2250-1 7:30am Lecture Record Week 12 S2010

Last Modified: April 11, 2010, 18:59 MDT.    Today: October 19, 2017, 15:54 MDT.

Week 12, Apr 5 to 9: Sections 10.5, 5.5, EPbvp3.7, 7.1, 7.2, 7.3

05 Apr: Sections 10.5, 5.5

 Piecewise Functions
   Unit Step: step(t)=1 for t>=0, step(t)=0 for t<0.
   Pulse: pulse(t,a,b)=step(t-a)-step(t-b)
   Ramp: ramp(t-a)=(t-a)step(t-a)
 Periodic function theorem
    Laplace of the square wave. Problem 10.5-25. Done earlier.
      Answer: (1/s)tanh(as/2)
    Laplace of the sawtooth wave. Problem 10.5-26.
      Answer: (1/s^2)tanh(as/2)
      Method: (d/dt) sawtooth = square wave
               The use the parts theorem.
    Laplace of the staircase function. Problem 10.5-27.
      This is floor(t/a). The Laplace answer is
          L(floor(t/a))=1/(s(exp(as)-1))
      This answer can be verified in maple by the code
         with(inttrans):
         laplace(floor(t/a),t,s);
    Problem 10.5-28. Details in class.
         f(t)=t on 0 <= t <= a,
         f(t)=0 on a <= t <= 2a
      According to the periodic function theorem, the answer is
      found from maple integration:
        int(t*exp(-s*t),t=0..a)/(1-exp(-2*a*s));
        # answer == (-1+exp(-s*a)+exp(-s*a)*s*a)/s^2/(-1+exp(-2*s*a))
      A better way to solve the problem is to write a formula for
      f'(t) and use the s-differentiation rule. We  get for a=1
                f'(t) = (1/2)(1+sqw(t))
      and then
               sL(f(t)) = (1/(2s))(1+tanh(s/2))
                L(f(t)) = (1/(2s^2))(1+tanh(s/2))
    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.
                1.Find all distinct atoms that appear in f, f', f'', ...
                  The number k of atoms found is exactly the number
                  of atoms that appear in the shortesttrial solution.
                2. Test each group of atoms having the same base atom.

                   2(a). If group contains a solution of the homogeneous DE,
                         then multiply every atom in the group by x.
                   2(b). Repeat 2(a) until that group contains NO CONFLICTS with
                         solutions of the homogeneous DE.

                   This STRIKE-OUT RULE removes atoms that appear in the
                   homogeneous solution y_h(x). Removed atoms are  REPLACED
                   by atoms with the same base atom, but higher powers of x.
                   At all correction stages, the number k of atoms remains constant.

                3. The conflict-free list of k atoms is the desired list of atoms that
                   appear in the shortest trial solution.

             Relation between the non-redundant 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 adding related lower-power atoms.
             Then the shortest trial solution has exactly k atoms.

     EXAMPLE. How to find a shortest trial solution using a
              shortcut correction rule.

              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?

06 Apr: Sections 5.5, EPbvp3.7, 7.1

Undetermined coefficients
  Examples continued from Monday.
  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.

07 Apr: 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.
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([[2,3],[0,4]]);
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([[2,3],[0,4]]).
   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.

09 Apr: Sections 6.1, 6.2, 7.2, 7.3

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([[2,3],[0,4]]).
     Dynamical system scalar form is
         x' = 2x + 3y,
         y' = 4y,
         x(0)=1, y(0)=2.
      Laplace resolvent method
        The shortcut equations
        Solving for L(x), L(y)
        Backward table and Lerch's theorem
        Answers: x(t) = -2 e^{2t} + 3 e^{4t}, 
                 y(t) = 2 e^{4t}.
      Chapter 1+5 Method
        Solve w'+p(t)w=0 as w = constant / integrating factor.
        Then  y(t) = 2 exp(4t)
        Stuff y(t) into the first DE to get the linear DE
           x' - 2x = 6 exp(4t)
        Superposition: x(t)=x_h(t)+x_p(t),
           x_h(t)=c exp(2t),
           x_p(t) = d1 exp(4t) = 3 exp(4t) by undetermined coeff.
        Then x(t)= -2 exp(2t) + 3 exp(4t).
Cayley-Hamilton Theorem
   A matrix satisfies its own characteristic equation.
   ILLUSTRATION: det(A-r I)=0 for the previous example
     is (2-r)(4-r)=0 or r^2 -6r + 8=0. Then C-H says
        A^2 - 6A + 8I = 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)(4-lambda)=0
          with roots lambda = 2,4
        Euler's theorem implies the atoms are exp(2t), exp(4t).
        Ziebur's Lemma says that
           u(t) = exp(2t) u_1 + exp(4t) 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 and set t=0 in both relations. Then u'=Au
        implies
             u_0 =   u_1 +   u_2,
            Au_0 = 2 u_1 + 4 u_2.
        These equations can be solved by elimination.
        The answer:
            u_1 = -(Au_0 - 4 u_0)/2, u_2 = (Au_0 - 2 u_0)/2
                = vector([-2,0])          = vector([3,2])
        These are recognized as eigenvectors of A for lambda=2
        and lambda=4, 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(4t).
        Use the first DE to solve for y(t):
           y(t) = (1/3)(x'(t) - 2x(t))
                =  (1/3)(2 k1 exp(2t) + 4 k2 exp(4t) -
                         2 k1 exp(2t) - 2 k2 exp(4t))
                =  (2/3) k2 exp(4t)
        For example, x(0)=1, y(0)=2 implies k1 and k2 are
        defined by
           k1 + k2 = 1,
           (2/3) k2 = 2,
        which implies k1 = -2, k2 = 3, agreeing with a previous
        solution formula.
Google Algorithm
   Lawrence Page's pagerank algorithm, google web page rankings.
  
    Variation of Parameters and Undetermined Coefficients references
    Slides: Basic undetermined coefficients, draft 4 (104.9 K, pdf, 07 Nov 2009)
    Slides: Variation of parameters (109.8 K, pdf, 07 Nov 2009)
    Slides: Resonance and undetermined coefficients (143.3 K, pdf, 07 Nov 2009)
    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 method for solving vector-matrix system u'=Au. (111.4 K, pdf, 30 Nov 2009)
    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 (0.0 K, pdf, 31 Dec 1969)
    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 (1968.3 K, pdf, 13 Nov 2003)
    Text: Laplace theory problem notes F2008 (8.9 K, txt, 31 Dec 2009)
    Text: Final exam study guide (7.6 K, txt, 12 Dec 2009)