Newton and the early triumphs of calculus

Equipped with his calculus, Issac Newton, with his laws of dynamics was able to explain:

The motion of planets.

The precession of equinoxes.

The formation of tides.

Astronomy hitherto an empirical science transformed into a dynamical science. A Monumental Achievement ! In essence Newton had dealt with success the differential equations governing the motion of two celestial bodies deriving the Kepler's laws of planetary motion.

The very long list has been truncated to cover only those parts that are relevant to the course.

Loan I Vrabie: Differential equation: An introduction to basic concepts, results and applications, 2nd Edition, World Scientific, 2011

Origin and scope of differential equations

Laws of physics when expressed in mathematical terms give rise to differential equations !

The law of mass action in chemistry - Differential equations of chemical kinetics and enzyme kinetics.

Models arising in Ecology and demography. Striking similiarities with equations modelling chemical kinetics.

Mathematical Biology: Spread of diseases, Models of heart-beat, Mathematical Physiology...

(i) E. C. Zeeman: Modelling of heart beat

(ii) Hodgkin and Huxley: Nerve impulses (1952) Received Nobel Prize for their work.

Problems arising in geometry (applications within the field of mathematics)

$m l^2 \frac{d^2 y}{d t^2}=-m \lg \sin (y(t))$

which simplifies to the differential equation

$\frac{d^2 y}{d t^2}+\frac{g}{l} \sin (y)=0$

We have derived the differential equation governing the motion of a simple pendulum. This is a second order differential equation.

A particle moving along a straight line executes simple harmonic oscillations. The force acting on it is proportional to the displacement and in the opposite direction. Thus if $y(t)$ is the displacement form the origin at time $t$ and $m$ is the mass of the particle then Newton's second law of motion gives

m $\frac{d^2 y}{d t^2}=-k y$

where $k$ is a positive constant (namely the constant of proportionality).

Denoting $k / m$ by $\omega^2$ we get the differential equation for harmonic oscillations as:

$\frac{d^2 y}{d t^2}+\omega^2 y=0$

$\quad$ (1.2)

which is again a second order differential equation

Let us take up the differential equation of SHM:

$\frac{d^2 y}{d t^2}+\omega^2 y=0$ $\quad$ (1.2), $\frac{d^2y} {dt^2} + \frac{g} {l} \sin y$

$y(t)=A \cos \omega t+B \sin \omega t$ $\quad$ (1.3)

also satisfies the equation (1.2). Thus we have enlisted many solutions of (1.2).

Question: Have we enlisted all the solutions? How do we know that if $z(t)$ is a solution of (1.2) then $z(t)=A \cos \omega t+B \sin \omega t$ for certain constants $A$ and $B$ ?We see that we are led to the problem of describing the class of ALL the solutions.

Analogue of equation (1.2) also appears in the theory of electric circuits the LC circuits. For a discussion on this see p. 845 (equation 38.5) in D. Halliday and R. Resnick, Physics, Vol 2, Third Edition, John Wiley and Sons, 1978. On p. 848 of this important book one finds the following differential equation for an LCR circuit:

$\frac{d^2 q}{d t^2}+\frac{R}{L} \frac{d q}{d t}+\frac{q}{L C}=0$

Malthus (in 1798 but earlier and independently L. Euler) had proposed the following model for population growth in an ecology consisting of a single species. If $y(t)$ is the population at time $t$ then the rate of change of population is proportional to $y(t)$ which means

$\frac{d y}{d t}=k y$ $\quad$ (1.4)

where $k$ is the constant of proportionality. Observe that (1.4) implies $y(t)=A \exp (k t)$ showing that the population $y(t)$ must grow exponentially fast !

Is this practical?

Would nature allow such a exponential growth of population??

Would there not be some inhibiting factors caused by limitations of natural resources preventing this exponential growth?

Verhulst (1836) proposed a modification of the Malthusian/Eulerian model taking into account the limitations of natural resources. Paucity of resources would produce social friction which would check the rate of growth of population by introducting the term $-k y^2 / R$ in the differential equation:

$\text { of } \frac{d y}{d t}=k y\left(1-\frac{y}{R}\right)$ $\quad$(1.5)

Let us consider a two species ecology consisting of a predator with population $x(t)$ and a prey with population $y(t)$. For the predators, the source of food is the availability of prey and the latter are herbivorous living on say algae.

In the absence of prey $y(t)$ the population of the predators $x(t)$ would decrease exponentially:

$\frac{d x}{d t}=-a x, \quad a>0$ $\quad$(1.6)

whereas the population of prey would increase:

$\frac{d y}{d t}=k y, \quad k>0$ $\quad$(1.7)

However when the two are put together, the rate of decrease indicated in the RHS of (1.6) would have to be modified as

$-a x+b x y, \quad b>0 .$

and the rate of increase indicated on the right hand side of (1.7) would be checked by the addition of a term $-cxy$ namely ky $- cxy$.

Thus we get the pair of differential equations:

$\frac{d x}{d t}=-a x+b x y, \quad \frac{d y}{d t}=ky-cxy$ $\quad$(1.8) .

where the constants $a, b, c$ and $k$ are all positive.

This model is known as the Volterra-Lotka system of equations.

It is a system of simultaneous differential equations and is a coupled

systems since the unknown $y$ appears in the $x$ equation and vice-versa. We shall not undertake a systematic analysis of this system.

This was mainly as an illustration of how differential equations arise in the study of Ecology

From now on we shall only deal with differential equations of first order. that is involving only one derivative with respect to time. The differential equations of the pendulum, SHM and LCR circuits involve two time derivatives and so are second order equations. The general first order equations that we shall deal with are of the form

$\frac{d y}{d x}=f(x, y)$ $\quad$(1.9)

where $f(x, y)$ is a function of two variables. We shall look at three types of equations

Variable separable equations

Homogeneous equations

Linear equations and its close cousin namely the Bernoulli equation.

This is a case when the function $f(x, y)$ appearing in (1.9) is a product of $g(x) h(y)$ in which case we rewrite the equation in the form

$\frac{1}{h(y)} \frac{d y}{d x}=g(x) .\quad (1.10)$

This is permissible as long as $h(y)$ is different from zero. Infact we shall assume that neither $h(y)$ nor $g(x)$ is zero. Such cases as always would have to be dealt with separately working on intervals where they are non-zero to begin with. Integrating both sides of (1.10) with respect to $x$ we get

$\int \frac{1}{h(y)}\left(\frac{d y}{d x}\right) d x=\int g(x) d x$

Disclaimer: We shall assume that the functions $g(x)$ and $h(y)$ are continuous on the interval in question.

$\frac{1}{h(y)} \frac{d y}{d x}=g(x), \quad h(y) \neq 0$

Integrate both sides with respect to $x$ :

$\int \frac{1}{h(y)} \frac{d y}{d x} d x=\int g(x) d x$

Let us look at the innocent differential equation

$\frac{d y}{d t}=y^2$

and prescribe initial condition $y(0)=c$ where $c>0$. Then proceeding as above,

$\int \frac{d y}{y^2}=\int d t$

Integrate (and remember the const. of integration !) $y(t)=-1 /(t+b)$. Put $t=0$ and the corresponding $y=c$. We get $b=-1 / c$ and so the solution is:

$y(t)=\frac{c}{1-c t}$

As $t \longrightarrow 1 / c$ from the left, the solution $y(t)$ escapes to infinity !

$\frac{d y}{d t}=y^2, \quad y(0)=c$

Divide by $y^2$ :

$\frac{1}{y^2} \frac{d y}{d t}=1$

Integrate with respect to $t$ :

$\int \frac{1}{y^2} \frac{d y}{d t} d t=\int d t$

Simplifying the LHS.

$\int \frac{d y}{y^2}=\int d t$

$\int \frac{d y}{y^2}=\int d t$

$-\frac{1}{y}=t+b,$

where, $b$ is the constant of integration.

$y(t)=\frac{-1}{t+b}$

Put in the initial conditions: When $t=0, y(0)=c$. We get

$c=\frac{-1}{b}, \quad \text { or } b=\frac{-1}{c}$

Feedback into the previous equation:

$y(t)=\frac{-1}{t+b}=\frac{-1}{t-\frac{1}{c}}$

Solution of the differential equation is

$y(t)=\frac{c}{1-c t}$

Notice what happens when $t \rightarrow \frac{1}{c}-$ ?

The RHS $c /(1-c t)$ blows up to infinity!

Let us look at the innocent differential equation

$\frac{d y}{d t}=y^2$

and prescribe initial condition $y(0)=c$ where $c>0$. Then proceeding as above,

$\int \frac{d y}{y^2}=\int d t$

Integrate (and remember the const. of integration !) $y(t)=-1 /(t+b)$. Put $t=0$ and the corresponding $y=c$. We get $b=-1 / c$ and so the solution is:

$y(t)=\frac{c}{1-c t}$