Page 264 - Determinants and Their Applications in Mathematical Physics

6.4 The Kay–Moses Equation 249
2
= −4c(c − 1)α . (6.3.20)
1
Referring to (6.3.16),

y +4n (c − 1)xA + n
2
11 2
=4
1+ x 1+ x
2
= −4cλ . (6.3.21)
Differentiating (6.3.19) and using (6.3.17),

y =8c(c − 1)λα 1 β 1 . (6.3.22)
The theorem follows from (6.3.19) and (6.3.22).

6.4 The Kay–Moses Equation

Theorem. The Kay–Moses equation, namely
2 2 2
D + ε +2D (log A) y = 0 (6.4.1)
is satisfied by the equation

 
n
e (c i +c j )ωεx A ij 2
y = e −ωεx  1 −  , ω = −1,
c j − 1
i,j=1
where
A = |a rs | n ,
e (c r +c s )ωεx
a rs = δ rs b r + c r + c s .

The b r , r ≥ 1, are arbitrary constants and the c r , r ≥ 1, are constants such
that c j =1, 1 ≤ j ≤ n and c r + c s =0, 1 ≤ r, s ≤ n, but are otherwise
arbitrary.
The analysis which follows differs from the original both in the form of
the solution and the method by which it is obtained.
Proof. Let A = |a rs (u)| n denote the symmetric determinant in which

e (c r +c s )u
a rs = δ rs b r + c r + c s = a sr ,
a = e (c r +c s )u . (6.4.2)
rs
Then the double-sum relations (A)–(D) in Section 3.4 with f r = g r = c r
become

(log A) = e (c r +c s )u A , (6.4.3)
rs
r,s