= Solution
Use the <radial reduction of the three-dimensional wave equation> $w=r\phi$ and extend $w$ oddly in $r$. Write $w=w_h+w_F$, where $w_h$ is the homogeneous <wave equation> solution with the prescribed <Cauchy data>, and $w_F$ has zero <Cauchy data>. With $w_0(r)=r\phi_0(r)$ and $w_1(r)=r\phi_1(r)$, the <D'Alembert formula> gives
$$
w_h(t,r)=\frac{w_0(r-t)+w_0(r+t)}2+\frac12\int_{r-t}^{r+t}w_1(\rho)\,d\rho.
$$
Both odd initial profiles have <compact support>, so
$$
|w_h(t,r)|\leq A,\qquad A=\|w_0\|_\infty+\frac12\|w_1\|_1.
$$
The <Duhamel principle>, applied to $w_{tt}-w_{rr}=-rF$, gives
$$
w_F(t,r)=-\frac12\int_0^t\int_{r-(t-s)}^{r+(t-s)}\rho F(s,\rho)\,d\rho\,ds.
$$
Now write $r=t+a$ with $1/2\leq a\leq1$. The lower integration limit is $s+a>0$, so this <domain of dependence> never crosses the axis. Intersecting its integration interval with the <support> of the source leaves a subset of $[s+a,s+1]$, of length at most one. On this interval,
$$
\rho |F(s,\rho)|\leq\frac{s+1}{(1+s)^2}=\frac1{1+s}.
$$
The <outgoing shell source estimate for a radial wave> therefore yields
$$
|w_F(t,t+a)|\leq\frac12\int_0^t\frac{ds}{1+s}=\frac12\log(1+t).
$$
Finally $r\geq t+1/2\geq(1+t)/2$, and thus
$$
|\phi(t,t+a)|\leq\frac{2A+\log(1+t)}{1+t}
\leq\boxed{\frac{C\log(2+t)}{1+t}},\qquad C=1+\frac{2A}{\log2}.
$$
This constant depends only on the <Cauchy data> and the fixed unit bound for the source. The logarithm comes from integrating the shell's accumulated forcing, whereas the factor $(1+t)^{-1}$ comes from division by $r$.
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