On the Muskat problem: global in time results in 2D and 3D

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1 On the Muskat problem: global in time results in 2D and 3D Peter Constantin, Diego Córdoba, Francisco Gancedo Luis Rodríguez-Piazza and Robert M. Strain Februar 4, 24 Abstract This paper considers the three dimensional Muskat problem in the stable regime. We obtain a conservation law which provides an L 2 maximum principle for the fluid interface. We also show global in time existence for strong and weak solutions with initial data controlled b explicit constants. Furthermore we refine the estimates from our paper [5] to obtain global existence and uniqueness for strong solutions with larger initial data than we previousl had in 2D. Finall we provide global in time results in critical spaces, giving solutions with bounded slope and time integrable bounded curvature. Kewords: Porous media, incompressible flows, fluid interface, global existence. Mathematics subject classification: 35A, 76S5, 76B3 Introduction We consider the dnamics of the interface in between two incompressible fluids in porous media in three dimensional space. This is the Muskat problem (see [3]. We assume that both fluids are immiscible and have the same constant viscosit but different constant densities. We simplif matters b taking gravit g =, the permeabilit of the medium κ = and the viscosit ν =. Then the motion of the fluids satisf: ρ t + (uρ = u + P = (,, ρ u = ( where ρ = ρ(x, x 2, x 3, t is the densit, P = P (x, x 2, x 3, t is the pressure, u = (u (x, x 2, x 3, t, u 2 (x, x 2, x 3, t, u 3 (x, x 2, x 3, t is the incompressible velocit field, x i R, i =, 2, 3 and t. The first equation is the conservation of mass and the second equation is Darc s law, where the velocit is proportional to the driving forces, the pressure gradient and the buoanc force. We denote the interface that separates the space in two domains Ω and Ω 2 b x 3 = f(x, x 2, t. We consider the densit ρ = ρ(x, x 2, x 3, t to be the following step function: { ρ ρ(x, x 2, x 3, t =, (x, x 2, x 3 Ω (t = {x 3 > f(x, x 2, t}, ρ 2, (x, x 2, x 3 Ω 2 (t = {x 3 < f(x, x 2, t}. (2

2 Then the interface satisfies the equation f t (x, t = ρ 2 ρ ( f(x, t f(x, t P V 4π [ 2 + (f(x, t f(x, t 2 d, ] 3/2 f(x, = f (x, x = (x, x 2, (3 in order to be a solution of the sstem ( (see [9] for a detail derivation. A first approach is to linearize equation (3 around the stead state f (x =, which ields f t = ρ ρ 2 Λf, 2 f(x, = f (x, (4 where the operator Λf is defined in Fourier variables b Λf(ξ = ξ f(ξ. The case ρ < ρ 2 gives a stable regime, and for ρ > ρ 2 the sstem is unstable. Stabilit versus instabilit is determined b the normal component of the pressure gradient jump at the interface having a distinguished sign. This is known as the Raleigh-Talor condition which implies local existence in H s when the heavier fluid is below the lighter one, and ill-posedness in the unstable regime (see [9] for a proof of both statements. Earlier works on the well-posedness in Sobolev spaces for the 3D Muskat problem, where both fluids have also different viscosities, include [], [4], [] and [8]. Our goal is to prove global in time existence results for the stable regime. Our main concern is about the size of the initial data needed to reach this conclusion. Global existence for large slopes turns out to be false. There exist initial data that turn to the unstable regime; in finite time the interface becomes no longer a graph (see [3]. Moreover, there exist smooth initial data in the stable regime that in finite time turn to the unstable regime and at a later time the are no longer C 4 (see [4]. In our previous work [5], we studied the two dimensional Muskat equation; we showed global existence of Lipschitz continuous solutions for initial data that satisf f L < and x f L <. We also proved global existence for unique strong solutions if the initial data is smaller than a constant c ; f c where def f s = dξ ξ s ˆf(ξ, s. We have checked numericall that c is not small; it is greater than 5. Recentl, in [2], global results are obtained in a confined domain for initial data satisfing smallness conditions relating the amplitude, the slope and the depth. We also point out a new work [2] where instant analticit is proved for small initial data represented on the Fourier side b positive measures. In this paper we show that in 3D it is possible to obtain similar global existence results but with different constants. First, in Section 2 we prove the following identit for the evolution of the L 2 norm of the contour f 2 L 2 (t+ ρ π t ( [ 2 +(f(x, s f(x, s 2 ] /2 dxdds = f 2 L 2, (5 where ρ = (ρ 2 ρ /2. We further explain using this this formula that there is no parabolic behavior in the contour equation at the level of f. In Section 3 we prove global existence 2

3 of unique C([, T ]; H k ( solutions for k 3 if initiall f is controlled b f < k where k /5 (see (9 for the exact size of k. We also use the calculations in Section 3 to improve the size of the initial data in our global existence and uniqueness theorem for smooth solutions in [5] for 2D. In Section 4 we show that if a strong solution has the propert f L < /3, then it will be preserved in time. In Section 5 we prove global in time existence of Lipschitz continuous solutions in the stable case for initial data satisfing f L < and f L < /3. Finall in Section 6 we use the parabolicit of the problem to show global in time solutions in critical spaces with f < k, f (t L ([, T ] and f 2 (t L ([, T ] for an T >. This result gives in particular that f C (t L ([, T ] and f C 2(t L ([, T ]. 2 L 2 maximum principle This section is devoted to the proof of the identit (5. In order to simplif the exposition we take (ρ 2 ρ /( = and we write f(x, t = f(x for a fixed t. Then, the contour equation (3 is given b: where f t (x = P V x Integration b parts allows us to observe that d 2 dt f 2 L (t = P V d dx f(x 2 R 2 = P V d dx f(x ( f(x [ + ( f(x 2 ] /2 2 d, f(x = (f(x f(x /. (6 2 f(x [ + ( f(x 2 ] /2 x x f(x x 2 [ + ( x f(x 2 ] /2. Next we split this into two terms, d dx ( x f(x 2 2 dt f 2 L (t = d 2 x [ + ( x f(x 2 ] /2 P V d dx x x x([ + ( x f(x 2 ] /2 = I + I 2. With these computations, a further integration b parts provides I 2 = x ([ + ( x f(x 2 ] /2 dxd, and this equalit gives d 2 dt f 2 L = 2 From above (5 follows easil. x ( [ + ( x f(x 2 ] /2 dxd. 3

4 Next we show the bound J def = ( [ + ( f(x 2 ] /2 dxd 4π 2 f L, which controls the integral J with zero derivatives. This expresses the fact that identit (5 does not give a gain of regularit at the level of f. Besides the linearization (4, the nonlinear structure of the equation does not ield a parabolic dissipation for large initial data. In order to deal with J we observe that J ( [ + ( f(x + f(x 2 2 ] /2 dxd. Using the function H(z = ( + z 2 /2 and the fact that H( z + z 2 H( 2 z + H( 2 z 2 it is eas to get ( J [+2 f(x 2 2 ] /2 + [+2 f(x 2 2 ] /2 dxd, and therefore ( J 2 [ + 2 f(x 2 2 ] /2 dxd = K. B an eas change of variable one finds K = 2 2 f(x dx ( z z 2 + dz, so that K = 4π 2 f L. This provides the desired bound. 3 A global existence result for data less than 5 In this section we give a global existence result for classical solutions of the Muskat contour equation. We consider the norm def f s = dξ ξ s ˆf(ξ, s, (7 which allows us to use Fourier techniques for small initial data. theorem: We prove the following Theorem 3.. Suppose that initiall f H l ( for l 3 and f < k, where k is a constant such that π +δ (2n +! (2n + (2 n n! 2 k2n, (8 n for some < δ <. Then there is a unique solution f of (3 with initial data f that satisfies f C([, T ]; H l ( for an T >. 4

5 Remark 3.2. Computing the limit case δ =, so that π (2n +! (2n + (2 n n! 2 k2n < n one finds < k In particular, this holds if k /5. Remark 3.3. Analogous estimations allow us to obtain a better size for f than in [5] in order to have a global existence and uniqueness result in 2D (D interface. In fact, if initiall f H l (R for l 2 and f < c, where c is a constant such that 2 (2n + +δ c 2n, (9 n for some < δ < /2, then there exists a unique solution f of the two dimensional Muskat contour equation with initial data f that satisfies f C([, T ]; H l (R for an T >. In the limit case δ = we find < c (4 3/ , and the result is true if for example f /3. The remainder of this section is devoted to the proof of Theorem 3.. We point out that the argument used in [5] does not work directl here. It is valid in 2D onl. To overcome the difficult for 3D we need to smmetrize the operators involved in the equation to find an extra cancellation. We define f(x as in (6 and we take (ρ 2 ρ /2 = for the sake of simplicit. The contour equation for the Muskat problem (3 can be written as f t (x, t = Λf N(f, ( where the operator Λ is the square root of the negative Laplacian and we have N(f = 2 x f(xr( f(xd, with R(z = /( + z 2 3/2. A change of variable allows us to obtain N(f = ( 4π 2 x f(xr( f(x x f(xr( f(x d. ( We consider the norm f (7 as follows: d dt f (t = dξ ξ ( ˆf t (ξ ˆf(ξ + ˆf(ξ ˆf t (ξ/(2 ˆf(ξ R 2 dξ ξ 2 ˆf(ξ + dξ ξ F(N(f(ξ. We will show that the first term controls the evolution in such a wa that f is decreasing if initiall f < k, where k

6 Since f(x f < we can use the Talor expansion R(z = ( n a n z 2n, with a n = n (2n +! (2 n, z <, n! 2 to obtain Recall that N(f = 4π ( n a n n 2 ( x ( f ( f 2n x ( f ( f 2n d. (2 F( f = ˆf(ξm(ξ,, F( x f = iξ ˆf(ξm(ξ,, where m(ξ, = ( e iξ /. (3 Therefore F( x ( f ( f 2n = ((iξ ˆfm ( ˆfm ( ˆfm(ξ, α, with 2n convolutions, one with iξ ˆfm and 2n with ˆfm. Using (2 F(N(ξ = i ( n a n d dξ dξ 2n 4π 2 (ξ ξ n ˆf(ξ ξ ( 2n j= where M n ( = M n (ξ, ξ,..., ξ 2n, is given b M n ( = m(ξ ξ, ˆf(ξ j ξ j+ ˆf(ξ2n (M n ( M n (, ( 2n j= m(ξ j ξ j+, m(ξ 2n,. We then use Fubini theorem to obtain F(N(ξ = a n dξ dξ 2n n (ξ ξ ˆf(ξ ξ ( 2n ˆf(ξ j ξ j+ ˆf(ξ2n I n, j= (4 where the integral I n = I n (ξ, ξ,..., ξ 2n reads def I n = i 4π ( n 2 (M n( M n ( d. Polar coordinates, = ru with u = (cos θ, sin θ, provide I n = i π 4π ( n udθ π + dr(m n (r, u M n (r, u, 6

7 where we redefine and M n (r, u = m(ξ ξ, r, u m(ξ, r, u = ( e irξ u /r, ( 2n j= m(ξ j ξ j+, r, u m(ξ 2n, r, u. Since we have m(ξ, r, u = m(ξ, r, u and m(ξ, r, u = m(ξ, r, u, the change of variable r = s ields I n = i π 4π ( n u dθ ds(m n (s, u M n (s, u, π and therefore I n = i π 8π ( n u dθ dr(m n (r, u M n (r, u. π R The identit m(ξ, r, u = iξ u ds eir(s ξ u allows us to obtain 2n M n (r, u = ( n ds (ξ j ξ j+ u ξ 2n u e ir(ξ ξ u r which is simplified b writing with ds 2n ( exp ir M n (r, u = ( n ds ( exp(ira u r A = 2n ( 2n j= (s j (ξ j ξ j+ +(s 2n ξ 2n u, j= ds 2n 2n (ξ j ξ j+ u ξ 2n u j= exp(irb u r, (s j (ξ j ξ j+ + (s 2n ξ 2n, j= and 2n B = (ξ ξ + (s j (ξ j ξ j+ + (s 2n ξ 2n. j= It follows that I n = i π 2n ud θ ds ds 2n (ξ j ξ j+ u ξ 2n u 8π π j= ( exp(ira u exp(irb u exp( ira u exp( irb u dr + R r r r r 7

8 and the equalit P V R dr exp(irα/r = πisgn α ields I n = 4 π π ud θ ds At this point it is eas to bound I n : ( 2n ds 2n (ξ j ξ j+ u j= 2n I n π ξ j ξ j+ ξ 2n. j= ξ 2n u(sgn (A u sgn (B u. The above estimate and (4 allow us to get dξ ξ F(N(ξ π a n dξ dξ dξ 2n ξ n ξ ξ ˆf(ξ 2n ξ ξ j ξ j+ ˆf(ξ j ξ j+ ξ 2n ˆf(ξ 2n. The inequalit ξ ξ ξ + ξ ξ ξ 2n ξ 2n + ξ 2n ields dξ ξ F(N(ξ π ( (2n + a n dξ ξ 2 ˆf(ξ ( dξ ξ ˆf(ξ n j= 2n, and therefore ( dξ ξ F(N(ξ dξ ξ 2 ˆf(ξ π (2n + a n f 2n n ( dξ ξ 2 ˆf(ξ ( + 2 f 2 π ( f 2. 5/2 For x < k one finds (+2x 2 /( x 2 5/2 < /π. Therefore if f < k this inequalit will be maintained when we propagate forward in time because of d dt f (t, and f (t f < k. Considering a higher order norm, with s > in (7, we aim to obtain f +δ (t + µ t ds f 2+δ (s f +δ, (5 for some < δ < and < µ <. Let us recall that f +δ C f H 3 for < δ <. We use the inequalit ξ +δ (2n + δ ( ξ ξ +δ + ξ ξ 2 +δ + + ξ 2n ξ 2n +δ + ξ 2n +δ, to obtain as before ξ +δ F(N(ξ dξ ξ 2+δ ˆf(ξ dξ π (2n + +δ a n f 2n n 8.

9 Due to > π (2n + +δ a n f 2n = µ π (2n + +δ a n f 2n (t, n n for some < µ <, we find ξ +δ F(N(ξ dξ ( µ ξ 2+δ ˆf(ξ dξ, for δ small enough. Since d dt f +δ(t µ f 2+δ (t, integration in time provides (5. From previous work [9], one could find the following a priori bound: d 2 dt 3 x f 2 L P ( f 2 L ( 2 f L f C δ + f L 2 f C δ f 2 H, 3 where P is a polnomial function and C δ is the homogeneous Hölder norm. The terms that appear in the evolution can be handled as in [9] (see Section 4 except for a couple of low order terms: L.O.T. = 3 x f(x ( f(x 3 ( x f(x 3 x f(x 2 [ + ( f(x 2 ] 9/2 ddx, L.O.T. 2 = 3 x f(x ( f(x( x f(x 3 x f(x 2 [ + ( f(x 2 ] 7/2 ddx. We bound L.O.T. + L.O.T x f(x x f(x x f(x 3 ddx = J. Splitting J for > and < it is eas to find J = dx d + dx d > < C 3 x f L 2 f L 2 f L f C δ 2 f L + C 3 x f L 2 2 f 2 L 4 f C δ 2 f L < > δ 2 d. δ 4 d Interpolation inequalit 2 f 2 L 4 f L 3 f L 2 allows us to obtain J C f L f C δ 2 f L f 2 H 2, as desired. Proceeding in a similar wa for 3 x 2 f L 2 we find d dt f 2 H 3 P ( f L ( 2 f L f C δ + f L 2 f C δ f 2 H 3. 9

10 Fourier transform ields k f L f k and k f C δ(t f k+δ for k =, 2 and b interpolation it is eas to obtain f 2 f +δ f f 2+δ. We find d dt f 2 H P ( f 3 f 2+δ f 2 H, 3 which together with the a priori bound provides f H 3(t f H 3 exp(cp (k after integration in time. Using (5 we get finall t f 2+δ (sds, f H 3(t f H 3 exp(cp (k f +δ /µ. We finish with the conclusion that the solution can be continued in H 3 for all time if f is initiall smaller than k defined b (9. An analogous calculation gives getting the result for an H k for k > 3. f H k(t f H k exp(cp (k f +δ /µ, 4 Initial data smaller than /3 In this section our goal is to prove the following maximum principle for the evolution of f L (t assuming that f L < /3. Theorem 4.. Let f H s with s 4 and f L < /3. Then the unique solution of the sstem (3 satisfies f L (t < /3, for t >. Proof: We consider (ρ 2 ρ /2 = without loss of generalit. We take one derivative in x i in (3 to find xi f t (x, t = I i (x, t + I i 2(x, t + I i 3(x, t, where and with I i = P V I i 2 = P V xi f(x, t [ 2 + (f(x, t f(x, t 2 d, ] 3/2 xi f(x, t [ 2 + (f(x, t f(x, t 2 d ] 3/2 I3 i = xi f(x, t xi f(x, t [ 2 +(f(x, t f(x, t 2 A(x, d, ] 3/2 (f(x, t f(x, t( f(x, t f(x, t A(x, = (f(x f(x 2.

11 Integration b parts ields where I i 2 = P V P V 2( xi f(x, t xi f(x, t [ 2 + (f(x, t f(x, t 2 d ] 3/2 ( xi f(x, t xi f(x, t [ 2 + (f(x, t f(x, t 2 B(x, d, ] 3/2 B(x, = (f(x, t f(x, t f(x 2 + (f(x f(x 2. Adding I2 i and Ii 3 one finds I2 i + I3 i = P V xi f(x, t xi f(x, t [ 2 + (f(x, t f(x, t 2 C(x, d, (6 ] 3/2 where C = A + B 2. Consider M(t = max x { ( x f(x, t 2 + ( x2 f(x, t 2} = ( x f(x t, t 2 + ( x2 f(x t, t 2. Next we follow the time derivative of M(t to find that M (t for almost ever t > if M( < /9. This will ield the desired result. We obtain M (t = 2( x f(x t, t x f t (x t, t + x2 f(x t, t x2 f t (x t, t for almost ever t (see [] for more details. It gives M (t = 2 xi f(x t, t(i2(x i t, t + I3(x i t, t, i=,2 due to the fact that at the maximum we have x f(x t, ti (x t, t + x2 f(x t, ti 2 (x t, t =. Equation (6 shows that it remains to check that C(x t,. We write C(x, t = + 3 f(x( f(x u f(x + ( f(x 2, for u = /. It is eas to check that it is positive if f L < /3. 5 Global existence for initial data smaller than /3 Here we prove the existence of weak solutions for the Muskat contour equation. First we provide the notion of weak solution. It is possible to rewrite (3 as follows: f t = ρ x P V 2 f(x [ + ( f(x 2 d, (7 ] /2

12 where ρ and f(x are defined as before. Then integrating b parts in the nonlinear term, it is eas to find that for an η(x, t Cc ([, T, a weak solution f should satisf T η t (x, tf(x, tdxdt + η(x, f (xdx R 2 T ρ = x η(x, t P V The main result we prove below is the following: f(x 2 [ + ( f(x 2 ddxdt. (8 ] /2 Theorem 5.. Suppose that f L < and f L < /3. Then there exists a weak solution of (8 that satisfies f(x, t C([, T ] L ([, T ]; W, (, for an T >. In particular f is a global in time Lipschitz continuous solution. We split the proof of Theorem 5. in several sections. A regularized model is defined below in (9 with solutions f ε (x, t; here the model will be defined for a sufficientl small ε >. In Section 5. we prove some necessar a priori bounds for f ε (x, t. The are used in Section 5.2 to give global in time existence of classical solutions to the regularized model. Then, in Section 5.3 we explain how to obtain the weak solution as a limit as ε + ; to this end we will establish to a strong convergence result. The regularized model is given b ft ε (x, t = εcλ ε f ε + ε f ε + ρ x P V d 2 ε f ε (x [ + ( f ε (x 2 ] /2, (9 where C > is an universal constant fixed below, the operator Λ ε is a Fourier multiplier given b Λ ε f(ξ = ξ ε f(ξ or equivalentl using its integral from b Λ ε f(x f(x f(x = c ε 3 ε d, with ε small enough. We define f(x = 2 x f(x + 2 x 2 f(x, and f(x is given in (6. In the next two subsections we write f = f ε for the solution to (9 for the sake of simplicit of notation. 5. A priori bounds For solutions of the regularized sstem (9 we get the following two a priori bounds f L f L, f L f L < /3. The first one is obtained b checking the evolution of M(t = max f(x, t = f(x t, t. x Here x t is thought of as the point where the maximum is attained. 2

13 with Since For almost ever t we find it is eas to find M (t = f t (x t, t = εcλ ε f(x t + ε f(x t + ρ I(x t, I(x = x P V d 2 ε f(x [ + ( f(x 2 ] /2. x x f(x I(x = x P V d x 2 ε [ + ( x f(x 2, (2 ] /2 I(x = εp V d R 2 + P V d f(x f( x 3 ε [ + ( x f(x 2 ] /2 f(x (x (f(x f( x 3 ε [ + ( x f(x 2 ] 3/2. The previous formula shows that for C large enough εcλ ε f(x t + ρ I(x t. Then M (t for a.e. t (, T ] because f(x t and therefore M(t M(. Analogousl m(t m(. Next we consider the evolution of L(t = max x ( x f(x, t 2 + ( x2 f(x, t 2 = ( x f(x t, t 2 + ( x2 f(x t, t 2. We can proceed as in the previous section, but in this case more terms will appear. In xi f t we have analogous terms that can be handled as before. Terms with the correct sign, that appear due to ελ ε f and ε f in (9. And a new element J i (x has terms which are given b That is J i (x = ε d x i f(x xi f(x 3 ε [ + ( f(x 2 ] 3/2. xi f t (x = εcλ ε xi f(x + ε xi f(x + ρ J i (x + Analogous terms. In checking the time derivative of L L (t = 2 xi f(x t, t xi f t (x t, t, i=,2 all the terms are handled as before but for ρ xi f(x tj i (x t. i=,2 But at this point it is eas to check that εc xi f(x tλ ε xi f(x t + ρ xi f(x tj i (x t, i=,2 for C big enough. Therefore L (t if L(t < /3 for almost ever t. This ields the desired maximum principle. i=,2 (2 3

14 5.2 Global existence for the regularized model We consider regular initial data f H 4 for the sstem (9. Local existence can easil be proved using the energ method following the arguments for the non-regularized Muskat problem (3, as in [9]. As we did for (3, it follows that d dt f L 2(t = ρ + ε π x ε ( [ + ( f(x 2 ] /2 dxd Therefore f L 2(t f L 2. 2Cε Λ ( ε/2 f L 2(t 2ε f L 2(t. Remark 5.2. The global existence theorem for weak solutions can also be found with f L 2 < instead of f L <. We chose the version above because it is more general. We see that if the solution satisfies initiall a L 2 bound then f(x, t L ([, T ]; L 2 (. where Next, we consider the evolution of 3 x f 3 x f t dx Cε Λ ( ε/2 3 x f L 2(t ε 3 x f 2 L 2 + I + I 2, I = ρ ( x 3 f(x x 3 P V ( f(x f(x 3 ε d dx, I 2 = ρ ( x 3 f(x x 3 ( f(x f(x [ + ( f(x 2 ] 3/2 P V 3 ε [ + ( f(x 2 ] 3/2 d dx. The term f(x cancels out in I due to the PV and an integration b parts shows that I = ρ C(ε x 3 f(xλ ε x 3 f(xdx. For I 2 one finds I 2 = ρ ( x 4 f(x x 2 ( f(x f(x [ + ( f(x 2 ] 3/2 P V 3 ε [ + ( f(x 2 ] 3/2 d dx, and the splitting I 2 = J + J 2 + J 3 + J 4 gives J = ρ x 4 f(x ( 2 x f(x 2 x f(x [ + ( f(x 2 ] 3/2 3 ε [ + ( f(x 2 ] 3/2 ddx, J 2 = 3ρ π x 4 f(x ( x f(x x f(x 3 ε x f(x f(x [ + ( f(x 2 ddx, ] 5/2 4

15 J 3 = 3ρ 4 ( f(x f(x x f(x 3 ε x 2 f(x f(x [ + ( f(x 2 ddx, ] 5/2 J 4 = 3ρ 4 ( f(x f(x x f(x 3 ε ( x f(x 2 4( f(x 2 [ + ( f(x 2 ddx. ] 7/2 For J we proceed as follows ( J = ρ dx dx + d dx C(ε( f L + x 2 f L 2 x 4 f L 2. > < The identit ields J 2 3ρ π and therefore xi x f(x xi x f(x = ds + 3ρ π d < 2 ε ds > xi x f(x + (s ds, dx xf(x 4 2 x f(x+(s ( x f(x + x f(x d 3 ε dx x 4 f(x 2 x f(x+(s ( x f(x + x f(x ( f(x + f(x, J 2 C(ε( + f L f L 4 x f L 2 2 x f L 2. In J 3 we use the splitting J 3 = K + K 2 where K = 3ρ d dx, K 2 = 3ρ > and then The equalit < d dx, K 3ρ π f d L > 3 ε dx x 4 f(x ( x 2 f(x + x 2 f(x allows us to obtain C f L 4 x f L 2 2 x f L 2 C f L 4 x f L 2( f L x f L 2. 2 x f(x 2 x f(x = K 2 3ρ π f L ds < C f L 4 x f L 2 2 x f L 2. 2 x f(x + (s ds, d 2 ε dx xf(x 4 x 2 f(x + (s 5

16 In J 4 we use the splitting J 4 = K 3 + K 4 where K 3 = 3ρ d dx, > K 4 = 3ρ d dx, < and then K 3 C f 2 L > d 4 ε dx x 4 f(x ( x f(x + x f(x The equalit allows us to obtain C f 2 L 4 x f L 2 x f L 2 C f 2 L 4 x f L 2( f L x f L 2. 2 x f(x 2 x f(x = K 4 C f L The following estimate ds dr < C f L 4 x f L 2 x f 2 L 4. 2 x f(x + (s ds, ε 2 d dx x 4 f(x ( x f(x + (s x f(x + (r xi x f 4 L 4 3 f L xi x f 2 L 4 xi 2 x f L 2, ields Using Young s inequalit K 4 C f 2 L 4 x f L 2 2 x f L 2. d dt 3 x f 2 L 2 C(ε( f 2 L + f 2 L + f 4 L + f 2 H 3. Proceeding in a similar manner, at this point it is eas to find and therefore d dt 3 x 2 f 2 L 2 C(ε( f 2 L + f 2 L + f 4 L + f 2 H 3, d dt f 2 H 3 C(ε( f 2 L + f 2 L + f 4 L + f 2 H 3. The Gronwall inequalit then ields for ( f 2 H (t f 3 2 t H exp 3 C(εG(sds, G(s = ( f 2 L (s + f 2 L (s + f 4 L (s +. We find f C([, T ]; H 3 (R for an T > b the a priori bounds. 6

17 For the argument in next sections we will need f C([, T ]; H 4 (R for an T >. Therefore we consider the evolution of four derivatives. Most of the terms can be controlled as before. We will show how to deal with the rest using the estimate of the H 3 norm. Since 4 x f 4 x f t dx Cε Λ ( ε/2 4 x f L 2(t ε 4 x f 2 L 2 + L + L 2, where L = ρ ( x 4 f(x x 4 P V ( f(x f(x 3 ε d dx, L 2 = ρ ( x 4 f(x x 4 ( f(x f(x [ + ( f(x 2 ] 3/2 P V 3 ε [ + ( f(x 2 ] 3/2 d dx. The term L has the correct sign as I. For L 2 one finds L 2 = ρ ( x 5 f(x x 3 ( f(x f(x [ + ( f(x 2 ] 3/2 P V 3 ε [ + ( f(x 2 ] 3/2 d dx, and the splitting L 2 = M + M 2 + M 3 + M 4 gives M = ρ x 5 f(x ( 3 x f(x 3 x f(x [ + ( f(x 2 ] 3/2 3 ε [ + ( f(x 2 ] 3/2 ddx, M 2 = 9ρ x 5 f(x ( 2 x f(x 2 x f(x x f(x f(x 3 ε [ + ( f(x 2 ddx, ] 5/2 M 3 = 9ρ x 5 f(x M 4 = 3ρ x 5 f(x For M and M 2 we obtain as before ( x f(x x f(x 3 ε ( f(x f(x 3 ε ( x f(x f(x x [ + ( f(x 2 ] 5/2 ddx, ( x 2 x f(x f(x [ + ( f(x 2 ] 5/2 ddx. M + M 2 C(ε( + f L ( f L + 5 x f L 2 f H 4. In M 3 we use the splitting M 3 = N + N 2 where N = 9ρ 5 ( x f(x x f(x x f(x 3 ε x 2 f(x f(x [ + ( f(x 2 ddx, ] 5/2 N 2 = 9ρ 5 ( x f(x x f(x x f(x 3 ε ( x f(x 2 4( f(x 2 [ + ( f(x 2 ] 7/2 ddx, 7

18 We take to find as before N = 9ρ d dx + 9ρ d dx, > < B Sobolev embedding N C(ε 5 x f L 2 x f L 4( 2 x f L x f L 4. N C(ε 5 x f L 2 f H 3( f H 3 + f H 4. Similarl for N 2 therefore N 2 = 9ρ d dx + 9ρ d dx, > < N 2 C(ε 5 x f L 2 x f L ( x f L x f L x f L 4 x f L 4, which ields N 2 C(ε 5 x f L 2 f L f H 3( f L + f H 4, For M 4 we split further M 4 = N 3 + N 4 + N 5 N 3 = 3ρ 5 ( f(x f(x x f(x 3 ε x 3 f(x f(x [ + ( f(x 2 ddx, ] 5/2 For N 3 one finds N 4 = 9ρ 5 ( f(x f(x x f(x 3 ε x 2 f(x x f(x 4( f(x 2 [ + ( f(x 2 ddx, ] 7/2 N 5 = 3ρ 5 ( f(x f(x x f(x 3 ε and similarl for N 4 ( x f(x 3 2( f(x 3 8 f(x 7 [ + ( f(x 2 ] 9/2 ddx, N 3 C(ε 5 x f L 2 f L ( 3 x f L x f L 2 C(ε 5 x f L 2 f L ( f H 3 + f H 4, N 4 C(ε 5 x f L 2 f L ( 2 x f L 2 x f L + 2 x f L 4 x f L 4 C(ε 5 x f L 2 f L f H 3( f L + f H 4. 8

19 Finall, for N 5 we conclude that N 5 C(ε 5 x f L 2 f L ( x f L 2 x f 2 L + x f 3 L 6 C(ε 5 x f L 2 f L ( f H 3 f 2 L + f 3 H 3, b Sobolev embedding. If we gather all the estimates above and use Young s inequalit, it is not difficult to check that 4 x f(x 4 x f t (xdx C(ε(+ f 2 L (+ f 2 L (+ f 2 H 3 f 2 H 4 A repetition of the argument for x 4 2 gives Therefore + C(ε( + f 6 L ( + f 6 H 3. 4 x 2 f(x 4 x f t (xdx C(ε(+ f 2 L (+ f 2 L (+ f 2 H 3 f 2 H 4 + C(ε( + f 6 L ( + f 6 H 3. d dt f 2 H 4 C(ε(+ f 2 L (+ f 2 L (+ f 2 H 3 f 2 H 4 + C(ε( + f 6 L ( + f 6 H 3. We use the Gronwall inequalit and additionall the control of the H 3 norm to obtain the desired global estimate for H Taking ε + This section ends the proof of Theorem 5. b showing that solutions of the regularized sstem converge to a weak solution. First we approximate the initial data to have a global solution of the regularized sstem. An approximation to the identit ζ Cc ( is defined as follows: dx ζ(x =, ζ, ζ(x = ζ( x, where ζ ε (x = ζ(x/ε/ε 2. (22 Then, for an f W, ( and f L < /3, we define the initial data for the regularized sstem as follows f ε (x = (ζ ε f (x + ε 2 x 2. Notice that f ε Hs (R for an s >, and f ε L f L. More importantl, f ε L < /3 if ε is sufficientl small (ε depends upon the size of f L. Therefore global existence of the regularized sstem (9 holds with initial data f ε under the condition that ε > is small enough. Now consider the solutions {f ε } to the regularized sstem (9 with initial data given b the f ε as described above. Integration b parts provides T η t f ε dxdt + ε(cλ ε η f ε dxdt + η(x, f(xdx ε R 2 T ρ = x η(x, t P V f ε (23 (x 2 ε [ + ( f ε (x 2 ddxdt, ] /2 9

20 for an η Cc ([, T. Now we send ε + to in order to obtain (8. The third integral above converges as a result of the properties of the the approximation to the identit which was previousl introduced. The second integral converges to because of the bound f ε L (t f L. Together with the other bound ( f ε L (t < /3, we find the existence of a subsequence (denoted again b f ε that converges in the weak* topolog to a function f L ([, T ]; W, ( b the Banach-Alaoglu theorem. This provides the solution f and implies the convergence of the first integral in (29. It remains to check that as ε + we have T dt ρ dx η(x, t P V T dt f ε (x d 2 ε [ + ( f ε (x 2 ] /2 ρ dx η(x, t P V d f(x 2 [ + ( f(x 2 ] /2. R We let B R denote the open ball of radius R and center (,, then we claim that there is a subsequence (denoted again b f ε such that f ε f L ([,T ] B R, as ε. (24 2 δ L B δ We will prove this at the end of the section b using a strong convergence theorem. Since f ε C([, T ] R for an ε > and, up to a subsequence, f ε converges to f on compact sets, we obtain f C([, T ] R. Choose M > so that supp(η B M. For an small δ > and an large L, with L > M + we split the integral as R B B BL c d = d + d + d. (25 The first and last integrals separatel are arbitraril small independent of ε for L > sufficientl large and for δ > sufficientl small: The bound f ε (x [ + ( f ε (x 2 ] /2. ields T ρ dt dx η(x, t P V B δ d 2 ε f ε (x [ + ( f ε (x 2 ] /2 ρ η L ([,T ] δ. For the integral on B c L we note that and therefore z [ + z 2 ] /2 = z ( [ + z 2 ] /2 = z + d sz ds [ + (sz 2 ] ds = z /2 [ + (sz 2 ds, ] 3/2 [ + (sz 2 ] 3/2 [ + (sz 2 ] 3/2 ds = z ( z 2 s 2 h(szds, 2

21 where h(sz = (3 + 3(sz 2 + (sz 4 /([ + (sz 2 ] 3/2 ( + [ + (sz 2 ] 3/2. This expression allows us to split i f ε (x P V d 2 ε [ + ( f ε (x 2 = Rε,L ] /2 i (f ε B c L for i =, 2. Here R ε,l i P V has the form R ε,l i (f ε def = P V B c L B c L d i 2 ε ( f ε (x 3 d f ε (x i 3 ε, s 2 h(s f ε (xds, with the principal value at infinit. On the other hand in the second term in the left hand side the principal value is not necessar and we obtain d ds C f ε 3 d L 4 ε C f 3. L B c L It remains to show a similar bound for I L def = dx η(x, t (R ε,l (f ε, R ε,l 2 (f ε. B M The principal value ields I L = lim n IL n where In L = dx η(x, t d f ε (x B M B n\b L 3 ε. Integration b parts provides ( In L ( f(x nu = dx η(x, t B M n ε + ( ε B c L f ε (x B n\b L 3 ε for u =, which allows us to bound the following term as ( In L C η L f L L ε + n ε. Hence I L C η L f L /L /2 f(x Lu L ε dθ d, and we conclude that I L is arbitraril small if L is arbitraril large. For the last integral we recall that we have uniform convergence on compact sets. Due to B L B δ and x B M we have T dt dx η(x, t T ρ P V dt R B L \B δ d 2 ε dx η(x, t f ε (x [ + ( f ε (x 2 ] /2 d B L \B δ 2 ρ P V f(x [ + ( f(x 2 ] /2, 2

22 as ε +. For L sufficientl large and δ > sufficientl small, we conclude b taking ε +. It remains to prove the strong convergence in L ([, T ]; L (B R for an R > which was claimed in (24. The idea is to use the weak space W 2, (B R to obtain bounds for ft ε (x, t which are uniform: sup ft ε W 2, (B R (t C f L (, t [,T ] (26 where C does not depend on R or ε. For v L (B R we consider the norm W 2, as follows: v W 2, (B R = sup φ(xv(xdx, φ W 2, B R (B R : φ 2, (B R where W 2, (B R = Cc (B R W 2,. Now the Banach space W 2, (B R is defined to be the completion of L (B R with respect to this norm W 2, (B R. We have the following result for convergence in this space (see [5] Lemma 4.3: Lemma 5.3. Consider a sequence {u m } in C([, T ] B R that is uniforml bounded in the space L ([, T ]; W, (B R. Assume further that the weak derivative t u m is in L ([, T ]; L (B R (not necessaril uniform and is uniforml bounded in L ([, T ]; W 2, (B R. Finall suppose that xi u m C([, T ] B R for i =, 2 and an m (not necessaril uniform. Then there exists a subsequence of u m that converges strongl in L ([, T ]; L (B R. B appling this lemma the strong convergence claimed in (24 is obtained. It onl remains to check the hpothesis of the lemma. For an regularized solution f ε to (9 we need ft ε in L ([, T ]; L (B R (but not uniforml and (26. Due to f ε C([, T ]; H 4 (R, in (9 it is eas to bound the linear terms. The nonlinear term can be written as and therefore N(f = C ε Λ ε f ε + ρ P V ( f(x f(x [ + ( f(x 2 ] 3/2 d 2 ε [ + ( f(x 2 ] 3/2, b Sobolev embedding. The norm of ft ε W 2, (B R is given b N(x, t C(ε f ε H 4(t, ft ε W 2, (B R (t = sup φ W 2, (B R : φ W 2, R dx ft ε (x, tφ(x, since φ vanishes on the boundar of B R. Then we have I = Λ ε f(xφ(xdx = B R Λ ε f(xφ(xdx = f(xλ ε φ(xdx, and therefore I f L (t Λ ε φ L. 22

23 We split so that Λ ε φ(x φ(x φ(x = c 3 ε d = d + d = J (x + J 2 (x, > < J (x dx > We rewrite J 2 as follows J 2 (x = c d 3 ε dx( φ(x + φ(x C φ L (B R. < We also consider the following identities φ(x φ(x = φ(x φ(x φ(x = φ(x φ(x φ(x 3 ε d. (s ds φ(x + (s ds, dr ( 2 φ(x + r(s α, The expression for J 2 and these identities together ield J 2 (x dx ε ds dr dx 2 φ(x + r(s < C 2 φ L (B R. We obtain Λ ε f ε W 2, (B L + f ε W 2, (B L C f ε L ( C f L (. For the last term in (9 we integrate b parts dx φ(x P V ρ R P V d 2 ε f(x [ + ( f(x 2 ] /2 to realize that the splitting from (25 with L = δ = allows us to conclude that the integral above is bounded b C φ W, f L (R. 6 Global existence for initial data in critical spaces This section is devoted to show global existence results for strong solutions of the Muskat contour equation in critical spaces. Theorem 6.. Suppose that f L 2 and f < k ( f < c for the 2D case. Then there is a unique solution f of Muskat with initial data f that satisfies f L 2(t f L 2, f (t + µ t ds f 2 (s f, for µ >, a.e. t [, T ] and an T >. The time derivative of f satisfies where C = C( f. f t (t C, T 23 ds f t (s C,

24 Remark 6.2. The scale invariance for Muskat solutions f λ (x, t = λf(λx, λt makes the following norms critical: ess sup t [,T ] f (t, T ds f 2 (s. The control of these norms gives in particular solutions such that T ess sup t [,T ] f C (t + µ ds 2 f C (s f, where C is the space of continuous functions vanishing at infinit. Proof: For f such that f < k we proceed as before to obtain the following a priori bound d dt f (t, f (t < k. Due to > π (2n + a n f 2n = µ π (2n + a n f 2n (t n n for < µ <, we find d dt f (t µ f 2 (t, (see Section 3 for details and time integration gives the desired a priori bound We also find dξ ˆf t (ξ dξ ξ f(ξ + f (t + µ t f 2 (sds f. dξ F(N(f(ξ f (+π n a n f 2n C( f, (27 and similarl T dtdξ ξ ˆf t (ξ T T ( dtdξ ξ 2 f(ξ + ξ F(N(f(ξ dt f 2 (t(+π (2n+a n f 2n C( f. n (28 Next we would like to find a bona fide solution of Muskat satisfing those bounds. We consider the following regularized model f ε t = ζ ε (T (f ε, f ε (x, = (ζ ε f (x, where T (f ε (x = x P V d 2 (ζ ε f ε (x [ + ( (ζ ε f ε (x 2 ] /2, 24

25 (we take ρ = for the sake of simplicit and ζ ε given b (22. Local existence can be shown as in [9] for regular initial data, since f L 2 it is eas to find ζ ε f H k for an k. Then, as in Section 2, it is possible to obtain an L 2 maximum principle: d 2 dt f ε 2 L = 2 ( [ + ( (ζ ε f ε (x 2 ] /2 dxd. Due to ζ ε f ε C 2,δ ζ ε f ε H 4 C(ε f ε L 2 C(ε f L 2 it is possible to get global in time bounds and therefore global existence for f ε C([, T ]; H k for an k 3 and an T > (see Section 2. Proceeding as before we find f ε (t + µ t ds ζ ε f ε 2 (s f. Next, we will take the limit as ε. We will find strong and weak limits so most of the time the argument will be up to various subsequences. All of them will be denoted b f εn b abuse of notation. In particular f ε is uniforml bounded in L ([, T ]; L 2 so that there exists a subsequence {f εn } which converges in the weak* topolog of L ([, T ]; L 2 to f. The subsequence { f εn } is also uniforml bounded in L 2 ([, T ] so there exists a subsequence { f εn } that converges weakl to f L 2 ([, T ]. Then it is eas to check that (ζ εn f εn (ξ, t = ζ(ε n ξ f εn (ξ, t converges weakl to f L 2 ([, T ]. We use Mazur s lemma to conclude that a convex combination G n (ξ, t = (G n(ξ, t, G 2 n(ξ, t = N(n k=n of ( f εn (ξ, t, ζ(ε n ξ f εn (ξ, t with (, denoting a vector and λ k ( f ε k (ξ, t, ζ(ε k ξ f ε k (ξ, t, λ k, N(n k=n λ k =, converges strongl to ( f, f in (L 2 ([, T ] 2. We extract a subsequence (denoted b G n to get that G n (ξ, t converges to ( f(ξ, t, f(ξ, t pointwise for almost ever (ξ, t [, T ]. Therefore for t [, T ] Ω with Ω = we find that G n(ξ, t converges to f(ξ, t pointwise for almost ever ξ. We use Fatou s lemma to conclude that for t [, T ] Ω and the following holds M(t lim inf n lim inf n lim inf n M(t = f (t + µ ( dξ ξ G n(ξ, t + µ t N(n k=n N(n k=n t ds λ k ( dξ ξ f ε k (ξ, t + µ λ k ( f ε k (t + µ t ds f 2 (s, t dξ ξ 2 G 2 n(ξ, s ds dξ ξ 2 ζ(ε k ξ f ε k (ξ, s ds ζ εk f ε k 2 (s f. 25

26 Therefore T ess sup t [,T ] f (t + µ ds f 2 (s f. In order to find that the limit function f satisfies Muskat equation we claim that f is a weak solution. Then the regularit of f allows to conclude that it is in fact a strong solution. We will follow the arguments in Section 5 and Lemma 5.3 to get strong convergence in L. We just need to bound ft εn uniforml in L ([, T ]; W 2, (B R. But f εn t W 2, (B R εn (t ft L (B R (t ft εn (t C( f, since the last inequalit can be obtained as we did in the a priori bound (27. Since {f εn } satisfies T η t (x, tf εn (x, tdxdt + η(x, (ζ εn f (xdx R 2 T = x (ζ εn η(x, t P V (ζ εn f εn (x, tddxdt 2 [ + ( (ζ εn f εn (x, t 2, (29 ] /2 we can pass to the limit as ε n and the strong convergence gives f as a weak Muskat solution. Now we have f a strong Muskat solution due to its regularit and we can find bounds (27 and (28. In order to end the result we just need to get uniqueness. We consider two Muskat solutions f and f 2 with the above properties and f (x, = f 2 (x, = f (x. Then for the difference f = f f 2 we find where and III = I = II = f(x d 2 dt f 2 L (t = I + II + III, 2 f(x f(x P V f(xp V [ 2 + (f (x f (x 2 ddx, ] 3/2 f(x [ 2 + (f (x f (x 2 ddx, ] 3/2 ( f 2 (x 2 [ + ( f (x 2 ] 3/2 [ + ( f 2 (x 2 ] 3/2 ddx. We integrate b parts in I to get I = 3 f(x 2 A(xdx, for A(x = 4π P V Next, we bound as follows A(x dξ F(A(ξ. 3 x f (x f (x [ + ( f (x 2 ] 5/2 d. In order to deal with F(A(ξ we proceed as for N(f in (. Since z( + z 2 5/2 = n b nz 2n+ for z < we can obtain dξ F(A(ξ π 4 f 2 (t b n ( f (t 2n+ C( f f 2 (t. n 26

27 This ields I C( f f 2 (t f 2 L 2 (t. In the term II we write f(x = (f(x f(x and integrate b parts in to find II = II + II 2 where II = (f(x f(x f(xp V [ 2 + (f (x f (x 2 ddx, ] 3/2 and II 2 = 3 f(xp V (f(x f(x (f (x f (x (f (x f (x f (x [ 2 + (f (x f (x 2 ] 5/2 ddx. One could smmetrize II to get II = (f(x f( 2 4π [ x 2 + (f (x f ( 2 ddx. ] 3/2 For II 2 we split further II 2 = II2 + II2 2 where II2 = 3 f(x 2 B(xdx, for B(x = P V and Since we denote to rewrite B as follows B(x, d, B(x, = f (x( f (x f (x 3 [ + ( f (x 2 ] 5/2, II2 2 = 3 f(xp V f (x f (x = x s f (x = B(x = P V f(x B(x, ddx. f (x + (s ds, f (x + (s ds f (x 3 x s f (x f (x [ + ( f (x 2 ] 5/2 d. At this point it is eas to find that B and A are similar in such a wa that an analogous analsis allows us to obtain B(x π 2 f 2 (t n b n ( f (t 2n+ C( f f 2 (t. It is possible to smmetrize II2 2 as follows II2 2 = 3 f(xp V [f(x B(x, + f(x + B(x, ]ddx, 4π 27

28 and to use Parseval s identit in order to obtain II2 2 = dξ f(ξ n b n dξ f(ξ ξ dξ 2 dξ 2n+2 ( 2n+ (ξ ξ 2 The integral J n = J n (ξ, ξ,..., ξ 2n+2 reads J n = 3i 4π P V 3 (M n( M n ( d, where M n ( = M n (ξ, ξ,..., ξ 2n+2, is given b j= M n ( = e i(ξ ξ e i(ξ ξ 2 eis(ξ ξ 2 m(ξ 2 ξ 3, ˆf (ξ j ξ j+ ˆf (ξ 2n+2 J n. m(ξ 3 ξ 4,... m(ξ 2n+ ξ 2n+2, m(ξ 2n+2,, using the operator x s in the B(x, formula. With the PV cancelation we get J n 3π 2 2n+ j= ξ j ξ j+ ξ 2n+2, and therefore II2 2 3π dξ 2 f(ξ b n [ ˆf ( 2 ˆf ( ˆf... ( }{{} ˆf ] (ξ n 2n+2 convolutions 3π 2 f 2 L f 2 2 b n f 2n+ C( f f 2 (t f 2 L (t. 2 n Above we use Schwarz s and Young s inequalities. It ields the desired estimate for II: II C( f f 2 (t f 2 L 2 (t. We expand III to obtain III = f(x ( n a n n 2 x f 2 (x[( f (x 2n ( f 2 (x 2n ]ddx. Since III = f(x ( n a n n 2 x f 2 (x f(x 2n j= ( f (x 2n j ( f 2 (x j ddx, 28

29 we split further III = III + III 2 to find III = f(x 2 n ( n a n P V 3 x f 2 (x 2n j= ( f (x 2n j ( f 2 (x j ddx, and III 2 = f(x n ( n a n P V f(x 3 x f 2 (x We can proceed as before to get III f 2 L 2 (t f 2 2 (t π 2 We deal with III 2 as with II 2 2 : We obtain finall III 2 f 2 L 2 (t f 2 2 (t π 2 n n 2n j= ( f (x 2n j ( f 2 (x j ddx. 2na n f 2n C( f f 2 2 (t f 2 L 2 (t. 2na n f 2n C( f f 2 2 (t f 2 L 2 (t. d 2 dt f 2 L (t C( f 2 ( f 2 (t + f 2 2 (t f 2 L (t, 2 and time integration provides ( t f 2 L (t f 2 2 L exp C( f 2 ( f 2 (s + f 2 2 (sds, to find f =. This completes the proof of uniqueness. Next we consider the following norms for s, p given b f p s,p = ξ sp f(ξ p dξ, with the homogeneous space We provide the following result: R F s,p = {T S (R : T is a function and T s,p < }. Theorem 6.3. Suppose that f L 2 F, F 2 p,p with p > and f = f, < k, ( f < c in 2D. Then there is a unique solution f of Muskat with initial data f that satisfies f L ([, T ]; L 2 F, F 2 p,p L ([, T ]; F 2, L p ([, T ]; F 2,p for an T >. The time derivative of f satisfies f t L ([, T ]; F, F p,p L ([, T ]; F, L p ([, T ]; F,p. 29

30 Remark 6.4. We would like to point out that all the homogeneous norms ( T /p sup f 2,p, and f p 2,p (tdt p [,T ] are critical in 2D under the scale invariant for Muskat contour equation. In 3D these norms are supercritical due to the fact that, for example, sup f 2,2 (t [,T ] is critical. In particular, we give a new result in Sobolev spaces taking p = 2. In the case < p < 2 it is possible to use Hausdorff-Young inequalit to find f L ([, T ], W 2 p, p p L p ([, T ]; W 2, p p Proof: For f such that f < k we proceed as before to obtain a priori estimates. Next we check the evolution of d p dt f p = dξ ξ 2p f(ξ p ( f 2 p,p t (ξ f(ξ + f(ξ f t (ξ/(2 f(ξ R dξ ξ 2p f(ξ p + dξ ξ 2p f(ξ p F(N(f(ξ R R = dξ ξ 2p f(ξ p + I R We bound as follows I dξ ξ 2(p f(ξ p ξ π a n dξ dξ 2n R n ξ ξ ˆf(ξ 2n ξ ξ j ξ j+ ˆf(ξ j ξ j+ ξ 2n ˆf(ξ 2n. j= The inequalit ξ ξ ξ + ξ ξ ξ 2n ξ 2n + ξ 2n gives I dξ ξ 2(p f(ξ p π [ (2n + a n ( 2 ˆf ( ˆf... ( R }{{} ˆf ] (ξ. n 2n convolutions Hölder and Young s inequalities ield I f p 2,p π n (2n + a n ( 2 ˆf ( ˆf... ( }{{} ˆf L p 2n convolutions f p 2,p π (2n + a n f 2,p f 2n. n Due to > π (2n + a n f 2n = µ π (2n + a n f 2n (t n n 3

31 for < µ <, we find d p dt f p 2,p(t µ f p 2,p (t, p and time integration gives the following a priori bound We also find t f p 2,p(t + pµ ds f p 2,p (s f p 2,p. p p dξ ˆf t (ξ R dξ ξ f(ξ + dξ F(N(f(ξ R a n f 2n 2. f ( + π n For g L p p and g p we find L p dξg(ξ ξ p ft (ξ dξ g(ξ ( ξ 2 p f(ξ + ξ p F(N(f(ξ R R f 2,p + J. p Using that ξ p ( ξ ξ p + ξ ξ 2 p ξ 2n ξ 2n p + ξ 2n p we get J dξ g(ξ 2 (2n + [ ( 2 p ˆf ( ˆf... ( R }{{} ˆf ] (ξ n 2n convolutions Therefore dualit provides π (2n + + p a n f 2n f 2 p,p C( f f 2,p. p n An analogous approach provides T f t p,p C( f f 2 p,p. dt f t p,p (t C( f f p pµ 2,p. p In order to find bona fide solutions of the sstem we regularize the initial data as in the previous theorem. We then obtain the same a priori bounds for the regularized solution as above. We pass to the limit to find a global-in-time solution. We find weak* and weak convergence of the regularized sstem to the Muskat solution in the weak* and weak topolog of L ([, T ]; F 2 p,p and L p ([, T ]; F 2,p respectivel using that for p > the spaces L p are reflexive. Acknowledgments PC was partiall supported b NSF grants DMS and DMS DC and FG were partiall supported b MCINN grant MTM (Spain. FG acknowledges support from the Ramón Cajal program. LRP was partiall supported b MCINN grant MTM (Spain. RMS was partiall supported b NSF grants DMS-2747, DMS- 9463, and an Alfred P. Sloan Foundation Research Fellowship. FG wish to thank Juan Casado-Díaz and David Ruiz for helpful discussions. 3

32 References [] D. Ambrose. Well-posedness of two-phase Darc flow in 3D. Quart. Appl. Math. 65, no., (27. [2] T. Beck, P. Sosoe and P. Wong. Duchon-Robert solutions for the Raleigh-Talor and Muskat problems. Journal of Differential Equations, 256, no., (23. [3] A. Castro, D. Córdoba, Ch. Fefferman, F. Gancedo and M. López-Fernández. Raleigh- Talor breakdown for the Muskat problem with applications to water waves. Annals of Math. 75 (2: (22. [4] A. Castro, D. Córdoba, Ch. Fefferman and F. Gancedo. Breakdown of smoothness for the Muskat problem. Arch. Rat. Mech. Anal. 28, no. 3, (23. [5] P. Constantin, D. Córdoba, F. Gancedo and R. M. Strain. On the global existence for the Muskat problem. J. Eur. Math. Soc. (JEMS 5, (23. [6] P. Constantin and M. Pugh. Global solutions for small data to the Hele-Shaw problem. Nonlinearit, 6, (993. [7] A. Córdoba, D. Córdoba and F. Gancedo. Interface evolution: the Hele-Shaw and Muskat problems. Annals of Math. 73, (, (2. [8] A. Córdoba, D. Córdoba and F. Gancedo. Porous media: the Muskat problem in 3D. Analsis & PDE, Vol. 6, No. 2, (23. [9] D. Córdoba and F. Gancedo. Contour dnamics of incompressible 3-D fluids in a porous medium with different densities. Comm. Math. Phs., 273, no. 2, (27. [] D. Córdoba and F. Gancedo. A maximum principle for the Muskat problem for fluids with different densities. Comm. Math. Phs., 286, no. 2, (29. [] J. Escher and B.V. Matioc. On the parabolicit of the Muskat problem: Well-posedness, Fingering, and stabilit results. Z. Anal. Anwend. 3, no. 2, (2. [2] R. Granero-Belinchon. Global existence for the confined Muskat problem. Preprint, arxiv: (23. [3] M. Muskat. The flow of homogeneous fluids through porous media. New York, 937. [4] M. Siegel, R. Caflisch and S. Howison. Global Existence, Singular Solutions, and Ill- Posedness for the Muskat Problem. Comm. Pure and Appl. Math., 57: 374-4,

33 Peter Constantin Department of Mathematics Princeton Universit 2 Fine Hall, Washington Rd, Princeton, NJ 8544, USA const@math.princeton.edu Diego Córdoba Instituto de Ciencias Matemáticas Consejo Superior de Investigaciones Científicas C/ Nicolas Cabrera, 3-5, 2849 Madrid, Spain dcg@icmat.es Francisco Gancedo Departamento de Análisis Matemático & IMUS Universidad de Sevilla C/ Tarfia s/n, Campus Reina Mercedes, 42 Sevilla, Spain fgancedo@us.es Luis Rodríguez-Piazza Departamento de Análisis Matemático & IMUS Universidad de Sevilla C/ Tarfia s/n, Campus Reina Mercedes, 42 Sevilla, Spain piazza@us.es Robert M. Strain Department of Mathematics Universit of Pennslvania David Rittenhouse Lab 29 South 33rd Street, Philadelphia, PA 94, USA strain@math.upenn.edu 33