Structure of soliton equations
José de Jesús Martı́nez
Department of Mathematics, Kobe University,
Rokko, Kobe 657-8501, Japan
Contents
1 Introduction
1
2 The KP equation
2.1 Wronskian Solutions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
2.2 Grammian Solutions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
1
1
5
3 The coupled KP equation
3.1 Wronski-type pfaffian solutions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
3.2 Gramm-type pfaffian solutions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
8
9
13
4 The BKP equation
15
5 The two-dimensional Toda lattice equation
5.1 Wronskian solution . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
5.2 Grammian solution . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
18
19
20
6 The two-dimensional Toda molecule equation
6.1 Bi-directional wronskian solution . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
6.2 Double wronskian solution . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
21
22
25
1
Introduction
In this report we discuss the structure of various soliton equations. We make use of wronskian forms and
pfaffian representation to construct N -soliton solutions to equations such as the KP equation, BKP equation
and the Toda lattice equation. The first step in constructing these new solutions, is to bilinearize the soliton
equation. We recall that the bilinear form is useful in that the Hirota-type equation is a polynomial of the
D-operator acting on a pair of functions. On many occasions we see the D-operator acting on a pair of identical
functions. This functions is commonly known as the tau function. How we represent the latter is the key to
understand the structure of soliton equations. Once we have bilinearized the equation, we make use of either
wronskian, grammian or pfaffian forms to express the tau function. We then compute the derivatives of this
functions and plug them into the bilinear equation. The last step is proving that the new solution satisfies the
bilinear equation. It was Freeman and Nimmo who discovered that by writing the tau function in the forms
mentioned above, what we obtain is some determinantal identity. We discuss the cases of pfaffian identities,
plücker relations, and the Jacobi identity.
2
2.1
The KP equation
Wronskian Solutions
The KP equation or two-dimensional KdV equation is a two-dimensional space and one-dimensional time
nonlinear partial differential equation.
(ut + uxxx + 6uux )x + uyy = 0
By multiplying variables t, x and y by the appropriate constant scaling factors we obtain,
1
(−4ut + uxxx + 6uux )x + 3uyy = 0
Under the logarithmic transformation u = 2(log τ )xx , we derive the KP bilinear equation
(D14 − 4D1 D3 + 3D22 )τ · τ = 0
where x = x1 , y = x2 and t = x3 . Since this is a KdV-type equation, the N -soliton solution is given by,
τN =
X
exp
N
hX
µi ηi +
i=1
(N )
X
Aij µi µj
i
i<j
Here ηi = Pi x1 + Qi x2 + Ωi x3 + c and the dispersion relation is expressed as,
Pi4 − 4Pi Ωi + 3Q2i = 0
We want to know if there is a wronskian expression that satisfies the bilinear equation. Assume that we
may express τN as the N × N wronskian,
τN
(0)
f
1
(0)
f2
= .
..
(0)
f
N
(1)
f1
(1)
f2
..
.
(1)
fN
(N −1) . . . f1
(N −1) . . . f2
.. . (N −1) . . . fN
where each function fi satisfies the differential equation,
∂m
∂
fi =
fi
∂xm
∂xm
(1)
For τN to hold, we check that the wronskian expression satisfies the bilinear equation. First we expand the
D-operator and simply write the equation in terms of normal derivatives.
(D14 − 4D1 D3 + 3D22 )τ · τ
= (2τ τxxxx + 6τxx τxx − 8τx τxxx ) − 4(−2τx3 τx + 2τx3 x τ ) + 3(2τx2 x2 τ − 2τx2 τx2 )
h
i
= 2 τ (τxxxx − 4τx3 x + 3τx2 x2 ) − 4τx (τxxx − τx3 ) + 3(τxx − τx2 )(τxx + τx2 )
By doing so, we notice that all we have to do is compute the derivatives of the wronskian τN . Let us use the
following notation for τN ,
τN = [0, 1, . . . , N − 2, N − 1]
The virtue of the wronskian determinant is that its derivative is very easy to compute. To calculate the
derivative of a determinant we
2
∂
τN = [0, 1, . . . , N − 2, N ]
∂x
To calculate the derivative of τN with respect to another variable, we recall equation (1). For example,
τx2 = [0, 1, . . . , N − 3, N, N − 1] + [0, 1, . . . , N − 3, N − 2, N + 1]
= [0, 1, . . . , N − 3, N − 2, N + 1] − [0, 1, . . . , N − 3, N − 1, N ]
Hence,
τ = [0, . . . , N − 4, N − 3, N − 2, N − 1]
τx = [0, . . . , N − 4, N − 3, N − 2, N ]
τxx = [0, . . . , N − 4, N − 3, N − 2, N + 1] + [0, . . . , N − 4, N − 3, N − 1, N ]
τxxx = [0, . . . , N − 4, N − 3, N − 2, N + 2] + 2[0, . . . , N − 4, N − 3, N − 1, N + 1] + [0, . . . , N − 4, N − 3, N − 2, N ]
τxxxx = [0, . . . , N − 4, N − 3, N − 2, N + 3] + 3[0, . . . , N − 4, N − 3, N − 1, N + 2] + 2[0, . . . , N − 4, N − 3, N, N + 1]
+ 3[0, . . . , N − 4, N − 2, N − 1, N + 1] + [0, . . . , N − 3, N − 2, N − 1, N ]
τx2 = [0, 1, . . . , N − 3, N − 2, N + 1] − [0, 1, . . . , N − 3, N − 1, N ]
τx2 x2 = [0, . . . , N − 4, N − 3, N − 2, N + 3] − [0, . . . , N − 4, N − 3, N − 1, N + 2] + 2[0, . . . , N − 4, N − 3, N, N + 1]
− [0, . . . , N − 4, N − 2, N − 1, N + 1] + [0, . . . , N − 3, N − 2, N − 1, N ]
τx3 = [0, . . . , N − 4, N − 3, N − 2, N + 2] − [0, . . . , N − 4, N − 3, N − 1, N + 1] + [0, . . . , N − 4, N − 2, N − 1, N ]
τx3 x = [0, . . . , N − 4, N − 3, N − 2, N + 3] − [0, . . . , N − 4, N − 3, N, N + 1] + [0, . . . , N − 3, N − 2, N − 1, N ]
This way it becomes really easy to compute the following and represent them using Maya diagrams.
N -2
[0, . . . , N − 3, N − 2, N − 1]
=
h
τx =
[0, . . . , N − 3, N − 2, N ]
=
h
τxx − τx2 =
[0, . . . , N − 3, N − 1, N ]
=
τxx + τx2 =
[0, . . . , N − 3, N − 2, N + 1]
=
τxxx + τx3 =
[0, . . . , N − 3, N − 1, N + 1]
=
τxxxx − 4τx3 x + 3τx2 x2 =
[0, . . . , N − 3, N, N + 1]
=
τ
=
N -1
N +1
N
h
h
h
h
h
h
h
h
h
h
Substituting these results into the Bilinear KP equation we get,
N -2
(D14
− 4D1 D3 +
3D22 )τ
·τ
h
=
+
h
−
h
N -1
N
h
N -2
N -1
N
h
×
h
3
N +1
×
h
h ×
h
N +1
h
h
h
= 0
We recognize this to be the Maya Diagram for the plücker relation
(0 1)(2 3) − (0 2)(1 3) + (0 3)(1 2) = 0 .
and thus confirm that τN satisfies the KP bilinear equation. Next, by using Young Diagrams, the Bilinear KP
equation becomes,
τ∅ τ
Note
−τ τ
+τ
τ =0.
The pfaffian expression for a wronskian determinant is given by,
τN = (d0 , . . . , dN −1 , 1, . . . , N ) ,
where
(dn , j) =
∂n
fj , (dm , dn ) = 0 .
∂xn
Now consider the vanishing pfaffian,
N −2 N −1
N
N +1
(d0 , . . . , dN −1 , dN , dN +1 , 1, . . . , N ) = [][][][] = 0
Then the Bilinear KP equation also reduces to the pfaffian identity
(1, 2, . . . , 2n)(a1 , a2 , . . . , a2m , 1, 2, . . . , 2n)
=
2m
X
(−1)k (a1 , ak , 1, 2, . . . , 2n)(a2 , a3 , . . . , a
bk , . . . , a2m 1, 2, . . . , 2n)
k=2
when m = 2,
(a1 , a2 , a3 , a4 , •)(•) = (a1 , a2 , •)(a3 , a4 , •)
− (a1 , a3 , •)(a2 , a4 , •)
+ (a1 , a4 , •)(a2 , a3 , •)
Employing Maya diagrams the above pfaffian identity is expressed as
a1
=
a2
h
h
h
h
−
h
+
h
a3
h
a4
a1
a2
a3
a4
h ×
h
×
h
×
h
h ×
h
4
h
h
h
2.2
Grammian Solutions
A Grammian is a determinant G = det(gi,j )1≤i,j,≤N where the entries are given by
b
Z
gi,j ≡
fi fj dx
a
Previously we learned to calculate the derivative of the pfaffian corresponding to the N × N determinant,
τN = det(ai,j )1≤i,j,≤N
= (1, 2, . . . , N, N ∗ , . . . , 2∗ , 1∗ )
where
Z
x
ai,j ≡ constant +
fi gj dx .
Seeing that G and τN are so similar, we also call τN a grammian. Now we want to know if τN satisfies the
Bilinear KP equation.
(D14 − 4D1 D3 + 3D22 )τ · τ
h
i
= 2 τ (τxxxx − 4τx3 x + 3τx2 x2 ) − 4τx (τxxx − τx3 ) + 3(τxx − τx2 )(τxx + τx2 )
=0.
We keep in mind the following regarding the pfaffian entries.
Z
∗
x
(i, j ) = ai,j = c +
fi gj dx
(i, j) = (i∗ , j ∗ ) = 0
∂n
gj ,
(dn , j ∗ ) =
∂xn
(dm , d∗n ) = 0
(d∗n , i) =
∂n
fi
∂xn
(dn , i) = (d∗m , j ∗ ) = 0
for m, n ≥ 0. Also, the functions fi and gj satisfy the linear differential equations below,
∂n
∂
fi =
fi
∂xn
∂xn
∂
∂n
gj = (−1)n+1 n gj
∂xn
∂x
Previously we learned how to compute
∂n
∂xn τN .
τ = (1, . . . , N, N ∗ , . . . , 1∗ )
τx = (d0 , d∗0 , τ)
τxx = (d1 , d∗0 , τ) + (d0 , d∗1 , τ)
τxxx = (d2 , d∗0 , τ) + 2(d1 , d∗1 , τ) + (d0 , d∗2 , τ)
τxxxx = (d3 , d∗0 , τ) + (d0 , d∗3 , τ) + 3(d2 , d∗1 , τ) + 3(d1 , d∗2 , τ) + 2(d0 , d∗0 , d1 , d∗1 , τ)
5
In order to compute the derivative of τ with respect to another variable we first make the following calculations.
∂
ai,j = fi gj = (d0 , d∗0 , i, j ∗ )
∂x
Z x
Z x
∂
∂
ai,j =
fi gj dx =
fi x2 gj + fi gj x2 dx
∂x2
∂x
Z x 2
fi xx gj − fi gj xx dx = fi x gj − fi gj x
=
∂
ai,j
∂x3
= −(d1 , d∗0 , i, j ∗ ) + (d0 , d∗1 , i, j ∗ )
Z x
Z x
∂
=
fi gj dx =
fi xxx gj − fi gj xxx dx
∂x3
= fi xx gj + fi gj xx − fi x gj x
= (d2 , d∗0 , i, j ∗ ) − (d1 , d∗1 , i, j ∗ ) + (d0 , d∗2 , i, j ∗ )
Noting the above and using the definition for the derivative of a pfaffian , we obtain
∂
τ
∂x2
∂2
τ
∂x22
∂
τ
∂x3
∂2
τ
∂x3 x
= −(d1 , d∗0 , τ ) + (d0 , d∗1 , τ )
= (d3 , d∗0 , τ ) − (d2 , d∗1 , τ ) + 2(d0 , d∗0 , d1 , d∗1 , τ ) − (d1 , d∗2 , τ ) + (d0 , d∗3 , τ )
= (d2 , d∗0 , τ ) − (d1 , d∗1 , τ ) + (d0 , d∗2 , τ )
= (d3 , d∗0 , τ ) − (d0 , d∗0 , d1 , d∗1 , τ ) + (d0 , d∗3 , τ )
By simply adding and subtracting pfaffians, we get the expressions below pertaining to the Bilinear KP equation,
τxxxx − 4τx3 x + 3τx2 x2 = 12(d0 , d∗0 , d1 , d∗1 , τ )
τ = (τ )
τxxx − τx3 = 3(d1 , d∗1 , τ )
τx = (d0 , d∗0 , τ )
τxx − τx2 = 2(d1 , d∗0 , τ ) = −2(d∗0 , d1 , τ )
τxx + τx2 = 2(d0 , d∗1 , τ )
Next we rewrite the latter using Maya Diagrams,
d0
τxxxx − 4τx3 x + 3τx2 x2 =
12
h
d∗
0
h
d1
d∗
1
h
h
h
h
τ =
τxxx − τx3 =
3
h
τx =
h
τxx + τx2 = −2
τxx + τx2 =
By substituting into the bilinear equation we that,
6
3
h
h
h
h
d∗
0
d0
(D14 − 4D1 D3 + 3D22 )τ · τ
d1
h
h
−
h
h
−
h
=
d∗
0
h
d∗
0
d0
d1
d∗
0
h ×
h
×
h ×
h
h
h
= 0
We know this expression holds since it is identical to the Maya Diagram for the Jacobi Identity,
D
n−1
n
n−1
n
n−1
D
−D
D
=D
n−1
n
n
n−1
n−1
n
D
n
Furthermore, it is a special case of the pfaffian identity,
(1, 2, . . . , 2n)(a1 , a2 , . . . , a2m , 1, 2, . . . , 2n)
=
2m
X
(−1)k (a1 , 1, 2, . . . , 2n, ak )(a2 , a3 , . . . , a
bk , . . . , a2m 1, 2, . . . , 2n)
k=2
Next, let d1 → d0 , d2 → d∗0 , d3 → d1 and d4 → d∗1 . Thus, when m = 2 we have that,
(d0 , d∗0 , d1 , d∗1 , •)(•) =
(d0 , d∗0 , •)(d1 , d∗1 , •)
− (d0 , d1 , •)(d∗0 , d∗1 , •)
+ (d0 , d∗1 , •)(d∗0 , d1 , •)
which we may then express using Maya Diagrams,
d∗
0
d0
=
h
h
h
h
−
h
+
h
d∗
1
d1
h
d∗
0
d0
d∗
1
d1
h ×
h
×
h
×
h
h ×
h
h
h
h
However, note that the term
(d0 , d1 , •)(d∗0 , d∗1 , •)
vanishes since there is an uneven-number of indices with and without the superscript ∗. Hence,
d0
d∗
0
h
h
=
h
h
+
h
d1
h
d∗
0
d0
d∗
0
d1
d∗
0
h ×
h
×
h ×
7
h
h
h
Now we present an example. Consider the two-soliton bilinear solution of the KP equation,
τ2 = 1 + exp(η1 ) + exp(η2 ) + a12 exp(η1 + η2 )
By parametrizing the variable η and introducing new phase factors, the solution above becomes
τ2 = 1 +
1
1
(p1 − p2 )(q1 − q2 )
exp(ξ1 − ξb1 ) +
exp(ξ2 − ξb2 ) +
exp(ξ1 + ξ2 − ξb1 − ξb2 )
p1 − q1
p2 − q2
(p1 − q2 )(q1 − p2 )(p1 − q1 )(p2 − q2 )
where
ξi = pi x1 + p2i x2 + p3i x3 + c
ξbi = q x1 + q 2 x2 + q 3 x3 + c
i
i
i
Notice how this is equivalent to the following Grammian determinant,
1 + 1 exp(ξ − ξb )
1
b 1
1
p1 −q1
p1 −q2 exp(ξ1 − ξ2 ) τ2 = 1
1
b1 )
b2 )
p −q
exp(ξ
−
ξ
1
+
exp(ξ
−
ξ
2
2
p2 −q2
2
1
The elements of the Grammian τ2 = det(ai,j )1≤i,j≤2 are written as,
Z
ai,j = δi,j +
x
exp(ξi ) exp(−ξbj ) dx
Furthermore, we see that indeed exp(ξi ) and exp(−ξbj ) satisfy the linear differential equations,
∂
∂n
exp(ξi ) =
exp(ξi )
∂xn
∂xn
∂
∂n
exp(−ξbj ) = (−1)n+1 n exp(−ξbj ) .
∂xn
∂x
3
The coupled KP equation
We have been studying the bilinear KP equation
(D14 − 4D1 D3 + 3D22 )τ · τ = 0
It was shown that the τ function may be expressed as a wronskian or grammian determinant. Either way, the
bilinear KP equation reduces to a known special case of the pfaffian identity. We saw that from the pfaffian
identity
(a1 ,a2 , . . . , a2m , 1, 2, . . . , 2n)(1, 2, . . . , 2n)
2n
X
=
(−1)j (a1 , aj , 1, 2, . . . , 2n) (a2 , a3 , . . . , abj , . . . , a2m , 1, 2, . . . , 2n)
j=2
8
we let m = 2 and thus obtain
(a1 , a2 , a3 , a4 , •)(•)
= (a1 , a2 , •)(a3 , a4 , •)
−(a1 , a3 , •)(a2 , a4 , •)
+(a1 , a4 , •)(a2 , a3 , •)
Specifically, when solution τ is expressed as the wronskian
τN = [0, 1, . . . , N − 2, N − 1] ,
the bilinear KP equation reduces to the plücker relation
(f 0 1)(f 2 3) − (f 0 2)(f 1 3) + (f 0 3)(f 1 2) = 0 .
Then we learned that by expressing the solution τ as a grammian
τN = (1, 2, . . . , N, N ∗ , . . . , 2∗ , 1∗ ) ,
the bilinear KP equation reduces to the Jacobi Identity for determinants
n−1
n
n−1
n
n−1
D
D
−D
D
=D
n−1
n
n
n−1
n−1
n
D.
n
From this observation, Professor Hirota realized that it could be possible to find new hierarchies. This new
hierarchy is called the coupled KP equation. The lowest order equations are
(D14 − 4D1 D3 + 3D22 )τ · τ = 24 σ
bσ
(2)
(D13
(D13
+ 2D3 + 3D1 D2 )σ · τ = 0
(3)
+ 2D3 − 3D1 D2 )b
σ·τ =0
(4)
These functions τ , σ and σ
b can be expressed in terms of what is called a wronski-type pfaffian and gram-type
pfaffian.
3.1
Wronski-type pfaffian solutions
We define the pfaffian entries (l, m) satisfying the following differential rules with respect to some variable xn
such that
∂
(l, m) = (l + n, m) + (l, m + n) .
∂xn
Let us consider a special wronski-type pfaffian,
(l, m) =
M h
X
(l)
(m)
Φk Ψk
k=1
9
(m)
(l)
− Φ k Ψk
i
.
(5)
Assume that functions Φ and Ψ satisfy the following linear differential equations,
∂
(n)
Φk = Φk
∂xn
,
∂
(n)
Ψk = Ψk .
∂xn
Then, the differential rule (5) for entry (l, m) holds. In general, the differential formula for pfaffian (i0 , i1 , . . . , i2N −1 )
is given by
2N
−1
X
∂
(i0 , i1 , . . . , i2N −1 ) =
(i0 , . . . , ik+n , . . . , i2N −1 ) .
∂xn
k=0
The reason why this is called a wronski-type pfaffian is because the differential rule for this pfaffian is exactly
the same as that of a wronskian determinant,
2N
−1
X
∂
[0, . . . , 2N − 1] =
[0, . . . , k + n, . . . , 2N − 1] .
∂xn
k=0
We now denote the wronski-type pfaffian as
τ W = (0, 1, . . . , 2N − 1) .
Since this pfaffian behaves like a normal wronskian, we see that by substituting τ W into the left-hand side of
the bilinear KP equation yields
2N -2 2N -1
1
4
24 (D1
− 4D1 D3 + 3D22 )τ W · τ W
h
=
−
h
+
h
h
=
2N
2N +1
h
×
h
h
2N -2 2N -1
h
h
2N
2N +1
h
×
h
h ×
h
h
h
h ×
Note The difference between the wronskian solution for solution τ and that of the wronski-type pfaffian is
that here the right-hand side is not necessarily zero. In fact, if it is equal to zero, then we say that solution τ
is a wronskian.
We introduce two new pfaffians,
2N -2 2N -1
σW
=
(0, 1, . . . , 2N − 3)
=
σ
bW
=
(0, 1, . . . , 2N + 1)
=
10
h
h
2N
h
2N +1
h
We check if equations (3) and (4) are satisfied by τ , σ and σ
b. As we have done previously with the bilinear KP
equation, we start by expressing equations (3) and (4) using normal differential operators.
(D13 + 2D3 + 3D1 D2 )σ · τ =(σxxx + 2σx3 + 3σx2 x )τ
− 3(σxx + σx2 )τx
+ 3σx (τxx − τx2 )
− σ(τxxx + 2τx3 − 3τx2 x ) ,
and
(D13 + 2D3 − 3D1 D2 )b
σ· τ =τ (b
σxxx + 2b
σx3 − 3b
σx2 x )
− 3τx (b
σxx − σ
bx2 )
+ 3(τxx + τx2 )b
σx
− (τxxx + 2τx3 + 3τx2 x )b
σ,
Next, we only need to compute the necessary derivatives for functions τ , σ and σ
b. Employing the differential
formula,
2N
X
∂
(0, 1, . . . , 2N ) =
(0, . . . , k + n, . . . , 2N ) ,
∂xn
k=0
we obtain the following formulae:
τ
τ W =(0, . . . , 2N − 4, 2N − 3, 2N − 2, 2N − 1) ,
τxW =(0, . . . , 2N − 4, 2N − 3, 2N − 2, 2N ) ,
W
τxx
=(0, . . . , 2N − 4, 2N − 3, 2N − 1, 2N ) + (0, . . . , 2N − 4, 2N − 3, 2N − 2, 2N + 1) ,
W
τxxx
=(0, . . . , 2N − 4, 2N − 2, 2N − 1, 2N ) + 2(0, . . . , 2N − 4, 2N − 3, 2N − 1, 2N + 1)
+ (0, . . . , 2N − 4, 2N − 3, 2N − 2, 2N + 2) ,
τxW2
τxW2 x
τxW3
= − (0, . . . , 2N − 3, 2N − 1, 2N ) + (0, . . . , 2N − 3, 2N − 2, 2N + 1) ,
= − (0, . . . , 2N − 4, 2N − 2, 2N − 1, 2N ) + (0, . . . , 2N − 4, 2N − 3, 2N − 2, 2N + 2) ,
=(0, . . . , 2N − 4, 2N − 2, 2N − 1, 2N ) − (0, . . . , 2N − 4, 2N − 3, 2N − 1, 2N + 1)
+ (0, . . . , 2N − 4, 2N − 3, 2N − 2, 2N + 2) .
σ
σ W =(0, . . . , 2N − 6, 2N − 5, 2N − 4, 2N − 3) ,
σxW =(0, . . . , 2N − 6, 2N − 5, 2N − 4, 2N − 2) ,
W
σxx
=(0, . . . , 2N − 6, 2N − 5, 2N − 3, 2N − 2) + (0, . . . , 2N − 6, 2N − 5, 2N − 4, 2N − 1) ,
W
σxxx
=(0, . . . , 2N − 6, 2N − 4, 2N − 3, 2N − 2) + 2(0, . . . , 2N − 5, 2N − 3, 2N − 1)
+ (0, . . . , 2N − 5, 2N − 4, 2N ) ,
σxW2
σxW2 x
σxW3
= − (0, . . . , 2N − 5, 2N − 3, 2N − 2) + (0, . . . , 2N − 5, 2N − 4, 2N − 1) ,
= − (0, . . . , 2N − 6, 2N − 4, 2N − 3, 2N − 2) + (0, . . . , 2N − 6, 2N − 5, 2N − 4, 2N ) ,
=(0, . . . , 2N − 6, 2N − 4, 2N − 3, 2N − 2) − (0, . . . , 2N − 6, 2N − 5, 2N − 3, 2N − 1)
+ (0, . . . , 2N − 6, 2N − 5, 2N − 4, 2N ) .
11
σ
b
σ
bW =(0, . . . , 2N − 2, 2N − 1, 2N, 2N + 1)
σ
bxW =(0, . . . , 2N − 2, 2N − 1, 2N, 2N + 2)
W
σ
bxx
=(0, . . . , 2N − 2, 2N − 1, 2N + 1, 2N + 2) + (0, . . . , 2N − 2, 2N − 1, 2N, 2N + 3)
W
σ
bxxx
=(0, . . . , 2N − 2, 2N, 2N + 1, 2N + 2) + 2(0, . . . , 2N − 2, 2N − 1, 2N + 1, 2N + 3)
+ (0, . . . , 2N − 2, 2N − 1, 2N, 2N + 4)
σ
bxW2
σ
bxW2 x
σ
bxW3
= − (0, . . . , 2N − 1, 2N + 1, 2N + 2) + (0, . . . , 2N − 1, 2N, 2N + 3)
= − (0, . . . , 2N − 2, 2N, 2N + 1, 2N + 2) + (0, . . . , 2N − 2, 2N − 1, 2N + 4)
=(0, . . . , 2N − 2, 2N, 2N + 1, 2N + 2) − (0, . . . , 2N − 2, 2N − 1, 2N + 1, 2N + 3)
+ (0, . . . , 2N − 2, 2N − 1, 2N, 2N + 4)
Just as we have done before, we substitute these into the bilinear equations and express them as Maya diagrams.
2N -3 2N -2 2N -1
(D13
+ 2D3 + 3D1 D2 )σ · τ =
h
h
×
h
h
×
h
h
+
h
−
2N -1
−
h
×
2N
2N +1 2N +2
h
+ 2D3 − 3D1 D2 )b
σ·τ =
2N -3 2N -2 2N -1
h ×
h
−
(D13
2N
2N -1
h
×
h
h
+
−
2N
×
h
×
h
h
h ×
h
h
2N
h
h
h
h
h
h =0
2N +1 2N +2
h
h
h
h
h
h
=0
We point out that the wronski-type pfaffians τ W , σ W and σ
b do satisfy the set of equations since they reduce
to a known pfaffian identity:
(1, 2, . . . , 2n − 1, 2n)(1, 2, . . . , 2n − 1, a1 , a2 , a3 )
= (1, 2, . . . , 2n − 1, a1 )(1, 2, . . . , 2n, a2 , a3 )
−(1, 2, . . . , 2n − 1, a2 )(1, 2, . . . , 2n, a1 , a3 )
+(1, 2, . . . , 2n − 1, a3 )(1, 2, . . . , 2n, a1 , a2 )
The identity above is a special case of the identity
(1, 2, . . . , 2n)(a1 , a2 , . . . , am , 1, 2, . . . , 2n − 1) ,
m
X
=
(−1)k+1 (ak , 1, 2, . . . , 2n − 1)(a1 , a2 , . . . , b
ak , . . . , am , 1, 2, . . . , 2n) .
k=1
12
3.2
Gramm-type pfaffian solutions
In the previous section we saw that the coupled KP equation
(D14 − 4D1 D3 + 3D22 )τ · τ = 24b
σσ ,
(D13 + 2D3 + 3D1 D2 )σ · τ = 0 ,
(D13 + 2D3 − 3D1 D2 )b
σ·τ =0,
is satisfied by the Wronski-type pfaffian τ W , σ W and σ
bW ,
τ W = (0, . . . , 2N − 1) ,
σ W = (0, . . . , 2N − 3) ,
σ
bW = (0, . . . , 2N + 1) ,
where the differential rule was given by,
2N
X
∂
(0, . . . , 2N ) =
(0, . . . , k + n, . . . , 2N ) .
∂xn
k=0
Now we check if the coupled KP equation is satisfied by the following Gramm-type pfaffians,
τ G = (1, 2, . . . , 2N ) ,
σ G = (c1 , c0 , 1, 2, . . . , 2N ) ,
σ
bG = (d0 , d1 , 1, 2, . . . , 2N ) .
We define the pfaffian entries as,
Z
(i, j) ≡ cij +
x
(fi gj − fj gi )dx ,
∂n
∂n
f
,
(c
,
i)
≡
gi ,
i
n
∂xn
∂xn
(dm , dn ) = (cm , cn ) = (cm , dn ) = 0 ,
(dn , i) ≡
where cij = −cji . We also require functions fi and gj to satisfy the linear differential equations below,
∂n
∂
fi =
fi ,
∂xn
∂xn
∂n
∂
gj = (−1)n+1
gj .
∂xn
∂xn
Same as with the KP equation, we make the following computations to construct a differential rule.
∂
(i, j) =fi gj − fj gi = (d0 , i)(c0 , j) − (d0 , j)(c0 , i)
∂x
=(c0 , d0 , i, j) ,
Z x
∂
∂
(i, j) =
(fi gj − fj gi )dx = fix gj − fi gjx − fjx gi + fj gix
∂x2
∂x2
=(c0 , d1 , i, j) − (c1 , d0 , i, j)
Z x
∂
∂
(i, j) =
(fi gj − fj gi )dx = fixx gj − fjxx gi − fix gjx + fjx gix + fi gjxx − fj gixx
∂x3
∂x3
=(c2 , d0 , i, j) + (c0 , d2 , i, j) − (c1 , d1 , i, j)
13
Let the indices 1, 2, . . . , 2n be denoted by •. Then, from the method on the section of the derivative formula of
a pfaffian, we get the following differential rules:
∂
(cm , dn , •) = (cm+1 , dn , •) + (cm , dn+1 , •) + (c0 , d0 , cm , dn , •) ,
∂x
∂
(cm , dn , •) = −(cm+2 , dn , •) + (cm , dn+2 , •) + (c0 , d1 , cm , dn , •) − (c1 , d0 , cm , dn , •) ,
∂x2
∂
(cm , dn , •) = (cm+3 , dn , •) + (cm , dn+3 , •) + (c0 , d2 , cm , dn , •) + (c2 , d0 , cm , dn , •) − (c1 , d1 , cm , dn , •) ,
∂x3
∂
(cm , cn , •) = (cm+1 , cn , •) + (cm , cn+1 , •) + (c0 , d0 , cm , cn , •) ,
∂x
∂
(cm , cn , •) = −(cm+2 , cn , •) − (cm , cn+2 , •) + (c0 , d1 , cm , cn , •) − (c1 , d0 , cm , cn , •) ,
∂x2
∂
(cm , cn , •) = (cm+3 , cn , •) + (cm , cn+3 , •) + (c0 , d2 , cm , cn , •) + (c2 , d0 , cm , cn , •) − (c1 , d1 , cm , cn , •) ,
∂x3
∂
(dm , dn , •) = (dm+1 , dn , •) + (dm , dn+1 , •) + (c0 , d0 , dm , dn , •) ,
∂x
∂
(dm , dn , •) = (dm+2 , dn , •) + (dm , dn+2 , •) + (c0 , d1 , dm , dn , •) + (c1 , d0 , dm , dn , •) ,
∂x2
∂
(dm , dn , •) = (dm+3 , dn , •) + (dm , dn+3 , •) + (c0 , d2 , dm , dn , •) + (c2 , d0 , dm , dn , •) − (c1 , d1 , dm , dn , •) ,
∂x3
Now we can easily compute the derivatives of τ G , σ G and σ
bG .
τ
τ G = (•)
τxG = (c0 , d0 , •)
G
τxx
= (c1 , d0 , •) + (c0 , d1 , •)
G
τxxx
= (c2 , d0 , •) + 2(c1 , d1 , •) + (c0 , d2 , •)
G
τxxxx
= (c3 , d0 , •) + (c0 , d3 , •) + 3(c2 , d1 , •) + 3(c1 , d2 , •) + 2(c0 , d0 , c1 , d1 , •)
τxG2 = −(c1 , d0 , •) + (c0 , d1 , •)
τxG2 x2 = −(c2 , d1 , •) + (c0 , d3 , •) + (c3 , d0 , •) − (c1 , d2 , •)
τxG3 = (c2 , d0 , •) + (c0 , d2 , •) − (c1 , d1 , •)
τxG3 x = (c3 , d0 , •) + (c0 , d3 , •) − (c0 , d0 , c1 , d1 , •)
σ
σ G = −(c0 , c1 , •)
σxG = −(c0 , c2 , •)
G
σxx
= (c3 , c1 , •) − (c1 , c2 , •)
G
σxxx
= (c4 , c0 , •) + 2(c3 , c1 , •) − (c0 , c1 , c2 , d0 , •)
σxG2 = −(c3 , c0 , •) − (c1 , c2 , •)
σxG2 x = −(c4 , c0 , •) − (c0 , c1 , c2 , d0 , •)
σxG3 = (c4 , c0 , •) + (c1 , c3 , •) − (c0 , c1 , c2 , d0 , •)
14
σ
b
G
σ
b = (d0 , d1 , •)
σ
bxG = (d0 , d2 , •)
G
σ
bxx
= (d1 , d2 , •) − (d0 , d3 , •)
G
σ
bxxx
= (d0 , d4 , •) + 2(d1 , d3 , •) − (d0 , d1 , d2 , c0 , •)
σ
bxG2 = −(d0 , d3 , •) − (d1 , d2 , •)
σ
bxG2 x = (d0 , d4 , •) + (d0 , d1 , d2 , c0 , •)
σ
bxG3 = −(d1 , d3 , •) + (d0 , d4 , •) − (d0 , d1 , d2 , c0 , •)
We substitute these results into the bilinear equations and get
c0
1
4
24 (D1
− 4D1 D3 +
3D22 )τ G
·τ
G
h
=
d0
h
−
h
−
=
−
h
c0
1
3
6 (D1
G
+ 2D3 + 3D1 D2 )σ · τ
G
=
−
h
+
=
+
h
−
h
+ 2D3 − 3D1 D2 )b
σG · τ G
=
=
c0
d1
h
h ×
h
h ×
h
h
×
h
h
×
c2
h
h ×
h
h
×
h
h
c1
h
c2
h
d0
h
h
h
×
d1
h
×
h
c1
h
c0
d0
h
d0
h
h
0
d0
1
3
6 (D1
c1
c1
d1
d2
−
c0
d0
×
+
h
+
h
−
h
h ×
h
h
d1
h
h
h
h
h
×
h
×
d2
h
c0
h
h
h
0
These are simply pfaffian identities. Hence, we confirm that τ G , σ G and σ
bG do indeed satisfy the coupled KP
equation.
4
The BKP equation
In this section we will study another type of soliton equation. Unlike the KP equation, whose solution is
expressed in determinant form, the solution to the BKP equation is best suited to be written in pfaffian form.
15
The BKP hierarchy, introduced by Date, Jimbo, Kashiwara and Miwa, includes the two following equations in
the bilinear form:
h
i
(D3 − D31 )D−1 + 3D12 τ · τ = 0
(6)
(D16 − 5D13 D3 − 5D32 + 9D1 D5 )τ · τ = 0
..
.
(7)
The reduction of these equations yields various well known soliton equations: Sawada-Kotera equation, Ramani
equation, KdV+ Sawada-Kotera equation, shallow water waves model equation, etc.
Now we want to show that the KBP equation is satisfied by the pfaffian
τN = (1, 2, . . . , 2N ) ,
where the pfaffian entry (i, j) has the form
Z
(i, j) = cij +
x
Dx fi (x) · fj (x)dx .
Note that, function fi (x) satisfy the linear differential equations
∂n
∂
fi (x) =
fi (x) ,
∂xn
∂xn
(n = −1, 1, 3, 5, . . . ) ,
where n = −1 means integrating function fi with respect to x once. Now we check to see if τN satisfies the
BKP equations (6) and (7). We start by expressing equation (6) in normal derivatives,
h
i
(D3 − D13 )D−1 + 3D12 τ · τ = ∂−1 (∂3 − ∂13 )τ + 3∂12 τ τ
+ 3 (∂−1 ∂12 − ∂1 )τ ∂1 τ
− 3 ∂−1 ∂1 τ ∂12 τ
− (∂3 − ∂13 )τ ∂−1 τ
In the case of equation (7), we have that,
(D16 − 5D13 D3 − 5D32 + 9D1 D5 )τ · τ = − 5 (∂13 − ∂3 )τ (2∂13 + ∂3 )τ
− 15 (−∂14 + ∂1 ∂3 )τ ∂12 τ
+ 3 (−3∂5 + 5∂12 ∂3 − 2∂15 )τ ∂1 τ
+ (−5∂32 + 9∂1 ∂5 − 5∂13 ∂3 + ∂16 )τ τ
Using the definition for the pfaffian entries and the Hirota derivative, we differentiate entry (i, j) with respect
to variables x1 , x−1 , x3 and x5 ,
16
∂
(i, j) = D1 fi · fj = fi x fj − fi fj x
∂x
= (d0 , d1 , i, j)
Z x
∂
fi x fj x−1 − fi x−1 fj x dx = fi fj x−1 − fi x−1 fj
(i, j) =
∂x−1
= (d−1 , d0 , i, j)
Z x
∂
fi (4) fj + fi x fj (3) − fi (3) fj x − fi fj (4) dx
(i, j) =
∂x3
= −2(d1 , d2 , i, j) + (d0 , d3 , i, j)
Z x
∂
(i, j) =
fi (6) fj + fi x fj (5) − fi (5) fj x − fi fj (6) dx
∂x5
= 2(d2 , d3 , i, j) − 2(d1 , d4 , i, j) + (d0 , d5 , i, j)
Let • represent the indices 1, . . . , 2N . Then the differential rule for τN is given by,
∂
(dm , dn , •) = (dm+1 , dn , •) + (dm , dn+1 , •) + (d0 , d1 , dm , dn , •)
∂x
∂
(dm , dn , •) = (dm−1 , dn , •) + (dm , dn−1 , •) + (d−1 , d0 , dm , dn , •)
∂x−1
∂
(dm , dn , •) = (dm+3 , dn , •) + (dm , dn+3 , •) + (d0 , d3 , dm , dn , •) − 2(d1 , d2 , dm , dn , •)
∂x3
∂
(•) = 2(d2 , d3 , •) − 2(d1 , d4 , •) + (d0 , d5 , •)
∂x5
Next we compute the necessary derivatives,
∂
(•) = (d0 , d1 , •)
∂x
∂2
(•) = (d0 , d2 , •)
∂x 2
∂3
(•) = (d1 , d2 , •) + (d0 , d3 , •)
∂x 3
∂4
(•) = 2(d1 , d3 , •) + (d0 , d4 , •)
∂x 4
∂5
(•) = 2(d2 , d3 , •) + 3(d1 , d4 , •) + (d0 , d5 , •)
∂x 5
∂6
(•) = 5(d2 , d4 , •) + 2(d0 , d1 , d2 , d3 , •) + 4(d1 , d5 , •) + (d0 , d6 , •)
∂x 6
∂
(•) = (d−1 , d0 , •)
∂x−1
∂2
(•) = (d−1 , d1 , •)
∂x−1 ∂x
∂3
(•) = (d0 , d1 , •) + (d−1 , d2 , •)
∂x−1 ∂x2
∂4
(•) = 2(d0 , d2 , •) + (d−1 , d3 , •) + (d−1 , d0 , d1 , d2 , •)
∂x−1 ∂x3
∂2
(•) = (d−1 , d3 , •) − (d0 , d2 , •) − 2(d−1 , d0 , d1 , d2 , •)
∂x−1 ∂x3
17
∂
(•) = (d0 , d3 , •) − 2(d1 , d2 , •)
∂x3
∂2
(•) = (d0 , d6 , •) − 2(d1 , d5 , •) + 2(d2 , d4 , •) − 4(d0 , d1 , d2 , d3 , •)
∂x3 2
∂2
(•) = (d0 , d4 , •) − (d1 , d3 , •)
∂x∂x3
∂3
(•) = (d0 , d5 , •) − (d2 , d3 , •)
∂x2 ∂x3
∂4
(•) = (d1 , d5 , •) + (d0 , d6 , •) − (d2 , d4 , •) − (d0 , d1 , d2 , d3 , •)
∂x3 ∂x3
∂
(•) = 2(d2 , d3 , •) − 2(d1 , d4 , •) + (d0 , d5 , •)
∂x5
∂2
(•) = (d0 , d6 , •) − (d1 , d5 , •) + 2(d0 , d1 , d2 , d3 , •)
∂x∂x5
Substituting the corresponding expressions into equation (6), we obtain
d−1 d0
d1
d2
d−1 d0
−
+
−
+
d2
×
×
d1
×
×
d1
d2
d3
=
0
=
0
Similarly, equation (7) reduces to
d0
d1
d2
d3
d0
−
+
−
×
×
×
×
These are simply pfaffian identities. Therefore, solution τN does indeed satisfy the BP equations.
5
The two-dimensional Toda lattice equation
Now we discuss the two-dimensional Toda lattice equation (2DTL) of the form
∂2
Qn = Vn+1 − 2Vx + Vn−1 ,
∂s∂x
Qn = log(1 + Vn ) ,
(8)
where n ∈ Z. We note here that this equation is an extension to the KP hierarchy. There are many ways
to obtain solutions of the 2DTL. Here we show that the solution τn of the 2DTL in the bilinear form, may
be expressed in terms of a wronskian and a grammian determinant. Let us bilinearize equation (8) using the
dependent variable transformation
18
Vn =
∂2
log(τn )
∂s∂x
After integrating with respect to x and s we obtain the bilinear 2DTL
Ds Dx τn · τn = 2(τn+1 τn−1 − τn2 )
5.1
(9)
Wronskian solution
The solution τn to equation (9) may be written as the Casorati determinant
φ1 (n) φ1 (n + 1)
φ2 (n) φ2 (n + 1)
τn = .
..
..
.
φN (n) φN (n + 1)
...
...
...
φ1 (n + N − 1) φ2 (n + N − 1) ..
.
φN (n + N − 1)
where functions φi (n) satisfy the linear differential equation
∂
φi (n) = φi (n + 1)
∂x
∂
φi (n) = −φi (n − 1)
∂s
We compute the derivatives and shifts of τn .
-1
τn
=
∂
∂s τn
=
∂
∂x τn
=
0
h
−
h
h
τn+1 =
h
τn−1 =
∂2
=
∂x∂s τn
−
−
h
h
h
1
N -3
N -2
N -1
h ...
h
h
h
h ...
h
h
h
h ...
h
h
h ...
h
h
h ...
h
h
h ...
h
h
h ...
h
h
h
h
h
h
h
By expanding the Hirota operator on equation (9) we have that
∂ 2 τn
∂
∂
τn −
τn τn = τn+1 τn−1 − τn2
∂s∂x
∂s ∂x
Substituting the results obtained previously into the above equation gives
19
N
-1
0
−
h
+
h
=
N -1
-1
N
0
h ×
h
h
×
h
h
h ×
h
N -1
N
h
h
h
This is simply the Plücker relation for determinants. Therefore solution τn satisfies the two-dimensional
toda lattice equation
Ds Dx τn · τn = τn+1 τn−1 − τn2 .
5.2
Grammian solution
Next we discuss the grammian solution to the Toda lattice equation. The 2DTL equation
∂ ∂ ∂
∂
τn τn −
τn τn = τn+1 τn−1 − τn2 .
∂s ∂x
∂s ∂x
(10)
This equation as a solution τn of the grammian form
Z
n
τn = cij + (−1)
x
(n) (−n)
fi gj
dx 1≤i,j≤N
we require functions fi and gi to satisfy the linear differential equations
∂ (n)
(n+k)
f
= fi
,
∂xk i
∂ (n)
(n−k)
f
= −fi
,
∂sk i
∂ (−n)
(−n+k)
g
= (−1)k−1 gi
∂xk i
∂ (−n)
(−n−k)
g
= (−1)k gi
∂sk i
where k = 1, 2, . . . . Here variables x1 and s1 , correspond to variables x and s, respectively. We express the
above grammian using the pfaffian
τn = (1, 2, . . . , N, N ∗ , . . . , 2∗ , 1∗ )n
Z x
(n) (−n)
∗
n
(i, j )n = cij + (−1)
fi gj dx
(i, j)n = (i∗ , j ∗ )n = 0
Moreover, the following equalities hold for pfaffian entries.
(n)
(d∗n , i)n = fi ,
(−n)
(d−n , j ∗ )n = (−1)n gi
(d−n , d∗n )n = (d∗−n , d∗n )n = (d−n , dn )n = 0
We compute the necessary derivatives and shifts in n of the pfaffian entry (i, j ∗ )n .
20
∂
(n) (−n)
(i, j ∗ )n = (−1)n fi gj
∂x
= (d−n , d∗n , i, j ∗ )n
∂
(−n−1)
(i, j ∗ )n = (−1)n−1 f (n−1 )i gj
∂s
= (d−n−1 , d∗n−1 , i, j ∗ )n
(n)
(i, j ∗ )n+1 = (i, j ∗ )n + (−1)n+1 fi
(−n−1)
gj
= (i, j ∗ )n + (d−n−1 , d∗n , i, j ∗ )n
(n−1)
(−n)
(i, j ∗ )n−1 = (i, j ∗ )n + (−1)n−1 fi
gj
= (i, j ∗ )n − (d−n , d∗n−1 , i, j ∗ )n
With the above results, we are able to obtain the following derivatives and shifts with respect to τn .
∂
(•)
∂x n
∂
(•)
∂s n
∂2
(•)
∂s∂x n
∗
(i, j )n+1
= (d−n , d∗n , •)n
= (d−n−1 , d∗n−1 , •)n
= −(d−n , d∗n−1 , •)n + (d−n−1 , d∗n , •)n + (d−n , d∗n , d−n−1 , d∗n−1 , •)n
= (•)n + (d−n−1 , d∗n , •)n
(i, j ∗ )n−1 = (•)n − (d−n , d∗n−1 , •)n
By substituting the above expressions into the 2DTL equation (10), we obtain the following expression in
Maya diagrams
-n-1
n-1∗
h
h
−
h
h
+
h
-n
h
n∗
-n-1
n-1∗
-n
n∗
h ×
h
×
h ×
h
h
h
= 0
which shows that τn is a solution to the two-dimensional Toda lattice equation
Ds Dx τn · τn = τn+1 τn−1 − τn2 .
6
The two-dimensional Toda molecule equation
The two-dimensional Toda molecule equation (2DTM) is given by
∂2
Qn = Vn+1 − 2Vn + Vn−1 ,
∂x∂y
Qn = log(Vn ) ,
(11)
21
where n = 1, 2, . . . N . Hence, n represents the position in the Toda molecule. Notice that if Vn = 0, then
Qn = ∞. Thus, the boundary conditions for the 2DTM equation are different to that of the two-dimensional
Toda molecule equation. These new boundary conditions are given as
V0 = VN +1 = 0 .
The difference in the boundary conditions affects the structure of its solution. Equation (11) may be bilinearized
through the same dependent variable transformation
Vn =
∂2
log(τn )
∂x∂y
as the 2DTL equation. After integrating equation (11) with respect to variables x and y we have that
∂ 2 τn
∂
∂
τn −
τn τn = τn+1 τn−1 .
∂x∂y
∂x ∂y
(12)
Employing the bilinear operator, equation (12) becomes
Dx Dy τn · τn = 2 τn+1 τn−1 .
We consider two solutions to the two-dimensional Toda molecule equation: the Bi-directional wronskian and
Double wronskian solutions.
6.1
Bi-directional wronskian solution
Consider equation (12)
∂
∂ 2 τn
∂
τn −
τn τn = τn+1 τn−1 .
∂x∂y
∂s ∂x
Dividing both sides by τn2 we see that the left-hand side of the equation may be rewritten as,
∂2
τn+1 τn−1
log(τn ) =
∂x∂y
τn2
It follows from the dependent variable transformation that
Vn =
∂2
τn+1 τn−1
log(τn ) =
∂x∂y
τn2
Now let
τ0 = 1 ,
τN +1 = Φ(x)Ω(y)
where functions Φ(x) and Ω(y) are arbitrary functions. Then by means of the convention
τ−1 = 0 ,
τN +2 = 0
the boundary conditions V0 = VN +1 = 0 are satisfied.
22
The solution τn of the bilinear two-dimensional Toda molecule equation is expressed by means of an n × n
wronskian
∂ i−1 ∂ j−1
τn = i−1 j−1 Ψ(x, y)
∂x
∂y
1≤i,j≤n
For now the function Ψ(x, y) is an arbitrary function. Later on we show how this function satisfies the
boundary conditions for the 2DTM equation. By ignoring the y derivative in τn , we see that this solution is a
wronskian in the row direction. Similarly, by ignoring the x derivative, solution τn is also a wronskian in the
column direction. Thus, this solution τn is called a bi-directional wronskian solution. Hereafter, we employ the
following notation to denote the derivatives of function Ψ(x, y)
∂m ∂n
Ψ(x, y) = Ψm,n
∂xm ∂y n
We check that the case for n = 1 satisfies equation (12). If n = 1, then equation (12) becomes
∂ 2 τ1
∂
∂
τ1 −
τ1 τ1 = τ2 τ0 .
∂x∂y
∂x ∂y
We know that τ0 = 1. Then by computing the derivatives of τ1 we have that
Ψ1,1 Ψ0,0 − Ψ1,0 Ψ0,1
Ψ
= 0,0
Ψ1,0
Ψ0,1 .
Ψ1,1 Hence τn is a solution if n = 1. Through a similar calculation we see that τn is also a solution for τ2 . Then
bilinear 2DTM equation for n = 2
∂ 2 τ2
∂
∂
τ2 −
τ2 τ2 = τ3 τ1 .
∂x∂y
∂x ∂y
yields
Ψ0,0
Ψ2,0
Ψ0,2 Ψ0,0
Ψ2,2 Ψ1,0
Ψ0,1 Ψ0,0
−
Ψ1,1 Ψ2,0
Ψ0,0
Ψ0,2 = Ψ
Ψ1,2 1,0
Ψ2,0
Ψ0,1 Ψ0,0
Ψ2,1 Ψ1,0
Ψ0,1
Ψ1,1
Ψ2,1
Ψ0,2 Ψ1,2 Ψ0,0
Ψ2,2 Now we show that solution τn satisfies the bilinear 2DTM equation for any n. In order to do so, we will
show that equation (12) is in fact the Jacobi Identity for determinants. Let
∂ i−1 ∂ j−1
= τn+1
D ≡ i−1 j−1 Ψ(x, y)
∂x
∂y
1≤i,j≤n+1
We make use of the minors of D, using the notation
a1,1
.
..
i
D
= i
j
.
..
aN,1
a1,2
..
.
...
j
...
..
.
aN,2
a1,N
..
.
..
.
...
...
aN,N
.
to represent the determinant of D where the ith row and the j th column have been deleted. Similarly, the
determinant
23
D
i
k
j
l
is the determinant obtained by eliminating the ith and j th rows and the k th the lth columns of D. Using this
notation, we see that
n+1
τn = D
n+1
τn−1
n
=D
n
n+1
n+1
∂
n
τn = D
n+1
∂x
∂
n+1
τn = D
n
∂y
Substituting the above results into equation (12) gives
n−1
n
n−1
n
n−1
D
D
−D
D
=D
n−1
n
n
n−1
n−1
n
D
n
As mentioned before, this is simply the Jacobi Identity for determinants. Therefore, τn is a solution of the
bilinear two-dimensional Toda molecule equation. Previously we said that function Ψ(x, y) is an arbitrary
function. Next we confirm that it satisfies the boundary conditions. Now let
Ψ(x, y) =
N
+1
X
uk (x)vk (y)
k=1
where u(x) and v(y) are arbitrary functions. By choosing Ψ(x, y) to take the form above, we have that solution
τn becomes
∂ i−1 ∂ j−1
τn = i−1 j−1 Ψ(x, y)
∂x
∂y
1≤i,j≤n
+1
∂ i−1 ∂ j−1 NX
uk (x)vk (y)
= i−1 j−1
∂x
∂y
1≤i,j≤n
k=1
+1
NX
∂ i−1
∂ j−1
=
u
(x)
v
(y)
k
k
∂xi−1
∂y j−1
1≤i,j≤n
k=1
We note that τn can then be expressed as the determinant of the product of an n × (N + 1) matrix and an
(N + 1) × n matrix. We let An be this n × (N + 1) matrix such that
(An )ik =
∂ i−1
uk (x)
∂xi−1
Similarly, we let Bn be the (N + 1) × n matrix such that
(Bn )kj =
∂ j−1
vk (y)
∂y j−1
24
Then τn becomes
τn = An × Bn In the case where n = N + 1, An and Bn are square matrices. Therefore, we rewrite τN +1 as
τn = AN +1 × BN +1 and choose functions Φ(x) and Ω(y) for AN +1 and BN +1 respectively such that
τN +1 = Φ(x)Ω(y)
This is a boundary condition for τn . Hence we have confirmed that τn is indeed a solution for the Toda
molecule equation satisfying the boundary conditions.
6.2
Double wronskian solution
We consider another solution to the 2DTM equation
Dx Dy τn · τn = 2τn+1 τn−1
The boundary conditions for this new solution are given by
τ0 = Ω(y) ,
τM = Φ(x)
Next, we introduce functions fi (x) and gi (y), where i = 1, . . . , M .Let us consider the double wronskian
(0)
f
1
(0)
f2
τn = .
..
(0)
f
M
(1)
. . . f1
(n−1)
. . . f2
..
.
(1)
...
f1
(1)
f2
..
.
fM
(n−1)
g1
(0)
g2
..
.
(n−1)
gM
fM
(0)
g1
(1)
g2
..
.
(0)
gM
(M −n−1)
(1)
. . . g1
(M −n−1)
. . . g2
..
.
(1)
...
where functions fi and gi satisfy the linear differential equations
(n)
fi
(n)
gi
∂n
∂
fi =
fi
∂xn
∂xn
∂
∂n
=
gi =
gi
n
∂x
∂xn
=
Consider the case n = 0.
(0)
g1
τ0 = ...
(0)
gM
(M −1)
...
g1
...
gM
Likewise, the case for n = M gives
25
(M −1)
(M −n−1)
gM
τM
(0)
f1
= ...
(0)
fM
(M −1)
...
f1
..
.
(M −1)
...
fM
We may assign function Ω(y) to determinant τ0 since gi is a function of y. Similarly we assign function Φ(x)
to determinant τM since fi is just a function of x. Hence, the boundary conditions are satisfied. Employing
Maya diagrams and computing the necessary derivatives and shifts of τn , we have that
-n
0
1
n-2
=
h
h ···
h
τnx =
h
h ···
h
τny
=
h
h ···
h
τnxy =
h
h ···
h
τn+1 =
h
h ···
h
τn−1 =
h
h ···
h
τn
n
n-1
n+1
-2
M
1
h
h
h
h ···
h
h
h
h ···
h
h
h
h
h ···
h
h
h
h
h ···
h
h
h ···
h
h
h
We rewrite the Toda molecule equation as
∂
∂ 2 τn
∂
+
τn τn = 0
∂x∂y ∂x ∂y
Then using Maya diagrams the above equation becomes
-1
-n
n-1
n
h
−
h
+
h
M
-1
-n
-n
M
h
n-1
n
M
×
h
h ×
h
-n
M
h
×
h
-n
M
h ···
h
τn+1 τn−1 − τn
1
-1
-n
M
h
h
h
0
h
h
h
= 0
This is equivalent to the plücker relation
(f 0 1)(f 2 3) − (f 0 2)(f 1 3) + (f 0 3)(f 1 2) = 0
and therefore τn satisfies the two-dimensional Toda molecule equation.
26
h
+
-n
M