# Construction of 2-Peakon Solutions and Nonuniqueness for a Generalized mCH Equation

Authors: Hao Yu; Aiyong Chen; Kelei Zhang
Journal: Advances in Mathematical Physics
 (2021)

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## Abstract

For the generalized mCH equation, we construct a 2-peakon solution on both the line and the circle, and we can control the size of the initial data. The two peaks at different speeds move in the same direction and eventually collide. This phenomenon is that the solution at the collision time is consistent with another solitary peakon solution. By reversing the time, we get two new solutions with the same initial value and different values at the rest of the time, which means the nonuniqueness for the equation in Sobolev spacesHs is proved for s<3/2.


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## Body

## 1. Introduction

The Camassa-Holm (CH) equation [1–3] is an integrable system with a bi-Hamiltonian structure, which is derived by Camassa and Holm using the asymptotic expansion in the Hamiltonian for Euler’s equation. A special kind of weak solution for this equation describes the solitary wave at the peak, called peakons [4, 5], whose wave slope is discontinuous at the peak. The interactions between any number of peakons were described by the multipeakon solutions [6, 7], in the form of a linear superposition of peakons whose amplitude and velocity change with time.In recent years, people’s great interest in the research of Camassa-Holm (CH) equation has inspired people to explore the CH-type equation, especially the equations that admit peakons and mutipeakons. The CH, Degasperis-Procesi (DP) [8–13], modified CH (mCH) [14–19], and Novikov (NE) [20–22] equations are all integrable systems that admit peakons and mutipeakons. Of course, there are also some nonintegrable systems that admit peakons and mutipeakons, such as the b-family of equations [23], the modified b-family of equations [24], and the cubic ab-family of equations [25]. It is worth noting that the b-family of equations includes the CH equation and the DP equation, the modified b-family of equations includes the NE equation, and the cubic ab-family of equations includes the mCH equation and the NE equation.With the development of research, great interest has been aroused in the uniqueness or posedness of solutions, setting initial valueu0x∈Hsℝ. The study of Li and Olver [26] shows that the CH equation is locally well posed in Hs for s>3/2, and Byers [27] proved the ill-posedness for the CH equation in Hs when s<3/2. Himonas, Grayshan, and Holliman [28] studied the ill-posedness for the DP equation. Himonas and Holliman [29] proved that the NE equation is well posed in Hs for s>3/2. Himonas, Kenig, and Holliman [30] demonstrated the nonuniqueness for the NE equation in Hs when s<3/2 by studying the collision of the peakons. Guo et al. [31] studied the ill-posedness for the CH, DP, and NE equations in critical spaces. Himonas and Mantzavinos [32] proved that the FORQ equation (also called mCH) is well posed in Hs for s>5/2. The nonuniqueness results of Himonas and Holliman [33] show that solutions to the Cauchy problem for the FORQ equation are not unique in Hs when s<3/2. At present, there is no theory to show the uniqueness for the FORQ equation in Hs when 3/2≤s≤5/2. Holmes and Puri [34] discussed the nonuniqueness for the ab-family of equations. Himonas, Grayshan, and Holliman [35] considered the ill-posedness for the b-family of equations in Hs for s<3/2 when b>1. On this basis, Novruzov [36] studied the ill-posedness for the b-family of equations when b<1.In this paper, we consider the Cauchy problem for a generalized mCH (gm- CH) equation which has the following form(1)mt+u2−ux22mx=0,m=u−uxx,(2)ux,0=u0x∈Hs,t>0,x∈X=ℝorT.This equation is obtained by Anco and Recio [37], by extending a Hamiltonian structure of the CH equation. Substituting m=u−uxx into the first equation of (1), it infers the following partial differential equation
(3)ut+u4ux+15ux5−23u2ux3−1−∂x2−115ux5−23u2ux3+1−∂x2−1∂x45u5−13uux4+2u3ux2=0.The results of Anco and Recio [37] show that the gmCH equation admits peakon traveling wave solutions and multipeakon solutions. They studied the existence of the single peakon travelling solutions with c≠0 and classification of 2-peakon solutions. Recio and Anco [38] considered the conservation laws (energy, momentum, H1-norm, etc.) of the gmCH equation, by modifying the general multiplier method combined with some tools from variational calculus. They also discussed the Hamiltonian structure and solitary traveling waves of the gmCH equation, by using the conservation laws. One remark is that the Hamiltonian structure for the family (1) corresponds to an energy conservation law that has a local density but a nonlocal flux.Based on the conservation laws in [38], the Cauchy problem and nonuniqueness of the peakon solutions in this paper are studied. Under this premise, we obtain our main result, and its proof is closely related to the conservation of norms. And based on the existence of peakons in [37], we conduct the research on the peakon solutions. The difference is that we obtain the peakon traveling wave solutions by verifying the weak solution. The peakon traveling wave solutions on the line are given by
(4)ux,t=ae−x−ct,wherec=1118a4.On the circle, they are given by(5)ux,t=b·chζ,ζ=π−x−ctp,b4=c8/15sh4π+4/3sh2π+1,where .p is defined by
(6)xp=x−2πx2π.On the other hand, the classification of 2-peakon solutions in [37] helps us construct 2-peakon solutions. In contrast to this, we construct a special 2-peakon solution based on the characteristics of the ODE system and study the collision of peakons. The result is summarized in the following theorem.Theorem 1.
Solutions to the Cauchy problem for the gmCH equation (1) are not unique in Sobolev spaces Hs when s<3/2.The rest is organized as follows. In Section2, we study the ODE systems that the 2-peakon solutions of the gmCH Equation (1) need to satisfy. In Section 3, we give the proof of Theorem 1 on the line by constructing a 2-peakon solution. In Section 4, we prove Theorem 1 on the circle.


## 2. 2-Peakon on the Line and the Circle

In [37], Recio and Anco studied the multipeakon solutions on the line, and they proved the following result.Theorem 2 (see [37]).
The nonperiodic 2-peakon(7)ux,t=p1te−x−q1t+p2te−x−q2t,x,t∈ℝ×ℝ,is a solution to Equation (1) if its positions q1,q2 and momenta p1,p2 satisfy
(8)p1′=0,q1′=815p12p12+5p1p2e−q1−q2+10p22e−2q1−q2,p2′=0,q2′=815p22p22+5p1p2e−q1−q2+10p12e−2q1−q2.
Now, we consider the 2-peakon system on the circle, based on the methods in [25].Theorem 3.
The periodic 2-peakon(9)ux,t=p1tchx−q1tp−π+p2tchx−q2tp−πis a solution to Equation (1) if its positions q1,q2. and momenta p1,p2 satisfy
(10)q1′=45p121+23sh2πp12ch2π+5p1p2chπchq1−q2p−π+15p14+p24+4p1p23chπchq1−q2p−π+2p12p2283ch2πch2q1−q2p−π−23ch2q1−q2p−π+1,(11)q2′=45p221+23sh2πp22ch2π+5p1p2chπchq2−q1p−π+15p24+p14+4p13p2chπchq2−q1p−π+2p12p2283ch2πch2q2−q1p−π−23ch2q2−q1p−π+1,(12)p1′=0,p2′=0,where ·p is defined as in (6).Proof.
We can rewrite the equation (1) as the following equivalent form
(13)1−∂x2ut+1−15∂x2∂xu5+ux5+∂x223u2ux3+∂x2u3ux2−13uux4−ux5=0.
Letφ∈C∞T be any smooth periodic test function on T, and
(14)ujx,t=pjtchx−qjtp−π,j=1,2,which causes the periodic 2-peakon solution (9) to be rewritten as ux,t=u1x,t+u2x,t. So, we have
(15)1−∂x2ut,φ=1−∂x2u1t,φ+1−∂x2u2t,φ.
Firstly, we calculate∂xuj,φ. Note that when qj∈2kπ,2k+1π,k∈ℕ, we have qj/2π=k and −qj/2π=−k+1, which leads to 2π−qj/2π=−k. Let qj∗=qj−2πqj/2π. Obviously, qj∗∈0,2π. Since x−qj/2π=−k+1 for 0<x<qj∗ and x−qj/2π=−k for qj∗<x<2π, it obtains by integrating by parts
(16)∂xuj,φ=−∫0qj∗pjchx−qj+2k+1πdφ−∫qj∗2πpjchx−qj+2k−1πdφ=−pjchπφqj∗+pjch−qj+2k+1πφ0+pjch−πφqj∗−pjch−qj+2k+1πφ2π+pjshx−qjp−π,φ.
Sincechx=ch−x and φx=φx+2π,
(17)∂xuj,φ=pjshx−qjp−π,φ.
On the other hand, whenqj=2kπ,k∈ℕ, we find x−qj/2π=−k for x∈0,2π. It follows that
(18)∂xuj,φ=−∫02πpjchx−qjp−πφ′dx=pjshx−qjp−π,φ,along with (17) leads to ∂xuj,φ=pjshx−qjp−π,φ, for all qj≥0. Analogously, since sh· is odd, ∂x2uj,φ=−2pjshπφqj∗+uj,φ, which means that 1−∂x2uj,φ=2pjshπφqj∗. Moreover, we find
(19)1−∂x2ut,φ=2shπp1′φq1∗+p1q1′φ′q1∗+p2′φq2∗+p2q2′φ′q2∗.
Now, we conclude−∂x2u4ux,φ. We use ux,t=u1x,t+u2x,t to get
(20)−∂x2u4ux,φ=−u4ux,φ″=−u14u1x+4u13u1xu2+6u12u1xu22+4u1u1xu23+u1xu24+u24u2x+4u1u23u2x+6u12u22u2x+4u13u2u2x+u14u2x,φ″.
Since−qj/2π=−k+1 for qj∈2kπ,2k+1π, where k∈ℕ, we find 0−qjp−π=−qj+2k+1π and 2π−qjp−π=−qj+2k+1π, which combined with integration by parts give that
(21)u14u1x,φ″=p15ch4x−q1p−πshx−q1p−π,φ″=p15∫0q1∗ch4x−q1+2k+1πshx−q1+2k+1πφ″dx+∫q1∗2πch4x−q1+2k−1πshx−q1+2k−1πφ″dx=2p15ch4πshπφ′q1∗+12u12u1x3+12u13u1xu1xx+u14u1xxx,φ,u24u2x,φ″=p25ch4x−q2p−πshx−q2p−π,φ″=2p25ch4πshπφ′q2∗+12u22u2x3+12u23u2xu2xx+u24u2xxx,φ.
Without loss of generality, we assumeq1≤q2. So, we have
(22)4u13u1xu2,φ″=4p14p2∫0q1∗+∫q1∗q2∗+∫q2∗2πch3x−q1p−π⋅shx−q1p−πchx−q2p−πφ″dx=8p14p2shπch3πchq1∗−q2p−πφ′q1∗−ch3πshq1∗−q2p−πφq1∗−ch3q2∗−q1p−πshq2∗−q1p−πφq2∗+24u1u1x3u2+36u12u1xu1xxu2+24u12u1x2u2x+8u13u1xxu2x+4u13u1xu2xx+4u13u1xxxu2,φ,(23)6u12u1xu22,φ″=6p13p22∫0q1∗+∫q1∗q2∗+∫q2∗2πch2x−q1p−π⋅shx−q1p−πch2x−q2p−πφ″dx=12p13p22shπch2πch2q1∗−q2p−πφ′q1∗−2ch2πchq1∗−q2p−πshq1∗−q2p−πφq1∗−2chπch2q2∗−q1p−πshq2∗−q1p−πφq2∗+12u1x3u22+36u1u1xu1xxu22+48u1u1x2u2u2x+24u12u1xxu2u2x+12u12u1xu2x2+12u12u1xu2u2xx+6u12u1xxxu22,φ.
Similar to (22), we obtain
(24)4u1u1xu23,φ″=4p12p23∫0q1∗+∫q1∗q2∗+∫q2∗2πchx−q1p−π⋅shx−q1p−πch3x−q2p−πφ″dx=8p12p23shπchπch3q1∗−q2p−πφ′q1∗−3chπch2q1∗−q2p−πshq1∗−q2p−πφq1∗−3ch2πchq2∗−q1p−πshq2∗−q1p−πφq2∗+12u1xu1xxu23+24u1x2u22u2x+24u1u1xxu22u2x+24u1u1xu2u2x2+12u1u1xu22u2xx+4u1u1xxxu23,φ,u1xu24,φ″=p1p24∫0q1∗+∫q1∗q2∗+∫q2∗2πshx−q1p−πch4x−q2p−πφ″dx=2p1p24shπch4q1∗−q2p−πφ′q1∗−4ch3q1∗−q2p−πshq1∗−q2p−πφq1∗−4ch3πshq2∗−q1p−πφq2∗+u1xxxu24+8u1xxu23u2x+12u1xu22u2x2+4u1xu23u2xx,φ.
Although the above calculation is made by assumingq1≤q2, the result is also valid to any q1 and q2. We next compute
(25)4u1u23u2x,φ″=8p1p24shπch3πchq2∗−q1p−πφ′q2∗−ch3πshq2∗−q1p−πφq2∗−ch3q1∗−q2p−πshq1∗−q2p−πφq1∗+24u1u2u2x3+36u1u22u2xu2xx+24u1xu22u2x2+8u1xu23u2xx+4u1xxu23u2x+4u1u23u2xxx,φ,(26)6u12u22u2x,φ″=12p12p23shπch2πch2q2∗−q1p−πφ′q2∗−2ch2πchq2∗−q1p−πshq2∗−q1p−πφq2∗−2chπch2q1∗−q2p−πshq1∗−q2p−πφq1∗+12u12u2x3+36u12u2u2xu2xx+48u1u1xu2u2x2+24u1u1xu22u2xx+12u1x2u22u2x+12u1u1xxu22u2x+6u12u22u2xxx,φ.(27)4u13u2u2x,φ″=8p13p22shπchπch3q2∗−q1p−πφ′q2∗−3chπch2q2∗−q1p−πshq2∗−q1p−πφq2∗−3ch2πchq1∗−q2p−πshq1∗−q2p−πφq1∗+12u13u2xu2xx+24u12u1xu2x2+24u12u1xu2u2xx+24u1u1x2u2u2x+12u12u1xxu2u2x+4u13u2u2xxx,φ,(28)u14u2x,φ″=2p14p2shπch4q2∗−q1p−πφ′q2∗−4ch3q2∗−q1p−πshq2∗−q1p−πφq2∗−4ch3πshq1∗−q2p−πφq1∗+u14u2xxx+8u13u1xu2xx+12u12u1x2u2x+4u13u1xxu2x,φ.
It follows from (20)-(28) that
(29)−∂x2u4ux,φ=−2p1shπφ′q1∗p1chπ+p2chq1∗−q2p−π4−2p2shπφ′q2∗p2chπ+p1chq2∗−q1p−π4+16p1p2shπφq1∗shq1∗−q2p−πp1chπ+p2chq1∗−q2p−π3+16p1p2shπφq2∗shq2∗−q1p−πp2chπ+p1chq2∗−q1p−π3−u4uxxx+12u3uxuxx+12u2ux3,φ.
Similar to (29), it infers
(30)−∂x215ux5,φ=−2p1shπφ′q1∗−45p14sh4π+p12sh2π+p22sh2q1∗−q2p−π2−2p2shπφ′q2∗−45p24sh4π+p22sh2π+p12sh2q2∗−q1p−π2+8p1p2shπφq1∗shq1∗−q2p−πp1chπ+p2chq1∗−q2p−π⋅p12sh2π+p22sh2q1∗−q2p−π+8p1p2shπφq2∗shq2∗−q1p−πp2chπ+p1chq2∗−q1p−π·p22sh2π+p12sh2q2∗−q1p−π−ux4uxxx+4ux3uxx2,φ,∂x223u2ux3,φ=2p1shπφ′q1∗·23p1chπ+p2chq1∗−q2p−π2p12sh2π+3p22sh2q1∗−q2p−π+2p2shπφ′q2∗·23p2chπ+p1chq2∗−q1p−π2p22sh2π+3p12sh2q2∗−q1p−π−8p1p2shπφq1∗shq1∗−q2p−πp1chπ+p2chq1∗−q2p−π3+43p1chπ+p2chq1∗−q2p−πp12sh2π+p22sh2q1∗−q2p−π−8p1p2shπφq2∗shq2∗−q1p−πp2chπ+p1chq2∗−q1p−π3+43p2chπ+p1chq2∗−q1p−πp22sh2π+p12sh2q2∗−q1p−π+43ux5+283uux3uxx+2u2ux2uxxx+4u2uxuxx2,φwith
(31)∂x2u3ux2,φ−8p1p2shπφq1∗shq1∗−q2p−πp1chπ+p2chq1∗−q2p−π3−8p1p2shπφq2∗shq2∗−q1p−πp2chπ+p1chq2∗−q1p−π3+4u3uxuxx+6u2ux3,φ,(32)∂x13uux4,φ=83p1p2shπφq1∗shq1∗−q2p−πp1chπ+p2chq1∗−q2p−π·p12sh2π+p22sh2q1∗−q2p−π+83p1p2shπφq2∗shq2∗−q1p−πp2chπ+p1chq2∗−q1p−π·p22sh2π+p12sh2q2∗−q1p−π−13ux5+43uux3uxx,φ.
Substituting (29)-(32) into the equation (13), also noting q1∗−q2p=q1−q2p and q2∗−q1p=q2−q1p, which is caused by the fact that qj∗/2π=0 for qj∗∈0,2π, we obtain the system (10).


## 3. Nonuniqueness on the Line

In this section, we use the ODE system (8) to prove Theorem 1 on the line. To do this, we take a 2-peakon solution of the form (7). From the first two items of the system (8), p1t=p10 and p2t=p20 are obvious. At the same time, we have q1′=q2′ if we take the symmetric initial data, p10=p20. The two peaks move at the same speed which means there is no collision. Therefore, we choose the following initial data
(33)p10=−b+δ,q10=0,p20=b,q20=a,with a>0 and b2+b+δ2>5bb+δ, where b+δ>b>0. The selection of these initial data is summarized in Figure 1.Figure 1
Initial profileu0x.According to (33), p1t=−b+δ and p2t=b are obtained. We introduce the symbol q to represent the difference between the positions of the two peakons, in other words, q≐q2−q1. It follows from the ODE system (8) that
(34)q′=815p24−p141+αe−q=815b4−b+δ41+αe−q<0,where −1<α=5p1p2/p12+p22=−5bb+δ/b2+b+δ2<0. Integrating (34), we calculate
(35)lneq+αea+α=815b4−b+δ4t.Sinceq′<0 and q0=q20−q10=a>0, we obtain a collision and a positive collision time when qt=0. Using the symbol τ for the collision time, from (35), we find
(36)τ=158b4−b+δ4ln1+αea+α.Applying expressionsqτ≐limt⟶τ−q1t=limt⟶τ−q2t and vx≐−δe−x−qτ to define the collision location qτ and the collision function v, we get the following proposition.Proposition 4.
TheHs limit of u as t approaches τ is v, or
(37)limt⟶τ−ut−vHs=0.Proof.
We compute the Fourier transform ofu and v, which is denoted by
(38)u^ξ,t=2p1e−iξq11+ξ2+2p2e−iξq21+ξ2=−2b+δe−iξq11+ξ2+2be−iξq21+ξ2,(39)v^ξ=−2δe−iξqτ1+ξ2.
Combining (38) and (39), we have
(40)limt⟶τ−ut−vHs2=limt⟶τ−4∫ℝ1+ξ2s−2be−iξq2−b+δe−iξq1+δe−iξqτ2dξ.
Notice that the equation inside the absolute value can be scaled up to(41)be−iξq2−b+δe−iξq1+δe−iξqτ≤b+δe−iξq1+be−iξq2+δe−iξqτ=2b+δ.
Letfξ=4b+δ21+ξ2s−2. There is no doubt that f is integrable when s<3/2, and f dominates the original integrand, which means, we can apply the dominated convergence theorem and put the limit inside the integral. So, we get
(42)limt⟶τ−4∫ℝ1+ξ2s−2be−iξq2−b+δe−iξq1+δe−iξqτ2dξ=4∫ℝ1+ξ2s−2be−iξq2τ−b+δe−iξq1τ+δe−iξqτ2dξ=0.
Proposition4 is proven.Proof of Theorem 1. (On the line).
In view of the 2-peakon solution we constructed, we need to construct a traveling wave solutionwx,t that satisfies
(43)wx,τ=ux,τ=vx.
Reviewing the system (8), we take the following data
(44)p10=−δ,q10=β,p2=0,q20=0.
The system is simplified and easy to be solved as(45)p1t=−δ,q1t=815δ4t+β.
We introduce the symbolq1u to represent q1 corresponding to u and q1w to represent q1 corresponding to w. It follows from (43) that q1uτ=q1wτ. So, we find β=q1uτ−δ4/b4−b+δ4ln1+α/ea+α,which makes the construction complete. In order to prove the nonuniqueness of the solution, we define two new solutions u~x,t≐u−x,−t+τ,w~x,t≐w−x,−t+τ. Since ux,t and wx,t are two solutions to (1), u−x,−t and w−x,−t are also solutions to (1), which can be obtained by reversing time. These mean u~x,t and u~x,t solve (1). Moreover, through Proposition 4, we have limt⟶0+u~x,t=v−x=limt⟶0+w~x,t. Finally, we note that the initial data for these nonunique solutions can be made arbitrarily small. Since the new initial data is the collision function vx=−δe−x−qτ, we have
(46)vHs2=4μδ2,whereμ=∫ℝ1+ξ2s−2dξ.
Therefore, for anyε>0, we can find a δ to make vHs<ε.


## 4. Nonuniqueness on the Circle

In this section, we use the ODE system (10) to prove the Theorem 1 on the circle. To do this, we take a 2-peakon solution of the form (9). From the first two items of the system (10), p1t=p10 and p2t=p20 are obvious. At the same time, we have q1′=q2′ if we take the symmetric initial data, p10=p20. The two peaks move at the same speed which means there is no collision. Therefore, we choose the same initial data as the line case
(47)p10=−b+δ,q10=0,p2=b,q20=a,with 0<a<π and b2+b+δ2>10ch2π/2ch2π+3bb+δ, where b+δ>b>0. The selection of these initial data is summarized in Figure 2.Figure 2
Initial profileu0x.According to (47), p1t=−b+δ and p2t=b are obtained. We introduce the symbol q to represent the difference between the positions of the two peakons, in other words, q≐q2−q1. It follows from the ODE system (10) that
(48)q′=815p24−p14sh2πch2π+321+α2chπ−q=−815α11+α2chπ−q,where
(49)α1=b+δ4−b4sh2πch2π+32,(50)α2=5p1p2p12+p22·chπch2π+3/2=−5bb+δb2+b+δ2·chπch2π+3/2.These are easy to get from (49) and (50) that α1>0 and −1/chπ<α2<0. So, we have 1+α2chπ−q>0, which leads to q′<0. Integrating (48), we calculate
(51)gq−gq0=415α1α2t,where
(52)gx=ln1−α22+1+α2eπ−x1−α22−1−α2eπ−x.Differentiating (52), we have
(53)g′x=−2α21−α22eπ−x1−α22+1+α2eπ−x1−α22−1−α2eπ−x.Since−1−α22<1+α2eπ−x<1−α22, we find g′x>0. It follows that gx increases in 0,π. Combined with q0=q20−q10=a>0, we obtain a collision and a positive collision time when qt=0. Using the symbol τ for the collision time, from (51), we find
(54)τ=154α1α2ln1−α22+1+α2eπ1−α22−1−α2eπ−ln1−α22+1+α2eπ−a1−α22−1−α2eπ−a.Applyingqτ≐limt⟶τ−q1t=limt⟶τ−q2t and vx≐−δ·chx−qτtp−πto define the collision location qτ and the collision function v, we get the following proposition.Proposition 5.
TheHs limit of u as t approaches τ is v, or
(55)limt⟶τ−ut−vHs=0.Proof.
We compute the Fourier transform ofu and v, which is denoted by
(56)u^n,t=shπ2p1e−inq11+n2+2p2e−inq21+n2=shπ−2b+δe−inq11+n2+2be−inq21+n2,(57)v^n=−shπ·2δe−inqτ1+n2.
Combining (56) and (57), we have
(58)limt⟶τ−ut−vHs2=limt⟶τ−4sh2π∑n∈ℤ1+n2s−2be−inq2−b+δe−inq1+δe−inqτ2.
Notice that the equation inside the absolute value can be scaled up to(59)be−inq2−b+δe−inq1+δe−inqτ≤b+δe−inq1+be−inq2+δe−inqτ=2b+δ.
Lethn≐4b+δ21+n2s−2. h is addable when s<3/2 and h dominate the original integrand, which means that we can apply the dominated convergence theorem and put the limit inside the integral. So, we get
(60)limt⟶τ−4sh2π∑n∈ℤ1+n2s−2be‐inq2−b+δe−inq1+δe−inqτ2=4sh2π∑n∈ℤ1+n2s−2be−inq2τ−b+δe−inq1τ+δe‐inqτ2=0.Proof of Theorem 1 (On the circle).
In view of the 2-peakon solution we constructed, we need to construct a traveling wave solutionwx,t that satisfies
(61)wx,τ=ux,τ=vx.
Reviewing the system (10), we take the following data
(62)p1=−δ,q10=β,p2=0,q20=0.
The system is simplified and easy to be solved as(63)p1t=−δ,q1t=815ch4π+415ch2π+15δ4t+β.
We introduce the symbolq1u to represent q1 corresponding to u and q1w to represent q1 corresponding to w. It follows from (61) that q1uτ=q1wτ. So, we get
(64)β=q1uτ−2δ4α1α2ch4π+12ch2π+38·ln1−α22+1+α2eπ1−α22−1−α2eπ−ln1−α22+1+α2eπ−a1−α22−1−α2eπ−a,which makes the construction complete. In order to prove the nonuniqueness of the solution, we define two new solutions u~x,t≐u−x,−t+τ,w~x,t≐w−x,−t+τ. Since ux,t and wx,t are two solutions to (1), u−x,−t and w−x,−t are also solutions to (1), which can be obtained by reversing time. These mean u~x,t and u~x,t solve (1). In addition, similar to the line case, through Proposition 5, we have limt⟶0+u~x,t=v−x=limt⟶0+w~x,t. Finally, we note that the initial data for these nonunique solutions can be made arbitrarily small. Since the new initial data is the collision function vx=−δ·chx−qτtp−π, we have
(65)vHs2=4μδ2sh2π,whereμ=∑n∈ℤ1+n2s−2.
Therefore, for anyε>0, we can find a δ to make vHs<ε.



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*Source: 9363673-2021-11-10.xml*