arXiv:2104.12073v2 [nlin.SI] 29 Jun 2021

DOUBLE AND TRIPLE POLES SOLUTIONS FOR THE GERDJIKOV-IVANOV TYPE OF

DERIVATIVE NONLINEAR SCHRODINGER EQUATION WITH ZERO/NONZERO

BOUNDARY CONDITIONS

WEIQI PENG AND YONG CHEN∗

Abstract. In this work, the double and triple poles soliton solutions for the Gerdjikov-Ivanov(GI) type of derivative

nonlinear Schrodinger equation with zero boundary conditions(ZBCs) and nonzero boundary conditions(NZBCs) arestudied via Riemann-Hilbert (RH) method. Though spectral problem analysis, we first give out the Jost function and

scattering matrix under ZBCs and NZBCs. Then according to the analyticity, symmetry and asymptotic behaviorof Jost function and scattering matrix, the Riemann-Hilbert problem(RHP) with ZBCs and NZBCs are constructed.

Further, the obtained RHP with ZBCs and NZBCs can be solved in the case that reflection coefficients have double or

triple poles. Finally, we derive the general precise formulae of N -double and N -triple poles solutions corresponding toZBCs and NZBCs, respectively. The dynamical behaviors for these solutions are further discussed by image simulation.

1. Introduction

The nonlinear Schrodinger (NLS) equation is well known to be one of the most vital integrable systems, which hasbeen constantly reported in various fields such as nonlinear optics and water waves [1,2]. To describe more abundantphysical effects and research the effects of high-order perturbations, the Gerdjikov-Ivanov (GI) type of the derivativeNLS equations was also proposed as follows [3]

iut + uxx − iu2u∗x +1

2|u|4u = 0, (1.1)

where the asterisk ∗ denotes the complex conjugation, u denotes transverse magnetic field perturbation function withspatial variable x and temporal variable t. As a extended form of the NLS equation, the GI equation(1.1) is used todescribe Alfve’n waves propagating parallel to the ambient magnetic field in plasma physics. The explicit soliton-likesolutions for Eq.(1.1) are constructed via applying its Darboux transformation [4], and its algebro-geometric solutionsare established according to the Riemann theta functions [5]. By the DT method, the breather wave and rogue wavesolutions of the GI equation are presented in Refs. [6, 7]. The soliton molecules and dynamics of the smooth positonsfor the GI equation are discussed in Ref. [8]. Besides, N -soliton for Eq.(1.1) under ZBCs was given through usingRH method [9]. By using the nonlinear steepest descent method, the long-time asymptotic behavior for Eq.(1.1) wasdiscussed in Refs. [10–12]. Recently, the RH method and Dbar-dressing method were used to construct simple polesolutions of the GI equation with NZBCs in Ref. [13] and Ref. [14], respectively. Higher-order pole soliton solutions ofthe GI equation with ZBCs are derived by the dressing method based on the technique of regularization [15]. Multiplehigher-order pole solutions were also obtained by the Laurent’s series and generalization of the residue theorem [16].

In soliton theory, the inverse scattering transform (IST) method is the most powerful tool for analysing initialvalue problems of integrable nonlinear evolution equations (NLEEs). The method was raised first by Gardner et al.in 1967 for the KdV equation [17]. As is well-known, the classic IST method was based on the Gel’fand-Levitan-Marchenko(GLM) integral equations. Subsequently, Zakharov et al. developed a RH formulation to replace theGLM equation, which simplifies the IST method [18]. After decades of development, the RH formulation has beensuccessfully applied to numerous integrable equations, and it is still a hot topic today [19–29]. In terms of the directscattering, the corresponding RHP was constructed and soliton of the can be represented by solution of the RHP. Thenthrough solving the RHP under reflection-less case, we can obtain the formulae of multiple solitons. However, since weassume that the reflection coefficient had multiple poles, the PHP will be non-regular, which can not be solved directlyby Plemelj formula. To solve this obstacle, we need to regularize the RHP via eliminating singularity. Different fromthe process in Refs. [15, 22, 30] which transformed the non-regular RHP into the regular problem via multiplying adressing operator, the method, pioneered by Ablowitz, Biondini, Demontis, Prinary, et al., regularized the RHP bysubtracting the asymptotic behavior and the pole contribution [2, 31–37]. Inspired by this idea, the research on thesoliton solution of the integrable system under ZBCs and NZBCs has been complected in recent years [38–43]. Inthis work, we apply this idea to derive the double and triple poles solutions for GI equation with ZBCs and NZBCs,which is different with the simple pole solutions obtained in Refs. [9, 13, 14]. It is worth mentioning that the methodwe used is different from computing soliton solutions with higher-order poles in Refs. [15, 16], because the techniquesin dealing with severe spectral singularities are disparate. Moreover, for the case of NZBCs, a general double polesand triple poles solutions of GI equation with mixed discrete spectra, i.e., 2N1-breather-2N2-soliton solutions and3N1-breather-3N2-soliton solutions, are analysed in this paper, which does not know yet for GI equation. Therefore,

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2 WEIQI PENG AND YONG CHEN∗

to our knowledge, for the GI equation, the research using this idea for double and triple poles solutions under ZBCsand NZBCs has not been reported yet.

It is well known that the Lax pair of GI equation (1.1) can be given by

Φx = XΦ, Φt = TΦ, (1.2)

where

X = −ik2σ3 + kQ− i

2Q2σ3,

T = −2ik4σ3 + 2k3Q− ik2Q2σ3 − ikQxσ3 +1

2(QxQ−QQx) +

i

4Q4σ3, (1.3)

and

σ3 =

(1 00 −1

), Q =

(0 u−u∗ 0

), (1.4)

where k ∈ C is a spectral parameter, and Φ = Φ(x, t, k) is the 2 × 2 matrix-valued eigenfunction. Eq.(1.1) can bederived from the compatibility condition Xt − Tx + [X,T ] = 0 of system (1.2).

The outline of this paper is organized as follows: In section 2, we establish the RHP for the GI equation underZBCs by spectral analysis, and the potential function is shown by the solution of the RHP. In section 3, throughsolving the RHP, we obtain the explicit N -double poles soliton solutions for the reflection coefficients with doublepoles under ZBCs. Additionally, the general N -triple poles soliton solutions for the reflection coefficients with triplepoles under ZBCs are shown in section 4. In section 5, the RHP for the GI equation under NZBCs is presented viamore miscellaneous spectral problem analysis. Following that, the accurate N -double and N -triple poles solutionsunder NZBCs are given in section 6 and section 7, respectively. Finally, some summaries are given in the last section.

2. The construction of Riemann-Hilbert problem with ZBCs

The direct scattering problem for the GI equation (1.1) with ZBCs has been discussed in Ref. [16]. In this section,we will recall some results for the direct scattering problem for the targeted GI equation with following ZBCs atinfinity

limx→±∞

u(x, t) = 0. (2.1)

2.1. Spectral analysis. Let x→ ±∞, the Lax pair (1.2) under ZBCs (2.1) changes into

Φx = X0Φ = −ik2σ3, Φt = T0Φ = 2k2X0Φ, (2.2)

which admits the following fundamental matrix solution

Φbg(x, t; k) = e−iθ(x,t;k)σ3 , θ(x, t; k) = k2(x+ 2k2t). (2.3)

Defining Σ := R ∪ iR, the Jost solutions Φ±(x, t; k) can be written as

Φ±(x, t; k) = e−iθ(x,t;k)σ3 + o(1), k ∈ Σ, as x→ ±∞. (2.4)

Through the variable transformation

µ±(x, t; k) = Φ±(x, t; k)eiθ(x,t;k)σ3 , (2.5)

the modified Jost solutions µ±(x, t; k) will tend to I as x→ ±∞, and it can be solved as

µ±(x, t; k) = I +

∫ x

±∞e−ik

2(x−y)σ3

[(kQ− i

2Q2σ3)(y, t)µ±(y, t; k)

]dy. (2.6)

Proposition 2.1. Suppose u ∈ L1(R±), then µ±(x, t; k) given in Eq.(2.5) are unique solutions for the Jost integralequation (2.6) in Σ, and they satisfy the following characteristics:• µ−1(x, t; k) and µ+2(x, t; k) become analytical for D+ and continuous in D+ ∪ Σ;• µ+1(x, t; k) and µ−2(x, t; k) become analytical for D− and continuous in D− ∪ Σ;• µ±(x, t; k)→ I as k →∞;• detµ±(x, t; k) = 1, x, t ∈ R, k ∈ Σ.

DOUBLE AND TRIPLE POLES SOLUTIONS FOR THE GI TYPE OF DERIVATIVE NLS EQUATION 3

Rek

Imk

0

kn

k∗n

−k∗n

−kn

Figure 1. (Color online) Distribution of the discrete spectrum and jumping curves for the RHP on complex k-plane, Region

D+ = {k ∈ C|RekImk > 0} (gray region), region D− = {k ∈ C|RekImk < 0} (white region).

Since the Jost solutions Φ±(x, t; k) are the simultaneous solutions of spectral problem (1.2), which satisfies followinglinear relation by the constant scattering matrix S(k) = (sij(k))2×2

Φ+(x, t; k) = Φ−(x, t; k)S(k), k ∈ Σ, (2.7)

where S(k) = σ2S∗(k∗)σ2, S(k) = σ1S

∗(−k∗)σ1, and σ2 =

(0 −ii 0

), σ1 =

(0 11 0

). The scattering coefficients

can be shown in what follows by Wronskians determinant

s11(k) = Wr(Φ+,1,Φ−,2), s12(k) = Wr(Φ+,2,Φ−,2),

s21(k) = Wr(Φ−,1,Φ+,1), s22(k) = Wr(Φ−,1,Φ+,2), (2.8)

where Wr(·, ·) denotes the Wronskian determinant. From these representations, it is not hard to get following propo-sition.Proposition 2.2. The scattering matrix S(k) satisfies:• detS(k) = 1 for k ∈ Σ;• s22(k) becomes analytical for D+ and continuous in D+ ∪ Σ;• s11(k) becomes analytical for D− and continuous in D− ∪ Σ;• S(x, t, k)→ I as k →∞.

2.2. The Riemann-Hilbert problem. In terms of the analytic properties of Jost solutions µ±(x, t; k) in Proposition2.1, we have the following sectionally meromorphic matrices

M−(x, t; k) = (µ+,1

s11, µ−,2), M+(x, t; k) = (µ−,1,

µ+,2

s22), (2.9)

where superscripts ± represent analyticity in D+ and D−, respectively. Naturally, a matrix RHP is proposed:Riemann-Hilbert Problem 1 M(x, t; k) solves the following RHP: M(x, t; k) is analytic in C \ Σ,

M−(x, t; k) = M+(x, t; k)(I −G(x, t; k)), k ∈ Σ,M(x, t; k)→ I, k →∞,

(2.10)

of which the jump matrix G(x, t; k) is

G =

(ρ(k)ρ(k) e−2iθ(x,t;k)ρ(k)

−e2iθ(x,t;k)ρ(k) 0

), (2.11)

where ρ(k) = s21(k)s11(k) , ρ(k) = s12(k)

s22(k) .

Taking

M(x, t; k) = I +1

kM (1)(x, t; k) +O(

1

k2), k →∞, (2.12)

then the potential u(x, t) of the GI equation (1.1) with ZBCs is denoted by

u(x, t) = 2iM(1)12 (x, t; k) = 2i lim

k→∞kM12(x, t; k). (2.13)

3. The solution of GI equation under ZBCs with double poles

In this section, we will discuss the inverse scattering problem with double poles discrete spectrum for the GIequation(1.1) under ZBCs, and present the general N -double poles solutions.


3.1. Inverse scattering problem with ZBCs and double poles. We suppose that s22(k) has N double zeroskn (n = 1, 2, · · · , N) in D0 = {k ∈ C : Rek > 0, Imk > 0}, which means s22(kn) = s′22(kn) = 0 and s′′22(kn) 6= 0.According to the symmetries relation of the scattering matrix, one has{

s22(kn) = s22(−kn) = s11(k∗n) = s11(−k∗n) = 0,s′22(kn) = s′22(−kn) = s′11(k∗n) = s′11(−k∗n) = 0.

(3.1)

Thus, the corresponding discrete spectrum can be collected as

Γ = {kn, k∗n,−k∗n,−kn}Nn=1 , (3.2)

whose distributions are displayed in Fig. 1.Since s22(k0) = 0 (k0 ∈ Γ∩D+), we easily know that Φ−1(x, t; k0) and Φ+2(x, t; k0) are linearly dependent. Similarly,

Φ+1(x, t; k0) and Φ−2(x, t; k0) are linearly dependent due to s11(k0) = 0 for k0 ∈ Γ ∩D−. That is to say

Φ+2(x, t; k0) = b[k0]Φ−1(x, t; k0), k0 ∈ Γ ∩D+,

Φ+1(x, t; k0) = b[k0]Φ−2(x, t; k0), k0 ∈ Γ ∩D−, (3.3)

where b[k0] is a norming constant. As well as, due to s′22(k0) = 0 in k0 ∈ Γ ∩ D+, we find that Φ′+2(x, t; k0) −b[k0]Φ′−1(x, t; k0) and Φ−1(x, t; k0) are linearly dependent. Analogously, Φ′+1(x, t; k0)−b[k0]Φ′−2(x, t; k0) and Φ−2(x, t; k0)are linearly dependent for k0 ∈ Γ ∩D−. Then, we obtain

Φ′+2(x, t; k0)− b[k0]Φ′−1(x, t; k0) = d[k0]Φ−1(x, t; k0), k0 ∈ Γ ∩D+,

Φ′+1(x, t; k0)− b[k0]Φ′−2(x, t; k0) = d[k0]Φ−2(x, t; k0), k0 ∈ Γ ∩D−, (3.4)

where d[k0] is also a norming constant. Therefore, one has

L−2k=k0

[Φ+2(x, t; k)

s22(k)

]= A[k0]Φ−1(x, t; k0), k0 ∈ Γ ∩D+,

L−2k=k0

[Φ+1(x, t; k)

s11(k)

]= A[k0]Φ−2(x, t; k0), k0 ∈ Γ ∩D−,

Resk=k0

[Φ+2(x, t; k)

s22(k)

]= A[k0]

[Φ′−1(x, t; k0) +B[k0]Φ−1(x, t; k0)

], k0 ∈ Γ ∩D+,

Resk=k0

[Φ+1(x, t; k)

s11(k)

]= A[k0]

[Φ′−2(x, t; k0) +B[k0]Φ−2(x, t; k0)

], k0 ∈ Γ ∩D−, (3.5)

where L−2[f(x, t; k)] means the coefficient of O((k−k0)−2) term in the Laurent series expansion of f(x, t; k) at k = k0.

A[k0] =

2b[k0]s′′22(k0) , k0 ∈ Γ ∩D+

2b[k0]s′′11(k0) , k0 ∈ Γ ∩D−.

(3.6)

B[k0] =

d[k0]b[k0] −

s′′′22(k0)3s′′22(k0) , k0 ∈ Γ ∩D+

d[k0]b[k0] −

s′′′11(k0)3s′′11(k0) , k0 ∈ Γ ∩D−.

(3.7)

Proposition 3.1. Let k0 ∈ Γ, then the following symmetry relations are satisfied:• The first symmetry relation A[k0] = −A[k∗0 ]∗, B[k0] = B[k∗0 ]∗;• The second symmetry relation A[k0] = A[−k∗0 ]∗, B[k0] = −B[−k∗0 ]∗.

In order to solve the RHP conveniently, we take

ξn =

{kn, n = 1, 2, · · ·N−kn−N , n = N + 1, N + 2, · · · 2N. (3.8)

Then the residue and the coefficient L−2 of M(x, t; k) can be expressed as

L−2k=ξn

M+ =(

0, A[ξn]e−2iθ(x,t;ξn)µ−1(x, t; ξn)),

L−2k=ξ∗n

M− =(A[ξ∗n]e2iθ(x,t;ξ∗n)µ−2(x, t; ξ∗n), 0

),

Resk=ξn

M+ =(

0, A[ξn]e−2iθ(x,t;ξn)[µ′−1(x, t; ξn) + [B[ξn]− 2iθ′(x, t; ξn)]µ−1(x, t; ξn)

]),

Resk=ξ∗n

M− =(A[ξ∗n]e2iθ(x,t;ξ∗n)

[µ′−2(x, t; ξ∗n) + [B[ξ∗n] + 2iθ′(x, t; ξ∗n)]µ−2(x, t; ξ∗n)

], 0). (3.9)


By subtracting out the residue, the coefficient L−2 and the asymptotic values as k →∞ from the original non-regularRHP, one can obtain the following regular RHP

M− − I −2N∑n=1

L−2k=ξn

M+

(k − ξn)2+

Resk=ξn

M+

k − ξn+

L−2k=ξ∗n

M−

(k − ξ∗n)2+

Resk=ξ∗n

M−

(k − ξ∗n)

=

M+ − I −2N∑n=1

L−2k=ξn

M+

(k − ξn)2+

Resk=ξn

M+

k − ξn+

L−2k=ξ∗n

M−

(k − ξ∗n)2+

Resk=ξ∗n

M−

(k − ξ∗n)

−M+G, (3.10)

which can be solved by the Plemelj’s formulae, given by

M(x, t; k) =I +

2N∑n=1

L−2k=ξn

M+

(k − ξn)2+

Resk=ξn

M+

k − ξn+

L−2k=ξ∗n

M−

(k − ξ∗n)2+

Resk=ξ∗n

M−

(k − ξ∗n)

+

1

2πi

∫Σ

M+(x, t; ζ)G(x, t; ζ)

ζ − kdζ, k ∈ C\Σ, (3.11)

where

L−2k=ξn

M+

(k − ξn)2+

Resk=ξn

M+

k − ξn+

L−2k=ξ∗n

M−

(k − ξ∗n)2+

Resk=ξ∗n

M−

(k − ξ∗n)=(

Cn(k)

[µ′−2(ξ∗n) +

(Dn +

1

k − ξ∗n

)µ−2(ξ∗n)

], Cn(k)

[µ′−1(ξn) +

(Dn +

1

k − ξn

)µ−1(ξn)

]), (3.12)

and

Cn(k) =A[ξn]e−2iθ(ξn)

k − ξn, Dn = B[ξn]− 2iθ′(ξn),

Cn(k) =A[ξ∗n]e2iθ(ξ∗n)

k − ξ∗n, Dn = B[ξ∗n] + 2iθ′(ξ∗n). (3.13)

Furthermore, according to (2.12), one has

M (1)(x, t; k) = − 1

2πi

∫Σ

M+(x, t; ζ)G(x, t; ζ)dζ+

2N∑n=1

(A[ξ∗n]e2iθ(ξ∗n)

(µ′−2(ξ∗n) + Dnµ−2(ξ∗n)

), A[ξn]e−2iθ(ξn)

(µ′−1(ξn) +Dnµ−1(ξn)

)). (3.14)

The potential u(x, t) with double poles for the GI equation with ZBCs is redefined into

u(x, t) = 2iM(1)12 = − 1

π

∫Σ

(M+(x, t; ζ)G(x, t; ζ))12dζ

+2i

2N∑n=1

A[ξn]e−2iθ(ξn)(µ′−11(ξn) +Dnµ−11(ξn)

). (3.15)

3.2. Double poles soliton solutions with ZBCs. To derive the explicit double poles soliton solutions of the GIequation with ZBCs, we take ρ(k) = ρ(k) = 0 called the reflectionless. Then, the second column of Eq.(3.11) yields

µ−2(ξ∗j ) =

(01

)+

2N∑n=1

Cn(ξ∗j )

[µ′−1(ξn) +

(Dn +

1

ξ∗j − ξn

)µ−1(ξn)

],

µ′−2(ξ∗j ) = −2N∑n=1

Cn(ξ∗j )

ξ∗j − ξn

[µ′−1(ξn) +

(Dn +

2

ξ∗j − ξn

)µ−1(ξn)

],

µ−1(ξj) =

(10

)+

2N∑n=1

Cn(ξj)

[µ′−2(ξ∗n) +

(Dn +

1

ξj − ξ∗n

)µ−2(ξ∗n)

],

µ′−1(ξj) = −2N∑n=1

Cn(ξj)

ξj − ξ∗n

[µ′−2(ξ∗n) +

(Dn +

2

ξj − ξ∗n

)µ−2(ξ∗n)

]. (3.16)


Then, we have

µ−12(ξ∗j ) =

2N∑n=1

Cn(ξ∗j )

[µ′−11(ξn) +

(Dn +

1

ξ∗j − ξn

)µ−11(ξn)

],

µ′−12(ξ∗j ) = −2N∑n=1

Cn(ξ∗j )

ξ∗j − ξn

[µ′−11(ξn) +

(Dn +

2

ξ∗j − ξn

)µ−11(ξn)

],

µ−11(ξj) = 1 +

2N∑n=1

Cn(ξj)

[µ′−12(ξ∗n) +

(Dn +

1

ξj − ξ∗n

)µ−12(ξ∗n)

],

µ′−11(ξj) = −2N∑n=1

Cn(ξj)

ξj − ξ∗n

[µ′−12(ξ∗n) +

(Dn +

2

ξj − ξ∗n

)µ−12(ξ∗n)

], (3.17)

Theorem 3.1 The explicit formula of the double poles soliton solution for the GI equation (1.1) with ZBCs (2.1) isexpressed as

u(x, t) = −2i

det

(I −H βαT 0

)det(I −H)

, (3.18)

where β,H, α are given in (3.20), (3.21) and (3.23).

Proof. We can rewrite the linear system (3.17) in the matrix form:

ψ −Hψ = β, (3.19)

where

ψ =

(ψ(1)

ψ(2)

), ψ(1) = (µ−11(ξ1), · · · , µ−11(ξ2N ))T , ψ(2) = (µ′−11(ξ1), · · · , µ′−11(ξ2N ))T ,

β =

(β(1)

β(2)

), β(1) = (1)2N×1, β

(2) = (0)2N×1. (3.20)

The 4N × 4N matrix H =

(H(11) H(12)

H(21) H(22)

)with H(im) =

(H

(im)jk

)2N×2N

(i,m = 1, 2) given by

H(11)jk =

2N∑n=1

Cn(ξj)Ck(ξ∗n)

[− 1

ξ∗n − ξk(Dk +

2

ξ∗n − ξk) + (Dn +

1

ξj − ξ∗n)(Dk +

1

ξ∗n − ξk)

]

H(12)jk =

2N∑n=1

Cn(ξj)Ck(ξ∗n)

[− 1

ξ∗n − ξk+ (Dn +

1

ξj − ξ∗n)

]

H(21)jk =

2N∑n=1

Cn(ξj)Ck(ξ∗n)

ξj − ξ∗n

[1

ξ∗n − ξk(Dk +

2

ξ∗n − ξk)− (Dn +

2


1

ξ∗n − ξk)

]

H(22)jk =

2N∑n=1

Cn(ξj)Ck(ξ∗n)

ξj − ξ∗n

[1

ξ∗n − ξk− (Dn +

2

ξj − ξ∗n)

]. (3.21)

According to the reflectionless potential, Eq.(3.15) can be rewritten as

u = 2iαTψ, (3.22)

where

α =

(α(1)

α(2)

), α(2) = (A[ξ1]e−2iθ(ξ1), A[ξ2]e−2iθ(ξ2), · · · , A[ξ2N ]e−2iθ(ξ2N ))T ,

α(1) = (A[ξ1]e−2iθ(ξ1)D1, A[ξ2]e−2iθ(ξ2)D2, · · · , A[ξ2N ]e−2iθ(ξ2N )D2N )T . (3.23)

Combining Eqs. (3.19), the expression of the double poles soliton solution can be derived. �

We will analyse the dynamical behaviors of the double poles soliton solution for GI equation with ZBCs. At thecase of N = 1, we display the following figures by choosing suitable parameters. From Fig.2 (a)(b), we easily find thatthe one-double poles soliton solution is actually a kind of bound-state soliton solution which represents the interactionof two bright soliton. Besides, the interaction between two bright soliton is an elastic collision with the shape and sizeof the soliton unchange. Fig. 2(c) presents the wave propagation along the x-axis at t = −5, 0, 5. Fig.3 displays the


interaction of two pairs of one-double soliton solutions,i.e., two-double soliton solutions, at the case of N = 2.

(a) (b) (c)Figure 2. (Color online) The one-double pole soliton solution for Eq.(1.1) with ZBCs and N = 1. The parameters are

A[k1] = 1, B[k1] = i, k1 = 13+ 1

2i. (a) Three dimensional plot; (b) The density plot; (c) The wave propagation along the x-axis

at t = −5(longdash), t = 0(solid), t = 5(dashdot).

(a) (b) (c)Figure 3. (Color online) The two-double pole soliton solution for Eq.(1.1) with ZBCs and N = 2. The parameters are

A[k1] = 1, B[k1] = 1, A[k2] = 1, B[k2] = 1, k1 = 12+ 1

2i, k2 = 1

3+ 1

2i. (a) Three dimensional plot; (b) The density plot; (c) The

wave propagation along the x-axis at t = −10(longdash), t = 0(solid), t = 10(dashdot).

4. The solution of GI equation under ZBCs with triple poles

In this section, we devote to derive the general N -triple poles solutions through discussing the inverse scatteringproblem with triple poles discrete spectrum for the GI equation(1.1) under ZBCs.

4.1. Inverse scattering problem with ZBCs and triple poles. The triple zeros kn (n = 1, 2, · · · , N) in D0 ={k ∈ C : Rek > 0, Imk > 0} mean s22(kn) = s′22(kn) = s′′22(kn) = 0 and s′′′22(kn) 6= 0. Be similar to the expressions(3.3), (3.4), and due to s′′22(k0) = 0 in k0 ∈ Γ ∩ D+, the Φ′′+2(x, t; k0) − b[k0]Φ′′−1(x, t; k0) − 2d[k0]Φ′−1(x, t; k0) andΦ−1(x, t; k0) are linearly dependent. Besides, Φ′′+1(x, t; k0) − b[k0]Φ′′−2(x, t; k0) − 2d[k0]Φ′−2(x, t; k0) and Φ−2(x, t; k0)are linearly dependent for k0 ∈ Γ ∩D−. Naturally, we get

Φ′′+2(x, t; k0)− b[k0]Φ′′−1(x, t; k0)− 2d[k0]Φ′−1(x, t; k0) = h[k0]Φ−1(x, t; k0), k0 ∈ Γ ∩D+,

Φ′′+1(x, t; k0)− b[k0]Φ′′−2(x, t; k0)− 2d[k0]Φ′−2(x, t; k0) = h[k0]Φ−2(x, t; k0), k0 ∈ Γ ∩D−, (4.1)


where h[k0] is also a norming constant. Thus, one has

L−3k=k0

[Φ+2(x, t; k)

s22(k)

]= A[k0]Φ−1(x, t; k0), k0 ∈ Γ ∩D+,

L−3k=k0

[Φ+1(x, t; k)

s11(k)

]= A[k0]Φ−2(x, t; k0), k0 ∈ Γ ∩D−,

L−2k=k0

[Φ+2(x, t; k)

s22(k)

]= A[k0]

[Φ′−1(x, t; k0) + B[k0]Φ−1(x, t; k0)

], k0 ∈ Γ ∩D+,

L−2k=k0

[Φ+1(x, t; k)

s11(k)

]= A[k0]

[Φ′−2(x, t; k0) + B[k0]Φ−2(x, t; k0)

], k0 ∈ Γ ∩D−,

Resk=k0

[Φ+2(x, t; k)

s22(k)

]= A[k0]

[1

2Φ′′−1(x, t; k0) + B[k0]Φ′−1(x, t; k0) + C[k0]Φ−1(x, t; k0)

], k0 ∈ Γ ∩D+,

Resk=k0

[Φ+1(x, t; k)

s11(k)

]= A[k0]

[1

2Φ′′−2(x, t; k0) + B[k0]Φ′−2(x, t; k0) + C[k0]Φ−2(x, t; k0)

], k0 ∈ Γ ∩D−, (4.2)

where L−3[f(x, t; k)] denotes the coefficient of O((k − k0)−3) term in the Laurent series expansion of f(x, t; k) atk = k0.

A[k0] =

6b[k0]s′′′22(k0) , k0 ∈ Γ ∩D+

6b[k0]s′′′11(k0) , k0 ∈ Γ ∩D−.

(4.3)

B[k0] =

d[k0]b[k0] −

s′′′′22 (k0)4s′′′22(k0) , k0 ∈ Γ ∩D+

d[k0]b[k0] −

s′′′′11 (k0)4s′′′11(k0) , k0 ∈ Γ ∩D−.

(4.4)

C[k0] =

h[k0]2b[k0] −

d[k0]s′′′′22 (k0)4b[k0]s′′′22(k0) +

(s′′′′22 )2(k0)16(s′′′22)2(k0) , k0 ∈ Γ ∩D+

h[k0]2b[k0] −

d[k0]s′′′′11 (k0)4b[k0]s′′′11(k0) +

(s′′′′11 )2(k0)16(s′′′11)2(k0) , k0 ∈ Γ ∩D−.

(4.5)

Proposition 4.1. Let k0 ∈ Γ, then the following symmetry relations are satisfied:• The first symmetry relation A[k0] = −A[k∗0 ]∗, B[k0] = B[k∗0 ]∗, C[k0] = C[k∗0 ]∗;

• The second symmetry relation A[k0] = −A[−k∗0 ]∗, B[k0] = −B[−k∗0 ]∗, C[k0] = C[−k∗0 ]∗.Then the residue, the coefficient L−2 and the coefficient L−3 of M(x, t; k) can be written as

L−3k=ξn

M+ =(

0, A[ξn]e−2iθ(ξn)µ−1(ξn)),

L−3k=ξ∗n

M− =(A[ξ∗n]e2iθ(ξ∗n)µ−2(ξ∗n), 0

),

L−2k=ξn

M+ =(

0, A[ξn]e−2iθ(ξn)[µ′−1(x, t; ξn) +

[B[ξn]− 2iθ′(ξn)

]µ−1(ξn)

]),

L−2k=ξ∗n

M− =(A[ξ∗n]e2iθ(ξ∗n)

[µ′−2(ξ∗n) +

[B[ξ∗n] + 2iθ′(ξ∗n)

]µ−2(ξ∗n)

], 0),

Resk=ξn

M+ =

(0, A[ξn]e−2iθ(ξn)

[1

2µ′′−1(ξn) +

[B[ξn]− 2iθ′(ξn)

]µ′−1(ξn) + [C[ξn]−Θ1(ξn)]µ−1(ξn)

]),

Resk=ξ∗n

M− =


[1

2µ′′−2(ξ∗n) +

[B[ξ∗n] + 2iθ′(ξ∗n)

]µ′−2(ξ∗n) + [C[ξ∗n]−Θ2(ξ∗n)]µ−2(ξ∗n)

], 0

), (4.6)

where

Θ1(ξn) = 2(θ′(ξn))2 + iθ′′(ξn) + 2B[ξn]iθ′(ξn),

Θ2(ξ∗n) = 2(θ′(ξ∗n))2 − iθ′′(ξ∗n)− 2B[ξ∗n]iθ′(ξ∗n). (4.7)


Subtracting out the residue, the coefficient L−2, L−3 and the asymptotic values as k →∞ from the original non-regularRHP, the regular RHP is derived

M− − I−

2N∑n=1

L−3k=ξn

M+

(k − ξn)3+

L−2k=ξn

M+

(k − ξn)2+

Resk=ξn

M+

k − ξn+

L−3k=ξ∗n

M−

(k − ξ∗n)3+

L−2k=ξ∗n

M−

(k − ξ∗n)2+

Resk=ξ∗n

M−

(k − ξ∗n)

=

M+ − I−

2N∑n=1

L−3k=ξn

M+

(k − ξn)3+

L−2k=ξn

M+

(k − ξn)2+

Resk=ξn

M+

k − ξn+

L−3k=ξ∗n

M−

(k − ξ∗n)3+

L−2k=ξ∗n

M−

(k − ξ∗n)2+

Resk=ξ∗n

M−

(k − ξ∗n)

−M+G, (4.8)

which can be solved as follows by the Plemelj’s formulae

M(x, t; k) = I +1

2πi

∫Σ

M+(x, t; ζ)G(x, t; ζ)

ζ − kdζ

+

2N∑n=1

L−3k=ξn

M+

(k − ξn)3+

L−2k=ξn

M+

(k − ξn)2+

Resk=ξn

M+

k − ξn+

L−3k=ξ∗n

M−

(k − ξ∗n)3+

L−2k=ξ∗n

M−

(k − ξ∗n)2+

Resk=ξ∗n

M−

(k − ξ∗n)

. (4.9)

where

L−3k=ξn

M+

(k − ξn)3+

L−2k=ξn

M+

(k − ξn)2+

Resk=ξn

M+

k − ξn+

L−3k=ξ∗n

M−

(k − ξ∗n)3+

L−2k=ξ∗n

M−

(k − ξ∗n)2+

Resk=ξ∗n

M−

(k − ξ∗n)=(

Cn(k)

[1

2µ′′−2(ξ∗n) +

(Dn +

1

k − ξ∗n

)µ′−2(ξ∗n) + (

1

(k − ξ∗n)2+

Dn

k − ξ∗n+ Fn)µ−2(ξ∗n)

],

Cn(k)

[1

2µ′′−1(ξn) +

(Dn +

1

k − ξn

)µ′−1(ξn) + (

1

(k − ξn)2+

Dn

k − ξn+ Fn)µ−1(ξn)

]), (4.10)

and

Cn(k) =A[ξn]e−2iθ(ξn)

k − ξn, Dn = B[ξn]− 2iθ′(ξn), Fn = C[ξn]−Θ1(ξn)

Cn(k) =A[ξ∗n]e2iθ(ξ∗n)

k − ξ∗n, Dn = B[ξ∗n] + 2iθ′(ξ∗n), Fn = C[ξ∗n]−Θ2(ξ∗n). (4.11)

Furthermore, according to (2.12), we have

M (1)(x, t; k) = − 1

2πi

∫Σ

M+(x, t; ζ)G(x, t; ζ)dζ+

2N∑n=1


(1

2µ′′−2(ξ∗n) + Dnµ

′−2(ξ∗n) + Fnµ−2(ξ∗n)

), A[ξn]e−2iθ(ξn)

(1

2µ′′−1(ξn) +Dnµ

′−1(ξn) + Fnµ1(ξn)

)).

(4.12)

The potential u(x, t) with triple poles for the GI equation with ZBCs (2.1) is reconstructed as

u(x, t) = 2iM(1)12 = − 1

π

∫Σ

(M+(x, t; ζ)G(x, t; ζ))12dζ

+2i

2N∑n=1

A[ξn]e−2iθ(ξn)

(1

2µ′′−11(ξn) +Dnµ

′−11(ξn) + Fnµ11(ξn)

). (4.13)


4.2. Triple poles soliton solutions with ZBCs. Taking ρ(k) = ρ(k) = 0, we can derive the explicit triple polessoliton solutions of the GI equation with ZBCs. Then, the second column of Eq.(4.9) yields

µ−2(ξ∗j ) =

(01

)+

2N∑n=1

Cn(ξ∗j )

[1

2µ′′−1(ξn) +

(Dn +

1

ξ∗j − ξn

)µ′−1(ξn) + (

1

(ξ∗j − ξn)2+

Dn

ξ∗j − ξn+ Fn)µ−1(ξn)

],

µ′−2(ξ∗j ) = −2N∑n=1

Cn(ξ∗j )

ξ∗j − ξn

[1

2µ′′−1(ξn) +

(Dn +

2

ξ∗j − ξn

)µ′−1(ξn) + (

3

(ξ∗j − ξn)2+

2Dn


],

µ′′−2(ξ∗j ) =

2N∑n=1

2Cn(ξ∗j )

(ξ∗j − ξn)2

[1

2µ′′−1(ξn) +

(Dn +

3

ξ∗j − ξn

)µ′−1(ξn) + (

6

(ξ∗j − ξn)2+

3Dn


],

µ−1(ξj) =

(10

)+

2N∑n=1

Cn(ξj)

[1

2µ′′−2(ξ∗n) +

(Dn +

1

ξj − ξ∗n

)µ′−2(ξ∗n) + (

1

(ξj − ξ∗n)2+

Dn

ξj − ξ∗n+ Fn)µ−2(ξ∗n)

],

µ′−1(ξj) = −2N∑n=1

Cn(ξj)

ξj − ξ∗n

[1

2µ′′−2(ξ∗n) +

(Dn +

2

ξj − ξ∗n

)µ′−2(ξ∗n) + (

3

(ξj − ξ∗n)2+

2Dn


],

µ′′−1(ξj) =

2N∑n=1

2Cn(ξj)

(ξj − ξ∗n)2

[1

2µ′′−2(ξ∗n) +

(Dn +

3

ξj − ξ∗n

)µ′−2(ξ∗n) + (

6

(ξj − ξ∗n)2+

3Dn


]. (4.14)

Furthermore, we have

µ−12(ξ∗j ) =

2N∑n=1

Cn(ξ∗j )

[1

2µ′′−11(ξn) +

(Dn +

1

ξ∗j − ξn

)µ′−11(ξn) + (

1

(ξ∗j − ξn)2+

Dn


],

µ′−12(ξ∗j ) = −2N∑n=1

Cn(ξ∗j )

ξ∗j − ξn

[1

2µ′′−11(ξn) +

(Dn +

2

ξ∗j − ξn

)µ′−11(ξn) + (

3

(ξ∗j − ξn)2+

2Dn


],

µ′′−12(ξ∗j ) =

2N∑n=1

2Cn(ξ∗j )

(ξ∗j − ξn)2

[1

2µ′′−11(ξn) +

(Dn +

3

ξ∗j − ξn

)µ′−11(ξn) + (

6

(ξ∗j − ξn)2+

3Dn


],

µ−11(ξj) = 1 +

2N∑n=1

Cn(ξj)

[1

2µ′′−12(ξ∗n) +

(Dn +

1

ξj − ξ∗n

)µ′−12(ξ∗n) + (

1

(ξj − ξ∗n)2+

Dn


],

µ′−11(ξj) = −2N∑n=1

Cn(ξj)

ξj − ξ∗n

[1

2µ′′−12(ξ∗n) +

(Dn +

2

ξj − ξ∗n

)µ′−12(ξ∗n) + (

3

(ξj − ξ∗n)2+

2Dn


],

µ′′−11(ξj) =

2N∑n=1

2Cn(ξj)

(ξj − ξ∗n)2

[1

2µ′′−12(ξ∗n) +

(Dn +

3

ξj − ξ∗n

)µ′−12(ξ∗n) + (

6

(ξj − ξ∗n)2+

3Dn


]. (4.15)

Theorem 4.1 The explicit formula of the triple poles soliton solution for the GI equation (1.1) with ZBCs (2.1) isexpressed as

u(x, t) = −2i

det

(I − H βαT 0

)det(I − H)

, (4.16)

where β, H, α are given in (4.18), (4.19) and (4.23).

Proof. We rewrite the linear system (4.15) in the matrix form:

ψ − Hψ = β, (4.17)

where

ψ =

ψ(1)

ψ(2)

ψ(3)

, ψ(1) = (µ′′−11(ξ1), · · · , µ′′−11(ξ2N ))T , ψ(2) = (µ′−11(ξ1), · · · , µ′−11(ξ2N ))T ,

ψ(3) = (µ−11(ξ1), · · · , µ−11(ξ2N ))T , β =

β(1)

β(2)

β(3)

, β(1) = (0)2N×1, β(2) = (0)2N×1, β

(3) = (1)2N×1. (4.18)


The 6N × 6N matrix H =(H(im)

)3×3

with H(im) =(H

(im)jk

)2N×2N

(i,m = 1, 2, 3) given by

H(11)jk =

2N∑n=1

Cn(ξj)Ck(ξ∗n)

(ξj − ξ∗n)2

[1

(ξ∗n − ξk)2− 1

ξ∗n − ξk(Dn +

3

ξj − ξ∗n) + (

6

(ξj − ξ∗n)2+

3Dn

ξj − ξ∗n+ Fn)

],

H(12)jk =

2N∑n=1

2Cn(ξj)Ck(ξ∗n)

(ξj − ξ∗n)2

[1

(ξ∗n − ξk)2(Dk +

3

ξ∗n − ξk)− 1

ξ∗n − ξk(Dn +

3


2

ξ∗n − ξk)

+(6

(ξj − ξ∗n)2+

3Dn

ξj − ξ∗n+ Fn)(Dk +

1

ξ∗n − ξk)

],

H(13)jk =

2N∑n=1

2Cn(ξj)Ck(ξ∗n)

(ξj − ξ∗n)2

[1

(ξ∗n − ξk)2(

6

(ξ∗n − ξk)2+

3Dk

ξ∗n − ξk+ Fk)

− 1

ξ∗n − ξk(Dn +

3

ξj − ξ∗n)(

3

(ξ∗n − ξk)2+

2Dk

ξ∗n − ξk+ Fk)

+(6

(ξj − ξ∗n)2+

3Dn

ξj − ξ∗n+ Fn)(

1

(ξ∗n − ξk)2+

Dk

ξ∗n − ξk+ Fk)

], (4.19)

H(21)jk = −

2N∑n=1

Cn(ξj)Ck(ξ∗n)

2(ξj − ξ∗n)

[1

(ξ∗n − ξk)2− 1

ξ∗n − ξk(Dn +

2

ξj − ξ∗n) + (

3

(ξj − ξ∗n)2+

2Dn

ξj − ξ∗n+ Fn)

],

H(22)jk = −

2N∑n=1

Cn(ξj)Ck(ξ∗n)

ξj − ξ∗n

[1

(ξ∗n − ξk)2(Dk +

3

ξ∗n − ξk)− 1

ξ∗n − ξk(Dn +

2


2

ξ∗n − ξk)

+(3

(ξj − ξ∗n)2+

2Dn


1

ξ∗n − ξk)

],

H(23)jk = −

2N∑n=1

Cn(ξj)Ck(ξ∗n)

ξj − ξ∗n

[1

(ξ∗n − ξk)2(

6

(ξ∗n − ξk)2+

3Dk

ξ∗n − ξk+ Fk)

− 1

ξ∗n − ξk(Dn +

2

ξj − ξ∗n)(

3

(ξ∗n − ξk)2+

2Dk

ξ∗n − ξk+ Fk)

+(3

(ξj − ξ∗n)2+

2Dn


1

(ξ∗n − ξk)2+

Dk

ξ∗n − ξk+ Fk)

], (4.20)

H(31)jk =

2N∑n=1

Cn(ξj)Ck(ξ∗n)

2

[1

(ξ∗n − ξk)2− 1

ξ∗n − ξk(Dn +

1

ξj − ξ∗n) + (

1

(ξj − ξ∗n)2+

Dn

ξj − ξ∗n+ Fn)

],

H(32)jk =

2N∑n=1

Cn(ξj)Ck(ξ∗n)

[1

(ξ∗n − ξk)2(Dk +

3

ξ∗n − ξk)− 1

ξ∗n − ξk(Dn +

1


2

ξ∗n − ξk)

+(1

(ξj − ξ∗n)2+

Dn


1

ξ∗n − ξk)

],

H(33)jk =

2N∑n=1

Cn(ξj)Ck(ξ∗n)

[1

(ξ∗n − ξk)2(

6

(ξ∗n − ξk)2+

3Dk

ξ∗n − ξk+ Fk)

− 1

ξ∗n − ξk(Dn +

1

ξj − ξ∗n)(

3

(ξ∗n − ξk)2+

2Dk

ξ∗n − ξk+ Fk)

+(1

(ξj − ξ∗n)2+

Dn


1

(ξ∗n − ξk)2+

Dk

ξ∗n − ξk+ Fk)

], (4.21)

At the case of reflectionless potential, Eq.(4.13) can be redefined as

u = 2iαT ψ, (4.22)


where

α =

α(1)

α(2)

α(3)

, α(1) = (1

2A[ξ1]e−2iθ(ξ1),

1

2A[ξ2]e−2iθ(ξ2), · · · , 1

2A[ξ2N ]e−2iθ(ξ2N ))T ,

α(2) = (A[ξ1]e−2iθ(ξ1)D1, A[ξ2]e−2iθ(ξ2)D2, · · · , A[ξ2N ]e−2iθ(ξ2N )D2N )T ,

α(3) = (A[ξ1]e−2iθ(ξ1)F1, A[ξ2]e−2iθ(ξ2)F2, · · · , A[ξ2N ]e−2iθ(ξ2N )F2N )T . (4.23)

Combining Eqs. (4.17), the triple poles soliton solution (4.16) can be given out. �

As a matter of convenience, we take N = 1 as a example to illustrate the correlative dynamic behavior for theone-triple poles soliton solution for GI equation with ZBCs (2.1). As we can see in Fig. 4, it displays the bright-bright-bright soliton solutions, which stands for the interaction of three bright soliton waves.

(a) (b) (c)Figure 4. (Color online) The one-triple pole soliton solution for Eq.(1.1) with ZBCs and N = 1. The parameters are

A[k1] = B[k1] = C[k1] = 1, k1 = 12+ 1

2i. (a) Three dimensional plot; (b) The density plot; (c) The wave propagation along

the x-axis at t = −5(longdash), t = 0(solid), t = 5(dashdot).

5. The construction of Riemann-Hilbert problem with NZBCs

The direct scattering problem for the GI equation (1.1) with NZBCs has been studied in Ref. [13,16]. In this section,we first recall some results for the direct scattering problem for the targeted GI equation with following NZBCs atinfinity

limx→±∞

u(x, t) = u±e− 3

2 iu40t+iu

20x, (5.1)

where |u±| = u0 > 0, and u± are constant.

5.1. Spectral analysis. As x→ ±∞, the Lax pair (1.2) under the boundary (5.1) becomes

Φx = X±Φ = −ik2σ3 + kQ±, Φt = T±Φ = (2k2 − u20)X±, (5.2)

where

Q± =

(0 u±−u∗± 0

). (5.3)

The fundamental matrix solution of this lax pair is

Φbg± (x, t; k) =

{Y±(k)e−iθ(x,t;k)σ3 , k 6= ±iu0,I + (x− 3u2

0t)X±(k), k = ±iu0,(5.4)

where

Y± =

(1 − iu±

λ+k

− iu∗±λ+k 1

), θ(x, t; k) = kλ[x+ (2k2 − u2

0)t], λ2 = k2 + u20. (5.5)

Introducing a uniformization variable z = k + λ, we obtain

k =1

2(z − u2

0

z), λ =

1

2(z +

u20

z), (5.6)

which means that the scattering problem can be analysed on a standard z-plane instead of the two-sheeted Riemannsurface.

Defining D+, D− and Σ on z-plane as Σ = R∪ iR \ {0}, D± = {z ∈ C|RezImz ≷ 0} , the Jost solutions Φ±(x, t, ; z)are given by

Φ±(x, t; z) = Y±e−iθ(x,t;z)σ3 + o(1), z ∈ Σ, as x→ ±∞. (5.7)


After the variable transformation

µ±(x, t; z) = Φ±(x, t; z)eiθ(x,t;z)σ3 , (5.8)

the modified Jost solutions µ±(x, t; z) tend to Y±(z) as x → ±∞, and it also satisfy the following Volterra integralequations

µ±(x, t, z) =

{Y± +

∫ x±∞ Y±e

−ikλ(x−y)σ3[Y −1± ∆X±(y, t)µ±(y, t; z)

]dy, z 6= ±iu0,

Y± +∫ x±∞ [I + (x− y)X±(z)] ∆X±(y, t)µ±(y, t; z)dy, z = ±iu0,

(5.9)

where ∆X± = X −X±.Proposition 5.1. Suppose u − u± ∈ L1(R±), then µ±(x, t; z) given in Eq.(5.8) are unique solutions for the Jostintegral equation (5.9) in Σ0 = Σ \ {±iu0}, µ±(x, t; z) have the following characteristics:• µ−1(x, t; z) and µ+2(x, t; z) become analytical for D+ and continuous in D+ ∪ Σ0;• µ+1(x, t; z) and µ−2(x, t; z) become analytical for D− and continuous in D− ∪ Σ0;• µ±(x, t; z)→ I as z →∞;• µ±(x, t; z)→ − i

zσ3Q± as z → 0;

• detµ±(x, t; z) = detY± = γ = 1 +u20

z2 , x, t ∈ R, z ∈ Σ0.

Rez

Imz

0

zn

z∗n

−z∗n

−zn

−u20

zn

−u20

z∗n

u20

z∗n

u20

zn

−ω∗m

−ωm

ωm

ω∗m

Figure 5. (Color online) Distribution of the discrete spectrum and jumping curves for the RHP on complex z-plane, Region

D+ = {z ∈ C|RezImz > 0} (gray region), region D− = {z ∈ C|RezImz < 0} (white region).

Due to the Jost solutions Φ±(x, t; z) are the simultaneous solutions of spectral problem (1.2), which satisfies followinglinear relation by the constant scattering matrix S(z) = (sij(z))2×2

Φ+(x, t; z) = Φ−(x, t; z)S(z), z ∈ Σ0, (5.10)

where S(z) = σ2S∗(z∗)σ2, S(z) = σ1S

∗(−z∗)σ1, S(z) = (σ3Q−)−1S(−u20

z )σ3Q+ for z ∈ Σ. The scattering coefficientscan be expressed as Wronskians determinant in the following form:

s11(z) =Wr(Φ+,1,Φ−,2)

γ(z), s12(z) =

Wr(Φ+,2,Φ−,2)

γ(z),

s21(z) =Wr(Φ−,1,Φ+,1)

γ(z), s22(z) =

Wr(Φ−,1,Φ+,2)

γ(z). (5.11)

Proposition 5.2. The scattering matrix S(z) satisfies:• detS(z) = 1 for z ∈ Σ0;• s22(z) becomes analytical D+ and continuous in D+ ∪ Σ0;• s11(z) becomes analytical D− and continuous in D− ∪ Σ0;• S(x, t; z)→ I as z →∞;

• S(x, t; z)→ diag(u−u+, u+

u−

)as z → 0.

5.2. The Riemann-Hilbert problem. Based on the analytic properties of Jost solutions µ±(x; t; z) in Proposition5.1, we get the following sectionally meromorphic matrices

M−(x, t; z) = (µ+,1

s11, µ−,2), M+(x, t; z) = (µ−,1,

µ+,2

s22), (5.12)

where superscripts ± imply analyticity in D+ and D−, respectively. Then, a matrix RHP is raised:


Riemann-Hilbert Problem 2 M(x, t; z) solves the following RHP:M(x, t; z) is analytic in C \ Σ,M−(x, t; z) = M+(x, t; z)(I −G(x, t; z)), z ∈ Σ,M(x, t; z)→ I, z →∞,M(x, t; z)→ − i

zσ3Q−, z → 0,

(5.13)

of which the jump matrix G(x, t; z) is

G =

(ρ(z)ρ(z) e−2iθ(x,t;z)ρ(z)

−e2iθ(x,t;z)ρ(z) 0

), (5.14)

where ρ(z) = s21(z)s11(z) , ρ(z) = s12(z)

s22(z) . Taking

M(x, t; z) = I +1

zM (1)(x, t; z) +O(

1

z2), z →∞, (5.15)

then the potential u(x, t) of the GI equation (1.1) with NZBCs is given by

u(x, t) = iM(1)12 (x, t; z) = i lim

z→∞zM12(x, t; z). (5.16)

6. The solution of GI equation under NZBCs with double poles

In this section, the inverse scattering problem with double poles discrete spectrum for the GI equation (1.1) underNZBCs will be considered, and the general N -double poles solutions will be given.

6.1. Inverse scattering problem with NZBCs and double poles. We first suppose that s22(z) has N1 doublezeros zn (n = 1, 2, · · · , N1) in D+ ∩ {z ∈ C : Rez > 0, Imz > 0, |z| > u0}, and N2 double zeros ωm in {z = u0e

iϑ :0 < ϑ < π

2 }, which denotes s22(z0) = s′22(z0) = 0 and s′′22(z0) 6= 0 when z0 is the double zero of s22(z). From thesymmetries of the scattering matrix, the corresponding discrete spectrum is summed up as(see Fig. 5)

Υ =

{±zn,±z∗n,±

u20

zn,±u

20

z∗n

}N1

n=1

∪ {±ωm,±ω∗m}N2m=1. (6.1)

Due to s22(z0) = 0 (z0 ∈ Υ ∩D+), we can find that Φ−1(x, t; z0) and Φ+2(x, t; z0) are linearly dependent. Analo-gously, since s11(z0) = 0 (z0 ∈ Υ ∩D−), Φ+1(x, t; z0) and Φ−2(x, t; z0) can be linearly dependent. Then, we have

Φ+2(x, t, z0) = b[z0]Φ−1(x, t, z0), z0 ∈ Υ ∩D+,

Φ+1(x, t, z0) = b[z0]Φ−2(x, t, z0), z0 ∈ Υ ∩D−, (6.2)

where b[z0] is a norming constant. Due to s′22(z0) = 0 (z0 ∈ Υ∩D+), we find that Φ′+2(x, t; z0)− b[z0]Φ′−1(x, t; z0) andΦ−1(x, t; z0) are linearly dependent. Similarly, when z0 ∈ Υ ∩D−, Φ′+1(x, t; z0) − b[z0]Φ′−2(x, t; z0) and Φ−2(x, t; z0)are linearly dependent. Then, we have

Φ′+2(x, t; z0)− b[z0]Φ′−1(x, t; z0) = d[z0]Φ−1(x, t; z0), z0 ∈ Υ ∩D+,

Φ′+1(x, t; z0)− b[z0]Φ′−2(x, t; z0) = d[z0]Φ−2(x, t; z0), z0 ∈ Υ ∩D−, (6.3)

where d[z0] is a norming constant. Therefore, one obtains

L−2z=z0

[Φ+2(x, t; z)

s22(z)

]= A[z0]Φ−1(x, t; z0), z0 ∈ Υ ∩D+,

L−2z=z0

[Φ+1(x, t; z)

s11(z)

]= A[z0]Φ−2(x, t; z0), z0 ∈ Υ ∩D−,

Resz=z0

[Φ+2(x, t; z)

s22(z)

]= A[z0]

[Φ′−1(x, t; z0) +B[z0]Φ−1(x, t; z0)

], z0 ∈ Υ ∩D+,

Resz=z0

[Φ+1(x, t; z)

s11(z)

]= A[z0]

[Φ′−2(x, t; z0) +B[z0]Φ−2(x, t; z0)

], z0 ∈ Υ ∩D−, (6.4)

where L−2[f(x, t; z)] is the coefficient of O((z − z0)−2) term in the Laurent series expansion of f(x, t; z) at z = z0.

A[z0] =

{2b[z0]s′′22(z0) , z0 ∈ Υ ∩D+

2b[z0]s′′11(z0) , z0 ∈ Υ ∩D−.

B[z0] =

{d[z0]b[z0] −

s′′′22(z0)3s′′22(z0) , z0 ∈ Υ ∩D+

d[z0]b[z0] −

s′′′11(z0)3s′′11(z0) , z0 ∈ Υ ∩D−.

(6.5)

Proposition 6.1. Let k0 ∈ Υ, then the following symmetry relations are satisfied:• The first symmetry relation A[z0] = −A[z∗0 ]∗, B[z0] = B[z∗0 ]∗;


• The second symmetry relation A[z0] = A[−z∗0 ]∗, B[z0] = −B[−z∗0 ]∗;

• The Third symmetry relation A[z0] =z40u−u40u∗−A[−u

20

z0], B[z0] =

u20

z20B[−u

20

z0] + 2

z0.

As a matter of convenience, we set ζn = −u20

ζnand

ζn =

zn, n = 1, 2, · · · , N1

−zn−N1, n = N1 + 1, N1 + 2, · · · , 2N1

u20

z∗n−2N1

, n = 2N1 + 1, 2N1 + 2, · · · , 3N1

− u20

z∗n−3N1

, n = 3N1 + 1, 3N1 + 2, · · · , 4N1

ωn−4N1, n = 4N1 + 1, 4N1 + 2, · · · , 4N1 +N2

−ωn−4N1−N2, n = 4N1 +N2 + 1, 4N1 +N2 + 2, · · · , 4N1 + 2N2

(6.6)

Then the residue and the coefficient L−2 of M(x, t; z) can be written as

L−2z=ζn

M+ =(

0, A[ζn]e−2iθ(x,t;ζn)µ−1(x, t; ζn)),

L−2

z=ζn

M− =(A[ζn]e2iθ(x,t;ζn)µ−2(x, t; ζn), 0

),

Resz=ζn

M+ =(

0, A[ζn]e−2iθ(x,t;ζn)[µ′−1(x, t; ζn) + [B[ζn]− 2iθ′(x, t; ζn)]µ−1(x, t; ζn)

]),

Resz=ζn

M− =(A[ζn]e2iθ(x,t;ζn)

[µ′−2(x, t; ζn) +

[B[ζn] + 2iθ′(x, t; ζn)

]µ−2(x, t; ζn)

], 0). (6.7)

By subtracting out the asymptotic values as z → ∞, z → 0, the residue, and the coefficient L−2 from the originalnon-regular RHP, one can obtain the following regular RHP

M− +i

zσ3Q− − I−

4N1+2N2∑n=1

L−2z=ζn

M+

(z − ζn)2+

Resz=ζn

M+

z − ζn+

L−2

z=ζn

M−

(z − ζn)2+

Resz=ζn

M−

(z − ζn)

=

M+ +i

zσ3Q− − I−

4N1+2N2∑n=1

L−2z=ζn

M+

(z − ζn)2+

Resz=ζn

M+

z − ζn+

L−2

z=ζn

M−

(z − ζn)2+

Resz=ζn

M−

(z − ζn)

−M+G. (6.8)

Via the Plemelj’s formulae, the above RHP can be solved as

M(x, t; z) = I − i

zσ3Q− +

1

2πi

∫Σ

M+(x, t; ζ)G(x, t; ζ)

ζ − zdζ

+

4N1+2N2∑n=1

L−2z=ζn

M+

(z − ζn)2+

Resz=ζn

M+

z − ζn+

L−2

z=ζn

M−

(z − ζn)2+

Resz=ζn

M−

(z − ζn)

. (6.9)

where

L−2z=ζn

M+

(z − ζn)2+

Resz=ζn

M+

z − ζn+

L−2

z=ζn

M−

(z − ζn)2+

Resz=ζn

M−

(z − ζn)=(

Cn(z)

[µ′−2(ζn) +

(Dn +

1

z − ζn

)µ−2(ζn)

], Cn(z)

[µ′−1(ζn) +

(Dn +

1

z − ζn

)µ−1(ζn)

]), (6.10)

and

Cn(z) =A[ζn]e−2iθ(ζn)

z − ζn, Dn = B[ζn]− 2iθ′(ζn),

Cn(z) =A[ζn]e2iθ(ζn)

z − ζn, Dn = B[ζn] + 2iθ′(ζn). (6.11)


Moreover, according to (5.15), we obtain

M (1)(x, t; z) = − 1

2πi

∫Σ

M+(x, t; ζ)G(x, t; ζ)dζ − iσ3Q−

+

4N1+2N2∑n=1

(A[ζn]e2iθ(ζn)

(µ′−2(ζn) + Dnµ−2(ζn)

), A[ζn]e−2iθ(ζn)

(µ′−1(ζn) +Dnµ−1(ζn)

)). (6.12)

The potential u(x, t) with double poles for the GI equation with NZBCs is given by

u(x, t) = iM(1)12 = u− −

1

2π

∫Σ

(M+(x, t; ζ)G(x, t; ζ))12dζ

+i

4N1+2N2∑n=1

A[ζn]e−2iθ(ζn)(µ′−11(ζn) +Dnµ−11(ζn)

). (6.13)

6.2. Trace formulae and theta condition. Since ζn and ζn are double zeros of the scattering coefficients s22(z)and s11(z), respectively, we can set

β+(z) = s22(z)

4N1+2N2∏n=1

(z − ζnz − ζn

)2, β−(z) = s22(z)

4N1+2N2∏n=1

(z − ζnz − ζn

)2, (6.14)

such that β+(z) is analytic and has no zeros in D+, and β−(z) is analytic and has no zeros in D−. They both tendto o(1) as z →∞. Through using Cauchy projectors and the Plemelj’s formulae, β±(z) can be expressed as

log β±(z) = ∓ 1

2πi

∫Σ

log(1− ρρ)

s− zds, z ∈ D±. (6.15)

According to Eq.(6.14), the trace formulae is given as

s22(z) = exp

[− 1

2πi

∫Σ

log(1− ρρ)

s− zds

] 4N1+2N2∏n=1

(z − ζnz − ζn

)2,

s11(z) = exp

[1

2πi

∫Σ

log(1− ρρ)

s− zds

] 4N1+2N2∏n=1

(z − ζnz − ζn

)2. (6.16)

Furthermore, let z → 0, one has

u+

u−= exp

[i

2π

∫Σ

log(1− ρρ)

sds

] N1∏n=1

(znz∗n

)8N2∏n=1

(ωnu0

)8. (6.17)

Therefore, we easily obtain the theta condition for Eq.(6.17), given by

arg(u+

u−) =

1

2π

∫Σ

log(1− ρρ)

sds+ 16

N1∑n=1

arg(zn) + 8

N2∑n=1

arg(ωn). (6.18)

6.3. Double poles soliton solutions with NZBCs. To get double poles soliton solution, we let ρ(z) = ρ(z) = 0.Then, the second column of Eq.(6.9) yields

µ−2(z) =

(− izu−1

)+

4N1+2N2∑n=1

Cn(z)

[µ′−1(ζn) +

(Dn +

1

z − ζn

)µ−1(ζn)

], (6.19)

µ′−2(z) =

(iz2u−

0

)−

4N1+2N2∑n=1

Cn(z)

z − ζn

[µ′−1(ζn) +

(Dn +

2

z − ζn

)µ−1(ζn)

]. (6.20)

According to the symmetric relation, one can obtain

µ−2(z) = − iu−zµ−1(−u

20

z). (6.21)

Taking the first-order derivative about z in above formula, we get

µ′−2(z) =iu−z2

µ−1(−u20

z)− iu−u

20

z3µ′−1(−u

20

z). (6.22)

Putting Eqs.(6.21) and (6.22) into Eqs. (6.19) and (6.20), and letting z = ζj , j = 1, 2, · · · , 4N1 + 2N2, we obtain a8N1 + 4N2 linear system, given by

4N1+2N2∑n=1

{Cn(ζj)µ

′−1(ζn) +

[Cn(ζj)

(Dn +

1

ζj − ζn

)+iu−

ζjδj,n

]µ−1(ζn)

}=

(iζju−

−1

), (6.23)


4N1+2N2∑n=1

{(Cn(ζj)

ζj − ζn− iu−u

20

ζ3j

δj,n

)µ′−1(ζn)

+

[Cn(ζj)

ζj − ζn

(Dn +

2

ζj − ζn

)+iu−

ζ2j

δj,n

]µ−1(ζn)

}=

(iζ2ju−

0

). (6.24)

Theorem 6.1 The general formula of the double poles solution for the GI equation (1.1) with NZBCs (5.1) is expressedas

u(x, t) = u− − idet

(H ϕχT 0

)det(H)

, (6.25)

where ϕ,H, χ are given by (6.29), (6.30), (6.32), respectively.

Proof. From Eqs.(6.23) and (6.24), we get a 8N1 + 4N2 linear system with respect to µ−11(ζn), µ′−11(ζn)

4N1+2N2∑n=1

{Cn(ζj)µ

′−11(ζn) +

[Cn(ζj)

(Dn +

1

ζj − ζn

)+iu−

ζjδj,n

]µ−11(ζn)

}=iu−

ζj(6.26)

4N1+2N2∑n=1

{(Cn(ζj)


20

ζ3j

δj,n

)µ′−11(ζn)

+

[Cn(ζj)

ζj − ζn

(Dn +

2

ζj − ζn

)+iu−

ζ2j

δj,n

]µ−11(ζn)

}=iu−

ζ2j

, (6.27)

which can be rewritten in the matrix form:

Hγ = ϕ, (6.28)

where

ϕ =

(ϕ(1)

ϕ(2)

), ϕ(1) = (

iu−

ζ1,iu−

ζ2, · · · , iu−

ζ4N1+2N2

)T , ϕ(2) = (iu−

ζ21

,iu−

ζ22

, · · · , iu−

ζ24N1+2N2

)T ,

γ =

(γ(1)

γ(2)

), γ(1) = (µ−11(ζ1), µ−11(ζ2), · · · , µ−11(ζ4N1+2N2

))T ,

γ(2) = (µ′−11(ζ1), µ′−11(ζ2), · · · , µ′−11(ζ4N1+2N2))T ,H =

(H(11) H(12)

H(21) H(22)

), (6.29)

with H(im) =(H(im)jn

)4N1+2N2×4N1+2N2

(i,m = 1, 2) given by

H(11)jn = Cn(ζj)

(Dn +

1

ζj − ζn

)+iu−

ζjδj,n, H(12)

jn = Cn(ζj),

H(21)jn =

Cn(ζj)

ζj − ζn

(Dn +

2

ζj − ζn

)+iu−

ζ2j

δj,n, H(22)jn =

Cn(ζj)


20

ζ3j

δj,n. (6.30)

According to the reflectionless potential, Eq.(6.13) can be rewritten as

u = u− + iχT γ, (6.31)

where

χ =

(χ(1)

χ(2)

), χ(2) = (A[ζ1]e−2iθ(ζ1), A[ζ2]e−2iθ(ζ2), · · · , A[ζ4N1+2N2

]e−2iθ(ζ4N1+2N2))T ,

χ(1) = (A[ζ1]e−2iθ(ζ1)D1, A[ζ2]e−2iθ(ζ2)D2, · · · , A[ζ4N1+2N2]e−2iθ(ζ4N1+2N2

)D4N1+2N2)T . (6.32)

From Eqs. (6.28), the expression of the double poles soliton solution is derived. �

As a example, through choosing some appropriate parameters, we discuss the dynamical behaviors for one-doubleand two-double poles soliton solution in the case of N1 = 0, N2 = 1, N1 = 1, N2 = 0, and N1 = 1, N2 = 1, respectively.For N1 = 0, N2 = 1, we choose the parameters u− = 1, A[ω1] = 1, B[ω1] = 1, ω1 = e

π4 i in which the theta condition is

arg(u+

u−) = 2π such that u+ = u−. For this case, the one-double poles soliton solution exhibits the interaction of dark

and bright solitons, which can be verified in Fig.6. Additionally, from Fig.6(c), it is easily to find that the collisionis an elastic collision, due to the shape and size of the dark and bright solitons remain unchanged after the collision.On the other hand, when N1 = 1, N2 = 0, we select the parameters u− = 1, A[z1] = 1, B[z1] = i, z1 = 2e

π6 i and the

theta condition is arg(u+

u−) = 8

3π which leads to u+ = e23πi. As shown in Fig.7, the one-double poles soliton solution

stands for the interaction of two breather waves. Furthermore, when N1 = N2 = 1 i.e., a mixed discrete spectra, the


two-breather-two-soliton solutions can be obtained, which displays the interaction of two-breather and two-soliton.

(a) (b) (c)Figure 6. (Color online) The one-double pole soliton solution for Eq.(1.1) with NZBCs and N1 = 0, N2 = 1. The parameters

are u± = 1, A[ω1] = 1, B[ω1] = 1, ω1 = eπ4i. (a) Three dimensional plot; (b) The density plot; (c) The wave propagation along

the x-axis at t = −5(longdash), t = 0(solid), t = 5(dashdot).

(a) (b) (c)Figure 7. (Color online) The one-double pole soliton solution for Eq.(1.1) with NZBCs and N1 = 1, N2 = 0. The parameters

are u− = 1, u+ = e23πi, A[z1] = 1, B[z1] = i, z1 = 2e

π6i. (a) Three dimensional plot; (b) The density plot; (c) The wave

propagation along the x-axis at t = −15(longdash), t = 0(solid), t = 15(dashdot).

7. The solution of GI equation under NZBCs with triple poles

In this section, we aim to derive the general N -triple poles solutions via analysing the inverse scattering problemwith triple poles discrete spectrum for the GI equation (1.1) under NZBCs.

7.1. Inverse scattering problem with NZBCs and triple poles. Be similar to the expressions (6.2), (6.3), andsince s′′22(z0) = 0 in z0 ∈ Υ∩D+ for the case of triple poles, it is not hard to find that Φ′′+2(x, t; z0)−b[z0]Φ′′−1(x, t; z0)−2d[z0]Φ′−1(x, t; z0) and Φ−1(x, t; z0) are linearly dependent, and Φ′′+1(x, t; z0) − b[z0]Φ′′−2(x, t; z0) − 2d[z0]Φ′−2(x, t; z0)and Φ−2(x, t; z0) are linearly dependent for z0 ∈ Υ ∩D−. That is to say

Φ′′+2(x, t; z0)− b[z0]Φ′′−1(x, t; z0)− 2d[z0]Φ′−1(x, t; z0) = h[z0]Φ−1(x, t; z0), z0 ∈ Υ ∩D+,

Φ′′+1(x, t; z0)− b[z0]Φ′′−2(x, t; z0)− 2d[z0]Φ′−2(x, t; z0) = h[z0]Φ−2(x, t, z0), z0 ∈ Υ ∩D−, (7.1)


where h[z0] is also a norming constant. Therefore, one has

L−3z=z0

[Φ+2(x, t; z)

s22(z)

]= A[z0]Φ−1(x, t; z0), z0 ∈ Υ ∩D+,

L−3z=z0

[Φ+1(x, t; z)

s11(z)

]= A[z0]Φ−2(x, t; z0), z0 ∈ Υ ∩D−,

L−2z=z0

[Φ+2(x, t; z)

s22(z)

]= A[z0]

[Φ′−1(x, t; z0) + B[z0]Φ−1(x, t; z0)

], z0 ∈ Υ ∩D+,

L−2z=z0

[Φ+1(x, t; z)

s11(z)

]= A[z0]

[Φ′−2(x, t; z0) + B[z0]Φ−2(x, t; z0)

], z0 ∈ Υ ∩D−,

Resz=z0

[Φ+2(x, t; z)

s22(z)

]= A[z0]

[1

2Φ′′−1(x, t; z0) + B[z0]Φ′−1(x, t; z0) + C[z0]Φ−1(x, t; z0)

], z0 ∈ Υ ∩D+,

Resz=z0

[Φ+1(x, t; z)

s11(z)

]= A[z0]

[1

2Φ′′−2(x, t; z0) + B[z0]Φ′−2(x, t; z0) + C[z0]Φ−2(x, t; z0)

], z0 ∈ Υ ∩D−, (7.2)

where L−3[f(x, t; z)] means the coefficient of O((z− z0)−3) term in the Laurent series expansion of f(x, t; z) at z = z0,and

A[z0] =

6b[z0]s′′′22(z0) , z0 ∈ Υ ∩D+

6b[z0]s′′′11(z0) , z0 ∈ Υ ∩D−.

(7.3)

B[z0] =

d[z0]b[z0] −

s′′′′22 (z0)4s′′′22(z0) , z0 ∈ Υ ∩D+

d[z0]b[z0] −

s′′′′11 (z0)4s′′′11(z0) , z0 ∈ Υ ∩D−.

(7.4)

C[z0] =

h[z0]2b[z0] −

d[z0]s′′′′22 (z0)4b[z0]s′′′22(z0) +

(s′′′′22 )2(z0)16(s′′′22)2(z0) , z0 ∈ Υ ∩D+

h[z0]2b[z0] −

d[z0]s′′′′11 (z0)4b[z0]s′′′11(z0) +

(s′′′′11 )2(z0)16(s′′′11)2(z0) , z0 ∈ Γ ∩D−.

(7.5)

Proposition 7.1. Let k0 ∈ Υ, then the following symmetry relations are satisfied:• The first symmetry relation A[z0] = −A[z∗0 ]∗, B[z0] = B[z∗0 ]∗, C[z0] = C[z∗0 ]∗;

• The second symmetry relation A[z0] = −A[−z∗0 ]∗, B[z0] = −B[−z∗0 ]∗, C[z0] = C[−z∗0 ]∗;

• The Third symmetry relation A[z0] =z60u−u60u∗−A[−u

20

z0], B[z0] =

u20

z20B[−u

20

z0] + 3

z0, C[z0] =

u40

z40C[−u

20

z0] +

2u30

z30B[−u

20

z0] + 3

z20.

The residue and the coefficient L−2, L−3 of M(x, t; z) are

L−3z=ζn

M+ =(

0, A[ζn]e−2iθ(ζn)µ−1(ζn)),

L−3

z=ζn

M− =(A[ζn]e2iθ(ζn)µ−2(ζn), 0

),

L−2z=ζn

M+ =(

0, A[ζn]e−2iθ(ζn)[µ′−1(x, t; ζn) +

[B[ζn]− 2iθ′(ζn)

]µ−1(ζn)

]),

L−2

z=ζn

M− =(A[ζn]e2iθ(ζn)

[µ′−2(ζn) +

[B[ζn] + 2iθ′(ζn)

]µ−2(ζn)

], 0),

Resz=ζn

M+ =

(0, A[ζn]e−2iθ(ζn)

[1

2µ′′−1(ζn) +

[B[ζn]− 2iθ′(ζn)

]µ′−1(ζn) + [C[ζn]−Θ1(ζn)]µ−1(ζn)

]),

Resz=ζn

M− =

(A[ζn]e2iθ(ζn)

[1

2µ′′−2(ζn) +

[B[ζn] + 2iθ′(ζn)

]µ′−2(ζn) + [C[ζn]−Θ2(ζn)]µ−2(ζn)

], 0

), (7.6)

where

Θ1(ζn) = 2(θ′(ζn))2 + iθ′′(ζn) + 2B[ζn]iθ′(ζn),

Θ2(ζn) = 2(θ′(ζn))2 − iθ′′(ζn)− 2B[ζn]iθ′(ζn). (7.7)


By subtracting out the residue, the coefficient L−2, L−3 and the asymptotic values as z → ∞ from the originalnon-regular RHP, the following regular RHP is derived

M− +i

zσ3Q− − I−

4N1+2N2∑n=1

L−3z=ζn

M+

(z − ζn)3+

L−2z=ζn

M+

(z − ζn)2+

Resz=ζn

M+

z − ζn+

L−3

z=ζn

M−

(z − ζn)3+

L−2

z=ζn

M−

(z − ζn)2+

Resz=ζn

M−

(z − ζn)

=

M+ +i

zσ3Q− − I−

4N1+2N2∑n=1

L−3z=ζn

M+

(z − ζn)3+

L−2z=ζn

M+

(z − ζn)2+

Resz=ζn

M+

z − ζn+

L−3

z=ζn

M−

(z − ζn)3+

L−2

z=ζn

M−

(z − ζn)2+

Resz=ζn

M−

(z − ζn)

−M+G, (7.8)

which can be solved by the Plemelj’s formulae, given by

M(x, t; z) = I − i

zσ3Q− +

1

2πi

∫Σ

M+(x, t; ζ)G(x, t; ζ)

ζ − zdζ

+

4N1+2N2∑n=1

L−3z=ζn

M+

(z − ζn)3+

L−2z=ζn

M+

(z − ζn)2+

Resz=ζn

M+

z − ζn+

L−3

z=ζn

M−

(z − ζn)3+

L−2

z=ζn

M−

(z − ζn)2+

Resz=ζn

M−

(z − ζn)

, (7.9)

where

L−3z=ζn

M+

(z − ζn)3+

L−2z=ζn

M+

(z − ζn)2+

Resz=ζn

M+

z − ζn+

L−3

z=ζn

M−

(z − ζn)3+

L−2

z=ζn

M−

(z − ζn)2+

Resz=ζn

M−

z − ζn=(

Cn(z)

[1

2µ′′−2(ζn) +

(Dn +

1

z − ζn

)µ′−2(ζn) + (

1

(z − ζn)2+

Dn

z − ζn+ Fn)µ−2(ζn)

],

Cn(z)

[1

2µ′′−1(ζn) +

(Dn +

1

z − ζn

)µ′−1(ζn) + (

1

(z − ζn)2+

Dn

z − ζn+ Fn)µ−1(ζn)

]), (7.10)

and

Cn(z) =A[ζn]e−2iθ(ζn)

z − ζn, Dn = B[ζn]− 2iθ′(ζn), Fn = C[ζn]−Θ1(ζn),

Cn(z) =A[ζn]e2iθ(ζn)

z − ζn, Dn = B[ζn] + 2iθ′(ζn), Fn = C[ζn]−Θ2(ζn). (7.11)

Furthermore, according to (5.15), we get

M (1)(x, t; z) = − 1

2πi

∫Σ

M+(x, t; ζ)G(x, t; ζ)dζ − iσ3Q−

4N1+2N2∑n=1

(A[ζn]e2iθ(ζn)

(1

2µ′′−2(ζn) + Dnµ

′−2(ζn) + Fnµ−2(ζn)

),

A[ζn]e−2iθ(ζn)

(1

2µ′′−1(ζn) +Dnµ

′−1(ζn) + Fnµ1(ζn)

)). (7.12)

The potential u(x, t) with triple poles for the GI equation with NZBCs turns into

u(x, t) = iM(1)12 = u− −

1

2π

∫Σ

(M+(x, t; ζ)G(x, t; ζ))12dζ

+i

4N1+2N2∑n=1

A[ζn]e−2iθ(ζn)

(1

2µ′′−11(ζn) +Dnµ

′−11(ζn) + Fnµ11(ζn)

). (7.13)

7.2. Trace formulae and theta condition. In consideration of ζn and ζn being trace zeros of the scatteringcoefficients s22(z) and s11(z), respectively, we introduce following functions

β+(z) = s22(z)

4N1+2N2∏n=1

(z − ζnz − ζn

)3, β−(z) = s22(z)

4N1+2N2∏n=1

(z − ζnz − ζn

)3, (7.14)


of which β+(z) is analytic and has no zeros in D+, and β−(z) is analytic and has no zeros in D−. As z → ∞, theyboth tend to o(1). Furthermore, β±(z) can be shown as

log β±(z) = ∓ 1

2πi

∫Σ

log(1− ρρ)

s− zds, z ∈ D±, (7.15)

and the trace formulae is

s22(z) = exp

[− 1

2πi

∫Σ

log(1− ρρ)

s− zds

] 4N1+2N2∏n=1

(z − ζnz − ζn

)3,

s11(z) = exp

[1

2πi

∫Σ

log(1− ρρ)

s− zds

] 4N1+2N2∏n=1

(z − ζnz − ζn

)3. (7.16)

Let z → 0, it arrives in

u+

u−= exp

[i

2π

∫Σ

log(1− ρρ)

sds

] N1∏n=1

(znz∗n

)12N2∏n=1

(ωnu0

)12. (7.17)

Ultimately, the theta condition for Eq.(7.17) is given by

arg(u+

u−) =

1

2π

∫Σ

log(1− ρρ)

sds+ 24

N1∑n=1

arg(zn) + 12

N2∑n=1

arg(ωn). (7.18)

7.3. Triple poles soliton solutions with NZBCs. We take ρ(z) = ρ(z) = 0 to generate the explicit triple polessoliton solutions of the GI equation with NZBCs. Then, the second column of Eq.(7.9) yields

µ−2(z) =

(− izu−1

)+

4N1+2N2∑n=1

Cn(z)

[1

2µ′′−1(ζn) +

(Dn +

1

z − ζn

)µ′−1(ζn) +

(1

(z − ζn)2+

Dn

z − ζn+ Fn

)µ−1(ζn)

],

µ′−2(z) =

(iz2u−

0

)−

4N1+2N2∑n=1

Cn(z)

z − ζn

[1

2µ′′−1(ζn) +

(Dn +

2

z − ζn

)µ′−1(ζn) +

(3

(z − ζn)2+

2Dn

z − ζn+ Fn

)µ−1(ζn)

]µ′′−2(z) =

( −2iz3 u−

0

)+

2N∑n=1

2Cn(z)

(z − ξn)2

[1

2µ′′−1(ζn) +

(Dn +

3

z − ζn

)µ′−1(ζn) +

(6

(z − ζn)2+

3Dn

z − ζn+ Fn

)µ−1(ζn)

]. (7.19)

Taking the second-order derivative about z in above formula (6.21), we get

µ′′−2(z) = −2iu−z3

µ−1(−u20

z) +

4iu−u20

z4µ′−1(−u

20

z)− iu−u

40

z5µ′′−1(−u

20

z). (7.20)

Putting Eqs.(6.21),(6.22) and (7.20) into Eqs. (7.19), and letting z = ζj , j = 1, 2, · · · , 4N1 + 2N2, we obtain a12N1 + 6N2 following linear system

4N1+2N2∑n=1

{Cn(ζj)

(1

2µ′′−1(ζn) +

(Dn +

1

ζj − ζn

)µ′−1(ζn)

)

+

[Cn(ζj)

(1

(ζj − ζn)2+

Dn

ζj − ζn+ Fn

)+iu−

ζjδj,n

]µ−1(ζn)

}=

(iζju−

−1

), (7.21)

4N1+2N2∑n=1

{Cn(ζj)

2(ζj − ζn)µ′′−1(ζn) +

[Cn(ζj)

ζj − ζn

(Dn +

2

ζj − ζn

)− iu2

0u−

ζ3j

δj,n

]µ′−1(ζn)

+

[Cn(ζj)

ζj − ζn

(3

(ζj − ζn)2+

2Dn

ζj − ζn+ Fn

)+iu−

ζ2j

δj,n

]µ−1(ζn)

}=

(iζ2ju−

0

), (7.22)


4N1+2N2∑n=1

{[Cn(ζj)

(ζj − ζn)2+iu4

0u−

ζ5j

δj,n

]µ′′−1(ζn) +

[2Cn(ζj)

(ζj − ζn)2

(Dn +

3

ζj − ζn

)− 4iu2

0u−

ζ4j

δj,n

]µ′−1(ζn)

+

[2Cn(ζj)

(ζj − ζn)2

(6

(ζj − ζn)2+

3Dn

ζj − ζn+ Fn

)+

2iu−

ζ3j

δj,n

]µ−1(ζn)

}=

(2iζ3ju−

0

), (7.23)

Theorem 7.1 The general formula of the triple poles solution for the GI equation (1.1) with NZBCs (5.1) is expressedas

u(x, t) = u− − idet

(H ϕχT 0

)det(H)

, (7.24)

where ϕ, H, χ are given by (7.29), (7.30), (7.32), respectively.

Proof. From Eq.(7.21), (7.22) and (7.23), we get a 12N1 + 6N2 linear system with respect to µ−11(ζn), µ′−11(ζn)

4N1+2N2∑n=1

{Cn(ζj)

(1

2µ′′−11(ζn) +

(Dn +

1

ζj − ζn

)µ′−11(ζn)

)

+

[Cn(ζj)

(1

(ζj − ζn)2+

Dn

ζj − ζn+ Fn

)+iu−

ζjδj,n

]µ−11(ζn)

}=

i

ζju−, (7.25)

4N1+2N2∑n=1

{Cn(ζj)

2(ζj − ζn)µ′′−11(ζn) +

[Cn(ζj)

ζj − ζn

(Dn +

2

ζj − ζn

)− iu2

0u−

ζ3j

δj,n

]µ′−11(ζn)

+

[Cn(ζj)

ζj − ζn

(3

(ζj − ζn)2+

2Dn

ζj − ζn+ Fn

)+iu−

ζ2j

δj,n

]µ−11(ζn)

}=

i

ζ2j

u−, (7.26)

4N1+2N2∑n=1

{[Cn(ζj)

(ζj − ζn)2+iu4

0u−

ζ5j

δj,n

]µ′′−11(ζn) +

[2Cn(ζj)

(ζj − ζn)2

(Dn +

3

ζj − ζn

)− 4iu2

0u−

ζ4j

δj,n

]µ′−11(ζn)

+

[2Cn(ζj)

(ζj − ζn)2

(6

(ζj − ζn)2+

3Dn

ζj − ζn+ Fn

)+

2iu−

ζ3j

δj,n

]µ−11(ζn)

}=

2i

ζ3j

u−, (7.27)

which can be rewritten in the matrix form:

Hγ = ϕ, (7.28)

where

ϕ =

ϕ(1)

ϕ(2)

ϕ(3)

, ϕ(1) = (iu−

ζ1, · · · , iu−

ζ4N1+2N2

)T , ϕ(2) = (iu−

ζ21

, · · · , iu−

ζ24N1+2N2

)T , ϕ(3) = (2iu−

ζ31

, · · · , 2iu−

ζ34N1+2N2

)T ,

γ =

γ(1)

γ(2)

γ(3)

, γ(1) = (µ−11(ζ1), · · · , µ−11(ζ4N1+2N2))T , γ(2) = (µ′−11(ζ1), · · · , µ′−11(ζ4N1+2N2

))T ,

γ(3) = (µ′′−11(ζ1), · · · , µ′′−11(ζ4N1+2N2))T , H =

H(11) H(12) H(13)

H(21) H(22) H(23)

H(31) H(32) H(33)

, (7.29)

with H(im) =(H(im)jn

)4N1+2N2×4N1+2N2

(i,m = 1, 2, 3) given by

H(11)jn = Cn(ζj)

(1

(ζj − ζn)2+

Dn

ζj − ζn+ Fn

)+iu−

ζjδj,n, H(12)

jn = Cn(ζj)

(Dn +

1

ζj − ζn

), H(13)

jn =1

2Cn(ζj),

H(21)jn =

Cn(ζj)

ζj − ζn

(3

(ζj − ζn)2+

2Dn

ζj − ζn+ Fn

)+iu−

ζ2j

δj,n, H(22)jn =

Cn(ζj)

ζj − ζn

(Dn +

2

ζj − ζn

)− iu2

0u−

ζ3j

δj,n,

H(23)jn =

Cn(ζj)

2(ζj − ζn), H(31)

jn =2Cn(ζj)

(ζj − ζn)2

(6

(ζj − ζn)2+

3Dn

ζj − ζn+ Fn

)+

2iu−

ζ3j

δj,n,

H(32)jn =

2Cn(ζj)

(ζj − ζn)2

(Dn +

3

ζj − ζn

)− 4iu2

0u−

ζ4j

δj,n, H(33)jn =

Cn(ζj)

(ζj − ζn)2+iu4

0u−

ζ5j

δj,n. (7.30)


For the case of reflectionless potential, Eq.(7.13) is denoted as

u = u− + iχT γ, (7.31)

where

χ =

χ(1)

χ(2)

χ(3)

, χ(1) = (A[ζ1]e−2iθ(ζ1)F1, · · · , A[ζ4N1+2N2]e−2iθ(ζ4N1+2N2

)F4N1+2N2)T ,

χ(2) = (A[ζ1]e−2iθ(ζ1)D1, · · · , A[ζ4N1+2N2]e−2iθ(ζ4N1+2N2

)D4N1+2N2)T ,

χ(3) = (1

2A[ζ1]e−2iθ(ζ1), · · · , 1

2A[ζ4N1+2N2

]e−2iθ(ζ4N1+2N2))T . (7.32)

Using Eqs. (7.28), the expression of the triple poles soliton solution can be derived finally. �

In particular, we exhibit the obtained solutions with N1 = 0, N2 = 1 and N1 = 1, N2 = 0, respectively. In the caseof N1 = 0, N2 = 1, the one-triple poles soliton become dark-bright-dark solutions of the GI equation with NZBCs

(5.1) in Fig.8 for the parameters u− = 1, A[ω1] = B[ω1] = C[ω1] = 1, ω1 = eπi6 in which the theta condition become

arg(u+

u−) = 2π such that u+ = u−. On the other hand, when selecting N1 = 1, N2 = 0, the one-triple poles soliton

displays breather-breather-breather in Fig. 9 for the parameters u− = 1, A[ω1] = B[ω1] = C[ω1] = 1, ω1 = 2eπi6 in

which the theta condition is arg(u+

u−) = 4π such that u+ = u−. Besides, when N1 = N2 = 1 i.e., a mixed discrete

spectra, it turns into the three-breather-three-soliton solutions, meaning the interaction of three-breather and three-soliton.

(a) (b) (c)Figure 8. (Color online) The one-triple pole soliton solution for Eq.(1.1) with NZBCs and N1 = 0, N2 = 1. The parameters

are u± = 1, A[ω1] = B[ω1] = C[ω1] = 1, ω1 = eπi6 . (a) Three dimensional plot; (b) The density plot; (c) The wave propagation

along the x-axis at t = −5(longdash), t = 0(solid), t = 5(dashdot).

(a) (b) (c)Figure 9. (Color online) The one-triple pole soliton solution for Eq.(1.1) with NZBCs and N1 = 1, N2 = 0. The parameters are

u± = 1, A[ω1] = B[ω1] = C[ω1] = 1, ω1 = 2eπi6 . (a) Three dimensional plot; (b) The density plot; (c) The wave propagation

along the x-axis at t = −10(longdash), t = 0(solid), t = 10(dashdot).

8. Conclusion

In this paper, we have applied the RH method to discuss the GI type of derivative NLS equation with ZBCs andNZBCs. Through solving the RHP at the case of double and triple poles, we have given out the N -double and N -triple poles soliton solutions under ZBCs and NZBCs. The critical technique shown in this work is to eliminate theproperties of singularities via subtracting the residue and the coefficient L−2 from the original non-regular RHP when


reflection coefficients have double poles. For the case of triple poles, we have to subtract another the coefficient L−3.Additionally, the asymptotic value of jump matrix is subtracted from the original non-regular RHP. Then the regularRHP can be displayed, which can be solved by Plemelj formula. Finally, the N -double and N -triple poles solitonsolutions can be derived by using the solution of RHP to reconstruct potential function. Also, through choosingsuitable parameters, the dynamic behaviors of one-double poles soliton, two-double poles soliton, one-triple polessoliton corresponding to ZBCs, one-double poles soliton, and one-triple poles soliton corresponding to NZBCs areanalysed. In the near future, more works remain to be solved for other integrable systems via the technique shown inthis paper.

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(WP) School of Mathematical Sciences, Shanghai Key Laboratory of Pure Mathematics and Mathematical Practice,

East China Normal University, Shanghai 200062, People’s Republic of China

(YC) School of Mathematical Sciences, Shanghai Key Laboratory of Pure Mathematics and Mathematical Practice, East

China Normal University, Shanghai 200062, People’s Republic of China

(YC) College of Mathematics and Systems Science, Shandong University of Science and Technology, Qingdao 266590,

People’s Republic of China

Email address: [email protected](Corresponding author).


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arXiv:2104.12073v2 [nlin.SI] 29 Jun 2021