Page 72 - INTRODUCTION TO THE CALCULUS OF VARIATIONS

Second form of the Euler-Lagrange equation 59

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Exercise 2.2.9 Let X = u ∈ C ([0, 1]) : u (0) = 0,u (1) = 1 and
½ Z 1 ¾
(P) inf I (u)= |u (x)| dx = m.
0
u∈X
0
Show that (P) has infinitely many solutions.
0
Exercise 2.2.10 Let p ≥ 1 and a ∈ C (R),with a (u) ≥ a 0 > 0.Let A be
defined by
1/p
A (u)= [a (u)] .
0
Show that a minimizer (which is unique if p> 1)of
( )
Z
b
p
(P) inf I (u)= a (u (x)) |u (x)| dx
0
u∈X
a
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1
where X = u ∈ C ([a, b]) : u (a)= α, u (b)= β is given by
∙ ¸
A (β) − A (α)
−1
u (x)= A (x − a)+ A (α) .
b − a
2.3 Second form of the Euler-Lagrange equation
The next theorem gives a different way of expressing the Euler-Lagrange equa-
tion, this new equation is sometimes called DuBois-Reymond equation.It turns
out to be useful when f does not depend explicitly on x, as already seen in some
of the above examples.
2
Theorem 2.7 Let f ∈ C ([a, b] × R × R), f = f (x, u, ξ),and
( )
Z b
0
(P) inf I (u)= f (x, u (x) ,u (x)) dx = m
u∈X a
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2
1
where X = u ∈ C ([a, b]) : u (a)= α, u (b)= β .Let u ∈ X ∩ C ([a, b]) be a
minimizer of (P) then for every x ∈ [a, b] the following equation holds
d
0
0
0
[f (x, u (x) ,u (x)) − u (x) f ξ (x, u (x) ,u (x))] = f x (x, u (x) ,u (x)) . (2.3)
0
dx
Proof. We will give two different proofs of the theorem. The first one is
very elementary and uses the Euler-Lagrange equation . The second one is more
involved but has several advantages that we do not discuss now.