1-13 One example: mapping from an ellipsoid-of-revolution to the sphere 79
In the light of the equivalence theorem 1.14 (areomorphism), we are now prepared to solve the following
problems. (i) Can we prove the first equivalence given by (1.285) and (ii) can we prove the third
equivalence given by (1.286)?
det[C l ] = det[G l ] or det[C r ] = det[G r ] ,
(1.285)
u U v V − u V u U =
det[G l ]
det[G r ]
=
G 11 G 22 − G 2
12
g 11 g 22 − g 2
12
.
(1.286)
Solution (the first problem).
Start from the equiareal map of Box 1.41 in order to prove det[C l ] = det[G l ], where the left Cauchy–
Green matrix C l as well as the left matrix G l of the metric is given by means of Box 1.40. Here, again
we collect all deviational items in Box 1.43. As soon as we implement f
(Φ) into the determinantal
identity, the proof is closed.
End of Solution (the first problem).
Solution (the second problem).
By means of Box 1.44, let us work out the partial differential equation which governs an equiareal
mapping. Note that we here specify {u = λ, v = φ} and {U = Λ, V = Φ} subject to the “Ansatz”
{λ = Λ, φ = f (Φ)}. Indeed, we find f (Φ) as given already in Box 1.42.
End of Solution (the second problem).
Note that the second or canonical equivalence Λ 1 Λ 2 = 1 has already been used to construct the
equiareal map E
2
A 1 ,A 1 ,A 2
→ S
2
r .
Box 1.43 (Equiareal mapping: E
2
A 1 ,A 1 ,A 2 → S
2
r , det[C l ] = det[G l ]).
C l =
2
4
r
2 cos
2 φ
0
0
r
2 f
2 (φ)
3
5 , G l =
2
6
6
6
4
A
2
1 cos
2 Φ
1 − E 2 sin
2 Φ
0
0
A
2
1 (1 − E
2 )
2
(1 − E 2 sin
2 Φ) 3
3
7
7
7
5
,
2
6
6
4
det [C l ] = r
4 cos
2 φ f
2 (φ)
f
2 (Φ) =
1
r 4 cos 2 φ
A
4
1 (1 − E
2 )
2
(1 − E 2 sin
2 Φ) 4 cos
2 Φ
3
7
7
5 ⇒ det [C l ] =
A
4
1 (1 − E
2 )
2
(1 − E 2 sin
2 Φ) 4 cos
2 Φ = det [G l ]
q. e. d.
(1.287)
Box 1.44 (Equiareal mapping: E
2
A 1 ,A 1 ,A 2 → S
2
r , partial differential equation).
u U v V − u V u U =
s
det[G l ]
det[G r ]
, u = λ , v = φ = f (Φ) , U = Λ , V = Φ ⇔
⇔ λ Λ φ Φ − λ Φ φ Λ =
r
G 11 G 22
g 11 g 22
⇔ f
2 (Φ) =
1
r 2 cos φ
A
2
1 (1 − E
2 )
(1 − E 2 sin
2 Φ) 2 cos Φ ,
f
(Φ) : see Box 1.42
q. e. d.
(1.288)
In the light of the equivalence theorem 1.14 (areomorphism), we are now prepared to solve the following
problems. (i) Can we prove the first equivalence given by (1.285) and (ii) can we prove the third
equivalence given by (1.286)?
det[C l ] = det[G l ] or det[C r ] = det[G r ] ,
(1.285)
u U v V − u V u U =
det[G l ]
det[G r ]
=
G 11 G 22 − G 2
12
g 11 g 22 − g 2
12
.
(1.286)
Solution (the first problem).
Start from the equiareal map of Box 1.41 in order to prove det[C l ] = det[G l ], where the left Cauchy–
Green matrix C l as well as the left matrix G l of the metric is given by means of Box 1.40. Here, again
we collect all deviational items in Box 1.43. As soon as we implement f
(Φ) into the determinantal
identity, the proof is closed.
End of Solution (the first problem).
Solution (the second problem).
By means of Box 1.44, let us work out the partial differential equation which governs an equiareal
mapping. Note that we here specify {u = λ, v = φ} and {U = Λ, V = Φ} subject to the “Ansatz”
{λ = Λ, φ = f (Φ)}. Indeed, we find f (Φ) as given already in Box 1.42.
End of Solution (the second problem).
Note that the second or canonical equivalence Λ 1 Λ 2 = 1 has already been used to construct the
equiareal map E
2
A 1 ,A 1 ,A 2
→ S
2
r .
Box 1.43 (Equiareal mapping: E
2
A 1 ,A 1 ,A 2 → S
2
r , det[C l ] = det[G l ]).
C l =
2
4
r
2 cos
2 φ
0
0
r
2 f
2 (φ)
3
5 , G l =
2
6
6
6
4
A
2
1 cos
2 Φ
1 − E 2 sin
2 Φ
0
0
A
2
1 (1 − E
2 )
2
(1 − E 2 sin
2 Φ) 3
3
7
7
7
5
,
2
6
6
4
det [C l ] = r
4 cos
2 φ f
2 (φ)
f
2 (Φ) =
1
r 4 cos 2 φ
A
4
1 (1 − E
2 )
2
(1 − E 2 sin
2 Φ) 4 cos
2 Φ
3
7
7
5 ⇒ det [C l ] =
A
4
1 (1 − E
2 )
2
(1 − E 2 sin
2 Φ) 4 cos
2 Φ = det [G l ]
q. e. d.
(1.287)
Box 1.44 (Equiareal mapping: E
2
A 1 ,A 1 ,A 2 → S
2
r , partial differential equation).
u U v V − u V u U =
s
det[G l ]
det[G r ]
, u = λ , v = φ = f (Φ) , U = Λ , V = Φ ⇔
⇔ λ Λ φ Φ − λ Φ φ Λ =
r
G 11 G 22
g 11 g 22
⇔ f
2 (Φ) =
1
r 2 cos φ
A
2
1 (1 − E
2 )
(1 − E 2 sin
2 Φ) 2 cos Φ ,
f
(Φ) : see Box 1.42
q. e. d.
(1.288)