Y p problem explain why the largest and smallest

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Unformatted text preview: (cos t − sin t). Y P The system is unstable because Re (λi ) > 0 for each eigenvalue. Indeed, u1 (t) and u2 (t) both become unbounded as t → ∞. However, a population cannot become negative–once it’s zero, it’s extinct. Figure 7.4.2 shows that the graph of u2 (t) will cross the horizontal axis before that of u1 (t). O C 400 300 u1(t) 200 u2(t) 100 t 0 0.2 0.4 0.6 0.8 1 -100 -200 Figure 7.4.2 Copyright c 2000 SIAM Buy online from SIAM http://www.ec-securehost.com/SIAM/ot71.html Buy from AMAZON.com 546 Chapter 7 Eigenvalues and Eigenvectors http://www.amazon.com/exec/obidos/ASIN/0898714540 Therefore, the prey species will become extinct at the value of t for which u2 (t) = 0 —i.e., when 100et (cos t − sin t) = 0 =⇒ cos t = sin t =⇒ t = π . 4 It is illegal to print, duplicate, or distribute this material Please report violations to meyer@ncsu.edu Exercises for section 7.4 7.4.1. Suppose that An×n is diagonalizable, and let P = [x1 | x2 | · · · | xn ] be a matrix whose columns are a complete set of linearly independent eigenvectors corresponding to eigenvalues λi . Show that the solution to u = Au, u(0) = c, can be written as D E T H u(t) = ξ1 eλ1 t x1 + ξ2 eλ2 t x2 + · · · + ξn eλn t xn in which the coefficients ξi satisfy the algebraic system Pξ = c. 7.4.2. Using only the eigenvalues, determine the long-run behavior of the solution to u = Au, u(0) = c for each of the following matrices. −1 −2 1 −2 1 −2 (a) A = . (b) A = . (c) A = . 0 −3 0 3 1 −1 IG R 7.4.3. Competing Species. Consider two species that coexist in the same environment but compete for the same resources. Suppose that the population of each species increases proportionally to the number of its own kind but decreases proportionally to the number in the competing species—say that the population of each species increases at a rate equal to twice its existing number but decreases at a rate equal to the number in the other population. Suppose that there are initially 100 of species I and...
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