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For antiplane displacement $u=(0,0,w)$ the only nonzero strain components are $e_{xz}=w_x/2$ and $e_{yz}=w_y/2$. Commutation of mixed partial derivatives gives the necessary local compatibility conditionThe supplied stress is symmetric, independent of $z$ and trace-free. Its first two equilibrium equations are automatic; its third givesaway from the origin. Antiplane shear has zero trace, so isotropic Hooke's law reduces to $\boxed{e=\tau /(2\mu )}$. HenceLocally this integrates to $w=\arg (x+iy)+C$, represented on a suitable branch by $\arctan (y/x)+C$. The strain condition is satisfied at every point of the punctured plane, but the displacement cannot be made globally single-valued there:A gradient of a globally single-valued smooth displacement would instead integrate to zero around any closed curve. Local compatibility is therefore not sufficient on this multiply connected region.

A physical construction is a Volterra screw dislocation: cut the cylinder along a radial half-plane ending on its axis, shift one cut face relative to the other parallel to the axis, and rejoin. The axial mismatch is the Burgers vector. In the given normalization its magnitude is $2\pi$; with a dimensional mismatch $b$ one uses $w=b\arg (x+iy)/(2\pi )$. An intact pristine solid cannot reach this state by a smooth single-valued elastic deformation alone. The axis is a dislocation core where the continuum elastic field is singular and must be regularized by microscopic structure or plastic deformation. On the punctured domain the differential condition holds locally; globally its nonzero circulation detects the defect. In a distributional description, that incompatibility is concentrated on the axis. This is local strain compatibility versus screw-dislocation circulation.