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Let $H$ be any antiderivative of the initial velocity, $H^{\prime }=g$. One choice of profiles isOpposite additive constants in $F$ and $G$ cancel, so the displacement is independent of the chosen antiderivative. Even for compact support of $g$, these separate profiles can approach nonzero constants; cancellation in their sum is essential to support widening for the wave equation.

For a short derivation, the characteristic coordinates $\xi =x-ct$, $\eta =x+ct$ give $u_{tt}-c^2{u}_{xx}=-4c^2{u}_{\xi \eta }$ by the chain rule. Thus $u=F(\xi )+G(\eta )$, and the Cauchy data imply $F+G=f$ and $-c{F}^{\prime }+c{G}^{\prime }=g$, giving the profiles above.

For constant wave speed $c>0$, the whole-line Cauchy problemhas the unique classical solution, for $f\in C^2(\mathbb{R})$ and $g\in C^1(\mathbb{R})$,The first term evolves displacement released from rest; the second evolves initial velocity. The factor $1/(2c)$ converts an integral of velocity over distance into displacement. On a finite string the formula applies before the backward characteristic curves reach an endpoint, or globally after an extension that enforces the boundary conditions. ^[ref]

For a completely smooth example, choose a bump function with compact support:where $a>0$ sets the half-width and $A$ sets the initial peak. ThenAt $t=0$ the two copies coincide, recovering peak $A$. At $t=a/(2c)$, $u(0,t)=A{e}^{-1/3}\simeq 0.716531A$. At $t=3a/(2c)$ the pulses are separated, centred at $x=\pm 3a/2$, each of width $2a$ and peak $A/2$. Zero initial velocity forces equal left and right contributions; a purely right-moving initial profile instead requires $g=-c{f}^{\prime }$. Thus “released from rest” and “launched to the right” describe different Cauchy data.

The domain of dependence of $(x_{\ast },t_{\ast })$, for $t_{\ast }\ge 0$, is contained in $[x_{\ast }-c{t}_{\ast },x_{\ast }+c{t}_{\ast }]$ on the initial line. More precisely, displacement contributes only through the two endpoints, while initial velocity contributes throughout the interval. With $c=1$, $x_{\ast }=0.5$ and $t_{\ast }=1.5$, that interval is $[-1,2]$. Changing either initial datum outside it cannot change $u(0.5,1.5)$. If both initial data vanish outside $[-a,a]$, the solution vanishes outside $[-a-ct,a+ct]$: this is finite propagation speed. The expanding interval bounds possible influence; it need not be filled by nonzero displacement, as the separated bumps demonstrate.

Historically, d'Alembert studied the vibrating-string problem in “Recherches sur la courbe que forme une corde tendue mise en vibration”, in the Berlin Academy's volume for the year 1747, pages 214–219. The memoir introduces the differential equation and integrates it through two arbitrary one-variable functions. The original geometric problem makes the formula's enduring appeal clear: arbitrary initial shapes can be handled without first decomposing them into normal modes. ^[ref]

For a brief derivation, use characteristic coordinates $\xi =x-ct$ and $\eta =x+ct$. The chain rule gives $u_{tt}-c^2{u}_{xx}=-4c^2{u}_{\xi \eta }$, so $u=F(\xi )+G(\eta )$. The initial conditions yield $F+G=f$ and $-c{F}^{\prime }+c{G}^{\prime }=g$. Integrating these relations gives the displayed formula; the arbitrary integration constants cancel in their sum.

Put $v(r,t)=ru(r,t)$. Multiplication of the radial wave equation by $r$ gives $v_{tt}=c^2{v}_{rr}$ for $r>0$. In the characteristic coordinates $\xi =r+ct$, $\eta =r-ct$, this becomes $v_{\xi \eta }=0$. Integrating first in $\xi$ and then in $\eta$ gives $v=f(\xi )+g(\eta )$. ConsequentlyThe two terms are incoming and outgoing radial disturbances. If the field is finite at the origin, their numerators must cancel there; arbitrary independent choices generally produce a $1/r$ singularity.

Take the wave speed $c>0$ and introduce characteristic coordinates $\xi =x+ct$, $\eta =x-ct$. The chain rule givesThe wave equation therefore says $y_{\xi \eta }=0$. Integrating first with respect to $\eta$ gives $y_{\xi }=F(\xi )$, and integrating with respect to $\xi$ gives $y=f(x+ct)+g(x-ct)$ with twice differentiable $f,g$. Conversely substitution shows that every such sum solves the wave equation. The two terms are travelling waves of unchanged shape, moving respectively to the left and right at speed $c$.

The zero initial displacement gives $g(x)=-f(x)$, and the initial velocity gives $2c{f}^{\prime }(x)=\psi (x)$. Hence the D'Alembert formula with initial velocity isFor a classical twice differentiable solution one may take $\psi \in C^1$. Differentiation gives $y_t(x,0)=\psi (x)$ and $y(x,0)=0$ directly. The coordinate calculation also works for any nonzero signed $c$, with physical speed $|c|$. If the printed constant is allowed to be $c=0$, the equation instead gives $y=A(x)+tB(x)$ and these initial data give $y=t\psi (x)$; the displayed travelling-wave representation requires $c\ne 0$.

The chain rule gives ${\partial }_x={\partial }_{\xi }+\beta {\partial }_{\eta }$ and ${\partial }_y=\alpha {\partial }_{\xi }+{\partial }_{\eta }$. The coefficients of $u_{\xi \xi }$ and $u_{\eta \eta }$ are $2+3{\alpha }^2-7\alpha$ and $2{\beta }^2+3-7\beta$. Their integer roots are $\alpha =2$ and $\beta =3$. Thus choose characteristic coordinates $\xi =x+2y$, $\eta =3x+y$, whose Jacobian determinant is $-5\ne 0$. The coefficient of $u_{\xi \eta }$ is $4\beta +6\alpha -7(1+\alpha \beta )=-25$, so the equation becomes $-25u_{\xi \eta }=0$, equivalently $u_{\xi \eta }=0$.

Integrating the transformed hyperbolic partial differential equation gives $u=F(\xi )+G(\eta )$. Along $x=-2y$ the coordinates are $(0,-5y)$, so the zero boundary condition gives $G(\eta )=-F(0)$ for every real $\eta$. Along $x=0$ they are $(2y,y)$, so $F(2y)-F(0)=4y^2$, hence $F(\xi )={\xi }^2+F(0)$. The constants cancel and the unique resulting solution isIt directly satisfies both boundary conditions and $2u_{xx}+3u_{yy}-7u_{xy}=4+24-28=0$.

For the inflow transport boundary condition, the weak formulation isHere $\phi$ is a smooth compactly supported test function on the closed quadrant. Put $z=x-at$ and $\psi (t,z)=\phi (t,z+at)$. The transformed region is $t>\tau (z):=\max (0,-z/a)$. The formula in the preceding solution is $v(t,z)=b(z)$, with $b(z)=u_0(z)$ for $z\ge 0$ and $b(z)=f(-z/a)$ for $z<0$. Consequently the first integral equals $-\int b(z)\psi (\tau (z),z)dz$. Its positive-$z$ portion cancels the initial integral; the substitution $z=-at$ in its negative-$z$ portion cancels the inflow integral, including the factor $a$. This proves weak existence for arbitrary bounded data without corner compatibility.

For uniqueness, subtract two weak solutions. Interior test functions in these characteristic coordinates give ${\partial }_{t}v=0$ distributionally, so $v(t,z)=b(z)$ on each vertical ray. This follows first on rectangles compactly contained in $t>\tau (z)$ using tensor test functions, and then on the whole region by overlapping rectangles. In the full weak formulation the remaining lower-boundary term is $-\int b(z)\psi (\tau (z),z)dz=0$. Arbitrary smooth traces supported separately on $z>0$ and $z<0$ force $b=0$ there. The single point $z=0$ has zero measure. Thus the two-branch formula is the unique bounded weak solution.

Expanding the equation givesFor the hyperbolic partial differential equation discriminant, $A=1$, $B=\rho$, $C={\rho }^2-1$, soFor a curve $\rho =\rho (t)$, the characteristic hypersurface equation is $({\rho }^{\prime }{)}^2-2\rho {\rho }^{\prime }+{\rho }^2-1=0$, or ${\rho }^{\prime }=\rho \pm 1$. Therefore the two families of characteristic curves areThe values $C=0$ include the characteristic lines $\rho =-1$ and $\rho =1$. Every curve in the first family approaches $\rho =-1$ as $t\to -\infty$; every curve in the second approaches $\rho =1$. The two curves through $(\rho ,t)=(0,0)$ are $\rho =e^t-1$ and $\rho =1-e^t$, shown as the thick curves below. Equivalently, the characteristic coordinates are $e^{-t}(\rho +1)$ and $e^{-t}(\rho -1)$.

Introduce the null coordinatesThen $\square =-4{\partial }_{\xi }{\partial }_{\eta }$, so the equation becomesWrite the compatible boundary values asTwice integrating the equation gives the equivalent Volterra integral equation

Let $T$ denote the double-integral operator including the factor $-1/4$, and put $g=a+b-c$. Successive approximation gives the Neumann seriesOn a rectangle $|\xi |\le R$, $|\eta |\le R$,The series and its differentiated series converge locally uniformly. Since $a$ and $b$ are analytic, the sum is analytic and solves the equation and data near the origin.

If two solutions have the same data, their difference $w=Tw$. Iterating and using the same factorial estimate gives $\Vert w{\Vert }_{\infty }=0$ on every sufficiently small rectangle. This proves uniqueness. The argument is the Analytic Goursat problem for a Klein--Gordon equation.

A characteristic hypersurface of the form $\varphi (t,x)=p(t)\pm q(x)$ must satisfyOn either connected component of $tx>0$, separation givesOne choice valid on both components isIndeed, $p^{\prime }(t)=|t{|}^{-1/2}$ and $q^{\prime }(x)=|x{|}^{1/2}$ away from the axes, so $tx{p}^{\prime 2}=q^{\prime 2}$. Thusare the two families of characteristics, and $p\pm q$ are characteristic coordinates.