global conservative solutions of the two-component -hunter

Research ArticleGlobal Conservative Solutions of the Two-Componentμ-HunterndashSaxton System

Xiayang Shi 1 Jingjing Liu2 and Hongyang Zhang3

1College of Software Engineering Zhengzhou University of Light Industry 450002 Zhengzhou China2Department of Mathematics and Information Science Zhengzhou University of Light Industry 450002 Zhengzhou China3College of Water Conservancy North China University of Water Resources and Electric Power Zhengzhou 450046 China

Correspondence should be addressed to Xiayang Shi aryang123163com

Received 22 June 2021 Accepted 20 September 2021 Published 16 October 2021

Academic Editor Mostafa M A Khater

Copyright copy 2021 Xiayang Shi et al (is is an open access article distributed under the Creative Commons Attribution Licensewhich permits unrestricted use distribution and reproduction in any medium provided the original work is properly cited

In this paper we establish global conservative solutions of the two-component μ-HunterndashSaxton system by the methods de-veloped in ldquoA Bressan A Constantin Global Conservative Solutions of the Camassa-Holm Equation Arch Ration Mech Anal183 (2) 215ndash239 (2007)rdquo and ldquoH Holden X Raynaud Periodic Conservative Solutions of the Camassa-Holm Equation Ann InstFourier (Grenoble) 58(3) 945ndash988 (2008)rdquo

1 Introduction

If a solution remains bounded pointwise and its slope be-comes unbounded in finite time we say this solution breaksdown in finite time Blow-up is a highly interesting propertyexhibited in a lot of nonlinear dispersive-wave equationseg the CamassandashHolm equation [1ndash3]

ut minus utxx + 3uux minus 2uxuxx minus uuxxx 0 (1)

(e μ-HunterndashSaxton equation [4] is as follows

utxx minus 2μ(u)ux + 2uxuxx + uuxxx 0 (2)

(e HunterndashSaxton equation [5] is as follows

utxx + 2uxuxx + uuxxx 0 (3)

(en what will happen after wave breaking is an interestingproblem which has received considerable attention in the pastdecade Several methods have been developed to study this issueincluding the vanishing viscosity approach initial data molli-fication and coordinate transformation [6ndash15] Among all themethods two of them will be used in this paper One wasintroduced by Bressan and Constantin in [6] and the other wasproposed by Holden and Raynaud in [10] Both methodsconverted the problem to solving a corresponding semilinear

system by the application of new variables (eir mutual dif-ference lies in the fact that Holden and Raynaud used a differentset of variables and constructed a bijectivemap between Eulerianand Lagrangian coordinates for (CH) In addition if the H1

energy 1113938(u2 + u2x) dx remains constant except for the exact

time of break down we call the conservative solutions if 1113938(u2 +

u2x) dx decreases to zero at the breakdown time we call the

solutions dissipativeIn this paper we discuss the conservative solutions of the

following periodic two-component μ-HunterndashSaxton sys-tem [16]

At + 2Aux + uAx + ρρx 0

ρt +(uρ)x 0

u(0 x) u0(x)

ρ(0 x) ρ0(x)

u(t x + 1) u(t x)

ρ(t x + 1) ρ(t x)

⎧⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎨

⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎩

(4)

where A(t x) μ(u) minus uxx μ(u) 111393810 u(t x) dx where

tgt 0 is the time vector and x isin R is a space vector (issystem is of a bi-Hamilton structure and it can also beviewed as a bivariational equation set (erefore equation(4) can be rewritten as

HindawiMathematical Problems in EngineeringVolume 2021 Article ID 1878991 10 pageshttpsdoiorg10115520211878991

u

ρ⎛⎝ ⎞⎠

t

J1

δH2

δu

δH2

δρ

⎛⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎝

⎞⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎠ J2

δH1

δu

δH1

δρ

⎛⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎝

⎞⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎠ (5)

where

H1 12

11139461

0uf + ρ21113872 1113873 dx H2 1113946

1

0μ(f)f

2+12

ff2x +

12

fρ21113874 1113875 dx

(6)

Also

J1 zxA 0

0 zx

1113888 1113889 J2 uzx + zxu ρzx

zxρ 01113888 1113889 (7)

with f Aminus 1u (μ minus z2x)minus 1u It is a generalization of μHSequation (where ρ 0) If A(t x) u minus uxx then equation (4)becomes a two-component CH system which has been studiedin [17ndash21] (e system exhibits local well-posedness and it hasfinite-time blowup solutions and global strong solutions in timeGlobal conservative weak solutions can be obtained by co-ordinate transformation in [19 20] and admissible weak so-lutions can be found in [21] by mollifying the initial data IfA(t x) minus uxx then equation (4) turns to a two-componentHS system which has been looked into in [22 23] Hence wecan say 2-μHS system equation (4) lies in an intermediatebetween 2-CH and 2-HS systems (e Cauchy problem forequation (4) has been studied extensively in [24ndash27] In additionresearch studies have shown that this system is locally well-posed[26] for (u0 ρ0) isin Hs times Hsminus 1 sgt (32) besides its globalclassical solutions [26] and finite-time blowup solutions [25 27]

have also been found and its geometric background has beencomprehensively given by Escher in [24] (e global admissibleweak solution of system equation (4) has been obtained in [28]by mollifying the initial date Here we will follow previousresearch studies [6 10 14] and demonstrate the existence ofglobal conservativeweak solutionsHowever compared to the 2-CH system the existing μ(u) in the 2-μHS system brings somedifficulties to the calculation of equation (28) Fortunately weovercame it Because A(t x) minus uxx in the 2-HS system the 2-μHS system is structurally more complex than the 2-HS systemIn [23] the author gave the specific expression ofy U H and r (one can find them in equations (14)ndash(17))which is very helpful to the proof of themain theoremHoweverthis practice is almost impossible for the 2-μHS system so ourproof is a little bitmore difficult To sumup althoughwe refer tothe methods in [6 10 14] our results and the proofs are quitedifferent

Our paper is organized as follows In Section 2 we refor-mulate system equation (4) and give an equivalent system inLagrangian coordinates We also try to illustrate the existenceand uniqueness of solutions to the equivalent system to Banachcontraction arguments In Section 3 we establish maps betweenLagrangian and Eulerian coordinates which can connect con-servative weak solutions of equation (4) and solutions ofa semilinear system together In Section 4 we give the existenceof global conservative weak solutions to equation (4)

2 Preliminaries

Firstly we reformulate system equation (4) AssumeA(t x) μ(u) minus uxx in equation (4) we have

μ ut( 1113857 minus utxx minus uuxxx + 2μ(u)ux minus 2uxuxx + ρρx 0 tgt 0 x isin R

ρt + uρx + uxρ 0 tgt 0 x isin R

u(0 x) u0(x) x isin R

ρ(0 x) ρ0(x) x isin R

u(t x + 1) u(t x) tgt 0 x isin R

ρ(t x + 1) ρ(t x) tgt 0 x isin R

⎧⎪⎪⎪⎪⎪⎪⎪⎪⎨

⎪⎪⎪⎪⎪⎪⎪⎪⎩

(8)

which is equivalent to

ut + uux + Aminus 1

zx 2μ(u)u +12u2x +

12ρ21113874 1113875 0 tgt 0 x isin R






⎧⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎨

⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎩

(9)

2 Mathematical Problems in Engineering

where A μ minus z2x and Aminus 1f (μ minus z2x)minus 1f glowastf for allf isin L2 with g(x) (12)(x2 minus |x|) + (1312) By differen-tiating the first equation in equation (9) we obtain

utx + uuxx + u2x + glowast 2μ(u)uxx + u

2xx + uxuxxx + ρ2x + ρρxx1113872 1113873 0 (10)

Based on the second equation in equations (9) and (10)a direct computation implies

u2x + ρ21113872 1113873

t+ u u

2x + ρ21113872 11138731113872 1113873

x u minus μ u

2x + ρ21113872 1113873 minus 4μ(u)

2+ 2μ(u)u1113872 11138731113872 1113873

x (11)

For smooth solutions we combine the first equation inequations (8) and (11) and we find the following conser-vation laws

μ(u) 11139461

0u(t x) dx 1113946

1

0u0(x) dx ≔ μ

11139461

0u2x + ρ21113872 1113873 dx 1113946

1

0u20x + ρ201113872 1113873 dx ≔ e

(12)

Since system equation (4) is periodic with period 1 wedefine a space

V1 f isin H1loc(R) ∣ f(ξ + 1) f(ξ) + 1)1113966 1113967 (13)

However V1 is not a Banach space We definey R⟶ V1 t⟶ y(t middot) as the solution of

yt(t ξ) u(t y(t ξ)) (14)

And then we define

U(t ξ) u(t y(t ξ)) (15)

H(t ξ) 1113946y(tξ)

y(t0)u2x + ρ21113872 1113873 dx (16)

r(t ξ) ρ(t y(t ξ))yξ(t ξ) (17)

Taking the derivative of both sides of equations (15)ndash(17)with respect to t and using equation (11) we can obtaina result Combining this result with equation (14) we havethe following semilinear system of (y U H r)

yt(t ξ) U

Ut(t ξ) minus Q

Ht(t ξ) U minus e minus 4μ2 + 2μU1113872 11138731113966 1113967(t y(t ξ)) minus U minus e minus 4μ2 + 2μU1113872 11138731113966 1113967(t y(t 0))

≔ U minus e minus 4μ2 + 2μU1113872 11138731113966 1113967 ∣y(t ξ)

y(t 0)

μt et 0

rt(t ξ) 0

⎧⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎨

⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎩

(18)

where

Q μ minus z2x1113872 1113873

minus 1zx 2μ(u)u +

12u2x +

12ρ21113874 1113875(t y(t ξ))

glowast zx 2μ(u)u +12u2x +

12ρ21113874 1113875(t y(t ξ))

11139461

0(y(t ξ) minus z) minus

12sgn(y(t ξ) minus z)1113874 1113875 2μ(u)u +

12u2x +

12ρ21113874 1113875(t z) dz

(19)

Mathematical Problems in Engineering 3

After we define a new variable z y(t ξprime) we have

Q 11139461

0y(t ξ) minus y t ξprime( 1113857( 1113857 minus

12sgn y(t ξ) minus y t ξprime( 1113857( 11138571113874 1113875 2μU t ξprime( 1113857yξprime +

12Hξprime1113874 1113875 dξprime (20)

Here we will make some explanation about H Since(u ρ) is periodic with period 1 and y isin V1 we can obtainH(t ξ + 1) minus H(t ξ) H(t 1) minus H(t 0) By equation (11)

a direct computation implies that(ddt)(H(t 1) minus H(t 0)) 0 which follows thatH(t 1) minus H(t 0) H(0 1) minus H(0 0) Define space V as

V f isin H1loc(R) ∣ there exists α isin R such thatf(ξ + 1) f(ξ) + α for all ξ isin R1113966 1113967 (21)

with norm fV fH1([01]) as a Banach space [10] andH isin V Moreover we introduce the Banach space

H1per f isin H

1loc(R) ∣ f(ξ + 1) f(ξ) for all ξ isin R1113966 1113967

L2per f isin H


(22)

with the norm fH1per

fH1([01]) andfL2

per fL2([01])

Next we will give the existence and uniqueness of so-lution to equation (18) based on the Banach contractionargument However two important issues are noteworthyabout y and H One is that the space V1 in which y belongsto is not a Banach space and the other is that H is notperiodic with period 1 Hence we let ζ y minus Id and σ

H minus eId to be transient in order to use Banach contractionargument And equation (18) becomes

ζt(t ξ) U

Ut(t ξ) minus Q

σt(t ξ) U minus e minus 4μ2 + 2μU1113872 11138731113966 1113967 ∣y(t ξ)

y(t 0)

μt et 0

rt(t ξ) 0

⎧⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎨

⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎩

(23)

Since the first four equations in both equations (18) and(23) are independent of r and r is preserved with respect totime t by following closely the proofs of(eorems 3 and 4 in[14] we have the following results Let E H1

per times H1per times

H1per times R times R times L2

per be equipped with the norm

(ζ U σ μ e r)E ζH1per

+UH1per

+σH1per

+|μ| +|e| +rL2per

(24)

Theorem 1 (local existence and uniqueness) For initialdata X0 (ζ0 U0 σ0 μ0 e0 r0) isin E there exists a time T

T(X0E)gt 0 such that system equation (23) has a uniquesolution in C1([0 T] E)

In order to obtain global existence and uniqueness weneed to make more hypotheses on initial data so let G bea space consisting of all (ζ U σ μ e r) in Ecap (W1infin

per )3 times

R2 times Linfinper such that

yξ ge 0 Hξ ge 0 yξ + Hξ gt 0 ae (25)

yξHξ U2ξ + r

2 ae (26)

11139461

0Uyx dx μ ae (27)

Theorem 2 (global existence and uniqueness) For initialdata X0 (y0 U0 H0 μ0 e0 r0) isin G system equation (18)has a unique global solution X(t) isin C1(R+ E) MoreoverX(t) isin G is satisfied at all times Furthermore the mapS G times R+⟶ G defined as St(X0) X(t) which is a con-tinuous semigroup

Proof (e proof follows the same clue as(eorem 4 in [14]so we prove only equation (26) here Firstly by equation(18) we have


ytξ Uξ

Utξ minus μ minus z2x1113872 1113873

minus 1z2x 2μ(u)u +

12u2x +

12ρ21113874 1113875(t y(t ξ))yξ

μ minus z2x1113872 1113873

minus 1μ minus z

2x1113872 1113873 minus 2μ2 minus

12

e + 2μ(u)u +12u2x +

12ρ21113874 1113875(t y(t ξ))yξ

minus 2μ2 minus12

e + 2μ(u)U1113874 1113875yξ +12Hξ

Htξ minus e minus 4μ21113872 1113873Uξ + 4μUUξ

⎧⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎨

⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎩

(28)

It satisfies that

yξHξ1113872 1113873t

ytξHξ + yξHtξ UξHξ + yξ minus e minus 4μ2 + 4μU1113872 1113873Uξ

U2ξ + r

21113872 1113873

t 2UξUtξ minus e minus 4μ2 + 4μU1113872 1113873Uξyξ + UξHξ

(29)

(us (yξHξ minus U2ξ minus r2)t 0 Since initial data X0 isin G

equation (26) can be obtained

3 Bijective Maps between Eulerian andLagrangian Coordinates

Since the energy is concentrated on the zero measure setswhen wave breaking occurs we must consider a periodic

positive Radon measure Consequently we make the fol-lowing definition

Definition 1 D is the set of all triplets (u ρ η) such thatu isin H1

per ρ isin L2per and η is a positive and periodic Radon

measure whose absolute continuous part isηac (u2

x + ρ2) dxNote that the variables in Eulerian space are (u ρ η) and

those in the Lagrangian space are (y U H r) As we preferto get one-to-one correspondence between Eulerian andLagrangian coordinates we define equivalence of the latterby establishing an equivalence class map onG Let us start byrelabeling invariance first

Let

G f isinW1infinloc ∣ f is invertible f(x + 1) f(x) + 1 forx isin R andf minus Id f

minus 1minus Id isinW

1infinper1113966 1113967 (30)

and

Gs f isin G ∣ f minus IdW1infin + fminus 1

minus Id

W1infin le s1113966 1113967 (31)

with sge 1 As is described in many references (for exampleLemma 32 in [29]) if f isin Gs then (11 + s)lefx le 1 + s aeand if f isinW1infin

loc f is invertible and f(x + 1) f(x) + 1 forx isin R and there is a cge 1 such that (1c)lefx le c ae andthen f isin Gs for some s is dependent only on c

Define subsets F and Fs of G as

F (y U H μ e r) isin G1

1 + e(y + H) isin G1113882 1113883 (32)

and

Fs (y U H μ e r) isin G1

1 + e(y + H) isin Gs1113882 1113883 (33)

Let 1113957G G times R be a group with its operation defined by(f1 c1)(f2 c2) (f2degf1 c1 + c2) (e mapΦ 1113957G times F⟶ F defined as

Φ((f c)(y U H μ e r)) y deg f U deg f Hdeg f + c μ e r deg ffx( 1113857

(34)

is an equivalence class map on 1113957G Based on the proof of(eorem 42 in [14] we have the following theorem bya slight modification

Theorem 3 Define 1113957St on F 1113957G as 1113957St([X]) [StX] then 1113957St

generates a continuous semigroup

Theorem 4 For any (u ρ η) isin D let

y(ξ) sup y ∣Fη(y) + y

1 + elt ξ1113896 1113897

H(ξ) (1 + e)ξ minus y(ξ)

U(ξ) u(t y(t ξ))

r(ξ) ρ(t y(t ξ))yξ

μ 11139461

0u(t x) dx

e η([0 1])

(35)

where


Fη(x)

η([0 x)) if xgt 0

0 if x 0

minus η([x 0)) if xlt 0

⎧⎪⎪⎨

⎪⎪⎩(36)

en (y U H μ e r) isin F0 We denoteL D⟶ (F 1113957G) and let L(u ρ η) isin (F 1113957G) denote theequivalence class of (y U H μ e r)

Before giving the proof of eorem 4 we give a criticallemma Define a set

B x isin R ∣ limε⟶0

12ε

η(x minus ε x + ε) u2x + ρ21113882 1113883 (37)

Note that (u2x + ρ2) dx here is the absolute continuous

part of η By Besicovitchrsquos derivation theorem one can obtainmeas (Bc) 0

Lemma 1 For ξ isin yminus 1(B) we have

yξ(ξ) u2x(y(ξ)) + ρ2(y(ξ))1113872 1113873 + yξ 1 + e (38)

Proof Firstly we claim that for all i isin N there isa 0lt εlt (1i) such that x minus ε and x + ε are in supp(ηs)

cwhere ηs is the singular part of Radon measure η and itssupport supp (ηs) is a point set with a countable number ofelements If not then there exists i isin N such that for any0lt εlt (1i) (x minus ε) isin supp (ηs) or (x + ε) isin supp (ηs) andthen for any z isin (x minus ε x + ε)supp (ηs)(2x minus z) isin supp (ηs) Consequently we may construct aninjection between (x minus (1i) x + (1i))supp (ηs) andsupp (ηs) which is rather impossible because (x minus (1i) x +

(1i))supp (ηs) is uncountable and supp (ηs) is countable(en we can construct sequences y(ξi) and y(ξi

prime) suchthat

12

y ξi( 1113857 + y ξiprime( 1113857( 1113857 y(ξ) andy ξi( 1113857 minus y ξi

prime( 1113857le1i (39)

By the definition of Fη we have

η y ξi( 1113857 y ξiprime( 11138571113858 1113857( 1113857 + y ξi

prime( 1113857 minus y ξi( 1113857 (1 + e) ξiprime minus ξi( 1113857 (40)

Dividing equation (40) by ξiprime minus ξi and taking i⟶infin we

obtain equation (38)

Proof of eorem 4 By Lemma 1 and slight modifications of(eorem 43 in [14] we will establish the map from La-grangian coordinates to Eulerian ones which is a general-ization of(eorem 47 in [14] We only state the results hereas this proof and that of (eorem 47 in [14] are verysimilar

Theorem 5 Given any [X] isin F 1113957G we define (u ρ η) by

u(x) U(ξ) for any ξ such thatx y(ξ)

ρ(x) dx y♯(r dξ)

η y♯ Hξdξ1113872 1113873

(41)

belonging to D where f♯ξ(B) ξ(fminus 1(B)) for any Borel setB is called the push forward element of ξ by f en (u ρ η)

belongs to D and is independent of the representative X from[X] We denote M F 1113957G⟶ D

Next we will clarify the relation between L and M

Theorem 6 e maps M F 1113957G⟶ D and L D⟶ F 1113957G

are invertible and

L degM Id ∣F1113957G

Mdeg

L Id ∣ D (42)

Proof (e proof follows the same lines as in(eorem 48 in[14] so we do not present it here

Now we obtain the solution map Tt Mdeg1113957St deg L that isD⟶L F1113957G⟶St F1113957G⟶M D

4 Weak Solutions

Definition 2 Let u R+ times R⟶ R and ρ R+ times R⟶ RAssume that u and ρ satisfy the following

(i) u isin Linfin([0infin) H1per) and ρ isin Linfin([0infin) L2

per)(ii) If the equations

1113946 1113946R+times[01]

minus u(t x)φt(t x) + uux + Px( 1113857φ(t x)( 1113857 dxdt 1113946[01]

u0(x)φ0(x) dx (43)

where P (μ minus z2x)minus 1(2μ(u)u + (12) u2x + (12)ρ2)

(t y(t ξ))

1113946 1113946R+timesR

2μ2 +12e2

minus 2μu minus12u2x minus

12ρ21113874 1113875φ(t x) + Px(t x)φx(t x)1113876 1113877 dxdt 0 (44)


and

1113946 1113946R+times[01]

minus ρ(t x)φt(t x) minus u(t x)ρ(t x)φx(t x)( 1113857 dxdt 1113946[01]

ρ0(x)φ0(x) dx (45)

hold for all spatial periodic functionsφ isin Cinfin0 ([0infin)R) then we say (u ρ) is a globalweak solution of equation (8)

Moreover if this solution (u ρ) satisfies

1113946[01]

u2x + ρ21113872 1113873 dx 1113946

[01]u20x + ρ201113872 1113873 dx ae for tge 0 (46)

then we say it is a global conservative solution of equation(8)

Theorem 7 Given (u0 ρ0 η0) isin D if Tt(u0 ρ0 η0) (u(t)

ρ(t) η(t)) then (u ρ) is a global conservative solution ofequation (8)

Proof (eorem 2 and Definition 1 imply that ( _1) in Def-inition 2 holds In the following section we will proveequations (43)ndash(46) one by one for any spatial periodicfunction φ isin Cinfin0 ([0infin)R) Let x y(t ξ) and we havedx yξ dξ Since Uξ ux(t y(t ξ)) andyξ ytξ we obtain

1113946 1113946R+times[01]

minus u(t x)φt(t x) + uuxφ(t x)( 1113857 dxdt

1113946 1113946R+times[01]

minus u(t y(t ξ))φt(t y(t ξ))yξ + u(t y(t ξ))ux(t y(t ξ))φ(t y(t ξ))yξ1113960 1113961 dξdt

1113946 1113946R+times[01]

minus U(t ξ)φt(t y(t ξ))yξ + U(t ξ)Uξ(t y(t ξ))φ(t y(t ξ))1113960 1113961 dξdt

(47)

By

Uyξφ degy1113872 1113873t

Utyξφ degy + UUξφ degy + Uyξφtdegy + U2yξφ degy (48)

and Ut ut + uux minus Q minus Px we have

minus Uyξφt deg y + UUξφ degy

minus Uyξφ degy1113872 1113873t+ Utyξφ degy + 2UUξφ degy + U

2yξφ degy

minus Uyξφ degy1113872 1113873tminus Pxyξφ degy + U

2φ degy1113872 1113873ξ

(49)

Integrating this formula into equation (47) we obtain

1113946 1113946R+times[01]


1113946[01]

1113946infin

0minus Uyξφ degy1113872 1113873

tdt1113874 1113875 dx minus 1113946 1113946

R+times[01]Pxyξφ degy1113872 1113873 dξdt

1113946[01]

u0(x)φ0(x) dx minus 1113946 1113946R+times[01]

Px(t x)φ(t x) dxdt

(50)


And the proof for equation (43) completes here Usingequation (28) a direct computation implies that

1113946 1113946R+timesR

Px(t x)φx(t x) dxdt

1113946 1113946R+timesR

Px(t y(t ξ))φx(t y(t ξ))yξ dξdt

1113946 1113946R+timesR

Q(t ξ)φξ(t y(t ξ)) dξdt

minus 1113946 1113946R+timesR

Qξ(t ξ)φ(t y(t ξ)) dξdt


minus 2μ2 minus12

e + 2μ(u)U(t ξ) +12u2x(t y(t ξ)) +

12ρ2(t y(t ξ))1113874 1113875yξφ((t y(t ξ)) dξdt


minus 2μ2 minus12

e + 2μ(u)u(t x) +12u2x(t x) +

12ρ2(t x)1113874 1113875φ(t x) dxdt

(51)

(is completes the proof for equation (44) By rt 0 wehave

(r(t ξ)φ(t y(t ξ)))t r(t ξ)φt(t y(t ξ)) + r(t ξ)U(t ξ)φx(t y(t ξ)) (52)

It satisfies that

1113946 1113946R+times[01]

minus ρ(t x)φt(t x) minus u(t x)ρ(t x)φx(t x)( 1113857 dxdt

1113946 1113946R+times[01]

minus ρ(t y(t ξ))φt(t y(t ξ)) minus u(t y(t ξ))ρ(t y(t ξ))φx(t y(t ξ))1113858 1113859yξ dξdt

1113946 1113946R+times[01]

minus r(t ξ)φt(t y(t ξ)) minus U(t ξ)r(t ξ)φx(t y(t ξ))1113858 1113859 dξdt

1113946 1113946R+times[01]

minus (r(t ξ)φ(t y(t ξ)))t dξdt 1113946[01]

ρ0(x)φ0(x) dx

(53)

And this completes the proof of equation (45) Similarlylet x y(t ξ) in the left side of equation (46) we have


1113946[01]

u2x + ρ21113872 1113873 dx

1113946[01]

u2x(t y(t ξ)) + ρ2(t y(t ξ))1113960 1113961yξ dξ

1113946[01]

Hξ dξ

H(t 1) minus H(t 0) H(0 1) minus H(0 0) 1113946[01]

u20x + ρ201113872 1113873 dx

(54)

(is completes the proof of equation (46)

Data Availability

(e computation data used to support the findings of thisstudy are included within the article

Conflicts of Interest

(e authors declare that they have no conflicts of interest

Acknowledgments

(is work was partially supported by the National NaturalScience Foundation of China (Nos 11701525 11971446 and51609087)

References

[1] R Camassa and D D Holm ldquoAn integrable shallow waterequation with peaked solitonsrdquo Physical Review Lettersvol 71 no 11 pp 1661ndash1664 1993

[2] R Camassa D D Holm and J M Hyman ldquoA new integrableshallow water equationrdquo Advances in Applied Mechanicsvol 31 pp 1ndash33 1994

[3] A Constantin and H P McKean ldquoA shallow water equationon the circlerdquo Communications on Pure and Applied Math-ematics vol 52 no 8 pp 949ndash982 1999

[4] B Khesin J Lenells and G Misiołek ldquoGeneralized Hunter-Saxton equation and the geometry of the group of circlediffeomorphismsrdquo Mathematische Annalen vol 342 no 3pp 617ndash656 2008

[5] J K Hunter and R Saxton ldquoDynamics of director fieldsrdquoSIAM Journal on Applied Mathematics vol 51 no 6pp 1498ndash1521 1991

[6] A Bressan and A Constantin ldquoGlobal conservative solutionsof the Camassa-Holm equationrdquo Archive for Rational Me-chanics and Analysis vol 183 no 2 pp 215ndash239 2007

[7] A Bressan and A Constantin ldquoGlobal dissipative solutions ofthe Camassa-Holm equationrdquo Analysis and Applicationsvol 5 no 1 pp 1ndash27 2007

[8] G M Coclite H Holden and K H Karlsen ldquoGlobal weaksolutions to a generalized hyperelastic-rod wave equationrdquoSIAM Journal on Mathematical Analysis vol 37 no 4pp 1044ndash1069 2005

[9] G Gui Y Liu and M Zhu ldquoOn the wave-breaking phe-nomena and global existence for the generalized periodicCamassa-Holm equationrdquo International Mathematics Re-search Notices vol 2012 no 21 pp 4858ndash4903 2012

[10] H Holden and X Raynaud ldquoPeriodic conservative solutionsof the Camassa-Holm equationrdquo Annales de lrsquoInstitut Fouriervol 58 no 3 pp 945ndash988 2008

[11] H Holden X Raynaud and X Raynaud ldquoDissipative solu-tions for the Camassa-Holm equationrdquo Discrete amp Contin-uous Dynamical SystemsmdashA vol 24 no 4 pp 1047ndash11122009

[12] J K Hunter and Y Zheng ldquoOn a nonlinear hyperbolicvariational equation I global existence of weak solutionsrdquoArchive for Rational Mechanics and Analysis vol 129 no 4pp 305ndash353 1995

[13] J K Hunter and Y Zheng ldquoOn a nonlinear hyperbolicvariational equation II the zero-viscosity and dispersionlimitsrdquo Archive for Rational Mechanics and Analysis vol 129no 4 pp 355ndash383 1995

[14] F Tiglay ldquoConservative weak solutions of the periodicCauchy problem for μ HS equationrdquo Journal of Mathematicsand Physics vol 56 Article ID 021504 2015

[15] Z Xin and P Zhang ldquoOn the weak solutions to a shallowwater equationrdquo Communications on Pure and AppliedMathematics vol 53 no 11 pp 1411ndash1433 2000

[16] D Zuo ldquoA two-component μ-Hunter-Saxton equationrdquo In-verse Problems vol 26 no 8 Article ID 085003 2010

[17] A Constantin and R I Ivanov ldquoOn an integrable two-component Camassa-Holm shallow water systemrdquo PhysicsLetters A vol 372 no 48 pp 7129ndash7132 2008

[18] M Chen S-Q Liu and Y Zhang ldquoA two-component gen-eralization of the camassa-holm equation and its solutionsrdquoLetters in Mathematical Physics vol 75 no 1 pp 1ndash15 2006

[19] K Grunert H Holden and X Raynaud ldquoGlobal solutions forthe two-component camassa-holm systemrdquo Communicationsin Partial Differential Equations vol 37 no 12 pp 2245ndash2271 2012

[20] K Grunert H Holden and X Raynaud ldquoPeriodic conser-vative solutions for the two-component Camassa-Holmsystemrdquo in Spectral Analysis Differential Equations andMathe-Matical Physics A Festschrift for Fritz Gesztesy on theOccasion of His 60th Birthday H Holden B Simon andG Teschl Eds pp 165ndash182 American Mathematical So-ciety 2013

[21] C Guan and Z Yin ldquoGlobal weak solutions for a two-component Camassa-Holm shallow water systemrdquo Journal ofFunctional Analysis vol 260 no 4 pp 1132ndash1154 2011

[22] C Guan and Z Yin ldquoGlobal weak solutions and smoothsolutions for a two-component Hunter-Saxton systemrdquoJournal of Mathematical Physics vol 52 no 10 Article ID103707 2011

[23] A Nordli ldquoA lipschitz metric for conservative solutions of thetwo-component Hunter-Saxton systemrdquo Methods and Ap-plications of Analysis vol 23 no 3 pp 215ndash232 2016


[24] J Escher ldquoNon-metric two-component Euler equations onthe circlerdquo Monatshefte fur Mathematik vol 167 no 3-4pp 449ndash459 2012

[25] Y Guo and T Xiong ldquoBlow-up analysis for the periodic two-component μ-Hunter-Saxton systemrdquo Mathematical Prob-lems in Engineering vol 2018 Article ID 5374180 11 pages2018

[26] J Liu ldquo(e Cauchy problem of a periodic 2-componentμ-Hunter-Saxton system in Besov spacesrdquo Journal of Mathe-matical Analysis and Applications vol 399 no 2 pp 650ndash666 2013

[27] Y Zhang Y Liu and C Qu ldquoBlow up of solutions andtraveling waves to the two-component μ-Camassa-Holmsystemrdquo International Mathematics Research Noticesvol 2013 no 15 pp 3386ndash3419 2013

[28] J Liu and Z Yin ldquoGlobal weak solutions for a periodic two-component μ-Hunter-Saxton systemrdquo Monatshefte furMathematik vol 168 no 3-4 pp 503ndash521 2012

[29] H Holden and X Raynaud ldquoGlobal conservative solutions ofthe Camassa-Holm equation-a Lagrangian point of viewrdquoCommunications in Partial Differential Equations vol 32no 10 pp 1511ndash1549 2007


u

ρ⎛⎝ ⎞⎠

t

J1

δH2

δu

δH2

δρ

⎛⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎝

⎞⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎠ J2

δH1

δu

δH1

δρ

⎛⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎝

⎞⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎠ (5)

where

H1 12

11139461

0uf + ρ21113872 1113873 dx H2 1113946

1

0μ(f)f

2+12

ff2x +

12

fρ21113874 1113875 dx

(6)

Also

J1 zxA 0

0 zx

1113888 1113889 J2 uzx + zxu ρzx

zxρ 01113888 1113889 (7)

with f Aminus 1u (μ minus z2x)minus 1u It is a generalization of μHSequation (where ρ 0) If A(t x) u minus uxx then equation (4)becomes a two-component CH system which has been studiedin [17ndash21] (e system exhibits local well-posedness and it hasfinite-time blowup solutions and global strong solutions in timeGlobal conservative weak solutions can be obtained by co-ordinate transformation in [19 20] and admissible weak so-lutions can be found in [21] by mollifying the initial data IfA(t x) minus uxx then equation (4) turns to a two-componentHS system which has been looked into in [22 23] Hence wecan say 2-μHS system equation (4) lies in an intermediatebetween 2-CH and 2-HS systems (e Cauchy problem forequation (4) has been studied extensively in [24ndash27] In additionresearch studies have shown that this system is locally well-posed[26] for (u0 ρ0) isin Hs times Hsminus 1 sgt (32) besides its globalclassical solutions [26] and finite-time blowup solutions [25 27]

have also been found and its geometric background has beencomprehensively given by Escher in [24] (e global admissibleweak solution of system equation (4) has been obtained in [28]by mollifying the initial date Here we will follow previousresearch studies [6 10 14] and demonstrate the existence ofglobal conservativeweak solutionsHowever compared to the 2-CH system the existing μ(u) in the 2-μHS system brings somedifficulties to the calculation of equation (28) Fortunately weovercame it Because A(t x) minus uxx in the 2-HS system the 2-μHS system is structurally more complex than the 2-HS systemIn [23] the author gave the specific expression ofy U H and r (one can find them in equations (14)ndash(17))which is very helpful to the proof of themain theoremHoweverthis practice is almost impossible for the 2-μHS system so ourproof is a little bitmore difficult To sumup althoughwe refer tothe methods in [6 10 14] our results and the proofs are quitedifferent

Our paper is organized as follows In Section 2 we refor-mulate system equation (4) and give an equivalent system inLagrangian coordinates We also try to illustrate the existenceand uniqueness of solutions to the equivalent system to Banachcontraction arguments In Section 3 we establish maps betweenLagrangian and Eulerian coordinates which can connect con-servative weak solutions of equation (4) and solutions ofa semilinear system together In Section 4 we give the existenceof global conservative weak solutions to equation (4)

2 Preliminaries

Firstly we reformulate system equation (4) AssumeA(t x) μ(u) minus uxx in equation (4) we have

μ ut( 1113857 minus utxx minus uuxxx + 2μ(u)ux minus 2uxuxx + ρρx 0 tgt 0 x isin R






⎧⎪⎪⎪⎪⎪⎪⎪⎪⎨

⎪⎪⎪⎪⎪⎪⎪⎪⎩

(8)

which is equivalent to

ut + uux + Aminus 1

zx 2μ(u)u +12u2x +

12ρ21113874 1113875 0 tgt 0 x isin R






⎧⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎨

⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎩

(9)




2xx + uxuxxx + ρ2x + ρρxx1113872 1113873 0 (10)


u2x + ρ21113872 1113873

t+ u u

2x + ρ21113872 11138731113872 1113873

x u minus μ u

2x + ρ21113872 1113873 minus 4μ(u)

2+ 2μ(u)u1113872 11138731113872 1113873

x (11)


μ(u) 11139461

0u(t x) dx 1113946

1

0u0(x) dx ≔ μ

11139461

0u2x + ρ21113872 1113873 dx 1113946

1

0u20x + ρ201113872 1113873 dx ≔ e

(12)




yt(t ξ) u(t y(t ξ)) (14)

And then we define

U(t ξ) u(t y(t ξ)) (15)

H(t ξ) 1113946y(tξ)

y(t0)u2x + ρ21113872 1113873 dx (16)



yt(t ξ) U

Ut(t ξ) minus Q



y(t 0)

μt et 0

rt(t ξ) 0

⎧⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎨

⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎩

(18)

where

Q μ minus z2x1113872 1113873

minus 1zx 2μ(u)u +

12u2x +

12ρ21113874 1113875(t y(t ξ))


12ρ21113874 1113875(t y(t ξ))

11139461



12u2x +

12ρ21113874 1113875(t z) dz

(19)



Q 11139461



12Hξprime1113874 1113875 dξprime (20)





H1per f isin H


L2per f isin H


(22)


fH1([01]) andfL2

per fL2([01])



ζt(t ξ) U

Ut(t ξ) minus Q


y(t 0)

μt et 0

rt(t ξ) 0

⎧⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎨

⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎩

(23)






+UH1per

+σH1per

+|μ| +|e| +rL2per

(24)




per )3 times



yξHξ U2ξ + r

2 ae (26)

11139461

0Uyx dx μ ae (27)




ytξ Uξ



12u2x +

12ρ21113874 1113875(t y(t ξ))yξ

μ minus z2x1113872 1113873

minus 1μ minus z

2x1113872 1113873 minus 2μ2 minus

12

e + 2μ(u)u +12u2x +

12ρ21113874 1113875(t y(t ξ))yξ

minus 2μ2 minus12

e + 2μ(u)U1113874 1113875yξ +12Hξ


⎧⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎨

⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎩

(28)

It satisfies that

yξHξ1113872 1113873t


U2ξ + r

21113872 1113873


(29)











Let



1infinper1113966 1113967 (30)

and


minus Id

W1infin le s1113966 1113967 (31)





1 + e(y + H) isin G1113882 1113883 (32)

and


1 + e(y + H) isin Gs1113882 1113883 (33)



(34)






1 + elt ξ1113896 1113897


U(ξ) u(t y(t ξ))


μ 11139461

0u(t x) dx

e η([0 1])

(35)

where


Fη(x)

η([0 x)) if xgt 0

0 if x 0


⎧⎪⎪⎨

⎪⎪⎩(36)




12ε

η(x minus ε x + ε) u2x + ρ21113882 1113883 (37)




yξ(ξ) u2x(y(ξ)) + ρ2(y(ξ))1113872 1113873 + yξ 1 + e (38)




prime) suchthat

12


prime( 1113857le1i (39)










η y♯ Hξdξ1113872 1113873

(41)





are invertible and


Mdeg

L Id ∣ D (42)



4 Weak Solutions




1113946 1113946R+times[01]


u0(x)φ0(x) dx (43)


(t y(t ξ))

1113946 1113946R+timesR

2μ2 +12e2




and

1113946 1113946R+times[01]


ρ0(x)φ0(x) dx (45)



1113946[01]

u2x + ρ21113872 1113873 dx 1113946

[01]u20x + ρ201113872 1113873 dx ae for tge 0 (46)





1113946 1113946R+times[01]


1113946 1113946R+times[01]


1113946 1113946R+times[01]


(47)

By

Uyξφ degy1113872 1113873t





2yξφ degy


2φ degy1113872 1113873ξ

(49)


1113946 1113946R+times[01]


1113946[01]

1113946infin


tdt1113874 1113875 dx minus 1113946 1113946


1113946[01]


Px(t x)φ(t x) dxdt

(50)



1113946 1113946R+timesR


1113946 1113946R+timesR


1113946 1113946R+timesR





minus 2μ2 minus12

e + 2μ(u)U(t ξ) +12u2x(t y(t ξ)) +



minus 2μ2 minus12

e + 2μ(u)u(t x) +12u2x(t x) +

12ρ2(t x)1113874 1113875φ(t x) dxdt

(51)



It satisfies that

1113946 1113946R+times[01]


1113946 1113946R+times[01]


1113946 1113946R+times[01]


1113946 1113946R+times[01]


ρ0(x)φ0(x) dx

(53)



1113946[01]

u2x + ρ21113872 1113873 dx

1113946[01]


1113946[01]

Hξ dξ


u20x + ρ201113872 1113873 dx

(54)


Data Availability




Acknowledgments


References


































2xx + uxuxxx + ρ2x + ρρxx1113872 1113873 0 (10)


u2x + ρ21113872 1113873

t+ u u

2x + ρ21113872 11138731113872 1113873

x u minus μ u

2x + ρ21113872 1113873 minus 4μ(u)

2+ 2μ(u)u1113872 11138731113872 1113873

x (11)


μ(u) 11139461

0u(t x) dx 1113946

1

0u0(x) dx ≔ μ

11139461

0u2x + ρ21113872 1113873 dx 1113946

1

0u20x + ρ201113872 1113873 dx ≔ e

(12)




yt(t ξ) u(t y(t ξ)) (14)

And then we define

U(t ξ) u(t y(t ξ)) (15)

H(t ξ) 1113946y(tξ)

y(t0)u2x + ρ21113872 1113873 dx (16)



yt(t ξ) U

Ut(t ξ) minus Q



y(t 0)

μt et 0

rt(t ξ) 0

⎧⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎨

⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎩

(18)

where

Q μ minus z2x1113872 1113873

minus 1zx 2μ(u)u +

12u2x +

12ρ21113874 1113875(t y(t ξ))


12ρ21113874 1113875(t y(t ξ))

11139461



12u2x +

12ρ21113874 1113875(t z) dz

(19)



Q 11139461



12Hξprime1113874 1113875 dξprime (20)





H1per f isin H


L2per f isin H


(22)


fH1([01]) andfL2

per fL2([01])



ζt(t ξ) U

Ut(t ξ) minus Q


y(t 0)

μt et 0

rt(t ξ) 0

⎧⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎨

⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎩

(23)






+UH1per

+σH1per

+|μ| +|e| +rL2per

(24)




per )3 times



yξHξ U2ξ + r

2 ae (26)

11139461

0Uyx dx μ ae (27)




ytξ Uξ



12u2x +

12ρ21113874 1113875(t y(t ξ))yξ

μ minus z2x1113872 1113873

minus 1μ minus z

2x1113872 1113873 minus 2μ2 minus

12

e + 2μ(u)u +12u2x +

12ρ21113874 1113875(t y(t ξ))yξ

minus 2μ2 minus12

e + 2μ(u)U1113874 1113875yξ +12Hξ


⎧⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎨

⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎩

(28)

It satisfies that

yξHξ1113872 1113873t


U2ξ + r

21113872 1113873


(29)











Let



1infinper1113966 1113967 (30)

and


minus Id

W1infin le s1113966 1113967 (31)





1 + e(y + H) isin G1113882 1113883 (32)

and


1 + e(y + H) isin Gs1113882 1113883 (33)



(34)






1 + elt ξ1113896 1113897


U(ξ) u(t y(t ξ))


μ 11139461

0u(t x) dx

e η([0 1])

(35)

where


Fη(x)

η([0 x)) if xgt 0

0 if x 0


⎧⎪⎪⎨

⎪⎪⎩(36)




12ε

η(x minus ε x + ε) u2x + ρ21113882 1113883 (37)




yξ(ξ) u2x(y(ξ)) + ρ2(y(ξ))1113872 1113873 + yξ 1 + e (38)




prime) suchthat

12


prime( 1113857le1i (39)










η y♯ Hξdξ1113872 1113873

(41)





are invertible and


Mdeg

L Id ∣ D (42)



4 Weak Solutions




1113946 1113946R+times[01]


u0(x)φ0(x) dx (43)


(t y(t ξ))

1113946 1113946R+timesR

2μ2 +12e2




and

1113946 1113946R+times[01]


ρ0(x)φ0(x) dx (45)



1113946[01]

u2x + ρ21113872 1113873 dx 1113946

[01]u20x + ρ201113872 1113873 dx ae for tge 0 (46)





1113946 1113946R+times[01]


1113946 1113946R+times[01]


1113946 1113946R+times[01]


(47)

By

Uyξφ degy1113872 1113873t





2yξφ degy


2φ degy1113872 1113873ξ

(49)


1113946 1113946R+times[01]


1113946[01]

1113946infin


tdt1113874 1113875 dx minus 1113946 1113946


1113946[01]


Px(t x)φ(t x) dxdt

(50)



1113946 1113946R+timesR


1113946 1113946R+timesR


1113946 1113946R+timesR





minus 2μ2 minus12

e + 2μ(u)U(t ξ) +12u2x(t y(t ξ)) +



minus 2μ2 minus12

e + 2μ(u)u(t x) +12u2x(t x) +

12ρ2(t x)1113874 1113875φ(t x) dxdt

(51)



It satisfies that

1113946 1113946R+times[01]


1113946 1113946R+times[01]


1113946 1113946R+times[01]


1113946 1113946R+times[01]


ρ0(x)φ0(x) dx

(53)



1113946[01]

u2x + ρ21113872 1113873 dx

1113946[01]


1113946[01]

Hξ dξ


u20x + ρ201113872 1113873 dx

(54)


Data Availability




Acknowledgments


References

































Q 11139461



12Hξprime1113874 1113875 dξprime (20)





H1per f isin H


L2per f isin H


(22)


fH1([01]) andfL2

per fL2([01])



ζt(t ξ) U

Ut(t ξ) minus Q


y(t 0)

μt et 0

rt(t ξ) 0

⎧⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎨

⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎩

(23)






+UH1per

+σH1per

+|μ| +|e| +rL2per

(24)




per )3 times



yξHξ U2ξ + r

2 ae (26)

11139461

0Uyx dx μ ae (27)




ytξ Uξ



12u2x +

12ρ21113874 1113875(t y(t ξ))yξ

μ minus z2x1113872 1113873

minus 1μ minus z

2x1113872 1113873 minus 2μ2 minus

12

e + 2μ(u)u +12u2x +

12ρ21113874 1113875(t y(t ξ))yξ

minus 2μ2 minus12

e + 2μ(u)U1113874 1113875yξ +12Hξ


⎧⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎨

⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎩

(28)

It satisfies that

yξHξ1113872 1113873t


U2ξ + r

21113872 1113873


(29)











Let



1infinper1113966 1113967 (30)

and


minus Id

W1infin le s1113966 1113967 (31)





1 + e(y + H) isin G1113882 1113883 (32)

and


1 + e(y + H) isin Gs1113882 1113883 (33)



(34)






1 + elt ξ1113896 1113897


U(ξ) u(t y(t ξ))


μ 11139461

0u(t x) dx

e η([0 1])

(35)

where


Fη(x)

η([0 x)) if xgt 0

0 if x 0


⎧⎪⎪⎨

⎪⎪⎩(36)




12ε

η(x minus ε x + ε) u2x + ρ21113882 1113883 (37)




yξ(ξ) u2x(y(ξ)) + ρ2(y(ξ))1113872 1113873 + yξ 1 + e (38)




prime) suchthat

12


prime( 1113857le1i (39)










η y♯ Hξdξ1113872 1113873

(41)





are invertible and


Mdeg

L Id ∣ D (42)



4 Weak Solutions




1113946 1113946R+times[01]


u0(x)φ0(x) dx (43)


(t y(t ξ))

1113946 1113946R+timesR

2μ2 +12e2




and

1113946 1113946R+times[01]


ρ0(x)φ0(x) dx (45)



1113946[01]

u2x + ρ21113872 1113873 dx 1113946

[01]u20x + ρ201113872 1113873 dx ae for tge 0 (46)





1113946 1113946R+times[01]


1113946 1113946R+times[01]


1113946 1113946R+times[01]


(47)

By

Uyξφ degy1113872 1113873t





2yξφ degy


2φ degy1113872 1113873ξ

(49)


1113946 1113946R+times[01]


1113946[01]

1113946infin


tdt1113874 1113875 dx minus 1113946 1113946


1113946[01]


Px(t x)φ(t x) dxdt

(50)



1113946 1113946R+timesR


1113946 1113946R+timesR


1113946 1113946R+timesR





minus 2μ2 minus12

e + 2μ(u)U(t ξ) +12u2x(t y(t ξ)) +



minus 2μ2 minus12

e + 2μ(u)u(t x) +12u2x(t x) +

12ρ2(t x)1113874 1113875φ(t x) dxdt

(51)



It satisfies that

1113946 1113946R+times[01]


1113946 1113946R+times[01]


1113946 1113946R+times[01]


1113946 1113946R+times[01]


ρ0(x)φ0(x) dx

(53)



1113946[01]

u2x + ρ21113872 1113873 dx

1113946[01]


1113946[01]

Hξ dξ


u20x + ρ201113872 1113873 dx

(54)


Data Availability




Acknowledgments


References
































ytξ Uξ



12u2x +

12ρ21113874 1113875(t y(t ξ))yξ

μ minus z2x1113872 1113873

minus 1μ minus z

2x1113872 1113873 minus 2μ2 minus

12

e + 2μ(u)u +12u2x +

12ρ21113874 1113875(t y(t ξ))yξ

minus 2μ2 minus12

e + 2μ(u)U1113874 1113875yξ +12Hξ


⎧⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎨

⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎩

(28)

It satisfies that

yξHξ1113872 1113873t


U2ξ + r

21113872 1113873


(29)











Let



1infinper1113966 1113967 (30)

and


minus Id

W1infin le s1113966 1113967 (31)





1 + e(y + H) isin G1113882 1113883 (32)

and


1 + e(y + H) isin Gs1113882 1113883 (33)



(34)






1 + elt ξ1113896 1113897


U(ξ) u(t y(t ξ))


μ 11139461

0u(t x) dx

e η([0 1])

(35)

where


Fη(x)

η([0 x)) if xgt 0

0 if x 0


⎧⎪⎪⎨

⎪⎪⎩(36)




12ε

η(x minus ε x + ε) u2x + ρ21113882 1113883 (37)




yξ(ξ) u2x(y(ξ)) + ρ2(y(ξ))1113872 1113873 + yξ 1 + e (38)




prime) suchthat

12


prime( 1113857le1i (39)










η y♯ Hξdξ1113872 1113873

(41)





are invertible and


Mdeg

L Id ∣ D (42)



4 Weak Solutions




1113946 1113946R+times[01]


u0(x)φ0(x) dx (43)


(t y(t ξ))

1113946 1113946R+timesR

2μ2 +12e2




and

1113946 1113946R+times[01]


ρ0(x)φ0(x) dx (45)



1113946[01]

u2x + ρ21113872 1113873 dx 1113946

[01]u20x + ρ201113872 1113873 dx ae for tge 0 (46)





1113946 1113946R+times[01]


1113946 1113946R+times[01]


1113946 1113946R+times[01]


(47)

By

Uyξφ degy1113872 1113873t





2yξφ degy


2φ degy1113872 1113873ξ

(49)


1113946 1113946R+times[01]


1113946[01]

1113946infin


tdt1113874 1113875 dx minus 1113946 1113946


1113946[01]


Px(t x)φ(t x) dxdt

(50)



1113946 1113946R+timesR


1113946 1113946R+timesR


1113946 1113946R+timesR





minus 2μ2 minus12

e + 2μ(u)U(t ξ) +12u2x(t y(t ξ)) +



minus 2μ2 minus12

e + 2μ(u)u(t x) +12u2x(t x) +

12ρ2(t x)1113874 1113875φ(t x) dxdt

(51)



It satisfies that

1113946 1113946R+times[01]


1113946 1113946R+times[01]


1113946 1113946R+times[01]


1113946 1113946R+times[01]


ρ0(x)φ0(x) dx

(53)



1113946[01]

u2x + ρ21113872 1113873 dx

1113946[01]


1113946[01]

Hξ dξ


u20x + ρ201113872 1113873 dx

(54)


Data Availability




Acknowledgments


References
































Fη(x)

η([0 x)) if xgt 0

0 if x 0


⎧⎪⎪⎨

⎪⎪⎩(36)




12ε

η(x minus ε x + ε) u2x + ρ21113882 1113883 (37)




yξ(ξ) u2x(y(ξ)) + ρ2(y(ξ))1113872 1113873 + yξ 1 + e (38)




prime) suchthat

12


prime( 1113857le1i (39)










η y♯ Hξdξ1113872 1113873

(41)





are invertible and


Mdeg

L Id ∣ D (42)



4 Weak Solutions




1113946 1113946R+times[01]


u0(x)φ0(x) dx (43)


(t y(t ξ))

1113946 1113946R+timesR

2μ2 +12e2




and

1113946 1113946R+times[01]


ρ0(x)φ0(x) dx (45)



1113946[01]

u2x + ρ21113872 1113873 dx 1113946

[01]u20x + ρ201113872 1113873 dx ae for tge 0 (46)





1113946 1113946R+times[01]


1113946 1113946R+times[01]


1113946 1113946R+times[01]


(47)

By

Uyξφ degy1113872 1113873t





2yξφ degy


2φ degy1113872 1113873ξ

(49)


1113946 1113946R+times[01]


1113946[01]

1113946infin


tdt1113874 1113875 dx minus 1113946 1113946


1113946[01]


Px(t x)φ(t x) dxdt

(50)



1113946 1113946R+timesR


1113946 1113946R+timesR


1113946 1113946R+timesR





minus 2μ2 minus12

e + 2μ(u)U(t ξ) +12u2x(t y(t ξ)) +



minus 2μ2 minus12

e + 2μ(u)u(t x) +12u2x(t x) +

12ρ2(t x)1113874 1113875φ(t x) dxdt

(51)



It satisfies that

1113946 1113946R+times[01]


1113946 1113946R+times[01]


1113946 1113946R+times[01]


1113946 1113946R+times[01]


ρ0(x)φ0(x) dx

(53)



1113946[01]

u2x + ρ21113872 1113873 dx

1113946[01]


1113946[01]

Hξ dξ


u20x + ρ201113872 1113873 dx

(54)


Data Availability




Acknowledgments


References
































and

1113946 1113946R+times[01]


ρ0(x)φ0(x) dx (45)



1113946[01]

u2x + ρ21113872 1113873 dx 1113946

[01]u20x + ρ201113872 1113873 dx ae for tge 0 (46)





1113946 1113946R+times[01]


1113946 1113946R+times[01]


1113946 1113946R+times[01]


(47)

By

Uyξφ degy1113872 1113873t





2yξφ degy


2φ degy1113872 1113873ξ

(49)


1113946 1113946R+times[01]


1113946[01]

1113946infin


tdt1113874 1113875 dx minus 1113946 1113946


1113946[01]


Px(t x)φ(t x) dxdt

(50)



1113946 1113946R+timesR


1113946 1113946R+timesR


1113946 1113946R+timesR





minus 2μ2 minus12

e + 2μ(u)U(t ξ) +12u2x(t y(t ξ)) +



minus 2μ2 minus12

e + 2μ(u)u(t x) +12u2x(t x) +

12ρ2(t x)1113874 1113875φ(t x) dxdt

(51)



It satisfies that

1113946 1113946R+times[01]


1113946 1113946R+times[01]


1113946 1113946R+times[01]


1113946 1113946R+times[01]


ρ0(x)φ0(x) dx

(53)



1113946[01]

u2x + ρ21113872 1113873 dx

1113946[01]


1113946[01]

Hξ dξ


u20x + ρ201113872 1113873 dx

(54)


Data Availability




Acknowledgments


References

































1113946 1113946R+timesR


1113946 1113946R+timesR


1113946 1113946R+timesR





minus 2μ2 minus12

e + 2μ(u)U(t ξ) +12u2x(t y(t ξ)) +



minus 2μ2 minus12

e + 2μ(u)u(t x) +12u2x(t x) +

12ρ2(t x)1113874 1113875φ(t x) dxdt

(51)



It satisfies that

1113946 1113946R+times[01]


1113946 1113946R+times[01]


1113946 1113946R+times[01]


1113946 1113946R+times[01]


ρ0(x)φ0(x) dx

(53)



1113946[01]

u2x + ρ21113872 1113873 dx

1113946[01]


1113946[01]

Hξ dξ


u20x + ρ201113872 1113873 dx

(54)


Data Availability




Acknowledgments


References
































1113946[01]

u2x + ρ21113872 1113873 dx

1113946[01]


1113946[01]

Hξ dξ


u20x + ρ201113872 1113873 dx

(54)


Data Availability




Acknowledgments


References
































global conservative solutions of the two-component -hunter

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