Page 213 - INTRODUCTION TO THE CALCULUS OF VARIATIONS

200 Solutions to the Exercises

and the divergence theorem to get

ZZ Z
¡ ¢
det ∇udxdy = ϕψ ν 1 − ϕψ ν 2 dσ
y x
Ω ∂Ω
where ν =(ν 1 ,ν 2 ) is the outward unit normal to ∂Ω.Since ϕ = α on ∂Ω,we
have, applying twice the divergence theorem,
ZZ ZZ
h i
¡ ¢
det ∇udxdy = αψ − (αψ ) dxdy
y x x y
Ω Ω
ZZ Z
h i
= (α x ψ) − (α y ψ) dxdy = (α x ψν 2 − α y ψν 1 ) dσ .
y x
Ω ∂Ω
Since ψ = β on ∂Ω, we obtain, using again the divergence theorem, that
ZZ ZZ ZZ
h i
det ∇udxdy = (α x β) − (α y β) dxdy = det ∇vdxdy.
y x
Ω Ω Ω

Step 2. We first regularize v, meaning that for every > 0 we find v ∈
¡ ¢
C 2 Ω; R 2 so that

kv − v k 1,p ≤ .
W
1,p ¡ 2 ¢ ¡ 2 ¢
Since u − v ∈ W Ω; R ,wecan find w ∈ C ∞ Ω; R so that
0 0

k(u − v) − w k W 1,p ≤ .
¡ 2 ¢

2
Define u = v + w and observe that u ,v ∈ C Ω; R ,with u = v on ∂Ω,
and

ku − u k 1,p = k(u − v) − w +(v − v )k 1,p ≤ 2 .
W W
Using Exercise 3.5.4 below, we deduce that there exists α 1 (independent of )so
that

kdet ∇u − det ∇u k L p/2 , kdet ∇v − det ∇v k L p/2 ≤ α 1 .
Combining Step 1 with the above estimates we obtain that there exists a constant
α 2 (independent of ) such that
¯ZZ ¯ ¯ZZ ¯
¯ ¯ ¯ ¯
¯
¯ (det ∇u − det ∇v) dxdy ≤ ¯ (det ∇u − det ∇v ) dxdy ¯
¯ ¯ ¯ ¯
Ω Ω
¯ZZ ¯ ¯ZZ ¯
¯ ¯ ¯ ¯

+ ¯ (det ∇u − det ∇u ) dxdy + ¯ (det ∇v − det ∇v) dxdy ≤ α 2 .
¯
¯
¯ ¯ ¯ ¯
Ω Ω
Since is arbitrary we have indeed the result.