Global existence of dissipative solutions to the Hunter–Saxton equation via vanishing viscosity

J. Differential Equations 250 (2011) 1427–1447

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Journal of Differential Equations

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Global existence of dissipative solutions to theHunter–Saxton equation via vanishing viscosity

Jingyu Li, Kaijun Zhang ∗

School of Mathematics and Statistics, Northeast Normal University, Changchun 130024, PR China

a r t i c l e i n f o a b s t r a c t

Article history:Received 12 May 2010Revised 26 August 2010Available online 15 September 2010

Keywords:Hunter–Saxton equationDissipative solutionVanishing viscosityLp Young measure

Using the vanishing viscosity method, we prove the global exis-tence of dissipative weak solutions to the Hunter–Saxton equationthat describes the propagation of waves in a massive director fieldof a nematic liquid crystal. Our main tool is the Lp Young measuretheory. We also derive the upper bound on the convergence ratefor the vanishing viscosity approximations.

© 2010 Elsevier Inc. All rights reserved.

1. Introduction

In this paper we study the initial–boundary value problem for the Hunter–Saxton equation

⎧⎪⎪⎨⎪⎪⎩

(ut + uux)x = 1

2u2

x , t > 0, x > 0,

u(t,0) = 0,

u(0, x) = u(x),

(1.1)

which models the propagation of waves in a massive director field of a nematic liquid crystal [6],with the orientation of the molecules described by the director field n(t, x) = (cos u(t, x), sin u(t, x)),x being the space variable in a reference frame moving with the unperturbed wave speed, and t be-ing a slow time variable. Eq. (1.1) is a high-frequency limit of the Camassa–Holm equation [4] andis a completely integrable system with a bi-Hamiltonian structure [7]. An interesting geometric inter-pretation of the Hunter–Saxton equation [10,11] is that, for spatially periodic functions, it describes

* Corresponding author.E-mail address: [email protected] (K. Zhang).

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1428 J. Li, K. Zhang / J. Differential Equations 250 (2011) 1427–1447

the geodesic flow on the homogeneous space Vir(S)/Rot(S) of the Virasoro group Vir(S) modulo therotations Rot(S), with respect to the right-invariant homogeneous H1 metric: 〈 f , g〉 = ∫

Sfx gx dx.

Using the method of characteristics, Hunter and Saxton [6] showed that, if u(x) is not monotoneincreasing, then the first derivative of the smooth solution to (1.1), ux , blows up in finite time. More-over, employing Kato’s theory on abstract quasi-linear evolution equations, Yin [14] established thelocal well-posedness of strong solutions to the spatially periodic Hunter–Saxton equation and provedthat all strong solutions except the space-independent solutions blow up in finite time. It is thereforenecessary to study the global existence of admissible weak solutions. Among the many weak solu-tions that (1.1) may have, the most natural ones are the dissipative and conservative solutions [8,9].For smooth u(x), the dissipative and conservative solutions coincide while they remain smooth, butthey are distinct after the blowup time. The dissipative solution loses all the kinetic energy whereasthe conservative solution still preserves the kinetic energy after the blowup time. For example [8],consider the initial datum

u(x) ={−x, 0 � x � 1,

−1, x > 1,

then the dissipative weak solution related to u(x) is

udis(t, x) ={

2xt−2 , 0 < x < (1 − 1

2 t)2, 0 < t < 2,

t2 − 1, otherwise;

and the conservative weak solution is

ucon(t, x) ={

2xt−2 , 0 < x < (1 − 1

2 t)2, t > 0,

t2 − 1, otherwise.

Therefore, after the blowup time t∗ = 2, the kinetic energy of the dissipative solution is

E(udis(t)

) =∫

R+

∣∣∂xudis(t, x)∣∣2

dx = 0,

while the kinetic energy of the conservative solution is

E(ucon(t)

) =∫

R+

∣∣∂xucon(t, x)∣∣2

dx = 1 = E(u).

In this paper we are interested in the dissipative weak solution. To state our results more precisely,we introduce a new variable v := ∂xu, then we can write (1.1) as the system

⎧⎪⎪⎪⎪⎨⎪⎪⎪⎪⎩

∂t v + u∂x v = −1

2v2, t > 0, x > 0,

∂xu = v, t > 0, x > 0,

u(t,0) = 0,

v(0, x) = v0(x).

(1.2)

Recall that Hunter and Zheng [8] proved the global existence of both dissipative and conservativeweak solutions to (1.2) with compactly supported v0 ∈ BV(R+), and they also presented many quali-tative properties of (1.1) that are analogous or in contrast with those of the inviscid Burgers equation.

J. Li, K. Zhang / J. Differential Equations 250 (2011) 1427–1447 1429

Zhang and Zheng [16] established the global existence of dissipative weak solutions to (1.2) withcompactly supported and nonnegative v0 ∈ L p(R+) for any p > 2. Note that the space L2 is a suit-able space for studying the existence of weak solutions, in the sense that the L2 norm is preservedby smooth solutions to (1.2). From this point of view, Zhang and Zheng [17] established the globalexistence and uniqueness of dissipative weak solutions to (1.2) with compactly supported and non-negative v0 ∈ L2(R+). Later, they [18] proved the global existence and uniqueness of both dissipativeand conservative weak solutions to (1.2) with general compactly supported v0 ∈ L2(R+). It is worth-while to point out that Bressan and Constantin [2] constructed a continuous semigroup of globaldissipative weak solutions and introduced a distance functional demonstrating the locally Lipschitzcontinuous dependence of solutions on the initial data.

Recall that the approaches in [8,16–18] are all based on constructing approximate solution se-quence by the method of characteristics. More precisely, they approximate the initial datum v0(x) by asequence of step functions {vn

0(x)}, then applying the characteristic method, they construct the explicitsolutions (vn(t, x), un(t, x)) to (1.2) with initial data vn

0(x), the sequence {(vn(t, x), un(t, x))} is finallytaken as the approximate solution sequence. In this paper, we construct the approximate solution se-quence by a vanishing viscosity method. In fact, we take the viscous approximations (vε(t, x), uε(t, x))that solve the following viscous problem

⎧⎪⎪⎪⎪⎪⎪⎪⎨⎪⎪⎪⎪⎪⎪⎪⎩

∂t vε + uε∂x vε = −1

2

(vε

)2 + ε∂2x vε, t > 0, x > 0,

∂xuε = vε, t > 0, x > 0,

uε(t,0) = 0,

∂x vε(t,0) = 0,

vε(0, x) = vε0(x),

(1.3)

where vε0(x) is a smooth approximation of v0 ∈ L2(R+). Similar viscous approximations were also

used by Hunter and Zheng [9] in a special case that

v0 ={−1, 0 � x � 1,

0, x > 1.

In the current paper, we further investigate the works of [9] on the existence of dissipative solutionsin various respects. Firstly, we study the case of general initial data v0 ∈ L2(R+). Secondly, [9] dis-cussed the case of a finite space interval x ∈ [0, l] for a constant l � 10; nevertheless we consider thecase of half space x ∈ [0,+∞). Thirdly, to obtain the strong compactness for the approximate solutionsequence in L2, we employ the theory of L p Young measure, which is quite different from the toolused in [9].

Before the statement of our main results, let us recall the concept of the dissipative weak solutionsintroduced by Zhang and Zheng [18]. Denote Q T := [0, T ) × [0,+∞).

Definition 1.1. We call (v(t, x), u(t, x)) an admissible dissipative weak solution to (1.2) if

(a) the functions (v, u) have the regularity that

(v, u)(t, x) ∈ L∞(R

+, L2(R

+)) ⊗ C(Q T ); (1.4)

(b) (v, u) satisfy in the sense of distributions the equations

⎧⎨⎩ ∂t v + ∂x(uv) = 1

2v2,

∂xu = v;(1.5)


(c) the energy∫

R+ v2(t, x)dx is non-increasing in t ∈ [0,+∞);(d) u(t, x) is equal to zero at x = 0 as a continuous function, and v(t, x) takes on the initial value in

the sense of C([0,+∞), L1(R+));(e) the entropy condition holds:

v(t, x) � 2

t, for a.e. (t, x) ∈ Q ∞. (1.6)

We now state the first result of this paper concerning the global existence of weak solutions.

Theorem 1.1. Assume that v0(x) ∈ L2(R+) has compact support. Then (1.2) has a global dissipative weaksolution (v, u) in the sense of Definition 1.1. Moreover, the solution satisfies v ∈ L p

loc(Q ∞) for all p < 3,

v ∈ C([0,+∞), Lq(R+)) for all q < 2 and v ∈ C+([0,+∞), L2(R+)); u ∈ W 1,ploc (Q ∞) satisfies

∥∥u(t, ·)∥∥L∞ �∞∫

0

∣∣v0(x)∣∣dx + t

2

∞∫0

∣∣v0(x)∣∣2

dx.

Here C+([0,+∞)) denotes the set of right continuous functions on [0,+∞).

To establish Theorem 1.1, note that (1.3) contains a quadratically nonlinear term, thus we need toderive the strong compactness of the approximate solution sequence {vε} in L2

loc(Q ∞). However, thea priori estimates only give the weak compactness. To overcome this difficulty, we adopt the Youngmeasure theory spirited by the works of [13,15,18–20]. Based on a one-sided supernorm estimate(see (2.4)), we show that the Young measure associated with the sequence {vε} deduces to a Diracmeasure, which gives the strong compactness of the approximations in Lq

loc(R+, Lr

loc(R+)) for q < ∞

and r < 2. Then in combination with a local space–time higher integrability estimate (see (2.5)), weobtain the strong compactness for {vε} in L2

loc(Q ∞). For the sequence {uε}, thanks to a uniform W 1,ploc

estimate for p < 3 (see (2.6)), we easily get its compactness in Cloc(Q ∞).Comparing with the work of Zhang and Zheng [18], we also need the Young measure theory

to deduce the strong compactness of the approximate solution sequence. To use the frame work ofYoung measure theory, we should establish some necessary estimates for the approximate solutions.Nevertheless, because we utilize the viscous approximation, unlike the work of Zhang and Zheng[18] where their approximate solutions are explicit, we adopt a different approach to derive thesenecessary estimates (see Lemma 2.1). Among these estimates, the L∞ estimate for uε and the uniformlocal space–time higher integrability estimate for vε (see (2.5)) are the main new difficulties. Toestablish the former estimate, we first show that uε exists globally in the space of L∞ by provingthe well-posedness of system (1.3). This L∞ estimate depends on ε. By studying the behavior ofuε(t, x) as x → ∞, we then obtain the uniform L∞ estimate for uε (see (2.3)). This estimate alsoplays an important role in the proofs of the uniform one-sided supernorm estimate (see (2.4)) andthe uniform W 1,p

loc estimate (see (2.6)). Estimate (2.5), which is quite important to obtain the strongcompactness for {vε} in L2, is the other difficulty. We establish this estimate by using a truncationtechnique.

Our second result gives an upper bound on the convergence rate for the vanishing viscosity ap-proximations.

Theorem 1.2. Assume that v0(x) � 0 has bounded total variation and compact support. Suppose that vε0(x) =

j√ε ∗ v0(x), where j√ε(x) is the standard mollifier j√ε(x) = 1√ε

j( x√ε) and we have taken the even extension

of v0(x). Then there exists a positive constant C = C(TV(v0)) such that

∥∥uε(t, ·) − u(t, ·)∥∥ ∞ + � Cε1/2 exp(Ct) (1.7)
L (R )


and

∥∥vε(t, ·) − v(t, ·)∥∥L2(R+)� Cε1/4 exp(Ct). (1.8)

Remark 1.1. Hunter and Zheng [8, Theorem 3.2] proved that if v0(x) � 0 has bounded total variationand compact support, then (1.2) has a global dissipative solution that also has bounded total vari-ation. For the works on the vanishing viscosity approximations and their convergence rate for thestrictly hyperbolic, genuinely nonlinear system of conservation laws with BV initial data, we refer theinterested readers to [1,3].

Remark 1.2. Based on the method of characteristics, Zhang and Zheng [18] proved that system (1.2)has a unique global dissipative weak solution for general compactly supported v0 ∈ L2. In this paper,we revisit this problem from the view point of viscous approximation. We obtain that our dissipativeweak solution also exists globally for general compactly supported v0 ∈ L2, and by a slight modifica-tion of the proof of our Theorem 1.2, one can prove that our dissipative solution is also unique if thecompactly supported v0 has bounded total variation. Furthermore, we obtain the convergence rate forsuch bounded variation approximate sequence. If v0 ∈ L2, it is not very clear whether our vanishingviscosity solution is unique or not, and we leave this issue for the future work.

The organization of the paper is as follows. In Section 2, we prove the global existence of smoothsolutions (vε, uε) to the viscous system (1.3) and establish some a priori estimates for the approx-imate solution sequence {(vε, uε)}. Section 3 is devoted to the proof of Theorem 1.1 employing thenotion of Young measure. Finally, we present the proof of Theorem 1.2 in Section 4.

2. Preliminaries

In this section, we derive some basic a priori estimates for the solutions to the viscous system(1.3). It is convenient to extend system (1.3) to the whole space R

+t × Rx . To do so we take the odd

extension in x of uε(t, x) and the even extensions of vε(t, x) and v0(x) to obtain⎧⎪⎪⎪⎪⎪⎨⎪⎪⎪⎪⎪⎩

∂t vε + uε∂x vε = −1

2

(vε

)2 + ε∂2x vε, t > 0, x ∈ R,

∂xuε = vε, t > 0, x ∈ R,

vε(0, x) = vε0(x),

vε0(x) = vε

0(−x),

(2.1)

where vε0(x) = j√ε ∗ v0(x), and j√ε(x) is the standard mollifier j√ε(x) = 1√

εj( x√

ε).

Lemma 2.1. If vε0 ∈ Hk(R) for some integer k � 1 is compactly supported, then the following estimates hold.

(1) (2.1) has a unique global solution (vε, uε) such that vε ∈ C([0, T ], Hk(R)) ∩ L2((0, T ), Hk+1(R)),uε ∈ L∞(Q T ) for all T > 0 and

∫R+

∣∣vε(T , x)∣∣2

dx + 2ε

T∫0

∫R+

∣∣∂x vε(s, x)∣∣2

dx ds =∫

R+

∣∣vε0(x)

∣∣2dx. (2.2)

(2)

∥∥uε(t, ·)∥∥L∞ �∞∫

0

∣∣vε0(x)

∣∣dx + t

2

∞∫0

∣∣vε0(x)

∣∣2dx. (2.3)


(3) For a.e. (t, x) ∈ Q ∞ ,

vε(t, x) � 2

t. (2.4)

(4) For all 2 � p < 3, T > 0 and R > 0,

T∫0

∫0�x�R

∣∣vε(t, x)∣∣p

dx dt � C, (2.5)

with a constant C = C(R, T , p,‖v0‖L2 ) independent of ε.(5) For all 2 � p < 3,

∥∥uε∥∥

W 1,ploc (Q ∞)

� C, (2.6)

with C = C(R, T , p,‖v0‖L2).

Proof. Proof of (1). It is standard to establish the local well-posedness for system (2.1). In other words,for any vε

0 ∈ Hk(R) with k � 1, there exists a positive constant T such that (2.1) has a unique solution(vε, uε) satisfying vε ∈ C([0, T ], Hk(R)) ∩ L2((0, T ); Hk+1(R)) and uε ∈ L∞((0, T ) × R). Moreover, ifTmax is the life span of the solution and Tmax < +∞, then it holds that

limt→Tmax

(∥∥vε(t, ·)∥∥Hk(R)+ ∥∥uε(t, ·)∥∥L∞(R)

) = +∞.

Thus to derive the global existence of solution to (2.1), it suffices to bound ‖vε(t, ·)‖Hk + ‖uε(t, ·)‖L∞on any finite time interval. Multiplying (2.1)1 by vε , integrating the product on (0, t) × R, and per-forming an integration by parts, we get

∫R

∣∣vε(t, x)∣∣2

dx + 2ε

t∫0

∫R

∣∣∂x vε(s, x)∣∣2

dx ds =∫R

∣∣vε0(x)

∣∣2dx. (2.7)

Thanks to the symmetry of the solution to (2.1) with respect to x = 0, we derive (2.2) and for anyt > 0

∂x vε(t, x = 0) = 0 and uε(t, x = 0) = 0. (2.8)

We differentiate (2.1)1 with respect to x, multiply the resulting equation by ∂x vε , and integrate on R

to obtain

d

dt

∫R

∣∣vεx(t, x)

∣∣2dx + 2ε

∫R

∣∣vεxx(t, x)

∣∣2dx = −3

∫R

vε∣∣vε

x

∣∣2(t, x)dx. (2.9)

By Gagliardo–Nirenberg’s inequality and (2.7), we get

∣∣∣∣∫

vε∣∣vε

x

∣∣2(t, x)dx

∣∣∣∣ � C∥∥vε(t, ·)∥∥7/4

L2

∥∥vεxx(t, ·)

∥∥5/4L2 � ε

4

∫ ∣∣vεxx(t, x)

∣∣2dx + C

(‖v0‖L2

)ε−5/3,

R R


which in combination with (2.9) leads to

∫R

∣∣vεx(t, x)

∣∣2dx + ε

t∫0

∫R

∣∣vεxx

∣∣2dx dt �

∫R

∣∣vε0x

∣∣2dx + C

(‖v0‖L2

)ε−5/3t. (2.10)

We now estimate ‖uε‖L∞(Q T ) . By (2.7) and (2.10), we obtain

∥∥vε(t, ·)∥∥L∞ � C∥∥vε(t, ·)∥∥1/2

L2

∥∥vεx(t, ·)

∥∥1/2L2 � C

(∥∥vε0

∥∥H1

) + C(ε,‖v0‖L2

)t1/2 (2.11)

and

∥∥vεx(t, ·)

∥∥L∞ � C

∥∥vε(t, ·)∥∥1/4L2

∥∥vεxx(t, ·)

∥∥3/4L2 � C

(ε,‖v0‖L2

) + ε

∫R

∣∣vεxx(t, ·)

∣∣2dx. (2.12)

Integrating (2.1)1 on [0, t) × [0, x), in view of (2.8), we get

uε(t, x) = −t∫

0

uε vε(s, x)ds + 1

2

t∫0

x∫0

(vε

)2(s, y)dy ds + ε

t∫0

vεx(s, x)ds +

x∫0

vε0(y)dy. (2.13)

Because vε0 is compactly supported, by (2.11), (2.12) and (2.10), we have

∣∣uε(t, x)∣∣ �

(C + Ct1/2) t∫

0

∣∣uε(s, x)∣∣ds + Ct + C,

where C = C(ε,‖vε0‖H1 ). It then follows from Gronwall’s inequality that

∥∥uε∥∥

L∞(Q T )� C

(ε,

∥∥vε0

∥∥H1

)exp

(C T 3/2). (2.14)

We next turn to estimate ‖vε‖C([0,T ],Hk(R)) . By means of energy estimates,

1

2

d

dt

∥∥vε(t, ·)∥∥2Hk + ε

∥∥vεx(t, ·)

∥∥2Hk

=k∑

i=0

∫R

[−1

2uε∂x

((∂ i

x vε)2)

(t, x)

+ (uε∂ i+1

x vε − ∂ ix

(uε∂x vε

))∂ i

x vε(t, x) − 1

2∂ i

x

((vε

)2)∂ i

x vε(t, x)

]dx.

Applying Gagliardo–Nirenberg’s inequality, we obtain

∣∣∣∣∫

uε∂x((

∂ ix vε

)2)(t, x)dx

∣∣∣∣ =∣∣∣∣∫

vε(∂ i

x vε)2

(t, x)dx

∣∣∣∣ �∥∥vε(t, ·)∥∥L∞

∥∥∂ ix vε(t, ·)∥∥2

L2 ,

R R


∣∣∣∣∫R

(uε∂ i+1

x vε − ∂ ix

(uε∂x vε

))∂ i

x vε(t, x)dx

∣∣∣∣ �∥∥uε∂ i+1

x vε − ∂ ix

(uε∂x vε

)(t, ·)∥∥L2

∥∥∂ ix vε(t, ·)∥∥L2

� C∥∥vε(t, ·)∥∥L∞

∥∥∂ ix vε(t, ·)∥∥2

L2 ,

∣∣∣∣∫R

∂ ix

((vε

)2)∂ i

x vε(t, x)dx

∣∣∣∣ �∥∥vε(t, ·)∥∥L∞

∥∥∂ ix vε(t, ·)∥∥2

L2 .

Together with (2.11), we have

1

2

d

dt

∥∥vε(t, ·)∥∥2Hk �

(C + Ct1/2)∥∥vε(t, ·)∥∥2

Hk .

Thanks to Gronwall’s inequality, we conclude that

∥∥vε∥∥

Hk(Q T )� C

(ε,

∥∥vε0

∥∥H1

)exp

(C T 3/2). (2.15)

The global well-posedness for (2.1) follows from (2.14) and (2.15) immediately.

Proof of (2). Integrating (2.1)1 on (0, x) for x > 0, we get the equation of uε ,

⎧⎪⎪⎪⎪⎪⎪⎪⎪⎪⎨⎪⎪⎪⎪⎪⎪⎪⎪⎪⎩

uεt − εuε

xx + uεuεx − 1

2

x∫0

(vε

)2(t, y)dy = 0, t > 0, x > 0,

uε(t,0) = 0,

uε(0, x) =x∫

0

vε0(y)dy.

(2.16)

Thanks to (2.13), we get

uε(t,+∞) =+∞∫0

vε0(x)dx + 1

2

t∫0

+∞∫0

∣∣vε(s, x)∣∣2

dx ds.

Denote u1 := ∫ ∞0 |vε

0(x)|dx + t2

∫ ∞0 |vε

0(x)|2 dx, u2 := − ∫ ∞0 |vε

0(x)|dx − t2

∫ ∞0 |vε

0(x)|2 dx. Let P =uε − u1, then we have

⎧⎪⎪⎨⎪⎪⎩

Pt = εP xx − uε P x + 1

2

x∫0

(vε

)2dy − 1

2

∞∫0

∣∣vε0(y)

∣∣2dy,

P (t,0) < 0, P (t,+∞) � 0, P (0, x) � 0.

(2.17)

We claim that P � 0 for all (t, x) ∈ R+ × R

+ . Otherwise, assume that P (t, x) has a strictly positivemaximum on the domain [0, T ]×[0,+∞] at a point (t, x) = (t0, x0). Clearly, it follows from the initialand boundary conditions of (2.17) that 0 < t0 � T , 0 < x0 < +∞. Then we have

P xx(t0, x0) � 0, P x(t0, x0) = 0, Pt(t0, x0) � 0. (2.18)


On the other hand, by Eqs. (2.17) and (2.2),

Pt(t0, x0) � 1

2

x0∫0

(vε

)2dy − 1

2

∞∫0

∣∣vε0(y)

∣∣2dy < 0,

which contradicts (2.18). Therefore uε � u1. A similar argument gives uε � u2.

Proof of (3). Consider the following ordinary differential equation

⎧⎨⎩ wt = −1

2w2,

w(0) = w0 = max{

0, vε0(x)

}.

A simple calculation yields

w = 2w0

t w0 + 2� 2

t.

Because uε ∈ L∞(Q T ), applying the comparison principle, we have

vε � w � 2

t.

Proof of (4). This uniform local higher integrability estimate is similar to the estimates in [12,13].Let η ∈ C∞(R+), η � 0, η ≡ 1 on [0, R] and η ≡ 0 on [R + 1,+∞). For two positive integers mand l satisfying m � l, take α = 2m/(2l + 1). Set ϕ(y) = (1 + α)

∫ y0 max{1, sα}ds for y ∈ R

+ . Clearly,ϕ′(y) = (1 + α)max{1, yα}. Multiplying (2.1)1 by η(x)ϕ′(vε), integrating the product on [0, T ] × R

+and performing an integration by parts, we get

T∫0

∫R+

vεηϕ(

vε)

dx dt − 1

2

T∫0

∫R+

(vε

)2ηϕ′(vε

)dx dt

=∫

R+ηϕ

(vε(T , x)

)dx −

∫R+

ηϕ(

vε0(x)

)dx −

T∫0

∫R+

uε∂xηϕ(

vε)

dx dt

+ ε

T∫0

∫R+

∂x vε∂xηϕ′(vε)

dx dt + ε

T∫0

∫R+

∣∣∂x vε∣∣2

ηϕ′′(vε)

dx dt. (2.19)

From (2.2), it follows that

∣∣∣∣∫

R+ηϕ

(vε(T , x)

)dx

∣∣∣∣ = (1 + α)

∣∣∣∣∣∫

R+η

vε(T ,x)∫0

max{

1, sα}

ds dx

∣∣∣∣∣= (1 + α)

∫|vε |�1

ηdx +∫

|vε |>1

η(∣∣vε

∣∣1+α + α)

dx

� C(R,α) + C(R,α)‖v0‖1+α2 .

L


In view of the second equation of (2.1), uε(x) = ∫ x0 vε(y)dy, thus

∣∣uε∣∣ � |x|1/2

∥∥vε∥∥

L2 . (2.20)

Then by Cauchy’s inequality and (2.2), we have

∣∣∣∣∣T∫

0

∫R+

uε∂xηϕ(

vε)

dx dt

∣∣∣∣∣ � C(R,α, T ) + C(R,α, T )‖v0‖2+αL2 ,

∣∣∣∣∣εT∫

0

∫R+

∂x vε∂xηϕ′(vε)

dx dt

∣∣∣∣∣ � ε1/2∥∥vε

0

∥∥L2

(C(R,α, T ) + (1 + α)

T∫0

∫|vε |>1

|∂xη|2∣∣vε∣∣2α

)1/2

� ε1/2‖v0‖L2

(C(R,α, T ) + C(R,α, T )‖v0‖2α

L2

)1/2,∣∣∣∣∣ε

T∫0

∫R+

∣∣∂x vε∣∣2

ηϕ′′(vε)

dx dt

∣∣∣∣∣ � C(α)ε

T∫0

∫R+

∣∣∂x vε(t, x)∣∣2

dx dt � C(α)‖v0‖2L2 .

The left-hand side of (2.19) can be estimated as

T∫0

∫R+

vεηϕ(

vε)

dx dt − 1

2

T∫0

∫R+

(vε

)2ηϕ′(vε

)dx dt

� 1 + α

2

T∫0

∫|vε |�1

∣∣vε∣∣2

ηdx dt − α

T∫0

∫|vε |>1

∣∣vε∣∣ηdx dt + 1 − α

2

T∫0

∫|vε |>1

∣∣vε∣∣2+α

ηdx dt.

Therefore, we conclude that

T∫0

∫|vε |>1

∣∣vε∣∣2+α

ηdx dt � C(

R,α, T ,‖v0‖L2

),

which implies (2.5).

Proof of (5). (2.16) gives that

uεt = εvε

x − uε vε + 1

2

x∫0

(vε

)2dy.

Because of (2.3), (2.2) and (2.5), we only need to estimate ‖εvεx‖Lp(Q T ) . By Gagliardo–Nirenberg’s

inequality, (2.2) and Cauchy’s inequality,

ε∥∥vε

x

∥∥L p(Q T )

� C(T , p,‖v0‖L2

)ε

( t∫ ∫ ∣∣vεxx

∣∣2dx ds

) 3p−28p

.

0 R


Note that p < 3, ‖vε0x‖L2 ∼ ε−1/2, due to (2.10), we get

ε∥∥vε

x

∥∥L p(Q T )

� Cε2

3p .

Therefore for all p < 3,

∥∥uε∥∥

W 1,ploc (Q T )

� C(

p, T ,‖v0‖L2

). �

3. Proof of Theorem 1.1

Lemma 3.1. There exist a subsequence {(vε j , uε j )} and (v, u) with v ∈ L∞(R+, L2(R)) and u ∈W 1,p

loc (R+t × Rx) for all p < 3, such that as j → +∞,

uε j → u uniformly on each compact subset of R+t × Rx,

vε j = ∂xuε j ⇀ ∂xu =: v weakly in Lploc

(R

+t × Rx

)for all p < 3. (3.1)

Moreover,

∥∥u(t, ·)∥∥L∞ �∞∫

0

∣∣v0(x)∣∣dx + t

2

∞∫0

∣∣v0(x)∣∣2

dx. (3.2)

Proof. By (2.6) and the compact imbedding theorem, there exist a subsequence {uε j } and a functionu ∈ W 1,p

loc (R+t × Rx) such that {uε j } converges to u uniformly on each compact subset of R

+ × R asj → ∞. (3.1) is due to (2.5). (3.2) is the result of (2.3). �

Next let us prove the strong compactness of the sequence {vε} in L2loc(R

+t ×Rx). To do so we need

to investigate the structure of the Young measure associated with {vε}. Let us start with the existenceof the Young measure.

Lemma 3.2. There exist a subsequence of the solution sequence {vε(t, x)} that is still denoted by {vε(t, x)}, anda family of Young measures μt,x(λ), such that for any continuous functions f (t, x, λ) = O (|λ|q), ∂λ f (t, x, λ) =O (|λ|q−1) as λ → ∞ for q < 2, and for any ψ(x) ∈ Lr

c(R) with r−1 + q/2 = 1, we have

limε→0

∫R

f(t, x, vε(t, x)

)ψ(x)dx =

∫R

f (t, x, v)ψ(x)dx (3.3)

uniformly on each compact subset of [0,∞), where

f (t, x, v) :=∫

λ∈R

f (t, x, λ)dμt,x(λ) ∈ C([0,∞); Lq′/q(R)

)(3.4)

for any q′ ∈ (q,2). Moreover, for any T > 0,

limε→0

T∫ ∫g(t, x, vε(t, x)

)ϕ dx dt =

T∫ ∫g(t, x, v)ϕ dx dt,

0 R 0 R


where the continuous function g(t, x, λ) = O (|λ|p) as λ → ∞ for some p < 3 and ϕ(t, x) ∈ Ls(Q T ) with1/p + 1/s = 1. In addition,

λ ∈ L∞(R

+, L2(R × R,dx ⊗ dμt,x(λ)

)) ∩ Lploc

(R

+ × R × R,dt ⊗ dx ⊗ dμt,x(λ))

(3.5)

for any p < 3. Also comparing the notations in (3.1) and (3.4), we have

v ≡ v. (3.6)

Proof. The proof is similar to that of Lemma 3 in [18]. �The following lemma gives the structure of the Young measures μt,x(λ).

Lemma 3.3. μt,x(λ) = δv(t,x)(λ) for a.e. (t, x) ∈ R+ × R.

Proof. Step 1: Let ε → 0 in (2.1); we get from Lemmas 3.1, 3.2 and estimate (2.2) that

∂t v + ∂x(uv) = 1

2v2,

which is

∂t v + u∂x v = 1

2v2 − (v)2, (3.7)

in the weak sense. By applying the standard Friedrichs mollifier jδ to regularize Eq. (3.7), we get

∂t vδ + u∂x vδ = jδ ∗(

1

2v2 − (v)2

)+ rδ, (3.8)

where vδ := v ∗ jδ and rδ := u∂x vδ − (u∂x v) ∗ jδ . Take

T +ρ (ζ ) =

⎧⎪⎨⎪⎩

0, if ζ < 0,12 ζ 2, if 0 � ζ � ρ,

ρζ − 12ρ2, if ζ > ρ.

It then follows from (3.8) that

∂t T +ρ (vδ) + ∂x

(uT +

ρ (vδ)) = vT +

ρ (vδ) + DT +ρ (vδ)

(jδ ∗

(1

2v2 − (v)2

))+ DT +

ρ (vδ)rδ. (3.9)

By Lemma II.1 of [5], rδ → 0 in L1loc(R

+, L1(R)) ∩ L p/2loc (R+ × R) for all p < 3 as δ → 0. Thus passing

to the limit δ → 0 in (3.9), we have

∂t T +ρ (v) + ∂x

(uT +

ρ (v)) = vT +

ρ (v) + DT +ρ (v)

(1

2v2 − (v)2

). (3.10)

On the other hand, we obtain from (2.1) that


∂t T +ρ

(vε

) + ∂x(uεT +

ρ

(vε

))= vεT +

ρ

(vε

) − 1

2DT +

ρ

(vε

)(vε

)2 + ε∂x(

DT +ρ

(vε

)∂x vε

) − εD2T +ρ

(vε

)(∂x vε

)2.

Passing to the limit as ε → 0, observing that T +ρ (ζ ) ∈ W 2,∞(R1) is a convex function, by Lem-

mas 3.1, 3.2 and estimate (2.2), we get

∂t T +ρ (v) + ∂x

(uT +

ρ (v))� vT +

ρ (v) − 1

2DT +

ρ (v)v2. (3.11)

Subtracting (3.11) from (3.10) gives that

∂t(T +ρ (v) − T +

ρ (v)) + ∂x

(uT +

ρ (v) − uT +ρ (v)

)� ρ

2

∫R

λ(λ − ρ)1λ>ρ dμt,x(λ) − 1

2DT +

ρ (v)(

v2 − (v)2) + 1

2ρv(ρ − v)1v>ρ, (3.12)

where 1A denotes the characterization function of the set A. It follows from (2.4) that, both vε

and v are bounded by 2/t . Thus by choosing specific functions f vanishing on (−∞,2/t], we havesuppμt,x(·) ⊆ (−∞,2/t]. It then follows from Jensen’s inequality that, for t > 2/ρ ,

∂t(T +ρ (v) − T +

ρ (v)) + ∂x

(uT +

ρ (v) − uT +ρ (v)

)� 0. (3.13)

Step 2: Take

T −ρ (ζ ) =

⎧⎪⎨⎪⎩

0, if ζ > 0,12ζ 2, if −ρ � ζ � 0,

−ρζ − 12ρ2, if ζ < −ρ.

Similar to the argument of (3.12), we have

∂t(T −ρ (v) − T −

ρ (v)) + ∂x

(uT −

ρ (v) − uT −ρ (v)

)� −ρ

2

∫R

λ(λ + ρ)1λ<−ρ dμt,x(λ) + 1

2ρv(v + ρ)1v<−ρ − 1

2DT −

ρ (v)(

v2 − (v)2). (3.14)

For any h, define h+ = max{h,0} and h− = min{h,0}. Then v2 = (v+)2 + (v−)2 and (v)2 = ((v)+)2 +((v)−)2. By the definition of T −

ρ (ζ ),

T −ρ (v) − T −

ρ (v)

= 1

2

((v−)2 − (

(v)−)2) − 1

2

∫R

(λ + ρ)21λ<−ρ dμt,x(λ) + 1

2(v + ρ)21v<−ρ, (3.15)

in combination with (3.14), this gives that


∂t(T −ρ (v) − T −

ρ (v)) + ∂x

(uT −

ρ (v) − uT −ρ (v)

)� ρ

(T −ρ (v) − T −

ρ (v)) + ρ2

2

∫R

(λ + ρ)1λ<−ρ dμt,x(λ) − ρ2

2(v + ρ)1v<−ρ

− ρ + v

2

((v−)2 − (

(v)−)2)

1−ρ�v�0 − ρ + v

2

((v+)2 − (

(v)+)2)

1−ρ�v�0

+ ρ

2

[(v+)2 − (

(v)+)2]

.

Because (v±)2 � ((v)±)2 and (λ + ρ)1λ<−ρ is concave, it follows that

∂t(T −ρ (v) − T −

ρ (v)) + ∂x

(uT −

ρ (v) − uT −ρ (v)

)� ρ

(T −ρ (v) − T −

ρ (v)) + ρ

2

[(v+)2 − (

(v)+)2]

. (3.16)

Step 3: Our goal is to derive that

∫R

(v2 − (v)2)dx = 0, for all t ∈ R

+, (3.17)

which implies μt,x(λ) = δv(t,x)(λ). We first prove that

∫R

((v+)2 − (

(v)+)2)

(t, x)dx = 0, ∀t ∈ R+. (3.18)

By (3.13), we have for t > 2/ρ ,

0 �∫R

(T +ρ (v) − T +

ρ (v))(t, x)dx �

∫R

(T +ρ (v) − T +

ρ (v))(2/ρ, x)dx. (3.19)

By the definition of T +ρ (ζ ),

T +ρ (v) − T +

ρ (v) = 1

2

((v+)2 − (

(v)+)2) − 1

2

∫R

(λ − ρ)21λ>ρ dμt,x(λ) + 1

2(v − ρ)21v>ρ. (3.20)

Because λ ∈ L∞ , by the Lebesgue dominated convergence theorem, we have

limρ→∞

∫R

(T +ρ (v) − T +

ρ (v))(t, x)dx = 1

2

∫R

((v+)2 − (

(v)+)2)

(t, x)dx, ∀t ∈ R+. (3.21)

Note that (λ − ρ)21λ>ρ is convex; by Jensen’s inequality, we get from (3.20) that

∫R

(T +ρ (v) − T +

ρ (v))(2/ρ, x)dx � 1

2

∫R

(v2 − (v)2)(2/ρ, x)dx. (3.22)

We claim that


limt→0+

∫R

(v)2(t, x)dx =∫R

v20(x)dx. (3.23)

Assume that (3.23) holds; it then follows from (3.3), Fatou’s Lemma and (2.2) that

limρ→∞

∫R

(v2 − (v)2)(2/ρ, x)dx �

∫R

v20(x)dx − lim

ρ→∞

∫R

(v)2(2/ρ, x)dx = 0. (3.24)

Summing up (3.19), (3.21), (3.22) and (3.24) gives (3.18).We now turn to show (3.23). Actually, by (3.7) and observing that v ∈ L∞(R+, L2(R)), we have

v(t, x) → v0(x) weakly in L2(R) as t → 0+ . Thus

∫R

v20(x)dx � lim

t→0+

∫R

(v)2(t, x)dx.

On the other hand, by (2.2) and Fatou’s Lemma,

∫R

(v)2(t, x)dx �∫R

v20(x)dx, ∀t ∈ R

+.

The desired (3.23) follows from these two inequalities.To get (3.17), it is left to show

∫R

((v−)2 − (

(v)−)2)

(t, x)dx = 0, ∀t ∈ R+. (3.25)

By (3.16) and (3.18),

d

dt

∫R

(T −ρ (v) − T −

ρ (v))(t, x)dx � ρ

∫R

(T −ρ (v) − T −

ρ (v))(t, x)dx.

Observe that (3.22) still holds for T −ρ , which together with (3.24) yields

limt→0+

∫R

(T −ρ (v) − T −

ρ (v))(t, x)dx = 0.

Thus by Gronwall’s inequality, we get

∫R

(T −ρ (v) − T −

ρ (v))(t, x)dx = 0, ∀t ∈ R

+.

Consequently, it follows from (3.5), (3.15) and the Lebesgue dominated convergence theorem that

1

2

∫R

((v−)2 − (

(v)−)2)

(t, x)dx = limρ→∞

∫R

(T −ρ (v) − T −

ρ (v))(t, x)dx = 0,

which gives the desired (3.25). �


End of the proof of Theorem 1.1. It follows from Lemmas 3.2, 3.3 and the Lebesgue dominated con-vergence theorem that, for any p1 < ∞, p2 < 2,

vε → v in Lp1loc

(R

+, Lp2loc

(R

+)).

By (2.5) and the interpolation theorem, we get

vε → v in L2loc

(R

+ × R+)

.

Thus by Lemma 3.1, we can pass to the limit in (2.1), and we conclude that (u, v) is a global weaksolution of (1.2). By (2.5), v ∈ L p

loc(Q ∞) for all p < 3. By (3.4), v ∈ C([0,+∞), Lq(R+)) for all q < 2.v ∈ C+([0,+∞), L2(R+)) is due to (3.23). �4. Proof of Theorem 1.2

Let us first derive the upper bound on the distance ‖vε(t, ·) − v(t, ·)‖L1 .

Theorem 4.1. Assume that v0(x) � 0 has bounded total variation and compact support. Then

∥∥vε(t, ·) − v(t, ·)∥∥L1(R+)� Cε1/2 exp(Ct) (4.1)

and

∥∥uε(t, ·) − u(t, ·)∥∥L∞(R+)� Cε1/2 exp(Ct), (4.2)

with a positive constant C = C(TV(v0)).

Proof. Observing that uε(x) − u(x) = ∫ x0 (vε(y) − v(y))dy, (4.2) follows from (4.1) immediately. Thus

we only need to show (4.1). Multiplying (1.2)1 by sgn(v(τ , ξ) − vε(s, x)), we have

∣∣v(τ , ξ) − vε(s, x)∣∣τ

+ u∣∣v(τ , ξ) − vε(s, x)

∣∣ξ

= −1

2v2(τ , ξ) sgn

(v(τ , ξ) − vε(s, x)

). (4.3)

Observe that if φ is Lipschitz and convex, then (φ(vε))xx � φ′(vε)vεxx . Hence similar to (4.3), multi-

plying (2.1)1 by sgn(vε(s, x) − v(τ , ξ)), we get

∣∣vε(s, x) − v(τ , ξ)∣∣s + uε(s, x)

∣∣vε(s, x) − v(τ , ξ)∣∣x

� ε∣∣vε(s, x) − v(τ , ξ)

∣∣xx − 1

2

(vε(s, x)

)2sgn

(vε(s, x) − v(τ , ξ)

). (4.4)

Take the standard mollifier jδ(z) = 1δ

j( zδ), where j(z) ∈ D(R), 0 � j(z) � 1 and

∫R

j(z)dz = 1. Define

g(s, τ , x, ξ) := jα(s − τ ) jβ(x − ξ).

Taking the odd extension in x of u(t, x) and the even extension of v(t, x) to the whole space, it thenfollows from (4.3) that


∫g(s, t, x, ξ)

∣∣v(t, ξ) − vε(s, x)∣∣dξ −

∫g(s,0, x, ξ)

∣∣v0(ξ) − vε(s, x)∣∣dξ

−∫ ∫

0<τ<t

gτ

∣∣v(τ , ξ) − vε(s, x)∣∣dξ dτ −

∫ ∫0<τ<t

gξ u∣∣v(τ , ξ) − vε(s, x)

∣∣dξ dτ

−∫ ∫

0<τ<t

gv∣∣v(τ , ξ) − vε(s, x)

∣∣dξ dτ

= −1

2

∫ ∫0<τ<t

gv2(τ , ξ) sgn(

v(τ , ξ) − vε(s, x))

dξ dτ .

We now integrate this equation with respect to s and x on [0, t)× R, note that ∂ g∂τ = − ∂ g

∂s , ∂ g∂ξ

= − ∂ g∂x ;

due to (4.4), we have

∫ ∫ ∫0<s<t

g(s, t, x, ξ)∣∣v(t, ξ) − vε(s, x)

∣∣dx dξ ds −∫ ∫ ∫0<s<t

g(s,0, x, ξ)∣∣v0(ξ) − vε(s, x)

∣∣dx dξ ds

+∫ ∫ ∫0<τ<t

g(t, τ , x, ξ)∣∣vε(t, x) − v(τ , ξ)

∣∣dx dξ dτ

−∫ ∫ ∫0<τ<t

g(0, τ , x, ξ)∣∣vε

0(x) − v(τ , ξ)∣∣dx dξ dτ

+∫ ∫ ∫ ∫0<s,τ<t

gx∣∣vε(s, x) − v(τ , ξ)

∣∣(u(τ , ξ) − uε(s, x))

dx dξ ds dτ

� −ε

∫ ∫ ∫ ∫0<s,τ<t

gx∣∣vε(s, x) − v(τ , ξ)

∣∣x dx dξ ds dτ

+ 1

2

∫ ∫ ∫ ∫0<s,τ<t

g(

vε(s, x) + v(τ , ξ))∣∣vε(s, x) − v(τ , ξ)

∣∣dx dξ ds dτ .

Letting α → 0, we get

2∫ ∫

jβ(x − ξ)∣∣v(t, ξ) − vε(t, x)

∣∣dx dξ − 2∫ ∫

jβ(x − ξ)∣∣v0(ξ) − vε

0(x)∣∣dx dξ

+∫ ∫ ∫0<s<t

∂x jβ(x − ξ)∣∣vε(s, x) − v(s, ξ)

∣∣(u(s, ξ) − uε(s, x))

dx dξ ds

� −ε

∫ ∫ ∫0<s<t

∂x jβ(x − ξ)∣∣vε(s, x) − v(s, ξ)

∣∣x dx dξ ds

+ 1

2

∫ ∫ ∫0<s<t

jβ(x − ξ)(

vε(s, x) + v(s, ξ))∣∣vε(s, x) − v(s, ξ)

∣∣dx dξ ds. (4.5)

For the first term on the left-hand side of (4.5),


∫ ∫jβ(x − ξ)

∣∣v(t, ξ) − vε(t, x)∣∣dx dξ

�∫ ∫

jβ(x − ξ)∣∣v(t, ξ) − vε(t, ξ)

∣∣dx dξ −∫ ∫

jβ(x − ξ)∣∣vε(t, ξ) − vε(t, x)

∣∣dx dξ

= ∥∥vε(t, ·) − v(t, ·)∥∥L1 −∫ ∫

j(z)∣∣vε(t, ξ + βz) − vε(t, ξ)

∣∣dξ dz

�∥∥vε(t, ·) − v(t, ·)∥∥L1 − CβTV

(vε(t)

).

Claim. TV(vε(t)) � TV(vε0) � CTV(v0).

We first assume that this claim holds, then

∫ ∫jβ(x − ξ)

∣∣v(t, ξ) − vε(t, x)∣∣dx dξ �

∥∥vε(t, ·) − v(t, ·)∥∥L1 − CβTV(v0). (4.6)

Because vε0(x) = j√ε ∗ v0(x), the second term of (4.5) satisfies

∫ ∫jβ(x − ξ)

∣∣v0(ξ) − vε0(x)

∣∣dx dξ �∫ ∫

jβ(x − ξ)(∣∣v0(ξ) − v0(x)

∣∣ + ∣∣v0(x) − vε0(x)

∣∣)dx dξ

� CβTV(v0) +∫R

∣∣v0(x) − vε0(x)

∣∣dx

� CβTV(v0) + Cε1/2TV(v0).

For the third term of (4.5),

∣∣∣∣∫ ∫ ∫0<s<t

∂x jβ(x − ξ)∣∣vε(s, x) − v(s, ξ)

∣∣(u(s, ξ) − uε(s, x))

dx dξ ds

∣∣∣∣�

∣∣∣∣∫ ∫ ∫0<s<t

jβ(x − ξ)∣∣vε(s, x) − v(s, ξ)

∣∣vε(s, x)dx dξ ds

∣∣∣∣+

∣∣∣∣∫ ∫ ∫0<s<t

jβ(x − ξ)∣∣vε(s, x) − v(s, ξ)

∣∣x

(u(s, ξ) − uε(s, x)

)dx dξ ds

∣∣∣∣:= I1 + I2.

Similar to (4.6), in view of that ‖vε(t, ·)‖L∞ � ‖v0‖L∞ , we get

I1 �t∫

0

∥∥vε(s, ·) − v(s, ·)∥∥L1 ds + CβtTV(v0).

I2 can be estimated as follows:


I2 �∫ ∫ ∫0<s<t

jβ(x − ξ)∣∣u(s, ξ) − uε(s, ξ)

∣∣∣∣vεx(s, x)

∣∣dx dξ ds

+∫ ∫ ∫0<s<t

jβ(x − ξ)∣∣uε(s, ξ) − uε(s, x)

∣∣∣∣vεx(s, x)

∣∣dx dξ ds

�t∫

0

∥∥uε(s, ·) − u(s, ·)∥∥L∞

∫ ∫jβ(x − ξ)

∣∣vεx(s, x)

∣∣dξ dx ds

+ ‖v0‖L∞∫ ∫ ∫0<s<t

jβ(x − ξ)|x − ξ |∣∣vεx(s, x)

∣∣dx dξ ds

�t∫

0

∥∥uε(s, ·) − u(s, ·)∥∥L∞

∫ ∫j(z)dz

∣∣vεx(s, x)

∣∣dx ds + Cβ‖v0‖L∞

t∫0

TV(

vε(s))

ds

� CTV(v0)

t∫0

∥∥vε(s, ·) − v(s, ·)∥∥L1 ds + Cβ‖v0‖L∞ TV(v0)t.

Similar to I2, we see that the first term on the right-hand side of (4.5) is less than or equal to

Cε

βTV(v0)t.

Similar to (4.6), the last term of (4.5) is less than

t∫0

∥∥vε(s, ·) − v(s, ·)∥∥L1 ds + CβTV(v0)t.

Therefore, taking β = ε1/2, we conclude that

∥∥vε(t, ·) − v(t, ·)∥∥L1 �(1 + TV(v0)

) t∫0

∥∥vε(s, ·) − v(s, ·)∥∥L1 ds + Cε1/2(t + 1)TV(v0).

By Gronwall’s inequality, we have

∥∥vε(t, ·) − v(t, ·)∥∥L1 � Cε1/2 exp(Ct),

where C = C(TV(v0)).It is left to show the claim. Differentiating (2.1)1 with respect to x, multiplying the resulting

equation by sgn vεx , and integrating the product on R, we get

d

dt

∫R

∣∣vεx

∣∣dx = −∫R

vε∣∣vε

x

∣∣dx � 0.

Thus


∫R

∣∣vεx(t, x)

∣∣dx �∫R

∣∣vε0x

∣∣dx.

Because vε0 ∈ C∞

0 (R), we then have

TV(

vε(t)) =

∫R

∣∣vεx(t, x)

∣∣dx �∫R

∣∣vε0x

∣∣dx = TV(

vε0

)� CTV(v0),

from which the claim follows. �Proof of Theorem 1.2. It is easy to see that (vε − v) satisfies

(vε − v

)t − εvε

xx + (uε vε

x − uvx) = −1

2

[(vε

)2 − v2].Multiplying this equation by vε − v , integrating on [0, t] × R

+ , note that

(uε vε

x − uvx)(

vε − v) = 1

2uε

((vε

)2)x − uε vε

x v − uvx vε + 1

2u(

v2)x,

we have

∫R+

∣∣vε − v∣∣2

dx =∫

R+

∣∣vε0 − v0

∣∣2dx +

t∫0

∫R+

[2εvε

xx

(vε − v

) + 2vεx v

(uε − u

) + v vε(

vε − v)]

dx ds

�∥∥vε

0 − v0∥∥

L1

∥∥vε0 − v0

∥∥L∞ + ε2

t∫0

∫R+

∣∣vεxx

∣∣2dx ds +

t∫0

∫R+

∣∣vε − v∣∣2

dx ds

+ C

t∫0

∥∥uε(s, ·) − u(s, ·)∥∥L∞∥∥v(s, ·)∥∥L∞ TV

(vε(s)

)ds

+t∫

0

∥∥v(s, ·)∥∥L∞∥∥vε(s, ·)∥∥L∞

∥∥vε(s, ·) − v(s, ·)∥∥L1 ds. (4.7)

By the comparison principle, 0 � vε(t, x) � ‖v0‖L∞ for any (t, x) ∈ R+ × R

+ . Due to (2.9),

ε

t∫0

∫R+

∣∣vεxx

∣∣2dx �

∫R+

∣∣vε0x

∣∣2dx �

∥∥vε0x

∥∥L∞

∥∥vε0x

∥∥L1 � ε−1/2

∥∥vε0

∥∥L1 TV

(vε

0

).

Substituting this inequality into (4.7), thanks to (4.1) and (4.2), we get

∫+

∣∣vε − v∣∣2

dx � Cε1/2(1 + exp(Ct)) +

t∫ ∫+

∣∣vε − v∣∣2

dx ds,

R 0 R


where C = C(TV(v0)). Therefore, it follows from Gronwall’s inequality that

∫R+

(vε(t, x) − v(t, x)

)2dx � Cε1/2 exp(Ct),

with C = C(TV(v0)). �Acknowledgments

The authors would like to thank Professor Ping Zhang very much for many useful discussions. Theauthors are grateful to the anonymous referee for the helpful and important suggestions. This workis partially supported by the Chinese NSF (Grant No. 11071034) and the Fundamental Research Fundsfor the Central Universities.

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Global existence of dissipative solutions to the Hunter–Saxton equation via vanishing viscosity