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For two semi-infinite cracks separated by a ligament, choose $\chi (z)=\sqrt{a^2-z^2}>0$ on the ligament. Its upper crack values are $-i\sqrt{t^2-a^2}$ on the right and $+i\sqrt{t^2-a^2}$ on the left. Multiplication by $\chi$ converts the Hilbert problem for an antiplane crack to an additive jump; the Sokhotski–Plemelj theorem gives the displayed solution, assuming convergence. A homogeneous term $C/\chi$ transfers an extra resultant $\pi C$ through the ligament and has logarithmic displacement at infinity. The usual correction with no additional remote resultant excludes it. The far-field normalization is essential to uniqueness.

Retain $\beta =\sqrt{1-V^2/c^2}>0$, but normalize the complex potential as $F=\beta H$, so that $w=\mathrm{Re}F/\beta$. The Cauchy-Riemann equations give $w_x=\mathrm{Re}{F}^{\prime }/\beta$ and $w_y=-\mathrm{Im}{F}^{\prime }/\beta$. Since ${\partial }_{x_2}=\beta {\partial }_y$, the two stresses satisfyUsing 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 $Q=F^{\prime }$. For the antisymmetric crack-opening response, reflection gives $Q^{-}(s)=-\overline{Q^{+}(s)}$ on the faces. The prescribed equal face stress gives $\mathrm{Im}{Q}^{\pm }(s)=f(s)/\mu$. Multiplication by the square-root factor converts this face condition to an additive jump. In fact, with $G=\sqrt{z}Q$ and $r=\sqrt{-s}$,The Sokhotski–Plemelj theorem applied to this jump giveswhere $P$ 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, $G\to 0$ at infinity and Liouville theorem sets $P=0$. The resulting Cauchy solution for a steadily moving semi-infinite antiplane crack isIntegrals 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 $I_{\mathrm{pv}}(s)=\mathrm{PV}{\int }_{-\infty }^0\sqrt{-t}f(t)/(s-t)dt$. The Cauchy boundary limits giveBoth have $-\mu \mathrm{Im}{Q}^{\pm }=-f(s)$, 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 isNo $\beta$ appears: 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.

For the finite crack use the branchcut on $[-a,a]$. For $-a<t<a$, its limiting values are ${\chi }^{+}(t)=i\sqrt{a^2-t^2}$ and ${\chi }^{-}(t)=-i\sqrt{a^2-t^2}$. Consequently $H=\chi G$ converts the sum condition of the Hilbert problem for an antiplane crack into an additive jump:By the Sokhotski–Plemelj theorem, the Cauchy integral with this jump isThus the finite antiplane crack Cauchy solution is $G=H/\chi$. For regular integrable loading, $H=O(z^{-1})$ and $G=O(z^{-2})$. A homogeneous term $C/\chi$ would instead produce a logarithmic potential. In the reflected real-loading class $C$ is real, and its $1/z$ term gives a nonzero displacement period around the crack. It is excluded for the induced, dislocation-free correction. A nonvanishing constant stress term would change the remote applied field and is likewise excluded.

For $x=x_1>a$, $\chi (x)=\sqrt{x^2-a^2}>0$, soThis is the additional stress, not the total stress; the applied field must still be added. The integral is real at these intact-axis points, so ${\sigma }_{13}(x,0)=0$. On the crack itself, the upper Cauchy principal value formula gives $H^{+}={\pi }^{-1}\mathrm{PV}\int \sqrt{a^2-t^2}p(t)/(x-t)dt-i\sqrt{a^2-x^2}p(x)$. Division by $i\sqrt{a^2-x^2}$ verifies $\mathrm{Re}{G}^{+}=-p$, checking the sign of the face traction cancellation.

For example, constant applied shear $p_0$ gives $H(z)=p_0(z-\chi (z))$ and $G(z)=p_0(z/\chi (z)-1)$. The total shear ahead of the crack is then $p_{0}x/\sqrt{x^2-a^2}$, providing a direct check of the normalization and the crack-tip singularity.

For the two semi-infinite cracks, choose insteadwith cuts on the two crack segments. Approaching from above givesThese opposite signs on the two cracks are essential. Multiplying the Hilbert problem for an antiplane crack by $\chi$ givesThe Sokhotski–Plemelj theorem therefore yields the antiplane ligament Cauchy solutionFor $-a<x=x_1<a$, $\chi (x)=\sqrt{a^2-x^2}$ and the kernel has no pole on either crack. HenceAgain this is the additional stress, and ${\sigma }_{13}=0$ on the intact ligament. The integrals have the Cauchy principal value interpretation wherever required, and their convergence at infinity is assumed. For a point on the right crack, the upper value of $H$ has imaginary part $+\sqrt{t^2-a^2}p(t)$; division by $-i\sqrt{t^2-a^2}$ gives real part $-p(t)$. On the left crack both signs reverse and the same cancellation follows.

The homogeneous ambiguity here is physically different from the finite-crack displacement period. A real termhas zero real traction on the crack faces but adds ligament shear with resultantIts potential is $(C/\mu )\arcsin (z/a)$, producing logarithmic displacement growth at infinity. Thus it represents a separately imposed load transfer between the two sides, even though its stress tends to zero pointwise at infinity. The induced correction with no added remote resultant selects $C=0$, giving the boxed formula. Merely requiring pointwise decay of stress would not fix this constant; the far-field load normalization is needed.

For antiplane shear, the infinitesimal displacement is $(0,0,w(x_1,x_2))$. The isotropic linear elasticity constitutive law gives ${\sigma }_{13}=\mu w_{,1}$ and ${\sigma }_{23}=\mu w_{,2}$, together with the symmetric entries ${\sigma }_{31},{\sigma }_{32}$. All other stress tensor components vanish. Static force balance without body force becomes $\mu (w_{,11}+w_{,22})=0$, so $w$ satisfies the Laplace equation.

Locally choose a harmonic conjugate $\varphi$ by ${\varphi }_{,1}=w_{,2}$ and ${\varphi }_{,2}=-w_{,1}$. Compatibility follows from the Laplace equation. These are the Cauchy-Riemann equations for $F=\varphi +iw$, and henceThus the complex stress potential for antiplane shear supplies both stresses and, by integration, the displacement. A global potential must have the displacement periods appropriate to the cracked body; these periods also help fix homogeneous solutions.

For the additional crack-induced field, set $G=\mu F^{\prime }$ and $p(t)={\sigma }_{23}^A(t,0)$. Only $p$ loads the prospective faces, since ${\sigma }_{13}^A$ is tangential to their normal. Denote upper and lower boundary values by $G^{+},G^{-}$. The imposed cancellation of face tractions is $\mathrm{Re}{G}^{\pm }=-p$. The traction data and geometry are invariant under the displacement reflection $w(x_1,x_2)\mapsto -w(x_1,-x_2)$. For the induced correction with no independently added loading, uniqueness up to a rigid translation allows the odd displacement choice. Then $w_{,2}$ is even, $w_{,1}$ is odd, and $G^{-}=\overline{G^{+}}$ on the real axis. Therefore the Hilbert problem for an antiplane crack isThe second condition expresses continuation through uncracked material. There $G$ is real and the additional ${\sigma }_{13}$ vanishes. This scalar Riemann-Hilbert problem is supplemented by at most inverse-square-root stress singularities at the crack tips and the usual correction with no additional remote resultant or imposed dislocation. Without a far-field or period normalization, the face conditions alone permit homogeneous terms. The formulas below select this normalized induced field and explicitly identify the excluded terms.