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2280 12:55pm Lectures Week 9 S2015

Last Modified: March 22, 2015, 20:58 MDT.    Today: November 20, 2017, 23:22 MST.
 Edwards-Penney, sections 7.1 to 7.6 and 4.1 to 4.3
  The textbook topics, definitions and theorems
Edwards-Penney 7.1 to 7.5 (20.1 K, txt, 17 Apr 2015)
Edwards-Penney 4.1 to 4.3 (10.4 K, txt, 05 Jan 2015)

Monday: Topics from last week

Maple Example to find roots of the characteristic equation
 Consider the recirculating brine tank example:
    20 x' = -6x + y,
    20 y' = 6x - 3y
 The maple code to solve the char eq:
   # Answer: -9/40+(1/40)*sqrt(33), -9/40-(1/40)*sqrt(33)
Applications of Laplace's method from 7.3, 7.4, 7.5
Convolution theorem 
    DEF. Convolution of f and g = f*g(t) = integral of f(x)g(t-x) from x=0 to x=t
    THEOREM. L(f(t))L(g(t))=L(convolution of f and g)
    Application:   L(cos t)L(sin t) = L(0.5 t sin(t))
 Second shifting Theorems
   e^{-as}L(f(t))=L(f(t-a)u(t-a)) Backward table
   L(g(t)u(t-a))=e^{-as}L(g(t+a)) Forward table


 EXAMPLES: Second Shift Theorem.
   Forward table
   L(sin(t)u(t-Pi)) = e^{-Pi s} L(sin(t)|t->t+Pi)
                    = e^{-Pi s} L(sin(t+Pi))
                    = e^{-Pi s} L(sin(t)cos(Pi)+sin(Pi)cos(t))
                    = e^{-Pi s} L(-sin(t))
                    = e^{-Pi s} ( -1/(s^2+1))
   Backward table
   L(f(t)) = e^{-2s}/s^2
           = e^{-2s} L(t)
           = L(t u(t)|t->t-2)
           = L((t-2)u(t-2))
   Therefore f(t) = (t-2)u(t-2) = ramp at t=2.

Laplace Resolvent Method.
Slides: Laplace resolvent method (88.1 K, pdf, 03 Mar 2012) The general vector-matrix DE Model u'=Au Laplace of u(t) = Resolvent times u(0) Resolvent = inverse(sI - A) Chapter 1 methods for solving 2x2 systems 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). Laplace Resolvent Method Consider problem 7.2-16 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 Then u:=Vector([x,y,z]); 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 (sI - B) L(u(t)) = u0 which is called the Resolvent Equation. For the example, the resolvent equation is Matrix([[s-1,0,-1],[-1,s-1,0],[2,0,s+1]]).Vector([L(x),L(y),L(z)]) = Vector([1,0,0]); DEF. 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 = L(u(t)) = inverse(sI - B) times vector u0 Briefly, Laplace of VECTOR u(t) = RESOLVENT MATRIX times VECTOR u(0) ADVICE: 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.

Monday: Resolvent. Dirac Impulse. Delta and hammer hits.

 Hammer hits and the Delta function
   Definition of delta(t)
     delta(t) = idealized injection of energy into a system at
                   time t=0 of impulse=1.
   A hammer hit model in mechanics:
      Camshaft impulse in a car engine
   How delta functions appear in circuit calculations
      Start with Q''+Q=E(t) where E is a switch. Then differentiate to get
      I''+I=E'(t). Term E'(t) is a Dirac impulse.
   Paul Dirac (1905-1985) and impulses
   Laurent Schwartz (1915-2002) and distribution theory
   Riemann Stieltjes integration theory: making sense of the Dirac delta.
      Def: RS-integral equals the limit of RS-sums as N-->infinity and mesh-->zero.
      RS-sum = sum of terms  f(x_i)(alpha(x_i)-alpha(x_{i-1})) where alpha(x) is
      the monotonic RS integrator.
   Why int( f(t) delta(t-a), t=-infinity .. infinity) = f(a)
   The symbol delta(t-a) makes sense only under an integral sign.

Engineering models
   Short duration impulses: Injection of energy into a mechanical or electrical model.
     Definition: The impulse of force f(t) on interval [a,b] equals the
     integral of f(t) over [a,b]
     An example for f(t) with impulse 5 is defined by
        f(t) = (5/(2h))pulse(t,-h,h)
     EXAMPLE. The Laplace integral of f(t) and its limit as h --> 0.
              Answer is the Dirac impulse.
     EXAMPLE. The delta function model x''(t) + 4x(t) = 5 delta(t-t0),
              x(0)=0, x'(0)=0. The model is a mass on a spring with no
              damping. It is at rest until time t=t0, when a short duration
              impulse of 5 is applied. This starts the mass oscillation.
     EXAMPLE. The delta impulse model from EPbvp 7.6,
              x''(t) + 4x(t) = 8 delta(t-2 pi), x(0)=3, x'(0)=0.
              The model is a mass on a spring with no damping. The mass is moved
              to position x=3 and released (no velocity). The mass oscillates until
              time t=2Pi, when a short duration impulse of 8 is applied. This
              alters the mass oscillation, producing a piecewise output x(t).
        # How to solve it with dsolve in maple.
        ic:=x(0)=3,D(x)(0)=0; dsolve({de,ic},x(t));
      Details of the Laplace calculus in maple: inttrans package.
         with(inttrans): f:=x->cos(omega*t)+8*Dirac(t-2*Pi);
         G:=laplace(f(t),t,s); invlaplace(G,s,t);
   CALCULATION. Phase amplitude conversion [see EP 5.4]
      x(t) = 3 cos(2t) until hammer hit at t=2Pi. It has amplitude 3.
      x(t) = 3 cos(2t)+4 sin(2t) after the hit. It has amplitude 5.
           = 5 cos(2t - arctan(4/3))

   An RLC circuit model
      Q'' + 110 Q' + 1000 Q = E(t)
      Differentiate to get [see EPbvp 3.7]
      I'' + 100 I' + 1000 I = E'(t)
      When E(t) is a switch, then E'(t) is a Dirac delta.

Tuesday-Wednesday: Problem session. Sections 7.1 to 7.5.

 Forward and Backward Table Applications
   Review of previously solved problems.
     Problem 7.1-18. Trig identity 2 sin (theta) cos(theta) = sin(2 theta)
       used for L(sin(3t)cos(3t)).
     Problem 7.1-28. Splitting a fraction into backward table entries.
  Partial Fractions and Backward Table Applications
    Problem 7.2-24. L(f)=1/(s(s+1)(s+2)) solved by the three methods for
    partial fractions: sampling, atom method, Heaviside cover-up.
    Problem 7.2-9. Solve x''+3x'+2x=t, x(0)=0, x'(0)=2. Get the resolvent
      L(x)=A/s + B/s^2 + C/(s+2) + D(s+1)
      L(x)=L(A+Bt+C e^{-2t} +D e^{-t})
     Solve for A,B,C,D by the sampling method (partial fraction method).
   Shifting Theorem and u-substitution Applications
    Problem 7.3-8. L(f)=(s-1)/(s+1)^3
      See #18 details for a similar problem.
     Problem 7.3-18. L(f)=s^3/(s-4)^4.
       L(f) = (u+4)^3/u^4  where u=s-4
       L(f) = (u^3+12u^2+48u+64)/u^4
       L(f) = (1/s + 12/s^2 + 48/s^3 + 64/s^4) where s --> (s-4)
       L(f)=L(e^{4t}(1+12t+48t^2/2+64t^3/6)) by shifting thm
     Problem 7.3-8. L(f)=(s+2)/(s^2+4s+5)
       L(f) = (s+2)/((s+2)^2 + 1)
       L(f) = u/(u^2 + 1)  where u=s+2
       L(f) = s/(s^2 + 1) where s --> s+2
       L(f) = L(e^{-2t} cos(t))  by shifting thm
   S-differentiation theorem
     Problem 7.4-21. Similar to Problem 7.4-22.
     Clear fractions, multiply by (-1), then:
     (-t)f(t) = -exp(3t)+1
     L((-t)f(t)) = -1/(s-3) + 1/s
     (d/ds)F(s) = -1/(s-3) + 1/s
     F(s) = ln(|s|/|s-3|)+c
     To show c=0, use this theorem:
     THEOREM. The Laplace integral has limit zero at s=infinity.
   Convolution theorem
     THEOREM. L(f(t)) L(g(t)) = L(convolution of f and g)
     Example. L(cos t)L(sin t) = L(0.5 t sin t)
     Example: 7.4-36. x''+4x=f(t), x(0)=x'(0)=0 has solution
        x(t)=0.5 int(sin(2u)f(t-u),u=0..t)

 Periodic function theorem application
    Problem 7.5-28.
     Find L(f(t)) where f(t) = t on 0 <= t < a and f(t)=0 on a <= t < 2a,
     with f(t) 2a-periodic [f(t+2a)=f(t)].
      According to the periodic function theorem, the answer is
      found from maple integration:
       L(f) = 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))

 Piecewise Functions
   Unit Step: u(t)=1 for t>=0, u(t)=0 for t<0.
   Pulse: pulse(t,a,b)=u(t-a)-u(t-b)
   Ramp: ramp(t-a)=(t-a)u(t-a)
 Problem Session: Periodic function theorem
    Laplace of the square wave. Problem 7.5-25. Done earlier.
      Answer: (1/s)tanh(as/2)
    Laplace of the sawtooth wave. Problem 7.5-26. Done earlier.
      Answer: (1/s^2)tanh(as/2)
      Method: (d/dt) sawtooth = square wave
               The use the parts theorem.
               Or, use the Integral theorem.
    Laplace of the staircase function. Problem 7.5-27. Done earlier.
      This is floor(t/a). The Laplace answer is
      This answer can be verified by maple code
    Laplace of the sawtooth wave, revisited.
       Identity: floor(t) = staircase with jump 1.
       Identity: t - floor(t) = saw(t) = sawtooth wave
       General:  t - a*floor(t/a) = a*saw(t/a) = sawtooth wave of period a.
    Problem 7.5-28. Details revisited.
         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:
        # 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))
        Use the Laplace integral theorem, which says the answer is (1/s)
        times the Laplace answer for the 2a-periodic function g(t)=1 on
        [0,a], g(t)=0 on [a,2a]. We check that g(t)=(1/2)(1+sqw(t/a)).
  Problem Session: Second Shifting Theorem Applications
     Second shifting Theorems
       e^{-as}L(f(t))=L(f(t-a)u(t-a))  Requires a>=0.
       L(g(t)u(t-a))=e^{-as}L(g(t+a))  Requires a>=0.
     Problem 7.5-3. L(f)=e^{-s}/(s+2)
     Problem 7.5-4. L(f) = (e^{-s} - e^{2-2s})/(s-1)
     Problem 7.5-22. f(t)=t^3 pulse(t,1,2)
     Problem 7.5-4.
      F_1(s) = exp(-s)/(s-1) = exp(-as)L(exp(t)) with a=1
           = L(exp(t-1)u(t-1)) by the second shifting theorem
      F_2(s) = exp(2-2s)/(s-1) = exp(-2s)/s with shift s --> s-1
             = L(1 u(t-2)) [2nd shifting theorem] shift s --> s-1
             = L( exp(t) 1 u(t-2)) by the first shifting theorem
      F=F_1 - F_2 = L(exp(t-1)u(t-1)-exp(t)u(t-2))
      f(t) =  exp(t-1)u(t-1)-exp(t)u(t-2)


   Problem 7.5-22.
     f(t)=t^3 pulse(t,1,2)
         = t^3 u(t-1) - t^3 u(t-2)
     L(t^3 u(t-1)) = exp(-s)L((t+1)^3) 2nd shifting theorem
     L(t^3 u(t-2)) = exp(-2s)L((t+2)^3) 2nd shifting theorem
     Details to be finished in class. Pascal's triangle and (a+b)^3.
     Function notation and dummy variables.

   Dirac Applications
     x''+x=5 Delta(t-1), x(0)=0,x'(0)=1
       The Laplace integral has limit zero at t=infinity, provided
       f(t) is of exponential order. The Laplace of the Dirac impulse
       violates this theorem's hypothesis, because L(delta(t))=1.

  Cayley-Hamilton-Ziebur Method: Ch4, sections 4.1, 4.2

 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
        (2I-A)(3I-A)=0 or A^2 - 5A + 6I = 0.
Cayley-Hamilton-Ziebur Method
        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)
Slides: Cayley-Hamilton-Ziebur method for solving vector-matrix system u'=Au. (137.7 K, pdf, 28 Mar 2015) PROBLEM: Solve by Cayley-Hamilton-Ziebur the 2x2 dynamical system x' = 2x + y, y' = 3y, x(0)=1, y(0)=2. 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 Theorem says that u(t) = exp(2t) vec(v_1) + exp(3t) vec(v_2) where vectors v_1, uv_2 are to be determined from the matrix A = matrix([[2,1],[0,3]]) and initial conditions x(0)=1, y(0)=2. ZIEBUR ALGORITHM. To solve for v_1, v_2 in the example, differentiate the equation u(t) = exp(2t) v_1 + exp(3t) v_2 and set t=0 in both relations. Then u'=Au implies u_0 = v_1 + v_2, Au_0 = 2 v_1 + 3 v_2. These equations can be solved by elimination. The answer: v_1 = (3 u_0 -Au_0), v_2 = (Au_0 - 2 u_0) = vector([-1,0]) = vector([2,2]) Vectors v_1, v_2 are recognized as eigenvectors of A for lambda=2 and lambda=3, respectively. ZIEBUR SHORTCUT [Edwards-Penney textbook method, Example 5 in 4.1] 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. Cayley-Hamilton topics, Section 6.3. Cayley-Hamilton Theorem A matrix satisfies its own characteristic equation. Proof of the Ziebur Lemma for 2x2 matrices.
 Numerical Methods for Systems: Section 4.3

 Review of Scalar methods 
   Euler, Heun, RK4

 Change the methods to the vector case, based upon the statement of
 the IVP in Picard's Theorem. It amounts to putting arrows over y and f.

 Vector Methods
   Euler, Heun, RK4
   Slides: Vector Methods

 EXAMPLE. Solve the vector problem u'=Au, u(0)=u_0 with A=Matrix([[1,2],[0,3]]) and
           u_0 = <1,-1> using Euler's method and Heuns's method with step size h=0.1.