In this section the local quadratic converges for the iterative procedure 4 is proved, the truth of the inequality
is shown. For t h e proof we use the same approach as for Newton's method [5].
Let
11 1)
be the vector 11-norm and the induced matrix norm onR'
[2]. Let S,(x) be the open domain with radius E>
0 and with center in x, i.e.The main result is given in
Proposition 4.3 (Sobolev, 1989 [ll]). Let the function
be given i n a n open convex set
D
implicitly by the systemAssume there ezists the vector
8
EB
CR1
such that equality ~ ( 8 ) = 0 holds. JacobianJ
of functionF
evaluated i n8
i s nonsingular and the equalityholds.
T h e n there exists a small scalar E
>
0 such that for allPo
E ~ ~ ( 8 ) the sequence derived by the iterative procedure4 {Pn,
n = 0, l , 2 , ..
.} i s defined correctly and converges t o the desiredp
quadratically.Proof. First, let us show that the iterative process is defined correctly and the eval- uation
holds for all n, i.e., Pn+1 E n = 1,2,.
. . .
Consider the iterative scheme 4 for k = 0
PI
=Po
-[
Ao]-'
FO.Vector
pl
is defined correctly if matrix A. is nonsingular. If any matrixB
is nonsingular and for any other matrix A the inequalityholds then the matrix A is nonsingular too and the evaluation
is valid [5].
Consider matrix A0 as A and matrix J. =
~ ( 8 )
asB
as above. Then the following evaluationholds. The theorem about high-order derivatives for implicit function guarantees that there exist continuous derivatives for function F at least second order with apprecia- ble smoothness of function
U.
Therefore function F and its Jacobian J are Lipschitz continuous inD,
i.e. there are constants p, q such that VP, P' ED
are valid.
Consider the norm of difference
. .
Fi(Po
+
Foej) - F i ( P o ) - J?Fj(po)F j ( P o ) ( 7 4 )
The Jacobian definition and Newton's theorem implies immediately for all vectors x and x
+
y the equalitywhere it means the element-by-element integration of the matrix-valued function. Conse- quently, the elements of matrix Jo - A. are evaluated as
Then (74) is transformed into the form
(1
[Ji'(po+
tFoej) -J?] 1)
dt.0 1x1
Using (72) and (73) we have
Therefore for any
1
we obtain
11
J r l ( J * - Ao)((I
r(ll J*-
Joll+ ll
Jo - Aoll)So the matrix A. is nonsingular, the inequality
holds and
Dl
is defined correctly. Now let us show thata
belongs tos C ( j ) .
The equality implies the other equalityLet GnA be the notation for the difference
A - 8
and F, be the evaluation of the function F atp.
Rewrite the last equality in the formAOGl = AoG0
-
Fo+
F,.Adding the vector JOGo, where Jo = J ( P o ) , we receive
Then
61 = [AO]-'(AO - Jo)Go
+
[Ao]-I(-&+
F*+
JOGo).Using (75) we have
Taking into account that Jacobian of the function F is Lipschitz continuous in
D
with ( 7 3 ) we haveI[(-Fo
+
F*+
Jo6o)Il<
112~11~011~<
1/2~&116011.Then for the norm llG1
11
we can writeit means that
PI
belongs to~ ~ ( 6 ) .
The inductive consideration implies that matrix An is nonsingular, the evaluations
for n = 1 , 2 , .
. .
are valid. It means the convergence of the sequenceto vector
8.
Consequently the sequenceconverges to zero.
To prove the quadratic convergence it is sufficient t o show that the sequence ( 7 7 ) satisfies the inequality [6]
Ilhn+~
I1 5
cl16nll,2- For n>
0 rewrite ( 7 6 ) in the formAs far as for n the matrix
An
is nonsingular and the expression6n+1 = [An]-'(An
-
Jn)6n+
[An]-'(-Fn+
F*+
J n 6 n )is defined correctly. It implies the evaluation
II6n+l
II 5
2r(lIAn - JnI((16nII+
Il(-Fn+
J'*+
Jn6n:1I().With ( 7 5 ) the last inequality can be transformed to
Il6n+lII
I
2r(1/2~q116n11~+
Il(-Fn+
F*+
Jn6n)Il).The implicit function F is continuously differentiable. This implies the evaluation Il(-Fn
+
F*+
Jn6n)llI
1 / 2 ~ 1 1 6 ~ ) ( ~ .As a result we receive the inequality
l16n+111
5
C116n1I2, whereReferences
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