1-9 Equivalence theorem of conformal mapping 47
Proof (third part).
(ii) ⇒ (iv).
In order to derive a linear system of partial differential equations for {u U , u V , v U , v V }, we depart from
the inverse right Cauchy–Green deformation tensor C r since it contains just the above quoted partials.
For an inverse portrait involving the partials {U u , U v , V u , V v }, in previous sections, we start from the
inverse left Cauchy–Green deformation tensor, a procedure we are not following further.
1st step:
C
−1
r = J l G
−1
l J
T
l =
G
−1
r
λ 2 ⇔
⎡
⎢
⎢
⎢
⎣
u U u V
v U v V
G
−1
l
u U v U
u V v V
=
G
−1
r
λ 2
x 1 :=
u U
u V
, x 2 :=
v U
v V
⎤
⎥
⎥
⎥
⎦
⇒
⎡
⎢
⎢
⎢
⎢
⎢
⎢
⎢
⎢
⎣
(α) x
T
1 G
−1
l x 1 = +
g 22
λ 2
(β) x
T
2 G
−1
l x 2 = +
g 11
λ 2
(γ) x
T
1 G
−1
l x 2 = −
g 12
λ 2
(δ) x
T
2 G
−1
l x 1 = −
g 12
λ 2
⎤
⎥
⎥
⎥
⎥
⎥
⎥
⎥
⎥
⎦
. (1.164)
Without loss of generality – see through the remark that follows after the proof – let us here assume
that the right two-dimensional Riemann manifold (i. e. the right parameterized surface) M
2
r is charted
by orthogonal parameters (orthogonal coordinates) such that g 12 = 0 holds. Such a parameterization
of a surface can always be achieved though it might turn out to be a difficult numerical procedure.
2nd step:
x
T
2 G
−1
l x 1 = 0 (δ)
“Ansatz” x 1 = G l Xx 2 (X = unknown matrix)
⇔ x
T
2 G
−1
l x 2 = 0 ∀ x 2 ∈ R
2×1 (ε) . (1.165)
A quadratic form over the field of real numbers can only be zero (“isotropic”) if and only if X is
antisymmetric, i. e. X = −X
T (for a proof, we refer to A. Crumeyrolle (1990), Proposition 1.1.3):
“Ansatz” X = Ax ∀ A = −A
T , A :=
0
1
−1 0
∈ R
2×2 , x ∈ R .
(1.166)
3rd step:
⎡
⎣
1
g 22
x
T
1 G
−1
l x 1 =
1
g 11
x
T
2 G
−1
l x 2 =
1
λ 2
x 1 = G l Axx 2
⎤
⎦ ⇒
⇒
1
g 22
x
T
1 G
−1
l x 1 =
1
g 22
x
T
2 A
T G l Ax 2 x =
1
g 11
x
T
2 G
−1
l x 2 ⇔
⇔
g 11
g 22
A
T G l AG l x = I ⇔
⇔
⎡
⎢
⎣
x =
1
G 11 G 22 − G 2
12
g 22
g 11
x 1 = G l Axx 2
⎤
⎥
⎦ ⇒
⇒ x 1 = G l A
1
G 11 G 22 − G 2
12
g 22
g 11
x 2 .
(1.167)
The converse (iv) ⇒ (ii) is obvious.
End of Proof (third part).
Proof (third part).
(ii) ⇒ (iv).
In order to derive a linear system of partial differential equations for {u U , u V , v U , v V }, we depart from
the inverse right Cauchy–Green deformation tensor C r since it contains just the above quoted partials.
For an inverse portrait involving the partials {U u , U v , V u , V v }, in previous sections, we start from the
inverse left Cauchy–Green deformation tensor, a procedure we are not following further.
1st step:
C
−1
r = J l G
−1
l J
T
l =
G
−1
r
λ 2 ⇔
⎡
⎢
⎢
⎢
⎣
u U u V
v U v V
G
−1
l
u U v U
u V v V
=
G
−1
r
λ 2
x 1 :=
u U
u V
, x 2 :=
v U
v V
⎤
⎥
⎥
⎥
⎦
⇒
⎡
⎢
⎢
⎢
⎢
⎢
⎢
⎢
⎢
⎣
(α) x
T
1 G
−1
l x 1 = +
g 22
λ 2
(β) x
T
2 G
−1
l x 2 = +
g 11
λ 2
(γ) x
T
1 G
−1
l x 2 = −
g 12
λ 2
(δ) x
T
2 G
−1
l x 1 = −
g 12
λ 2
⎤
⎥
⎥
⎥
⎥
⎥
⎥
⎥
⎥
⎦
. (1.164)
Without loss of generality – see through the remark that follows after the proof – let us here assume
that the right two-dimensional Riemann manifold (i. e. the right parameterized surface) M
2
r is charted
by orthogonal parameters (orthogonal coordinates) such that g 12 = 0 holds. Such a parameterization
of a surface can always be achieved though it might turn out to be a difficult numerical procedure.
2nd step:
x
T
2 G
−1
l x 1 = 0 (δ)
“Ansatz” x 1 = G l Xx 2 (X = unknown matrix)
⇔ x
T
2 G
−1
l x 2 = 0 ∀ x 2 ∈ R
2×1 (ε) . (1.165)
A quadratic form over the field of real numbers can only be zero (“isotropic”) if and only if X is
antisymmetric, i. e. X = −X
T (for a proof, we refer to A. Crumeyrolle (1990), Proposition 1.1.3):
“Ansatz” X = Ax ∀ A = −A
T , A :=
0
1
−1 0
∈ R
2×2 , x ∈ R .
(1.166)
3rd step:
⎡
⎣
1
g 22
x
T
1 G
−1
l x 1 =
1
g 11
x
T
2 G
−1
l x 2 =
1
λ 2
x 1 = G l Axx 2
⎤
⎦ ⇒
⇒
1
g 22
x
T
1 G
−1
l x 1 =
1
g 22
x
T
2 A
T G l Ax 2 x =
1
g 11
x
T
2 G
−1
l x 2 ⇔
⇔
g 11
g 22
A
T G l AG l x = I ⇔
⇔
⎡
⎢
⎣
x =
1
G 11 G 22 − G 2
12
g 22
g 11
x 1 = G l Axx 2
⎤
⎥
⎦ ⇒
⇒ x 1 = G l A
1
G 11 G 22 − G 2
12
g 22
g 11
x 2 .
(1.167)
The converse (iv) ⇒ (ii) is obvious.
End of Proof (third part).