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These characteristic curves are straight when wave speed $c$ is constant. A right-moving component is constant on $x-ct=s$, and a left-moving component on $x+ct=s$. The full superposition generally is not constant on either family, because the other component can vary along it. Their backwards intersections with the initial line select the endpoints used in D'Alembert's formula.

For the wave equation $u_{tt}=c^2{u}_{xx}$ with zero initial displacement and initial velocity $g$, D'Alembert's formula becomesFor $c=1$ this is the original speed-one form $u(t,x)=\frac{1}{2}{\int }_{x-t}^{x+t}f(s)ds$ when the initial velocity is named $f$. The integral records the total initial kick inside the domain of dependence.

For a computed example, set $g(x)=V$ on $[-a,a]$ and zero elsewhere, with $V>0$. This discontinuous datum gives a weak solution:At the centre, $u(0,t)=V\min (t,a/c)$. For $a=1\mathrm{m}$, $c=2\mathrm{m}{\mathrm{s}}^{-1}$ and $V=0.4\mathrm{m}{\mathrm{s}}^{-1}$, this is $0.1\mathrm{m}$ at $t=0.25\mathrm{s}$ and $0.2\mathrm{m}$ at $t=1\mathrm{s}$. Once $ct>a$, a central plateau $u=Va/c$ occupies $|x|\le ct-a$. In contrast, central zero region for odd compactly supported initial velocity follows when the full integral of $g$ vanishes. Finite propagation speed permits an interior tail; it does not imply the Strong Huygens principle.

For real constants $\gamma ,c$, the substitution $v=e^{\gamma t}u$ turns the displayed equation into the wave equation $v_{tt}=c^2{v}_{xx}$. If $u(x,0)=f(x)$ and $u_t(x,0)=g(x)$, then $v(x,0)=f(x)$ and $v_t(x,0)=g(x)+\gamma f(x)$. For $v_0=|c|>0$, D'Alembert's formula consequently givesThe extra $\gamma f$ term is necessary even when the original initial velocity is zero. For $c=0$, take the limit to obtain $u=e^{-\gamma t}[f+t(g+\gamma f)]$. The formula gives classical solutions for suitably smooth data and weak solutions for less regular data. Positive $\gamma$ damps the transformed wave, whereas negative $\gamma$ amplifies it.

For the unforced, constant-$c$ one-dimensional wave equation, finite propagation speed says that changing Cauchy data outside $[x_{\ast }-c{t}_{\ast },x_{\ast }+c{t}_{\ast }]$ cannot change the solution at $(x_{\ast },t_{\ast })$, where $c>0$ and $t_{\ast }\ge 0$. This interval is the initial-data domain of dependence. Its endpoints are found by following the two characteristic lines for the one-dimensional wave equation backwards from the observation. D'Alembert's formula makes the dependence sharper: initial displacement is sampled at the endpoints, while initial velocity is integrated over the entire interval.^[ref]

For the synthetic choice $c=1$, $(x_{\ast },t_{\ast })=(0.5,1.5)$ has initial-data domain of dependence $[-1,2]$. A change confined near $x=3$ is irrelevant to that observation, however large it is. If both Cauchy data vanish outside $[-a,a]$, their compact support can influence only $[-a-ct,a+ct]$ at $t\ge 0$. Thus an observer at $x>a$ sees exactly zero displacement for $0\le t<(x-a)/c$. There is no nonzero precursor beyond this causal cone for the constant-speed wave equation in the ideal model. This is an exact statement about the partial differential equation, stronger than an experimental failure to detect a small signal.

Support widening for the wave equation is a bound on possible influence, not a claim that every point inside the expanding interval moves. A bump function released with zero initial velocity splits into two separated travelling waves; the middle can become empty. An initial kick can instead leave an interior tail, as in the D'Alembert formula with initial velocity. Consequently finite propagation speed does not imply the Strong Huygens principle. A finite string needs boundary conditions, and the whole-line formula applies locally only until its backward causal cone for the constant-speed wave equation encounters an endpoint.

An applied mathematics educator whose partial differential equation notes derive the wave equation on a string from a physical model and then connect its solutions with initial and boundary data. D'Alembert's formula decomposes an initial disturbance into profiles traveling in opposite directions. The construction lets a reader predict how an initially plucked string evolves before accounting for reflections at its ends.^[ref]

For zero initial velocity and initial displacement $h(x)$ on $x\ge 0$ with $h(0)=0$, let $H$ be its odd extension to the real line. The displayed D'Alembert's formula solves the wave equation with $y(0,t)=0$, since $H(-ct)+H(ct)=0$. A pulse travelling toward the endpoint reflects with reversed displacement. For piecewise linear data the equation holds classically between moving corners and in the sense of weak derivatives globally.

On the whole line, the homogeneous one-dimensional wave equation with constant wave speed $c>0$ isHere $u$ is displacement in the string interpretation; $f$ is initial displacement and $g$ is initial velocity. Its two directions of travel giveThis is D'Alembert's formula, including both Cauchy data. The countertravelling decomposition of the one-dimensional wave equation identifies $F(x-ct)$ as the right-moving travelling wave and $G(x+ct)$ as the left-moving travelling wave. Their sum is permitted by the superposition principle. For $f\in C^2$ and $g\in C^1$, it is a classical solution. More generally, travelling waves with profiles $F$ and $G$ that are locally integrable functions solve the equation as distributions, with derivatives interpreted distributionally.^[ref]

The synthetic diagram uses the smooth bump function $b$ defined in D'Alembert's formula, with $f(x)=b(x)$ and $c=1$. Releasing this displacement from rest, $g=0$, gives $F=G=f/2$. Each emerging travelling wave has half the original peak. Launching the identical initial displacement solely to the right instead requires $g=-c{f}^{\prime }$ and gives $F=f$, $G=0$. At $t=1.5$, the released profiles have peaks $1/2$ at $x=\pm 1.5$, while the launched profile has peak $1$ at $x=1.5$. These are calculated model values. A purely left-moving launch requires $g=c{f}^{\prime }$. The initial velocity therefore selects the balance of directions; the initial shape alone does not.

For a separated solution $y=X(x)Q(t)$, the wave equation gives $X^{\prime \prime }/X=\ddot{Q}/(v^{2}Q)$. Choose the separation constant $-k^2$. ThenThese are the oscillatory separated building blocks. Since no endpoint conditions have yet been imposed, the general solution is a superposition over the allowed continuum of $k$, rather than one product with four constants. Equivalently, combining the left- and right-travelling separated components gives the D'Alembert's formulawith arbitrary twice differentiable functions $F,G$ on the relevant characteristic intervals. Indeed, in characteristic coordinates $\xi =x-vt$, $\eta =x+vt$, the equation becomes $y_{\xi \eta }=0$, whose general solution is exactly this sum. This also includes zero-frequency and polynomial limits that would be omitted by restricting to one nonzero-$k$ sine mode.

Take the constant speed parameter $v\ne 0$, so $r=x+vt$ and $s=x-vt$ are independent characteristic coordinates. The chain rule givesApplying these operators twice to a twice differentiable function yieldsConsequently the wave equation reduces to $4F_{rs}=0$. Integrate first with respect to $s$, obtaining $F_r=A(r)$, and then with respect to $r$, allowing a separate integration function of $s$. The general local solution iswith arbitrary sufficiently differentiable functions $G$ and $H$. For $v>0$, the first profile travels towards negative $x$ and the second towards positive $x$. Substitution directly checks the equation. Initial or boundary conditions select the two profiles; none are specified here. This is D'Alembert's formula in characteristic form.

Assume $v\ne 0$ so that the change of variables is invertible. The chain rule gives ${\partial }_x={\partial }_r+{\partial }_s$ and ${\partial }_t=v({\partial }_r-{\partial }_s)$. Commuting the mixed partial derivatives for a twice continuously differentiable solution givesThus the wave equation is $F_{rs}=0$. Integrating once says that $F_r$ depends only on $r$; integrating again givesThe two arbitrary twice differentiable functions are waves moving in opposite directions at speed $|v|$. This is D'Alembert's formula, on a product domain in $(r,s)$ or globally on the usual full line.

Let the mass per unit length be $m$ and the constant tension be $T$. Small transverse displacement obeys the wave equationWith characteristic coordinates $\xi =x+ct$ and $\eta =x-ct$, the equation becomes $y_{\xi \eta }=0$. Integrating once in each variable gives D'Alembert's formulaFor a smooth solution with vanishing endpoint energy flux, multiplication by $y_t$ and integration by parts giveThe fixed endpoint has $y_t(0,t)=0$. Decay of displacement alone does not guarantee either a pointwise derivative-flux limit or finiteness of the energy integral; the conservation statement is understood in the usual finite-energy class. The profile formula below proves it directly there without requiring such a pointwise limit at infinity.

The endpoint condition gives $g(-ct)=-f(ct)$, allowing the reflected wave to be expressed through one profile:For forward time one may define the negative-argument part of this profile using the original $g$; no restriction on the independent initial waves is lost. Differentiation gives $y_t=c[f^{\prime }(x+ct)-f^{\prime }(ct-x)]$ and $y_x=f^{\prime }(x+ct)+f^{\prime }(ct-x)$. HenceThe two substitutions partition the real line at $ct$, making conservation of energy explicit.

Enforce the fixed-end Dirichlet boundary condition by taking the odd extension of the initial velocity to the whole line:This agrees with the specified initial condition for $s\ge 0$. To derive the relevant form of D'Alembert's formula, the zero initial displacement in b gives $G=-F$, up to a cancelling constant. The initial velocity then gives $-2c{F}^{\prime }=v_o$. HenceAn antiderivative, valid across $s=0$, isEvaluation therefore gives the solution throughout $x>0,t>0$:For $ct\le x$, replace $|x-ct|$ by $x-ct$; for $ct\ge x$, replace it by $ct-x$. Thus the same expression includes the reflected wave after it reaches the fixed end.

At $x=0$ the two terms agree, so $y(0,t)=0$. At $t=0$ they again agree, so $y(x,0)=0$. Differentiating the integral with respect to $t$ giveswhich recovers $y_t(x,0)=\kappa x{e}^{-x/L}$. Since $v_o$ is a continuously differentiable function, the formula is twice continuously differentiable even across $x=ct$, and satisfies the wave equation there as well.

Write $u(x,t)=U(\xi ,\eta )$, where $\xi =x-t$ and $\eta =x+t$. The chain rule givesFor a twice continuously differentiable classical solution, mixed partial derivatives commute, so the wave equation becomesHence $U_{\xi }$ depends only on $\xi$. Integrating first in $\xi$ yields the countertravelling decomposition of the one-dimensional wave equation:On a rectangular domain in $(\xi ,\eta )$ this follows immediately; on the full line for $t\ge 0$, the connected characteristic slices give the same representation. The two terms travel at wave speed one in opposite directions.

Impose the initial conditions. At $t=0$,Differentiate the first identity and combine:If $H(x)={\int }_{x_{\ast }}^{x}g(s)ds$, then $F=(f-H)/2+C$ and $G=(f+H)/2-C$. Substitution cancels $C$ and the arbitrary base point $x_{\ast }$, giving D'Alembert's formula:For example, $f\in C^2$ and $g\in C^1$ make this a classical solution of the wave equation satisfying both initial conditions. The same formula defines a weak solution for less regular data, including the continuous piecewise smooth data of part (b).

Since the initial velocity vanishes, D'Alembert's formula reduces toEach translated copy has half the original amplitude. To identify each requested cross-section, track the translated support intervals and add the values where they overlap.

At $t=\pi$, $f(x-\pi )$ is supported on $[-\pi ,0]\cup [2\pi ,3\pi ]$ and $f(x+\pi )$ on $[-3\pi ,-2\pi ]\cup [0,\pi ]$. On these intervals $\sin (x\pm \pi )=-\sin x$, soThe pieces meet continuously at zero, where both contributions vanish.

At $t=3\pi /2$, the support of $f(x-3\pi /2)$ is $[-\pi /2,\pi /2]\cup [5\pi /2,7\pi /2]$; that of $f(x+3\pi /2)$ is $[-7\pi /2,-5\pi /2]\cup [-\pi /2,\pi /2]$. In the central overlap,so the two contributions cancel. The remaining wave profile isFor the time cross-section at the origin, the initial profile is an odd function, soFor the time cross-section at $x=\pi /2$, the left-travelling contribution is nonzero for $\pi /2\le t\le 3\pi /2$ and equals $\cos t/2$. The right-travelling contribution is nonzero for $3\pi /2\le t\le 5\pi /2$ and also equals $\cos t/2$. At the common endpoint both vanish. HenceThe following graphs show all four cross-sections. In particular, the last trace reaches $-1/2$ at $t=\pi$ and $1/2$ at $t=2\pi$.

Because $u(0,t)=0$, restriction to the half-line solves the following initial-boundary value problem with a homogeneous Dirichlet boundary condition:This is wave reflection by odd extension. If a condition at infinity is desired, take the solution with finite wave energy; here the profile has bounded support at every finite time.

The initial profile is continuous and piecewise continuously differentiable, but its first derivative jumps at the support endpoints. Thus the formula gives a weak solution globally and a piecewise classical solution away from the characteristic lines for the one-dimensional wave equation issuing from those endpoints. The initial velocity is zero in the square-integrable function sense and pointwise wherever the initial profile is differentiable; at the kink points a pointwise right time derivative need not equal zero. This regularity qualification is needed for the given data.

The printed arguments $x-t$ and $x+t$ assume unit wave speed. For the stated wave equation with general $c>0$, the correct arguments are $x-ct$ and $x+ct$, as in the paper's subsequent formula. Equivalently, use rescaled time $\tilde{t}=ct$ to obtain the printed unit-speed form.

Introduce coordinates $\xi =x-ct$, $\eta =x+ct$ along the characteristic lines for the one-dimensional wave equation, and write $u(x,t)=U(\xi ,\eta )$. The chain rule givesso the wave equation becomesThus $U_{\xi }$ is independent of $\eta$, and integration gives the countertravelling decomposition of the one-dimensional wave equationConversely, differentiating this expression twice verifies the wave equation for any sufficiently differentiable $F,G$.

The two initial conditions giveDifferentiating the first and solving these two equations for the derivatives givesFor a fixed reference point $x_0$, their antiderivatives can be writtenSubstituting the two travelling arguments cancels the constants and combines the integrals. This proves D'Alembert's formula with initial velocity:At $t=0$ the integral vanishes, giving $u=f$; differentiating by the fundamental theorem of calculus gives $u_t=g$ there. These checks also fix the sign and the factor $1/(2c)$.

Here $f=0$ and $ct=1$, so D'Alembert's formula givesOnly the overlap $[1/2,1]$ with the support of the initial velocity contributes. ThereforeThe relevant interval is determined by the two characteristic lines for the one-dimensional wave equation through the observation point.

For $c>0$ and $L>0$, D'Alembert's formula for the wave equation on the full real line isFor the specified initial data, define the cumulative initial-velocity integralThen $\boxed{y(x,t)=[A(x+ct)-A(x-ct)]/(2c)}$. The initial velocity jumps at $\pm L$, so the resulting solution is interpreted piecewise classically, or as a weak wave-equation solution across the corresponding characteristic curves.

If $\psi =f(\xi )+g(\eta )$ with twice differentiable $f,g$, its mixed partial derivative vanishes because each summand depends on only one coordinate. Conversely ${\psi }_{\xi \eta }=0$ means ${\psi }_{\xi }$ is independent of $\eta$; integration in $\xi$ supplies $f(\xi )$ and an integration function $g(\eta )$. Hence, on a characteristic rectangle, the general solution isThese terms travel in opposite directions without changing profile, as in D'Alembert's formula.

Expanding gives $u=2x^2+2c^2{t}^2$, so $u_{tt}=4c^2$ and $u_{xx}=4$. Hence $u_{tt}=c^2{u}_{xx}$, proving that it solves the one-dimensional wave equation. Equivalently its two summands are oppositely travelling profiles, as in D'Alembert's formula.

Use separation of variables, $f=X(x)T(t)$, and setFor $\kappa =-{\lambda }^2$, with $\lambda >0$, the separated solutions areFor $\kappa =0$, they areFor $\kappa =+{\lambda }^2$, they areArbitrary admissible superpositions of separated solutions are again solutions. An individual product is only a separated family, not the entire general solution. Without prescribed boundaries, the general twice-differentiable solution is equivalently D'Alembert's formula,For example, changing to $\xi =x-ct$, $\eta =x+ct$ makes the differential equation $f_{\xi \eta }=0$, whose two integrations give this form.