Splash singularities for the one-phase Muskat problem in stable regimes

Splash singularities for the one-phase Muskat problem in stable regimes Angel Castro, Diego C´ordoba, Charles Fefferman and Francisco Gancedo Abstract This paper shows finite time singularity formation for the Muskat problem in a stable regime. The framework we exhibit is with a dry region, where the density and the viscosity are set equal to 0 (the gradient of the pressure is equal to (0,0)) in the complement of the fluid domain. The singularity is a splash-type: a smooth fluid boundary collapses due to two different particles evolve to collide at a single point. This is the first example of a splash singularity for a parabolic problem. 1 Introduction This paper establishes some scenarios where the 2D Muskat problem produces splash singularities; that is to say, we prove that a free boundary evolving by the Muskat problem collapses at a single point while the interface prevails smooth. The situation is stable; we show geometries for initial data where the Rayleigh-Taylor condition holds. The singularities we construct are “splash” singularities in which the interface self-intersects at a single point at the time of breakdown T∗as in Fig. 1. Our previous papers [6] and [7] showed the existence of a splash singularity for the water wave problem. The strategy there is to start with a “splash” singularity at the time T∗then solve water wave equation backwards in time. This yields a solution to the water wave equation in a time interval [T∗−, T∗] that is well behaved at any time [T∗−, T∗) but exhibits a splash at time T∗. In our present setting, we cannot use that strategy because the Muskat problem in the stable regime is parabolic and therefore cannot be solved backwards in time. The importance of this issue is made clear by the fact that water waves can form a “splat” singularity [7] whereas Muskat solution cannot [14]. (A “splat” occurs when, at the time of breakdown, the interface self-intersects along an arc). On the other hand, an analysis of the Muskat problem has in common with our previous work on the water waves a conformal map to the “tilde domain”, see [7]. Recall the Muskat problem, which describes the evolution of two fluids of different nature in porous media. Both fluids are assumed to be immiscible and incompressible, been the most common example for applications the dynamics of water and oil [3]. In two dimensions, the two fluids occupy the connected open set D(t) and R2rD(t) respectively. The characteristics of the fluids are their constant densities and viscosities. Then the step functions ρ(x, t) and µ(x, t) represent the density and viscosity respectively on the media given by: ρ(x, t) = ρ0, x ∈D(t), ρ0, x ∈R2rD(t), 1 arXiv:1311.7653v2 [math.AP] 7 Feb 2015 Figure 1: Splash singularity µ(x, t) = µ0, x ∈D(t), µ0, x ∈R2rD(t), for x∈R2,t≥0 and ρ0,ρ0,µ0,µ0constant values. The main concern is about the dynamics of the common free boundary ∂D(t), which is given by using the experimental Darcy’s law: µ(x, t)v(x, t) = −∇p(x, t)−(0, ρ(x, t)).(1) Here v(x, t)=(v1(x, t), v2(x, t)) is the incompressible velocity ∇·v(x, t)=0,(2) and p(x, t) is the scalar pressure. Above, the permeability of the media and the gravity constant are set equal to one without loss of generality. The Muskat problem is a long standing matter [24] of recognized importance, specially because of its analogies with the evolutions of fluids in Hele-Shaw cells. In that setting the fluids are confined inside two closely parallel flat surfaces in such a way that the dynamics is essentially bidimensional. The Hele-Shaw evolution law is given by 12 b2µ(x, t)v(x, t) = −∇p(x, t)−(0, ρ(x, t)), where bis distance among the surfaces. Therefore, it is possible to observe that for both different scenarios comparable phenomena and properties hold [27]. A main feature of the problem is the appearance of instabilities, which have been shown in different situations [25],[28]. From a contour dynamics point of view, the system of equations for the free boundary is essentially ill-possed from a Hadamard point of view [27],[11]. 2 Although taking into account surface tension effects the system becomes well-possed [16][2], still it shows fingering [17] and exponential growing modes [22]. On the other hand, the Muskat problem is well-possed in stable regimes without surface tension [11],[10],[13]. This situation is reached for the problem when the difference of the gradient pressure jump at the free interface is positive [1]. Then it is said that the RayleighTaylor condition holds. In such case, linearizing the contour equation it is possible to get the following [11]: fL t(α, t) = −σΛfL(α, t),(3) where (α, fL(α, t)) represents the free boundary (α∈R), σis the Rayleigh-Taylor constant and the operator Λ is the square root of the negative Laplacian. Then, the fact that σ > 0 turns the Muskat problem into a parabolic system at the linear level. This fact has been used to prove global in time regularity and instant analyticity for small initial data in different situations [27],[11],[17],[9],[4],[21]. For the case of equal viscosities (µ0=µ0), the Rayleigh-Taylor condition holds when the more dense fluid lies below the interface and the less dense fluid lies above it [11]. In this situation, the regime is stable if the free boundary ∂D is represented by the graph of a function (α, f(α, t)). In particular, it is possible to get a decay of the L∞norm [12] as follows: kf−1 2πZπ −π f0dαkL∞(t)≤ kf0−1 2πZT f0dαkL∞e−Ct, for f(α+ 2π, t) = f(α, t) and with f(α, t)∈L2(R) kfkL∞(t)≤ kf0kL∞(1 + Ct)−1, C =C(f0)>0. It is easy to check that above formulas provide the same rate of decay than equation (3) for fLat the linear level. On the other hand, the L2norm evolution allows to control half derivative for fLdue to the identity kfLk2 L2(t)+2σZt 0kΛ1/2fLk2 L2(s)ds =kfL 0k2 L2, meanwhile at the nonlinear level the following equality kfk2 L2(t) + σ πZt 0ZRZR ln 1 + f(α, s)−f(β, s) α−β2dαdβds =kf0k2 L2, does not give a chance of gaining any regularity [9]. The case of a drop on a solid substrate in porous media have been studied in [23]. This case considers the dynamics of one fluid, also known as the one-phase Muskat problem. The authors show local well-posedness on the problem with estimates independent of the contact angle. In [8] it is shown solutions of the Muskat equation for initial smooth stable graphs with precise geometries which enter in unstable regime becoming non-graph in finite time. The pattern is far from trivial and recently it has been shown to be richer for the inhomogeneous and confined problems (see [20] and references therein). In particular the significance of a 3 turnover (non-graph scenario) is that the Rayleigh-Taylor condition breaks down. Furthermore, [5] there exist smooth initial data in the stable regime for the Muskat problem such that the solutions turn to the unstable regime and later the regularity breaks down. Therefore global-existence is false for some large initial data in the stable regime as the time evolution solutions develop singularities. In this paper we show that the Muskat problem can develop singularities in stable regimes. The singularity is a splash, where for the free boundary given by ∂D(t) = {z(α, t) = (z1(α, t), z2(α, t)) : α∈R},(4) there exist a blow-up time Ts>0 and a point xs∈R2such that xs=z(α1, Ts) = z(α2, Ts) for α16=α2. In particular the curve is regular, and satisfies the chord-arc condition up to the time Ts: |z(α, t)−z(β, t)| ≥ Cca(t)|α−β|,∀α, β ∈R, Cca(t)>0, t ∈[0, Ts). Free boundary incompressible fluid equations can develop splash singularities. This scenario have been shown for the incompressible Euler equations in the water waves form [6],[7] which considers the evolution of a free boundary given by air, with density 0, and water, with density 1, and irrotational velocity. This type of singularities can also be shown for the case with vorticity [15]. Although for the case of two incompressible fluids with positive densities, this setting has been recently ruled out [18]. For Muskat this type of singularities does not also hold in the case in which µ0=µ0and ρ06=ρ0[19]. In this work we show finite time splash singularities with ρ0=µ0= 0: (ρ(x, t), µ(x, t)) = (ρ0, µ0)x∈D(t), (0,0), x ∈R2rD(t),(5) considering the one fluid dynamics with R2rD(t) a dry region. We also yield some geometries for the interface where the Rayleigh-Taylor condition is satisfied, getting rid of unstable situations. The main theorem of the paper is the following: Theorem 1.1 There exist an open set of curves O ⊂ H3, satisfying the chord-arc and Rayleigh-Taylor condition, such that for any z0∈ O the solution of Muskat (1,2,4,5) with z(α, 0) = z0(α)violates the chord-arc condition at a finite time Ts=Ts(z0)>0. In addition, this holds in such a way that z(α1, Ts) = z(α2, Ts)with α16=α2. At the time Tsthe Muskat system (1,2,4,5) breaks down. In the rest of the paper we show the proof of above result splitting it in several sections. In section 2 we construct a family of curves zlfor which there is a unique self-intersection point xswhere xs=zl(α1) = zl(α2) with α16=α2and ∂αzl 1(α1) = ∂αzl 1(α2) = 0. Plugging these curves in Darcy’s law, we get that the Rayleigh-Taylor condition holds. Furthermore, the velocity indicates that the self-intersection point is going to disappear going backward in time. A more general scenario can be found in section 7. In section 3 we show how to make sense the problem with a self-intersecting point, transforming the Muskat problem into a new contour dynamics equation we call P(Muskat). Up to the time of the splash we can 4 recover Muskat from P(Muskat), but at the time of splash P(Muskat) makes sense and it is possible to go further in time. In section 4 we prove local existence of the P(Muskat) system. In section 5 we show a stability result for P(Muskat). Finally, in section 6 we show how the family of curves zl(α) together with the existence and stability for P(Muskat) allow us to conclude the proof of theorem 1.1. 2 Self-intersecting stable curves with suitable sign of velocity In this section we show that there exits a family of splash curves such that the Rayleigh-Taylor condition hold and with velocities which separate the splash point running backward-in-time. First we use Hopf’s lemma to achieve the Rayleigh-Taylor condition. Taking divergence in Darcy’s law (1) we have ∆p(x, t)=0, for any x∈D(t). In addition, the continuity of the pressure on the free boundary [10] and the fact that −∇p(x, t) = (0,0) for any xin the interior of R2rD(t) allow us to get p(z(α, t), t) = 0. On the other hand we consider velocities with mean zero vorticity ∂x1v2−∂x2v1which provides v∈L2(D(t)) and finite energy settings. Approaching to infinity in D(t) yields lim x2→−∞ v(x, t) = 0, and therefore Darcy’s law gives lim x2→−∞ ∂x1p(x, t) = 0, lim x2→−∞ ∂x2p(x, t) = −ρ0. It is possible to find that p(x, y)∼ −ρ0x2+c(t) when x2→ −∞ and to conclude that the pressure is positive in D(t) by the maximum principle for harmonic functions. In this situation we can apply Hopf’s lemma to obtain that −∇p(z(α, t), t)·∂⊥ αz(α, t)≥c(t)>0,(6) where ∂⊥ αz(α, t)=(−∂αz2(α, t), ∂αz1(α, t)) is the normal vector pointing out the domain D(t). Next we deal with curves zl(α) with a splash point xs=zl(α1) = zl(α2) for α16=α2 where ∂αzl 1(α1) = ∂αzl 1(α2)=0. We show that this configuration provides a sign for the velocity at xs. Taking the trace of Darcy’s law to the surface and multiplying by ∂⊥ αzl(α) we have that µ0v(zl(α)) ·∂⊥ αzl(α) = −∇p(zl(α)) ·∂⊥ αzl(α)−ρ0∂αzl 1(α). 5 Thanks to our choice of the splash curve it must be satisfied v(zl(αi)) ·∂⊥ αzl(αi) = −µ−1 0∇p(zl(αi)) ·∂⊥ αzl(αi)≥c > 0, i = 1,2,(7) where again we have used Hopf’s lemma (6). It is clear that (7) implies that the velocity separates the splash point backwards in time. In figure 1 we give a graphic sketch of the kind of splash singularities we are considering. Theses splash curves yield the simplest scenario we can consider. In section 7 we show the existence of different geometries that give rise to a splash singularity for the one-phase Muskat problem. 3 Transformation to a non-splash scenario This section is devoted to transform the system into a new contour evolution equation where we handle the splash singularity. We consider solutions of Muskat satisfying (1,2,4,5) for regular z(α, t) satisfying the chord-arc condition. Taking limit as x→z(α, t) from D(t) we find v(z(α, t), t) = u(α, t), where u(α, t) = BR(z, ω)(α, t) + ω(α, t) 2 zα(α, t) |zα(α, t)|2. BR stands for the Birkhoff-Rott integral, which is given by BR(α, t) = BR(z, ω)(α, t) = 1 2πPV ZR (z(α, t)−z(α−β, t))⊥ |z(α, t)−z(α−β, t)|2ω(α−β, t)dβ, (8) and ωis the amplitude of the vorticity concentrated on the free boundary: (∂x1v2−∂x2v1)(x, t) = ω(β, t)δ(x=z(β, t)). By approaching to the contour in Darcy’s law and taking the dot product with zα(α, t) it is easy to relate the amplitude of the vorticity and the free boundary by an elliptic implicit equation: ω(α, t) = −2BR(z, ω)(α, t)·∂αz(α, t)−2ρ0 µ0 ∂αz2(α, t).(9) We have the dynamics given by the following contour equation zt(α, t) = u(α, t) + c(α, t)∂αz(α, t)(10) where crepresents reparameterization freedom. See [10] for a detail derivation of the system. In a periodic setting in the x1direction, we will transform the system with the conformal map: P(w) = tan(w/2)1/2, w ∈C. 6 Above, the branch of the square root is chosen in such a way that crosses the self-intersecting point of the zl(α) curve from before section. Therefore P(zl(α)) becomes a one-to-one curve. We then consider by this new transformation the curve ˜z(α, t) = P(z(α, t)). This provides easily ˜zα(α, t) = ∇P(z(α, t))zα(α, t), and ˜zt(α, t) = ∇P(z(α, t))zt(α, t) = ∇P(z(α, t))(u(α, t) + c(α, t)zα(α, t)) = =∇P(z(α, t))u(α, t) + c(α, t)˜zα(α, t). For the potential φ(x, t) (∇φ(x, t) = v(x, t)) we define in the tilde domain ˜ φ(˜x, t) = φ(x, t). Then v(x, t) = ∇φ(x, t)=(∇˜ φ)(P(x), t)∇P(x) = ∇P(x)T(∇˜ φ)(P(x), t). Taking limit we find u(α, t) = ∇P(z(α, t))T(∇˜ φ)(P(z(α, t)), t) = ∇P(z(α, t))T˜u(α, t), where ˜u(α, t) = ∇˜ φ(˜z(α, t), t). It yields ˜zt(α, t) = Q2(α, t)˜u(α, t) + c(α, t)˜zα(α, t),(11) where Q2is given by ∇P(z(α, t))∇P(z(α, t))T=Q2(α, t)I, and Iis the 2 ×2 identity matrix. In other words Q2(α, t) = dP dw (z(α, t)) 2=dP dw (P−1(˜z(α, t))) 2.(12) Next we consider the velocity ˜vdefined on the whole space by ˜v(˜x, t) = ∇˜ φ(˜x, t) = 1 2πPV ZR (˜x−˜z(α−β, t))⊥ |˜x−˜z(α−β, t)|2˜ω(α−β, t)dβ, where (∂˜x1˜v2−∂˜x2˜v1)(˜x, t) = ˜ω(β, t)δ(˜x= ˜z(β, t)), in a distributional sense. Approaching to the free boundary it is possible to obtain ˜u=BR(˜z, ˜ω) + ˜ω 2|˜zα|2˜zα.(13) In order to close the system we integrate Darcy’s law to find µ0φ(z(α, t), t) = −p(z(α, t), t)−ρ0z2(α, t) = −ρ0z2(α, t), due to the continuity of the pressure at the free boundary and the vacuum state. The conformal map Pprovides µ0˜ φ(˜z(α, t), t) = −ρ0P−1 2(˜z(α, t)).(14) 7 Taking one derivative and identity (13) allow us to find µ0(BR(˜z, ˜ω)·˜zα+˜ω 2) = −ρ0∂α(P−1 2(˜z)). We rewrite above identity as ˜ω(α, t) = −2BR(˜z, ˜ω)(α, t)·˜zα(α, t)−2ρ0 µ0 ∂α(P−1 2(˜z(α, t))).(15) Next we will pick a tangential component to get |˜zα|depending only on the variable t. Identities (11) and (13) give ˜zt(α, t) = Q2(α, t)BR(˜z, ˜ω)(α, t) + ˜c(α, t)˜zα(α, t),(16) for ˜c=Q2˜ω/(2|˜zα|2) + c. This provides ˜c(α, t) = α+π 2πZπ −π ∂β(Q2BR)(β, t)·˜zβ(β, t) |˜zβ(β, t)|2dβ −Zα −π ∂β(Q2BR)(β, t)·˜zβ(β, t) |˜zβ(β, t)|2dβ. (17) We end up with a contour equation given by (15,16,17). Finally we will find the Rayleigh-Taylor condition in terms of ˜z. We define ˜p(˜x, t) = p(x, t) to obtain with Darcy’s law −∇˜p(˜x, t) = µ0∇˜ φ(˜x, t) + ρ0∇P−1 2(˜x). Approaching to the free boundary, we find easily ˜σ(α, t) = −∇˜p(˜z(α, t), t)·˜z⊥ α=µ0BR(˜z, ˜ω)·˜z⊥ α+ρ0∇P−1 2(˜z(α, t)) ·˜z⊥ α.(18) 4 Local-existence in the tilde domain This section is devoted to prove local existence for ˜zsolutions of (15,16,17) with ˜z∈C([0, T]; Hk) with k≥3. We show the proof for k= 3 with the rest of the cases being analogous. In order to simplified the exposition we suppress the time variable and the tilde in the equation. We follow the same strategy as in [10]. We define q0= (0,0), q1= ( 1 √2,1 √2), q2= (−1 √2,1 √2), q3= (−1 √2,−1 √2), q4= ( 1 √2,−1 √2), which are the singular points of the P−1conformal map. We set z(α, t) to hold ˜z(α, t)6=ql for l= 0, ..., 4. In order to get this we fix D(0) so that dP dw (w)6= 0 for any w∈D(0) without loss of generality. We will check that this property remains true for short time. Next we define the quantity Ek(z, t) = Ek(t) = kzk2 Hk(t) + kF(z)k2 L∞(t) + 1 m(Q2σ)(t)+ 4 X l=0 1 m(ql)(t),(19) 8 where F(z) = |β| |z(α)−z(α−β)|, α, β ∈[−π, π], and m(Q2σ)(t) = min α∈TQ2(α, t)σ(α, t), m(ql)(t) = min α∈T|z(α, t)−ql|. We shall show a proof of the following result: Proposition 4.1 Let z(α, t)be a solution of (15,16,17). Then, the following estimate holds: d dtEk(t)≤C(Ek(t))p for k≥3. The constants Cand pdepend only on k. Below we will show the proof for k= 3, being the rest of the cases analogous. These a priori estimates will lead to a local existence result for the contour equation in the tilde domain. We refer the reader to [10] in order to obtain d dtkzk2 L2(t) + kF(z)k2 L∞(t) + 1 m(Q2σ)(t)+ 4 X l=0 1 m(ql)(t)≤C(Ek(t))p, as a similar approach can be made. Next we check d dtk∂3 αzk2 L2(t)=2Z∂3 αz(α)·∂3 αzt(α)dα. We can estimate most of the terms as in [10]. We also quote [6] for dealing with the Q2 factor. This term do not introduce any unbounded character as kQ2kHk≤C(Ek(t))p. We will show how to deal with the unbounded and therefore singular terms. We find d dtk∂3 αzk2 L2(t)≤C(Ek(t))p+I, for I=Z∂3 αz(α)·Q2(α)1 πZ(z(α)−z(α−β))⊥ |z(α)−z(α−β)|2∂3 αω(α−β)dβdα. We get I≤C(Ek(t))p+II where II =Z∂3 αz(α)·z⊥ α(α) |zα(α)|2Q2(α)H(∂3 αω)(α)dα. Identity H(∂α) = Λ allows us to rewrite II as follows II =1 |zα(α)|2ZΛ(∂3 αz·z⊥ αQ2)(α)∂2 αω(α)dα. 9 At this point it is easy to get I4≤Ckzk2 H1. If one gathers above inequalities the following is obtained: d dtkzk2 L2≤Ckzk2 H1. Next step is to analyzed d dtkzαk2 L2= 2 Z∂αz·∂αztdγ =I5+I6+I7+I8+I9, where I5= 2 Zzα·∂αQ2 x(BRx−BRy)dα, I6= 2 Zzα·Q2 x∂α(BRx−BRy)dα I7= 2 Zzα·∂α(Q2 x−Q2 y)BRydα, I8= 2 Zzα·∂α(cx−cy)xαdα, I9= 2 Zzα·∂αcyzαdα. It is easy to get I5≤Ckzk2 H1. For I6we consider I6=I6,1+I6,2+I6,3+I6,4+I6,5+I6,6where I6,1=1 πZzα·Q2 xZx⊥ − |x−|2ω0 αdβdα, I6,2=1 πZzα·Q2 xZx⊥ − |x−|2−y⊥ − |y−|2ζ0 αdβdα, I6,3=1 πZzα·Q2 xZ∂αz⊥ − |x−|2γ0dβdα, I6,4=1 πZzα·Q2 xZ∂αy⊥ −γ0 |x−|2−ζ0 |y−|2dβdα, I6,5=−2 πZzα·Q2 xZx⊥ − |x−|4x−·∂αz−ζ0dβdα, and I6,6=−2 πZzα·Q2 xZx⊥ − |x−|4x−·∂αy−γ0−y⊥ − |y−|4y−·∂αy−ζ0dβdα. It is easy to get I6,2+I6,4+I6,6≤Ckzk2 H1. We split further: I6,3=I6,3,1+I6,3,2to find I6,3,1=1 2πZzα·Z∂αz⊥ − |x−|2[Q2 xγ0−(Q2 x)0γ]dβdα, and I6,3,2=1 2πZzα·Z∂αz⊥ − |x−|2[Q2 xγ0+ (Q2 x)0γ]dβdα. 16 We find as before I6,3,1≤Ckzk2 H1. Changing variables one could obtain I6,3,2=1 4πZ Z ∂αz−·∂αz⊥ − |x−|2[Q2 xγ0+ (Q2 x)0γ]dβdα = 0. We are done with I6,3. The term I6,5is decomposed as follows I6,5,1=−2 πZzα·Q2 xZhζ0x⊥ − |x−|4x−−ζx⊥ α |xα|44 sin2(β/2)xαi·∂αz−dβdα, I6,5,2=−2Zzα·Q2 xζx⊥ α |xα|4xα·Λ(zα)dα. Therefore it yields I6,5,1≤Ckzk2 H1. We use the fact that Λ = H(∂α) and the identity xα·∂2 αz=−xα·∂2 αy=−zα·∂2 αy to rewrite I6,5,2=2 Zzα·Q2 xζx⊥ α |xα|4[Λ(xα·zα)−xα·Λ(zα)] −2Zzα·Q2 xζx⊥ α |xα|4H(∂2 αx·zα)dα + 2 Zzα·Q2 xζx⊥ α |xα|4H(zα·∂2 αy)dα Since the commutator estimate for Λ gives I6,5,2≤Ckzk2 H1, we are done with I6,5. It remains to deal with I6,1where we have to find the Rayleigh-Taylor condition. We decompose further I6,1,1=1 πZzα·Q2 xZx⊥ − |x−|2−x⊥ α |xα|22 tan(β/2)ω0 αdβdα, I6,1,2=ZQ2 xzα·x⊥ α |xα|2H(ωα)dα. As before I6,1,1≤Ckzk2 H1. Next we will decompose ωα, pointing out first the bounded terms, and dealing later with the unbounded. We take ωα=G3+G4+G5+G6where G3=−2BRx·∂2 αx+ 2BRy·∂2 αy, G4=−2∂αBRx·xα+ 2∂αBRy·yα G5=−2ρ0 µ0 (∂α(∇P−1 2(x))·xα−∂α(∇P−1 2(y))·yα), G6=−2ρ0 µ0 (∇P−1 2(x)·∂2 αx−∇P−1 2(y)·∂2 αy). 17 We split further G3=G3,1+G3,2 G3,1=−2BRx·∂2 αz, G3,2= 2(BRy−BRx)·∂2 αy to obtain as before kG3,2kL2≤CkzkH1. The term G3,1is part of the unbounded characters. We continue by taking G4=G4,1+ G4,2+G4,3+G4,4+G4,5+G4,6where G4,1=−1 πZ∂αz⊥ − |x−|2γ0dβ ·xα, G4,2=−1 πZ∂αy⊥ −·γ0 |x−|2xα−ζ0 |y−|2yαdβ, G4,3=2 πZx⊥ − |x−|4·xαx−·∂αz−γ0dβ, G4,4=2 πZx⊥ − |x−|4·xαx−γ0−y⊥ − |y−|4·yαy−ζ0·∂αy−dβ, G4,5=−1 πZx⊥ − |x−|2ω0 αdβ ·xα, G4,6=−1 πZx⊥ − |x−|2·xα−y⊥ − |y−|2·yαζ0 αdβ. Next, G4,1joints the unbounded terms and kG4,2kL2+kG4,4kL2+kG4,6kL2≤CkzkH1. It is possible to obtain a kernel of degree −1 applied to ∂αzin G4,3as follows: G4,3=2 πZx⊥ −−x⊥ αβ |x−|4·xαx−·∂αz−γ0dβ. Therefore kG4,3kL2≤CkzkH1. Since G4,5=−1 πZx⊥ −−x⊥ αβ |x−|2ω0 αdβ ·xα we obtain in an analogous way kG4,5kL2≤CkωkL2≤CkzkH1. For G5it is easy to get kG5kL2≤CkzkH1, but the term G6has to be decomposed as follows: G6,1=−2ρ0 µ0∇P−1 2(x)·∂2 αz, G6,2=−2ρ0 µ0 (∇P−1 2(x)−∇P−1 2(y)) ·∂2 αy. G6,1remains unbounded and kG6,2kL2≤CkzkH1, 18 easily. Thanks to all this decomposition we find I6,1,2≤Ckzk2 H1+I1 6,1,2+I2 6,1,2+I3 6,1,2, where I1 6,1,2=ZQ2 xzα·x⊥ α |xα|2H(G3,1)dα, I2 6,1,2=ZQ2 xzα·x⊥ α |xα|2H(G4,1)dα, and I3 6,1,2=ZQ2 xzα·x⊥ α |xα|2H(G6,1)dα. In I1 6,1,2and I3 6,1,2the Rayleigh-Taylor condition will show up. For I2 6,1,2we consider the splitting I2,1 6,1,2=ZH(Q2 xzα·x⊥ α |xα|2)1 πZ∂αz⊥ −γ0 |x−|2−γ |xα|24 sin2(β/2)dβ ·xαdα, I2,2 6,1,2=ZH(Q2 xzα·x⊥ α |xα|2)γ |xα|2Λ(∂αz⊥)·xαdα, using that His skew-adjoint. First term satisfies I2,1 6,1,2≤Ckzk2 H1. For the second one we use the commutator estimates to find I2,2 6,1,2≤Ckzk2 H1+ZH(Q2 xzα·x⊥ α |xα|2)Λ( γ |xα|2∂αz⊥·xα)dα. The fact that H2=−Iyields I2,2 6,1,2≤Ckzk2 H1+ZQ2 xzα·x⊥ α |xα|2∂α(γ |xα|2∂αz⊥·xα)dα. In the integral above we expand the derivative, to find out that it is possible to integrate by parts in ∂α(zα·x⊥ α). This yields I2,2 6,1,2≤Ckzk2 H1,and therefore I2 6,1,2≤Ckzk2 H1. Next we consider I1 6,1,2=−ZQ2 xzα·x⊥ α |xα|2H(2BRx·∂2 αz)dα, for which we use the commutator for the Hilbert transform kH(g∂αf)−gH(∂αf)kL2≤CkgkC1,1 3kfkL2, to find I1 6,1,2≤ − 2 |xα|2ZQ2 xzα·x⊥ αBRx·H(∂2 αz)dα, 19 Next we split above integral by components: I1,1 6,1,2=2 |xα|2ZQ2 x∂αz1∂αx2BRx1·H(∂2 αz1)dα, (25) I1,2 6,1,2=2 |xα|2ZQ2 x∂αz1∂αx2BRx2·H(∂2 αz2)dα, I1,3 6,1,2=−2 |xα|2ZQ2 x∂αz2∂αx1BRx1·H(∂2 αz1)dα, I1,4 6,1,2=−2 |xα|2ZQ2 x∂αz2∂αx1BRx2·H(∂2 αz2)dα. (26) The commutator for the Hilbert transform allows us to obtain I1,2 6,1,2≤Ckzk2 H1+2 |xα|2ZQ2 x∂αz1BRx2·H(∂αx2∂2 αz2)dα, and together with identity ∂αx2∂2 αz2=−∂αx1∂2 αz1−∂αz·∂2 αy provides I1,2 6,1,2≤Ckzk2 H1−2 |xα|2ZQ2 x∂αz1BRx2·H(∂αx1∂2 αz1)dα. The commutator estimate yields I1,2 6,1,2≤Ckzk2 H1−2 |xα|2ZQ2 x∂αz1∂αx1BRx2·H(∂2 αz1)dα. (27) In a similar manner we find I1,3 6,1,2≤Ckzk2 H1+2 |xα|2ZQ2 x∂αz2∂αx2BRx1·H(∂2 αz2)dα. (28) Adding (25), (27), (28) and (26) we find I1 6,1,2≤Ckzk2 H1−2 |xα|2ZQ2 xBRx·x⊥ αzαΛ(zα)dα. (29) Next I3 6,1,2=−2ρ0 µ0ZQ2 xzα·x⊥ α |xα|2H(∇P−1 2(x)·∂2 αz)dα, and a decomposition in components as before provides I3 6,1,2≤Ckzk2 H1−2ρ0 µ0|xα|2ZQ2 x∇P−1 2(x)·x⊥ αzα·Λ(zα)dα. (30) Adding (29) and (30) we find I6,1,2≤Ckzk2 H1+I1 6,1,2+I3 6,1,2≤Ckzk2 H1−21 µ0|xα|2ZQ2 xσxzα·Λ(zα)dα. 20 The positivity of the Rayleigh-Taylor condition for the curve xgives I6,1,2≤Ckzk2 H1, I6,1≤Ckzk2 H1,and finally I6≤Ckzk2 H1. We find easily I7≤Ckzk2 H1. For I8we consider I8= 2 Zzα·(cx−cy)∂2 αxdα + 2 Zzα·xα∂α(cx−cy)dα, and integrate by parts to find I8=−2Z∂2 αz·xα(cx−cy)dα. The fact that I8= 2 Z∂2 αy·zα(cx−cy)dα, allows us to deal with I8as for I3to get I8≤Ckzk2 H1. Finally, integration by parts provides I9=Z|zα|2∂αcydα ≤Ckzk2 H1. 6 Applying perturbative argument and concluding the proof Finally, we will applied a perturbative argument to conclude the proof of theorem 1.1. Consider a curve zl(α) as in section 2 and P(zl(α)) an initial datum for P(Muskat). Then we get a solution ezl(α, t)∈C([0, T], H3) given by using section 3. Next we consider a perturbation of zl(α), the curve z0(α), for which the Rayleigh-Taylor and chord-arc conditions holds. Furthermore, the velocity given by Darcy’s law for z0(α) shows that two different branches of the interface are going to approach as time goes forward. Next we take P(z0(α)) as an initial datum for P(Muskat), getting a solution ez(α, t)∈C([0, T], H3). The stability result in section 5 gives kez−ezlkH1(t)≤C(sup [0,T] E3(ez, t) + sup [0,T] E3(ezl, t))kP(z0(α)) −P(zl(α))kH1 and therefore kez−ezlkH1(t)≤C(sup [0,T] E3(ez, t) + sup [0,T] E3(ezl, t))kz0(α)−zl(α)kH1. Here we point out that the time of existence in section 3 is independent of the smallness of kz0(α)−zl(α)kH1. Since the transformation P−1is well define for ˜zand the fact that zl self-intersect at a point allows us to conclude that in the evolution of z=P−1(ez) there exists a finite time such that zhas to break down with a splash singularity. 21 7 A remark on the family of splash singularities The scenario in section 2 is the simplest one to obtain a splash singularity. However the one-phase Muskat problem can develop this kind of point-wise collapse for more geometries. In order to check that we proceed as follows. Let z(α) be a splash curve with α16=α2such that z(α1) = z(α2) and |∂αz(α)|>0 for every α. To consider a different situation than in section 2, we also assume that ∂αz1(α1)6= 0. We make the following distinction between α1 and α2: There exist a neighborhood Uα1of α1and a neighborhood Uα2of α2such that, if z1(β1) = z1(β2) for β1∈Uα1and β2∈Uα2, then z2(β1)≤z2(β2). Roughly speaking, we just mean that z(α2) is the upper splash point and z(α1) is the lower splash point. Let us analyze the normal velocity at α2and α1. For α2we have µ0u(α2)·n(α2) = −∇p(z(α2)) ·n(α2)−ρ0 ∂αz1(α2) |∂αz(α2)|, where n(α) = ∂⊥ αz(α)/|∂αz(α)|. As in section 2, Hopf’s lemma yields −∇p(z(α2)) ·n(α2)>0 and we consider ∂αz1(α2) |∂αz(α2)|<0. On the other hand, we have µ0u(α1)·n(α1) = −∇p(z(α1)) ·n(α1)−ρ0 ∂αz1(α1) |∂αz(α1)|, with −∇p(z(α1)) ·n(α1)>0, and ∂αz1(α1) |∂αz(α1)|>0. Then the sign of ∂αz(α1) is bad for our purpose. However we can notice that ∂αz1(α1) |∂αz(α1)|=−∂αz1(α2) |∂αz(α2)|, and therefore u(α2)·n(α2)>−u(α1)·n(α1). The last inequality is enough to show that the velocity separates the splash points backward in time. Unfortunately this is not enough to assure that we can produce a splash singularity by using the previous analysis. It is possible to find u(α1)·n(α1) negative. Then the solution would cross the branch of Pbackward in time. This is a mere technical problem that we can solve as follow. 22 Let’s define a velocity v(x1, x2, t) = v(x1, x2−ρ0 µ0 t, t) + (0,ρ0 µ0 ), a density ρ(x1, x2, t) = ρ(x1, x2−ρ0 µ0 t, t), and a viscosity µ(x1, x2, t) = µ(x1, x2−ρ0 µ0 t, t). Therefore ∂tρ(x1, x2, t) =(∂tρ)(x1, x2−ρ0 µ0 t, t)−ρ0 µ0 (∂x2ρ)(x1, x2−ρ0 µ0 t, t) =−v(x1, x2−ρ0 µ0 , t)·(∇ρ)(x1, x2−ρ0 µ0 t, t)−ρ0 µ0 (∂x2ρ)(x1, x2−ρ0 µ0 t, t) =−v(x1, x2−ρ0 µ0 , t)·∇ρ(x1, x2, t)−ρ0 µ0 ∂x2ρ(x1, x2, t) =−v(x1, x2−ρ0 µ0 t, t) + (0,ρ0 µ0 )·∇ρ(x1, x2, t). Thus we have that ρsatisfies ∂tρ+v·∇ρ= 0, and in a similar manner it is easy to get ∂tµ+v·∇µ= 0. On the other hand µ v =−∇p, where we consider p(x1, x2, t) = p(x1, x2−ρ0 µ0 t, t). Then, by using v,ρ,µand p, we can write our Muskat problem as the system ∂tρ+v·∇ρ=0, ∂tµ+v·∇µ=0, µ v =−∇p, ∇·v=0, with the boundary condition lim x2→−∞ v(x1, x2, t)−(0,ρ0 µ0 )= 0. 23 In this new system we find the following: If z(α) is a splash curve such that z(α1) = z(α2) then µ0v(z(α1)) ·n(α1) = −∇p(z(α1)) ·n(α1), µ0v(z(α2)) ·n(α2) = −∇p(z(α2)) ·n(α2), and again we can invoke Hopf’s lemma to obtain that −∇p(z(α1)) ·n(α1)>0,−∇p(z(α2)) ·n(α2)>0. Then, the velocity separates the splash point and u(α1)·n(α1)n(α1) points in the opposite direction to u(α2)·n(α2)n(α2). Therefore we can carry out the same analysis we did for the simpler case of section 2. Acknowledgements AC, DC and FG were partially supported by the grant MTM2011-26696 (Spain). AC was partially supported by the ERC grant 307179-GFTIPFD. CF was partially supported by NSF and ONR grants DMS 09-01040 and N00014-08-1-0678. FG acknowledges support from the Ram´on y Cajal program. References [1] D.M. Ambrose. 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