# Analyze This!

Consider the following chemical reaction

$\underbrace{A + A + ... + A}_{n} \underset{k_2}{\stackrel{k_1}{\rightleftharpoons}} A_n$

where $n$ molecule of $A$ combine reversibly to form $A_n$ and, $k_1, k_2$ are the reaction rates.

If $x, y$ are the concentrations of $A, A_n$ respectively, then according to the Law of Mass Action, the reaction is governed by

$\begin{cases}\frac{dx}{dt}=n k_2 y - n k_1 x^n \quad\quad(0-1)\\ \frac{dy}{dt}=k_1 x^n-k_2 y\quad\quad\quad(0-2)\\x(0)=x_0, y(0)=y_0\;\quad(0-3)\end{cases}$

Without solving this initial-value problem quantitatively, the future state of system can be predicted through qualitatively analyzing how the value of $(x, y)$ changes over the course of time.

To this end, we solve (0-1) for $y$ first:

$y=\frac{1}{n k_2}(\frac{dx}{dt} +n k_1 x^n).$

Substitute it in (0-2),

$\frac{d}{dt} (\frac{1}{n k_2}(\frac{dx}{dt} +n k_1 x^n)) =k_1 x^n -k_2 \cdot \frac{1}{n k_2}(\frac{dx}{dt} +n k_1 x^n).$

It simplifies to

$\frac{d^2x}{dt^2} + (n^2 k_1 x^{n-1} + k_2)\frac{dx}{dt} =0.$

Let

$p=\frac{dx}{dt},$

we have

$\frac{d^2x}{dt^2}=\frac{d}{dt}(\frac{dx}{dt})=\frac{dp}{dt}=\frac{dp}{dx}\frac{dx}{dt}=\frac{dx}{dt}\frac{dp}{dx}=p\cdot\frac{dp}{dx}.$

Substituting $p, p\frac{dp}{dx}$ for $\frac{dx}{dt}, \frac{d^2x}{dt^2}$ respectively in (1-1) gives

$p\frac{dp}{dx}+(n^2 k_1 x^{n-1}+k_2)p=0.$

It means $p=0$ or

$\frac{dp}{dx}=-n^2k_1x^{n-1}-k_2.$

Integrate it with respect to $x$,

$p = \frac{dx}{dt}=-n k_1 x^n - k_2 x +c_0.$

Let

$f(x) = -n k_1 x^n - k_2 x +c_0$,

we have

$\frac{df(x)}{dx} = -n^2 k_1 x^{n-1} - k_2 < 0 \implies f(x) = -n k_1 x^{n-1} - k_2 x + c_0$ is a monotonically decreasing function.

In addition, Descartes’ rule of signs reveals that

$f(x)=0$ has exactly one real positive root.

By definition, this root is the $x_*$ in an equilibrium point $(x_*, y_*)$.

Fig. 1

Hence,

As time advances, $x\uparrow$ if $x_0 < x_*$. Otherwise $(x_0>x_*)$, $x\downarrow \quad\quad\quad(1-1)$

Dividing (0-2) by (0-1) yields

$\frac{dy}{dx} = -\frac{1}{n}.$

That is,

$y=-\frac{1}{n} x + c_1.$

By (0-3),

$c_1 = y_0 + \frac{1}{n}x_0$.

And so,

$y=-\frac{1}{n} x + y_0 + \frac{1}{n}x_0.$

Since $y$ is a line with a negative slope,

$y$ is a monotonically decreasing function of $x.\quad\quad\quad(1-2)$

Moreover, from (0-1) and (0-2), we see that

$\forall x > 0, (x, \frac{k_1}{k_2}x^n)$ is an equilibrium point.

i.e.,

All points on the curve $y = \frac{k_1}{k_2}x^n$ in the first quadrant of x-y plane are equilibriums $\quad\quad\quad(1-3)$.

Based on (1-1), (1-2) and (1-3), for a initial state $(x_0, y_0)$,

$x_0 < x_* \implies x\uparrow, y\downarrow$.

Similary,

$x_0 > x_* \implies x\downarrow, y\uparrow$.

Fig. 2

A phase portrait of the system is shown in Fig. 3.

Fig. 3

It shows that $(x, y)$ on the trajectory approaches the equilibrium point $(x_*, y_*)$ over the course of time. Namely, the system is asymptotically stable.