Page 201 - INTRODUCTION TO THE CALCULUS OF VARIATIONS

188 Solutions to the Exercises

We start by observing that for any > 0 and u ∈ X piec ,wecan find v ∈ X such
that
ku − vk W 1,2 ≤ .
It is an easy matter (exactly as above) to show that if
Z 1
2
I (u)= x (u (x)) dx
0
0
then we can find a constant K so that
0 ≤ I (v) ≤ I (u)+ K .

Taking the infimum over all elements v ∈ X and u ∈ X piec we get that
0 ≤ m ≤ K

whichisthe desiredresultsince is arbitrary.
1
Exercise 2.2.7. Let u ∈ C ([0, 1]) with u (0) = u (1) = 0. Invoking Poincaré
inequality (cf. Theorem 1.47), we can find a constant c> 0 such that

Z Z
1 1
02
2
u dx ≤ c u dx .
0 0
2
We hence obtain that m λ =0 if λ is small (more precisely λ ≤ 1/c). Observe
that u 0 ≡ 0 satisfies I λ (u 0 )= m λ =0. Furthermore it is the unique solution
2
2
of (P λ ) since, by inspection, it is the only solution (if λ <π )ofthe Euler-
Lagrange equation
½ 2
u + λ u =0
00
u (0) = u (1) = 0 .
The claim then follows.
Exercise 2.2.8. Let
© 1 ª
X piec = u ∈ C ([−1, 1]) : u (−1) = 0,u (1) = 1
piec
and ½
0 if x ∈ [−1, 0]
u (x)=
x if x ∈ (0, 1] .
It is then obvious to see that
½ Z ¾
1
inf I (u)= f (u (x) ,u (x)) dx = I (u)= 0 .
0
u∈X pie c
−1
Note also that the only solution in X piec is u.