254 IV Steady Stokes Flow in Bounded Domains
Thus, i n particular, by Lemma II.6.1,
w ∈ W
m+2,q
(C), s ∈ W
m+1,q
(C),
for all (open) cubes C ⊂ R
n
+
. Final ly, w and s satisfy (IV.3.20), and, if q ∈
(1, n), (IV.3.21) for all ` ∈ [0, m], and w has zero trace at the boundary.
We have therefore proved
Lemma IV.3.2 Fo r any f, g in the class defined by (IV.3.22), there exists a
solution w, s a.e. to the nonhomogeneous Stokes system (IV.3.14) such that
w ∈ W
m+2,q
(C), s ∈ W
m+1,q
(C) ,
for all open cubes C ⊂ R
n
+
. Moreover, w, s satisfy (IV.3.23) and, if q ∈ (1 , n),
also (IV.3.24). In addi tion, for every ` ∈ [0, m], the f ollowing inequality holds:
|w|
`+2,q
+ |s|
`+1,q
≤ c (|f|
`,q
+ |g|
`+1,q
) . (IV.3.25)
If q ∈ (1, n), we also have:
|w|
`+1,nq/(n−q)
+ |w|
`+2,q
+ |s|
`+1,q
≤ c (|f|
`,q
+ |g|
`+1,q
) . (IV.3.26)
In the above inequalities, c = c(n, `, q).
The next step is to prove uniqueness for such solutions. In this respect,
the general result proved in the fol lowing theorem is appropriate.
Theorem IV.3.1 Let u ∈ W
m+2,q
(C), π ∈ W
m+1,q
(C) (m ≥ 0, C arbitrary
open cube in R
n
+
, n ≥ 2) be a solution a.e. to the Stokes system (IV.3.14)
with f ≡ g ≡ 0. Assume |u|
`+2,q
finite fo r some ` ≥ 0 and some q ∈ (1, ∞).
Then
|u|
`+2,q
= |π|
`+1,q
= 0.
In particular, if ` = 0, then
u = a x
n
, π = const.
with a = (a
1
, . . . , a
n−1
, 0) constant vector.
Proof. To fix the ideas, we shall consider the case n = 3 and ` = 0, the general
case being handled in a completely analogous way. As in the proof of Theorem
IV.2. 1, we obtain at once that π is harmonic througho ut the half-space. Let
u
ε
0
and π
ε
0
be the regularizatio ns of u a nd π, respectively, with respect to
x
0
= (x
1
, x
2
). It is readily shown that, fo r all ε
0
> 0, u
ε
0
and π
ε
0
satisfy the
same boundary-value problem as u, π and that
w ≡ D
02
u
ε
0
, s ≡ D
02
π
ε
0
with D
0
defined in (IV.3 .8), are solutions to the following problem