Riemann-Hilbert problem for the sextic nonlinear Schrödinger equation with non-zero boundary conditions
aa r X i v : . [ n li n . S I] A ug Riemann-Hilbert problem for the sextic nonlinearSchr ¨odinger equation with non-zero boundary conditions
Xin Wu, Shou-Fu Tian ∗ , Jin-Jie Yang and Zhi-Qiang Li School of Mathematics and Institute of Mathematical Physics, China University of Mining and Technology,Xuzhou 221116, People’s Republic of China
Abstract
We consider a matrix Riemann-Hilbert problem for the sextic nonlinear Schr¨odingerequation with a non-zero boundary conditions at infinity. Before analyzing the spec-trum problem, we introduce a Riemann surface and uniformization coordinate variablein order to avoid multi-value problems. Based on a new complex plane, the directscattering problem perform a detailed analysis of the analytical, asymptotic and sym-metry properties of the Jost functions and the scattering matrix. Then, a generalizedRiemann-Hilbert problem (RHP) is successfully established from the results of the di-rect scattering transform. In the inverse scattering problem, we discuss the discretespectrum, residue condition, trace formula and theta condition under simple poles anddouble poles respectively, and further solve the solution of a generalized RHP. Finally,we derive the solution of the equation for the cases of di ff erent poles without reflec-tion potential. In addition, we analyze the localized structures and dynamic behaviorsof the resulting soliton solutions by taking some appropriate values of the parametersappeared in the solutions. Keywords:
The sextic nonlinear Schr¨odinger equation, Non-zero boundaryconditions, A generalized Riemann-Hilbert problem, Simple poles and double poles,Breathers and soliton solutions. ✩ Project supported by the Fundamental Research Fund for the Central Universities under the grant No.2019ZDPY07. ∗ Corresponding author.
E-mail addresses : [email protected], [email protected] (S. F. Tian)
Preprint submitted to Journal of L A TEX Templates August 19, 2020 . Introduction
The nonlinear Schr¨odinger (NLS) equation iq t + q xx + | q | q = , (1.1)is a classic physical model, which is completely integrable. It has significant physicalsignificance in many aspects of research such as deep water waves, fluid mechanics,plasma physics, nonlinear optics, Bose-Einstein condensation. For instance, it candescribe the grouping evolution of quasi-monochromatic waves slowly changing inweakly nonlinear dispersion media. Because of the rich physical meaning of the NLSequation, it has attracted a lot of attention. Especially after Zakharov and Shabat’sresearch about the NLS equation[1]-[3], there are more and more researches to extendit [4]-[16].Furthermore, on the basis of the previous results, it is found that high-order disper-sion terms and other profound physical e ff ects have more in-depth significance in thephysical research of equations. For example, pulses of shorter duration propagate inoptics along the fiber [17]-[19]. Therefore, the classic NLS equations can no longermeet the requirement of the research and the study of the higher-order NLS equationis more worthy of being put forward and carried out. In this work, we mainly considerthe sextic nonlinear Schr¨odinger (sNLS) equation [20] iq t +
12 ( q xx + | q | q ) + δ q xxxxxx + δ h q ∗ | q x | + q ∗ ) q xx + q ∗ xxxx i q + δ q h q ∗ q xxxx + q x q ∗ xxx + | q xx | i + δ q h q xxx q ∗ x + q ∗ ) q x i + δ ( q x ) q ∗ xx + δ q x (cid:2) q xx q ∗ x + q ∗ q xxx (cid:3) + δ q ∗ q xx + δ q h ( q ∗ x ) + q ∗ q xx i + δ q | q | = , (1.2)where q ( x , t ) is the complex variable about x and t , the asterisk denotes the complexconjugation, and δ is a real coe ffi cient. In addition, (1.2) will be reduced to the classicnonlinear Schr¨odinger equation when δ =
0. Some results of the sNLS equation havealso been studied in various ways such as analytic breather solutions have been ob-tained through the Darboux transformation in [21], and the soliton solutions also havebeen obtained through inverse scattering transform in [22].In this work, we are dedicated to study the inverse scattering of sNLS equation withnon-zero boundary conditions at infinity, i.e., q ( x , t ) → q ± e i ( q + δ q ) t , x → ±∞ , (1.3)here | q ± | = q ,
0. Di ff erent from the case of zero boundary value, it will be moredi ffi cult to establish and solve the Riemann-Hilbert problem.2t is well known that the inverse scattering transform (IST) has developed into apowerful analysis tool for solving a large class of pure and applied mathematics, whichplays an indispensable important role in the field of nonlinear science. IST was firstapplied to solve the Korteweg-de Vries equation by Garderner, Greene, Kruskal andMiurra (GGKM) in 1967 [23], and then it was found that IST can also be appliedto the NLS equation after the research of Zakharov and Shabat [1]. After IST wasfurther developed, it was widely used in various equations [24]-[49]. According towhat we know, the research of the sNLS equation with NZBCs (1.2) by using theinverse scattering transform has not been reported. Therefore, the main purpose ofour work is to find the soliton solution of the sNLS equation and study its physicalmeanings, so as to enrich the research of Schr¨odinger equation.The structure of this work is given as follows. In section 2, we mainly study thedirect scattering. Riemann surface and uniformization variables are introduced to avoiddouble-valued problems. Based on this result, we then discuss the analytical, symme-try and asymptotic properties of the Jost function and the scattering matrix. In section3, we establish a generalized Riemann-Hilbert problem from the basis of direct scat-tering, and then we also solve the sNLS equation by solving the generalized RHP. Insection 4, the inverse scattering problem with simple poles is discussed. To start with,we recover the potential by reconstructing the formula based on the discrete spectrumand residue conditions. Furthermore, trace formulate and theta condition are derived.Finally, the soliton solutions are given under reflection-less potential and its propaga-tion behavior is also illustrated graphically when selecting appropriate parameters. Insection 5, similar to simple poles, the results of the inverse scattering problem withdouble poles are obtained. In the end, some conclusions and discussions are presented.
2. Direct scattering transform
In this section, similar but di ff erent from the calculation of the zero boundary con-ditions, we will make some transformations and introduce new definitions under thenon-zero boundary conditions. In this subsection, Riemann surface will be introduced to solve the problem ofmulti-valued functions. The corresponding uniformization variable will also be intro-duced to solve the complexity. 3he Lax pair of (1.2) reads ψ x = X ψ, X = ik σ + Q ,ψ t = T ψ, T = T T T − T , (2.1)here ψ = ( ψ ( x , t ) , ψ ( x , t )) T , σ = − , Q = iq ∗ ( x , t ) iq ( x , t ) 0 , T = i " − | q | − δ | q | − δ (cid:16) q x ( q ∗ ) (cid:17) − δ | q xx | + δ (cid:0) q x q ∗ xxx − q ∗ q xxxx − qq ∗ xxxx (cid:1) − δ | q | (cid:0) q xx q ∗ + q ∗ xx q (cid:1)i + ik h i δ | q | (cid:0) q x q ∗ − q ∗ x q (cid:1) + i δ (cid:0) q x q ∗ xx − q ∗ x q xx + q ∗ q xxx − q ∗ xxx q (cid:1)i + ik h + δ | q | − δ | q x | + δ (cid:0) q ∗ xx q + q xx q ∗ (cid:1)i − δ k (cid:0) qq ∗ x − q ∗ q x (cid:1) − i δ k | q | + i δ k , T = i δ k q − δ k q x + ik (cid:16) − δ | q | q − δ q xx (cid:17) + ik (cid:16) − i δ | q | q x − i δ q xxx (cid:17) ik h q + δ q ∗ q x + δ | q | q xx + δ q q ∗ xx + δ q xxxx + δ | q | q + δ q | q x | i + i (cid:20) i q x + i δ q xxxxx + i δ (cid:16) qq ∗ x q xx + qq ∗ xx q x + | q | q xxx + | q | q x + q x | q x | q ∗ q x q xx (cid:17)(cid:21) , T = i δ k q ∗ − δ k q ∗ x + ik (cid:16) − δ | q | q ∗ − δ q ∗ xx (cid:17) + ik (cid:16) i δ | q | q ∗ x + i δ q ∗ xxx (cid:17) ik h q ∗ + δ qq x + δ | q | q ∗ xx + δ q q xx + δ q ∗ xxxx + δ | q | q ∗ + δ q ∗ | q x | i − i (cid:20) i q ∗ x + i δ q ∗ xxxxx + i δ (cid:16) q ∗ q x q ∗ xx + q ∗ q xx q ∗ x + | q | q ∗ xxx + | q | q ∗ x + q ∗ x | q x | + qq ∗ x q ∗ xx (cid:17)(cid:21) , where k is the spectral parameter. Furthermore, (1.2) satisfies zero curvature equation U t − V x + [ U , V ] = t into a non-zero boundaryindependent of time t , we make the following transformation q → qe i ( q + δ q ) t ,ψ → ψ e − i ( q + δ q ) t σ . (2.2)Then (1.2) develops into iq t + q xx + q ( | q | − q ) + δ q ( | q | − q ) + δ q xxxxxx + δ q h ( q ∗ x ) + q ∗ q xx i + δ q h q ∗ q xxxx + q x q ∗ xxx + | q xx | i + δ q h q xxx q ∗ x + q ∗ ) q x i + δ ( q x ) q ∗ xx + δ q x (cid:2) q xx q ∗ x + q ∗ q xxx (cid:3) + δ q ∗ q xx + δ h q ∗ | q x | + q ∗ ) q xx + q ∗ xxxx i q = , (2.3)4nd the Lax pair also becomes ψ x = X ψ,ψ t = T ψ, (2.4)here X = ik σ + Q , T = T − i (cid:16) q + δ q (cid:17) T T − T + i (cid:16) q + δ q (cid:17) . The new boundary conditions islim x →±∞ q ( x , t ) = q ± , | q ± | = q , , (2.5)so when x → ±∞ , we obtain the limit spectral problem ψ x = X ± ψ,ψ t = T ± ψ, (2.6)where X ± = ik σ + Q ± , Q ± = iq ∗± iq ± , T ± = i δ k σ + δ k Q ± + i δ k Q ± σ + δ k Q ± + ik σ + i δ k Q ± σ + kQ ± + i δ kQ ± + iQ ± σ + i δ Q ± σ − i q + δ q ! σ = (cid:16) δ k + δ k q + δ kq + k (cid:17) X ± . From the calculation, we can know that X ± has two eigenvalues i λ and − i λ , where λ = k + q = ( k + iq )( k − iq ) , the Riemann surface determined by this is formed by bonding complex k -planes S and S cut along secant [ − iq , iq ], here the branch points are k = ± iq . On S , introducinglocal polar coordinates k + iq = r e i θ , k − iq = r e i θ , − π < θ , θ < π , (2.7)5hen the single-valued analytical branch function can be written on the Riemann surfaceas λ ( k ) = ( r r ) e θ + θ , on S , − ( r r ) e θ + θ , on S . (2.8)Meanwhile, it has the following properties • Imk > S and Imk < S are mapped into Im λ > • Imk < S and Imk > S are mapped into Im λ < • The branch [ − iq , iq ] of S and S is mapped into the branch [ − q , q ] of λ -plane.Defining the uniformization variable z = k + λ, (2.9)so we get the following relations k ( z ) = z − q z , λ ( z ) = z + q z . (2.10)Considering the second equation of (2.10), namely, Joukowsky transformation, λ ( z ) = | z | h ( | z | − q ) z + q Rez i , one obtains Im λ ( z ) = | z | ( | z | − q ) Imz . Therefore, Joukowsky transformation has the following properties: • Im λ > D + = n z ∈ C : ( | z | − q ) Imz > o ; • Im λ < D − = n z ∈ C : ( | z | − q ) Imz < o ; • The branch [ − q , q ] is mapped into the circle of z -plane C = {| z | = q , z ∈ C } . The transformation relationship from k -plane to z -plane is shown in the figure be-low. 6 ekImkiq − iq S Imk > Imk < RekImkiq − iq S Imk > Imk < RekImkiq − iq λ − plane 0 Im λ > Im λ < RezImz iq − iq Figure 1.
Transformation relationship from k -Riemann plane to λ -plane and z -plane. In addition, for k → ∞ , z has two asymptotic states: on S , k → ∞ ⇒ z → ∞ ; on S , k → ∞ ⇒ z → . In this subsection, some results of Jost functions are obtained.Through the previous related calculations, we know that X ± has two eigenvalues ± i λ and T ± also has two eigenvalues ± i λ (32 δ k + δ k q + δ kq + k ). Accordingto the relation of X ± and T ± , it is obviously that they can be diagonalized by the samematrix, i.e., X ± ( x , t ; z ) = Y ± ( z )( i λσ ) Y − ± ( z ) , T ± ( x , t ; z ) = Y ± ( z )( i λ (32 δ k + δ k q + δ kq + k ) σ ) Y − ± ( z ) , (2.11)where Y ± ( z ) = − q ∗± zq ± z = I + iz σ Q ± . ( Y − ± ψ ) x = i λσ Y − ± ψ, ( Y − ± ψ ) t = i λ (32 δ k + δ k q + δ kq + k ) σ Y − ± ψ. (2.12)Then, the solution of asymptotic spectral problem (2.6) is ψ = Y ± e i θ ( z ) σ , (2.13)where θ ( z ) = λ ( z ) h x + (32 δ k + δ k q + δ kq + k ) t i .Therefore, the Lax pair (2.4) has a solution asymptotically to ψψ ± ∼ Y ± e i θ ( z ) σ , x → ±∞ . (2.14)For (2.4), making the following transformation µ ± = ψ ± e − i θ ( z ) σ , (2.15)we obtain µ ± ∼ Y ± , x → ±∞ . (2.16)The equivalent Lax pair of (2.4) is ( Y − ± µ ± ) x + i λ h Y − ± µ ± , σ i = Y − ± △ Q ± µ ± , ( Y − ± µ ± ) t + i λ (32 δ k + δ k q + δ kq + k ) h Y − ± µ ± , σ i = Y − ± △ T ± µ ± , (2.17)with △ Q ± = Q − Q ± , △ T ± = T − T ± .Furthermore, the full derivative form of the equivalent Lax pair (2.17) reads d (cid:16) e i θ ( z ) ˆ σ Y − ± µ ± (cid:17) = e i θ ( z ) ˆ σ h Y − ± ( △ Q ± dx − △ T ± dt ) µ ± i , (2.18)where e ˆ σ A = e σ Ae − σ . By integrating along two special paths ( −∞ , t ) → ( x , t ) and( + ∞ , t ) → ( x , t ), the following two Volterra integral equations can be derived as µ − ( x , t ; z ) = Y − + Z x −∞ Y − e i λ ( x − y ) ˆ σ h Y − − △ Q − ( y , t ; z ) µ − ( y , t ; z ) i dy ,µ + ( x , t ; z ) = Y + − Z + ∞ x Y + e i λ ( x − y ) ˆ σ h Y − + △ Q + ( y , t ; z ) µ + ( y , t ; z ) i dy . (2.19)If q − q ± ∈ L ( R ), then from the first of the Volterra integral equations, we get Y − − µ − , = + Z x −∞ G ( x − y , z ) △ Q − ( y ) µ − , dy , (2.20)8ere x − y > G ( x − y , z ) = γ q ∗− z − q − z e − i λ ( x − y ) e − i λ ( x − y ) , with γ = det ( Y ± ) = + q z . Noticing e − i λ ( x − y ) = e − i ( x − y ) Re λ e x − y ) Im λ , it is obviously that µ − , is analytic in D − . In the same way, we obtain µ + , is also analyticin D − , while µ + , and µ − , are analytic in D + . Furthermore, they can be recorded as µ + = (cid:16) µ ++ , , µ − + , (cid:17) ,µ − = (cid:16) µ −− , , µ + − , (cid:17) . In this section, some results of scattering matrix are obtained.Before the beginning of this section, we first introduce Abel’s theorem.
Theorem 2.1.
Assuming A ( x ) ∈ C n × n , forY x = A ( x ) Y , then ( detY ) x = trAdetY , therefore detY ( x ) = detY ( x ) e R xx trA ( t ) dt . Obviously, from the expressions of X and T in the Lax pair, we know trX = trT = . (2.21)Again according to Abel’s theorem, it can be deduced that( det ψ ± ) x = ( det ψ ± ) t = . (2.22)Thus, through (2.14) we know det ψ ± = detY ± = γ. (2.23)9ue to ψ ± are the solution of Lax pair (2.4), they have a linear relationship eachother, there exists S ( z ) satisfying ψ + ( x , t ; z ) = ψ − ( x , t ; z ) S ( z ) , (2.24)where S ( z ) = ( s i j ) × does not depend on x and t .Furthermore, it can be written in component form as ψ + , = s ψ − , + s ψ − , , (2.25) ψ + , = s ψ − , + s ψ − , . (2.26)Defining Wronskian as W ( φ, ψ ) = φ ψ − φ ψ , it is obviously that Wronskiansatisfies the following properties W ( φ, ψ ) = − W ( ψ, φ ) , W ( c φ, c ψ ) = c c W ( φ, ψ ) . By the component form of (2.24), the follow results can be obtained by s = W ( ψ + , , ψ − , ) /γ, s = W ( ψ + , , ψ − , ) /γ, s = W ( ψ − , , ψ + , ) /γ, s = W ( ψ − , , ψ + , ) /γ. (2.27)Combining (2.15) and (2.24) deduces µ + = µ − e i θ ˆ σ S ( z ) . (2.28)According to det ( µ ± ) x = det ( µ ± ) t =
0, then det ( µ ± ) = lim x →±∞ det ( µ ± ) = det ( Y ± ) = γ , . (2.29)Thus, µ ± is reversible. Direct calculation shows that S ( z ) can be represented by µ ± .Assuming q − q ± ∈ L ( R ), according to the analyticity of µ ± , we know that s isanalytic on D + , s is analytic on D − , and s and s are continue to Σ .In addition, we define the reflection coe ffi cients as ρ ( z ) = s ( z ) s ( z ) , ˜ ρ ( z ) = s ( z ) s ( z ) . (2.30) According to the first of the equivalent lax pair (2.17), one obtains h Y − ± ( z ) µ ± ( z ) i x + i λ ( z ) h Y − ± ( z ) µ ± ( z ) σ − σ Y − ± ( z ) µ ± ( z ) i = Y − ± ( z ) △ Q ± ( z ) µ ± ( z ) . (2.31)10or the above equation, we replace z with z ∗ and take the conjugate, and multiply σ onboth sides. Then By simple calculation one can obtain h Y − ± ( z ) σµ ∗± ( z ∗ ) σ i x + i λ ( z ) h Y − ± ( z ) σµ ∗± ( z ∗ ) σσ − σ Y − ± ( z ) σµ ∗± ( z ∗ ) σ i = Y − ± ( z ) △ Q ± ( z ) σµ ∗± ( z ∗ ) σ, (2.32)here σ = − . Due to (2.16), it is easy to see that − σµ ∗± ( z ∗ ) σ ∼ Y ± , x → ±∞ . (2.33)Thus, µ ± ( z ) and − µ ∗± ( z ∗ ) satisfy the same equation and have the same asymptotic be-havior, i.e., they are equal. From this, we get the symmetry relation µ ± ( z ) = − σµ ∗± ( z ∗ ) σ. (2.34)Similarly, the following symmetry relation can be obtained as µ ± ( z ) = iz µ ± − q z σ Q ± . (2.35)Expanding the symmetry relationship by column yields µ ± , ( z ) = σµ ∗± , ( z ∗ ) , µ ± , ( z ) = − σµ ∗± , ( z ∗ ) ,µ ± , ( z ) = q ± z µ ± , − q z , µ ± , ( z ) = − q ∗± z µ ± , − q z . (2.36)Because of the symmetry of µ ± ( z ) and the relationship between S ( z ) and µ ± ( z ), weget − σ S ∗ ( z ∗ ) σ = S ( z ) . (2.37)Applying the same way, the following relation also can be obtained as S ( z ) = ( σ Q − ) − S (cid:16) − q / z (cid:17) σ Q + . (2.38)By column, we have s ( z ) = s ∗ ( z ∗ ) , s ( z ) = − s ∗ ( z ∗ ) , s ( z ) = q + q − s − q z , s ( z ) = q ∗ + q − s − q z . Furthermore, the reflection coe ffi cient has the following symmetry ρ ( z ) = − ˜ ρ ∗ ( z ∗ ) = − q − q ∗− ˜ ρ − q z . (2.39)11 .5. Asymptotic behavior of µ ± and scattering matrix In this subsection, we give the asymptotic properties of µ ± ( x , t ; z ) and S ( z ) as fol-lows. Proposition 2.2.
The asymptotic properties of µ ± ( x , t ; z ) and S ( z ) are µ ± ( x , t ; z ) = I + iz σ Q + O z ! , z → ∞ , iz σ Q ± + O (1) , z → , (2.40) and S ( z ) = I + O z ! , z → ∞ , diag q + q − , q − q + ! + O ( z ) , z → . (2.41) Proof.
When z = ∞ , the asymptotic expansion of µ ± is µ ± = µ ± + µ ± z + · · · . (2.42)Combining Y − ± = γ (cid:18) I − iz Q ± σ (cid:19) , (2.43)we reconsider the equivalent Lax pair and compare the coe ffi cients of the power termof z to obtain µ ± = I , µ ± = i σ Q . Thus, we have µ ± ( x , t ; z ) = I + iz σ Q + O z ! , z → ∞ . (2.44)In the same way, we also can obtain the asymptotic properties of µ ± ( x , t ; z ) when z → µ ± ( x , t ; z ), the asymptotic properties of S ( z ) can be derived. (cid:3)
3. Generalized Riemann-Hilbert problem
In this section, we will construct a generalized Riemann-Hilbert problem based onthe previous results. 12ccording to (2.28), one have µ + , = s µ − , + s e − i θ µ − , , (3.1) µ + , = s e i θ µ − , + s µ − , . (3.2)According to the analysis of the Jost function µ ± and the scattering matrix, defininga sectionally meromorphic matrix as M ( x , t ; z ) = M + ( x , t ; z ) = µ + , ( x , t ; z ) s ( z ) , µ − , ( x , t ; z ) ! , M − ( x , t ; z ) = µ − , ( x , t ; z ) , µ + , ( x , t ; z ) s ( z ) ! . (3.3)Thus, by the asymptotic of µ ± and the scattering coe ffi cient, we know M ± ( x , t ; z ) ∼ I + O z ! , z → ∞ , (3.4) M ± ( x , t ; z ) ∼ iz σ Q − + O (1) , z → . (3.5)So far, the following theorem can be concluded. Definition 3.1.
The generalized Riemann-Hilbert problem is constructed as • M ( x , t ; z ) is meromorphic in C \ Σ ; • The jump condition is M − ( x , t ; z ) = M + ( x , t ; z ) ( I − G ( x , t ; z )) , (3.6)where G ( x , t ; z ) = − ˜ ρ ( z ) e i θ ( x , t ; z ) ρ ( z ) e i θ ( x , t ; z ) ρ ( z ) ˜ ρ ; • M ± ( x , t ; z ) ∼ I + O (cid:16) z (cid:17) , z → ∞ ; • M ± ( x , t ; z ) ∼ iz σ Q − + O (1) , z →
4. The inverse scattering transform with the simple poles
In this section, we will discuss and solve the soliton solution under the simple poles.
The discrete spectrum of the scattering problem is composed of all values z ∈ C \ Σ satisfying eigenfunctions in L ( R ). These values make s ( z ) = z ∈ D + and13 ( z ) = z ∈ D − respectively. Assuming z n ∈ D + T { z ∈ C : Imz > } is the simplepoles of s ( z ), then s ( z ) = s ( z ) , n = , , · · · , N .From the symmetry of the scattering data, one obtains s ( z n ) = ⇔ s ( z ∗ n ) = ⇔ s ( − q / z n ) = ⇔ s ( − q / z ∗ n ) = , (4.1)which yields a quartet of discrete eigenvalues, namely, Z = n z n , z ∗ n , − q / z n , − q / z ∗ n o , n = , , · · · , N . (4.2)According to (2.27), When s ( z n ) =
0, there is a constant b n and ˜ b n that do notdepend on x , t and z such that ψ + , ( x , t ; z n ) = b n ψ − , ( x , t ; z n ) , (4.3a) ψ + , ( x , t ; z ∗ n ) = ˜ b n ψ − , ( x , t ; z ∗ n ) . (4.3b)The transformation (2.15) yields ψ ± , = µ ± , e i θ ( z ) ,ψ ± , = µ ± , e − i θ ( z ) , (4.4)then (4.3a) and (4.3b) derive µ + , ( z n ) = b n e − i θ ( z n ) µ − , ( z n ) , (4.5) µ + , ( z ∗ n ) = ˜ b n e i θ ( z ∗ n ) µ − , ( z ∗ n ) . (4.6)Therefore, we get the residue condition as Res z = z n " µ + , ( z ) s ( z ) = µ + , ( z n ) s ′ ( z ) = b n s ′ ( z ) e − i θ ( z n ) µ − , ( z n ) , (4.7) Res z = z ∗ n " µ + , ( z ) s ( z ) = µ + , ( z ∗ n ) s ′ ( z ∗ ) = ˜ b n s ′ ( z ∗ ) e i θ ( z ∗ n ) µ − , ( z ∗ n ) . (4.8)For the convenience of calculation, introducing C n = b n s ′ ( z ) , ˜ C n = ˜ b n s ′ ( z ∗ ) , then (4.7) and (4.8) become Res z = z n " µ + , ( z ) s ( z ) = C n e − i θ ( z n ) µ − , ( z n ) , (4.9) Res z = z ∗ n " µ + , ( z ) s ( z ) = ˜ C n e i θ ( z ∗ n ) µ − , ( z ∗ n ) . (4.10)14oing some simple calculations based on symmetry after substituting (4.4) into(4.3a) and (4.3b), it is easy to get the relation between b n and ˜ b n by compare the results,i.e., − b ∗ n = ˜ b n . (4.11)Combining s ′ ( z n ) = (cid:16) s ′ ( z ∗ n ) (cid:17) ∗ , we have˜ C n = − C ∗ n . (4.12)Applying the symmetry of µ ± for µ + , ( z n ) = b n µ − , ( z n ) e − i θ ( z n ) , it can be see ψ + , − q z n = − q ∗− q + b n ψ − , − q z n . (4.13)Similarly, ψ + , − q z ∗ n = − q − q ∗ + ˜ b n ψ − , − q z ∗ n . (4.14)Next, we need to deal with s ( z ) = q + q − s (cid:18) − q z (cid:19) as follows. Taking the derivative of z ,substitute z = z ∗ , and making the conjugate on both sides, we can get s ′ − q z ∗ n = q − q + ! ∗ z n q ! (cid:0) s ′ ( z n ) (cid:1) ∗ . (4.15)In the same way, one obtains s ′ − q z n = q − q + z n q ! (cid:0) s ′ ( z ∗ n ) (cid:1) ∗ . (4.16)Therefore, the residue conditions on z = − q z ∗ n and z = − q z n are Res z = − q z ∗ n " µ + , ( z ) s ( z ) = − q − q ∗− z n q ! ˜ C n e − i θ ( − q / z ∗ n ) µ − , − q z ∗ n = C N + n e − i θ ( − q / z ∗ n ) µ − , − q z ∗ n , (4.17) Res z = − q zn " µ + , ( z ) s ( z ) = − q ∗− q − z n q ! C n e i θ ( − q / z n ) µ − , − q z n = ˜ C N + n e − i θ ( − q / z n ) µ − , − q z n , (4.18)15here C N + n = − q − q ∗− z n q ! ˜ C n , ˜ C N + n = − q ∗− q − z n q ! C n . For convenience, numbering a quartet of discrete eigenvalues uniformly and defin-ing ξ n = z n and ξ N + n = − q z n for n = , , · · · , N , the residue condition of M + ( z ) and M − ( z ) are Res z = ξ n M + ( z ) = (cid:16) C n e − i θ ( ξ n ) µ − , ( ξ n ) , (cid:17) , (4.19) Res z = ξ ∗ n M − ( z ) = (cid:16) , ˜ C n e i θ ( ξ ∗ n ) µ − , ( ξ ∗ n ) (cid:17) , (4.20)here n = , , · · · , N . To solve the Riemann-Hilbert(RH) problem, we need to transform the original RHproblem into a regular RH problem by subtracting the asymptotic properties and thepole contributions. Thus (3.6) can be rewritten as M − ( z ) − I − iz σ Q − − N X n = Res z = ξ ∗ n M − ( z ) z − ξ ∗ n − N X n = Res z = ξ n M + ( z ) z − ξ n = M + ( z ) − I − iz σ Q − − N X n = Res z = ξ ∗ n M − ( z ) z − ξ ∗ n − N X n = Res z = ξ n M + ( z ) z − ξ n − M + ( z ) G ( z ) . (4.21)For (4.21), we know that the first five terms on the left side are analyzed on D − , whilethe first five terms on the right side are analyzed on D + . Introducing the projectionoperators P ± on Σ as P ± [ f ]( z ) = π i Z Σ f ( ζ ) ζ − ( z ± i d ζ, (4.22)where R Σ means the integral along the oriented contour shown in Fig.1 and z ± i / right of z ( z ∈ Σ ) respectively. Via applying Plemelj’sformulae, (4.21) can be solved as M ( z ) = I + iz σ Q − + N X n = Res z = ξ ∗ n M − ( z ) z − ξ ∗ n + N X n = Res z = ξ n M + ( z ) z − ξ n − π i Z Σ M + ( s ) G ( s ) s − z ds , z ∈ C \ Σ . (4.23)16onsidering the residue condition, the second column of (4.23), in z = z n and z = − q / z ∗ n , i.e., z = ξ n , is solved as µ − , ( ξ n ) = − q ∗− /ξ n + N X k = ˜ C k e i θ ( ξ ∗ k ) ξ n − ξ ∗ k µ − , ( ξ ∗ k ) + π i Z Σ ( M + G ) ( s ) s − ξ n ds . (4.24)Similarly, the first column of (4.23) in z = z ∗ n and z = − q / z n , i.e., z = ξ ∗ n , is µ − , ( ξ ∗ n ) = q − /ξ ∗ n + N X j = C j e − i θ ( ξ j ) ξ ∗ n − ξ j µ − , ( ξ j ) + π i Z Σ ( M + G ) ( s ) s − ξ ∗ n ds . (4.25)When z → ∞ , asymptotic expansion of (4.23) is M ( z ) = I + z i σ Q − + N X n = Res z = ξ ∗ n M − ( z ) + Res z = ξ n M + ( z ) ! − π i Z Σ M + ( s ) G ( s ) ds ) + O z ! . (4.26)Taking M = M − , we obtain the follow result by combining the (2, 1)-element of asymp-totic behavior of µ ± as q = q − + N X n = C n e − i θ ( ξ n ) µ − , ( ξ n ) − π i Z Σ ( M + ( s ) G ( s )) ds . (4.27) Recall that s and s are analytic on D + and D − respectively, and the discretespectrum is Z = { z n , z ∗ n , − q / z n , − q / z ∗ n } for n = , , · · · , N . Constructing the followfunctions β + ( z ) = s ( z ) N Y n = ( z − z ∗ n )( z + q / z n )( z − z n )( z + q / z ∗ n ) ,β − ( z ) = s ( z ) N Y n = ( z − z n )( z + q / z ∗ n )( z − z ∗ n )( z + q / z n ) , (4.28)here β + ( z ) and β − ( z ) are analytic in D + and D − respectively, and there is no zero point.In addition, β ± ( z ) → z → ∞ via the asymptotic behavior of the scattering data.For all z ∈ Σ , we have β + ( z ) β − ( z ) = s ( z ) s ( z ). According to detS ( z ) =
1, it canbe seen that 1 s ( z ) s ( z ) = − ρ ( z ) ˜ ρ ( z ) = + ρ ( z ) ρ ∗ ( z ∗ ) , β + ( z ) β − ( z ) = + ρ ( z ) ρ ∗ ( z ∗ ) , z ∈ Σ . (4.29)By taking the logarithm of both sides of the above formula and using the Plemelj’sformulae we knowlog β ± ( z ) = ∓ π i Z Σ log[1 + ρ ( ζ ) ρ ∗ ( ζ ∗ )] ζ − z d ζ, z ∈ D ± . (4.30)Substituting (4.30) into (4.28), we obtain trace formulate as s ( z ) = exp " − π i Z Σ log[1 + ρ ( ζ ) ρ ∗ ( ζ ∗ )] ζ − z d ζ N Y n = ( z − z n )( z + q / z ∗ n )( z − z ∗ n )( z + q / z n ) , z ∈ D + , s ( z ) = exp " π i Z Σ log[1 + ρ ( ζ ) ρ ∗ ( ζ ∗ )] ζ − z d ζ N Y n = ( z − z ∗ n )( z + q / z n )( z − z n )( z + q / z ∗ n ) , z ∈ D − . (4.31)Taking z → arg q + q − ! = π Z Σ log[1 + ρ ( ζ ) ρ ∗ ( ζ ∗ )] ζ − z d ζ + N X n = argz n . (4.32) Now, we study the potential q ( x , t ) when the reflection coe ffi cient ρ ( z ) graduallydisappears. At this point, without the jump from M + to M − , the inverse problem willalso degenerate into an algebraic system. According to (4.24) and (4.25), it is easy tosee that µ − , ( ξ n ) = + N X k = ˜ C k ξ n − ξ ∗ k e i θ ( ξ ∗ k ) µ − , ( ξ ∗ k ) = − N X k = c ∗ j ( ξ ∗ k ) µ − , ( ξ ∗ k ) , (4.33) µ − , ( ξ ∗ k ) = q − ξ ∗ k + N X j = C j ξ ∗ k − ξ j e − i θ ( ξ j ) µ − , ( ξ j ) = q − ξ ∗ k + N X j = c j ( ξ ∗ k ) µ − , ( ξ j ) , (4.34)here c j ( z ) = C j z − ξ j e − i θ ( ξ j ) , with j = , , · · · , N . Substituting (4.34) into (4.33) get µ − , ( ξ n ) = − N X k = c ∗ k ( ξ ∗ n ) q − ξ ∗ k − N X k = N X j = c ∗ k ( ξ ∗ n ) c j ( ξ ∗ k ) µ − , ( ξ j ) . (4.35)18ntroducing X n = µ − , ( ξ n ) , X = ( X , · · · , X N ) T (4.36) B n = − N X k = c ∗ k ( ξ ∗ n ) q − ξ ∗ k , B = ( B , · · · , B N ) T , (4.37) A n , j = N X j = c ∗ k ( ξ ∗ n ) c j ( ξ ∗ k ) , A = ( A n , j ) N × N , (4.38)then (4.35) is converted into component form MX = B , (4.39)where M = I + A = ( M , · · · , M N ).Using Cramer’s Rule, we can calculate that X n = detM extn detM , (4.40)here M extn = ( M , · · · , M n − , B , M n + , · · · , M N ). Thus q ( x , t ) = q − − detM aug detM , (4.41)where M aug = Y T B M , Y n = C n e − i θ ( ξ n ) , Y = ( Y , · · · , Y N ) . In the last subsection, we specifically obtain the exact expression of the solitonsolution of the sNLS equation and the images of the solutions by selecting appropriateparameters to study the propagation behavior and dynamic behavior of the solution.Let N =
1, we discuss the figures under the appropriate parameters.( a ) ( b ) ( c ) ( d ) Figure 2.
The breather wave solution for | q | with the parameters selection δ = , ξ = i . (a) the oliton solution with q − = (b) the density plot corresponding to (a), (c) the soliton solutionwith q − = . i , (d) the density plot corresponding to (c). In Fig.2, we have discussed the image when q − is a real number and an imaginarynumber, and found that the breathing wave solution can be obtained in both cases.( a ) ( b ) ( c ) Figure 3.
The breather wave solution for | q | with the parameters selection δ = , ξ = i . (a) thesoliton solution with q − = . (b) the soliton solution with q − = (c) the soliton solution with q − = . In Fig.3, when the value of q − is di ff erent, the shape of the wave is also di ff er-ent. When the value of q − gets smaller and smaller and gradually approaches 0, thebreathing wave gradually becomes a bright soliton wave.Next, we discuss the figures of the soliton when N =
2. In Fig.4, it is Obvious toseen that the two solitons are side by side and do not a ff ect each other.( a ) ( b ) ( c ) Figure 4.
The breather wave solution for | q | with the parameters selection δ = . , ξ = + . i , ξ = − + . i . (a) the soliton solution with q − = (b) density plot correspondingto (a), (c) the contour line of the soliton solution corresponding to (a). . The inverse scattering transform with the double poles In this section, we will discuss the case where z n is a double pole of s , i.e., s ( z n ) = s ′ ( z n ) =
0, but s ′′ ( z n ) ,
0. Therefore, our calculations in this section will be di ff erent from thesimple poles case. According to the symmetry of the scattering data S ( z ), if s ( z n ) = s ( z ∗ n ) = s ( − q / z n ) = s ( − q / z ∗ n ) = . Then discrete spectrum is Z = { Z n , z ∗ n , − q / z n , − q / z ∗ n } , n = , · · · , N . (5.1)For convenience, defining ˆ z n = − q / z n and ˇ z n = − q / z ∗ n .From the simple poles, we know µ + , ( z n ) = b n e − i θ n µ − , ( z n ) . (5.2)Because z n is the double pole of s ( z ), s ′ ( z n ) =
0. For s = γ W ( ψ + , , ψ − , ), there are0 = s ′ ( z n ) = " γ W ( ψ ′ + , , ψ − , ) + W ( ψ + , , ψ ′− , ) − γ ′ γ W ( ψ + , , ψ − , ) ! z = z n , i.e., 0 = W ( ψ ′ + , ( z n ) − b n ψ ′− , ( z n ) , ψ − , ( z n )) . Thus, there is a constant d n that satisfies ψ ′ + , ( z n ) − b n ψ ′− , ( z n ) = d n ψ − , ( z n ) , namely, ψ ′ + , ( z n ) = d n ψ − , ( z n ) + b n ψ ′− , ( z n ) . Using (4.4) and (5.2), the follow result can be obtained µ ′ + , ( z n ) = e − i θ n h ( d n − i θ ′ n b n ) µ − , ( z n ) + b n µ ′− , ( z n ) i , (5.3)here θ n = θ ( z n ), θ ′ n = θ ′ ( z n ).In the same way, we also get the follow relations µ + , ( z ∗ n ) = ˜ b n e i θ ∗ n µ − , ( z ∗ n ) , (5.4) µ ′ + , ( z ∗ n ) = e i θ ∗ n h ( ˜ d n + i θ ′∗ n ˜ b n ) µ − , ( z ∗ n ) + ˜ b n µ ′− , ( z ∗ n ) i , (5.5) µ + , (ˇ z n ) = ˇ b n e − i ˇ θ n µ − , (ˇ z n ) , (5.6) µ ′ + , (ˇ z n ) = e − i ˇ θ n h ( ˇ d n − i ˇ θ ′ n ˇ b n ) µ − , (ˇ z n ) + ˇ b n µ ′− , (ˇ z n ) i , (5.7) µ + , (ˆ z n ) = ˆ b n e i ˆ θ n µ − , (ˆ z n ) , (5.8) µ ′ + , (ˆ z n ) = e i ˆ θ n h ( ˆ d n + i ˆ θ ′ n ˆ b n ) µ − , (ˆ z n ) + ˆ b n µ ′− , (ˆ z n ) i , (5.9) here θ ∗ n = θ ( z ∗ n ) , ˇ θ n = θ (ˇ z n ) , ˆ θ n = θ (ˆ z n ) ,θ ′ ∗ n = θ ′ ( z ∗ n ) , ˇ θ ′ n = θ ′ (ˇ z n ) , ˆ θ ′ n = θ ′ (ˆ z n ) , and ˜ b n , ˜ d n , ˇ b n , ˇ d n , ˆ b n and ˆ d n are constants.Similar to the proof of (4.11), we also have the following relationshipˆ b n = − q ∗− q + b n , − ˇ b n = ˆ b ∗ n , d ∗ n = − ˜ d n , ˆ d n = − q ∗− z n q + q d n , ˇ d n = − ˆ d ∗ n . (5.10)If f and g is analyzed in Ω ∈ C , and g has a double zero at z ∈ Ω and f ( z ) ,
0, then
Res z = z " fg = f ′ ( z ) g ′′ ( z ) − f ( z ) g ′′′ ( z )3( g ′′ ( z )) , P − z = z " fg = f ( z ) g ′′ ( z ) . Thus P − z = z n " µ + , s = µ + , ( z n ) s ′′ ( z n ) = b n s ′′ ( z n ) e − i θ n µ − , ( z n ) , (5.11) Res z = z n " µ + , s = b n s ′′ ( z n ) e − i θ n " µ ′− , ( z n ) + d n b n − i θ ′ n − s ′′′ ( z n )3 s ′′ ( z n ) ! µ − , ( z n ) . (5.12)Let A n = b n s ′′ ( z n ) and B n = d n b n − s ′′′ ( z n )3 s ′′ ( z n ) , then the above equations change into P − z = z n " µ + , s = A n e − i θ n µ − , ( z n ) , (5.13) Res z = z n " µ + , s = A n e − i θ n h µ ′− , ( z n ) + (cid:0) B n − i θ ′ n (cid:1) µ − , ( z n ) i . (5.14)In the same way, we also get the similar relations as follows P − z = z ∗ n " µ + , s = ˜ A n e i θ ∗ n µ − , ( z ∗ n ) , (5.15) Res z = z ∗ n " µ + , s = ˜ A n e i θ ∗ n h µ ′− , ( z ∗ n ) + (cid:16) ˜ B n + i θ ′∗ n (cid:17) µ − , ( z ∗ n ) i , (5.16) P − z = ˇ z n " µ + , s = ˇ A n e − i ˇ θ n µ − , (ˇ z n ) , (5.17) Res z = ˇ z n " µ + , s = ˇ A n e − i ˇ θ n h µ ′− , (ˇ z n ) + (cid:16) ˇ B n − i ˇ θ ′ n (cid:17) µ − , (ˇ z n ) i , (5.18) P − z = ˆ z n " µ + , s = ˆ A n e i ˆ θ n µ − , (ˆ z n ) , (5.19) Res z = ˆ z n " µ + , s = ˆ A n e i ˆ θ n h µ ′− , (ˆ z n ) + (cid:16) ˆ B n + i ˆ θ ′ n (cid:17) µ − , (ˆ z n ) i , (5.20) here ˜ A n = b n s ′′ ( z ∗ n ) , ˜ B n = ˜ d n ˜ b n − s ′′′ ( z ∗ n )3 s ′′ ( z ∗ n ) , ˇ A n = b n s ′′ (ˇ z n ) , ˇ B n = ˇ d n ˇ b n − s ′′′ (ˇ z n )3 s ′′ (ˇ z n ) , ˆ A n = b n s ′′ (ˆ z n ) , ˆ B n = ˆ d n ˆ b n − s ′′′ (ˆ z n )3 s ′′ (ˆ z n ) . According to the symmetry in (4.11) and (5.10), the following relation also can be obtained˜ A n = − A ∗ n , ˜ B n = B ∗ n , ˆ A n = − q q ∗− z n q − A n , ˆ B n = q ˆ z n B n + z n , ˇ A n = − ˆ A ∗ n , ˇ B n = ˆ B ∗ n . For convenience, let ξ n : = z n , ξ N + n : = ˇ z n , ξ ∗ n : = ˆ z n , ξ ∗ N + n : = z ∗ n . Similar to the case of the simple poles, to get a regular RHP, we need to subtract the asymptoticproperties and the pole contributions. The jump condition becomes M − ( z ) − I − iz σ Q − − N X n = Res z = ξ ∗ n M − ( z ) z − ξ ∗ n + P − z = ξ ∗ n M − ( z )( z − ξ ∗ n ) Res z = ξ n M + ( z ) z − ξ n + P − z = ξ n M + ( z )( z − ξ n ) = M + ( z ) − I − iz σ Q − − N X n = Res z = ξ ∗ n M − ( z ) z − ξ ∗ n + P − z = ξ ∗ n M − ( z )( z − ξ ∗ n ) Res z = ξ n M + ( z ) z − ξ n + P − z = ξ n M + ( z )( z − ξ n ) − M + ( z ) G ( z ) . (5.21)During the defining of M ± ( z ), we have Res z = ξ ∗ n M − ( z ) = , Res z = ξ ∗ n " µ + , s , P − z = ξ ∗ n M − ( z ) = , P − z = ξ ∗ n " µ + , s , Res z = ξ n M + ( z ) = Res z = ξ n " µ + , s , ! , P − z = ξ n M + ( z ) = P − z = ξ n " µ + , s , ! . (5.22)It is easy to see that the left side of (5.22) is analytic in D − , and the right side except the last itemis analytic in D + . Resorting to Plemelj’ formulae, the solution of RHP is M ( z ) = I + iz σ Q − + N X n = Res z = ξ ∗ n M − ( z ) z − ξ ∗ n + P − z = ξ ∗ n M − ( z )( z − ξ ∗ n ) Res z = ξ n M + ( z ) z − ξ n + P − z = ξ n M + ( z )( z − ξ n ) + π i Z Σ M + ( s ) G ( s ) s − z ds , z ∈ C \ Σ . (5.23) valuating the first column of (5.23) at z = ξ ∗ k , we get µ − , ( ξ ∗ k ) = q − ξ ∗ k + π i Z Σ (( M + G )( s )) s − ξ ∗ k ds + N X n = " C n ( ξ ∗ k ) µ ′− , ( ξ n ) + C n ( ξ ∗ k ) D n + ξ ∗ k − ξ n ! µ − , ( ξ n ) , (5.24)here C n ( z ) = A n e − i θ n z − ξ n and D n = B n − i θ ′ n .By the symmetry of (2.36), we know that µ − , ( ξ ∗ k ) = − ξ k q ∗− µ − , ( ξ k ), then (5.24) can be rewrittenas 0 = q − ξ ∗ k + π i Z Σ (( M + G )( s )) s − ξ ∗ k ds + N X n = ( C n ( ξ ∗ k ) µ ′− , ( ξ n ) + C n ( ξ ∗ k ) " D n + ξ ∗ k − ξ n ! + δ kn ξ k q ∗− µ − , ( ξ n ) ) , (5.25)where δ kn is Kronecker delta. In order to obtain further results at the derivative of the eigenfunc-tion, we derive the derivative of (5.25) with respect to z at z = ξ ∗ k and obtain0 = − q − ( ξ ∗ k ) + π i Z Σ (( M + G )( s )) ( s − ξ ∗ k ) ds − N X n = ( C n ( ξ ∗ k ) ξ ∗ k − ξ n − ξ ∗ k δ kn q q ∗− ! µ ′− , ( ξ n ) + " C n ( ξ ∗ k ) ξ ∗ k − ξ n D n + ξ ∗ k − ξ n ! − ξ k q q ∗− µ − , ( ξ n ) ) . (5.26)Let M = M − , using the same method as in the simple poles get q = q − + i π Z Σ (( M + G )( s )) ds − N X n = A n e − i θ n h µ ′− , ( ξ n ) + D n µ − , ( ξ n ) i . (5.27) According to analyticity of s and s and the discrete spectrum, it is obvious to knowthat β + ( z ) has the double poles in z n and − q / z ∗ n , and β − ( z ) in z ∗ n and − q / z n respectively for n = , , · · · , N . For β + ( z ) = s ( z ) N Y n = ( z − z ∗ n ) ( z + q / z n ) ( z − z n ) ( z + q / z ∗ n ) ,β − ( z ) = s ( z ) N Y n = ( z − z n ) ( z + q / z ∗ n ) ( z − z ∗ n ) ( z + q / z n ) , (5.28) β + ( z ) and β − ( z ) are analytic in D + and D − respectively and have no zero point. Furthermore,their asymptotic properties are the same as s ( z ) and s ( z ) when z → ∞ . Thus there have β + ( z ) β − ( z ) = s ( z ) s ( z ) for z ∈ Σ . Again by detS ( z ) =
1, we know β + ( z ) β − ( z ) = + ρ ( z ) ρ ∗ ( z ∗ ) , z ∈ Σ . (5.29) et’s take the logarithms for the above equation and apply the Plemelj’s formulae, it is easy toobtain that log β ± ( z ) = − π i Z Σ log[1 + ρ ( ζ ) ρ ∗ ( ζ ∗ )] ζ − ( z ± i d ζ, z ∈ D ± . (5.30)Similar to the calculation in the simple pole, trace formulate and theta condition are s ( z ) = exp " − π i Z Σ log[1 + ρ ( ζ ) ρ ∗ ( ζ ∗ )] ζ − z d ζ N Y n = ( z − z n ) ( z + q / z ∗ n ) ( z − z ∗ n ) ( z + q / z n ) , z ∈ D + , s ( z ) = exp " − π i Z Σ log[1 + ρ ( ζ ) ρ ∗ ( ζ ∗ )] ζ − z d ζ N Y n = ( z − z ∗ n ) ( z + q / z n ) ( z − z n ) ( z + q / z ∗ n ) , z ∈ D − , (5.31)and arg q + q − ! = π Z Σ log[1 + ρ ( ζ ) ρ ∗ ( ζ ∗ )] ζ − z d ζ + N X n = argz n . (5.32) In this section, we will consider the case of gradual disappearance of the reflection coe ffi -cient, i.e., G ( z ) =
0. From (5.27), the solution of the equation is q = q − − N X n = A n e − i θ n h µ ′− , ( ξ n ) + D n µ − , ( ξ n ) i . (5.33)In the following, we will focus on solving µ ′− , ( ξ n ) and µ − , ( ξ n ).When G ( z ) =
0, through (5.25) and (5.26) we have0 = q − ξ ∗ k + N X n = ( C n ( ξ ∗ k ) µ ′− , ( ξ n ) + C n ( ξ ∗ k ) " D n + ξ ∗ k − ξ n ! + δ kn ξ k q ∗− µ − , ( ξ n ) ) , (5.34)and 0 = q − ( ξ ∗ k ) + N X n = ( C n ( ξ ∗ k ) ξ ∗ k − ξ n − ξ ∗ k δ kn q q ∗− ! µ ′− , ( ξ n ) + " C n ( ξ ∗ k ) ξ ∗ k − ξ n D n + ξ ∗ k − ξ n ! − ξ k q q ∗− µ − , ( ξ n ) ) . (5.35)Defining X n = µ − , ( ξ n ) , X N + n = µ ′− , ( ξ n ) , V n = q − ( ξ n ) , V N + n = q − ( ξ n ) , M k , n = C n ( ξ ∗ k ) D n + ξ ∗ k − ξ n ! + δ kn ξ k q ∗− , M k , N + n = C n ( ξ ∗ k ) , M N + k , n = C n ( ξ ∗ k ) ξ ∗ k − ξ n D n + ξ ∗ k − ξ n ! − ξ k q q ∗− , M N + k , N + n = C n ( ξ ∗ k ) ξ ∗ k − ξ n − ξ ∗ k δ kn q q ∗− , ith n = , , · · · , N .Therefore, (5.34) and (5.35) are expressed in component form as MX = V . (5.36)By Cramer’s Rule, the solution of (5.36) is X n = detM extn detM , (5.37)where M extn = ( M , · · · , M n − , V , M n + , · · · , M N + n ). In this subsection, we mainly discuss the exact solution images obtained in the previous sec-tion and analyze their dynamic behavior. Let N =
1, the image of the solution of the equation isas follows. ( a ) ( b ) ( c )( d ) ( e ) ( f )( g ) ( h ) ( i ) Figure 5.
The breather wave solution for | q | with the parameters selection δ = , ξ = i . (a) he soliton solution with q − = . (b) density plot corresponding to (a), (c) the contour lineof the soliton solution corresponding to (a), (d) the soliton solution with q − = . (e) densityplot corresponding to (d), (f) the contour line of the soliton solution corresponding to (d), (g) thesoliton solution with q − = . (h) density plot corresponding to (g), (i) the contour line of thesoliton solution corresponding to (g).In Fig.5, we separately discuss the images of q − at di ff erent values, and find that when q − approaches 0, the breathing wave approaches a bright soliton.
6. Conclusions and discussions
In this work, we mainly solve the exact solution of the sixtic Schr¨odinger equation basedon the generalized Riemann-Hilbert problem with non-zero boundary values. Since the non-zero boundary considerations is considered here, the sixtic Schr¨odinger equation needs to betransformed first, so that the non-zero boundary does not depend on t . In order to avoid the multi-value problem, Riemann surface and uniformization coordinate are introduced to transform theoriginal spectrum problem into a new spectrum problem. On this basis, we discuss the analysis,symmetry and asymptotic properties of the Jost functions and the scattering matrix, which willprepare for the establishment of a generalized Riemann-Hilbert problem. In inverse scattering, itis divided into two cases: simple pole and double pole. Doing the same process, we reconstructthe formula for potential according to the discrete spectrum and residue conditions, and discussthe trace formula and theta condition. Finally, the soliton solutions of the equation are solvedwithout reflection potential. In addition, we have made vivid image descriptions of the exactsolutions in simple poles and double poles respectively by selecting appropriate parameters,which is of great help to the study of the propagation of the equation and its dynamic behavior.We also hope that our results will be of great significance to the study of Schr¨odinger equation. Acknowledgements
This work was supported by the National Natural Science Foundation of China under GrantNo. 11975306, the Natural Science Foundation of Jiangsu Province under Grant No. BK20181351,the Six Talent Peaks Project in Jiangsu Province under Grant No. JY-059, the Fundamental Re-search Fund for the Central Universities under the Grant Nos. 2019ZDPY07 and 2019QNA35,and the Assistance Program for Future Outstanding Talents of China University of Mining andTechnology under Grant No.2020WLJCRCZL074.
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