Fixed points of a system of differential equations

Question

I'm wondering about how to find the fixed points for the following system:

$$ \dot{x} = \frac{xr_1}{k_1}\left(k_1 - c_1 x - i_1 y \right) $$

$$ \dot{y} = \frac{y r_2}{k_2}\left(k_2 - c_2 y - i_2 x \right) $$

I think the approach would be;

For $\dot{x}$ I can state that either $x=0$ or the term in the parenthesis is zero. For the term the parenthesis, consider $x=0$ and $y=0$ separately. This gives the points $(0, k_1/i_1)$ when $x=0$ and $(k_1/c_1 , 0)$ when $y=0$.

The same approach is taken for $\dot{y}$ which gives $(0, k_2/c_2)$ when $x=0$ and $(k_2/i_2, 0)$ when $y=0$.

This gives the fixed points

• $(0 , 0) $
• $(0 , \frac{k_1}{i_1}) $, (from $\dot{x}$, where $x=0$)
• $(\frac{k_1}{c_1} , 0) $, (from $\dot{x}$, where $y=0$)
• $(0 , \frac{k_2}{c_2}) $, (from $\dot{y}$, where $x=0$)
• $(\frac{k_2}{c_2} , 0) $, (from $\dot{y}$, where $y=0$)

Answer 1

No, that's not correct.

For example, if $x=0$ then $\dot x=0$ already, so you shouldn't look at the other factor in $\dot x$ in that case.

The correct cases are:

• $x = 0$ and $y = 0$

or

• $x = 0$ and $k_2 - c_2 y - i_2 x = 0$

or

• $k_1 - c_1 x - i_1 y = 0$ and $y = 0$

or

• $k_1 - c_1 x - i_1 y = 0$ and $k_2 - c_2 y - i_2 x = 0$

Each of these cases gives you one fixed point $(x,y)$, so there are four fixed points in total (the origin, one on each axis, plus a nontrivial one).