# Global Conservative Solutions of the Two-Componentμ-Hunter–Saxton System

Authors: Xiayang Shi; Jingjing Liu; Hongyang Zhang
Journal: Mathematical Problems in Engineering
 (2021)

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## Abstract

In this paper, we establish global conservative solutions of the two-componentμ-Hunter–Saxton system by the methods developed in “A. Bressan, A. Constantin, Global Conservative Solutions of the Camassa-Holm Equation, Arch. Ration. Mech. Anal. 183 (2), 215–239 (2007)” and “H. Holden, X. Raynaud, Periodic Conservative Solutions of the Camassa-Holm Equation, Ann. Inst. Fourier (Grenoble) 58(3), 945–988 (2008).”


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## Body

## 1. Introduction

If a solution remains bounded pointwise and its slope becomes unbounded in finite time, we say this solution breaks down in finite time. Blow-up is a highly interesting property exhibited in a lot of nonlinear dispersive-wave equations, e.g., the Camassa–Holm equation [1–3]:(1)ut−utxx+3uux−2uxuxx−uuxxx=0.Theμ-Hunter–Saxton equation [4] is as follows:(2)utxx−2μuux+2uxuxx+uuxxx=0.The Hunter–Saxton equation [5] is as follows:(3)utxx+2uxuxx+uuxxx=0.Then, what will happen after wave breaking is an interesting problem, which has received considerable attention in the past decade. Several methods have been developed to study this issue, including the vanishing viscosity approach, initial data mollification, and coordinate transformation [6–15]. Among all the methods, two of them will be used in this paper. One was introduced by Bressan and Constantin in [6], and the other was proposed by Holden and Raynaud in [10]. Both methods converted the problem to solving a corresponding semilinear system by the application of new variables. Their mutual difference lies in the fact that Holden and Raynaud used a different set of variables and constructed a bijective map between Eulerian and Lagrangian coordinates for (CH). In addition, if the H1 energy ∫u2+ux2dx remains constant except for the exact time of break down, we call the conservative solutions; if ∫u2+ux2dx decreases to zero at the breakdown time, we call the solutions dissipative.In this paper, we discuss the conservative solutions of the following periodic two-componentμ-Hunter–Saxton system [16]:(4)At+2Aux+uAx+ρρx=0,ρt+uρx=0,u0,x=u0x,ρ0,x=ρ0x,ut,x+1=ut,x,ρt,x+1=ρt,x,where At,x=μu−uxx, μu=∫01ut,xdx, where t>0 is the time vector, and x∈ℝ is a space vector. This system is of a bi-Hamilton structure, and it can also be viewed as a bivariational equation set. Therefore, equation (4) can be rewritten as(5)uρt=J1δH2δuδH2δρ=J2δH1δuδH1δρ,where(6)H1=12∫01uf+ρ2dx,H2=∫01μff2+12ffx2+12fρ2dx.Also,(7)J1=∂xA00∂x,J2=u∂x+∂xuρ∂x∂xρ0,with f=A−1u=μ−∂x2−1u. It is a generalization of μHS equation (where ρ=0). If At,x=u−uxx, then equation (4) becomes a two-component CH system, which has been studied in [17–21]. The system exhibits local well-posedness, and it has finite-time blowup solutions and global strong solutions in time. Global conservative weak solutions can be obtained by coordinate transformation in [19, 20], and admissible weak solutions can be found in [21] by mollifying the initial data. If At,x=−uxx, then equation (4) turns to a two-component HS system, which has been looked into in [22, 23]. Hence, we can say 2-μHS system equation (4) lies in an intermediate between 2-CH and 2-HS systems. The Cauchy problem for equation (4) has been studied extensively in [24–27]. In addition, research studies have shown that this system is locally well-posed [26] for u0,ρ0∈Hs×Hs−1, s>3/2; besides, its global classical solutions [26] and finite-time blowup solutions [25, 27] have also been found, and its geometric background has been comprehensively given by Escher in [24]. The global admissible weak solution of system equation (4) has been obtained in [28] by mollifying the initial date. Here, we will follow previous research studies [6, 10, 14] and demonstrate the existence of global conservative weak solutions. However, compared to the 2-CH system, the existing μu in the 2-μHS system brings some difficulties to the calculation of equation (28). Fortunately, we overcame it. Because At,x=−uxx in the 2-HS system, the 2-μHS system is structurally more complex than the 2-HS system. In [23], the author gave the specific expression of y,U,H,andr (one can find them in equations (14)–(17)), which is very helpful to the proof of the main theorem. However, this practice is almost impossible for the 2-μHS system, so our proof is a little bit more difficult. To sum up, although we refer to the methods in [6, 10, 14], our results and the proofs are quite different.Our paper is organized as follows. In Section2, we reformulate system equation (4) and give an equivalent system in Lagrangian coordinates. We also try to illustrate the existence and uniqueness of solutions to the equivalent system to Banach contraction arguments. In Section 3, we establish maps between Lagrangian and Eulerian coordinates, which can connect conservative weak solutions of equation (4) and solutions of a semilinear system together. In Section 4, we give the existence of global conservative weak solutions to equation (4).


## 2. Preliminaries

Firstly, we reformulate system equation (4). Assume At,x=μu−uxx in equation (4); we have(8)μut−utxx−uuxxx+2μuux−2uxuxx+ρρx=0,t>0,x∈ℝ,ρt+uρx+uxρ=0,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,which is equivalent to(9)ut+uux+A−1∂x2μuu+12ux2+12ρ2=0,t>0,x∈ℝ,ρt+uρx+uxρ=0,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 and A−1f=μ−∂x2−1f=g∗f for all f∈L2 with gx=1/2x2−x+13/12. By differentiating the first equation in equation (9), we obtain(10)utx+uuxx+ux2+g∗2μuuxx+uxx2+uxuxxx+ρx2+ρρxx=0.Based on the second equation in equations (9) and (10), a direct computation implies(11)ux2+ρ2t+uux2+ρ2x=u−μux2+ρ2−4μu2+2μuux.For smooth solutions, we combine the first equation in equations (8) and (11), and we find the following conservation laws:(12)μu=∫01ut,xdx=∫01u0xdx≔μ,∫01ux2+ρ2dx=∫01u0x2+ρ02dx≔e.Since system equation (4) is periodic with period 1, we define a space(13)V1=f∈Hloc1ℝ∣fξ+1=fξ+1.However,V1 is not a Banach space. We define y:ℝ⟶V1,t⟶yt,⋅ as the solution of(14)ytt,ξ=ut,yt,ξ.And then, we define(15)Ut,ξ=ut,yt,ξ,(16)Ht,ξ=∫yt,0yt,ξux2+ρ2dx,(17)rt,ξ=ρt,yt,ξyξt,ξ.Taking the derivative of both sides of equations (15)–(17) with respect to t and using equation (11), we can obtain a result. Combining this result with equation (14), we have the following semilinear system of y,U,H,r:(18)ytt,ξ=U,Utt,ξ=−Q,Htt,ξ=U−e−4μ2+2μUt,yt,ξ−U−e−4μ2+2μUt,yt,0≔U−e−4μ2+2μU∣yt,ξyt,0,μt=et=0,rtt,ξ=0,where(19)Q=μ−∂x2−1∂x2μuu+12ux2+12ρ2t,yt,ξ=g∗∂x2μuu+12ux2+12ρ2t,yt,ξ=∫01yt,ξ−z−12sgnyt,ξ−z2μuu+12ux2+12ρ2t,zdz.After we define a new variablez=yt,ξ′, we have(20)Q=∫01yt,ξ−yt,ξ′−12sgnyt,ξ−yt,ξ′2μUt,ξ′yξ′+12Hξ′dξ′.Here, we will make some explanation aboutH. Since u,ρ is periodic with period 1 and y∈V1, we can obtain Ht,ξ+1−Ht,ξ=Ht,1−Ht,0. By equation (11), a direct computation implies that d/dtHt,1−Ht,0=0, which follows that Ht,1−Ht,0=H0,1−H0,0. Define space V as(21)V=f∈Hloc1ℝ∣there existsα∈ℝsuch thatfξ+1=fξ+αfor allξ∈ℝ,with norm fV=fH10,1 as a Banach space [10] and H∈V. Moreover, we introduce the Banach space(22)Hper1=f∈Hloc1ℝ∣fξ+1=fξfor allξ∈ℝ,Lper2=f∈Hloc2ℝ∣fξ+1=fξfor allξ∈ℝ,with the norm fHper1=fH10,1 and fLper2=fL20,1.Next, we will give the existence and uniqueness of solution to equation (18) based on the Banach contraction argument. However, two important issues are noteworthy about y and H. One is that the space V1 in which y belongs to is not a Banach space, and the other is that H is not periodic with period 1. Hence, we let ζ=y−Id and σ=H−eId to be transient in order to use Banach contraction argument. And, equation (18) becomes(23)ζtt,ξ=U,Utt,ξ=−Q,σtt,ξ=U−e−4μ2+2μU∣yt,ξ,yt,0,μt=et=0,rtt,ξ=0.Since the first four equations in both equations (18) and (23) are independent of r and r is preserved with respect to time t, by following closely the proofs of Theorems 3 and 4 in [14], we have the following results. Let E=Hper1×Hper1×Hper1×ℝ×ℝ×Lper2 be equipped with the norm(24)ζ,U,σ,μ,e,rE=ζHper1+UHper1+σHper1+μ+e+rLper2.Theorem 1 (local existence and uniqueness).
For initial dataX0=ζ0,U0,σ0,μ0,e0,r0∈E, there exists a time T=TX0E>0 such that system equation (23) has a unique solution in C10,T,E.In order to obtain global existence and uniqueness, we need to make more hypotheses on initial data, so letG be a space consisting of all ζ,U,σ,μ,e,r in E∩Wper1,∞3×ℝ2×Lper∞ such that(25)yξ≥0,Hξ≥0,yξ+Hξ>0,a.e.,(26)yξHξ=Uξ2+r2,a.e.,(27)∫01Uyxdx=μ,a.e..Theorem 2 (global existence and uniqueness).
For initial dataX0=y0,U0,H0,μ0,e0,r0∈G, system equation (18) has a unique global solution Xt∈C1ℝ+,E. Moreover, Xt∈G is satisfied at all times. Furthermore, the map S:G×ℝ+⟶G defined as StX0=Xt which is a continuous semigroup.Proof.
The proof follows the same clue as Theorem4 in [14], so we prove only equation (26) here. Firstly, by equation (18), we have(28)ytξ=Uξ,Utξ=−μ−∂x2−1∂x22μuu+12ux2+12ρ2t,yt,ξyξ=μ−∂x2−1μ−∂x2−2μ2−12e+2μuu+12ux2+12ρ2t,yt,ξyξ=−2μ2−12e+2μuUyξ+12Hξ,Htξ=−e−4μ2Uξ+4μUUξ.
It satisfies that(29)yξHξt=ytξHξ+yξHtξ=UξHξ+yξ−e−4μ2+4μUUξ,Uξ2+r2t=2UξUtξ=−e−4μ2+4μUUξyξ+UξHξ.
Thus,yξHξ−Uξ2−r2t=0. Since initial data X0∈G, equation (26) can be obtained.


## 3. Bijective Maps between Eulerian and Lagrangian Coordinates

Since the energy is concentrated on the zero measure sets when wave breaking occurs, we must consider a periodic positive Radon measure. Consequently, we make the following definition.Definition 1.
D is the set of all triplets u,ρ,η such that u∈Hper1, ρ∈Lper2, and η is a positive and periodic Radon measure whose absolute continuous part is ηac=ux2+ρ2dx.
Note that the variables in Eulerian space areu,ρ,η and those in the Lagrangian space are y,U,H,r. As we prefer to get one-to-one correspondence between Eulerian and Lagrangian coordinates, we define equivalence of the latter by establishing an equivalence class map on G. Let us start by relabeling invariance first.
Let(30)G=f∈Wloc1,∞∣fis invertible,fx+1=fx+1forx∈ℝ,andf−Id,f−1−Id∈Wper1,∞and(31)Gs=f∈G∣f−IdW1,∞+f−1−IdW1,∞≤s.with s≥1. As is described in many references (for example, Lemma 3.2 in [29]), if f∈Gs, then 1/1+s≤fx≤1+sa.e. and if f∈Wloc1,∞, f is invertible and fx+1=fx+1 for x∈ℝ, and there is a c≥1 such that 1/c≤fx≤ca.e. and then f∈Gs for some s is dependent only on c.
Define subsetsℱ and ℱs of G as(32)ℱ=y,U,H,μ,e,r∈G:11+ey+H∈Gand(33)ℱs=y,U,H,μ,e,r∈G:11+ey+H∈Gs.
LetG˜=G×ℝ be a group with its operation defined by f1,γ1f2,γ2=f2°f1,γ1+γ2. The map Φ:G˜×ℱ⟶ℱ defined as(34)Φf,γy,U,H,μ,e,r=y°f,U°f,H°f+γ,μ,e,r°ffx,is an equivalence class map on G˜. Based on the proof of Theorem 4.2 in [14], we have the following theorem by a slight modification.Theorem 3.
DefineS˜t on ℱ/G˜ as S˜tX=StX; then, S˜t generates a continuous semigroup.Theorem 4.
For anyu,ρ,η∈D, let(35)yξ=supy∣Fηy+y1+e<ξ,Hξ=1+eξ−yξ,Uξ=ut,yt,ξ,rξ=ρt,yt,ξyξ,μ=∫01ut,xdx,e=η0,1,where(36)Fηx=η0,x,ifx>0,0,ifx=0,−ηx,0,ifx<0.
Theny,U,H,μ,e,r∈ℱ0. We denote L:D⟶ℱ/G˜, and let Lu,ρ,η∈ℱ/G˜ denote the equivalence class of y,U,H,μ,e,r.
Before giving the proof of Theorem4, we give a critical lemma. Define a set(37)B=x∈ℝ∣limε⟶012εηx−ε,x+ε=ux2+ρ2.
Note thatux2+ρ2dx here is the absolute continuous part of η. By Besicovitch’s derivation theorem, one can obtain measBc=0.Lemma 1.
Forξ∈y−1B, we have(38)yξξux2yξ+ρ2yξ+yξ=1+e.Proof.
Firstly, we claim that for alli∈ℕ, there is a 0<ε<1/i such that x−ε and x+ε are in suppηsc, where ηs is the singular part of Radon measure η and its support suppηs is a point set with a countable number of elements. If not, then there exists i∈ℕ, such that for any 0<ε<1/i, x−ε∈suppηs or x+ε∈suppηs and then for any z∈x−ε,x+ε/suppηs, 2x−z∈suppηs. Consequently, we may construct an injection between x−1/i,x+1/i/suppηs and suppηs, which is rather impossible because x−1/i,x+1/i/suppηs is uncountable and suppηs is countable.
Then, we can construct sequencesyξi and yξi′ such that(39)12yξi+yξi′=yξandyξi−yξi′≤1i.
By the definition ofFη, we have(40)ηyξi,yξi′+yξi′−yξi=1+eξi′−ξi.
Dividing equation (40) by ξi′−ξi and taking i⟶∞, we obtain equation (38).Proof of Theorem4.
By Lemma1 and slight modifications of Theorem 4.3 in [14], we will establish the map from Lagrangian coordinates to Eulerian ones, which is a generalization of Theorem 4.7 in [14]. We only state the results here, as this proof and that of Theorem 4.7 in [14] are very similar.Theorem 5.
Given anyX∈ℱ/G˜, we define u,ρ,η by(41)ux=Uξfor anyξsuch thatx=yξ,ρxdx=y♯rdξ,η=y♯Hξdξ,belonging to D, where f♯ξB=ξf−1B for any Borel set B is called the push forward element of ξ by f. Then, u,ρ,η belongs to D and is independent of the representative X from X. We denote M:ℱ/G˜⟶D.
Next, we will clarify the relation betweenL and M.Theorem 6.
The mapsM:ℱ/G˜⟶D and L:D⟶ℱ/G˜ are invertible and(42)L°M=Id∣ℱ/G˜,M°L=Id∣D.Proof.
The proof follows the same lines as in Theorem 4.8 in [14], so we do not present it here.
Now, we obtain the solution mapTt=M°S˜t°L, that is, D⟶Lℱ/G˜⟶Stℱ/G˜⟶MD.


## 4. Weak Solutions

Definition 2.
Letu:ℝ+×ℝ⟶ℝ and ρ:ℝ+×ℝ⟶ℝ. Assume that u and ρ satisfy the following:(i)
u∈L∞0,∞;Hper1 and ρ∈L∞0,∞;Lper2.(ii)
If the equations(43)∫∫ℝ+×0,1−ut,xφtt,x+uux+Pxφt,xdxdt=∫0,1u0xφ0xdx,where P=μ−∂x2−12μuu+1/2ux2+1/2ρ2t,yt,ξ,(44)∫∫ℝ+×ℝ2μ2+12e2−2μu−12ux2−12ρ2φt,x+Pxt,xφxt,xdxdt=0,and(45)∫∫ℝ+×0,1−ρt,xφtt,x−ut,xρt,xφxt,xdxdt=∫0,1ρ0xφ0xdx,hold for all spatial periodic functions φ∈C0∞0,∞,ℝ, then we say u,ρ is a global weak solution of equation (8).
Moreover, if this solutionu,ρ satisfies(46)∫0,1ux2+ρ2dx=∫0,1u0,x2+ρ02dx,a.e. fort≥0,then we say it is a global conservative solution of equation (8).Theorem 7.
Givenu0,ρ0,η0∈D, if Ttu0,ρ0,η0=ut,ρt,ηt, then u,ρ is a global conservative solution of equation (8).Proof.
Theorem2 and Definition 1 imply that 1˙ in Definition 2 holds. In the following section, we will prove equations (43)–(46) one by one for any spatial periodic function φ∈C0∞0,∞,ℝ. Let x=yt,ξ and we have dx=yξdξ. Since Uξ=uxt,yt,ξandyξ=ytξ, we obtain(47)∫∫ℝ+×0,1−ut,xφtt,x+uuxφt,xdxdt=∫∫ℝ+×0,1−ut,yt,ξφtt,yt,ξyξ+ut,yt,ξuxt,yt,ξφt,yt,ξyξdξdt=∫∫ℝ+×0,1−Ut,ξφtt,yt,ξyξ+Ut,ξUξt,yt,ξφt,yt,ξdξdt.
By(48)Uyξφ°yt=Utyξφ°y+UUξφ°y+Uyξφt°y+U2yξφ°yand Ut=ut+uux=−Q=−Px, we have(49)−Uyξφt°y+UUξφ°y=−Uyξφ°yt+Utyξφ°y+2UUξφ°y+U2yξφ°y=−Uyξφ°yt−Pxyξφ°y+U2φ°yξ.
Integrating this formula into equation (47), we obtain(50)∫∫ℝ+×0,1−ut,xφtt,x+uuxφt,xdxdt=∫0,1∫0∞−Uyξφ°ytdtdx−∫∫ℝ+×0,1Pxyξφ°ydξdt=∫0,1u0xφ0xdx−∫∫ℝ+×0,1Pxt,xφt,xdxdt.
And, the proof for equation (43) completes here. Using equation (28), a direct computation implies that(51)∫∫ℝ+×ℝPxt,xφxt,xdxdt=∫∫ℝ+×ℝPxt,yt,ξφxt,yt,ξyξdξdt=∫∫ℝ+×ℝQt,ξφξt,yt,ξdξdt=−∫∫ℝ+×ℝQξt,ξφt,yt,ξdξdt=−∫∫ℝ+×ℝ−2μ2−12e+2μuUt,ξ+12ux2t,yt,ξ+12ρ2t,yt,ξyξφt,yt,ξdξdt=−∫∫ℝ+×ℝ−2μ2−12e+2μuut,x+12ux2t,x+12ρ2t,xφt,xdxdt.
This completes the proof for equation (44). By rt=0, we have(52)rt,ξφt,yt,ξt=rt,ξφtt,yt,ξ+rt,ξUt,ξφxt,yt,ξ.
It satisfies that(53)∫∫ℝ+×0,1−ρt,xφtt,x−ut,xρt,xφxt,xdxdt=∫∫ℝ+×0,1−ρt,yt,ξφtt,yt,ξ−ut,yt,ξρt,yt,ξφxt,yt,ξyξdξdt=∫∫ℝ+×0,1−rt,ξφtt,yt,ξ−Ut,ξrt,ξφxt,yt,ξdξdt=∫∫ℝ+×0,1−rt,ξφt,yt,ξtdξdt=∫0,1ρ0xφ0xdx.
And, this completes the proof of equation (45). Similarly, let x=yt,ξ in the left side of equation (46); we have(54)∫0,1ux2+ρ2dx∫0,1ux2t,yt,ξ+ρ2t,yt,ξyξdξ,=∫0,1Hξdξ=Ht,1−Ht,0=H0,1−H0,0=∫0,1u0,x2+ρ02dx.
This completes the proof of equation (46).



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