advanced30 min studyLesson 887 in Mathematics

Saddle-point problems and Stokes elements

Choose velocity and pressure spaces that control incompressibility and eliminate the unavoidable pressure gauge.

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What you’ll understand

  • Formulate homogeneous Dirichlet Stokes in H¹₀ velocity and mean-zero L² pressure spaces.
  • Relate the block system and pressure Schur complement to inf-sup stability.
  • Qualify element-pair stability by domain and mesh assumptions.

Incompressible Stokes flow couples a velocity field to a pressure that enforces zero divergence. The pressure is not an independent energy-minimizing field: it acts as a multiplier, producing a saddle-point matrix. A good approximation therefore needs a compatible velocity-pressure pair, not just accurate scalar interpolation in each component. We use a bounded connected Lipschitz domain, homogeneous velocity on its full boundary, a positive constant viscosity, and a bounded load functional on the velocity space.

Mean-zero pressure fixes a genuine null mode

Set V to the vector-valued zero-trace H¹ space and Q to square-integrable pressures with zero spatial mean. Use viscosity times the gradient inner product for a, and negative pressure times divergence for b. The velocity energy is coercive in the gradient norm. The continuous divergence inf-sup property on this connected domain gives the other stability ingredient. Pressure need only be L² in this weak formulation, so requiring its classical gradient pointwise would impose unnecessary regularity.

Adding a constant to pressure changes none of its pairings with zero-trace velocity tests, because the integral of their divergence vanishes. The mean-zero condition removes that gauge. On a disconnected domain one global mean condition would not remove all componentwise constant modes, so connectedness matters. Nonzero velocity or divergence data need appropriate flux compatibility and a lifting; the homogeneous full-Dirichlet setting here avoids silently introducing incompatible inflow or outflow.

V=[H01(Ω)]d,Q=L02(Ω),a(u,v)=ν∫Ω∇u:∇v,b(v,q)=−∫Ωq ∇⋅v.V=[H_0^1(\Omega)]^d,\quad Q=L_0^2(\Omega),\quad a(u,v)=\nu\int_\Omega\nabla u:\nabla v,\quad b(v,q)=-\int_\Omega q\,\nabla\cdot v.
The weak equations are a(u,v)+b(v,p)=F(v) and b(u,q)=0; pressure is normalized on a connected domain.↔ Scroll wide equations horizontally.

The Schur complement exposes pressure control

For a conforming discrete pair, the matrix has a positive velocity block A and a rectangular discrete divergence block B. The zero pressure block records that pressure enforces a constraint rather than carrying its own positive energy. Eliminating velocity gives a pressure Schur complement B A inverse B transpose. Its positivity on the normalized pressure space depends on B detecting every remaining pressure direction. A pressure null mode makes this Schur complement singular even if A is perfectly well conditioned.

Uniform spectral control of a suitably norm-scaled Schur complement is closely related to the discrete inf-sup constant, rather than to its unscaled entries alone. This also guides block preconditioners: a pressure mass matrix may approximate the correct scale under a stable pair and the relevant viscosity convention, but that is an additional spectral statement. Counting more velocity coefficients than pressure coefficients cannot exclude a checkerboard mode or a stability margin that tends to zero under refinement.

(ABTB0)(UP)=(FG),S=BA−1BT,SP=BA−1F−G.\begin{pmatrix}A&B^T\\B&0\end{pmatrix}\begin{pmatrix}U\\P\end{pmatrix}=\begin{pmatrix}F\\G\end{pmatrix},\qquad S=BA^{-1}B^T,\qquad SP=BA^{-1}F-G.
A is invertible after velocity boundary constraints; pressure normalization and inf-sup control determine whether S is positive definite.↔ Scroll wide equations horizontally.

An element name is not a complete stability theorem

Taylor–Hood uses continuous quadratic velocity and continuous linear pressure on simplices, with the appropriate boundary and pressure constraints. Its stability must be attached to an admissible mesh theorem. For example, the cited result in dimensions two and three assumes an exact polygonal or polyhedral domain with a fitted conforming shape-regular affine mesh family, and each simplex must have at least d internal edges. Those hypotheses are not supplied merely by choosing polynomial degrees. Other stable pairs and stabilizations have different assumptions and sometimes different discrete kernels.

Equal-order continuous linear velocity and pressure are not generally inf-sup stable without suitable modification, even though both variables have familiar approximation estimates. Conversely, eliminating the constant pressure gauge does not eliminate every possible spurious pressure mode. A verification should identify the exact spaces, physical norms, geometry, mesh family and boundary conditions, then supply uniform stability and approximation results. For an actual flow computation, algebraic solver accuracy and pressure normalization are additional checks after the variational pair has been justified.

Solve a small constrained velocity-pressure system

Use the Euclidean block model A=2I₂, B=[1,−1], F=(2,0), G=0. Compute U,P directly and through the Schur complement. This is an algebraic illustration, not a claimed mesh discretization.

  1. The constraint gives U₁=U₂. The primal equations are 2U₁+P=2 and 2U₂−P=0. Adding them yields U₁=U₂=1/2.

  2. The second equation then gives P=1. The pressure Schur complement is [1,−1](I/2)[1,−1]ᵀ=1.

  3. Its right-hand side is B A inverse F−G=[1,−1](1,0)ᵀ=1, so the reduced equation again gives P=1. Back substitution recovers U=(1/2,1/2), satisfying both blocks.

The result

The velocity block and divergence coupling play distinct roles; the pressure is fixed by the coupled constraint system.

Explore the model

Interactive lab

Interactive lab · matrix

A determinant does not certify a coupling margin

Reset, set a=1,b=0; predict determinant one and smallest singular value one. Reset, set a=1,b=2; the determinant remains one but predict smallest singular value √2−1, about 0.414, and condition number 3+2√2. This illustrates why a volume or dimension check cannot replace a norm-aware coupling bound. It is a triangular matrix family, not a Stokes element or pressure solver. Readouts describe the final matrix; changes retain time and Reset restores both settings and zero time.

A determinant does not certify a coupling marginThe selected matrix is A = [[a,b],[0,1]]. Playback interpolates geometry between the identity and A; readouts always describe A. Its determinant is a and its singular values come from AᵀA. Positive a makes every supported matrix invertible. This two-dimensional family is not a general eigensolver or a learned network. Current readings: Area scale · det A: 1.4; Singular values · σmax, σmin: 1.597, 0.876; Condition number · κ₂(A): 1.823. All parameters are adjustable with the labeled controls below.Matrix A with first row 1.4, 0.6 and second row zero, one; display interpolates between identity and A while final determinant remains 1.4.−4−2024−2−1012x · equal geometric scaleyA = [ 1.4 0.6 ; 0 1 ]Display blend: 0% · final values below
Interpolated geometryOriginal unit squareSelected final transform
Area scale · det A
1.4
Singular values · σmax, σmin
1.597, 0.876
Condition number · κ₂(A)
1.823
t = 0 s

The selected matrix is A = [[a,b],[0,1]]. Playback interpolates geometry between the identity and A; readouts always describe A. Its determinant is a and its singular values come from AᵀA. Positive a makes every supported matrix invertible. This two-dimensional family is not a general eigensolver or a learned network. Parameter changes pause playback and recompute the model at the current elapsed time. Reset restores both parameters and returns time to zero. Playback starts paused. Adjust either parameter, then choose Play or Step.

Look forThe smallest singular value can change substantially while determinant and dimensions remain fixed.

Check your understanding

Why is one mean-zero pressure constraint necessary in the connected full-Dirichlet Stokes setting?

Try it yourself

For full zero velocity boundary data, show that b(v,1)=0 for every sufficiently smooth test v and extend the conclusion to H¹₀ by density. What happens if a discrete pressure space retains the constant function?

Take a moment to make a prediction or work it out on paper. Open a hint when you need one.

Give me a hint

Integrate divergence over the domain and use the boundary trace.

Compare with the solution

The divergence theorem gives b(v,1)=−∫Ωdiv v=−∫∂Ωv·n=0 for smooth zero-trace tests. Continuity of the divergence pairing and density extend this to every H¹₀ velocity test. If the constant pressure remains in the discrete space, it is a nonzero multiplier with zero pairing against all discrete tests, so the inf-sup constant on that unnormalized pressure space is zero. Removing that mode is necessary, but does not prove that all other discrete pressure directions are stably detected.

Take it with you

  • Stokes uses a coercive velocity block and a multiplier pressure with an additive gauge.
  • The pressure Schur complement is controlled by the divergence coupling in the correct norms.
  • Stable element pairs require explicit domain, boundary, and mesh hypotheses beyond degree labels.

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