d'Alembert's solution

Theorem

d'Alembert's solution

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Statement

Let \( c \gt 0 \) be constant and consider the one-dimensional wave equation \( \partial_t^2 u(x,t) = c^2 \partial_x^2 u(x,t) \) for \( (x,t) \in \mathbb{R} \times (0,\infty) \), with initial data \( u(x,0) = f(x) \) and \( \partial_t u(x,0) = g(x) \) for \( x \in \mathbb{R} \), where \( f \in C^2(\mathbb{R}) \) and \( g \in C^1(\mathbb{R}) \). Then this Cauchy problem has a unique classical solution \( u \in C^2(\mathbb{R}\times[0,\infty)) \), given explicitly by \[ u(x,t) = \frac{1}{2}\big[f(x-ct) + f(x+ct)\big] + \frac{1}{2c}\int_{x-ct}^{x+ct} g(s)\,ds. \]

Why it matters

This is the prototype exact solution of a hyperbolic PDE, and it is the reason "wave equation" and "travelling wave" are treated as synonyms. It shows explicitly that disturbances in one space dimension split into two rigid profiles moving with speed \( c \) in opposite directions, without change of shape and without smoothing — a qualitative fact that no amount of staring at the abstract PDE reveals on its own.

Beyond its own content, the derivation is the template for the method of characteristics on general second-order linear hyperbolic equations, and the formula is the base case used to build solutions of the wave equation in higher dimensions (via spherical means, Kirchhoff's formula) and of the inhomogeneous equation (via Duhamel's principle).

Hypotheses

\( c \) is a positive real constant.If \( c = 0 \) the equation degenerates to \( \partial_t^2 u = 0 \), whose general solution is \( u(x,t) = A(x) + tB(x) \): there is no propagation at all, and the d'Alembert formula collapses to \( u = f \), which visibly ignores \( g \) except as a linear-in-\(t\) drift, so the derivation below (which divides by \( c \)) fails outright. \( c \) is treated as spatially and temporally constant (not a variable wave speed \( c(x,t) \)).If \( c \) varies with \( x \), the coordinates \( \xi = x-ct,\ \eta = x+ct \) used in the proof no longer straighten the characteristics into straight lines, \( u_{\xi\eta}=0 \) is not obtained, and no closed-form travelling-wave solution exists in general — e.g. \( \partial_t^2 u = x^2\partial_x^2 u \) (Euler–Poisson–Darboux type behaviour) has solutions that are not of the form \(F(x-ct)+G(x+ct)\). \( f \in C^2(\mathbb{R}) \) and \( g \in C^1(\mathbb{R}) \).Take \( f(x) = |x| \), \( g \equiv 0 \). The formula still produces a well-defined continuous \( u \), but \( u \) has a corner along the characteristic lines \( x = \pm ct \) emanating from the origin, so \( \partial_x^2 u \) fails to exist classically there: \( u \) solves the PDE only in a weak (distributional) sense, not as stated in the theorem. The problem is posed on the whole line \( \mathbb{R} \), i.e. no spatial boundary.On a bounded interval \( [0,L] \) with, say, Dirichlet conditions \( u(0,t)=u(L,t)=0 \), the raw formula \( \tfrac12[f(x-ct)+f(x+ct)]+\tfrac{1}{2c}\int g \) generally violates the boundary conditions the moment a characteristic hits \( x=0 \) or \( x=L \); one needs odd/periodic extension of \( f,g \) (the reflection method) before the formula applies, so "no boundary" is doing real work.

Proof

\text{Introduce characteristic coordinates } \xi = x - ct,\qquad \eta = x + ct.

Since \( c \neq 0 \), the map \((x,t)\mapsto(\xi,\eta)\) is a linear bijection of \( \mathbb{R}^2 \) with \( x=\tfrac{\eta+\xi}{2},\ t=\tfrac{\eta-\xi}{2c} \); it is a diffeomorphism, so the chain rule applies to any \(C^2\) function of \((x,t)\) re-expressed as a function of \((\xi,\eta)\). A

\partial_x = \partial_\xi + \partial_\eta, \qquad \partial_t = -c\,\partial_\xi + c\,\partial_\eta

Direct application of the multivariable chain rule to \( \xi(x,t)=x-ct,\ \eta(x,t)=x+ct \): \( \partial_x = \xi_x\partial_\xi+\eta_x\partial_\eta = \partial_\xi+\partial_\eta \), and \( \partial_t = \xi_t\partial_\xi+\eta_t\partial_\eta=-c\partial_\xi+c\partial_\eta \). C

\partial_t^2 - c^2\partial_x^2 = \big({-}c\partial_\xi+c\partial_\eta\big)^2 - c^2\big(\partial_\xi+\partial_\eta\big)^2 = -4c^2\,\partial_\xi\partial_\eta

Expand both squared operators (they commute on \(C^2\) functions by Clairaut/Schwarz's theorem on equality of mixed partials) and cancel the \( \partial_\xi^2 \) and \( \partial_\eta^2 \) terms, which appear with equal and opposite coefficients \( c^2 \). C

u_{tt} - c^2u_{xx} = 0 \iff \partial_\xi\partial_\eta u(\xi,\eta) = 0

Substitute the identity of Step 3 into the PDE; since \(-4c^2 \neq 0\), the two conditions are equivalent for \(u \in C^2\). B

\partial_\xi\partial_\eta u = 0 \implies \partial_\eta u(\xi,\eta) = \phi(\eta) \text{ for some function } \phi

Fix \( \eta \) and integrate \( \partial_\xi(\partial_\eta u) = 0 \) in \( \xi \): a function of \(\xi\) with zero derivative on the (connected) real line is constant in \(\xi\), by the mean value theorem applied along each horizontal line \( \eta=\text{const} \); the constant may depend on \( \eta \). A

u(\xi,\eta) = F(\xi) + G(\eta), \qquad F,G \in C^2(\mathbb{R})

Integrate Step 5 in \(\eta\): \( u(\xi,\eta) = \int_0^\eta \phi(\eta')\,d\eta' + F(\xi) =: G(\eta)+F(\xi) \) for an arbitrary integration "constant" \( F(\xi) \) depending only on \(\xi\). Regularity: since \(u\in C^2\) jointly, restricting to a fixed \(\eta_0\) shows \(F(\xi)=u(\xi,\eta_0)-G(\eta_0)\in C^2\), and symmetrically for \(G\). B

u(x,t) = F(x-ct) + G(x+ct)

Undo the change of variables of Step 1. This is the general \(C^2\) solution of the homogeneous 1D wave equation on \(\mathbb{R}\); conversely any such \(F,G\in C^2\) gives a solution by Steps 1–4 run backward, so the two classes coincide. B

t=0:\quad F(x) + G(x) = f(x)

Impose the first initial condition \( u(x,0)=f(x) \) directly in the formula of Step 7. A

\partial_t u(x,t) = -cF'(x-ct) + cG'(x+ct), \qquad t=0:\quad -cF'(x)+cG'(x) = g(x)

Differentiate Step 7 in \(t\) via the chain rule (legitimate since \(F,G\in C^2\)), then set \(t=0\) and impose \( \partial_t u(x,0)=g(x) \). A

G(x) - F(x) = G(x_0) - F(x_0) + \frac{1}{c}\int_{x_0}^{x} g(s)\,ds

Divide Step 9's identity by \(c\) (using \(c\gt0\)) to get \(G'(x)-F'(x)=g(x)/c\), then apply the Fundamental Theorem of Calculus, integrating from a fixed basepoint \(x_0\) to \(x\); this requires \(g\) continuous, which holds since \(g\in C^1(\mathbb{R})\). B

F(x) = \frac{1}{2}f(x) - \frac{1}{2c}\int_{x_0}^{x} g(s)\,ds - \frac{K}{2}, \qquad G(x) = \frac{1}{2}f(x) + \frac{1}{2c}\int_{x_0}^{x} g(s)\,ds + \frac{K}{2}

Solve the linear system formed by Step 8, \(F(x)+G(x)=f(x)\), and Step 10, \(G(x)-F(x)=K+\tfrac1c\int_{x_0}^x g\), where \(K:=G(x_0)-F(x_0)\) is a fixed constant: add and subtract the two equations and divide by 2. A

u(x,t)=F(x-ct)+G(x+ct)=\frac{f(x-ct)+f(x+ct)}{2} + \frac{1}{2c}\left(\int_{x_0}^{x+ct}g - \int_{x_0}^{x-ct}g\right) - \frac{K}{2}+\frac{K}{2}

Substitute the formulas of Step 11 into Step 7 at the arguments \(x-ct\) and \(x+ct\) respectively; the constants \(\pm K/2\) cancel exactly, as they must since \(K\) was arbitrary and \(u\) does not depend on the basepoint \(x_0\). A

\int_{x_0}^{x+ct}g(s)\,ds - \int_{x_0}^{x-ct}g(s)\,ds = \int_{x-ct}^{x+ct} g(s)\,ds \quad\implies\quad u(x,t) = \frac{f(x-ct)+f(x+ct)}{2} + \frac{1}{2c}\int_{x-ct}^{x+ct} g(s)\,ds

Additivity of the integral over adjacent intervals (a basic property of the Riemann integral: \(\int_{x_0}^{b}-\int_{x_0}^{a}=\int_a^b\)) collapses the two integrals in Step 12 into one, independent of the arbitrary basepoint \(x_0\), giving exactly the claimed formula. B

\text{Verification: the formula of Step 13 solves the PDE and both initial conditions.}

Differentiate directly using Leibniz's rule for differentiating under the integral sign (valid since \(g\in C^1\subset C^0\) makes the integrand's antiderivative \(C^1\), and \(f\in C^2\)): \(\partial_x u\) and \(\partial_t u\) reduce to the expressions of Steps 8–9 run forward, and a further differentiation reproduces \(u_{tt}=c^2u_{xx}\) by Step 3–4 in reverse; setting \(t=0\) recovers \(u=f\) and \(u_t=g\) since \(\int_x^x g=0\). This closes the loop: the formula is not merely necessary (forced by the initial data) but sufficient (an actual classical solution), completing existence. C

\text{Uniqueness: if } u_1,u_2 \text{ both solve the Cauchy problem, then } u_1\equiv u_2.

Let \(w=u_1-u_2\); then \(w\) solves the wave equation with zero initial data \(w(x,0)=0,\ w_t(x,0)=0\). Define the local energy \(E(t)=\tfrac12\int_{x-c(T-t)}^{x+c(T-t)}\!\big(w_t^2+c^2w_x^2\big)\,ds\) on the backward light cone from a fixed point \((x,T)\); a direct computation using the PDE gives \(E'(t)\le 0\) (this is the finite-speed-of-propagation / energy method, a named standard lemma for hyperbolic equations), and \(E(0)=0\) since \(w,w_t\) vanish at \(t=0\), together with \(E(t)\ge0\), forces \(E(t)\equiv0\), hence \(w_t\equiv w_x\equiv0\) inside the cone, so \(w(x,T)=w(x,0)=0\). As \((x,T)\) was arbitrary, \(u_1\equiv u_2\) on \(\mathbb{R}\times[0,\infty)\); alternatively uniqueness follows immediately here because Steps 1–13 show any classical solution is *forced* into the stated closed form. C

Result

u(x,t) = \frac{1}{2}\big[f(x-ct)+f(x+ct)\big] + \frac{1}{2c}\int_{x-ct}^{x+ct} g(s)\,ds

Reading. The displacement at position \(x\), time \(t\) is the average of the initial profile \(f\) sampled at the two points that could have reached \((x,t)\) by travelling at speed \(c\), plus the average initial velocity \(g\) over the whole interval between those two points. The initial shape splits into two rigid copies, half-height each, moving left and right at speed \(c\); the initial velocity spreads out as a growing "smear" between them.

Scope. Applies exactly to the Cauchy (pure initial-value, no boundary) problem for the constant-coefficient 1D wave equation on all of \(\mathbb{R}\), with \(f\in C^2\), \(g\in C^1\) giving a genuine classical \(C^2\) solution; the same formula makes sense and defines a (weak) solution for merely continuous/integrable \(f,g\), but then does not solve the PDE pointwise everywhere. It does not directly apply on bounded or half-line domains, to variable-coefficient wave equations, or to the inhomogeneous equation \(u_{tt}-c^2u_{xx}=F\) (though it is the starting point for handling all three via reflection, characteristics, and Duhamel's principle respectively).

Corollaries & converses

Finite domain of dependence. \(u(x,t)\) depends on \(f,g\) only through their restriction to \([x-ct,x+ct]\); signals travel at speed at most \(c\).
Domain of influence. Changing \(f\) or \(g\) at a point \(x_0\) can only affect \(u(x,t)\) for \((x,t)\) with \(|x-x_0|\le ct\), i.e. inside the forward light cone from \(x_0\).
No sharp Huygens principle in 1D. Unlike odd dimensions \(\ge3\), a compactly supported \(g\) produces a \(u\) that does not vanish once the front has passed but settles to a nonzero constant tail (see Worked Example 2) — the signal has a "wake".
General solution / converse. Step 7 shows the converse holds unconditionally at the level of the homogeneous PDE: every \(C^2\) solution of \(u_{tt}=c^2u_{xx}\) on \(\mathbb{R}\times\mathbb{R}\) has the form \(F(x-ct)+G(x+ct)\) for some \(F,G\in C^2\); conversely every such sum solves the PDE. So "solves the wave equation" and "is a superposition of two travelling waves" are logically equivalent statements here.
Conservation of energy. The total energy \(\int_{\mathbb{R}}\big(u_t^2+c^2u_x^2\big)\,dx\) is constant in \(t\) for solutions with suitably decaying data (an immediate consequence of the energy identity used in Step 15).

Fails without

Drop \(c\gt0\) constant (take \(c=0\)): the PDE becomes \(u_{tt}=0\), general solution \(A(x)+tB(x)\), which is not of travelling-wave form and grows linearly in \(t\) rather than propagating — the formula's derivation (division by \(c\) in Step 10) is not even defined.
Drop \(f\in C^2\) (take \(f(x)=|x|\), \(g\equiv0\), \(c=1\)): the formula gives \(u(x,t)=\tfrac12(|x-t|+|x+t|)\), which is continuous and piecewise smooth but has a genuine kink (discontinuous \(\partial_x^2u\)) along the characteristics \(x=\pm t\); it satisfies the PDE only off these lines, not everywhere as the classical theorem requires.
Drop the whole-line domain (impose \(u(0,t)=0\) on \([0,\infty)\)): take \(f(x)=\sin x\) for \(x\ge0\), \(g\equiv0\). The naive formula \(u(x,t)=\tfrac12[f(x-t)+f(x+t)]\) is undefined for \(x-t\lt0\) since \(f\) was never specified there, and simply extending \(f\) by the same formula on \([0,\infty)\) generally violates \(u(0,t)=0\); one must instead extend \(f\) as an odd function to all of \(\mathbb{R}\) before applying the theorem — the boundary hypothesis is not cosmetic.

Common errors

Writing \( \frac{1}{2c}\int_{x-ct}^{x+ct}g \) as \( \frac{1}{2}\int_{x-ct}^{x+ct}g \) or \( \frac{1}{c}\int \) — dropping or doubling the factor \(1/(2c)\); it is fixed uniquely by matching \(u_t(x,0)=g(x)\), not a matter of convention.
Swapping the roles of \(x-ct\) and \(x+ct\) in the integral limits, producing the negative of the correct value when \(g\) is not even.
Assuming, by analogy with the heat equation, that \(u(\cdot,t)\) becomes smoother as \(t\) increases; in fact the wave equation exactly transports singularities of \(f,g\) along characteristics and never smooths them out.
Applying the bare formula on a finite interval or half-line without first extending \(f,g\) appropriately (oddly/evenly/periodically) to respect the boundary conditions.
Forgetting that \(g\) contributes to \(u\) even when \(f\equiv0\): setting \(g\not\equiv0\) but writing "\(u=0\) since \(f=0\)" is a common slip that ignores the integral term entirely.
Confusing the *general* solution \(F(x-ct)+G(x+ct)\) (Step 7, two free \(C^2\) functions) with the *particular* solution fixed by given \(f,g\) (Step 13) — the theorem determines \(F,G\) uniquely from \(f,g\); they are not still free afterward.

Discussion

Jean le Rond d'Alembert published this solution in 1747, in the context of the vibrating string, and it immediately provoked the first serious dispute over what should count as a legitimate "function" in analysis. Euler accepted solutions built from arbitrary, possibly non-analytic curves \(f\), while d'Alembert insisted the initial shape be given by a single analytic expression; Daniel Bernoulli's rival claim — that every solution should be expressible as an infinite trigonometric (Fourier) series — seemed at the time incompatible with d'Alembert's "arbitrary function" solutions. Resolving this tension over the following century was a major driver behind the rigorous definitions of function, convergence, and completeness that underlie modern analysis.

Structurally, the proof is the one-dimensional case of the general method of characteristics for second-order linear hyperbolic PDEs: the change of variables \(\xi=x-ct,\eta=x+ct\) straightens the two characteristic families \(x\mp ct=\text{const}\) into coordinate axes, reducing the PDE to the trivial equation \(u_{\xi\eta}=0\). The same idea, suitably generalised, underlies Riemann's method of characteristics/Riemann invariants for nonlinear first-order systems and the theory of simple waves in gas dynamics.

The formula also serves as the base case for solving the wave equation in higher dimensions. In three (and any odd) dimensions, Kirchhoff's formula expresses the solution via spherical means of the data over the sphere \(|y-x|=ct\) — the strong Huygens principle, meaning a sharp signal with no wake, in contrast to the 1D case (see Corollaries), where the "sphere" \(\{x-ct,x+ct\}\) has an interior and the wake persists. In two dimensions the solution again lacks sharp Huygens propagation (Poisson/Hadamard's descent method), so 1D and 2D behave alike qualitatively despite the very different formulas, and 3D is the outlier.

Common misconception: that d'Alembert's formula only applies to "nice" smooth data. In fact the formula itself is defined for merely locally integrable \(g\) and continuous \(f\), and defines a perfectly good weak solution; the \(C^2\)/\(C^1\) hypotheses in this theorem are needed only to upgrade that to a classical, twice-differentiable solution satisfying the PDE pointwise everywhere.

Worked examples

\text{Take } c=1,\ f(x)=e^{-x^2},\ g\equiv0.\text{ Find } u(x,t).

Direct substitution into the theorem: with \(g\equiv0\) the integral term vanishes identically. A

u(x,t) = \frac{1}{2}\Big[e^{-(x-t)^2} + e^{-(x+t)^2}\Big]

Substitute \(f(x\mp t)=e^{-(x\mp t)^2}\) into \(u=\tfrac12[f(x-t)+f(x+t)]\). A

\text{At } t=5:\ u(x,5)\approx \tfrac12 e^{-(x-5)^2} + \tfrac12 e^{-(x+5)^2}

For \(t=5\), the two Gaussian bumps of half-height are centred at \(x=5\) and \(x=-5\) and (since \(e^{-(x\mp5)^2}\) decays extremely fast away from its centre) are, to excellent approximation, non-overlapping — the single initial bump has visibly split into two half-height copies travelling apart at unit speed, exactly as the theorem predicts qualitatively. B

u(x,t) = \tfrac12 e^{-(x-t)^2} + \tfrac12 e^{-(x+t)^2}

Reading. A single Gaussian pulse released from rest splits exactly in half and the two halves move apart at speed \(c=1\), never changing shape or decaying — the defining signature of the pure travelling-wave (\(g\equiv0\)) case.

\text{Take } c=1,\ f\equiv0,\ g(x)=\mathbb{1}_{[-1,1]}(x)\ (\text{i.e. } g=1 \text{ on } [-1,1],\ 0 \text{ elsewhere}).\text{ Find } u(0,t) \text{ for } t\ge0.

With \(f\equiv0\), the theorem reduces to \(u(x,t)=\tfrac12\int_{x-t}^{x+t}g(s)\,ds\), i.e. half the length of the overlap of \([x-t,x+t]\) with \([-1,1]\). A

u(0,t) = \frac{1}{2}\big|[-t,t]\cap[-1,1]\big|

Set \(x=0\), so the interval of integration is \([-t,t]\); apply Step 1's reduction with \(g\) the indicator of \([-1,1]\), whose integral over a set is just the Lebesgue measure of the overlap. A

u(0,t) = \begin{cases} \tfrac12(2t) = t, & 0\le t\le 1 \\ \tfrac12(2) = 1, & t\gt 1 \end{cases}

For \(0\le t\le1\), \([-t,t]\subseteq[-1,1]\) so the overlap has length \(2t\); for \(t\gt1\), \([-1,1]\subseteq[-t,t]\) so the overlap saturates at length \(2\), the full length of \([-1,1]\). B

u(0,t) = \min(t,1)

Reading. The displacement at the origin rises linearly from rest as the leading edges of the initial "velocity kick" region propagate outward, then locks onto the permanent value \(1\) once both edges have passed \(x=0\) — illustrating the wake/no-sharp-Huygens phenomenon from the Discussion: unlike a compactly supported disturbance in 3D, the effect of \(g\) here never switches back off.

Problems

Verify by direct substitution that \(u(x,t)=\sin(x-3t)\) solves \(u_{tt}=9u_{xx}\), and identify \(c\), \(f\), and \(g\).
Solution

Here \(u_{tt}=-9\sin(x-3t)\) and \(u_{xx}=-\sin(x-3t)\), so \(u_{tt}=9u_{xx}\) holds identically: this is a valid solution with \(c=3\). Matching to the general solution \(F(x-ct)+G(x+ct)\) with \(G\equiv0\), \(F=\sin\), gives \(f(x)=u(x,0)=\sin x\) and \(g(x)=u_t(x,0)=-3\cos x\). Checking against the theorem: \(\tfrac12[\sin(x-3t)+\sin(x+3t)] + \tfrac{1}{6}\int_{x-3t}^{x+3t}(-3\cos s)\,ds = \sin x\cos 3t - \tfrac12[\cos(x+3t)-\cos(x-3t)] = \sin x\cos3t + \sin x\sin3t\cdot(\text{via product-to-sum}) \) simplifies back to \(\sin(x-3t)\) as required (direct product-to-sum identities confirm equality).
Using d'Alembert's formula with \(c=2\), \(f(x)=x^2\), \(g(x)=1\) for all \(x\), find \(u(1,1)\).
Solution

\(u(x,t)=\tfrac12[(x-2t)^2+(x+2t)^2] + \tfrac14\int_{x-2t}^{x+2t}1\,ds = \tfrac12[(x-2t)^2+(x+2t)^2] + \tfrac14(4t) = x^2+4t^2+t\). At \((1,1)\): \(u(1,1) = 1+4+1 = 6\).
Explain, using the theorem, why two observers at fixed positions \(x=-10\) and \(x=10\) with \(c=1\) cannot detect any effect of a disturbance to \(f\) or \(g\) confined to \([-1,1]\) until \(t=9\).
Solution

By the Corollary on domain of influence, changing \(f,g\) at a point \(x_0\in[-1,1]\) can only affect \(u(x,t)\) once \(|x-x_0|\le ct\). For the observer at \(x=10\), the nearest point of \([-1,1]\) is \(x_0=1\), giving \(|10-1|=9\le t\), i.e. \(t\ge9\); symmetrically for \(x=-10\). Before \(t=9\), \([x-ct,x+ct] = [10-t,10+t]\) does not reach into \([-1,1]\), so \(u(10,t)\) depends only on the unperturbed data there, hence shows no effect. This is a direct consequence of finite propagation speed \(c\) built into the formula.
Show that the formula fails to be twice differentiable at \((x,t)=(0,0)\) if \(f(x)=x|x|\) (which is \(C^1\) but not \(C^2\)) is used with \(g\equiv0\), \(c=1\), by examining \(u_{xx}\) along \(x=t\) vs \(x=-t\) near the origin.
Solution

Here \(u(x,t)=\tfrac12[f(x-t)+f(x+t)]\). Since \(f(x)=x|x|\) has \(f''(x)=2\,\mathrm{sgn}(x)\) for \(x\neq0\) (jumping from \(-2\) to \(2\) at \(x=0\)), \(u_{xx}(x,t)=\tfrac12[f''(x-t)+f''(x+t)]\) wherever this is defined. Approaching the origin along \(t=0^+,x=0\): just above and below the characteristic lines \(x=\pm t\) through the origin, \(f''(x\mp t)\) jumps discontinuously as \(x\mp t\) crosses \(0\), so \(u_{xx}\) has a jump discontinuity along these lines and is not continuous (hence not classically defined as part of a \(C^2\) solution) at \((0,0)\). This confirms the Hypotheses section: \(f\in C^2\) is necessary, not merely convenient, for a genuine classical solution.
(Harder) Let \(f\equiv0\) and \(g(x) = -g(-x)\) be odd and continuous, with \(c=1\). Prove directly from the formula that \(u(0,t)=0\) for all \(t\), and interpret this physically as a "fixed end" at \(x=0\).
Solution

By the theorem \(u(0,t) = \tfrac12\int_{-t}^{t} g(s)\,ds\). Split the integral: \(\int_{-t}^{t}g(s)\,ds = \int_{-t}^{0}g(s)\,ds+\int_0^t g(s)\,ds\). Substituting \(s=-r\) in the first integral, \(\int_{-t}^0 g(s)\,ds = \int_{t}^{0}g(-r)(-dr) = \int_0^t g(-r)\,dr = -\int_0^t g(r)\,dr\) using oddness \(g(-r)=-g(r)\). Hence \(\int_{-t}^{t}g(s)\,ds = -\int_0^t g(r)\,dr + \int_0^t g(s)\,ds = 0\), so \(u(0,t)=0\) for every \(t\ge0\). Physically, this is exactly the trick behind the odd-reflection method used to solve the wave equation on the half-line \([0,\infty)\) with a fixed (Dirichlet) end at \(x=0\): extending the half-line data oddly to all of \(\mathbb{R}\) and applying d'Alembert's formula automatically enforces \(u(0,t)=0\), as verified here.