Hopefully a simple question however it has me stumped...
I have the following system:
$\frac{dN_1}{dt} = r_1N_1(1- \frac{N_1}{k_1} + b_{12} \frac{N_2}{k_1})$
$\frac{dN_2}{dt} = r_2N_2(1- \frac{N_2}{k_2} + b_{21} \frac{N_1}{k_2})$
Transforming it to its dimensionless form (skipped workings since not relevant to question):
$\frac{du}{d \tau} = u(1-u+ \alpha v)$
$\frac{dv}{d \tau} = \rho v(1-v + \beta u)$
Setting both equal 0 we get the following steady states:
$(0,0), (0,1), (1,0)$ and $(1+ \alpha v, 1+ \beta u)$
QUESTION:
How was the steady state
$(1+ \alpha v, 1+ \beta u)$
transformed into
$(\frac{1+ \alpha}{1-\alpha \beta}, \frac{1+ \beta}{1-\alpha \beta})$ ?
Thanks in advance.
The steady state is when the derivatives are both zero. This occurs when $u(1-u+\alpha v)=0$ and $\rho v (1-v + \beta u)=0$. If $u=0$ then we have either $v=0$ or $v=1$ and similarly if $v=0$ we have $u=0$ or $u=1$. That accounts for your first three cases.
The final case is the equations $1-u+\alpha v=0$ and $1-v + \beta u=0$ (which is equivalent to your fourth solution). Just solve these two simultaneous equations and you get the solution you mentioned. That is, just substitute $u=1 +\alpha v$ into the equation $1-v + \beta u=0$ to solve for $v$ and then substitute the value or $v$ back into the first equation to find $u$.