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].


==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>.
'''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==
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.
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>. 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.
'''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 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>
'''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.
Using these three claims, we have shown that the solution <math>\Phi(x)</math> exists.


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

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>

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


Proof of Claim 2:
'''Proof of Claim 2''':


<math> |\Phi_n(x) - \Phi_{n-1}(x)|</math>
<math> |\Phi_n(x) - \Phi_{n-1}(x)|</math>
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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.
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]
'''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>

Revision as of 21:07, 25 October 2012

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

Lipschitz

Def. [math]\displaystyle{ f: \mathbb{R}_y \rightarrow \mathbb{R} }[/math] is called Lipschitz if [math]\displaystyle{ \exists \epsilon \gt 0, k \gt 0 }[/math] (a Lipschitz constant of f) such that [math]\displaystyle{ |y_1 - y_2| \lt \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.

Statement of Existence and Uniqueness Theorem

Thm. Existence and Uniqueness Theorem for ODEs

Let [math]\displaystyle{ 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]\displaystyle{ \Phi' = f(x, \Phi) }[/math] with [math]\displaystyle{ \Phi(x_0) = y_0 }[/math] has a unique solution [math]\displaystyle{ \Phi : [x_0 - \delta, x_0 + \delta] \rightarrow \mathbb{R} }[/math] where [math]\displaystyle{ \delta = min(a, ^b/_M) }[/math] where M is a bound of f on [math]\displaystyle{ \mathbb{R} }[/math].

Proof of Existence

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

Let [math]\displaystyle{ \Phi_0(x) = y_0 }[/math] and let [math]\displaystyle{ \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]\displaystyle{ \Phi_n }[/math] is well-defined. More precisely, [math]\displaystyle{ \Phi_n }[/math] is continuous and [math]\displaystyle{ \forall x \in [x_0 - \delta, x_0 + \delta] }[/math], [math]\displaystyle{ |\Phi_n(x) - y_0| \leq b }[/math] where b is as referred to above.

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

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

Using these three claims, we have shown that the solution [math]\displaystyle{ \Phi(x) }[/math] exists.

Proofs of Claims

Proof of Claim 1:

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

[math]\displaystyle{ |\Phi_n - y_0| }[/math]

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

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

[math]\displaystyle{ \leq | \int_{x_0}^x M dt | = M |x_0 - x| }[/math]

[math]\displaystyle{ \leq M \delta }[/math]

[math]\displaystyle{ \leq M \cdot \frac{b}{M} }[/math]

[math]\displaystyle{ = b }[/math]

[math]\displaystyle{ \Box }[/math]

Proof of Claim 2:

[math]\displaystyle{ |\Phi_n(x) - \Phi_{n-1}(x)| }[/math]

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

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

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

[math]\displaystyle{ \leq |\int_{x_0}^x k \frac{M k^{n-2}}{(n-1)!} |t-x_0|^{n-1}dt| }[/math]

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

[math]\displaystyle{ = \frac{M k^{n-1}}{n!} |x-x_0|^n }[/math]

[math]\displaystyle{ \Box }[/math]

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

Proof of Claim 3: Assigned in Homework 3, Task 1, see page for solutions.

Proof of Uniqueness

Suppose [math]\displaystyle{ \Phi }[/math] and [math]\displaystyle{ \Psi }[/math] are both solutions. Let [math]\displaystyle{ \Chi(x) = |\Phi(x) - \Psi(x)| }[/math].

[math]\displaystyle{ \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]\displaystyle{ \Chi \leq k \int_{x_0}^x \Chi(x) dx }[/math] for some constant k, which means [math]\displaystyle{ \Chi' \leq k\Chi }[/math], and that [math]\displaystyle{ \Chi(x) \geq 0 }[/math].

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

[math]\displaystyle{ 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]\displaystyle{ U(x_0) = 0 \and U'(x) = 0 \implies U(x) \leq 0 }[/math], and [math]\displaystyle{ 0 \leq U(x) \leq 0 \implies U(x) \equiv 0 \implies \Chi(x) \equiv 0 \implies \Phi(x) \equiv \Psi(x) }[/math].

[math]\displaystyle{ \Box }[/math]