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{
g
1
(
α,r
)
,
g
2
(
α,r
)
}
is called
contravariant
. The properly posed question can be answered imme-
diately. The second-order tensor C
r
(
α, r
) is represented in the contravariant or two-co-basis
{
g
1
}
, in general. Due to the diagonal structure of the right defor-
mation tensor C
r
(
r
), contains only components
g
1
⊗
g
1
,
g
1
⊗
g
2
,
g
2
⊗
g
1
,
g
2
⊗
g
2
⊗
g
1
and
g
2
⊗
g
2
,or
g
1
⊗
g
1
and
g
2
⊗
g
2
,
respectively.”
End of Solution (the first problem).
Solution (the second problem).
are collected in Boxes
1.15
and
1.16
, exclusively. In particular, we aim at computing the right eigenvalues, eigencolumns, and
eigenvectors, namely in Box
1.15
in Cartesian coordinates
The results on the right eigenspace analysis of the matrix pair
{
C
r
,
G
r
}
{
x, y
}
along the fixed orthonormal
frame
{
e
1
,e
2
}
andinBox
1.16
in polar coordinates
{
α,r
}
along the moving orthogonal frame
{
g
1
,
g
2
|
p
}
. First, we solve the right general eigenvalue problem, both in Cartesian representa-
tion
{
λ
1
(
x, y
)
,λ
2
=1
}
and in polar representation
{
λ
1
(
r
)
,λ
2
=1
}
. The characteristic equation
λ
2
G
r
|
|
= 0 is solved in Box
1.16
if both C
r
and G
r
are diagonal. The determinantal identity is
factorized directly into the right eigenvalues
λ
1
and
λ
2
, a result we take advantage from in a follow-
ing section. Second, we derive the simple structure of the eigencolumns
C
r
−
{
f
11
(
x, y
)
,f
21
(
x, y
)
}
and
{
f
12
(
x, y
)
,f
22
(
x, y
)
}
in case of Cartesian coordinates as well as of the eigencolumns
{
f
11
(
r
)
,f
21
(
r
)
}
and
in polar coordinates. Third, let us derive the right eigenvectors. In Box
1.15
,
we succeed to represent the orthonormal right eigenvectors in the Cartesian basis
{
f
12
(
r
)
,f
22
(
r
)
}
.In
contrast, in Box
1.16
, we are able to compute the first right eigenvector as a tangent vector of
the image of the parallel circle, while the second right eigenvector “radial” as a tangent vector of
the image (straight line) of the meridian. Such a beautiful result is illustrated by Fig.
1.11
.
{
e
1
,e
2
|
p
}
End of Solution (the second problem).
Box 1.13 (Orthogonal projection
S
2
R
+
onto
P
2
O
, Cartesian coordinates, the first problem).
x
=
r
cos
α, y
=
r
sin
α, Λ
(
x, y
) = arctan
Y
X
= arctan
y
x
=
α,
(1.91)
= arccos
x
2
+
y
2
R
Φ
(
x, y
) = arccos
√
X
2
+
Y
2
R
= arccos
r
R
.
Right Jacobi matrix:
⎡
⎤
J
r
:=
D
x
ΛD
y
Λ
=
√
x
2
+
y
2
y
√
x
2
+
y
2
x
−
−
1
⎣
⎦
,
x
2
+
y
2
(1.92)
√
R
2
−
(
x
2
+
y
2
)
x
√
R
2
−
(
x
2
+
y
2
)
y
D
x
ΦD
y
Φ
−
y
x
2
+
y
2
,D
y
Λ
=+
x
x
2
+
y
2
,
D
x
Λ
=
−
x
x
2
+
y
2
y
x
2
+
y
2
1
1
D
x
Φ
=
−
R
2
−
(
x
2
+
y
2
)
,D
y
Φ
=
−
R
2
−
(
x
2
+
y
2
)
.
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