Page 227 - INTRODUCTION TO THE CALCULUS OF VARIATIONS

214 Solutions to the Exercises

we find that
ϕ xx = f, ϕ xy = g, ϕ yy = h
and hence that
ϕ ϕ − ϕ 2 xy =1 .
xx yy
The fact that ϕ is convex follows from the above identity, ϕ > 0, ϕ > 0 and
xx yy
Theorem 1.50.

7.6 Chapter 6: Isoperimetric inequality

7.6.1 The case of dimension 2
Exercise 6.2.1. One can consult Hardy-Littlewood-Polya [55], page 185, for
more details. Let u ∈ X where
½ Z 1 ¾
X = u ∈ W 1,2 (−1, 1) : u (−1) = u (1) with u =0 .
−1
Define
z (x)= u (x +1) − u (x)

and note that z (−1) = −z (0),since u (−1) = u (1). Wededucethatwecan find
α ∈ (−1, 0] so that z (α)=0, which means that u (α +1) = u (α).We denote
this common value by a (i.e. u (α +1) = u (α)= a). Since u ∈ W 1,2 (−1, 1) it
2
is easy to see that the function v (x)= (u (x) − a) cot (π (x − α)) vanishes at
x = α and x = α +1 (this follows from Hölder inequality, see Exercise 1.4.3).
We therefore have (recalling that u (−1) = u (1))

Z
1 n 2 2 o
02
2
u − π (u − a) − (u − π (u − a)cot π (x − α)) dx
0
−1
h i 1
2
= π (u (x) − a) cot (π (x − α)) =0 .
−1
R 1
Since u =0,weget from theabove identity that
−1
Z Z
1 1
¡ 02 2 2 ¢ 2 2 2
0
u − π u dx =2π a + (u − π (u − a)cot π (x − α)) dx
−1 −1
and hence Wirtinger inequality follows. Moreover we have equality in Wirtinger
inequality if and only if a =0 and,c denoting a constant,
0
u = πu cot π (x − α) ⇔ u = c sin π (x − α) .