Let $\square u = 0$ be the wave equation in $\mathbb R^3$, where $\square$ is the wave operator $\square = \partial_t^2 - \Delta_x$. Let $u \in C^2(\mathbb R^3 \times [0, \infty))$ be a solution of this wave equation that is radial with respect to $x \in \mathbb R^3$ (i.e. knowing $|x|$ and $t$, we already know $u(x, t)$). I know want to show that there exists a function $F \in C^2(\mathbb R)$ so that
$$u(x, t) = \frac{F(t + |x|) - F(t - |x|)}{|x|} \qquad \text{for } x \in \mathbb R^3 \backslash \{0\}, t \geq 0$$
Now I was given the hint that I might consider $\tilde u(|x|, t) := u(x, t)$ and consider the fact that $\tilde u(r, t)$ can be extended to an even function $\tilde u \in C^2(\mathbb R \times [0, \infty))$ (even with respect to $r$).
Now I assuming that I've extended $\tilde u$ in such a prescribed way, I'm not sure how to continue. I know I somehow want to find an $F$ as above and I (probably) need $\tilde u$ to construct it. I've been thinking that there's maybe a clever trick of writing $u$, i.e. something like $u(x, t) = \frac{\tilde u(|x|, t)}{2} + \frac{\tilde u(-|x|, t)}{2}$ (which already "looks" somewhat like the desired equation), just something more general? But I'm not entirely sure.
Edit: Maybe I need to apply d'Alembert's formula to $\tilde u$? If a radial $u$ satisfies the wave equation $\square u = 0$ in the $3+1$-dimensional case, does $\tilde u$ as defined above actually satisfy $\square \tilde u = 0$ (in the 1+1-dimensional case)? Because that would give me some function $f$ that might help?