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The global conservative solutions for the generalized camassa-holm equation

  • Received: 01 August 2019 Revised: 01 September 2019
  • Primary: 35A01, 35A02; Secondary: 35D30, 35G25, 35G25

  • This paper deals with the continuation of solutions to the generalized Camassa-Holm equation with higher-order nonlinearity beyond wave breaking. By introducing new variables, we transform the generalized Camassa-Holm equation to a semi-linear system and establish the global solutions to this semi-linear system, and by returning to the original variables, we obtain the existence of global conservative solutions to the original equation. We introduce a set of auxiliary variables tailored to a given conservative solution, which satisfy a suitable semi-linear system, and show that the solution for the semi-linear system is unique. Furthermore, it is obtained that the original equation has a unique global conservative solution. By Thom's transversality lemma, we prove that piecewise smooth solutions with only generic singularities are dense in the whole solution set, which means the generic regularity.

    Citation: Li Yang, Chunlai Mu, Shouming Zhou, Xinyu Tu. The global conservative solutions for the generalized camassa-holm equation[J]. Electronic Research Archive, 2019, 27: 37-67. doi: 10.3934/era.2019009

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  • This paper deals with the continuation of solutions to the generalized Camassa-Holm equation with higher-order nonlinearity beyond wave breaking. By introducing new variables, we transform the generalized Camassa-Holm equation to a semi-linear system and establish the global solutions to this semi-linear system, and by returning to the original variables, we obtain the existence of global conservative solutions to the original equation. We introduce a set of auxiliary variables tailored to a given conservative solution, which satisfy a suitable semi-linear system, and show that the solution for the semi-linear system is unique. Furthermore, it is obtained that the original equation has a unique global conservative solution. By Thom's transversality lemma, we prove that piecewise smooth solutions with only generic singularities are dense in the whole solution set, which means the generic regularity.



    In this paper we consider the continuation of solutions for the generalized Camassa-Holm (g-CH) equation

    {utuxxt+(m+2)(m+1)2umux=(m2u(m1)u2x+umuxx)x,xR,t>0,u(x,0)=u0(x),xR, (1)

    where m is a positive integer.

    The equation (1) was first proposed by Hakkaev and Kirchev in [18], and the local well-posedness of the Cauchy problem (1) was studied for the Sobolev spaces Hs with s>32. Under suitable assumptions and energy conservations, the orbital stability and instability of solitary wave solutions were considered. In [20], the authors established the local well-posedness to (1) for a range of Besov spaces and proved that its solutions are analytic in both variables. The persistence property of strong solutions for (1) was investigated in weighted Lp spaces [23]. And it was shown that the equation is well-posed in Sobolev spaces Hs (s>32) for both the periodic and the nonperiodic case in the sense of Hadamard [21]. Moreover, the nonuniform dependence and H¨older continuous to (1)were discussed.

    In fact, the equation (1) is a natural generalization of the famous Camassa-Holm (CH) model

    ut+uxxt+3uux=uuxxx+2uxuxx. (2)

    The Camassa-Holm equation first arisen in the context of hereditary symmetries was studied by Fokas and Fuchssteiner [14], but did not receive much attention until Camassa and Holm [9,7] derived it as a model of shallow water waves over a flat bottom. It has bi-Hamiltonian structure, infinitely many conservation laws and is completely integrable [8,7,14,19]. In addition, the stability of the smooth solitons and the orbital stability of the peaked solitons to (1) were established in [10] and [11] respectively. Particularly, the Camassa-Holm equation possesses solutions with presence of wave breaking (that is, the solution remains bounded while its slope becomes unbounded in finite time [6,12]). When these two waves collide at some time, the combined wave forms an infinite slope. After the collision, there are two things that happen: either two waves pass through each other with total energy preserved; or annihilate each other with a lose of energy. The solutions in the first case is called conservative, and the second case is called dissipative.

    So far, the continuation of the solutions after wave breaking has been studied widely. Bressan and Constantin proved that the solution of the Camassa-Holm equation can be continued as either global conservative or global dissipative solutions [2,3]. Notice that, the conservative solutions are about preservation of the H1 norm, while dissipative solutions are characterized by a sudden drop in the norm at blow-up. Afterwards, the uniqueness of the conservative solution and the dissipative solution for the Camassa-Holm equation were obtained [4,16]. Recently, the generic regularity of conservative solutions to Camassa-Holm equation was discussed in [17]. It is worth mentioning that the H1(R) norm conserved quantity plays a key role in the process of studying the conservative and dissipative solutions.

    In the case of a more general Camassa-Holm equation, the global existence and uniqueness to the solution are established in [22,25]. Moreover, for the Camassa-Holm equation with a forcing term ku, Zhu obtained the global existence and uniqueness [26].

    Similar to the well-known Camassa-Holm equation, the system (1) also models the peculiar wave breaking phenomena [24]. Therefore, in this paper, we still focus on the conservative case of equation (1.1) in H1 space, including the existence, uniqueness and generic regularity problems. Setting G(x)=12e|x|, xR, then (12x)1f=Gf for all fL2(R). Thus the equation (1.1) can be rewritten as the following integral-differential form:

    {ut+umux=Px,xR,t>0,u(x,0)=u0(x),xR, (3)

    where

    P=G(m2u(m1)u2x+m(m+3)2(m+1)). (4)

    Motivated by [2,4,17], in this paper, we consider the global weak conservative solutions defined by as follows.

    Definition 1.1. Let u0H1(R), there exists a family of Radon measure {μ(t),tR}, depending continuously on time w.r.t. the topology of weak convergence of measures, such that the following properties hold.

    (ⅰ) The map tu(.,t) is Lipschitz continuous from [0,T] into L2(R) with the initial data u0H1(R).

    (ⅱ) The solution u=u(x,t) satisfies the initial data u0L2(R). For any test function ϕCc1(Ω) with Ω={(x,t)|xR,t[0,+)}, one has

    Ω(ux(ϕt+umϕx)+(m2umu2xm(m+3)2(m+1)um+1+P)ϕ)dxdt+Ru0xϕ(0,x)dx=0. (5)

    For a solution u=u(t,x), we say that u is conservative, which means that the balance law (10) is satisfied in the following sense.

    There exists a family of Radon measure {μ(t),tR+}, depending continuously on time w.r.t. the topology of weak convergence of measures. For any tR+, the absolutely continuous measure μ(t) has density u2x(t,) w.r.t. Lebesgue measure. Moreover, for any test function ϕCc1(Ω), the family {μ(t);tR} supplies a measure-valued solution to the balance law

    0(R(ϕt+umϕx)dμ(t)+R2ux(m(m+3)2(m+1)um+1P)dx)dtRu20xϕ(0,x)dx=0. (6)

    Based on the characteristic, by introducing a new variables, we transform the equation (1) to a semi-linear system, and prove the semi-linear system has global solutions. Then by a reverse transformation, one can get the conservative solutions for equation (1). Our results are stated as follows.

    Theorem 1.2. Let u0H1(R). Then the generalized Camassa-Holm equation (3) has global conservative solution u=u(x,t) defined on R×(0,+). Moreover, the solution has the following properties.

    (i) u(x,t) is 1/2-H¨older continuous on both t and x.

    (ii) The function u provides a solution to the Cauchy problem (3) in the sense of Definition 1.1.

    (iii) There exists a null set NR with measN=0 such that for any tN, the measure μ(t) is absolutely continuous and has density u2x(t,) w.r.t. the Lebesgue measure.

    (iv) The energy u2+u2x coincides a.e. with a constant, that is,

    E(t)=E(0)fortN,E(t)<E(0)fortN.

    (v) The continuous dependence of solutions to system (3) holds with the initial data belongs to H1(R). More precisely, given a sequence of initial data {u0n} satisfy ||u0nu0||H1(R)0, then the corresponding solutions un(t,x) converge to u(t,x) uniformly for (t,x)[0,T]×R.

    Theorem 1.3. Given any initial data u0H1(R), the Cauchy problem (3) has a unique conservative solution.

    Remark 1. In fact, we know the process for the proof of the existence is an inverse, but the method here is an irreversible.

    By virtue of the analysis of solutions along characteristic, we show that piecewise smooth solutions with only generic singularities are dense in the whole solution set. Using the Thom's transversality Lemma [1,15], we give the following generic regularity result.

    Theorem 1.4. For any T>0, there exists an open dense set of initial data DC3(R)H1(R), such that for any u0D, the conservative solution u=u(t,x) of the equation (3) is twice continuously differentiable in the complement of finitely many characteristic curves, within the domain [0,T]×R.

    Remark 2. The generic regularity is very interesting since it reflects the structure of singularities. Similar issue was first established for the variational wave equation[5], and later this method was applied to the Camassa-Holm equation [17].

    This paper is organized as follows. In Section 2, we give the energy conservation laws and introduce a new set of independent and dependent variables. In Section 3, we first obtain a global conservative solution of the semi-linear system (27), and then by inverse transformation we prove the existence of the global conservative solution to equation (1). In Section 4, we establish the uniqueness of the characteristic curve through each initial point, and by considering the dynamics of a conservative solution along a characteristic, we obtain the proof of the uniqueness for the global conservation solution. In Section 5, the generic regularity of conservative solutions to equation (1) is investigated.

    For smooth solutions, we claim that the total energy

    E(t)=R(u2+u2x)dx (7)

    is constant in time. In fact, by using 2xGf=Gff and differentiating the equation (3) with respect to x, we have

    uxt+umuxx+mum1u2x=m2um1u2x+m(m+3)2(m+1)um+1P. (8)

    Multiplying (3) by u, and (8) by ux, one get

    (u22)t+(1m+1um+1)x+uPx=0, (9)
    (u2x2)t+(umux2)x=(m(m+3)2(m+1)um+1P)ux. (10)

    It follows from (9)-(10) that

    ddtE(t)=ddtR(u2+u2x)dx=0. (11)

    Therefore, the conservation law is given by

    E(t):=R(u2+u2x)dx=E(0). (12)

    Since P, Px are both defined as convolutions, by Young's inequality and Sobolev's inequality uL(R)uH1(R)=E(0)12, it implies that

    P(t)L,Px(t)L12e|x|Lm2um1u2x+m(m+3)2(m+1)um+1L1m(m+3)4(m+1)E(0)m+12, (13)

    and

    P(t)L2,Px(t)L212e|x|L2m2um1u2x+m(m+3)2(m+1)um+1L1m(m+3)2(m+1)E(0)m+12. (14)

    Let ˜u=u0(x)H1(R) be the initial data. Considering the energy variable ξR, the non-decreasing map ξ˜y(ξ) is defined by

    ˜y(ξ)0(1+˜u2x)dx=ξ. (15)

    Then the characteristic map ty(t,ξ) satisfies

    ty(t,ξ)=um(t,y(t,ξ)),y(0,ξ)=˜y(ξ). (16)

    And the new variables θ=θ(t,ξ) and h=h(t,ξ) are introduced as

    θ2arctanux,h1+u2xyξ, (17)
    h(0,ξ)1, (18)
    11+u2x=cos2θ2,ux1+u2x=12sinθ,u2x1+u2x=sin2θ2, (19)
    yξ=h(1+u2x)=cos2θ2h, (20)
    y(t,ˉξ)y(t,ξ)=ˉξξcos2θ(t,s)2h(t,s)ds. (21)

    Furthermore, we get

    P(t,ξ)=12exp{|y(t,ξ)x|}(m(m+3)2(m+1)um+1+m2um1u2x)dx,Px(t,ξ)=12(y(t,ξ)y(t,ξ))exp{|y(t,ξ)x|}(m(m+3)2(m+1)um+1+m2um1u2x)dx. 

    It follows from identities (18) to (21) that an expression for P and Px in terms of the new variable ξ

    P(t,ξ)=12exp{|ˉξξ(hcos2θ2)(s)ds|}[h(m(m+3)2(m+1)um+1cos2θ2+m2um1sin2θ2)](ˉξ)dˉξ, (22)
    Px(t,ξ)=12(ξξ)exp{|ˉξξ(hcos2θ2)(s)ds|}[h(m(m+3)2(m+1)um+1cos2θ2+m2um1sin2θ2)](ˉξ)dˉξ, (23)

    By (4) and (16), the evolution equation for u takes the form

    tu(t,ξ)=ut+uyyt=ut+umux=Px(t,ξ), (24)

    where Px is given in (23).

    By the definition of variable h, it follows that

    ξ2ξ1h(t,ξ)dξ=y(t,ξ2)y(t,ξ1)(1+u2x(t,x))dx. 

    By using (16) and Pxx=Pm2um1u2xm(m+3)2(m+1)um+1, we have

    ddtξ2ξ1h(t,ξ)dξ=y(t,ξ2)y(t,ξ1){[(1+u2x)]t+[u(1+u2x)]x}dx=y(t,ξ2)y(t,ξ1)(m(m+3)m+1um+1+mum12P)uxdx. 

    Differentiating with respect to ξ, we obtain

    th(t,ξ)=(m(m+3)m+1um+1+mum12P)ux1+u2xh=(m(m+3)m+1u2(m+1)+m2um1P)sinθh. (25)

    Using (17) and (19), we see

    tθ(t,ξ)=21+u2x(uxt+umuxx)=21+u2x(m2um1(ux)2+m(m+3)2(m+1)um+1P)=(m(m+3)m+1um+12P)cos2θ2mum1sin2θ2, (26)

    where P is defined by (22).

    According to (24)-(26), we obtain the following semi-linear system

    {ut=Px,θt=(m(m+3)m+1um+12P)cos2θ2mum1sin2θ2,ht=(m(m+3)2(m+1)um+1+m2um1P)sinθh (27)

    with the initial data

    {u(0,ξ)=˜u(˜y(ξ))θ(0,ξ)=2arctan˜ux(˜y(ξ))h(0,ξ)=1, (28)

    where P and Px are give by (22)-(23). System (27) can be regarded as an ODE in the Banach space

    XH1(R)×[L2(R)L(R)]×L(R), (29)

    with (u,θ,h)uH1+θL2+θL+hL.

    In light of the standard theory of ODE in the Banach space, we can establish that all functions on the right-hand side of (27) are locally Lipschitz continuous, this implies the local existence of solutions to the system (27)-(28).

    Lemma 3.1. Given initial data ˜uH1(R), the Cauchy problem (27)-(28) has a unique solution defined on any given time interval [0,T] with T0.

    Proof. Set any bounded domain ΛX defined by

    Λ=(u,θ,h)={uH1γ,θL2δ,θL3π2,hh(x)h+a.e.xR}, (30)

    for any positive constants γ,δ,h,h+. In view of the Sobolev's inequality

    uLuH1, (31)

    and the uniform boundedness of θ,h, it is easy to see that

    m(m+3)m+1um+1cos2θ2,mum1sin2θ2,(m(m+3)m+1um+1+m2um1)sinθh

    are Lipschitz continuous from Λ into L2L. The next aim is to prove the maps

    (u,θ,h)(P,Px) (32)

    are Lipschitz continuous from Λ into L2L. Actually, we only need to show that these maps are Lipschitz continuous from Λ into H1. To this, we first observe that

    measure{ξR;|θ(ξ)2|π4}measure{ξR;sin2θ(ξ)214}4{ξR;sin2θ(ξ)214}sin2θ(ξ)2dξ14{ξR;sin2θ(ξ)214}θ(ξ)2dξ14δ2

    for (u,θ,h)Λ. Then we have

    ˉξξcos2θ(ξ)2h(ξ)dξ{ξ[ξ,ˉξ]θ(ξ)2π4}h2dξh2(ˉξξ14δ2)for anyξˉξ, (33)

    which guarantees that exponential term in the (22)-(23) for P andPx decreases quickly as |ˉξξ|. Taking

    Γ(ϵ)min{1,exp(18δ2hh2|ϵ|)}, (34)

    we see

    ΓL1=(ϵ14δ2+ϵ 14δ2)Γ(ϵ)dϵ=12δ2+h4. (35)

    Next we show that P,PxH1, namely,

    P,Pξ,Px,PxξL2. (36)

    Since the estimates for P and Px are similar, we only need consider a priori bounds on Px. From the definition of Px in (23), it follows that

    |Px(ξ)|h+2|Γ(m(m+3)2(m+1)um+1cos2θ2+m2um1sin2θ2)|. (37)

    A standard properties of convolutions ensures that

    ||Px||L2h+2||Γ||L1(m(m+3)2(m+1)um+1L2+m8||um1θ2||L2)Ch+2||Γ||L1(||u||mL||u||L2+||u||m2L||u||L2||θ2||L2)<, (38)

    where C=m(m+3)2(m+1). Next we observe that

    Pxξ=[h(m(m+3)2(m+1)um+1cos2θ2+m2um1sin2θ2)](ξ)+12(ξξ)exp{|ˉξξ(hcos2θ2)(s)ds|}[cos2θ2h(ξ)]sign(ξˉξ)[h(m(m+3)2(m+1)um+1cos2θ2+m2um1sin2θ2)](ˉξ)dˉξ. (39)

    Since

    |Pxξ(ξ)|h+|m(m+3)2(m+1)um+1(ξ)+m8um1(ξ)θ2(ξ)|+h+2|Γ(m(m+3)2(m+1)um+1cos2θ2+m2um1sin2θ2)(ξ)|, (40)

    this implies

    ||Pxξ||L2=h+(m(m+3)2(m+1)||um+1||L2+m8||um1θ2||L2)+h+2||Γ||L1(m(m+3)2(m+1)um+1L2+m8||um1θ2||L2)Ch+2||Γ||L1(||u||mL||u||L2+||u||m2L||u||L2||θ2||L2)C(h++h+2)||Γ||L1<. (41)

    A similar argument leads to P,PξL2.

    To show that the maps given in (32) are Lipschitz continuous. It suffices to verify that partial derivatives

    Pu,Pθ,Ph,Pxu,Pxθ,Pxh, (42)

    are uniformly bounded for (u,θ,h)Λ. We observe that above derivatives are bounded linear operators from appropriate spaces into H1(R). For the sake of illustration, we just give a detailed estimate for Pxu, the boundedness of other derivatives can be obtained in a similar way.

    For (u,θ,h)Λ, the partial derivative Pxu and (ξPx)u are linear operator, defined by

    [Px(u,θ,h)uu](ξ)=12(ξξ)exp{|ˉξξ(hcos2θ2)(s)ds|}[hu(m(m+3)2umcos2θ2+m(m1)2um2sin2θ2)](ˉξ)dˉξ (43)

    and

    [(ξPx)(u,θ,h)uu](ξ)=[hu(m(m+3)2umcos2θ2+m(m1)2um2sin2θ2)](ξ)+12(ξξ)exp{|ˉξξ(hcos2θ2)(s)ds|}(hcos2θ2)sign(ˉξξ)(m(m+3)2umcos2θ2+m(m1)2um2sin2θ2)u(ˉξ)dˉξ. (44)

    In view of ||u||L||u||H1, the above operators norm can be estimated as follows:

    PxuuL2h+2Γ(m(m+3)2umcos2θ2+m(m1)2um2sin2θ2)L2||u||Lh+2||Γ||1L(m(m+3)2umL2+m(m1)8||um2θ2||L2)||u||H1 (45)

    and

    (ξPx)uuL2h+2m(m+3)2umcos2θ2+m(m1)2um2sin2θ2L2||u||H1+(h+)22Γ(m(m+3)2umcos2θ2+m(m1)2um2sin2θ2)L2||u||H1. (46)

    By (45) and (46), we obtain Pxu is a bounded linear operator from H1(R) into H1(R). And the boundedness of other partial derivatives in (42) can be proved by the same arguments. which means that the maps (32) are Lipschitz continuous.

    Next we show that the local solutions of the system (27) can be extended globally in time.

    Lemma 3.2. Given initial data ˜uH1(R), the Cauchy problem (27))-(28) has a unique solution defined for all time T>0.

    Proof. To extend the local solutions of the system (27) to global solutions, we only need to prove that

    ||u||H1+||θ||L2+||θ||L+||h||L+1hL< (47)

    for all T<. We first claim

    uξ=h2sinθ. (48)

    In fact, recalling (27), (22) and (23), one can get

    uξt=h(m(m+3)2(m+1)um+1cos2θ2+m2um1sin2θ2Pcos2θ2)=utξ. (49)

    Moreover, from (19)) and ((20) we have

    uξ=ux1+ux2=12sinθ,h=1

    at t=0. This implies that (48) holds for all t, as long as the solution defined.

    Next we prove

    ddtR(u2cos2θ2+sin2θ2)hdξ=0. (50)

    Using (27), a direct calculation yields that

    ddtR(u2cos2θ2+sin2θ2)hdξ=Rh{(u2cos2θ2+sin2θ2)(m(m+3)m+1um+1+mum12P)cosθ2sinθ22uPxcos2θ2+cosθ2sinθ2(1u2)[(m(m+3)m+1um+12P)cos2θ2mum1sin2θ2]}dξ=Rh{2m(m+2)m+1um+1cosθ2sinθ22Pcosθ2sinθ22uPxcos2θ2}dξ. (51)

    By (20), we have

    Pξ=hPxcos2θ2,

    which implies

    (uP)ξ=(Psinθ2cosθ2+uPxcos2θ2)h (52)

    and

    2m(m+2)m+1um+1cosθ2sinθ2h=2m(m+2)m+1um+1uξ=(2mm+1um+2)ξ. (53)

    From (52) and (53), one has

    ddtR(u2cos2θ2+sin2θ2)dξ=0,

    namely,

    R(u2cos2θ2+sin2θ2)dξ=E(0)=E0. (54)

    If the solution is well defined, we obtain a priori bound on ||u(t)||L(R) as follows

    supξR|u2(t,ξ)|2R|uuξ|dξ2R|u||sinθ2cosθ2|hdξE0. (55)

    According to (22), (23) and (54) we know

    ||P(t)||L,||Px(t)||L12h(m(m+3)2(m+1)um+1cos2θ2+m2um1sin2θ2)L1m(m+3)4(m+1)Em+120. (56)

    Using the third equation in (27), together with (55) and (56), we conclude

    |ht|(3m(m+3)8(m+1)Em+120+m4Em120)h.

    Therefore, it follows that

    e(3m(m+3)8(m+1)Em+120+m4Em120)th(t)e(3m(m+3)8(m+1)Em+120+m4Em120)t. (57)

    Also, we observe that there exists a suitable constant A=A(E0) such that

    θLeAt. (58)

    Multiplying the first equation of (27) by u and integrating, one has

    |ddt(12u(t)L22)|u(t)LPxL1.

    Similarly, we deduce

    |ddt(12uξ(t)L22)|uξ(t)LξPxL1.

    Therefore, we obtain that u and uξ are uniformly bound on [0,T] by (48), (55)and (57). A bound on the L1 norms of Px and xPx yield that uH1 is bounded for any T. Therefore, we only consider estamates xPxL1, PxL1. In fact, for ξ>ˉξ, we find that

    ξˉξhcos2θ2(s)ds{s[ˉξ,ξ],|θ2(ˉξ)|π4}hcos2θ2(s)ds{s[ˉξ,ξ],|θ2(ˉξ)|π4}h2(s)dsh2(ξˉξ){s[ˉξ,ξ],|θ2(ˉξ)|π4}h2(s)dsh2(ξˉξ){s[ˉξ,ξ],|θ2(ˉξ)|π4}hsin2θ2(s)dsh2(ξˉξ)2E0, (59)

    where h=e(3m(m+3)8(m+1)Em+120+m4Em120)t=eBt. Denoting

    Γ1(ϵ)min{1,e(2E0h|ξ|2)},

    it follows from (39) that

    ξ(xP(t))L1|Pxξ(ξ)|=h(m(m+3)2(m+1)um+1cos2θ2+m2um1sin2θ2)(ξ)L1+h+2LΓ1(m(m+3)2(m+1)um+1cos2θ2+m2um1sin2θ2)(ξ)L1(1+12hLΓ1L1)h(m(m+3)2(m+1)um+1cos2θ2+m2um1sin2θ2)(ξ)L1m(m+3)2(m+1)[1+2e2Bt(2E0+1)]Em+120,

    where

    Γ1L1=4eBt(2E0+1).

    Multiplying the second equation (27) by θ, we see that

    |ddt(12θ(t)L22)|R|(m(m+3)m+1um+12P)θ|dξ+Rm8um1|θ|3dξθL(m(m+3)m+1u2L2um1L+2PL1)+m8um1LθLθ2L2.

    From this, we prove that θL2 remains bounded on bounded intervals of time. This complete the proof of Lemma 3.2.

    Let (u,θ,h) be a global solution to (27), and

    y(t,ξ)˜y(ξ)+t0um(τ,ξ)dτ. (60)

    Then for each fixed ξ, the functions ty(t,ξ) provides a solution to the Cauchy problem

    ty(t,ξ)=um(t,ξ),y(0,ξ)=˜y(ξ). (61)

    Now we claim that

    u(t,x)u(t,ξ) if y(t,ξ)=x, (62)

    is a weak solutions of equation (3).

    Proof of Theorem 1.2. The proof is divided into the following steps.

    Step 1. It is clearly that we have the uniform bound

    |u(t,ξ)|E120.

    From (60), we have the estimate

    ˜y(ξ)Em20ty(t,ξ)˜y(ξ)+Em20t,t0.

    The definition of ξ in (15) implies

    limξ±y(t,ξ)=±.

    Hence the image of the map (t,ξ)(t,y(t,ξ)) is the entire plane R2.

    Step 2. We check

    yξ=hcos2θ2 for all t0 and a.e ξR. (63)

    Indeed, thanks to system (27), by a straightforward computation we have

    t(hcos2θ2)(t,ξ)=hθtsinθ2cosθ2+htcos2v2=mum1hsinθ2cosθ2=(um)ξ(t,ξ).

    On the other hand, (60) implies

    tyξ(t,ξ)=(um)ξ(t,ξ).

    Since the function x2arctan˜ux(x) is measurable, hence (63) is true for almost every ξ at t=0, then the above calculation (63) remains true for all t0, and y(t,ξ) is non-deceasing. Moreover, if ξ<ˉξ but y(t,ξ)=y(t,ˉξ), then

    ˉξξyξ(t,s)ds=ˉξξh(t,s)cos2θ2ds=0.

    Hence cosθ20 throughout the interval of the integration. By (48), we have

    u(t,ˉξ)u(t,ξ)=ˉξξh(t,s)2sinθ(t,s)cosθ2ds=0.

    This shows that the map (t,x)u(t,y(ξ)) is well defined for all t0 and xR.

    Step 3. Recalling the basic relations, we have

    u(t,ξ)ξ=h(t,ξ)2sinθ(t,ξ),y(t,ξ)ξ=h(t,ξ)cos2θ(t,ξ)2. (64)

    In addition, if x=y(t,ξ),cosθ(t,ξ)1, and

    ux(t,x)=sinθ(t,ξ)1+cosθ(t,ξ), (65)

    then for any time t, it follows from (64) and (65) that

    R(u2(t,x)+u2x(t,x))dx=Rcosθ21(u2(t,ξ)cos2θ(t,ξ)2+sin2θ(t,ξ)2)h(t,ξ)dξE0, (66)

    which implies E(t)=E0. Since the boundedness of PxL, we obtain

    du(t,y(t,ξ))dt=ut.

    On the basis of the Sobolev inequality, u(t,x) is Hölder continuous with exponent 12 on both x and x.

    Step 4. We are ready to show that the Lipschitz continuity of u(t,x) with values in L2(R). Consider any interval [τ,τ+h], given a point x, we choose ξR such that the characteristic ty(t,ξ) passes through the point (τ,x). By (27) and (55), it follows that

    |u(τ+s,x)u(τ,x)||u(τ+s,x)u(τ+s,y(τ+s,ξ))|+|u(τ+s,y(τ+s,ξ))u(τ,x)|sup|yx|Em20s|u(τ+s,y)u(τ+s,x)|+τ+sτ|Px(t,ξ)|dt. 

    Integrating over R, using the boundedness of ||Px||L2(R) and uxL2, we deduce that

    R|u(τ+s,y)u(τ,y)|2dx2R(x+Em20sxEm20s|ux(τ+s,y)|dy)2dx+2R(τ+sτ|Px(t,ξ)|dt)2h(τ,ξ)cos2θ(τ,ξ)2dξ4Em20sRx+Em20sxEm20s|ux(τ+s,y)|2dydx+2shLRτ+sτ|Px(t,ξ)|2dtdξ8Em0s2ux(τ+s)||2L2(R)+2shL||Px(t)||2L2(R)dtCs,

    where the constant C depending only on T. The above inequality implies that the map tu(t) is Lipschitz continuous for the variable x.

    Step 5. Define Ω=[0,)×R and Ω=Ω{(t,y)|cos2θ(τ,ξ)20}, for any text function ϕ(x,t)C1c(Ω), we have the following weak form

    0=Ω{uξtϕt+hϕ(m(m+3)2(m+1)um+1cos2θ2+m2um1sin2θ2Pcos2θ2)}ξdt=Ω{uξϕtϕt+hϕ(m(m+3)2(m+1)um+1cos2θ2+m2um1sin2θ2Pcos2θ2)}dξdt=Ω{uξϕtϕt+hϕ(m(m+3)2(m+1)um+1cos2θ2+m2um1sin2θ2Pcos2θ2)}ξdt=Ω{ux(ϕt+umϕx)+ϕ(m(m+3)2(m+1)um+1m2um1u2x+P)}dxdt, (67)

    which proves that (5) holds. Let us introduce the Radon measures {μ(t),tR+}, for any Lebesgue measurable set {xA} in R, assuming the corresponding pre-image set of transformation is {ξF(A)}, one has

    μt(A)=F(A)sin2θ2(t,ξ)dξ.

    By (63), the measure μ(t) is absolutely continuous and has density u2x(t,) w.r.t. Lebesgue measure. It is easy to check that (6) is right. Indeed, from (3.1) we have

    R+{(ϕt+umϕx)dν(t)}dt=Ωsin2θ2hϕtdξdt=Ω(sin2θ2h)tϕdξdt=Ω(m(m+3)m+1um+12P)sinθ2cosθ2hϕdξdt (68)
    =Ω(m(m+3)m+1um+12P)uxϕdxdt.

    Step 6. Ultimately, we show that for almost every tR+, the singular part of vt is concentrated on the set where u=0. The proof is similar to the argument in [2]. Note that when blow up occurs, cosθ2=0, it follows that θt=mum1, which implies θt0 only when m0 or u0. Moreover, the proof in the seventh step is different from the Camassa-Holm equation.

    Let u=u(t,x) be a conservative solution of equation (1). We introduce the new coordinates (t,β), and define x(t,β) is the unique point x such that

    x(t,β)+μ(t){(,x)}βx(t,β)+μ(t){(,x]} (69)

    for any time t and βR. When the measure μ(t) is absolutely continuous with density u2x w.r.t Lebesgue measure, the above definition gives that

    x(t,β)+x(t,β)u2x(t,ξ)dξ=β. (70)

    Next, we will give following Lemma which is helpful to prove the Lipschitz continuity of x and u as functions of the variables t,β.

    Lemma 4.1. Assume that u=u(t,x) is a conservative solution of (1). For every t0, then the maps βx(t,β) and βu(t,β)u(t,x(t,β)), which are implicitly defined by (69), are Lipschitz continuous. Moreover, The map tx(t,β) is also Lipschitz continuous with a constant depending only on u0H1.

    Proof of Lemma 4.1. We split the proof into three steps.

    Step 1. For any time t0, the map

    xβ(t,x)

    is right continuous and strictly increasing. Thus, the inverse βx(t,β) is well defined, continuous, nondecreasing. If β1<β2, we see that

    x(t,β2)x(t,β1)+μ(t){(x(t,β2),x(t,β1))}β2β1,

    which implies

    x(t,β2)x(t,β1)β2β1, (71)

    and the map βx(t,β) is Lipschitz continuous.

    Step 2. For the map βu(t,β), as β1<β2, it follows from (69) and (71) that

    |u(t,x(t,β2))u(t,x(t,β1))|x(t,β2)x(t,β1)|ux|dxx(t,β2)x(t,β1)12(1+u2x)dy12[x(t,β2)x(t,β1)+μ(t){(x(t,β2),x(t,β1))}]12(β2β1). (72)

    Hence, the map βu(t,β) is Lipschitz continuous.

    Step 3. Now we claim the Lipschitz continuity of the map tx(t,β). Assume x(τ,β)=y, since the family of measure μ(t) satisfies the balance law (6), we infer the source term 2ux(m(m+3)2(m+1)um+1P) satisfies

    2ux(m(m+3)2(m+1)um+1P)L12(m(m+3)2(m+1)umLuL2+PL2)uxL2Cs. (73)

    For tτ, it follows from (73) that

    μ(t){(,yC(tτ))}μ(τ){(,y)}+Cs(tτ),

    where the constant Cs depending only on the H1(R) norm of u and m. Denoting y(t)y(C+Cs)(tτ), we have

    y(t)+μ(t){(,y(t)]}y(C+Cs)(tτ)+μ(τ){(,y)}+Cs(tτ)y+μ(τ){(,y)}β,

    which implies x(t,β)y(t) for all t>τ. A similar argument yields

    x(t,β)y+(t)y+(C+Cs)(tτ).

    This proves the uniform Lipschitz continuity of the map tx(t,β).

    Lemma 4.2. Let u=u(t,x) be the conservative solution of equation (1). Then there exists a unique Lipschitz continuous map tx(t) for any ˜yR, where the map satisfies

    ddtx(t)=um(t,x),x(0)=˜y, (74)

    and

    ddtx(t,β)u2x=x(t,β)2ux(m(m+3)2(m+1)um+1P)dx. (75)

    Moreover, we have

    u(t,x(t))u(τ,x(τ))=tτPx(s,x(s))ds, (76)

    for any 0τt.

    Proof. Step 1. According to the adapted coordinates (t,β), we write the characteristic beginning with ˜y in the form tx(t)=x(t,β(t)). β() is a map to be determined. Together with (74) and (75), we obtain

    x(t)+x(t)u2x(t,y)dy=˜y+˜yu20,xdy+t0(um(s,x(s))+x(s)2ux(m(m+3)2(m+1)um+1P)(s,y)dy)ds. (77)

    For convenience, let

    G(t,β)x(s)2ux(m(m+3)2(m+1)um+1P)dx (78)

    and

    ˜β=˜y+˜yu20,x(y)dy. (79)

    Therefore, we can rewrite the equation (77) as follow

    β(t)=˜β+t0G(s,β(s))ds,for allt>0. (80)

    Step 2. For every fix t0, in view of the maps xu(t,x),xP(t,x)H1(R), and the function βG(t,β) defined by (78) is uniformly bounded and absolutely continuous. Furthermore, we have

    Gβ=2ux(m(m+3)2(m+1)um+1P)xβ=2ux(m(m+3)2(m+1)um+1P)1+u2x[C,C] (81)

    for some constant C, which depends only on the H1 norm of u. Consequently, the function G in (80) is uniformly Lipschitz continuous w.r.t. β.

    Step 3. Based on the Lipschitz continuity of the function G, applying the standard fixed point theory, one can get the existence of a unique solution for the integral (80). More details can refer to [4].

    Step 4. Owing to the previous construction, we conclude that the map tx(t)x(t,β(t)) is a unique solution for equation (77). β(t) and x(t) are differentiable almost everywhere because of the Lipschitz continuity of β(t) and x(t)=x(t,β(t)), so we only consider the time where x(t) is differentiable. We prove that (74) holds at almost every time. Suppose, on the contrary, ˙x(τ)um(τ,x(τ)). Without loss of generality, let

    ˙x(τ)um(τ,x(τ))+2ε0 (82)

    for some ε0>0. The case ε0<0 can be proved by similar approach. For t(τ,τ+δ], choosing δ>0 small enough, we find

    x+(t)x(τ)+(tτ)[um(τ,x(τ))+ε0]<x(t). (83)

    We also observe that (6) still holds for any test ϕ with compact support.

    For any ϵ>0, we consider the test functions as

    ρϵ(s,y)={0ifyϵ1,(y+ϵ1)ifϵ1y1ϵ1,1if1ϵ1yx+(s),1ϵ1(yx(s))ifx+(s)yx+(s)+ϵ,0ifyx+(s)+ϵ,
    χϵ(s,y)={0ifsτϵ,ϵ1(sτ+ϵ)ifτϵsτ,1ifτst,1ϵ1(st)ifts<t+ϵ,0ifst+ϵ. (84)

    We define

    ψϵ(s,y)min{ρϵ(s,y),χϵ(s)}. (85)

    Using ψϵ as a test function in (6), it follows that

    [u2xψϵt+umu2xψϵx+2ux(m(m+3)2(m+1)um+1P)ψϵ]dxdt=0. (86)

    If tτ, we have

    limϵ0tτx+(s)+ϵx+(s)ϵu2x(ψϵt+umψϵx)dyds0. (87)

    Actually, for s[τ+ϵ,tϵ], one has

    0=ψϵt+[um(τ,x(τ))+ε0]ψϵxψϵt+um(s,x)ψϵx, (88)

    where we use the fact that um(s,x)<um(τ,x(τ))+ε0 and ψϵx0.

    Due to the family of measures μ(t) depend continuously on t in the topology of weak convergence, taking the limit of (86) as ϵ0, we have

    0=x(τ)u2x(τ,y)dyx+(t)u2x(t,y)dy+tτx+(s)2ux(m(m+3)2(m+1)um+1P)dyds+limϵ0tτx+(s)+ϵx+(s)ϵu2x(ψϵt+umψϵx)dydsx(τ)u2x(τ,y)dyx+(t)u2x(t,y)dy+tτx+(s)2ux(m(m+3)2(m+1)um+1P)dyds,

    which yields

    x+(t)u2x(t,y)dyx(τ)u2x(τ,y)dy+tτx+(s)2ux(m(m+3)2(m+1)um+1P)dyds=x(τ)u2x(τ,y)dy+tτx(s)2ux(m(m+3)2(m+1)um+1P)dyds+o1(tτ).

    Note that the last term is higher order infinitesimal, satisfying o1(tτ)tτ0 as tτ. Indeed,

    |o1(tτ)|=|tτx+(s)+ϵx+(s)2ux(m(m+3)2(m+1)um+1P)dxdyds|2(m(m+3)2(m+1)um+1P)Ltτx+(s)+ϵx+(s)|ux|dyds2(m(m+3)2(m+1)um+1P)Ltτ(x(s)x+(s))12ux(s,)L2dsC(tτ)32.

    On the other hand, together with (78) and (80), we see

    β(t)=β(τ)+(tτ)[um(τ,x(τ)+x(τ)2ux(m(m+3)2(m+1)um+1P)dy]+o2(tτ), (89)

    with o2(tτ)tτ as tτ. For t sufficiently close to τ, we have

    β(t)=x(t)+x(t)(u2x(t,y)dy>x(τ)+(tτ)[um(τ,x(τ))+ε0]+x+(t)u2x(t,y)dyx(τ)+(tτ)[um(τ,x(τ))+ε0]+x(τ)u2x(τ,y)dy+tτx(s)2ux(m(m+3)2(m+1)um+1P)dyds+o1(tτ). (90)

    By (89) and (90), we have

    β(t)+(tτ)[um(τ,x(τ))+x(τ)2ux(m(m+3)2(m+1)um+1P)dy]+o2(tτ)[x(τ)+x(τ)u2x(τ,y)dy]+(tτ)[um(τ,x(τ))+ε0]+tτx(s)2ux(m(m+3)2(m+1)um+1P)dyds+o1(tτ). (91)

    Dividing both sides by tτ and letting tτ, we get a contradiction, namely, (74) holds.

    Step 5. Now we prove (75). By (3), one has

    0[uϕt+um+1m+1ϕx+Pxϕ]dxdt+u0(x)ϕ(0,x)dx=0, (92)

    for every test function ϕCc(R). Let ϕ=ψx, where ψCc. Due to the fact that the map xu(t,x) is absolutely continuous, integrating by part w.r.t. x, then we get

    0[uxψt+umuxψx+Pxψx]dxdt+u0,x(x)ψ(0,x)dx=0. (93)

    By an approximation argument, we find the identity (93) still holds for any test function ψ which is Lipschitz continuous with compact support. consider the function

    η(s,y)={0ifyϵ1,(y+ϵ1)ifϵ1y1ϵ1,1if1ϵ1yx(s),1ϵ1(yx(s))ifx(s)yx(s)+ϵ,0ifyx(s)+ϵ,

    for any ϵ>0 sufficiently small. We define

    φϵ(s,y)=min{ηϵ(s,y),χϵ(s)}, (94)

    with χϵ(s) defined in (84). We use the test function ψ=φϵ in (93). And let ϵ0. Since the function Px is continuous, we have

    x(t)ux(t,y)dy=x(τ)ux(τ,y)dytτPx(s,x(s))ds+limϵ0t+ϵτϵx(s)+ϵx(s)u2x(φϵt+umφϵx)dyds. (95)

    It suffices to prove that the last term of the limit in (95) is zero. The Cauchy's inequality implies

    |tτx(s)+ϵx(s)ux(φϵt+umφϵx)dyds|tτ(x(s)+ϵx(s)|ux|2dy)12(x(s)+ϵx(s)(φϵt+umφϵx)2dy)12ds, (96)

    where uxL2. For each ϵ>0, denoting

    ςϵ(s)(supxRx+ϵxu2x(s,y)dy)12, (97)

    we see that all functions ςϵ are uniformly bounded and ςϵ(t)0 pointwise at a.e. time t as ϵ0. Therefore, it follows from the dominated convergence theorem that

    limϵ0tτ(x(s)+ϵx(s)|ux(s,y)|2dx)12dslimϵ0tτςϵ(s)ds=0. (98)

    On the other hand, for every time s[τ,t], we obtain

    φϵx(s,y)=ϵ1,φϵt(s,y)+um(s,x(s))φϵx(s,y)=0,

    for x(s)<y<x(s)+ϵ. This yields

    x(s)+ϵx(s)|φϵt(s,y)+um(s,x(s))φϵx(s,y)|2dy=ϵ2x(s)+ϵx(s)|um(s,y)um(s,x(s))|2dyϵ1(maxx(s)yx(s)+ϵ|um(s,y)um(s,x(s))|)2m2ϵ1(x(s)+ϵx(s)|um1ux(s,y)|dy)2ϵ1m2(um1Lϵ12ux(s)L2)2m2u(s)2mH1. (99)

    By (98) and (99), one has the integral in (96) approaches to zero as ϵ0. We now estimate the integral near the corners of the domain,

    |(ττϵ+t+ϵt)x(s)+kϵx(s)ux(ψϵt+umφϵx)dxds|(ττϵ+t+ϵt)(x(s)+ϵx(s)|ux|2dx)12(x(s)+ϵx(s)(ψϵt+umφϵx)2dx)12ds2ϵu(s)H1(x(s)+ϵx(s)4ϵ2uL)2mdx)12Cϵ120 (100)

    as ϵ0. We conclude

    limϵ0t+ϵτϵx(s)+ϵx(s)u2x(ψϵt+umuφϵx)dxds=0. (101)

    Therefore, using (95), we deduce (76).

    Step 6. Finally, the uniqueness of the solution x(t) is clear.

    Lemma 4.3. If u=u(t,x) is a conservative solution of equation (3), then the map (t,β)u(t,β)u(t,x(t,β)) is Lipschitz continuous, where the Lipschitz constant depending only on the norm u0H1.

    Proof. By (72), (76) and (80), we have

    |u(t,x(t,˜β))u(τ,˜β)||u(t,x(t,˜β))u(t,x(t,β(t)))|+|u(t,x(t,β(t)))u(t,x(τ,β(τ)))|12|β(t)˜β|+C(tτ)C(tτ), (102)

    where C is a constant depending only on u0H1.

    Lemma 4.4. Let u be a conservative solution to the equation (3). If tβ(t;τ,˜β) is the solution to the integral equation

    β(t)=˜β+tτG(τ,β(τ))dτ, (103)

    where the G is defined in (78), then there exists a constant C, such that for any two initial data ~β1,~β2 and any t,τ0 the corresponding solutions satisfy

    |β(t;τ,~β1)β(t;τ,~β2)|eC|tτ||~β1~β2|. (104)

    Proof. Using the Lipschitz continuity of G with respect to β, the lemma can be proved. We omit here for brevity.

    Lemma 4.5. Suppose uH1(R). Then Px is absolutely continuous and satisfies

    Pxx=Pm2um1u2xm(m+3)2(m+1)um+1. (105)

    Proof. The function ϕ(x)=12e|x| satisfies the distributional identity

    D2xϕ=ϕδ0.

    Here δ0 denotes a unit Dirac mass at the origin. For every function fL1(R), the convolution satisfies

    D2x(ϕf)=ϕff.

    Choosing f=m2um1u2x+m(m+3)2(m+1)um+1, we obtain the result.

    In this subsection, we mainly prove the uniqueness of conservative solutions for equation (1).

    Proof of Theorem 1.3. The proof is divided into following steps.

    Step 1. It follows from Lemma 4.1 and Lemma 4.3 that the map (t,β)(x,u)(t,β) is Lipschitz continuous. By a similar approach, we find the maps βG(t,β)G(t,x(β)) and βPx(t,β)Px(t,x(t,β)) are also Lipschitz continuous. Thanks to the Rademacher's theorem, the partial derivatives xt,xβ,ut,uβ and Px,β exist almost everywhere. And for these derivatives, a.e. point (t,β) is a Lebesgue one. Recalling that tβ(t,˜β) the unique solution of the equation (80), for a.e ˜β, from Lemma 4.4 we can draw the following conclusion.

    (GC) For a.e. t>0, except a measure zero set NR+, the point (t,β(t,˜β)) is a Legbesgue point with respect to the partial derivatives xt,xβ,ut,uβ,Gβ,Px,β. And xβ(t,β(t,˜β))>0 for a.e. t>0.

    If the above condition is true, we say that tβ(t,˜β) is a good characteristic.

    Step 2. We find an ODE to describe a change in these two quantities uβ and xβ along a good characteristic. Denote tβ(t;τ,˜β) to be the solution of (103). For τ,tN, let β(;τ,˜β) be a good characteristic. Differentiating (103) w.r.t ˜β we obtain

    β(t)˜β=1+tτGβ(s,β(s;τ,˜β))˜ββ(s;τ,˜β)ds. (106)

    Next, differentiating w.r.t. ˜β the identity

    x(t,β(t;τ,˜β))=x(τ,˜β)+tτum(s,x(s,β(t;τ,˜β)))ds,

    we have

    xβ(t,β(t;τ,˜β))˜ββ(t;τ,˜β)=xβ(τ,˜β)+tτumβ(s,β(s;τ,˜β))˜ββ(t;s,˜β)ds. (107)

    Finally, by (76), differentiating w.r.t. ˜β, we have

    uβ(t,β(t;τ,˜β))˜ββ(t;τ,˜β)=uβ(τ,˜β)+tτPx,β(s,β(s;τ,˜β))˜ββ(t;s,˜β)ds. (108)

    Together with (106), (107) and (108) yield the following ODE system:

    {ddt[˜ββ(t;τ,˜β)]=Gβ(t,β(t;τ,˜β))˜ββ(t;τ,˜β),ddt[xβ(t,β(t;τ,˜β))˜ββ(t;τ,˜β)]=(um)β(t,β(t;τ,˜β))˜ββ(t;τ,˜β),ddt[uβ(t,β(t;τ,˜β))˜ββ(t;τ,˜β)]=Px,β(β(t;τ,˜β))β(t;τ,˜β). (109)

    The quantities within square brackets on the left hand sides of (109) are absolutely continuous. By the above system and using Lemma 4.5, along a good characteristic, we deduce

    {ddtxβ+Gβxβ=mum1uβ,ddtuβ+Gβuβ=(m2um1u2x+m(m+3)2(m+1)um+1P)xβ=(m(m+3)2(m+1)um+1+m2um1(1xβ1)P)xβ=(m(m+3)2(m+1)um+1Pm2um1)xβ+m2um1. (110)

    Step 3. Now we return to the original coordinates (t,x) and deduce an evolution equation for ux along a ``good" characteristic curve.

    Fixed a point (τ,˜x) for τN. Suppose that ˜x is a Lebesgue point for the map xux(τ,x). Let ˜β be such that ˜x=x(τ,˜β). Assume that tβ(t;τ,˜β) is a good characteristic, so that (GC) holds. Notice that

    u2x(τ,x)=1xβ(τ,˜β)10,xβ(τ,˜β)>0.

    If xβ>0, along the characteristic though (τ,˜x), it follows that

    ux(t,x(t,β(t;τ,˜β)))=uβ(t,β(t;τ,˜β))xβ(t,β(t;τ,˜β)). (111)

    From (110), we obtain that the map tux(t,x(t,β(t;τ˜β))) is absolutely continuous (as long as xβ0) and satisfies

    ddtux(t,x(t,β(t;τ˜β)))=ddt(uβxβ)=xβ{(m(m+3)2(m+1)um+1Pm2um1)xβ+m2um1xβuβGβ}uβ{mum1uβxβGβ}x2β=m(m+3)2(m+1)um+1Pm2um1+mum12xβuβGβxβmum1u2βx2β+uβGβxβ=m(m+3)2(m+1)um+1Pm2um1+mum12xβmum1u2βx2β. (112)

    Thus, for xβ>0, one has

    ddtarctanux(t,x(t,β(t;τ˜β)))=11+u2xddtux=(m(m+3)2(m+1)um+1Pm2um1+mum12xβmum1u2βx2β)xβ=(m(m+3)2(m+1)um+1Pm2um1)xβ+mum12mum1u2βxβ=(m(m+3)2(m+1)um+1mum1u2xPm2um1)xβ+mum12. (113)

    Step 4. Introduce the function

    θ{2arctanuxif0<xβ1,πifxβ=0. (114)

    This yields

    xβ=11+u2x=cos2θ2,ux1+u2x=12sinθ,u2x1+u2x=sin2θ2. (115)

    where θ can be seen as a map taking values in the unit circle Ω[π,π] with endpoints identified. We say that this map tθ(t)θ(t,x(t,β(t;τ˜β))) is absolutely continuous and satisfies

    ddtθ(t)=θt=(m(m+3)m+1um+12P)cos2θ2m2um1sin2θ2, (116)

    along each good characteristic. In fact, for simplicity, denote by xβ(t),uβ(t) and ux(t)=uβ(t)xβ(t) the values of xβ,uβ and ux along this particular characteristic. From (GC), for a.e. t>0, we have xβ(t)>0. Assume that τ is any time where xβ(τ)>0, we find a neighborhood I=[τδ,τ+δ] satisfies xβ(τ)>0 on I. It follows from (113) and (115) that v=2arctan(uβxβ) is absolutely continuous restricted to I and satisfies (116). To prove our previous conclusion, we need to prove that tv(t) is continuous on the null set N of times at xβ(t)=0. Let xβ(t0)=0. By the following identity

    u2x(t)=1xβ(t)xβ(t), (117)

    which is valid as long as xβ>0, we have u2x as tt0 and xβ(t)0, which denotes θ(t)=2arctanux(t)±π. Since we identify the points ±π in Ω, so we establish the continuity of θ for all t0. This completes our conclusion.

    Step 5. If u=u(t,x) is a conservation solution, in terms of the variables t,β, the quantities x,u,θ, we deduce

    {ddtβ(t,˜β)=G(t,β(t,˜β)),ddtx(t,β(t,˜β))=um(t,β(t,˜β)),ddtu(t,β(t,˜β))=Px(t,β(t,˜β)),ddtv(t,β(t,˜β))=(m(m+3)m+1um+12P)cos2θ2m2um1sin2θ2. (118)

    Recalling the definition of P and G, in term of the variable β, the function P and Px have representations as follows

    P(x(β))=12exp{|ββcos2v(s)2ds|}[(m(m+3)2(m+1)um+1cos2θ2+m2um1sin2θ2)(β)]dβ, (119)
    Px(x(β))=12(ξξ)exp{|βξcos2v(s)2ds|}[(m(m+3)2(m+1)um+1cos2θ2+m2um1sin2θ2)(β)]dβ. (120)

    For every ˜βR, the initial condition is

    {β(0,˜β)=˜β,x(0,˜β)=x(0,˜β),u(0,˜β)=u0(x(0,˜β)),θ(0,˜β)=2arctanu0,x(x(0,˜β)). (121)

    Since the Lipschitz continuity of coefficients, the Cauchy problem (118), (121) has a unique solution with initial data condition (121), which is globally defined for all t0,xR.

    Step 6. Suppose that u,ˉu are two conservative solutions of equation (3) with the same initial data u0H1(R). For a.e. t>0, the Lipschitz continuous maps βx(t,β),βˉx(t,β) are strictly increasing. Therefore, the above maps have continuous inverses, i.e. xβ(t,x),xˉβ(t,x). In summary, the map (t,β)↦↦(x,u,θ)(t,β) is uniquely resolved by the initial data u0. Accordingly,

    x(t,β)=ˉx(t,β),u(t,β)=ˉu(t,β).

    In turn, for a.e. t>0, we conclude

    u(t,x)=u(t,β(t,x))=ˉu(t,ˉβ(t,x))=ˉu(t,x).

    To prove the singularities of the solution for u to (3) in tx plane, we need to consider the level sets {θ(t,ξ)=π}. According to the fact that u,θ and h are smooth, the generic structure of these level sets can be studied by Thom's transversality theorem [1,15]. Our aim is to establish several families of perturbations for a given solution of (27). To this, we introduce the following lemma.

    Lemma 5.1. Let (u,θ,h) be a smooth solution of the semilinear system for (27). Given a point (t0,ξ0)R+×R.

    (1) If (θ,θξ,θξξ)(t0,ξ0)=(π,0,0), then there exists a 3parameter family of smooth solutions (uλ,θλ,hλ), depending smoothly on λR3, such that the following holds.

    i) When λ=0R3, one goes back to the original solution, namely, (u0,θ0)=(u,θ).

    ii) At the point (t0,ξ0), when λ=0 one can obtain

    rankDλ(vλ,θλξ,θλξξ)=3.

    2) If (θ,θξ,θt)=(π,0,0), then there exists a 3parameter family of smooth solutions (uλ,θλ,hλ), depending smoothly on λR3, satisfying (i)-(ii).

    Proof. Let (u,θ,h) be a smooth solution of the semilinear system (27). Given a point (t0,ξ). Taking derivatives to the equation of θ in the semilinear system (27), one has

    tθξ(t,ξ)=(m(m+3)2umuξPξ)(1+cosθ)(m(m+3)2(m+1)um+1P+m2um1)sinθθξm(m1)2um2uξ(1cosθ), (122)
    tθt(t,ξ)=(m(m+3)2umPxPt)(1+cosθ)(m(m+3)2(m+1)um+1P+m2um1)sinθ[(m(m+3)m+1um+12P) (123)
    cos2θ2mum1sin2θ2]+m(m1)2um2Px(1cosθ),
    tθξξ(t,ξ)=(m2(m+3)2um1u2ξ+m(m+3)2umuξξPξξ)(1+cosθ)(m(m+3)umuξ2Pξ+m(m1)2um2uξ)sinθθξ(m(m+3)2(m+1)um+1P+m2um1(cosθθ2ξ+sinθθξξ))m(m1)2um2uξsinθθξm(m1)2um2u2ξm(m1)(m2)2um3u2ξ, (124)
    thξ(t,ξ)=(m(m+3)2(m+1)um+1+m2um1P)sinθhξ)(m(m+3)2(m+1)um+1+m2um1P)cosθθξh(m(m+3)2umuξ+m(m1)2um2uξPξ)sinθh, (125)

    with

    uξ=12sinθh,uξξ=12sinθhξ+12cosθhθξ. (126)

    We consider the families (ˉuλ,ˉθλ,ˉhλ) of perturbations of the initial data as

    ˉuλ=ˉu(ξ)+i=1,2,3λiUi(ξ), (127)
    ˉθλ=ˉθ(ξ)+i=1,2,3λiΘi(ξ), (128)
    ˉhλ=ˉh(ξ)+i=1,2,3λiHi(ξ). (129)

    Together with (27), ((122), (124) and (125) form a complete system. Next, the following lemma will be used to get the rank which we desired.

    Lemma 5.2 ([5]). Consider the following ODE system

    ddtuϵ=g(uϵ),uϵ(0)=u0+ϵ1v1++ϵkvk, (130)

    where uϵ(t):RRn and g is a Lipschitz continuous function. The system is well-posed in [0,T). If the matrix

    Dϵuϵ0=(v1,v2,,vk)Rn×k, (131)

    and the rank of this matrix is

    rank(Dϵuϵ0)=l. (132)

    Then rank(Dϵuϵ(t))=l for any t[0,T).

    Proof Lemma 5.1. By (27), (122) and (124), we achieve an ODE system as follows,

    t(uθhθξθξξ)=(Px(m(m+3)m+1um+12P)cos2θ2mum1sin2θ2(m(m+3)2(m+1)um+1+m2um1P)sinθh(m(m+3)2umuξPξ)(1+cosθ)(m(m+3)2(m+1)um+1P+m2um1)sinθθξm(m1)2um2uξ(1cosθ)(m2(m+3)2um1u2ξ+m(m+3)2umuξξPξξ)(1+cosθ)(m(m+3)umuξ2Pξ+m(m1)2um2uξ)sinθθξ(m(m+3)2(m+1)um+1P+m2um1(cosθθ2ξ+sinθθξξ))m(m1)2um2uξsinθθξm(m1)2um2u2ξm(m1)(m2)2um3u2ξ). (133)

    Then we establish a family of solutions (ˉuλ,ˉvλ,ˉhλ) of perturbations with the initial data given in (127)-(129). Differentiating w.r.t. λ, one has

    t(DλˉuλDλˉθλDλˉhλDλˉθλξDλˉθλξξ)=(Dλgλ1Dλgλ2Dλgλ3Dλgλ4Dλgλ5), (134)

    where gλ1,gλ5 are the perturbation of the right-hand-side of (134). Therefore, we have

    t(DλˉuλDλˉθλDλˉhλDλˉθλξDλˉθλξξ)=(Dugλ1Dθgλ1Dhgλ1Dθξgλ1Dθξξgλ1Dugλ2Dθgλ2Dhgλ2Dθξgλ2Dθξξgλ2Dugλ3Dθgλ3Dhgλ3Dθξgλ3Dθξξgλ3Dugλ4Dθgλ4Dhgλ4Dθξgλ4Dθξξgλ4Dugλ5Dθgλ5Dhgλ5Dθξgλ5Dθξξgλ5)(Dλ1ˉuλDλ2ˉuλDλ3ˉuλDλ1ˉθλDλ2ˉθλDλ3ˉθλDλ1ˉhλDλ2ˉhλDλ3ˉhλDλ1ˉθλξDλ2ˉθλξDλ3ˉθλξDλ1ˉθλξξDλ2ˉθλξξDλ3ˉθλξξ). (135)

    Based on Lemma 5.2, we need to explain the Lipschitz continuity of gλi(i=15). Since the function (u,θ,h) is smooth, we only need to prove the Lipschitz continuity of the nonlocal term of P and Px. In Section 3, we have obtained the boundedness of |Pu|,|Pθ|,|Ph|,|Pxu|,|Pxθ|,|Pxh|. Choosing suitable perturbation Θi(i=1,2,3), when λ=0, we have

    rankDλ(θθξθξξ)=3. (136)

    The system (27) combine with (122), (123) forms a complete system. By choosing suitable perturbation Θi(i=1,2,3) observe that

    rankDλ(θθξθt)=3, (137)

    while λ=0.

    Next, we are going to investigate smooth solutions to the semi-linear system (27), and determine the generic structure of level sets {θ(t,ξ)=π}. We give the key lemma to prove Theorem 1.4.

    Lemma 5.3 ([5]). Given a compact domain

    D:={(t,ξ);0tT,|ξ|M},

    let W be the family of all C2 solutions (u,θ,h) for the semi-linear system (27), with h>0 for all (t,ξ)[0,T]×R+. Moreover, let WW be the subfamily of all solutions (u,θ,h), such that for (t,ξ)D, the value

    (θ,θξ,θξξ)=(π,0,0),(θ,θξ,θt)=(π,0,0) (138)

    cannot be obtained. Then W is a relatively open and dense subset of W, in the topology induced by C2(D).

    Proof of Theorem 1.4. Step 1. For convenience, we denote the space

    M:=C3(R+)H1(R+),

    with the norm

    u0M:=u0C3+u0H1.

    Given a initial data u0M, and we introduce the open ball

    Bδ:={u0M;u0u0M<δ}.

    By the definition of the space of M, it follows that u0(x)0 and u0,x(x)0. Therefore, we choose κ>0 big enough such that u0(x) and u0x(x) are uniformly bounded for |x|>κ. By a standard comparison argument on the domain {(t,x);t[0,T],|x|κ+umL}, we see that the partial derivative ux is uniformly bounded. This implies the singularity of u(t,x) in set [0,T]×R only appears on the compact set Δ:=[0,T]×[rumLT,r+umLT], where uL:=max{u(t,x),(t,x)[0,T]×R}. In (t,x) plane, we take a domain D such that ΔJ(D), where J is a map from (t,ξ) to (t,x(t,ξ)).

    We define the subset ΓBδ as follows: u0Γ if u0Bδ and for the corresponding solution (u,θ,h) of (27), the values (138) are never attained for (t,x)Δ.

    Step 2. We now claim the set Γ is open, in the topology of C3. Take a sequence of initial data (un0)n1 such that the sequence converges to u0. From the definition of Γ, there exists a point (tn,ξn) such that

    (θn,θnξ,θnξξ)(tn,θn)=(π,0,0),(tn,xn(tn,ξn))Δ

    for all n1. Since the domain Δ is compact, we can take a subsequence (tn,ξn), which converges to some point (t,ξ). It follows from continuity that

    (θ,θξ,θξξ)(t,ξ=(π,0,0),(t,x(t,ξ))Δ.

    This denotes u0Γ. Using similar procedure, other case (θ,θξ,θt)=(π,0,0) can be proved. So Γ is open.

    Step 3. We explain that Γ is dense in Bδ. Let u0Bδ, by a small perturbation, we assume u0C. From Lemma 5.3, we construct a sequence of solutions (un,θn,hn) of (3.1), such that

    ⅰ) for every n1, the values in (138) are never attained for any (t,ξ)D.

    ⅱ) The Ck(k>1) norm of the difference satisfies

    limn(unu,θnθ,hnh,xnx)Ck(I)=0,

    for every bounded set I[0,T]×R+. When t=0, the corresponding sequence of initial value satisfies

    limnun0u0Ck[a,b]=0

    for every bounded set [a,b]R+.

    Introducing a cutoff function

    p(x)={1,if|x|r,0,if|x|r+1, (139)

    where rκ+um1LT is large enough. For every n1, let the initial data

    ˜uv0:=pun0+(1p)u0.

    We obtain

    limn˜un0u0M=0.

    Furthermore, choosing r>0 sufficiently large for any (t,x)Δ, we have

    ˜un(t,x)=un(t,x).

    It is obvious that ˜un(t,x) is C2 on the outer domain. Therefore, ˜un(t,x)Γ for every n1 sufficiently large. Thus Γ is dense in Bδ.

    Step 4. Finally, we prove that, for every initial data u0Γ, the solution of (3) is piecewise C2 on the domain [0,T]×R+. By previous argument, we only study the singularity of u on the inner domain Δ. For every point (t0,ξ0)D, two cases can appear.

    Case Ⅰ. θ(t0,ξ0)π. By the coordinate change xξ=hcos2θ2, we know that the map (t,ξ)(t,x) is locally invertible in a neighborhood of (t0,ξ0). Then we conclude that the function u is C2 in a neighborhood of the point (t0,x(t0,ξ0)).

    Case Ⅱ. θ(t0,ξ0)=π. From (138), θt(t0,ξ0)0 or θξ(t0,ξ0)0. Thanks to the implicit function theorem, we derive that the set

    Wθ:={(t,ξ)Δ;θ(t,ξ)=π}

    is the union of finitely many C2 curves.



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