Page 229 - INTRODUCTION TO THE CALCULUS OF VARIATIONS

216 Solutions to the Exercises

We next let
© 02 02 ª
O = x ∈ (a, b): u (x)+ v (x) > 0 .
Thecasewhere O =(a, b) hasbeenconsideredinStep1of Theorem6.4. If O
is empty the result is trivial, so we will assume from now on that this is not the
case. Since the functions u and v are continuous, the set O is open. We can
0
0
then find (see Theorem 6.59 in [57] or Theorem 9 of Chapter 1 in [37])
a ≤ a i <b i <a i+1 <b i+1 ≤ b, ∀i ≥ 1

∞
O = ∪ (a i ,b i ) .
i=1
02
c
02
In the complement of O, O ,wehave u + v =0, and hence
L (b i ,a i+1 )= M (b i ,a i+1 )= 0 . (7.26)
Step 2. We then change the parametrization on every (a i ,b i ).We choose a
multiple of the arc length, namely

⎧
L (a, x)
⎨ y = η (x)= −1+2
⎪
⎪
L (a, b)
⎪
⎪ ¡ ¢ ¡ ¢
ϕ (y)= u η (y) and ψ (y)= v η (y) .
⎩ −1 −1
Note that this is well defined, since (a i ,b i ) ⊂ O.We then let
L (a, a i ) L (a, b i )
α i = −1+2 and β = −1+ 2
i
L (a, b) L (a, b)
so that
L (a i ,b i )
β − α i =2 .
i
L (a, b)
Furthermore, since L (b i ,a i+1 )= 0,weget
∞
β = α i+1 and ∪ [α i ,β ]=[−1, 1] .
i i
i=1
We also easily find that, for y ∈ (α i ,β ),
i
q L (a, b) L (a i ,b i )
02
02
ϕ (y)+ ψ (y)= =
2
β − α i
i
ϕ (α i )= u (a i ) ,ψ (α i )= v (a i ) ,ϕ (β )= u (b i ) ,ψ (β )= v (b i ) .
i i