We now take up the pair of Volterra-Lotka differential equations for the two species ecology consisting of a predator and a prey.

$\frac{d x}{d t}=-a x+b x y, \quad \frac{d y}{d t}=k y-c x y$

Each solution is a pair of functions $(x(t), y(t))$ and so can be thought of as a parametrized curve $C$ in the plane. We now try to determine the cartesian equation of this curve $C$. Using the chain rule (and the dots denoting the derivative with respect to time $t$ )

$\frac{d x}{d y}=\frac{\dot{x}}{\dot{y}}=\frac{-a x+b x y}{k y-c x y}=\frac{x(b y-a)}{y(k-c x)}$

This is a variable separable equation !

Since $x$ and $y$ denote populations, they are assumed to be positive. Also we assume that by $-a \neq 0$ and $k-c x \neq 0$. Proceeding as usual,

$\frac{k-c x}{x} \frac{d x}{d y}=\frac{b y-a}{y}$

Integrating with respect to $y$ we get

$k \log x-c x+a \log y-b y=E$

where $E$ is the constant of integration. Note that $(1.15)$ is the cartesian equation of the parametrized curve $(x(t), y(t))$ - called a phase curve.

Note that (1.15) is a very interesting equation. Although the individual $x(t)$ and $y(t)$ keep varying with time, the combination

$k \log x(t)-c x(t)+a \log y(t)-b y(t)$

is always constant in time. This should be compared with the familiar situation in physics with conservative mechanical systems namely the law of conservation of energy. Thus equation (1.15) can be termed as the "law of conservation of ecological energy"

It is natural to ask for a picture of the phase curves represented by (1.15). Unfortunately the curve sketching would involve calculus of several variables and so is beyond the scope of the present lecture series. I would only mention that the curves are closed curves in the plane.

The student whose curiosity has been aroused may consult: P. Glendinning: Stability, instability and chaos, Cambridge University Press, 1999, pp 130-132.

On page 3 of the following article in the website you can find a beautiful sketch of the phase curves for the Volterra-Lotka model:

Let us discuss the phase diagram of some systems of differential equations:

$\frac{d x}{d t}=y, \quad \frac{d y}{d t}=-x$

This is really the equation of SHM rewritten as a system. We see that

$\frac{d y}{d x}=\frac{-x}{y}$

and integrating this last equation gives

$x^2+y^2=c$

The phase curves are concentric circles. Note that equation (1.17) carries LESS information than (1.16). Solving (1.16) amounts to finding $x$ and $y$ INDIVIDUALLY as functions of time. Whereas ALL that we get out of (1.17) is a relationship (1.18) between $x(t)$ and $y(t)$.

Consider the pair of differential equations

$\frac{d y}{d t}=2 x y, \quad \frac{d x}{d t}=1+x^2$

Work in the first quadrant $x>0, y>0$ and obtain the first order differential equations for the phase curves

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

Find the solutions of (1.20) and so determine the phase curves for (1.19). Sketch these curves. Is it possible to find the solutions $x(t), y(t)$ of the original system (1.19)?

Again consider the pair of differential equations

$\frac{d x}{d t}=\frac{1}{1+y^2}, \quad \frac{d y}{d t}=\frac{1}{1+x^2}$

Find the phase curves of (1.21) by obtaining a differential equation of the form (1.20) and solving it. Sketch the phase curves. Here one can solve the two equations in (1.21) independently. However, you would only get $x$ and $y$ implicitly as functions of $t$.

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

The solutions are

$\tan ^{-1} x=\tan ^{-1} y+C$

$x=\tan \left(\tan ^{-1} y+C\right)=\frac{y+\tan C}{1-y \tan C}$

Sketch these curves. These are a family of rectangular hyperbolas In the next exercise we shall see a whole family of related equations ALL of them having the SAME phase curves!

Let us consider the system

$\frac{d x}{d t}=y \phi(x, y), \quad \frac{d y}{d t}=-x \phi(x, y)$

where $\phi(x, y)$ is a function of $x$ and $y$. For every $\phi(x, y)$ the system (1.22) has the circles $x^2+y^2=c^2$ as phase curves. In particular, for the choice $\phi(x, y)=1$, the solutions to (1.22) are:

$x(t)=c \sin t, \quad y(t)=c \cos t$

These specific systems (1.16), (1.19) and (1.22) are RARE instances where the pair of differential equations can be solved explicitly for $x(t)$ and $y(t)$ as for instance in the case of SHM

$\frac{d x}{d t}=y, \quad \frac{d y}{d t}=-x$

we have the explicit solution

$x(t)=c \sin t, \quad y(t)=c \cos t$

(and similarly for the system (1.19)).

However in the case of the Volterra-Lotka (reproduced here for convenience)

$\frac{d x}{d t}=-a x+b x y, \frac{d y}{d t}=k y-c x y$

It is NOT possible to find $x(t)$ and $y(t)$ individually and we have to be content with the equation

$k \log x-c x+a \log y-b y=E$

spelling out the law of conservation of ecological energy! But equation (1.25) is adequate for most practical purposes.

Recall that the equation of the pendulum is

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

Let us introduce the (angular) velocity variable $z(t)=\frac{d y}{d t}$ so that $\frac{d z}{d t}=\frac{d^2 y}{d t^2}=-\frac{g}{1} \sin y$ and finally,

$\frac{d y}{d t}=z(t), \quad \frac{d z}{d t}=-\frac{g}{l} \sin y$

for the parametrized curves $(y(t), z(t))$.

Obtain a combination of $z(t)$ and $y(t)$ that is constant in time and interpret it physically. Have you seen the corresponding phase diagram somewhere? If not, can you use your physical intuition to sketch these curves?

Again finding the individual functions $y(t)$ and $z(t)$ is not easy - they involve elliptic functions! All we get easily is a relationship between $y(t)$ and $z(t)$.

In books you often see a differential equation being written as

$M(x, y) d x+N(x, y) d y=0$

What is the meaning of this? Although the expression $M d x+N d y$ can be defined with mathematical precision, we shall not do so. So what do we do with (1.28)? We proceed to clarify this now.

We have seen that differential equations arising naturally in science and technology are pairs (or triples etc., but we shall restrict to pairs) of simultaneous ordinary differential equations such as (1.16), (1.19) or (1.22) and they are all of the form

$\frac{d x}{d t}=N(x, y), \quad \frac{d y}{d t}=-M(x, y)$

with $t$ representing time. For a mechanical system, the pair (1.29) spells out the components $N(x, y)$ and $-M(x, y)$ of the instantaneous velocity of a moving point and the problem of solving the pair differential equations amounts to finding the two functions $x(t)$ and $y(t)$.

Bringing in the opprobrious $M d x+N d y=0$ !

Since in practice it is RARELY possible to obtain from (1.29) the individual functions $x(t)$ and $y(t)$ (e.g Volterra-Lotka), we should settle for a relation between $x(t)$ and $y(t)$. This is given by the differential equation

$\frac{d y}{d x}=\frac{\dot{y}}{\dot{x}}=-\frac{M(x, y)}{N(x, y)}$

However nothing prevents us from proceeding as

$\frac{d x}{d y}=\frac{\dot{x}}{\dot{y}}=-\frac{N(x, y)}{M(x, y)}$

In view of the equal roles played by $x$ and $y$ in the system, it is convenient think of the equation

$M(x, y) d x+N(x, y) d y=0$

as representing either (1.30) or (1.31).

So from a practical point of view interesting first order differential equations (1.28) actually arise out of a study of pairs of equations (1.29) and the purpose of solving (1.28) (that is to say $(1.30)$ or (1.31)) is to determine the phase curves for (1.29).

Stated differently, the pair of differential equations

$\frac{d x}{d t}=N(x, y), \quad \frac{d y}{d t}=-M(x, y)$

is to be regarded as the MORE fundamental object of study and the equation

$M(x, y) d x+N(x, y) d y=0$

as arising out of (1.29). The interest lies in the family of curves represented by $(1.28)$

It should be evident that for a non-vanishing function $\mu(x, y)$, the pair

$\frac{d x}{d t}=\mu(x, y) N(x, y), \quad \frac{d y}{d t}=-\mu(x, y) M(x, y)$

has the SAME phase curves as the pair (1.29). The solutions $x((t), y(t))$ of (1.29) and (1.32) differ in parametrizations.

We have seen many examples of phase curves already and an instance of reparametrization mentioned above.

We have seen that a differential equation

$M(x, y) d x+N(x, y) d y=0$

gives a one parameter family of curves in the plane (we have seen many examples). Conversely, given a one parameter family of curves can we construct a differential equation of the type (1.28)? One parameter families of curves are VERY pretty objects. They appear naturally in Electrostatics as equipotential curves for distribution of charges (you could start with a three dimensional configuration where these would be surfaces in $\mathbb{R}^3$ and slice the picture along the $x-y$ plane to get curves). As a starter, look at the pictures on page 635 in Halliday-Resnick's book (cited earlier and the same edition!).

Let us now look at some interesting examples of such families. Among the simplest examples are the family of concentric circles

$x^2+y^2=c$

Differentiating and clearing the factor of 2 :

$x+y \frac{d y}{d x}=0$

Consider the family of circles touching the $y$-axis at the origin. If the center of the circle is $(c, 0)$ then the equation of the circle is

$x^2+y^2-2 c x=0$

Differentiating (1.34) with respect to $x$ we get (treating $y$ as an implicit function of $x$ )

$x+y \frac{d y}{d x}=c$

Feeding back the value of $c$ from (1.35) into (1.34) and simplifying we get

$\left(x^2-y^2\right) d x+2 x y d y=0$

In the next slide you have an exercise to work out ! But it is not difficult :-)

which is the equation we want! Very simple eh?

Let us now proceed to a not so friendly case :-)

Consider the family of circles touching the $y$-axis at the origin. If the center of the circle is $(c, 0)$ then the equation of the circle is

$x^2+y^2-2 c x=0$

Differentiating (1.34) with respect to $x$ we get (treating $y$ as an implicit function of $x$ )

$x+y \frac{d y}{d x}=c$

Feeding back the value of $c$ from (1.35) into (1.34) and simplifying we get

$\left(x^2-y^2\right) d x+2 x y d y=0$

In the next slide you have an exercise to work out ! But it is not difficult :-)