Consider a system of two genes whose dynamics are described by the following two equations: $$\dot{X}_1=\frac{K}{1+X_2^n}-X_1, \quad \dot{X}_2=\frac{K}{1+X_1^n}-X_2. $$ Find the critical value of $K$ at which the equilibrium state with $X_1 = X_2$ loses stability, and is replaced by the two stable symmetrical equilibria $(X_1, X_2)$ and $(X_2, X_1).$
Because of symmetry, the steady state solutions are either $x_1 = x_2$, or a pair $(x_1, x_2)$ and $(x_2, x_1)$. If $x_1 = x_2 = x_0$, then $x_0^{n+1} + x_0 +K = 0$. For this eq, we have one positive root $x_0$. Linearising around it, and using $x_{i} = x_0 +\epsilon_{i}$, where $i=1,2$, we have the following differential equation:
$$\dot{\epsilon_1}=\frac{K}{1+(x_0+\epsilon_2)^n}-\epsilon_1-x_0.$$
Linearising, we get
$$\dot{\epsilon_1}=-\frac{n}{K}x_0^{n+1}\epsilon_2-\epsilon_1.$$
I don't really understand how they linearised either time.
I tried to combine the $x_0$ terms to get
$$ \frac{K-x_0-x_0^{n+1}-\ldots-n\epsilon_2^{n-1}x_0-\epsilon_2^n}{1+x_0^n+\ldots+n\epsilon_2^{n-1}x_0+\epsilon_2^n}-\epsilon_1.$$
If I linearise the first term, I get
$$\frac{K-x_0(1+n\epsilon_2^{n-1})-\epsilon_2^n}{1+n\epsilon_2^{n-1}x_0-\epsilon_2^n}-\epsilon_1.$$
How do I proceed?
I'm not sure what version of linearization you are using, but I believe I have a solution. First compute the Jacobian about the point $(X_1,X_2) = (x_0,x_0)$: $$ J = \left.\begin{pmatrix} -1 & \frac{-KnX_2^{n-1}}{(1+X_2^n)^2} \\ \frac{-KnX_1^{n-1}}{(1+X_1^n)^2} & -1 \end{pmatrix}\right|_{(x_0,x_0)} = \begin{pmatrix} -1 & -K\alpha \\ -K\alpha & -1 \end{pmatrix}, $$ where $\alpha = \frac{nx_0^{n-1}}{(1+x_0^n)^2}$. We then compute the eigenvalues to find $\lambda_1 = -K\alpha -1$ and $\lambda_2 = K\alpha-1$. Stability occurs when both of these eigenvalues are strictly negative. Assuming that $x_0$ and $K$ are both positive, $\lambda_1<0$, so we must find the value of $K$ where $\lambda_2$ changes sign. This occurs at $$ K = \frac{1}{\alpha} = \frac{(1+x_0^n)^2}{nx_0^{n-1}}. $$