Singularity formations for a surface wave model

arXiv:1004.3975v1 [math.AP] 22 Apr 2010 Singularity Formation in a Surface Wave Model Angel Castro, Diego C´ordoba and Francisco Gancedo April 21, 2010 Abstract In this paper we study the Burgers equation with a nonlocal term of the form Hu where His the Hilbert transform. This system has been considered as a quadratic approximation for the dynamics of a free boundary of a vortex patch (see [6] and [2]). We prove blow up in finite time for a large class of initial data with finite energy. Considering a more general nonlocal term, of the form ΛαHu for 0 < α < 1, finite time singularity formation is also shown. 1 Introduction. We shall study the formation of singularities for the equation ut+uux= ΛαHu, (1) u(x, 0) = u0(x), with 0 ≤α < 1, where His the Hilbert transform [9] defined by Hf(x) = 1 πP.V. ZR f(y) x−ydy and Λα≡(−∆)α/2is given by the following expression Λαf(x) = kαZR f(x)−f(y) |x−y|1+αdy, kα=Γ(1 + α) cos((1 −α)π/2) π. The case α= 0 ut+uux=Hu (2) was introduced by J. Marsden and A. Weinstein [6] as a second order approximation for the dynamics of a free boundary of a vortex patch (see [3] and [1]). Recently J. Biello and J.K. Hunter [2] proposed it as a model for waves with constant nonzero linearized frequency. They gave a dimensional argument to show that it models nonlinear Hamiltonian waves with constant frequency. In addition, an asymptotic equation from (2) is derived, describing surface waves on a planar discontinuity in vorticity for a two-dimensional inviscid incompressible fluid. They also carried out numerical analysis showing evidence of singularity formation in finite 1 time. Let us point out that the Hamiltonian structure of the equation (1) (in particular for α= 0) comes from the representation ut+∂xδH δu = 0,where H(u) = ZR1 2uΛα−1u+1 6u3dx. (3) In section 2 we show that the linear term in the equation (2) is too weak to prevent the singularity formation of the Burgers equation. In fact, we show that, if the L∞norm of the initial data is large enough compare with the L2norm, the maximum of the solution has a singular behavior during the time of existence. One of the ingredients in the proof is to use the following pointwise inequality u(x)4≤16||u||2 L2(R)ZR (u(x)−u(y))2 (x−y)2dy, (4) (see lemma 2.2 below) which can be understood as the local version of the well-known bound ||u||4 L4≤C||u||2 L2||Λ1/2u||2 L2=C 2π||u||2 L2ZRZR (u(x)−u(y))2 (x−y)2dydx. In the appendix we provide a generalized pointwise inequality (n−dimensional) in terms of fractional derivatives. In section 3 we consider the more general family of equations, with a higher order term in derivatives, given by (1). By a different method, we prove that the blow up phenomena still arises. Let us note that, since ΛHu =−ux, the case α= 1 trivializes. Using the same approach as in section 2, it is possible to obtain blow up for 0 < α < 1/3. Inspired by the method used in [5], we check the evolution of the following quantity Jp qu(x) = ZR wp q(x−y)u(y)dy, where wp q(x) = |x|−qsign (x) if |x|<1 |x|−psign (x) if |x|>1, with 0 < q < 1 and p > 2 to find a singular behavior. Let us note that a similar approach was used by H. Dong, D. Lu and D. Li (see [7]) to show blow up for the Burgers equation with fractional dissipation in the supercritical case (0 < α < 1): ut+uux=−Λαu. (5) A different method to show singularities can be found in [8]. It is well known that the Lpnorms of the solutions of equation (5) are bounded for all 1≤p≤ ∞. However, to the best to the authors knowledge, two quantities are conserved by equation (1). The orthogonality property of the Hilbert transform provides the conservation of the L2norm, i.e. ||u(·, t)||L2(R)=||u0||L2(R). Since the equation is given by (3), we have that ZR1 3u3(x, t) + Λα−1 2u(x, t)2dx =ZR1 3u3 0(x) + Λα−1 2u0(x)2dx. 2 2 Blow up for the Burgers-Hilbert equation. The purpose of this section is to show finite time singularity formation in solutions of the equation (2). The result we shall prove is the following: Theorem 2.1 Let u0∈L2(R)∩C1+δ(R), with 0< δ < 1, satisfying the following condition: There exists a point β0∈Rwith Hu0(β0)>0,(6) such that u0(β0)≥32π||u0||2 L2(R)1/3.(7) Then there is a finite time Tsuch that lim t→T||u(·, t)||C1+δ(R)=∞, where u(x, t)is the solution to the equation (2). Proof: Let us assume that there exist a solution of the equation (2) u(x, t)∈C([0, T), C1+δ(R)), for all time T < ∞and with u0satisfying the hypotheses. Now, we shall define the trajectories x(β, t) by the equation dx(β, t) dt =u(x(β, t), t), x(β, 0) = β. Considering the evolution of the solution along trajectories, it is easy to get the identity du(x(β, t), t) dt =ut(x(β, t), t) + dx(β, t) dt ux(x(β, t), t) = Hu(x(β, t), t), and taking a derivative in time we obtain d2u(x(β, t), t) dt2=Hut(x(β, t), t) + u(x(β, t), t)Hux(x(β, t), t) =−H(uux)(x(β, t), t)−u(x(β, t), t) + u(x(β, t), t)Hux(x(β, t), t). Since H(uux)(x) = 1 2H((u2)x) = 1 2Λ(u2)(x), we can write 1 2Λ(u2)(x) = 1 2πP.V ZR u(x)2−u(y)2 (x−y)2dy =u(x)Λu(x)−1 2πZR (u(x)−u(y))2 (x−y)2dy, 3 and therefore it follows that d2u(x(β, t), t) dt2=1 2πZR (u(x(β, t), t)) −u(y, t)))2 (x(β, t)−y)2dy −u(x(β, t), t).(8) In order to continue with the proof we will prove the lemma below (for similar approach see [4]): Lemma 2.2 Let u∈L2(R)∩C1+δ(R), for 0< δ < 1. Then 1 2πZR (u(x)−u(y))2 (x−y)2dy ≥Cu(x)4, where C=1 32πE and E=||u||2 L2(R). Proof of lemma 2.2: Let us assume that u(x)>0 (a similar proof holds for u(x)<0). Let Ω be the set Ω = {y∈R:|x−y|<∆}, where ∆ will be given below. And let Ω1and Ω2be the subsets Ω1={y∈Ω : u(x)−u(y)≥u(x) 2}, Ω2={y∈Ω : u(x)−u(y)<u(x) 2}={y∈Ω : u(y)>u(x) 2}. Then 1 2πZR (u(x)−u(y))2 (x−y)2dy ≥u(x)2 8π∆2|Ω1|. On the other hand E=ZR u(y)2dy ≥ZΩ2 u(y)2dy ≥u(x)2 4|Ω2|, and therefore |Ω2| ≤ 4E u(x)2. Since |Ω1|=|Ω| − |Ω2|and |Ω|= 2∆, we have that 1 2πZR (u(x)−u(y))2 (x−y)2dy ≥u(x)2 8π∆2(2∆ −4E u(x)2). We achieve the conclusion of lemma 2.2 by taking ∆ = 4E u(x)2. 4 Next, let us define J(t) = u(x(β0, t), t). Thus, applying lemma 2.2 to the expression (8), we obtain the inequality Jtt(t)≥CJ(t)4−J(t).(9) Since Hu0(β0)>0 and Jt(t) = Hu(x(β0, t), t), we obtain that Jt(t)>0 and J(t)> J(0) for t∈(0, t∗) and t∗small enough. Therefore, multiplying (9) by Jt(t) we have that 1 2(Jt(t)2)t≥C 5(J(t)5)t−1 2(J(t)2)t,∀t∈[0, t∗). Integrating this inequality in time from 0 to twe get Jt(t)≥Jt(0)2+2C 5(J(t)5−J(0)5)−(J(t)2−J(0)2)1 2,∀t∈[0, t∗).(10) Now, since CJ(0)4−J(0) ≥0, by the statements of the theorem we obtain that Jtt(t)> Jtt(0) ≥0 for t∈(0, t∗). Therefore, Jt(t) is an increasing function [0, t∗). Thus, the inequality (10) holds for all time tand we have a contradiction. Remark 2.3 It is easy to check that there exists a large class of functions satisfying the requirement of the theorem (2.1). For example, we can consider the function u0(x) = −ax 1 + (bx)2, Hu0(x) = a 1 + (bx)2, where a, b > 0. Choosing aand bin a suitable way we can have the norm ||u0||L2(R)as small as we want and the norm ||u0||L∞(R)as large as we want. Remark 2.4 We note that the requirements (6) and (7) in theorem 2.1 can be replaced by Hu0(β0)≥0, u0(β0)>32π||u0||2 L2(R)1/3, attaining the same conclusion. 3 Blow up for the whole range 0< α < 1. In this section we shall show formation of singularities for the equation (1), with 0 < α < 1. The aim is to prove the following result: Theorem 3.1 There exist initial data u0∈L2(R)∩C1+δ(R), with 0< δ < 1, and a finite time T, depending on u0, such that lim t→T||u(·, t)||C1+δ(R)=∞ where u(x, t)is the solution to the equation (1). 5 Proof: Let us assume that there exists a solution of the equation (1), u(x, t)∈C([0, T), C1+δ(R)), for all time T < ∞. Let Jp qube the convolution Jp qu(x) = ZR wp q(x−y)u(y)dy where wp q(x) = (1 |x|qsign (x) if |x|<1 1 |x|psign (x) if |x|>1, with 0 < q < 1 and p > 2. In order to prove theorem 3.1 we shall need the following two lemmas. Lemma 3.2 Let fin C1+δ(R)∩L2(R)and 0< α < 1. Then ΛαHf(x) = kαZR f(x)−f(y) |x−y|1+αsign (x−y)dy where kα=−Γ(1 + α) sin((1 + α)π/2) π. Proof: Let fbe a function on the Schwartz class. The inverse Fourier transform formula yields ΛαHf(x) = 1 2πZR −isign (k)|k|αˆ f(k) exp(ikx)dk. We will understand the above identity as the following limit ΛαHf(x) = lim ε→0+ 1 2πZR −isign (k)|k|αexp(−ε|k|) exp(ikx)ZR f(y) exp(−iky)dydk = lim ε→0+ 1 πZR f(y)Z∞ 0 kαexp(−εk) sin(k(x−y))dkdy. Next, we can compute that ΛαHf(x) = lim ε→0+ Γ(1 + α) πZR f(y) (ε2+ (x−y)2)(1+α)/2sin (1 + α) arctan x−y εdy =−lim ε→0+ Γ(1 + α) πZR f(x)−f(y) (ε2+ (x−y)2)(1+α)/2sin (1 + α) arctan x−y εdy =−Γ(1 + α) sin((1 + α)π/2) πZR f(x)−f(y) |x−y|1+αsign (x−y)dy. We achieve the conclusion of lemma 3.2 by the classical density argument. Lemma 3.3 Let Ip q(x)be the integral Ip q(x) = ZR wp q(x)−wp q(y) |x−y|1+αsign (x−y)dy 6 where 0< q < 1and p > 2. Then |Ip q(x)| ≤      K1 |x|q+αif 0<|x|<1 2 K2 |x|2+αif 2<|x|<∞ K3if 1 2≤ |x| ≤ 2 where K1,K2and K3are universal constants depending on qand p. Proof: Since the function Ip q(x) is even, we can assume that x > 0. The constant values of K1and K2can be different along the estimates below. First, let us consider the case 0 < x < 1/2. We split as follows Ip q(x) = Z|y|<1 dy +Z|y|>1 dy =I1(x) + I2(x). It yields I1(x) = Z|y|<1 1 xq−sign (y)1 |y|q |x−y|1+αsign (x−y)dy =Z1 01 xq−1 yq |x−y|1+αsign (x−y) + 1 xq+1 yq |x+y|1+αdy, and a change of variables allow us to split further I1(x) = 1 xq+αZ1 x 01−1 ηq |1−η|1+αsign (1 −η) + 1 + 1 ηq |1 + η|1+αdη =1 xq+αZ1 0 +Z1 x 1=1 xq+α(F1(x) + F2(x)). For F1(x) we find the bound |F1(x)| ≤ Z1 0 1−1 ηq |1−η|1+αdη +Z1 0 1 + 1 ηq |1 + η|1+αdη ≤K1. On the other hand F2(x) = Z1 x 11 ηq−1 |1−η|1+α+ 1 ηq+ 1 |1 + η|1+αdη =Z3 2 1 +Z1 x 3 2 =j1(x) + j2(x). For j1(x) it is easy to obtain |j1(x)| ≤ Z3 2 1 1 ηq−1 |1−η|1+αdη +Z3 2 1 1 ηq+ 1 |1 + η|1+αdη ≤K1. For j2(x) we decompose as follows j2(x) = Z1 x 3 2 1 ηq1 |1−η|1+α+1 |1 + η|1+αdη +Z1 x 3 21 |1−η|1+α−1 |1 + η|1+αdη. 7 Thus, since 0 < q < 1 and  1 |1−η|1+α+1 |1 + η|1+α≤K1 |η|1+α,for η∈[3/2,∞) we have that |j2(x)| ≤ K1Z∞ 3 2 1 ηq+1 dη +Z∞ 3 2 1 |1−η|1+α−1 |1 + η|1+αdη ≤K1. Let us continue with I2which can be written in the form I2(x) = Z|y|>1 1 xq−sign (y)1 |y|p |x−y|1+αsign (x−y)dy =Z∞ 1− 1 xq−1 |y|p |x−y|1+α+ 1 xq+1 |y|p |x+y|1+αdy =1 xq+αZ∞ 1 xxq−p ηp−1 |1−η|1+α+1 + xq−p ηp |1−η|1+αdη. The following decomposition I2(x) = 1 xq+αZ∞ 1 x1 |1 + η|1+α−1 |1−η|1+αdη +1 xp+αZ∞ 1 x 1 ηp1 |1−η|1+α+1 |1 + η|1+αdη yields |I2(x)| ≤ K1 xq+α+1 xp+αZ∞ 1 x 1 ηp 1 |1−η|1+α+1 |1 + η|1+αdη ≤K1 xq+α+K1 xp+αZ∞ 1 x 1 ηp+1 dη ≤K11 |x|α+q+1 |x|α≤K1 xq+α. Next, we consider the case 2 < x < ∞taking Ip q(x) = ZR 1 xp−w(y) |x−y|1+αsign (x−y)dy =Z|y|<1 dy +Z|y|>1 dy =J1(x) + J2(x). For J2(x) we have that J2(x) = Z|y|>1 1 xp−sign (y)1 |y|p |x−y|1+αsign (x−y)dy =Z∞ 11 xp−1 |y|p |x−y|1+αsign (x−y) + 1 xp+1 |y|p |x+y|1+αdy 8 and a change of variables provides J2(x) = 1 xp+αZ∞ 1 x1−1 ηp |1−η|1+αsign (1 −η) + 1 + 1 ηp |1 + η|1+αdη =1 xα+pZ1 1 x +Z∞ 1=1 xα+p(H1(x) + H2(x)). For H2(x) one could bound as follow |H2(x)| ≤ Z∞ 1 1−1 ηp |1−η|1+αdη +Z∞ 1 1 + 1 ηp |1 + η|1+αdη ≤K2. On the other hand, in H1(x) we split further H1(x) = Z1 1 x1−1 ηp |1−η|1+α+1 + 1 ηp |1 + η|1+αdη =Z2 3 1 x dη +Z1 2 3 dη =h1(x) + h2(x). The term h2(x) is bounded by |h2(x)| ≤ Z1 2 3 1−1 ηp |1−η|1+αdη +Z1 2 3 1 + 1 ηp |1 + η|1+αdη ≤K2. We reorganize h1(x) so that h1(x) = Z2 3 1 x1 |1−η|1+α+1 |1 + η|1+αdη +Z2 3 1 x 1 ηp1 |1−η|1+α−1 |1 + η|1+αdη. Since p > 2 and  1 |1−η|1+α−1 |1 + η|1+α ≤K2ηfor η∈[0,2/3], we obtain that |h1(x)| ≤ Z2 3 0 1 |1−η|1+α+1 |1 + η|1+αdη +K2Z2 3 1 x 1 ηp−1dη ≤K2(1 + xp−2). Therefore |J2(x)| ≤ K21 xp+α+1 x2+α≤K2 x2+α. Next, we deal with J1given by J1(x) = Z|y|<1 1 xp−sign (y)1 |y|q |x−y|1+αdy =Z1 0 1 xp−1 |y|q |x−y|1+α+ 1 xp+1 |y|q |x+y|1+α!dy =1 xp+αZ1 x 0 1−xp−q ηq |1−η|1+α+1 + xp−q ηq |1 + η|1+α!dη. 9