Evolution Navier Stokes Equations
For U Ð R n , n ² 4, a bounded open set with a smooth boundary, @, consider the full Navier
Stokes system for f 5 L 2 ÝU T Þ and u 0 5 L 2 ÝUÞ,
3 6 4Þu
3 + 4p = 3f
/ t3
uÝx, tÞ ? X4 2 3
uÝx, tÞ + Ýu
div 3
u=0
3
uÝx, tÞ = 0
3
uÝx, 0Þ = 3
u0
in U T
in U T
on @ T
in U
We define a weak solution for this system to consist of a vector valued function
3
v 5 V and
uÝx, tÞ 5 L 2 ß0, T : Và and a scalar valued pÝx, tÞ 5 L 2 ß0, T : Hà such that for all 3
almost all t 5 Ý0, TÞ,
3ÝtÞ, 3
3 v ÝtÞ, 3
3ÝtÞ, 3
vÞ H
vÞ = Ýf3ÝtÞ, 3
uÝtÞ, 3
v àà + bÝu
v Þ + Xßßu
Ýu
H
3Ý0Þ ? 3
u0, 3
Ýu
vÞ H = 0
and
3ÝtÞà ? 3
3ÝtÞ ? Bßu
u v ÝtÞ
4pÝtÞ = 3fÝtÞ ? XAu
where A, Bß6à are defined by
3ÝtÞ, 3
3ÝtÞ, 3
3ÝtÞ, 3
v × V v v ×V = ÝAu
v àà = ÖAu
vÞH
F 3u Ýv3Þ = ßßu
3ÝtÞà, 3
3ÝtÞ, 3
vÞ
v × V v v ×V = bÝu
uÝtÞ, 3
ÖBßu
-v3 5 V
-v3 5 V
With respect to the last definition, recall that
n
3, 3
3 Þ = > i,j=1
XU u i / i v j w j dx
bÝu
v, w
3, 3
3 5 H 1o ÝUÞ n
-u
v, w
Since the extended Holder inequality implies
XU u i / i v j w j dx ² |u i | 4 |/ i v j | 2 |w j | 4
and H 1o ÝUÞ n is continuously embedded in L 4 ÝUÞ n when n ² 4, it follows that
| X u i / i v j w j dx| ² C||u i || H 1o ÝUÞ ||v j || H 1o ÝUÞ ||w j || H 1o ÝUÞ .
U
This gives,
3 || 2V ||v3|| V
3ÝtÞ, 3
v Þ| ² C ||u
uÝtÞ, 3
|bÝu
1
and so
3ÝtÞà, 3
3ÝtÞ, 3
vÞ
v × V v v ×V = bÝu
uÝtÞ, 3
ÖBßu
-v3 5 V
3ÝtÞà of V’. Then the weak equation can be written
defines an element Bßu
as
3
3ÝtÞà = Pf3ÝtÞ,
3ÝtÞ + Bßu
u v ÝtÞ + XAu
a.e. in Ý0, TÞ
3o
3
uÝ0Þ = Pu
3ÝtÞà that 3
3ÝtÞ ? Bßu
u v ÝtÞ 5 L 2 ß0, T : V v à
We would now like to conclude from 3
u v ÝtÞ = Pf3ÝtÞ ? XAu
3ÝtÞà belongs. Here, P denotes the
but we do not, at this point, know to what space Bßu
projection from L 2 ßUà to H. From this point on, we do not distinguish between Pf3ÝtÞ and 3fÝtÞ,
3 o and 3
nor Pu
uo.
3ÝtÞà 5 L 1 ß0, T : V v à
Lemma 1 If 3
uÝtÞ 5 L 2 ß0, T : Và then Bßu
Proof- For 3
uÝtÞ 5 L 2 ß0, T : Và,
T
T
T
X 0 ||Bßu3ÝtÞà|| V v dt = X 0 sup |ÖBßu3ÝtÞà, v×|dt = X 0 sup | bÝu3ÝtÞ, 3uÝtÞ, 3v Þ| dt
||v|| V =1
||v|| V =1
T
3ÝtÞ|| 2V dt = C ||u|| 2L 2 ß0,T:Và < K
² C X ||u
0
3ÝtÞà5 L 1 ß0, T : V v à.
This proves that Bßu
As a result of the lemma, we have that if 3
uÝtÞ 5 L 2 ß0, T : Và solves the weak NS equation,
then 3
u v ÝtÞ 5 L 1 ß0, T : V v à. Since L 2 ß0, T : Và is continuously embedded in L 1 ß0, T : V v à
uÝtÞ u ũÝtÞ 5
we have 3
uÝtÞ 5 W 1,1 ß0, T : V v à and it follows then from a previous lemma that 3
v
Cß0, T : V à
not Cß0, T : Hà ! . We can improve on this when n ² 4.
Lemma 2 Suppose 3
uÝtÞ 5 L 2 ß0, T : Và solves the weak NS system with n ² 4. Then
3
uÝtÞ 5 L 2p ß0, T : L 4 ÝUÞ n à
and
3
u v ÝtÞ 5 L p ß0, T : V v à
p ² 4/n.
In particular, we have the following cases when n = 2, 3, 4 :
(i)
n=2
and
(ii)
n=3
3
uÝtÞ 5 L 4 ß0, T : L 4 ÝUÞ n à p L 4 ÝU T Þ n
3
Ýthen 3
uÝtÞ u ũÝtÞ 5 Cß0, T : Hà Þ
u v ÝtÞ 5 L 2 ß0, T : V v à
3
uÝtÞ 5 L 8/3 ß0, T : L 4 ÝUÞ n à
3
u v ÝtÞ 5 L 4/3 ß0, T : V v à
2
(iii)
n=4
Lemma 2 = Lemma 1
Proof- If 3
uÝtÞ 5 L 2 ß0, T : Và solves the weak NS equation, then
3ÝtÞ, 3
3 v ÝtÞ, 3
3ÝtÞ, 3
uÝtÞÞ H
uÝtÞÞ = Ýf3ÝtÞ, 3
uÝtÞ, 3
uÝtÞàà + bÝu
uÝtÞÞ H + Xßßu
Ýu
3ÝtÞ|| 2H + X||u
3ÝtÞ|| 2V ² ||f|| V v ||u
3ÝtÞ|| V
1/2 d/dt ||u
3ÝtÞ, 3
uÝtÞÞ = 0
uÝtÞ, 3
recall bÝu
Then
3ÝtÞ|| 2H + X||u
3ÝtÞ|| 2V ² 1/2 1/X||fÝtÞ|| 2V v + X||u
3ÝtÞ|| 2V
1/2 d/dt ||u
and
3ÝtÞ|| 2V ² 1/X||fÝtÞ|| 2V v
3ÝtÞ|| 2H + X||u
d/dt ||u
(1)
(a) First energy estimate- Integrate (1) from t = 0 to t = b ² T,
3 0 || 2H + 1/X||f|| 2L 2 ß0T:V v à
3ÝbÞ|| 2H ² ||u
||u
Since this holds for all b ² T,
3Ý6Þ|| 2Cß0,T:Hà ² ||u
3 0 || 2H + 1/X||f|| 2L 2 ß0T:V v à
3ÝbÞ|| 2H = ||u
max ||u
0² b²T
(b) Second energy estimate- Integrate (1) from t = 0 to t = T,
T
3 0 || 2H + 1/X||f|| 2L 2 ß0T:V v à
3ÝtÞ|| 2V dt ² ||u
X X ||u
0
i.e.,
3Ý6Þ|| 2L 2 ß0,T:Và ² 1/X ||u
3 0 || 2H + 1/X 2 ||f|| 2L 2 ß0T:V v à
||u
Now recall,
3 || Vp ||u
3 || 1?V
3 || q ² C||4u
||u
r
-u 5 C K0 ÝUÞ
where
0 ² V ² 1,
1/q = VÝ1/p ? 1/nÞ + Ý1 ? VÞ 1/r
Choose q=4 and p=r=2 so that V = n/4. Then
3ÝtÞ|| n/4
3ÝtÞ|| 1?n/4
3ÝtÞ|| 4 ² C ||4u
||u
||u
2
2
If 3
uÝtÞ 5 L 2 ß0, T : Và solves the weak NS equation, then by (a)
3ÝbÞ|| H ² Ý||u
3 0 || 2H + 1/X||f|| L2 2 ß0T:V v à Þ 1/2 = C 1
3ÝbÞ|| 2 = ||u
||u
3
and
3ÝtÞ|| 4 ² C ||4u
3ÝtÞ|| n/4
3ÝtÞ|| n/4
C 1?n/4
= C 2 ||u
||u
1
2
V
In addition,
3 || 2L 4 ÝUÞ ||v3|| V
3ÝtÞ, 3
v Þ| ² ||u
uÝtÞ, 3
|bÝu
3 || 2L 4 ÝUÞ
3ÝtÞà|| V v v ² ||u
||Bßu
implies
0 ² t ² T,
and it follows that
T
T
T
X 0 ||Bßu3ÝtÞà|| pV vv dt ² X 0 ||u3|| L2p4 ÝUÞ dt ² C 2 X 0 ||u3ÝtÞ|| np/2
dt
V
This last integral is finite for np/2 ² 2 and 3
uÝtÞ 5 L 2 ß0, T : Và; then we
have proved that
3
uÝtÞ 5 L 2p ß0, T : L 4 ÝUÞ n à
and
3ÝtÞà, 3
u v ÝtÞ 5 L p ß0, T : V v à
Bßu
p ² 4/n.
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;
3ÝtÞ, vÞ H
i.e., t n ¸ t 5 Ý0, TÞ implies Ý 3
uÝt n Þ, vÞ H ¸ Ýu
-v 5 H
3 j â denote the eigenfunctions of the Stokes operator A. Then
Proof- Let áw
3 j, w
3 j, w
3 j, w
3 k àà = ÝAw
3 k Þ H = V j Ýw
3 k Þ H = 0 if j ® k
ßßw
3 j || H = 1 for all j.
3 vk s are orthogonal in both V and in H. Let ||w
Then the w
sequence of approximate solutions
For N = 1, 2, ... let
N
3j
3
u N ÝtÞ = > j=1 c j Ýt; NÞ w
satisfy the approximate N-S equation,
4
3 vN ÝtÞ, w
3 N ÝtÞ, 3
3 k Þ H + Xßßu
3 N ÝtÞ, w
3 k àà + bÝu
3 k Þ = Ýf3, w
3 kÞH
u N ÝtÞ, w
Ýu
3 vk s implies that
for 1 ² k ² N. Then the orthogonality of the w
c vk Ýt, NÞ + XV k c k Ýt, NÞ + > i,j=1 B ijk c i Ýt; NÞ c j Ýt; NÞ = 3f k ÝtÞ
30 , w
3 k ÞH
c k Ý0; NÞ = Ýu
N
where
3 i 6 4w
3j 6 w
3 k dx,
B ijk = X w
U
3f k ÝtÞ = Ýf3, w
3 kÞH.
This is a system of N nonlinear ordinary differential equations for the unknown
coefficients, c j Ýt; NÞ. This initial value problem has a solution on ß0, T N Þ for each
N, and, since we are going to prove a-priori bounds on 3
u N ÝtÞ that are uniform in
N on Ý0, TÞ, it follows that T N = T for every N.
a-priori estimates
For each N, 3
u N ÝtÞ solves the approximate N-S equation, hence it follows by the
same arguments used to prove lemma 2, that for every N,
3 0 || 2H + 1/X||f|| L2 2 ß0T:V v à
3 N Ý6Þ|| 2Cß0,T:Hà ² ||u
(a) ||u
3 N Ý6Þ|| 2L 2 ß0,T:Và ² 1/X ||u
3 0 || 2H + 1/X 2 ||f|| L2 2 ß0T:V v à
(b) ||u
3 N Ý6Þâ is a bounded infinite set in L K ß0, T : Hà and in L 2 ß0, T : Và. In addition
i.e., áu
the lemma 2 arguments imply that for p ² 4/n,
3 N Ý6Þâ is a bounded infinite set in L 2p ß0, T : L 4 ÝUÞ n à
(c) áu
3 vN Ý6Þâ is a bounded infinite set in L p ß0, T : V v à
(d) áu
3 N Ý6Þâ) such that
It follows that there exists a subsequence (which we also denote by áu
i)
uÝtÞ 5 L 2 ß0, T : Và
(b) implies that 3
u N Ý6Þ ¸ 3
ii)
(a) implies that 3
u N Ý6Þ ¸ 3
uÝtÞ 5 L K ß0, T : Hà
weakly in L 2 ß0, T : Và
weak-* in L K ß0, T : Hà
u v ÝtÞ 5 L p ß0, T : V v à weakly in L p ß0, T : V v à
iii) (d) implies that 3
u vN Ý6Þ ¸ 3
This last assertion relies on the fact that p = 4/n > 1 for n=2,3, so L p ß0, T : V v à is
reflexive with dual space L q ß0, T : V v à. Finally, it follows from i) and iii) and a previous
3 N Ý6Þâ) such that
lemma, that there exists a subsequence (which we also denote by áu
5
uÝtÞ 5 L 2 ß0, T : Và
iv) 3
u N Ý6Þ ¸ 3
strongly in L 2 ß0, T : Hà p L 2 ßU T à
passing to the limit
Now for 3
vÝtÞ 5 Cß0, T : Và Ð L 2 ß0, T : Và we have
T
T
X 0 Ýf3ÝtÞ, 3vÝtÞÞ H dt = X 0 3fÝtÞ, 3vÝtÞ
dt
is well defined,
X 0 ßßu3N ÝtÞ, 3vÝtÞàà dt ¸ X 0 ßßu3ÝtÞ, 3vÝtÞàà dt
(follows from i)
T
V v ×V
T
and, since Cß0, T : Và Ð L q ß0, T : Và
T
T
T
X 0 Ýu3vN ÝtÞ, 3vÝtÞÞ H dt = X 0 Öu3vN ÝtÞ, 3vÝtÞ× V v ×V dt ¸ X 0 Ýu3v ÝtÞ, 3vÝtÞÞ H dt
N
3 i , a i 5 Cß0, Tà
a i ÝtÞw
> i=1
Cß0, T : V N à =
Then, for
(follows from iii)
it follows from the approximate N-S equation that for every 3
vÝtÞ 5 Cß0, T : V N à
X 0 Ýu3vN ÝtÞ, 3vÝtÞÞ H + Xßßu3N ÝtÞ, 3vÝtÞàà + bÝu3N ÝtÞ, 3u N ÝtÞ, 3vÝtÞÞ ? Ýf3ÝtÞ, 3vÝtÞÞ H dt = 0
T
and it only remains to show that we can pass to the limit in the nonlinear term in
order to show that 3
uÝtÞ is a solution of the weak N-S equation.
Note that
T
T
X 0 ÝBßu3N ÝtÞà, 3vÝtÞÞ H dt = X 0 bÝu3N ÝtÞ, 3u N ÝtÞ, 3vÝtÞÞdt
T
3 N ÝtÞ, 3
= ? X bÝu
vÝtÞ, 3
u N ÝtÞÞdt
0
T
= ? X > i,j=1 X 3
u i,N ÝtÞ / i3
v j ÝtÞ 3
u j,N ÝtÞ dx dt
0
U
n
where 3
u j,N ÝtÞ denotes the j-th component of 3
u N ÝtÞ. Now it follows from
u j ÝtÞ strongly in L 2 ÝU T Þ which implies, in turn, that
iv) that 3
u j,N ÝtÞ ¸ 3
3
u 2j ÝtÞ
u 2j,N ÝtÞ ¸ 3
and
3
u j,N ÝtÞ 3
u i,N ÝtÞ ¸ 3
u j ÝtÞ 3
u i ÝtÞ as N ¸ K
where the convergence is strong convergence in L 1 ÝU T Þ. i.e.,
T
2
u 2j ÝtÞ || L 1 ÝU T Þ = X X | 3
u j,N
ÝtÞ ? 3
u 2j ÝtÞ | dx dt
|| 3
u 2j,N ÝtÞ ? 3
0 U
6
T
= X X |3
u j,N ÝtÞ ? 3
u j ÝtÞ | | 3
u j,N ÝtÞ + 3
u j ÝtÞ | dx dt
0 U
u j || L 2 ÝU T Þ || 3
u j,N ? 3
u j || L 2 ÝU T Þ ² C|| 3
u j,N ? 3
u j || L 2 ÝU T Þ
² || 3
u j,N + 3
and,
3 j,N ÝtÞ 3
||u
u i,N ÝtÞ ¸ 3
u j ÝtÞ 3
u i ÝtÞ || L 1 ÝU T Þ ²
² || 3
u i,N || L 2 ÝU T Þ || 3
u j,N ? 3
u j || L 2 ÝU T Þ + || 3
u j || L 2 ÝU T Þ || 3
u i,N ? 3
u i || L 2 ÝU T Þ
Now we combine these results to conclude that as N ¸ K,
T
X 0 ÝBßu3N ÝtÞà ? Bßu3ÝtÞà, 3vÝtÞÞ H dt =
T
3 i ÝtÞ 3
= X > i,j=1 X ßu
u j,N ÝtÞ ? 3
u i,N ÝtÞ 3
u j,N ÝtÞà/ i3
v j ÝtÞ dx dt
0
U
n
n
3 i ÝtÞ 3
² > i,j=1 ||u
u j,N ÝtÞ ? 3
u i,N ÝtÞ 3
u j,N ÝtÞ|| L 1 ÝU T Þ ||/ i3
v j ÝtÞ || K ¸ 0
It follows now that
T
X 0 Ýu3v ÝtÞ, 3vÝtÞÞ H + XÝÝu3ÝtÞ, 3vÝtÞÞÞ + bÝu3ÝtÞ, 3uÝtÞ, 3vÝtÞÞ ? Ýf3ÝtÞ, 3vÝtÞÞ H dt = 0
N
3 i , a i 5 Cß0, Tà for every N, hence it
This holds for every 3
v 5 Cß0, T : V N à = > i=1 a i ÝtÞw
must hold for every 3
v 5 Cß0, T : Và, and by continuity, for every 3
v 5 L 2 ß0, T : Và.
This completes the proof that the weak limit of the sequence of approximate solutions is a
solution of the weak N-S equation. Using a test function 3
v 5 C 1 ß0, T : Và, with vÝTÞ = 0 and
integrating the time derivative term by parts leads to the result that the weak solution must
u o by the same argument that we used for the linear
satisfy the initial condition, 3
uÝ0Þ = 3
parabolic problems. This completes the proof that the N-S equation has at least one weak
solution.
The weak continuity of 3
uÝtÞ is a result of the facts:
uÝtÞ weak-* in L K ß0, T : Hà so 3
uÝtÞ 5 L K ß0, T : Hà
(i) 3
u N Ý6Þ ¸ 3
(ii) 3
uÝtÞ 5 L 2 ß0, T : Và
and
3
u v ÝtÞ 5 L 4/n ß0, T : V v à
so
3
uÝtÞ 5 Cß0, T : V v à
(iii) Lemma- If X and Y are two Banach spaces with X continuously embedded in Y
and if u 5 L K ß0, T : Xà is weakly continuous t with values in Y, then u is weakly
continuous t with values in X.
7
In the case n=2 we can show that the weak solution is unique and belongs to Cß0, T : Hà .
In the case n=3 the weak solution has not been shown to be unique and the weak solution
is only weakly continuous from ß0, Tà to H.
8