Edwards-Penney, sections 4.2, 5.1 to 5.6, 6.1, 6.2 The textbook topics, definitions and theorems

Edwards-Penney 5.1 to 5.6 (22.6 K, txt, 04 Jan 2015)

Edwards-Penney 6.1 to 6.4 (11.8 K, txt, 05 Apr 2015)

Maple Example to find roots of the characteristic equationConsider the recirculating brine tank example, section 5.2: 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)EIGENANALYSIS WARNINGReading Edwards-Penney Chapter 5 may deliver the wrong ideas about how to solve for eigenpairs. The examples emphasize a clever shortcut, which does not apply in general to solve for eigenpairs.Solving DE System u' = Au by EigenanalysisExample: Solving a 2x2 dynamical system Study of u'=Au, u(0)=vector([1,2]), A=matrix([[2,3],[0,4]]). Dynamical system scalar form is x' = 2x + 1y, y' = 3y, x(0)=1, y(0)=2. Find the eigenpairs (2, v1), (3,v2) where v1=vector([1,0]) and v2=vector([1,1]). THEOREM. The solution of u' = Au in the 2x2 case is u(t) = c1 exp(lambda1 t) v1 + c2 exp(lambda2 t) v2 APPLICATION: u(t) = c1 exp(2t) v1 + c2 exp(4t) v2 [ 1 ] [ 1 ] u(t) = c1 e^{2t} [ ] + c2 e^4t} [ ] [ 0 ] [ 1 ] which means x(t) = c1 exp(2t) + 3 c2 exp(4t), y(t) = 2 c2 exp(4t).Section 5.1: Topics from linear systems:Systems of two differential equationsSolving a system from Chapter 1 methods The Laplace resolvent method for systems. Cramer's Rule, Matrix inversion methods. EXAMPLE: Solving a 2x2 dynamical system using Laplace's resolvent method. Study of u'=Au, u(0)=vector([1,2]), A=matrix([[2,1],[0,3]]). EXAMPLE: Problem 10.2-16, This problem is a 3x3 system for x(t), y(t), z(t) solved by Laplace theory methods. The resolvent formula (sI - A) L(u(t)) = u(0) with u(t) the fixed 3-vector with components x(t), y(t), z(t), amounts to a shortcut to obtain the equations for L(x(t)), L(y(t)), L(z(t)). After the shortcut is applied, in which Cramer's Rule is the method of choice, to find the formulas, there is no further shortcut: we have to find x(t), for example, by partial fractions and the backward table, followed by Lerch's theorem. EXAMPLE. Recirculating brine tanks 20 x' = -6x + y, 20 y' = 6x - 3y x(t)=pounds of salt in tank 1 (100 gal) y(t)=pounds of salt in tank 2 (200 gal) x(0), y(0) = initial salt amounts in each tank t=minutes 20=inflow rate=outflow rate 0=inflow salt concentration EXAMPLE. Solve x'=-2y, y'=x/2. ANSWER: x(t)=A cos(t) + B sin(t), y(t) = (-A/2) cos(t) + (B/2) sin(t).Conversion Methods to Create a First Order SystemThe position-velocity substitution. How to convert second order systems. EXAMPLE. Transform to a first order system 2x'' = -6x + 2y, y'' = 2x - 2y + 40 sin(3t) ANSWER: u1=x,u2=x',u3=y,u4=y' ==> u1' = u2, u2' = -3u1 + u3, [a division by 2 needed] u3' = u4, u4' = 2u1 - 2u3 + 40 sin(3t) How to convert nth order scalar differential equations. EXAMPLE. x''' + 2x'' + x = 0 Use u1=x(t), u2=x'(t), u3=x''(t) Non-homogeneous terms and the vector matrix system u' = Au + F(t) Non-linear systems and the vector-matrix system u' = F(t,u) Answer checks for u'=Au Example: The system u'=Au, A=matrix([[2,1],[0,3]]);Systems of two differential equationsThe Laplace resolvent method for systems. Solving the resolvent equation for L(x), L(y). Cramer's Rule Matrix inversion Elimination Example: Solving a 2x2 dynamical system Study of u'=Au, u(0)=vector([1,2]), A=matrix([[2,1],[0,3]]). Dynamical system scalar form is x' = 2x + y, y' = 3y, x(0)=1, y(0)=2. The equations for L(x), L(y) (s-2)L(x) + (-1)L(y)=1, (0)L(x) + (s-3)L(y)=2 REMARK: Laplace resolvent method shortcut. How to solve the [resolvent] equations for L(x), L(y). Cramer's Rule Matrix inversion Elimination Answers: L(x) = delta1/delta, L(y)=delta2/delta delta=(s-2)(s-3), delta1=s-1, delta2=2(s-2) L(x) = -1/(s-2)+2/(s-3), L(y)=2/(s-3) Backward table and Lerch's theorem Answers: x(t) = - e^{2t} + 2 e^{3t}, y(t) = 2 e^{3t}. Edwards-Penney Shortcut Method in Example 1, 4.2. Uses Chapter 1+3 methods. This is the Cayley-Hamilton-Ziebur method. See below. Solve w'+p(t)w=0 as w = constant / integrating factor. Then y' -2y=0 ==> y(t) = 2 exp(3t) Stuff y(t) into the first DE to get the linear DE x' - 2x = 2 exp(3t) Superposition: x(t)=x_h(t)+x_p(t), x_h(t)=c exp(2t), x_p(t) = d1 exp(t) = 2 exp(3t) by undetermined coeff. Then x(t)= - exp(2t) + 2 exp(3t).Cayley-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 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. ZIEBUR SHORTCUT [Edwards-Penney textbook method, Example 1 in 4.2] 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.Slides

Exam 2 ReviewSample exam 2 Eigenvalues A 4x4 matrix. Block determinant theorem. Eigenvectors for a 4x4. B:=matrix([[5,0,0,0],[0,5,0,0],[0,0,0,3],[0,0,-3,0]]); lambda=5,5,3i,-3i v1=[1,0,0,0], v2=[0,1,0,0], v3=[0,0,i,-1], v4=[0,0,i,1] One panel for lambda=5 First frame is A-5I with 0 appended Find rref Apply last frame algorithm Scalar general solution Take partials on t1, t2 to find v1,v2 Eigenpairs are (5,v1), (5,v2) One panel for lambda=3i Same outline as lambda=5 Get one eigenpair (3i,v3) Other eigenpair=(-3i,v4) where v4 is the conjugate of v3.Second Order SystemsHow to convert mx''+cx'+kx=F0 cos (omega t) into a dynamical system u'=Au+F(t). Electrical systems u'=Au+E(t) from LRC circuit equations. Electrical systems of order two: networks Mechanical systems of order two: coupled systems Second order systems u''=Au+F Examples are railway cars, earthquakes, vibrations of multi- component systems, electrical networks.Second Order Vector-Matrix Differential EquationsThe model u'' = Ax + F(t) Coupled Spring-Mass System. Problem 5.3-6 A:=matrix([[-6,4],[2,-4]]); eigenvals(A); lambda1= -2, lambda2= -8Ziebur's Methodroots for Ziebur's theorem are plus or minus sqrt(lambda) Roots = sqrt(2)i, sqrt(8)i, -sqrt(2)i, -sqrt(8)i Atoms = cos (sqrt(2)t), sin(sqrt(2)t), cos(sqrt(8)t), sin(sqrt(8)t) Vector x(t) = vector linear combination of the above 4 atomsMaple routines for second orderde1:=diff(x(t),t,t)=-6*x(t)+4*y(t); de2:=diff(y(t),t,t)=2*x(t)-4*y(t); dsolve({de1,de2},{x(t),y(t)}); x(t) = _C1*sin(sqrt(2)*t)+_C2*cos(sqrt(2)*t)+_C3*sin(2*sqrt(2)*t)+_C4*cos(2*sqrt(2)*t), y(t) = _C1*sin(sqrt(2)*t)+_C2*cos(sqrt(2)*t)-(1/2)*_C3*sin(2*sqrt(2)*t)-(1/2)*_C4*cos(2*sqrt(2)*t)}Eigenanalysis method section 5.3u(t) = (a1 cos(sqrt(2)t) + b1 sin(sqrt(2)t)) v1 + (a2 cos(sqrt(8)t) + b2 sin(sqrt(8)t)) v2 where (-2,v1), (-8,v2) are the eigenpairs of A. The two vector terms in u(t) are called the natural modes of oscillation. The natural frequencies are sqrt(2), sqrt(8). Eigenanalysis of A gives v1=[1,1], v2=[2,-1].Railway cars. Problem 5.3-24Cayley-Hamilton-Ziebur method Laplace Resolvent method for second order Eigenanalysis method section 5.2

Some Essentail Topics Putzer's method for the 2x2 matrix exponential. Solution of u'=Au is: u(t) = exp(A t)u(0) THEOREM: exp(A t) = r1(t) I + r2(t) (A-lambda_1 I), Lambda Symbols: lambda_1 and lambda_2 are the roots of det(A-lambda I)=0. The DE System: r1'(t) = lambda_1 r1(t), r1(0)=0, r2'(t) = lambda_2 r2(t) + r1(t), r2(0)=0 See the slides and manuscript on systems for proofs and details. THEOREM. The formula can be used as e^{r1 t} - e^{r2 t} e^{At} = e^{r1 t} I + ------------------- (A-r1 I) r1 - r2 where r1=lambda_1, r2=lambda_2 are the eigenvalues of A. EXAMPLE. Solve a homogeneous system u'=Au, u(0)=vector([1,2]), A=matrix([[2,3],[0,4]]) using the matrix exponential, Zeibur's method, Laplace resolvent and eigenanalysis. EXAMPLE. Solve a non-homogeneous system u'=Au+F(t), u(0)=vector([0,0]), A=matrix([[2,3],[0,4]]), F(t)=vector([3,1]) using variation of parameters.

Extra Credit Maple Project: Tacoma narrows. Explore an alternative explanation for what caused the bridge to fail, based on the hanging cables.: Maple Lab 9. Tacoma Narrows (0.0 K, pdf, 31 Dec 1969)MAPLEExtra Credit Maple Project: Earthquakes. Explore a 5-story or 7-story building and the resonant frequencies of oscillation of the building which might make it destruct during an earthquake. See Edwards-Penney, application section in 5.3.

Dynamical Systems TopicsEquilibria. Stability. Instability. Asymptotic stability. Classification of equilibria for u'=Au when det(A) is not zero, for the 2x2 case. Impact of Cayley-Hamilton-Ziebur on classification

Slides on Dynamical Systems: Systems theory and examples (730.9 K, pdf, 10 Apr 2014)Manuscript: Laplace second order systems, spring-mass, boxcars, earthquakes (288.1 K, pdf, 03 Mar 2012)Slides: Introduction to dynamical systems (145.4 K, pdf, 05 Apr 2015)Slides: Phase Portraits for dynamical systems (221.3 K, pdf, 05 Apr 2015)Slides: Stability for dynamical systems (154.4 K, pdf, 05 Apr 2015)Slides: Nonlinear classification spiral, node, center, saddle (98.0 K, pdf, 05 Apr 2015)Slides: Matrix Exponential, Putzer Formula, Variation Parameters (123.5 K, pdf, 29 Mar 2015)Slides

Systems of Differential Equations references: Systems of DE examples and theory (730.9 K, pdf, 10 Apr 2014)Manuscript: Laplace resolvent method (88.1 K, pdf, 03 Mar 2012)Slides: Laplace second order systems (288.1 K, pdf, 03 Mar 2012)Slides: Home heating, attic, main floor, basement (99.3 K, pdf, 09 Apr 2014)Slides: Cable hoist example (73.2 K, pdf, 21 Aug 2008)Slides: Sliding plates example (105.8 K, pdf, 21 Aug 2008)Slides

References for Eigenanalysis and Systems of Differential Equations.: Algebraic eigenanalysis (187.6 K, pdf, 03 Mar 2012)Sildes: What's eigenanalysis 2008 (174.2 K, pdf, 03 Mar 2012)Slides: Cayley-Hamilton-Ziebur method for solving vector-matrix system u'=Au. (137.7 K, pdf, 28 Mar 2015)Slides: Systems of DE examples and theory (730.9 K, pdf, 10 Apr 2014)Manuscript: Final exam study guide (7.6 K, txt, 28 Apr 2015)Text