d'Alembert's General Solution

Derivation

d'Alembert's General Solution

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Statement

For the one-dimensional wave equation u_tt = c² u_xx on the whole line, every twice-differentiable solution can be written as u(x,t) = f(x − ct) + g(x + ct) for two arbitrary profiles f and g: a rigid right-moving shape superposed on a rigid left-moving shape, each translating at speed c without change of form. Conversely, any such sum solves the equation.

Why it matters

This is the complete general solution of the archetypal hyperbolic PDE. It says the wave equation does not merely permit travelling waves — it admits nothing else. Every disturbance on an ideal string, in an ideal acoustic tube, or along a lossless transmission line is exactly a pair of counter-propagating profiles, so understanding a single arbitrary pulse suffices to understand all one-dimensional wave motion.

It also gives the operational meaning of the "wave speed" c that emerged from the loaded-string equation: c is literally the translation speed of an undistorted shape. From here follow the characteristic coordinates, the domain of dependence, and d'Alembert's initial-value formula — the foundation of causality in classical field theory.

Assumptions

The medium obeys the linear, constant-coefficient wave equation.If the string tension or density varies with position, c becomes c(x), the equation no longer factors into constant-speed operators, and shapes distort as they propagate — no rigid f, g exist.

One spatial dimension, unbounded domain.In 2D or 3D the analogous decomposition fails (Huygens' principle differs, wakes appear); on a finite interval, boundary reflections mix left- and right-movers so the profiles are no longer independent.

No dispersion and no dissipation: ω = ck exactly for all k.Any k-dependence of phase speed (stiffness, resistance, a mass term as in Klein–Gordon) spreads a pulse out; the "undistorted profile" statement is the strongest thing that fails first.

The solution is C² (twice continuously differentiable).If one only wants u_tt = c²u_xx in a weak/distributional sense, then f and g may be merely continuous or even distributions; the classical derivation below then needs the weak-solution machinery to remain valid.

Derivation

u_tt − c² u_xx = 0

Starting point: the 1D wave equation, taken as given from the loaded-string result. A

u_tt − c² u_xx = ( ∂_t − c ∂_x )( ∂_t + c ∂_x ) u

The scalar wave operator factors like a difference of squares. Legal because c is constant, so ∂_t and c∂_x commute and the cross terms ±c∂_x∂_t cancel exactly. B

ξ = x − ct , η = x + ct

Introduce characteristic coordinates — the two families of lines along which each factor is constant. This is an invertible linear change of variables (Jacobian ∂(ξ,η)/∂(x,t) = −2c ≠ 0), so nothing is lost. B

∂_x = ∂_ξ + ∂_η , ∂_t = −c ∂_ξ + c ∂_η

Chain rule applied to the map of step 3, using ∂ξ/∂x = 1, ∂η/∂x = 1, ∂ξ/∂t = −c, ∂η/∂t = c. B

∂_t + c ∂_x = 2c ∂_η , ∂_t − c ∂_x = −2c ∂_ξ

Substitute step 4 into each first-order factor and collect. The two directional derivatives collapse onto the two characteristic variables individually — that is exactly what characteristics are chosen to do. B

( ∂_t − c ∂_x )( ∂_t + c ∂_x ) u = ( −2c ∂_ξ )( 2c ∂_η ) u = −4c² u_ξη

Compose the two operators from step 5. Legal because the coefficients ±2c are constants and pass through the derivatives. B

u_ξη = 0

The wave equation of step 1 equals −4c²u_ξη, and −4c² ≠ 0, so the mixed second derivative must vanish. This is the wave equation in its canonical (normal) form. A

∂_ξ ( u_η ) = 0 ⇒ u_η = Φ(η)

Integrate step 7 once in ξ. A quantity whose ξ-derivative is zero everywhere is independent of ξ, so u_η is an arbitrary function of η alone. Requires u ∈ C² so the mixed partials are equal and the integration is valid. C

u(ξ,η) = ∫ Φ(η) dη + f(ξ) = g(η) + f(ξ)

Integrate step 8 in η. The antiderivative of Φ defines an arbitrary g(η); the "constant" of this integration may depend on ξ, giving the arbitrary f(ξ). Two integrations of a second-order equation yield exactly two arbitrary functions. B

u(x,t) = f(x − ct) + g(x + ct)

Return to physical coordinates via step 3. Since the change of variables is invertible, this is the general solution — no solution has been excluded and, by direct substitution, every such sum satisfies step 1. A

Result

u(x,t) = f(x − ct) + g(x + ct)

Reading. The field at position x and time t is the sum of two independent shapes. The term f(x − ct) keeps its argument constant along x = ct + const, i.e. it slides rigidly to the right at speed c; g(x + ct) slides rigidly to the left at speed c. Neither profile changes height or width — only position. The two are fixed by the initial displacement and velocity of the medium.

Units check. The arguments x − ct and x + ct have units of length ([c][t] = (m s⁻¹)(s) = m), so f and g are ordinary functions of a length. Each returns whatever u measures — metres for a transverse string displacement, pascals for acoustic pressure — so the sum carries the units of u, as required.

Limiting cases

Pure right-mover: if g ≡ 0, then u = f(x − ct) and u_t = −c u_x — a single advection equation; the disturbance travels one way only.
Standing wave: f(ξ) = ½A cos kξ and g(η) = ½A cos kη sum to A cos kx cos ωt with ω = ck — two equal counter-movers make a stationary pattern.
Static limit (c → 0 or t = 0): u(x,0) = f(x) + g(x), the frozen initial shape; time-translation switches off.
Single harmonic: taking f = ½A e^ik(x−ct) recovers a plane wave e^i(kx−ωt), showing the Fourier modes are the special sinusoidal case of the general profile.

Breaks when

Dispersive media. If the dispersion relation is not ω = ck (e.g. a stiff string with ω² = c²k² + αk⁴, or Klein–Gordon ω² = c²k² + m²), different Fourier components move at different speeds. A pulse spreads and no fixed profiles f, g exist.
Variable or non-constant coefficients. With c = c(x) the operator does not factor into two constant-speed derivatives, characteristics curve, and a shape both distorts and partially reflects as it propagates through the gradient.
Higher dimensions. In 3D the general solution is not a sum of two rigid profiles; expanding spherical waves decay as 1/r and, in even dimensions, leave a trailing wake (failure of Huygens' principle).
Dissipation / driving. Adding a damping term γu_t or a source makes the equation inhomogeneous or non-conservative; amplitudes decay and the clean left/right split no longer holds.

Failure modes

Sign confusion in the argument. Writing f(x + ct) for the right-mover. The right-moving shape needs the argument x − ct so that increasing t requires increasing x to hold the profile fixed.
Thinking f and g must be sinusoidal. They are arbitrary C² functions — a square-ish bump, a ramp, anything. Sinusoids are just the eigen-basis, not a restriction.
Dropping one of the two arbitrary functions. Integrating u_ξη = 0 and keeping only g(η) loses the ξ-dependent integration "constant" f(ξ) — and with it half of every general solution.
Assuming the split is unique. One can add any constant to f and subtract it from g; the decomposition is unique only up to such a constant, fixed once initial data are imposed.
Confusing phase speed with signal speed here. For the non-dispersive wave equation they coincide at c; students carry over the "group vs phase" distinction where there is none, or fail to invoke it where dispersion is later added.

Discussion

The heart of the result is the factorisation ∂_tt − c²∂_xx = (∂_t − c∂_x)(∂_t + c∂_x). Each first-order factor is a transport (advection) operator: (∂_t + c∂_x)h = 0 is solved by any h = f(x − ct), a shape swept rightward at speed c. The wave equation is thus, quite literally, two transport equations composed, and its solution space is the direct sum of the two transport solution spaces. Characteristic coordinates are the change of variables that makes each factor act on one variable alone.

Physically, the two profiles carry the two pieces of initial data. The displacement u(x,0) and the velocity u_t(x,0) together determine f and g uniquely (up to the trivial constant), which is d'Alembert's formula u = ½[φ(x−ct)+φ(x+ct)] + (1/2c)∫_x−ct^x+ctψ(s)ds. The integral limits expose the domain of dependence: the value at (x,t) depends only on initial data in [x−ct, x+ct] — nothing outside that interval can have reached the point, which is causality made geometric.

The result also underwrites the method of images and normal modes on bounded domains. A boundary condition (fixed or free end) is satisfied by choosing g to be a reflected, possibly inverted, copy of f; the reflection is exactly the constraint that couples the otherwise-independent left- and right-movers. Standing waves are then the resonant superpositions selected by two such boundaries.

More abstractly, d'Alembert's solution is the statement that the 1D wave operator is hyperbolic with two distinct real characteristic families, and that its canonical form u_ξη=0 integrates in closed form. This is special to one space dimension: in n dimensions the characteristics form a cone, the solution is given by Kirchhoff/Poisson surface integrals rather than a two-function sum, and sharp propagation (Huygens' principle) holds only for odd n ≥ 3. The clean "two rigid profiles" picture is the lowest-dimensional, most degenerate — and most transparent — case of the general theory.

Common misconceptions. The formula is not an approximation or a special ansatz — it is the complete general solution, exhaustive for the ideal 1D equation. And "wave speed c" is not the speed of any particle of the medium (that is u_t, generally different and much smaller); c is the speed of the pattern.

Worked examples

Example 1 — A Gaussian pulse released from rest.

u(x,0) = φ(x) = A e^−x²/a² , u_t(x,0) = 0

Initial data: a symmetric bump of height A and width a, released with zero velocity. A

f(ξ) + g(ξ) = φ(ξ) , −c f′(ξ) + c g′(ξ) = 0

Impose the two initial conditions on u = f(x−ct)+g(x+ct) at t=0; the second gives f′ = g′, so f and g differ by a constant. B

f(ξ) = g(ξ) = ½φ(ξ)

Zero initial velocity splits the shape equally between the two movers. A

u(x,t) = ½A e^{−(x−ct)²/a²} + ½A e^{−(x+ct)²/a²}

Substitute back — the bump splits into two half-height Gaussians moving apart. A

A = 3 cm, a = 0.20 m, c = 12 m s⁻¹, t = 0.10 s ⇒ ct = 1.2 m

Insert numbers. Each half-pulse (height 1.5 cm, width 0.20 m) is now centred at x = ±1.2 m; their centres are 2.4 m apart, far larger than the width, so they are cleanly separated. A

u(x, 0.10 s) = 1.5 cm · [ e^{−(x−1.2)²/0.04} + e^{−(x+1.2)²/0.04} ] (x in m)

Reading. The single 3 cm bump has become two 1.5 cm bumps, one at x = +1.2 m and one at x = −1.2 m, each keeping the original 0.20 m width. Total displacement is conserved; the amplitude halved because energy went into two pulses.

Example 2 — A struck string (initial velocity, zero displacement).

u(x,0) = 0 , u_t(x,0) = ψ(x) = V for |x| < b, else 0

A hammer gives a length 2b of string a uniform transverse velocity V, with no initial displacement. A

u(x,t) = (1/2c) ∫_x−ct^x+ct ψ(s) ds

d'Alembert's formula with φ = 0; this is the f,g pair fixed by pure velocity data. Legal because ψ is integrable. B

u(x,t) = (V / 2c) × ( length of overlap of [x−ct, x+ct] with [−b, b] )

Evaluate the integral of the box function: it just measures how much of the moving window lies inside the struck region. B

u_max = (V / 2c)(2b) = Vb / c (reached once ct ≥ 2b at the centre)

The plateau value: once the characteristic window fully contains the struck segment, the displacement saturates. A

V = 0.80 m s⁻¹, b = 0.05 m, c = 40 m s⁻¹

Numbers for a lightly struck steel string. A

u_max = Vb/c = (0.80)(0.05)/40 = 1.0 × 10⁻³ m = 1.0 mm

Reading. The struck patch rises to a flat-topped plateau 1.0 mm high whose edges spread outward at c = 40 m s⁻¹ — a growing rectangular displacement, the velocity-driven counterpart of the splitting bump. Units: (m s⁻¹)(m)/(m s⁻¹) = m. ✓

Problems

Show by direct substitution that u = f(x − ct) satisfies u_tt = c²u_xx for any twice-differentiable f.
Solution
Let ξ = x − ct. Then u_x = f′(ξ), u_xx = f″(ξ). For time, u_t = −c f′(ξ), u_tt = c² f″(ξ). Hence u_tt = c² f″ = c² u_xx identically, for any f ∈ C². The identical argument with η = x + ct (which gives u_t = +c g′) confirms the left-mover, and linearity gives the sum.
Initial data u(x,0) = φ(x), u_t(x,0) = ψ(x). Derive f and g explicitly.
Solution
At t=0: f(x) + g(x) = φ(x) and −c f′(x) + c g′(x) = ψ(x). Integrate the second: −f(x) + g(x) = (1/c)∫₀^xψ(s)ds + K. Solve the pair: f(x) = ½φ(x) − (1/2c)∫₀^xψ − K/2 and g(x) = ½φ(x) + (1/2c)∫₀^xψ + K/2. Substituting into u = f(x−ct)+g(x+ct) the constants K cancel, giving d'Alembert's formula u = ½[φ(x−ct)+φ(x+ct)] + (1/2c)∫_x−ct^x+ctψ(s)ds.
A string has c = 50 m s⁻¹. A pulse of shape u(x,0) = h e^−x²/a² with h = 2 cm, a = 0.10 m is released from rest. Where are the two half-pulses at t = 20 ms, and what is each height?
Solution
Released from rest ⇒ each mover carries half: height h/2 = 1.0 cm, width unchanged at a = 0.10 m. Displacement of each centre: ct = 50 × 0.020 = 1.0 m. So one half-pulse is centred at x = +1.0 m, the other at x = −1.0 m, each 1.0 cm high. They are 2.0 m apart, hugely larger than 0.10 m, so fully separated.
Two identical right- and left-moving sinusoids f = ½A cos k(x−ct), g = ½A cos k(x+ct) superpose. Show the result is a standing wave and give its nodal spacing.
Solution
Sum: u = ½A[cos k(x−ct) + cos k(x+ct)]. Use cos(P) + cos(Q) = 2 cos(&frac{P+Q}{2}) cos(&frac{P−Q}{2}) with P = k(x−ct), Q = k(x+ct): u = A cos(kx) cos(ωt), ω = ck. This is separable in x and t — a standing wave. Nodes where cos kx = 0, i.e. kx = (n+½)π; adjacent nodes differ by Δx = π/k = λ/2.
The Klein–Gordon equation u_tt = c²u_xx − μ²u (μ ≠ 0). Show that a single undistorted travelling profile u = f(x−ct) is not in general a solution, and find its dispersion relation.
Solution
Try u = f(x−ct): u_tt = c²f″, c²u_xx = c²f″, so the equation demands c²f″ = c²f″ − μ²f, i.e. μ²f = 0. Since μ ≠ 0 this forces f ≡ 0 — no non-trivial rigid profile exists; the factorisation of the operator fails because of the −μ²u term. Instead try u = e^i(kx−ωt): −ω² = −c²k² − μ², giving ω² = c²k² + μ². The phase speed ω/k = c√(1 + μ²/c²k²) depends on k, so the medium is dispersive and pulses spread — exactly why d'Alembert's undistorted-profile picture is special to the μ = 0 wave equation.