Finite Elements in Solid Mechanics - Finite Elements: Theory, Fast Solvers, and Applications in Elasticity Theory - page 317

The PEERS Element

Mixed methods have not been heavily used for problems in plane elasticity, since
in the finite element approximation of the Hellinger-Reissner principle (3.22), two
stability problems occur simultaneously. The bilinear form a(σ, τ) : = ( C − 1 σ, τ) 0

= H( div ,) , but only on the kernel V . Here
we are using the notation of the general theory in Ch. III, §4.
In order to ensure the ellipticity on V h , for two decades the best possibility
was considered to choose V h ⊂ V which assures that the condition in III.4.7
is satisfied. As Brezzi and Fortin [1991, p. 284] have shown via a dimensional
argument, the symmetry of the stress tensor σ is a major difficulty. For this reason,
Arnold, Brezzi, and Douglas [1984] have developed the PEERS element (p
¯

is not elliptic on the entire space X :

lane
¯ lasticity ¯ lement with ¯ educed ¯ ymmetry). It has been studied further by Stenberg

[1988], among others. The so-called BDM elements of Brezzi, Douglas, and Marini
[1985] are constructed in a similar way. All of these elements are also useful for
nearly incompressible materials.
Finite element spaces for stresses with full symmetry have been established
by Arnold and Winther [2002]. They can be understood as extensions of the
Raviart-Thomas element, and the adaptation of the commuting diagram prop-
erties in Ch. III, §5 play an important role. A disadvantage is however that there
are 24 local degrees of freedom. There are no satisfactory mixed methods with
less degrees of freedom and full symmetry. Often there exist so-called zero en-
ergy modes , which must be filtered out, since otherwise we get the (hour-glass)
instabilities discussed in Ch. III, §7, due to the violation of the inf-sup condition.
In discussing the PEERS elements, for simplicity we restrict ourselves to pure
displacement boundary conditions. In this case (3.22) simplifies to
( C − 1 σ, τ) 0 + ( div τ,u) 0 = 0 for all τ ∈ H( div , ),

( 4 . 7 )

for all v ∈ L 2 () 2 . ( div σ, v) 0 =− (f, v) 0

Since we allow unsymmetric tensors in the following, the antisymmetric part

= τ − τ T ∈ L 2 () 2 × 2 , as(τ ) :

i.e. as(τ ) ij = τ ij − τ ji , plays a role. Since as(τ ) is already completely determined
by its (2,1)-component, we will refer to this component.
We consider the following saddle point problem:
Find σ H( div ,) 2 × 2 L 2 () 2 ∈ X : = and (u, γ ) ∈ M : = × L 2 () such that

( C − 1 σ, τ) 0 + ( div τ,u) 0 + (as(τ ), γ ) = τ ∈ H( div ,) 2 × 2 , 0 ,

=− (f, v) 0 ,v ∈ L 2 () 2 , ( div σ, v) 0 ( 4 . 8 )

= 0 ,

η ∈ L 2 (). (as(σ ), η)

= (τ 12 − τ 21 )ηdx . Here (as(τ ), η) :