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)

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 entriesPiecewise 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) Applications of Laplace's method from 10.3, 10.4, 10.5Convolution 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(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.

Forward and Backward Table ApplicationsReview 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 ApplicationsProblem 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 ApplicationsProblem 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 thmS-differentiation theoremProblem 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 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: 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 applicationProblem 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 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 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 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 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 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.

Maple Example to find roots of the characteristic equationConsider the recirculating brine tank example: 20 x' = -6x + y, 20 y' = 6x - 3y Themaple codeto 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 WARNINGReading 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 calledFourier'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 ModelTHEOREM. 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: 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: Optional Maple Lab 7. Laplace applications (0.0 K, pdf, 31 Dec 1969)MAPLE: DE systems, examples, theory (730.9 K, pdf, 10 Apr 2014)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 (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)Slides: Heaviside's method 2008 (352.3 K, pdf, 07 Jan 2014)Manuscript: Laplace theory 2008 (500.9 K, pdf, 16 Mar 2014)Manuscript: Ch10 Laplace solutions 10.1 to 10.4 (1068.7 K, pdf, 28 Nov 2010)Transparencies: Laplace theory problem notes for Chapter 10 (17.7 K, txt, 18 Mar 2014)Text: Final exam study guide (8.0 K, txt, 20 Apr 2014)TextSystems of Differential Equations references: Systems of DE examples and theory (730.9 K, pdf, 10 Apr 2014)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 (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)Slides

References for Eigenanalysis and Systems of Differential Equations.: Algebraic eigenanalysis (187.6 K, pdf, 04 Mar 2012)Sildes: 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)Slides: Systems of DE examples and theory (730.9 K, pdf, 10 Apr 2014)Manuscript: Home heating, attic, main floor, basement (99.3 K, pdf, 10 Apr 2014)Slides: 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: Cable hoist example (73.2 K, pdf, 21 Aug 2008)Slides: Sliding plates example (105.8 K, pdf, 21 Aug 2008)Slides