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

Last Modified: November 21, 2010, 17:38 MST.    Today: December 16, 2017, 08:05 MST.

Week 12, Nov 15 to 19: Sections 5.5, 5.6, EPbvp3.7, 7.1, 7.2, 7.3

16 Nov: Sections 5.5, 5.6

Exam 3 Review
Michal and Laura will do exam 3 review this week Monday, Wednesday, Thursday.
Locations are 335 jwb at 12:55 Mon-Wed and WEB 103 at 7:30 on Thursday.

Exam 3 part I: 22, 23, 24 Nov.
The Tuesday exam is during the regular 7:25am classtime in WEB 103.
The Mon-Wed exams are in JWB 335 at 12:50pm.
As usual, you may write the exam on any one of the three days.
The problems are 1,2 on the sample exam below, which matches 3,4 on the S2010 exam key.

Exam 3 part II: 29 Nov, 1, 2 Dec.
The Thursday exam is at the usual 7:25am time in WEB 103.
The Mon-Wed exams are inJWB 335 at 12:50pm.
As usual, you may write exam 3, part II, on any one of the three days.
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 F2010
    PDF: sample exam 3, all problems. (182.4 K, pdf, 14 Nov 2010)
    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: Relation between the Rule I + II trial solution and
             the book's table that uses the mystery factor x^s.
      EXAMPLES.
        y'' + y = 1 + 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),
                   by 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),
              to obtain 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?

Undetermined coefficients
  Examples continued from the previous lecture.
  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.
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.
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.

16 Nov: Resonance, Section 5.6

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 (152.1 K, pdf, 19 Nov 2010)
Slides: Variation of parameters (109.8 K, pdf, 07 Nov 2009)
Slides: Resonance and undetermined coefficients (199.3 K, pdf, 20 Nov 2010)

18 Nov: 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.

18 Nov: 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).
Cayley-Hamilton Theorem
   A matrix satisfies its own characteristic equation.
   ILLUSTRATION: det(A-r I)=0 for the previous example
     is (2-r)(3-r)=0 or r^2 -5r + 6=0. Then C-H says
        A^2 - 5A + 6I = 0.
Cayley-Hamilton-Ziebur Method
  ZIEBUR'S LEMMA.
        The components of u in u'=Au are linear combinations of
        the atoms created by Euler's theorem applied to the
        roots of the characteristic equation det(A-rI)=0.
  THEOREM. Solve u'=Au without complex numbers or eigenanalysis.
        The solution of u'=Au is a linear combination of atoms
        times certain constant vectors [not arbitrary vectors].
             u(t)=(atom_1)vec(c_1)+ ... + (atom_n)vec(c_n)

  PROBLEM: Solve by Ziebur's Lemma the 2x2 dynamical system above.
        The characteristic equation is (2-lambda)(3-lambda)=0
          with roots lambda = 2,3
        Euler's theorem implies the atoms are exp(2t), exp(3t).
        Ziebur's Lemma says that
           u(t) = exp(2t) u_1 + exp(3t) u_2
        where vectors u_1, U_2 are to be determined from A and
        the initial conditions x(0)=1, y(0)=2.

  ZIEBUR ALGORITHM.
        To solve for u_1, u_2 in the example, differentiate the
        equation u(t) = exp(2t) u_1 + exp(3t) u_2 and set t=0
        in both relations. Then u'=Au implies
             u_0 =    u_1  +   u_2,
            Au_0 = 2 u_1 + 3 u_2.
        These equations can be solved by elimination.
        The answer:
            u_1 = (3 u_0 -Au_0), u_2 = (Au_0 - 2 u_0)
                = vector([-1,0])          = vector([2,2])
        These are recognized as eigenvectors of A for lambda=2
        and lambda=3, respectively, after studying chapter 6.

  ZIEBUR SHORTCUT [a textbook method]
        Start with Ziebur's theorem, which implies that
           x(t) = k1 exp(2t) + k2 exp(3t).
        Use the first DE to solve for y(t):
           y(t) = x'(t) - 2x(t)
                =  2 k1 exp(2t) + 3 k2 exp(3t)
                         - 2 k1 exp(2t) - 2 k2 exp(3t))
                =   k2 exp(3t)
        For example, x(0)=1, y(0)=2 implies k1 and k2 are
        defined by
           k1 + k2 = 1,
                  k2 = 2,
        which implies k1 = -1, k2 = 2, agreeing with a previous
        solution formula.
  Lecture: Fourier's Model. Intro to Eigenanalysis, Ch6.
  Examples and motivation.
     The ellipse and eigenpairs
      The eigenpairs in the solution of a differential equation u'=Au.
EIGENANALYSIS WARNING
  Reading Edwards-Penney Chapter 6 may deliver the wrong ideas
  about how to solve for eigenpairs.

     HISTORY. Chapter 6 originally appeared in the 2280 book
     as a summary, which assumes a linear algebra course. The
     chapter was copied without changes into the Edwards-Penney
     Differential Equations and Linear Algebra textbook, which you
     currently own. The text contains only shortcuts. There is
     no discussion of a general method for finding eigenpairs.
     You will have to fill in the details by yourself. The online
     lecture notes and slides were created to fill in the gap.
History of Eigenanalysis
    J.B.Fourier's 1822 treatise on the theory of heat.
    The rod example.
      Physical Rod: a welding rod of unit length, insulated on the
                    lateral surface and ice packed on the ends.
    Define f(x)=thermometer reading at loc=x along the rod at t=0.
    Define u(x,t)=thermometer reading at loc=x and time=t>0.
Algebraic Eigenanalysis Section 6.2.
  Calculation of eigenpairs     
   THEOREM. The eigenvalues of A are found from the determinant
    equation
                        det(A -lambda I)=0,
    which is called the characteristic equation.
  THEOREM. The eigenvectors of A are found from the frame
    sequence which starts with B=A-lambda I [lambda a root of
    the characteristic equation], ending with last frame rref(B).

    The eigenvectors for lambda are the partial derivatives of
    the general solution obtained by the Last Frame Algorithm,
    with respect to the invented symbols t1, t2, t3, ...
 
  Conversion Methods to Create a First Order System
    The position-velocity substitution.
    How to convert second order systems.
    How to convert nth order scalar differential equations.
    Non-homogeneous terms and the vector matrix system
         u' = Au + F(t)
    Non-linear systems and the vector-matrix system
         u' = F(t,u)
    Answer checks for u'=Au
      Example: The system u'=Au, A=matrix([[2,1],[0,3]]);
  
    Variation of Parameters and Undetermined Coefficients references
    Slides: Basic undetermined coefficients, draft 5 (152.1 K, pdf, 19 Nov 2010)
    Slides: Variation of parameters (109.8 K, pdf, 07 Nov 2009)
    Slides: Resonance and undetermined coefficients (199.3 K, pdf, 20 Nov 2010)
    References for Eigenanalysis and Systems of Differential Equations.
    Manuscript: Algebraic eigenanalysis (127.8 K, pdf, 23 Nov 2009)
    Manuscript: What's eigenanalysis 2008 (126.8 K, pdf, 11 Apr 2010)
    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. (152.6 K, pdf, 23 Nov 2010)
    Slides: Laplace resolvent method (56.4 K, pdf, 01 Nov 2009)
    Slides: Laplace second order systems (248.9 K, pdf, 01 Nov 2009)
    Manuscript: Systems of DE examples and theory (785.8 K, pdf, 16 Nov 2008)
    Slides: Home heating, attic, main floor, basement (73.8 K, pdf, 30 Nov 2009)
    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. (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 (186.8 K, pdf, 20 Oct 2009)
    Manuscript: Laplace theory (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 (8.9 K, txt, 18 Nov 2010)
    Text: Final exam study guide (8.3 K, txt, 09 Dec 2010)