I work on geological problems and I use the Finite Element Method. But this question can be applied on other classical mechanical problems.I work on implicit 3D surfaces (which represent the limits between two geological layers aka two media).I have to impose displacement on these surfaces (Dirichlet Condition). The condition that I impose on surfaces are transfered on the surrounding nodes. I make the assumption that I can use the  classical way to apply Dirichlet Condition (penalty method in my case) on these nodes, that are NOT on the the boundaries.My question is: do you think this assumption is valid? Have you got some references about the subject? As I understand the problem I try to apply Boundary Conditions not on the boundaries...Thank you in advance.Best.EDITWe try to solve this problem:$$  \left\{    \begin{aligned}      \sigma_{ij,j} + F_{i} &= 0 & &\text{in the domain $\Omega$}&\\      u_i&= q_{i}& &\text{on the boundary $\Gamma_{q}$}&\\      \sigma_{ij}n_{j} &= h_{i}& &\text{on the boundary $\Gamma_{h}$}&\\      u_i &= b_{i}& &\text{punctually in the domain $\Omega$}&\\    \end{aligned}  \right.$$We run the FEA on a mesh. We basically try to impose Dirichlet Condition on the nodes which are inside the domain $\Omega$ (fourth line). We constrain the displacement value of $u_i$ to $b_i$ for these nodes.
FEM: Possible to have boundary conditions inside the domain?
finite element;boundary conditions
The solution of the equation you are looking for is in the space $H^1$ of functions that have one weak derivative, but in 2d and 3d, this does not imply that the solution is in fact continuous. As a consequence, it is not possible to define the value of a solution at individual points, and equally consequentially, it is mathematically not possible to prescribe (boundary) values at individual points.But I suspect that you're not actually interested in prescribing values for the displacement at individual points, but for all points along a line (in 2d) or surface (in 3d), and this is perfectly valid. (Mathematically speaking, this is so because the trace operator is well defined on $H^1$). At the discrete level, this is then equivalent to prescribing the values on all nodes that lie on this line/surface.Conceptually, you can think of this as a crack in your domain that is occupied by an infinitely thin device whose displacement you can manipulate and that attached to the two sides of the crack.