So far, we’ve tried to give some fresh insights into things that are often covered in schools. Now we start to properly bridge the gap between school and university science…
Why are ‘differential equations’ the next place to go? Well, because they are EVERYWHERE in science and engineering (and a bunch of other places too, like economics…). So let’s answer the question in this section title and explain what they are by introducing some in this video…
So, to summarise the video, a differential equation is an equation that contains a ‘rate of change’ (i.e. a derivative). Since we spent the whole of Section 1 explaining that rates of change are really important, you shouldn’t be that surprised that this kind of equation keeps cropping up…
Differential equations often arise in situations involving feedback. A (by now familiar) example is given in the video below.
We used a ‘model’ for the air resistance here, in which we assumed that it is proportional to the square of the velocity, so that [latex]D=kv^2[/latex]. This assumption is qualitatively reasonable – as the object falls, it has to push ‘air molecules’ (apologies to chemists for using this lazy term!) out of the way. The faster the object, the greater the rate of collisions with molecules, and so the greater the air resistance.
It is important to remember when you have used a ‘model’, and the conditions for which it is appropriate. If you start to consider conditions for which your model is unsuitable, you may need to find a new model! In this case, if the falling object falls very slowly, then a linear assumption for the air resistance ([latex]D=kv[/latex] rather than [latex]D=kv^2[/latex]) might be more appropriate. This might be the case if:
– The object is falling through a viscous liquid, like treacle, rather than through air
– The object has a density similar to the medium through which it is falling (and in this case, a further change to the underlying equation would be needed, because it would no longer be reasonable to ignore the buoyancy force)
So now you can recognise a differential equation, but it doesn’t mean you can solve one (to be fair, there aren’t that many we can solve either…). So let’s start solving them by asking:
“What does it even mean to ‘solve’ a differential equation?”
A simple differential equation involves a rate of change of one quantity (call it [latex]A[/latex]) with respect to another ([latex]B[/latex]). Solving the differential equation means finding a relationship between [latex]A[/latex] and [latex]B[/latex], from the information about the rate of change.
There are a limited number of ways differential equations can be more complicated than this. Two important ones are:
– There can be second, and higher-order derivatives, such as [latex]\frac{d^2A}{dB^2}[/latex]. If the highest order derivative present is [latex]\frac{d^nA}{dB^n}[/latex], then it is an ‘nth-order’ differential equation (which only matters because it will affect which text book you go and borrow to solve it!)
– [latex]A[/latex] can be a function of more than one variable, e.g. [latex]A=A(B,C)[/latex], and the differential equation can have derivatives with respect to [latex]B[/latex] and to [latex]C[/latex]. These would be partial derivatives (see Section 2) and the equation would be called a ‘partial differential equation’. A differential equation that is not a partial one is called an ‘ordinary differential equation’. The examples we solve in this section will all be ‘ordinary’ in this sense…
We’ll show you in two stages what we mean about solving differential equations:
- In the first stage below, we’ll revisit the constant acceleration motion from Section 1. This is motion that is so simple (scientifically speaking, not in terms of intellectual effort necessarily), that you don’t even need to use differential equations to solve it. But we will anyway, and this will help make the links between uniform and non-uniform motion
- In the second, we’ll revisit the object falling through air, but we’ll treat it numerically (that means rather than finding a general relationship between [latex]A[/latex] and [latex]B[/latex] that holds for all values of [latex]B[/latex], we find a value of [latex]A[/latex] for the particular value of [latex]B[/latex] that we are interested in)
Solving differential equations in a fairly trivial case
Let’s look at the example of the motion from Section 1, and calculate a bunch of stuff that we have already calculated…
There’s one other place where we’ll put you through the pain of watching us solve a differential equation analytically, and that’s when we discuss all the science surrounding the ‘harmonic oscillator’ in sections 14 and 15 .
Solving differential equations in a slightly less trivial case, by a numerical method
Here, once again, is the motion of an object through air, described by the differential equation
[latex]\displaystyle mg-kv^2=m \frac{dv}{dt}[/latex]
So what does it mean when we say we are “solving a differential equation” in this instance? Well, ideally we would find an equation that links [latex]v[/latex] to [latex]t[/latex] without reference to a derivative. Let’s pretend that that’s too hard to do for this equation, so that we are forced to use a ‘numerical method’.
We are not going to solve this equation here analytically. This site, and this page in particular, does not aim to be a course in how to solve differential equations; there are enormous text books out there whose sole purpose is to help you solve all the differential equations your imagination can deal with. Instead, we are aiming to provide some insight into what these equations are and what they mean, so that when struggling to solve them at some point in the future, you remember what you are trying to do rather than getting bogged down in the techniques. If you do want to solve this equation analytically, you can find an undergraduate mathematics or ‘mathematics for science students’ textbook, or use an ‘online differential equation solver’.
So what can we do if the differential equation is too hard? What would this ‘numerical method’ consist of?
Well, we can split the motion into a series of tiny steps (time intervals, each of duration [latex]\Delta t[/latex]), and assume the acceleration, [latex]a[/latex], is constant in each one. Then, the velocity [latex]v(t)[/latex] will become a new velocity [latex]v(t+\Delta t)[/latex] according to:
[latex]\displaystyle v(t+\Delta t)=v(t)+a\Delta t[/latex]
If you need to, take a minute to make sure you understand what this bit of maths is telling you – it’s a really important concept. It means:
“The velocity a bit later is equal to the velocity now, plus an extra bit of velocity due to the acceleration.”
(And we assume that acceleration is constant, which may require the “bit later” to be only a tiny bit later for that approximation to work…)
We can use our new value of [latex]v[/latex] to calculate a new value for [latex]a[/latex], which we can use to calculate the next value for [latex]v[/latex] and so on, and so on. The method will break down if our time intervals are too big, because then the approximation that [latex]a[/latex] is constant across the time interval will not be valid. But if the acceleration is approximately constant over each [latex]\Delta t[/latex], then it should work.
Richard Feynman explains this approach brilliantly in the Feynman Lecture in Physics Volume 1, chapter 9
Now, this approach is not one to attempt with a calculator and a pencil. Many iterations are needed to get the behaviour of [latex]v[/latex] over sensible timescales (because [latex]\Delta t[/latex] is in general quite small…). And ‘many iterations’ is the kind of situation that computers were invented for… So we’ll show you how it can be done with a spreadsheet in the next video:
The animation below allows you to change the mass, gravitational field strength and drag coefficient and see the effect…
INSERT INTERACTIVE ELEMENT
So where should you go next? Well, there are loads of places…
Link to Section 5 – What exponential means (and what it doesn’t)
It would make sense to go here, because it is a very important and common class of scientific situation. It will introduce the number ‘e’, and we’ll finally solve the falling ball problem. We promise not to mention the falling ball again afterwards…
Link to Home Page
And you can always return here…