Page 230 - INTRODUCTION TO THE CALCULUS OF VARIATIONS

Chapter 6: Isoperimetric inequality 217

In particular we have that ϕ, ψ ∈ W 1,2 (−1, 1),with ϕ (−1) = ϕ (1) and ψ (−1) =
ψ (1),and
Z
2 β i ¡ 02 02 ¢
L (a i ,b i )= ϕ (y)+ ψ (y) dy (7.27)
L (a, b)
α i
Z
β i
0
M (a i ,b i )= ϕ (y) ψ (y) dy . (7.28)
α i
We thus obtain, using (7.26), (7.27) and (7.28),
∞ Z 1
X 2 ¡ 02 ¢
02
L (a, b)= L (a i ,b i )= ϕ (y)+ ψ (y) dy
L (a, b) −1
i=1
∞ Z 1
X
M (a, b)= M (a i ,b i )= ϕ (y) ψ (y) dy .
0
−1
i=1
We therefore find, invoking Corollary 6.3, that
2
2
[L (u, v)] − 4πM (u, v)= [L (a, b)] − 4πM (a, b)
Z 1 Z 1
¡ 02 02 ¢
=2 ϕ + ψ dy − 4π ϕψ dy ≥ 0
0
−1 −1
as wished.
7.6.2 The case of dimension n
Exercise 6.3.1. We clearly have
¡ ¢
C =(a + B) ∪ b + A ⊂ A + B.

¡ ¢ © ª
It is also easy to see that (a + B) ∩ b + A = a + b . Observe then that
¡ ¢ £ ¡ ¢¤
M (C)= M (a + B)+ M b + A − M (a + B) ∩ b + A = M (A)+ M (B)

and hence
M (A)+ M (B) ≤ M (A + B) .
Exercise 6.3.2. (i) We adopt the same notations as those of Exercise 5.2.4.
2
By hypothesis there exist a bounded smooth domain Ω ⊂ R and a map v ∈
¡
¢
¡ ¢
C 2 Ω; R 3 (v = v (x, y),with v x × v y 6=0 in Ω)sothat ∂A 0 = v Ω .
From the divergence theorem it follows that
ZZ
1
M (A 0 )= hv; v x × v y i dxdy . (7.29)
3 Ω