Section8.13How Euler Did It: Harmonic Oscillators, and Complex Numbers
After exponential quantities the circular functions, sine and cosine, should be considered because they arise when imaginary quantities are involved in the exponential.
was one of the all-time great mathematicians. In any list of stellar mathematicians, he would stand out among the best. Indeed, Pierre-Simon Laplace 20
(1749–1827), an outstanding mathematician in his own right, was once asked by a novice how best to learn Calculus. Laplace told him “Read Euler, read Euler, he is the master of us all.” The title of this section was borrowed from the book of the same name 21
https://bookstore.ams.org/spec-52/
which was itself compiled from a series of articles, also of the same name, written by the late Prof. Ed Sandifer, of Western Connecticut State University.
In Section 6.2.5 we saw that one model for the motion of an object having mass \(m\text{,}\) bouncing without resistance on a spring is the simple harmonic oscillator (SHO) equation:
Equation (8.34) is unrealisitic because the oscillations described never vary. In the real world, as a result of friction, the spring’s oscillations will get shorter and shorter, eventually stopping altogether. How could we modify this model in order to capture the effects of, say, air resistance?
A relatively simple way to model air resistance is to assume that the resistive force is proportional to the speed of the object. This sounds imposing but you are quite familiar with this sort of resistance. If you put your hand out the window of a moving car you’ll notice that as the car moves faster there is more force (wind) pushing back on your hand. This simple model breaks down at high speeds where other factors such as turbulence will also affect the resistance, but it can be shown experimentally that the assumption that the resistive force is proportional to velocity works well for our spring.
Putting the words “resistance is proportional to velocity” into symbols means we need to add the term \(-b\dfdx{y}{t}\) to equation (8.33) where \(b\gt0\) is a constant which depends on the viscosity of the ambient medium (air, in this case). (Why must this proportionality be negative?)
Thus our model for harmonic oscillation with resistance becomes
satisfies equation (8.36) for any constants \(A\) and \(B\text{.}\)
(b)
Let \(A=1, B=0\) and plot this solution for \(0\le t\le 15\) and \(-0.5\le y \le 0.5.\) Does this seem to model a damped oscillation?
It is one thing to confirm a given solution as in Problem 8.13.2, but quite another to find the solution in the first place. A natural question to ask is “How was this problem solved the first time?”
Leonhard Euler (1707–1783) solved equation (8.36) by guessing that for some carefully chosen value of \(c\)
\begin{equation*}
y(t)=e^{ct}
\end{equation*}
would be a solution. In view of the results of Problem 8.13.2 this guess looks completely crazy. Where is the sine? Where is the cosine? Nothing seems to be oscillating!
It is difficult to imagine what might have led him to make such a guess. But Euler knew that if his guess didn’t pan out he didn’t have to tell anyone about it. He could just toss all of his notes into his fireplace and pretend that he’d never made his crazy–seeming guess.
The really crazy thing though, is that Euler’s guess worked. Here’s how.
The key is the constant \(c\text{.}\) By leaving it unspecified Euler gave himself a little “wiggle room.” All he had to do now was find a value of \(c\) that would work.
Problem8.13.3.
Show that if \(y=e^{ct}\) satisfies the differential equation
Put the first two derivatives of \(y\) into the differential equation.
Solving equation (8.37) we see that \(c=\frac{-1}{2}\pm\frac{\sqrt{-3}}{2},\) so we conclude that if \(c=-\frac12+\frac{\sqrt{-3}}{2}\) or \(c=-\frac12-\frac{\sqrt{-3}}{2}\) then \(y=e^{ct}\) satisfies equation (8.36).
The appearance of \(\sqrt{-3}\) is troubling since there is no real number which is the square root of a negative number. But Euler knew that such “imaginary” numbers had been effectively used in the previous century, although no one could yet explain what they really were. Knowing that if nothing useful came of his efforts he could discard them and try something else, Euler decided to treat the square roots of negative numbers, like differentials, as “convenient fictions”
Since he had assumed that \(c\) was a constant Euler wanted to see where that assumption would lead. In particular he decided to treat \(\sqrt{-3}\) as the constant it is. To make things “easier on the eyes,” Euler made the substitution \(\sqrt{-1}=i\) so that
Show that \(y=e^{ct}\) is a solution of the differential equation (8.40) when \(c=i\) or \(c=-i\text{.}\)
Drill 8.13.5 shows that \(y=e^{it}\) and \(y=e^{-it}\) are both solutions of equation (8.40). But Euler knew (as we do, see Problem 6.2.21) that another solution is \(y(t)=a\cos(t)+b\sin(t)\) where \(a\) and \(b\) are constants.
Since he now had very different looking solutions of equation (8.40). Euler guessed that somehow these must be the same, so he set
Euler’s Identity is foundational to electrical engineering, and physics (especially Quantum Mechanics). For mathematicians it is the key to understanding the geometry of the Complex Numbers (numbers of the form \(a+bi\text{,}\) where \(a\) and \(b\) are real numbers and \(i^2=-1\text{.}\)
For Euler, it was a way to translate trigonometric functions to exponential (and hyperbolic trigonometric) functions and back. For us, it is the key to our dampened oscillation problem.
Before we get to that, notice the following curious consequences of Euler’s Identity.
Drill8.13.7.
Show that \(e^{i\pi}+1=0\text{.}\)
We call this curious because it relates the numbers \(0,
1, i, e\) , and \(\pi\) all in one simple formula. If you want to learn more about the complex numbers you’ll need to become a mathematics major and take a course in Complex Analysis. It’s fun. You should do that.
Another curious consequence of Euler’s Identity is that the trigonometric functions and the hyperbolic trigonometric functions are related via the constant \(i=\sqrt{-1}\text{.}\)
Problem8.13.8.
(a)
Show that \(\cosh(it)=\frac{e^{it}+e^{-it}}{2}=\cos(t)\)
(b)
Show that \(\frac{1}{i}\sinh(it)=\frac{e^{it}-e^{-it}}{2i}=\sin(t)\text{.}\)
