Page 225 - INTRODUCTION TO THE CALCULUS OF VARIATIONS

212 Solutions to the Exercises

2
has no C solution satisfying w (0) = w (1) = α and hence (P α )alsohas no C 2
minimizer).
Exercise 5.2.4. By hypothesis there exist a bounded smooth domain Ω ⊂ R 2
¡
¢
¡ ¢
and a map v ∈ C 2 Ω; R 3 (v = v (x, y),with v x ×v y 6=0 in Ω)sothat Σ 0 = v Ω .
Let e 3 =(v x × v y ) / |v x × v y |.We then let for ∈ R and ϕ ∈ C 0 ∞ (Ω)

v (x, y)= v (x, y)+ ϕ (x, y) e 3 .

Ω .Since Σ 0 is of minimal area and ∂Σ = ∂Σ 0 ,weshould
Finally let Σ = v ¡ ¢
have ZZ ZZ
¯ ¯
v × v
|v x × v y | dxdy ≤ ¯ ¯ dxdy . (7.23)
x y
Ω Ω
¯ ¯
®
2
­
2
2

Let E = |v | , F = v ; v , G = v y 2 , E = |v x | , F = hv x ; v y i and G = |v y | .
¯ ¯
y
x
x
We therefore get
¡ ¢
2
E = |v x + ϕe 3x + ϕ e 3 | = E +2 [ϕ hv x ; e 3 i + ϕ hv x ; e 3x i]+ O 2
x x
¤
£
¡ ¢
F = F + ϕ hv y ; e 3 i + ϕ hv x ; e 3 i + ϕ hv y ; e 3x i + ϕ hv x ; e 3y i + O 2
x y
¤
£
¡ ¢
G = G +2 ϕ hv y ; e 3 i + ϕ hv y ; e 3y i + O 2
y
where O (t) stands for a function f so that |f (t) /t| is bounded in a neighborhood
of t =0.Appealing to the definition of L, M, N, Exercise 5.2.1 and to the fact
that hv x ; e 3 i = hv y ; e 3 i =0,weobtain
¡ ¢
2
2

E G − (F ) =(E − 2 Lϕ)(G − 2 ϕN) − (F − 2 ϕM) + O 2
¡ ¢
2
= EG − F − 2 ϕ [EN − 2FM + GL]+ O 2
¡ 2 ¢ ¡ ¢
2
= EG − F [1 − 4 ϕH]+ O .
We therefore conclude that
¯ ¯ ¡ ¢
¯ ¯ = |v x × v y | (1 − 2 ϕH)+ O
2
v × v
x y
and hence
ZZ
¡ ¢

Area (Σ )= Area (Σ 0 ) − 2 ϕH |v x × v y | dxdy + O 2 . (7.24)
Ω
Using (7.23) and (7.24) (i.e., we perform the derivative with respect to )weget
ZZ
ϕH |v x × v y | dxdy =0, ∀ϕ ∈ C ∞ (Ω) .
0
Ω
Since |v x × v y | > 0 (due to the fact that Σ 0 is a regular surface), we deduce
from the fundamental lemma of the calculus of variations (Theorem 1.24) that
H =0.