KdV 類型雙線性方程式的廣義Hirota 方法

 

國 立 交 通 大 學

應用數學系

碩 士 論 文

KdV

類型雙線性方程式的廣義Hirota 方法

Generalized Hirota method of KdV type bilinear

equation

研 究 生 : 沈中柱

指導老師 : 邵錦昌 教授

中 華 民 國 九 十 八 年 一 月

 

 

 

KdV  類型雙線性方程式的廣義Hirota  方法

Generalized Hirota method of KdV type bilinear 

equation 

研 究 生 :  沈中柱

Student : Chun‐Chue Shen

指導教授 :  邵錦昌

Advisor : Dr. Jiin‐Chang Shaw

國 立 交 通 大 學 應用數學系 碩 士 論 文

A thesis

Submitted to Department of Applied Mathematics

College of Science

National Chiao Tung University

In partial Fulfillment of the Requirements

For the Degree of

Master

In

Applied Mathematics

January 2009

KdV  類型雙線性方程式的廣義Hirota  方法

研 究 生 :  沈中柱 指導教授 :  邵錦昌

國立交通大學應用數學系 (研究所)  碩士班

摘 要

很多偏微分方程式都可以轉換成F(D)f‧f= 0 這種雙線性方程式 。這篇論文

我們試著得到這雙線性方程式的廣義解 , 由於這廣義解擁有Fredholm 行列式值

的結構 , 從中我們發展出一個 GLM 積分方程式 , 相較於逆散射方法所得到的

GLM 積分方程式 , 我們所得到的積分方程擁有更寬廣的應用空間

 

Generalized Hirota method of 

 

KdV type bilinear equation 

Student : Chun‐Chue Shen      Advisor : Dr. Jiin‐Chang Shaw 

 

 

Department of Applied Mathematics 

National Chiao Tung University 

Hsinchu, Taiwan, R.O.C 

Abstract

 

F(D)f‧f= 0

is an important bilinear equation into which many PDEs can be transformed. In this thesis we try to derive the generalized soliton solutions for this bilinear equation. Owing to the structure of Fredholm's determinant of generalized soliton solution we can develop a GLM integral equation whose application is wider than GLM equation produced in inverse scattering method.

誌謝

這真是一個很漫長的過程 , 不管怎樣 , 我總算畢業了 , 感謝

邵老師對我的悉心指導 , 感謝口試委員的不吝指正 , 以及所有被我

麻煩過的人 , 我想呢 , 要報答這些人的最好方法就是 , 在社會上

做一個頂天立地 , 堂堂正正的男子漢

Contents

1 Introduction 2

1.1 History . . . 2

1.2 Three major methods . . . 2

2 D operator and Bilinearization 5 2.1 The properties of D operator . . . 5

2.2 Bilinearization . . . 7

3 Generalized Soliton Solution 10 3.1 Introduction . . . 10

3.2 The pure N-soliton solutions . . . 11

3.3 Construction of generalized soliton solutions . . . 14

3.4 Two Examples . . . 20

4 Relation with GLM equation 24 5 Perspective 30 A 31 A.1 . . . 31

A.2 . . . 32

Chapter 1

Introduction

1.1

History

The story of soliton starts in 1834 from the findings of Russel[1] , but his discovery didn’t evoke many ripples around the scientific circle in British and suffered from attacks. It was not until the 1870’s that Russel’s work was finally vindicated and its scientific importance can be measured by the eminence of the men who did the job[2]. After 25 years Korteweg and de Vries were to derive their famous equation

ut+ 6uux+ uxxx (1.1)

,where subscript denote partial derivatives. However , the KdV equation didn’t draw much attention until the remarkable discoveries of Zabusky and Kruskal , who were investigating Fermi-Pasta-Ulam problem , in 1965[3]. They observe the particle-like nature of those interacting solitay waves in their numarical experiment. The name ”soliton” was born. In a later analysis of the interaction Lax[4] verified their observation rigorously. Afterwards, analytical methods of soliton theory were developed prosperously.

1.2

Three major methods

We introduce briefly below three major methods of soliton theory

Inverse scattering transform

Gardner et al[5] were able to relate equation (1.1) to the eigenvalue problem

ϕxx+ uϕ = λϕ (1.2)

They proved that (1.2) is isospectral in time when u satisfies (1.1) Using inverse scattering they were able to find a GLM equation for the initial value problem of KdV equation and to derive a number of important results , including the

explicit solution for the interaction of any number of solitary waves. The success of inverse scattering was explained by a deeper and more general argument by Lax[4], opening the way for more equations to be solved. In 1972, Zakharov and Shabat[6] found an eigenvalue problem with which they were able to solve the nonlieanr Schrodinger equation

iϕt+ ϕxx+ |ϕ|2ϕ = 0 (1.3)

Also in 1972, Waditi[7,8] applied essentially the same eigenvalue problem to solve the modified KdV equation

ut+ 6u2ux+ uxxx= 0 (1.4)

Backlund transformation

Consider a pair of partial differential equations

A(u) = 0 (1.5)

B(˜u) = 0 (1.6)

in which u and ˜u denote the unknown functions, while A and B represent dif-ferential operators in m independent variables. Then the set of relations

Rj((u), (˜u), (ξ)) = 0 j = 1, . . . , n

where (u) and (˜u) denote a finite sequence of partial derivatives of u and ˜u respectively.The number of each sequence need not be equal. (ξ) represent parameters. In reality, A and B are not restricted to the average differential op-erator. Hirota[9] created a new form of Backlund transformations,in which case A and B are specialized differential operator, the combinations of D opera-tor. For example, let f and f0 _{be two solutions of the bilinearized KdV equation}

, in symbols

Dx(Dt+ D3x)f · f = 0 (1.7)

Dx(Dt+ D3x)f0· f0= 0 (1.8)

,where Dx and Dt will be explained later.

(Dt+ 3λDx+ D3x)f

0_{· f = 0 (1.9)}

D2_xf0· f = λf0· f (1.10) constitute the Backlund transformation of the KdV equation in the bilinear fornalism

Hirota direct method

A perturbational series is a commonly used technique to solve a PDE. The same technique applies to Hirota bilinear equation. For example if we put ϕ = G/F then nonlinear Schrodinger equation iϕt+ ϕxx− 2|ϕ|2ϕ = 0 can be

transformed into

(iDt+ D2x− λ)G · F = 0 (1.11)

(D_x2− λ) = −2G · G∗ (1.12) (1.11) and (1.12) are nonlinear Schrodinger equation in the bilinear form. λ is an parameter to be determined. F and G can be set equal to:

F = 1 + εf1+ ε2f2+ · · · (1.13)

G = g0(1 + εg1+ εg2+ · · · ) (1.14)

Substituting (1.13) and (1.14) into (1.11) and (1.12) and colleting terms with the same power of ε. In this case, the calculated result indicates that both F and G turn out to have finite terms , i.e , F = 1 + εf1+ ε2f2+ · · · + mfmand

G = g0(1 + εg1+ εg2+ · · · + εngn)(say), provided the solutions to f1, g0 and g1

are properly selected.

Generally speaking , a perturbational series applied to a PDE may not turn out to have finite terms , perhaps not even be convergent. The advantage of Hirota method is able to truncate the series after a number of finite terms.

The marvel of math is that different approaches may lead to the same con-clusion. Hirota[9] used the concept of Backlund transformations to show that a new forms of the Backlund transformations lead to the known inverse scatter-ing methods of solutions of the initial-value problem for the respective nonlinear evolution equations.

The drawback of bilinear method is unable to solve a initial value problem for a soliton equation. Oishi studied using bilinear method to solve initial-value problems whose solutions may be expressed as a determinant. The thesis is primarily a review on Oishi’s papers[11][12].

Chapter 2

D operator and

Bilinearization

2.1

The properties of D operator

The D operator is defined by

DtmDnxa(t, x) · b(t0, x0) = (∂t− ∂t0)m(∂_x− ∂_x0)na(t, x)b(t0, x0)|_t0_=t,x0_=x (2.1)

For example:

Dxa · b = axb − abx

D2_xa · b = axxb − 2axbx+ abxx

D3_xa · b = axxxb − 3axxbx+ 3axbxx− abxxx

From the definition (2.1) we have the following lemma

Lemma 2.1 Dm_xa(x) · 1 = ∂_xma(x) (2.2) Dm1 x1 D m2 x2 . . . D mn xn a · b = (−1) m1+m2+...+mn_Dm1 x1 D m2 x2 . . . D mn xnb · a (2.3) Dm_xa · a = 0 for odd m (2.4) DxDta · 1 = DxDt1 · a = ∂x∂ta (2.5) Proof i(2.3) Dm1 x1 D m2 x2 . . . D mn xna · b = (∂x1− ∂x01) m1_(∂ x2− ∂x02) m2_{· · · (∂} xn− ∂x0n) mn_{a · b|} x0 1=x1,x02=x2...x0n=xn = (−1)m1+m2+...+mn_(∂ x0 1− ∂x1) m1_(∂ x0 2−∂x2) m2_{· · · (∂} x0 n− ∂xn) mn_{a · b|} x0 1=x1,x02=x2...x0n=xn = (−1)m1+m2+...+mn_Dm1 x1 D m2 x2 . . . D mn xn b · a

ii(2.4) By (2.3) D_xma · a = (−1)mD_xma · a = −D_xma · a ⇒

Dmxa · a = 0

QED

Also using definition (2.1) we have

Dm1 x1D m2 x2 . . . D mn xnexp(Φ1· x) · exp(Φ2· x) = (φ1₁− φ2 1) m1_(φ1 2− φ 2 2) m2_{· · · (φ}1 n− φ 2 n) mn_exp[(Φ 1+ Φ2) · x] (2.6)

with Φi and x being vectors

Φi= (φi1, φ i 2, . . . , φ i n) i = 1, 2 x = (x1, x2, . . . xn) Φi· x ≡ n X k=1 φi_kxk In particular , we have Dm1 x1D m2 x2 . . . D mn xnexp(Φ · x) · exp(Φ · x) = 0 (2.7)

Generally speaking , if F is a multipolynomial in Dx1, Dx2. . . Dxm , then

F (D) exp(Φ1· x) · exp(Φ2· x) = F (Φ1− Φ2) exp[(Φ1+ Φ2) · x] (2.8)

F (D)f · 1 = F (∂x1, ∂x2, . . . , ∂xn)f (2.9)

The following lemma are essential to bilinearize an evolution equation

Lemma 2.2

exp(δDx)a(x) · b(x) = a(x + δ)b(x − δ) (2.10)

Proof

exp(δDx)a(x) · b(x) = exp[δ(∂x− ∂x0)]a(x)b(x0)|_x0_=x

= exp(δ∂x)a(x) exp(−δ∂x0)b(x0)|x0=x

= [1 + δ∂x+ (δ2∂2x)/2! + · · · ]a(x)×

[1 − δ∂x+ (δ2∂2x)/2! − · · · ]b(x0)|x0_=x

From Taylor series we know the right hand side

= a(x + δ)b(x0− δ)|x0=x= a(x + δ)b(x − δ)

2.2

Bilinearization

There are several techniques to transform nonlinear partial differential equation

L(u, ut, ux, utt, uxx, utx, . . .) = 0

into bilinear forms. We introduce some of them here. Let us do some calcula-tions ∂x[ a(x) b(x)] = axb − bxa b2 = (∂x− ∂x0)a(x)b(x0) b2 |x0=x= Dxa · b b2 ∂_x2[a(x) b(x)] = b2(axb − bxa)x− 2bbx(axb − bxa) b4 =axxb − 2axbx+ abxx b2 − a b 2 b2(bxxb − b 2 x) =D 2 xa · b b2 − a b D2xb · b b2 Therefore a change of dependent variable u = a

b plays a natural role in bilin-earization. To avoid tedious calculations in transforming ∂n

x for n ≥ 3 we need a lemma Lemma 2.3 exp(δ∂x) a b = exp(δDx)a · b cosh(δDx)b · b (2.11) proof

By making use of (2.10) we have

cosh(δDx)b · b = [exp(δDx) + exp(−δDx)] 2 b · b = b(x + δ)b(x − δ) 2 +b(x − δ)b(x + δ) 2 = b(x + δ)b(x − δ) Therefore exp(δ∂x) a b = a(x + δ) b(x + δ) = a(x + δ)b(x − δ) b(x + δ)b(x − δ) = exp(δDx)a · b cosh(δDx)b · b QED

Expanding Taylor series on both sides of (2.11) with respect to the parameter δ , we have (1 + δ∂x+ δ2 2∂ 2 x+ δ3 6 ∂ 3 x+ · · · ) a b = (1 + δDx+ 1/2δ2D2x+ 1/6δ 3_D3 x+ · · · )a · b (1 + 1/2δ2D_x2+ 1/24δ4D_x4+ · · · )b · b = (a b + δ Dxa · b b2 + δ 2_/2D 2 xa · b b2 + · · · ) × (1 + δ 2_/2D 2 xb · b b2 + δ 4_/24D 4 xb · b b2 + · · · ) −1

Expanding the denominator using (1 + X)−1 = 1 − X + X2+ · · · , and col-lecting terms in powers of δ , we can obtain formulae which express derivatives of u = a/b in terms of the D-operator. The change of dependent variable u = 2(log f )xx also plays a natural role in bilinearization. Likewise , we also

need a lemma to avoid those messy calculations

Lemma 2.4

2 cosh(δ∂x) log f (x) = log[cosh(δDx)f (x) · f (x)] (2.12)

proof

2 cosh(δ∂x) log f (x) = 2[

exp(δ∂x) + exp(−δ∂x)

2 ] log f (x) = log f (x + δ) + log f (x − δ) = log f (x + δ)f (x − δ) = log[f (x + δ)f (x − δ) 2 + f (x − δ)f (x + δ) 2 ] = log[exp(δDx)f · f 2 + exp(−δDx)f · f 2 ] = log[cosh(δDx)f (x) · f (x)] QED

Expanding (2.12) with respect to δ , we have

2∂x2log f = D2xf · f f2 (2.13) 2∂x∂tlog f = DxDtf · f f2 (2.14) 2∂4_xlog f = D 4 xf · f f2 − 3( D_x2f · f f2 ) 2 _(2.15) 2∂_x6log f = D 6 xf · f f2 − 15 D_x4f · f f2 D_x2f · f f2 + 30( D2_xf · f f2 ) 3 _(2.16)

Now we apply these formulae to some equations. Making the dependent variable transformation u = 2(log f )xxto KdV equation (1.1) we obtain

2(log f )xxt+ 6(2 log f )xx(2 log f )xxx+ (2 log f )xxxxx= 0

Integrate with respect to x once

2(log f )xt+ 3[(2 log f )xx]2+ (2 log f )xxxx= 0

By making use of (2.13) , (2.14) , (2.15) we have

DxDtf · f f2 + 3( D2xf · f f2 ) 2₊D 4 xf · f f2 − 3( D2xf · f f2 ) 2_{= 0}

Multiplying f2 on both sides we have

Dx(Dt+ D3x)f · f = 0 (2.17)

Now we proceed to apply the same transformation of dependent variable to K-P equation

(−4ut+ uxxx+ 6uux)x+ 3uyy = 0

Set u = 2(log f )xxand integrate with respect to x twice we obtain

−4(2 log f )xt+ (2 log f )xxxx+ 3[(2 log f )xx]2+ 3(2 log f )yy= 0

Use (2.13) , (2.14) , and (2.15). Then

−4DxDtf · f f2 + Dx4f · f f2 − 3( Dx2f · f f2 ) 2_{+ 3(}D 2 xf · f f2 ) 2_{+ 3}D 2 yf · f f2 = 0 Multiplying f2 _{on both side we obtain the KP equation in the bilinear form}

Chapter 3

Generalized Soliton

Solution

3.1

Introduction

Many soliton equations can be transformed into bilinear equations such as:

Ex1 The KdV equation : ut+ 6uux+ uxxx= 0 u = 2(log f )xx

⇒ Dx(Dt+ D3x)f · f = 0

Ex2 The Sawada-Kotera equation:

ut+ 15(u3+ uuxx) + uxxxxx= 0 u = 2(log f )xx

⇒ Dx(Dt+ Dx5)f · f = 0

Ex3 The K-P equation:(−4ut+ uxxx+ 6uux)x+ 3uyy = 0 u = 2(log f )xx

⇒ (D4

x− 4DxDt+ 3Dy2)f · f = 0

These equations , which can be written in the form

F (D) = 0

and F is a multipolynomial in D = (Dx1, Dx2, . . . Dxm) with x = x1 , y = x2 ,

t = x3....etc. , have three conditions in common.

F (−D) = F (D) (3.1) F (0) = 0 (3.2) X σ=1,−1 F ( N X i=1 σiPi) (N ) Y i<j F (σiPi− σjPj)σiσj= 0 , f or N = 1, 2, . . . (3.3)

(3.3) is well known as Hirota condition. In this thesis we try to solve bilinear equation

under these three conditions. In what follows we tacitly assume these conditions

Lemma 3.1 If F is a multipolynomial in Dx1, Dx2, . . . Dxm satisfying

condi-tion (3.1) , then F (D)f · g = F (D)g · f (3.4) pf : Since F (D) = F (Dx1, Dx2, . . . , Dxm) is a multipolynomial in Dx1, Dx2, . . . Dxm , then by using (2.3) F (D)f · g = X 0≤n1,n2...nm≤N αn1n2...nmD n1 x1D n2 x2 . . . D nm xmf · g = X 0≤n1,n2...nm≤N (−1)n1+n2+...+nm_α n1n2...nmD n1 x1D n2 x2. . . D nn xmg · f = F (D)g · f , if n1+ n2+ . . . + nm is even

and the evenness of n1+ n2+ . . . + nmis exactly what (3.1) assures

QED

3.2

The pure N-soliton solutions

Now we proceed to solve

F (D)f · f = F (Dx1, Dx2, . . . , Dxm)f · f = 0

What we tackle this problem is making use of perturbation method , so we put

f = 1 + ε1f1+ ε2f2+ ε3f3+ · · ·

Substituting this perturbational series of f into the above bilinear equation and collecting like powers of ε , we have

ε0

F (D)1 · 1 = 0 , this is why (3.2) is required (3.5)

ε1 F (D)(f1· 1 + 1 · f1) = 0 (3.6) ε2 F (D)(f2· 1 + f1· f1+ 1 · f2) = 0 (3.7) ε3 F (D)(f3· 1 + f2· f1+ f1· f2+ 1 · f3) = 0 (3.8) ε4 F (D)(f4· 1 + f3· f1+ f2· f2+ f1· f3+ 1 · f4) = 0 (3.9)

Running through the above procedures with a proper solution to f1 , for

ex-ample , f1 = exp(φ1x1+ φ2x2+ · · · + φmxm) ≡ exp(Φ · x) , we can derive a

truncated series , thus an explicit solution is obtained without resorting to sum-ming the series. Now we set f1= exp(Φ · x) then ε1

F (D)(f1· 1 + 1 · f1) = 2F (D)f1· 1 = 2F (∂x1, ∂x2, . . . ∂xm) exp(Φ · x)

= 2F (φ1, φ2. . . φm) exp(Φ · x) = 0

⇒ F (φ1, φ2. . . φm) = 0 (3.10)

the above equalities are on account of (2.9) and (3.4). (3.10) is a confinement implying that each φi, i = 1, 2 . . . m can be expressed by (m − 1) independent

variables such as φi= φi(p, q, . . . , z) ε2 2F (D)f2· 1 = −F (D)f1· f1 = −F (D)[exp(Φ · x)] · [exp(Φ · x)] = −F (Φ − Φ) exp[(Φ + Φ) · x] = −F (0) exp(2Φ · x) = 0

⇒ f2 can be assigned to zero

The above equalities are due to (2.8) and (3.2)

ε3

2F (D)f3· 1 = −2F (D)f2· f1= 0

⇒ f3 can be assigned to zero Going on the calculation for steps εn, n > 3 it is

readily to have the conclusion that fn = 0 for n > 3. Therefore we obtained

a truncated series as formerly promised. Collectively , u = 2(log f )xx , where

f = 1 + exp(φ1x1+ φ2x2+ · · · + φmxm) with condition F (φ1, φ2, . . . φm) = 0 ,

is a one-soliton solution

If we put f1 = exp(Φ1· x) + exp(Φ2· x) , where Φ1 = (φ11, φ12, . . . φ1m) and

Φ2= (φ21, φ22, . . . φ2m) , we have (note that φ2i does not mean square)

ε1

F (D)(f1· 1 + 1 · f1) = 2F (D)f1· 1 = 2F (∂x1, ∂x2, . . . , ∂xm)[exp(Φ1· x)

+ exp(Φ2· x)]

= 2[F (Φ1) exp(Φ1· x) + F (Φ2) exp(Φ2· x)] = 0

ε2

2F (D)f2· 1 = −F (D)f1· f1

= −2F (D)[exp(Φ1· x)] · [exp(Φ2· x)] − F (D)[exp(Φ1· x)] · [exp(Φ1· x)]

− F (D)[exp(Φ2· x)] · [exp(Φ2· x)]

= −2F (Φ1− Φ2)exp[(Φ1+ Φ2) · x]

⇒ To meet the above equation we can use (2.9) and put f2= − F (Φ1− Φ2) F (Φ1+ Φ2) exp[(Φ1+ Φ2) · x] ε3 2F (D)f3· 1 = −2F (D)f2· f1

= (const)F (D){exp[(Φ1+ Φ2) · x]} · {exp(Φ1· x) + exp(Φ2· x)}

= (const){F (Φ1+ Φ2− Φ1)exp[(2Φ1+ Φ2) · x]

+ F (Φ1+ Φ2− Φ2)exp[(Φ1+ 2Φ2) · x]}

= 0 by (3.11)

⇒ f3 can be assigned zero to meet the above equation

ε4

2F (D)f4· 1 = −2F (D)f3· f1− F (D)f2· f2= 0

⇒ f4 can be assigned zero by f3= 0 , (2.7) and (3.2)

Going on the step for εn _{it is readily to conclude that f}

n= 0 f or n > 4

Collectively , u = 2(log f )xx , where f = 1 + εf1 + ε2f2 = 1 + exp(Φ1 ·

x + η0) + exp(Φ2· x + η0) + a12exp[(Φ1+ Φ2) · x + 2η0] , exp η0 = ε , and

a12= −

F (Φ1− Φ2)

F (Φ1+ Φ2) , is a two-soliton solution

If we put f1=P N

i=1exp Φi· x , then we obtain N-soliton solution? The answer

is yes. Here we directly quote N-soliton solution derived by Hirota[10]

f = X µ=0,1 exp[ N X i=1 µiηi+ (N ) X i<j Aijµiµj] (3.12) where exp Aij = − F (Φi− Φj) F (Φi+ Φj) and ηi = Φi· x + (const). P µ=0,1 means a

summation over all possible combinations of µ1= 0, 1, µ2= 0, 1, . . . , µN = 0, 1

, P(N )

i<j means a summation over all possible pairs (i, j) chosen from the set

3.3

Construction of generalized soliton solutions

As formerly stated , if we put f1=PN_i=1exp(Φi·x) , where Φi= (φi1, φi2, . . . , φim)

and x = (x1, x2, . . . , xm) , we have a N-soliton solution for arbitrary integer N

The starting point of generalized soliton solutions is to express f1 as a m − 1

multiple integral , i.e. ,

f1= Z Γ1 Z Γ2 · · · Z Γm−1 exp Φ · xdτ (p, q . . . z) (3.13)

, which can be abbreviated asR

Γ(m−1)exp Φ · xdτm−1 with the definitions

Φ · x ≡ m X i=1 φi(p, q, . . . , z)xi and Z Γ(m−1) dτ (p, q, . . . , z) ≡ Z Γ1 Z Γ2 · · · Z Γm−1 c(p, q, . . . , z)dpdq · · · dz

The integration represents a multiple complex integral along possibly differ-ent path Γi, i = 1, 2, . . . , m − 1 with a properly selected complex function

c(p, q, . . . , z)

With f1 specified by (3.13) now we go through the same perturbational

ap-proach as before ε1 F (D)f1· 1 = F (D)[ Z Γ(m−1) exp Φ · xdτm−1] · 1 = Z Γ_(m−1) F (D)[exp(Φ · x)] · 1dτm−1 = Z Γ(m−1) F (Φ)[exp(Φ · x)]dτm−1= 0 by (2.9)

To meet the above equation

⇒ F (Φ) = 0 (3.14) (3.14) explains why each component function φi = φi(p, q, . . . z) has m − 1

in-dependable variables. This is also the reason for m−1 multiple complex integral.

F (D)f2· 1 = − 1 2F (D)f1· f1 = −1 2F (D)[ Z Γ(m−1) exp Φ · xdτm−1] · [ Z Γ(m−1) exp Φ · xdτm−1] = −1 2[ Z Γ2 (m−1) F (D)[exp(Φ1· x)] · [exp(Φ2· x)]dτm−12 = −1 2 Z Γ2 (m−1) F (Φ1− Φ2) exp[(Φ1+ Φ2) · x]dτm−12 by (2.8)

where Φi (i = 1, 2) means a vector function

(φ1(pi, qi, . . . zi), φ2(pi, qi, . . . zi), . . . , φm(pi, qi, . . . zi)) for i=1,2

A simple example to clarify the above equation is that

[ Z Γ f (z)dz] × [ Z Γ f (z)dz] = Z Γ2 f (z1)f (z2)dz1dz2

Making the substitution f2 = cR_Γ2 (m−1)

exp[(Φ1+ Φ2) · x]dτm−12 into the above

equation to meet the identity corresponding to ε2_{, we have}

c Z Γ2 (m−1) F (D){exp[(Φ1+ Φ2) · x]} · 1dτm−12 = c Z Γ2 (m−1) F (Φ1+ Φ2){exp[(Φ1+ Φ2) · x]}dτm−12 by (2.9) = −1 2 Z Γ2 (m−1) F (Φ1− Φ2) exp[(Φ1+ Φ2) · x]dτm−12

Deriving c by equating the above equation leads to

f2= Z Γ2 (m−1) −1 2 F (Φ1− Φ2) F (Φ1+ Φ2) exp[(Φ1+ Φ2) · x]dτm−12

For convenience , we define I3≡ 2F (D)f3· 1 , as in (3.8) , we have

I3= −2F (D)f1· f2 = −2F (D)[ Z Γ(m−1) exp (Φ1· x)dτm−1] · [ Z Γ2 (m−1) −1 2 F (Φ2− Φ3) F (Φ2+ Φ3) exp [(Φ2+ Φ3) · x]dτm−12 ] = Z Γ3 (m−1) F (Φ2− Φ3) F (Φ2+ Φ3) F (D) exp(Φ1· x) · exp[(Φ2+ Φ3) · x]dτm−13 = Z Γ3 (m−1) F (Φ2− Φ3) F (Φ2+ Φ3) F (Φ1− Φ2− Φ3) exp[(Φ1+ Φ2+ Φ3) · x]dτm−13

Since Φi , i = 1, 2, 3 are dummy variables we can write I3 in such a form

I3= 1 3 Z Γ3 (m−1) F (Φ1− Φ2) F (Φ1+ Φ2) F (Φ3− Φ1− Φ2) exp[(Φ1+ Φ2+ Φ3) · x]dτm−13 + 1 3 Z Γ3 (m−1) F (Φ1− Φ3) F (Φ1+ Φ3) F (Φ2− Φ1− Φ3) exp[(Φ1+ Φ2+ Φ3) · x]dτm−13 + 1 3 Z Γ3 (m−1) F (Φ2− Φ3) F (Φ2+ Φ3) F (Φ1− Φ2− Φ3) exp[(Φ1+ Φ2+ Φ3) · x]dτm−13

Using Hirota condition (3.3) we have(see Appendix A.1)

I3= − 1 3 Z Γ3 (m−1) F (Φ1+ Φ2+ Φ3) F (Φ1− Φ2)F (Φ1− Φ3)F (Φ2− Φ3) F (Φ1+ Φ2)F (Φ1+ Φ3)F (Φ2+ Φ3) × exp[Φ1+ Φ2+ Φ3]dτm−13

To satisfy the above equation f3can be assigned

f3= − 1 3! Z Γ3 (m−1) F (Φ1− Φ2)F (Φ1− Φ3)F (Φ2− Φ3) F (Φ1+ Φ2)F (Φ1+ Φ3)F (Φ2+ Φ3) exp[Φ1+Φ2+Φ3]dτm−13

The mathematical form of f2 , f3 leads to a conjecture

fn= 1 n! Z Γn (m−1) Y 1≤j<k≤n [−F (Φj− Φk) F (Φj+ Φk) ] exp( n X i=1 Φi· x)dτm−1n n ≥ 2

To justify this conjecture such fn must be compatible in the original

perturba-tional series , i.e. ,

F (D)(

n

X

l=0

fl· fn−l) = 0 (3.15)

define F (D)(Pn l=0fl· fn−l) ≡ In Then In = F (D)[1 · fn+ f1· fn−1+ n−2 X l=2 fl· fn−l+ fn−1· f1+ fn· 1] = 1 n! Z Γn (m−1) Y 1≤j<k≤n [−F (Φj− Φk) F (Φj+ Φk) ]F (D)1 · exp[ n X i=1 Φi· x]dτm−1n + 1 (n − 1)! Z Γn (m−1) Y 2≤j<k≤n [−F (Φj− Φk) F (Φj+ Φk) ]F (D) exp(Φ1· x) · exp( n X i=2 Φi· x)dτm−1n + n−2 X l=2 1 l!(n − l)! Z Γn (m−1) Y 1≤j<k≤l [−F (Φj− Φk) F (Φj+ Φk) ] Y l+1≤j<k≤n [−F (Φj− Φk) F (Φj+ Φk) ]× F (D) exp( l X i=1 Φi· x) · exp( n X i=1+1 Φi· x)dτm−1n + 1 (n − 1)! Z Γn (m−1) Y 1≤j<k≤n−1 [−F (Φj− Φk) F (Φj+ Φk) ]F (D) exp( n−1 X i=1 Φi· x) · exp(Φn· x)dτm−1n + 1 n! Z Γn (m−1) Y 1≤j<k≤n [−F (Φj− Φk) F (Φj+ Φk) ]F (D) exp[ n X i=1 Φi· x] · 1dτm−1n

Using (2.8) and we obtain

In = 1 n! Z Γn (m−1) Y 1≤j<k≤n [−F (Φj− Φk) F (Φj+ Φk) ]F (− n X i=1 Φi) exp( n X i=1 Φi· x)dτm−1n + 1 (n − 1)! Z Γn (m−1) Y 2≤j<k≤n [−F (Φj− Φk) F (Φj+ Φk) ]F (Φ1− n X i=2 Φi) exp( n X i=1 Φi· x)dτm−1n + n−2 X l=2 1 l!(n − l)! Z Γn (m−1) Y 1≤j<k≤l [−F (Φj− Φk) F (Φj+ Φk) ] Y l+1≤j<k≤n [−F (Φj− Φk) F (Φj+ Φk) ]× F ( l X i=1 Φi− n X i=l+1 Φi) exp( n X i=1 Φi· x)dτm−1n + 1 (n − 1)! Z Γn (m−1) Y 1≤j<k≤n−1 [−F (Φj− Φk) F (Φj+ Φk) ]F ( n−1 X i=1 Φi− Φn) exp( n X i=1 Φi· x)dτm−1n + 1 n! Z Γn (m−1) Y 1≤j<k≤n [−F (Φj− Φk) F (Φj+ Φk) ]F ( n X i=1 Φi) exp( n X i=1 Φi· x)dτ(m−1)n ≡ In(1)+ In(2)+ In(3)+ In(4)+ In(5)

In(1) can be written as In(1)= 1 n!C n 0 X σ=1,−1 (1 − σ1)(1 − σ2) · · · (1 − σn) 2n Z Γn (m−1) Y 1≤j<k≤n [−F (σjΦj− σkΦk) F (Φj+ Φk) σjσk]× F ( n X i=1 σiΦi) exp( n X i=1 Φi· x)dτm−1n , Cln= n! l!(n − l)! (3.16) P

σ=1,−1means the summation over all possible combinations of σ1= 1, −1 σ2=

1, −1 . . . σn= 1, −1 Since σi= −1 i = 1, 2, . . . n is the only combination

which contributes , (3.16) is valid Similarly , we have In(2)= 1 n!C n 1 X σ=1,−1 (1 + σ1)(1 − σ2) · · · (1 − σn) 2n Z Γn (m−1) Y 1≤j<k≤n [−F (σjΦj− σkΦk) F (Φj+ Φk) σjσk]× F ( n X i=1 σiΦi) exp( n X i=1 Φi· x)dτm−1n In(3)= 1 n! n−2 X l=2 C_ln X σ=1,−1 (1 + σ1) · · · (1 + σl)(1 − σl+1) · · · (1 − σn) 2n × Z Γn (m−1) Y 1≤j<k≤n [−F (σjΦj− σkΦk) F (Φj+ Φk) σjσk]F ( n X i=1 σiΦi) exp( n X i=1 Φi· x)dτm−1n In(4)= 1 n!C n n−1 X σ=1,−1 (1 + σ1) · · · (1 + σn−1)(1 − σn) 2n Z Γn (m−1) Y 1≤j<k≤n [−F (σjΦj− σkΦk) F (Φj+ Φk) σjσk]× F ( n X i=1 σiΦi) exp( n X i=1 Φi· x)dτm−1n In(5)= 1 n!C n n X σ=1,−1 (1 + σ1) · · · (1 + σn) 2n Z Γn (m−1) Y 1≤j<k≤n [−F (σjΦj− σkΦk) F (Φj+ Φk) σjσk]× F ( n X i=1 σiΦi) exp( n X i=1 Φi· x)dτm−1n =⇒ In = In(1)+ In(2)+ In(3)+ In(4)+ In(5) = 1 n! X σ=1,−1 n X l=0 C_ln(1 + σ1) · · · (1 + σl)(1 − σl+1) · · · (1 − σn) 2n × Z Γn (m−1) Y 1≤j<k≤n [−F (σjΦj− σkΦk) F (Φj+ Φk) σjσk]F ( n X i=1 σiΦi) exp( n X i=1 Φi· x)dτm−1n

Due to the dummy variables Φi , we can choose (r1, r2, . . . , rl) from the set

{1, . . . , n} with the relation (r1 < r2 < . . . < rl) and the left was denoted by

(rl+1, rl+2, . . . , rn) with the relation (rl+1< rl+2< . . . < rn) Then

In = 1 n! X σ=1,−1 n X l=0 C_ln(1 + σr1) · · · (1 + σrl)(1 − σrl+1) · · · (1 − σrn) 2n × Z Γn (m−1) Y 1≤j<k≤n [−F (σrjΦrj − σrkΦrk) F (Φrj + Φrk) σrjσrk]F ( n X i=1 σriΦri) exp( n X i=1 Φi· x)dτm−1n We denote Z Γn (m−1) Y 1≤j<k≤n [−F (σrjΦrj− σrkΦrk) F (Φrj + Φrk) σrjσrk]F ( n X i=1 σriΦri) exp( n X i=1 Φi·x)dτm−1n

by C Because C is independent of the selection of (r1, . . . , rl) from the set

{1, . . . , N } we have In = 1 n! X σ=1,−1 n X l=0 C_ln(1 + σr1) · · · (1 + σrl)(1 − σrl+1) · · · (1 − σrn) 2n × C = 1 n! X σ=1,−1 n X l=0 X Cn l (1 + σr1) · · · (1 + σrl)(1 − σrl+1) · · · (1 − σrn) 2n × C where X Cn l

means the summation over all the selection possibilities

From observation we have

In = 1 n! X σ=1,−1 2−n X 1...n=1,−1 (1 + 1σ1)(1 + 2σ2) · · · (1 + nσn) × C = 1 n! X σ=1,−1 2−n X 2...n=1,−1 [(1 + σ1) + (1 − σ1)](1 + 2σ2) · · · (1 + nσn) × C = 1 n! X σ=1,−1 2−n+1 X 2...n=1,−1 (1 + 2σ2) · · · (1 + nσn) × C = · · · = 1 n! X σ=1,−1 ×C = 1 n! Z Γn (m−1) X σ=1,−1 Y 1≤j<k≤n [−F (σjΦj− σkΦk) F (Φj+ Φk) σjσk]F ( n X i=1 σiΦi) exp( n X i=1 Φi· x)dτm−1n = 0 by Hirota condition (3.3)

We summarize the discussion of this section into the following theorem

Theorem 3.1 If F (D) , a bilinear operator in a multipolynomial form of Dx1, . . . Dxm , satisfies (3.1)(3.2) and (3.3) , equation F (D)f · f = 0 has a

solution f = 1 + ∞ X n=1 εnfn (3.17)

where f1= Z Γ(m−1) exp(Φ · x)dτm−1 and fn= 1 n! Z Γn (m−1) Y 1≤j<k≤n [−F (Φj− Φk) F (Φj+ Φk) ] exp( n X i=1 Φi· x)dτm−1n n ≥ 2

3.4

Two Examples

(i)

KdV equation ut+ 6uux+ uxxx= 0 can be transformed into

F (Dx1, Dx2)f ·f = Dx1(Dx2+D

3

x1)f ·f = 0 with x1= x, x2= t, u = 2(log f )xx

From (3.13) and (3.14) , in this case m = 2 , we have

f1= Z Γ exp[φ1(p1)x1+ φ2(p1)x2]dτ (p1) (3.18) and F (φ1(p1), φ2(p1)) = φ1(φ2+ φ31) = 0 (3.19)

From (3.19) we can assign

φ1(p1) = p1, φ2(p1) = −p31

Using theorem3.1 we have

f1= Z Γ exp(p1x1− p31x2)dτ (p1) fn= 1 n! Z Γn Y 1≤j<k≤n [−F (φ1(pj) − φ1(pk), φ2(pj) − φ2(pk)) F (φ1(pj) + φ1(pk), φ2(pj) + φ2(pk)) ]× exp[ n X i=1 φ1(pi)x1+ φ2(pi)x2]dτn = 1 n! Z Γn Y 1≤j<k≤n {−(pj− pk)[−p 3 j+ p3k+ (pj− pk)3] (pj+ pk)[−p3j− p3k+ (pj+ pk)3] }× exp( n X i=1 pix1− n X i=1 p3_ix2)dτn = 1 n! Z Γn Y 1≤j<k≤n (pj− pk pj+ pk )2exp( n X i=1 pix − p3it) n Y i=1 dτ (pi)

for n ≥ 2 , where x1 and x2 have been replaced back by x and t respectively. Then f = 1 + ε Z Γ exp(p1x − p31t)dτ (p1) + ∞ X n=2 εn n! Z Γn Y 1≤j<k≤n (pj− pk pj+ pk )2 × exp( n X i=1 pix − p3it) n Y i=1 dτ (pi) (3.20) Now we set Z Γ dτ (pi) = Z ∞ −∞ N X l=1 αlδ(pi− p0l)dpi

where αl and p0l (l = 1, 2, . . . , N ) are real constants with p0j 6= p0k for j 6= k

(1 ≤ j, k ≤ N ) and δ is Dirac’s delta function. Then from (3.20) f becomes

f = 1 + N X l=1 ε Z ∞ −∞ exp(p1x − p31t)αlδ(p1− p0l)dp1 + ∞ X n=2 εn n! Z ∞ −∞ · · · Z ∞ −∞ Y 1≤j<k≤n (pj− pk pj+ pk )2exp[ n X i=1 (pix − p3it)] n Y i=1 N X li=1 αliδ(pi− p0li)dpi = 1 + N X l=1 εαlexp(p0lx − p30lt) + ∞ X n=2 εn n! N X l1···ln=1 αl1· · · αln Y 1≤j<k≤n (p0lj− p0lk p0lj+ p0lk )2× exp[ n X i=1 (p0lix − p 3 0lit)] We observe thatQ 1≤j<k≤n( p0lj − p0lk p0lj + p0lk

)2makes l1· · · ln having different values

respectively and if n > N it is impossible to make l1· · · lnhaving different values

respectively thus the terms for n > N contribute nothing. Therefore f can be written as f = 1 + N X l=1 εαlexp(p0lx − p30lt) + N X n=2 εn X (l1...ln)∈CnN αl1· · · αln× Y 1≤j<k≤n (p0lj − p0lk p0lj + p0lk )2exp[ n X i=1 (p0lix − p 3 0lit)] = 1 + N X l=1 exp(p0lx − p30lt + c0l) + N X n=2 X (l1...ln)∈CnN Y 1≤j<k≤?n (p0lj− p0lk p0lj+ p0lk )2 exp[ n X i=1 (p0lix − p 3 0lit + c0li)]

where c0li ≡ log(εαli) and

P

(l1...ln)∈CnN denotes a summation over l1, . . . , ln

which are chosen from a set {1, . . . , N } as well as l1< l2< · · · < ln

Now with p0ix − p30it + c0i≡ ηi (p0i− p0j p0i+ p0j )2≡ exp Aij we have f = 1 + N X i=1 exp ηi+ N X n=2 X (l1,...,ln)∈CNn exp[ n X j=1 ηlj+ X 1≤j<k≤n Aljlk] (3.21)

Then f expressed in (3.21) equals to f in (3.12) , in other words , we can derive pure N-soliton solution from generalized soliton solution.

(ii)

K-P equation (−4ut+ uxxx+ 6uux)x+ 3uyy = 0 can be transformed into

F (Dx1, Dx2, Dx3) = (D

4

x1− 4Dx1Dx3+ 3D

2

x2)f · f = 0

where x1= x ,x2= y , x3= t and u = 2(log f )xx. From (3.13) and (3.14) , in

this case m = 3 , we have

f1= Z Γ Z Γ2 exp[φ1(p, q)x1+ φ2(p, q)x2+ φ3(p, q)x3]dτ (p, q) (3.22) and F (φ1(p, q), φ2(p, q), φ3(p, q)) = φ41− 4φ1φ3+ 3φ22= 0 (3.23)

From (3.23) we can assign

φ1(p, q) = p − q , φ2(p, q) = p2− q2and φ3(p, q) = p3− q3 (3.24)

Hence from theorem 3.1 and Appendix A.2 we obtain

fn= 1 n! Z Γn (2) Y 1≤j<k≤n [−F (φ1(pj, qj) − φ1(pk, qk), φ2(pj, qj) − φ2(pk, qk), φ3(pj, qj) − φ3(pk, qk)) F (φ1(pj, qj) + φ1(pk, qk), φ2(pj, qj) + φ2(pk, qk), φ3(pj, qj) + φ3(pk, qk)) ] × exp[ n X i=1 φ1(pi, qi)x1+ φ2(pi, qi)x2+ φ3(pi, qi)x3] n Y i=1 dτ (pi, qi) = 1 n! Z Γn (2) Y 1≤j<k≤n [(pj− pk)(qj− qk) (pj− qk)(qj− pk) ] exp[ n X i=1 (pi− qi)x1+ (p2i − q 2 i)x2+ (p3i − q 3 i)x3] n Y i=1 dτ (pi, qi) Now we set Z Γ₍₂₎ dτ (p, q) = Z Γ1 Z Γ2 dτ (p, q) = Z ∞ −∞ Z ∞ −∞ N X l=1 αlδ(p − p0l)δ(q − q0l)dpdq

Then f = 1 + ε N X l=1 αlexp[(p0l− q0l)x1+ (p20l− q 2 0l)x2+ (p30l− q 3 0l)x3] +ε n n! N X l1,...,ln=1 αl1. . . αln Y 1≤j<k≤n (p0lj − p0lk)(q0lj − q0lk) (p0lj − q0lk)(q0lj − p0lk) × exp[ n X i=1 (p_0l i− q0li)x1+ (p 2 0li− q 2 0li)x2+ (p 3 0li− q 3 0li)x3] Now with εαiexp[(p0li− q0li)x1+ (p 2 0li− q 2 0li)x2+ (p 3 0li− q 3 0li)x3] ≡ exp ηli (p0j− p0k)(q0j− q0k) (p0j− q0k)(q0j− p0k) ≡ exp Ajk

Chapter 4

Relation with GLM

equation

In this chapter , we relate the generalized soliton solution for KdV equation to the GLM integral equation. To begin , we need the following lemma

Lemma 4.1Q

1≤i<j≤n(

pi− pj

pi+ pj)

2_{= det4}

n , where 4n is an n×n matrix with

i-j element 2pi

pi+ pj

pf see Appendix A.3

Using Lemma4.1 and letting ε = 1 , (3.20) becomes

f = 1 + Z Γ exp(px − p3t)dτ (p) + ∞ X n=2 1 n! Z Γn det(4n) exp[ n X i=1 (pix − p3it)] n Y i=1 dτ (pi)

Using the definition of determinant

A = (aij) , det A =

X

σ

sgn(σ)a1σ(1)a2σ(2)· · · anσ(n)

, a summation over all permutations , f becomes

f = 1 + Z Γ exp(px − p3t)dτ (p) + ∞ X n=2 1 n! X σ sgn(σ) Z Γn 2p1 p1+ pσ(1) · · · 2pn pn+ pσ(n) × exp[ n X i=1 (pix − p3it)] n Y i=1 dτ (pi)

Meanwhile X σ sgn(σ) Z Γn 2p1 p1+ pσ(1) · · · 2pn pn+ pσ(n) exp[ n X i=1 (pix − p3it)] n Y i=1 dτ (pi) =X σ sgn(σ) Z Γn p1· · · pnexp[ n X i=1 (−p3_it)] 2 p1+ pσ(1) · · · 2 pn+ pσ(n) exp[ n X i=1 pi+ pσ(i) 2 x] n Y i=1 dτ (pi) =X σ sgn(σ)(−1)n Z Γn p1· · · pnexp[ n X i=1 (−p3_it)] Z ∞ x · · · Z ∞ x exp[ n X i=1 pi+ pσ(i) 2 si] n Y i=1 dsi n Y i=1 dτ (pi) =X σ sgn(σ)(−1)n Z ∞ x · · · Z ∞ x Z Γn p1· · · pnexp n X i=1 [(pi+ pσ(i)) 2 si− p 3 it] n Y i=1 dτ (pi) n Y i=1 dsi =X σ sgn(σ)(−1)n Z ∞ x · · · Z ∞ x Z Γn p1· · · pnexp[ n X i=1 (pisi 2 + pisσ−1_(i) 2 − p 3 it)] n Y i=1 dτ (pi) n Y i=1 dsi

rename σ−1= ρ. The above equation becomes = Z ∞ x · · · Z ∞ x X ρ−1 sgn(ρ−1)(−1)n Z Γn p1· · · pnexp n X i=1 [(si+ sρ(i)) 2 pi− p 3 it] n Y i=1 dτ (pi) n Y i=1 dsi

It is noteworthy that we can restrict the path Γ on the left complex plane to avoid the explosion of improper integral at infinity. Now let F (t, s) = −R

Γp exp(

sp

2 − p3t)dτ (p) and we know that P σ = P σ−1 = P ρ as well as sgn(ρ) = sgn(ρ−1)

Then the above equation becomes

= Z ∞ x · · · Z ∞ x X ρ sgn(ρ)F (t, s1+ sρ(1))F (t, s2+ sρ(2)) · · · F (t, sn+ sρ(n)) n Y i=1 dsi = Z ∞ x · · · Z ∞ x detΨn n Y i=1 dsi

where Ψn is a matrix with i − j element F (t, si+ sj)

And Z Γ exp(px − p3t)dτ (p) = − Z Γ Z ∞ x p exp(ps − p3t)dsdτ (p) = − Z ∞ x Z Γ p exp(ps1+ s1 2 − p 3_{t)dτ (p)ds} 1 = Z ∞ x F (t, s1+ s1)ds1

which can be written asR∞

x detΨ1ds1 with F (t, s1+ s1) ≡ detΨ1

Therefore f becomes f = 1 + ∞ X n=1 1 n! Z ∞ x · · · Z ∞ x detΨn(t; s1. . . sn) Y i=1 dsi (4.1)

(4.1) has a form of Fredholm’s determinant of a certain integral equation. In-spired by Fredholm’s trick[13] we introduce Fredholm’s first minor

detΩn(t, x, z; s1. . . sn) ≡            F (t, x + z) F (t, x + s1) F (t, x + s2) · · · F (t, x + sn) F (t, s1+ z) F (t, s1+ s1) F (t, s1+ s2) · · · F (t, s1+ sn) F (t, s2+ z) F (t, s2+ s1) F (t, s2+ s2) · · · F (t, s2+ sn) .. . ... ... . .. ... F (t, sn+ z) F (t, sn+ s1) F (t, sn+ s2) · · · F (t, sn+ sn)            (4.2) And particularly detΩ0(t, x, z) ≡ F (t, x + z) (4.3)

Take cofactor expansion along the first row of (4.2)

detΩn= F (t, x + z)detΨn+ n X j=1 (−1)jF (t, sj+ z)×                 F (t, x + s1) F (t, x + s2) · · · F (t, x + sn) F (t, s1+ s1) F (t, s1+ s2) · · · F (t, s1+ sn) .. . ... . .. ... F (t, sj−1+ s1) F (t, sj−1+ s2) · · · F (t, sj−1+ sn) F (t, sj+1+ s1) F (t, sj+1+ s2) · · · F (t, sj+1+ sn) .. . ... . .. ... F (t, sn+ s1) F (t, sn+ s2) · · · F (t, sn+ sn)                 Rearrange the j-th column to the first one

detΩn= F (t, x + z)detΨn+ n X j=1 (−1)jF (t, sj+ z)(−1)j−1×                 F (t, x + sj) F (t, x + s1) · · · F (t, x + sj−1) F (t, x + sj+1) · · · F (t, x + sn) F (t, s1+ sj) F (t, s1+ s1) · · · F (t, s1+ sj−1) F (t, s1+ sj+1) · · · F (t, s1+ sn) .. . ... . .. ... ... . .. ... F (t, sj−1+ sj) F (t, sj−1+ s1) · · · F (t, sj−1+ sj−1) F (t, sj−1+ sj+1) · · · F (t, sj−1+ sn) F (t, sj+1+ sj) F (t, sj+1+ s1) · · · F (t, sj+1+ sj−1) F (t, sj+1+ sj+1) · · · F (t, sj+1+ sn) .. . ... . .. ... ... . .. ... F (t, sn+ sj) F (t, sn+ s1) · · · F (t, sn+ sj−1) F (t, sn+ sj+1) · · · F (t, sn+ sn)                 = F (t, x + z)detΨn(t; s1. . . sn) − n X j=1 F (t, sj+ z)detΩn−1(t, x, sj; s1, . . . , sj−1sj+1, . . . sn)

ImposingP∞ n=1_n!1 R∞ x · · · R∞ x Qn

i=1dsi on both sides of the above equation ∞ X n=1 1 n! Z ∞ x · · · Z ∞ x detΩn(t, x, z; s1. . . sn) n Y i=1 dsi = F (t, x + z) ∞ X n=1 1 n! Z ∞ x · · · Z ∞ x detΨn(t; s1, . . . , sn) n Y i=1 dsi − ∞ X n=1 1 n! Z ∞ x · · · Z ∞ x n X j=1 F (t, sj+ z)detΩn−1(t, x, sj; s1, . . . , sj−1sj+1, . . . , sn) n Y i=1 dsi

Renaming sj = s, sj+1= sj, sj+2= sj+1, . . . , sn= sn−1 for the second

summa-tion integral and using (4.1) , the above equasumma-tion becomes

∞ X n=1 1 n! Z ∞ x · · · Z ∞ x detΩn(t, x, z; s1. . . sn) n Y i=1 dsi = F (t, x + z)(f − 1) − Z ∞ x F (t, s + z)[ ∞ X n=1 1 (n − 1)! Z ∞ x · · · Z ∞ x det Ωn−1(t, x, s; s1. . . sn−1) n−1 Y i=1 dsi]ds

Now define E(t, x, z) ≡P∞

n=1 1 n! R∞ x · · · R∞ x detΩn(t, x, z; s1. . . sn) Qn i=1dsiand

recall (4.3). Then from the above we obtain

E(t, x, z) = F (t, x + z)(f − 1) − Z ∞ x F (t, s + z)[Ω0(t, x, s) + E(t, x, s)]ds = F (t, x + z)(f − 1) − Z ∞ x F (t, s + z)[F (t, x + s) + E(t, x, s)]ds ⇒ E(t, x, z) + F (t, x + z) = F (t, x + z)f (t, x) − Z ∞ x F (t, s + z)[F (t, x + s) + E(t, x, s)] It is natural to define D(t, x, z) = −E(t, x, z) − F (t, x + z) (4.4)

Then the above equation becomes

D(t, x, z) + F (t, x + z)f (t, x) + Z ∞

x

F (t, s + z)D(t, x, s)ds

Divide f on both side(of course we assume f is not identical to zero) and put K(t, x, z) = D(t,x,z)_{f (t,x)} Then we obtain the GLM equation

K(t, x, z) + F (t, x + z) + Z ∞

x

Recall F (t, s) = −R Γp exp( sp 2 − p 3_{t)dτ (p) and} R Γdτ (p) ≡ R Γc(p)dp with an

properly selected c(p) as well as suitable path Γ. Now we set Γ lie in the real-axis and imaginary-real-axis such that

F (t, s) = − Z ∞ −∞ p[exp(sp 2 − p 3_t)][−1 p N X n=1 c2_n(0)δ(p + 2pn)]dp − Z ∞ −∞

2ip[exp(isp + 8ip3t)][ir(p, 0) 4πp ]dp

where i is imaginary number pn are different real values and the first integral of

right hand side involves the path on real axis while the second integral on the imaginary. Then F (t, s) = N X n=1 c2_n(0) exp(8p3_nt)exp(−pns) + 1 2π Z ∞ −∞ r(p, 0) exp(8ip3t)exp(ips)dp which can be abbreviated as

F (t, s) = N X n=1 c2_n(t) exp(−pns) + 1 2π Z ∞ −∞ r(p, t) exp(ips)dp

with cn(t) = cn(0) exp(4p3nt) and r(p, t) = r(p, 0) exp(8ip3t)

GLM equation (4.5) with F defined as above entirely coincides with GLM equa-tion described in [14]. Those familiar with[5] would aware that cn(t) and r(p, t)

are nothing but scattering data. [15] also has a result that

u = 2∂x[K(t, x, x)]

Now we prove the same result in our approach. Firstly , we need the following lemma Lemma 4.2 fx(t, x) = D(t, x, x) pf ∂x Z ∞ x · · · Z ∞ x detΨn(t; s1, . . . , sn) n Y i=1 dsi = − n X i=1 Z ∞ x · · · Z ∞ x

detΨn(t; s1, . . . , si−1, x, si+1, . . . , sn)ds1. . . dsi−1dsi+1. . . dsn

(4.6)

From the structure of matrix Ψn we know we can construct a matrix

by interchanging i-th and (i+1)-th columns as well as i-th and (i+1)-th rows of Ψn(s1, . . . , sn) respectively. Hence we have

detΨn(s1, . . . , si−1, si+1, si, si+2, . . . , sn) = (−1)2detΨn(s1, . . . , sn)

Therefore the r.h.s of (4.6) becomes

= − n X i=1 Z ∞ x · · · Z ∞ x

(−1)2(i−1)detΨn(t; x, s1, . . . , si−1, si+1, . . . , sn)ds1· · · dsi−1dsi+1· · · dsn

Rename si+1= si, si+2= si+1, . . . , sn= sn−1 the above equation becomes

= −n Z ∞ x · · · Z ∞ x detΨn(t; x, s1, . . . , sn−1) n−1 Y i=1 dsi

From the structure of Ωn it is readily to derive

detΨn(t; x, s1, . . . , sn−1) = detΩn−1(t, x, x; s1, . . . , sn−1)

Now imposing ∂x on both sides of (4.1)

∂xf = ∞ X n=1 1 n!∂x Z ∞ x · · · Z ∞ x detΨn(t; s1, . . . , sn) n Y i=1 dsi = − ∞ X n=1 1 (n − 1)! Z ∞ x · · · Z ∞ x detΩn−1(t, x, x; s1, . . . , sn−1) n−1 Y i=1 dsi = D(t, x, x) by(4.3) and (4.4) QED

And we know u = 2(log f )xx= 2(f_fx)xThen by lemma 4.2

u(t, x) = 2(D(t, x, x)

Chapter 5

Perspective

In chapter 4 we derive GLM equation from generalized soliton solutions

K(t, x, z) + F (t, x + z) + Z ∞ x K(t, x, s)F (t, s + z)ds (5.1) with F (t, s) = − Z Γ p exp(sp 2 − p 3_{t)dτ (p) (5.2)}

As previously stated if the integral path Γ is restricted along the real and imag-inary axis , F (t, s) will be transformed into

F (t, s) = N X n=1 c2_n(t) exp(−pns) + 1 2π Z ∞ −∞ r(p, t) exp(ips)dp (5.3)

The first term of r.h.s of (5.3) represents solitons and the second one represents an oscillatory wave train. Segur[15] has gained the result that an initial dis-turbance , in general , evolves into solitons as well as an oscillatory wave train and there appears to be no permanent effect on the solitons from the interaction with the oscillatory wave train. Apparently F (t, s) in (5.2) owns more flexibility for us to investigate the interaction between solitons and oscillatory wave train than F (t, s) in (5.3). This is the value of (5.2).

Appendix A

A.1

Here we shall prove

F (Φ1− Φ2) F (Φ1+ Φ2) F (Φ3− Φ1− Φ2) + F (Φ1− Φ3) F (Φ1+ Φ3) F (Φ2− Φ1− Φ3) +F (Φ2− Φ3) F (Φ2+ Φ3) F (Φ1− Φ2− Φ3) = −F (Φ1+ Φ2+ Φ3) F (Φ1− Φ2)F (Φ1− Φ3)F (Φ2− Φ3) F (Φ1+ Φ2)F (Φ1+ Φ3)F (Φ2+ Φ3)

this relation in p16 We recall Hirota condition

X σ=1,−1 F ( N X i=1 σiPi) (N ) Y i<j F (σiPi− σjPj)σiσj= 0 , f or N = 1, 2, . . .

For N = 3 the summation has eight terms. They are

F (Φ1+ Φ2+ Φ3)F (Φ1− Φ2)F (Φ1− Φ3)F (Φ2− Φ3) (A.1) F (Φ1+ Φ2− Φ3)F (Φ1− Φ2)F (Φ1+ Φ3)F (Φ2+ Φ3) (A.2) F (Φ1− Φ2+ Φ3)F (Φ1+ Φ2)F (Φ1− Φ3)F (−Φ2− Φ3) (A.3) F (Φ1− Φ2− Φ3)F (Φ1+ Φ2)F (Φ1+ Φ3)F (−Φ2+ Φ3) (A.4) F (−Φ1+ Φ2+ Φ3)F (−Φ1− Φ2)F (−Φ1− Φ3)F (Φ2− Φ3) (A.5) F (−Φ1+ Φ2− Φ3)F (−Φ1− Φ2)F (−Φ1+ Φ3)F (Φ2+ Φ3) (A.6) F (−Φ1− Φ2+ Φ3)F (−Φ1+ Φ2)F (−Φ1− Φ3)F (−Φ2− Φ3) (A.7) F (−Φ1− Φ2− Φ3)F (−Φ1+ Φ2)F (−Φ1+ Φ3)F (−Φ2+ Φ3) (A.8)

Because F (−Φ) = F (Φ) then (A.1) = (A.8), (A.2) = (A.7), (A.3) = (A.6), (A.4) = (A.5) Therefore Hirota condition for N = 3 reduces to (A.5) + ((A.6) + ((A.7) + ((A.8) = 0 i.e.

F (−Φ1+ Φ2+ Φ3)F (−Φ1− Φ2)F (−Φ1− Φ3)F (Φ2− Φ3)

+ F (−Φ1+ Φ2− Φ3)F (−Φ1− Φ2)F (−Φ1+ Φ3)F (Φ2+ Φ3)

+ F (−Φ1− Φ2+ Φ3)F (−Φ1+ Φ2)F (−Φ1− Φ3)F (−Φ2− Φ3)

+ F (−Φ1− Φ2− Φ3)F (−Φ1+ Φ2)F (−Φ1+ Φ3)F (−Φ2+ Φ3) = 0

Now divide F (Φ1+ Φ2)F (Φ1+ Φ3)F (Φ2+ Φ3)) on both sides. Then

F (Φ1− Φ2) F (Φ1+ Φ2) F (Φ3− Φ1− Φ2) + F (Φ1− Φ3) F (Φ1+ Φ3) F (Φ2− Φ1− Φ3) +F (Φ2− Φ3) F (Φ2+ Φ3) F (Φ1− Φ2− Φ3) = −F (Φ1+ Φ2+ Φ3) F (Φ1− Φ2)F (Φ1− Φ3)F (Φ2− Φ3) F (Φ1+ Φ2)F (Φ1+ Φ3)F (Φ2+ Φ3)

A.2

This is a proof for p22

φ1(p, q) = p − q , φ2(p, q) = p2− q2 , φ3(p, q) = p3− q3 F (a, b, c) = a4− 4ac + 3b2 F (φ1(pj, qj) − φ1(pk, qk), φ2(pj, qj) − φ2(pk, qk), φ3(pj, qj) − φ3(pk, qk)) = [pj− qj− (pk− qk)]4− 4[pj− qj− (pk− qk)][p3j− q 3 j − (p 3 k− q 3 k)] + 3[p2 j− qj2− (p2k− q 2 k)] 2 [pj− qj− (pk− qk)]4 = p4 j+qj4+p4k+q 4 k−4p 3 jqj−4pjqj3−4p3jpk−4pjp3k+4p 3 jqk+4pjqk3+4q 3 jpk+qjp3k− 4q3_jqk− 4qjqk3− 4p 3 kqk− 4pkq3k+ 6p 2 jq 2 j+ 6p 2 jp 2 k+ 6p 2 jq 2 k+ 6q 2 jp 2 k+ 6q 2 jq 2 k+ 6p 2 kqk+ 12p2 jqjpk− 12p2jqjqk− 12p2jpk− 12q2jpjpk+ 12qj2pjqk− 12qj2pkqk− 12p2kpjqj+ 12p2 kpjqk− 12p2kqjqk− 12qk2pjpj− 12qk2pjpk+ 12qk2qjpk+ 24pjqjpkqk −4[pj− qj− (pk− qk)][p3j− q 3 j − (p 3 k− q 3 k)] = −4(p4 j− pjqj3− pjp3k+ pjq3k− qjp3j+ qj4+ qjp3k− qjqk3− pkp3j + pkq3j + p4k − pkqk3+ qkp3j− qkqj3− qkp3k+ qk4) 3[p2_j− q2 j − (p 2 k− q 2 k)] 2 = 3(p4 j+ q4j+ p4k+ q 4 k− 2p 2 jqj2− 2p2jp2k+ 2p 2 jq2k+ 2q 2 jp2k− 2q 2 jqk2− 2p 2 kq 2 k)

Adding the three terms we obtain

pk)(pjq2k+ pjqjpk− pjqjqk− pjpkqk− q2jpk+ q2jqk− qjqk2+ qjpkqk) = 12(pj− pk)(qj− qk)(qjqk− pjqk+ pjpk− qjpk) Likewise F (φ1(pj, qj) + φ1(pk, qk), φ2(pj, qj) + φ2(pk, qk), φ3(pj, qj) + φ3(pk, qk)) = [pj− qj+ (pk− qk)]4− 4[pj− qj+ (pk− qk)][p3j− qj3+ (p3k− q 3 k)] + 3[p2 j− qj2+ (p2k− q2k)]2 = [pj−qj−(qk−pk)]4−4[pj−qj−(qk−pk)][p3j−q 3 j−(q 3 k−p 3 k)]+3[p 2 j−q 2 j−(q 2 k−p 2 k)] 2 = 12(pj− qk)(qj− pk)(qjpk− pjpk+ pjqk− qjqk) = −12(pj− qk)(qj− pk)(qjqk− pjqk+ pjpk− qjpk) Therefore −F (φ1(pj, qj) − φ1(pk, qk), φ2(pj, qj) − φ2(pk, qk), φ3(pj, qj) − φ3(pk, qk)) F (φ1(pj, qj) + φ1(pk, qk), φ2(pj, qj) + φ2(pk, qk), φ3(pj, qj) + φ3(pk, qk)) =(pj− pk)(qj− qk) (pj− qk)(qj− pk)

A.3

This is a proof for lemma 4.1 The following proof can be readily extended to n × n matrix. Now we consider a 3 × 3 matrix

D3(x1, x2, x3; α1, α2, α3) =         1 x1− α1 1 x1− α2 1 x1− α3 1 x2− α1 1 x2− α2 1 x2− α3 1 x3− α1 1 x3− α2 1 x3− α3         (A.9)

If we set xi = xj or αi = αj for i 6= j , then D3= 0. Hence D3 is divisible by

ζ12(x₁, x₂, x₃)ζ 1 2(α₁, α₂, α₃) where ζ12_{(x1, x2, x3) ≡}       1 x₁ x2 1 1 x₂ x2 2 1 x₃ x2 3       = Y 1≤j<k≤3 (xk− xj) = (−1) 3(3−1) 2 Y 1≤j<k≤3 (xj− xk)

Now multiplying the i-th row of D3 with ui = (xi− α1)(xi− α2)(xi− α3) we

obtain u1u2u3D3 =       (x1− α2)(x1− α3) (x1− α1)(x1− α3) (x1− α1)(x1− α2) (x2− α2)(x2− α3) (x2− α1)(x2− α3) (x2− α1)(x2− α2) (x3− α2)(x3− α3) (x3− α1)(x3− α3) (x3− α1)(x3− α2)      

Owing to D3being divisible by ζ

1

2(x₁, x₂, x₃)ζ 1

2(α₁, α₂, α₃) and u₁u₂u₃D₃

hav-ing the same polynomial degree on x1, x2, x3, α1, α2, α3 as ζ

1 2(x₁, x₂, x₃)ζ12(α₁, α₂, α₃) we have u1u2u3D3= (const)ζ 1 2_{(x1, x2, x3)ζ}12_{(α1, α2, α3)}

Now put αi= xi for i = 1, 2, 3 then the above equation becomes       (x1− x2)(x1− x3) 0 0 0 (x2− x1)(x2− x3) 0 0 0 (x3− x1)(x3− x2)       = (const)ζ(x1, x2, x3) ⇒ (−1)3(3−1)2 Y 1≤j<k≤3 (xj− xk)2= (const) Y 1≤j<k≤3 (xj− xk)2 (const) = (−1)3(3−1)2 Therefore D3= (−1)3(3−1)2 ζ12(x₁, x₂, x₃)ζ12(α₁, α₂, α₃) u1u2u3

Now replace αi= −xi for i = 1, 2, 3 then

        1 x1+ x1 1 x1+ x2 1 x1+ x3 1 x2+ x1 1 x2+ x2 1 x2+ x3 1 x3+ x1 1 x3+ x2 1 x3+ x3         = D3(x1, x2, x3; −x1, −x2, −x3) =(−1) 3(3−1) 2 ζ12(x₁, x₂, x₃)ζ12(−x₁, −x₂, −x₃) 23_x 1x2x3Q1≤j<k≤3(xj+ xk)2 while ζ12(−x₁, −x₂, −x₃) = (−1) 3(3−1) 2 Y 1≤j<k≤3 (xk−xj) = (−1) 3(3−1) 2 (−1) 3(3−1) 2 Y 1≤j<k≤3 (xj−xk) = Y 1≤j<k≤3 (xj− xk) ⇒         1 x1+ x1 1 x1+ x2 1 x1+ x3 1 x2+ x1 1 x2+ x2 1 x2+ x3 1 x3+ x1 1 x3+ x2 1 x3+ x3         = 1 23_x 1x2x3 Y 1≤j<k≤3 (xj− xk xj+ xk )2 ⇒         2x1 x1+ x1 2x1 x1+ x2 2x1 x1+ x3 2x2 x2+ x1 2x2 x2+ x2 2x2 x2+ x3 2x3 x3+ x1 2x3 x3+ x2 2x3 x3+ x3         = Y 1≤j<k≤3 (xj− xk xj+ xk )2

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