# Inverse Scattering Transform for the Generalized Derivative Nonlinear Schrödinger Equation via Matrix Riemann–Hilbert Problem

Authors: Fang Fang; Beibei Hu; Ling Zhang
Journal: Mathematical Problems in Engineering
 (2022)

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## Abstract

The inverse scattering transformation for a generalized derivative nonlinear Schrödinger (GDNLS) equation is studied via the Riemann–Hilbert approach. In the direct scattering process, we perform the spectral analysis of the Lax pair associated with a2×2 matrix spectral problem for the GDNLS equation. Then, the corresponding Riemann–Hilbert problem is constructed. In the inverse scattering process, we obtain an N-soliton solution formula for the GDNLS equation by solving the Riemann–Hilbert problem with the reflection-less case. In addition, we express the N-soliton solution of the GDNLS equation as determinant expression.


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## Body

## 1. Introduction

The nonlinear Schrödinger (NLS) equation,(1)iqt+qxx+2q2q=0,is an important integrable model in physics which was first derived by Zakharov in his study of modulation stability of deep waves [1]. Also, it makes sense to describe soliton propagation in nonlinear fiber optics, water waves, plasma physics, etc. So far, a number of research areas on the NLS equation have achieved remarkable results. For example, soliton solutions of the NLS equation have been obtained via the mathematical and physical methods, such as the inverse scattering transformation (IST) [2], the Bäcklund transformation (BT), and the Darboux transformation (DT). In addition, the dynamic properties of COVID-19 analyzed by mathematical and physical methods have also been widely concerned by scholars [3–7]. With the development of the Riemann–Hilbert (RH) problem, soliton solutions of the NLS equation can be derived from IST via a particular matrix RH problem. Researchers found that it was simpler and more effective than the original IST. On the one hand, scholars have used the RH approach to obtain the solutions to many integrable equations, such as the coupled derivative NLS (DNLS) equation [8], the coupled higher-order NLS equation [9], the generalized Camassa–Holm equation [10], the short pulse (SP) equation [11], and other integrable equations [12–25]. On the other hand, the RH method has been considered to be an effective way to work the initial-boundary value (IBV) problems of the integrable nonlinear evolution PDEs [26–29].In 1987, Clarkson and Cosgrove [30] proposed a generalized derivative NLS (GDNLS) equation in the form of(2)iqt=qxx+iαq2qx+iβq2qx∗+κq4q,where q is the amplitude of the complex field envelope and ∗ denotes the complex conjugation, α and β are the real parameters and α≠β. Equation (2) has several applications in optical fibers [31–33], nonlinear optics [34, 35], weakly nonlinear dispersion water waves [36–38], quantum field theory, plasma physics, etc. However, it is well known that equation (2) has Painlevé property only if κ=1/4β2β−α holds. At this time, equation (2) is reduced to an integrable GDNLS model as follows:(3)iqt=qxx+iαq2qx+iβq2qx∗+14β2β−αq4q,α≠β.Givenα=2β≠0, equation (3) becomes the DNLS-I (Kaup–Newell) equation,(4)iqt+qxx=iq2qx,and if α≠0andβ=0, equation (3) becomes the DNLS-II (Chen–Lee–Liu) equation,(5)iqt=qxx+iαq2qx.Recently, the conservation laws [39] and the IBV problems [40] of equation (3) on the half-line have been reported. In this paper, based on the IST idea, we use the RH method to solve the initial value problem of the GDNLS equation, and then, the scattering data from the scattering problem is obtained, and the N-soliton solution of the potential function qx,t is constructed.This paper is organized as follows. In Section2, based on the Lax pair of the GDNLS equation, we perform spectral analysis, and construct the RH problem in terms of the Jost solutions and scattering data Sλ. In Section 3, we apply the Plemelj formula to solve the regular and the nonregular RH problem, respectively. In the inverse scattering process, we reconstruct the potential qx,t and derive the determinant expression of the N-soliton solution for the GDNLS equation. Furthermore, we display explicitly one-soliton solution in the section. Finally, we summarise the content of the paper.


## 2. The Construction of Riemann–Hilbert Problem

In this section, we perform direct scattering transformation for the GDNLS equation. According to Hu et al. [40], the Lax pair of equation (2) is(6a)Φx=UΦ,(6b)Φt=VΦ,where(7)U=iλ2β−ασ+λQ−i4α−2βQ2σ=iλ2β−ασ+U1,V=2iλ4α−β2σ−iλ2Q2+i8α2−αβ−2β2Q4+14α−2βQQx−QxQσ+2λ3α−βQ+α2Q3−iQxσ=2iλ4α−β2σ+V1,with(8)σ=100−1,Q=0q∗q0,where Φ=Φx,t;λ is a matrix function of the complex isospectral parameter, λ is the spectral parameter, Q is named potential function, and ∗ denotes complex conjugation.It will be convenient to expressΦ as(9)Φ=Jeiλ2/β−ασx+2iλ4/α−β2σt,where J=Jx,t,λ is introduced as a new matrix spectral function. Inserting equation (9) into equations (6a) and (6b), the spectral problem equations (6a) and (6b) becomes(10a)Jx=iλ2β−ασ,J+U1J,(10b)Jt=2iλ4α−β2σ,J+V1J,where σ,J=σJ−Jσ is the commutator.In the following direct scattering process, we first treat equation (10a) as a spatial scattering problem at a fixed time t. Now, we concentrate on Jost solutions J±=J±x,λ for the x part of the Lax pair equation (10a) with the following asymptotic property at large distance:(11a)J+⟶I,x⟶+∞,(11b)J−⟶I,x⟶−∞.Applying Abel’s formula,TrU1=0 and using the boundary conditions equations (11a) and (11b), we see that(12)detJ±=1,for all x,λ.Based on Volterra integral equations,J±x,λ can be uniquely cast as(13a)J+x,λ=I−∫x+∞eiλ2/β−ασ^x−ξU1J+ξ,λdξ,J−x,λ=I+∫−∞xeiλ2/β−ασ^x−ξU1J−ξ,λdξ,where σ^ represents a matrix operator acting on 2×2 matrix X by(14)σ^X=σ,X,eσ^X=eσXe−σ.It is important to analyze the analytic properties of the Jost solutionsJ±, thus we partition J± into columns, namely, J±=J±1,J±2. Now, we assume β>α (in fact, result is the same thing if β<α), then, it is easy to find that J−1andJ+2 are continuous for λ∈D+∪∂D and analytic for λ∈D+; J+1andJ−2 are continuous for λ∈D−∪∂D and analytic λ∈D−; here,(15)D+=λ|argλ∈π2,π∪3π2,2π,D−=λ|argλ∈0,π2∪π,3π2,and ∂D=ℝ∪iℝ.In order to formulate a Riemann–Hilbert problem for the GDNLS equation, we construct a matrix functionP1=P1x,λ which is analytic for λ∈D+:(16)P1=J−1,J+2=J−H1+J+H2,where H1≡diag1,0andH2≡diag0,1.In addition, we see that the large-λ asymptotic of P1 is(17)P1⟶I,λ⟶+∞,λ∈D+.We can check that the inverse matricesJ±−1 satisfy this adjoint equation. On the other hand, we obtain an analytic counterpart of P1 in D− by the adjoint scattering equation of equation (11a):(18)Kx=iλ2β−ασ,K−KU1.If we expressJ±−1 as a collection of rows(19)J+−1=J+−11J+−12,J−−1=J−−11J−−12.Then, by applying the same techniques as above, we can show that(20)P2=J−−11J+−12=H1J−−1+H2J+−1,where J−−11andJ+−12 are continuous for λ∈D−∪∂D and analytic for λ∈D−; meanwhile, J+−11 and J−−12 are continuous for λ∈D+∪∂D and analytic for λ∈D+. Also, the large-λ asymptotic behavior of P2 is(21)P2⟶I,λ⟶−∞,λ∈D−.In addition, definingE=eiλ2/β−ασx, we can get that J−E and J+E are both solutions of the linear equations (6a) and (6b) and are linearly related by a scattering matrix Sλ=skj2×2.(22)J−E=J+E·Sλ,λ∈∂D.In view of equation (12) and equation (21), we see that Sλ satisfies(23)detSλ=1,λ∈∂D.Furthermore, from the properties ofJ− and J+−1, Sλ has the following analyticity structures:(24)Sx,λ=s11+s12s21s22−,where the superscripts ± refer to which λ∈D± the scattering elements are analytic in. Elements without superscripts indicate that such elements do not allow analytical extensions to D± in general.Denoting the limit ofP1 taken from the left-hand side of ∂D as P+, and the limit of P2 from the right-hand side of ∂D as P−, respectively, we arrive at a Riemann–Hilbert problem for the GDNLS equation.(25)P−x,λP+x,λ=Gx,λ,λ∈∂D,where the jump matrix G=Gx,λ is as follows:(26)Gx,λ=1−s12e2iλ2/β−αxs21e−2iλ2/β−αx1,and the canonical normalization condition for this Riemann–Hilbert problem is(27)P1⟶I,λ⟶+∞,λ∈D+,P2⟶I,λ⟶−∞,λ∈D−.To finish the direct scattering transformation, we determine the time evolution of the scattering matrixSλ. By inserting J+ES into equation (10b), taking the limit x⟶+∞, and recalling the boundary condition equation (11a) for J+ as well as the fact that V1⟶0, as x⟶±∞, we obtain(28)St=2iλ4α−β2σ,S.Equation (28) leads to(29a)∂s11∂t=∂s22∂t=0,(29b)∂s12∂t=4iλ4α−β2s12,∂s21∂t=−4iλ4α−β2s21.Equation (29a) shows that s11 and s22 are time independent, and the time evolution for the scattering data s12 and s21 are given by(30)s12t,λ=s120,λe4iλ4/α−β2,s21t,λ=s210,λe−4iλ4/α−β2.


## 3. Solution for the Riemann–Hilbert Problem

In this section, we discuss how to solve a regular Riemann–Hilbert problem and a nonregular Riemann–Hilbert problem. The Riemann–Hilbert problem (25) constructed in above section is regular when detP1≠0 and detP2≠0 in the respective planes of analyticity and is nonregular when detP1=0 and detP2=0 at certain discrete locations of λ.


### 3.1. Solution for the Regular Riemann–Hilbert Problem

Recalling the definitions ofP1 and P2, we see that(31)detP1=s11λ,λ∈D+,detP2=s22λ,λ∈D−.IfdetP1≠0 in their analytic domain, we can solve this Riemann–Hilbert problem by Plemelj formula. Under the canonical normalization equation (17) and equation (21), the solution to the regular Riemann–Hilbert problem is provided by the following integral equation:(32)P+−1λ=I+12πi∫−∞+∞G^ξP+−1ξξ−λdξ,λ∈D,where(33)G^λ=I−G=0s12e2iλ2/β−αx−s21e−2iλ2/β−αx0.Moreover, we can prove that the solution to the regular Riemann–Hilbert problem under the boundary condition is unique. This can be easily proved as follows. Suppose (25) has two sets of solutions P± and P˜±, then, P−λP+λ=P˜−P˜+, thus(34)P+λP˜+−1λ=P−−1λP˜−λ,λ∈∂D.SinceP+P˜+−1 is analytic in D+ and P−−1P˜− is analytic in D−. On the ∂D, they are equal to each other. Thus, they together define a matrix function which is analytic in the whole ℂ plane. Due to the boundary condition equation (17) and equation (21), this analytic matrix function is equal to the unit matrix I2×2 everywhere by Liouville theorem, i.e.,(35)P+λP˜+−1λ=P−−1λP˜−λ,λ∈ℂ.Therefore,P±=P˜±, which means that the solution to the regular Riemann–Hilbert problem (25) is unique.


### 3.2. Solution for the Nonregular Riemann–Hilbert Problem

In the more general case, the Riemann–Hilbert problem (25) is not regular and we can transform a nonregular Riemann–Hilbert problem into a regular one. In view of equation (6b), we can see that zeros of detP1 and detP2 are determined by s11λ=0 and s22λ=0, respectively.To specify these zeros, we easily find that there is a symmetry relation forU1 in equation (10a):(36)U1†−λ∗=−U1λ,where the superscript † means the Hermitian conjugation. According to equation (18), we get(37)J±†−λ∗=J±−1λ.Then, it follows that(38)P1†−λ∗=P2λ,λ∈D−.FromJ−E=J+ES, we also verify that Sλ satisfies the involution property:(39)S†−λ∗=S−1λ.Obviously, equation (39) leads to(40)s11∗−λ∗=s22λ,λ∈D−,s22∗−λ∗=s11λ,λ∈∈D+,s12∗−λ∗=−s21λ,λ∈∂D,s21∗−λ∗=−s12λ,λ∈∂D.In addition, there is one more symmetry relation,(41)U1−λ=σU1λσ.Owing to equation (10a), it is easy to see that(42)J±−λ=σJ±λσ.Applying equations (41) and (42) to equation (46), we can get(43)S−λ=Sλ.Equation (43) implies that(44)s11−λ=s11λ,λ∈D+,s22−λ=s22λ,λ∈D−,s12−λ=s12λ,λ∈∂D,s21−λ=s21λ,λ∈∂D.Furthermore, we deduce a property forP1 and P2:(45)σPi−λσ=Piλ,i=1,2.The solution to the Riemann–Hilbert problem is associated to the zeros of detP1 and det P2 as well as the kernel structures of P1 and P2 at the zeros. So we assume all zeros are simple, and see that if λj is a zero of det P1, then λj=λj∗ is a zero of det P2, and −λj is also a zero of det P1. Firstly, we suppose that det P1 has 2N zeros λj12N,λj∈D+, which satisfy λN+j=−λj,1≤j≤N, thus det P2 possesses 2N zeros λ^j12Nλ^j∈D− which satisfy λ^j=λj∗,1≤j≤2N. In this case, each ker P1λ and ker P2λ^j contains only a single column vector vj and row vector v^j, respectively(46)P1λjvj=0,v^jP2λ^j=0.Now, we derive the column vectorsvj and the row v^j. Using equations (38) and (46), we deduce that(47)v^j=vj†σ,1≤j≤2N,v^j+N=v^jσ,1≤j≤N,(48)vj+N=σvj1≤j≤N.On the other hand, we get the spatial evolutions of the vectorsvj1≤j≤2N. By taking the x-derivative to the first equations (46) and (10a), we get(49)vjx=iλ2β−ασvj.Similarly, we have(50)vjt=2iλ4α−β2σvj.By solving equations (48) and (49) explicitly, we get(51)vj=eiλ2/β−ασx+2iλ4/α−β2σtvj0=eθjσvj0,1≤j≤N,where θj=iλ2/β−αx+2iλ4/α−β2t and vj0=vj|x=0=cj,1T,1≤j≤N is a nonzero complex constant vector. Therefore, according to the above relationship between each pair of vj and v^j, the column vector vj and row vector v^j are determined as follows:(52)vj=eθjσvj0,1≤j≤N,σeθj−Nσvj−N,0,N+1≤j≤2N,and(53)v^j=vj0†eθj∗σσ,1≤j≤N,vj−N,0†eθj−N∗σ,N+1≤j≤2N.Based on above results, we reduce the nonregular Riemann–Hilbert problem to the regular Riemann–Hilbert problem by the following theorem.Theorem 1.
The solution to the nonregular Riemann–Hilbert problem (25) with zeros equation (46) under the canonical normalization condition equations (17) and (21) is(54)P+λ=P^+λΓλ,P−λ=Γ−1λP^−λ,where(55)Γλ=I+∑j=12N∑k=12NvjM−1jkv^kλ−λ^k,Γ−1λ=I−∑j=12N∑k=12NvjM−1jkv^kλ−λj,where M is an 2N×2N matrix with its j,k the element given by(56)mjk=v^jvkλ^j−λk,1≤j,k≤2N,detΓλ=∏k=12Nλ−λkλ−λ^k,and P^± is the unique solution to the following regular Riemann–Hilbert problem.(57)P^−λP^+λ=ΓλGλΓ−1λ,λ∈∂D,where P^± are analytic in D± and P^±⟶I as λ⟶∞.This theorem was obtained by Zakharov and Shabat in 1979 (see [41]).In order to obtain N-soliton solutions for GDNLS, we choose the jump matrixG=I, i.e., s12=s21=0. In this case, the corresponding scattering problem is called reflection-less. This solution for this particular the Riemann–Hilbert problem is written out explicitly as follows.(58a)P1λ=I−∑j=12N∑k=12NvjM−1jkv^kλ−λ^k,(58b)P2λ=I+∑j=12N∑k=12NvjM−1jkv^kλ−λj,where M=mjk2N×2N is a matrix whose entries are as follows:(59)mjk=v^kvjλj−λ^k,1≤k,j≤2N.


## 3.1. Solution for the Regular Riemann–Hilbert Problem

Recalling the definitions ofP1 and P2, we see that(31)detP1=s11λ,λ∈D+,detP2=s22λ,λ∈D−.IfdetP1≠0 in their analytic domain, we can solve this Riemann–Hilbert problem by Plemelj formula. Under the canonical normalization equation (17) and equation (21), the solution to the regular Riemann–Hilbert problem is provided by the following integral equation:(32)P+−1λ=I+12πi∫−∞+∞G^ξP+−1ξξ−λdξ,λ∈D,where(33)G^λ=I−G=0s12e2iλ2/β−αx−s21e−2iλ2/β−αx0.Moreover, we can prove that the solution to the regular Riemann–Hilbert problem under the boundary condition is unique. This can be easily proved as follows. Suppose (25) has two sets of solutions P± and P˜±, then, P−λP+λ=P˜−P˜+, thus(34)P+λP˜+−1λ=P−−1λP˜−λ,λ∈∂D.SinceP+P˜+−1 is analytic in D+ and P−−1P˜− is analytic in D−. On the ∂D, they are equal to each other. Thus, they together define a matrix function which is analytic in the whole ℂ plane. Due to the boundary condition equation (17) and equation (21), this analytic matrix function is equal to the unit matrix I2×2 everywhere by Liouville theorem, i.e.,(35)P+λP˜+−1λ=P−−1λP˜−λ,λ∈ℂ.Therefore,P±=P˜±, which means that the solution to the regular Riemann–Hilbert problem (25) is unique.


## 3.2. Solution for the Nonregular Riemann–Hilbert Problem

In the more general case, the Riemann–Hilbert problem (25) is not regular and we can transform a nonregular Riemann–Hilbert problem into a regular one. In view of equation (6b), we can see that zeros of detP1 and detP2 are determined by s11λ=0 and s22λ=0, respectively.To specify these zeros, we easily find that there is a symmetry relation forU1 in equation (10a):(36)U1†−λ∗=−U1λ,where the superscript † means the Hermitian conjugation. According to equation (18), we get(37)J±†−λ∗=J±−1λ.Then, it follows that(38)P1†−λ∗=P2λ,λ∈D−.FromJ−E=J+ES, we also verify that Sλ satisfies the involution property:(39)S†−λ∗=S−1λ.Obviously, equation (39) leads to(40)s11∗−λ∗=s22λ,λ∈D−,s22∗−λ∗=s11λ,λ∈∈D+,s12∗−λ∗=−s21λ,λ∈∂D,s21∗−λ∗=−s12λ,λ∈∂D.In addition, there is one more symmetry relation,(41)U1−λ=σU1λσ.Owing to equation (10a), it is easy to see that(42)J±−λ=σJ±λσ.Applying equations (41) and (42) to equation (46), we can get(43)S−λ=Sλ.Equation (43) implies that(44)s11−λ=s11λ,λ∈D+,s22−λ=s22λ,λ∈D−,s12−λ=s12λ,λ∈∂D,s21−λ=s21λ,λ∈∂D.Furthermore, we deduce a property forP1 and P2:(45)σPi−λσ=Piλ,i=1,2.The solution to the Riemann–Hilbert problem is associated to the zeros of detP1 and det P2 as well as the kernel structures of P1 and P2 at the zeros. So we assume all zeros are simple, and see that if λj is a zero of det P1, then λj=λj∗ is a zero of det P2, and −λj is also a zero of det P1. Firstly, we suppose that det P1 has 2N zeros λj12N,λj∈D+, which satisfy λN+j=−λj,1≤j≤N, thus det P2 possesses 2N zeros λ^j12Nλ^j∈D− which satisfy λ^j=λj∗,1≤j≤2N. In this case, each ker P1λ and ker P2λ^j contains only a single column vector vj and row vector v^j, respectively(46)P1λjvj=0,v^jP2λ^j=0.Now, we derive the column vectorsvj and the row v^j. Using equations (38) and (46), we deduce that(47)v^j=vj†σ,1≤j≤2N,v^j+N=v^jσ,1≤j≤N,(48)vj+N=σvj1≤j≤N.On the other hand, we get the spatial evolutions of the vectorsvj1≤j≤2N. By taking the x-derivative to the first equations (46) and (10a), we get(49)vjx=iλ2β−ασvj.Similarly, we have(50)vjt=2iλ4α−β2σvj.By solving equations (48) and (49) explicitly, we get(51)vj=eiλ2/β−ασx+2iλ4/α−β2σtvj0=eθjσvj0,1≤j≤N,where θj=iλ2/β−αx+2iλ4/α−β2t and vj0=vj|x=0=cj,1T,1≤j≤N is a nonzero complex constant vector. Therefore, according to the above relationship between each pair of vj and v^j, the column vector vj and row vector v^j are determined as follows:(52)vj=eθjσvj0,1≤j≤N,σeθj−Nσvj−N,0,N+1≤j≤2N,and(53)v^j=vj0†eθj∗σσ,1≤j≤N,vj−N,0†eθj−N∗σ,N+1≤j≤2N.Based on above results, we reduce the nonregular Riemann–Hilbert problem to the regular Riemann–Hilbert problem by the following theorem.Theorem 1.
The solution to the nonregular Riemann–Hilbert problem (25) with zeros equation (46) under the canonical normalization condition equations (17) and (21) is(54)P+λ=P^+λΓλ,P−λ=Γ−1λP^−λ,where(55)Γλ=I+∑j=12N∑k=12NvjM−1jkv^kλ−λ^k,Γ−1λ=I−∑j=12N∑k=12NvjM−1jkv^kλ−λj,where M is an 2N×2N matrix with its j,k the element given by(56)mjk=v^jvkλ^j−λk,1≤j,k≤2N,detΓλ=∏k=12Nλ−λkλ−λ^k,and P^± is the unique solution to the following regular Riemann–Hilbert problem.(57)P^−λP^+λ=ΓλGλΓ−1λ,λ∈∂D,where P^± are analytic in D± and P^±⟶I as λ⟶∞.This theorem was obtained by Zakharov and Shabat in 1979 (see [41]).In order to obtain N-soliton solutions for GDNLS, we choose the jump matrixG=I, i.e., s12=s21=0. In this case, the corresponding scattering problem is called reflection-less. This solution for this particular the Riemann–Hilbert problem is written out explicitly as follows.(58a)P1λ=I−∑j=12N∑k=12NvjM−1jkv^kλ−λ^k,(58b)P2λ=I+∑j=12N∑k=12NvjM−1jkv^kλ−λj,where M=mjk2N×2N is a matrix whose entries are as follows:(59)mjk=v^kvjλj−λ^k,1≤k,j≤2N.


## 4. Inverse Scattering Transform

Now we know that the Riemann–Hilbert problem (25) can be solved from the given scattering data λjλ^jvjov^jo by applying the formulas (58a) and (58b). In this section, we reconstruct the potential q by means of the inverse scattering transform of the GDNLS equation. Indeed, due to that P1λ is solution of the scattering problem (10a), we expand P1λ at large λ as(60)P1x,λ=I+P11λ+P12λ2+Oλ−3,λ⟶∞,and substitute it into equation (10a). By comparing O1 terms, we can get(61)QP11−i4α−2βQ2σ=0.Then, the solutionqx,t can be obtained as(62)qx,t=4i2β−αP1121,where P11=P112×2 and P11ij is the i;j-entry of P11,i,j=1,2. Here, the matrix function P11 to the Riemann–Hilbert problem which is reflection-less case, one can be derived in equation (58a):(63)P11=∑k=12N∑j=12NvjM−1jkv^k.The corresponding solutionqx,t is(64)q=4i2β−α∑k=12N∑j=12NvjM−1jkv^k21.


## 5. N-Soliton Solutions

In what follows, we substitute equations (52) and (53) into equation (64), then, a N-soliton solution formula for the GDNLS equation is given by(65)q=4i2β−α∑j=1N∑k=1Nck∗e−θj+θk∗M−1jk−∑j=N+12N∑k=1Nck∗e−θj−N+θk∗M−1jk−∑j=1N∑k=N+12Nck−N∗e−θj+θk−N∗M−1jk−∑j=N+12N∑k=N+12Nck−N∗e−θj−N+θk−N∗M−1jk,where M=mjk2N×2N with(66)mjk=ckcj∗eθj∗+θk−e−θj∗−θkλk−λj∗,1≤j,k≤N,ck−Ncj∗eθj∗+θk−N+e−θj∗−θk−N−λj∗−λk−N,1≤j≤N,N+1≤k≤2N,ckcj−N∗eθj−N∗+θk+e−θj−N∗−θkλj−N∗+λk,1≤k≤N,N+1≤j≤2N,ck−Ncj−N∗eθj−N∗+θk−N−e−θj−N∗−θk−Nλj−N∗−λk−N,N+1≤j,k≤2N.Recalling that the determinant of a matrix can be expressed as the sum of its elements along a row or column multiplying their corresponding cofactor, we rewrite the solution (66) as(67)qx,t=4iα−2β·detM˜21detM,where(68)M˜21=0f1g^1M,and f1=v1,v2…,v2N and g^1=v^1,v^2…,v^2NT. When N=1, the single-soliton solution is(69)qx,t=8c1∗ξ2ηeiYsinhX2β−αc1ξ2sinh2X+η2cosh2X,with(70)c1=eξ1,ξ1=lnc1,λ=ξ+iη,X=σ0+θ1+θ1∗=4ξηx−16ξ2−η2ξηtβ−α+lnc1,Y=−θ1+θ1∗=4ξ4−6ξ2η2+η4t−2ξ2−η2xβ−α,θ1=2ξηx−8ξ2−η2ξηtα−β+i2ξ4−6ξ2η2+η4t−ξ2−η2xα−β.By choosing different parameters, we can obtain some plots which are displayed to particularly describe the dynamic behaviors for single-soliton solution equation (69). Figure 1 illustrates that the single-soliton solution qx,t is a M-type soliton.Figure 1
Profiles ofqx,t in (69) with the parameters chosen as ξ=0.45,η=−0.05,c1=1,α=1,β=8. (a) M-type soliton solution of Real qx,t; (b) M-type soliton solution of Imaginary rx,t; (c) the envelope surface of qx,t; (d–f) are the propagation pattern along the x-axis at t=0 in (a–c), respectively.
(a)(b)(c)(d)(e)(f)


## 6. Conclusions

In this paper, we started with the spectral analysis of the Lax pair of the GDNLS equation and formulated a Riemann–Hilbert problem on theR∪iR, which admits simple zero points generated by the roots of s11λ. By taking spectral analysis, we found that s11λ is an odd function. So, the zero points are paired. According to the symmetry relations of zero points, a transformation was established, which eliminates the zero points and makes the Riemann–Hilbert problem be regular. By solving particular Riemann–Hilbert problem with vanishing scattering coefficients, we successfully constructed as a simple representation for the N-soliton solution for the GDNLS equation. Moreover we reduced the N-soliton solution for the GDNLS equation to a ratio of 2N+1×2N+1 determinant and 2N×2N determinant.



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