Up to now if I gave you an equation, and asked you to solve it for x you would be, in general, looking for a value of x which solved the equation. Given:

 

x^2+3x+2=0

 

You can solve this equation to find two values of x.

I could also give you an equation which linked x and y explicitly, and you could find a relationship between the two which, given a value of x would give you a value of y. You’ve been doing this now for many years. Now we’re going to add a hugely powerful tool to our mathematical arsenal. We’re going to allow our equations to include information about gradients of the function…let’s see what this means…

We’re going to take everything that you learnt about integration and turn it into a way to model and understand the world around us. This is a very powerful statement and indeed differential equations are without a doubt the most powerful mathematical tool we have to understand the behaviour of everything from fundamental particles to populations, economies, weather, flow of wealth, heat, fluids, the motion of planets, the life of stars, the flight of an aircraft, the trajectory of a meteor, the way a pendulum swings, the way a ponytail swings (see paper on this here), the way fish move, the way algae grow, the way a neuron fires, the way a fire spreads…and so much more.

So what is a differential equation? It is an equation which contains one or more derivatives of a function.
Let’s look at a very simple example, of population growth. We might want to ask the question, how fast does a population grow? We will take the most naive model we can think of…which will be horribly wrong, but it will give us an idea of how to deal with differential equations and what they can tell us.

We have a population and we want to ask how fast it grows. What might the rate of change of a population be related to? Well, presumably if you already have more members of a population then the population can grow faster. If you only have two members of a population then it can’t grow very fast but if you have 100 members, there will be more babies added to the population more regularly.

Let’s say that a couple will have a baby every 30 years, on average – of course in reality the rate won’t be so smooth, but when we have a large sample, this won’t matter. If we have a single couple then the rate of change of the population will be one every 30 years:

 

\frac{dP(t)}{dt}\approx 1/30

 

the rate of change of the population is 1 every 30 years. You can think of this as the velocity of the population, but rather than being the distance moved over time, it’s the rate of the increase in population over time.

Note that we can choose to use whatever units we want, ie. years, seconds, millennia, as long as we are consistent throughout and use the same unit for every measurement).

If we have 10 couples, then on average there will be one baby every 3 years (Note that we’re making a gross approximation here that we are mixing the number of couples and the number of individuals but to get a general idea of what’s going on, this won’t matter very much – basically factors of 2 are not very important for this discussion).

We’re not yet taking into account the fact that these babies will grow and start having babies of their own. This will be the crucial step in a moment. So it seems that the higher the population, the higher the rate of baby production. It seems reasonable that as we double the population, we’ll double the rate of change of the population so let’s postulate that:

 

\frac{dP(t)}{dt}\propto P(t)

 

That is, the rate of change of the population at time t is proportional to the population that you have at time t. There is a constant of proportionality, and that is the rate of baby production per individual. Let’s call that constant k. It will depend on the type of population, be it people, or rabbits, or bacteria. They clearly have very different reproductive rates. So now we have:

 

\frac{dP(t)}{dt}= kP(t)

 

This is our first differential equation – it’s an equation, and it has a differential in it. It’s called first order because it contains the first derivative. If we had the second derivative anywhere in the equation it would be called a second order differential equation, and so on. The aim is going to be to find a function P(t) that satisfies this equation. We want to find some function P(t) whose derivative is proportional to the function itself.

You’ll see what that is when we’ve found it! Let’s rearrange the equation a little bit:

 

\frac{dP(t)}{P(t)}= kdt

 

This links the way P(t) changes (i.e. dP(t)) with small changes in time (dt). We can now simply put an integral sign on both sides (as long as we do the same thing to both sides, we’re good!). This gives us:

 

\int \frac{dP(t)}{P(t)}= \int kdt

 

But that’s fine, we know how to work with this because we’ve spent an age looking at integrals by now! Let’s just integrate both sides to get:

 

\ln|P(t)|+C_1= kt+C_2

 

we see here that we have two constants of integration, but they are just constants, so let’s add them together to get:

 

\ln|P(t)|= kt+C

 

where C is a new constant, having absorbed C_1 into C_2. Whenever we have two constants of integration like this we can just add them together to get a single constant. Soon you will see exactly WHY we have constants of integration and what they really mean. They are very very important, and related to the fact that there are a whole family of functions which solve our original differential equation. Everything you did before will suddenly become oh-so-very clear!

Ok, let’s do a bit more rearrangement but first we’re going to postulate that the population will always be positive and so we can remove the modulus about \ln|P(t)|. The equation we have now rearranges to:

 

P(t)= e^{kt+C}=e^Ce^{kt}

 

Now e^C is also just a constant, so let’s rename this whole thing in a rather suggestive way. Let’s call this the constant P_0:

 

P(t)= P_0e^{kt}

 

remember, P_0 is just a constant, and k is also a constant which we have to give because it tells us about the average rate of baby production per member of the population. This type of behaviour is called exponential growth, and models many things, from populations of bacteria, to compound interest, to radioactive decay.

What does P_0 represent? Well, let’s put in t=0 into the equation and we get:

 

P(0)= P_0

 

$P_0$ is the size of the population at time t=0. It is the initial condition of our population. Note that we are also free to choose when we start to measure time, but once we decide upon it, we must always measure time relative to that point. That point will be t=0. This is a really profound moment – all of a sudden we see that the integration constant that we’ve been throwing around for the last few months is really related to being able to define an initial condition of a system which solves a differential equation! This will take some time to sink in…

Let’s say that at the start of our experiment the population of bacteria in our dish is 1000, then P_0=1000. Or if we look at the population of people in South Africa, and we choose to start measuring from now, when the population is around 50 million, then P_0=5\times10^7. The value of P_0 depends on the system that you’re looking at.

Note that there is something really really important that we’ve missed out. We are not taking into account that there is another way for the population to change than simply new people being born/bacteria being produced. That is death. This is a fairly important part of the way a population works, but for now we are leaving it out. We will see that because we have left it out, we will get some pretty crazy results, but we will at least get an idea of what a differential equation can give us.

Let’s work our way through an example.

Let’s take rabbits. If you look it up online, you will find that for rabbits k (the rate of production of rabbits per rabbit) is somewhere around 13 per year. So we have:

 

P(t)=P_0e^{13t}

 

Let’s say we start with 2 rabbits. What happens after a year? Well, according to our calculations this will give us:

 

P(1 year)=2e^{13\times 1}=884827

 

That’s already a lot of rabbits, but let’s see what happens after 10 years…

 

P(10 years)=2e^{13\times 10}\approx 5.7\times 10^{56}

 

damn, that’s a lot of rabbits!!! How much rabbit is that?

Well, let’s say that a rabbit is about 2 kilos. This is 10^{57} kilos of rabbit. The sun weighs around 10^{30} kilos, so this is around 10^{27} suns, which is more than there are stars in all the galaxies in the observable universe. After 10 years we already have a universe of rabbits. Let’s keep going with this calculation, just because it’s fun.

Let’s say that our population of rabbits which is increasing rapidly in size is in the form of a giant ball of rabbits, all reproducing and making more rabbits (it’d be strange if they started making lemurs).

Let’s also say that you can fit roughly 100 rabbits into a cubic meter. This is probably an overestimate in normal conditions, but in the centre of the sphere of rabbits the pressure will be pretty high, so this is probably a vast underestimate. If that’s the case then the volume of the population is

 

\frac{P(t)}{100} m^3.

 

What’s the radius of this sphere of rabbits?

 

V_{sphere}=\frac{4}{3}\pi r^3

 

so:

 

r=({\frac{3V_{sphere}}{4\pi}})^{\frac{1}{3}}

 

so the radius is:

 

r_{\text{sphere of rabbits}}(t)=\left({\frac{3\frac{P(t)}{100}}{4\pi}}\right)^{\frac{1}{3}}

 

let’s plug in our known rabbit population growth and we have that the radius at time t is given by:

 

r_{\text{sphere of rabbits}}(t)=\left({\frac{3\frac{2e^{13t}}{100}}{4\pi}}\right)^{\frac{1}{3}}=\left(\frac{3\times 2}{100 \times 4\pi}\right)^\frac{1}{3}e^{\frac{13t}{3}}

 

assuming that we start with 2 rabbits at time t=0.

What’s the rate of change of the radius over time? Well, we can think of this as the velocity of the surface of the sphere:

 

velocity_{\text{surface of sphere of rabbits}}=\frac{dr(t)}{dt}=\left(\frac{3\times 2}{100 \times 4\pi}\right)^\frac{1}{3}\frac{13}{3}e^{\frac{13t}{3}}

 

The speed of light is roughly 9.5\times 10^{15} meters per year. So we can ask when the surface of our shell of rabbits starts expanding faster than the speed of light:

 

9.5\times 10^{15}=\frac{dr(t)}{dt}=\left(\frac{3\times 2}{100 \times 4\pi}\right)^\frac{1}{3}\frac{13}{3}e^{\frac{13t}{3}}

 

Solving this we get that this happens at roughly t=8.5 years.

If we took an idealised rabbit population that reproduced in proportion to its population, then after 8.5 years this ball of rabbits would be expanding faster than the speed of light. In fact we’ve not taken a few crucial things into account, including the fact that the pressure of the inner part of the sphere of rabbits would start nuclear fusion and our sphere would become star, which might make reproduction hard, and even if it didn’t, the sphere would have become a black hole far before its outer limb accelerated past the speed of light.

Phew! ok, that was all very silly but quite fun and we have learnt several things:

  • We have solved our first differential equation
  • The assumptions that went into our initial differential equation probably weren’t very realistic.
  • If you ever find perfectly idealised rabbits, don’t let them breed.
  • Never let an applied mathematician with an overactive imagination set your tut questions.

What we did learn that is very important is that starting off with knowledge of the rate of change of a quantity, we were able to find out how it increased over time.

 

How clear is this post?