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$$\dot{x} = x(1-\mu x-y), \quad \quad \dot{y} = y(-1+x-y)$$

where $\mu > 0$ is a constant. The number and type of equilibria depend on the value of μ; in fact there are three essentially different cases. Find the three corresponding ranges of μ and the number, type and stability of the equilibria in each case. (Ignore a borderline case with a zero eigenvalue, but look carefully at any other borderline cases.)

I've already worked out the following:

$$(x_0, y_0) = (0,0), \quad (0,-1), \quad (\frac{1}{\mu},0), \quad (\frac{2}{\mu +1},\frac{1-\mu}{\mu+1})$$

$\therefore$ If $\mu = -1$ or $\mu = 0$ then there exist only 3 equilibrium points. However, if $[\mu \in \mathbb{R} : \mu \ne 0, -1]$ then 4 equilibria exists.

Thus from here, assume $\mu \ne 0,-1$

Jacobian:

$$J = \begin{bmatrix} 1-2\mu x -y & -x\\y & -1+x-2y\end{bmatrix}$$

I've also worked out the Jacobian for $(\frac{1}{\mu},0)$ and found the polynomial for the EVal's to be $\frac{-b \pm \sqrt{b^2 - 4b}}{2}$ where $b = \frac{\mu -1}{\mu}$.

Now where do I go from here to complete the question? Also please do keep in mind that I'm studying Maths not Physics.

Many thanks.

  • 2
    Shouldn't the critical points be $(x, y) = (0, 0), (0, -1), (2-\mu, 1-\mu)$? If you substitute each of those into the system, what do you get? Now, try that with the ones you show. – Moo Mar 18 '17 at 13:14
  • 2
    Why you spend time on the cases $\mu=0$ and $\mu=-1$ when $\mu>0$ is assumed, escapes me. And you are missing that the case $\mu=1$ is quite special since then, every $(x,0)$ is a fixed point. – Did Mar 18 '17 at 13:21
  • @Moo My apologies, $\dot{x} = x(1-\mu x -y)$. –  Mar 18 '17 at 14:03
  • Strategy. – Cosmas Zachos Mar 18 '17 at 21:57
  • http://math.stackexchange.com/q/2191385/ – Did Mar 19 '17 at 11:19

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