Page 215 - INTRODUCTION TO THE CALCULUS OF VARIATIONS

202 Solutions to the Exercises

ν
(i) We have seen (Exercise 1.4.6) that the sequence u 0 in W 1,2 (we have
shown this only up to a subsequence, but it is not difficult to see that the whole
sequence has this property). An elementary computation gives

ν
det ∇u = −ν (1 − y) 2ν−1 .
ν
1
Let us show that det ∇u 0 in L does not hold. Indeed, let ϕ ≡ 1 ∈ L ∞ (Ω).
It is not difficult to see that
ZZ
ν
lim det ∇u (x, y) dxdy 6=0,
ν→∞
Ω
and thus the result.
ν
(ii) Note first that by Rellich theorem (Theorem 1.43) we have that if u
q
u in W 1,p then u ν → u in L , ∀q ∈ [1, 2p/ (2 − p)) provided p< 2 and
∀q ∈ [1, ∞) if p =2 (the case p> 2 has already been considered in Lemma
ν
ν
4
3.23). Consequently if p> 4/3,we have u → u in L . Let therefore u =
ν
ν
ν
ν
ν
ν
u (x, y)=(ϕ (x, y) ,ψ (x, y)) and v ∈ C ∞ (Ω).Since det ∇u = ϕ ψ −
y
x
0
¡
ν
ν
2
ν
ν
ϕ ψ ν = ϕ ψ ν ¢ − (ϕ ψ ) (this is allowed since u ν ∈ C )wehave, after
y x y x x y
integrating by parts,
ZZ ZZ
¡ ν ν ν ν ¢
ν
det ∇u vdxdy = ϕ ψ v y − ϕ ψ v x dxdy .
y
x
Ω Ω
ν
ν
However we know that ψ ψ in W 1,4/3 (since u u in W 1,p and p> 4/3)
¡ ν ν ¢ ¡ ¢
ν
ν
ν
4
and ϕ → ϕ in L , we therefore deduce that ϕ ψ ,ϕ ψ ϕψ ,ϕψ in L 1
x y x y
(Exercise 1.3.3). Passing to the limit and integrating by parts once more we get
the claim, namely
ZZ ZZ ZZ
ν
lim det ∇u vdxdy = (ϕψ v y − ϕψ v y ) dxdy = det ∇uv dxdy .
x
x
ν→∞
Ω Ω Ω
Exercise 3.5.6. (i) Let x =(x 1 ,x 2 ),wethen find
⎛ 2 ⎞
x 2 x 1 x 2
3 − 3
⎜ |x| |x| ⎟ 2 1
∇u = ⎜ 2 ⎟ ⇒ |∇u| = .
⎝ x 1 x 2 x 1 ⎠ |x| 2
−
3 3
|x| |x|
Wethereforededuce(cf. Exercise1.4.1)that u ∈ L ∞ and u ∈ W 1,p provided
0
p ∈ [1, 2),but, however, u/∈ W 1,2 and u/∈ C .
(ii) Since
ZZ Z Z
1 r 2π ν+1 s − 1
q
ν
|u (x) − u (x)| dx =2π q dr = 2 q ds
Ω 0 (νr +1) ν 1 s