Ladyzhenskaya–Babuška–Brezzi condition

Remarks

The same two conditions with $V$, $Q$ and the kernel $Z = \set{v \in V : b(v, q) = 0 \text{ for all } q \in Q}$ give existence and uniqueness for the continuous problem; this is Brezzi's theorem. The inf-sup condition says that the operator $B \colon V \to Q'$ given by $b$ maps onto $Q'$ with a bounded right inverse. So one first finds $u_g$ with $b(u_g, q) = g(q)$ and $\norm{u_g} \le \norm{g} / \beta$. Then one solves for $u - u_g$ in $Z$ by the Lax–Milgram theorem. Finally one recovers $p$ from the first equation, using the inf-sup condition again. The point of the discrete theorem is that $\beta$ does not depend on $h$; otherwise the constant $C$ can blow up as the mesh is refined.

Example: for the Stokes equations, $V = H^1_0(\Omega)^d$, $Q = L^2_0(\Omega)$, $a(u, v) = \int_\Omega \nabla u : \nabla v$ and $b(v, q) = -\int_\Omega q \, \nabla \cdot v$. The Taylor–Hood pair, continuous piecewise quadratic velocities with continuous piecewise linear pressures, satisfies the inf-sup condition uniformly on suitable meshes.

Necessity: the pair of continuous piecewise linear velocities and pressures on a triangle mesh violates the condition. The discrete problem then has spurious pressure modes or a constant $\beta_h$ that tends to $0$, and the computed pressure oscillates.

Going deeper

For a student

The condition is the key stability requirement of mixed finite element methods. It replaces the coercivity of the Lax–Milgram setting, which a saddle-point problem never has on the whole space $V \times Q$. Checking it for a given pair of spaces is often done with a Fortin operator: a bounded projection $\Pi_h \colon V \to V_h$ with $b(v - \Pi_h v, q) = 0$ for all $q \in Q_h$. Stable pairs include Taylor–Hood elements for Stokes flow and Raviart–Thomas elements with piecewise constants for the mixed Poisson problem.