<s>[INST] fixed points for the following system I'm trying to find the fixed point for the system (see document attached) but it seems so hard and I don't know what Im doing wrong. Can somebody help me with this. I need to find the to look for the value where the bifurcation occur (for that process i need to evaluate the jacobian matrix for the system on the fixed points and the looks for the delta, which is really easy) but this is the only part where I'm stuck. Please see document attached 
 [/INST] We need to simultaneously solve:
$$\begin{align*} a(1 - u) - u v^2 = 0 \\ u v^2 -(a + k) v = 0 \end{align*}$$
From the first equation we get:
$$u = \dfrac{a}{a+v^2}$$
Substituting this into the second equation yields:
$$-(a+k) v + \dfrac{a v^2}{a + v^2} = -(a+k)(v)(a + v^2) + av^2 = 0$$
Expanding and simplifying, yields:
$$ -a^2 v-a k v-a v^3+a v^2-k v^3=0$$
The three roots (critical or fixed points) to this cubic are:
$$v = 0, \dfrac{a-\sqrt{-4 a^3-8 a^2 k+a^2-4 a k^2}}{2 (a+k)}, \dfrac{a+\sqrt{-4 a^3-8 a^2 k+a^2-4 a k^2}}{2 (a+k)}$$
Now, we need to find the corresponding $u$ values using either equation. We have:
$$u = \dfrac{a}{v^2 + a}$$


*

*At $v = 0$, we get $u = 1$.

*At $v = \dfrac{a-\sqrt{-4 a^3-8 a^2 k+a^2-4 a k^2}}{2 (a+k)}, u = \dfrac{a+\sqrt{a \left(-4 a^2-8 a k+a-4 k^2\right)}}{2 a}$

*At $v = \dfrac{a+\sqrt{-4 a^3-8 a^2 k+a^2-4 a k^2}}{2 (a+k)}, u = \dfrac{a-\sqrt{a \left(-4 a^2-8 a k+a-4 k^2\right)}}{2 a}$

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