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Homework Statement
Find all the positive equilibria for the system of the difference equations
[tex]x(t+1) = \frac{ax(t)y(t)}{1+x(t)};
y(t+1) = \frac{bx(t)y(t)}{1+y(t)}[/tex]
a, b > 0.
Then determine conditions on parameters so the equilibriums are locally asymptotically stable.
Homework Equations
An equilibrium point will have corresponding eigenvalues < 1 iff [tex]|Trace(J)| < 1 + det(J) < 2[/tex],
where J is the Jacobian matrix of the system evaluated at the equilibrium point.
In this case [tex]J = \left(
\begin{array}{cc}
\frac{ay(t)}{1+x(t)} - \frac{ax(t)y(t)}{(1+x(t))^2} & \frac{ax(t)}{1+x(t)}\\
\frac{by(t)}{1+y(t)} & \frac{bx(t)}{1+y(t)} - \frac{bx(t)y(t)}{(1+y(t))^2}
\end{array}
\right)
[/tex]
The Attempt at a Solution
I found the following equilibria
(0,0) and [tex](\frac{1+a}{ab-1},\frac{1+b}{ab-1}); ab > 1[/tex]
The Jacobian Matrix evaluated at (0,0) gives me a zero matrix. So that means the trivial solution is always stable for any choice of parameters a,b, right? Or do I have to look at H.O.T.?
The Jacobian matrix for the second point is:
[tex]J = \left(
\begin{array}{cc}
\frac{ab-1}{ab+a} & \frac{1+a}{1+b}\\
\frac{1+b}{1+a} & \frac{ab-1}{ab+b}
\end{array}
\right)[/tex]
or, if I let x and y denote the equilibrium points
[tex]J = \left(
\begin{array}{cc}
\frac{1}{ay} & \frac{x}{y}\\
\frac{y}{x} & \frac{1}{bx}
\end{array}
\right)[/tex]
And since the trace is always positive, we have the following requirements
[tex]ax + by < 1 < 2aybx[/tex]
Or [tex](ab)(xy) > \frac{1}{2}[/tex]
and [tex]ay + bx < 1[/tex]
Does that make any sense?
If we substitute x and y back in though, we get nonsense (something like, pos number < -1).
What am I doing wrong?
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