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

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Disclamer: This is a student prepared note based on [http://drorbn.net/dbnvp/12-267-120928.php the lecure of Monday September 21st].


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].
Def. <math>f: \mathbb{R}_y \rightarrow \mathbb{R}</math> is called Lipschitz if <math>\exists \epsilon > 0, k > 0</math> (a Lipschitz constant of f) such that <math>|y_1 - y_2| < \epsilon \implies |f(y_1) - f(y_2)| \leq k |y_1 = y_2|</math>.

==Lipschitz==
'''Def.''' <math>f: \mathbb{R}_y \rightarrow \mathbb{R}</math> is called Lipschitz if <math>\exists \epsilon > 0, k > 0</math> (a Lipschitz constant of f) such that <math>|y_1 - y_2| < \epsilon \implies |f(y_1) - f(y_2)| \leq k |y_1 - y_2|</math>.


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.
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
==Statement of Existence and Uniqueness Theorem==
'''Thm.''' Existence and Uniqueness Theorem for ODEs


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>.


==Proof of Existence==
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>.
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.


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


'''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.
Proof of Claim 1:

'''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.

==Proofs of Claims==
'''Proof of Claim 1''':


The statement is trivially true for <math>\Phi_0</math>. Assume the claim is true for <math>\Phi_{n-1}</math>. <math>\Phi_n</math> is continuous, being the integral of a continuous function.
The statement is trivially true for <math>\Phi_0</math>. Assume the claim is true for <math>\Phi_{n-1}</math>. <math>\Phi_n</math> is continuous, being the integral of a continuous function.


<math>|\Phi_n - y_0|</math>
<math>|\Phi_n - y_0| = |\int_{x_0}^x f(t, \Phi_{n-1}(t))dt| \leq |\int_{x_0}^x |f(t, \Phi_{n-1}(t))|dt| \leq | \int_{x_0}^x M dt | = M |x_0 - x| \leq M \delta \leq M \cdot \frac{b}{M} = b.</math>

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

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

<math> \leq | \int_{x_0}^x M dt | = M |x_0 - x|</math>

<math> \leq M \delta</math>

<math> \leq M \cdot \frac{b}{M}</math>

<math> = b</math>


<math> \Box </math>
<math> \Box </math>

'''Proof of Claim 2''':

<math> |\Phi_n(x) - \Phi_{n-1}(x)|</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 |</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 \frac{M k^{n-2}}{(n-1)!} |t-x_0|^{n-1}dt|</math>

<math> = \frac{M k^{n-1}}{(n-1)!} \int_0^{|x-x_0|} t^{n-1} dt</math>

<math> = \frac{M k^{n-1}}{n!} |x-x_0|^n </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], see page for solutions.

==Proof of Uniqueness==
Suppose <math>\Phi</math> and <math>\Psi</math> are both solutions. Let <math>\Chi(x) = |\Phi(x) - \Psi(x)|</math>.

<math>\Chi(x) = |\Phi(x) - \Psi(x)| = |\int_{x_0}^x(f(x, \Phi(x)) - f(x, \Psi(x))) dx | \leq \int_{x_0}^x k|\Phi(x) - \Psi(x)| dx</math>

We have that <math>\Chi \leq k \int_{x_0}^x \Chi(x) dx</math> for some constant k, which means <math>\Chi' \leq k\Chi</math>, and that <math>\Chi(x) \geq 0</math>.

Let <math>U(x) = e^{-kx}\int_{x_0}^x \Chi(x) dx</math>. Note that <math>U(x_0) = 0</math> as in this case we are integrating over an empty set, and that U thus defined has <math>U(x) \geq 0</math>. Then

<math>U'(x) = -ke^{-kx}\int_{x_0}^x\Chi(x) dx + e^{-kx} \Chi(x) = e^{-kx}(\Chi(x) - k\int_{x_0}^x\Chi(x) dx) \leq 0</math>

Then <math>U(x_0) = 0 \and U'(x) = 0 \implies U(x) \leq 0</math>, and <math> 0 \leq U(x) \leq 0 \implies U(x) \equiv 0 \implies \Chi(x) \equiv 0 \implies \Phi(x) \equiv \Psi(x)</math>.

<math>\Box</math>

Latest revision as of 19:39, 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 .