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Proof (third part).
(iv)
.
In order to derive a linear system of partial differential equations for
(ii)
⇒
,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
}
,inprevious
sections, we start from the inverse left Cauchy-Green deformation tensor, a procedure we are not
following further.
{
U
u
,U
v
,V
u
,V
v
}
1st step:
⎡
v
U
u
V
u
U
v
V
G
−
1
u
U
v
U
u
V
v
V
=
G
−
1
⎤
λ
2
⎣
⎦
⇒
C
−
r
=J
l
G
−
l
J
l
=
G
−
1
l
x
1
:=
u
U
,
x
2
:
u
U
r
λ
2
⇔
u
V
u
V
⎡
⎤
(
α
)
x
1
G
−
l
x
1
=+
g
22
λ
2
(
β
)
x
2
G
−
l
x
2
=+
g
11
⎣
⎦
λ
2
.
(1.164)
(
γ
)
x
1
G
−
l
x
2
=
g
12
λ
2
−
g
12
λ
2
(
δ
)
x
2
G
−
l
x
1
=
−
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
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 di
cult
numerical procedure.
2nd step:
x
2
G
−
l
x
1
=0 (
δ
)
“Ansatz”
x
1
=G
l
X
x
2
(X = unknown matrix)
x
2
G
−
l
x
2
=0
2
×
1
⇔
∀
x
2
∈
R
(
)
(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
Crumeyrolle 1990
, Proposition 1.1.3):
−
A
T
,
A:=
01
2
×
2
,
x
∈
R
“Ansatz” X = A
x
∀
A=
−
∈
R
.
(1.166)
−
10
3rd step:
g
22
x
1
G
−
l
x
1
=
x
2
G
−
l
x
2
=
λ
2
x
1
=G
l
A
x
x
2
1
g
11
⇒
1
1
1
g
22
x
1
G
−
l
x
1
=
g
22
x
2
A
T
G
l
A
x
2
x
=
g
11
x
2
G
−
l
x
2
⇔
⇒
g
11
g
22
A
T
G
l
AG
l
x
=I
⇔
⇔
(1.167)
x
=
g
22
g
11
√
G
11
G
22
−G
12
1
⇔
⇒
x
1
=G
l
A
x
x
2
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