# Global Existence and Blow-Up for a Weakly Dissipative Modified Two-Component Camassa-Holm System

Authors: Xuanxuan Han; Tingting Wang; Yibin Lu
Journal: Advances in Mathematical Physics
 (2022)

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## Abstract

In this paper, we consider a weakly dissipative modified two-component Camassa-Holm system. We demonstrate a simple sufficient condition on initial data to guarantee blow-up of solutions in finite time, and the solutions exist globally in time.


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## Body

## 1. Introduction

The well-known Camassa-Holm (CH) equation(1)mt+umx+2uxm=0,m=u−uxx,which was first derived by Fokas and Fuchssteiner [1] and later derived as a model for unidirectional propagation of shallow water over a flat bottom by Camassa and Holm [2]. After the birth of the CH equation, a lot of works have been carried out to it. For example, the CH equation has travelling wave solutions of the form ce−x−ct, called peakons, which describes an essential feature of the travelling waves of largest amplitude [3–6]. It is shown in [7] that the blow-up occurs in the form of breaking waves; namely, the solution remains bounded but its slope becomes unbounded in finite time.In general, it is difficult to avoid energy dissipation mechanisms in our real world. Ott and Sudan [8] studied how the KdV equation was modified by the presence of dissipation and the effect of such dissipation on solitary solution of the KdV equation. Ghidaglia [9] investigated the long-time behavior of solutions to a weakly dissipative KdV equation as a finite-dimensional dynamical system. Inspired by the above works, Wu and Yin consider the following weakly dissipative CH equation [10, 11]:
(2)mt+umx+2uxm+λm=0,m=uxx,where λm is the weakly dissipative term and λ>0 is a dissipative parameter. Wu and Yin show that if the initial momentum m0=u0−u0xx at some point x0∈ℝ satisfies some sign condition, then the corresponding solution to Equation (2) exists globally in time and blows up in finite time. Novruzov and Hagverdiyev [12] derived a condition of the changing of the sign of m0 at some point x0∈ℝ to guarantee blow-up in finite time.The two-component Camassa-Holm system is as follows:(3)mt+umx+2uxm+ρρx=0,ρt+ρux=0,where m=u−uxx, ut,x describes the horizontal velocity of the fluid, and ρt,x describes the horizontal deviation of the surface from equilibrium. The system (3) appears initially in [13]; then, Constantin and Ivanov [14] give a demonstration about its derivation in view of the shallow water theory from the hydrodynamic point of view. Local well-posedness, blow-up, global existence, stability, and other mathematical properties can be seen in [15–22] and references therein.The modified two-component Camassa-Holm system is as follows:(4)mt+umx+2uxm+ρρ¯x=0,ρt+ρux=0,where m=u−uxx, ρ=1−∂x2ρ¯−ρ¯0, u denotes the velocity field, and ρ¯0 is taken to be a constant. The system (4) does admit peaked solutions in the velocity and average density; we refer this to Ref. [23] for details. In Ref. [23], the authors analytically identified the steepening mechanism that allows the singular solutions to emerge from smooth spatially confined initial data. They found that wave breaking in the fluid velocity does not imply singularity in the pointwise density ρ at the point of vertical slope. Some related work can be found in [24–29]. Let γ=ρ¯−ρ¯0; then, the system (4) is equivalent to the following one, where m=u−uxx and ρ=γ−γxx.In this paper, we are interested in the effect of the weakly dissipative term on the system (5) as follows:
(5)mt+umx+2uxm+λm+ργx=0,ρt+ρux+λρ=0,where m=u−uxx, ρ=γ−γxx, and λ>0 is a dissipative parameter. The main difference between the systems (4) and (5) is that the system (5) does not have conservation law.
(6)∫ℝu2+ux2+γ2+γx2dx.In fact, for the system (5), Et decays to zero as time goes to infinity (see Lemma 7). Recently, a blow-up result for the system (5) is presented in [30]. Similar to [10, 11], Ref. [30] shows that if the initial momentum m0=u0−u0xx at some point x0∈ℝ satisfies some sign condition, then the corresponding solution to the system (5) blows up in finite time.The main goal of the present paper is to demonstrate a simple condition guaranteeing blow-up of solutions in finite time and guaranteeing the solutions exist globally in time by using some properties of the solution generated by initial data. Our results could be stated as follows:Theorem 1.
Suppose thatu0,γ0∈Hsℝ with s≥5/2, if
(7)m0L22+ρ0L221/2<4λ3.
Then, the corresponding solutionut,x to the system (5) exists globally in time.Theorem 2.
Suppose thatu0,γ0∈Hsℝ×Hs−1ℝ with s≥5/2. Assume that ρ0x0=0, ρ0x≥0 on −∞,x0, and ρ0x≤0 on x0,∞, and assume further that
(8)2λ+12u0H12+γ0H121/2<u0x0−u0xx0,for some point x0. Then, the corresponding solution to the system (5) blows up in finite time.Remark 3.
Note that our theorems do not need to assume that the initial momentumm0=u0−u0xx at some point x0∈ℝ satisfies some sign condition, so our theorems improve the results in [30]. If ρ=0, our theorems improve the global existence and blow-up results in [11] and cover the results in [12].The rest of this paper is organized as follows. In Section2, we recall several useful results which are crucial in the proof of Theorem 1 and Theorem 2. In Section 3 and Section 4, we complete the proof of our results.


## 2. Preliminaries

In this section, we recall several useful results to pursue our goal. First, we recall local well-posedness for the system (5).Theorem 4 (see [30]).
Givenu0,γ0∈Hsℝ×Hs−1ℝ with s≥5/2, there exist a maximal existence time T=Tu0Hsℝ,γ0Hs−1ℝ and a unique solution u,γ to the system (5) such that
(9)u,γ∈C0,T;Hsℝ×Hs−1ℝ∩C10,T;Hs−1ℝ×Hs−2ℝ.
Moreover, the solution depends continuously on the initial data, i.e., the mappingu0,γ0↦u,γ:
(10)Hsℝ×Hs−1ℝ↦C0,T;Hsℝ×Hs−1ℝ∩C10,T;Hs−1ℝ×Hs−2ℝis continuous.Next, we state the following precise blow-up scenario.Theorem 5 (see [30]).
Letu0,γ0∈Hsℝ×Hs−1ℝ with s≥5/2 and T>0 be the maximal existence time of the solution u,γ to the system (5) with initial data u0,γ0. Then, the corresponding solution blows up in a finite time T<∞ if and only if
(11)liminft⟶Tinfx∈ℝuxt,x=−∞.Consider the following initial value problem of ordinary differential equation (ODE):(12)qt=ut,qt,x,t,x∈0,T×ℝ,q0,x=x,x∈ℝ.The following lemma will be used to prove our theorem.Lemma 6 (see [30]).
Letu0,γ0∈Hsℝ×Hs−1ℝ with s≥5/2 and T>0 be the maximal existence time of the solution u,γ to the system (5) with initial data u0,γ0. Then, we have
(13)ρt,qt,xqxt,x=ρ0xe−λt,t,x∈0,T×ℝ.
Moreover, if there existx0∈ℝ such that ρ0x0=0, then ρt,qt,x0=0 for all t∈0,T.Lemma 7 (see [30]).
Letu0,γ0∈Hsℝ×Hs−1ℝ with s≥5/2 and T>0 be the maximal existence time of the solution u,γ to the system (5) with initial data u0,γ0. Then, we have
(14)Et=∫ℝu2+ux2+γ2+γx2dx=e−2λt∫ℝu02+u0x2+γ02+γ0x2dxor
(15)Et=uH12+γH12=e−2λtu0H12+γ0H12=e−2λtE0.Lemma 8.
Ifρ0x0=0, ρ0x≥0 on −∞,x0, and ρ0x≤0 on x0,∞, then for any fixed t,
(16)γx2t,x−γ2t,x≤γx2t,qt,x0−γ2t,qt,x0,for all x∈ℝ.Proof.
Ifx≤qt,x0, then
(17)γx2t,x−γ2t,x=−∫−∞qt,x0eξρt,ξdξ−∫xqt,x0eξρt,ξdξ×∫qt,x0∞e−ξρt,ξdξ+∫xqt,x0e−ξρt,ξdξ=γx2t,qt,x0−γ2t,qt,x0−∫−∞xeξρt,ξdξ∫xqt,x0e−ξρt,ξdξ+∫xqt,x0eξρt,ξdξ∫qt,x0∞e−ξρt,ξdξ≤γx2t,qt,x0−γ2t,qt,x0.
Similarly, ifx≥qt,x0, we also have
(18)γx2t,x−γ2t,x≤γx2t,qt,x0−γ2t,qt,x0.
Therefore, we complete the proof of Lemma8.


## 3. Proof of Theorem1

First, multiplying the first equation of the system (5) by 2m and integrating by parts, we get
(19)ddt∫ℝm2dx=−2∫ℝummxdx−4∫ℝm2uxdx−2λ∫ℝm2dx−2∫ℝmργxdx=−3∫ℝm2uxdx−2λ∫ℝm2dx−2∫ℝmργxdx.Similarly,(20)ddt∫ℝρ2dx=−∫ℝρ2uxdx−2λ∫ℝρ2dx.Adding (19) and (20), we have
(21)ddt∫ℝm2+ρ2dx+2λ∫ℝm2+ρ2dx=−3∫ℝm2uxdx−∫ℝρ2uxdx−2∫ℝmργxdx.Multiplying (21) by e2λt yields
(22)ddte2λt∫ℝm2+ρ2dx=−3e2λt∫ℝm2uxdx−e2λt∫ℝρ2uxdx−2e2λt∫ℝmργxdx.Note thatpx=Δ1/2e−x; we have 1−∂x2−1f=p∗f for all f∈L2ℝ and p∗m=u, where we denote by $∗$ the convolution. Then, taking the Young inequality, one gets
(23)uxL∞=px∗mL∞≤pxL2mL2≤12mL2,∫ℝρ2uxdx≤uxL∞∫ℝρ2dx≤12mL2∫ℝρ2dx,∫ℝmργxdx≤mL∞∫ℝργxdx=0.Hence, we have(24)ddte2λt∫ℝm2+ρ2dx≤32e2λt∫ℝm2dx1/2∫ℝm2dx+12e2λt∫ℝm2dx1/2∫ℝρ2dx≤32e2λt∫ℝm2+ρ2dx3/2=32e−λte2λt∫ℝm2+ρ2dx3/2.It is easy to derive that(25)ddte2λt∫ℝm2+ρ2dx−1/2≥−34e−λt.Integrating from0 to t yields
(26)e2λt∫ℝm2+ρ2dx−1/2−∫ℝm02+ρ02dx−1/2≥34λe−λt−1≥−34λ.Thus, it follows that(27)mL22+ρL221/2≤e−λtm0L22+ρ0L22−1/2−34λ−1,due to m0L22+ρ0L221/2<4λ/3 and
(28)uxL∞≤12mL2<mL22+ρL221/2<e−λtm0L22+ρ0L22−1/2−34λ−1.According to Theorem5, we have that the solution exists globally in time.


## 4. Proof of Theorem2

First, we introduce the following notation:(29)It=∫−∞qt,x0eξmt,ξdξ.DifferentiatingIt with respect to x, we get
(30)dItdt=∫−∞qt,x0eξmtt,ξdξ+eqt,x0mt,qt,x0qtt,x0.By using the first equation of the system (5) and integrating by parts, we have
(31)∫−∞qt,x0eξmtt,ξdξ=∫−∞qt,x0eξ−umx−uxm−λm−ργxdξ=−eξum−∞qt,x0+∫−∞qt,x0eξumdξ−12eξu2−ux2−∞qt,x0+12∫−∞qt,x0eξu2−ux2dξ−λ∫−∞qt,x0eξmdξ+12eqt,x0γx2t,qt,x0−12eqt,x0γ2t,qt,x0−12∫−∞qt,x0eξγx2−γ2dξ=−eξum−∞qt,x0+∫−∞qt,x0eξu2+12ux2dξ−λ∫−∞qt,x0eξmdξ+12eξux2−∞qt,x0−eξuux−∞qt,x0+12eqt,x0γx2t,qt,x0−12eqt,x0γ2t,qt,x0−12∫−∞qt,x0eξγx2−γ2dξ.Note that(32)∫−∞qt,x0eξu2+12ux2dξ=12∫−∞qt,x0eξu2dξ+12∫−∞qt,x0eξu2+ux2dξ≥12∫−∞qt,x0eξu2dξ+12∫−∞qt,x02eξuuxdξ≥12∫−∞qt,x0eξu2dξ+12∫−∞qt,x0eξu2xdξ=12eqt,x0u2t,qt,x0.Inserting (31), (32), and (12) into (30), we have
(33)dItdt+λ∫−∞qt,x0eξmdξ≥12eqt,x0u2t,qt,x0+12eξux2−∞qt,x0−eξuux−∞qt,x0−eξum−∞qt,x0+eqt,x0mt,qt,x0qtt,x0+12eqt,x0γx2t,qt,x0−12eqt,x0γ2t,qt,x0−12∫−∞qt,x0eξγx2−γ2dξ≥12eqt,x0u2t,qt,x0−eqt,x0ut,qt,x0uxt,qt,x0+12eqt,x0ux2t,qt,x0=12eqt,x0ut,qt,x0−uxt,qt,x02.Next, by integrating by parts, we have(34)It=∫−∞qt,x0eξmt,ξdξ=∫−∞qt,x0eξudξ+∫−∞qt,x0eξuxdξ−eqt,x0uxt,qt,x0=eqt,x0ut,qt,x0−uxt,qt,x0.Thus, from (33) and (34), it is obvious that
(35)dItdt≥12e−qt,x0It2−λIt.Multiplying (35) by e−qt,x0, we have
(36)e−qt,x0dItdt≥12e−2qt,x0It2−λe−qt,x0It.Due toqtt,x0=ut,qt,x0, adding −qtt,x0e−qt,x0It to the left side of (36) and −ut,qt,x0e−qt,x0It to the right side of (36) yields
(37)ddte−qt,x0It≥12e−qt,x0It2−λ+ue−qt,x0It=12e−qt,x0It−λ−ue−qt,x0It.Hence,d/dte−qt,x0It>0 holds as long as 1/2e−qt,x0It>λ+u and e−qt,x0It>0. In view of Lemma 7, we have
(38)uL∞≤12uH1≤12uH12+γH121/2≤12e−λtE0.Note that in (34) and the condition of Theorem 2, we get
(39)12e−q0x0I0=12u0x0−u0xx0>λ+12E0≥λ+u0L∞≥λ+u0x0.Therefore, we can conclude thatd/dte−qt,x0It=d/dtut,qt,x0−uxt,qt,x0>0 holds. Since ut,qt,x0−uxt,qt,x0 is a continuous function, so we have
(40)12ut,qt,x0−uxt,qt,x0>12u0x0−u0xx0>λ+12E0.Lemma7 implies that ut,qt,x0⟶0 as t⟶∞, so there exist a T0>0 such that for t>T0(41)−12uxt,qt,x0>λ+12E0.Thus, we have(42)−12ux2t,qt,x0−λuxt,qt,x0=uxt,qt,x0−12uxt,qt,x0−λ=−uxt,qt,x0−12uxt,qt,x0−λ<−2λ+12E012E0=−2λ+12E0E0.Note thatpx=Δ1/2e−x, we have 1−∂x2−1f=p∗f for all f∈L2ℝ and p∗m=u, where we denote by ∗ the convolution. Then, we can rewrite the first equation of the system (5) as follows:
(43)ut+uux+∂xp∗u2+12ux2+12γ2−12γx2+λu=0.Differentiating (43) with respect to x, we get
(44)utx+uuxx=u2−12ux2+12γ2−12γx2−p∗u2+12ux2+12γ2−12γx2−λux,then
(45)ddtuxt,qt,x0=utxt,qt,x0+ut,qt,x0uxxt,qt,x0=u2t,qt,x0−12ux2t,qt,x0+12γ2t,qt,x0−12γx2t,qt,x0−p∗u2+12ux2+12γ2−12γx2−λuxt,qt,x0≤u2t,qt,x0−12ux2t,qt,x0−p∗u2+12ux2−λuxt,qt,x0≤12u2t,qt,x0−12ux2t,qt,x0−λuxt,qt,x0,where we used Lemma 6 and the inequality p∗u2+1/2ux2≥1/2u2. Then, by (42) and (45), we can infer that
(46)duxt,qt,x0dt<12u2t,qt,x0−2λ+12E0E0.Further, by Lemma7, for sufficiently large t, we have
(47)12u2t,qt,x0<2−12λ+12E0E0.Then, there exists aT1 such that for t>T1>T0,
(48)duxt,qt,x0dt<−12λ+12E0E0.Thus, integrating fromT1 to t, we get
(49)uxt,qt,x0<uxT1,qT1,x0−12λ+12E0E0t−T1.This means that we can choose aT2>T1 such that for all t>T2,
(50)−λuxt,qt,x0<18ux2t,qt,x0,and by Lemma 7, we have
(51)12u2t,qt,x0<18ux2t,qt,x0,for t>T2.Then, substituting (50) and (51) into (45) yields
(52)ddtuxt,qt,x0≤12u2t,qt,x0−12ux2t,qt,x0−λuxt,qt,x0<18ux2t,qt,x0−12ux2t,qt,x0+18ux2t,qt,x0=−14ux2t,qt,x0,which leads to
(53)−ddt1uxt,qt,x0<−14,and integrating from T2 to t gives
(54)0<−1uxt,qt,x0<−1uxT2,qT2,x0−14t−T2.Thus, if we suppose that the solution exists globally, then for sufficiently larget∗>T2,
(55)0<−1uxt,qt,x0<−1uxT2,qT2,x0−14t∗−T2<0,which contradicts that the solution blows up in finite time. Therefore, we complete the proof of Theorem 2.



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