d'Alembert's Solution and Characteristics of the Wave Equation

The one-dimensional wave equation $u_{tt}=c^2u_{xx}$ is the canonical hyperbolic partial differential equation, and d'Alembert's solution is the most transparent window ever opened onto it. The trick is algebraic: the wave operator factors as $(\partial_t-c\partial_x)(\partial_t+c\partial_x)$ , a product of two first-order transport operators. That single factorisation carries all of the physics. It says every solution is a superposition of two disturbances, one gliding right and one gliding left, each at the fixed speed $c$ and each holding its shape forever — no spreading, no dispersion, no decay. A plucked string does not blur; it splits. This rigidity is special to the lossless, non-dispersive, one-dimensional case, and it is the cleanest demonstration in all of mathematical physics that the type of a PDE dictates its qualitative behaviour. The natural coordinates are the characteristics $\xi=x-ct$ and $\eta=x+ct$ , in which the equation collapses to $u_{\xi\eta}=0$ and integrates by inspection to $u=f(\xi)+g(\eta)$ . Impose an initial displacement $\phi$ and velocity $\psi$ and the arbitrary profiles are pinned down uniquely, giving d'Alembert's closed-form formula — no eigenfunction expansion, no infinite series. Read geometrically, that formula hands you the two deepest ideas in PDE theory: the domain of dependence, the finite slice of initial data that can reach a given spacetime point, and the domain of influence, the wedge a data point can affect. Both are bounded by characteristics, which is the mathematical statement that signals travel at finite speed — the seed of causality and the light cone. Bounded domains fold back onto the infinite line through odd or even reflection of the data, reuniting the travelling-wave picture with the standing-wave modes of separation of variables.

Key takeaways

The 1D wave equation factors into two transport operators, so every solution is a sum $u=f(x-ct)+g(x+ct)$ of a right- and a left-moving wave of fixed profile travelling at speed $c$ .
Characteristic coordinates $\xi=x-ct,\ \eta=x+ct$ reduce $u_{tt}=c^2u_{xx}$ to $u_{\xi\eta}=0$ , whose general solution is manifestly the sum of two arbitrary twice-differentiable functions.
Cauchy data $u(x,0)=\phi(x),\ u_t(x,0)=\psi(x)$ fix $f$ and $g$ , giving d'Alembert's formula $u=\tfrac12[\phi(x-ct)+\phi(x+ct)]+\tfrac1{2c}\int_{x-ct}^{x+ct}\psi\,\mathrm ds$ .
The field at $(x_0,t_0)$ depends only on data in the interval $[x_0-ct_0,\,x_0+ct_0]$ — the domain of dependence — encoding finite propagation speed.
Characteristics $x\mp ct=\text{const}$ are the curves along which information rides and across which derivatives may jump; kinks propagate undistorted forever.
Bounded domains are handled by odd (Dirichlet) or even (Neumann) reflection of the data, turning finite-domain problems back into the infinite-line formula.
Key idea: factor the wave operator into two transport operators; the characteristic coordinates $\xi=x-ct,\ \eta=x+ct$ turn $u_{tt}=c^2u_{xx}$ into $u_{\xi\eta}=0$ , whose general solution $u=f(x-ct)+g(x+ct)$ encodes finite-speed propagation along straight characteristics.

Where this lives today

d'Alembert's finite-speed propagation is not a historical curiosity — it is the working backbone of hyperbolic solvers and the physical statement of causality. The 1928 Courant–Friedrichs–Lewy condition, which every explicit numerical scheme for wave-like equations must obey, is d'Alembert's domain of dependence made computational: a stencil is stable only if its own domain of dependence contains the true one, so the time step must satisfy $c\,\Delta t\le\Delta x$ and information cannot outrun the grid. Seismologists invert exactly this structure, reading arrival times along ray-traced characteristics to image the Earth's interior; time-domain reflectometry sends a step down a cable and times the inverted or upright echo — the fixed-end / free-end reflection derived above — to locate faults to the centimetre. In relativistic field theory the same operator with $c$ the speed of light has characteristics that trace the light cone, so the domain of dependence is the past light cone: d'Alembert's finite-speed propagation is the classical seed of microcausality. And in nonlinear gas dynamics, characteristics that cross are the birth of shock waves — the weak-solution theory that begins precisely where a d'Alembert kink steepens.

The physical picture

Pluck an infinitely long string, hold the shape, and let go. Nothing about the equation tells the hump which way to move, so it does the only symmetric thing possible: it splits into two identical copies of half the amplitude, one sliding right and one sliding left, each at the speed $c=\sqrt{T/\rho}$ set by tension $T$ and linear density $\rho$ . Neither copy ever changes shape, because the equation contains no length scale and no time scale with which to spread them. The first figure freezes this split — one pulse at release, two half-height pulses a time $t$ later, their peaks separated by exactly $2ct$ .

Figure

A localized pulse released from rest splits into two copies of half the amplitude, A divided by 2 each, one moving right and one left at speed c; after a time t their peaks sit a distance 2ct apart while each keeps its exact shape

The characteristics are the rails of spacetime. A disturbance can only ride a $+c$ rail or a $-c$ rail, never leave them, and never smear across them. To learn the field at a point you trace both rails backward in time until they strike the initial line $t=0$ ; the data collected between those two intercepts is the only data that matters — nothing outside that interval could have reached you in time. The second figure draws this in the $x$ – $t$ plane as a bowtie: the backward cone below the point is its domain of dependence, the forward cone above is the region the point will later influence, and both are cut out by characteristics of slope $\pm 1/c$ .

Figure

Spacetime bowtie for c = 2 at the point x₀ = 5, t₀ = 3: the two backward characteristics x − ct = −1 and x + ct = 11 strike t = 0 at x = −1 and x = 11, so only initial data between −1 and 11 can determine the field at P; the forward cone above P is the region P can later affect

Displacement data and velocity data leave different fingerprints. An initial shape with no speed splits cleanly into two travelling copies; an initial velocity — a hammer blow — is different, because the field at a point accumulates all the velocity that has passed through it, so a localized kick leaves a permanent residual displacement stretched between the two outgoing fronts. That is the physical meaning of the $\int\psi$ term. And a wall behaves like a mirror folded into the string: send a pulse in and an inverted (fixed end) or upright (free end) phantom pulse comes back out, reproducing reflection without ever leaving the infinite-line formula.

The core equations

The starting point is the lossless, non-dispersive wave equation in one space dimension. It is second order in both time and space, and its single material parameter is the propagation speed $c$ , assumed constant and positive.

u_{tt}=c^2\,u_{xx} \qquad (1)

where: $u(x,t)$ — the wave field (string displacement, acoustic pressure, a field component); $c$ — propagation speed, constant and positive; $x,t$ — space and time. Read it as curvature in space drives acceleration in time, with $c^2$ the exchange rate between them. Because $\partial_t$ and $\partial_x$ commute and $c$ is constant, the operator on the left is a difference of squares and factors exactly into two first-order transport operators. This is the whole secret of the equation compressed into one line.

\big(\partial_t-c\,\partial_x\big)\big(\partial_t+c\,\partial_x\big)\,u=0 \qquad (2)

where: $\partial_t-c\,\partial_x$ annihilates any right-moving profile $f(x-ct)$ ; $\partial_t+c\,\partial_x$ annihilates any left-moving profile $g(x+ct)$ . A solution need only be killed by one factor or the other. The two factors motivate a change to the coordinates that are constant along each family of transport lines — the characteristic coordinates.

\xi=x-ct,\qquad \eta=x+ct \qquad (3)

where: $\xi$ — label of a right-moving characteristic ( $\xi=\text{const}$ ); $\eta$ — label of a left-moving characteristic ( $\eta=\text{const}$ ). These straight lines are the rails along which signals travel. In these variables the wave equation loses every term except the mixed derivative, reaching its hyperbolic canonical form.

u_{\xi\eta}=0 \qquad (4)

where: $u_{\xi\eta}$ — mixed second derivative in the characteristic variables. Vanishing mixed derivative is the simplest possible hyperbolic equation, and it integrates twice by inspection. Integrating $u_{\xi\eta}=0$ once in $\eta$ makes $u_\xi$ a function of $\xi$ alone; integrating again gives the general solution as a sum of two arbitrary twice-differentiable profiles.

u(x,t)=f(x-ct)+g(x+ct) \qquad (5)

where: $f$ — an arbitrary $C^2$ profile carried rigidly rightward; $g$ — an arbitrary $C^2$ profile carried rigidly leftward. Every classical solution of (1) has exactly this shape. To select the two profiles from a physical problem we supply Cauchy data: the initial shape and the initial velocity of the field.

u(x,0)=\phi(x),\qquad u_t(x,0)=\psi(x) \qquad (6)

where: $\phi$ — initial displacement, taken $C^2$ ; $\psi$ — initial velocity, taken $C^1$ . Two functions of one variable are needed because the equation is second order in time. Solving for $f$ and $g$ and reassembling gives d'Alembert's formula — an exact, closed-form solution requiring no series.

u(x,t)=\tfrac12\big[\phi(x-ct)+\phi(x+ct)\big]+\frac1{2c}\!\int_{x-ct}^{x+ct}\!\psi(s)\,\mathrm ds \qquad (7)

where: the bracket splits the initial shape into two half-amplitude travelling copies; the integral accumulates the initial velocity swept between the two backward characteristic intercepts $x\pm ct$ ; $s$ is a dummy variable along the initial line. Reading the arguments of (7) geometrically gives the two causal regions. Only data in a bounded interval reaches a spacetime point, and a single data point influences only a bounded wedge.

D(x_0,t_0)=[\,x_0-ct_0,\;x_0+ct_0\,],\qquad I(x_0)=\{(x,t):|x-x_0|\le c\,t\} \qquad (8)

where: $D$ — the domain of dependence, the data that determines $u(x_0,t_0)$ ; $I$ — the domain of influence, the region a disturbance at $x_0$ can reach. Both boundaries are characteristics; both encode finite propagation speed.

Derivation AFactoring the wave operator and reaching canonical form

The goal is to turn the second-order wave equation $u_{tt}=c^2u_{xx}$ into the trivially integrable form $u_{\xi\eta}=0$ , and in doing so to prove that its general solution is the travelling-wave sum $u=f(x-ct)+g(x+ct)$ . The route is a change of variables to the characteristic coordinates $\xi=x-ct,\ \eta=x+ct$ , executed with the chain rule; the payoff is that the two arbitrary functions $f$ and $g$ fall out by two elementary integrations, with no assumption about their form beyond twice-differentiability.

Step 1. Treat the operator as a difference of squares. Since $\partial_t$ and $\partial_x$ commute and $c$ is constant, $\partial_t^2-c^2\partial_x^2=(\partial_t-c\partial_x)(\partial_t+c\partial_x)$ , the two cross terms $\mp c\,\partial_t\partial_x$ cancelling exactly. Why: this exhibits the wave operator as a product of two first-order transport operators, already telling us solutions come in a right-moving and a left-moving family.

Step 2. Change variables with $\xi=x-ct,\ \eta=x+ct$ , so that $\xi_x=\eta_x=1,\ \xi_t=-c,\ \eta_t=c$ . The chain rule then gives $\partial_x=\partial_\xi+\partial_\eta$ and $\partial_t=-c\,\partial_\xi+c\,\partial_\eta$ . Why: characteristic coordinates are the ones held constant along each transport line, so they are exactly the variables in which the two factors decouple.

Step 3. Square each operator: $u_{xx}=u_{\xi\xi}+2u_{\xi\eta}+u_{\eta\eta}$ and $u_{tt}=c^2(u_{\xi\xi}-2u_{\xi\eta}+u_{\eta\eta})$ , the sign flip on the cross term coming from the $-c$ in $\partial_t$ . Why: only the mixed second derivative carries opposite signs in the two expressions, so it is the term that will survive the subtraction.

Step 4. Form $u_{tt}-c^2u_{xx}=-4c^2\,u_{\xi\eta}$ . Setting this to zero and dividing by the nonzero $-4c^2$ leaves $u_{\xi\eta}=0$ , the hyperbolic canonical form. Why: the pure $\xi\xi$ and $\eta\eta$ derivatives cancel identically, and canonical form is where the equation is simplest to integrate.

Step 5. Integrate. $u_{\xi\eta}=0$ means $\partial_\eta(u_\xi)=0$ , so $u_\xi$ depends on $\xi$ alone; integrating in $\xi$ gives $u=f(\xi)+g(\eta)=f(x-ct)+g(x+ct)$ with $f,g$ arbitrary $C^2$ functions. Why: two integrations of a mixed derivative produce exactly two arbitrary functions of one variable each — the general solution, not a special case. *Why this works: substituting $f(x-ct)$ gives $u_{tt}=c^2f''=c^2u_{xx}$ , and likewise for $g(x+ct)$ , so the residual $u_{tt}-c^2u_{xx}$ is identically zero for both travelling-wave profiles.*

Go deeper · Derivation B · 1 minFixing the profiles from initial data — d'Alembert's formula

Derivation BFixing the profiles from initial data — d'Alembert's formula

The goal is to determine the two arbitrary profiles $f$ and $g$ in $u=f(x-ct)+g(x+ct)$ from the initial displacement $\phi$ and velocity $\psi$ , arriving at d'Alembert's closed-form solution. The method is to impose the Cauchy data at $t=0$ , obtaining one algebraic and one differential relation between $f$ and $g$ , then to integrate and solve the resulting linear system; the reference constants introduced along the way must and do cancel, leaving a symmetric formula that depends only on the data.

Step 1. Set $t=0$ in $u=f(x-ct)+g(x+ct)$ to get $f(x)+g(x)=\phi(x)$ . Why: the initial displacement is the sum of the two profiles evaluated on the initial line, one relation between the unknowns.

Step 2. Differentiate in time, $u_t=-c\,f'(x-ct)+c\,g'(x+ct)$ , and set $t=0$ to get $-c\,f'(x)+c\,g'(x)=\psi(x)$ . Why: the initial velocity constrains the slopes of the profiles, supplying the second, independent relation.

Step 3. Divide by $c$ and integrate: $-f(x)+g(x)=\tfrac1c\int_{x_0}^{x}\psi(s)\,\mathrm ds+K$ , with $x_0$ and $K$ constants of integration. Why: converting the derivative relation into $f$ and $g$ themselves lets us pair it with Step 1 as a two-by-two linear system.

Step 4. Add and subtract the relations from Steps 1 and 3 to solve for each profile: $f(x)=\tfrac12\phi(x)-\tfrac1{2c}\int_{x_0}^{x}\psi-\tfrac K2$ and $g(x)=\tfrac12\phi(x)+\tfrac1{2c}\int_{x_0}^{x}\psi+\tfrac K2$ . Why: with both profiles known as functionals of the data, the solution is fully determined up to the bookkeeping constant $K$ .

Step 5. Reassemble by putting the argument $x-ct$ into $f$ and $x+ct$ into $g$ . The half- $\phi$ terms give $\tfrac12[\phi(x-ct)+\phi(x+ct)]$ ; the $K$ terms cancel; and the two velocity integrals combine, $\int_{x_0}^{x+ct}\psi-\int_{x_0}^{x-ct}\psi=\int_{x-ct}^{x+ct}\psi$ , erasing the reference point $x_0$ . Why: a physical solution cannot depend on an arbitrary reference $x_0$ or constant $K$ , and indeed they vanish, which is a strong internal check.

Step 6. The result is d'Alembert's formula $u(x,t)=\tfrac12[\phi(x-ct)+\phi(x+ct)]+\tfrac1{2c}\int_{x-ct}^{x+ct}\psi(s)\,\mathrm ds$ . Why: this is an exact solution of the initial-value problem for any $\phi\in C^2,\ \psi\in C^1$ , with the causal structure of dependence built directly into its arguments. *Why this works: for $\phi=\cos kx,\ \psi=0$ the formula gives $\tfrac12[\cos k(x-ct)+\cos k(x+ct)]=\cos kx\,\cos kct$ , whose residual $u_{tt}-c^2u_{xx}$ is identically zero.*

Go deeper · Derivation C · 1 minReflection at a fixed end by odd extension

Derivation CReflection at a fixed end by odd extension

The goal is to solve the wave equation on the half-line $x\ge0$ with a fixed (Dirichlet) end $u(0,t)=0$ using nothing but the infinite-line d'Alembert formula, by extending the initial data as odd functions about $x=0$ . The pay-off is a proof that the boundary condition is satisfied automatically for all time, together with the physical picture that a pulse reflects off a clamped end inverted — the mirror-image, phase-reversed wave that guitarists and transmission-line engineers rely on.

Step 1. Extend $\phi$ and $\psi$ , defined for $x\ge0$ , to odd functions $\tilde\phi,\tilde\psi$ on the whole line, so $\tilde\phi(-x)=-\tilde\phi(x)$ and $\tilde\psi(-x)=-\tilde\psi(x)$ . Solve the infinite-line problem with this extended data. Why: the whole-line d'Alembert formula already exists, so if the extension enforces the boundary condition we get the half-line solution for free.

Step 2. Evaluate the formula at the boundary $x=0$ : $u(0,t)=\tfrac12[\tilde\phi(-ct)+\tilde\phi(ct)]+\tfrac1{2c}\int_{-ct}^{ct}\tilde\psi(s)\,\mathrm ds$ . Why: the boundary condition is a statement about $x=0$ , so we test it there directly.

Step 3. The displacement bracket vanishes because $\tilde\phi$ is odd: $\tilde\phi(-ct)+\tilde\phi(ct)=0$ . Why: an odd function takes equal-and-opposite values at $\pm ct$ , so the two travelling copies exactly cancel at the wall.

Step 4. The velocity integral vanishes because a symmetric integral of an odd integrand is zero: $\int_{-ct}^{ct}\tilde\psi=0$ . Why: the two halves of the integration range contribute equal and opposite areas.

Step 5. Hence $u(0,t)=0$ for all $t$ , and for $x\ge0$ the solution equals the physical half-line motion. A right-moving pulse $f(x-ct)$ approaching the wall reappears as $-f(-x-ct)$ , an upside-down left-mover. Why: the odd image supplies exactly the inverted reflected wave, so clamping the end flips the pulse. A Neumann (free) end uses an even extension instead, giving an upright reflection. *Why this works: with odd $\tilde\phi,\tilde\psi$ both terms are identically zero at $x=0$ , and the reflected profile $-f(-x-ct)$ is exactly the inverted fixed-end reflection.*

Limiting cases

The clean d'Alembert picture is a statement about one non-dispersive space dimension. The table tracks what survives and what breaks as each assumption is relaxed.

Case	Result	Meaning
Pure right-mover	$\psi=c\,\phi'\ \Rightarrow\ u=\phi(x-ct)$	Tying the initial velocity to the slope cancels $g$ , leaving a single rigid wave — a one-way wave.
Point velocity impulse	$\phi\equiv0,\ \psi=\delta(x)\ \Rightarrow\ u=\tfrac1{2c}$ between the fronts	A hammer blow leaves a permanent step of height $1/(2c)$ spreading between $\pm ct$ : lasting displacement, no clean tail.
Three dimensions	$u=\partial_t\!\big(\tfrac1{4\pi c^2t}\!\int_{\|y-x\|=ct}\phi\,\mathrm dS\big)+\cdots$	Kirchhoff's formula keeps sharp fronts (Huygens' principle); in 2D a wake trails behind — the rigid 1D profile is dimension-specific.
Variable speed / dispersion	$u_{tt}=c(x)^2u_{xx}$ or $\omega^2=c^2k^2+\Omega^2$	Characteristics curve and packets spread; rigid profiles no longer exist, though WKB still tracks the rays.
Damping	$u_{tt}+\gamma u_t=c^2u_{xx}$	Amplitude decays as it propagates; the factorisation into pure transport operators is lost.

Common trapd'Alembert's rigid travelling wave is a one-dimensional, non-dispersive miracle — it does not generalize for free

The exact "splits into two shape-preserving copies" picture is special to one space dimension and the constant-coefficient, non-dispersive operator. In three dimensions Kirchhoff's formula still gives sharp fronts, but in two dimensions the solution develops a trailing wake — Huygens' principle holds cleanly only in odd dimensions $\ge 3$ , which is a deep and often-forgotten fact. Any $x$ -dependence of $c$ , any dispersion relation beyond $\omega=ck$ , and any damping all destroy the rigid-profile picture: characteristics bend, packets broaden, amplitudes fade. And even in 1D the classical formula presumes $\phi\in C^2$ and $\psi\in C^1$ ; a corner in the initial data is not a classical solution but a weak solution, with the kink riding a characteristic forever rather than smoothing out as it would under the heat equation.

Worked examples

1 · The struck string — a hammer blow leaves a permanent dent (fully worked)

A long taut string carries waves at $c=300\ \text{m/s}$ . A felt hammer of contact half-width $a=0.01\ \text{m}$ strikes it, imparting an initial velocity $\psi(x)=V_0=2\ \text{m/s}$ for $|x|\le a$ and zero elsewhere, with no initial displacement, $\phi\equiv0$ . Find the field at the centre for late times, the profile it settles into, and when the plateau forms.

Step 1. With $\phi\equiv0$ , d'Alembert's formula reduces to the velocity integral alone, $u(x,t)=\tfrac1{2c}\int_{x-ct}^{x+ct}\psi(s)\,\mathrm ds$ . Because $\psi$ is a rectangle of height $V_0$ over $[-a,a]$ , the integral is just $V_0$ times the overlap length of the window $[x-ct,\,x+ct]$ with $[-a,a]$ . The field measures how much of the struck region the backward characteristics have swept.

Step 2. Evaluate at the centre $x=0$ once the window fully contains the strike, i.e. for $ct>a$ . Then the overlap is the whole interval, length $2a$ , so $u(0,t)=\tfrac1{2c}\,V_0\,(2a)=\dfrac{aV_0}{c}$ . Numerically, $u(0,t)=\dfrac{(0.01)(2)}{300}=6.67\times10^{-5}\ \text{m}\approx\mathbf{67\ \mu m}$ . This value is constant in time — a permanent residual displacement, the hallmark of velocity data.

Step 3. The plateau forms when the two inner edges of the window clear the strike, at $t=a/c=0.01/300=\mathbf{3.33\times10^{-5}\ s}\ (33.3\ \mu\text{s})$ . For later times the profile is a symmetric trapezoid: flat at height $aV_0/c$ over $|x|<ct-a$ , ramping linearly down over the two edge zones of width $2a$ , and zero beyond the outrunning fronts at $|x|=ct+a$ . The struck string does not spring back — it lifts a lasting plateau between two fronts racing apart at $\pm c$ . Residual displacement $= aV_0/c \approx 67\ \mu\text{m}$ , permanent, bounded by fronts at $|x|=ct+a$ .

2 · The plucked string — split, and read off the domain of dependence

The same string ( $c=2\ \text{m/s}$ for easy arithmetic) is released from rest, $\psi\equiv0$ , with an initial displacement $\phi$ . (a) Describe the motion. (b) At the spacetime point $x_0=5,\ t_0=3$ , which initial data can possibly affect the field?

Step 1. With $\psi\equiv0$ , d'Alembert collapses to $u(x,t)=\tfrac12[\phi(x-ct)+\phi(x+ct)]$ . The initial shape splits into two half-amplitude copies, one riding right at $+c$ , one riding left at $-c$ , each rigid. This is exactly the first figure. No integral term, so displacement data produces a clean split with no residual.

Step 2. For the causal question, trace the two backward characteristics from $(5,3)$ : $x-ct=5-2\cdot3=-1$ and $x+ct=5+2\cdot3=11$ . They strike $t=0$ at $x=-1$ and $x=11$ . These intercepts are the arguments $x\pm ct$ appearing in the formula.

Step 3. Therefore $u(5,3)$ depends only on $\phi$ (and $\psi$ , were it nonzero) on the interval $\mathbf{[-1,\,11]}$ — the domain of dependence, width $2ct_0=12$ . Data outside this interval, no matter how large, cannot influence the field at $P$ : signals travelling at $c=2$ simply have not had time to arrive. The field at $(5,3)$ is set entirely by data on $[-1,11]$ .

3 · Standing wave and the pitch of a guitar string

A steel high-E guitar string has tension $T=73\ \text{N}$ , linear density $\rho=4.0\times10^{-4}\ \text{kg/m}$ , and vibrating length $L=0.648\ \text{m}$ , clamped at both ends. Find the wave speed and the fundamental frequency, and connect the travelling-wave and standing-wave pictures.

Step 1. The speed follows from tension and density: $c=\sqrt{T/\rho}=\sqrt{73/(4.0\times10^{-4})}=\sqrt{1.825\times10^{5}}=\mathbf{427.2\ \text{m/s}}$ . This is the single parameter that scales every characteristic slope.

Step 2. Repeated odd reflection about the two fixed ends makes the motion $2L$ -periodic; the fundamental has half a wavelength on the string, $\lambda_1=2L$ , so $f_1=c/\lambda_1=c/(2L)=427.2/(2\cdot0.648)=\mathbf{329.6\ \text{Hz}}$ — the pitch E4, with harmonics $f_n=n f_1$ at $659.3\ \text{Hz},\ 988.9\ \text{Hz},\dots$

Step 3. The same result comes from d'Alembert directly. Take displacement data $\phi(x)=A\cos kx$ (with $\psi=0$ ) as a stand-in for a single mode; the formula gives $u=A\cos kx\,\cos kct$ , a standing wave oscillating at angular frequency $\omega=ck$ . For the lowest fixed-fixed mode $k=\pi/L$ , $\omega=c\pi/L=427.2\cdot\pi/0.648=2071\ \text{rad/s}$ , i.e. $f=\omega/2\pi=329.6\ \text{Hz}$ — identical to Step 2. The travelling-wave sum and the normal-mode sum are two languages for one motion; $c=427\ \text{m/s}$ gives $f_1=329.6\ \text{Hz}$ (E4).

Physics of the portrait

d'Alembert's solution is where a musical instrument and a differential equation first met. The vibrating string was the eighteenth century's great test problem, and the argument over what its solutions could look like — smooth functions or arbitrary shapes — ran for decades and seeded the modern theory of functions, Fourier series, and partial differential equations.

Origins · 1 minJean le Rond d'Alembert and how this was discovered

Jean le Rond d'Alembert

1717–1783 · the wave equation and the vibrating-string controversy

Abandoned on the steps of the Paris church of Saint-Jean-le-Rond as an infant, Jean le Rond d'Alembert rose to become a towering figure of the French Enlightenment — mathematician, physicist, and co-editor with Diderot of the Encyclopédie. In a 1747 memoir to the Berlin Academy, Recherches sur la courbe que forme une corde tendue mise en vibration, he wrote down the one-dimensional wave equation for a vibrating string and gave its general solution as the sum of two arbitrary functions, $u=f(x-ct)+g(x+ct)$ — the first general solution of a partial differential equation in the history of mathematics. He insisted these functions be given by single analytic expressions. Euler objected that a plucked string has a corner, and that reasonable initial shapes need not be analytic at all; Daniel Bernoulli then argued that every motion could instead be written as a sum of sinusoidal normal modes. The three-way dispute — travelling waves, arbitrary profiles, trigonometric series — could not be settled with the mathematics of the day and was resolved only when Fourier showed, decades later, that arbitrary functions really can be expanded in trigonometric series. d'Alembert's factorisation and his causal reading of the characteristics remain the cleanest entry point to hyperbolic PDEs, and his name still labels the wave operator $\Box=\partial_t^2-c^2\nabla^2$ .

History

1713

— Brook Taylor computes the fundamental mode of a vibrating string

1747

— d'Alembert publishes the wave equation and its solution u = f(x−ct) + g(x+ct)

1748

— Euler recasts the solution and argues arbitrary initial shapes are admissible

1753

— Daniel Bernoulli proposes the trigonometric normal-mode superposition

1822

— Fourier's analytic theory of heat vindicates the trigonometric-series view

1928

— Courant, Friedrichs, and Lewy formalize the domain of dependence and finite propagation

Going further

Texts. Stone and Goldbart, Mathematics for Physics, chapter on PDEs (wave equation and characteristics) · Arfken, Weber and Harris, Mathematical Methods for Physicists, 7th ed., PDE chapter · Riley, Hobson and Bence, Mathematical Methods for Physics and Engineering, 3rd ed., §20 · Courant and Hilbert, Methods of Mathematical Physics, Vol. II · Evans, Partial Differential Equations, §2.4. Key results to keep in mind: the chain-rule reduction yields the identity $u_{tt}-c^2u_{xx}=-4c^2u_{\xi\eta}$ ; the integral-combination step erases the reference point $x_0$ and constant $K$ from d'Alembert's formula; the check $\phi=\cos kx,\ \psi=0\Rightarrow u=\cos kx\cos kct$ confirms the formula solves the PDE; the struck-string residual is $aV_0/c$ and the guitar values are $c=427\ \text{m/s},\ f_1=329.6\ \text{Hz}$ ; odd/even extensions enforce Dirichlet/Neumann conditions respectively.

Prerequisites. The Wave Equation · Classification of Second-Order Linear PDEs · Method of Characteristics · Boundary Conditions and Well-Posedness. Leads to. Separation of Variables in Cartesian Coordinates · Fundamental Solutions and Green's Functions for PDEs · The Heat (Diffusion) Equation · Sturm–Liouville Theory and Eigenfunction Expansions.

Ask the tutor: Ask me to trace a specific pulse through a reflection off a fixed or free end, or to show how repeated images rebuild a string's normal modes.

Why does the wave equation have two arbitrary functions in its general solution, not two constants?

Because it is a partial differential equation, not an ordinary one. A second-order ODE like $\ddot y=-\omega^2 y$ has solutions fixed by two numbers (initial position and velocity). A second-order PDE is fixed by two functions — here the initial displacement $\phi(x)$ and velocity $\psi(x)$ , each a whole profile along the line. The general solution $u=f(x-ct)+g(x+ct)$ accordingly carries two arbitrary functions, one per characteristic family, and d'Alembert's formula reads $f$ and $g$ off from $\phi$ and $\psi$ . That is the structural difference between solving for a curve and solving for a field.

What exactly is a characteristic, and why do discontinuities travel along them?

A characteristic is a curve in the $x$ – $t$ plane along which the PDE degenerates — for the wave equation, the straight lines $x\mp ct=\text{const}$ of slope $\pm 1/c$ . They are the only curves across which a solution can be non-smooth while still solving the equation in a weak sense: a jump in a derivative can sit on a characteristic and propagate, but it cannot cross into the smooth region on either side. Physically they are the trajectories of signal fronts. This is why a corner in the initial string shape stays a corner forever, riding its characteristic undistorted, whereas the heat equation — which has no real characteristics — instantly smooths every kink.

Why does a localized velocity leave a permanent displacement, but a localized displacement does not?

It is the difference between the two terms of d'Alembert's formula. Displacement data enters through $\tfrac12[\phi(x-ct)+\phi(x+ct)]$ , which simply carries the shape away in two directions and leaves nothing behind. Velocity data enters through $\tfrac1{2c}\int_{x-ct}^{x+ct}\psi$ , an integral that accumulates all the velocity the expanding window has swept up. Once the window fully contains a localized kick, the integral stops changing and locks in a constant, nonzero value — a permanent step of height (total impulse) $/2c$ stretched between the outgoing fronts. A pluck springs back to two travelling copies; a struck string lifts a lasting plateau.

How can the infinite-line formula solve a finite string with boundaries?

By the method of images. A boundary condition can be enforced by reflecting the initial data across the boundary with the right symmetry and then applying the whole-line d'Alembert formula. A fixed (Dirichlet) end $u=0$ needs an odd extension, so the two travelling copies cancel at the wall and the reflected pulse comes back inverted; a free (Neumann) end $u_x=0$ needs an even extension, giving an upright reflection. For a string clamped at both $x=0$ and $x=L$ you extend oddly about both ends, which forces a $2L$ -periodic pattern of images — and summing them reproduces exactly the standing-wave normal modes of separation of variables. Two pictures, one motion.

Is d'Alembert's clean "no spreading" picture true in three dimensions?

No, and the way it fails is subtle. In three space dimensions Kirchhoff's formula still gives sharp wavefronts with no wake — a flash of light or a clap of sound reaches you as a clean pulse, which is Huygens' principle. But in two dimensions (ripples on a pond) the solution develops a trailing wake: the disturbance arrives, then decays slowly rather than switching off. Remarkably, sharp Huygens propagation holds cleanly only in odd spatial dimensions $\ge 3$ . The one-dimensional rigid-profile result is thus special, and even the three-dimensional case is "spherical spreading with a clean back edge," not truly "no spreading." The dimension of space genuinely changes the qualitative physics.

What does d'Alembert's solution have to do with relativity and causality?

Everything, structurally. Take the wave operator with $c$ the speed of light. Its characteristics are the light cones of spacetime, and the domain of dependence of an event is precisely its past light cone — the earlier events that could have sent it a signal. Finite propagation speed, which in the string is just "you cannot feel a pluck before it reaches you," becomes the relativistic statement that no influence outruns light. In quantum field theory it reappears as microcausality: field operators commute at spacelike separation, outside each other's light cones. d'Alembert's vibrating string already held the geometry Einstein would make law.