Breakdown of smoothness for the Muskat problem
Abstract
In this paper we show that there exist analytic initial data in the stable regime for the Muskat problem such that the solution turns to the unstable regime and later breaks down, i.e., no longer belongs to C4.
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Breakdown of smoothness for the Muskat problem Ángel Castro, Diego Córdoba, Charles Fefferman, and Francisco Gancedo. 27-1-2012 Abstract In this paper we show that there exist analytic initial data in the stable regime for the Muskat problem such that the solution turns to the unstable regime and later breaks down, i.e., no longer belongs to C4. Contents 1 Introduction 2 2 Preliminary estimates 9 2.1 Basic Properties of Λ................................ 9 2.2 Contour Integrals Analogous to Λ......................... 10 2.3 Applying results derived above . . . . . . . . . . . . . . . . . . . . . . . . . . . 14 2.4 GårdingInequalities................................. 16 2.5 Applications of Gårding’s Inequality . . . . . . . . . . . . . . . . . . . . . . . . 18 3 The unperturbed solution zand the height function h20 3.1 The Rayleigh Taylor Function . . . . . . . . . . . . . . . . . . . . . . . . . . . 20 3.2 TwoHeightFunctions................................ 23 3.3 Rayleigh Taylor in the Complex Domain . . . . . . . . . . . . . . . . . . . . . . 28 3.4 Sobolev Embedding in Ω(t)............................ 29 4 Perturbing the Muskat solution 31 4.1 Differentiating The Muskat Equation . . . . . . . . . . . . . . . . . . . . . . . . 32 4.2 Perturbing Easy and Safe terms . . . . . . . . . . . . . . . . . . . . . . . . . . . 39 4.3 AuxiliaryFunctions................................. 45 4.4 Perturbing Auxiliary Functions . . . . . . . . . . . . . . . . . . . . . . . . . . . 47 4.5 Perturbing the Dangerous Term . . . . . . . . . . . . . . . . . . . . . . . . . . . 52 4.6 TheMainEnergyEstimate............................. 54 4.6.1 The main energy estimate I . . . . . . . . . . . . . . . . . . . . . . . . . 58 4.6.2 The main energy estimate II . . . . . . . . . . . . . . . . . . . . . . . . 63 4.7 Conclusion ..................................... 65 1 arXiv:1201.2525v2 [math.AP] 29 Jan 2012
5 Galerkin approximations and main theorem 65 5.1 Conformal Maps Depending on a Parameter . . . . . . . . . . . . . . . . . . . . 66 5.2 ChangingCoordinates ............................... 69 5.3 GeneralizedMuskat................................. 78 5.4 GalerkinApproximation .............................. 80 5.5 Local Existence for Muskat Solutions in the Complex Domain . . . . . . . . . . 88 5.6 Continuing Back to Negative Time . . . . . . . . . . . . . . . . . . . . . . . . . 89 5.7 Proof of the Main Theorem 1.1 . . . . . . . . . . . . . . . . . . . . . . . . . . . 91 1 Introduction The Muskat equation governs the motion of an interface separating two fluids in a porous medium (e.g. oil and water in sand). See [10]. The same equation governs an interface separating two fluids trapped between two closely spaced parallel vertical plates (a "Helle Shaw cell"). See [9]. In this paper, we exhibit a solution of the Muskat equation for which the interface is initially smooth and stable, but later enters an unstable regime, and still later develops a singularity. Let us briefly recall the Muskat equation and its derivation from physical laws. Imagine the plane R2filled with two incompressible fluids that cannot mix. At time t, let Ω1(t)be the region occupied by FLUID 1, and let Ω2(t)be the region occupied by FLUID 2. These two regions are separated by their common boundary ∂Ω1(t) = ∂Ω2(t)(the "interface"), which we represent as a parametrized curve {z(α, t) : α∈R}, where z(α, t)=(z1(α, t), z2(α, t)) ∈R2. We suppose that all points (x1, x2)∈R2with large positive x2belong to Ω1(t)and that all (x1, x2)with large negative x2belong to Ω2(t). Let pi(x, t)and ui(x, t)denote (respectively) the pressure and fluid velocity in region Ωi(t), for i= 1,2. Then the piand uisatisfy the following conditions. 1. divui= 0 in Ωi(t), (i= 1,2), i.e., the fluids are incompressible. 2. µi κui=−∇pi−g(0, ρi)in Ωi(t)(i= 1,2); this is Darcy’s law [10] for fluid flow in a porous medium. (Here κis a physical constant associated to the porous medium, µiand ρi are physical constants associated to FLUID i, and gis the acceleration due to gravity. In particular, ρiis the density of FLUID i and µiis the viscosity of FLUID i.) Moreover, on the interface, 3. p1=p2 and 4. u1−u2is tangent to the interface. Finally, the interface moves with the fluid, i.e., 5. ∂tz(α, t) = u1(z(α, t), t) + c1(α, t)∂αz(α, t). Here the function c1(α, t)may be chosen arbitrarily; it affects only the parametrization of the interface, which has no intrinsic physical meaning. (We could have used u2in 5, in place of u1.) See, e.g., [7] where the Muskat problem is interpreted in terms of weak solutions. We may study two scenarios, namely 2
ASYMPTOTICALLY FLAT INTERFACES: limα→±∞[z(α, t)−(α, 0)] = 0 and PERIODIC INTERFACES: z(α, t)−(α, 0) is 2π−periodic in αfor fixed t. We restrict attention to the case of two fluids with the same viscosity but different densities. In that case, µ1=µ2but ρ16=ρ2in 2. Equations 1,...,5 take pressure and gravity into account but neglect surface tension. Once we know the interface z(α, t), we can derive explicit formulas for u1,u2,p1and p2 from 1,...,4 and elementary potential theory. Substituting these formulas into 5, and making a convenient choice of the function c1(α, t), we obtain the Muskat equation for z(α, t). For asymptotically flat interfaces, the Muskat equation takes the form ∂tzµ(α, t) = ρ2−ρ1 2πP.V. Z∞ −∞ z1(α, t)−z1(β, t) |z(α, t)−z(β, t)|2[∂αzµ(α, t)−∂βzµ(β, t)]dβ (µ= 1,2), (1.1) where "P.V." denotes a principal value integral (the integral does not converge absolutely at infinity). For periodic interfaces, the Muskat equation is ∂tzµ(α, t) = ρ2−ρ1 2πZR/2πZ sin(z1(α, t)−z1(β, t))[∂αzµ(α, t)−∂βzµ(β, t)] cosh(z2(α, t)−z2(β, t)) −cos(z1(α, t)−z1(β, t))dβ (µ= 1,2). (1.2) Given an initial interface z0(α)=(z0 1(α), z0 2(α)) (α∈R), we try to solve (1.1) or (1.2), with the initial condition z(α, 0) = z0(α) (all α∈R).(1.3) There is no significant difference between the two flavors (1.1), (1.2) of the Muskat equation. Except for the introduction, we will work in the periodic setting. For the moment, we work with asymptotically flat interfaces, since the relevant formulas are then a bit simpler. Suppose first that the interface is the graph of a function, x2=f(x1). Then it is well-known that the initial-value problem (1.1), (1.3) is linearly stable for positive time, if and only if the heavy fluid lies below the interface and the light fluid lies above it (i.e., ρ2> ρ1). If instead the heavy fluid lies above the light fluid (ρ1> ρ2), then the initial-value problem (1.1), (1.3) is linearly stable for negative time. See [7]. If the interface is not the graph of a function, as in Figure 1, then the heavy fluid lies below the light fluid locally near point A, and the reverse holds locally near point B. In that case, the initial-value problem (1.1), (1.3) is linearly unstable, both for positive time (thanks to point B), and for negative time (thanks to point A). Therefore, one asks whether an interface may begin in a (presumably) stable configuration as in Figure 2, and then "turn over" to a (presumably) highly unstable configuration as in Figure 3. For "very small" initial data, Constantin-Pugh [4], Yi [20], Siegel-Caflish-Howison [15], Córdoba-Gancedo [7] and Escher-Matioc [8], prove that such a "turnover" cannot occur. It is shown in [3] that a turnover cannot occur if the initial interface has the form {(x1, x2) : x2=f(x1)} with |f0|<1everywhere. Nevertheless, Castro et al. [1], [2] show that a turnover from Figure 2 to Figure 3 can occur. Moreover, the Muskat solution exists and remains smooth for some time after the turnover, despite the instability of interfaces that look like Figure 3. The reason why the Muskat solution continues to exist in a highly unstable regime is that, by the time the turnover occurs, the interface has become real-analytic. In fact, starting from a 3
Figure 1: Non-graph interface. Figure 2: Initially stable interface. smooth, stable configuration as in Figure 2, the Muskat solution becomes instantly real-analytic. For t > 0, the function z(α, t)defining the interface ∂Ω1(t) = ∂Ω2(t) = {z(α, t) : α∈R} continues analytically to a strip {α∈C:|=α|< h(t)}, with h(t)> ct for small t. As t approaches the turnover time, the function h(t)may decrease, but it remains strictly positive up to the moment when the turnover occurs (i.e. up to the appearance of a vertical tangent to the interface). Therefore, the variants [11], [12] of the Cauchy-Kowalewski theorem show that a real-analytic Muskat solution continues to exist for a short time after the turnover. The interface becomes (presumably) more and more unstable as the turnover progresses. For details see [2]. 4
Figure 3: Turnover. Figure 4: Breakdown. In [7] (following the scheme of [15] for a scenario of two fluids with different viscosities and the same density) the construction of global-in-time solutions leads to ill-posedness (in Sobolev spaces in the Hadamard sense) for the unstable regime by going backwards in time. In [18] and [19] it was proposed, in the unstable regime, a relaxation of the phase distribution that leads to a well-posed problem and captures a mixing profile. From a completely different approach, a convex integration framework, unstable weak solutions are constructed for porous media in [17, 14, 6]. 5
It is natural to believe that at some finite time T, the Muskat solution will break down. However, the technique in [1], [2] is not strong enough to prove that such a breakdown occurs. In this paper, we show that a breakdown occurs, by proving the following result. Theorem 1.1 (Main Theorem). There exists a solution z(x, t)of the Muskat equation (1.2), defined for t∈[t0, t2],α∈R/2πZ, such that the following hold: A. At time t0, the interface is a graph. B. At some time t1∈(t0, t2),the interface is no longer a graph. C. For each time t∈[t0, t2), the interface is real analytic. D. At time t2, the interface is C3smooth but not C4smooth. Thus, at time t2, there is no longer a C4solution of the Muskat equation. Note that (D) includes the assertion that the curve α→z(α, t2)cannot be made C4-smooth by reparametrization. It is not enough merely to show that z(α, t2)/∈C4 loc. In our proof of the Main Theorem, the failure of C4smoothness in (D) will occur at a single point, as shown in Figure 4. To start to understand why Muskat solutions behave as described above, let us linearize equation (1.1) about the trivial solution z(α, t)=(α, 0),a stationary horizontal interface. Perturbing z(α, t)to z(α, t) + δz(α, t), and keeping only terms up to first order in δz in (1.1), we obtain the linear equation ∂t[δzµ(α)] = ρ2−ρ1 2πP.V. Z∞ −∞ ∂α[δzµ(α, t)] −∂β[δzµ(β, t)] α−βdβ (1.4) with µ= 1,2. The operator f→Λf(x) := −1 πP.V. Z∞ −∞ f0(x)−f0(y) x−ydy (1.5) is the square root of the Laplacian in one dimension, i.e., c Λf(ξ) = |ξ|b f(ξ)(all ξ∈R), where b denotes the Fourier transform. The equation (1.4) takes the form ∂tF(x, t) =(ρ1−ρ2) 2ΛF(x, t)(1.6) F(x, 0) =f(x)(given). The operator Λhas a well-known connection with harmonic functions (see [16]). In fact, let u(x+iy)denote the Poisson integral of f(x), defined for y > 0. Similarly, for each fixed t, let U(x+iy, t)denote the Poisson integral of F(x, t). Then (1.6) is equivalent to the initial-value problem ∂tU(x+iy, t) = −(ρ1−ρ2) 2∂yU(x+iy, t) U(x+iy, 0) =u(x+iy), which (formally) admits the solution U(x+iy, t) = u(x+i[y−(ρ1−ρ2) 2t]). In particular setting y= 0, we find that F(x, t) = u(x−i(ρ1−ρ2) 2t).(1.7) 6
This is the formal solution of our linearized Muskat equation (1.6). It is now obvious that the sign of ρ1−ρ2plays a crucial role, and that solutions are realanalytic in the stable case. Indeed the harmonic function u(x+iy)is real-analytic in the upper half-plane y > 0but undefined in the lower half-plane y < 0and unless fis real-analytic. Therefore, if (ρ2−ρ1)t > 0, then (1.7) yields a real-analytic function of x; more precisely, F(x, t)continues analytically to the strip |=x|<(ρ2−ρ1) 2t. On the other hand, if (ρ2−ρ1)t < 0, then (1.7) makes no sense unless the initial datum f(x)continues analytically to the strip |=x|< |(ρ2−ρ1) 2t|. Note that the same “Rayleigh-Taylor" condition that predicts stability as time flows forward also predicts instability as time flows backward. More generally, if we linearize the Muskat equation about any given solution z(α, t) = (z1(α, t), z2(α, t)), we arrive at a linearized Muskat equation of the form Ft(x, t) = −σ(x, t)ΛF(x, t) + HARMLESS TERMS, F(x, 0) = f(x), where σ(x, t) = (ρ2−ρ1) 2 ∂xz1(x, t) (∂xz1(x, t))2+ (∂xz2(x, t))2the “Rayleigh-Taylor function." We therefore expect that our Muskat solution will be real-analytic for t > 0provided σ(x, t)> 0everywhere, and highly unstable or non-existent for t > 0if σ(x, t)<0everywhere. If we pass from initial conditions at time zero to solutions at times t < 0, then the roles of σ > 0and σ < 0are reversed. Observe that (assuming ρ2> ρ1) the function σ(x, t)is positive if and only if the interface (z1(α, t), z2(α, t)) is the graph of a function. In particular the significance of a turnover is that the Rayleigh-Taylor condition σ > 0for linear stability breaks down. In [2], we made a rigorous analysis of the full nonlinear problem and established analytic continuation of Muskat solutions to the time-varying strip {|=x|< h(t)}. This allowed us to construct real-analytic Muskat solutions z0(x, t)defined for t∈[−T, T]that “turn over" at time 0. Here, we prove our Main Theorem 1.1 by constructing Muskat solutions z(x, t)analytic on a carefully chosen time-varying domain Ω(t) = {|=x|< h(<x, t)}. We take as initial datum the interface z(x, τ)at some small positive time τ, and solve the Muskat equation backwards in time until we reach time −τ2. We take our initial datum z(x, τ)to be a small perturbation of z0(x, τ), where z0is the real-analytic “turnover" solution constructed in [2]. Thus, we obtain Muskat solutions zanalytic on Ω(t)and close to z0in a suitable Sobolev norm. For all times t∈[−τ2, τ), we take the domain Ω(t)to be a neighborhood of the real axis, as in Figure 5. However, at time t=τ, we take Ω(t)to be a domain that pinches to a point as shown in Figure 6. In particular an initial datum z(x, τ)may continue analytically to Ω(τ), yet fail to belong to C4as a function of x∈R. We take z(x, τ)to belong to C3but not C4. Solving the Muskat equation (backwards in time) on Ω(t) (t∈[−τ2, τ]) for such an initial datum, we obtain a Muskat solution as in the statement of our Main Theorem 1.1. In particular, assertions (A) and (B) of the Main Theorem hold because zis a small perturbation of z0, which is known to be a graph at time t0=−τ2, but not at time t1=τ2.(Recall that z0turns over at time zero.) Assertion (C) holds because z(x, t)continues analytically to Ω(t), with Ω(t)as in Figure 5 for tless than t2=τ. Finally, assertion (D) holds because we picked our initial datum z(·, τ)to be in C3but not in C4. 7
Figure 5: Domain Ω(t)at time t < τ. Figure 6: Domain Ω(t)at time t=τ. 8
This concludes our summary of the proof of our main theorem. We warn the reader that the summary is somewhat oversimplified. The correct version of our proofs appears in the sections below. 2 Preliminary estimates This section is devoted to proving crucial estimates which will be used in sections 4 and 5. 2.1 Basic Properties of Λ Let T=R/2πZand let f∈C2(T). For x∈Twe define Λf(x) = −1 πZy∈T 1 2cot x−y 2(f0(x)−f0(y))dy. (2.1) For x∈T, let ||x|| =distance(x, 2πZ).(2.2) Then a carefully justified integration by parts yields Λf(x) = 1 4πlim ε→0Z y∈T ||x−y||>ε csc2x−y 2(f(x)−f(y))dy (2.3) where the limit converges uniformly in x. Consequently, Zx∈T f(x)Λf(x)dx =1 4πlim ε→0ZZ x,y∈T ||x−y||>ε f(x) csc2x−y 2(f(x)−f(y))dydx =1 8πlim ε→0ZZ x,y∈T ||x−y||>ε (f(x)−f(y)) csc2x−y 2(f(x)−f(y))dydx i.e., Zx∈T f(x)Λf(x)dx =1 8πZ Zx,y∈T csc2x−y 2|f(x)−f(y)|2dydx. (2.4) More generally, let A∈C2(T), and suppose that |A00| ≤ Con T. We write C0,C00, etc. to denote constants depending only on C. Then the expression (2.3) gives Zx∈T A(x)f(x)Λf(x)dx =1 4πlim ε→0ZZ x,y∈T ||x−y||>ε A(x)f(x) csc2x−y 2(f(x)−f(y))dydx =1 8πlim ε→0ZZ x,y∈T ||x−y||>ε (A(x)f(x)−A(y)f(y)) csc2x−y 2(f(x)−f(y))dydx 9
2.4 Gårding Inequalities First we work on R. Later on, we will switch over to T. Here ΛRwill be the square root of (−∆) in one dimension, D=1 i d dx on Rand Πσ:L2(R)→L2(R) for σ=±1is the orthogonal projection onto either positive frequencies (σ= 1) or negative frequencies (σ=−1). Note Πσ=1 2(I+iσH), where His the Hilbert transform ( ˆ H= −i sign(ξ)). Commutator estimates: Let abe a real-valued function on R. We suppose a0∈L∞(R), and write afor the operator f7→ a·fon L2(R). Then [a, H]D:L2(R)→L2(R) is bounded, with norm at most C||a0||L∞(R), where Cis a universal constant. That follows from Calderón’s theorem on the commutator integral (see [16]). Since ΛR=iDH, it follows that [a, H]ΛR:L2(R)→L2(R) is bounded, with norm at most C||a0||L∞(R). Since Πσ=1 2(I+iσH), it follows that ||[a, Πσ]D||,||[a, Πσ]ΛR|| ≤ C||a0||L∞(R). Next, recall that 2<f·ΛRf≥ΛR|f|2 pointwise (see [5]). If a≥0everywhere and Λa∈L∞, then <ZR a(x)f(x)ΛRf(x)dx≥1 2ZR a(x)ΛR|f|2dx =1 2ZR (ΛRa) (x)|f(x)|2dx ≥ −1 2||ΛRa||L∞||f||2 L2. Now suppose a,bare real, a0,b0∈L∞;ΛRa,ΛRb∈L∞;a≥ |b|pointwise. Note that ΛR=Pσ=±1ΠσΛRΠσand D=Pσ=±1ΠσDΠσ. Therefore, aΛR+bD =X σ=±1 {[a, Πσ]ΛRΠσ+ [b, Πσ]DΠσ} +X σ Πσ(aΛR+bD)Πσ. The terms in curly brackets are operators with norm at most C||a0||L∞+C||b0||L∞on L2(R), with Ca universal constant. Therefore, <h(aΛR+bD)f, fi=Error +X σ <hΠσ(aΛR+bD)Πσf, fi =Error +X σ <h(aΛR+bD)Πσf, Πσfi =Error +X σ <h(a+σb)ΛRΠσf, Πσfi, 16
since DΠσ=σΛRΠσ; here |Error| ≤ C(||a0||L∞+||b0||L∞)||f||2 L2. Recalling that a+σb ≥0on R, we see that <h(aΛR+bD)f, fi≥−C{||a0||L∞+||b0||L∞+||ΛRa||L∞+||ΛRb||L∞} ||f||2 L2, where Cis a universal constant. Now suppose a, b real, a≥ |b|on R,|a|,|b|,|a0|,|b0|,|a00|,|b00| ≤ 1. Then ΛRa(x) = 1 πP.V. ZR a(x)−a(y) (x−y)2dy. Now Z|x−y|>1 a(x)−a(y) (x−y)2dy≤Z|x−y|>1 |a(x)|+|a(y)| (x−y)2dy ≤C, and P.V. Z|x−y|≤1 a(x)−a(y) (x−y)2dy=P.V. Z|x−y|≤1 a(x)−a(y) + a0(x)(x−y) (x−y)2dy≤C. Similarly for b. Therefore, a0,b0,ΛRa,ΛRbare bounded on R. Therefore, we obtain the following result: Suppose a, b are real functions on R, such that |a|,|b|,|a0|,|b0|,|a00|,|b00| ≤ 1; and suppose a≥ |b|everywhere on R. Then <h(aΛR+bD)f, fi≥−C||f||2 L2 for f∈C2 0(R), where Cis a universal constant. Rescaling, we obtain the following result: Lemma 2.4 Let 0< δ < 1, and let a, b be real functions on R, such that |a| ≤ δ,|b| ≤ δ, |a0| ≤ 1,|b0| ≤ 1,|a00| ≤ δ−1,|b00| ≤ δ−1everywhere, and also a≥ |b|everywhere. Then <h(aΛR+bD)f, fi≥−C||f||2 L2 for f∈C2 0(R), where Cis a universal constant. Next, we transfer our result from Rto T. Lemma 2.5 Let 0< δ < 1, and let a, b be real-valued functions on T. Suppose that d dxj a(x),d dxj b(x)≤δ1−j for j= 0,1,2and x∈T. Suppose also that |b| ≤ afor all x∈T. Then, for any smooth function fon T, we have <ZT f(x) (a(x)Λf(x) + b(x)if0(x)) dx≥ −CZT |f(x)|2dx, (2.24) where Cis a universal constant. 17
Proof: Suppose first that supp(a),supp(b)⊂[−1,1]. We identify Twith [−π, π], and let f= 0 on R\[−π, π]. Thus, we may regard fas a function on R. For x∈[−1,1], we have Λf(x)−ΛRf(x) = ZR K(x−y)f(y)dy for a kernel K, where ||K||L∞(R)≤C. Hence, in proving (2.24), we may replace Λby ΛR. Estimate (2.24) now follows from lemma (2.4). We have established (2.24) in the case where supp(a),supp(b)⊂I, where Iis any interval of length 2 in T. Conclusion (2.24) in the general case now follows by using a partition of unity. 2.5 Applications of Gårding’s Inequality In this section, we are given functions h,ht∈C3(T)(with norms bounded a priori). As usual, we assume that h > 0, and we take Γ+={x+ih(x) : x∈T},Γ−={x−ih(x) : x∈T}. We let Fbe analytic on the region of T+iRbounded by Γ+and Γ−and we suppose that Fis smooth on the closure of this region. As usual, we set f+(x) = F(x+ih(x)),f−(x) = F(x−ih(x)). Let ˜σ1(x)and θ(x)be functions on T. As in section (2.3), we want a lower bound for X≡ < Zx∈T f+(x)θ(x)−˜σ1(x) πZu∈T 1 2cot (x+ih(x)) −(u+ih(u)) 2 ×(F0(x+ih(x)) −F0(u+ih(u)))(1 + ih0(u))du +iht(x)F0(x+ih(x)) odx. (2.25) We assume that the C2norm of ˜σ1is bounded a priori and that 0≤θ≤1on T. According to (2.7) and (2.10) we have X=<Zx∈T f+(x)θ(x)˜σ1(x)(1 + ih0(x))−1Λf+(x) + iht(x)F0(x+ih(x))dx +ˇ Err, with |ˇ Err| ≤ C||f+||2 L2(T). Also, F0(x+ih(x)) = (1 + ih(x))−1f0 +(x), as noted before. Therefore, X=<Zx∈T f+(x)θ(x)(1 + ih0(x))−1˜σ1(x)Λf+(x) + iht(x)f0 +(x)dx +ˇ Err, (2.26) with |ˇ Err| ≤ C||f+||2 L2(T). We now suppose that: The C2norms of θ(x)˜σ1(x)and θ(x)ht(x)are bounded a priori. (2.27) Then the same holds for θ(x)(1 + ih0(x))−1˜σ1(x)and θ(x)(1 + ih0(x))−1ht(x). Therefore, formula (2.6) gives <Zx∈T f+(x)θ(x)(1 + ih0(x))−1˜σ1(x)Λf+dx 18
=<Zx∈T f+(x)<θ(x)(1 + ih0(x))−1˜σ1(x)Λf+(x)dx +ˇ ˇ Err, (2.28) with |ˇ ˇ Err| ≤ C||f+||2 L2(T). Also, if iB(x)is real, then <Zx∈T f+(x)B(x)if0 +(x)dx =Zx∈T iB(x)<(f+(x)f0 +(x)) =Zx∈T iB(x)1 2 d dx|f+(x)|2dx =−i 2Zx∈T B0(x)|f+(x)|2dx. Applying this with B(x) = i={θ(x)(1 + ih0(x))−1ht(x)}, we see that <Zx∈T f+(x)θ(x)(1 + ih0(x))−1ht(x)if0 +(x)dx =<Zx∈T f+(x)<{θ(x)(1 + ih0(x))−1ht(x)}if0 +(x)dx +ˇ ˇ ˇ Err, (2.29) with |ˇ ˇ ˇ Err| ≤ C||f+||2 L2(T). Putting (2.28) and (2.29) into (2.26) we see that X=<Zx∈T f+(x)A(x)Λf+(x) + B(x)if0 +(x)dx +Err, (2.30) where |Err| ≤ C||f+||2 L2(T) and A(x) = <{θ(x)(1 + ih0(x))−1˜σ1(x)},(2.31) B(x) = <{θ(x)(1 + ih0(x))−1ht(x))}.(2.32) We assume that d dxj A(x),d dxj B(x)≤Cδ1−j(2.33) for 0≤j≤2; and that |B(x)| ≤ A(x)for all x∈T.(2.34) From (2.30) and Gårding’s inequality we obtain the estimate X≥ −C||f+||2 L2(T).(2.35) Thus, we have proven the following result: Lemma 2.6 Let θ,h,ht,˜σ1be functions on T, and let 0< δ < 1be a number. We make the following assumptions: 1. h > 0. 2. The norm of ˜σ1in C2and the norms of hand htin C3are bounded a priori. 3. The C2norms of θ(x)˜σ1(x)and θ(x)ht(x)are bounded a priori. 19
4. supp θ ⊂[−δ, δ]. 5. d dx j<{θ(x)(1 + ih0(x))−1˜σ1(x)},d dx j<{θ(x)(1 + ih0(x))−1ht(x)}≤Cδ1−j for 0≤j≤2,x∈T. 6. |<{θ(x)(1 + ih0(x))−1ht(x)}| ≤ <{θ(x)(1 + ih0(x))−1˜σ1(x)}for all x∈T. 7. 0≤θ≤1. Then <Zx∈T f+(x)θ(x)−˜σ1(x) πZu∈T 1 2cot (x+ih(x)) −(u+ih(u)) 2 ×(F0(x+ih(x)) −F0(u+ih(u)))(1 + ih0(u))du +iht(x)F0(x+ih(x)) odx ≥ −C0||f+||2 L2(T)2, where f+(x) = F(x+ih(x)). Here, C0depends only on the a priori bounds in 2, 3 and on the constant Cin 5. Remark 2.7 An analogous result holds for f−(x) = F(x−ih(x)). 3 The unperturbed solutions zand the height function h In this section we study the Rayleigh-Taylor function associated to the unperturbed solutions. We construct our time-varying domain of analyticity Ω(t) = {|=x| ≤ h(<x, t)}adapted to the Rayleigh-Taylor function. 3.1 The Rayleigh Taylor Function We consider the analytic solution z(x, t) = (z1(x, t), z2(x, t)), for t∈[−T, T ](T > 0), found in [2]. We will call z(x, t)the unperturbed solution and we recall that we can choose the origin of time so that z(x, t)satisfies the following properties: •z1(x, t)−xand z2(x, t)are 2π−periodic functions and real for xreal. •z1(x, 0) and z2(x, 0) are odd functions. •z1(0,0) = 0, •∂xz1(0,0) = 0, •∂2 xz1(0,0) = 0, •∂3 xz1(0,0) >˜c1>0, •∂t∂xz1(0,0) <−˜c2<0. •∂xz2(0,0) > c > 0. Then since |∂4 xz1(x, 0)| ≤ C3it follows from Taylor’s theorem that c4x2≤∂xz1(x, 0) ≤C5x2,(3.1) for |x| ≤ c6. Similarly, since ∂t∂xz1(0,0) <−˜c2<0 20
and |∂2 x∂tz1(x, 0)| ≤ C7 it follows that ∂t∂xz1(x, 0) <−c8<0 for |x|< c9. Since also |∂2 t∂xz1(x, t)| ≤ C10 for |x| ≤ c9,|t| ≤ c11, it follows in turn that ∂xz1(x, 0) −C12t≤∂xz1(x, t)≤∂xz1(x, 0) −c13t(3.2) for |x| ≤ c9,0≤t≤c14. Combining our estimates (3.1) and (3.2), we find that c4x2−C12t≤∂xz1(x, t)≤C5x2−c13t(3.3) for |x| ≤ c15,0≤t≤c16. Similarly we have that c0 4x2−c0 12t≤∂xz1(x, t)≤C0 5x2−C0 13t(3.4) for |x| ≤ c15,−c16 ≤t≤0. Also, since ∂2 xz1(0,0) = 0 and z1(x, t)has C3−norm at most Con {|x| ≤ c, |t| ≤ c}, it follows that |∂2 xz1(x, t)| ≤ C17|t|+C17|x|(3.5) for |x| ≤ c18,|t| ≤ c19. We have also |∂3 xz1(x, t)| ≤ C20 (3.6) for |x|,|t| ≤ c. Finally we notice that ∂xz1(x, −t)> c20tfor all x∈Tand 0< t < c21.(3.7) Now we define the Rayleigh-Taylor function associated to the unperturbed solution as σ0 1(x, t) = −2π∂xz1(x, t) (∂xz1(x, t))2+ (∂xz2(x, t))2(3.8) Therefore σ0 1(x, t)satisfies the following: 1. σ0 1(·, t)is analytic on {x+iy :x∈T,|y| ≤ cb}with |σ0 1(x+iy, t)| ≤ C, for all x+iy as above and for all t∈[−τ, τ]. 2. σ0 1(x, t)is real for x∈T,t∈[−τ, τ]. 3. σ0 1has a priori bounded C2norm as a function of (x, t)∈T×[−τ, τ]. 4. σ0 1(0,0) = 0. 5. ∂xσ0 1(0,0) = 0. 6. ∂2 xσ0 1(0,0) = −c2<0. 7. ∂tσ0 1(0,0) = c1>0. 21
For complex x(|=x|< c0), we have σ0 1(x, t) = σ0 1(0, t) + x∂xσ0 1(0, t) + 1 2∂2 xσ0 1(0, t)x2+O(||x||3), where ||x|| =||<x|| +|=x| ≈ distance(x, 2πZ). Moreover, σ0 1(0, t) = c1t+O(t2), and we have ∂2 xσ0 1(0, t) = −c2+O(t) ∂xσ0 1(0, t) = O(t). Hence, for complex xsuch that |x| ≤ 10−3and |=x|< c0we have σ0 1(x, t) = c1t+O(t2) + O(xt)−c2 2x2+O(tx2) + O(x3). Since also |O(xt)| ≤ c2 1000 x2+O(t2), we conclude that |σ0 1(x, t)−[c1t−c2 2x2]| ≤ 10−2[c1t+c2 2x2], if |x|< c,t∈[0, τ]. In particular, c1(1 −10−2)t−c2 2(1 + 10−2)x2≤σ0 1(x, t)≤c1(1 + 10−2)t−c2 2(1 −10−2)x2(3.9) for real x, if |x|< c,t∈[0, τ]. Similarly, we have that c0 1(1 + 10−2)t−c2 2(1 + 10−2)x2≤σ0 1(x, t)≤c0 1(1 −10−2)t−c2 2(1 −10−2)x2(3.10) for real x, if |x|< c,t∈[−τ, 0]. On the other hand, for the Muskat solution in [2], (3.8) gives σ0 1(x, 0) <−c0if c 100 ≤ |x| ≤ π,xreal.(3.11) Then, by taking τsmall enough, we achieve σ0 1(x, t)<−c0 2if c 100 ≤ |x| ≤ π,xreal, t∈[0, τ].(3.12) If h(x, t)is a positive functions defined for x∈T,0≤t≤τ, and if 0< h(x, t)<1 2cbfor all such (x, t)(see property 1 above),(3.13) then |σ0 1(x±ih(x, t), t)−σ0 1(x, t)| ≤ Ch(x, t).(3.14) We will work with a perturbed Rayleigh-Taylor function σ1(x, t), defined and analytic on {|=x|< h(<x, t)}for each fixed t∈[0, τ]; and smooth on the closure of that region. We assume that ∂j x(σ1(x, t)−σ0 1(x, t))≤τ10 (3.15) for 0≤j≤2, and for |=x| ≤ h(<x, t),0≤t≤τ. We have not yet picked the function h(x, t). This we do in the next section. 22
Figure 7: Red: Function h(x, t)at time t=τ2, with κ= 0. Blue: Function h(x, t)at time t=τ, with κ= 0. 3.2 Two Height Functions In this section, we suppose we are given a Rayleigh-Taylor function σ0 1as in the section (3.1). We pick A >> 1and 0< τ,κ,λ << 1real numbers. We first pick Alarge enough, then pick τsmall enough (depending on A). Having picked Aand τ, we then take 0<λ<τ100 and 0< κ < τ100. We fix such A,τ,λuntil the end of the paper. The parameter κwill be held fixed until section 5.7 below, in which will let κtend to zero. From now on, we suppose A,τ,λand κ have been picked as above. For x∈Tand τ2≤t≤τ, we define h(x, t) = A−1(τ2−t2)+(A−1−A(τ−t)) sin2x 2+κ. (3.16) See Figures (7) and (8). Thus, ∂th(x, t) = −2A−1t+Asin2x 2.(3.17) See Figures (9) and (10). If ||x|| ≥ 10A−1t1 2, then sin2(x/2) ≥10−2||x||2≥A−2t, hence (3.17) yields |∂th(x, t)| ≤ 2A−1t+Asin2x 2for all x∈T,t∈[τ2, τ], (3.18) and |∂th(x, t)| ≤ 3Asin2x 2if ||x|| ≥ 10A−1t1 2,t∈[τ2, τ].(3.19) On the other hand, (3.16) yields h(x, t)≥1 2A−1sin2(x/2) for all x∈T,t∈[τ2, τ). Hence, (3.19) implies |∂th(x, t)| ≤ 6A2h(x, t)for ||x|| ≥ 10A−1t1 2,t∈[τ2, τ].(3.20) Next, let ||x|| ≤ c 2with cas in inequality (3.9). Applying that estimate we learn that σ0 1(x, t) + ∂th(x, t)−A1 2h(x, t) 23
Figure 8: Zoom of Figure 1 around the point (0,0). Figure 9: Blue: R-T function at time t=τ. Red: Function ∂th(x, t)at time t=τ. Green: Function ∂th(x, t)at time t=τwith a larger value of A. Figure 10: Zoom of Figure 3 around the point (0,0). 24
≥˜c1t−˜ C1sin2x 2+−2A−1t+Asin2x 2−A1 2hA−1(τ2−t2)+(A−1−A(τ−t)) sin2x 2+κi =h˜c1t−2A−1t−A−1 2(τ2−t2)−A1 2κi+h−˜ C1+A−(A−1 2−A3 2(τ−t))isin2x 2. The second term in square brackets is positive (since Ais large but A3/2τis small), and the first term in square brackets is greater than cτ2. Hence, σ0 1(x, t) + ∂th(x, t)−A1 2h(x, t)≥cτ2if ||x|| ≤ 1 2c,t∈[τ2, τ]. Remark 3.1 The reason we restrict to t≥τ2is that, if we allowed t << τ2, then the first term in square brackets above might be negative. On the other hand, suppose that ||x|| >1 2c,t∈[τ2, τ]. Then |σ0 1(x, t)| ≤ C, and (3.17) yields ∂th(x, t)≥cA, while (3.16) implies that 0≤h(x, t)≤C0A−1. Hence, in this case, σ0 1(x, t) + ∂th(x, t)−A1 2h(x, t)≥cτ2. Thus, in all cases, we have σ0 1(x, t) + ∂th(x, t)−A1 2h(x, t)≥cτ2for x∈T,t∈[τ2, τ](3.21) Next, suppose ||x|| ≤ 20A−1t1 2,t∈[τ2, τ]. Then inequality (3.9) gives σ0 1(x, t)≥ct, while (3.16) gives 0≤h(x, t)≤CA−1tand (3.17) gives |∂th(x, t)| ≤ CA−1t. Hence, σ0 1(x, t)− |∂th(x, t)| − A1 2h(x, t)≥c0t≥c0τ2for ||x|| ≤ 20A−1t1 2,t∈[τ2, τ].(3.22) Next, we investigate the x−derivatives of h(x, t),ht(x, t)and σ0 1(x, t). From (3.16) and (3.17), we see that ∂j xh(x, t)≤CA−1for x∈T,t∈[τ2, τ],0≤j≤3; (3.23) and ∂j xht(x, t)≤CA for x∈T,t∈[τ2, τ],0≤j≤3.(3.24) Also, ∂j xσ0 1(x, t)≤Cfor x∈T,t∈[τ2, τ],0≤j≤2.(3.25) (See section (3.1)). Now suppose that ||x|| ≤ 20A−1t1 2. According to estimate (3.9), we have ct ≤σ0 1(x, t)≤Ct for ||x|| ≤ 20A−1t1 2,t∈[τ2, τ].(3.26) Since (3.26) holds for all x∈[−20A−1t1 2,20A−1t1 2], the mean-value theorem implies that |∂xσ0 1(x, t)| ≤ CAt1 2somewhere in [−20A−1t1 2,20A−1t1 2]. Hence, by (3.25), we have ∂xσ0 1(x, t)≤CAt1 2,for all ||x|| ≤ 20A−1t1 2,t∈[τ2, τ].(3.27) From (3.25), (3.26), (3.27), we see that ∂j xσ0 1(x, t)≤CA t1 22−jfor all ||x|| ≤ 20A−1t1 2,t∈[τ2, τ],0≤j≤2.(3.28) 25
4.1 Differentiating The Muskat Equation In this section, we suppose we are given a pair of functions zµ(x, t)(µ= 1,2), z1(x, t)−xand z2(x, t)periodic of period 2πin x, analytic in the complex variable xin the region Ω(t) = {|=x|< h(<x, t)}, and smooth in (x, t)on the closure of this region. We write Γ±(t) = {x±ih(x, t) : x∈T} for the two contours bounding the above region. Later, we will take h(x, t)(of this section) to be either our previous h(x, t)or our previous ~(x, t), depending on whether we work with t∈[τ2, τ]or with t∈[−τ2, τ2]. For now, we simply suppose that h(x, t)is a positive, smooth function of xfor each fixed t. We suppose that our zµsatisfy the Muskat equation ∂tzµ(x, t) = Zy∈T sin (z1(x, t)−z1(x+y)) (∂xzµ(x, t)−∂xzµ(x+y, t)) cosh (z2(x, t)−z2(x+y)) −cos (z1(x, t)−z1(x+y))dy. (4.1) Note the sign in (4.1) and note that there is no factor 1 2πin front of the integral. In (4.1), xand y∈Tare real. Later we will use the analyticity of zµ(x, t)in xto move into the complex plane. We will assume the following COMPLEX CHORD-ARC CONDITION: Let x,y∈Ω(t). Then |cosh (z2(x, t)−z2(y, t)) −cos (z1(x, t)−z1(y, t)) | ≥ cCA [||<(x−y)|| +|=(x−y)|]2. (4.2) Let kbe a large enough positive integer constant (here, k= 4 is enough). We apply the k−th derivative ∂k xto both sides of (4.1). The goal of this section is to see what results. We obtain ∂t∂k xzµ(x, t) =X k+k0=k c(k, k)Zy∈T ∂k xsin (z1(x, t)−z1(x+y)) cosh (z2(x, t)−z2(x+y)) −cos (z1(x, t)−z1(x+y)) ×∂k0 x{∂xzµ(x, t)−∂xzµ(x+y, t)}dy, (4.3) where c(k, k)are harmless coefficients and c(0, k)=1. We fix t,yand write ˆzµ(x) = zµ(x, t)−zµ(x+y, t). The integrand in (4.3) is then (with k+k0=k): ∂k xsin(ˆz1(x)) cosh(ˆz2(x)) −cos(ˆz1(x))h∂k0+1 xˆzµ(x)i =X k]+kb=k ˜c(k], kb)h∂k] x{sin(ˆz1(x))}i∂kb x1 cosh(ˆz2(x)) −cos(ˆz1(x))h∂k0+1 xˆzµ(x)i, where ˜c(k], kb)are harmless coefficients. Now ∂k] x{sin(ˆz1(x))}is a linear combination (with harmless coefficients) of terms hsin(m])(ˆz1(x))im] Y j=1 ∂k] j xˆz1(x), 32
where k] j≥1for each j,k] 1+···+k] m]=k]and sin(m])denotes the m]th derivative of the sine function. Also, ∂kb x1 cosh(ˆz2(x)) −cos(ˆz1(x)) is a linear combination, with harmless coefficients, of terms [cosh(ˆz2(x)) −cos(ˆz1(x))]−(m+1) m Y j=1 ∂kj x{cosh(ˆz2(x)) −cos(ˆz1(x))}, with kj≥1for each j, and with k1+···+km=kb. Therefore, the integrand in (4.3) is a linear combination, with harmless coefficients, of terms hz(k0+1) µ(x)ihsin(m])(ˆz1(x))im] Y j=1 ˆz(k] j) 1(x) ×[cosh(ˆz2(x)) −cos(ˆz1(x))]−(m+1) m Y j=1 ∂kj x{cosh(ˆz2(x)) −cos(ˆz1(x))}, with kj,k] j≥1for each j, and with k] 1+···+k] m]+k1+···+km+k0=k. Furthermore, for kj≥1, ∂kj x{cosh(ˆz2(x))} is a sum with harmless coefficients of terms cosh(mj)(ˆz2(x)) mj Y i=1 ˆzkji 2(x), with mj≥1, with kji ≥1, and with Pmj i=1 kji =kj. Similarly, ∂kj x{cos(ˆz1(x))} is a sum with harmless coefficients of terms cos(mj)(ˆz1(x)) mj Y i=1 ˆz(kji) 1(x), with mj≥1, with each kji ≥1, and with Pmj i=1 kji =kj. Consequently, the integrand in (4.3) is a linear combination with harmless coefficients, of terms hz(k0+1) µ(x)i hsin(m])(ˆz1(x))im] Y j=1 ˆz(k] j) 1(x) ×[cosh(ˆz2(x)) −cos(ˆz1(x))]−(m+1) m0 Y j=1 hcosh(m0 j)(ˆz2(x))im0 j Y i=1 hˆz(k0 ji) 2(x)i × m00 Y j=1 hcos(m00 j)(ˆz1(x))im00 j Y i=1 hˆz(k00 ji) 1(x)i ,(4.4) 33
where all the k] j,k0 ji,k00 ji ≥1, all the m0 j,m00 j≥1, (4.5) and k0+ m] X j=1 k] j+ m0 X j=1 m0 j X i=1 k0 ji + m00 X j=1 m00 j X i=1 k00 ji =k, m0+m00 =m. (4.6) Let us call each expression in square brackets in (4.4) a BASIC FACTOR. Thus, the integrand in (4.3) is a linear combination, with harmless coefficients, of products of basic factors. We assign to each basic factor a weight, as follows. •[ˆz(k) 1(x)] has weight 1. •[ˆz(k) 2(x)] has weight 1. •[sin(m])(ˆz1(x))] has weight 1 if m]is even, 0 if m]is odd. •[cos(m00 j)(ˆz1(x))] has weight 1 if m00 jis odd, 0 if m00 jis even. •[cosh(m0 j)(ˆz2(x))] has weight 1 if m0 jis odd, 0 if m0 jis even. •[cosh(ˆz2(x)−cos(ˆz1(x))]−(m+1) has weight −2(m+ 1). The significance of this notion is that if φ(x)is a basic factor with weight w, then |φ(x)|= O(||y||w)when ||y|| is small, uniformly in x. That is because ˆz(k) µ=∂k xzµ(x, t)−∂k xzµ(x+y, t) and because of the COMPLEX CHORD-ARC CONDITION. We define the TOTAL WEIGHT of a product of basic factors to be the sum of the weights of those basic factors. We now check that (4.5) and (4.6) implies that the sum of the weight of the basic factors in (4.4) is greater than or equal to zero. To see this, we note that the total weight of cosh(m0 j)(ˆz2(x)) m0 j Y i=1 ˆzk0 ji 2(x) is 1m0 jodd +m0 j≥2, since m0 j≥1by (4.5). Similarly, the total weight of cos(m00 j)(ˆz1(x)) m00 j Y i=1 ˆz(k00 ji) 1(x) is 1m00 jodd +m00 j≥2, since m00 j≥1by (4.6). Also, the total weight of hsin(m])(ˆz1(x))im] Y j=1 ˆz(k] j) 1(x) , is 1m]even +m]≥1, since m]≥0. Consequently, the total weight of the terms (4.4) is greater than or equal to 1+1−2(m+ 1) + 2m0+ 2m00 = 0, since m0+m00 =mby (4.6). This completes the verification that the total weight of (4.4) is bigger than or equal to zero. We classify the terms (4.4) as follows. 34
•Suppose [ˆz(k+1) 1(x)] or [ˆz(k+1) 2(x)] appears in the product (4.4). Then we say that the term (4.4) is dangerous. •If [ˆz(k) 1(x)] or [ˆz(k) 2(x)] appears in a term (4.4), then we say that the term (4.4) is safe. •If the only [ˆz(k) 1(x)],[ˆz(k) 2(x)] appearing in a term (4.4) have k≤k−1, then we call term (4.4) easy. Thus, any term (4.4) (for which (4.5) and (4.6) holds) is dangerous, safe, or easy. These conditions are mutually exclusive thanks to (4.5) and (4.6) and because we will take k≥4. Let us identify the dangerous and the safe terms of the form (4.4). Since k0+ m] X j=1 k] j+ m0 X j=1 m0 j X i=1 k0 ji + m00 X j=1 m00 j X i=1 k00 ji =k, we see that the only possible dangerous term (4.4) is the term arising from k0=k, all k] j,k0 ji, k00 ji = 0. Since all the k0 ji,k00 ji ≥1by (4.5) and (4.6), we must have k0=k,m]= 0,m0= 0, m00 = 0. That is, the only dangerous term of the form (4.4) is hˆz(k+1) µ(x)i[sin(ˆz1(x))] 1 cosh(ˆz2(x)) −cos(ˆz1(x)) This term appears in the integrand of (4.3) with a coefficient 1. Next, we look for all the safe terms (4.4). We distinguish several cases. Case 1: The basic factor [ˆz(k0+1) 1(x)] may have k0=k−1. In this case (4.5) and (4.6) imply that there is exactly one nonzero number among the k] j,k0 ji,k00 ji; and that number is 1. Since all the k] j,k0 ji,k00 ji are at least 1 by (4.5) and (4.6), there remain only the following possibilities: •m]= 1,k] 1= 1,m0= 0,m00 = 0 •m]= 0,m0= 1,m0 1= 1,k0 11 = 1,m00 = 0 •m]= 0,m0= 0,m00 = 1,m00 1= 1,k00 11 = 1. These possibilities give rise to the following term of the form (4.4): •hˆz(k) µ(x)i[cos(ˆz1(x))] 1 cosh(ˆz2(x)) −cos(ˆz1(x))[ˆz0 1(x)] •hˆz(k) µ(x)i[sin(ˆz1(x))] 1 cosh(ˆz2(x)) −cos(ˆz1(x))2 [sinh(ˆz2(x))][ˆz0 2(x)] •hˆz(k) µ(x)i[sin(ˆz1(x))] 1 cosh(ˆz2(x)) −cos(ˆz1(x))2 [sin(ˆz1(x))][ˆz0 1(x)] Case 2: A basic factor [ˆz(k] j) 1(x)] in (4.4) has k] j=k. Then (4.5) and (4.6) show that k0= 0, m]= 1,m0= 0,m00 = 0,m= 0. This gives rise to a term (4.4) of the form •ˆz0 µ(x)[cos(ˆz1(x))] hˆz(k) 1(x)i1 cosh(ˆz2(x)) −cos(ˆz1(x)) 35
Case 3: A basic factor [ˆz(k0 ji) 2(x)] in (4.4) has k0 ji =k. Then (4.5) and (4.6) show that k0= 0, m]= 0,m0= 1,m0 1= 1,k0 11 =k,m00 = 0, and m= 1. This gives rise to a term (4.4) of the form •ˆz0 µ(x)[sin(ˆz1(x))] 1 cosh(ˆz2(x)) −cos(ˆz1(x))2 [sinh(ˆz2(x))][ˆz(k) 2(x)]. Case 4: A basic factor [ˆz(k00 ji) 1(x)] in (4.4) has k00 ji =k. Then (4.5) and (4.6) show that k0= 0, m]= 0,m0= 0,m00 = 1,m00 1= 1,k00 11 =k,m= 1. This give rise to a term (4.4) of the form •ˆz0 µ(x)[sin(ˆz1(x))] 1 cosh(ˆz2(x)) −cos(ˆz1(x))2 [sin(ˆz1(x))][ˆz(k) 1(x)]. These are all the safe terms. We have now identified all the dangerous and the safe terms. All the remaining terms are easy. We now substitute into (4.3) the results of the above discussion of the integrand of (4.3). Recalling our definition of ˆz1(x),ˆz1(x), and making the change of variable u=x+yin each of our integrals, we obtain the formula: ∂t∂k xzµ(x, t) =Zu∈T sin(z1(x, t)−z1(u, t)) cosh(z2(x, t)−z2(u, t)) −cos(z1(x, t)−z1(u, t)) ∂k+1 xzµ(x, t)−∂k+1 uzµ(u, t)du +(coeff) Zu∈T [∂xz1(x, t)−∂uz1(u, t)] cos(z1(x, t)−z1(u, t)) cosh(z2(x, t)−z2(u, t)) −cos(z1(x, t)−z1(u, t)) ∂k xzµ(x, t)−∂k uzµ(u, t)du +(coeff) Zu∈T [∂xz2(x, t)−∂uz2(u, t)] sinh(z2(x, t)−z2(u, t)) sin(z1(x, t)−z1(u, t)) (cosh(z2(x, t)−z2(u, t)) −cos(z1(x, t)−z1(u, t)))2 ×∂k xzµ(x, t)−∂k uzµ(u, t)du +(coeff) Zu∈T [∂xz1(x, t)−∂uz1(u, t)][sin(z1(x, t)−z1(u, t))]2 (cosh(z2(x, t)−z2(u, t)) −cos(z1(x, t)−z1(u, t)))2∂k xzµ(x, t)−∂k uzµ(u, t)du +(coeff) Zu∈T [∂xzµ(x, t)−∂uzµ(u, t)] cos(z1(x, t)−z1(u, t)) cosh(z2(x, t)−z2(u, t)) −cos(z1(x, t)−z1(u, t)) ∂k xz1(x, t)−∂k uz1(u, t)du +(coeff) Zu∈T [∂xzµ(x, t)−∂uzµ(u, t)] sinh(z2(x, t)−z2(u, t)) sin(z1(x, t)−z1(u, t)) (cosh(z2(x, t)−z2(u, t)) −cos(z1(x, t)−z1(u, t)))2 ×∂k xz2(x, t)−∂k uz2(u, t)du +(coeff) Zu∈T [∂xzµ(x, t)−∂uzµ(u, t)][sin(z1(x, t)−z1(u, t))]2 (cosh(z2(x, t)−z2(u, t)) −cos(z1(x, t)−z1(u, t)))2∂k xz1(x, t)−∂k uz1(u, t)du + νmax X ν=1 cµ νEasy Termµ ν(x, t),(4.7) where Easy Termµ ν(x, t) 36
=Zu∈T m1 Y i=1 ∂ki xz1(x, t)−∂ki uz1(u, t)m2 Y i=1 h∂k0 i xz2(x, t)−∂k0 i uz2(u, t)i ×[sin(z1(x, t)−z1(u, t))]m3[cos(z1(x, t)−z1(u, t))]m4 ×[sinh(z2(x, t)−z2(u, t))]m5[cosh(z2(x, t)−z2(u, t))]m6 ×(cosh(z2(x, t)−z2(u, t)) −cos(z1(x, t)−z1(u, t)))−m7du, (4.8) where the mi,ki,k0 imay depend on µ,ν, although we do not indicate that in the notation, and where all the kiand k0 iare ≥1, and m7≥1, (4.9) and m1+m2+m3+m5−2m7≥0; (4.10) and all kiand k0 iare ≤k−1. (4.11) Now we recall that we assumed zµ(x, t)are analytic in Ω(t), and that (z1(x, t), z2(x, t)) are assumed to satisfy the COMPLEX CHORD-ARC CONDITION. The integrands (4.7), (4.8) have no singularities; they are holomorphic in (x, u)∈Ω(t)×Ω(t). Hence, we may deform the contour in (4.7), (4.8) from Tto Γ±(t), and the formulas holds, not merely for xreal, but for all xin the closure of Ω(t). We record this result as a lemma. Lemma 4.1 Let k≥4. Let h(x, t)be a positive smooth function of x∈Tfor each fixed t. Set Ω(t) = {x∈C/2πZ:|=x| ≤ h(<x, t)},Γ±={x±ih(x, t) : x∈T}. Let z1(x, t),z2(x, t) be smooth functions of (x, t)on {x∈Ω(t)closure,t∈Time interval}. For each fixed t, suppose z1(x, t),z2(x, t)are analytic in Ω(t). Assume that z1(x, t),z2(x, t)satisfy the Muskat equation (ME) : ∂tzµ(x, t) = Zy∈T sin (z1(x, t)−z1(x+y)) (∂xzµ(x, t)−∂xzµ(x+y, t)) cosh (z2(x, t)−z2(x+y)) −cos (z1(x, t)−z1(x+y))dy, for µ= 1,2. Assume also that z1(x, t),z2(x, t)satisfy the COMPLEX CHORD-ARC CONDITION (CCA) : |cosh (z2(x, t)−z2(y, t)) −cos (z1(x, t)−z1(y, t)) | ≥cCA [||<(x−y)|| +|=(x−y)|]2. for x,y∈Ω(t). Then, for ζ∈Γ+(t), we have that (I)∂th∂k ζzµ(ζ, t)i=Dangerousµ(ζ, t) + 6 X j=1 Safeµ j(ζ, t) + νmax X ν=1 cµ νEasyµ ν(ζ, t), where (II)Dangerousµ(ζ, t) =Z w∈Γ+(t) sin(z1(ζ, t)−z2(w, t)) cosh(z2(ζ, t)−z2(w, t)) −cos(z1(ζ, t)−z1(w, t)) ×(∂k+1 ζzµ(ζ, t)−∂k+1 wzµ(w, t))dw 37
(III)Safeµ 1(ζ, t) = (coeff)Z w∈Γ+(t) (∂ζz1(ζ, t)−∂wz1(w, t)) cos(z1(ζ, t)−z1(w, t)) cosh(z2(ζ, t)−z2(w, t)) −cos(z1(ζ, t)−z1(w, t)) ×(∂k ζzµ(ζ, t)−∂k wzµ(w, t))dw (IV )Safeµ 2(ζ, t) = (coeff)Z w∈Γ+(t) (∂ζz2(ζ, t)−∂wz2(w, t)) sinh(z2(ζ, t)−z2(w, t)) sin(z1(ζ, t)−z1(w, t)) (cosh(z2(ζ, t)−z2(w, t)) −cos(z1(ζ, t)−z1(w, t)))2 ×(∂k ζzµ(ζ, t)−∂k wzµ(w, t))dw (V)Safeµ 3(ζ, t) =(coeff) Z w∈Γ+(t) (∂ζz1(ζ, t)−∂wz1(w, t)) sin2(z1(ζ, t)−z1(w, t)) (cosh(z2(ζ, t)−z2(w, t)) −cos(z1(ζ, t)−z1(w, t)))2 ×(∂k ζzµ(ζ, t)−∂k wzµ(w, t))dw (V I)Safeµ 4(ζ, t) =(coeff) Z w∈Γ+(t) (∂ζzµ(ζ, t)−∂wzµ(w, t)) cos(z1(ζ, t)−z1(w, t)) (cosh(z2(ζ, t)−z2(w, t)) −cos(z1(ζ, t)−z1(w, t))) ×(∂k ζz1(ζ, t)−∂k wz1(w, t))dw (V II)Safeµ 5 =(coeff) Z w∈Γ+(t) (∂ζzµ(ζ, t)−∂wzµ(w, t)) sin(z1(ζ, t)−z1(w, t)) sinh(z2(ζ, t)−z2(w, t)) (cosh(z2(ζ, t)−z2(w, t)) −cos(z1(ζ, t)−z1(w, t)))2 ×(∂k ζz2(ζ, t)−∂k wz2(w, t))dw (V III)Safeµ 6(ζ, t) =(coeff) Z w∈Γ+(t) (∂ζzµ(ζ, t)−∂wzµ(w, t)) sin2(z1(ζ, t)−z1(w, t)) (cosh(z2(ζ, t)−z2(w, t)) −cos(z1(ζ, t)−z1(w, t)))2 ×(∂k ζz1(ζ, t)−∂k wz1(w, t))dw and each term Easyµ ν(ζ, t)has the form (IX)Easyµ ν(ζ, t) =Z w∈Γ+(t) m1 Y i=1 (∂ki ζz1(ζ, t)−∂ki wz1(w, t)) m2 Y i=1 (∂k0 i ζz2(ζ, t)−∂k0 i wz2(w, t)) ×[sin(z1(ζ, t)−z1(w, t))]m3[cos(z1(ζ, t)−z1(w, t))]m4 ×[sinh(z2(ζ, t)−z2(w, t))]m5[cosh(z2(ζ, t)−z2(w, t))]m6 ×[(cosh(z2(ζ, t)−z2(w, t)) −cos(z1(ζ, t)−z1(w, t)))]−m7dw, 38
where (X)m7≥1,1≤ki≤k−1,1≤k0 i≤k−1, and (XI)m1+m2+m3+m5−2m7≥0; also (XII)X i ki+X i k0 i=k+ 1. Here, the mi,ki,k0 imay depend on µand ν, though we have suppressed this dependence in the notation. 4.2 Perturbing Easy and Safe terms Let z= (zµ(x, t))µ=1,2and z= (zµ(x, t))µ=1,2be two solutions of the Muskat equation (ME) as in the lemma (4.1). We assume that zµ(x, t)and zµ(x, t)are defined on T×T, where the time interval Tcontains some given t0∈[τ2, τ]. We make the following assumptions on zµand zµ(x, t): 1. zµ(x, t0)and zµ(x, t0)extend to functions analytic on Ω = {|=x|< h(<x, t0)}and smooth on the closure of Ω. 2. Zζ∈Γ±∂14 ζzµ(ζ, t0)2d<ζ≤C, where Γ±={=ζ=±h(<ζ, t0)}. Here, h(x, t)is as in section 3.2. 3. |cosh (z2(ζ, t0)−z2(w, t0)) −cos (z1(ζ, t0)−z1(w, t0)) | ≥cCA [||<(ζ−w)|| +|=(ζ−w)|]2, for ζ,w∈Ω. 4. ||z(·, t)−z(·, t)||H4(Ω(t)) ≤λ, with λas in section 3.2 . Here we recall that ||f||2 H4(Ω(t)) =X ±Zζ∈Γ±(t) |f(ζ)|2d<ζ+X ±Zζ∈Γ±(t) |∂4 ζf(ζ)|2d<ζ. Note that our unperturbed solution zis assumed to have 4 + 10 controlled derivatives in hypothesis 2, whereas the perturbation z−zis assumed to have only 4controlled derivatives in the hypothesis 4. In this section, we write c,C,C0, etc. to denote constants determined by the constants Cand cCA in hypotheses 2 and 3. These symbols may denote different constants in different occurrences. From the previous hypotheses, we conclude that Zζ∈Γ±∂4 ζzµ(ζ, t0)2d<ζ≤C, (4.12) 39
∂j ζzµ(ζ, t0),∂j ζzµ(ζ, t0)≤Cfor 0≤j≤3,ζ∈Ωclosure,(4.13) |cosh (z2(ζ, t0)−z2(w, t0)) −cos (z1(ζ, t0)−z1(w, t0)) | ≥cCA [||<(ζ−w)|| +|=(ζ−w)|]2,for ζ,w∈Ωclosure.(4.14) Let Safeµ j(ζ, t0)(j= 1, ..., 6)) and Easyµ ν(ζ, t0)be as in the lemma (4.1), arising from the Muskat solution (z1(ζ, t), z2(ζ, t)), taking t=t0in that lemma. Similarly, let Safeµ j(ζ, t0)and Easyµ ν(ζ, t0)arise from the Muskat solution (z1(ζ, t), z2(ζ, t)). Our goal in this section is to estimate Easyµ ν(ζ, t0)−Easyµ ν(ζ, t0)and Safeµ j(ζ, t0)−Safeµ j(ζ, t0). We begin with the Easy terms. From (1)...(5), we obtain the estimates ∂j ζ(zµ−zµ)(ζ, t0)≤Cλ 0≤j≤3,ζ∈Γ±;(4.15) and ∂3 ζ(zµ−zµ)(ζ, t0)−∂3 w(zµ−zµ)(w, t0)≤C||<(ζ−w)||M∂4 ζ(zµ−zµ)(ζ)(4.16) for ζ,w∈Γ+, where Mis a Hardy-Littlewood maximal function. We prepare to estimate |[cosh (z2(ζ, t0)−z2(w, t0)) −cos (z1(ζ, t0)−z1(w, t0))] −[cosh (z2(ζ, t0)−z2(w, t0)) −cos (z1(ζ, t0)−z1(w, t0))]| for ζ,w∈Γ+. To do so, let ξ,η,ξ,ηbe complex numbers, with |ξ|,|η|,|ξ|,|η| ≤ C. We set ξs= sξ + (1 −s)ξand ηs=sη + (1 −s)ηfor 0≤s≤1. We then have d ds [cosh(ξs)−cos(ηs)]=sinh(ξs)(ξ−ξ) + sin(ηs)(η−η) ≤C0|ξs||ξ−ξ|+C0|ηs||η−η| ≤ C0(|ξ|+|ξ|)|ξ−ξ|+C0(|η|+|η|)|η−η|. We integrate in s. We take ξ=z2(ζ, t0)−z2(w, t0), ξ=z2(ζ, t0)−z2(w, t0), η=z1(ζ, t0)−z1(w, t0), η=z1(ζ, t0)−z1(w, t0) for |ζ−w| ≤ c, where cis a small constant and thus obtain: |[cosh (z2(ζ, t0)−z2(w, t0)) −cos (z1(ζ, t)−z1(w))] −[cosh (z2(ζ, t0)−z2(w, t0)) −cos (z1(ζ, t)−z1(w))]| ≤C0[|z2(ζ, t0)−z2(w, t0)|+|z2(ζ, t0)−z2(w, t0)|]×|[z2(ζ, t0)−z2(w, t0)] −[z2(ζ, t0)−z2(w, t0)]| +C0[|z1(ζ, t0)−z1(w, t0)|+|z1(ζ, t0)−z1(w, t0)|]×|[z1(ζ, t0)−z1(w, t0)] −[z1(ζ, t0)−z1(w, t0)]| (4.17) for ζ,w∈Γ+,|ζ−w| ≤ c. 40
Since |z1(ζ, t0)−z1(w, t0)|,|z1(ζ, t0)−z1(w, t0)|,|z2(ζ, t0)−z2(w, t0)|,|z2(ζ, t0)−z2(w, t0)| ≤C||<(ζ−w)||, for ζ,w∈Γ+; and since |[z1(ζ, t0)−z1(w, t0)]−[z1(ζ, t0)−z1(w, t0)]|=|[z1(ζ, t0)−z1(ζ, t0)]−[z1(w, t0)−z1(w, t0)]| ≤max ξ∈Ω|∂ξ(z1−z1)(ξ, t0)| · |<(ζ−w)| ≤ C0λ|<(ζ−w)| for ζ,w∈Γ+,|ζ−w| ≤ c, thanks to (4.15); and since, similarly, |[z2(ζ, t0)−z2(w, t0)]−[z2(ζ, t0)−z2(w, t0)]|=|[z2(ζ, t0)−z2(ζ, t0)]−[z2(w, t0)−z2(w, t0)]| ≤max ξ∈Ω|∂ξ(z2−z2)(ξ, t0)| · |<(ζ−w)| ≤ C0λ|<(ζ−w)| for ζ,w∈Γ+,|ζ−w| ≤ c, we conclude from (4.17) that |[cosh (z2(ζ, t0)−z2(w, t0)) −cos (z1(ζ, t0)−z1(w, t0))] −[cosh (z2(ζ, t0)−z2(w, t0)) −cos (z1(ζ, t0)−z1(w, t0))]| ≤Cλ sin ζ−w 2 2 (4.18) for ζ,w∈Γ+. We have proven this for |ζ−w| ≤ c. For |ζ−w|> c,ζ,w∈Γ+the estimate (4.18) is trivial from (4.15). Together with the chord-arc condition (4.14), this yields the following estimate [cosh (z2(ζ, t0)−z2(w, t0)) −cos (z1(ζ, t0)−z1(w, t0))]−1 −[cosh (z2(ζ, t0)−z2(w, t0)) −cos (z1(ζ, t0)−z1(w, t0))]−1 ≤Cλ sin ζ−w 2 −2 ,(4.19) for ζ,w∈Γ+. Next, from (4.15) we obtain easily the estimates for ζ,w∈Γ+: |cos (z1(ζ, t0)−z1(w, t0)) −cos (z1(ζ, t0)−z1(w, t0))| ≤ Cλ (4.20) |cosh (z2(ζ, t0)−z2(w, t0)) −cosh (z2(ζ, t0)−z2(w, t0))| ≤ Cλ. (4.21) Next, for ζ,w∈Γ+, we have |sinh (z2(ζ, t0)−z2(w, t0)) −sinh (z2(ζ, t0)−z2(w, t0))| ≤C|[z2−z2](ζ, t0)−[z2−z2](w, t0)| ≤C0λ||<(ζ−w)|| thanks to (4.15). Because of the periodicity we conclude that |sinh (z2(ζ, t0)−z2(w, t0)) −sinh (z2(ζ, t0)−z2(w, t0))| 41
=G1(ζ, w, t)·(ζ−w), with the C2−norm of G1(ζ, w, t)at most Cfor fixed t. Examining the asymptotics as w→ζ, yields G1(ζ, ζ, t) = ∂ζz1(ζ, t), hence sin (z1(ζ, t)−z1(w, t)) = [∂ζz1(ζ, t) + G2(ζ, w, t)(ζ−w)] (ζ−w),(4.60) with the C1−norm of G2(ζ, w, t)at most Cfor fixed t. Similarly, cosh (z2(ζ, t)−z2(w, t)) −1 = cosh (z2(ζ, t)−z2(w, t)) −1 (z2(ζ, t)−z2(w, t))2(z2(ζ, t)−z2(w, t))2 =G3(ζ, w, t)(ζ−w)2, with the C2−norm of G3(ζ, w, t)at most Cfor fixed t. Examining the asymptotics as w→ζ, gives G3(ζ, ζ, t) = 1 2(∂ζz2(ζ, t))2, hence cosh (z2(ζ, t)−z2(w, t)) −1 = 1 2(∂ζz2(ζ, t))2+G4(ζ, w, t)(ζ−w)(ζ−w)2, with the C1−norm of G4(ζ, w, t)at most Cfor fixed t. Similarly, 1−cos (z1(ζ, t)−z1(w, t)) = 1 2(∂ζz1(ζ, t))2+G5(ζ, w, t)(ζ−w)(ζ−w)2, with the C1−norm of G5(ζ, w, t)at most Cfor fixed t. Adding, and recalling the COMPLEX CHORD-ARC CONDITION (4.46), we see that cosh (z2(ζ, t)−z2(w, t)) −cos (z1(ζ, t)−z1(w, t)) =1 2h(∂ζz2(ζ, t))2+ (∂ζz1(ζ, t))2i·[1 + G6(ζ, w, t)·(ζ−w)] (ζ−w)2,(4.61) with the C1−norm of G6(ζ, w, t)at most Cfor fixed t. Dividing (4.60) by (4.61), and noting that 1 1 + G6·(ζ−w)= 1 −G6 1 + G6·(ζ−w)·(ζ−w), we obtain sin (z1(ζ, t)−z1(w, t)) cosh (z2(ζ, t)−z2(w, t)) −cos (z1(ζ, t)−z1(w, t)) ="2∂ζz1(ζ, t) (∂ζz2(ζ, t))2+ (∂ζz1(ζ, t))2+G7(ζ, w, t)·(ζ−w)#1 (ζ−w) =2∂ζz1(ζ, t) (∂ζz2(ζ, t))2+ (∂ζz1(ζ, t))2 1 (ζ−w)+G7(ζ, w, t), with the C1−norm of G7(ζ, w, t)at most Cfor fixed t.. 48
Since also 2∂ζz1(ζ, t) (∂ζz2(ζ, t))2+ (∂ζz1(ζ, t))2cot ζ−w 2 =2∂ζz1(ζ, t) (∂ζz2(ζ, t))2+ (∂ζz1(ζ, t))2 1 ζ−w+G8(ζ, w, t), with the C1norm of G8(ζ, w, t)on Ωbounded by C, we conclude that a(ζ, w, t)belongs to C1(Ω), with norm at most C. Again recalling that there is no problem outside Ω(i.e., when |ζ−w|> c even after ζ,ware translated by multiples of 2π), we conclude (4.59). Next, we notice that ˜a(ζ, t) = Z=w=h(<w,t) a(ζ, w, t)dw =Zw∈T a(ζ, w, t)dw (4.62) if t∈[τ2, τ]; ˜a(ζ, t) = Z=w=~(<w,t) a(ζ, w, t)dw =Zw∈T a(ζ, w, t)dw (4.63) if t∈[−τ2, τ2]. We shall study the regularity of the function ˜a(ζ)for ζ∈Γ+and Γ+={ζ∈C:ζ=x+ih(x), x ∈T}. We recall that h(x)is a smooth function and we will prove that ˜ahas C2-norm bounded by a controlled constant. (We replace hby ~if we are in the interval [−τ2, τ2].) Remark 4.2 Notice that since z(ζ, t)∈C3(Ω(t)closure), from (4.62) and (4.63) we see that ˜ais aC2-function. However, this argument is not enough to control the norm ||˜a(·+ih(·, t))||C2(T) by a controlled constant. We would obtain a constant which depends on 1 minx∈Th(x,t)and we can not permit it. In fact, we will prove that ˜ahas bounded norm in C2+δ(Ω(t)) for any δ < 1/2. To see this, it is enough to show that ˜a|Γ+(t)∈C2+δwith norm at most C(δ), for any δ < 1 2.(4.64) (The reduction from Ω(t)to Γ+(t)is possible, thanks to X1in section 3.4). To prove (4.64), we first note that ˜a(ζ, t) = P.V. ZΓ+(t) sin(z1(ζ, t)−z1(w, t)) cosh(z2(ζ, t)−z2(w, t)) −cos(z1(ζ, t)−z1(w, t))dζ (4.65) for ζ∈Γ+(t), since P.V. ZΓ+(t) cot ζ−w 2dw = 0, as we noted in section 2. 49
It is useful to make the change of variable ζ=x+ih(x, t), w=x+y+ih(x+y, t), and to set ˆzµ(x)≡zµ(x+ih(x, t), t)(4.66) for µ= 1, 2. Then ˆz1(x)−xand ˆz2(x)belong to C3.5(T), with norm less or equal than C. (4.67) . Hence also ˆzµ(x)−ˆzµ(x+y) y=−Z1 0 ˆz0 µ(x+sy)ds (4.68) belongs to C2.5(T×[−π, π]) with norm less or equal than C. Our change of variable (4.66) gives ˜a(x+ih(x, t), t) = P.V. Zπ −π sin(ˆz1(x)−ˆz1(x+y)) (1 + i∂xh(x+y, t)) cosh(ˆz2(x)−ˆz2(x+y)) −cos(ˆz1(x)−ˆz1(x+y))dy (4.69) for x∈T. It is convenient to introduce the analytic functions gsn(ζ) = sin(ζ) ζ, gcsh(ζ) = cosh(ζ)−1 ζ2, gcs(ζ) = 1−cos(ζ) ζ2(4.70) for ζ∈Cand g] sn(ζ, ζ) = gsn(ζ)−gsn(ζ) ζ−ζ, g] csh(ζ, ζ) = gcsh(ζ)−gcsh(ζ) ζ−ζ, g] cs(ζ, ζ) = gcs(ζ)−gcs(ζ) ζ−ζ,(4.71) for ζ,ζ∈C. We can then write y−1sin(ˆz1(x)−ˆz1(x+y)) = gsn(ˆz1(x)−ˆz1(x+y)) ˆz1(x)−ˆz1(x+y) y(4.72) and y−2[cosh(ˆz2(x)−ˆz2(x+y)) −cos(ˆz1(x)−ˆz1(x+y))] gcsh(ˆz2(x)−ˆz2(x+y)) ˆz2(x)−ˆz2(x+y) y2 +gcs(ˆz1(x)−ˆz1(x+y)) ˆz1(x)−ˆz1(x+y) y2 (4.73) 50
Thanks to (4.67) and (4.68), the right-hand sides of (4.72) and (4.73) belong to C2.5(T×[−π, π]), with norm at most C. Also, the left-hand side of (4.73) has absolute value larger or equal than c, thanks to the chord-arc condition. Therefore, we may express the integrand in (4.69) in the form Q(x,y) y, where Q(x, y) = [LHS of (4.72)] (1 + i∂xh(x+y, t)) [LHS of (4.73)] belongs to C2.5(T×[−π, π]), with norm at most C. That is, ˜a(x+ih(x, t), t) = P.V. Zπ −π Q(x, y) ydy =Zπ −π Q(x, y)−Q(x, 0) ydy. (4.74) with ||Q||C2.5(T×[−π, π]) ≤C. It follows easily that the function x7→ ˜a(x+ih(x, t), t)belongs to C2+δ(T)with norm less or equal than C(δ)for each δ < 1 2. This completes the proof of (4.64). To conclude this section, we prove that sin(z1(ζ, t)−z1(w, t)) cosh(z2(ζ, t)−z2(w, t)) −cos(z1(ζ, t)−z1(w, t)) −sin(z1(ζ, t)−z1(w, t)) cosh(z2(ζ, t)−z2(w, t)) −cos(z1(ζ, t)−z1(w, t)) =λΘ(ζ, t) cot ζ−w 2+ Θ](ζ, w, t),(4.75) for ζ,w∈Γ+(t), where Θ∈C0(Γ+(t)) and Θ]∈C0(Γ+(t)×Γ+(t)) for fixed t, with norms less or equal than C. To see this, we again use the change of variable ζ=x+ih(x, t), w=x+y+ih(x+y, t); and we introduce ˆzµ(x),ˆzµ(x)for µ= 1,2, where ˆzµ(x)is given by (4.66) and ˆzµ(x)is the analogous expression with zµin place of zµ. From (4.71) and (4.72) and the analogue of (4.72) for the ˆzµ, we see that y−1[sin(ˆz1(x)−ˆz1(x+y)) −sin(ˆz1(x)−ˆz1(x+y))] = [gsn(ˆz1(x)−ˆz1(x+y)) −gsn(ˆz1(x)−ˆz1(x+y))] (ˆz1(x)−ˆz1(x+y)) y + [gsn(ˆz1(x)−ˆz1(x+y))] (ˆz1(x)−ˆz1(x+y)) y−(ˆz1(x)−ˆz1(x+y)) y =g] sn(ˆz1(x)−ˆz1(x+y),ˆz1(x)−ˆz1(x+y))[(ˆz1−ˆz1)(x)−(ˆz1−ˆz1)(x+y)] ׈z1(x)−ˆz1(x+y) y + [gsn(ˆz1(x)−ˆz1(x+y))] (ˆz1−ˆz1)(x)−(ˆz1−ˆz1)(x+y) y.(4.76) Since ˆzµ−ˆzµbelongs to C3.5with norm less or equal than Cλ, we know that ˆzµ(x)−ˆzµ(x+y) y belongs to C2.5with norm less or equal than Cλ, as in the proof of (4.68). Therefore, (4.76) 51
shows that y−1[sin(ˆz1(x)−ˆz1(x+y)) −sin(ˆz1(x)−ˆz1(x+y))] belongs to C2.5(T×[−π, π]), with norm less or equal Cλ. (4.77) A similar argument involving the function g] csh and g] cs in (4.71) shows that y−2[{cosh(ˆz2(x)−ˆz2(x+y)) −cos(ˆz1(x)−ˆz1(x+y))} − {cosh(ˆz2(x)−ˆz2(x+y)) −cos(ˆz1(x)−ˆz1(x+y))}] belongs to C2.5(T×[−π, π]), with norm less or equal than Cλ. (4.78) We have seen that y−1sin(ˆz1(x)−ˆz1(x+y)) and y−2[cosh(ˆz2(x)−ˆz2(x+y)) −cos(ˆz1(x)−ˆz1(x+y))] belong to C2.5(T×[−π, π]) with norms less or equal than C, and that y−2{cosh(ˆz2(x)−ˆz2(x+y)) −cos(ˆz1(x)−ˆz1(x+y))}≥con T×[−π, π].(4.79) From (4.77), (4.78), (4.79) it now follows easily that sin(ˆz1(x)−ˆz1(x+y)) cosh(ˆz2(x)−ˆz2(x+y)) −cos(ˆz1(x)−ˆz1(x+y))− sin(ˆz1(x)−ˆz1(x+y)) cosh(ˆz2(x)−ˆz2(x+y)) −cos(ˆz1(x)−ˆz1(x+y)) =λQ](x, y) y(4.80) for a function Q]∈C2.5(T×[−π, π]) with norm less or equal than C. Our result (4.80) is much sharper than our assertion (4.75) In particular, we have proven (4.75). 4.5 Perturbing the Dangerous Term Let z,zbe as in the section (4.2). Suppose that zsatisfies the additional assumptions made in the section (4.3). Let a(ζ, w, t),a(ζ, w, t),˜a(ζ, t),˜a(ζ, t)be as in section (4.3). We recall from section (4.1) that Dangerousµ(ζ, t) = Z w∈Γ+(t) sin (z1(ζ, t)−z1(w, t)) cosh (z2(ζ, t)−z2(w, t)) −cos (z1(ζ, t)−z1(w, t))[∂4+1 ζzµ(ζ, t)−∂4+1 wzµ(w, t)]dw (4.81) for ζ∈Γ+(t). For the Muskat solution zin place of z, the analogous expression is Dangerousµ(ζ, t) = Z w∈Γ+(t) sin (z1(ζ, t)−z1(w, t)) cosh (z2(ζ, t)−z2(w, t)) −cos (z1(ζ, t)−z1(w, t))[∂4+1 ζzµ(ζ, t)−∂4+1 wzµ(w, t)]dw (4.82) 52
for ζ∈Γ+(t). Our goal in this section is to compute Dangerousµ(ζ, t)−Dangerousµ(ζ, t), modulo small errors. Let ζ∈Γ+(t)be given. From (4.81) and (4.82), we have Dangerousµ(ζ, t)−Dangerousµ(ζ, t) = Z w∈Γ+(t) [sin (z1(ζ, t)−z1(w, t))][∂4+1 ζ(zµ−zµ)(ζ, t)−∂4+1 w(zµ−zµ)(w, t)] cosh (z2(ζ, t)−z2(w, t)) −cos (z1(ζ, t)−z1(w, t)) dw +Z w∈Γ+(t)sin (z1(ζ, t)−z1(w, t)) cosh (z2(ζ, t)−z2(w, t)) −cos (z1(ζ, t)−z1(w, t)) −sin (z1(ζ, t)−z1(w, t)) cosh (z2(ζ, t)−z2(w, t)) −cos (z1(ζ, t)−z1(w, t)) ×[∂4+1 ζzµ(ζ, t)−∂4+1 wzµ(w, t)]dw. Substituting the definition of a(ζ, w, t)and applying (4.75), we conclude that Dangerousµ(ζ, t)−Dangerousµ(ζ, t) = ∂ζz1(ζ, t) (∂ζz1(ζ, t))2+ (∂ζz2(ζ, t))2Z w∈Γ+(t) cot ζ−w 2h∂4+1 ζ(zµ−zµ)(ζ, t)−∂4+1 w(zµ−zµ)(w, t)idw +Z w∈Γ+(t) a(ζ, w, t)h∂4+1 ζ(zµ−zµ)(ζ, t)−∂4+1 w(zµ−zµ)(w, t)idw +Z w∈Γ+(t) λΘ(ζ, t) cot ζ−w 2+ Θ](ζ, w, t)[∂4+1 ζzµ(ζ, t)−∂4+1 wzµ(w, t)]dw. (4.83) Recall that we define ΛΓ+(t)F(ζ) = −1 2πZ w∈Γ+(t) cot ζ−w 2[F0(ζ)−F0(w)]dw (4.84) for holomorphic function F. Also, in the terms involving a(ζ, w, t)in (4.83), we recall (4.62), and we integrate by parts in w. Thus, (4.83) becomes Dangerousµ(ζ, t)−Dangerousµ(ζ, t) = −2π∂ζz1(ζ, t) (∂ζz1(ζ, t))2+ (∂ζz2(ζ, t))2ΛΓ+(t)(∂4 µzµ(·, t)−∂4 µzµ(·, t)) +˜a(ζ, t)∂4+1 ζ(zµ(ζ, t)−zµ(ζ, t)) +Z w∈Γ+(t) [∂wa(ζ, w, t)]∂4 w(zµ−zµ)(w, t)dw +λZ w∈Γ+(t)Θ(ζ, t) cot ζ−w 2+ Θ](ζ, w, t)[∂4+1 ζzµ(ζ, t)−∂4+1 wzµ(w, t)]dw. (4.85) 53
Recall from assumption 2 in section (4.2) that |∂4+1 ζzµ(ζ, t)−∂4+1 ζzµ(w, t)| ≤ min{C|ζ−w|, C}. (Here, it is important that we refer to z, not to z.) Since also |Θ(ζ, t)|,|Θ](ζ, w, t)| ≤ C by (4.75), we have λZ w∈Γ+(t)Θ(ζ, t) cot ζ−w 2+ Θ](ζ, w, t)·[∂4+1 ζzµ(ζ, t)−∂4+1 ζzµ(w, t)]dw ≤Cλ, (4.86) for any ζ∈Γ+(t). Also, since |∂wa(ζ, w, t)| ≤ C by (4.59) and assumption 4 in section 4.2 (recall, we assume λ < 1), we have Z w∈Γ+(t) [∂wa(ζ, w, t)]∂4 w(zµ−zµ)(w, t)dw ≤Cλ (4.87) for all ζ∈Γ+(t), thanks to assumption 4 in section (4.2). Substituting (4.86) and (4.87) into (4.85), we see that Dangerousµ(ζ, t)−Dangerousµ(ζ, t) = −2π∂ζz1(ζ, t) (∂ζz1(ζ, t))2+ (∂ζz2(ζ, t))2ΛΓ+(t)(∂4 µzµ(·, t)−∂4 µzµ(·, t)) +˜a(ζ, t)∂4+1 ζ(zµ(ζ, t)−zµ(ζ, t)) +ErrorP DT (ζ, t)(4.88) for ζ∈Γ+(t), with |ErrorP DT (ζ, t)| ≤ Cλ. (4.89) The basic results of this section are (4.88), (4.89). 4.6 The Main Energy Estimate In this subsection, we suppose that zis an unperturbed solution of Muskat problem, while zis a perturbed solution, as in the last several sections. Let σ0 1,σ1,σ±,hand ~as in the sections 3.1, 3.2 and 3.3. Our goal is to estimate (from below) the quantity d dt Zζ∈Γ+(t)∂4 ζ(zµ−zµ)(ζ, t)2d<ζ. To do so, we first recall (I) from lemma (4.1). Applying that result to both our Muskat solutions, we see that ∂t[∂4 ζ(zµ−zµ)(ζ, t)] = 54
[Dangerousµ(ζ, t)−Dangerousµ(ζ, t)] + 6 X j=1 [Safeµ j(ζ, t)−Safeµ j(ζ, t)] +X 1≤ν≤νmax cµ ν[Easyµ ν(ζ, t)−Easyµ ν(ζ, t)] for ζ∈Γ+(t). According to (4.27), (4.41) and (4.88), (4.89), this implies that ∂t[∂4 ζ(zµ−zµ)(ζ, t)] = −2π∂ζz1(ζ, t) (∂ζz1(ζ, t))2+ (∂ζz2(ζ, t))2ΛΓ+(t)(∂4 µzµ(·, t)−∂4 µzµ(·, t))+˜a(ζ, t)∂4+1 ζ(zµ(ζ, t)−zµ(ζ, t)) +Error(ζ, t)(4.90) for ζ∈Γ+(t), where Zw∈Γ+(t) |Error(ζ, t)|2d<ζ≤Cλ2.(4.91) Recall our assumption Zζ∈Γ+(t)∂4 ζ(zµ−zµ)(ζ, t)2d<ζ≤λ2(4.92) and note that 1 2 d dt Zζ∈Γ+(t)∂4 ζ(zµ−zµ)(ζ, t)2d<ζ =<Z ζ∈Γ+(t) [∂4 ζ(zµ−zµ)(ζ, t)]·n∂t[∂4 ζ(zµ−zµ)(ζ, t)] + iht(<ζ, t)·∂4+1 ζ(zµ−zµ)(ζ, t)od<ζ, where Γ+(t) = {x+ih(x) : x∈T} if t∈[τ2, τ]and Γ+(t) = {x+i~(x) : x∈T} if t∈[−τ2, τ2]. (In this range we use ~in place of hin the previous expression.) We conclude that 1 2 d dt Zζ∈Γ+(t)∂4 ζ(zµ−zµ)(ζ, t)2d<ζ =<Z ζ∈Γ+(t) F(ζ)·−2πz0 1(ζ, t) (z0 1(ζ, t))2+ (z0 2(ζ, t))2ΛΓ+(t)F(ζ)+(i∂th(<ζ, t) + ˜a(ζ, t)) F0(ζ)d<ζ +Error, (4.93) where F(ζ) = ∂4 ζ(zµ−zµ)(ζ, t),(4.94) and |Error| ≤ Cλ2.(4.95) Recall (4.54),...,(4.57), (4.49) and (4.64). Our assumptions on A,λ,τ,κin section 3.2 imply 0< λ < τ2.(4.96) 55
Recall from section 4.3, equations (4.49), (4.55), (4.56) and (4.57), that ˜a(ζ, t)is a C2−function of ζ∈Γ+(t), with norm at most C; (4.97) |=˜a(ζ, t)| ≤ Ch(<ζ, t)for ζ∈Γ+(t); (4.98) |∂xh(x, t)˜a(ζ, t)| ≤ Ch(x, t) + CA−1τ2for ζ=x+ih(x, t)∈Γ+(t). (4.99) Estimates (4.98), (4.99) hold for t∈[τ2, τ]. When t∈[−τ2, τ2], we use ~in place of h. Note that ={˜a(x+ih(x, t), t)·(1 + i∂xh(x, t))−1} = [=˜a(x+ih(x, t), t)−∂xh(x, t)· <˜a(x+ih(x, t), t)] ·1+(∂xh(x, t))2−1. Hence (4.97),...,(4.99) show that ˜a(x+ih(x, t), t)·(1 + i∂xh(x, t))−1=aR(x) + iaI(x),(4.100) where aR(x), aI(x)are real, C2−functions on T, with norm ≤C(4.101) and |aI(x)| ≤ Ch(x, t) + CA−1τ2 1+(∂xh(x, t))2for all x∈T.(4.102) If t∈[−τ2, τ2], then we use ~in place of h. Let f(x) = F(x+ih(x, t)) = ∂4 ζ(zµ−zµ)(ζ, t)ζ=x+ih(x,t).(4.103) Thus, (4.92) gives Zx∈T |f(x)|2dx ≤Cλ2.(4.104) Note that f0(x) = (1 + i∂xh(x, t))F0(x+ih(x, t)).(4.105) We have <Zζ∈Γ+(t) F(ξ)˜a(ξ)F0(ξ)d<ξ =<Zζ∈Γ+(t) f(x)· {[aR(x) + iaI(x)] ·(1 + i∂xh(x, t))}(1 + i∂xh(x, t))−1f0(x)dx =<Zx∈T f(x)·[aR(x) + iaI(x)]f0(x)dx =Zx∈T aR(x)·1 2 d dx |f(x)|2dx +<Zx∈T f(x)aI(x)if0(x)dx =−Zx∈T a0 R(x)·1 2|f(x)|2dx +<Zx∈T f(x)aI(x)if0(x)dx. (4.106) The first term on the right is dominated by Cλ2in absolute value, thanks to (4.101), (4.104). By (4.101), (4.102) and the sharp Gårding inequality, the second term on the right is bounded below by −< Z x∈T f(x)Ch(x, t) + CA−1τ2(1 + (∂xh(x, t))2)−1Λf(x)dx −CZ x∈T |f(x)|2. 56
Putting the above remarks and (4.104) into (4.106), we see that <Zζ∈Γ+(t) F(ζ)˜a(ζ, t)F0(ζ)dζ ≥ −Cλ2−C<Zx∈T f(x)h(x, t) + A−1τ2(1 + (∂xh(x, t))2)−1Λf(x)dx. (4.107) It is convenient to use (2.6) to conclude that <Zx∈T f(x)h(x, t) + A−1τ2i∂xh(x, t) 1+(∂xh(x, t))2Λf(x)dx ≤CZx∈T |f(x)|2dx ≤C0λ2. Together with (4.107), this yields the estimate <Zζ∈Γ+(t) F(ζ)˜a(ζ, t)F0(ζ)d<ζ ≥ −Cλ2−C<Zx∈T f(x)h(x, t) + A−1τ2(1 −i∂xh(x, t)) 1+(∂xh(x, t))2Λf(x)dx. Recalling that Λf(x) = (1 + i∂xh(x, t))ΛΓ+(t)F(x+ih(x, t)) + Error(x), with ||Error(x)||L2(T)≤C||f||L2(T) (see (2.10)), we conclude that <Zζ∈Γ+(t) F(ζ)˜a(ζ, t)F0(ζ)d<ζ ≥ −Cλ2−C<Zx∈T f(x)h(x, t) + A−1τ2(1 −i∂xh(x, t))(1 + i∂xh(x, t)) 1+(∂xh(x, t))2ΛΓ+(t)F(x+ih(x, t))dx, where again we use (4.104) to control the term involving Error(x). Recalling the definition (4.103) of f, we can conclude that <Zζ∈Γ+(t) F(ζ)˜a(ζ, t)F0(ζ)d<ζ ≥ −Cλ2−C<Zζ∈Γ+(t) F(ζ)h(x, t) + A−1τ2ΛΓ+(t)F(ζ)d<ζ. Substituting this inequality into (4.93), we see that 1 2 d dt Zζ∈Γ+(t)∂4 ζ(zµ−zµ)(ζ, t)2d<ζ ≥ < Z ζ∈Γ+(t) F(ζ)·σ+(<ζ, t)−Ch(x, t)−CA−1τ2ΛΓ+(t)F(ζ) + (i∂th(<ζ, t)) F0(ζ)od<ζ−Cλ2,(4.108) where we recall that σ+(<(ζ), t) = −2π∂ζz1(ζ, t) (∂ζz1(ζ, t))2+ (∂ζz2(ζ, t))2for ζ∈Γ+(t). 57
The above remarks imply that, for all x∈T, we have <σ+(x, t) + ∂t~(x, t)−C~(x, t)−CA−1τ2 +∂x~(x, t)=σ+(x, t) ≥σ0 1(x, t) + ∂t~(x, t)−C~(x, t)−CA−1τ2.(4.133) The right-hand side of (4.133) is positive; see (3.45). Hence, (4.133) implies (4.132), completing the proof of hypothesis 6. Thus, hypotheses 1,...,6 hold, with constants C(A). Hypothesis 7 holds trivially, since θ= 1. We may now apply the lemma in section (2.3). We obtain the estimate <Z ζ∈Γ+(t) F(ζ)· −2πz0 1(ζ, t) (z0 1(ζ, t))2+ (z0 2(ζ, t))2−C~(<ζ, t)−CA−1τ2ΛΓ+(t)F(ζ) + (i∂t~(<ζ, t)) F0(ζ)od<ζ ≥ −C(A)τ−1X ±Zζ∈Γ±(t) |F(ζ)|2d<ζ. (4.134) This estimate holds for t∈[−τ2, τ2]. Moreover, we have estimate (4.108), in which hshould be replaced by ~since we are working with t∈[−τ2, τ2]. Hence, (4.108) and (4.134) yield the estimate 1 2 d dt Zζ∈Γ+(t)∂4 ζ(zµ−zµ)(ζ, t)2d<ζ ≥ −Cλ2−C(A)τ−1X ±Zζ∈Γ±(t) |∂4 ζ(zµ−zµ)(ζ)|2d<ζ(4.135) for t∈[−τ2, τ2], with Γ±(t) = {x±i~(x, t) : x∈T}. Recalling from (4.94), and from assumption 4 in section (4.2) that X ±Zζ∈Γ±(t) |∂4 ζ(zµ−zµ)(ζ)|2d<ζ≤λ2, we conclude from (4.135) that 1 2 d dt Zζ∈Γ+(t)∂4 ζ(zµ−zµ)(ζ, t)2d<ζ≥ −C(A)τ−1λ2,(4.136) if t∈[−τ2, τ2]. This estimate holds provided zµis a Muskat solution satisfying X ±Zζ∈Γ±(t) |(zµ−zµ)(ζ)|2d<ζ+X ±Zζ∈Γ±(t) |∂4 ζ(zµ−zµ)(ζ)|2d<ζ≤λ2(4.137) and A,τ,λ,κare as assumed in section 3.2.(4.138) Here, Γ±(t) = {x±i~(x, t) : x∈T}. Our basic energy estimates are (4.127),..., (4.129) for t∈[τ2, τ]and (4.136),...,(4.138) for t∈ [−τ2, τ2]. 64
4.7 Conclusion The previous estimates allows us to achieve the following result. Theorem 4.3 Let z(x, t)be a solution of the Muskat equation in the interval t∈[tleast, τ]⊂ [−τ2, τ]and let z(x, t)be the unperturbed solution. Assume that z(x, t)satisfies •z1(x, t)−xand z2(x, t)are periodic with period 2π. •z(ζ, t)is real for ζreal. •z(ζ, t)is analytic in ζ∈Ω(t). •z(ζ, t)−(ζ, 0) ∈H4(Ω(t)). •Complex Arc-Chord condition. |cosh(z2(ζ, t)−z2(w, t)) −cos(z1(ζ, t)−z1(w, t))| ≥ cCA[||<(ζ−w)|| +|=(ζ−w)|]2, for ζ,w∈Ω(t). Here in the definition of Ω(t)we use h(x, t)if t∈[τ2, τ]and ~(x, t)if t∈[−τ2, τ2].Then d dt||z(·, t)−z(·, t)||2 H4(Ω(t)) ≥ −C(A)λ2, if t∈[τ2, τ]∩[tleast, τ], and if also ||z(·, t)−z(·, t)||H4(Ω(t)) ≤λ. In addition, d dt||z(·, t)−z(·, t)||2 H4(Ω(t)) ≥ −C(A)τ−1λ2, if t∈[−τ2, τ2]∩[tleast, τ], and if also ||z(·, t)−z(·, t)||H4(Ω(t)) ≤λ and λ≤τ2. Proof: We have to estimate the quantity: d dt||z(·, t)−z(·, t)||2 H4(Ω(t)),(4.139) where ||z(·, t)−z(·, t)||2 H4(Ω(t)) =X ±Zζ∈Γ±(t) |z(ζ, t)−z(ζ, t)|2d<ζ+X ±Zζ∈Γ±(t) |∂4 ζ(z(ζ, t)−z(ζ, t))|2d<ζ. (4.140) The term in (4.139) coming from the second term on the right hand side on (4.140) is bounded using (4.127) and (4.136). The rest of terms are lower order terms which are easy to control. 5 Galerkin approximations and main theorem We prove existence of solution of the Muskat equation in H4(Ω(t)) and the main theorem 1.1. 65
5.1 Conformal Maps Depending on a Parameter Let m≥100 and h0>0be given, and let h(x, t)be a real-valued function on T×[0,1] with |∂l t∂j xh(x, t)| ≤ 1 for l, j ≤100m+ 100. We write c,C, etc. to denote constants determined by mand h0alone. Let V0(x+iy)be the solution of the Dirichlet problem ∆V0(z) = 0 for −2h0<=z < 0 V0(x) = 0 for x∈R V0(x−2ih0) = 1 2π=cot x−2ih0 2 for x∈R. Then V0is 2π−periodic and real-analytic; hence we may regard V0as harmonic, real analytic and 2π−periodic on a strip H0={−2h0−c0<=z < c0}. We will solve the Dirichlet problem on the region Ω(t;δ) = {x+iy :−2h0−δh(x, t)< y < δh(x, t)} for 0< δ < c1(small enough c1). As an approximate Poisson kernel, we try P(x+iy, x+) = 1 2π=cot ζ 2·(1 + (δh0(x+, t))2)−V0(ζ)·(1 + (δh0(x+, t))2), for x+iy ∈Ω(t;δ),x+∈R, where ζ= (x+iy)−(x++iδh(x+, t)) −iδh0(x+, t) sin[(x+iy)−(x++iδh(x+, t))]. Since δ < c1for small enough c1, we know that ζ∈H0, so P(x+iy, x+)is well-defined. Suppose f∈Cm(T). Then ut(x+iy) = Zx+∈T P(x+iy, x+)f(x+)dx+ is harmonic in Ω(t, δ), and one checks that lim ε→0+ut(x+iδh(x, t)−iε) = f(x) + ZT K1(x, x+, t)dx+(5.1) lim ε→0+ut(x+iδh(x, t)−2h0i−iε) = ZT K2(x, x+, t)dx+(5.2) where |∂l t∂j x∂k x+Ki(x, x+, t)| ≤ Cδ for l,j,k≤mand i= 1,2,(5.3) and |∂j x∂k x+Ki(x, x+, t)| ≤ Cδ for j,k≤mand i= 1,2.(5.4) 66
We indicate the main steps in verifying the above. We introduce a new variable p, and we study how ζbehaves when we take y=δh(x, t)−p. We work on the domain {(x, x+, p, δ, t)|x−x+| ≤ π, δ ∈[0, c], t ∈[0,1]} for a small constant c. Let us write G1,G2, etc. to denote functions on this domain having at least 10mderivatives of absolute value at most C. Then ζ=Z−iδh0(x, t) sin(Z)(5.5) with Z= [x+iδh(x, t)−ip]−[x++iδh(x+, t)].(5.6) First of all, note that ζ=Z·1−iδh0(x+, t)sin(Z) Z = [(x−x+)−ip]·1 + iδ h(x, t)−h(x+, t) (x−x+) + ip ·1−iδh0(x+, t)sin(Z) Z, and therefore |ζ|=|(x−x+)−ip| · (1 + O(δ)) (5.7) uniformly in x,x+,p,δ,t. Next, by writing sin(Z) = Z+Z3sin(Z)−Z Z3, we find that ζ=Z−iδh0(x+, t)Z+δZ3G1; moreover, Z= (x−x+)·[1 + iδh0(x+, t)] −ip +δ(x−x+)2G2. Hence, ζ= [1 + (δh0(x+, t))2]·(x−x+)−δph0(x+, t)−ip +δ(x−x+)2G3+δp2G4.(5.8) Next, we refine (5.7) in the case |x−x+|< p2 3. In that case, we have |δ(x−x+)2G3+δp2G4| ≤ Cδp4 3 and |δph0(x+, t)| ≤ Cδp, while |[1 + (δh0(x+, t))2](x−x+)−ip| ≥ cp. Therefore, (5.8) implies that |ζ|=|[1 + (δh0(x+, t))2](x−x+)−δph0(x+, t)−ip| · (1 + O(δp1 3)),(5.9) uniformly in x,x+,p,δ,t, under the assumptions |x−x+|< p2 3. 67
Now we are ready to analyze =cot(ζ/2). For |<ζ| ≤ 3 2π, we have 1 2cot ζ 2=1 ζ+H(ζ) with Hanalytic and real-valued for real ζ. On the other hand, ζis a C90m+90−smooth function of x,x+,p,δ,t, equal to (x−x+)when p=δ= 0. Therefore, 1 2cot ζ 2=−=ζ |ζ|2+pG5+δG6.(5.10) From (5.7),...,(5.10) and the properties of V0discussed above, it is now a routine task to check (5.1),...,(5.4). Thus, ut(x+iy)is approximately equal to fon the upper boundary of Ω(t;δ), approximately zero on the lower boundary of Ω(t;δ)and (exactly) harmonic in Ω(t, δ). Moreover, Ki(x, x+, t)(i= 1,2) are given by explicit formulas. There are analogous explicit formulas for a harmonic function on Ω(t;δ)approximately equal to a given function gon the lower boundary of Ω(t;δ)and approximately equal to zero on the upper boundary. Our explicit formulas, together with a Neumann series, show that the Dirichlet problem V(z, t)harmonic in Ω(t;δ)for each fixed t V(x+iδh(x, t), t)=1 V(x−iδh(x, t)−2ih0, t) = −1 admits a solution Vthat is Cmon the region Ξ(δ) = {(z, t) : z∈Ω(t;δ)closure, t ∈[0,1]}. In particular, the above Neumann series converges, since we take δ < c1for a small enough c1. Taking a harmonic conjugate Uof V, we may suppose Ualso belongs to Cmon the region Ξ(δ). The analytic function Φt=U+iV is a conformal map of Ω(t;δ)to the strip {|=z|<1}. Moreover, Φtdepends Cm−smoothly on t, and Φt(z+ 2π)=Φt(z) + λ(t)(all z∈Ω(t;δ) for a smooth, real-valued function λ(t). A trivial rescaling now shows that Ω(t;δ)maps conformally to a strip {|=z| ≤ h](t)}by a conformal map Φ] t, that depends Cm−smoothly on tand satisfies Φ] t(z+ 2π) = Φ] t(z) + 2π; moreover, h](t)is Cm−smooth. We can now easily remove our small δassumptions. Let Ω(t) = {x+iy :h−(x, t)< y < h+(x, t)} for t∈[0,1]; where h+and h−are real, smooth, and 2π-periodic. We assume that h−(x, t)< h+(x, t)for all x∈R,t∈[0,1]; and we suppose that h−(x, t) and h+(x, t)are real-analytic in xfor each fixed t. (The real-analyticity assumptions ought to be removed, but our h+and h−happen to be real-analytic. See section 3.2.) Given t0∈[0,1], a conformal mapping defined on a neighborhood of Ω(t0)closure carries Ω(t0)to a strip. This same conformal transformation carries Ω(t)to a domain close to that strip, whenever tis close enough to t0. Rescaling in the t−variable we reduce matters to the case 68
considered above. Hence, Ω(t)is mapped conformally to {|=z|< h](t)}by a conformal map z7→ Φ(z, t), that depends Cm−smoothly on (z, t)for z∈Ω(t)closure,tnear t0. We have also Φ(z+ 2π, t) = Φ(z, t)+2π. Thus, we have our smooth family of conformal maps, defined in a small neighborhood of any given t0. These maps carry the upper boundary (respectively, the lower boundary) of Ω(t) to the upper boundary (respectively, the lower boundary) of H(t). Since such Φtare uniquely determined up to translations, it is trivial to patch together our results on small t-intervals. Thus, we obtain the following result. Lemma 5.1 Let m≥100 be given. For t∈[0,1], let Ω(t) = {x+iy :h−(x, t)< y < h+(x, t)} where h+,h−are smooth, real-valued functions on R×[0,1]. We suppose that h−(x, t)< h+(x, t)for all (x, t)∈R×[0,1], and that h+(x, t),h−(x, t)are 2π−periodic and real-analytic in xfor each fixed t. Then there exist a Cm−smooth, positive function h](t)defined on [0,1], and a Cm−smooth map Φ : {(z, t)∈C×[0,1] : z∈Ωclosure(t)} → C, such that, for each fixed t∈[0,1],z7→ Φ(z, t)maps Ω(t)conformally to {|=ζ| ≤ h](t)}. Moreover, Φ(z+ 2π, t) = Φ(z, t)+2πfor each (z, t)in the domain of φ. 5.2 Changing Coordinates Let h(x, t)be a real-analytic, positive function on T×[t0−δ, t0+δ], for h(x+ 2π, t) = h(x, t) for all x,t. Let Γ±(t) = {ζ∈C:=ζ=±h(<ζ, t)} Ω(t) = {ζ∈C:|=ζ|< h(<ζ, t)}.(5.11) Then, by lemma 5.1 there exist a smooth positive function ˆ h(ˆ t)on [t0−δ, t0+δ]and a smooth map Φ : {(ζ, t) : ζ∈Ωclosure(t),|t−t0| ≤ δ} 7→ {(ˆ ζ, ˆ t) : |=ˆ ζ| ≤ ˆ h(ˆ t),|ˆ t−t0| ≤ δ}(5.12) with the following properties: Φ(ζ, t)=(ˆ ζ, ˆ t)=(φ(ζ, t), t)(5.13) for each (ζ, t). The map ζ7→ φ(ζ, t)(5.14) is a biholomorphic map from Ω(t)to the strip {|=ˆ ζ|<ˆ h(ˆ t)}, for each t. φ(ζ+ 2π, t) = φ(ζ, t)+2π(5.15) φ: Γ+(t)7→ {=ˆ ζ=ˆ h(t)}(5.16) for each fixed t. φ(ζ, t)is real for real ζ. (5.17) The inverse of Φis the map Ψ(ˆ ζ, ˆ t) = (ψ(ˆ ζ, ˆ t),ˆ t)=(ζ, t).(5.18) 69
We would like to solve the Muskat equation ∂tzµ(ζ, t) = Z w∈Γ+(t) sin(z1(ζ, t)−z1(w, t))[∂ζzµ(ζ, t)−∂wzµ(w, t)] cosh(z2(ζ, t)−z2(w, t)) −cos(z1(ζ, t)−z1(w, t))dw (5.19) for ζ∈Ω(t),t∈[t0−δ0, t0], with zµ(ζ, t)holomorphic in ζ∈Ω(t)for fixed t, (5.20) with zµ(ζ, t)−ζδ1, µ ∈H4(Γ±(t)) for fixed t, (5.21) where δi, j is the Kronecker delta, with z1(ζ, t)−ζand z2(ζ, t)periodic with period 2π, for fixed t. (5.22) zµ(ζ, t)is real for real ζ, any t∈[t0−δ0, t0], (5.23) and satisfying the initial condition zµ(ζ, t0) = z0 µ(ζ),(5.24) where z0 µ(·)satisfy (5.20),...,(5.23), as well as the chord-arc condition |cosh(z2(ζ, t0)−z2(w, t0)) −cos(z1(ζ, t0)−z1(w, t0))| ≥cCA [||<(ζ−w)|| +|=(ζ−w)|]2(5.25) for ζ,w∈Ωclosure(t0). We assume that z0 µ(ζ)satisfy a Rayleigh-Taylor condition, to be explained below. Our goal is to transform the Muskat problem (5.19),...,(5.25) to an equivalent problem in (ˆ ζ, ˆ t)coordinates, and to show that the Rayleigh-Taylor condition for (5.19),...,(5.25) implies an analogous Rayleigh-Taylor condition in (ˆ ζ, ˆ t)coordinates. For (5.19),...,(5.25), we introduce the ’Generalized Rayleigh-Taylor function’ RT(ζ, t) = <−2π∂ζz1(ζ, t) (∂ζz1(ζ, t))2+ (∂ζz2(ζ, t))2[1 + i∂xh(<ζ, t)]−1 += P.V. Z w∈Γ+(t) sin(z1(ζ, t)−z1(w, t)) cosh(z2(ζ, t)−z2(w, t)) −cos(z1(ζ, t)−z1(w, t)))dw +i∂th(<ζ, t)o[1 + i∂xh(<ζ, t)]−1(5.26) for ζ∈Γ+(t). Our Rayleigh-Taylor assumption regarding (5.19),...,(5.25) is as follows RT(ζ, t0)>0(5.27) (strict inequality) for ζ∈Γ+(t). Let us check that assumption (5.27) holds for zµ(ζ, t)as in the previous sections, using either of our previous functions h(x, t)or ~(x, t)for the function h of this section. (If we use h(x, t), then we assume that t∈[τ2, τ]; if we use ~(x, t), then we assume 70
that t∈[−τ2, τ2]). We will also assume that λ≤A−1τ2. After we check this, we return to the task of transforming (5.19),...,(5.27) into (ˆ ζ, ˆ t)−coordinates. Let’s start with the case of the function h(x, t)(rather than ~(x, t)). Then, for ζ=x+ih(x, t)∈Γ+(t), we have −2π∂ζz1(ζ, t) (∂ζz1(ζ, t))2+ (∂ζz2(ζ, t))2−−2π∂xz1(x, t) (∂xz1(x, t))2+ (∂xz2(x, t))2≤Ch(x, t), and −2π∂xz1(x, t) (∂xz1(x, t))2+ (∂xz2(x, t))2−−2π∂xz1(x, t) (∂xz1(x, t))2+ (∂xz2(x, t))2≤CA−1τ2. Therefore, <−2π∂ζz1(ζ, t) (∂ζz1(ζ, t))2+ (∂ζz2(ζ, t))2[1 + i∂xh(<ζ, t)]−1 ≥ −Ch(x, t)−CA−1τ2+<−2π∂xz1(x, t) (∂xz1(x, t))2+ (∂xz2(x, t))2[1 + i∂xh(x, t)]−1 =−Ch(x, t)−CA−1τ2+−2π∂xz1(x, t) (∂xz1(x, t))2+ (∂xz2(x, t))2(1 + (∂xh(x, t))2)−1.(5.28) Recall that ˜a(ζ, t)≡Z w∈Γ+(t)sin(z1(ζ, t)−z1(w, t)) cosh(z2(ζ, t)−z2(w, t)) −cos(z1(ζ, t)−z1(w, t)) −∂ζz1(ζ, t) (∂ζz1(ζ, t))2+ (∂ζz2(ζ, t))2cot ζ−w 2odw. (5.29) Thanks to our observation in section (2.2) that P.V. Zw∈Γ+(t) cot ζ−w 2dw = 0, we have that ˜a(ζ, t)≡P.V. Z w∈Γ+(t) sin(z1(ζ, t)−z1(w, t)) cosh(z2(ζ, t)−z2(w, t)) −cos(z1(ζ, t)−z1(w, t))dw (5.30) Comparing (5.29) with (4.58) and (4.62), we see that ˜ais as in section (4.4). Hence, (4.75) together with (4.54), (4.55), tell us that |∂xh(x, t)˜a(ζ, t)| ≤ Ch(x, t) + CA−1τ2 with x=<ζ,ζ∈Γ+(t). Thanks to (4.56), (4.57) and (4.75), analogous estimates hold also for ~(x, t)in place of h(x, t). Consequently, =˜a(ζ, t)[1 + i∂xh(x, t)]−1≥ −Ch(x, t)−CA−1τ2. 71
Recalling (5.30), we conclude that = P.V. Z w∈Γ+(t) sin(z1(ζ, t)−z1(w, t)) cosh(z2(ζ, t)−z2(w, t)) −cos(z1(ζ, t)−z1(w, t)))dw[1 + i∂xh(x, t)]−1 ≥ −Ch(x, t)−CA−1τ2(5.31) for x=<ζ,ζ∈Γ+(t). Analogous estimates hold for ~(x, t)in place of h(x, t). We note that =i∂th(x, t)[1 + i∂xh(x, t)]−1=∂th(x, t) 1+(∂xh(x, t))2.(5.32) Putting (5.28), (5.30), (5.31), (5.32) into definition (5.26), we see that, for ζ∈Γ+(t),x=<ζ, we have: RT(ζ, t)≥(1 + (∂xh(x, t))2)−1 ×−Ch(x, t)−CA−1τ2+−2π∂xz1(x, t) (∂xz1(x, t))2+ (∂xz2(x, t))2+∂th(x, t)(5.33) The term in square brackets in (5.33) is strictly positive, thanks to (3.21). Thus, (5.27) holds for t∈[τ2, τ], under the assumptions of the previous sections. Similarly, (5.27) holds also for t∈[−τ2, τ2], with ~(x, t)in place of h(x, t), under the assumptions of the previous sections. We have succeeded in verifying the Rayleigh-Taylor condition (5.27) for our Muskat solutions from section 4. We now return to the business of transforming our Muskat problem (5.19),...,(5.25), into (ˆ ζ, ˆ t)−coordinates. We make the change of variable ˆ ζ=φ(ζ, t),ˆw=φ(w, t), for fixed t, (5.34) so that also ζ=ψ(ˆ ζ, ˆ t),w=ψ( ˆw, ˆ t), for fixed ˆ t. (5.35) Recall, ˆ t=t. We define ˆzµ(ˆ ζ, ˆ t) = zµ(ζ, t)for (ζ, t) = Ψ(ˆ ζ, ˆ t).(5.36) Note ∂ˆ ζˆzµ(ˆ ζ, ˆ t) = ∂ˆ ζψ(ˆ ζ, t)∂ζzµ(ζ, t)for (ζ, t) = Ψ(ˆ ζ, ˆ t).(5.37) Similarly, ∂ˆwˆzµ( ˆw, ˆ t) = ∂ˆwψ( ˆw, t)∂wzµ(w, t)for (w, t) = Ψ( ˆw, ˆ t).(5.38) Thus ∂ζzµ(ζ, t) = ∂ζφ(ζ, t)∂ˆ ζˆzµ(ˆ ζ, ˆ t)for (ˆ ζ, ˆ t) = Φ(ζ, t); (5.39) and ∂wzµ(w, t) = ∂wφ(w, t)∂ˆwˆzµ( ˆw, ˆ t)for ( ˆw, ˆ t) = Φ(ζ, t).(5.40) Also, dw =∂ˆwψ( ˆw, ˆ t)dˆw(5.41) 72
for fixed t. The chain rule shows that for (ˆ ζ, ˆ t) = Φ(ζ, t), we have ∂tzµ(ζ, t) = ∂t[ˆzµ◦Φ] (ζ, t) = ∂t{ˆzµ(φ(ζ, t), t)}=∂ˆ tˆzµ(ˆ ζ, ˆ t)+[∂tφ(ζ, t)]∂ˆ ζˆzµ(ˆ ζ, ˆ t) =∂ˆ tˆzµ(ˆ ζ, ˆ t) + [(∂tφ)◦Ψ(ˆ ζ, ˆ t)]∂ˆ ζˆzµ(ˆ ζ, ˆ t).(5.42) In view of the above remarks, our Muskat equation (5.19) is equivalent to the following equation in (ˆ ζ, ˆ t)−coordinates ∂ˆ tˆzµ(ˆ ζ, ˆ t) + [(∂tφ)◦Ψ(ˆ ζ, ˆ t)]∂ˆ ζˆzµ(ˆ ζ, ˆ t) = Z =ˆw=ˆ h(ˆ t) sin(ˆz1(ˆ ζ, ˆ t)−ˆz1( ˆw, ˆ t)) cosh(ˆz2(ˆ ζ, ˆ t)−ˆz2( ˆw, ˆ t)) −cos(ˆz1(ˆ ζ, ˆ t)−ˆz1( ˆw, ˆ t)) ×n[(∂ζφ)◦Ψ(ˆ ζ, ˆ t)]∂ˆ ζˆzµ(ˆ ζ, ˆ t)−[(∂wφ)◦Ψ( ˆw, ˆ t)]∂ˆwˆzµ( ˆw, ˆ t)o∂ˆwψ( ˆw, ˆ t)dˆw(5.43) Since φ(·, t)and ψ(·, t)are inverse functions, we see that [(∂wφ)◦Ψ( ˆw, ˆ t)] ·[∂ˆwψ( ˆw, ˆ t)] = 1. Hence, defining ˆ B(ˆ ζ, ˆw, ˆ t)≡[(∂ζφ)◦Ψ(ˆ ζ, ˆ t)][∂ˆwψ( ˆw, ˆ t)] −1,(5.44) we see that ˆ B(ˆ ζ, ˆw, ˆ t)is smooth on {|=ˆ ζ|,|= ˆw| ≤ ˆ h(ˆ t),ˆ t∈[t0−δ, t0+δ]}(5.45) ˆ B(ˆ ζ, ˆw, ˆ t)is holomorphic in (ˆ ζ, ˆw)∈ {|=ˆ ζ|,|= ˆw|<ˆ h(ˆ t)}for fixed ˆ t(5.46) and ˆ B(ˆ ζ, ˆ ζ, ˆ t)=0.(5.47) Remark 5.2 Note that ˆ Bdoes not depend on our Muskat solution; it is entirely determined by the function h(x, t)and the coordinate change (ˆ ζ, ˆ t) = Φ(ζ, t). Thanks to (5.44),...,(5.47), we can write (5.43) in the equivalent form ∂ˆ tˆzµ(ˆ ζ, ˆ t) = −[(∂tφ)◦Ψ(ˆ ζ, ˆ t)] + Z =ˆw=ˆ h(ˆ t) sin(ˆz1(ˆ ζ, ˆ t)−ˆz1( ˆw, ˆ t)) ˆ B(ˆ ζ, ˆw, ˆ t) cosh(ˆz2(ˆ ζ, ˆ t)−ˆz2( ˆw, ˆ t)) −cos(ˆz1(ˆ ζ, ˆ t)−ˆz1( ˆw, ˆ t))dˆw ∂ˆ ζˆzµ(ˆ ζ, ˆ t) +Z =ˆw=ˆ h(ˆ t) sin(ˆz1(ˆ ζ, ˆ t)−ˆz1( ˆw, ˆ t)) cosh(ˆz2(ˆ ζ, ˆ t)−ˆz2( ˆw, ˆ t)) −cos(ˆz1(ˆ ζ, ˆ t)−ˆz1( ˆw, ˆ t))[∂ˆ ζˆzµ(ˆ ζ, ˆ t)−∂ˆwˆzµ( ˆw, ˆ t)]dˆw. (5.48) This holds for |=ˆ ζ| ≤ ˆ h(ˆ t). We define ˆ A(ˆ ζ, ˆ t) = −[(∂tφ)◦Ψ(ˆ ζ, ˆ t)] (5.49) 73
+=A(ζ, t) + h0(t).(5.106) Then RT(ζ, t)>0for t∈Itime,ζ∈Γ+(t). (5.107) For (ζ, t)∈Ω, the following equation holds: ∂tzµ(ζ, t) = A(ζ, t) + Z w∈Γ+(t) sin(z1(ζ, t)−z1(w, t))B(ζ, w, t) cosh(z2(ζ, t)−z2( ˆw, t)) −cos(z1(ζ, t)−z1(w, t))dw ∂ζzµ(ζ, t) +Z w∈Γ+(t) sin(z1(ζ, t)−z1(w, t)) cosh(z2(ζ, t)−z2(w, t)) −cos(z1(ζ, t)−z1(w, t))[∂ζzµ(ζ, t)−∂wzµ(w, t)]dw. (5.108) In the above integrals, the integrands are holomorphic in w∈Ω(t)for fixed ζand t; in particular, the integrands have no poles. Hence, we may change the contours in these integrals from Γ+(t) to T, without changing the values of the integrals. In the next section, we prove existence of solutions to the above GENERALIZED MUSKAT PROBLEM. Thanks to the results in section (5.2) this yields the desired local existence of solutions of our original Muskat equation, with desired holomorphic and H4properties, etc. 5.4 Galerkin Approximation Let h(t),Γ±(t),t0,δ,Ω(t),Ω,Ω+,A(ζ, t),B(ζ, w, t),z0 µ(ζ)be as in section (5.3).(5.109) For a large enough positive integer N, we define z[N] µ(ζ)from z0 µ(ζ), by using the projection ΠN: ∞ X −∞ Akeikζ 7→ N X −N Akeikζ.(5.110) We define z[N] µ(ζ)by stipulating that z[N] 1(ζ)−ζ= ΠN[z0 1(ζ)−ζ](5.111) and z[N] 2(ζ)=ΠN[z0 2(ζ)].(5.112) For Nlarge enough, the functions z[N] µsatisfy the chord-arc and Rayleigh-Taylor condition (5.105), (5.106), (5.107) with controlled constants. Also, since z[N] 2(ζ)and z[N] 2(ζ)−ζare trigonometric polynomials, those function are holomorphic on Ω(t0), they extend to smooth functions on the closure of Ω(t0), and their H4−norms on Γt0are at most the H4−norms of z0 2and z0 1(ζ)−ζ, respectively, and are thus dominated by a controlled constant. Consequently, the conditions imposed in section (5.3) for z0 µhold also for z[N] µ, with controlled constants, provided Nis large enough. From now on, we suppose that Nis large enough that this holds. In place of the Generalized Muskat equation (5.108), we will solve its Galerkin approximation. That is, we will solve the equation ∂tzµ(ζ, t)=ΠN[J(·, t)](ζ, t),(5.113) 80
where J(ζ, t) = A(ζ, t) + Z w∈Γ+(t) sin(z1(ζ, t)−z1(w, t))B(ζ, w, t) cosh(z2(ζ, t)−z2(w, t)) −cos(z1(ζ, t)−z1(w, t))dw ∂ζzµ(ζ, t) +Z w∈Γ+(t) sin(z1(ζ, t)−z1(w, t)) cosh(z2(ζ, t)−z2(w, t)) −cos(z1(ζ, t)−z1(w, t))[∂ζzµ(ζ, t)−∂wzµ(w, t)]dw (5.114) We impose the initial condition zµ(ζ, t0) = z[N] µ(ζ).(5.115) Here, zµ(ζ, t)depend on N, though our notation does not indicate that. We hope to solve (5.113), (5.114), (5.115) for t∈Itime. Note that (5.113), (5.114), (5.115) is simply a system of ODE’s. We solve these ODE’s, starting at time t0, and proceeding backwards in time, until either we reach time t0−δ(the left endpoint of the interval Itime), or else the conditions (5.99),...,(5.107) (with suitable controlled constants) are not satisfied. We will see that we get all the way back to time t0−δ. Thus, we obtain a solution of (5.113), (5.114), (5.115), satisfying conditions (5.99),...,(5.107) with controlled constants; these controlled constants are independent of N, so by (5.103), we may pick a sequence Nν→ ∞ for which the corresponding solutions of (5.113), (5.114), (5.115) tend weakly to a limit zµ(ζ, t)in H4(Γ±(t)) for fixed t(rational). The resulting zµ(ζ, t)is then easily seen to solve the GENERALIZED MUSKAT PROBLEM in section (5.3). It all comes down to an a priori estimate d dt Z ζ∈Γ±(t)∂4 ζzµ(ζ, t)2dζ ≥ −C0,(5.116) to be proven as long as our solution to (5.113), (5.114), (5.115) satisfies (5.99),...,(5.107). Note that zµ(ζ, t) = zµ(ζ, t)so the integrals over Γ+(t)and Γ−(t)in (5.116) are equal. We sketch a proof of (5.116). We have 1 2 d dt Z ζ∈Γ±(t)∂4 ζzµ(ζ, t)2dζ =<Z ζ∈Γ+(t) ∂4 ζzµ(ζ, t)n∂t∂4 ζzµ(ζ, t) + ih0(t)∂4+1 ζzµ(ζ, t)odζ =<Z ζ∈Γ+(t) ∂4 ζzµ(ζ, t)n∂4 ζΠN[J(·, t)](ζ, t) + ih0(t)∂4+1 ζzµ(ζ, t)odζ =<Z ζ∈Γ+(t) ∂4 ζzµ(ζ, t)nΠN[∂4 ζJ(·, t)](ζ, t) + ih0(t)∂4+1 ζzµ(ζ, t)odζ =<Z ζ∈Γ+(t) ∂4 ζzµ(ζ, t)n∂4 ζJ(ζ, t) + ih0(t)∂4+1 ζzµ(ζ, t)odζ (5.117) 81
since ∂4 ζzµ(ζ, t), is a trigonometric polynomial in the range of ΠN. Thus, ΠNhas disappeared from the right-hand side of (5.117). The above manipulations are made possible by the fact that the contours Γ±(t)are horizontal straight lines. That is why we took the trouble to make the conformal mapping in sections 5.1 and 5.2. Using ideas from section 4.1, we can compute ∂4 ζJ(ζ, t)modulo errors with controlled L2−norm on Γ+(t). To do so, we first recall our assumptions that the H4−norms of z1(ζ, t)−ζand z2(ζ, t) on Γ+(t)are bounded by a controlled constant C, and that the chord-arc condition holds with a controlled constant c. Consequently, for ζ∈Γ+(t)and y∈[−π, π], the following estimates hold: |sin(z1(ζ, t)−z1(ζ+y, t))| ≤ C|y|,|cos(z1(ζ, t)−z1(ζ+y, t)| ≤ C(5.118) |sinh(z1(ζ2, t)−z2(ζ+y, t))| ≤ C|y|,|cosh(z2(ζ, t)−z2(ζ+y, t))| ≤ C(5.119) |∂k ζzµ(ζ, t)−∂k ζzµ(ζ+y, t)| ≤ C|y|for 0≤k≤2,µ= 1,2(5.120) |∂3 ζzµ(ζ, t)−∂3 ζzµ(ζ+y, t)| ≤ C|y|Mµ(ζ)for µ= 1,2(5.121) where Mµ(ζ)is the Hardy-Littlewood maximal function of ∂4 ζzµ|Γ+(t), and consequently ZΓ+(t) |Mµ(ζ)|2dζ ≤C. (5.122) Also, our assumptions on the function B(ζ, w, t)tell us that ˜ B(ζ, y, t)≡B(ζ, ζ +y, t)for ζ∈Γ+(t),y∈[−π, π], (5.123) satisfies |∂l ζ˜ B(ζ, y, t)| ≤ C|y|for 0≤l≤4.(5.124) We now proceed to the computation of ∂4 ζJ(ζ, t), for ζ∈Γ+(t). By the proof of lemma 4.1, we have ∂4 ζZw∈Γ+(t) sin(z1(ζ, t)−z1(w, t)) cosh(z2(ζ, t)−z2(w, t)) −cos(z1(ζ, t)−z1(w, t)) [∂ζzµ(ζ, t)−∂wzµ(w, t)] dw =Dangerousµ(ζ, t) + 6 X j=1 Safeµ j(ζ, t) + νmax X ν=1 cµ νEasyµ ν(ζ, t),(5.125) with Dangerousµ, Safeµ jand Easyµ νdefined as in lemma 4.1, with k= 4. Using our estimates (5.118),...,(5.122), one shows easily that |Easyµ ν(ζ, t)| ≤ Cfor each µ,ν. (5.126) To see how to estimate the "Safe" terms, we take for example Safeµ 2(ζ, t) = (coeff)Zπ −π [∂ζz2(ζ, t)−∂ζz2(ζ+y, t)] sinh(z2(ζ, t)−z2(ζ+y, t)) sin(z1(ζ, t)−z1(ζ+y, t)) (cosh(z2(ζ, t)−z2(ζ+y, t), t)−cos(z1(ζ, t)−z1(ζ+y, t)))2 ×∂4 ζzµ(ζ, t)−∂4 ζzµ(ζ+y, t)dy (5.127) 82
Since z1(ζ, t)−ζand z2(ζ, t)have H4−norms at most C, and since the chord-arc condition holds with a controlled constant c, one shows easily that the expression in curly brackets in (5.127) has the form θ(ζ,y,t) y, with estimates |θ(ζ, y, t)| ≤ C and |∂yθ(ζ, y, t)| ≤ C. Consequently, (5.127) and the L2−norm boundedness of the Hilbert transform tell us that ZΓ+(t) |Safeµ 2(ζ, t)|2dζ ≤C. We can treat the other "Safe" terms in (5.125) in the same way. Together with (5.126) and the definition of the "Dangerous" term in (5.125), this now implies that ∂4 ζZw∈Γ+(t) sin(z1(ζ, t)−z1(w, t)) cosh(z2(ζ, t)−z2(w, t)) −cos(z1(ζ, t)−z1(w, t)) [∂ζzµ(ζ, t)−∂ζzµ(w, t)] dw Zw∈Γ+(t) sin(z1(ζ, t)−z1(w, t)) cosh(z2(ζ, t)−z2(w, t)) −cos(z1(ζ, t)−z1(w, t)) ∂5 ζzµ(ζ, t)−∂5 ζzµ(w, t)dw +Errorµ(ζ, t),(5.128) with ZΓ+(t) |Errorµ(ζ, t)|2dζ ≤C. (5.129) Thus, modulo errors with bounded L2−norms, we have computed the fourth ζ−derivative of the last integral in (5.114). We now turn our attention to the integral involving B(ζ, w, t)in (5.114). Let 0≤k≤4. As in the proof of lemma 4.1, we find that ∂k ζZw∈Γ+(t) sin(z1(ζ, t)−z1(w, t)) cosh(z2(ζ, t)−z2(w, t)) −cos(z1(ζ, t)−z1(w, t))B(ζ, w, t)dw (5.130) may be written as a linear combination (with harmless coefficients) of terms of the form Zπ −π [sin(z1(ζ, t)−z1(ζ+y, t))]k0 [cos(z1(ζ, t)−z1(ζ+y, t))]m0 [sinh(z1(ζ, t)−z1(ζ+y, t))]k00 [cosh(z1(ζ, t)−z1(ζ+y, t))]m00 ∂l ζ˜ B(ζ, y, t) ν0 max Y ν=1 h∂k0 ν ζz1(ζ, t)−∂k0 ν ζz1(ζ+y, t)iν00 max Y ν=1 h∂k00 ν ζz2(ζ, t)−∂k00 ν ζz2(ζ+y, t)i [cosh(z2(ζ, t)−z2(ζ+y, t)) −cos(z1(ζ, t)−z1(ζ+y, t))]−mdy (5.131) where m≥1, each k0 ν,k00 ν≥1, and also k0+k00 +1+ν0 max +ν00 max −2m≥0, and l+X ν k0 ν+X ν k00 ν=k. 83
These terms (5.131) can be estimated using (5.118),...,(5.124), as we have just done for the terms Easyµ ν(ζ, t)and Safeµ 2in (5.125). If each k0 νand k00 νis at most 3, then we can prove that the integral (5.131) has absolute value at most C(this is the analogue of the "Easy" terms). If one of the k0 νor k00 νis equal to 4, then we can prove that (5.131) has L2−norm at most Cas a function of ζ∈Γ+(this is the analogue of the "Safe" terms). Note that the analogue of the Danguerous term from lemma 4.1 is an Easy term in the present context, because four derivatives fall on the known, harmless function B. Consequently, for (5.130), we learn that ∂k ζZw∈Γ+(t) sin(z1(ζ, t)−z1(w, t)) cosh(z2(ζ, t)−z2(w, t)) −cos(z1(ζ, t)−z1(w, t))B(ζ, w, t)dw≤C for ζ∈Γ+(t),0≤k≤3; and ∂4 ζZw∈Γ+(t) sin(z1(ζ, t)−z1(w, t)) cosh(z2(ζ, t)−z2(w, t)) −cos(z1(ζ, t)−z1(w, t))B(ζ, w, t)dw has L2−norm at most Con Γ+(t). Therefore, ∂4 ζ(Zw∈Γ+(t) sin(z1(ζ, t)−z1(w, t))B(ζ, w, t) cosh(z2(ζ, t)−z2(w, t)) −cos(z1(ζ, t)−z1(w, t))dw∂ζzµ(ζ, t)) =Zw∈Γ+(t) sin(z1(ζ, t)−z1(w, t))B(ζ, w, t)dw cosh(z2(ζ, t)−z2(w, t)) −cos(z1(ζ, t)−z1(w, t))∂5 ζzµ(ζ, t) + ˜ Errorµ(ζ, t) (5.132) with ZΓ+(t)˜ Errorµ(ζ, t)2dζ ≤C. (5.133) We have succeeded in computing the fourth ζ−derivative of another of the terms in (5.114). Finally, since the H4-norms of z1(ζ, t)−ζand z2(ζ, t)on Γ+(t)are less than a controlled constant C, and since A(ζ, t)has 4 derivatives bounded by C, we have ∂4 ζ{A(ζ, t)∂ζzµ(ζ, t)}= A(ζ, t)∂5 ζzµ(ζ, t) + Error] µ(ζ, t),(5.134) for ζ∈Γ+(t), with ZΓ+(t)Error] µ(ζ, t)2dζ ≤C. (5.135) Now, comparing (5.128), (5.129), (5.132), (5.133), and (5.134), (5.135) with the definition (5.114) of J, we learn that ∂4 ζJ(ζ, t) = (A(ζ, t) + Zw∈Γ+(t) sin(z1(ζ, t)−z1(w, t))B(ζ, w, t) cosh(z2(ζ, t)−z2(w, t)) −cos(z1(ζ, t)−z1(w, t))dw)∂5 ζzµ(ζ, t) +Zw∈Γ+(t) sin(z1(ζ, t)−z1(w, t)) cosh(z2(ζ, t)−z2(w, t)) −cos(z1(ζ, t)−z1(w, t)) ∂5 ζzµ(ζ, t)−∂5 ζzµ(w, t)dw +Errµ(ζ, t)for ζ∈Γ+(t),(5.136) 84
where ZΓ+(t) |Errµ(ζ, t)|2dζ ≤C. (5.137) We have succeeded in computing ∂4 ζJ(ζ, t)modulo an error with L2−norm at most Con Γ+(t). We may rewrite ∂4 ζJ(ζ, t)in the form ∂4 ζJ(ζ, t) = (A(ζ, t) + P.V. Zw∈Γ+(t) sin(z1(ζ, t)−z1(w, t)) (B(ζ, w, t) + 1) cosh(z2(ζ, t)−z2(w, t)) −cos(z1(ζ, t)−z1(w, t))dw)∂5 ζzµ(ζ, t) +Zw∈Γ+(t)(∂ζz1(ζ, t) (∂ζz1(ζ, t))2+ (∂ζz2(ζ, t))2cot ζ−w 2 −sin(z1(ζ, t)−z1(w, t)) cosh(z2(ζ, t)−z2(w, t)) −cos(z1(ζ, t)−z1(w, t))∂5 wzµ(w, t)dw −P.V. Zw∈Γ+(t) ∂ζz1(ζ, t) (∂ζz1(ζ, t))2+ (∂ζz2(ζ, t))2cot ζ−w 2∂5 wzµ(w, t)dw +Errµ(ζ, t)for ζ∈Γ+(t),(5.138) with Errµ(ζ, t)as in (5.137). Let us estimate the second-to-last integral in (5.138). The expression in curly brackets in that integral has C1−norm at most C, as a function of w∈Γ+(t); this follows by arguments in sections 4.3 and 4.4. Therefore, integrating by parts in w, we obtain an integral of the form Zw∈Γ+(t) [Quantity of absolute value at most C]·∂4 wzµ(w, t)dw; such an integral has absolute value at most C. Consequently, (5.138) may be rewritten in the form ∂4 ζJ(ζ, t) = (A(ζ, t) + P.V. Zw∈Γ+(t) sin(z1(ζ, t)−z1(w, t)) (B(ζ, w, t) + 1) dw cosh(z2(ζ, t)−z2(w, t)−cos(z1(ζ, t)−z1(w, t))))∂5 ζzµ(ζ, t) −∂ζz1(ζ, t) (∂ζz1(ζ, t))2+ (∂ζz2(ζ, t))2P.V. Zw∈Γ+(t) cot ζ−w 2∂5 wzµ(w, t) +Err]] µ(ζ, t)for ζ∈Γ+(t),(5.139) with ZΓ+(t)Err]] µ2dζ ≤C. (5.140) Now let Fµ(ζ) = ∂4 ζzµ(ζ, t)(5.141) Then Zζ∈Γ+(t) |Fµ(ζ)|2dζ ≤C. (5.142) 85
We have P.V. Zw∈Γ+(t) cot ζ−w 2∂5 wzµ(w, t)dw =−P.V. Zw∈Γ+(t) cot ζ−w 2F0 µ(ζ)−F0 µ(w)dw = 2πΛFµ(ζ)for ζ∈Γ+(t),(5.143) thanks to (2.1). Also, applying (2.16) (with our present h(t)in place of h(x)in (2.16)) we see that ∂5 ζzµ(ζ, t) = F0 µ(ζ, t) = −iΛFµ(ζ) + Goofµ(ζ)(5.144) for ζ∈Γ+(t), where ZΓ+(t) |Goofµ(ζ)|2dζ ≤C1 h(t)2X ±ZΓ±(t) |Fµ(ζ)|2dζ. (5.145) We recall that h(t)> c; see our definition of controlled constants in section 5.3. Also, since Fµ(ζ) = Fµ(ζ), we have ZΓ+(t) |Fµ(ζ)|2dζ =ZΓ−(t) |Fµ(ζ)|2dζ ≤C by (5.142). Hence, (5.145) yields ZΓ+(t) |Goofµ(ζ)|2dζ ≤C. (5.146) Now, putting (5.143), (5.144) and (5.146) into (5.139), and recalling (5.140), we see that ∂4 ζJ(ζ, t) = −i(A(ζ, t) + P.V. Zw∈Γ+(t) sin(z1(ζ, t)−z1(w, t)) (B(ζ, w, t) + 1) dw cosh(z2(ζ, t)−z2(w, t)−cos(z1(ζ, t)−z1(w, t))))ΛFµ(ζ) −2π∂ζz1(ζ, t) (∂ζz1(ζ, t))2+ (∂ζz2(ζ, t))2ΛFµ(ζ) +Erreurµ(ζ, t)for ζ∈Γ+(t),(5.147) with ZΓ+(t) |Erreurµ(ζ, t)|2dζ ≤C. (5.148) (Here, we have also used the fact that the quantity in curly brackets in (5.139) is at most C, thanks to our assumptions on A(ζ, t),B(ζ, t)and on zµ(ζ, t).) Also, from (5.144) and (5.146), we have ih0(t)∂5 ζzµ(ζ, t) = ih0(t)F0 µ(ζ) = h0(t)ΛFµ(ζ) + Goofextra µ(ζ, t)(5.149) for ζ∈Γ+(t), with ZΓ+(t)Goofextra µ(ζ, t)2dζ ≤C. (5.150) 86
From (5.147),..., (5.150), we see that ∂4 ζJ(ζ, t) + ih0(t)∂5 ζzµ(ζ, t) =(−iA(ζ, t)−iP.V. Zw∈Γ+(t) sin(z1(ζ, t)−z1(w, t)) (B(ζ, w, t) + 1) dw cosh(z2(ζ, t)−z2(w, t)−cos(z1(ζ, t)−z1(w, t))) −2π∂ζz1(ζ, t) (∂ζz1(ζ, t))2+ (∂ζz2(ζ, t))2+h0(t))ΛFµ(ζ) + Errrµ(ζ)(5.151) for ζ∈Γ+(t), with ZΓ+(t) |Errrµ(ζ, t)|2dζ ≤C. (5.152) From (5.141), (5.142) and (5.151), (5.152), we find that <Zζ∈Γ+(t) ∂4 ζzµ(ζ, t)∂4 ζJ(ζ, t) + ih0(t)∂5 ζzµ(ζ, t)dζ ≥ < Zζ∈Γ+(t) Fµ(ζ)R(ζ)ΛFµ(ζ)dζ −C, (5.153) where R(ζ) = −iA(ζ, t)−iP.V. Zw∈Γ+(t) sin(z1(ζ, t)−z1(w, t)) (B(ζ, w, t) + 1) dw cosh(z2(ζ, t)−z2(w, t)−cos(z1(ζ, t)−z1(w, t))) −2π∂ζz1(ζ, t) (∂ζz1(ζ, t))2+ (∂ζz2(ζ, t))2+h0(t).(5.154) We note that R(ζ)has C2−norm at most Con Γ+(t). (Compare with (4.58), (4.62) and (4.64). From (2.6) and our present estimate (5.142), we see that <Zζ∈Γ+(t) Fµ(ζ)R(ζ)ΛFµ(ζ)dζ ≥ < Zζ∈Γ+(t) Fµ(ζ)<(R(ζ)) ΛFµ(ζ)dζ −C. (5.155) Moreover, <(R(ζ)) ≥0on Γ+(t), thanks to the Rayleigh-Taylor hypothesis (5.97) and (5.98). (Recall: We are solving backwards in time, and we stop as soon as RT fails.) Since also the C2−norm of <(R(ζ)) on Γ+(t)is at most C, the inequality (2.6) now tell us that <ZΓ+(t) Fµ(ζ)<(R(ζ)) ΛFµ(ζ)dζ ≥ −CZΓ+(t) |Fµ(ζ)|2dζ ≥ −C, (5.156) by (5.142). From (5.153), (5.155), (5.156), we learn that <ZΓ+(t) ∂4 ζzµ(ζ, t)∂4 ζJ(ζ, t) + ih0(t)∂5 ζzµ(ζ, t)dζ ≥ −C. (5.157) Recalling (5.117), we conclude that d dt Zζ∈Γ+(t)∂4 ζzµ(ζ, t)2dζ ≥ −C. (5.158) Since we are solving backwards in time, this is our desired energy inequality. 87
We conclude that our Galerkin problem can be solved in a suitable time interval [t0−δ, t0], maintaining control of the H4−norm, the chord-arc constant, and a positive lower bound for the Rayleigh-Taylor function. This control is uniform in the parameter N used to define our Galerkin approximation. Here, δmay be taken to be a small enough controlled constant. It is now a routine exercise to deduce the following local existence theorem for the Generalized Muskat Problem, Lemma 5.4 Let h(t),Itime,Ω(t),Ω,Ω+,A(ζ, t),B(ζ, w, t),z0 1(ζ),z0 2(ζ), and RT0(ζ)be as in (5.76),...,(5.98). Then there exist functions z1(ζ, t),z2(ζ, t)on Ω, such that (5.99),...,(5.108) hold. This lemma provides only local existence, since hypothesis (5.98) demands that the length of our time interval is less than a small enough controlled constant. 5.5 Local Existence for Muskat Solutions in the Complex Domain In this section, we give the following local-in-time result. Theorem 5.5 1. Let h(x, t)be a positive real-analytic function on T×[t0−δ, t0+δ]; set Γ±(t) = {ζ∈C:=ζ=±h(<ζ, t)} and Ω(t) = {ζ∈C:|=ζ|< h(<ζ, t)}. Let z0(ζ) = (z0 1(ζ), z0 2(ζ)) be continuous on Ω(t0)closure and holomorphic on Ω(t0). Suppose that z0 1(ζ)−ζand z0 2(ζ)are 2π−periodic and take real values for real ζ. Also, suppose that z0 1(ζ)−ζand z0 2(ζ)belong to H4(Γ±(t0)). 2. Assume the chord-arc condition cosh(z0 2(ζ)−z0 2(w)) −cos(z0 1(ζ)−z0 1(w))≥c[||<(ζ−w)|| +|=(ζ−w)|]2, for ζ,w∈Ω(t0)closure. 3. Finally, assume the Rayleigh-Taylor condition RT (ζ)>0for all ζ∈Γ+(t), where RT is defined by (5.26) in terms of (z0 1(ζ), z0 2(ζ)). Then there exist positive numbers δ0,C0,c0and, functions (z1(ζ, t), z2(ζ, t)), continuous on (ζ, t)∈C×[t0−δ0, t0] : ζ∈Ω(t)closure, for which the following hold. A. For each fixed t∈[t0−δ0, t0], the functions z1(ζ, t)−ζand z2(ζ, t)are 2π−periodic, holomorphic on Ω(t), real-valued for real ζ, and have H4−norm at most C0on Γ±(t). Moreover, |cosh(z2(ζ, t)−z2(w, t)) −cos(z1(ζ, t)−z1(w, t))| ≥ c0[||<(ζ−w)|| +|=(ζ−w)|]2, for ζ,w∈Ωclosure,t∈[t0−δ0, t0]. Also, RT(ζ, t)>0(strict inequality) for all ζ∈Γ+(t),t∈[t0−δ0, t0]. (Here, RT(ζ, t)is defined from z1(ζ, t),z2(ζ, t)by (5.26)). 88
B. The functions z1(ζ, t),z2(ζ, t)satisfy the Muskat equation ∂tzµ(ζ, t) = Zw∈Γ+ sin(z1(ζ, t)−z1(w, t))[∂ζzµ(ζ, t)−∂wzµ(w, t)] cosh(z2(ζ, t)−z2(w, t)) −cos(z1(ζ, t)−z1(w, t))dw, for µ= 1,2,ζ∈Ω(t)closure,t∈[t0−δ0, t0], with initial conditions zµ(ζ, t0) = z0 µ(ζ) for µ= 1,2and ζ∈Ω(t0)closure. C. The constants δ0,C0,c0above may be taken to depend only on the following: •δ(see 1.); •The function h(x, t); •Bounds for the H4−norms of z0 1(ζ)−ζand z0 2(ζ)on Γ+(t0); •The constant cin the cord-arc condition 2; and •A positive lower bound for the Rayleigh-Taylor function in 3. Proof: In section 5.2, we showed that the Muskat equation on a time-varying domain Ω(t)is equivalent to a generalized Muskat equation on a time-varying strip. In section 5.3 and 5.4, we proved a local existence theorem for generalized Muskat. The present result follows at once. 5.6 Continuing Back to Negative Time Let z= (z1(ζ, t), z2(ζ, t)), and A,τ,λ,κ,h(x, t),~(x, t),Ω(t),Γ±(t)be as in section 3. Recall that ~(x, τ2)< h(x, τ2)for all x. We suppose we are given initial data z0(ζ) = (z0 1(ζ), z0 2(ζ)) on Ω(τ)closure, holomorphic in Ω(τ), and real-valued for real ζ. We assume that the functions z0 µ(ζ)−zµ(ζ, τ)(µ= 1,2) are 2π−periodic, and that their H4(Ω(τ))−norms are less than 10−2λ. Starting from the initial data z0at time t=τ, we solve the Muskat problem backwards in time, until we reach a time tleast at which either the assumptions of theorem 4.3 break down, or else we reach tleast =τ2. We can produce such a Muskat solution, thanks to a routine argument that combines the local existence result theorem 5.5 with the a-priori estimates given by theorem 4.3. For t∈[tleast, τ]⊂[τ2, τ], we have d dt||z(·, t)−z(·, t)||2 H4(Ω(t)) ≥ −C(A)λ2. Since also ||z(·, τ)−z(·, τ)||2 H4(Ω(τ)) ≤10−4λ2, it follows that ||z(·, tleast)−z(·, tleast)||2 H4(Ω(tleast)) ≤10−4λ2+C(A)τλ2≤10−3λ2. Consequently, the assumptions of theorem 4.3 cannot break down at t=tleast, so we have tleast =τ2. Thus, we have solved the Muskat equation back to time t=τ2. Our Muskat solution z(ζ, t) satisfies: 89