Steady states of a system

Question

How can I find the steady states? I am aware that the condition is to equal 0 but I am not able to say how many steady states there are...

$$\begin{cases} \dot x=x-y^2 \\ \dot y= -x+2y-z^2 \\ \dot z= -y+z^2 \end{cases} $$

Answer 1

The "steady states" are normally given when the time-derivatives are all set to zero, so that here we have:

$$ 0=x-y^2 \\ 0= -x+2y-z^2 \\ 0= -y+z^2 $$

so that the first and third equations give $x=y^2$ and $y=z^2$. Putting these into the middle equation we get

$$0=-y^2+2y-y\implies y^2-y=0 \implies y(y-1)=0\implies y=0 \textrm{ or } y=1$$.

$y=0$ gives $x=z=0$. $y=1$ gives $x=1$ and $z=\pm1$, so there are 3 solutions: $$(0,0,0), (1,1,1) \textrm{ and } (1,1,-1)$$