# This autonomous problem has two equilibrium points y1

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Unformatted text preview: s a co-existence of both species. The stability of 0 0]T may be analyzed by using the linear system y1 = a 0 y1 : y 0 ;c y 0 2 2 Since the system matrix is diagonal, its eigenvalues are 1 = a, 2 = ;c. Thus, 1 > 0 and 0 0]T is unstable. In order to analyze the stability of the second equilibrium point, we introduce the change of variables z1 = y1 ; c z2 = y2 ; a to move the equilibrium point to the origin. Substituting this transformation into the predator-prey equations gives z = (z + c ) a ; (z + a )] = ; z ( c + z ) 0 1 1 2 2 1 z2 = (z2 + a ) ;c + (z1 + c )] = z1 ( a + z2 ): 0 The linearized system is now z1 = 0 ; c= z1 : z2 a= 0 z2 p The eigenvalues of this system are 1 2 = i c=a. Since the eigenvalues are imaginary, the linear system is stable however, the stability of the nonlinear system is undetermined without additional analysis. 0 Problems 1. Suppose @[email protected] is continuous on a closed convex domain D. Show that f (t y) satis es a Lipschitz condition on D with (t L = (tmax j @[email protected] y) j: y) D (Hint: use t...
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