The Wave Equation in One, Two and Three Dimensions

One bump, struck or plucked, in a line, on a sheet and in space. Watch it spread, and read what a listener at distance r records in each, long after the disturbance has gone by.

Worked example — The Wave Equation in One, Two and Three Dimensions

The disturbance reaches the listener, at ct/a	4
Its back has passed the listener, at ct/a	6
String's highest reading	0.5333

These are the values the calculator shows when the page opens. Change any input above to recompute them.

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Lesson

The theory — The Wave Equation in One, Two and Three Dimensions

The wave equation, u_tt = c²∇²u, says that the second time derivative of u at each point (its acceleration, when u is a displacement) is c² times its Laplacian, which is proportional to how far the average of u over the point's immediate surroundings sits above u itself. In one dimension every solution is two shapes travelling at speed c in opposite directions (d'Alembert, 1747). In three, a disturbance confined to a region reaches a distant point only while the sphere of radius ct centred on that point is crossing the region, and leaves it at rest afterwards. In two dimensions it does not: a residual tail remains, and after a struck start it falls only as 1/t.

What each symbol means

u: the wave's value: the displacement of the string or the sheet, or in space a quantity such as the excess air pressure
c: the wave speed
a: the radius of the starting bump; every length on the page is in units of a, every time in units of a/c
g: the shape of the starting bump, (1 − s²)² at distance s from its centre (in units of a), zero for s ≥ 1

Assumes: A line, sheet or space with no ends or walls, so nothing is reflected back to the listener. Small displacements, so the equation is linear: a bump struck twice as hard gives twice every number on the page.

A struck string never settles back 🖖

Press Struck, heard at 5a and watch the teal string. The kick sends a ramp out each way, and behind the ramps the string is left lifted, flat, by 0.5333 v₀a/c. The listener's point climbs while the disturbance passes and then stays there, at ct = 20a and at 80a alike. Switch to Plucked and the same bump, pushed up and let go, splits into two copies of half its height that run apart: the point rises to 0.5000 and falls back to exactly 0. D'Alembert showed in 1747 that every motion of a string is two shapes travelling in opposite directions. Written in terms of the starting shape f and starting speed g it reads u = ½[f(x − ct) + f(x + ct)] + (1/2c)∫_x−ct^x+ctg. A shape leaves with the two travelling halves. A speed is summed over the stretch of string within ct of the point, and once that stretch has grown to take in the whole patch, the sum is the whole kick, ½∫g = 8/15. A real string has fixed ends whose reflections pull it back. This one has no ends, so nothing ever comes back.

On a sheet, a kick never finishes arriving 🖖

Press Struck, heard at 5a and watch the two surfaces. Each carries a ring outwards, and both rings have passed the listener by ct = 6a. Inside the dotted circle, where the back of the wave has already been, space is flat: its listener reads exactly 0 from then on, and the table's last column is zero to every place. The sheet inside its circle is still raised. At ct = 20a its listener reads 0.00861, 9.93% of the 0.0867 it peaked at; press Struck, 5a, listen at 80a and it is 0.00209, about a quarter of that for four times the wait. The tail falls as 1/t and never reaches zero. The two solution formulas say why. In space the answer at a point is built from the start's average over the surface of a sphere of radius ct centred on it, and once that sphere has grown past the bump there is nothing on it. On a sheet the answer is an integral over the whole disc inside the circle of radius ct, so the bump stays inside it for ever, weighted by 1/√(c²t² − ρ²). Hadamard, in the Yale lectures he published in 1923, called the leftover a residual integral and wrote that cylindrical waves, unlike spherical ones, diffuse.

In space a released bump arrives as a bump and a dip 🖖

Press Plucked, heard at 5a and watch the blue trace. Nothing in the bump was ever below zero, yet the listener in space records a rise to 0.02862 followed by a fall just as deep, with areas of 0.01667 and −0.01667 that cancel exactly. For r > a the answer is u = (r − ct) g(ct − r)/(2r): the starting shape multiplied by a ramp that passes through zero just as the bump's centre arrives. Sound pressure obeys this equation, so a ball of compressed air released at rest reaches a listener as a compression followed by an equal rarefaction, with no net push left over. The string carries the same release without changing its sign. The sheet records a bump, then a tail below zero that falls as 1/t², which is the sign Hadamard worked out for a residual in two dimensions.

Problem solved in full

How long does a clap linger in a flat world? 4 steps
On the sheet, struck, a listener 5a away records a largest reading of 0.0867 v₀a/c. How long until the tail has fallen to 1% of that? Press Struck, heard at 5a: the page prints the peak, and the reading at ct = 20a beside the tail law.

Open this problem in the tool
1. Long after the wave has passed, the sheet's reading follows the tail law for a point-sized kick. At ct = 20a the two already agree to three figures:
  
  $\frac{1}{6 2 0 ^{2} - 5 ^{2}} = \frac{1}{6 375} = 0.00861$
2. One per cent of the peak is
  
  $0.01 \times 0.0867 = 0.000867$
3. Set the tail law equal to it and solve for ct/a:
  
  $(c t / a)^{2} - 25 = \frac{1}{6 \times 0 . 000867} = 192.2 \Rightarrow \frac{c t}{a} = 192 . 2^{2} + 25 = 192.3$
4. In space the same listener is silent from ct = 6a. For a disturbance about 10 cm across, a = 0.05 m, clapping in air at c = 343 m/s, the unit of time a/c is
  
  $\frac{a}{c} = \frac{0 . 05 m}{343 m/s} = 0.146 ms, 6 \frac{a}{c} = 0.87 ms, 192.3 \frac{a}{c} = 28 ms$
Answer

At about ct = 192a, thirty-two times the 6a at which space falls silent. For a clap 10 cm across that is 28 milliseconds against less than one. In a flat world every sound would drag a tail behind it into whatever came next. A 2% threshold would roughly halve the wait, because the tail falls as 1/t.

$\frac{c t}{a} \approx 192 \approx 32 \times 6$

Learning path

The heat, wave and Laplace equations

References (2)

The vibrating string's equation solved as the sum of two arbitrary functions, y = Ψ(t + s) + Γ(t − s): every motion of the string is two shapes travelling in opposite directions (art. VII), followed by a long continuation working out which shapes satisfy fixed ends: J. le Rond d'Alembert, "Recherches sur la courbe que forme une corde tenduë mise en vibration", Histoire de l'Académie royale des sciences et belles lettres de Berlin 3 (1747, published 1749), 214–219; "Suite des recherches…", same volume, 220–249.
Huygens' principle in its narrow sense, a disturbance localised at the start stays localised on a thin spherical shell (Book II, §33), and its failure in two dimensions: the 'residual integral' left after the wave has passed, so that 'cylindrical waves, unlike spherical ones, diffuse' (Book III, §111), and the sign of that residual, negative for a positive starting displacement when there are three variables (§112): J. Hadamard, Lectures on Cauchy's Problem in Linear Partial Differential Equations, Yale University Press, New Haven, 1923 (Mrs. Hepsa Ely Silliman Memorial Lectures), Book II §33, pp. 53–55, and Book III §§111–112, pp. 175–178.

By the 42Math team · Reviewed against published sources. Corrections welcome.