For printing - MSP

184 JAIME RIPOLL
where a is a constant. In the points where x = x(z), we therefore get
x ′2
x
=
a − Hx2 2 − 1.
We then have
for r ≤ x ≤ x1, and
z ′ (x) =
z ′ (x) = −
1
x
a−Hx 2
1
x
a−Hx 2
2 − 1
2 − 1
for x1 ≤ x ≤ R, where x1 ∈ (r, R) is such that z ′ (x1) = 0. Since z ′ (r) = ∞,
we have
r
a − Hr2 2 − 1 = 0
and a = r(−1 + Hr) or a = r(1 + Hr). In the case of the nodoids, we
know moreover that x ′ (z1) = ∞, where z1 = z(x1). Therefore we have
a = Hx 2 1 ≥ Hr2 , and this implies that a = r(1 + rH). It follows that
r(1 + rH)
x1 =
.
H
Now, a computation shows that
z ′ (x1 − x) ≥ −z ′ (x1 + x),
for all x ∈ [0, x1 − r]. It follows then that
proving the proposition.
R ≥ x1 + (x1 − r) ≥
H
1 +
2
1 + 1
rH
Proof of Theorem 2. Considering the family of Dirichlet’s problems
∇u
div
1 + |∇u| 2 = −2h, u|∂Ω = 0, u ∈ C 2 (9)
(Ω), h ∈ [0, H0)
we first prove that
S := {h ∈ [0, H0) | (9) has a solution}
coincides with [0, H0). Since (9) has the trivial solution for h = 0, we have
S = ∅. From the implicit function theorem, it follows that S is open in
[0, H0). For proving that S is closed in [0, H0), let us consider a sequence
{hn} ⊂ S converging to H1 ∈ [0, H0). If H1 = 0 then H1 ∈ S and we are
done. Thus, let us assume that H1 > 0. By contradiction, assume that