Examples Using Energy Estimates
We will use energy estimates for solutions to the N-S equations to derive existence results
for solutions having various types of regularity. Of course the various types of regularity for
the solution are determined by the regularity assumptions on the data in the problem and it
will be our aim to show precisely how the energy estimates are modified to correspond to
the data regularity in such a way as to lead to the various kinds of regularity for the solution.
Recall first, that in the basic existence proof we derived the following estimate for the
nonlinear term
3, 3
u, 3
vÞ| ² C || 3
u || 2V || 3
v || V
|bÝu
for 3
u, 3
v 5 H 10 ÝUÞ n
3ÝtÞÞ : L 2 ß0, T : Và ¸ L p ß0, T : V v à for p ² 4/n.
and this eventually led to the result that BÝu
u 0 5 H, on the data lead to a weak solution
Then the assumptions, 3fÝtÞ 5 L 2 ß0, T : V v à, 3
2
K
v
3
u ÝtÞ 5 L p ß0, T : V v à for p ² 4/n.
uÝtÞ 5 L ß0, T : Và V L ß0, T : Hà, with 3
Example 1 Here, we assume that n = 2, and
3fÝtÞ and 3f v ÝtÞ 5 L 2 ß0, T : V v à, 3fÝ0Þ 5 H
3
u 0 5 H 2 ÝUÞ n V V.
and
3 j â a family of functions in K (the divergence free test functions in U) that
Then for áw
spans a dense subset of V, let
N
3j
3
u N ÝtÞ = > j=1 C j,N ÝtÞ w
satisfy
3 k Þ H + cßßu
3 k àà + bÝu
3 N ÝtÞ, 3
3 k Þ = 3fÝtÞ, w
3 vN ÝtÞ, w
3 N ÝtÞ, w
3k
u N ÝtÞ, w
Ýu
for 1 ² k ² N.
H
,
Ý1Þ
3 0 à, where P N denotes the
u N Ý0Þ = P N ßu
Since 3
u 0 5 H 2 ÝUÞ n V V, we can also let 3
2
n
3 1 , ..., w
3 N à. Then
orthogonal projection in H ÝUÞ V V onto the subspace V N = spanßw
clearly,
3 0 à || H 2 ² || 3
30 à ? 3
u 0 || H 2 ¸ 0 as N ¸ K
u 0 || H 2 and ||P N ßu
|| 3
u N Ý0Þ|| H 2 = || P N ßu
The initial value problem for (1) has a unique solution for each N and by multiplying
(1) by C vj,N ÝtÞ and forming linear combinations, we can infer that
3 vN ÝtÞ, 3
3 N ÝtÞ, 3
3 N ÝtÞ, 3
u vN ÝtÞÞ H + cßßu
u vN ÝtÞàà + bÝu
u N ÝtÞ, 3
u vN ÝtÞÞ = 3fÝtÞ, 3
u vN ÝtÞ
Ýu
H
,
1
At t=0 in particular, we have
u vN Ý0Þ
u vN Ý0Þ|| 2H = 3fÝ0Þ, 3
|| 3
= 3fÝ0Þ, 3
u vN Ý0Þ
H
3 N Ý0Þ, 3
3 N Ý0Þ, 3
u vN Ý0Þàà ? bÝu
u N Ý0Þ, 3
u vN Ý0ÞÞ
? cßßu
H
3 N Ý0Þ, 3
3 N Ý0Þ, 3
u vN Ý0ÞÞ H ? bÝu
u N Ý0Þ, 3
u vN Ý0ÞÞ
? cÝAu
and, since we assumed that 3fÝ0Þ 5 H,
2
3 N Ý0Þ|| H || 3
3 N Ý0Þ| | H || 3
u vN Ý0Þ|| H + c || Au
u vN Ý0Þ|| H + ||Bu
u vN Ý0Þ|| H
u vN Ý0Þ|| H ² ||3fÝ0Þ|| H || 3
|| 3
i.e.,
3 N Ý0Þ|| H + ||Bu
3 N Ý0Þ| | H
u vN Ý0Þ|| H ² ||3fÝ0Þ|| H + c || Au
|| 3
Now observe that,
3 N Ý0Þ = ?4 2 3
ÝiÞ Au
u N Ý0Þ hence
3 N Ý0Þ|| H = ||4 2 3
|| Au
u N Ý0Þ|| H ² C|| 3
u N Ý0Þ|| H 2 ² C|| 3
u 0 || H 2
3 || 4 || 3
3, 3
v || 2 (by the generalized Holder inequality)
u, 3
vÞ| ² C|| 3
u || 4 || 4u
ÝiiÞ |bÝu
The Sobolev inequality for n=2 implies,
3 || 1/2
u || 1/2
u || 4 ² || 3
|| 4u
|| 3
2
2
and
3 || 4 ² ||4 3
u || 1/2
u || 1/2
|| 4u
|| 4 2 3
2
2
Then the Cauchy inequality gives,
1/2
1/2
3 || 2 ² || 3
u || 2 || 4u
u || V ,
|| 3
1/2
1/2
u || 2 || 4 2 3
u || 2 ² || 3
u || H 2
||4 3
3, 3
and, finally, |bÝu
u, 3
vÞ| ² C || 3
u || V || 3
u 5 H 2 ÝUÞ n V V
u || H 2 ||v3|| H for all 3
and all v 5 H. It follows that for each 3
u 5 H 2 ÝUÞ n V V there is a
3Þ 5 H such that |bÝu
3, 3
3 Þ, 3
v Þ H -v 5 H.
unique BÝu
u, 3
vÞ| = ÝBÝu
Then
3 N Ý0Þ|| H ² C || 3
u N Ý0Þ|| V || 3
u N Ý0Þ|| H 2
||Bu
2
² C || 3
u N Ý0Þ|| H 2 ² C || 3
u 0 || H 2
These results lead to,
u 0 || H 2 + C || 3
u 0 || H 2 = M 0 Ý3fÝ0Þ, 3
u0Þ
u vN Ý0Þ|| H ² ||3fÝ0Þ|| H + cC || 3
|| 3
2
u vN Ý0Þâ is a bounded sequence in H.
á3
i.e,
By differentiating the approximate equation with respect to t , and then
forming linear combinations of the results, we get
3 ”N ÝtÞ, 3
3 vN ÝtÞ, 3
3 vN ÝtÞ, 3
u vN ÝtÞÞ H + cßßu
u vN ÝtÞàà + bÝu
u N ÝtÞ, 3
u vN ÝtÞÞ
Ýu
3 N ÝtÞ, 3
u vN ÝtÞ, 3
u vN ÝtÞÞ = 3f v ÝtÞ, 3
u vN ÝtÞ
+ bÝu
H
,
Note that since we have assumed 3f v ÝtÞ 5 L 2 ß0, T : V v à, then
3f v ÝtÞ, 3
u vN ÝtÞ
and
1 d
2 dt
V v ×V
.
2
u vN ÝtÞ|| H
|| 3
3, 3
3 Þ| ² C ||u
3 || H ||u
3 || V
v, w
|bÝu
and since
for n=2,
Now
= 3f v ÝtÞ, 3
u vN ÝtÞ
3 ”N ÝtÞ, 3
u vN ÝtÞÞ H =
Ýu
In addition,
d
dt
H
1/2
3 || H ||w
3 || V
||v3|| V ||w
1/2
2
3 vN ÝtÞ|| H ||u
3 vN ÝtÞ|| V || 3
u vN ÝtÞ|| 2V ² 2 ||3f v ÝtÞ|| V v || 3
u vN ÝtÞ|| V + 2C||u
u N ÝtÞ|| V
u vN ÝtÞ|| H + 2c|| 3
|| 3
3 vN ÝtÞ|| H ||u
3 vN ÝtÞ|| 2H ||u
3 vN ÝtÞ|| V || 3
3 N ÝtÞ|| 2V
2C||u
u N ÝtÞ|| V ² c|| 3
u vN ÝtÞ|| 2V + CÝcÞ ||u
2 ||3f v ÝtÞ|| V v || 3
u vN ÝtÞ|| V ²
therefore,
d
dt
2
u vN ÝtÞ|| H +
|| 3
1
2
1
2
c|| 3
u vN ÝtÞ|| 2V + c/2 ||3f v ÝtÞ|| 2V v
3 vN ÝtÞ|| 2H ||u
3 N ÝtÞ|| 2V
c|| 3
u vN ÝtÞ|| 2V ² C ||3f v ÝtÞ|| 2V v + ||u
Ý2Þ
The approximate equation has been manipulated into the form of an energy
estimate from which the basic a-priori estimates can be derived.
Note first that (2) implies
d
dt
2
u vN ÝtÞ|| 2H ² C ||3f v ÝtÞ|| 2V v
u vN ÝtÞ|| H ? a N ÝtÞ || 3
|| 3
where
a N ÝtÞ = C || 3
u N ÝtÞ|| 2V .
Since we showed in the course of proving the basic existence theorem that
(under weaker assumptions on the data) || 3
u N || L 2 ß0,T:Và is bounded, uniformly in N,
3
it follows that a N ÝtÞ 5 L 1 ß0, Tà. Then the Gronwall inequality implies
t
X a N ÝtÞdt
3 vN Ý0Þ|| 2H e 0
|| 3
u vN ÝtÞ|| 2H ² C ||3f v ÝtÞ|| 2V v + ||u
² C ||3f v ÝtÞ|| 2V v + M 0
Since this holds for all t, 0 ² t ² T, we have proved
|| 3
u vN || L K ß0,T:Hà ² M 1
u vN ÝtÞâ is a bounded sequence in L K ß0, T : Hà
á3
i.e.,
A second a-priori estimate is obtained by integrating (2) from 0 to T. Then
1
2
T
T
T v
3 N ÝtÞ|| 2H ||u
3 N ÝtÞ|| 2V dt
c X || 3
u vN ÝtÞ|| 2V dt ² C X ||3f v ÝtÞ|| 2V v dt + X ||u
0
0
0
3 vN || 2L K ß0,T:Hà ||u
3 N || 2L 2 ß0,T:Và
|| 3
u vN || 2L 2 ß0,T:Và ² C ||3f v || L2 2 ß0,T:V v à + ||u
i.e.,
² M2
u vN ÝtÞâ is a bounded sequence in L 2 ß0, T : Và
á3
These a-priori estimates now imply the existence of a subsequence converging
weakly in L 2 ß0, T : Và and weak-* in L K ß0, T : Hà to a limit vÝtÞ 5 L 2 ß0, T : VàVL K ß0, T : Hà
In addition, we can use distributional arguments to show that vÝtÞ = u v ÝtÞ where
3 N ÝtÞâ.
uÝtÞ denotes the limit in L 2 ß0, T : Và of the sequence áu
Then these hypotheses on the data have led to the existence of a weak solution 3
uÝtÞ
of the N-S equations with
3
uÝtÞ 5 L 2 ß0, T : VàVCß0, T : Hà
and
u v ÝtÞ 5 L 2 ß0, T : VàVL K ß0, T : Hà
Example 2 Here we assume n=2,3 and U Ð R n has a smooth (i.e., C 2 Þ boundary.
We will suppose the data satisfies,
3fÝtÞ 5 L 2 ß0, T : Hà,
and
3
u 0 5 V.
For 3
u 5 H 2 ÝUÞ n V V
2
2
3 || 2H = || 4 2 3
u || H ² > |J|²2 ||D J 3
u || H = || 3
u || 2H 2
ÝiÞ || Au
3 5 L 2 ÝUÞ n but then the regularity theorem for the Stokes’ problem implies
ÝiiÞ 3f = Au
3
3 || 2H
u 5 H 2 ÝUÞ n and || 3
u || 2H 2 ² C|| 3f || 2H = C || Au
4
3 || H defines a norm on H 2 ÝUÞ n V V that is equivalent
Then it is clear that || 3
u || A = || Au
to the H 2 ÝUÞ n ? norm.
Before proceeding with the energy estimates we need an estimate for the nonlinear
term from the N-S equation.
3Þ 5 H and
Lemma 1 For 3
u 5 H 2 ÝUÞ n V V, BÝu
ÝiÞ
ÝiiÞ
1/2
3 || 1/2
3 || V
3 Þ|| H ² C 1 || 3
u || H || Au
||u
||BÝu
H
3/2
3 || 1/2
3 Þ|| H ² C 2 || 3
u || V || Au
||BÝu
H
if n = 2
if n = 3
Proof of lemma- When n=2, apply the generalized Holder inequality with p=q=4, and
r=2 so that
XU u i / i u j v j dx ² || u i || 4 || / i u j || 4 || v j || 2
Now the Sobolev inequality for n=2 can be used to obtain
XU u i / i u j v j dx ² C Ý|| u i || 2 ||4 u i || 2 Þ 1/2 Ý || / i u j || 2 || 4/ i u j || 2 Þ 1/2 || v j || 2
1/2
² C || u i || 1/2
Ý || 4/ i u j || 2 Þ 1/2 || v j || 2
H Ý || / i u j || 2 || 4u i || 2 Þ
and it follows that,
3, 3
u|| 2V Þ
u, 3
vÞ| ² C || 3
u|| 1/2
|bÝu
H Ý || 3
1/2
u|| H 2 Þ
Ý || 3
1/2
|| 3
v||| H ² C || 3
u|| 1/2
u|| V || 3
u|| 1/2
v||| H
H || 3
A || 3
3, 3
This estimate implies that for each 3
u 5 H 2 ÝUÞ n V V, bÝu
u, 6Þ defines a bounded
1/2
3
3
3
3
3
linear functional BÝuÞ on H with ||BÝuÞ|| H ² C || u|| H || u|| V || u|| 1/2
A .
When n=3, apply the generalized Holder inequality with four functions and p=6, q=4,
r=12 and s=2, so that
XU u i / i u j v j dx = XU u i Ý/ i u j Þ 1/2 Ý/ i u j Þ 1/2 v j dx
² || u i || 6 || Ý/ i u j Þ 1/2 || 4 || Ý/ i u j Þ 1/2 || 12 || v j || 2
1/2
² || u i || 6 || Ý/ i u j Þ|| 1/2
2 || Ý/ i u j Þ|| 6 || v j || 2
Now the Sobolev embedding theorem asserts that for n=3, H 1 ÝUÞ Ð L 6 ÝUÞ and so
5
this last estimate implies
1/2
XU u i / i u j v j dx ² C|| u i ||| 3/2
H 1 || Ý/ i u j Þ|| H 1 || v j || 2
3 5 H 2 ÝUÞ n V V,
-u
hence
3, 3
u|| H 2 Þ 1/2 || 3
u, 3
vÞ| ² C || 3
u|| 3/2
v||| H
|bÝu
V Ý || 3
which leads to the result ÝiiÞ.n
3 j â to be the eigenfunctions of the Stokes’ operator
In this example, we take the basis áw
A.
Then
3 j, 3
3 j, 3
3 j, 3
v ÞÞ = ÝAw
v Þ H = V j Ýw
vÞH
-v3 5 V
ÝÝw
and
3 j, w
3 k ÞÞ = ÝAw
3 k ÞH =
3 j, w
ÝÝw
V j if j = k
0 if j ® k
If
N
3j
3
u N ÝtÞ = > j=1 C j,N ÝtÞ w
solves
3 j Þ H + ÝcAu
3 N ÝtÞÞ, w
3 j Þ H = 3fÝtÞ, w
3 vN ÝtÞ, w
3 N ÝtÞ + BÝu
3j
Ýu
H
1 ² j ² N,
N
30 , w
3 j,
3 j ÞÞw
3
u N Ý0Þ = > j=1 ÝÝu
3 N Ý0Þ ? 3
u 0 || V ¸ 0, as N ¸ K since we have assumed that u 0 5 V.
then clearly, ||u
Multiplying the approximate equation by V j leads to
3 j ÞÞ + ÝcAu
3 N ÝtÞ + BÝu
3 N ÝtÞÞ, Aw
3 j Þ H = 3fÝtÞ, Aw
3 vN ÝtÞ, w
3j
ÝÝu
H
1 ² j ² N,
and, forming linear combinations of these equations, we get
3 N ÝtÞ + BÝu
3 N ÝtÞÞ, Au
3 N ÝtÞÞ H = 3fÝtÞ, Au
3 N ÝtÞ
3 vN ÝtÞ, 3
u N ÝtÞÞÞ + ÝcAu
ÝÝu
H
1 ² j ² N.
It follows from lemma 1 and the hypothesis that 3fÝtÞ 5 L 2 ß0, T : Hà, that we have
1 d
2 dt
2
1/2
3 N ÝtÞ|| 2H ² ||f3ÝtÞ|| H ||Au
3 N ÝtÞ|| H + C 1 || 3
3 N ÝtÞ|| 3/2
3 N ÝtÞ|| V
u N ÝtÞ|| V + c||Au
u N ÝtÞ|| H || Au
|| 3
||u
H
² C c ||f3ÝtÞ|| 2H +
1
4
3 N ÝtÞ|| 2H +
c||Au
1
4
3 N ÝtÞ|| 2H + CÝcÞ|| 3
3 N ÝtÞ|| 4V
c||Au
u N ÝtÞ|| 2H ||u
Here we have used both the Cauchy and the Young’s inequalities. Now we have
6
d
dt
2
2
3 N ÝtÞ|| 2H ² 2C c ||f3ÝtÞ|| 2H + 2CÝcÞ|| 3
3 N ÝtÞ|| 4V
u N ÝtÞ|| H ||u
u N ÝtÞ|| V + c||Au
|| 3
Ý3Þ
3 N ÝtÞ|| 2V 5 L 1 ß0, Tà and then
u N ÝtÞ|| 2H ||u
As in the previous example, a N ÝtÞ = 2CÝcÞ|| 3
d
dt
2
3 N ÝtÞ|| 2V ² 2C c ||f3ÝtÞ|| 2H
u N ÝtÞ|| V + a N ÝtÞ||u
|| 3
Then the Gronwall inequality is used to conclude that á 3
u N ÝtÞâ is a bounded sequence
in L K ß0, T : Và. Alternatively, if we integrate Ý3Þ from 0 to T, we get
T
T
T
2
2
3 N ÝtÞ|| 2H dt ² || 3
3 N ÝtÞ|| 4V dt
u N Ý0Þ|| V + 2C c X ||f3ÝtÞ|| 2H dt + 2CÝcÞ X || 3
u N ÝtÞ|| H ||u
c X ||Au
0
0
0
and since á 3
u N ÝtÞâ is a bounded sequence in both L K ß0, T : Và and in L K ß0, T : Hà, it
follows that
T
3 N ÝtÞ|| 2H dt ² M 3 .
c X ||Au
0
Finally, since ||6|| A is equivalent to ||6|| H 2 , we conclude that á 3
u N ÝtÞâ is a bounded
2
2
n
sequence in L ß0, T : H ÝUÞ à . and in the usual way, we obtain the existence of a
weak solution 3
uÝtÞ 5 L 2 ß0, T : H 2 ÝUÞ n à. Note that the (n=2) lemma implies
T
T
X 0 ||BÝu3ÝtÞÞ|| 4H dt ² C X 1 || 3uÝtÞ|| 2H || Au3ÝtÞ|| 2H ||u3N ÝtÞ|| 4V dt < K
3ÝtÞ 5 L 2 ß0, T : Hà, we have
3ÝtÞÞ 5 L 4 ß0, T : Hà and since Au
Then BÝu
3ÝtÞÞ 5 L 2 ß0, T : Hà.
3ÝtÞ ? BÝu
3
u v ÝtÞ = 3fÝtÞ ? cAu
In the n=3 case, the equation
3 N ÝtÞ + BÝu
3 N ÝtÞÞ, Au
3 N ÝtÞÞ H = 3fÝtÞ, Au
3 N ÝtÞ
3 vN ÝtÞ, 3
u N ÝtÞÞÞ + ÝcAu
ÝÝu
H
becomes,
1 d
2 dt
2
3/2
3 N ÝtÞ|| 2H ² ||f3ÝtÞ|| H ||Au
3 N ÝtÞ|| H + C 1 || 3
3 N ÝtÞ|| 3/2
u N ÝtÞ|| V + c||Au
u N ÝtÞ|| V || Au
|| 3
H
²
1
4
3 N ÝtÞ|| 2H +
c||Au
1
c
||f3ÝtÞ|| 2H +
1
4
6
3 N ÝtÞ|| 2H + CÝcÞ 1 || 3
c||Au
u N ÝtÞ|| V
where we have used again, the Cauchy and Young’s inequalities. Then
d
dt
3 N ÝtÞ|| 2H ²
u N ÝtÞ|| 2V + c||Au
|| 3
2
c
6
||f3ÝtÞ|| 2H + 2CÝcÞ|| 3
u N ÝtÞ|| V .
7
6
The term || 3
u N ÝtÞ|| V . appearing on the right side of this estimate presents an
obstacle which is not easily overcome. We are only able to obtain local in time
versions of the results that we found in the case n=2.
8