Page 69 - INTRODUCTION TO THE CALCULUS OF VARIATIONS

56 Classical methods

3. Minimal surfaces of revolution.
This example will be treated in Chapter 5. Let us briefly present it here.
p 2
The function under consideration is f (u, ξ)=2πu 1+ ξ and the min-
imization problem (which corresponds to minimization of the area of a
surface of revolution) is
( )
Z
b
(P) inf I (u)= f (u (x) ,u (x)) dx = m
0
u∈X
a
© 1 ª
where X = u ∈ C ([a, b]) : u (a)= α, u (b)= β, u > 0 and α, β > 0.
The Euler-Lagrange equation and its first integral are
∙ ¸ 0
u u p 02
0
00
√ = 1+ u 02 ⇔ u u =1 + u
1+ u 02
0
p u u
02 0
u 1+ u − u √ = λ = constant.
1+ u 02
This leads to
u 2
02
u = 2 − 1.
λ
The solutions, if they exist (this depends on a, b, α and β, see Exercise
5.2.3), are of the form (µ being a constant)
³ x ´
u (x)= λ cosh + µ .
λ
We conclude this section with a generalization of a classical example.
p 2
Example 2.6 Fermat principle. The function is f (x, u, ξ)= g (x, u) 1+ ξ
and
( )
b
Z
0
(P) inf I (u)= f (x, u (x) ,u (x)) dx = m
u∈X
a
© ª
1
where X = u ∈ C ([a, b]) : u (a)= α, u (b)= β . Therefore the Euler-Lagrange
equation is
∙ ¸
d u 0 p
02
g (x, u) √ = g u (x, u) 1+ u .
dx 1+ u 02
Observing that
∙ ¸
d u 0 u 00
√ =
dx 1+ u 02 (1 + u ) 3/2
02
we get
¡ 02 ¢
g (x, u) u +[g x (x, u) u − g u (x, u)] 1+ u =0.
00
0