294 6. Applications of Determinants in Mathematical Physics
Hence, the application of transformationγtoPngivesP
′
n
.
In order to prove that the application of transformationβtoP
′
n
gives
Pn+1, it is required to prove that
φn+1=
ρ
φ′
n
,
which is obviously satisfied, and
∂ψn+1
∂ρ
=−
ωρ
(φ′
n
)^2
∂ψ
′
n
∂z
∂ψn+1
∂z
=
ωρ
(φ
′
n)
2
∂ψ
′
n
∂ρ
, (6.10.19)
that is,
∂
∂ρ
[
(−1)
n+1
ωρ
n− 1
E
n 1
]
=−ωρ
[
ρ
n− 2
A
11
] 2
∂
∂z
[
(−1)
n
ωA
1 n
ρ
n− 2
]
,
∂
∂z
[
(−1)
n+1
ωρ
n− 1
E
n 1
]
=ωρ
[
ρ
n− 2
A
11
] 2
∂
∂ρ
[
(−1)
n
ωA
1 n
ρ
n− 2
]
(ω
2
=−1). (6.10.20)
But when the derivatives of the quotients are expanded, these two relations
are found to be identical with the two identities in Lemma 6.10.4 which
have already been proved. Hence, the application of transformationβto
P
′
ngivesPn+1and the theorem is proved.
The solutions of (6.10.1) and (6.10.2) can now be expressed in terms of
the determinantBand its cofactors. Referring to Lemmas 6.17 and 6.18,
φn=
ρ
n− 2
Bn− 1
Bn− 2
,
ψn=−
(−ω)
n
ρ
n− 2
B 1 n
Bn− 2
(ω
2
=−1),n≥ 3 , (6.10.21)
φ
′
n
=
Bn− 1
ρ
n− 2
Bn
,
ψ
′
n
=
(−ω)
n
B 1 n
ρ
n− 2
Bn
,n≥ 2. (6.10.22)
The first few pairs of solutions are
P
′
1 (φ, ψ)=
(
ρ
u 0
,
−ωρ
u 0
)
,
P 2 (φ, ψ)=(u 0 ,−u 1 ),
P
′
2
(φ, ψ)=
(
u 0
u
2
0
+u
2
1
,
u 1
u
2
0
+u
2
1
)
,
P 3 (φ, ψ)=
(
ρ(u
2
0 +u
2
1 )
u 0
,
ωρ(u 0 u 2 −u
2
1 )
u 0