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)

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)

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(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: Second Shift Theorem. 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.: Laplace resolvent method (81.0 K, pdf, 14 Mar 2016) The general vector-matrix DE Model u'=Au Laplace of u(t) = Resolvent times u(0) Resolvent = inverse(sI - A)SlidesChapter 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 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. 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. 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 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 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 impulse. 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 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. 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 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).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 to be 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 Dirac impulse violates this theorem's hypothesis, because L(delta(t))=1.

Cayley-Hamilton-Ziebur Method: Ch4, sections 4.1, 4.2Cayley-Hamilton TheoremA matrix satisfies its own characteristic equation. ILLUSTRATION: det(A-r I)=0 for the previous example is (2-r)(3-r)=0 or r^2 -5r + 6=0. Then C-H says (2I-A)(3I-A)=0 or A^2 - 5A + 6I = 0.Cayley-Hamilton-Ziebur MethodZIEBUR'S LEMMA. The components of u in u'=Au are linear combinations of the atoms created by Euler's theorem applied to the roots of the characteristic equation det(A-rI)=0. THEOREM. Solve u'=Au without complex numbers or eigenanalysis. The solution of u'=Au is a linear combination of atoms times certain constant vectors [not arbitrary vectors]. u(t)=(atom_1)vec(c_1)+ ... + (atom_n)vec(c_n): Cayley-Hamilton-Ziebur method for solving vector-matrix system u'=Au. (137.4 K, pdf, 14 Mar 2016) PROBLEM: Solve by Cayley-Hamilton-Ziebur the 2x2 dynamical system x' = 2x + y, y' = 3y, x(0)=1, y(0)=2. The characteristic equation is (2-lambda)(3-lambda)=0 with roots lambda = 2,3 Euler's theorem implies the atoms are exp(2t), exp(3t). Ziebur's Theorem says that u(t) = exp(2t) vec(v_1) + exp(3t) vec(v_2) where vectors v_1, uv_2 are to be determined from the matrix A = matrix([[2,1],[0,3]]) and initial conditions x(0)=1, y(0)=2. ZIEBUR ALGORITHM. To solve for v_1, v_2 in the example, differentiate the equation u(t) = exp(2t) v_1 + exp(3t) v_2 and set t=0 in both relations. Then u'=Au implies u_0 = v_1 + v_2, Au_0 = 2 v_1 + 3 v_2. These equations can be solved by elimination. The answer: v_1 = (3 u_0 -Au_0), v_2 = (Au_0 - 2 u_0) = vector([-1,0]) = vector([2,2]) Vectors v_1, v_2 are recognized as eigenvectors of A for lambda=2 and lambda=3, respectively. ZIEBUR SHORTCUT [Edwards-Penney textbook method, Example 5 in 4.1] Start with Ziebur's theorem, which implies that x(t) = k1 exp(2t) + k2 exp(3t). Use the first DE to solve for y(t): y(t) = x'(t) - 2x(t) = 2 k1 exp(2t) + 3 k2 exp(3t) - 2 k1 exp(2t) - 2 k2 exp(3t)) = k2 exp(3t) For example, x(0)=1, y(0)=2 implies k1 and k2 are defined by k1 + k2 = 1, k2 = 2, which implies k1 = -1, k2 = 2, agreeing with a previous solution formula.SlidesCayley-Hamilton topics, Section 6.3.Cayley-Hamilton TheoremA matrix satisfies its own characteristic equation. Proof of the Ziebur Lemma for 2x2 matrices.