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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
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=
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 @f=@y is continuous on a closed convex domain D. Show that f (t y)
satis es a Lipschitz condition on D with
L = (tmax j @f@y y) j:
(Hint: use t...
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- Spring '14