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2250-1 7:30am Lecture Record Week 11 S2012

Last Modified: April 10, 2012, 12:34 MDT.    Today: October 19, 2017, 01:17 MDT.

Week 11, March 26 to April 2: Sections 10.5,EPbvp7.6,5.5,5.6,,EPbvp3.7,7.1

Thu, Apr 1:Patrick

Exam 2, problems 4,5.

Mon, Mar 26: Piecewise Functions. Section 10.5 and EPbvp supplement 7.6.

Laplace Resolvent Method.
  --> This method is considered a shortcut for solving systems by Laplace's method.
  --> It is also a convenient way to solve systems with maple.
  Consider problem 10.2-16
       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.
 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
   vector([L(x),L(y),L(z)]) = L(u(t)) = inverse(sI - B) u0
Briefly,
    Laplace of u(t) = RESOLVENT x u(0)
Intro to the Laplace resolvent method for 2x2 systems
  Remarks on problem 10.2-16, a 3x3 system that can be solved with
     the resolvent equation shortcut.
   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).
    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.

Mar 27,28: 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))
 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)
     delta(t) = idealized injection of energy into a system at
                   time t=0 of impulse=1.
   Hammer hit models
   How delta functions enter into circuit calculations
   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 [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))  Previously 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).
 later lecture, with slides and video:
    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.

Mar 27,28: 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 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: 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))
 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)
 Periodic function theorem
    Laplace of the square wave. Problem 10.5-25. Done earlier.
      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
          L(floor(t/a))=1/(s(exp(as)-1))
      This answer can be verified in maple by the code
         with(inttrans):
         laplace(floor(t/a),t,s);
    Laplace of the sawtooth wave, revisited.
       Identity: sqw(t) = floor(t) = staircase with jump 1.
       Identity: t - floor(t) = saw(t) = sawtooth wave
       General:  t - a*floor(t/a) = saw(t,a) = sawtooth wave of period a.
    Problem 10.5-28. Details
         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))

Fri, Mar 30: Laplace theory, Dirac, Variation of parameters, 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)
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.
    Undetermined Coefficients
   Which equations can be solved
   Intro to the basic trial solution method
      Laplace solution of x'' + 9x = 30 sin(2t)
      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 (164.5 K, pdf, 03 Mar 2012)
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. (200.2 K, pdf, 04 Mar 2012)
    Slides: Intro to Laplace theory. Calculus assumed. (163.0 K, pdf, 19 Mar 2012)
    Slides: Laplace rules (160.3 K, pdf, 04 Mar 2012)
    Slides: Laplace table proofs (169.6 K, pdf, 04 Mar 2012)
    Slides: Laplace examples (149.1 K, pdf, 04 Mar 2012)
    Slides: Piecewise functions and Laplace theory (108.1 K, pdf, 04 Mar 2012)
    MAPLE: Maple Lab 7. Laplace applications (60.6 K, pdf, 09 Dec 2011)
    Manuscript: DE systems, examples, theory (785.8 K, pdf, 16 Nov 2008)
    Slides: Laplace resolvent method (88.1 K, pdf, 04 Mar 2012)
    Slides: Laplace second order systems (288.1 K, pdf, 04 Mar 2012)
    Slides: Home heating, attic, main floor, basement (109.8 K, pdf, 04 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 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 (8.2 K, txt, 11 Dec 2011)
    Variation of Parameters and Undetermined Coefficients references
    Slides: Basic undetermined coefficients, draft 4 (104.9 K, pdf, 07 Nov 2009)
    Slides: Variation of parameters (164.5 K, pdf, 03 Mar 2012)
    Systems of Differential Equations references
    Manuscript: Systems of DE examples and theory (785.8 K, pdf, 16 Nov 2008)
    Slides: Laplace resolvent method (88.1 K, pdf, 04 Mar 2012)
    Slides: Laplace second order systems (288.1 K, pdf, 04 Mar 2012)
    Slides: Home heating, attic, main floor, basement (109.8 K, pdf, 04 Mar 2012)
    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 (124.8 K, pdf, 04 Mar 2012)
    Slides: Forced damped vibrations (280.7 K, pdf, 04 Mar 2012)
    Slides: Forced vibrations and resonance (240.3 K, pdf, 04 Mar 2012)
    Slides: Forced undamped vibrations (214.2 K, pdf, 03 Mar 2012)
    Slides: Resonance and undetermined coefficients (199.4 K, pdf, 03 Mar 2012)
    Slides: Unforced vibrations 2008 (667.9 K, pdf, 03 Mar 2012)
    Eigenanalysis and Systems of Differential Equations.
    Manuscript: Eigenanalysis 2010, 46 pages (345.3 K, pdf, 31 Mar 2010)
    Manuscript: Algebraic eigenanalysis 2008 (187.6 K, pdf, 04 Mar 2012)
    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 (174.2 K, pdf, 04 Mar 2012)
    Slides: Cayley-Hamilton-Ziebur method for solving vector-matrix system u'=Au. (152.9 K, pdf, 04 Mar 2012)