= Solution
Away from the line source, insert a leading <wavefront> form $u\sim a(\mathbf x)W(t-T(\mathbf x))$ into the <shear-horizontal wave> equation and match the two most singular orders. The first gives the <eikonal equation> $|\nabla T|=1/\beta$. The next gives
$$
2\nabla T\cdot\nabla a+a\Delta T=0,
\qquad\nabla\cdot(a^2\nabla T)=0.
$$
Equivalently, <SH ray-tube amplitude transport> conserves $\rho\beta a^2J$, where $J$ is the transverse width of a two-dimensional ray tube. A line source has straight radial <elastic rays> and $J=r\,d\theta$ for fixed angular width, so
$$
\boxed{a(r)=C r^{-1/2}}.
$$
The same geometrical spreading applies to the first nonzero derivative discontinuity or leading <wavefront> singularity; the source history determines which waveform $W$ and normalization $C$ occur. In three dimensions a point source instead has $J\propto r^2$ and amplitude proportional to $r^{-1}$.
This does not contradict the <SH line-source near-field singularity>. Ray transport describes an arrival-front limit with $t-r/\beta$ small, or the high-<frequency> field outside the immediate source neighborhood. The logarithm describes $r\downarrow0$ at fixed elapsed time and includes the tail behind the two-dimensional <wavefront>. For example, a causal step force $F_0H(t)$ has the exact radial displacement
$$
u(r,t)=\frac{F_0}{2\pi\mu}\operatorname{arcosh}\frac{\beta t}{r}\ H(t-r/\beta).
$$
For $r\ll\beta t$, this is $F_0\log(2\beta t/r)/(2\pi\mu)$, the near-source logarithm. For $t=r/\beta+\tau$ with $0<\tau\ll r/\beta$, it is
$$
\boxed{u\sim\frac{F_0\sqrt{2\beta}}{2\pi\mu}\,r^{-1/2}\sqrt\tau}.
$$
The coefficient of the leading arrival singularity therefore has exactly the ray-theory dependence. A force impulse gives the corresponding $r^{-1/2}H(\tau)/\sqrt\tau$ leading field. \b[The limits describe different regions of one causal solution; neither is a uniform approximation to the other].
Back to article page
