TopicsSections 7.1 to 7.6 The textbook topics, definitions, examples and theorems

Edwards-Penney Ch 7, 7.1 to 7.5 (21.0 K, txt, 22 Feb 2015)

Laplace Methods for SystemsWe'll use the textbook, section 4.1, as a reference for applications in Laplace theory to systems of equations.Intro to the Laplace resolvent shortcut for 2x2 systemsTheory: Picard's vector theorem on existence-uniqueness 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 systemsSolve 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).

Forward and Backward Table ApplicationsReview of previously solved problems. Problem 7.1-18. Trig identity 2 sin (theta) cos(theta) = sin(2 theta) used for L(sin(3t)cos(3t)). Problem 7.1-28. Splitting a fraction into backward table entries.Partial Fractions and Backward Table ApplicationsProblem 7.2-24. L(f)=1/(s(s+1)(s+2)) solved by the three methods for partial fractions: sampling, atom method, Heaviside cover-up. Problem 7.2-9. 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). Forward and backward table entries for sine and cosine. Why cosh(t) and sinh(t) are not in the tables.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.: Laplace resolvent method (81.0 K, pdf, 14 Mar 2016) First example: x'=x-y, y'=x+y, x(0)=1, y(0)=2 Solve with the resolvent equation shortcut. Second example: x'=-x-y, y'=x-y, x(0)=1, y(0)=2 Solve with the resolvent equation shortcut. 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 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. TheSlidesRESOLVENTis 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.

Piecewise FunctionsUnit 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 TheoremL(int(g(x),x=0..t)) = (1/s) L(g(t)) Applications to computing ramp(t-a) L(ramp(t-a)) = (1/s^2) exp(-as) [for a >= 0 only] This formula is dupicated by the Second shifting Theorem, applied to f(t)=t.Piecewise defined periodic wavesSquare 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 wavePeriodic function theoremProof 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, 7.5Convolution theoremDEF. 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(int(g(x),x=0..t)) = L(1) L(g(t)) = (1/s)L(g(t)), which is the Integral Theorem Application: L(cos t)L(sin t) = L(0.5 t sin(t))

Piecewise FunctionsUnit 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]Piecewise defined periodic wavesSquare 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 wavePeriodic function theoremProof details Laplace of the square wave. Problem 7.5-25. Answer: (1/s)tanh(as/2)Second shifting Theoremse^{-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.

Hammer hits and the Dirac ImpulseDefinition 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 modelsShort 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 impulse.

Shifting Theorem and u-substitution ApplicationsProblem 7.3-8. L(f)=(s-1)/(s+1)^3 See #18 details for a similar problem. Problem 7.3-18. 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 7.3-8. 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 thmS-differentiation theoremProblem 7.4-21. Similar to Problem 7.4-22. 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 theoremTHEOREM. 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: 7.4-36. 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 applicationProblem 7.5-28. 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 FunctionsUnit 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 theoremLaplace of the square wave. Problem 7.5-25. Done earlier. Answer: (1/s)tanh(as/2) Laplace of the sawtooth wave. Problem 7.5-26. 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 7.5-27. 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 7.5-28. 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 ApplicationsSecond shifting Theoremse^{-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 7.5-3. L(f)=e^{-s}/(s+2) Problem 7.5-4. L(f) = (e^{-s} - e^{2-2s})/(s-1) Problem 7.5-22. f(t)=t^3 pulse(t,1,2) Problem 7.5-4. 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 7.5-22. 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 Applicationsx''+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.

Laplace theory references: Laplace and Newton calculus. Photos. (188.3 K, pdf, 14 Mar 2016)Slides: Intro to Laplace theory. Calculus assumed. (144.9 K, pdf, 13 Mar 2016)Slides: Laplace theory (497.3 K, pdf, 19 Mar 2014)Manuscript: Laplace rules (144.9 K, pdf, 14 Mar 2016)Slides: Laplace table proofs (151.9 K, pdf, 14 Mar 2016)Slides: Laplace examples (133.7 K, pdf, 14 Mar 2016)Slides: Piecewise functions and Laplace theory (98.5 K, pdf, 14 Mar 2016)Slides: Laplace resolvent method (81.0 K, pdf, 14 Mar 2016)Slides: Cable hoist example (73.2 K, pdf, 21 Aug 2008)Slides: Sliding plates example (105.8 K, pdf, 21 Aug 2008)Slides: Heaviside's method (352.3 K, pdf, 07 Jan 2014)Manuscript: DE systems, examples, theory (730.9 K, pdf, 10 Apr 2014)Manuscript: Laplace theory 2008 (497.3 K, pdf, 19 Mar 2014)Manuscript: Ch7 Laplace solutions 7.1 to 7.4 (1068.7 K, pdf, 22 Feb 2015)Transparencies: Laplace second order systems (273.7 K, pdf, 14 Mar 2016)Slides