Darboux Transformations in Integrable Systems: Theory and their Applications to Geometry (Mathematical Physics Studies, 26)

DARBOUX TRANSFORMATIONS IN INTEGRABLE SYSTEMS MATHEMATICAL PHYSICS STUDIES Editorial Board: Maxim Kontsevich, IHES, ...

Author: Chaohao Gu | Hesheng Hu | Zixiang Zhou

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DARBOUX TRANSFORMATIONS IN INTEGRABLE SYSTEMS

MATHEMATICAL PHYSICS STUDIES Editorial Board:

Maxim Kontsevich, IHES, Bures-sur-Yvette, France Massimo Porrati, New York University, New York, U.S.A. Vladimir Matveev, Université Bourgogne, Dijon, France Daniel Sternheimer, Université Bourgogne, Dijon, France

VOLUME 26

Darboux Transformations in Integrable Systems Theory and their Applications to Geometry

by

Chaohao Gu Fudan University, Shanghai, China

Hesheng Hu Fudan University, Shanghai, China and

Zixiang Zhou Fudan University, Shanghai, China

A C.I.P. Catalogue record for this book is available from the Library of Congress.

ISBN 1-4020-3087-8 (HB) ISBN 1-4020-3088-6 (e-book)

Published by Springer, P.O. Box 17, 3300 AA Dordrecht, The Netherlands. Sold and distributed in North, Central and South America by Springer, 101 Philip Drive, Norwell, MA 02061, U.S.A. In all other countries, sold and distributed by Springer, P.O. Box 322, 3300 AH Dordrecht, The Netherlands.

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All Rights Reserved © 2005 Springer No part of the material protected by this copyright notice may be reproduced or utilized in any form or by any means, electronic or mechanical, including photocopying, recording or by any information storage and retrieval system, without written permission from the copyright owner. Printed in the Netherlands.

Contents

Preface

ix

1. 1+1 DIMENSIONAL INTEGRABLE SYSTEMS 1.1

1

KdV equation, MKdV equation and their Darboux transformations 1.1.1 Original Darboux transformation 1.1.2 Darboux transformation for KdV equation 1.1.3 Darboux transformation for MKdV equation 1.1.4 Examples: single and double soliton solutions 1.1.5 Relation between Darboux transformations for KdV equation and MKdV equation

10

1.2

AKNS system 1.2.1 2 × 2 AKNS system 1.2.2 N × N AKNS system

11 11 16

1.3

Darboux transformation 1.3.1 Darboux transformation for AKNS system 1.3.2 Invariance of equations under Darboux transformations 1.3.3 Darboux transformations of higher degree and the theorem of permutability 1.3.4 More results on the Darboux matrices of degree one

18 18

30

KdV hierarchy, MKdV-SG hierarchy, NLS hierarchy and AKNS system with u(N ) reduction 1.4.1 KdV hierarchy 1.4.2 MKdV-SG hierarchy 1.4.3 NLS hierarchy 1.4.4 AKNS system with u(N ) reduction

34 35 40 46 48

1.4

1 1 2 3 6

23 25

vi


1.5

Darboux transformation and scattering, inverse scattering theory 1.5.1 Outline of the scattering and inverse scattering theory for the 2 × 2 AKNS system 1.5.2 Change of scattering data under Darboux transformations for su(2) AKNS system

2. 2+1 DIMENSIONAL INTEGRABLE SYSTEMS

51 51 58 65

2.1

KP equation and its Darboux transformation

65

2.2

2+1 dimensional AKNS system and DS equation

68

2.3

Darboux transformation 2.3.1 General Lax pair 2.3.2 Darboux transformation of degree one 2.3.3 Darboux transformation of higher degree and the theorem of permutability

70 70 71

2.4

Darboux transformation and binary Darboux transformation for DS equation 2.4.1 Darboux transformation for DSII equation 2.4.2 Darboux transformation and binary Darboux transformation for DSI equation

75 78 78 82

2.5

Application to 1+1 dimensional Gelfand-Dickey system

84

2.6

Nonlinear constraints and Darboux transformation in 2+1 dimensions

87

3. N + 1 DIMENSIONAL INTEGRABLE SYSTEMS

103

3.1

n + 1 dimensional AKNS system 3.1.1 n + 1 dimensional AKNS system 3.1.2 Examples

103 103 106

3.2

Darboux transformation and soliton solutions 3.2.1 Darboux transformation 3.2.2 u(N ) case 3.2.3 Soliton solutions

108 108 110 111

3.3

A reduced system on Rn

116

4. SURFACES OF CONSTANT CURVATURE, ¨ BACKLUND CONGRUENCES

121 3

4.1

Theory of surfaces in the Euclidean space R

4.2

Surfaces of constant negative Gauss curvature, sine-Gordon equation and B¨ acklund transformations

122 126

vii

Contents

4.2.1

Relation between sine-Gordon equation and surface of constant negative Gauss curvature in R3 4.2.2 Pseudo-spherical congruence 4.2.3 B¨ äcklund transformation 4.2.4 Darboux transformation 4.2.5 Example 4.3 Surface of constant Gauss curvature in the Minkowski space R2,1 and pseudo-spherical congruence 4.3.1 Theory of surfaces in the Minkowski space R2,1 4.3.2 Chebyshev coordinates for surfaces of constant Gauss curvature 4.3.3 Pseudo-spherical congruence in R2,1 4.3.4 B¨ äcklund transformation and Darboux transformation for surfaces of constant Gauss curvature in R2,1 4.4 Orthogonal frame and Lax pair 4.5 Surface of constant mean curvature 4.5.1 Parallel surface in Euclidean space 4.5.2 Construction of surfaces 4.5.3 The case in Minkowski space 5. DARBOUX TRANSFORMATION AND HARMONIC MAP 5.1 Definition of harmonic map and basic equations 5.2 Harmonic maps from R2 or R1,1 to S 2 , H 2 or S 1,1 5.3 Harmonic maps from R1,1 to U (N ) 5.3.1 Riemannian metric on U (N ) 5.3.2 Harmonic maps from R1,1 to U (N ) 5.3.3 Single soliton solutions 5.3.4 Multi-soliton solutions 5.4 Harmonic maps from R2 to U (N ) 5.4.1 Harmonic maps from R2 to U (N ) and their Darboux transformations 5.4.2 Soliton solutions 5.4.3 Uniton 5.4.4 Darboux transformation and singular Darboux transformation for unitons 6. GENERALIZED SELF-DUAL YANG-MILLS AND YANG-MILLS-HIGGS EQUATIONS 6.1 Generalized self-dual Yang-Mills flow

126 129 132 135 139 141 141 144 149

154 174 179 179 181 184 189 189 192 199 199 201 207 210 213 213 219 220 225 237 237

viii


6.1.1 Generalized self-dual Yang-Mills flow 6.1.2 Darboux transformation 6.1.3 Example 6.1.4 Relation with AKNS system 6.2 Yang-Mills-Higgs field in 2+1 dimensional Minkowski space-time 6.2.1 Yang-Mills-Higgs field 6.2.2 Darboux Transformations 6.2.3 Soliton solutions 6.3 Yang-Mills-Higgs field in 2+1 dimensional anti-de Sitter space-time 6.3.1 Equations and their Lax pair 6.3.2 Darboux transformations 6.3.3 Soliton solutions in SU (2) case 6.3.4 Comparison with the solutions in Minkowski space-time 7. TWO DIMENSIONAL TODA EQUATIONS AND LAPLACE SEQUENCES OF SURFACES 7.1 Signed Toda equations 7.2 Laplace sequences of surfaces in projective space Pn−1 7.3 Darboux transformation 7.4 Su chain (Finikoff configuration) 7.5 Elliptic version of Laplace sequence of surfaces in CPn 7.5.1 Laplace sequence in CPn 7.5.2 Equations of harmonic maps from R2 to CPn in homogeneous coordinates 7.5.3 Cases of indefinite metric 7.5.4 Harmonic maps from R1,1 7.5.5 Examples of harmonic sequences from R2 to CPn or R1,1 to CPn References

237 242 245 247 248 248 249 251 256 256 257 260 263 267 267 271 277 281 291 291 292 295 296 296 299

Preface

GU Chaohao The soliton theory is an important branch of nonlinear science. On one hand, it describes various kinds of stable motions appearing in nature, such as solitary water wave, solitary signals in optical fibre etc., and has many applications in science and technology (like optical signal communication). On the other hand, it gives many effective methods of getting explicit solutions of nonlinear partial differential equations. Therefore, it has attracted much attention from physicists as well as mathematicians. Nonlinear partial differential equations appear in many scientific problems. Getting explicit solutions is usually a difficult task. Only in certain special cases can the solutions be written down explicitly. However, for many soliton equations, people have found quite a few methods to get explicit solutions. The most famous ones are the inverse scattering method, Backlund ¨ transformation etc.. The inverse scattering method is based on the spectral theory of ordinary differential equations. The Cauchy problem of many soliton equations can be transformed to solving a system of linear integral equations. Explicit solutions can be derived when the kernel of the integral equation is degenerate. The B¨ a¨cklund transformation gives a new solution from a known solution by solving a system of completely integrable partial differential equations. Some complicated “nonlinear superposition formula” arise to substitute the superposition principle in linear science. However, if the kernel of the integral equation is not degenerate, it is very difficult to get the explicit expressions of the solutions via the inverse scattering method. For the B¨ a¨cklund transformation, the nonlinear superposition formula is not easy to be obtained in general. In

x


late 1970s, it was discovered by V. B. Matveev that a method given by G. Darboux a century ago for the spectral problem of second order ordinary differential equations can be extended to some important soliton equations. This method was called Darboux transformation. After that, it was found that this method is very effective for many partial differential equations. It is now playing an important role in mechanics, physics and differential geometry. V. B. Matveev and M. A. Salle published an important monograph [80] on this topic in 1991. Besides, an interesting monograph of C. Rogers and W. K. Schief [90] with many recent results was published in 2002. The present monograph contains the Darboux transformations in matrix form and provides purely algebraic algorithms for constructing explicit solutions. Consequently, a basis of using symbolic calculations to obtain explicit exact solutions for many integrable systems is established. Moreover, the behavior of simple and multi-solutions, even in multi-dimensional cases, can be elucidated clearly. The method covers a series of important topics such as varies kinds of AKNS systems in acklund congruences and surfaces of constant Rn+1 , the construction of B¨ Gauss curvature in R3 and R2,1 , harmonic maps from two dimensional manifolds to the Lie group U (n), self-dual Yang-Mills fields and the generalizations to higher dimensional case, Yang-Mills-Higgs fields in 2 + 1 dimensional Minkowski and anti-de Sitter space, Laplace sequences of surfaces in projective spaces and two dimensional Toda equations. All these cases are stated in details. In geometric problems, the Lax pair is not only a tool, but also a geometric object to be studied. Many results in this monograph are obtained by the authors in recent years. This monograph is partially supported by the Chinese Major State Basic Research Program “Frontier problems in nonlinear sciences”, the Doctoral Program Foundation of the Ministry of Education of China, National Natural Science Foundation of China and Science Foundation of Shanghai Science Committee. Most work in this monograph was done in the Institute of Mathematics of Fudan University.

Chapter 1 1+1 DIMENSIONAL INTEGRABLE SYSTEMS

Starting from the original Darboux transformation, we first discuss the classical form of the Darboux transformations for the KdV and the MKdV equation, then discuss the Darboux transformations for the AKNS system and more general systems. The coefficients in the evolution equations discussed here may depend on t. The Darboux matrices are constructed algebraically and the algorithm is purely algebraic and universal to whole hierarchies. The Darboux transformations for reduced systems are also concerned. We also present the relations between Darboux transformation and the inverse scattering theory, and show that the number of solitons (the number of eigenvalues) increases or decreases after the action of a Darboux transformation.

1.1

KdV equation, MKdV equation and their Darboux transformations 1.1.1 Original Darboux transformation In 1882, G. Darboux [18] studied the eigenvalue problem of a linear partial differential equation of second order (now called the onedimensional Schr¨ odinger equation) −φxx − u(x)φ = λφ.

(1.1)

Here u(x) is a given function, called potential function; λ is a constant, called spectral parameter. He found out the following fact. If u(x) and φ(x, λ) are two functions satisfying (1.1) and f (x) = φ(x, λ0 ) is a solution of the equation (1.1) for λ = λ0 where λ0 is a fixed constant,

2


then the functions u and φ defined by u = u + 2(ln f )xx , satisfy

φ (x, λ) = φx (x, λ) −

fx φ(x, λ) f

−φxx − u φ = λφ ,

(1.2)

(1.3)

which is of the same form as (1.1). Therefore, the transformation (1.2) transforms the functions (u, φ) to (u , φ ) which satisfy the same equations. This transformation (u, φ) −→ (u , φ ),

(1.4)

is the original Darboux transformation, which is valid for f = 0.

1.1.2

Darboux transformation for KdV equation

In 1885, the Netherlandish applied mathematicians Korteweg and de Vries introduced a nonlinear partial differential equation describing the motion of water wave, which is now called the Korteweg-de Vries equation (KdV equation) ut + 6uux + uxxx = 0.

(1.5)

In the middle of 1960’s, this equation was found out to be closely related to the Schrodinger ¨ equation mentioned above [87]. KdV equation (1.5) is the integrability condition of the linear system − φxx − uφ = λφ, φt = −4φxxx − 6uφx − 3ux φ

(1.6)

which is called the Lax pair of the KdV equation. Here u and φ are functions of x and t. (1.6) is the integrability condition of (1.5). In other words, (1.5) is the necessary and sufficient condition for (φxx )t = (φt )xx being an identity for all λ, where (φxx )t is computed from φxx = (−λ − u)φ (the first equation of (1.6)) and (φt )xx is given by the second equation of (1.6). Since the first equation of the Lax pair of the KdV equation is just the Schrodinger ¨ equation, the Darboux transformation (1.2) can also be applied to the KdV equation, where the functions depend on t. Obviously the transformation keeps the first equation of (1.6) invariant, i.e., (u , φ ) satisfies (1.7) −φxx − u φ = λφ . Moreover, it is easily seen that (u , φ ) satisfies the second equation of (1.6) as well. Therefore, u satisfies the KdV equation, which is the

3

1+1 dimensional integrable systems

integrability condition of (1.6). In summary, suppose one knows a solution u of the KdV equation, solving the linear equations (1.6) one gets φ(x, t, λ). Take λ to be a special value λ0 and let f (x, t) = φ(x, t, λ0 ), then u = u + 2(ln f )xx gives a new solution of the KdV equation, and φ given by (1.2) is a solution of the Lax pair corresponding to u . This gives a way to obtain new solutions of the KdV equation. This process can be done successively as follows. For a known solution u of (1.5), first solve a system of linear differential equations (1.6) and get φ. Then explicit calculation from (1.2) gives new special solutions of the KdV equation. Since φ is known, it is not necessary to solve any linear differential equations again to obtain (u , φ ). That is, we only need algebraic calculation to get (u , φ ) etc.: (u, φ) −→ (u , φ ) −→ (u , φ ) −→ · · · .

(1.8)

Therefore, we have extended the Darboux transformation for the Schr¨ odinger equation to the KdV equation. The basic idea here is to get the new solutions of the nonlinear equation and the corresponding solutions of the Lax pair simultaneously from a known solution of the nonlinear equation and a solution of its Lax pair by using algebraic and differential computation. Note that the formula is valid only for f = 0. If f = 0, the Darboux transformation will have singularities. Remark 1 Let ψ1 = φ, ψ2 = φx , Ψ = (ψ1 , ψ2 )T , then the Lax pair (1.6) can be written in matrix form as ⎛

Ψx = ⎝ ⎛

Ψt = ⎝

⎞

0

1

−λ − u 0

⎠ Ψ,

ux

4λ − 2u

−4λ2 − 2λu + uxx + 2u2

−ux

⎞

(1.9)

⎠ Ψ.

The transformation φ → φ in (1.2) can also be rewritten as a transformation of Ψ, which can be realized via algebraic algorithm only. We shall discuss this Darboux transformation in matrix form later.

1.1.3

Darboux transformation for MKdV equation

The method of Darboux transformation can be applied to many other equations such as the MKdV equation, the sine-Gordon equation etc. [105]. We first take the MKdV equation as an example. General cases will be considered in the latter sections. MKdV equation (1.10) ut + 6u2 ux + uxxx = 0

4


is the integrability condition of the over-determined linear system [2, 119] ⎛

Φx = U Φ = ⎝

⎞

λ

u

−u −λ

⎠ Φ,

Φt = V Φ ⎛

=⎝

−4λ3 − 2u2 λ

−4uλ2 − 2ux λ − 2u3 − uxx

4uλ2 − 2ux λ + 2u3 + uxx

4λ3 + 2u2 λ

⎞ ⎠ Φ,

(1.11) that is, (1.10) is the necessary and sufficient condition for Φxt = Φtx being an identity. The system (1.11) is called a Lax pair of (1.10) and λ a spectral parameter. Here Φ may be regarded as a column solution or a 2 × 2 matrix solution of (1.11). There are several ways to derive the Darboux transformation for the MKdV equation. Here we use the Darboux matrix method. For a given solution u of the MKdV equation, suppose we know a fundamental solution of (1.11) ⎛

Φ(x, t, λ) = ⎝

⎞

Φ11 (x, t, λ) Φ12 (x, t, λ)

⎠

(1.12)

Φ21 (x, t, λ) Φ22 (x, t, λ)

which composes two linearly independent column solutions of (1.11). Let λ1 , µ1 be arbitrary real numbers and σ=

Φ22 (x, t, λ1 ) + µ1 Φ21 (x, t, λ1 ) Φ12 (x, t, λ1 ) + µ1 Φ11 (x, t, λ1 )

(1.13)

be the ratio of the two entries of a column solution of the Lax pair (1.11). Construct the matrix ⎛

⎞

2σ λ1 ⎝ 1 − σ 2 ⎠ D(x, t, λ) = λI − 2 2 1+σ 2σ σ −1

(1.14)

and let Φ (x, t, λ) = D(x, t, λ)Φ(x, t, λ). Then it is easily verified that Φ (x, t, λ) satisfies Φx = U Φ ,

Φt = V Φ ,

(1.15)

5


where ⎛

U = ⎝ ⎛

V =⎝

λ

u

−u −λ

⎞ ⎠,

−4λ3 − 2u2 λ

−4u λ2 − 2ux λ − 2u3 − uxx

4u λ2 − 2ux λ + 2u3 + uxx

⎞ ⎠

4λ3 + 2u2 λ (1.16)

with u = u +

4λ1 σ . 1 + σ2

(1.17)

(1.15) and (1.16) are similar to (1.11). The only difference is that the u in (1.11) is replaced by u defined by (1.17). For any solution Φ of (1.11), DΦ is a solution of (1.15), hence (1.15) is solvable for any given initial data (the value of Φ at some point (x0 , t0 )). In other words, (1.15) is integrable. The integrability condition of (1.15) implies that u is also a solution of the MKdV equation. Using this method, we obtain a new solution of the MKdV equation together with the corresponding fundamental solution of its Lax pair from a known one. The above conclusions can be summarized as follows. Let u be a solution of the MKdV equation and Φ be a fundamental solution of its Lax pair. Take λ1 , µ1 to be two arbitrary real constants, and let σ be defined by (1.13), then (1.17) gives a new solution u of the MKdV equation, and the corresponding solution to the Lax pair can be taken as DΦ. The transformation (u, Φ) → (u , Φ ) is the Darboux transformation for the MKdV equation. This Darboux transformation in matrix form can be done successively and purely algebraically as (u, Φ) −→ (u , Φ ) −→ (u , Φ ) −→ · · · .

(1.18)

Remark 2 Both (1.9) and (1.11) are of the form Φx = U Φ,

Φt = V Φ

(1.19)

where U and V are N × N matrices and independent of Φ. The integrability condition of (1.19) is Ut − Vx + [U, V ] = 0

([U, V ] = U V − V U ).

(1.20)

According to the elementary theory of linear partial differential equations, the solution of (1.19) exists uniquely for given initial data Φ(x0 , t0 ) = Φ0 if and only if (1.20) holds identically. Here Φ is an N × N matrix, or a

6


vector with N entries. In this case Φ(x, t) is determined by the ordinary differential equation dΦ = (U dx + V dy)Φ (1.21) along an arbitrary path connecting (x0 , t0 ) and (x, t). (1.20) is also called a zero-curvature condition.

1.1.4

Examples: single and double soliton solutions

Starting with the trivial solution u = 0 of the MKdV equation, one can use the Darboux transformation to obtain the soliton solutions. For u = 0, the fundamental solution of the Lax pair can be obtained as ⎛

Φ(x, t, λ) = ⎝

⎞

exp(λx − 4λ3 t)

0

0

exp(−λx + 4λ3 t)

⎠

(1.22)

by integrating (1.11). Take λ1 = 0 and µ1 = exp(2α1 ) > 0, then (1.13) gives (1.23) σ = σ1 = exp(−2λ1 x + 8λ31 t − 2α1 ), hence

⎛

⎞

1 λ1 ⎝ sinh v1 ⎠, D = λI − cosh v1 1 − sinh v1

(1.24)

v1 = 2λ1 x − 8λ31 t + 2α1 .

(1.25)

where (1.17) gives the single soliton solution u = 2λ1 sech(2λ1 x − 8λ31 t + 2α1 ),

(1.26)

of the MKdV equation. (1.26) is called the single soliton solution because it has the following properties: (i) It is a travelling wave solution, i.e., it is in the form u = f (x − ct); (ii) For any t, limx→±∞ u = 0. Speaking intuitively, u is near 0 outside a small region, i.e., |u| < 2|λ1 | sech K when |2λ1 x − 8λ31 t + 2α1 | > K. The solution of the corresponding Lax pair is Φ (x, t, λ) = (Φij (x, t, λ)) ⎛

= D(x, t, λ) ⎝

exp(λx − 4λ3 t) 0

⎞

0 exp(−λx + 4λ3 t)

⎠ (1.27)

7


where D is given by (1.24). If we take u as a seed solution, a new Darboux matrix can be constructed from Φ and a series of new solutions of the MKdV equation can be obtained. We write down the second Darboux transformation explicitly. Suppose u is a solution of the MKdV equation (1.10), Φ is a fundamental solution of the corresponding Lax pair (1.11). Construct the Darboux matrix D = (Dij ) according to (1.13) and (1.14) and let σ = σ1 . Moreover, take constants λ2 = 0 (λ2 = λ1 ) and µ2 = exp(2α2 ). According to (1.13), Φ (x, t, λ2 ) + µ2 Φ21 (x, t, λ2 ) . (1.28) σ2 = 22 Φ12 (x, t, λ2 ) + µ2 Φ11 (x, t, λ2 ) Substituting Φ = DΦ into it, we have σ2 =

D21 (Φ12 + µ2 Φ11 ) + D22 (Φ22 + µ2 Φ21 ) D11 (Φ12 + µ2 Φ11 ) + D12 (Φ22 + µ2 Φ21 ) λ=λ2

(1.29)

D21 (λ2 ) + D22 (λ2 )σ2 = , D11 (λ2 ) + D12 (λ2 )σ2 where σ2 =

Φ22 (x, t, λ2 ) + µ2 Φ21 (x, t, λ2 ) . Φ12 (x, t, λ2 ) + µ2 Φ11 (x, t, λ2 )

(1.30)

Starting from u = 0, (1.26) and (1.27) are the single soliton solution and the corresponding fundamental solution of the Lax pair. Substituting (1.24), the expression of D, into (1.27), we have ⎛

Φ (x, t, λ) = ⎝

3

(λ − λ1 tanh v1 )eλx−4λ t − λ1 sech v1 e−λx+4λ 3

−λ1 sech v1 eλx−4λ t (λ + λ1 tanh v1 )e−λx+4λ

3t

3t

⎞ ⎠,

(1.31) hence σ2 =

−λ1 sech v1 + (λ2 + λ1 tanh v1 ) exp(−v2 ) , (λ2 − λ1 tanh v1 ) − λ1 sech v1 exp(−v2 )

(1.32)

v2 = 2λ2 x − 8λ32 t + 2α2 ,

(1.33)

where (i = 1, 2).

According to (1.17), 4λ2 σ2 u = 4λ1 σ12 + 1 + σ1 1 + σ22 2(λ22 − λ21 )(λ2 cosh v1 − λ1 cosh v2 ) = 2 2 (λ1 + λ2 ) cosh v1 cosh v2 − 2λ1 λ2 (1 + sinh v1 sinh v2 )

(1.34)

8


is a new solution of the MKdV equation. This is called the double soliton solution of the MKdV equation. This name follows from the following asymptotic property of the solutions. We shall show that a double soliton solution is asymptotic to two single soliton solutions as t → ∞. Suppose λ2 > λ1 > 0, M is a fixed positive number. Let v1 be bounded by |v1 | ≤ M , then x ∼ ∞ as t ∼ ∞. Since v2 =

λ2 2λ2 α1 v1 − 8λ2 (λ22 − λ21 )t + 2α2 − , λ1 λ1

(1.35)

v2 ∼ +∞ as t ∼ −∞, and u ∼ −2λ1 sech(v1 − v0 ) as t → −∞ where v0 = tanh−1

2λ1 λ2 . λ21 + λ22

(1.36)

(1.37)

If t ∼ +∞, then v2 ∼ −∞, and u ∼ −2λ1 sech(v1 + v0 ).

(1.38)

Hence, for fixed v1 (i.e., the observer moves in the velocity 4λ21 ), the solution is asymptotic to one single soliton solution (corresponding to the parameter λ1 ) as t ∼ −∞ or t ∼ +∞. However, there is a phase shift between the asymptotic solitons as t ∼ −∞ and t ∼ +∞. That is, the center of the soliton (the peak) moves from v1 = v0 to v1 = −v0 . Similarly, if |v2 | ≤ M , then v1 =

λ1 2λ1 α2 v2 + 8λ1 (λ22 − λ21 )t + 2α1 − λ2 λ2

(1.39)

implies that v1 ∼ ±∞ as t ∼ ±∞, and u ∼ 2λ2 sech(v2 + v0 ),

t ∼ −∞,

u ∼ 2λ2 sech(v2 − v0 ),

t ∼ +∞.

(1.40)

Finally, if t ∼ ±∞ and both v1 , v2 tend to ±∞ (i.e., the observer moves in the velocity = 4λ21 , 4λ22 ), then u ∼ 0. Therefore, whenever t ∼ +∞ or t ∼ −∞, u is asymptotic to two single soliton solutions (see Figures 1.1 – 1.3. This fact means that: (i) a double soliton solution is asymptotic to two single soliton solutions as t → ±∞; (ii) if two single solitons (the asymptotic behavior as t → −∞) interact, they will almost recover later


Figure 1.1.

Double soliton solutions of the MKdV equation, t = −1

Figure 1.2.

Double soliton solutions of the MKdV equation, t = 0.1

Figure 1.3.

9

Double soliton solutions of the MKdV equation, t = 1

(t → +∞). Both the shape and the velocity do not change. The only change is the phase shift. Physically speaking, there is elastic scattering between solitons. This is the most important character of solitons. The discovery of this property (first to the KdV equation) greatly promotes the progress of the soliton theory. Remark 3 Starting from the trivial solution u = 0, we can also obtain the single and double soliton solutions of the KdV equation by using the original Darboux transformation mentioned at the beginning of this section. The computation is simpler and is left for the reader.

10


The Darboux transformation for the MKdV equation can be used to get not only the single and double soliton solutions, but also the multisoliton solutions. Moreover, this method can be applied to many other nonlinear equations. We shall discuss the general problem in the next section.

1.1.5

Relation between Darboux transformations for KdV equation and MKdV equation

The Darboux transformation for the MKdV equation can also be derived from the “complexification” of the Schr¨ odinger equation (1.1) directly. That is why the transformation given by the matrix D is also called a Darboux transformation. ⎛ ⎞ φ1 ⎠ of (1.11), then the first equation of (1.11) Take a solution Φ = ⎝ φ2 is φ1,x = λφ1 + uφ2 , (1.41) φ2,x = −uφ1 − λφ2 . Let ψ = φ1 + iφ2 and suppose λ is a real parameter, u is a real function, then ψ satisfies ψxx = λ2 ψ − (iux + u2 )ψ. (1.42) This is a complex Schrödinger equation with potential (iux + u2 ). It can be checked directly that if u is a solution of the MKdV equation, then w = iux + u2 is a complex solution of the KdV equation wt + 6wwx + wxxx = 0. The transformation from the solution u of the MKdV equation to the solution w of the KdV equation is called a Miura transformation. Remark 4 Let v = iu, then (1.10) is vt − 6v 2 vx + vxxx = 0,

(1.10)

and the Miura transformation becomes w = vx −v 2 . If v is a real solution of (1.10) , then w is a real solution of the KdV equation. f2 of the equation Take a real number λ ⎛0 and⎞a solution f = f1 + if f1 ⎠ is a solution of (1.41) for λ = λ0 . Using (1.42) for λ = λ0 , then ⎝ f2 the conclusion to the KdV equation, we know that ψ = ψx − (ffx /f )ψ,

w = w + 2(ln f )xx

(1.43)

satisfy (1.42) and w is a solution of the KdV equation. Moreover, there is a corresponding u satisfying the MKdV equation. Now we write down the explicit expression of u .

11


Considering (1.41), the first equation of (1.43) can be rewritten in terms of the components as f¯ φ1 + iφ2 = λφ1 − iλφ2 − λ0 (φ1 + iφ2 ). f

(1.44)

If λ and λ0 are real numbers, then φ1 and φ2 can be chosen as real functions. (1.44) becomes ⎛

⎞

⎛

⎜ ⎝ ⎠=⎜ ⎜ ⎝ φ

φ1 2

f 2 − f22 λ − λ0 12 f1 + f22 2f1 f2 λ0 2 f1 + f22

2f1 f2 −λ0 2 f1 + f22 f 2 − f22 −λ − λ0 12 f1 + f22

⎞

⎛ ⎞ ⎟ φ ⎟⎝ 1 ⎠ . ⎟ ⎠ φ2

(1.45)

It should be noted that the matrix in the right hand side of (1.45) is the counterpart of the Darboux ⎛ matrix ⎞ defined by (1.24). It can be checked that ⎝

φ1 φ2

⎠ satisfies

φ1,x = λφ1 + u φ2 , φ2,x = −u φ1 − λφ2 where u = −u −

4λ0 f1 f2 . f12 + f22

(1.46)

(1.47)

The integrability condition of (1.46) implies that u is a solution of the MKdV equation. Remark 5 The matrix given by (1.14) and ⎞ that given by (1.45) are dif⎛ 1 0 ⎠. Notice that if (u, φ1 , φ2 ) ferent by a left-multiplied factor ⎝ 0 −1 is a solution of (1.41), then (−u, φ1 , −φ2 ) is also a solution of (1.41). Therefore, the solution (1.17) given by (1.14) is the minus of the solution given by (1.47). Both u and −u satisfy the MKdV equation. We can take any one transformation as the Darboux transformation. The matrix D is very important hereafter. It is called a Darboux matrix.

1.2 AKNS system 1.2.1 2 × 2 AKNS system In order to generalize the Lax pair of the MKdV equation, V. E. Zakharov, A. B. Shabat and M. J. Ablowitz, D. J. Kaup, A. C. Newell,

12


H. Segur introduced independently a more general system [2, 119] which is now called the AKNS system. For simplicity, we first discuss the 2 × 2 AKNS system (i.e., the AKNS system of 2 × 2 matrices), and then the more general N × N AKNS system. 2 × 2 AKNS system is the linear system of differential equations Φx = U Φ = λJΦ + P Φ, Φt = V Φ =

n

(1.48)

Vj λn−j Φ,

j=0

where

⎛

J =⎝

⎞

1

0

0 −1

⎠,

⎛

P =⎝

A= B= C=

⎞

0

p

q

0

n j=0 n j=0 n

⎠,

⎛

V =⎝

⎞

A

B

C −A

⎠

(1.49)

aj (x, t)λn−j , bj (x, t)λn−j ,

(1.50)

cj (x, t)λn−j ,

j=0

p, q, aj , bj , cj are complex or real functions of x and t, λ is a real or complex parameter, called the spectral parameter. As mentioned in Remark 2, the integrability condition of (1.48) Ut − Vx + [U, V ] = 0

(1.51)

should hold for all λ. In terms of the components, (1.51) becomes Ax = pC − qB, Bx = pt + 2λB − 2pA,

(1.52)

Cx = qt − 2λC + 2qA. Both sides of the above equations are polynomials of λ. Expanding them in terms of the powers of λ, we have b0 = c0 = 0, aj,x = pccj − qbj (0 ≤ j ≤ n), 1 bj+1 = bj,x + paj (0 ≤ j ≤ n − 1), 2 1 cj +1 = − cj,x + qaj (0 ≤ j ≤ n − 1), 2

(1.53)


13

and the evolution equations pt = bn,x + 2pan , qt = cn,x − 2qan .

(1.54)

(1.53) can be regarded as the equations to determine A, B, C, and (1.54) is a system of evolution equations of p and q. In (1.53), aj , bj , cj can be derived through algebraic calculation, differentiation and integration. We can see later that they are actually polynomials of p, q and their derivatives with respect to x (without any integral expressions of p and q), the coefficients of which are arbitrary functions of t. After solving aj , bj , cj from (1.53), we get the system of nonlinear evolution equations of p and q from (1.54). For j = 0, 1, 2, 3, a0 = α0 (t),

b0 = c0 = 0,

a1 = α1 (t), b1 = α0 (t)p, c1 = α0 (t)q, 1 a2 = − α0 (t)pq + α2 (t), 2 1 b2 = α0 (t)px + α1 (t)p, 2 1 c2 = − α0 (t)qx + α1 (t)q, 2 1 1 a3 = α0 (t)(pqx − qpx ) − α1 (t)pq + α3 (t), 4 2 1 1 2 b3 = α0 (t)(pxx − 2p q) + α1 (t)px + α2 (t)p, 4 2 1 1 2 c3 = α0 (t)(qxx − 2pq ) − α1 (t)qx + α2 (t)q. 4 2

(1.55)

Here α0 (t), α1 (t), α2 (t), α3 (t) are arbitrary functions of t, which are the integral constants in integrating a0 , a1 , a2 , a3 from the second equation of (1.53). Here are some simplest and most important examples.

Example 1.1 n = 3, p = u, q = −1, α0 = −4, α1 = α2 = α3 = 0. In this case, a3 = −ux , b3 = −uxx − 2u2 , c3 = 2u. (1.54) becomes the KdV equation (1.56) ut + uxxx + 6uux = 0. Example 1.2 n = 3, p = u, q = −u, α0 = −4, α1 = α2 = α3 = 0, then a3 = 0, b3 = −uxx − 2u3 , c3 = uxx + 2u3 . The equation becomes the MKdV equation (1.57) ut + uxxx + 6u2 ux = 0.

14


Example 1.3 n = 2, p = u, q = −¯ u, α0 = −2i, α1 = α2 = 0, a2 = ux . (1.54) is the nonlinear Schr¨dinger ¨ −i|u|2 , b2 = −iux , c2 = −i¯ equation (1.58) iut = uxx + 2|u|2 u. We have seen that for j = 0, 1, 2, 3, aj , bj , cj are differential polynomials of p and q, i.e., polynomials of p, q and their derivatives with respect to x, whose coefficients are constants or arbitrary functions of t.

Lemma 1.4 aj , bj , cj given by (1.53) are differential polynomials of p and q. Proof. Use induction. The conclusion is obviously true for j = 0. Suppose aj , bj , cj are differential polynomials of p and q for j < l, we will prove that al , bl , cl are also differential polynomials of p and q. (1.53) implies that bl , cl are differential polynomials of p and q. Hence it is only necessary to prove that al is a differential polynomial of p and q. For 1 ≤ j ≤ l − 1, bj cl+1−j − cj bl+1−j 1 1 = bj (qal−j − cl−j,x ) − cj (pal−j + bl−j,x ) 2 2 1 = (qbj − pccj )al−j − (bj cl−j,x + cj bl−j,x ) 2 1 1 = −(aj al−j + bj cl−j + cj bl−j )x + aj al−j,x 2 2 1 + (bj,x cl−j + cj,x bl−j ) 2 1 1 1 = −(aj al−j + bj cl−j + cj bl−j )x + (paj + bj,x )cl−j 2 2 2 1 −(qaj − cj,x )bl−j 2 1 1 = −(aj al−j + bj cl−j + cj bl−j )x + bj+1 cl−j − cj +1 bl−j . 2 2 Summarize for j from 1 to l − 1, we have b1 cl − c1 bl = −

(1.59)

l−1

1 1 (aj al−j + bj cl−j + cj bl−j )x − (b1 cl − c1 bl ), (1.60) 2 2 j=1

i.e., pcl − qbl = −

l−1 1

(2aj al−j + bj cl−j + cj bl−j )x .

(1.61)

(2aj al−j + bj cl−j + cj bl−j ) + αl (t)

(1.62)

j=1

4a0

Hence al = −

l−1 1 j=1

4a0

15


is a differential polynomial of p and q. The lemma is proved. Since {aj , bj , cj } are differential polynomials of p and q, we can define {a0j [[p, q ]}, {b0j [[p, q ]}, {cc0j [[p, q ]} recursively so that they satisfy the recursion relations (1.53) and the conditions a00 [0, 0] = 1, a0j [0, 0] = 0 (1 ≤ j ≤ n). Clearly, these {a0j , b0j , c0j } are uniquely determined as certain polynomials of p, q and their derivatives with respect to x. From (1.53), we have

Lemma 1.5 b0j [0, q] = 0,

c0j [p, [ 0] = 0, (1 ≤ j ≤ n)

a0j [p, [ 0] = a0j [0, q] = 0

(1.63)

for any p and q. Moreover, for any {aj , bj , cj } satisfying (1.53), there exist αj (t) (0 ≤ j ≤ n) such that ak [[p, q ] =

k

αk−j (t)a0j [[p, q ],

j=0

bk [[p, q ] = ck [[p, q ] =

k

j=0 k

αk−j (t)b0j [[p, q ],

(1.64)

αk−j (t)cc0j [[p, q ].

j=0

Remark 6 For any positive integer n, the first equation of (1.48) (x ( equation) is fixed, but the second one depends on the choice of α0 (t), · · ·, αn (t). Therefore, the evolution equations (1.54) also depend on the choice of α0 (t), · · ·, αn (t). This means that (1.54) is a series of equations, which is called the AKNS hierarchy. If α0 (t), · · ·, αn (t) are all constants, then the evolution equations have the coefficients independent of t and form a series of infinite dimensional dynamical systems. Especially, if α0 = · · · = αn−1 = 0, αn = 1, then we obtain the normalized AKNS hierarchy, written as ⎛ ⎝

⎞

p q

⎡

⎠ = Kn ⎣ t

⎤

p q

⎦,

(1.65)

where Kn is a nonlinear differential operator defined by ⎡

Kn ⎣

⎤

p q

⎛

⎦=⎝

b0n,x + 2pa0n c0n,x − 2qa0n

⎞

⎠.

(1.66)

16


From this definition and (1.53), we know that Kn is derived from Kn−1 by recursive algorithm.

1.2.2

N × N AKNS system

In the last subsection, we introduced the 2 × 2 AKNS system. In order to obtain more nonlinear partial differential equations, the 2 × 2 Lax pair should be generalized naturally to the problems of N × N matrices. Therefore, we discuss the Lax pair Φx = U Φ = λJΦ + P (x, t)Φ, Φt = V Φ =

n

Vj (x, t)λn−j Φ

(1.67)

j=0

where J is an N × N constant diagonal matrix, P (x, t), Vj (x, t) are N × N matrices and P (x, t) is off-diagonal (i.e., its diagonal entries are all zero), λ is a spectral parameter. We assume that all the entries of J are distinct, though the assumption can be released with restrictions on P and V . The integrability condition of (1.67) is still Ut − Vx + [U, V ] = 0.

(1.68)

Using the expressions of U and V in (1.67), we have Pt −

n

Vj,x λ

n−j

j=0

+

n−1

[J, Vj +1 ]λ

j=−1

n−j

+

n

[P, Vj ]λn−j = 0.

(1.69)

j=0

The coefficients of each power of λ on the left hand side should be zero. This leads to [J, V0 ] = 0, [J, Vj +1 ] − Vj,x + [P, Vj ] = 0

(0 ≤ j ≤ n − 1),

(1.70)

Pt − Vn,x + [P, Vn ] = 0. For any N × N matrix M , we divide it as M = M diag + M off , where is the diagonal part of M and M off = M − M diag (hence M off is M off-diagonal). Since J is diagonal with distinct diagonal entries and P diag

17


is off-diagonal, (1.70) is divided into V0off = 0, (0 ≤ j ≤ n),

diag Vj,x = [P, Vjoff ]diag

[J, Vj +1 ] = Vj,x − [P, Vj ] off

off

(1.71)

(0 ≤ j ≤ n − 1),

off

and off − [P, Vn ]off . Pt = Vn,x

(1.72)

We can solve Vj (j = 0, · · · , n) from (1.71) by differentiation and integration. In fact, similar to the 2 × 2 case, Vj can be obtained by differentiation and integration for n = 0, 1, 2, 3. They are differential polynomials of the entries of P . For general n, it can be proved by induction that each entry of Vj is a differential polynomial of the entries of P whose coefficients may depend on t. (The proof is omitted here. See [111]). Therefore, (1.72) gives a system of partial differential equations of the entries of P . We shall write Vj [P ] for the Vj to specify the dependence on P .

Example 1.6 Let n = 1, J = A = diag(a1 , · · · , aN ), V0 = B = diag(b1 , · · · , bN ) with ai = aj and bi = bj (i = j). Take V1 = Q(x, t) whose diagonal entries are all 0, then, from (1.71), Qij =

bi − b j Pij ai − aj

(i = j),

(1.73)

and the equation (1.72) becomes Pt = Qx − [P, Q]off .

(1.74)

Written in terms of the components, it becomes

Pij,t = cij Pij,x +

(cik − ckj )P Pik Pkj

(1.75)

k= i,j

where cij =

bi − b j . ai − aj

(1.76)

This is a system of nonlinear partial differential equations of Pij (i = j), called the N wave equation. Similar to the discussion in Lemma 1.5, let Vj0 [P ] be the solution of (1.71) satisfying V0 [0] = I, Vl [0] = 0 (1 ≤ l ≤ n), then the following lemma holds.

18


Lemma 1.5 . The general solution of (1.71) can be expressed as Vk [P ] =

k

αj Vk0−j [P ].

(1.77)

j=0

Vj [P ]}, where α0 , · · · , αn are the corresponding integral constants of {V which are diagonal matrices independent of x but may depend on t.

1.3 Darboux transformation 1.3.1 Darboux transformation for AKNS system Let (1.78) F (u, ux , ut , uxx , · · ·) = 0, be a system of partial differential equations where u is a function or a vector valued function. Consider the AKNS system Φx = U Φ = (λJ + P )Φ, Φt = V Φ =

n

Vj λn−j Φ,

(1.79)

j=0

where J, P , Vj satisfy the condition in the last section and P is a differential polynomial of u. Suppose (1.78) is equivalent to (1.68), the integrability condition of (1.79), then (1.79) is called the Lax pair of (1.78). In this case, (1.78) is the evolution equation (1.72). The non-degenerate N × N matrix solution of (1.79) is called a fundamental solution of the Lax pair. In this section, we suppose that the (off-diagonal) entries of P are independent and (1.78) is the system of differential equations (1.72) in which all off-diagonal entries of P are unknown functions. This system is called unreduced. We first discuss the Darboux transformation for the unreduced AKNS system.

Definition 1.7 Suppose D(x, t, λ) is an N × N matrix. If for given P and any solution Φ of (1.79), Φ = DΦ satisfies a linear system Φx = U Φ = (λJ + P )Φ , Φt = V Φ =

n

Vj λn−j Φ ,

(1.80)

j=0

where P is an off-diagonal N ×N matrix function, then the transformation (P, Φ) → (P , Φ ) is called a Darboux transformation for the unreduced AKNS system, D(x, t, λ) is called a Darboux matrix. A Darboux matrix is of degree k if it is a polynomial of λ of degree k.

19


According to this definition, P satisfies the equation off Ptoff − Vn,x + [P , Vn ]off = 0,

(1.72)

where the entries of Vn are differential polynomials of P . Later, we will see that Vn = Vn [P ] when the Darboux matrix is a polynomial of λ. In this case (1.72) and (1.72) are the same partial differential equations. Substituting Φ = DΦ into (1.80), we get U = DU D−1 + Dx D−1 , V = DV D−1 + Dt D−1 .

(1.81)

Proposition 1 If D is a Darboux matrix for (1.79) and D is a Darboux matrix for (1.80), then D D is a Darboux matrix for (1.79). Proof. Since D is a Darboux matrix for (1.80), there exists U = λJ + P (P is off-diagonal) and V = nj=0 Vj λn−j such that Φ = D Φ = D DΦ satisfies Φx = U Φ ,

Φt = V Φ .

(1.82)

Hence, by definition, D D is a Darboux matrix for (1.79). Remark 7 Any constant diagonal matrix K independent of λ is a Darboux matrix of degree 0, since under its action according to (1.81), λJ + P → λJ + KP K −1 , n j=0

Vj λn−j →

n

KV Vj K −1 λn−j .

(1.83)

j=0

If we do not consider the relations among the entries of P , this kind of Darboux matrices are trivial. Now we first consider the Darboux matrix of degree one, which is linear in λ. Suppose it has the form λI − S where S an N × N matrix function, I is the identity matrix. According to Proposition 1 and Remark 7, the discussion on the Darboux matrix K(λI − S) (K is a non-degenerate constant matrix which must be diagonal in order to get the first equation of (1.80)) can be reduced to the discussion on the Darboux matrix λI − S. Therefore, to construct the Darboux matrix, it is only necessary to construct S.

20


The differential equations of S are derived as follows. From the first equation of (1.80), (λJ + P )(λI − S)Φ = ((λI − S)Φ)x = (λI − S)(λJ + P )Φ − Sx Φ. (1.84) It must hold for any solution of (1.79). Comparing the coefficients of the powers of λ, we have P = P + [J, S],

(1.85)

This is the expression of P . The term independent of λ in (1.84) gives Sx = P S − SP = P S − SP + JS 2 − SJS,

(1.86)

Sx + [S, JS + P ] = 0.

(1.87)

i.e., This is the first equation which S satisfies. The second equation of (1.80) leads to n

Vj λn−j (λI − S)Φ = ((λI − S)Φ)t

j=0

(λI − S)

=

n

Vj λn−j Φ − St Φ.

(1.88)

j=0

Comparing the coefficients of λn+1 , λn , · · ·, λ, we can determine {V Vj } recursively by V0 = V0 ,

Vj+1 = Vj +1 + Vj S − SV Vj ,

(1.89)

and get the second equation of S Vn . St = Vn S − SV

(1.90)

From (1.89) Vj ’s can be expressed as V0 = V0 , Vj

= Vj +

j

[V Vj −k , S]S k−1

(1 ≤ j ≤ n),

(1.91)

k=1

and (1.90) becomes St + [S,

n j=0

Vj S n−j ] = 0.

(1.92)

21


Theorem 1.8 λI − S is a Darboux matrix for (1.79) if and only if S satisfies Sx + [S, JS + P ] = 0, St + [S,

n

Vj S n−j ] = 0.

(1.93)

j=0

Moreover, under the action of the Darboux matrix λI −S, P = P +[J, S]. Proof. Suppose λI − S is a Darboux matrix, then (1.93) is just (1.87) and (1.92) derived above. Conversely, if (1.87) and (1.92) hold, then for any solution Φ of (1.79), there are the relations (1.84) and (1.88). Hence Vj } determined (1.80) holds for the P determined by (1.85) and the {V by (1.89), which means that λI − S is a Darboux matrix. This theorem implies that we need to solve S from the system of nonlinear partial differential equations (1.93) to get the Darboux matrix. Fortunately, most of the solutions of (1.93) can be constructed explicitly. The following theorem gives the explicit construction of the Darboux matrix of degree one. Suppose P is a solution of (1.72). Take complex numbers λ1 , · · ·, λN such that they are not all the same. Let Λ = diag(λ1 , · · · , λN ). Let hi be a column solution of (1.79) for λ = λi . H = (h1 , · · · , hN ). When det H = 0, let (1.94) S = HΛH −1 , then we have the following theorem.

Theorem 1.9 The matrix λI −S defined by (1.94) is a Darboux matrix for (1.79). Proof. hi is a solution of (1.79) for λ = λi , that is, it satisfies hi,x = λi Jhi + P hi , hi,t =

n

Vj λn−j hi .

(1.95)

j=0

By taking the derivatives of H = (h1 , h2 , · · · , hN ) with respect to x and t, (1.95) is equivalent to Hx = JHΛ + P H, Ht =

n j=0

Vj HΛn−j .

(1.96)

22


Hence Sx = Hx ΛH −1 − HΛH −1 Hx H −1 = [H Hx H −1 , S] = [JS + P, S] Ht H −1 , S] St = Ht ΛH −1 − HΛH −1 Ht H −1 = [H =[

n

(1.97)

Vj S n−j , S].

j=0

Therefore, the matrix S defined by (1.94) is a solution of (1.93). Theorem 1.8 implies that λI −S is a Darboux matrix for (1.79). The theorem is proved.

Theorem 1.10 For any given (x0 , t0 ) and the matrix S0 , (1.93) has a solution satisfying S(x0 , t0 ) = S0 . That is, the system is integrable. Proof. First suppose the Jordan form of S0 is a diagonal matrix. Suppose its eigenvalues are λ1 , · · ·, λN and the corresponding eigenvectors are h0i . Let hi be a solution of (1.95) satisfying hi (x0 , t0 ) = h0i . Then these hi are linearly independent in a neighborhood of (x0 , t0 ). Theorem 1.9 implies that the Darboux matrix exists, i.e., (1.93) has a solution. If the Jordan form of S0 is not diagonal, then there is a series of matrices (k) (k) (k) {S S0 } such that the Jordan form of S0 is diagonal and S0 → S0 as (k) k → ∞. Construct S according to (1.94) with initial value S0 , then S (k) solves (1.93). The smooth dependency of the solution of (1.93) to (k) (k) the initial value implies that S (k) → S and Sx , St converge in a neighborhood of (x0 , t0 ). Therefore S is a solution of (1.93) with initial value S0 . Thus (1.93) is solvable for any given initial value, which means that it is integrable. The theorem is proved. We can also prove this theorem by direct but tedious calculation. This theorem implies that a Darboux matrix of degree one can be obtained either by (1.94) or the limit of such Darboux matrices. Remark 8 hi can be expressed as hi = Φ(λi )li (i = 1, 2, · · · , N ), where l1 , l2 , · · ·, lN are N linearly independent constant column matrices. Hence H in (1.94) can be written as H = (Φ(λ1 )l1 , Φ(λ2 )l2 , · · · , Φ(λN )lN ) .

(1.98)

This construction of Darboux matrix was given by [33, 94]. Theorem 1.9 and 1.10 implies that (1.94) contains all the matrices S which are similar to diagonal matrices and λI − S are Darboux matrices of

23


degree one. A Darboux matrix expressed by (1.94) is called a diagonalizable Darboux matrix or Darboux matrix with explicit expressions. It is useful in constructing the solutions because it is expressed explicitly. Hereafter, we mostly use the diagonalizable Darboux matrices and the word “diagonalizable” is omitted. The “single soliton solution” can be obtained by the Darboux transformation from the seed solution P = 0. For P= 0, the fundamental solution of (1.67) is Φ = eλJx+Ω(λ,t) where n n−j dt is a diagonal matrix. For any constants Ω(t) = j=0 Vj [0](t)λ λ1 , · · ·, λN and column matrices l1 , · · ·, lN , let

then

H = eλ1 Jx+Ω(λ1 ,t) l1 , · · · , eλN Jx+Ω(λN ,t) lN ,

(1.99)

P = [J, HΛH −1 ]

(1.100)

is a solution of (1.72) and the fundamental solution of the Lax pair (1.80) is Φ = (λI − HΛH −1 )Φ. Remark 9 If Vj [0] depends on t, then Ω(λi , t) is not a linear function of t, hence the velocities of the solitons are not constants [45]. The double soliton solution can be obtained from P by applying further Darboux transformation. Since Φ is known in this process, P and Φ can be obtained by a purely algebraic algorithm. The multisoliton solutions are obtained similarly. For general AKNS system, det H = 0 may not hold for all (x, t). Therefore, the solutions given by Darboux transformations may not be regular for all (x, t).

1.3.2

Invariance of equations under Darboux transformations

We have known that (P , Vj ) and (P, Vj ) satisfy the same recursion relations (1.71) and (1.72) holds true for the two sets of functions. Vj is a differential polynomial of P which is expressed by a similar equality as (1.77), i.e., Vk [P ] =

k

αj (t)V Vk0−j [P ].

(1.101)

j=0

Here we prove that actually the coefficients α0 (t), · · · , αn (t) are the same as α0 (t), · · · , αn (t) respectively. Therefore P and P satisfy the same evolution equation (1.72).

24


Theorem 1.11 Suppose Vj ’s are differential polynomials of P satisfying (1.71). S is a matrix satisfying (1.87), Vi ’s are defined by (1.89) and P = P + [J, S]. Then Vi ’s are differential polynomials of P and Vj [P ] = Vj [P ]

(j = 1, 2, · · · , n).

(1.102)

Proof. At first we see that the equation P = P + [J, S] is equivalent to P = P − [J, S] and the equation (1.87) is equivalent to Sx = [P + SJ, S].

(1.103)

Moreover, for arbitrary x-function P . This equation admits solutions in a neighborhood around any given point x = x0 . Thus we may consider P as an arbitrary off-diagonal matrix-valued function of x. From (1.89) we have + [P , Vj ] [J, Vj+1 ] − Vj,x

= [J, Vj +1 ] − Vj,x + [P, Vj ] + ([J, Vj ] − Vj−1,x + [P , Vj−1 ])S

(1.104)

−S([J, Vj ] − Vj −1,x + [P, Vj −1 ]) (j = 0, · · · , n − 1). Using induction, we know that + [P , Vj ] = 0 (j = 0, · · · , n − 1) [J, Vj+1 ] − Vj,x

(1.105)

from the equations (1.71) for Vj [P ]. Moreover, we can prove diag − [P , Vn ]diag = 0. Vn,x

(1.106)

This means that Vj and P also satisfy (1.71). Therefore, as mentioned above, Vj can be expressed as a differential polynomial of P : Vj = Vj [P ]. Let (1.107) ∆j [P ] = Vj [P ] − Vj [P ], (1.89) implies ∆0 = 0. Suppose ∆k = 0, then (1.71) leads to off = 0, [J, ∆k+1 ] = ∆off k,x − [P , ∆k ]

(1.108)

hence ∆off k+1 = 0. From (1.71), off = 0, ∆diag k+1,x = [P , ∆k+1 ]

(1.109)

which means that ∆diag k+1 [P ] is independent of x. We should prove that (α)

∆diag k+1 [P ] is independent of P . Denote Pij be the entries of P , Pij

be

25


the αth derivative of Pij with respect to x. Suppose the order of the highest derivatives of P in ∆diag k+1 is r, then 0=

r ∂∆diag ∂∆diag k+1 k+1 (α+1) = P . (α) ij ∂x P i,j α=0 ∂P

(1.110)

ij

(r+1)

In this equation, the coefficient of Pij should be 0. Hence ∆diag k+1 does not contain the rth derivative of P , which means that it is independent of P . Especially, let S = 0, P = P , then (1.89) implies ∆diag k+1 = 0. Thus (1.102) is proved Theorem 1.11 implies that for the evolution equations (1.72) in the AKNS system, the Darboux transformation transforms a solution of an equation to a new solution of the same equation. Note that the Darboux transformation (P, Φ) −→ (P , Φ ) defined by

P = P + [J, S],

(1.111)

Φ = (λI − S)Φ

(1.112)

can be taken successively in a purely algebraic algorithm and leads to an infinite series of solutions of the AKNS system: (P, Φ) −→ (P , Φ ) −→ (P , Φ ) −→ · · ·

1.3.3

(1.113)

Darboux transformations of higher degree and the theorem of permutability

The Darboux matrices discussed above are all of degree one. In this subsection, we construct Darboux matrices with explicit expressions which are the polynomials of λ of degree > 1. Then we derive the theorem of permutability from the Darboux matrices of degree two. Clearly, the composition of r Darboux transformations of degree one gives a Darboux transformation of degree r. On the other hand, we can also construct the Darboux transformations of degree r directly. As known above, a Darboux matrix of degree one is D(x, t, λ) = λI −S where S is given by (1.94) if it can be diagonalized (Theorem 1.9). Then, SH = HΛ is equivalent to D(x, t, λi )hi = 0 where hi is a column solution of the Lax pair for λ = λi such that det H = det(h1 , · · · , hN ) = 0. This fact can be generalized to the Darboux matrix of degree r, that is, we can consider an N × N Darboux matrix in the form D(x, t, λ) =

r j=0

Dr−j (x, t)λj ,

D0 = I.

(1.114)

26


Take N r complex numbers λ1 , λ2 , · · ·, λN r and the column solution hi of the Lax pair for λ = λi (i = 1, · · · , N r). Let ⎛

⎜ ⎜ ⎜ Fr = ⎜ ⎜ ⎜ ⎝

h1

h2

···

hN r

λ1 h1 .. .

λ2 h2 .. .

··· .. .

λN r hN r .. .

λ1r−1 h1 λ2r−1 h2 · · · λr−1 N r hN r

⎞ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎠

(1.115)

which is an N r × N r matrix. The system D(x, t, λi )hi = 0 is equivalent to r−1

Dr−j (x, t)λji hi = −λri hi

(i = 1, · · · , N r)

(1.116)

Fr = −(λr1 h1 , · · · , λrN r hN r ). (Dr , Dr−1 , · · · , D1 )F

(1.117)

j=0

and can be written as

This is a system of linear algebraic equations for (Dr , Dr−1 , · · · , D1 ). When det Fr = 0, it has a unique solution (Dr , Dr−1 , · · · , D1 ). Therefore, when det Fr = 0, there exists a unique N ×N matrix D(x, t, λ) satisfying D(x, t, λi )hi = 0 (i = 1, · · · , N r). We write it as D(h1 , · · · , hN r , λ) to indicate that D is constructed from h1 , · · · , hN r . The next theorem shows that it is a Darboux matrix and decomposable as a product of two Darboux matrices of lower degree [52, 74].

Theorem 1.12 Given N r complex numbers λ1 , · · · , λN r . Let hi be a column solution of the Lax pair (1.79) for λ = λi (i = 1, · · · , N r), Fr be defined by (1.115). Suppose det Fr = 0, then the following conclusions hold. (1) There exists a unique matrix D(h1 , · · · , hN r , λ) in the form (1.114) such that D(h1 , · · · , hN r , λi )hi = 0

(i = 1, 2, · · · , N r).

(1.118)

In this case, D(h1 , · · · , hN r , λ) is a Darboux matrix of degree r for (1.79). (2) If det Fr−1 = 0, then the above Darboux matrix of degree r can be decomposed as D(h1 , · · · , hN r , λ)

= D D(h1 , · · · , hN (r−1) , λN (r−1)+1 )hN (r−1)+1 , · · · ,

D(h1 , · · · , hN (r−1) , λN r )hN r , λ · ·D(h1 , · · · , hN (r−1) , λ).

(1.119)

27


On the right hand side of this equality, the first term is a Darboux matrix of degree one and the second term is a Darboux matrix of degree (r − 1). (3) The Darboux matrix D(h1 , · · · , hN r , λ) of degree r can be decomposed into the product of r Darboux matrices of degree one. (4) P = P − [J, D1 ] is a solution of (1.72). Proof. We first prove (2). Let

then

⎛ ⎜ ⎜ ⎜ Fr = ⎜ ⎜ ⎜ ⎝

Since

Λk = diag(λN (k−1)+1 , · · · , λN k ),

(1.120)

Hk = (hN (k−1)+1 , · · · , hN k ),

(1.121)

H1

H2

···

Hr

H1 Λ1 .. .

H2 Λ2 .. .

··· .. .

H r Λr .. .

H2 Λ2r−1 · · · Hr Λrr−1 H1 Λr−1 1

⎛ ⎜ ⎜ ⎜ Fr−1 = ⎜ ⎜ ⎜ ⎝

···

H1

H2

H1 Λ1 .. .

H2 Λ2 .. .

⎞ ⎟ ⎟ ⎟ ⎟. ⎟ ⎟ ⎠

Hr−1

· · · Hr−1 Λr−1 .. .. . .

r−2 H1 Λ1r−2 H2 Λ2r−2 · · · Hr−1 Λr−1

(1.122)

⎞ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎠

(1.123)

is non-degenerate, there is a matrix D(h1 , · · · , hN (r−1) , λ) of degree (r−1) with respect to λ such that D(h1 , · · · , hN (r−1) , λi )hi = 0

(i = 1, 2, · · · , N (r − 1)).

(1.124)

Let hi = D(h1 , · · · , hN (r−1) , λi )hi

(i = N (r − 1) + 1, · · · , N r). (1.125)

Construct a Darboux matrix D(hN (r−1)+1 , · · · , hN r , λ) from hi and let D (λ) = D(hN (r−1)+1 , · · · , hN r , λ)D(h1 , · · · , hN (r−1) , λ),

(1.126)

then D (λi )hi = 0 (i = 1, 2, · · · , N (r − 1)). Moreover, for i = N (r − 1) + 1, · · · , N r, D (λi )hi = D(hN (r−1)+1 , · · · , hN r , λi )hi = 0. Hence D (λ) = D(λ).

(1.127)

28


Thus, we have decomposed the matrix of degree r determined by (1.116) to the product of a matrix of degree (r − 1) and a matrix of degree one, expressed by (1.119). This proves (2). For D(h1 , · · · , hN (r−1) , λ), if all the determinants of Fr−2 , Fr−3 , · · · are non-zero, then D(h1 , · · · , hN , λ) can be decomposed to r matrices of degree one by repeating the above procedure. For r = 1, D(h1 , · · · , hN , λ) is a Darboux matrix. Hence D(h1 , · · · , hN r , λ) is also a Darboux matrix and it can be decomposed to the product of r Darboux matrices of degree one: (1.128) D = (λI − Sr ) · · · (λI − S1 ). Since det Fr = 0, we can always permute the subscripts of Λi and Hi so that all the determinants of Fr−2 , Fr−3 , · · · are non-zero. (3) is proved. Since D is the product of r Darboux matrices of degree one, D itself is a Darboux matrix. Hence (1) holds. Note that (1.129) D1 = −(S1 + · · · + Sr ). After the transformation of λI − S1 , P → P = P + [J, S1 ]. Then after the transformation of λI −S2 , P → P = P +[J, S2 ], and so on. Hence, after the transformation of D, P → P + [J, S1 + · · · + Sr ] = P − [J, D1 ].

(1.130)

Therefore, P − [J, D1 ] is a solution of (1.72). (4) is proved. This proves the lemma. Darboux transformation has an important property — the theorem of permutability. This theorem originated from the B¨ a¨cklund transformation of the sine-Gordon equation and there are a lot of generalizations and various proofs. The proof here is given by [52] (2 × 2 case) and [33] (N × N case). This proof does not depend on any boundary conditions and the permutation of the parameters is expressed definitely. From the solution (P, Φ(λ)), we can construct the Darboux transfor(1) (1) (1) (1) mation with parameters λ1 , · · ·, λN and the solutions hi = Φ(λi )li (1) of the Lax pair. Then the solution (P (1) , Φ(1) (λ)) is obtained. Here li ’s are N constant vectors. Next, construct a Darboux matrix for (P (1) , (2) (2) (2) Φ(1) (λ)) with parameters λ1 , · · ·, λN and li to get (P (1,2) , Φ(1,2) (λ)). On the other hand, construct the Darboux transformation for (P, Φ(λ)) (2) (2) with parameters λi and li to get (P (2) , Φ(2) (λ)). Then construct the (1) (1) Darboux transformation with parameters λi and li to get (P (2,1) , Φ(2,1) (λ)). The following theorem holds.

29


Theorem 1.13 (Theorem of permutability) Suppose ⎛

det ⎝

⎞

H1

H2

H1 Λ1 H2 Λ2

⎠ = 0,

(1.131)

then (P (1,2) , Φ(1,2) (λ)) = (P (2,1) , Φ(2,1) (λ)).

(1.132)

Proof. Theorem 1.12 implies that Φ(1,2) (λ) and Φ(2,1) (λ) are both obtained from Φ(λ) by the action of the Darboux transformation of degree two, and are expressed by (1)

(1)

(2)

(2)

(2)

(2)

(1)

(1)

Φ(1,2) (λ) = D(h1 , · · · , hN , h1 , · · · , hN , λ), φ(2,1) (λ) = D(h1 , · · · , hN , h1 , · · · , hN , λ).

(1.133)

From (1) of Theorem 1.12, we know that the right hand side of the above equations are equal. Hence the theorem of permutability holds. The theorem of permutability can be expressed by the following Bianchi diagram:

*

Λ(1) , L(1)

(P, Φ)

HH

Λ(2) , L(2) H j H

(P (1) , Φ(1) )H

(2) , L(2) HΛ H j H

(P (1,2) , Φ(1,2) ) = (P (2,1) , Φ(2,1) ) * (1)

(P (2) , Φ(2) ) Λ

, L(1)

(1.134) (1)

(2)

Here L(1) and L(2) denote the sets {li } and {li } respectively. Remark 10 The Darboux transformation of higher degree is much more complicated than the Darboux transformation of degree one. The theorem of decomposition implies that Darboux transformations of degree one can generate Darboux transformations of higher degree. Therefore, we can use Darboux transformations of degree one successively instead of a Darboux transformation of higher degree so as to avoid the calculation of the determinant of a matrix of very high order (of order N r). Since the algorithm for the Darboux transformation of degree one is purely algebraic and independent of the seed solution P , it is quite convenient to calculate the solutions using symbolic calculation with computer. However, some solutions, e.g., multi-solitons can be expressed by an explicit formulae by using Darboux transformations of higher degree [80].

30


Remark 11 The proof for Theorem 1.13 is for the Darboux transformations with explicit expressions. Since any Darboux transformation without explicit expression is a limit of Darboux transformations with explicit expressions, the theorem of permutability also holds for the Darboux transformations without explicit expressions. Now we compute the more explicit expression of the Darboux matrix of degree two. Suppose it is constructed from (Λ1 , H1 ) and (Λ2 , H2 ) which satisfy (1.131). Let Sj = Hj Λj Hj−1 and denote (α)

(α)

Λα = diag(λ1 , · · · , λN ),

(α)

(α)

Hα = (h1 , · · · , hN ).

(2)

(2)

(2)

After the action of λI −S1 , hj is transformed to (λj I −S1 )hj . Hence 2 = H2 Λ2 − S1 H2 = (S2 − S1 )H H2 . The second H2 is transformed to H Darboux matrix of degree one is λI − S2 where 2 Λ2 H −1 = (S2 − S1 )S2 (S2 − S1 )−1 . S2 = H 2

(1.135)

According to (1.131), S2 − S1 is non-degenerate. The Darboux matrix of degree two is D(λ) = (λI − S2 )(λI − S1 ) = λ2 I − λ(S22 − S12 )(S2 − S1 )−1 + (S2 − S1 )S2 (S2 − S1 )−1 S1 . (1.136) It is easy to check that D(λ) is symmetric to S1 and S2 . Therefore, we can also obtain the theorem of permutability by this symmetry.

1.3.4

More results on the Darboux matrices of degree one

In this subsection, we show that the Darboux matrix method in Theorem 1.9 can be applied not only to the AKNS system, but also to many other evolution equations, especially to the Lax pairs whose U and V are polynomials of λ. On the other hand, we also show that those Darboux matrices include all the diagonalizable Darboux matrices of the form λI − S, and any non-diagonalizable Darboux matrix can be obtained from the limit of diagonalizable Darboux matrices. We generalize the Lax pair (1.79) to Φx = U Φ, Φt = V Φ,

(1.137)

31


where U and V are polynomials of the spectral parameter λ: U (x, t, λ) = V (x, t, λ) =

m j=0 n

Uj (x, t)λm−j , (1.138) n−j

Vj (x, t)λ

,

j=0

Uj ’s and Vj ’s are N × N matrices. Clearly, the integrability condition of (1.137) is Ut − Vx + [U, V ] = 0.

(1.139)

In this subsection, we still discuss the Darboux matrices for the Lax pairs without reductions. That is, we suppose that all the entries of Uj ’s and Vj ’s are independent except for the partial differential equations (1.139). This is to say that apart from the integrability condition (1.139), there is no other constraint. Therefore, the nonlinear partial differential equation to be studied is just (1.139), i.e., the equations given by the coefficients of each power of λ in (1.139) and the unknowns are the N ×N matrices Uj and Vj (j = 0, 1, · · · , n). Compared with Subsection 1.3.1, D = λI − S is a Darboux matrix if and only if there exist

U (x, t, λ) =

V (x, t, λ) =

m j=0 n

Uj (x, t)λm−j , (1.140) Vj (x, t)λn−j

j=0

such that

Φ

= (λI − S)Φ satisfies Φx = U Φ ,

Φt = V Φ

(1.141)

where Φ is a fundamental solution of (1.137). Clearly, U and V have the expressions U = DU D−1 + Dx D−1 , V = DV D−1 + Dt D−1 and they satisfy

Ut − Vx + [U , V ] = 0.

(1.142)

(1.143)

The remaining problem is to obtain S so that (1.141) holds. If S is obtained, we have the Darboux transformation (U, V, Φ) → (U , V , Φ ).

(1.144)

32


Comparing to Theorem 1.8, 1.9 and 1.10, we have Theorem 1.8 . λI − S is a Darboux matrix of degree one for (1.137) if and only if S satisfies Sx + [S, U (S)] = 0, Here U (S) =

m j=0

Uj S m−j ,

St + [S, V (S)] = 0.

V (S) =

n

Vj S n−j .

(1.145)

(1.146)

j=0

Now suppose (U, V ) satisfies the integrability condition (1.139). For given constant diagonal matrix Λ = diag(λ1 , · · · , λN ), let hi be a column solution of (1.137) for λ = λi , H = (h1 , · · · , hN ). If det H = 0, let S = HΛH −1 , then the following theorems holds. Theorem 1.9 . The matrix λI − S is a Darboux matrix for (1.137). Theorem 1.10 . The system (1.145) is integrable. The proofs are omitted since they are similar to the proofs for the corresponding theorems above. Note that for the AKNS system, we can solve Vi [P ]’s from a system of differential equations by choosing “integral constants” and these Vi [P ]’s are differential polynomials of P . The remaining equation is only the equation (1.72) for P . In the present case, all the entries of Ui and Vi are regarded as independent unknowns satisfying the partial differential equations (1.139). The inverse of Theorem 1.9 also holds.

Theorem 1.14 (1) If λI − S is a Darboux matrix for (1.137) and S is diagonalized at one point, then there exists a constant diagonal matrix Λ = diag(λ1 , · · · , λN ) and column solutions hi ’s of the Lax pair (1.137) for λ = λi (i = 1, 2, · · · , N ) such that H = (h1 , · · · , hN ) and S = HΛH −1 . (2) If λI − S is a Darboux matrix for (1.137) but it can not be diagonalized at any points, then there exist a series of Darboux matrices λI − Sk such that Sk ’s and their derivatives with respect to x and t converge to S and its derivatives respectively. The proof is similar to that for Theorem 1.10.

Example 1.15 An example of a Darboux matrix which is not diagonalizable everywhere.

33


Consider the Lax pair ⎛

Φx = ⎝ ⎛

Φt = ⎝

⎞

λ

p

q

−λ

−2iλ2

⎠ Φ, ⎞

+ iipq −2iλp − ip i x

−2iλq + iqx

(1.147)

⎠Φ

2iλ2 − ipq i

whose integrability condition leads to the nonlinear evolution equations i t = pxx − 2p2 q, ip

−iqt = qxx − 2pq 2 .

(1.148)

This system of equations has a solution 2

2

p = α sech(αx)e−iα t ,

q = −α sech(αx)e iα t ,

(1.149)

which is derived from the trivial solution p = q = 0 by the Darboux matrix D = λI − HΛH −1 with ⎛

⎞

1 0 α ⎠, Λ= ⎝ 2 0 −1 ⎛

H=⎝

2 t/2

e−iα

0

⎞⎛

0

⎠⎝

2 t/2

e iα

eαx/2

−e−αx/2

e−αx/2

eαx/2

⎞

(1.150)

⎠.

Now we take (1.149) as a seed solution, whose corresponding fundamental solution of the Lax pair (1.147) is ⎛

(λI − HΛH −1 ) ⎝

eλx−2iλ

Take

2t

e−λx+2iλ

⎛

⎛

H () = ⎝

0

0

Λ() = ⎝ then we can choose

⎞

2t

⎠.

(1.151)

⎞

0 0 0

⎠,

h11

()

h12

()

h22

h21

() ()

(1.152)

⎞ ⎠,

(1.153)

34


where ()

h11 = () h12 () h21

22

α

− tanh(αx) eθ −

α 2 sech(αx)e−iα t−θ , 2 2

= − tanh(αx) + sech(αx)e−iα t , α 2 = − sech(αx)e iα t+θ + + tanh(αx) e−θ , 2

(1.154)

2

()

h22 = − sech(αx)e iα t − tanh(αx), θ = x − 2i2 t. When → 0, ⎛

H () Λ() (H () )−1 → S, 2

2

(1.155) ⎞

sinh(αx)e−iα t e−2iα t α ⎠, S= ⎝ 2 ∆ − sinh2 (αx) − sinh(αx)e−iα t 2

2αe−2iα t , p = ∆

where

q =

2α sinh2 (αx) ∆ 2

∆ = (α + 2 − 2 sech(αx)e−iα t ) cosh2 (αx).

(1.156)

(1.157)

(1.158)

Note that both eigenvalues of S are zero, but S = 0. Hence S is not diagonalizable. However, from the construction of S, we know that λI − S is a Darboux matrix, i.e., it satisfies (1.145). Finally, the conclusions for the Darboux transformations of higher degree and the theorem of permutability in Subsection 1.3.3 also hold for the general Lax pair (1.137). Moreover, when U and V in (1.137) are generalized to rational functions of λ, similar conclusions hold [121].

1.4

KdV hierarchy, MKdV-SG hierarchy, NLS hierarchy and AKNS system with u(N ) reduction

In the last section, we discussed the Darboux transformations for the AKNS system and more general systems. In those cases, we supposed that there were no reductions. In particular, there were no restrictions among the off-diagonal entries of P . However, in many cases, there are constraints on P and the Darboux transformation should keep those constraints. This problem is solved in many cases. Nevertheless, it should be very interesting to establish a systematic method to treat with reduced problems.

35


In this section, we first discuss some equations when N = 2 and there is certain relation between p and q. They are the important special cases of the 2 × 2 AKNS system: (1) KdV hierarchy: p is real and q = −1; (2) MKdV-SG hierarchy: q = −p is real; (3) Nonlinear Schr¨ o¨dinger hierarchy: q = −¯ p. These special cases were studied widely (e.g. [82, 88, 91, 105, 117, 118]). Here we use a unified method to deal with the whole hierarchy, and the coefficients may depend on t [32, 45]. At the end of this section, we discuss the general AKNS system with u(N ) reduction. This is a generalization of the nonlinear Schr¨ odinger hierarchy and has many applications to other problems.

1.4.1

KdV hierarchy

Consider the Lax pair [45] Φx = U Φ, where

⎛

U =⎝

Φt = V Φ

⎞

0

1

ζ −u 0

(1.159)

⎛

⎠,

V =⎝

⎞

A

B

C −A

⎠,

(1.160)

A, B and C are polynomials of the spectral parameter ζ. Compared with Section 1.2, the integrability condition (1.68) leads to − Ax + C − B(ζ − u) = 0,

Bx + 2A = 0,

−ut − Cx + 2A(ζ − u) = 0.

(1.161)

The first two equations imply 1 A = − Bx , 2 C = ζB − uB − 12 Bxx .

(1.162)

Substituting (1.162) into (1.161) we get 1 ut = −2(ζ − u)Bx + ux B + Bxxx . 2 Let B=

n

bj (x, t)ζ n−j ,

(1.163)

(1.164)

j=0

then (1.163) leads to b0,x = 0, bj+1,x = ubj,x + 12 ux bj + 14 bj,xxx

(0 ≤ j ≤ n − 1),

(1.165)

36


1 ut = 2ubn,x + ux bn + bn,xxx . (1.166) 2 (1.166) is the equation of u. When n ≥ 2, it is called a KdV equation of higher order. Similar to Lemma 1.5, (1.165) leads to bk =

k

αk−j (t)b0j [u],

(1.167)

j=0

where b0j [u]’s satisfy the recursion relations (1.165) and b00 [0] = 1, b0j [0] = 0 (j ≥ 1). Clearly b0j [u]’s are determined by (1.165) uniquely. The first few b0j ’s are 1 b01 = u, 2 1 3 2 0 b2 = uxx + u , · · · . 8 8 b00 = 1,

(1.168)

The corresponding equations are n = 0: Linear equation ut = α0 (t)ux . n = 1:

(1.169)

1

3 uxxx + uux + α1 (t)ux . (1.170) 4 2 If α0 = constant, α1 = 0, (1.170) is the standard KdV equation. n = 2:

ut = α0 (t)

1

5 5 15 uxxxxx + uuxxx + ux uxx + u2 ux 16 8 4 8 1 3 +α1 (t) uxxx + uux + α2 (t)ux , 4 2

ut = α0 (t)

(1.171)

which is called the KdV equation of 5th-order. Next we discuss the Darboux transformation for the KdV hierarchy by using the general results for the AKNS system. It seems that the calculation is tedious. However, we can see the application of the general results more clearly. The method here is valid to the whole hierarchy comparing to the special method in Section 1.1. The U and V given by (1.160) are different from those of the AKNS system. However, the Lax pair can be transformed to a Lax pair in the AKNS system by a similar transformation given by a constant matrix depending on ζ.

37


Let

⎛

R=⎝

⎞

λ

1

−1 0

⎠

Ψ = RΦ,

(λ2 = ζ),

⎛

⎞

λ

= RU R−1 = ⎝ U ⎛

V = RV R−1 = ⎝ then Ψ satisfies

u

−1 −λ

(1.172)

⎠,

λB − A λ2 B − 2λA − C −B

A − λB

Ψ, Ψx = U

⎞ ⎠,

Ψt = V Ψ.

(1.173)

This is the Lax pair for the KdV equation in the AKNS form. The Darboux transformation can be constructed based on the discussion in Section 1.3. Take two constants λ1 , λ2 and column solutions h1 , h2 of the Lax pair when λ = λ1 , λ2 respectively. Moreover, we want that the matrices given by the Darboux transformation are still of the form of (1.172). That is, ⎛

= ⎝ U ⎛

V =⎝

u

λ

−1 −λ

⎞

⎠,

λB[u ] − A[u ] λ2 B[u ] − 2λA[u ] − C[u ] −B[u ]

A[u ] − λB[u ]

⎞

(1.174)

⎠.

is −1) holds only This condition (especially that the (2, 1) entry of U , h2 and when λ2 , λ1⎛ ⎞ h1 are specified. α ⎠ is a solution of the Lax pair (1.173) for λ = λ0 , then Suppose ⎝ β ⎛ ⎝

⎞

α + 2λ0 β β

⎠ is a solution of (1.173) for λ = −λ0 . Thus we choose ⎛

Λ=⎝

⎞

λ0

0

0

−λ0

⎛

⎠,

Let

H=⎝ ⎛

S = HΛH −1

⎞

α α + 2λ0 β β

1 ⎜ −λ0 − τ =⎝ −1

β

⎠.

(1.175)

⎞

1 2λ0 + 2 τ τ ⎟ ⎠ 1 + λ0 τ

(1.176)

38


where τ = β/α, and ⎛ =⎝ D

⎞

1

0

0 −1

⎠ (λI − S),

(1.177)

then after the action of the Darboux transformation given by the Dar boux matrix D, ⎛ = D U D −1 + D xD −1 = ⎝ U

where u = −u − 2

λ

u

−1 −λ

⎞ ⎠

(1.178)

1 2λ0 + . 2 τ τ

(1.179)

According to the general discussion to the AKNS system, V is given by the second equation of (1.174). Therefore, the Darboux transformation in (1.177) is a Darboux transformation from any equation in given by D the KdV hierarchy to the same equation. ⎛ ⎞ α ⎠ Next we compare the results here with those in Section 1.1. If ⎝ β is a solution of (1.173) for λ = λ0 , then the corresponding solution of (1.159) is ⎛

R−1 (λ0 ) ⎝

⎞

α β

⎛

⎠=⎝

−β α + λ0 β

⎞

⎠.

(1.180)

Let σ be the ratio of the second and the first components, i.e., σ=

1 α + λ0 β = − − λ0 , −β τ

(1.181)

then σ satisfies σx = λ20 − u − σ 2 , and

⎛

S=⎝

σ

σ 2 − λ20

−1

−σ

(1.182) ⎞ ⎠.

(1.183)

39


In order to get the Darboux matrix for the Lax pair in the form (1.159), let ⎛

D = R−1 ⎝ ⎛

=⎝

⎞

1

0

0 −1

⎛

⎠ (λI − S)R = ⎝ ⎞

−σ

1

ζ − ζ0 +

−σ

σ2

⎠

−σ

⎞

1

λ2 − λ20 + σ 2 −σ

⎠

(ζζ0 = λ20 ). (1.184)

Then U = DU D−1 + Dx D−1 ⎛

=⎝

⎞

0

1

ζ − 2ζζ0 + u + 2σ 2 0

⎠.

(1.185)

Hence the action of D keeps the x part of the Lax pair invariant, and transforms u to (1.186) u = 2ζζ0 − u − 2σ 2 (the same as (1.179)). Theorem 1.11 implies that V [u] = V [u ], i.e., the Darboux transformation keeps the t part invariant. Therefore, we have

Theorem 1.16 u is a solution of (1.166), ζ0 is a non-zero ⎛ Suppose ⎞ a real constant, ⎝ ⎠ is a solution of the Lax pair (1.159) for ζ = ζ0 , b σ = b/a, then ⎛

D=⎝

−σ

⎞

1

ζ − ζ0 + σ 2 −σ

⎠

(1.187)

is a Darboux matrix for (1.159). It transforms a solution u of (1.166) to a new solution (1.188) u = 2ζζ0 − u − 2σ 2 of the same equation. and U have the same (2, 1)-entry −1, the Remark 12 In order to let U ⎛ ⎞ 1 0 ⎠ (λI − S)R, Darboux matrix (1.184) is chosen as D = R−1 ⎝ 0 −1 not R−1 (λI − S)R. This guarantees that the transformation transforms a solution of (1.166) to a solution of the same equation (1.166).

40

DARBOUX TRANSFORMATIONS IN INTEGRABLE SYSTEMS ⎛

Remark 13 Let Φ = ⎝

⎞

φ ψ

⎠, then φ satisfies

φxx = (ζ − u)φ,

(1.189)

φt = Aφ + Bφx . It is similar to the system in Section 1.1. However, here A and B can be polynomials of ζ of arbitrary degrees, whose coefficients are differential polynomials of u. The problem discussed in Section 1.1 was a special case. In Theorem 1.16, b = ax , hence σ = ax /a. The transformation D in (1.184) gives ax (1.190) φ → φ = φx − σφ = φx − φ, a and (1.182), (1.186) give the original Darboux transformation u = u + 2(ln a)xx .

(1.191)

Remark 14 From (1.165), we can get b0 , b1 , · · · recursively, whose integral constants can be functions of t. Therefore, the coefficients of the nonlinear equations can be functions of t, as in the examples (1.170) and (1.171). The solutions of the equations whose coefficients depending on t differ a lot from the solutions of the equations whose coefficients independent of t. In the latter case, each soliton moves in a fixed velocity and the soliton with larger amplitude moves faster. However, in the former case, each soliton can have varying velocity (e.g. oscillates), and the soliton with larger amplitude may move slower.

1.4.2

MKdV-SG hierarchy

Consider the Lax pair [32]

⎛

Φx = U Φ = ⎝ ⎛

Φt = V Φ = ⎝

⎞

λ

p

−p −λ

⎠ Φ,

⎞

A

B

C −A

(1.192)

⎠ Φ,

where A, B and C are polynomials of λ and λ−1 satisfying A(−λ) = −A(λ),

B(−λ) = −C(λ).

(1.193)

Moreover, suppose A=

n+m j=0

aj λ2n−2j+1

(1.194)

41


(m ≥ 0, n ≥ 0). Unlike the AKNS system, here A, B and C are not restricted to the polynomials of λ, but the negative powers of λ are allowed. The term with lowest power of λ in (1.194) is an+m λ−2m+1 . The integrability condition Ut − Vx + [U, V ] = 0 leads to Ax = p(B + C), pt − Bx − 2pA + 2λB = 0,

(1.195)

pt + Cx + 2pA + 2λC = 0. Hence aj,x Ax n+m = λ2n−2j+1 , p p j=0 (B + C)x + 44pA B−C = 2λ n+m 1 aj,x = + 2paj λ2n−2j , 2 p x j=0

B+C =

(1.196)

(thus B(−λ) = −C(λ) holds automatically) and 1 pt = (B − C)x − λ(B + C). 2

(1.197)

Comparing the coefficients of λ in (1.197), we can obtain the recursion relations among aj ’s. They include two parts. The first part a0,x = 0, 1 aj+1,x = p 4

aj,x p

(j = 0, 1, · · · , n − 1)

+ 4aj p x

(1.198)

x

are obtained from the coefficients of positive powers of λ and the second part an+m,x + 4an+m p = 0, x x p aj+1,x aj,x (1.199) + 4aj p = 4 p x p x (j = n + m − 1, · · · , n + 1) are obtained from the coefficients of negative powers of λ. Moreover, the term without λ leads to the equation 1 pt − 4

an,x p

+ 4an p x

+ x

an+1,x = 0. p

(1.200)

42


The first few aj ’s (0 ≤ j ≤ n) are a0 = α0 (t), 1 a1 = α0 (t)p2 + α1 (t), 2 1 1 2 3 4 1 a2 = α0 (t) ppxx − px + p + α1 (t)p2 + α2 (t), 4 8 8 2 ···.

(1.201)

If V does not contain negative powers of λ, i.e., m = 0, then from the general conclusion to the AKNS system, all aj ’s are differential polynomials of p. The equation (1.200) becomes pt −

1 4

an,x p

+ 4an p x

= 0.

(1.202)

x

This is called the MKdV hierarchy. By using the notion a0j in Section 1.2, these equations can be written as pt +

n

αj (t)M Mn−j [p [ ] = 0,

(1.203)

j=0

where 1 [ ]=− Ml [p 4

0 a l,x

p

+

(l = 0, 1, · · · , n).

4a0l p

x

(1.204)

x

Especially, if n = 1 and α0 = −4, α1 = 0, then (1.202) becomes the MKdV equation (1.205) pt + pxxx + 6p2 px = 0. Next, we consider the negative powers of λ in V . Take p = −ux /2 and suppose it satisfies the boundary condition: u − kπ and its derivatives tend to 0 fast enough as x → −∞ (k is an integer). The first equation of (1.199) gives

an+m,x ux

x

+ an+m ux

= 0.

(1.206)

x

Write an+m as a function of u, then the above equation becomes ((an+m,uu + an+m )ux )x = 0.

(1.207)

The boundary condition as x → −∞ gives an+m,uu + an+m = 0. Hence an+m = α cos(u + β)

(1.208)

43


where α and β are constants. Now take a special an+m : a0n+m = 14 cos u. an+j (j = 1, 2, · · · , m − 1) can be determined as follows. Let gn+j = a0n+j,x /p, then gn+m = 12 sin u, and a0j−1 =

x

−∞

pggj −1 dx + a− j−1

(n + 2 ≤ j ≤ n + m)

(1.209)

0 where a− j−1 is the limit of aj−1 as x → −∞. From the boundary condition lim (ggj −1 )x = 0, the recursion relations (1.199) become x→−∞

1 (ggj −1 )x + p a− j−1 + 4

x

pggj −1 dx =

−∞

x −∞

gj dx.

(1.210)

Moreover, suppose lim gj −1 = 0,

(1.211)

x→−∞

then

x

gj −1 + 4 x

=4

−∞ ξ

−∞ −∞

p(ξ)

a− j−1

ξ

+

−∞

p(ζ)ggj −1 (ζ) dζ

dξ (1.212)

gj (ζ) dζ dξ.

This is an integral equation of Volterra type. It has a unique solution in the class of functions which tend to zero fast enough together with its derivatives as x → −∞. Take a− j−1 = 0 and write the solution of (1.212) as 1 gj −1 = Q(ggj ) = Q2 (ggj +1 ) = · · · = Qn+m−j+1 [sin u]. 2

(1.213)

Here Q is the operator to determine gj −1 from gj defined by (1.212). gj −1 is not a differential polynomial of gj . If n = 0, α0 = 0, then we obtain the SG hierarchy pt +

1 m−1 βj (t)Qm−j−1 [sin u] = 0 2 j=0

(1.214)

where βj (t)’s are arbitrary functions of t. Generally, we have the compound MKdV-SG hierarchy pt + (p

n

αj (t)M Mn−j [p [ ]+

j=0 = − u2x ).

1 m−1 βj (t)Qm−j−1 [sin u] = 0, 2 j=0

(1.215)

44


Example 1.17 n = 0, m = 2, β0 = 0, β1 = 1, then, g2 = the equation becomes the sine-Gordon equation

1 2

sin u, and

uxt = sin u.

(1.216)

Example 1.18 n = 1, m = 2, α0 = −4, α1 = 0, β0 = 0, β1 = 1, then the equation becomes the equation describing one-dimensional nonlinear lattice of atoms [70] 3 uxt + u2x uxx + uxxxx − sin u = 0. 2

(1.217)

⎛

Now we consider the Darboux transformation. If ⎝ ⎛

−β

of (1.192) for λ = λ0 , then ⎝

α

⎞

⎞

α β

⎠ is a solution

⎠ is a solution of (1.192) for λ = −λ0 .

Therefore, we can choose ⎛

Λ=⎝ ⎛

where ⎝

⎞

λ0

0

0

−λ0

⎠,

⎛

H=⎝

α −β β

α

⎞ ⎠,

(1.218)

⎞

α β

⎠ is a solution of (1.192) for λ = λ0 . Let σ = β/α, ⎛

S = HΛH −1

and denote tan

⎞

2σ λ0 ⎝ 1 − σ 2 ⎠ = 2 2 1+σ 2σ σ −1

θ = σ, then 2

⎛

S = λ0 ⎝

(1.219)

⎞

cos θ

sin θ

sin θ − cos θ

⎠.

(1.220)

From σx = −p(1 + σ 2 ) − 2λ0 σ, we have θx = −2p − 2λ0 sin θ.

(1.221)

By direct calculation, ⎛

(λI − S)U (λI − S)−1 − Sx (λI − S)−1 = ⎝

λ

p

−p −λ

⎞ ⎠

(1.222)

45


where p = p + 2λ0 sin θ = −p − θx ,

(1.223)

u = −u + 2θ

(1.224)

or equivalently,

for suitable choice of the integral constant. It remains to prove that the Darboux matrix λI − S keeps the reduction of MKdV-SG hierarchy. This includes (1) the transformed A , B and C still satisfy A (−λ) = −A (λ) and B (−λ) = −C(λ); (2) the coefficients αj (t)’s keeps invariant. Since V T (−λ) = −V (λ), S T = S and (λI + S)T (λI − S) = λ2 I − S 2 = 2 (λ −λ20 )I, it can be verified by direct calculation that V T (−λ) = −V (λ) holds. This proves (1). (2) is proved as follows. For aj (j ≤ n), this has been proved for the AKNS system; for aj (j ≥ n + 1), the conclusion follows from the boundary condition at infinity. Therefore, the following theorem holds.

Theorem ⎛ 1.19 ⎞ Suppose u is a solution of (1.200), λ0 is a non-zero real a number, ⎝ ⎠ is a solution of the Lax pair (1.192) for λ = λ0 . Let b −1 θ = 2 tan (b/a), ⎛

S = λ0 ⎝

⎞

cos θ

sin θ

sin θ − cos θ

⎠,

(1.225)

then λI − S is a Darboux matrix for (1.192). It transforms a solution p of (1.200) to the solution p = p + 2λ0 sin θ of the same equation. Moreover, u = −u + 2θ with suitable boundary condition, where p = −ux /2, p = −ux /2. Remark 15 For the sine-Gordon equation, the Backlund ¨ transformation is a kind of method to get explicit solutions, which was known in the nineteenth century. In that method, to obtain a new solution from a known solution, there is an integrable system of differential equations to be solved (moreover, one can obtain explicit expression by using the theorem of permutability and the nonlinear superposition formula). Using Darboux transformation, that explicit expression can be obtained directly. This will be discussed in Chapter 4 together with the related geometric problems.

46


1.4.3

NLS hierarchy

The Lax pair for the nonlinear Schrodinger ¨ hierarchy (NLS hierarchy) is

⎛

Φx = U Φ = ⎝ ⎛

Φt = V Φ = ⎝

⎞

λ

p

−¯ p −λ

⎠ Φ,

⎞

A

B

C −A

(1.226)

⎠ Φ,

where A, B and C are polynomials of λ (λ, p, A, B and C are complexvalued) satisfying ¯ = −C(λ) B(−λ)

¯ = −A(λ), A(−λ)

(1.227)

¯ = −V (λ) where ∗ refers to the complex conjugate trans(i.e., V ∗ (−λ) pose of a matrix). This is also a special case of the AKNS system. We shall construct a Darboux matrix keeping this reduction. The integrability condition Ut − Vx + [U, V ] = 0 is ¯ Ax = pC + pB, Bx = pt + 2λB − 2pA,

(1.228)

Cx = −¯ pt − 2λC − 2pA. ¯ We can use (1.55) to write down the coefficients of the powers of λ in A, B and C. They depend on p, px , · · · and the integral constants αj (t). Moreover, there is a nonlinear evolution equation pt = bn,x + 2pan .

(1.229)

Especially, for n = 2, α0 = −2i, α1 = α2 = 0, the equation is the nonlinear Schr¨ odinger equation i t = pxx + 2|p|2 p. ip

(1.230)

The Darboux transformation for the nonlinear Schr¨ o¨dinger ⎛ hierarchy ⎞ α ⎠ is a is also constructed from the choice of Λ and H. Suppose ⎝ β ⎛

solution of (1.226) for λ = λ0 , then ⎝

−β¯ α ¯

⎞

⎠ is a solution of (1.226) for

47


¯ 0 . Hence, we can choose λ = −λ ⎛

Λ=⎝

⎞

λ0 0

0 ⎠, ¯0 −λ

⎛

H=⎝

α −β¯ β

α ¯

⎞ ⎠,

⎛

S = HΛH −1

(1.231) ⎞

¯ 0 |σ|2 (λ0 + λ ¯ 0 )¯ λ −λ σ 1 ⎝ 0 ⎠, = 2 ¯ ¯ 1 + |σ| (λ0 + λ0 )σ −λ0 + λ0 |σ|2

(1.232)

where σ = β/α. Since det H = 0, S can be defined globally. It can be checked that H satisfies H ∗ H = |α|2 + |β|2 ,

(1.233)

hence S satisfies S ∗ S = |λ0 |2 ,

(1.234)

¯0. S − S ∗ = λ0 − λ

Therefore, under the action of the Darboux matrix λI − S, U is transformed to ⎛

U = (λI − S)U (λI − S)−1 − Sx (λI − S)−1 = ⎝ where p = p + 2S12 = p +

λ

p

−¯ p −λ

¯ 0 )¯ 2(λ0 + λ σ . 1 + |σ|2

⎞ ⎠ (1.235)

(1.236)

From the discussion on the AKNS system (Theorem 1.11), we know that V = (λI − S)V (λI − S)−1 − St (λI − S)−1 is also a polynomial ∗ ¯ = −V (λ) holds. Moreover, λI − S gives a Darboux of λ and V (−λ) transformation from an equation in the nonlinear Schr¨ o¨dinger hierarchy to the same equation. This leads to the following theorem.

Theorem 1.20 Suppose ⎛ ⎞ p is a solution of (1.229), λ0 is a non-real a complex number, ⎝ ⎠ is a solution of the Lax pair (1.226) for λ = λ0 . b Let σ = b/a, ⎛

⎞

¯ 0 |σ|2 (λ0 + λ ¯ 0 )¯ λ −λ σ 1 ⎝ 0 ⎠, S= ¯ 0 )σ −λ ¯ 0 + λ0 |σ|2 1 + |σ|2 (λ0 + λ

(1.237)

48


then λI − S is a Darboux matrix for (1.226). It transforms a solution p of (1.229) to a solution p = p +

¯ 0 )¯ 2(λ0 + λ σ 2 1 + |σ|

(1.238)

of the same equation.

1.4.4

AKNS system with u(N ) reduction

For the nonlinear Schrodinger ¨ hierarchy, (1.226) and (1.227) imply ¯ = −U (λ)∗ , U (−λ)

¯ = −V (λ)∗ . V (−λ)

(1.239)

Here we generalize it to the AKNS system. For the AKNS system (1.226), if U and V satisfy (1.239), then we say that (1.226) has u(N ) reduction, because U (λ) and V (λ) are in the Lie algebra u(N ) when λ is purely imaginary. This is a very popular reduction. We want to construct Darboux matrix which keeps u(N ) reduction. That is, after the action of the Darboux matrix, the derived potentials U (λ) and V (λ) must satisfy ¯ = −U (λ)∗ , U (−λ)

¯ = −V (λ)∗ . V (−λ)

(1.240)

With this additional condition, Λ and H in (1.94) can not be arbitrary. They should satisfy the following two conditions: µ where µ is a complex number (µ (1) λ1 , · · · , λN can only be µ or −¯ is not real). (2) If λj = λk , then (1.241) h∗j hk = 0 holds at one point (x0 , t0 ). In fact, if (1.241) holds at one point, then it holds everywhere. This is proved as follows. ¯ j , hence When λj = λk , λk = −λ hk,x = U (λk )hk ,

hk,t = V (λk )hk , ¯ j ) = −h∗ U (λk ), h∗j,x = h∗j U (λj )∗ = −h∗j U (−λ j ¯ j ) = −h∗ V (λk ). h∗ = h∗ V (λj )∗ = −h∗ V (−λ j,t

This implies that

j

j

(h∗j hk )x = 0,

(1.242)

j

(h∗j hk )t = 0.

Therefore, h∗j hk = 0 holds everywhere if it holds at one point.

(1.243)

49


Theorem 1.21 If λj ’s, hj ’s satisfy the above conditions (1) and (2), H = (h1 , · · · , hN ), then det H = 0 holds everywhere if it holds at one point. Moreover, U and V given by (1.81) satisfy ¯ = −U (λ)∗ , U (−λ)

¯ = −V (λ)∗ . V (−λ)

(1.244)

Proof. Let (x0 , t0 ) be a fixed point. Then by the property of linear differential equation, all {hα } with λα = µ are linearly independent if ¯ they are linearly independent at (x0 , t0 ). Likewise, all {hα } with λα = µ are also linearly independent if they are linearly independent at (x0 , t0 ). Moreover, (1.241) implies that all {h1 , · · · , hN } are linearly independent if they are linearly independent at (x0 , t0 ). Therefore, det H = 0 and S = HΛH −1 is globally defined. According to the definition of S, Shj = λj hj , Hence

¯k . h∗k S ∗ = h∗k λ

¯ k )h∗ hj . h∗k (S − S ∗ )hj = (λj − λ k

(1.245) (1.246)

µ, then If λj = µ, λk = −¯ h∗k (S − S ∗ )hj = 0.

(1.247)

If λj = λk = µ (or λj = λk = −¯ µ), then ¯)h∗k hj . h∗k (S − S ∗ )hj = (µ − µ

(1.248)

¯)I. S − S ∗ = (µ − µ

(1.249)

Hence

On the other hand, from (1.245), we have ¯ k h∗ hj . h∗k S ∗ Shj = λj λ k

(1.250)

h∗k S ∗ Shj = 0.

(1.251)

If λj = µ, λk = −¯ µ, then

If λj = λk = µ (or λj = λk = −¯ µ), then

Therefore,

h∗k S ∗ Shj = |µ|2 h∗k hj ,

(1.252)

S ∗ S = |µ|2 I.

(1.253)

From (1.249) and (1.253), we obtain ¯ + S)∗ (λI − S) = (λ − µ)(λ + µ ¯)I. (λI

(1.254)

50


According to the action of the Darboux transformation on Vj , m

Vj λm−j

j=0

= (λI − S)

m

(1.255) Vj λm−j (λI − S)−1 + (λI − S)t (λI − S)−1 ,

j=0

(

m

Vj λm−j )∗

j=0

= (λI − S)∗−1

m

¯ m−j (λI − S)∗ + (λI − S)∗−1 (λI − S)∗ Vj∗ λ t

j=0

¯ + S) = −(λI =−

m

m

¯ m−j (λI ¯ + S)−1 − (λI ¯ + S)t (λI ¯ + S)−1 Vj (−λ)

j=0

¯ m−j . Vj (−λ)

j=0

(1.256) ¯ = −V (λ)∗ . Likewise, U (−λ) ¯ = −U (λ)∗ . The theorem Hence V (−λ) is proved. As in Section 1.3, a Darboux transformation of higher degree can be derived by the composition of Darboux transformations of degree one. However, with the u(N ) reduction, we have also the following special and more direct construction [117, 17]. Suppose we take l times of Darboux transformations of degree one. Each Darboux transformation is constructed from Λα , Hα (α = 1, · · · , l). (α) (α (α) (α) In each Λα = diag(λ1 , · · · , λk ), suppose λ1 = · · · = λk = µα , (α) (α) µα . Here k is the same for all α. For each λk+1 = · · · = λN = −¯ (α)

(α)

λj , solve the Lax pair and get a solution hj satisfying the orthogonal relations (1.241). (α) (α) ˚ = (h(α) , · · · , h(α) ). Let Denote Hα = (h1 , · · · , hN ), H α 1 k Γαβ =

D(λ) =

l γ=1

⎛

(λ + µ ¯ γ ) ⎝1 −

˚∗ H ˚ H β α , µβ + µ ¯α l ˚ (Γ−1 ) H ˚∗ H α αβ β

α,β=1

¯β λ+λ

Now we prove that D(λ) is a Darboux matrix.

(1.257) ⎞ ⎠.

(1.258)

51


˚ ,H ˚∗ H ˆ α = (h(α) , · · · , h(α) ). Then Hα = ( H ˆ α ) and H ˆ Let H α α α = 0 for N k+1 all α = 1, · · · , l. Hence ˚ = 0, D(µα ) H α

ˆ α = 0. D(−¯ µα )H

(1.259)

According to Theorem 1.12, D(λ) is a Darboux matrix. Moreover, the inverse of D(λ) can be written out explicitly as D(λ)−1 =

l

⎛

⎞ l ˚ (Γ−1 ) H ˚∗ H α αβ β⎠

(λ + µ ¯γ )−1 ⎝1 +

γ=1

λ − λα

α,β=1

.

(1.260)

(1.258) gives a compact form of Darboux matrix of higher degree. Although it is special, it is very useful.

1.5

Darboux transformation and scattering, inverse scattering theory

The scattering and inverse scattering theory is an important part of the soliton theory. It transforms the problem of solving the Cauchy problem of a nonlinear partial differential equation to the problem of describing the spectrum and eigenfunctions of the Lax pair. Here we consider the 2 × 2 AKNS system as an example to show the outline of the scattering and inverse scattering theory (see [23] for details). Moreover, we discuss the change of the scattering data under Darboux transformation for su(2) reduction. For the KdV equation, the problem can be solved similarly, but the scattering and inverse scattering theory is simpler.

1.5.1

Outline of the scattering and inverse scattering theory for the 2 × 2 AKNS system

First, we give the definition of the scattering data for the 2×2 complex AKNS system. In order to coincide with the usual scattering theory, let λ = −iζ, then the first equation of the 2×2 AKNS system (1.48) becomes ⎛

Φx = ⎝

⎞

−iζ

p

q

iζ

⎠ Φ.

(1.261)

Suppose p, q and their derivatives with respect to x decay fast enough at infinity. Let C be the complex plane and R be the real line. Besides C+ and C− are the upper and lower half plane of C respectively, i.e., C+ = {z ∈ C | Im ζ > 0}, C− = {z ∈ C | Im ζ < 0}.

52


Property 1. For each one of the following boundary conditions, the equation (1.261) has a unique column solution ⎛

(1)

ψr (x, ζ) = R(x, ζ)e−iζx ,

lim R(x, ζ) = ⎝

x→−∞

⎞

1 0

⎠,

(1.262)

( Im ζ ≥ 0), ⎛

(2)

ψr (x, ζ) = R(x, ζ)e iζx ,

lim R(x, ζ) = ⎝

x→−∞

⎞

0 1

⎠,

(1.263)

( Im ζ ≤ 0), ⎛

(3)

ψl (x, ζ) = L(x, ζ)e

iζx

,

lim L(x, ζ) = ⎝

x→+∞

⎞

0 1

⎠,

(1.264)

( Im ζ ≥ 0), ⎛

(4)

ψl (x, ζ) = L(x, ζ)e−iζx ,

lim L(x, ζ) = ⎝

x→+∞

⎞

1 0

⎠,

(1.265)

( Im ζ ≤ 0). Moreover, ψr , ψl (resp. ψr , ψl ) are continuous for ζ ∈ C+ ∪ R (resp. z ∈ C− ∪ R), and holomorphic with respect to ζ in C+ (resp. C− ). These solutions are called Jost solutions. If ζ ∈ R, then ψl and ψl are linearly independent. Hence, there exist functions r+ (ζ), r− (ζ), r+ (ζ) and r− (ζ) such that ψr = r+ ψl + r− ψl , ψr = r+ ψl + r− ψl .

(1.266)

Considering the Wronskian determinant between ψr , ψl and the Wronskian determinant between ψr , ψl , we have Property 2. For ζ ∈ R, r− (ζ) = R1 (x, ζ)L2 (x, ζ) − R2 (x, ζ)L1 (x, ζ), 2 (x, ζ)L 1 (x, ζ) − R 1 (x, ζ)L 2 (x, ζ). r+ (ζ) = R

(1.267)

r− (ζ) can be holomorphically extended to C+ ∪ R, and r+ (ζ) can be holomorphically extended to C− ∪ R. Here R1 and R2 are two compo1, R 2, L 1, L 2 have nents of the vector R, i.e., R = (R1 , R2 )T . L1 , L2 , R the similar meanings.

53


The asymptotic properties of the four Jost solutions in Property 1 as x → ±∞ are listed in the next property. Property 3. (1) The following limits hold uniformly for ζ: ⎛

lim R(x, ζ) = ⎝

x→−∞

⎞

1

⎛

0

lim R(x, ζ) = ⎝

0

x→−∞

⎠, ⎞ ⎠,

⎛

1

⎞

lim L(x, ζ) = ⎝

0

⎠,

x→+∞

⎛

1

⎞

lim L(x, ζ) = ⎝

1

⎠,

x→+∞

0

ζ ∈ C+ ∪ R, ζ ∈ C− ∪ R, (1.268) ζ ∈ C+ ∪ R, ζ ∈ C− ∪ R.

(2) The following limits hold uniformly for ζ in a compact subset: ⎛

lim R(x, ζ) = ⎝

x→+∞

⎞

r− (ζ)

⎛

0

lim R(x, ζ) = ⎝

0

x→+∞

⎛

lim L(x, ζ) = ⎝

x→−∞

⎛

lim L(x, ζ) = ⎝

x→−∞

r+ (ζ) 0 r− (ζ) r+ (ζ) 0

⎠,

ζ ∈ C+ ,

⎞ ⎠,

ζ ∈ C− ,

⎞ ⎠,

(1.269) ζ ∈ C+ ,

⎞ ⎠,

ζ ∈ C− .

(3) The following limits hold uniformly for real ζ ∈ R: ⎛ ⎞ r− (ζ) = 0, ⎝ ⎠ lim R(x, ζ) − x→+∞ r+ (ζ)e2iζx ⎛ ⎞ − (ζ)e−2iζx r ⎠ = 0, ζ) − ⎝ lim R(x, x→+∞ r+ (ζ) ⎛ ⎞ − (ζ)e−2iζx − r ⎠ = 0, lim L(x, ζ) − ⎝ x→−∞ r− (ζ)

ζ ∈ R,

ζ ∈ R,

ζ ∈ R,

54

DARBOUX TRANSFORMATIONS IN INTEGRABLE SYSTEMS ⎛ ⎞ + (ζ) r ⎠ = 0, lim L(x, ζ) − ⎝ x→−∞ −r+ (ζ)e2iζx

ζ ∈ R.

Rewrite (1.261) as LΦ = ζΦ, where

⎛

⎞⎛

i

(1.270) ⎛

⎞⎞

0

0 p d ⎠⎝ ⎠⎠ , −⎝ L=⎝ dx 0 −i q 0

(1.271)

then (1.261) becomes a spectral problem of a linear ordinary differential operator. We consider its spectrum in L2 (R) × L2 (R). If ζ ∈ C+ and r− (ζ) = 0, then (1.267) implies that ψr and ψl are linearly dependent. Hence ψr → 0 as x → ±∞. Similarly, if ζ ∈ C− and r+ (ζ) = 0, then ψr and ψl are linearly dependent. Hence ψr → 0 as x → ±∞. Since r− (ζ) and r+ (ζ) are holomorphic in C+ and C− respectively, their zeros are discrete. These zeros are the eigenvalues of L. The set of all eigenvalues of L is denoted by IP σ(L). If ζ ∈ R, then it can be proved that (1.270) has a nontrivial bounded solution. σ(L) = R ∪ IP σ(L) is called the spectrum of the operator L. Its compliment C − σ(L) is called the regular set of L. Property 4. If r− (ζ) = 0 and r+ (ζ) = 0 hold for ζ ∈ R, then IP σ(L) is a finite set. Hereafter, we always suppose r− (ζ) = 0 and r+ (ζ) = 0 when ζ ∈ R. First we consider the eigenvalues. If ζ ∈ IP σ(L), then ψr and ψl are linearly dependent. Suppose ψr (x, ζ) = α(ζ)ψl (x, ζ)

(ζ ∈ C+ ∩ IP σ(L)),

ψl (x, ζ) ψr (x, ζ) = α(ζ)

(ζ ∈ C− ∩ IP σ(L)).

(1.272)

Denote IP σ(L) ∩ C+ = {ζ1 , · · · ζd } and IP σ(L) ∩ C− = {ζ1 , · · · ζd} to be the set of eigenvalues in C+ and C− respectively. Moreover, suppose ζ1 , · · ·, ζd are all simple zeros. Corresponding to each eigenvalue, there is a constant

dr− (ζk ) dζ d r (ζk ) + k) Ck = α(ζ dζ Ck = α(ζk )

(k = 1, · · · , d), (k = 1, · · · , d).

(1.273)

55


Using these data, we define the functions Bd (y) = −i

d

Ck e

iζk y

,

d (y) = i B

k=1

d

Ck e−iζk y .

(1.274)

k=1

Next, we consider the continuous spectrum ζ ∈ R. As is known, the Fourier transformation of a Schwarz function φ is 1 F (φ)(k) = √ 2π

+∞ −∞

φ(s)e−iks ds.

(1.275)

It can be extended to L2 (R) and becomes a bounded map from L2 (R) to L2 (R). Property 5. ⎛

L(x, ·) − ⎝

⎞

0 1

⎛

⎠ ∈ L2 (R),

Denote

L(x, ·) − ⎝

⎛

⎛

⎞

1 0

⎠ ∈ L2 (R).

⎞⎞

0 1 N (x, s) = √ F ⎝L(x, ·) − ⎝ ⎠⎠ (s), 2π 1 ⎛

(x, s) = N

then

√1 F 2π

⎛

L(x, ζ) = ⎝ ⎛ L(x, ζ) = ⎝

⎞

0 1 1 0

⎠+

⎛

⎝L(x, ·) − ⎝

+∞

⎞⎞

1 0

⎠⎠ (s),

N (x, s)e iζs ds,

0

⎞ ⎠+

+∞ 0

(1.276)

(1.277) (s ≥ 0),

∀ζ ∈ C+ ∪ R, (1.278)

(x, s)e−iζs ds, N

∀ζ ∈ C− ∪ R,

and the above integrals converge absolutely. Moreover, p(x) = −2N N1 (x, 0),

2 (x, 0), q(x) = −2N

(1.279)

where the subscripts refer to the components. For ζ ∈ R, denote b(ζ) =

r+ (ζ) , r− (ζ)

r− (ζ) . b(ζ) = r+ (ζ)

(1.280)

56


It can be proved that b, b ∈ L2 (R) ∩ L1 (R) ∩ C 0 (R). Hence, we can define

(1.281)

1 +∞ b(ζ)e iζy dζ, Bc (y) = 2π −∞ 1 +∞ Bc (y) = b(ζ)e−iζy dζ. 2π −∞

(1.282)

The data b(ζ), {ζk , Ck (k = 1, · · · , d), ζk , Ck (k = 1, · · · , d), b(ζ) (ζ ∈ R)} (1.283)

are called the scattering data corresponding to (p, q), denoted by Σ(p, q). We can also call {r− (ζ) (ζ ∈ C+ ∪ R), r+ (ζ) (ζ ∈ C− ∪ R)}

(1.284)

the scattering data, since the data in (1.283) can be obtained from the data in (1.284). =B c + B d according to (1.274) and (1.282). Define B = Bc +Bd and B satisfy the follow system of linear integral Property 6. N and N equations (Gelfand-Levitan-Marchenko equations) ⎛

⎞

1 + s) ⎝ ⎠ + N (x, s) + B(2x ⎛ (x, s) + B(2x + s) ⎝ N

0

0 1

⎞ ⎠ +

+∞

(x, σ)B(2x N + s + σ) dσ = 0,

0

+∞

N (x, σ)B(2x + s + σ) dσ = 0. 0

(1.285) If the scattering data {ζk , Ck , ζk , Ck , b(ζ), b(ζ)} are known, N and N are solved from the above integral equations and (1.279) gives (p, q). The process to get scattering data from (p, q) is called the scattering process. It needs to solve the spectral problem of ordinary differential equations. The process to get (p, q) from the scattering data is called the inverse scattering process. It needs to solve linear integral equations. Now we consider the evolution of the scattering data. In the AKNS system, p and q are functions of (x, t). Therefore, we should consider the full AKNS system (with time t) ⎛

Φx = ⎝

⎞

−iζ

p

q

iζ

⎠ Φ,

⎛

Φt = ⎝

⎞

A

B

C −A

⎠ Φ,

(1.286)

57


where A, B and C are polynomials of ζ, A= B= C=

n j=0 n j=0 n

aj (−iζ)n−j , bj (−iζ)n−j ,

(1.287)

cj (−iζ)n−j .

j=0

We also suppose A|p=q=0 = iω(ζ, t).

(1.288)

Lemma 1.5 implies B|p=q=0 = C|p=q=0 = 0. Property 7. Suppose (p, q) satisfies the equations qt = cn,x − 2qan

pt = bn,x + 2pan ,

(1.289)

given by the integrability condition, then the evolution of the corresponding scattering data is given by r− (ζ, t) = r− (ζ, 0)

ζ ∈ C+ ∪ R,

r+ (ζ, t) = r+ (ζ, 0)

ζ ∈ C− ∪ R, t

r+ (ζ, t) = r+ (ζ, 0) exp(−2i r− (ζ, t) = r− (ζ, 0) exp(2i

t0

ω(ζ, τ ) dτ )

ω(ζ, τ ) dτ ) 0

ζ ∈ R,

(1.290)

ζ ∈ R,

and ζk (t) = ζk (0), ζk (t) = ζk (0),

t

Ck (t) = Ck (0) exp(−2i Ck (t) = Ck (0) exp(2i

ω(ζ, τ ) dτ ),

t0

ω(ζ, τ ) dτ ),

t

0

ω(ζ, τ ) dτ ),

b(ζ, t) = b(ζ, 0) exp(−2i b(ζ, t) = b(ζ, 0) exp(2i

(1.291)

t

0

ω(ζ, τ ) dτ ). 0

(1.290) or (1.291) gives the evolution of the scattering data explicitly.

58


In summary, the process of solving the initial value problem of nonlinear evolution equations (1.289) of (p, q) is as follows. Here the initial condition is t = 0 : p = p0 , q = q0 . For given (p0 , q0 ), first solve the x-part of the Lax pair (1.286) for p = p0 , q = q0 and get the scattering data corresponding to p0 and q0 . Then, using the evolution of the scattering data (1.291), the scattering data corresponding to (p(t), q(t)) are obtained. Finally, solve the integral equations (1.285) to get (p(t), q(t)). Therefore, the inverse scattering method changes the initial value problem of nonlinear partial differential equations to the problem of solving systems of linear integral equations. This gives an effective way to solve the initial value problem. Especially, c = 0, (1.285) has a degenerate kernel. Hence when br = br = 0, Bc = B it can be solved algebraically and the soliton solutions can be obtained. Please see [23] for details. Remark 16 Denote Σ(p, q) to be the scattering data corresponding to (p(x, t), q(x, t)), p0 (x) and q0 (x) to be the initial values of p and q at t = 0, then the procedure of inverse scattering method can be shown in the following diagram: t = 0 : (p0 , q0 )

t=t:

(p, q)

scattering

-

Σ(p0 , q0 )

inverse scattering

?

Σ(p, q) (1.292)

For a linear equation, if “scattering” is changed to “Fourier transformation” and “inverse scattering” is changed to “inverse Fourier transformation” in the above diagram, then it becomes the diagram for solving the initial value problem by Fourier transformations which has been used extensively for linear problems. Therefore, the scattering and inverse scattering method can be regarded as a kind of Fourier method for nonlinear problems.

1.5.2

Change of scattering data under Darboux transformations for su(2) AKNS system

For the AKNS system, the scattering data include {ζk , Ck , ζk , Ck , b(ζ), b(ζ)}.

(1.293)

The su(2) AKNS system means that U, V ∈ su(2) for ζ ∈ R, i.e., q = −¯ p, ¯ Therefore, it is just the nonlinear Schr¨ A¯ = −A, C = −B. odinger

59


hierarchy. The Lax pair is

⎛

Φx = ⎝ ⎛

Φt = ⎝

⎞

−iζ

p

−¯ p

iζ

A

B

C −A

⎠ Φ, ⎞

(1.294)

⎠Φ

¯ = −A(ζ), B(ζ) ¯ = −C(ζ). Here we consider the su(2) AKNS where A(ζ) system instead of general 2 × 2 AKNS system because the Darboux transformation ⎛ ⎞ may exist globally in this case.⎛ ⎞ ¯ ζ) ¯ α(ζ) −β( ⎠ is a solution of (1.294), then ⎝ ⎠ is also its soIf ⎝ ¯ β(ζ) α ¯ (ζ) lution. This leads to the following property. Property 8. For the Lax pair (1.294), if ζ ∈ R, then there are following relations among the Jost solutions and the scattering data: 1 = −R ¯2, R 1 = L ¯ 2, L

r+ (ζ) = ¯− (ζ),

2 = R ¯1, R

(1.295)

2 = −L ¯ 1, L

r− (ζ) = ¯+ (ζ).

(1.296)

b(ζ) = ¯b(ζ)

(1.297)

By reordering the eigenvalues, d = d,

ζk = ζ¯k ,

Ck = −C¯k ,

(ζ ∈ R).

Therefore, for the su(2) AKNS system, the scattering data can be reduced to ζk ∈ C+ , Ck (k = 1, 2, · · · , d) and b(ζ) (ζ ∈ R). Now we consider the change of the scattering data under Darboux transformations. From the discussion on the nonlinear Schr¨ o¨dinger hierarchy, we know that if p is defined globally on (−∞, +∞), so is the Darboux matrix. In order to use the scattering theory, we want that p and its derivatives tend to 0 fast enough at infinity. Take a constant µ and a column solution of the Lax pair ⎛

ψr (ζζ0 ) − µψl (ζζ0 ) = ⎝

R1 (ζζ0 )e−iζ0 x − µL1 (ζζ0 )e iζ0 x R2 (ζζ0 )e−iζ0 x − µL2 (ζζ0 )e iζ0 x

⎞ ⎠

(1.298)

(ζζ0 ∈ C+ ). Let σ=

R2 (ζζ0 ) − µL2 (ζζ0 )e2iζ0 x R1 (ζζ0 ) − µL1 (ζζ0 )e2iζ0 x

(1.299)

60


be the ratio of the second and the first components. Then the Darboux matrix is − iζI − S

⎛

⎞

−iζζ0 − iζ¯0 |σ|2 (−iζζ0 + iζ¯0 )¯ σ 1 ⎝ ⎠, = −iζI − 2 1 + |σ| (−iζζ0 + iζ¯0 )σ −iζ¯0 − iζζ0 |σ|2 (1.300) and the solution is transformed by p = p + 2i

(ζ¯0 − ζ0 )¯ σ . 2 1 + |σ|

(1.301)

The change of the scattering data under Darboux transformation is given by the following theorem [75].

Theorem 1.22 If the scattering data for (1.294) are r− (ζ) (ζ ∈ C+ ∪ R), r+ (ζ) (ζ ∈ R) and α(ζk ) (k = 1, · · · , d), then, under the action of the Darboux matrix (1.300) (µ ( = 0, ζ0 ∈ C+ ), the scattering data are changed as follows: (1) If ζ0 is not an eigenvalue, then, after the action of the Darboux transformation, the number of eigenvalues increase one. All the original eigenvalues are not changed, and ζ0 is a unique additional eigenvalue. Moreover, ζ − ζ0 r− (ζ) = r− (ζ) (ζ ∈ C+ ∪ R), ζ − ζ¯0 (ζ) = r+ (ζ) r+

α (ζk ) = α(ζk )

(ζ ∈ R),

(1.302)

(k = 1, · · · , d),

α (ζζ0 ) = 1/µ, hence

ζ − ζ¯0 b(ζ) (ζ ∈ R), H ζ − ζ0 ζk − ζ¯0 Ck (k = 1, · · · , d), H Ck = ζk − ζ0 ζ0 − ζ¯0 . C0 = µr− (ζζ0 ) b (ζ) =

(1.303)

(2) If ζ0 is an eigenvalue: ζ0 = ζj , and µ = α(ζζj ), then, after the action of the Darboux transformation, ζ0 is no longer an eigenvalue.

61


Moreover, r− (ζ) =

and

ζ − ζ¯0 r− (ζ) ζ − ζ0

(ζ ∈ C+ ∪ R), (1.304)

(ζ) = r+ (ζ) r+

(ζ ∈ R),

α (ζk ) = α(ζk )

(k = 1, · · · , d, k = j),

ζ − ζ0 b(ζ) (ζ ∈ R), H ζ − ζ¯0 ζk − ζ0 Ck = Ck (k = 1, · · · , d, k = j). ζk − ζ¯0 b (ζ) =

(1.305)

Proof. (1) ζ0 ∈ IP σ(L). Then, both the numerator and denominator of (1.299) are not 0. Property 3 implies lim σ = ∞,

lim σ = 0.

x→−∞

Hence

⎛

lim (−iζI − S) = ⎝

x→−∞

⎛

lim (−iζI − S) = ⎝

⎞

−iζ + iζ¯0

0

0

−iζ + iζζ0

−iζ + iζζ0

0

0

−iζ + iζ¯0

x→+∞

(1.306)

x→+∞

⎠, ⎞

(1.307)

⎠.

Under the action of the Darboux transformation, the Jost solutions are changed to 1 (−iζI − S)ψr (x, t, ζ), −iζ + iζ¯0 1 (−iζI − S)ψl (x, t, ζ). ψl (x, t, ζ) = −iζ + iζ¯0 ψr (x, t, ζ) =

Hence R = If ζ ∈ C+ ,

1 (−iζI − S)R. −iζ + iζ¯0 ⎛

ζ − ζ0 ⎜ ζ − ζ¯ lim R = ⎝ 0 x→+∞ 0

Thus (ζ) = r−

⎞⎛

(1.309) ⎞

0 ⎟ r− (ζ) ⎠. ⎠⎝ 0 1

ζ − ζ0 r− (ζ), ζ − ζ¯0

(1.308)

(1.310)

(1.311)

62


(ζ) has an additional zero ζ than r (ζ). This means that ζ is a new r− 0 − 0 eigenvalue. For ζ ∈ R,

⎛

⎞⎛

ζ − ζ0 ⎜ ζ − ζ¯ R ∼⎝ 0 0

⎞

0 ⎟ r− (ζ) ⎠. ⎠⎝ 2iζx (ζ)e r + 1

(1.312)

ζ − ζ¯0 b(ζ). ζ − ζ0

(1.313)

(ζ) = r (ζ), and Hence r+ +

b (ζ) =

If ζk is a zero of r− (ζ), then (1.308) implies α (ζk ) = α(ζk ), and

Ck

= α (ζk )

(ζ ) dr− ζk − ζ¯0 k Ck . = dζ ζk − ζ0

(1.314)

When ζ = ζ0 , ⎛

⎞

|σ|2 −¯ σ 1 ⎝ ⎠ ψr (x, t, ζ0 ), ψr (x, t, ζ0 ) = 2 1 + |σ| −σ 1 ⎛

⎞

|σ|2 −¯ σ 1 ⎝ ⎠ ψl (x, t, ζ0 ), ψl (x, t, ζ0 ) 2 1 + |σ| −σ 1 α (ζζ0 ) =

σL1 exp(iζζ0 x) − L2 exp(iζζ0 x) 1 = , σR1 exp(−iζζ0 x) − R2 exp(−iζζ0 x) µ dr (ζζ0 ) ζ0 − ζ¯0 C0 = α (ζζ0 )/ − = . dζ µr− (ζζ0 )

(1) is proved. (2) ζ0 = ζj ∈ IP σ(L), µ = α(ζζj ). Now L2 (ζζj ) R2 (ζζj ) = , σ= R1 (ζζj ) L1 (ζζj )

(1.315)

(1.316) (1.317)

(1.318)

hence lim σ = ∞,

lim σ = 0,

x→−∞

⎛

lim (−iζI − S) = ⎝

x→−∞

⎛

lim (−iζI − S) = ⎝

x→+∞

x→+∞

(1.319) ⎞

−iζ + iζζ0

0

0

−iζ + iζ¯0

−iζ + iζ¯0

0

0

−iζ + iζζ0

⎠, ⎞ ⎠.

(1.320)

63


Under the action of the Darboux transformation, the Jost solutions become 1 (−iζI − S)ψr (x, t, ζ), −iζ + iζζ0 1 ψl (x, t, ζ) = (−iζI − S)ψl (x, t, ζ). −iζ + iζζ0 ψr (x, t, ζ) =

For ζ ∈ C+ ,

⎛

ζ − ζ¯0 ⎜ lim R = ⎝ ζ − ζ0 x→+∞ 0 and for ζ ∈ R,

⎛

ζ − ζ¯0 ⎜ R ∼ ⎝ ζ − ζ0 0 Hence (ζ) = r− (ζ) r+

⎞

0 ⎟ r− (ζ) ⎠, ⎠⎝ 0 1

⎞⎛

(1.322)

⎞

0 ⎟ r− (ζ) ⎠. ⎠⎝ 2iζx (ζ)e r + 1

ζ − ζ¯0 r− (ζ) ζ − ζ0

= r+ (ζ)

⎞⎛

(1.321)

(ζ ∈ C+ ∪ R),

(1.323)

(1.324)

(ζ ∈ R),

and

ζ − ζ0 b(ζ). (1.325) ζ − ζ¯0 From (1.324) we know that the Darboux transformation removes the eigenvalue ζ0 (= ζj ). If ζ = ζk (k = j), then ψr = α(ζk )ψl , hence b (ζ) =

α (ζk ) = α(ζk ), dr (ζ ) ζk − ζ0 − k = Ck . Ck = α (ζk ) dζ ζk − ζ¯0

(1.326)

The theorem is proved. We have given the formulae for the change of the scattering data under Darboux transformation in the su(2) case. A Darboux transformation increase or decrease the number of eigenvalues (number of solitons). However, it does not affect the scattering data related to the continuous spectrum. Thus we can use the Darboux transformation to change a general inverse scattering problem to an inverse scattering problem without eigenvalues.

64


Remark 17 For the KdV equation, q = 1 (or −1) in the Lax pair (1.286). Since q does not tend to zero at infinity, the above conclusions can not be applied directly. However, the inverse scattering theory for the KdV equation is actually simpler than the AKNS system (see [23]). The conclusions similar to Theorem 1.22 for the KdV equation holds true as well [29].

Chapter 2 2+1 DIMENSIONAL INTEGRABLE SYSTEMS

This chapter is devoted to the Darboux transformations of 2+1 dimensional integrable systems. Starting from the KP equation, we discuss the Darboux transformation for 2+1 dimensional AKNS system and more general systems. Unlike the Darboux matrices in 1+1 dimensions, the Darboux transformations here are given by differential operators (called Darboux operators). The construction of the Darboux operators is uniform to all the equations in the system, as in the 1+1 dimensional case. The binary Darboux transformation, which is a kind of Darboux transformation in integral form, is introduced briefly. Explicit solutions of the DSI equation can be obtained by the combination of Darboux transformation and binary Darboux transformation. Moreover, the nonlinear constraint method is used to separate the differentials in the 2+1 dimensional AKNS system so that the Darboux transformation in 1+1 dimensions can be used to get the localized soliton solutions.

2.1

KP equation and its Darboux transformation

A 2+1 dimensional integrable system has three independent variables (x, y, t) where x and y usually refer to space variables and t refers to time variable. A typical 2+1 dimensional integrable partial differential equation is the Kadomtsev-Petviashvili equation (KP equation) [68] uxt = (uxxx + 6uux )x + 3α2 uyy ,

(2.1)

where α = ±1 or ±i. (2.1) is called the KPI equation if α = ±1, and the KPII equation if α = ±i. The KP equation is the natural generalization

66


of the KdV equation, which describes the motion of two dimensional water wave. (2.1) can also be written as vxt = vxxxx + 6vx vxx + 3α2 vyy

(2.2)

where v satisfies vx = u. The KP equation has a Lax pair φy = α−1 φxx + α−1 uφ,

(2.3)

φt = 4φxxx + 6uφx + 3(αvy + ux )φ.

(2.4)

We first derive φyt by differentiating (2.3) with respect to t and inserting the expression of φt . Also, we can derive φty by differentiating (2.4) with respect to y and inserting the expression of φy . The equality φyt = φty is equivalent to (2.1) when φ = 0. The proof of this fact is direct, which is left for the reader. Therefore, (2.1) is the integrability condition of the overdetermined system (2.3) and (2.4). The Darboux transformation for the KP equation is similar with that for the KdV equation. It can be constructed as follows. Let h be a solution of the Lax pair (2.3) and (2.4). For any solution φ of (2.3) and (2.4), define (2.5) φ = φx − (hx /h)φ, then φ is a solution of φy = α−1 φxx + α−1 u φ , φt = 4φxxx + 6u φx + 3(αvy + ux )φ where

u = u + 2(hx /h)x ,

v = v + 2hx /h.

(2.6)

(2.7)

Comparing (2.6) with (2.3) and (2.4), the only difference is that (u, φ) is changed to (u , φ ). Hence (2.7) gives a new solution u of the KP equation [77]. Similar to 1+1 dimensions, if the seed solution u is simple enough, we can solve the Lax pair (2.3) and (2.4) to get h, then (2.5) gives a more complicated solution of the KP equation. Especially, if u = v = 0, then (2.3) and (2.4) becomes φy = α−1 φxx (2.8) φt = 4φxxx . Therefore, for any solution h of (2.8) with h = 0, u = 2(hx /h)x gives a solution of the KP equation.

67


Example 2.1 For α = 1, h can be chosen as 2 y+4λ3 t

h = eλx+λ

+ 1,

(2.9)

where λ is a real constant, then u =

1 λ2 sech2 (λx + λ2 y + 4λ3 t) 2 2

(2.10)

is a solution of the KPI equation.

Example 2.2 For α = −i, let 2 y+4λ3 t

h = eλx+iλ

¯

¯ 2 y−4λ ¯3 t

+ e−λx+iλ

,

(2.11)

where λ = a + bi is a complex constant, then we obtain a solution of the KPII equation: u = 2a2 sech2 (ax − 2aby + 4(a3 − 3ab2 )t).

(2.12)

These two solutions are both travelling waves, i.e., they are of form = f (t + a1 x + a2 y) and u is invariant along the line t + a1 x + a2 y = constant on the (x, y) plane. For fixed t, u is a non-zero constant along certain lines (for KPI, they are λx+λ2 y = constant, while for KPII, they are ax − 2aby = constant), and u tends to zero exponentially at infinity along other lines. Hence the region where u is far from zero forms a band on the (x, y) plane. This kind of solutions are call “line-solitons”. This does not happen in 1+1 dimensions. Suppose we have known a solution u of the KP equation and a set of solutions {φ} of the corresponding Lax pair. Let h be a special φ, then u = u + 2(ln h)xx is a solution of the KP equation. Moreover, φ = φx − (hx /h)φ gives the set of solutions of the Lax pair for u . Now we take a special φ as h , then we can obtain another solution u = u + 2(ln h )xx of the KP equation and the solution φ = φx − (hx /h )φ of the corresponding Lax pair by constructing Darboux transformation with h . Continuing this procedure, we obtain a series of solutions of the KP equation without solving differential equations. Except the first step, this algorithm can be realized by algebraic computation and differentiations. Therefore, it can be done by symbolic calculation. The solutions are global for all (x, y, t) if h, h , h · · · do not equal zero. This process can be expressed as u

(u, φ) −→ (u , φ ) −→ (u , φ ) −→ · · · .

(2.13)

The differential operator of order three on the right hand side of (2.4) can be changed to differential operators of arbitrary order, then we get

68


the KP hierarchy φy = α−1 φxx + α−1 uφ, φt =

n

(2.14)

vn−j ∂ j φ,

j=0

(∂ = ∂/∂x). Computing the integrability condition of (2.14) and letting all the coefficients of the derivatives of φ with respect to x be zero, we have 2vj+1,x = αvj,y − vj,xx +

j−1

j−k Cnn−j u, −k vk ∂

(2.15)

k=0

ut = αvn,y − vn,xx +

n−1

vk ∂ n−k u.

(2.16)

k=0

In (2.15), vj+1 can be solved by integration. Unlike the 1+1 dimensional systems such as the KdV hierarchy, here, in general, vj ’s can not be expressed as differential polynomials of u. Therefore, (u, v1 , · · · , vn ) are regarded as a set of unknowns of (2.15) – (2.16). The Darboux transformation is still valid for this system. In practical problems, some additional relations among (v1 , · · · , vn , u) should be satisfied. This is called a reduction of the original one. In that case, we should choose proper h so that the relations among v1 , · · · , vn , u keeps after the Darboux transformation. Usually this is a difficult problem and some special cases can be solved by certain techniques.

2.2

2+1 dimensional AKNS system and DS equation

2+1 dimensional AKNS system is Φy = JΦx + P Φ,

Φt =

n

Vn−j ∂ j Φ,

(2.17)

j=0

where J is an N × N constant diagonal matrix, P (x, y, t) is an offdiagonal N × N matrix, Vj (x, y, t)’s are also N × N matrices, ∂ = ∂/∂x. For simplicity, we assume that the diagonal entries of J are distinct. Moreover, we consider the non-degenerate Φ only. The integrability condition of (2.17) leads to off off Vj,x − [P, Vj ]off + [J, Vjoff +1 ] = Vj,y − JV

j−1 k=0

Cnn−j Vk ∂ j−k P )off ,(2.18) −k (V

69

2+1 dimensional integrable systems j−1

diag diag Vj,y − JV Vj,x = [P, Vjoff ]diag −

off off − JV Vn,x − [P, Vn ]off + Pt = Vn,y

Cnn−j Vk ∂ j−k P )diag , −k (V

k=0 n−1

(V Vk ∂ n−k P )off .

(2.19) (2.20)

k=0

Here the superscripts “diag” and “off” refer to the diagonal and offdiagonal part of a matrix respectively. Usually Vj ’s are not differential polynomials of P . But they can be generated from P by differentiation and integration with respect to x. (2.19) and (2.20) are regarded as a system of partial differential equations for P and Vjdiag ’s (j = 0, 1, · · · , n) where Vjoff ’s (j = 1, · · · , n) are determined by (2.18). (2.17) is the Lax pair of this system of equations. A typical equation in 2+1 dimensional AKNS system is the DaveyStewartson equation (DS equation), which is the natural generalization of the nonlinear Schr¨ odinger equation in 2+1 dimensions. Take N = 2, n = 2 in (2.17) and let ⎛

⎞

1

J = α−1 ⎝

0

0 −1

⎠,

α = ±1 or ± i,

⎛

P =⎝

⎞

0

−¯ u 0

⎠,

= ±1, (2.21)

V0 = 2iαJ,

V1 = 2iαP,

V2 = iα ⎝

w1

ux + αuy

−¯ ux + α¯ uy

w2

⎛

u

⎞ ⎠.

where u, w1 , w2 are complex-valued functions, u ¯ is the complex conjugate of u. Then, (2.17) becomes ⎛

Φy = α−1 ⎝ ⎛

Φt = 2i ⎝

⎞

1

0

0 −1

⎠ Φx + ⎝

⎞

1

0

0 −1

⎛

⎞

0

−¯ u 0

⎠ Φ,

⎛

⎠ Φxx + 2iα ⎝

⎛

+iα ⎝

u 0

⎞

u

−¯ u 0

⎠ Φx

⎞

w1

ux + αuy

−¯ ux + α¯ uy

w2

⎠ Φ.

(2.22)

70


(2.19) and (2.20) lead to iut = −uxx − α2 uyy − αu(w1 − w2 ), w1,y − α−1 w1,x = (|u|2 )x + α(|u|2 )y ,

(2.23)

w2,y + α−1 w2,x = (|u|2 )x − α(|u|2 )y , and w2 − w1 = α−2 (w2 − w1 ). Denote v = −|u|2 +

1 (w1 − w2 ), 2α

(2.24)

then (2.23) becomes iut = −uxx − α2 uyy − 2α2 |u|2 u − 2α2 uv,

(2.25)

vxx − α2 vyy + 2(|u|2 )xx = 0. (2.25) is called the DSI equation if = 1, α = ±1, and DSII equation if = 1, α = ±i. They describe the motion of long wave and short wave in the water of finite depth [20].

2.3 Darboux transformation 2.3.1 General Lax pair Similar to the KP equation, we also want to construct the Darboux transformation for the AKNS system. Here we first discuss the Darboux transformation for the following more general Lax pair without any reductions. Consider Lax pair Φy = U (x, y, t, ∂)Φ, where U (x, y, t, ∂) = V (x, y, t, ∂) =

Φt = V (x, y, t, ∂)Φ, m j=0 n

(2.26)

Um−j (x, y, t)∂ j , (2.27) Vn−j (x, y, t)∂

j

j=0

are differential operators with respect to x whose coefficients Uj ’s and Vj ’s are N × N matrices. For simplicity, we write U (x, y, t, ∂) = U (∂), V (x, y, t, ∂) = V (∂). Φyt = Φty can be obtained by differentiating the first equation of (2.26) with respect to t or by differentiating the second equation with

71


respect to x. Let these two equal, we get Ut (∂) − Vy (∂) + [U (∂), V (∂)] = 0.

(2.28)

(2.26) is called integrable if (2.28) holds. (2.28) is the generalization of the zero-curvature equations in 1+1 dimensions. It gives a system of partial differential equations by equating all the coefficients of ∂ to be zero. Remark 18 The existence and uniqueness of the solutions of a system of partial differential equations are very difficult problems. The local solvability of a system of linear partial differential equations have been studied by many authors. In the present case, even (2.28) holds, local solution of (2.26) near t = t0 , y = y0 with initial data Φ(t0 , x, y0 ) = Φ0 (x) may not exist. However, if each set of equations in (2.26) is locally solvable and the solutions are smooth enough with respect to the parameters y and t, U (∂) and V (∂) satisfy (2.28), then (2.26) is locally solvable. This follows from the following consideration. Suppose that the initial data (x0 , y0 , t0 , Φ0 (x)) are given. First solve the first set of equations of (2.26) at t = t0 with initial value Φ1 (x, y0 ) = Φ0 (x) and get the solution Φ1 (x, y). Using Φ1 (x, y) as the initial value, solve the second set of equations of (2.26) for fixed y and get the solution Φ(x, y, t). Using (2.28) and the second equation of (2.26), we have (Φy − U (∂)Φ)t = V (∂)(Φy − U (∂)Φ).

(2.29)

Therefore, Φy = U (∂)Φ holds identically near (x0 , y0 , t0 ) by the uniqueness of the solution. No matter whether the existence and uniqueness hold, (2.28) is called the integrability condition of (2.26). It gives a system of nonlinear partial differential equations of U (∂) and V (∂). (2.26) is called the Lax pair of this system of nonlinear partial differential equations. It is interesting to see that we can apply Darboux transformation as well provided that the set of solutions of (2.26) is not empty.

2.3.2

Darboux transformation of degree one

Similar to 1+1 dimensional case, we can define Darboux operator for the integrable nonlinear partial differential equations (2.28) and there Lax pair (2.26).

Definition 2.3 A differential operator D(x, y, t, ∂) with respect to x is called a Darboux operator for (2.26) if there exist differential operators U (∂) and V (∂) with respect to x such that for any solution Φ of (2.26), Φ = D(∂)Φ satisfies Φy = U (∂)Φ ,

Φt = V (∂)Φ .

(2.30)

72


The transformation (Φ, U (∂), V (∂)) → (Φ , U (∂), V (∂)) given by D(∂) is called a Darboux transformation. Substituting Φ = DΦ into (2.30), we have Dy (∂) = U (∂)D(∂) − D(∂)U (∂), Dt (∂) = V (∂)D(∂) − D(∂)V (∂), and

Ut (∂) − Vy (∂) + [U (∂), V (∂)] = 0.

(2.31)

(2.32)

(2.31) is the necessary and sufficient condition for D(∂) being a Darboux operator. Hence, if (U (∂), V (∂)) is a solution of (2.28), so is (U (∂), V (∂)). This means that the Darboux transformation gives a new solution of (2.28). Our main task is to construct the solution D of (2.31). We first discuss the most fundamental Darboux operator, the Darboux operator of degree one. This is the Darboux operator in the form D(x, y, t, ∂) = ∂ − S(x, y, t). The Darboux operator of higher degree will be discussed later. In order to get the general construction of S, we first derive the equations that S should satisfy. For a matrix M (x), we define a sequence of matrices M (j) by M (0) = I and (2.33) M (j+1) = Mx(j) + M (j) M, then, for any solution Φ of the equation Φx = M Φ, ∂ j Φ = M (j) Φ holds. For any differential operator U (∂) =

k

Uk−j ∂ j ,

V (∂) =

j=0

k

Vk−j ∂ j

(2.34)

Vk−j S (j) .

(2.35)

j=0

and an N × N matrix S, we define U (S) =

k j=0

Uk−j S (j) ,

V (S) =

k j=0

Suppose that Φ satisfies Φx = SΦ, then U (∂)Φ = U (S)Φ, V (∂)Φ = V (S)Φ. Notice that U (S) and V (S) are not given by replacing ∂ in U (∂) and V (∂) with S. Actually they are obtained by replacing ∂ j with S (j) .

Theorem 2.4 ∂ − S is a Darboux operator for (2.26) if and only if S satisfies Sy + [S, U (S)] = (U (S))x , (2.36) St + [S, V (S)] = (V (S))x .

73


Proof. Suppose ∂ − S is a Darboux operator for (2.26), then the first equation of (2.31) is Sy − (∂ − S)U (∂) + U (∂)(∂ − S) = 0.

(2.37)

Let Ψ be the fundamental solution of Ψx = SΨ, then Sy Ψ = (∂ − S)U (S)Ψ = (U (S))x Ψ − [S, U (S)]Ψ.

(2.38)

This gives the first equation of (2.36). The second equation is derived similarly. The necessity of (2.36) is proved. Conversely, suppose S is a solution of (2.36). Define U (∂) =

m

j Um −j ∂ ,

(2.39)

j=0

where Uj ’s are determined recursively by U0 = U0 , Uj +1

= Uj +1 + Uj,x − SU Uj +

j

m−j j−k Cm S. −k Uk ∂

(2.40)

k=0

Then

Sy − (∂ − S)U (∂) + U (∂)(∂ − S)

(2.41)

does not contain any terms with ∂, i.e., it is a matrix-valued function of x, y and t. On the other hand, for any fundamental solution Φ of Ψx = SΨ, (2.36) leads to (S Sy − (∂ − S)U (∂) + U (∂)(∂ − S))Ψ = 0.

(2.42)

Hence, as a matrix, Sy − (∂ − S)U (∂) + U (∂)(∂ − S) = 0.

(2.43)

This shows that ∂ − S satisfies the first equation of (2.31). The second one can be proved similarly. Therefore, ∂ − S is a Darboux operator for (2.26). The theorem is proved.

Theorem 2.5 ∂ −S is a Darboux operator for (2.26) if and only if there exists an N × N non-degenerate matrix solution H of (2.26) such that S = Hx H −1 [122].

74


Proof. First we prove the sufficiency, i.e., to show the S satisfies (2.36). From (2.26), Hy H −1 = (U (S)H)x H −1 − SU (S) Sy = Hxy H −1 − SH

(2.44)

= [U (S), S] + (U (S))x , which is the first equation of (2.36). The second one is derived in the same way. Now suppose S satisfies (2.36). We shall show that the system of equations Hx = SH,

Hy = U (∂)H,

Ht = V (∂)H,

(2.45)

Ht = V (S)H.

(2.46)

has a solution. Clearly (2.45) is equivalent to Hx = SH,

Hy = U (S)H,

Hence, we only need to verify the integrability conditions of (2.46). Let Ψ be a fundamental solution of Ψx = SΨ. (2.37) implies (Ψy − U (∂)Ψ)x = (SΨ)y − ∂U (∂)Ψ = S(Ψy − U (∂)Ψ).

(2.47)

Hence (V Vy (∂) + V (∂)U (∂))Ψ = (V (∂)Ψ)y − V (∂)(Ψy − U (∂)Ψ) = (V (S)Ψ)y − V (S)(Ψy − U (S)Ψ)

(2.48)

= V (S)y Ψ + V (S)U (S)Ψ. Similarly, (U Ut (∂) + U (∂)V (∂))Ψ = U (S)t Ψ + U (S)V (S)Ψ.

(2.49)

Since det Ψ = 0, the integrability condition (2.28) gives U (S)t − V (S)y + [U (S), V (S)] = 0.

(2.50)

Hence the integrability condition Hyt = Hty for (2.46) holds. Theorem 2.4 gives the other two integrability conditions Hxy = Hyx and Hxt = Htx . Hence (2.46) is integrable. For given initial value H = H0 at (t, x, y) = (t0 , x0 , y0 ), (2.46) has a solution H. If H0 is nondegenerate, H is also non-degenerate in a neighborhood of (t0 , x0 , y0 ). That is, (2.46) has a non-degenerate matrix solution H such that S = Hx H −1 . The theorem is proved.

75


If there is no reduction, this theorem shows that any Darboux operator in the form ∂ − S can be expressed explicitly by the solutions of the Lax pair. Darboux transformation exists as long as the Lax pair has a nondegenerate N × N matrix solution. Under the Darboux transformation, Uj is transformed to Uj given by (2.40). Vj ’s have similar expressions. Thus, we have constructed the Darboux transformation (U, V, Φ) −→ (U , V , Φ ).

(2.51)

This process can be continued by algebraic and differential operations to get infinite number of solutions provided that the set of solutions of the Lax pair for the seed solution is big enough. For the AKNS system (2.17), the action of the Darboux operator ∂ −S gives (2.52) (∂ − S)(J∂ + P ) − Sx = (J∂ + P )(∂ − S). The coefficients of ∂ 2 on both sides are equal. Comparing the coefficient of ∂, we have P = P + [J, S]. (2.53) For practical problems, the entries of U and V often have some constraint relations. In that case, H in the theorem should also satisfy certain conditions so that (U , V ) and (U, V ) satisfy the same constraints. If so, we can obtain a transformation from a solution of a nonlinear partial differential equation to a solution of the same equation. Remark 19 For the KP equation, the construction for the Darboux operator is completely the same as in Section 2.1. However, for the DaveyStewartson equation, it is more difficult because we should consider the relations among the entries of P . We shall discuss it in Section 2.4. Similar with the 1+1 dimensional case, we can also compose several Darboux transformations of degree one to a Darboux transformation of higher degree. However, they can be constructed directly with explicit formulae.

2.3.3

Darboux transformation of higher degree and the theorem of permutability

Now we discuss a Darboux operator of higher degree. It is a differential operator in the form D(∂) =

r j=0

Dr−j ∂ j ,

D0 = I

(2.54)

76


such that

Dy (∂) = U (∂)D(∂) − D(∂)U (∂), Dt (∂) = V (∂)D(∂) − D(∂)V (∂).

(2.55)

Here U (∂) and V (∂) are differential operators with respect to x. For simplicity, we only discuss the Darboux operator of degree two. When r > 2, the Darboux operator can also be written down explicitly, but is more complicated.

Theorem 2.6 Let H1 and H2 be two N × N non-degenerate matrix solutions of (2.26). Let F be the block matrix ⎛ ⎝

⎞

H1

H2

∂H1 ∂H H2

⎠.

(2.56)

Suppose det F = 0, then the following conclusions hold: (1) There is a unique differential operator of degree two D(H1 , H2 , ∂) = ∂ 2 + D1 ∂ + D2

(2.57)

satisfying Hi = 0 D(H1 , H2 , ∂)H

(i = 1, 2).

(2.58)

It is a Darboux operator. (2) The theorem of permutability holds: H2 , H1 , ∂). D(H1 , H2 , ∂) = D(H

(2.59)

(3) There is a decomposition H2 , ∂)D(H1 , ∂). D(H1 , H2 , ∂) = D(D(H1 , ∂)H

(2.60)

Proof. Since det F = 0, the linear algebraic system D1 ∂H1 + D2 H1 = −∂ 2 H1 ,

D1 ∂H H2 + D2 H2 = −∂ 2 H2

(2.61)

for D1 , D2 has a unique solution, which determines D(∂) uniquely and D(∂) satisfies (2.58). Since (2.61) is symmetric with respect to H1 and H2 , (2) holds. By the definitions of D(D(H1 , ∂)H H2 , ∂) and D(H1 , ∂), D(D(H1 , ∂)H H2 , ∂)D(H1 , ∂)H1 = 0, D(D(H1 , ∂)H H2 , ∂)D(H1 , ∂)H H2 = 0.

(2.62)

77


Hence (2.60) holds. From (2.60) it is seen that D(H1 , H2 , ∂) is a Darboux operator because it is the composition of two Darboux operators of degree one. Similar to (1.134), the theorem of permutability can be expressed by the following diagram: (U (1) , V (1) , Φ(1) ) H HH2 * H1 H j H (1,2) (U , V (1,2) , Φ(1,2) ) (U, V, Φ) (U (2,1) , V (2,1) , Φ(2,1) ) H HH j H

H2

*

(U (2) , V (2) , Φ(2) ) H1

(2.63)

Example 2.7 For the KP equation, N = 1, we can get the expression of u after the Darboux transformation. Denote Hi = hi . Suppose the Darboux operator is r

Dr−j ∂ j ,

(2.64)

Dr−j ∂ j hi = −∂ r hi ,

(2.65)

j=0

then Theorem 2.6 implies r−1 j=0

i.e., Fr = −(∂ r h1 , · · · , ∂ r hr ). (Dr , · · · , D1 )F

(2.66)

Solving this system, we have ⎛

D1

h ∂h1 ⎜ 1 .. ⎜ .. = − det ⎜ . . ⎝ ⎛

⎛

hr ∂hr

h ∂h1 ⎜ ⎜ 1 .. ⎜ ⎜ .. · ⎜det ⎜ . . ⎝ ⎝ hr ∂hr = −(ln det Fr )x .

· · · ∂ r−2 h1 ∂ r h1 .. .. .. . . . · · · ∂ r−2 hr ∂ r hr · · · ∂ r−2 h1 ∂ r−1 h1 .. .. .. . . . · · · ∂ r−2 hr ∂ r−1 hr

⎞ ⎟ ⎟ ⎟· ⎠ ⎞⎞−1 ⎟⎟ ⎟⎟ ⎟⎟ ⎠⎠

(2.67)

78


Therefore, for the KP equation, the transformation between two solutions is u = u − 2D1,x = u + 2(ln det Fr )xx .

(2.68)

Many solutions can be obtained in this way [77, 80, 89].

2.4

Darboux transformation and binary Darboux transformation for DS equation 2.4.1 Darboux transformation for DSII equation In Section 2.2 we introduced the DSI and DSII equations (2.25) and their Lax pairs (2.17) and (2.21). Since the reductions in DSI equation and in DSII equation are different, the method of solving these two equations are also quite different. First, consider the DSII equation, i.e., = 1, α = −i [120]. i ¯1 . Hence v = −|u|2 + (w1 − w ¯1 ), In this case, we should have w2 = w 2 and J, P , Vj (j = 0, 1, 2) are ⎛

J = i⎝ V0 = 2J,

⎞

1

0

0 −1

⎠,

V1 = 2P,

⎛

P =⎝

⎞

0

u

−¯ u 0

⎠,

⎛

V2 = ⎝

w1

ux − iuy

−¯ ux − i¯ uy

w ¯1

⎞

(2.69)

⎠.

J, P and Vj have the properties J¯ = σJσ −1 , where

P¯ = σP σ −1 ,

V¯j = σV Vj σ −1 ⎛

σ=⎝

(j = 0, 1, 2),

(2.70)

⎞

0

1

−1 0

⎠,

(2.71)

P¯ is the matrix each of whose entry is the complex conjugate of the corresponding entry of P . Now (2.25) becomes iut = −uxx + uyy + 2|u|2 u + 2uv, vxx + vyy + 2(|u|2 )xx = 0.

(2.72)

79


Its Lax pair is ⎛

Φy =

⎞

1

i⎝

0

0 −1

⎠ Φx + ⎝

⎛

Φt = 2i ⎝ ⎛

+⎝

1

⎛

0

0 −1

⎞

⎞

0

u

−¯ u 0

⎠ Φ,

⎛

⎠ Φxx + 2 ⎝

w1

ux − iuy

−¯ ux − i¯ uy

w ¯1

0

⎞

u

−¯ u 0

⎠ Φx

(2.73)

⎞

⎠ Φ,

with i v = −|u|2 + (w1 − w ¯1 ). 2

(2.74)

The Darboux operator for (2.73) is constructed as follows. ¯ T is also its soluSuppose (ξ, η)T is a solution of (2.73), then (−¯ η , ξ) tion. Hence we can choose ⎛

⎞

ξ −¯ η ⎠, H=⎝ η ξ¯ ⎛

S = Hx H −1

(2.75) ⎞

¯ x + η η¯x ηξ ξξ ¯ x − ξ η¯x 1 ⎝ ⎠. = 2 2 ¯ x − η ξ¯x ξ ξ¯x + η¯ηx |ξ| + |η| ξη

(2.76)

¯ = σHσ −1 , we have S¯ = σSσ −1 . The equations Since H U (∂)(∂ − S) = (∂ − S)U (∂) − Sy , V (∂)(∂ − S) = (∂ − S)V (∂) − St

(2.77)

imply ¯ = σU σ −1 , U V¯ = σV σ −1 .

(2.78)

This means that the Darboux transformation keeps the reduction relations (2.70) invariant. After the action of the Darboux operator ∂ − S, P = P + [J, S], V2 = V2 + V1,x + 2V V0 Sx + [V V0 , S]S + [V V1 , S].

(2.79)

80


Figure 2.1.

Single line-soliton, t = 0

Hence the new solution of the DSII equation is u = u + 2iS12 = u + 2i

ηξ ¯ x − ξ η¯x , |ξ|2 + |η|2

(2.80)

v = v − 2( Re S11 )x = v − (ln(|ξ| + |η| ))xx . 2

2

Example 2.8 Take the seed solution u = 0, then we can choose v = 0 ((w1 = 0), ξ = ξ(x + iy, t), η = η(x − iy, t) (i.e., ξ is analytic with respect to x + iy and η is analytic with respect to x − iy) satisfying ξt = 2iξxx , ηt = −2iηxx . For these (ξ, η), (u, v) given by (2.80) are all solutions of 2 ¯ ¯ ¯2 DSII equation. Especially, let ξ = eαx+iαy+2iα t , η = eβx−iβy−2iβ t , then 2

2

2i(α − β)e(α+β)x+i(α+β)y+2i(α +β )t , u = 2Reαx−2Imαy−2Im(α2 )t e + e2Reβx−2Imβy−2Im(β 2 )t 2 2 4( Re α − Re β)2 e2Re(α+β)x−2Im(α+β)y−2Im(α +β )t v = − 2Reαx−2Imαy−2Im(α2 )t . (e + e2Reβx−2Imβy−2Im(β 2 )t )2

(2.81)

When t is fixed, the solution u is a constant along the line with slope x : y = Im(β − α) : Re(β − α), and tends to zero in any other directions. This kind of solution also belongs to “line-soliton”. Multi-line-solitons can be obtained by successive Darboux transformations. They tend to zero at infinity except for finitely many directions. Figures 2.1 – 2.4 show the single line-soliton and multi-line-solitons, where the parameters are α1 = 3 + 2i, β1 = 1 + i, α2 = i, β2 = (2 + i)/4. (For the single line-soliton, only (α1 , β1 ) is used.)


Figure 2.2.

Figure 2.3.

Figure 2.4.

Double line-soliton, t = 0

Double line-soliton, t = 0.5

Double line-soliton, t = 1

81

82


Remark 20 Comparing to the general method discussed in the last section, the key point in the construction of Darboux transformation for the DSII equation is the choice of H in (2.75). Although it is successful to the DSII equation, it can not be applied to the DSI equation.

2.4.2

Darboux transformation and binary Darboux transformation for DSI equation

When = 1 and α = 1, (2.25) becomes iut + uxx + uyy + 2|u|2 u + 2uv = 0, vxx − vyy + 2(|u|2 )xx = 0,

(2.82)

and its Lax pair is ⎛

Φy = ⎝

⎞

1

0

0 −1

Φt = 2i ⎝ ⎛

+i ⎝

⎞

0

⎠ Φx + ⎝

⎛

1

⎛

0 −1

⎠ Φ,

−¯ u 0

⎞

0

u ⎛

⎠ Φxx + 2i ⎝

⎞

0

u

−¯ u 0

⎠ Φx

(2.83)

⎞

w1

ux + uy

−¯ ux + u ¯y

w2

⎠ Φ,

where

1 (2.84) v = −|u|2 + (w1 − w2 ), 2 w1 and w2 are real functions. Since we can not find the solution H of (2.83) like that of (2.75) to construct the Darboux transformation for (2.82), we should use the binary Darboux transformation. The binary Darboux transformation was first introduced by V. B. Matveev et al and has many applications [4, 81, 80, 72, 125, 126]. Here we show its application to DSI equation for constructing new solutions. For the general case, please refer to [80]. For simplicity, rewrite (2.83) as Φy = JΦx + P Φ, V2 Φ. Φt = 2iJΦxx + 2iP Φx + iV

(2.85)

Apart from this Lax pair, consider its adjoint equations Ψy = Ψx J − ΨP, Ψt = −2iΨxx J + 2i(ΨP )x − iΨV V2 .

(2.86)

83


With P ∗ = −P and V2∗ = V2 − 2P Px , we know that if Φ is a solution of ∗ (2.85), then Ψ = Φ is a solution of (2.86), and vice versa. Therefore, as soon as a solution of (2.85) or (2.86) is known, a solution of its adjoint equations is also known. Similar to Section 2.3, we first take Darboux transformation for the adjoint equation (2.86): Ψ = Ψx − ΨS, P = P + [J, S], V2 where

(2.87)

= V2 − 2P Px + 2[P, S] − 2S[J, S] + 2(JS Sx + Sx J) S = Ψ−1 0 Ψ0,x ,

(2.88)

Ψ0 is a non-degenerate 2 × 2 matrix solution of (2.86). Notice that P does not satisfy P ∗ = −P , and Ψ∗ is not a solution of (2.85) with P replaced by P . In order to preserve the reduction, the binary Darboux transformation is a useful tool. To get a new solution of the DSI equation, it needs the following steps: Step 1: For a solution Φ of (2.85) and a solution Ψ of the adjoint equations (2.86), define 1-form ω(Ψ, Φ) = ΨΦ dx + ΨJΦ dy + 2i(ΨP Φ + ΨJΦx − Ψx JΦ) dt.

(2.89)

It can be verified that ω(Ψ, Φ) is a closed 1-form, that is, its exterior differential dω(Ψ, Φ) = 0. Hence, in a simply connected region, the integral of ω along any closed curve is zero. In R2,1 , define (x,y,t)

Ω(Ψ, Φ)(x, y, t) =

ω(Ψ, Φ),

(2.90)

(x0 ,y0 ,t0 )

which is independent of the path of integration, and ω(Ψ, Φ) = dΩ(Ψ, Φ). Step 2: Let Φ = Ψ−1 0 Ω(Ψ0 , Φ), then we can verify that Φ is a solution of (2.85) with (P, V2 ) replaced by (P , V2 ). ∗ Step 3: Let Φ0 = Ψ−1 0 Ω(Ψ0 , Ψ0 ). Acting the Darboux operator ∂ − −1 Φ0 Φ0 on Φ , we get the Darboux transformation ∗ ∗ −1 Φ = Φx − Φ0,x Φ−1 0 Φ = Φ − Ψ0 Ω(Ψ0 , Ψ0 ) Ω(Ψ0 , Φ),

∗ ∗ −1 P = P + [J, Ψ0,x Ψ−1 0 ] = P + [J, Ψ0 Ω(Ψ0 , Ψ0 ) Ψ0 ].

(2.91)

P ∗ = −P leads to P ∗ = −P . Therefore, we get a new solution of the DSI equation.

84


The process in Step 1 and Step 2 is called a binary Darboux transformation. For the DSI equation, a new solution is obtained by the composition of a Darboux transformation and a binary Darboux transformation. It needs differentiation and integration in this procedure.

2.5

Application to 1+1 dimensional Gelfand-Dickey system

In this section, we use Theorem 2.5 to discuss the Darboux transformation for the (1+1 dimensional) Gelfand-Dickey system λΦ = U (x, t, ∂)Φ,

(2.92)

Φt = V (x, t, ∂)Φ where U (∂) = V (∂) =

m j=0 n

Um−j (x, t)∂ j , (2.93) j

Vn−j (x, t)∂ .

j=0

From the first equation of (2.92), we can compute Φt and it should be the same as that given by the second equation of of (2.92). This gives the integrability condition Ut (∂) + [U (∂), V (∂)] = 0

(2.94)

of (2.92). Let D(x, t, ∂) be a differential operator. If for any solution Φ of (2.92), Φ = D(∂)Φ satisfies λΦ = U (x, t, ∂)Φ , (2.95) Φt = V (x, t, ∂)Φ , where U and V are differential operators of the form

U (∂) =

V (∂) =

m j=0 n

j Um −j (x, t)∂ ,

(2.96) Vn −j (x, t)∂ j ,

j=0

then D(x, t, ∂) is called a Darboux operator for (2.92). For a differential operator D(x, t, ∂) = ∂ − S(x, t), we have the following theorem.


85

Theorem 2.9 ∂ − S(x, t) is a Darboux operator for (2.92) if and only if S = Hx H −1 , where H is an N × N non-degenerate matrix solution of HΛ = U (∂)H, Ht = V (∂)H,

(2.97)

and Λ is a constant upper-triangular matrix. Proof. Introduce a new variable y and consider the system Ψy = U (x, t, ∂)Ψ, Ψt = V (x, t, ∂)Ψ.

(2.98)

If ∂ − S(x, t) is a Darboux operator for (2.92), then there exist U (∂) and V (∂) such that 0 = (∂ − S)U (∂) − U (∂)(∂ − S), St = (∂ − S)V (∂) − V (∂)(∂ − S).

(2.99)

Since S is independent of y and (2.37) holds, ∂ −S is a Darboux operator for (2.98) which is independent of y. According to Theorem 2.5, there exists an N × N non-degenerate matrix solution H0 of (2.98) such that S = H0,x H0−1 . Here H0 may depend on y. Let L0 = H0−1 H0,y , (2.36) leads to L0 = H0−1 U (S)H H0 , H0−1 H0,x H0−1 U (S)H H0 + H0−1 (U (S))x H0 L0,x = −H +H H0−1 U (S)H H0,x = H0−1 {(U (S))x − [S, U (S)]}H H0 = 0, and (2.50) leads to H0−1 H0,y H0−1 U (S)H H0 + H0−1 U (S)H H0,y = 0, L0,y = −H H0−1 H0,t H0−1 U (S)H H0 + H0−1 (U (S))t H0 L0,t = −H +H H0−1 U (S)H H0,t = H0−1 {(U (S))t + [U (S), V (S)]}H H0 = 0. Hence L0 is a constant matrix. Therefore, there exists a constant uppertriangular matrix Λ and a constant matrix T such that L0 = T ΛT −1 . According to the definition of L0 , H0,y = H0 T ΛT −1 .

86


Hence

H0 (x, y, t) = H(x, t) exp(Λy)T −1

where H satisfies (2.97) and S = Hx H −1 . Conversely, if H is a solution (2.97) and S = Hx H −1 , then S satisfies (2.99), i.e., ∂ − S is a Darboux operator for (2.92). The theorem is proved. Remark 21 (1) When N = 1, H satisfies the Lax pair (2.92). (2) If L0 in the above theorem is diagonalizable, then each column in H satisfies the Lax pair (2.92) for specific λ.

Example 2.10 The original Darboux transformation for the KdV equation can also be deduced from the above theorem. The KdV equation ut + 6uux + uxxx = 0

(2.100)

has the Lax pair λφ = −φxx − uφ, φt = 2(2λ − u)φx + ux φ.

(2.101)

From Theorem 2.9, the Darboux transformation is φ = φx −

fx φ f

(2.102)

where f is a solution of the Lax pair for λ = λ0 . The new solution given by this Darboux transformation is u = u + 2(ln f )xx .

(2.103)

Example 2.11 The Boussinesq equation (uxxx + 6uux )x + 3utt = 0

( = ±1)

has the Lax pair 3u φx + wφ, 2 φt = σφxx + σuφ

λφ = φxxx +

(2.104)

((σ 2 = , wx = 3(σuxx + ut )/4σ). Theorem 2.9 also gives the Darboux transformation [73] fx φ = φx − φ (2.105) f


87

where f is a solution of the Lax pair for λ = λ0 . The new solution of the Boussinesq equation is u = u + 2(ln f )xx , 3 w = w + ux + 3((ln f )xxx + (ln f )x (ln f )xx ). 2

2.6

(2.106)

Nonlinear constraints and Darboux transformation in 2+1 dimensions

Now we come back to the 2+1 dimensional AKNS system. In this section we will use the nonlinear constraint method and the Darboux transformation method to solve this system. The basic idea of the nonlinear constraint method is: (1) Find a suitable nonlinear relation between U and Ψ and express U as a nonlinear matrix function of Ψ: U = f (Ψ). (2) Substitute U = f (Ψ) into the Lax pair so that the original Lax pair becomes a system of nonlinear partial differential equations of Ψ. In each equation, the derivative with respect to only one of x, y, t is concerned. (3) The constraint U = f (Ψ) is suitable so that the new system of nonlinear equations has a Lax set (generalized Lax pair). Then by solving the new system of nonlinear equations and its Lax set, we can get solutions of the original problem. This idea was first applied in 1+1 dimensional integrable systems [11] and was generalized to the (2+1 dimensional) KP equation [14, 71]. Here we pay our attention to the 2+1 dimensional AKNS system so that we can get localized soliton solutions. With this method, we can also get a lot of non-localized solutions [123, 124]. However, since localized solutions are more interesting, here we only consider localized solutions [127, 128]. In order to use the nonlinear constraint method, here we add some conditions on the 2+1 dimensional AKNS system. As in Section 2.2, the 2+1 dimensional AKNS system is Ψy = JΨx + U (x, y, t)Ψ, Ψt =

n

Vj (x, y, t)∂ n−j Ψ

(2.107)

j=0

where ∂ = ∂/∂x, J = diag(J J1 , · · · , JN ) is a constant diagonal N × N matrix with distinct entries. U (x, y, t) is off-diagonal. Moreover, here we want that all Jj ’s are real and U ∗ = −U . In this case, we call (2.107) a hyperbolic u(N ) AKNS system. The condition U ∗ = −U will imply

88


that the solutions are globally defined, and the condition Jj ’s are real will guarantee that there exist localized solutions. As in Section 2.2, the integrability conditions of (2.107) are given by (2.20). Now we introduce a new linear system ⎛

Φx = ⎝ ⎛

Φt = ⎝

⎞

iλI

iF

iF ∗

0

⎠ Φ,

⎞

W

X

Y

Z

⎛

⎠Φ =

Φy = ⎝ n

⎛ ⎝

j=0

⎞

iλJ + U

iJF

iF ∗ J

0

⎞

Wj

Xj

−X Xj∗

Zj

⎠λ

n−j

⎠ Φ,

(2.108)

Φ

where F , Wj , Xj , Zj are N × K, N × N , N × K, K × K matrices Wj , Zj∗ = −Z Zj . respectively (K ≥ 1) and satisfy Wj∗ = −W The integrability conditions of (2.108) consists of Fy = JF Fx + U F, Wn F − iF Zn , iF Ft = Xn,x + iW iX Xj +1 = Xj,x + iW Wj F − iF Zj Xj F ∗ Wj,x = −iF Xj∗ − iX Xj∗ F (j Zj,x = iF ∗ Xj + iX

(2.109)

(j = 0, 1, · · · , n − 1)

(j = 0, 1, · · · , n) = 0, 1, · · · , n)

i[J, Wj +1 ] = Wj,y − [U, Wj ] + iJF Xj∗ + iX Xj F ∗ J

(2.110)

(j = 0, 1, · · · , n − 1) Xj + iX Xj∗ JF Zj,y = iF ∗ JX

(j = 0, 1, · · · , n),

Ux = [J, F F ∗ ],

(2.111)

Xn F ∗ J. Ut = Wn,y − [U, Wn ] + iJF Xn∗ + iX

(2.112)

For U = 0, F = 0, (2.110) implies that Wj (λ) = iΩj (t), Xj = 0, Zj0 (t) where Ωj (t)’s are real diagonal matrices and Zj0 (t)’s are Zj = iZ real matrices. When Zj0 (t) = ζj (t)IIK (IIK is the K × K identity matrix) where ζj (t) is a real function of t, (2.109) is just the Lax pair (2.107) for n = 1, 2, 3. (2.110) and (2.112) give the recursion relations to determine Wj , Xj , Zj , together with the evolution equations corresponding to (2.18)–(2.20), which are the integrability conditions of (2.107). (2.111) gives a nonlinear constraint between U and F .

89


This system includes the DSI equation and the 2+1 dimensional Nwave equation as special cases. In order to consider the asymptotic behavior of the solution U , here we suppose Ωj is independent of t and ζj = 0. Moreover, denote Ω =

n n−j and write Ω = diag(ω , · · · , ω ). 1 N j=0 Ωj λ Remark 22 (2.108) is a special case of the high-dimensional generalized AKNS system (3.1). Here we only consider this special system to find localized solutions. The general theory will be discussed in the next chapter. The soliton solutions are obtained by Darboux transformations from U = 0, F = 0. In the present case, the Darboux transformation can be constructed as in Subsection 1.4.4 with u(n) reduction. However, in order to get localized solutions, there should be more restrictions on the parameters of Darboux transformations. Let λα (α = 1, 2, · · · , r) be r non-real complex numbers such that ¯ β for all α, β. Let λα = λβ for α = β and λα = λ ¯α, · · · , λ ¯ α ). Λα = diag(λα , · · · , λα , λ

N

(2.113)

K

Considering the orthogonal relation (1.241), we can always take ⎛

Hα = ⎝

exp(Qα ) − exp(−Q∗α )C Cα∗ Cα

IK

⎞ ⎠,

(2.114)

where Cα ’s are K × N constant matrices, Qα = diag(q1 , · · · , qN ),

qj = iλα x + iλα Jj y + iωj (λα , t).

(2.115)

According to Section 1.4, the derived solutions are always global. However, in order to get localized solutions, we choose special Cα = (0, · · · , 0, κα , 0, · · · , 0)

(2.116)

lα

where κα is a constant K × 1 non-zero vector being the lα ’s column of Cα .

90


The Darboux matrices for such {Λα , Hα } can be constructed as follows. Let D(1) (λ) = λ − H1 Λ1 H1−1 ,

(1)

Hα = D(1) (λα )H Hα

(α = 2, 3, · · · , r), (1)

(1)−1

D(2) (λ) = λ − H2 Λ2 H2

,

(2)

(1)

Hα = D(2) (λα )H Hα

(α = 3, 4, · · · , r),

(2.117)

··· (r−1)

D(r) (λ) = λ − Hr

(r−1)−1

Λr Hr

,

D(λ) = D(r) (λ)D(r−1) (λ) · · · D(1) (λ),

(2.118)

then D(λ) is a polynomial of λ of degree r. The permutability (Theorem 1.12) implies that if (Λα , Hα ) and (Λβ , Hβ ) are interchanged, D(λ) is invariant. Let mj = #{ α | 1 ≤ α ≤ r, lα = j }

m = (m1 , · · · , mN )

then m1 + · · · + mN = r. Suppose D(λ) = λr − D1 λr−1 + · · · + (−1)r Dr .

(2.119)

(2.120)

The solution given by this Darboux matrix is U [m] = i[J, (D1 )BN ].

(2.121)

Here (D1 )BN denotes the first N × N principal submatrix of D1 . In order to consider the localization, the asymptotic behavior as t → ∞ and the asymptotic behavior as the phase difference tends to infinity uniformly, we write (2.122) qj = aαj s + bαj where aαj and bαj are independent of s. Here s can be a linear parameter of a straight line in (x, y) plane, or time t, or any other parameters. Moreover, denote ρα = Re(aα,lα ),

φα = Im(aα,lα ),

πα = Re(bα,lα ),

ψα = Im(bα,lα ).

(2.123)

In order to prove the following theorem, we need some symbols and simple facts. . If M1 , M2 are j × k matrices, we write M1 = M2 if there is a nondegenerate diagonal k × k matrix A such that M2 = M1 A.

91


If L is a k × k diagonal matrix, M1 and M2 are k × k matrices with . M1−1 = M2 LM M2−1 . M1 = M2 and det M1 = 0, then M1 LM Let ⎛ ⎞ a −v ∗ /¯ a ⎠ M =⎝ (2.124) v IK where v = 0 is an K × 1 vector, a = 0 is a number. Let ⎛

Λ=⎝ Then we have

⎞

λ0 ¯ 0 IK λ

⎠.

⎛

M −1

(2.125) ⎞

a ¯ v∗ 1 ⎠, = ⎝ ∆ −¯ av ∆IIK − vv ∗

(2.126)

⎛

⎞

¯ 0 )|a|2 ¯ 0 )av ∗ ¯ 0 ∆ + (λ0 − λ (λ0 − λ λ 1 ⎠ M ΛM −1 = ⎝ ¯ 0 )¯ ¯ 0 ∆IIK + (λ0 − λ ¯ 0 )vv ∗ ∆ (λ0 − λ av λ (2.127) ∗ 2 where ∆ = v v + |a| . Moreover, ⎛

lim M ΛM −1 = ⎝

a→∞

⎞

λ0 ¯ 0 IK λ

⎠,

⎛

¯ ⎜ λ0

lim M ΛM −1 = ⎝

a→0

(2.128) ⎞

∗

¯ 0 IK + (λ0 − λ ¯ 0 ) vv λ v∗v

⎟ ⎠.

(2.129)

Theorem 2.12 (1) If there is at most one α (1 ≤ α ≤ r) such that ρα = 0, then lims→∞ U [m] = 0. (2) If ραj = 0 (j = 1, 2, · · · , q) with αj = αk for j = k, ργ = 0 for all γ = αj (j = 1, · · · , q) and lα1 = · · · = lαq , then lims→∞ U [m] = 0. (3) If ρα = 0, ρβ = 0 (α = β), ργ = 0 for all γ = α, β, and lα = lβ , then [m] (2.130) lim Uab = 0 for (a, b) = (lα , lβ ) s→∞

and as s → ∞, [m]

Ulα ,lβ ∼

Bαβ exp(i(ψα − ψβ ) + i(φβ − φα )s) (1)

(2)

Aαβ cosh(πα + πβ − δαβ ) + cosh(πα − πβ − δαβ )

(2.131)

92

DARBOUX TRANSFORMATIONS IN INTEGRABLE SYSTEMS (1)

(2)

where Aαβ , δαβ , δαβ are real constants, Aαβ > 0, and Bαβ are complex constants. Moreover, if K = 1, then Bαβ = 0 if and only if κα = 0 and κβ = 0. Proof. First suppose ρα = 0. By (2.128) and (2.129), lim

ρα s→±∞

Hα Λα Hα−1 = Sα±∞

(2.132)

where ⎛

Sα+∞ = ⎝

⎞

λα IN

⎠, ¯ α IK λ ⎛ ¯ α − λα )El l λα IN + (λ α α

⎞

⎜

⎟

Sα−∞ = ⎝

∗ , ¯ α ) κα κα , ⎠ ¯ α IK + (λα − λ λ ∗ κα κα (2.133) Ejk is an N × N matrix whose (j, k)th entry is 1 and the rest entries are zero. For β = α,

⎛

⎞

exp(Qβ (s)) − exp(−Qβ (s)∗ )Cβ±∗ . ⎠ Hβ = ⎝ (λβ − Sα±∞ )H Cβ± IK

(2.134)

where ± Cβ± = (0, · · · , 0, κ β , 0, · · · , 0), lβ

¯α λβ − λ + κ = κβ , β λβ − λ α ⎧ ¯α ¯ α κ∗ κβ λ −λ λα − λ ⎪ α ⎪ ⎪ β κβ − κα if lβ = lα , ⎨ λβ − λ α λβ − λα κ∗α κα − β = κ ¯ α κ∗ κβ ⎪ λα − λ α ⎪ ⎪ κα if lβ = lα . ⎩ κβ − ∗ ¯ κ λβ − λα α κα

(2.135)

Therefore, if ρα = 0, the action of the limit Darboux matrix λ − Sα±∞ on Hβ (β = α) does not change the form of Hβ , but only changes the constant vector κβ . ± ±∗ ± If K = 1, then κ±∗ γ = 0. When K > 1, this β κγ = 0 implies κ β κ does not hold in general.

93


Now suppose ρα = 0. Without loss of generality, suppose lα = 1. Then ⎛

⎜ ⎜ ⎜ ⎜ . ⎜ Hα = ⎜ ⎜ ⎜ ⎜ ⎝

− exp(−¯ πα )κ∗α

exp(πα ) 1 ..

κα

0 .. .

. 1

0

0 ··· 0

IK

⎞

⎟ ⎟ ⎟ ⎟ ⎟ ⎟. ⎟ ⎟ ⎟ ⎠

(2.136)

By (2.127), Hα Λα Hα−1 = ⎛

⎜ ⎜ ⎜ ⎜ ·⎜ ⎜ ⎜ ⎝

1 · ∆

¯ α ) exp(πα )κ∗α (λα − λ

¯ α ) exp(πα + π ¯ α ∆ + (λα − λ ¯α ) λ λα ..

0 .. .

. λα

¯ α ) exp(¯ πα )κα (λα − λ

0

···

0

0 ¯ α ∆IK + (λα − λ ¯ α )κα κ∗α λ

⎞ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎠

(2.137) ¯α ) + κ∗α κα . where ∆ = exp(πα + π Part (1) of the theorem is derived as follows. Owing to the permutability of Darboux transformations, we can suppose ρ1 = 0, · · ·, ρr−1 = 0, ρr = 0. Then, as s → ∞, D(α) tends to a diagonal matrix for α ≤ r − 1. Considering (2.137), the limit of (D(r) (λ))BN is also diagonal, hence U [m] = i[J, (D1 )BN ] → 0.

(2.138)

Now we turn to prove part (2). We use the construction of Darboux matrices in (1.258). However, the λ in (1.258) should be replaced by iλ because of its appearance in (2.108). Let ⎛ ⎞ ˚∗ H ˚ exp(Q (s)) H α β α ˚ =⎝ ⎠ , Γαβ = (2.139) H α ¯α , λβ − λ Cα then the Darboux matrix is D(λ) =

r

⎛

r ˚ (Γ−1 ) H ˚∗ H α αβ β

¯ α ) ⎝1 − (λ − λ

α=1

and the new solution is

α,β=1

⎡

U [m] = i ⎣J,

r α,β=1

¯β λ−λ

˚ (Γ−1 ) H ˚∗ H α αβ β

⎞ ⎠

(2.140)

⎤

BN

⎦.

(2.141)

94


First, suppose q = r and αj = j (j = 1, 2, · · · , r). Since ⎛ 1 ⎜ .. ⎜ . ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ . ˚ =⎜ H ⎜ j ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎝

1 exp(π πj ) 1 .. 0 ··· 0

κj

.

⎞ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ 1 ⎟ ⎟ ⎠

(2.142)

0 ··· 0

lj

and l1 = · · · = lq , we have ⎛

Γ · · · Γ1q ⎜ 11 .. ⎜ . .. Γ = ⎜ .. . . ⎝ Γq1 · · · Γqq

⎞ ⎟ ⎟ ⎟ ⎠

(2.143)

where Γjk ’s are N × N diagonal matrices. Therefore, ⎛

Γ−1

Σ · · · Σ1q ⎜ 11 .. ⎜ .. .. =⎜ . . . ⎝ Σq1 · · · Σqq

⎞ ⎟ ⎟ ⎟ ⎠

(2.144)

where Σjk ’s are also N × N diagonal matrices. This implies that U [m] = 0. When r > q, we use the permutability of Darboux transformations and suppose ρ1 = 0, · · ·, ρr−q = 0, ρr−q+1 = · · · = ρr = 0. Then, after (r−q) (r−q) have the action of D(r−q) (λ) · · · D(1) (λ), the derived Hr−q+1 and Hr the same asymptotic form as Hr−q+1 and Hr respectively, provided that (r−q) (r−q) . the constant vectors κr−q+1 and κr are changed to κr−q+1 and κr [m] are Therefore, as in the case r = q, the limits of the components of U all zero. This proved part (2). ˚ is Now we prove part (3). First consider the case r = 2. Suppose H j given by (2.142) (j = 1, 2) and l1 = l2 . Denote θ12 = #

κ∗1 κ2 , κ∗1 κ1 κ∗2 κ2

(2.145)


g12 = 1 −

4 Im λ1 Im λ2 2 ¯ 1 |2 |θ12 | > 0, |λ2 − λ

95 (2.146)

then, by direct calculation, we have 2i(J Jl2 − Jl1 ) Im λ1 Im λ2 θ12 · ¯ 2 − λ1 λ exp(i(ψ1 − ψ2 )) ·√ , g12 cosh(π1 + π2 − δ1 ) + cosh(π 1 − π2 − δ2 ) λ2 − λ 1 1 1 δ1 = ln g12 + ln(κ∗1 κ1 κ∗2 κ2 ) + 2 ln ¯1 , 2 2 λ − λ 2 1 κ∗ κ1 δ2 = ln 1∗ , 2 κ2 κ2 [m]

lim Ul1 ,l2 exp(i(φ2 − φ1 )s) =

s→∞

(2.147)

[m]

and Uµν → 0 if (µ, ν) = (l1 , l2 ). When r > 2, we still use the permutability of Darboux transformations and suppose ρ1 = 0, · · ·, ρr−2 = 0, ρr−1 = ρr = 0. As in the proof (r−2) of part (2), after the action of D(r−2) (λ) · · · D(1) (λ), the derived Hr−1 (r−2)

and Hr have the same asymptotic form as Hr−1 and Hr respectively, (r−2) provided that the constant vectors κr−1 and κr are changed to κr−1 (r−2)

and κr . Therefore, as in the case r = 2, the limit of Ulr−1 ,lr has the desired form, and the limits of the other components of U [m] are all zero. The theorem is proved. Now we can discuss the properties of the solution U [m] . (1) Localization of the solutions For the Lax pair (2.108), Qα = iλα (x + Jy) + iω(λα )t.

(2.148)

Consider the limit of the solution as (x, y) → ∞ along a straight line x = ξ + vx s, y = η + vy s (vx2 + vy2 > 0), then Qα = iλα (ξ + Jη) + iω(λα )t + iλα (vx + Jvy )s.

(2.149)

ρα = Re (iλα (vx + Jlα vy )) = − Im λα (vx + Jlα vy ).

(2.150)

Now

If there is at most one ρα = 0, then part (1) of Theorem 2.12 implies that U [m] → 0 as s → ∞. If ρα = 0, ρβ = 0 (α = β), then lα = lβ since Jlα = Jlβ if lα = lβ . Hence, part (2) of Theorem 2.12 also implies U [m] → 0 as s → ∞. Therefore, we have

96


Theorem 2.13 U [m] → 0 as (x, y) → ∞ in any directions. (2) Asymptotic behavior of the solutions as t → ∞ Now we use a frame (ξ, η) which moves in a fixed velocity (vx , vy ), that is, let x = ξ + vx t, y = η + vy t, then Qα = iλα (ξ + Jη) + (iλα (vx + Jvy ) + iω(λα ))t,

(2.151)

ρα = − Im λα (vx + Jlα vy ) − Im(ωlα (λα )).

(2.152)

Suppose that for distinct α, β, γ, ⎛ ⎜

1 Jlα

det ⎜ ⎝ 1 Jlβ 1 Jlγ

σα

⎞ ⎟

σβ ⎟ ⎠ = 0

(2.153)

σγ

where σα = Im(ωlα (λα ))/ Im(λα ).

(2.154)

Then there are at most two ρα = 0 (α = 1, · · · , r). By Theorem 2.12, [m] Ulα ,lβ → 0 only if ρα = 0, ρβ = 0. This leads to vx =

Jlα σβ − Jlβ σα , Jlβ − Jlα

vy =

σα − σ β . Jlβ − Jlα

(2.155)

[m]

For Ujk → 0, α, β can take mj and mk values respectively, hence [m]

there are at most mj mk velocities (vx , vy ) such that Ujk → 0. Therefore, we have

Theorem 2.14 Suppose (2.153) is satisfied. Then as t → ∞, the [m] asymptotic solution of Ujk has at most mj mk peaks whose velocities are given by (2.155) (l ( α = j, lβ = k). If a peak has velocity (vx , vy ), then, in the coordinate ξ = x − vx t, η = y − vy t, limt→∞ Uab = 0 for all (a, b) = (j, k), and as t → ∞ Bαβ exp(i Re(λα − λβ )ξ + i(λα Jj − λβ Jk )η + i(φα − φβ )t) , ∆ (1) ∆ = Aαβ cosh( Im(λα + λβ )ξ + Im(λα Jj + λβ Jk )η + δαβ ) [m]

Ujk ∼

(2)

+ cosh( Im(λα − λβ )ξ + Im(λα Jj − λβ Jk )η + δαβ ) where Aαβ , constants,

(1) δαβ ,

(2) δαβ

are real constants, Aαβ > 0, and Bαβ

φγ = Re λγ (vx + Jlγ vy ) + Re(ωlγ (λγ ))

(γ = α, β).

(2.156) are complex (2.157)

97


Remark 23 The condition (2.153) implies that the velocities of the solitons are all different. This is true for the DSI equation. However, for the 2+1 dimensional N-wave equation, all the solitons move in the same velocity. We shall discuss this problem later.

Example 2.15 DSI equation Let n = 2, N = 2, ⎛

J =⎝

⎞

1

0

0 −1

⎛

⎠,

U =⎝

⎞

0

u

−¯ u 0

⎠,

ω = −2iJλ2 ,

(2.158)

then we have Fx + U F, Fy = JF

⎛

Fxx + 2iU Fx + i ⎝ Ft = 2iJF

⎞

|u|2 + 2q1

ux + uy

−¯ ux + u ¯y

−|u|2 − 2q2

−iut = uxx + uyy + 2|u|2 u + 2(q1 + q2 )u, q1,x − q1,y = q2,x + q2,y = −(|u|2 )x , ⎛

⎠ F,

(2.159)

(2.160)

⎞

q1 0 1 ⎠, (F F ∗ )D = ⎝ 2 0 q2

[J, F F ∗ ] = Ux .

(2.161)

(2.160) is the DSI equation. If we construct the solution U [m] as above, then Theorem 2.13 implies that U [m] → 0 as (x, y) → ∞ in any directions. If Re λα = Re λβ for α = β and lα = lβ , then, Theorem 2.14 implies that as t → ∞, ( 1 + m2 = r). From the derived solution u has at most m1 m2 peaks (m (2.154), σα = −4J Jlα Re λα , hence (2.155) implies that each peak has the velocity vx = 2 Re(λα − λβ ), vy = 2 Re(λα + λβ ) (lα = 1, lβ = 2). This is the (m1 , m2 ) solitons [30]. When K = 1, these peaks do not vanish if and only if all κα ’s are non-zero. Figures 2.5 – 2.7 show the solitons u[1,3] , u[2,3] and u[3,3] respectively. The parameters are K = 1, t = 2, λ1 = 1 − 2i, λ2 = −3 − i, λ3 = 2 + i, λ4 = −1 + 3i, λ5 = 2 + 1.5i, λ6 = −0.5 − 1.5i, C1 = (1, 0), C2 = (0, 1), C3 = (0, 2), C4 = (0, −2), C5 = (2, 0), C6 = (−2, 0). (3) Asymptotic solutions as the phases differences tend to infinity For the equations whose solitons move in the same speed, like the 2+1 dimensional N-wave equation, the peaks do not separate as t → ∞.

98


Figure 2.5.

u[1,3] of the DSI equation

Figure 2.6.


However, we can still see some peaks in the figures. Here we will get the corresponding asymptotic properties of the solitons.

Theorem 2.16 Let pα (α = 1. · · · , r) be constant real numbers satisfying ⎛ ⎜

1 Jlα

det ⎜ ⎝ 1 Jlβ 1 Jlγ

pα / Im λα

⎞ ⎟

pβ / Im λβ ⎟ ⎠ = 0

(2.162)

pγ / Im λγ

for distinct α, β, γ. Let dα be complex constant K × 1 vectors, κα = dα exp(pα τ ) and construct the Darboux transformations as above. Let

99


Figure 2.7.


x = ξ + vx τ , y = η + vy τ , then, for any j, k with 1 ≤ j, k ≤ N , j = k, [m] limτ →∞ Ujk = 0 only if (vx , vy ) takes specific mj mk values. Proof. Here Qα = iλα (ξ + Jη) + iω(λα )t + iλα (vx + Jvy )τ. Hence

⎛

. ⎝ ˚ = H α

α (τ )) exp(Q

(2.163)

⎞ ⎠

Dα

(2.164)

where Dα = (0, · · · , 0, dα , 0, · · · , 0),

(2.165)

lα

α (τ ) = iλα (ξ + Jη) + iω(λα )t + (iλα (vx + Jvy ) − pα )τ. Q

(2.166)

α (τ ) is The real part of the coefficient of τ in Q

ρα = − Im λα (vx + Jvy ) − pα .

(2.167)

Condition (2.162) implies that there are at most two ρα ’s such that [m] ρα = 0. According to Theorem 2.12, as τ → ∞, Ujk → 0 only if there exist ρα = 0, ρβ = 0, α = β, lα = j, lβ = k. Therefore, the theorem is verified. When the condition (2.153) holds, this theorem is useless, because the evolution will always separate the peaks. However, when (2.153) does not hold, especially when it is never satisfied, this theorem reveals a fact of the separation of the peaks.

100


Example 2.17 2+1 dimensional N-wave equation Let n = 1, ω = Lλ where L = diag(L1 , · · · , LN ) is a constant real diagonal matrix such that Lj = Lk for j = k. Then, the integrability conditions (2.109) – (2.112) imply Fx + U F, Ft = LF Fx + V F, (2.168) Fy = JF [J, V ] = [L, U ], Ut − Vy + [U.V ] + JV Vx − LU Ux = 0, (2.169) ∗ (2.170) Ux = [J, F F ]. (2.169) is just the 2+1 dimensional N-wave equation. Suppose U [m] is constructed as above, then Theorem 2.13 implies that [m] → 0 as (x, y) → ∞ in any directions. Theorem 2.14 cannot be U applied here. The reason is: the condition (2.153) holds only if lα = lβ for α = β. Hence for any j, mj = 0 or 1. This implies that (2.153) does not hold generally unless mj = 0 or 1 for all 1 ≤ j ≤ N . Therefore, we apply Theorem 2.16 to the previous problem. Theorem 2.16 implies that if we choose {pα } such that (2.162) is satisfied, then, for each (j, k), [m] limτ →∞ Ujk has at most mj mk peaks. When K = 1, these peaks do not vanish if and only if all κα ’s are non-zero. Remark 24 Here τ → ∞ means that the phase differences of different peaks tend to infinity. Therefore, the peaks are separated by enlarging the phase differences. Here are the figures describing the solutions U [0,1,2] and U [1,1,2] of the 3-wave equation. The vertical axis is (|u12 |2 + |u13 |2 + |u23 |2 )1/4 so that all the components are shown in one figure. The parameters are ⎛ ⎜

J =⎜ ⎝

1 0 −1

⎞

⎛

⎟ ⎟ ⎠

L=⎜ ⎝

⎜

⎞

2

⎟ ⎟ ⎠

−1 1

K = 1, t = 10, λ1 = 1 − 2i, λ2 = −3 − i, λ3 = 2 + i, λ4 = −1 + 3i, C1 = (0, 1, 0), C2 = (0, 0, 1), C3 = (0, 0, 4096), C4 = (1, 0, 0). Note that for U [0,1,2] , only U23 has two peaks, and for U [1,1,2] , U12 , U13 , U23 have one, two, two peaks respectively.


Figure 2.8.

U [0,1,2] of the 3-wave equation

Figure 2.9.

U [1,1,2] of the 3-wave equation

101

Chapter 3 N + 1 DIMENSIONAL INTEGRABLE SYSTEMS

In this chapter, we discuss the soliton theory in higher dimensions and generalize the AKNS system in 1 + 1 dimensions to the AKNS system in n + 1 dimensions. The Darboux transformation method can still be used and the algorithm is universal and purely algebraic. The collision of solitons in higher dimensions is also elastic as in 1 + 1 dimensions. For the Cauchy problems, the initial data of this generalized AKNS system is given on a straight line. This theory covers many problems, especially some differential geometric problems in higher dimensions. It can also be applied to 2 + 1 dimensional problems (like the KP, DSI and 2+1 dimensional N-wave equations) as in Section 2.6. However, for n = 2, the contents do not cover those in Chapter 2.

3.1 n + 1 dimensional AKNS system n + 1 dimensional AKNS system 3.1.1 The physical space-time is 3 + 1 dimensional. On the other hand, the soliton theory is developed extensively in 1 + 1 and 2 + 1 dimensions. Here we will pay attention to the integrable systems in higher dimensions and introduce a kind of higher dimensional solitons with the property of elastic collision [34, 35, 39, 53]. First, we generalize the AKNS system in Section 1.2 to arbitrary n + 1 dimensions. When n = 2, the system here is different from that in Chapter 2. Here the system still has spectral parameter while in the generalized AKNS system the spectral parameter is replaced by a differential operator. Suppose that (t, x1 , x2 , · · · , xn ) are the coordinates of Rn+1 where (x1 , · · · , xn ) denotes spatial coordinates and t denotes the time coordi-

104


nate. Consider the following linear system: ∂i Ψ = Ui Ψ = (λJ Ji + Pi )Ψ ∂t Ψ = V Ψ =

m

(i = 1, · · · , n),

Vα λm−α Ψ,

(3.1)

α=0

where ∂i are ∂t are the partial derivatives with respect to xi and t respectively, Pi ’s and Vα ’s are N × N matrix functions, Pi ’s are off-diagonal, Ji ’s are constant diagonal matrices. Since (3.1) contains several equations, we call it Lax set. The integrability conditions of the first part of (3.1) (its spatial part) are Ui , Uj ] = 0. (3.2) ∂j Ui − ∂i Uj + [U Considering the coefficients of the powers of λ, we have [J Ji , Jj ] = 0, Jj , Pi ], [J Ji , Pj ] = [J ∂j Pi − ∂i Pj + [P Pi , Pj ] = 0.

(3.3) (3.4) (3.5)

Since Ji is diagonal, the equations in (3.3) are identities. In what follows, we need two more assumptions: (1) Ji ’s are linearly independent. (2) If a matrix A satisfies [A, Ji ] = 0 for all i, then A is a diagonal matrix. The first condition implies that the system is really an n + 1 dimensional problem and cannot be reduced to a lower dimensional system in some sense. In fact, if not, there would be a linear combination of Ji ’s and a linear transformation of spatial variables so that Jn = 0, that is, the essential part of the system is (n − 1) + 1 dimensional. The second condition is equivalent to that there is a linear combination J = ai Ji such that the diagonal entries of J are distinct. Therefore, there is a linear transformation in xi ’s so that the diagonal entries of J1 are distinct. From (3.4) and (3.5), we have: (i) There is an N × N off-diagonal matrix function P such that Pi = [P, Ji ]

(i = 1, · · · , n).

(3.6)

(ii) P satisfies the spatial constraints ∂j Pi − ∂i Pj + [P Pi , Pj ] = 0.

(3.7)

n + 1 dimensional integrable systems

105

(iii) The integrability conditions of (3.1) give the equations for Vα ’s: [J Ji , V0 ] = 0, ∂i V0 = 0, off Pi , Vα ]off , [J Ji , Vαoff +1 ] = ∂i Vα − [P diag ∂i Vαdiag Pi , Vαoff , +1 ] +1 = [P (α = 0, 1, · · · , m − 1),

(3.8) (3.9) (3.10)

and the evolution equations ∂t Pi − ∂i Vmoff + [P Pi , Vm ]off = 0.

(3.11)

There are many redundant equations in (3.7) – (3.11). In fact, with the assumptions on J, we can always assume that all the diagonal entries of J1 are distinct, which can be realized by a linear transformation of (x1 , · · · , xn ). (3.6) implies that the off-diagonal entries of P are determined by P1 . The following lemma shows that (3.7) – (3.10) can be replaced by part of the equations in it.

Lemma 3.1 If the diagonal entries of J1 are distinct, then (3.7) – (3.11) are equivalent to ∂1 Pi − ∂i P1 + [P Pi , P1 ] = 0, off Pi , Vα ]off , [J Ji , Vα+1 ] = ∂i Vαoff − [P Pi , Vαoff ]diag , ∂i Vαdiag = [P ∂t P1 − ∂1 Vmoff + [P P1 , Vm ]off = 0.

(3.12) (3.13) (3.14) (3.15)

Proof. (3.13) and (3.14) are the same as (3.9) and (3.10). Hence we only need to prove that (3.7) and (3.11) are consequence of (3.12) and (3.15) with the help of (3.13) and (3.14). Let Pi , Pj ]. ∆ij = ∂j Pi − ∂i Pj + [P

(3.16)

By Jacobi identity, we have [J J1 , ∆ij ] = ∂j [J Ji , P1 ] − ∂i [J Jj , P1 ] + [J J1 , [P Pi , Pj ]] Jj , ∆1i ]. = [J Ji , ∆1j ] − [J

(3.17)

Pi = [P, Ji ] and Pj = [P, Jj ] imply that all the diagonal entries of [P Pi , Pj ] are 0. Hence all the diagonal entries of ∆ij are 0. Using ∆1i = 0 (i ≥ 2), we know that [J J1 , ∆ij ] = 0 for any i and j. Hence ∆off ij = 0, i.e., ∆ij = 0. (3.7) is proved.

106


(3.11) is proved as follows. For any i > 1, using [P Pi , Jj ] = [P Pj , Ji ], we have Pi , Vm ], J1 ] [∂ ∂t Pi − ∂i Vm + [P = [∂1 Vm − [P P1 , Vm ], Ji ] − [∂ ∂i Vm − [P Pi , Vm ], J1 ]

(3.18)

Vm , Ji ] − ∂i [V Vm , J1 ] + [[P Pi , Vm ], J1 ] − [[P P1 , Vm ], Ji ]. = ∂1 [V With (3.7), (3.9), (3.10) and the Jacobi identity, the last equality equals Pi , Vm−1 ] − ∂i [P P1 , Vm−1 ] − [[V Vm , J1 ], Pi ] + [[V Vm , Ji ], P1 ] ∂1 [P Pi , [P P1 , Vm−1 ]] − [[P P1 , [P Pi , Vm−1 ]] = 0. = [∂1 Pi − ∂i P1 , Vm−1 ] + [P (3.19) Hence (3.11) holds. The lemma is proved. From this lemma, the spatial constraints are simplified to (3.12) and the evolution equations are simplified to (3.15). In general, (3.12) and (3.15) are still over-determined partial differential equations. In Chapter 1, we have known that Vα ’s do exist when n = 1, and they are differential polynomials of P . When n > 1, Vα ’s should satisfy many more equations (3.13) and (3.14). Hence we need to show that they do exist and are still differential polynomials of P .

Lemma 3.2 (1) There exist {V Vα }’s satisfying (3.13) and (3.14). These {V Vα }’s are polynomials of P and its derivatives with respect to x, and are denoted as Vα = Vα [P ]. (2) For given diagonal matrices Vα0 (t)’s which Vα [P ]} such that Vα [0] = are independent of xi ’s, there exist unique {V Vα0 (t). Sketch of proof. V0 and V1 can be obtained from (3.13) and (3.14) by direct calculation. For general Vα , this can be derived inductively. First, from the result in [111], Vα ’s are polynomials of P , ∂1 P , · · ·, ∂1α P if x2 , · · ·, xn are regarded as parameters. Then from the spatial constraints (3.12) we can prove that these differential polynomials satisfy all the equations in (3.13) and (3.14). The complete proof is formal and tedious. For the details, see [53]. By this lemma, all the equations derived from the Lax set (3.1) are partial differential equations of x1 , · · ·, xn and t.

3.1.2

Examples

In (3.1), m, n, N (N ≥ n) and Ji are quite arbitrary. Hence (3.1) leads to a lot of nonlinear partial differential equations. Here we show two simple examples.

107


Example 3.3 n = N = 2, m = 2, ⎛

J1 = ⎝

⎞

1 0

⎛

⎠,

0 0

J2 = ⎝

⎞

0

0

0 −i

⎠,

⎛

P =⎝

⎞

0 p q 0

⎠,

p, q are complex-valued functions. Then, ⎛

P1 = [P, J1 ] = ⎝

0 −p q

0

⎞ ⎠,

⎛

P2 = [P, J2 ] = i ⎝

0 −p q

0

⎞ ⎠,

and the Lax set is J1 + P1 )Ψ, ∂1 Ψ = (λJ V0 ∂t Ψ = (V

λ2

∂2 Ψ = (λJ J2 + P2 )Ψ,

+ V1 λ + V2 )Ψ.

The spatial constraints ∂2 P1 − ∂1 P2 = 0 becomes ∂1 p + i∂2 p = 0,

∂1 q + i∂2 q = 0,

i.e., p, q are holomorphic functions of the complex variable z = x1 + ix2 , and ∂1 p = pz , ∂1 q = qz . From the recursion relations, ⎛

V0 = V00 = ⎝

⎞

a1

0

0

a2

⎛

V1 = ⎝ ⎛

V2 = ⎝

⎠,

b1

(a2 − a1 )p

(a1 − a2 )q

b2

⎞ ⎠,

(a1 − a2 )pq + c1

(a2 − a1 )px + (b2 − b1 )p

(a2 − a1 )qx + (b1 − b2 )q

(a2 − a1 )pq + c2

⎞ ⎠,

where a1 , b1 , c1 , a2 , b2 , c2 are functions of t only. Moreover, we get the nonlinear evolution equations ∂p + (a2 − a1 )pzz + (b2 − b1 )pz ∂t +2(a2 − a1 )p2 q + (c2 − c1 )p = 0, ∂q + (a1 − a2 )qzz + (b2 − b1 )qz ∂t +2(a1 − a2 )pq 2 + (c1 − c2 )q = 0.

108


Example 3.4 n = N = 3, m = 2, ⎛ ⎜

J1 = ⎜ ⎝

⎞

i

⎛

⎟ ⎟, ⎠

0

⎜

J2 = ⎜ ⎝

⎞

0

⎟ ⎟, ⎠

i

0

⎛ ⎜

J3 = ⎜ ⎝

⎞

0

⎟ ⎟. ⎠

0

0

i

Now V0 [P ] = V00 (t), V1 [P ] = [P, V0 ] + V10 (t), $ % ∂P Jj , V0 + [P, [P, V0 ]]off + [P, V10 ]off V2 [P ] = − ∂xj −([P, V0 ]P )diag + V20 (t). We get a system of second order quasi-linear partial differential equations. Its soliton solutions will be discussed later.

3.2 Darboux transformation and soliton solutions 3.2.1 Darboux transformation In this subsection we discuss the Darboux transformation for the higher dimensional AKNS system. We have known in Chapter 1 that in all Darboux matrices those of degree one are the most fundamental ones. The Darboux matrices of higher degree can be produced from the Darboux matrices of degree one inductively. Hence we only discuss the Darboux matrix in the form D(x, t, λ) = λI − S(x, t).

(3.20)

Let Ψ = (λI − S)Ψ and substitute it into

we get

∂i Ψ = (λJ Ji + Pi )Ψ ,

(3.21)

Pi = Pi + [J Ji , S],

(3.22)

Pi = Pi S. ∂i S + SP

(3.23)

From (3.22) and Pi = [P , Ji ], we have P = P − S + arbitrary diagonal matrix. For simplicity, let

P = P − S.

(3.24)

109

n + 1 dimensional integrable systems Then (3.23) becomes Pi , S] + [J Ji , S]S. ∂i S = [P

(3.25)

Substitute Ψ = (λI − S)Ψ into ∂t Ψ =

m

Vα λm−α Ψ ,

(3.26)

α=0

and expand both sides in terms of the powers of λ, we get V0 = V0 , Vα + Vα S Vα +1 = Vα+1 − SV and ∂t S =

$ m

(α = 0, 1, · · · , m − 1),

(3.27)

%

Vα S

m−α

,S .

(3.28)

α=0

(3.23) and (3.28) are partial differential equations for S. It can be verified directly that they are completely integrable. Therefore, for the initial value of S at t = t0 , xi = xi0 , the solution S exists uniquely in some region around t = t0 , xi = xi0 . What is more, we need to prove Vα = Vα [P ].

(3.29)

so that P and P satisfy the same evolution equation (3.11) (or (3.15)). For α = 0, 1, 2 · · ·, (3.29) can be verified directly. For general α, the proof is quite tedious. It is similar to that for Theorem 1.11 and can be found in [53]. From the above discussion and the general method of constructing Darboux matrix, we have

Theorem 3.5 Suppose P is a solution of (3.15). For given constant diagonal matrix Λ = diag(λ1 , · · · , λN ), let hi be a column solution of (3.1) with λ = λi , H = (h1 , · · · , hN ). When det H = 0, let S = HΛH −1 . Then λI − S is a Darboux matrix for (3.1). Proof. The proof is similar to that for Theorem 1.9. Therefore, with Darboux transformation, we can get a series of solutions (P (0) , Ψ(0) ) −→ (P (1) , Ψ(1) ) −→ (P (2) , Ψ(2) ) −→ · · · from a given solution P (0) and the corresponding solution Ψ(0) of the Lax set (3.1). We have also seen that the formula S = HΛH −1 appears in many cases.

(3.30)

110

3.2.2


u(N ) case

If P satisfies some extra constraints, usually the Darboux transformation constructed above can not keep these constraints. In this case, corresponding constraints should be added on the Darboux transformation. This would be quite complicated in general. Here we will discuss an interesting and simple case in which Jj = iej = i diag(0, · · · , 0, 1, 0, · · · , 0), j

(3.31)

and P ∈ u(N ) (i.e. P ∗ = −P ). Suppose a solution P ∈ u(N ) of (3.5) is known, let Φ be the fundamental solution of its Lax set. Now Ji∗ , P ∗ ] = [J Ji , P ] = −P Pi . Pi∗ = [P, Ji ]∗ = [J

(3.32)

In solving (3.8) – (3.10), take the “integral constant” (function of t) for Vα to be a diagonal matrix Vα0 (t) in u(N ), i.e., Vα0 (t) are purely imaginary diagonal matrices, then Vα ∈ u(N ),

α = 0, 1, · · · , m.

(3.33)

In fact, V0 = V00 (t) ∈ u(N ). Suppose Vl ∈ u(N ), then from (3.9) for off α = l, we have [J Jj , Vloff +1 ] ∈ u(N ). This implies Vl+1 ∈ u(N ). From (3.10) diag for α = l, we have ∂i Vl+1 ∈ u(N ). Therefore, Vldiag +1 ∈ u(N ) if it holds at one point. (3.33) is proved by induction. Now we construct the Darboux transformation (P, Φ) → (P , Φ ) so that P ∈ u(N ). As in Subsection 1.4.4, let λα (α = 1, · · · , N ) take only two mutually conjugate complex numbers: λα = µ Let h0α ’s satisfy

or

0 h0∗ β hα = 0

µ ¯ (µ = µ). ¯ (µβ = µα ).

Solve ∂i hα = (λα Ji + Pi )hα ,

(3.34)

∂t hα =

m

Vl λm−l α hα

l=0

for initial data hα = h0α at some point to get the column solution hα . ¯ β = λα ), ∂i (h∗ hα ) = 0 and ∂t (h∗ hα ) = 0 Then when λα = λβ (i.e. λ β β hold. Hence (3.35) h∗β hα = 0 holds everywhere. Moreover, suppose that all the solutions {h0α } corresponding to λα = µ are linearly independent, and all the solutions {h0β }

111


corresponding to λβ = µ ¯ are linearly independent. Then all these solutions are linearly independent. Hence (3.35) implies that {h1 , · · · , hN } are linearly independent. This implies det H = 0, and S = HΛH −1 is defined globally on Rn+1 . According to the discussion in Subsection 1.4.4, we have S + S ∗ = (µ + µ ¯)I,

(3.36)

S ∗ S = |µ|2 I.

(3.37)

and We have also

¯ − S)∗ (λI − S) = (λ − µ)(λ − µ (λI ¯)I. Moreover, from Subsection 1.4.4, Vα ∈ u(N ). Therefore, we get the construction of Darboux transformation for the AKNS system with u(N ) reduction. The Darboux transformations for many other systems with u(N ) reduction can be constructed similarly.

3.2.3

Soliton solutions

Now we construct single and multi-soliton solutions. For simplicity, only the case n = 3, N = 3 and m = 2 will be considered. It will also be proved that the soliton interaction in multi-soliton solution is elastic. Take ⎛ ⎜

a1 i

V00 = ⎜ ⎝ 0 0 ⎛

V20

0 a2 i

c1 i

⎜ =⎜ ⎝ 0

a3 i

0

0

0

⎞

⎛

⎟

0 ⎟ ⎠,

0 c2 i

0

0

⎞

⎜

b1 i

0

V10 = ⎜ ⎝ 0 0

b2 i 0

0

⎞ ⎟

0 ⎟ ⎠, b3 i

⎟

0 ⎟ ⎠, c3 i

where ai , bi , ci (i = 1, 2, 3) are real constants. As before, take the seed solution P = 0, then ⎛ ⎜

Ψ0 = ⎜ ⎝

e i(λx1 +φ1 (λ)t)

0

0

0


0

0


0

⎞ ⎟ ⎟, ⎠

(3.38)

where φi (λ) = ai λ2 + bi λ + ci

(i = 1, 2, 3).

(3.39)

112


Let λ1 = µ,

(µ = µ), ¯

λ2 = λ3 = µ ¯ ζi = µxi + φi (µ)t.

Let

⎛ ⎜

¯

−¯ ae iζ1

e iζ1

iζ2 H=⎜ ⎝ ae be iζ3

(3.41)

¯ −¯be iζ1

¯

(3.40)

e iζ2

0

0

¯ e iζ3

⎞ ⎟ ⎟, ⎠

(3.42)

then λi ’s and H satisfy the conditions in (3.34) and (3.35), and ¯

¯

det H = e i(ζ1 +ζ2 +ζ3 ) ∆, ¯

¯

(3.43)

¯

¯

∆ = 1 + |a|2 e i[(ζ2 −ζ2 )−(ζ1 −ζ1 )] + |b|2 e i[(ζ3 −ζ3 )−(ζ1 −ζ1 )] > 0.

(3.44)

The derived solution consists of three waves. They are ¯

¯

¯)e i[(ζ2 −ζ1 )−(ζ3 −ζ1 )] /∆, p = S23 = a¯b(µ − µ ¯ ¯ ¯)¯be i(ζ1 −ζ3 ) /∆, q = S13 = (µ − µ r = S12 = (µ −

(3.45)

¯ ¯ µ ¯)¯ ae i(ζ1 −ζ2 ) /∆.

According to (3.41) and (3.45), |p|, |q| and |r| are travelling waves with velocities φi (¯ µ) − φi (µ) (i = 1, 2, 3) (3.46) vi = − µ ¯−µ respectively. In fact, |p| = |ab| |µ − µ ¯|e i(µ−¯µ)[(ξ2 −ξ1 )+(ξ3 −ξ1 )]/2 /∆, |q| = |b| |µ − µ ¯|e i(µ−¯µ)(ξ3 −ξ1 )/2 /∆,

(3.47)

|r| = |a| |µ − µ ¯|e i(µ−¯µ)(ξ2 −ξ1 )/2 /∆, where

∆ = 1 + |a|2 e i(µ−¯µ)(ξ2 −ξ1 ) + |b|2 e i(µ−¯µ)(ξ3 −ξ1 ) , ξi = xi − vi t

(i = 1, 2, 3).

(3.48) (3.49)

For fixed t, |p|, |q| and |r| are all bounded. Moreover, when |x| → ∞, they tend to zero rapidly along all directions except the following directions: (i)

for |p| :

|x2 − x3 | → +∞, 2x1 − x2 − x3 → +∞,

(ii)

for |q| :

|x3 − x1 | → +∞, 2x2 − x1 − x3 → +∞,

(iii)

for |r| :

|x2 − x1 | → +∞, 2x3 − x1 − x2 → +∞.

(3.50)

113


Here we suppose i(µ − µ ¯) > 0. The solution we derived here is called a single-soliton solution. Each solution consists of three waves p, q and r. If i(µ − µ ¯) < 0, the inequalities 2x1 − x2 − x3 → +∞ · · · in (3.50) should be replaced by 2x1 − x2 − x3 → −∞ · · ·. These solutions are not localized, since p, q and r are constants along the straight line x2 − x1 = constant, x3 − x1 = constant. They are a kind of line solitons appeared in Chapter 2. The other components of S are S11 = S22 = S33 =

1 ∆ (µ 1 µ ∆ (¯ 1 µ ∆ (¯

+µ ¯|a|2 e i(µ−¯µ)(ξ2 −ξ1 ) + µ ¯|b|2 e i(µ−¯µ)(ξ3 −ξ1 ) ), + µ|a|2 e i(µ−¯µ)(ξ2 −ξ1 ) + µ ¯|b|2 e i(µ−¯µ)(ξ3 −ξ1 ) ), +

µ ¯|a|2 e i(µ−¯µ)(ξ2 −ξ1 )

+

(3.51)

µ|b|2 e i(µ−¯µ)(ξ3 −ξ1 ) ).

These expressions will be useful in considering multi-soliton solutions. Remark 25 If ai ’s, bi ’s and ci ’s are functions of t rather than constants, then the soliton solution has variable velocities. The term φi (λ)t in (3.41) should be changed to t

ωi (λ, t) =

t0

t

φi (λ, τ ) dτ =

t0

[ai (τ )λ2 + bi (τ )λ + ci (τ )] dτ.

(3.52)

Especially, |p|, |q| and |r| are constants along the curve xi +

µ ¯, t) − ωi (µ, t) ωi (¯ = constant. µ ¯−µ

Different choice of the functions ai ’s, bi ’s and ci ’s leads to different behavior of the waves. For instance, if the integrals of ai ’s, bi ’s and ci ’s are periodic functions, we get the waves which oscillate periodically. If ai ’s, bi ’s and ci ’s increase fast at infinity, the waves have large acceleration and will move apart soon. k-multi-soliton solution can be obtained by k times of Darboux transformation of degree one from the seed solution P = 0. Take the param¯b when a = b), then eters µ0 , µ1 , · · ·, µk−1 (µa = µb , µa = µ Ψl (λ, x, t) = (λI − Sl−1 ) · · · (λI − S0 )Ψ0

(k = 1, · · · , l − 1).

(3.53)

Here the matrices Sk (k = 0, 1, · · · , l − 1) are constructed from Φk by using (3.30) with (3.34) and (3.35) successively. Then we obtain the solution S0 + S1 + · · · + Sl−1 ) (3.54) P l = −(S

114


of the nonlinear equation. Moreover, for considering the behavior of the multi-soliton solutions, we assume (i = 1, 2, · · · , n; l = 0, 1, · · · , k − 1)

vi (µl )

(3.55)

are distinct. Multi-soliton solutions have the following property.

Theorem 3.6 When t → ±∞, a k-multi-soliton is asymptotic to k single solitons. These single solitons for t → −∞ and t → +∞ have the same amplitude. Proof. We first consider the case k = 2. Choose the moving coordinate with velocity {vi (µ1 )}, that is, keep ξi1 = xi −vi (µ1 )t finite and t → ±∞. Then ξi0 − ξj0 = xi − xj − (vi (µ0 ) − vj (µ0 ))t → ±∞ ⎛ ⎜

(3.56)

⎞

σ1±

S0 ∼ ⎜ ⎝

(i = j),

⎟ ⎟ ⎠

σ2± σ3±

(t → ±∞)

(3.57)

where σi± = µ0 or µ ¯0 . Hence, ⎛ ⎜

⎞

(λ − σ1± )e i(λx1 +φ1 (λ)t)

Ψ1 ∼ ⎜ ⎝

(λ − σ2± )e i(λx2 +φ2 (λ)t) (λ − σ3± )e i(λx3 +φ3 (λ)t)

⎟ ⎟. ⎠

(3.58)

Choose H1 so that ⎛ ⎜

1

¯1

−¯ a1 e iζ1

e iζ1

iζ H1 ∼ ⎜ ⎝ a1 e 2 1 b1 e iζ3

where

1

¯1 −¯b1 e iζ1

¯1

e i ζ2

0

0

¯1 e i ζ3

⎞ ⎟ ⎟, ⎠

ζi1 = µ1 xi + φi (µ1 )t − ln(µ1 − σi± )i.

(3.59)

(3.60)

(3.59) and (3.60) are similar to (3.42) and (3.41). For the asymptotic behavior of S1 , the equalities similar to (3.45), (3.47) and (3.51) hold, provided that a, b, µ and ζi ’s are changed to a1 , b1 , µ1 and ζi1 ’s respectively. Hence, when t → ±∞ and ξi1 ’s keep finite, S0off − S1off P 2 = −S


Figure 3.1.

Figure 3.2.

115

p wave of the double soliton solution, t = −1

p wave of the double soliton solution, t = −0.5

is asymptotic to a single soliton solution. The asymptotic behavior of p1 , q1 and r1 are similar when t → +∞ and t → −∞ respectively. However, the extra term − ln[(µ1 − σi± )i in (3.60) are different for t → +∞ and t → −∞. The difference of these two terms represents the phase shift. This means that the “center” of the wave ξi1 = 0, xi − vi (µ1 )t = 0 shifts differently for t → +∞ and t → −∞. When t → ±∞, ξi0 = xi − v(µ0 )t keep finite (then ξi1 → ±∞), the same result can be obtained by the theorem of permutability. Hence the theorem is proved for k = 2. Similar to Chapter 1, multi-soliton solution can be obtained by several Darboux transformations of degree one. Using induction, we can prove that when t → ±∞, a multi-soliton solution splits up into several single soliton solutions. The interactions of these single solitons are elastic. That is, the shape and velocity of the norm of these solitons will not change from t → −∞ to t → +∞, but there is a phase shift. The phase shift is given by the difference of σi± in (3.60) for t → −∞ and t → +∞. In summary, the interactions of the solitons are elastic, and the phase shifts of the solitons can be computed explicitly. This implies that the higher dimensional solitons also have this basic property of 1+1 dimensional solitons. However, these solitons are not local. They do not approach to zero along every direction. Figures 3.1 – 3.5 describe the p wave of the double soliton solution.

116


Figure 3.3.

Figure 3.4.

p wave of the double soliton solution, t = 0

p wave of the double soliton solution, t = 0.5

Figure 3.5.

3.3

p wave of the double soliton solution, t = 1

A reduced system on Rn

As an example, we consider a special AKNS system on Rn . Let ⎛ ⎜ ⎜ ⎜ ⎜ e1 = ⎜ ⎜ ⎜ ⎝

⎞

1

⎟ ⎟ ⎟ ⎟ ⎟, ⎟ ⎟ ⎠

0 0 ..

. 0

⎛ ⎜ ⎜ ⎜ ⎜ e2 = ⎜ ⎜ ⎜ ⎝

⎞

0

⎟ ⎟ ⎟ ⎟ ⎟, ⎟ ⎟ ⎠

1 0 ..

. 0

117

n + 1 dimensional integrable systems ⎛

···

⎜ ⎜ ⎜ ⎜ en = ⎜ ⎜ ⎜ ⎝

⎞

0

⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎠

0 0 ..

.

(3.61)

1

be n diagonal matrices. Let ⎛

Ei = ⎝

⎞

ei

0

0

−ei

⎠

(3.62)

be n 2n × 2n matrices. Consider the system ∂Ψ = (λE Ei + Pi )Ψ, ∂xi

(3.63)

· · · , n) are where Pi ’s are real 2n × 2n matrices. Note that if ai (i = 1, 2, n real numbers so that |ai | are distinct, then the entries of ai Ei are distinct. Hence there exists 2n × 2n matrix P such that Pi = [P, Ei ].

(3.64)

What is more, we want that P satisfies the following extra conditions. This is a reduction of (3.63). Let P be written as ⎛

P =⎝

⎞

P11 P12

⎠

P21 P22

(3.65)

where P11 , P12 , P21 and P22 are n × n matrices. It is required that P22 is symmetric, P11 = −P P21 is anti-symmetric. P12 = −P

(3.66)

This is an AKNS system with strong reduction, which is derived from a geometric problem [7]. The Darboux transformation will be constructed as follows [47]. By the general theory (Section 3.2), let Λ = diag(λ1 , · · · , λ2n ), ψa be a column solution of (3.63) for the eigenvalue λ = λa . Let H = (ψ1 , · · · , ψ2n ),

(3.67)

S = HΛH −1 .

(3.68)

118


From the general method for constructing the Darboux matrix, we know that λI − S is a Darboux matrix, and the Darboux transformation (P, Ψ) → (P , Ψ ) is given by Ψ = (λI − S)Ψ,

P = P − S.

(3.69)

In order that P still satisfies the conditions (3.65) and (3.66), we need special choice of λa ’s and ψa ’s. First, we prove the following lemma.

Lemma 3.7 Suppose P satisfies (3.66), and ⎛

ψ(λ0 ) = ⎝

ψ 1 (λ0 )

⎞ ⎠

ψ 2 (λ0 )

(3.70)

is a column solution of (3.63) with λ = λ0 where ψ 1 , ψ 2 are n-vectors, then, ⎛ ⎞ 2 (λ ) ψ 0 0) = ⎝ ⎠ ψ(λ (3.71) ψ 1 (λ0 ) is a column solution of (3.63) with λ = −λ0 . Proof. Let P11 = F , P21 = G, then ⎛

Pi = ⎝

[F, ei ]

{G, ei }

{G, ei }

[F, ei ]

⎞

⎛

⎠=⎝

⎞

Fi

Gi

Gi

Fi

⎠,

(3.72)

where Fi = [F, ei ],

Gi = {G, ei } = Gei + ei G.

(3.73)

(3.63) can be written as ∂ψ 1 = λei ψ 1 + Fi ψ 1 + Gi ψ 2 , ∂xi ∂ψ 2 = −λei ψ 2 + Gi ψ 1 + Fi ψ 2 ∂xi

(3.74)

which is invariant under the transformation ⎛

ψ=⎝

ψ1 ψ2

⎞

⎛

⎠ −→ ψ = ⎝

ψ2 ψ1

⎞ ⎠,

λ → −λ

(3.75)

The lemma is proved. Now take a non-zero real number µ. Let λi = µ,

λn+i = −µ,

(i = 1, 2, · · · , n).

(3.76)

119


Let ψi = ψi (µ) (i = 1, 2, · · · , n) be n linearly independent column solutions of (3.63) with λ = µ, and ψn+i = ψi

(3.77)

where ψi (i = 1, 2, · · · , 2n) are defined by (3.72). Using (3.74) and Pi∗ = −P Pi we know ∂ (ψ ∗ ψβ ) = 0 (λα = λβ ) (α, β = 1, 2, · · · , 2n). ∂xi α Hence if

ψα∗ ψβ = 0 (λα = λβ )

(3.78)

(3.79)

hold at one point, they hold everywhere. We shall prove later that there are linearly independent ψα ’s so that (3.77) and (3.79) hold. From the construction of S, SH − HΛ = 0, i.e., Sψβ = λβ ψβ , Hence

ψγ∗ S ∗ = λγ ψγ∗ .

ψγ∗ S ∗ Sψβ = λβ λγ ψγ∗ ψβ .

(3.80) (3.81)

When λβ = λγ , ψγ∗ ψβ = 0. When λβ = λγ , λβ λγ = µ2 . Thus, ψγ∗ S ∗ Sψβ = µ2 ψγ∗ ψβ

(3.82)

for all β and γ. Therefore, S ∗ S = µ2 I. On the other hand, ψγ∗ (S ∗ − S)ψβ = (λγ − λβ )ψγ∗ ψβ = 0. So S satisfies Rewrite S as ⎛

S=⎝

S11 S12 S21 S22

S ∗ = S,

S ∗ S = µ2 I.

(3.83) (3.84)

⎞ ⎠

(S Sab (a, b = 1, 2) are n × n matrices).

(3.85)

∗ . Since S ∗ = S, S11 and matrices, and S12 = S21 ⎛ S22 are symmetric ⎞ µI 0 ⎠, by the construction of H we know Moreover, since Λ = ⎝ 0 −µI that (3.86) S11 = −S22 , S12 = −S21 .

This implies that the matrix S satisfies the conditions (3.65) and (3.66) for P . Hence P = P − S also satisfies these conditions. It is remained

120


to prove that there are linearly independent column vectors ψα (λ) satisfying (3.79). In fact, it is sufficient to choose their initial value to be linearly independent and to satisfy (3.79). From S ∗ S = µ2 I, 2 2 − S12 = µ2 I, S11

Let then

[S12 , S11 ] = 0.

(3.87)

σ = µ−1 (S11 + S12 ),

(3.88)

σ ∗ = µ−1 (S11 − S12 ),

(3.89)

σ∗σ

= I, i.e., σ is an n × n orthogonal matrix. and Take σ 0 to be an arbitrary constant orthogonal matrix. Let µ µ 0 0 S11 = (σ 0 + σ 0∗ ), S12 = (σ 0 − σ 0∗ ), 2 2 then 0 2 0 2 0 0 ) − (S12 ) = µ2 I, [S12 , S11 ] = 0. (S11 Let

⎛

S0 = ⎝

0 S11

0 S12

0 0 −S12 −S11

(3.90) (3.91)

⎞ ⎠.

(3.92)

S 0 is a symmetric matrix. Since S 0∗ S 0 = (S 0 )2 = µ2 I, the eigenvalues of S 0 are ±µ. If ψ 0 is the eigenvector with eigenvalue µ, then ψ0 is the eigenvector with eigenvalue −µ. Therefore, ψi0 ’s and ψi0 (i = 1, · · · , n) form a set of 2n linearly independent vectors ψα0 (α = 1, · · · , 2n) with ψn+i = ψi0 (i = 1, · · · n). These are the initial values of ψα ’s to be required and the Darboux matrix S which keeps the reduction is constructed.

Theorem 3.8 The Darboux transformation which keeps the reduction (3.66) exists and is obtained from the above construction of S. Remark 26 (3.63) can be generalized to Qi ∂Ψ )Ψ. = (λE Ei + Pi + ∂xi λ

(3.93)

In this case the above construction of Darboux transformation is still valid. The transformation of Qi is given by Qi = SQi S −1 .

(3.94)

Using the Darboux transformation in (3.66) and (3.93), we can solve the geometric problem in [7] and give the explicit solution of the B¨cklund ¨ transformation presented there.

Chapter 4 SURFACES OF CONSTANT CURVATURE, ¨ BACKLUND CONGRUENCES AND DARBOUX TRANSFORMATION There are many important partial differential equations originating from classical differential geometry. The famous sine-Gordon equation is one of them. The non-Euclidean geometry was initiated in the early nineteenth century. Afterwards, it was found that the surfaces of constant negative Gauss curvature realize the non-Euclidean geometry locally. Therefore, these surfaces were studied extensively. The sine-Gordon equation and its Bäcklund transformation appeared in that period. A surface of constant negative Gauss curvature corresponds to a non-zero solution of the sine-Gordon equation, and the B¨ a¨cklund transformation provides a way to construct a new solution of the sine-Gordon equation from a known one, and a way to construct a new surface of constant negative Gauss curvature from a known surface of constant negative Gauss curvature too. Since the middle of the twentieth century, the transformations of the solutions of some partial differential equations become effective methods in the soliton theory. The role of differential geometry in the soliton theory becomes more and more important. Since the B¨ acklund transformation depends on solving a system of integrable nonlinear partial differential equations, usually its solutions can not be expressed explicitly, except for some very special cases. However, as in the previous chapters, the Darboux transformation is a way to obtain explicit expressions. In this chapter, we combine the Darboux transformation and the classical B¨ acklund transformation to realize the construction of the B¨ acklund congruences and the surfaces of constant negative Gauss curvature. Hence a serious of such congruences and surfaces can be obtained explicitly by purely algebraic algorithm. Besides, we also discuss the surfaces of constant Gauss curvature in the Minkow-

122


ski space R2,1 . The existence and explicit construction of the generalized Backlund ¨ congruence are studied as well. However, the situation becomes more complicated since a surface or a line congruence may be time-like or surface-like. Especially, we give the geometric meaning of the Bäcklund transformation between α = sin α and α = sinh α, the explicit construction of the time-like and space-like surfaces of constant negative Gauss curvature and the B¨ acklund congruences. The Darboux transformations are also used to construct surfaces of constant mean curvature. In summary, in these geometric problems, the Lax pair is not only a tool for solving related partial differential equations, but also the geometric objects that we want to study. With the idea of Darboux transformation, some of the geometric objects can be constructed by purely algebraic algorithm. So it becomes much easier to constructing them by computer. There are also other geometric problems related with integrable systems which we will not discuss in this book. For example, the soliton surfaces introduced by A. Sym [96, 97, 17] and the topics in [90]. For the convenience of the readers we give a sketch of the basic facts of the theory of surfaces in R3 and R2,1 respectively by using the differential forms.

4.1

Theory of surfaces in the Euclidean space R3

We use the moving frame method to introduce the basic theory of the position vector of surfaces in Euclidean space. In R3 , r represents # a point. The length of r is r =| r |= x2 + y 2 + z 2 . As is known, a surface in R3 is a two dimensional differential manifold embedded (or immersed) in R3 . It can be covered by some open subsets (surface charts) which are homeomorphic to connected open regions of a plane. In each surface chart, r can be represented by a vector-valued function of two parameters as r = r(u1 , u2 ) and in the intersection of two surface charts, the parametric representations (u1 , u2 ) and (v 1 , v 2 ) are linked ∂(φ1 , φ2 ) = by differentiable relations v a = φa (u1 , u2 ) (a = 1, 2) with ∂(u1 , u2 ) ∂r (a = 1, 2) form a basis of the tangent plane at r, and 0. ra = ∂ua r1 × r2 n = is the normal vector of the surface. {r1 , r2 , n} form | r1 × r2 | a frame at r. As a basis of the global theory of surfaces, the local theory of surfaces discusses the properties of the surface charts with the above-mentioned parametrization. We use the term surface to indicate

Surfaces of constant curvature, B¨cklund ¨ congruences

123

a surface chart, since we only consider the local theory. The frame (r1 , r2 , n) is called a natural frame because it is obtained naturally from the parametrization of the surface. Differentiate r1 , r2 , n and write them as linear combinations of ra , n, we obtained the fundamental equations of a surface dr = dua ra , dra = ωab rb + ωa3 n, dn =

ω3b rb ,

(4.1)

(a, b = 1, 2),

where ωab , ωa3 , ω3b are 1-forms of u1 , u2 . In (4.1) and hereafter, the summation convention is used, i.e., the symbol Σ is omitted for double indices. The first fundamental form of the surface is I = ds2 = dr · dr = gab dua dub ,

(4.2)

gab = ra · rb = gba .

(4.3)

where Since n · dra + ra · dn = 0, ω3b and ω3a are related by ωa3 = −gab ω3b .

(4.4)

ωa3 = bab dub .

(4.5)

Let 2

From d r = 0,

ωa3

∧

dua

= 0. We have bab = bba . The quadratic form

II = −dr · dn = −gab ω3b dua = bab dua dub

(4.6)

is the second fundamental form of the surface with coefficients bab = −

∂r ∂n ∂2r · = · n. ∂ua ∂ub ∂ua ∂ub

(4.7)

Two principal curvatures are eigenvalues of the second fundamental form with respect to the first fundamental form, i.e., they are two roots of det(bαβ − λgαβ ) = 0. Since the first fundamental form is positive definite, two principal curvatures are both real. Now ωab can be written as ωab = Γbac duc ,

(4.8)

124


where Γbac ’s are the Christoffel symbols of the surface. From d2 r = 0, we obtain dua ∧ ωab = 0, hence Γcab = Γcba . Moreover, ra · rb = gab and ra · drb + dra · rb = dgab imply dgab = gac ωbc + gcb ωac . Therefore,

1 ∂gad ∂gbd ∂gab Γcab = g cd + − . (4.9) 2 ∂ub ∂ua ∂ud Exterior differentiating the fundamental equations (4.1) gives the integrability conditions of (4.1). Using d2 r = 0 and the exterior differentiations of the second and third equations of (4.1), we obtain dωab + ωcb ∧ ωac = ωa3 ∧ ω3b , dωa3 + ωb3 ∧ ωab = 0.

(4.10) (4.11)

(4.10) and (4.11) are called Gauss equations and Codazzi equations of the surface respectively. The left hand side of (4.10) are usually written as 1 (4.12) dωab + ωcb ∧ ωac = Rb acd duc ∧ dud , 2 where ∂Γbad ∂Γbac Rb acd = − + Γbec Γead − Γbed Γeac (4.13) ∂uc ∂ud is the Riemannian curvature tensor. It is fully determined by the coefficients of the fundamental form gab and their derivatives of first and second order. From (4.10), (4.4) and (4.5), 1 b R acd duc ∧ dud = −bac duc ∧ g be bed dud . 2

(4.14)

Rb acd = −g be (bac bed − bad bec ).

(4.15)

Rbacd = gbe Re acd ,

(4.16)

Hence Denote then Rbacd = bbc bad − bac bbd , where the subscripts a, b, c, d take the values 1 or 2. In fact, there is only one independent equation, i.e., R1212 = b11 b22 − b212 .

(4.17)


125

This is another form of the Gauss equation. It can also been written as R1212 b11 b22 − b212 = 2 2 =K g11 g22 − g12 g11 g22 − g12

(4.18)

where K is called the total curvature or Gauss curvature of the surface. The left hand side of (4.18) is determined by the first fundamental form. Before Gauss’ work, K is expressed by the first fundamental form together with the second fundamental form (second equality of (4.18)). The first equality in (4.18) is the famous Gauss Theorem. It implies that the Gauss curvature is actually determined by the first fundamental form of the surface. The properties which can be determined by the first fundamental form are call intrinsic properties. The above discussion is summarized as follows. For a given surface S, its first and second fundamental forms satisfy the Gauss-Codazzi equations. Conversely, given two differential forms I = gab dua dub (a, b = 1, 2, gab is positive definite), II = bab dua dub , and suppose that the Gauss-Codazzi equations hold, then there exists a surface chart whose first and second fundamental forms are I and II respectively. This surface chart is uniquely determined in a simply connected region up to rigid motions and reflections. This is the fundamental theorem of surfaces. In fact, the surface chart r = r(u, v) is determined by solving the system of linear equations (4.1). Since the integrability condition of (4.1) is just the Gauss-Codazzi equations, (4.1) is completely integrable. Hence the fundamental theorem of surfaces holds. The tangent vectors r1 and r2 can be replaced by their linear combinations. Suppose e1 and e2 are linear combinations of r1 , r2 and e1 , e2 are orthogonal with each other, then {e1 , e2 , n} form an orthogonal frame of the surface at r. The fundamental equations of the surface can be written as dr = ω a ea , dea = ωab eb + ωa3 n,

(4.19)

dn = ω3a ea . There are also the following relations among ω a and ωab : dω a + ωba ∧ ω b = 0 ωij

+

ωji

=0

(a, b = 1, 2),

(i, j = 1, 2, 3).

(4.20) (4.21)

ωba (a, b = 1, 2) are uniquely determined by (4.20) and the relation ωba + ωab = 0. In the orthogonal frame, the Gauss-Codazzi equations are simplified as (4.22) dω21 = ω23 ∧ ω31

126


and dωa3 + ωb3 ∧ ωab = 0.

(4.23)

Especially, the Gauss equation can also be written as dω21 = R1212 ω 1 ∧ ω 2 = Kω 1 ∧ ω 2 .

(4.24)

Hereafter, we will mainly use the orthogonal frames. Moreover, the orientation of the frame {e1 , e2 , n} will not always be fixed, that is, n may be replaced by −n.

4.2

Surfaces of constant negative Gauss curvature, sine-Gordon equation and Backlund ¨ transformations 4.2.1 Relation between sine-Gordon equation and surface of constant negative Gauss curvature in R3

Suppose S is a surface of constant negative Gauss curvature in R3 . By using a scaling transformation of S, we can also suppose K = −1. Take the lines of curvature as coordinate curves and let e1 and e2 be the unit tangent vectors of the lines of curvature. Denote dr =

∂r ∂r du + dv = A due1 + B dve2 , ∂u ∂v ω 1 = A du,

ω 2 = B dv.

(4.25) (4.26)

The first fundamental form of the surface is ds2 = A2 du2 + B 2 dv 2 ,

(4.27)

and the second fundamental form is II = k1 A2 du2 + k2 B 2 dv 2 = k1 (ω 1 )2 + k2 (ω 2 )2 ,

(4.28)

where k1 and k2 are principal curvatures. The Gauss curvature K = k1 k2 (= −1). On the other hand, (4.6) implies II = ω13 ω 1 + ω23 ω 2 .

(4.29)

Comparing with (4.28), we get ω13 = k1 ω 1 = k1 A du,

ω23 = k2 ω 2 = k2 B dv.

From (4.20), dω a + ωba ∧ ω b = 0,

(4.30)


hence ω12 = −ω21 = −

Av Bu du + dv. B A

127

(4.31)

The Codazzi equation dω13 + ω23 ∧ ω12 = 0 leads to (k1 A)v = k2 Av , i.e., (4.32) (k1 − k2 )Av + k1v A = 0. α α Since K = k1 k2 = −1, we can set k1 = tan , k2 = − cot , (0 < α < π), 2 2 then 1 k1 − k2 = α. α sin cos 2 2 Substituting it into (4.32), we get

(log A)v − log cos Hence A = cos

α = 0. 2 v

α U (u). 2

Similarly,

α V (v). 2 Here U (u) and V (v) are functions depending on u and v respectively. Let du1 = U (u) du, dv1 = V (v) dv, then u1 and v1 become new parameters and will still be written as u, v. Hence B = sin

α α , B = sin , 2 2 α α 1 2 ω = cos du, ω = sin dv, 2 2 α α 3 3 ω1 = sin du, ω2 = − cos dv, 2 2 1 2 ω1 = (αv du + αu dv) = −ω21 . 2 A = cos

(4.33)

Substituting them into Gauss equation (4.10), we get dω21 = ω23 ∧ ω21 = k1 k2 ω 1 ∧ ω 2 = −ω 1 ∧ ω 2 , from which it is seen that α satisfies the sine-Gordon equation αuu − αvv = sin α.

(4.34)

128


The coordinates we are using are called the Chebyshev coordinates, and the corresponding frame is called the Chebyshev frame. It can be checked directly that the Codazzi equations are consequence of the sine-Gordon equation (4.34). Hence the fundamental theorem of surfaces leads to the following theorem.

Theorem 4.1 For any solution α of the sine-Gordon equation (4.34) (0 < α < π), one can construct a surface of constant negative Gauss curvature by solving the fundamental equation of surfaces (4.19), in which the coefficients ω a , ωba , ωb3 are given by (4.33). Note that under the Chebyshev coordinates, the first fundamental form of the surface is ds2 = cos2

α 2 α du + sin2 dv 2 , 2 2

(4.35)

and the second fundamental form is II = cos

α α sin (du2 − dv 2 ). 2 2

(4.36)

Hence two systems of asymptotic curves are real and the directions du : dv = 1 : ±1 are the asymptotic directions of the surface. The cosine of the angle between these two directions is cos θ = cos2

α α − sin2 = cos α. 2 2

(4.37)

Therefore, the geometric meaning of α is the angle between two asymptotic curves of the surface. Remark 27 Theorem 4.1 is a local result. On the other hand, the classical Hilbert Theorem says that there is no complete surface of constant negative Gauss curvature in R3 [56]. Here a complete surface means a two dimensional open manifold on which each geodesic can be extended infinitely. Remark 28 On a surface of constant positive Gauss curvature (not a sphere), there are also Chebyshev coordinates. In this case, α α du, ω 2 = sinh dv, 2 2 α α 3 3 ω1 = sinh du, ω2 = cosh dv, 2 2 1 ω12 = −ω21 = (−αv du + αu dv). 2 ω 1 = cosh

(4.38)


129

The construction of the surface depends on the solution of the negative sinh-Laplace equation

α = − sinh α. (4.39) The proof of these facts is similar with the case of constant negative Gauss curvature. Remark 29 Suppose a solution of the sine-Gordon equation or the negative sinh-Laplace equation is known. The construction of surface of constant negative Gauss curvature or constant positive Gauss curvature is reduced to solving the fundamental equations of the surface. These are completely integrable systems of linear partial differential equations. The solutions of this system can be obtained by solving ordinary differential equations. However, the construction of explicit solutions is still not easy. The Darboux transformation will provide an efficient method for the explicit construction.

4.2.2

Pseudo-spherical congruence

Line congruences originated from the study of refraction and reflection of light. A two-parametric family of straight lines is called a line congruence. For example, all the normal lines of a surface constitute a line congruence which is called the normal congruence of the surface. However, in general, a line congruence may not be a normal congruence, that is, there may not exist a surface which is orthonormal to all the straight lines in the congruence. Locally, a line congruence can be expressed as follows. Suppose S is a surface expressed as X = X(u, v). Given a unit vector ξ(u, v) at each point of the surface, let Y(u, v, λ) = X(u, v) + λξ(u, v).

(4.40)

When (u, v) are fixed and λ changes, Y forms a straight line. Hence Y represents a line congruence. The surface S is called the reference surface. Clearly, the reference surface of a line congruence is quite arbitrary. For a curve C on the reference surface S, the straight lines in the congruence passing through C form a ruled surface. Suppose the equations of C are u = u(t), v = v(t). Substituting them into (4.40), we can get the equations of the ruled surface parametrized by t and λ. If there exists dY and ξ are parallel, then the ruled surface becomes λ = λ(t) such that dt a developable surface, and the curve Y(u, v, λ) = Y(u(t), v(t), λ(t)) is the line of regression of the developable surface. Now Y = Y(t) = X(u(t), v(t)) + λ(t)ξ(u(t), v(t)),

130


and

dY dX dλ dξ = + ξ + λ(t) = µξ, dt dt dt dt dX dξ that is, ξ, and are linearly dependent. Hence, the ruled surface dt dt is developable if and only if u = u(t), v = v(t) satisfy det

dX dξ

,

dt dt

,ξ

= 0.

(4.41)

The differential form of this condition is det(Xu du + Xv dv, ξu du + ξv dv, ξ ) = 0. Write this equation as A du2 + 2B du dv + C dv 2 = 0.

(4.42)

This is the quadratic equation of du : dv. When B 2 − AC > 0, there are two different real roots for du : dv which correspond to two developable surfaces in the line congruence. Thus for each line in the congruence there are two developable surfaces passing through it. As is known, a developable surface is generated by the tangent lines of a spatial curve named line of regression. When B 2 − AC > 0, each line in the congruence are tangent to a line of regression of each developable surface. The tangent points are called focal points. The set of all focal points form the focal surfaces. A focal surface is also obtained from the lines of regression. Therefore a line congruence can be regarded as the set of all common tangent lines of two focal surfaces. We only consider the case B 2 − AC > 0. Suppose S and S ∗ are two focal surfaces of a line congruence. Suppose P P ∗ is a line in the congruence, which is the common tangent line of two focal surfaces, and P , P ∗ are the tangent points. The correspondence between P and P ∗ leads to the correspondence between S and S ∗ . In differential geometry, this is called a Laplace transformation between the surfaces S and S ∗ (it is different from the terminology Laplace transformation in analysis). Suppose n and n∗ are unit normal vectors of S at P and of S ∗ at P ∗ respectively. Let τ be the angle between n and n∗ , and l be the distance between P and P ∗ , i.e., n · n∗ = cos τ dpp∗ = l

(sin τ = 0), (l = 0).

(4.43) (4.44)

When τ and l are both constants, the congruence is called a pseudospherical congruence.

131


Historically, a surface of constant negative Gauss curvature is called a pseudo-sphere. This is the origin of the name pseudo-spherical congruence. A pseudo-spherical congruence is also called a B¨äcklund congruence.

Theorem 4.2 (B¨ a ¨cklund Theorem) Two focal surfaces S and S ∗ of a pseudo-spherical congruence are both the surfaces of constant negative Gauss curvature with the same K = − sin2 τ /l2 . Proof. Let P be a general point on S, with position vector r(u, v). Its −→

corresponding point P ∗ has position vector r ∗ (u, v). P P ∗ is a common −→

tangent line of S and S ∗ . Let e1 be the unit vector along P P ∗ . Let {e1 , e2 , n} be the orthogonal frame of S at P and {e1∗ , e2∗ , n ∗ } be the orthogonal frame of S ∗ at P ∗ with e1∗ = e1 . From the above assumptions, r ∗ = r + le1 , e2∗ = cos τ e2 + sin τ n, n ∗ = − sin τ e2 + cos τ n.

(4.45)

(4.46)

The fundamental equations of S and S ∗ are dr = ω a ea , dea = ωab eb + ωa3 n, dn = and

(4.47)

ω3a ea ,

dr ∗ = ω ∗a ea∗ , dea∗ = ωa∗b eb∗ + ωa∗3 n,

(4.48)

dn ∗ = ω3∗a ea

respectively. From (4.45) – (4.48), we get dr ∗ = ω a ea + lde1 = (ω a + lω1a )ea + lω13 n = ω ∗a ea∗ = ω ∗1 e1 + ω ∗2 (cos τ e2 + sin τ n). Hence

l ω3 . (4.49) sin τ 1 Rewrite (4.45) and (4.46) so that r, e1 , e2 are expressed by r∗ , e∗1 , e∗2 , then we have l ω ∗3 . ω2 = (4.50) sin τ 1 ω ∗1 = ω 1 ,

ω ∗2 =

132


By (4.46) and (4.48), de2∗ = cos τ (ω21 e1 + ω23 n) + sin τ (ω31 e1 + ω32 e2 ) = ω2∗1 e1∗ + ω2∗3 n ∗ = ω2∗1 e1 + ω2∗3 (− sin τ e2 + cos τ n), which leads to

ω2∗3 = ω23 .

The Gauss curvature K ∗ of S ∗ is determined by dω1∗2 = ω1∗3 ∧ ω3∗2 = −K ∗ ω ∗1 ∧ ω ∗2 .

(4.51)

Denote ω2∗3 = ω23 = bω 1 + cω 2 , then ω1∗3 ∧ ω3∗2 =

sin τ 2 sin τ 1 ω ∧ (−bω 1 ) = b ω ∧ ω2 . l l

(4.52)

On the other hand, ωa3 ∧ ω a = 0 implies that ω13 has the form ω13 = aω 1 + bω 2 . Hence

1 1 sin τ ∗1 ω ∧ ω ∗2 . ω 1 ∧ ω 2 = ω 1 ∧ ω13 = b b l

Substituting (4.52) and (4.53) into (4.51), we get K ∗ = − larly, K = −

4.2.3

sin2 τ holds. The theorem is proved. l2

(4.53) sin2 τ . Simil2

B¨ acklund transformation

Theorem 4.1 implies that a solution α (α = 0) of the sine-Gordon equation leads to a surface of constant negative Gauss curvature. The Backlund ¨ theorem implies that two focal surfaces of a pseudo-spherical congruence have the same constant negative Gauss curvature, and these two focal surfaces correspond to two solutions of the sine-Gordon equation. However, the above proof of the B¨ acklund theorem does not imply the existence of the Bäcklund congruences. We are going to construct the Bäcklund congruence from a given surface S of constant negative Gauss curvature with K = −1. Suppose (e1 , e2 , n) is the Chebyshev frame of S at r, and (u, v) are the Chebyshev coordinates. Let r ∗ = r + le = r + l(cos θe1 + sin θe2 )

(4.54)

be the transformation from the surface S to the surface S ∗ . Let P and P ∗ be the corresponding points on S and S ∗ with position vectors r and


133

Figure 4.1.

r ∗ respectively. It is required that the lines P P ∗ form a pseudo-spherical congruence. Here e = cos θe1 + sin θe2 , or θ, is to be determined. Suppose S corresponds to a solution α of the sine-Gordon equation. Differentiating (4.54) and using (4.33), we have dr ∗ = dr + l(cos θde1 + sin θde2 ) + l(− sin θe1 + cos θe2 )dθ & α αu ' α v du + dv e1 cos du − l sin θdθ − l sin θ = 2 2 2 & α αv αu ' + sin dv + l cos θdθ + l cos θ du + dv e2 2 2' 2 & α α + l sin cos θ du − l cos sin θ dv n. 2 2

(4.55)

Since the angle τ between the normal vector n ∗ of S ∗ at P ∗ and the normal vector n of S at P is a constant, and P P ∗ is tangent to S ∗ at P ∗ , we have n ∗ = sin τ sin θe1 − sin τ cos θe2 + cos τ n.

(4.56)

n ∗ is the normal vector of S ∗ , hence it must satisfy dr ∗ · n ∗ = 0. Thus α α l sin τ dθ − sin τ cos sin θ du − sin cos θ dv 2 α α2u v du + dv +l sin τ 2 2 α α −l cos τ sin cos θ du − cos sin θ dv = 0. 2 2

(4.57)

If (4.57) can be solved for θ, then P P ∗ generates exactly a pseudospherical congruence. From the B¨ äcklund theorem, the surfaces S and sin2 τ ∗ S ∗ have the same K = K = − 2 . Without loss of generality, l ∂α1 1 ∂α1 dv . (4.57) suppose l = sin τ . Let θ = α1 /2, then dθ = du + 2 ∂u ∂v

134


can be written as a system of partial differential equations: ∂α ∂α α1 α α1 α 1 1 sin τ + = sin τ sin cos + cos τ cos sin 2 ∂u ∂v 2 2 2 2 ∂α ∂α α α1 α α1 1 1 sin τ + sin − cos τ sin cos . = − sin τ cos 2 ∂v ∂u 2 2 2 2 (4.58)

This system is the original form of the Bäcklund transformation of sine-Gordon equation. Regarded as a system of α1 , its integrability condition is the sine-Gordon equation αuu − αvv = sin α. Since it holds already, the solution α1 of (4.58) exists uniquely for a given value α10 (0 < α10 < π) of α1 at (u0 , v0 ). From αuv = αvu , we see that α1 also satisfies the sine-Gordon equation. This transformation a¨cklund transformation between two solutions from α to α1 is called a B¨ of the sine-Gordon equation. Thus we have obtained

Theorem 4.3 Let S be a surface of constant negative Gauss curvature K = − sin2 τ /l2 in R3 , l and sin τ be non-zero constants and e0 be a unit tangent vector of S at P ∈ S. Then there exists a surface S ∗ of K = − sin2 τ /l2 such that the common tangent lines of SS ∗ constitute a pseudo-spherical congruence with parameters (l, sin τ ) and the line from P in the direction e0 belongs to the congruence. This theorem indicates the method of the construction of pseudospherical congruences. In differential geometry, the surface S ∗ is called a B¨äcklund transformation of the surface S. For S ∗ , (4.55) can be rewritten as dr ∗ = A∗ due1∗ + B ∗ dve2∗ .

(4.59)

We shall prove that the Chebyshev coordinates (u, v) of S are also the Chebyshev coordinates of S ∗ . Since e1∗ and e2∗ are unit vectors, (4.48) and (4.55) lead to α1 α1 B ∗ = sin , A∗ = cos , 2 2 α1 α α1 α ∗ e1 = (cos cos − cos τ sin sin )e1 2 2 2 2 α α1 α α1 α +(cos sin + sin τ sin cos )e2 + sin τ sin n, (4.60) 2 2 2 2 2 α1 α α1 α e2∗ = (sin cos + cos τ cos sin )e1 2 2 2 2 α1 α α1 α α − cos τ cos cos )e2 − sin τ cos n. +(sin sin 2 2 2 2 2


Hence

135

α1 α1 du, ω2∗ = sin dv. (4.61) 2 2 Computing dn ∗ from (4.56) and expanding it in terms of e1∗ , e2∗ , we get dn ∗ = ω3∗1 e1∗ + ω3∗2 e2∗ , (4.62) α1 α1 du, ω3∗2 = cos dv. ω3∗1 = − sin 2 2 Comparing with (4.33), we conclude that (u, v) are the Chebyshev coordinates of S ∗ . The difference of sign can be eliminated by using −n∗ instead of n∗ . ω1∗ = cos

Theorem 4.4 Suppose S ∗ is the Backlund ¨ transformation of a surface S of constant negative Gauss curvature, and (u, v) are the Chebyshev coordinates of S, then (u, v) are also the Chebyshev coordinates of S ∗ . Thus one can apply the Bä¨cklund transformation successively to obtain a sequence of pseudo-spherical congruences and surfaces of constant negative Gauss curvature.

4.2.4

Darboux transformation

In the last subsection, we have got the following well-known facts. Suppose α is a solution of the sine-Gordon equation and S is the corresponding surface of constant negative Gauss curvature. If α1 is a solution of (4.58), then (4.54) gives an explicit expression of the B¨ äcklund transformation of S where θ = α1 /2. In order to get the explicit expression of S ∗ , we need to use Darboux transformation to get the explicit expression of α1 . Let u−v u+v , η= , (4.63) ξ= 2 2 then (4.58) becomes α1 − α , 2 α1 + α 2 , (α1 − α)η = sin β 2 (α1 + α)ξ = 2β sin

(4.64)

where

1 − cos τ = 0. sin τ The sine-Gordon equation becomes β=

αξη = sin α.

(4.65)

(4.66)

136


In Chapter 1, the Darboux transformation for the sine-Gordon equation has already been introduced. The construction is as follows. Let α be a solution of the sine-Gordon equation, then the Lax pair ⎛ αη ⎞ λ − 2 ⎠ Φ, Φη = ⎝ αη −λ 2 (4.67) ⎛

⎞

1 ⎝ cos α sin α ⎠ Φ Φξ = 4λ sin α − cos α ⎛

is completely integrable. It can be checked that if ⎝ ⎛

solution of the Lax pair (4.67), then ⎝

⎞

φ2 (λ) −φ1 (λ)

⎞

φ1 (λ) φ2 (λ)

⎠ is a column

⎠ is a column solution ⎛

of the Lax pair with λ replaced by −λ. Hence ⎝

⎞

φ2 (−λ)

⎠ is also a

−φ1 (−λ) column solution of the Lax pair (4.67). Φ(λ) can be chosen as ⎛

Φ(λ) = ⎝ ⎛

Take λ = λ1 = 0. Let ⎝

⎞

φ1 (λ)

φ2 (−λ)

φ2 (λ) −φ1 (−λ)

⎠.

(4.68)

⎞

h1 h2

⎠ be a column solution of the Lax pair

(4.67) with λ = λ1 , and h2 = φ2 (λ1 ) − b1 φ1 (−λ1 ),

h1 = φ1 (λ1 ) + b1 φ2 (−λ1 ), ⎛

where b1 is a constant. Then ⎝ (4.67) with λ = −λ1 . Let

⎛

H=⎝

−h2 h1

⎞

⎠ is a solution of the Lax pair

h1 −h2 h2

h1

⎞ ⎠, ⎛

then det H = h21 + h22 , which is not zero provided that ⎝

⎞

h1

⎠ is not h2 a trivial solution. (From the property of linear ordinary differential

137


equations, h1 = h2 = 0 holds everywhere if it holds at one point.) Let ⎛

S =H⎝

⎞

λ1

0

0

−λ1

⎛

⎛

2h1 h2

λ1 ⎝ ⎠ h21 + h22 2h1 h2 h22 − h21 ⎛ ⎞ ⎞ ψ ψ sin cos 2σ 2 2 ⎟ ⎠ = λ1 ⎜ ⎝ ψ ψ ⎠, 2 −1 + σ − cos sin 2 2

⎠ H −1 =

λ1 ⎝ 1 − σ 2 = 1 + σ2 2σ where

h2 , h1

σ= cos

⎞

h21 − h22

ψ = 4 tan−1 σ,

1 − σ2 ψ = , 2 1 + σ2

sin

(4.69)

(4.70)

ψ 2σ = . 2 1 + σ2

(4.71)

The matrix D(λ) = λI − S

(4.72)

Φ1 (λ) = D(λ)Φ(λ),

(4.73)

is the Darboux matrix. Let

then Φ1 (λ) satisfies the Lax pair (4.67) when α is replaced by certain α1 . This can be verified directly as follows. From Φ1η = −S Sη Φ + (λI − S)Φη

⎛

− Sη + (λI − S) ⎝

= ⎛

=⎝

λ

−α1η /2

α1η /2

−λ

and det Φ(λ) = 0, we obtain ⎛

−S Sη + (λI − S) ⎝

λ

−αη /2

αη /2

−λ

⎞

⎞

λ

−αη /2

αη /2

−λ

⎞ ⎠ Φ(λ)

⎠ (λI − S)Φ

⎛

⎠=⎝

λ

−α1η /2

α1η /2

−λ

⎞ ⎠ (λI − S).

Now compare the coefficients of the powers of λ in both sides. The terms with λ2 are equal identically. The terms with λ lead to α1η − αη = −4λ1 sin

ψ . 2

(4.74)

The term with λ0 leads to α1η = −αη + ψη .

(4.75)

138


Similarly, considering the part of the Lax pair with derivative to ξ, we get (4.76) α1 = ψ − α, 1 α − α1 . (4.77) sin λ1 2 (4.76) gives the new solution of the sine-Gordon equation explicitly by 1 Darboux transformation. Let β = − , then (4.75) – (4.77) lead to 2λ1 (4.64). Therefore, α1 is a solution of the partial differential equations (4.64) in the Bäcklund transformation. More precisely, we have 1 Theorem 4.5 Take λ1 = − , then α1 derived by the Darboux trans2β formation (Φ, α) → (Φ1 , α1 ) is a solution of (4.64). ψξ =

The explicit construction of the surface S ∗ of constant negative Gauss curvature is as follows.

Theorem 4.6 Suppose α (0 < α < π) is a solution of the sine-Gordon equation, and Φ is the fundamental solution of its Lax pair. Let S be the surface of constant negative Gauss curvature corresponding to α (expressed in Chebyshev coordinates). Then the Darboux transformation gives a new solution α1 of the sine-Gordon equation, and the transformation (4.54) gives the surface of constant negative Gauss curvature corresponding to α1 . According to this theorem, we can get a series of surfaces of constant negative Gauss curvature by applying Darboux transformation (4.54) to a solution α of the sine-Gordon equation, the fundamental solution Φ of its Lax pair and the corresponding surface S of constant negative Gauss curvature with Chebyshev coordinates. This can be illustrated by the following diagram.

α

solution Φ of Lax pair −→ (α1 , Φ1 ) −→ (α2 , Φ2 ) −→ ↓ surface S

−→

S∗

···

↓ −→

S ∗∗

−→

In the diagram, the arrows from α to Φ and from α to S are realized by solving a system of linear integrable partial differential equations. Other arrows can be realized by purely algebraic operations. Here the algebraic operations include exponential, trigonometric and inverse trigonometric functions. In the construction of B¨ acklund transformation, we only need α1 α α α1 and sin in terms of σ, cos and sin . These to express cos 2 2 2 2 expressions are purely algebraic.

139


4.2.5

Example

Starting from α = 0, a series of non-trivial solutions of the sineGordon equation can be obtained by using Darboux transformation. With α = 0, the Lax pair is ⎛

Φη = ⎝

⎞

λ

0

0 −λ

⎛

⎠ Φ,

hence the fundamental solution is ⎛

Φ=⎝

⎞

1 ⎝ 1 0 ⎠ Φξ = Φ, 4λ 0 −1 ⎞

ξ

eλη+ 4λ

0

0

e−λη− 4λ

ξ

⎠.

Take λ = λ1 . From (4.70) we have σ = −be

−2λ1 η− 2λξ

1

.

For simplicity, take b = −1, then α1 = 4 tan−1 σ = 4 tan−1 (e

−2λ1 η− 2λξ

1

).

The Darboux matrix is ⎛

D(λ) = ⎝

⎞

λ + λ1 tanh γ

−λ1 sech γ

−λ1 sech γ

λ − λ1 tanh γ

γ = −2λ1 η −

ξ . 2λ1

where

⎠,

Then Φ1 (λ) = D(λ)Φ(λ) ⎛

=⎝

ξ

ξ

⎞

ξ

−λ1 sech γe−λη− 4λ

(λ + λ1 tanh γ)eλη+ 4λ

ξ

(λ − λ1 tanh γ)e−λη− 4λ

−λ1 sech γeλη+ 4λ

⎠.

When λ = λ2 , φ1 (λ2 ) = (λ2 + λ1 tanh γ)e φ2 (λ2 )

σ =

= (−λ1 sech γ)e

(−λ1 sech γ)e

λ2 η+ 4λξ

(λ2 + λ1 tanh γ)e

2

λ2 η+ 4λξ

λ2 η+ 4λξ 2

2

,

,

+ b(λ2 − λ1 tanh γ)e

λ2 η+ 4λξ 2

+ b(−λ1 sech γ)e

−λ2 η− 4λξ

2

−λ2 η− 4λξ 2

140


and the new solution α2 = 4 tan−1 σ − α1 . On the other hand, by the fundamental equations of the surface for α = 0, dr = e1 du,

de1 = 0,

de2 = −n dv,

dn = e2 dv,

we have r = (u, 0, 0),

e1 = (1, 0, 0),

e2 = (0, cos v, sin v),

n = (0, sin v, − cos v).

It is not a surface, but a line together with a two-parametric family of orthonormal frames along the line. However, since the fundamental equations hold, we can still use it to construct surface of constant negative Gauss curvature. Now α1 = 2 tan−1 σ, θ= 2 cos θ = − tanh γ, sin θ = sech γ. According to (4.54), the equation for the surface S ∗ is α1 α1 r ∗ = r + l cos e1 + sin e2 , 2 2 or x1 = u − l tanh γ, x2 = l sech γ cos v, x3 = l sech γ sin v, −4λ1 ξ l= , γ = −2λ η − . 1 2λ1 1 + λ21 In order to construct the second surface S ∗∗ , we first write down the expressions of e1∗ and e2∗ . From (4.60) and α = 0, α1 α1 e1∗ = cos e1 + sin e2 , 2 α1 2 α1 e2∗ = cos τ sin e1 − cos e2 − sin τ n. 2 2 Hence the equation for the second surface is 2λ1 α1 α1 e e2 cos + sin r ∗∗ = ue1 − 1 2 2 1 + λ21 2λ2 α2 ∗ α2 ∗ − cos e + sin e . 1 2 2 2 1 + λ22 These surfaces of constant negative Gauss curvature can be plotted by computer. In [85], there are figures for some interesting surfaces including several surfaces of constant negative Gauss curvature.


141

4.3

Surface of constant Gauss curvature in the Minkowski space R2,1 and pseudo-spherical congruence 4.3.1 Theory of surfaces in the Minkowski space R2,1

As the Euclidean space R3 , the Minkowski space R2,1 is a three dimensional flat space. A vector l in R2,1 has three coordinates l1 , l2 and l3 . In an orthonormal coordinate system, the inner product of two vectors l = (l1 , l2 , l3 ) and m = (m1 , m2 , m3 ) in R2,1 is given by l · m = l1 m1 + l2 m2 − l3 m3 ,

(4.78)

and the square of the norm of a vector l is l2 = l12 + l22 − l32 ,

(4.79)

which is not positive definite. According to the sign of l2 , there are three types of non-zero vectors: l2 > 0

space-like,

l2 < 0

time-like,

l2 = 0

light-like.

(4.80)

A light-like vector is also called a null vector. All the null vectors form a light cone l12 + l22 − l32 = 0. A space-like vector points to the exterior of the light cone, whereas a time-like vector points to the interior of the light cone. Moreover, according the sign of l3 , time-like and light-like vectors are divided into past-oriented and future-oriented vectors. All the vectors orthogonal to a time-like vector are space-like. A vector orthogonal to a space-like vector may be space-like, time-like or lightlike. The vectors orthogonal to a light-like vector may be light-like or space-like, and all these vectors form the tangent planes of the light cone. A light-like vector is always orthogonal to itself. A point (x, y, z) in R2,1 can be expressed by a position vector r. In the special theory of relativity, the space and time are expressed uniformly by a four dimensional Minkowski space-time. The study of the geometry of R2,1 may help understanding the space-time in the relativity. The terms space-like, time-like, light-like etc. also originate from the relativity. For a surface in R2,1 , normal vector n can also be defined as the unit vector orthogonal to the tangent plane. When n is time-like, all the vectors on the tangent plane are space-like, and the surface is called space-like. When n is space-like, the vectors on the tangent plane may

142


be space-like, time-like or light-like, and the surface is called time-like. There is also light-like surface, whose normal vector is light-like. In this case, the normal vector locates in the tangent plane. A mixed surface is a connected surface which includes space-like, time-like and light-like parts. In this book, we only consider the space-like surfaces and time-like surfaces. Similar to R3 , the fundamental equation of a space-like or time-like surface can be written as dr = ω a ea , dea = ωab eb + ωa3 n,

(4.81)

dn = ω3a ea . The integrability conditions are dω a + ωba ∧ ω b = 0,

ωa3 ∧ ω a = 0,

dωba + ωca ∧ ωac = −ω3a ∧ ωb3

(Gauss equation),

(4.82)

dωa3 + ωb3 ∧ ωab = 0 (Codazzi equation). Space-like and time-like surfaces should be considered separately. (1) space-like surface For the orthogonal frame {e1 , e2 , n} at r, ea · eb = δab ,

ea · n = 0,

n2 = −1,

(4.83)

ω11 = ω22 = 0,

ω21 = −ω21 ,

ωa3 = ω3a .

(4.84)

I = dr · dr = (ω 1 )2 + (ω 2 )2 = gab ωa ωb ,

(4.85)

and The first fundamental form of the surface is

where g11 = g22 = 1,

g12 = g21 = 0.

Let ωa3 = bac ω c

(bac = bca ),

(4.86)

then the second fundamental form of the surface is II = −dr · dn = −ωa3 ω a = −bac ω a ω c . The Gauss equation is dωba

= 12 Ra bcd ω c ∧ ω d = −ω3a ∧ ωb3 = − 12 (bac bbd − bad bbc )ω c ∧ ω d .

(4.87)


143

As before, it contains only one independent equality R1212 = −(b11 b22 − b212 )

(R1212 = R1 212 ).

(4.88)

The intrinsic definition of Gauss curvature is K=

R1212 2 = R1212 . g11 g22 − g12

(4.89)

Hence the Gauss equation can be written as K = −(b11 b22 − b212 ),

(4.90)

dω21 = −ω31 ∧ ω23 = R1212 ω 1 ∧ ω 2 = Kω 1 ∧ ω 2 .

(4.91)

while

(4.90) differs from the corresponding formula in Euclidean space by a sign. The principal curvatures of the surface are still the eigenvalues of the second fundamental form with respect to the first fundamental form. (2) Time-like surface Take the orthogonal frame {e1 , e2 , n} at r, so that e1 is space-like, e2 is time-like, and (4.92) e12 = 1, e22 = −1, n 2 = 1. Then ω11 = ω22 = 0,

ω21 = ω12 ,

ω31 = −ω13 ,

ω23 = ω23 .

(4.93)

The first fundamental form ds2 = dr · dr = (ω 1 )2 − (ω 2 )2 = gab ω a ω b

(4.94)

is not positive definite. Here g11 = 1, g22 = −1, g12 = 0. Write ωa3 = bac ω c

(bac = bca ),

(4.95)

then the second fundamental form is II = −dr · dn = −ω 1 ω31 + ω 2 ω32 = ω 1 ω13 + ω 2 ω23 = bac ω a ω c .

(4.96)

The intrinsic definition of Gauss curvature is K=

R1212 2 = −R1212 , g11 g22 − g12

(4.97)

144


The Gauss equation is 1 ω1 ∧ ω = R 1 2 1 3 dω21 = R212 2 1212 ω ∧ ω = −ω3 ∧ ω2

= (b11 b22 − b212 )ω 1 ∧ ω 2 = −Kω 1 ∧ ω 2 .

(4.98)

The right hand side of the above formula differs from (4.91) by a sign. Note that if the frame satisfies e21 = −1, e22 = 1, then (4.98) will be changed to dω21 = R1 212 ω 1 ∧ ω2 = −R1212 ω 1 ∧ ω 2 = −ω31 ∧ ω23 = −(b11 b22 −

b212 )ω 1

∧

ω2

=

Kω 1

∧

ω2 .

(4.98 )

Comparing the formulae for the surfaces in Euclidean space with the space-like or time-like surfaces in Minkowski space, there are differences on sign (especially for the expressions of K).

4.3.2

Chebyshev coordinates for surfaces of constant Gauss curvature

The surfaces of constant Gauss curvature in R2,1 may be space-like, time-like or light-like, also may have positive or negative curvature (K = ±1). Moreover, for time-like surface of constant positive Gauss curvature, the eigenvalues of the second fundamental form with respect to the first fundamental form (i.e., the principal curvatures) may be real, imaginary, or be repeated eigenvalues with only one dimensional eigenspace. Therefore, we need to consider these cases separately. In each case, appropriate frame and coordinates are chosen and the surfaces are constructed with the help of the solutions of the corresponding integrable equations. First of all, we write down the fundamental equations and the Gauss equations under Chebyshev coordinates in each case. (1) space-like surface of constant positive Gauss curvature (K = +1) e1 , e2 and n satisfy (4.83), and e1 , e2 are tangent to the lines of curvature. Take α α (4.99) ω 1 = cos du, ω 2 = sin dv. 2 2 Similar to the surface of constant negative Gauss curvature in the Euclidean space, we still have 1 ω12 = −ω21 = (αv du + αu dv). 2 Take ω13 = sin

α du, 2

ω23 = − cos

α dv, 2

(4.100) (4.101)


145

then it can be verified easily that the Codazzi equation dω13 + ω23 ∧ ω12 = 0, holds, and b11 = tan

α , 2

dω23 + ω13 ∧ ω21 = 0

b22 = − cot

α , 2

b12 = 0.

(4.102)

(4.103)

Hence K = −b11 b22 = 1.

(4.104)

The first fundamental form is ds2 = (ω 1 )2 + (ω 2 )2 = cos2

α 2 α du + sin2 dv 2 , 2 2

and the second fundamental form is α α II = cos sin (du2 − dv 2 ). 2 2

(4.105)

(4.106)

The Gauss equation becomes the negative sine-Gordon equation αuu − αvv = − sin α.

(4.107)

By exchanging the parameters u and v, it is the same as the usual sineGordon equation. (2) space-like surface of constant negative Gauss curvature (K = −1) Under the assumption that there are no umbilic points, e1 , e2 and n satisfy (4.83), and e1 , e2 are tangent to the lines of curvature. Take ω 1 = cosh then

α du, 2

ω 2 = sinh

α dv, 2

1 ω21 = −ω12 = (αv du − αu dv). 2

Take ω13 = sinh then b11 = tanh

α du, 2

α , 2

ω23 = cosh

b22 = coth

α , 2

(4.108)

(4.109)

α dv, 2

(4.110)

b12 = 0.

(4.111)

Hence K = −b11 b22 = −1.

(4.112)

The first fundamental form is ds2 = (ω 1 )2 + (ω 2 )2 = cosh2

α 2 α du + sinh2 dv 2 , 2 2

(4.113)

146


and the second fundamental form is α α II = cosh sinh (du2 + dv 2 ). 2 2

(4.114)

The Gauss-Codazzi equations become

α = sinh α.

(4.115)

(3) Time-like surface of constant positive Gauss curvature (K = +1) Under the assumption that there are no umbilic points, this case is divided into the following three subcases. (3a) Principal curvatures are real Let e1 , e2 and n satisfy (4.92). Take ω 1 = cosh then

α du, 2

ω 2 = sinh

α dv, 2

1 ω21 = ω12 = (αv du + αu dv). 2

Let ω13 = sinh then

α du, 2

ω23 = − cosh

α dv, 2

α α , b22 = − coth , 2 2 K = −b11 b22 = +1.

b11 = tanh

(4.116)

(4.117) (4.118) (4.119) (4.120)

The Codazzi equation holds. The first fundamental form is ds2 = (ω 1 )2 − (ω 2 )2 = cosh2

α 2 α du − sinh2 dv 2 , 2 2

and the second fundamental form is α α II = cosh sinh (du2 − dv 2 ). 2 2

(4.121)

(4.122)

The Gauss equation becomes αuu − αvv = − sinh α. (3b) Principal curvatures are imaginary Take the frame so that e21 = 0,

e22 = 0,

n · ea = 0 (a = 1, 2),

(4.123)


and

147

1 e1 · e2 = eα > 0. 2

From the structure equations (4.81), we get ω21 = ω12 = 0, 1 ω13 + eα ω32 = 0, 2

1 ω23 + eα ω31 = 0. 2

Choose coordinates (u, v) [51] so that ω 1 = du − e−α dv, and

Then

ω 2 = −e−α du − dv,

1 ω31 = (du − eα dv), 2

1 ω23 = (eα du + dv). 2

1 ω13 = (du − eα dv), 2

1 ω23 = (eα du + dv). 2

d2 r = 0 leads to ω11 = αu du,

ω22 = αv dv.

Hence the first fundamental form and the second fundamental form are I = dr2 = −du2 − 2 sinh α du dv + dv 2 , II = −dr · dn = −2 cosh α du dv. Therefore, K = 1 and two principal curvatures are imaginary. From the Gauss equation dωba + ωca ∧ ωbc = −ω3a ∧ ωb3 , we get αuv = cosh α. All the Codazzi equations are also the consequences of the cosh-Gordon equation. Therefore, from a solution of the cosh-Gordon, a time-like surface with Gauss curvature 1 and two imaginary principal curvatures can be obtained by solving the fundamental equations of the surface. The parameters (u, v) are called asymptotic Chebyshev coordinates, since u and v are asymptotic lines. (3c) Repeated principal curvature with only one principal direction Take e1 , e2 satisfying e21 = 0,

e22 = 0,

1 e1 · e2 = eα . 2

148


Take the parameters (u, v) so that ω 1 = du − e−α dv,

ω 2 = −dv,

ω31 = −du − e−α dv,

ω32 = dv.

The first and second fundamental forms are I = −eα du dv + dv 2 , II = −eα du dv. The Gauss-Codazzi equations become the Liouville equation 1 αuv = eα . 2 From any solution of this equation, one can get a surface of this kind by solving the fundamental equations of surface. (4) Time-like surface of constant negative Gauss curvature (K = −1) Let e1 , e2 and n satisfy (4.92), α dv, 2

(4.124)

1 ω21 = ω12 = (−αv du + αu dv). 2

(4.125)

ω 1 = cos then

α du, 2

ω 2 = sin

Take

α du, 2 then the first fundamental form is ω13 = sin

ds2 = (ω 1 )2 − (ω 2 )2 = cos2

ω23 = cos

α dv, 2

α 2 α du − sin2 dv 2 , 2 2

and the second fundamental form is α α II = sin cos (du2 + dv 2 ), 2 2

(4.126)

(4.127)

(4.128)

and

α α , b22 = cot . 2 2 The Gauss curvature K = −1. The Gauss-Codazzi equation b11 = tan

dω21 = −ω31 ∧ ω23 becomes

α = sin α.

(4.129)


149

In summary, we have the following theorem.

Theorem 4.7 The Gauss-Codazzi equations of various kinds of surfaces (space-like or time-like) of constant Gauss curvature ((K = ±1) can be sine-Gordon equation, sinh-Gordon equation, cosh-Gordon equation or Liouville equation in a suitable parametrization. The construction of these surfaces is reduced to solve these equations and to integrate the fundamental equations of surfaces. This result can be listed as follows: First fundamental form: cos2 K=1

Second fundamental form:

α du2 + sin2 α2 dv 2 2 α cos 2 sin α2 (du2 − dv 2 )

Gauss equation: αuu − αvv = − sin α

Space-

First fundamental form: cosh2

like K = −1


α du2 + sinh2 α2 dv 2 2 cosh α2 sinh α2 (du2 + dv 2 )

Gauss equation: α = sinh α (a) First fundamental form: cosh2 K=1


α du2 − sinh2 α2 dv 2 2 cosh α2 sinh α2 (du2 − dv 2 )

Gauss equation: αuu − αvv = − sinh α (b) First fundamental form: −du2 − 2 sinh αdu dv + dv 2 Second fundamental form: −2 cosh α du dv Time-

Gauss equation: αuv = cosh α (c) First fundamental form: −eα dudv + dv 2

like

Second fundamental form: −e−α du dv Gauss equation: αuv = 12 eα First fundamental form: cos2 K = −1


α du2 − sin2 α2 dv 2 2 cos α2 sin α2 (du2 + dv 2 )

Gauss equation: α = sin α

There are still two problems left. (1) How to get explicit solutions of these partial differential equations? (2) How to get the explicit expressions of these surfaces. In order to answer these problems, we need to consider the Bäcklund transformation and pseudo-spherical congruences in R2,1 and use the Darboux transformation.

4.3.3

Pseudo-spherical congruence in R2,1

In R2,1 , a line congruence may be space-like, time-like, light-like or mixed. Here we want to study time-like line congruences (i.e., all the lines in it are time-like) and space-like line congruences (i.e., all the

150


lines in it are space-like). Under the assumption that they have two focal surfaces, these line congruences can be classified into following four types. (a) The line congruence is space-like, and two focal surfaces are spacelike; (b) The line congruence is space-like, and two focal surfaces are timelike; (c) The line congruence is time-like, and two focal surfaces are also time-like; (d) The line congruence is space-like, and one focal surface is spacelike, another focal surface is time-like. Let S and S ∗ be two focal surfaces of a line congruence Σ, P P ∗ be the lines in Σ which are common tangent lines of S and S ∗ , and P ∈ S, P ∗ ∈ S ∗ are the tangent points. Let n and n ∗ be the normal vectors of S and S ∗ at P and P ∗ respectively. If |P P ∗ | = l,

n · n∗ = k

(4.130)

are non-zero constants, then Σ is called a pseudo-spherical congruence. In [66, 84], the B¨ a¨cklund transformation for various pseudo-spherical congruences has been discussed. However, the discussion there is to be completed and the method of explicit construction is insufficient. Here we shall consider all the possible cases and use the Darboux transformation to get explicit expressions of the surfaces and congruences.

Theorem 4.8 (Generalized B¨ acklund theorem) Two focal surfaces of a pseudo-spherical congruence in R2,1 are surfaces of the same constant Gauss curvature K. In cases (a), (b) and (c), K is positive. In case (d), K is negative. Proof. The theorem should be proved for all four cases. Case (a): The congruence is space-like, and two focal surfaces S and S ∗ are space-like Take the orthogonal frames {e1 , e2 , n} and {e1∗ , e2∗ , n ∗ } of S at r and of S ∗ at r∗ respectively so that e1 = −e1∗ is the unit vector parallel to P P ∗ . Here r and r∗ are the position vectors of P and P ∗ respectively. Then e12 = e22 = e1∗2 = e2∗2 = 1, n 2 = n ∗2 = −1, and e2∗ = (cosh τ )e2 + (sinh τ )n, n ∗ = (sinh τ )e2 + (cosh τ )n.

(4.131)


151

Here τ = constant = 0. Moreover, r ∗ = r + le1

(l = constant = 0).

(4.132)

Differentiating this equation and comparing it with dr ∗ = ω ∗1 e1∗ +ω ∗2 e2∗ , we get ω 1 = −ω ∗1 , ω 2 + lω12 = (cosh τ )ω ∗2 ,

(4.133)

lω13 = (sinh τ )ω ∗2 . Hence (cosh τ )ω13 − (sinh τ )ω12 = From

sinh τ 2 ω . l

ω2 sinh τ, l = (−de2∗ ) · n ∗ = ω23 ,

ω1∗3 = (−de1∗ ) · n ∗ = − ω2∗3

(4.134)

(4.135)

we have −ω3∗1 ∧ ω2∗3 = ω23 ∧ (−

ω2 sinh τ 3 ω1 ∧ ω 1 sinh τ ) = l l

sinh2 τ ∗1 = ω ∧ ω ∗2 , l2 Hence the Gauss curvatures of S and S ∗ are the same constant

(4.136)

sinh2 τ . l2 Case (b): The congruence is space-like, and two focal surfaces S and ∗ S are time-like Now K∗ = K =

e12 = e1∗2 = 1,

e22 = e2∗2 = −1,

n 2 = n ∗2 = 1,

(4.137)

and e1∗ = −e1 , e2∗ = cosh τ e2 + sinh τ n, n ∗ = sinh τ e2 + cosh τ n,

(4.138) (n · n ∗ = cosh τ = constant).

From r ∗ = r + lr1 (l = constant), ω 1 = −ω ∗1 , ω 2 + lω12 = cosh τ ω ∗2 , lω13 = sinh τ ω ∗2 .

(4.139)

152


(4.138) and (4.139) are the same as (4.131) and (4.133). Moreover, ω1∗3 = (de1∗ ) · n ∗ = −

ω2 sinh τ, l

ω2∗3 = (de2∗ ) · n = ω23 .

(4.140)

Hence sinh τ 3 sinh τ 3 ω2 ∧ ω 2 = − ω1 ∧ ω 1 l 2 l sinh τ ∗1 =− ω ∧ ω ∗2 . l2

−ω3∗1 ∧ ω2∗3 =

(4.141)

This means that S and S ∗ are surfaces of the same constant positive Gauss curvature sinh2 τ K∗ = K = . (4.142) l2 Case (c): The congruence is time-like, and two focal surfaces S and S ∗ are time-like Now e12 = e1∗2 = −1,

e22 = e2∗2 = 1,

n 2 = n ∗2 = 1,

(4.143)

and e1∗ = −e1 , e2∗ = (cos τ )e2 + (sin τ )n, n ∗ = −(sin τ )e2 + (cos τ )n

(4.144)

(n · n ∗ = cos τ = constant = 0). Differentiating r ∗ = r + le1 , we get ω 1 = −ω ∗1 , ω 2 + lω12 = (cos τ )ω ∗2 ,

(4.145)

lω13 = (sin τ )ω ∗2 . (4.144) implies ω2 sin τ, l = ω23 ,

ω1∗3 = − ω2∗3

(4.146)

which leads to −ω3∗1 ∧ ω2∗3 = ω23 ∧ (− =

sin τ 3 ω2 sin τ ) = ω1 ∧ ω 1 l l

sin2 τ ∗1 ω ∧ ω ∗2 . l2

(4.147)

153


(4.98) implies that sin2 τ , l2

K∗ =

K=

sin2 τ . l2

(4.148)

Hence S and S ∗ are surfaces of the same constant positive Gauss curvature. Case (d): The congruence is space-like, and one focal surface S is time-like, another focal surface S ∗ is space-like Now e12 = e1∗2 = 1,

e22 = −1,

e2∗2 = 1,

n 2 = 1,

n ∗2 = −1,

(4.149)

and e1∗ = −e1 , e2∗ = (sinh τ )e2 + (cosh τ )n, n ∗ = (cosh τ )e2 + (sinh τ )n

(4.150)

(n · n ∗ = sinh τ = constant). Differentiating r ∗ = r + le1 (l = constant), we get ω 1 = −ω ∗1 , ω 2 + lω12 = (sinh τ )ω ∗2 ,

(4.151)

lω13 = (cosh τ )ω ∗2 . Moreover, cosh τ 2 ω , l

ω1∗3 =

ω2∗3 = ω23

(4.152)

leads to −ω3∗1 ∧ ω2∗3 =

cosh τ 3 cosh τ 3 ω2 ∧ ω 2 = − ω1 ∧ ω 1 l l cosh2 τ ∗1 =− ω ∧ ω ∗2 . l2

(4.153)

cosh2 τ . l2

(4.154)

Hence, from (4.91), K∗ = − On the other hand, −ω31 ∧ ω23 = −ω2∗3 ∧ =

cosh τ ∗2 cosh τ ∗3 ω = ω1 ∧ ω ∗1 l l

cosh2 τ 1 ω ∧ ω2. l

(4.155)

154


Hence, from (4.98), K=−

cosh2 τ . l2

(4.156)

The theorem is proved. This theorem does not confirm that all four cases can occur, because it does not contain the proof of existence. This will be completed in the next subsection.

4.3.4

B¨ acklund transformation and Darboux transformation for surfaces of constant Gauss curvature in R2,1

Now we shall give the method for constructing new surfaces of constant Gauss curvature in R2,1 from a known one by using pseudo-spherical congruence [61, 63]. This also gives the proof of existence of various pseudo-spherical congruences. It is the generalization of the classical Backlund ¨ transformation. Using Darboux transformation, this analytical process can be realized algebraically. The method is similar with that in Section 4.2. However, in each case the method has some special feature. (1) S is a space-like surface of constant positive Gauss curvature, the congruence is space-like a¨cklund Let {e1 , e2 , n} be the Chebyshev frames at r. If there exists a B¨ transformation such that its another focal surface S ∗ is space-like, then r ∗ = r + l(cos θe1 + sin θe2 ),

(l = constant = 0).

(4.157) −→

The normal vector n ∗ of S ∗ is time-like and orthogonal to P P ∗ . Hence n ∗ = sinh τ (− sin θe1 + cos θe2 ) + cosh τ n,

(τ = constant = 0). (4.158) Differentiating r ∗ , using the fundamental equations of surface, and using the equations (4.99) – (4.101) on space-like surfaces of constant positive Gauss curvature, we get dr ∗ =

(&

α αv ' αv − l sin θ θu + e1 + l cos θ θu + e2 2 2 (2 α ) αu +l cos θ sin n du + − l sin θ θv + e1 2 2 & ' α αu α ) + sin + l cos θ θv + e2 − l sin θ cos n dv. 2 2 2

cos

(4.159)


155

Let θ = α1 /2 and suppose K = 1, then l = sinh τ . The condition dr ∗ · n ∗ = 0 implies α1 α α1 α cos + 2 cosh τ cos sin , 2α 2α 2α 2α 1 1 sin − 2 cosh τ sin cos . l(αv + αu ) = −2 cos 2 2 2 2

l(αu + αv ) = 2 sin

(4.160)

(4.160) is a system of partial differential equations for α1 . Its integrability condition is exactly (4.107) αuu − αvv = − sin α.

(4.161)

Therefore, (4.160) is solvable. This gives the following theorem.

Theorem 4.9 For a given space-like surface S of constant positive Gauss curvature, there exists a space-like pseudo-spherical congruence whose focal surfaces are S and another space-like surface S ∗ of the same constant Gauss curvature. Using the method in Section 4.2, we can prove that (u, v) are also the Chebyshev coordinates of S ∗ . Using the Darboux transformation for the sine-Gordon equation, a series of surfaces of constant positive Gauss curvature can be obtained. If a solution α of the sine-Gordon equation corresponding to the given surface of constant positive Gauss curvature together with a fundamental solution of its Lax pair are known, then the construction is purely algebraic. Note that in this case u and v should be u+v u−v ,η= ) so that (4.161) can be changed interchanged (i.e. ξ = 2 2 to the standard sine-Gordon equation (4.66). (2) S is a time-like surface of constant positive Gauss curvature There are following three subcases. (2a) Two principal curvatures of S are real and distinct Take the Chebyshev coordinates, then α α du, ω 2 = sinh dv, 2 2 1 ω21 = ω12 = (αv du + αu dv), 2 α α 3 ω1 = sinh du, ω23 = − cosh dv, 2 2 ω 1 = cosh

and e21 = 1,

e22 = −1,

n2 = 1.

(4.162)

156


(2a1 ) Space-like B¨ äcklund transformation The equation of S ∗ is r ∗ = r + l(cosh θe1 + sinh θe2 )

(4.163)

and the normal vector is n ∗ = sinh τ (sinh θe1 + cosh θe2 ) + cosh τ n.

(4.164)

Suppose K = 1, then the proof of Theorem 4.8 leads to l = sinh τ . We have (& α αv ' αv cosh + l sinh θ θu + e1 + l cosh θ θu + e2 dr ∗ = 2 2 (2 α ) αu +l cosh θ sinh n du + l sinh θ θv + e1 2 2 & α αu ' α ) + sinh + l cosh θ θv + e2 − l sinh θ cosh n dv. 2 2 2 (4.165) ∗ ∗ From dr · n = 0, we have

α αv α = sinh θ cosh + cosh τ cosh θ sinh , 2 2 2 αu α α = − cosh θ sinh − cosh τ sinh θ cosh . l θv + 2 2 2 l θu +

(4.166)

Regarded as a system of partial differential equations for θ = α1 /2, its integrability condition is αvv − αuu = sinh α.

(4.167)

This equation holds because it is the Gauss equation of S. It is easy to see that S ∗ is also time-like. This proves the existence of the pseudospherical congruence. By Theorem 4.8, the curvature of S ∗ is also +1. Therefore, we have the following theorem.

Theorem 4.10 There exist space-like pseudo-spherical congruences with two time-like focal surfaces of constant positive Gauss curvature. Similarly, we can prove that (u, v) are also the Chebyshev coordinates of S ∗ , and can write down the explicit expressions of e1∗ , e2∗ . In order to construct S ∗ explicitly in terms of the Darboux transformation, we want to discuss the Darboux transformation for the sinh-Gordon equation. Let v−u u+v , η= , (4.168) ξ= 2 2 then (4.123) becomes (4.169) αξη = sinh α.

157


It has a Lax pair ⎛

Φξ =

λ⎝ 0 2 eα

e−α

⎛

⎞

⎞

1 1 ⎜ −αη λ ⎟ Φη = ⎝ 1 ⎠ Φ, 2 αη λ

⎠ Φ,

0

(4.170)

that is,⎛the integrability condition of (4.170) is just (4.169). ⎞ Let ⎝ ⎛

then ⎝ ⎛

H=⎝

h1

⎠ be a column solution of the Lax pair (4.170) with λ = λ1 ,

h2

⎞

h1 −h2 h1

⎠ is a column solution of (4.170) with λ = −λ1 .

Let

⎞

h1

h2 −h2

⎠ and suppose h1 , h2 = 0, then the Darboux matrix is ⎛

1 ⎜ λ 1 D(λ) = I − λH ⎜ ⎝ 0

⎞

⎛

0 ⎟ ⎜ 0 ⎟ H −1 = I − λ ⎜ 1 ⎠ λ1 ⎝ h2 − λ1 h1

⎞

h1 h2 ⎟ ⎟ . (4.171) ⎠ 0

It can be checked directly that (α1 , Φ1 ) defined by eα1 = e−α

h 2 2

h1

,

Φ1 = D(λ)Φ

(4.172)

still satisfy the Lax pair (4.170). Hence α1 is a solution of (4.169), and (α, Φ) −→ (α1 , Φ1 ) is the Darboux transformation for (4.169). Now we prove that α1 obtained by (4.172) provides a solution θ = α1 /2 of (4.166). Let θ = α1 /2. Adding and subtracting (4.168) and (4.166), we get αξ α1 − α = (1 − cosh τ ) sinh , 2 2 2 α αη α1 + α 1η − . = −(1 + cosh τ ) sinh l 2 2 2 l

⎛

Since ⎝

α

1ξ

+

(4.173)

⎞

h1

⎠ satisfies the Lax pair (4.170),

h2 1 h2 λ1 −α h2 αη e + , (ln |h1 |)η = − , 2 h1 2 2λ1 h1 λ1 α h1 1 h1 1 e , (ln |h2 |)η = + αη . (ln |h2 |)ξ = 2 h2 2λ1 h2 2 (ln |h1 |)ξ =

(4.174)

158


Choose ⎝

⎞

h1

⎠ so that

h2 is positive (if it is negative, the discussion is h1

h2 similar), then (4.172) leads to

α1ξ + αξ = (2 ln |h2 |)ξ − (2 ln |h1 |)ξ h h2 α1 − α 1 . = λ1 eα − e−α = −2λ1 sinh h2 h1 2

(4.175)

Likewise, α1η + αη = 2αη +

1 −(α1 +α)/2 (e − e(α+α1 )/2 ), λ1

or equivalently α1η − αη = −

1 α1 + α , sinh λ1 2

(4.176)

−1 + cosh τ 1 + cosh τ , (4.175) . Since l = sinh τ , λ−1 1 = l sinh τ and (4.176) become (4.173). Thus by using Darboux transformation we can obtain the solution α1 of (4.176) explicitly. Besides, in realizing α1 the Bäcklund transformation (4.163), we only need to use cosh and 2 α1 sinh , hence the algorithm is purely algebraic. 2 Now we turn to examples. Starting from α = 0 and solving the Lax pair ⎛ ⎞ ⎛ ⎞ λ⎝ 0 1 ⎠ 1 ⎝ 0 1 ⎠ Φ, Φη = Φ, (4.177) Φξ = 2 2λ 1 0 1 0

where λ1 =

we get ⎛

Φ(λ) = ⎝

⎞

cosh γ

sinh γ

sinh γ

cosh γ

⎠,

1 λ η . γ= ξ+ 2 2λ

(4.178)

Let

h1 = cosh γ1 ,

h2 = sinh γ1 ,

λ1 1 γ1 = η . ξ+ 2 2λ1

(4.179)

According to (4.171), the Darboux matrix is ⎛

D(λ) = I −

⎞

0

λ ⎝ λ1 tanh γ1

coth γ1 0

⎠.

(4.180)

159


Hence Φ1 (λ) = D(λ)Φ(λ) ⎛ λ ⎜ cosh γ − λ coth γ1 sinh γ sinh γ − 1 =⎜ ⎝ λ tanh γ1 cosh γ cosh γ − sinh γ − λ1 Take λ2 > 0 (λ2 = λ1 ), γ2 =

⎞

λ coth γ1 cosh γ ⎟ λ1 ⎟. ⎠ λ tanh γ1 cosh γ λ1 (4.181)

λ2 1 η, and let ξ+ 2 2λ2

λ2 coth γ1 sinh γ2 , λ1 2 = sinh γ2 − λ2 tanh γ1 cosh γ2 , h λ1 1 = cosh γ2 − h

then eα2 = e−α1

(4.182)

h 2 2 1 h

gives a new solution α2 . This process can be done successively to get α1 , α2 , α3 , · · ·. The surfaces are constructed as follows. Starting from α = 0 and solving the fundamental equations of the surface, we obtain a family of frames along a line. In suitable coordinates, they are r = (u, 0, 0),

e1 = (1, 0, 0),

e2 = (0, sinh v, cosh v),

n = (0, − cosh v, − sinh v).

(4.183)

Then by Theorem 4.8 and formula (4.163), α1 gives a time-like surface of Gauss curvature +1. Using α2 , α3 , · · ·, we can construct a series of surfaces in the same way. (2a2 ) Time-like B¨äcklund transformation From the same S, the time-like Darboux transformation is α1 α1 r ∗ = r + l(sinh e1 + cosh e2 ), 2 2 (4.184) α1 α1 ∗ n = sin τ (cosh e1 + sinh e2 ) + cos τ n, 2 2 where l and τ are constants, sin τ = l = 0. Computing dr∗ and using the condition n∗ · dr∗ = 0, we get the equations α αv α1 α α1 α 1u l + = − cosh cosh + cos τ sinh sinh , 2 2 2 2 2 2 (4.185) α αu α α1 α α1 1v + sinh + cos τ cosh cosh , = sinh l ( 2 2 2 2 2 2

160


which is similar with (4.166). This is the Bäcklund transformation from α to α1 . Its integrability condition αuu − αvv = − sinh α

(4.186)

already holds. Hence the B¨ acklund transformation from S to S ∗ can be realized by solving this integrable system. If the expression of S and the solution of the Lax pair corresponding to the sinh-Gordon equation are known, then S ∗ can be obtained by Darboux transformation in algebraic algorithm. (2b) Two principal curvatures of S are imaginary When two principal curvatures of S are imaginary, the frame {e1 , e2 , n}, parameters (u, v) and ω 1 , ω 2 , ωji (i, j = 1, 2, 3) have been chosen in the case (3b) of Subsection 4.3.2. (2b1 ) Time-like B¨äcklund transformation Suppose that the expression of S ∗ is r∗ = r + l(ae1 + be2 ), where l is a non-zero constant and abeα = −1. The last condition means that P P ∗ is a time-like straight line and (r∗ − r)2 = −l2 . If we want that n∗ = (ae1 − be2 ) cos τ + n sin τ is the normal vector of S ∗ where sin τ = n · n∗ = constant, then by n∗ · dr∗ = 0, we get bu l )e1 + (−e−α + lbu )e2 + (a − a−1 )n, b 2 av l r∗v = (−e−α + lav )e1 + (−1 − lb )e2 + (b + b−1 )n a 2

r∗u = (1 − la

(4.187)

where l = sin τ . From n∗ · dr∗ = 0, 2b−1 bu = −µ(a − a−1 ),

1 2a−1 av = − (b + b−1 ), µ

(4.188)

with µ = sec τ − tan τ . Let α1 be a function such that a = exp

α1 − α , 2

b = − exp

−α1 − α , 2

then (α1 + α)u = 2µ sinh

α1 − α , 2

(α1 − α)v =

2 α1 − α cosh . µ 2

(4.189)

161


This is the Bäcklund transformation for the cosh-Gordon equation. The integrability condition for α1 is that α satisfies the cosh-Gordon equation which is the Gauss equation of S and hence is satisfied by α. Therefore, α1 exists and S ∗ is obtained via α1 . From the Bäcklund theorem we know that S ∗ is time-like and has constant positive Gauss curvature K = 1. Moreover, by tedious calculation we can verify that (u, v) are the asymptotic Chebyshev coordinates of S ∗ . Now we want to get the expression of α1 by using Darboux transformation.

Lemma 4.11 The Lax pair of cosh-Gordon equation is ⎛

⎞

0 e−α λ ⎠ Φ, Φu = U Φ = ⎝ α 2 e 0

⎛

⎞

−αv λ−1 1 ⎠ Φ. Φv = V Φ = ⎝ −1 2 λ αv (4.190)

Proof. The proof follows from direct calculations, showing that the integrability condition of (4.190) Uv − Vu + [U, V ] = 0 is equivalent to the cosh-Gordon equation. ⎛

Lemma 4.12 (i) If ⎝ ⎛

then ⎝

−h1

⎞

⎞

h1 h2

⎠ is a solution of the Lax pair for λ = λ0 ,

⎠ is a solution of the Lax pair for λ = −λ0 . h2 (ii) Suppose λ is purely imaginary and h2 /h1 is purely imaginary at one point (u0 , v0 ), then h2 /h1 is purely imaginary in a neighborhood of (u0 , v0 ).

Proof. (i) is obvious. Now we prove (ii). From the Lax pair (4.190),

h2 = h1 u h2 = h1 v

λ0 α h2 2 e + e−α , 2 h1 2 1 h2 1 h2 + αv − . 2λ0 h1 2λ0 h1

162


Let A =

¯2 h2 h + ¯ , then h1 h1 ¯0 λ Au = e−α 2

¯ 2 h2 h ¯ 1 − h1 A, h ¯2 1 h2 h Av = αv A + ¯ − ¯ A. 2λ0 h1 h 1 Hence if A = 0 at (u0 , v0 ), then it holds in a neighborhood of (u0 , v0 ). The lemma is proved. By the general formulae for constructing Darboux matrix, we have ⎛

Φ1

⎛

⎞

⎞

0 λ ⎝ λ−1 0 ⎝ ⎠ H −1 ⎠ Φ = I− H λ0 0 −λ−1 0 ⎛

⎛

⎞⎞

0 h1 /h2 λ ⎝ ⎠⎠ Φ, = ⎝I + λ0 −h2 /h1 0 ⎛

where

H=⎝

h1 −h1 h2

⎛

From Φ1u

0 λ = ⎝ 2 e α1

we get

α1

e

h2 =− h1

⎞ ⎠.

h2 −eα1 0 2

⎞ ⎠ Φ1 ,

e−α .

This is the explicit expression of α1 . The explicit expression of S ∗ can be obtained from r∗ = r + l(ae1 + be2 ) by using this α1 . It can also be proved that α1 is a solution of the Bä¨cklund transformation. Therefore, the explicit formula for the B¨ acklund transformation of S is obtained. äcklund transformation (2b2 ) Space-like B¨ The expression of S ∗ is still r∗ = r + l(ae1 + be2 ), but abeα = 1 now. The normal vector of S ∗ is n∗ = (ae1 − be2 ) sinh τ + n cosh τ.


163

Then bu l )e1 + (−e−α + lbu )e2 + (a + a−1 )n, b 2 av l r∗v = (−e−α + lav )e1 + (−1 − lb )e2 + (b − b−1 )n. a 2 r∗u = (1 − la

(4.191)

Let l = sinh τ . n∗ · dr∗ = 0 leads to 1 2au = − (b − b−1 ), a µ

2bu = µ(a + a−1 ), b where µ = cosech τ − coth τ . Let a = exp

−α1 − α , 2

b = exp

α1 − α , 2

then we get (α1 − α)u = 2µ cosh

α1 + α , 2

(α1 + α)v =

2 α1 − α sinh , µ 2

(4.192)

which is the Backlund ¨ transformation for the cosh-Gordon equation. If µ −1 is changed to µ , this is just (4.189). Using Darboux transformation, we can get the explicit expression of α1 , which gives the explicit Bä¨cklund transformation from S to S ∗ . (2c) Two principal curvatures of S are equal and there is only one principal direction Choose the frame (e1 , e2 , n) and the parameters (u, v) as the case (3c) in Subsection 4.3.2, then 1 e1 · e 2 = e α , 2 ω 1 = du − e−α dv, ω 2 = −dv, e21 = e22 = 0,

ω31 = −du − e−α dv,

ω32 = dv.

Hence I = −eα du dv + dv 2 ,

II = −eα du dv.

The Gauss-Codazzi equations become the Liouville equation 1 αuv = eα . 2 Take the time-like B¨ äcklund transformation for S as r∗ = r + l(ae1 + be2 ),

abeα = −1,

164


then

n∗ = (ae1 − be2 ) cos τ + n sin τ.

From n∗ · dr∗ = 0, we get (α1 + α)u = −µ exp

α − α1 , 2

Considering the Lax pair ⎛

⎞

0 0 λ ⎠ Φ, Φu = ⎝ 2 eα 0

(α1 − α)v =

α1 + α 2 cosh . µ 2

⎛

⎞

−αv λ−1 1 ⎠Φ Φv = ⎝ 2 λ−1 αv

(4.193)

of the Liouville equation, we can use Darboux transformation to get the explicit expression of S ∗ . The space-like Bäcklund transformation is given by r∗ = r + l(ae1 + be2 ),

abeα = 1,

and S ∗ can be obtained by Darboux transformation in a similar way. In summary, we have the following theorem.

Theorem 4.13 Suppose S is a time-like surface of constant positive Gauss curvature without umbilic points, then there are time-like and space-like Backlund ¨ congruences such that S is one of its focal surface, and another focal surface is also a time-like surface of constant positive Gauss curvature. (3) Space-like pseudo-spherical congruence with one spacelike focal surface and one time-like focal surface Suppose that S is a time-like surface of constant negative Gauss curvature −1. Choose the Chebyshev coordinates and the corresponding frame as (4) in Subsection 4.3.2, then (4.92) and (4.124) – (4.126) hold. Apply the Bäcklund transformation r ∗ = r + l(cosh θe1 + sinh θe2 ), ∗

(l = cosh τ ),

(4.194)

n = cosh τ (sinh θe1 + cosh θe2 ) + sinh τ n. By the fundamental equations of S, the integrability condition, together with (4.124) – (4.126), we have dr ∗ =

(&

α αv ' αv + l sinh θ θu − e1 + l cosh θ θu − e2 2 2 (2 α ) αu +l cosh θ sin n du + l sinh θ(θv + e1 2 2 & α αu ' α ) + sin + l cosh θ θv + e2 + l sinh θ cos n dv. 2 2 2 (4.195)

cos


From n ∗ · dr ∗ = 0, we have α αv sinh θ cos + l sinh θ θu − 2 2 α αv ) + sinh τ cosh θ sin = 0, −l cosh2 θ θu − 2 αu 2 α αu l sinh2 θ θv + − cosh θ sin + l cosh θ θv + 2 2 2 α + sinh τ sinh θ cos = 0, 2 i.e., αv α α l θu − = sinh θ cos + sinh τ cosh θ sin , 2α 2α α2u = − cosh θ sin + sinh τ sinh θ cos , l θv + 2 2 2 where l = cosh τ . The integrability condition for θ is

α = sin α,

165

(4.196)

(4.197)

(4.198)

which is satisfied already. Hence for any given initial condition θ = θ0 at u = u0 and v = v0 , the solution θ of (4.197) exists, and it satisfies

α1 = sinh α1 , where α1 = θ/2. This implies that the corresponding pseudo-spherical congruence exists, and r ∗ is a space-like surface of constant negative Gauss curvature.

Theorem 4.14 There exists a space-like pseudo-spherical congruence with one time-like and one space-like focal surfaces of constant negative Gauss curvature. We shall prove that (u, v) are also the Chebyshev coordinates of the space-like surface of constant negative Gauss curvature S ∗ . In fact, according to (4.195), dr ∗ = ω ∗1 e1∗ + ω ∗2 e2∗ , ω ∗1 = cosh θ,

ω ∗2 = sinh θ,

(4.199) (4.200)

where 1 (& α αv cos + l sinh θ θu − e1 cosh θ 2 α 2 α ) v +l sinh θ θu − e2 + l sinh θ sin n , 2 ( 2 1 αu e2∗ = l sinh θ θv + e1 sinh & θα 2 α ' α ) u + sin + l cosh θ θv + e2 + l sinh θ cos n . 2 2 2

e1∗ =

(4.201)

166


From (4.197), Hence and

e1∗2 = 1,

e2∗2 = 1,

e1∗ · e2∗ = 0.

dn ∗ = ω1∗3 e1∗ + ω2∗3 e2∗ , ω1∗3 = sinh θ du,

ω2∗3 = cosh θ dv.

(4.202) (4.203) (4.204)

Let θ = α1 /2, then (u, v) are the Chebyshev coordinates of S ∗ . Moreover, α1 satisfies (4.205)

α1 = sinh α1 . This can also be derived from the differentiation of (4.197). We can do the same procedure conversely, i.e., starting from a spacelike surface of constant negative Gauss curvature S ∗ , the reverse procedure leads to a time-like surface of constant negative Gauss curvature S. The Bäcklund transformations described above are not transformations between two solutions of the same sine-Gordon equation or the same sine-Laplace equation. In fact, they transform a solution of the sine-Laplace equation to a solution of the sinh-Laplace equation, and vice versa (the differential form of these transformations has been given in [91]). Applying the transformations twice, we can get a transformation of solutions of the same equation. Darboux transformation will give an explicit form of these transformations. Introduce the complex variables ζ=

u + iv , 2

u − iv ζ¯ = , 2

(4.206)

then

∂ ∂ ∂ ∂ ∂ ∂ = −i , = +i , ¯ ∂ζ ∂u ∂v ∂u ∂v ∂ζ ∂2 ∂2 ∂2 + = . = ∂ζ∂ ζ¯ ∂u2 ∂v 2 The integrability condition of the Lax pair ⎛

⎞

0 −e−α λ ⎠ Φ, Φζ = ⎝ 2 eα 0 ⎛

(4.207)

⎞

−αζ¯ 1/λ 1 ⎠Φ Φζ¯ = ⎝ 2 −1/λ αζ¯

(4.208)

is

α = αζ ζ¯ = sinh α.

(4.209)

167


If α can be complex, this equation is called the complex sinh-Laplace equation. If α is real, it is the usual sinh-Laplace equation. If α is purely imaginary (α = iβ, β is real), it becomes the sine-Laplace equation

β = βζ ζ¯ = sin β. For the complex sinh-Laplace equation, the Darboux transformation (α, Φ) −→ (α1 , Φ1 ) can be constructed similarly as the real sinh-Laplace equation. However, the Lax pair (4.208) is slightly different from (4.170) for simplifying the calculations. ⎛ ⎞ h1 ⎠ be a column solution of the Lax pair for Take λ1 = 0. Let ⎝ h2 ⎛

λ = λ1 , then ⎝

−h1 h2

⎞

⎠ is a column solution of the Lax pair for λ = −λ1 .

Let

⎛

H=⎝

⎞

h1 −h1 h2

⎠,

h2

then ⎛

S =H⎝

⎛

⎞

1/λ1

0

0

−1/λ1

1

⎜

−

0 1 ⎜ ⎜ ⎝ h2 λ1 ⎞h1 λ h1 − λ1 h2 ⎟ ⎟.

⎠ H −1 =

⎛

D(λ) = I − λS = ⎜ ⎝

⎞

λ h2 λ1 h1

h1 h2 ⎟ ⎟, ⎠ 0

⎠

1

Hence the Darboux transformation is Φ1 (λ) = D(λ)Φ(λ). Here α1 is determined by α1

e

= −e

−α

h2 h1

2

.

(4.210)

Suppose α is real, then α1 is purely imaginary if and only if 2 h2 −α e = 1. h 1

(4.211)

168


It is remained to find h1 and h2 such that (4.211) is satisfied. Take λ = λ1 with |λ1 | = 1. From the Lax pair, λ1 −α λ1 α e h2 , h2ζ = e h1 , 2 2 αζ¯ 1 1 1 h1ζ¯ = − αζ¯h1 + h2 . h2 , h2ζ¯ = − h1 + 2 2λ1 2λ1 2

h1ζ = −

(4.212)

Take the complex conjugation, we get ¯ 1ζ = − 1 αζ h ¯1 + 1 h ¯ h ¯1 2, 2 2λ ¯ ¯ 2, h ¯ ¯= ¯ ¯ = − λ1 e−α h h 1ζ 2ζ 2 Let

¯ 2ζ = − 1 h ¯ 2, ¯ 1 + αζ h h ¯ 2 2λ1 ¯1 λ ¯ 1. eα h 2

2 ¯2 h2 h2 h A = e−α − 1 = ¯ e−α − 1, h1 h1 h1

(4.213)

(4.214)

then $

%

¯ 2ζ + h2ζ h ¯ 2 h2 h ¯ 2 (h1 h ¯ 1ζ + h1ζ h ¯ 1) ¯2 h2 h h2 h Aζ = − ¯ e−α αζ + e−α − . ¯1 ¯ 1 )2 h1 h1 h1 h (h1 h Using (4.212) and (4.213), λ1 Aζ = 2 Similarly, ¯1 λ Aζ¯ = 2

¯2 h2 −α h e −¯ h1 h1

A.

(4.215)

¯2 h2 h − ¯ e−α A. h1 h1

(4.216)

If we choose h1 and h2 so that A = 0 holds at one point (initial value), then A = 0 holds identically. Hence the following theorem is true.

Theorem 4.15 Suppose α is real, λ1 is a complex number with |λ1 | = 1. Take h1 and h2 satisfying (4.211) at some point. Then α1 derived by the Darboux transformation is purely imaginary. This is the Darboux transformation from a solution of the sinh-Laplace equation to a solution of the sine-Laplace equation. Now suppose α is purely imaginary. From (4.211), α1 is real if and only if 2 2 h2 h2 −α e = e−α < 0, h1 h1

169


since eα1 > 0. From the above equality,

hence

¯1 h2 h −α ¯2 e h1 h

2

= 1,

¯1 h2 h −α ¯ 2 e = ±1. h1 h

On the other hand,

h2 h1

2

e−α =

¯ 2 h2 h ¯1 h2 h −α ¯ 1 h1 h ¯2 e . h1 h

Hence we should choose h1 and h2 such that ¯1 h2 h −α ¯ 2 e = −1. h1 h

(4.217)

Similar to the proof of Theorem 4.15, by taking λ1 such that |λ1 | = 1, (4.217) holds everywhere if it holds at one point. This leads to the following theorem.

Theorem 4.16 Suppose α is a purely imaginary solution of (4.209), λ1 is a complex number with |λ1 | = 1. Take h1 and h2 satisfying (4.217) at some point. Then the Darboux transformation provides a real α1 which is a solution of the sinh-Laplace equation. Using the above two kinds of Darboux transformations, we get the following series of Darboux transformations (α, Φ) −→ (α1 , Φ1 ) −→ (α2 , Φ2 ) −→ · · · . If α is purely imaginary and β = −iα is a solution of the sine-Laplace equation, then α1 , α3 , · · ·, α2n+1 , · · · are solutions of the sinh-Laplace equation, and β2 = −iα2 , β4 = −iα4 , · · ·, β2n = −iα2n , · · · are solutions of the sine-Laplace equation. On the other hand, if α is a solution of the sinh-Laplace equation, then all α2n (n = 1, 2, 3, · · ·) are solutions of the sinh-Laplace equation, and all β2n+1 = −iα2n+1 (n = 0, 1, 2, · · ·) are solutions of the sine-Laplace equation. This can be shown by the following figure: 1 α

α0 @ @ 0 α

@ @

α1

3 α

α2 @ @ 2 α

@ @

α3

···

α4 @ @ 4 α

···

170


i−1 are real for odd i and purely imaginary for even i. αi and α Now we prove that this Darboux transformation is consistent with the Backlund ¨ transformation (4.197). Rewrite (4.210) as h2 . (4.218) e(α1 +α)/2 = h1

Differentiating (4.218) with respect to ζ and using (4.212), we get (α1 + α)ζ = λ1 sinh(α − α1 ). 2

(4.219)

Similarly, differentiating (4.218) with respect to ζ¯ and using (4.212), we get (α1 − α)ζ 1 α + α1 = sinh . (4.220) 2 λ1 2 1 + i sinh τ α1 , λ1 = , then from (4.197), we obtain (4.219) and 2 cosh τ (4.220). Therefore, the function α1 derived from the Darboux transformation is a solution of (4.219) and the construction of the B¨ a¨cklund congruence from a time-like surface of constant negative Gauss curvature is completed. This gives the following theorem.

Let θ =

Theorem 4.17 Starting from one focal surface, the B¨ acklund congruence of type (d) together with its another focal surface can be constructed by using Darboux transformation (4.210) and B¨ acklund transformation (4.194), Example: Take the trivial solution α = 0. The Lax pair is ⎛

⎞

0 −1 λ ⎠ Φ, Φζ = ⎝ 2 1 0

⎛

⎞

0 1/λ 1 ⎠ Φ. Φζ¯ = ⎝ 2 −1/λ 0

(4.221)

Its fundamental solution is ⎛

Φ(λ) = ⎝

eγ

e−γ

−ieγ

ie−γ

⎞ ⎠,

γ=

λ

2

ζ−

1 ¯ ζ i. 2λ

(4.222)

Using the Darboux transformation, we can get α1 . Since α = 0 can be regarded as a real or a purely imaginary solution, we can get purely imaginary or real α1 from α = 0.

171


First we want to find real α1 . Let λ = λ1 with |λ1 | = 1, then γ1 = ⎛

is real. Let ⎝

1

2

ζ−

1 ¯ ζ i 2λ1

(4.223)

⎞

h1 h2

i.e.,

λ

⎠ be a column solution of the Lax pair for λ = λ1 ,

h1 = eγ1 + be−γ1 ,

h2 = −ieγ1 + bie−γ1 .

(4.224)

Take b to be real, then

¯1 h2 h ¯ 2 = −1. h1 h Suppose b is positive, then it can be taken as 1 by adding a constant on γ1 . From (4.210), h2 eγ1 − e−γ1 = γ1 = tanh γ1 , h1 e + e−γ1

eα1 /2 = i and

(γ1 > 0),

e−α1 /2 = coth γ1 , α1 α1 = coth(2γ1 ), sinh = − cosech(2γ1 ). cosh 2 2

(4.225)

When γ1 < 0, the right hand sides of (4.225) should change signs. α = 0 does not correspond to a time-like surface, but only a system of orthogonal frames along a straight line. In fact, from (4.124) – (4.126), dr = due1 ,

de1 = 0,

de2 = dvn,

dn = dve2 ,

(4.226)

hence r = (u, 0, 0),

e1 = (1, 0, 0),

e2 = (0, sinh v, cosh v),

(4.227)

n = (0, cosh v, sinh v).

According to (4.194), the space-like surface of constant negative Gauss curvature is r ∗ = (u + l coth(2γ1 ), −l cosech(2γ1 ) sinh v, −l cosech(2γ1 ) cosh v).

(4.228)

According to the general theory, α1 = −2 tanh−1 ( sech(2γ1 )) is a solution of the sinh-Laplace equation. It is defined on the (u, v) plane except for

172


the straight line γ1 = 0. When γ1 → 0, the point on the surface tends to infinity. Now we apply the Darboux transformation again. Φ1 (λ) = D(λ)Φ(λ) ⎛

⎛

⎞⎞

⎛ ⎞ h1 γ −γ ⎜ ⎜ 0 ⎟ ⎟ e e λ h2 ⎟⎟ ⎝ ⎜ ⎠ =⎜ ⎝I − λ ⎝ h2 ⎠⎠ 1 −ieγ ie−γ 0 h1 ⎛ λ λ γ −γ 1 + coth γ e 1 − coth γ 1 1 e ⎜ λ λ 1 1 =⎜ ⎝ λ λ −i 1 + tanh γ1 eγ i 1 − tanh γ1 e−γ λ1 λ1

⎞ ⎟ ⎟. ⎠

(4.229) Take λ2 with |λ2 | = 1 and let λ2 /λ1 = e iµ (µ is real), then h1 = (1 + e iµ coth γ1 )eγ2 + b(1 − e iµ coth γ1 )e−γ2 , h2 = i(−1 − e iµ tanh γ1 )eγ2 + ib(1 − e iµ tanh γ1 )e−γ2 , where

1 ¯ ζ i (4.231) 2 2λ2 is real. In order that the derived solution α2 is purely imaginary and β = −iα2 satisfies the sine-Laplace equation, (4.211) should hold. This can be done if b is chosen to be a real number. In fact, when b is real, ¯ = ih coth γ1 e iµ , (4.232) h γ2 =

λ

(4.230)

1

2

ζ−

2

2 h2 −α 1 e 1 = coth2 γ1 = 1. h 2 coth γ 1 1

h2 h is replaced by 2 , then β = −iα2 h1 h1 is a solution of the sine-Laplace equation. From the explicit expressions β β of cos , sin , the Bäcklund transformation can be constructed explic2 2 itly and the time-like surface of constant negative Gauss curvature is obtained. Now we construct purely imaginary α1 from α = 0. By the fundamental solution (4.222) ⎛ ⎞of the Lax pair corresponding to α = 0, take the

In (4.210), if α is replaced by α1 ,

column solution ⎝

h1

⎠ such that

h1 2 h2 = 1, h 1

(4.233)

173


i.e.,

e2γ1 + |b|2 e−2γ1 − (b + ¯b) = 1. e2γ1 + |b|2 e−2γ1 + (b + ¯b)

(4.234)

Hence b must be purely imaginary. Since |b| can be transformed to 1 by adding a constant to γ1 , we can choose b = ±i. Suppose b = i, then h1 = eγ1 + ie−γ1 , Hence

h2 = eγ1 − ie−γ1 .

(4.235)

h2 = sech(2γ1 ) + i tanh(2γ1 ), h1 h1 = −i = sech(2γ1 ) − i tanh(2γ1 ). h2

eα1 /2 = i e−α1 /2 Let α1 = iβ1 , then

β1 β1 = sech(2γ1 ), sin = tanh(2γ1 ), cos 2 2 (4.236) β1 = cos−1 ( sech(2γ1 )), 2 β1 is a solution of the sine-Laplace equation. The Darboux matrix is ⎛

0 λ ⎝ D(λ) = I − λ1 −ieα1 /2

ie−α1 /2

⎞ ⎠

(4.237)

0

and the fundamental solution is Φ1 (λ) = D(λ)Φ(λ) ⎛

⎞

λ −α1 /2 λ γ ) e−γ (1 + e−α1 /2 ) ⎟ ⎜ e (1 − λ e λ1 ⎟ 1 =⎜ ⎝ λ α1 /2 λ α1 /2 ⎠ . γ −γ ) ie (1 + e ) −ie (1 − e λ1 λ1

(4.238)

Take λ2 such that |λ2 | = 1. Let λ2 /λ1 = e iµ (µ is real) and γ2 as before. Let

h1 = eγ2 1 − e iµ e−α1 /2 + be−γ2 1 + e iµ e−α1 /2 ,

(4.239)

¯ = h e−iµ e−α1 /2 , h 2 2

(4.240)

h2 = −ieγ2 1 − e iµ eα1 /2 + ibe−γ2 1 + e iµ eα1 /2 . Take b to be purely imaginary, then ¯ = −h e−iµ eα1 /2 , h 1 1 hence

¯ h2 h 1 −α1 = −1. ¯ e h1 h 2

174


This implies that eα2 /2 = i

h2 −α1 /2 e , h1

e−α2 /2 = i

h1 α1 /2 e h2

(4.241)

are real positive functions, and α2 satisfies the sinh-Laplace equation. From α1 and α, one can use the similar method as in the last subsection to construct Backlund ¨ congruences as well as the space-like and time-like surfaces of constant negative Gauss curvature.

4.4

Orthogonal frame and Lax pair

In this section, we consider the relation between the orthogonal frame of a surface of constant negative Gauss curvature in R3 and the Lax pair of the sine-Gordon equation. The geometric meaning of the Lax pair is elucidated clearly. The group SU (2) consists of all 2 × 2 matrices A satisfying A∗ A = I and det A = 1. Its general element is of form ⎛

A=⎝

⎞

a0 + a1 i

a2 + a3 i

−a2 + a3 i a0 − a1 i

⎠

(4.242)

where a0 , a1 , a2 , a3 are real numbers with a20 + a21 + a22 + a23 = 1. The Lie algebra su(2) consists of all 2 × 2 matrices A satisfying A∗ + A = 0 and tr A = 0. Its general element is of form ⎛

A=⎝

⎞

βi

γ + δi

−γ + δ i

−β i

⎠

(4.243)

where β, γ, δ are real numbers. The group SO(3) consists of all 3 × 3 real orthogonal matrices with determinant 1. Its Lie algebra so(3) consists of all 3 × 3 anti-symmetric real matrices. The correspondence ⎛

σ:

⎞

⎛

⎞

0

β

δ

⎜ ⎟ βi γ + δi 1⎝ ⎜ ⎟ ⎠ ←− −→ ⎜ −β 0 γ ⎟ ⎝ ⎠ 2 −γ + δ i −β i −δ −γ 0

(4.244)

is an isomorphism between su(2) and so(3), i.e., [σA, σB] = σ[A, B]

(4.245)

for all A, B ∈ su(2). This isomorphism of Lie algebras can be lifted to an isomorphism of the group SU (2) to the double covering of SO(3) (see Lemma 4.18 and Lemma 4.19).

175


We turn to discuss the geometrical meaning of the fundamental solution Φ to the Lax pair of the sine-Gordon equation. According to (4.33), under the Chebyshev coordinates, there exist orthogonal frames of a surface S of constant negative Gauss curvature satisfying ⎛

⎞

⎛

e ⎜ ⎜ 1 ⎟ ⎜ ⎜ ⎟ ⎜ e2 ⎟ = ⎜ ⎜ ⎝ ⎠ ⎝ n

⎛

u

⎞

⎛

0

e ⎜ ⎜ 1 ⎟ ⎜ αu ⎜ ⎟ ⎜ e2 ⎟ = ⎜ ⎜ − 2 ⎝ ⎠ ⎝ n

v

0

⎞ α ⎞⎛ e 1 2 ⎟ ⎟ ⎟⎜ ⎜ ⎟ 0 ⎟ ⎜ e2 ⎟ , ⎟

αv 2 0

0 αv − 2α − sin 2

sin

0

⎠⎝

0

αu 2 0 α cos 2

⎠

n

⎞⎛

0 α − cos 2 0

⎞

(4.246)

⎟ ⎜ e1 ⎟ ⎟⎜ ⎟⎜ e ⎟ . ⎟⎝ 2 ⎟ ⎠ ⎠

n

These frames are called Chebyshev frames. Using the isomorphism σ −1 , we derive a system of equations in 2 × 2 matrices ⎛

Ψu =

1⎜ ⎝ 2 ⎛

1⎜ Ψv = ⎝ 2

α αv i i sin 2 α αv 2 − i i sin 2 2 αu α i − cos 2α αu 2 − i cos 2 2

⎞ ⎟ ⎠ Ψ, ⎞

where Ψ is a 2 × 2 matrix in SU (2). Let u−v u+v , η= . ξ= 2 2 We have ⎛

Ψξ = Ψu + Ψv =

1⎜ ⎝ 2

⎛

Ψη = Ψu − Ψv = Define

1⎜ ⎝ 2 ⎛

Φ(ξ, η) = ⎝

(4.247)

⎟ ⎠ Ψ,

αξ i −e−αi/2 2 αξ eαi/2 − i 2 αη − i eαi/2 2 αη i −e−αi/2 2

(4.248) ⎞ ⎟ ⎠ Ψ, ⎞

(4.249)

⎟ ⎠ Ψ.

⎞

e−αi/4

0

0

eαi/4

⎠ Ψ,

(4.250)

176


(4.249) becomes

⎛

⎞

⎛

⎞

0 −e−αi 1 ⎠ Φ, Φξ = ⎝ 2 eαi 0

(4.251)

−αη i 1 1 ⎠ Φ. Φη = ⎝ 2 −1 αη i

1 η1 and denote ξ1 and η1 by ξ and η again, we µ can introduce a spectral parameter µ in (4.251) formally. Thus we have an alternative form of the Lax pair of the sine-Gordon equation with spectral parameter µ: Let ξ = µξ1 , η =

⎛

⎞

⎛

⎞

0 −e−αi µ ⎠ Φ, Φξ = ⎝ αi 2 e 0

(4.252)

−αη i 1/µ 1 ⎠ Φ. Φη = ⎝ 2 −1/µ αη i Remark 30 By using the isomorphism of su(2) ⎛ ⎝ ⎛ ⎝

⎞

i

0

0 −i

⎠ −→ ⎝

⎞

0

i

i 0

⎛ ⎛

⎠ −→ ⎝

and changing λ to

⎞

0

1

−1 0 0

−i

−i

0

⎠, ⎞

⎛ ⎝

⎞

0

1

−1 0

⎛

⎠ −→ ⎝

⎞

i

0

0 −i

⎠,

⎠,

i , the Lax pair (4.67) becomes (4.252). 2µ

In applying Darboux transformation we can use the Lax pair (4.252) as well as (4.67), but one should note that the λ in (4.67) and µ in i , i.e., if we use real λ1 to construct the (4.252) are related by λ = 2µ Darboux matrix for (4.67), then we should use purely imaginary µ1 for (4.252). In order to find out the relation between the Chebyshev frames and the fundamental solution of the Lax pair, we need the map from SU (2) to SO(3) corresponding to the isomorphism (4.244) between their Lie algebras.

177


Lemma 4.18 Let A ∈ SU (2) be defined by (4.242). The map ⎛

⎞ a20 +a22 −a23 −a21

⎜ ⎜ −2(a0 a1 +a2 a3 ) ⎝

τ (A) = ⎜

2(−a0 a3 +a1 a2 )

2(a0 a1 −a2 a3 )

2(a0 a3 +a1 a2 )

a20 +a23 −a22 −a21

2(a0 a2 −a1 a3 )

2(−a0 a2 −a3 a1 )

a20 −a23 −a22 +a21

⎟ ⎟ ⎟ ⎠

(4.253)

is an isomorphism from SU (2) to the double covering of SO(3). If Ψ satisfies (4.249), then τ (Ψ) satisfies (4.246). Proof. It is necessary to prove: (i) τ (A) ∈ SO(3); (ii) τ is a homomorphism, i.e., τ (AB) = τ (A)τ (B); (iii) τ (A) = τ (B) if and only if B = ±A; (iv) τ : SU (2) → SO(3) induces an isomorphism σ between their Lie algebras. Moreover, if A is a matrix function of certain parameters, then (dτ (A))(τ (A))−1 = σ((dA)A−1 ). These are the basic facts on the relationship between SU (2) and SO(3). They can all be verified by direct calculations.

Lemma 4.19 Suppose τ (A) = (aij ) ∈ SO(3), then A is given by 1 a20 = (1 + a11 + a22 + a33 ), 4 1 a22 = (1 + a11 − a22 − a33 ), 4 1 a0 a1 = (a12 − a21 ), a0 a2 = 4 1 a0 a3 = (a13 − a31 ). 4

1 a21 = (1 + a33 − a11 − a22 ), 4 1 a23 = (1 + a22 − a11 − a33 ), 4 1 (a23 − a32 ), 4

(4.254)

If the sign of any one of ai (say a0 ) is fixed, the signs of the other ai ’s are also fixed. Hereafter, unless otherwise stated, τ −1 (A) is referred to one branch of this double covering. Now we can discuss the relation between the solution of the Lax pair related to the solution α of the sine-Gordon equation and the Chebyshev frame of the surface S of constant negative Gauss curvature corresponding to α. Let S be the surface with K = −1 and related with the solution α(ξ, η) of the sine-Gordon equation, (e1 , e2 , n) be the Chebyshev frames and Φ(1, ξ, η) be the fundamental solution of the Lax pair with µ = 1.

178


From the above discussion, we have ⎛

Φ(1, ξ, η) = ⎝

and

⎛

⎞

e1

⎜ ⎟ ⎜ ⎟ ⎜ e2 ⎟ = τ ⎝ ⎠

⎞

e−αi/4

0

0

eαi/4

⎛ ⎝

⎠ τ −1

⎛

⎞

⎜ ⎟ ⎜ ⎟ ⎜ e2 ⎟ . ⎝ ⎠

e1

(4.255)

n ⎞

e−αi/4

0

0

eαi/4

n

⎠ Φ(1, ξ, η) .

(4.256)

(4.255) and (4.256) give the algebraic relation between Φ(1, ξ, η) and the Chebyshev frame, so we can say Φ(1, ξ, η) is an SU (2) representation of the Chebyshev frame. We turn to the geometrical meaning of the fundamental solution = Φ µ, ξ , µη (µ is Φ(µ, ξ, η). From (4.252) it is easily seen that Φ µ real and non-zero) satisfies ⎛

⎞

0 −e−αµ i ξ = 1 ⎝ ⎠ Φ, Φ 2 eαµ i 0

⎛

⎞

−αµη i 1 (4.257) η = 1 ⎝ ⎠ Φ. Φ 2 −1 αµη i

η is another solution of the sine-Gordon equation. µ The corresponding surface Sµ−1 with K = −1 is called the Lie transformation of S [26]. Comparing (4.257) with (4.251) we obtain Here αµ = α µξ,

ξ Theorem 4.20 Φ µ, , µη is an SU (2) representation of the surface µ Sµ−1 .

Remark 31 When α and Φ(1, ξ, η), or equivalently the Chebyshev frame, is known, the surface r = r(ξ, η) can be determined by direct integration. In fact, the right hand side of the equation dr = cos

α α due1 + sin dve2 , 2 2

(4.258)

is known already and its exterior derivative is zero. Hence r can be determined by the integration of (4.258). In conclusion, if a solution α of the sine-Gordon equation is known, then a surface S with K = −1 and the fundamental solution of the Lax

179


pair together with the Lie transformation Sµ can be obtained by solving a system of linear partial differential equations, or simply by integration. Having these data, by using Darboux transformation, a series of surfaces of K = −1 together with their Lie transformations can be obtained by purely algebraic algorithm. Similarly, for various surfaces of constant Gauss curvature in R2,1 , we can get the relation between the Chebyshev frames and the solutions of the Lax pair, i.e. the relation between the Lax pair and the Chebyshev frames of the family of surfaces obtained by the Lie transformation from the seed surface. In this case, we need the isomorphism between the Lie algebras su(1, 1) and so(2, 1). Here su(1, 1) is the set of 2 × 2 matrix A satisfying ⎛ ⎞ ⎛ ⎞ 1 0 1 0 ⎠+⎝ ⎠ A∗ = 0, A⎝ 0 −1 0 −1 and so(2, 1) is the Lie algebra of the Lorentz group on R2,1 with determinant 1.

4.5

Surface of constant mean curvature

Recently, the surface of constant mean curvature is studied widely. For example, the surface of constant mean curvature immersed in Euclidean space and homeomorphic to torus is constructed which gives an answer to the Hopf conjecture [110]. In this section we will use Darboux transformation to construct a series of surfaces of constant mean curvature from a known surface of constant mean curvature by purely algebraic algorithm.

4.5.1

Parallel surface in Euclidean space

Suppose S is a surface in R3 . By moving each point on S a distance l along the normal direction, we obtain another surface, which is a parallel surface of S. Take the orthogonal frame such that e1 , e2 are tangent to the lines of curvature, then the fundamental equations of S are dr = ω a ea

(a = 1, 2),

dea = ωab eb + dn = ω3a ea .

ωa3 n,

(4.259)

Since e1 and e2 are tangent to the lines of curvature, ω13 = b11 ω 1 ,

ω23 = b22 ω 2 ,

(4.260)

where b11 and b22 are principal curvatures of the lines of curvature.

180


Suppose S(l) is the parallel surface of S whose distance to S is l, then its position vector is (4.261) r ∗ = r + ln. By differentiation, we get

where

dr ∗ = ω ∗a ea ,

(4.262)

ω ∗a = ω a + lω3a = (1 − lbaa )ω a .

(4.263)

Hence e1 and e2 are still the tangent vectors of S(l), and n is the normal vector. When 1 − lb11 = 0 and 1 − lb22 = 0 hold, the surface S(l) is regular, and {e1 , e2 , n} is an orthonormal frame of S(l) at r∗ . Moreover, ω2∗1 = ω21 ,

ω1∗3 = ω13 = b11 ω 1 ,

ω2∗3 = ω23 = b22 ω 2 .

(4.264)

On the other hand, ω1∗3 and ω2∗3 can be written as the linear combinations of ω ∗1 and ω ∗2 : ω1∗3 = b∗11 ω ∗1 + b∗12 ω ∗2 = b∗11 ω 1 − lb∗11 ω13 + b∗12 ω ∗2 , ω2∗3 = b∗22 ω 2 − lb∗22 ω23 + b∗21 ω ∗1 .

(4.265)

Hence b∗11 − lb∗11 b11 = b11 ,

b∗22 − lb∗22 b22 = b22 ,

b∗12 = b∗21 = 0,

(4.266)

which gives b11 b22 , b∗22 = . (4.267) 1 − lb11 1 − lb22 This means that e1 and e2 are still unit vectors tangent to the lines of curvature on S(l), and the principal curvatures are b∗11 and b∗22 . The Gauss curvature and the mean curvature of the parallel surface S(l) are b∗11 =

K∗ =

b11 b22 K = , (1 − lb11 )(1 − lb22 ) 1 − 2lH + l2 K H∗ =

H − lK , 1 − 2lH + l2 K

(4.268)

(4.269)

respectively, where H = 12 (b11 + b22 ) is the mean curvature of the surface S. The most simple surfaces of constant mean curvature are spheres and cylinders. We shall not consider these trivial cases. (4.269) implies H(−2lH ∗ − 1) = −lK − H ∗ (1 + l2 K).

(4.270)

181


Suppose that K is a constant, H is not a constant. We want to find S(l) such that H ∗ is a constant. The right hand side of (4.270) is a constant. Hence 1 H ∗ = − , 1 − l2 K = 0. (4.271) 2l *

1 . Conversely, if the mean curvature H of K the surface S is a constant, then from (4.268),

Therefore, K > 0, l = ±

K(1 − l2 K ∗ ) = K ∗ (1 − 2lH).

(4.272)

If K is not a constant and K ∗ is a constant, then the right hand side of (4.272) is a constant. However, since K is not a constant, we should 1 1 , K ∗ = 2 . This gives the proof of the well-known facts: have l = 2H l

Theorem 4.21 If the surface S of constant positive Gauss curvature 1 K = 2 is not a sphere, then on each side of S, there is a parallel l 1 surface of constant mean curvature H ∗ = ± whose distance to S is l. 2l 1 If the surface S of constant mean curvature H = is neither sphere 2l nor cylinder, then there is a parallel surface of constant positive Gauss 1 curvature K = 2 whose distance to S is l. l Therefore, in the Euclidean space, the construction of the surface of constant mean curvature and the construction of surface of constant positive Gauss curvature are equivalent in local sense.

4.5.2

Construction of surfaces

We shall consider the construction of surfaces of constant positive Gauss curvature. Although there is no B¨ acklund congruence, Darboux transformation can still be used. As mentioned before, if we choose the Chebyshev coordinates of a surface of constant positive Gauss curvature, then (4.38) holds, hence ⎛

⎛

⎞

e ⎜ ⎜ 1 ⎟ ⎜ ⎜ ⎟ ⎜ e2 ⎟ = ⎜ ⎜ ⎝ ⎠ ⎝ n

u

0 αv − 2 α − sinh 2

αv 2 0 0

⎞ α ⎞⎛ e 1 2 ⎟ ⎟ ⎟⎜ ⎟ ⎟⎜ 0 ⎜ e2 ⎟ , ⎟

sinh

0

⎠⎝

⎠

n

(4.273)

182


⎞

e1

⎛

⎜ ⎜ ⎟ ⎜ ⎜ ⎟ ⎜ e2 ⎟ = ⎜ ⎜ ⎝ ⎠ ⎝

n

v

−

0 αu 2 0

αu 2 0

− cosh

⎞⎛

0 cosh α 2

⎞

e1

⎟⎜ ⎟ ⎟⎜ ⎟⎜ e ⎟ . ⎟⎝ 2 ⎟ ⎠ ⎠

α 2

(4.274)

n

0

Its integrability condition is

α = − sinh α. Using the isomorphism (4.244) between so(3) and su(2), the system (4.273) and (4.274) is equivalent to ⎛

Ψu =

1⎜ ⎝ 2 ⎛

1⎜ Ψv = ⎝ 2

⎞

α iαv i sinh ⎟ − 2 2 iαv ⎠ Ψ, α i sinh 2 2 ⎞ α iαu cosh ⎟ 2 2 α iαu ⎠ Ψ. − cosh − 2 2

Let ζ=

u + iv , 2

then

⎛

Ψζ = Ψu − iΨv =

1⎜ ⎝ 2 ⎛

1⎜ Ψζ¯ = Ψu + iΨv = ⎝ 2 Write

⎛

Φ=⎝ then

u − iv ζ¯ = , 2

αζ −ie−α/2 2 αζ ieα/2 − 2 αζ¯ ieα/2 − 2 αζ¯ −ie−α/2 2

(4.276) ⎞ ⎟ ⎠ Ψ, ⎞

(4.277)

⎟ ⎠ Ψ.

⎞

e−α/4

0

0

eα/4

⎠ Ψ,

⎛

(4.278)

⎞

0 −ie−α 1 ⎠ Φ, Φζ = ⎝ 2 ieα 0 ⎛

(4.275)

−αζ¯

1 Φζ¯ = ⎝ 2 −i

⎞

i αζ¯

⎠ Φ.

(4.279)

183


¯ Rewrite ζ as ζ Take the constant λ such that |λ| = 1, then 1/λ = λ. λ ¯ then we get the Lax pair with spectral parameter and ζ¯ as λζ, ⎛

⎞

0 −ie−α λ ⎠ Φ, Φζ = ⎝ 2 ieα 0 ⎛

−αζ¯ 1 Φζ¯ = ⎝ 2 −i/λ

⎞

i/λ

(4.280)

⎠ Φ,

αζ¯

whose integrability condition is still the negative sinh-Laplace equation (4.39). Thus we have the Lax pair for the sinh-Laplace equation. The surface of constant positive Gauss curvature can be constructed explicitly as follows. For any complex α, sinh(π i + α) = − sinh α. By changing λ to −iλ, α to π i + α, the Lax pair (4.208) of the sinh-Laplace equation becomes the Lax pair of the negative sinh-Laplace equation. Thus the solution of the sinh-Laplace equation (and the fundamental solution of its Lax pair) and the solution of the negative sinh-Laplace equation (and the fundamental solution of its Lax pair) can be changed with each other. We shall use the Darboux transformation for the sinh-Laplace equation to construct the solution of the negative sinh-Laplace equation. Suppose α is a solution of the negative sinh-Laplace equation and Φ is the fundamental solution of its Lax pair, the algorithm is shown in the following diagram. (α2 , Φ2 (λ))

(α, Φ(λ))

⏐ ⏐ α=iπ+α ⏐ ⏐ , λ=iλ

(α , Φ (λ ))

⏐ λ =−iπ ⏐ ⏐ ⏐ α =α−iπ

twice Darboux transformations

− (α2 , Φ2 (λ )) −→

In this diagram, (α , Φ (λ )) and (α2 , Φ2 (λ )) are solutions of the sinhLaplace equation and the fundamental solutions of the corresponding Lax pairs, (α, Φ(λ)) and (α2 , Φ2 (λ)) are solutions of the negative sinhLaplace equation and the fundamental solutions of the corresponding Lax pairs, the “twice Darboux transformations” are (α , Φ (λ )) −→ (α1 , Φ1 (λ )) −→ (α2 , Φ2 (λ )) where α1 is purely imaginary while α and α2 are real. This process has been explained in detail in Section 4.3.

184


Following the inverse of ⎛ the procedure from (4.273) and (4.274) to ⎞ e1 ⎜ ⎜ ⎝

⎟ ⎟ ⎠

(4.280), we get the frame ⎜ e2 ⎟ from Φ2 . In this case there is no n Backlund ¨ congruence, hence we can not get the expression of the surface algebraically. However, α2 α2 du, ω 2 = sinh dv (4.281) ω 1 = cosh 2 2 are known, and the integrability condition of dr = ω 1 e1 + ω 2 e2

(4.282) r

holds. Hence the surface of constant positive Gauss curvature can be obtained by a direct integration (without solving differential equations). Therefore, the surfaces of constant mean curvature are parallel with the surfaces of constant positive Gauss curvature. The new surface of constant mean curvature is derived from suitable parallel surface of the surface of constant positive Gauss curvature (K = 1, l = ±1). From b11 = b22 , b11 b22 = 1, we have 1 − b11 = 0, 1 − b22 = 0, hence the parallel surface exists and is regular.

Theorem 4.22 Suppose the expression of a non-spherical surface of constant positive Gauss curvature (resp. surface of constant mean curvature) under the Chebyshev coordinates is known and its Lie transformation is also known, then new surface of constant positive Gauss curvature (resp. surface of constant mean curvature) and its Lie transformation can be obtained by algebraic computation together with an integral. This process can be continued successively. Whenever we know a family of non-spherical surfaces of constant positive Gauss curvature (resp. surfaces of constant mean curvature) which are Lie transformation with each other, a series of such kinds of surfaces can be obtained. Algebraic computation together with an integral is needed in this process.

4.5.3

The case in Minkowski space

In the Minkowski space, we shall consider both space-like and timelike surfaces. (a) Space-like surface Suppose S is a space-like surface of constant Gauss curvature. Let e1 and e2 be the unit vectors tangent to the lines of curvature and n be the unit normal vector. (e1 , e2 , n) forms an orthogonal frame, and ω13 = b11 ω 2 ,

ω23 = b22 ω 2 .

(4.283)

185


From e12 = e22 = 1, n 2 = −1, we have ω13 = ω31 , ω23 = ω32 . The position vector of the parallel surface S(l) is r ∗ = r + ln.

(4.284)

dr ∗ = ω a ea + lω3a ea = ω ∗a ea .

(4.285)

By differentiation,

Hence ω ∗1 = ω 1 + lω13 = (1 + lb11 )ω 1 , ω ∗2 = ω 2 + lω23 = (1 + lb22 )ω 2 .

(4.286)

Since ωa∗3 = ωa3 , b∗11 = we have

b11 , 1 + lb11

b∗22 =

b22 , 1 + lb22

b∗12 = 0,

(4.287)

K , 1 + 2lH − l2 K 1 H − lK H ∗ = (b∗11 + b∗22 ) = , 2 1 + 2lH − l2 K

(4.288)

H(2lH ∗ − 1) = −lK − (1 − l2 K)H ∗ .

(4.289)

K ∗ = −b∗11 b∗22 =

and Suppose that K, H ∗ are constants and H is not a constant, then H∗ =

1 , 2l

1 K = − 2. l

(4.290)

Alternatively, if H, K ∗ are constants and K is not a constant, then from K(1 + l2 K ∗ ) = K ∗ (1 + 2lH), we have

1 1 , K∗ = − 2 . 2l l This leads to the following theorem. H=−

(4.291)

Theorem 4.23 For each space-like surface of constant Gauss curvature 1 − 2 , two parallel surfaces with distance ±l have constant mean curvature l 1 respectively. For each space-like surface of constant mean H = ± 2l

186


1 and non-constant Gauss curvature, the parallel surface 2l 1 with distance l is of constant negative Gauss curvature − 2 . l curvature −

Therefore, the construction of a space-like surface of constant mean curvature is equivalent to the construction of a space-like surface of constant negative Gauss curvature (see Section 4.2). The following theorem holds.

Theorem 4.24 From a known space-like surface of constant mean curvature, a series of space-like surfaces of constant mean curvature can be obtained via constructing parallel surfaces, applying B¨ a ¨cklund transformation and Darboux transformation. Remark 32 In this case, the B¨ acklund congruence belongs to the case (d) in Subsection 4.3.3. In the construction of B¨cklund ¨ transformation, time-like surface of constant negative Gauss curvature will appear as an intermediate configuration. (b) Time-like surface Now suppose S is a time-like surface, e12 = 1, e22 = −1. Similar to the space-like case, ω13 = −ω31 , ω23 = ω32 , and b∗11 =

b11 , 1 − lb11

b∗22 =

b22 , 1 + lb22

b12 = 0.

(4.292)

1 K = −b11 b22 , H = (b11 − b22 ) leads to 2 K∗ =

K , 1 − 2lH + l2 K

H∗ =

H − lK . 1 − 2lH + l2 K

(4.293)

If K and H ∗ are constants, then H∗ = −

1 , 2l

K=

1 , l2

(4.294)

while K ∗ and H are constants, we have H=

1 , 2l

K∗ =

1 . l2

(4.295)

This gives the following theorem.

Theorem 4.25 Two parallel surfaces of a time -like surface of constant 1 positive Gauss curvature K = 2 with “distance” ±l are surfaces of l


187

1 . On one side of a time-like surface of 2l 1 constant mean curvature H = , there is a parallel surface with distance 2l 1 l and with constant positive Gauss curvature 2 . l constant mean curvature ±

Therefore, the construction of time-like surface of constant mean curvature and the construction of time-like surface of constant positive Gauss curvature are equivalent.

Theorem 4.26 From a known time-like surface of constant mean curvature, a series of time-like surfaces of constant mean curvature can be obtained via constructing parallel surfaces, B¨ a ¨cklund transformation and Darboux transformation. In this case the related Backlund ¨ congruences belong to the case (2) in Subsection 4.3.4.

Chapter 5 DARBOUX TRANSFORMATION AND HARMONIC MAP

Harmonic map is an important subject in differential geometry [25, 24, 115] and is closely related with mathematical physics and the soliton theory [24, 57, 46, 48]. At the beginning of this chapter, we introduce the notion of the harmonic map. Then the harmonic maps from Euclidean plane R2 or Minkowski plane R1,1 to Euclidean sphere S 2 in R3 , and H 2 or S 1,1 in Minkowski space R2,1 are elucidated. We shall show that all these harmonic maps can be obtained from the construction of the surfaces of constant Gauss curvature in Chapter 4, and show the relations of these harmonic maps with some special soliton equations. Therefore, we can construct new harmonic maps from known harmonic maps and their extended solutions by using purely algebraic algorithm. Using Darboux transformation, we can also construct the harmonic maps from R1,1 or R2 to the Lie group U (N ) and get explicit expressions of the solutions. The solitons interact elastically. The unitons are also considered by using Darboux transformation. Comparing with [102], the construction here is more explicit with purely algebraic algorithm by using Darboux transformation.

5.1

Definition of harmonic map and basic equations

Riemannian manifold and Lorentzian manifold are the generalizations of the Euclidean space and Minkowski space. A Riemannian manifold or a Lorentzian manifold is an n dimensional differential manifold M with a metric g. In the local coordinates (x1 , · · · , xn ) of M , the metric g is

190


expressed as ds2 =

n

gij (x)dxi dxj = gij (x)dxi dxj .

i,j=1

If (gij ) is positive definite, then M is a Riemannian manifold and ds is the differential of the arc length of a curve. If the eigenvalues of (gij ) have the signs (+, · · · , +, −), then M is a Lorentzian manifold and g is called a Lorentzian metric. Let M and N be Riemannian manifolds or Lorentzian manifolds, φ : M → N be a C 2 -map. The energy (or the action if M is a Lorentzian manifold) of the map φ is

E(φ) = M

e(φ) dV VM

(5.1)

where dV VM is the volume element of M . e(φ) is called the energy (action) density. In local coordinates, e(φ) = gαβ (φ) N

∂φα ∂φβ ij g (x) ∂xi ∂xj M

(5.2)

(α, β = 1, · · · , n; i, j = 1, · · · , m). Here n = dim N , m = dim M . g ij ’s and gαβ ’s are contravariant compoM

N

nents of g and covariant components of g respectively, and (g ij ) is the M

N

M

inverse of (gij ). M

The Euler equation of the functional E(φ) is derived directly as fol√ lows. In local coordinates, dV VM = g dm x where g = | det(gij )|. Hence M the Euler equation of E(φ) is √ √ ∂e(φ) g ∂ ∂e(φ) g − k = 0. (5.3) ∂φγ ∂x ∂(φγ,k ) For simplicity, we write gij for gij , and gαβ for gαβ . The partial derivative M N ∂φγ γ is denoted by φ,k . Then, the first term of (5.3) is ∂xk ∂gαβ (φ) α β ij √ g, φ,i φ,j g ∂φγ and the second term is ∂g ∂ √ αγ ki α β √ g φ,i φ,k g 2 k (gαγ g ki φα,i g ) = 2 ∂x ∂φβ √ ∂ g ∂g ki √ ∂ 2 φα √ + gαγ k φα,i g + gαγ g ki i k g + gαγ g ki φα,i k . ∂x ∂x ∂x ∂x

Darboux transformation and harmonic map

191

Using the Christoffel symbols Γkij and Γγαβ of M , N and the well-known M N formulae ∂gαγ = Γλαβ gλγ + Γλγβ gαλ , (5.4) N N ∂φβ ∂g ij = −g lj Γilk − g il Γjlk , M M ∂xk √ ∂ g √ = Γiik g, M ∂xk we obtain the Euler equation g ik

∂ 2 φγ

∂xi ∂xk

− Γjik M

α β ∂φγ γ ∂φ ∂φ + Γ = 0. ∂xj N αβ ∂xi ∂xk

(5.5) (5.6)

(5.7)

Definition 5.1 A C 2 map φ from M to N is called a harmonic map if and only if it satisfies (5.7) [24]. (5.7) is the system of partial differential equations for the harmonic map φ from M to N . If M is a Riemannian manifold, (5.7) is a system of nonlinear elliptic equations. If M is a Lorentzian manifold, (5.7) is a system of nonlinear hyperbolic equations. The harmonic map from a Lorentzian manifold is also called a wave map. Harmonic map is a very important subject both in mathematics and in physics. In mathematics, there are many known special cases. If N is the line R, a harmonic map is a harmonic function on M (especially, when M = Rn , it is the ordinary harmonic function); if M is the line R or the circle S 1 , the harmonic map is a geodesic or a closed geodesic respectively. A minimal surface is a conformal harmonic map which keeps the angles between two lines invariant. In physics, there are quite a lot of applications: (1) Nonlinear σ-model (or chiral field) is a harmonic map M → N where M is a Minkowski space and N is usually a homogeneous space. Especially, if N is a Lie group, it is called a principal chiral field [86]. (2) Ernst equation describes the static axially symmetric solution of Einstein gravitation in vacuum [27]. It is the Euler equation of the Energy 3 & 1 ∂φ 2 ∂ψ 2 ' 3 + d x φ2 i=1 ∂xi ∂xi under the axially symmetric constraint. Therefore, the Ernst equation is an axially symmetric harmonic map from R3 to the hyperbolic plane

192


H 2 which is equipped with the Poincar´ é metric 1 (dφ2 + dψ 2 ). φ2

ds2 =

(3) In particle physics, string is used as a model of hadron. In four dimensional Lorentzian space-time V , the world surface describing the motion of a classical string is a two dimensional time-like surface. It is determined by the equation φαττ − φασσ + Γαβγ (φβτ φγτ − φβσ φγσ ) = 0, V

This is the equation for the harmonic map from two dimensional Minkowski plane R1,1 to V [36]. (4) Some solutions of the Yang-Mills equation in R4 under R-gauge can also be obtained from harmonic maps. (5) The simplest model for liquid crystal is a harmonic map from R2 to S 2 . In this book, we only discuss harmonic maps which can be constructed explicitly by Darboux transformation. The starting manifold M is the Euclidean plane R2 = {(x, y)} or Minkowski plane R1,1 = {(t, x)} (or a part of them). In these two cases, the equations for the harmonic maps are (5.8) φγxx + φγyy + Γγαβ (φαx φβx + φαy φβy ) = 0 N

and φγtt − φγxx + Γγαβ (φαt φβt − φαx φβx ) = 0 N

(5.9)

respectively.

5.2

Harmonic maps from R2 or R1,1 to S 2 , H 2 or S 1,1

Let S 2 be the sphere of Euclidean space R3 , which consists of all the points l = (l1 , l2 , l3 ) with l 2 = 1. The coordinates in Euclidean plane R2 are represented by (x, y), and the differential form of the Euclidean metric is ds2 = dx2 + dy 2 . A map from R2 to S 2 can be written as l = l(x, y) with l 2 = 1. From (5.7) or (5.8), the equation of l for the harmonic map R2 → S 2 is given by (5.10) lxx + lyy + (lx2 + ly2 )l = 0. It can be derived as follows: E(l) =

& ∂l 2 Ω

∂x

+

∂l 2 '

∂y

dx dy,

193


where Ω is a region of R2 , l satisfies l 2 = 1. Introduce the Lagrangian undetermined multipliers λ, we rewrite E(l) as E(l) =

& ∂l 2 Ω

∂x

+

∂l 2

∂y

'

+ λ(l 2 − 1) dx dy,

then the Euler equation becomes λl =

∂ ∂l ∂ ∂l + , ∂x ∂x ∂y ∂y

or

l = λl. With = 1, we have l · lx = 0 and l · ly = 0, hence l · lxx = −lx2 , l · lyy = −ly2 . This leads to λ = −(lx2 + ly2 ), and (5.10) follows. As is known, the Minkowski plane R1,1 is the simplest two dimensional Lorentzian manifold, whose metric is ds2 = dt2 − dx2 . Instead of (5.7), the equation for the harmonic map from R1,1 to S 2 is l2

ltt − lxx + (lt2 − lx2 )l = 0.

(5.11)

The proof is similar to the Euclidean space. In the Minkowski space R2,1 , we can define two kinds of “spheres”: 2 H and S 1,1 . H 2 is defined by l 2 = −1,

l3 > 0.

(5.12)

It is the upper branch of the biparted rotational hyperboloid in R2,1 and realizes the Lobachevsky geometry globally, while the surface of constant negative Gauss curvature in R3 only realizes a certain part of the Lobachevsky plane. S 1,1 is given by (5.13) l 2 = 1. It is a uniparted rotational hyperboloid. It is a time-like surface in the Minkowski space. Its metric is indefinite. Similar to the above discussions, the equations for the harmonic maps from R2 and H 2 to S 1,1 are lxx + lyy − (lx2 + ly2 )l = 0

(l 2 = −1, l3 > 0)

(5.14)

and lxx + lyy + (lx2 + ly2 )l = 0 (l 2 = 1),

(5.15)

respectively. The equations for the harmonic maps from R1,1 to H 2 and S 1,1 are ltt − lxx − (lt2 − lx2 )l = 0 (l 2 = −1, l3 > 0) (5.16)

194


and ltt − lxx + (lt2 − lx2 )l = 0

(l 2 = 1)

(5.17)

respectively. A harmonic map may be defined in a region of R2 or R1,1 instead of the whole plane. The harmonic maps from R2 (or R1,1 ) are conformal invariant. That is, suppose h : Ω → N is a harmonic map from a region Ω in R2 (or R1,1 ) to a Riemannian (or Lorentzian) manifold N , φ is a conformal map from a region Ω1 in R2 (or R1,1 ) to Ω, then the map h ◦ φ is a harmonic map from Ω1 to N . This fact can be derived from the general equation or the energy integral. In fact, when n = 2, the energy integral (5.1) is invariant under the conformal transformation of R2 (or R1,1 ). Hence its Euler equation has the same property. Here we explain it by two cases. Case 1: Conformal invariance of the harmonic map from R2 to S 2 (or H 2 , S 1,1 ) First we consider R2 as a one dimensional complex space C1 with complex coordinate {z = u + iv} and metric ds2 = dz d¯. Suppose z = f (w) = u(u1 , v1 ) + iv(u1 , v1 ) is a conformal map from a region Ω1 to Ω, l is a harmonic map from Ω to S 2 , then l◦f : l = l(x(u1 , v1 ), y(u1 , v1 )) is a harmonic map from Ω1 to S 2 (or H 2 , S 1,1 ). In fact, (5.10) can be written as (5.18) lz z¯ + (lz · lz¯)l = 0. Under the transformation z = z(w), (5.18) is transformed to lww¯ + (lw · lw¯ )l = 0. If S 2 is changed to H 2 or S 1,1 , the proof is similar. Case 2: Conformal invariance of the harmonic map from R1,1 to S 2 (or H 2 , S 1,1 ) Let t−x t+x , η= (5.19) ξ= 2 2 be the characteristic coordinates (or light-cone coordinates) of R1,1 , then ds2 = 4dξ dη. A conformal map φ of R1,1 is of form ξ = f (ξ1 ),

η = g(η1 ).

(5.20)

(5.11) can be written as lξη + (lξ · lη )l = 0 (l 2 = 1). l ◦ φ is also a harmonic map, since it satisfies lξ1 η1 + (lξ1 · lη1 )l = 0.

(5.21)

195


Similar result holds if S 2 is changed to H 2 or S 1,1 . Using the conformal invariance, we can define normalized harmonic map. First we show that a harmonic map l from R1,1 to S 2 satisfies (lξ2 )η = 0,

(lη2 )ξ = 0,

l 2 = 1.

(5.22)

Suppose (5.21) holds, then take the inner product with lξ , we obtain (lξ2 )η = 0. Similarly, (lη2 )ξ = 0 holds. This leads to (5.22). Conversely, suppose (5.22) holds and lξ ’s, lη ’s are linearly independent, then lξ · lξη = 0,

lη · lξη = 0.

(5.23)

Hence lξη = σl. The inner product with l leads to σ = l · lξη = (l · lξ )η − lξ · lη = −lξ · lη , Hence (5.21) holds. From (5.22), lξ2 = f (ξ),

lη2 = g(η).

When f (ξ) = 0 and g(η) = 0, we can define the transformation ξ = ξ(ξ1 ) dξ 2 dη 2 1 1 and η = η(η1 ) such that = f (ξ), = g(η). Rewrite (ξ1 , η1 ) dξ dη as (ξ, η), we obtain (5.24) lξ2 = 1, lη2 = 1. The harmonic map satisfying condition (5.24) is called a “normalized harmonic map” [38]. When lξ2 = 0 and lη2 = 0, a harmonic map can always be transformed to a normalized harmonic map by the transformation ξ = ξ(ξ1 ), η = η(η1 ). Thus, some problems on harmonic maps can be simplified by using conformal transformations. For a normalized harmonic map, lx2 + lt2 = 1,

lt · l x = 0

(5.25)

holds [58]. In fact, (5.24) is equivalent to (5.25). Hence a normalized harmonic map can also be defined by (5.25). A harmonic map from R2 to S 2 (or H 2 , S 1,1 ) can also be normalized. From (5.18) and l 2 = 1, we have (lζ2 )ζ¯ = 0 and (lζ¯2 )ζ = 0. Hence lζ2 is a holomorphic function of ζ. If lζ2 = (lu2 − lv2 ) − 2ilu · lv = 1,

(5.26)

196


i.e., lu2 − lv2 = 1,

lu · lv = 0,

(5.27)

this harmonic map is called normalized. Suppose l = l(u, v) is a harmonic map defined in a simply connected region Ω and lζ2 = 0, then l can be normalized. In fact, take the conformal map ζ = ζ(w), dζ 2 . (5.28) lw2 = lζ2 dw dw 2 ) = lζ2 has a single-valued solution When lζ2 = 0, the equation ( dζ w = f (ζ) in the simply connected region Ω. w maps Ω to Ω1 . The map l ◦ f −1 from Ω1 to S 2 (or H 2 , S 1,1 ) has been normalized. We have seen in Chapter 4 that there are many kinds of surfaces of constant Gauss curvature on which Chebyshev coordinates (u, v) and the corresponding Chebyshev frames (e1 , e2 , n) exist (under the assumption that there are no umbilical points). The map from a region of R2 (or R1,1 ) to the surface defined by Chebyshev coordinates is called a Chebyshev map. In this case, the normal vector n defines the Gauss map of S 2 (or H 2 , S 1,1 ).

Theorem 5.2 The composition of the Chebyshev map and the Gauss map is a normalized harmonic map, and vise versa [58]. Proof. We prove this theorem in several cases. (a) R1,1 → S 2 Suppose l is a normalized harmonic map. Take e1 and e2 so that e1 is the unit vector of lt , e2 is the unit vector of lx , then (5.25) implies that e1 is orthogonal to e2 . Moreover, there is a function α(x, t) such that lt = − sin

α e1 , 2

lx = cos

α e2 . 2

(5.29)

Since l, e1 and e2 form a orthonormal frame, e1t = sin α2 l + σe2 , e1x = τ e2 ,

(5.30)

e2t = −σe1 , e2x = − cos α2 l − τ e1 . Their integrability condition leads to 1 σ = αx , 2

1 τ = αt 2

(5.31)

197


and αtt − αxx = sin α.

(5.32)

Denote l = n, we have ⎛

⎞

e1

⎜ ⎜ ⎜ e2 ⎝

n ⎛

e

⎜ 1 ⎜ ⎜ e2 ⎝

n

⎛

αx 0 ⎜ 2 ⎟ ⎜ αx ⎟ − 0 ⎟ =⎜ ⎜ ⎠ 2α ⎝ 0 − sin t 2 α ⎛ ⎞ t 0 ⎜ 2 ⎟ ⎜ αt ⎟ − 0 ⎟ =⎜ ⎜ ⎠ 2 ⎝ α 0 cos x 2

⎞ α ⎞⎛ e 1 ⎟ 2 ⎟ ⎟⎜ ⎜ ⎟ ⎟ 0 ⎟ ⎜ e2 ⎟ ,

sin

0

⎠⎝

0 α − cos 2 0

⎠

n

⎞⎛

⎞

(5.33)

⎟ ⎜ e1 ⎟ ⎟⎜ ⎟⎜ e ⎟ , ⎟⎝ 2 ⎟ ⎠ ⎠

n

This is just the equation of Chebyshev frame for a surface of constant negative Gauss curvature, l = n is its normal vector, l(u, v) is the Gauss map. Therefore, a normalized harmonic map R1,1 → S 2 is the composition of a Chebyshev map and a Gauss map. Converse conclusion follows from α α nξ2 = (nt + nx )2 = sin2 + cos2 = 1, 2 2 nη2 = (nt − nx )2 = 1 and the fact that nξ and nη are linearly independent. (b) R2 → H 2 Suppose l(x, y) is a normalized harmonic map, then (5.25) holds. Let e1 and e2 be the unit vectors of lu and lv respectively. We can choose function α(x, t) so that α e1 , 2 α lv = sinh e2 . 2 lu = cosh

Similar to the case (a), they satisfy a system of different equations. The equations are just the equations of the Chebyshev frame for the surface of constant negative Gauss curvature in R2,1 (Section 4.3), n is the Gauss map from this surface to H 2 . Other cases can be proved similarly. Here we list the results in the following table.

198


From

To

Constant curvature

S 2 ⊂ R3 R

2

2

H ⊂R

S 1,1 ⊂ R2,1 2

3

S ⊂R

− (space-like)

α = sinh α

− (time-like)

α = sin α ∂2 α ∂u2

−

H 2 ⊂ R2,1 R

α = − sinh α

+

2,1

Differential equation

+ (space-like)

2

∂ α ∂u2 2

1,1

∂ α ∂u2

S

1,1

2,1

⊂R

+ (time-like)

∂2 α ∂u2 2

∂ α ∂u2

−

∂2 α ∂v 2

−

2

∂ α ∂v 2

= sin α

−

∂2 α ∂v 2

= sinh α

−

∂2 α ∂v 2

= cosh α

−

2

∂ α ∂v 2

= sin α

= eα

This theorem implies that the construction of various normalized harmonic maps is closely related to the construction of the Chebyshev frames of various surfaces of constant Gauss curvature. Therefore, the following theorem holds [58–60].

Theorem 5.3 Suppose a normalized harmonic map from R2 or R1,1 to various two dimensional “sphere” is known. Then, using Darboux transformation, we can obtain an infinite series of normalized harmonic maps of the same type. Moreover, normalized harmonic map from R2 to S 1,1 (resp. H 2 ) can also be constructed by Darboux transformation from a known normalized harmonic map from R2 → H 2 (resp. S 1,1 ). Remark 33 It is know that for a surface in R3 with constant positive Gauss curvature, the Gauss map is a harmonic map [92]. This can be derived from Theorem 5.3 as a corollary. For example, for a surface of constant Gauss curvature 1 in R3 , the corresponding surface of constant mean curvature (see Section 4.5) is ω ∗1 = ω 1 ± ω13 ,

ω ∗2 = ω 2 ± ω23 ,

α α ± sinh )du, 2 2

ω ∗2 = (cosh

i.e., ω ∗1 = (cosh

α α ± sinh )dv. 2 2

Its first fundamental form is ds∗2

= (cosh

α α ± sinh )2 (du2 + dv 2 ) 2 2

= e±α (du2 + dv 2 ).

199


Hence its Chebyshev coordinates are isothermal coordinates. This means that the Chebyshev coordinates give a conformal correspondence of surfaces of constant mean curvature and the Euclidean plane. The above conclusion follows from the conformal invariance of the harmonic map in two dimensional cases. Remark 34 We can prove that there is a similar conclusion as Remark 33 for space-like and time-like surfaces of constant mean curvature in R2,1 . In Section 4.5 we have known that a space-like surface of constant mean curvature is a parallel surface of a space-like surface of constant negative Gauss curvature, while a time-like surface of constant mean curvature is a parallel surface of a time-like surface of constant positive Gauss curvature. Suppose their Gauss curvatures are −1 and +1 respectively, then ω ∗1 = ω 1 ± ω13 , ω ∗2 = ω 2 ± ω23 . For space-like surface, α α ω ∗1 = (cosh ± sinh )du, 2 2 hence

ω ∗2 = (sinh

α α ± cosh )dv, 2 2

ds∗2 = (ω ∗1 )2 + (ω ∗2 )2 = e±α (du2 + dv 2 )

where (u, v) are isothermal coordinates. In the time-like case, ω ∗1 = (cosh hence

α α ± sinh )du, 2 2

ω ∗2 = (sinh

α α ± cosh )dv, 2 2

ds∗2 = (ω ∗1 )2 − (ω ∗2 )2 = e±α (du2 − dv 2 )

where (u, v) are also isothermal coordinates. From the above theorem, we know that a Gauss map of a surface of constant mean curvature is a harmonic map, and a Chebyshev map from (u, v) plane to the surface is a conformal map. Remark 35 The harmonic map from the Minkowski plane R1,1 was first studied in [37]. The solution of the Cauchy problem for harmonic map exists globally if the target manifold is a complete Riemannian manifold. In [38], it was pointed out that when the target manifold is S 1,1 , global solution of the Cauchy problem may not exist.

5.3 Harmonic maps from R1,1 to U (N ) 5.3.1 Riemannian metric on U (N ) The group U (N ) is composed of all N × N unitary matrices, i.e., it consists of all N × N matrices g satisfying gg ∗ = I. On U (N ), there is

200


a Riemannian metric ds2 = − tr(dg g −1 dg g −1 ) = − tr(dg g ∗ dg g ∗ ).

(5.34)

Here g ∈ U (N ), dg is its differential, the trace (tr) of a matrix is the sum of all its diagonal entries. This metric ds2 is invariant under the left and right translation (i.e., the transformation g → g1 = ag and g → g2 = ga for any fixed element a of U (N )) of the group. In fact, dg1 g1−1 = a dg g −1 a−1 , dg2 g2−1 = dg g −1 imply tr(dg g −1 dg g −1 ) = tr(dg1 g1−1 dg1 g1−1 ) = tr(dg2 g2−1 dg2 g2−1 ). Next we prove that ds2 is actually a Riemannian metric, i.e., (5.34) is positive definite. Since U (N ) is a differential manifold, we take the local coordinate {uα } of (N ) so that g = g(uα )

(α = 1, 2, · · · , r),

where r = N 2 is the dimension of U (N ). Then ∂g duα , ∂uα ∂g ∗ ∂g ∗ g g duα duβ = gαβ duα duβ , ds2 = − tr ∂uα ∂uβ ∂g ∗ ∂g ∗ g g . gαβ = − tr ∂uα ∂uβ dg =

(5.35)

Suppose {ξ α } are real and not all zero. Let Q=

∂g −1 α g ξ . ∂uα

Substituting ξ α for duα into ds2 , we obtain gαβ ξ α ξ β = − tr(Q2 ). On the other hand, by differentiating gg ∗ = I, we have ∂g ∗ ∂g ∗ g + g = 0, ∂uα ∂uα which gives Q + Q∗ = 0. Hence − tr(Q2 ) = tr(QQ∗ ) ≥ 0. Clearly tr(QQ∗ ) = 0 if and only if Q = 0, which means ξ α = 0. Therefore, gαβ ξ α ξ β ≥ 0, and the equality holds if and only if ξ α = 0. This proves that ds2 is a Riemannian metric.

201


5.3.2

Harmonic maps from R1,1 to U (N )

Suppose (x, t) are orthonormal coordinates of R1,1 . Let ξ = x + t, η = x−t, then (ξ, η) are characteristic coordinates (light-cone coordinates) of R1,1 , and the metric of R1,1 is ds2 = dξ dη. The action of the harmonic map g(ξ, η) from R1,1 to U (N ) is A[g] =

gαβ uαξ uβη dξ dη ∂g ∗ ∂g ∗ = − tr g g uαξ uβη dξ dη ∂uα ∂uβ ∂g ∂g g ∗ g ∗ dξ dη. = − tr ∂ξ ∂η

Denote

A = gη g ∗ ,

B = gξ g ∗ ,

(5.36)

(5.37)

then A and B are valued in the Lie algebra u(N ) of the Lie group U (N ). We can write down the Euler equation of A[g] by using the standard procedure of variation. Denote AΩ [g] = −

Ω

tr(gξ g ∗ gη g ∗ )dξ dη,

where Ω is a bounded region. Suppose g depends on another parameter ∂g τ , i.e., g = g(τ, ξ, η) and g(0, ξ, η) = g. Let = h and h is 0 at ∂τ τ =0 the boundary of Ω. Then

dAΩ [g] =− dτ τ =0

Ω

tr(hξ g ∗ A + gξ h∗ A + Bhη g ∗ + Bgη h∗ )dξ dη.

Take the partial integration for the first and the third terms, we have

dAΩ [g] = − dτ τ =0

Ω

tr(−hg ∗ Aξ − hgξ∗ A + gξ h∗ A

−Bη hg ∗ − Bhgη∗ + Bgη h∗ )dξ dη. Differentiating gg ∗ = I and g ∗ g = I leads to hg ∗ = −gh∗ and h∗ g = −g ∗ h. Hence gξ h∗ A = gξ h∗ gg ∗ gη g ∗ = −gξ g ∗ hg ∗ gη g ∗ , Bgη h∗ = gξ g ∗ gη h∗ gg ∗ = −gξ g ∗ gη g ∗ hg ∗ , hgξ∗ A = hgξ∗ gg ∗ gη g ∗ = −hg ∗ gξ g ∗ gη g ∗ , Bhgη∗ = gξ g ∗ hg ∗ ggη∗ = −gξ g ∗ hg ∗ gη g ∗ .

202


With the above equation and using the equality tr(A1 A2 · · · An ) = tr(An A1 A2 · · · An−1 ), we obtain

dAΩ [g] = dτ τ =0

Ω

tr(hg ∗ (Aξ + Bη ))dξ dη = 0.

Since hg ∗ can be an arbitrary matrix in the Lie algebra u(N ) and Aξ +Bη is also a matrix in the Lie algebra u(N ), the Euler equation Aξ + Bη = 0

(5.38)

is derived. By direct calculation, A and B defined by (5.37) also satisfy Aξ − Bη + [A, B] = 0.

(5.39)

Remark 36 (5.37) is actually gξ = Bg and gη = Ag, and (5.39) is the integrability condition of (5.37). Hence, in R1,1 (or its simply connected region), A and B which satisfy (5.39) can determine g(ξ, η) uniquely up to a right-multiplier of a fix element of U (N ). Therefore, we can consider A and B as unknown functions and study the partial differential equations (5.38) and (5.39) for A and B. As in many cases considered before, we need to find the Lax pair with spectral parameter for (5.38) and (5.39). Notice that the integrability condition of (5.40) Φη = λAΦ, Φξ = λ(2λ − 1)−1 BΦ is

λAξ − λ(2λ − 1)−1 Bη + λ2 (2λ − 1)−1 [A, B] = 0,

i.e., λ2 (Aξ − Bη + [A, B]) + (λ2 − λ)(Aξ + Bη ) = 0. It should hold true for arbitrary λ, so we get (5.38) and (5.39). Hence, the following theorem holds.

Theorem 5.4 The partial differential equations (5.38) and (5.39) have Lax pair (5.40). A non-degenerate N ×N matrix solution Φ(λ) of the Lax pair is called its fundamental solution. A fundamental solution is determined by its value at one point, say (0, 0). When λ is real, if Φ(λ)|(0,0) ∈ U (N ),


203

then Φ(λ) ∈ U (N ) holds everywhere. Hereafter, we suppose Φ(λ)|(0,0) ∈ U (N ) for real λ. Then Φ(1)K is a harmonic map for any K ∈ U (N ). Remark 37 We can replace U (N ) by any matrix Lie group G. If we use A = gη g −1 ,

B = gξ g −1

to replace (5.37), then (5.38) and (5.39) are extended as the equations for harmonic maps to G, and the Lax pair is still (5.40). In this case G may not have an invariant positive definite metric. Now we use the Darboux transformation to construct explicit solutions of (5.38) and (5.39). First, suppose the group G is a complex (or real) general linear group GL(N, C) (or GL(N, R)). A(ξ, η) and B(ξ, η) are real (or complex) N × N matrix functions. Suppose Φ(ξ, η, λ) satisfies det Φ = 0, and A, B, Φ are solutions of (5.38)–(5.40). We want to construct a Darboux matrix D in the form D(λ) = I − λS

(5.41)

Φ1 = D(λ)Φ

(5.42)

so that with suitable A1 and B1 satisfying (5.38)–(5.39). That is, Φ1 , A1 , B1 satisfy (5.40) Φ1η = λA1 Φ1 , Φ1ξ = λ(2λ − 1)−1 B1 Φ1 . Substituting (5.42) into the above equations and using (5.40), we have A1 = A − Sη ,

B1 = B + Sξ

(5.43)

and Sη S = AS − SA,

Sξ S − 2S Sξ = SB − BS.

(5.44)

Therefore, as soon as we get a nontrivial solution S(ξ, η) of (5.44), then, A1 , B1 are obtained from (5.43) and Φ1 is obtained from (5.41), (5.42). Hence, Φ1 satisfies (5.40), and A1 , B1 satisfy (5.38), (5.39). This means that A1 and B1 give a new solution. The explicit expression of S is constructed as follows. Take N real (or complex) constants λ1 , · · ·, λN , which are not all the same and λα = 0, 1/2, 1; α = 1, 2, · · · , N . Take N constant column vectors l1 , · · · , lN so that the N × N matrix H = (Φ(λ1 )l1 , · · · , Φ(λN )lN )

(5.45)

204


is non-degenerate. (Each column Φ(λα )lα in H is a column solution of the Lax pair when λ = λα .) Let Λ = diag(λ1 , λ2 , · · · , λN ),

(5.46)

then we have the following theorem.

Theorem 5.5 The matrix S = HΛ−1 H −1

(5.47)

satisfies (5.44). Proof. hα = Φ(λα )lα is a column solution of (5.40) for λ = λα , i.e., it satisfies hαη = λα Ahα , hαξ = λα (2λα − 1)−1 Bhα , (5.48) (α = 1, 2, · · · , N ). Hence Hη = AHΛ,

Hξ = BHΛ(2Λ − 1)−1 ,

and Sη = AHΛΛ−1 H −1 − HΛ−1 H −1 AHΛH −1 = A − SAS −1 . This is the first equation of (5.44). On the other hand, Sξ

= BHΛ(2Λ − 1)−1 Λ−1 H −1 − HΛ−1 H −1 BHΛ(2Λ − 1)−1 H −1 = BH(2Λ − 1)−1 H −1 − SBHΛ(2Λ − 1)−1 H −1 .

It is easy to check that the second equation of (5.44) is satisfied. Therefore, we have the following theorem.

Theorem 5.6 Suppose (A, B, Φ) is a solution of (5.38)–(5.40), S is defined by (5.47), then (A1 , B1 , Φ1 ) defined by (5.43) and (5.42) is also a solution of (5.38)–(5.40). Thus, we have the Darboux transformation (A, B, Φ) −→ (A1 , B1 , Φ1 ). Moreover, the algorithm is purely algebraic since S, Sξ , Sη can be obtained from A, B and Φ with explicit algebraic expressions.


205

In this construction, we cannot guarantee that the condition det H = 0 holds globally. That is, if the seed solution (A, B, Φ) is defined on R1,1 (or its simply connected region Ω), it is not always possible that A1 , B1 and Φ1 can be defined on R1,1 (or Ω) globally. What we can be sure is that det H = 0 holds in a small region since it holds at some point by construction. There is another problem. If G is a group, A and B are valued in the Lie algebra G of G, whether will A1 and B1 be still in G ? These are two related difficult problems. We shall give their answers for G = U (N ). As we have known, the group U (N ) is composed of all the N × N matrices g satisfying g ∗ = g −1 . A matrix A belongs to the Lie algebra u(N ) of U (N ) if and only if A∗ + A = 0.

(5.49)

Now suppose (A, B, Φ) is a solution of (5.38)–(5.40) which is defined on whole R1,1 and A, B ∈ u(N ). (If they are defined in a simply connected region of R1,1 , the statements below hold as well.) The Darboux transformation is constructed following (5.43) and (5.42). In order to have A1 , B1 ∈ u(N ), we need (S + S ∗ )η = 0,

(S + S ∗ )ξ = 0.

(5.50)

Hence we want to make specific S to satisfy (5.50). Let λ0 be a non-zero complex number and ¯0 λα = λ0 or λ Choose lα so that

(α = 1, 2, · · · , N ).

h∗α hβ = 0 (if λα = λβ )

(5.51) (5.52)

holds at one point (say ξ = η = 0) and hα ’s are linearly independent. We shall show that the S constructed from these λα and lα satisfies (5.50). First, we prove that (5.52) holds everywhere on R1,1 if it holds at one point. In fact, (5.48) leads to ¯ α h∗ A∗ hβ + λβ h∗ Ahβ , (h∗α hβ )η = λ α α ¯ α (2λ ¯ α − 1)−1 h∗ B ∗ hβ + λβ (2λβ − 1)−1 h∗ Bhβ . (h∗α hβ )ξ = λ α α Hence

(h∗α hβ )η = (h∗α hβ )ξ = 0

¯ α = λβ ). This means that (5.52) holds everywhere when λα = λβ (i.e. λ if it holds at one point.

206


¯ 0 ), hα ’s are linearly independent Next, for the same λα = λ0 (or λ everywhere if they are linearly independent at one point, since (5.48) is linear. Choose the initial value of {hα } so that they are linearly independent and satisfy (5.52), then {hα } are linearly independent and (5.52) holds everywhere. This means that the solution is globally defined on R1,1 . (5.47) leads to SH − HΛ−1 = 0, i.e., Shβ − λ−1 β hβ = 0. Its complex conjugate transpose gives ¯ −1 h∗ = 0. h∗β S ∗ − λ β β Hence

¯ −1 + λ−1 )h∗ hγ . h∗β (S ∗ + S)hγ = (λ γ β β

¯ 0 ), it equals It is zero if λβ = λγ . When λβ = λγ = λ0 (or λ 1 ∗ ¯ 0 hβ hγ . Hence λ 1 1 h∗β (S ∗ + S)hγ = h∗β 0 + ¯ 0 Ihγ . λ λ

1

λ0

+

Since {hγ } (γ = 1, 2, · · · , N ) are linearly independent, we have S∗ + S =

1

1 + ¯ 0 I. λ0 λ

(5.53)

Hence (5.50) holds, and A1 , B1 ∈ u(N ). Therefore, we have the following theorem.

Theorem 5.7 Let G = U (N ), A, B ∈ u(N ). Suppose (A, B, Φ) is a solution of (5.38)–(5.40) on R1,1 , S is defined by (5.47) in which λα and lα ’s are chosen so that (5.51), (5.52) and det H = 0 are satisfied. Then (A1 , B1 , Φ1 ) defined by (5.42) is also a global solution of (5.38)– (5.40) on R1,1 . Moreover, A1 , B1 ∈ u(N ). Now we turn to the Lie group SU (N ). A matrix g ∈ SU (N ) if and only if g ∈ U (N ) and det g = 1. The Lie algebra su(N ) of SU (N ) composes of all matrices A such that A ∈ u(N ) and tr A = 0. The above Theorem 5.7 also holds for the case of SU (N ). In this case, there are additional conditions tr A = tr B = 0 and tr A1 = tr B1 = 0. Thus, we must have tr Sη = 0 and tr Sξ = 0. In fact, the first equation of (5.44) can be written as Sη = −SAS −1 + A

207


and tr Sη = 0 follows. From the definition of S, −S + 2I = −H(Λ−1 − 2I)H −1 . Since λ0 is not real, 2I − S is non-degenerate. The second equation of (5.44) can be written as Sξ (S − 2I) = (S − 2I)B − B(S − 2I), which implies tr Sξ = 0. Remark 38 Let λ = 1, then Φ(1) satisfies Φ(1)η = AΦ(1),

Φ(1)ξ = BΦ(1).

Comparing with (5.37), we know that g = Φ(1)g0 is a harmonic map where g0 is a constant matrix. This kind of harmonic map has been studied in [8]. Here we use Darboux transformation to get more explicit conclusions.

5.3.3

Single soliton solutions

Take the trivial solution as a seed solution, single soliton and multisoliton solutions can be obtained by Darboux transformations as mentioned above. In this process, only algebraic algorithm is necessary. For simplicity, we only consider the harmonic map R1,1 → SU (2). However, the following discussions are essentially the same for U (N ). An element of SU (2) can be written as ⎛

⎞

γ β ⎝ ⎠, −β¯ γ¯ where β and γ are complex numbers satisfying γ¯ γ + β β¯ = 1. Take A and B to be two non-zero constant elements of su(2) with [A, B] = 0. Clearly they satisfy the equation of harmonic map (5.38) and (5.39). Suppose ⎛ ⎞ ⎛ ⎞ i ip 0 iq 0 ⎠, B = ⎝ ⎠, A=⎝ 0 −ip i 0 −iq where p and q are non-zero real numbers. The solution of dg0 = (Adη + Bdξ)g0 is ⎛ ⎞ g0 = ⎝

e i(pη+qξ)

0

0

e−i(pη+qξ)

⎠,

which is a harmonic map. The corresponding Φ0 (λ) is ⎛

Φ0 (λ) = ⎝

⎞

l(λ)

0

0

l−1 (λ)

⎠,

208


where

iλ qξ . 2λ − 1 g0 , A, B and Φ0 constitute the seed solution of the Darboux transformation. ¯0, Take λ1 = λ0 , λ2 = λ

l(λ) = exp iλpη +

⎛

l(λ0 )

H = (h1 , h2 ) = ⎝ Then

¯0) bl(λ

¯0) al−1 (λ0 ) l−1 (λ

⎞ ⎠.

¯ 0 )l(λ0 ) + al−1 (λ ¯ 0 )l−1 (λ0 ). h∗2 h1 = ¯bl(λ

Since

¯ 0 ) = exp iλ ¯ 0 pη + iλ¯ 0 qξ = l−1 (λ0 ), l(λ ¯ l(λ0 ) =

¯ l−1 (λ

2λ0 −1

0 ),

the equality h∗2 h1 = 0 holds if and only if ¯b = −a. By direct calculations, we have det H = er + |a|2 e−r > 0, S = HΛ−1 H −1 =

where

⎛

1 er + |a|2 e−r

⎜ ⎜ ⎜ ⎝

1 er e−r 1 is + ¯ |a|2 −¯ a ¯e λ0 λ0 λ0 λ0 1 1 er e−r 2 − ¯ ae−is ¯ + |a| λ0 λ0 λ0 λ0

⎞ ⎟ ⎟ ⎟, ⎠

¯0 λ0 λ qξ, − ¯ 2λ0 − 1 2λ0 − 1 λ ¯0 λ 0 ¯ 0 )pη + + ¯ s = (λ0 + λ qξ 2λ0 − 1 2λ0 − 1 are real linear functions of ξ and η. The new fundamental solution and the new harmonic map are ¯ 0 )pη + i r = i(λ0 − λ

Φ1 (λ) = (I − λS)Φ(λ), λ0 g1 = Φ1 (1) λ0 − 1

λ0 is used to keep g1 ∈ respectively. The above right-multiplier λ0 − 1

SU (2).

209


Figure 5.1.

Remark 39 If we use r − ln |a| and s + i ln(a/|a|) instead of r and s, then a in S can be changed to 1. g1 can be written as ⎛

g1 = ⎝

with

1−

1 λ0

γ1

β1

−β¯1 γ¯1

er + 1 −

1 ¯0 λ

e−r

⎞ ⎠

λ0 , r −r e +e λ0 − 1 1 1 1 i(s−pη−qξ) λ0 β1 = − −¯ e . r −r λ0 λ0 e + e λ0 − 1 γ1 =

e i(pη+qξ)

Write β1 = ρ1 e iθ1 ,

γ1 = σ1 e iτ1 ,

where ρ1 > 0, σ1 > 0, θ1 , τ1 is real, then

¯ 0 1 λ0 − λ ρ1 = sech r, 2 λ0 (λ0 − 1) σ1 = (1 − ρ21 )1/2 . The figures for ρ1 and σ1 with respect to r are shown in Figure 5.1. Both ρ1 and σ1 are of the shape of solitons. As is known, SU (2) is a three dimensional manifold, which is actually the three dimensional sphere S 3 = {(x1 , x2 , x3 , x4 ) | x21 + x22 + x23 + x24 = 1} ⎛

⎞

γ β ⎠ is given by in R4 . The relation between (x1 , x2 , x3 , x4 ) and ⎝ ¯ −β γ¯ γ = x1 + ix2 ,

β = x3 + ix4 .

210


Unless r depends on x or t only, the image of the map tends to two circles x3 = x4 = 0 and x21 + x22 = 1 as x → ±∞ for fixed t, or as t → ±∞ for fixed x. Hence these two circles are limiting circles. Remark 40 For SU (2), the invariant metric is the standard metric of the sphere S 3 . Therefore, a harmonic map to SU (2) is just a harmonic map to the sphere S 3 .

5.3.4

Multi-soliton solutions

Multi-soliton solutions can be obtained by composition of successive Darboux transformations. Let λ0 , λ1 , · · ·, λk be k + 1 non-real complex numbers. Suppose that they are distinct and satisfy (i) |2λl − 1| (l = 0, 1, · · · , k) are distinct, λ ¯l λ l ¯ l )p ± (ii) (λl − λ − ¯ q = 0. 2λl − 1 2λl − 1 Here p and q are the constants appeared in the seed solution ⎛

g0 = ⎝ Let

⎞

e i(pη+qξ)

0

0

e−i(pη+qξ)

⎠.

¯l λ λl − ¯ qξ, 2λl − 1 2λl − 1 λ ¯l λ l ¯ l )pη + + ¯ qξ, sl = (λl + λ 2λl − 1 2λl − 1 (l = 0, 1, 2, · · · , k). ¯ l )pη + i rl = i(λl − λ

They are real linear functions of ξ and η. It can be verified that rl /rj ’s (l = j) are not constants if and only if (i) holds. On the other hand, (ii) means that rl really depends on x and t. Here r0 and s0 are just r and s in the construction of the single soliton solutions. Since we have g0 and Φ0 , we can define Sj −1 ) · · · (I − λS S0 )Φ0 (λ), Φj (λ) = (I − λS λ λ0 j−1 gj = (I − Sj −1 ) · · · (I − S0 )g0 · · · λj−1 − 1 λ0 − 1 (j = 1, 2, · · · , l), recursively. Let S0 be the matrix S in the construction of the single soliton solutions, S1 , S2 , · · · be defined by (5.47) for Φ = Φ1 , Φ2 , · · ·.

211


According to the expression of S0 , ⎛

⎞

lim S0 = ⎝

r0 →±∞

where

⎧ ⎪ ⎨ 1/λ0 µ0 = ⎪ ⎩ ¯

1/λ0

µ0

0

0

µ ¯0

⎠

as r0 → +∞, as r0 → −∞.

Lemma 5.8 When rl is bounded and t → +∞, ⎛

lim Sl −→ ⎝

⎞

µl

0

⎠

t→±∞

0 µ ¯l ¯ l ; l = 0, 1, 2, · · · , k − 1). (µl = 1/λl or 1/λ Proof. We use induction to prove it. Suppose it holds for l = 0, 1, 2, · · · , i − 1. Now we consider the case l = i. Since ri is finite, r0 , · · ·, ri−1 → ±∞ when t → +∞, hence ⎛

Φi ∼ ⎝

⎞

A(λ)l(λ)

0

0

−1 (λ) A(λ)l

⎠,

where ∼ represents asymptotic value, A(λ) =

i−1

(1 − λµh ),

A(λ) =

h=0

i−1

(1 − λ¯ µk ).

h=0

Since λ), ¯ A(λ) = A(

we have

⎛

Hi ∼ ⎝

A(λi )l(λi )

¯ i )l(λ ¯i) bi A(λ

i )l−1 (λi ) ai A(λ

λ ¯ i )l−1 (λ ¯i) A(

where bi = −¯ ai .

⎞ ⎠

212


As before, ai and bi can be put into ri and si so that ai = 1, bi = −1. By direct calculation, −1 ∼ Si = Hi Λ−1 i Hi

⎛ ⎜ ⎜ ⎝

·⎜

|A(λi

)|2 eri

1 ¯ i )|2 e−ri + |A(λ

¯ i )|2 |A(λi )|2 ri |A(λ −ri e + ¯i e λi λ 1 1 ¯ −isi − ¯ A(λ i )A(λi )e λi λi

When t → +∞ and ri → ±∞, ⎛

Si ∼ ⎝

1

1 ¯ i )e isi − ¯ A(λi )A(λ λi λi ¯ i )|2 |A(λi )|2 ri |A(λ e + e−ri ¯i λi λ

gk ∼ (I − Sk−1 ) ⎝ ⎛

·⎝

e i(pη+qξ)

⎟ ⎟ ⎟. ⎠

⎞

µi

0

0

µ ¯i

⎠,

µi =

1 1 or ¯ . λi λi

The lemma is proved. When rk−1 is finite and t → +∞, r0 , · · · , rk−2 → ±∞, hence ⎛

⎞

⎞

1 − µk−2 1−µ ¯k−2 ⎞

⎛

⎠···⎝

1 − µ0 1−µ ¯0

⎞ ⎠

λ0 λk−1 ⎠ λ − 1 · · · λ . 0 k−1 − 1 e−i(pη+qξ)

According to the asymptotic expression of Sk−1 , we know that the asymptotic behavior of gk as rk−1 bounded and t → +∞ is the same as a single soliton solution. If one of r0 , · · · , rk−2 does not tend to ±∞ (say, ri → ±∞ with certain i < k − 1) as t → +∞, then the other rj → ±∞. According to the theorem of permutability of Darboux transformation, we can change the Darboux matrix with λi to be the last one. The factors should have some change, but the result is still gk . The asymptotic behavior of gk as t → +∞ along the direction ri → ±∞ is still a single soliton. If t → −∞, there are similar conclusions, but µi should be changed to its complex conjugate. This means that the soliton has “phase shift”. Therefore, we have the following theorem.

Theorem 5.9 A harmonic map from R1,1 to U (N ) (or SU (N )) derived from the trivial solution by k Darboux transformations has the following property. When t → ±∞, the asymptotic solution split up into k single solitons. When t → +∞ and t → −∞, these k single solitons are arranged in opposite order and have their own phase shifts.


213

Thus, the behavior of the multi-soliton solutions of a harmonic map from R1,1 to U (N ) (or SU (N )) is quite similar to that of the multisolitons of KdV equations etc., i.e., all of them have the property of elastic scattering.

5.4 Harmonic maps from R2 to U (N ) 5.4.1 Harmonic maps from R2 to U (N ) and their Darboux transformations Let (x, y) be the orthonormal coordinates of R2 , then the energy of φ(x, y) from R2 (or a region Ω in R2 ) to U (N ) is S[φ] = −

tr(φx φ−1 φx φ−1 + φy φ−1 φy φ−1 )dx dy.

(5.54)

φ is called a harmonic map from R2 to U (N ) if the Euler equation for S[φ] holds. We suppose Ω to be simply connected. Denote ζ = x + iy, ζ¯ = x − iy (5.55) and

1 ∂ ∂ ∂ = + i , ∂y ∂ ζ¯ 2 ∂x

∂ 1 ∂ ∂ = −i . ∂ζ 2 ∂x ∂y

(5.56)

B = φζ φ−1 .

(5.57)

Let A = φζ¯φ−1 ,

Similar to Section 5.3, φ is a harmonic map if and only if A and B satisfy Aζ − Bζ¯ + [A, B] = 0,

(5.58)

Aζ + Bζ¯ = 0.

(5.59)

Since φ ∈ U (N ), i.e., φ∗ = φ−1 , A∗ = φ∗−1 φ∗ζ = φ(φ−1 )ζ = −φφ−1 φζ φ−1 = −B. Hence the constraint on A and B is A∗ = −B.

(5.60)

Conversely, if A and B satisfy the constraint (5.60) and the equations (5.58), (5.59), then φ can be solved from φζ¯ = Aφ,

φζ = Bφ.

(5.61)

214


Moreover, φ satisfies (φ∗ φ)ζ

= φ∗ζ φ + φ∗ φζ = (φζ¯)∗ φ + φ∗ φζ = (Aφ)∗ φ + φ∗ Bφ = φ∗ (A∗ + B)φ = 0.

Similarly,

(φ∗ φ)ζ¯ = 0.

(5.62)

(5.63)

If φ∗ φ = I holds at one point, it holds identically. Hence, (5.58), (5.59) and (5.60) can be regarded as the equations of the harmonic map from R2 to U (N ). These equations uniquely determine the harmonic map up to a right-multiplied constant matrix in U (N ). Now consider the Lax pair Φζ¯ = λAΦ,

Φζ =

λ BΦ 2λ − 1

(5.64)

(λ = 1/2) whose integrability conditions are (5.58) and (5.59). Suppose the initial condition Φ|ζ=0 ∈ U (N ), then g = Φ(1)

(5.65)

is a harmonic map. g is unique up to a right-multiplied constant matrix in U (N ). Here Φ(λ) is called an extended solution of the harmonic map ¯= λ and Φ(λ) ∈ U (N ). g [102]. Notice that if |2λ − 1| = 1, then λ 2λ − 1 The harmonic maps from R2 to U (N ) have been discussed in [102]. Here we mainly consider their properties related to the Darboux transformation. Remark 41 The symbols used here differ from those in [102]. The main differences are: (i) The A and B are equivalent to 2Aζ¯ and 2Aζ in [102]. (ii) In the definition of A, B and the Lax pair (5.64), the order of multiplication of matrices is different from those used in [102]. Therefore, left multiplications in [102] are changed to right multiplications in many places. 1 1 λ were written as (1−λ) and (1− (iii) The parameters λ and 2λ − 1 2 2 λ−1 ) in [102]. In our notation we introduce µ = 1−2λ which corresponds to the λ in [102]. As in Section 5.3, we can construct the Darboux matrix D = I − λS and the Darboux transformation (A, B, Φ) −→ (A1 , B1 , Φ1 )

(5.66)

215


where Φ1 = (I − λS)Φ,

(5.67)

A1 = A − Sζ¯,

(5.68)

B1 = B + Sζ .

In order to make (A1 , B1 , Φ1 ) satisfy the Lax pair Φ1ζ¯ = λA1 Φ1 ,

Φ1ζ =

λ B1 Φ1 , 2λ − 1

(5.69)

S should satisfy Sζ¯S = AS − SA,

Sζ (S − 2I) = SB − BS.

(5.70)

Moreover, A∗1 = −B1 must hold. Take N complex numbers λ1 , λ2 , · · ·, λN (= 0, 1/2). Let hρ = Φ(λρ )lρ

(ρ = 1, 2, · · · , N )

(5.71)

where lρ ’s are constant column vectors so that H = (h1 , h2 , · · · , hN ) is non-degenerate, then (5.72) S = HΛ−1 H −1 with Λ = diag(λ1 , λ2 , · · · , λN ) satisfies (5.70). The proof for this is the same as in R1,1 case. It is necessary to choose λρ and lρ so that A∗1 = −B1 holds. This can be done as follows. Take ω1 to be a non-real complex number and ω2 =

ω ¯1 . 2¯ ω1 − 1

(5.73)

We choose

⎧ ⎨ ω1 λρ = ⎩

ω2

(ρ = 1, · · · , k) (ρ = k + 1, · · · , N )

(0 < k < N ).

(5.74)

Moreover, suppose |2ω1 − 1| = 1 so that ω1 = ω2 . Let h1 , · · · , hk be the column solutions of the Lax pair for λ = ω1 , hk+1 , · · · , hN be the column solutions of the Lax pair for λ = ω2 , i.e., ha = Φ(ω1 )la ,

hp = Φ(ω2 )lp

(a = 1, 2, · · · , k; p = k + 1, · · · , N )

(5.75)

where l1 , · · · , lN are linearly independent constant column vectors. If h1 , · · · , hN are chosen to be orthogonal with each other at one point (say (0, 0)), then h∗p ha = 0 (a = 1, 2, · · · , k; p = k + 1, · · · , N )

(5.76)

216


hold everywhere. This follows from (h∗p )ζ¯ = (hpζ )∗ =

∗ ω2 Bhp = −ω1 h∗p A, 2ω2 − 1

(h∗p ha )ζ¯ = −ω1 h∗1 Aha + h∗1 ω1 Aha = 0, and similarly,

(h∗p ha )ζ = 0.

(5.77)

(5.78)

From the definition of S, we have

Hence

Sha =

1 ha , ω1

Shp =

h∗a S ∗ =

1 ∗ h , ω ¯1 a

h∗p S ∗ =

1 ∗ h . ω ¯2 p

1 ∗ h hb , ω ¯ 1 ω1 a 1 1 ∗ h∗p (S ∗ − S)hq = − h hq , ω ¯ 2 ω2 p h∗a (S ∗ − S)hb =

1

1 hp , ω2

(5.79) (5.80)

−

h∗p (S ∗ − S)ha = 0,

(5.81)

h∗a (S ∗ − S)hp = 0 (a, b = 1, · · · , k; p, q = k + 1, · · · , N ). According to (5.73), 1 1 1 1 − = − . ω1 ω ¯1 ω2 ω ¯2 So h∗β (S ∗ − S)hα = h∗β

1

ω ¯1

−

(5.82)

1 Ihα ω1

(α, β = 1, 2, · · · , N ) 1 1 − is real. Since {hα } consists of N linearly independent ω1 ω2 vectors, 1 1 S∗ − S = − I. (5.83) ω ¯ 1 ω1 Then and

A∗1 + B1 = (A1 − Sζ¯)∗ + B + Sζ = A∗ + B − Sζ∗ + Sζ = 0, i.e., A1 and B1 satisfy the constraint (5.60) for U (N ). Therefore, we have the following theorem.

217


Theorem 5.10 Suppose that λρ ’s and hρ ’s satisfy (5.74) and (5.76), hρ ’s are linearly independent. Then the Darboux transformation (5.66) gives a new harmonic map from R2 to U (N ), which is expressed by g1 = Φ1 (1)(Φ1 |0 )−1 where (Φ1 |0 ) is the value of Φ1 at a fixed point ζ = 0. This topic has also been discussed by several authors such as [112] and [54]. Now we use the Darboux transformation to derive the projective operators π and π ⊥ in CN given by [102]. According to (5.83), S∗ − so S −

1 1 I = S − I, ω ¯1 ω1

1 I is Hermitian. On the other hand, ω1 S−

1 1 I = H Λ−1 − I H −1 , ω1 ω1

1 1 1 I has k eigenvalues 0 and N − k real eigenvalues − . ω1 ω2 ω1 Moreover, so S −

π

=

1

ω2

−

1 −1 1 1 1 −1 S− I = − · ω1 ω1 ω2 ω1

⎡⎛

⎞

1 ⎢⎜ ω Ik ⎜ 1 ·H ⎢ ⎣⎝ 0 ⎛

=H⎝

0 1 IN −k ω2

⎛

1 ⎟ ⎜ ω Ik ⎟−⎜ 1 ⎠ ⎝ 0

⎞⎤

0

⎟⎥ ⎟⎥

⎠⎦ H 1 IN −k ω1

−1

⎞

0

0

⎠ H −1

0 IN −k (5.84) is an Hermitian matrix with k eigenvalues 0 and N − k eigenvalues 1. π is a projective matrix, and ⎛

π⊥ = I − π = H ⎝

⎞

Ik 0 0

0

⎠ H −1

(5.85)

218


is the complement of π. The invariant subspace of π ⊥ is orthogonal to that of π. From (5.84) and (5.85), S=

1 1 π + π⊥ , ω2 ω1

(5.86)

and the Darboux matrix

D = I − λS = 1 −

λ λ ⊥ λ π+ 1− π = (π + γπ ⊥ ) 1 − (5.87) ω2 ω1 ω2

where γ=

ω2 (ω1 − λ) . ω1 (ω2 − λ)

(5.88)

Remark 42 S can also be written as S=

1 1 1 I− − π. ω1 ω1 ω2

(5.89)

Substituting it into (5.70), we get πζ¯ = −ω1 πA + ω2 Aπ + (ω1 − ω2 )πAπ, πζ = ω ¯ 1 Bπ − ω ¯ 2 πB + (¯ ω2 − ω ¯ 1 )πBπ,

(5.90)

These equations are first derived in [102] by the decomposition of loop group. Here we use Darboux transformation to give the explicit expressions of the solutions of these equations. Moreover, the new harmonic map is given by g1 = Φ1 (1)

1−

1 = (π + γ1 π ⊥ )Φ(1) ω2

(5.91)

1 − 1/ω1 1 − 1/ω2

(5.92)

where γ1 =

satisfies γ¯1 γ1 = 1. If we are considering the harmonic map to SU (N ), then the new harmonic map is g1 = Φ1 (1)(I − Λ−1 )−1 .

(5.93)

219


5.4.2

Soliton solutions

Similar to Section 5.3, the soliton solutions of the harmonic map from R → U (N ) can be constructed explicitly. For simplicity, we choose N = 2, so the solution is a special harmonic map from R2 → SU (2). Take the seed solution 2

⎛

⎞

¯

g0 = ⎝

eτ ζ−τ¯ζ

0

0

e−(τ ζ−τ¯ζ )

¯

⎠

where τ is a non-zero complex number, then ⎛

A = g0ζ¯g0−1 = ⎝

⎞

τ

0

⎛

B = g0ζ g0−1 = ⎝

⎠,

0 −τ

−¯ τ 0 0

⎞ ⎠.

τ¯

The solution of the Lax pair (5.64) is ⎛

Φ0 = ⎝ where

⎞

l(λ)

0

0

l−1 (λ)

⎠

(5.94)

λ τ¯ζ . 2λ − 1 Take a complex number ω1 satisfying |2ω1 − 1| = 1. Let

l(λ) = exp λτ ζ¯ −

ω2 = then

ω ¯1 , 2¯ ω1 − 1

l−1 (ω1 ) = l(ω2 ),

Let

⎛

H=⎝ then

(5.95)

l−1 (ω2 ) = l(ω1 ).

l(ω1 )

−al ¯ (ω2 )

al−1 (ω1 )

l−1 (ω2 )

(5.96)

⎞ ⎠,

(5.97)

det H = |l(ω1 )|2 + |a|2 |l(ω2 )|−2 .

(5.98)

By (5.72), ⎛

S=

1 r e + |a|2 e−r

⎜ ⎜ ⎜ ⎝

1 er e−r 1 is + |a|2 − a ¯e ω1 ω2 ω1 ω2 1 1 −is er e−r − ae + |a|2 ω1 ω2 ω2 ω1

⎞ ⎟ ⎟ ⎟, ⎠

(5.99)

220


where r = (ω1 − ω2 )τ ζ¯ + (¯ ω1 − ω ¯ 2 )¯ τ¯ζ, s = −i((ω1 + ω2 )τ ζ¯ − (¯ ω1 + ω ¯ 2 )¯ τ¯ζ )

(5.100)

¯ can all be eliminated by adding are real valued functions. |a|2 , a and a some constants to r and s. Therefore, we can suppose a = 1. The harmonic map is ⎛

g1

⎛

=⎝ where

⎞

ω1 ⎜ = Φ1 (1) = (I − S) ⎝ 1 − ω1 0

0 ⎟ ¯ ω2 ⎠ g0 (ζ, ζ) 1 − ω2

⎞

(5.101)

α β ⎠, −β¯ α ¯

1 ¯ α = e + e−r eτ ζ−τ¯ζ , γ1 r

β=

ω2 − ω1 is−(τ ζ−τ ¯ ¯ζ ) e , ω1 (1 − ω2 )

(5.102)

γ1 is defined by (5.92). The asymptotic behavior of g1 is similar to that in R1,1 case. Moreover, we can also construct multi-soliton solutions explicitly, and can prove that a k-soliton solution is asymptotic to k single solitons as y → ±∞. The proof is similar to the R1,1 case and is omitted here.

5.4.3

Uniton

Uniton is a special harmonic map from R2 (or a region Ω ⊂ R2 ) to U (N ) and was introduced by K. Uhlenbeck in [102]. It is an important notion because each harmonic map from S 2 to U (N ) is a uniton. In order to be compatible with [102], we first change some parameters which are used before. For Lax pair (5.64), let µ = 1 − 2λ, then λ= Let

λ 1 − µ−1 = . 2λ − 1 2

1−µ , 2

1−µ Φ(λ) = Φ 2

(5.103)

= Ψ(µ),

(5.104)

1 − µ−1 BΨ. 2

(5.105)

(5.64) becomes Ψζ¯ =

1−µ AΨ, 2

Ψζ =

221


Definition 5.11 Suppose g is a harmonic map from R2 (or Ω) to U (N ). If its extended solution Ψ(µ) satisfies the following four conditions: (a) Ψ(µ) =

n

Ta µa

(a polynomial of µ),

(5.106)

a=0

(b) Ψ(1) = I, (c) Ψ(−1) = g, µ−1 )−1 (d) Ψ(µ)∗ = Ψ(¯

(µ = 0),

(5.107) (5.108) (5.109)

then g is called a uniton, and Ψ(µ) is called an extended solution of uniton. The conditions (b), (c) and (d) above are constraints to the initial values. (b) means that Φ(µ)|0 = I for µ = 1. (c) holds if Ψ(−1)|0 ∈ U (N ). (d) holds everywhere if it holds at z = 0. The last statement µ−1 ) and using the Lax pair can be proved by differentiating Ψ(µ)∗ Ψ(¯ ∗ together with the relation A = −B. (a) is an essential condition, based on the conditions (b), (c) and (d). For a uniton, the extended solution Ψ(µ) and its degree are not unique. The minimum degree of the extended solutions is called the degree of the uniton, or uniton number [102].

Theorem 5.12 Suppose an N × N matrix valued polynomial Ψ(µ) =

n

Ta µa

a=0

satisfies the conditions (b) and (d), then Ψ(µ) can be written as Ψ(µ) = (π1 + µπ1⊥ ) · · · (πn + µπn⊥ ),

(5.110)

where πa (a = 1, · · · , n) are Hermitian projections, and πa⊥ ’s are their orthogonal complements. Proof. First, we consider the case n = 1, i.e., Ψ(µ) = T0 + T1 µ. By Ψ(1) = I, T0 + T1 = I. The condition (d) is T0 + (I − T0 )¯ µ−1 ) = I, (T T0 + (I − T0 )µ)∗ (T

222


which leads to T0∗ (I − T0 ) = (I − T0 )∗ T0 = 0, T0∗ T0 + (I − T0∗ )(I − T0 ) = I. Hence

T0∗ T0 = T0∗ = T0 .

This means that T0 is the Hermitian projective operator π0 and T1 = I − T0 = π ⊥ . Now we use mathematical induction to prove the general result. Assume (5.110) is true for n, and Ψ(µ) =

n+1

Ta µa

a=0

satisfies conditions (b) and (d). Without loss of generality, suppose T0 = 0, Tn+1 = 0. (Otherwise, the degree of Ψ(µ) is less than n + 1 or Ψ(µ) can be written as µΨ1 (µ) with deg(Ψ1 (µ)) ≤ n. The assumption of induction implies that the theorem is true.) det Ψ(µ) is a polynomial of µ. If it has a non-zero root µ = µ0 = 0, then Ψ(µ0 ) is degenerate, which is contradict to condition (d). Hence det Ψ(µ) = 0 only when µ = 0. It follows that det T0 = 0 and T0 l = 0 has non-zero solutions for l. All ⊥ be the the solutions l of T0 l = 0 form a proper subspace P . Let πn+1 Hermitian projective operator to P . It is non-trivial and its orthogonal ⊥ = 0). complement is denoted by πn+1 (πn+1 = 0, πn+1 Let −1 ⊥ Ψ(µ) = Ψ(µ)(π n+1 + µ πn+1 ). ⊥ Since T0 πn+1 = 0, Ψ(µ) is a polynomial of µ of degree n and condition (b) holds. The condition (d) holds for Ψ(µ) too, since ⊥ )(π ⊥ )Ψ(µ) ∗ µ−1 )(πn+1 + µ ¯πn+1 ¯−1 πn+1 Ψ(¯ µ−1 )Ψ(µ)∗ = Ψ(¯ n+1 + µ ∗ = I. µ−1 )Ψ(µ) = Ψ(¯

By the assumption of induction, Ψ(µ) =

n

(πa + µπa⊥ ), hence

a=1

Ψ(µ) =

n+1

(πa + µπa⊥ ).

a=1

The theorem is proved. This theorem shows that the extended solution of uniton can be expressed as a product (5.110). However, this decomposition is point-wise.


223

We have not proved the smoothness of πa and πa⊥ . This will be obtained later. In [102], the uniton of degree one is constructed as follows. Suppose Ψ(µ) = π + µπ ⊥ is an extended solution of uniton. Let Ψζ¯(µ)Ψ−1 (µ) = (1 − µ)πζ¯(π + µ−1 π ⊥ ) 1−µ = (1 − µ)πζ¯π + πζ¯π ⊥ , µ Ψζ (µ)Ψ−1 (µ) = (1 − µ)πζ (π + µ−1 π ⊥ ) = (1 − µ−1 )(−πζ π ⊥ ) + (1 − µ)πζ π. Comparing with the Lax pair (5.105), we have A = 2πζ¯π,

B = −2πζ π ⊥

(5.111)

πζ π = 0.

(5.112)

and the conditions for π πζ¯π ⊥ = 0,

The last two conditions are equivalent, since ∗ ⊥ ∗ πζ π = (ππζ¯)∗ = (−ππζ⊥ ¯ ) = (πζ¯π ) .

From π ⊥ π = 0, we have 0 = (π ⊥ π)ζ = π ⊥ πζ − πζ π. Hence (5.112) is equivalent to (5.112) π ⊥ πζ = 0. The projective operator π satisfying this condition is constructed as follows. If u1 , · · · , uk are k linearly independent vector functions satisfying

uaζ =

cab ub ,

(5.113)

then for any invertible linear combination va =

qab ub

(det(qab ) = 0),

whose coefficients are arbitrary functions, vaζ ’s are also linear combinations of v1 , · · · , vk . Now suppose v1 , · · · , vk are Schmidt orthogonalization of u1 , · · · , uk , i.e., they are linear combinations of u1 , · · ·, uk satisfying (5.114) vb∗ va = δba , then π=

va va∗

(5.115)

224


is an Hermitian projective operator of rank k: π ∗ = π,

ππ = π,

(5.116)

and v1 , · · ·, vk (also u1 , · · ·, uk ) are its invariant vectors. Moreover, from (5.113), vaζ = cab vb where cab ’s are suitable functions, and πζ =

cab vb va∗ +

∗ va vaζ .

From π ⊥ va = (I − π)va = va − va = 0 we know that (5.112) holds. Conversely, suppose π is an Hermitian projective operator of rank k satisfying (5.112) , then for any k linearly independent invariant vectors ua (k = 1, · · · , k), πua = ua , π ⊥ ua = 0. By (5.112) , π ⊥ uaζ = π ⊥ (πua )ζ = π ⊥ πζ ua + π ⊥ πuaζ = 0. Hence uaζ is a linear combination of u1 , · · ·, uk . Moreover, if ua ’s satisfy (5.113), then there exist linearly independent vectors w1 , · · ·, wk which are linear combinations of ua ’s and satisfy waζ = 0. In fact, this can be done as follows. Let U = (u1 , · · · , uk ), C = (cab ), then U satisfies Uζ = U C. Let W = (w1 , · · · , wk ) = U B where B is a k × k matrix, then Wζ = Uζ B + U Bζ = U (CB + Bζ ). B can be determined by solving the linear equation Bζ + CB = 0. Therefore, to construct π, we only need a set of linearly independent vectors wα whose entries are holomorphic functions of ζ¯ (i.e. antiholomorphic functions of ζ). This gives the construction of unitons of degree one. Remark 43 Let the set of all k dimensional subspaces of CN be attached at each point of R2 (or Ω). Then we get a k dimensional plane bundle on R2 . A section of this plane bundle is to assign a k dimensional subspace at each point of R2 (or Ω). It can also be obtained by assigning ¯ at each point. k dimensional Hermitian projective operator π = π(ζ, ζ)

225


In [102], the section was called holomorphic if π satisfies the condition π ⊥ πζ¯ = 0. Because of the difference of the symbols (See Remark 41), the condition here should be πζ¯π ⊥ = 0, or equivalently, π ⊥ πζ = 0, i.e., the section is anti-holomorphic. As mentioned above, this kind of section can be constructed by k linearly independent anti-holomorphic vectors.

Example 5.13 Take N = 2. Let f , g be two holomorphic functions of ζ¯ without common zero. Let ⎛

u=

⎞

f

1 ⎝ ⎠, (|f |2 + |g|2 )1/2 g

then ⎛

⎞

⎛

⎞

f |f |2 f g¯ 1 1 ⎝ ⎠ (f¯, g¯) = ⎝ ⎠, π= |f |2 + |g|2 |f |2 + |g|2 g g f¯ |g|2 ⎛

⎞

|g|2 −f g¯ 1 ⎝ ⎠. π⊥ = I − π = |f |2 + |g|2 −g f¯ |f |2 The expression of the uniton is ⎛

⎞

|f |2 − |g|2 2f g¯ 1 ⎝ ⎠, g = π − π⊥ = 2 2 |f | + |g| 2g f¯ |g|2 − |f |2 and the extended solution is ⎛

⎞

|f |2 + µ|g|2 (1 − µ)f g¯ 1 ⎝ ⎠. Ψ(µ) = |f |2 + |g|2 (1 − µ)g f¯ |g|2 + µ|f |2

5.4.4

Darboux transformation and singular Darboux transformation for unitons

We construct the Darboux matrix by using the algorithm in Section 5.1. Suppose g is a uniton and Ψ(µ) is its extended solution satisfying the µ−1 ) = Ψ(µ)−1 . Let normalizing condition: Ψ∗ (¯ ω1 =

1− , 2

ω2 =

1 − ¯−1 . 2

(5.117)

226


This is equivalent to putting µ to be . Choose the basis of Cn such that L1 = L⊥ 1

l1 · · · l k

= L2 =

⎛

=⎝

⎞

Ik

⎠,

0

lk+1 · · · lN

⎛

⎞

0

=⎝

⎠.

IN −k

Then H =

Ψ()L1 Ψ(¯−1 )L2 ⎛

π = H ⎝

⎞

0

0

⎠ H −1 ,

0 IN −k

,

⎛

π⊥ = H ⎝

⎞

Ik 0 0

⎠ H −1 .

0

(5.118)

The Darboux transformation of the extended uniton Ψ(µ) is Ψ1 (µ) = (π + γπ⊥ )Ψ(µ)(σ + γ −1 σ ⊥ ),

(5.119)

where (π + γπ⊥ ) is the Darboux matrix, (σ + γ −1 σ ⊥ ) is an additional factor in which σ and σ ⊥ are the constant Hermitian projective operators to L⊥ 1 and L1 respectively. Moreover,

γ = γ (µ) =

ω2 ω1 −

1−µ 2

ω1 ω2 −

1−µ 2

=

(µ − )(¯ − 1) . (¯µ − 1)(1 − )

(5.120)

Clearly, µ−1 ) = γ (¯

(¯ µ−1 − )(¯ − 1) (1 − ¯ µ)(¯ − 1) = = (γ (µ))−1 . −1 (¯¯ µ − 1)(1 − ) (¯ − µ ¯)(1 − )

(5.121)

Hence µ−1 ) = (π + γ¯ −1 π⊥ )Ψ(¯ µ−1 )(σ + γσ ⊥ ), Ψ1 (¯ µ−1 )−1 = (π + γπ⊥ )Ψ(µ)(σ + γ −1 σ ⊥ ) = Ψ1 (µ). Ψ∗1 (¯

(5.122)

This means that Ψ1 also satisfies the normalizing condition. Now consider Ψ1 (µ) = π Ψ(µ)σ + π Ψ(µ)γ −1 (µ)σ ⊥ + π⊥ γ(µ)Ψ(µ)σ + π⊥ Ψ(µ)σ ⊥ . (5.123) On the right hand side, the first and fourth terms are polynomials of µ, the second and third terms are rational functions of µ. µ = might be a pole of the second term and µ = ¯−1 might be a pole of the third term.

227


However, we can prove that the denominators µ − and ¯µ − 1 in these two terms can be cancelled with the corresponding enumerators. Hence all the terms are polynomials of µ. This fact is proved as follows. Notice that ⎛ ⎞ ∗ Ψ()∗ C ()L 1 1 ⎠, (5.124) H−1 = ⎝ ∗ C2 ()L2 Ψ(¯−1 )∗ where C1 () is a k × k matrix and C2 () is an (N − k) × (N − k) matrix, which are determined by C1 ()L∗1 Ψ()∗ Ψ()L1 = Ik , C2 ()L∗2 Ψ(¯−1 )∗ Ψ(¯−1 )L2 = IN −k

(5.125)

respectively. In (5.125), L∗1 Ψ()∗ Ψ()L1 is invertible. In fact, suppose l is a k dimensional vector such that L∗1 Ψ()∗ Ψ()L1 l = 0, then from l∗ L∗1 Ψ()∗ Ψ()L1 l = 0 we obtain Ψ()L1 l = 0. It follows that L1 l = 0 and hence l = 0. Consequently, C1 () is determined by (5.125) uniquely. Similarly, C2 () also exists uniquely. Therefore, ⎛

π = H ⎝ ⎛

π⊥ = H ⎝

⎞

0

0

0 IN −k

⎠ H −1 = Ψ(¯−1 )L2 C2 ()L∗ Ψ(¯−1 )∗ , 2

⎞

Ik 0 0

0

(5.126)

⎠ H −1 = Ψ()L1 C1 ()L∗ Ψ()∗ , 1

and π Ψ(µ)γ −1 (µ)σ ⊥ = Ψ(¯−1 )L2 C2 ()L∗2 Ψ(¯−1 )∗ Ψ(µ)γ −1 σ ⊥ .

(5.127)

Denote the right hand side of the last equation by γ −1 F (µ). When µ = , (5.128) F () = Ψ(¯−1 )L2 C2 ()L∗2 Ψ(¯−1 )∗ Ψ()σ ⊥ . Using the normalizing condition Ψ(¯−1 )∗ Ψ() = I and ⎛

σ⊥ = ⎝

⎞

Ik 0 0

⎠,

0

we get F () = 0. This means that µ = is actually not a pole of the second term of (5.123). Similar conclusion holds for the third term. Thus, we have the following theorem.

Theorem 5.14 A Darboux transformation with right-multiplied normalizing factor (σ + µ−1 σ ⊥ ) transforms an extended solution Ψ(µ) to

228


an extended solution Ψ1 (µ). The degree of Ψ1 (µ) with respect to µ cannot exceed the degree of Ψ(µ) with respect to µ. We have mentioned before that the extended solutions of a uniton are not unique, and their degrees are not definite either. However, the uniton number is defined as the minimum of the degrees of all possible extended solutions. The above theorem implies that a Darboux transformation cannot increase uniton number. In [102], singular B¨ a¨cklund transformation was introduced. For a set of usual Backlund ¨ transformations with parameter , let → 0, then the derived transformation is called a singular B¨ a¨cklund transformation. Since Backlund ¨ transformation can be solved explicitly by Darboux transformation, singular Darboux transformation (limit of Darboux transformations with parameter as → 0) should give singular Backlund ¨ transformation explicitly. Clearly, when → 0, γ → µ. Hence if π → π, π⊥ → π ⊥ , the Darboux transformations will converge to a singular Darboux transformation Ψ1 (µ) = (π + µπ ⊥ )Ψ(µ). Here the factor (σ + µ−1 σ ⊥ ) is neglected.

Lemma 5.15 Suppose Ψ(µ) is an extended solution of a uniton, then Ψ1 (µ) = (π + µπ ⊥ )Ψ(µ) is an extended solution of a uniton if and only if (2πζ¯ + πA)π ⊥ = 0,

(5.129)

π ⊥ Aπ = 0.

(5.130)

Proof. By direct calculation, 1−µ (2πζ¯π + πAπ + π ⊥ Aπ ⊥ ) 2 1−µ µ(1 − µ) ⊥ (2πζ¯π ⊥ + πAπ ⊥ ) + π Aπ + 2µ 2 1−µ = A1 . 2 This leads to (5.129) and (5.130). Alternatively, = Ψ1ζ¯Ψ−1 1

1 − µ−1 (−2πζ π ⊥ + πBπ + π ⊥ Bπ ⊥ ) 2 1 − µ−1 1−µ (2πζ π − π ⊥ Bπ) + πBπ ⊥ + 2 2µ 1 − µ−1 B1 . = 2

= Ψ1ζ Ψ−1 1


Hence we have

229

2πζ π − π ⊥ Bπ = 0,

(5.129)

πBπ ⊥ = 0.

(5.130)

However, (5.129) and (5.130) are equivalent to (5.129) and (5.130) respectively. In fact, complex conjugate of (5.130) is (5.130) . Taking the complex conjugate of (5.129) , we have ⊥ ⊥ ⊥ 0 = 2ππζ¯ + πAπ ⊥ = −2ππζ⊥ ¯ + πAπ = 2πζ¯π + πAπ .

This is just (5.129). The lemma is proved. Remark 44 Let → 0 in the Backlund ¨ transformation (5.90), then we obtain (5.129) and (5.130). Hence (5.129) and (5.130) are called the equations of singular B¨ acklund transformation in [102]. Notice that when → 0, C1 () and C2 () may not converge, and L∗1 Ψ()∗ Ψ()L1 and L∗2 Ψ(¯−1 )∗ Ψ(¯−1 )L2 may not converge to invertible matrices. However, by the following method we can eliminate the term which tends to infinity as → 0 so that π tends to π. This method can be regarded as a method of renormalization [49]. The rank of Ψ()L1 is k. From the second equation of (5.126) and the first equation of (5.125), we have π⊥ Ψ()L1 = Ψ()L1 . Hence each column of Ψ()L1 is an invariant vector of π⊥ and its entries are polynomials of whose degrees do not exceed n. Consider the terms of the zeroth order of in the columns of Ψ()L1 and choose a maximal linearly independent set in it. The corresponding column vectors of Ψ()L1 constitute an -dependent matrix χ0 = χ00 + χ01 + · · · + χ0n n , where χ00 is of full rank. The coefficients of 0 in other columns of Ψ()L1 are linear combinations of the columns in χ00 . Subtract these other columns by suitable linear combination of the columns in χ0 so that all the coefficients of 0 are eliminated. After divided by , we obtain a set of columns in the form b + b1 + · · · + bn−1 n−1 . Combining these columns to the matrix χ0 , we obtain a matrix M1 (). Evidently, the columns of M1 span the same bundle of subspaces P over

230


R2 (or Ω) as the columns of Ψ()L1 . We can transform the matrix M1 () to M2 () in the same way and so on. Finally we obtain a matrix M () = τ0 + τ1 + · · · + τn n

(5.131)

such that τ0 is of rank k. In fact, if rankττ0 < k, we can make the above transformation further, since the columns of M span the bundle P . Let (5.132) W = M ()(M ∗ ()M ())−1 M ∗ (). It is easily seen that W is Hermitian and W2 = W . Moreover, W M () = M (), i.e., W () is the Hermitian projective operator to P . Hence W = π⊥ .

(5.133)

Let → 0, we have π⊥ → π ⊥ with π ⊥ = τ0 (ττ0∗ τ0 )−1 τ0∗

(5.134)

and π → π = I − π ⊥ . Thus we have proved the convergence of π (and τ⊥ ) as → 0 and obtained the explicit formulae for π ⊥ and π. The convergence of π,ζ and π,ζ¯ as → 0 can be obtained from some properties of harmonic maps (see [50]). We have the following theorem.

Theorem 5.16 Let π ⊥ be defined by (5.134), then Ψ1 = (π + µπ ⊥ )Ψ

(5.135)

is a new extended solution of uniton. Moreover, π and π ⊥ satisfy (5.129) and (5.130). Thus we have established the singular Darboux transformation which realizes the singular B¨äcklund transformation. According to the Theorem 5.14, the extended solution of uniton (π + µπ ⊥ )Ψ(σ + µ−1 σ ⊥ )

(5.136)

is of degree ≤ n. Hence the uniton number cannot be increased by singular B¨ a¨cklund transformation. We shall take n = 1 as the example to see the limiting process of π and π⊥ as → 0. When n = 1, Ψ(µ) = π1 + µπ1⊥ ,

231


π = (π1 + ¯−1 π1⊥ )L2 C2 ()L∗2 (π1 + −1 π1⊥ ), π⊥ = (π1 + π1⊥ )L1 C1 ()L∗1 (π1 + ¯π1⊥ ),

(5.137) (5.138)

C1 ()L∗1 (π1 + ||2 π1⊥ )L1 = Ik , C2 ()L∗2 (π1 + ||−2 π1⊥ )L2 = IN −k .

(5.139) (5.140)

The existence of the limits of π and π⊥ as → 0 is discussed as follows. By using a transformation of the basis, we have ⎛

L1 = ⎝

⎞

Ik

⎛

⎠,

0

L2 = ⎝

⎛

Let

π1 = ⎝

⎞

0

⎠.

(5.141)

IN −k ⎞

π11 π12

⎠

(5.142)

π21 π22

as a block matrix where π11 , π12 , π21 , π22 are k × k, k × (N − k), (N − k) × k, (N − k) × (N − k) matrices respectively. Since π1∗ = π1 , we have ∗ π11 = π11 ,

∗ π12 = π21 ,

∗ π21 = π12 ,

∗ π22 = π22 .

(5.143)

From π12 = π1 , 2 +π π π11 12 21 = π11 ,

π21 π11 + π22 π21 = π21 ,

π11 π12 + π12 π22 = π12 ,

2 =π . π21 π12 + π22 22

(5.144)

Since π11 is an Hermitian matrix, there exists a k × k unitary matrix β such that π11 = β diag(0, · · · , 0, λr+1 , · · · , λk )β ∗

(λr+1 , · · · , λk = 0).

(5.145)

Hence (5.139) becomes

C1 ()β diag(0, · · · , 0, λr+1 , · · · , λk )

+||2 diag(1, · · · , 1, 1 − λr+1 , · · · , 1 − λk ) β ∗ = Ik , or equivalently,

C1 () = β

diag ||−2 , · · · , ||−2 ,

1 1 , · · · , β∗, λr+1 + ||2 (1 − λr+1 ) λk + ||2 (1 − λk )

(5.146)

232


From (5.126), ⎛

π⊥ = ⎝

π11 + (1 − π11 ) π21 − π21

From (5.145),

⎞ ⎠ C1 ()(π11 + ¯(1 − π11 ), π12 − ¯π12 ). (5.147)

π11 β diag(IIr , 0)β ∗ = 0.

(5.148)

On the other hand, suppose a k dimensional vector l satisfies π11 l = 0, then by (5.144), π12 π21 l = 0. Hence 2 (π21 l)∗ (π21 l) = l∗ π12 π21 l = l∗ (π11 − π11 )l = 0,

which implies π21 l = 0. Therefore, π21 β diag(IIr , 0, · · · , 0) = 0.

(5.149)

Substituting these relations into (5.147) and letting → 0, we get lim π⊥ = π ⊥ ⎛

→0

=⎝

⎞

π11 π

⎠ β diag 0, · · · , 0,

⎛ 21

+⎝

β diag(IIr , 0)β ∗ 0 0

⎞

1 λr+1

,···,

1 ∗ β π11 π12 λk

⎠,

0 (5.150)

and

⎛

lim π = π = ⎝

−π12

⎞

⎠α 1 − π22 1 1 · diag ,···, , 0, · · · , 0 α∗ −π21 1 − π22 λk+1 λk+s ⎛ ⎞ 0 0 ⎠. +⎝ 0 α diag(0, IN −k−s )α∗

→0

(5.151)

Here α is a suitable (N − k) × (N − k) unitary matrix which can make π22 to be diagonal. Finally, we prove the following theorem.

Theorem 5.17 (Theorem of factorization) Any extended solution of uniton Ψ(µ) of degree n can be factorized as ⊥ ) · · · (π1 + µπ1⊥ )C C0 , Ψ(µ) = (πn + µπn⊥ )(πn−1 + µπn−1

(5.152)


233

where πi and πi⊥ = I − πi (i = 1, 2, · · · , n) are Hermitian projective operators, being analytic functions of (x, y), C0 is a constant matrix in U (N ) which may be chosen as I. Before proving this theorem, we first give the following two lemmas.

Lemma 5.18 Suppose Ψ(µ) is an extended solution of uniton of degree n, then there exists an Hermite projective operator π such that Ψ1 (µ) = (π + µπ ⊥ )Ψ(µ) is an extended solution of uniton defined on a dense open subset of Ω and the degree of Ψ1 (µ) is at most n. Proof. Suppose Ψ(µ) = T0 + T1 µ + · · · + Tn µn . Since the equations for harmonic map is elliptic and U (N ) is a real analytic manifold, A and B are both analytic functions of the real variables x and y, so is T0 . Suppose the maximum rank of T0 is k, then the points at which the rank of T0 is k form a dense open subset Ω1 of Ω. Moreover, suppose L1 is a constant N × (k + a) matrix (a ≥ 0) such that the maximum rank of T0 L1 is also k, then the points at with the rank of T0 L1 is k also form a dense open subset Ω2 of Ω. We use L = (L1 , L2 ) to construct the Darboux transformation. From (5.125) and (5.126), π⊥ = Ψ()L1 C1 ()L∗1 Ψ()∗ , C1 ()L∗1 Ψ1 ()∗ Ψ1 ()L1 = Ik . When → 0, C1 (0) = (L∗1 T0∗ T0 L1 )−1 , ∗ π ⊥ = T0 L1 C(0)L−1 1 T0

holds in Ω2 . By Ψ(µ)∗ Ψ(¯ µ−1 ) = I, we have T0∗ Tn = 0. Hence π ⊥ Tn = 0, and the degree of Ψ1 (µ) on Ω2 does not exceed n. The lemma is proved.

Lemma 5.19 Suppose Ψ(µ) is an extended solution of uniton of degree n, then there exists an Hermitian projective operator τ defined on a dense open subset of Ω such that Ψ(µ) = (τ + µτ ⊥ )Ψ−1 (µ),

234


where Ψ−1 (µ) is an extended solution of uniton whose degree does not exceed n − 1. Proof. Take τ as π ⊥ in Lemma 5.18, then τ ⊥ = π. Let

Ψ−1 (µ) = τ ⊥ +

1 τ Ψ(µ), µ

then Ψ1 (µ) = (π + µπ ⊥ )Ψ(µ) = (π + µπ ⊥ )(π ⊥ + µπ)Ψ−1 (µ) = µΨ−1 (µ). Hence the degree of Ψ−1 (µ) does not exceed n − 1. On the other hand, it can be verified directly that Ψ−1 (µ) is an extended solution of uniton if and only if τ ⊥ Aτ − 2τ ⊥ τζ¯ = 0, τ Aτ ⊥ = 0. Since τ = π ⊥ , τ ⊥ = π, these are just the equations (5.129) and (5.130) for π and π ⊥ . The lemma is proved. Proof of Theorem 5.17. Use mathematical induction. Suppose the theorem is true for all extended solutions of uniton of degree n − 1. According to Lemma 5.19, Ψ(µ) = (ττn + µττn⊥ )Ψ−1 (µ) holds in a dense open subset. By the assumption of induction, ⊥ ) · · · (π1 + µπ1⊥ ) Ψ(µ) = (πn + µπn⊥ )(πn−1 + µπn−1

holds in a dense open subset where τn and τn⊥ are denoted by πn and πn⊥ . The left hand side is an analytic function of x and y on Ω, and the right hand side is an analytic function of x and y in a dense open subset of Ω. By analytic continuation, we know that the above equality holds in whole Ω. The theorem is proved. Remark 45 This theorem was obtained in [102]. Here the construction of factorization is more general. Especially, the rank of each πi is not fixed. The only restriction is 1 ≤ rank πi ≤ N − ki where ki is given by Lemma 5.18. Since singular Darboux transformation can be constructed algebraically, the factorization here can also be realized algebraically. Take µ = −1, then Ψ(−1) can be factorized as ⊥ ) · · · (π1 − π1⊥ ). Ψ(−1) = (πn − πn⊥ )(πn−1 − πn−1


235

Moreover, the unitons to the Grassmannian have the similar factorization (see [44]). Remark 46 In the proof of Lemma 5.19, π and π ⊥ (or τ ⊥ and τ ) are constructed from Ψ(µ). But we have not constructed them from Ψ−1 (µ) algebraically. There was some negligence in [49]. It is still not clear whether one can obtain all unitons by purely algebraic algorithm from one unitons. Partial solution has been obtained in [55].

Chapter 6 GENERALIZED SELF-DUAL YANG-MILLS EQUATIONS AND YANG-MILLS-HIGGS EQUATIONS

6.1

Generalized self-dual Yang-Mills flow

Yang-Mills equations are one of the most important equations in theoretical physics to describe the fundamental interactions in nature. Selfdual Yang-Mills equations are special case of the Yang-Mills equations. They have also great significance in differential topology. Moreover, a lot of soliton equations can be reduced from self-dual Yang-Mills equations. Instead of the general theory of Yang-Mills fields, here we only consider a kind of their generalization in the point view of soliton theory. The Darboux transformation can be applied to this generalized self-dual Yang-Mills flow [40, 41]. Furthermore, a reduction of the generalized self-dual Yang-Mills flow leads to the AKNS system.

6.1.1

Generalized self-dual Yang-Mills flow

First we introduce briefly the self-dual Yang-Mills fields. Let G be a matrix Lie group, g be its Lie algebra. The Euclidean space R4 has metric ds2 = η ij dxi dxj

(with summation convention, η ij = δ ij ).

(6.1)

Gauge potential is a set of functions Ai ’s valued in the Lie algebra g. The strength of the gauge field is the anti-symmetric tensor Fij = ∂i Aj − ∂j Ai + [Ai , Aj ].

(6.2)

A Yang-Mills field is defined by the gauge potential satisfying the YangMills equations ∂F Fij + [Ak , Fij ] = 0. (6.3) η jk ∂xk

238


For any G-valued function G(x), the transformation Ai = G(x)Ai G(x)−1 + (∂ ∂i G(x))G(x)−1 is called a gauge transformation. A gauge transformation does not change the physical essence of the gauge field. Especially, it does not change the Yang-Mills equations. The dual of the gauge intensity F is a tensor ∗F defined by 1 (∗F )ij = ijkl η ka η lb Fab , 2 where

ijkl =

⎧ ⎪ 1, ⎪ ⎪ ⎨

−1,

⎪ ⎪ ⎪ ⎩ 0,

(6.4)

(i, j, k, l) is an even permutation of (1, 2, 3, 4) (i, j, k, l) is an odd permutation of (1, 2, 3, 4)

(6.5)

otherwise,

that is, (∗F )12 = F34 ,

(∗F )23 = F14 ,

(∗F )31 = F24 ,

(∗F )34 = F12 ,

(∗F )14 = F23 ,

(∗F )24 = F31 .

(6.6)

If ∗F = F , i.e., F12 = F34 ,

F23 = F14 ,

F31 = F24 ,

(6.7)

this gauge field is called a self-dual Yang-Mills field. An self-dual YangMills field always satisfies the Yang-Mills equation. Similarly, the gauge field satisfying ∗F = −F is called anti-self-dual. The self-duality and anti-self-duality can be interchanged by changing the orientation of the space-time. Therefore, it is enough for us to discuss only the self-dual Yang-Mills field. The self-dual Yang-Mills field on R4 can be complexified to the gauge field on C4 . Besides, R4 can be replaced by R2,2 with the metric ds2 = dx21 + dx22 − dx23 − dx24 .

(6.8)

In [98], the self-duality was extended to R4n . Here we will extend it to R2n . Suppose (x, p) = (x1 , · · · , xn ; p1 , · · · , pn ) are the real coordinates of R2n (n ≥ 2), Ai (x1 , · · · , xn , p1 , · · · , pn )’s are a set of N × N (real or complex) matrix functions of (x, p). Consider the Lax set

Li Ψ ≡

∂ ∂ −λ Ψ = −Ai Ψ, ∂pi ∂xi

(6.9)

Generalized self-dual Yang-Mills and Yang-Mills-Higgs equations

239

where Ψ is an N × N matrix function, λ is the spectral parameter which ∂ . appears as the coefficient of ∂xi The integrability conditions of (6.9) are ∂Ai ∂Aj − − [Ai , Aj ] = 0 ∂pj ∂pi

(6.10)

∂Ai ∂Aj − = 0. ∂xj ∂xi

(6.11)

and

When n = 2, this is just the self-dual Yang-Mills equation [116]. In [98], self-dual Yang-Mills field was generalized to the current form with “space” dimension 4n (n ≥ 1). We call {Ai } satisfying (6.10) and (6.11) a generalized Yang-Mills potential. Remark 47 We have mentioned that the self-duality depends on the metric of the space. For four dimensional space, here we only consider R2,2, with the metric ds2 = d dp1 dx1 + d dp2 dx2 . From (6.10), there exists an N × N matrix J such that Ai = −

∂J −1 J . ∂pi

(6.12)

Hence, (6.11) becomes ∂ ∂xi

∂J −1 ∂ J − ∂pj ∂xj

∂J −1 J = 0. ∂pi

(6.13)

Clearly, (6.13) and (6.12) are equivalent to (6.10) and (6.11). We shall consider a flow of the generalized self-dual Yang-Mills field by introducing a “time” variable so that Ai ’s, J and Ψ depend on t as well as the “space” variables (x, p). Moreover, suppose that Ψ satisfies the evolution equation m+q ∂Ψ =VΨ= Va λm−a Ψ (q ≥ 0), ∂t a=0

(6.14)

where Va ’s are N × N -matrix valued functions which are independent of the spectral parameter λ.

240


(6.14) and (6.9) form a Lax set in R2n+1 . Its integrability conditions consist of the recursive relations among Va ’s ∂V V0 = 0, ∂xi ∂V Va ∂V Va−1 = + [Ai , Va−1 ], ∂xi ∂pi

(6.15) (a = 1, 2, · · · , m),

and ∂V Vm+q = [V Vm+q , Ai ], ∂pi ∂V Va ∂V Va−1 = + [V Va−1 , Ai ], ∂pi ∂xi

(6.16) (a = m + q, · · · , m + 2)

together with the evolution equations Vm ∂V Vm+1 ∂Ai ∂V + − + [Ai , Vm ] = 0. ∂t ∂pi ∂xi

(6.17)

A solution of (6.10), (6.11) and (6.17) is called a generalized self-dual Yang-Mills flow. The gauge potentials Ai ’s satisfy not only the self-dual Yang-Mills equation, but also an additional evolution equation (6.17). Remark 48 There is no direct physical meaning of the generalized selfdual Yang-Mills flow. However, it contains a one parametric family of self-dual Yang-Mills solutions and it will be seen later that almost all known soliton equations are the reductions of the generalized self-dual Yang-Mills equation.

Theorem 6.1 If Ai ’s satisfy the generalized self-dual Yang-Mills equation (6.10) and (6.11), then the equations (6.16) and (6.17) for Vi ’s are completely integrable and all Vi ’s are expressed by differentiations and integrations of Ai ’s. Proof. From the first equation of (6.15), V0 = V0 (p, t). The other equations in (6.15) lead to ∂ 2 Va = ∂xi ∂xj

$

%

$

(6.18) %

∂Ai ∂ 2 Va−1 ∂V Va−1 + , Va−1 + Ai , ∂pi ∂xj ∂xj ∂xj 0 1 0 1 ∂ 2 Va−2 ∂Aj ∂V Va−2 = + , Va−2 + Aj , ∂p$i ∂pj ∂p %i $ % ∂pi ∂Ai ∂V Va−2 + , Va−1 + Ai , + [Ai , [Aj , Va−2 ]]. ∂xj ∂pj

(6.19)


241

Using (6.10), (6.11) and the Jacobi identity, we know that the right hand side of (6.19) is symmetric for i and j. Hence (6.15) is completely integrable. Take pi ’s as parameters, Vn ’s can be determined recursively as V0 = V00 (p, t), Va =

(x,p,t) ∂V Va−1 i

∂pi

(0,p,t)

+ [Ai , Va−1 ] dxi + Va0 (p, t),

(6.20)

where Va0 (p, t)’s are integral “constants” and the integrals are taken along the path in the n-dimensional subspace pi = const. In fact, these integrals are independent of the path because of the integrability conditions. Moreover, we can choose Va0 ’s as arbitrary functions of p and t. If they are independent of t, then the evolution equations (6.17) are a system of integro-differential equations which do not depend on t explicitly. Let (6.21) Wa = J −1 Va J (a = m + q, · · · , m + 1), then (6.16) implies

0

∂W Wa ∂J ∂W Wa−1 = + J −1 , Wa ∂pi ∂xi ∂xi and

1

(a = m + q, · · · , m + 2),

(6.22)

∂W Wm+q = 0. ∂pi

Hence 0 Wm+q = Wm +q (x, t).

Moreover, according to (6.22), Wa−1 =

(x,p,t) ∂W Wa i

(x,0,t)

∂xi

0

+ J −1

∂J , Wa ∂xi

1

d i + Wa0−1 (x, t) dp

(6.23)

(a = m + q, · · · , m + 2). The integrals are now taken along a path in the n-dimensional subspace xi = constant, and independent of the path. Here xi ’s are regarded as parameters. Therefore, when Ai ’s satisfy (6.10) and (6.11), we can write down Va (a = 0, · · · , m; a = m + q, m + q − 1, · · · , m + 1) recursively so that (6.15) and (6.16) are satisfied. The theorem is proved. (6.10) and (6.11) can be regarded as “spatial constraints” among Ai ’s, and (6.17) are their evolution equations. The meaning of “spatial constraints” is that at any moment Ai ’s satisfy the generalized self-dual Yang-Mills equation.

242

6.1.2



Let D be an N × N matrix D = λI − S,

(6.24)

where S is independent of λ. Let Ψ = DΨ and we want

Li Ψ =

(6.25)

∂ ∂ −λ Ψ = −Ai Ψ , ∂pi ∂xi

m+q ∂Ψ = Va λm−a Ψ . ∂t a=0

(6.26)

As in Section 3.2, direct calculation implies Ai = Ai −

∂S , ∂xi

(6.27)

∂S ∂S = SAi − Ai S = [S, Ai ] + S, ∂pi ∂xi V0 = V0 , Va−1 Va = Va + Va−1 S − SV

(a = 1, 2, · · · , m),

(6.28)

(6.29)

Vm +q = SV Vm+q S −1 , Vm+k S −1 − (V Vm+k+1 − Vm +k+1 )S −1 , Vm +k = SV

(6.30)

(k = q − 1, · · · , 1), and

∂S + SV Vm − Vm S − Vm+1 + Vm +1 = 0. (6.31) ∂t Here we assume that S is non-degenerate. It can be verified that (6.28) and (6.31) are completely integrable, that is, for any given initial data S(t0 , p0 , x) at t = t0 , p = p0 , there exists S(t, p, x) satisfying (6.28), (6.31) and the initial condition.

Theorem 6.2 Suppose the matrix S is non-degenerate and satisfies the equations (6.28) and (6.31), then D = λI − S is a Darboux matrix for (6.9) and (6.14). Now we turn to the explicit construction of the matrix S, which is essentially the same as that for the AKNS system. Take λ1 , · · ·, λN to


243

be N complex numbers such that at least two of them are different. Let hα (α = 1, 2, · · · , N ) be a column solution of (6.9) and (6.14) for λ = λα , i.e., hα satisfies m+q ∂hα = Va λm−a hα . α ∂t a=0

∂hα ∂hα = λα − Ai hα , ∂pi ∂xi

(6.32)

Let H = (h1 , h2 , · · · , hN ),

Λ = diag(λ1 , · · · , λN ).

(6.33)

Suppose H is non-degenerate. Let S = HΛH −1 .

(6.34)

Theorem 6.3 Suppose S is defined by (6.34), then D = λI − S is a Darboux matrix. Proof. From the first equation of (6.32), we have ∂H ∂H = Λ − Ai H. ∂pi ∂xi

(6.35)

Hence ∂S ∂H ∂H 2 −1 = Λ H − HΛH −1 ΛH −1 − Ai S + SAi , ∂pi ∂xi ∂xi ∂H ∂H −1 ∂S = ΛH −1 − HΛH −1 H . ∂xi ∂xi ∂xi

(6.36)

Therefore, ∂S ∂S = [S, Ai ] + S. ∂pi ∂xi

(6.37)

This is just (6.28). On the other hand, it is easy to see that m+q ∂H = Va HΛm−a , ∂t a=0

(6.38)

which implies ∂S ∂t

= =

m+q a=0 m+q a=0

Va HΛm−a+1 H −1 − Va S

m−a

S−S

m+q a=0

m+q

HΛH −1 Va HΛm−a H −1

a=0

Va S

m−a

(6.39) .

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From (6.29) and (6.30), we know that the right hand side of (6.39) is −SV Vm S +V Vm+1 −V Vm +1 . Hence (6.31) holds. The theorem is proved. Vm +V If Ψ(λ) is a fundamental solution of the Lax set, then a column solution of the Lax set can be written as Ψ(λ)l where l = l(λx + p) is a column and can be an arbitrary function of λx1 + p1 , · · · , λxn + pn . It is noted that in the case of AKNS system, l should be a constant column. Hence there is more freedom for constructing the Darboux matrix in the present case. Thus we have constructed explicitly the Darboux transformation (Ψ, Ai ) → (Ψ , Ai )

(6.40)

provided that the gauge group G is GL(N, C). In the construction, there are only algebraic and differential operations. Moreover, the Darboux transformations can be continued successively with the same algorithm. In many cases, the gauge potentials Ai ’s should belong to certain subalgebra of gl(N ), say Ai ∈ u(N ), i.e., A∗i + Ai = 0. We also want Va ∈ u(N ). In the construction of Darboux transformation, the equalities A∗ i + Va = 0 should also be satisfied. This can be guaranteed Ai = 0 and Va∗ +V by the following two constraints: (i) Take ¯ (µ is not real), (6.41) λα = µ or µ (ii) Take hα ’s such that

h∗α hβ = 0

(6.42)

whenever λα = λβ . ¯ β ), the Lax set leads to In fact, when λα = λβ (i.e. λα = λ

∂ ∂ −µ (h∗α hβ ) = 0, ∂pi ∂xi ∂ ∗ (h hβ ) = 0. ∂t α

(6.43)

Hence, h∗α hβ is a holomorphic function of xi + µpi and independent of t. If we take the initial values of hα , hβ such that h∗α hβ = 0 at t = 0 and xi + µpi = 0, then h∗α hβ = 0 holds everywhere. Similar to the discussion in Section 1.4, we have ¯)I, S ∗ + S = (µ + µ

S ∗ S = |µ|2 .

∗ (6.27) implies that A∗ i + Ai = 0 holds. Moreover, we have Va + Va = 0 inductively by (6.29) and (6.30). Therefore, the above construction of the Darboux matrix keeps the u(N ) reduction.


245

Remark 49 If Ai ’s are independent of t, then the above system is reduced to the generalized self-dual Yang-Mills system, and the Darboux transformation is still valid.

6.1.3

Example

Starting from the trivial solution Ai = 0, we want to construct solutions by Darboux transformation. We have to determine V0 , V1 , · · ·, Vm , Vm+1 , · · ·, Vm+q first. When Ai = 0, (6.15) and (6.16) become V0 = V0 (p), ∂V Va ∂V Va−1 = (a = 1, 2, · · · , m; m + q, · · · , m + 2), ∂xi ∂pi Vm+q = Vm0 +q (x).

(6.44)

Hence, the entries of Vm are polynomials of x of degree ≤ m, the entries of Vm+1 are polynomials of p of degree ≤ q−1. Moreover, (6.17) becomes ∂V Vm+1 ∂V Vm = . ∂pi ∂xi

(6.45)

Thus the entries of Vm are polynomials of p of degree ≤ q, the entries of Vm+1 are polynomials of x of degree ≤ m + 1. Therefore, the entries of V0 are polynomials of p of degree ≤ m + q and the entries of Vm+q are polynomials of x of degree ≤ m + q. From the expression of V and (6.44), we have ∂V ∂V λ − = 0. (6.46) ∂xi ∂pi Hence the entries of V are functions of λpi + xi (i = 1, · · · , n). V can be expressed as 1 V = q P (λp + x). (6.47) λ Here the entries of P (λp + x) are polynomials of λpi + xi (i = 1, · · · , n) of degree ≤ m + q. Therefore, the fundamental solution of (6.9) and (6.14) is 1 P (λp + x)t . (6.48) Ψ = exp λq The Darboux transformation (0, Ψ) → (A , Ψ ) → (A , Ψ ) → · · · ,

(6.49)

can be constructed successively so that a series of explicit solutions are obtained. We still call them soliton solutions.

246


To write down the solutions more explicitly, take N = 2, ⎛

V =⎝

⎞

iv(λ)

0

0

−iv(λ)

⎠,

(6.50)

where

1 u(λp + x), λq u is a real polynomial of degree m + q. Then

(6.51)

v(λ) = ⎛

Ψ=⎝

⎞

e iv(λ)t

0

0

e−iv(λ)t

⎠.

(6.52)

Let µ = σ + iτ be a complex number where σ, τ are real and τ = 0. Take ⎛ ⎞ e iv(µ)t −g ∗ (¯ µ)e iv(¯µ)t ⎠, H=⎝ (6.53) g(µ)e−iv(µ)t e−iv(¯µ)t where g(µ) = g(µp + x), g ∗ (¯ µ) = g(µ). For any holomorphic function g, the two columns of H are solutions of the Lax set for λ = µ and µ ¯ respectively. The condition (6.42) is also satisfied. Moreover, ∆ = det H = ewt + |g(µ)|2 e−wt = 0,

(6.54)

where w = i(v(µ) − v(¯ µ)) is a real-valued function. Let g(µ) = eρ(µ)+iθ(µ) ,

(6.55)

v(µ) + v(¯ µ) = κ(µ), where ρ(µ), θ(µ) are real-valued functions. Then S = HΛH −1 ⎛

=⎝

σ − iτ tanh(ρ − wt) iτ sech(ρ −

iτ sech(ρ − wt)e i(−θ+κt)

wt)e i(θ−κt)

σ + iτ tanh(ρ − wt)

⎞ ⎠

(6.56)

and we get the single soliton solution Ai = −

∂S . ∂xi

(6.57)

This kind of soliton solutions may have very complicated behavior because the choice of v and g has large freedom. The appearance of the functions tanh and sech in S implies that if v and g are chosen suitably,


247

the entries of S tend to constants and the entries of Ai ’s tend to 0 when ρ − wt → ∞. However, these solutions are not travelling waves. Multisoliton solutions can be obtained by successive Darboux transformations with long expressions.

6.1.4

Relation with AKNS system

Let Ji ’s be n N × N constant matrices and Φ(p, t) = exp(−

Pi (p, t) = − exp(− Ua (p, t) = exp(−

xj Jj )Ψ,

xj Jj )Ai exp(

xj Jj )V Va exp(

xj Jj ),

(6.58)

xj Jj ).

Here Ψ, Ai ’s and Va ’s are the matrix valued functions in the Lax set (6.9) and (6.14). Suppose that Φ, Pi ’s and Ua ’s are all independent of xi ’s. Thus, (6.58) gives constraints for the Lax set (6.9) and (6.14). Under these constraints, (6.9) and (6.14) become ∂Φ = (λJ Ji + Pi )Φ, ∂pi m+q ∂Φ = Ua λm−a Φ. ∂t a=0

(6.59)

This is the AKNS system on Rn+1 where the space-time variables are (p1 , · · · , pn , t). Therefore we have

Theorem 6.4 AKNS system is a kind of constraints of the self-dual Yang-Mills flow. R. S. Ward points out that all known soliton equations in lower dimensions can be the constraints of the self-dual Yang-Mills equation on R4 . This means that the self-dual Yang-Mills equation has profound content. Although the high dimensional AKNS system in Chapter 3 cannot become a constraint of the self-dual Yang-Mills equation on R4 , it is a constraint of a high dimensional generalized self-dual Yang-Mills flow.

248


6.2

Yang-Mills-Higgs field in 2+1 dimensional Minkowski space-time 6.2.1 Yang-Mills-Higgs field The Yang-Mills-Higgs equations in R2,1 are important partial differential equations in mathematical physics. The system of equations is a reduction of the self-dual Yang-Mills equation on R2,2 and the Higgs field appears as one of the reduced gauge potential. This means that the Higgs field occurs inside the framework of Yang-Mills theory. This equation is called Bogomolny equation or monopole equation. There were a series of work on the Bogomolny equations [76]. The case of R2,1 has the advantage that there is a real time variable and the metric is of Minkowski. The system is also integrable. A systematical sketch with beautiful results on these equations can be found in [108]. Let R2,1 be the Minkowski space-time {(x, y, t)} with the metric ds2 = dx2 + dy 2 − dt2 , G be a matrix Lie group, g be its Lie algebra. Let A = (At , Ax , Ay ) be a gauge potential valued in g and Φ a g−valued function called Higgs field. We consider the Yang-Mills-Higgs field satisfying the following Bogomolny equations (∗F )i = Di Φ (i = 1, 2, 3, x1 = x, x2 = y, x3 = t).

(6.60)

Here F is the field strength, Di Φ is the gauge-derivative of Φ, i.e., Fij = ∂j Ai − ∂i Aj + [Ai , Aj ],

(6.61)

Di Φ = ∂i Φ + [Ai , Φ],

(6.62)

and ∗ is the Hodge operator. For the Minkowski space-time R2,1 with oriented coordinate (t, y, x) and metric ds2 = dx2 + dy 2 − dt2 , the Bogomolny equation becomes Fxy = Dt Φ,

Fty = Dx Φ,

Fxt = Dy Φ.

(6.63)

It admits a Lax pair. In fact, let ∂t − ∂y + λ∂ ∂x , L1 = −∂

L2 = λ∂ ∂t − λ∂ ∂ y − ∂x ,

(6.64)

where λ is a parameter. The Lax pair is L1 Ψ = (At + Ay − λAx − λΦ)Ψ, L2 Ψ = (−λAt + λAy + Ax − Φ)Ψ,

(6.65)


249

where Ψ is an N ×N matrix-valued function. The integrability condition of the Lax pair (6.65) is just the Bogomolny equation (6.63) [76]. The Lax pair (6.65) can also be written down in the covariant form (Dt + Dy − λDx )Ψ = λΦΨ, (λDt − λDy − Dx )Φ = −ΦΨ

(6.66)

where Di Ψ = ∂i Ψ + Ai Ψ (i = t, x, y). If Ψ(t, x, y, λ) is an N × N matrix solution of Lax pair (6.65) with det Ψ = 0, and (L1 Ψ)Ψ−1 , (L2 Ψ)Ψ−1 are of degree one with respect to λ, then Ψ is called a Lax representation of the Yang-Mills-Higgs field, since At , Ax , Ay and Φ can be determined by Ψ.

6.2.2

Darboux Transformations

At first we consider the case G = GL(N, C). Let (Ψ, A, Φ) satisfy the Lax pair. It is required to find a matrix S(t, x, y) such that Ψ = (λI − S)Ψ,

(6.67)

together with some A , Φ , satisfy the Lax pair (6.65) too. Hence, A = (At , Ax , Ay ) and Φ give a solution of the Bogomolny equation [42]. Substituting Ψ in (6.67) into the Lax pair (6.65) with (Ai , Φ) replaced by (Ai , Φ ) (i = t, x, y), we obtain At = At − (Φ S − SΦ),

Ay = Ay − (Φ S − SΦ),

Ax = Ax − (Φ − Φ)

(6.68)

and −S St − Sy = At S − SAt + Ay S − SAy − 2(Φ S − SΦ)S, St − Sy = SAt − At S + Ay S − SAy + 2(Φ − Φ),

(6.69)

Sx = SAx − Ax S + (Φ − Φ)S + (Φ S − SΦ). From the last equation of (6.69), we get 1 Φ = (S Sx S −1 − SAx S −1 + Ax + Φ + SΦS −1 ). 2

(6.70)

Substituting (6.70) into the first two equations of (6.69), we have (∂ ∂t + ∂y )S = Sx S − (At S − SAt ) − (Ay S − SAy ) +Ax S 2 − SAx S + ΦS 2 − SΦS,

(6.71)

250


(∂ ∂t − ∂y )S = Sx S −1 − (At S − SAt ) + (Ay S − SAy ) +Ax − SAx S −1 − Φ + SΦS −1 .

(6.72)

The problem is reduced to find S such that (6.71) and (6.72) are satisfied. This is a complicated system of nonlinear partial differential equations. Fortunately, as before, the solution S can be constructed explicitly by a “universal” formula. Let λ1 , · · · , λN be N complex numbers so that they are not all equal, ha be a column solution of the Lax pair for λ = λa (a = 1, 2, · · · , N ). Construct (6.73) H = (h1 , h2 , · · · , hN ) (det H = 0), and

S = HΛH −1 ,

Λ = diag(λ1 , · · · , λN ).

(6.74)

It is easily seen that (∂ ∂t + ∂y )H = Hx Λ − At H − Ay H + Ax HΛ + ΦHΛ

(6.75)

(∂ ∂t − ∂y )H = Hx Λ−1 − At H + Ay H + Ax HΛ−1 − ΦHΛ−1 .

(6.76)

and

Consequently, direct calculation shows that (6.71) and (6.72) are satisfied. Thus, we have

Theorem 6.5 Let S be defined by (6.74). Then λI − S is a Darboux matrix for the Lax pair (6.65), and (At , Ax , Ay , Φ ) given by (6.68) and (6.70) satisfies the Bogomolny equation (6.63). The Darboux transformation (Ψ, A, Φ) → (Ψ , A , Φ ) can be applied successively to obtain a series of solutions by using the same purely algebraic algorithm. Remark 50 In general, if G = GL(N, R), the solution (Ψ , A , Φ ) is only a local solution, since we cannot make sure that the condition det H = 0 holds globally. Remark 51 ha (a = 1, · · · , N ) may be expressed as ha = Ψ(λa )la ,

(6.77)

where la ’s are columns satisfying L1 la = 0,

L2 la = 0

(6.78)

251


for λ = λa . Moreover, if λa is not real, then each entry of la is a holomorphic function of ω(λa ) = λ2a (t + y) + 2λa x + (t − y).

(6.79)

When G is not GL(N, C), but its subgroup, some reductions should be added in the construction of Darboux transformation. An interesting case is G = U (N ) (or SU (N )). In this case, (A, Φ) should be valued in the Lie algebra u(N ), i.e., Ai + A∗i = 0,

Φ + Φ∗ = 0.

(6.80)

After Darboux transformation we should have Ai + A∗ i = 0,

Φ + Φ∗ = 0.

(6.81)

This can be realized by choosing λa ’s and ha ’s such that ¯ (µ is not real), λa = µ or µ ∗ ha hb = 0 if λa = λb .

(6.82) (6.83)

Theorem 6.6 Suppose Ai ’s and Φ are valued in u(N ). If we choose λa ’s and ha ’s such that (6.82) and (6.83) hold, then after the Darboux transformation, Ai ’s and Φ are still valued in u(N ). Proof. As in Section 3.2, with the conditions (6.82) and (6.83), we have ¯)I, S + S ∗ = (µ + µ

S ∗ S = µµ∗ I.

(6.84)

Thus µ1 S is valued in U (N ), Sx and S − µI are valued in u(N ). From (6.70) it is seen that Φ ∈ u(N ) and 1 Φ S − SΦ = (S Sx − SAx + Ax S − ΦS + SΦ) ∈ u(N ). 2

(6.85)

Hence Ax , Ay and At are valued in u(N ). The proof is completed. Remark 52 The solution (Ψ , A , Φ ) is global if (Ψ, A, Φ) is global. Remark 53 The theorem also holds for the case of SU (N ).

6.2.3

Soliton solutions

Starting with the trivial solution Ai = Φ = 0,

252


the first Darboux transformation gives single solitons. Now the Lax pair (6.65) becomes ∂x )ψ = 0, (−∂ ∂t − ∂t + λ∂

(6.86)

∂t − ∂x )ψ = 0. (λ∂ ∂t − λ∂ The general solution of (6.86) is ψ = f (ω(λ))

(6.87)

where each entry of f is an arbitrary holomorphic function, ω is defined by (6.79). The Darboux transformation is constructed in the last subsection and the expressions for Φ , Ai are Φ = 12 Sx S −1 , At = − 12 Sx ,

Ax = − 12 Sx S −1 ,

(6.88)

Ay = − 12 Sx .

It is obvious that all solutions are SU (N ) solutions. If µ = i, the solution is static. If µ = i, the line ω(µ) = c can be expressed in the form x = βt + β0 ,

y = γt + γ0 ,

(6.89)

and the solution (Ψ , A , Φ ) is a travelling wave. That is, as functions of (x, y, t), (Ψ , A , Φ ) depends on x − βt and y − γt only. In the case N = 2, we write down the single solitons more explicitly. Let f1 , f2 be two holomorphic functions without common zero. We have ⎛

H=⎝

f1 (ω(µ)) −ff2 (ω(µ)) f2 (ω(µ))

f1 (ω(µ))

⎞ ⎠,

⎛

Λ=⎝

⎞

µ 0

⎠.

(6.90)

0 µ ¯

Then ⎛

⎞

µ|F F1 |2 + µ ¯|F F2 |2 (µ − µ ¯)F F1 F¯2 1 ⎝ ⎠. S = HΛH −1 = |F F1 |2 + |F F2 |2 (µ − µ ¯)F¯1 F2 µ|F F2 |2 + µ ¯|F F1 |2 (6.91) Here F1 and F2 denote f1 (ω(µ)) and f2 (ω(µ)) respectively. Hence 1 1 4( Im µ)2 |F F1 F2 − F2 F1 |2 tr Φ2 = − tr(S Sx S −1 Sx S −1 ) = 2 8 (|F F1 |2 + |F F2 |2 )2 (6.92) where Fj denotes fj (ω(µ)). It is noted that if µ = i, along any straight line on the plane t = constant, ω(µ) → ∞ when x2 + y 2 → ∞. It ||Φ ||2 = −


Figure 6.1.

Example (1), ||Φ||2 at t = 10

Figure 6.2.

Example (2), ||Φ||2 at t = 10

253

was shown in [108] that if F1 = P1 (ω(µ)), F2 = P2 (ω(µ)) where P1 and P2 are certain polynomials of ω, then the solution is localized. More generally, if F1 = P1 (ω(µ)), F2 = P2 (ω(µ)) exp(Q(w(µ))) where Q is another polynomial of ω with deg Q ≤ max{deg P1 , deg P2 }, then along any straight line on the plane t = t0 , we still have Sx → 0 as x2 +y 2 → ∞. However, the solutions may not be completely localized (see example (3)). This class of single solitons is slightly wider than those constructed in [108], where Q(ω) = 1. We draw the pictures of ||Φ||2 at t = 10 for the following examples (1) µ = i/2, f1 = 1, f2 = ω. (2) µ = i/2, f1 = 1, f2 = (ω − 1)(ω + 6)(ω − 4i). (3) µ = i/2, f1 = 1, f2 = ωeω/10 . The solution Φ does not approach to zero if t = t0 and (x, y) → ∞ along the curves |ω|2 eRe ω/10 = constant.

254


Figure 6.3.

Example (3), ||Φ||2 at t = 10

By applying Darboux transformations p times to the trivial solution, we obtain p-soliton solutions. Let µ1 , µ2 , · · · , µp be p complex numbers (µj = real, µj = i, µj = µk for j, k = 1, 2, · · · , p). The Lax representation of a p-soliton solution is Ψp = (λI − Sp )(λI − Sp−1 ) · · · (λI − S1 )

(6.93)

At first we consider the asymptotic behavior of the double soliton Ψ2 = (λI − S2 )(λI − S1 )

(6.94)

as t → ±∞. Let x = a1 t + b1 ,

y = a2 t + b2

(6.95)

be a straight line which is different from ω(µ1 ) = c1 . Then along (6.95), ω(µ1 ) → ∞ as t → ±∞. Hence λI − S1 ∼ λI − S0

(6.96)

where S0 is the limit of S as t → ±∞. Hence the asymptotic behavior of Ψ2 is the Darboux transformation of the constant matrix λI − S0 by using µ = µ2 . Therefore, Ψ2 behaves as the Lax representation of a single-soliton along the straight line ω(µ2 ) = constant. Besides, the potential Ai ’s and Higgs field Φ approach to zero along each straight line if it is neither ω(µ1 ) = constant nor ω(µ2 ) = constant.. It is remaining to consider the asymptotic behavior of Ψ2 along the straight line ω(µ1 ) = constant.. From the theorem of permutability for Darboux transformation, Ψ2 = (λI − S2 )(λI − S1 ) = (λI − S1 )(λI − S2 ),

(6.97)

255


Figure 6.4.

Splitting of ||Φ||2 for a 2-soliton (t = 0)

where S1 , S2 are defined in the same way as S1 , S2 but the order of (µ1 , l1 ) and (µ2 , l2 ) is changed. Thus the above argument implies that Ψ2 behaves as the Lax representation of the single-soliton defined by (µ1 , l1 ) along the straight line ω(µ1 ) = constant. as t → ±∞. The double soliton is splitting up into two single solitons as t → ±∞. For general p we have following splitting theorem as well.

Theorem 6.7 A p-soliton splits up into p single solitons asymptotically as t → ±∞. We draw a picture for ||Φ||2 showing the splitting of a 2-soliton with µ1 = i/2, µ2 = 2 − i/2 and for both µ1 and µ2 , f1 = 1, f2 = ω. Remark 54 The expression (6.93) can be written in the form [108] Ψp = I +

p

Mk . λ − µk k=1

(6.98)

Remark 55 Consider the relationship between gauge transformation and Darboux transformation. Let g be a function of (x, y, t) valued in the group G and Ψ(x, y, t, λ) be a Lax representation of a solution (A, Φ) of the Bogomolny equation (6.63). Then = gΨ Ψ→Ψ

(6.99)

is a gauge transformation. In fact, from the Lax pair it is easily seen that = gΦg −1 . Ai = gAi g −1 − gx g −1 , Φ (6.100)

256


Figure 6.5.

Splitting of ||Φ||2 for a 2-soliton (t = 10)

Moreover, Ψ gΨ = g(λI − S)Ψ = (λI − S)gΨ = (λI − S)

with

S = gSg −1 .

(6.101) (6.102)

(6.101) means that the gauge transformations commute with the Darboux transformation.

6.3

Yang-Mills-Higgs field in 2+1 dimensional anti-de Sitter space-time 6.3.1 Equations and their Lax pair The Bogomolny equation in 2+1 dimensional anti-de Sitter spacetime is also known to be integrable in the sense that it has a Lax pair [108, 107, 109]. Here we use the Darboux transformation method to derive its explicit solutions and discuss the asymptotic behavior of the solutions as time t → ∞. Moreover, the solutions are compared with those in the Minkowski space-time when the curvature of the anti-de Sitter space-time tends to zero [129, 130]. The 2+1 dimensional anti-de Sitter space-time is the hyperboloid U 2 + V 2 − X 2 − Y 2 = ρ2

(6.103)

(ρ > 0) in R2,2 with the metric ds2 = −dU 2 − dV 2 + dX 2 + dY 2 .

(6.104)

257


It has constant Gauss curvature −1/ρ2 . Define the local coordinate r=

ρ − ρ + 1, U +X

x=

Y , U +X

t=−

V , U +X

(6.105)

then a part U + X > 0 on the 2+1 dimensional anti-de Sitter space-time is represented by the Poincaré coordinates (r, x, t) with r > −ρ + 1 and the metric is ρ2 ρ2 2 2 2 (−dt + dr + dx ) = (dr2 + du dv) (r + ρ − 1)2 (r + ρ − 1)2 (6.106) where u = x + t, v = x − t. Clearly, when ρ → +∞, the metric on the region r > −ρ + 1 tends to the flat metric of the Minkowski space-time. ds2 =

Remark 56 Here we take the half space as r > −ρ + 1 rather than r > 0 so that the solitons can keep in a bounded region as ρ → +∞. This will be shown in Subsection 6.3.4. As in the Minkowski space-time, the Bogomolny equation is also (∗F )i = Di Φ

(i = u, v, r).

(6.107)

With the metric (6.106) and the orientation (v, u, r), (6.107) becomes r+ρ−1 Fur , ρ r+ρ−1 Dv Φ = − Fvr , ρ 2(r + ρ − 1) Dr Φ = − Fuv . ρ Du Φ =

(6.108)

It has a Lax pair ((r + ρ − 1)Dr + ρΦ − 2(ρλ − u)Du )ψ = 0, ρλ − u ρ(ρλ − u) Dr − 2Dv + Φ ψ=0 r+ρ−1 (r + ρ − 1)2

(6.109)

where the action of Dµ on ψ is Dµ ψ = ∂µ ψ + Aµ ψ. This Lax pair was first proposed by [107] to ρ = 1, which can be easily generalized to arbitrary ρ. With the help of this Lax pair, we can construct Darboux transformation in next subsection.

6.3.2

Darboux transformations

First we consider the case G = GL(N, C). This is a case free of reduction. Let (λ − u/ρ)I − S(u, v, r) (6.110)

258


be a Darboux matrix, then there exists (Au , Av , Ar , Φ ) such that for any solution ψ of (6.109), ψ = ((λ − u/ρ)I − S)ψ satisfies ((r + ρ − 1)Dr + ρΦ − 2(ρλ − u)Du )ψ = 0, ρλ − u ρ(ρλ − u) 2Dv + D − Φ ψ = 0 r + ρ − 1 r (r + ρ − 1)2

(6.111)

with Dµ = ∂µ +Aµ . Here the term u/ρ in (6.110) is used only to simplify the calculation. To determine the Darboux transformation, it is sufficient to find the matrix function S. For given (A, Φ), (A , Φ ) and an arbitrary matrix function Q, we define ∆i Q = ∂i Q + Ai Q − QAi (i = u, v, r), (6.112) δQ = Φ Q − QΦ. Substituting ψ = ((λ − u/ρ)I − S)ψ into (6.111) and considering (6.109), we get (r + ρ − 1)∆r I + 2ρ∆u S + ρδI + 2I = 0, ρ (r + ρ − 1)∆r S + ρδS = 0, ∆r I − δI = 0, r+ρ−1 ρ ρ2 ∆r S + δS = 0, ∆v S = 0. 2∆v I − r+ρ−1 (r + ρ − 1)2

∆u I = 0,

(6.113)

This system is equivalent to ∆u I = 0, ∆v S = 0,

(6.114) (6.115)

∆v I =

(6.116)

ρ ∆r S, r+ρ−1 ρ 1 ∆u S + I = 0, ∆r I + r+ρ−1 r+ρ−1 ρ ∆r I − δI = 0, r+ρ−1 ρ ∆r S + δS = 0. r+ρ−1

(6.117) (6.118) (6.119)

The new solution (Au , Av , Ar , Φ ) is solved from (6.114), (6.115), (6.118) and (6.119) as Au = Au ,

(6.120)

259


Av = SAv S −1 − (∂ ∂v S)S −1 , 1 1 ρ (SΦS −1 − Φ), Ar = (SAr − ∂r S)S −1 + Ar + 2 2 r+ρ−1 r+ρ−1 r+ρ−1 (SAr − ∂r S)S −1 − Ar Φ = 2ρ 2ρ 1 + (SΦS −1 + Φ), 2

(6.121) (6.122)

(6.123)

while (6.116) and (6.117) give the equations that S should satisfy. Remark 57 The Bogomolny equation is gauge invariant. We choose the form of the Darboux matrix (6.110) so that Au keeps unchanged in Darboux transformation. Using the standard procedure of constructing Darboux transformation, we have

Theorem 6.8 Let Λ = diag(λ1 , · · · , λN ) where λ1 , · · · , λN are complex numbers. Let hj be a column solution of the Lax pair (6.109) with λ = λj , H = (h1 , · · · , hN ). If det H = 0, then S = HΛH −1 − (u/ρ)I satisfies (6.116) and (6.117). Hence (λ − u/ρ)I − S is a Darboux matrix for (6.109). This theorem can also be generalized as follows.

Theorem 6.9 Let Λ = diag(λ1 , · · · , λN ) where λj (u, v, r) is a solution of ∂r λ −

2(ρλ − u) ∂u λ = 0, r+ρ−1

∂v λ +

ρλ − u ∂r λ = 0. 2(r + ρ − 1)

(6.124)

Let hj be a column solution of the Lax pair (6.109) with λ = λj (u, v, r), H = (h1 , · · · , hN ). If det H = 0 and define S = HΛH −1 − (u/ρ)I, then (λ − u/ρ)I − S is a Darboux matrix for (6.109). The general non-constant solution of (6.124) is given by the implicit function (r + ρ − 1)2 ρ − 1 + = constant. (6.125) v− ρλ − u λ Remark 58 Clearly Theorem 6.9 is a generalization of Theorem 6.8 because constant λ is a solution of (6.124). We shall show later that constant Λ always leads to global solutions on the 2+1 dimensional anti-de Sitter space-time. However, usually a non-constant Λ only leads to local solutions.

260


According to Theorem 6.8 and Theorem 6.9, we can construct explicit solutions of (6.108) from a known solution of (6.108) and the corresponding solution of the Lax pair (6.109). When the group G = U (N ), the Lie algebra consists of all antiHermitian matrices. Hence A∗µ = −Aµ , Φ∗ = −Φ. To keep this reduction, there should be more constraints on λj ’s and hj ’s in the construction of the Darboux matrix. They are ¯ 0 for certain non-real λ0 , λj = λ0 or λ h∗j hk = 0 if λj = λk .

(6.126)

The second condition holds identically if it holds on a line u = u0 = constant, v = v0 = constant. In fact, h∗j hk satisfies ∂u (h∗j hk ) = 0, (r + ρ − 1)∂ ∂r (h∗j hk ) − 2(ρλk − u)∂ ρλk − u 2∂ ∂v (h∗j hk ) + ∂r (h∗j hk ) = 0. r+ρ−1

(6.127)

The general solution of this system is h∗j hk = fjk (ω(λk )) where ω(λ) = v −

(r + ρ − 1)2 ρ − 1 + ρλ − u λ

(6.128)

and fjk (z) is a holomorphic function of z. Hence fjk (ω(λk )) ≡ 0 if fjk (ω(λk )) = 0 on the line u = u0 , v = v0 . Therefore, h∗j hk = 0 identically. With the condition (6.126), the Darboux transformation gives Φ ∈ u(N ), Aµ ∈ u(N ), provided that Φ ∈ u(N ), Aµ ∈ u(N ).

6.3.3

Soliton solutions in SU (2) case

(1) Expressions of the solutions Single soliton solutions are given by Darboux transformations from the trivial seed solution Ai = 0 (i = u, v, r), Φ = 0. Here we only consider the case where all λj ’s are constants. In this case ∂r hj −

2(ρζζj − u) ∂u hj = 0, r+ρ−1

∂v hj +

ρζζj − u ∂r hj = 0. (6.129) 2(r + ρ − 1)

The general solution is hj = fj (ω(ζζj ))

(6.130)

where fj is a column matrix whose entries are all holomorphic functions and ω is defined by (6.128). Then the Darboux matrix is λI − HΛH −1 with H = (h1 , · · · , hN ).


261

Now suppose G = SU (2). Considering the conditions (6.126), we take ¯ 0 ) where λ0 ∈ C is not real, τ = ω(λ0 ), Λ = diag(λ0 , λ ⎛

H=⎝

α(τ ) −β(τ ) β(τ )

⎞ ⎠

α(τ )

where α, β are holomorphic functions of τ . Let σ(τ ) = β(τ )/α(τ ). Then ⎛

⎞

¯0 1 σ ¯ λ0 − λ ¯0 − u . ⎝ ⎠+λ S= 2 1 + |σ| ρ σ |σ|2 ⎛

(6.131)

⎞

¯0 (|σ|2 )u σ ¯ 2 σu − σ ¯u λ0 − λ ⎝ ⎠, Φ = 2 2 (1 + |σ| ) σ2σ ¯u − σu −(|σ|2 )u

(6.132)

1 4( Im λ0 )2 tr Φ2 = |∂ ∂u σ|2 . 2 (1 + |σ|2 )2

(6.133)

||Φ ||2 = −

Therefore, a single soliton solution depends on an arbitrary meromorphic function σ(τ ) and ||Φ || is given by (6.133). Multi-soliton solutions can be constructed by successive actions of Darboux transformations on trivial solution. (2) Globalness of the solutions Till now, the solutions are only defined by the local coordinate (u, v, r), or on the part with U + X > 0 in the 2+1 dimensional anti-de Sitter space-time (6.103). However, these solutions can be extended to the whole 2+1 dimensional anti-de Sitter space-time in a natural way. According to (6.105) and (6.128), τ

ρλ0 (Y + V )(U + X) − 1 − Y 2 + V 2 ρ − 1 + (ρλ0 (U + X) − Y + V )(U + X) λ0 ρλ0 (Y + V ) + X − U ρ−1 = + . ρλ0 (U + X) − Y + V λ0

=

(6.134)

Denote ξ = ρλ0 (Y + V ) + X − U,

η = ρλ0 (X + U ) − Y + V,

(6.135)

then both ξ and η can never be zero on (6.103) when λ0 is not real. Hence τ is a smooth function of U, V, X, Y on (6.103), so are (r + ρ − 1)2 ρ2 = − , ∂v τ = 1, (ρλ0 − u)2 η2 2(r + ρ − 1)2 2ρ =− . ∂r τ = − ρλ0 − u η

∂u τ = −

(6.136)

262


Therefore, all the solutions which are obtained by Darboux transformation from trivial solution can be extended naturally to whole 2+1 dimensional anti-de Sitter space-time, provided that λ0 is chosen to be a non-real constant. (3) Localization of the solutions The infinity of the 2+1 dimensional anti-de Sitter space-time is at r → −ρ + 1. We shall verify that ||Φ || → 0 as r → −ρ + 1. In fact, 4( Im λ0 )2 |σ (τ )|2 |∂ ∂ u τ |2 (1 + |σ(τ )|2 )2 4 4( Im λ0 )2 2 (r + ρ − 1) = |σ (τ )| . (1 + |σ(τ )|2 )2 |ρλ0 − u|4

||Φ ||2 =

(6.137)

When r → −ρ + 1, τ → v + (ρ − 1)/λ0 . Hence ||Φ || → 0. This means that the solutions are all localized on the 2+1 dimensional anti-de Sitter space-time when λ0 is a constant. (4) Asymptotic behavior of the solutions when t → ∞ Without loss of generality, suppose λ0 = i. For general λ0 , a transformation of the coordinates will lead to this case.

Example 6.10 σ(τ ) = τ , this is just the localized solution (25) of [107], and 4r4 ||Φ ||2 = . (6.138) 2 2 2 2 ((r + x − t ) + 2x2 + 2t2 + 1)2 Let x = tR cos θ, r = tR sin θ. (6.139) When t and θ are√fixed, ||Φ ||2 is a function of R only. Its maximum appears at R = ± t2 + 1/t. Hence as t → ∞, the solution has a ridge which is located on the circle x2 + r2 = t2 + 1. This is quite different from the usual solitons which have only some peaks. Figure 6.6 shows this solution at t = 10. The vertical axis is ||Φ ||1/2 .

Example 6.11 σ(τ ) is a polynomial of τ of degree k (k ≥ 1) and all the roots τ1 , · · · , τk of σ(τ ) are simple. The asymptotic behavior of the solution is roughly as follows [129]. (1) If | Im τj | > 1, then ||Φj || is not small only near one point. The shape is a peak. (3) If Im τj −ρ + 1 with metric (6.106) tends to whole R2,1 with the Minkowski metric ds2 = dr2 + du dv. By taking the limit ρ → +∞ in (6.108) and (6.109), the Bogomolny equation and

264


Figure 6.8.

its Lax pair tend to Du Φ = Fur ,

Dv Φ = −F Fvr ,

Dr Φ = −2F Fuv ,

(6.141)

and (Dr + Φ − 2λDu )ψ = 0, (2Dv + λDr − λΦ)ψ = 0,

(6.142)

which are the Bogomolny equation (6.63) and the Lax pair (6.66) in the Minkowski space-time R2,1 , with the correspondence (λ, r, u, v) → ( λ1 , x, y + t, y − t). Then the Darboux transformation (λ − u/ρ)I − S = λI − HΛH −1 also tends to the Darboux transformation in R2,1 , so are the solutions (Au , Av , Ar , Φ ). The choice of the half space r > −ρ + 1 keeps the soliton solutions in a bounded region as ρ → +∞. However, the asymptotic behavior of the solutions are quite different for finite ρ and ρ → +∞. In the Minkowski space-time, the asymptotic behavior of the solutions is quite simple when σ(τ ) is a polynomial with only simple roots. In this case, each root of σ(τ ) corresponds to a peak in its graph. However, for finite ρ, we have shown that different kind of the root of σ(τ ) creates different shape in the graph. For σ(τ ) = (τ − 2)(τ − 6)(τ + 6)(τ − 2i)(τ − 6i)(τ + 6i), Figure 6.7 has shown the solution for ρ = 1. Figure 6.9 and Figure 6.10 show the solution for ρ = 2 and ρ → +∞ respectively. In Figure 6.9, the ridges are shorter than those in Figure 6.7, and in Figure 6.10, each root of σ(τ ) corresponds to a peak. For σ(τ ) = sin(τ /20), Figure 6.8 has shown the solution for ρ = 1. Here Figure 6.11 and Figure 6.12 show the solution for ρ = 30 and ρ → +∞ respectively.


Figure 6.9.

Figure 6.10.

Figure 6.11.

265

266


Figure 6.12.

Chapter 7 TWO DIMENSIONAL TODA EQUATIONS AND LAPLACE SEQUENCES OF SURFACES IN PROJECTIVE SPACE 7.1

Signed Toda equations

Toda equations are a class of important integrable systems [100]. Besides the original one-dimensional Toda equations, the two dimensional Toda equations also attract many authors in recent years [80, 31, 54, 83, 10]. In particular, the elliptic version of two dimensional Toda equations has many applications in differential geometry, such as minimal surfaces, surfaces of constant mean curvature, harmonic maps, etc. The hyperbolic two dimensional Toda equations can be traced back to G. Darboux [19]. Consider the hyperbolic partial differential equation zxt + azx + bzt + cz = 0.

(7.1)

z1 = zt + bz,

(7.2)

Define then z1 satisfies a hyperbolic equation z1,xt + a1 z1,x + b1 z1,t + c1 z1 = 0,

(7.3)

which is called a Laplace transformation of (7.1) in t-direction. Similarly, the function (7.4) z−1 = zx + az, satisfies z−1,xt + a−1 z−1,x + b−1 z−1,t + c−1 z−1 = 0,

(7.5)

and is called a Laplace transformation of (7.1) in x-direction. Applying these Laplace transformations successively, we get a series of hyperbolic systems zi,xt + ai zi,x + bi zi,t + ci zi = 0 (i = 0, ±1, ±2, · · ·),

(7.6)

268


which is called a Laplace sequence of equations starting from (7.1) G. Darboux derived the equations (ln hi )xt = −hi−1 + 2hi − hi+1

(hi =

∂ai + ai bi − ci ). ∂t

(7.7)

He set hi = eωi and obtained the equations ωi,xt = −eωi−1 + 2eωi − eωi+1

(i = 1, 2, · · · , n)

(7.8)

which are called hyperbolic Toda equations now. The original Toda equations are a system of ordinary differential equations in the form d2 ωi = eωi−1 − 2eωi + eωi+1 dt2

(i = 1, 2, · · · , n)

(7.9)

(in the periodic case, ω0 = ωn , ωn+1 = ω1 ). It was established by M. Toda [100] in the study of the motion of a chain of particles in which two neighboring particles attract each other in a nonlinear way. Consider a system of n particles of the same kind with equal mass arranged along a line and jointed by identical springs. Let the position of the n particles be q1 , q2 , · · ·, qn and q1 < q2 < · · · < qn . Suppose the ith particle is acted by the i − 1th particle and the i + 1th particle only. If the force is dependent of the distance exponentially, then we have a system of Toda equations (7.9). In [64], we noted that hi may be negative. We should set hi = αi eωi .

(7.10)

Here αi = ±1 as hi > 0 or hi < 0 respectively. Thus (7.8) is changed to ωi,xt = −αi−1 eωi−1 + 2αi eωi − αi+1 eωi+1 .

(7.11)

We call (7.11) two dimensional signed Toda equations. Accordingly, the one dimensional signed Toda equations should be d2 ωi = αi−1 eωi−1 − 2αi eωi + αi+1 eωi+1 dt2

(i = 1, 2, · · · , n).

(7.12)

The physical interpretation is that the system describes the motion of a chain of two kinds of particles and the interacting force between two neighboring particles is attractive or repulsive according to that they belong to the different kinds or the same kind respectively [65]. This is a system consisting of two different kinds of particles, positive and negative, with equal mass and the interaction between two neighboring particles of the same (resp. different) kind is repulsive (resp. attractive). The magnitudes of the forces are the same exponential function of

Two dimensional Toda equations and Laplace sequences of surfaces

269

the relative displacement qi+1 − qi . Consequently, the acting force from the i + 1th particle to the ith particle is m

d2 qi = ±eqi+1 −qi . dt2

(7.13)

Here we have + sign (resp. − sign) if two particles are of different (resp. same) kinds. Define αi = 1 if the ith particle and the i+1th particle are of different kinds, αi = −1 if the ith particle and the i + 1th particle are of the same kinds, then the force acting on the ith particle should be m

d2 q i = αi eqi+1 −qi − αi−1 eqi −qi−1 dt2

d2 q 1 m 2 = α1 eq2 −q1 , dt

(i = 2, · · · , n − 1),

d2 q n m 2 = −αn−1 eqn −qn−1 . dt

(7.14)

If the particles lie on a closed chain and let q1 = qn+1 , q0 = qn , we have m

d2 qi = αi eqi+1 −qi − αi−1 eqi −qi−1 dt2

(i = 1, · · · , n),

d2 qi+1 m = αi+1 eqi+2 −qi+1 − αi eqi+1 −qi dt2

(7.15) (i = 1, · · · , n).

Then m

d2 (qi+1 − qi ) = αi+1 eqi+2 −qi+1 − 2αi eqi+1 −qi + αi−1 eqi −qi−1 . (7.16) dt2

Let qi+1 − qi = ωi ,

(7.17)

then m

d2 ωi = αi+1 ewi+1 − 2αi ewi + αi−1 ewi−1 . dt2

(7.18)

Thus we get the signed Toda equations. Consequently, we see that (i) The arrangement of the particles of the two kinds on the line determine the values of αi (i = 1, 2, · · · , n). (ii) The values αi (= ±1) determine the arrangement of the two kinds of particles on the line. (iii) If the chain is a closed one (i.e. q1 = qn+1 ), then the number of αi which equals 1 should be even.

270


We return to the two dimensional case and treat the signed Toda equations from the point of view of integrable system. It is known that the two dimensional Toda equations can be deduced from their Lax pair ⎛ ⎛ ⎜ ⎜ ⎜ ⎝

⎜ ⎜ ⎜ ⎜ ⎟ ⎜ ⎟ ⎟ = λ⎜ ⎜ ⎠ ⎜ ⎜ ψn ⎝ t

ψ1 .. .

⎞

⎛ ⎛ ⎜ ⎜ ⎜ ⎝

⎜ ⎜ ⎜ ⎜ ⎟ ⎜ ⎟ ⎟ =⎜ ⎜ ⎠ ⎜ ⎜ ψn ⎝ x

ψ1 .. .

⎞

⎞

0

0

···

0

p1

p2

0

···

0

0 ⎟ ⎟

0 .. .

p3 · · · .. . . . .

0 .. .

0 .. .

0

0

· · · pn

0

σ1

1 λ

0

···

0

0

σ2

1 λ

···

0 .. .

0 .. .

σ3 · · · .. . . . .

1 λ

0

0

⎟⎛

ψ1 .. .

⎟⎜ ⎟⎜ ⎟⎜ ⎟⎝ ⎟ ⎟ ψn ⎠

⎞ ⎟ ⎟ ⎟, ⎠

(7.19)

⎞ ⎟⎛

0 ⎟ ⎟

⎟⎜

⎜ 0 ⎟ ⎟⎜

ψ1 .. .

⎟⎝ 1 ⎟ ⎟ ψn λ ⎠

⎞ ⎟ ⎟ ⎟, ⎠

(7.20)

· · · σn

i.e., the integrability conditions of (7.19) and (7.20) are pi,x = pi (σi − σi−1 ),

(7.21)

σi,t = pi − pi+1 .

In the case of pi > 0, we can write pi = eωi and (7.21) is equivalent to the two dimensional Toda equations (7.8). However, in the general cases, we should put pi = αi eωi in (7.21) (αi = ±1), and the signed Toda equations (7.12) are derived as well. In particular, if p1 = p3 = · · · = p2k−1 = eω , p2 = p4 = · · · = p2k = e−ω ,

(7.22) n = 2k,

the equations (7.21) are reduced to sinh-Gordon equation ωxt = 4 sinh ω.

(7.23)

When p1 = p3 = · · · = p2k−1 = e−ω , p2 = p4 = · · · = p2k = −eω ,

(7.24)

we have the cosh-Gordon equation ωxt = −4 cosh ω.

(7.25)

271


¯ the complex coordinates of the Euclidean If x, t are replaced by ζ, ζ, plane, then (7.11) becomes the elliptic version of the signed Toda equations (7.26) ωi,ζ ζ¯ = −αi−1 eωi−1 + 2αi eωi − αi+1 eωi+1 . Here the sign on the right hand side of the Toda equations (7.8) and the signed Toda equations (7.11) is different from that in the ordinary differential equations (7.9) and (7.12). This is due to our calculation is according to Darboux’s original equation (7.8). If we change t to −t, then the sign of (7.8) and (7.11) will be the same as that in (7.9) and (7.12).

7.2

Laplace sequences of surfaces in projective space Pn−1

The Laplace sequences of surfaces as an important subject in classical projective differential geometry, have been studied extensively [28, 95, 63, 62]. In this section, we elucidate the relationship between Laplace sequences of surfaces of period n in Pn−1 and the Toda equations of period n. Different from Darboux, we consider this problem from the point of view of integrable system. We start from the fundamental equations of the periodic Laplace sequences of surfaces which are written in the form of first-order partial differential equations. Multiplying suitable factors to the homogeneous coordinates of the points of the surfaces and changing an independent variable of the fundamental equations, we can simplify the fundamental equations of the periodic Laplace sequences of surfaces quite significantly. We find that there are two types of n periodic Laplace sequences of surfaces in Pn−1 . It is noted that type II occurs only for even n and was not mentioned by Darboux and the researchers in this field. Moreover, both types have the integrability conditions of the same form ∂ 2 ωi = −αi−1 eωi−1 + 2αi eωi − αi+1 eωi+1 , ∂x∂t

(i = 1, 2, · · · , n; ωn+1 = ω1 ) (7.27)

where αi = ±1. Let Pn−1 be n − 1 dimensional projective space, (x1 , x2 , · · · , xn ) be the homogeneous coordinates of a point N ∈ Pn−1 and N = N (t, x) (i.e. xa = xa (t, x), a = 1, 2, · · · , n)

(7.28)

be the equations of a surface of Pn−1 in homogeneous coordinates. The straight line determined by N and Nt is the tangent line of the t curve (i.e. x = constant) and the straight line passing through N and Nx is the tangent lines of the x-curve. Suppose that N , Nt , Nx are linearly

272


independent, then two tangent lines at N do not coincide and N , Nt , Nx determine the tangent plane of the surface N . This also means that the surface N is regular. Suppose that there exist parameters (t, x) of the surface N (t, x) such that along each x-curves the tangent lines T (t) of the t curves form a developable surface, i.e., there exists a point λN + Nt on each T (t) generating a curve C(t) parametrized by x such that the tangent line of C(t) is just T (t). Analytically, N x + λx N (λN + Nt )x = Ntx + λN

(7.29)

is a linear combination of N and Nt . Hence N (t, x) should satisfy a hyperbolic equation Nt + bN Nx + cN = 0. Ntx + aN

(7.30)

Conversely, suppose that a surface N (t, x) satisfies hyperbolic equation (7.30). Let (7.31) N = Nt + bN which is a point on the tangent line T (t). From Nt + bN )x = Ntx + bN Nx + bx N = −aN Nt + (bv − c)N, Nx = (N

(7.32)

it is seen that when x varies N generates a curve whose tangent line is T (t), i.e., along each x-curve the tangent lines of the t-curves form a developable surface. If a surface N (t, x) has the above property, we say that the surface N (t, x) admits a conjugate net and (t, x) are called conjugate parameters. We have

Theorem 7.1 A surface N (t, x) in Pn−1 admits a conjugate net with conjugate parameters (t, x) if and only if it satisfies a hyperbolic equation (7.30). Suppose that N (t, x) satisfies (7.30). The surface N = Nt + bN

(7.33)

is called the Laplace transformation of the surface N in the t-direction. It can be verified by calculation that N satisfies a hyperbolic equation in the form + a Nt + b Nx + c N = 0. (7.30) Ntx Hence N admits a conjugate net with conjugate parameters (t, x) too. The tangent lines T (t) to the surface N are tangent lines of surface N


273

too, i.e., the straight lines N N constitute a line congruence with focal surfaces N and N . Similarly, we can define the Laplace transformation of N in the xdirection (7.34) N = Nx + aN and N satisfies a hyperbolic equation in the form + a Nt + b Nx + c N = 0. Ntx

Any regular surface in P3 always admits a conjugate net as seen in the elementary theory of surfaces in Euclidean space. In Pn , a surface N (t, x) satisfying (7.30) or equivalently admitting a conjugate net is called a Laplace surface. Starting from a given Laplace surface S0 , the Laplace transformations in t and x directions give a sequence of surfaces · · · , N−l , N−l+1 , · · · , N0 , N1 , · · · , Nm , · · · .

(7.35)

It can be easily shown that the Laplace transformation (7.33), (7.34) and the Laplace sequence (7.35) are independent of the change of homogeneous coordinates N (t, x) → µN (t, x) (µ = 0) geometrically. The Laplace sequences of period n in the projective space Pn−1 is of special interest. Let a system of surfaces Ni = Ni (t, x) (i = 1, 2, · · · , n)

(7.36)

satisfy Ni,t = µi Ni + pi Ni−1

(pi = 0),

Ni,x = σi Ni + qi Ni+1

(qi = 0),

(7.37)

(N Nn+1 = N1 , N0 = Nn ), then the system of surfaces constitute a Laplace sequence of period n, and vice versa. In fact, the line Ni Ni+1 is the common tangent line of surfaces Ni and Ni+1 . In other words, (7.37) means that the surfaces Ni and Ni+1 are the two focal surfaces of the Ni Ni+1 }, and hence Ni+1 (resp. Ni ) is the Laplace line congruence {N transformation of Ni (resp. Ni+1 ). We shall first simplify the fundamental equation (7.37) of Laplace sequences by multiplying a suitable factor on the homogeneous coordinates of each surface Ni = Ni (t, x). Let i = ki (x, t)N N Ni , (ki (x, t) = 0) (7.38) we have

i,t = µ i + pi N i−1 , i N N i,x = σ i + qi N i+1 , i N N

(7.37)

274


where

ki,t + µi , ki ki,x i = σ + σi , ki i = µ

ki pi , ki−1 ki qi = qi . ki+1

pi =

i , we choose From the expression µ

ki = ki0 (x)e−

µi dt

,

(7.39)

(7.40)

i = 0 holds true. Here ki0 (x) is an arbitrary function of x. then µ i − µ i+1 ) = From the integrability condition of (7.37) , we get qi,t = qi (µ 0. Hence qi ’s depend on x only. i (mi (x) = 0), we have Furthermore, let Ni = mi (x)N = p N , Ni,t i i−1 Ni,x

where

=

σi Ni

+

(7.37)

qi Ni+1

mi mi,x i , pi , σi = σ mi−1 mi (mn+1 = m1 , m0 = mn ). pi =

qi =

mi qi , mi+1

(7.41)

From the last equation, we see that q1 q2 · · · qn = q1 q2 · · · qn ≡ Q(x)

(7.42)

is independent of the choice of mi . When Q > 0 or Q < 0 and n is odd, let (7.43) q = Q1/n and take arbitrary m1 = 0, mi+1 = mi qi q −1

(i = 1, 2, · · · , n),

(7.44)

then from (7.41) we have qi = q

(i = 1, 2, · · · , n).

(7.45)

We should note that the choice of mn+1 is consistent with mn+1 = m1 , since mn+1 = mn qn q −1 = mn−1 qn−1 qn q −2 = · · · = m1 qn qn−1 · · · q1 q −n = m1 . (7.46) When Q < 0 and n is even, let q = (−Q)1/n

(7.47)


275

and take arbitrary m1 = 0, mi+1 = mi qi q −1

(i = 1, 2, · · · , n − 1),

mn+1 = −mn qn q −1 .

(7.48)

Then we have qi (x) = q(x)

(i = 1, 2, · · · , n − 1),

qn = −q(x).

(7.49)

The choice of mn+1 is still consistent with mn+1 = m1 . Thus we have the following theorem.

Theorem 7.2 There are two types of Laplace sequences of surfaces with period n in projective space Pn−1 . Their fundamental equations can be written as Type I: and Type II:

⎧ ⎨ N =pN i,t i i−1 (pi = 0), ⎩ N = σ N + qN Ni+1 (q = 0) i,x i i

(7.50)

⎧ ⎨ N =pN i,t i i−1 (pi = 0), ⎩ N = σ N + c qN i,x i i i Ni+1 (q = 0)

(7.51)

respectively. Here ci = 1 for i = n and cn = −1. However, the Laplace sequence of Type II can occur only for even n. By a transformation of the variable x, (7.50) and (7.51) can be reduced to Type I: and Type II:

⎧ ⎨ N =pN , i,t i i−1 ⎩ N =σ N +N i,x i i i+1

(7.52)

⎧ ⎨ N =pN , i,t i i−1 ⎩ N =σ N +c N . i,x i i i i+1

(7.53)

Remark 59 In the case n being even, sgn(Q) is a projective invariant of Laplace sequences of surfaces. This is the reason why there are two types of Laplace sequences of surfaces of period n for even n. We write

⎛

N1

⎞

⎜ ⎟ ⎜ ⎟ ⎜ N2 ⎟ ⎜ Ψ=⎜ . ⎟ ⎟. ⎜ .. ⎟ ⎝ ⎠

Nn

(7.54)

276


x λt, , λσi , pi , we can introduce the λ spectral parameter in (7.50) and (7.51). Thus we obtain the Lax pair.

By the rescaling (t, x, σi , pi ) →

Theorem 7.3 The fundamental equations (7.52) of the Laplace sequence of type I with period n in Pn−1 are the Lax pair of two dimensional signed Toda equations. Note that the matrix form of (7.52) is (7.19) and (7.20). The Laplace sequence of surface of type II should correspond to the Lax pair ⎛ ⎛ ⎜ ⎜ ⎜ ⎝

⎜ ⎜ ⎜ ⎜ ⎟ ⎜ ⎟ ⎟ = λ⎜ ⎜ ⎠ ⎜ ⎜ ψn ⎝ t

ψ1 .. .

⎞

⎛ ⎛ ⎜ ⎜ ⎜ ⎝

⎜ ⎜ ⎜ ⎜ ⎟ ⎜ ⎟ ⎟ =⎜ ⎜ ⎠ ⎜ ⎜ ψn ⎝ x

ψ1 .. .

⎞

⎞

0

0

···

0

p1

p2

0

···

0

0 ⎟ ⎟

0 .. .

p3 · · · .. . . . .

0 .. .

0 .. .

0

0

· · · pn

0 0

σ1

1 λ

0

···

0

σ2

1 λ

···

0 .. .

0 .. .

σ3 · · · .. . . . .

− λ1

0

0

⎟⎛

ψ1 .. .

⎟⎜ ⎟⎜ ⎟⎜ ⎟⎝ ⎟ ⎟ ψn ⎠

⎞ ⎟ ⎟ ⎟, ⎠

(7.55)

⎞ ⎟⎛

0 ⎟ ⎟

⎟⎜

⎜ 0 ⎟ ⎟⎜

ψ1 .. .

⎟⎝ 1 ⎟ ⎟ ψn λ ⎠

⎞ ⎟ ⎟ ⎟, ⎠

(7.56)

· · · σn

(7.56) differs from (7.20) slightly. However, (7.55) and (7.56) can serve as the Lax pair of the signed Toda equations too. The integrability conditions of (7.55) and (7.56) are slightly different from (7.21) too. In fact, they are pa,x = pa (σa − σa−1 ), p1,x = p1 (σ1 − σn ),

σa,t = pa − pa+1

(a = 2, · · · , n − 1),

σ1,t = −p1 − p2 ,

pn,x = pn (σn − σn−1 ),

(7.57)

σn,t = p1 + pn .

Comparing with (7.21), it is seen that the only change is the expressions for σ1,t and σn,t . If we set pa = αa eωa

(a = 2, · · · , n),

−p1 = α1 eω1

(7.58)

α1 = sgn(−p1 ),

(7.59)

where αa = sgn(pa )

(a = 2, · · · , n),


277

then (7.57) is reduced to the signed Toda equations (7.11). Theorem 7.2 The fundamental equations of the Laplace sequences of type II with period n are the Lax pair of the two dimensional signed Toda equations too. Remark 60 If the Laplace sequences of surfaces is non-periodic, there are infinite number of surfaces corresponding to the infinite number of Toda equations and the Lax pair contains matrices of infinite order.

7.3


In the previous chapters we have applied Darboux transformation to construct new solutions of many integrable systems from some seed solutions. More precisely, the Darboux transformation is an algorithmic method to accomplish the transformation (u, Φ) → (u , Φ ). Here u is a known solution and Φ is a fundamental solution of the Lax pair corresponding to u, u is the new solution and Φ is the fundamental solution of the Lax pair corresponding to u . In some geometrical problems, the fundamental solutions are the geometrical objects to be identified. In this chapter, (u, Φ) are solutions of the two dimensional signed Toda equations and the Laplace sequences of surfaces of period n in Pn−1 . It is very interesting to see how to apply the Darboux transformation to these objects. Matveev has provided the Darboux transformation successively already [79]. We can use his method to construct periodic Laplace sequences of type I. Besides, Hu has modified Matveev’s method to fit the case of type II. The Darboux matrix of the two cases can be derived by using the general formula D = I − λS in a unified way too. At first, let (Ψ, pi , σi ) be the functions appeared in the Lax pair (7.19) and (7.20) in which Ψ = (N N1 , · · · , Nn )T is the Laplace sequence of type ω I in Pn−1 and pi = αi e i . The Darboux transformation for Laplace sequence of surfaces of type I is Na = Na −

Ψ0a Na−1 Ψ0a−1

(7.60)

where Ψ0a = Nai li ,

(7.61)

278


Nai is the ith homogeneous coordinate of Na and li is a set of constants. The corresponding pa = pa −

Ψ0a , Ψ0a−1

σa = σa +

Ψ0a−1 Ψ0a − Ψ0a Ψ0a−1

(7.62) (a = 1, · · · , n).

(7.63)

To construct the Darboux transformation for the Laplace sequence of type II we can still use (7.62) and (7.63) for a = 2, · · · , n − 2. However, and σn should be changed slightly as σ1 , σn−1 Ψ02 Ψ01 + , Ψ01 Ψ0n Ψ0 Ψ0 = σn−1 + 0 n − n−1 , σn−1 Ψn−1 Ψ0n−2 Ψ0 Ψ0 σn = σn − 01 − 0 n . Ψn Ψn−1

σ1 = σ1 +

(7.64)

The validity of all these formulae can be verified by direct calculations. On the other hand, let λ0 be a special value of the spectral parameter, Ψ0 be a solution of the Lax pair (7.19) and (7.20) (resp. (7.55) and 2π i

(7.56)) with λ = λ0 . Let ω = e n and Ω = diag(1, ω, ω 2 , · · · , ω n−1 ). The Darboux matrix is I − λS with S = HΛ−1 H −1 .

(7.65)

H = (Ψ0 , ΩΨ0 , · · · , Ωn−1 Ψ0 ).

(7.66)

Here Λ = λ0 Ω,

The Darboux transformation is Ψ = (λ − λS)Ψ.

(7.67)

Direct calculations show that the formulae (7.60) and (7.62)–(7.64) hold. Now we turn to some examples, i.e., use the above Darboux transformation to construct Laplace sequences. (1) Laplace sequence of type I Take the trivial solution of two dimensional Toda equation σa = 0,

pa = 1.

(7.68)

The fundamental equations (7.52) are reduced to N1,t = N4 ,

N2,t = N1 ,

N1,x = N2 ,

N2,x = N3 ,

N3,t = N2 , N3,x = N4 ,

N4,t = N3 , N4,x = N1 .

(7.69)

279


Solving the equations, we obtain the Laplace sequence of period 4: N1 cosh u sinh u

cos v

sin v

sinh u cosh u

sin v

− cos v

N2

N3 cosh u sinh u − cos v N4

− sin v

sinh u cosh u − sin v

(7.70)

cos v

Here u = t + x, v = t − x. In (7.70), the rows are homogeneous coordinates of the point Na , and each column is a solution of the Lax pair ∂Ψa = Ψa−1 , ∂t

∂Ψa = Ψa+1 ∂x

(a = 1, 2, 3, 4; Ψ0 = Ψ4 , Ψ5 = Ψ1 ). (7.71)

It is seen that N1 and N3 generate the surface S1 : x21 − x22 = x23 + x24

or x2 + y 2 + z 2 = 1

(7.72)

if we use the inhomogeneous coordinates x = x1 /x4 , y = x2 /x4 , z = x3 /x4 . Similarly, N2 and N4 generate the surface S2 : x22 − x21 = x23 + x24

or y 2 − x2 − z 2 = 1.

(7.73)

A typical quadrilateral together with S1 and S2 is shown in Figure 7.1. By using Darboux transformation, we obtain a new Laplace sequence of period 4 Ψ01 N4 , Ψ04 Ψ0 N3 = N3 − 30 N2 , Ψ2 N1 = N1 −

Ψ02 N1 , Ψ01 Ψ0 N4 = N4 − 40 N3 . Ψ3 N2 = N2 −

(7.74)

Here Ψ0a is a linear combination of the 4 columns in (7.70), i.e., Ψ01 = a cosh u + b sinh u + c cos v + d sin v, Ψ02 = a sinh u + b cosh u + c sin v − d cos v, Ψ03 = a cosh u + b sinh u − c cos v − d sin v,

(7.75)

Ψ04 = a sinh u + b cosh u − c sin v + d cos v. By a long calculation, we obtain N1

b − cz3 + dz2

−a − cz1 − dz4

az3 + bz1 + d

−az2 + bz4 − c

N2

−b − cz4 + dz1

a + cz2 + dz3

az4 − bz2 + d

−az1 − bz3 − c

N3

b + cz3 − dz2

−a + cz1 + dz4

−az3 − bz1 + d

az2 − bz4 − c

N4

−b + cz4 − dz1

a − cz2 − dz3

−az4 + bz2 + d

az1 + bz3 − c.

(7.76)

280


Figure 7.1. Example of Laplace sequences of surfaces of period 4 (Type I): N1 and N3 are on the surface x2 + y 2 + z 2 = 1, N2 and N4 are on the surface y 2 − x2 − z 2 = 1

Here z1 = cosh u cos v + sinh u sin v, z2 = cosh u cos v − sinh u sin v,

(7.77)

z3 = cosh u sin v + sinh u cos v, z4 = cosh u sin v − sinh u cos v. It is not difficult to prove that N1 , N2 , N3 , N4 generate four algebraic varieties in P3 . In fact, among the parameters z1 , z2 , z3 , z4 , there are algebraic relations z1 z2 + z3 z4 = 0,

z12 − z22 − z32 + z42 = 0.

(7.78)

From these relations and the parametric representation of Ni , we can eliminate these parameters and find that x = x1 /x4 , y = x2 /x4 , z = x3 /x4 satisfy an algebraic equation Fi (x, y, z) = 0. Hence each Ni lies on an algebraic variety. On can apply Darboux transformations successively to get an infinite sequence of Laplace sequences of surfaces of period 4 in P3 . (2) Laplace sequences of Type II Now we construct the Laplace sequences of surfaces of type II of period 4 in P3 explicitly. Here we also take λ0 = 1.


281

For the trivial solution of (7.57), we can take σ1 = σ2 = σ3 = σ4 = 0,

p1 = −1,

p2 = p3 = p4 = 1.

(7.79)

So the fundamental equations for the Laplace sequences are N1,t = −N N4 , N1,x = N2 ,

N2,t = N1 , N2,x = N3 ,

N3,t = N2 , N3,x = N4 ,

N4,t = N3 , N4,x = −N N1 .

(7.80)

Solving these equations we obtain N1 N2 N3 N4

ev cos u − sin u) √ 2 −ev sin u ev (sin u + cos u) − √ 2 ev (cos u

ev sin u + sin u) √ 2 ev cos u ev (cos u − sin u) √ 2 ev (cos u

e−v cos v + sin u) − √ 2 e−v sin u e−v (cos u − sin u) √ 2 e−v (cos u

e−v sin v − sin u) √ 2 −e−v cos u e−v (cos u + sin u) √ 2

e−v (cos u

(7.81) √ √ Here u = (x − t)/ 2, v = (x + t)/ 2, N1 and N3 generate the surface S1 : x1 x4 = x2 x3 or x = yz, N2 and N4 generate the surface S2 : x1 x3 = −x2 x4 or y = −xz. A typical quadrilateral N1 , N2 , N3 , N4 together with the surfaces S1 and S2 is shown in Figure 7.2. By using the formula (7.60), we can get Na = Na −

Ψ0a Na−1 Ψ0a−1

(7.82)

as in the case of Type I. Hence the Darboux transformation gives a new Laplace sequence of period 4. These four surfaces are algebraic surfaces too. Remark 61 The surfaces (7.72), (7.73) do not contain straight lines, while the surfaces S1 and S2 appeared in Laplace sequences of surfaces of type II contain two systems of straight lines. This fact reflects that the Laplace sequences of surfaces of type I and those of type II are not equivalent in the real projective geometry. The projective invariant sgn(Q) of type I and type II are +1 and −1 respectively.

7.4

Su chain (Finikoff configuration)

There are a class of special Laplace sequences of period 4 in the projective space P3 . They were introduced by P. S. Finikoff and B. Q. Su around 1930 with a plenty of geometrical properties but without explicit examples [28, 95]. Su showed that the integrability condition of the fundamental equations is the sinh-Gordon equation. In this section, we

282


Figure 7.2. Example of Laplace sequences of surfaces of period 4 (Type II): N1 and N3 are on the surface z = xy, N2 and N4 are on the surface y = −xz

simplify the fundamental equations of the Su chain and introduce the spectral parameter. Then the Darboux transformations are constructed and a series of explicit examples are obtained [62]. Following Su’s notations, the Laplace sequences of the surfaces is N1 N3 N2 N4 and the fundamental equations are reduced to ⎛

⎞

⎛

N1

0

⎜ ⎟ ⎜ ⎜ ⎟ ⎜ ⎜ N2 ⎟ ⎜ 0 ⎜ ⎟ = 1⎜ ⎜ ⎟ 2⎜ ⎜ N3 ⎟ ⎜ 0 ⎝ ⎠ ⎝

N4 ⎛

eφ

u

⎞

⎛

N1

−φv

⎜ ⎟ ⎜ ⎜ ⎟ ⎜ ⎜ N2 ⎟ ⎜ 0 1 ⎜ ⎟ = ⎜ ⎜ ⎟ 2⎜ ⎜ N3 ⎟ ⎜ −1 ⎝ ⎠ ⎝

N4

v

0

⎞⎛

−e−φ

0

0

0

e−φ

eφ

0

0

0

0

0

0

⎞

N1

⎟⎜ ⎟ ⎟⎜ ⎟ ⎟ ⎜ N2 ⎟ ⎟⎜ ⎟, ⎟⎜ ⎟ ⎟ ⎜ N3 ⎟ ⎠⎝ ⎠

(7.83)

N4 ⎞⎛

⎞

0

0

1

−φv

1

0

φv

⎜ N ⎟ 0 ⎟ ⎟⎜ 2 ⎟. ⎟⎜ ⎟ 0 ⎟ ⎜ N3 ⎟

1

0

φv

⎟⎜ ⎟⎜

N1

⎠⎝

⎟ ⎟ ⎠

N4

(7.84)


283

To introduce the spectral parameter, we use the transformation u → u , v → λv and the equations for Su chain lead to the Lax pair λ ⎛

⎞

0

0

−e−φ

λ⎜ 0 Φu = λU Φ = ⎜ 2⎜ ⎜ 0

0

0

eφ

0

0 ⎟

eφ

0

0

0

⎜ ⎜ ⎝

⎛

⎟ ⎟

e−φ ⎟ ⎟

⎟ Φ, ⎠

(7.85)

⎞

−φv

⎜ ⎜

0

0

0

1/λ

−φv 1/λ 1⎜ 0 Φv = V Φ = ⎜ ⎜ 2 ⎜ −1/λ 0 φv

⎟ Φ. ⎠

0 ⎟

⎝

0

1/λ

⎟ ⎟

0 ⎟ ⎟

0

(7.86)

φv

The integrability condition of the Lax pair is φuv = sinh φ.

(7.87)

We have already known the way of constructing the solutions of (7.87) together with the solution of 2 × 2 matrix Lax pair. Now we should construct the solution of (7.87) together with 4 × 4 matrix Lax pair (7.85) and (7.86). Introduce 2 × 2 matrices ⎛

a=⎝ ⎛

I=⎝

−1 0 0

⎞

⎛

⎠,

b=⎝

1

1 0

⎞ ⎠,

O=⎝

⎞

(7.88)

⎠.

⎞

0 e−φ a 1 ⎠, U= ⎝ 2 0 eφ b ⎛

0 0

⎠,

0 0

Write U and V to be block matrices

where

1 0

⎛

0 1 ⎛

⎞

0 1

V =C+

⎞

1 D, λ

⎛

−φv I 0 1 ⎠, C= ⎝ 2 0 φv I

(7.89)

⎞

0 b 1 ⎠, D= ⎝ 2 a 0

(7.90)

then the Lax pair (7.85) and (7.86) take the form Φu = λU Φ,

Φv = C +

1 D Φ. λ

(7.91)

284


Consider the 2 × 2 Lax pair Ψ, Ψu = λU ⎛

where

⎞

0 e−φ = 1⎝ ⎠, U 2 eφ 0

Ψv = V Ψ,

(7.92)

⎛

⎞

−φv 1/λ 1 ⎠. V = ⎝ 2 1/λ φv

(7.93)

This Lax pair has appeared in Section 4.3. Let (h1 , h2 )T be a column solution of (7.92) for λ = λ1 , then the Darboux matrix is ⎛ D(λ) =I−

⎞

⎛

⎞

h1 1 ⎜ h2 ⎟ ⎟=⎜ ⎠ ⎝ λ h2 0 − λ1 h1

0 λ ⎜ ⎜ λ1 ⎝ h2 h1

λ h1 − λ1 h2 ⎟ ⎟, ⎠ 1

(7.94)

and the new solution φ1 of the sinh-Gordon equation is defined by eφ1 = e−φ

h 2 2

h1

.

(7.95)

Now introduce the map E from 2 × 2 matrices to 4 × 4 matrices: ⎛

E: ⎝

⎞

α β γ

δ

⎛

⎠ −→ ⎝

⎞

αI βa γb

⎠.

(7.96)

δI

is mapped to The Darboux matrix D(λ) ⎛

⎞

λ1 h1 I − a ⎜ λ h2 ⎟ ⎟. D(λ) = ⎜ ⎝ λ1 h2 ⎠ b I − λ h1

(7.97)

By using the differential equations for h1 and h2 , we can see that Φ1 (λ) = D(λ)Φ(λ) satisfies U1 Φ1 (λ), Φ1u (λ) = λU

Φ1v (λ) = V1 Φ1 (λ),

(7.98)

where U1 and V1 are derived from replacing φ in U and V by φ1 in (4.175). Φ1 is a Su chain obtained from the known Su chain by Darboux transformation. In fact, ⎛

⎞

λ h1 I − a ⎜ λ1 h2 ⎟ ⎟ Φ(λ). Φ1 (λ) = ⎜ ⎝ ⎠ λ h2 b I − λ1 h1

(7.99)

285


Differentiating it with respect to u, the first equation of (7.98) is ⎛ ⎜ ⎜ ⎝

⎞

λ h1 − a λ1 h2 u ⎟ ⎟

0 −

λ h2 λ1⎛ h1

u

b

⎠

0

⎞

⎛ λ h1 −φ I − a ⎟ λ⎜ λ1 h2 ⎟ ⎝ 0 e a + ⎜ ⎠ 2 ⎝ − λ h2 b 0 eφ b I λ1 h1 ⎛ ⎛ ⎞ λ h1 I − a e−φ1 a ⎜ λ⎝ 0 λ1 h2 ⎠⎜ = ⎝ λ h2 2 0 eφ1 b b I − λ1 h1

⎞ ⎠

(7.100)

⎞ ⎟ ⎟. ⎠

Equating the terms with λ2 , we get the first equation of (4.172) eφ1 =

h 2 2

h1

e−φ .

The terms with λ give 1 h1 + λ1 h2 u 1 h2 − + λ1 h1 u

−

e−φ e−φ1 − = 0, 2 2 −φ −φ e e 1 − = 0, 2 2

(7.101)

h1 h2 and in 2 × 2 Darboux matrix should h2 h1 satisfy. Hence the first equation of (7.98) holds. The second equation of (7.98) can be verified in a similar way. Hence the matrix D(λ) derived from the Darboux matrix D(λ) for the 2×2 Lax pair is exactly a Darboux matrix for the Su chain. Thus the following theorem holds. which are the equations that

Theorem 7.4 Suppose a solution of the sinh-Gordon equation and a fundamental solution of its Lax pair are known. Then a series of Su chains can be constructed by the map E and the Darboux transformation successively. The algorithm is purely algebraic.

286


Now we give some geometric properties. Let µ = ⎛

1 ⎜ ⎜ ⎜ ⎜ 0 ⎜ D(λ) = ⎜ ⎜ ⎜ −λ 1 ⎜ λ µ 1 ⎜ ⎝

λ µ λ1

0

h1 , then h2

⎞

0

1

0

λ − µ λ1

0

1

0

⎟ ⎟ ⎟ ⎟ ⎟ ⎟. ⎟ ⎟ ⎟ ⎟ ⎠

(7.102)

λ 1 0 1 λ1 µ Su chain obtained from a known one via Darboux transformation is 0

⎛

N1

⎞

⎛

1

⎜ ⎜ ⎟ ⎜ 0 ⎜ N2 ⎟ ⎜ ⎜ ⎟ ⎜ ⎜ N ⎟ = ⎜ − 1 1 ⎝ 3 ⎠ ⎜ λ µ 1 ⎝ N4

0

1

1 µ λ1 0

0

1

0 1 − µ λ1 0

0

1

0

−

−

1 1 λ1 µ

⎞

⎛

⎞

⎛

N1 ⎟ ⎜ ⎟⎜ ⎟ ⎜ ⎟ ⎜ N2 ⎟ ⎜ ⎟⎜ ⎟ ⎜ ⎟⎜ N ⎟ = ⎜ ⎟⎝ 3 ⎠ ⎜ ⎠ ⎝ N4

µ N3 λ1 µ N2 − N4 λ1 1 N3 − N1 λ1 µ 1 N4 − N2 λ1 µ N1 +

⎞ ⎟ ⎟ ⎟ ⎟. ⎟ ⎟ ⎠

(7.103) When u and v are fixed, N1 N3 N2 N4 forms a spatial quadrilateral. After the Darboux transformation, N1 , N3 , N2 and N4 locate on four sides of the quadrilateral N1 N3 N2 N4 . We say that the quadrilateral N1 N3 N2 N4 ’ is incident to the quadrilateral N1 N3 N2 N4 . The following theorem holds.

Theorem 7.5 Under the Darboux transformation, the spatial quadrilateral N1 N3 N2 N4 is incident to N1 N3 N2 N4 . In the construction of Darboux transformation of the sinh-Gordon equation, we want two parameters. One is assigned value of the spectral parameter λ = µ, another is a constant column vector l such that h = C0 be a given Su chain. Φ(µ)l. Now let (µA , lA ) be n sets of parameters, SC By applying the Darboux transformation with parameters (µA , lA ) (A = 1, 2, · · · , n) successively, we get a series of Su chains Cn−1 → SC Cn . SC C0 → SC1 → · · · → SC If the order of the parameters (µA , lA ) is changed, we have another series of Su chains Cn −1 → SC Cn . SC C0 → SC1 → · · · → SC From the theorem of permutability, SC Cn = SC Cn . Since the inverse of a Darboux transformation is a Darboux transformation too, we have a periodic chain of Su chains of period 2n SC C0

SC1 → SC C2 → · · · → SC Cn−1 C2 ← · · · ← SC Cn −1 SC1 ← SC

SC Cn .

287


Therefore, we get a series of solutions with period 2n (n = 2, 3, 4, · · ·), i.e., with period 4, 6, 8, · · ·.

Theorem 7.6 There exists a periodic chain of Su chains of period 2n connected by Darboux transformations and each quadrilateral of these Su chains is incident to the quadrilateral of the neighboring Su chain. Especially, the series of Su chains of period 4 has four Su chains. The quadrilaterals of their congruences are incident one another. Example: The simplest Su chain corresponds to the solution φ = 0. In this case, ⎛

⎞

⎛

N1

0 0 −1 0

⎜ ⎟ ⎜ ⎜ ⎟ ⎜ ⎜ N2 ⎟ ⎜ 0 0 λ ⎜ ⎟ = ⎜ ⎜ ⎟ 2⎜ ⎜ N3 ⎟ ⎜ 0 1 ⎝ ⎠ ⎝ ⎛

N4

⎞u

⎛

1 0

N1

⎞⎛ ⎟⎜ ⎟⎜

⎞

N1

⎟ ⎟

0

⎜ N ⎟ 1 ⎟ ⎟⎜ 2 ⎟, ⎟⎜ ⎟ 0 ⎟ ⎜ N3 ⎟

0

0

0

⎠⎝

⎞⎛

⎠

N4

0

0 0 1

N1

0

1 0 0

N4

⎞

(7.104)

⎜ ⎟ ⎜ ⎟⎜ ⎟ ⎜ ⎟ ⎜ ⎟⎜ ⎟ ⎜ N2 ⎟ ⎜ 0 0 1 0 ⎟ ⎜ N2 ⎟ 1 ⎜ ⎟ = ⎜ ⎟⎜ ⎟. ⎜ ⎟ ⎟⎜ ⎟ 2λ ⎜ ⎜ N3 ⎟ ⎜ −1 0 0 0 ⎟ ⎜ N3 ⎟ ⎝ ⎠ ⎝ ⎠⎝ ⎠

N4

v

The first equation can be written as λ N 1u = − N 3 , 2

N 3u =

λ N2 , 2

N 2u =

λ N4 , 2

N 4u =

λ N1 , 2

hence λ 4

N1uuuu = −

2

N1 .

(7.105)

Solving this ordinary differential equation, N1 = A1 cosh β cos β + A2 cosh β sin β +A3 sinh β cos β + A4 sinh β sin β,

(7.106)

288


√

2λ u, and A1 , A2 , A3 , A4 are four dimensional row vectors where β = 4 which are independent of u. Then, √ 2 2 − (A2 + A3 ) cosh β cos β N 3 = − N 1u = λ 2 +(A1 − A4 ) cosh β sin β − (A1 + A4 ) sinh β cos β

+(A3 − A2 ) sinh β sin β , 2 N3u = −A4 cosh β cos β N2 = λ +A3 cosh β sin β − A2 sinh β cos β

(7.107)

+A1 sinh β sin β, √ 2 2 N4 = N 2u = (A3 − A2 ) cosh β cos β λ 2 +(A1 + A4 ) cosh β sin β + (A1 − A4 ) sinh β cos β

+(A2 + A3 ) sinh β sin β . From the second equation of (7.104),

and hence A1v A3v

1 N4 , 2λ λ 1 = N3 , 2λ

1 N2 , 2λ 1 = − N1 , 2λ

N 1v =

N 4v =

N 2v

N 3v

√ 1 2 = (A3 − A2 ), 2λ √2 1 2 (A1 − A4 ), = 2λ 2

A2v A4v

√ 1 2 = (A4 + A1 ), 2λ √2 1 2 (A2 + A3 ). = 2λ 2

(7.108)

λ 4 By differentiation, we have Ai,vvvv = − Ai (i = 1, 2, 3, 4). Hence 2 each component of

Ai = (ai , bi , ci , di ) (i = 1, 2, 3, 4) are linear combinations of cosh α cos α, cosh α sin √ α, sinh α cos α and 2 λv. sinh α sin α with constant coefficients. Here α = 4 When the initial condition is taken as N1 = (1, 0, 0, 0),

N2 = (0, 1, 0, 0),

N3 = (0, 0, 1, 0),

N4 = (0, 0, 0, 1)

(7.109)


289

at u = v = 0, then ⎛

⎞T

cosh α cos α

⎜ ⎜ ⎜ sinh α sin α A1 = ⎜ ⎜ √2 ⎜ 2 (− sinh α cos α + cosh α sin α) ⎝ √ 2 2 (cosh α sin α

+ sinh α cos α)

⎛

⎞T

cosh α sin α

⎜ ⎜ ⎜ − sinh α cos α A2 = ⎜ ⎜ √2 ⎜ 2 (− cosh α cos α − sinh α sin α) ⎝ √ 2 2 (− cosh α cos α

⎞T

sinh α cos α

⎜ ⎜ ⎜ cosh α sin α A3 = ⎜ ⎜ √2 ⎜ 2 (− cosh α cos α + sinh α sin α) ⎝ √

⎟ ⎟ ⎟ ⎟ , ⎟ ⎟ ⎠

+ sinh α sin α)

⎛

sinh α sin α

⎞T

⎜ ⎜ ⎜ − cosh α cos α A4 = ⎜ ⎜ √2 ⎜ − 2 (cosh α sin α + sinh α cos α) ⎝ √ 2 2 (cosh α sin α

⎟ ⎟ ⎟ ⎟ , ⎟ ⎟ ⎠

+ sinh α sin α)

⎛

2 2 (cosh α cos α

⎟ ⎟ ⎟ ⎟ , ⎟ ⎟ ⎠

− sinh α cos α)

⎟ ⎟ ⎟ ⎟ . ⎟ ⎟ ⎠

Substituting into (7.107) and letting λ = 1, we can get the √ first Su 2 (u + v), chain. Let α + β = σ, α − β = τ . When λ = 1, σ = 4 √ 2 (u − v). τ= 4 ⎛

cosh σ cos τ

⎜ ⎜ ⎜ sinh σ sin τ N1 = ⎜ ⎜ √2 ⎜ 2 (cosh σ sin τ − sinh σ cos τ ) ⎝ √ 2 2 (cosh σ sin τ

+ sinh σ cos τ )

⎞T ⎟ ⎟ ⎟ ⎟ , ⎟ ⎟ ⎠

290


⎞T

− sinh σ sin τ

⎜ ⎜ ⎜ cosh σ cos τ N2 = ⎜ ⎜ √2 ⎜ 2 (cosh σ sin τ + sinh σ cos τ ) ⎝ √ 2 2 (− cosh σ sin τ

⎛ √ ⎜ ⎜ ⎜ N3 = ⎜ ⎜ ⎜ ⎝

2 (− cosh σ sin τ √2 2 2 (− cosh σ sin τ

+ sinh σ cos τ )

− sinh σ cos τ )

⎞T ⎟ ⎟

+ sinh σ cos τ ) ⎟ ⎟

⎟ , ⎟ ⎠

cosh σ cos τ − sinh σ sin τ

⎛ √

2 2√ (− cosh σ sin τ + sinh σ cos τ ) 2 2 (cosh σ sin τ + sinh σ cos τ )

⎜ ⎜ ⎜ N4 = ⎜ ⎜ ⎜ ⎝

⎟ ⎟ ⎟ ⎟ , ⎟ ⎟ ⎠

sinh σ sin τ

⎞T ⎟ ⎟ ⎟ ⎟ . ⎟ ⎟ ⎠

cosh σ cos τ It is easy to see that N1 and N2 are the same quadratic surface x23 − x24 = −2x1 x2 ,

(7.110)

N3 and N4 are the same quadratic surface x21 − x22 = −2x3 x4 .

(7.111)

The second Su chain can be constructed as follows. From (7.94), the 2 × 2 Darboux matrix is ⎛

⎞

0 tanh γ1 λ ⎝ ⎠, I− λ1 coth γ1 0 where γ1 =

λ1 1 u+ v. Hence the Darboux matrix for the Su chain is 2 2λ1 ⎛

I−

⎞

0

λ ⎝ λ1 b coth γ1

a tanh γ1 0

⎠.


Let λ = 1, then ⎛

N

⎜ 1 ⎜ ⎜ N2 ⎜ ⎜ ⎜ N3 ⎝

N4

A series of solutions can

7.5

⎛

⎞

⎜ N1 +

⎟ ⎟ ⎟ ⎟ ⎟ ⎟. ⎟ ⎟ ⎟ ⎟ ⎠

tanh γ1 N3 λ1 ⎜ ⎟ ⎜ ⎟ ⎜ N2 − tanh γ1 N4 ⎟ ⎜ λ1 ⎟=⎜ ⎟ ⎜ ⎟ ⎜ N3 − coth γ1 N1 ⎠ ⎜ λ1 ⎜ ⎝ coth γ1 N4 − N2 λ1 be obtained successively ⎞

291

(7.112)

in this way.

Elliptic version of Laplace sequence of surfaces in CPn

In this section, we define the elliptic Laplace sequences and obtain their relations with harmonic sequences. We also get explicit examples of harmonic sequences [21]. Using Darboux transformation, a series of explicit harmonic sequences can be constructed.

7.5.1

Laplace sequence in CPn

CPn is the n dimensional complex projective space. Let Z = (Z 1 , · · · , be homogeneous coordinates of a point in CPn where Z i (i = ¯ 1, 2, · · · , n + 1) are complex numbers which are not all zero. Z = f (ζ, ζ) 2 2 n is a map from R (or a region of R ) to CP . Here ζ = z + iy is the complex coordinate of R2 . If f satisfies Z n+1 )

fζ ζ¯ + affζ + bffζ¯ + cf = 0

(7.113)

and fζ , fζ¯, f are linearly independent, then f is called an elliptic Laplace ¯ We use [f ] to denote the surface surface in CPn parametrized by (ζ, ζ). or the class of functions {λf | λ = 0}. As for the case of real projective space, we can define two kinds of Laplace transformations LI ([f ]) and LII ([f ]) of an elliptic Laplace surface [f ] by (7.114) f1 = fζ + bf, f−1 = fζ¯ + af respectively. Geometrically, the transformations LI and LII are independent of the choice of f in [f ], i.e., if we replace f by λf (λ = 0), then the surface LI ([f ]) and LII ([f ]) are unchanged. A sequence of elliptic Laplace surfaces · · · , [f−2 ], [f−1 ], [f ], [f1 ], [ff2 ], · · ·

(7.115)

can be obtained such that [ffi+1 ] = LI ([ffi ]),

[ffi−1 ] = LII ([ffi ]).

(7.116)

292


The sequence (7.115) is called an elliptic Laplace sequence, which is determined by [f ] completely, provided that fi,ζ , fi,ζ¯ and fi are linearly independent. An elliptic Laplace sequence of surfaces can be represented in the canonical form fi,ζ ζ¯ = σi fi,ζ¯ + pi fi , (7.117)

fi,ζ = fi+1 + σi fi , fi+1,ζ¯ = pi+1 fi and pi+1 = pi − σi,ζ¯,

σi+1 =

pi+1,ζ + σi . pi+1

(7.118)

In fact, by changing f to f0 = λf , (7.113) can be written as (p0 = 0)

(7.119)

f0,ζ¯ = p0 f−1 .

(7.120)

f0,ζ ζ¯ = σ0 f0,ζ¯ + p0 f0 and (7.114) takes the form f0,ζ = f1 + σ0 f0 ,

All other equations in (7.117) can be obtained by acting LI and LII successively. Remark 62 If pi = αi eωi , (7.118) is equivalent to the signed Toda equations of elliptic version ωi,ζ ζ¯ = −αi−1 eωi−1 + 2αi eωi − αi+1 eωi+1 .

7.5.2

(7.121)

Equations of harmonic maps from R2 to CPn in homogeneous coordinates

In Chapter 5 we have seen that the energy of the map from a Riemannian manifold (M, g) to another Riemannian manifold (N, a) is defined by E(φ) =

E(φ) dV VM .

(7.122)

Here E(φ) = g ij aαβ

∂φα ∂φβ , ∂xi ∂xj

dV VM =

√

g dx1 · · · dxn

(7.123)

in local coordinates. The map φ is called a harmonic map if it is a critical point of the energy integral and the Euler-Lagrange equations √ √ ∂(E(φ) g ) ∂(E(φ) g ) ∂ − i =0 (7.124) ∂y α ∂x ∂yiα

293


are the equations of harmonic maps. By introducing the Hermitian metric (Z, W ) =

n+1

α

Z αW ,

|Z|2 =

α=1

n+1

Z αZ

α

(7.125)

α=1

in Cn+1 , CPn becomes a Riemannian manifold with the Fubini-Study metric [69] |Z|2 (dZ, dZ) − (Z, dZ)(dZ, Z) . (7.126) ds2 = |Z|4

Lemma 7.7 The map [f ] : R2 → CPn is harmonic if and only if fζ ζ¯ = σffζ¯ + pf

(7.127)

(ffζ¯, f ) = 0,

(7.128)

(|f |2 )ζ |f |2

(7.129)

(ffζ¯, fζ¯) . |f |2

(7.130)

and

σ= hold true for some f ∈ [f ]. Here p=−

The proof can be done by writing out and simplifying the EulerLagrange equations based on the Fubini-Study metric. By using this lemma and the canonical form of elliptic Laplace sequences, we obtain the following result.

Theorem 7.8 Let {[ffi ]} be an elliptic Laplace sequence. If one of [ffi ] (say [ff0 ]) is harmonic, then all [ffi ] are harmonic maps. Proof. It is sufficient to prove that f1 and f−1 are harmonic if f0 = f is harmonic. At first, it is seen that the condition (7.128) means that f ⊥ f−1 and condition (7.129) means that f ⊥ f1 . From f1 = fζ − σf , it follows that f1,ζ¯ = fζ ζ¯ − σζ¯f − σffζ¯ = (p − σζ¯)f.

(7.131)

Hence (f1,ζ¯, f1 ) = (p − σζ¯)(f, f1 ) = 0. Besides, p1 =

(f1,ζ ζ¯, f1 ) (f1,ζ¯, f1,ζ¯) |p1 |2 |f |2 = − = − . |f1 |2 |f1 |2 |f1 |2

(7.132)

294


Hence p1 = − From σ1 =

|f1 |2 . |f |2

p1,ζ (|f1 |2 )ζ (|f |2 )ζ +σ = − , p1 |f1 |2 |f |2

(7.133)

(7.134)

it follows that condition (7.129) is satisfied. Hence [f1 ] is harmonic. Now turn to f−1 . From fζ¯ = pf−1 , it follows (f−1 , f ) = 0. Hence σ−1 =

(f−1,ζ , f−1 ) . (f−1 , f−1 )

(7.135)

Since f is harmonic, we have p|f |2 = −

(ffζ¯, fζ¯) = −|p|2 |f−1 |2 , |f |2

and hence

|f |2 . |f−1 |2

(7.137)

pζ + σ−1 , p

(7.138)

(f−1 , f−1 )ζ . (f−1 , f−1 )

(7.139)

p=− From σ= we have σ−1 =

(7.136)

Hence (f−1,ζ , f−1 ) = (f−1 , f−1 )ζ

(7.140)

(f−1 , f−1,ζ¯) = 0.

(7.141)

and it follows Hence f−1 is harmonic. The theorem is proved. Thus, the Laplace sequences of surfaces becomes the harmonic sequences in [16].

Theorem 7.9 Let {[ffi ]} be an elliptic Laplace sequence. If f−1 ⊥ f0 and f0 ⊥ f1 , the {[ffi ]} is a harmonic sequence. From the properties of harmonic sequences or by direct calculation, there are following facts. (1) Let |ffi+1 |2 . (7.142) eωi = |ffi |2


295

The elliptic Toda equations ωi,ζ ζ¯ = eωi−1 − 2eωi + eωi+1

(7.143)

are satisfied. Hence from a harmonic sequence, a solution of two-dimensional elliptic Toda equations can be constructed. (2) If f−1 ⊥ f0 , f0 ⊥ f1 and f−1 ⊥ f1 , then fi−1 ⊥ fi , fi ⊥ fi+1 and fi−1 ⊥ fi+1 holds true for all i, and these fi ’s are minimal surfaces in CPn . For the proof of fi−1 ⊥ fi+1 , it is sufficient to verify f ⊥ f2 and f ⊥ f−2 . From 0 = (f, f−1 )ζ = (ffζ , f−1 )+(f, f−1,ζ¯) = (f1 , f−1 )−(σf, f−1 )+ ¯−1 (f, f−2 ), (7.144) it is seen that (f, f−2 ) = 0. From 0 = (f1 , f )ζ = (f1,ζ , f ) + (f1 , fζ¯) = (ff2 , f ) + (σ1 f1 , f−1 ) + ¯(f1 , f−1 ), (7.145) we can see (ff2 , f ) = 0. Moreover, the metric on the surface [ffi ] induced ¯ i.e., from the Fubini-Study metric is proportional to the metric dζ dζ, the map [ffi ] is conformal. It is well-known that a harmonic map is minimal if it is conformal [114].

7.5.3

Cases of indefinite metric

If we use the indefinite metric (Z, W )J = |Z|2J

=

J a=1

J a=1

a

Z Z − a

n+1

a

Z aW − n+1

b

Z bW ,

b=J+1 b

Z Z

(7.146) b

b=J+1

in Cn+1 to replace the positive definite metric (Z, W ), we obtain the submanifolds CPnJ+ = {[Z] | |Z|2J > 0}, CPnJ− = {[Z] | |Z|2J < 0},

(7.147)

CPnJ0 = {[Z] | |Z|2J = 0}, and the Fubini-Study metric can be extended to CPnJ+ and CPnJ− . The above results can be extended to CPnJ+ and CPnJ− with some modifications, provided that |ffi |2J = 0. In these cases the two-dimensional signed Toda equations of elliptic version ωp,ζ ζ¯ = αp−1 eωp−1 − 2αp eωp + αp+1 eωp+1 can be obtained.

(7.148)

296

7.5.4


Harmonic maps from R1,1

Instead of R2 (or S 2 ), we take the Minkowski plane R1,1 = {(ξ, η)} with ds2 = dξ dη, then the Laplace sequence can be written in the form fi,ξη = σi fi,η + pi fi , fi,ξ = fn+1 + σi fi ,

(7.149)

fi+1,η = pi+1 fi and pi+1 = pi − σi,η , pi+1,η + σi . σi+1 = pi+1

(7.150)

Moreover, it is seen that if fi−1 ⊥ fi and fi ⊥ fi+1 , then fi is a harmonic map (wave map) from R1,1 to CPn (or CPnJ+ , CPnJ− ). However, in general fi−1 ⊥ fi and fi ⊥ fi+1 do not imply fi+1 ⊥ fi+2 , i.e., starting with a harmonic map, the Laplace sequence may not be a harmonic map.

7.5.5

Examples of harmonic sequences from R2 to CPn or R1,1 to CPn

Example 1. Let f

= (f 1 , f 2 , · · · , f n+1 ) ξ¯ ξ¯ = α1 exp λ1 ξ − , · · · , αn+1 exp λn+1 ξ − . λ1 λn+1

(7.151)

We take λk (k = 1, 2, · · · , n + 1) such that |λk |2 = 1 and |α1 |2 λ1 + · · · + |αn+1 |2 λn+1 = 0.

(7.152)

· · · , fz¯z¯, fz¯, f, fz , fzz , · · ·

(7.153)

Then

is a harmonic sequence in CPn since fζ ζ¯ + f = 0. In particular, if we take |αk |2 = 1,

2kπ

λk = e n+1 i ,

then the harmonic sequence is of period n.

(7.154)


297

Example 2. For a harmonic sequence from R1,1 to CPn−1 (n = 2k) f = (f 1 , f 2 , · · · , f n ), η η , f 2 = α1 sin λ1 ξ − , f 1 = α1 cos λ1 ξ − λ1 λ1 η η , f 4 = α2 sin λ2 ξ − , f 3 = α2 cos λ2 ξ − λ2 λ2 ··· η η , f 2k = αk sin λk ξ − , f 2k−1 = αk cos λk ξ − λk λk

(7.155)

where λ1 , · · ·, λk are real numbers, we have (f, f ) = |α1 |2 + · · · + |αn |2 = constant, (f, fη ) = 0,

(7.156)

· · · , fηη , fη , f, fξ , fξξ , · · · .

(7.157)

(f, fξ ) = 0, fξη + f = 0. The Laplace sequence is

It is easily verified that (ffk , fk+1 ) = 0, Hence fk is a harmonic map.

(ffk−1 , fk ) = 0.

(7.158)

Index

2+1 dimensional AKNS system, 65, 68, 69, 87 2+1 dimensional N-wave equation, 89, 97, 100 AKNS hierarchy, 15 AKNS system, 1, 12, 16, 18, 23, 25, 30, 32, 34–36, 38, 41, 42, 45–48, 51, 56, 58, 59, 64, 70, 75, 103, 108, 111, 117, 237, 242, 244, 247 Backlund ¨ congruence, 121, 122, 131, 132, 164, 170, 174, 181, 184, 186, 187 Backlund ¨ transformation, 28, 45, 120– 122, 134, 135, 138, 149, 150, 154, 156, 158–164, 166, 170, 172, 186, 187, 228–230 binary Darboux transformation, 65, 82– 84 Bogomolny equation, 248–250, 255, 257 Boussinesq equation, 86

Darboux operator, 65, 71–73, 75–77, 79, 83–86 Darboux transformation, 1–7, 9–11, 18, 23, 25, 28–31, 34, 36–40, 44– 47, 50, 51, 59–61, 63, 65–68, 70–72, 75, 77, 79, 80, 82–84, 86, 87, 89, 93–95, 98, 103, 108–111, 113, 115, 117, 118, 120–122, 129, 135, 136, 138, 139, 149, 150, 154–161, 163, 164, 166–170, 172, 176, 179, 181, 183, 186, 187, 189, 192, 198, 203–205, 207, 208, 210, 212, 214, 217, 218, 226–228, 233, 237, 244, 245, 247, 250– 252, 254–260, 262, 264, 277– 281, 284–287 Davey-Stewartson equation, 69, 75 differential polynomial, 14, 15, 17–19, 23, 24, 32, 40, 42, 43, 68, 69, 106 DS equation, 69 DSII equation, 70, 78, 80, 82 DSI equation, 65, 70, 78, 82–84, 89, 97

Chebyshev coordinates, 128, 132, 134, 135, 138, 144, 147, 155, 156, 161, 164–166, 175, 181, 184 Chebyshev frame, 128, 132, 154, 175–179, 196–198 continuous spectrum, 55, 63 cosh-Gordon equation, 147, 149, 161

elliptic Laplace sequence, 292–294 elliptic Laplace surface, 291 elliptic Toda equations, 295

Darboux matrix, 1, 4, 7, 11, 18, 19, 21–23, 25–28, 30–34, 38, 39, 45–48, 50, 51, 59, 60, 65, 90, 92, 93, 108, 109, 118, 120, 137, 139, 157, 158, 162, 173, 176, 203, 212, 214, 218, 225, 226, 242– 244, 250, 258–260, 277, 278, 284, 285, 290

gauge transformation, 238, 255, 256

focal surface, 130–132, 150–155, 164, 165, 170

harmonic map, 189, 191–196, 198, 199, 201, 203, 207, 208, 210, 212– 214, 217–221, 230, 233, 292, 293, 295–297 harmonic sequence, 294–297 Higgs field, 248, 254

300


integrability condition, 2–5, 11, 12, 16, 18, 31–33, 35, 41, 46, 57, 66, 68, 71, 74, 84, 88, 100, 104, 105, 124, 125, 134, 142, 155– 157, 160, 161, 164–166, 182– 184, 196, 202, 214, 239–241, 249, 270, 274, 276, 283 inverse scattering, 1, 51, 56, 58, 63, 64 Jost solution, 52, 53, 59, 61, 63 KdV equation, 2, 3, 9, 10, 36, 37, 51, 64, 66, 86, 213 KdV hierarchy, 35, 36, 38, 68 KP equation, 65–67, 70, 75, 77, 78, 87 KP hierarchy, 68 KPII equation, 67 KPI equation, 65, 67 Laplace sequence, 268, 273, 275–282, 294, 296, 297 Laplace transformation, 130, 267, 272, 273 Lax pair, 2–7, 11, 16, 18, 23, 25, 26, 28, 30–37, 39, 40, 45–47, 50, 51, 58, 59, 64, 66, 67, 69–71, 75, 78, 79, 82, 86–88, 95, 122, 136– 139, 155, 157, 158, 160, 161, 164, 166–168, 170–172, 174– 177, 179, 183, 202–204, 214, 215, 219–221, 223, 248–250, 252, 255–257, 259, 260, 264, 270, 276–279, 283–285 Lax set, 87, 104, 106, 107, 109, 110, 238, 240, 244, 246, 247 line congruence, 122, 129, 130, 149, 150 Miura transformation, 10 MKdV equation, 3–8, 10, 11, 13, 42 MKdV hierarchy, 42 MKdV-SG hierarchy, 35, 43, 45 negative sine-Gordon equation, 145 negative sinh-Laplace equation, 129, 183 nonlinear constraint, 65, 87, 88 nonlinear Schr¨ ¨ odinger equation, 14, 46, 69 nonlinear Schr¨ ¨ odinger hierarchy, 35, 46– 48, 59 normalized harmonic map, 195–198

pseudo-spherical congruence, 130–135, 149, 150, 154–156, 164, 165 scattering data, 51, 56–60, 63 self-dual Yang-Mills equation, 237, 239– 241, 247 self-dual Yang-Mills field, 237–239 self-dual Yang-Mills flow, 237, 240, 247 signed Toda equations, 268–271, 276, 277 signed Toda equations of elliptic version, 292, 295 sine-Gordon equation, 3, 28, 44, 45, 121, 127–129, 132–136, 138, 139, 145, 149, 155, 166, 174–178 sine-Laplace equation, 166–169, 172, 173 singular Darboux transformation, 228, 230, 234 sinh-Gordon equation, 149, 156, 160, 284– 286 sinh-Laplace equation, 166, 167, 169, 171, 174, 183 spectral parameter, 1, 4, 12, 16, 31, 35, 103, 176, 183, 202, 239 Su chain, 283–287, 289, 290 surface of constant Gauss curvature, 121, 144, 154, 179, 184, 185, 189, 196, 198 surface of constant mean curvature, 122, 179–181, 184, 186, 187, 198, 199 surface of constant negative Gauss curvature, 121, 126, 128, 129, 131, 132, 134, 135, 138, 140, 144, 145, 148, 164–166, 170–172, 174, 186, 193 surface of constant positive Gauss curvature, 128, 144, 146, 154–156, 164, 181, 183, 184, 186, 187 theorem of permutability, 25, 28–30, 34, 45, 76, 77, 115, 212, 254, 286 Toda equation, 267, 268, 270, 277 uniton, 189, 220–226, 228, 230, 232–235 Yang-Mills equation, 237, 238 Yang-Mills field, 237, 238 Yang-Mills-Higgs equation, 248 Yang-Mills-Higgs field, 248, 249

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