Edwards-Penney, sections 10.4, 10.5, EPbvp7.6, 5.5, 5.6 The textbook topics, definitions and theorems

Edwards-Penney 10.1, 10.2, 10.3, 10.4, 10.5 (20.5 K, txt, 03 Mar 2013)

Edwards-Penney 5.5, 5.6 (15.5 K, txt, 27 Dec 2012)

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)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)).

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.

PREVIEW: Undetermined CoefficientsWhich equations can be solved by undetermined coefficients. 1. Constant coefficients 2. Forcing term a linear combination of Euler solution atoms. Intro to the basic trial solution method Laplace solution of x'' + 9x = 30 sin(2t) x(t)=c1 cos 3t + c2 sin 3t + a cos 2t + b sin 2t = sum of two harmonics, of frequencies 3 and 2 = BEATS example xh(t) = c1 cos 3t + c2 sin 3t, the first harmonic xp(t) = a cos 2t + b sin 2t, the second harmonic Laplace solution of y'' + y = 1+x [use x''(t)+x(t) = 1+t, x(0)=x'(0)=0] x = xh + xp, xh = c1 cos t + c2 sin t, xp = a + b t TRIAL SOLUTION The equation y'' + y = 1 + x should have a solution yp = a + bx for some constants a, b. We find them by substitution of yp into the DE. How to find the Euler solution atoms in y_p(x). 1. Use Laplace theory (a bit slow). 2. Use rules from undetermined coefficient theory. Faster. How to find the Euler solution atoms in y_h(x) Use the characteristic equation roots. 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

: Basic undetermined coefficients, draft 5 (156.3 K, pdf, 29 Mar 2013)Slides: Variation of parameters (164.5 K, pdf, 03 Mar 2012)SlidesExtra Credit Maple Project: Tacoma narrows. Explore an alternative explanation for what caused the bridge to fail, based on the hanging cables.

REVIEW: Undetermined CoefficientsWhich equations can be solved Laplace solution of x'' + 9x = 30 sin(2t) Laplace solution of y'' + y = 1+x [use x''(t)+x(t) = 1+t, x(0)=x'(0)=0] 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: Basic undetermined coefficients, draft 5 (156.3 K, pdf, 29 Mar 2013)Slides: Variation of parameters (164.5 K, pdf, 03 Mar 2012) 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_p(t)=1+t. BASIC 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, ... METHODS to FIND the SOLUTION. Laplace solution of y''(t) + y(t) = f(t) when f(t)=linear combination of atoms We get L(y) = polynomial fraction. By partial fractions and the backward table, L(y)=L(g(t)), where g(t)=linear combination of atoms! This proves the general result: 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 shortest trial solution for y_p(x) METHOD 1. Laplace theory. It works, but it is slow. Central to this method is dividing L(f(t)) by the characteristic polynomial, then writing this polynomial fraction as L(g(t)). The homogeneous solution atoms are shrunk from g(t) to obtain the trial solution. METHOD 2. A rule for finding y_p(x) from f(x) and the DE. We get rid of Laplace theory's t-variable and use x. Finding a trial solution with fewest symbols.SlidesRule I. Assume the right side f(x) of the differential equation is a linear combination of atoms. Make a list of all distinct atoms that appear in the derivatives f(x), f'(x), f''(x), ... . Multiply these k atoms byundetermined coefficientsd_1, ... , d_k, then add to define atrial solution y. This ruleFAILSif one or more of the k atoms is a solution of the homogeneous differential equation.Rule II. If Rule IFAILS, then break the k atoms into groups with the samebase atom. Cycle through the groups, replacing atoms as follows. If the first atom in the group is a solution of the homogeneous differential equation, then multiply all atoms in the group by factor x. Repeat until the first atom is not a solution of the homogeneous differential equation. Multiply the constructed k atoms by symbols d_1, ... , d_k and add to define trial solution y.Explanation: The relation between the Rule I + II trial solution and the book's table that uses the mystery factor x^s. EXAMPLES. y'' = x y'' + y = x exp(x) y'' - y = x exp(x) y'' + y = cos(x) y''' + y'' = 3x + 4 exp(-x) THEOREM. Suppose a list of k atoms is generated from the atoms in f(x), using Rule I. Then the shortest trial solution has exactly k atoms. EXAMPLE. How to find a shortest trial solution using Rules I and II. Details for x''(t)+x(t) = t^2 + cos(t), obtaining the shortest trial solution x(t)=d1+d2 t+d3 t^2+d4 t cos(t) + d5 t sin(t). How to use dsolve() in maple to check the answer. EXAMPLE. Suppose the DE has order n=4 and the homogeneous equation has solution atoms cos(t), t cos(t), sin(t), t sin(t). Assume f(t) = t^2 + cos(t). What is the shortest trial solution? EXAMPLE. Suppose the DE has order n=2 and the homogeneous equation has solution atoms cos(t), sin(t). Assume f(t) = t^2 + t cos(t). What is the shortest trial solution? EXAMPLE. Suppose the DE has order n=4 and the homogeneous equation has solution atoms 1, t, cos(t), sin(t). Assume f(t) = t^2 + t cos(t). What is the shortest trial solution?

Undetermined coefficientsExamples continued from the previous lecture. EXAMPLES. y'' = x y'' + y = x exp(x) y'' - y = x exp(x) y'' + y = cos(x) y''' + y'' = 3x + 4 exp(-x) Shortest trial solution. Two Rules to find the shortest trial solution. 1. Compute the atoms in f(x). The number k of atoms found is the number needed in the shortest trial solution. 2. Correct groups with the same base atom, by multiplication by x until the group contains no atom which is a solution of the homogeneous problem [eliminate homogeneous DE conflicts].The x^s mystery factorin the book's table. The number s is the multiplicity of the root in the homogenous DE characteristic equation, which constructed the base atom of the group.Reference: Edwards-Penney, Differential Equations and Boundary Value Problems, 4th edition, section 3.7 [math 2280 textbook]. Extra pages supplied by Pearson with bookstore copies of the 2250 textbook. Also available as a xerox copy in case your book came from elsewhere. Check-out the 2280 book in the math center or the Math Library. All editions of the book have identical 3.7 and 7.6 sections.Transform TerminologyConvolution theorem and x'' + 4x = cos(t), x(0)=x'(0)=0. Input Output Transfer FunctionCircuits EPbvp3.7: Electrical resonance. Derivation from mechanical problems 5.6. THEOREM: omega = 1/sqrt(LC). Impedance, reactance. Steady-state current amplitude Transfer function. Input and output equation.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. Wine glass experiment. Tacoma narrows.: Wine glass breakage (QuickTime MOV) (96.8 K, mov, 21 Mar 2013)Video: Wine glass experiment (12mb mpg, 2min) (12493.8 K, mpg, 01 Apr 2008)Video: Tacoma Narrows Bridge Nov 7, 1940 (18mb mpg, 4min) (18185.8 K, mpg, 01 Apr 2008)Video: Basic undetermined coefficients, draft 5 (156.3 K, pdf, 29 Mar 2013)Slides: Variation of parameters (164.5 K, pdf, 03 Mar 2012)Slides: Resonance and undetermined coefficients (199.4 K, pdf, 03 Mar 2012)Slides

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: Maple Lab 7. Laplace applications (156.0 K, pdf, 04 Dec 2012)MAPLE: DE systems, examples, theory (785.8 K, pdf, 16 Nov 2008)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 (109.8 K, pdf, 04 Mar 2012)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 (186.8 K, pdf, 20 Oct 2009)Manuscript: Laplace theory 2008 (351.3 K, pdf, 09 Apr 2013)Manuscript: Ch10 Laplace solutions 10.1 to 10.4 (1068.7 K, pdf, 28 Nov 2010)Transparencies: Laplace theory problem notes for Chapter 10 (0.0 K, txt, 31 Dec 1969)Text: Final exam study guide (8.2 K, txt, 05 Dec 2012)TextVariation of Parameters and Undetermined Coefficients references: Basic undetermined coefficients, draft 5 (156.3 K, pdf, 29 Mar 2013)Slides: Variation of parameters (164.5 K, pdf, 03 Mar 2012)SlidesSystems of Differential Equations references: Systems of DE examples and theory (785.8 K, pdf, 16 Nov 2008)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 (109.8 K, pdf, 04 Mar 2012)Slides: Cable hoist example (73.2 K, pdf, 21 Aug 2008)Slides: Sliding plates example (105.8 K, pdf, 21 Aug 2008)SlidesOscillations. Mechanical and Electrical.: Electrical circuits (124.8 K, pdf, 04 Mar 2012)Slides: Forced damped vibrations (280.7 K, pdf, 04 Mar 2012)Slides: Forced vibrations and resonance (240.3 K, pdf, 04 Mar 2012)Slides: Forced undamped vibrations (214.2 K, pdf, 03 Mar 2012)Slides: Resonance and undetermined coefficients (199.4 K, pdf, 03 Mar 2012)Slides: Unforced vibrations 2008 (667.9 K, pdf, 03 Mar 2012)SlidesEigenanalysis and Systems of Differential Equations.: Eigenanalysis 2010, 46 pages (345.3 K, pdf, 31 Mar 2010)Manuscript: Algebraic eigenanalysis 2008 (187.6 K, pdf, 04 Mar 2012)Manuscript: 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: What's eigenanalysis, draft 1 (152.2 K, pdf, 01 Apr 2008)Manuscript: What's eigenanalysis, draft 2 (124.0 K, pdf, 14 Nov 2007)Manuscript: What's eigenanalysis 2008 (174.2 K, pdf, 04 Mar 2012)Manuscript: Cayley-Hamilton-Ziebur method for solving vector-matrix system u'=Au. (152.9 K, pdf, 04 Mar 2012)Slides