Topics will be chapter 5, 10, EPbvp7.6. Problems 1,2 center on Laplace theory, but there is some contact with chapter 5, sections 5.1 to 5.4. See the online sample exam for details. Problems 3,4,5 will be the week after Thanksgiving. Exam 3 part I: 22, 23, 24 Nov. The Mon-Wed exams are in JWB 335 at 12:50pm. The Tuesday exam is during the regular 7:25am classtime in WEB 103. As usual, you may write the exam on any one of the three days. The problems are 1,2 on the sample exam below, which matches 3,4 on the S2010 exam key. Exam 3 part II: 29 Nov, 1, 2 Dec. The Thursday exam is at the usual 7:25am time in WEB 103. The Mon-Wed exams are in JWB 335 at 12:50pm. As usual, you may write exam 3, part II, on any one of the three days. The problems are 3,4,5 on the sample exam below, which matches 1,2,5 on the S2010 exam key.

Review from last week: Systems of two differential equationsExample: 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}. Chapter 1+5 Shortcut Method [Ziebur§s method. Appears in Edwards-Penney] 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).Conversion Methods to Create a First Order SystemThe position-velocity substitution. How to convert second order systems. How to convert nth order scalar differential equations. 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]]);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) PROBLEM: Solve by Ziebur's Lemma 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 Lemma says that u(t) = exp(2t) u_1 + exp(3t) u_2 where vectors u_1, U_2 are vectors to be determined from A = matrix([[2,1],[0,3]]) and initial conditions x(0)=1, y(0)=2. ZIEBUR ALGORITHM. To solve for u_1, u_2 in the example, differentiate the equation u(t) = exp(2t) u_1 + exp(3t) u_2 and set t=0 in both relations. Then u'=Au implies u_0 = u_1 + u_2, Au_0 = 2 u_1 + 3 u_2. These equations can be solved by elimination. The answer: u_1 = (3 u_0 -Au_0), u_2 = (Au_0 - 2 u_0) = vector([-1,0]) = vector([2,2]) Vectors u_1, u_2 are recognized as eigenvectors of A for lambda=2 and lambda=3, respectively, after studying chapter 6. ZIEBUR SHORTCUT [Edwards-Penney textbook method] 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.

EIGENANALYSIS WARNINGReading Edwards-Penney Chapter 6 may deliver the wrong ideas about how to solve for eigenpairs. HISTORY. Chapter 6 originally appeared in the 2280 book as a summary, which assumed a linear algebra course. The chapter was copied without changes into the Edwards-Penney Differential Equations and Linear Algebra textbook, which you currently own. The text contains only shortcuts. There is no discussion of a general method for finding eigenpairs. You will have to fill in the details by yourself. The online lecture notes and slides were created to fill in the gap.Lecture: Fourier's Model. Intro to Eigenanalysis, Ch6.Examples and motivation. Ellipse, rotations, eigenpairs. General solution of a differential equation u'=Au and eigenpairs. Fourier's model. History. J.B.Fourier's 1822 treatise on the theory of heat. The rod example. Physical Rod: a welding rod of unit length, insulated on the lateral surface and ice packed on the ends. Define f(x)=thermometer reading at loc=x along the rod at t=0. Define u(x,t)=thermometer reading at loc=x and time=t>0. Problem: Find u(x,t). Fourier's solution assume that f(x) = 17 sin (pi x) + 29 sin(5 pi x) = 17 v1 + 29 v2 Packages v1, v2 are vectors in a vector space V of functions on [0,1]. Fourier computes u(x,t) by re-scaling v1, v2 with numbers Lambda_1, Lambda_2 that depend on t. This idea is calledFourier's Model.u(x,t) = 17 ( exp(-pi^2 t) sin(pi x)) + 29 ( exp(-25 pi^2 t) sin (5 pi x)) = 17 (Lambda_1 v1) + 29 (Lambda_2 v2) Eigenanalysis of u'=Au is the identical idea. u(0) = c1 v1 + c2 v2 implies u(t) = c1 exp(lambda_1 t) v1 + c2 exp(lambda_2 t) v2 Fourier's re-scaling idea from 1822, applied to u'=Au, replaces v1 and v2 in the expression c1 v1 + c2 v2 by their re-scaled versions to obtain the answer c1 (Lambda1 v1) + c2 (Lambda2 v2) where Lambda1 = exp(lambda_1 t), Lambda2 = exp(lambda_2 t).

Main Theorem on Fourier's ModelTHEOREM. Fourier's model A(c1 v1 + c2 v2) = c1 (lambda1 v1) + c2 (lambda2 v2) with v1, v2 a basis of R^2 holds [for all constants c1, c2] if and only if the vector-matrix system A(v1) = lambda1 v1, A(v2) = lambda2 v2, has a solution with vectors v1, v2 independent if and only if the diagonal matrix D=diag(lambda1,lambda2) and the augmented matrix P=aug(v1,v2) satisfy 1. det(P) not zero [then v1, v2 are independent] 2. AP=PD THEOREM. The eigenvalues of A are found from the determinant equation det(A -lambda I)=0, which is called the characteristic equation. THEOREM. The eigenvectors of A are found from the frame sequence which starts with B=A-lambda I [lambda a root of the characteristic equation], ending with last frame rref(B). The eigenvectors for lambda are the partial derivatives of the general solution obtained by the Last Frame Algorithm, with respect to the invented symbols t1, t2, t3, ...

Algebraic Eigenanalysis Section 6.2.Calculation of eigenpairs to produce Fourier's model. Connection between Fourier's model and a diagonalizable matrix. How to find the variables lambda and v in Fourier's model using determinants and frame sequences. Solved in class: examples similar to the problems in 6.1 and 6.2. Web slides and problem notes exist for the 6.1 and 6.2 problems. Examples where A has an eigenvalue of multiplicity greater than one.

23 Nov: First Order Systems. Sections 7.1-7.3Solving 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).Drill ProblemsIn the case of a 2x2 matrix A, FOURIER'S MODEL is A(c1 v1 + c2 v2) = c1(lambda1 v1) + c2(lambda2 v2) where v1,v2 are a basis for the plane equivalent to DIAGONALIZATION AP=PD, where D=diag(lamba1,lambda2), P=augment(v1,v2), where det(P) is not zero equivalent to EIGENPAIR EQUATIONS A(v1)=lambda1 v1, A(v2)=lambda2 v2, where vectors v1,v2 are independent 1. Problem: Given P and D, find A in the relation AP=PD. 2. Problem: Given Fourier's model, find A. 3. Problem: Given A, find Fourier's model. 4. Problem: Given A, find all eigenpairs. 5. Problem: Given A, find packages P and D such that AP=PD. 6. Problem: Give an example of a matrix A which has no Fourier's model. 7. Problem: Give an example of a matrix A which is not diagonalizable. 8. Problem: Given 2 eigenpairs, find the 2x2 matrix A.Cayley-Hamilton topics, Section 6.3.Computing powers of matrices. Stochastic matrices. Example of 1984 telecom companies ATT, MCI, SPRINT with discrete dynamical system u(n+1)=A u(n). Matrix A is stochastic. EXAMPLE: [ 6 1 5 ] [ a(t) ] 10 A = [ 2 7 1 ] u(t) = [ m(t) ] [ 2 2 4 ] [ s(t) ] Meaning: 60% stay with ATT and 20% switch to MCI, 20% switch to SPRINT. 70% stay with MCI and 20% switch to SPRINT, 10% switch to ATT. 40% stay with SPRINT and 50% switch to ATT, 10% switch to MCI. Determinant problem from chapter 3: B(n+1)=2B(n)-B(n-1). This is a second order difference equation.Google AlgorithmLawrence Page's pagerank algorithm, google web page rankings. Methods to solve dynamical systems like x'=x-5y, y'=x-y, x(0)=1, y(0)=2. Cayley-Hamilton-Ziebur method. Laplace resolvent. Eigenanalysis method. Exponential matrix using maple Putzer's method Spectral methods [ch8; not studied in 2250]Survey of Methods for solving a 2x2 dynamical system1. Cayley-Hamilton-Ziebur method for u'=Au Solution: u(t)=(atom_1)vec(c_1)+ ... + (atom_n)vec(c_n) Atoms: They are constructed by Euler's theorem from roots of det(A-rI)=0 Vectors: Symbols vec(c_1), ..., vec(c_n) are not arbitrary. They are determined from A and u(0). Algorithm outlined above for 2x2. 2. Laplace resolvent L(u)=(s I - A)^(-1) u(0) 3. Eigenanalysis u(t) = exp(lambda_1 t) v1 + exp(lambda_2 t) v2 4. 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 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 eigenavalues of A.