# Is steady linear elasticity inherently ill-conditioned?

Compared to the transient PDE for linear elasticity, the steady equations appear to less well-conditioned. Are they inherently ill-conditioned without the transient term?

The condition number for the operator matrix seems to be fairly large (at least $$O(10^4)$$) for all the problems that I have solved. Is this fairly normal for linear elasticity? I don't usually see condition numbers this large for some other computational science problems.

I have not tried a FEM discretization for linear elasticity; would FEM yield a better conditioned system than, say, FVM/FD?

The condition number for the stiffness matrix of any method (finite elements, finite volumes, finite differences) applied to a second order differential operator always grows as $${\cal O}(h^{-2})$$ where the exponent equals the order of the differential operator. As a consequence, if you just make the mesh fine enough, you can make the condition number arbitrarily large -- and in fact, $$10^4$$ is a pretty small value that probably results from the fact that you have a relatively small number of cells or mesh points.
The reason why you don't observe this issue is because for time dependent (dynamic) problems, the matrix that needs to be inverted is either of the form $$M+\Delta t\, A$$ (parabolic problems with one time derivative) or $$M+\Delta t^2\, A$$ (hyperbolic problems with two time derivatives). The mass matrix has condition number $${\cal O}(1)$$ and so the total matrix you need to invert has condition number of either $${\cal O}(1+\Delta t\,h^{-2})$$ or $${\cal O}(1+\Delta t^2\,h^{-2})$$ in the two cases above. In other words, the multiplication of the bad matrix by a small number $$\Delta t$$ makes the problem better conditioned.
• The real estimate for the condition number is ${\cal O}(rh^{-2})$ where $r$ is the ratio between the maximum and minimum of the coefficients (material properties). – Wolfgang Bangerth Dec 6 '18 at 23:10
• Thanks! Is $h$ the smallest spatial step size in the mesh? – anonuser01 Dec 7 '18 at 20:44