Geography Reference
In-Depth Information
r
cos
φ
.
p
λ
=
r, q
λ
=
p
φ
=0
,q
φ
=
(1.195)
Solution (the first problem).
Start from the conformal map that is defined in Box
1.24
. The three conditions to be fulfilled are
given in Box
1.25
for “left UMP” and in Box
1.26
for “right
UMP”. F
irst
, we writ
e down
the left
an
d right
specified Korn-Lichtenstein equations, namely
G
11
/G
22
,
G
22
/G
11
and
g
11
/g
22
,
g
22
/g
11
, respectively. Indeed, by transforming
left
and right, KL 1st and KL 2nd are satisfied. Second, we specialize the left and right integrability
conditions of the vector-valued Korn-Lichtenstein equations, namely the left and right Laplace-
Beltrami equations, by
{
Q
Λ
,Q
Φ
,P
Λ
,P
Φ
}
as well as
{
q
λ
,q
φ
,p
λ
,p
φ
}
2
A
1
,A
1
,A
2
{
G
11
,G
22
,P
Λ
,P
Φ
,Q
Λ
,Q
Φ
}
of
E
as well as
{
g
11
,g
22
,p
λ
,p
φ
,q
λ
,q
φ
}
2
of
are
right harmonic
.
Third, we prove left and right orientation by computing the left and right
Jacobians
which turn
out positive.
S
r
. Indeed, we have succeeded that
{
P,Q
}
are
left harmonic
and
{
p, q
}
End of Solution (the first problem).
Solution (the second problem).
In any textbook of Differential Geometry, you will find the representation of the Gaussian curva-
ture of a surface in terms of conformal coordinates (isometric, isothermal). Let the left and the
right matrix of the metric be equipped with a conformally flat structure
{
G
l
=
λ
l
I
2
,
G
r
=
λ
r
I
2
}
,
which is generated by a left and a right conformal coordinate representation. Then the left
and the right Gaussian curvature are given by
k
l
=
−
(1
/
2
λ
l
)
Δ
l
ln
λ
l
=
−
(1
/λ
l
)
Δ
l
ln
λ
l
and
k
r
=
(1
/λ
r
)
Δ
r
ln
λ
r
as well as
Δ
l
:=
D
PP
+
D
QQ
=
D
P
+
D
Q
and
D
p
+
D
q
=
D
pp
+
D
qq
:=
Δ
r
,where
Δ
l
and
Δ
r
represent the left and the right Laplace-Beltrami
operator. Let us apply this result in solving the second problem. By means of Boxes
1.27
,
1.28
,
and
1.29
, we have outlined how to generate a conformally flat metric of an ellipsoid-of-revolution
and of a sphere. It is the classical Gauss factorization.
−
(1
/
2
λ
r
)
Δ
r
ln
λ
r
=
−
End of Solution (the second problem).
Solution (the third problem).
By means of the left and the right mapping equations, i.e. by means of
{
P
=
A
1
Λ, Q
(
Φ
=0)=0
}
2
A
1
,A
1
,A
2
and
{
p
=
rλ,q
(
φ
=0)=0
}
, respectively, it is obvious that an equatorial arc of
E
and
2
of
r
is mapped equidistantly. Similarly, the factor of conformality derived in Box
1.27
amounts
to
λ
l
(
Φ
= 0) = 1 for the left manifold and to
λ
r
(
φ
= 0) = 1 for the right manifold. Such a
configuration on the ellipsoidal as well as the spherical equator we call an
isometry
. Indeed, the
postulate of an equidistant mapping of the left or right equator constitutes a boundary condition
for the left and right Korn-Lichtenstein equations, namely to make their solution unique.
S
End of Solution (the third problem).
Search WWH ::
Custom Search