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2250-4 12:25pm Lecture Record Week 11 F2010

Last Modified: November 14, 2010, 20:42 MST.    Today: December 14, 2017, 12:09 MST.

Week 11, Nov 8 to 12: Sections 10.5,EPbvp7.6,5.5,5.6,,EPbvp3.7,7.1

Nov 8, 10, 11:Laura and Michal

Review starts for Exam 3, using the 7:30 exam key from S2010. Solved laplace theory problems from chapter 10 dailies.

9 Nov: Piecewise Functions. Shifting. Section 10.5 and EPbvp supplement 7.6. Delta function and hammer hits.

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)
   L(step(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]
 Second shifting Theorems
   e^{-as}L(f(t))=L(f(t-a)step(t-a))
   L(g(t)step(t-a))=e^{-as}L(g(t+a))
   Used for switches and hammer hits.
 Piecewise defined periodic waves
   Square wave
   Triangular wave
   Sawtooth
   Rectified sine
   Half-wave rectified sine
   Parabolic wave
 Periodic function theorem
    Laplace of the square wave, tanh function.
    Periodic function theorem
     Proof
     Application to the square wave.
   Convolution theorem 
    Application:   L(cos t)L(sin t) = L(0.5 t sin(t))
 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.

9 Nov: Problem session. Sections 10.4, 10.5.

 Forward and Backward Table Applications
   Review of previously solved problems.
     Problem 10.1-18. Trig identity 2 sin (theta) cos(theta) = sin(2 theta)
       used for L(sin(3t)cos(3t)).
     Problem 10.1-28. Splitting a fraction into backward table entries.
  Partial Fractions and Backward Table Applications
    Problem 10.2-24. L(f)=1/(s(s+1)(s+2)) solved by the three methods for
    partial fractions: sampling, atom method, Heaviside cover-up.
    Problem 10.2-9. Solve x''+3x'+2x=t, x(0)=0, x'(0)=2. Get 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.
   Shifting Theorem and u-substitution Applications
    Problem 10.3-8. L(f)=(s-1)/(s+1)^3
      See #18 details for a similar problem.
     Problem 10.3-18. 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 10.3-8. 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 10.4-21. Similar to Problem 10.4-22.
     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 t=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: 10.4-36. 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
    Laplace of the square wave. Problem 10.5-25.
      Answer: (1/s)tanh(as/2)
    Problem 10.5-28.
     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))

10 Nov: Undetermined Coefficients. Sections 10.5,7.1,5.5

  Second Shifting Theorem Applications
     Problem 10.5-3. L(f)=e^{-s}/(s+2)
     Problem 10.5-4. L(f) = (e^{-s} - e^{2-2s})/(s-1)
     Problem 10.5-22. f(t)=t^3 pulse(t,1,2)
     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)) [2nd shifting theorem] 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) 2nd shifting theorem
     L(t^3 step(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.

   Piecewise Applications
     Staircase or floor function
     Sawtooth wave
     Square wave
   Dirac Applications
     x''+x=5 Delta(t-1), x(0)=0,x'(0)=1
     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)

Nov 11: Laplace Resolvent. Resonance. Variation of parameters. Sections 5.5, 10.5, EPbvp7.6

 More Laplace Examples
   Continuing 10.5 examples from last lecture.
 Transform Terminology
   Input
   Output
   Transfer Function
Engineering models
   Short duration impulses: Injection of energy into a mechanical or electrical model.
     Definition: The 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
           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));
   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.
 Resonance examples
    x'' + x = cos(t)
    Pure resonance, unbounded solution x(t) = 0.5 t sin(t)
    mx'' + cx' + kx = F_0 cos(omega t)
    Practical resonance, all solutions bounded, but x(t)
      can have extremely large amplitude when omega is tuned
      to the frequency omega = sqrt(k/m - c^2/(2m^2))
    LQ'' + RQ' + (1/C)Q = E_0 sin(omega t)
    Practical resonance, all solutions bounded, but the current
      I(t)=dQ/dt can have large amplitude when omega is tuned
      to the resonant frequency omega = 1/sqrt(LC).
 To do later:
    Soldiers marching in cadence, Tacoma narrows bridge,
    Wine Glass Experiment. Theodore Von Karman and vortex shedding.
    Cable model of the Tacoma bridge, year 2000. Resonance explanations.
 Theory of Practical Resonance
   The equation is
     mx''+cx'+kx=F_0 cos(omega t)
   THEOREM. The limit of x_h(t) is zero at t=infinity
   THEOREM. x_p(t) = C(omega) cos(omega t - phi)
            C(omega) = F_0/Z, Z^2 = A^2+B^2, A and B are the
            undetermined coefficient answers for trial solution
            x(t) = A cos(omega t) + B sin(omega t).
   THEOREM. The output x(t) = x_h(t) + x_p(t) is graphically
            just x_p(t) = C(omega) cos(omega t - phi) for large t.
            Therefore, x_p(t) is the OBSERVABLE output.
   THEOREM. The amplitude C(omega) is maximized over all possible
            input frequencies omega>0 by the single choice
                omega = sqrt(k/m - c^2/(2m^2)).
   DEFINITION. The practical resonance frequency is the number omega
               defined by the above square root expression.
 
 Intro to the Laplace resolvent method 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 x u(0)
      Resolvent = inverse(sI - A)
   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).
   Remarks on problem 10.2-16, a 3x3 system that can be solved with
     the resolvent equation shortcut.
   RULE: 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.
 Transform Terminology
   Input
   Output
   Transfer Function
 Variation of parameters
   The second order formula.
   Application to y''=1+x
   Application to y''+y=sec(x) [see also slides]
   How to calculate y_p(x) from the five parameters
    y1(x)
    y2(x)
    W(x) = y1(x)y2'(x)-y1'(x)y2(x)
    A(x) = coefficient in the DE of y''
    f(x) = input or forcing term, the RHS of the DE
   See (33) in section 5.5.
    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
 

Slides: Basic undetermined coefficients, draft 4 (104.9 K, pdf, 07 Nov 2009)
Slides: Variation of parameters (109.8 K, pdf, 07 Nov 2009)
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 (155.7 K, pdf, 27 Nov 2010)
    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 (1068.7 K, pdf, 28 Nov 2010)
    Text: Laplace theory problem notes for Chapter 10 (8.9 K, txt, 18 Nov 2010)
    Text: Final exam study guide (8.3 K, txt, 09 Dec 2010)
    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)
    Systems of Differential Equations references
    Manuscript: Systems of DE examples and 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)
    Oscillations. Mechanical and Electrical.
    Slides: Electrical circuits (87.1 K, pdf, 11 Oct 2009)
    Slides: Forced damped vibrations (235.0 K, pdf, 11 Oct 2009)
    Slides: Forced vibrations and resonance (185.3 K, pdf, 11 Oct 2009)
    Slides: Forced undamped vibrations (174.7 K, pdf, 11 Oct 2009)
    Slides: Resonance and undetermined coefficients (199.3 K, pdf, 20 Nov 2010)
    Slides: Unforced vibrations 2008 (620.4 K, pdf, 11 Oct 2009)
    Eigenanalysis and Systems of Differential Equations.
    Manuscript: Eigenanalysis 2010, 46 pages (345.3 K, pdf, 31 Mar 2010)
    Manuscript: Algebraic eigenanalysis 2008 (127.8 K, pdf, 23 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)
    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)
    Manuscript: What's eigenanalysis 2008 (126.8 K, pdf, 11 Apr 2010)
    Slides: Cayley-Hamilton method for solving vector-matrix system u'=Au. (152.6 K, pdf, 23 Nov 2010)