186
2 Variational Problems with Fixed Boundaries
d
dx
(yy
) =
d
dx
1
2
d
dx
y
2
=
1
2
d
2
dx 2 y
2
= −x
2
(9)
Integrating the above equation twice, the result of Eq. (5) can be obtained, then
making use of the boundary condition, the result is still Eq. (6).
Example 2.10.6 Prove that the extremal curve of the functional J [r ] =
θ 1
θ 0
f (r sin θ)
√
r 2 + r 2 dθ can often be obtained through integration.
Proof Let x = r cos θ , y = r sin θ , then the functional can be rewritten as
J [y] =
x 1
x 0
f (y)
1 + y 2 dx
Because the integrand does not contain x, so there is the first integral
f (y)
1 + y 2 − y
f (y)
y
1 + y 2
= c 1
or
f (y)
1 + y 2
= c 1
The above integral can be written as
dy
f 2 (y) − c
2
1
=
x
c 1
+ c 2
If the integral on the left-hand side of the equation can be solved, then the extremal
curve can be obtained. Quod erat demonstrandum.
Example 2.10.7 Prove the Clairaut theorem: On a surface of revolution, the multiplicative product of the parallel radius at every point of any geodesic line and the
sine of included angle between the geodesic line through the point and the meridian
is a constant.
Proof Let the equation of a surface of revolution under cylindrical coordinate system
have the following form
x = r cos ϕ, y = r sin ϕ, z = f (r )
2 Variational Problems with Fixed Boundaries
d
dx
(yy
) =
d
dx
1
2
d
dx
y
2
=
1
2
d
2
dx 2 y
2
= −x
2
(9)
Integrating the above equation twice, the result of Eq. (5) can be obtained, then
making use of the boundary condition, the result is still Eq. (6).
Example 2.10.6 Prove that the extremal curve of the functional J [r ] =
θ 1
θ 0
f (r sin θ)
√
r 2 + r 2 dθ can often be obtained through integration.
Proof Let x = r cos θ , y = r sin θ , then the functional can be rewritten as
J [y] =
x 1
x 0
f (y)
1 + y 2 dx
Because the integrand does not contain x, so there is the first integral
f (y)
1 + y 2 − y
f (y)
y
1 + y 2
= c 1
or
f (y)
1 + y 2
= c 1
The above integral can be written as
dy
f 2 (y) − c
2
1
=
x
c 1
+ c 2
If the integral on the left-hand side of the equation can be solved, then the extremal
curve can be obtained. Quod erat demonstrandum.
Example 2.10.7 Prove the Clairaut theorem: On a surface of revolution, the multiplicative product of the parallel radius at every point of any geodesic line and the
sine of included angle between the geodesic line through the point and the meridian
is a constant.
Proof Let the equation of a surface of revolution under cylindrical coordinate system
have the following form
x = r cos ϕ, y = r sin ϕ, z = f (r )