Week 11: 09 Nov, |
10 Nov, | 11 Nov, | 12 Nov, | 13 Nov, |

Undetermined CoefficientsWhich equations can be solved Intro to the basic trial solution method Laplace solution of y'' + y = 1+x [use x''(t)+x(t) = 1+t, x(0)=x'(0)=0] How to find the atoms in y_p(x). How to find the atoms in y_h(x) THEOREM. Solution y_h(x) is a linear combination of atoms. THEOREM. Solution y_p(x) is a linear combination of atoms. THEOREM. (superposition) y = y_h + y_p EXAMPLE. How to find a shortest expression for y_p(x) using Laplace's method. Details for x''(t)+x(t) = 1+t, to obtain the trial solution x(t)=A+Bt and the answer x(t)=1+t.Intro to the Laplace resolvent method for 2x2 systemsConverting a dynamical system to vector-matrix form u'=Au. Position-velocity substitution and the harmonic oscillator. Solve the systems by ch1 methods x' = 2x, x(0)=100, y' = 3y, y(0)=50. Answer: x=100exp(2t), y=50 exp(3t) x' = 2x+y, x(0)=1, y' = 3y, y(0)=2. Answer: y=2 exp(3t) and x(t) is the solution of the linear integrating factor problem x'=2x+2 exp(3t).

Engineering modelsThe job-site cable hoist example [delayed] Sliding plates example [delayed] Home heating example [delayed]

Forward and Backward Table ApplicationsProblem 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 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.Shifting Theorem and u-substitution ApplicationsProblem 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+1 L(f) = s/(s^2 + 1) where s --> s+1 L(f) = L(e^{-t} cos(t)) by shifting thmSecond Shifting Theorem ApplicationsProblem 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)Piecewise ApplicationsStaircase or floor function Sawtooth wave Square waveDirac Applicationsx''+x=5 Delta(t-1), x(0)=0,x'(0)=1

More Laplace ExamplesContinuing 10.3, 10.5 examples from last lecture.Transform TerminologyInput Output Transfer FunctionVariation of parametersThe second order formula. Application to y''=1+x Application to y''+y=sec(x) [slides]Undetermined CoefficientsBASIC METHOD. Given a trial solution with undetermined coefficients, find a system of equations for d1, d2, ... and solve it. Report y_p as the trial solution with substituted answers d1, d2, d3, ... METHOD to FIND the SOLUTION. Laplace solution of y'' + y = f(x) when f(x)=linear combination of atoms THEORY. y = y_h + y_p, and each is a linear combination of atoms. How to find the homogeneous solution y_h(x) from the characteristic equation. How to determine the form of the trial solution for y_p(x) Laplace theory method. A rule for finding y_p(x) without using Laplace theory. Finding the redundant trial solution from g(x) = x^n f(x). Finding a trial solution with fewest symbols [non-redundant trial sol]. Finding the non-redundant trial solution from g(x) = f(x) and thecross-out correction rule. Relation between the non-redundant trial solution and the book's table that uses the mystery factor x^s. EXAMPLES.

Laplace resolvent method for 2x2 systemsModel u'=Au, u(0)=u_0 x' = 2x+y, x(0)=1, y' = 3y, y(0)=2. How to solve it by Laplace's method. The resolvent equations, before the answers are found for L(x), L(y) (s-2)L(x) + (-1)L(y) = 1, (0)L(x) + (s-3)L(y) = 2 A shortcut to the resolvent equations. Laplace resolvent formula (sI-A)L(u)=u(0) Problem 10.2-16. x'=x+z, y'=x+y, z'= -2x -z, x(0)=1, y(0)=z(0)=0. How to find the resolvent equations. Backward table solution. Problem 10.2-6. x''+4x=cos(t), x(0)=x'(0)=0. Resolvent equation of the dynamical system Solving the equation by standard Laplace methods.Wine Glass ExperimentThe lab table setup Speaker. Frequency generator with adjustment knob. Amplifier with volume knob. Wine glass. x(t)=deflection from equilibrium of the radial component of the glass rim, represented in polar coordinates, orthogonal to the speaker front. mx'' + cx' + kx = F_0 cos(omega t) The model of the wine glass m,c,k are properties of the glass sample itself F_0 = volume knob adjustment omega = frequency generator knob adjustmentTheory of Practical ResonanceThe 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. Thepractical resonance frequencyis the number omega defined by the above square root expression.

Projection: glass-breaking video [delayed] Next week: More on resonance, details of practical resonance theory. Next week: Calculations from 5.6.