Contour dynamics of incompressible 3-D fluids in a porous medium with different densities

Contour dynamics of incompressible 3-D fluids in a porous medium with different densities Diego C´ordoba and Francisco Gancedo Abstract We consider the problem of the evolution of the interface given by two incompressible fluids through a porous medium, which is known as the Muskat problem and in two dimensions it is mathematically analogous to the two-phase Hele-Shaw cell. We focus on a fluid interface given by a jump of densities, being the equation of the evolution obtained using Darcy’s law. We prove local well-posedness when the smaller density is above (stable case) and in the unstable case we show ill-posedness. 1 Introduction The evolution of a fluid in a porous medium is an important and interesting topic of fluid mechanics (see [3]). This phenomena is based on an experimental physical principle given by H. Darcy in 1856. Darcy’s law for a 3-D fluid is given by the momentum equation µ κv=−∇p−(0,0,gρ), where vis the incompressible velocity, pis the pressure, µis the dynamic viscosity, κis the permeability of the medium, ρis the liquid density and g is the acceleration due to gravity. A different problem is the motion of a 2-D fluid in a Hele–Shaw cell (see [12]). In this case the fluid is set between two fixed parallel plates. These plates are close enough in such a way that the mean velocity is described by 12µ b2v=−∇p−(0,gρ), where bdenotes the distance between the plates. Considering that the fluid in the porous medium only moves in two directions suppressing one of the variables in the horizontal plane, these two different physical phenomena of fluid dynamics become nevertheless mathematically analogous if we identify the permeability of the medium κand the constant b2/12. The Muskat problem (see [14]) and the two-phase Hele-Shaw flow (see [17]) model the evolution of an interface between two fluids (in a porous medium and in a Hele–Shaw cell respectively) with different viscosities and densities. A lot of information can be found in the literature about both problems (see references in [9] and [13]). These free boundary problems are considered with surface tension using the Laplace–Young condition and also without surface tension in which case the pressures are equal on the interface. With surface tension, in the two dimensional case has been proven that the problems have classical solutions (see [11]). Without surface tension, Siegel, Caflisch and Howison [18] proved ill-posedness in an unstable 2-D case, namely when the higher-viscosity fluid contracts, and they show global-in-time existence of small initial data in the stable case when the higher-viscosity fluid expands. The results rely on the assumption that the Atwood number Aµ=µ1−µ2 µ1+µ2 1 X2X1 X3 Ω2(t) Ω1(t) is nonzero where µ1and µ2are the viscosities of the fluids. In the same year, Ambrose [1] treated the 2-D problem with an initial data fulfilling (ρ2−ρ1)g cos(θ(α, 0)) + 2AµU(α, 0) >0, and the following condition (x(α, 0) −x(α0,0))2+ (y(α, 0) −y(α0,0))2 (α−α0)2>0,(1) where the interface is the curve (x(α, t), y(α, t)), ρ1and ρ2are the densities of the fluids, θis the angle that the tangent to the curve forms with the horizontal and Uis the normal velocity (given by the Birkhoff-Rott integral). We are interested in the case Aµ= 0 that presents the evolution of the interface for different densities. This case, for example, models moist and dry regions in a porous medium. Meanwhile the work of Ambrose is based on the arclength and the tangent angle formulation used by Hou, Lowengrub and Shelley [13], due to the particular form of the vorticity in the case Aµ= 0, we get to parameterize the curve in the two dimensional problem getting the condition (1) for any time (see equation (16)). The free boundary problems given by fluids with different densities are been intensely studied. Notice the classical paper of Taylor [22] and the works of Wu [23] and [24] where the full water wave problem is solved considering the water with positive density and the air with zero density. A study of the two-dimensional case can be found in [2] due to Ambrose and Masmoudi. In order to simplify the notation, we consider µ/k = 12µ/b2= 1 and g = 1. Thus, the 3-D system is written as v(x1, x2, x3, t) = −∇p(x1, x2, x3, t)−(0,0, ρ(x1, x2, x3, t)),(2) where (x1, x2, x3)∈R3are the spatial variables and t≥0 denotes the time. Here ρis defined by ρ(x1, x2, x3, t) = ½ρ1in Ω1(t) ρ2in Ω2(t), with ρ1, ρ2≥0 constants and ρ16=ρ2. We show in section 2 that in this case it is not necessary to assume any condition on the pressure along the interface to obtain the contour equation. Furthermore, we illustrate below that the solutions to this model are weak solutions to the following conservation of mass equation Dρ Dt =ρt+v· ∇ρ= 0,(3) 2 where div v= 0. We notice the similarity with the 2-D vortex patch problem given by the two-dimensional Euler equation where the vorticity is conserved along trajectories in a weak sense. The vorticity is considered to be a characteristic function of a domain. Chemin [8] proved global-in-time existence using paradifferential calculus. A simpler proof can be found in [4] due to Bertozzi and Constantin. In section 2 we show that due to (2) the velocity can be determined from the density by singular integral operators (see [21]). It makes the equation more singular than the 2-D vortex patch problem where the velocity is given by the Biot-Savart law. A singular problem, more analogous to (3), is the evolution of the 2-D quasi-geostrophic (QG) equation for sharp fronts. The QG equation models the dynamics of cold and hot air and the formation of fronts. Here the temperature θis conserved along particle trajectories and the velocity is given by singular integral operators in the following form v= (−R2θ, R1θ), where R1and R2are the Riesz transforms (see [10] for more details of the QG equation). Rodrigo [19] proposed the contour equation of the sharp fronts where the temperature is concentrated in a domain and proved local existence and uniqueness. The paper is organized as follows. In section 2 we derive the contour equation. We show that this equation fulfills the conservation of mass equation in section 3. In section 4 we prove local existence and uniqueness of the stable case. In section 5 we get a family of global solutions of the 2-D stable case with small initial data. Finally, as a consequence of the previous section, in 6 we prove ill-posedness for the 3-D unstable case. 2 The Contour Equation We consider the equation with (x1, x2, x3)∈R3where the fluid has different densities, that is ρis represented by ρ(x1, x2, x3, t) = ½ρ1,{x3> f(x1, x2, t)} ρ2,{x3< f(x1, x2, t)},(4) being fthe interface. Using Darcy’s Law we get curl curl v= (−∂x1∂x3ρ, −∂x2∂x3ρ, (∂2 x1+∂2 x2)ρ). Since div v= 0 we have curl curl v=−∆v, therefore it follows v= (∂x1∆−1∂x3ρ, ∂x2∆−1∂x3ρ, −(∂2 x1+∂2 x2)∆−1ρ).(5) The integral operators ∂x1∆−1and ∂x2∆−1are given by the kernels K1(x1, x2, x3) = 1 4π x1 (x2 1+x2 2+x2 3)3/2, K2(x1, x2, x3) = 1 4π x2 (x2 1+x2 2+x2 3)3/2, respectively, thus the velocity can be expressed by v= (K1∗∂x3ρ, K2∗∂x3ρ, −K1∗∂x1ρ−K2∗∂x2ρ).(6) Since ρsatisfies (4) we have ∇ρ= (ρ2−ρ1)(∂x1f(x1, x2, t), ∂x2f(x1, x2, t),−1)δ(x3−f(x1, x2, t)),(7) 3 where δis the Dirac distribution. Using (6) we obtain v(x1, x2, x3, t) = −ρ2−ρ1 4πPV ZR2 (y1, y2,∇f(x−y, t)·y) [|y|2+ (x3−f(x−y, t))2]3/2dy, (8) where we note x= (x1, x2), y= (y1, y2) and ∇f(x−y, t)·y=∂x1f(x−y, t)y1+∂x2f(x−y, t)y2. In (8) x36=f(x, t) and the principal value is taken at infinity (see [21]). When x3approaches f(x, t) in the normal direction, we get a discontinuity on the velocity due to the fact that the vorticity is concentrated on the interface. Thus, for ε > 0 we define v1(x, f(x, t), t) = lim ε→0v(x1−ε∂x1f(x, t), x2−ε∂x2f(x, t), f(x, t) + ε, t), and v2(x, f(x, t), t) = lim ε→0v(x1+ε∂x1f(x, t), x2+ε∂x2f(x, t), f(x, t)−ε, t). It follows v1(x, f(x, t), t) = −ρ2−ρ1 4πPV ZR2 (y1, y2,∇f(x−y, t)·y) [|y|2+ (f(x, t)−f(x−y, t))2]3/2dy +ρ2−ρ1 2 ∂x1f(x, t)(1,0, ∂x1f(x, t)) 1+(∂x1f(x, t))2+ (∂x2f(x, t))2 +ρ2−ρ1 2 ∂x2f(x, t)(0,1, ∂x2f(x, t)) 1+(∂x1f(x, t))2+ (∂x2f(x, t))2, (9) v2(x, f(x, t), t) = −ρ2−ρ1 4πPV ZR2 (y1, y2,∇f(x−y, t)·y) [|y|2+ (f(x, t)−f(x−y, t))2]3/2dy −ρ2−ρ1 2 ∂x1f(x, t)(1,0, ∂x1f(x, t)) 1+(∂x1f(x, t))2+ (∂x2f(x, t))2 −ρ2−ρ1 2 ∂x2f(x, t)(0,1, ∂x2f(x, t)) 1+(∂x1f(x, t))2+ (∂x2f(x, t))2. (10) The velocity in the tangential directions only moves the particles on the surface f(x, t); i.e., if we rewrite the velocity in the tangential directions, we only make a change on the parametrization and do not alter the shape of the interface. Thus, it follows that v(x, f(x, t), t) = −ρ2−ρ1 4πPV ZR2 (y1, y2,∇f(x−y, t)·y) [|y|2+ (f(x, t)−f(x−y, t))2]3/2dy, (11) due to the fact that the terms ±ρ2−ρ1 2 ∂x1f(x, t)(1,0, ∂x1f(x, t)) 1 + (∂x1f(x, t))2+ (∂x2f(x, t))2, ±ρ2−ρ1 2 ∂x2f(x, t)(0,1, ∂x2f(x, t)) 1 + (∂x1f(x, t))2+ (∂x2f(x, t))2, are in the tangential directions. Moreover, if we add the following tangential terms to (11) ρ2−ρ1 4πPV ZR2 y1 [|y|2+ (f(x, t)−f(x−y, t))2]3/2dy(1,0, ∂x1f(x, t)), 4 ρ2−ρ1 4πPV ZR2 y2 [|y|2+ (f(x, t)−f(x−y, t))2]3/2dy(0,1, ∂x2f(x, t)), we obtain v(x, f(x, t), t) = ρ2−ρ1 4π(0,0, PV ZR2 (∇f(x, t)− ∇f(x−y, t)) ·y [|y|2+ (f(x, t)−f(x−y, t))2]3/2dy).(12) Finally we have the contour equation given by df dt (x, t) = ρ2−ρ1 4πPV ZR2 (∇f(x, t)− ∇f(x−y, t)) ·y [|y|2+ (f(x, t)−f(x−y, t))2]3/2dy, f(x, 0) = f0(x). (13) In the periodic case, we can obtain an equivalent equation to (13) due to the integral operators ∂x1∆−1and ∂x2∆−1can be presented by the kernels Kp 1(x1, x2, x3) = 1 4π(x1 (x2 1+x2 2+x2 3)3/2L(x1, x2, x3) + M(x1, x2, x3)), Kp 2(x1, x2, x3) = 1 4π(x2 (x2 1+x2 2+x2 3)3/2L(x1, x2, x3) + M(x1, x2, x3)), respectively for (x1, x2, x3)∈T2×Rwith T2= [−π, π]2and the functions L, M ∈C∞(T2×R) (see [20] for the kernel of the Riesz potentials on the n-torus). Adding an appropiate function to the singular part of Kp 1and KP 2, we can choose L∈C∞ c(T2×R), L ≥0,supp L⊂ {x2 1+x2 2+x2 3≤4}, L= 1 in {x2 1+x2 2+x2 3≤1}and L(−x1,−x2,−x3) = L(x1, x2, x3). (14) The function Mbelongs to C∞ b(T2×R) and M(0,0,0) = 0. The velocity can be expressed by v= (Kp 1∗∂x3ρ, Kp 2∗∂x3ρ, −Kp 1∗∂x1ρ−Kp 2∗∂x2ρ), and due to (7) it follows (suppressing the dependence on t) v(x1, x2, x3) = −ρ2−ρ1 4πPV ZT2 (y1, y2,∇f(x−y)·y) [|y|2+ (x3−f(x−y))2]3/2L(y, x3−f(x−y))dy −ρ2−ρ1 4πZT2 (1,1,∇f(x−y)·(1,1))M(y, x3−f(x−y))dy, if x36=f(x). Adding a term in the tangential direction we obtain v(x, f(x)) =ρ2−ρ1 4π(0,0,ZT2 (∇f(x)− ∇f(x−y)) ·y [|y|2+ (f(x)−f(x−y))2]3/2L(y, f(x)−f(x−y))dy +ZT2 (∇f(x)− ∇f(x−y)) ·(1,1))M(y, f(x)−f(x−y))dy). Finally we have the contour equation in the periodic case given by df dt (x, t) =ρ2−ρ1 4πZT2 (∇f(x, t)− ∇f(x−y, t)) ·y [|y|2+ (f(x, t)−f(x−y, t))2]3/2L(y, f(x, t)−f(x−y, t))dy +ρ2−ρ1 4πZT2 (∇f(x, t)− ∇f(x−y, t)) ·(1,1))M(y, f(x, t)−f(x−y, t))dy, f(x, 0) =f0(x). (15) 5 We use both formulations throughout the paper. Suppose that the function f(x) only depends on x1 in equation (13). Then the contour equation in the 2-D case (with a 1-D interface) follows df dt (x, t) = ρ2−ρ1 2πPV ZR (∂xf(x, t)−∂xf(x−α, t))α α2+ (f(x, t)−f(x−α, t))2dα, f(x, 0) = f0(x); x∈R. (16) This equation can be obtained in a similar way that (13) using the stream function. Performing a two-dimensional analysis using the stream function, we obtain an equivalent equation to (16) in the two dimensional periodic case as follows df dt (x, t) =ρ2−ρ1 2πZT (∂xf(x, t)−∂xf(x−α, t))α α2+ (f(x, t)−f(x−α, t))2P(α, f(x, t)−f(x−α, t))dα +ρ2−ρ1 2πZT (∂xf(x, t)−∂xf(x−α, t))Q(α, f(x, t)−f(x−α, t))dα, f(x, 0) =f0(x), (17) with P(x1, x2)∈C∞ c(T×R), P ≥0,supp P⊂ {x2 1+x2 2≤4}, P= 1 in {x2 1+x2 2≤1}and P(−x1,−x2) = P(x1, x2). The function Q(x1, x2) belongs to C∞ b(T×R) and Q(0,0) = 0. If we consider the linearized equation of the motion, we obtain a dissipative equation when ρ1< ρ2 (the greater density is below) and an unstable equation when ρ1> ρ2. The unstable linearized equation presents an instability similar to the Kelvin-Helmholtz’s (see [6]). As usual, we note the Riesz transforms in R2(see [21]) R1f(x) = 1 2πP.V. ZR2 y1 |y|3f(x−y)dy, R2f(x) = 1 2πP.V. ZR2 y2 |y|3f(x−y)dy, and the operator Λsfdefined by the fourier transform d Λsf(ξ) = |ξ|sb f(ξ). Suppose that f(x) is uniformly small and we can neglect the terms of order grater than one in (13), then it reduces to the following linear equation ft=ρ1−ρ2 2(R1∂x1f+R2∂x2f) = ρ1−ρ2 2Λf, f(x, 0) = f0(x). (18) Applying the fourier transform we get ˆ f(ξ) = ˆ f0(ξ)eρ1−ρ2 2|ξ|t, and therefore (18) is a dissipative equation when ρ1< ρ2and an ill posed problem in the case ρ1> ρ2 with a general initial data in the Schwartz class. We need an analytic initial data in order to get a well posed problem for ρ1> ρ2. 6 3 The conservation of mass equation We show that if ρis defined by (4) and f(x, t) is convected by the velocity (12) then ρis a weak solution of the conservation of mass equation (3) and conversely. From now on, Ω is equal to R2or T2and ex= (x1, x2, x3). Definition 3.1 The density ρis a weak solution of the conservation of mass equation if for any ϕ∈C∞(Ω ×R×(0, T)),ϕwith compact support in the real case and periodic in (x1, x2)otherwise, we have ZT 0ZΩZR (ρ(ex, t)∂tϕ(ex, t) + v(ex, t)ρ(ex, t)∇ϕ(ex, t))dexdt = 0,(19) where the incompressible velocity vis given by Darcy’s law. Then Proposition 3.2 If f(x, t)satisfies (13) and ρ(ex, t)is defined by (4), then ρis a weak solution of the conservation of mass equation. Furthermore, if ρis a weak solution of the conservation of mass equation given by (4), then f(x, t)satisfies (13). Proof: Let ρbe a weak solution of (3) defined by (4). Integrating by parts we have I=ZT 0ZΩZR ρ∂tϕdexdt =ρ1ZT 0Z{x3>f} ∂tϕdexdt +ρ2ZT 0Z{x3<f} ∂tϕdexdt = (ρ1−ρ2)ZT 0ZΩ ϕ(x, f(x, t), t)∂tf(x, t)dxdt. On the other hand, due to (9) and (10) we obtain J=ZT 0ZΩZR ρv ∇ϕ dexdt =ρ1ZT 0Z{x3>f} v∇ϕdexdt +ρ2ZT 0Z{x3<f} v∇ϕdexdt =ZT 0ZΩ ϕ(x, f(x, t), t)(ρ1v1(x, f(x, t), t)−ρ2v2(x, f(x, t), t))·(∂x1f(x, t), ∂x2f(x, t),−1)dxdt = (ρ1−ρ2)ZT 0ZΩ ϕ(x, f(x, t), t)v(x, f(x, t), t)·(∂x1f(x, t), ∂x2f(x, t),−1)dxdt, where v(x, f(x, t), t) is given by (11). We get J=(ρ1−ρ2)2 4πZT 0ZΩ ϕ(x, f(x, t), t)PV ZR2 (∇f(x, t)− ∇f(x−y, t)) ·y [|y|2+ (f(x, t)−f(x−y, t))2]3/2dydxdt. Then I+J= 0 due to (19). Thus, if we choose ϕ(ex, t) = ϕ(x, t) for x3∈[−kfkL∞,kfkL∞] follows that f(x, t) fulfils (13). Following the same arguments it is easy to check that if f(x, t) satisfies (13), then ρis a weak solution given by (4). ¤ Remark 3.3 Note that due to (5), the velocity satisfies v= (R1(R3ρ), R2(R3ρ),−(R2 1+R2 2)(ρ)), where the operators R1,R2and R3are the Riesz transforms in three dimensions (see [21]). Since ρ∈L∞(Ω ×R)then v belongs to BMO (bounded mean oscillation) and therefore vis in L2(Ω ×R) locally (see [21] for the definitions and properties of the BMO space). 7 4 Local well-posedness for the stable case In this section we prove local existence and uniqueness for the stable case using energy estimates. First we study the case Ω = R2and at the end of the section we give the main differences with the periodic domain. Denote the Sobolev spaces by Hk, the Holder spaces by Ck,δ with 0 ≤δ < 1 the Holder continuity and the hessian matrix of a function f(x) by ∇2f(x). The norms of Hkand Ck,δ are defined as follows kfk2 Hk=kfk2 L2+kΛkfk2 L2, kfkCk,δ =kfkCk+ max i+j=kmax x6=y |∂i x1∂j x2f(x)−∂i x1∂j x2f(y)| |x−y|δ. 4.1 Case Ω = R2 The main theorem in this section is the following Theorem 4.1 Let f0(x)∈Hk(R2)for k≥4and ρ2> ρ1. Then there exists a time T > 0so that there is a unique solution to (13) in C1([0, T]; Hk(R2)) with f(x, 0) = f0(x). We choose ρ2−ρ1= 4πwithout loss of generality, then df dt (x, t) = PV ZR2 (∇f(x, t)− ∇f(x−y, t)) ·y [|y|2+ (f(x, t)−f(x−y, t))2]3/2dy, f(x, 0) = f0(x). (20) We show the proof with k= 4 being analogous for k > 4. We apply energy methods (see [5] for more details). Then 1 2 d dtkfk2 L2(t) = ZR2 f(x)PV ZR2 (∇f(x)− ∇f(x−y)) ·y [|y|2+ (f(x)−f(x−y))2]3/2dydx =ZR2 f(x)Z|y|<1 (∇f(x)− ∇f(x−y)) ·y [|y|2+ (f(x)−f(x−y))2]3/2dydx +ZR2 f(x)PV Z|y|>1 ∇f(x)·y [|y|2+ (f(x)−f(x−y))2]3/2dydx −ZR2 f(x)PV Z|y|>1 ∇f(x−y)·y [|y|2+ (f(x)−f(x−y))2]3/2dydx =I1+I2+I3. The identity ∂xif(x)−∂xif(x−y) = Z1 0 ∇∂xif(x+ (s−1)y)·y ds, yields I1≤CZ1 0 ds Z|y|<1 |y|−1ZR2 |f(x)||∇2f(x+ (s−1)y)| [1 + ((f(x)−f(x−y))2|y|−2]3/2dxdy ≤CZ1 0 ds Z|y|<1 |y|−1dykfkL2X i+j=2 k∂i x1∂j x2fkL2≤Ckfk2 H2. 8 Integrating by parts, the term I2is written I2=3 2Z|y|>1ZR2 |f(x)|2(f(x)−f(x−y))(∇f(x)− ∇f(x−y)) ·y [|y|2+ ((f(x)−f(x−y))2]5/2dxdy ≤CZ|y|>1 |y|−3ZR2 |f(x)|2|f(x)−f(x−y)||y|−1|∇f(x)− ∇f(x−y)| [1 + ((f(x)−f(x−y))2|y|−2]5/2dxdy ≤CkfkL∞kfk2 H1. Integrating by parts in I3, it follows I3=Z|y|>1ZR2 f(x)f(x−y)|y|2−2(f(x)−f(x−y))2 [|y|2+ (f(x)−f(x−y))2]5/2dxdy + 3 Z|y|>1ZR2 f(x)f(x−y)(f(x)−f(x−y))∇f(x−y)·y [|y|2+ (f(x)−f(x−y))2]5/2dxdy −Z|y|=1 ZR2 f(x)f(x−y)|y|2 [|y|2+ (f(x)−f(x−y))2]3/2dxdσ(y) ≤C(kfkL∞+ 1)kfk2 H1. Using Sobolev inequalities, we get finally d dtkfk2 L2(t)≤C(kfk3 H2(t) + 1).(21) We consider the quantity 1 2 d dtk∂4 x1fk2 L2(t) = I4+I5+I6+I7+I8, where I4=ZR2 ∂4 x1f(x)PV ZR2 (∇∂4 x1f(x)− ∇∂4 x1f(x−y)) ·y [|y|2+ (f(x)−f(x−y))2]3/2dydx, I5= 4 ZR2 ∂4 x1f(x)ZR2 (∇∂3 x1f(x)− ∇∂3 x1f(x−y)) ·y ∂x1A(x, y)dydx, I6= 6 ZR2 ∂4 x1f(x)ZR2 (∇∂2 x1f(x)− ∇∂2 x1f(x−y)) ·y ∂2 x1A(x, y)dydx, I7= 4 ZR2 ∂4 x1f(x)ZR2 (∇∂x1f(x)− ∇∂x1f(x−y)) ·y ∂3 x1A(x, y)dydx, I8=ZR2 ∂4 x1f(x)ZR2 (∇f(x)− ∇f(x−y)) ·y ∂4 x1A(x, y)dydx, and A(x, y) = [|y|2+ (f(x)−f(x−y))2]−3/2. The most singular term is I4. In order to estimate it I4=ZR2 ∂4 x1f(x)PV ZR2 ∇∂4 x1f(x, t)·y [|y|2+ (f(x, t)−f(x−y, t))2]3/2dydx −ZR2 ∂4 x1f(x)PV ZR2 ∇∂4 x1f(y, t)·(x−y) [|x−y|2+ (f(x, t)−f(y, t))2]3/2dydx =J1+J2. 9 Due to |L(x1, x2, x3)−1| ≤ C|(x1, x2, x3)|we have K1≤C(1 + kfk4 C2,δ )kfk2 H4+Ck∂4 x1fk2 L2max x¯¯¯PV ZT2 (∇f(x)·y)(y· ∇2f(x)·y) [|y|2+ (∇f(x)·y))2]5/2dy¯¯¯ ≤C(1 + kfk4 C2,δ )kfk2 H4. Using that |f(x)−f(x−y)| ≤ kfkC1|y|and Lx3= 0 in {x2 1+x2 2+x2 3≤4}, we have that K2=−1 2ZT2 |∂4 x1f(x)|2Z|y|>2 1+kfkC1 Lx3(y, f(x)−f(x−y))(∇f(x)− ∇f(x−y)) ·y [|y|2+ (f(x)−f(x−y))2]3/2dydx ≤C(kfk4 C2,δ + 1)kfk2 H4. In order to estimates J2, we integrate by parts getting J2=−ZT2 ∂4 x1f(x)PV ZT2 ∇y(∂4 x1f(y)−∂4 x1f(x)) ·(x−y) [|x−y|2+ (f(x)−f(y))2]3/2L(x−y, f(x)−f(y))dydx =K3+K4+K5+K6 with K3=−ZT2 ∂4 x1f(x)PV ZT2 ∂4 x1f(x)−∂4 x1f(y) [|x−y|2+ (f(x)−f(y))2]3/2L(x−y, f(x)−f(y))dydx, K4=ZT2 ∂4 x1f(x)ZT2 (∂4 x1f(x)−∂4 x1f(y))B(x, y)L(x−y, f(x)−f(y))dydx, K5=ZT2ZT2 ∂4 x1f(x)(∂4 x1f(y)−∂4 x1f(x))C(x, y)Lx3(x−y, f(x)−f(y))dydx, K6=ZT2ZT2 ∂4 x1f(x)(∂4 x1f(y)−∂4 x1f(x))D(x, y)dydx, and B(x, y) = 3(f(x)−f(y))(f(x)−f(y)− ∇f(y)·(x−y)) [|x−y|2+ (f(x)−f(y))2]5/2, C(x, y) = −∇f(y)·(x−y) [|x−y|2+ (f(x)−f(y))2]3/2, D(x, y) = −Lx1(x−y, f(x)−f(y))(x1−y1) + Lx2(x−y, f(x)−f(y))(x2−y2) [|x−y|2+ (f(x)−f(y))2]3/2, Lx1(x1, x2, x3) = ∂x1L(x1, x2, x3), Lx2(x1, x2, x3) = ∂x2L(x1, x2, x3). Exchanging the variables xand ywe can obtain K3≤0. The terms K4, K5and K6can be estimated in a similar way as K1. Therefore, we obtain d dtk∂4 x1fkL2(t)≤C(kfk5 H4(t) + 1), and analogously d dtk∂4 x2fkL2(t)≤C(kfk5 H4(t) + 1). This proves local existence. The proof of the uniqueness is similar to the case Ω = R2. 16 4.3 2-D case Using the equation (16) in Rand the equation (17) in the periodic case we obtain the following theorem Theorem 4.3 Let f0(x)∈Hkfor k≥3and ρ2> ρ1. Then there exists a time T > 0so that there is a unique solution to (16) in C1([0, T ]; Hk)with f(x, 0) = f0(x). The proof is similar to the theorem’s 4.1 and 4.2. 5 Global solution for the 2-D stable case In this section we obtain a family of global solutions for the 2-D stable case with a small initial data with respect to a fixed norm. Indeed, we can get the result with an initial data with the property kf0kHs=∞for s > 3/2. In this section we consider x∈Rand kfka=X|ˆ f(k)|ea|k|. For a > 0, if kfka<∞, then the function fcan be extended analytically on the strip |=z|< a. Furthermore k∂xfka≤Ckfkb b−a,(30) for b > a. The main result of this section is Theorem 5.1 Let f0(x)be a function such that RTf0(x)dx = 0,k∂xf0k0≤εfor εsmall enough and k∂2 xf0kb(t)≤εeb(t)(1 + |b(t)|γ−1),(31) with 0< γ < 1,b(t) = a−(ρ2−ρ1)t/2,ρ2> ρ1and a≤(ρ2−ρ1)t/2. Then, there exists a unique solution of (16) with f(x, 0) = f0(x)and ρ2> ρ1satisfying k∂xfka(t)≤C(ε) exp((2σa −(ρ2−ρ1)t)/4),(32) and k∂2 xfka(t)≤C(ε)(1 + |σa −ρ2−ρ1 2t|γ−1) exp((2σa −(ρ2−ρ1)t)/4),(33) for a≤ρ2−ρ1 2σt,σ= 1 + δand 0< δ < 1. The condition (31) can be satisfied for example if kΛ1+γf0k0< ε and ˆ f0(0) = ˆ f0(1) = ˆ f0(−1) = 0 since k∂2 xf0kb(t)≤eb(t)kΛ1+γf0k0max k≥2|k|1−γeb(t)(|k|−1). In order to prove the theorem, we use the Cauchy-Kowalewski method (see [15] and [16]) in a similar way as Caflisch and Orellana [7] and Siegel, Caflisch and Howison [18]. We show the proof with ρ2−ρ1= 2 without loss of generality. Let g(x, t) and h(x, t) be functions satisfying gt=−Λg, g(x, 0) = f0(x), ht=−Λh+T(g+h), h(x, 0) = 0, (34) 17 with T(f) = −π−1ZR ∂xf(x)−∂xf(x−α) α¡f(x)−f(x−α) α¢2 1 + ¡f(x)−f(x−α) α¢2dα. (35) Then the function f(x, t) = g(x, t) + h(x, t) is a solution of (16). First, we show some properties of the nonlinear operator T. Lemma 5.2 If k∂xfka,k∂xgka<1for a≥0then [ T(f)(0) = 0,(36) k∂xT(f)ka≤C1k∂2 xfkak∂xfka,(37) and k∂xT(f)−∂xT(g)ka≤C2(k∂2 xfka+k∂2 xgka)k∂xf−∂xgka +C2(k∂xfka+k∂xgka)k∂2 xf−∂2 xgka,(38) with C1= 4(1 − k∂xfk2 a)−2and C2= 4(1 − k∂xfk2 a)−2+ (1 − k∂xgk2 a)−2. Proof of the Lemma: Due to the inequality |∂xf(x)| ≤ k∂xfka<1 and by (35) we obtain T(f) = π−1X n≥1 (−1)nZR ∂xf(x)−∂xf(x−α) αµf(x)−f(x−α) α¶2n dα, (39) and T(f) = π−1∂xX n≥1 (−1)n 2n+ 1 ZRµf(x)−f(x−α) α¶2n+1 dα. Thus [ T(f)(0) = 0. Using (39) [ T(f)(k) = π−1X n≥1 (−1)nZRX k0,...,k2n δ( 2n X j=0 kj, k)ik0 2n Y j=0 ˆ f(kj)1−e−iαkj αdα =X n≥1 (−1)nX k0,...,k2n δ( 2n X j=0 kj, k)Mn(k0, ..., k2n)ik0 2n Y j=0 ˆ f(kj), where Mn(k0, ..., k2n) = π−1ZR 2n Y j=0 1−e−iαkj αdα. (40) We get Mn(k0, ..., k2n) = (−1)nmn(k0, ..., k2n) 2n Y j=1 kj, 18 with mn(k0, ..., k2n) = π−1Z1 0 ds1... Z1 0 ds2nZR 1−e−iαk0 αexp³iα 2n X j=1 (sj−1)kj´dα =π−1Z1 0 ds1... Z1 0 ds2nPV ZR exp³iα 2n X j=1 (sj−1)kj´dα α −π−1Z1 0 ds1... Z1 0 ds2nPV ZR exp³−iαk0+iα 2n X j=1 (sj−1)kj´dα α =iZ1 0 ds1... Z1 0 ds2n(sing A−sing B), and A= 2n X j=1 (sj−1)kj, B =−k+ 2n X j=1 sjkj. It follows [ T(f)(k) = X n≥1X k0,...,kn δ( 2n X j=0 kj, k)mn(k0, ..., k2n) 2n Y j=0 kjˆ f(kj), with |mn(k0, ..., k2n)| ≤ 2. We have X k ea|k||k||[ T(f)(k)| ≤ 2X kX n≥1X k0,...,kn ea|k||k|δ( 2n X j=0 kj, k) 2n Y j=0 |kj|| ˆ f(kj)| ≤2X n≥1 (2n+ 1) X k0,...,kn ea|k0||k0|2|ˆ f(k0)| 2n Y j=1 ea|kj||kj|| ˆ f(kj)|, and therefore k∂xT(f)ka≤2k∂2 xfkaX n≥1 (2n+ 1)k∂xfk2n a= 2k∂2 xfka 3k∂xfk3 a− k∂xfk4 a (1 − k∂xfk2 a)2. We get (37) for k∂xfka<1. In a similar way we obtain (38).¤ From (34) gcan be expressed as follows ˆg(k, t) = e−|k|tˆ f0(k), and by the hypothesis of the initial data we have k∂xgka(t)≤εea−t,(41) k∂2 xgka(t)≤εea−t(1 + (t−a)γ−1),(42) for t≥a. We will prove the existence of hby an induction argument on the iterative equation: ∂thn+1 =−Λhn+1 +T(g+hn), hn+1(x, 0) = 0, h0= 0, 19 or [ hn+1(k, t) = Zt 0 e−|k|(t−s)(T(g+hn)) b (k, s)ds, h0= 0. For h1we obtain the following estimates k∂xh1ka(t)≤Zt 0 kT(g)ka+s−t(s)ds =Zt−a 0 +Zt t−a =I1+I2. Using (37), (41) and (42) we get I1≤ea−tZt−a 0 esk∂xT(g)k0(s)ds ≤Cea−tZt−a 0 esk∂2 xgk0(s)k∂xgk0(s)ds ≤Cε2ea−t Zt−a 0 e−s(1 + sγ−1)ds ≤Cε2(1 + 2γ) γea−t. By (41) and (42) we have I2≤CZt t−a k∂2 xgka+s−t(s)k∂xgka+s−t(s)ds ≤Cε2e2(a−t)a(1 + (t−a)γ−1)≤2Cε2 δea−t, due to the inequalities (aδ)γ−1>(t−a)γ−1and aea−t≤δ−1for σa < t. Then k∂xh1ka(t)≤5Cε2 δγ ea−t. Choosing b=a+s−t+t−a 2we have k∂2 xh1ka(t)≤Zt 0 k∂2 xT(g)ka+s−t(s)ds ≤Zt 0 k∂xT(g)kb(s) b−(a+s−t)ds ≤2Zt 0 k∂xT(g)kb t−a ≤³Zt−a 2 0 +Zt t−a 2´=I3+I4, where I3≤2Cea−t 2 t−aZt−a 2 0 esk∂2 xgk0(s)k∂xgk0(s)ds ≤2Cε2ea−t 2 t−aZt−a 2 0 e−s(1 + sγ−1)ds ≤2Cε2 γea−t 2(1 + (t−a)γ−1), and I4≤2C t−aZt t−a 2 k∂2 xgkb(s)k∂xgkb(s)ds ≤2Cε2 t−aea−t(1 + (t−a 2)γ−1)( t 2+a 2) ≤3Cε2 δea−t(1 + (t−a)γ−1). Therefore 20 k∂xh1ka(t)≤5Cε2 δγ ea−t,(43) k∂2 xh1ka(t)≤5Cε2 δγ ea−t 2(1 + (t−a)γ−1).(44) Define rn+1 =hn+1 −hn, Rn= sup 0≤a < ∞ σa < t ³k∂xrnka+k∂2 xrnka 1+(t−σa)γ−1´et−σa 2, and Mn= sup 0≤a < ∞ σa < t ³k∂xhnka+k∂2 xhnka 1 + (t−σa)γ−1´et−σa 2. Take M1=R1≤5Cε2 δγ ≤ε0 2and suppose that Mj, Rj≤ε0 2for any j= 2, ..., n, then k∂xrn+1ka≤Zt 0 k∂xT(g+hn)−∂xT(g+hn−1)ka+s−t(s)ds =Zt−a 0 +Zt t−a =I7+I8. Using (38) we have I7≤Cea−tZt−a 0 es(k∂xrnk0(s)(k∂2 xg+∂2 xhnk0(s) + k∂2 xg+∂2 xhn−1k0(s)) +Cea−tZt−a 0 es(k∂2 xrnk0(s)(k∂xg+∂xhnk0(s) + k∂xg+∂xhn−1k0(s))ds ≤2Cε0Rnea−tZt−a 0 (1 + sγ−1)ds ≤2Cε0 γRneσa−t 2, and I8≤CZt t−a (k∂xrnka+s−t(s)(k∂2 xg+∂2 xhnka+s−t(s) + k∂2 xg+∂2 xhn−1ka+s−t(s))ds +CZt t−a (k∂2 xrnka+s−t(s)(k∂xg+∂xhnka+s−t(s) + k∂xg+∂xhn−1ka+s−t(s))ds ≤2Cε0RnZt t−a eδs−σ(t−a)(1 + (σ(t−a)−δs)γ−1)ds ≤2Cε0 δRnZt−a t−σa e−x(1 + xγ−1)dx ≤6Cε0 γδ Rneσa−t. We obtain for b=a+s−t+σ(t−a)−δs 2σ k∂2 xrn+1ka(t)≤Zt 0 k∂2 x(T(g+hn)−T(g+hn−1))ka+s−t(s)ds ≤Zt 0 k∂x(T(g+hn)−T(g+hn−1)kb(s) b−(a+s−t)ds ≤2σZt 0 k∂x(T(g+hn)−T(g+hn−1)kb σ(t−a)−δs ≤³Zσ σ+1 (t−a) 0 +Zt σ σ+1 (t−a)´=I9+I10. 21 We have σ(t−a)−δs > 2σ σ+1 (t−a) for 0 ≤s≤σ σ+1 (t−a) and therefore we obtain I9≤(σ+ 1)C t−aZσ σ+1 (t−a) 0 eb(k∂xrnk0(s)(k∂2 xg+∂2 xhnk0(s) + k∂2 xg+∂2 xhn−1k0(s)) +(σ+ 1)C t−aZσ σ+1 (t−a) 0 eb(k∂2 xrnk0(s)(k∂xg+∂xhnk0(s) + k∂xg+∂xhn−1k0(s))ds ≤2(σ+ 1)Cε0 t−aRnZσ σ+1 (t−a) 0 eb−s(1 + sγ−1)ds ≤4σCε0 γRnea−t 2(1 + (t−a)γ−1). Using (38) and the induction hypothesis we get I10 ≤4σCε0RnZt σ σ+1 (t−a) eσb−s(1 + (s−σb)γ−1) σ(t−a)−δs ds ≤4σCε0RnZt σ σ+1 (t−a) eδs−σ(t−a) 21+(σ(t−a)−δs 2)γ−1 σ(t−a)−δs ds ≤4σCε0 δRnZσ σ+1 (t−a) t−σa 2 e−x(x−1+xγ−2)dx ≤8σCε0 δ(1 −γ)Rneσa−t 2(1 + (σa −t)γ−1). Due to the estimates for I7, I8, I9and I10 we obtain Rn+1 ≤Cσε0 δγ(γ−1)Rn.(45) Choosing ε0small enough we get Rn+1 ≤1 2Rn≤... ≤1 2nR1≤ε0 2n+1 , and Mn+1 ≤ n+1 X j=1 Rn+1 ≤ε0. Therefore, we obtain the function h= lim n→∞ hnsatisfying k∂xhka(t)≤X n Rneσa−t 2≤ε0eσa−t 2. Taking f(x, t) = g(x, t) + h(x, t), we get (32) for ρ2−ρ1= 2. In order to show the uniqueness, we write the equation (13) for ρ2−ρ1= 2 in the following form ft=−Λf+T(f), f(x, 0) = f0(x). Suppose that there exist two solutions f1and f2with f1(x, 0) = f2(x, 0). Define Rby R= sup 0≤a < ∞ σa < t ³k∂xf1−∂xf2ka+k∂2 xf1−∂2 xf2ka 1+(t−σa)γ−1´et−σa 2. It follows that R≤C(ε)σ δγ(γ−1) Rand for εsmall enough it yields C(ε)σ δγ(γ−1) <1 and therefore f1=f2. 22 6 Ill-posedness for the unstable case Here we show ill-posedness for the unstable case ρ1> ρ2. We use the global solution for the 2-D stable case f(x1, t) satisfying (32) with kΛ1+γf0k0< C and kΛ1+γ+ζf0k0=∞for γ, ζ > 0. Making a change of variables, we define fλ(x1, t) = λ−1f(λx1,−λt +λ1/2) obtaining {fλ}λ>0a family of solutions to the unstable case. 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