= Solution
Let $u(x,z,t)$ be displacement along the source line and let $F(t)$ be its force per unit length. The <shear-horizontal wave> equation is
$$
\rho u_{tt}-\mu\Delta u=F(t)\delta^{(2)}(x,z),\qquad \mu=\rho\beta^2.
$$
Rotational symmetry about the line makes $u=u(r,t)$. Integrate over a disk of radius $r$ and use the <divergence theorem>:
$$
\rho\int_{D_r}u_{tt}\,dA-2\pi\mu r u_r=F(t).
$$
At a fixed regular time, a logarithmic singularity and its time derivatives are locally integrable; the inertial integral is $O(r^2|\log r|)$ and vanishes as $r\downarrow0$. The leading balance is consequently $r u_r=-F(t)/(2\pi\mu)$, giving the <SH line-source near-field singularity>
$$
\boxed{u(r,t)=-\frac{F(t)}{2\pi\mu}\log\frac r{r_*}+O(1)\qquad(r\downarrow0)}.
$$
The reference length $r_*$ only changes the bounded term; boundary conditions and source history determine that term. This is a local fixed-time statement, interpreted distributionally in time if the source force has jumps. It establishes logarithmic spatial dependence directly, without extrapolating far-field ray spreading into the source.
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