Page 65 - INTRODUCTION TO THE CALCULUS OF VARIATIONS

52 Classical methods

contrary to (P piec ), has no solution. Indeed I (u)= 0 implies that |u | =1
0
almost everywhere and no function u ∈ X can satisfy |u | =1 (since by
0
continuity of the derivative we should have either u =1 everywhere or
0
0
u = −1 everywhere and this is incompatible with the boundary data).
We now show that m =0. We will give a direct argument now and a more
elaborate one in Exercise 2.2.6. Consider the following sequence
⎧ £ 1 1 ¤
x if x ∈ 0, −
⎪ 2 n
⎪
⎪
⎪
⎨ ¡ ¢ 3 ¡ ¢ 2 ¡ ¤
1
u n (x)= −2n 2 x − 1 2 − 4n x − 1 2 − x +1 if x ∈ 1 2 − , 1
n 2
⎪
⎪
⎪
⎪
⎩ ¡ 1 ¤
1 − x if x ∈ , 1 .
2
Observe that u n ∈ X and
1
Z Z 1
2 4
0
0
I (u n )= f (u (x)) dx = f (u (x)) dx ≤ → 0 .
n
n
0 1 − 1 n
2 n
This implies that indeed m =0.
We can also make the further observation that the Euler-Lagrange equation
is
d h ³ 0 2 ´i
u 0 (u ) − 1 =0.
dx
It has u ≡ 0 as a solution. However, since m =0, it is not a minimizer
(I (0) = 1).
Case 2.4 f (x, u, ξ)= f (x, ξ).
The Euler-Lagrange equation is
d
0
[f ξ (x, u )] = 0, i.e. f ξ (x, u )= constant.
0
dx
The equation is already harder to solve than the preceding one and, in general,
it has not as simple a solution as in (2.2).
We now give an important example known as Weierstrass example.Let
2
f (x, ξ)= xξ (note that ξ → f (x, ξ) is convex for every x ∈ [0, 1] and even
strictly convex if x ∈ (0, 1])and
½ Z ¾
1
0
(P) inf I (u)= f (x, u (x)) dx = m
u∈X
0