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

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1 Contour dynamics of incompressible 3-D fluids in a porous medium with different densities Diego Córdoba Francisco Gancedo September 4, 6 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 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) in the unstable case we show ill-posedness. 1 Introduction The evolution of a fluid in a porous medium is an important interesting topic of fluid mechanics (see [3]). This phenomena is based on an experimental physical principle given by H. Darcy in Darcy s law for a 3-D fluid is given by the momentum equation µ v = p (,, g ρ), κ where v is the incompressible velocity, p is the pressure, µ is the dynamic viscosity, κ is the permeability of the medium, ρ is the liquid density g is the acceleration due to gravity. A different problem is the motion of a -D fluid in a Hele Shaw cell (see [1]). 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 1µ v = p (, g ρ), b where b denotes 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 κ the constant b /1. The Muskat problem (see [14]) the two-phase Hele-Shaw flow (see [17]) model the evolution of an interface between two fluids (in a porous medium in a Hele Shaw cell respectively) with different viscosities densities. A lot of information can be found in the 1

2 literature about both problems (see references in [9] [13]). These free boundary problems are considered with surface tension using the Laplace Young condition 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 Howison [18] proved ill-posedness in an unstable -D case, namely when the higher-viscosity fluid contracts, they show global-intime existence of small initial data in the stable case when the higher-viscosity fluid exps. The results rely on the assumption that the Atwood number A µ = µ 1 µ µ 1 + µ is nonzero where µ 1 µ are the viscosities of the fluids. In the same year, Ambrose [1] treated the -D problem with an initial data fulfilling the following condition (ρ ρ 1 )g cos(θ(α, )) + µ µ 1 µ + µ 1 U(α, ) >, (x(α, ) x(α, )) + (y(α, ) y(α, )) (α α ) >, (1) where the interface is the curve (x(α, t), y(α, t)), ρ 1 ρ are the densities of the fluids, θ is the angle that the tangent to the curve forms with the horizontal U is the normal velocity (given by the Birkhoff-Rott integral). We are interested in the case A µ = that presents the evolution of the interface for different densities. This case, for example, models moist dry regions in a porous medium. Meanwhile the work of Ambrose is based on the arclength the tangent angle formulation used by Hou, Lowengrub Shelley [13], due to the particular form of the vorticity in the case A µ =, we get to parameterize the curve in the two dimensional problem getting the condition (1) for any time (see equation (15)). The free boundary problems given by fluids with different densities are been intensely studied. Notice the classical paper of Taylor [1] the works of Wu [] [3] where the full water wave problem is solved considering the water with positive density the air with zero density. A study of the two-dimensional case can be found in [] due to Ambrose Masmoudi. In order to simplify the notation, we consider µ/k = 1µ/b = 1 g = 1. Thus, the 3-D system is written as v(x 1, x, x 3, t) = p(x 1, x, x 3, t) (,, ρ(x 1, x, x 3, t)), () where (x 1, x, x 3 ) R 3 are the spatial variables t denotes the time. Here ρ is defined by { ρ1 in Ω ρ(x 1, x, x 3, t) = 1 (t) ρ in Ω (t), with ρ 1, ρ constants ρ 1 ρ.

3 Ω 1 (t) X 3 Ω (t) X X 1 We show in section 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 = ρ t + v ρ =, (3) where div v =. We notice the similarity with the -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 globalin-time existence using paradifferential calculus. A simpler proof can be found in [4] due to Bertozzi Constantin. In section we show that due to () the velocity can be determined from the density by singular integral operators (see []). It makes the equation more singular than the -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 -D quasi-geostrophic (QG) equation for sharp fronts. The QG equation models the dynamics of cold hot air the formation of fronts. Here the temperature is conserved along particle trajectories the velocity is given by singular integral operators in the following form v = ( R θ, R 1 θ), where R 1 R are the Riesz transforms (see [1] for more details of the QG equation). Rodrigo [19] proposed the contour equation of the sharp fronts where the temperature θ is concentrated in a domain proved local existence uniqueness. The paper is organized as follows. In section 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 uniqueness of the stable case. In section 5 we get a family of global solutions of the -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. 3

4 The Contour Equation We consider the equation with (x 1, x, x 3 ) R 3 where the fluid has different densities, that is ρ is represented by { ρ1, {x ρ(x 1, x, x 3, t) = 3 > f(x 1, x, t)} (4) ρ, {x 3 < f(x 1, x, t)}, being f the interface. Using Darcy s Law we get curl curl v = ( x1 x3 ρ, x x3 ρ, ( x 1 + x )ρ). Since div v = we have curl curl v = v, therefore it follows v = ( x1 1 x3 ρ, x 1 x3 ρ, ( x 1 + x ) 1 ρ). (5) The integral operators x1 1 x 1 are given by the kernels K 1 (x 1, x, x 3 ) = 1 4π x 1 (x 1 + x + x 3 )3/, K (x 1, x, x 3 ) = 1 4π respectively, thus the velocity can be expressed by Since ρ satisfies (4) we have x (x 1 + x + x 3 )3/, v = (K 1 x3 ρ, K x3 ρ, K 1 x1 ρ K x ρ). (6) ρ = (ρ ρ 1 )( x1 f(x 1, x, t), x f(x 1, x, t), 1)δ(x 3 f(x 1, x, t)), (7) where δ is the Dirac distribution. Using (6) we obtain v(x 1, x, x 3, t) = ρ ρ 1 (y 1, y, f(x y, t) y) 4π R [ y + (x 3 f(x y, t)) dy, ] 3/ where x 3 f(x, t), x, y R the principal value is taken at infinity (see []). When x 3 approaches 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 > we define It follows v 1 (x, f(x, t), t) = lim ε v(x 1 ε x1 f(x, t), x ε x f(x, t), f(x, t) + ε, t), v (x, f(x, t), t) = lim ε v(x 1 + ε x1 f(x, t), x + ε x f(x, t), f(x, t) ε, t). v 1 (x, f(x, t), t) = ρ ρ 1 (y 1, y, f(x y, t) y) 4π R [ y + (f(x, t) f(x y, t)) ] + ρ ρ 1 + ρ ρ 1 x1 f(x, t)(1,, x1 f(x, t)) 1 + ( x1 f(x, t)) + ( x f(x, t)) x f(x, t)(, 1, x f(x, t)) 1 + ( x1 f(x, t)) + ( x f(x, t)), 3/ dy (8) 4

5 v (x, f(x, t), t) = ρ ρ 1 (y 1, y, f(x y, t) y) 4π R [ y + (f(x, t) f(x y, t)) ] ρ ρ 1 ρ ρ 1 x1 f(x, t)(1,, x1 f(x, t)) 1 + ( x1 f(x, t)) + ( x f(x, t)) x f(x, t)(, 1, x f(x, t)) 1 + ( x1 f(x, t)) + ( x f(x, t)). 3/ dy (9) 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 of the interface. Thus, it follows that v(x, f(x, t), t) = ρ ρ 1 4π due to the fact that the terms ± ρ ρ 1 ± ρ ρ 1 (y 1, y, f(x y, t) y) R [ y + (f(x, t) f(x y, t)) dy, (1) ] 3/ x1 f(x, t)(1,, x1 f(x, t)) 1 + ( x1 f(x, t)) + ( x f(x, t)), x f(x, t)(, 1, x f(x, t)) 1 + ( x1 f(x, t)) + ( x f(x, t)), do not alter the shape of the interface. If we add the following tangential terms to (1) ρ ρ 1 y 1 4π R [ y + (f(x, t) f(x y, t)) ] 3/ dy(1,, x 1 f(x, t)), ρ ρ 1 y 4π R [ y + (f(x, t) f(x y, t)) ] 3/ dy(, 1, x f(x, t)), we obtain v(x, f(x, t), t) = ρ ρ 1 (,, 4π R ( f(x, t) f(x y, t)) y [ y + (f(x, t) f(x y, t)) dy). (11) ] 3/ Finally we have the contour equation given by df dt (x, t) = ρ ρ 1 ( f(x, t) f(x y, t)) y 4π [ y + (f(x, t) f(x y, t)) dy, ] 3/ f(x, ) = f (x). R (1) In the periodic case, we can obtain an equivalent equation to (1) due to the integral operators x1 1 x 1 can be presented by the kernels K p 1 (x 1, x, x 3 ) = 1 4π ( x 1 (x 1 + x + x 3 )3/ L(x 1, x, x 3 ) + M(x 1, x, x 3 )), K p (x 1, x, x 3 ) = 1 4π ( x (x 1 + x + x 3 )3/ L(x 1, x, x 3 ) + M(x 1, x, x 3 )), 5

6 respectively for (x 1, x, x 3 ) T R with T = [ π, π] the functions L, M C (T R). We can choose L C c (T R), L, supp L {x 1 + x + x 3 4}, L = 1 in {x 1 + x + x 3 1} L( x 1, x, x 3 ) = L(x 1, x, x 3 ). The function M belongs to Cb (T R) M(,, ) =. The velocity can be expressed by v = (K p 1 x 3 ρ, K p x 3 ρ, K p 1 x 1 ρ K p x ρ), due to (7) it follows (suppressing the dependence on t) v(x 1, x, x 3 ) = ρ ρ 1 (y 1, y, f(x y) y) 4π T [ y + (x 3 f(x y)) ] 3/ L(y, x 3 f(x y))dy ρ ρ 1 (1, 1, f(x y) (1, 1))M(y, x 3 f(x y))dy, 4π T if x 3 f(x). Adding a term in the tangential direction we obtain v(x, f(x)) = ρ ρ 1 ( f(x) f(x y)) y (,, 4π T [ y + (f(x) f(x y)) L(y, f(x) f(x y))dy ] 3/ + ( f(x) f(x y)) (1, 1))M(y, f(x) f(x y))dy). T Finally we have the contour equation in the periodic case given by df dt (x, t) =ρ ρ 1 4π + ρ ρ 1 4π f(x, ) =f (x). T (13) ( f(x, t) f(x y, t)) y [ y + (f(x, t) f(x y, t)) L(y, f(x, t) f(x y, t))dy ] 3/ (14) ( f(x, t) f(x y, t)) (1, 1))M(y, f(x, t) f(x y, t))dy, T We use both formulations throughout the paper. Suppose that the function f(x) only depends on x 1 in equation (1). Then the contour equation in the -D case (with a 1-D interface) follows df dt (x, t) = ρ ρ 1 ( x f(x, t) x f(x α, t))α π R α + (f(x, t) f(x α, t)) dα, (15) f(x, ) = f (x); x R. This equation can be obtained in a similar way that (1) using the stream function. Performing a two-dimensional analysis using the stream function, we obtain an equivalent equation to (15) in the two dimensional periodic case as follows df dt (x, t) =ρ ρ 1 π T + ρ ρ 1 π f(x, ) =f (x), ( x f(x, t) x f(x α, t))α α P (α, f(x, t) f(x α, t))dα + (f(x, t) f(x α, t)) ( x f(x, t) x f(x α, t))q(α, f(x, t) f(x α, t))dα, T (16) 6

7 with P (x 1, x ) C c (T R), P, supp P {x 1 + x 4}, P = 1 in {x 1 + x 1} P ( x 1, x ) = P (x 1, x ). The function Q(x 1, x ) belongs to Cb (T R) Q(, ) =. If we consider the linearized equation of the motion, we obtain a dissipative equation when ρ 1 < ρ (the greater density is below) an unstable equation when ρ 1 > ρ. The unstable linearized equation presents an instability similar to the Kelvin-Helmholtz s (see [6]). As usual, we note the Riesz transforms in R (see []) R 1 f(x) = 1 π P.V. y 1 f(x y)dy, R y 3 R f(x) = 1 π P.V. y f(x y)dy, R y 3 the operator Λf defined by the fourier transform Λf(ξ) = ξ f(ξ). Suppose that f(x) is uniformly small we can neglect the terms of order grater than one in (1), then it reduces to the following linear equation f t = ρ 1 ρ (R 1 x1 f + R x f) = ρ 1 ρ Λf, f(x, ) = f (x). (17) Applying the fourier transform we get ˆf(ξ) = ˆf (ξ)e ρ 1 ρ ξ t, therefore (17) is a dissipative equation when ρ 1 < ρ an ill posed problem in the case ρ 1 > ρ with a general initial data in the Schwartz class. We need an analytic initial data in order to get a well posed problem for ρ 1 > ρ. 3 The conservation of mass equation We show that if ρ is defined by (4) f(x, t) is convected by the velocity (11) then ρ is a weak solution of the conservation of mass equation (3) conversely. From now on, Ω is equal to R or T x = (x 1, x, x 3 ). Definition 3.1 The density ρ is a weak solution of the conservation of mass equation if for any ϕ C (Ω R (, T )), ϕ with compact support in the real case periodic in (x 1, x ) otherwise, we have T Ω R (ρ( x, t) t ϕ( x, t) + v( x, t)ρ( x, t) ϕ( x, t))d xdt =, (18) where the incompressible velocity v is given by Darcy s law. 7

8 Then Proposition 3. If f(x, t) satisfies (1) ρ( x, t) is defined by (4), then ρ is a weak solution of the conservation of mass equation. Furthermore, if ρ is a weak solution of the conservation of mass equation given by (4), then f(x, t) satisfies (1). Proof: Let ρ be a weak solution of (3) defined by (4). Integrating by parts we have T T T I = ρ t ϕd xdt = ρ 1 t ϕd xdt + ρ t ϕd xdt Ω = (ρ 1 ρ ) R T Ω {x 3 >f} ϕ(x, f(x, t), t) t f(x, t)dxdt. On the other h, due to (8) (9) we obtain T T T J = ρv ϕ d xdt = ρ 1 v ϕd xdt + ρ = T Ω Ω = (ρ 1 ρ ) R {x 3 >f} {x 3 <f} {x 3 <f} v ϕd xdt ϕ(x, f(x, t), t)(ρ 1 v 1 (x, f(x, t), t) ρ v (x, f(x, t), t)) ( x1 f(x, t), x f(x, t), 1)dxdt T Ω ϕ(x, f(x, t), t)v(x, f(x, t), t) ( x1 f(x, t), x f(x, t), 1)dxdt, where v(x, f(x, t), t) is given by (1). We get J = (ρ 1 ρ ) T ϕ(x, f(x, t), t) 4π Ω R ( f(x, t) f(x y, t)) y [ y + (f(x, t) f(x y, t)) dydxdt. ] 3/ Then I + J = due to (18). Thus, if we choose ϕ( x, t) = ϕ(x, t) for x 3 [ f L, f L ] follows that f(x, t) fulfils (1). Following the same arguments it is easy to check that if f(x, t) satisfies (1), then ρ is a weak solution given by (4). Remark 3.3 Note that due to (5), the velocity satisfies v = (R 1 (R 3 ρ), R (R 3 ρ), (R 1 + R )(ρ)), where the operators R 1, R R 3 are the Riesz transforms in three dimensions (see []). Since ρ L (Ω R) then v belongs to BMO (bounded mean oscillation) therefore v is in L (Ω R) locally (see [] for the definitions properties of the BMO space). 4 Local well-posedness for the stable case In this section we prove local existence uniqueness for the stable case using energy estimates. First we study the case Ω = R at the end of the section we give the main differences with the periodic domain. Denote the Sobolev spaces by H k, the Holder spaces by C k,δ with δ < 1 the fractional derivative the hessian matrix of a function f(x) by f(x). 8

9 4.1 Case Ω = R The main theorem in this section is the following Theorem 4.1 Let f (x) H k (R ) for k 4 ρ > ρ 1. Then there exists a time T > so that there is a unique solution to (1) in C 1 ([, T ]; H k (R )) with f(x, ) = f (x). We choose ρ ρ 1 = 4π without loss of generality, then df ( f(x, t) f(x y, t)) y (x, t) = dt [ y + (f(x, t) f(x y, t)) dy, ] 3/ f(x, ) = f (x). R (19) We show the proof with k = 4 being analogous for k > 4. We apply energy methods (see [5] for more details). Then The identity yields 1 d ( f(x) f(x y)) y dt f L (t) = f(x) R R [ y + (f(x) f(x y)) dydx ] 3/ ( f(x) f(x y)) y = f(x) R y <1 [ y + (f(x) f(x y)) dydx ] 3/ f(x) y + f(x) R y >1 [ y + (f(x) f(x y)) dydx ] 3/ f(x y) y f(x) R [ y + (f(x) f(x y)) dydx ] 3/ I 1 C C = I 1 + I + I 3. xi f(x) xi f(x y) = 1 1 ds ds y <1 y <1 y 1 R y >1 1 xi f(x + (s 1)y) y ds, f(x) f(x + (s 1)y) [1 + ((f(x) f(x y)) y dxdy () ] 3/ y 1 dy f H C f H. (1) Integrating by parts, the term I is written I = 3 (f(x) f(x y))( f(x) f(x y)) y f(x) y >1 R [ y + ((f(x) f(x y)) ] 5/ dxdy C y 3 f(x) f(x) f(x y) y 1 f(x) f(x y) y >1 R [1 + ((f(x) f(x y)) y ] 5/ dxdy C f L f H 1. 9

10 Integrating by parts in I 3, it follows R I 3 = f(x)f(x y) y (f(x) f(x y)) y >1 [ y + (f(x) f(x y)) dxdy ] 5/ (f(x) f(x y)) f(x y) y + 3 f(x)f(x y) y >1 R [ y + (f(x) f(x y)) ] 5/ dxdy y f(x)f(x y) y =1 R [ y + (f(x) f(x y)) dxdσ(y) ] 3/ We get finally C( f L + 1) f H 1. We consider the quantity d dt f L (t) C( f 3 H (t) + 1). () where 1 d dt 4 x 1 f L (t) = I 4 + I 5 + I 6 + I 7 + I 8, I 4 = x 4 1 f(x) ( 4 x1 f(x) 4 x1 f(x y)) y R R [ y + (f(x) f(x y)) dydx, ] 3/ I 5 = 4 x 4 1 f(x) ( x 3 1 f(x) x 3 1 f(x y)) y x1 A(x, y)dydx, R R I 6 = 6 x 4 1 f(x) ( x 1 f(x) x 1 f(x y)) y x 1 A(x, y)dydx, R R I 7 = 4 x 4 1 f(x) ( x1 f(x) x1 f(x y)) y x 3 1 A(x, y)dydx, R R I 8 = x 4 1 f(x) ( f(x) f(x y)) y x 4 1 A(x, y)dydx, R R A(x, y) = [ y + (f(x) f(x y)) ] 3/. The most singular term is I 4. In order to estimate it I 4 = x 4 1 f(x) 4 x1 f(x, t) y R R [ y + (f(x, t) f(x y, t)) dydx ] 3/ x 4 1 f(x) 4 x1 f(y, t) (x y) R R [ x y + (f(x, t) f(y, t)) dydx ] 3/ = J 1 + J. 1

11 Integrating by parts J 1 = 3 x 4 1 f(x) (f(x) f(x y))( f(x) f(x y)) y R R [ y + ((f(x) f(x y)) ] 5/ dydx = + C( f C 1 + M(f)) x 4 1 f L, where y >1 y <1 M(f) = max x y <1 (f(x) f(x y))( f(x) f(x y)) y [ y + ((f(x) f(x y)) ] 5/ dy. We estimate this maximum in the following form (f(x) f(x y) f(x) y)( f(x) f(x y)) y M(f) max x y <1 [ y + ((f(x) f(x y)) ] 5/ dy ( f(x) y)(( f(x) f(x y)) y y f(x) y) + max x y <1 [ y + ((f(x) f(x y)) ] 5/ dy + max ( f(x) y)(y f(x) y)(b(x, y) C(x, y))dy x y <1 ( f(x) y)(y f(x) y) + max x [ y + ( f(x) y) ] 5/ dy, y <1 (3) (4) where B(x, y) = [ y + ((f(x) f(x y)) ] 5/, C(x, y) = [ y + ( f(x) y) ] 5/. Making the change of variables y = z, we obtain that the last integral in (4) is null, then we can estimate M(f) by M(f) f y 1 C max x y <1 [1 + ((f(x) f(x y)) y 1 ) ] 5/ dy + f C 1 f C,δ y +δ dy + f C f 1 C y 1 dy with < δ < 1, having finally y <1 C( f C + f C 1 f C,δ + f C 1 f C ), In order to estimate J, we integrate by parts getting y <1 J 1 C( f 4 C,δ + 1) 4 x 1 f L. (5) J = 4 y x 1 f(x) ( 4 x1 f(x) 4 x1 f(y)) (x y) R R [ x y + (f(x) f(y)) ] 3/ dydx = K 1 + K, 11

12 with K 1 = x 4 1 f(x) R R 4 x1 f(x) 4 x1 f(y) [ x y + (f(x) f(y)) dydx, ] 3/ K = x 4 1 f(x) ( x 4 1 f(x) 4 3(f(x) f(y))(f(x) f(y) f(y) (x y)) x 1 f(y)) R R [ x y + (f(x) f(y)) ] 5/ dydx. Making a change of variables we can obtain K 1 = = = 1. R R 4 x 4 x 1 f(x) 1 f(x) x 4 1 f(y) R R [ x y + (f(x) f(y)) dydx ] 3/ 4 x 4 x 1 f(y) 1 f(x) x 4 1 f(y) R [ x y + (f(x) f(y)) dydx ] 3/ R ( 4 x1 f(x) 4 x1 f(y)) [ x y + (f(x) f(y)) dydx ] 3/ Here we observe the main difference with the unstable case in which we obtain the opposite sign. Now we consider K = L 1 + L + L 3, being L 1 = 3 x 4 1 f(x) (f(x) f(y))(f(x) f(y) f(y) (x y)) R R [ x y + (f(x) f(y)) ] 5/ dydx, L = 3 4 x 4 1 f(x) x 4 1 f(y) (f(x) f(y))(x y) ( f(x) + f(y)) (x y) R R [ x y + (f(x) f(y)) ] 5/ dydx, L 3 = 3 x 4 1 f(x) x 4 1 f(y)(f(x) f(y))d(x, y)dydx, R R with D(x, y) = (f(x) f(y) f(y) (x y) 1 4 (x y) ( f(x) + f(y)) (x y)) [ x y + (f(x) f(y)) ] 5/. The term L 1 can be estimated like J 1 in (3) one finds that Exchanging x for y we obtain that L =. L 1 C(1 + f 4 C,δ ) 4 x 1 f L. 1

13 For the last term L 3, it follows L 3 C x 4 1 f(x) f(x) f(y) D(x, y) dydx R R + C x 4 1 f(y) f(x) f(y) D(x, y) dxdy R R C x 4 1 f(x) f(x) f(x y) D(x, x y) dydx R R + C x 4 1 f(y) f(x + y) f(y) D(x + y, y) dxdy R R C x 4 1 f(x) dx( dy + dy) + x 4 1 f(y) dy( R y <1 y >1 R Finally C f C 1 f C,δ 4 x 1 f L. x <1 dx + dx) x >1 J = K 1 + K K = L 1 + L + L 3 = L 1 + L 3 C( f 4 C,δ + 1) 4 x 1 f L, due to (5) we obtain I 4 C( f 4 C,δ + 1) 4 x 1 f L. (6) Now we estimate the integral I 5. We have I 5 = J 3 + J 4 where J 3 = 4 x 4 1 f(x) x 3 1 f(x) y (f(x) f(x y))( x 1 f(x) x1 f(x y)) R R [ y + (f(x) f(x y)) ] 5/ dydx, J 4 = 4 = 4 R R x 4 1 f(x) x 3 1 f(y) (x y) (f(x) f(y))( x 1 f(x) x1 f(y)) R [ x y + (f(x) f(y)) ] 5/ dydx x 4 1 f(x) x x 3 1 f(y)(x y ) (f(x) f(y))( x 1 f(x) x1 f(y)) R [ x y + (f(x) f(y)) ] 5/ dydx. The way to estimate J 3 is similar to the term J 1 in (3), we find that J 3 C(1 + f 4 C,δ ) f H 4. We decompose the term J 4 = K 3 + K 4 + K 5 + K 6 as follows K 3 = 4 R x 4 1 f(x) x x 3 1 f(x y)y E(x, y)dydx, R K 4 = 4 R x 4 1 f(x) x x 3 1 f(x y)y F (x, y)dydx, R K 5 = 4 x 4 1 f(x) x x 3 1 f(x y)y ( f(x) y)( x1 f(x) y)(b(x, y) C(x, y))dydx R y <1 13

14 K 6 = 4 R y <1 x 4 1 f(x) x x 3 ( f(x) y)( x1 f(x) y) 1 f(x y)y [ y + ( f(x) y) ] 5/ dydx, where E(x, y) = (f(x) f(x y) f(x) y)( x 1 f(x) x1 f(x y)) [ y + (f(x) f(x y)) ] 5/, F (x, y) = ( f(x) y)( x 1 f(x) x1 f(x y) x1 f(x) yx { y <1} ) [ y + (f(x) f(x y)) ] 5/. The terms K 3, K 4, K 5 are estimated as J 1. Then we find that K 3 C(1 + f C ) f H 4, K 4 C(1 + f C 1 f C,δ) f H 4 K 5 C(1 + f C 1 f C ) f H 4. We rewrite K 6 we get K 6 = 4 x 4 1 f(x)s( x x 3 1 f)(x)dx, R with the operator S defined by S(g)(x) = The function Σ(x, y) satisfies that y <1 Σ(x, y) y g(x y)dy, Σ(x, y) = y ( f(x) y y )( x 1 f(x) y y ) y [1 + ( f(x) y y ) ] 5/. (7) (i) Σ(x, λy) = Σ(x, y), λ >, (ii) (iii) Σ(x, y) = Σ(x, y), sup Σ(x, y) x1 f L, x therefore S is a bounded linear map on L p (R ) for 1 < p < S p C x1 f L (see [] references therein for more details). Then K 6 C f C 4 x 1 f L x 3 x 1 f L. We obtain finally I 5 C(1 + f 4 C,δ ) f H 4. 14

15 In order to estimate the term I 6 we take 1 I 6 = 6 ds dy x 4 1 f(x) y ( x 1 f(x + (s 1)y)) y x 1 A(x, y)dx R R 1 ds( dy + dy) x 4 1 f(x) x 1 f(x + (s 1)y) x 1 A(x, y) y dx y <1 y >1 R C( y +δ dy + y 3 dy)(1 + f 4 C ) f,δ H. 4 y <1 y >1 The most singular term of I 7 is K 7 K 7 = 1 x 4 1 f(x)( x1 f(x) x1 f(x y)) y G(x, y)dydx, R where R G(x, y) = (f(x) f(x y))( 3 x 1 f(x) 3 x 1 f(x y)) [ y + (f(x) f(x y)) ] 5/. Due to x1 f(x) x1 f(x y) f C,δ y δ writing 3 x 1 f(x) 3 x 1 f(x y) = 1 3 x 1 f(x + (s 1)y) y ds, we obtain K 7 C( f + 1) f C,δ H I 4 7 C(1 + f 4 ) f C,δ H. 4 The most singular term of I 8 is K 8 K 8 = 1 x 4 1 f(x)( x 4 1 f(x) x 4 1 f(x y)) H(x, y)dydx, R R where H(x, y) = (f(x) f(x y))( f(x) f(x y)) y [ y + (f(x) f(x y)) ] 5/. Then K 8 = 1 x 4 1 f(x) (f(x) f(x y))( f(x) f(x y)) y R R [ y + (f(x) f(x y)) ] 5/ dydx, is controlled as before. We obtain K 8 C(1+ f 4 C,δ ) 4 x 1 f L I 8 C(1+ f 4 C,δ ) f H 4. Finally, we have d dt 4 x 1 f L (t) C(1 + f 4 C,δ ) f H 4, using Sobolev inequalities we get In a similar way we can obtain d dt 4 x 1 f L (t) C( f 6 H 4 + 1). d dt 4 x f L (t) C( f 6 H 4 + 1), 15

16 due to the identity f H = f 4 L + 4 x 1 f L + 4 x f L we get d dt f H 4(t) C( f 5 H 4 (t) + 1). Using Gronwall s inequality we get that the quantity f H 4 is bounded up to a time T = T ( f H 4). Then, applying energy methods the local existence result follows. Let the functions f 1 (x, t), f (x, t) be two solutions of equation (1) with f 1 (x, ) = f (x, ) = f (x), f = f 1 f. Then with with d dt f L (t) = I 9 + I 1 + I 11, I 9 = f(x) f(x) R y[ y + (f 1 (x) f 1 (x y)) ] 3/ dydx, R I 1 = f(x) R f(y, t) (x y)[ x y + (f 1 (x) f 1 (y)) ] 3/ dydx R I 11 = f(x) R ( f (x) f (x y)) yn(x, y)dydx, R N(x, y) = [ y + (f 1 (x) f 1 (x y)) ] 3/ [ y + (f (x) f (x y)) ] 3/. Integrating by part in I 9 we have I 9 C( f 1 H 4) f L, I 11 C( f 1 H 4, f H 4) f L. The term I 1 = f(x) y (f(y) f(x)) (x y)[ x y + (f 1 (x) f 1 (y)) ] 3/ dydx R R = f(x)(f(x) f(y))[ x y + (f 1 (x) f 1 (y)) ] 3/ dydx R R + f(x)(f(x) f(y)) 3(f 1(x) f 1 (y))(f 1 (x) f 1 (y) f 1 (x)(x y)) R R [ x y + (f 1 (x) f 1 (y)) ] 5/ dydx. Then we have that I 1 J 5 + J 6 where J 5 = f(x) 3(f 1 (x) f 1 (y))(f 1 (x) f 1 (y) f 1 (x)(x y)) R R [ x y + (f 1 (x) f 1 (y)) ] 5/ dydx J 6 = f(x) f(x y) 3(f 1(x) f 1 (x y))(f 1 (x) f 1 (x y) f 1 (x) y) R R [ y + (f 1 (x) f 1 (x y)) ] 5/ dydx. The term J 5 is estimated as J 1 obtaining J 5 C( f 1 H 4) f L. The term J 6 can be expressed as J 6 = K 9 + K 1 with (f 1 (x) f 1 (x y))g(x, y) K 9 = 3 f(x) f(x y) R R [ y + (f 1 (x) f 1 (x y)) dydx, ] 5/ 16

17 where the function G(x, y) is given by G(x, y) = f 1 (x) f 1 (x y) f 1 (x) y 1 4 y ( f 1 (x) + f 1 (x y)) y. One finds that the principal value K 1 is null. Therefore, we obtain finally J 6 C( f 1 H 4) f L. Applying Gronwall s inequality we get uniqueness. 4. Case Ω = T In the periodic case we give the theorem of local well-posedness the differences with Ω = R. Theorem 4. Let f (x) H k (T ) for k 4 ρ > ρ 1. Then there exists a time T > so that there is a unique solution to (14) in C 1 ([, T ]; H k (T )) with f(x, ) = f (x). The proof is similar to the theorem 4.1 but we must use the properties of the function L in (13). We consider without loss of generality ρ ρ 1 = 4π. In order to control the evolution of the quantity x 4 1 f L, the most singular term is I = x 4 1 f(x) ( 4 x1 f(x) 4 x1 f(x y)) y T T [ y + (f(x) f(x y)) L(y, f(x) f(x y))dydx ] 3/ = x 4 1 f(x) 4 L(y, f(x) f(x y)) x 1 f(x) y T T [ y + (f(x) f(x y)) dydx ] 3/ x 4 1 f(x) 4 x1 f(y) (x y) T T [ x y + (f(x) f(y)) L(x y, f(x) f(y))dydx ] 3/ = J 1 + J. Integrating by parts J 1 = 3 x 4 1 f(x) A(x, y)l(y, f(x) f(x y))dydx T T 1 x 4 1 f(x) L x3 (y, f(x) f(x y))( f(x) f(x y)) y T T [ y + (f(x) f(x y)) ] 3/ dydx = K 1 + K, (8) where A(x, y) = (f(x) f(x y))( f(x) f(x y)) y [ y + ((f(x) f(x y)) ] 5/ L x3 (x 1, x, x 3 ) = x3 L(x 1, x, x 3 ). Due to L(x 1, x, x 3 ) 1 C (x 1, x, x 3 ) we have K 1 C(1 + f 4 C,δ ) f H 4 + C 4 x 1 f L max x C(1 + f 4 C,δ ) f H 4. ( f(x) y)(y f(x) y) T [ y + ( f(x) y)) ] 5/ dy 17

18 Using that f(x) f(x y) f C 1 y L x3 = in {x 1 + x + x 3 4}, we have that K = 1 x 4 1 f(x) L x3 (y, f(x) f(x y))( f(x) f(x y)) y T y > [ y + (f(x) f(x y)) ] 3/ dydx C( f 4 C,δ + 1) f H f C 1 In order to estimates J, we integrate by parts getting J = 4 y x 1 f(x) ( 4 x1 f(y) 4 x1 f(x)) (x y) T T [ x y + (f(x) f(y)) ] 3/ L(x y, f(x) f(y))dydx = K 3 + K 4 + K 5 + K 6 with K 3 = x 4 1 f(x) T T 4 x1 f(x) 4 x1 f(y) [ x y + (f(x) f(y)) L(x y, f(x) f(y))dydx, ] 3/ K 4 = x 4 1 f(x) T ( x 4 1 f(x) x 4 1 f(y))b(x, y)l(x y, f(x) f(y))dydx, T K 5 = T x 4 1 f(x)( x 4 1 f(y) x 4 1 f(x))c(x, y)l x3 (x y, f(x) f(y))dydx, T K 6 = x 4 1 f(x)( x 4 1 f(y) x 4 1 f(x))d(x, y)dydx, T T B(x, y) = 3(f(x) f(y))(f(x) f(y) f(y) (x y)) [ x y + (f(x) f(y)) ] 5/, f(y) (x y) C(x, y) = [ x y + (f(x) f(y)) ] 3/, D(x, y) = L x 1 (x y, f(x) f(y))(x 1 y 1 ) + L x (x y, f(x) f(y))(x y ) [ x y + (f(x) f(y)) ] 3/, L x1 (x 1, x, x 3 ) = x1 L(x 1, x, x 3 ), L x (x 1, x, x 3 ) = x L(x 1, x, x 3 ). Exchanging the variables x y we can obtain K 3. The terms K 4, K 5 K 6 can be estimated in a similar way as K 1. Therefore, we obtain d dt 4 x 1 f L (t) C( f 5 H 4 (t) + 1), analogously d dt 4 x f L (t) C( f 5 H 4 (t) + 1). This proves local existence. The proof of the uniqueness is similar to the case Ω = R. 18

19 4.3 -D case Using the equation (15) in R the equation (16) in the periodic case we obtain the following theorem Theorem 4.3 Let f (x) H k for k 3 ρ > ρ 1. Then there exists a time T > so that there is a unique solution to (15) in C 1 ([, T ]; H k ) with f(x, ) = f (x). The proof is similar to the theorem s Global solution for the -D stable case In this section we obtain a family of global solutions for the -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 f H s = for s > 3/. In this section we consider x R, the operator Λ s = ( ) s/ f a = ˆf(k) e a k. For a >, if f a <, then the function f can be extended analytically on the strip Iz < a. Furthermore for b > a. The main result of this section is x f a C f b b a, (9) Theorem 5.1 Let f (x) be a function such that T f (x) dx =, x f ε for ε small enough xf b(t) εe b(t) (1 + b(t) γ 1 ), (3) with < γ < 1, b(t) = a (ρ ρ 1 )t/, ρ > ρ 1 a (ρ ρ 1 )t/. Then, there exists a unique solution of (15) with f(x, ) = f (x) ρ > ρ 1 satisfying x f a (t) C(ε) exp((σa + (ρ ρ 1 )t)/4), (31) xf a (t) C(ε)(1 + σa ρ ρ 1 t γ 1 ) exp((σa + (ρ ρ 1 )t)/4), (3) for a ρ ρ 1 σ t, σ = 1 + δ < δ < 1. The condition (3) can be satisfied for example if Λ 1+γ f < ε ˆf (1) = ˆf ( 1) =. In order to prove the theorem, we use the Cauchy-Kowalewski method (see [15] [16]) in a similar way as Caflisch Orellana [7] Siegel, Caflisch Howison [18]. We show the proof with ρ ρ 1 = without loss of generality. 19

20 Let g(x, t) h(x, t) be functions satisfying g t = Λg, g(x, ) = f (x), h t = Λh + T (g + h), h(x, ) =, (33) with T (f) = π 1 R x f(x) x f(x α) α ( f(x) f(x α) ) α 1 + ( f(x) f(x α) α Then the function f(x, t) = g(x, t) + h(x, t) is a solution of (15). properties of the nonlinear operator T. Lemma 5. If x f a, x g a < 1 for a then ) dα. (34) First, we show some T (f)() =, (35) x T (f) a C 1 xf a x f a, (36) x T (f) x T (g) a C ( xf a + xg a ) x f x g a + C ( x f a + x g a ) xf xg a, (37) with C 1 = 4(1 x f a) C = 4(1 x f a) + (1 x g a). Proof of the Lemma: Due to the inequality x f(x) x f a < 1 by (34) we obtain T (f) = π ( ) 1 ( 1) n x f(x) x f(x α) f(x) f(x α) n dα, (38) n 1 R α α T (f) = π 1 x n 1 ( 1) n n + 1 R ( ) f(x) f(x α) n+1 dα. α Thus T (f)() =. Using (38) T (f)(k) = ( 1) n n 1 n δ( k,...,k n j= k j, k) ik M n (k,..., k n ) n j= ˆf(k j ), where M n (k,..., k n ) = π 1 R dα n α (1 e iαk 1 e iαk j ). (39) α j=1

21 We get M n (k,..., k n ) = ( 1) n m n (k,..., k n ) n j=1 k j, with m n (k,..., k n ) = π 1 = i 1 1 ds 1... ds ds n R 1 e iαk α ds n (sing A sing B), ( exp iα n j=1 (s j 1)k j )dα A = α n (s j 1)k j, B = αk + α n j=1 j=1 s j k j. It follows T (f)(k) = n 1 with m n (k,..., k n ). We have therefore e a k k T (f)(k) k k n δ( k,...,k n j= n 1 n 1(n + 1) k j, k) m n (k,..., k n ) k,...,k n e a k k δ( x T (f) a xf a (n + 1) x f n n 1 n j= k j, k) n j= n k,...,k n e a k k ˆf(k ) We get (36) for x f a < 1. In a similar way we obtain (37). From (33) g can be expressed as follows ĝ(k, t) = e k t ˆf (k), by the hypothesis of the initial data we have a j= n k j ˆf(kj ), k j ˆf(k j ) j=1 e a k j k j ˆf(k j ), = xf a 3 x f 3 a x f 4 a (1 x f a). x g a (t) εe a t, (4) xg a (t) εe a t (1 + (t a) γ 1 ), (41) 1

22 for t a. We will prove the existence of h by an induction argument on the iterative equation: or ĥ n+1 (k, t) = h =. t h n+1 = Λh n+1 + T (g + h n ), h n+1 (x, ) =, h =, t e k (t s) (T (g + h n )) (k, s)ds, For h 1 we obtain the following estimates x h 1 a (t) t Using (36), (4) (41) we get T (g) a+s t (s)ds = t a t + = I 1 + I. t a t a t a I 1 e a t e s x T (g) (s)ds Ce a t e s xg (s) x g (s)ds By (4) (41) we have I C t t a t a Cε e a t e s (1 + s γ 1 )ds Cε (1 + γ) e a t. γ xg a+s t (s) x g a+s t (s)ds Cε e (a t) a(1 + (t a) γ 1 ) Cε e a t, δ due to the inequalities (aδ) γ 1 > (t a) γ 1 ae a t δ 1 for σa < t. Then Choosing b = a + s t + t a we have where xh 1 a (t) I 3 a t Ce t a t x h 1 a (t) 5Cε δγ ea t. xt (g) a+s t (s)ds ( t a t a t t ) + = I 3 + I 4, t a Cε γ e a t (1 + (t a) γ 1 ), x T (g) b (s) b (a + s t) ds e s xg (s) x g (s)ds Cε e a t t a t a t x T (g) b t a e s (1 + s γ 1 )ds

23 Therefore I 4 C t t a xg b (s) x g b (s)ds Cε t a t a ea t (1 + ( t a )γ 1 )( t + a ) 3Cε e a t (1 + (t a) γ 1 ). δ x h 1 a (t) 5Cε δγ ea t, (4) xh 1 a (t) 5Cε δγ e a t (1 + (t a) γ 1 ). (43) Define r n+1 = h n+1 h n, R n = M n = sup a < σa < t sup a < σa < t ( x r n a + xr n ) a 1 + (t σa) γ 1 e t σa, ( x h n a + xh n ) a 1 + (t σa) γ 1 e t σa. Take M 1 = R 1 5Cε δγ ε suppose that M j, R j ε for any j =,..., n, then x r n+1 a t Using (37) we have x T (g + h n ) x T (g + h n 1 ) a+s t (s)ds = t a t + = I 7 + I 8. t a t a I 7 Ce a t e s ( x r n (s)( xg + xh n (s) + xg + xh n 1 (s)) t a + Ce a t e s ( xr n (s)( x g + x h n (s) + x g + x h n 1 (s))ds t a Cε R n e a t (1 + s γ 1 )ds Cε γ R ne σa t, 3

24 I 8 C t + C t a t ( x r n a+s t (s)( xg + xh n a+s t (s) + xg + xh n 1 a+s t (s))ds t a ( xr n a+s t (s)( x g + x h n a+s t (s) + x g + x h n 1 a+s t (s))ds t Cε R n e δs σ(t a) (1 + (σ(t a) δs) γ 1 )ds t a Cε t a R n e x (1 + x γ 1 )dx 6Cε δ t σa γδ R ne σa t. We obtain for b = a + s t + σ(t a) δs σ xr n+1 a (t) t t x(t (g + h n ) T (g + h n 1 )) a+s t (s)ds x (T (g + h n ) T (g + h n 1 ) b (s) ds b (a + s t) t x (T (g + h n ) T (g + h n 1 ) ( σ b σ σ(t a) δs σ+1 (t a) t + σ σ+1 (t a) ) = I 9 + I 1. We have σ(t a) δs > σ σ σ+1 (t a) for s σ+1 (t a) therefore we obtain σ (σ + 1)C σ+1 (t a) I 9 e b ( x r n (s)( t a xg + xh n (s) + xg + xh n 1 (s)) + (σ + 1)C t a (σ + 1)Cε t a σ σ+1 (t a) R n σ σ+1 (t a) Using (37) the induction hypothesis we get e b ( xr n (s)( x g + x h n (s) + x g + x h n 1 (s))ds t I 1 4σCε R n e σb s (1 + (s σb)γ 1 ) ds σ σ+1 (t a) σ(t a) δs t 4σCε R n e δs σ(t a) 1 + ( σ(t a) δs ) γ 1 σ σ+1 (t a) σ(t a) δs σ σ+1 (t a) R n 4σCε δ t σa e b s (1 + s γ 1 )ds 4σCε R n e a t (1 + (t a) γ 1 ). γ ds e x (x 1 + x γ )dx 8σCε δ(1 γ) R ne σa t Due to the estimates for I 7, I 8, I 9 I 1 we obtain (1 + (σa t) γ 1 ). R n+1 Cσε δγ(γ 1) R n. (44) 4

25 Choosing ε small enough we get R n+1 1 R n... 1 n R 1 ε n+1, n+1 M n+1 R n+1 ε. j=1 Therefore, we obtain the function h = lim n hn satisfying x h a (t) n R n e σa t ε e σa t. Taking f(x, t) = g(x, t) + h(x, t), we get (31) for ρ ρ 1 =. In order to show the uniqueness, we write the equation (1) for ρ ρ 1 = in the following form sup a < σa < t f t = Λf + T (f), f(x, ) = f (x). Suppose that there exist two solutions f 1 f with f 1 (x, ) = f (x, ). Define R by ( R = x f 1 x f a + xf 1 xf ) a 1 + (t σa) γ 1 e t σa. It follows that R f 1 = f. C(ε)σ C(ε)σ δγ(γ 1) R for ε small enough it yields δγ(γ 1) < 1 therefore 6 Ill-posedness for the unstable case Here we show ill-posedness for the unstable case ρ 1 > ρ. We use the global solution for the -D stable case f(x 1, t) satisfying (31) with Λ 1+γ f < C Λ 1+γ+ζ f = for γ, ζ >. Making a change of variables, we define f λ (x 1, t) = λ 1 f(λx 1, λt+λ 1/ ) obtaining {f λ } λ> a family of solutions to the unstable case. Then it follows f λ H s() = λ s 3 f H s(λ 1/ ) C λ s 3 f 1 (λ 1/ ) C λ s 3 e λ 1/, f λ H s(λ 1/ ) = λ s 3 f H s() λ s 3 C k 1+γ+ζ ˆf (k) =, for s > 3/ γ, ζ small enough. We obtain an ill posed problem for s > 3/. Theorem 6.1 Let s > 3/, then for any ε > there exists a solution f of (15) with ρ 1 > ρ < δ < ε such that f H s() ε f H s(δ) =. Remark 6. Due to the fact that the solutions of the -D case are solutions of the 3-D problem we obtain ill-posedness for (1) with ρ 1 > ρ. k 5

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