Edwards-Penney, sections 5.4, 10.1, 10.2, 10.3 The textbook topics, definitions and theorems

Edwards-Penney 10.1, 10.2, 10.3, 10.4, 10.5 (20.5 K, txt, 03 Mar 2013)

ApplicationsHow to solve differential equations Solving y''+y=0, y(0)=0, y'(0)=1 Solving y''+y=1, y(0)=y'(0)=0 Solving y''+y=cos(t), y(0)=y'(0)=0 (resonance) Solving y''+y=cos(2t), y(0)=y'(0)=0 (beats) LRC Circuit Specialized mechanical models. Pure Resonance x''+x=cos(t), frequency mnatching Solution explosion, unbounded solution x=(1/2) t sin t. Practical Resonance: x'' + x = cos(omega t) with omega near 1 Large amplitude harmonic oscillations: Pure resonance y = x sin(x) (74.7 K, pdf, 18 Mar 2013) Resonance examples: 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. Millenium Foot-Bridge London Beats x''+x=cos(2t) Graphics for beats [x=sin(10 t)sin(t/2)], slowly-oscillating envelope, rapidly oscillating harmonic with time-varying amplitude.: Beats y=sin(10x)sin(x/2) (68.9 K, pdf, 18 Mar 2013)Theory of Practical Resonance: Forced vibrations and resonance (240.3 K, pdf, 04 Mar 2012) 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. TheSlidespractical resonance frequencyis the number omega defined by the above square root expression.Chapter 5 references: 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)Slides: Water glass shattering due to resonant sound waves. (96.8 K, mov, 21 Mar 2013)MOVIE

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)) = 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 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 10.5-25. Answer: (1/s)tanh(as/2)Laplace theory references: 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)Slides: Maple Lab 7. Laplace applications (156.0 K, pdf, 04 Dec 2012)MAPLE: DE systems, examples, theory (785.8 K, pdf, 16 Nov 2008)Manuscript: 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)Slides: Heaviside's method 2008 (186.8 K, pdf, 20 Oct 2009)Manuscript: Laplace theory 2008 (351.3 K, pdf, 15 Apr 2009)Manuscript: Ch10 Laplace solutions 10.1 to 10.4 (1068.7 K, pdf, 28 Nov 2010)Transparencies: Laplace theory problem notes F2008 (17.2 K, txt, 03 Dec 2012)Text: Final exam study guide (8.2 K, txt, 05 Dec 2012)Text

Convolution 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(cos t)L(sin t) = L(0.5 t sin(t))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.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 (88.1 K, pdf, 04 Mar 2012)SlidesIntro to the Laplace resolvent shortcut for 2x2 systemsProblem: 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).

Laplace Resolvent MethodConsider 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. TheRESOLVENTis 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 functionDefinition 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 delta.Resonance examplesx'' + 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). Future 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.Laplace theory references: 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)Slides: Maple Lab 7. Laplace applications (156.0 K, pdf, 04 Dec 2012)MAPLE: DE systems, examples, theory (785.8 K, pdf, 16 Nov 2008)Manuscript: 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)Slides: Heaviside's method 2008 (186.8 K, pdf, 20 Oct 2009)Manuscript: Laplace theory 2008 (351.3 K, pdf, 15 Apr 2009)Manuscript: Ch10 Laplace solutions 10.1 to 10.4 (1068.7 K, pdf, 28 Nov 2010)Transparencies: Laplace theory problem notes for Chapter 10 (17.2 K, txt, 03 Dec 2012)Text: Final exam study guide (8.2 K, txt, 05 Dec 2012)TextVariation of Parameters and Undetermined Coefficients references: Basic undetermined coefficients, draft 4 (104.9 K, pdf, 07 Nov 2009)Slides: Variation of parameters (164.5 K, pdf, 03 Mar 2012)SlidesSystems of Differential Equations references: Systems of DE examples and theory (785.8 K, pdf, 16 Nov 2008)Manuscript: 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)SlidesOscillations. Mechanical and Electrical.: 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)SlidesEigenanalysis and Systems of Differential Equations.: Eigenanalysis 2010, 46 pages (345.3 K, pdf, 31 Mar 2010)Manuscript: Algebraic eigenanalysis 2008 (187.6 K, pdf, 04 Mar 2012)Manuscript: Lawrence Page's pagerank algorithm (0.7 K, txt, 06 Oct 2008)Text: History of telecom companies (1.9 K, txt, 04 Apr 2013)Text: 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)Manuscript: Cayley-Hamilton-Ziebur method for solving vector-matrix system u'=Au. 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