advanced30 min studyLesson 880 in Mathematics

A posteriori error estimation and adaptive meshes

Estimate discretization error with element residuals and flux jumps while preserving the assumptions behind reliability.

On this page

What you’ll understand

  • Define a residual estimator for a conforming Poisson discretization in an energy norm.
  • Distinguish global reliability, local efficiency and data oscillation.
  • Compute a one-dimensional residual indicator and explain the roles of marking, refinement and algebraic error.

An algebraic residual can be tiny because a discrete system was solved accurately, even when its mesh approximation is poor. A posteriori analysis uses the computed function and the PDE data to estimate that discretization error. Its guarantees depend on the equation, boundary data, mesh regularity and error norm. An indicator is not automatically a certified upper bound merely because it becomes smaller under refinement.

Interior equations and interface balance leave different residuals

Consider minus Laplacian u=f on a bounded polygonal domain, with homogeneous Dirichlet boundary data, f in L2, and a conforming piecewise linear Galerkin solution on a shape-regular simplicial mesh. The error norm is the L2 norm of its gradient. Inside an element, the strong residual is f plus the elementwise Laplacian of the approximation; for linear elements that Laplacian vanishes. Across interior faces, the outward-normal derivative contributions fail to cancel pointwise, leaving a flux jump.

A residual estimator combines squared element residuals weighted by element diameter squared and squared interior normal-flux jumps weighted by face diameter. Count each interior face once in the global estimator, or split it equally between adjacent local indicators. No Neumann boundary residual is included in this homogeneous Dirichlet model; different boundary conditions require their corresponding terms. For variable or discontinuous diffusion coefficients, both the energy norm and robust residual weights must be reconsidered rather than copied from the unit-coefficient Poisson case.

η2=∑KhK2∥f+Δuh∥L2(K)2+∑e interiorhe∥[∇uh⋅ne]∥L2(e)2\eta^2=\sum_K h_K^2\|f+\Delta u_h\|_{L^2(K)}^2+\sum_{e\ \mathrm{interior}}h_e\|[\nabla u_h\cdot n_e]\|_{L^2(e)}^2
This residual estimator is for conforming linear Poisson elements with unit coefficient; each interior face is counted once and its jump uses a fixed orientation.↔ Scroll wide equations horizontally.

Reliability and efficiency are different inequalities

Under these assumptions, global reliability bounds the energy error by a mesh-independent constant times the estimator, with the constant depending on the domain and mesh shape regularity. It does not say that the constant is one, nor that each local indicator individually bounds the error on its own element. An elliptic error can be affected by residuals elsewhere, so a purely elementwise upper-bound interpretation is generally false. A rigorous proof uses Galerkin orthogonality and a stable quasi-interpolant to turn residuals into an energy dual estimate.

Efficiency runs in the other direction: an indicator is controlled by error on a neighboring patch plus unresolved variation in the data. The latter data oscillation is typically the norm of h times the difference between f and an appropriate local polynomial approximation. Omitting it can make an efficiency assertion false for rapidly varying data. Reliability shows that the estimator does not miss the error, while efficiency prevents excessive overestimation up to patch and data terms; neither identifies the exact error without constants.

∥∇(u−uh)∥L2(Ω)≤Crelη,ηK≤Ceff(∥∇(u−uh)∥L2(ωK)+osc⁡ωK(f))\|\nabla(u-u_h)\|_{L^2(\Omega)}\le C_{\mathrm{rel}}\eta,\qquad \eta_K\le C_{\mathrm{eff}}\left(\|\nabla(u-u_h)\|_{L^2(\omega_K)}+\operatorname{osc}_{\omega_K}(f)\right)
These are global reliability and patchwise efficiency statements for shape-regular conforming Poisson meshes; oscillation depends on local data projection.↔ Scroll wide equations horizontally.

Adaptive refinement must also control the solve

An adaptive loop solves, estimates, marks and refines. A bulk marking rule selects elements whose squared indicators account for a prescribed fraction of the total. Refinement must maintain conformity and shape regularity, often by refining additional neighbors. Convergence and optimal work rates require more than computing this loop once: the estimator, refinement strategy, marking and approximation class must satisfy the hypotheses of the chosen adaptive theorem. A shrinking global indicator on a few experiments is useful evidence but not that theorem.

The estimator assumes the discrete Galerkin problem has been solved accurately enough. A large algebraic error can contaminate the residual and the Galerkin orthogonality used in its proof, so stopping criteria should balance solver error against estimated discretization error. For the explicit one-dimensional illustration below, interior faces are points and the jump contribution is defined as h_e times the squared derivative jump. This is the one-dimensional counterpart of the face weighting, not an unannounced replacement of a two-dimensional face integral.

A residual estimator on two linear elements

Solve−u″=1 on[0,1] with zero endpoints, using two equal conforming linear elements. The exact solution is u=x(1−x)/2. Use h_e=1/2 at the midpoint and the stated one-dimensional jump weighting to compute eta and the exact energy error.

  1. The Galerkin nodal midpoint value is 1/8, so the two element slopes are 1/4 and−1/4. Each element residual is one, and the derivative jump has magnitude 1/2.

  2. The volume contribution is two times h² times integral of 1 over a half-cell:2·(1/4)·(1/2)=1/4. The jump contribution is(1/2)(1/2)²=1/8, so eta²=3/8 and eta≈0.612372.

  3. The derivative error is 1/4−x on the first half and 3/4−x on the second. Its squared integral is 2∫01/2(1/4−x)2 dx=1/482\int_0^{1/2}(1/4-x)^2\,dx=1/48, giving energy error 1/48≈0.1443381/\sqrt{48}\approx0.144338.

The result

Eta is larger than the exact energy error here, but their values are not equal and no universal reliability constant of one has been inferred.

Explore the model

Interactive lab

Interactive lab · function

A squared gradient error before integration

Pause and Reset, then set coefficient a = 1 and position x = 0.25. Interpret x as a local derivative mismatch: exact y=0.0625, rounded 0.063 in the readout, is its squared energy-error density. Reset and set both a = 1 and x = 0.50: the density becomes 0.25, four times larger. The instrument evaluates a quadratic density; it does not integrate over a mesh, compute flux jumps or certify an a posteriori estimator.

A squared gradient error before integrationThis lab uses only y = ax². Playback traces the curve within the fixed integration interval from 0 to x; it does not estimate numerical error or change the chosen point. Current readings: y at chosen x: 0.608; Tangent slope · 2ax: 1.35; Signed area · ax³/3: 0.182. All parameters are adjustable with the labeled controls below.−2.4−1.201.22.4−9−4.504.59xy(x, ax²)
y = ax²TangentSigned integral
y at chosen x
0.608
Tangent slope · 2ax
1.35
Signed area · ax³/3
0.182
t = 0 s

This lab uses only y = ax². Playback traces the curve within the fixed integration interval from 0 to x; it does not estimate numerical error or change the chosen point. 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 forEnergy error aggregates squared gradient mismatches over space, whereas a residual indicator needs PDE-specific reliability analysis.

Check your understanding

What does the stated Poisson estimator’s reliability guarantee mean?

Try it yourself

Repeat the one-dimensional example with four uniform elements. The Galerkin solution interpolates the same quadratic at the nodes. Use h_e=1/4 and count all three interior derivative jumps once. Find eta² and the exact energy error.

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

Give me a hint

The element width and derivative-jump magnitude are both 1/4. On a uniform mesh, the squared derivative-interpolation error for this quadratic is h²/12.

Compare with the solution

The volume contribution is ∑Kh2∫K1 dx=h2=1/16\sum_K h^2\int_K1\,dx=h^2=1/16. The three jumps each have magnitude 1/4, so their contribution is 3(1/4)(1/4)2=3/643(1/4)(1/4)^2=3/64. Thus eta²=7/64 and eta=sqrt(7)/8≈0.330719. The exact squared energy error is h2/12=1/192h^2/12=1/192, giving energy error about 0.0721688, half the two-element error. The estimator also decreases but by a different factor. These explicit values check the chosen one-dimensional normalization and do not replace a shape-regularity-dependent reliability or efficiency proof for general multidimensional adaptive meshes.

Take it with you

  • Residuals inside elements and normal-flux jumps describe different discretization defects.
  • Reliability is global, while efficiency is local up to patches and data oscillation.
  • Adaptive accuracy also depends on mesh admissibility, marking and algebraic solver tolerances.

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