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