12-267/Existence And Uniqueness Theorem - Revision history

Twine: Corrected error in proof of claim 2

2012-12-17T00:39:14Z

Corrected error in proof of claim 2

← Older revision		Revision as of 20:39, 16 December 2012
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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>		<math> = \|\int_{x_0}^x f(t, \Phi_{n-1}(t))dt - \int_{x_0}^x f(t, \Phi_{n-2}(t))dt\|</math>

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

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

Twine: Added structure and headings to page

2012-10-26T02:07:28Z

Added structure and headings to page

← Older revision		Revision as of 22:07, 25 October 2012
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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].

			==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>.		'''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~~. The proofs of the claims are below~~.		Using these three claims, we have shown that the solution <math>\Phi(x)</math> exists.

			==Proofs of Claims==
⚫	Proof of ~~Uniqueness:~~
		⚫	'''Proof of Claim 1''':

⚫	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>

⚫	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>

		⚫	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>

Twine: Added navigation

2012-10-24T21:54:36Z

Added navigation

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			{{12-267/Navigation}}

	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].

Twine: Typo edits

2012-10-24T20:32:19Z

Typo edits

← Older revision		Revision as of 16:32, 24 October 2012
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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>.

	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>.

	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>.

Twine: Corrected typo in definition of lipschitz

2012-10-24T19:34:21Z

Corrected typo in definition of lipschitz

← Older revision		Revision as of 15:34, 24 October 2012
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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>.		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.

Twine: Corrected dates and links

2012-10-13T00:35:55Z

Corrected dates and links

← Older revision		Revision as of 20:35, 12 October 2012
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	Disclamer: This is a student prepared note based on [http://drorbn.net/dbnvp/12-267-120928.php ~~the~~ ~~lecure~~ of Monday ~~September~~ ~~21st~~].		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>.		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>.

Twine: Inserted proof of uniqueness

2012-10-13T00:33:32Z

Inserted proof of uniqueness

← Older revision		Revision as of 20:33, 12 October 2012
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	Using these three claims, we have shown that the solution <math>\Phi(x)</math> exists. The proofs of the claims are below.		Using these three claims, we have shown that the solution <math>\Phi(x)</math> exists. The proofs of the claims are below.

			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>

	Proof of Claim 1:		Proof of Claim 1:

Twine: Added Claim 3 and finished existence proof

2012-10-12T23:58:05Z

Added Claim 3 and finished existence proof

← Older revision		Revision as of 19:58, 12 October 2012
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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>.

			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>.

	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 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.

	Proof of Claim 1:		Proof of Claim 1:
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	<math> \Box </math>		<math> \Box </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>.

	Proof of Claim 2:		Proof of Claim 2:
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	<math>\Box</math>		<math>\Box</math>

			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]

Twine: Added Claim 2 and proof.

2012-10-12T23:45:01Z

Added Claim 2 and proof.

← Older revision		Revision as of 19:45, 12 October 2012
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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.

			<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>

			<math> = \|\int_{x_0}^x f(t, \Phi_{n-1}(t))dt\|</math>

			<math> \leq \|\int_{x_0}^x \|f(t, \Phi_{n-1}(t))\|dt\|</math>

			<math> \leq \| \int_{x_0}^x M dt \| = M \|x_0 - x\|</math>

			<math> \leq M \delta</math>

			<math> \leq M \cdot \frac{b}{M}</math>

			<math> = b</math>

	<math> \Box </math>		<math> \Box </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>.

			Proof of Claim 2:

			<math> \|\Phi_n(x) - \Phi_{n-1}(x)\|</math>

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

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

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

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

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

			<math> = \frac{M k^{n-1}}{n!} \|x-x_0\|^n </math>

			<math>\Box</math>

Twine: Added Claim 1 and proof. Also fixed some formatting.

2012-10-12T23:32:55Z

Added Claim 1 and proof. Also fixed some formatting.

← Older revision		Revision as of 19:32, 12 October 2012
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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>.

			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>.

			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.

			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.

			<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>

			<math> \Box </math>

← Older revision		Revision as of 20:39, 16 December 2012
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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>		<math> = \|\int_{x_0}^x f(t, \Phi_{n-1}(t))dt - \int_{x_0}^x f(t, \Phi_{n-2}(t))dt\|</math>

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

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