Determinants and Their Applications in Mathematical Physics

(Chris Devlin) #1
6.5 The Toda Equations 257

whereAnandBnare Hankelians defined as


An=|φm|n, 0 ≤m≤ 2 n− 2 ,

Bn=|φm|n, 1 ≤m≤ 2 n− 1 ,

φ


m=(m+1)φm+1.

Proof.


B

(n)
1 n

=(−1)

n+1
A

(n)
11

,

A

(n+1)
1 ,n+1

=(−1)

n
Bn. (6.5.10)

It follows from Theorems 4.35 and 4.36 in Section 4.9.2 on derivatives of


Turanians that


D(An)=−(2n−1)A

(n+1)
n+1,n,

D(Bn)=− 2 nB

(n+1)
n,n+1

,

D(A

(n)
11 )=−(2n−1)A

(n+1)
1 ,n+1;1n,

D(B

(n)
11
)=− 2 nB

(n+1)
1 ,n+1;1,n

. (6.5.11)

The algebraic identity in Theorem 4.29 in Section 4.8.5 on Turanians is


satisfied by bothAnandBn.


B

2
n
y


2 n− 1
=BnD(B

(n)
11

)−B

(n)
11
D(Bn)

=2n

[

B

(n)
11

B

(n+1)
n,n+1
−BnB

(n+1)
1 ,n+1;1n

]

=2nB

(n)
1 nB

(n+1)
1 ,n+1

=− 2 nA

(n)
11

A

(n+1)
11

.

Applying the Jacobi identity,


A

(n)
11

A

(n+1)
11
(y 2 n−y 2 n− 2 )=An+1A

(n)
11
−AnA

(n+1)
11

=An+1A

(n+1)
1 ,n+1;1,n+1

−A

n+1
n+1,n+1

A

(n+1)
11

=−

[

A

(n+1)
1 ,n+1

] 2

=−B

2
n.

Hence,


y


2 n− 1
(y 2 n−y 2 n− 2 )=2n,

which proves the theorem whennis odd.


[

A

(n+1)
1 ,n+1

] 2

y


2 n=A

(n+1)
11
D(An+1)−An+1D(A

(n+1)
11

)

=(2n+1)

[

An+1A

(n+2)
1 ,n+2;1,n+1−A

(n+1)
11 A

(n+2)
n+2,n+1

]

=−(2n+1)A

(n+1)
1 ,n+1

A

(n+2)
1 ,n+2

.
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