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2280 12:55pm Lectures Week 9 S2016

Last Modified: March 12, 2016, 11:40 MST.    Today: November 18, 2017, 20:13 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 (21.0 K, txt, 22 Feb 2015)
Edwards-Penney 4.1 to 4.3 (10.7 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:
   A:=(1/20)*Matrix([[-6,1],[6,-3]]);
   linalg[charpoly](A,r);
   solve(%,r);
   # 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

second-shift-theorem

 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 (81.0 K, pdf, 14 Mar 2016) 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.
        de:=diff(x(t),t,t)+4*x(t)=f(t);f:=t->8*Dirac(t-2*Pi);
        ic:=x(0)=3,D(x)(0)=0; dsolve({de,ic},x(t));
        convert(%,piecewise,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);
         de:=diff(x(t),t,t)+4*x(t)=f(t);
         laplace(de,t,s);
         subs(ic,%);
         solve(%,laplace(x(t),t,s));
   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
    equation
     (s^2+3s+2)L(x)=2+L(t)
      L(x)=(1+2s^2)/(s^2(s+2)(s+1))
      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)].
    Details
      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
          L(floor(t/a))=(1/s)/(exp(as)-1))
      This answer can be verified by maple code
         inttrans[laplace](floor(t/a),t,s);
    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:
        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))
      ALTERNATIVE
        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)

second-shift-theorem-7.5-22

   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
     THEOREM.
       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
  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)
Slides: Cayley-Hamilton-Ziebur method for solving vector-matrix system u'=Au. (137.4 K, pdf, 14 Mar 2016) 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.