# Well-posedness of Elasticity Boundary Conditions

For geotechnical engineering problems, it is common to fix a single component of displacement along a boundary as a Dirichlet boundary condition (roller boundary condition). However, I'm having trouble seeing why this leads to a well-posed problem.

The number of unknowns in an elastic problem is equal to the dimension of the problem (i.e. one unknown for each displacement component for the Navier equations). I was under the impression you needed to specify a boundary condition for each unknown in a boundary value problem. Why can we get away with a single boundary condition for a single component of displacement? Is there an implicit stress boundary condition implied when we do this? I've included an example sketch with a common configuration. My intuition says this should have a unique solution but I can't see why mathematically. • This question is probably better suited for Physics.SE. – nicoguaro Mar 28 '17 at 19:25

For this problem you have mixed boundary conditions. Then, you really have 8 boundary conditions for the problem that you present in your sketch. Although, 4 of them are not explicitly written.

These are:

• Top side:
• Non-homogeneous Neumann BC (normal traction): $\sigma_n=C$
• Homogeneous Neumann BC (tangent traction): $\sigma_t = 0$
• Left side:
• Homogeneous Dirichlet BC: $u_x=0$
• Homogeneous Neumann BC (tangent traction): $\sigma_t = 0$
• Right side:
• Homogeneous Dirichlet BC: $u_x=0$
• Homogeneous Neumann BC (tangent traction): $\sigma_t = 0$
• Bottom side:
• Homogeneous Dirichlet BC: $u_y=0$
• Homogeneous Neumann BC (tangent traction): $\sigma_t = 0$

I think that this might answer your question regarding the number of boundary conditions.

• It's been a while since you answered this, but I have a question regarding the bottom side's boundary condition. Is there an additional constraint on the bottom side that $\sigma_n =C$ to obtain an explicit balance of forces? If this constraint does not exist, then $\sigma_n$ at the bottom side may not necessarily be $C$, but I believe force balance will still be satisfied. – David Mar 21 '19 at 19:23
• @David, no, you don't impose that on the solution. In this case, you would obtain that due to the symmetry of the problem. In general, what you obtain is a global balance as a sum of forces. – nicoguaro Mar 21 '19 at 22:20
• Ah I see! So if you solved the system, and then checked the traction at the bottom surface, would it be $C$? – David Mar 22 '19 at 0:00
• @David, in this problem, yes. In general, no. – nicoguaro Mar 22 '19 at 0:04
• The reason why I ask is I was solving the simply supported beam problem, where there are 2 equidistant and equal loads on the top surface, and 2 supports on the bottom. The 2 supports on the bottom are also equidistant, and I was checking the solution for the force at those bottom supports, and they were not equal to the applied load on the top surface. I was quite puzzled as to why this occurred. – David Mar 22 '19 at 0:17