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 (20.1 K, txt, 17 Apr 2015)

Edwards-Penney 4.1 to 4.3 (10.4 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 (88.1 K, pdf, 03 Mar 2012) 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. (137.7 K, pdf, 28 Mar 2015) 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.

Numerical Methods for Systems: Section 4.3 Review of Scalar methods Euler, Heun, RK4 Change the methods to the vector case, based upon the statement of the IVP in Picard's Theorem. It amounts to putting arrows over y and f. Vector Methods Euler, Heun, RK4 Slides: Vector Methods EXAMPLE. Solve the vector problem u'=Au, u(0)=u_0 with A=Matrix([[1,2],[0,3]]) and u_0 = <1,-1> using Euler's method and Heuns's method with step size h=0.1.