Global weak solutions for a two-component Camassa–Holm shallow water system

Journal of Functional Analysis 260 (2011) 1132–1154 www.elsevier.com/locate/jfa

Global weak solutions for a two-component Camassa–Holm shallow water system Chunxia Guan, Zhaoyang Yin ∗ Department of Mathematics, Sun Yat-sen University, 510275 Guangzhou, China Received 24 February 2010; accepted 30 November 2010

Communicated by J. Bourgain

Abstract In this paper, we prove the existence of global weak solution for an integrable two-component Camassa– Holm shallow water system provided the initial data satisfying some certain conditions. © 2010 Elsevier Inc. All rights reserved. Keywords: An integrable two-component Camassa–Holm shallow water system; Global existence; Weak solution

1. Introduction In this paper we consider the following integrable two-component Camassa–Holm shallow water system: 

mt + 2ux m + umx + σρρx = 0, ρt + (uρ)x = 0,

t > 0, x ∈ R, t > 0, x ∈ R,

(1.1)

where m = u − uxx , σ = ±1. Eq. (1.1) was recently derived by Constantin and Ivanov [18] in the context of shallow water theory. The variable u(t, x) describes the horizontal velocity of the fluid and the variable ρ(t, x) is in connection with the horizontal deviation of the surface from equilibrium, all measured in dimensionless units [18]. The case σ = 1 (σ = −1) corresponds to the situation in which the gravity acceleration points downwards (upwards) [18]. Eq. (1.1) * Corresponding author.

E-mail addresses: [email protected] (C. Guan), [email protected] (Z. Yin). 0022-1236/$ – see front matter © 2010 Elsevier Inc. All rights reserved. doi:10.1016/j.jfa.2010.11.015

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with σ = −1 was originally proposed by Chen et al. in [6] and Falqui in [32]. Eq. (1.1) with σ = −1 is identified with the first negative flow of the AKNS hierarchy and has peakon and multi-kink solutions [6]. The extended N = 2 supersymmetric Camassa–Holm equation was presented recently by Popowicz in [43]. The mathematical properties of Eq. (1.1) have been studied in many works, cf. [6,18,29,36]. For ρ ≡ 0, Eq. (1.1) becomes the Camassa–Holm equation, modeling the unidirectional propagation of shallow water waves over a flat bottom. Here u(t, x) stands for the fluid velocity at time t in the spatial x direction [4,21,28,35,37,38]. The Camassa–Holm equation is also a model for the propagation of axially symmetric waves in hyperelastic rods [24,25]. It has a bi-Hamiltonian structure [8,33] and is completely integrable [4,10]. Also there is a geometric interpretation of Eq. (1.1) in terms of geodesic flow on the diffeomorphism group of the circle [20,39]. Its solitary waves are peaked [5]. They are orbitally stable and interact like solitons [1,23]. The peaked traveling waves replicate a characteristic for the waves of great height – waves of largest amplitude that are exact solutions of the governing equations for water waves, cf. [11,17,45]. Recently, it was claimed in the paper [40] that the equation might be relevant to the modeling of tsunami, see also the discussion in [19]. The Cauchy problem and initial–boundary value problem for the Camassa–Holm equation have been studied extensively [13,14,26,30,31,41,44,48]. It has been shown that this equation is locally well-posed [13,14,26,41,44] for initial data u0 ∈ H s (R), s > 32 . More interestingly, it has global strong solutions [9,13,14] and also finite time blow-up solutions [9,12–14,16,26,41,44]. On the other hand, it has global weak solutions in H 1 (R) [2,3,15,22,47]. The advantage of the Camassa–Holm equation in comparison with the KdV equation lies in the fact that the Camassa–Holm equation has peaked solitons and models wave breaking [5,12] (by wave breaking we understand that the wave remains bounded while its slope becomes unbounded in finite time [46]). For ρ ≡ 0, the Cauchy problems of Eq. (1.1) with σ = −1 and with σ = 1 have been discussed in [29] and [18], respectively. Recently, a new global existence result and several new blow-up results of strong solutions for the Cauchy problem of Eq. (1.1) with σ = 1 were obtained in [34]. However, the existence and uniqueness of global weak solutions to Eq. (1.1) have not been discussed yet. Note that Eq. (1.1) with σ = −1 or with σ = 1 has peakon solutions which are not strong solutions. Our aim of this paper is to prove the existence of global week solutions to Eq. (1.1) with σ = 1 provided the initial data satisfying some certain conditions. We hope that our result sheds some light on important physical phenomena of Eq. (1.1) with σ = 1 such as wave breaking. Up to now, we have no uniqueness result for the obtained global weak solutions to Eq. (1.1) with σ = 1. This problem will be discussed later on. Since Eq. (1.1) with σ = 1 is a system, there are more difficulties in analyzing it than a single equation. The main difficulties are the mutual effect between two components ρ and u and the estimates of ux and ρ. One cannot follow directly the same argument as in [22] or in [7,47] to deal with this problem. To solve the problem, we combine the method in [22] with the method in [7,47] to show the existence of global weak solution to Eq. (1.1) with σ = 1 provided the initial data satisfying some certain conditions. The paper is organized as follows. In Section 2, we recall and present some useful lemmas which will be used in the sequel. In Section 3, we prove the existence of global weak solution to Eq. (1.1) with σ = 1. Notation. In the following, we denote by ∗ the spatial convolution. Given a Banach space Z, we denote its norm by  · Z .

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2. Preliminaries In this section, we will recall and present some useful lemmas which will be used in the sequel. Notice that in the hydrodynamical derivation of Eq. (1.1) [18] it is required that u(t, x) → 0 and ρ(t, x) → 1 as |x| → ∞, at any instant t. Then, letting ρ¯ = ρ − 1, we have ρ¯ → 0 as |x| → ∞. With m = u − uxx and ρ¯ = ρ − 1, we can rewrite Eq. (1.1) with σ = 1 as follows: ⎧ ut − utxx + 3uux = 2ux uxx + uuxxx − ρ¯ ρ¯x − ρ¯x , ⎪ ⎪ ⎨ ρ¯ + (uρ) ¯ x + ux = 0, t ⎪ u(0, x) = u0 (x), ⎪ ⎩ ρ(0, ¯ x) = ρ¯0 (x),

t > 0, x ∈ R, t > 0, x ∈ R, x ∈ R, x ∈ R.

(2.1)

Note that if p(x) := 12 e−|x| , x ∈ R, then (1 − ∂x2 )−1 f = p ∗ f for all f ∈ L2 (R) and p ∗ m = u. Using this identity, Eq. (2.1) takes the form of a quasi-linear evolution equation of hyperbolic type:   ⎧ 1 2 1 2 2 ⎪ ⎪ ut + uux = −∂x p ∗ u + ux + ρ¯ + ρ¯ , ⎪ ⎪ 2 2 ⎨ ρ¯t + uρ¯x = −ux ρ¯ − ux , ⎪ ⎪ ⎪ ⎪ u(0, x) = u0 (x), ⎩ ρ(0, ¯ x) = ρ¯0 (x),

t > 0, x ∈ R, t > 0, x ∈ R, x ∈ R, x ∈ R.

(2.2)

Consider now the following initial value problem 

qt = u(t, q), q(0, x) = x,

t ∈ [0, T ), x ∈ R,

(2.3)

where u denotes the first component of the solution to Eq. (2.1). Applying classical results in the theory of ordinary differential equations, one can obtain the two following results on q which are crucial in studying global existence of strong solutions to Eq. (2.1). Lemma 2.1. (See [18,29].) Let u ∈ C([0, T ); H s (R)) ∩ C 1 ([0, T ); H s−1 (R)), s  2. Then Eq. (2.3) has a unique solution q ∈ C 1 ([0, T ) × R; R). Moreover, the map q(t, ·) is an increasing diffeomorphism of R with t qx (t, x) = exp

 ux s, q(s, x) ds > 0,

∀(t, x) ∈ [0, T ) × R.

0

u Lemma 2.2. (See [34].) Let z0 = ρ¯00 ∈ H s (R) × H s−1 (R), s  2 and let T > 0 be the maximal existence time of the corresponding solution z = (u, ρ) ¯ to Eq. (2.1). Then we have    ρ¯ t, q(t, x) + 1 qx (t, x) = ρ¯0 (x) + 1 , ∀(t, x) ∈ [0, T ) × R. Then, we give a useful conservation law for strong solutions to Eq. (2.1).

(2.4)

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u Lemma 2.3. (See [34].) Let z0 = ρ¯00 ∈ H s (R) × H s−1 (R), s  2 and let T > 0 be the maximal existence time of the corresponding solution z = (u, ρ) ¯ to Eq. (2.1). Then we have

E(t) =



2

u

+ u2x

+ ρ¯

2



R

dx =

 2 u0 + u20,x + ρ¯02 dx,

∀t ∈ [0, T ).

R

Moreover, we have   u(t, ·)2 ∞

L (R)



1 u0 2H 1 (R) + ρ¯0 2L2 (R) , 2

∀t ∈ [0, T ).

Next, we present a very useful lemma on global strong solutions to Eq. (2.1). u Lemma 2.4. Let z0 = ρ¯00 ∈ H 2 (R) × H 1 (R). If ρ¯0 (x) = −1 for all x ∈ R, then the correu sponding strong solution z = ρ¯ to Eq. (2.1) exists globally in time, i.e. z ∈ C(R+ ; H 2 (R) × H 1 (R)) ∩ C 1 (R+ ; H 1 (R) × L2 (R)). Moreover, there exists β > 0 such that for all t ∈ R,    32 (u0 2 1 +ρ¯0 2 2 +1)t ux (t, ·) ∞  1 3 + 2ρ¯0 2 ∞ + u0,x 2 ∞ H (R) L (R) e , L L (R) (R) L (R) 2β where β = infx∈R |ρ¯0 (x) + 1|. Proof. Differentiating the first equation in (2.2) with respect to x and using the identity ∂x2 p ∗ f = p ∗ f − f , we have   1 1 1 1 utx + uuxx + u2x = u2 + ρ¯ 2 + ρ¯ − p ∗ u2 + u2x + ρ¯ 2 + ρ¯ . 2 2 2 2

(2.5)

Set M(t, x) := ux (t, q(t, x)), γ (t, x) =: ρ(t, ¯ q(t, x)) + 1 and 

 1 2 1 2 f (t, x) := u (t, q) − p ∗ u + ux + ρ¯ + ρ¯ (t, q). 2 2 2

2

Using Lemma 2.3, we get    f (t, x)  3 u0 2 1 + ρ¯0 2 2 + 1, H L (R) (R) 2

∀(t, x) ∈ [0, T ) × R.

(2.6)

∂γ = −γ M. ∂t

(2.7)

By Lemma 2.1 and Eq. (2.2), we have  ∂M = (utx + uuxx ) t, q(t, x) ∂t

and

Then by (2.5), we have 1 1 Mt = − M 2 + γ 2 + f (t, x), 2 2

(t, x) ∈ [0, T ) × R.

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By Lemmas 2.1–2.2, we know that γ (t, x) has the same sign with γ (0, x) = ρ0 (x) for all x ∈ R. Since ρ¯0 (x) ∈ H 1 (R), it follows that ρ¯0 (x) ∈ C(R) and there exists R0 such that |ρ¯0 (x)| < 12 for all |x|  R0 . Note that ρ¯0 (x) ∈ C(R) and ρ¯0 (x) = −1 for all x ∈ R. Then we have     inf γ (0, x) = inf ρ¯0 (x) + 1 > 0.

|x|R0

|x|R0

Set β := min{ 12 , inf|x|∈R |γ (0, x)|}. Then |γ (0, x)|  β > 0 for all x ∈ R. Therefore γ (t, x)γ (0, x) > 0 for all x ∈ R. Next, we consider the following Lyapunov positive function first introduced in [18], w(t, x) = γ (0, x)γ (t, x) +

 γ (0, x)  1 + M 2 (t, x) , γ (t, x)

(t, x) ∈ [0, T ) × R.

(2.8)

By Sobolev imbedding theorem, we have 0 < w(0, x) = γ 2 (0, x) + 1 + M 2 (0, x)  2ρ¯02 (x) + 3 + u20,x (x)   3 + 2ρ¯0 2L∞ (R) + u0,x 2L∞ (R) .

(2.9)

Differentiating Eq. (2.8) with respect to t, in view of (2.7), we obtain   ∂w γ (0, x) 1 (t, x) = 2 M(t, x) f (t, x) + ∂t γ (t, x) 2  γ (0, x)  3  u0 2H 1 (R) + ρ0 2L2 (R) + 1 1 + M2 2 γ (t, x)  3  u0 2H 1 (R) + ρ0 2L2 (R) + 1 w(t, x) 2 for all (t, x) ∈ [0, T ) × R. Thus by Gronwall’s inequality and (2.9), we get 3

(u0 2

+ρ0 2

+1)t

H 1 (R) L2 (R) w(t, x)  w(0, x)e 2 3 (u0 2 1 +ρ¯0 2 2 +1)t  H (R) L (R)  3 + 2ρ¯0 2L∞ (R) + u0,x 2L∞ (R) e 2

for all (t, x) ∈ [0, T ) × R. On the other hand,     w(t, x)  2 γ 2 (0, x) 1 + M 2  2β M(t, x),

∀(t, x) ∈ [0, T ) × R.

Therefore    32 (u0 2 1 +ρ¯0 2 2 +1)t M(t, x)  1 w(t, x)  1 3 + 2ρ¯0 2 ∞ + u0,x 2 ∞ H (R) L (R) L (R) L (R) e 2β 2β for all (t, x) ∈ [0, T ) × R. By Lemma 2.1 and the above inequality, we have

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  ux (t, ·)

L∞ (R)

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   = ux t, q(t, ·) L∞ (R) 3 (u0 2 1 +ρ¯0 2 2 +1)t 1  H (R) L (R) 3 + 2ρ¯0 2L∞ (R) + u0,x 2L∞ (R) e 2 2β



for all (t, x) ∈ [0, T ) × R. By Lemma 2.3 in [34], we have T = ∞. This completes the proof of the lemma. 2 We finally recall the two following useful lemmas. Lemma 2.5. (See Appendix C of [42].) Let X be a separable reflexive Banach space and let f n be bounded in L∞ (0, T ; X) for some T ∈ (0, ∞). We assume that f n ∈ C([0, T ]; Y ) where Y is a Banach space such that X → Y , Y  is separable and dense in X  . Furthermore, (φ, f n (t))Y  ×Y is uniformly continuous in t ∈ [0, T ] and uniformly in n  1. Then f n is relatively compact in C w ([0, T ]; X), the space of continuous functions from [0, T ] with values in X when the latter space is equipped with its weak topology. Remark 2.1. If the conditions which f n satisfies in Lemma 2.5 are replaced by the following conditions: f n ∈ L∞ (0, T ; X),

∂t f n ∈ Lp (0, T ; Y )

for some p ∈ (1, ∞),

and  n f 

L∞ (0,T ;X)

,

 n ∂t f 

Lp (0,T ;Y )

 C,

∀n  1,

then the conclusion of Lemma 2.5 holds true. Throughout this paper, we will denote by {φn }n1 the mollifiers 

φn (x) :=

−1 φ(ξ ) dξ

nφ(nx),

x ∈ R, n  1,

R

where φ ∈ Cc∞ (R) is defined by 

1/(x φ(x) = e 0,

2 −1)

, |x| < 1, |x|  1.

Lemma 2.6. (See [42].) Let f ∈ W 1,α (R) and g ∈ Lp (R) with 1  p  ∞. Then   f g ∗ φn − f (g ∗ φn ) → 0, in Lβ (R), where

1 β

=

1 α

+ p1 .

Remark 2.2. Note that if the domain R is replaced by a bounded domain, then Lemma 2.6 has already been shown in [42]. We observe that Lemma 2.6 holds also true for an unbounded domain R. In the following, we would like to write down the whole proof although it might be similar to the one in [42].

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Proof of Lemma 2.6. We first observe that

    f g ∗ φn − f (g ∗ φn ) = − g(y) f (y)φn (x − y) y + f (x)φn (x − y) dy

=

R

  g(y) f (y) − f (x) φn (x − y) dy − gf  ∗ φn .

R

By standard results on convolutions, the second term converges in Lβ (R) as n → ∞ to gf  . Next, we estimate the first term as follows for n large enough 

     g(y) f (y) − f (x) φ  (x − y) dy  n   R





 CgLp (R) R

Lβ (R)

 α   nf (x) − f (y) dy dx

1/α ,

|x−y| Cn

where C denotes various constants independent of n, f and g. Then we deduce that 



 α   nf (x) − f (y) dy dx

1/α

R |x−y| C n



=

1/α  α

1        f x + tz  dt dz dx  C f  Lα (R) .   n

R |z|C

0

To this end, we just need to observe that it is now enough to show that

 g(y) f (y) − f (x) φn (x − y) dy → gf  in Lβ (R), R

when f and g are smooth. Indeed, the general case follows by density using the above bounds. But, this convergence is clear if f and g are smooth since

  g(y) f (y) − f (x) φn (x − y) dy → −g(x) −f  (x) = g(x)f  (x). 2 R

3. The existence and regularity of global weak solution In this section, we first establish the global existence of approximate solutions. By acquiring the precompactness of approximate solutions, we then prove the existence of the global weak solutions to Eq. (2.1). We finally show the regularity of the obtained global weak solutions. Before giving the precise statements of the main result, we first introduce the definition of an admissible weak solution to the Cauchy problem (2.2).

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Definition 3.1. Give z0 =

 u0 ρ¯0

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∈ H 1 (R) × L2 (R). If

z(t, x) =

   u ∈ L∞ (0, ∞); H 1 (R) × L2 (R) ρ¯

satisfies Eq. (2.2), z(t, ·) → z0 as t → 0+ in the sense of distribution and zH 1 (R)×L2 (R)  z0 H 1 (R)×L2 (R) , then z =

u ρ¯

is called an admissible weak solution to Eq. (2.2).

The main result of this paper can be stated as follows: u Theorem 3.1. Let z0 = ρ¯00 ∈ (H 1 (R) ∩ W 1,∞ (R)) × (L2 (R) ∩ L∞ (R)). If there exists α > −1 such that ρ¯0 (x)  α for all x ∈ R, then Eq. (2.2) has an admissible weak solution z=

    u ∈ C R+ ; H 1 (R) × L2 (R) ∩ L∞ R+ ; H 1 (R) × L2 (R) . ρ¯

Moreover, E(z(t, ·)) = z(t, ·)2H 1 (R)×L2 (R) is a conservation law. Furthermore,  1,∞ (R) u ∈ L∞ loc R+ ; W

 ∞ and ρ¯ ∈ L∞ loc R+ ; L (R) .

Proof. The proof of the theorem is divided into the four following steps. Step 1. The global existence of approximate solutions. u Let z0 = ρ¯00 ∈ (H 1 (R) ∩ W 1,∞ (R)) × (L2 (R) ∩ L∞ (R)) and let there exist α > −1 such that ρ¯0 (x)  α > −1 for all x ∈ R.  φ ∗u  un Define z0n := φnn ∗ρ¯00 = ρ¯0n ∈ H 2 (R) × H 1 (R) for n  1. Note that ρ¯0 (x)  α > −1 and 0 φn (x)  0 for all x ∈ R. Then we get

ρ¯0n (x) = φn

∗ ρ¯0 (x)  α

φn (y) dy = α > −1,

∀x ∈ R.

R

Obviously, we have z0n → z0

in H 1 (R) × L2 (R) as n → ∞,

(3.1)

and  n z 

0 H 1 (R)×L2 (R)

 z0 H 1 (R)×L2 (R) .

(3.2)

By Lemma 2.4, we obtain that the corresponding solution zn ∈ C(R+ ; H 2 (R) × H 1 (R)) to Eq. (2.1) with initial data z0n exists globally in time.

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Remark 3.1. By Lemma 2.3 and (3.2), we have  n  u (t, ·)2

H 1 (R)

 2  2 + ρ¯ n (t, ·)L2 (R) = z0n H 1 (R)×L2 (R)  z0 2H 1 (R)×L2 (R) ,

∀t ∈ R+ .

(3.3)

Then, we obtain   n u (t, ·)

√

√  2 2 n   u (t, ·) H 1 (R)  z0 H 1 (R)×L2 (R) ,  L∞ (R) 2 2

∀t ∈ R+ .

(3.4)

Step 2. The precompactness of approximate solutions. Let us denote P n (t, x) =: p ∗ ((un )2 + 12 (unx )2 + 12 (ρ¯ n )2 + ρ¯ n )(t, x) in the following text. Lemma 3.1. There exist a pair of subsequences {znk (t, x)} ⊂ {zn (t, x)} and {P nk (t, x)} ⊂ {P n (t, x)} and a pair of functions z(t, x) ∈ L∞ (R+ ; H 1 (R) × L2 (R)) and P (t, x) ∈ L∞ (R+ ; H 1 (R)) such that znk  z nk

u

  weakly in H 1 (0, T ) × R × L2 (0, T ) × R as nk → ∞, ∀T > 0, → u uniformly on each compact subset of R+ × R as nk → ∞

(3.5) (3.6)

and P nk → P

uniformly on each compact subset of R+ × R as nk → ∞.

(3.7)

Proof. By Lemma 2.3, we can easily obtain that {zn (t, x)} is uniformly bounded in L∞ (R+ ; H 1 (R) × L2 (R)). We claim that the sequence {un } is uniformly bounded in H 1 ((0, T ) × R) for fixed T > 0. Indeed, unt ∈ L2 ((0, T ) × R), in view of (3.3)–(3.4), we have  n n u u 

x L2 ((0,T )×R)

√      un L∞ ((0,T )×R) unx L2 ((0,T )×R)  z0 2H 1 (R)×L2 (R) T

and  2       px ∗ un 2 + 1 un 2 + 1 ρ¯ n 2 + ρ¯ n   2  2 x 2 L ((0,T )×R)  px 2L2 (R)

2 

T     n 2 1  n 2 1  n 2  u + u ρ¯ (t, ·) dt x +  1  2 2 L ((0,T )×R) 0

T + px 2L1 (R)

  n ρ¯ (t, ·)2 2

L ((0,T )×R)

dt

0

 T z0 4H 1 (R)×L2 (R)

+ T z0 2H 1 (R)×L2 (R) ,

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where we used the inequalities px 2L2 (R)  1 and px 2L1 (R)  1. Then by the first equation of (2.2), we obtain that {unt } is uniformly bounded in L2 ((0, T ) × R). From Lemma 2.3, we  u can easily get that {ρ¯ n } is uniformly bounded in L2 ((0, T ) × R). Therefore, there exist z = ρ¯ ∈ L∞ (R+ ; H 1 (R) × L2 (R)) and a subsequence {znk (t, x)} such that   weakly in H 1 (0, T ) × R × L2 (0, T ) × R as nk → ∞.

znk  z

By the classical Lions–Aubin lemma, {unk (t, x)} is weakly compact in L∞ (R+ ; H 1 (R)). Furthermore, {unk } converges to u(t, x) uniformly on each compact subset of R+ × R as nk → ∞. Next, we turn to the compactness of {P n }. For fixed t ∈ (0, T ), in view of p2L2 (R)  p2L1 (R) = 1, we have   n P (t, ·)2 2 L

     2    2 p ∗ un 2 + 1 un 2 + 1 ρ¯ n 2   + p ∗ ρ¯ n L2 (R) x   (R) 2 2 L2 (R) 2      n 2 1  n 2 1  n 2 2  (t, ·)  pL2 (R)  u + ux + ρ¯  1 2 2 L (R)   2 + p2L1 (R) ρ¯ n (t, ·)L2 (R)  4  2  zn (t, ·) 1 + zn (t, ·) 1 2 2 H (R)×L (R)

 z0 4H 1 (R)×L2 (R)

H (R)×L (R)

+ z0 2H 1 (R)×L2 (R) .

In a similar way, in view of ∂x p2L2 (R)  ∂x p2L1 (R)  1, we can get   ∂x P n (t, ·)2 2

L (R)

 z0 4H 1 (R)×L2 (R) + z0 2H 1 (R)×L2 (R) .

(3.8)

Therefore, {P n } is uniformly bounded in L∞ (R+ ; H 1 (R)). Since u0 ∈ W 1,∞ (R) and un0 = φn ∗ u0 , we have  n  u  ∞  φn  1 u0,x L∞ (R) = u0,x L∞ (R) . L (R) 0,x L (R) By Lemma 2.4, for fixed T > 0, the sequence unx is uniformly bounded in L∞ ((0, T ) × R). Since ρ¯0 ∈ L∞ (R), then we have  n ρ¯  ∞  φn  1 ρ¯0 L∞ (R) = ρ¯0 L∞ (R) . L (R) 0 L (R) Note that unx is uniformly bounded in L∞ ((0, T ) × R). Then by Lemmas 2.1–2.2, we obtain that {ρ¯ n } is uniformly bounded in L∞ ((0, T ) × R). Thus, there exists M(T ) > 0 such that         n u (t, x), ux (t, x), ρ¯ n (t, x), ρ(t, ¯ x)  M(T ), x

∀(t, x) ∈ (0, T ) × R.

Moreover, by Eq. (2.2), we can obtain untx + un unxx +

1  n 2  n 2 1  n 2 u = u + ρ¯ + ρ¯ n − P n 2 x 2

(3.9)

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and  ρ¯tn + un ρ¯ n x + unx = 0. Therefore, we have  ∂P n = p ∗ 2un unt + unx untx + ρ¯ n ρ¯tn + ρ¯tn ∂t    = p ∗ 2un −un unx − ∂x P n + ρ¯ n −unx ρ¯ n − un ρ¯xn − unx    1  2  2 1  2 + p ∗ unx −un unxx − unx + un + ρ¯ n + ρ¯ n − P n 2 2   n n n + p ∗ − u ρ¯ x − ux = I1 + I2 + I3 .

(3.10)

Next, we estimate I1 , I2 and I3 , respectively.    I1 = p ∗ 2un −un unx − ∂x P n + ρ¯ n −unx ρ¯ n − un ρ¯xn − unx

   2 1 = p ∗ 2un −un unx − ∂x P n − unx ρ¯ n − ρ¯ n unx − e−|x−y| un ρ¯ n ρ¯yn dy 2 R

   2 1 = p ∗ 2un −un unx − ∂x P n − unx ρ¯ n − ρ¯ n unx + 4 

 1  2 = p ∗ 2un −un unx − ∂x P n − unx ρ¯ n − ρ¯ n unx 2

 2 1 + e−|x−y| un sign(x − y) ρ¯ n dy. 4





 −|x−y| n  n 2 e u y ρ¯ dy

R

R

Using (3.3)–(3.4), (3.8)–(3.9) and Hölder’s inequality, we get

I1 2L2 ((0,T )×R)

 p2L2 (R)

2

T   n n n   2u −u u − ∂x P n − 1 un ρ¯ n 2 − ρ¯ n un  dt x x x  2 L1 (R) 0

T

  + un ρ¯ n 

p2L1 (R) L∞ (0,T )×R

  n ρ¯ (t, ·)2 2

L (R)

dt

0

T   0

    n 2  n n     u + u u + ∂x P n 2 + 1 M(T ) + 1 ρ¯ n 2 + un 2  dt x x   2 L1 (R)

  + un ρ¯ n 

T

L∞ (0,T )×R

  n ρ¯ (t, ·)2 2

L (R)

0

dt

C. Guan, Z. Yin / Journal of Functional Analysis 260 (2011) 1132–1154

1143

  C T , z0 H 1 (R)×L2 (R) , M(T ) , here we used pL2 (R)  pL1 (R)  1. Since −un unx unxx − 12 (unx )3 = − 12 (un (unx )2 )x , it follows that    1  2  2 1  2 I2 = p ∗ unx −un unxx − unx + un + ρ¯ n + ρ¯ n − P n 2 2     2 1  2   2 1 = p ∗ unx un + ρ¯ n + ρ¯ n − P n − p ∗ un unx x 2 2     2 1  2   2 1 − ∂x p ∗ un unx . = p ∗ unx un + ρ¯ n + ρ¯ n − P n 2 2 Using (3.3)–(3.4), (3.8)–(3.9) and Hölder’s inequality, we obtain I2 2L2 ((0,T )×R)

T 

2      n  n 2 1  n 2 n n ux u + ρ¯ (t, ·) + ρ¯ − P  1 dt 2 L (R)

p2L2 (R)  

0

 1 + un unx L∞ (0,T )×R 2

T

2  p2L1 (R) unx (t, ·)L2 (R) dt

0

2   M(T ) + 1 + z0 H 1 (R)×L2 (R)   C T , z0 H 1 (R)×L2 (R) , M(T ) ,

T

 n 2  n 2  n 2  n 2   u + u + ρ¯ + P  x

L1 (R)

dt

0

here we used pL2 (R) , pL1 (R)  1. Next, we estimate the last term. Using Hölder’s inequality, (3.3)–(3.4) and (3.9), in view of ∂x pL2 (R) , pL1 (R)  1, we get

T I3 2L2 ((0,T )×R)



2   2  2 2 p2L1 (R) unx (t, ·)L2 (R) + ∂x p2L2 (R)  un + ρ¯ n L1 (R) dt

0

  C T , z0 H 1 ×L2 , M(T ) . n

2 By (3.10) and the above estimates, we deduce that { ∂P ∂t } is uniformly bounded in L ((0, T )×R). Thus, by the classical Lions–Aubin lemma again, it has a subsequence (we denote it again by {P nk }) and P ∈ L∞ (R+ ; H 1 (R)) such that {P nk (t, x)} is weakly compact in L∞ (R+ ; H 1 (R)). Moreover, {P nk (t, x)} converges to P (t, x) uniformly on each compact subset of R+ × R as nk → ∞. This completes the proof of the lemma. 2

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By (3.4)–(3.6) and (3.9), we have that for any T > 0, unk ρ¯ nk  uρ¯

 weakly in L2 (0, T ) × R

(3.11)

unk unx k  uux

 weakly in L2 (0, T ) × R .

(3.12)

and

Remark 3.2. By the above argument, we see that for any fixed T > 0, there exists a pair of subsequences {(unx k )2 } ⊂ {(unx )2 } and {(ρ¯ n )2 } ⊂ {(ρ¯ nk )2 } converging weakly in Lr ((0, T ) × R) for all 1 < r < ∞, i.e. there exists a pair of functions u2x ∈ Lr ((0, T ) × R) and ρ¯ 2 ∈ Lr ((0, T ) × R) such that  n 2 ux k  u2x

and

 n 2 ρ¯ k  ρ¯ 2

 weakly in Lr (0, T ) × R .

Moreover, we have   weakly in Lp (0, T ) × R and unx k  ux weakly* in L∞ R+ ; L2 (R) ,    ρ¯ weakly in Lp (0, T ) × R and ρ¯ nk  ρ¯ weakly* in L∞ R+ ; L2 (R) ,

unx k  ux ρ¯ nk

where p  2. Furthermore, we have u2x (t, x)  u2x (t, x)

and ρ¯ 2 (t, x)  ρ¯ 2 (t, x)

a.e. on (R+ × R).

(3.13)

Note that unx k  ux weakly in Lp ((0, T ) × R), unx k  ux weakly* in L∞ (R+ ; L2 (R)),  ρ¯ weakly in Lp ((0, T ) × R) and ρ¯ nk  ρ¯ weakly* in L∞ (R+ ; L2 (R)). Then, we conclude that for any η ∈ C 1 (R) with η bounded, Lipschitz continuous on R and η(0) = 0, we have ρ¯ nk

  η unx k  η(ux ) weakly in Lp (0, T ) × R ,   η unx k  η(ux ) weakly* in L∞ R+ ; L2 (R) and   ¯ weakly in Lp (0, T ) × R , η ρ¯ nk  η(ρ)   η ρ¯ nk  η(ρ) ¯ weakly* in L∞ R+ ; L2 (R) . Lemma 3.2. There hold ∂ ∂ux 1 1 + (uux ) = u2x + ρ¯ 2 + ρ¯ + u2 − P ∂t ∂x 2 2

(3.14)

∂ ρ¯ ∂ + (uρ) ¯ + ux = 0 ∂t ∂x

(3.15)

and

in the sense of distributions on R+ × R.

C. Guan, Z. Yin / Journal of Functional Analysis 260 (2011) 1132–1154

1145

Proof. Note that znk is the solution of (2.2). Differentiating the first equation in (2.2) with respect to x and using the relation ∂x2 p ∗ f = p ∗ f − f , we have  1  2  2 1  2 untxk + unk unx k x = unx k + unk + ρ¯ nk + ρ¯ nk − P nk . 2 2

(3.16)

Then by the second equation of (2.2), we get  n ρ¯t k + unk ρ¯ nk x + unx k = 0.

(3.17)

By (3.5)–(3.7), (3.11)–(3.12) and Remark 3.2, we obtain (3.14) and (3.15) immediately from (3.16) and (3.17) as nk → ∞, respectively. 2 The next lemma contains renormalized formulations of (3.14) and (3.15). Lemma 3.3. For any η ∈ C 1 (R) with η bounded, Lipschitz continuous on R and η(0) = 0, we have   ∂  1 2 ∂η(ux ) + uη(ux ) = ux η(ux ) + ux − u2x η (ux ) + u2 η (ux ) ∂t ∂x 2 1 + ρ¯ 2 η (ux ) + ρη ¯  (ux ) − P η (ux ) 2

(3.18)

and ∂η(ρ) ¯ ∂  + uη(ρ) ¯ = ux η(ρ) ¯ − ux ρη ¯  (ρ) ¯ − ux η (ρ) ¯ ∂t ∂x

(3.19)

in the sense of distributions on R+ × R. Proof. Denote un,x (t, x) := (ux (t, ·) ∗ φn )(x) and ρ¯n (t, x) := (ρ(t, ¯ ·) ∗ φn )(x). According to Lemma II.1 of [27], we may deduce from (3.14)–(3.15) that un,x and ρ¯n solve    ∂un,x ∂un,x 1 2 2 2 +u = −ux + ux + ρ¯ + ρ¯ ∗ φn + u2 − P ∗ φn + τn ∂t ∂x 2

(3.20)

∂ ρ¯n ∂ ρ¯n +u + (ux ρ) ¯ ∗ φn + ux ∗ φn = σn , ∂t ∂x

(3.21)

and

where the errors τn and σn tend to zero in L1loc (R+ × R). Multiplying (3.20) and (3.21) by η (un,x ) and η (ρ¯n ) respectively, we get ∂  ∂η(un,x ) + uη(un,x ) = ∂t ∂x

  1 2 1 2 2 u + ρ¯ − ux + ρ¯ ∗ φn η (un,x ) 2 x 2  + ux η(un,x ) + u2 − P ∗ φn + τn η (un,x )



(3.22)

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and  ∂  ∂η(ρ¯n ) + uη(ρ¯n ) = − (ux ρ¯ + ux ) ∗ φn η (ρ¯n ) + ux η(ρ¯n ) + τn η (ρ¯n ). ∂t ∂x

(3.23)

Using the boundedness of η and η , we can send n → ∞ in (3.22) and (3.23) to obtain (3.18) and (3.19). 2 Multiplying (3.16) and (3.17) by η (unx k ) and η (ρ¯ nk ) respectively, we get  1  2  ∂  nk ∂  nk  nk η ux + u η ux = unx k η unx k − unx k η unx k ∂t ∂x 2    n 2 1  n 2  nk nk k k η unx k + u + ρ¯ + ρ¯ − P 2 and    ∂  nk ∂  nk  nk η ρ¯ + u η ρ¯ = unx k η ρ¯ nk − unx k ρ¯ nk (η) ρ¯ nk − unx k (η) ρ¯ nk . ∂t ∂x Then we have ∂η(ux ) 1 ∂  uη(ux ) = ux η(ux ) − u2x η (ux ) + ∂t ∂x 2 1 2  ¯ η (ux ) + ρη ¯  (ux ) − P η (ux ) + u2 η (ux ) + (ρ) 2

(3.24)

and ∂ ∂  uη(ρ) η(ρ) ¯ + ¯ = ux η(ρ) ¯ − ux ρη ¯  (ρ) ¯ − ux η (ρ) ¯ ∂t ∂x

(3.25)

in the sense of distributions on R+ × R. Here f is the limit of f nk in the sense of distributions on R+ × R. Lemma 3.4. There hold

lim

t→0+



u2x (t, x) dx

R

= lim

t→0+

u2x (t, x) = R

u20,x (x) dx

(3.26)

R

and

ρ¯ (t, x) dx = lim

lim

t→0+

R



ρ¯ 2 (t, x) dx

2

t→0+

R

=

ρ¯0 2 (x) dx. R

(3.27)

C. Guan, Z. Yin / Journal of Functional Analysis 260 (2011) 1132–1154

1147

Proof. By Lemma 3.1 and Lemma 2.4, for any T > 0, we have that un ∈ L∞ ((0, T ); H 1 (R)), {unt } is uniformly bounded in L∞ ((0, T ); L2 (R)), and un ∈ C([0, T ]; H 1 (R)). Then in view of Lemma 2.5, Remark 2.1 and the proof of Lemma 3.1, we have that {un } contains a subsequence denoted again by {unk }, which converges to u weakly in H 1 (R) uniformly in t. This implies that u is weakly continuous from (0, T ) into H 1 (R), i.e.  u ∈ C w [0, T ]; H 1 (R) .

(3.28)

Similarly, since ρ¯ n ∈ L∞ ((0, T ); L2 (R)) and for all t ∈ (0, T ),  n  ρ¯ (t, ·) t

= H −1 (R)

sup

φH 1 (R)

   n n u ρ¯ + un φx dx  un ρ n + un L2 (R)

R

 2z0 2H 1 (R)×L2 (R) + 2z0 H 1 (R)×L2 (R) in view of (3.3) and (3.4), it follows that {ρ¯tn } is uniformly bounded in L∞ ((0, T ); H −1 (R)). Then by Remark 2.1, we have that {ρ¯ n } contains a subsequence denoted again by {ρ¯ nk }, which converges to ρ¯ weakly in L2 (R) uniformly in t. This implies that ρ¯ is weakly continuous from (0, T ) into L2 (R), i.e.  ρ¯ ∈ C w [0, T ]; L2 (R) .

(3.29)

Then by (3.28) and (3.29), we get ρ(t, ¯ ·)  ρ¯0

weakly in L2 (R) as t → 0+ .

and ux (t, ·)  u0,x

Thus, we have



ρ¯ (t, x) dx 

lim inf t→0+

ρ¯0 2 (x) dx

2

R

R

and



u2x (t, x) dx

lim inf t→0+



R

u20,x (x) dx. R

Therefore, we deduce that

lim inf t→0+

u2x (t, x) dx

+ ρ¯ (t, x) dx  lim inf 2

t→0+

R



+ lim inf

R

u20,x (x) + ρ¯0 2 (x) dx.

 R

Moreover, from Lemma 2.3 we have

u2x (t, x) dx

t→0+

ρ¯ 2 (t, x) dx R

(3.30)

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 2 u (t, x) + u2x (t, x) + ρ¯ 2 (t, x) dx

R

 lim inf nk →∞

R



= lim inf nk →∞

=

 n 2  2  2 u k (t, x) + unx k (t, x) + ρ¯ nk (t, x) dx  nk 2  nk 2  n 2 u0 (x) + u0,x (x) + ρ¯0 k (x) dx

R

 2 u0 (x) + u20,x (x) + ρ¯02 (x) dx.

R

Using the continuity of u and limt→0+

lim sup t→0+



2 R u (t, x) dx

 u2x (t, x) + ρ¯ 2 (t, x) dx 

R

=



2 R u0 (x) dx,

we obtain

 2 u0,x (x) + ρ¯02 (x) dx.

(3.31)

R

In view of (3.13), (3.30) and (3.31), we get (3.26) and (3.27).

2

Step 3. The existence of global weak solution. Define  1 2 ξ , |ξ |  R, ηR (ξ ) := 2 1 2 R|ξ | − 2 R , |ξ | > R.  is bounded. Then ηR (0) = 0 and ηR

Remark 3.3. Let R > 0. Then for each ξ ∈ R 1 1 ηR (ξ ) = ξ 2 − R − |ξ |2 χ(−∞,−R)∪(R,∞) (ξ ), 2 2   ηR (ξ ) = ξ + R − |ξ | sign(ξ )χ(−∞,−R)∪(R,∞) (ξ ). Lemma 3.5. For almost all t ∈ (0, T ) with T > 0, we have

R



u2x

− u2x

(t, x) dx 

t



ρ¯ 2 ux



− ρ¯ 2 ux (s, x)

0 R

t

dx ds + 2

 ρu ¯ x (s, x) − ρu ¯ x (s, x) dx ds

0 R

t

+ M(T )



u2x (s, x) − u2x (s, x) dx ds.

(3.32)

0 R

Proof. Let R > M(T ) (see (3.9)). Subtracting (3.18) from (3.24) and using the entropy ηR , we get

C. Guan, Z. Yin / Journal of Functional Analysis 260 (2011) 1132–1154

1149

 ∂ ∂   u ηR (ux ) − ηR (ux ) ηR (ux ) − ηR (ux ) + ∂t ∂x  1  ux ηR (ux ) − ux ηR (ux ) − u2x (ηR ) (ux ) − u2x (ηR ) (ux ) 2   1 2 − ux − u2x (ηR ) (ux ) + u2 − P (ηR ) (ux ) − (ηR ) (ux ) 2  1 ¯ R ) (ux ) − ρ(η ¯ R ) (ux ) . + ρ¯ 2 (ηR ) (ux ) − ρ¯ 2 (ηR ) (ux ) + ρ(η 2 From Remark 3.3, we know that 1 ux ηR (ux ) − u2x (ηR ) (ux ) = 0, ∀(t, x) ∈ (0, T ) × R, 2 1 ux ηR (ux ) − u2x (ηR ) (ux ) = 0, ∀(t, x) ∈ (0, T ) × R 2 and (ηR ) (ux ) = (ηR ) (ux ) = ux , ηR (ux ) = 12 u2x , ηR (ux ) = 12 u2x . Then the following inequality holds in (0, T ) × R, i.e.    ∂ 2 ∂   2 u ux − u2x  ρ¯ 2 ux − ρ¯ 2 ux + 2(ρu ux − u2x + ¯ x − ρu ¯ x ) − u2x − u2x ux . ∂t ∂x Integrating the above inequality over (ε, t) × R, we obtain

 u2x − u2x (t, x) dx 

R

t



ρ¯ 2 ux



− ρ¯ 2 ux (s, x)

t

dx ds + 2

ε R

+



ρu ¯ x (s, x) − ρu ¯ x (s, x) dx ds

ε R





u2x (ε, x) − u2x (ε, x)

R

t

dx −



u2x − u2x ux

ε R

for almost all t ∈ (0, T ). Sending ε → 0 and using Lemma 3.4 and (3.9), we get (3.32).

2

We now estimate the second component of z. Lemma 3.6. For any T > 0, we have

R



ρ¯ 2 − ρ¯ 2 (t, x) dx =

t

 ux ρ¯ 2 − ux ρ¯ 2 (s, x) dx ds + 2

0 R

t

(ux ρ¯ − ux ρ)(s, ¯ x) dx ds 0 R

(3.33) in the sense of distributions on (0, T ) × R.

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Proof. Let R > M(T ) (see (3.9)). Subtracting (3.19) from (3.25) and using the entropy ηR , in ¯ = ρ¯ 2 , 2ηR (ρ) ¯ = ρ¯ 2 , (ηR ) (ρ) ¯ = ρ, ¯ and (ηR ) (ρ) ¯ = ρ. ¯ Thus, view of Remark 3.3, we have 2ηR (ρ) we deduce that  ∂ 2 ∂   2 u ρ¯ − ρ¯ 2 = ux ρ¯ 2 − ux ρ¯ 2 + 2ux ρ¯ 2 − 2ux ρ¯ 2 + 2ux ρ¯ − 2ux ρ¯ ρ¯ − ρ¯ 2 + ∂t ∂x ¯ = ux ρ¯ 2 − ux ρ¯ 2 + 2ux ρ¯ − 2ux ρ. Integrating the above equality over (ε, t) × R, we obtain



ρ¯ 2 − ρ¯ (t, x) dx =



2

R



ρ¯ 2 − ρ¯ (ε, x) dx +

t

2

R

 ux ρ¯ 2 − ux ρ¯ 2 (s, x) dx ds

ε R

t

+2

(ux ρ¯ − ux ρ)(s, ¯ x) dx ds. ε R

Letting ε → 0 and using Lemma 3.4, we get (3.33). This completes the proof of Lemma 3.6.

2

Lemma 3.7. There hold u2x (t, x) = u2x (t, x)

and ρ¯ 2 (t, x) = ρ¯ 2 (t, x)

a.e. on R+ × R.

Proof. Adding (3.32) and (3.33), we have that for fixed T > 0 and all t ∈ (0, T )

 u2x − u2x + ρ¯ 2 − ρ¯ 2 (t, x) dx

R

t



 ux ρ¯ 2 − ρ¯ 2 (s, x) ds dx + M(T )

t

0 R

 u2x (s, x) − u2x (s, x) dx ds.

0 R

Using (3.9) again, we get



u2x − u2x

+ ρ¯ 2 − ρ¯ (t, x) dx  M(T )

t

2

R

 u2x − u2x + ρ¯ 2 − ρ¯ 2 (s, x) dx ds.

0 R

Using Gronwall’s inequality and Lemma 3.4, we conclude that

R



u2x − u2x + ρ¯ 2 − ρ¯ 2 (t, x) dx  0.

(3.34)

C. Guan, Z. Yin / Journal of Functional Analysis 260 (2011) 1132–1154

1151

By (3.13), it yields

0



u2x − u2x + (ρ) ¯ 2 − ρ¯ 2 (t, x) dx  0,

R

that is



u2x − u2x (t, x) dx =

R





(ρ) ¯ 2 − ρ¯ 2 (t, x) dx = 0.

R

This implies that (3.34) holds true.

2

By Lemma 3.7, we have  p∗

 n 2 1  n 2 1  n 2 u k + ux k + ρ¯ k + ρ¯ nk 2 2



  1 1 → p ∗ u2 + u2x + ρ¯ 2 + ρ¯ . 2 2

(3.35)

From the relations (3.3)–(3.5), (3.11)–(3.12) and (3.35), we infer that z satisfies Eq. (2.2) in D  ((0, T ) × R) and z ∈ C w ([0, T ]; H 1 (R) × L2 (R)) for any T > 0. Step 4. The regularity of the obtained global weak solution. w (R ; H 1 (R) × L2 (R)). In order to conclude that z ∈ C(R ; H 1 (R) × Note that z ∈ Cloc + + 2 L (R)), it suffices to show that the functional 2   E z(t) = z(t, ·)H 1 (R)×L2 (R) is conserved in time. Indeed, if this holds, then   z(t) − z(s)2

H 1 (R)×L2 (R)

2 2    = z(t)H 1 (R)×L2 (R) − 2 z(t), z(s) H 1 (R)×L2 (R) + z(s)H 1 (R)×L2 (R)  = 2z0 2H 1 (R)×L2 (R) − 2 z(t), z(s) H 1 (R)×L2 (R) , ∀t, s ∈ R+ .

Thus, the scalar product in the last line converges to   z(t)2

H 1 (R)×L2 (R)

= z0 2H 1 (R)×L2 (R)

as s → t.

The conservation of E(z(t)) in time can be proved by a regularization technique. Denote fn := f ∗ φn . Since z solves Eq. (2.2) in the sense of distribution, we see that for a.e. t ∈ R+ ,   ∂un 1 1 + (uux ) ∗ φn + px ∗ u2 + u2x + ρ¯ 2 + ρ¯ ∗ φn = 0, ∂t 2 2

n  1.

(3.36)

Multiplying (3.36) by un and integrating by parts, we have 1 d 2 dt



(un )2 dx + R

R

 un (uux ) ∗ φn dx +

R

    1 1 un px ∗ u2 + u2x + ρ¯ 2 + ρ¯ ∗ φn dx = 0. 2 2 (3.37)

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By (3.20), we get   ∂un,x ∂un,x 1 1 +u − u2 − u2x + ρ¯ 2 + ρ¯ ∗ φn ∂t ∂x 2 2   1 1 + p ∗ u2 + u2x + ρ¯ 2 + ρ¯ ∗ φn + τn = 0, 2 2

(3.38)

here the error τn tends to zero in L1 (R) by applying Lemma 2.6 with α = 2, p = 2 and β = 1. Multiplying (3.38) by un,x and integrating by parts, we obtain 1 d 2 dt

(un,x )2 dx + R





−





+

un,x R



 u u2n,x x dx

R

  1 2 1 2 u − ux + ρ¯ + ρ¯ ∗ φn dx 2 2 2

un,x R

1 2



 

1 2 1 2 p ∗ u + ux + ρ¯ + ρ¯ ∗ φn dx + un,x τn dx = 0. 2 2 2

(3.39)

R

By the second equation in (2.2), we have ∂ ρ¯n ∂ ρ¯n ¯ ∗ φn + ux ∗ φn + σn = 0, +u + (ux ρ) ∂t ∂x

(3.40)

here again the error σn tends to zero in L1 (R) by applying Lemma 2.6 with α = 2, p = 2 and β = 1. Multiplying (3.40) by ρ¯n and integrating by parts, we deduct that 1 d 2 dt

(ρ¯n )2 dx + R

1 2



 u ρ¯n2 x dx +

R



ρ¯n (ux ρ) ¯ ∗ φn + ux ∗ φn dx +

R

ρ¯n σn dx = 0. R

(3.41) Adding (3.37), (3.39) and (3.41), and then integrating by parts, we infer that 1 d 2 dt

R

 2 un + u2n,x + ρ¯n2 dx

=− R

1 + 2 1 + 2

 un (uux ) ∗ φn dx +

 ux u2n,x dx −

R



R

 ux ρ¯n2 dx −







un,x R

  1 2 1 2 u − ux + ρ¯ + ρ¯ ∗ φn dx 2 2 2

ρ¯n (ux ρ) ¯ ∗ φn + ux ∗ φn dx R



un,x τn dx − R

ρ¯n σn dx. R

C. Guan, Z. Yin / Journal of Functional Analysis 260 (2011) 1132–1154

1153

Note that for fixed T > 0, u, ux and ρ¯ are bounded in (0, T ) × R. An application of Lebesgue’s dominated convergence theorem yields 1 d 2 dt



 2 u + u2x + ρ¯ 2 dx = 0.

R

This completes the proof of Theorem 3.1.

2

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