Part 1 · Oscillators and waves

1.5 The wave equation

Source: Postorino, week 1 slides, pp. 21–23, and week 2 slides, p. 1

Objectives

Picture. A disturbance passed from mass to mass along a chain of springs travels without changing shape: longitudinal or transverse.
Theorem. Every solution of the wave equation is $f (x - v t) + g (x + v t)$ ; $f$ slides towards $+ x$ at speed $v$ , $g$ towards $- x$ .
Recipe. Prove a function of $x \mp v t$ solves the wave equation with the chain rule, via $h = x \mp v t$ .
Vocabulary. Wave equation (d’Alembert’s), progressive and regressive waves, longitudinal and transverse.

A chain of oscillators

Take a chain of masses $1, 2, \dots, i - 1, i, i + 1, \dots, N$ joined by springs, running from $- \infty$ to $+ \infty$ along $x$ . Move one mass a little from equilibrium and let go. Its springs push and pull its neighbours, which push and pull theirs. The disturbance (perturbazione) travels along the chain towards both increasing and decreasing $x$ .

Call $Y (x, t)$ the displacement from equilibrium of the mass at position $x$ at time $t$ . The disturbance can be:

longitudinal: the masses move along the direction of propagation (compressions and rarefactions, like sound);
transverse: the masses move across the direction of propagation (like a shaken rope).

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Figure 1.5.1 A chain of masses carrying a pulse. Switch between transverse and longitudinal: in both, each mass only moves a little back and forth about its own place, while the pulse travels along the chain. Press Play.

Theorem 1.5.1The wave equationslides w1, p. 21

In both cases the disturbance obeys $\frac{\partial ^{2} Y ( x , t )}{\partial x ^{2}} = \frac{1}{v ^{2}} \frac{\partial ^{2} Y ( x , t )}{\partial t ^{2}},$ the wave equation, or d’Alembert’s equation. It is a linear differential equation, so the superposition principle holds.

Deriving it for the chainbeyond the lectures

Masses $m$ , springs of constant $K$ , spacing $a$ at rest; mass $i$ sits at $x = ia$ and is displaced by $Y_{i}$ . Its left spring is stretched by $Y_{i} - Y_{i - 1}$ and its right one by $Y_{i + 1} - Y_{i}$ , so Newton’s law reads $m \ddot{Y}_{i} = K (Y_{i + 1} - Y_{i}) - K (Y_{i} - Y_{i - 1}) = K (Y_{i + 1} - 2 Y_{i} + Y_{i - 1}) .$ When $Y$ changes little from one mass to the next (wavelengths much longer than $a$ ), Taylor expansion gives $Y_{i \pm 1} = Y \pm a \partial_{x} Y + \frac{a ^{2}}{2} \partial_{x}^{2} Y \pm \dots$ . The odd terms cancel in the sum, so $Y_{i + 1} - 2 Y_{i} + Y_{i - 1} \approx a^{2} \partial^{2} Y / \partial x^{2}$ , and $m \frac{\partial ^{2} Y}{\partial t ^{2}} = K a^{2} \frac{\partial ^{2} Y}{\partial x ^{2}} ⟹ v = a \frac{K}{m} .$ Stiffer springs and lighter masses carry the disturbance faster. In a solid this is the speed of sound in the spring–ball model (The harmonic approximation).

The wave equation is very general: $Y (x, t)$ can be a position, an electric field (§2.2 Electromagnetic waves in vacuum), the surface of the sea, the density of a gas, a density of matter…

$v$ is a speed

Example 1.5.2Dimensional analysisslides w1, p. 22

With $[Y]$ the units of $Y$ , $[L]$ length and $[t]$ time: $[\frac{Y}{L ^{2}}] = [\frac{1}{v ^{2}}] [\frac{Y}{t ^{2}}] \Rightarrow [v^{2}] = [\frac{L ^{2}}{t ^{2}}] \Rightarrow [v] = [\frac{L}{t}] .$

The general solution

Theorem 1.5.3General solution of the wave equationslides w1, p. 22

$Y (x, t) = f (x - v t) + g (x + v t),$ with $f$ and $g$ practically any functions, regular enough to be differentiated twice. The equation is linear, so each part, the progressive wave $f$ and the regressive wave $g$ , satisfies it on its own.

Proof

Show that $f (x - v t)$ is a solution. Define $h = x - v t$ , so $f$ depends on $x$ and $t$ only through $h$ , with $\partial h / \partial x = 1$ and $\partial h / \partial t = - v$ . By the chain rule:

first derivative in $x$ : $\frac{\partial f}{\partial x} = \frac{\partial f}{\partial h} \frac{\partial h}{\partial x} = \frac{\partial f}{\partial h}$ ;
second in $x$ : $\frac{\partial ^{2} f}{\partial x ^{2}} = \frac{\partial}{\partial h} (\frac{\partial f}{\partial h}) \frac{\partial h}{\partial x} = \frac{\partial ^{2} f}{\partial h ^{2}}$ ;
first in $t$ : $\frac{\partial f}{\partial t} = \frac{\partial f}{\partial h} \frac{\partial h}{\partial t} = - v \frac{\partial f}{\partial h}$ ;
second in $t$ : $\frac{\partial ^{2} f}{\partial t ^{2}} = \frac{\partial}{\partial t} (- v \frac{\partial f}{\partial h}) = - v \frac{\partial ^{2} f}{\partial h ^{2}} \frac{\partial h}{\partial t} = v^{2} \frac{\partial ^{2} f}{\partial h ^{2}}$ .

Substitute into the wave equation: $\frac{\partial ^{2} f}{\partial x ^{2}} = \frac{1}{v ^{2}} \frac{\partial ^{2} f}{\partial t ^{2}} ⟹ \frac{\partial ^{2} f}{\partial h ^{2}} = \frac{1}{v ^{2}} v^{2} \frac{\partial ^{2} f}{\partial h ^{2}} . Q.E.D.$ For $g (x + v t)$ the same steps with $h = x + v t$ give $\partial h / \partial t = + v$ , and again $v^{2}$ , since the sign is squared. That every solution has this form is d’Alembert’s theorem; the slide states it.

Why every solution has this formbeyond the lectures

Change variables to $p = x - v t$ and $q = x + v t$ . By the chain rule, $\partial_{x} = \partial_{p} + \partial_{q}$ and $\partial_{t} = v (\partial_{q} - \partial_{p})$ , so $\partial_{x}^{2} - \frac{1}{v ^{2}} \partial_{t}^{2} = (\partial_{p} + \partial_{q})^{2} - (\partial_{q} - \partial_{p})^{2} = 4 \partial_{p} \partial_{q} .$ The wave equation becomes $\partial_{p} \partial_{q} Y = 0$ . So $\partial_{q} Y$ does not depend on $p$ : it is some function of $q$ alone. Integrating in $q$ , $Y = g (q) + f (p)$ .

Which way does $f (x - v t)$ move? Follow the peak of a bell-shaped pulse: at $t = t_{0}$ it is at $x_{0}$ , at a later time $t_{1}$ at $x_{1}$ .

Proposition 1.5.4Direction of travelslides w1, p. 23

$f (x - v t)$ keeps its shape and slides towards increasing $x$ at speed $v$ . Likewise $g (x + v t)$ slides towards decreasing $x$ .

Proof

The peak has the same height, $f (x_{0}, t_{0}) = f (x_{1}, t_{1})$ , so the argument is the same: $x_{0} - v t_{0} = x_{1} - v t_{1} \Rightarrow v (t_{1} - t_{0}) = x_{1} - x_{0} .$ With $t_{1} > t_{0}$ and $v > 0$ , $x_{1} > x_{0}$ .

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Figure 1.5.2 A

progressive

pulse

f (x - v t)

and a

regressive

one

g (x + v t)

, and their

sum

. Drag the time, or press Play. Each pulse keeps its shape; where they meet they add, then pass through each other unchanged.

Superposition, both ways?

The superposition principle holds for linear differential equations. With $Y = Y_{1} + Y_{2}$ , the week 2 slide states it in both directions:

if $Y_{1}$ and $Y_{2}$ are solutions, then $Y$ is a solution;
if $Y$ is a solution, then $Y_{1}$ and $Y_{2}$ are solutions as well.

“Easy proof, do it yourselves”: the first direction is Theorem 1.3.3.

WarningSlide slip

The second direction is not true as stated. Any function can be split into two pieces that are not solutions: $Y = (Y + ϕ) + (- ϕ)$ . What holds is the first direction. In the general solution $f (x - v t) + g (x + v t)$ it also happens that each piece separately is a solution, as proved above; that is the sense in which the slide uses it.

Exercises

Exercise 1.5.1beyond the lectures

A pulse $Y (x, 0) = e^{- x^{2}}$ is released as $Y (x, t) = e^{- (x - v t)^{2}}$ , with $v = 2$ m/s. Where is its peak at $t = 3$ s?

Solution

The peak is where the argument vanishes: $x - 2 \cdot 3 = 0$ , so $x = 6$ m. The shape is unchanged.

Exercise 1.5.2beyond the lectures

Which of these solve the wave equation with speed $v$ ? (a) $(x - v t)^{3}$ ; (b) $sin x cos v t$ ; © $e^{- (x^{2} - v^{2} t^{2})}$ ; (d) $ln (x + v t)$ .

Solution

(a) yes, a function of $x - v t$ . (b) yes: $sin x cos v t = \frac{1}{2} [sin (x - v t) + sin (x + v t)]$ . © no: $x^{2} - v^{2} t^{2} = (x - v t) (x + v t)$ is not a sum of a function of each; check $\partial_{x}^{2} \neq = \partial_{t}^{2} / v^{2}$ at $t = 0$ , $x = 0$ : $\partial_{x}^{2} = - 2$ , $\partial_{t}^{2} / v^{2} = + 2$ . (d) yes, for $x + v t > 0$ .

Exercise 1.5.3beyond the lectures

A chain has atoms of mass $4.7 \times 1 0^{- 26}$ kg, 0.3 nm apart, with bonds of $K = 10$ N/m. Estimate the speed of sound along it.

Solution

$v = a K / m = 0.3 \times 1 0^{- 9} 10/4.7 \times 1 0^{- 26} = 0.3 \times 1 0^{- 9} \times 1.46 \times 1 0^{13} = 4.4 \times 1 0^{3}$ m/s: the right order for sound in solids.

Exercise 1.5.4beyond the lectures

At $t = 0$ a string is at rest with shape $Y (x, 0) = F (x)$ . Show that $Y (x, t) = \frac{1}{2} [F (x - v t) + F (x + v t)]$ is the solution, and describe what happens.

Solution

It is of d’Alembert’s form, so it solves the equation. At $t = 0$ it gives $F (x)$ . Its time derivative at $t = 0$ is $\frac{1}{2} [- v F^{'} (x) + v F^{'} (x)] = 0$ : at rest. The initial bump splits into two half-height copies running in opposite directions.

1.5 The wave equation

Objectives

A chain of oscillators

v is a speed

The general solution

Superposition, both ways?

Exercises

$v$ is a speed