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

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(Added Claim 2 and proof.)
(Added Claim 3 and finished existence proof)
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Let <math>f:\mathbb{R} = [x_0 - a, x_0 + a] \times [y_0 - b, y_0 + b] \rightarrow \mathbb{R}</math> be continuous and uniformly Lipschitz relative to y. Then the equation <math>\Phi' = f(x, \Phi)</math> with <math> \Phi(x_0) = y_0</math> has a unique solution <math>\Phi : [x_0 - \delta, x_0 + \delta] \rightarrow \mathbb{R}</math> where <math>\delta = min(a, ^b/_M)</math> where M is a bound of f on <math>\mathbb{R}</math>.
Let <math>f:\mathbb{R} = [x_0 - a, x_0 + a] \times [y_0 - b, y_0 + b] \rightarrow \mathbb{R}</math> be continuous and uniformly Lipschitz relative to y. Then the equation <math>\Phi' = f(x, \Phi)</math> with <math> \Phi(x_0) = y_0</math> has a unique solution <math>\Phi : [x_0 - \delta, x_0 + \delta] \rightarrow \mathbb{R}</math> where <math>\delta = min(a, ^b/_M)</math> where M is a bound of f on <math>\mathbb{R}</math>.

This is proven by showing the equation <math>\Phi(x) = y_0 | \int_{x_0}^x f(t, \Phi(t))dt</math> exists, given the noted assumptions.


Let <math>\Phi_0(x) = y_0</math> and let <math>\Phi_n(x) = y_0 + \int_{x_0}^x f(t, \Phi_{n-1}(t))dt</math>.
Let <math>\Phi_0(x) = y_0</math> and let <math>\Phi_n(x) = y_0 + \int_{x_0}^x f(t, \Phi_{n-1}(t))dt</math>.


Claim 1: <math>\Phi_n</math> is well-defined. More precisely, <math>\Phi_n</math> is continuous and <math>\forall x \in [x_0 - \delta, x_0 | \delta]</math>, <math>|\Phi_n(x) - y_0| \leq b</math> where b is as referred to above.
Claim 1: <math>\Phi_n</math> is well-defined. More precisely, <math>\Phi_n</math> is continuous and <math>\forall x \in [x_0 - \delta, x_0 | \delta]</math>, <math>|\Phi_n(x) - y_0| \leq b</math> where b is as referred to above.

Claim 2: For <math>n \geq 1</math>, <math>|\Phi_n(x) - \Phi_{n-1}(x)| \leq \frac{Mk^{n-1}}{n!} |x-x_0|^n</math>.

Claim 3: if <math> \Phi_n(x)</math> is a series of functions such that <math>|\Phi_n(x) - \Phi_{n-1}(x)| < c_n</math>, with <math>\sum_{n=1}^{\infty} c_n</math> equal to some finite number, then <math>\Phi_n</math> converges uniformly to some function <math>\Phi</math>

Using these three claims, we have shown that the solution <math>\Phi(x)</math> exists. The proofs of the claims are below.


Proof of Claim 1:
Proof of Claim 1:
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<math> \Box </math>
<math> \Box </math>

Claim 2: For <math>n \geq 1</math>, <math>|\Phi_n(x) - \Phi_{n-1}(x)| \leq \frac{Mk^{n-1}}{n!} |x-x_0|^n</math>.


Proof of Claim 2:
Proof of Claim 2:
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<math>\Box</math>
<math>\Box</math>

Note that the sequence <math> c_n = \frac{M k^{n-1}}{n!} |x-x_0|^n</math> has <math>\sum_{n=1}^{\infty} c_n</math> equal to some finite number.

Proof of Claim 3: Assigned in [http://drorbn.net/index.php?title=12-267/Homework_Assignment_3 Homework 3, Task 1]

Revision as of 18:58, 12 October 2012

Disclamer: This is a student prepared note based on the lecure of Monday September 21st.

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 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