Page 64 - INTRODUCTION TO THE CALCULUS OF VARIATIONS

Euler-Lagrange equation 51

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where X = u ∈ C ([0, 1]) : u (0) = u (1) = 0 . We have from (2.2) that
u ≡ 0 (and it is clearly a maximizer of I in the class of admissible functions
X), however (P) has no minimizer as we will now show. Assume for a
moment that m =0, then, clearly, no function u ∈ X can satisfy
Z 1 2
0
e −(u (x)) dx =0
0
and hence (P) has no solution. Let us now show that m =0.Let n ∈ N
and define
µ ¶ 2
1 n
u n (x)= n x − −
2 4
then u n ∈ X and

Z 1 1 Z n
2
I (u n )= e −4n (x−1/2) 2 dx = e −y 2 dy → 0,as n →∞.
0 2n −n
Thus m =0, as claimed.

1
3. Solutions of (P) are not necessarily C .
1
We now show that solutions of (P) are not necessarily C even in the
present simple case (another example with a similar property will be given
¡ 2 ¢ 2
in Exercise 2.2.8). Let f (ξ)= ξ − 1
1
½ Z ¾
0
(P) inf I (u)= f (u (x)) dx = m
u∈X
0
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where X = u ∈ C ([0, 1]) : u (0) = u (1) = 0 . We associate to (P) the
following problem
½ Z 1 ¾
0
(P piec ) inf I (u)= f (u (x)) dx = m piec
u∈X pie c 0
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X piec = u ∈ C ([0, 1]) : u (0) = u (1) = 0 .
piec
This last problem has clearly
½
x if x ∈ [0, 1/2]
v (x)=
1 − x if x ∈ (1/2, 1]
1
as a solution since v is piecewise C and satisfies v (0) = v (1) = 0 and
I (v)= 0;thus m piec =0. Assume for a moment that we already proved
that not only m piec =0 but also m =0. This readily implies that (P),