A plucked string, a dropped stone in a pond, and a beam of light all obey the same partial differential equation. Two families of solutions capture everything: waves that travel, and waves that stand still and merely oscillate in place.
IntuitionA plucked string, frozen in time
Pluck a guitar string and record its height at every point, at every instant: you get a function u(x,t), the displacement at position x and time t. Drop a stone in a still pond and the ripples spreading outward are governed by the same kind of law, just in two spatial dimensions instead of one. In every case, the key physical fact is local: how a point accelerates depends only on how curved the wave is right there — a taut string pulls a dip back up harder the sharper the dip is. That single idea, "acceleration proportional to curvature," is the wave equation.
A 3D surface plot of a radially symmetric ripple function, peaking at the center and decaying in concentric rings outward, used here as a stand-in for a spreading circular wave.
A ripple spreading outward, like circular waves on a pond after a stone is dropped at the center — height decays with distance while the wave pattern expands outward at constant speed.
UndergraduateFrom Newton's second law to a partial differential equation
Model an idealized string under tension T with mass density ρ per unit length, vibrating with small vertical displacement u(x,t). Newton's second law applied to a short segment [x,x+dx], using the small-angle approximation for the tension's vertical components, gives ρutt=Tuxx. Writing c2=T/ρ (a speed, since T has units of force and ρ mass per length) turns this into the standard form.
On the whole line R, change to characteristic coordinates ξ=x−ct, η=x+ct. The chain rule turns utt=c2uxx into the far simpler uξη=0, whose general solution is u=F(ξ)+G(η)=F(x−ct)+G(x+ct) for arbitrary (twice-differentiable) functions F,G — a right-moving wave plus a left-moving wave, both traveling at speed c without changing shape. Matching this to initial position u(x,0)=φ(x) and initial velocity ut(x,0)=ψ(x) pins down F and G exactly.
For utt=c2uxx on R×(0,∞) with u(x,0)=φ(x), ut(x,0)=ψ(x), the unique solution is u(x,t)=21(φ(x−ct)+φ(x+ct))+2c1∫x−ctx+ctψ(s)ds.
Why is it true?
Anything happening at (x,t) can only depend on initial data in [x−ct,x+ct]: information cannot travel faster than c. This finite propagation speed is the defining feature that separates the wave equation from, say, the heat equation, where an initial disturbance is felt everywhere instantly.
Proof
Change to characteristic coordinates as above to get u=F(x−ct)+G(x+ct); imposing u(x,0)=φ(x) and ut(x,0)=ψ(x) gives two equations for F and G in terms of φ and an antiderivative of ψ, which combine into the stated formula.
Example: A plucked pulse splits in two
An infinite string starts at rest (ψ(x)=0) with a triangular pulse φ(x)=1−∣x∣ for ∣x∣≤1 and φ(x)=0 otherwise. Describe u(x,t) for t>0.
Solution
Since ψ=0, the integral term vanishes and u(x,t)=21(φ(x−ct)+φ(x+ct)): the original pulse, at half its height, splits into two identical copies moving apart at speed c, one to the right and one to the left. Wherever the two copies still overlap (for small t), the heights simply add, momentarily reconstructing part of the original shape; once ct>1 they are fully separated.
Example: Seismic wave arrival time and epicenter distance
A seismograph station detects a P-wave (compressional, speed cP=6 km/s) arriving at time tP=0 and an S-wave (shear, speed cS=3.5 km/s) arriving Δt=45 seconds later. Assuming both waves are generated simultaneously at the same source and travel in straight lines, find the distance d to the earthquake epicenter.
Solution
Step 1 — Set up the timing equations. The P-wave travels distance d in time d/cP and the S-wave takes time d/cS. The measured lag is Δt=d/cS−d/cP.
**Step 2 — Solve for d.** Factoring out d: Δt=d(cS1−cP1), so d=cS1−cP1Δt.
Step 3 — Substitute values.cS1−cP1=3.51−61≈0.2857−0.1667=0.1190 s/km, so d=0.119045≈378 km. This is the standard seismological S-P method used to locate earthquakes with a single station.
UndergraduateStanding waves and Fourier series on a finite string
Now pin the string down at both ends: u(0,t)=u(L,t)=0 for a string of length L. Separating variables, u(x,t)=X(x)T(t), forces X to satisfy X′′+λX=0 with X(0)=X(L)=0 — an eigenvalue problem whose solutions are exactly Xn(x)=sin(nπx/L), λn=(nπ/L)2, for n=1,2,3,…. Each such Xn pairs with Tn(t)=cos(nπct/L+phase) to give a normal mode: the whole string oscillating in place, all points crossing zero at the same instants. These are exactly the shapes seen in [Fourier series](/chuoi-fourier) — the wave equation is where that topic's sine basis was born, in Daniel Bernoulli's 1753 analysis of exactly this problem.
Graph of a triangle wave together with its approximation by five sine harmonics, illustrating how a plucked string's initial shape decomposes into normal modes.
A triangle wave (the shape of a string plucked at its midpoint) built from 5 sine harmonics — exactly the normal modes sin(nπx/L) that the plucked string decomposes into.
AdvancedEnergy conservation and higher dimensions
The total energy E(t)=21∫0L(ut2+c2ux2)dx (kinetic plus elastic potential) is conserved: differentiating under the integral, integrating by parts, and using utt=c2uxx together with the fixed-end boundary conditions makes E′(t)=0. This both confirms the physics (a frictionless string never runs out of energy) and gives a clean proof that d'Alembert's and the normal-mode solutions are the only solutions: if two solutions shared the same initial data, their difference would have zero energy, hence be identically zero.
Let u(x,t) solve the wave equation utt=c2uxx on [0,L] with fixed-end boundary conditions u(0,t)=u(L,t)=0. The total energy E(t)=21∫0L(ut2+c2ux2)dx is constant: E′(t)=0 for all t>0.
Why is it true?
Energy has two parts: kinetic 21∫ut2dx (mass moving) and elastic potential 21c2∫ux2dx (string stretched). The wave equation turns kinetic energy into elastic energy and back, at the same rate, keeping the total constant — exactly as a pendulum trades kinetic and potential energy without loss.
Proof
Differentiate E(t) under the integral sign: E′(t)=∫0L(ututt+c2uxuxt)dx.
Substitute utt=c2uxx from the wave equation to get E′(t)=∫0L(c2utuxx+c2uxuxt)dx=c2∫0L(utuxx+uxuxt)dx. Recognize the integrand as the derivative ∂x∂(utux), so E′(t)=c2[utux]0L.
Apply the fixed-end boundary conditions u(0,t)=u(L,t)=0, which force ut(0,t)=ut(L,t)=0 as well (differentiating the boundary conditions with respect to t). Therefore E′(t)=c2(ut(L,t)ux(L,t)−ut(0,t)ux(0,t))=0, so E(t)=E(0) is constant for all t≥0.
In two spatial dimensions, a rectangular drum membrane pinned at its edges has normal modes sin(mx)sin(ny); the picture below shows the related pattern sin(x)cos(y), a standing wave rippling across a surface rather than along a line. Higher dimensions also bring a striking qualitative difference known as Huygens' principle: in odd spatial dimensions 3,5,7,…, a localized disturbance produces a sharp wavefront with silence afterward (this is why sound is intelligible), while in even dimensions 2,4,… (and in dimension 1) a disturbance leaves a persistent "wake" that never fully dies out.
A 3D surface plot of the function sin(x)cos(y), showing an alternating checkerboard pattern of peaks and troughs, used here to depict a two-dimensional standing-wave mode of a membrane.
A standing-wave mode z=sinxcosy on a 2D membrane: a snapshot at one instant, with the whole checkerboard pattern of peaks and troughs rising and falling in place rather than traveling sideways.
The wave equation utt=c2uxx needs how many pieces of initial data to determine a unique solution?
In d'Alembert's formula, the value at (x,t) depends only on the initial data on which interval?
For a string of length L fixed at both ends, the normal modes (standing waves) have spatial shape
d'Alembert's formula u(x,t)=21[φ(x−ct)+φ(x+ct)]+2c1∫x−ctx+ctψ(s)ds reduces to just u(x,t)=21[φ(x−ct)+φ(x+ct)] precisely when