# Difference between using displacement and current configuration as unknown?

We could use either the current configuration $x$ or the displacement $u$ as unknown while solving for the deformation, for example, of a solid object. I want to know what's the difference between them? Is it that there is no difference, or one is numerically better than the other?

When you study a deformable continuum, the current configuration $x$ means very little, unless you also know what the original configuration $X$ was. Interesting things only happen when the displacement $x-X$ is different from point to point of the continuum. More precisely, when the displacement is not the sum of a rotation and a traslation, and thus the continuum is not simply undergoing rigid body motion. So in the end, you are going to be interested not in the displacement, but in the gradient of it.
In the Lagrangian description of deformation, you define the current configuration as a function of the original configuration, $x = \phi(X)$, and you track what happens to 'particles' in the original configuration as the body deforms. The displacement under such description takes the form $U(X) = x-X = \phi(X)-X$, You could compute the gradient of the displacement, which is closely related to deformation, at the point that was originally at position $X$ as $\nabla U(X) = \nabla \phi(X)-I(x)$.
In the Eulerian description, you look at a fixed point in space, and consider what is going on to the particle that now is at that position. In this description, displacement is $u(x) = x-X=x - \phi^{-1}(x)$, and the gradient of the displacement for the point currently at position $x$ is $\nabla u(x) =I - \nabla \phi^{-1}(x)$.