Difference between revisions of "12-267/Existence And Uniqueness Theorem"

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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].
  
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>.
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==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
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==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>.
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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.
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'''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>.
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'''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>
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'''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.
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Using these three claims, we have shown that the solution <math>\Phi(x)</math> exists.
  
Proof of Uniqueness:
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==Proofs of Claims==
 
+
'''Proof of Claim 1''':
Suppose <math>\Phi</math> and <math>\Psi</math> are both solutions. Let <math>\Chi(x) = |\Phi(x) - \Psi(x)|</math>.
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<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>.
+
 
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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>
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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>.
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<math>\Box</math>
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Proof of Claim 1:
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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:
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'''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]
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'''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 22:07, 25 October 2012

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

Contents

Lipschitz

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

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

Proof of Existence

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

Let \Phi_0(x) = y_0 and let \Phi_n(x) = y_0 + \int_{x_0}^x f(t, \Phi_{n-1}(t))dt. IF we can prove the following three claims, we have proven the theorem. The proofs of these claims will follow below.

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

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

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

Using these three claims, we have shown that the solution \Phi(x) exists.

Proofs of Claims

Proof of Claim 1:

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

|\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

 \Box

Proof of Claim 2:

 |\Phi_n(x) - \Phi_{n-1}(x)|

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

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

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

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

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

 = \frac{M k^{n-1}}{n!} |x-x_0|^n

\Box

Note that the sequence  c_n = \frac{M k^{n-1}}{n!} |x-x_0|^n has \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 \Phi and \Psi are both solutions. Let \Chi(x) = |\Phi(x) - \Psi(x)|.

\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

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

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

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

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

\Box