= Solution
An <antiplane shear>[antiplane displacement] has the form $v(x_1,x_2,x_3)=(0,0,u(x_1,x_2))$, independent of the out-of-plane coordinate. Only $e_{13}=e_{31}=u_{x_1}/2$ and $e_{23}=e_{32}=u_{x_2}/2$ are nonzero, so its strain energy per unit out-of-plane thickness is $\mu|\nabla u|^2/2$.
The equilibrium minimizes total potential energy
$$
\boxed{\mathcal E_s(u)=\int_{\omega\setminus K_s}
\left(\frac\mu2|\nabla u|^2-fu\right)dx,\qquad
u\in H^1(\omega\setminus K_s),\quad u=u_D\text{ on }\Gamma_D.}
$$
Here $f$ is the out-of-plane body-force component; the two <crack> faces have independent <traces>. For a variation $\eta$ with zero outer-boundary <trace>, stationarity gives
$$
\mu\int\nabla u\cdot\nabla\eta=\int f\eta.
$$
Integrating by parts first for interior <test functions> and then for variations free on either <crack> face yields
$$
\boxed{-\mu\Delta u=f\text{ in }\omega\setminus K_s,\qquad
u=u_D\text{ on }\Gamma_D,\qquad\mu\partial_nu=0\text{ on both crack faces}.}
$$
These are the weak equilibrium equation, prescribed <displacement>, and zero <traction>. The tip behavior is selected by finite energy; it is not an additional prescribed tip value.
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