Consider the dynamical system. x' = -ax + xy y' = 1 - By - x2 Show that the system has a unique critical point when aß > 1 and that this critical point is stable when B > 0. From x' = -ax + xy = 0, either x = or y = . If x = , then 1 - øy = 0 and so y = . The case y = a Implies that 1 - pa - x2 = 0. If aß > 1, this equation has no real solutions. It follows that (x, y) = 0, is the unique critical point. The Jacobian matrix is the following. 1 B- o g'(X) = -B (ав — 1) +B Therefore if B > 0, r = O and A = aß - 1 0. It follows that (x, y) =| 0, is a stable critical point.
Consider the dynamical system. x' = -ax + xy y' = 1 - By - x2 Show that the system has a unique critical point when aß > 1 and that this critical point is stable when B > 0. From x' = -ax + xy = 0, either x = or y = . If x = , then 1 - øy = 0 and so y = . The case y = a Implies that 1 - pa - x2 = 0. If aß > 1, this equation has no real solutions. It follows that (x, y) = 0, is the unique critical point. The Jacobian matrix is the following. 1 B- o g'(X) = -B (ав — 1) +B Therefore if B > 0, r = O and A = aß - 1 0. It follows that (x, y) =| 0, is a stable critical point.
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Transcribed Image Text:Consider the dynamical system.
x' = -ax + xy
y' = 1- By - x2
Show that the system has a unique critical point when aß > 1 and that this critical point is stable when f > 0.
a.
If x = 0
1
then 1 - Øy = 0 and so y =
From x' = -ax + xy = 0, either x =
or y =
The case
y =
a.
implies that 1 - Ba - x? = 0. If aß > 1, this equation has no real solutions. It follows that (x, y) =
0,
is the unique critical point. The Jacobian matrix
is the following.
1
В — а
g'(X) =
-B
(ав — 1)
+B
>v 0. It follows that (x, y) =(|0,
Therefore if B > 0, r =
O and A = aß – 1
is a stable critical point.
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