Contents (I): The model: The physical scenario and the derivation of the model: Muskat and the confined Hele-Shaw cell problems.

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2 Contents (I): The model: The physical scenario and the derivation of the model: Muskat and the confined Hele-Shaw cell problems. First ideas and results: Existence and uniqueness But, are the boundaries playing an important role? Let see some numerics. Smoothing effect Ill-posedness in the R-T unstable case * -Note: * means that the proof will be treated in a more detailed way.

3 Contents (II): Qualitative theory: Maximum principle for f L (R) * Decay estimate for f L (R) * Maximum principle for x f L (R) * Singularities in finite time: Existence of such singularities * Finite vs. infinite depth Numerical simulations and final remarks -Note: * means that the proof will be treated in a more detailed way.

4 The Muskat problem The problem that we want to consider is the evolution of two different fluids in a bounded porous medium (like water and oil in an oil well). Figure: Water and air in a porous medium.

5 We do these hypothesis: The lower fluid is homogeneous with density ρ 2. The upper fluid has density equal to ρ 1. Both fluid are incompressible, inviscid and irrotational. The unique external force is gravity.

6 The interface between both fluid (the wave) initially is a graph and we don t consider surface tension. Due to friction of the fluid particles with the porous grid we have a friction force proportional with velocity: µ κu. µ and κ came from physical properties from the fluid and the medium. The fluid velocities are small and they don t changes too much. Thus we neglect acceleration terms. This is what is known as the Muskat problem.

7 At this point we need to translate these hypothesis into equations. We start with Euler equation ρ( t u + (u )u) = p. Now, we use the hipothesis concerning gravity ρ( t u + (u )u) = p gρ(0, 1) t,

8 At this point we need to translate these hypothesis into equations. We start with Euler equation ρ( t u + (u )u) = p. Now, we use the hipothesis concerning gravity ρ( t u + (u )u) = p gρ(0, 1) t,

9 ρ( t u + (u )u) = p Using the hypothesis about friction ρ( t u + (u )u) = p gρ(0, 1) t µ κ u,

10 ρ( t u + (u )u) = p Using the hypothesis about friction ρ( t u + (u )u) = p gρ(0, 1) t µ κ u,

11 ρ( t u + (u )u) = p Neglecting accelerations we obtain ρ 0 ( t u + (u )u) = p gρ(0, 1) t µ κ u.

12 ρ( t u + (u )u) = p Neglecting accelerations we obtain ρ 0 ( t u + (u )u) = p gρ(0, 1) t µ κ u.

13 The previous conservation of momentum equation is known as Darcy s law. This law was formulated by a french engineer, Henry Philibert Gaspard Darcy (June 10, 1803 January 3, 1858), who derived it from experiments. Figure: Henry Darcy. Mathematically can be derived using homogenization from the Stokes system (see works by Tartar and Allaire and references therein).

14 The Hele-Shaw cell problem The problem that we want to consider is the evolution of a Stokes 3D flow between two parallel flat plates separated by an infinitesimal (when compared with the other dimensions in the problem) gap. Figure: Rhin river (Mainz, Germany)

15 The Hele-Shaw cell problem The problem that we want to consider is the evolution of a Stokes 3D flow between two parallel flat plates separated by an infinitesimal (when compared with the other dimensions in the problem) gap.

16 To obtain the equation for the Hele-Shaw cell problem we start with the Stokes system. Looking how we write the coordinates (see the pictures before) we have x,y,z p + gρ(0, 1, 0) = µ x,y,z u; x,y,z u = 0. As the walls are very close (write b for the distance) the flow only moves in other the directions, i.e. if the walls are in the z axis, u 3 = 0. Inserting this ansatz in the equations we obtain p(x, y, z) = p(x, y).

17 Now we have that the derivatives in the z direction of u 1, u 2 are bigger that the derivatives in the other directions, so we arrive to the following system x p = µ 2 z u 1 ; y p + gρ = µ 2 z u 2. The solution of thus system can be explicitly obtained (using the boundary conditions in z) as u = ( p + gρ(0, 1)) z2 zb 2µ

18 Define ū the mean velocity in the z direction. Then we have that ū = ( p + gρ(0, 1)) 2bµ b 0 z 2 zbdz = b2 ( p + gρ(0, 1)), 12µ being this new averaged velocity also incompressible. The Hele-Shaw cell problem receives its name in honour of Henry Selby Hele-Shaw ( ), an english engineer.

19 We observe that, mathematically, both physical situations (the flow in a porous medium and the Hele-Shaw cell) are equivalent. Indeed, we have the following system of equations (where I normalized the constants that don t play any role) The equations v = p (0, ρ), Conservation of momentum v = 0, Incompressibility (1) t ρ + v ρ = 0, Conservation of mass. Recall that ρ is a jump function with values ρ 1 and ρ 2.

20 We have the equations, thus, only remains to define the spatial domain. As we are interested in the effects of the boundaries in the evolution of the interface between both fluids we consider our porous medium (or Hele-Shaw cell) bounded. Then, we consider the strip R ( l, l) for some l > 0 as our spatial domain. Figure: Scheme of the problem.

21 We have the system (1) posed in the correct domain with impermeable boundary conditions. Now, the next step is Step 0 to translate this system to a unique equation for the interface in such a way that, if we solve the problem for the interface, we recover weak solutions to the original system.

22 Using the fundamental solution for the Poisson equation on the strip S = (0, 2l) (so the wave is in l + f (η)) and Darcy s Law, we have that if Ψ denotes the stream function Ψ(x, y) = 1 2π log R 2l 0 n= [ ) log ( (x η) 2 + (y (4nl + ν)) 2 ( (x η) 2 + (y (4nl ν)) 2 ) ] x ρ(η, ν)dνdη.

23 ... After some painful computations, and, for the sake of simplicity, taking l = π 2... We obtain the velocity on the curve v(x, f (x)) = ρ2 ρ 1 P.V. Ξ 1 (x, x η, f ) + Ξ 2 (x, x η, f )dη 4π R i ρ2 ρ 1 P.V. x f (η)(ξ 1 (x, x η, f ) Ξ 2 (x, x η, f ))dη. 4π R where the kernels are Ξ 1 = Ξ 2 = sinh (η) cosh (η) cos(f (x) f (x η)), sinh (η) cosh (η) + cos(f (x) + f (x η))

24 A delicate issue is the tangential velocity... Let us recall that, due to the definition of ρ, we have ρ ρ = α z α z = ( xf, 1) ( x f, 1) using that it is a graph. From here we obtain that ρ ρ = αz α z = (1, xf ) (1, x f ) using that it is a graph.

25 The conservation of mass equation reads Now observe that t ρ + v ρ = 0. t ρ + v ρ = t ρ + (v + c ρ) ρ = 0, where c is any smooth scalar function. Following this reasoning we can change the tangential velocity of the curve and we will obtain a weak solution of the system (1). Moreover, as the tangential velocity only changes the parametrization of the curve, we can alter the tangential velocity without changing the shape of the wave.

26 Important idea As we said, we can change the tangential velocity of the curve without changing the evolution of the curve. Let us summarize the (equivalent) arguments leading to this claim: The conservation of mass equation (which is the relevant one for the densities) does not see terms with ρ. The tangential velocity in the evolution of a curve only changes the parametrization and not the shape. Using lagrangian coordinates we obtain that the curve is not advected by the velocity, but by the normal component of the velocity.

27 Thus, in order to ensure that our initial graph continues, at least for a short time interval, being a graph we add the term (1, x f (x)) ρ2 ρ 1 4π P.V. Ξ 1 (x, x η, f ) + Ξ 2 (x, x η, f )dη, R and we obtain that the system (1) posed (with impermeable boundary conditions) on the flat strip S is equivalent to...

28 The equation for the interface; l = π/2 t f (x) = ρ2 ρ 1 ( x f (x) x f (x η)) sinh (η) P.V. 4π R cosh (η) cos(f (x) f (x η)) + ( xf (x) + x f (x η) sinh (η) dη. (2) cosh (η) + cos(f (x) + f (x η)) Remark: The second kernel becames singular if f reaches the boundaries.

29 Or, if we consider general depth l, The equation for the interface; 0 < l < t f (x) = ρ2 ρ 1 8l P.V. R ( x f (x) x f (x η)) sinh ( π cosh ( π 2l η) cos( π 2l + ( xf (x) + x f (x η) sinh ( π cosh ( π 2l η) + cos( π 2l 2l η) (f (x) f (x η))) 2l η) (f (x) + f (x η)))dη. (3) Remark: The second kernel becames singular if f reaches the boundaries. Now we have our equation for the case with finite depth.

30 D. Córdoba and F. Gancedo (CMP-2007) proved that the system (1) posed in the plane is equivalent to The equation for the interface; l = t f = ρ2 ρ 1 ( x f (x) x f (x η))η P.V. 2π η 2 dη. (4) + (f (x) f (x η)) 2 This is the equation for infinite depth. R

31 Given an initial data, f 0, and a depth, l, we define A = f 0 L (R). l By analogy with the case of the water waves (see works by Lannes and references therein), we call it the amplitude (or nonlinearity) parameter. As we said before there are three different regimes: A = 0, which corresponds with infinite depth and has been studied by Córdoba and co-workers. We call it the deep water regime. 0 < A < 1, which is the case with finite depth. This is the topic of this talk. A = 1. We call it the large amplitude regime and it won t be treated here.

32 Ok, we have the equation. The next step is... Step 1 given an initial data and parameters ρ i 0, is there a solution corresponding to this data? And the answer is... it depends on the data.

33 Let us define Rayleigh-Taylor condition RT (x, t) = ( p 2 (z(x, t)) p 1 (z(x, t))) x z(x, t). We say that we are in the R-T stable regime if RT > 0. Then, if we use the incompressibility of the fluids and Darcy s law, we obtain RT (x, t) = (ρ 2 ρ 1 ) x z 1 (x, t) > 0 This condition is satisfied for graphs if the denser fluid is below the lighter one.

34 There are different pictures here: Consider ρ 1 < ρ 2 (the lighter fluid above) and initial data in H k (R), k 3, and being a graph. Then there is a local in time classical solution.

35 Theorem: R-T stable, graph Let f 0 H 3 be an initial data and consider the case ρ 2 ρ 1 > 0. Then there exists a unique classical solution, f, at least for a small time, T = T (f 0 ).

36 (Idea of the proof:) Using that the linear operator is like Λ =, we obtain some a priori estimates for the following energy E[f ](t) = f (t) 2 H 3 + d[f ](t) L, where d[f ] : R 2 R + R + is defined as d[f ](x, η, t) = 1 cosh(η) + cos(f (x) + f (x η)).

37 There are different pictures here: Consider ρ 1 < ρ 2 (the lighter fluid above) and initial data in H k (R), k 3, and being a graph. Then there is a local in time classical solution. Consider ρ 1 > ρ 2 (the heavier fluid above) and initial data in H k (R), k 3, and being a graph. Then the problem is ill-posed in Sobolev spaces. For the moment we forget about our claim of ill-posedness. We will return to this issue in few minutes.

38 There are different pictures here: Consider ρ 1 < ρ 2 (the lighter fluid above) and initial data in H k (R), k 3, and being a graph. Then there is a local in time classical solution. Consider ρ 1 > ρ 2 (the heavier fluid above) and initial data in H k (R), k 3, and being a graph. Then the problem is ill-posed in Sobolev spaces. Consider an analytic initial data which is a graph. Then there is a local in time analytic solution. We remark that in this result, there is not any hypothesis on the RT stablity or even on the finiteness of some integral norms for the initial data like H k (R). This result is not so important and can be obtained following the techniques in the next result, so we skip it.

39 Well.. the previous result are for the case where the initial data is a graph... Is there any result for a general curve? Sure. If you have a general curve wich is analytic then there is a local in time analytic solution corresponding to this initial data.

40 Theorem: R-T stable or R-T unstable, analytic curve Let z 0 = (z0 1, z2 0 ) be an anaytic initial curve flat at infinity. Then there exists a unique analytic solution, z, in the time interval [ T, T ] for a small time, T = T (z 0 ). (Idea of the proof:) Use that we have a first order parabolic equation and apply a Cauchy-Kovalevsky Theorem. In this result we do not require any hypothesis on the RT condition.

41 Skepticism appearing... Wait a moment! You said that the case with infinite depth has been studied by Córdoba and co-workers. Maybe your equation considering the effects coming from the bottom and the top of the spatial domain are only some kind of perturbation that can be handled being patient. Well, I have some simulations to show the evolution of the same initial data with the same physical parameters. Of course the numerical parameters like the size of the spatial grid and so on are the same. Let see some videos...

42 Figure: The evolution of f (x, t) with finite depth

43 Figure: The evolution of f (x, t) with finite depth

44 Figure: The evolution of f (x, t) with infinite depth

45 Step 2 Now, the question is how smooth is the solution?

46 We said in the first five minutes of the talk that there is some kind of Smoothing effect. Indeed, using that the linear term is some kind of Λ = with the correct sign if the Rayleigh-Taylor condition holds, we have Theorem: Smoothing effect Assume that the R-T condition holds and consider an initial data f 0 H 3 (graph). Then the solution, which exists due to the previous Theorem becames analytic for all times t > 0. Moreover it continues analytically on the following growing complex strip B = {x + iξ, ξ < k(f 0 )t}.

47 Figure: The strip of analitycity for f.

48 (Idea of the proof:) Take R a big constant. We define f 2 L 2 (B) = f (x + ikt) 2 dx + f (x ikt) 2 dx, R D[f ](γ) = d + [f ](x + iξ, η) = d [f ](x + iξ, η) = 1 cosh(r) 2 cosh(2 f (γ) ), R cosh 2 (η/3) cosh(η) + cos(f (x + iξ) + f (x + iξ η)), sinh 2 (η/3) cosh(η) cos(f (x + iξ) f (x + iξ η)).

49 We consider the following energy: E B [f ] = f 2 H 3 (B) + d + [f ] L (B)) + d [f ] L (B) + D[f ] L (B). If this energy is finite the function is analytic. Due to the linear part the solution gains half derivative in L 2. As the boundary of the complex strip depends on time you need the previous gain of derivatives to compensates the derivatives that get lost.

50 Lets go with the ill-posedness Theorem. First of all observe that the common approach to this theorem, which is based in some global solutions starting with some small initial data and the correct scaling (like in Córdoba-Gancedo, (CMP-2007)) will not work here. First of all because we (at least today) don t have global solutions, even for small data. Also the spatial operator is too nonlinear to obtain some scaling.

51 Theorem: Ill-posedness There exists solutions f of (2) with ρ 2 < ρ 1 (R-T unstable) such that f 0 H s (R) < ɛ and f (δ) H s (R) =, for any 4 s, 0 < ɛ and small enough δ > 0.

52 (Idea of the proof:) Take g 0 (x) H 3 (R) but g 0 / H 4 (R) and ρ = 2. Now consider the segment λg 0, with 0 λ 1. We obtain f λ (x, t) the solutions to the problem (2), in the RT stable case, with initial data f λ (x, 0) = λg 0 (x). The key point is that the time interval of existence, [0, δ (g 0 )], and the constant k in the growing complex strip can be take uniform in λ. We consider 0 < δ < δ (g 0 ).

53 Now define f λ,δ (x, t) = f λ (x, t + δ). We have f λ,δ (δ) H 4 (R) = λ g 0 H 4 (R) =. Applying Cauchy s integral formula to control x L 2 (R) by some expression involving L 2 (B) and an uniform bound of this last quantity (which holds due to the Theorem about smoothing effect) we get x 4 f λ,δ (0) L 2 (R) c(g 0) λ. δ { } δɛ Now, given ɛ > 0 take 0 < λ = min 1, c(g 0 ) to conclude

54 We finish the part entitled First ideas and results. Let us summarize what we have and let us also compare with the deep water regime (infinite depth). Existence and uniqueness of solutions in different pictures: The results are the same in both regimes. The case with finite depth is harder in the sense that one control more quantities like d +, this is the main difference. Smoothing effect: The results are, again, identical, with the same main difference. Ill-posedness: The proofs of the results are completely different and the technique that works for the infinite depth doesn t work in the case with finite depth (at least up to our knowledge).

55 Now we deal with the qualitative theory: Step 3 We know that, in some cases, there are a smooth local solution, how does the solution behave? Maximum Principle for f L (R) Let f be the unique classical solution of (2) in the R-T stable regime. Then, f satisfies that f (t) L (R) f 0 L (R).

56 Now we deal with the qualitative theory: Step 3 We know that, in some cases, there are a smooth local solution, how does the solution behave? Maximum Principle for f L (R) Let f be the unique classical solution of (2) in the R-T stable regime. Then, f satisfies that f (t) L (R) f 0 L (R).

57 (Idea of the proof:) Use Rademacher Theorem to obtain the ODE for f (x t, t) = f (t) L (R): Define The ODE is θ = f (x t) f (x t η) 2 d dt f (x t) = 4f (x t ) + P.V. Let see the sign of each term... R R, θ = f (x t) + f (x t η). 2 cot( θ) 1 cosh 2 (η/2) tanh 2 (η/2) + cot 2 ( θ) dη tan(θ) 1 cosh 2 (η/2) tanh 2 (η/2) + tan 2 (θ) dη,

58 d dt f (x t) = Good sign cot( θ) 1 4f (x t ) + R cosh 2 (η/2) tanh 2 (η/2) + cot 2 ( θ) dη tan(θ) 1 P.V. cosh 2 (η/2) tanh 2 (η/2) + tan 2 (θ) dη, R

59 Bad sign d dt f (x cot( θ) 1 t) = 4f (x t ) + R cosh 2 (η/2) tanh 2 (η/2) + cot 2 ( θ) dη tan(θ) 1 P.V. cosh 2 (η/2) tanh 2 (η/2) + tan 2 (θ) dη, R

60 d dt f (x cot( θ) 1 t) = 4f (x t ) + R cosh 2 (η/2) tanh 2 (η/2) + cot 2 ( θ) dη Good sign tan(θ) 1 P.V. cosh 2 (η/2) tanh 2 (η/2) + tan 2 (θ) dη. R We need to balance them... The key point is that arctan(tan(f (x t )) tanh(η/2)) = 2f (x t ), thus 4f (x t ) = R 1 cosh 2 ( η ) 2 tan( π 2 f (x t)) tan 2 ( π 2 f (x t)) + tanh 2 ( η 2 )dη,

61 Just compute now... and you obtain d dt f (t) L (R) 0 a.e. t. Now integrate to conclude the result.

62 Once we have that the interface decays, it is interesting to know how fast it does: Decay estimate Let f 0 0, f 0 L 1 H k l (R). Then the L norm satisfies the following inequality d dt f L (t) c( f 0 L 1, f 0 L, ρ 2, ρ 1, l)e Recall that there is dependence on the depth l. π f 0 L 1 l f L (t),

63 Figure: The evolution of f (t) L (R) in the case of infinite depth (left) and with finite depth (right)

64 In the proof of the decay we use that the mean is conserved, i.e. f (x, t)dx = f 0 (x)dx, R and for positive (or negative) initial datas which are integrable this is equivalent to the conservation of the L 1 (R) norm. Recall that, due to the maximum principle the sign is preserved. R

65 To obtain the conservation of this quantity just observe that the equation can be rewritten (taking l = π/2 to simplify notation) as ( ) t f = ρ2 ρ 1 P.V. x arctan tan f (x) f (x η) 2 2π R tanh ( η ) 2 ( ( ) f (x) + f (x η) ( η ) + x (arctan )) tan tanh dη. 2 2 Now integrate to obtain...

66 d f dx = ρ2 ρ 1 dt R 2π ( + arctan ( tan ( ) x P.V. arctan tan f (x) f (x η) 2 R R tanh ( η ) 2 ( ) f (x) + f (x η) ( η ) )) tanh dηdx. 2 2 Where we used that here P.V. R means lim ɛ 0,R B c (0,ɛ) B(0,R).

67 (Idea of the proof:) In order to show the decay we recall that (due to the maximum principle Theorem) we have d dt f (x 1 t) = cosh 2 Π(x, η, t)dη, (η/2) with R Π tan(θ) 1 (tan 2 ( f 0 L (R)) + 1) 2 + tan 2 ( f 0 L (R)) 1 + tan 2 ( f 0 L (R)),

68 Fix the interval [ r, r]. We consider the sets { U 1 = η : η [ r, r], θ f (x } t) 4 and U 2 = { η : η [ r, r], θ < f (x } t). 4 The Conservation of the total mass Lemma gives us a control for the measure of these sets. Now we use the set U 1 to bound the integral and the set U 2 to control the measure of the previous set.

69 R 1 cosh 2 (η/2) Π(x, η, t)dη 1 Π cosh 2 dη (r/2) U 1 tan(f (x t )/4) 1 Π = (tan 2 ( f 0 L (R)) + 1) 2 + tan 2 ( f 0 L (R)) 1 + tan 2 ( f 0 L (R)). Now we have f 0 L 1 (R) = Then we choose r = 2 f 0 L 1 f (x t). R f (x η)dη f (x η)dη (f (x t )/2) U 2, U 2 2r U 2 = U 1.

70 D. Córdoba and F. Gancedo proved that for a (positive) classical solution in the case of infinite depth the following inequality holds d dt f (t) L c(ρ2, ρ 1, f 0 ) f 2 L. Thus, the decay of our system is slower. Corollary There are not non-trivial, one-signed, integrable, steady state solutions.

71 Now we study the last part in this Qualitative theory section: The evolution of x f L (R). What we can show is that if the initial data satisfies some hypothesis on the size and the shape then the derivative of this initial data is depleted. First hypothesis x f 0 L 1, there are no units in this hypothesis! (5) Second hypothesis ( ) π f0 L tan 2l ( π ) x f 0 L tanh, (6) 4l

72 Now we study the last part in this Qualitative theory section: The evolution of x f L (R). What we can show is that if the initial data satisfies some hypothesis on the size and the shape then the derivative of this initial data is depleted. First hypothesis x f 0 L 1, there are no units in this hypothesis! (5) Second hypothesis ( ) π f0 L tan 2l ( π ) x f 0 L tanh, (6) 4l

73 Third hypothesis ( ( π ) ( x f 0 L + 2(cos 2) sec 4 π ( 2l ( 4l 1 + x f 0 L x f 0 L + tan + 4 tan ( π 2l f 0 L ) x f 0 3 L ) π 3 ( π x f 0 L 2l 2 tanh( π 4l ) ) )) 8l 3 6 tanh ( ) π 4l ) ( π 4 x f 0 L cos l f 0 L π 2 4l 2 ) 0 (7) This hypothesis hold if the slope is small and also if the curve has small amplitude.

74 Maximum principle for x f L Let f 0 Hl 3 (R) be a smooth initial data such that conditions (5),(6) and (7) hold, then x f (t) L x f 0 L. (8) Moreover, there exists (x(l), y(l)) such that if the following conditions hold x f 0 L < y(l) and f 0 L < x(l) we have that x f L 1.

75 The previous result gives us conditions on the smallness of A and x f 0 L. Thus, the Physical meaning is that if we are in the long wave regime (small amplitude and large wavelenght) then there is not turning effect, i.e. there are not shocks. The result in the case of infinite depth is that, if initially x f 0 L 1, then the maximum principle holds. We remark that if we take the limit A 0 we recover the result for the deep water regime (see Córdoba-Gancedo CMP-2009)

76 Figure: Different regions in ( f 0 L, x f 0 L ) for the behaviour of x f L when π = 2l.

77 (Idea of the proof:) The idea is the same as for the Maximum principle for f L (R). We apply Rademacher Theorem and we obtain the ODE d dt xf (t) L (R) = t x f (x t ) = 8 x f (x t )+P.V. I 1 +P.V. I 2, R R for some explicit, but intrincate, integrands I i.

78 Now, we can show that P.V. I 1 0 1st and the 2nd hypothesis. P.V. B(0,1) B c (0,1) I 1 + P.V. R I 2 8 x f (x t ) 0 3rd hypothesis. Thus, if initially the curve satisfies all the hypothesis the slope decays locally in time (this is due to the fact that we can not control the evolution of this hypothesis!).

79 From local to global decay What can we do to obtain the global in time decay? Assume now that the initial data satisfies the hypothesis. Then there exist a time 0 < t 1 where the slope decays. If this time interval is finite then, in the worst case, there exist another time t 1 < t 2 where x f (t 2 ) L (R) = x f 0 L (R)... but now the maximum principle for f L (R) ensures that at t 2 the hypothesis hold!, so we obtain decay up to time t 2 + δ... and so on. We conclude the result.

80 We finish the part entitled Qualitative theory. Let us summarize what we have and let us also compare with the deep water regime (infinite depth). Maximum principle for f L (R): The proofs are completely different due to the fact that some bad-signed terms coming from the boundaries appear. Decay for f (t) L (R): The results are completely different and with finite depth the decay that can be shown is slower. This is consistent with the numerical simulations. Maximum principle for x f L (R): The result and the proof are completely different. Now it is important how big is the slope but also where is located the curve while in the infinite depth only the size of the slope plays a role.

81 Step 4 We have local existence but maybe all these solutions exists globally in time... Singularities at finite time Take ρ 2 ρ 1 > 0 (RT stable). Then there exists initial datas f 0 Hl 3(R) such that they develop a blow up for xf (t) L at time t = t 1 and, for t > t 1, the curve is no longer a graph.

82 Step 4 We have local existence but maybe all these solutions exists globally in time... Singularities at finite time Take ρ 2 ρ 1 > 0 (RT stable). Then there exists initial datas f 0 Hl 3(R) such that they develop a blow up for xf (t) L at time t = t 1 and, for t > t 1, the curve is no longer a graph.

83 Figure: The evolution at times t = t 1, 0, t 1 respectively.

84 (Idea of the proof:) We conclude the result by showing that there exists curves z(α) = (z 1 (α), z 2 (α)) such that the following holds: 1. z i are analytic, odd functions. 2. α z 1 (α) > 0, α 0, α z 1 (0) = 0, and α z 2 (0) > α v 1 (0) = α t z 1 (0) < 0. Now take this curve the initial data in our well posedness result without taking into account RT condition. Forward in time we have a curve that can not be parametrized as a graph, z(α, δ). Backward in time we obtain a graph, f (α, 0) = z(α, δ), with finite H 3 norm. Clearly x f (δ) L (R) =.

85 Step 5 And these singularities how they works with the finite depth? Singularities at finite time There exists initial datas z 0 (α) = (z 1 (α), z 2 (α)) such that they achieve the (Rayleigh-Taylor) unstable case only when the depth is finite. If the depth is infinite the same curves are depleted.

86 Step 5 And these singularities how they works with the finite depth? Singularities at finite time There exists initial datas z 0 (α) = (z 1 (α), z 2 (α)) such that they achieve the (Rayleigh-Taylor) unstable case only when the depth is finite. If the depth is infinite the same curves are depleted.

87 Figure: A curve that realizes the Theorem

88 (Idea of the proof:) The proof follows from the curves constructed in the previous result and the correct choice of parameters. For the correct choice of parameters we obtain Figure: Comparison of x v 1 (0) in different regimes

89 Numerical simulations In the numerical simulations we used a collocation method. We consider the cubic spline approximating the data and we desingularize the integrals by removing any zeros. In time we use the classic explicit Runge-Kutta scheme of fourth order.

90