In section 9.2 on page 241, question #12 is given as follows: "Solve the three-dimensional wave equation in $\{r\ne0,t>0\}$ with zero initial conditions and with the limiting condition \begin{equation*} \lim_{r\to 0}4\pi r^{2}u_{r}(r,t) = g(t). \end{equation*} Assume that $g(0) = g^{\prime}(0) = g^{\prime\prime}(0) = 0$. The text also provides the following solution: \begin{equation*} u = \begin{cases} -\frac{1}{4\pi r}g\left(t-\frac{r}{c}\right) &\mbox{if } t \ge \frac{r}{c} \\ 0 & \mbox{if } 0\le t\le \frac{r}{c}. \end{cases} \end{equation*}
For simplicity, let $c=1$. Now, \begin{alignat*}{2} u_{r} &= \frac{4\pi r^{2}g^{\prime}(t-r) + 8\pi rg(t-r)}{16\pi^{2}r^{4}} \\ &= \frac{g^{\prime}(t-r)}{4\pi r^{2}} + \frac{g(t-r)}{2\pi r^{3}} \Rightarrow \\ u_{rr} &= \frac{-4\pi r^{2}g^{\prime\prime}(t-r)-8\pi rg^{\prime}(t-r)}{16\pi^{2}r^{4}} - \left[ \frac{2\pi r^{3}g^{\prime}(t-r)+6\pi r^{2}g(t-r)}{4\pi^{2}r^{6}}\right] \\ &= -\frac{g^{\prime\prime}(t-r)}{4\pi r^{2}} - \frac{g^{\prime}(t-r)}{\pi r^{3}} - \frac{3g(t-r)}{2\pi r^{4}}. \end{alignat*} However, \begin{alignat*}{2} u_{rr} + \frac{2}{r}u_{r} &= -\frac{g^{\prime\prime}(t-r)}{4\pi r^{2}} - \frac{g^{\prime}(t-r)}{2\pi r^{3}} - \frac{g(t-r)}{2\pi r^{4}} \\ &\ne -\frac{g^{\prime\prime}(t-r)}{4\pi r^{2}} \\ &= u_{tt}. \end{alignat*} So, \begin{equation*} u_{tt} \ne u_{rr} + \frac{2}{r}u_{r} = \Delta u \Rightarrow \end{equation*} $u$ does not solve the wave equation in 3D. What am I missing in my reasoning or calculation that can make the given solution for $u$ correct?
