Linear Systems - Study Guide

Solve the linear first order system x0 = Ax Given x01 = ax1 + bx2 x02 = cx1 + dx2  ⇔ x01 x2   = a b c d  x1 x2  ⇔ x0 = Ax Then we could solve the system in three ways: • Convert the system to 2d order DE of the form ay 00 + by 0 + cy = 0, and solve. • Could try to write the system in implicit form: dx2 /dt cx1 + dx2 dx2 = = dx1 dx1 /dt ax1 + bx2 Then we could try to use a first order method. • Use the eigenvalues and eigenvectors from the matrix A. This is summarized below for 2 × 2 matrices. Eigenvalues and Eigenvectors • Definition: Given an n × n matrix A, if there is a constant λ and a non-zero vector v so that Av = λv then λ is an eigenvalue, and v is an associated eigenvector for matrix A. Note that an eigenvector is not uniquely determined; we usually choose the simplest vector (integer valued if possible). • If we try to solve our equation: Av − λv = 0 or (A − λI)v = 0 (1) This system has a non-trivial (non-zero) solution for v1 , v2 only if the determinant is zero (so that A − λI is not invertible): |A − λI| = 0 And this is the characteristic equation, and simplifying it will give you an nth degree polynomial in λ, which we solve for the eigenvalues.   a b If A = , then the characteristic equation becomes: c d λ2 − (a + d)λ + (ad − bc) = 0 ⇔ λ2 − Tr(A)λ + det(A) = 0 where Tr(A) is the trace of A (which we defined as a + d). For each λ, we must go back and solve Equation (1). 1 • Given a λ and v, the generalized eigenvector w is computed as the solution to (A − λI)w = v and in the 2 × 2 case, it is used when λ has algebraic multiplicity 2, but geometric multiplicity 1 (a double root with only one eigenvector). Summary To solve x0 = Ax, find the trace, determinant and discriminant for the matrix A. The eigenvalues are found by solving the characteristic equation: √ ∆ Tr(A) ± λ2 − Tr(A)λ + det(A) = 0 λ= 2 The solution is one of three cases, depending on ∆: • Real λ1 , λ2 give two eigenvectors, v1 , v2 : x(t) = C1 eλ1 t v1 + C2 eλ2 t v2 • Complex λ = a + ib, v (we only need one):

x(t) = C1 Real eλt v + C2 Imag eλt v • One eigenvalue, one eigenvector v. Compute the generalized eigenvector w, (A − λI)w = v. Then x(t) = eλt (C1 v + C2 (tv + w)) You might find this helpful- Below there is a chart comparing the solutions from Chapter 3 to the solutions in Chapter 7: Chapter 3 Form: ay 00 + by 0 + cy = 0 Ansatz: y = ert Char Eqn: ar2 + br + c = 0 Real Solns y = C1 er1 t + C2 er2 t Complex y = C1 Re(ert ) + C2 Im(ert ) Repeated Root y = ert (C1 + C2 t) Chapter 7 x0 = Ax x = eλt v λ2 − Tr(A)λ + det(A) = 0 x(t) = C1 eλ1 t v
1 + C2 eλ2 t v2
x(t) = C1 Re eλt v + C2 Im eλt v x(t) = eλt (C1 v + C2 (tv + w)) Exercise Set 3 1. Verify that the following function solves the given system of DEs:       1 2 3 −2 −t 2t 0 x(t) = C1 e + C2 e x = x 2 1 2 −2 2. For each matrix, find the eigenvalues and eigenvectors (these are selected from 16-23, p. 384 in the textbook). Note that they could be complex, and the matrix A may have complex numbers. Try the last one to see if you can do it! 2  5 −1 3 1   3 −2 4 −1   −2 1 1 −2 (a) A = (b) A = (c) A =   1 i (d) A = −i 1 √   3 1 (e) A = √ 3 −1   3 2 2 4 1  (f) A =  1 −2 −4 −1  3. For each given λ and v, find an expression for Re(eλt v) and Im(eλt v): (a) λ = 3i, v = [1 − i, 2i]T (c) λ = 2 − i, v = [1, 1 + 2i]T (b) λ = 1 + i, v = [i, 2]T (d) λ = i, v = [2 + 3i, 1 + i]T 4. Given the eigenvalues and eigenvectors for some matrix A, write the general solution to x0 = Ax. Furthermore, classify the origin as a sink, source, saddle, or none of the above.     1 1 (a) λ = −2, 3 v1 = , v2 = 2 −1     1 3 (b) λ = −2, −2 v= , w= 1 1     −1 2 (c) λ = 2, −3 v1 = v2 = 2 −1 5. Give the general solution to each system x0 = Ax using eigenvalues and eigenvectors, and sketch a phase plane (solutions in the x1 , x2 plane). Identify the origin as a sink, source or saddle:     1 5 −6 10 (a) A = (c) A = 5 1 −2 3     7 2 8 6 (b) A = (d) A = −4 1 −15 −11 6. (Extra Practice) For each system below, find y as a function of x by first writing the differential equation as dy/dx. (a) x0 = −2x y0 = y (c) x0 = −(2x + 3) y 0 = 2y − 2 (b) x0 = y + x3 y y 0 = x2 (d) x0 = −2y y 0 = 2x 3