Cervical Pain 1 Day Before Period

integral equation (differential) Lipschitz and the type of Picard iterations.

Exercise

was repeated under the same theoreme1

in the message of 28/11/2006

Let (E, II) a real Banach space. <= kIx-yI sur ExE.

C denotes the space of continuous functions

defined on [0, T] with values in E.

space C is equipped with the standard

u, denoted II.II

Let K be a continuous mapping of [0, T] x [0, T) xE

in E satisfies the Lipschitz condition: There

L> 0: IK (t, s, x)-K (t, s, y) I

for all (t, s) [0, T] x [0, T] and x , y

E. It defines the same application S

on C (see the proof of theoreme1) by:

S (x) (t) = f (t) + \\ int_ {0} ^ {t} K (t, s, x (s)) ds;

. where f is a fixed element x in C and C

i) Verify that C is stable in S.

ii) for x, y in C, show that

II (S ^ {n} x) - (S ^ {n} y) II

ii) Deduce that S admits a unique fixed point z C.

iii) Show that for each x in C, we have:

IIS ^ {n} x-ZII or

R_ {n) is a constant to be determined.
<= LIx-yI
iv) Search the benefits of this approach.

(Compare with theoreme1)
Proof

i) obvious

ii) Demonstration by recurrence:

for n = 1, the result is évident.Supposons

true has order n, then:

I (S ^ {n +1} x) (t) - (S ^ { n +1} y) (t) I

<= [(LT)^{n}/n!] IIx-yII





= (Lt) ^ {n +1} / (n +1)! IIx-YII <=R_{n}ISx-zI

Taking the sup over [0, T], we obtain

IIS ^ {n +1} xS ^ {n +1} YII

ii) Taking into account the previous inequality (* ) apply

Assertion i) or the assertion ii) the exercise of the message

20/10/2007.

iii) For any x in C, we have:

IIS ^ {n} x-ZII



or R_ {n} = Sigma (i = 0 to infinity) [(LT) ^ {n + i } / (n + i)]
<=\int_{0}^{t} IK(t,s,(S^{n}x)(s))-K(t,s,(S^{n}y)(s))I ds
is the rest of the series expansion of exp (LT).
<= L\int_{0}^{t} I(Sx)(s)-(Sy)(s)Ids
iv) leaves the reader.
<= \int(L^{n+1} s^{n})/n!)Ix(s)-y(s)I ds



(

<= (LT)^{n+1}/(n+1)! IIx-yII (*)