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>. | + | ==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 | + | ==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 | + | Using these three claims, we have shown that the solution <math>\Phi(x)</math> exists. |
− | + | ==Proofs of Claims== | |
− | + | '''Proof of Claim 1''': | |
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− | 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> | ||
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+ | 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 | ||
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+ | <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> |
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.
Contents |
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 .