On the global existence for the Muskat problem

On the global existence for the Muskat problem Peter Constantin, Diego C´ordoba, Francisco Gancedo and Robert M. Strain July 21, 2010 Abstract The Muskat problem models the dynamics of the interface between two incompressible immiscible fluids with different constant densities. In this work we prove three results. First we prove an L2(R) maximum principle, in the form of a new “log” conservation law (3) which is satisfied by the equation (1) for the interface. Our second result is a proof of global existence of Lipschitz continuous solutions for initial data that satisfy kf0kL∞<∞and k∂xf0kL∞<1. We take advantage of the fact that the bound k∂xf0kL∞<1 is propagated by solutions, which grants strong compactness properties in comparison to the log conservation law. Lastly, we prove a global existence result for unique strong solutions if the initial data is smaller than an explicitly computable constant, for instance kfk1≤1/5. Previous results of this sort used a small constant 1 which was not explicit [5, 16, 7, 12]. Keywords: Porous media, incompressible flows, fluid interface, global existence. Mathematics subjet classification: 35A01, 76S05, 76B03 Contents 1 Introduction 2 2L2maximum principle 5 3 A global existence result for data less than 1/57 4 Global existence for initial data smaller than 1 12 1 1 Introduction The Muskat problem models the dynamics of an interface between two incompressible immiscible fluids with different characteristics, in porous media. The phenomena have been described using the experimental Darcy’s law that is given in two dimensions by the following momentum equation: µ κu=−∇p−g(0, ρ). Here µis viscosity, κpermeability of the isotropic medium, uvelocity, p pressure, ggravity and ρdensity. Saffman and Taylor [15] related this problem with the evolution of an interface in a Hele-Shaw cell since both physical scenarios can be modeled analogously (see also [5] and reference therein). Recently, the well-posedness has been shown without surface tension in [6] (for previous work on the topic see [1], [18] and [7]) using arguments that rely upon the boundedness properties of the Hilbert transforms associated to C1,γ curves. Precise estimates are obtained with arguments involving conformal mappings, the Hopf maximum principle and Harnack inequalities. The initial data have to satisfy the Rayleigh-Taylor condition initially, otherwise the problem has been shown to be ill-posed [16], [7]. With surface tension, the initial value problem becomes more regular, and instabilities do not appear [11]. The case of more than one free boundary has been treated in [10] and [12]. In this paper we consider an interface given by fluids of different constant densities ρi, with the same viscosity and without surface tension. The step function ρis represented by ρ(x, t) = ρ1, x ∈Ω1(t), ρ2, x ∈Ω2(t) = R2\Ω1(t), for Ωi(t) connected regions. As the density ρis transported by the flow ρt+u·∇ρ= 0, the free boundary evolves with the two dimensional velocity u= (u1, u2). The Biot-Savart law recovers ufrom the vorticity given by ω=∂x1u2−∂x2u1, via the integral operator u(x, t) = ∇⊥∆−1ω(x, t). Darcy’s law then provides the relation ω=−∂x1ρwhere µ/κ and gare taken equal to 1 for the sake of simplicity. Then the velocity field can be obtained 2 in terms of the density as follows: u(x, t) = PV ZR2 K(x−y)ρ(y, t)dy −1 2(0, ρ(x, t)). Here the kernel Kis of Calder´on-Zygmund type: K(x) = 1 π−x1x2 |x|2,x2 1−x2 2 2|x|2, (see [17]). As a consequence of ρ∈L∞(R2×R+) it follows that the velocity belongs to BMO. Moreover, as Kis an even kernel, it has the property that the mean of K(in the principal value sense) are zero on hemispheres [3], and this yields a bound of the velocity u(x, t) in terms of C1,γ norms (0 < γ < 1) of the free boundary [10]. In order to have a well-posed problem we need to consider initially an interface parameterized as a graph of a function with the denser fluid below: ρ2> ρ1as in [7]. The interface is characterized as a graph of the function (x, f(x, t)). This characterization is preserved by the system and fsatisfies ft(x, t) = ρ2−ρ1 2πPV ZR dα (∂xf(x, t)−∂xf(x−α, t))α α2+ (f(x, t)−f(x−α, t))2, f(x, 0) = f0(x), x ∈R. (1) The above equation can be linearized around the flat solution to find the following nonlocal partial differential equation ft(x, t) = −ρ2−ρ1 2Λf(x, t), f(α, 0) = f0(α), α ∈R, (2) where the operator Λ is the square root of the Laplacian. This linearization shows the parabolic character of the problem in the stable case (ρ2> ρ1), as well as the ill-posedness in the unstable case (ρ2< ρ1). The nonlinear equation (1) is ill-posed in the unstable situation and locally well-posed in Hk(k≥3) for the stable case [7]. Furthermore the stable system gives a maximum principle kfkL∞(t)≤ kfkL∞(0), see [8]; decay rates are obtained for the periodic case as: kfkL∞(t)≤ kf0kL∞e−Ct, and also for the case on the real line (flat at infinity) as: kfkL∞(t)≤kf0kL∞ 1 + Ct . 3 Numerical solutions performed in [9] further indicate a regularizing effect. The decay of the slope and the curvature is stronger than the rate of decay of the maximum of the difference between fand its mean value. Thus, the irregular regions in the graph are rapidly smoothed and the flat regions are smoothly bent. It is shown analytically in [8] that, if the initial data satisfy k∂xf0kL∞<1, then there is a maximum principle that shows that this derivative remains in absolute value smaller than 1. The three main results we present in this paper are the following: 1) In Section 2, we prove that a solution of (1) satisfies kfk2 L2(t) + ρ2−ρ1 2πZt 0 ds ZR dα ZR dx ln 1 + f(x, s)−f(α, s) x−α2 =kf0k2 L2.(3) Furthermore, we have the the inequality ZR dx ZR dα ln 1 + f(x, s)−f(α, s) x−α2≤CkfkL1(s). This identity shows a major difference with the linear equation (2) where the evolution of the L2norm provides a gain of half derivative for ρ2> ρ1: kfk2 L2(t) + ρ2−ρ1Zt 0 ds kΛ1/2fk2 L2(s) = kf0k2 L2,(4) or equivalently kfk2 L2(t) + ρ2−ρ1 2πZt 0 ds ZR dx ZR dα f(x, s)−f(α, s) x−α2 =kf0k2 L2. Notice that this linear energy balance (4) directly implies compactness, whereas compactness does not follow from the nonlinear L2energy (3). 2) In Section 4 we prove global in time existence of Lipschitz continuous solutions in the stable case. We understand the solution of (1) using its weak formulation: ZT 0 dt ZR dx ηt(x, t)f(x, t) + ZR dx η(x, 0)f0(x) =ZT 0 dt ZR dx ηx(x, t)ρ2−ρ1 2πPV ZR dα arctan f(x, t)−f(α, t) x−α. (5) 4 This equality holds ∀η∈C∞ c([0, T)×R). For initial data f0satisfying kf0kL∞<∞and k∂xf0kL∞<1 we prove that there exists a solution of (5) that remains in the spaces f(x, t)∈C([0, T]×R)∩L∞([0, T ]; W1,∞(R)) for any T > 0. We point out that, because of the condition f∈L∞(R), the nonlinear term in (5) has to be understood as a principal value for the integral of two functions, one in H1and the other in BMO [17]. There are several results of global existence for small initial data (small compared to 1 or 1) in several norms (more regular than Lipschitz) [5, 16, 7, 12] taking advantage of the parabolic character of the equation for small initial data. Here we show that we just need k∂xf0kL∞<1, therefore  f0(x)−f0(α) x−α<1. Notice that considering the first order term in the Taylor series of ln(1+ y2) for |y|<1, then the identity (3) becames (4). 3) Our third result, discussed in Section 3, proves global existence of unique C([0, T]; H3(R)) solutions if initially the norm (6) of f0is controlled as kf0k1< c0where kf0k1=ZR dξ |ξ||ˆ f0(ξ)|. The key point here, in comparison to previous work [5, 16, 7, 12], is that the constant c0can be easily explicitly computed. We have checked numerically that c0is not that small; it is greater than 1/5. 2L2maximum principle In this section we provide a proof of the identity (3). As we are in the stable case, we take without loss of generality (ρ2−ρ1)/(2π) = 1 to simplify the exposition. The contour equation (1) can be written as follows: ft(x, t) = PV ZR ∂xarctan f(x, t)−f(x−α, t) αdα. We multiply by f, integrate over dx, and use integration by parts to observe 1 2 d dtkfk2 L2(t) = −ZR dx ZR dα fx(x) arctan f(x, t)−f(x−α, t) α =−ZR dx ZR dα fx(x) arctan f(x, t)−f(α, t) x−α. 5 We use the splitting 1 2 d dtkfk2 L2(t) = −ZRZRf(x, t)−f(α, t) x−αarctan f(x, t)−f(α, t) x−αdxdα −ZRZRfx(x)(x−α)−(f(x, t)−f(α, t)) x−αarctan f(x, t)−f(α, t) x−αdxdα =I1+I2. We also use the function Gdefined by G(x) = xarctan x−ln p1 + x2=Zx 0 dy arctan y. With these, it is easy to observe that I2=−ZRZR (x−α)∂xGf(x, t)−f(α, t) x−αdxdα. The identity below lim |x|→∞(x−α)Gf(x, t)−f(α, t) x−α= 0, allows us to integrate by parts to obtain I2=ZRZR Gf(x, t)−f(α, t) x−αdxdα =−I1−ZRZR ln r1 + f(x, t)−f(α, t) x−α2dxdα. This equality gives 1 2 d dtkfk2 L2(t) = −ZRZR ln r1 + f(x, t)−f(α, t) x−α2dxdα, and integrating in time we get the desired identity. The above equality indicates that for large initial data, the system is not parabolic at the level of f. We prove below the inequality ZRZR ln 1 + f(x, t)−f(α, t) x−α2dxdα ≤4π√2kfkL1(t) which shows that there is no gain of derivatives for the stable case. If the initial data are positive, then kfkL1(t)≤ kf0kL1follows from [8], so that the dissipation is bounded in terms of the initial data with zero derivatives. 6 For the proof of the inequality, we denote by Jthe integral Jdef =ZRZR ln 1 + f(x)−f(x−α) α2dxdα. We now use that the function ln(1 + y2) is increasing to observe that J≤ZRZR ln 1 + 2|f(x)|2 α2+2|f(x−α)|2 α2dxdα. The inequality ln(1 + a2+b2)≤ln(1 + a2) + ln(1 + b2) yields J≤ZRZR ln 1 + 2|f(x)|2 α2dxdα +ZRZR ln 1 + 2|f(x−α)|2 α2dxdα, and therefore J≤2ZRZR ln 1 + 2|f(x)|2 α2dxdα =K. For Kit is easy to get K= 2 Z{x:|f(x)|6=0} dx ZR dα ln 1 + 2|f(x)|2 α2, so that an easy integration in αprovides K= 4π√2Z{x:|f(x)|6=0} dx |f(x)|= 4π√2kfkL1. This concludes our discussion of the L2maximum principle (3) for (1). 3 A global existence result for data less than 1/5 In this section we prove global existence of C([0, T]; H3(R)) small data solutions. A key point is to consider the norm kfks def =ZR dξ |ξ|s|ˆ f(ξ)|, s ≥1.(6) This norm allows us to use Fourier techniques for small initial data that give rise to a global existence result for classical solutions. 7 Theorem 3.1. Suppose that initially f0∈H3(R)and kf0k1< c0, where c0 is a constant such that 2X n≥1 (2n+ 1)2+δc2n 0≤1 for 0< δ < 1/2. Then there is a unique solution fof (1) that satisfies f∈C([0, T]; H3(R)) for any T > 0. Remark 3.2. We compute the limit case δ= 0, so that 2X n≥1 (2n+ 1)2c2n 0≤1 for 0≤c0≤1 3s7−14 ×52/3 3 p9√39 −38 + 2 3 q5(9√39 −38) ≈0.2199617648835399. In particular, 2X n≥1 (2n+ 1)2.1c2n 0<1, if say c0≤1/5. The remainder of this section is devoted to the proof of Theorem 3.1. The contour equation for the stable Muskat problem (1) can be written as ft(x, t) = −ρ(Λf+T(f)),(7) where we recall that ρ=ρ2−ρ1 2>0 and we have T(f) = 1 πZR ∂xf(x)−∂xf(x−α) αf(x)−f(x−α) α2 1 + f(x)−f(x−α) α2dα, (8) We define ∆αf(x)def =f(x)−f(x−α) α. We consider the evolution of the norm kfk1(6): d dtkfk1(t) = ZR dξ |ξ|sgn ( ˆ f(ξ)) ˆ ft(ξ) =ρZR dξ |ξ|sgn ( ˆ f(ξ)) (−|ξ|ˆ f(ξ)−F(T)(ξ)). 8 We will show that the first term dominates the second term if initially kf0k1<q(4 −√13)/6,where q(4 −√13)/6>1/4. The key point, again, is that the constant is given explicitly. Notice that under the local existence theorem of [7], this bound will be propagated for a short time. Then we may use the Taylor expansion x2 1 + x2=∞ X n=1 (−1)n+1x2n, to obtain T(f) = −1 πX n≥1 (−1)nZR ∂x(∆αf) (∆αf)2ndα. (9) Notice that F(∆αf) = ˆ f(ξ)m(ξ, α),F(∂x∆αf) = −iξ ˆ f(ξ)m(ξ, α), m(ξ, α) = 1−e−iξα α. Therefore F(∂x(∆αf) (∆αf)2n) = ((−iξ ˆ fm)∗(ˆ fm)∗···∗(ˆ fm))(ξ, α), with 2nconvolutions, one with −iξ ˆ fm and 2n−1 with ˆ fm. Using (9) F(T)(ξ) = i πX n≥1 (−1)nZR dα ZR dξ1···ZR dξ2n(ξ−ξ1)ˆ f(ξ−ξ1)m(ξ−ξ1, α) ׈ f(ξ1−ξ2)m(ξ1−ξ2, α)··· ˆ f(ξ2n−1−ξ2n)m(ξ2n−1−ξ2n, α)ˆ f(ξ2n)m(ξ2n, α) =X n≥1ZR dξ1···ZR dξ2n(ξ−ξ1)ˆ f(ξ−ξ1) 2n−1 Y i=1 ˆ f(ξi−ξi+1)!ˆ f(ξ2n)Mn, where Mn=Mn(ξ, ξ1, . . . , ξ2n) is given by Mn def =i π(−1)nZR m(ξ−ξ1, α) 2n−1 Y i=1 m(ξi−ξi+1, α)!m(ξ2n, α)dα. (10) Since m(ξ, α) = iξ R1 0ds eiα(s−1)ξwe obtain Mn(ξ, ξ1, . . . , ξ2n) = mn(ξ, ξ1, . . . , ξ2n) (ξ1−ξ2)···(ξ2n−1−ξ2n)ξ2n, 9 using (17) and that −εC 2Λ1−εfx(xt) + ρ πJ2(xt)≤0, by (16). In fact −εC 2Λ1−εfx(xt) + ρ πK2(xt) =−εPV ZR fx(xt)−f(xt)−f(xt−α) α |α|2−ε Cc2(ε) 2(∆ε αf(xt))2+Cc2(ε) 2−ρ π 1 + (∆ε αf(xt))2dα. The mean value theorem gives |f(x)−f(x−α)|/|α|≤kfxkL∞. Thus if we take C≥2ρ cmπwe obtain the first inequality. Also −εC 2Λ1−εfx(xt) + ρ πJ2(xt) = −εPV ZR fx(xt)−fx(xt−α) |α|2−ε Cc1(ε) 2(∆ε αf(xt))2+Cc1(ε) 2−ρ π 1 + (∆ε αf(xt))2dα −εPV ZR fx(xt)−fx(xt−α) |α|2−ε 2(∆ε αf(xt))2dα (1 + (∆ε αf(xt))2)2≤0. We find fxxx(xt)≤0 and K1(xt) = 0. We still have to deal with K3and K4. Considering K3(xt) + K4(xt), we realize that if P(α) = 2 + 2(∆ε αf(xt))2+ 2(∆ε αf(xt))(fx(xt)|α|ε−(1−ε)∆ε αf(xt)) ≥0, we are done. We rewrite P(α) = 2 + 2ε(∆ε αf(xt))2+ 2(∆ε αf(xt))fx(xt)|α|ε, and therefore we need |(∆ε αf(xt))fx(xt)|α|ε| ≤ 1. This fact holds if |f|C1−2ε= sup α6=0 |f(xt)−f(xt−α)| |α|1−2ε<1. 16 Now we will check that if kfkL∞≤ kf0kL∞and kfxkL∞<1 then |f|C1−2ε< 1 for εsmall enough uniformly. We replace 2εby εwithout lost of generality. If kf0kL∞= 0 or kfxkL∞= 0 all done. Otherwise |f(xt)−f(xt−α)| |α|1−ε≤ kfxkL∞δε, for 0 <|α| ≤ δand |f(xt)−f(xt−α)| |α|1−ε≤2kf0kL∞ δ1−ε, for |α| ≥ δ. We take δ1−ε= 2kf0kL∞/kfxkL∞and therefore |f|C1−ε≤max{kfxkL∞,kfxk1−ε 1−ε L∞(2kf0kL∞)ε 1−ε}. Now it is clear that given kf0kL∞, if kfxkL∞<1 there exists ε0>0 such that, for any 0 ≤ε≤ε0it follows that |f|C1−ε≤1. 4.2 Global existence for the regularized model We use here the a priori bounds to show global existence. Local existence can be easily proving using the local existence proof for the non-regularized Muskat problem (1), as in [7]. We use energy estimates and the Gronwall inequality. As we did for (1), it follows that d dtkfkL2(t) = −ρ πZRZR 1−ε |x−α|εln 1 + f(x, t)−f(α, t) x−α2dxdα −2CεkΛ(1−ε)/2fkL2(t)−2εkfxkL2(t). Therefore kfkL2(t)≤ kf0kL2. Remark 4.2. The theorem of this section can also be found with kf0kL2<∞instead of kf0kL∞<∞. We picked the version above because it is more general. We see that if the solution satisfies initially a L2bound then f(x, t)∈L∞([0, T]; L2(R)). Next, we consider the evolution of ZR ∂3 xf∂3 xftdx ≤ −CεkΛ(1−ε)/2∂3 xfkL2(t)−εk∂4 xfxk2 L2+L1+L2, 17 where L1=ρ πZR ∂3 xf(x)∂3 xPV ZR fx(x)−fx(x−α) φ(α)dαdx, L2=−ρ πZR ∂3 xf(x)∂3 xPV ZR fx(x)−fx(x−α) φ(α) (∆ε αf(x))2dα 1 + (∆ε αf(x))2dx. The term fx(x) cancel out in L1due to the PV and an integration by parts shows that L1=−ρ πC(ε)ZR ∂3 xf(x)Λ1−ε∂3 xf(x)dx ≤0. For L2one finds L2=ρ πZR ∂4 xf(x)∂2 xPV ZR fx(x)−fx(x−α) φ(α) (∆ε αf(x))2dα 1 + (∆ε αf(x))2dx, and the splitting L2=M1+M2+M3gives M1=ρ πZR ∂4 xf(x)ZR ∂3 xf(x)−∂3 xf(x−α) φ(α) (∆ε αf(x))2dα 1 + (∆ε αf(x))2dx, M2=3ρ πZR ∂4 xf(x)ZR ∂2 xf(x)−∂2 xf(x−α) φ(α) fx(x)−fx(x−α) φ(α) ×2(∆ε αf(x))dα (1 + (∆ε αf(x))2)2dx, M3=ρ πZR ∂4 xf(x)ZRfx(x)−fx(x−α) φ(α)3(2 −6(∆ε αf(x))2)dα (1 + (∆ε αf(x))2)3dx. For M1we proceed as follows |M1|=ρ π Z|α|>1 dα ZR dx +Z|α|<1 dα ZR dx! ≤C(ε)(kfkL∞+ 1)k∂3 xfkL2k∂4 xfkL2. The identity ∂2 xf(x)−∂2 xf(x−α) = Z1 0 ∂3 xf(x+ (s−1)α)αds, 18 yields |M2| ≤ 6ρ πZ1 0 ds Z|α|<1 dα |α|1−2εZR dx|∂4 xf(x)||∂3 xf(x+ (s−1)α)| ×(|fx(x)|+|fx(x−α)|) +6ρ πZ1 0 ds Z|α|>1 dα |α|2−3εZR dx |∂4 xf(x)||∂3 xf(x+(s−1)α)| ×(|fx(x)|+|fx(x−α)|)(|f(x)|+|f(x−α)|), and therefore |M2| ≤ C(ε)(1 + kfkL∞)k∂3 xfkL2k∂4 xfkL2kfxkL∞. In M3we use the splitting M3=N1+N2where N1=ρ πZ|α|>1 dα ZR dx, N2=Z|α|<1 dα ZR dx, and then |N1| ≤ 16ρ πkfxk2 L∞Z1 0 ds Z|α|>1 dα |α|3−3εZR dx|∂4 xf(x)|(|fx(x)|+|fx(x−α)|) ≤Ckfxk2 L∞kfxkL2k∂4 xfkL2 ≤Ckfxk2 L∞(kfkL2+k∂3 xfkL2)k∂4 xfkL2. To finish, the equality fx(x)−fx(x−α) = Z1 0 ∂2 xf(x+ (s−1)α)α ds, allows us to obtain (since 1 2+1 4+1 4= 1): |N2| ≤ 16ρ πkfxkL∞ Z1 0 dr Z1 0 ds Z|α|<1 dα |α|1−3εZR dx ×|∂4 xf(x)||∂2 xf(x+ (r−1)α)||∂2 xf(x+ (s−1)α)| ≤CkfxkL∞k∂4 xfkL2k∂2 xfk2 L4. The following estimate k∂2 xfk4 L4=ZR (∂2 xf)3∂2 xfdx =−3ZR (∂2 xf)2∂3 xf∂xfdx ≤3kfxkL∞k∂2 xfk2 L4k∂3 xfkL2, 19 yields |N2| ≤ Ckfxk2 L∞k∂4 xfkL2k∂3 xfkL2. Using Young’s inequality d dtk∂3 xfk2 L2≤C(ε)(kfk4 L∞+kfk2 L∞+kfxk4 L∞+kfxk2 L∞+ 1) ×(kfk2 L2+k∂3 xfk2 L2), and therefore the Gronwall inequality yields kfk2 L2(t) + k∂3 xfk2 L2(t)≤(kf0k2 L2+k∂3 xf0k2 L2) exp Zt 0 C(ε)G(s)ds, for G(s) = kfk4 L∞+kfk2 L∞+kfxk4 L∞(s) + kfxk2 L∞(s)+1. We find f∈C([0, T]; H3(R)) for any T > 0 by the a priori bounds. 4.3 Approximation of the initial data The approximation of the initial data described below is needed in order to construct a weak solution. First consider a common approximation to the identity ζ∈C∞ c(R) satisfying ZR dx ζ(x)=1, ζ ≥0, ζ(x) = ζ(−x). Now we denote the standard mollifier ζε(x) = ζ(x/ε)/ε so that ζε(x) continues to satisfy the normalization condition above. For any f0∈W1,∞(R) and k∂xf0kL∞<1, we define the initial data for the regularized system as follows fε 0(x) = (ζε∗f0)(x) 1 + εx2. Notice that fε 0∈Hs(R) for any s > 0, and kfε 0kL∞≤ kf0kL∞. More importantly, k∂xfε 0kL∞≤ k∂xf0kL∞if εis sufficiently small (here ε will generally depend upon the size of kf0kL∞). In particular ∂xfε 0(x) = (ζε∗∂xf0)(x) 1 + εx2−2εx(ζε∗f0)(x) (1 + εx2)2, 20 and clearly  (ζε∗∂xf0)(x) 1 + εx2≤ k(ζε∗∂xf0)kL∞(R)≤ k∂xf0kL∞(R). On the other hand, by splitting into |x| ≤ ε−2/3and |x|> ε−2/3we have that 2εx(1 + εx2)−2≤2 max{ε1/3, ε}. On the unbounded region we have x(1 + εx2)−2=1 √x+εx3/2−2 ≤1. Thus, the desired bound follows if εis small enough. Therefore global existence of the regularized system (13) is true with fε 0if εis small enough. Now consider the solution to the regularized system (13) with initial data given by the fε 0just described above. For ε > 0 sufficiently small, we decompose ZR η(x, 0)fε 0(x)dx =ZR η(x, 0)(ζε∗f0)(x) 1 + εx2dx =Iε 1+Iε 2, where Iε 1=ZR η(x, 0)(ζε∗f0)(x)1 1 + εx2−1dx, and Iε 2=ZR η(x, 0)(ζε∗f0)(x)dx. We apply the dominated convergence theorem to find that as ε↓0 it holds that Iε 1→0. For Iε 2we write Iε 2=ZR ζε∗(η(·,0))f0(x)dx. The L1approximation of the identity property shows that Iε 2→ZR η(x, 0)f0(x)dx. Thus, it remains to check the convergence of the rest of the terms in (5). 21 4.4 Weak solution In this section we prove that solutions of the regularized system converge to a weak solution satisfying the bounds kfkL∞(t)≤ kf0kL∞,k∂xfkL∞(t)≤ k∂xf0kL∞<1.(18) Given a collection of regularized solutions {fε}to (13), we have the uniform (in ε > 0) bound kfεkL∞(t)≤ kf0kL∞,k∂xfεkL∞(R)(t)≤1, ε > 0.(19) This implies that there is a subsequence (denoted again by fε) such that ZT 0ZR fε(x, t)g(x, t)dxdt −→ ZT 0ZR f(x, t)g(x, t)dxdt, ZT 0ZR ∂xfε(x, t)g(x, t)dxdt −→ ZT 0ZR ∂xf(x, t)g(x, t)dxdt, for f∈L∞([0, T]; W1,∞(R)) and any g∈L1([0, T]×R) by the Banach- Alaoglu theorem. We find weak* convergence in L∞([0, T]; W1,∞(R)). We denote BN= [−N, N], then we claim that there is a subsequence (denoted again by fε) such that kfε−fkL∞([0,T]×BN)→0,as ε→0. We will prove this fact in the next Section 4.5. Then, up to a subsequence, we find uniform convergence of fεto fon compact sets. Since fε∈C([0, T]×R) we find that fis continuous. The only thing to check is that as ε↓0 we have ZT 0 dt ZR dx ηx(x, t)ρ πPVZR dα arctan fε(x)−fε(x−α) φε(α) →ZT 0 dt ZR dx ηx(x, t)ρ πPVZR dα arctan f(x)−f(x−α) α, where φε(α) = α/|α|ε. The rest of the terms will converge in the usual obvious way (since they are linear). Choose M > 0 so that supp(η)⊆BM. For any small δ > 0 and any large R1, with R > M + 1 we split the integral as ZR dα =ZBδ dα +ZBR−Bδ dα +ZBc R dα. 22 We begin with a proof that the first and last integrals separately are arbitrarily small independent of εfor R > 0 sufficiently large and for δ > 0 sufficiently small. One finds that  arctan fε(x)−fε(x−α) φε(α)≤π 2. Here we don’t need any regularity for fε, and we conclude ZT 0 dt ZR dx ηx(x, t)ρ πPV ZBδ dα arctan fε(x)−fε(x−α) φε(α) ≤ρkηxkL1([0,T]×R)δ. Therefore this term can clearly be chosen arbitrarily small, depending upon the smallness of δ. We now estimate the term integrated over Bc R. We note that arctan y=Z1 0 d ds(arctan(sy))ds =yZ1 0 1 1 + s2y2ds, and therefore arctan y=y1−Z1 0 s2y2 1 + s2y2ds. This is morally the first order Taylor expansion for arctan with remainder in integral form. With this expression we have that PVZBc R dα arctan fε(x)−fε(x−α) φε(α)=−Hε R(fε) −PV ZBc R dα fε(x)−fε(x−α) φε(α)3Z1 0 s2ds 1 + s2fε(x)−fε(x−α) φε(α)2. Here Hε Ris a (Hilbert-type) Transform, which has the form Hε R(fε)def =PVZBc R dα fε(x−α) φε(α). The principal value is evaluated at infinity (if necessary). For the second term in the left hand side notice that the integral is over Bc Rand the principal value is not necessary. In particular, we have ZBc R dα Z1 0 ds≤Ckfεk3 L∞Z∞ R dα α3−3ε≤Ckf0k3 ∞ R. 23 This term is therefore arbitrarily small if Ris chosen sufficiently large. We are going to show the same for IR def =ZM −M dx ηx(x, t)Hε R(fε). We write IR=JR+KRwhere JR def = lim n→∞ZM −M dx ηx(x, t)Z−R −n dα fε(x−α) φε(α), KR def = lim n→∞ZM −M dx ηx(x, t)Zn R dα fε(x−α) φε(α). We shall show how to control JR, the same follows for KR. We write JR= lim n→∞ZM −M dx ηx(x, t)Zn x+R dα fε(α) φε(x−α). An integration by parts yields JR= lim n→∞ZM −M dx η(x, t)f(x+R)|R|ε −R+ (1 −ε)Zn x+R dα fε(α) |x−α|2−ε. Hence |JR| ≤ 2kηkL1kf0kL∞/R1/2. Since the same estimate holds for |KR|, one finds that IRis arbitrarily small if Ris arbitrarily large. It remains to prove the convergence of ZT 0ZBR−Bδ dα ηx(x, t) arctan fε(x)−fε(x−α) φε(α). Recall that we have uniform convergence on compact sets. Lets consider Gε=fε(x)−fε(x−α) φε(α), where x∈BMand α∈BR−Bδ. Since arctan is a continuous function then arctan(Gε)→arctan(G0) converges uniformly. Thus also the integral over 24 a bounded region of arctan(Gε) also converges. Then for any R > M + 1 and any small δ > 0 as ε↓0 we have ZT 0 dt ZR dx ηx(x, t)ρ πZBR−Bδ dα arctan fε(x)−fε(x−α) φε(α) →ZT 0 dt ZR dx ηx(x, t)ρ πZBR−Bδ dα arctan f(x)−f(x−α) α. We conclude by first choosing Rsufficiently large and δ > 0 sufficiently small and then sending ε↓0. Note that Rand δwill generally depend upon the size of kf0k∞, but this has no effect on our argument. 4.5 Strong convergence in L∞([0, T]; L∞(BR)) In order to prove the strong convergence in L∞([0, T]; L∞(BR)), the idea is to use the non-standard weak space W−2,∞ ∗(BR) which will be defined below. Crucially, we will have the uniform bounds: sup t∈[0,T]kfε(t)kW1,∞(BR)≤Ckf0kL∞(R), sup t∈[0,T]    ∂fε ∂t (t)   W−2,∞ ∗(BR)≤Ckf0kL∞(R), (20) where Cdoes not depend on Ror ε. From here we will conclude that for any finite R > 0 there exists a subsequence such that fε→fstrongly in L∞([0, T]; L∞(BR)). We define the space W−2,∞ ∗(BR) as follows. For v∈L∞(BR) we consider the norm kvk−2,∞= sup φ∈W2,1 0(BR) : kφk2,1≤1ZBR φ(x)v(x)dx . Here W2,1 0(BR) is the usual set of functions in W2,1(BR) which vanish on the boundary of BRtogether with their first two weak derivatives. Now the Banach space W−2,∞ ∗(BR) is defined to be the completion of L∞(BR) with respect to the norm k·k−2,∞. In general this is all we need for our convergence study. This will be explained after the proof of Lemma 4.3 below. The full space W−2,∞ ∗may be a large space, but because we are going to deal with fεand dfε dt , both in L∞([0, T ]; L∞(BR)), it is not difficult to find the norms of both functions in W−2,∞ ∗(BR). 25