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2250-4 12:25pm Lecture Record Week 10 F2010

Last Modified: November 09, 2010, 21:20 MST.    Today: October 19, 2017, 21:29 MDT.

Week 10, Nov 1 to 5: Sections 10.2, 10.3, 10.4, 10.5, EPbvp7.6

1,2 and 4 Nov: Laura and Michal

Exam 2, problems 4,5 at 12:50pm in 335 JWB on Monday, 1 Nov or Wednesday 3 Nov. Also possible is 7:25am in WEB 103 on Thursday, 4 Nov. Study the S2010 exam solution key for similar problems.

2 Nov: Tables, Rules of Laplace's Calculus 10.2, 10.3.

  Partly solved 5.4-34.
    The DE is 3.125 x'' + cx' + kx=0. The characteristic equation
    is 3.125r^2 + cr + kr=0 which factors into 3.125(r-a-ib)(r-a+ib)=0
    having complex roots a+ib, a-ib. Problems 32, 33 find the numbers
    a, b from the given information. This is an inverse problem, one
    in which experimental data is used to discover the differential
    equation model. The book uses its own notation for the symbols
    a,b: a ==> -p and b ==> omega1. Because the two roots a+ib, a-ib
    determine the quadratic equation, then c and k are known in terms
    of symbols a,b. See also the web site FAQ for more details.
  Partly solved 5.4-20
   The problem breaks into two distinct initial value problems:
    (1)   2x'' + 16x' + 40x=0, x(0)=5, x'(0)=4
           Characteristic equation  2(r^2+8r+20)=0. Roots r=-4+2i,r=-4-2i.
           Solution Atoms=e^{-4t}cos 2t, e^{-4t}sin 2t. Underdamped.

    (2)   2x'' + 0x' + 40x=0, x(0)=5, x'(0)=4
           Characteristic equation 2(r^2+0+20)=0. Roots r=sqrt(20)i,r=-sqrt(20)i.
           Solution Atoms=cos( sqrt(20)t), sin( sqrt(20)t).
   Each system has general solution a linear combination of the solution atoms.
   Evaluate the constants in the linear combination, in each of the two
   cases, using the initial conditions x(0)=5, x'(0)=4. There are two linear algebra
   problems to solve.
   Answers: (1)  Coefficients 5, 2  for 2x'' + 16x' + 40x=0
                      Amplitude sqrt(5^2 + 12^2) = 13
                (2)  Coefficients 5, 2/sqrt(5) for 2x'' + 0x' + 40x=0
                      Amplitude sqrt(5^2 + 4/5) = sqrt(129/5)
   Plots can be made from these answers directly.
   Write each solution in phase-amplitude form, a trig problem. See section
   5.4 for specific instructions. The book's answers:
     (1) tan(alpha) = 5/12   (2) tan(alpha) = 5 sqrt(5)/2
Laplace Theory
 A brief Laplace table.
   Forward table.
   Backward table.
   Extensions of the Table.
 Laplace rules.
   The s-differentiation theorem (d/ds)L(f(t))=L((-t)f(t)).
   Shift theorem.
   Parts theorem.
 Finding Laplace integrals using Laplace calculus.
 Solving differential equations by Laplace's method.
 Basic Theorems of Laplace Theory
   Lerch's theorem
   The s-differentiation theorem (d/ds)L(f(t))=L((-t)f(t)).
   Shift theorem L(exp(at)f(t)) = L(f(t))|s->(s-a)
   Parts theorem L(y')=sL(y)-y(0)

2 Nov: Applications of Laplace's method from 10.3, 10.4, 10.5.

 Solving y' = -1, y(0)=2
 Solving y''+y=0, y(0)=0, y'(0)=1
 Solving y''+y=1, y(0)=y'(0)=0
 Solving y''+y=cos(t), y(0)=y'(0)=0
 Computing Laplace integrals
 Solving an equation L(y(t))=expression in s for y(t)
 Dealing with complex roots and quadratic factors
 Partial fraction methods
 Using the s-differentiation theorem
 Using the shifting theorem
 Harmonic oscillator y''+a^2 y=0

4 Nov: Mechanical oscillators. Resonance. Beats. Sections 10.4, 10.5

Basic Theorems of Laplace Theory
   Periodic function theorem
     Some engineering functions
     unit step
     sawtooth wave
     other periodic waves next Monday
   Convolution theorem application
     L(cos t)L(sin t) = L(0.5 t sin(t))
   How to solve differential equations
   LRC Circuit
   Second shifting rule
   Specialized models.
     Pure Resonance x''+x=cos(t)
       Solution explosion, unbounded solution x=(1/2) t sin t.
       Resonance examples: Soldiers marching in cadence, Tacoma narrows bridge,
              Wine Glass Experiment. Theodore Von Karman and vortex shedding.
              Cable model of the Tacoma bridge, year 2000. Resonance explanations.
     Beats x''+x=cos(2t)
        Graphics for beats [x=sin(10 t)sin(t/2)], slowly-oscillating envelope,
        rapidly oscillating harmonic with time-varying amplitude.

4 Nov: Piecewise Functions. Shifting. Section 10.5 and EPbvp supplement 7.6.

 Piecewise Functions
   Unit Step
 Second shifting Theorems
 Piecewise defined periodic waves
   Square wave
   Triangular wave
   Rectified sine
   Half-wave rectified sine
   Parabolic wave
 Periodic function theorem
    Laplace of the square wave, tanh function.
 Laplace Resolvent Method.
  --> This method is considered a shortcut for systems by Laplace's method.
  --> It is also a convenient way to solve systems with maple.
  Consider problem 10.2-16
       x'=x+z, y'=x+y, z'=-2x-z, x(0)=1, y(0)=0, z(0)=0
  Write this as a matrix differential equation
       u'=Bu, u(0)=u0
    B:=matrix([[1,0,1],[1,1,0],[-2,0,-1]]); u0:=vector([1,0,0]);
  If we think of the matrix differential equation as a scalar equation, then
  its Laplace model is
      -u(0) + s L(u(t)) = BL(u(t))
  or equivalently
      sL(u(t)) - B L(u(t)) = u0
 Write s = sI where I is the 3x3 identity matrix. Then the Laplace model is
      (aI - B) L(u(t)) = u0
 which is called the Resolvent Equation.
 The RESOLVENT is the inverse of the matrix multiplier on the left:
    Resolvent == inverse(sI - B)
 It is so-named because the vector of Laplace answers is
   vector([L(x),L(y),L(z)]) = L(u(t)) = inverse(sI - B) u0
    Laplace of u(t) = RESOLVENT x u(0)
Intro to the Laplace resolvent method for 2x2 systems
  Remarks on problem 10.2-16, a 3x3 system that can be solved with
     the resolvent equation shortcut.
   Problem: Write a 2x2 dynamical system as a vector-matrix equation u'=Au.
   Problem: Solve a 2x2 dynamical system in vector-matrix form u'=Au.
   The general vector-matrix DE Model u'=Au
      Laplace of u(t) = Resolvent x u(0)
      Resolvent = inverse(sI - A)
   Solve the systems by ch1 methods for x(t), y(t):
     x' = 2x, x(0)=100,
     y' = 3y, y(0)=50.
       Answer: x = 100 exp(2t), y = 50 exp(3t)
     x' = 2x+y, x(0)=1,
     y' = 3y, y(0)=2.
       Answer: y(t) = 2 exp(3t) and x(t) is the solution of the linear
               integrating factor problem x'(t)=2x(t)+2 exp(3t).
    RULE: Use Cramer's rule or matrix inversion to solve the resolvent
         equation for the vector of components L(x), L(y), L(z). Any
         linear algebra problem Bu=c where B contains symbols should
         be solved this way, unless B is triangular.
    Laplace theory references
    Slides: Laplace and Newton calculus. Photos. (145.3 K, pdf, 01 Nov 2009)
    Slides: Intro to Laplace theory. Calculus assumed. (109.5 K, pdf, 01 Nov 2009)
    Slides: Laplace rules (112.2 K, pdf, 01 Nov 2009)
    Slides: Laplace table proofs (130.3 K, pdf, 01 Nov 2009)
    Slides: Laplace examples (101.2 K, pdf, 07 Nov 2009)
    Slides: Piecewise functions and Laplace theory (64.7 K, pdf, 01 Nov 2009)
    MAPLE: Maple Lab 7. Laplace applications (155.7 K, pdf, 27 Nov 2010)
    Manuscript: DE systems, examples, theory (785.8 K, pdf, 16 Nov 2008)
    Slides: Laplace resolvent method (56.4 K, pdf, 01 Nov 2009)
    Slides: Laplace second order systems (248.9 K, pdf, 01 Nov 2009)
    Slides: Home heating, attic, main floor, basement (73.8 K, pdf, 30 Nov 2009)
    Slides: Cable hoist example (73.2 K, pdf, 21 Aug 2008)
    Slides: Sliding plates example (105.8 K, pdf, 21 Aug 2008)
    Manuscript: Heaviside's method 2008 (186.8 K, pdf, 20 Oct 2009)
    Manuscript: Laplace theory 2008 (350.5 K, pdf, 06 Mar 2009)
    Transparencies: Ch10 Laplace solutions 10.1 to 10.4 (1068.7 K, pdf, 28 Nov 2010)
    Text: Laplace theory problem notes F2008 (8.9 K, txt, 18 Nov 2010)
    Text: Final exam study guide (8.3 K, txt, 09 Dec 2010)