Page 157 - INTRODUCTION TO THE CALCULUS OF VARIATIONS

144 Minimal surfaces

we then choose λ and µ so that

⎧
⎨ λ x − µ = σ
y
⎩
λ y + µ = τ
x
(thisisalwayspossible; find first λ satisfying ∆λ = σ x + τ y then choose µ such
¡ ¢
that µ ,µ y =(τ − λ y ,λ x − σ)). Returning to (5.13) we find
x
ZZ
{(E − G) σ +2Fτ} dxdy =0, ∀σ, τ ∈ C 0 ∞ (Ω) .
Ω
The fundamental lemma of the calculus of variations (Theorem 1.24) implies
then E = G and F =0. Thus, up to the condition v x ×v y 6=0, Plateau problem
is solved (cf. Theorem 5.17).
We have still to prove (iii) in the statement of the theorem. However this
follows easily from (i), (ii) and (S2) of the theorem, cf. Dierkes-Hildebrandt-
Küster-Wohlrab [39] page 248.
¡ ¢
Step 4.We let Σ 0 = v Ω where v is the element that has been found in the
previous steps and satisfies in particular

d =inf {D (v): v satisfies (S1), (S2), (S3)} = D (v) .

To conclude the proof of the theorem it remains to show that

a =inf {Area (Σ): Σ ∈ S} =Area (Σ 0 )= D (v)= d.
We already know, from the previous steps, that

a ≤ Area (Σ 0 )= D (v)= d
and we therefore wish to show the reverse inequality.
A way of proving this claim is by using a result of Morrey (see Dierkes-
Hildebrandt-Küster-Wohlrab [39] page 252 and for a slightly different approach
see Courant [24] and Nitsche [78]) which asserts that for any > 0 and any v

satisfying (S1), (S2), (S3), we can find v verifying (S1), (S2), (S3) so that

D (v ) − ≤ Area (Σ)
¡ ¢
where Σ = v Ω .Since d ≤ D (v ) and is arbitrary, we obtain that d ≤ a;the
other inequality being trivial we deduce that a = d.
This concludes the proof of the theorem.