Drorbn - User contributions [en]

12-267/Existence And Uniqueness Theorem

2012-12-17T00:39:14Z

Twine: Corrected error in proof of claim 2

{{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].

==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.

==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>.

==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.

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 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.

==Proofs of Claims==
'''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|</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>

'''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 |</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>

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], 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>

12-267/Topic List

2012-12-17T00:04:55Z

Twine: Added links to two handouts

Here is a (incomplete) list of subjects covered in MAT267 this semester.

==First-order linear equations==

[http://imgur.com/a/OSx1U#0 Solution techniques]

The brachistochrone

Separable equations

Escape velocities

Changing source and target coordinates

Homogeneous equations

Reverse-engineering separable and exact equations

Solving exact equations with and without integration factors

[http://drorbn.net/index.php?title=12-267/Existence_And_Uniqueness_Theorem Fundamental Theorem and proof]

==Calculus of Variations==

Chain law

Euler-Lagrange, [http://drorbn.net/index.php?title=12-267/Derivation_of_Euler-Lagrange derivation] ([http://graphics.ethz.ch/teaching/former/vc_master_06/Downloads/viscomp-varcalc_6.pdf alternate]) and reductions

Lagrange multipliers

Isoperimetric Inequality

==Numerical methods==

[http://drorbn.net/index.php?title=Numerical_Methods Examples] and derivations

Evaluating local error

==Higher Order Constant Coefficient Homogeneous Linear ODEs==

Multiple roots

Reduction of order

Method of Undetermined Coefficients

==Systems of First-Order Linear Equations==

[http://i.imgur.com/uTugV.jpg Solution techniques]

Matrix exponetiation

Phase portraits

Non-homogeneous systems

==Power-series solutions==

[http://imgur.com/a/sZSYx#0 Quick guide]

Global existence for linear ODEs

Wronskian

Series solutions for y' = f(x,y)

Radius of convergence

==Qualitative Analysis==

[http://drorbn.net/AcademicPensieve/Classes/12-267/QualitativeAnalysis/QualitativeAnalysis.pdf Handout]

Airy's equation

Fuch's Theorem

Regular singular points

[http://drorbn.net/images/0/03/12-267-Frobenius.pdf Frobenius series and Frobenius Method]

The basic oscillation theorem

Non-oscillation theorem

Sturm comparison theorem

Changing the independent variable

Amplitudes of oscillations

12-267/Topic List

2012-12-11T02:03:12Z

Twine: Added some links, improved formatting

Here is a (incomplete) list of subjects covered in MAT267 this semester.

==First-order linear equations==

[http://imgur.com/a/OSx1U#0 Solution techniques]

The brachistochrone

Separable equations

Escape velocities

Changing source and target coordinates

Homogeneous equations

Reverse-engineering separable and exact equations

Solving exact equations with and without integration factors

[http://drorbn.net/index.php?title=12-267/Existence_And_Uniqueness_Theorem Fundamental Theorem and proof]

==Calculus of Variations==

Chain law

Euler-Lagrange, [http://drorbn.net/index.php?title=12-267/Derivation_of_Euler-Lagrange derivation] ([http://graphics.ethz.ch/teaching/former/vc_master_06/Downloads/viscomp-varcalc_6.pdf alternate]) and reductions

Lagrange multipliers

Isoperimetric Inequality

==Numerical methods==

[http://drorbn.net/index.php?title=Numerical_Methods Examples] and derivations

Evaluating local error

==Higher Order Constant Coefficient Homogeneous Linear ODEs==

Multiple roots

Reduction of order

Method of Undetermined Coefficients

==Systems of First-Order Linear Equations==

[http://i.imgur.com/uTugV.jpg Solution techniques]

Matrix exponetiation

Phase portraits

Non-homogeneous systems

==Power-series solutions==

[http://imgur.com/a/sZSYx#0 Quick guide]

Global existence for linear ODEs

Wronskian

Series solutions for y' = f(x,y)

Radius of convergence

==Qualitative Analysis==

Airy's equation

Fuch's Theorem

Regular singular points

Frobenius series and Frobenius Method

The basic oscillation theorem

Non-oscillation theorem

Sturm comparison theorem

Changing the independent variable

Amplitudes of oscillations

12-267/Topic List

2012-12-11T01:48:03Z

Twine:

Here is a (incomplete) list of subjects covered in MAT267 this semester.

The brachistochrone

First-order linear equations

Separable equations

Escape velocities

Changing source and target coordinates

Homogeneous equations

Reverse-engineering separable and exact equations

Solving exact equations with and without integration factors

[http://drorbn.net/index.php?title=12-267/Existence_And_Uniqueness_Theorem Fundamental Theorem and proof]

Chain law

Variational problems

Euler-Lagrange, derivation and reductions

Lagrange multipliers

Isoperimetric Inequality

[http://drorbn.net/index.php?title=Numerical_Methods Numerical methods]

Evaluating local error

Higher order constant coefficient homogeneous linear ODEs

Multiple roots

Reduction of order

Method of Undetermined Coefficients

Systems of equations

Matrix exponetiation

Phase portraits

Non-homogeneous systems

Power-series solutions

Global existence for linear ODEs

Wronskian

Series solutions for y' = f(x,y)

Radius of convergence

Airy's equation

Fuch's Theorem

Regular singular points

Frobenius series and Frobenius Method

The basic oscillation theorem

Non-oscillation theorem

Sturm comparison theorem

Changing the independent variable

Amplitudes of oscillations

12-267

2012-12-11T01:46:10Z

Twine:

__NOEDITSECTION__
__NOTOC__
{{12-267/Navigation}}
==Advanced Ordinary Differential Equations==
===Department of Mathematics, University of Toronto, Fall 2012===

{{12-267/Crucial Information}}

===Text===
Boyce and DiPrima, [http://ca.wiley.com/WileyCDA/WileyTitle/productCd-EHEP002451.html Elementary Differential Equations and Boundary Value Problems] (current edition is 9th and 10th will be coming out shortly. Hopefully any late enough edition will do).

===Further Resources===

* [http://www.math.toronto.edu/undergrad/ Undergraduate Information] at the [http://www.math.toronto.edu/ UofT Math Department]

* [http://www.artsandscience.utoronto.ca/ofr/calendar/crs_mat.htm Undergraduate Course Descriptions].

* Vitali Kapovitch's 2007 classes: [http://www.math.toronto.edu/vtk/267/ Spring], [http://www.math.toronto.edu/vtk/267fall07/ Fall].

* Also previously taught by T. Bloom, C. Pugh, D. Remenik.

* My {{Pensieve Link|Classes/12-267/|12-267 notebook}}.

{{Template:12-267:Dror/Students Divider}}

[[User:Drorbn|Drorbn]] 06:36, 12 September 2012 (EDT): Material by [[User:Syjytg|Syjytg]] moved to [[12-267/Tuesday September 11 Notes]].

[http://imgur.com/a/OSx1U#0 Summary of techniques to solve differential equations ] [[User:Vsbdthrsh|Vsbdthrsh]]

[http://drorbn.net/index.php?title=12-267/Existence_And_Uniqueness_Theorem Fundamental Theorem and Proof from Lecture] [[User:Twine|Twine]]

[http://drorbn.net/index.php?title=12-267/Derivation_of_Euler-Lagrange Derivation of Euler-Lagrange from Lecture] [[User:Twine|Twine]]

[http://graphics.ethz.ch/teaching/former/vc_master_06/Downloads/viscomp-varcalc_6.pdf Useful PDF: proof of Euler-Lagrange equation, explanation, examples ] [[User:Vsbdthrsh|Vsbdthrsh]]

12-267 [http://math.hunter.cuny.edu/mbenders/cofv.pdf In-depth coverage of Calculus of Variations][[User:Simon1|Simon1]]

12-267 [http://highered.mcgraw-hill.com/sites/dl/free/007063419x/392340/Calculus_of_Variations.pdf A good summary of Calculus of Variations]
[[User:Simon1|Simon1]]

12-267 [http://imgur.com/a/1vy11#0 All class notes from September 10th to October 5th] [[User:Simon1|Simon1]]

12-267 [http://imgur.com/a/uLSlM Summary of Numerical Methods] [[User:Simon1|Simon1]]

[http://drorbn.net/index.php?title=Numerical_Methods Numerical Methods (wiki)] [[User:Twine|Twine]]

12-267 [http://imgur.com/a/tliYg Summary of Chapter 3 from the Textbook on Constant Coefficient Second Order ODEs] [[User:Simon1|Simon1]]

[http://drorbn.net/images/e/e6/Geometric_Interpretation_of_Lagrange_Multiplier.pdf Geometric Interpretation of Lagrange Multiplier] [[User:Mathstudent|Mathstudent]]

Past exams from [http://exams.library.utoronto.ca.myaccess.library.utoronto.ca/handle/exams/4429 December 2009] and [http://exams.library.utoronto.ca.myaccess.library.utoronto.ca/handle/exams/4429 December 2010] [[User:Twine|Twine]]

Handwritten notes by [[User:Ktnd3|Ktnd3]]:

* September: [[Media:Mat267_-_lecture_1(sep.10).PDF|10th]], [[Media:Mat267_-_lecture_2%28sep.11%29.PDF|11th]], [[Media:Mat267_-_lecture_3%28sep.14%29.PDF|14th]], [[Media:Mat267_-_lecture_4%28sep.17%29.PDF|17th]], [[Media:12-267%28lecture5%29.PDF|18th]], [[Media:12-267%28lecture6%29.PDF|21st]], [[Media:12-267%28lecture7%29.PDF|24th]], [[Media:12-267%28lecture8%29.PDF|25th]], [[Media:12-267%28lecture9%29.PDF|28th]]

* October: [[Media:12-267%28lecture10%29.PDF|1st]], [[Media:12-267%28lecture11%29.PDF|2nd]], [[Media:12-267%28lecture12%29.PDF|5th]], [[Media:12-267%28lecture13%29.PDF|9th]] [[Media:12-267%28lecture14%29.PDF|12th]]

[http://i.imgur.com/uTugV.jpg Quick guide: system of 1st order linear equations] [[User:Vsbdthrsh|Vsbdthrsh]]

Help of Inverse Matrix [http://mathworld.wolfram.com/MatrixInverse.html Matrix Inverse] [[User:Dongwoo.kang|Dongwoo.kang]]

12-267 [http://imgur.com/a/50sRR All class notes from October 5th to October 30th] [[User:Simon1|Simon1]]

[http://imgur.com/a/sZSYx#0 Quick guide: Power Series + ODE] [[User:Vsbdthrsh|Vsbdthrsh]]

[https://lh4.googleusercontent.com/-V1WnvvzXRlE/TxTGwl_BBjI/AAAAAAAAE9k/yVuy1zRmnWs/Gatos-cute-cute-32-800x500.jpg Mirrors do actually flip top-to-bottom, depending on how you look at them] [[User:jonathanrlove|jonathanrlove]]

[http://www.guardian.co.uk/notesandqueries/query/0,5753,-19877,00.html Actually, mirrors don't flip top-to-bottom *or* left-to-right] [[User:Twine|Twine]]

[http://drorbn.net/index.php?title=12-267/Topic_List Topics covered this semester] [[User:Twine|Twine]]

12-267

2012-12-11T01:45:53Z

Twine:

12-267/Topic List

2012-12-11T01:45:25Z

Twine: Created first version of page

Here is a (incomplete) list of subjects covered in MAT267 this semester.

The brachistochrone

First-order linear equations

Separable equations

Escape velocities

Changing source and target coordinates

Homogeneous equations

Reverse-engineering separable and exact equations

Solving exact equations with and without integration factors

Fundamental Theorem and proof

Chain law

Variational problems

Euler-Lagrange, derivation and reductions

Lagrange multipliers

Isoperimetric Inequality

Numerical methods

Evaluating local error

Higher order constant coefficient homogeneous linear ODEs

Multiple roots

Reduction of order

Method of Undetermined Coefficients

Systems of equations

Matrix exponetiation

Phase portraits

Non-homogeneous systems

Power-series solutions

Global existence for linear ODEs

Wronskian

Series solutions for y' = f(x,y)

Radius of convergence

Airy's equation

Fuch's Theorem

Regular singular points

Frobenius series and Frobenius Method

The basic oscillation theorem

Non-oscillation theorem

Sturm comparison theorem

Changing the independent variable

Amplitudes of oscillations

12-267

2012-12-11T01:33:12Z

Twine:

12-267/Existence And Uniqueness Theorem

2012-10-26T02:07:28Z

Twine: Added structure and headings to page

{{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].

==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.

==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>.

==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.

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 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.

==Proofs of Claims==
'''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|</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>

'''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>

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], 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>

12-267/Numerical Methods

2012-10-26T01:55:32Z

Twine: Added derivation of local error of Improved Euler method

==Summary of Numerical Methods==

Based largely off of a note available [http://imgur.com/a/uLSlM here] posted by [[User:Simon1|Simon1]] --[[User:Twine|Twine]] 20:55, 25 October 2012 (EDT)

Numerical methods: <math>\frac{dy}{dt} = f(t, y)</math> and <math>y(t_0) = y_0</math>, <math>y = \Phi(t)</math> is a solution.

'''1. Using the proof of Picard's Theorem:'''

<math>\Phi_0(x) = y_0</math>

<math>\Phi_n(x) = y_0 + \int_{x_0}^x f(x, \Phi_{n-1}(x)) dx</math>

<math>\Phi_n(x) \rightarrow \Phi(x)</math>

'''2. The Euler Method:'''

<math>y_{n+1} = y_n + f(t_n, y_n)(t_{n+1} - t_n) = y_n + f_n h</math> if h is constant

Backward Euler formula: <math>y_{n+1} = y_n + h f(t_{n+1}, y_{n+1})</math>

Local truncation error: <math>e_{n+1} = \frac{1}{2} \Phi''(t_n)h^2 \leq \frac{Mh^2}{2}</math> where <math>m = \mathrm{max} |\Phi''(t)|</math>

Local error is proportional to <math>h^2</math>.

Global error is proportional to h.

'''3. Improved Euler Formula (or Heun Formula):'''

<math>y_{n+1} = y_n + \frac{f_n + f(t_n + h, y_n + hf_n)}{2} h</math>

To determine local error we took the taylor expansions of <math>y</math> and <math>\Phi</math> and compared them.

<math>\Phi(x_1) = \Phi(x_0 + h) = \Phi(x_0) + h \Phi'(x_0) + \frac{h^2}{2} \Phi''(x_0) + O(h^3)</math>

<math>\Phi'(x) = f(x, \Phi(x)) \quad \Phi''(x) = \Phi'(x) = f_x(x, \Phi(x)) + f_y(x, \Phi(x))\Phi'(x)</math>

<math>\Phi(x_0 + h) = y_0 + h f(x_0, y_0) + \frac{h^2}{2}(f_x + f_y f) + O(h^3)</math>

We compare this to the computed value <math>y_1</math>

<math>y_1 = y_0 +\frac{h}{2}(f(x_0, y_0) + f(x_0 + h, y_0 + hf(x_0, y_0)))</math>

Taking the taylor expansion we get

<math>y_1 = y_0 + h f(x_0, y_0) + \frac{h^2}{2}(f_x + f_y f) + O(h^3)</math>

So we have shown that the Improved Euler Formula is accurate up to an error term proportional to <math>h^3</math>

Local truncation error is proportional to <math>h^3</math>

Global truncation error is proportional to <math>h^2</math>

'''4. The Runge-Kutta Method:'''

<math>y_{n+1} = y_n + \frac{k_{n1} + 2k_{n2} + 2k_{n3} + k_{n4}}{6} h</math>

where

<math>k_{n1} = f(t_n, y_n) \quad k_{n2} = f(t_n + \frac{1}{2} h, y_n + \frac{1}{2}hk_{n1})</math>

<math>k_{n3} = f(t_n + \frac{1}{2} h, y_n + \frac{1}{2}hk_{n2}) \quad k_{n4} = f(t_n + h, y_n + hk_{n3})</math>

Local truncation error is proportional to <math>h^5</math>.

Global truncation error is proportional to <math>h^4</math>.

==Python Example of Euler's Method==
In class on October 15th we discussed Euler's Method to numerically compute a solution to a differential equation. <math>x_0</math> and <math>y_0</math> are given as well as an increment amount <math>h</math>, <math>x_{n+1} = x_n + h</math>, and we use the guess <math>y_{n+1} = y_n + f(x_n, y_n)*h</math> where f computes the derivative as a function of x and y.

Here is an example of code (written in Python) which carries out Euler's Method for the example we discussed in class, <math>y' = -y</math>:

def f(x, y):
return -y

def euler(x, y, f, h, x_max):
"""Take in coordinates x and y, a function f(x, y) which calculates
dy/dx at (x, y), an increment h, and a maximum value of x.

Return a list containing coordinates in the Euler's Method computation
of the solution to Phi' = f(x, Phi(x)), Phi(x) = y, with the x
values of those coordinates separated by h, and not exceeding x_max.
"""
if x > x_max: # we have already calculated all our values
return []
x_next, y_next = (x + h, y + f(x, y)*h) # calculate the next x, y values
# return the current coordinates, and every coordinates following it, in a list
return [(x_next, y_next)] + euler(x_next, y_next, f, h, x_max)

if __name__ == '__main__':
print euler(0, 1, f, 0.01, 1)[-1]

12-267/Numerical Methods

2012-10-26T01:00:44Z

Twine:

==Summary of Numerical Methods==

Based largely off of a note available [http://imgur.com/a/uLSlM here] posted by [[User:Simon1|Simon1]] --[[User:Twine|Twine]] 20:55, 25 October 2012 (EDT)

Numerical methods: <math>\frac{dy}{dt} = f(t, y)</math> and <math>y(t_0) = y_0</math>, <math>y = \Phi(t)</math> is a solution.

'''1. Using the proof of Picard's Theorem:'''

<math>\Phi_0(x) = y_0</math>

<math>\Phi_n(x) = y_0 + \int_{x_0}^x f(x, \Phi_{n-1}(x)) dx</math>

<math>\Phi_n(x) \rightarrow \Phi(x)</math>

'''2. The Euler Method:'''

<math>y_{n+1} = y_n + f(t_n, y_n)(t_{n+1} - t_n) = y_n + f_n h</math> if h is constant

Backward Euler formula: <math>y_{n+1} = y_n + h f(t_{n+1}, y_{n+1})</math>

Local truncation error: <math>e_{n+1} = \frac{1}{2} \Phi''(t_n)h^2 \leq \frac{Mh^2}{2}</math> where <math>m = \mathrm{max} |\Phi''(t)|</math>

Local error is proportional to <math>h^2</math>.

Global error is proportional to h.

'''3. Improved Euler Formula (or Heun Formula):'''

<math>y_{n+1} = y_n + \frac{f_n + f(t_n + h, y_n + hf_n)}{2} h</math>

Local truncation error is proportional to <math>h^3</math>

Global truncation error is proportional to <math>h^2</math>

'''4. The Runge-Kutta Method:'''

<math>y_{n+1} = y_n + \frac{k_{n1} + 2k_{n2} + 2k_{n3} + k_{n4}}{6} h</math>

where

<math>k_{n1} = f(t_n, y_n) \quad k_{n2} = f(t_n + \frac{1}{2} h, y_n + \frac{1}{2}hk_{n1})</math>

<math>k_{n3} = f(t_n + \frac{1}{2} h, y_n + \frac{1}{2}hk_{n2}) \quad k_{n4} = f(t_n + h, y_n + hk_{n3})</math>

Local truncation error is proportional to <math>h^5</math>.

Global truncation error is proportional to <math>h^4</math>.

==Python Example of Euler's Method==
In class on October 15th we discussed Euler's Method to numerically compute a solution to a differential equation. <math>x_0</math> and <math>y_0</math> are given as well as an increment amount <math>h</math>, <math>x_{n+1} = x_n + h</math>, and we use the guess <math>y_{n+1} = y_n + f(x_n, y_n)*h</math> where f computes the derivative as a function of x and y.

Here is an example of code (written in Python) which carries out Euler's Method for the example we discussed in class, <math>y' = -y</math>:

def f(x, y):
return -y

def euler(x, y, f, h, x_max):
"""Take in coordinates x and y, a function f(x, y) which calculates
dy/dx at (x, y), an increment h, and a maximum value of x.

Return a list containing coordinates in the Euler's Method computation
of the solution to Phi' = f(x, Phi(x)), Phi(x) = y, with the x
values of those coordinates separated by h, and not exceeding x_max.
"""
if x > x_max: # we have already calculated all our values
return []
x_next, y_next = (x + h, y + f(x, y)*h) # calculate the next x, y values
# return the current coordinates, and every coordinates following it, in a list
return [(x_next, y_next)] + euler(x_next, y_next, f, h, x_max)

if __name__ == '__main__':
print euler(0, 1, f, 0.01, 1)[-1]

12-267/Numerical Methods

2012-10-26T00:59:41Z

Twine: Added python example of Euler's method, headings

==Summary of Numerical Methods==

Based largely off of a note available [http://imgur.com/a/uLSlM here] [[User:Simon1|Simon1]] --[[User:Twine|Twine]] 20:55, 25 October 2012 (EDT)

Numerical methods: <math>\frac{dy}{dt} = f(t, y)</math> and <math>y(t_0) = y_0</math>, <math>y = \Phi(t)</math> is a solution.

'''1. Using the proof of Picard's Theorem:'''

<math>\Phi_0(x) = y_0</math>

<math>\Phi_n(x) = y_0 + \int_{x_0}^x f(x, \Phi_{n-1}(x)) dx</math>

<math>\Phi_n(x) \rightarrow \Phi(x)</math>

'''2. The Euler Method:'''

<math>y_{n+1} = y_n + f(t_n, y_n)(t_{n+1} - t_n) = y_n + f_n h</math> if h is constant

Backward Euler formula: <math>y_{n+1} = y_n + h f(t_{n+1}, y_{n+1})</math>

Local truncation error: <math>e_{n+1} = \frac{1}{2} \Phi''(t_n)h^2 \leq \frac{Mh^2}{2}</math> where <math>m = \mathrm{max} |\Phi''(t)|</math>

Local error is proportional to <math>h^2</math>.

Global error is proportional to h.

'''3. Improved Euler Formula (or Heun Formula):'''

<math>y_{n+1} = y_n + \frac{f_n + f(t_n + h, y_n + hf_n)}{2} h</math>

Local truncation error is proportional to <math>h^3</math>

Global truncation error is proportional to <math>h^2</math>

'''4. The Runge-Kutta Method:'''

<math>y_{n+1} = y_n + \frac{k_{n1} + 2k_{n2} + 2k_{n3} + k_{n4}}{6} h</math>

where

<math>k_{n1} = f(t_n, y_n) \quad k_{n2} = f(t_n + \frac{1}{2} h, y_n + \frac{1}{2}hk_{n1})</math>

<math>k_{n3} = f(t_n + \frac{1}{2} h, y_n + \frac{1}{2}hk_{n2}) \quad k_{n4} = f(t_n + h, y_n + hk_{n3})</math>

Local truncation error is proportional to <math>h^5</math>.

Global truncation error is proportional to <math>h^4</math>.

==Python Example of Euler's Method==
In class on October 15th we discussed Euler's Method to numerically compute a solution to a differential equation. <math>x_0</math> and <math>y_0</math> are given as well as an increment amount <math>h</math>, <math>x_{n+1} = x_n + h</math>, and we use the guess <math>y_{n+1} = y_n + f(x_n, y_n)*h</math> where f computes the derivative as a function of x and y.

Here is an example of code (written in Python) which carries out Euler's Method for the example we discussed in class, <math>y' = -y</math>:

def f(x, y):
return -y

def euler(x, y, f, h, x_max):
"""Take in coordinates x and y, a function f(x, y) which calculates
dy/dx at (x, y), an increment h, and a maximum value of x.

Return a list containing coordinates in the Euler's Method computation
of the solution to Phi' = f(x, Phi(x)), Phi(x) = y, with the x
values of those coordinates separated by h, and not exceeding x_max.
"""
if x > x_max: # we have already calculated all our values
return []
x_next, y_next = (x + h, y + f(x, y)*h) # calculate the next x, y values
# return the current coordinates, and every coordinates following it, in a list
return [(x_next, y_next)] + euler(x_next, y_next, f, h, x_max)

if __name__ == '__main__':
print euler(0, 1, f, 0.01, 1)[-1]

12-267

2012-10-26T00:57:36Z

Twine: Added link to numerical methods page, removed link to python example of euler method (plan to amalgamate)

__NOEDITSECTION__
__NOTOC__
{{12-267/Navigation}}
==Advanced Ordinary Differential Equations==
===Department of Mathematics, University of Toronto, Fall 2012===

{{12-267/Crucial Information}}

===Text===
Boyce and DiPrima, [http://ca.wiley.com/WileyCDA/WileyTitle/productCd-EHEP002451.html Elementary Differential Equations and Boundary Value Problems] (current edition is 9th and 10th will be coming out shortly. Hopefully any late enough edition will do).

===Further Resources===

* [http://www.math.toronto.edu/undergrad/ Undergraduate Information] at the [http://www.math.toronto.edu/ UofT Math Department]

* [http://www.artsandscience.utoronto.ca/ofr/calendar/crs_mat.htm Undergraduate Course Descriptions].

* Vitali Kapovitch's 2007 classes: [http://www.math.toronto.edu/vtk/267/ Spring], [http://www.math.toronto.edu/vtk/267fall07/ Fall].

* Also previously taught by T. Bloom, C. Pugh, D. Remenik.

* My {{Pensieve Link|Classes/12-267/|12-267 notebook}}.

{{Template:12-267:Dror/Students Divider}}

[[User:Drorbn|Drorbn]] 06:36, 12 September 2012 (EDT): Material by [[User:Syjytg|Syjytg]] moved to [[12-267/Tuesday September 11 Notes]].

[http://imgur.com/a/OSx1U#0 Summary of techniques to solve differential equations ] [[User:Vsbdthrsh|Vsbdthrsh]]

[http://drorbn.net/index.php?title=12-267/Existence_And_Uniqueness_Theorem Fundamental Theorem and Proof from Lecture] [[User:Twine|Twine]]

[http://drorbn.net/index.php?title=12-267/Derivation_of_Euler-Lagrange Derivation of Euler-Lagrange from Lecture] [[User:Twine|Twine]]

[http://graphics.ethz.ch/teaching/former/vc_master_06/Downloads/viscomp-varcalc_6.pdf Useful PDF: proof of Euler-Lagrange equation, explanation, examples ] [[User:Vsbdthrsh|Vsbdthrsh]]

[http://math.hunter.cuny.edu/mbenders/cofv.pdf In-depth coverage of Calculus of Variations][[User:Simon1|Simon1]]

[http://highered.mcgraw-hill.com/sites/dl/free/007063419x/392340/Calculus_of_Variations.pdf A good summary of Calculus of Variations]
[[User:Simon1|Simon1]]

[http://imgur.com/a/1vy11#0 All class notes from September 10th to October 5th] [[User:Simon1|Simon1]]

[http://imgur.com/a/uLSlM Summary of Numerical Methods] [[User:Simon1|Simon1]]

[http://drorbn.net/index.php?title=Numerical_Methods Numerical Methods (wiki)] [[User:Twine|Twine]]

[http://imgur.com/a/tliYg Summary of Chapter 3 from the Textbook on Constant Coefficient Second Order ODEs] [[User:Simon1|Simon1]]

[http://drorbn.net/images/e/e6/Geometric_Interpretation_of_Lagrange_Multiplier.pdf Geometric Interpretation of Lagrange Multiplier] [[User:Mathstudent|Mathstudent]]

Past exams from [http://exams.library.utoronto.ca.myaccess.library.utoronto.ca/handle/exams/4429 December 2009] and [http://exams.library.utoronto.ca.myaccess.library.utoronto.ca/handle/exams/4429 December 2010] [[User:Twine|Twine]]

Handwritten notes by [[User:Ktnd3|Ktnd3]]:

* September: [[Media:Mat267_-_lecture_1(sep.10).PDF|10th]], [[Media:Mat267_-_lecture_2%28sep.11%29.PDF|11th]], [[Media:Mat267_-_lecture_3%28sep.14%29.PDF|14th]], [[Media:Mat267_-_lecture_4%28sep.17%29.PDF|17th]], [[Media:12-267%28lecture5%29.PDF|18th]], [[Media:12-267%28lecture6%29.PDF|21st]], [[Media:12-267%28lecture7%29.PDF|24th]], [[Media:12-267%28lecture8%29.PDF|25th]], [[Media:12-267%28lecture9%29.PDF|28th]]

* October: [[Media:12-267%28lecture10%29.PDF|1st]], [[Media:12-267%28lecture11%29.PDF|2nd]], [[Media:12-267%28lecture12%29.PDF|5th]], [[Media:12-267%28lecture13%29.PDF|9th]] [[Media:12-267%28lecture14%29.PDF|12th]]

12-267/Numerical Methods

2012-10-26T00:55:19Z

Twine: Created page, based largely off of http://imgur.com/a/uLSlM posted by Simon1

Numerical methods: <math>\frac{dy}{dt} = f(t, y)</math> and <math>y(t_0) = y_0</math>, <math>y = \Phi(t)</math> is a solution.

'''1. Using the proof of Picard's Theorem:'''

<math>\Phi_0(x) = y_0</math>

<math>\Phi_n(x) = y_0 + \int_{x_0}^x f(x, \Phi_{n-1}(x)) dx</math>

<math>\Phi_n(x) \rightarrow \Phi(x)</math>

'''2. The Euler Method:'''

<math>y_{n+1} = y_n + f(t_n, y_n)(t_{n+1} - t_n) = y_n + f_n h</math> if h is constant

Backward Euler formula: <math>y_{n+1} = y_n + h f(t_{n+1}, y_{n+1})</math>

Local truncation error: <math>e_{n+1} = \frac{1}{2} \Phi''(t_n)h^2 \leq \frac{Mh^2}{2}</math> where <math>m = \mathrm{max} |\Phi''(t)|</math>

Local error is proportional to <math>h^2</math>.

Global error is proportional to h.

'''3. Improved Euler Formula (or Heun Formula):'''

<math>y_{n+1} = y_n + \frac{f_n + f(t_n + h, y_n + hf_n)}{2} h</math>

Local truncation error is proportional to <math>h^3</math>

Global truncation error is proportional to <math>h^2</math>

'''4. The Runge-Kutta Method:'''

<math>y_{n+1} = y_n + \frac{k_{n1} + 2k_{n2} + 2k_{n3} + k_{n4}}{6} h</math>

where

<math>k_{n1} = f(t_n, y_n) \quad k_{n2} = f(t_n + \frac{1}{2} h, y_n + \frac{1}{2}hk_{n1})</math>

<math>k_{n3} = f(t_n + \frac{1}{2} h, y_n + \frac{1}{2}hk_{n2}) \quad k_{n4} = f(t_n + h, y_n + hk_{n3})</math>

Local truncation error is proportional to <math>h^5</math>.

Global truncation error is proportional to <math>h^4</math>.

Based largely off of a note available [http://imgur.com/a/uLSlM here] [[User:Simon1|Simon1]] --[[User:Twine|Twine]] 20:55, 25 October 2012 (EDT)

12-267

2012-10-24T23:36:51Z

Twine: Added links to past exams in U of T old exams repository

__NOEDITSECTION__
__NOTOC__
{{12-267/Navigation}}
==Advanced Ordinary Differential Equations==
===Department of Mathematics, University of Toronto, Fall 2012===

{{12-267/Crucial Information}}

===Text===
Boyce and DiPrima, [http://ca.wiley.com/WileyCDA/WileyTitle/productCd-EHEP002451.html Elementary Differential Equations and Boundary Value Problems] (current edition is 9th and 10th will be coming out shortly. Hopefully any late enough edition will do).

===Further Resources===

* [http://www.math.toronto.edu/undergrad/ Undergraduate Information] at the [http://www.math.toronto.edu/ UofT Math Department]

* [http://www.artsandscience.utoronto.ca/ofr/calendar/crs_mat.htm Undergraduate Course Descriptions].

* Vitali Kapovitch's 2007 classes: [http://www.math.toronto.edu/vtk/267/ Spring], [http://www.math.toronto.edu/vtk/267fall07/ Fall].

* Also previously taught by T. Bloom, C. Pugh, D. Remenik.

* My {{Pensieve Link|Classes/12-267/|12-267 notebook}}.

{{Template:12-267:Dror/Students Divider}}

Handwritten notes by [[User:Ktnd3|Ktnd3]]:

September: [http://drorbn.net/images/7/72/Mat267_-_lecture_1%28sep.10%29.PDF 10th], [http://drorbn.net/images/b/ba/Mat267_-_lecture_2%28sep.11%29.PDF 11th], [http://drorbn.net/images/5/5a/Mat267_-_lecture_3%28sep.14%29.PDF 14th], [http://drorbn.net/images/8/82/Mat267_-_lecture_4%28sep.17%29.PDF 17th], [http://drorbn.net/images/d/df/12-267%28lecture5%29.PDF 18th], [http://drorbn.net/images/3/34/12-267%28lecture6%29.PDF 21st], [http://drorbn.net/images/a/a4/12-267%28lecture7%29.PDF 24th], [http://drorbn.net/images/a/a8/12-267%28lecture8%29.PDF 25th], [http://drorbn.net/images/f/fe/12-267%28lecture9%29.PDF 28th]

October: [http://drorbn.net/images/9/97/12-267%28lecture10%29.PDF 1st], [http://drorbn.net/images/c/cc/12-267%28lecture11%29.PDF 2nd], [http://drorbn.net/images/4/43/12-267%28lecture12%29.PDF 5th], [http://drorbn.net/images/4/4d/12-267%28lecture13%29.PDF 9th] [http://drorbn.net/images/7/73/12-267%28lecture14%29.PDF 12th]

[[User:Drorbn|Drorbn]] 06:36, 12 September 2012 (EDT): Material by [[User:Syjytg|Syjytg]] moved to [[12-267/Tuesday September 11 Notes]].

[http://imgur.com/a/OSx1U#0 Summary of techniques to solve differential equations ] [[User:Vsbdthrsh|Vsbdthrsh]]

[http://drorbn.net/index.php?title=12-267/Existence_And_Uniqueness_Theorem Fundamental Theorem and Proof from Lecture] [[User:Twine|Twine]]

[http://drorbn.net/index.php?title=12-267/Derivation_of_Euler-Lagrange Derivation of Euler-Lagrange from Lecture] [[User:Twine|Twine]]

[http://graphics.ethz.ch/teaching/former/vc_master_06/Downloads/viscomp-varcalc_6.pdf Useful PDF: proof of Euler-Lagrange equation, explanation, examples ] [[User:Vsbdthrsh|Vsbdthrsh]]

[http://drorbn.net/index.php?title=12-267/Euler%27s_Method Python example for Euler's Method] [[User:Twine|Twine]]

[http://math.hunter.cuny.edu/mbenders/cofv.pdf In-depth coverage of Calculus of Variations][[User:Simon1|Simon1]]

[http://highered.mcgraw-hill.com/sites/dl/free/007063419x/392340/Calculus_of_Variations.pdf A good summary of Calculus of Variations]
[[User:Simon1|Simon1]]

[http://imgur.com/a/1vy11#0 All class notes from September 10th to October 5th] [[User:Simon1|Simon1]]

[http://imgur.com/a/uLSlM Summary of Numerical Methods] [[User:Simon1|Simon1]]

[http://imgur.com/a/tliYg Summary of Chapter 3 from the Textbook on Constant Coefficient Second Order ODEs] [[User:Simon1|Simon1]]

Past exams from [http://exams.library.utoronto.ca.myaccess.library.utoronto.ca/handle/exams/4429 December 2009] and [http://exams.library.utoronto.ca.myaccess.library.utoronto.ca/handle/exams/4429 December 2010] [[User:Twine|Twine]]

12-267

2012-10-24T23:31:32Z

Twine: Gathered links to Ktnd3's notes from lecture video pages, put them in student link section

__NOEDITSECTION__
__NOTOC__
{{12-267/Navigation}}
==Advanced Ordinary Differential Equations==
===Department of Mathematics, University of Toronto, Fall 2012===

{{12-267/Crucial Information}}

===Text===
Boyce and DiPrima, [http://ca.wiley.com/WileyCDA/WileyTitle/productCd-EHEP002451.html Elementary Differential Equations and Boundary Value Problems] (current edition is 9th and 10th will be coming out shortly. Hopefully any late enough edition will do).

===Further Resources===

* [http://www.math.toronto.edu/undergrad/ Undergraduate Information] at the [http://www.math.toronto.edu/ UofT Math Department]

* [http://www.artsandscience.utoronto.ca/ofr/calendar/crs_mat.htm Undergraduate Course Descriptions].

* Vitali Kapovitch's 2007 classes: [http://www.math.toronto.edu/vtk/267/ Spring], [http://www.math.toronto.edu/vtk/267fall07/ Fall].

* Also previously taught by T. Bloom, C. Pugh, D. Remenik.

* My {{Pensieve Link|Classes/12-267/|12-267 notebook}}.

{{Template:12-267:Dror/Students Divider}}

Handwritten notes by [[User:Ktnd3|Ktnd3]]:

September: [http://drorbn.net/images/7/72/Mat267_-_lecture_1%28sep.10%29.PDF 10th], [http://drorbn.net/images/b/ba/Mat267_-_lecture_2%28sep.11%29.PDF 11th], [http://drorbn.net/images/5/5a/Mat267_-_lecture_3%28sep.14%29.PDF 14th], [http://drorbn.net/images/8/82/Mat267_-_lecture_4%28sep.17%29.PDF 17th], [http://drorbn.net/images/d/df/12-267%28lecture5%29.PDF 18th], [http://drorbn.net/images/3/34/12-267%28lecture6%29.PDF 21st], [http://drorbn.net/images/a/a4/12-267%28lecture7%29.PDF 24th], [http://drorbn.net/images/a/a8/12-267%28lecture8%29.PDF 25th], [http://drorbn.net/images/f/fe/12-267%28lecture9%29.PDF 28th]

October: [http://drorbn.net/images/9/97/12-267%28lecture10%29.PDF 1st], [http://drorbn.net/images/c/cc/12-267%28lecture11%29.PDF 2nd], [http://drorbn.net/images/4/43/12-267%28lecture12%29.PDF 5th], [http://drorbn.net/images/4/4d/12-267%28lecture13%29.PDF 9th] [http://drorbn.net/images/7/73/12-267%28lecture14%29.PDF 12th]

[[User:Drorbn|Drorbn]] 06:36, 12 September 2012 (EDT): Material by [[User:Syjytg|Syjytg]] moved to [[12-267/Tuesday September 11 Notes]].

[http://imgur.com/a/OSx1U#0 Summary of techniques to solve differential equations ] [[User:Vsbdthrsh|Vsbdthrsh]]

[http://drorbn.net/index.php?title=12-267/Existence_And_Uniqueness_Theorem Fundamental Theorem and Proof from Lecture] [[User:Twine|Twine]]

[http://drorbn.net/index.php?title=12-267/Derivation_of_Euler-Lagrange Derivation of Euler-Lagrange from Lecture] [[User:Twine|Twine]]

[http://graphics.ethz.ch/teaching/former/vc_master_06/Downloads/viscomp-varcalc_6.pdf Useful PDF: proof of Euler-Lagrange equation, explanation, examples ] [[User:Vsbdthrsh|Vsbdthrsh]]

[http://drorbn.net/index.php?title=12-267/Euler%27s_Method Python example for Euler's Method] [[User:Twine|Twine]]

[http://math.hunter.cuny.edu/mbenders/cofv.pdf In-depth coverage of Calculus of Variations][[User:Simon1|Simon1]]

[http://highered.mcgraw-hill.com/sites/dl/free/007063419x/392340/Calculus_of_Variations.pdf A good summary of Calculus of Variations]
[[User:Simon1|Simon1]]

[http://imgur.com/a/1vy11#0 All class notes from September 10th to October 5th] [[User:Simon1|Simon1]]

[http://imgur.com/a/uLSlM Summary of Numerical Methods] [[User:Simon1|Simon1]]

[http://imgur.com/a/tliYg Summary of Chapter 3 from the Textbook on Constant Coefficient Second Order ODEs] [[User:Simon1|Simon1]]

Notes for 12-267-120928

2012-10-24T23:19:37Z

Twine: Added link

Proof of the Fundamental Theorem.

Proof and statement of the theorem based on this lecture are available [http://drorbn.net/index.php?title=12-267/Existence_And_Uniqueness_Theorem here].

12-267/Homework Assignment 2

2012-10-24T22:44:46Z

Twine: Added solution to task 2

{{12-267/Navigation}}

This assignment is due at the tutorial on Tuesday October 2. Here and everywhere, '''neatness counts!!''' You may be brilliant and you may mean just the right things, but if the your readers will be having hard time deciphering your work they will give up and assume it is wrong.

'''Task 0.''' Identify yourself in the [[12-267/Class Photo|Class Photo]]!

'''Task 1.''' Solve the following differential equations:
# <math>x^2y^3+x(1+y^2)y'=0</math> (hint: try <math>\mu=x^\alpha y^\beta</math>).
# <math>dx+(\frac{x}{y}-\sin y)dy=0</math>.
# <math>(x^2+3xy+y^2)dx-x^2dy=0</math>.
# <math>\frac{dy}{dx}=\frac{2y-x+5}{2x-y-4}</math> (hint: consider trying <math>x_1=x+\alpha</math> and <math>y_1=y+\beta</math> for good <math>\alpha</math>, <math>\beta</math>).
# <math>y'=\frac{y^3}{1-2xy^2}</math> with <math>y(0)=1</math>.
# <math>\frac{dy}{dx}=\frac{2y+\sqrt{x^2-y^2}}{2x}</math>.

'''Task 2.''' Let <math>M</math> and <math>N</math> be differentiable functions of <math>x</math> and <math>y</math>.
# Show that if <math>\frac{N_x-M_y}{xM-yN}</math> depends only on <math>xy</math>, then the differential equation <math>Mdx+Ndy=0</math> has an integrating factor of the form <math>\mu(xy)</math>, where <math>\mu</math> is a function of a single variable.
# Find a condition on <math>M</math> and <math>N</math> that would imply that the differential equation <math>Mdx+Ndy=0</math> would have an integrating factor of the form <math>\mu(x+y)</math>, where <math>\mu</math> is a function of a single variable.

'''Task 3.''' The equation <math>y'+p(x)y=q(x)y^n</math> is called a "Bernoulli Equation".
# Explain why you already know how to solve the Bernoulli equation when <math>n=0</math> and when <math>n=1</math>.
# Show that if <math>n\neq0,1</math>, then the substitution <math>v=y^{1-n}</math> reduces the Bernoulli equation to an equation you already know how to solve.
# Solve the equation <math>x^2y'+2xy-y^3=0</math> (for <math>x>0</math>).

'''Task 4.''' Find an example of a non-differentiable function which is nevertheless Lipschitz.

{{Template:12-267:Dror/Students Divider}}

'''Solution to Task 2''' --[[User:Twine|Twine]] 18:44, 24 October 2012 (EDT)

'''Part 1'''

What we want is an integration factor <math>\mu</math> such that <math>(\mu M)_y = (\mu N)_x</math>. Let <math>xy = z</math>

When <math>\mu</math> is a function of z, <math>\frac{dv}{dx} = \frac{dv}{dz} \frac{dz}{dx} = \frac{dv}{dz} y</math>.

Similarly <math>\frac{dv}{dy} = \frac{dv}{dz} x</math>.

Then we have

<math>(\mu M)_y = (\mu N)_x</math>

<math>\iff \mu_{z}xM - \mu_{xy}yN = \mu (N_x - M_y)</math>

<math>\iff \frac{\mu_{z}}{mu} = \frac{N_x - M_y}{xM - yN}</math>

If the right hand side depends only on xy, we can get

<math>\mu (xy) = e^{\int \frac{N_x - M_y}{xM - yN} d(xy)}</math>

which satisfies the requirements of an integrating factor.

'''Part 2'''

As in Part 1, we need <math>(\mu M)_y = (\mu N)_x</math>. Let <math>z = x + y</math>

When <math>\mu</math> is a function of <math>x + y</math>, <math>\frac{d \mu}{dx} = \frac{d\mu}{dz} \frac{dz}{dx} = \frac{d\mu}{dz}</math>.

Similarly <math>\frac{d \mu}{dy} = \frac{d\mu}{dz}</math>

Then we have

<math>(\mu M)_y = (\mu N)_x</math>

<math>\iff \frac{\mu_{z}}{\mu} = \frac{N_x - M_y}{M-N}</math>

If the right hand side of this equation depends only on z (that is, only on (x+y)), then we have

<math>\mu (x + y) = e^{\int \frac{N_x - M_y}{M - N} d(x + y)}</math>

which satisfies the requirements of an integrating factor.

Solution to HW2: [[User:Mathstudent|Mathstudent]]
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12-267/Derivation of Euler-Lagrange

2012-10-24T22:18:13Z

Twine: Added special cases (without derivations)

{{12-267/Navigation}}

Disclamer: This is a student prepared note based on [http://drorbn.net/dbnvp/12-267-121002-2.php the lecure of Tuesday October 2nd].

For a function <math>y(x)</math> defined on <math>[a, b]</math> to be an extremum of <math>J(y) = \int_a^b F(x, y, y') dx</math>, it must be that for any function <math>h(x)</math> defined on <math>[a, b]</math> that preserves the endpoints of <math>y</math> (that is, <math>h(a) = 0</math> and <math>h(b) = 0</math>), we have <math> \frac{d}{d \epsilon } J(y + \epsilon h) </math><math>|_{\epsilon = 0} = 0 </math>.

<math> \frac{d}{d \epsilon } J(y + \epsilon h) </math><math>|_{\epsilon = 0} </math>

<math> = \frac{d}{d \epsilon } \int_a^b F(x, y + \epsilon h, y' + \epsilon h') dx |_{\epsilon = 0} </math>

Let <math>F_n</math> signify F differentiated with respect to its nth variable.

<math> = \int_a^b (F_1 \cdot 0 + F_2 \cdot h + F_3 \cdot h') dx |_{\epsilon = 0} </math>

<math> = \int_a^b (F_2(x, y, y') \cdot h + F_3(x, y, y') \cdot h') dx </math>

<math> = \int_a^b (F_2 \cdot h - [\frac{d}{dx} F_3] \cdot h) dx + F_3 \cdot h |_a^b</math> (integrating by parts)

Due to the constraints of <math>h(a) = 0</math> and <math>h(b) = 0</math>, <math>F_3 \cdot h |_a^b = 0</math>.

<math> = \int_a^b (F_2 - \frac{d}{dx} F_3) \cdot h dx</math>

As this must be equal to 0 for all h satisfying the endpoint constraints, we must have that <math>F_2 = \frac{d}{dx} F_3</math>, or in other terms, <math>F_y - \frac{d}{dx} F_y' = 0</math>.

Special cases (without derivations):

In the case that F does not depend on y', we have <math>F_y = 0</math>

In the case that F does not depend on y, we have <math>F_{y'} = c</math>

In the case that F does not depend on x, we have <math>F - y'F_{y'} = c</math>

12-267/Homework Assignment 3

2012-10-24T22:11:30Z

Twine: Added solution to 1

{{12-267/Navigation}}

This assignment is due at the tutorial on Tuesday October 9. Here and everywhere, '''neatness counts!!''' You may be brilliant and you may mean just the right things, but if your readers have a hard time deciphering your work they will give up and assume it is wrong.

'''Task 0.''' Identify yourself in the [[12-267/Class Photo|Class Photo]]!

'''Task 1.''' Let <math>\phi_n\colon X\to{\mathbb R}</math> be a sequence of functions defined on some set <math>X</math>, and suppose that some sequence <math>c_n</math> of non-negative reals is given such that for every <math>x\in X</math>, <math>|\phi_n(x)-\phi_{n+1}(x)|\leq c_n</math>. Suppose also that <math>\sum_{n=1}^\infty c_n</math> is finite. Prove that the sequence <math>\phi_n</math> is uniformly convergent.

'''Task 2.''' Find the extrema of the following functionals:
# <math>y\mapsto \int_0^1y'dx</math> subject to <math>y(0)=0</math> and <math>y(1)=1</math>.
# <math>y\mapsto \int_0^1yy'dx</math> subject to <math>y(0)=0</math> and <math>y(1)=1</math>.
# <math>y\mapsto \int_0^1xyy'dx</math> subject to <math>y(0)=0</math> and <math>y(1)=1</math>.
# <math>y\mapsto\int_a^b\frac{y'^2}{x^3}dx</math>.
# <math>y\mapsto\int_a^b(y^2+y'^2+2ye^x)dx</math>.
# Postponed! <math>y\mapsto\int_0^1(y'^2+x^2)dx</math> subject to <math>\int_0^1y^2dx=2</math> and <math>y(0)=0</math> and <math>y(1)=1</math>.

'''Task 3.''' A roach <math>R</math> I once met was mortally afraid of walls, and so when it walked on my kitchen's floor, its speed was exactly proportional to its distance from the nearest wall (that is, very near a wall it crawled very slowly, while in the centre of the room it run around quickly and happily). As a step towards simplifying <math>R</math>'s life, help it find the fastest path from one point in the upper half plane <math>\{y>0\}</math> to another point in the upper half plane, assuming there is only one wall around, built along the <math>x</math>-axis <math>y=0</math>.

{{Template:12-267:Dror/Students Divider}}

'''Solution to Task 1.''' --[[User:Twine|Twine]] 18:11, 24 October 2012 (EDT)

Let <math>\sum_{n=1}^\infty c_n = S</math>. We have that <math>\lim_{n \rightarrow \infty} c_n = 0</math>.

Let <math>\epsilon > 0</math>.

<math>\exists N \in \mathbb{Z}^+</math> such that <math>\sum_{n=1}^N c_n > S - \epsilon</math>.

Choose such an N. Then <math>\sum_{n=N}^{\infty} c_n < \epsilon</math>.

Then <math>\forall x \in X, \forall m > N, n>N</math>, we have <math>|\Phi_n(x) - \Phi_m(x)| < \epsilon</math>.

As this result is independent of our choice of x, we have by the cauchy criterion that <math>\Phi_n</math> is uniformly convergent and <math>\lim_{n \rightarrow \infty} \Phi_n(x) = \Phi(x)</math> exists.

Solutions to HW3: [[User:Mathstudent|Mathstudent]]
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12-267/Derivation of Euler-Lagrange

2012-10-24T21:54:59Z

Twine: Added navigation

12-267/Existence And Uniqueness Theorem

2012-10-24T21:54:36Z

Twine: Added navigation

{{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].

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.

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>.

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>.

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 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:

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> = |\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>

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>

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]

12-267/Homework Assignment 4

2012-10-24T21:54:12Z

Twine: Added solutions to 4 and 5

{{12-267/Navigation}}

This assignment is due at the tutorial on Tuesday October 16. Here and everywhere, '''neatness counts!!''' You may be brilliant and you may mean just the right things, but if your readers have a hard time deciphering your work they will give up and assume it is wrong.

'''Task 0.''' Identify yourself in the [[12-267/Class Photo|Class Photo]]!

'''Task 1.''' Find the general solution of the Euler-Lagrange equation corresponding to the functional <math>J(y)=\int_a^bf(x)\sqrt{1+y'^2}dx</math>, and investigate the special cases <math>f(x)=\sqrt{x}</math> and <math>f(x)=x</math>.

'''Task 2.''' Find the extrema the following functional <math>y\mapsto\int_0^1(y'^2+x^2)dx</math> subject to <math>\int_0^1y^2dx=2</math> and <math>y(0)=0</math> and <math>y(1)=0</math>. (An earlier version of this assignment had by mistake <math>y(1)=1</math>, which leads to much uglier numbers. If you already solved the problem with <math>y(1)=1</math>, you may submit either solution).

'''Task 3.''' Solve the "power line problem": Of all the curves <math>y</math> with <math>y(a)=A</math> and <math>y(b)=B</math> and with total arc-length <math>l</math>, find the one with the least potential energy <math>\int_a^by\sqrt{1+y'^2}dx</math>.

'''Task 4.''' Find a necessary condition for a function <math>y</math> satisfying <math>y(a)=A</math>, <math>y'(a)=A'</math>, <math>y(b)=B</math>, and <math>y'(b)=B'</math> to be an extremal of a functional of the form <math>y\mapsto\int_a^bF(x,y,y',y'')dx</math>.

'''Task 5.''' Find the curve <math>y</math> joining the points <math>(0,0)</math> and <math>(1,0)</math> and for which the integral <math>\int_0^1y''^2dx</math> is minimal, if <math>y'(0)=a</math> and <math>y'(1)=b</math>.

{{Template:12-267:Dror/Students Divider}}

'''Solution to Task 4.''' --[[User:Twine|Twine]] 17:54, 24 October 2012 (EDT)

Let <math>h(x)</math> be any function defined on <math>[a, b]</math> with <math>h'(a) = h(a) = h'(b) = h(b) = 0</math>. For y to be an extremal of the functional with the boundary constraints given, we must have that <math>\frac{d}{d\epsilon} J(y + \epsilon h) |_{\epsilon = 0} = 0</math> for any such <math>h(x)</math>.

<math>\frac{d}{d\epsilon} J(y + \epsilon h) |_{\epsilon = 0}</math>

<math>= \frac{d}{d\epsilon} \int_a^b F(x, y+\epsilon h, y'+\epsilon h', y'' +\epsilon h'')dx|_{\epsilon = 0}</math>

<math>= \int_a^b (F_1 \cdot 0 + F_2 \cdot h + F_3 \cdot h' + F_4 \cdot h'')dx|_{\epsilon = 0}</math>

<math>= \int_a^b (F_2 \cdot h - F_3' \cdot h - F_4' \cdot h')dx + F_3 \cdot h|_a^b + F_4\cdot h'|_a^b</math>

<math>= \int_a^b (F_2 \cdot h - F_3' \cdot h + F_4'' \cdot h)dx + F_4'\cdot h|_a^b</math>

<math>= \int_a^b (F_2 - F_3' + F_4'') \cdot h dx</math>

For this to be equivalent to 0 for any h defined above, we must have

<math>F_y - \frac{d}{dx}F_{y'} + \frac{d^2}{dx^2}F_{y''} = 0</math>

'''Solution to Task 5.''' --[[User:Twine|Twine]] 17:54, 24 October 2012 (EDT)

We use the result of 4. As in this case F is independent of x, y, and y', the equation reduces to

<math>\frac{d^2}{dx^2}2y'' = 0</math>

<math>y'''' = 0</math>

This has the solution <math>y = c_0 + c_1 x + c_2 x^2 + c_3 x^3</math>. We can use the constraint equations <math>y(0) = 0</math>, <math>y(1) = 0</math>, <math>y'(0) = a</math>, <math>y'(1) = b</math> to show that <math>y = (b+a)x^3 - (2a + b)x^2 + ax</math>. Hence, This is the only y for which the functional is extremal.

12-267/Derivation of Euler-Lagrange

2012-10-24T21:24:31Z

Twine: Corrected typo

Disclamer: This is a student prepared note based on [http://drorbn.net/dbnvp/12-267-121002-2.php the lecure of Tuesday October 2nd].

For a function <math>y(x)</math> defined on <math>[a, b]</math> to be an extremum of <math>J(y) = \int_a^b F(x, y, y') dx</math>, it must be that for any function <math>h(x)</math> defined on <math>[a, b]</math> that preserves the endpoints of <math>y</math> (that is, <math>h(a) = 0</math> and <math>h(b) = 0</math>), we have <math> \frac{d}{d \epsilon } J(y + \epsilon h) </math><math>|_{\epsilon = 0} = 0 </math>.

<math> \frac{d}{d \epsilon } J(y + \epsilon h) </math><math>|_{\epsilon = 0} </math>

<math> = \frac{d}{d \epsilon } \int_a^b F(x, y + \epsilon h, y' + \epsilon h') dx |_{\epsilon = 0} </math>

Let <math>F_n</math> signify F differentiated with respect to its nth variable.

<math> = \int_a^b (F_1 \cdot 0 + F_2 \cdot h + F_3 \cdot h') dx |_{\epsilon = 0} </math>

<math> = \int_a^b (F_2(x, y, y') \cdot h + F_3(x, y, y') \cdot h') dx </math>

<math> = \int_a^b (F_2 \cdot h - [\frac{d}{dx} F_3] \cdot h) dx + F_3 \cdot h |_a^b</math> (integrating by parts)

Due to the constraints of <math>h(a) = 0</math> and <math>h(b) = 0</math>, <math>F_3 \cdot h |_a^b = 0</math>.

<math> = \int_a^b (F_2 - \frac{d}{dx} F_3) \cdot h dx</math>

As this must be equal to 0 for all h satisfying the endpoint constraints, we must have that <math>F_2 = \frac{d}{dx} F_3</math>, or in other terms, <math>F_y - \frac{d}{dx} F_y' = 0</math>.

12-267/Homework Assignment 3

2012-10-24T21:05:56Z

Twine: Added student divider

{{12-267/Navigation}}

This assignment is due at the tutorial on Tuesday October 9. Here and everywhere, '''neatness counts!!''' You may be brilliant and you may mean just the right things, but if your readers have a hard time deciphering your work they will give up and assume it is wrong.

'''Task 0.''' Identify yourself in the [[12-267/Class Photo|Class Photo]]!

'''Task 1.''' Let <math>\phi_n\colon X\to{\mathbb R}</math> be a sequence of functions defined on some set <math>X</math>, and suppose that some sequence <math>c_n</math> of non-negative reals is given such that for every <math>x\in X</math>, <math>|\phi_n(x)-\phi_{n+1}(x)|\leq c_n</math>. Suppose also that <math>\sum_{n=1}^\infty c_n</math> is finite. Prove that the sequence <math>\phi_n</math> is uniformly convergent.

'''Task 2.''' Find the extrema of the following functionals:
# <math>y\mapsto \int_0^1y'dx</math> subject to <math>y(0)=0</math> and <math>y(1)=1</math>.
# <math>y\mapsto \int_0^1yy'dx</math> subject to <math>y(0)=0</math> and <math>y(1)=1</math>.
# <math>y\mapsto \int_0^1xyy'dx</math> subject to <math>y(0)=0</math> and <math>y(1)=1</math>.
# <math>y\mapsto\int_a^b\frac{y'^2}{x^3}dx</math>.
# <math>y\mapsto\int_a^b(y^2+y'^2+2ye^x)dx</math>.
# Postponed! <math>y\mapsto\int_0^1(y'^2+x^2)dx</math> subject to <math>\int_0^1y^2dx=2</math> and <math>y(0)=0</math> and <math>y(1)=1</math>.

'''Task 3.''' A roach <math>R</math> I once met was mortally afraid of walls, and so when it walked on my kitchen's floor, its speed was exactly proportional to its distance from the nearest wall (that is, very near a wall it crawled very slowly, while in the centre of the room it run around quickly and happily). As a step towards simplifying <math>R</math>'s life, help it find the fastest path from one point in the upper half plane <math>\{y>0\}</math> to another point in the upper half plane, assuming there is only one wall around, built along the <math>x</math>-axis <math>y=0</math>.

{{Template:12-267:Dror/Students Divider}}

Solutions to HW3: [[User:Mathstudent|Mathstudent]]
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12-267/Homework Assignment 2

2012-10-24T21:05:39Z

Twine: Added student divider

12-267/Homework Assignment 1

2012-10-24T21:05:13Z

Twine: Added student divider

{{12-267/Navigation}}

This assignment is due at the tutorial on Tuesday September 25. Here and everywhere, '''neatness counts!!''' You may be brilliant and you may mean just the right things, but if the your readers will be having hard time deciphering your work they will give up and assume it is wrong.

'''Question 1.''' Show that if <math>y=y_1(x)</math> is a solution of <math>y'+p(x)y=0</math>, and <math>y=y_2(x)</math> is a solution of <math>y'+p(x)y=g(x)</math>, then for any constant <math>c</math>, <math>y=cy_1+y_2</math> is a solution of <math>y'+p(x)y=g(x)</math>.

'''Question 2.''' Solve the following differential equations
# For <math>x>0</math>, <math>xy'+2y=\sin x</math>.
# <math>\frac{dy}{dx}=\frac{1}{e^y-x}</math> with <math>y(1)=0</math>; you may want to solve for <math>x</math> first.
# <math>xy'=\sqrt{1-y^2}</math>.
# <math>\frac{dy}{dx}=\frac{x-e^{-x}}{y+e^y}</math>.
# <math>xdx+ye^{-x}dy=0</math>, with <math>y(0)=1</math>.
# <math>\frac{dy}{dx}=\frac{ay+b}{cx+d}</math>, where <math>a,b,c,d</math> are arbitrary constants.
# <math>\frac{dy}{dx}=-\frac{ax+by}{bx+cy}</math>, where <math>a,b,c</math> are arbitrary constants.
# <math>0=(e^x\sin y + 3y)dx + (3(x+y)+e^x\cos y)dy</math>.

{{Template:12-267:Dror/Students Divider}}

'''Solution to Question 1.''' [[User:Twine|Twine]]

Take y defined by <math>y=cy_1+y_2</math> and plug it into <math>y'+p(x)y=g(x)</math>. We get

<math>(cy_1 + y_2)' + p(x)(cy_1 + y_2) = g(x)</math>

<math>c(y_1' + p(x)y_1) + (y_2' + p(x)y_2) = g(x)</math>

Based on our assumptions about <math>y_1</math> and <math>y_2</math>, we have <math>y_1' + p(x)y_1 = 0</math> and we have <math>y_2' + p(x)y_2 = g(x)</math>, and so the above equation holds <math>\forall c \in \mathbb{R}</math>.

Hence, <math>\forall c \in \mathbb{R}</math>, <math>cy_1 + y_2</math> is a solution of <math>y'+p(x)y=g(x)</math>.

Solutions to HW1: [[User:Mathstudent|Mathstudent]]
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12-267/Homework Assignment 3

2012-10-24T21:01:39Z

Twine: Changed format of Mathstudent's solutions to a gallery

{{12-267/Navigation}}

This assignment is due at the tutorial on Tuesday October 9. Here and everywhere, '''neatness counts!!''' You may be brilliant and you may mean just the right things, but if your readers have a hard time deciphering your work they will give up and assume it is wrong.

'''Task 0.''' Identify yourself in the [[12-267/Class Photo|Class Photo]]!

'''Task 1.''' Let <math>\phi_n\colon X\to{\mathbb R}</math> be a sequence of functions defined on some set <math>X</math>, and suppose that some sequence <math>c_n</math> of non-negative reals is given such that for every <math>x\in X</math>, <math>|\phi_n(x)-\phi_{n+1}(x)|\leq c_n</math>. Suppose also that <math>\sum_{n=1}^\infty c_n</math> is finite. Prove that the sequence <math>\phi_n</math> is uniformly convergent.

'''Task 2.''' Find the extrema of the following functionals:
# <math>y\mapsto \int_0^1y'dx</math> subject to <math>y(0)=0</math> and <math>y(1)=1</math>.
# <math>y\mapsto \int_0^1yy'dx</math> subject to <math>y(0)=0</math> and <math>y(1)=1</math>.
# <math>y\mapsto \int_0^1xyy'dx</math> subject to <math>y(0)=0</math> and <math>y(1)=1</math>.
# <math>y\mapsto\int_a^b\frac{y'^2}{x^3}dx</math>.
# <math>y\mapsto\int_a^b(y^2+y'^2+2ye^x)dx</math>.
# Postponed! <math>y\mapsto\int_0^1(y'^2+x^2)dx</math> subject to <math>\int_0^1y^2dx=2</math> and <math>y(0)=0</math> and <math>y(1)=1</math>.

'''Task 3.''' A roach <math>R</math> I once met was mortally afraid of walls, and so when it walked on my kitchen's floor, its speed was exactly proportional to its distance from the nearest wall (that is, very near a wall it crawled very slowly, while in the centre of the room it run around quickly and happily). As a step towards simplifying <math>R</math>'s life, help it find the fastest path from one point in the upper half plane <math>\{y>0\}</math> to another point in the upper half plane, assuming there is only one wall around, built along the <math>x</math>-axis <math>y=0</math>.

Solutions to HW3: [[User:Mathstudent|Mathstudent]]
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Image:012.jpg|page 1
Image:013.jpg|page 2
Image:014.jpg|page 3
Image:015.jpg|page 4
</gallery>

12-267/Homework Assignment 1

2012-10-24T21:00:32Z

Twine: Changed format of Mathstudent's solutions to a gallery

{{12-267/Navigation}}

This assignment is due at the tutorial on Tuesday September 25. Here and everywhere, '''neatness counts!!''' You may be brilliant and you may mean just the right things, but if the your readers will be having hard time deciphering your work they will give up and assume it is wrong.

'''Question 1.''' Show that if <math>y=y_1(x)</math> is a solution of <math>y'+p(x)y=0</math>, and <math>y=y_2(x)</math> is a solution of <math>y'+p(x)y=g(x)</math>, then for any constant <math>c</math>, <math>y=cy_1+y_2</math> is a solution of <math>y'+p(x)y=g(x)</math>.

'''Question 2.''' Solve the following differential equations
# For <math>x>0</math>, <math>xy'+2y=\sin x</math>.
# <math>\frac{dy}{dx}=\frac{1}{e^y-x}</math> with <math>y(1)=0</math>; you may want to solve for <math>x</math> first.
# <math>xy'=\sqrt{1-y^2}</math>.
# <math>\frac{dy}{dx}=\frac{x-e^{-x}}{y+e^y}</math>.
# <math>xdx+ye^{-x}dy=0</math>, with <math>y(0)=1</math>.
# <math>\frac{dy}{dx}=\frac{ay+b}{cx+d}</math>, where <math>a,b,c,d</math> are arbitrary constants.
# <math>\frac{dy}{dx}=-\frac{ax+by}{bx+cy}</math>, where <math>a,b,c</math> are arbitrary constants.
# <math>0=(e^x\sin y + 3y)dx + (3(x+y)+e^x\cos y)dy</math>.

Disclamer: The solutions below are by students, for students.

'''Solution to Question 1.''' [[User:Twine|Twine]]

Take y defined by <math>y=cy_1+y_2</math> and plug it into <math>y'+p(x)y=g(x)</math>. We get

<math>(cy_1 + y_2)' + p(x)(cy_1 + y_2) = g(x)</math>

<math>c(y_1' + p(x)y_1) + (y_2' + p(x)y_2) = g(x)</math>

Based on our assumptions about <math>y_1</math> and <math>y_2</math>, we have <math>y_1' + p(x)y_1 = 0</math> and we have <math>y_2' + p(x)y_2 = g(x)</math>, and so the above equation holds <math>\forall c \in \mathbb{R}</math>.

Hence, <math>\forall c \in \mathbb{R}</math>, <math>cy_1 + y_2</math> is a solution of <math>y'+p(x)y=g(x)</math>.

Solutions to HW1: [[User:Mathstudent|Mathstudent]]
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</gallery>

12-267/Homework Assignment 2

2012-10-24T20:58:13Z

Twine: Changed format of Mathstudent's solutions to a gallery

12-267/Homework Assignment 1

2012-10-24T20:50:53Z

Twine: Corrected some formatting, added solution for question 1

{{12-267/Navigation}}

This assignment is due at the tutorial on Tuesday September 25. Here and everywhere, '''neatness counts!!''' You may be brilliant and you may mean just the right things, but if the your readers will be having hard time deciphering your work they will give up and assume it is wrong.

'''Question 1.''' Show that if <math>y=y_1(x)</math> is a solution of <math>y'+p(x)y=0</math>, and <math>y=y_2(x)</math> is a solution of <math>y'+p(x)y=g(x)</math>, then for any constant <math>c</math>, <math>y=cy_1+y_2</math> is a solution of <math>y'+p(x)y=g(x)</math>.

'''Question 2.''' Solve the following differential equations
# For <math>x>0</math>, <math>xy'+2y=\sin x</math>.
# <math>\frac{dy}{dx}=\frac{1}{e^y-x}</math> with <math>y(1)=0</math>; you may want to solve for <math>x</math> first.
# <math>xy'=\sqrt{1-y^2}</math>.
# <math>\frac{dy}{dx}=\frac{x-e^{-x}}{y+e^y}</math>.
# <math>xdx+ye^{-x}dy=0</math>, with <math>y(0)=1</math>.
# <math>\frac{dy}{dx}=\frac{ay+b}{cx+d}</math>, where <math>a,b,c,d</math> are arbitrary constants.
# <math>\frac{dy}{dx}=-\frac{ax+by}{bx+cy}</math>, where <math>a,b,c</math> are arbitrary constants.
# <math>0=(e^x\sin y + 3y)dx + (3(x+y)+e^x\cos y)dy</math>.

Disclamer: The solutions below are by students, for students.

Solutions to HW1: [http://drorbn.net/index.php?title=Image:001.jpg page 1], [http://drorbn.net/index.php?title=Image:002.jpg page 2], [http://drorbn.net/index.php?title=Image:003.jpg page 3], [http://drorbn.net/index.php?title=Image:004.jpg page 4] [[User:Mathstudent|Mathstudent]]

'''Solution to Question 1.''' [[User:Twine|Twine]]

Take y defined by <math>y=cy_1+y_2</math> and plug it into <math>y'+p(x)y=g(x)</math>. We get

<math>(cy_1 + y_2)' + p(x)(cy_1 + y_2) = g(x)</math>

<math>c(y_1' + p(x)y_1) + (y_2' + p(x)y_2) = g(x)</math>

Based on our assumptions about <math>y_1</math> and <math>y_2</math>, we have <math>y_1' + p(x)y_1 = 0</math> and we have <math>y_2' + p(x)y_2 = g(x)</math>, and so the above equation holds <math>\forall c \in \mathbb{R}</math>.

Hence, <math>\forall c \in \mathbb{R}</math>, <math>cy_1 + y_2</math> is a solution of <math>y'+p(x)y=g(x)</math>.

12-267/Existence And Uniqueness Theorem

2012-10-24T20:32:19Z

Twine: Typo edits

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>.

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 <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>.

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 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:

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> = |\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>

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>

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]

12-267/Existence And Uniqueness Theorem

2012-10-24T19:34:21Z

Twine: Corrected typo in definition of lipschitz

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>.

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 <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>.

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 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:

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> = |\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>

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>

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]

User:Twine

2012-10-24T19:33:08Z

Twine: Corrected latex errors

Week 6, Lecture 3

Constant Coefficient Homogeneous High Order ODEs

Ex <math>Ly = a y'' + b y' + c y = 0</math>, <math>a \in \mathbb{R}, b \in \mathbb{R}, c \in \mathbb{R}</math>

Or generally <math>Ly = \sum_k=0^n a_k y^{(k)} = 0, a_k \in \mathbb{R}</math>

<math>L:\{f: \mathbb{R} \rightarrow \mathbb{R}\} \rightarrow \{f: \mathbb{R} \rightarrow \mathbb{R}\}</math> is a linear transformation ("linear operator").

What do we expect from <math>\{y: Ly = 0\} = ker(L)</math>? We expect an n-dimensional vector space.

Take <math> y''+y'-6y = 0</math>, guess <math> y = c, y' = \alpha e^{\alpha x}, y'' = \alpha^2e^{\alpha x}</math>

<math> \alpha^2 e^{\alpha x} + \alpha e^{\alpha x} - 6 e^{\alpha x} = 0</math>

<math> (\alpha^2 +\alpha - 6) e^{\alpha x} = 0</math>

<math> (\alpha +3)(\alpha - 2) = 0</math>

So we have <math> y = c_1e^{-3x} +c_2 e^{2x}</math> as the general solution.

Say we have complex <math>\alpha</math>. Then what?

User:Twine

2012-10-19T14:02:24Z

Twine: Notes from Oct 19, temporary placement

Week 6, Lecture 3

Constant Coefficient Homogeneous High Order ODEs

Ex <math>L(y) = ay" + by' +cy = 0, a, b, c \in \mathbb{R}</math>

Or generally <math>L(y) = \sum_k=0^n a_k y^{(k)} = 0, a_k \in \mathbb{R}</math>

<math>L:{functions on \mathbb{R} \rightarrow {functions on \mathbb{R}</math> is a linear transformation ("linear operator").

What do we expect from <math>{y:L(y) = 0} = ker(L)</math>? We expect an n-dimensional vector space.

Take <math> y"+y'-6y = 0</math>, guess <math> y = c, y' = \alpha e^{\alpha x}, y" = \alpha^2e^{\alpha x}</math>

<math> \alpha^2 e^{\alpha x} + \alpha e^{\alpha x} - 6 e^{\alpha x} = 0</math>

<math> (\alpha^2 +\alpha - 6) e^{\alpha x} = 0</math>

<math> (\alpha +3)(\alpha - 2) = 0</math>

So we have <math> y = c_1e^{-3x} +c_2 e^{2x}</math> as the general solution.

Say we have complex <math>\alpha</math>. Then what?

12-267

2012-10-15T15:43:53Z

Twine: corrected formatting error in previous edit

12-267

2012-10-15T15:43:26Z

Twine: Added link for Euler's method

12-267

2012-10-13T00:37:37Z

Twine: Added link to fundamental theorem proof

12-267/Existence And Uniqueness Theorem

2012-10-13T00:35:55Z

Twine: Corrected dates and links

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>.

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 <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>.

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 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:

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> = |\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>

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>

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]

12-267/Existence And Uniqueness Theorem

2012-10-13T00:33:32Z

Twine: Inserted proof of uniqueness

Disclamer: This is a student prepared note based on [http://drorbn.net/dbnvp/12-267-120928.php the lecure of Monday September 21st].

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.

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>.

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>.

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 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:

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> = |\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>

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>

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]

12-267/Existence And Uniqueness Theorem

2012-10-12T23:58:05Z

Twine: Added Claim 3 and finished existence proof

Disclamer: This is a student prepared note based on [http://drorbn.net/dbnvp/12-267-120928.php the lecure of Monday September 21st].

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.

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>.

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>.

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:

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> = |\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>

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>

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]

12-267/Existence And Uniqueness Theorem

2012-10-12T23:45:01Z

Twine: Added Claim 2 and proof.

12-267/Existence And Uniqueness Theorem

2012-10-12T23:32:55Z

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

12-267/Existence And Uniqueness Theorem

2012-10-12T23:20:41Z

Twine: Created page with statement of theorem, proofs to follow

12-267/Derivation of Euler-Lagrange

2012-10-12T23:07:00Z

Twine: corrected date of lecture

Disclamer: This is a student prepared note based on [http://drorbn.net/dbnvp/12-267-121002-2.php the lecure of Tuesday October 2nd].

For a function <math>y(x)</math> defined on <math>[a, b]</math> to be an extremum of <math>J(y) = \int_a^b F(x, y, y') dx</math>, it must be that for any function <math>h(x)</math> defined on <math>[a, b]</math> that preserves the endpoints of <math>y</math> (that is, <math>h(a) = 0</math> and <math>h(b) = 0</math>), we have <math> \frac{d}{d \epsilon } J(y + \epsilon h) </math><math>|_{\epsilon = 0} = 0 </math>.

<math> \frac{d}{d \epsilon } J(y + \epsilon h) </math><math>|_{\epsilon = 0} </math>

<math> = \frac{d}{d \epsilon } \int_a^b F(x, y + \epsilon h, y' + \epsilon h') dx |_{\epsilon = 0} </math>

Let <math>F_n</math> signify F differentiated with respect to its nth variable.

<math> = \int_a^b (F_1 \cdot 0 + F_2 \cdot h + F_3 \cdot h') dx |_{\epsilon = 0} </math>

<math> = \int_a^b (F_2(x, y, y') \cdot h + F_3(x, y, y') \cdot h') dx </math>

<math> = \int_a^b (F_2 \cdot h + [\frac{d}{dx} F_3] \cdot h) dx + F_3 \cdot h |_a^b</math> (integrating by parts)

Due to the constraints of <math>h(a) = 0</math> and <math>h(b) = 0</math>, <math>F_3 \cdot h |_a^b = 0</math>.

<math> = \int_a^b (F_2 - \frac{d}{dx} F_3) \cdot h dx</math>

As this must be equal to 0 for all h satisfying the endpoint constraints, we must have that <math>F_2 = \frac{d}{dx} F_3</math>, or in other terms, <math>F_y - \frac{d}{dx} F_y' = 0</math>.

12-267

2012-10-12T23:03:10Z

Twine:

12-267/Derivation of Euler-Lagrange

2012-10-12T23:01:07Z

Twine: Created page

Disclamer: This is a student prepared note based on [http://drorbn.net/dbnvp/12-267-121002-2.php the lecure of Tuesday October 1st].

For a function <math>y(x)</math> defined on <math>[a, b]</math> to be an extremum of <math>J(y) = \int_a^b F(x, y, y') dx</math>, it must be that for any function <math>h(x)</math> defined on <math>[a, b]</math> that preserves the endpoints of <math>y</math> (that is, <math>h(a) = 0</math> and <math>h(b) = 0</math>), we have <math> \frac{d}{d \epsilon } J(y + \epsilon h) </math><math>|_{\epsilon = 0} = 0 </math>.

<math> \frac{d}{d \epsilon } J(y + \epsilon h) </math><math>|_{\epsilon = 0} </math>

<math> = \frac{d}{d \epsilon } \int_a^b F(x, y + \epsilon h, y' + \epsilon h') dx |_{\epsilon = 0} </math>

Let <math>F_n</math> signify F differentiated with respect to its nth variable.

<math> = \int_a^b (F_1 \cdot 0 + F_2 \cdot h + F_3 \cdot h') dx |_{\epsilon = 0} </math>

<math> = \int_a^b (F_2(x, y, y') \cdot h + F_3(x, y, y') \cdot h') dx </math>

<math> = \int_a^b (F_2 \cdot h + [\frac{d}{dx} F_3] \cdot h) dx + F_3 \cdot h |_a^b</math> (integrating by parts)

Due to the constraints of <math>h(a) = 0</math> and <math>h(b) = 0</math>, <math>F_3 \cdot h |_a^b = 0</math>.

<math> = \int_a^b (F_2 - \frac{d}{dx} F_3) \cdot h dx</math>

As this must be equal to 0 for all h satisfying the endpoint constraints, we must have that <math>F_2 = \frac{d}{dx} F_3</math>, or in other terms, <math>F_y - \frac{d}{dx} F_y' = 0</math>.

12-267/Tuesday September 11 Notes

2012-10-04T22:20:10Z

Twine: Corrected misspelling of Brachistochrone

{{12-267/Navigation}}

===Solving the complicated integral in the Brachistochrone integral===

<math>\int \sqrt{\frac{d-y}{y}} dy </math>

<math> = \int \sqrt{\frac{d y-y^2}{y}} dy </math>

Complete the square in the integrand:

<math> = \int \sqrt{\frac{\frac{d^2}{4} - (y - \frac{d}{2})^2}{y}} </math>

Substitute <math>u = y-\frac{d}{2}</math> and <math> du = dy </math>:

<math> = \int 2 \sqrt{\frac{\frac{d^2}{4} - u^2}{d+2 u}} du </math>

Assuming all variables are positive, substitute <math>u = \frac{1}{2} d \sin{s}</math> and <math>du = \frac{1}{2} d \cos{s} ds</math>. Then <math>\sqrt{\frac{d^2}{4} - u^2} = \sqrt{\frac{d^2}{4} - \frac{1}{4} d^2 \sin^2{s}} = \frac{1}{2} d \cos{s}</math> and <math>s = \sin^{-1}{\frac{2u}{d}}</math>:

<math> = \frac{d^2}{2} \int \frac{\cos^2{s}}{d \sin{s} + d} ds</math>

For the integrand substitute <math> p=\tan{\frac{s}{2}}</math> and <math>dp = \frac{1}{2} \sec^2{\frac{s}{2}} ds</math>. Then transform the integrand using the substitutions <math>\sin{s} = \frac{2p}{p^2 + 1}</math>, <math>\cos{s} = \frac{1-p^2}{p^2 + 1}</math> and <math>ds = \frac{2 dp}{p^2 + 1}</math>:

<math> = \frac{d^2}{2} \int 2 \frac{(1-p^2)^2}{(p^2 + 1)^3 (\frac{2 d p}{p^2 + 1} + d)} dp</math>

Simplify the integrand <math>\frac{2(1-p^2)^2}{(p^2 + 1)^3 (\frac{2 d p}{p^2 + 1} + d)}</math> to get <math>\frac{2 (p-1)^2}{d p^4 + 2 d p^2 + d}</math>:

<math> = \frac{d^2}{2} \int \frac{2(p-1)^2}{d p^4+2 d p^2+d} dp </math>

<math> = d^2 \int \frac{(p-1)^2}{d p^4+2 d p^2+d} dp </math>

<math> = d^2 \int \frac{(p-1)^2}{d (p^2+1)^2} dp </math>

<math> = d \int \frac{(p-1)^2}{(p^2+1)^2} dp </math>

For the integrand <math>\frac{(p-1)^2}{(p^2+1)^2}</math> use partial fractions:

<math> = d \int (\frac{1}{p^2+1}-\frac{2 p}{(p^2+1)^2}) dp</math>

<math> = d \int \frac{1}{(p^2+1)} dp - 2 d \int \frac{p}{(p^2+1)^2} dp</math>

For the integrand <math>\frac{p}{(p^2+1)^2}</math>, substitute <math>w = p^2+1</math> and <math>dw = 2 p dp</math>:

<math> = d \int \frac{1}{p^2+1} dp - d \int \frac{1}{w^2} dw </math>

The integral of <math>\frac{1}{p^2+1}</math> is <math>\tan^{-1}{p}</math>:

<math> = d \tan^{-1}{p}-d \int \frac{1}{w^2} dw</math>

<math> = d \tan^{-1}{p}+ \frac{d}{w}+constant</math>

Substitute back for <math>w = p^2+1</math>:

<math> = \frac{d ((p^2+1) tan^{-1}{p}+1)}{p^2+1}+C </math>

Substitute back for <math>p = \tan{\frac{s}{2}}</math>:

<math> = \frac{1}{2} d (\cos{s}+2 \tan^{-1}{\tan{\frac{s}{2}}}+1)+C </math>

Substitute back for <math>s = \sin^{-1}{\frac{2 u}{d}}</math>:

<math> = 1/2 (\sqrt{d^2-4 u^2}+2 d \tan^{-1}(\frac{2 u}{d \sqrt{1-\frac{4 u^2}{d^2}}+1)})+d)+C </math>

Substitute back for <math>u = y-\frac{d}{2}</math>:

<math> = d (-\tan^{-1}{\frac{d-2 y}{2 d \sqrt{\frac{y (d-y)}{d^2}}+d}})+\sqrt{y (d-y)}+\frac{d}{2}+C</math>

Factor the answer a different way:

<math> = \frac{1}{2} (-2 d \tan^{-1}{\frac{d-2 y}{2 d \sqrt{\frac{y (d-y)}{d^2}}+d}}+2 \sqrt{y (d-y)}+d)+C</math>

Which is equivalent for restricted y and d values to:

<math> = y \sqrt{\frac{d}{y}-1}-\frac{1}{2} d \tan^{-1}{\frac{\sqrt{\frac{d}{y}-1} (d-2 y)}{2 (d-y)}}+C</math>

[[User:Syjytg|Syjytg]] 23:00, 11 September 2012 (EDT)

12-267/Tuesday September 11 Notes

2012-10-04T22:19:41Z

Twine: Converted math to latex notation (verbatum)

{{12-267/Navigation}}

===Solving the complicated integral in the Brachistochroe integral===

<math>\int \sqrt{\frac{d-y}{y}} dy </math>

<math> = \int \sqrt{\frac{d y-y^2}{y}} dy </math>

Complete the square in the integrand:

<math> = \int \sqrt{\frac{\frac{d^2}{4} - (y - \frac{d}{2})^2}{y}} </math>

Substitute <math>u = y-\frac{d}{2}</math> and <math> du = dy </math>:

<math> = \int 2 \sqrt{\frac{\frac{d^2}{4} - u^2}{d+2 u}} du </math>

Assuming all variables are positive, substitute <math>u = \frac{1}{2} d \sin{s}</math> and <math>du = \frac{1}{2} d \cos{s} ds</math>. Then <math>\sqrt{\frac{d^2}{4} - u^2} = \sqrt{\frac{d^2}{4} - \frac{1}{4} d^2 \sin^2{s}} = \frac{1}{2} d \cos{s}</math> and <math>s = \sin^{-1}{\frac{2u}{d}}</math>:

<math> = \frac{d^2}{2} \int \frac{\cos^2{s}}{d \sin{s} + d} ds</math>

For the integrand substitute <math> p=\tan{\frac{s}{2}}</math> and <math>dp = \frac{1}{2} \sec^2{\frac{s}{2}} ds</math>. Then transform the integrand using the substitutions <math>\sin{s} = \frac{2p}{p^2 + 1}</math>, <math>\cos{s} = \frac{1-p^2}{p^2 + 1}</math> and <math>ds = \frac{2 dp}{p^2 + 1}</math>:

<math> = \frac{d^2}{2} \int 2 \frac{(1-p^2)^2}{(p^2 + 1)^3 (\frac{2 d p}{p^2 + 1} + d)} dp</math>

Simplify the integrand <math>\frac{2(1-p^2)^2}{(p^2 + 1)^3 (\frac{2 d p}{p^2 + 1} + d)}</math> to get <math>\frac{2 (p-1)^2}{d p^4 + 2 d p^2 + d}</math>:

<math> = \frac{d^2}{2} \int \frac{2(p-1)^2}{d p^4+2 d p^2+d} dp </math>

<math> = d^2 \int \frac{(p-1)^2}{d p^4+2 d p^2+d} dp </math>

<math> = d^2 \int \frac{(p-1)^2}{d (p^2+1)^2} dp </math>

<math> = d \int \frac{(p-1)^2}{(p^2+1)^2} dp </math>

For the integrand <math>\frac{(p-1)^2}{(p^2+1)^2}</math> use partial fractions:

<math> = d \int (\frac{1}{p^2+1}-\frac{2 p}{(p^2+1)^2}) dp</math>

<math> = d \int \frac{1}{(p^2+1)} dp - 2 d \int \frac{p}{(p^2+1)^2} dp</math>

For the integrand <math>\frac{p}{(p^2+1)^2}</math>, substitute <math>w = p^2+1</math> and <math>dw = 2 p dp</math>:

<math> = d \int \frac{1}{p^2+1} dp - d \int \frac{1}{w^2} dw </math>

The integral of <math>\frac{1}{p^2+1}</math> is <math>\tan^{-1}{p}</math>:

<math> = d \tan^{-1}{p}-d \int \frac{1}{w^2} dw</math>

<math> = d \tan^{-1}{p}+ \frac{d}{w}+constant</math>

Substitute back for <math>w = p^2+1</math>:

<math> = \frac{d ((p^2+1) tan^{-1}{p}+1)}{p^2+1}+C </math>

Substitute back for <math>p = \tan{\frac{s}{2}}</math>:

<math> = \frac{1}{2} d (\cos{s}+2 \tan^{-1}{\tan{\frac{s}{2}}}+1)+C </math>

Substitute back for <math>s = \sin^{-1}{\frac{2 u}{d}}</math>:

<math> = 1/2 (\sqrt{d^2-4 u^2}+2 d \tan^{-1}(\frac{2 u}{d \sqrt{1-\frac{4 u^2}{d^2}}+1)})+d)+C </math>

Substitute back for <math>u = y-\frac{d}{2}</math>:

<math> = d (-\tan^{-1}{\frac{d-2 y}{2 d \sqrt{\frac{y (d-y)}{d^2}}+d}})+\sqrt{y (d-y)}+\frac{d}{2}+C</math>

Factor the answer a different way:

<math> = \frac{1}{2} (-2 d \tan^{-1}{\frac{d-2 y}{2 d \sqrt{\frac{y (d-y)}{d^2}}+d}}+2 \sqrt{y (d-y)}+d)+C</math>

Which is equivalent for restricted y and d values to:

<math> = y \sqrt{\frac{d}{y}-1}-\frac{1}{2} d \tan^{-1}{\frac{\sqrt{\frac{d}{y}-1} (d-2 y)}{2 (d-y)}}+C</math>

[[User:Syjytg|Syjytg]] 23:00, 11 September 2012 (EDT)

12-267/Tuesday September 11 Notes

2012-10-04T21:05:30Z

Twine: Minor formatting edit

{{12-267/Navigation}}

===Solving the complicated integral in the Brachistochroe integral===

integral sqrt((d-y)/y) dy
= integral sqrt(d y-y^2)/y dy
For the integrand sqrt(d y-y^2)/y, complete the square:
= integral sqrt(d^2/4-(y-d/2)^2)/y dy
For the integrand sqrt(d^2/4-(y-d/2)^2)/y, substitute u = y-d/2 and du = dy:
= integral (2 sqrt(d^2/4-u^2))/(d+2 u) du
= 2 integral sqrt(d^2/4-u^2)/(d+2 u) du
For the integrand sqrt(d^2/4-u^2)/(d+2 u), (assuming all variables are positive) substitute u = 1/2 d sin(s) and du = 1/2 d cos(s) ds. Then sqrt(d^2/4-u^2) = sqrt(d^2/4-1/4 d^2 sin^2(s)) = 1/2 d cos(s) and s = sin^(-1)((2 u)/d):
= d^2/2 integral (cos^2(s))/(d sin(s)+d) ds
For the integrand (cos^2(s))/(d sin(s)+d), substitute p = tan(s/2) and dp = 1/2 sec^2(s/2) ds. Then transform the integrand using the substitutions sin(s) = (2 p)/(p^2+1), cos(s) = (1-p^2)/(p^2+1) and ds = (2 dp)/(p^2+1):
= d^2/2 integral (2 (1-p^2)^2)/((p^2+1)^3 ((2 d p)/(p^2+1)+d)) dp
Simplify the integrand (2 (1-p^2)^2)/((p^2+1)^3 ((2 d p)/(p^2+1)+d)) to get (2 (p-1)^2)/(d p^4+2 d p^2+d):
= d^2/2 integral (2 (p-1)^2)/(d p^4+2 d p^2+d) dp
= d^2 integral (p-1)^2/(d p^4+2 d p^2+d) dp
= d^2 integral (p-1)^2/(d (p^2+1)^2) dp
= d integral (p-1)^2/(p^2+1)^2 dp
For the integrand (p-1)^2/(p^2+1)^2, use partial fractions:
= d integral (1/(p^2+1)-(2 p)/(p^2+1)^2) dp
= d integral 1/(p^2+1) dp-2 d integral p/(p^2+1)^2 dp
For the integrand p/(p^2+1)^2, substitute w = p^2+1 and dw = 2 p dp:
= d integral 1/(p^2+1) dp-d integral 1/w^2 dw
The integral of 1/(p^2+1) is tan^(-1)(p):
= d tan^(-1)(p)-d integral 1/w^2 dw
= d tan^(-1)(p)+d/w+constant
Substitute back for w = p^2+1:
= (d ((p^2+1) tan^(-1)(p)+1))/(p^2+1)+C
Substitute back for p = tan(s/2):
= 1/2 d (cos(s)+2 tan^(-1)(tan(s/2))+1)+C
Substitute back for s = sin^(-1)((2 u)/d):
= 1/2 (sqrt(d^2-4 u^2)+2 d tan^(-1)((2 u)/(d (sqrt(1-(4 u^2)/d^2)+1)))+d)+C
Substitute back for u = y-d/2:
= d (-tan^(-1)((d-2 y)/(2 d sqrt((y (d-y))/d^2)+d)))+sqrt(y (d-y))+d/2+C
Factor the answer a different way:
= 1/2 (-2 d tan^(-1)((d-2 y)/(2 d sqrt((y (d-y))/d^2)+d))+2 sqrt(y (d-y))+d)+C
Which is equivalent for restricted y and d values to:
= y sqrt(d/y-1)-1/2 d tan^(-1)((sqrt(d/y-1) (d-2 y))/(2 (d-y)))+C
[[User:Syjytg|Syjytg]] 23:00, 11 September 2012 (EDT)