12-267/Existence And Uniqueness Theorem: Difference between revisions
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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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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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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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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. |
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<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> |
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<math> \Box </math> |
Revision as of 18:32, 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 .
Let and let .
Claim 1: is well-defined. More precisely, is continuous and , where b is as referred to above.
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.