# Considering the predator prey model to find the range of values to be a spiral

I have the following problem: Consider the predator-prey model: $$\frac{du}{dt}=u(1-\alpha(u)-v), \frac{dv}{dt}=\rho(-1-\alpha(v)+u),$$

where $\rho$ and $\alpha$ are positive parameters with $0\leq\alpha<0$.

The First bit asks that at the non-trivial fixed point, there is a centre at $\alpha=0$ which changes into a stable point for $0<\alpha<0$. I have shown this. However it then asks to find the range of values for the non-trivial fixed point to be a spiral?

I have that the det(A) at our non-trivial fixed point will be always greater than zero, hence I have to show that $\Delta<0$.

$\Rightarrow {(-\alpha u+\alpha\rho v)^2}-4({\alpha}^{2} \rho uv+u\rho v)<0$, then i tried using the our fixed point (u,v)= $(\frac{1+\alpha}{{\alpha}^{2}+1}, \frac{1-\alpha}{{\alpha}^{2}+1})$ implies u>1 and v<1. Eventually simplifying down i get an expression for $\rho$ and $\alpha$ but I am unable to solve for $\alpha$!

Is this the correct approach to take?

Do you mean to assume $0\le \alpha < \rho$? –  MSEoris Nov 10 '12 at 15:05
@MSEoris it says to assume $\rho$ is positive. However it asks find the range of values for the non-trivial fixed point to be a spiral, therefore thing it may be interested in looking at $\rho$ in conjunction with $\alpha$? –  user24930 Nov 10 '12 at 15:14
From my understanding, your steady states are stable if all of the eigenvalues of A satisfy RE$(r)<0$. Thus you need to solve $\det(A-rI)=0$ and for all values of r, you must also have that $\lim e^{rt} =0$ as $t\rightarrow \infty$ –  MSEoris Nov 10 '12 at 15:26