Uniqueness for Weak Solutions to Navier Stokes
We proved the following existence result for Navier-Stokes:
Theorem (Existence of a weak solution, n ² 4) For every 3f 5 L 2 ß0, T : V v à and each
u 5 L 2 ß0, T : Và that is a weak solution of the N-S
u 0 5 H there exists at least one 3
system. In addition, this weak solution satisfies:
(a) 3
u 5 L K ß0, T : Hà and 3
u v 5 L 4/n ß0, T : V v à
(b) 3
uÝtÞ is weakly continuous in H;
Uniqueness in the Case n = 2
When n=2, the weak solution has additional regularity and is, in fact, unique.
Theorem (Uniqueness of the weak solution, n = 2) In the case n = 2, the weak solution
of the Navier Stokes equations satisfies:
u v 5 L 2 ß0, T : V v à hence 3
u ~ ũÝtÞ 5 Cß0, T : Hà
i) 3
u 5 L 2 ß0, T : Và and 3
ii) 3
uÝtÞ is unique
Proof- We begin by repeating an estimate for the nonlinear term in the N-S system.
3 5 H 10 ÝUÞ n , we have (by the extended Holder ineqality followed by
For arbitrary 3
u, 3
v, w
the C-S inequality)
2
2
3, 3
3 Þ| ² > i,j=1
XU |u i / i v j w j | dx ² > i,j=1
v, w
|| u i || 4 || / i v j || 2 || w j || 4
|bÝu
²
2
|| / i v j || 22
> i,j=1
1/2
2
|| u i || 24
> i=1
1/2
2
|| w j || 24
> j=1
1/2
In addition, the Sobolev inequality implies
|| u|| q ² C|| 4u|| Vp || u|| 1?V
r
-u 5 C K0 ÝUÞ
where
0 ² V ² 1,
1/q = VÝ1/p ? 1/nÞ + Ý1 ? VÞ 1/r
If we choose q=4 and p=r=2 so that V = n/4, then with n = 2, we have
1/2
|| u || 4 ² C || 4u|| 1/2
2 || u || 2
-u 5 H 10 ÝUÞ
and, combined with the previous estimate,
1
2
|| u i || 24
> i=1
2
² C > i=1 || u i || 2 || 4u i || 2 ² C || 3
u || H || 3
u || V .
3, 3
3 Þ|
This leads to the following estimate on |bÝu
v, w
3, 3
3 Þ| ² C
v, w
|bÝu
1/2
|| 3
u || H || 3
u || V
.||v|| V
3 || H || w
3 || V .
|| w
1/2
and
3, 3
3, 3
u, 3
vÞ| = |? bÝu
v, 3
uÞ| ² C || 3
u || H || 3
u || V .||v|| V
|bÝu
3, 3
-u
v5V
This implies that
3ÝtÞÞ|| V v ² C || 3
u || H || 3
u || V
||BÝu
Since the weak solution, 3
uÝtÞ, belongs to L K ß0, T : Hà V L 2 ß0, T : Và, we have
T
T
X 0 ||BÝu3ÝtÞÞ|| 2V v dt ² C X 0
|| 3
uÝtÞ || H || 3
uÝtÞ|| V
2
3 || 2K ||u
3 || 2L 2 ß0,T:Và
dt ² C||u
3Ý6ÞÞ|| V v ² C||u
3 || K ||u
3ÝtÞÞ 5 L 2 ß0, T : V v à
3 || L 2 ß0,T:Và and BÝu
hence
||BÝu
Then 3
u v ÝtÞ also belongs to L 2 ß0, T : V v à and then it follows from Theorem 3, p287
in Evans that
3
u 5 L 2 ß0, T : Và
and
Now, for 3
u 5 L 2 ß0, T : Và
3
u v 5 L 2 ß0, T : V v à implies
and
3 v ÝtÞ, 3
uÝtÞ× V×V v =
Öu
3
uÝtÞ ~ ũÝtÞ 5 Cß0, T : Hà
3
u v 5 L 2 ß0, T : V v à we have
1 d
2 dt
2
uÝtÞ|| H
|| 3
Then, if 3
u 1 ÝtÞ, 3
u 2 ÝtÞ are two weak solutions for N-S, let 3
uÝtÞ = 3
u 1 ÝtÞ ? 3
u 2 ÝtÞ . Then
3ÝtÞ = Bßu
3 2 ÝtÞà?Bß 3
3
u 1 ÝtÞà ,
u v ÝtÞ + cAu
and
But
d
dt
3
uÝ0Þ = 0,
3 2 ÝtÞ, 3
3 1 ÝtÞ, 3
uÝtÞ|| 2V = 2bÝu
u 2 ÝtÞ, 3
uÝtÞÞ ? 2bÝu
u 1 ÝtÞ, 3
uÝtÞÞ.
uÝtÞ|| 2H + 2c|| 3
|| 3
3 1 ÝtÞ, 3
3 1 ÝtÞ, 3
3 2 ÝtÞ, 3
3 2 ÝtÞ, 3
u 2 ÝtÞ, 3
uÝtÞÞ ? bÝu
u 1 ÝtÞ, 3
uÝtÞÞ = ?ßbÝu
u 1 ÝtÞ, 3
uÝtÞÞ ? bÝu
u 2 ÝtÞ, 3
uÝtÞÞà
bÝu
3 2 ÝtÞ, 3
3 2 ÝtÞ, 3
3 2 ÝtÞ, 3
3 1 ÝtÞ, 3
u 1 ÝtÞ, 3
uÝtÞÞ ? bÝu
u 1 ÝtÞ, 3
uÝtÞÞ + bÝu
u 1 ÝtÞ, 3
uÝtÞÞ ? bÝu
u 2 ÝtÞ, 3
uÝtÞÞà
= ?ßbÝu
and
3 2 ÝtÞ, 3
3ÝtÞ, 3
3 1 ÝtÞ, 3
u 1 ÝtÞ, 3
uÝtÞÞ ? bÝu
u 1 ÝtÞ, 3
uÝtÞÞ = bÝu
u 1 ÝtÞ, 3
uÝtÞÞ
bÝu
3 2 ÝtÞ, 3
3 2 ÝtÞ, 3
3ÝtÞ, 3
bÝu
u 1 ÝtÞ, 3
uÝtÞÞ ? bÝu
u 2 ÝtÞ, 3
uÝtÞÞ = bÝu
uÝtÞ, 3
uÝtÞÞ = 0
Then
d
dt
2
2
3ÝtÞ, 3
uÝtÞ|| V = ?2bÝu
u 1 ÝtÞ, 3
uÝtÞÞ.
uÝtÞ|| H + 2c|| 3
|| 3
2
But as we showed previously, (and using abc ²
c
C
a2 +
C
c
b2c2 )
3ÝtÞ, 3
uÝtÞÞ| ² C|| 3
uÝtÞ|| H || 3
uÝtÞ|| V || 3
u 1 ÝtÞ|| V
u 1 ÝtÞ, 3
| bÝu
2
2
2
² Cß c || 3
uÝtÞ|| V + C
uÝtÞ|| H || 3
u 1 ÝtÞ|| V à
|| 3
c
C
and therefore,
d
dt
2
uÝtÞ|| H ²
|| 3
2C 2
c
2
2
uÝtÞ|| H || 3
u 1 ÝtÞ|| V
|| 3
2
Since || 3
u 1 ÝtÞ|| V 5 L 1 ß0, Tà, we get
d
dt
2
uÝtÞ|| H exp ? 2Cc
|| 3
2
t
X 0 || 3u 1 ÝsÞ|| 2V ds
²0
2
2
uÝ0Þ|| H = 0, or 3
u 1 ÝtÞ = 3
u 2 ÝtÞ.
uÝtÞ|| H ² || 3
|| 3
i.e.,
Summarizing,
when n = 2 the NS system has a unique solution 3
uÝtÞ 5 Cß0, T : Hà V L 2 ß0, T : Và,
with 3
u v ÝtÞ 5 L 2 ß0, T : V v à. In addition, 3
uÝtÞ 5 L 4 ßU T à
Since the solution is unique in the case n = 2 and is not unique in the case n ³ 3, it must be
that the assumptions under which a physical flow can be treated as a 2-dimensional flow
are also the conditions that preclude turbulence and the consequent loss of uniqueness.
Thus a flow which can be treated as 2-dimensional under some conditions (e.g., shallow
flow with low flow velocity ) may under other conditions behave in ways that cannot be
viewed as 2-dimensional, even though the flow may still be a shallow flow.
Uniqueness in the Case n=3
Recall the corollary to the Sobolev inequality implies
1?n/4
|| u || 4 ² C || 4u|| n/4
2 || u || 2
-u 5 H 10 ÝUÞ
so that, in the case n = 3, we have
1/4
|| u || 4 ² C || 4u|| 3/4
2 || u || 2
-u 5 H 10 ÝUÞ
This leads to the result that the weak solution of N-S satisfies,
3
uÝtÞ 5 L 8/3 ß0, T : L 4 ßUà n à,
and
3
u v ÝtÞ 5 L 4/3 ß0, T : V v à
Note that this is strictly weaker than what was true of the solution in the case n = 2.
3
Theorem Suppose n = 3 and the weak solution of the N-S system satisfies:
3
uÝtÞ 5 L K ß0, T : Hà V L 2 ß0, T : Và
3
uÝtÞ 5 L 8 ß0, T : L 4 ßUà n à,
and
Then, under these conditions, 3
uÝtÞ is unique and 3
uÝtÞ 5 Cß0, T : Hà
Proof- Supppose 3
uÝtÞ is a weak solution for N-S with the additional regularity indicated
above.
Then
2
3ÝtÞ, 3
-3
uÝtÞ, 3
v 5 V,
v || V
uÝtÞ, 3
v Þ| ² C || 3
uÝtÞ|| 4 || 3
|bÝu
implies
3ÝtÞÞ|| V v ² C || 3
uÝtÞ|| 24
||BÝu
and
T
T
T
np/2
X 0 ||BÝu3ÝtÞÞ|| pV v dt ² C X 0 || 3uÝtÞ|| 2p
dt ² C X || 3
uÝtÞ|| V dt
4
0
3ÝtÞÞ 5 L p ß0, T : V v à if 3
It follows that BÝu
uÝtÞ 5 L 2 ß0, T : Và and np/2 ² 2; i.e.,
3ÝtÞÞ 5 L 4/n ß0, T : V v à = L 4/3 ß0, T : V v à.
BÝu
On the other hand, if
3
uÝtÞ 5 L 8 ß0, T : L 4 ßUà n à, then
T
X 0 || 3uÝtÞ|| 2p
dt < K for p ² 4
4
and
3ÝtÞÞ 5 L 2 ß0, T : V v à
BÝu
(at least)
Then it follows from the N-S equation that 3
u v ÝtÞ 5 L 2 ß0, T : V v à and this, together
uÝtÞ 5 Cß0, T : Hà.
with 3
uÝtÞ 5 L 2 ß0, T : Và imply 3
Now observe that
3ÝtÞ, 3
uÝtÞ, 3
v Þ| ² C 0 || 3
uÝtÞ|| 4 || 3
uÝtÞ|| V || 3
v || 4
|bÝu
1/4
3/4
² C 1 || 3
uÝtÞ|| 2 || 3
uÝtÞ|| V
1/4
uÝtÞ|| V || 3
v || 4
|| 3
7/4
² C 1 || 3
uÝtÞ|| H || 3
uÝtÞ|| V || 3
v || 4
Now suppose 3
u 1 ÝtÞ, 3
u 2 ÝtÞ are two weak solutions for N-S both of which have the additional
u 2 ÝtÞ , and note that as in the n = 2 proof
regularity of the hypotheses. Let 3
uÝtÞ = 3
u 1 ÝtÞ ? 3
4
d
dt
2
2
3ÝtÞ, 3
uÝtÞ|| V = 2bÝu
uÝtÞ, 3
u 2 ÝtÞÞ.
uÝtÞ|| H + 2c|| 3
|| 3
Also
1/4
7/4
2
2
8
3ÝtÞ, 3
uÝtÞ|| V + C 2 || 3
| bÝu
uÝtÞ, 3
u 2 ÝtÞÞ.| ² C 1 || 3
uÝtÞ|| H || 3
uÝtÞ|| V || 3
u 2 ÝtÞ|| 4 ² c|| 3
uÝtÞ|| H || 3
u 2 ÝtÞ|| 4
d
dt
and so
2
8
2
uÝtÞ|| H ² C 2 || 3
u 2 ÝtÞ|| 4 || 3
uÝtÞ|| H .
|| 3
8
Since || 3
u 2 ÝtÞ|| 4 5 L 1 ß0, Tà, we can finish the proof as in the n = 2 case.
In the step,
7/4
2
8
3
uÝtÞ|| 1/4
uÝtÞ|| 2V + C 2 || 3
uÝtÞ|| V || 3
u 2 ÝtÞ|| 4 ² c|| 3
uÝtÞ|| H || 3
u 2 ÝtÞ|| 4
|| 3
H ||
we used the following version of Young’s inequality,
ab ²
ap
p
+
bq
q
² Oa p + CÝOÞb q ,
1
p
+
1
q
= 1, a, b > 0
5