NoteNextra-origin/content/Swap/Math4501/Math4501_L3.md

Math4501 Lecture 3

Review from last lecture

Bisection method for finding root

P1. Let f be a continuous function on [a,b]\to \mathbb{R}, find \xi \in [a,b] such that f(\xi)=0.

P2. Let g be a continuous function on [a,b]\to \mathbb{R}, find \xi \in [a,b] such that g(\xi)=\xi.

Theorem 1:

solution to P1 exists if f(a)f(b)<0.

Theorem 2:

Fixed point theorem:

If solution to P2 exists, then g:[a,b]\to [a,b].

Bijection method

Obtain two sequence (a_n)_{n=1}^\infty and (b_n)_{n=1}^\infty.

Initially, we set a_0=a and b_0=b.

If |a_n-b_n|<2^{-n}|a_0-b_0|, then a_n and b_n are Cauchy sequence. So their limit exists.

\lim_{n\to\infty} a_n=\lim_{n\to\infty} b_n=\xi.

Simple iteration method

Definition of Simple Iteration

Given x_0\in [a,b], we define a sequence (x_n)_{n=0}^\infty by

x_{n+1}=g(x_n)

If a simple iteration converges, the it converges to a fixed point of

Proof

Let c\coloneqq \lim_{n\to\infty} x_n=g(c).

g:[a,b]\to \mathbb{R} is continuous if and only if g is continuous at x_0\in [a,b].

Definition of Lipschitz continuous

A function g:[a,b]\to \mathbb{R} is Lipschitz continuous if for all x,y\in [a,b], there exists a constant L>0 such that

|g(x)-g(y)|\leq L|x-y|

for some L>0.

Note

Lipschitz continuous is a stronger condition than continuous.

If a function is Lipschitz continuous, then it is continuous.

However, the converse is not true.

Definition of contraction mapping

A function g:[a,b]\to \mathbb{R} is a contraction mapping if it is Lipschitz continuous with L<1.

Theorem of simple iteration

Let g:[a,b]\to [a,b] is a contraction mapping (Lipschitz continuous with L<1, with pointwise ontinuous g).

Then

• g has a unique fixed point \xi \in [a,b]
• Simple iteration x_{n+1}=g(x_n) converges to \xi for any x_0\in [a,b].

Proof

Uniqueness:

Suppose \xi_1 and \xi_2 are two fixed points of g.

Then |x_1-x_2|=|g(\xi_1)-g(\xi_2)|\leq L|\xi_1-\xi_2|.

Thus, (1-L)|\xi_1-\xi_2|\leq 0, which implies |\xi_1-\xi_2|=0.

A more general result:

Brouwer's fixed point theorem

Convergence:

Let \xi\in [a,b] be the unique fixed point of g.

Then,

\begin{aligned}
|x_n-\xi|&=|g(x_{n-1})-g(\xi)|\\
&\leq L|x_{n-1}-\xi|\\
&=L|g(x_{n-2})-g(\xi)|\\
&\leq L^2|x_{n-2}-\xi|\\
&\vdots\\
&\leq L^n|x_0-\xi|

Thus, we can always find N such that L^N|x_0-\xi|<\epsilon for any \epsilon>0. Choose N=\log(\frac{\epsilon}{|x_0-\xi|})/\log(L).

Therefore, x_n converges to \xi.