Monthly Archives: June 2020

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.

A Delightful Piece of Mathematics

Prove: \sqrt[3]{\sqrt{108} +10} - \sqrt[3]{\sqrt{108} - 10} = 2.

We will proceed as follows:

For all a, b,

a^3-b^3=(a-b)(a^2+ab+b^2)=(a-b)(a^2-2ab+b^2+3ab)=(a-b)((a-b)^2+3ab).

That is,

a^3-b^3 = (a-b)((a-b)^2+3ab).\quad\quad\quad(1)

When

a = \sqrt[3]{\sqrt{108}+10}, \quad b=\sqrt[3]{\sqrt{108}-10},

we have

a^3 = (\sqrt[3]{\sqrt{108}+10})^3 = \sqrt{108}+10, \quad b^3=(\sqrt[3]{\sqrt{108}-10})^3=\sqrt{108}-10.

It means

a^3-b^3= 20.

We also have

ab = \sqrt[3]{\sqrt{108}+10}\cdot\sqrt[3]{\sqrt{108}-10}=\sqrt[3]{(\sqrt{108})^2-10^2}=\sqrt[3]{8}=2.

As a result, (1) becomes

20 = (a-b)((a-b)^2+3\cdot 2).

Rewrite it as

20 = x (x^2+6)

where x denotes a-b, we see that

a-b=\sqrt[3]{\sqrt{108}+10}-\sqrt[3]{\sqrt{108}-10} is a positive root of 20=x(x^2+6).\quad\quad\quad(\star)

Clearly,

2 is also a positive root of 20=x(x^2+6)\quad\quad\quad(\star\star)

since 20=2\cdot(2^2+6).

Moreover (see Exercise 1),

20=x(x^2+6) has only one positive root \quad\quad\quad(\star\star\star)

Therefore, it must be true that

\sqrt[3]{\sqrt{108} +10} - \sqrt[3]{\sqrt{108} - 10} = 2.

Exercise-1 Prove: 20=x(x^2+6) has only one positive root.

Exercise-2 Prove: \sqrt[3]{8+3\sqrt{21}} + \sqrt[3]{8-3\sqrt{21}} = 1

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.

Qualitative Analysis of a Simple Chemical Reaction

We will study a simple chemical reaction described by

A + A \underset{k_2}{\stackrel{k_1}{\rightleftharpoons}} A_2

where two molecules of A are combined reversibly to form A_2 and, k_1, k_2 are the reaction rates.

If x is the concentration of A, y of A_2, then according to the Law of Mass Action,

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

or equivalently,

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

We seek first the equilibrium points that represent the steady state of the system. They are the constant solutions where \frac{dx}{dt}=0 and \frac{dy}{dt}=0, simultaneously.

From \frac{dx}{dt}=2 k_2 y - 2 k_1 x^2=0 and \frac{dy}{dt}=k_1 x^2-k_2 y=0, it is apparent that

\forall  x \ge 0, (x_*, y_*) = (x, \frac{k_1}{k_2}x^2)\quad\quad\quad(0-4)

is an equilibrium point.

To find the value of x_*, we solve for y from (0-1),

y = \frac{1}{2k_2}(\frac{dx}{dt}+2 k_1x^2).\quad\quad\quad(0-5)

Substitute it in (0-2),

\frac{d}{dt}(\frac{1}{2k_2}(\frac{dx}{dt}+2k_1x^2))=k_1x^2-k_2\frac{1}{2k_2}(\frac{dx}{dt}+2k_1x^2)

\implies  \frac{d}{dt}(\frac{dx}{dt}+2k_1x^2)=2k_2k_1x^2-k_2(\frac{dx}{dt}+2k_1x^2),

i.e.,

\frac{d^2x}{dt^2} + \frac{dx}{dt}(4k_1x+k_2)=0.\quad\quad\quad(0-6)

This is a 2nd order nonlinear differential equaion. Since it has no direct dependence on t, we can reduce its order by appropriate substitution of its first order derivative.

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}

so that (0-6) is reduced to

p\cdot\frac{dp}{dx}+p\cdot(4k_1x+k_2)=0,\quad\quad\quad(0-7)

a 1st order differential eqution. It follows that either p=\frac{dx}{dt}=0 or \frac{dp}{dx}+(4k_1x+k_2)=0.

The second case gives

\frac{dp}{dx}=-4k_1x-k_2.

Integrate it with respect to x,

p=\frac{dx}{dt}=-2k_1x^2-k_2x + c_0.\quad\quad\quad(0-8)

Hence, the equilibrium points of (0-1) and (0-2) can be obtained by solving a quadratic equation

-2k_1x^2-k_2x + c_0=0.

Notice in order to have x_* \ge 0 as a solution, c_0 must be non-negative .

Fig. 1

The valid solution is

x_* = \frac{\sqrt{k_2^2+8c_0k_1}-k_2}{4k_1}.

Fig. 2

By (0-4),

y_* =\frac{k_1}{k_2}x_*^2=\frac{(\sqrt{k_2^2+8c_0k_1}-k2)^2}{16k_1 k_2}.

and so, the equilibrium point is

(x_*, y_*) = (\frac{\sqrt{k_2^2+8c_0k_1}-k_2}{4k_1}, \frac{(\sqrt{k_2^2+8c_0k_1}-k2)^2}{16k_1 k_2}).

Next, we turn our attentions to the phase-plan trajectories that describe the paths traced out by the (x, y) pairs over the course of time, depending on the initial values.

For (x, y) \ne (x_*, y_*), \frac{dx}{dt} \ne 0. Dividing (0-2) by (0-1) yields

\frac{\frac{dy}{dt}}{\frac{dx}{dt}}= \frac{k_1 x^2-k_2 y}{2 k_2 y- 2 k_1 x^2}=-\frac{1}{2}

i.e.,

\frac{dy}{dx} = -\frac{1}{2}.

Integrating it with respect to x,

y = -\frac{1}{2} x + c_1.

By (0-3),

c_1= y_0 +\frac{1}{2}x_0.

Therefore,

y = -\frac{1}{2}x +y_0 + \frac{1}{2}x_0.\quad\quad\quad(1-1)

Moreover, by (0-5)

y=\frac{1}{2k_2}(\frac{dx}{dt} + 2 k_1 x^2) \overset{(0-8)}{=} \frac{1}{2k_1}(-2k_1 x^2-k_2 x + x_0 +2 k_1 x^2)=\frac{1}{2k_2}(-k_2x +x_0).

As a result,

y_0 = \frac{1}{2k_2}(-k_2 x_0 + x_0).

Substitute y_0 in (1-1), we have

y = -\frac{1}{2} x + \frac{1}{2k_2}(-k_2x_0+x_0) + \frac{1}{2}x_0.

This is the trajectory of the system. Clearly, all trajectories are monotonically decreaseing lines.

At last, let us examine how the system behaves in the long run.

If x < x_* then \frac{dx}{dt}>0 (see Fig. 2) and x will increase. As a result, y will decrease. Similarly, if x > x_*, \frac{dx}{dt}<0 ensures that x will decrease. Consequently, y will increase.

Fig. 3 Trajectories and Equilibriums

It is evident that as time t advances, (x, y) on the trajectory approaches the equilibrium point (x_*, y_*).

A phase portrait of the system is illustrated in Fig. 4.

Fig. 4

It shows that the system is asymptotically stable.