Global existence of smooth solutions to two-dimensional compressible isentropic Euler equations for Chaplygin gases

Transcription

1 Global existence of smooth solutions to two-dimensional compressible isentropic Euler equations for Chaplygin gases De-Xing Kong a Yu-Zhu Wang b a Center of Mathematical Sciences, Zhejiang University Hangzhou 31007, China b Department of Mathematics, Shanghai Jiao Tong University Shanghai 0040, China Abstract In this paper we investigate the two-dimensional compressible isentropic Euler equations for Chaplygin gases. Under the assumption that the initial data is close to a constant state the vorticity of the initial velocity vanishes, we prove the global existence of the smooth solution to the Cauchy problem for two-dimensional flow of Chaplygin gases. Key words phrases: Chaplygin gases, compressible Euler equations, irrotational flow, null condition, Cauchy problem, smooth solution. 000 Mathematics Subject Classification: 76N10; 35Q35; 35L99. 1

2 1 Introduction This paper concerns the following two-dimensional compressible isentropic Euler equations t ρ + u ρ + ρ u = 0, t u + u u ρ P = 0, where ρx, t > 0 is the density, ux, t = u 1 x, t, u x, t T denotes the pressure the state equation reads sts for the velocity, P P = P 0 Aρ 1, in which P 0, A are two positive constants. We are interested in the global existence on the smooth solution of the Cauchy problem for the system 1.1 with the initial data t = 0 : ρ = ρ + ερ 0 x, u = εu 0 x, 1. where ε is a positive small parameter, ρ sts for a positive constant satisfies P 0 A ρ 1 > 0, ρ 0 x is a given smooth function with compact support, u 0 x = u 1 0 x, u 0 xt is a given vector-valued smooth function with compact support, without loss of generality, we may assume that there exists a positive number R such that supp{ρ 0, supp{u 1 0, supp{u 0 {x R x R. In 1.1 here after, we use the following notations: x = x 1, x R, = x1, x 0 = t = t, i = xi = x i i = 1,. The compressible Euler equations comprise a nonlinear symmetric hyperbolic system of PDEs that model the ideal fluid flow. The characteristic wave speeds of this system are given by the fluid velocity the local sound speed. The two wave families associated to these speeds may be considered separately. Smooth Eulerian flows of ideal polytropic gases with initial data close to a constant state do not, in gereral, exist for all positive times see [6, 17, 18]. Godin [6] obtained the precise information on the asymptotic behavior of the life-span of the smooth solution to three-dimensional spherically symmetric Eulerian flows of ideal polytropic gases with variable entropy, provided that the initial data is a smooth

3 small perturbation with compact support to a constant state. Sideris [17] showed that the classical solution to the three-dimensional Euler equations for a polytropic, ideal fluid must blow up in finite time under some assumption on the initial data. In Sideris [18], the estimates on life-span of smooth solutions in two space dimensions were obtained under suitable assumptions. In particular, we would like to mention that Grassin [8] proved that the suitable initial data, which force particles to spread out, may lead to the global existence in time for ideal polytropic gases. When the adiabatic constant γ of the gas is equal to 1 in this case, the equation of state is suitably modified to keep the pressure its derivative with respect to the density positive, the gas is named Chaplygin gas, or called Karman-Tsien gas see [0, 1], in the isentropic case, see [4]. Godin [7] obtained the global existence of a class of smooth three-dimensional spherically symmetric flows of Chaplygin gases with the variable entropy, provided that the initial data is a smooth small perturbation with compact support to a constant state. By the light-cone gauge, Bordemann Hoppe [] simplified the description of relativistic membrane moving in the Minkowski space. By performing variables transformations, they established the relationship between the dynamics of relativistic membrane two-dimensional fluid dynamics. The equations for the two-dimensional fluid dynamics derived in [] are nothing but the system 1.1. This implies that the system 1.1 can be used to describe the motion of relativistic membrane moving in the Minkowski space R 1+3. On this research topic, we refer Huang Kong [1]. In the present paper, we investigate the global existence of the smooth solution of the two-dimensional flow of Chaplygin gases. More precisely, under the assumption that the initial data is close to a constant state the vorticity of the initial velocity vanishes, we show the global existence of the smooth solution of the Cauchy problem The main result in this paper is the following theorem. Theorem 1.1 Suppose that the initial velocity is irrotational, i.e., u 0 satisfies u 0 1 u 0 u 1 0 = 0. Then there exists ε 0 > 0 such that the Cauchy problem has a unique global smooth solution for any ε [0, ε 0 ]. The key idea to prove Theorem 1.1 is as follows: i write the velocity as the gradient of a potential function; ii reduce the system 1.1 to a quasilinear wave equation for 3

4 the potential function; iii verify that this quasilinear wave equation satisfies the null condition in two space dimensions; iv by null condition, L -L estimates energy estimates, obtain the global existence on the smooth solution under consideration. Remark 1.1 Theorem 1.1 still holds if the initial data ux, 0 = εu 0 x is replaced by ux, 0 = ū + εu 0 x, where ū R is a constant vector. In fact, as is well known, replacing x by x tū u by u ū leads to a new solution of with the same pressure law, so we reduce to the assumptions of Theorem 1.1 immediately. The paper is organized as follows. In Section we reduce the Cauchy problem to a Cauchy problem for a quasilinear wave equation see below. In Section 3, we introduce some notations prove some lemmas, these lemmas play an important role in the proof of Theorem 1.1. Sections 4-5 are devoted to establishing the L -L estimates energy estimates of the smooth solution, respectively. Theorem 1.1 is proved in Section 6 by the bootstrap argument. Two-dimensional irrotational flow Let then =, 1, ω = ω 1, ω, ω = 1 ω ω 1. As mentioned before, the key idea of the proof of Theorem 1.1 is to write the velocity u as the gradient of a potential function. To do so, we first prove the following lemma which plays an important roles in the present paper. Lemma.1 Let T be a given positive real number assume that ρ, u is a smooth solution to the Cauchy problem in the domain R [0, T. If the initial velocity u 0 x satisfies u 0 = 0,.1 then u = 0, x, t R [0, T.. 4

5 Proof. Noting u = u 1, u, by a direct calculation we have u u = 1 u 1 1 u + u 1 1 u + 1 u u + u 1 u u 1 1 u 1 u 1 1 u 1 u u 1 u u1,.3 u u = u 1 1u u 1 1 u 1 + u 1 u u u 1.4 u u = 1 u 1 1 u 1 u 1 u 1 + u 1 u u 1 u gives u u = u u + u u..6 Noting.6 taking the curl of the second equation in 1.1 gives t u + u u + u u = 0..7 Combining.1.7 yields. immediately. completed. Thus, the proof of Lemma.1 is Remark.1 Lemma.1 shows that, for two-dimensional compressible isentropic Euler equations for Chaplygin gases, if the flow is irrotational at the initial time, then so is it at any time t [0, T, where T sts for the maximal existence time of the smooth solution. Under the assumptions of Theorem 1.1, for system 1.1 there exists a scalar function v = vx, t such that u = v.8 where the scalar function vx, t is a smooth function has a compact support in x since u has compact support by the finite propagation speed. Substituting u = v into the second equation in 1.1 noting Lemma.1 leads to t v + 1 v + fρ = 0,.9 where the function f is defined by f ρ = 1 dp ρ dρ.10 f ρ =

6 It follows from that where Noting.9, we have fρ = ρ ρ c = 1 dp λ dλ dλ = c A ρ, { 1 dp dρ ρ..1 t v + 1 u + fρ = Thus, by.8.13, we obtain u = v, t v = u t u f ρ t ρ,.14 t v = u u f ρ ρ. It follows from.14 that u = v, t ρ = 1 f ρ t + v t v, ρ = 1 f ρ tv + v v. Substituting.15 into the first equation in 1.1 yields.15 t v c v + i v t i v t v v + i v j v i j v v v = i, j=1 On the other h, noting 1.,.8.13, we get t = 0 : v = ε x1 R u 1 0y, x dy, v t = 1 ε u 0 x f ρ + ερ 0 x..17 Remark. For quasilinear wave equations in two space dimensions satisfying the null condition resp. both null conditions, Alinhac [1] proved the quasi-global existence resp. global existence for their Cauchy problems. His proof relies on the construction of an approximate solution, combined with an energy integral method which displays the null conditions. The method used in the present paper is different from one in Alinhac [1]. By.17, it is easy to see that vx, 0, t x, 0 are smooth functions of x R has compact support, i.e., supp{vx, 0, supp{ t x, 0 {x R x R. 6

7 Hence by the finite propagation speed, we obtain vx, t = 0, x {x R x R + ct..18 where Obviously,.16 can be rewritten as t v c v + g k ij k v i j v + ij k v l v i j v = 0,.19 g k ij k v i j v = 1 v 0 1 v + v 0 v 0 v 1v 0 v v.0 ij k v l v i j v = 1 v v 1 v v 1v 1 v v..1 For any given vector X 0, X 1, X R 3 with X0 = c X1 + X, we have g k ijx i X j X k = X 0 X 1 + X 0 X X 0 X 1 X 0 X = 0. ij X i X j X k X l = X 1 X X 1 X X 1X X 1X = 0..3 This shows that the wave equation.16 satisfies the null condition in two space dimensions see[1]. The concept of null condition was introduced by Christodoulou Klainerman for nonlinear wave equations in three space dimensions see [3] [14]. Therefore, we can rewrite the terms as where Q ij ϕ, ψ are defined by g k ij k v i j v ij k v l v i j v g k ij k v i j v = Q 01 1 v, v + Q 0 v, v.4 ij k v l v i j v = vq 1 1 v, v + 1 vq 1 v, v,.5 Q ij ϕ, ψ = i ϕ j ψ i ψ j ϕ 0 i < j, which are null forms. 7

8 3 Preliminaries In order to prove Theorem 1.1, in this section we give some preliminaries. Following Klainerman [13], we introduce the following vector fields Γ = Γ 0,, Γ 4 = 0, 1,, Ω, S, 3.1 where Ω = x 1 x 1, S = t t + r r. 3. It is easy to show that [Γ i, ] = δ 4i i = 0,, 4, [ i, j ] = 0 i, j = 0, 1,, [S, i ] = i i = 0, 1,, 3.3 [Ω, 1 ] =, [Ω, ] = 1, [Ω, 0 ] = 0, [S, Ω] = 0, where [, ] denotes the usual commutator of linear operators δ ij symbol. In what follows, we shall use the following notations vx, t κ = a κ Γ a vx, t, vt κ = sup x R vx, t κ, is the Kronecker [vx, t] κ = 1 + x x ct Γ a vx, t, [vt] κ = sup [vx, t] κ x R a κ vx, t = 1 + x + t 1 Γ a vx, t, a κ vt = sup x R vx, t, vt κ = Γ a v, t L, where κ is a nonnegative integer, a = a 0,, a 4 is a multi-index a = a a 4. Moreover, we also use the following notations a κ v κ,t = sup vs κ, 0 s<t [v] κ,t = sup [vs] κ 0 s<t Let v κ,t = sup vs κ, 0 s<t c 0 = 1 c v κ,t = sup vs κ. 0 s<t 8

9 define ΛT = {x, t R [0, T x ct c 0 t. 3.4 Thanks to the null condition, we will be able to show that the null form possesses a good decay in the domain ΛT. In the our argument, the following operators Z = Z 1, Z = ct x 1, + x 1, x S + x, x 1 Ω t x t 3.5 will play an important role. It follows from 3.5 that Z i = c i + x i x t i = 1, 3.6 x ct Zϕx, t C ϕx, t + 1 t t ϕx, t 1, 3.7 here hereafter C sts for a common absolute constant. We have Lemma 3.1 Assume that h k ij are constants it holds that h k ijx i X j X k = for any vector X 0, X 1, X satisfying X0 = c X1 + X. If ϕ, ψ C R [0, T, then h k ij k ψ i j ϕx, t C [ Zψx, t ϕx, t + ψx, t Z ϕx, t ]. 3.9 Lemma 3. Assume that h kl ij are constants it holds that h kl ij X i X j X k X l = for any vector X 0, X 1, X satisfying X 0 = c X 1 + X. If ϕ, ψ, φ C R [0, T, then ij k ψ l φ i j ϕx, t h kl C [ ψx, t φx, t Z ϕx, t + ψx, t Zφx, t ϕx, t + Zψx, t φx, t ϕx, t ] In what follows, we only prove Lemma 3., because the proof of Lemma 3.1 is of the same type even somewhat simpler. 9

10 Proof. In fact, it follows from 3.10 that where = h kl ij k ψ l φ i j ϕ = i,j,k.l=0 + 1 c hkl ij c k + ω k t ψ l φ i j ϕ h kl ω k ij c kψ c l + ω l t φ i j ϕ + h kl ω k ω l ω i ij c 4 k ψ l φ c j + ω j t t ϕ + h kl 1 ij c c k + ω k t ψ l φ i j ϕ h kl ij ω k ω l c 3 kψ l φ c i + ω i t j ϕ ω 0, ω 1, ω = c, h kl ij ω k ω l c 3 kψ l φ c i + ω i t j ϕ h kl ij ω k ω l ω i ω j c 4 t ψ t φ t ϕ h kl ω k ij c kψ c l + ω l t φ i j ϕ x 1 x, x. x h kl ω k ω l ω i ij c 4 k ψ l φ c j + ω j t t ϕ, 3.1 Noting , we obtain 3.11 immediately. This completes the proof of Lemma 3.. Lemma 3.3 There exist constants g k ij such that i g k ij k v i j v satisfies the null condition, that is, for any vector X 0, X 1, X satisfying X0 = c X1 + X, it holds that ii the following equality holds Γ gij k k v i j v = gij k k Γv i j v + Lemma 3.4 There exist constants ij g k ijx i X j X k = 0; 3.13 such that I g k ij k v i j Γv + g k ij k v i j v ij k v l v i j v satisfies the null condition, that is, for any vector X 0, X 1, X satisfying X0 = c X1 + X, it holds that ij X i X j X k X j = 0;

11 II the following equality holds Γ gij kl k v l v i j v = ij k Γv l v i j v + ij k v l Γv i j v + ij k v l i j Γv + ij k v l v i j v As before, in what follows we only prove Lemma 3.4, because the proof of Lemma 3.3 is of the same type even somewhat simpler. Proof. It follows from 3.3 that m gij kl k v l v i j v = Ω S = ij k v l m v i j v + ij k v l v i j v = ij k v l i j Sv 4 ij k m v l v i j v+ ij k v l v i j m v, ij k Sv l v i j v + ij k v l v i j v m = 0, 1,, 3.17 ij k v l Sv i j v+ ij k v l v i j v = 1 Ωv v 1 v + 1 v Ωv 1 v + 1 v v 1 Ωv ij k Ωv l v i j v Ωv 1 v v 1 v Ωv v Ωv 1v v 1Ωv ij k v l Ωv i j v + ij k v l i j Ωv The desired follows from immediately. This proves Lemma 3.4. By Lemmas , we can obtain the following lemma. Lemma 3.5 Let v = vx, t be a smooth solution of the Cauchy problem on the domain R [0, T κ be a positive integer, where T is a given positive real number. 11

12 Then there exists a positive constant C κ depending on κ but independent of T such that the following inequalities hold for every x, t ΛT {y, s s 1 gij k k v i j vx, t C κ ZΓ a vx, t Γ b vx, t, 3.0 κ a κ Γa where the terms gij k k v i j v x, t C κ ij k v l v i j vx, t C κ κ g k ij k v i j v a+b b+c, b, c κ a+b+c gij k k v i j Γ a vx, t ZΓ b vx, t Γ c vx, t 3.1 ZΓ a vx, t Γ b vx, t Γ c vx, t, 3. ij k v l v i j v are defined by.0.1, respectively. Moreover, it holds that gij k k v i j vx, t x ct C κ 1 + x + t vx, t [ ] vx, t + κ 1 C κ vx, t 1 + x + t [ ]+1 vx, t + vx, t [ ] vx, t κ+, 3.3 ij k v l v i j vx, t x ct C κ 1 + x + t vx, t [ ] vx, t + κ a κ Γa 1 C κ 1 + x + t vx, t [ ] vx, t [ ]+1 vx, t + vx, t [ ] vx, t κ+ gij k k v i j v x, t C κ x ct 1 + x + t vx, t [ ] vx, t κ+ C κ x + t gij k k v i j Γ a vx, t vx, t [ ]+1 vx, t κ + vx, t [ ] vx, t

13 Proof. We only prove 3.0, 3.1, , because the proof of is similar. where By Lemma 3.3, we obtain Γ a gij k k v i j v = b+c a g ka ijbc kγ b v i j Γ c v, 3.6 g ka ijbc kγ b v i j Γ c v satisfies the null condition, that is, for any vector X 0, X 1, X satisfying X0 = c X1 + X, it holds that g ka ijbc X ix j X k = 0. Noting 3.3, , we get 3.0 immediately. By Lemma 3.3 again, we have gij k k v i j v Γ a where g k ij k v i j Γ a v = b+c a,c<a g ka ijbc kγ b v i j Γ c v, 3.7 g ka ijbc kγ b v i j Γ c v satisfies the null condition, that is, for any vector X 0, X 1, X satisfying X0 = c X1 + X, it holds that g ka ijbc X ix j X k = 0. Thus, using 3.3, , we obtain 3.1. Finally, noting 3.7, 3.0 using 3.1, we get immediately. Thus, the proof of Lemma 3.5 is completed. 4 L -L estimates The L -L estimates play an important role in the proof of Theorem 1.1. This section is devoted to establishing these estimates. Theorem 4.1 Let v = vx, t be a smooth solution of the Cauchy problem on the domain R [0, T, where T is a given positive real number, let κ be a positive integer, δ be a positive small constant. Suppose that [ v] [ κ+6 ],T + v [ κ+6 ],T δ,

14 then there exist positive constants C 0, C 1 ν such that it holds that [ vt] κ + vt C 0 ε+c 1 [ v] [ κ+6 ],t + v [ κ+6 for every t [0, T. Proof. For t [0, 1], we have On the other h, noting the fact that using.18, we obtain ],t sup {1+τ ν vτ κ τ<t t vx, t κ C vx, t C 1 + t vx, t. 4.3 t = S t x t t vx, t κ C vx, t κ + C 1 + t vx, t for t Thus, it follows from that, for every t [0, T t vx, t κ C vx, t κ + C 1 + t vx, t. 4.5 Thanks to 4.5, in what follows, we only need to prove 4. with replaced by i i = 1, in the left-h side. To do so, we can rewrite the solution v = vx, t of the Cauchy problem as v = v 1 + ΦF, where v 1 is the solution of the following Cauchy problem t v 1 c v 1 = 0, t = 0 : v 1 = ε x1 M u 1 0y, x dy, while ΦF is the solution of the following Cauchy problem t ΦF c ΦF = F, t v 1 = 1 ε u 0 x f ρ + ερ 0 x, t = 0 : ΦF = 0, t ΦF = 0, in which F = g k ij k v i j v On the other h, it follows from 4.6 that see [5] [16] ij k v l v i j v. 4.8 [ v 1 t] κ + v 1 t Cε for t

15 In order to prove 4., it suffices to show [ ΦF t] κ + ΦF t Cε + C 1 [ v] [ κ+6 ], t + v [ κ+6 ], t sup 0 τ<t { 1 + τ ν vτ κ for t [0, T. Here hereafter, we interpret [ ΦF t] κ ΦF t as [ ΦF t] κ = sup x R a κ ΦF t κ = sup 1 + x x ct Γ a ΦF x, t x R a κ To prove 4.10, we need the following lemmas. 1 + x + t 1 Γ a ΦF x, t. Lemma 4.1 Suppose that F C R [0, T. Then for any given positive real number µ > 0 there exists a positive constant C = Cµ such that the following inequalities holds ΦF x, t k CL µ, κ F x, t 4.11 [ ΦF x, t] k CL µ, F x, t, 4.1 where L µ, κ F x, t = sup y,τ Dx,t T Λt in which sup y,τ Dx,t T Λt c { y y + τ 1+µ 1 + y cτ F y, τ κ + { y y + τ 1+µ 1 + y F y, τ κ, 4.13 Dx, t = { y, τ R [0, t] y x ct τ Λt c denotes the complementary set of Λt. Lemma 4.1 can be found in [10] see also [11]. The following Lemma comes from [15]. Lemma 4. Suppose that h C R [0, T satisfies ht <. Then there exits a positive constant C independent of T such that the following estimate holds: x 1 hx, t C ht, t [0, T

16 In what follows, we prove a more general lemma see Lemma 4.3 below, whose special case is nothing but 4.10 more precisely, the case θ = 0 in Lemma 4.3 gives After the proof of Theorem 4.1, we shall prove Lemma 4.3. Lemma 4.3 Let v = vx, t be the smooth solution of the Cauchy problem on the domain R [0, T. Suppose that the following estimate holds [ v] [ κ+6 ] θ, T + v [ κ+6 ] θ, T δ, 4.16 where the positive constants κ δ are defined in Theorem 4.1. Then there exists positive constant ν such that for any t [0, T, it holds that 1 + x + t 1 +µθ [ ΦF x, t] κ + ΦF x, t C [ v] [ κ+6 ] θ, t + 1 δ θ v [ κ+6 ] θ, t ε + sup {1 + τ ν vτ κ+8 θ, 0 τ<t 4.17 where µ is a positive constant satisfying µ 0, 1 6, θ takes its value in {0, 1, δθ is the Kronecker symbol. We will prove Lemma 4.3 after the proof of Theorem 4.1. Proof of Theorem 4.1. Obviously, taking θ = 0 in 4.17 gives the desired 4.10 immediately. This proves Theorem 4.1. We now prove Lemma 4.3. Proof of Lemma 4.3. We distinguish three cases. Case 1: θ = In this case, it follows from that, for every y, τ Dx, t Λt it holds that y y + τ 3µ 1 + y cτ F y, τ C y y + τ 3µ 1 + y cτ vy, τ [ κ+ ] vy, τ κ+ C y y + τ 3µ 1 + y y y cτ vy, τ [ κ+ ] vy, τ κ+ C[ v] [ κ+ ], t sup {1 + τ 1 +3µ vτ κ+4. 0 τ<t

17 Similarly, we obtain from.18, that y y + τ 3µ 1 + y F y, τ C y y + τ 3µ 1 + y vy, τ [ κ+ ] vy, τ κ C[ v] [ κ+ ] t sup {1 + τ 1 +3µ vτ κ+4 0 τ<t for y, τ Dx, t Λt c. On the other h, by the fact that we have 1 + y + τ C1 + x + t, y, τ Dx, t, x + t 1+µ {[ ΦF x, t] κ + ΦF x, t { C { y y + τ 3µ 1 + y cτ F y, τ sup y,τ Dx,t T Λt { + sup y,τ Dx,t T y y + τ 3µ 1 + y F y, τ. Λt c 4.1 Combining , 4.1 choosing ν 0, 1 3µ gives 4.17 in the case θ =. where Case : θ = 1 In the present situation, it follows from 4.8, 4.11, that C 1 + x + t 1 +µ [ ΦF x, t] κ + ΦF x, t { sup y,τ Dx,t T Λt By, τ = y y +τ +µ 1+ y Ay, τ + sup y,τ Dx,t T By, τ Λt c Ay, τ = y y + τ +µ 1 + y cτ gij k k v i j vy, τ + ij k v l v i j vy, τ gij k k v i j vy, τ +, 4. ij k v l v i j vy, τ When τ [0, 1], it follows from that, for every y, τ Dx, t Λt 1 + y + τ 1 +µ 1 + y cτ y 1 gij k k v i j vy, τ + ij k v l v i j vy, τ. C[ v] [ κ+ ],t vτ κ+4 C[ v] [ κ+ ],t1 + τ ν vτ κ+4, ν R

18 Similarly, when τ [0, 1], it follows from.18, that, for every y, τ Dx, t Λt c 1 + y 1 + y + τ 1 +µ y 1 gij k k v i j vy, τ + C[ v] [ κ+ ],t1 + τ ν vτ κ+4, ν R. Thus, in what follows, it suffices to investigate the case τ 1. ij k v l v i j vy, τ 4.4 By 3.3, 4.5, with θ =, for y, τ Dx, t Λt {z, s s 1 we have 1 + y + τ 1 +µ y y cτ gij k k v i j vy, τ { C y y + τ 1 +µ 1 + y cτ vy, τ [ κ+ ] vy, τ κ y cτ vy, τ [ κ+ ] vy, τ κ+3 + vy, τ [ κ+4 ] vy, τ κ+ { C [ v] [ κ+ ], t + v [ κ+4 ], t y y + τ 1 +µ 1 + y 1 vy, τ κ y y cτ vy, τ κ y cτ 1 + y + τ 1 vy, τ κ+ { C [ v] [ κ+ ], t + v [ κ+4 ], t y y + τ 1 +µ 1 + y y cτ vy, τ κ τ 1 vy, τ κ y 1 vy, τ κ y cτ 1 + y + τ 1 vy, τ κ τ 1 vy, τ κ+3 { C [ v] [ κ+ ], t + v [ κ+4 ], t ε + y y + τ 1 +µ 1 + y y cτ ΦF y, τ κ+ + y y + τ 1 +µ 1 + y 1 ΦF y, τ κ+3 {ε C [ v] [ κ+ ], t + v [ κ y + τ ], t 1+µ [ ΦF y, τ] κ+ + ΦF y, τ κ+3 C [ v] [ κ+ ], t + v [ κ+4 ], t ε + [ v] [ κ+4 ], t sup {1 + τ ν vτ κ+6. 0 τ<t

19 Making use of.18, 4.5, with θ = gives 1 + y + τ 1 +µ y y gij k k v i j vy, τ C y y + τ 1 +µ 1 + y vy, τ [ κ+ ] vy, τ κ+ C[ v] 1 1 [ κ+ ], t y 1 + y + τ +µ 1 + y y cτ 1 vy, τ κ+ C[ v] 1 1 [ κ+ ], t y 1 + y + τ +µ 1 + y y cτ 1 vy, τ κ τ 1 vy, τ κ+3 { C[ v] [ κ+ ], t ε + y y + τ +µ 1 + y cτ [ ΦF y, τ] κ+ + y y + τ µ 1 + y y cτ 1 ΦF y, τ κ+3 { C[ v] [ κ+ ε y + τ ], t 1+µ [ ΦF y, τ] κ+ + ΦF y, τ κ+3 C[ v] [ κ+ ], t ε + [ v] [ κ+4 ], t sup {1 + τ ν vτ κ+6 0 τ<t 4.6 for y, τ Dx, t Λt c {z, s s 1. On the other h, by 4.15, for y, τ Dx, t Λt {z, s s 1 we have 1 + y + τ 1 +µ y y cτ ij k v l v i j vy, τ C[ v] 1 1 [ κ+ ],t y 1 + y + τ +µ 1 + y cτ y 1 vy, τ κ+ C[ v] [ κ+ ],t sup {1 + τ ν vτ κ+4. 0 τ<t It follows from that, for y, τ Dx, t Λt c {z, s s y + τ 1 +µ y y gij kl k v l v i j vy, τ C[ v] 1 1 [ κ+ ], t y 1 + y + τ +µ 1 + y cτ vy, τ κ+ C[ v] [ κ+ ], t sup {1 + τ ν vτ κ+4. 0 τ<t yields 4.17 in the case θ = 1. Case 3: θ = 0 19

20 For y, τ Dx, t Λt {z, s s 1, we obtain from that 1 + y + τ 1+µ y y cτ gij k k v i j vy, τ + ij k v l v i j vy, τ C[ v] [ κ+ ], t y y + τ 1+µ 1 + y 1 vy, τ κ+ C[ v] [ κ+ ], t sup {1 + τ ν vτ κ+4. 0 τ<t 4.9 Similarly, it follows from.18, that, for y, τ Dx, t Λt c {z, s s y + τ 1+µ y y gij k k v i j vy, τ + ij k v l v i j vy, τ C[ v] [ κ+ ], t y y + τ 1+µ 1 + y y cτ 1 vy, τ κ+ C[ v] [ κ+ ], t sup {1 + τ ν vτ κ+4. 0 τ<t We next consider the case τ

21 Noting , 4.5, 4.9, 4.16 using 4.17 in the case θ = 1, for y, τ Dx, t Λt {z, s s 1 we have 1 + y + τ 1+µ 1 + y cτ y 1 gij k k v i j vy, τ + C y y + τ µ { 1 + y cτ vy, τ [ κ+ ] vy, τ κ++ C C C C C ij k v l v i j vy, τ 1 + y cτ vy, τ [ κ+ ] vy, τ κ+3 + vy, τ [ κ+4 ] vy, τ κ+ { [ v] [ κ+ ], t + v [ κ+4 ], t y y + τ µ 1 + y y cτ vy, τ κ y 1 vy, τ κ y cτ 1 + y + τ 1 vy, τ κ+ { [ v] [ κ+ ], t + v [ κ+4 ], t y y + τ µ 1 + y y cτ vy, τ κ τ 1 vy, τ κ y 1 vy, τ κ y cτ 1 + y + τ 1 vy, τ κ τ 1 vy, τ κ+3 { [ v] [ κ+ ], t + v [ κ+4 ], t ε + y y + τ µ 1 + y y cτ ΦF y, τ κ+ + y y + τ µ 1 + y 1 ΦF y, τ κ+3 { [ v] [ κ+ ], t + v [ κ+4 ], t { [ v] [ κ+ ], t + v [ κ+4 ], t ε + ε y + τ 1 +µ [ ΦF y, τ] κ+ + ΦF y, τ κ+3 [ v] [ κ+6 ], t + v [ κ+6 ], t sup {1 + τ ν vτ κ+8 0 τ<t

22 Using.18, 4.5, 4.9, 4.16 noting 4.17 with θ = 1, for y, τ Dx, t Λt c {z, s s 1 we obtain 1 + y + τ 1+µ 1 + y y 1 gij k k v i j vy, τ + ij k v l v i j vy, τ C[ v] [ κ+ ], t y y + τ 1+µ 1 + y y cτ 1 vy, τ κ+ C[ v] [ κ+ ], t y y + τ 1+µ 1 + y y cτ 1 vy, τ κ τ 1 vy, τ κ+3 C[ v] [ κ+ ], t { ε + y y + τ 1+µ 1 + y cτ [ ΦF y, τ] κ+ + y y + τ 1 +µ 1 + y y cτ 1 ΦF y, τ κ+3 C[ v] [ κ+ ], t { C[ v] [ κ+ ], t C ε y + τ 1 +µ [ ΦF y, τ] κ+ + ΦF y, τ κ+3 { ε + [ v] [ κ+6 ], t + v [ κ+6 ], t ε + sup {1 + τ ν vτ κ+8 0 τ<t [ v] [ κ+6 ], t + v [ κ+6 ], t ε + sup {1 + τ ν vτ κ+8 0 τ<t. 4.3 Therefore, combining gives 4.17 in the case θ = 0 immediately. Thus, the proof of Lemma 4.3 is completed. 5 Energy estimate This section is devoted to establishing the energy estimates, which play an important role in the proof of Theorem 1.1. Theorem 5.1 Let v = vx, t be a smooth solution of the Cauchy problem on the domain R [0, T. If there exists a positive small constant δ such that then, for every t [0, T, it holds that [ v] [ ],T + v [ ]+1,T δ, 5.1 vt κ C ε1 + t C 3 [ v] [ ],T + v [ ]+1, T where C C 3 are two positive constants independent of T. «, 5. Proof. Define qλ = λ 1 dϱ ϱ

23 px, t = q x ct, ṗx, t = q x ct. Obviously, Moreover, q C 1 R, p C 1 R R + ṗ C 0 R R +. 0 px, t 5.4 for any x, t R 0, +. ṗx, t = x ct 5.5 On the other h, we introduce the following energy functions { E 0 vt = t vx, t + c vx, t 1 dx, 5.6 R E κ vt = E 0 Γ a vt, 5.7 a κ { Ẽ 0 vt = e p x, t t vx, t + c vx, t dx R Thus, it follows from 5.4, that Ẽ κ vt = Ẽ 0 Γ a vt. 5.9 a κ 1 Ĉ vt κ 1 C E κ vt Ẽκvt CE κ vt Ĉ vt κ 5.10 for some positive constant Ĉ C which are independent of t. Therefore, in order to prove 5., it suffices to show «Ẽ κ vt Cε1 + t C [ v] [ ],T + v [ ]+1,T for t [0, T

24 To do so, noting.16, we have t Γ a v c Γ a v + = b a i v t i Γ a v t v Γ a v + C b Γ b i v t i v + t v v i v t i Γ a v t v Γ a v + i, j=1 i v j v i j Γ a v v Γ a v i,j=1 i, j=1 i v j v i j v + v v + i v j v i j Γ a v v Γ a v 5.1 for any given multi-index a, where C b is a constant which is determined by 3.3. Notice that C a = 1. Multiplying e p t Γ a v to 5.1 then integrating with respect to x on R leads to 1 d { e p t Γ a vx, t + c Γ a vx, t dx+ dt R c v + i v t i Γ a v t Γ a v t v Γ a v t Γ a v+ ṗ ZΓa R e p i, j=1 i v j v i j Γ a v t Γ a v v Γ a v t Γ a v dx = I 1 t, 5.13 where I 1 t = e p t Γ a v C R b Γ b i v t i v + t v v i v j v i j v + v v + b a i, j=1 i v t i Γ a v t v Γ a v + i v j v i j Γ a v v Γ a v dx. i, j=

25 Notice that e p i v t i Γ a v t Γ a v t v Γ a v t Γ a v + = i e p i v t Γ a v t Γ a v j=1 i, j=1 i e p t v i Γ a v t Γ a v + i e p i v j v j Γ a v t Γ a v 1 i e p v i Γ a v t Γ a v + 1 ṗ ep 1 ep ω i i v t Γ a v t Γ a v 4 j=1 j=1 i v j v i j Γ a v t Γ a v v Γ a v t Γ a v t e p i v i Γ a v i Γ a v+ t e p i v j v j Γ a v i Γ a v t e p v i Γ a v i Γ a v ω i t v i Γ a v t Γ a v c ω i i v j v j Γ a v t Γ a v + c ω i v i Γ a v t Γ a v c i v t Γ a v t Γ a v 4 j=1 j=1 t v i Γ a v i Γ a v+ i v j v j Γ a v i Γ a v v i Γ a v i Γ a v i t v i Γ a v t Γ a v + i i v j v j Γ a v t Γ a v i v i Γ a v t Γ a v + j=1 t v i Γ a v i Γ a v+ t i v j v j Γ a v i Γ a v t v i Γ a v i Γ a v, where ω i = x i i = 1,. It follows from that x I 1 t = 1 d { e p t Γ a vx, t + c Γ a vx, t dx + dt R d e p i v i Γ a v i Γ a v 1 dt R j=1 + 1 e p v i Γ a v i Γ a v R e p i v j v j Γ a v i Γ a v dx + I t + I 3 t, 5.15 c ep ṗ ZΓ a v dx

26 where ṗ I t = R ep 1 I 3 t = R ep ω i i v t Γ a v t Γ a v 4 j=1 ω i t v i Γ a v t Γ a v c ω i i v j v j Γ a v t Γ a v + c ω i v i Γ a v t Γ a v c i v t Γ a v t Γ a v 4 j=1 j=1 t v i Γ a v i Γ a v+ i v j v j Γ a v i Γ a v v i Γ a v i Γ a v dx i t v i Γ a v t Γ a v + i i v j v j Γ a v t Γ a v i v i Γ a v t Γ a v + Combining yields Ẽ κ vt + t a κ for t [0, T, where Now we claim that for t [0, T. 0 C It 1 j= t v i Γ a v i Γ a v+ t i v j v j Γ a v i Γ a v t v i Γ a v i Γ a v dx. R e p ṗ ZΓ a vx, τ dxdτ CẼκv0 + C t Iτ dτ 5.19 It = I 1 t + I t + I 3 t. 5.0 a κ C 1 + t R a κ ṗe p ZΓ a vx, t dx+ [ v] [ ],T + v [ ]+1,T R e p vx, t κdx 5.1 For the time being, we assume that 5.1 is true. We will prove it later. Thus, it follows from that Ẽ κ vt + 1 a κ t 0 CẼκv0 + C R e p ṗ ZΓ a vx, τ dxdτ t 0 1 [ v] 1 + τ [ ],T + v [ ]+1,T Ẽκ vτ dτ. 6 5.

27 Noting.17, 5. using Gronwall inequality gives 5.11 immediately. It suffices to prove 5.1 in the rest of this section. To prove 5.1, we distinguish two cases. Case I: t [0, 1] In this case, 5.1 comes from 5.14, immediately. Case II: t [1, T We first estimate I 1 t. Noting.18,.0,.1, 3.7, 3.0, 3.1, 5.1, 5.14 using Cauchy inequality, we have I 1 t x,t ΛT Γa b<a b a b a C C x,t ΛT x,t ΛT c R gij k k v i j v gij k k v i j Γ a v tγ a v e p dx+ C bγ b gij k k v i j v tγ a v e p dx+ C bγ b gij k k v i j v gij k k v i j Γ a v tγ a v e p dx+ C bγ b gij kl k v l v i j v gij kl k v l v i j Γ a v tγ a v e p dx x ct 1 + x + t v [ ] x + t v [ ]+1 e p v κdx+ x,t ΛT b κ b κ x,t ΛT x,t ΛT e p v [ ] ZΓb v v κ dx + C x,t ΛT c e p v [ ] v κdx+ C e p v R [ ] v κdx x ct C x,t ΛT 1 + x + t v [ ] x + t v [ ]+1ep v κdx+ C e p v [ ] ZΓb v v κ dx+ e p 1 R 1 + t [ v] [ ],T + v [ ] v κdx ɛ R e p 1 + x ct ZΓb v dx + C ɛ [ v] 1 + t [ ],T + v [ ]+1,T e p vx, t κdx R b κ ɛ b κ R ṗe p ZΓ b v dx + C ɛ 1 + t [ v] [ ],T + v [ ]+1,T R e p vx, t κdx, 7 5.3

28 where ɛ is a positive small constant. We now estimate I t. Noting.18, 3.7, 5.1, 5.17 using the null condition Cauchy inequality, we obtain I t x,t ΛT ṗe p ω i i v t Γ a v t Γ a v ω i t v i Γ a v t Γ a v c t v i Γ a v i Γ a v dx + ṗe p ω i i v t Γ a v t Γ a v x,t ΛT c ω i t v i Γ a v t Γ a v c t v i Γ a v i Γ a v dx+ ṗ R ep ω i i v j v j Γ a v t Γ a v + c i v j v j Γ a v i Γ a v j=1 j=1 ω i v i Γ a v t Γ a v c v i Γ a v i Γ a v dx c x,t ΛT ṗe p ω i i v t Γ a v t Γ a v ω i t v i Γ a v t Γ a v t v i Γ a v i Γ a v dx + C v v κdx + v v κdx x,t ΛT c R x,t ΛT ṗe p ω 1 c c 1 + ω 1 t v t Γ a v + ω c c + ω t v t Γ a v 1 c tv c 1 + ω 1 t Γ a v c 1 + ω 1 t Γ a v 1 c tv c + ω t Γ a v c + ω t Γ a v ω 1 c tv c 1 + ω 1 t Γ a v t Γ a v ω c tv c + ω t Γ a v t Γ a v dx+ C 1 + t [ v] 0,T e p v κdx R C ṗe p Zv Γ a v + v ZΓ a v Γ a v dx + C x,t ΛT 1 + t [ v] 0,T e p vx, t κdx R ɛ ṗe p ZΓ a vx, t dx + C ɛ R 1 + t [ v] 0,T + v 1,T e p vx, t κdx. R a κ where ω i = x i x i = 1, ɛ is a positive small constant. 5.4 Similarly, using.18, 3.7, 5.1, 5.18, the null condition Cauchy inequality, we have I 3 t ɛ a κ R ṗe p ZΓ a vx, t dx + C ɛ 1 + t [ v] 1,T 8 R e p vx, t κdx. 5.5

29 Choosing ɛ suitably small, by we see that 5.1 holds for t [1, T. Thus the proof of Theorem 5.1 is completed. 6 Proof of Theorem 1.1 This section is devoted to the proof of Theorem 1.1. In order to prove Theorem 1.1, it suffices to show the following theorem. Theorem 6.1 Under the assumptions of Theorem 1.1, there exists a positive constant ε 0 such that, for every ε [0, ε 0 ], the smooth solution v = vx, t of the Cauchy problem exists globally in time. Proof. The local existence of classical solutions has been proved by the method of Picard iteration see Sogge [19] Hörmer [9]. It suffices to show that if v = vx, t is a smooth solution of the Cauchy problem on the domain R [0, T, then α vx, t L R [0, T. α 4 In what follows, we prove Theorem 6.1 by the method of continuous induction, or say, by the bootstrap argument. Taking an integer κ 8, we have [ ] κ + 9 κ. According to the bootstrap argument, we first assume that [ v] κ, T + v, T Mε, 6.1 then we prove that, by choosing M sufficiently large ε suitably small we have [ v] κ, T + v, T 1 Mε. 6. To do so, let δ be a positive small constant. Choosing ε 1 [0, δ M ], we have [ v] [ κ+6 ], T + v [ κ+6 ], T [ v] [ κ+9 ], T + v [ κ+9 ]+1, T [ v] κ, T + v, T Mε δ 6.3 9

30 for ε [0, δ M ]. Thus, it follows from Theorems that [ v] κ, T + v, T C 0 ε + C 1 [ v] [ κ+6 ], T + v [ κ+6 ], T sup 0 τ<t C 0 ε + C 1 C Mε sup {1 + τ C3Mε ν 0 τ<t {1 + τ ν vτ κ+8 for ε [0, ε ], provided that C 0 ε ε [ { C0 0, min C 1 C M, ] ν, C 3 M 6.4 where ν is the constant appearing in Theorem 4.1. Therefore, taking ε 0 = min{ε 1, ε M 4C 0, we obtain 6. from 6.4 immediately. This implies that α vx, t L R [0, T. α 4 Thus, the proof of Theorem 6.1 is completed. Acknowledgements. This work was supported by the NNSF of China Grant No the Qiu-Shi Chair Professor Fellowship from Zhejiang University. References [1] S. Alinhac, The null condition for quasilinear wave equations in two space dimensions I, Invent. Math , [] M. Bordeman J. Hoppe, The dynamics of relativistic membranes: Reduction to -dimensional fluid dynamics, Phys. Lett. B , [3] D. Christodoulou, Global solutions of nonlinear hyperbolic equations for small initial data, Comm. Pure Appl. Math , [4] R. Courant K. O. Friedrichs, Supersonic Flow Shock Waves, Appl. Math. Sci. 1, Spring-Verlag, [5] R. T. Glassey, Existence in the large for u = F u in two space dimensions, Math. Z ,

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