Page 217 - INTRODUCTION TO THE CALCULUS OF VARIATIONS

204 Solutions to the Exercises

So let > 0 be arbitrary and let ϕ ∈ C ∞ (Ω). We can therefore find δ = δ ( ) > 0
such that
x ∈ Ω δ ⇒ |ϕ (x) − ϕ (0)| < . (7.16)

We then combine (7.14), (7.15) and (7.16) to get the result. Indeed let ϕ ∈
C ∞ (Ω) and obtain
0
ZZ ZZ ZZ
ν
ν
ν
det ∇u ϕdx = ϕ (0) det ∇u dx + det ∇u (ϕ (x) − ϕ (0)) dx
Ω Ω Ω δ
ZZ
ν
+ det ∇u (ϕ (x) − ϕ (0)) dx .
Ω−Ω δ
This leads to the following estimate
¯ZZ ZZ ¯
¯ ν ν ¯
det ∇u ϕdx − ϕ (0) det ∇u dx
¯ ¯
¯ ¯
Ω Ω
ZZ ZZ
ν
ν
≤ sup [|ϕ (x) − ϕ (0)|] |det ∇u | dx +2 kϕk |det ∇u | dx .
L ∞
Ω
x∈Ω δ Ω−Ω δ
Keeping first fixed , and thus δ,welet ν →∞ and obtain
¯ZZ ¯
¯ ν ¯
lim ¯ det ∇u ϕdx − πϕ (0) ≤ π ,
¯
¯ ¯
ν→∞
Ω
being arbitrary, we have indeed obtained the result.
7.3.5 Relaxation theory
Exercise 3.6.1. (i) Let
½
ax + α if x ∈ [0,λ]
u (x)=
b (x − 1) + β if x ∈ [λ, 1] .
Note that u (0) = α, u (1) = β and u is continuous at x = λ since β − α =
λa+(1 − λ) b, hence u ∈ X.Since f ∗∗ ≤ f and f ∗∗ is convex, we have appealing
to Jensen inequality that, for any u ∈ X,
1 1
Z Z
I (u)= f (u (x)) dx ≥ f ∗∗ (u (x)) dx
0
0
0 0
1
µZ ¶
≥ f ∗∗ u (x) dx = f ∗∗ (β − α)= λf (a)+(1 − λ) f (b)= I (u) .
0
0
Hence u is a minimizer of (P).