Nonconforming and Other Methods - Finite Elements: Theory, Fast Solvers, and Applications in Elasticity Theory - page 181

§ 9. A Posteriori Error Estimates via the Hypercircle Method

The a posteriori error estimators in the preceding section provide a bound of the
error up to a generic constant; cf. Theorem 8.1. The theorem of Prager and Synge
(Theorem 5.1) admits the computation of an error bound without such a generic
constant . The essential idea is that a comparison of an approximate solution of the
primal variational problem with a feasible function of the dual mixed problem (cf.
(5.5) v ) yields an estimate. An elaborated theory was provided by Neittaanmaki
and Repin [2004]. Since Theorem 5.1 looks like Pythagoras' rule in an infinite
dimensional space, the term hypercircle method is frequently found.
We consider the Poisson equation (Example II.2.10) as the simplest case.
Given a finite element solution u h of the primal problem, for applying Theorem
5.1 an auxiliary function σ ∈ H( div ) with

div σ =− f ( 9 . 1 )

is required. Following Braess and Sch oberl [2006] we demonstrate that such a
function σ can be constructed by the solution of cheap local problems based on
the knowledge of u h .
Let

2 . A crucial step is
the evaluation of the error with respect to the solution for the differential equation
with a right-hand side f h that is piecewise constant on the triangulation. We know
from (8.8) that there is only an extra term ch f − f h

T h be a triangulation of a polygonal domain ⊂ R

if we approximate a given

0 . Obviously, the extra term is a term of higher
order. [We suggest for a first reading to assume that f is piecewise constant. In
this case the data oscillation vanishes and (9.5) below can be replaced by a simpler
expression.]
The mixed method by Raviart-Thomas yields a bound that is optimal in a
certain sense.

function f ∈ L 2 () by f h ∈ M

0 ( T h ). Moreover, let (σ h ,w h ) be
a solution of the mixed variational problem with the Raviart-Thomas element of
lowest order in RT 0 ( T h ) × M

1

9.1 Lemma. Let u h ∈ M

0 ( T h ) and f h ∈ M

0 ( T h ). Then

min 0 . 9 . 2 ) ∇ u h − σ h = ∇ u h − τ h ; τ h ∈ RT 0 ( T h ), div τ h + f h =

0 0

Proof. The Lagrange function for the minimization problem (9.2) is

1
2 τ

2
0

L (τ, v) = − ( ∇ u h ,τ) 0 + (v, div τ + f h ) 0 ,