# An Ellipse in Its Polar Form

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

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

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).

# From Dancing Planet to Kepler’s Laws

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.

# Evaluate a Definite Integral without FTC

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

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

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

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 $00, 00, c_2>0$.

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 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 then $\frac{d}{da}v(a) >0 \quad\quad\quad(\star\star)$

For all $a \ge 0$, if $a 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 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 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

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,

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

$\begin{cases} \alpha_1+\alpha_2+\alpha_3=0\\-\alpha_1+\alpha_3=0\\(\frac{\alpha_1}{2}+\frac{\alpha_3}{2})h^2=1 \end{cases}$

Fig. 2

whose solution (see Fig. 2) is

$\alpha_1 = \frac{1}{h^2}, \alpha_2 = -\frac{2}{h^2}, \alpha_3=\frac{1}{h^2}$.

Hence,

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

i.e.,

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

# A Primer on Mathematical Epidemiology

In Memory of Dr. Li WenLiang (1986-2020)

This post is an introduction to deterministic models of infectious diseases and their Computer Algebra-Aided mathematical analysis.

We assume the followings for the simplistic SI model:

(A1-1) The population $N$ under consideration remains a constant.

(A1-2) The population is divided into two categories: the infectious and the susceptible. Their percentages are denoted by $i(t)$ and $s(t)$ respectively. At $t=0, i(0) = i_0$.

(A1-3) The infectious’ unit time encounters with other individuals is $\lambda$. Upon an encounter with the infectious, the susceptible becomes infected.

When a infectious host have $\lambda$ encounters with the population, $\lambda\cdot s(t)$ susceptible become infected. There are $N\cdot i(t)$ infectious in total at time $t$. It means that within any time interval $[t, t+\Delta t]$, the infectious will increase by $N\cdot i(t) \cdot \lambda s(t)\cdot \Delta t$. i.e.,

$N\cdot i(t+\Delta t) -N\cdot i(t) = N\cdot i(t)\cdot\lambda s(t)\cdot\Delta t$.

Cancelling out the $N$‘s,

$\frac{i(t+\Delta t)-i(t)}{\Delta t} = \lambda\cdot i(t) s(t)$

and so

$\lim\limits_{\Delta t\rightarrow 0}\frac{i(t+\Delta t)-i(t)}{\Delta t} = \lambda i(t) s(t)$.

That is,

$\frac{d i(t)}{dt} = \lambda\cdot i(t) s(t)$.

Deduce further from (A1-2) ($i+s=1$) is that

$\begin{cases} \frac{d i}{dt} = \lambda i (1-i) \\ i(0) = i_0 \end{cases}\quad\quad\quad(1-1)$

Let’s examine (1-1) qualitatively first.

We see that the SI model has two critical points:

$i_1^* = 0, \quad i_2^* = 1$.

So

$\forall i \in (0, 1), \frac{d i}{dt} = i(1-i) > 0 \implies \lim\limits_{t \rightarrow \infty} i(t)=1$.

This indicates that in the presence of any initial infectious hosts, the entire population will be infected in time. The rate of infection is at its peak when $i = 0.5$.

The qualitative results can be verified quantitatively by Omega CAS Explorer.

Fig. 1-1

From Fig. 1-1, we see that

$i(t) = \frac{1}{1-(1-\frac{1}{i_0})e^{-\lambda t}}$.

Therefore,

$\lim\limits_{t \rightarrow \infty} i(t) = \lim\limits_{t \rightarrow \infty} \frac{1}{1-(1-\frac{1}{i_0})e^{-\lambda t}} = 1$.

Fig. 1-2 confirms that the higher the number of initial infectious hosts($i_0$), the sooner the entire population becomes infected ($i = 1$)

Fig. 1-2

The SI model does not take into consideration any medical practice in combating the spread of infectious disease. It is pessimistic and unrealistic.

An improved model is the SIR model. The assumptions are

(A2-1) See (A1-1)

(A2-2) See (A1-2)

(A2-3) See (A1-3)

and,

(A2-4) Number of individuals recovered from the disease in unit time is $\mu$. The recovered are without immunity. They can be infected again.

By (A2-1) – (A2-4), the modified model is

$N \frac{d i}{dt} = \lambda N s i - \mu N i$

or,

$\begin{cases} \frac{d i}{dt} = \lambda i \cdot(1-i)-\mu\cdot i \\ i(0) = i_0 \end{cases}$

Let

$\sigma = \frac{\lambda}{\mu}$,

we have

$\begin{cases} \frac{d i}{dt} = -\lambda i \cdot(i - (1-\frac{1}{\sigma}))\\ i(0) = i_0 \end{cases}\quad\quad\quad(2-1)$

The new model has two critical points:

$i_1^*=0, \quad i_2^* = 1-\frac{1}{\sigma}$.

And,

$\frac{d^2i}{dt^2} =\frac{d}{dt}(\frac{di}{dt})=\frac{d}{di}(-\lambda i (i-(1-\frac{1}{\sigma})))\cdot \frac{di}{dt}=-2\lambda (i-\frac{1}{2}(1-\frac{1}{\sigma}))\cdot \frac{di}{dt}$

i.e.,

$\frac{d^2i}{dt^2} = 2\lambda^2 i\cdot (i-\frac{1}{2}(1-\frac{1}{\sigma}))\cdot (i-(1-\frac{1}{\sigma}))$

Without solving (2-1), we extract from it the following qualitative behavior:

Case-1 $\sigma > 1$

[2-1-1] $\lim\limits_{t \rightarrow \infty} i(t) = 1-\frac{1}{\sigma}$

[2-1-2] $\forall i>1-\frac{1}{\sigma}, \frac{di}{dt} < 0$.

[2-1-3] $\forall i < 1-\frac{1}{\sigma}, \frac{di}{dt} > 0$.

[2-1-4] $i > 1-\frac{1}{\sigma}, \frac{d^2i}{dt^2} > 0$.

[2-1-5] $\forall (\frac{1}{2}(1-\frac{1}{\sigma}) < i < (1-\frac{1}{\sigma})), \frac{d^2i}{dt^2} < 0$.

[2-1-6] $\forall(0 < i < \frac{1}{2}(1-\frac{1}{\sigma})), \frac{d^2i}{dt^2} > 0$.

Case-2 $\sigma < 1$

[2-2-1] $\lim\limits_{t\rightarrow \infty} i(t) = 0$.

[2-2-2] $\forall i, \frac{di}{dt} < 0$.

[2-2-3] $\forall i, \frac{d^2i}{dt^2} >0$.

Case-3 $\sigma = 1$

[2-3-1] $\lim\limits_{t\rightarrow \infty} i(t) = 0$.

[2-3-2] $\forall i, \frac{di}{dt} < 0$.

[2-3-3] $\forall i, \frac{d^2i}{dt^2} >0$.

The cases are illustrated by solving (2-1) analytically using Omega CAS Explorer (see Fig. 2-1,2-2,2-3)

Fig. 2-1 $\sigma> 1, i$ approaches $1-\frac{1}{\sigma}$ asymptotically.

Fig. 2-2 $\sigma < 1, i$ approaches $0$ asymptotically.

Fig. 2-3 $\sigma = 1, i$ approaches $0$ asymptotically.

Fig. 2-4 $\sigma > 1, i$‘s monotonicity depends on $i_0$.

Fig. 2-4 shows that for $\sigma > 1$, if $\frac{1}{2}(1-\frac{1}{\sigma})< i_0 < 1-\frac{1}{\sigma}$, then $i$ increases on a convex curve. Otherwise, $\forall i_0 < \frac{1}{2}(1-\frac{1}{\sigma}), i$ increases on a concave curve first. The curve turns convex after $i$ reaches $\frac{1}{2}(1-\frac{1}{\sigma})$. However, $\forall i_0 > (1-\frac{1}{\sigma}), i$ monotonically decreases along a concave curve.

Fig. 2-5 $\sigma < 1, i$ monotonically decrease.

Fig 2-5 illustrates the case $\sigma >1$.

We also have:

Fig. 2-6 $\sigma =1, i$ monotonically decrease.

From these results we may draw the following conclusion:

If $\sigma >1$, the monotonicity of $i(t)$ depends on the level of $i_0$. Otherwise ($\sigma = \frac{\lambda}{\mu} \le 1$), $i(t)$ will decrease and approach to $0$ since the rate of recovery from medical treatment $\mu$ is at least on par with the rate of infection $\lambda.$

This model is only valid for modeling infectious disease with no immunity such as common cold, dysentery. Those who recovered from such disease become the susceptible and can be infected again.

However, for many disease such as smallpox, measles, the recovered is immunized and therefore, falls in a category that is neither infectious nor susceptible. To model this type of disease, a new mathematical model is needed.

Enter the Kermack-McKendrick model of infectious disease with immunity.

There are three assumptions:

(A3-1) The total population $N$ does not change.

(A3-2) Let $i(t), s(t)$ and $r(t)$ denote the percentage of the infectious, susceptible and recovered respectively. At $t=0, i(0) = i_0, s(0) = s_0, r(0) = r_0 = 0$. The recovered are the individuals who have been infected and then recovered from the disease. They will not be infected again or transmit the infection to others.

(A3-3) $\lambda$ is the unit time number of encounters with the infectious, $\mu$ the unit time recoveries from the disease.

For the recovered, we have

$N\frac{dr}{dt} = \mu N i$.

Hence,

$\begin{cases} \frac{d i(t)}{dt} = \lambda s(t) i(t) - \mu i(t),\;i(0) = i_0\quad\quad\; (3-1-1)\\ \frac{d s(t)}{dt} = -\lambda s(t)i(t),\;s(0) = s_0\quad\quad\quad\quad\;\;(3-1-2) \\ \frac{d r(t)}{dt} = \mu i(t),\;r(0)=r_0=0\quad\quad\quad\quad\;\;\;\;(3-1-3)\\i_0, s_0 \in (1, 0), i_0+s_0=1\quad\quad\quad\quad\quad\;\;\;\;(3-1-4)\end{cases}$

This system of differential equations appears to defy any attempts to obtain an analytic solution (i.e., no solution can be expressed in terms of known function).

Numerical treatments for two sets of given $\lambda, \mu$ and $(s_0, i_0)$ are depicted in Fig. 3-1 and Fig. 3-2.

Fig. 3-1 $\lambda = 0.25, \mu=0.1, i_0= 0.15, s_0=0.85$

Fig. 3-2 $\lambda = 0.25, \mu=0.1, i_0=0.75, s_0=0.25$

However, it is only the rigorous analysis in general terms gives the correct interpretations and insights into the model.

To this end, we let

$\rho = \frac{\mu}{\lambda}$.

For $\frac{ds}{dt} \ne 0 (i.e., i \ne 0, s \ne 0)$,

$\frac{\frac{di}{dt}}{\frac{ds}{dt}} = \frac{di}{ds} = -1+\rho\frac{1}{s}$

or,

$\frac{di}{ds} = -1+\rho\frac{1}{s}.$

It has the following qualitatives:

[3-1] $\forall s < \rho, \frac{di}{ds} > 0 \implies i$ increases.

[3-2] $s = \rho, \frac{di}{ds} = 0\implies i$ reaches its maximum.

[3-3] $\forall s>\rho, \frac{di}{ds} < 0\implies i$ decreases.

The analytical solution to

$\begin{cases} \frac{di}{ds} = -1+\frac{\rho}{s}\\ i(s_0) = i_0\end{cases}\quad\quad\quad(3-2)$

(see Fig. 3-3) is

$i(s) = \rho\log(\frac{s}{s_0}) -s +i_0+s_0 \overset{[3-1-4]}{=} \rho\log(\frac{s}{s_0}) -s +1\quad\quad\quad(3-3)$

Fig. 3-3

Notice that

$\forall s, (s, 0)$ is a critical point of (3-1-1) and (3-1-2).

or,

all points on the s-axis of the s-i phase plane are critical points of (3-1-1) and (3-1-2).

By a theorem of qualitative theory of ordinary differential equations (see Fred Brauer and John Nohel: The Qualitative Theory of Ordinary Differential Equations, p. 192, Lemma 5.2),

$\forall t>0, i(t)>0\quad\quad\quad(3-4)$

Moreover, let $s' = s_0e^{-\frac{1}{\rho}},$

$i(s')=1-s' + \rho\log(\frac{s'}{s_0}) <1 + \rho\log(\frac{s'}{s_0}) = 1+ \rho\log(\frac{s_0e^{-\frac{1}{\rho}}}{s_0})=1+\rho\cdot(-\frac{1}{\rho})=0$

i.e.,

$i(s') < 0\quad\quad\quad(3-5)$

Since $s' = s_0e^{-\frac{1}{\rho}} = \frac{s_0}{e^{\frac{1}{\rho}}} \overset{\frac{1}{\rho}>0\implies e^{\frac{1}{\rho}}>e^0=1}{<} s_0$, together, (3-5) and $i(s_0) = i_0 >0$ implies

$\exists s_{\infty} : s' < s_{\infty}

or,

$\exists s_{\infty} : 0 \overset{s' = s_0e^{-\frac{1}{\rho}}> 0}{<} s_{\infty} < s_0 \ni 1-s_{\infty} + \rho\log(\frac{s_{\infty}}{s_0})=0.$

Clearly, $(s_{\infty}, 0)$ is a critical point of (3-1-1) and (3-1-2). Lemma 5.2 thus ensures

$\forall t>0, s(t) > s_{\infty} > 0\quad\quad\quad(3-6)$

It follows that

$\forall t>0, \frac{ds(t)}{dt} = -\lambda i(t)\cdot s(t) \overset{(3-4), (3-6)}{<} 0\quad\quad\quad(3-7)$

To the list ([3-1]-[3-3]) , we now add:

[3-4] $\forall t_2> t_1>0, s(t_2) \overset{(3-7)}{<} s(t_1)$

[3-5] $\lim\limits_{t \rightarrow \infty} i(t) = 0, \lim\limits_{t \rightarrow \infty} s(t) = s_{\infty}$

[3-6] $\forall t>0, \frac{dr(t)}{dt} = \mu i(t) \overset{(3-4)}{>}0$

And so, for all $(s_0, i_0) : 0 < s_0 <1, 0,

if $\rho < s_0 <1$ then

$\forall s_+: \rho \le s_+ < s_0, i_+ = i(s_+) \overset{[3-1], [3-4], (3-1-4)}{>} i(s_0)=i_0.$

Since

$i_{max} \overset{[3-2]}{=} i(\rho) \overset{(3-3)}{=} \rho\log(\frac{\rho}{s_0})-\rho+1\overset{s_0 > \rho \implies \frac{\rho}{s_0} < 1 \implies \log(\frac{\rho}{s_0}) < 0}{<}1\quad\quad\quad(3-8),$

we have

$\forall s_+: 0 < s_+ \le \rho, i_+ = i(s_+) \le i_{max} < 1.$

Fig. 3-4

If $s_0=\rho$ then,

$i_+ = i(s_+ < \rho) \overset{[3-2], [3-4]}{<} i_{max} \overset{(3-8)}{<} 1.$

Fig. 3-5

If $0 then

$i_+=i(s_+ < s_0) \overset{[3-1], [3-4]}{<} i(s_0)=i_0 < 1.$

Fig. 3-6

In fact, for all finite $t>0$,

$\forall s_+ : 0.

Thus, the orbits of (3-1-1) and (3-1-2) have the form illustrated in Fig. 3-7.

Fig. 3-7

For example,

Fig. 3-8 $\rho=0.3$

What we see is that as time $t$ advances, $(i(t), s(t))$ moves along the curve (3-3) in the direction of decreasing $s.$ Consequently, if $s_0$ is less than $\rho$, then $i(t)$ decreases monotonically to $0$, and $s(t)$ decreases to $s_{\infty}$. Therefore, if a small group of infectious $i_0$ is introduced into the population with the susceptibles $s_0$, with $s_0 < \rho$, the disease will die out rapidly. On the other hand, if $s_0$ is greater than $\rho$, then $i(t)$ increases as $s(t)$ decreases to $\rho$. Only after attaining its maximum at $s=\rho, i(t)$ starts to decrease when the number of susceptibles falls below the threshold value $\rho.$

We therefore conclude:

An epidemic will occur only if the number of susceptibles in a population exceeds the threshold value $\rho$.

It means a larger $\rho$ is preferred.

To increase $\rho = \frac{\mu}{\lambda}$, the recovery rate $\mu$ is boosted through adequate medical care and treatment. Meanwhile, $\lambda$ is reduced by quarantine and social distancing.

In addition to increase $\rho$, we can also decrease $s_0$ through immunizing the population.

If the number of susceptibles $s_0$ is initially greater than, but close to the threshold value $\rho$:

$\rho < s_0 \approx \rho\quad\quad\quad(3-9)$,

and $i_0$ is very small compared to $s_0$:

$i_0 \ll s_0\quad\quad\quad(3-10)$,

we can estimate the number of individuals who ultimately contracted the disease.

From (3-1-4), we have

$s_0+i_0+0=1\quad\quad\quad(3-11)$

and $\lim\limits_{t\rightarrow \infty}(s(t) + i(t) + r(t))$ yields

$s_{\infty} + 0 + r_{\infty} = 1\quad\quad\quad(3-12)$

(3-11)-(3-12),

$s_0-s_{\infty} + i_0 - r_{\infty} =0 \overset {(3-10)}{\implies} s_0 - s_{\infty} - r_{\infty} = 0$

or,

$r_{\infty} = s_0 - s_{\infty}\quad\quad\quad(3-13)$

Given (3-9), we deduce from it that

$1< \frac{s_0}{\rho} \approx 1 \implies 0 < \frac{s_0}{\rho}-1 \approx 0 \implies 0 <\frac{s_0-\rho}{\rho} \approx 0\implies 0 <\frac{s_0-\rho}{\rho} \ll 1$

i.e.,

$\rho < s_0 \approx \rho \implies 0<\frac{s_0-\rho}{\rho} \ll 1\quad\quad\quad(3-14)$

(3-1-2) / (3-1-3) gives

$\frac{\frac{ds}{dt}}{\frac{dr}{dt}}=\frac{ds}{dr} = -\frac{\lambda}{\mu}s=-\frac{1}{\rho}s$

Solving

$\begin{cases} \frac{ds}{dr} =-\frac{1}{\rho}s\\s(r_0=0)=s_0\end{cases}$

yields

$s = s_0 e^{-\frac{r}{\rho}}\quad\quad\quad(3-15)$

After substituting (3-15) in (3-1-3),

$\frac{dr}{dt} = \mu(1-r-s) \overset{(3-14)}{=}\mu(1 - r -s_0 e^{-\frac{r}{\rho}})\quad\quad\quad(3-16)$

In view of the fact that

$\frac{r}{\rho} \overset{[3-6]}{<} \frac{s_0-s_{\infty}}{\rho} \overset{(3-9)}{\approx} \frac{s_0-s_{\infty}}{s_0}<1\quad\quad\quad(3-17)$,

we approximate the term $e^{-\frac{r}{\rho}}$ in (3-16) with a Taylor expression up to the second order (see Fig. 3-9)

Fig. 3-9

The result is an approximation of equation (3-16):

$\frac{dr}{dt}=\mu(1-r-s_0(1-\frac{r}{\rho}+\frac{r^2}{2\rho^2}))$

i.e.,

$\frac{dr}{dt}=\mu(1-s_0+(\frac{s_0}{\rho}-1)r-\frac{s_0}{2\rho^2}r^2)$

It can be solved analytically (see Fig. 3-10).

Fig. 3-10

As a result,

$\lim\limits_{t \rightarrow \infty} r(t) = 2\rho(1-\frac{\rho}{s_0})\quad\quad\quad(3-18)$

It follows from

$2\rho(1-\frac{\rho}{s_0}) = 2\rho (\frac{s_0-\rho}{s_0}) =2(s_0-\rho)(\frac{\rho}{s_0-\rho+\rho})=2(s_0-\rho)(\frac{1}{1+\frac{s_0-\rho}{\rho}}) \overset{(3-13)}{\approx} 2(s_0-\rho)$

that (3-18) yields

$r_{\infty} \approx 2(s_0-\rho)$

Namely, the size of the epidemic is roughly $2(s_0-\rho)$. Consequently, by (3-13),

$s_{\infty} = s_0 - 2(s_0-\rho)$.

The above analysis leads to the following threshold theorem of epidemiology:

(a) An epidemic occurs if and only if $s_0$ exceeds the threshold $\rho$.

(b) If $\rho and $i_0 \ll s_0$, then after the epidemic, the number of susceptible individuals is reduced by an amount approximately $2(s_0-\rho)$, namely, $s_{\infty} \approx s_0-2(s_0-\rho)$.

We can also obtain (b) without solving for $r(t)$:

From (3-3), as $t\rightarrow \infty$,

$0 =\rho \log(\frac{s_{\infty}}{s_0})-s_{\infty} + i_0 + s_0$

$\overset{i_0 \ll s_0}{=} s_0 - s_{\infty} + \rho\log(\frac{s_0-s_0+s_{\infty}}{s_0})$

$= s_0 -s_{\infty} + \rho\log(\frac{s_0-(s_0-s_{\infty})}{s_0})$

$= s_0 -s_{\infty} + \rho\log(1-\frac{s_0-s_{\infty}}{s_0})$

When $s_0-s_{\infty}$ is small compared to $s_0$ (see (3-17)), we approximate $\log(1-\frac{s_0-s_{\infty}}{s_0})$ with a truncated Taylor series after two terms. Namely,

$\log(1-\frac{s_0-s_{\infty}}{s_0}) = -\frac{s_0-s_{\infty}}{s_0} - \frac{1}{2}(\frac{s_0-s_{\infty}}{s_0})^2+ ...$

Then,

$0 = s_0-s_{\infty}+ \rho(-\frac{s_0-s_{\infty}}{s_0}-\frac{1}{2}(\frac{s_0-s_{\infty}}{s_0})^2)$

$=(s_0-s_{\infty})(1-\frac{\rho}{s_0} - \frac{\rho}{2s_0^2}(s_0-s_{\infty}))$

Solving for $s_0-s_{\infty}$ yields

$s_0-s_{\infty} = 2s_0(\frac{s_0}{\rho}-1)$

$= 2(s_0-\rho+\rho)(\frac{s_0-\rho}{\rho})$

$= 2(s_0-\rho)(\frac{s_0-\rho+\rho}{\rho})$

$= 2(s_0-\rho)(1+\frac{s_0-\rho}{\rho})$

$\overset{(3-14)}{\approx} 2(s_0-\rho)$

Exercise-1 For the Kermack-McKendrick model, show that $\forall t>0$,

1. $s(t)+i(t)+r(t)=1$.
2. $0.

# Restate Feynman’s “Great Identity”

In a letter dated July 7, 1948, Richard Feynman, a Nobel laureate in physics (1965) made a claim:

I am the possessor of a swanky new scheme to do each problem in terms of one with one less energy denominator. It is based on the great identity $\frac{1}{ab} = \int \limits_{0}^{1} \frac{1}{(ax+b(1-x))^2}\;dx$.

(The Beat of a Different Drum: The Life and Science of Richard Feynman, by Jagdish Mehra, page 262)

Assuming non-zero constants $a, b$ are both real but otherwise arbitrary, let’s check the validity of Feynman’s “great identity”.

If $a \ne b$,

$\int \frac{1}{(ax+b(1-x))^2}\;dx$

$= \int \frac{1}{((a-b)x+b)^2}\;dx$

$= \int \frac{1}{a-b}\cdot \frac{a-b}{((a-b)x+b)^2}\;dx$

$= \frac{1}{a-b}\int \frac{((a-b)x+b)'}{((a-b)x+b)^2}\;dx$

$= \frac{1}{a-b}\cdot \frac{-1}{(a-b)x+b}$.

Using Leibniz’s rule,

$\int\limits_{0}^{1}\frac{1}{(ax+b(1-x))^2}\;dx = \frac{1}{a-b}\cdot\frac{-1}{(a-b)x+b}\bigg| _{0}^{1} =\frac{1}{a-b}(\frac{-1}{a}-\frac{-1}{b})=\frac{1}{ab}$.

When $a=b$,

$\int \frac{1}{(ax+b(1-x))^2}\;dx = \int \frac{1}{b^2}\;dx =\frac{1}{b^2}x$

and,

$\int \limits_{0}^{1}\frac{1}{(ax+b(1-x))^2}\;dx = \frac{1}{b^2}x\bigg|_{0}^{1} = \frac{1}{b^2}=\frac{1}{ab}$.

All is as Feynman claimed:

$\int\limits_{0}^{1}\frac{1}{(ax+b(1-x))^2}\;dx = \frac{1}{ab}\quad\quad\quad(\star)$

There is something amiss:

If $a$ and $b$ have opposite signs i.e., $ab<0$ then the right hand side of ($\star$) is negative. But the integrand is squared so the integral on the left hand side of ($\star$) is never negative, no matter what $a$ and $b$ may be.

Let’s figure it out !

In its full glory, Leibniz’s rule we used to obtain $(\star)$ is

If the real-valued function $F$ on an open interval $I$ in $R$ has the continuous derivative $f$ and $a, b \in I$ then $\int\limits_{a}^{b} f(x)\; dx = F(b)-F(a)$.

Essentially, the rule requires the integrand $f$ to be a continuous function on an open interval that contains $a$ and $b$.

Solving $ax+b(1-x)=0$ for $x$ yields

$x = \frac{b}{b-a}$,

the singularity of integrand $\frac{1}{(ax+(1-x))^2}$ in $(\star)$.

For $ab<0$, we consider the following two cases:

Case (1-1) $(a>0, b<0) \implies (a>0, b<0, b-a<0)\implies (\frac{b}{b-a}>0$,

$\frac{b}{b-a} = \frac{b-a+a}{b-a} = 1+\frac{a}{b-a}<1)\implies 0<\frac{b}{b-a}<1$

Case (1-2) $(a<0, b>0) \implies (a<0, b>0, b-a>0) \implies (\frac{b}{b-a}>0$,

$\frac{b}{b-a} = \frac{b-a+a}{b-a} = 1+\frac{a}{b-a}<1) \implies 0<\frac{b}{b-a}<1$

From both cases, we see that

when $ab<0, \frac{1}{(ax+b(1-x))^2}$ has a singularity in $(0, 1) \implies \frac{1}{(ax+b(1-x))^2}$ is not continuous in $(0, 1)$.

Applying Leibniz’s rule to $\int\limits_{0}^{1} \frac{1}{(ax+b(1-x))^2}\;dx$ regardless of integrand’s singularity thus ensured an outcome of absurdity.

However, $ab>0$ paints a different picture.

We have

Case (2-0) $a=b \implies \frac{1}{(ax+b(1-x))^2}=\frac{1}{b^2} \implies$ no singularity

Case (2-1) $(a>b, a>0, b>0) \implies (b-a<0, a>0, b>0) \implies \frac{b}{b-a}<0$

Case (2-2) $(a0, b>0) \implies (b-a>0, a>0, b>0)$

$\implies \frac{b}{b-a} = \frac{b-a+a}{b-a}=1+\frac{a}{b-a}>1$

Case (2-3) $(a>b, a<0, b<0) \implies (b-a<0, a<0, b<0) \implies \frac{b}{b-a}=1+\frac{a}{b-a}>1$

Case (2-4) $(a0, a<0, b<0) \implies \frac{b}{b-a} <0$

All cases show that when $ab>0$, the integrand has no singularity in $(0,1)$.

It means that $\frac{1}{(ax+b(1-x))^2}$ is continuous in $(0, 1)$ and therefore, Leibniz’s rule applies.

So let’s restate Feynman’s “great identity”:

$a\cdot b > 0 \iff \frac{1}{ab} = \int\limits_{0}^{1}\frac{1}{(ax+b(1-x))^2}\;dx$

Exercise-1 Evaluate $\int\limits_{0}^{1}\frac{1}{(ax+b(1-x))^2}\;dx$ using Omega CAS Explorer. For example,

(hint : for $ab>0$, specify $a > b$ or $b>a$)

# Oh! Matryoshka!

Given polynomial $f(x) = a_0 + a_1 x+a_2 x^2 + ... + a_{n-1}x^{n-1}+a_n x^n$, we wish to evaluate integral

$\int \frac{f(x)}{(x-a)^p}\;dx, \quad p \in N^+\quad\quad\quad(1)$

When $p = 1$,

$\int \frac{f(x)}{x-a} \;dx= \int \frac{f(x)-f(a)+f(a)}{x-a}\;dx$

$= \int \frac{f(x)-f(a)}{x-a}\;dx + \int \frac{f(a)}{x-a}\;dx$

$=\int \frac{f(x)-f(a)}{x-a}\;dx + f(a)\cdot \log(x-a)$.

Since

$f(x) = a_0 + a_1x + a_2x^2 + ... + a_{n-1}x^{n-1} + a_n x^n$

and

$f(a) = a_0 + a_1 a + a_2 a^2 + ... + a_{n-1}a^{n-1} + a_n a^n$

It follows that

$f(x)-f(a) = a_1(x-a) + a_2(x^2-a^2) + ... + a_{n-1}(x^{n-1}-a^{n-1}) + a_n (x^n-a^n)$.

That is

$f(x)-f(a) = \sum\limits_{k=1}^{n}a_k(x^k-a^k)$

By the fact (see “Every dog has its day“) that

$x^k-a^k =(x-a)\sum\limits_{i=1}^{k}x^{k-i}a^{i-1}$,

we have

$f(x)-f(a) = \sum\limits_{k=1}^{n}a_k(x-a)\sum\limits_{i=1}^{k}x^{k-i}a^{i-1}=(x-a)\sum\limits_{k=1}^{n}(a_k\sum\limits_{i=1}^{k}x^{k-i}a^{i-1})$

or,

$\frac{f(x)-f(a)}{x-a}= \sum\limits_{k=1}^{n}(a_k\sum\limits_{i=1}^{k}x^{k-i}a^{i-1})\quad\quad\quad(2)$

Hence,

$\int\frac{f(x)}{x-a}\;dx = \int \sum\limits_{k=1}^{n}(a_k \sum\limits_{i=1}^{k}x^{k-i}a^{i-1})\;dx + f(a)\log(x-a)$

$=\sum\limits_{k=1}^{n}(a_k \sum\limits_{i=1}^{k}\int x^{k-i}a^{i-1}\; dx)+ f(a)\log(x-a)$

i.e.,

$\int \frac{f(x)}{x-a} = \sum\limits_{k=1}^{n}(a_k\sum\limits_{i=1}^{k}\frac{x^{k-i+1}}{k-i+1}a^{i-1})+ f(a)\log(x-a)$

Let us now consider the case when $p>1$:

$\int \frac{f(x)}{(x-a)^p}\; dx$

$=\int \frac{f(x)-f(a)+f(a)}{(x-a)^p}\;dx$

$=\int \frac{f(x)-f(a)}{(x-a)^p} + \frac{f(a)}{(x-a)^p}\;dx$

$=\int \frac{f(x)-f(a)}{(x-a)}\cdot\frac{1}{(x-a)^{p-1}} + \frac{f(a)}{(x-a)^p}\;dx$

$= \int \frac{f(x)-f(a)}{x-a}\cdot\frac{1}{(x-a)^{p-1}}\;dx + \int\frac{f(a)}{(x-a)^p}\; dx$

$\overset{(2)}{=}\int \frac{g(x)}{(x-a)^{p-1}}\;dx + \frac{f(a)(x-a)^{1-p}}{1-p}$

where

$g(x) = \frac{f(x)-f(a)}{x-a}=\sum\limits_{k=1}^{n}(a_k\sum\limits_{i=1}^{k}x^{k-i}a^{i-1})$, a polynomial of order $n-1$.

What emerges from the two cases of $p$ is a recursive algorithm for evaluating (1):

Given polynomial $f(x) = \sum\limits_{k=0}^{n} a_k x^k$,

$\int \frac{f(x)}{(x-a)^p} \;dx, \; p \in N^+= \begin{cases}p=1: \sum\limits_{k=1}^{n}(a_k\sum\limits_{i=1}^{k}\frac{x^{k-i+1}}{k-i+1}a^{i-1})+ f(a)\log(x-a) \\p>1: \int \frac{g(x)}{(x-a)^{p-1}}\;dx + \frac{f(a)(x-a)^{1-p}}{1-p}, \\ \quad\quad\quad g(x) = \frac{f(x)-f(a)}{x-a}=\sum\limits_{k=1}^{n}(a_k\sum\limits_{i=1}^{k}x^{k-i}a^{i-1}). \end{cases}$

Exercise-1 Optimize the above recursive algorithm (hint: examine how it handles the case when $f(x)=0$)

# Integration of Trigonometric Expressions

We will introduce an algorithm for obtaining indefinite integrals such as

$\int \frac{(1+\sin(x))}{\sin(x)(1+\cos(x))}\;dx$

or, in general, integral of the form

$\int R(\sin(x), \cos(x))\;dx\quad\quad\quad(1)$

where $R$ is any rational function $R(p, q)$, with $p=\sin(x), q=\cos(x)$.

Let

$t = \tan(\frac{x}{2})\quad\quad(2)$

Solving (2) for $x$, we have

$x = 2\cdot\arctan(t)\quad\quad\quad(3)$

which provides

$\frac{dx}{dt} = \frac{2}{1+t^2}\quad\quad\quad(4)$

and,

$\sin(x) =2\sin(\frac{x}{2})\cos(\frac{x}{2})\overset{\cos^(\frac{x}{2})+\sin^2(\frac{x}{2})=1}{=}\frac{2\sin(\frac{x}{2})\cos(\frac{x}{2})}{\cos^2(\frac{x}{2})+\sin^2(\frac{x}{2})}=\frac{2\frac{\sin(\frac{x}{2})}{\cos(\frac{x}{2})}}{1+\frac{\sin^2(\frac{x}{2})}{\cos^2(\frac{x}{2})}}=\frac{2\tan(\frac{x}{2})}{1+\tan^2(\frac{x}{2})}$

yields

$\sin(x) = \frac{2 t}{1+t^2}\quad\quad\quad(5)$

Similarly,

$\cos(x) = \cos^2(\frac{x}{2})-\sin^2(\frac{x}{2})=\frac{\cos^2(\frac{x}{2})-\sin^2(\frac{x}{2})}{\cos^2(\frac{x}{2})+\sin^2(\frac{x}{2})}=\frac{1+\frac{\sin^2(\frac{x}{2})}{\cos^2(\frac{x}{2})}}{1+\frac{\sin^2(\frac{x}{2})}{\cos^2(\frac{x}{2})}}=\frac{1-\tan^2(\frac{x}{2})}{1+\tan^2(\frac{x}{2})}$

gives

$\cos(x)=\frac{1-t^2}{1+t^2}\quad\quad\quad(6)$

We also have (see “Finding Indefinite Integrals” )

$\int f(x)\;dx \overset{x=\phi(t)}{=} \int f(\phi(t))\cdot\frac{d\phi(t)}{dt}\;dt$.

Hence

$\int R(\cos(x), \sin(x))\;dx \overset{(2), (4), (5), (6)}{=} \int R(\frac{1-t^2}{1+t^2}, \frac{2t}{1+t^2})\cdot\frac{2}{1+t^2}\;dt$,

and (1) is reduced to an integral of rational functions in $t$.

Example-1 Evaluate $\int \csc(x)\;dx$.

Solution: $\csc(x) = \frac{1}{\sin(x)}\implies \int \csc(x)\;dx = \int \frac{1}{\sin(x)}\;dx$

$= \int \frac{1}{\frac{2t}{1+t^2}}\cdot\frac{2}{1+t^2}\;dt=\int\frac{1}{t}\;dt = \log(t) = \log(\tan(\frac{x}{2}))$.

Example-2 Evaluate $\int \sec(x)\;dx$.

Solution: $\sec(x) = \frac{1}{\cos(x)}\implies \int \sec(x)\; dx =\int \frac{1}{\cos(x)}\;dx$

$= \int \frac{1}{\frac{1-t^2}{1+t^2}}\cdot \frac{2}{1+t^2}\; dt=\int \frac{2}{1-t^2}\;dt=\int \frac{2}{(1+t)(1-t)}\;dt=\int \frac{1}{1+t} + \frac{1}{1-t}\;dt$

$=\int \frac{1}{1+t}\;dt - \int \frac{-1}{1-t}\;dt$

$=\log(1+t) -\log(1-t) =\log\frac{1+t}{1-t}=\log(\frac{1+\tan(\frac{x}{2})}{1-\tan(\frac{x}{2})})$.

According to CAS (see Fig. 1),

Fig. 1

However, the two results are equivalent as a CAS-aided verification (see Fig. 2) confirms their difference is a constant (see Corollary 2 in “Sprint to FTC“).

Fig. 2

Exercise-1 According to CAS,

Show that it is equivalent to the result obtained in Example-1

Exercise-2 Try

$\int \frac{1}{\sin(x)+1}\;dx$

$\int \frac{1}{\sin(x)+\cos(x)}\;dx$

$\int \frac{1}{(2+\cos(x))\sin(x)}\;dx$

$\int \frac{1}{5+4\sin(x)}\;dx$

$\int \frac{1}{2\sin(x)-\cos(x)+5}\;dx$

and of course,

$\int \frac{1+\sin(x)}{\sin(x)(1+\cos(x))}\;dx$