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)

DEF:Piecewise Continuous FunctionExistence of the Laplace integral. One-sided and two-sided Laplace integral Freeway example, suspension collides with a ramp. One-sided and two-sided Laplace integral Freeway example, suspension collides with a ramp. DEF. Gamma function Gamma(t) = integral x=0 to x=infinity x^{t-1} e^{-x} dx 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) dx DEF. Two-sided Laplace transform {Bf}(s) = {Mf(-ln(x))}(s) = integral x=0 to x=infinity x^{s-1}f(-ln x) dx 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 entries

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).Brine Tank CascadesThe three-tank problem has a special form called a cascade because the system can be solved by Ch1 methods: Linear Integrating Factor method and shortcuts. x' = b0 - b1 x, Tank 1 amount x(t) y' = b1 x - b2 y, Tank 2 amount y(t), z' = b2 y - b3 z, Tank 3 amount z(t). In system form: u'=Au, A=3x3 band matrix. All such systems can be solved by the same method. Laplace also works, not much worse for details.Coupled Mechanical OscialltionsEXAMPLE. Forced system of undamped oscillators m1 x'' = - k1 x + k2 (y-x), m2 y'' = -k2(y-x) + f(t) INSTANCE m1=2, m2=1, k1=4, k2=2, f(t)=40sin(3t) 2 x'' = - 6x + 2y, y'' = 2x - 2y + 40 sin(3t) Can be solved by Laplace's method, but tedious. This can be converted to a 4x4 system solving for four variables x,x',y,y'. The first order system can be solved by a variety of methods, Laplace being just one method.Recirculating Brine Tank: 2 Lakes Pollution ExampleConsider the recirculating brine tank example: 20 x' = -6x + y, 20 y' = 6x - 3y The is u'=Au in vector-matrix form. We have to solve for roots of the characteristic equationMaple ExampleThe maple code to solve the char eq: A:=(1/20)*Matrix([[-6,1],[6,-3]]); chareq:=linalg[charpoly](A,r); solve(chareq,r); # Answer: -9/40+(1/40)*sqrt(33), -9/40-(1/40)*sqrt(33) We will also solve the system by Laplace methods, getting the answers X:=c1*exp(-(9+alpha)*t/40)+c2*exp((-9+gamma)*t/40); Y:=c3*exp((-9+gamma)*t/40)+c4*exp(-(9+gamma)*t/40); [gamma]=[sqrt(33)]; The method: Resolvement method below.Maple codeeq1:=20*diff(x(t),t)=-6*x(t) + y(t); eq2:=20*diff(y(t),t)=6*x(t) - 3*y(t); dsolve([eq1,eq2],[x(t),y(t)]);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. The determinants must be evaluated by the cofactor method, due to symbols.

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 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) dx 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)) This is a BEATS example from 3.6, solved by Laplace theory.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 (81.0 K, pdf, 14 Mar 2016)Slides

Hammer hits and the Delta impulsenDefinition 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 impulses 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 impulse 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 impulse.