Solution to the wave equation in $\mathbb{R}^{3}$ with certain initial data Suppose $f$ is a smooth function satisfying $f(0) = f'(0) = 0$. The question I am working on is to determine the solution $u$ to $u_{tt} - \Delta u = 0$ in $\mathbb{R}^{3}$ with $u(x, 0) = f(|x|)/|x|$ and $u_{t}(x, 0) = f'(|x|)/|x|$.
By (essentially) the Kirchhoff formula, I get
$$u(x, t) = \frac{\partial}{\partial t}\left(\frac{1}{4\pi t}\int_{\partial B(x, t)}\frac{f(|y|)}{|y|}\, d\sigma_{y}\right) + \frac{1}{4\pi t}\int_{\partial B(x, t)}\frac{f'(|y|)}{|y|}\, d\sigma_{y}.$$
However, this does not make use of $f(0) = f'(0) = 0$ which I think will simplify the above expression. I was wondering if one could simplify the above expression further?

The initial conditions together with the (second order) differential equation determine $u(t)$ for all later $t$. One way of doing this is the following:
\begin{eqnarray*}
&&\partial _{t}^{2}u-\partial _{\mathbf{x}}^{2}u-0 \\
u_{1} &=&u,\;u_{2}=\partial _{t}u \\
\partial _{t}u_{1} &=&u_{2} \\
\partial _{t}u_{2} &=&\partial _{\mathbf{x}}^{2}u_{1}
\end{eqnarray*}
Set
\begin{eqnarray*}
\mathbf{u} &=&\left(
\begin{array}{c}
u_{1} \\
u_{2}
\end{array}
\right)  \\
\partial _{t}\mathbf{u} &=&\left(
\begin{array}{cc}
0 & 1 \\
\partial _{\mathbf{x}}^{2} & 0
\end{array}
\right) \mathbf{u} \\
\mathbf{u}(t) &=&\exp [\left(
\begin{array}{cc}
0 & 1 \\
\partial _{\mathbf{x}}^{2} & 0
\end{array}
\right) t]\mathbf{u}(0)
\end{eqnarray*}
Then
\begin{eqnarray*}
\left(
\begin{array}{cc}
0 & 1 \\
\partial _{\mathbf{x}}^{2} & 0
\end{array}
\right) ^{2} &=&\left(
\begin{array}{cc}
0 & 1 \\
\partial _{\mathbf{x}}^{2} & 0
\end{array}
\right) \left(
\begin{array}{cc}
0 & 1 \\
\partial _{\mathbf{x}}^{2} & 0
\end{array}
\right) =\left(
\begin{array}{cc}
\partial _{\mathbf{x}}^{2} & 0 \\
0 & \partial _{\mathbf{x}}^{2}%
\end{array}
\right) =\partial _{\mathbf{x}}^{2}\left(
\begin{array}{cc}
1 & 0 \\
0 & 1
\end{array}
\right)  \\
\left(
\begin{array}{cc}
0 & 1 \\
\partial _{\mathbf{x}}^{2} & 0
\end{array}
\right) ^{3} &=&\partial _{\mathbf{x}}^{2}\left(
\begin{array}{cc}
0 & 1 \\
\partial _{\mathbf{x}}^{2} & 0
\end{array}
\right) ,\;\mathrm{etc}
\end{eqnarray*}
Hence
\begin{eqnarray*}
\exp [\left(
\begin{array}{cc}
0 & 1 \\
\partial _{\mathbf{x}}^{2} & 0
\end{array}
\right) t] &=&\left(
\begin{array}{cc}
1 & 0 \\
0 & 1
\end{array}
\right) +\left(
\begin{array}{cc}
0 & 1 \\
\partial _{\mathbf{x}}^{2} & 0
\end{array}
\right) t+\frac{1}{2!}\partial _{\mathbf{x}}^{2}\left(
\begin{array}{cc}
1 & 0 \\
0 & 1
\end{array}
\right) +\frac{1}{3!}\partial _{\mathbf{x}}^{2}\left(
\begin{array}{cc}
0 & 1 \\
\partial _{\mathbf{x}}^{2} & 0
\end{array}
\right) \cdots  \\
&=&\cos (\partial _{\mathbf{x}}^{2}t)\left(
\begin{array}{cc}
1 & 0 \\
0 & 1
\end{array}
\right) +\frac{\sin (\partial _{\mathbf{x}}^{2}t)}{\partial _{\mathbf{x}}^{2}%
}\left(
\begin{array}{cc}
0 & 1 \\
\partial _{\mathbf{x}}^{2} & 0
\end{array}
\right)
\end{eqnarray*}
If you do not feel comfortable with ${∂_x}^2$ use the Fourier transform so ${∂_x}^2$ is replaced by $-k^2$.
Your initial conditions are tricky, since ${f_t}(x,0)$ need not vanish.
