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2280 Lectures Week 8 S2015

Last Modified: March 09, 2015, 08:46 MDT.    Today: November 23, 2017, 09:40 MST.
Topics
  Sections 7.1 to 7.6
  The textbook topics, definitions, examples and theorems
Edwards-Penney Ch 7, 7.1 to 7.5 (20.1 K, txt, 17 Apr 2015)

Week 8: Sections 7.2, 7.3, 7.4, 7.5, 7.6

Monday: Solving differential equations, Resolvent


Laplace Resolvent Method.
  --> This method is a shortcut for solving systems by Laplace's method.
  --> It is also a convenient way to solve systems with maple.
Slides: Laplace resolvent method (88.1 K, pdf, 03 Mar 2012) Intro to the Laplace resolvent shortcut for 2x2 systems Problem: Write a 2x2 dynamical system as a vector-matrix equation u'=Au. Problem: Solve a 2x2 dynamical system in vector-matrix form u'=Au. 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).

Tuesday: Piecewise Functions, Convolution, Resolvent

   Forward and backward table entries for sine and cosine.
    Why cosh(t) and sinh(t) are not in the tables.

Laplace Resolvent Method

  First example: x'=x-y, y'=x+y, x(0)=1, y(0)=2
    Solve with the resolvent equation shortcut.
  Second example: x'=-x-y, y'=x-y, x(0)=1, y(0)=2
    Solve with the resolvent equation shortcut.
 
  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.
 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.
 
 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)
   L(u(t-a)) = (1/s) exp(-as) [for a >= 0 only]
Integral Theorem
   L(int(g(x),x=0..t)) = s L(g(t))
   Applications to computing ramp(t-a)
    L(ramp(t-a)) = (1/s^2) exp(-as) [for a >= 0 only]
 Piecewise defined periodic waves
   Square wave: f(t)=1 on [0,1), f(t)=-1 on [1,2), 2-periodic
   Triangular wave: f(t)=|t| on [-1,1], 2-periodic
   Sawtooth wave: f(t)=t on [0,1], 1-periodic
   Rectified sine: f(t)=|sin(kt)|
   Half-wave rectified sine: f(t)=sin(kt) when positive, else zero.
   Parabolic wave
 Periodic function theorem
      Proof details
      Laplace of the square wave. Problem 7.5-25.
      Answer: (1/s)tanh(as/2)

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(int(g(x),x=0..t)) = L(ramp(t)) L(g(t)) = (1/s)L(g(t)), which is the Integral Theorem
    Application:   L(cos t)L(sin t) = L(0.5 t sin(t))
 

Wednesday: Second Shift Theorem, Dirac Impulse

 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)
   L(u(t-a)) = (1/s) exp(-as) [for a >= 0 only]
Piecewise defined periodic waves
   Square wave: f(t)=1 on [0,1), f(t)=-1 on [1,2), 2-periodic
   Triangular wave: f(t)=|t| on [-1,1], 2-periodic
   Sawtooth wave: f(t)=t on [0,1], 1-periodic
   Rectified sine: f(t)=|sin(kt)|
   Half-wave rectified sine: f(t)=sin(kt) when positive, else zero.
   Parabolic wave
 Periodic function theorem
      Proof details
      Laplace of the square wave. Problem 7.5-25.
      Answer: (1/s)tanh(as/2)

 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.
   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.

  Hammer hits and the Dirac Impulse
   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 Delta.
   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 delta.
     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 function 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 impulse.

Friday: 7.5, 7.6 Problem Session

 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)
   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 were 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 delta function
       violates this theorem's hypothesis, because L(delta(t))=1.
Laplace theory references
 
Slides: Laplace and Newton calculus. Photos. (200.2 K, pdf, 03 Mar 2012)
Slides: Intro to Laplace theory. Calculus assumed. (163.0 K, pdf, 18 Mar 2012)
Manuscript: Laplace theory (497.3 K, pdf, 19 Mar 2014)
Slides: Laplace rules (160.3 K, pdf, 03 Mar 2012)
Slides: Laplace table proofs (169.6 K, pdf, 03 Mar 2012)
Slides: Laplace examples (149.1 K, pdf, 03 Mar 2012)
Slides: Piecewise functions and Laplace theory (108.5 K, pdf, 03 Mar 2013)
Slides: Laplace resolvent method (88.1 K, pdf, 03 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 (352.3 K, pdf, 07 Jan 2014)
Manuscript: DE systems, examples, theory (730.9 K, pdf, 10 Apr 2014)
Manuscript: Laplace theory 2008 (497.3 K, pdf, 19 Mar 2014)
Transparencies: Ch7 Laplace solutions 7.1 to 7.4 (1068.7 K, pdf, 22 Feb 2015)
Slides: Laplace second order systems (288.1 K, pdf, 03 Mar 2012)