# Relationship between displacement field and dislocation density tensor

Let $u$ denote the displacement field in a solid body $\Omega \subset \mathbb{R}^3$ in the realm of continuum mechanics. Suppose we know that the restriction of $u$ to the boundary $\partial \Omega$ is discontinuous i.e. $u|_{\partial \Omega}$ is not continuous. Does it imply that the dislocation density tensor within the body is not zero? In other words, does the information about the displacement of boundary points (only) tell us anything about the dislocation density tensor?

You may want to restrict $$u$$ to $$u$$ modulo $$b$$, where $$b$$ is the size of the Burger's vector, in which case, if a dislocation went through your body, and there are no other dislocations, $$u$$ is constant. But now you say $$u$$ is not constant and has a discontinuity.
If you integrate $${\mathbf{u}}$$ over a loop you get the (signed sum of the dislocation's) Burger's vector. $${\mathbf{b}} = - \oint {d{\mathbf{u}}} = \int_A {{\mathbf{\alpha }}d{\mathbf{A}}} ,$$ where $${\mathbf{\alpha }}$$ is the dislocation tensor. Because dislocation lines must end and at the surface or in another dislocations fulfilling the conservation of Burger's vector, the dislocation must go inside the body, therefore, it is not $$0$$ inside.