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2250-1 7:30am Lecture Record Week 12 S2012

Last Modified: April 04, 2012, 05:01 MDT.    Today: October 22, 2017, 07:57 MDT.

Week 12, Apr 2 to 6: Sections 5.5, 5.6, EPbvp3.7, 7.1, 7.2, 7.3

Apr 2 and 3: Sections 5.5, 5.6

Exam 3 Review
Patrick will do exam 3 review this week on Thursday, JTB 140 7:30 to 9:25am.

Exam 3 part I: April 12
The problems are 1,2 on the sample exam below, which matches 3,4 on the S2010 exam key.
See the Fall 2010 exam 3 solution key for another sample exam, in which the problem
numbering is the same as your exam.

Exam 3 part II: April 19
The Thursday exam is at the usual 7:30am time in JTB 140.
The problems are 3,4,5 on the sample exam below, which matches 1,2,5 on the S2010 exam key.
    Sample Exam 3 for S2012
    PDF: sample exam 3, all problems. (182.4 K, pdf, 14 Nov 2010)
    HTML: All sample exams and solution keys. (18.9 K, html, 02 May 2012)

Mon, Apr 2: Variation of parameters. Sections 5.5, 10.4

 More Laplace Examples
   Continuing 10.3, 10.5 examples from last lecture.
 Review of the Laplace resolvent method for 2x2 systems
   Problem: Write a 2x2 dynamical system as a vector-matrix equation u'=Au.
            x' = 2x + y, y' = 3y ==> u'=Au
   Problem: Solve a 2x2 dynamical system in vector-matrix form u'=Au.
            u'=Au where A=<2,0|1,3> and u(0)=<1,2>
   The general vector-matrix DE Model u'=Au
      BASIC RESULT: Laplace of u(t) = Resolvent matrix times vector u(0)
      DEFINITION: Resolvent matrix = 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).
   Review of problem 10.2-16, a 3x3 system that can be solved with
     the resolvent equation shortcut.
   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.
 

 Variation of parameters
   The second order formula.
   Application to y''=1+x
   Application to y''+y=sec(x) [see also slides]
   How to calculate y_p(x) from the five parameters
    y1(x)
    y2(x)
    W(x) = y1(x)y2'(x)-y1'(x)y2(x)
    A(x) = coefficient in the DE of y''
    f(x) = input or forcing term, the RHS of the DE
   See (33) in section 5.5.
    Undetermined Coefficients
   Which equations can be solved
   Intro to the basic trial solution method
      Laplace solution of y'' + y = 1+x
         [use x''(t)+x(t) = 1+t, x(0)=x'(0)=0]
      How to find the atoms in y_p(x).
      How to find the atoms in y_h(x)
   THEOREM. Solution y_h(x) is a linear combination of atoms.
   THEOREM. Solution y_p(x) is a linear combination of atoms.
   THEOREM. (superposition)  y = y_h + y_p
   EXAMPLE. How to find a shortest expression for y_p(x) using
            Laplace's method.
            Details for x''(t)+x(t) = 1+t, to
            obtain the trial solution x(t)=A+Bt
            and the answer x_p(t)=1+t.
   BASIC METHOD. Given a trial solution with undetermined coefficients,
                 find a system of equations for d1, d2, ... and solve it.
                 Report y_p as the trial solution with substituted
                 answers d1, d2, d3, ...
   METHODS to FIND the SOLUTION.
     Laplace solution of y''(t) + y(t) = f(t) when
       f(t)=linear combination of atoms
       We get L(y) = polynomial fraction. By partial fractions and
       the backward table, L(y)=L(g(t)), where g(t)=linear combination
       of atoms! This proves the general result:

  THEORY. y = y_h + y_p, and each is a linear combination of atoms.

      How to find the homogeneous solution y_h(x) from the characteristic equation.
      How to determine the form of the shortest trial solution for y_p(x)
         METHOD 1. Laplace theory. It works, but it is slow. Central to this
                   method is dividing L(f(t)) by the characteristic polynomial,
                   then writing this polynomial fraction as L(g(t)). The homogeneous
                   solution atoms are shrunk from g(t) to obtain the trial solution.
         METHOD 2. A rule for finding y_p(x) from f(x) and the DE.
             We get rid of Laplace theory's t-variable and use x.
             Finding a trial solution with fewest symbols.
             Rule I. Assume the right side f(x) of the differential equation is a
             linear combination of atoms. Make a list of all distinct atoms that appear in
             the derivatives f(x), f'(x), f''(x), ... . Multiply these k atoms by
             undetermined coefficients d_1, ... , d_k, then add to define a trial solution y.

            This rule FAILS if one or more of the k atoms is a solution of
            the homogeneous differential equation.

            Rule II. If Rule I FAILS, then break the k atoms into groups
            with the same base atom. Cycle through the groups, replacing atoms
            as follows. If the first atom in the group is a solution of the homogeneous
            differential equation, then multiply all atoms in the group by factor x. Repeat
            until the first atom is not a solution of the homogeneous differential equation.
            Multiply the constructed k atoms by symbols d_1, ... , d_k and add to define trial solution y.

            Explanation: The relation between the Rule I + II trial solution and
             the book's table that uses the mystery factor x^s.
      EXAMPLES.
        y'' = x
        y'' + y = x exp(x)
        y'' - y = x exp(x)
        y'' + y = cos(x)
        y''' + y'' = 3x + 4 exp(-x)

    THEOREM. Suppose a list of k atoms is generated from the
             atoms in f(x), using Rule I. Then the shortest trial
             solution has exactly k atoms.

     EXAMPLE. How to find a shortest trial solution using
                   Rules I and II.

              Details for x''(t)+x(t) = t^2 + cos(t), obtaining
              the shortest trial solution
                 x(t)=d1+d2 t+d3 t^2+d4 t cos(t) + d5 t sin(t).
              How to use dsolve() in maple to check the answer.
     EXAMPLE. Suppose the DE has order n=4 and the homogeneous
              equation has solution atoms cos(t), t cos(t), sin(t),
              t sin(t). Assume f(t) = t^2 + cos(t). What is the
              shortest trial solution?
     EXAMPLE. Suppose the DE has order n=2 and the homogeneous
              equation has solution atoms cos(t), sin(t). Assume
                  f(t) = t^2 + t cos(t).
              What is the shortest trial solution?
     EXAMPLE. Suppose the DE has order n=4 and the homogeneous
              equation has solution atoms 1, t, cos(t), sin(t).
              Assume
                  f(t) = t^2 + t cos(t).
              What is the shortest trial solution?

Wed Apr 4: Resonance, Section 5.6

Undetermined coefficients
  Examples continued from the previous lecture.
      EXAMPLES.
        y'' = x
        y'' + y = x exp(x)
        y'' - y = x exp(x)
        y'' + y = cos(x)
        y''' + y'' = 3x + 4 exp(-x)

  Shortest trial solution.
  Two Rules to find the shortest trial solution.
     1. Compute the atoms in f(x). The number k of atoms found
        is the number needed in the shortest trial solution.
     2. Correct groups with the same base atom, by
        multiplication by x until the group contains no atom
        which is a solution of the homogeneous problem
        [eliminate homogeneous DE conflicts].
  The x^s mystery factor in the book's table. The number s is the
    multiplicity of the root in the homogenous DE characteristic
    equation, which constructed the base atom of the group.
Reference:
     Edwards-Penney, Differential Equations and Boundary Value
     Problems, 4th edition, section 3.7 [math 2280 textbook].
     Extra pages supplied by Pearson with bookstore copies of
     the 2250 textbook. Also available as a xerox copy in case
     your book came from elsewhere. Check-out the 2280 book in
     the math center or the Math Library. All editions of the
     book have identical 3.7 and 7.6 sections.
Transform Terminology
   Convolution theorem and x'' + 4x = cos(t), x(0)=x'(0)=0.
   Input
   Output
   Transfer Function
Circuits EPbvp3.7:
  Electrical resonance.
    Derivation from mechanical problems 5.6.
    THEOREM: omega = 1/sqrt(LC).
  Impedance, reactance.
  Steady-state current
  amplitude
  Transfer function.
  Input and output equation.

Wine Glass Experiment
   The lab table setup
      Speaker.
      Frequency generator with adjustment knob.
      Amplifier with volume knob.
      Wine glass.
   x(t)=deflection from equilibrium of the radial component of the
      glass rim, represented in polar coordinates, orthogonal to
      the speaker front.
   mx'' + cx' + kx = F_0 cos(omega t)  The model of the wine glass
      m,c,k are properties of the glass sample itself
      F_0 = volume knob adjustment
      omega = frequency generator knob adjustment
Theory of Practical Resonance
   The equation is
     mx''+cx'+kx=F_0 cos(omega t)
   THEOREM. The limit of x_h(t) is zero at t=infinity
   THEOREM. x_p(t) = C(omega) cos(omega t - phi)
            C(omega) = F_0/Z, Z^2 = A^2+B^2, A and B are the
            undetermined coefficient answers for trial solution
            x(t) = A cos(omega t) + B sin(omega t).
   THEOREM. The output x(t) = x_h(t) + x_p(t) is graphically
            just x_p(t) = C(omega) cos(omega t - phi) for large t.
            Therefore, x_p(t) is the OBSERVABLE output.
   THEOREM. The amplitude C(omega) is maximized over all possible
            input frequencies omega>0 by the single choice
                omega = sqrt(k/m - c^2/(2m^2)).
   DEFINITION. The practical resonance frequency is the number omega
               defined by the above square root expression.
 Projection: glass-breaking video. Wine glass experiment. Tacoma narrows.

Video: Wine glass breakage (QuickTime MOV) (96.8 K, mov, 31 Mar 2008)
Video: Wine glass experiment (12mb mpg, 2min) (12493.8 K, mpg, 01 Apr 2008)
Video: Tacoma Narrows Bridge Nov 7, 1940 (18mb mpg, 4min) (18185.8 K, mpg, 01 Apr 2008)
Slides: Basic undetermined coefficients, draft 5 (153.6 K, pdf, 03 Mar 2012)
Slides: Variation of parameters (164.5 K, pdf, 03 Mar 2012)
Slides: Resonance and undetermined coefficients (199.4 K, pdf, 03 Mar 2012)

Fri Apr 6: Sections 7.1, 7.2

Topics from linear systems:
    Brine tank models.
    Recirculating brine tanks.
    Pond pollution.
    Home heating.
    Earthquakes.
    Railway cars.
    All are 2x2 or 3x3 or nxn system applications that can be solved by Laplace methods.
Systems of two differential equations
   Solving a system from Chapter 1 methods
   The Laplace resolvent method for systems.
    Cramer's Rule,
    Matrix inversion methods.
   EXAMPLE: Solving a 2x2 dynamical system using Laplace's resolvent method.
     Study of u'=Au, u(0)=vector([1,2]), A=matrix([[2,1],[0,3]]).
   EXAMPLE: Problem 10.2-16,
     This problem is a 3x3 system for x(t), y(t), z(t) solved
     by Laplace theory methods. The resolvent formula
                     (sI - A) L(u(t)) = u(0)
     with u(t) the fixed 3-vector with components x(t), y(t), z(t),
     amounts to a shortcut to obtain the equations for L(x(t)),
     L(y(t)), L(z(t)). After the shortcut is applied, in which
     Cramer's Rule is the method of choice, to find the formulas,
     there is no further shortcut: we have to find x(t), for example,
     by partial fractions and the backward table, followed by Lerch's
     theorem.

Mon Apr 9: Sections 6.1, 7.1

Systems of two differential equations
   The Laplace resolvent method for systems.
        Solving the resolvent equation for L(x), L(y).
          Cramer's Rule
          Matrix inversion
          Elimination
   Example: Solving a 2x2 dynamical system
     Study of u'=Au, u(0)=vector([1,2]), A=matrix([[2,1],[0,3]]).
     Dynamical system scalar form is
         x' = 2x + y,
         y' = 3y,
         x(0)=1, y(0)=2.
     Chapter 1: Linear integrating factor method.
      Laplace resolvent method
        The shortcut equations
        Solving for L(x), L(y)
        Backward table and Lerch's theorem
        Answers: x(t) = - e^{2t} + 2 e^{3t},
                     y(t) = 2 e^{3t}.
      Chapter 1+5 Shortcut Method [appears in Edwards-Penney]
        Solve w'+p(t)w=0 as w = constant / integrating factor.
        Then  y(t) = 2 exp(3t)
        Stuff y(t) into the first DE to get the linear DE
           x' - 2x = 6 exp(3t)
        Superposition: x(t)=x_h(t)+x_p(t),
           x_h(t)=c exp(2t),
           x_p(t) = d1 exp(t) = 2 exp(3t) by undetermined coeff.
        Then x(t)= - exp(2t) + 2 exp(3t).
  
    Variation of Parameters and Undetermined Coefficients references
    Slides: Basic undetermined coefficients, draft 5 (153.6 K, pdf, 03 Mar 2012)
    Slides: Variation of parameters (164.5 K, pdf, 03 Mar 2012)
    Slides: Resonance and undetermined coefficients (199.4 K, pdf, 03 Mar 2012)
    References for Eigenanalysis and Systems of Differential Equations.
    Manuscript: Algebraic eigenanalysis (187.6 K, pdf, 04 Mar 2012)
    Manuscript: What's eigenanalysis 2008 (174.2 K, pdf, 04 Mar 2012)
    Manuscript: What's eigenanalysis, draft 1 (152.2 K, pdf, 01 Apr 2008)
    Manuscript: What's eigenanalysis, draft 2 (124.0 K, pdf, 14 Nov 2007)
    Slides: Cayley-Hamilton-Ziebur method for solving vector-matrix system u'=Au. (152.9 K, pdf, 04 Mar 2012)
    Slides: Laplace resolvent method (88.1 K, pdf, 04 Mar 2012)
    Slides: Laplace second order systems (288.1 K, pdf, 04 Mar 2012)
    Manuscript: Systems of DE examples and theory (785.8 K, pdf, 16 Nov 2008)
    Slides: Home heating, attic, main floor, basement (109.8 K, pdf, 04 Mar 2012)
    Text: Lawrence Page's pagerank algorithm (0.7 K, txt, 06 Oct 2008)
    Text: History of telecom companies (1.4 K, txt, 30 Dec 2009)
    Slides: Cable hoist example (73.2 K, pdf, 21 Aug 2008)
    Slides: Sliding plates example (105.8 K, pdf, 21 Aug 2008)

Extra Credit Maple Project: Tacoma narrows. Explore an alternative explanation for what caused the bridge to fail, based on the hanging cables.
    Laplace theory references
    Slides: Laplace and Newton calculus. Photos. (200.2 K, pdf, 04 Mar 2012)
    Slides: Intro to Laplace theory. Calculus assumed. (163.0 K, pdf, 19 Mar 2012)
    Slides: Laplace rules (160.3 K, pdf, 04 Mar 2012)
    Slides: Laplace table proofs (169.6 K, pdf, 04 Mar 2012)
    Slides: Laplace examples (149.1 K, pdf, 04 Mar 2012)
    Slides: Piecewise functions and Laplace theory (108.1 K, pdf, 04 Mar 2012)
    MAPLE: Maple Lab 7. Laplace applications (60.6 K, pdf, 09 Dec 2011)
    Manuscript: DE systems, examples, theory (785.8 K, pdf, 16 Nov 2008)
    Slides: Laplace resolvent method (88.1 K, pdf, 04 Mar 2012)
    Slides: Laplace second order systems (288.1 K, pdf, 04 Mar 2012)
    Slides: Home heating, attic, main floor, basement (109.8 K, pdf, 04 Mar 2012)
    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 (8.2 K, txt, 11 Dec 2011)