Page 224 - INTRODUCTION TO THE CALCULUS OF VARIATIONS

Chapter 5: Minimal surfaces 211

p 2
(ii) Let f (w, ξ)= w 1+ ξ . Observe that the function f is not convex over
(0, +∞)×R; although the function ξ → f (w, ξ) is strictly convex, whenever w>
0. We will therefore only give necessary conditions for existence of minimizers
of (P α ) and hence we write the Euler-Lagrange equation associated to (P α ),
namely

∙ ¸
d d ww 0 p
02
0
0
[f ξ (w, w )] = f w (w, w ) ⇔ √ = 1+ w . (7.21)
dx dx 1+ w 02
Invoking Theorem 2.7, we find that any minimizer w of (P α )satisfies
∙ ¸
d d w
0 0 0
[f (w, w ) − w f ξ (w, w )] = 0 ⇔ √ =0
dx dx 1+ w 02
which implies, if we let a> 0 be a constant,
w 2
02
w = − 1 . (7.22)
a 2
Before proceeding further, let us observe the following facts.
1) The function w ≡ a is a solution of (7.22) but not of (7.21) and therefore
it is irrelevant for our analysis.
2) To a =0 corresponds w ≡ 0, which is also not a solution of (7.21) and
moreover does not satisfy the boundary conditions w (0) = w (1) = α> 0.
2
2
3) Any solution of (7.22) must verify w ≥ a and, since w (0) = w (1) = α>
0,thusverifies w ≥ a> 0.
We can therefore search for solutions of (7.22) the form
f (x)
w (x)= a cosh
a
where f satisfies, when inserted into the equation, f 02 =1, which implies that
1
either f ≡ 1 or f ≡−1,since f is C . Thus the solution of the differential
0
0
equation is of the form
x + µ
w (x)= a cosh .
a
Since w (0) = w (1), we deduce that µ = −1/2. Finally since w (0) = w (1) = α,
2
every solution C of (P α )mustbeofthe form
µ ¶
2x − 1 1
w (x)= a cosh and a cosh = α.
2a 2a
Summarizing, we see that depending on the values of α, the Euler-Lagrange
equation (7.21) may have 0, 1 or 2 solutions (in particular for α small, (7.21)