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]. |
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==Lipschitz== |
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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>. |
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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. |
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==Statement of Existence and Uniqueness Theorem== |
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'''Thm.''' Existence and Uniqueness Theorem for ODEs |
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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>. |
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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. |
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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>. |
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. |
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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 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 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>. |
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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> |
'''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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Using these three claims, we have shown that the solution <math>\Phi(x)</math> exists |
Using these three claims, we have shown that the solution <math>\Phi(x)</math> exists. |
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==Proofs of Claims== |
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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> |
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Proof of Claim 2: |
'''Proof of Claim 2''': |
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<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. |
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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] |
'''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. |
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==Proof of Uniqueness== |
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Revision as of 22:07, 25 October 2012
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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 (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 .