The Stokes Operator
A Summary of the Function Spaces
Suppose U Ð R n is open, bounded and has a smooth boundary /U. Then we have defined
the following spaces of functions
3 5 L 2 ÝUÞ n : div 3
EÝUÞ = áu
u 5 L 2 ÝUÞâ.
Then EÝUÞ is a Hilbert space and supports a trace operator
T : EÝUÞ í H ?1/2 Ý/UÞ
ÝontoÞ
for which the following integration by parts formula then holds
3, 4wÞÞ 0 + Ýdiv 3
3, T 0 w×
u, wÞ 0 = ÖTu
ÝÝu
3 5 EÝUÞ, w 5 H 1 ÝUÞ.
-u
The fact that the divergence and gradient operators are transposes of one another implies
that the annihilator of the subspace
3 5 DÝUÞ n : div d
3 = 0â
K = ád
is composed of gradients. A typical result asserts that
3f 5 K 0 V H ?1 ÝUÞ n
If we define
é 3f = 4p for some p 5 L 2 ÝUÞ.
H = the completion of K in the norm of L 2 ÝUÞ n
V = the completion of K in the norm of H 1 ÝUÞ n ,
then
3=0
u 5 L 2 ÝUÞ n : div 3
u = 0, Tu
H= 3
n
1
3 5 H 0 ÝUÞ : div 3
V = áu
u = 0â,
and
where
L 2 ÝUÞ n = H ã H e = H ã H 1 ã H 2
H1 =
H2 =
3
u 5 L 2 ÝUÞ n : 3
u = 4p, p 5 H 1 ÝUÞ, 4 2 p = 0
3
u 5 L 2 ÝUÞ n : 3
u = 4p, p 5 H 10 ÝUÞ .
Then
K Ð V Ð H Ð EÝUÞ Ð L 2 ÝUÞ n Ð H ?1 ÝUÞ n Ð D v ÝUÞ n
.
The Hilbert space projection theorem implies the existence of projections
1
P : L 2 ÝUÞ n ¸ H
Q : L 2 ÝUÞ n ¸ H e
and
where
3=3
Pu
u0 = 3
u?3
u1 ? 3
u2.
In addition, P maps H 10 ÝUÞ n continuously into H 1 ÝUÞ n . To see this, suppose 3
u 5 H 10 ÝUÞ n .
Then
4 2 q = div 3
u 5 L 2 ÝUÞ
q 5 H 10 ÝUÞ,
u 2 = 4q 5 H 1 ÝUÞ n .
has a unique solution q 5 H 2 ÝUÞ, and 3
In addition,
3?3
3
u 2 Þ 5 H 1/2 Ý/UÞ
u?3
u 2 5 H 1 ÝUÞ n , so TÝu
and
in U
42p = 0
3
u 2 Þ,
/ N p = TÝu ? 3
has a solution p 5 H 2 ÝUÞ, and 3
u 1 = 4p 5 H 1 ÝUÞ n . Then
and
3
u?3
u1 ? 3
u 2 5 H 1 ÝUÞ n
u0 = 3
P : H 10 ÝUÞ n ¸ H 1 ÝUÞ n
is bounded since
3 || 2H 1 ÝUÞ n = || 3
u 0 || 2H 1 ÝUÞ n = || 3
u || 2H 1 ÝUÞ n ? || 3
u1 + 3
u 2 || 2H 1 ÝUÞ n ² || 3
u || H 1 ÝUÞ n
|| Pu
The Stokes Operator
To say 3
u 5 V satisfies
?X4 2 3
u + 4p = 3f
means
3, 3
Xßßu
v àà= ÝÝf3, 3
vÞÞ 0
More precisely, since 4 2 3
u 5 L 2 ÝUÞ n ,
for 3f 5 L 2 ÝUÞ n
for all 3
v 5 V.
3 + 3z, for w
3 5 H and 3z 5 H e ; i.e.
4 23
u=w
3, 3
3, 3
u, 3
v ÞÞ 0 = ÝÝw
v ÞÞ 0 + ÝÝz3, 3
v ÞÞ 0 = ÝÝw
v ÞÞ 0 ,
ÝÝ4 2 3
since
v ÞÞ 0 = ÝÝ4q, 3
v ÞÞ 0 = ÖTv3, T 0 q× ? Ýdiv 3v, qÞ 0 = 0
ÝÝz3, 3
Of course,
v ÞÞ 0 = 0,
ÝÝ4p, 3
-v3 5 V.
and thus
ÝÝ?X4 2 3
u + 4p ? 3f, 3
vÞÞ 0 = ÝÝ?XP4 2 3
u ? Pf3, 3
vÞÞ 0 = 0
-v3 5 V
2
3 = Pf3, where Au
3 = ?XP4 2 3
This means Au
u and P : L 2 ÝUÞ n í H denotes the
projection into H. Note that P does not commute with ?4 2 , in general. Now let
3 5 V : Au
3 5 Hâ = áu
3 5 H 2 ÝUÞ n V Vâ
D A = áu
=
3
u 5 H 2 ÝUÞ n V H 10 ÝUÞ n : div 3
u=0
Then we can show that A is self adjoint as an operator from D A to H.
Lemma 1-
3, Av3ÞÞ 0
3, 3
A is symmetric; i.e., ÝÝAu
v ÞÞ 0 = ÝÝu
3, 3
-u
v 5 DA
3, f
3=d
3, and Pf
3 5 K = á the divergence free test functionsâ, Pd
3 =f
3 , so
Proof- For d
3, f
3, f
3àà= ÝÝAf
3ÞÞ 0
3, f
3 ÞÞ 0 = X ßßd
3 àà = X ßßf
3, d
3, d
3 ÞÞ 0 = ÝÝ?X4 2 d
ÝÝAd
Now K is dense in V and hence then also dense in D A and so we can extend the
equality here to the following cases
3, 3
3, 3
ÝÝAu
vÞÞ 0 = ßßu
vàà for 3
u 5 DA, 3
v5V
3, 3
3, Av3ÞÞ 0 = ßßu
vàà for 3
v 5 DA, 3
u5V
ÝÝu
3, Av3ÞÞ 0
3, 3
It follows from this that ÝÝAu
vÞÞ 0 = ÝÝu
3, 3
-u
v 5 DA n
Lemma 2- A is self adjoint
3, Av3ÞÞ 0 for 3
Proof- Fix 3
u 5 H, and define FÝv3Þ = ÝÝu
v 5 D A . Then F is clearly linear but is
u in H such that F is
not necessarily bounded. However, let D A D denote the set of all 3
D
bounded. It is evident that zero belongs to D A and it follows from lemma 1 that D A Ð D A D ;
i.e.,
3, Av3ÞÞ 0 = ÝÝAu
3, 3
FÝv3Þ = ÝÝu
vÞÞ 0 if both
3
u, 3
v 5 DA.
3, 3
3, 3
Then 3
vn í 3
v, in H implies FÝv3n Þ = ÝÝAu
v n ÞÞ 0 í ÝÝAu
vÞÞ 0 in R, so F is continuous.
Now it follows from the Riesz theorem that for each 3
u 5 D A D there is a unique 3f u in H such
vÞÞ 0 for 3
v 5 D A . Note that since D A is dense in H, 3f u 5 H is uniquely
that FÝv3Þ = ÝÝf3u , 3
determined by 3
u 5 D A D . If we denote this correspondence between 3
u and 3f u by
u for all 3
u 5 D A D , then
writing 3f u = A D 3
3 5 DAD
-u
3, Av3ÞÞ 0 = ÝÝA D 3
ÝÝu
u, 3
vÞÞ 0
-3
v 5 DA.
3
For any 3
u 5 D A D we have A D 3
u 5 H, so we can use the existence theorem for the Stokes
3
problem to find a unique w 5 D A D such that
3 + 4p = A D 3
?X4 2 w
u
3
div w = 0
3 =0
w
in U
in U
on @
i.e.,
3 5 DAD
w
is such that
3 = A D3
Aw
u5H
3 =3
3 ÞÞ 0 for 3z 5 H, arbitrary. Using the
Now we claim that w
u. To see this, consider ÝÝ 3z, 3
u?w
existence theorem for the Stokes problem once again, we find 3
v 5 D A such that Av3 = 3z.
Then
3 ÞÞ 0 = ÝÝz3, 3
3 ÞÞ 0 = ÝÝAv3, 3
3 ÞÞ 0
ÝÝz3, 3
u?w
u ÞÞ 0 ? ÝÝ 3z, w
u ÞÞ 0 ? ÝÝAv3, w
But
and
Then
ÝÝAv3, 3
u ÞÞ 0 = ÝÝv3, A D 3
u ÞÞ 0
3 ÞÞ 0 = ÝÝv3, Aw
3 ÞÞ 0
ÝÝAv3, w
since 3
u 5 DAD
3, 3
since w
v 5 DA.
3 ÞÞ 0 = 0 -z3 5 H
3 ÞÞ 0 = ÝÝv3, A D 3
u ÞÞ 0 ? ÝÝv3, Aw
ÝÝz3, 3
u?w
3 5 D A and Au
3 = A D3
i.e., for every 3
u 5 DAD , 3
u=w
u, so A is self adjoint.n
Theorem 1- The inverse of A : D A í H is compact
3 = 3f is uniquely solvable for every 3f 5 H, the inverse of A exists. From the
Proof- Since Au
regularity results for the Stokes problem, we have
3 || H 2 + ||p|| H 1 ² C||f|| L 2 i.e.,
||u
||A ?1 f|| H 2 ² C||f|| L 2 .
which implies A ?1 : H í D A = H 2 ÝUÞ n V V is bounded. But the inclusion, V Ð H is
compact (by Rellich’s lemma) hence A ?1 : H í H is compact.n
Note that K = A ?1 is compact and self-adjoint since A is self adjoint. Then K has an
3 j â that is complete in H (and in V). Then
orthonormal (in H) family of eigenfunctions áw
3 j = Vjw
3 j , where 0 < V 1 ² V 2 ² 6 6 6 í K
Aw
3 j 5 H m ÝUÞ n , m ³ 2.
It can be shown that if /U 5 C m , then w
Interpolation Spaces
3 j â for K to build a sequence
We can use the orthonormal (in H) family of eigenfunctions áw
of Hilbert spaces all contained in H. For 3
u 5 H,
4
3
3, w
3 j ÞÞ 0 w
3 j = > j³1 U j w
3j
u = > j³1 ÝÝu
and
3, w
3 j ÞÞ 0 | 2 = > j³1 | U j | 2
3 || 2H = > j³1 |ÝÝu
||u
For u 5 D A
3, w
3 j ÞÞ 0 Aw
3 j = > j³1 U j V j w
3j
A3
u = > j³1 ÝÝu
and
3 || 2H = > j³1 V 2j | U j | 2 .
||Au
For J > 0, define
and
3 5 HJ
-u
Evidently,
H J = DÝA J Þ =
2
3
u 5 H : > j³1 V 2J
j |U j | < K
3, w
3 j ÞÞ 0 A J w
3 j = > j³1 U j V Jj w
3 j.
A J3
u = > j³1 ÝÝu
u || 2H + ||A J 3
u || 2H
u || 2J = || 3
|| 3
is the graph norm on D A , and thus it follows from the closed graph theorem that H J is a
Banach space for this norm. Of course the norm then supports the inner product
3, 3
3, 3
v Þ J = Ýu
v Þ H + ÝA J/2 3
u, A J/23
v Þ H = > j³1 Ý1 + |V j | 2J Þ|U j V j |
Ýu
and H J becomes a Hilbert space for this inner product. In addition, since V j > 0 for all j,
this inner product is equivalent to
3, 3
vÞÞ J = ÝA J/2 3
u, A J/23
v Þ H = > j³1 V 2J
ÝÝu
j |U j V j |.
It is clear from the definitions that H 0 = H and H 1 = D A , and in addition, we can show that
u 5 V,
H 1/2 = V. To see this, note that for 3
2
3, 3
3, 3
uàà= ÝÝAu
u ÞÞ 0 =
u || V = ßßu
|| 3
3 j , > k³1 U k w
3k
> j³1 U j V j w
0
2
3 j, w
3 k ÞÞ 0 = > j³1 |U j | 2 V j = ÝÝu
3, 3
= > j³1 > k³1 U j V j U k ÝÝw
u ÞÞ 1/2 = || 3
u || 1/2
The spaces H J are a continuously distributed scale of Hilbert spaces for 0 ² J ² 1;
i.e.,
0 ² J ² 1.
H 10 ÝUÞ V H 2 ÝUÞ Ð D A = H 1 Ð H J Ð H 0 = H 0 ÝUÞ,
We can use arguments similar to those used in the development of the spaces H s ÝR n Þ to
show that
5
H J is continuously embedded in W p,q ÝUÞ n if
2J > p
2J ? n/2 > p ? n/q
H J is continuously embedded in C m ÝŪÞ n if 2J ? n/2 > m
In addition, we have the following interpolation results:
Lemma 3- For a ² J ² b, and 3
u 5 Hb
3 || 1?S
3 || J ² ||u
3 S
||u
b ||u || a
S=
b?J
b?a
Proof- For J ² b, and 3
u 5 H b , it follows that 3
u 5 H J and
2
2bÝ1?SÞ
2S
3 || 2J = > j³1 V 2J
|U j | 2Ý1?SÞ Þ V 2aS
||u
j |U j | = > j³1 ÝV j
j |U j |
Now use the Hölder inequality with p = 1/Ý1 ? SÞ and q = 1/S, to get
3 || 2J =
||u
2
> j³1 V 2b
j |U j |
1?S
2
> j³1 V 2a
j |U j |
S
3 || J ² ||u
3 || 1?S
3 S .n
from which it follows that ||u
b ||u || a
The particular choice a=0, b=1 leads to
3 || 1?S
3 || 1?S
3 || S0 = ||Au
3 || SH
3 || J ² ||u
||u
||u
||u
1
H
3 5 DA
-u
6