Page 231 - INTRODUCTION TO THE CALCULUS OF VARIATIONS

218 Solutions to the Exercises

Let then ∈ R, ϕ ∈ C 0 ∞ (Ω) and

v (x, y)= v (x, y)+ ϕ (x, y) e 3
where e 3 =(v x × v y ) / |v x × v y |.
© ª ¡ ¢

We next consider ∂A = v (x, y):(x, y) ∈ Ω = v Ω . Wehavetoevalu-

ate M (A ) and we start by computing
¡
¡

v × v y =(v x + (ϕ e 3 + ϕe 3x )) × v y + ϕ e 3 + ϕe 3y ¢¢
x
y
x
= v x × v y + [ϕ (e 3x × v y + v x × e 3y )]
£ ¤ ¡ ¢
2
+ ϕ e 3 × v y + ϕ v x × e 3 + O
y
x
(where O (t) stands for a function f so that |f (t) /t| is bounded in a neighbor-
hood of t =0)which leadsto
­ ® ­ ®
v ; v × v = v + ϕe 3 ; v × v
x y x y
= hv; v x × v y i + ϕ he 3 ; v x × v y i + hv; ϕ (e 3x × v y + v x × e 3y )i
­ ® ¡ ¢
2
+ v; ϕ e 3 × v y + ϕ v x × e 3 + O .
x y
Observing that he 3 ; v x × v y i = |v x × v y | and returning to (7.29), we get after
integration by parts that (recalling that ϕ =0 on ∂Ω).
ZZ

M (A ) − M (A 0 )= ϕ {|v x × v y | + hv; e 3x × v y + v x × e 3y i
3
Ω
o
¡ ¢
2
− (hv; e 3 × v y i) − (hv; v x × e 3 i) dxdy + O
x y
ZZ
= ϕ {|v x × v y | − hv x ; e 3 × v y i
3 Ω
¡ ¢
2
− hv y ; v x × e 3 i} dxdy + O .
Since hv x ; e 3 × v y i = hv y ; v x × e 3 i = − |v x × v y |, we obtain that
ZZ
¡ ¢

M (A ) − M (A 0 )= ϕ |v x × v y | dxdy + O 2 . (7.30)
Ω
(ii) We recall from (7.24) in Exercise 5.2.4 that we have
ZZ
¡ ¢

L (∂A ) − L (∂A 0 )= −2 ϕH |v x × v y | dxdy + O 2 . (7.31)
Ω
Combining (7.30), (7.31), the minimality of A 0 and a Lagrange multiplier α,we
get
ZZ
(−2ϕH + αϕ) |v x × v y | dxdy =0, ∀ϕ ∈ C 0 ∞ (Ω) .
Ω
The fundamental lemma of the calculus of variations (Theorem 1.24) implies
then that H = constant (since ∂A 0 is a regular surface we have |v x × v y | > 0).