Page 66 - INTRODUCTION TO THE CALCULUS OF VARIATIONS

Euler-Lagrange equation 53

© 1 ª
where X = u ∈ C ([0, 1]) : u (0) = 1,u (1) = 0 . We will show that (P) has
1
1
no C or piecewise C solution (not even in any Sobolev space). The Euler-
Lagrange equation is
c
0 0
0
(xu ) =0 ⇒ u = ⇒ u (x)= c log x + d, x ∈ (0, 1)
x
where c and d are constants. Observe first that such a u cannot satisfy simulta-
neously u (0) = 1 and u (1) = 0.
We associate to (P) the following problem
½ Z ¾
1
0
(P piec ) inf I (u)= f (x, u (x)) dx = m piec
u∈X pie c
0
© 1 ª
X piec = u ∈ C ([0, 1]) : u (0) = 1,u (1) = 0 .
piec
We next prove that neither (P) nor (P piec ) have a minimizer. For both cases
it is sufficient to establish that m piec = m =0. Let us postpone for a moment the
1
proof of these facts and show the claim. If there exists a piecewise C function
v satisfying I (v)= 0, this would imply that v =0 almost everywhere in (0, 1).
0
Since the function v ∈ X piec , it should be continuous and v (1) should be equal
to 0, we would then deduce that v ≡ 0, which does not verify the other boundary
condition, namely v (0) = 1. Hence neither (P) nor (P piec ) have a minimizer.
We first prove that m piec =0.Let n ∈ N and consider the sequence
⎧ £ 1 ¤
1 if x ∈ 0,
⎨ n
u n (x)=
⎩ − log x ¡ 1 ¤
if x ∈ , 1 .
log n n
1
Note that u n is piecewise C , u n (0) = 1, u n (1) = 0 and
1
I (u n )= → 0,as n →∞,
log n
hence m piec =0.
We finally prove that m =0. This can be done in two different ways. A more
sophisticated argument is given in Exercise 2.2.6 and it provides an interesting
continuity argument. A possible approach is to consider the following sequence
⎧
−n 2 2 n £ 1 ¤
⎪ x + x +1 if x ∈ 0,
⎨ log n log n n
u n (x)=
⎪ − log x ¡ 1 ¤
⎩ if x ∈ , 1 .
log n n
We easily have u n ∈ X and since
⎧ £ ¤
n (1 − 2nx) if x ∈ 0, 1
⎨ log n n
0
u (x)=
n
⎩ −1 ¡ 1 ¤
x log n if x ∈ n , 1