2.10 Invariance of the Euler Equation
187
The differential relationships are
dx = cos ϕdr − r sin ϕdϕ, dy = sin ϕdr + r cos ϕdϕ, dz = f r dr
Then the differential of the arc length on a surface of revolution is
ds =
(dx) 2 + (dy) 2 + (dz) 2 =
(dr ) 2 + (r dϕ) 2 + ( f r dr ) 2 =
r 2 + (1 + f 2
r )r 2 dϕ
The geodesic line on a surface of revolution is the following functional
J [r ] =
ϕ 1
ϕ 0
r 2 + (1 + f 2
r )r 2 dϕ
Because the functional is only the function of r and r
, so the Euler equation of
the functional is
r 2 + (1 + f 2
r )r 2 − r
(1 + f
2
r )r
r 2 + (1 + f 2
r )r 2
= c
or
r
2
r 2 + (1 + f 2
r )r 2
= c
Using the differential relationship of the arc length, the above equation can be
written as r
2 dϕ
ds
= c, note that r dϕ = sin θ ds, where, θ is the included angle
between the geodesic line and the meridian, thus r sin θ = c can be obtain. Quod
erat demonstrandum.
Example 2.10.8 Find the extremal curve of the functional J [θ ] =
r 1
r 0
ln r
r
√
1 + r 2 θ 2 dr .
Solution Since the integrand does not contain θ , so the Euler equation has the first
integral
r θ
ln r
√
1 + r 2 θ 2
= c 1
or
θ
=
c 1
r
ln
2 r − c
2
1
187
The differential relationships are
dx = cos ϕdr − r sin ϕdϕ, dy = sin ϕdr + r cos ϕdϕ, dz = f r dr
Then the differential of the arc length on a surface of revolution is
ds =
(dx) 2 + (dy) 2 + (dz) 2 =
(dr ) 2 + (r dϕ) 2 + ( f r dr ) 2 =
r 2 + (1 + f 2
r )r 2 dϕ
The geodesic line on a surface of revolution is the following functional
J [r ] =
ϕ 1
ϕ 0
r 2 + (1 + f 2
r )r 2 dϕ
Because the functional is only the function of r and r
, so the Euler equation of
the functional is
r 2 + (1 + f 2
r )r 2 − r
(1 + f
2
r )r
r 2 + (1 + f 2
r )r 2
= c
or
r
2
r 2 + (1 + f 2
r )r 2
= c
Using the differential relationship of the arc length, the above equation can be
written as r
2 dϕ
ds
= c, note that r dϕ = sin θ ds, where, θ is the included angle
between the geodesic line and the meridian, thus r sin θ = c can be obtain. Quod
erat demonstrandum.
Example 2.10.8 Find the extremal curve of the functional J [θ ] =
r 1
r 0
ln r
r
√
1 + r 2 θ 2 dr .
Solution Since the integrand does not contain θ , so the Euler equation has the first
integral
r θ
ln r
√
1 + r 2 θ 2
= c 1
or
θ
=
c 1
r
ln
2 r − c
2
1
