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.

2 thoughts on “Analyze This!

  1. Pingback: An Epilogue of “Analyze This!” | Vroom

  2. Pingback: An Edisonian Moment | Vroom

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