Determinants and Their Applications in Mathematical Physics

(Chris Devlin) #1
A.12 B ̈acklund Transformations 339

∇·

(

∇ψ

φ

)

=

1

φ^2

(φ∇

2
ψ−∇φ·∇ψ), (A.12.10)

∇·

(

∇ψ

φ
2

)

=

1

φ
3

(φ∇

2
ψ− 2 ∇φ·∇ψ), (A.12.11)


2
(logφ)=

1

φ
2

[φ∇

2
φ−(∇φ)

2
], (A.12.12)


2
(logρ)=0. (A.12.13)

Applying (A.12.12) and (A.12.11), the coupled equations (A.12.2) and


(A.12.3) become


φ

2

2
(logφ)+(∇ψ)

2
=0, (A.12.14)

∇·

(

∇ψ

φ
2

)

=0. (A.12.15)

Transformationβ(Ehlers)


If the pairP(φ, ψ) is a solution of (A.12.4) and (A.12.5), andφ



andψ


are

functions which satisfy the relations


a.φ


=

ρ

φ

,

b.


∂ψ


∂ρ

=−

ωρ

φ
2

∂ψ

∂z

,

c.

∂ψ


∂z

=

ωρ

φ
2

∂ψ

∂ρ

,(ω
2
=−1),

then the pairP







) is also a solution.

Proof. Applying (A.12.6) and (A.12.7) to (A.12.15),


∇·

(

1

φ
2

∂ψ

∂ρ

,

1

φ
2

∂ψ

∂z

)

=0,


∂ρ

(

ρ

φ
2

∂ψ

∂ρ

)

+


∂z

(

ρ

φ
2

∂ψ

∂z

)

=0,

which is satisfied by (b) and (c). Eliminatingψfrom (b) and (c),



∂ρ

(

φ
2

ρ

∂ψ

∂ρ

)

+


∂z

(

φ
2

ρ

∂ψ

∂z

)

=0,


2
ψ


∂ρ
2


1

ρ

∂ψ


∂ρ

+


2
ψ


∂z
2

=−

2

φ

(

∂φ

∂ρ

∂ψ


∂ρ

+

∂φ

∂z

∂ψ


∂z

)

.

Hence, referring to (A.12.8) and (a),



2
ψ


=

2 φ

ρ

[(

1

φ


ρ

φ
2

∂φ

∂ρ

)

∂ψ

∂ρ


ρ

φ
2

∂φ

∂z

∂ψ

∂z

]

=

2 φ

ρ

[


∂ρ

(

ρ

φ

)

∂ψ


∂ρ

+


∂z

(

ρ

φ

)

∂ψ


∂z

]
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