# Equilibria analysis for Biomath Problem

1. Sep 30, 2007

### end3r7

1. The problem statement, all variables and given/known data
Find all the positive equilibria for the system of the difference equations
$$x(t+1) = \frac{ax(t)y(t)}{1+x(t)}; y(t+1) = \frac{bx(t)y(t)}{1+y(t)}$$
a, b > 0.
Then determine conditions on parameters so the equilibriums are locally asymptotically stable.

2. Relevant equations
An equilibrium point will have corresponding eigenvalues < 1 iff $$|Trace(J)| < 1 + det(J) < 2$$,
where J is the Jacobian matrix of the system evaluated at the equilibrium point.
In this case $$J = \left( \begin{array}{cc} \frac{ay(t)}{1+x(t)} - \frac{ax(t)y(t)}{(1+x(t))^2} & \frac{ax(t)}{1+x(t)}\\ \frac{by(t)}{1+y(t)} & \frac{bx(t)}{1+y(t)} - \frac{bx(t)y(t)}{(1+y(t))^2} \end{array} \right)$$

3. The attempt at a solution
I found the following equilibria
(0,0) and $$(\frac{1+a}{ab-1},\frac{1+b}{ab-1}); ab > 1$$
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:
$$J = \left( \begin{array}{cc} \frac{ab-1}{ab+a} & \frac{1+a}{1+b}\\ \frac{1+b}{1+a} & \frac{ab-1}{ab+b} \end{array} \right)$$

or, if I let x and y denote the equilibrium points

$$J = \left( \begin{array}{cc} \frac{1}{ay} & \frac{x}{y}\\ \frac{y}{x} & \frac{1}{bx} \end{array} \right)$$

And since the trace is always positive, we have the following requirements
$$ax + by < 1 < 2aybx$$
Or $$(ab)(xy) > \frac{1}{2}$$
and $$ay + bx < 1$$

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?

Last edited: Sep 30, 2007
2. Sep 30, 2007

### end3r7

Sorry about the confusion fellas, first time using latex in a long time.
Any questions about my work, I'll be gladly to answer. And if possible, don't tell me the answer, but rather around where my thinking goes wrong.

3. Sep 30, 2007

### end3r7

Does the zero matrix mean that I have to consider the nonlinear terms of the Jacobian? Cuz that would suck =P

4. Sep 30, 2007

### end3r7

Oops, I'm thinking Differential Equations, where the nonhyperbolic/stationary case is when the Jacobian Matrix is the zero Matrix. For difference, I would have to consider H.O.T. if the Jacobian was the identity, which it clearly isn't.
So I guess (0,0) is asymptotically stable, which makes sense if you look a the equations I guess. For x, y small it should converge.

Am I right?

5. Sep 30, 2007

### EnumaElish

What is the formal definition of asymptotically stable?

6. Sep 30, 2007

### end3r7

Well, locally asymptotically stability implies that it's both stable and attracting.

That is, given a certain distance from equilibrium point, it will not ever exceed that distance (stable), and will eventually converge to the equilibrium (attracting).

For the vector function X(t+1) = F(X(t)), where X(t) is a vector depending on x,y, whatever...
An equilibrium T is locally stable if for any E > 0 there exists D > 0 such that if |X(0) - T| < D, then |X(t) - T| < E
It's attracting if lim X(t) = T.

7. Sep 30, 2007

### EnumaElish

Can you show formally that either of your equilibria satisfy the definition?

8. Sep 30, 2007

### end3r7

Yes. But it really doens't matter, I know I"m supposed to use it, cuz that's what the book tells me. =P But yea, I can show it does.

9. Sep 30, 2007

### end3r7

Anyway, the identity that's giving me problems is the following

$$ay + bx < 1$$

which is equivalent to

a + 2ab + b < ab - 1
or a + ab + b + 1 < 0, which is impossible since a,b>0