= Solution
{c}
\b[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 $\widetilde t=ct$ 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> gives
$$
\partial_x=\partial_\xi+\partial_\eta,\qquad
\partial_t=-c\partial_\xi+c\partial_\eta,
$$
so the <wave equation> becomes
$$
u_{tt}-c^2u_{xx}=-4c^2U_{\xi\eta}=0.
$$
Thus $U_\xi$ is independent of $\eta$, and <integration> gives the <countertravelling decomposition of the one-dimensional wave equation>
$$
\boxed{u(x,t)=F(x-ct)+G(x+ct)}.
$$
Conversely, differentiating this expression twice verifies the <wave equation> for any sufficiently differentiable $F,G$.
The two <initial conditions> give
$$
F(x)+G(x)=f(x),\qquad -cF'(x)+cG'(x)=g(x).
$$
Differentiating the first and solving these two equations for the <derivatives> gives
$$
F'(x)=\frac12f'(x)-\frac1{2c}g(x),\qquad
G'(x)=\frac12f'(x)+\frac1{2c}g(x).
$$
For a fixed reference point $x_0$, their <antiderivatives> can be written
$$
F(s)=\frac12f(s)-\frac1{2c}\int_{x_0}^s g(q)\,dq+C,\qquad
G(s)=\frac12f(s)+\frac1{2c}\int_{x_0}^s g(q)\,dq-C.
$$
Substituting the two travelling arguments cancels the constants and combines the <integrals>. This proves <D'Alembert's formula> with initial velocity:
$$
\boxed{u(x,t)=\frac12\bigl[f(x-ct)+f(x+ct)\bigr]
+\frac1{2c}\int_{x-ct}^{x+ct}g(s)\,ds}.
$$
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 $1/(2c)$.
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