Unformatted text preview: 324 CHAPTER 5 NONLINEAR SYSTENLS (b) y y
y
4 4 \v‘ 32
2 2
x x x
2 4 8 12 2 4 (c) We could compute the eigenvalues and eigenvectors for each of the linear systems. how
ever. we can determine much of the qualitative information about behavior of the solutions
from the ﬁgures in part (b). It is interesting to note the size of the eigenvalues (which are quite large in absolute value). so solutions move toward or away from the equilibria very
quickly. 9. (a) The equilibrium points are (0, O), (0, 25), (100, 0) and (75, 12.5). We can classify these
equilibrium points by computing the Jacobian matrix 100  2x  2y 2x
—y 150  x — 12y at each of the equilibrium points and computing the eigenvalues. At (75, 12.5), the Jaco bian matrix is
—'75 150
—12.5 —75 and the eigenvalues are approximame ~32 and — l 18 so this point is a sink. At (0. 0) the Jacobian matrix is
100 O
0 [50 and the eigenvalues are 100 and ISO. so this point is a source. At (0. 25) the Jacobian matrix is
50 0
—25 —150 and the eigenvalues are 50 and —lSO. so this point is a saddle. Finally, at (100, 0) the Jacobian matrix is
—100 200
O 50 and the eigenvalues are — 100 and 50, so this point is a saddle. ...
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 Spring '08
 JU

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