The Stokes System
Settings for the Stokes Problem
As a first step to considering the Navier-Stokes equations, we will consider various settings
for the Stokes problem of finding ⃗
u, p which satisfy
−ν∇ 2 ⃗
u + ∇p = ⃗f
div ⃗
u = 0.
in U ⊂ R n
1
We could consider this system of equations on:
1.
a.
U = R n : in this case we would have to impose some ”behavior at ∞"
⃗ as | ⃗
conditions such as ⃗
u→ψ
x| → ∞
b.
U ⊂ R n , open bounded with smooth boundary ∂U. In this case we would
impose conditions on ∂U such as ⃗
u=⃗
0 on ∂U. The more smoothness
we require of ∂U the easier the proofs become but the less realistic the
problem becomes.
c.
U = 0, L n = n − dim coordinate cell. In this case we impose periodic
boundary conditions
⃗
ux⃗ + Le⃗i  = ⃗
ux⃗
px⃗ + Le⃗i  = px⃗
∀x⃗ 1 ≤ i ≤ n,
∀x⃗.
Setting (c) is simpler than (b) but maintains most of the relevant features of a problem on a
bounded set. For considering the problem in this setting we let
Hn =
⃗
u ∈ L 2 U n : ⃗
u satisfies the periodic boundary conditions
⃗, ⃗
v  Hn = ∫ ⃗
ux⃗ ⋅ ⃗
vx⃗ dx
u
U
and we let H denote H n in the case n = 1. For convenience, we let L = 2π, n = 2. Then an
orthonormal basis for H is the family
φ mn x, y = e i mx e iny
where the normalization constants have been suppressed for convenience. Then
⃗fx, y = ∑ f mn⃗
e 1 + g mn⃗
e 2  φ mn x, y
m,n
px, y = ∑ m,n p mn φ mn x, y
⃗
ux, y = ∑ m,n u mn⃗
e 1 + v mn⃗
e 2  φ mn x, y
where
p mn = p, φ mn  H
and
⃗, φ mn  H . ,
e 1 + v mn⃗
e 2  = u
u mn⃗
e 1 + g mn⃗
e 2  = f⃗, φ mn  H .
f mn⃗
Then
1
∇p = ∑ m,n p mn ∇φ mn x, y = ∑ m,n p mn i m ⃗
e1 + n ⃗
e 2 φ mn x, y
div ⃗
u=⃗
ux, y = ∑ m,n u mn ∂ x φ mn x, y + v mn ∂ y φ mn x, y
= ∑ m,n i m u mn + n v mn  φ mn x, y,
and
∇ 2⃗
u = ∑ m,n u mn⃗
e 1 + v mn⃗
e 2  ∇ 2 φ mn x, y
= − ∑ m,n m 2 + n 2 u mn⃗
e 1 + v mn⃗
e 2  φ mn x, y.
The Stokes system then becomes (letting ν = 1 for convenience),
−∇ 2 ⃗
u + ∇p = ∑ m,n m 2 + n 2 u mn⃗
e 1 + v mn⃗
e 2  + p mn i m ⃗
e1 + n ⃗
e 2  φ mn x, y
= ∑ m,n f mn⃗
e 1 + g mn⃗
e 2  φ mn x, y,
and
∑ m,n i m u mn + n v mn  φ mn x, y = 0.
This reduces then to the following algebraic system
m 2 + n 2  u mn + i m p mn = f mn
m 2 + n 2  v mn + i n p mn = g mn
m u mn + n v mn = 0.
It follows that
im 2 + n 2 p mn = m f mn + n g mn
or
p mn =
m f mn + n g mn
im 2 + n 2 
u mn =
f mn
− 2m 2
2
m +n
m +n
and then
2
m f mn + n g mn
im 2 + n 2 
.
It is clear that u 00 = ⃗
u, ⃗
1 , and p 00 = p, 1 H are indeterminate from these equations,
H
which is to say, ⃗
u and p are not uniquely determined but are unique only up to additive
constants. Then, in the quotient spaces
̃ = p ∈ H : p, 1 = 0 and H
̃n =
H
H
⃗
u ∈ Hn : ⃗
u, ⃗
1
H
=0
the solutions of the Stokes system (1) are uniquely determined.
In addition, using the isomorphism
H 2 ∋ ⃗f  f mn , g mn  ∈ ℓ 2  2
⃗, p such that
we can show that for every f ∈ H 2 , there is a ”unique” pair u
2⃗
2
2
̃2×H
̃
∇ u ∈ H and ∇p ∈ H . This means that there exists a one to one mapping from H
2
onto H .
2
On the basis of this example, we now feel justified is seeking an abstract setting in
which to prove the existence of a weak solution to the Stokes system in the more general
situation of case (b). If we are successful in this attempt it will hopefully lead to the correct
approach for dealing with the more difficult issues of the nonlinear steady and unsteady
equations.
Weak Formulation of the Stokes System
Let U ⊂ R n n = 2, 3 be open and bounded with a smooth boundary ∂U. Let ⃗f ∈ L 2 U n be
given and consider
− ν∇ 2 ⃗
u + ∇p = ⃗f
div ⃗
u = 0,
⃗
u=0
in U ⊂ R n
in U,
S
on ∂U.
u, p is called the Stokes system. We
This set of n + 1 equations for the n + 1 unknowns ⃗
will now define several function spaces and develop their properties in preparation for giving
a weak formulation for the Stokes system.
Function Spaces
Let
EU =
⃗
u ∈ L 2 U n : div ⃗
u ∈ L 2 U
with
⃗, ⃗
vE = ∫ ⃗
ux ⋅ ⃗
vx dx + ∫ div ⃗
ux div ⃗
vx dx
u
U
U
and
⃗ || 2E = ||u
⃗ || 2L 2 U n + ||div ⃗
u || L2 2 U .
||u
Then we have the following results about EU
Lemma 1 (a) EU is a Hilbert space
(b) C ∞ U n ⊂ EU is dense
Evidently EU is a vector valued version of H 1 U and the proof of lemma 1 is nearly
identical to the proof of analogous results about H 1 U.
Recall that the ”trace operator” T 0 : H 1 U → L 2 ∂U had the interpretation of a
restriction to the boundary operator. In the simplest case, U = R n+ , ∂U = R n−1 , we have for
any φ ∈ C 1 Ū and ⃗
x = x⃗′ , x n , ⃗
x ′ ∈ R n−1 , x n > 0,
∞
2
2
|φx⃗′ , 0| = − ∫ 0 ∂ n |φx⃗′ , x n | dx n .
Then
∞
∫R n−1 |φx⃗′ , 0| 2 dx ′ = − ∫R n−1 ∫ 0 ∂ n |φx⃗′ , x n | 2 dx n dx ′
and
3
2
||φx⃗′ , 0|| L 2
R n−1
||T 0 φ|| 2L 2
i.e.,
R n−1
≤ 2 ||∂ n φ|| L 2 R n+  || φ|| L 2 R n+  ≤ ||∂ n φ|| L2 2 R n+  + || φ|| L2 2 R n+ 
≤ || φ|| H2 1 R n+  .
This proves that the restriction operator is continuous from H 1 U to L 2 ∂U in this simple
case. Extension to more general sets U is accomplished using the previously discussed
local charts for "flattening the boundary"..
Recall further that T 0 satisfied
i) ∀φ ∈ C 1 Ū
T 0 φ = φ| ∂U
1
ii) ker T 0 = H 0 U
iii) Rng T 0 = H 1/2 ∂U
H 1/2 ∂U ⊂ L 2 ∂U ⊂ H −1/2 ∂UV
Here
⟨f, g⟩ H −1/2 ×H 1/2 = f, g L 2 ∂U = ∫∂U f g ds;
where
i.e., the duality pairing on H −1/2 ∂U × H 1/2 ∂U is just the extension of the L 2 ∂U inner
product. This makes H −1/2 ∂U a realization for the dual of H 1/2 ∂U. Our aim now is to
define a similar restriction operator for the space EU.
Trace Theorem for EU Suppose U ⊂ R n has smooth boundary ∂U with ⃗
n x =outward unit
normal to ∂U at x. Then there is a continuous linear map
T : EU → H −1/2 ∂U onto
⃗ =w
⃗ ⋅⃗
Tw
n x | x∈∂U
with
⃗ ∈ C 1 Ū n
∀w
⃗ ∈ EU, w ∈ H 1 U
∀u
Moreover,
⃗, ∇w L 2 U n + div ⃗
u, w
u
L 2 U
⃗, T 0 w⟩ H −1/2 ×H 1/2
= ⟨Tu
n
To see the meaning of this last equation suppose we have ⃗
φ n , ψ n ∈ C 1 Ū × C 1 Ū ,
and note that a version of the divergence theorem asserts that for each n, we have the
following integration by parts formula,
∫U ⃗φ n ⋅ ∇ψ n dx = ∫∂U ⃗φ n ⋅ ⃗n x ψ n ds − ∫U div ⃗φ n ψ n dx
∀n
Suppose also that as n tends to infinity,
∫U ⃗φ n ⋅ ∇ψ n dx + ∫U div ⃗φ n ψ n dx = ∫∂U ⃗φ n ⋅ ⃗n x ψ n ds
↓
↓
↓
∫U ⃗u ⋅ ∇w dx + ∫U div ⃗u w dx = ∫∂U ⃗u ⋅ ⃗n x w ds
Note that by the continuity of the inner products
⃗
φn  ⃗
u in L 2 U n
∇ψ n  ∇w in L 2 U
implies
∫U ⃗φ n ⋅ ∇ψ n dx  ∫U ⃗u ⋅ ∇w dx
Similarly,
4
div ⃗
φ n  div ⃗
u in L 2 U
ψ n  w in L 2 U
implies
∫U divφ⃗ n ψ n dx  ∫U div ⃗u w dx
In addition,
⃗
φn  ⃗
u in L 2 U n and div ⃗
φ n  div ⃗
u in L 2 U
⃗
implies
φn − ⃗
u
 0, and ⃗
u ∈ EU
EU
and
ψ n  w in L 2 U and ∇ψ n  ∇w in L 2 U
implies ‖ψ − w‖ 1  0 and w ∈ H 1 U
Together, these results imply that ⃗
φn → ⃗
u, in EU, and ψ n → w
in H 1 U and
∫∂U ⃗φ n ⋅ ⃗n x ψ n ds  ⟨Tu⃗, T 0 w⟩ H −1/2 ×H 1/2 = ∫∂U ⃗u ⋅ ⃗n x w ds
where the trace operator T is defined in terms of this limit.
To show that T maps EU onto H −1/2 ∂U, pick an arbitary ψ ∈ H −1/2 ∂U, and let
φ = ψ−
⟨ψ, 1⟩
⟨1, 1⟩
Then φ ∈ H −1/2 ∂U and ⟨φ, 1⟩ = 0; i.e.,we subtract from ψ its average over ∂U to obtain a φ
whose integral over ∂U is zero. This condition on φ is sufficient to imply the existence of a
(nonunique) p ∈ H 1 U satisfying
∇2p = 0
in U
∂ N p = ∇p ⋅ ⃗
n x = φ.
Denote the solution of this Neumann problem by p = pφ, and let
⃗
u = ∇pφ +
⟨ψ, 1⟩
⃗
u0
⟨1, 1⟩
where
n
⃗
u 0 ∈ C 1 Ū is such that
Then
⃗0 = 1
Tu
⟨ψ, 1⟩
⃗
u = ∇2p +
div ⃗
u 0 ∈ L 2 U
u ∈ L 2 U n and div ⃗
⟨1, 1⟩
which means ⃗
u ∈ EU. Moreover,
1⟩
⟨ψ, 1⟩
⃗ = ∂ N pφ + ⟨ψ,
Tu
= φ+
= ψ.
1⟩
⟨1,
⟨1, 1⟩
⃗ = ψ.
Thus for every ψ ∈ H −1/2 ∂U there is a ⃗
u ∈ EU such that Tu
Now let E 0 U = the completion of C ∞c U n in the norm of EU. Then E 0 U = ker T.
To see that this is true, let ⃗
u ∈ E 0 U with ⃗
φ n ∈ C ∞c U n such that
⃗ n = 0 ∀n which implies Tu
⃗ = 0 since T is continuous.
|| ⃗
φn − ⃗
u|| E → 0 as n → ∞. Then Tφ
5
⃗ = 0. For F ∈ C ∞c R n , let f = the restriction
Conversely, suppose ⃗
u ∈ EU is such that Tu
⃗ = 0, we have
of F to U. Since Tu
⃗, T 0 F⟩ = 0.
⟨Tu
But
⃗, T 0 F⟩ = ∫ f divu
⃗+⃗
u ⋅ ∇f dx.
⟨Tu
U
Then
∫R n F div ⃗u ~ + u⃗ ~ ⋅ ∇F dx = 0
∀F ∈ C ∞c R n ,
⃗ ~ denote the extensions by zero from U to all of R n . But
where div ⃗
u ~ and u
∫R n F div u⃗ ~ dx = − ∫R n u⃗ ~ ⋅ ∇F dx
∀F ∈ C ∞c R n ,
⃗ ~ . Then it follows
which, combined with the previous result, implies that div ⃗
u ~ = div u
~
n
⃗ ∈ ER . It now has to be shown that this forces ⃗
that u
u ∈ E 0 U but the key part of the
⃗ = 0 we can extend ⃗
u to
result resides in what we have just shown. More precisely, when Tu
the whole space by letting ⃗
u = 0 outside U and the extended function is still an element of
H 1 R n  n . In general, when extending a function in H 1 U n by zero, the extended function is
no longer in H 1 R n  n .
Summarizing, we have a continuous linear map
T : EU → H −1/2 ∂U onto
⃗ =w
⃗ ⋅⃗
Tw
n x | x∈∂U
with
⃗ ∈ C 1 Ū n
∀w
ker T = E 0 U
⃗, ∇w L 2 U n + div ⃗
u, w
u
and
L 2 U
⃗, T 0 w⟩ H −1/2 ×H 1/2
= ⟨Tu
⃗ ∈ EU, w ∈ H 1 U
∀u
In order to attack the Stokes system, our plan is going to be to incorporate the continuity
equation div ⃗
u = 0 into the definition of the solution space. To this end, we begin by
considering
K=
⃗
φ ∈ C ∞c U n : div ⃗
φ=0
.
This is a linear space but it cannot support Hilbert space, Banach space or even metric
space topology without destroying the character of the space. Note, however, that for any
⃗
φ ∈ K, and any distribution p ∈ D ′ U, we have
∇p, ⃗
φ
D ′n ×D n
i.e., this shows that
and
n
n
i=1
i=1
= ∑ ⟨∂ i p, φ i ⟩ D ′ ×D = − ∑ ⟨p, ∂ i φ i ⟩ D ′ ×D = − p, div ⃗
φ
D ′ ×D
= 0;
div : DU n ∋ ⃗
u → div ⃗
u ∈ DU
grad : D ′ U n ∋ ∇p  p ∈ D ′ U
are transposes of one another. A consequence of this fact is that every gradient annihilates
K but the converse, that every annihilator of K must be a gradient, is much more difficult to
prove. The following result will be stated without proof.
6
Theorem 1 Suppose U ⊂ R n is open, bounded and has Lipschitz smooth boundary ∂U.
a.
For ⃗f ∈ D ′ U n
b.
For ⃗f ∈ H −1 U n
⃗f, ⃗
φ =0
⃗ ∈ K ⇔ ⃗f = ∇p some p ∈ L 2 U
∀φ
c.
For ⃗f ∈ L 2loc U n
⃗f, ⃗
φ =0
⃗ ∈ K ⇔ ⃗f = ∇p some p ∈ H 1 U
∀φ
Note that
⃗f, ⃗
φ =0
⃗ ∈ K ⇔ ⃗f = ∇p some p ∈ D ′ U
∀φ
grad : L 2 U  H −1 U n
is not an isomorphism since the gradient is not injective on any space containing the
constants. However, the gradient is an isomorphism on the quotient space
L 2∗ U = L 2 U modR ≊
p ∈ L 2 U : ∫ p dx = 0
U
Now define
H = completion of K in the norm of L 2 U n
V = completion of K in the norm of H 1 U n
where
⃗
φ
n
2
H 1 U n
n
n
= ∑||φ i || L2 2 U + ∑ ∑ ||∂ j φ i || L2 2 U
i=1
i=1 j=1
Since K consists of the divergence free vector valued test functions, we expect the
elements of H and V to continue to have some of the properties enjoyed by elements of K.
Theorem 2 Suppose U ⊂ R n is open, bounded and has Lipschitz smooth boundary ∂U, and
let
H∗ =
Ĥ=
Then
⃗
⃗=0
u ∈ L 2 U n : div ⃗
u = 0, and Tu
⃗
u ∈ L 2 U n : ⃗
u = ∇p, for some p ∈ H 1 U .
H = H ∗ and Ĥ = H  .
Proof- (a) Ĥ ⊂ H 
Suppose ⃗
u ∈ Ĥ. Then
⃗
u, ⃗
φ
0
=
∇p, ⃗
φ
0
= − p, div ⃗
φ
0
⃗ ∈ K.
= 0 ∀φ
Here we use ⋅, ⋅ 0 and ⋅, ⋅ 0 to denote the inner products in
L 2 U n and L 2 U, respectively. Since K is dense in H, it follows from this result and the
continuity of the inner product, that
⃗, ⃗
v  0 = 0 ∀v⃗ ∈ H,
u
hence Ĥ ⊂ H  .
(b) H  ⊂ Ĥ.
Suppose ⃗
u ∈ H  . Then
⃗
u, ⃗
φ
0
⃗ ∈ K ⊂ H,
= 0 ∀φ
and it follows from theorem 1 that ⃗
u = ∇p for some p ∈ D ′ U. But ∇p = ⃗
u ∈ L 2 U n so
p ∈ H 1 U, and it follows that ⃗
u ∈ Ĥ.
7
The results (a) and (b) together imply Ĥ = H  .
(c) H ⊂ H ∗ .
⃗ n  ⊂ K. Then
u is the limit in L 2 U n of a sequence φ
Suppose ⃗
u ∈ H. Then ⃗
2
⃗
div φ n = 0 ∀n implies div ⃗
u = 0 in L U. Moreover, since div ⃗
u = div ⃗
φ n = 0 ∈ L 2 U, we
have ⃗
u, ⃗
φ ∈ EU and hence
⃗
φn
u−⃗
i.e.,
EU
=
⃗
φn
u−⃗
L 2 U n
→ 0 as n → ∞
⃗
φn → ⃗
u in EU.
⃗ n → Tu
⃗ = 0, since T is continuous on EU. This proves
But this implies that 0 = Tφ
⃗
u ∈ H implies ⃗
u ∈ H ∗ ; i. e. , H ⊂ H ∗ .
To see that H = H ∗ , suppose H is not all of H ∗ . then H, since it is the L 2 U n closure of K, is
a closed subspace of H ∗ .Then H has an orthogonal complement, H 0 , in H ∗ . But if ⃗
u ∈ H0,
then ⃗
u = ∇p for some p ∈ H 1 U by the argument used to prove (b). In this case,
⃗
u ∈ H 0 ⊂ H ∗ satisfies
∇ 2 p = div ⃗
u=0
⃗ = ∂Np = 0
and Tu
Then p =constant and ⃗
u = ∇p = 0, so H 0 = 0 and H = H ∗ . ■
A more precise characterization of H  is possible when ∂U has additional smoothness.
Theorem 3 Suppose U ⊂ R n is open, bounded and has a smooth boundary ∂U. Then
L 2 U n = H ⊕ H  = H ⊕ H 1 ⊕ H 2
where
H1 =
⃗
u ∈ L 2 U n : ⃗
u = ∇p, p ∈ H 1 U, ∇ 2 p = 0
H2 =
⃗
u ∈ L 2 U n : ⃗
u = ∇p, p ∈ H 10 U
Proof- It is clear that H 1 , H 2 ⊂ H  . To see that H 1  H 2 , let
⃗
u = ∇p ∈ H 1 and ⃗
v = ∇q ∈ H 2 . Then
⃗, ⃗
⃗, ∇q 0 = ⟨Tu
⃗, T 0 q⟩ − div ⃗
v  0 = u
u, q 0 .
u
But
⃗, T 0 q⟩ = 0 for q ∈ H 10 U and div ⃗
u, q 0 = 0 as div ⃗
u = ∇ 2 p = 0.
⟨Tu
Now, to show L 2 U n = H ⊕ H 1 ⊕ H 2 , let ⃗
u ∈ L 2 U n . Then div ⃗
u ∈ H −1 U so our results on
the existence of weak solutions to elliptic boundary value problems imply that there exists a
unique p ∈ H 10 U such that ∇ 2 p = div ⃗
u ∈ H −1 U.
Let ⃗
u 2 = ∇p, and note that ⃗
u 2 ∈ H 2 . Then
⃗ − ∇p = 0,
div ⃗
u − ∇ 2 p = divu
⃗
u − ∇p ∈ EU,
so
and
⃗ − ∇p ∈ H −1/2 ∂U.
Tu
Next, let q denote the weak solution of
8
∇ 2 q = 0 in U,
⃗ − ∇p ∈ H −1/2 ∂U.
∂ N q = Tu
Since the Green’s formula implies,
⃗ − ∇p, 1⟩ = ∫ 1divu
⃗ − ∇p = 0,
⟨Tu
U
it follows that the Neumann problem has a (non-unique) solution q ∈ H 1 U. Finally, we let
⃗
u 1 ∈ H 1 and ⃗
u0 = ⃗
u−⃗
u1 − ⃗
u 2 ∈ H; i. e. ,
u 1 = ∇q so that ⃗
⃗−⃗
⃗ − ∇p = 0
div ⃗
u 0 = divu
u1 − ⃗
u 2  = div ⃗
u − ∇ 2 q − ∇ 2 p = divu
⃗ 0 = Tu
⃗−⃗
⃗ − ∇p − ∂ n q = 0.
Tu
u1 − ⃗
u 2  = Tu
This shows that every ⃗
u ∈ L 2 U n can be written ⃗
u=⃗
u0 + ⃗
u1 + ⃗
u2. ■
The Hilbert space projection theorem implies the existence of projections
P H : L 2 U n → H
and
Q H : L 2 U n → H 
where
P H⃗
u=⃗
u0 = ⃗
u−⃗
u1 − ⃗
u2.
In addition, P H maps H 10 U n continuously into H 1 U n . To see this, suppose ⃗
u ∈ H 10 U n .
Then
∇ 2 q = div ⃗
u ∈ L 2 U
q ∈ H 10 U,
u 2 = ∇q ∈ H 1 U n .
has a unique solution q ∈ H 2 U, and ⃗
In addition,
⃗
⃗−⃗
u−⃗
u 2 ∈ H 1 U n , so Tu
u 2  ∈ H 1/2 ∂U
so
∇2p = 0
in U
⃗−⃗
∂ N p = Tu
u 2 ,
has a solution p ∈ H 2 U, and ⃗
u 1 = ∇p ∈ H 1 U n . Then
⃗
u0 = ⃗
u−⃗
u1 − ⃗
u 2 ∈ H 1 U n
and
P H : H 10 U n → H 1 U n
is bounded since
u || H 1 U n = || ⃗
u 0 || H 1 U n ≤ || ⃗
u || H 1 U n
|| P H⃗
Next, we will give a characterization for the space V obtained by completing K in the
H 1 U − norm.
Theorem 4 Suppose U ⊂ R n is open, bounded and has Lipschitz smooth boundary ∂U,
9
and let
⃗ ∈ H 10 U n : div ⃗
V ∗ = u
u = 0.
Then
⃗
φ ∈ C ∞c U n : div ⃗
φ=0
V ∗ = V = H 1 U n completion of K =
Proof- (a) V ⊂ V ∗
⃗ n  ⊂ K. This implies
Suppose ⃗
u ∈ V. Then ⃗
u is the limit in H 1 U n of a sequence φ
⃗
φn − ⃗
u
H 1 U n
→0
0 = div ⃗
φ n → div ⃗
u in L 2 U,
and
hence
⃗
u ∈ H 10 U n
i.e.,
div ⃗
u = 0 in L 2 U.
and
⃗
u ∈ V implies ⃗
u ∈ V ∗ so V ⊂ V ∗ .
To now show that V is, in fact, all of V ∗ , we will show that any continuous linear functional L
on V ∗ which vanishes on V must then vanish on all of V ∗ . So, suppose L is such a linear
functional on V ∗ . Since V ∗ is, by its definition, a closed subspace of H 10 U n , it follows that
any linear functional that is continuous on V ∗ can be extended to H 10 U n as a continuous
linear functional. Of course this extension is not unique but, since the dual of H 10 U n is
H −1 U n , the extended functional has a representation of the form
n
L⋅ = ∑⟨λ i , ⋅⟩
λ i ∈ H −1 U.
i=1
Since we have assumed that L vanishes on V = H 1 U n completion of K, it follows that
n
φ = ∑⟨λ i , φ i ⟩ = 0
L ⃗
⃗ ∈ K.
∀φ
i=1
But then theorem 1 implies that ⃗
λ = λ 1 , … , λ n  = ∇p for some p ∈ L 2 U. Then this
implies, in turn, that
n
v ⟩ = −p, div ⃗
v0 = 0
Lv⃗ = ∑⟨λ i , v i ⟩ = −⟨p, div ⃗
∀v⃗ ∈ V ∗ . ;
i=1
i.e., we have succeeded in showing that an L that vanishes on V necessarily also vanishes
on V ∗ . ■
If U is bounded but the boundary is not Lipschitz smooth, then it is not known whether
V = V ∗ . If U is not bounded, then there are examples where dimV\V ∗  = 1 or even larger.
When V is not equal to V ∗ . this can lead to serious difficulties in the weak formulation of the
Stokes problem.
Weak Solutions of the Stokes System
⃗, p is a
We now return to considering the weak formulation of the Stokes system (S). If u
solution for (S), then for any ⃗
φ ∈ K,
−ν
∇ 2⃗
u, ⃗
φ
0
= −ν ∫ ∇ 2 ⃗
u⋅⃗
φ dx
U
10
⃗ dx =: ν
⃗ ∗ ∇φ
= ν ∫ ∇u
U
⃗
u, ⃗
φ
;
i.e.,
n
− ∫ ∇ 2⃗
u⋅⃗
φ dx = − ∫ ∑ φ j ∇ 2 u j dx
U
U
j=1
n
⃗ ∗ ∇u
⃗ dx =
= ∫ ∑ ∇φ j ⋅ ∇u j dx = ∫ ∇φ
U
Also
∇p, ⃗
φ
0
= ∫ ∇p ⋅ ⃗
φ dx = − ∫ p div ⃗
φ dx + ∫
U
U
ν ⃗
u, ⃗
φ  = ⃗f, ⃗
φ  0
Thus
⃗
⃗
u, φ
U
j=1
∂U
.
⃗ dS = 0.
T 0 p Tφ
⃗ ∈ K,
∀φ
⃗ ∈ H 10 U n : div ⃗
and ⃗
u ∈ V = u
u = 0 is a weak solution of the Stokes system if
= ⃗f, ⃗
φ  0
⃗
u, ⃗
φ
ν
⃗ ∈ V.
∀φ
Conversely, suppose ⃗
u ∈ V is a weak solution of the Stokes system. Then
⃗
u∈V
Moreover,
Then
⃗  0
= ⃗f, φ
⃗
⃗
u, φ
ν
implies
u i ∈ H 10 U, 1 ≤ i ≤ n, hence T 0 u i = u| ∂U = 0.
div ⃗
u = 0.
implies
and
 − ν∇ 2 ⃗
u − ⃗f, ⃗
φ  0 = 0
⃗ ∈ V.
∀φ
⃗ ∈ K.
∀φ
−ν∇ 2 ⃗
u − ⃗f ∈ K 0 and it follows from theorem 1 that, − ν∇ 2 ⃗
u − ⃗f = ∇p
for some p ∈ H 1 U if − ν∇ 2 ⃗
u − ⃗f ∈ L 2loc U n
or for some p ∈ L 2 U if − ν∇ 2 ⃗
u − ⃗f ∈ H −1 U n .
Here p is uniquely determined in L 2∗ U but otherwise there is an arbitrary additive
constant. We seem to have proved,
Theorem 5 (Weak Formulation of Stokes System)
Suppose U ⊂ R n is open, bounded and has Lipschitz smooth boundary ∂U. Then the
following are equivalent:
⃗
u, ⃗
φ
= ⃗f, ⃗
φ  0
⃗ ∈ V.
∀φ
1.
⃗
u ∈ V satisfies
2.
i) ⃗
u ∈ H 10 U n so ⃗
u vanishes on ∂U in the trace sense
ν
ii) ∃ p ∈ L 2∗ U such that − ν∇ 2 ⃗
u − ⃗f = ∇p in D ′ U n
iii) div ⃗
u=0
For U ⊂ R n open, bounded with a Lipschitz smooth boundary ∂U, the Poincare inequality
holds for H 10 U n and it follows that
11
n
∫U ∑ ∇v j ∇u j dx = ∫U ∇v⃗ ⋅ ∇u⃗ dx =
⃗
u, ⃗
φ
j=1
defines an inner product on V, and that V is a Hilbert space for the associated norm. Then
we have
Theorem 6 (Existence of a Weak Solution of Stokes System)
Suppose U ⊂ R n is open, bounded and has Lipschitz smooth boundary ∂U. Then for each
⃗f ∈ L 2 U n there exists a unique ⃗
u ∈ V such that ⃗
u is a weak solution of the Stokes system.
2
In addition there exists a p ∈ L U satisfying ∇p = −ν∇ 2 ⃗
u − ⃗f and p is unique up to an
additive constant.
Proof- For ⃗f ∈ L 2 U n , let Fv⃗ = f⃗, ⃗
v 0 ∀v⃗ ∈ V. Then
| Fv⃗| ≤ || ⃗f || 0 ||v⃗|| 0 ≤ || ⃗f || 0 ||v⃗|| V
∀v⃗ ∈ V.
Then F is a bounded linear functional on V and it follows that there exists a unique ⃗z F ∈ V
such that
Fv⃗ =  ⃗z F , ⃗
v 
Then
implies
∀v⃗ ∈ V.
⃗, ⃗
νu
v  = Fv⃗ =  ⃗z F , ⃗
v 
∀v⃗ ∈ V,
⃗
u= 1
ν ⃗z F ∈ V
is the unique weak solution of the Stokes system. It is clear that div ⃗
u = 0 by virtue of the
⃗, ⃗
fact that ⃗
u ∈ V and νu
v  = Fv⃗ ∀v⃗ ∈ V, implies
−ν∇ 2 ⃗
u − ⃗f, ⃗
φ
=0
⃗ ∈ K,
∀φ
which in turn implies that −ν∇ 2 ⃗
u − ⃗f = ∇p for p ∈ D ′ U. Additionally, ⃗
u ∈ V implies
n
2⃗ ⃗
−1
2
−ν∇ u − f ∈ H U which is sufficient to conclude that p ∈ L U ■
Corollary to Theorem 5 Assertions 1 and 2 are equivalent to
3. ⃗
u ∈ V minimizes
⃗ =
Ju
1
2
⃗, ⃗
νu
u  − ⃗f, ⃗
u  0
Proof- For ⃗
u, ⃗
v∈V
⃗ + t⃗
⃗  + t νu
⃗, ⃗
Ju
v  = Ju
v  − ⃗f, ⃗
v  0 +
1
2
t 2 ν v⃗, ⃗
v .
⃗ + t⃗
⃗  ∀v⃗ ∈ V; i. e. , 1 implies 3. Conversely, 3 implies the term in
Then 1 implies Ju
v  ≥ Ju
⃗ + t⃗
Ju
v  which is linear in t must vanish, and this implies 1.■
Regularity of the Weak Solution
We have seen in the past that if u = ux is a weak solution of the elliptic boundary value
problem
Lux = fx
in U ⊂ R n
12
ux = 0
on ∂U
then
au, v = f, v 0
∀v ∈ V = H 10 U
determines a unique u ∈ V for each f ∈ H −1 U. For ∂U sufficiently regular, we can show
further that
f ∈ H 0 U  u ∈ H 10 U ∩ H 2 U
here ∂U ∈ C 2 
f ∈ H m U  u ∈ H 10 U ∩ H m+2 U
here ∂U ∈ C m .
For the Stokes system, a weak solution ⃗
u ∈ V, p ∈ L 2 U exists for all ⃗f ∈ L 2 U n provided
∂U is smooth. In fact, the solution possesses additional regularity, similar to that found in
the elliptic problem discussed above. We can show that
u || H 2 U n + || p|| H 1 U ≤ C || ⃗f || L 2 U n
|| ⃗
which implies
⃗f ∈ L 2 U n  ⃗
u ∈ V ∩ H 2 U n and p ∈ H 1 U
here ∂U ∈ C 2 .
13