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2.3 The Stokes Equations 77

and the right–hand side f is given by

1 1
f =( I + K st )ϕ − ϕ · ndsV ∆ n
2
β ∆ Γ
# $
1 1 1
+c ( I + K ∆ + L 1 )ϕ − {( I + K ∆ + L 1 )ϕ}· ndsV ∆ n .
2 2
β ∆ Γ
Since for c = 0 the equation (2.3.45) is uniquely solvable, the regularly per-
turbed equation (2.3.45) for small c but c = 0 is still uniquely solvable.
With σ 0 available, α can be found from (2.3.44) and, finally, the bound-
ary traction σ is given by (2.3.40). Then the representation formula (2.2.6)
provides us with the elastic displacement field u and the solution’s behavior
for the elastic, but almost incompressible materials, which one may expand
with respect to small c ≥ 0, as well. In particular we see that, for the almost
incompressible material

u e = u st + 1 ϕ · ndsV ∆ n + O(c)as c → 0
β ∆
Γ
where u st is the unique solution of the Stokes problem with

1
u st | Γ = ϕ − ϕ · ndsV ∆ n + O(c) .
β ∆
Γ
We also have the relation

ϕ · nds = divudx = −c pdx .
Γ Ω Ω

This shows that only if the given datum ϕ · nds = O(c) then we have
Γ
u e = u st + O(c) .

Next, we consider the interior traction problem for the almost incompress-
ible material. For simplicity, we now employ Equation (2.2.46),
1

D e u =( I − K )ψ on Γ (2.3.46)
2 e
where now ψ, the boundary stress, is given on Γ satisfying the compatibility
conditions (2.2.47), and the boundary displacement u is the unknown.
With (2.3.38), i.e.,

1 c 1
E e (x, y)= E st (x, y)+ γ n (x, y)I (2.3.47)
1+ c 1+ c 2(n − 1)πµ
and with Lemma 2.3.1 we obtain for the hypersingular operators

D e ϕ = D st ϕ + cL 2 ϕ (2.3.48)