# An Epilogue of “Analyze This!”

In “Analyze This!“, we examined system $\begin{cases}\frac{dx}{dt}=n k_2 y - n k_1 x^n \\ \frac{dy}{dt}=k_1 x^n-k_2 y\\x(0)=x_0, y(0)=y_0\;\end{cases}$

qualitatively.

Now, let us seek its equilibrium $(x_*, y_*)$ quantitatively.

In theory, one may first solve differential equation $\frac{dx}{dt} = -nk_1x^n-k_2x+c_0$

for $x(t)$, using a popular symbolic differential equation solver such as ‘ode2’. Then compute $x_*$ as $\lim\limits_{t \rightarrow \infty} x(t)$, followed by $y_* = \frac{k_1}{k_2}x_*^n$.

However,in practice, such attempt meets a deadend rather quickly (see Fig. 1).

Fig. 1

Bring in a more sophisticated solver is to no avail (see Fig. 2)

Fig. 2

An alternative is getting $x_*$ directly from polynormial equation $-nk_1x^n-k_2x+c_0=0\quad\quad\quad(1)$

We can solve (1) for $x$ if $n \le 4$. For example, when $n=3$, $-3k_1 x^3-k_2x+c_0=0$ has three roots (see Fig. 3).

Fig. 3

First two roots are complex numbers. By Descartes’ rule of signs, the third root $\frac{(\sqrt{4k_2^3+81c_0^2k_1}+9c_0\sqrt{k_1})^{\frac{2}{3}}-2^{\frac{2}{3}}k_2}{3 \cdot 2^{\frac{1}{3}}\sqrt{k_1}(\sqrt{4k_2^3+81c_0^2k_1}+9c_0\sqrt{k_1})^{\frac{1}{3}}}$

is the $x_*$ of equilibrium $(x_*, y_*)$ (see Exercise-1).

Exercise-1 Show the third root is real and positive.

Exercise-2 Obtain $x_*$ from $-4k_1x^4-k_2x+c_0=0.$