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2280 12:55pm Lectures Week 9 S2018

Last Modified: March 07, 2018, 12:17 MST.    Today: June 23, 2018, 17:48 MDT.
 Edwards-Penney, sections 7.1 to 7.6 and 4.1 to 4.3
  The textbook topics, definitions and theorems
Edwards-Penney 7.1 to 7.5 (21.0 K, txt, 22 Feb 2015)
Edwards-Penney 4.1 to 4.3 (10.7 K, txt, 05 Jan 2015)

Monday-Tuesday: Piecewise Functions, Convolution, Resolvent

DEF: Piecewise Continuous Function
   Existence of the Laplace integral.
   One-sided and two-sided Laplace integral
      Freeway example, suspension collides with a ramp.

   One-sided and two-sided Laplace integral
      Freeway example, suspension collides with a ramp.

   DEF. Gamma function
      Gamma(t) = integral x=0 to x=infinity  x^{t-1} e^{-x} dx
      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) dx
   DEF. Two-sided Laplace transform
      {Bf}(s) = {Mf(-ln(x))}(s) = integral x=0 to x=infinity x^{s-1}f(-ln x) dx

 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
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).

 Brine Tank Cascades

   The three-tank problem has a special form called
   a cascade because the system can be solved by Ch1
   methods: Linear Integrating Factor method and shortcuts.

   x' = b0 - b1 x,  Tank 1 amount x(t)
   y' = b1 x - b2 y, Tank 2 amount y(t),
   z' = b2 y - b3 z, Tank 3 amount z(t).

   In system form: u'=Au, A=3x3 band matrix. All such
   systems can be solved by the same method. Laplace 
   also works, not much worse for details.

 Coupled Mechanical Oscialltions
    EXAMPLE. Forced system of undamped oscillators
    m1 x'' = - k1 x + k2 (y-x),
    m2 y'' = -k2(y-x) + f(t)
    INSTANCE m1=2, m2=1, k1=4, k2=2, f(t)=40sin(3t)
       2 x'' = - 6x + 2y,
         y'' = 2x - 2y + 40 sin(3t)
    Can be solved by Laplace's method, but tedious.
    This can be converted to a 4x4 system solving for
    four variables x,x',y,y'. The first order system
    can be solved by a variety of methods, Laplace being
    just one method.

Recirculating Brine Tank: 2 Lakes Pollution Example
 Consider the recirculating brine tank example:
    20 x' = -6x + y,
    20 y' = 6x - 3y
 The is u'=Au in vector-matrix form.
 We have to solve for roots of the characteristic equation

 Maple Example
 The maple code to solve the char eq:
   # Answer: -9/40+(1/40)*sqrt(33), -9/40-(1/40)*sqrt(33)

 We will also solve the system by Laplace methods, getting the answers
 The method: Resolvement method below.

 Maple code
  eq1:=20*diff(x(t),t)=-6*x(t) + y(t); 
  eq2:=20*diff(y(t),t)=6*x(t) - 3*y(t);
Laplace Resolvent Method
  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
    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. 

 For the example, the resolvent equation is

    Matrix([[s-1,0,-1],[-1,s-1,0],[2,0,s+1]]).Vector([L(x),L(y),L(z)]) = Vector([1,0,0]);

 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
    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. The determinants
        must be evaluated by the cofactor method, due to symbols.

 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) dx 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))
       This is a BEATS example from 3.6, solved by Laplace theory.
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
   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 (81.0 K, pdf, 14 Mar 2016)

Friday: Dirac Impulse. Energy injection like hammer hits.

 Hammer hits and the Delta impulsen
   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 impulses 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 impulse.
   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 impulse.
     EXAMPLE. The delta impulse 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 impulse 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.
        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);
   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.