Page 100 - INTRODUCTION TO THE CALCULUS OF VARIATIONS

A general existence theorem 87

© 1,2 ª
where X = u ∈ W (0, 1) : u (0) = 1,u (1) = 0 . All the hypotheses of the
theorem are verified with the exception of (H2) that is satisfied only with α 1 =0.
This is enough to show that (P) has no minimizer in X. Indeedwehaveseen
1
in Exercise 2.2.6 that (P) has no solution in Y = X ∩ C ([0, 1]) and that the
corresponding value of the infimum, letusdenoteitby m Y ,is 0. Since trivially
0 ≤ m X ≤ m Y we deduce that m X =0. Now assume, by absurd hypothesis,
that (P) has a solution u ∈ X, we should then have I (u)= 0,but since the
0
integrand is non negative we deduce that u =0 a.e. in (0, 1). Since elements of
X are continuous we have that u is constant, and this is incompatible with the
boundary data. Hence (P) has no solution.
Example 3.9 The present example (cf. Poincaré-Wirtinger inequality) shows
that we cannot allow, in general, that q = p in (H2). Let n =1, λ>π and
1 ¡ 2 2 2 ¢
f (x, u, ξ)= f (u, ξ)= ξ − λ u .
2
We have seen in Section 2.2 that if

½ Z 1 ¾
1,2
(P) inf I (u)= f (u (x) ,u (x)) dx : u ∈ W (0, 1) = m
0
0
0
then m = −∞, which means that (P) has no solution.
Example 3.10 (Bolza example). Wenowshowthat, asageneralrule, one
cannot weaken (H1) either. One such example has already been seen in Section
2.2 where we had f (x, u, ξ)= f (ξ)= e −ξ 2 (which satisfies neither (H1) nor
(H2)). Let n =1,
¡ 2 ¢ 2 4
f (x, u, ξ)= f (u, ξ)= ξ − 1 + u
½ Z 1 ¾
0
(P) inf I (u)= f (u (x) ,u (x)) dx : u ∈ W 1,4 (0, 1) = m.
0
0
Assume for a moment that we already proved that m =0 and let us show that
1,4
(P) has no solution, using an argument by contradiction. Let u ∈ W (0, 1) be
0
0
a minimizer of (P), i.e. I (u)= 0. This implies that u =0 and |u | =1 a.e. in
(0, 1).Since the elements of W 1,4 are continuous we have that u ≡ 0 and hence
0
u ≡ 0 which is clearly absurd.
So let us show that m =0 by constructing an appropriate minimizing se-
1,4
quence. Let u ν ∈ W (ν ≥ 2 being an integer) defined on each interval
0
[k/ν, (k +1) /ν], 0 ≤ k ≤ ν − 1,asfollows
⎧ £ ¤
k 2k 2k+1
x − if x ∈ ,
⎨
ν 2ν 2ν
u ν (x)=
⎩ k+1 ¡ 2k+1 2k+2 ¤
−x + if x ∈ , .
ν 2ν 2ν