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Week 10: 02 Nov,  03 Nov,  04 Nov,  05 Nov,  06 Nov,

2250 Lecture Record Week 10 F2009

Last Modified: November 07, 2009, 08:06 MST.    Today: October 23, 2017, 00:22 MDT.

Week 10, Nov 2 to 6: Sections 10.5,EPbvp7.6,5.5,7.1,,EPbvp3.7

02 Nov: Piecewise Functions. Shifting. Delta function. Section 10.5 and EPbvp supplement 7.6.

 Piecewise Functions
   Unit Step
 Second shifting Theorems
 Piecewise defined periodic waves
   Square wave
   Triangular wave
   Rectified sine
   Half-wave rectified sine
   Parabolic wave
 Periodic function theorem
    Laplace of the square wave, tanh function.
 Hammer hits and the Delta function
   Definition of delta(t)
   Hammer hit models
   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.

03 Nov: Delta function and hammer hits. Periodic piecewise functions. Second shifting theorem.

Engineering models
   Short duration impulses
     Definition: impulse of force f(t) on [a.b] equals the integral of f(t) over [a,b]
     An example when f(t) has impulse 5, defined by
        f(t) = (5/(2h))pulse(t,-h,h)
     Laplace integral of f(t) and its limit as h --> 0.  Answer is the Dirac delta.
     The delta function model
      x''(t) + 4x(t) = 5 delta(t-t0),
      x(0)=0, x'(0)=0
   The delta function model from EPbvp 7.6
      x''(t) + 4x(t) = 8 delta(t-2 pi),
      x(0)=3, x'(0)=0
   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))  Discussed how to convert in class. See 5.4.
   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 step, then E'(t) is a Dirac delta.

04 Nov: Intro Laplace Resolvent. Undetermined Coefficients. Section 5.5, EPbvp7.6, 3.7.

Engineering models
   The delta function model
      x''(t) + 4x(t) = 8 delta(t-2 pi),
      x(0)=3, x'(0)=0
      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);
   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

 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

 Intro to the Laplace resolvent method for 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).

05nov: Fusi and Richins

Exam 2 at 6:50am to 8:30am in WEB 103 on 05 Nov

06 Nov: Problem session. Sections 10.4, 10.5.

 S-differentiation theorem
   Problem 10.4-22. Due next week.
   Problem 10.4-21.
     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 t=infinity. 
 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)
 Second shifting Theorems
   e^{-as}L(f(t))=L(f(t-a)step(t-a))  Requires a>=0.
   L(g(t)step(t-a))=e^{-as}L(g(t+a))  Requires a>=0.
   Problem 10.5-4.
     F_1(s) = exp(-s)/(s-1) = exp(-as)L(exp(t)) with a=1
          = L(exp(t-1)step(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 step(t-2)) [by the second shifting theorem] with shift s --> s-1 
            = L( exp(t) 1 step(t-2)) by the first shifting theorem
     F=F_1 - F_2 = L(exp(t-1)step(t-1)-exp(t)step(t-2))
     f(t) =  exp(t-1)step(t-1)-exp(t)step(t-2)
   Problem 10.5-22.
     f(t)=t^3 pulse(t,1,2)
         = t^3 step(t-1) - t^3 step(t-2)
     L(t^3 step(t-1)) = exp(-s)L((t+1)^3) by the second shifting theorem
     L(t^3 step(t-2)) = exp(-2s)L((t+2)^3) by the second shifting theorem  
     Details were finished in class. Pascal's triangle and (a+b)^3.
     Function notation and dummy variables. 
 Periodic function theorem
    Laplace of the square wave. Problem 10.5-25.
      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
      This answer can be verified in maple by the code
    Problem 10.5-28. Details in class.
      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))
    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 (84.3 K, pdf, 19 Jul 2009)
    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 (8.1 K, txt, 21 Nov 2008)
    Text: Final exam study guide (7.6 K, txt, 12 Dec 2009)