Page 228 - INTRODUCTION TO THE CALCULUS OF VARIATIONS

Chapter 6: Isoperimetric inequality 215

Exercise 6.2.2. Since the minimum in (P) is attained by u ∈ X,wehave, for
any v ∈ X ∩ C 0 ∞ (−1, 1) and any ∈ R,that
I (u + v) ≥ I (u) .

Therefore the Euler-Lagrange equation is satisfied, namely
Z 1
¡ 2 ¢
0 0
u v − π uv dx =0, ∀v ∈ X ∩ C ∞ (−1, 1) . (7.25)
0
−1
Let us transform it in a more classical way and choose a function f ∈ C 0 ∞ (−1, 1)
R 1
with f =1 and let ϕ ∈ C ∞ (−1, 1) be arbitrary. Set
−1 0
1 Z 1
µZ ¶
1 ¡ 2 ¢
0 0
v (x)= ϕ (x) − ϕdx f (x) and λ = − 2 u f − π uf dx .
π
−1 −1
R 1 R 1
Observe that v ∈ X ∩ C ∞ (−1, 1). Use (7.25), the fact that f =1, v =0
0 −1 −1
and the definition of λ to get, for every ϕ ∈ C 0 ∞ (−1, 1),
1
Z
£ 2 ¤
0 0
u ϕ − π (u − λ) ϕ dx
−1
Z ∙ µ Z ¶ µ Z ¶¸ Z
2
2
= u 0 v + f 0 ϕ − π u v + f ϕ + π λ ϕ
0
Z ∙Z ¸ ∙ Z ¸
¡ 2 ¢ 2 ¡ 2 ¢
0 0
= u v − π uv + ϕ π λ + u f − π uf =0 .
0 0
The regularity of u (which is a minimizer of (P) in X) then follows (as in
Proposition4.1) atoncefromtheaboveequation. Sinceweknow(fromThe-
orem 6.1) that among smooth minimizers of (P) the only ones are of the form
u (x)= α cos πx + β sin πx,wehavetheresult.
Exercise 6.2.3. We divide the proof into two steps.
Step 1. We start by introducing some notations. Since we will work with
fixed u, v, we will drop the dependence on these variables in L = L (u, v) and
M = M (u, v). However we will need to express the dependence of L and M on
the intervals (α, β),where a ≤ α< β ≤ b, and we will therefore let
Z
β p
02
02
L (α, β)= u + v dx
α
Z β
M (α, β)= uv dx .
0
α
So that in these new notations
L (u, v)= L (a, b) and M (u, v)= M (a, b) .