Finite Elements in Solid Mechanics - Finite Elements: Theory, Fast Solvers, and Applications in Elasticity Theory - page 337

2 is simply connected. Then every function η ∈
H − 1 ( div ,)is uniquely decomposable in the form

6.1 Lemma. Assume that ⊂ R

η =∇ ψ + curl p ( 6 . 5 )

with ψ ∈ H 0 () and p ∈ L 2 ()/ R . Moreover, the norms

1
2

2
1 ,

2

η H − 1 ( div ,) and ( ψ + p

0 )

( 6 . 6 )

are equivalent, where p is the representer satisfying pdx = 0 .
Proof. By hypothesis, χ :

div η ∈ H − 1 () . Let ψ ∈ H 0 () be the solution
of the equation ψ = χ . Then div (η −∇ ψ) =

=

0. By classical
estimates, every divergence-free function in can be represented as a rotation, i.e.,
η −∇ ψ =

div η − ψ =

curl p with a suitable p ∈ L 2 ()/ R . This establishes the decomposition.

We also observe that

div η − 1 = ψ − 1 =| ψ | 1 , ( 6 . 7 )

and

η − 1 =∇ ψ + curl p − 1 ≤∇ ψ − 1 + curl p − 1 ≤ ψ 0 + p 0 .

2 2

1

2

After summation, it follows that

0 .
In view of (6.7), to complete the proof we need only show that

η

H − 1 ( div ,) ≤

2 ψ + 2 p

p ≤ 0

c η H − 1 ( div ,) . Note that pdx =

0. It is known from the Stokes problem (see
Problem III.6.7) that there exists a function v ∈ H 0 () 2 with

div v = p and v 1 ≤ c p 0 . ( 6 . 8 )

= ( − v 2 ,v 1 ) , it clearly follows that ξ ∈ H 0 () 2 , Then for ξ = (ξ 1 ,ξ 2 ) :

rot ξ = p and ξ 1 ≤ c p 0 .

Moreover, taking account of (6.7), we see that the decomposition (6.5) satisfies

2
0

p = (p, rot ξ) = ( curl p, ξ) = (η −∇ ψ, ξ)

≤ η − 1 ξ 1 +| ψ | 1 ξ 0
≤ c( η − 1 + div η − 1 ) p 0 .

Supplement. If η lies in L 2 () 2 and not just in H − 1 ( div ,) , then we even have

p ∈ H 1 ()/ R

for the second component of the Helmholtz decomposition (6.5),