vert f_n\right\Vert _{\dot{H}^{s+\frac{1}{2}-\frac{\varepsilon }{2}}}\left\Vert f_n\right\Vert _{\dot{H}^{s+\frac{1}{2}}}\\&\quad \leqq \frac{C}{2(1+K^2)}\left\Vert f_n\right\Vert _{\dot{H}^{s+\frac{1}{2}}}^2+\frac{1+K^2}{2C} \mathcal {F}(\left\Vert f_n\right\Vert _{\dot{H}^1\cap \dot{H}^{s}})^2\left\Vert f_n\right\Vert _{\dot{H}^{s+\frac{1}{2}-\frac{\varepsilon }{2}}}^2. \end{aligned}$$

Since \(K\lesssim \left\Vert f_n\right\Vert _{\dot{H}^1\cap \dot{H}^{s}}\) by Sobolev embedding, up to modifying the value of the function \(\mathcal {F}\), by inserting the above inequality in (5.11), we get

$$\begin{aligned} \frac{1}{2} \frac{{{\,\mathrm{d}\,}}}{{{\,\mathrm{d}\,}} t}\Vert f_n \Vert ^{2}_{\dot{H}^{s}} +\frac{C}{2(1+K^2)}\left\Vert f_n\right\Vert _{\dot{H}^{s+\frac{1}{2}}}^2 \leqq \mathcal {F}(\left\Vert f_n\right\Vert _{\dot{H}^1\cap \dot{H}^{s}})\left\Vert f_n\right\Vert _{\dot{H}^{s+\frac{1}{2}-\frac{\varepsilon }{2}}}^2. \end{aligned}$$
(5.12)

To conclude, it will suffice to replace in the right side the norm \(\left\Vert f_n\right\Vert _{\dot{H}^{s+\frac{1}{2}-\frac{\varepsilon }{2}}}^2\) by \(\left\Vert f_n\right\Vert _{\dot{H}^{s}}^2\). To do this, we begin by using the interpolation inequality

$$\begin{aligned} \left\Vert f_n\right\Vert _{\dot{H}^{s+\frac{1}{2}-\frac{\varepsilon }{2}}}^2\leqq \left\Vert f_n\right\Vert _{\dot{H}^{s}}^{2\theta } \left\Vert f_n\right\Vert _{\dot{H}^{s+\frac{1}{2}}}^{2-2\theta } \end{aligned}$$
(5.13)

for some \(\theta \in (0,1)\). Next, because of Young’s inequality,

$$\begin{aligned} xy\leqq \frac{1}{p}x^p+\frac{1}{p'}y^{p'}\quad \text {with}\quad \frac{1}{p}+\frac{1}{p'}=1, \end{aligned}$$
(5.14)

and applied with \(p=2/(2-2\theta )\), we infer that

$$\begin{aligned} \frac{1}{2} \frac{{{\,\mathrm{d}\,}}}{{{\,\mathrm{d}\,}} t}\Vert f_n \Vert ^{2}_{\dot{H}^{s}} +\frac{C}{4(1+K^2)}\left\Vert f_n\right\Vert _{\dot{H}^{s+\frac{1}{2}}}^2 \leqq \mathcal {F}(\left\Vert f_n\right\Vert _{\dot{H}^1\cap \dot{H}^{s}})\left\Vert f_n\right\Vert _{\dot{H}^{s}}^2, \end{aligned}$$
(5.15)

where as above we modified the value of the function \(\mathcal {F}\). From this, it is now an easy matter to obtain the conclusion of the proposition. Firstly, integration of the above estimate gives

$$\begin{aligned}&\frac{1}{2}\Vert f_n(t) \Vert ^{2}_{\dot{H}^{s}} +\frac{C}{4(1+K^2)}\int \nolimits _0^t \left\Vert f_n(t')\right\Vert _{\dot{H}^{s+\frac{1}{2}}}^2{{\,\mathrm{d}\,}} t'\\&\quad \leqq \frac{1}{2}\Vert f_n(0) \Vert ^{2}_{\dot{H}^{s}}+t\sup _{t'\in [0,t]} \mathcal {F}(\left\Vert f_n(t')\right\Vert _{\dot{H}^1\cap \dot{H}^{s}})\left\Vert f_n(t')\right\Vert _{\dot{H}^{s}}^2. \end{aligned}$$

Modifying \(\mathcal {F}(\cdot )\) and C, we deduce that

$$\begin{aligned}&\Vert f_n(t) \Vert ^{2}_{\dot{H}^{s}} +\frac{C}{1+K^2}\int \nolimits _0^t \left\Vert f_n(t')\right\Vert _{\dot{H}^{s+\frac{1}{2}}}^2{{\,\mathrm{d}\,}} t' \\&\quad \leqq \Vert f_n(0) \Vert ^{2}_{\dot{H}^{s}} +T\mathcal {F}\Big (\sup _{t\in [0,T]}\left\Vert f_n(t)\right\Vert _{\dot{H}^1\cap \dot{H}^s}^2\Big ), \end{aligned}$$

for any \(t\in [0,T]\). By taking the supremum over \(t\in [0,T]\), we deduce an estimate for \(\sup _{t\in [0,T]}\left\Vert f_n(t)\right\Vert _{\dot{H}^s}^2\). Now, the desired estimate for \(\sup _{t\in [0,T]}\left\Vert f_n(t)-f_0\right\Vert _{\dot{H}^s}^2\) follows from the triangle inequality and the fact that \(\Vert f_n(0) \Vert _{\dot{H}^{s}}\leqq \Vert f_0 \Vert _{\dot{H}^{s}}\). \(\quad \square \)

We will also need another energy estimate to compare two different solutions \(f_1\) and \(f_2\). The main difficulty here will be to find the optimal space in which one can perform an energy estimate. The most simpler way to do so would be to estimate their difference \(f_1-f_2\) in the biggest possible space and to use an interpolation inequality to control the latter in a space of smoother function. This suggests that we should to estimate \(f_1-f_2\) in \(C^0([0,T];L^2(\mathbb {R}))\). On the other hand, by thinking of the fluid problem, we might think that it is compulsory to control the difference between the two functions parametrizing the two free surfaces in a space of smooth functions. We will see later that, somewhat unexpectedly, that it is enough to estimate \(f_1-f_2\) in \(C^0([0,T];\dot{H}^{1/2}(\mathbb {R}))\). In this direction, we will use the following proposition:

Proposition 5.3

  1. (i)

    For all s in (3/2, 2), there exists a non-decreasing function \(\mathcal {F}:\mathbb {R}_+\rightarrow \mathbb {R}_+\) such that, for any \(n\in \mathbb {N}{\setminus }\{0\}\), any \(T>0\), and any functions

    $$\begin{aligned} f&\in C^0([0,T];\dot{H}^1(\mathbb {R})\cap \dot{H}^s(\mathbb {R})), \\ g&\in C^1([0,T];H^\frac{1}{2}(\mathbb {R}))\quad \text {with }J_ng=g,\\ F&\in C^0([0,T];L^2(\mathbb {R})), \end{aligned}$$

    satisfying the equation

    $$\begin{aligned} \partial _tg-J_n\big (V(f)\partial _x g)+J_n\Big (\frac{1}{1+f_x^2}\Lambda g\Big )=F, \end{aligned}$$
    (5.16)

    where V(f) is as above, we have the estimate

    $$\begin{aligned} \frac{1}{2} \frac{{{\,\mathrm{d}\,}}}{{{\,\mathrm{d}\,}} t}\Vert g \Vert ^{2}_{\dot{H}^\frac{1}{2}} +\int \frac{(\Lambda g)^2}{1+f_x^2} {{\,\mathrm{d}\,}} x\leqq \mathcal {F}(\left\Vert f\right\Vert _{\dot{H}^1\cap \dot{H}^{s}})\left\Vert g\right\Vert _{\dot{H}^{1-\frac{\varepsilon }{2}}}\left\Vert g\right\Vert _{\dot{H}^1} +( F,\Lambda g)_{L^2}, \end{aligned}$$
    (5.17)

    where \(\varepsilon =s-3/2\) and \(C=\mathcal {F}\big (\left\Vert f\right\Vert _{L^\infty ([0,T];\dot{H}^1\cap \dot{H}^s)}\big )\).

  2. (ii)

    Moreover, the same result is true when one replaces \(J_n\) by the identity.

Proof

To prove (5.17) we take the \(L^2\)-scalar product of the Equation (5.16) with \(\Lambda g\). Since

$$\begin{aligned} \frac{1}{2} \frac{{{\,\mathrm{d}\,}}}{{{\,\mathrm{d}\,}} t}\Vert g \Vert ^{2}_{\dot{H}^\frac{1}{2}}=(\partial _t g,\Lambda g),\qquad \Big (J_n\Big (\frac{1}{1+f_x^2}\Lambda g\Big ),\Lambda g\Big )=\int \frac{( \Lambda g)^2}{1+f_x^2} {{\,\mathrm{d}\,}} x\end{aligned}$$

(where we used \(J_ng=g\)), we only have to estimate \(\big (J_n\big (V(f)\partial _x g),\Lambda g\big )\). As above, writing \(\partial _x=-\,\mathcal {H}\Lambda \), where \(\mathcal {H}\) is the Hilbert transform satisfying \(\mathcal {H}^*=-\,\mathcal {H}\), we obtain

$$\begin{aligned} \left|\big (J_n\big (V(f)\partial _x g),\Lambda g\big )\right|=\frac{1}{2}\left|\big ( \big [ \mathcal {H},V(f)\big ]\Lambda g,\Lambda g\big )\right|. \end{aligned}$$

Set \(\varepsilon =s-3/2\), \(\nu =2\varepsilon /3\) and \(\theta =\varepsilon /2\). We use Lemma 2.6 to obtain

$$\begin{aligned} \left|\big (J_n\big (V(f)\partial _x g),\Lambda g\big )\right|&\leqq \frac{1}{2}\left\Vert \big [\mathcal {H},V(f)\big ]\Lambda g\right\Vert _{L^2} \left\Vert \Lambda g\right\Vert _{L^2}\\&\leqq \left\Vert V(f)\right\Vert _{C^{0,\nu }}\left\Vert \Lambda g\right\Vert _{H^{-\theta }} \left\Vert \Lambda g\right\Vert _{L^2}\\&\leqq \mathcal {F}(\left\Vert f\right\Vert _{\dot{H}^1\cap \dot{H}^{s}}) \left\Vert \Lambda g\right\Vert _{H^{-\frac{\varepsilon }{2}}} \left\Vert \Lambda g\right\Vert _{L^2}\\&\leqq \mathcal {F}(\left\Vert f\right\Vert _{\dot{H}^1\cap \dot{H}^{s}})\left\Vert g\right\Vert _{\dot{H}^{1-\frac{\varepsilon }{2}}}\left\Vert g\right\Vert _{\dot{H}^1}. \end{aligned}$$

This completes the proof of (i) and the same arguments can be used to prove (ii). \(\quad \square \)

5.3 End of the proof

In this paragraph we complete the analysis of the Cauchy problem. We begin by proving the uniqueness part in Theorem 1.2.

Lemma 5.4

Assume that f and \(f'\) are two solutions of the Muskat equation with the same initial data and satisfying the assumptions of Theorem 1.4. Then \(f=f'\).

Proof

Set

$$\begin{aligned} g=f-f',\quad M=\left\Vert f\right\Vert _{L^\infty ([0,T];\dot{H}^1\cap \dot{H}^s)} +\left\Vert f'\right\Vert _{L^\infty ([0,T];\dot{H}^1\cap \dot{H}^s)}. \end{aligned}$$

We denote by C(M) various constants depending only on M.

We want to prove that \(g=0\). To do so, we use the energy estimate in \(\dot{H}^{1/2}(\mathbb {R})\). The key point is to write that g is a smooth function, in \(C^1([0,T];H^\frac{1}{2}(\mathbb {R}))\), satisfying

$$\begin{aligned} \partial _t g+\Lambda g =\mathcal {T}(f)g+F_1 \quad \text {with}\quad F_1=\mathcal {T}(f)f'-\mathcal {T}(f')f'. \end{aligned}$$

This term is estimated by means of point (ii) in Proposition 2.3 with \(\delta =\varepsilon =s-3/2\),

$$\begin{aligned} \left\Vert F_1\right\Vert _{L^2}=\left\Vert (\mathcal {T}(f)-\mathcal {T}(f'))f'\right\Vert _{L^2} \leqq C \left\Vert f-f'\right\Vert _{\dot{H}^{1-\varepsilon }}\left\Vert f'\right\Vert _{\dot{H}^{\frac{3}{2}+\varepsilon }}=C(M)\left\Vert g\right\Vert _{\dot{H}^{1-\varepsilon }}. \end{aligned}$$
(5.18)

Recall from (1.6) that

$$\begin{aligned} \mathcal {T}(f)g=\frac{f_x^2}{1+f_x^2}\Lambda g+V(f)\partial _x g+R(f,g), \end{aligned}$$

where R(fg) satisfies (setting \(\varepsilon =s-3/2\)) that

$$\begin{aligned} \left\Vert R(f,g)\right\Vert _{L^2}\leqq C\left\Vert f\right\Vert _{\dot{H}^{\frac{3}{2}+\varepsilon }}\left\Vert g\right\Vert _{\dot{B}^{1-\varepsilon }_{2,1}}\leqq C(M)\left\Vert g\right\Vert _{\dot{B}^{1-\varepsilon }_{2,1}}. \end{aligned}$$
(5.19)

Therefore, g satisfies

$$\begin{aligned} \partial _t g -V\partial _x g+\frac{1}{1+f_x^2}\Lambda g=F, \end{aligned}$$

where \(F=F_1+R(f)g\). In view of the estimates (5.18), (5.19) and the embedding \(\dot{H}^{1-3\varepsilon /2}(\mathbb {R})\cap \dot{H}^{1-\varepsilon /2}(\mathbb {R})\hookrightarrow \dot{B}^{1-\varepsilon }_{2,1}\) (see Lemma 2.2), we have

$$\begin{aligned} \left|(F,\Lambda g)\right|\leqq \left\Vert F\right\Vert _{L^2}\left\Vert g\right\Vert _{\dot{H}^1}\leqq C(M) \left\Vert g\right\Vert _{\dot{H}^{1-\frac{\varepsilon }{2}}\cap \dot{H}^{1-\frac{3\varepsilon }{2}}}\left\Vert g\right\Vert _{\dot{H}^1}. \end{aligned}$$

Hence, it follows from Proposition 5.3 [see point (ii)] that

$$\begin{aligned} \frac{1}{2} \frac{{{\,\mathrm{d}\,}}}{{{\,\mathrm{d}\,}} t}\Vert g \Vert ^{2}_{\dot{H}^\frac{1}{2}} +\int \frac{( \Lambda g )^2}{1+f_x^2} {{\,\mathrm{d}\,}} x\leqq C(M) \left\Vert g\right\Vert _{\dot{H}^{1-\frac{\varepsilon }{2}}\cap \dot{H}^{1-\frac{3\varepsilon }{2}}}\left\Vert g\right\Vert _{\dot{H}^1}. \end{aligned}$$

Next, we use interpolation inequalities as in the proof of Proposition 5.2. More precisely, by using arguments parallel to those used to deduce (5.15) from (5.13)-(5.14), we get

$$\begin{aligned} \frac{1}{2} \frac{{{\,\mathrm{d}\,}}}{{{\,\mathrm{d}\,}} t}\Vert g \Vert ^{2}_{\dot{H}^\frac{1}{2}} +\frac{1}{4(1+\left\Vert f_x\right\Vert _{L^\infty _{t,x}})}\int ( \Lambda g )^2 {{\,\mathrm{d}\,}} x\leqq C(M)\left\Vert g\right\Vert _{\dot{H}^\frac{1}{2}}^2. \end{aligned}$$

This obviously implies that

$$\begin{aligned} \frac{1}{2} \frac{{{\,\mathrm{d}\,}}}{{{\,\mathrm{d}\,}} t}\Vert g \Vert ^{2}_{\dot{H}^\frac{1}{2}} \leqq C(M)\left\Vert g\right\Vert _{\dot{H}^\frac{1}{2}}^2. \end{aligned}$$

Since \(g(0)=0\), the Gronwall’s inequality implies that \(\left\Vert g\right\Vert _{\dot{H}^\frac{1}{2}}=0\) so \(g=0\), which completes the proof. \(\quad \square \)

Having proved the uniqueness of solutions, we now study their existence. The key step will be here to apply the a priori estimates proved in Proposition 5.2. This will give us uniform bounds for the solutions \(f_n\) defined in Section 5.1.

Lemma 5.5

There exists \(T_0>0\) such that \(T_n\geqq T_0\) for all \(n\in \mathbb {N}{\setminus }\{0\}\) and such that \((f_n-f_0)_{n\in \mathbb {N}}\) is bounded in \(C^{0}([0,T_0]; H^{s}(\mathbb {R}))\).

Proof

We use the notations of Section 5.1 and Proposition 5.2. Given \(T<T_n\), we define

$$\begin{aligned} M_n(T)=\sup _{t\in [0,T]}\left\Vert f_n(t)-f_0\right\Vert _{L^2\cap \dot{H}^s}^2,\qquad N_n(T)=M_n(T)+\left\Vert f_0\right\Vert _{\dot{H}^1\cap \dot{H}^s}^2. \end{aligned}$$

Denote by \(\mathcal {F}\) the function whose existence is the conclusion of Proposition 5.2 and set

$$\begin{aligned} A=10\left\Vert f_0\right\Vert _{\dot{H}^1\cap \dot{H}^s}^2. \end{aligned}$$

We next pick \(0<T_{0}\leqq 1\) small enough such that

$$\begin{aligned} 3(1+T_0)^2\left\Vert f_0\right\Vert _{\dot{H}^1\cap \dot{H}^s}^2+T_0\mathcal {F}(A)< A, \end{aligned}$$

We claim the uniform bound

$$\begin{aligned} \forall n\in \mathbb {N}{\setminus }\{0\},~\forall T\in I_n\mathrel {:=}[0,\min \{T_{0},T_n\}),\quad N_n(T) <A. \end{aligned}$$

Let us prove this claim by contradiction. Assume that for some n there exists \(\tau _n\in I_n\) such that \(N_n(\tau _n)\geqq A\) and consider the smallest of such times (then \(\tau _n>0\) since \(T\mapsto N_n(T)\) is continuous and \(N_n(0)<A\) by construction). Then, by definition, for all \(0<T\leqq \tau _n\), one has \(N_n(T)\leqq A\) and \(N_n(\tau _n)=A\). Since \(\left\Vert \partial _xf_n(t)\right\Vert _{L^\infty (\mathbb {R})}\lesssim \left\Vert f_n(t)\right\Vert _{\dot{H}^1\cap \dot{H}^s}\) (see (3.4)), we have a uniform control of the \(L^\infty _{x}\)-norm of \(\partial _xf_n\) on [0, T] in terms of A, hence we are in position to apply the a priori estimate (5.5). Now, if we add \(\left\Vert f_0\right\Vert _{\dot{H}^1\cap \dot{H}^s}^2\) to both sides of (5.5) we deduce that

$$\begin{aligned}&N_n(\tau _n)+\frac{C}{1+C(A)^2}\int \nolimits _{0}^{\tau _n}\left\Vert f_n(t)\right\Vert _{\dot{H}^{s+\frac{1}{2}}}^2{{\,\mathrm{d}\,}} t\\&\quad \leqq 3(1+\tau _n)^2\left\Vert f_0\right\Vert _{\dot{H}^1\cap \dot{H}^s}^2+\tau _n\mathcal {F}(N_n(\tau _n)). \end{aligned}$$

We infer that

$$\begin{aligned} A=N_n(\tau _n)&\leqq 3(1+\tau _n)^2\left\Vert f_0\right\Vert _{\dot{H}^1\cap \dot{H}^s}^2+\tau _n\mathcal {F}(N_n(\tau _n))\\&\leqq 3(1+T_0)^2\left\Vert f_0\right\Vert _{\dot{H}^1\cap \dot{H}^s}^2+T_0\mathcal {F}(A)\\&<A, \end{aligned}$$

hence the contradiction. We thus have proved that, for all \(n\in \mathbb {N}\) and all \(T\leqq \min \{T_{0},T_n\}\), we have

$$\begin{aligned} \sup _{t\in [0,T]}\left\Vert f_n(t)-f_0\right\Vert _{L^2\cap \dot{H}^s}^2\leqq A. \end{aligned}$$

This obviously implies that

$$\begin{aligned} \sup _{t\in [0,T]}\left\Vert f_n(t)-f_0\right\Vert _{L^2}\leqq \sqrt{A}. \end{aligned}$$

Since

$$\begin{aligned} u_n=f_n-J_nf_0=f_n-f_0+(I-J_n)f_0, \end{aligned}$$

and since \(\left\Vert (I-J_n)f_0\right\Vert _{L^2}\leqq \left\Vert f_0\right\Vert _{\dot{H}^1}\), the previous bound implies that the norm \(\left\Vert u_n(t)\right\Vert _{L^2}\) is bounded for all \(t\leqq \min \{T_{0},T_n\}\). The alternative (5.2) then implies that the lifespan of \(f_n\) is bounded from below by \(T_0\), and the previous inequality shows that \((f_n-f_0)\) is bounded in \(C^0([0,T_0];H^s(\mathbb {R}))\). This completes the proof. \(\quad \square \)

At that point, we have defined a sequence \((f_n)\) of solutions to well-chosen approximate systems. The next task is to prove that this sequence converges. Here a word of caution is in order: \(\dot{H}^s(\mathbb {R})\) is not a Banach space when \(s>1/2\). To overcome this difficulty, we use the fact that \(u_n=f_n-f_0\) is bounded in \(C^0([0,T_0];H^s(\mathbb {R}))\), where \(H^s(\mathbb {R})\) is the nonhomogeneous space \(L^2(\mathbb {R})\cap \dot{H}^s(\mathbb {R})\), which is a Banach space. We claim that, in addition, \((u_n)\) is a Cauchy sequence in \(C^0([0,T_0];H^{s'}(\mathbb {R}))\) for any \(s'<s\). Let us assume this claim for the moment. This will imply that \((u_n)\) converges in the latter to some limit u. Now, setting \(f=f_0+u\) and using the continuity result for \(\mathcal {T}(f)f\) given by (ii) in Proposition 2.3, we verify immediately that f is a solution to the Cauchy problem for the Muskat equation. It would remain to prove that u is continuous in time with values in \(H^s(\mathbb {R})\) (instead of \(H^{s'}(\mathbb {R})\) for any \(s'<s\)). For the sake of shortness, this is the only point that we do not prove in details in this paper (referring to [2] for the proof of a similar result in a case with similar difficulties).

To conclude the proof of Theorem 1.4, it remains only to establish the following:

Lemma 5.6

For any real number \(s'\) in [0, s), the sequence \((u_n)\) is a Cauchy sequence in \(C^0([0,T_0];H^{s'}(\mathbb {R}))\).

Proof

The proof is in two steps. We begin by proving that \((f_n)\) is a Cauchy sequence in \(C^0([0,T_0];\dot{H}^{s'}(\mathbb {R}))\) for \(1/2\leqq s'<s\). Then, we use this result and an elementary \(L^2\)-estimate to infer that \((u_n)\) is a Cauchy sequence in \(C^0([0,T_0];L^2(\mathbb {R}))\).

By using estimates parallel to those used to prove Lemma 5.4, one obtains that \((f_n)\) is a Cauchy sequence in \(C^0([0,T_0];\dot{H}^\frac{1}{2}(\mathbb {R}))\). Now consider \(1/2<s'<s\). By interpolation, there exists \(\alpha \) in (0, 1) such that

$$\begin{aligned} \left\Vert u\right\Vert _{\dot{H}^{s'}}\lesssim \left\Vert u\right\Vert _{\dot{H}^\frac{1}{2}}^\alpha \left\Vert u\right\Vert _{\dot{H}^s}^{1-\alpha }. \end{aligned}$$

Consequently, since \((f_n)\) is bounded in \(C^0([0,T_0];\dot{H}^s(\mathbb {R}))\), we deduce that \((f_n)\) is a Cauchy sequence in \(C^0([0,T_0];\dot{H}^{s'}(\mathbb {R}))\) for any \(s'<s\).

It remains only to prove that \((u_n)\) is a Cauchy sequence in \(C^0([0,T_0];L^2(\mathbb {R}))\). To do so, we proceed differently. Starting from (see (5.7))

$$\begin{aligned} \partial _t f_n+\Lambda f_n=J_n\mathcal {T}(f_n)f_n, \end{aligned}$$

we obtain that \(u_n-u_p=f_n-f_p-(J_n-J_p)f_0\) satisfies

$$\begin{aligned} \partial _t (u_n-u_p)+\Lambda (u_n-u_p)=F_{np}+G_{np}, \end{aligned}$$
(5.20)

where

$$\begin{aligned} F_{np}=J_n(\mathcal {T}(f_n)f_n-\mathcal {T}(f_p)f_p),\quad G_{np}=(J_n-J_p)\big (-\Lambda f_0+\mathcal {T}(f_p)f_p\big ). \end{aligned}$$

We now use an elementary \(L^2\)-estimate. We take the \(L^2\)-scalar product of the Equation (5.20) with \(u_n-u_p\), to obtain, since \(u_n(0)-u_p(0)=0\),

$$\begin{aligned} \sup _{t\in [0,T]}\left\Vert u_n(t)-u_p(t)\right\Vert _{L^2}\leqq \left\Vert F_{np}\right\Vert _{L^1([0,T];L^2)} +\left\Vert G_{np}\right\Vert _{L^2([0,T];\dot{H}^{-\frac{1}{2}})}. \end{aligned}$$

So it remains only to prove that \(\left\Vert F_{np}\right\Vert _{L^1([0,T];L^2)}\) and \(\left\Vert G_{np}\right\Vert _{L^2([0,T];\dot{H}^{-\frac{1}{2}})}\) are arbitrarily small for np large enough. Here we use the result proved in the first part of the proof. Namely, since \((f_n)\) is a Cauchy sequence in \(C^0([0,T];\dot{H}^1(\mathbb {R})\cap \dot{H}^\frac{3}{2}(\mathbb {R}))\), we deduce from point (ii) in Proposition 2.3 that

$$\begin{aligned} \mathcal {T}(f_n)f_n-\mathcal {T}(f_p)f_p \end{aligned}$$

is small in \(C^0([0,T];L^2(\mathbb {R}))\) for np large enough. On the other hand, using the estimate

$$\begin{aligned} \left\Vert (J_n-J_p)u\right\Vert _{\dot{H}^{-\frac{1}{2}}}\leqq \frac{1}{\sqrt{\min (n,p)}}\left\Vert u\right\Vert _{L^2}, \end{aligned}$$

we verify that \(\left\Vert G_{np}\right\Vert _{L^2([0,T];\dot{H}^{-\frac{1}{2}})}\) is arbitrarily small for np large enough. This completes the proof. \(\quad \square \)