# Blow-Up Analysis for the Periodic Two-Componentμ-Hunter-Saxton System

Authors: Yunxi Guo; Tingjian Xiong
Journal: Mathematical Problems in Engineering
 (2018)

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## Abstract

The two-componentμ-Hunter-Saxton system is considered in the spatially periodic setting. Firstly, a wave-breaking criterion is derived by employing the localization analysis of the transport equation theory. Secondly, several sufficient conditions of the blow-up solutions are established by using the classic method. The results obtained in this paper are new and different from those in previous works.


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## Body

## 1. Introduction

In this article, we will consider the periodic two-componentμ-Hunter-Saxton system derived by Zuo [1](1)μut-utxx=-2μuux+2uxuxx+uuxxx+ρρx-γ1uxxx,t>0,x∈R,ρt=uρx+2γ2ρx,t>0,x∈R,u0,x=u0x,x∈R,ρ0,x=ρ0x,x∈R,ut,x+1=ut,x,t>0,x∈R,ρt,x+1=ρt,x,t>0,x∈R,where u(t,x) and ρ(t,x) are time-dependent functions on the unit circle S=R/Z,  μ(u)=∫Sudx denotes its mean, and γi∈R,  i=1,2. It is shown in [1] that system (1) is an Euler equation with bi-Hamilton structure(2)Γ1=∂xA00∂x,Γ2=Au∂x+∂xAu-γ1∂x3ρ∂x∂xρ2γ2∂x,where A(u)=μ(u)-uxx, and it also can be viewed as a bivariational equation. Moreover, for γi=0,  i=1,2, system (1) has a Lax pair given by(3)ψxx=λAu-λ2ρ2ψ,ψt=u-12λψx-12uxψ,where λ is a spectral parameter (see [1]).In fact, system (1) is a generalization of the generalized Hunter-Saxton equation [2, 3](4)μut-utxx=-2μuux+2uxuxx+uuxxx,which describes the geodesic flow on Ds(S) with the right-invariant metric given at the identity by the inner product 〈u,v〉=μ(u)μ(v)+∫Suxvxdx and models the propagation of weakly nonlinear orientation waves in a massive nematic liquid crystal with external magnetic nematic field and self-interaction. Here, the solution u(t,x) denotes the director field of a nematic liquid crystal. It was observed in [2–4] that the μ-Hunter-Saxton equation is formally integrable and has bi-Hamiltonian structure and infinite hierarchy of conservation laws. Further, the development of singularities in finite time and geometric descriptions of the system from nonstretching invariant curve flows in centroequiaffine geometries and pseudospherical surfaces and affine surfaces are described by Fu et al. [5].Forγi=0,  i=1,2,  μ(u)=0 and replacing t by -t, system (1) reduces to periodic two-component Hunter-Saxton equation(5)utxx+2uxuxx+uuxxx-ρρx=0,t>0,x∈R,ρt+uρx=0,t>0,x∈R,which is a generalization of the well-known Hunter-Saxton equation. It was also viewed as a particular case of the Gurevich-Zybin system [6] pertaining to nonlinear one-dimensional dynamics of dark matter as well as nonlinear ion-acoustic waves (cf. [7] and the references therein). The Hunter-Saxton system is formally integrable, possesses a bi-Hamiltonian structure, and admits peakon solutions [8]. The local well-posedness, global existence, blow-up phenomena, solitary wave solutions, and geometric properties of system (5) were recently discussed in literatures (see [9–17] and the references therein). It is worthwhile to mention that Moon and Liu [13] studied the Cauchy problem for the two-component Hunter-Saxton system in the periodic setting and gave some interesting results.Recently, Liu and Yin [18, 19] investigated the Cauchy problem for system (1). In [18], the local well-posedness and several precise blow-up criteria for the system were obtained. Under the conditions μ0=0 and μ0≠0, the sufficient conditions of blow-up solutions were presented. The global existence for strong solution for system (1) in the Sobolev space Hs(S)×Hs-1(S) with s=2 is also given [18], and in [19], existence of global weak solution is established for the periodic two-component μ-Hunter-Saxton system. The objective of the present paper is to focus mainly on wave-breaking criterion and several sufficient conditions of blow-up solutions.Motivated by the works in [13, 20], in the present paper, the localization analysis in the transport equation theory is employed to derive a new wave-breaking criterion of solutions for the system (1) in the Sobolev space Hs(S)×Hs-1(S) with s≥2. It implies that the wave-breaking criterion is determined only by the slope of the component u of solution definitely. Inspired by the work in [21, 22], we use the Lyapunov function of ∫Sux3dx to deduce several new blow-up results for the periodic two-component μ-Hunter-Saxton system (1), which are different from the ones obtained in [18]. These results obtained in this paper are new and different from those in Liu and Yin’s work [18].The rest of this paper is organized as follows. Section2 states several properties for the periodic two-component μ-Hunter-Saxton system and gives several lemmas. In Section 3, we employ the transport equation theory to prove a wave-breaking criterion in the Soblev space Hs(S)×Hs-1(S) with s≥2. Section 4 is devoted to the study of blow-up mechanism.


## 2. Preliminaries

Lemma 1 (see [20]).
The following estimates hold.
(i) Fors≥0,(6)fgHsR≤CfHsRgL∞R+fL∞RgHsR.
(ii) Fors>0,(7)f∂xgHsR≤CfHs+1RgL∞R+fL∞R∂xgHsR,where C are constants independent of f and g.Lemma 2 (see [20]).
Suppose thats>-d/2. Let ν be a vector filed such that ∇ν belongs to L1([0,T];Hs-1) if s>1+d/2 or to L1([0,T];Hd/2∩L∞) otherwise. Suppose also that f0∈Hs, F∈L1([0,T];Hs), and f∈L∞([0,T];Hs)∩C([0,T];S′) solves the d-dimensional linear transport equations(8)ft+v·∇f=F,ft=0=f0.Then f∈C([0,T];Hs). More precisely, there exists a constant C depending only on s,p, and d, such that the following statements hold:
(1) Ifs≠1+d/2,(9)fHs≤f0Hs+C∫0tFτHsdτ+C∫0tV′τfτHsdτ.Or hence(10)fHs≤eCVtf0Hs+∫0te-CVtFτHsdτwith V(t)=∫0t‖∇ν(τ)‖Hd/2∩L∞dτ if s<1+d/2 and V(t)=∫0t‖∇ν(τ)‖Hs-1dτ else.
(2) Iff=ν, then, for all s>0, the estimates (9) and (10) hold with V(t)=∫0t‖∂xu(τ)‖L∞dτ.Lemma 3 (see [20]).
Let0<σ<1. Suppose that f0∈Hσ,  g∈L1([0,T];Hσ),  ν,∂xν∈L1([0,T];L∞) and f∈L∞([0,T];Hσ)∩C([0,T];S′) solves the 1-dimensional linear transport equation(11)ft+ν∂xf=g,ft=0=f0.Then f∈C([0,T];Hσ). More precisely, there exists a constant C depending only on σ, and such that the following statement holds:(12)fHσ≤f0Hσ+C∫0tgτHσdτ+C∫0tV′τfτHσdτ.Or hence(13)fHσ≤eCVtf0Hσ+∫0tCgτHσdτwith V(t)=∫0t(‖ν(τ)‖L∞+‖∂xν(τ)‖L∞)dτ.Lemma 4 (see [18]).
Givenz0=(u0,ρ0)∈Hs(S)×Hs-1(S),  s≥2, then there exists a maximal T=T(‖z0‖Hs(S)×Hs-1(S))>0 and an unique solution z=(u,ρ) to system (1) such that(14)z=z·,z0∈C0,T;HsS⋂C10,T;Hs-1S.Lemma 5 (see [21]).
Iff∈H1(S) satisfies ∫Sf(x)dx=0, then it has(15)maxx∈S⁡f2x≤112∫Sfx2xdx.Lemma 6 (see [23]).
For everyf(x)∈H1(a,b) periodic and with zero average, i.e., such that ∫abf(x)dx=0, it holds that(16)∫abf2xdx≤b-a2π2∫abf′x2dx,and equality holds if and only if(17)fx=Acos⁡2πxb-a+Bsin⁡2πxb-a.Integrating the first equation of system (1) over the circle S=R/Z and noting the periodicity of u, we have μ(ut)=0. Making use of system (1), we have that ∫S(ux2+ρ2)dx is conserved in time (see [18]). In what follows we denote(18)μ0=μu0=μu=∫Sut,xdxand(19)μ1=∫Sux2t,x+ρ2t,xdx1/2=∫Sux20,x+ρ20,xdx1/2.Then μ0 and μ1 are constants and independent of time t.Notice that∫S(u(t,x)-μ0)dx=μ0-μ0=0. From Lemma 5, we get(20)maxx∈S⁡ut,x-μ02≤112∫Sux2t,xdx≤112∫Sux2t,x+ρ2t,xdx=112∫Sux20,x+ρ20,xdx=112μ12,which implies that the amplitude of wave remains bounded in any time. Namely, we have(21)ut,·L∞S-μ0≤ut,·-μ0L∞S≤36μ1,which results in(22)ut,·L∞S≤μ0+36μ1.In fact, the initial-value problem (1) can be recast in the following:(23)ut-u+γ1ux=A-1∂x2μ0u+12ux2+12ρ2,t>0,x∈R,ρt-u+2γ2ρx=ρux,t>0,x∈R,u0,x=u0x,x∈R,ρ0,x=ρ0x,x∈R,ut,x+1=ut,x,t>0,x∈R,ρt,x+1=ρt,x,t>0,x∈R,where A=μ-∂x2 is an isomorphism between Hs and Hs-2 with the inverse ν=A-1ω given explicitly by(24)νx=x22-x2+1312μω+x-12∫01∫0yωsdsdy-∫0x∫0yωsdsdy+∫01∫0y∫0sωrdrdsdy.Commuting A-1 and ∂x, we get(25)A-1∂xωx=x-12∫01ωx-∫0xωydy+∫01∫0xωydydxand(26)A-1∂x2ωx=-ωx+∫01ωxdx.Note that if f∈L2(S), we have A-1f=(μ-∂x2)-1f=g∗f, where we denotes by ∗ convolution and g is Green’s function of the operator A-1, given by(27)gx=12x-122+2324,and the derivative ofg can be assigned(28)gxx=0,x=0,x-12,x∈0,1.Now, consider the initial-value problem for the Lagrangian flow map:(29)ηt=ut,-η+2γ2,t∈0,T,η0,x=x,x∈R,where u denotes the first component of the solution z=(u,ρ) to system (1). Applying classical results from ordinary differential equations, one can obtain the result.Lemma 7 (see [18]).
Letu∈C([0,T);Hs(R))⋂C1([0,T);Hs-1(R)),  s≥2. Then (29) has a unique solution η∈C1([0,T)×R;R). Moreover, the map η(t,·) is an increasing diffeomorphism of R with(30)ηxt,x=exp⁡-∫0tuxs,-ηs,xds>0,t,x∈0,T×R.Lemma 8 (see [18]).
Letz0=(u0,ρ0)∈Hs(S)×Hs-1(S),  s≥2, and let T>0 be the maximal existence time of the corresponding solution z=(u,ρ) to system (1). Then it has(31)ρt,-ηt,xηxt,x=ρ0-x,t,x∈0,T×R.Lemma 9 (see [18]).
Assume thatz0=(u0,ρ0)∈Hs(S)×Hs-1(S),  s≥2, and let T be the maximal existence time of corresponding solution z=(u,ρ) to system (1) with the initial data z0. Then the corresponding solution blows up in finite time if and only if(32)limsupt→Tsupx∈Suxt,x=+∞orlimsupt→TρxL∞=+∞.


## 3. Wave-Breaking Criterion

Theorem 10.
Letz0=(u0,ρ0)∈Hs(S)×Hs-1(S) with s≥2, and z=(u,ρ) be the corresponding solution to (1). Assume that T>0 is the maximal existence time. Then(33)T<∞⇒∫0TuxL∞Sdτ=∞.Proof.
We shall complete the proof of the theorem by an inductive argument with respect to the indexs. Let us first give upper bound for ‖ρ‖L∞(S).
From (31), we derive(34)ρL∞=ρ0L∞e∫0tuxs,-ηs,xds≤ρ0L∞e∫0tuxs,·L∞ds.Next, we split four steps to finish the proof of Theorem 10.
Step 1. For s∈(2,3), applying Lemma 3 to the second equation of system (23), we have(35)ρHs-2S≤ρ0Hs-2+C∫0tρuxHs-2dτ+C∫0tρHs-2u+2γ2L∞+∂xuL∞dτ.Using (6), we get(36)uxρHs-2S≤CuxHs-2ρL∞+uxL∞ρHs-2=Cu+γ1xHs-2ρL∞+uxL∞ρHs-2.Therefore, it yields(37)ρHs-2S≤ρ0Hs-2+C∫0tu+γ1Hs-1ρL∞dτ+C∫0tρHs-2uL∞+1+uxL∞dτ.
Differentiating once the second equation of system (23) with respect to x, we have(38)∂tρx-u+2γ2∂xρx=2uxρx+uxxρ.Using Lemma 3, we get(39)ρxHs-2S≤ρ0xHs-2+C∫0t2ρxux+ρuxxHs-2dτ+C∫0tρxHs-2u+2γ2L∞+∂xuL∞dτ.From (7), we have(40)uxρxHs-2≤CuxHs-1ρL∞+uxL∞ρxHs-2=Cu+γ1xHs-1ρL∞+uxL∞ρxHs-2and(41)ρ∂xuxHs-2≤CρHs-1uxL∞+ρL∞uxxHs-2=CρHs-1uxL∞+ρL∞u+γ1xxHs-2.Therefore, we have(42)ρxHs-2S≤ρ0xHs-2+C∫0tu+γ1HsρL∞dτ+C∫0tρHs-1uL∞+1+uxL∞dτ.From (37) and (42), it has(43)ρHs-1S≤ρ0Hs-1+C∫0tu+γ1HsρL∞dτ+C∫0tρHs-1uL∞+1+uxL∞dτ.
On the other hand, the first equation of system (23) is equivalent to(44)∂t-u-γ1+-u-γ1∂x-u-γ1=-A-1∂x2μ0u+γ1+12u+γ1x2+12ρ2.
Therefore, using Lemma2, we get from (44)(45)u+γ1HsS≤C∫0tA-1∂x2μ0u+γ1+12u+γ1x2+12ρ2Hsdτ+u0+γ1Hs+C∫0tu+γ1Hs∂xuτL∞dτ.Note that(46)A-∂xfHsS≤A-∂xfL2S+A-∂x2fHs-1S≤gx∗fL2S+-f+∫SfdxHs-1S≤gxL1SfL2S+2fHs-1S≤CfHs-1S,where we have used (27), (28), and Young’s inequality.
Using (6) and (46), one has(47)A-1∂x2μ0u+γ1+12u+γ1x2+12ρ2Hs≤C2μ0u+γ1+12u+γ1x2+12ρ2Hs-1≤Cu+γ1Hs-1+u+γ1xHs-1uxL∞+ρHs-1ρL∞.
Thus, we reach(48)u+γ1HsS≤u0+γ1Hs+C∫0tu+γ1Hs1+uxL∞dτ+C∫0tρτHs-1ρτL∞dτ,which together with (43) reaches(49)u+γ1HsS+ρtHs-1S≤u0+γ1Hs+ρ0Hs-1+C∫0tu+γ1Hs+ρtHs-1×uxL∞+1+uL∞+ρL∞dτ.
Using Gronwall’s inequality, we have(50)ut+γ1HsS+ρtHs-1S≤u0+γ1Hs+ρ0Hs-1eC∫0tuL∞+uxL∞+1+ρL∞dτ.
From (22) and (34), we get(51)ut+γ1HsS+ρtHs-1S≤u0+γ1Hs+ρ0Hs-1eC1t+1exp⁡∫0tuxL∞dτ,where C1=C1(|μ0|,μ1,‖ρ0‖L∞).
Hence, if the maximal existence timeT<∞ satisfies ∫0t‖ux‖L∞dτ<∞, we obtain from (51) that(52)limsupt→Tut+γ1HsS+ρtHs-1S<∞,
which contradicts the assumption on the maximal existence timeT<∞. It completes the Theorem 10 for s∈(2,3).
Step 2. For s∈[2,5/2), applying Lemma 2 to the second equation of system (23), we get(53)ρHs-1S≤ρ0Hs-1+C∫0tuxρHs-1dτ+C∫0tρHs-1∂xu+γ1L∞∩H1/2dτ.From (6), we get(54)ρuxHs-1S≤CuxHs-1ρL∞+ρHs-1uxL∞=Cu+γ1xHs-1ρL∞+ρHs-1uxL∞,which ensures that(55)ρHs-1S≤ρ0Hs-1+C∫0tu+γ1HsρL∞dτ+C∫0tρHs-1uxL∞+∂xu+γ1L∞∩H1/2dτ,
which together with (48) gives rise to(56)ut+γ1Hs+ρtHs-1≤u0+γ1Hs+ρ0Hs-1+C∫0tu+γ1Hs+ρtHs-1×uxL∞+1+u+γ1xH1/2∩L∞+ρL∞dτ.
Using Gronwall’s inequality, we have(57)ut+γ1Hs+ρtHs-1≤u0+γ1Hs+ρ0Hs-1eC∫0t1+u+γ1H3/2+ε+ρL∞dτ,where ε∈(0,1/2) and we used the fact that H1/2+ε↪L∞∩H1/2.
Therefore, using Step 1 and arguing by induction assumption, we get that(58)u+γ1H3/2+ε+ρL∞is uniformly bounded. We obtain from (57) that(59)limsupt→Tut+γ1Hs+ρtHs-1<∞,
which contradicts the assumption on the maximal existence timeT<∞. It completes the Theorem 10 for s∈[2,5/2).
Step 3. For s=k∈N,  k≥3. Differentiating k-2 times the second equation of system (23) with respect to x, we obtain(60)∂t∂xk-2ρ-u+2γ2∂x∂xk-2ρ+∑l1+l2=k-3,l1,l2≥0Cl1,l2∂xl1+1u∂xl2+1ρ+ρ∂x∂xk-2u=0.Using Lemma 2, we get from (60) that(61)∂xk-2ρH1≤∂xk-2ρ0H1+C∫0t∂xk-2ρH1∂xuH1/2∩L∞dτ+C∫0t∑l1+l2=k-3,l1,l2≥0Cl1,l2∂xl1+1u∂xl2+1ρ+ρ∂xk-1uH1dτ.Since H1 is an algebra, we have(62)ρ∂xk-1uH1≤CρH1∂xk-1uH1≤CρH1u+γ1Hsand(63)∑l1+l2=k-3,l1,l2≥0Cl1,l2∂xl1+1u∂xl2+1ρH1≤CρHs-1u+γ1Hs-1.Therefore, we have(64)∂xk-2ρH1≤∂xk-2ρ0H1+C∫0tu+γ1Hs+ρHs-1×u+γ1Hs-1+ρH1dτ.Applying Lemma 3 to the second equation of system (23) yields(65)ρHσS≤ρ0Hσ+C∫0tu+γ1Hσ+1ρL∞dτ+C∫0tρHσuL∞+∂xuL∞+1dτ,where we used Lemma 1.
From (64) and (65), it yields(66)ρHs-1≤ρ0Hs-1+C∫0tu+γ1Hs+ρHs-1×u+γ1Hs-1+ρH1dτ,where we used the Gagliardo-Nirenberg inequality ‖ρ‖Hs-1≤C(‖ρ‖Hσ+‖∂xk-2ρ‖H1) for σ∈(0,1).
Using (48) implies that(67)ut+γ1Hs+ρtHs-1≤Cu0+γ1Hs+ρ0Hs-1+C∫0tu+γ1Hs+ρtHs-1×u+γ1Hs-1+ρH1+1dτ.Using Gronwall’s inequality, we get(68)ut+γ1Hs+ρtHs-1≤Cu0+γ1Hs+ρ0Hs-1eC∫0tu+γ1Hs-1+ρH1+1dτ.Therefore, if the maximal existence time T<∞ satisfies ∫0T‖ux‖L∞dτ<∞, using Step 2, we get that(69)u+γ1Hs-1+ρH1is uniformly bounded by the induction assumption. From (66), we get(70)limsupt→Tut+γ1Hs+ρtHs-1<∞,which contradicts the assumption that T<∞ is the maximal existence time. This completes the proof of Theorem 10 for s=k∈N and k≥3.
Step 4. For s∈(k,k+1),k∈N and k≥3. Differentiating k-1 times the second equation of system (23) with respect to x, we obtain(71)∂t∂xk-1ρ-u+2γ2∂x∂xk-1ρ+∑l1+l2=k-2,l1,l2≥0Cl1,l2∂xl1+1u∂xl2+1ρ+ρ∂x∂xk-1u=0.Applying Lemma 3 to (71) and noting s-k∈(0,1), we get(72)∂xk-1ρHs-k≤C∫0t∑l1+l2=k-2,l1,l2≥0Cl1,l2∂xl1+1u∂xl2+1ρ+ρ∂xkuHs-kdτ+C∫0t∂xk-1ρHs-kuL∞+∂xuL∞+1dτ+∂xk-1ρ0Hs-k.For each ε∈(0,1/2), using (7) and the fact that H1/2+ε(R)↪L∞(R), we have(73)ρ∂xkuHs-k≤CρHs-k+1∂xk-1uL∞+∂xkuHs-kρL∞≤CρHs-k+1u+γ1Hk-1/2+ε+u+γ1Hs-1ρL∞and(74)∑l1+l2=k-2,l1,l2≥0Cl1,l2∂xl1+1u∂xl2+1ρHs-k≤C∑l1+l2=k-2,l1,l2≥0Cl1,l2∂xl1+1uHs-k+1∂xl2ρL∞+∂xl1+1uL∞∂xl2+1ρHs-k≤Cu+γ1HsρHk-3/2+ε+u+γ1Hk-1/2+ερHs-1,where we used Lemma 1.
Therefore, from (72), (73), and (74), we get(75)∂xk-1ρHs-k≤C∫0tu+γ1Hs+ρHs-1u+γ1Hk-1/2+ε+ρHk-3/2+ε+1dτ+∂xk-1ρ0Hs-k,which together with (48) and (37) (where s-2 is replaced by s-k) gives rise to(76)ut+γ1Hs+ρtHs-1≤Cu0+γ1Hs+ρ0Hs-1+C∫0tu+γ1Hs+ρtHs-1×u+γ1Hk-1/2+ε+ρHk-3/2+ε+1dτ,from which we have(77)ut+γ1Hs+ρtHs-1≤Cu0+γ1Hs+ρ0Hs-1eC∫0tu+γ1Hk-1/2+ε+ρHk-3/2+ε+1dτ.Noting that k-1/2+ε<k,  k-3/2+ε<k-1 and k≥3. Therefore, if the maximal existence time T<∞ satisfies ∫0T‖ux‖L∞dτ<∞, using Step 3, we get that(78)u+γ1Hk-1/2+ε+ρHk-3/2+εis uniformly bounded by the induction assumption. From (77), we get(79)limsupt→Tut+γ1Hs+ρtHs-1<∞,which contradicts the assumption that T<∞ is the maximal existence time. This completes the proof of Theorem 10 for s=k∈N and k≥3.
Therefore, from Steps 1–4, we complete the proof of Theorem10.


## 4. Wave-Breaking Phenomena

In this section, we will present several blow-up results by using the Lyapunov function of∫Sux3dx (see [21]). Let z0=(u0,ρ0)∈Hs(S)×Hs-1(S) with s≥2 and let T be the maximal existence time of the corresponding solution to system (1) with the initial data z0.Now, we give the first blow-up result.Theorem 11.
Letz0=(u0,ρ0)∈Hs(S)×Hs-1(S) with s≥2 and N be a sufficient large positive real number and let T be the maximal existence time of the corresponding solution to system (1) with the initial data z0 and ∫Sρ0(-x)=0.
(i) Ifμ02<π2(μ12-N2), then the corresponding solution to system (1) must blow up in finite time T with(80)0≤T≤infα∈I⁡61-6αμ0∫Su0x31/3,where I=(|μ0|/π2(μ12-N2),1/6|μ0|), such that(81)limsupt→T-supx∈Suxt,x=+∞.(ii) If ∫Su0x3dx>(3|μ0|/2π2α+3N2)μ14/1-6α|μ0| then the corresponding solution to system (1) must blow up in finite time T with(82)0<T<μ121-6αμ0Kln⁡∫Su0x3dx+K∫Su0x3dx-K,where 0<α<1/6|μ0| and K=(3|μ0|/2π2α+3N2)μ14/1-6α|μ0|.Proof.
If the statement is not true, then from Lemma9 it implies that there exists a large enough real number N>0, such that ux(t,x)<N and ‖ρx(t,x)‖L∞≤N for all t∈[0,T). From the assumption ∫Sρ0(x)dx=0, we know(83)∫Sρt,xdx=∫Sρ0xdx=0.It then follows that, for any t∈[0,T), ρ(t,x) has a zero point ξ. Thus we have(84)ρt,x=∫ξxρxt,sds,x∈ξ,ξ+1,which results in(85)ρt,x=∫ξxρxt,sds≤N.(i) Differentiating the first equation of system (23) with respect to x yields(86)utx-ux2-u+γ1uxx=A-1∂x22μ0u+12ux2+12ρ2.Using (26), we get(87)utx=12ux2+u+γ1uxx-2μ0u+2μ02-12ρ2+12μ12.Multiplying (87) by 3ux2 and integrating on S with respect to x, we obtain for any t∈[0,T) that(88)ddt∫Sux3dx=∫S3ux2utxdx=32∫Sux4dx+∫S3ux2u+γ1uxxdx-6μ0∫Sux2udx+6μ02∫Sux2dx-32∫Sρ2ux2dx+32μ12∫Sux2dx=12∫Sux4dx-6μ0∫Su-μ0ux2dx-32∫Sρ2ux2dx+32μ12∫Sux2dx.On the other hand, it follows from Lemma 6 for any α>0 that(89)μ0∫Su-μ0ux2dx≤μ0∫Su-μ0ux2dx≤μ0∫Su-μ02dx1/2∫Sux4dx1/2≤μ0α2∫Sux4dx+12αμ0∫Su-μ02dx≤α2μ0∫Sux4dx+18π2αμ0∫Sux2dx.Therefore, we deduce that(90)ddt∫Sux3dx≥12-3αμ0∫Sux4dx+32μ12-3μ04π2α∫Sux2dx-32∫Sρ2ux2dx≥12-3αμ0∫Sux4dx+32μ12-3μ04π2α-32N2∫Sux2dx,where we have used Lemma 8 and (19). From the assumption of the theorem, we know that |μ0|/[π2(μ12-N2)]<1/(6|μ0|). Let α>0 satisfy(91)0≤μ0π2μ12-N2<α<16μ0.It implies that 1/2-3α|μ0|>0 and 3/2μ12-3|μ0|/4π2α-3/2N2>0. Define β1 and β2 by(92)β1=12-3αμ0>0,β2=32μ12-3μ04π2α-32N2∫Sux2dx>0.
We get(93)ddt∫Sux3dx≥β1∫Sux4dx+β2.
LettingV(t)=∫Sux3dx with t∈[0,T) derives(94)ddtVt≥β1Vt4/3,t∈t0,T,where we used the inequality (∫Sux3dx)4/3<∫Sux4dx, which leads to(95)Vt≥3V1/303-β1tV1/303→+∞,ast→3β1V1/30.On the other hand, we have(96)∫Sux3dx≤supx∈Suxt,x∫Sux2dx≤μ12supx∈Suxt,x,which implies that 0<T<3/β1(V0)1/3 satisfying(97)limsupt→T-supx∈Suxt,x=+∞.
( i i ) From (90), we get(98)ddt∫Sux3dx≥12-3αμ0∫Sux4dx-3μ04π2α+32N2μ12.Let α>0 satisfy(99)0<α<16μ0.it implies that 1/2-3α|μ0|>0. Define α1 and α2 by(100)α1=1/2-3αμ0μ12>0,α2=3μ04π2α+32N2μ12>0and let ∫Sux3dx=V(t); we have(101)ddtVt≥α1V2t-α2.Let α2/α1=K. If V(0)>K, then V(t)>K. Therefore, we solve (101) to obtain(102)1-V0-KV0+Ke2α1Kt>2KVt+K>0.It follows that there exists T satisfying(103)0<T<12α1Kln⁡V0+KV0-K,such that limt→TV(t)=+∞. This contradicts the assumption ux>N for all (t,x)∈[0,T)×S. From Lemma 9 we obtain that the solution z blows up in finite time.
It completes the proof of Theorem11.Before giving the second blow-up result, we firstly claim that, under the conditionγ1=2γ2, μ2=∫S(μ0u2+1/2uux2+1/2uρ2)dx is conserved in time. Indeed, like two-component Camassa-Holm equation (see [24–26]), we can easy obtain(104)ddt∫Sμ0u2+12uux2+12uρ2dx=∫S2μ0u+12ux2+12ρ2ut+uuxutx+uρρtdx.In view of (23) and (87), we get after some integration by parts(105)ddt∫Sμ0u2+12uux2+12uρ2dx=∫S2μ0u+12ux2+12ρ2Gxdx,whereG(x)=A-1(2μ0u+1/2ux2+1/2ρ2).Observe that(106)∫SGdx-Gxx=2μ0u+12ux2+12ρ2.Therefore, we have(107)ddt∫Sμ0u2+12uux2+12uρ2dx=0.Theorem 12.
Letz0=(u0,ρ0)∈Hs×Hs-1 with s≥2 and N be a sufficiently large positive real number and let T be maximal existence time of corresponding solution to system (1) with the initial data z0 and ∫Sρ0(x)dx=0.
(i) Assume thatθ=12μ0μ2+6μ02μ12+3|μ0|μ13+3/2N2μ12<0. Then the corresponding solution to system (1) must blow up in finite time T with(108)0≤T≤6-1+∫Su0x3dx12μ0μ2+6μ02μ12+3μ0μ13+3/2N2μ12.
(ii) Assume thatϕ=12μ0μ2+6μ02μ12+3|μ0|μ13+3/2N2μ12>0. Then the corresponding solution to system (1) must blow up in finite time T with(109)0≤T≤μ12Kln⁡∫Su0x3dx+K∫Su0x3dx-K,whereK=2μ12(12μ0μ2+6μ02μ12+3|μ0|μ13+3/2N2μ12).Proof.
Applying a simple density argument, we only need to consider the cases=3. From (88), one has(110)ddt∫Sux3dx=∫S3ux2utxdx=32∫Sux4dx+∫S3ux2u+γ1uxxdx-6μ0∫Sux2udx+6μ02∫Sux2dx-32∫Sρ2ux2dx+32μ12∫Sux2dx=12∫Sux4dx-6μ0∫Suux2dx+6μ02∫Sux2dx-32∫Sρ2ux2dx+32μ12∫Sux2dx=12∫Sux4dx-12μ0μ2+12μ02∫Su2dx+6μ0∫Suρ2dx+6μ02∫Sux2dx-32∫Sρ2ux2dx+32μ12∫Sux2dx.
Applying the assumption of the theorem and Hölder inequality give rise to(111)ddt∫Sux3dx≥12∫Sux3dx4/3-12μ0μ2-6μ02μ12-3μ0μ13-32N2μ12.Define V(t)=∫Sux3dx,  t∈[0,T). It is clear that(112)ddtVt≥12Vt4/3+θ,where θ=-(12μ0μ2+6μ02μ12+3|μ0|μ13+3/2N2μ12)>0. This implies that V(t) increase strictly in [0,T). Let t0=1+|V(0)|/θ; we assume t0<T. Otherwise, T≤t0<∞ and the theorem is proved.
From (112), we have(113)ddtVt≥θ.Now, integrating (113) over [0,t0) yields(114)Vt0≥V0+θt0≥-V0+θt0=1.
Using again (112), one has(115)ddtVt≥12Vt4/3,t∈t0,T,which leads to(116)Vt≥66-t-t03→+∞,ast→6+t0.On the other hand, we have(117)∫Sux3dx≤supx∈S⁡uxt,x∫Sux2dx≤μ12supx∈S⁡uxt,x,which implies that 0<T<6+t0 satisfying(118)limsupt→T-supx∈S⁡uxt,x=+∞.
(ii) Applying the assumption of the theorem, (110) and Hölder inequality give rise to(119)ddt∫Sux3dx≥12μ12∫Sux3dx2-12μ0μ2-6μ02μ12-3μ0μ13-32N2μ12.
DefineV(t)=∫Sux3dx,  t∈[0,T). It is clear that(120)ddtVt≥12μ12Vt2-ϕ,where ϕ=12μ0μ2+6μ02μ12+3|μ0|μ13+3/2N2μ12>0. Let 2μ12ϕ=K. If V(0)>K, then V(t)>K. Therefore, we solve (101) to obtain(121)1-V0-KV0+KeKt/μ12>2KVt+K>0.It follows that there exists T satisfying(122)0<T<μ12Kln⁡V0+KV0-K,such that limt→TV(t)=+∞. This contradicts the assumption ux>N for all (t,x)∈[0,T)×S. From Lemma 9 we obtain that the solution z blows up in finite time. It finishes the proof of the Theorem 12.Remark 13.
If we letρ0(x)=0, then we can obtain ρ(t,x)=0 easily. Then system (1) is degenerated into μ-version Camassa-Holm equation under some transform. The blow-up results related to μ-version Camassa-Holm equation; the reader is referred to [5].Remark 14.
It is worthwhile to mention that, comparing with the results in [18], our blow-up results are new and quite different from the ones in [18].



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