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2250-1 7:30am Lectures Week 12 S2014

Last Modified: April 24, 2014, 11:14 MDT.    Today: December 14, 2017, 23:17 MST.

Week 12: Sections 10.4, 10.5, EPbvp7.6, 6.1, 6.2

 Edwards-Penney, sections 10.4, 10.5, EPbvp7.6, 6.1, 6.2
  The textbook topics, definitions and theorems
Edwards-Penney 10.1, 10.2, 10.3, 10.4, 10.5 (20.5 K, txt, 19 Dec 2013)
Edwards-Penney 6.1, 6.2 (7.6 K, txt, 19 Dec 2013)

Monday and Tuesday: Piecewise Functions. Sections 10.4 and EPbvp supplement 7.6.

 DEF. Gamma function
    Gamma(t) = integral x=0 to x=infinity  x^{t-1} e^{-x}
    Gamma(n)=(n-1)!, generalizes the factorial function
 DEF. Mellin transform
   {Mf](s)= phi(s)=integral x=0 to x=infinity  x^{s-1} f(x)
 DEF. Two-sided Laplace transform
   {Bf}(s) = {Mf(-ln(x))}(s) = integral x=0 to x=infinity x^{s-1}f(-ln x)
 DEF. Unit step u(t-a)=1 for t>=a, else zero
 DEF. Ramp t->(t-a)u(t-a)
 Backward table problems: examples
 Forward table problems: examples
 Computing Laplace integrals L(f(t)) with rules
 Solving an equation L(y(t))=expression in s for y(t)
    Complex roots and quadratic factors
    Partial fraction methods
 Trig identities and their use in Laplace calculations
 Hyperbolic functions and Laplace calculations
   Why the forward and backward tables don't have cosh, sinh entries

 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 10.5-25.
      Answer: (1/s)tanh(as/2)

Applications of Laplace's method from 10.3, 10.4, 10.5
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
 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.

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, 04 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, Wednesday: Piecewise Functions. Shifting. Section 10.5 and EPbvp supplement 7.6. Delta function and hammer hits.

Laplace Resolvent Method
  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.
 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.

 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 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 delta.
 

Monday, Tuesday, Wednesday: Problem session. Sections 10.1 to 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 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 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 application
    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: 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 10.5-25. Done earlier.
      Answer: (1/s)tanh(as/2)
    Laplace of the sawtooth wave. Problem 10.5-26. 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 10.5-27. 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 10.5-28. 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 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)
     Problem 10.5-4.
      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 10.5-22.
     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.

Friday: Sections 6.1, 7.1

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)
EIGENANALYSIS WARNING
  Reading Edwards-Penney Chapter 6 may deliver the wrong ideas
  about how to solve for eigenpairs. The examples emphasize a
  clever shortcut, which does not apply in general to solve for
  eigenpairs.

     HISTORY. Chapter 6 originally appeared in the 2280 book
     as a summary, which assumed a linear algebra course. The
     chapter was copied without changes into the Edwards-Penney
     Differential Equations and Linear Algebra textbook, which you
     currently own. The text contains only shortcuts. There is
     no discussion of a general method for finding eigenpairs.
     You will have to fill in the details by yourself. The online
     lecture notes and slides were created to fill in the gap.

Lecture: Fourier's Model. Intro to Eigenanalysis, Ch6.
  Examples and motivation.
     Ellipse, rotations, eigenpairs.
     General solution of a differential equation u'=Au and eigenpairs.
  Fourier's model.
  History.
    J.B.Fourier's 1822 treatise on the theory of heat.
    The rod example.
      Physical Rod: a welding rod of unit length, insulated on the
                    lateral surface and ice packed on the ends.
    Define f(x)=thermometer reading at loc=x along the rod at t=0.
    Define u(x,t)=thermometer reading at loc=x and time=t>0.
    Problem: Find u(x,t).
      Fourier's solution. Let's assume that
      f(x) = 17 sin (pi x) + 29 sin(5 pi x)
           = 17 v1 + 29 v2
      Packages v1, v2 are vectors in a vector space V of functions on [0,1].
      Fourier computes u(x,t) by re-scaling v1, v2 with numbers Lambda_1,
      Lambda_2 that depend on t. This idea is called Fourier's Model.

      u(x,t) = 17 ( exp(-pi^2 t) sin(pi x)) + 29 ( exp(-25 pi^2 t) sin (5 pi x))
             = 17 (Lambda_1 v1) + 29 (Lambda_2 v2)

  Eigenanalysis of u'=Au is the identical idea.
     u(0) = c1 v1 + c2 v2  implies
     u(t) = c1 exp(lambda_1 t) v1 + c2 exp(lambda_2 t) v2
     Fourier's re-scaling idea from 1822, applied to u'=Au,
       replaces v1 and v2 in the expression
                c1 v1 + c2 v2
       by their re-scaled versions to obtain the answer
                c1 (Lambda1 v1) + c2 (Lambda2 v2)
       where
         Lambda1 = exp(lambda_1 t), Lambda2 = exp(lambda_2 t).

Main Theorem on Fourier's Model

  THEOREM. Fourier's model
     A(c1 v1 + c2 v2) = c1 (lambda1 v1) + c2 (lambda2 v2)
  with v1, v2 a basis of R^2 holds [for all constants c1, c2]
    if and only if
  the vector-matrix system
    A(v1) = lambda1 v1,
    A(v2) = lambda2 v2,
  has a solution with vectors v1, v2 independent
    if and only if
  the diagonal matrix D=diag(lambda1,lambda2) and
  the augmented matrix P=aug(v1,v2) satisfy
     1. det(P) not zero [then v1, v2 are independent]
     2. AP=PD

  THEOREM. The eigenvalues of A are found from the determinant
    equation
                        det(A -lambda I)=0,
    which is called the characteristic equation.
  THEOREM. The eigenvectors of A are found from the frame
    sequence which starts with B=A-lambda I [lambda a root of
    the characteristic equation], ending with last frame rref(B).

    The eigenvectors for lambda are the partial derivatives of
    the general solution obtained by the Last Frame Algorithm,
    with respect to the invented symbols t1, t2, t3, ...

Algebraic Eigenanalysis Section 6.2.
  Calculation of eigenpairs to produce Fourier's model.
    Connection between Fourier's model and a diagonalizable matrix.
    How to find the variables lambda and v in Fourier's model using
      determinants and frame sequences.
  Solved in class: examples similar to the problems in 6.1 and 6.2.
    Web slides and problem notes exist for the 6.1 and 6.2 problems.
  Examples where A has an eigenvalue of multiplicity greater than one.
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 (108.5 K, pdf, 03 Mar 2013)
MAPLE: Optional Maple Lab 7. Laplace applications (0.0 K, pdf, 31 Dec 1969)
Manuscript: DE systems, examples, theory (730.9 K, pdf, 10 Apr 2014)
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 (99.3 K, pdf, 10 Apr 2014)
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 (352.3 K, pdf, 07 Jan 2014)
Manuscript: Laplace theory 2008 (500.9 K, pdf, 16 Mar 2014)
Transparencies: Ch10 Laplace solutions 10.1 to 10.4 (1068.7 K, pdf, 28 Nov 2010)
Text: Laplace theory problem notes for Chapter 10 (17.7 K, txt, 18 Mar 2014)
Text: Final exam study guide (8.0 K, txt, 20 Apr 2014) Systems of Differential Equations references
Manuscript: Systems of DE examples and theory (730.9 K, pdf, 10 Apr 2014)
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 (99.3 K, pdf, 10 Apr 2014)
Slides: Cable hoist example (73.2 K, pdf, 21 Aug 2008)
Slides: Sliding plates example (105.8 K, pdf, 21 Aug 2008)
References for Eigenanalysis and Systems of Differential Equations.
 
Sildes: Algebraic eigenanalysis (187.6 K, pdf, 04 Mar 2012)
Slides: 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)
Slides: Laplace resolvent method (88.1 K, pdf, 04 Mar 2012)
Slides: Laplace second order systems (288.1 K, pdf, 04 Mar 2012)
Manuscript: Systems of DE examples and theory (730.9 K, pdf, 10 Apr 2014)
Slides: Home heating, attic, main floor, basement (99.3 K, pdf, 10 Apr 2014)
Text: Lawrence Page's pagerank algorithm (0.7 K, txt, 06 Oct 2008)
Text: History of telecom companies (1.9 K, txt, 04 Apr 2013)
Slides: Cable hoist example (73.2 K, pdf, 21 Aug 2008)
Slides: Sliding plates example (105.8 K, pdf, 21 Aug 2008)