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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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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|</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> |
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<math> = |\int_{x_0}^x f(t, \Phi_{n-1}(t))dt|</math> |
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<math> \leq |\int_{x_0}^x |f(t, \Phi_{n-1}(t))|dt|</math> |
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<math> \leq | \int_{x_0}^x M dt | = M |x_0 - x|</math> |
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<math> \leq M \delta</math> |
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<math> \leq M \cdot \frac{b}{M}</math> |
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<math> = b</math> |
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<math> \Box </math> |
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<math> \Box </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>. |
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Proof of Claim 2: |
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<math> |\Phi_n(x) - \Phi_{n-1}(x)|</math> |
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⚫ |
<math> = |\int_{x_0}^x f(t, \Phi_{n-1}(t))dt - \int_{x_0}^x f(t, \Phi_{n- 2}(t))dt|</math> |
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<math> \leq | \int_{x_0}^x (f(t, \Phi_{n-1}(t) - f(t, \Phi_{n-2}(t))dt )dt |</math> |
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<math> \leq |\int_{x_0}^x k|\Phi_{n-1}(t) - \Phi_{n-2}(t)|dt|</math> |
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<math> \leq |\int_{x_0}^x k \frac{M k^{n-2}}{(n-1)!} |t-x_0|^{n-1}dt|</math> |
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<math> = \frac{M k^{n-1}}{(n-1)!} \int_0^{|x-x_0|} t^{n-1} dt</math> |
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<math> = \frac{M k^{n-1}}{n!} |x-x_0|^n </math> |
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<math>\Box</math> |
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.
Claim 2: For , .
Proof of Claim 2: