We have the following dynamical system to linearize in order to find the critical points:
$$\dot{y_0}(t) = y_3(t) \\
\dot{y_1}(t) = y_4(t) \\
\dot{y_2}(t) = y_5(t) \\
\dot{y_3}(t) = Aa\mathrm{f}[y_1(t) – y_2(t)] – 2ay_3(t) – a^2y_0(t)\\
\dot{y_4}(t) = Aa\{p(t) + C_2\mathrm{f}[C_1y_0(t)]\} – 2ay_4(t) – a^2y_1(t) \\
\dot{y_5}(t) = BbC_4\mathrm{f}[C_3y_0(t)] – 2by_5(t) – b^2y_2(t)$$
where,
$$f$$ is a nonlinear function
and $a, A, b, B, C _{i}, \nu _{max}, u_{0},r$ are contants.
After setting the derivatives to zero, we end up with the following equations:
$$\frac{A}{a}\mathrm{f}[y_1(t) – y_2(t)] = y_0(t)\\
\frac{A}{a}\{p(t) + C_2\mathrm{f}[C_1y_0(t)]\} = y_1(t) \\
\frac{B}{b}C_4\mathrm{f}[C_3y_0(t)] = y_2(t)$$
which according to the paper I am reading, leads to this (implicit) equation of the equilibrium points:
$$y = \frac{A}{a}p + \frac{A}{a}C_{2}\mathrm{f}(\frac{A}{a}C_1\mathrm{f}(y)) – \frac{B}{b}C_{4}\mathrm{f}(\frac{A}{a}C_3\mathrm{f}(y))$$
No matter what I do, I receive totally different result when I substitute in the $y_0(t)$ equations, the expressions of $y_1(t)$ and $y_2(t)$, and I don't know where this result comes from. I assumed that they linearised the Sigmoid about its centre, but it still cannot be the case, since expressions such as:
$$\mathrm{f}(\mathrm{f}(y))$$
still exist in the result.
Best Answer
The result follows from the following definition: $y=y_1-y_2$. Simply subtract the third equation from the second and substitute $y_0$.