Algebro-geometric solutions of the coupled modified Korteweg–de Vries hierarchy

Advances in Mathematics 263 (2014) 123–153

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Advances in Mathematics www.elsevier.com/locate/aim

Algebro-geometric solutions of the coupled modified Korteweg–de Vries hierarchy Xianguo Geng a , Yunyun Zhai a,∗ , H.H. Dai b a

School of Mathematics and Statistics, Zhengzhou University, 100 Kexue Road, Zhengzhou, Henan 450001, People’s Republic of China b Department of Mathematics, City University of Hong Kong, 83 Tat Chee Avenue, Kowloon, Hong Kong, People’s Republic of China

a r t i c l e

i n f o

Article history: Received 28 January 2013 Accepted 23 June 2014 Available online xxxx Communicated by Charles Fefferman Keywords: cmKdV hierarchy Algebro-geometric solutions Baker–Akhiezer function Trigonal curve

a b s t r a c t Based on the stationary zero-curvature equation and the Lenard recursion equations, we derive the coupled modified Korteweg–de Vries (cmKdV) hierarchy associated with a 3 ×3 matrix spectral problem. Resorting to the Baker–Akhiezer function and the characteristic polynomial of Lax matrix for the cmKdV hierarchy, we introduce a trigonal curve with three infinite points and two algebraic functions carrying the data of the divisor. The asymptotic properties of the Baker–Akhiezer function and the two algebraic functions are studied near three infinite points on the trigonal curve. Algebro-geometric solutions of the cmKdV hierarchy are obtained in terms of the Riemann theta function. © 2014 Published by Elsevier Inc.

1. Introduction The study of soliton equations is one of the most prominent subjects in the field of nonlinear science [1]. Among so many soliton equations, the modified Korteweg–de Vries (mKdV) equations have been studied extensively because of their physical significance * Corresponding author. E-mail address: [email protected] (Y. Zhai). http://dx.doi.org/10.1016/j.aim.2014.06.013 0001-8708/© 2014 Published by Elsevier Inc.

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[4,33,41,42]. Generalization of the mKdV equation to a multi-component system or a matrix equation has been done in a number of references [4,23,41,42], which can reduced to the coupled mKdV (cmKdV) equation. Iwao and Hirota constructed multi-soliton solutions of the cmKdV equation by using the Hirota bilinear method [24]. In Ref. [40], Tsuchida and Wadati solved the initial value problem of the cmKdV equation with the aid of the inverse scattering transformation. It is shown that the cmKdV equation has an infinite number of conservation laws and multi-soliton solutions. The principal subject of the present paper is to construct algebro-geometric solutions of the cmKdV hierarchy associated with a 3 × 3 matrix spectral problem [40] on the basis of the theory of algebraic curves. The first nontrivial member in the hierarchy is the cmKdV equation ut1 = −uxxx + 3ux v 2 + 3uvvx + 6u2 ux , vt1 = −vxxx + 3u2 vx + 3uux v + 6v 2 vx ,

(1.1)

which is just reduced to the mKdV equation when v = u. We first review the related literature on the subject. Several systematic approaches have been developed to obtain algebro-geometric solutions of soliton equations such as the algebro-geometric method, the inverse scattering transformation for periodic problem, and others [3,6,9,10,12,13,16, 17,21,25,26,28,32]. These methods have the strong impact on the evolution of the modern mathematics and theoretical physics [3,6,9,10,12,13,16,17,20,21,25–28,32], that combine the spectral theory of differential and difference operators with periodic coefficients, the algebraic geometry of compact Riemann surfaces and their Jacobians, the Riemann theta functions and inverse problems. It has been applied extensively to 2 × 2 matrix spectral problem of continuous and semi-discrete cases, from which algebro-geometric solutions of many soliton equations are obtained such as the KdV, nonlinear Schrödinger, mKdV, sine-Gordon, Toda lattice, discrete Ablowitz–Ladik equations, and so on [3,6,9,10,12,13, 16,17,20,21,25–28,32]. However, for the case of 3 × 3, it is so difficult and complicated because of concerning the theory for trigonal curves [5,7,8,14,35] rather than hyperelliptic ones as for the 2 × 2 matrix spectral problems. Only a few of literature [2,29–31,36–39] studied algebro-geometric solutions of the Boussinesq equation related to a third-order differential operator by the reduction theory of Riemann theta functions. In Ref. [11], a unified framework is proposed which yields all quasi-periodic solutions of the entire Boussinesq hierarchy. On the basis of that, we give a systematical method to define the trigonal curve and the meromorphic functions on it, from which the algebro-geometric solutions of the entire soliton hierarchies are obtained by means of the algebro-geometric characteristic of the trigonal curve, the meromorphic functions and the Baker–Akhiezer function [18,19]. In the above known literature, there is only an infinite point on the three-sheeted Riemann surface, which is a branch point with the triple root. The present paper is devoted to study the trigonal curve with three infinite points, that are not branch points on the corresponding three-sheeted Riemann surface, from which algebro-geometric solutions of the cmKdV hierarchy associated with a 3 ×3 matrix

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spectral problem are constructed. So such an exploration has great significance both in theory and application. The outline of this paper is as follows. In Section 2, by introducing two sets of Lenard recursion equations, we derive the cmKdV hierarchy associated with the 3 × 3 matrix spectral problem. In Section 3, we introduce the Baker–Akhiezer functions, the meromorphic functions, and the trigonal curve Km−2 of arithmetic genus m − 2 with the help of the characteristic polynomial of Lax matrix for the cmKdV hierarchy. We list out some properties of the meromorphic functions φ2 and φ3 related to the Baker–Akhiezer function and obtain the Dubrovin-type equations. In Section 4, we give out the essential properties of φ2 , φ3 , and the Baker–Akhiezer function ψ1 , that concerns their asymptotic expansions and divisors. The last section is devoted to construct the algebro-geometric representation for ψ1 and solutions of the cmKdV hierarchy in terms of the Riemann theta function by employing three kinds of Abelian differentials. 2. The coupled mKdV hierarchy In this section, we shall derive the coupled mKdV hierarchy associated with a 3 × 3 matrix spectral problem ⎛

ψx = U ψ,

⎞ ψ1 ψ = ⎝ ψ2 ⎠ , ψ3

⎛

−λ u U =⎝ u λ v 0

⎞ v 0⎠, λ

(2.1)

where u, v are two potentials, and λ a constant spectral parameter. To this end, we first solve the stationary zero-curvature equation: Vx − [U, V ] = 0,

V = (Vij )3×3 ,

(2.2)

which is equivalent to V11,x + u(V12 − V21 ) + v(V13 − V31 ) = 0, V12,x + 2λV12 + u(V11 − V22 ) − vV32 = 0, V13,x + 2λV13 + v(V11 − V33 ) − uV23 = 0, V21,x − 2λV21 + u(V22 − V11 ) + vV23 = 0, V22,x + u(V21 − V12 ) = 0, V23,x + vV21 − uV13 = 0, V31,x − 2λV31 + v(V33 − V11 ) + uV32 = 0, V32,x + uV31 − vV12 = 0, V33,x + v(V31 − V13 ) = 0. In order to solve Eq. (2.2), we introduce the Lenard recursion equations:

(2.3)

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K11 hj + K12 gj = J11 hj+1 ,

K21 hj + K22 gj = 0,

j ≥ 0,

ˆ j + K12 gˆj = J11 h ˆ j+1 , K11 h

ˆ j + K22 gˆj = 0, K21 h

j ≥ 0,

(2.4)

with the starting points ⎛

  0 ˆ h0 = h0 = , 0

⎜ g0 = ⎜ ⎝

⎞

0 0 1

⎛

⎟ ⎟, ⎠

⎜ gˆ0 = ⎜ ⎝

−∂ −1 uv

⎞

0 1 0

⎟ ⎟, ⎠

(2.5)

∂ −1 uv

where ∂ −1 ∂ = ∂∂ −1 = 1 and the five operators are defined as  2    ∂ 0 ∂v −∂u 0 v K11 = , K12 = , 0 ∂2 ∂u 0 −∂v −u ⎛ ⎛ ⎞ v u ∂ 0 ⎜ −4u −2v ⎟ ⎜ 0 ∂ ⎟ K21 = ⎜ K22 = ⎜ ⎝ −2u −4v ⎠ , ⎝ 0 0 v∂ −u∂ v 2 − u2 −uv

 J11 = 4 0 0 ∂ uv

⎞ 0 0⎟ ⎟. 0⎠ ∂

1 0 0 1

 ,

(2.6)

ˆ j |(u,v)=0 = gj |(u,v)=0 = Here we give some accessional conditions, hj |(u,v)=0 = h ˆ j , gj , gˆj can be uniquely determined. It can gˆj |(u,v)=0 = 0, j ≥ 1, to insure that hj , h be calculated out that the first set of members read as     −1 1 −v∂ −1 uv ˆ 1 = 1 −ux + v∂ uv , h1 = , h −u∂ −1 uv 4 −vx + u∂ −1 uv 4 ⎛ g1 =

1⎜ ⎜ ⎜ 4⎝

∂ −1 uvx − ∂ −1 (u2 − v 2 )∂ −1 uv −v 2 − 2∂ −1 uv∂ −1 uv −2v + 2∂ 2

−1

uv∂

−1

uv

⎞ ⎟ ⎟ ⎟, ⎠

(4) g1

⎛ gˆ1 =

1⎜ ⎜ ⎜ 4⎝

∂ −1 ux v + ∂ −1 (u2 − v 2 )∂ −1 uv −2u2 + 2∂ −1 uv∂ −1 uv −u2 − 2∂ −1 uv∂ −1 uv

⎞ ⎟ ⎟ ⎟, ⎠

(4)

gˆ1 where

  1 3 1

(4) g1 = −∂ −1 uvxx + ∂ −1 u3 v + ∂ −1 uv 3 + ∂ −1 u2 − v 2 ∂ −1 uvx + u2 + v 2 ∂ −1 uv 2 2 2



 − 4∂ −1 uv∂ −1 uv∂ −1 uv − ∂ −1 u2 − v 2 ∂ −1 u2 − v 2 ∂ −1 uv,

  3 1 1

(4) gˆ1 = ∂ −1 uxx v − ∂ −1 u3 v − ∂ −1 uv 3 + ∂ −1 u2 − v 2 ∂ −1 ux v − u2 + v 2 ∂ −1 uv 2 2 2



 + 4∂ −1 uv∂ −1 uv∂ −1 uv + ∂ −1 u2 − v 2 ∂ −1 u2 − v 2 ∂ −1 uv.

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Now we define each entry Vij of the 3 × 3 matrix V as V11 =

2 λ(d + e), 3

V12 = ax + vc − ud − 2λa,

V13 = bx + uc − ve − 2λb,

2 λ(e − 2d), 3

V21 = ax + vc − ud + 2λa,

V22 =

V31 = bx + uc − ve + 2λb,

V32 = 2λc − f,

V23 = 2λc + f, V33 =

2 λ(d − 2e). 3

(2.7)

Substituting (2.7) into (2.3) yields the Lenard equations K11 H + K12 G = λ2 J11 H,

K21 H + K22 G = 0,

(2.8)

where H = (a, b)T , G = (c, d, e, f )T . Expanding H and G into the Laurent polynomials in λ:  aj  −2j λ−2j , Hj λ = H= bj j≥0

G=



j≥0

Gj λ−2j

j≥0

⎞ cj ⎜ dj ⎟ ⎜ ⎟ λ−2j , = ⎝ ej ⎠ j≥0

⎛

(2.9)

fj

(2.8) is equivalent to the following recursion equations K11 Hj + K12 Gj = J11 Hj+1 ,

K21 Hj + K22 Gj = 0,

J11 H0 = 0,

j ≥ 0. (2.10)

Since K22 G0 = 0 has a solution G0 = β0 g0 + δ0 gˆ0 ,

(2.11)

Eqs. (2.4) and (2.10) imply that functions j ˆ j−l ), Hj = (βl hj−l + δl h l=0

j Gj = (βl gj−l + δl gˆj−l )

(2.12)

l=0

satisfy Eqs. (2.10), where βl and δl are arbitrary constants. Let ψ satisfy the spectral problem (2.1) and the auxiliary problem ψtr = V (r) ψ,

(r)  V (r) = V ij 3×3 ,

(r) with each entry V ij = Vij (˜ a(r) , ˜b(r) , c˜(r) , d˜(r) , e˜(r) , f˜(r) ),

(2.13)

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⎛

⎞ ⎛ ⎞ c˜j c˜(r) r ⎜ d˜(r) ⎟ ⎜ d˜j ⎟ 2r−2j ⎜ ⎟ ⎜ ⎟λ , ⎝ e˜(r) ⎠ = ⎝ e˜j ⎠ j≥0 f˜(r) f˜j

  r  a ˜(r) a ˜j 2r−2j , ˜b(r) = ˜bj λ j≥0

(2.14)

where  ⎛

a ˜j ˜bj

 ⎞

j = =H

j ˆ j−l ), (β˜l hj−l + δ˜l h l=0

c˜j j ⎜ d˜j ⎟

j = ⎜ ⎟=G (β˜l gj−l + δ˜l gˆj−l ), ⎝ e˜j ⎠ l=0 f˜j

(2.15)

and β˜l , δ˜l are constants independent of βl , δl . Then the compatibility condition of (2.1) (r) and (2.13) yields the zero-curvature equation, Utr − V x + [U, V (r) ] = 0, which is equivalent to a hierarchy of nonlinear evolution equations   u

r + K12 G

r = J11 H

r+1 . = K11 H v tr

(2.16)

The first member in the hierarchy for r = 0 is as follows

 ut0 = −β˜0 v∂ −1 uv − δ˜0 ux − v∂ −1 uv , 

vt0 = −β˜0 vx − u∂ −1 uv − δ˜0 u∂ −1 uv.

(2.17)

For r = 1, the second member in the hierarchy (2.16) reads as  1 3 1 1 1 ut1 = (β˜0 − δ˜0 ) − ∂ 2 v∂ −1 uv + v∂ −1 u3 v + v∂ −1 uv 3 − ∂v∂ −1 ux v − v∂ −1 uxx v 4 8 8 4 4



  1 1 1 − ∂v∂ −1 u2 − v 2 ∂ −1 uv + ∂u∂ −1 uv∂ −1 uv + v u2 + v 2 ∂ −1 uv 4 2 8



 1 − v∂ −1 uv∂ −1 uv∂ −1 uv − v∂ −1 u2 − v 2 ∂ −1 u2 − v 2 ∂ −1 uv 4       −1 1 −1 2 3 1 3 3 2 2 2 ˜ ˜ − v∂ ux v + uvvx + δ0 − uxxx + u ux u − v ∂ ux v + β0 4 4 4 4 2

 − β˜1 v∂ −1 uv − δ˜1 ux − v∂ −1 uv ,

 1 3 1 1 1 vt1 = (δ˜0 − β˜0 ) − ∂ 2 u∂ −1 uv + u∂ −1 uv 3 + u∂ −1 u3 v − ∂u∂ −1 uvx − u∂ −1 uvxx 4 8 8 4 4

  1 1 1

− ∂u∂ −1 v 2 − u2 ∂ −1 uv + ∂v∂ −1 uv∂ −1 uv + u u2 + v 2 ∂ −1 uv 4 2 8



 1 − u∂ −1 uv∂ −1 uv∂ −1 uv − u∂ −1 u2 − v 2 ∂ −1 u2 − v 2 ∂ −1 uv 4

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 3 2 1 1 3 3 u vx + uvux + β˜0 − vxxx + v 2 vx − u∂ −1 v 2 − u2 ∂ −1 uvx + δ˜0 4 4 4 4 2

 − β˜1 vx − u∂ −1 uv − δ˜1 u∂ −1 uv, (2.18) which is reduced to the cmKdV equation (1.1) by choosing β˜1 = δ˜1 = 0, β˜0 = δ˜0 = 4. 3. The Baker–Akhiezer function and the Dubrovin-type equations In this section, we first introduce the Baker–Akhiezer function, a trigonal curve Km−2 of degree m, and two attendant meromorphic functions on it. Then the coupled mKdV hierarchy is decomposed into systems of solvable ordinary equations. Now we introduce the Baker–Akhiezer function ψ(P, x, x0 , tr , t0,r ) by

 ψx (P, x, x0 , tr , t0,r ) = U u(x, tr ), v(x, tr ); λ(P ) ψ(P, x, x0 , tr , t0,r ),

 ψtr (P, x, x0 , tr , t0,r ) = V (r) u(x, tr ), v(x, tr ); λ(P ) ψ(P, x, x0 , tr , t0,r ), 

V (n) u(x, tr ), v(x, tr ); λ(P ) ψ(P, x, x0 , tr , t0,r ) = y(P )ψ(P, x, x0 , tr , t0,r ), ψ1 (P, x0 , x0 , t0,r , t0,r ) = 1,

x, tr ∈ C,

(3.1)

n (n) (n) where V (n) = (λ2n V )+ = (Vij )3×3 , and Vij = k=0 Vij,k λ2n−2k . The compatibility conditions of the first three expressions in (3.1) yield that   Utr − V x(r) + U, V (r) = 0,   −Vx(n) + U, V (n) = 0,   (n) −Vtr + V (r) , V (n) = 0.

(3.2) (3.3) (3.4)

A direct calculation shows that yI − V (n) also satisfies the Lax equations (3.3) and (3.4), which implies that the characteristic polynomial Fm (λ, y) = det(yI − V (n) ) of the Lax matrix V (n) is a constant independent of variables x and tr with the expansion 

det yI − V (n) = y 3 + ySm (λ) − Tm (λ), where Sm and Tm are polynomials with constant coefficients of λ Sm

Tm

 (n) V  ii =  (n)  1≤i
 (n) Vij  (n)  V jj

  4 

= β0 δ0 − β02 − δ02 λ4n+2 + (β0 δ1 + β1 δ0 − 2β0 β1 − 2δ0 δ1 )λ4n + · · · , 3  (n) (n) (n)   V11 V12 V13    (n) (n) (n)  =  V21 V22 V23   (n)  (n) (n)  V V V 31

32

33

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130

 =λ +

8 (β0 + δ0 )(β0 − 2δ0 )(δ0 − 2β0 )λ6n+2 27

  8 2β0 δ0 (β1 + δ1 ) + δ02 (β1 − 2δ1 ) + β02 (δ1 − 2β1 ) λ6n + · · · . 9

(3.5)

Then one naturally leads to the trigonal curve Km−2 of degree m by Km−2 :

Fm (λ, y) = y 3 + ySm (λ) − Tm (λ) = 0,

(3.6)

where m = 6n+3 for β0 +δ0 = 0, β0 −2δ0 = 0, and δ0 −2β0 = 0. Assume β0 δ0 (β0 −δ0 ) = 0, then the discriminant of the local equation at infinity is non-zero. This implies that the curve has three different infinite points P∞1 , P∞2 and P∞3 , which are not the branch points. The compactification of the curve Km−2 is still denoted by the same symbol for the sake of convenience. Then Km−2 becomes a three-sheeted Riemann surface of arithmetic genus m − 2 if it is nonsingular or smooth. Here Km−2 is nonsingular means Fm ∂ Fm that for every point P ∈ Km−2 , ( ∂∂λ , ∂y ) = 0. Points P on Km−2 are represented as pairs P = (λ, y) satisfying Fm (λ, y) = 0 together with P∞j , j = 1, 2, 3, the points at infinity. The complex structure on Km−2 is defined by introducing local coordinates ζP0 : P → (λ − λ0 ) near points P0 ∈ Km−2 which are neither branch nor infinite points of Km−2 , ζP∞j : P → 1/λ near the infinite points P∞j ∈ Km−2 , j = 1, 2, 3, ζP0 : P → (λ − λ0 )1/3 near points P0 ∈ Km−2 which are triple points of Km−2 , and similarly at other branch points of Km−2 . For convenience, we define three points P, P ∗ , P ∗∗ on three different sheets of the same Riemann surface Km−2 . For a fixed λ, let yi (λ), i = 0, 1, 2, denote the three roots of polynomial Fm (λ, y) = 0, that is





 y − y0 (λ) y − y1 (λ) y − y2 (λ) = y 3 + ySm − Tm = 0.

(3.7)

Then points (λ, y0 (λ)), (λ, y1 (λ)) and (λ, y2 (λ)) are on the three different sheets of Riemann surface Km−2 , respectively. Let P = (λ, yi (λ)), i = 0, 1, 2, be an arbitrary point in the three points, then the other two points are defined as P ∗ and P ∗∗ , respectively. From (3.7), we can easily get y0 + y1 + y2 = 0, y0 y1 + y0 y2 + y1 y2 = Sm , y0 y1 y2 = Tm , y02 + y12 + y22 = −2Sm , y03 + y13 + y23 = 3Tm , 2 y02 y12 + y02 y22 + y12 y22 = Sm .

(3.8)

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Closely related to ψ(P, x, x0 , tr , t0,r ) are the following two meromorphic functions φ2 (P, x, tr ) and φ3 (P, x, tr ) on Km−2 defined by φ2 (P, x, tr ) =

ψ2 (P, x, x0 , tr , t0,r ) , ψ1 (P, x, x0 , tr , t0,r )

P ∈ Km−2 , x, tr ∈ C,

(3.9)

φ3 (P, x, tr ) =

ψ3 (P, x, x0 , tr , t0,r ) , ψ1 (P, x, x0 , tr , t0,r )

P ∈ Km−2 , x, tr ∈ C.

(3.10)

Using (3.1), a direct calculation shows that (n)

φ2 =

yV23 + Cm (n)

=

yV13 + Am

Fm−2 (n)

y 2 V23 − yCm + Dm

(n)

φ3 =

yV32 + Cm (n)

yV12 + Am

=

Fm−2 (n)

y 2 V32 − yCm + Dm

(n)

=

y 2 V13 − yAm + Bm , Em−2

=

y 2 V12 − yAm + Bm , −Em−2

(3.11)

(n)

(3.12)

where (n)

(n)

(n)

(n)

(n)

(n)

(n)

(n)

Am = V12 V23 − V13 V22 , (n) (n) (n) (n) (n)  (n) (n) (n) (n) (n)  Bm = V13 V11 V33 − V13 V31 + V12 V11 V23 − V13 V21 , Cm = V13 V21 − V11 V23 , (n) (n) (n) (n) (n)  (n) (n) (n) (n) (n)  Dm = V23 V22 V33 − V23 V32 + V21 V13 V22 − V12 V23 , (n)

(n)

(n)

(n)

(n)

(n)

(n)

(n)

(3.13)

Am = V13 V32 − V12 V33 , (n) (n) (n) (n) (n)  (n) (n) (n) (n) (n)  Bm = V12 V11 V22 − V12 V21 + V13 V11 V32 − V12 V31 , Cm = V12 V31 − V11 V32 , (n) (n) (n) (n) (n)  (n) (n) (n) (n) (n)  Dm = V32 V22 V33 − V23 V32 + V31 V12 V33 − V13 V32 ,

(3.14)

(n) 2 (n)

(n) 2 (n) (n) (n) (n) (n)  V32 + V12 V13 V22 − V33 − V12 V23 , Em−2 = V13

(n) 2 (n)



(n) (n) (n) (n) (n) (n) 2 V31 + V21 V23 V11 − V33 − V13 V21 , Fm−2 = V23





 (n) (n) 2 (n) (n) (n) (n) (n) (n) 2 + V31 V32 V11 − V22 − V12 V31 . Fm−2 = V21 V32

(3.15)

We can easily show that there exist various interrelationships among the polynomials Am , Bm , Cm , Dm , Am , Bm , Cm , Dm , Em−2 , Fm−2 , Fm−2 , Sm , Tm , some of which are listed below:

(n) 2 (n) (n) 2 V13 Fm−2 = V23 Dm − V23 Sm − Cm ,

(n) 2 Tm + Cm Dm , Am Fm−2 = V23

(3.16)

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(n) 2 (n) (n) V23 Em−2 = V13 Bm − V13 Sm − A2m ,

(n) 2 Cm Em−2 = V13 Tm + Am Bm , (n)

(n)

(n)

(3.17)

(n)

V23 Bm + V13 Dm − V13 V23 Sm + Am Cm = 0, (n)

(n)

(n)

(n)

V13 V23 Tm + V23 Am Sm + V13 Cm Sm − Bm Cm − Am Dm = 0, (n)

(n)

V23 Am Tm + V13 Cm Tm + Em−2 Fm−2 − Bm Dm = 0,

(n)

(3.18)

(n) 2 (n) (n) 2 V12 Fm−2 = V32 Dm − V32 S m − Cm ,

(n) 2 Am Fm−2 = V32 T m + Cm D m ,

(3.19)

(n) 2 (n) (n) −V32 Em−2 = V12 Bm − V12 Sm − A2m ,

(n) 2 −Cm Em−2 = V12 Tm + Am Bm ,

(3.20)

(n)

(n)

(n)

V32 Bm + V12 Dm − V12 V32 Sm + Am Cm = 0, (n)

(n)

(n)

(n)

V12 V32 Tm + V32 Am Sm + V12 Cm Sm − Bm Cm − Am Dm = 0, (n)

(n)

V32 Am Tm + V12 Cm Tm − Em−2 Fm−2 − Bm Dm = 0,

(n) 

(n)  Em−2,x = −4λEm−2 − u 2V13 Sm − 3Bm + v 2V12 Sm − 3Bm ,

(n)  Fm−2,x = 2λFm−2 − u 2V23 Sm − 3Dm ,

(n)  Fm−2,x = 2λFm−2 − v 2V32 Sm − 3Dm .

(3.21)

(3.22)

Then we have some further properties of φ2 (P, x, tr ) and φ3 (P, x, tr ) summarized in the following lemma. Lemma 3.1. Assume that (3.1), (3.9) and (3.10) hold and let P = (λ, y(P )) ∈ Km−2 \ {P∞1 , P∞2 , P∞3 }, and (λ, x, tr , ) ∈ C3 . Then φ2,x (P, x, tr ) + u(x, tr )φ22 (P, x, tr ) + v(x, tr )φ2 (P, x, tr )φ3 (P, x, tr ) − 2λφ2 (P, x, tr ) − u(x, tr ) = 0,

(3.23)

φ3,x (P, x, tr ) + v(x, tr )φ23 (P, x, tr ) + u(x, tr )φ2 (P, x, tr )φ3 (P, x, tr ) − 2λφ3 (P, x, tr ) − v(x, tr ) = 0,

(3.24)

(r)  (r) (r) φ2,tr (P, x, tr ) = V 21 (x, tr ) + V 22 (x, tr ) − V 11 (x, tr ) φ2 (P, x, tr ) + V 23 (x, tr )φ3 (P, x, tr ) − V 12 (x, tr )φ22 (P, x, tr ) (r)

(r)

(r) − V 13 (x, tr )φ2 (P, x, tr )φ3 (P, x, tr ),

(3.25)

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133

(r)  (r) (r) φ3,tr (P, x, tr ) = V 31 (x, tr ) + V 33 (x, tr ) − V 11 (x, tr ) φ3 (P, x, tr ) (r) (r) + V 32 (x, tr )φ2 (P, x, tr ) − V 13 (x, tr )φ23 (P, x, tr ) (r) − V 12 (x, tr )φ2 (P, x, tr )φ3 (P, x, tr ),

(3.26)



 Fm−2 (λ, x, tr ) φ2 (P, x, tr )φ2 P ∗ , x, tr φ2 P ∗∗ , x, tr = − , Em−2 (λ, x, tr )

(3.27)



 Fm−2 (λ, x, tr ) φ3 (P, x, tr )φ3 P ∗ , x, tr φ3 P ∗∗ , x, tr = , Em−2 (λ, x, tr )

(3.28)





  (n) (n) V12 (x, tr ) φ2 (P, x, tr ) + φ2 P ∗ , x, tr + φ2 P ∗∗ , x, tr + V13 (x, tr ) φ3 (P, x, tr )



 (n) + φ3 P ∗ , x, tr + φ3 P ∗∗ , x, tr = −3V11 (x, tr ), (3.29) 



  u(x, tr ) φ2 (P, x, tr ) + φ2 P ∗ , x, tr + φ2 P ∗∗ , x, tr + v(x, tr ) φ3 (P, x, tr )



 Em−2,x (λ, x, tr ) + φ3 P ∗ , x, tr + φ3 P ∗∗ , x, tr = + 4λ, Em−2 (λ, x, tr )

(3.30)



 φ2 (P, x, tr ) + φ2 P ∗ , x, tr + φ2 P ∗∗ , x, tr (n)

=−

2V13 (λ, x, tr )Sm (λ) − 3Bm (λ, x, tr ) Em−2 (λ, x, tr ) (n)

V13 (λ, x, tr ){ =

Em−2,x (λ,x,tr ) Em−2 (λ,x,tr )

(n)

(λ,x,tr ) + 4λ + 3v(x, tr ) V11 } (n) V13 (λ,x,tr )

(n)

(n)

u(x, tr )V13 (λ, x, tr ) − v(x, tr )V12 (λ, x, tr )

,

(3.31)



 φ3 (P, x, tr ) + φ3 P ∗ , x, tr + φ3 P ∗∗ , x, tr (n)

=

2V12 (λ, x, tr )Sm (λ) − 3Bm (λ, x, tr ) Em−2 (λ, x, tr ) (n)

=−

V12 (λ, x, tr ){

Em−2,x (λ,x,tr ) Em−2 (λ,x,tr )

(n) u(x, tr )V13 (λ, x, tr )

(n)

(λ,x,tr ) + 4λ + 3u(x, tr ) V11 } (n)

−

V12 (λ,x,tr ) (n) v(x, tr )V12 (λ, x, tr )

,

(3.32)

1 1 1 Fm−2,x (λ, x, tr ) − 2λFm−2 (λ, x, tr ) + + = , ∗ ∗∗ φ2 (P, x, tr ) φ2 (P , x, tr ) φ2 (P , x, tr ) u(x, tr )Fm−2 (λ, x, tr ) (3.33) 1 1 1 Fm−2,x (λ, x, tr ) − 2λFm−2 (λ, x, tr ) + + = , ∗ ∗∗ φ3 (P, x, tr ) φ3 (P , x, tr ) φ3 (P , x, tr ) v(x, tr )Fm−2 (λ, x, tr ) (3.34)

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134

φ2 (P, x, tr ) φ2 (P ∗ , x, tr ) φ2 (P ∗∗ , x, tr ) + + φ3 (P, x, tr ) φ3 (P ∗ , x, tr ) φ3 (P ∗∗ , x, tr ) (n)

=

(n)

(n)

−V31 (λ, x, tr )Fm−2,x (λ, x, tr ) + (2λV31 (λ, x, tr ) − 3vV33 (λ, x, tr ))Fm−2 (λ, x, tr ) (n)

,

vV32 (λ, x, tr )Fm−2 (λ, x, tr ) (3.35) φ3 (P, x, tr ) φ3 (P ∗ , x, tr ) φ3 (P ∗∗ , x, tr ) + + φ2 (P, x, tr ) φ2 (P ∗ , x, tr ) φ2 (P ∗∗ , x, tr ) (n)

=

(n)

(n)

−V21 (λ, x, tr )Fm−2,x (λ, x, tr ) + (2λV21 (λ, x, tr ) − 3uV22 (λ, x, tr ))Fm−2 (λ, x, tr ) (n)

.

uV23 (λ, x, tr )Fm−2 (λ, x, tr ) (3.36) Lemma 3.2. Assume that (3.1) holds and let (λ, x, tr ) ∈ C3 . Then Em−2,tr (λ, x, tr )

  (r) (r)

(n) (r)

(n) = 3V 11 Em−2 − V 12 2V13 Sm − 3Bm + V 13 2V12 Sm − 3Bm

= Em−2,x (λ, x, tr )

(n) (r) (n) (r) V13 V 12 − V12 V 13 (n)

(n)

uV13 − vV12

   (r) (r) (n) (r) (n) (r)  V13 V 12 − V12 V 13 uV 13 − v V 12 (r) (n) + 4λ , + Em−2 3 V 11 − V (n) (n) 11 (n) (n) uV13 − vV12 uV13 − vV12 Fm−2,tr (λ, x, tr ) = Fm−2,x (λ, x, tr )

(n) (r) (n) (r) V23 V 21 − V21 V 23 (n)

uV23

   (r) (n) (r) (n) (r)  V21 V 23 − V23 V 21 V (r) (n) + 2λ , + Fm−2 (λ, x, tr ) 3 V 22 − 23 V (n) 22 (n) V23 uV23 Fm−2,tr (λ, x, tr ) = Fm−2,x (λ, x, tr )

(n) (r) (n) (r) V32 V 31 − V31 V 32 (n)

vV32

   (r) (n) (r) (n) (r)  V31 V 32 − V32 V 31 V 32 (n) (r)

+ 2λ . + Fm−2 (λ, x, tr ) 3 V33 − (n) V33 (n) V32 vV32

(3.37)

Proof. The first two expressions in (3.1) imply that ψ1,x (P, x, x0 , tr , tr,0 ) = −λ + uφ2 (P, x, tr ) + vφ3 (P, x, tr ), ψ1 (P, x, x0 , tr , tr,0 )

(3.38)

ψ1,tr (P, x, x0 , tr , tr,0 ) (r) (r) (r) = V 11 + V 12 φ2 (P, x, tr ) + V 13 φ3 (P, x, tr ). ψ1 (P, x, x0 , tr , tr,0 )

(3.39)

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135

The compatibility condition of (3.38) and (3.39) means that 

(r) 

(r) (r) uφ2 (P, x, tr ) + vφ3 (P, x, tr ) t = V 11 + V 12 φ2 (P, x, tr ) + V 13 φ3 (P, x, tr ) x . (3.40) r

Differentiating (3.30) with respect to tr , we can get that 

Em−2,x Em−2

 tr





 = ∂x ∂tr (ln Em−2 ) = u φ2 + φ∗2 + φ∗∗ + v φ3 + φ∗3 + φ∗∗ 2 3 t

r

 (r)   (r)

(r)

= 3V 11 + V 12 φ2 + φ∗2 + φ∗∗ , + V 13 φ3 + φ∗3 + φ∗∗ 2 3 x

∗ ∗∗ here we take φ∗i and φ∗∗ i as an abbreviation for φi (P , x, tr ) and φi (P , x, tr ), i = 2, 3, respectively. Without loss of generality, we take the integration constant to be zero. This implies that

  (r) (r)

(r)

+ V 13 φ3 + φ∗3 + φ∗∗ ∂tr (ln Em−2 ) = 3V 11 + V 12 φ2 + φ∗2 + φ∗∗ 2 3 , which naturally yields the first expression in (3.37) by making use of (3.31) and (3.32). Differentiating (3.27) with respect to tr and using (3.25) and (3.27), we can deduce     φ∗2,tr φ∗∗ Fm−2 2,tr ∗ ∗∗ φ2,tr = φ2 φ2 φ2 + ∗ + ∗∗ − Em−2 tr φ2 φ2 φ2    Fm−2 (r) (r)  (r)

(r)

3 V22 − V11 − V12 φ2 + φ∗2 + φ∗∗ − V 13 φ3 + φ∗3 + φ∗∗ =− 2 3 Em−2     1 1 φ∗ φ∗∗ 1 (r) (r) φ3 . (3.41) + ∗ + ∗∗ + V 23 + 3∗ + 3∗∗ + V 21 φ2 φ2 φ2 φ2 φ2 φ2 Since   Em−2,tr (r) (r)

(r)

+ V 13 φ3 + φ∗3 + φ∗∗ = 3V 11 + V 12 φ2 + φ∗2 + φ∗∗ 2 3 , Em−2 we can prove the second expression of (3.37) by making use of (3.33) and (3.36). The third expression can be proved similarly. 2 Now we turn to consider Em−2 (λ, x, tr ), Fm−2 (λ, x, tr ), Fm−2 (λ, x, tr ) which can be inferred to be polynomial with respect to λ of degree m − 2 from (3.15) for β0 δ0 (β0 − δ0 ) = 0. Thus we can rewrite them in the following form: Em−2 (λ, x, tr ) = 2β0 δ0 (β0 − δ0 )uv

m−2 

 λ − μj (x, tr ) ,

(3.42)

j=1

Fm−2 (λ, x, tr ) = 2β0 δ0 (β0 − δ0 )u∂ −1 uv

m−2  j=1



 λ − νj (x, tr ) ,

(3.43)

136

X. Geng et al. / Advances in Mathematics 263 (2014) 123–153

Fm−2 (λ, x, tr ) = −2β0 δ0 (β0 − δ0 )v∂ −1 uv

m−2 

 λ − ξj (x, tr ) ,

(3.44)

j=1 m−2 m−2 are zeros of Em−2 (λ, x, tr ), where {μj (x, tr )}m−2 j=1 , {νj (x, tr )}j=1 , {ξj (x, tr )}j=1 Fm−2 (λ, x, tr ), Fm−2 (λ, x, tr ) respectively. Since

Am (μj (x, tr ), x, tr )

=

(n) V12 (μj (x, tr ), x, tr )

Am (μj (x, tr ), x, tr ) (n)

V13 (μj (x, tr ), x, tr )

which can be deduced from  (n) 2 (n)

(n) 2 (n)  (n) (n) (n) (n)  V13 V32 + V12 V13 V22 − V33 − V12 V23 λ=μ (x,t ) j r  (n)



 = V13 μj (x, tr ), x, tr Am μj (x, tr ), x, tr 

 (n)

− V12 μj (x, tr ), x, tr Am μj (x, tr ), x, tr = 0,

Em−2 |λ=μj (x,tr ) =

we can define the following points [11,18]   Am (μj (x, tr ), x, tr ) μj (x, tr ), − (n) V13 (μj (x, tr ), x, tr )   Am (μj (x, tr ), x, tr ) ∈ Km−2 , = μj (x, tr ), − (n) V12 (μj (x, tr ), x, tr )



 μ ˆj (x, tr ) = μj (x, tr ), y μj (x, tr ) =

(3.45)



 νˆj (x, tr ) = νj (x, tr ), y νj (x, tr ) =

 νj (x, tr ), −

Cm (νj (x, tr ), x, tr ) (n)

V23 (νj (x, tr ), x, tr )

 ∈ Km−2 , (3.46)



 ξˆj (x, tr ) = ξj (x, tr ), y ξj (x, tr ) = 1 ≤ j ≤ m − 2, (x, tr ) ∈ C2 .

  Cm (ξj (x, tr ), x, tr ) ξj (x, tr ), − (n) ∈ Km−2 , V32 (ξj (x, tr ), x, tr ) (3.47)

The dynamics of the zeros μj (x, tr ), νj (x, tr ) and ξj (x, tr ) of Em−2 (λ, x, tr ), Fm−2 (λ, x, tr ) and Fm−2 (λ, x, tr ) are then described in terms of Dubrovin-type equations in the following lemma. Lemma 3.3. Suppose that the zeros {μj (x, tr )}j=1,...,m−2 , {νj (x, tr )}j=1,...,m−2 and {ξj (x, tr )}j=1,...,m−2 of Em−2 (λ, x, tr ), Fm−2 (λ, x, tr ) and Fm−2 (λ, x, tr ) remain distinct for (x, tr ) ∈ Ωμ , (x, tr ) ∈ Ων and (x, tr ) ∈ Ωξ , respectively, where Ωμ , Ων , Ωξ ⊆ C2 are open and connected. Then {μj (x, tr )}j=1,...,m−2 , {νj (x, tr )}j=1,...,m−2 and {ξj (x, tr )}j=1,...,m−2 satisfy the system of differential equations

X. Geng et al. / Advances in Mathematics 263 (2014) 123–153 (n)

μj,x (x, tr ) =

137

(n)

[v(x, tr )V12 (μj (x, tr ), x, tr ) − u(x, tr )V13 (μj (x, tr ), x, tr )] 2β0 δ0 (β0 − δ0 )u(x, tr )v(x, tr ) ×

[3y 2 (μj (x, tr )) + Sm (μj (x, tr ))] , m−2 k=1 (μj (x, tr ) − μk (x, tr ))

1 ≤ j ≤ m − 2,

(3.48)

k=j

[V 13 (μj (x, tr ), x, tr )V12 (μj (x, tr ), x, tr ) − V 12 (μj (x, tr ), x, tr )V13 (μj (x, tr ), x, tr )] 2β0 δ0 (β0 − δ0 )u(x, tr )v(x, tr ) (r)

μj,tr (x, tr ) =

(n)

[3y 2 (μj (x, tr )) + Sm (μj (x, tr ))] , m−2 k=1 (μj (x, tr ) − μk (x, tr ))

×

(r)

(n)

1 ≤ j ≤ m − 2,

(3.49)

k=j

(n)

νj,x (x, tr ) =

[3y 2 (νj (x, tr )) + Sm (νj (x, tr ))] −V23 (νj (x, tr ), x, tr ) × , m−2 2β0 δ0 (β0 − δ0 )∂ −1 u(x, tr )v(x, tr ) k=1 (νj (x, tr ) − νk (x, tr )) k=j

1 ≤ j ≤ m − 2,

(3.50)

[V 23 (νj (x, tr ), x, tr )V21 (νj (x, tr ), x, tr ) − V23 (νj (x, tr ), x, tr )V 21 (νj (x, tr ), x, tr )] 2β0 δ0 (β0 − δ0 )u(x, tr )∂ −1 u(x, tr )v(x, tr ) (r)

νj,tr (x, tr ) =

×

(n)

[3y 2 (νj (x, tr )) + Sm (νj (x, tr ))] , m−2 k=1 (νj (x, tr ) − νk (x, tr ))

(n)

(r)

1 ≤ j ≤ m − 2,

(3.51)

k=j

(n)

ξj,x (x, tr ) =

[3y 2 (ξj (x, tr )) + Sm (ξj (x, tr ))] V32 (ξj (x, tr ), x, tr ) × m−2 , −1 2β0 δ0 (β0 − δ0 )∂ u(x, tr )v(x, tr ) k=1 (ξj (x, tr ) − ξk (x, tr )) k=j

1 ≤ j ≤ m − 2, ξj,tr (x, tr ) =

(3.52)

(r) (n) (n) (r) [V 31 (ξj (x, tr ), x, tr )V32 (ξj (x, tr ), x, tr ) − V31 (ξj (x, tr ), x, tr )V 32 (ξj (x, tr ), x, tr )] 2β0 δ0 (β0 − δ0 )v(x, tr )∂ −1 u(x, tr )v(x, tr )

×

[3y 2 (ξj (x, tr )) + Sm (ξj (x, tr ))] , m−2 k=1 (ξj (x, tr ) − ξk (x, tr ))

1 ≤ j ≤ m − 2.

(3.53)

k=j

Proof. From the first expression of (3.17) and (3.20), we have

 



 (n)

Bm μj (x, tr ), x, tr = V13 μj (x, tr ), x, tr y 2 μj (x, tr ) + Sm μj (x, tr ) ,

 



 (n)

Bm μj (x, tr ), x, tr = V12 μj (x, tr ), x, tr y 2 μj (x, tr ) + Sm μj (x, tr ) .

(3.54)

Inserting (3.54) into the first equation of (3.22) can naturally yields

    (n)

(n)

Em−2,x μj (x, tr ), x, tr = u(x, tr )V13 μj (x, tr ), x, tr − v(x, tr )V12 μj (x, tr ), x, tr 





(3.55) × 3y 2 μj (x, tr ) + Sm μj (x, tr ) .

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On the other hand, (3.42) implies that

 Em−2,x μj (x, tr ), x, tr = −2β0 δ0 (β0 − δ0 )u(x, tr )v(x, tr )μj,x (x, tr )

m−2 

 μj (x, tr ) − μk (x, tr ) .

(3.56)

k=1 k=j

Using (3.55) and (3.56), we can easily prove (3.48). Expressions (3.49)–(3.53) can be proved similarly. 2 4. Asymptotic expansions and divisors In this section we will investigate the asymptotic expansions of φ2 (P, x, tr ), φ3 (P, x, tr ) and ψ1 (P, x, x0 , tr , t0,r ) near P∞1 , P∞2 , P∞3 ∈ Km−2 by choosing the local coordinate ζ = λ−1 and obtain the divisors accordingly. Lemma 4.1. Suppose that u(x, tr ), v(x, tr ) satisfy the rth cmKdV equation. Moreover, let P ∈ Km−2 \ {P∞1 , P∞2 , P∞3 }, (x, tr ) ∈ C2 . Then ⎧ ⎪ ⎪ ⎪ ⎪ ⎪ ⎨ φ2 (P, x, tr ) =

2 −1 u(x,tr ) ζ

+

ux (x,tr )−v(x,tr )∂ −1 u(x,tr )v(x,tr ) u2 (x,tr )

as P → P∞1 ,

+ κ1,1 (x, tr )ζ + O(ζ 2 ),

ζ→0 ⎪ ⎪∂

−1

u(x,tr )v(x,tr ) + κ2,1 (x, tr )ζ + ⎪ v(x,tr ) ⎪ ⎪ ⎩ u(x,tr ) ux (x,tr ) 2 ζ + O(ζ 3 ), − 2 ζ− 4

⎧ ⎪ ⎪ ⎪ ⎪ ⎪ ⎨ φ3 (P, x, tr ) =

ζ→0 ⎪ ⎪

⎪ ⎪ ⎪ ⎩

∂ −1 u(x,tr )v(x,tr ) u(x,tr ) 2 −1 v(x,tr ) ζ

+

O(ζ ),

+ χ1,1 (x, tr )ζ + O(ζ 2 ),

as P → P∞2 ,

(4.1)

as P → P∞3 , as P → P∞1 ,

−1

vx (x,tr )−u(x,tr )∂ u(x,tr )v(x,tr ) v 2 (x,tr )

+ χ2,1 (x, tr )ζ + O(ζ 2 ), r) ζ− − v(x,t 2

2

vx (x,tr ) 2 ζ 4

+ O(ζ 3 ),

as P → P∞2 ,

(4.2)

as P → P∞3 ,

where 2 u2 − uuxx u2 + v 2 uvx − 2ux v −1 v v2

+ x 3 + ∂ uv + 2 ∂ −1 ux v + 3 ∂ −1 uv 3 2u 2u 2u 2u 2u  −1 v −1 2 2 − 2∂ v − u ∂ uv, 2u 2

 vx 1 u

1 = 2 ∂ −1 uv − ∂ −1 uvx − 2 ∂ −1 uv + ∂ −1 u2 − v 2 ∂ −1 uv, 2v 2v 2v 2v 2  ux 1 −1 v

1 −1 2 ∂ ux v − 2 ∂ −1 uv + ∂ = 2 ∂ −1 uv − v − u2 ∂ −1 uv, 2u 2u 2u 2u vx2 − vvxx u2 + v 2 ux v − 2uvx −1 u −1 u2 −1 2 + = + ∂ uv + ∂ uv + ∂ uv x 2v 2v 3 2v 3 2v 2 2v 3

 u − 2 ∂ −1 u2 − v 2 ∂ −1 uv. 2v

κ1,1 =

κ2,1 χ1,1 χ2,1

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139

Proof. We can insert the three sets of ansatzs

 (1) φ2 = κ1,−1 ζ −1 + κ1,0 + κ1,1 ζ + O ζ 2 ,

 φ3 = χ1,0 + χ1,1 ζ + O ζ 2 ;

ζ→0

(2) φ2 (3) φ2

 = κ2,0 + κ2,1 ζ + O ζ 2 ,

ζ→0

φ3 = χ2,−1 ζ

ζ→0

−1

ζ→0

 = κ3,1 ζ + κ3,2 ζ + O ζ 3 ,

 + χ2,0 + χ2,1 ζ + O ζ 2 ;

 φ3 = χ3,1 ζ + χ3,2 ζ 2 + O ζ 3 ;

2

ζ→0

ζ→0

into the Riccati-type equations (3.23) and (3.24). A comparison of the same powers of ζ then proves the lemma. 2 The divisor (f ) of a meromorphic function f (P ), not zero, on Km−2 is a map (f ) : Km−2 → Z, P → νf (P ), where νf (P ) is the order of f at P , that is, νf (P ) = n0 if f (P ) ∞ has the asymptotic expansion f (P ) = m=n0 cm ζPm under the local coordinate ζP [11]. Observing expression (3.11), we derive that νˆ1 (x, tr ), . . . , νˆm−2 (x, tr ) are m − 2 zeros and μ ˆ1 (x, tr ), . . . , μ ˆm−2 (x, tr ) are m−2 poles of meromorphic function φ2 (P, x, tr ). Combining the asymptotic expansions of φ2 (P, x, tr ) near P∞1 and P∞3 in Lemma 4.1, we finally obtain the divisor of φ2 (P, x, tr ) as follows

 φ2 (P, x, tr ) = DP∞3 ,ˆν1 (x,tr ),...,ˆνm−2 (x,tr ) (P ) − DP∞1 ,ˆμ1 (x,tr ),...,ˆμm−2 (x,tr ) (P ).

(4.3)

Similarly, one can derive the divisor of φ3 (P, x, tr ) 

φ3 (P, x, tr ) = DP∞

ˆ

ˆ

3 ,ξ1 (x,tr ),...,ξm−2 (x,tr )

(P ) − DP∞2 ,ˆμ1 (x,tr ),...,ˆμm−2 (x,tr ) (P ).

(4.4)

In order to derive the properties of ψ1 (P, x, x0 , tr , t0,r ), we shall do some preparations. ¯ (r) For the sake of convenience, we introduce the notation V ij to denote the corresponding (r) homogeneous case of V , that is to say ij

¯ (r) (r)  V ij = V ij β˜

˜

˜

˜

1 =···=βr =δ1 =···=δr =0

.

(4.5)

Accordingly, the homogeneous case of a defined function (r) (r) (r) Ir (P, x, tr ) = V 11 (λ, x, tr ) + V 12 (λ, x, tr )φ2 (P, x, tr ) + V 13 (λ, x, tr )φ3 (P, x, tr ),

(4.6) is defined as ¯ (r) ¯ (r) ¯ (r) I¯r (P, x, tr ) = V 11 (λ, x, tr ) + V 12 (λ, x, tr )φ2 (P, x, tr ) + V 13 (λ, x, tr )φ3 (P, x, tr ). (4.7)

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Lemma 4.2. Assuming (x, tr ) ∈ C2 , ζ = λ−1 denoting the local coordinate near P∞j , j = 1, 2, 3, then we have I¯r (P, x, tr ) ⎧ 2 ˜ ⎪ ¯˜r,xx + ∂v c¯˜r − ∂ud¯˜r + v f¯˜r ) + O(ζ), (β0 − 2δ˜0 )ζ −2r−1 + u1 (a ⎪ ⎪ ⎨3 = 2 (δ˜0 − 2β˜0 )ζ −2r−1 + 1 (¯˜br,xx + ∂uc¯˜r − ∂v e¯˜r − uf¯˜r ) + O(ζ), 3 v ⎪ ⎪ ⎪2 ˜ ⎩ −2r−1 ˜ + O(ζ), 3 (β0 + δ0 )ζ

P → P ∞1 , P → P ∞2 , P → P ∞3 . (4.8)

Proof. We use the inductive method to prove the first expression in this lemma while the ¯˜(0) = ¯˜b(0) = c¯˜(0) = 0, d¯˜(0) = δ˜0 , e¯˜(0) = β˜0 , other two can be proved similarly. For r = 0, a ¯ (0) −1 f˜ = (δ˜0 − β˜0 )∂ uv, then it is easy to see that

 (0)  2 ¯˜x + v c¯˜(0) − ud¯˜(0) − 2ζ −1 a ¯˜(0) φ2 I¯0 (P, x, tr ) = ζ −1 d¯˜(0) + e¯˜(0) + a 3

 ¯˜(0) − v e¯˜(0) − 2ζ −1¯˜b(0) φ3 + ¯˜b(0) x + uc =

2 ˜ 1 ¯ (β0 − 2δ˜0 )ζ −1 + (a ˜0,xx + ∂v c¯˜0 − ∂ud¯˜0 + v f¯˜0 ) + O(ζ), 3 u

P → P ∞1 . (4.9)

Suppose that I¯r (P, x, tr ) has the following expansion ∞ 2 I¯r (P, x, tr ) = (β˜0 − 2δ˜0 )ζ −2r−1 + σj (x, tr )ζ j , ζ→0 3 j=0

as P → P∞1 ,

(4.10)

for some coefficients {σj (x, tr )}j∈N0 to be determined. Observing (3.40), we arrive at σj,x = (uκ1,j + vχ1,j )tr ,

j = 0, 1, 2, 3 . . . .

(4.11)

Taking use of (2.10), (2.16) and Lemma 4.1, we get the first three expressions of (4.11)  ¯˜r+1 4a , u x  2¯ 2ux ¯ 2v 4 2 ˜r+1,x + 2 a = (uκ1,1 + vχ1,1 )tr = − a ˜r+1 − c¯˜r+1 + d¯˜r+1 − e¯˜r+1 u u u 3 3    2 ¯˜ 2v ¯ + ˜r+1 ∂ −1 uv , br+1 − 2 a u u x 

σ0,x = (κ1,0 + vχ1,0 )tr = σ1,x

σ2,x = (uκ1,2 + vχ1,2 )tr    2   ux ¯ uxx 1¯ ux ¯˜r+1 − 3v¯˜br+1 + vx − ux v c¯˜r+1 a ˜r+1,xx − 2 a = − − 3u ˜r+1,x + a u u u3 u2 u u2

X. Geng et al. / Advances in Mathematics 263 (2014) 123–153

141

    2  v ¯˜ 2ux v ¯ 1 ¯˜ ux ¯˜ v ¯ vx v + f r+1 2 a − 3 a − 1 c¯˜r+1 ˜r+1,x + ˜r+1 − br+1 + 2 br+1 + u u u2 u u u u2    2

2 v ¯˜ v ¯˜ v ¯ −1 ¯ − (dr+1 − e˜r+1 ) ∂ uv + a ˜r+1 − 2 br+1 ∂ −1 uv u u3 u     −1

 −1  v ¯ 1 ¯˜ −1 2 2 + v − u ∂ uv (4.12) a ˜r+1 − br+1 ∂ ux v − ∂ u2 u x from which it can be inferred ¯˜r+1 4a , u 2¯ 2ux ¯ 2v 4 2 σ1 (x, tr ) = − a ˜r+1,x + 2 a ˜r+1 − c¯˜r+1 + d¯˜r+1 − e¯˜r+1 u u u 3 3   2 ¯˜ 2v ¯ + ˜r+1 ∂ −1 uv, br+1 − 2 a u u   2   ux v ¯ 1¯ ux ¯ ux uxx vx ¯ ˜ ¯ σ2 (x, tr ) = a ˜r+1,xx − 2 a − 2 c˜r+1 − 2 − 3u a ˜r+1 − 3v br+1 + ˜r+1,x + u u u3 u u u    v˜ 2ux v ¯ 1 ux v ¯ vx ˜r+1 − ¯˜br+1 + 2 ¯˜br+1 + f¯ a ˜r+1,x + − 3 a u r+1 u2 u2 u u u   2    2

2 v v ¯˜ v ¯ v ¯˜ −1 ¯ ¯ + − 1 c˜r+1 − (dr+1 − e˜r+1 ) ∂ uv + a ˜r+1 − 2 br+1 ∂ −1 uv 2 3 u u u u   

  1 v ¯ + (4.13) a ˜r+1 − ¯˜br+1 ∂ −1 ux v − ∂ −1 v 2 − u2 ∂ −1 uv , 2 u u σ0 (x, tr ) =

where the integration constants are taken as zero because there are no arbitrary constants in the expansions of φ2 (P, x, tr ), φ3 (P, x, tr ) near P∞1 nor in the coefficients of the ¯ homogeneous polynomials V i,j with the condition ∂∂ −1 = ∂ −1 ∂ = 1. It is easy to see that ¯ (r+1) ¯ (r+1) ¯ (r+1) + V 12 φ2 + V 13 φ3 I¯r+1 (P, x, tr ) = V 11

 2 ¯˜r+1,x + v c¯˜r+1 − ud¯˜r+1 − 2ζ −1 a ¯˜r+1 φ2 = ζ −2 I¯r + ζ −1 (d¯˜r+1 + e¯˜r+1 ) + a 3

 + ¯˜b + uc¯˜ − v e¯˜ − 2ζ −1¯˜b φ r+1,x

=

r+1

r+1

r+1

3

2 ˜ 1 ¯ (β0 − 2δ˜0 )ζ −2r−3 + (a ˜r+1,xx + ∂v c¯˜r+1 − ∂ud¯˜r+1 + v f¯˜r+1 ) 3 u + O(ζ).

Thus I¯r (x, tr ) is proved to have the expansion as seen in (4.8) near P∞1 .

(4.14) 2

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From Lemma 4.2 and (2.15), we arrive at ⎧ r utr (x,tr ) 2 ˜ ⎪ (β − 2δ˜r−l )ζ −2l−1 + u(x,t + O(ζ), P → P∞1 , ⎪ r) ⎨ l=0 3 r−l r 2 ˜ v (x,t ) t r −2l−1 r Ir (P, x, tr ) = (δ − 2β˜r−l )ζ + v(x,t + O(ζ), P → P∞2 , r) ⎪ l=0 3 r−l ⎪ ⎩ r 2 ˜ −2l−1 ˜ + O(ζ), P → P ∞3 . l=0 3 (βr−l + δr−l )ζ

(4.15)

On the basis of the above preparation, we can give the asymptotic expansions of ψ1 (P, x, x0 , tr , t0,r ) near P∞j , j = 1, 2, 3 in the following lemma. Lemma 4.3. Suppose that u(x, tr ) and v(x, tr ) satisfy the rth coupled mKdV equation. Moreover, let P ∈ Km−2 \ {P∞1 , P∞2 , P∞3 }, (x, x0 , tr , t0,r ) ∈ C4 . Then ψ1 (P, x, x0 , tr , t0,r ) ⎧ u(x,tr ) r exp(ζ −1 (x − x0 ) + l=0 23 (β˜r−l − 2δ˜r−l )ζ −2l−1 ⎪ u(x ,t ) ⎪ 0 0,r ⎪ ⎪ ⎪ ⎪ × (tr − t0,r ) + O(ζ)), ⎪ ⎪ ⎪ ⎪ ⎨ v(x,tr ) exp(ζ −1 (x − x ) + r 2 (δ˜ − 2β˜ )ζ −2l−1 0 r−l l=0 3 r−l v(x0 ,t0,r ) = ζ→0 ⎪ ⎪ × (tr − t0,r ) + O(ζ)), ⎪ ⎪ ⎪ r ⎪ ⎪ exp(−ζ −1 (x − x0 ) + l=0 23 (β˜r−l + δ˜r−l )ζ −2l−1 ⎪ ⎪ ⎪ ⎩ × (tr − t0,r ) + O(ζ)),

as P → P∞1 , as P → P∞2 , as P → P∞3 . (4.16)

Proof. From (3.38) and (3.39), we obtain the expression of ψ1 (P, x, x0 , tr , t0,r )  x ψ1 (P, x, x0 , tr , t0,r ) = exp









 −λ + u x , tr φ2 P, x , tr + v x , tr φ3 P, x , tr dx

x0

tr +



 

 (r)

(r)

V 11 λ, x0 , t + V 12 λ, x0 , t φ2 P, x0 , t

t0,r

+



 (r)

V 13 λ, x0 , t φ3 P, x0 , t dt

 ,

(4.17)

which implies the lemma by making use of Lemma 4.1 and (4.15). 2 In the following, we shall inspect the zeros and poles of ψ1 (P, x, x0 , tr , t0,r ) on Km−2 \ {P∞1 , P∞2 , P∞3 }. By using (3.11), (3.12), (3.22), (3.37) and (3.45), one can compute −λ + u(x, tr )φ2 (P, x, tr ) + v(x, tr )φ3 (P, x, tr ) (n)

= −λ + u

(n)

y 2 V12 − yAm + Bm y 2 V13 − yAm + Bm −v Em−2 Em−2

X. Geng et al. / Advances in Mathematics 263 (2014) 123–153

=



(n) 1 (n)  uV13 − vV12 y 2 − (uAm − vAm )y + (Em−2,x + 4λEm−2 ) Em−2 3  2 (n) (n)  + uV13 − vV12 Sm − λ 3 1

(n)

=

(n)

2 (uV13 − vV12 )(3y 2 + Sm ) 1 Em−2,x + 3 Em−2 3 Em−2 (n)

−

uV13 y(y +

Am (n) ) V13

(n)

− vV12 y(y +

Am (n) ) V12

Em−2

=

λ→μj (x,tr )

=

λ→μj (x,tr )

1 + λ 3

μj,x (x, tr ) + O(1) λ − μj (x, tr )

 ∂x ln λ − μj (x, tr ) + O(1), −

and (r) (r) (r) V 11 (λ, x, tr ) + V 12 (λ, x, tr )φ2 (P, x, tr ) + V 13 (λ, x, tr )φ3 (P, x, tr ) (n)

(n)

2 2 (r) (r) y V13 − yAm + Bm (r) y V12 − yAm + Bm = V 11 + V 12 − V 13 Em−2 Em−2

=

(r) (n) (r) (n) (r) (r) y 2 (V 12 V13 − V 13 V12 ) − y(V 12 Am − V 13 Am ) + 13 Em−2,tr Em−2

− =

2 (r) (n) 3 (V13 V12

− V 12 V13 )Sm Em−2 (r)

(n)

(r) (n) (r) (n) 1 Em−2,tr 2 (V 12 V13 − V 13 V12 )(3y 2 + Sm ) + 3 Em−2 3 Em−2

−

(r) (n) V 12 V13 y(y +

Am (n) ) V13

(r) (n) − V 13 V12 y(y +

Am (n) ) V12

Em−2

μj,tr (x, tr ) + O(1) λ − μj (x, tr ) λ→μj (x,tr )

 = ∂tr ln λ − μj (x, tr ) + O(1). =

−

λ→μj (x,tr )

Consequently, ψ1 (P, x, x0 , tr , t0,r )  x 







 = exp −λ + u x , tr φ2 P, x , tr + v x , tr φ3 P, x , tr dx x0

143

X. Geng et al. / Advances in Mathematics 263 (2014) 123–153

144

tr +



 

 

 (r)

(r)

(r)

V 11 λ, x0 , t + V 12 λ, x0 , t φ2 P, x0 , t + V 13 λ, x0 , t φ3 P, x0 , t dt



t0,r

=

λ − μj (x, tr ) λ − μj (x0 , tr ) O(1) λ − μj (x0 , tr ) λ − μj (x0 , t0,r )

λ − μj (x, tr ) O(1) λ − μj (x0 , t0,r ) ⎧ for P near μ ˆj (x, tr ) = μ ˆj (x0 , t0,r ), ⎪ ⎨ (λ − μj (x, tr ))O(1) = O(1) for P near μ ˆj (x, tr ) = μ ˆj (x0 , t0,r ), ⎪ ⎩ −1 (λ − μj (x0 , t0,r )) O(1) for P near μ ˆj (x0 , t0,r ) = μ ˆj (x, tr ), =

(4.18)

where O(1) = 0. Therefore we have the following proposition: Proposition 4.4. Let P = (λ, y) ∈ Km−2 \ {P∞1 , P∞2 , P∞3 }, (x, x0 , tr , t0,r ) ∈ C4 . Then ψ1 (P, x, x0 , tr , t0,r ) on Km−2 \ {P∞1 , P∞2 , P∞3 } has m − 2 zeros and m − 2 poles which are μ ˆ1 (x, tr ), . . . , μ ˆm−2 (x, tr ) and μ ˆ1 (x0 , t0,r ), . . . , μ ˆm−2 (x0 , t0,r ) respectively. 5. Construction of the algebro-geometric solutions In this section, we shall construct the Riemann theta function representations for the Baker–Akhiezer function and solutions to the cmKdV hierarchy. To this end, we first give three kinds of Abelian differentials. Equip the Riemann surface Km−2 with homology basis {aj , bj }m−2 j=1 , which are independent and having intersection numbers as follows aj ◦ bk = δj,k ,

aj ◦ ak = 0,

bj ◦ bk = 0,

j, k = 1, . . . , m − 2.

For the present, we will choose as our basis the following set 1 l (P ) = 2 3y (P ) + Sm



1 ≤ l ≤ 4n + 1,

λl−1 dλ, y(P )λ

l−4n−2

dλ,

4n + 2 ≤ l ≤ m − 2,

(5.1)

which are m − 2 linearly independent holomorphic differentials on Km−2 . By using the m−2 homology basis {aj }m−2 j=1 and {bj }j=1 , the period matrices A = (Ajk ) and B = (Bjk ) can be constructed from  Ajk =

 j ,

ak

Bjk =

j .

(5.2)

bk

It is possible to show that the matrices A and B are invertible. Now we define the matrices C and τ by C = A−1 , τ = A−1 B. Then the matrix τ can be shown to be

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145

symmetric (τjk = τkj ) and has a positive-definite imaginary part (Im τ > 0) [15,22,34]. If we normalize l (P ) into new basis ω = (ω1 , · · · , ωm−2 ),

ωj =

m−2

Cjl l ,

(5.3)

l=1

then we have ak ωj = δjk , bk ωj = τjk , j, k = 1, . . . , m − 2. Calculation by using (4.1) and (4.2) reveals that the asymptotic behaviors of y(P ) near P∞1 , P∞2 , P∞3 are as follows: ⎧ −2n−1 2 ( 3 (β0 − 2δ0 ) + 23 (β1 − 2δ1 )ζ 2 + O(ζ 4 )), ⎪ ⎨ζ y(P ) = ζ −2n−1 ( 23 (δ0 − 2β0 ) + 23 (δ1 − 2β1 )ζ 2 + O(ζ 4 )), ζ→0 ⎪ ⎩ −2n−1 2 ζ ( 3 (β0 + δ0 ) + 23 (β1 + δ1 )ζ 2 + O(ζ 4 )),

as P → P∞1 , as P → P∞2 ,

(5.4)

as P → P∞3 .

Then we deduce the Laurent expansions of (5.3) near P∞1 , P∞2 , P∞3 : ⎧ 3Cj,4n+1 +2(β0 −2δ0 )Cj,6n+1 + O(ζ))dζ, ( ⎪ 12(β0 δ0 −δ02 ) ⎪ ⎨ 3Cj,4n+1 +2(δ0 −2β0 )Cj,6n+1 ωj = + O(ζ))dζ, ( 12(β0 δ0 −β02 ) ζ→0 ⎪ ⎪ ⎩ 3Cj,4n+1 +2(β0 +δ0 )Cj,6n+1 + O(ζ))dζ, (− 12β0 δ0

as P → P∞1 , as P → P∞2 ,

(5.5)

as P → P∞3 .

Furthermore, we could write ωj in the following form: ωk =

∞

as P → P∞j ,

k,l (P∞j )ζ l dζ,

(5.6)

l=0

where j,l (P∞j ) are constants, j = 1, 2, 3; k = 1, 2, · · · , m − 2. (2) Let ωP∞s ,j (P ), j ≥ 2, s = 1, 2, 3, denote the normalized Abelian differential of the second kind holomorphic on Km−2 \ {P∞s } satisfying 

(2)

ωP∞s ,j (P ) = 0,

k = 1, . . . , m − 2,

(5.7)

ak

 (2) ωP∞s ,j (P ) = ζ −j + O(1) dζ, ζ→0

as P → P∞s

(5.8)

and introduce the following Abelian differential by (4.16) (2)

(2)

(2)

Ω2 (P ) = ωP∞

1 ,2

(2) (P ) = Ω 2r+2

r 2(2l + 1) l=0

(2)

(P ) + ωP∞

3

2 ,2

(P ) − ωP∞

3 ,2

(P ),

(2)

(β˜r−l − 2δ˜r−l )ωP∞

1 ,2l+2

(P )

(5.9)

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146

+

r 2(2l + 1) l=0

+

r 2(2l + 1)

3

l=0

(2) (δ˜r−l − 2β˜r−l )ωP∞

2 ,2l+2

3

(2) (β˜r−l + δ˜r−l )ωP∞

3 ,2l+2

(P )

(P ).

(5.10)

Then we have P

(2) Ω2 (P )

⎧ (2) ⎪ −ζ −1 + e1 (Q0 ) + O(ζ), ⎪ ⎨ =

Q0

P Q0

as P → P∞1 ,

−1

(2) −ζ + e2 (Q0 ) + O(ζ), ⎪ ⎪ ⎩ −1 (2) ζ + e3 (Q0 ) + O(ζ),

as P → P∞2 ,

(5.11)

as P → P∞3 ,

⎧ r (2) 2 ˜ ⎪ − (β − 2δ˜r−l )ζ −(2l+1) + e˜1 (Q0 ) + O(ζ), ⎪ ⎨ l=0 3 r−l

(2) (P ) = − r 2 (δ˜r−l − 2β˜r−l )ζ −(2l+1) + e˜(2) (Q0 ) + O(ζ), Ω 2r+2 2 l=0 3 ⎪ ⎪ ⎩ r 2 ˜ (2) − l=0 3 (βr−l + δ˜r−l )ζ −(2l+1) + e˜3 (Q0 ) + O(ζ),

as P → P∞1 , as P → P∞2 , as P → P∞3 , (5.12)

(2)

(2)

(2)

(2)

(2)

(2)

where e1 (Q0 ), e2 (Q0 ), e3 (Q0 ), e˜1 (Q0 ), e˜2 (Q0 ), e˜3 (Q0 ) are integration constants with Q0 an appropriately chosen base point on Km−2 \ {P∞1 , P∞2 , P∞3 }. The b-periods (2)

(2) (P ) are denoted by of the differential Ω2 (P ) and Ω 2r+2

(2)  (2) (2) U 2 = U2,1 , · · · , U2,m−2 ,

(2)

U2,k =

1 2πi



(2)

Ω2 (P ), bk

(2) U 2r+2

(2)

= U



2r+2,1 , · · · , U2r+2,m−2 , (2)

1

(2) U 2r+2,j = 2πi



k = 1, . . . , m − 2,

(2) (P ), Ω 2r+2

(5.13)

j = 1, . . . , m − 2.

bj

(5.14) By the relationship between the normalized differential of the second kind and the normalized holomorphic differential ω, we can derive that (2)

U2,k = k,0 (P∞1 ) + k,0 (P∞2 ) − k,0 (P∞3 ),

(2) U 2r+2,k =

r 2 l=0

+

3

(β˜r−l − 2δ˜r−l )k,2l (P∞1 ) +

l=0 (3)

r 2 l=0

r 2

3

(β˜r−l + δ˜r−l )k,2l (P∞3 ),

k = 1, 2, · · · , m − 2,

3

(5.15)

(δ˜r−l − 2β˜r−l )k,2l (P∞2 )

k = 1, 2, · · · , m − 2.

(5.16)

Let ωQ1 ,Q2 denote the normalized Abelian differential of the third kind holomorphic on Km−2 \ {Q1 , Q2 } and having simple poles at Ql with residues (−1)l+1 , l = 1, 2, then

X. Geng et al. / Advances in Mathematics 263 (2014) 123–153



(3) ωQ1 ,Q2

 = 0,

ak

Q1 = 2πi ωk ,

(3) ωQ1 ,Q2

k = 1, . . . , m − 2,

147

(5.17)

Q2

bk

where the path of integration from Q2 to Q1 in the integral does not intersect the cycles a1 , · · · , am−2 , b1 , · · · , bm−2 . Let Tm−2 be the period lattice {z ∈ Cm−2 | z = N + Lτ, N , L ∈ Zm−2 }. The complex torus Jm−2 = Cm−2 /Tm−2 is called the Jacobian variety of Km−2 . An Abel map A : Km−2 → Jm−2 is defined by  P ω1 , · · · ,

A(P ) =



P

Q0

(mod Tm−2 )

ωm−2

(5.18)

Q0

with the natural linear extension to the factor group Div(Km−2 ) A

!

" nk Pk = nk A(Pk ).

(5.19)

Define

ρ

(1)

(x, tr ) = A

 6n+1

 μ ˆk (x, tr )

=

k=1

ρ

(2)

(x, tr ) = A

 6n+1

k=1

 νˆk (x, tr )

=

k=1

ρ

(3)

(x, tr ) = A

 6n+1

6n+1

6n+1 k=1

 ξˆk (x, tr )

=

k=1

6n+1 k=1

μ ˆ k (x,tr )

ω, Q0 ν ˆk (x,tr )

ω, Q0 (x,tr ) ξˆk

ω, Q0

where ρ (1) (x, tr ), ρ (2) (x, tr ), ρ (3) (x, tr ) can be linearized on Jm−2 in the following text. Let θ(z ) denote the Riemann theta function associated with Km−2 equipped with an appropriately fixed homology basis: θ(z ) =



# $ exp πi N τ, N  + 2πi N , z  .

(5.20)

N ∈Zm−2

Here z = (z1 , · · · , zm−2 ) ∈ Cm−2 is a complex vector. The diamond brackets denote the Euclidean scalar product:

N , z =

m−2 i=1

Ni zi ,

N τ, N  =

m−2 i,j=1

τij Ni Nj .

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148

Expression (5.20) implies that # $ θ(z + N + M τ ) = exp −πi M τ, M  − 2πi M , z  θ(z). For brevity, define the function z : Km−2 × σ m−2 Km−2 → Cm−2 by z (P, Q) = M − A(P ) +



  D Q A Q ,

P ∈ Km−2 ,

Q ∈Q

Q = (Q1 , · · · , Qm−2 ) ∈ σ m−2 Km−2 , where σ m−2 Km−2 denotes the (m − 2)th symmetric power of Km−2 and M = (M1 , · · · , Mm−2 ) is the vector of Riemann constant depending on the base point Q0 by the following expression  m−2  1 Mj = (1 + τjj ) − ωl (P ) ωj , 2 l=1 P

l=j al

j = 1, . . . , m − 2.

Q0

For the sake of convenience, we use μ ˆ (x, tr ) to denote (ˆ μ1 (x, tr ), · · · , μ ˆm−2 (x, tr )) ∈ m−2 σ Km−2 [19]. Then we have





 θ z P, μ ˆ (x, tr ) = θ M − A(P ) + ρ (1) (x, tr ) ,

P ∈ Km−2 .

Given these preparations, we can derive the theta function representations of ψ1 (P, x, x0 , tr , t0,r ), and in particular, that of solutions for the cmKdV hierarchy. Theorem 5.1. Let P = (λ, y) ∈ Km−2 \ {P∞1 , P∞2 , P∞3 } and let (x0 , t0,r ) ∈ C2 , (x, tr ) ∈ or Dνˆ (x,tr ) or Dξˆ(x,tr ) Ωμ ⊆ C2 , where Ωμ is open and connected. Suppose that Dμ(x,t ˆ r) is nonspecial for (x, tr ) ∈ Ωμ . Then ψ1 (P, x, x0 , tr , t0,r ) has the following representation ψ1 (P, x, x0 , tr , t0,r )   P θ(z(P, μ ˆ (x, tr )))θ(z (P∞3 , μ ˆ (x0 , t0,r ))) (2) (2) exp e3 (Q0 ) − Ω2 (P ) (x − x0 ) = θ(z(P∞3 , μ ˆ (x, tr )))θ(z (P, μ ˆ (x0 , t0,r ))) Q0

 +

(2) e˜3 (Q0 )

P −

  (2)

Ω2r+2 (P ) (tr − t0,r ) ,

(5.21)

Q0

where the paths of integration in the above integrals and in the Abel mapping (5.18) are same.

μ ⊆ Ωμ , Proof. Assume temporarily that μj (x, tr ) = μj  (x, tr ) for j = j and (x, tr ) ∈ Ω

where Ωμ is open and connected. Let Ψ1 denote the right hand side of (5.21). Observing

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149

Lemma 4.3, it can be easily seen that ψ1 and Ψ1 have the identical essential singularity at P∞1 , P∞2 and P∞3 . Proposition 4.4 shows that all zeros and poles of ψ1 and Ψ1 on Km−2 \ {P∞1 , P∞2 , P∞3 } are simple and coincident. The Riemann–Roch uniqueness Ψ1 results in that ψ = γ, where γ is a constant [11]. Using (4.16) and the right hand side 1 of (5.21), we get that Ψ1 (P, x, x0 , tr , t0,r ) ψ1 (P, x, x0 , tr , t0,r ) =

ζ→0

exp(−ζ −1 (x − x0 ) + exp(−ζ −1 (x −

= 1 + O(ζ),

ζ→0

r

2 ˜ −2l−1 ˜ (tr − t0,r ) + O(ζ))(1 + O(ζ)) l=0 3 (βr−l + δr−l )ζ r 2 ˜ x0 ) + l=0 3 (βr−l + δ˜r−l )ζ −2l−1 (tr − t0,r ) + O(ζ))

P → P∞3 .

(5.22)

Then we conclude that γ = 1. Observing the continuity of the Abel map A, one can

μ to (x, tr ) ∈ Ωμ , which completes the proof of the extend the result from (x, tr ) ∈ Ω theorem. 2 Theorem 5.2. Let (x, tr ), (x0 , t0,r ) ∈ C2 . Then (2)

ρ (1) (x, tr ) = ρ (1) (x0 , t0,r ) + U 2 (x − x0 ) + U 2r+2 (tr − t0,r )

(mod Tm−2 ),

(2)

(2) (tr − t0,r ) ρ (2) (x, tr ) = ρ (2) (x0 , t0,r ) + U 2 (x − x0 ) + U 2r+2

(mod Tm−2 ),

(2)

(2) (tr − t0,r ) ρ (3) (x, tr ) = ρ (3) (x0 , t0,r ) + U 2 (x − x0 ) + U 2r+2

(mod Tm−2 ).

(2)

(5.23)

Proof. Assume that Dμˆ (x,tr ) is nonspecial for (x, tr ) ∈ Ωμ ⊆ C2 , where Ωμ is open and connected. In order to prove the theorem, we introduce a meromorphic differential Ω(x, x0 , tr , t0,r ) =

 ∂

ln ψ1 (P, x, x0 , tr , t0,r ) dλ. ∂λ

(5.24)

From the representation (5.21) one infers

Ω(x, x0 , tr , t0,r ) = −(x − x0 )Ω2 − (tr − t0,r )Ω 2r+2 (2)

+

m−2

(2)

(3)

ωμˆj (x,tr ),ˆμj (x0 ,t0,r ) + ω

,

(5.25)

j=1

m−2

denotes a holomorphic differential on Km−2 , that is, ω

= j=1 hj ωj for some where ω hj ∈ C, j = 1, . . . , m − 2. Since ψ1 (P, x, x0 , tr , t0,r ) is single-valued on Km−2 , all a- and b-periods of Ω are integer multiples of 2πi and hence  2πilk =

 ω

= hk ,

Ω(x, x0 , tr , t0,r ) = ak

ak

k = 1, . . . , m − 2

(5.26)

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150

for some lk ∈ Z. Similarly, for some nk ∈ Z,  2πink =

Ω(x, x0 , tr , t0,r ) bk

 = −(x − x0 )



(2)

Ω2 − (tr − t0,r )

bk





(2)

bk

+ 2πi

m−2 j=1

m−2 



(3)

ω

ωμˆj (x,tr ),ˆμj (x0 ,t0,r ) +

j=1 b

bk

Ω2 − (tr − t0,r )

= −(x − x0 )

(2) + Ω 2r+2

bk

k

(2) Ω 2r+2

bk μ ˆ j (x,tr )

ωk + 2πi



m−2

lj

j=1

μ ˆ j (x0 ,t0,r )

ωj bk

= −2πi(x − x0 )U2,k − 2πi(tr − t0,r )U 2r+2,k (2)

%m−2 + 2πi j=1

(2)

μ ˆj (x,tr )

ωk − Q0

m−2 j=1

μ ˆ j (x 0 ,t0,r )

&

ωk + 2πi

m−2

lj τj,k .

(5.27)

j=1

Q0

So we have

N = −(x −

(2) x0 )U 2

(2) + − (tr − t0,r )U 2r+2

m−2 j=1

μ ˆ j (x,tr )

ω− Q0

m−2 j=1

μ ˆj (x 0 ,t0,r )

ω + Lτ, Q0

(5.28) where N = (n1 , . . . , nm−2 ), L = (l1 , . . . , lm−2 ) ∈ Zm−2 . Thus (5.28) is equivalent to (2)

(2) (tr − t0,r ) ρ (1) (x, tr ) = ρ (1) (x0 , t0,r ) + U 2 (x − x0 ) + U 2r+2

(mod Tm−2 ).

(5.29)

Using this equality and the linear equivalence DP∞1 μˆ (x,tr ) ∼ DP∞3 νˆ (x,tr ) and DP∞2 μˆ (x,tr ) ∼ DP∞ ξˆ(x,tr ) , that is, 3

A(P∞3 ) + ρ (2) (x, tr ) = A(P∞1 ) + ρ (1) (x, tr ), A(P∞3 ) + ρ (3) (x, tr ) = A(P∞2 ) + ρ (1) (x, tr ), then we can prove the other two equalities in (5.23). The result extends from (x, tr ) ∈ Ωμ to (x, tr ) ∈ C2 using the continuity of A and the fact that positive nonspecial divisors are dense in the space of the positive divisors. 2

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Noting Theorem 5.2, θ(z(P∞j , μ ˆ (x, tr ))) could be written briefly in the following form 





(2)

(2) tr , ˆ (x, tr ) = θ M (j) + U 2 x + U θ z P ∞j , μ 2r+2

(5.30)

where (2)

M (j) = M − A(P∞j ) + ρ (1) (x0 , t0,r ) − U 2 x0 − U 2r+2 t0,r , (2)

j = 1, 2, 3.

Theorem 5.3. Let P = (λ, y) ∈ Km−2 \ {P∞1 , P∞2 , P∞3 } and let (x0 , t0,r ) ∈ C2 , (x, tr ) ∈ Ωμ ⊆ C2 , where Ωμ is open and connected. Suppose that Dμ(x,t or Dνˆ (x,tr ) or Dξˆ(x,tr ) ˆ r) is nonspecial. Then the theta representations of u(x, tr ) and v(x, tr ) read

u(x, tr ) = u(x0 , t0,r )

θ(z (P∞1 , μ ˆ (x, tr )))θ(z (P∞3 , μ ˆ (x0 , t0,r ))) θ(z(P∞3 , μ ˆ (x, tr )))θ(z (P∞1 , μ ˆ (x0 , t0,r )))



(2)  

(2) (2) (2) × exp e3 (Q0 ) − e1 (Q0 ) (x − x0 ) + e˜3 (Q0 ) − e˜1 (Q0 ) (tr − t0,r ) , (5.31) v(x, tr ) = v(x0 , t0,r )

θ(z(P∞2 , μ ˆ (x, tr )))θ(z (P∞3 , μ ˆ (x0 , t0,r ))) θ(z(P∞3 , μ ˆ (x, tr )))θ(z (P∞2 , μ ˆ (x0 , t0,r )))



(2)  

(2) (2) (2) × exp e3 (Q0 ) − e2 (Q0 ) (x − x0 ) + e˜3 (Q0 ) − e˜2 (Q0 ) (tr − t0,r ) , (5.32) which are algebro-geometric solutions of the cmKdV hierarchy. Proof. Expanding ψ1 of (5.21) near P∞1 we can get ψ1 =

ζ→0

θ(z(P∞1 , μ ˆ (x, tr )))θ(z (P∞3 , μ ˆ (x0 , t0,r ))) θ(z(P∞3 , μ ˆ (x, tr )))θ(z (P∞1 , μ ˆ (x0 , t0,r ))) 

(2)  

(2) (2) (2) × exp e3 (Q0 ) − e1 (Q0 ) (x − x0 ) + e˜3 (Q0 ) − e˜1 (Q0 ) (tr − t0,r )   r

 2 × exp ζ −1 (x − x0 ) + (β˜r−l − 2δ˜r−l )ζ −(2l+1) × (tr − t0,r ) 1 + O(ζ) . 3 l=0

(5.33) Comparing with the asymptotic expansion near P∞1 in Lemma 4.3, we can get the expression of u(x, tr ) in (5.31) immediately. The expression (5.32) can be deduced similarly by expanding ψ1 near P∞2 . 2

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