Page 63 - INTRODUCTION TO THE CALCULUS OF VARIATIONS

50 Classical methods

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Since the integrand is, by strict convexity of f, positive unless u = v and u = v 0
we deduce that u ≡ v, as wished.
We now consider several particular cases and examples that are arranged in
an order of increasing difficulty.

Case 2.3 f (x, u, ξ)= f (ξ).

This is the simplest case. The Euler-Lagrange equation is
d
0
0
[f (u )] = 0, i.e. f (u )= constant.
0
0
dx
Note that
β − α
u (x)= (x − a)+ α (2.2)
b − a
is a solution of the equation and furthermore satisfies the boundary conditions
u (a)= α, u (b)= β. It is therefore a stationary point of I. It is not, however,
always a minimizer of (P) as will be seen in the second and third examples.
1. f is convex.
If f is convex the above u is indeed a minimizer. This follows from the
theorem but it can be seen in a more elementary way (which is also valid
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even if f ∈ C (R)). From Jensen inequality (cf. Theorem 1.51) it follows
1
that for any u ∈ C ([a, b]) with u (a)= α, u (b)= β
à ! µ ¶
1 Z b 1 Z b u (b) − u (a)
0
0
f (u (x)) dx ≥ f u (x) dx = f
b − a a b − a a b − a
µ ¶
β − α
0
= f = f (u (x))
b − a
Z b
1
= f (u (x)) dx
0
b − a
a
which is the claim.
2. f is non convex.
If f is non convex then (P) has, in general, no solution and therefore the
above u is not necessarily a minimizer (in the particular example below it
is a maximizer of the integral). Consider f (ξ)= e −ξ 2 and
½ Z ¾
1
(P) inf I (u)= f (u (x)) dx = m
0
u∈X
0