12-267/Tuesday September 11 Notes

DBN: Classes: 2012-13: 12-267 / Navigation Additions to this web site no longer count towards good deed points. # Week of... Notes and Links 1 Sep 10 About This Class.  Monday: Introduction and the Brachistochrone.  Tuesday: More on the Brachistochrone, administrative issues. Tuesday Notes.  Friday: Some basic techniques: first order linear equations. 2 Sep 17  Monday: Separated equations, escape velocities. HW1.  Tuesday: Escape velocities, changing source and target coordinates, homogeneous equations.  Friday: Reverse engineering separated and exact equations. 3 Sep 24  Monday: Solving exact equations, integration factors. HW2.  Tuesday: Statement of the Fundamental Theorem. Class Photo.  Friday: Proof of the Fundamental Theorem. 4 Oct 1  Monday: Last notes on the fundamental theorem. HW3.  Tuesday Hour 1: The chain law, examples of variational problems.  Tuesday Hour 2: Deriving Euler-Lagrange.  Friday: Reductions of Euler-Lagrange. 5 Oct 8 Monday is thanksgiving.  Tuesday: Lagrange multiplyers and the isoperimetric inequality. HW4.  Friday: More Lagrange multipliers, numerical methods. 6 Oct 15  Monday: Euler and improved Euler.  Tuesday: Evaluating the local error, Runge-Kutta, and a comparison of methods.  Friday: Numerical integration, high order constant coefficient homogeneous linear ODEs. 7 Oct 22  Monday: Multiple roots, reduction of order, undetermined coefficients.  Tuesday: From systems to matrix exponentiation. HW5. Term Test on Friday. 8 Oct 29  Monday: The basic properties of matrix exponentiation.  Tuesday: Matrix exponentiation: examples.  Friday: Phase Portraits. HW6. Nov 4 was the last day to drop this class 9 Nov 5  Monday: Non-homogeneous systems.  Tuesday: The Catalan numbers, power series, and ODEs.  Friday: Global existence for linear ODEs, the Wronskian. 10 Nov 12 Monday-Tuesday is UofT November break. HW7.  Friday: Series solutions for [math]\displaystyle{ y'=f(x,y) }[/math]. 11 Nov 19  Monday: [math]\displaystyle{ \pi }[/math] is irrational, more on the radius of convergence.  Tuesday (class): Airy's equation, Fuchs' theorem.  Tuesday (tutorial): Regular singular points. HW8.  Friday: Discussion of regular singular points.. 12 Nov 26  Monday: Frobenius series by computer. Qualitative Analysis Handout (PDF).  Tuesday: The basic oscillation theorem. Handout on the Frobenius Method. HW9.  Friday: Non-oscillation, Sturm comparison. 13 Dec 3  Monday: More Sturm comparisons, changing the independent variable.  Tuesday: Amplitudes of oscillations. Last class was on Tuesday! F1 Dec 10 F2 Dec 17 The Final Exam (time, place, style, office hours times) Register of Good Deeds Add your name / see who's in!

Solving the complicated integral in the Brachistochrone integral

[math]\displaystyle{ \int \sqrt{\frac{d-y}{y}} dy }[/math]

[math]\displaystyle{ = \int \sqrt{\frac{d y-y^2}{y}} dy }[/math]

Complete the square in the integrand:

[math]\displaystyle{ = \int \sqrt{\frac{\frac{d^2}{4} - (y - \frac{d}{2})^2}{y}} }[/math]

Substitute [math]\displaystyle{ u = y-\frac{d}{2} }[/math] and [math]\displaystyle{ du = dy }[/math]:

[math]\displaystyle{ = \int 2 \sqrt{\frac{\frac{d^2}{4} - u^2}{d+2 u}} du }[/math]

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

[math]\displaystyle{ = \frac{d^2}{2} \int \frac{\cos^2{s}}{d \sin{s} + d} ds }[/math]

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

[math]\displaystyle{ = \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]\displaystyle{ \frac{2(1-p^2)^2}{(p^2 + 1)^3 (\frac{2 d p}{p^2 + 1} + d)} }[/math] to get [math]\displaystyle{ \frac{2 (p-1)^2}{d p^4 + 2 d p^2 + d} }[/math]:

[math]\displaystyle{ = \frac{d^2}{2} \int \frac{2(p-1)^2}{d p^4+2 d p^2+d} dp }[/math]

[math]\displaystyle{ = d^2 \int \frac{(p-1)^2}{d p^4+2 d p^2+d} dp }[/math]

[math]\displaystyle{ = d^2 \int \frac{(p-1)^2}{d (p^2+1)^2} dp }[/math]

[math]\displaystyle{ = d \int \frac{(p-1)^2}{(p^2+1)^2} dp }[/math]

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

[math]\displaystyle{ = d \int (\frac{1}{p^2+1}-\frac{2 p}{(p^2+1)^2}) dp }[/math]

[math]\displaystyle{ = d \int \frac{1}{(p^2+1)} dp - 2 d \int \frac{p}{(p^2+1)^2} dp }[/math]

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

[math]\displaystyle{ = d \int \frac{1}{p^2+1} dp - d \int \frac{1}{w^2} dw }[/math]

The integral of [math]\displaystyle{ \frac{1}{p^2+1} }[/math] is [math]\displaystyle{ \tan^{-1}{p} }[/math]:

[math]\displaystyle{ = d \tan^{-1}{p}-d \int \frac{1}{w^2} dw }[/math]

[math]\displaystyle{ = d \tan^{-1}{p}+ \frac{d}{w}+constant }[/math]

Substitute back for [math]\displaystyle{ w = p^2+1 }[/math]:

[math]\displaystyle{ = \frac{d ((p^2+1) tan^{-1}{p}+1)}{p^2+1}+C }[/math]

Substitute back for [math]\displaystyle{ p = \tan{\frac{s}{2}} }[/math]:

[math]\displaystyle{ = \frac{1}{2} d (\cos{s}+2 \tan^{-1}{\tan{\frac{s}{2}}}+1)+C }[/math]

Substitute back for [math]\displaystyle{ s = \sin^{-1}{\frac{2 u}{d}} }[/math]:

[math]\displaystyle{ = 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]\displaystyle{ u = y-\frac{d}{2} }[/math]:

[math]\displaystyle{ = 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]\displaystyle{ = \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]\displaystyle{ = 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]

Syjytg 23:00, 11 September 2012 (EDT)