 Solution

For smooth rapidly decaying initial data, apply the Fourier transform in $x$ and differentiate under the integral. The Fourier transform of a derivative turns the wave equation intoSolving this constant-coefficient ordinary differential equation givesAt $k=0$ the quotient means its continuous limit $t$, so $\tilde{v}(0,t)=\tilde{f}(0)+t\tilde{g}(0)$.

Since $\cos kt=(e^{ikt}+e^{-ikt})/2$, the Translation property of the Fourier transform inverts the first term as $(f(x+t)+f(x-t))/2$. Also $K_t(x)=\frac{1}{2}{1}_{[-t,t]}(x)$ has Fourier transform $\sin kt/k$, including its limiting value at zero. The convolution theorem therefore inverts the second term asThis proves the D'Alembert formulaThe formula itself extends beyond the rapidly decaying class: $f\in C^2$ and $g\in C^1$ suffice for a classical solution, verified by differentiation and the two initial conditions. The decay assumption above justifies the ordinary transform derivation without distributional qualifications.