d'Alembert's Solution of the Wave Equation

How does a plucked guitar string's shape evolve after release, with no boundaries in sight? This question, posed on the infinite line, has an answer of striking simplicity. The disturbance splits into two identical copies of the initial shape. One travels left and one travels right, each at the wave speed. This article derives that formula rigorously from the The Wave Equation and Vibrating Strings model. It also proves this is the unique solution to the Cauchy problem on all of R\mathbb{R}. The idea matters because it is the first instance, in a linear PDE, of an explicit closed-form solution. This solution is built entirely from the equation's characteristic geometry. That is the same geometry used in First-Order Linear PDEs and the Method of Characteristics. Rather than separating variables or summing Fourier series, we factor the wave operator into two first-order transport operators. We then integrate each one directly. The result is not merely a formula to memorize. It is a lens revealing why waves propagate at finite speed. It shows why information travels along cones in spacetime. It also shows why the solution at a point depends only on a bounded interval of initial data. The big picture is this: the wave equation utt=c2uxxu_{tt} = c^2 u_{xx} admits a change of coordinates (x,t)↦(ξ,η)=(x−ct, x+ct)(x,t) \mapsto (\xi,\eta) = (x - ct,\, x+ct). This change diagonalizes it into uξη=0u_{\xi\eta} = 0. Solving that trivial equation and undoing the change of coordinates produces d'Alembert's formula. Every subsequent qualitative fact about the 1-D wave equation traces back to this one calculation. This includes finite propagation speed, domains of dependence, energy conservation, and well-posedness. This article delivers the full derivation via characteristic coordinates. It presents a rigorous existence-and-uniqueness theorem for the Cauchy problem, with careful hypotheses on smoothness. Two asymmetric worked examples follow: a non-symmetric pluck and a problem with nonzero initial velocity. A counterexample shows what fails without C2C^2 regularity. Finally, there is a discussion of how this circle of ideas foreshadows Domains of Dependence and Influence and Well-Posedness of PDE Problems.

Key takeaways
  • The 1-D wave equation utt=c2uxxu_{tt}=c^2u_{xx} factors into two transport operators along the characteristics x±ct=constx\pm ct=\text{const}.
  • The general solution is u(x,t)=F(x−ct)+G(x+ct)u(x,t)=F(x-ct)+G(x+ct): a right-mover plus a left-mover.
  • Matching initial position u(x,0)=ϕ(x)u(x,0)=\phi(x) and velocity ut(x,0)=ψ(x)u_t(x,0)=\psi(x) pins down FF and GG uniquely (up to a cancelling constant).
  • The result is d'Alembert's formula: u(x,t)=12[ϕ(x−ct)+ϕ(x+ct)]+12c∫x−ctx+ctψ(s) dsu(x,t)=\tfrac12[\phi(x-ct)+\phi(x+ct)]+\tfrac{1}{2c}\int_{x-ct}^{x+ct}\psi(s)\,ds.
  • The solution at (x,t)(x,t) depends only on data in [x−ct,x+ct][x-ct, x+ct]: this is finite propagation speed.
  • Key idea: diagonalize the wave operator via characteristic coordinates, then integrate twice.

Where this lives today

D'Alembert's traveling-wave decomposition is the prototype for finite propagation speed in hyperbolic PDE, a theme that reappears in the higher-dimensional wave equation, Maxwell's equations, and relativistic field theory, where signals cannot outrun a fixed maximal speed. In acoustics and music technology, the formula underlies physical models of string and wind instruments (digital waveguide synthesis decomposes a vibrating string's state into left- and right-traveling digital signals bouncing between simulated boundaries). In seismology and structural engineering, the same left/right (or outgoing/incoming) decomposition is used to model stress waves along cables, rods, and pipelines. In numerical analysis, characteristic-based schemes (e.g. upwind and Godunov-type methods for hyperbolic conservation laws) generalize the idea of tracking information along characteristic curves rather than differencing blindly across a grid. In control theory, boundary control of vibrating strings and transmission lines exploits the explicit traveling-wave representation to design feedback laws that absorb or reflect waves at endpoints. The formula is also the entry point to the broader study of Well-Posedness of PDE Problems, since it exhibits, in the simplest possible setting, existence, uniqueness, and continuous dependence on data — the three Hadamard conditions — and it is the direct ancestor of the sharper geometric statements found under Domains of Dependence and Influence, where the interval [x−ct,x+ct][x-ct,x+ct] generalizes to backward light cones in higher dimensions.

Ask the tutor: How does d'Alembert's formula change if the string has a forcing term, i.e. utt=c2uxx+f(x,t)u_{tt}=c^2u_{xx}+f(x,t)?

DefinitionThe Cauchy problem for the 1-D wave equation

Let c>0c>0 be a constant (the wave speed). The Cauchy problem on the whole line asks for a function u:R×[0,∞)→Ru : \mathbb{R}\times[0,\infty)\to\mathbb{R} satisfying

utt(x,t)=c2uxx(x,t),(x,t)∈R×(0,∞),u_{tt}(x,t) = c^2 u_{xx}(x,t), \qquad (x,t)\in\mathbb{R}\times(0,\infty),

subject to the initial conditions

u(x,0)=ϕ(x),ut(x,0)=ψ(x),x∈R.u(x,0)=\phi(x), \qquad u_t(x,0)=\psi(x), \qquad x\in\mathbb{R}.

A classical solution is a function u∈C2(R×[0,∞))u\in C^2(\mathbb{R}\times[0,\infty)) satisfying the PDE at every interior point and the initial conditions exactly. This requires ϕ∈C2(R)\phi\in C^2(\mathbb{R}) and ψ∈C1(R)\psi\in C^1(\mathbb{R}) for consistency of the boundary derivatives.

Intuition

The wave equation utt=c2uxxu_{tt}=c^2u_{xx} can be factored formally as an operator identity:

∂t2−c2∂x2=(∂t−c∂x)(∂t+c∂x).\partial_t^2 - c^2\partial_x^2 = (\partial_t - c\partial_x)(\partial_t + c\partial_x).

Each factor is a first-order transport operator, exactly the kind studied via characteristics for First-Order Linear PDEs and the Method of Characteristics. The operator ∂t+c∂x\partial_t + c\partial_x annihilates functions constant along lines x−ct=constx - ct = \text{const} (rightward characteristics), while ∂t−c∂x\partial_t - c\partial_x annihilates functions constant along x+ct=constx+ct=\text{const} (leftward characteristics). Applying one factor to the equation and using the method of characteristics on the result — which is exactly what the theorem below carries out — shows that any solution must have the form of a right-moving profile plus a left-moving profile, each rigidly translating at speed cc without changing shape. This is very different from, say, the heat equation, where an initial bump instantly smooths and spreads over all of space. Here, a bump splits into two half-height copies that simply glide apart, never distorting. The picture below shows the two families of characteristic lines in the (x,t)(x,t)-plane, and how a point of observation (x0,t0)(x_0,t_0) traces back to an interval of initial data — the geometric seed of finite propagation speed.

Figure
Characteristic lines through a point of observation
x t (x₀, t₀) x + ct = x₀ + ct₀ x − ct = x₀ − ct₀ x₀ − ct₀ x₀ + ct₀ domain of dependence

Central theorem

Theoremd'Alembert's Solution

Let c>0c>0. Suppose ϕ∈C2(R)\phi\in C^2(\mathbb{R}) and ψ∈C1(R)\psi\in C^1(\mathbb{R}). Then the Cauchy problem

utt=c2uxx  on R×(0,∞),u(x,0)=ϕ(x),ut(x,0)=ψ(x)u_{tt}=c^2u_{xx}\ \text{ on } \mathbb{R}\times(0,\infty), \qquad u(x,0)=\phi(x), \quad u_t(x,0)=\psi(x)

has a classical solution u∈C2(R×[0,∞))u\in C^2(\mathbb{R}\times[0,\infty)), and it is unique among classical solutions. It is given explicitly by

u(x,t)  =  12[ϕ(x−ct)+ϕ(x+ct)]  +  12c∫x−ctx+ctψ(s) ds.u(x,t) \;=\; \frac{1}{2}\bigl[\phi(x-ct)+\phi(x+ct)\bigr] \;+\; \frac{1}{2c}\int_{x-ct}^{x+ct}\psi(s)\,ds.
Proof

We proceed in three steps: (1) change coordinates to diagonalize the operator and derive the general solution of the PDE alone; (2) impose the initial conditions to solve for the two arbitrary profiles, obtaining the formula; (3) verify directly that the formula is a genuine C2C^2 classical solution, and separately prove uniqueness. The technique throughout is characteristic coordinates combined with direct integration and, for uniqueness, an energy method.

Step 1: General solution of the homogeneous equation. Suppose u∈C2(R×[0,∞))u \in C^2(\mathbb{R}\times[0,\infty)) solves utt=c2uxxu_{tt}=c^2u_{xx}. Introduce new coordinates

ξ=x−ct,η=x+ct,\xi = x - ct, \qquad \eta = x+ct,

so that x=ξ+η2x=\tfrac{\xi+\eta}{2}, t=η−ξ2ct=\tfrac{\eta-\xi}{2c}. Define v(ξ,η):=u(x(ξ,η),t(ξ,η))v(\xi,\eta) := u\bigl(x(\xi,\eta), t(\xi,\eta)\bigr); since (ξ,η)↦(x,t)(\xi,\eta)\mapsto(x,t) is an invertible linear (hence smooth, with smooth inverse) change of variables, v∈C2v\in C^2 wherever u∈C2u\in C^2, and derivatives transform by the chain rule:

ux=vξξx+vηηx=vξ+vη,ut=vξξt+vηηt=−cvξ+cvη.u_x = v_\xi \xi_x + v_\eta \eta_x = v_\xi + v_\eta, \qquad u_t = v_\xi \xi_t + v_\eta \eta_t = -c v_\xi + c v_\eta.

Differentiating again (using C2C^2 regularity to invoke equality of mixed partials, i.e. Clairaut's theorem, which is legitimate since v∈C2v\in C^2):

uxx=vξξ+2vξη+vηη,utt=c2(vξξ−2vξη+vηη).u_{xx} = v_{\xi\xi} + 2v_{\xi\eta} + v_{\eta\eta}, \qquad u_{tt} = c^2\bigl(v_{\xi\xi} - 2v_{\xi\eta}+v_{\eta\eta}\bigr).

Substituting into utt=c2uxxu_{tt}=c^2u_{xx}:

c2(vξξ−2vξη+vηη)=c2(vξξ+2vξη+vηη)⟹−4c2vξη=0.c^2(v_{\xi\xi}-2v_{\xi\eta}+v_{\eta\eta}) = c^2(v_{\xi\xi}+2v_{\xi\eta}+v_{\eta\eta}) \quad\Longrightarrow\quad -4c^2 v_{\xi\eta}=0.

Since c≠0c\neq 0, this gives vξη=0v_{\xi\eta}=0 identically. Now integrate: fixing η\eta and integrating ∂ξ(vη)=0\partial_\xi(v_\eta)=0 in ξ\xi shows vη(ξ,η)v_\eta(\xi,\eta) is independent of ξ\xi, i.e. vη(ξ,η)=g(η)v_\eta(\xi,\eta)=g(\eta) for some function gg. This step uses that v∈C2v\in C^2, so vηv_\eta is C1C^1 and the fundamental theorem of calculus applies along each line η=const\eta=\text{const}. Integrating vη=g(η)v_\eta = g(\eta) in η\eta (with ξ\xi fixed) gives

v(ξ,η)=G(η)+F(ξ)v(\xi,\eta) = G(\eta) + F(\xi)

for some function FF (a constant of integration depending on ξ\xi) and GG an antiderivative of gg. Since v∈C2(R2)v\in C^2(\mathbb{R}^2), one shows F,G∈C2(R)F, G \in C^2(\mathbb{R}): indeed F(ξ)=v(ξ,η0)−G(η0)F(\xi) = v(\xi,\eta_0) - G(\eta_0) for any fixed η0\eta_0, exhibiting FF as a C2C^2 function of ξ\xi directly, and symmetrically for GG. Translating back,

u(x,t)=F(x−ct)+G(x+ct),F,G∈C2(R).u(x,t) = F(x-ct) + G(x+ct), \qquad F,G\in C^2(\mathbb{R}).

This proves that every C2C^2 solution of the wave equation on R×[0,∞)\mathbb{R}\times[0,\infty) (or any open subset where the characteristic coordinates cover it, which for the full plane they do) has this traveling-wave form. Conversely, for any F,G∈C2(R)F,G\in C^2(\mathbb{R}), direct differentiation shows u(x,t)=F(x−ct)+G(x+ct)u(x,t)=F(x-ct)+G(x+ct) solves utt=c2uxxu_{tt}=c^2 u_{xx}: ut=−cF′(x−ct)+cG′(x+ct)u_t = -cF'(x-ct)+cG'(x+ct), utt=c2F′′(x−ct)+c2G′′(x+ct)u_{tt}=c^2F''(x-ct)+c^2G''(x+ct), while ux=F′(x−ct)+G′(x+ct)u_x = F'(x-ct)+G'(x+ct), uxx=F′′(x−ct)+G′′(x+ct)u_{xx}=F''(x-ct)+G''(x+ct), and these agree after multiplying by c2c^2. So the wave equation's classical solutions on R×[0,∞)\mathbb{R}\times[0,\infty) are exactly the functions of this form.

Step 2: Imposing initial data. Set t=0t=0 in u=F(x−ct)+G(x+ct)u=F(x-ct)+G(x+ct):

u(x,0)=F(x)+G(x)=ϕ(x).(i)u(x,0) = F(x)+G(x) = \phi(x). \tag{i}

Differentiate uu in tt and set t=0t=0:

ut(x,0)=−cF′(x)+cG′(x)=ψ(x)⟹−F′(x)+G′(x)=ψ(x)c.(ii)u_t(x,0) = -cF'(x) + cG'(x) = \psi(x) \quad\Longrightarrow\quad -F'(x)+G'(x) = \frac{\psi(x)}{c}. \tag{ii}

Since ψ∈C1\psi\in C^1, ψ/c\psi/c is continuous, so we may integrate (ii) from 00 to xx:

−F(x)+G(x)=−F(0)+G(0)+1c∫0xψ(s) ds.(ii’)-F(x)+G(x) = -F(0)+G(0) + \frac1c\int_0^x \psi(s)\,ds. \tag{ii'}

Adding and subtracting (i) and (ii'), and writing A:=F(0),B:=G(0)A := F(0), B:=G(0) with A+B=ϕ(0)A+B=\phi(0):

G(x)=ϕ(x)2+12c∫0xψ(s) ds+G(0)−F(0)2,G(x) = \frac{\phi(x)}{2} + \frac{1}{2c}\int_0^x\psi(s)\,ds + \frac{G(0)-F(0)}{2},
F(x)=ϕ(x)2−12c∫0xψ(s) ds−G(0)−F(0)2.F(x) = \frac{\phi(x)}{2} - \frac{1}{2c}\int_0^x\psi(s)\,ds - \frac{G(0)-F(0)}{2}.

The constant G(0)−F(0)2\tfrac{G(0)-F(0)}{2} cancels when we form u(x,t)=F(x−ct)+G(x+ct)u(x,t)=F(x-ct)+G(x+ct), because it appears with opposite signs in FF and GG. Explicitly,

u(x,t)=F(x−ct)+G(x+ct)=ϕ(x−ct)+ϕ(x+ct)2+12c[∫0x+ctψ(s) ds−∫0x−ctψ(s) ds].u(x,t) = F(x-ct)+G(x+ct) = \frac{\phi(x-ct)+\phi(x+ct)}{2} + \frac{1}{2c}\left[\int_0^{x+ct}\psi(s)\,ds - \int_0^{x-ct}\psi(s)\,ds\right].

The bracketed term equals ∫x−ctx+ctψ(s) ds\int_{x-ct}^{x+ct}\psi(s)\,ds by additivity of the integral. This yields exactly d'Alembert's formula, and confirms it is the only possible C2C^2 solution — this establishes uniqueness among classical solutions, since Step 1 showed every classical solution has the traveling-wave form, and Step 2 showed the initial data determine F,GF,G uniquely up to the cancelling constant.

Step 3: Direct verification that the formula is a classical solution. We must check the regularity and the two initial conditions directly, since Steps 1–2 assumed a solution exists and derived what it must be; existence requires exhibiting a genuine C2C^2 solution. Define Ψ(x):=∫0xψ(s) ds\Psi(x):=\int_0^x \psi(s)\,ds. Since ψ∈C1(R)\psi\in C^1(\mathbb{R}), the fundamental theorem of calculus gives Ψ∈C2(R)\Psi\in C^2(\mathbb{R}) with Ψ′=ψ∈C1\Psi'=\psi\in C^1. Then

u(x,t)=ϕ(x−ct)+ϕ(x+ct)2+Ψ(x+ct)−Ψ(x−ct)2c.u(x,t) = \frac{\phi(x-ct)+\phi(x+ct)}{2} + \frac{\Psi(x+ct)-\Psi(x-ct)}{2c}.

Since ϕ∈C2\phi\in C^2 and Ψ∈C2\Psi\in C^2, and (x,t)↦x±ct(x,t)\mapsto x\pm ct is smooth, uu is a C2C^2 function of (x,t)(x,t) on all of R×[0,∞)\mathbb{R}\times[0,\infty) (composition of C2C^2 functions with smooth linear maps, using the chain rule repeatedly). By the converse direction established in Step 1 (any function of the form F(x−ct)+G(x+ct)F(x-ct)+G(x+ct) with F,G∈C2F,G\in C^2 solves the PDE), and noting

F(y):=ϕ(y)2−Ψ(y)2c∈C2,G(y):=ϕ(y)2+Ψ(y)2c∈C2,F(y) := \frac{\phi(y)}{2} - \frac{\Psi(y)}{2c}\in C^2, \qquad G(y):=\frac{\phi(y)}{2}+\frac{\Psi(y)}{2c}\in C^2,

we conclude utt=c2uxxu_{tt}=c^2u_{xx} holds identically. Checking initial conditions: at t=0t=0,

u(x,0)=ϕ(x)+ϕ(x)2+Ψ(x)−Ψ(x)2c=ϕ(x).✓u(x,0) = \frac{\phi(x)+\phi(x)}{2} + \frac{\Psi(x)-\Psi(x)}{2c} = \phi(x). \checkmark

Differentiating in tt,

ut(x,t)=−cϕ′(x−ct)+cϕ′(x+ct)2+cΨ′(x+ct)+cΨ′(x−ct)2c,u_t(x,t) = \frac{-c\phi'(x-ct)+c\phi'(x+ct)}{2} + \frac{c\Psi'(x+ct)+c\Psi'(x-ct)}{2c},

so at t=0t=0,

ut(x,0)=0+ψ(x)+ψ(x)2=ψ(x).✓u_t(x,0) = 0 + \frac{\psi(x)+\psi(x)}{2} = \psi(x). \checkmark

This confirms existence. Combined with the uniqueness shown in Steps 1–2, the theorem is proved. ■\blacksquare

∎

Worked examples

Example 1 — an asymmetric pluck, zero initial velocity. Take c=2c=2 and initial displacement

ϕ(x)={x+1,−1≤x≤0,1−x/3,0<x≤3,0,otherwise,ψ(x)≡0.\phi(x) = \begin{cases} x+1, & -1\le x\le 0,\\ 1-x/3, & 0< x\le 3,\\ 0, & \text{otherwise},\end{cases} \qquad \psi(x)\equiv 0.

(Check continuity: at x=0x=0, both pieces give 11; at x=−1x=-1, ϕ=0\phi=0; at x=3x=3, ϕ=0\phi=0. Good — ϕ\phi is continuous and piecewise linear, a triangular bump of height 11, skewed with a short left ramp and a long right ramp.) Since ψ≡0\psi\equiv 0 the integral term vanishes, and d'Alembert's formula gives

u(x,t)=ϕ(x−2t)+ϕ(x+2t)2.u(x,t) = \frac{\phi(x-2t)+\phi(x+2t)}{2}.

Evaluate at a specific asymmetric point, say x=1x=1, t=1t=1: we need ϕ(x−2t)=ϕ(−1)\phi(x-2t)=\phi(-1) and ϕ(x+2t)=ϕ(3)\phi(x+2t)=\phi(3). From the piecewise formula, ϕ(−1)=−1+1=0\phi(-1) = -1+1 = 0 and ϕ(3)=1−3/3=0\phi(3) = 1-3/3 = 0. So u(1,1)=0+02=0u(1,1) = \tfrac{0+0}{2}=0. Now evaluate at x=0.5, t=0.5x=0.5,\ t=0.5: x−2t=−0.5∈[−1,0]x-2t = -0.5\in[-1,0], so ϕ(−0.5)=−0.5+1=0.5\phi(-0.5)=-0.5+1=0.5; x+2t=1.5∈(0,3]x+2t=1.5\in(0,3], so ϕ(1.5)=1−1.5/3=0.5\phi(1.5)=1-1.5/3=0.5. Thus

u(0.5,0.5)=0.5+0.52=0.5.u(0.5,0.5) = \frac{0.5+0.5}{2} = 0.5.

Note this is not obtainable by symmetry — the two arguments −0.5-0.5 and 1.51.5 lie on genuinely different pieces of ϕ\phi (left ramp vs. right ramp), yet happen to coincide numerically, illustrating that the formula mechanically averages whatever values ϕ\phi takes at the two characteristic feet, regardless of shape. As tt increases past 0.50.5 (with c=2c=2), the left-moving front x+2t=3x+2t=3 crosses x=0.5x=0.5 around t=1.25t=1.25, after which the right-hand term becomes permanently 00; the two humps (each of half-height, half-width copies of ϕ\phi) separate completely once 2t2t exceeds the support radius, here after t≥2t \ge 2.

Example 2 — zero initial displacement, nonzero (asymmetric) initial velocity. Take c=1c=1 and ϕ≡0\phi\equiv 0, with

ψ(x)={2,0≤x≤1,−1,1<x≤4,0,otherwise.\psi(x) = \begin{cases} 2, & 0\le x\le 1,\\ -1, & 1< x\le 4,\\ 0,& \text{otherwise.}\end{cases}

(This ψ\psi is only piecewise continuous, not C1C^1 — we use it here for a clean explicit computation of the formula, understanding the result as a weak/generalized solution; see the Scope table below for the regularity subtlety.) d'Alembert's formula reduces to

u(x,t)=12∫x−tx+tψ(s) ds.u(x,t) = \frac12\int_{x-t}^{x+t}\psi(s)\,ds.

Let's compute u(2,3)u(2,3), an asymmetric choice with the interval [x−t,x+t]=[−1,5][x-t,x+t]=[-1,5] straddling all three pieces of ψ\psi non-symmetrically. Split the integral:

∫−15ψ(s) ds=∫−100 ds+∫012 ds+∫14(−1) ds+∫450 ds=0+2−3+0=−1.\int_{-1}^{5}\psi(s)\,ds = \int_{-1}^0 0\,ds + \int_0^1 2\,ds + \int_1^4 (-1)\,ds + \int_4^5 0\,ds = 0 + 2 - 3 + 0 = -1.

So u(2,3)=−12=−0.5u(2,3) = \tfrac{-1}{2} = -0.5. Compare with u(0,0.5)u(0,0.5): interval is [−0.5,0.5][-0.5, 0.5], overlapping only the first piece for s∈[0,0.5]s\in[0,0.5]:

∫−0.50.5ψ(s) ds=∫00.52 ds=1,u(0,0.5)=0.5.\int_{-0.5}^{0.5}\psi(s)\,ds = \int_0^{0.5} 2\,ds = 1, \qquad u(0,0.5) = 0.5.

These two values (−0.5-0.5 and 0.50.5) differ in sign and magnitude relationship purely because of how the asymmetric velocity profile is sampled over each differently-positioned interval — exactly the point of choosing a non-symmetric ψ\psi: no shortcut symmetry argument could have predicted either value without carrying out the integral.

Counterexamplewhy C2C^2 initial data is needed for a classical solution

Drop the hypothesis ϕ∈C2(R)\phi \in C^2(\mathbb{R}) and instead take c=1c=1, ψ≡0\psi\equiv 0, and

ϕ(x)=∣x∣.\phi(x) = |x|.

This ϕ\phi is continuous and even Lipschitz, but fails to be differentiable at x=0x=0, hence certainly not C2C^2. d'Alembert's formula still makes sense as an expression:

u(x,t)=∣x−t∣+∣x+t∣2.u(x,t) = \frac{|x-t|+|x+t|}{2}.

Consider the point (0,1)(0,1) (with t>0t>0 fixed, looking at uu as a function of xx near x=0x=0): for xx near 00 and t=1t=1,

u(x,1)=∣x−1∣+∣x+1∣2.u(x,1) = \frac{|x-1|+|x+1|}{2}.

For ∣x∣<1|x|<1, x−1<0x-1<0 and x+1>0x+1>0, so u(x,1)=(1−x)+(1+x)2=1u(x,1) = \frac{(1-x)+(1+x)}{2} = 1, a constant — smooth here, no problem at this particular point. The failure instead shows up along the characteristic emanating from the singularity of ϕ\phi. Track the line x=tx = t (a right-moving characteristic from x=0x=0): along it, u(t,t)=∣t−t∣+∣t+t∣2=0+2t2=tu(t,t) = \frac{|t-t|+|t+t|}{2} = \frac{0+2t}{2}=t, which is smooth in tt. But approach the kink locus x=tx=t from the two sides in xx at fixed t=1t=1: for xx slightly greater than 11, u(x,1)=(x−1)+(x+1)2=xu(x,1) = \frac{(x-1)+(x+1)}{2} = x, so ux=1u_x = 1; for xx slightly less than 11 (but >−1>-1), u(x,1)=1u(x,1)=1 as computed above, so ux=0u_x=0. The one-sided derivatives ux(1−,1)=0≠1=ux(1+,1)u_x(1^-,1)=0 \ne 1 = u_x(1^+,1) disagree, so uxu_x has a jump discontinuity across the line x=1x=1 in the (x,t)(x,t)-plane (which is exactly the characteristic x−t=0x-t = 0 emanating from the original kink at (0,0)(0,0)). Consequently uu is continuous and even Lipschitz everywhere, and solves the wave equation in a distributional/weak sense, but u∉C2(R×[0,∞))u\notin C^2(\mathbb{R}\times[0,\infty)): the second derivatives uxxu_{xx} and uttu_{tt} fail to exist classically on the characteristic lines through the kink. So the formula's output is only a weak solution, not a classical one, precisely because the hypothesis ϕ∈C2\phi\in C^2 was dropped. This shows the regularity hypothesis in the theorem is not a technical nicety — it is exactly what is needed for utt=c2uxxu_{tt}=c^2u_{xx} to hold in the classical (pointwise, second-derivative) sense everywhere, and its absence produces exactly the phenomenon (persistent, propagating singularities along characteristics) that the theorem's hypotheses were designed to exclude.

Scope, edge cases & common traps

SituationWhat happens
ϕ∈C2\phi \in C^2, ψ∈C1\psi\in C^1Classical C2C^2 solution exists and is unique (theorem above).
ϕ\phi merely continuous / Lipschitz, or ψ\psi merely continuousFormula still defines a function, interpretable as a weak solution; singularities in the data propagate along characteristics forever (see counterexample) and never smooth out.
Bounded domain with boundary conditionsd'Alembert's formula in this exact closed form does not apply directly; one needs reflection/extension tricks or Domains of Dependence and Influence methods, or Fourier series methods for finite intervals.
ψ\psi not integrable on some interval (e.g. has a non-integrable singularity)The formula's integral term may fail to exist; hypotheses on ψ\psi (continuity, or at least local integrability) are essential.
Asking about uniqueness among all functions, not just C2C^2The theorem only claims uniqueness among classical (C2C^2) solutions; without that restriction, weak solutions may coincide with d'Alembert's formula but the proof method (Step 1) does not directly apply, since it used C2C^2 regularity essentially (Clairaut's theorem).

The mathematician behind it

Origins · 1 minJean-Baptiste le Rond d'Alembert and how this was discovered
JL
Jean-Baptiste le Rond d'Alembert
1717–1783 · Age of Enlightenment, Paris

D'Alembert was a French mathematician, physicist, and philosopher, one of the great polymaths of the Enlightenment and a principal editor of the Encyclopédie. In 1747 he presented to the Prussian Academy of Sciences his solution to the problem of the vibrating string, publishing the traveling-wave decomposition that now bears his name and thereby giving mathematics its first fully worked example of a partial differential equation solved in closed form. He derived the equation of motion for a vibrating string from mechanical first principles and observed that the equation is satisfied by any function of x−ctx-ct or x+ctx+ct, a fact that ignited a decades-long controversy with Euler and Daniel Bernoulli over what kinds of functions (in the 18th-century sense of "function") were legitimately admissible as initial data — a dispute that anticipated, by a century, the modern theory of function spaces and generalized solutions.

History
- 1743 — d'Alembert publishes his Traité de dynamique, laying groundwork in mechanics.
- 1747 — d'Alembert solves the vibrating string problem, presenting the traveling-wave decomposition to the Berlin Academy.
- 1749–1750s — Euler and Daniel Bernoulli dispute the class of admissible initial shapes, debating whether "arbitrary curves" (not given by a single analytic expression) can be legitimate solutions.
- 1807 — Fourier's work on heat conduction reframes the debate, eventually vindicating a much broader notion of "function" than d'Alembert or Euler had assumed.
- 20th century — the notion of weak/distributional solutions (Sobolev, Schwartz) finally resolves rigorously what kind of initial data the wave equation genuinely admits, closing the loop opened in 1747.
Why does the solution split into exactly two traveling waves, not three or more?

This follows from the algebraic factorization of the wave operator: ∂t2−c2∂x2\partial_t^2 - c^2\partial_x^2 factors as a product of exactly two commuting first-order operators, (∂t−c∂x)(∂t+c∂x)(\partial_t-c\partial_x)(\partial_t+c\partial_x). Each first-order factor, when set to annihilate a function via the method of characteristics, forces that function to be constant along one family of characteristic lines — there are only two independent characteristic directions for a second-order hyperbolic operator in one spatial dimension, namely slope +c+c and slope −c-c in the (x,t)(x,t)-plane. The proof shows any C2C^2 solution must equal F(x−ct)+G(x+ct)F(x-ct)+G(x+ct); no third independent term can appear because the general solution of vξη=0v_{\xi\eta}=0 has exactly two free functions, one for each integration in ξ\xi and one in η\eta.

Does d'Alembert's formula work if the wave speed cc is not constant?

No, not directly. The entire derivation relies on the characteristic lines x±ct=constx\pm ct=\text{const} being straight lines, which follows from cc being a fixed constant. If c=c(x)c=c(x) varies with position, the equation utt=c(x)2uxxu_{tt}=c(x)^2u_{xx} still has characteristics, but they become curves defined by the ODE dx/dt=±c(x)dx/dt=\pm c(x), generally not straight lines, and the simple closed-form averaging structure of d'Alembert's formula is lost. One can still use the method of characteristics to reduce the PDE to ODEs along these curves, but an explicit formula like the one here is rarely available in closed form.

How does this formula relate to conservation of energy for the vibrating string?

The energy of a vibrating string, E(t)=∫R12(ut2+c2ux2) dxE(t) = \int_{\mathbb{R}} \tfrac12(u_t^2 + c^2 u_x^2)\,dx, can be shown to be constant in time using d'Alembert's formula directly: writing u=F(x−ct)+G(x+ct)u=F(x-ct)+G(x+ct), one computes ut=−cF′(x−ct)+cG′(x+ct)u_t = -cF'(x-ct)+cG'(x+ct) and ux=F′(x−ct)+G′(x+ct)u_x = F'(x-ct)+G'(x+ct), and after expanding ut2+c2ux2u_t^2+c^2u_x^2 the cross terms cancel, leaving 2c2[F′(x−ct)2+G′(x+ct)2]2c^2[F'(x-ct)^2+G'(x+ct)^2]. Integrating over xx and substituting shows the total is time-independent (a change of variables y=x∓cty=x\mp ct leaves each integral unchanged). This is a first hint of the energy methods used more generally to prove uniqueness and well-posedness in higher-dimensional and bounded-domain wave problems.

What happens at points where the two characteristic feet x−ctx-ct and x+ctx+ct land in different smoothness regions of a piecewise-defined ϕ\phi?

Nothing pathological happens as long as ϕ\phi is globally C2C^2; the formula is evaluated pointwise using whatever values ϕ\phi and Ψ\Psi take at those two points, regardless of "which piece" they came from, since ϕ\phi itself is a single well-defined C2C^2 function. The subtlety in the worked examples and the counterexample arises only when ϕ\phi (or its derivatives) fails to be continuous or differentiable at the junction between pieces — then the formula is still computable as a number, but the resulting uu may fail to be twice differentiable at points on the characteristic lines through that junction, so it solves the PDE only weakly there.

Is the solution to the Cauchy problem stable under small perturbations of the initial data?

Yes, and this is a key part of well-posedness in the sense of Hadamard. If ϕ,ϕ~\phi,\tilde\phi are both C2C^2 with ∥ϕ−ϕ~∥∞<ε\|\phi-\tilde\phi\|_\infty < \varepsilon and ψ,ψ~\psi,\tilde\psi are C1C^1 with ∥ψ−ψ~∥∞<δ\|\psi-\tilde\psi\|_\infty <\delta, then directly from the formula, ∣u(x,t)−u~(x,t)∣≤ε+δ t|u(x,t)-\tilde u(x,t)| \le \varepsilon + \delta\, t for all (x,t)(x,t), since the formula is linear in the data and the averaging/integration operations are bounded by the sup-norms times fixed geometric factors. This continuous dependence on initial data — degrading only linearly in tt, not catastrophically — is exactly the third Hadamard condition explored more fully under Well-Posedness of PDE Problems.

Why is finite propagation speed considered a big deal, given it seems geometrically obvious from the formula?

It seems obvious in hindsight because we've already derived the closed-form solution, but it is a substantive physical and mathematical fact: it says information genuinely cannot travel faster than cc, in stark contrast to, say, the heat equation, where a localized initial disturbance influences every point in space instantaneously (for any t>0t>0, however small). The wave equation's finite propagation speed is what makes it compatible with relativistic causality, and the Domains of Dependence and Influence framework generalizes this exact observation — that u(x,t)u(x,t) depends only on ϕ,ψ\phi,\psi within [x−ct,x+ct][x-ct,x+ct] — to variable-coefficient and higher-dimensional hyperbolic equations, where it becomes a genuinely deep geometric statement about light cones.

Can this method be adapted to a bounded interval, like a string fixed at both ends?

Not directly with this exact formula, but the underlying idea survives via the method of images (reflection): one extends ϕ\phi and ψ\psi oddly (for fixed/Dirichlet ends) or evenly (for free/Neumann ends) across the boundary points to obtain data on the whole line, applies d'Alembert's formula there, and restricts back to the original interval. This works because the odd/even extensions automatically enforce the correct boundary conditions once one checks the reflected data still satisfies the necessary consistency (C2C^2 matching) at the boundary, a subtlety related to the domain-of-dependence ideas developed after this formula.

Before this

Where this leads