Page 68 - INTRODUCTION TO THE CALCULUS OF VARIATIONS

Euler-Lagrange equation 55

We will show in Exercise 2.2.7, Example 2.23 and Theorem 6.1 the follow-
ing facts.
-If λ ≤ π then m λ =0, which implies, in particular,
1 1
Z Z
2
02
u dx ≥ π 2 u dx .
0 0
Moreover if λ<π,then u 0 ≡ 0 is the only minimizer of (P λ ). If λ = π,
then (P λ )has infinitely many minimizers which are all of the form u α (x)=
α sin πx with α ∈ R.
-If λ>π then m λ = −∞, which implies that (P λ ) has no solution. To
see this fact it is enough to choose u α as above and to observe that since
λ>π,then
Z 1 £ ¤
2
2
2
2
I λ (u α )= α 2 π cos (πx) − λ sin (πx) dx →−∞ as α →∞.
0
2. Brachistochrone.
p 2 √
The function, under consideration, is f (u, ξ)= 1+ ξ / u, here (com-
pared with Chapter 0) we take g =1/2,and
( )
Z b
(P) inf I (u)= f (u (x) ,u (x)) dx = m
0
u∈X 0
© ª
1
where X = u ∈ C ([0,b]) : u (0) = 0,u (b)= β and u (x) > 0, ∀x ∈ (0,b] .
The Euler-Lagrange equation and its first integral are
√
∙ ¸ 0 02
u 0 1+ u
√ √ = − √
u 1+ u 02 2 u 3
√ ∙ ¸
1+ u 02 u 0
√ − u 0 √ √ = constant.
u u 1+ u 02
This leads (µ being a positive constant) to
¡ 02 ¢
u 1+ u =2µ.
The solution is a cycloid and it is given in implicit form by
¡ −1 ¢
u (x)= µ 1 − cos θ (x)
where
θ (t)= µ (t − sin t) .
Note that u (0) = 0. It therefore remains to choose µ so that u (b)= β.