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Let ρ> 0 be a parameter. Then ( A.22 ) is equivalent to
Find u
K
such that
(A.23)
ρf
ρBu
+
u
u, v
u
V
0
v
K
,
or to the fixed point problem
u = P K (ρf ρBu + u) in
K .
(A.24)
Define, for v
K
,
S
v
=
P
(v
ρ(Bv
f)) . Then, by ( A.6 ), for every
K
v 1 ,v 2 K
,
S
v 1 S
v 2 V
(v 1
v 2 )
ρB(v 1
v 2 )
V
,
and hence
2
2
ρ 2
2
V
S
v 1 S
v 2
V
v 1
v 2
V
2 ρ
B(v 1
v 2 ), v 1
v 2 V +
B(v 1
v 2 )
2
V
2 ρC 2 + ρ 2 C 1 ).
v 1 v 2
( 1
2 ρC 2 + ρ 2 C 1 ) 2 < 1. Then
We fix ρ> 0 such that κ := ( 1
S
is a contraction on
K
and ( A.24 ) has a unique solution.
Remark A.3.4 The proof is constructive—the contraction
S
converges with rate κ<
1. The rate is maximal, if
ρ 2 C 1 }=
C 2 /C 1 ,
ρ
=
ρ
=
arg min
{
1
2 ρC 2 +
(A.25)
then
C 2 /C 1 ) 2 < 1 .
κ
=
κ
=
( 1
(A.26)
In particular, if v ( 0 )
K
is arbitrary, the sequence
v (i + 1 )
v (i)
f)
v (i)
ρ(Bv (i)
= S
=
P
(A.27)
K
converges to u
K
, the solution of ( A.17 ), and
v (i)
i
v ( 0 )
u
V
u
V
,i
0 .
(A.28)
Corollary A.3.5 If
M V
is a closed , linear subspace , the problem
M
M
Find u
such that
b(u, v)
(v)
v
admits a unique solution .
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