12-267/Existence And Uniqueness Theorem

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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 Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \exists \epsilon > 0, k > 0} (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 Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \Phi(x_0) = y_0} has a unique solution where where M is a bound of f on .

Proof of Existence

This is proven by showing the equation Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \Phi(x) = y_0 + \int_{x_0}^x f(t, \Phi(t))dt} 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: Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \Phi_n} is well-defined. More precisely, is continuous and , where b is as referred to above.

Claim 2: For , .

Claim 3: if Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \Phi_n(x)} is a series of functions such that , with equal to some finite number, then Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \Phi_n} 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 Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \Phi_{n-1}} . Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \Phi_n} is continuous, being the integral of a continuous function.

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \leq | \int_{x_0}^x M dt | = M |x_0 - x|}

Proof of Claim 2:

Note that the sequence has Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \sum_{n=1}^{\infty} c_n} 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 .

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \Box}