Page 214 - INTRODUCTION TO THE CALCULUS OF VARIATIONS

Chapter 3: Direct methods 201

¢
2
Exercise 3.5.3. Let u ∈ C 2 ¡ Ω; R , u (x, y)=(ϕ (x, y) ,ψ (x, y)),be a mini-
¡ 2 ¢
mizer of (P) and let v ∈ C 0 ∞ Ω; R , v (x, y)= (α (x, y) ,β (x, y)), be arbitrary.
Since I (u + v) ≥ I (u) for every ,we musthave
¯
d ¯
0= I (u + v) ¯
d ¯
=0
½ZZ ¾¯
d £ ¡ ¢ ¡ ¢ ¤ ¯
= (ϕ + α x ) ψ + β y − ϕ + α y (ψ + β ) dxdy ¯
x
y
x
x
y
d Ω ¯ =0
ZZ
£¡ ¢ ¡ ¢¤
= ψ α x − ψ α y + ϕ β − ϕ β dxdy .
y x x y y x
Ω
Integrating by parts, we find that the right hand side vanishes identically. The
result is not surprising in view of Exercise 3.5.2, which shows that I (u) is in fact
constant.
Exercise 3.5.4. The proof is divided into two steps.
Step 1. It is easily proved that the following algebraic inequality holds
|det A − det B| ≤ α (|A| + |B|) |A − B| , ∀A, B ∈ R 2×2
where α is a constant.
Step 2. We therefore deduce that
p/2 p/2 p/2 p/2
|det ∇u − det ∇v| ≤ α (|∇u| + |∇v|) |∇u −∇v| .
Hölder inequality implies then

ZZ
p/2
|det ∇u − det ∇v| dxdy
Ω
µZZ ¶ 1/2 µZZ ¶ 1/2
p
p
≤ α p/2 (|∇u| + |∇v|) dxdy |∇u −∇v| dxdy .
Ω Ω
We therefore obtain that
2/p
µZZ ¶
kdet ∇u − det ∇vk L p/2 = |det ∇u − det ∇v| p/2 dxdy
Ω
µZZ ¶ 1/p µZZ ¶ 1/p
p p
≤ α (|∇u| + |∇v|) dxdy |∇u −∇v| dxdy
Ω Ω
and hence the claim.
Exercise 3.5.5. For more details concerning this exercise see [31] page 158.