Category Archives: Mathematics

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

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

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

$\frac{dp}{dt}=-4k_1x-k_2.$

Integrate it with respect to $t$ ,

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

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.

An Ellipse in Its Polar Form

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In this appendix to my previous post “From Dancing Planet to Kepler’s Laws“, we derive the polar form for an ellipse that has a rectangular coordinate system’s origin as one of its foci.

Fig. 1

We start with the ellipse shown in Fig. 1. Namely,

$\frac{x^2}{a^2}+\frac{y^2}{b^2}=1.$

Clearly,

$f^2=a^2-b^2\implies f< a.\quad\quad\quad(1)$

After shifting the origin $O$ to the right by $f$ , the ellipse has the new origin $O'$ as one of its foci (Fig. 2).

Fig. 2

Since $x = x' + f, y=y'$ , the ellipse in $x'O'y'$ is

$\frac{(x'+f)^2}{a^2}+\frac{y'^2}{b^2}=1.\quad\quad\quad(2)$

Substituting $x', y'$ in (2) by

$x'=r\cos(\theta), y'= r\sin(\theta)$

yields equation

$a^2r^2\sin^2(\theta)+b^2r^2\cos^2(\theta)+2b^2fr\cos(\theta)+b^2f^2-a^2b^2=0.$

Replacing $\sin^2(\theta), f^2$ by $1-\cos^2(\theta), a^2-b^2$ respectively, the equation becomes

$b^2r^2\cos^2(\theta)+a^2r^2(1-\cos^2(\theta)+2b^2fr\cos(\theta)-a^2b^2+b^2(a^2-b^2)=0.\quad\quad\quad(3)$

Fig. 3

Solving (3) for $r$ (see Fig. 3) gives

$r=\frac{b^2}{f cos(\theta)+a}$ or $r=\frac{b^2}{f \cos(\theta)-a}$ .

The first solution

$r=\frac{b^2}{f\cos(\theta)+a} \implies r= \frac{\frac{b^2}{a}}{\frac{f}{a}\cos(\theta)+1}$ .

Let

$p=\frac{b^2}{a}, e=\frac{f}{a}$ ,

we have

$r = \frac{p}{1+e\cdot\cos(\theta)}$ .

The second solution is not valid since it suggests that $r < 0$ :

$\cos(\theta)<1 \implies f\cdot\cos(\theta) < f \overset{(1)}{\implies} f\cos(\theta)<a \implies f\cos(\theta)-a<0$ .

From Dancing Planet to Kepler’s Laws

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“This most beautiful system of the sun, planets, and comets, could only proceed from the counsel and dominion of an intelligent powerful Being” – Sir. Issac Newton

When I was seven years old, I had the notion that all planets dance around the sun along a wavy orbit (see Fig. 1).

Fig. 1

Many years later, I took on a challenge to show mathematically the orbit of my ‘dancing planet’ . This post is a long overdue report of my journey.

Shown in Fig. 2 is the sun and a planet in a x-y-z coordinate system. The sun is at the origin. The moving planet’s position is being described by $x=x(t), y=y(t), z=z(t)$ .

Fig. 2 $r=\sqrt{x^2+y^2+z^2}, F=G\frac{Mm}{r^2}, F_z=-F\cos(c)=-F\cdot\frac{z}{r}$

According to Newton’s theory, the gravitational force sun exerts on the planet is

$F=-G\cdot M \cdot m \cdot \frac{1}{r^2}(\frac{x}{r},\frac{y}{r}, \frac{z}{r})=-\mu\cdot m \cdot \frac{1}{r^3}\cdot(x, y, z)$

where $G$ is the gravitational constant, $M, m$ the mass of the sun and planet respectively. $\mu = G\cdot M$ .

By Newton’s second law of motion,

$\frac{d^2x}{dt^2} = -\mu\frac{x}{r^3},\quad\quad\quad(0-1)$

$\frac{d^2y}{dt^2} = -\mu\frac{y}{r^3},\quad\quad\quad(0-2)$

$\frac{d^2z}{dt^2} = -\mu\frac{z}{r^3}.\quad\quad\quad(0-3)$

$y \cdot$ (0-3) $- z \cdot$ (0-2) yields

$y\frac{d^2z}{dt^2}-z\frac{d^2y}{dt^2} = -\mu\frac{yz}{r^3}+ \mu\frac{yz}{r^3}=0$ .

Since

$y\frac{d^2z}{dt^2}-z\frac{d^2y}{dt^2} = \frac{dy}{dt}\frac{dz}{dt}+y\frac{d^2z}{dt^2}-\frac{dz}{dt}\frac{dy}{dt}-z\frac{d^2y}{dt^2}=\frac{d}{dt}(y\frac{dz}{dt}-z\frac{dy}{dt})$ ,

it must be true that

$\frac{d}{dt}(y\frac{dz}{dt}-z\frac{dy}{dt}) = 0$ .

i.e.

$y\frac{dz}{dt}-z\frac{dy}{dt}=A\quad\quad\quad(0-4)$

where $A$ is a constant.

Similarly,

$z\frac{dx}{dt}-x\frac{dz}{dt}= B,\quad\quad\quad(0-5)$

$x\frac{dy}{dt}-y\frac{dx}{dt}= C\quad\quad\quad(0-6)$

where $B,C$ are constants.

Consequently,

$Ax=xy\frac{dz}{dt} - xz\frac{dy}{dt}$ ,

$By=yz\frac{dx}{dt} - xy\frac{dz}{dt}$ ,

$Cz=xz\frac{dy}{dt}-yz\frac{dx}{dt}$ .

Hence

$Ax + By +Cz=0.\quad\quad\quad(0-7)$

If $C \ne 0$ then by the following well known theorem in Analytic Geometry:

“If A, B, C and D are constants and A, B, and C are not all zero, then the graph of the equation Ax+By+Cz+D=0 is a plane“,

(0-7) represents a plane in the x-y-z coordinate system.

For $C=0$ , we have

$\frac{d}{dt}(\frac{y}{x})=\frac{x\frac{dy}{dt}-y\frac{dx}{dt}}{x^2}\overset{(0-6)}{=}\frac{C}{x^2}=\frac{0}{x^2}=0$ .

It means

$\frac{y}{x}=k$

where $k$ is a constant. Simply put,

$y=k x$ .

Hence, (0-7) still represents a plane in the x-y-z coordinate system (see Fig. 3(a)).

Fig. 3

The implication is that the planet moves around the sun on a plane (see Fig. 4).

Fig. 4

By rotating the axes so that the orbit of the planet is on the x-y plane where $z \equiv 0$ (see Fig. 3), we simplify the equations (0-1)-(0-3) to

$\begin{cases} \frac{d^2x}{dt^2}=-\mu\frac{x}{r^3} \\ \frac{d^2y}{dt^2}=-\mu\frac{y}{r^3}\end{cases}.\quad\quad\quad(1-1)$

It follows that

$\frac{d}{dt}((\frac{dx}{dt})^2 + (\frac{dy}{dt})^2)$

$= 2\frac{dx}{dt}\cdot\frac{d^2x}{dt^2} + 2 \frac{dy}{dt}\cdot\frac{d^2y}{dt^2}$

$\overset{(1-1)}{=}2\frac{dx}{dt}\cdot(-\mu\frac{x}{r^3})+ 2\frac{dy}{dt}\cdot(-\mu\frac{y}{r^3})$

$= -\frac{\mu}{r^3}\cdot(2x\frac{dx}{dt}+2y\frac{dy}{dt})$

$= -\frac{\mu}{r^3}\cdot\frac{d(x^2+y^2)}{dt}$

$= -\frac{\mu}{r^3}\cdot\frac{dr^2}{dt}$

$= -\frac{\mu}{r^3} \cdot 2r \cdot \frac{dr}{dt}$

$= -\frac{2\mu}{r^2} \cdot \frac{dr}{dt}$ .

i.e.,

$\frac{d}{dt}((\frac{dx}{dt})^2 + (\frac{dy}{dt})^2) = -\frac{2\mu}{r^2} \cdot \frac{dr}{dt}$ .

Integrate with respect to $t$ ,

$(\frac{dx}{dt})^2+(\frac{dy}{dt})^2 = \frac{2\mu}{r} + c_1\quad\quad\quad(1-2)$

where $c_1$ is a constant.

We can also re-write (0-6) as

$x\frac{dy}{dt}-y\frac{dx}{dt}=c_2\quad\quad\quad(1-3)$

where $c_2$ is a constant.

Using polar coordinates

$\begin{cases} x= r\cos(\theta) \\ y=r\sin(\theta) \end{cases},$

Fig. 5

we obtain from (1-2) and (1-3) (see Fig. 5):

$(\frac{dr}{dt})^2 + (r\frac{d\theta}{dt})^2-\frac{2\mu}{r} = c_1,\quad\quad\quad(1-4)$

$r^2\frac{d\theta}{dt} = c_2.\quad\quad\quad(1-5)$

If the speed of planet at time $t$ is $v$ then from Fig. 6,

Fig. 6

$v = \lim\limits_{\Delta t \rightarrow 0}\frac{\Delta l}{\Delta t} = \lim\limits_{\Delta t\rightarrow 0}\frac{l_r}{\Delta t}\overset{l_r=r\Delta \theta}{=}\lim\limits_{\Delta t \rightarrow 0}\frac{r\cdot \Delta \theta}{\Delta t}=r\cdot\lim\limits_{\Delta t\rightarrow 0}\frac{\Delta \theta}{\Delta t}=r\cdot\frac{d\theta}{dt}$

gives

$v = r\frac{d\theta}{dt}.\quad\quad\quad(1-6)$

Suppose at $t=0$ , the planet is at the greatest distance from the sun with $r=r_0, \theta=0$ and speed $v_0$ . Then the fact that $r$ attains maximum at $t=0$ implies $(\frac{dr}{dt})_{t=0}=0$ . Therefore, by (1-4) and (1-5),

$(\frac{dr}{dt})^2_{t=0} + (r\frac{d\theta}{dt})^2_{t=0}-\frac{s\mu}{r} = 0+ v_0^2-\frac{2\mu}{r}=c_1,$

$r (r\frac{d\theta}{dt})_{t=0}=r_0v_0=c_2.$

i.e.,

$c_1=v_0^2-\frac{2\mu}{r_0},\quad\quad\quad(1-7)$

$c_2=v_0 r_0.\quad\quad\quad(1-8)$

We can now express (1-4) and (1-5) as:

$\frac{dr}{dt} = \pm \sqrt{c_1+\frac{2\mu}{r}-\frac{c_2^2}{r^2}},\quad\quad\quad(1-9)$

$\frac{d\theta}{dt} = \frac{c_2}{r^2}.\quad\quad\quad(1-10)$

Let

$\rho = \frac{c_2}{r}\quad\quad\quad(1-11)$

then

$\frac{d\rho}{dr} = -\frac{c_2}{r^2},\quad\quad\quad(1-12)$

$r=\frac{c_2}{\rho}.\quad\quad\quad(1-13)$

By chain rule,

$\frac{d\theta}{dt} = \frac{d\theta}{d\rho}\cdot\frac{d\rho}{dr}\cdot\frac{dr}{dt}$ .

Thus,

$\frac{d\theta}{d\rho} = \frac{\frac{d\theta}{dt}}{ \frac{d\rho}{dr} \cdot \frac{dr}{dt}}$

$\overset{(1-10), (1-12), (1-9)}{=} \frac{\frac{c_2}{r^2}}{ (-\frac{c_2}{r^2})\cdot(\pm\sqrt{c_1+\frac{2\mu}{r}-\frac{c_2^2}{r^2}}) }$

$\overset{(1-11)}{=} \mp\frac{1}{\sqrt{c_1-\rho^2+2\mu(\frac{\rho}{c_2})}}$

$= \mp\frac{1}{\sqrt{c_1+(\frac{\mu}{c_1})^2-\rho^2+2\mu(\frac{\rho}{c_2}) -(\frac{\mu}{c_2})^2}}$

$= \mp\frac{1}{\sqrt{c_1+(\frac{\mu}{c_1})^2-(\rho^2-2\mu(\frac{\rho}{c_2}) +(\frac{\mu}{c_2})^2)}}$

$= \mp\frac{1}{\sqrt{c_1+(\frac{\mu}{c_1})^2-(\rho-\frac{\mu}{c_2})^2}}$ .

That is,

$\frac{d\theta}{d\rho} = \mp\frac{1}{\sqrt{c_1+(\frac{\mu}{c_1})^2-(\rho-\frac{\mu}{c_2})^2}}.\quad\quad\quad(1-14)$

Since

$c_1+(\frac{\mu}{c_2})^2\overset{(1-7)}{=}v_0^2-\frac{2\mu}{r_0}+(\frac{\mu}{v)0r_0})^2=(v_0-\frac{\mu}{v_0r_0})^2$ ,

we let

$\lambda = \sqrt{c_1 + (\frac{\mu}{c_2})^2}=\sqrt{(v_0-\frac{\mu}{v_0r_0})^2}=|v_0-\frac{\mu}{v_0r_0}|$ .

Notice that $\lambda \ge 0$ .

By doing so, (1-14) can be expressed as

$\frac{d\theta}{d\rho} =\mp \frac{1}{\sqrt{\lambda^2-(\rho-\frac{\mu}{c_2})^2}}$ .

Take the first case,

$\frac{d\theta}{d\rho} = -\frac{1}{\sqrt{\lambda^2-(\rho-\frac{\mu}{c_2})^2}}$ .

Integrate it with respect to $\rho$ gives

$\theta + c = \arccos(\frac{\rho-\frac{\mu}{c_2}}{\lambda})$

where $c$ is a constant.

When $r=r_0, \theta=0$ ,

$c = \arccos(1)=0$ or $c = \arccos(-1) = \pi$ .

And so,

$\theta = \arccos(\frac{\rho-\frac{\mu}{c_2}}{\lambda})$ or $\theta+\pi = \arccos(\frac{\rho-\frac{\mu}{c_2}}{\lambda})$ .

For $c = 0$ ,

$\lambda\cos(\theta) = \rho-\frac{\mu}{c_2}$ .

By (1-11), it is

$\frac{c_2}{r}-\frac{\mu}{c_2} = \lambda \cos(\theta).\quad\quad\quad(1-15)$

Fig. 7

Solving (1-15) for $r$ yields

$r=\frac{c_2^2}{c_2 \lambda \cos(\theta)+\mu}=\frac{\frac{c_2^2}{\mu}}{\frac{c_2}{\mu}\lambda \cos(\theta)+1}\overset{p=\frac{c_2^2}{\mu}, e=\frac{c_2 \lambda}{\mu}}{=}\frac{p}{e \cos(\theta) + 1}$ .

i.e.,

$r = \frac{p}{e \cos(\theta) + 1}.\quad\quad\quad(1-16)$

Studies in Analytic Geometry show that for an orbit expressed by (1-16), there are four cases to consider depend on the value of $e$ :

We can rule out parabolic and hyperbolic orbit immediately for they are not periodic. Given the fact that a circle is a special case of an ellipse, it is fair to say:

The orbit of a planet is an ellipse with the Sun at one of the two foci.

In fact, this is what Kepler stated as his first law of planetary motion.

Fig. 8

For $c=\pi$ ,

$\theta + \pi = \arccos(\frac{\rho-\frac{\mu}{c_2}}{\lambda})$

from which we obtain

$r=\frac{c_2^2}{c_2 \lambda \cos(\theta+\pi)+\mu}=\frac{\frac{c_2^2}{\mu}}{\frac{c_2}{\mu}\lambda\cos(\theta+\pi)+1}\overset{p=\frac{c_2^2}{\mu}, e=\frac{c_2 \lambda}{\mu}}{=}\frac{p}{e \cos(\theta+\pi) + 1}.\quad\quad(1-17)$

This is an ellipse. Namely, the result of rotating (1-16) by hundred eighty degrees or assuming $r$ attains its minimum at $t=0$ .

The second case

$\frac{d\theta}{d\rho} = +\frac{1}{\sqrt{\lambda^2-(\rho-\frac{\mu}{c_2})^2}}$

can be written as

$-\frac{d\theta}{d\rho} = -\frac{1}{\sqrt{\lambda^2-(\rho-\frac{\mu}{c_2})^2}}$ .

Integrate it with respect to $\rho$ yields

$-\theta + c = \arccos(\frac{\rho-\frac{\mu}{c_2}}{\lambda})$

from which we can obtain (1-16) and (1-17) again.

Fig. 9

Over the time duration $\Delta t$ , the area a line joining the sun and a planet sweeps an area $A$ (see Fig. 9).

$A = \int\limits_{t}^{t+\Delta t}\frac{1}{2}r\cdot v\;dt \overset{(1-6)}{=} \int\limits_{t}^{t+\Delta t}\frac{1}{2}r\cdot r\frac{d\theta}{dt}\;dt=\int\limits_{t}^{t+\Delta t}\frac{1}{2}r^2\frac{d\theta}{dt}\;dt\overset{(1-5)}{=}\int\limits_{t}^{t+\Delta t}\frac{1}{2}c_2\;dt = \frac{1}{2}c_2\Delta t$ .

It means

$A = \frac{1}{2}c_2\Delta t\quad\quad\quad(2-1)$

or that

$\frac{A}{\Delta t} = \frac{1}{2}c_2$

is a constant. Therefore,

A line joining the Sun and a planet sweeps out equal areas during equal intervals of time.

This is Kepler’s second law. It suggests that the speed of the planet increases as it nears the sun and decreases as it recedes from the sun (see Fig. 10).

Fig. 10

Furthermore, over the interval $T$ , the period of the planet’s revolution around the sun, the line joining the sun and the planet sweeps the enire interior of the planet’s elliptical orbit with semi-major axis $a$ and semi-minor axis $b$ . Since the area enlosed by such orbit is $\pi ab$ (see “Evaluate a Definite Integral without FTC“), setting $\Delta t$ in (2-1) to $T$ gives

${\frac{1}{2}c_2 T} = {\pi a b} \implies {\frac{1}{4}c_2^2 T^2}={\pi^2 a^2 b^2} \implies T^2 = \frac{4\pi^2 a^2 b^2}{c_2^2} \implies \frac{T^2}{a^3} = \frac{4\pi^2b^2}{c_2^2a}. (3-1)$

While we have $p = \frac{c_2^2}{\mu}$ in (1-16), it is also true that for such ellipse, $p=\frac{b^2}{a}$ (see “An Ellipse in Its Polar Form“). Hence,

$\frac{b^2}{a}=\frac{c_2^2}{\mu}\implies c_2^2=\frac{\mu b^2}{a}.\quad\quad\quad(3-2)$

Substituting (3-2) for $c_2^2$ in (3-1),

$\frac{T^2}{a^3} = \frac{4\pi^2 b^2}{(\frac{\mu b^2}{a})a}=\frac{4\pi^2}{\mu} \overset{\mu=GM}{=}\frac{4\pi^2}{GM}.\quad\quad\quad(3-3)$

Thus emerges Kepler’s third law of planetary motion:

The square of the orbital period of a planet is directly proportional to the cube of the semi-major axis of its orbit.

Established by (3-3) is the fact that the proportionality constant is the same for all planets orbiting around the sun.

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Evaluate a Definite Integral without FTC

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Problem: Find the area enclosed by ellipse

$\frac{x^2}{a^2} + \frac{y^2}{b^2}= 1$

Solution: For $0 \le x \le a$ ,

$y = b\cdot\sqrt{1-\frac{x^2}{a^2}} = \frac{b}{a}\sqrt{a^2-x^2}$ .

Clearly,

$A_{\;\frac{1}{4}ellipse} =\int\limits_{0}^{a}\frac{b}{a}\sqrt{a^2-x^2}\;dx =\frac{b}{a}\cdot\boxed{\int\limits_{b}^{a}\sqrt{a^2-x^2}\;dx}$ .

Since $\int\limits_{0}^{a}\sqrt{a^2-x^2}\;dx$ is a quarter of the area enclosed by a circle with radius $a$ ,

$\int\limits_{0}^{a}\sqrt{a^2-x^2}\;dx = \frac{\pi a^2}{4}$ .

It follows that

$A_{\; \frac{1}{4} ellipse} =\frac{b}{a}\cdot \frac{\pi a^2}{4}=\frac{\pi a b }{4} \implies A_{\;ellipse} = \pi a b$ .

Getting Animated!

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The above animation is produced by Omega CAS Explorer:

A Joint Work with David Deng on CAS-Aided Analysis of Rocket Flight Performance Optimization

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Research on rocket flight performance has shown that typical single-stage rockets cannot serve as the carrier vehicle for launching satellite into orbit. Instead, multi-stage rockets are used in practice with two-stage rockets being the most common. The jettisoning of stages allows decreasing the mass of the remaining rocket in order for it to accelerate rapidly till reaching its desired velocity and height.

Optimizing flight performance is a non-trivial problem in the field of rocketry. This post examines a two-stage rocket flight performance through rigorous mathematical analysis. A Computer Algebra System (CAS) is employed to carry out the symbolic computations in the process. CAS has been proven to be an efficient tool in carry out laborious mathematical calculations for decades. This post reports on the process and the results of using Omega CAS explorer, a Maxima based CAS to solve this complex problem.

A two-stage rocket consists of a payload $P$ propelled by two stages of masses $m_1$ (first stage) and $m_2$ (second stage), both with structure factor $1-e$ . The exhaust speed of the first stage is $c_1$ , and of second stage $c_2$ . The initial total mass, $m_1 + m_2$ is fixed. The ratio $b = \frac{P}{m_1+m_2}$ is small.

Based on Tsiolkovsky’s equation, we derived the multi-stage rocket flight equation $[1]$ . For a two-stage rocket, the final velocity can be calculated from the following:

$v = -c_1\log(1-\frac{em_1}{m_1+m_2+P}) - c_2\log(1-\frac{em_2}{m_2+P})\quad\quad\quad(1)$

Let $a = \frac{m_2}{m_1+m_2}$ , so that (1) becomes

$v=-c_1\log(1-\frac{e(1-a)}{1+b}) - c_2\log(1-\frac{ea}{a+b})\quad\quad\quad(2)$

where $0<a<1, b>0, 0<e<1, c_1>0, c_2>0$ .

We seek an appropriate value of $a$ that maximizes $v$ .

Consider $v$ as a function of $a$ , its derivative $\frac{d}{da}v$ is computed (see Fig. 1)

Fig. 1

We have $\frac{d}{da}v = -\frac{e(abc_2e+abc_1e+a^2c_1e+b^2c_2+bc_2-b^2c_1-2abc_1-a^2c_1)}{(b+a)(ae-b-a)(ae-e+b+1)}$ .

That is, $\frac{d}{da}v = \frac{e(abc_2e+abc_1e+a^2c_1e+b^2c_2+bc_2-b^2c_1-2abc_1-a^2c_1)}{(b+a)(b+a(1-e))(ae+b+1-e)}$ .

Fig. 2

As shown in Fig. 2, $\frac{d}{da}v$ can be expressed as

$\frac{d}{da}v = \frac{e(A_2a^2+B_2a+C_2)}{(b+a)(b+a(1-e))(ae+b+1-e)}\quad\quad\quad(3)$

Notice that $A_2 = c_1(e-1) = -c_1(1-e) < 0$ .

Solving $\frac{d}{da}v = 0$ for $a$ gives two solutions $a_1, a_2$ (see Fig. 3)

Fig. 3

We rewrite the expression under the square root in $a_1$ and $a_2$ as a quadratic function of $e$ : $Ae^2+Be+ C$ and compute $B^2-4AC$ (see Fig. 4)

Fig. 4

If $c_1 \ne c_2$ , $B^2-4AC< 0$ . It implies that $Ae^2+Be+C$ is positive since $A>0$ . When $c_1=c_2$ , $Ae^2+Be+C$ where $A>0, 0<e<1$ is still positive since as a result of $B^2-4AC=0$ , the zero point of function $Ax^2+Bx+C$ is $\frac{-B}{2A} = \frac{(8b^2+8b)c_1^2}{2(b^2c_1^2+(2b^2+4b)c_1^2+b^2c_1^2)} = 1$ .

The expression under the square root is positive means both $a_1$ and $a_2$ are real-valued and $a_1-a_2 >0$ (see Fig. 5), i.e., $a_1 > a_2$ .

Fig. 5

From (3) where $A_2 < 0$ , we deduce the following:

For all $a \ge 0$ , if $a>a_1$ then $\frac{d}{da}v(a) < 0 \quad\quad\quad(\star)$

For all $a \ge 0$ , if $a_2<a<a_1$ then $\frac{d}{da}v(a) >0 \quad\quad\quad(\star\star)$

For all $a \ge 0$ , if $a<a_2$ then $\frac{d}{da}v(a) < 0 \quad\quad\quad(\star\star\star)$

Fig. 6

Moreover, from Fig. 6,

$\frac{d}{da}v(0)\cdot \frac{d}{da}v(1) = -\frac{e^2(c_1e+bc_2-bc_1-c_1)(c_2e-bc_2-c_2+bc_1)}{b(b+1)(e-b-1)^2}=-\frac{e^2(c_1e+bc_2-bc_1-c_1)(c_2e-bc_2-c_2+bc_1)}{b(b+1)(b+1-e)^2}$ .

Since the expression in the numerator of $\frac{d}{da}v(0)\cdot \frac{d}{da}v(1)$ , namely

$(c_1e + bc_2-bc_1-c_1)(c_2e-bc_2-c_2 +bc_1)$

$= (c_1(e-1-b) + bc_2)(c_2(e-1-b)+bc_1)$

$=c_1c_2(e-1-b)^2+b^2c_1c_2+b(c_1^2+c_2^2)(e-1-b)$

$\ge c_1c_2(e-1-b)^2 + b^2c_1c_2+2bc_1c_2(e-1-b)$

$=c_1c_2(e-1-b+b)^2$

$=c_1c_2(e-1)^2>0$ ,

It follows that

$\frac{d}{da}v(0) \cdot \frac{d}{da}v(1) < 0\quad\quad\quad(4)$

The implication is that $\frac{d}{da}v$ has at least one zero point between $0$ and $1$ .

However, if both $a_1$ and $a_2$ , the two known zero points of $\frac{d}{da}v$ are between $0$ and $1$ , by ( $\star$ ) and ( $\star\star\star$ ), $\frac{d}{da}v(0)\cdot \frac{d}{da}v(1)$ must be positive, which contradicts (4). Therefore, $\frac{d}{da}v$ must have only one zero point between $0$ and $1$ .

We will proceed to show that the only zero lies between $0$ and $1$ is $a_1$ .

There are two cases to consider.

Case 1 ( $c_1 \le c_2$ ) $\frac{d}{da} v(0)=\frac{e(bc_2(1-e)+b^2(c_2-c1))}{b^2(b+1-e)} >0$ since $b>0, c_2>0, 0<e<1$ and $c_2-c_1>0$ . But this contradicts ( $\star\star\star$ ). Therefore, $a_2$ must not be positive.

Case 2 ( $c_1 > c_2$ ) The denominator of $a_2$ is negative since $c_1 > 0, e < 1$ . However, $(-bc_2-bc_1)e + 2bc_1$ , the terms not under the square root in the numerator of $a_2$ can be expressed as $bc_1(1-e) + b(c_1-c_2e)$ . This is a positive expression since $c_1 > c_2 > 0, 0<e<1$ implies that $c_1-c_2e > c_1 - c_1e = c_1(1-e) >0$ . Therefore, $a_2< 0$ .

The fact that only $a_1$ lies between $0$ and $1$ , together with ( $\star$ ) and ( $\star\star$ ) proves that $a_1$ is where the global maximum of $v$ occurs.

$a_1$ can be simplified to a Taylor series expansion (see Fig. 7)

Fig. 7

The result $-\frac{b((c_2+c_1)e-2c_1)}{2c_1e-2c_1} - \frac{\sqrt{b}\sqrt{c_2}}{\sqrt{c_1}}$ produced by CAS can be written as $-\sqrt{\frac{bc_2}{c_1}} + O(b)$ . However, it is incorrect as $\lim\limits_{b\rightarrow 0} \frac{-\sqrt{\frac{bc_2}{c_1}}+O(b)}{\sqrt{b}} = -\sqrt{\frac{c_2}{c_1}} < 0$ would suggest that $a_1$ is a negative quantity when $b$ is small.

To obtain a correct Taylor series expansion for $a_1$ , we rewrite $a_1$ as $\sqrt{b^2D+bE}+bF$ first where

$D=\frac{c_2^2e^2+2c_1c_2e^2+c_1^2e^2-8c_1c_2e+4c_1c_2}{(2c_1-2c_1e)^2}$

$E=\frac{4c_1c_2e^2-8c_1c_2e+4c_1c_2}{(2c_1-2c1e)^2}=\frac{c_2}{c_1}$

$F=\frac{c1e+c_2e-2c_1}{2c_1-2c_1e}$

Its first order Taylor series is then computed (see Fig. 8)

Fig. 8

The first term of the result can be written as $O(b)$ . Bring the value of $E$ into the result, we have:

$a_1= \sqrt{bE} + O(b) = \sqrt{\frac{bc_2}{c_1}} + O(b)\quad\quad\quad(6)$

To compute $v(a_1)$ from (6) , we substitute $\sqrt{Db^2+Eb} + Fb$ for $a$ in (2) and compute its Taylor series expansion about $b=0$ (see Fig. 9 )

Fig. 9

Writing its first term as $O(b)$ and substituting the value $E$ yields:

$v = -(c_1+c_2)\log(1-e)+\frac{\sqrt{b}(c_1E^{1/2}+c_2E^{-1/2})}{e-1} + O(b)$

$= -(c_1+c_2)\log(1-e) + \frac{\sqrt{b}(c_1e\sqrt{\frac{c_2}{c_1}}+c_2e\sqrt{\frac{c_1}{c_2}})}{e-1}+ O(b)$

$= -(c_1+c_2)\log(1-e)-2e\frac{\sqrt{c_1c_2b}}{1-e} + O(b)$

It is positive when $b$ is small.

We have shown the time-saving, error-reduction advantages of using CAS to aid manipulation of complex mathematical expressions. On the other hand, we also caution that just as is the cases with any software system, CAS may contain software bugs that need to be detected and weeded out with a well- trained mathematical mind.

References

$[1]$ M. Xue, Viva Rockettry! Part 2 https://vroomlab.wordpress.com/2019/01/31/viva-rocketry-part-2

$[2]$ Omega: A Computer Algebra System Explorer http://www.omega-math.com

Round Two on Finite Difference Approximations of Derivatives

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There is another way to obtain the results stated in “Finite Difference Approximations of Derivatives“.

Let $y'_i, y''_i$ denotes $y'(x_i)$ and $y''(x_i)$ respectively and,

$h=x_{i+1}-x_i=x_i-x_{i-1} >0$ .

We define

$y'_i = \alpha_1 y_{i-1} + \alpha_2 y_{i+1}\quad\quad\quad(1)$ .

By Taylor’s expansion around $x_i$ ,

$y_{i-1} \approx y_i + y'_i(-h)+\frac{y''_i}{2!}(-h)^2\quad\quad\quad(1-1)$

$y_{i+1} \approx y_i + y'_ih+\frac{y''_i}{2!}h^2\quad\quad\quad(1-2)$

Substituting (1-1), (1-2) into (1),

$y'_i \approx \alpha_1 y_i - \alpha y'_ih +\alpha_1\frac{y''_i}{2}h^2 + \alpha_2 y_i +\alpha_2y'_i h +\alpha_2\frac{y''_i}{2}h^2$ .

That is,

$y'_i \approx (\alpha_1+\alpha_2) y_i +(\alpha_2-\alpha_1)h y'_i + (\alpha_1+\alpha_2)\frac{y''_i}{2}h^2.$

It follows that

$\begin{cases} \alpha_1+\alpha_2=0 \\ (\alpha_2-\alpha_1)h=1\\ \alpha_1+\alpha_2=0 \end{cases}\quad\quad\quad(1-3)$

Fig. 1

Solving (1-3) for $\alpha_1, \alpha_2$ (see Fig. 1) yields

$\alpha_1=-\frac{1}{2h}, \alpha_2=\frac{1}{2h}.$

Therefore,

$y'_i \approx -\frac{1}{2h} y_{i-1} + \frac{1}{2h} y_{i+1} = \frac{y_{i+1}-y_{i-1}}{2h}$

or,

$\boxed{y'_i \approx \frac{y_{i+1}-y_i}{2h}}$

Now, let

$y''_i = \alpha_1y_{i-1}+\alpha_2 y_i + \alpha_3 y_{i+1}$ .

From

$y_{i-1} \approx y_i - y'_i h + \frac{y''_i}{2!} h^2$

and

$y_{i+1} \approx y_i +y'_i h + \frac{y''_i}{2!}h^2,$

we have,

$y''_i \approx \alpha_1 y_i - \alpha_1 y'_i h + \frac{\alpha_1}{2}y''_i h^2 + \alpha_2 y_i + \alpha_3 y_i + \alpha_3 y'_i h + \frac{\alpha_3}{2}y''_i h^2$ .