Mathematical Modelling - What is it?

People study mathematics for many reasons. There are those, like Dr Gavin Hitchcock, the editor of Zimaths, who believe that mathematics is worth studying just for its own sake. Generally, this is what most pure mathematicians believe - that there is a certain beauty about mathematics, enough reason to spend hours, days, or a life-time, doing mathematics just for the sake of maths.

My own ideas about mathematics closely resemble those of Richard Haberman, who is, in my opinion, one of the greatest mathematicians of this century. For me, the primary reason for studying mathematics lies in its applications. We study mathematics in order to apply it. Good mathematics is always applicable, even if we haven't yet found how it can be applied! Mathematics has been applied to virtually all facets of life.

The process of applying mathematics to a real life situation is often referred to as mathematical modelling. A model in this sense is a simplified, mathematical concept that represents a real life situation. Mathematical models are developed to help in the understanding of physical phenomena. Formulating a model usually involves making observations, collecting data or carrying out an experiment. The observations made, or the results of our experiment, are then put in a mathematical form, together with some assumptions. Often this means representing our observations or experimental results by an equation or set of equations involving some derivatives of one or more unknown functions. Such equations are called differential equations, and these constitute our mathematical model. However, before we can say anything to our friends about our model, we need to validate the model. To do this usually means solving the differential equations (it may sometimes be necessary to invent new mathematics to do this), then interpreting the solution and testing the predictions of the mathematical model against real-life situations. In most cases, it is then necessary to modify the initial model, possibly by removing some of the assumptions made in the formulation of the initial model, or applying different mathematical techniques to solve the set of differential equations that constitute the model.

Mathematical modelling is widely used in industry, commerce and government. Perhaps what most people might relate to, and what comes readily to mind when we speak of mathematical modelling, is the process of predicting the weather in meteorology. In an over-simplified way, this consists of collecting data on ocean currents, wind directions and speeds, cloud formations, etc., and putting together a mathematical model to predict future weather patterns.

Some areas in which mathematical modelling has been successfully used are: population dynamics (this includes predictions of population growths and the spread of epidemics, such as the AIDS virus), financial mathematics (for example, trading in stocks and derivatives), modelling traffic flow, modelling conflicts (such as political unrests and wars), etc. (See page 17, where the challenge is to model the degeneration of roads by potholes.)

Obviously, mathematical modelling is a vast subject. Looking at all the areas in which it has been applied might require more than the 32 volumes of the Encyclopedia Brittanica! We will look at two examples of mathematical models in our attempt to introduce some of the fundamental concepts to the reader. The first we will give now, and the second (modelling conflict situations) we will give in the next issue.

Example: Free fall

Let us consider the case of an object dropped from a certain height above the ground and falling under gravity. Newton's second law states that the acceleration of a particle times its mass equals the total force acting on it. And Newton's Law of Gravitation states that the force exerted on a particle of mass m by another particle of mass M is proportional (by a universal gravitational constant g) to the product of the masses divided by the square of the distance between them. These two mathematical equations together constitute a mathematical model of the way matter and space are related - and allow us to explain and predict, to a high degree of accuracy, the behaviour of both terrestrial projectiles and celestial planets. This grand and yet stunningly simple synthesis was arrived at by Isaac Newton in the seventeenth century, ``standing on the shoulders'' of others before him (like Kepler and Galileo), and based upon centuries of careful observation and measurement. How can we use these two laws to arrive at a reasonable mathematical model of our falling object?

A non-trivial piece of integration tells us that the attractive force exerted on a particle by a uniformly dense spherical object is the same as would be exerted on it were the whole mass of the sphere concentrated at its centre. Therefore, we assume that both our falling object and the earth are uniformly dense spheres of masses m and M, with the former small enough to behave like a particle, and R the radius of the earth. Now, let us take the origin O of our axis at the point where the object is falling from. Let z(t) be the distance from the origin to the centre of the object at time t, so that the acceleration (measured downward) of the object is [(d²z)/(dt²)]. Then the two laws lead, on our assumptions above, to the two model equations:

m d2z dt2 = F,  and  F = g mM (R+h-z)2

(1)

yielding the single differential equation

m d2z dt2 = g mM (R+h-z)2

(2)

In order to simplify the model still more we make one more assumption: that the height h is very small in comparison with the radius R of the earth, so that R+h-z » R; and so we obtain, finally, the differential equation we shall use to model the behaviour of our falling object:

m d2z dt2 = mg

(3)

where g = g[mM/(R²)] is called the acceleration of gravity. It is usually taken to be a constant, although in fact it varies from place to place on the Earth's surface.

Having obtained the differential equation we now need to solve it. Dividing by m and integrating with respect to time t leads to an equation for the velocity v(t) of the object at time t:

v = dz dt = gt+c1.

(4)

You may have met this equation in the form v = u+at. A further integration then gives an equation for the height of the object at time t:

z = 1 2 gt2+c1t+c2.

(5)

Clearly, the constants of integration c₁ and c₂ are the initial velocity and height of the object. You may have met this equation in the form s = ut+[1/2]at². By taking the origin of our axis at the point where the object is dropped from, this means that c₂ is zero.

Notice that you can use equation 5 to measure g in your classroom. All you need is a stop watch and a small object which you can drop from the top of a desk. Drop this object from different heights z and measure the time in seconds it takes to hit the floor. In each case the initial speed is zero and if you plot a graph of z against t², the slope should give you g/2. Compare your value of g with that from the expression g = (9.80616-0.025928cos2b+0.000069cos²2b-0.000003h) ms^-2, where b is the latitude and h is the height above sea-level measured in metres. What are the values of b and h in Harare, say? If you get an accurate value for g using this model, then it means that the model must be pretty good. This in general is how you validate a model, by comparing its predictions with reality.

However, thinking about the model just developed above, we realize that there is another factor which is not `true to life'. We have assumed that the only force acting on the object is that due to gravity. Surely, an object moving through air (or any other medium for that matter) must be subject to medium resistance - in this case, air resistance? Unfortunately, there is no formula that exactly determines the air resistance acting on the object. We suspect that the air resistance may depend on the air density, the velocity of the object and perhaps even the shape of the object! At this stage in order to make progress with our model we need to make another assumption. Let us assume that air resistance is proportional to the velocity (which means to the number of air particles the object bumps into each second). The model now becomes:

m d2z dt2 = mg-kv,

(6)

where k is a positive constant that depends on air density and body shape. Why have we put a negative sign? Because air resistance opposes the motion of the object, so the force is upward, in the opposite direction to the gravitational force and the direction in which we have measured z, [dz/dt], [(d²z)/(dt²)].

The new model equation can now be solved using a technique called ßeparation of variables" by first re-writing the equation in the form

dv mg-kv = dt m ,

(7)

and integrating once to obtain

- 1 k ln(mg-kv) = t m +c,

where c is a constant of integration. Let us multiply the last equation by -k to obtain

ln(mg-kv) = - kt m -kc.

Now taking exponentials of both sides and keeping in mind that for any arbitrary numbers x and y, e^lnx = x and e^x+y = e^xe^y, we obtain the general solution

v = mg k - 1 k e-kce-kt/m.

(8)

Let me leave it as an small exercise to show that if we now assume that at time t = 0, the speed of the particle is v₀, equation 8 gives the formula for the velocity at time t as

v = mg k +(v0- mg k )e-kt/m.

(9)

Notice from equation 9 that when an object has been falling for a long time (perhaps it is dropped from a high cliff and takes a long time to reach the ground!), the exponential term e^-kt/m becomes small, in fact, when t® ¥ then e^-kt/m® 0. In this case v = mg/k. In other words when an object has been in free fall fall for a long time, its velocity is approximately its weight mg divided by k. This is a constant. This constant mg/k is called the limiting or terminal velocity of the object.

So, what does our model tell us? It predicts that when dropped from the same height a heavier object would fall faster than a lighter one (since the velocity is directly proportional to the weight, mg). In addition, if somehow we were to reduce air resistance (which is equivalent to making k small), the model predicts that the object would fall faster. These predictions are broadly in line with our everyday experiences. This confirms the validity of our model.

- Dr Precious Sibanda. [Next Issue: Example 2 on modelling conflict]