= Cauchy solution for a steadily moving semi-infinite antiplane crack
{c}
{title2=$F'(z)=-\frac{i}{\pi\mu\sqrt z}\int_{-\infty}^0\frac{\sqrt{-s}\,f(s)}{z-s}\,ds$}
For the moving semi-infinite crack, take the <branch cut> of $\sqrt z$ on the negative real axis and $\sqrt x>0$ ahead of the tip. With <steadily moving antiplane shear>, face <tractions> $\sigma_{23}^{\pm}=-f(x)$ imply $\operatorname{Im}F'^{\pm}=f/\mu$. The antisymmetric <displacement> response gives $F'^-=-\overline{F'^+}$. Hence $G=\sqrt zF'$ has jump $G^+-G^-=-2\sqrt{-x}f(x)/\mu$. The <Sokhotski–Plemelj theorem> gives the displayed <Cauchy integral>, assuming convergence and no extra remote homogeneous loading. For $x>0$ it is purely imaginary, so $\sigma_{23}=\int_{-\infty}^0\sqrt{-s}f(s)/(x-s)\,ds/(\pi\sqrt x)$. At fixed co-moving loading this ahead-of-tip <stress> is independent of speed; the <displacement> still has the factor $1/\beta$.
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