12-267/Existence And Uniqueness Theorem: Difference between revisions

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Disclamer: This is a student prepared note based on the lecures of [http://drorbn.net/dbnvp/12-267-120928.php Friday, September 28th] and [http://drorbn.net/dbnvp/12-267-121001.php Monday October 1st].
Disclamer: This is a student prepared note based on the lecures of [http://drorbn.net/dbnvp/12-267-120928.php Friday, September 28th] and [http://drorbn.net/dbnvp/12-267-121001.php Monday October 1st].



Revision as of 17:54, 24 October 2012

Disclamer: This is a student prepared note based on the lecures of Friday, September 28th and Monday October 1st.

Def. is called Lipschitz if (a Lipschitz constant of f) such that .

Note that any function that is Lipschitz is uniformly continuous, and that if a function f and its derivative are both continous on a compact set then f is Lipschitz.

Thm. Existence and Uniqueness Theorem for ODEs

Let be continuous and uniformly Lipschitz relative to y. Then the equation with has a unique solution where where M is a bound of f on .

This is proven by showing the equation exists, given the noted assumptions.

Let and let .

Claim 1: is well-defined. More precisely, is continuous and , where b is as referred to above.

Claim 2: For , .

Claim 3: if is a series of functions such that , with equal to some finite number, then converges uniformly to some function

Using these three claims, we have shown that the solution exists. The proofs of the claims are below.

Proof of Uniqueness:

Suppose and are both solutions. Let .

We have that for some constant k, which means , and that .

Let . Note that as in this case we are integrating over an empty set, and that U thus defined has . Then

Then , and .

Proof of Claim 1:

The statement is trivially true for . Assume the claim is true for . is continuous, being the integral of a continuous function.

Proof of Claim 2:

Note that the sequence has equal to some finite number.

Proof of Claim 3: Assigned in Homework 3, Task 1