On the Cauchy Problem for the Muskat Equation. II: Critical Initial Data

Abstract

We prove that the Cauchy problem for the Muskat equation is well-posed locally in time for any initial data in the critical space of Lipschitz functions with three-half derivative in $L^2$. Moreover, we prove that the solution exists globally in time under a smallness assumption.

1 Introduction

The Muskat equation describes the dynamics of the interface separating two fluids in porous media whose velocities obey Darcy’s law [22, 30]. This equation belongs to the family of nonlocal parabolic equations that have attracted a lot of attention in recent years. Indeed, it has long been observed that one can reduce the Muskat equation to an evolution equation for the free surface parametrization (see [9, 24, 34, 35]). One interesting feature of the Muskat equation is that it admits a compact formulation in terms of finite differences, as observed by Córdoba and Gancedo [19]. More precisely, assume that the free surface is the graph of some function $f=f(t,x)$ with $x\in \mathbb {R}$. Then, Córdoba and Gancedo [19] showed that the Muskat equation reduces to

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

(1)

where $\Delta _\alpha f$ is the slope, defined by

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

(2)

It is easily verified that the Muskat equation is invariant by the change of unknowns:

$$\begin{aligned} f(t,x)\mapsto f_\lambda (t,x)\mathrel {:=}\frac{1}{\lambda }f\left( \lambda t,\lambda x\right) \qquad (\lambda \ne 0). \end{aligned}$$

(3)

Now, by a direct calculation,

$$\begin{aligned} \left\| f_\lambda \big \arrowvert _{t=0} \right\| _{\dot{W}^{1,\infty }}=\left\| f_0\right\| _{\dot{W}^{1,\infty }}; \quad \left\| f_\lambda \big \arrowvert _{t=0} \right\| _{ \dot{H}^{\frac{3}{2}}}=\left\| f_0\right\| _{ \dot{H}^{\frac{3}{2}}}. \end{aligned}$$

This means that the spaces $\dot{W}^{1,\infty }(\mathbb {R})$ and $\dot{H}^{\frac{3}{2}}(\mathbb {R})$ are critical for the study of the Cauchy problem. Let us clarify that we denoted by $\dot{W}^{1,\infty }(\mathbb {R})$ the space of Lipschitz functions, and by $H^s(\mathbb {R})$ (resp. $\dot{H}^{s}(\mathbb {R})$) the classical Sobolev (resp. homogeneous Sobolev) space of order s. They are equipped with the norm defined by

$$\begin{aligned} \left\| u\right\| _{\dot{W}^{1,\infty }}\mathrel {:=}\sup _{{\begin{array}{c} x,y\in \mathbb {R}\\ x\ne y \end{array}}}\frac{\left| u(x)-u(y)\right| }{\left| x-y\right| }, \end{aligned}$$

and

$$\begin{aligned} \left\| u\right\| _{\dot{H}^{s}}\mathrel {:=}\left( \int _\mathbb {R}\left| \xi \right| ^{2s}\big \vert \hat{u}(\xi )\big \vert ^2{{\,\mathrm{d}\,}} \xi \right) ^\frac{1}{2}, \quad \left\| u\right\| _{H^{s}}^2=\left\| u\right\| _{\dot{H}^s}^2+\left\| u\right\| _{L^2}^2. \end{aligned}$$

We are interested in the study of the Cauchy problem for the latter equation. Our main result states that the Cauchy problem for the Muskat equation is well-posed locally in time for any initial data in the critical space $\dot{W}^{1,\infty }(\mathbb {R})\cap {H}^{\frac{3}{2}}(\mathbb {R})$.

Our analysis is inspired by many previous works, and we begin by reviewing the literature on this problem. The first well-posedness results were established by Yi [38], Ambrose [4, 5], Córdoba and Gancedo [19], Córdoba, Córdoba and Gancedo [18], Cheng, Granero-Belinchón, Shkoller [14]. In recent years, these results were extended in several directions. In particular, the well-posedness of the Cauchy problem has been established in many sub-critical spaces: see Constantin, Gancedo, Shvydkoy and Vicol [17] for initial data in the Sobolev space $W^{2,p}(\mathbb {R})$ for some $p>1$, Deng, Lei and Lin [23] and Camerón [10] for initial data in Hölder spaces, and Matioc [29], Alazard and Lazar [2], Nguyen and Pausader [31] for initial data in $H^s(\mathbb {R})$ with $s>3/2$.

Special features of the Muskat equations were exploited to improve the analysis of the Cauchy problem in several directions. Constantin, Córdoba, Gancedo, Rodríguez-Piazza and Strain [15] (see also [17, 33]) proved a global well-posedness results assuming that the Lipschitz semi-norm is smaller than 1. Deng, Lei and Lin in [23] proved the existence of solutions whose slope can be arbitrarily large. Cameron [10] exhibited the existence of a modulus of continuity for the derivative (see also [1]) and obtained a global existence result assuming only that the product of the maximal and minimal slopes is bounded by 1. Córdoba and Lazar established in [21] the first global well-posedness result assuming only that the initial data is sufficiently smooth and that the critical $\dot{H}^{3/2}(\mathbb {R})$-norm is small enough (see also [25, 26, 28] for related global well-posedness results in Wiener spaces in the critical case, for small enough initial data). This result was extended to the 3D case by Gancedo and Lazar [27] for initial data in the critical Sobolev space $\dot{H}^2(\mathbb {R}^2)$. Eventually, in our companion paper [3], we initiated the study of the Cauchy problem for non-Lipschitz initial data.

For our subject matter, another fundamental component of the background is that the Cauchy problem is not well-posed globally in time: there are blow-up results for some large enough data by Castro, Córdoba, Fefferman, Gancedo and López-Fernández [11,12,13]. More precisely, they proved the existence of solutions such that at initial time $t=0$ the interface is a graph, at a later time $t_1>0$ the interface is no longer a graph and then at a subsequent time $t_2>t_1$, the interface is $C^3$ but not $C^4$.

Our main result in this paper is the following

Theorem 1.1

• (i) For any initial data $f_0$ in $\dot{W}^{1,\infty }(\mathbb {R})\cap H^{\frac{3}{2}}(\mathbb {R})$, there exists a time $T>0$ such that the Cauchy problem for the Muskat equation has a unique solution

  $$\begin{aligned} f\in L^\infty \big ([0,T];\dot{W}^{1,\infty }(\mathbb {R})\cap H^{\frac{3}{2}}(\mathbb {R})\big ) \cap L^2(0,T;\dot{H}^2(\mathbb {R})). \end{aligned}$$

• (ii) Moreover, there exists a positive constant $\delta$ such that, for any initial data $f_0$ in $\dot{W}^{1,\infty }(\mathbb {R})\cap H^{3/2}(\mathbb {R})$ satisfying

  $$\begin{aligned} \big (1+\left\| f_0\right\| _{\dot{W}^{1,\infty }}^4\big )\left\| f_0\right\| _{\dot{H}^{\frac{3}{2}}}\le \delta , \end{aligned}$$

  the Cauchy problem for the Muskat equation has a unique global solution

  $$\begin{aligned} f\in L^\infty \big ([0,+\infty );\dot{W}^{1,\infty }(\mathbb {R})\cap H^{\frac{3}{2}}(\mathbb {R})\big ) \cap L^2(0,+\infty ;\dot{H}^2(\mathbb {R})). \end{aligned}$$

Some remarks are in order.

• Let us discuss statement (ii) about the global well-posedness component of this result. This is a 2D analogous to the recent result by Gancedo and Lazar [27] for the 3D problem; it improves on a previous result by Córdoba and Lazar [21] which proves a similar global existence result for the 2D-problem with a similar smallness assumption, but under the extra assumption that the initial data belongs to $H^{5/2}(\mathbb {R})$.

• We now come to statement (i) about the local well-posedness result for arbitrary initial data. This is, in our opinion, the main new result in this paper. Since we are working in a critical space, this result is optimal in several directions.

Firstly, it follows from the results about singularity formation by Castro, Córdoba, Fefferman, Gancedo and López-Fernández [11,12,13] that one cannot solve the Cauchy problem for a time T which depends only on the norm of $f_0$ in $\dot{W}^{1,\infty }(\mathbb {R})\cap \dot{H}^{3/2}(\mathbb {R})$. Otherwise, one would obtain a global existence result for any initial data by an immediate scaling argument using (3). Notice that this argument does not contradict our main result: it means instead that the time of existence must depend on the initial data itself, and not only on its norm.

The previous discussion shows that one cannot prove statement (i) by using classical Sobolev energy estimates. This in turn poses new challenging questions since on the other hand the Muskat equation is a quasi-linear equation. To overcome this problem, we will estimate the solution for a norm whose definition depends on the initial data.

$\bullet$ We will also prove a result which elaborates on the previous discussion, stating that whenever one controls a bigger norm than the critical one, the time of existence is bounded from below on a neighborhood of the initial data.

To introduce this result, let us fix some notations.

Definition 1.2

Given a real number $s\ge 0$ and a function $\phi :[0,\infty )\rightarrow [1,\infty )$ satisfying the following assumptions:

(H1):

$\phi$ is increasing and $\lim \phi (r)=\infty$ when r goes to $+\infty$;

(H2):

there is a positive constant $c_0$ such that $\phi (2r)\le c_0\phi (r)$ for any $r\ge 0$;

(H3):

the function $r\mapsto \phi (r)/\log (4+r)$ is decreasing on $[0,\infty )$.

Then $|D|^{s,\phi }$ denotes the Fourier multiplier with symbol $|\xi |^s\phi (|\xi |)$, so that

$$\begin{aligned} \mathcal {F}( |D|^{s,\phi }f)(\xi )=|\xi |^s\phi (|\xi |) \mathcal {F}(f)(\xi ). \end{aligned}$$

Moreover, we define the space

$$\begin{aligned} \mathcal {X}^{s,\phi }(\mathbb {R})=\{ f\in \dot{W}^{1,\infty }(\mathbb {R})\cap L^2(\mathbb {R})\, :\, \left| D\right| ^s \phi (\left| D_x\right| )f\in L^2(\mathbb {R})\}, \end{aligned}$$

equipped with the norm

$$\begin{aligned} \left\| f\right\| _{\mathcal {X}^{s,\phi }}\mathrel {:=}\left\| f\right\| _{\dot{W}^{1,\infty }}+\left\| f\right\| _{L^2} +\left( \int _\mathbb {R}\left| \xi \right| ^{2s}(\phi (\left| \xi \right| ))^2\big \vert \hat{f}(\xi )\big \vert ^2{{\,\mathrm{d}\,}} \xi \right) ^\frac{1}{2}. \end{aligned}$$

Remark 1.3

The Fourier multiplier $|D|^{s,\phi }$ with $\phi (r)=\log (2+r)^a$ was introduced and studied in [6,7,8] for $s\in [0,1)$ (also see [32]).

Theorem 1.4

Consider a real number $M_0>0$ and a function $\phi$ satisfying assumptions $(\mathrm {H1})$–$(\mathrm {H3})$ in Definiton 1.2. Then there exists a time $T_0>0$ such that, for any initial data $f_0$ in $\mathcal {X}^{\frac{3}{2},\phi }(\mathbb {R})$ satisfying

$$\begin{aligned} \left\| f_0\right\| _{\mathcal {X}^{\frac{3}{2},\phi }}\le M_0, \end{aligned}$$

the Cauchy problem for the Muskat equation has a unique solution

$$\begin{aligned} f\in L^\infty \big ([0,T_0];\dot{W}^{1,\infty }(\mathbb {R})\cap {H}^{\frac{3}{2}}(\mathbb {R})\big ) \cap L^2(0,T_0;\dot{H}^2(\mathbb {R})). \end{aligned}$$

(4)

Remark 1.5

Statement (i) in Theorem 1.1 is a consequence of Theorem 1.4. Indeed, it is easily seen that (cf [3, Lemma 3.8]), for any $f_0$ in the critical space $\dot{W}^{1,\infty }(\mathbb {R})\cap \dot{H}^{\frac{3}{2}}(\mathbb {R})$, one may find a function $\phi$ such that $f_0$ belongs to $\mathcal {X}^{\frac{3}{2},\phi }(\mathbb {R})$ (and satisfying assumptions $(\mathrm {H1})$–$(\mathrm {H3})$ in Definiton 1.2).

Theorem 1.1 and Theorem 1.4 are proved in the next section.

2 Proof

2.1 Regularization

In order to rigorously justify the computations, we want to handle smooth functions (hereafter, a ‘smooth function’ is by definition a function that belongs to $C^1([0,T];H^\mu (\mathbb {R}))$ for any $\mu \in [0,+\infty )$ and some $T>0$). To do so, we must regularize the initial data and also consider an approximation of the Muskat equation. For our purposes, we further need to consider a regularization of the Muskat equation which will be compatible with the Sobolev and Lipschitz estimates. It turns out that this is a delicate technical problem.

Our strategy will consist in smoothing the equation in two different ways: (i) by introducing a cut-off function in the singular integral, removing wave-length shorter than some parameter $\varepsilon$ and (ii) by adding a parabolic term of order 2 with a small viscosity of size $\left| \log (\varepsilon )\right| ^{-1}$.

More precisely, we introduce the following Cauchy problem depending on the parameter $\varepsilon \in (0,1]$:

$$\begin{aligned} \left\{ \begin{aligned}&\partial _tf-|\log (\varepsilon )|^{-1}\partial _x^2 f =\frac{1}{\pi }\int _\mathbb {R}\frac{\partial _x\Delta _\alpha f}{1+\left( \Delta _\alpha f\right) ^2}\left( 1-\chi \left( \frac{\alpha }{\varepsilon }\right) \right) {{\,\mathrm{d}\,}} \alpha ,\\&f\arrowvert _{t=0}=f_0\star \chi _\varepsilon , \end{aligned} \right. \end{aligned}$$

(5)

where $\chi _\varepsilon (x)=\varepsilon ^{-1}\chi (x/\varepsilon )$ where $\chi$ is a smooth bump function satisfying $0\le \chi \le 1$ and

$$\begin{aligned} \chi (y)=\chi (-y),\quad \chi (y)=1 \quad \text {for}\quad |y|\le \frac{1}{4}, \quad \chi (y)=0 \quad \text {for }\left| y\right| \ge 2,\quad \int _\mathbb {R}\chi {{\,\mathrm{d}\,}} y=1. \end{aligned}$$

The equation (5) does not belong to a general class of parabolic equations. However, we will see that it can be studied by standard tools in functional analysis together with two estimates for the nonlinearity in the Muskat equation which plays a central role in our analysis.

Proposition 2.1

For any $\varepsilon$ in (0, 1] and any initial data $f_0$ in $H^{\frac{3}{2}}(\mathbb {R})$, there exists a unique global in time solution $f_\varepsilon$ satisfying

$$\begin{aligned} f_\varepsilon \in C^1([0,+\infty );H^\infty (\mathbb {R})). \end{aligned}$$

We postpone the proof of this proposition to §2.9.

2.2 An Estimate of the Lipschitz Norm

Lemma 2.2

For any real number $\beta _0$ in (0, 1/2), there exists a positive constant $C_0\ge 1$ such that, for any $\varepsilon \in (0,1]$ and any smooth solution $f\in C^1([0,T];H^{\infty }(\mathbb {R}))$ of the Muskat equation (5),

$$\begin{aligned} \frac{{{\,\mathrm{d}\,}}}{{{\,\mathrm{d}\,}} t}\left\| f(t)\right\| _{\dot{W}^{1,\infty }}\le C_0 \left\| f(t)\right\| _{\dot{H}^2}^2+C_0 \varepsilon ^{\beta _0}\left\| f(t)\right\| _{\dot{C}^{2,\beta _0}}, \end{aligned}$$

(6)

where

$$\begin{aligned} \left\| u\right\| _{\dot{C}^{2,\beta _0}}=\left\| \partial _{xx}u\right\| _{C^{0,\beta _0}} =\sup _{\begin{array}{c} x,y\in \mathbb {R}\\ x\ne y \end{array}} \frac{\left| (\partial _{xx}u)(x)-(\partial _{xx}u)(y)\right| }{\left| x-y\right| ^{\beta _0}}\cdot \end{aligned}$$

Proof

The proof is partially based on arguments from [10, 20, 27]. Firstly, it follows from the proof of [20, Lemma 5.1] that

$$\begin{aligned}&\partial _{x}\frac{1}{\pi }\int _\mathbb {R}\frac{\partial _x\Delta _\alpha f(x)}{1+\left( \Delta _\alpha f(x)\right) ^2}{{\,\mathrm{d}\,}} \alpha \\&\quad =\frac{\partial _x^2f(t,x)}{2\pi } \int \left( \frac{1}{1+(\Delta _\alpha f(t,x))^2}-\frac{1}{1+(\Delta _{-\alpha } f(t,x))^2}\right) \frac{{{\,\mathrm{d}\,}} \alpha }{\alpha }\\&\qquad -\frac{2}{\pi } \int \frac{\partial _x f(t,x)-\Delta _\alpha f(t,x)}{\alpha ^2} \frac{1+\partial _x f(t,x)\Delta _\alpha f(t,x)}{1+(\Delta _\alpha f(t,x))^2} {{\,\mathrm{d}\,}} \alpha . \end{aligned}$$

Moreover,

$$\begin{aligned} \begin{aligned}&\left| \partial _{x}\left( \frac{1}{\pi }\int _\mathbb {R}\frac{\partial _x\Delta _\alpha f}{1+\left( \Delta _\alpha f\right) ^2}\chi \left( \frac{|\alpha |}{\varepsilon }\right) {{\,\mathrm{d}\,}} \alpha \right) \right| \\&\quad \lesssim \int _{|\alpha |\le 2 \varepsilon }\left( |\Delta _\alpha f_{xx}| +|\Delta _\alpha f_{x}|^2\right) {{\,\mathrm{d}\,}} \alpha \\&\quad \lesssim \int _\mathbb {R}|\Delta _\alpha f_{x}|^2{{\,\mathrm{d}\,}} \alpha +\varepsilon ^{\beta _0}\left\| f_{xx}\right\| _{\dot{C}^{0,\beta _0}}, \end{aligned} \end{aligned}$$

(7)

where we used the notations $f_x=\partial _x f$ and $f_{xx}=\partial _{xx}f$. Thus, for any t and any x, we have

$$\begin{aligned} \begin{aligned}&(\partial _x \partial _t f)(t,x) -|\log (\varepsilon )|^{-1}\partial _{x}^2 f_x(t,x)\\&\quad \le \frac{\partial _x^2f(t,x)}{2\pi } \int \left( \frac{1}{1+(\Delta _\alpha f(t,x))^2}-\frac{1}{1+(\Delta _{-\alpha } f(t,x))^2}\right) \frac{{{\,\mathrm{d}\,}} \alpha }{\alpha }\\&\qquad -\frac{2}{\pi } \int \frac{\partial _x f(t,x)-\Delta _\alpha f(t,x)}{\alpha ^2} \frac{1+\partial _x f(t,x)\Delta _\alpha f(t,x)}{1+(\Delta _\alpha f(t,x))^2} {{\,\mathrm{d}\,}} \alpha \\&\qquad +C \int |\Delta _\alpha f_{x}(t,x)|^2{{\,\mathrm{d}\,}} \alpha +C\varepsilon ^{\beta _0}\left\| f_{xx}(t)\right\| _{\dot{C}^{0,\beta _0}}. \end{aligned} \end{aligned}$$

(8)

Consider the function $\varphi (t)=\left\| \partial _{x}f(t)\right\| _{L^\infty }$ and a function $t\mapsto x_t$ such that

$$\begin{aligned} \left\| \partial _{x}f(t)\right\| _{L^\infty }=(\partial _{x} f)(t,x_t). \end{aligned}$$

Then $(\partial _{x}^2f)(t,x_t)=0$ and $-(\partial _{xx} f_x)(t,x_t)\ge 0$. So, it follows from (8) that

$$\begin{aligned} \dot{\varphi }(t)&\le -\frac{2}{\pi }\int \frac{\partial _x f(t,x_t)-\Delta _\alpha f(t,x_t)}{\alpha ^2}{{\,\mathrm{d}\,}} \alpha \\&\quad -\frac{2}{\pi }\int \frac{(\partial _x f(t,x_t)-\Delta _\alpha f(t,x_t))^2}{\alpha ^2} \frac{\Delta _\alpha f(t,x_t)}{1+(\Delta _\alpha f(t,x_t))^2} {{\,\mathrm{d}\,}} \alpha \\&\quad +C \int |\Delta _\alpha f_{x}(t,x_t)|^2{{\,\mathrm{d}\,}} \alpha +C\varepsilon ^{\beta _0}\left\| f_{xx}(t)\right\| _{\dot{C}^{0,\beta _0}}. \end{aligned}$$

As already observed in [20] (see also [10, 27]), the first term in the right-hand side has a sign since $\partial _x f(t,x_t)\ge \Delta _\alpha f(t,x_t)$ for any $\alpha$. It follows that

$$\begin{aligned} \dot{\varphi }(t)&\le \frac{1}{\pi }\int \frac{(\partial _x f(t,x_t) -\Delta _\alpha f(t,x_t))^2}{\alpha ^2} {{\,\mathrm{d}\,}} \alpha +C \int |\Delta _\alpha f_{x}(t,x_t)|^2{{\,\mathrm{d}\,}} \alpha \\&\quad +C\varepsilon ^{\beta _0}\left\| f_{xx}(t)\right\| _{\dot{C}^{0,\beta _0}}. \end{aligned}$$

We now apply Hardy’s inequality to infer that

$$\begin{aligned} \int \frac{(\partial _x f(t,x_t) -\Delta _\alpha f(t,x_t))^2}{\alpha ^2} {{\,\mathrm{d}\,}} \alpha \lesssim \int |\Delta _\alpha f_{x}(t,x_t)|^2{{\,\mathrm{d}\,}} \alpha . \end{aligned}$$

Consequently, we end up with

$$\begin{aligned} \dot{\varphi }(t)\lesssim \int \left\| \Delta _\alpha f_{x}(t)\right\| _{L^\infty }^2{{\,\mathrm{d}\,}} \alpha +\varepsilon ^{\beta _0}\left\| f_{xx}(t)\right\| _{\dot{C}^{0,\beta _0}}. \end{aligned}$$

Introducing the difference operator $\delta _\alpha g(x)=g(x)-g(x-\alpha )$, the previous inequality is better formulated as follows:

$$\begin{aligned} \dot{\varphi }(t)\lesssim \int \left\| \delta _\alpha (\partial _xf)(t)\right\| _{L^\infty }^2 \frac{{{\,\mathrm{d}\,}} \alpha }{|\alpha |^{1+\frac{1}{2}2}}+\varepsilon ^{\beta _0}\left\| f_{xx}(t)\right\| _{\dot{C}^{0,\beta _0}}. \end{aligned}$$

Now the right-hand side is equivalent to the following homogeneous Besov norm: $\left\| \partial _x f(t)\right\| _{\dot{B}^{\frac{1}{2}}_{\infty ,2}}^2$ (see [36, 37] or Section 2 in [3]). Then it follows from Sobolev embeddings that

$$\begin{aligned} \dot{\varphi }(t)\lesssim \left\| f(t)\right\| _{\dot{H}^2}^2+\varepsilon ^{\beta _0}\left\| f_{xx}(t)\right\| _{\dot{C}^{0,\beta _0}} \end{aligned}$$

which is the wanted result. $\square$

2.3 Sobolev Estimates

In this paragraph we recall a generalized Sobolev energy estimate proved in our companion paper [3]. By generalized Sobolev energy estimate, we mean that, instead of estimating the $L^\infty _t(L^2_x)$-norm of $(-\Delta )^s f$, we shall estimate the $L^\infty _t(L^2_x)$-norm of $\left| D\right| ^{s,\phi }f$ for some function $\phi$ satisfying the assumptions in Definition 1.2.

There two technical results that we will borrow from [3]. The first result, which is Lemma 3.4 in [3], gives an energy estimate.

Lemma 2.3

There exists a positive constant C such that, for any $T>0$ and any smooth solution $f\in C^{1}([0,T];H^{\infty }(\mathbb {R}))$ to (1), there holds

$$\begin{aligned}&\frac{{{\,\mathrm{d}\,}}}{{{\,\mathrm{d}\,}} t}\big \Vert \left| D\right| ^{\frac{3}{2},\phi }f\big \Vert _{L^2}^2 + \int _\mathbb {R}\frac{\big \vert \left| D\right| ^{2,\phi }f\big \vert ^2}{1+(\partial _x f)^2}{{\,\mathrm{d}\,}} x+|\log (\varepsilon )|^{-1}\int _\mathbb {R}\big \vert \left| D\right| ^{\frac{5}{2},\phi }f\big \vert ^2{{\,\mathrm{d}\,}} x\nonumber \\&\quad \le C Q(f) \big \Vert \left| D\right| ^{2,\phi }f\big \Vert _{L^2}, \end{aligned}$$

(9)

where

$$\begin{aligned} Q(f)&= \left( \left\| f\right\| _{\dot{H}^2}+\left\| f\right\| _{\dot{H}^{\frac{7}{4}}}^2\right) \big \Vert \left| D\right| ^{\frac{3}{2},\phi }f\big \Vert _{L^2} +\big \Vert \left| D\right| ^{\frac{7}{4},\phi }f\big \Vert _{L^2} \left\| f\right\| _{{H}^{\frac{7}{4}}}\\&\quad +\left( \left\| f\right\| _{H^{\frac{19}{12}}}^{3/2}+\left\| f\right\| _{\dot{H}^{\frac{7}{4}}}^{1/2}\right) \big \Vert \left| D\right| ^{\frac{7}{4},\phi ^{2}}f\big \Vert ^{1/2}_{L^2} \left\| f\right\| _{\dot{H}^{\frac{7}{4}}}. \end{aligned}$$

Remark 2.4

Some explanations are in order since the reader may notice several modifications compared to our paper [3]. Firstly, in [3] we considered a function $\phi$ whose definition depends on an extra function $\kappa$. Here we ignore this point since it is irrelevant for the present analysis. Indeed, the functions $\phi$ and $\kappa$ are shown in [3] to be equivalent (such that $c\kappa (\lambda )\le \phi (\lambda )\le C \kappa (\lambda )$), and the distinction between them served only to organize the proof. Secondly, in [3] we also assume that $\phi (r)$ is bounded from below by $(\log (4+ r))^a$ for some $a\ge 0$. Here we will use that this property holds with $a=0$. Once the previous clarifications have been done, it remains to explain that in [3] we consider the equation (1) while here we work with (5). The elliptic term $(-\partial _{x}^2)$ is trivial to handle since in [3] we only applied an $L^2$-energy estimate and since the latter operator is positive. Eventually, the cut-off function $(1-\chi (\alpha /\varepsilon ))$ is also harmless in the various computations used to prove Lemma 3.4 in [3].

Secondly, we recall two interpolation inequalities from [3, Lemma 3.5]. Hereafter, we use the notations

$$\begin{aligned} \begin{aligned} A_\phi (t)&=\big \Vert \left| D\right| ^{\frac{3}{2},\phi }f(t)\big \Vert _{L^2}^2,\\ B_\phi (t)&=\big \Vert \left| D\right| ^{2,\phi }f(t)\big \Vert _{L^2}^2,\\ P_\phi (t)&=\big \Vert \left| D\right| ^{\frac{5}{2},\phi }f(t)\big \Vert _{L^2}^2, \end{aligned} \end{aligned}$$

(10)

and

$$\begin{aligned} \mu _\phi (t)=\left( \phi \left( \frac{B(t)}{A(t)}\right) \right) ^{-1} . \end{aligned}$$

Lemma 2.5

Consider a real number $7/4\le s\le 2$. Then, there exists a positive constant C such that, for any $T>0$, any smooth solution $f\in C^{1}([0,T];H^{\infty }(\mathbb {R}))$ to (5) and any $t\in [0,T]$,

$$\begin{aligned}&\left\| f(t)\right\| _{\dot{H}^s}\le C\mu _\phi (t) A_\phi (t)^{2-s} B_\phi (t)^{s-\frac{3}{2}}, \end{aligned}$$

(11)

$$\begin{aligned}&\big \Vert \left| D\right| ^{\frac{7}{4},\phi ^{2}}f(t)\big \Vert _{L^2}\le C \mu _\phi (t)A_\phi (t)^{\frac{1}{4}}B_\phi (t)^{\frac{1}{4}}. \end{aligned}$$

(12)

From these two lemmas, we get at once the following

Proposition 2.6

There exist two positive constants $C_1$ and $C_2$ such that, for any $T>0$ and any smooth solution $f\in C^1([0,T];H^\infty (\mathbb {R}))$ of the Muskat equation (5),

$$\begin{aligned}&\frac{{{\,\mathrm{d}\,}}}{{{\,\mathrm{d}\,}} t}A_\phi (t)+C_1\frac{B_\phi (t)}{1+\left\| f_x(t)\right\| _{L^\infty }^2}+|\log (\varepsilon )|^{-1} P_\phi (t)\nonumber \\&\quad \le C_2 \left( \sqrt{A_\phi (t)}+A_\phi (t) \right) \mu _\phi (t) B_\phi (t). \end{aligned}$$

(13)

We will also need an estimate for the $L^2$-norm.

Lemma 2.7

There holds

$$\begin{aligned} \frac{1}{2}\frac{{{\,\mathrm{d}\,}}}{{{\,\mathrm{d}\,}} t}\left\| f(t)\right\| _{L^2}^2\le C \varepsilon ^{\frac{1}{2}}\left\| f\right\| _{\dot{H}^{\frac{3}{2}}}\left\| f\right\| _{L^2}. \end{aligned}$$

In particular,

$$\begin{aligned} \left\| f(t)\right\| _{L^2}\le \left\| f_0\right\| _{L^2} +C\varepsilon ^{\frac{1}{2}}\int _0^t\left\| f(\tau )\right\| _{\dot{H}^{\frac{3}{2}}}{{\,\mathrm{d}\,}} \tau . \end{aligned}$$

(14)

Proof

Set

$$\begin{aligned} R_\varepsilon (f)=-\frac{1}{\pi }\int _\mathbb {R}\frac{\partial _x\Delta _\alpha f}{1+\left( \Delta _\alpha f\right) ^2} \chi \left( \frac{\alpha }{\varepsilon }\right) {{\,\mathrm{d}\,}} \alpha . \end{aligned}$$

(15)

We multiply the equation by f to obtain

$$\begin{aligned} \frac{1}{2}\frac{{{\,\mathrm{d}\,}}}{{{\,\mathrm{d}\,}} t}\left\| f(t)\right\| _{L^2}^2\le \frac{1}{\pi }\bigg \langle \int _\mathbb {R}\frac{\partial _x\Delta _\alpha f}{1+\left( \Delta _\alpha f\right) ^2}{{\,\mathrm{d}\,}} \alpha , f\bigg \rangle +\langle R_\varepsilon (f),f\rangle . \end{aligned}$$

Now, by [16, Section 2], the first term in the right-hand side has a sign. Indeed:

$$\begin{aligned}&\int _\mathbb {R}\left[ \int _\mathbb {R}\frac{\partial _x\Delta _\alpha f}{1+\left( \Delta _\alpha f\right) ^2}{{\,\mathrm{d}\,}} \alpha \right] f(x){{\,\mathrm{d}\,}} x\\&\quad =-\iint _{\mathbb {R}^2}\log \left[ \sqrt{1+\frac{(f(t,x)-f(t,x-\alpha ))^2}{\alpha ^2}}\right] {{\,\mathrm{d}\,}} x{{\,\mathrm{d}\,}} \alpha . \end{aligned}$$

It remains to estimate $R_\varepsilon (f)$. To do so, we use the estimate (19) to get

$$\begin{aligned} \begin{aligned} \left\| R_\varepsilon (f)\right\| _{L^2}&\lesssim \int _{|\alpha |\le 2\varepsilon }\left\| \Delta _\alpha f_{x}\right\| _{L^2}{{\,\mathrm{d}\,}} \alpha \\&\lesssim \varepsilon ^{\frac{1}{2}}\left( \int _{\mathbb {R}}\left\| \Delta _\alpha f_{x}\right\| _{L^2}^2{{\,\mathrm{d}\,}} \alpha \right) ^\frac{1}{2}\lesssim \varepsilon ^{\frac{1}{2}}\left\| f\right\| _{\dot{H}^{\frac{3}{2}}}, \end{aligned} \end{aligned}$$

(16)

which completes the proof. $\square$

2.4 Estimate of the Hölder Norm

To exploit the Sobolev energy estimate (13), the main difficulty is to estimate from above the factor $1+\left\| f_x(t)\right\| _{L^\infty }^2$. This is where we will apply Lemma 2.2. This in turn requires to estimate the Hölder norm $\left\| \cdot \right\| _{\dot{C}^{2,\beta _0}}$ of f. This is the purpose of the following result.

We will prove an estimate valid on arbitrary large time scale, which will be used later to prove a global existence result.

Proposition 2.8

For any $0<\beta <1/2$, there exist two positive constant $\varepsilon _0$ and $c_0$ such that, for any $\varepsilon \in (0,\varepsilon _0]$, any smooth solution $f\in C^1([0,T];H^{\infty }(\mathbb {R}))$ of the Muskat equation (5), and any time $t\le \min \{\varepsilon ^{-c_0},T\}$, there holds

$$\begin{aligned}&\varepsilon ^{\beta }\int _{0}^{t}\left\| f(\tau )\right\| _{C^{2,\beta }}d\tau \le \varepsilon ^{\frac{\beta }{2}}\left\| f_0\right\| _{\dot{H}^{\frac{3}{2}}} \\&~~~+ \varepsilon ^{\frac{\beta }{2}}\left( 1+\sup _{s\in [0,t]}\left\| f(s)\right\| _{H^{\frac{3}{2}}}\right) ^2 \log \left( 2+\int _{0}^{t} \left\| f(s)\right\| _{\dot{H}^{2}}^2{{\,\mathrm{d}\,}} s\right) ^{\frac{1}{2}} \left( \int _{0}^{t} \left\| f(s)\right\| _{\dot{H}^{2}}^2{{\,\mathrm{d}\,}} s\right) ^{\frac{1}{2}}. \end{aligned}$$

Proof

The classical Sobolev embeddings implies that

$$\begin{aligned} \left\| f(t)\right\| _{\dot{C}^{2,\beta }} \lesssim \left\| f(t)\right\| _{\dot{H}^{\frac{5}{2}+\beta }}. \end{aligned}$$

To estimate the latter Sobolev norm, the key point will be to apply the following interpolation inequality. $\square$

Lemma 2.9

Consider three real numbers

$$\begin{aligned} \gamma>0,\quad \beta _1>0\quad \text {and}\quad 0<\beta _2<2. \end{aligned}$$

Then, there exists a constant C such that, for any function $g=g(t,x)$,

$$\begin{aligned} \begin{aligned} \left\| g(t)\right\| _{\dot{H}^\gamma }&\lesssim \frac{1}{(\nu t)^{\frac{\beta _1}{2}}} \left\| g(0)\right\| _{\dot{H}^{\gamma -\beta _1}}\\&\quad +\int _0^{t} \frac{1}{(\nu (t-s))^{\frac{\beta _2}{2}}}\big \Vert \partial _tg(s)-\nu \partial _{xx} g(s)\big \Vert _{\dot{H}^{\gamma -\beta _2}}{{\,\mathrm{d}\,}} s. \end{aligned} \end{aligned}$$

(17)

Proof

Set $G\mathrel {:=}\partial _tg-\nu \partial _{xx} g$. Then, one has,

$$\begin{aligned} \hat{g}(t,\xi )=e^{-\nu t|\xi |^{2}}\hat{g}(0,\xi )+\int _{0}^{t}e^{-\nu (t-s)|\xi |^{2}}\hat{G}(s,\xi ){{\,\mathrm{d}\,}} s. \end{aligned}$$

The desired results then follows from Minkowski’s inequality. $\square$

Now, apply (17) with

$$\begin{aligned} \gamma =\frac{5}{2}+\beta , \quad \beta _1=1+\beta , \quad \beta _2=\frac{3}{2}+\beta ,\quad \nu =|\log (\varepsilon )|^{-1}, \end{aligned}$$

to get

$$\begin{aligned} \left\| f(t)\right\| _{\dot{C}^{2,\beta }}&\lesssim \left\| f(t)\right\| _{\dot{H}^{\frac{5}{2}+\beta }}\nonumber \\&\lesssim |\log (\varepsilon )|^{\frac{1+\beta }{2}} t^{-\frac{1+\beta }{2}}||f_0||_{\dot{H}^{\frac{3}{2}}}\nonumber \\&\quad +\int _0^{t} |\log (\varepsilon )|^{\frac{\frac{3}{2}+\beta }{2}} (t-s)^{-\frac{\frac{3}{2}+\beta }{2}} \big \Vert \partial _tf-|\log (\varepsilon )|^{-1}\partial _x^2f\big \Vert _{\dot{H}^1}{{\,\mathrm{d}\,}} s. \end{aligned}$$

(18)

It remains to estimate the $\dot{H}^1$-norm of $\partial _tf-|\log (\varepsilon )|^{-1}\partial _{xx} f$. In view of the equation (5), this is equivalent to bound the $\dot{H}^1$-norm of

$$\begin{aligned} \frac{1}{\pi }\int _\mathbb {R}\frac{\partial _x\Delta _\alpha f}{1+\left( \Delta _\alpha f\right) ^2}\left( 1-\chi \left( \frac{\alpha }{\varepsilon }\right) \right) {{\,\mathrm{d}\,}} \alpha . \end{aligned}$$

We will split the latter term into two pieces and estimate them separately.

Firstly, directly from (7) and Minkowski’s inequality, we obtain that

$$\begin{aligned} \left\| \frac{1}{\pi }\int _\mathbb {R}\frac{\partial _x\Delta _\alpha f}{1+\left( \Delta _\alpha f\right) ^2} \chi \left( \frac{\alpha }{\varepsilon }\right) {{\,\mathrm{d}\,}} \alpha \right\| _{\dot{H}^1}&\lesssim \int _{|\alpha |\le 2\varepsilon }\left( \left\| \Delta _\alpha f_{xx}\right\| _{L^2} +\left\| \Delta _\alpha f_{x}\right\| _{L^4}^2\right) {{\,\mathrm{d}\,}} \alpha \\&\lesssim \varepsilon ^{\frac{1}{2}+\beta }\left( \int _{\mathbb {R}}\left\| \Delta _\alpha f_{xx}\right\| _{L^2}^2|\alpha |^{-2\beta }{{\,\mathrm{d}\,}} \alpha \right) ^\frac{1}{2}\\&\quad + \int _{\mathbb {R}}\left\| \Delta _\alpha f_{x}\right\| _{L^4}^2{{\,\mathrm{d}\,}} \alpha . \end{aligned}$$

Now we use the following inequality:

$$\begin{aligned} \iint _{\mathbb {R}^2} \big \vert \Delta _\alpha \tilde{f}\big \vert ^2 |\alpha |^{-2\beta }{{\,\mathrm{d}\,}} \alpha {{\,\mathrm{d}\,}} x\sim \big \Vert \tilde{f}\big \Vert _{\dot{H}^{\frac{1}{2}+\beta }}^2. \end{aligned}$$

(19)

Indeed,

$$\begin{aligned} \iint _{\mathbb {R}^2} \big \vert \Delta _\alpha \tilde{f}\big \vert ^2 |\alpha |^{-2\beta }{{\,\mathrm{d}\,}} \alpha {{\,\mathrm{d}\,}} x= \iint _{\mathbb {R}^2} \left[ \frac{\big \vert \tilde{f}(x)-\tilde{f}(x-\alpha )\big \vert }{\left| \alpha \right| ^{1/2+\beta }}\right] ^2\frac{{{\,\mathrm{d}\,}} \alpha }{\left| \alpha \right| }{{\,\mathrm{d}\,}} x\sim \big \Vert \tilde{f}\big \Vert _{\dot{H}^{\frac{1}{2}+\beta }}^2. \end{aligned}$$

Similarly, using Sobolev embedding in Besov’s spaces, we get

$$\begin{aligned} \int _{\mathbb {R}}\left\| \Delta _\alpha f_{x}\right\| _{L^4}^2{{\,\mathrm{d}\,}} \alpha \lesssim \left\| f\right\| _{\dot{H}^{\frac{7}{4}}}^2. \end{aligned}$$

It follows that

$$\begin{aligned} \left\| \frac{1}{\pi }\int _\mathbb {R}\frac{\partial _x\Delta _\alpha f}{1+\left( \Delta _\alpha f\right) ^2} \chi \left( \frac{\alpha }{\varepsilon }\right) {{\,\mathrm{d}\,}} \alpha \right\| _{\dot{H}^1} \lesssim \varepsilon ^{\frac{1}{2}+\beta }\left\| f\right\| _{\dot{H}^{\frac{5}{2}+\beta }}+ \left\| f\right\| _{\dot{H}^{2}} \left\| f\right\| _{\dot{H}^{\frac{3}{2}}}, \end{aligned}$$

(20)

where we used an interpolation inequality in Sobolev spaces. On the other hand, it follows from the estimate (41) below that,

$$\begin{aligned} \begin{aligned} \left\| \int _\mathbb {R}\frac{\partial _x\Delta _\alpha f}{1+\left( \Delta _\alpha f\right) ^2}{{\,\mathrm{d}\,}} \alpha \right\| _{\dot{H}^1}&\lesssim \left\| \mathcal {T}(f)f\right\| _{\dot{H}^1}\\&\lesssim \left( 1+\left\| f\right\| _{H^{\frac{3}{2}}}\right) ^2 \log \left( 2+\left\| f\right\| _{\dot{H}^2}^2\right) ^{\frac{1}{2}} \left\| f\right\| _{\dot{H}^{2}}. \end{aligned} \end{aligned}$$

(21)

By gathering the two previous estimates, we conclude that

$$\begin{aligned}&\left\| \partial _tf-|\log (\varepsilon )|^{-1}\partial _x^2f\right\| _{\dot{H}^1}\\&\quad \lesssim \varepsilon ^{\frac{1}{2}+\beta }\left\| f\right\| _{H^{\frac{5}{2}+\beta }}+ \left( 1+\left\| f\right\| _{H^{\frac{3}{2}}}\right) ^2\log \left( 2+\left\| f\right\| _{\dot{H}^2}^2\right) ^{\frac{1}{2}} \left\| f\right\| _{\dot{H}^{2}}. \end{aligned}$$

Set

$$\begin{aligned} b=\frac{\frac{3}{2}+\beta }{2}\cdot \end{aligned}$$

By reporting this bound in (18), we find that

$$\begin{aligned}&\left\| f(t)\right\| _{\dot{H}^{\frac{5}{2}+\beta }}\lesssim |\log (\varepsilon )|^{\frac{1+\beta }{2}} t^{-\frac{1+\beta }{2 }}\left\| f_0\right\| _{\dot{H}^{\frac{3}{2}}}\\&\quad + \varepsilon ^{\frac{1}{2}+\beta }|\log (\varepsilon )|^{b}\int _0^{t} (t-s)^{-b} \left\| f(s)\right\| _{\dot{H}^{\frac{5}{2}+\beta }}{{\,\mathrm{d}\,}} s\\&\quad +|\log (\varepsilon )|^{b}\int _0^{t} (t-s)^{-b} \left( 1+\left\| f(s)\right\| _{H^{\frac{3}{2}}}\right) ^2\log \left( 2+\left\| f(s)\right\| _{\dot{H}^2}^2\right) ^{\frac{1}{2}} \left\| f(s)\right\| _{\dot{H}^{2}}{{\,\mathrm{d}\,}} s. \end{aligned}$$

So,

$$\begin{aligned}&\int _{0}^{t}\left\| f(\tau )\right\| _{\dot{H}^{\frac{5}{2}+\beta }}{{\,\mathrm{d}\,}} \tau \lesssim \left| \log (\varepsilon )\right| ^{\frac{1+\beta }{2 }} t^{1-\frac{1+\beta }{2 }}\left\| f_0\right\| _{\dot{H}^{\frac{3}{2}}}\\&\quad + \varepsilon ^{\frac{1}{2}+\beta }|\log (\varepsilon )|^{b} t^{1-b} \int _0^{t}\left\| f(s)\right\| _{\dot{H}^{\frac{5}{2}+\beta }}{{\,\mathrm{d}\,}} s\\&\quad +|\log (\varepsilon )|^{b} t^{1-b}\int _0^{t}\left( 1+\left\| f(s)\right\| _{H^{\frac{3}{2}}}\right) ^2\log \left( 2+\left\| f(s)\right\| _{\dot{H}^2}^2\right) ^{\frac{1}{2}} \left\| f(s)\right\| _{\dot{H}^{2}}{{\,\mathrm{d}\,}} s. \end{aligned}$$

As a result, there exists $c_0>0$ and $\varepsilon _0\le 1$ such that, if $t\le \varepsilon ^{-c_0}$ and $\varepsilon \le \varepsilon _0$,

$$\begin{aligned} \int _{0}^{t}\left\| f(\tau )\right\| _{\dot{H}^{\frac{5}{2}+\beta }}{{\,\mathrm{d}\,}} \tau&\le \varepsilon ^{-\frac{\beta }{2}}||f_0||_{\dot{H}^{\frac{3}{2}}} \\&\quad +|\log (\varepsilon )|^{b} t^{1-b} \mathcal {K}(t)\int _0^{t} \log \left( 2+\left\| f(s)\right\| _{\dot{H}^2}^2\right) ^{\frac{1}{2}} \left\| f(s)\right\| _{\dot{H}^{2}}{{\,\mathrm{d}\,}} s, \end{aligned}$$

where

$$\begin{aligned} \mathcal {K}(t)=\sup _{s\in [0,t]}\left( 1+\left\| f(s)\right\| _{H^{\frac{3}{2}}}\right) ^2. \end{aligned}$$

Now observe that

$$\begin{aligned}&\int _0^{t} \log \left( 2+\left\| f(s)\right\| _{\dot{H}^2}^2\right) ^{\frac{1}{2}} \left\| f(s)\right\| _{\dot{H}^{2}}{{\,\mathrm{d}\,}} s\\&\quad \le (t+1)^{\frac{1}{2}} \log \left( 2+ \int _{0}^{t} \left\| f(s)\right\| _{\dot{H}^{2}}^2ds\right) ^{\frac{1}{2}} \left( \int _{0}^{t} \left\| f(s)\right\| _{\dot{H}^{2}}^2ds\right) ^{\frac{1}{2}}. \end{aligned}$$

Therefore, up to modifying the values of $c_0>0$ and $\varepsilon _0$, we see that, for $t\le \varepsilon ^{-c_0}$ and $\varepsilon \le \varepsilon _0$, we have

$$\begin{aligned}&\varepsilon ^{\beta }\int _{0}^{t}\left\| f(\tau )\right\| _{\dot{C}^{2,\beta }}{{\,\mathrm{d}\,}} \tau \lesssim \varepsilon ^{\frac{\beta }{2}}\left\| f_0\right\| _{\dot{H}^{\frac{3}{2}}}\\&~~~+ \varepsilon ^{\frac{\beta }{2}}\mathcal {K}(t)\log \left( 2+ \int _{0}^{t} \left\| f(s)\right\| _{\dot{H}^{2}}^2{{\,\mathrm{d}\,}} s\right) ^{\frac{1}{2}} \left( \int _{0}^{t} \left\| f(s)\right\| _{\dot{H}^{2}}^2{{\,\mathrm{d}\,}} s\right) ^{\frac{1}{2}}. \end{aligned}$$

This completes the proof. $\square$

2.5 Global in Time Estimates, Under a Smallness Assumption

Proposition 2.10

Let $T>0$ and consider a smooth solution $f\in C^1([0,T],H^{\infty }(\mathbb {R}))$ of the Muskat equation (5). Set

$$\begin{aligned} K=1+16 \left( \frac{C_2}{C_1}\right) ^2 \end{aligned}$$

and assume that

$$\begin{aligned} 2\left( K+\frac{C_0}{C_1}\right) ^\frac{1}{2}\left( 2+\left\| \partial _x f_0\right\| _{L^\infty }\right) ^2\left\| f_0\right\| _{\dot{H}^{\frac{3}{2}}}\le 1, \end{aligned}$$

(22)

where the constants $C_0,C_1,C_2$ are as defined in the statements of Lemma 2.2 and Proposition 2.6. Then there exists $\varepsilon _0$ depending only on $C_0,C_1,C_2$ and $||f_0||_{L^2}$ such that, if $\varepsilon \le \varepsilon _0$, then

$$\begin{aligned} \sup _{0\le \tau \le T}\left\| f(\tau )\right\| _{{H}^{\frac{3}{2}}} \le \frac{1}{\sqrt{K}\big (2+\left\| \partial _xf_0\right\| _{L^\infty }\big )^2}\quad \text {and} \quad \int _{0}^{T}\left\| f(\tau )\right\| _{\dot{H}^2}^2{{\,\mathrm{d}\,}} \tau \le \frac{1}{C_0} \cdot \end{aligned}$$

(23)

Proof

We apply the previous a priori estimate (13) in the simplest case where $\phi =1$. With this choice, the quantities $A_\phi$ and $B_\phi$ defined by (10) simplify to

$$\begin{aligned} \begin{aligned} A(t)&=\big \Vert \left| D\right| ^{\frac{3}{2}}f(t)\big \Vert _{L^2}^2, \\ B(t)&=\big \Vert \left| D\right| ^{2}f(t)\big \Vert _{L^2}^2=\left\| f(t)\right\| _{\dot{H}^2}^2. \end{aligned} \end{aligned}$$

(24)

Introduce the set

$$\begin{aligned} I=\left\{ t\in [0,T]\,;\, \int _{0}^{t}B(\tau ){{\,\mathrm{d}\,}} \tau \le \frac{2}{3C_0} \text { and } \sup _{0\le \tau \le t}A(\tau )\le \frac{1}{K\big (2+\left\| \partial _xf_0\right\| _{L^\infty }\big )^4}\right\} . \end{aligned}$$

We want to prove that $I=[0,T]$. Since 0 belongs to I by assumption on the initial data, and since I is closed, it suffices to prove that I is open. To do so, we consider a time $t^*\in [0,T)$ which belongs to I. Our goal is to prove that

$$\begin{aligned} \int _{0}^{t^*}B(\tau ){{\,\mathrm{d}\,}} \tau \le \frac{1}{2C_0} \text { and } \sup _{0\le \tau \le t^*}A(\tau )\le \frac{1}{4K\big (2+\left\| \partial _xf_0\right\| _{L^\infty }\big )^4}. \end{aligned}$$

This will imply at once that $t^*$ belongs to the interior of I.

Since $\mu (t)=1$ for $\phi \equiv 1$, the estimate (13) implies that there are two positives constants $C_1,C_2$ such that

$$\begin{aligned} \frac{{{\,\mathrm{d}\,}}}{{{\,\mathrm{d}\,}} t}A(t)+C_1\frac{B(t)}{1+\left\| \partial _x f(t)\right\| _{L^\infty }^2}\le C_2\Big (A(t)+\sqrt{A(t)}\Big )B(t). \end{aligned}$$

(25)

By combining Proposition 2.8 with Lemma 2.2, we get, for any t,

$$\begin{aligned}&\left\| \partial _xf(t)\right\| _{L^\infty }- \left\| \partial _xf_0\right\| _{L^\infty }\le C_0\int _{0}^{t}B(\tau ) {{\,\mathrm{d}\,}} \tau +C_0 \varepsilon ^{\frac{\beta }{2}}\left\| f_0\right\| _{\dot{H}^{\frac{3}{2}}}\\&\quad + C_0 \varepsilon ^{\frac{\beta }{2}}\left[ \sup _{s\in [0,t]}\left( 1+\left\| f(s)\right\| _{H^{\frac{3}{2}}}\right) ^2\right] \log \left( 2+\int _{0}^{t} B(\tau ){{\,\mathrm{d}\,}} \tau \right) ^{\frac{1}{2}} \left( \int _{0}^{t} B(\tau ){{\,\mathrm{d}\,}} \tau \right) ^{\frac{1}{2}}. \end{aligned}$$

By (14),

$$\begin{aligned} \sup _{s\in [0,t]}\left\| f(s)\right\| _{L^{2}}\le \left\| f_0\right\| _{L^2}+C\varepsilon ^{\frac{1}{2}}t \sup _{s\in [0,t]}\left\| f(s)\right\| _{\dot{H}^{\frac{3}{2}}}. \end{aligned}$$

This implies

$$\begin{aligned} \sup _{s\in [0,t]}\left\| f(s)\right\| _{H^{\frac{3}{2}}}\le \left\| f_0\right\| _{L^2}+(1 +C\varepsilon ^{\frac{1}{2}}t) \sup _{s\in [0,t]}\left\| f(s)\right\| _{\dot{H}^{\frac{3}{2}}}. \end{aligned}$$

(26)

If $t\le t^*$, then the bound on the integral of B, $\sup _{0\le \tau \le t^*}||f||_{\dot{H}^{\frac{3}{2}}}\le 1$ and (26) imply that

$$\begin{aligned}&\left\| \partial _xf(t)\right\| _{L^\infty }- \left\| \partial _xf_0\right\| _{L^\infty }\le \frac{1}{2}+C_0 \varepsilon ^{\frac{\beta }{2}}+ C_0 \varepsilon ^{\frac{\beta }{2}}\left( 1+ \left\| f_0\right\| _{L^2}\right. \\&\quad \left. +(1 +C\varepsilon ^{\frac{1}{2}}t^*)\right) ^2\log \left( 3\right) ^{\frac{1}{2}}. \end{aligned}$$

For $\varepsilon$ small enough, we conclude that

$$\begin{aligned} \left\| \partial _xf(t)\right\| _{L^\infty }\le \frac{2}{3}+\left\| \partial _xf_0\right\| _{L^\infty }. \end{aligned}$$

On the other hand, if $t^*\in I$, then for any $t\le t^*$ we have

$$\begin{aligned} A(t)+\sqrt{A(t)}\le 2\sqrt{A(t)}\le \frac{2}{\sqrt{K}\big (2+\left\| \partial _xf_0\right\| _{L^\infty }\big )^2}\cdot \end{aligned}$$

Consequently, for any $t\le t^\star$, (25) gives

$$\begin{aligned} \frac{{{\,\mathrm{d}\,}}}{{{\,\mathrm{d}\,}} t}A(t)+C_1\frac{B(t)}{\big (2+\left\| \partial _{x}f_0\right\| _{L^\infty }\big )^2} \le \frac{2C_2}{\sqrt{K}\big (2+\left\| \partial _xf_0\right\| _{L^\infty }\big )^2}B(t). \end{aligned}$$

By definition of K, we have

$$\begin{aligned} K\ge \frac{16C_2^2}{C_1^2}, \end{aligned}$$

so, for any $t\le t^\star$,

$$\begin{aligned} \frac{{{\,\mathrm{d}\,}}}{{{\,\mathrm{d}\,}} t}A(t)+\frac{C_1}{2}\frac{B(t)}{\big (2 +\left\| \partial _{x}f_0\right\| _{L^\infty }\big )^2}\le 0. \end{aligned}$$

(27)

Integrate this on the time interval $[0,t^*]$, to infer that

$$\begin{aligned} \sup _{t\in [0,t^*]}A(t)+\frac{C_1}{2\big (2 +\left\| \partial _{x}f_0\right\| _{L^\infty }\big )^2} \int _{0}^{t^*}B(t){{\,\mathrm{d}\,}} t\le A(0). \end{aligned}$$

Using the smallness assumption (22), the previous inequality (27) implies at once that

$$\begin{aligned}&\sup _{t\in [0,t^*]}A(t)\le A(0)\le \frac{1}{4K\big (2+\left\| \partial _xf_0\right\| _{L^\infty }\big )^4},\\&\quad \int _{0}^{t^*}B(t){{\,\mathrm{d}\,}} t\le \frac{2\big (2+\left\| \partial _{x}f_0\right\| _{L^\infty }\big )^2}{C_1}A(0)\le \frac{1}{2C_0}. \end{aligned}$$

These are the wanted bootstrap inequalities. As explained above, by connexity, this proves that $I=[0,T]$, which implies the desired results in (23). $\square$

2.6 A Priori Estimates Locally in Time, for Arbitrary Initial Data

Proposition 2.11

Consider $\phi$ satisfying assumptions $(\mathrm {H1})$–$(\mathrm {H3})$ in Definiton 1.2. Let $T>0$ and consider a smooth solution $f\in C^1([0,T],H^{\infty }(\mathbb {R}))$ of the Muskat equation (5). For any $M_0>0$ there exists $\varepsilon _0>0$ and $T_0>0$ such that the following properties holds. If $\varepsilon \in (0,\varepsilon _0]$ and

$$\begin{aligned} \big \Vert \left| D\right| ^{\frac{3}{2},\phi }f(0)\big \Vert _{L^2}^2\le M_0, \end{aligned}$$

then, with $T^*=\min \{T,T_0\}$, there holds

$$\begin{aligned} \sup _{t\in [0,T^*]}A_\phi (t)\le 5M_0,\quad \int _0^{T^*}\mu _\phi (t)^2B_\phi (t){{\,\mathrm{d}\,}} t\le \frac{1}{C_0}, \end{aligned}$$

where $A_\phi$, $B_\phi$, $\mu _\phi$ are defined in (10) while $C_0$ is given by Lemma 2.2.

Proof

For this proof we skip the index $\phi$ and write simply $A,B,\mu$.

Since (see (11)),

$$\begin{aligned} \left\| f(t)\right\| _{\dot{H}^2}\le C\mu (t) B(t)^{\frac{1}{2}}. \end{aligned}$$

We then apply Proposition 2.8 for some fixed parameter $\beta >0$. Then, it follows from (13) that

$$\begin{aligned} \frac{{{\,\mathrm{d}\,}}}{{{\,\mathrm{d}\,}} t}A(t)+C_1\frac{B(t)}{\nu (t)^2}\le C_2 \left( \sqrt{A(t)}+A(t) \right) \mu (t) B(t), \end{aligned}$$

(28)

where

$$\begin{aligned} \nu (t)&= 1+ \left\| \partial _xf_0\right\| _{L^\infty }+ C_0\int _{0}^{t}\mu (\tau )^{2}B(\tau ) {{\,\mathrm{d}\,}} \tau +C_0 \varepsilon ^{\frac{\beta }{2}}\left\| f_0\right\| _{\dot{H}^{\frac{3}{2}}}\\&\quad + C_0 \varepsilon ^{\frac{\beta }{2}}\left[ \sup _{\tau \in [0,t]}\left( 1+\left\| f(\tau )\right\| _{H^{\frac{3}{2}}}\right) ^2\right] \log \left( 2+\int _{0}^{t} \mu (\tau )^{2}B(\tau ){{\,\mathrm{d}\,}} \tau \right) ^{\frac{1}{2}} \\&\qquad \left( \int _{0}^{t} \mu (\tau )^{2}B(\tau ){{\,\mathrm{d}\,}} \tau \right) ^{\frac{1}{2}}. \end{aligned}$$

Given a positive number $T_0$ to be determined, introduce the set

$$\begin{aligned} I(T_0)=\left\{ t\in [0,\min \{T,T_0\}]\,;\, \int _{0}^{t}\mu (\tau )^2B(\tau ){{\,\mathrm{d}\,}} \tau \le \frac{2}{3C_0} \text { and } \sup _{0\le \tau \le t}A(\tau )\le 5M_0\right\} . \end{aligned}$$

We want to prove that $I(T_0)=[0,\min \{T,T_0\}]$. Since 0 belongs to $I(T_0)$ by assumption on the initial data, and since $I(T_0)$ is closed, it suffices to prove that $I(T_0)$ is open. To do so, we consider a time $t^*\in [0,\min \{T,T_0\})$ which belongs to $I(T_0)$. Our goal is to prove that

$$\begin{aligned} \int _{0}^{t^*}\mu (\tau )^2B(\tau ){{\,\mathrm{d}\,}} \tau \le \frac{1}{2C_0} \text { and } \sup _{0\le \tau \le t^*}A(\tau )\le 4M_0. \end{aligned}$$

This will imply at once that $t^*$ belongs to the interior of $I(T_0)$.

As in the previous proof, we use (26) to write

$$\begin{aligned} \sup _{s\in [0,t]}\left\| f(s)\right\| _{H^{\frac{3}{2}}}\le \left\| f_0\right\| _{L^2}+(1 +C\varepsilon ^{\frac{1}{2}}t) \sup _{s\in [0,t]}\left\| f(s)\right\| _{\dot{H}^{\frac{3}{2}}}. \end{aligned}$$

(29)

It $t\le t^*$ with $t^*\in I(T_0)$, then

$$\begin{aligned} \nu (t)&\le 1+ \left\| \partial _xf_0\right\| _{L^\infty }+ \frac{2}{3} +C_0 \varepsilon ^{\frac{\beta }{2}}M_0\\&\quad + C_0 \varepsilon ^{\frac{\beta }{2}}\left( 1+\left\| f_0\right\| _{L^2}+6M_0\right) ^2 \log \left( 2+ \frac{2}{3C_0}\right) ^{\frac{1}{2}} \left( \frac{2}{3C_0}\right) ^{\frac{1}{2}}. \end{aligned}$$

Hence, one can define $\varepsilon _0$ small enough, depending only on $M_0$, $\left\| f_0\right\| _{L^2}$ and the fixed parameter $\beta$, such that if $\varepsilon \le \varepsilon _0$ and if $t^*\in I(T_0)$, then for any $t\in [0,t^*]$, we have

$$\begin{aligned} \nu (t)\le 2+\left\| \partial _xf_0\right\| _{L^\infty }. \end{aligned}$$

Consequently

$$\begin{aligned} \frac{{{\,\mathrm{d}\,}}}{{{\,\mathrm{d}\,}} t}A(t) +C_1 \frac{B(t)}{(2+\left\| \partial _{x}f_0\right\| _{L^\infty })^2} \le C_2\Big (A(t)+\sqrt{A(t)}\Big ) \mu (t)B(t). \end{aligned}$$

Introduce the function

$$\begin{aligned} \mathcal {E}(r,m)\mathrel {:=}\sup _{\rho \ge 0}\left\{ C_2 \big (\sqrt{r}+r\big )\left( \phi \left( \frac{\rho }{r}\right) \right) ^{-1} \rho -\frac{C_1}{2}\frac{\rho }{m}\right\} \cdot \end{aligned}$$

Then, for any $t\in [0,t^*]$, we have

$$\begin{aligned} \frac{{{\,\mathrm{d}\,}}}{{{\,\mathrm{d}\,}} t}A(t)+\frac{C_1}{2}\frac{B(t)}{\big (2+\left\| \partial _{x}f_0\right\| _{L^\infty }\big )^2} \le \mathcal {E}\Big (A(t),\left\| \partial _{x}f_0\right\| _{L^\infty }\Big ). \end{aligned}$$

Assume that the number $T_0$ satisfies

$$\begin{aligned} T_0\le \frac{A(0)}{4\mathcal {E}\big (4A(0),\left\| \partial _{x}f_0\right\| _{L^\infty }\big )}\cdot \end{aligned}$$

Then, for any $t\le t^*$, we get that

$$\begin{aligned} \sup _{\tau \le t} A(\tau )+\frac{C_1}{2}\frac{1}{\big (2+\left\| \partial _{x}f_0\right\| _{L^\infty }\big )^2} \int _{0}^{t}B(\tau ){{\,\mathrm{d}\,}} \tau \le 4A(0). \end{aligned}$$

In particular, for $t=t^*$, this gives

$$\begin{aligned} \sup _{t\le t^*} A(t)\le 4A(0),\quad \int _{0}^{t^*}B(t){{\,\mathrm{d}\,}} t\le \frac{8A(0)}{C_1}\big (2+\left\| \partial _{x}f_0\right\| _{L^\infty }\big )^2. \end{aligned}$$

(30)

To get the result, we must show that

$$\begin{aligned} C_0\int _{0}^{T}\mu (t)^2 B(t){{\,\mathrm{d}\,}} t\le \frac{1}{2}. \end{aligned}$$

(31)

Recall that

$$\begin{aligned} \mu (t)=\left( \phi \left( \frac{B(t)}{A(t)}\right) \right) ^{-1}. \end{aligned}$$

Since $\phi$ is increasing and since $A(t)\le 4A(0)$, we have

$$\begin{aligned} \mu (t)\le \left( \phi \left( \frac{B(t)}{4A(0)}\right) \right) ^{-1}. \end{aligned}$$

Now, we claim that the function $F:[0,+\infty )\rightarrow [0,+\infty )$, defined by

$$\begin{aligned} F(r)=\left( \phi \left( \frac{r}{4A(0)}\right) \right) ^{-1}r, \end{aligned}$$

is increasing. To see this decompose F(r) under the form $F(r)=F_1(r)\left( F_2\left( r\right) \right) ^2$ with

$$\begin{aligned} F_1(r)=\frac{r}{(\log (\lambda _0+r))^2}\quad F_2(r)=\frac{\log (\lambda _0+r)}{\phi (r/4A(0))}\cdot \end{aligned}$$

Then

$$\begin{aligned}&\int _{0}^{t^*}\mu (t)B(t){{\,\mathrm{d}\,}} t\le \int _{0}^{t^*}\left( \phi \left( \frac{B(t)}{4A(0)}\right) \right) ^{-2}B(t){{\,\mathrm{d}\,}} t\\&\quad \le \int _{0}^{t^*}\left( \phi \left( \frac{r}{4A(0)}\right) \right) ^{-2}r {{\,\mathrm{d}\,}} t+\int _{0}^{t^*}\left( \phi \left( \frac{r}{4A(0)}\right) \right) ^{-2}B(t){{\,\mathrm{d}\,}} t\\&\quad \overset{(30)}{\le }t^*\left( \phi \left( \frac{r}{4A(0)}\right) \right) ^{-2}r+ \left( \phi \left( \frac{r}{4A(0)}\right) \right) ^{-2}8A(0)\left( 2+\left\| \partial _{x}f_0\right\| _{L^\infty }\right) ^2, \end{aligned}$$

for any $r\ge 1$. Now we successively determine two numbers $r_0>1$ and $T_0>0$ such that

$$\begin{aligned} C_0\left( \phi \left( \frac{r_0}{4A(0)}\right) \right) ^{-2}8A(0)\left( 2+\left\| \partial _{x}f_0\right\| _{L^\infty }\right) ^2 =\frac{1}{4}, \end{aligned}$$

(32)

and

$$\begin{aligned} T_0\left( \phi \left( \frac{r_0}{4A(0)}\right) \right) ^{-2}r_0=\frac{1}{4}\cdot \end{aligned}$$

(33)

With this choice we get (31) and we obtain that $I(T_0)=[0,\min \{T,T_0\}]$, which is equivalent to the statement of the proposition. $\square$

2.7 Transfer of Compactness

Previously, we have proven a priori estimates for the spatial derivatives. In this paragraph, we gather results from which we will infer estimates for the time derivative as well as for the nonlinearity in the Muskat equation. These estimates serve to pass to the limit the equation (which is needed to regularize the solutions).

The Muskat equation (1) can be written under the form

$$\begin{aligned} \partial _tf+\left| D\right| f = \mathcal {T}(f)f, \end{aligned}$$

(34)

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

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

(35)

We recall the following result from Proposition 2.3 in [2] and from Remark 2.9 and Propositions 2.10 and 2.13 in [3].

Proposition 2.12

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

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

• (ii) One can decompose the nonlinearity under the form

  $$\begin{aligned} \mathcal {T}(f)g =\frac{(\partial _xf)^2}{1+(\partial _xf)^2}\left| D\right| g +V(f)\partial _x g+R(f,g), \end{aligned}$$

  (36)

  where the coefficient V(f) and the remainder term R(f, g) satisfy the following estimates:

  $$\begin{aligned}&\left\| V(f)\right\| _{L^\infty }\le C \int _\mathbb {R}\left| \xi \right| \big \vert \hat{f}(\xi )\big \vert {{\,\mathrm{d}\,}} \xi , \end{aligned}$$

  (37)
  $$\begin{aligned}&\left\| R(f,g)\right\| _{L^2}\le C\Vert g\Vert _{\dot{H}^{\frac{3}{4}}} \left\| f\right\| _{\dot{H}^{\frac{7}{4}}}, \end{aligned}$$

  (38)

  for some absolute constant C. Moreover,

  $$\begin{aligned} \left\| \mathcal {T}(f)f\right\| _{\dot{H}^1}\le C\left( \left\| f\right\| _{\dot{H}^{\frac{3}{2}}}+\left\| f\right\| _{\dot{H}^{\frac{3}{2}}}^2+1 +\left\| V(f)\right\| _{L^\infty }\right) \left\| f\right\| _{\dot{H}^{2}}, \end{aligned}$$

  (39)

  and,

  $$\begin{aligned} \left| \big (V(f)\partial _x g,\left| D\right| g\big )\right| \le C\Big ( \left\| f\right\| _{\dot{H}^2}+\left\| f\right\| _{\dot{H}^{\frac{7}{4}}}^2\Big ) \left\| g\right\| _{\dot{H}^\frac{1}{2}}\left\| g\right\| _{\dot{H}^1}. \end{aligned}$$

  (40)

For later purpose, we need a refinement of (39).

Proposition 2.13

There exists a positive constant $C>0$ such that, for all function $f\in H^2(\mathbb {R})$,

$$\begin{aligned} \left\| \mathcal {T}(f)f\right\| _{\dot{H}^1}\le C \left( 1+\left\| f\right\| _{H^{\frac{3}{2}}}\right) ^2 \log \left( 2+\left\| f\right\| _{\dot{H}^2}^2\right) ^{\frac{1}{2}} \left\| f\right\| _{\dot{H}^{2}}. \end{aligned}$$

(41)

Proof

In view of (39) and (37), it is sufficient to estimate the $L^1$-norm of $\left| \xi \right| \hat{f}$. Write,

$$\begin{aligned} \int _\mathbb {R}|\xi | |\hat{f}| {{\,\mathrm{d}\,}} \xi&=\int _{|\xi |>\lambda } |\xi |^{-1} |\xi |^{2}|\hat{f}| {{\,\mathrm{d}\,}} \xi +\int _{|\xi |\le \lambda } (|\xi |+1)^{-\frac{1}{2}} |\xi |(1+|\xi |)^{\frac{1}{2}}|\hat{f}| {{\,\mathrm{d}\,}} \xi \\&\lesssim \left( \int _{|\xi |>\lambda } \frac{1}{|\xi |^{2}} {{\,\mathrm{d}\,}} \xi \right) ^{\frac{1}{2}} \Vert f\Vert _{\dot{H}^2}\\&\quad + \left( \int _{|\xi |\le \lambda } \frac{1}{(|\xi |+1)} {{\,\mathrm{d}\,}} \xi \right) ^{\frac{1}{2}} \left( \Vert f \Vert _{\dot{H}^{\frac{3}{2}}}+ \Vert f\Vert _{L^2}\right) \\&\lesssim \lambda ^{-\frac{1}{2}} \Vert f\Vert _{\dot{H}^2}+ \log (1+\lambda )^{\frac{1}{2}} \left( \Vert f \Vert _{\dot{H}^{\frac{3}{2}}}+ \Vert f\Vert _{L^2}\right) . \end{aligned}$$

Choosing $\lambda =\left\| f\right\| _{\dot{H}^2}^2$, we obtain

$$\begin{aligned} \int _\mathbb {R}|\xi | |\hat{f}| {{\,\mathrm{d}\,}} \xi \lesssim 1+\log (1+\left\| f\right\| _{\dot{H}^2}^2)^{\frac{1}{2}} \left( \Vert f \Vert _{\dot{H}^{\frac{3}{2}}}+ \Vert f\Vert _{L^2}\right) . \end{aligned}$$

By reporting this in (37) and then using (39), we get the desired result (41). $\square$

By using the equation (34), we deduce at once the following bound.

Corollary 2.14

There exists a non-decreasing function $\mathcal {F}:\mathbb {R}^+\rightarrow \mathbb {R}^+$ such that, for any $T>0$, any $\varepsilon$ and any smooth solution f in $C^1([0,T];H^\infty (\mathbb {R}))$ of the Muskat equation (5), if one sets

$$\begin{aligned} M_\varepsilon (T)= & {} \sup _{t\in [0,T]}\left( \left\| f(t)\right\| _{\dot{H}^{\frac{3}{2}}}^2 +\left\| f(t)\right\| _{L^2}^2\right) \\&+\int _{0}^{T}\left\| f(t)\right\| _{\dot{H}^{2}}^2{{\,\mathrm{d}\,}} t+\left| \log (\varepsilon )\right| ^{-1} \int _0^T\left\| f(t)\right\| _{\dot{H}^{\frac{5}{2}}}^2{{\,\mathrm{d}\,}} t\end{aligned}$$

then,

$$\begin{aligned} \int _{0}^{T}\frac{\left\| \mathcal {T}(f)f\right\| _{\dot{H}^1}^2}{\log \big (2+\left\| \mathcal {T}(f)f\right\| _{\dot{H}^1}\big )}{{\,\mathrm{d}\,}} t\le \mathcal {F}(M_\varepsilon (T)), \end{aligned}$$

(42)

and

$$\begin{aligned} \int _{0}^{T} \frac{\left\| \partial _tf\right\| _{\dot{H}^1}^2}{\log \big (2+\Vert \partial _tf\Vert _{\dot{H}^1}^2)}{{\,\mathrm{d}\,}} t\le \mathcal {F}(M_\varepsilon (T)). \end{aligned}$$

(43)

Proof

Let C be the constant given by Proposition 2.13 and set $\tilde{C}=\max \{C,1\}$. We claim that

$$\begin{aligned} \frac{\left\| \mathcal {T}(f)f\right\| _{\dot{H}^1}^2}{\log \big (2+\left\| \mathcal {T}(f)f\right\| _{\dot{H}^1}\big )}\le \tilde{C}^2 \left( \left\| f\right\| _{\dot{H}^{\frac{3}{2}}} +\left\| f\right\| _{\dot{H}^{\frac{3}{2}}}^2+\left\| f\right\| _{\dot{H}^1}+1\right) ^2\left\| f\right\| _{\dot{H}^{2}}^2. \end{aligned}$$

If $\left\| \mathcal {T}(f)f\right\| _{\dot{H}^1}\le \left\| f\right\| _{\dot{H}^{2}}$, then this is obvious. Otherwise, this follows at once from (41). This implies (42).

The proof of (43) follows from similar argument, using the equation to estimate $\partial _tf$ in terms of $\mathcal {T}(f)f$. $\square$

It follows from the previous results that one can extract from the solutions of the approximate Cauchy problems (5) a sub-sequence converging to a solution of the Muskat equation (1). Since it is rather classical, we do not include the details and refer for instance to [19, 21].

2.8 Uniqueness

To prove the uniqueness of the solution to the Cauchy problem for rough initial data, we shall prove an estimate for the difference of two solutions.

Proposition 2.15

Let $T>0$ and consider two solutions $f_1,f_2$ of the Muskat equation, with initial data $f_{1,0},f_{2,0}$ respectively, satisfying

$$\begin{aligned} f_k\in C^0([0,T];\dot{W}^{1,\infty }(\mathbb {R})\cap \dot{H}^{\frac{3}{2}}(\mathbb {R}))\cap C^1([0,T];\dot{H}^{\frac{1}{2}}(\mathbb {R}))\cap L^2(0,T;\dot{H}^2(\mathbb {R})),\quad k=1,2. \end{aligned}$$

Assume that

$$\begin{aligned} \sup _{t\in [0,T]}\left( \left\| f_k(t)\right\| _{\dot{H}^{\frac{3}{2}}}^2+\left\| f_k(t)\right\| _{\dot{W}^{1,\infty }}^2\right) +\int _0^{T} \left\| f_k\right\| _{\dot{H}^2}^2{{\,\mathrm{d}\,}} t\le M<\infty ,~~k=1,2. \end{aligned}$$

(44)

Then the difference $g=f_1-f_2$ is estimated by

$$\begin{aligned} \sup _{t\in [0,T]} \Vert g(t) \Vert _{\dot{H}^\frac{1}{2}}\le \Vert g(0) \Vert _{\dot{H}^\frac{1}{2}}\exp \left( C(M+1)^{5} \int _0^T \left( \left\| f_1\right\| _{\dot{H}^2}^2+\left\| f_2\right\| _{\dot{H}^2}^2\right) {{\,\mathrm{d}\,}} t\right) . \end{aligned}$$

(45)

Proof

Since $\partial _tf_k+\left| D\right| f_k = \mathcal {T}(f_k)f_k$, it follows from the decomposition (36) of $\mathcal {T}(f_k)f_k$ that the difference $g=f_1-f_2$ satisfies

$$\begin{aligned} \partial _tg+\frac{\left| D\right| g}{1+(\partial _xf_1)^2}&= V(f_1)\partial _x g+R(f_1,g)+\left( \mathcal {T}(f_2+g)-\mathcal {T}(f_2)\right) f_2. \end{aligned}$$

Since g belongs to $C^1([0,T];\dot{H}^{\frac{1}{2}}(\mathbb {R}))$, we may take the $L^2$-scalar product of this equation with $\left| D\right| g$ to get

$$\begin{aligned} \frac{1}{2}\frac{{{\,\mathrm{d}\,}}}{{{\,\mathrm{d}\,}} t}\Vert g \Vert ^{2}_{\dot{H}^\frac{1}{2}}+\int \frac{( \left| D\right| g)^2}{1+(\partial _x f_{1})^2} {{\,\mathrm{d}\,}} x&\le \left| \big (V(f_1)\partial _x g,|D| g\big )\right| +\left\| R(f_1,g)\right\| _{L^2}\left\| g\right\| _{\dot{H}^1}\\&\quad +\left\| \left( \mathcal {T}(f_2+g)-\mathcal {T}(f_2)\right) f_2\right\| _{L^2}\left\| g\right\| _{\dot{H}^1}. \end{aligned}$$

It follows from Proposition 2.12 that

$$\begin{aligned} \frac{{{\,\mathrm{d}\,}}}{{{\,\mathrm{d}\,}} t}\Vert g \Vert ^{2}_{\dot{H}^\frac{1}{2}}+M^{-1}||g||_{\dot{H}^1}^2&\lesssim \left( \left\| f_1\right\| _{\dot{H}^2}+\left\| f_1\right\| _{\dot{H}^{\frac{7}{4}}}^2\right) \left\| g\right\| _{\dot{H}^{\frac{1}{2}}}\left\| g\right\| _{\dot{H}^{1}}\\&\quad +\left\| f_2\right\| _{\dot{H}^{\frac{7}{4}}} \Vert g\Vert _{\dot{H}^{\frac{3}{4}}}|| g||_{\dot{H}^1}. \end{aligned}$$

By Gagliardo-Nirenberg interpolation inequality

$$\begin{aligned} \frac{{{\,\mathrm{d}\,}}}{{{\,\mathrm{d}\,}} t}\Vert g \Vert ^{2}_{\dot{H}^\frac{1}{2}}+M^{-1} ||g||_{\dot{H}^1}^2&\lesssim \left\| f_1\right\| _{\dot{H}^2}\left( 1+\left\| f_1\right\| _{\dot{H}^{\frac{3}{2}}}\right) \left\| g\right\| _{\dot{H}^{\frac{1}{2}}}\left\| g\right\| _{\dot{H}^{1}}\\&\quad +\left\| f_2\right\| _{\dot{H}^{2}}^{\frac{1}{2}} \left\| f_2\right\| _{\dot{H}^{\frac{3}{2}}}^{\frac{1}{2}}\Vert g\Vert _{\dot{H}^{\frac{1}{2}}}^{\frac{1}{2}} \left\| g\right\| _{\dot{H}^1}^{\frac{3}{2}}\\&\lesssim \left\| f_1\right\| _{\dot{H}^2}\left( 1+M\right) \left\| g\right\| _{\dot{H}^{\frac{1}{2}}}\left\| g\right\| _{\dot{H}^{1}}\\&\quad +M^{\frac{1}{2}}\left\| f_2\right\| _{\dot{H}^{2}}^{\frac{1}{2}} \Vert g\Vert _{\dot{H}^{\frac{1}{2}}}^{\frac{1}{2}}\left\| g\right\| _{\dot{H}^1}^{\frac{3}{2}}. \end{aligned}$$

Thus, thanks to Holder’s inequality, one gets

$$\begin{aligned} \frac{{{\,\mathrm{d}\,}}}{{{\,\mathrm{d}\,}} t}\Vert g \Vert ^{2}_{\dot{H}^\frac{1}{2}}+\frac{1}{2M} ||g||_{\dot{H}^1}^2&\le C (M+1)^{5} \left( \left\| f_1\right\| _{\dot{H}^2}^2+\left\| f_2\right\| _{\dot{H}^2}^2\right) \left\| g\right\| _{\dot{H}^{\frac{1}{2}}}^2 \end{aligned}$$

which in turn implies (45). $\square$

2.9 The Cauchy Problem for the Approximate Equations

It remains to prove Proposition 2.1.

Rewrite the equation (5) under the form

$$\begin{aligned} \partial _tf-|\log (\varepsilon )|^{-1}\partial _x^2f =N_\varepsilon (f), \end{aligned}$$

(46)

with

$$\begin{aligned} N_\varepsilon (f)=\frac{1}{\pi }\int _\mathbb {R}\frac{\partial _x\Delta _\alpha f}{1+\left( \Delta _\alpha f\right) ^2}\left( 1-\chi \left( \frac{|\alpha |}{\varepsilon }\right) \right) {{\,\mathrm{d}\,}} \alpha . \end{aligned}$$

The next proposition shows that Equation (46) can be seen as a sub-critical parabolic equation.

Lemma 2.16

There holds

$$\begin{aligned} \begin{aligned} \left\| N_\varepsilon (f)\right\| _{\dot{H}^1} \lesssim \varepsilon ^{\frac{1}{2}}\left\| f\right\| _{\dot{H}^{\frac{5}{2}}}+\left( 1+\left\| f\right\| _{H^{\frac{3}{2}}}\right) ^2 \log \left( 2+\left\| f\right\| _{\dot{H}^2}^2\right) ^{\frac{1}{2}} \left\| f\right\| _{\dot{H}^{2}}, \end{aligned} \end{aligned}$$

(47)

and

$$\begin{aligned} \left\| N_\varepsilon (f)\right\| _{L^2}\le C \left( 1+\left\| f\right\| _{H^{\frac{3}{2}}}\right) ^2. \end{aligned}$$

(48)

Proof

The estimate (47) follows at once from (20) and (21). To prove (48), we decompose $N_\varepsilon (f)=-\left| D\right| f+\mathcal {T}(f)f+R_\varepsilon (f)$ where $\mathcal {T}(f)$ is the operator already introduced in §2.7 and the remainder $R_\varepsilon (f)$ is as defined by (15). Recall from Proposition 2.3 in [2] that

$$\begin{aligned} \left\| \mathcal {T}(f)f\right\| _{L^2}\lesssim \left\| f\right\| _{\dot{H}^1}\left\| f\right\| _{\dot{H}^{\frac{3}{2}}}. \end{aligned}$$

So the wanted conclusion follows from the estimate (16) for $R_\varepsilon (f)$. $\square$

Multiply the latter equation by $(I-\Delta )^{3/2} f$ and integrate in time, to obtain

$$\begin{aligned} \frac{1}{2}\frac{{{\,\mathrm{d}\,}}}{{{\,\mathrm{d}\,}} t}\left\| f\right\| _{H^{\frac{3}{2}}}^2 +|\log (\varepsilon )|^{-1}\left\| \left| D\right| f\right\| _{H^{\frac{3}{2}}}^2 \le \left\| N_\varepsilon (f)\right\| _{H^1}\left\| f\right\| _{H^2}. \end{aligned}$$

(49)

Recall that

$$\begin{aligned} \left\| N_\varepsilon (f)\right\| _{\dot{H}^1} \lesssim \varepsilon ^{\frac{1}{2}}\left\| f\right\| _{\dot{H}^{\frac{5}{2}}}+\left( 1+\left\| f\right\| _{H^{\frac{3}{2}}}\right) ^2 \log \left( 2+\left\| f\right\| _{\dot{H}^2}^2\right) ^{\frac{1}{2}} \left\| f\right\| _{\dot{H}^{2}}, \end{aligned}$$

(50)

Since $\varepsilon ^\frac{1}{2}\ll \left| \log (\varepsilon )\right| ^{-1}$ for $\varepsilon \ll 1$, we can absorb the contribution of $\varepsilon ^{\frac{1}{2}}\left\| f\right\| _{\dot{H}^{\frac{5}{2}}}$ in the right-hand side of (50) by the left-hand side of (49). On the other hand, since $5/2>2$, one can absorb the contribution of the other terms by using the Hölder’s inequality. This proves an a priori estimate for (46). We also get easily a contraction estimate similar to (but much simpler) the one given by Proposition 2.15. Then by using classical tools for semi-linear equations, we conclude that the Cauchy problem for (46) can be solved by standard iterative scheme.