= Solution
Retain $\beta=\sqrt{1-V^2/c^2}>0$, but normalize the complex potential as $F=\beta H$, so that $w=\operatorname{Re}F/\beta$. The <Cauchy-Riemann equations> give $w_x=\operatorname{Re}F'/\beta$ and $w_y=-\operatorname{Im}F'/\beta$. Since $\partial_{x_2}=\beta\partial_y$, the two stresses satisfy
$$
\boxed{\beta\sigma_{13}-i\sigma_{23}=\mu F'(z).}
$$
Using the same unscaled $H$ from part (a) would instead put an additional $\beta$ on its derivative; the potential normalization matters.
Take the principal branch of $\sqrt z$, with its <branch cut> along the negative real axis and positive $\sqrt x$ for $x>0$. Write $Q=F'$. For the antisymmetric crack-opening response, reflection gives $Q^-(s)=-\overline{Q^+(s)}$ on the faces. The prescribed equal face <stress> gives $\operatorname{Im}Q^\pm(s)=f(s)/\mu$. Multiplication by the square-root factor converts this face condition to an additive jump. In fact, with $G=\sqrt zQ$ and $r=\sqrt{-s}$,
$$
G^+=irQ^+,\qquad G^-=-irQ^-=ir\overline{Q^+},
\qquad
G^+-G^-=-\frac{2}{\mu}\sqrt{-s}\,f(s).
$$
The <Sokhotski–Plemelj theorem> applied to this jump gives
$$
G(z)=-\frac1{\pi i\mu}\int_{-\infty}^0\frac{\sqrt{-s}\,f(s)}{s-z}\,ds+P(z),
$$
where $P$ is entire after imposing the usual finite-energy crack-tip regularity. This is the <Hilbert problem for an antiplane crack>, with the stress-potential phase chosen explicitly above.
For localized face loading with no additional remote loading, the far <displacement> is bounded and the weighted Cauchy transform decays. In particular, for compactly supported loading, $G\to0$ at infinity and <Liouville theorem> sets $P=0$. The resulting <Cauchy solution for a steadily moving semi-infinite antiplane crack> is
$$
\boxed{F'(z)=-\frac{i}{\pi\mu\sqrt z}
\int_{-\infty}^0\frac{\sqrt{-s}\,f(s)}{z-s}\,ds.}
$$
Integrals and boundary values are understood for loadings with the corresponding convergence and regularity, or as limits of localized loading. The reflection used above is the pure antiplane opening response: reflection with a sign change of <displacement> leaves both prescribed face stresses unchanged, and the usual no-extra-load uniqueness condition eliminates an unforced symmetric addition.
One can check the face stresses directly. Put $I_{\mathrm{pv}}(s)=\operatorname{PV}\int_{-\infty}^0\sqrt{-t}f(t)/(s-t)\,dt$. The Cauchy boundary limits give
$$
Q^+(s)=-\frac{I_{\mathrm{pv}}(s)}{\pi\mu\sqrt{-s}}+\frac{i}{\mu}f(s),
\qquad
Q^-(s)=\frac{I_{\mathrm{pv}}(s)}{\pi\mu\sqrt{-s}}+\frac{i}{\mu}f(s).
$$
Both have $-\mu\operatorname{Im}Q^\pm=-f(s)$, verifying the prescribed <traction> and its sign on both faces. The opposite real parts also give the required reflection. Ahead of the tip, the integral is real and $Q$ is purely imaginary, so the bonded line has zero tangential <displacement> gradient and the two halves match.
For $x=x_1-Vt>0$, take $z=x$ in the boxed analytic solution. The ahead-of-tip <stress> is
$$
\boxed{\sigma_{23}(x_1,0,t)=\frac1{\pi\sqrt x}
\int_{-\infty}^0\frac{\sqrt{-s}\,f(s)}{x-s}\,ds.}
$$
No $\beta$ appears: \b[at fixed co-moving position and the specified co-moving face loading, this <stress> is independent of the speed]. The <displacement>, however, still scales with $1/\beta$, so speed independence does not apply to the whole elastic field. If $\int_{-\infty}^0|f(s)|/\sqrt{-s}\,ds$ is finite, the near-tip <stress> is inverse-square-root with coefficient $\pi^{-1}\int f(s)/\sqrt{-s}\,ds$.
The far-field condition is a genuine uniqueness requirement. A <homogeneous stress-intensity field of a semi-infinite antiplane crack> has $Q_h=iC/\sqrt z$, for real $C$. It adds no face <traction> but changes the <stress> ahead by $-\mu C/\sqrt x$. Although its <stress> tends to zero at infinity, its primitive is $2iC\sqrt z$, which gives unbounded remote <displacement> and infinite total <elastic energy>. Thus only asking for <stress> to vanish at infinity would leave the homogeneous constant undetermined; excluding that added remote stress-intensity loading selects the displayed face-loaded solution. The ordinary inverse-square-root tip singularity itself has finite energy in a bounded neighbourhood and is not being excluded.
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