# Darboux Transformations and Soliton-Like Solutions of a New System Associated with the Negative AKNS System

Authors: Shilong Huang; Hongmin Li
Journal: Advances in Mathematical Physics
 (2022)

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## Abstract

In this letter, we consider a new two-component system. By its reciprocal relation with the first negative flow in the AKNS hierarchy, we construct three Darboux-Bäcklund transformations for the new system and obtain some soliton-like solutions.


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## Body

## 1. Introduction

It is well known that Darboux transformation (DT) and Bäcklund transformation (BT) play important roles in establishing new exact solutions of integrable systems from old ones [1–4]. For example, it may be used to construct multisoliton solutions from a trivial seed. For some classical integrable systems, one may construct DT/BT via the gauge transformation. But for others such as the Camassa-Holm equation and the complex short pulse equation, one can not construct their DTs/BTs directly, because they involve the independent variable x. Fortunately, the reciprocal transformation provides a useful tool [5–8]. There are many other methods to study exact solutions of integrable systems, see Refs. [9–13].Recently, while discussing reciprocal transformations of negative flows of some classical integrable hierarchies, Li and Wu [14] proposed some new systems; one of them reads
(1)ut=2bu2−2aux,vt=−2cv2−2avx,bx=−2av,cx=2au,ax=cv−bu.A Lax pair of (1) has been derived by using its reciprocal relation with the first negative flow in the AKNS hierarchy. Based on this Lax pair and using the standard approach, one may present infinitely many conserved quantities for it. However, its DT/BT and exact solutions are still unknown.In this letter, we will construct three BTs and provide some soliton-like solutions for (1). The paper is arranged as follows. In Section 2, we will construct three BTs of (1) by using its reciprocal link with the first negative flow in the AKNS hierarchy. In Section 3, we will construct the multisoliton solutions of (1) by using the BTs discovered above.


## 2. Darboux Transformations for the Negative AKNS System

As pointed out in [14], the system (1) admits the following Lax pair
(2)ψx=λuvvu−λuvψ,ψt=aλ−2λauvbλ−2avcλ−2au−aλ+2λauvψ,withψ=ψ1,ψ2T. The system (1) possesses a conservation law of the form uvt=−2auvx, which allows a reciprocal transformation
(3)dy=uvdx−2auvdt,dτ=dt.Through this reciprocal transformation, it is easy to show that (1) is changed to the first negative flow in the AKNS hierarchy
(4)u~τ=−2b,v~τ=2c,by=−2au~,cy=2av~,ay=cu~−bv~,and the Lax pair (2) is converted to
(5)ψy=λu~v~−λψ,ψτ=1λabc−aψ.Now, we turn our attention to the negative AKNS system [15–17]. The famous DTs for the negative AKNS system may be summarized as the following proposition.Proposition 1.
The Lax pair (5) is covariant with respect to the following three DTs:
(1)
The first one is given by(6)ψ1=T1ψ,T1=λ−λ1−u~f22f1u~2−f2f11,where f1,f2T is a special solution of the Lax pair (5) at λ=λ1.
(2)
The second one reads(7)ψ1=T2ψ,T2=1−g1g2−v~2λ−λ1+v~g12g2,where g1,g2T is a special solution of the Lax pair (5) at λ=λ1.
(3)
The third one is provided by(8)ψ1=T3ψ,T3=λI−HΛH−1,where Λ=diagλ1,λ2, H=hi,j2×2, and h1i,h2iT is a special solution of the Lax presentation (5) at λ=λi.


## 3. DTs and Some Soliton-Like Solutions of the New System

In this section, we will use the three DTs for the negative AKNS system in Proposition1 to construct three BTs, which involve the independent variable x, for the new system (1) [8]. The key point of the procedure is to determine the coordinate transformation between x,t and y,τ via the inverse transformation of (3).Firstly, let us apply the first DT (6) to the Lax pair (5), which leads to the BT between u~1,v~1,a1,b1,c1 and u~,v~,a,b,c. In particular, we have
(9)a1=a−cu~2λ1+au~f2λ1f1+bf2f1+bu~f222λ1f12,u~1=−λ1u~+u~y2−u~2f22f1,v~1=−2f2f1,which leads to
(10)u~1v~1=u~v~−u~f2f1y,2a1=2a−u~f2f1τ.Considering the inverse of (3), we may obtain a BT of system (2) as
(11)x1=x−u~f2f1+l1,u1=−2f12λ1u~f1−u~yf1+u~2f2,v1=−f12f2,where x=∫u~v~dy+∫2adτ and l1 is an arbitrary constant.Hereafter, as an example of use of this BT, let us choose a seed of (4) as
(12)u~=c1ek1τ,v~=c2e−k1τ,a=0,b=−12c1k1ek1τ,c=−12c2k1e−k1τ,where k1≠0 is an arbitrary constant. It follows immediately that
(13)f1=α1ek1/2τ+ξ+β1ek1/2τ−ξ,f2=1c1α1η−λ1e−k1/2τ+ξ−β1η+λ1e−k1/2τ−ξ,with η=λ12+c1c2,ξ=ηy−k1/2λ1τ. Now, substituting them into (11) yields
(14)x1=c1c2y−c1ek1τf2f1+l1,u1=−2f12λ1c1f1ek1τ+c12f2e2k1τ,v1=−f12f2.A profile of 2-kink like solution is plotted in Figures1 and 2 at c1=1,c2=−1,k1=1,λ1=2,α1=1,β1=1,l1=−4.Figure 1
One antikink foru.Figure 2
One antikink forv.Secondly, applying the DT (7) to the Lax presentation (5) yields
(15)a1=a−bv~2λ1−av~g1λ1g2−cg1g2+cv~g122λ1g22,u~1=2g1g2,v~1=−λ1v~−v~y2+v~2g12g2,which leads to
(16)u~1v~1=u~v~−v~g1g2y,2a1=2a−v~g1g2τ.Using the inverse of (3), we may obtain the second BT of system (2) as
(17)x1=x−v~g1g2+l2,u1=g22g1,v1=2g2−2λ1v~g2−v~yg2+v~2g1,with x=∫u~v~dy+∫2adτ and l2 is an arbitrary constant. Choosing the same seed as (12), one may obtain
(18)x1=c1c2y−c2e−k1τf1f2+l2,u1=f22f1,v1=2f2−2λ1c2f2e−k1τ+c22f1e−2k1τ.The profile is similar to the first case, and hence, we omit it here.Finally, imposing the DT (8) to the Lax pair (5) and setting S=HΛH−1 leads to
(19)a1=a−s11τ,u~1=u~+2s12,v~1=v~−2s21,where
(20)s11=λ1h11h22−λ2h12h21h11h22−h12h21,s12=λ2−λ1h11h12h11h22−h12h21,s21=λ1−λ2h21h22h11h22−h12h21.Furthermore, a straightforward calculation shows that(21)s11y+v~s12−u~s21−2s12s21=0.Combining (19) and (21), we arrive at
(22)u~1v~1=u~v~−2s11y,2a1=2a−2s11τ.Hence, we acquire the third BT of system (2), which is
(23)x1=x−2s11+l3,u1=1u~+2s12,v1=1v~−2s21,where x=∫u~v~dy+∫2adτ and l3 is an arbitrary constant. Choosing the same initial solution as (12) gives rise to
(24)h1i=αiek1/2τ+ξi+βiek1/2τ−ξi,h2i=1c1αiηi−λie−k1/2τ+ξi−βiηi+λie−k1/2τ−ξi,whereηi=λi2+c1c2,ξi=ηiy−k1/2λiτ,i=1,2. Substituting them into (23), we get
(25)x1=c1c2y−2s11+l3,u1=1c1ek1τ+2s12,v1=1c2e−k1τ−2s21.A profile of 2-kink like solution is plotted in Figures3 and 4 at c1=5,c2=−2,k1=1/10,λ1=−4,λ2=5,α1=1,α2=1,β1=1,β2=1,l3=0.Figure 3
Two kinks foru (t=−200 for u/109, linestyle is dot; t=10 for u, linestyle is solid; t=200 for 108u, linestyle is dash).Figure 4
Two kinks forv (t=−200 for 107v, linestyle is dot; t=10 for v/10, linestyle is solid; t=200 for v/109, linestyle is dash.).



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*Source: 3276320-2022-04-15.xml*