 Solution

An 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 energyHere $f$ 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 givesIntegrating by parts first for interior test functions and then for variations free on either crack face yieldsThese 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.