# Wave Breaking for the Rotational Camassa–Holm Equation on the Circle

Authors: Rendong Chen
Journal: Journal of Mathematics
 (2022)

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## Abstract

In this paper, we consider the rotational Camassa–Holm equation on the circle. Sufficient conditions on the initial data to guarantee wave breaking are established.


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## Body

## 1. Introduction

Recently, a rotational Camassa–Holm equation was proposed in [1–4], which reads as(1)mt+umx+2uxm+cux−β0βuxxx+ω1α2u2ux+ω2α3u3ux=0,m=u−uxx,with(2)c=1+Ω2−Ω,α=c21+c2,β0=cc4+6c2−16c2+12,β=3c4+8c2−16c2+12,ω1=3cc2−1c2−22c2+13,ω2=c2−12c2−28c2−12c2+15,where Ω is a parameter related to the Coriolis effect. In the whole space, Gui et al. [3] established the local well-posedness in Hsℝ, s>3/2. Sufficient conditions to guarantee wave breaking phenomena were also studied in [3]. Tu et al. [4] established the global existence and uniqueness of the energy conservative weak solutions. In [5], they found some explicit solutions by elliptic integrals. They also classified the members of the equation describing pseudo-spherical surfaces. Generic regularity of conservative solutions was investigated in [6]. Moon [7] studied the question of nonexistence of periodic peaked traveling wave solution for rotational Camassa–Holm equation.IfΩ=0, it follows ω1=ω2=0, and then system (1) reduces to the Dullin–Gottwald–Holm equation [8]. Some mathematical studies can be found in [9–12].Another highly related model is the well-known Camassa–Holm equation [13] mt+umx+2uxm=0,m=u−uxx. This equation has a physical background with shallow water propagation. The Camassa–Holm equation [14,15] has infinitely many conservation laws. In [16,17], they established the local well-posedness. Wave breaking phenomena were widely studied in [16–19]. McKean [20] (see also [21] for a simple proof) established a necessary and sufficient condition on the initial datum u0, which depends on the shape of y=u−uxx. In [22], the orbital stability of the peakons was proved. In [23, 24], they studied persistence properties and unique continuation of solutions. The long-time behavior for the support of momentum density of the Camassa–Holm equation was discussed in [25]. Mathematical studies for the related models can been found in [26–28].For the convenience of research, in this paper, we consider the rotational Camassa–Holm equation as the following form on the circle:(3)mt+umx+2uxm=αux+βu2ux+γu3ux+Γuxxx,where m=u−uxx and α,β,γ,Γ are real constants. x∈S, where S denotes the unit circle, i.e., S=ℝ/ℤ.The paper is organized as follows. In Section2, we introduce some useful lemmas. The main result and its proof will be shown in Section 3.


## 2. Preliminaries

LetΛ=1−∂x21/2; then, the operator Λ−2 can be expressed by its associated Green’s function as(4)u=Λ−2mx,t=G∗mx,t,Gx=coshx−x−1/22sinh1/2.Then, (3) can be rewritten as(5)ut+u+Γux+∂xG∗u2+ux22=∂xG∗α+Γu+β3u3+γ3u4.By applying Kato’s semigroup theory [29] and similar arguments in [3], we can also have the local well-posedness on the circle.Theorem 1 (see [3]).
Givenux,t=0=u0∈HsS, s>3/2, then there exist a maximal T=Tu0>0 and a unique solution u to the rotational Camassa–Holm equation (3) such that(6)u=u·,u0∈C0,T;HsS∩C10,T;Hs−1S.Then, the precise blow-up scenario will been shown as follows.Theorem 2 (see [3]).
Assume thatu0∈H2S and let T be the maximal existence time of the solution ux,t to equation (3) with the initial data u0x. Then, the corresponding solution of the rotational Camassa–Holm equation (3) blows up in finite time if and only if(7)limt⟶Tinfx∈Suxx,t=−∞.Then, we introduce some useful inequality in the circle.Lemma 1 (see [30]).
For allf∈H1S, the following inequality holds:(8)maxx∈0,1f2x≤C0fH1S2,where C0=e1/2+e−1/2/2e1/2−e−1/2.Lemma 2 (see [30]).
Letf∈H1S. If ∫Sfdx=0, then the following inequalities hold:(9)fL∞S2≤112∫Sfx2dx≤112fH1S2,∫Sf2xdx≤112∫Sfx2xdx,∫Sf2xfx2xdx≤112∫Sfx2xdx2.


## 3. Main Results

In this section, we firstly establish a sufficient condition to guarantee the blow up of the solution to the rotational Camassa–Holm equation (3). We give the particle trajectory as(10)qtx,t=uqx,t,t+Γ,0<t<T,x∈S,qx,0=x,x∈S,where T is the lifespan of the solution. Taking derivative (10) with respect to x, we obtain(11)dqtx,tdx=qtx=uxqxqx,t,t,t∈0,T.Therefore,(12)qx=exp∫0tuxq,sds,0<t<T,x∈S,qxx,0=1,x∈S,which is always positive before the blow-up time.Theorem 3.
Assume thatu0∈H2S and there exists x0∈S such that(13)u0xx0<−2C∗,where C∗=1/2C0u0H12+2α+ΓC0u0H1+2β/3C03/2u0H13+γ/4u0H14; then, the corresponding solution ux,t to equation blows up at a finite time T bounded by(14)T≤1−1/2u0xx0+C∗/u0xx0.Proof.
Letℋu=α+Γu+β/3u3+γ/4u4. Differentiating (5) with respect to x yields(15)uxt+u+Γuxx+ux2+∂x2G∗u2+ux22=∂x2G∗ℋ.
Then, we haveuxt at the point qx,t,t as(16)uxtqx,t,t=u2−ux22−G∗u2+ux22−ℋu+G∗ℋqx,t,t.
Without loss of generality, we choose0≤qx,t≤1, and we have(17)G∗u2+ux22qx,t,t=12sinh1/2∫0qx,teqx,t−η−1/2+e1/2+η−qx,t2u2+ux22η,tdη+12sinh1/2∫qx,t1eqx,t−η+1/2+e−1/2+η−qx,t2u2+ux22η,tdη.
Note that(18)∫0qx,te−ηu2+ux2η,tdη≥−2∫0qx,te−ηuuxη,tdη=−e−ηu2η,t|0qx,t−∫0qx,te−ηu2η,tdη.
We have(19)∫0qx,te−ηu2+12ux2qx,tdη≥−12e−ηu2η,t|0qx,t.
Similar argument yields that(20)∫0qx,teηu2+12ux2η,tdη≥12eηu2η,t|0qx,t,∫qx,t1e−ηu2+12ux2η,tdη≥−12e−ηu2η,t|qx,t1,∫qx,t1eηu2+12ux2η,tdη≥12eηu2η,t|qx,t1.
Then, we have(21)G∗u2+ux22qx,t,t≥12u2qx,t,t.
Combining (21) into (16), we obtain(22)uxtqx,t,t≤u22−ux22−ℋu+G∗ℋqx,t,t.
A direct calculation gives(23)G∗ℋqx,t,t=12sinh1/2∫0qx,teqx,t−η−1/2+e1/2+η−qx,t2ℋη,tdη+12sinh1/2∫qx,t1eqx,t−η+1/2+e−1/2+η−qx,t2ℋη,tdη≤maxx∈Sℋ12sinh1/2∫0qx,teqx,t−η−1/2+e1/2+η−qx,t2dη+12sinh1/2∫qx,t1eqx,t−η+1/2+e−1/2+η−qx,t2dη=maxx∈Sℋ.
By Lemma1, we have(24)u22≤12C0u0H12.
Forγ≥0, we have(25)−ℋu≤−α+Γu−β3u3≤α+ΓC0u0H1+β3C03/2u0H13,G∗ℋ≤maxx∈Sℋ≤α+ΓC0u0H1+β3C03/2u0H13+γ4C02u0H14.
Forγ<0, we have(26)−ℋu≤α+ΓC0u0H1+β3C03/2u0H13+γ4C02u0H14,G∗ℋ≤G∗α+Γu+β3u3≤α+ΓC0u0H1+β3C03/2u0H13.
By (22) and the definition of C∗, we have(27)uxtqx,t,t≤−ux22qx,t,t+C∗.
This is a Riccati type inequality. By the fundamental ODE methods, the proof is completed by the initial condition.Theorem 4.
Assume thatu0∈H2S, ∫Su0dx=0, and there exists x0∈S such that(28)u0xx0<−2C⋆,where C⋆=1/24u0H12+2α+Γ/12u0H1+β/1812u0H13+γ/576u0H14; then, the corresponding solution ux,t to equation blows up at a finite time T bounded by(29)T≤1−1/2u0xx0+C⋆/u0xx0.Proof.
Recall that(30)uxtqx,t,t≤u22−ux22−ℋu+G∗ℋqx,t,t.
With the initial condition∫Su0dx=0, by Lemma 2, we have(31)u22≤124u0H12.
Forγ≥0, we have(32)−ℋu≤−α+Γu−β3u3≤α+Γ12u0H1+β3612u0H13,G∗ℋ≤maxx∈Sℋ≤α+Γ12u0H1+β3612u0H13+γ576u0H14.
Forγ<0, we have(33)−ℋu≤α+Γ12u0H1+β3612u0H13+γ576u0H14,G∗ℋ≤G∗α+Γu+β3u3≤α+Γ12u0H1+β3612u0H13.
Combining the above arguments into (30), we have(34)uxtqx,t,t≤−ux22qx,t,t+C⋆,where C⋆ is defined in Theorem 4. The proof of Theorem 4 is completed by the fundamental ODE methods.



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*Source: 9603850-2022-02-08.xml*