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Generalized Matrix Exponential Solutions to the AKNS Hierarchy

Jian-bing Zhang Jiangsu Normal University

Canyuan Gu Jiangsu Normal University

Wen-Xiu Ma University of South Florida, [email protected]

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Scholar Commons Citation Zhang, Jian-bing; Gu, Canyuan; and Ma, Wen-Xiu, "Generalized Matrix Exponential Solutions to the AKNS Hierarchy" (2018). Mathematics and Statistics Faculty Publications. 19. https://scholarcommons.usf.edu/mth_facpub/19

This Article is brought to you for free and open access by the Mathematics and Statistics at Scholar Commons. It has been accepted for inclusion in Mathematics and Statistics Faculty Publications by an authorized administrator of Scholar Commons. For more information, please contact [email protected]. Hindawi Advances in Mathematical Physics Volume 2018, Article ID 1375653, 9 pages https://doi.org/10.1155/2018/1375653

Research Article Generalized Matrix Exponential Solutions to the AKNS Hierarchy

Jian-bing Zhang ,1 Canyuan Gu,1 and Wen-Xiu Ma 2,3,4,5,6

1 School of Mathematics and Statistics, Jiangsu Normal University, Xuzhou, Jiangsu 221116, China 2Department of Mathematics and Statistics, University of South Florida, Tampa, FL 33620, USA 3College of Mathematics and Systems Science, Shandong University of Science and Technology, Qingdao, Shandong 266590, China 4Department of Mathematics, Zhejiang Normal University, Jinhua, Zhejiang 321004, China 5College of Mathematics and Physics, Shanghai University of Electric Power, Shanghai 200090, China 6Department of Mathematical Sciences, North-West University, Mafkeng Campus, Mmabatho 2735, South Africa

Correspondence should be addressed to Jian-bing Zhang; [email protected]

Received 4 December 2017; Revised 8 January 2018; Accepted 14 January 2018; Published 13 February 2018

Academic Editor: Zhijun Qiao

Copyright © 2018 Jian-bing Zhang et al. Tis is an open access article distributed under the Creative Commons Attribution License, which permits unrestricted use, distribution, and reproduction in any medium, provided the original work is properly cited.

Generalized matrix exponential solutions to the AKNS equation are obtained by the inverse scattering transformation (IST). Te resulting solutions involve six matrices, which satisfy the coupled Sylvester equations. Several kinds of explicit solutions including soliton, complexiton, and Matveev solutions are deduced from the generalized matrix exponential solutions by choosing diferent kinds of the six involved matrices. Generalized matrix exponential solutions to a general integrable equation of the AKNS hierarchy are also derived. It is shown that the general equation and its matrix exponential solutions share the same linear structure.

1. Introduction Since the discovery of scattering behavior of solitons [4] and the IST [5], solitons have received much attention. Many nonlinear models are studied and shown to possess Te IST has been also well developed and widely used to hierarchies, and recursion operators play a crucial role in solve nonlinear equations [1, 6]. It can be used to solve constructing hierarchies of soliton equations [1]. Associated not only normal soliton equations but also unusual soliton with the variational derivative, recursion operators have been equations such as equations with self-consistent sources [7], developed to formulate the Hamiltonian structures proving nonisospectral equations [8, 9], and equations with steplike the integrability of soliton hierarchies [2, 3]. Recursion oper- fnite-gap backgrounds [10] and on quasi-periodic back- ators also have a tight correlation with one-soliton solutions. For example, the Korteweg-de Vries (KdV) hierarchy grounds [11]. Recently, Ablowitz and Musslimani developed the IST for the integrable nonlocal nonlinear Schrodinger¨ � =��� , �=0,1,2,..., � � (1) (NLS) equation [12]. possesses the following one-soliton solution: Te Sylvester equation 2� �2 �(�+� �+�) �= 2 , �=0,1,2,..., (2) �� − �� = � (4) 2 sech 2 2 −1 is one of the most well-known matrix equations. It appears where �=� +4�+2��� is the recursion operator of the −1 −1 KdV hierarchy, � is a constant, and �=�/��, �� =� �= frequently in many areas of applied mathematics and plays a 1. Notice that the dispersion relation of (2) central role, in particular, in systems and control theory, sig- nal processing, fltering, model reduction, image restoration, ��+�2�+1� (3) and so on. In recent years, it has been used to solve soliton is linked closely with the order of the recursion operator �. equations [13, 14] successfully. Te method based on the 2 Advances in Mathematical Physics

Sylvester equation is also known as Cauchy matrix approach It is well known that the AKNS hierarchy has the [14–18], which is actually a by-product of direct linearization following Lax pairs [29]: approach frst proposed by Fokas and Ablowitz in 1981 [15] and developed to discrete integrable systems by Nijhof et al. −�� � �� =��, �=( ), (7a) in early of 1980s [16]. ��� In the process of solving soliton equations, it can yield fantastic results by using matrices properly. Successful exam- �� �� =��, �=( ), (7b) ples are the Wronskian technique and the IST. Ma et al. �−� had introduced the matrix element in Wronskian determi- nants when they used the Wronskian technique to solve �=(�,� )� � �=�(�,�) soliton equations [19–23]. Tey obtained various kinds of where 1 2 , is a spectral parameter, and , �=�(�,�)are potential functions. We assume that �(�, �) solutions such as soliton, rational, Matveev, and complexiton �(�, �) � � solutions. In 2006, Aktosun and Van Der Mee proposed and are smooth functions of variables and ,and their derivatives of any order with respect to � vanish rapidly a modifed inverse scattering transformation (MIST) [24]. �→∞ Tey expressed the scattering data of spectral problems by as . Te compatibility condition, zero curvature three matrices, �, �, �, and proved that the matrices �, �, equation � satisfy the Sylvester equation. Te advantage of the method � −� + [�, �] =0, is that it can get more kinds of solutions and its process is � � (8a) simpler than the traditional IST [24–27]. with boundary conditions Inthispaper,wewouldliketoconsidertheAKNS hierarchy 1 − (2��)� 0 2 �|(�,�)=(0,0) =( ), (8b) � −� 1 � � 0 (2��) ( ) =� ( ) (� = 0, 1, 2, . . .) , (5) � � 2 � can yield the isospectral AKNS hierarchy (5). Te frst two with the help of MIST, where nonlinear equations in the AKNS hierarchy are

2 � −1 0 � −��� +2� � −1 ( ) =( ), (9a) �=��+2( )� (�, �) , � = ( ). (6) � 2 −� 01 � ��� −2��

� ���� −6���� ( ) =( ). (9b) Many integrable systems can be reduced from it such as � � −6��� the modifed KdV, sine-Gordon, nonlinear Schrodinger,¨ and � ��� � nonlocal nonlinear Schrodinger¨ system. Tere have had already a lot of researches on AKNS hierarchy for it is a Next we mainly follow the notions and results given in representative integrable hierarchy [21, 22, 28]. In this paper, [1, 6]. we will construct the soliton, complexiton, and Matveev � Lemma 1. If the read potentials (�(�, �), �(�, �)) satisfy solutions to the frst nonlinear equation of (5). We will show that the linear relation (3) exists not only in one-soliton ∞ � � � � � solutions but also in multisoliton, complexiton, and Matveev ∫ �� � (�)� ��<+∞, solutions. −∞ ∞ � � Te paper is organized as follows: In Section 2, we will � � � (10) ∫ �� � (�)� ��<+∞, review the recovered potentials of the AKNS spectral problem −∞ byIST.InSection3,wewillobtainthecoupledSylvester equations and generalized matrix exponential solutions to the (� = 0, 1) , AKNS equation. In Section 4, some diferent types of explicit �(�, �) solutions will be constructed. In Section 5, generalized matrix the spectral problem (7a) has a group of Jost solutions , exponential solutions will be given to the AKNS hierarchy. �(�, �), �(�, �),and�(�, �) which are bounded for all values Te relationship between the recursion operator and the of � and also enjoy the following asymptotic behaviors: solutions of the AKNS hierarchy will be discussed also. We 0 conclude the paper in Section 6. � (�, �) ∼( )����, 1 2. Preparation 1 � (�, �) ∼( )�−���, To make the paper self-contained, we frst briefy recall the 0 Lax integrability of the isospectral AKNS hierarchy and its Gel’fand-Levitan-Marchenko (GLM) equations. (��→∞) , Advances in Mathematical Physics 3

1 Lemma 4. � (�, �) ∼( )�−���, Given the scattering data for the spectral problem 0 (7a) and

0 ��� � (�, �) ∼( )� , � (�) =�� (�) +�� (�) , −1 (18a) � (�) = � (�) + � (�) , (��→−∞) . � � (11) where In the usual manner, we defne the scattering coefcient by ∞ 1 ��� �� (�) = ∫ � (�) � ��, � (�, �) =�(�) � (�, �) +�(�) � (�, �) , (12a) 2� −∞ � (18b) � (�, �) =−� (�) � (�, �) + � (�) � (�, �) , (12b) 2 ���� �� (�) = ∑�� � , �=1 where 1 ∞ � (�) � (�) +�(�) � (�) =1. −��� (13) �� (�) = ∫ � (�) � ��, 2� −∞

Furthermore, one can give �(�, �), �(�, �) by the integral � (18c) 2 �� � representation � �� (�) =−∑��� , 0 ∞ �=1 ��� ��� � (�, �) =( )� + ∫ �(�,�)� ��, (14a) 1 � one has the GLM equations for the AKNS hierarchy 1 ∞ −��� −��� � (�, �) =( )� + ∫ �(�,�)� ��, (14b) 0 � 0 �(�,�)+( )�(�+�) � 1 where �(�, �) = (�1(�, �), �2(�, �)) and �(�, �) = (�1(�, � (19a) �), �2(�, �)) arecolumnvectors. ∞ In order for (14a) and (14b) to be valid, it is necessary that + ∫ � (�, �) �(�+�)��=0, � � (�) =−2� (�, �) , 1 1 (15) �(�,�)−( ) �(�+�) � (�) =−2�2 (�, �) . 0 (19b) � � �(�) �(�) ∞ Defnition 2. If � and � aresinglerootsof and , − ∫ � (�, �) �(�+�)��=0. respectively, there exist �� and �� such that � ∞ 2 2 ∫ �� �1 (�, ��)�2 (�, ��)=1, −∞ 3. Generalized Matrix Exponential ∞ (16) Solutions to the AKNS Equation 2 2 ∫ ���1 (�, ��) �2 (�, ��)=1. −∞ In this section, we will get generalized matrix exponential solutions to the AKNS equation (9a) via MIST. �� and �� are named the normalization constants for the From the previous section, we know that the potentials �(�, � ) �(�, � ) � �(�, � ) eigenfunctions � and � . Accordingly, � � of spectral problem (7a) can be recovered by (15). For � �(�, � ) and � � are named the normalization eigenfunctions. convenience, we denote �1(�, �) and �2(�, �) by �(�, �) and �(�, �), respectively, in this and the following sections. Defnition 3. Onenamedtheset Lemma 5. � (�) � (�) TeGLMequations(19a)and(19b)oftheAKNS {� ( �=0) ,�(�) = , � (�) = ; hierarchy have other forms: Im � (�) � (�)

�� (Im �� >0), �� (Im �� <0), ��, ��,�=1,2, (17) �(�,�)

...,�, �=1,2,...,�} = �(�+�) ∞ ∞ − ∫ ∫ �� ��� (�, �) � (�+�) �(�+�), to be the scattering data for the spectral problem (7a). � � 4 Advances in Mathematical Physics

�(�,�) Proof. We only prove the frst relation, and the second relationcanbeprovedsimilarly: =−�(�+�) ∞ ∞ ∞ ��−��=∫ (������−���−�������−��)�� − ∫ ∫ �� ��� (�, �) � (�+�) �(�+�). 0 � � (25) �∞ (20) =−��� ���−��� =��. �0 Proof. Rewrite (19a) and (19b) in their component forms as

∞ For convenience, we set �1 (�, �) + ∫ �1 (�, �) � (�+�) �� = 0, (21a) � Ω (�) =�−������, ∞ �2 (�, �) + � (� + �) + ∫ �2 (�, �) �(�+�)�� � (21b) Ω (�) =�����−��, =0, (26) �� ∞ Ξ (�) =� , �1 (�, �) − �1 (�+�)−∫ �1 (�, �) �(�+�)�� � (21c) Ξ (�) =�−��, =0, ∞ where �, � are �×�matrices, �, �, �, �, �, �, �, � are �2 (�, �) − ∫ �2 (�, �) �(�+�)��=0. (21d) � matrices defned in Lemma 6, and they enjoy the relations (24). Substituting (21a) and (21d) into (21c) and (21b) and replacing Setting � (�, �) � (�, �) �(�, �) �(�, �) 1 , 2 with , ,respectively,wehave Γ=�+Ω(�) Ξ (�) Ω (�) Ξ (�) , (27) �(�,�) Γ=�+Ω (�) Ξ (�) Ω (�) Ξ (�) , = �(�+�) if Γ and Γ are nondegenerate matrices, we have the following ∞ ∞ − ∫ ∫ �� ��� (�, �) � (�+�) �(�+�), theorem. � � (22) Teorem 7. If �� = �� and � �=� �,thepotentialsofthe �(�,�) AKNS spectral problem (7a) can be recovered as =−�(�+�) � (�, �) =−2��−��Ξ (�) Γ−1�−���, ∞ ∞ − ∫ ∫ �� ��� (�, �) � (�+�) �(�+�). (28) �� −1 �� � � � (�, �) =2�� Ξ (�) Γ � �.

Proof. Let ��(�) and ��(�) be zero and ��(�) and ��(�) be matrix exponential forms in (18a); that is, Lemma 6. Let � and � be �×�constant matrices, � and � be �×1constant column vectors, and � and � be 1×�constant � (�) =� (�) =�����, �� � row vectors. Assume that � and � satisfy lim�→∞� =0and (29) −�� −�� lim�→∞� =0,respectively,where0 is the �×�matrix with � (�) = �� (�) = �� �. all elements being zero. Ten upon taking Suppose that the time evolution of �(�) and that of �(�) are ∞ �=∫ ������−����, � (�, �) =����Ξ (�) �, 0 (23) (30) ∞ � (�, �) = ��−��Ξ (�) �. �=∫ �−���������, 0 We may take one has the following coupled Sylvester relations: �(�,�)=�(�, �) �−���, (31) ��−��=��, �(�,�)=� (�, �) ��� �, (24) � �−�� = ��. accordingly, where �(�, �), �(�, �) are 1×�row vectors. Advances in Mathematical Physics 5

Substituting (30), (31) into the frst equation of (20), we Proof. We only prove the frst equation of (9a), and the second get equationcanbeprovedsimilarly.

∞ ∞ On the one hand −�� 2 � (�, �) = �� Ξ (�) − ∫ ∫ �� ��� (�, �) −4� �−�� −1 2 2 �� =8�� Γ [Γ� +Ω(�) Ξ (�) � Ω (�) Ξ (�) � � (32) −�� �(�+�) −�� 2 ⋅� ��� Ξ (�) ��� Ξ (�) . −Ω(�) Ξ (�) Ω (�) Ξ (�) � ]Γ−1�−��� (38) Tat is, 2 2 =8��−4� �−��Γ−1 [� +Ω(�) Ξ (�) �2Ω (�) Ξ (�)] � (�, �) (� ⋅Γ−1�−���, ∞ ∞ −�� �� �� −�� (33) andontheotherhand + ∫ � ��� ��Ξ (�) ∫ � ��� ��Ξ (�)) 2 � � 2 −4� �−�� −1 2 2 −��� +2� �=2�� Γ [(Γ� + � Γ) = ��−��Ξ (�) . −1 +2(Γ� +Γ�) Γ (Γ� + �Γ) − Γ�� (39) In the light of 2 −�� 4�2�+�� −1 �� −4� �−�� −1 −�� ∞ +8� ��� Γ � ��� ]Γ � �. ∫ ������−���� = �����−��, � Trough proper simplifcation, we have ∞ (34) −�� �� −�� �� 2 2 ∫ � ��� �� = � �� , −� +2�2�=8��−4� �−��Γ−1 [� � �� (40) we have +Ω(�) Ξ (�) �2Ω (�) Ξ (�)]Γ−1�−���. � (�, �) = ��−��Ξ (�) Γ−1. (35a)

Similarly, we can arrive at 4. Exact Solutions to the AKNS Equation −1 � (�, �) =−����Ξ (�) Γ . (35b) In this section, we will construct diferent kinds of explicit solutions to the AKNS equation (9a) by taking diferent kinds Finally, we fnd that the potentials of the AKNS spectral of �, �, � and �, �, �. problem (7a) can be recovered as

(i) One-Soliton Solutions. Taking �=�1, �=�1, �=�1 and � (�) =−2�(�, �) =−2��−��Ξ (�) Γ−1�−���, �=�2, �=�2, �=�2,weget (36) �� −1 �� 2 −4�2�−2� � � (�) =−2� (�, �) =2�� Ξ (�) Γ � �, −2 (� −� ) � � � 2 2 �= 2 1 2 2 , 2 4(�2−�2)�+2(� −� )� (� −� ) +�� � � � 1 2 1 2 by taking advantage of (15). 2 1 1 2 1 2 (41) 2 4�2�+2� � 2(� −� ) � � � 1 1 NextwewillgetsolutionstotheAKNSequation(9a)from �= 2 1 1 1 , the recovered potentials (28). 2 4(�2−�2)�+2(� −� )� (�2 −�1) +�1�2�1�2� 1 2 1 2

Proposition 8. Te matrices Ω(�)Ξ(�) and Ω(�)Ξ(�) satisfy where ��,��,�� ∈ C (� = 1, 2) and Re �1 <0,Re�2 >0.

−1 Ω (�) Ξ (�) Γ =Γ−1Ω (�) Ξ (�) , (37a) (ii) Two-Soliton Solutions. Taking −2 0 −1 Γ Ω (�) Ξ (�) = Ω (�) Ξ (�) Γ−1, �=( ), (37b) 0−1

−1 −1 respectively. Tus, the two matrices Γ and Γ are similar. 1 �=( ), 2 Although Proposition 8 is simple, we will use it many (42) timesinthefollowingpartofthispaper. 2 2 2 �=( ), Teorem 9. If �=(2�) and �=(2�) in Ξ(�) and Ξ(�), 1 respectively, the recovered the potentials (28) are solutions to the AKNS equation (9a). �=�=(11), 6 Advances in Mathematical Physics we have

16�2(−8�+�) (9�12� + 50�2� + 1800�12�+8� + 3600�32�+10�) �=− , 1 + 1800�8� + 576�6(2�+�) + 6400�10(2�+�) + 1800�8(4�+�) (1 + 8�8�) (43) 16�4(4�+�) [9 + 50�20�+2� (1 + 72�8� +36�6(2�+�))] �= . 1 + 1800�8� + 576�6(2�+�) + 6400�10(2�+�) + 1800�8(4�+�) (1 + 8�8�)

(iii) Tree-Soliton Solutions. Taking 1 �=(−1), −1 0 0 2 �=(0−10), 00−2 �=(121),

200 �=(211), �=(010), (44) 002

1 �=(−1), 1 we have

−72�−4(4�+�) (�12� −4�2� − 144�6� + 36�12�+8�) �= , 720 cosh 12� − 648 cosh 2� − 1297 cosh 6� − 432 sinh 12� − 1295 sinh 6� (45) 72�4�−4� (1 + 36�8� −�2(6�+�) − 36�12�+6�) �= . 720 cosh 12� − 648 cosh 2� − 1297 cosh 6� − 432 sinh 12� − 1295 sinh 6�

� (iv) Matveev Solutions. Taking 2�4(4�+�) [4 + 64� + 8� + �2(6�+�) (3 + 16� + 4�)] 11 = . �=( ), 4+�4(6�+�) +�2(6�+�) [3 − 256�2 −96�(1+�) −2�(5+4�)] 01 (47) 21 �=( ), 02 (v) Complexiton Solutions. Taking 11 1 (46) �=( ), �=( ), 2 −1 1 22 2 �=( ), �=( ), −2 2 1 1 �=�=(11), �=( ), (48) 3 we have 3 � �=( ), 1 �−2(2�+�) [22 + 32� + 8� + �2(6�+�) (−6+32�+4�)] = , 4+�4(6�+�) +�2(6�+�) [3 − 256�2 −96�(1+�) −2�(5+4�)] �=�=(11), Advances in Mathematical Physics 7 we have

32 (8� + 2�) −16 (8� + 2�) −4�2� [2 (32� + 4�) + (32� + 4�)] �= cos sin cos sin , 4�2� +�6� −�4� [5 sin (24� + 2�) + sin (40� + 6�)] (49) 4�4� [2 (32� + 4�) − (32� + 4�)] +4�6� [2 (8� + 2�) + (8� + 2�)] �= cos sin cos sin . 4+�4� −�2� [5 sin (24� + 2�) + sin (40� + 6�)]

5. The Same Linear Relation That a General We consider a new general equation: Equation and Its Solutions Share � −� ( ) =�(�) ( ), (50) In this section, we will construct a new general equation � � � related to the AKNS hierarchy and get its generalized matrix exponential solutions. Furthermore, we fnd that the recur- where sion operator and the solutions of the AKNS hierarchy have � �−1 thesamelinearrelation. � (�) =�0� +�1� +⋅⋅⋅+��−1�+�� (51) is a polynomial operator on �. Let

� � 2�+1��−��Ξ (�) Γ−1 [� +Ω(�) Ξ (�) ��Ω (�) Ξ (�)]Γ−1�−��� ( �)=( ), �+1 �� −1 � � −1 �� �� 2 �� Ξ (�) Γ [� + Ω (�) Ξ (�) � Ω (�) Ξ (�)] Γ � �

−1 �−1 (52) � 2�����Ξ (�) Γ ��� (��−1�−�� ) Ξ (�) �−��Γ−1�−��� ( �)=( ); � −�� −1 −�� �−1 �−1 �� −1 �� V� 2 �� Ξ (�) Γ � (�� − � �) Ξ (�) � Γ � �

�−1 −1 we have the following Lemma. ⋅� Ω (�) Ξ (�) Γ Ω (�) Ξ (�) �Ω (�) Ξ (�) −Ω(�) ⋅Ξ(�) ��−1Ω (�) Ξ (�) Γ−1�+Ω(�) Ξ (�) ��Ω (�) Lemma 10. If �� = �� and � �=� �,oneobtains ⋅ Ξ (�)]Γ−1�−���, �� +���−1 =2�(�� + V�), (53) (54) �� −���−1 =−2�(�� + V�). we have Proof. We only prove the frst equation. For �+1 −(2�)��−�� −1 �� +���−1 =2 �� Γ {[�Ω (�) Ξ (�) � −�� �−1 �� =�{2 �� [� Ξ (�) −1 �−1 �−1 −Ω(�) Ξ (�) �] Γ Ω (�) Ξ (�) � −1 �−1 + Ξ (�) Γ Ω (�) Ξ (�) � Ω (�) Ξ (�) �−1 +[� Ω (�) Ξ (�) −Ω(�) Ξ (�) ��−1] �−1 − Ξ (�) Γ−1Ω (�) Ξ (�) Ω (�) Ξ (�) � ]Γ−1�−���} −1 ⋅ Γ [Ω (�) Ξ (�) �−�Ω (�) Ξ (�)] �+1 −�� −1 � −1 �−1 =2 �� Ξ (�) Γ [� − �Γ � −1 −[�Ω (�) Ξ (�) −Ω�Ξ(�) �] Γ ��−1Ω (�) Ξ (�)} �−1 �−1 − � Γ−1�−Ω(�) Ξ (�) �Ω (�) Ξ (�) Γ−1� ⋅Γ−1�−���=2�+1��−��Ξ (�) Γ−1�−�� [(� � �−1 − � Γ−1Ω (�) Ξ (�) �Ω (�) Ξ (�) −Ω(�) Ξ (�) −1 �−1 − ��) ���Ξ (�) Γ ��� (�� −��−1�) Ξ (�) ⋅�Ω (�) Ξ (�) Γ−1Ω (�) Ξ (�) ��−1Ω (�) Ξ (�) �−1 �−1 �� −1 �� − �Γ−1Ω (�) Ξ (�) ��−1Ω (�) Ξ (�) −Ω(�) Ξ (�) +(� �−�� )� Ξ (�) Γ � (�� 8 Advances in Mathematical Physics

� −��)Ξ (�)]�−��Γ−1�−���=2�+1��−��Ξ (�) Proof. When �(2�) = (2�) ,wehave

−1 �−1 �� � � ⋅Γ−1�−�� [�����Ξ (�) Γ ��� (�� −��−1�) =2�+1��−(2�) �−��� Γ−1�−��� ��

�−1 � ⋅ Ξ � +(� �−���−1)Ξ � −(2�) �−�� −1 −1 −�� ( ) ( ) +2�� Γ Γ�Γ � �,

−1 � ⋅���Γ ����� Ξ (�)]�−��Γ−1�−���. =2�+1��−��Ξ (�) Γ−1 [Γ�

(55) +Ω(�) Ξ (�) ��Ω (�) Ξ (�) (63)

� From (28), we know that the conclusion is right. −Ω(�) Ξ (�) Ω (�) Ξ (�) � ]Γ−1�−���,

Teorem 11. �� = �� � �=� � � If and ,onehas =2�+1��−��Ξ (�) Γ−1 [� � (�, �) −� (�, �) � � � −1 −�� ( )=� ( ). (56) +Ω(�) Ξ (�) � Ω (�) Ξ (�)]Γ � �. �� (�, �) � (�, �)

Tat is, ��/�� = �� =��.Inthesameway,wealsocanobtain � Proof. We use the mathematical induction to prove the �� =�� when �(2�) = (2�) .Tuswehave theorem. When �=1,wehave � � � −� ( ) =� ( �)+� ( �−1)+⋅⋅⋅+� ( ) � 0 � 1 � � � � � � � �−1 ( 1)=( �) � � (64) 1 � −� (57) =�(�) ( ). 4��−��Ξ (�) Γ−1 [�+Ω(�) Ξ (�) �Ω (�) Ξ (�)]Γ−1�−��� � =( ), −1 −1 4����Ξ (�) Γ [� + Ω (�) Ξ (�) �Ω (�) Ξ (�)] Γ ���� and (56) is established. Te corollary means that the operator and the solutions Supposing that of the AKNS hierarchy enjoy the same linear structure. � −� (�, �) �−1 �−1 ( )=� ( ), (58) 6. Conclusions ��−1 � (�, �) To sum up, we have solved the AKNS hierarchy by the we obtain MIST. Te MIST can determine solutions more directly and generate more diverse solutions than the traditional IST. �� ��−1 Actually, such solutions can be also obtained by applying ( )=�( ); (59) � � theWronskiantechnique[19–23].Tearbitrarinessofthe � �−1 matrices involved leads to the diversity of exact solutions. that is, A general integrable equation of the AKNS hierarchy was constructed and its matrix exponential solutions were −1 obtained. Te ways to generate the equation and its matrix �� −���−1 +2�� (���−1 +���−1) ( )=( ). (60) exponential solutions are the same. Tey share the same � −1 � ���−1 −2�� (���−1 +���−1) linear algebraic structure, not only in the case of one-soliton solutions but also in the case of other interesting solutions. Trough tedious calculation, we arrive at

�(� + V ) Conflicts of Interest � � =�� +�� . (61) �� �−1 �−1 Te authors declare that there are no conficts of interest regarding the publication of this paper. By Lemma 10, we know that the frst equality in (60) is right. Its second equality can be proved analogously. Acknowledgments Corollary 12. If �=�(2�)and �=�(2�) in Ξ(�) and Ξ(�), Te work was supported in part by NSFC under the Grants respectively, the recovered potentials (28) solve (50), where 11671177, 11771186, 11571079, 11371086, and 1371361; NSF under the Grant DMS-1664561; the Jiangsu Qing Lan Project (2014); � �−1 � (�) =�0� +�1� +⋅⋅⋅+��−1�+��� (62) SixTalentPeaksProjectofJiangsuProvince(2016-JY-08);and the Distinguished Professorships by Shanghai University of is a polynomial matrix on � and � is an unit matrix. Electric Power and Shanghai Second Polytechnic University. Advances in Mathematical Physics 9

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