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

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<math> = |\int_{x_0}^x f(t, \Phi_{n-1}(t))dt - \int_{x_0}^x f(t, \Phi_{n-2}(t))dt|</math>
<math> = |\int_{x_0}^x f(t, \Phi_{n-1}(t))dt - \int_{x_0}^x f(t, \Phi_{n-2}(t))dt|</math>


<math> \leq | \int_{x_0}^x (f(t, \Phi_{n-1}(t) - f(t, \Phi_{n-2}(t))dt )dt |</math>
<math> \leq | \int_{x_0}^x |f(t, \Phi_{n-1}(t)) - f(t, \Phi_{n-2}(t)) | dt |</math>


<math> \leq |\int_{x_0}^x k|\Phi_{n-1}(t) - \Phi_{n-2}(t)|dt|</math>
<math> \leq |\int_{x_0}^x k|\Phi_{n-1}(t) - \Phi_{n-2}(t)|dt|</math>

Latest revision as of 20:39, 16 December 2012

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

Lipschitz

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.

Statement of Existence and Uniqueness Theorem

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 .

Proof of Existence

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

Let and let . IF we can prove the following three claims, we have proven the theorem. The proofs of these claims will follow below.

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.

Proofs of Claims

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, see page for solutions.

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 .