The Navier Stokes Equations
In preparation for considering the N-S equations, we introduce some new definitions and
recall some old ones.
Vector Valued Functions of t
Let X denote a Banach space and, for T > 0, let Cß0, T; Xà denote continuous functions of t,
0 ² t ² T, with values in X. That is,
||fÝt n Þ ? fÝtÞ|| X ¸ 0 as t n ¸ t 5 ß0, Tà.
Cß0, T; Xà is complete for the norm ||f|| Cß0,T;Xà = max ||fÝtÞ|| X .
0²t²T
For 1 ² p < K, let L p ß0, T; Xà denote measurable functions of t, 0 ² t ² T, with values in
X such that
T
X 0 ||fÝtÞ|| X .dt
||f|| L p ß0,T;Xà =
1/p
< K.
L p ß0, T; Xà is complete for this norm and completing the space Cß0, T; Xà in this norm
produces L p ß0, T; Xà; i.e., Cß0, T; Xà is dense in L p ß0, T; Xà.
For 1 ² p < K, let W 1,p ß0, T; Xà denote functions fÝtÞ 5 L p ß0, T; Xà such that for some gÝtÞ
in L p ß0, T; Xà,
t
fÝtÞ = fÝ0Þ + X gÝsÞ ds,
0
t 5 ß0, Tà.
Then it is a classic measure theoretic result that the function f is strongly differentiable with
fÝt + hÞ ? fÝtÞ
h
f v ÝtÞ = lim
h¸0
= gÝtÞ,
and since weak (distributional) and strong derivatives coincide, W 1,p ß0, T; Xà consists of
functions fÝtÞ 5 L p ß0, T; Xà such that f v ÝtÞ 5 L p ß0, T; Xà.
We can define a norm on W 1,p ß0, T; Xà by
|| f|| W 1,p ß0,T;Xà =
T
X 0 á||fÝtÞ|| pX . + ||f v ÝtÞ|| pX âdt
1/p
and W 1,p ß0, T; Xà is complete in this norm and completing the space Cß0, T; Xà in this norm
produces W 1,p ß0, T; Xà; i.e., Cß0, T; Xà is dense in W 1,p ß0, T; Xà.
Let V denote a Banach space continuously,and densely embedded in a Hilbert space H
which is identified with its dual, H = H v . Then we have V Ð H Ð V v so that the duality
pairing on V × V v is realized by extending the inner product on H,
Öf, v× V×V v = Ýf, vÞ H
-v 5 V.
1
Now for 1 < p < K, we define
W p Ý0, TÞ =
fÝtÞ 5 L p ß0, T; Và such that f v ÝtÞ 5 L q ß0, T; V v à
with
|| f|| W p Ý0,TÞ = ||f|| L p + || f v || L q ,
1 + 1 = 1.
p
q
Then it is an extension of a previous result that,
i) W p Ý0, TÞ is contained in Cß0, T; Hà; i.e., there exists a constant C > 0
such that -f 5 W p Ý0, TÞ
|| f|| Cß0,T;Hà ² C|| f|| W p Ý0,TÞ
ii) || uÝ6Þ|| 2H is absolutely continuous on ß0, Tà and
a.e. t 5 ß0, Tà
|| uÝtÞ|| 2H = 2Ýu v ÝtÞ, uÝtÞÞ H
iii) -f, g 5 W p Ý0, TÞ ÝfÝ6Þ, gÝ6ÞÞ H is absolutely continuous on ß0, Tà and
d
dt
d
dt
ÝfÝtÞ, gÝtÞÞ H = Ýf v ÝtÞ, gÝtÞÞ H + Ýf ÝtÞ, g v ÝtÞÞ H
a.e. t 5 ß0, Tà
iv) If V, V v are reflexive, and the embeddings V Ð H, H Ð V v are compact
and continuous, respectively, then the inclusion of W p Ý0, TÞ into L p ß0, T; Hà
is compact.
We are going to consider the evolution N-S equations where the solutions will belong to
L p ß0, T; Và and the equation is understood to hold in the space L q ß0, T; V v à. It follows from the
result i) above that the solution uÝtÞ is absolutely continuous with values in H
not just in V’ so the initial conditions are attained in H.
Linearized N-S Equations
In order to prepare for considering the nonlinear evolution N-S equations, we will first
consider the linearized version.
uÝx, tÞ ? X4 2 3
uÝx, tÞ + 4p = 3fÝx, tÞ Ýx, tÞ 5 U × Ý0, TÞ = U T
/ t3
in U T
div 3
uÝx, tÞ = 0
3
u=3
0
Ýx, tÞ 5 /U × Ý0, TÞ = /U T
3
u 0 ÝxÞ x 5 U
uÝx, 0Þ = 3
2
Now recall
3 5 C KC ÝUÞ n : div d
3 = 0â = divergence free vector test functions
K = ád
3 5 L 2 ÝUÞ n : div 3
3 = 0â
H = completion of K in the norm of L 2 ÝUÞ n = áu
u = 0, Tu
n
n
1
1
3
3
V = completion of K in the norm of H 0 ÝUÞ = áu 5 H 0 ÝUÞ : div u = 0â
and
3, 3
-u
v 5 H,
3, 3
3, 3
u63
v dx = Ýu
vÞH
v ÞÞ 0 = X 3
ÝÝu
U
2
u63
u dx = || 3
u|| 2H .
u || 0 = X 3
|| 3
U
3, 3
-u
v 5 V,
n
3, 3
3, 3
ßßu
vàà= > X / i 3
u 6 / i3
v dx = Ýu
vÞV
U
i=1
n
3|| 2H .
u 6 / i3
u dx = || 4u
u || V = > X / i 3
|| 3
U
2
i=1
It follows from the Poincare inequality that for U bounded, || 3
u || V defines a norm on V that is
n
1
v
equivalent to the H ÝUÞ ? norm. Then V, V are reflexive, and the embeddings
V Ð H, H Ð V v = H ?1 ÝUÞ n are compact and continuous, respectively.
3, 3
v 5 V. Then F 5 V v and we can use the
v àà for 3
For 3
u 5 V, fixed, define FÝv3Þ = ßßu
3 to indicate the dependence of F on 3
notation F = Au
u; i.e., A : V ¸ V v . Then
3, 3
3, 3
3, 3
v × V×V v = ÝAu
v àà = FÝv3Þ = ÖAu
vÞH
ßßu
-v3 5 V.
Then we can say that 3
uÝtÞ 5 L 2 ß0, T; Và is a solution of the linearized N-S system if
3 v ÝtÞ, 3
3ÝtÞ, 3
vÞ H
v àà = Ý3fÝtÞ, 3
v Þ H + Xßßu
Ýu
3
u0
uÝ0Þ = 3
-v3 5 V, a.e.in Ý0, TÞ.
or, equivalently, if
3ÝtÞ = Pf3ÝtÞ,
3
u v ÝtÞ + X Au
3
u0
uÝ0Þ = 3
in V v a.e. in Ý0, TÞ
where
P : L 2 ÝUÞ n = H ã H e ¸ H ,
P : H 10 ÝUÞ n ¸ V ,
and
A = ?P4 2 : V ¸ V v .
u v ÝtÞ 5 L 2 ß0, T; V v à, so 3
uÝtÞ 5 W 2 Ý0, TÞ and the initial condition is
Evidently 3
uÝtÞ 5 L 2 ß0, T; Và, 3
3
satisfied in the sense of Cß0, T; Hà for u 0 5 H.
3
u 0 5 H. Then the linearized N-S system has a
Theorem Suppose 3fÝtÞ 5 L 2 ß0, T; V v à, and 3
unique weak solution 3
uÝtÞ 5 L 2 ß0, T; Và, with
3
u v ÝtÞ 5 L 2 ß0, T; V v à, so 3
uÝtÞ 5 W 2 Ý0, TÞ. The initial condition is satisfied in the sense of
Cß0, T; Hà.
Proof1) Define an approximate solution3 j â denote the orthonormal basis of eigenfunctions of A and, for N = 1, 2...
Let áw
N
3 j,
3
u N ÝtÞ = > c j,N ÝtÞ w
j=1
where the ác j,N â are chosen such that for 1 ² k ² N,
3 kÞH
3 k Þ H + Xßßu
3 k àà = Ýf3ÝtÞ, w
3 N ÝtÞ, w
Ý dtd 3
u N ÝtÞ, w
3 N Ý0Þ, w
3 k Þ H = Ýu
30 , w
3 k ÞH
Ýu
3 j, w
3 j, w
3 k àà = ÝAw
3 k Þ H = V j Ýw
3 j, w
3 k Þ H this approximate system is equivalent to
Since ßßw
d
dt
c j,N ÝtÞ + XV j c j,N ÝtÞ = f j ÝtÞ
30 , w
3 k ÞH
c j,N Ý0Þ = Ýu
1 ² j ² N,
and this linear system of ordinary differential equations has a unique solution for each N.
2) A-priori estimatesFor each N,
aÞ || 3
u N Ý6Þ|| Cß0,T;Hà
bÞ || 3
u N Ý6Þ|| L 2 ß0,T;Và
² C || u 0 || H + || 3f || L 2 ß0,T;V v à
cÞ || / t 3
u N Ý6Þ|| L 2 ß0,T;V v à
3) Existence
u N ÝtÞâ is bounded in L 2 ß0, T; Và by 2b)
á3
u N ÝtÞâ is bounded in L 2 ß0, T; V v à by 2c)
á/ t 3
Then there exists a subsequence á 3
u N v ÝtÞâ such that
3
weakly in L 2 ß0, T; Và
uÝtÞ
u N v ÝtÞ ¸ 3
weakly in L 2 ß0, T; V v à
vÝtÞ
/t 3
u N v ÝtÞ ¸ 3
4
Then the usual distributional argument implies / t 3
u=3
vÝtÞ, hence
3
uÝtÞ 5 W 2 ß0, Tà Ð Cß0, T; Hà.
It remains to be shown that the limit, 3
uÝtÞ, is a weak solution of the linearized N-S
system. Let
VN =
N
3 k , a k 5 C 1 ß0, Tà, -k
3
vÝtÞ = > a k ÝtÞw
k=1
Then it follows from the approximate equation that
X 0 Ýu3vN ÝtÞ, 3vÝtÞÞ H dt + X X 0 ßßu3N ÝtÞ, 3vÝtÞàà dt = X 0 Ýf3ÝtÞ, 3vÝtÞÞ H dt
T
T
T
¹
¹
2
v
weakly in L 2 ß0, T; V à
weakly in L ß0, T; V à
¹
¹
T v
T
T
X Ýu3 ÝtÞ, 3vÝtÞÞ H dt + X X ßßu3ÝtÞ, 3vÝtÞàà dt = X Ýf3ÝtÞ, 3vÝtÞÞ H dt
0
0
0
-v3 5 V N
-v3 5 4 V N
N>0
3 k â is an orthonormal basis for V, it follows that 4 V N is dense in L 2 ß0, T; V à and
Since áw
N>0
u 0 , we choose a
thus the limit is a weak solution for the pde. To see that 3
uÝ0Þ = 3
1
vÝTÞ = 0 and integrate by parts with respect to t in the approximate
v 5 C ß0, T; V à such that 3
equation before passing to the limit.
We can even show that the approximate solution 3
u N converges strongly in the norm of
2
uÝtÞ using the argument we applied to linear parabolic equations in the
L ß0, T; V à to 3
previous course. Uniqueness is proved as it was then as well.n
The Semigroup Approach to NL Navier Stokes
Consider a bounded open set U Ð R n Ýn = 2, 3Þ with smooth boundary /U. Then the
nonlinear N-S system is
3 6 4Þu
3 + 4p = 3fÝx, tÞ Ýx, tÞ 5 U T
/ t3
uÝx, tÞ ? X4 2 3
uÝx, tÞ + Ýu
in U T
div 3
uÝx, tÞ = 0
3
u=3
0
Ýx, tÞ 5 /U × Ý0, TÞ = /U T
3
u 0 ÝxÞ x 5 U
uÝx, 0Þ = 3
This can be expressed abstractly as
where
3 Ýu
3ÝtÞ, tÞ
3ÝtÞ = F
3
u v ÝtÞ + Au
3
3
uÝ0Þ = u 0
in Ý0, TÞ
A = ?PÝX4 2 Þ:D A í H
3 Ýu
3 6 4Þu
3ÝtÞ.
3ÝtÞ, tÞ = Pf3Ý6, tÞ ? PÝu
F
5
We must be able to show that
3 : H J × ß0, KÞ í H
F
such that for some 0 < W < 1,
3 Ýv3ÝbÞ, bÞ || H ² C R |t ? b| W + ||u
3 Ýu
3ÝtÞ, tÞ ? F
3?3
|| F
v || J
-t, b 5 ß0, Tà, 3
u, 3
v 5 B R Ý0Þ Ð H J
If 3fÝtÞ is Lipschitz continuous with values in H, then
Pf3ÝtÞ ? Pf3ÝbÞ
H
² C|t ? b|
-t, b 5 ß0, KÞ.
3 6 4Þu
3 we
3 Þ = Ýu
Now, H J Ð W 1,2 ÝUÞ if 2J > 1 and H J Ð C 0 ÝŪÞ if 4J > n. Then for NÝu
have
3 Þ|| H =
||NÝu
1/2
XU |Ýu3 6 4Þu3| 2 dx
3 || 0
² || 3
u || K ||4u
and
if 4J > n
u || J
u || K ² C 1 || 3
|| 3
3 || V ² C 2 || 3
3 || 0 = ||u
u|| J ,
if 2J > 1.
||4u
Therefore,
2
3 Þ|| H ² C|| 3
u || J
||NÝu
for
J>
J>
which implies
N : H J í H is bounded for J >
In addition,
3 6 4u
3?3
3 Þ ? NÝv3Þ|| H = ||u
v 6 4v3|| H
||NÝu
1
2
3
4
n
4
n=2
n=3
.
3 6 4ÞÝu
3?3
3?3
v Þ|| H + ||u
² ||Ýu
v || K ||4v3|| H
3?3
v Þ|| V + || 3
² || 3
u || J ||Ýu
u?3
v || J ||v3|| V
²
v || J || 3
u?3
v || J
u || J + || 3
u?3
v || J ² 2R || 3
|| 3
for all 3
u, 3
v 5 B R Ý0Þ Ð H J provided J >
3
4
. (when n = 2, J >
1
2
is sufficient). Then we have,
u 0 5 H J , and for 3fÝtÞ Lipschitz continuous in t with values in
Theorem For J > n4 and all 3
H, there is a positive T > 0 and 3
u 5 Cß0, T; H J à V C 1 ß0, T; H 0 à such that u solves the
abstract N-S initial value problem (in the stong sense since ?A generates an analytic
6
semigroup). In addition,
3?3
3 5 Cß0, T; H 0 à.
u ? Au
u 6 4u
4p = 3f ? / t 3
The solution is unique but there is no information about the size of T. We could try to
establish that
uÝtÞ|| J ² K
|| 3
for all t
by means of an energy estimate; i.e. let
EÝtÞ = X
U
1
2
3 Þ dx
3ÝtÞ| 2 ? ®Ýu
|4u
3 Þ.
3 Þ = FÝu
® v Ýu
Then under appropriate hypotheses on F, we may be able to show that E v ÝtÞ ² 0, and
3ÝtÞ|| 21/2 ² k 0 EÝtÞ + k 1 .
||u
3ÝtÞ|| 1/2 but since we need J > 12 to get existence of a
This gives a uniform bound on ||u
solution in H J , this approach fails. (i.e., energy estimates always involve the so-called
3ÝtÞ|| 1/2 = || 3
uÝtÞ|| V ).
”energy norm”, ||u
7