Determinants and Their Applications in Mathematical Physics

6.4 The Kay–Moses Equation 249

=− 4 c(c−1)α

2

1

. (6.3.20)

Referring to (6.3.16),

(

y+4n

2

1+x

)′

=4

[

(c−1)xA

11

+n

2

1+x

]′

=− 4 cλ

2

. (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

[

D

2

+ε

2

+2D

2

(logA)

]

y= 0 (6.4.1)

is satisfied by the equation

y=e

−ωεx



 1 −

n

∑

i,j=1

e

(ci+cj)ωεx

A

ij

cj− 1



,ω^2 =− 1 ,

where

A=|ars|n,

ars=δrsbr+

e

(cr+cs)ωεx

cr+cs

.

Thebr,r≥ 1 , are arbitrary constants and thecr,r≥ 1 , are constants such

thatcj=1, 1 ≤j≤nandcr+cs=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. LetA=|ars(u)|ndenote the symmetric determinant in which

ars=δrsbr+

e

(cr+cs)u

cr+cs

=asr,

a

′

rs=e

(cr+cs)u

. (6.4.2)

Then the double-sum relations (A)–(D) in Section 3.4 withfr=gr=cr

become

(logA)

′

=

∑

r,s

e

(cr+cs)u

A

rs

, (6.4.3)