On boundedness, existence and uniqueness of strong solutions of the Navier-Stokes Equations in 3 dimensions

0.19 MB·English

by  A. A. Ruzmaikina

On boundedness, existence and uniqueness of strong solutions of
the Navier-Stokes Equations in 3 dimenions.
A. A. Ruzmaikina.
Department of Statistics, Purdue University, West Lafayette, IN 47909.
9 [email protected]
0
0
Abstract.
2

n
In this paper we consider the Navier-Stokes Equations in R3 in the vorticity
a
J formulation in the absence of the external forces. We first derive the upper

3 bounds on |ω|6.302(t) for t ∈ [t1,t2] in terms of maxt∈[t1,t2]|ω|2(t), then derive
M] tahneduωppe6.r02b(ot1u)n;dtsheonnd|e∇¯riω¯v|e2(tth)efuorppter∈bo[tu1n,td2]onin ¯teω¯rm3.s01o(ft)mfoarxtt∈[t1,[tt21]|,ωt2|2](itn)
.G term|s o|f3maxt∈[t1,t2]|ω|2(t), |ω|6.302(t1) and |∇¯ω¯|2|(∇t1);| then we der∈ive the up-
th per bound on |ω|∞(t) for t ∈ [t1,t2] in terms of maxt∈[t1,t2]|ω|2(t), |ω|6.302(t1),
a ¯ω¯ (t ) and ¯ω¯ (t ). These estimates are used to improve the estimate
2 1 3.01 1
m |∇ | ∇ |
on ω (t) and to show that ω (t) max(ω (t ),ee), provided that cer-
∞ ∞ ∞ 1
[ | | | | ≤ | |
tain assumptions hold. Repeated use of Local Existence and Uniqueness

2 results together with the upper bounds on ω (t) gives the existence and
∞
v | |
uniqueness of the solution if ω (0) < or if ω (0) < .
8 | |∞ ∞ | |4 ∞
1
3 Introduction and statement of results.
0
.
0 The Navier-Stokes Equations for velocity can be written as
1
8 ∂ u¯(x¯,t) ν∆u¯(x¯,t)+(u¯(x¯,t) ¯)u¯(x¯,t)+ ¯p(x¯,t) = f(x¯,t), (1)
0 t − ·∇ ∇
: ¯ u¯(x¯,t)= 0, x¯ R3, t [0, ).
v ∇· ∈ ∈ ∞
i
X
We assume that f(x¯,t)= 0 and that lim u¯(x¯,t) = 0.
|x¯|→∞
r Introducingvorticity ω¯(x¯,t) = ¯ u¯(x¯,t)a|llows us| toobtain theNavier-
a
∇×
Stokes Equations without the pressure term.
The Navier-Stokes Equations for vorticity can be written as
∂ ω¯(x¯,t) ν∆ω¯(x¯,t)= (u¯(x¯,t) ¯)ω¯(x¯,t)+(ω¯(x¯,t) ¯)u¯(x¯,t).(2)
t
− − ·∇ ·∇
¯ ω¯(x¯,t) = 0, x¯ R3, t [0, ).
∇· ∈ ∈ ∞
We assume that lim ω¯(x¯,t) = 0.
|x|→∞
| |
The question of existence and uniqueness of a strong solution of these
equations in 3 dimensions is a longstanding problem. The solution of this
problem is described in this paper.
1
The main result of the paper are the following Theorems:
Theorem 1: Suppose that ω (t) < for t [t ,t ] and also that
2 1 2
| | ∞ ∈
ω (t ) < and ¯ω¯ (t ) < . Then ω (t) < for t [t ,t ] and
∞ 1 3.01 1 ∞ 1 2
| | ∞ |∇ | ∞ | | ∞ ∈
an upper bound (equation (43)) on ω (t) can be given as a function of t,
∞
| |
max ω (t), ω (t ) and ¯ω¯ (t ).
t∈[t1,t2]| |2 | |∞ 1 |∇ |3.01 1
Theorem2: Suppose that ω (t) < fort [t ,t ]andthat ω (t ) <
2 1 2 ∞ 1
| | ∞ ∈ | |
and ¯ω¯ (t ) < . Suppose also that there exists n such that for
3.01 1 0
∞ |∇ | ∞
all n > n , either ω (t) ee for all t [t ,t ] or ω (t) is a con-
0 n 1 2 n
| | ≥ ∈ | |
tinuous function of t for all t [t ,t ]. Then for all sufficiently large n,
1 2
ω n(t) max(en1eet2−t1 ω n(t1)∈,ee) and ω ∞(t) max(ω ∞(t1),ee) for
| | ≤ | | | | ≤ | |
t [t ,t ].
1 2
∈
Theorem 3: Suppose that ω (0) < . Then for any T > 0, there
∞
| | ∞
existsauniquesolutionof the 3D NavierStokesEquationsω¯(x¯,t)forx R3
∈
and t [0,T], such that ω (t) < for t [0,T].
∞
∈ | | ∞ ∈
Theorem 4: Suppose that ω (0) < . Then for any T > 0, there
4
| | ∞
existsauniquesolutionof the 3D NavierStokesEquationsω¯(x¯,t)forx R3
∈
and t [0,T], such that ω (t) < for t (0,T].
∞
∈ | | ∞ ∈
Proof of Theorem 1.
Part 1: Take Navier Stokes equation for ω¯ and take the inner product
with ω¯(x) integrate over space and by parts on the viscosity term to obtain
1 3 1
∂ ω¯(x¯)2d3x+ν ¯ω¯(x¯)2d3x= d3x d3y(ω¯(x¯) yˆ)(ω¯(x¯) ω¯(x¯+y¯) yˆ) . (3)
t
2 | | |∇ | 4π · × · y 3
Z Z Z Z | |
The absolute value of the integral on the right hand side of (3) can be
estimated as
1
d3x d3y(ω¯(x¯) yˆ)(ω¯(x¯) ω¯(x¯+y¯) yˆ) (4)
· × · y 3 ≤
(cid:12)Z Z | | (cid:12)
(cid:12) (cid:12)
(cid:12) (cid:12) d3y
(cid:12) d3x (ω¯(x¯) yˆ)(ω¯(x¯) (ω¯(x¯+y¯) ω¯((cid:12)x¯)) yˆ) +
≤ (cid:12)(cid:12)Z Z|y|<1 · × − · |y|3(cid:12)(cid:12)
(cid:12)(cid:12) (cid:12)(cid:12)d3y
(cid:12) + d3x (ω¯(x¯) yˆ)(ω¯(x¯) ω¯(x¯+y¯) yˆ)(cid:12)
(cid:12)(cid:12)Z Z|y|≥1 · × · |y|3(cid:12)(cid:12)
1 (cid:12) d3y (cid:12)
3√6 ds d(cid:12)(cid:12)3x ω¯(x¯)2 ¯ω¯(x¯+sy¯) (cid:12)(cid:12)
≤ Z0 Z Z|y|<1| | |∇ ||y|2
ω¯(x¯+y¯)
+3√2 d3xω¯(x¯)2 | |d3y
Z | | Z|y|≥1 |y|3
2
1 ω¯(x¯′ sy¯)2
3√6 ds d3x′ ¯ω¯(x¯′) d3y| − | +
≤ Z0 Z |∇ |Z|y|<1 |y|2
ω¯(x¯+y¯)
+3√2 d3xω¯(x¯)2 d3y| |
Z | | Z|y|≥1 |y|3
1
3√6 ds d3x′ ¯ω¯(x¯′)2
≤ Z0 sZ |∇ | ×
ω¯(x¯′ sy¯)2 ω¯(x¯′ sz¯)2
d3x′ | − | d3y | − | d3z+√24π d3xω¯(x¯)2( d3y ω¯(y¯)2)21
sZ Z|y|<1 |y|2 Z|z|<1 |z|2 Z | | Z | |
1 ω¯(x¯′ sy¯)2 ω¯(x¯′ sz¯)2
3√6 ds d3x ¯ω¯(x¯)2 d3x′ | − | d3y | − | d3z+
≤ Z0 sZ |∇ | sZ Z|y|≤1 |y|3−ǫ Z|z|≤1 |z|1+ǫ
+√24π ω 2 ω
| |2| |2
1 d3y ω¯(x¯+sy¯ sz¯)2
3√6 ds d3x ¯ω¯(x¯)2 d3xω¯(x¯)2 | − | d3z+
≤ Z0 sZ |∇ | sZ | | Z|y|≤1 |y|3−ǫ Z|z|≤1 |z|1+ǫ
+√24π ω 2 ω
| |2| |2
1 ds d3y ω¯(x¯+sy¯ z¯′)2
3√6 d3x ¯ω¯(x¯)2 d3xω¯(x¯)2 | − | d3z′+
≤ Z0 s1−2ǫ sZ |∇ | sZ | | Z|y|≤1 |y|3−ǫ Z|z′|≤1 |z′|1+ǫ
+√24π ω 2 ω
| |2| |2
1 ds
3√6 d3x ¯ω¯(x¯)2
ǫ
≤ Z0 s1−2sZ |∇ |
d3y 2(3−3ǫ) 2−3ǫ d3z′ 1
d3xω¯(x¯)2 ( d3z′ ω¯(x¯ z¯′+sy¯) 2−3ǫ )3−3ǫ( )3(1−ǫ) +
sZ | | Z|y|≤1 |y|3−ǫ Z | − | Z|z′|<1 |z′|3(1−ǫ2)
+√24π ω 2 ω
| |2| |2
≤ 3√6(4ǫπ)21(34ǫπ2)6(11−ǫ) Z01 s1d−s2ǫ sZ d3x|∇¯ω¯(x¯)|2sZ d3x|ω¯(x¯)|2(Z d3z|ω¯(z¯)|2(23−−33ǫǫ))32−−33ǫǫ
+√24π ω 2 ω
| |2| |2
≤ 6√ǫ6(4ǫπ)21(34ǫπ2)6(11−ǫ)( |∇¯ω¯(x¯)|2d3x)43(1−−2ǫǫ)( |ω¯(x¯)|2d3x)34−−44ǫǫ +√24π|ω|22|ω|2
Z Z
1−2ǫ(6√6)41−−42ǫǫ(4π)21−−22ǫǫ(4π)3(1−22ǫ)(20.33)31−−22ǫǫ( d3xω¯(x¯)2)13−−24ǫǫ +
≤ 4 4ǫ ǫ ǫ 3ǫ2 ν | |
− Z
ν
+ d3x ¯ω¯(x¯)2+√24π ω 2 ω .
20.33 |∇ | | |2| |2
Z
Notice that in the above estimate ǫ is arbitrarily small. Using this estimate
3
and introducing ε such that
2ǫ
ε =
1 2ǫ
−
we obtain the following inequality from (3)
1 1
∂ ω¯(x¯)2d3x+ν(1 ) ¯ω¯(x¯)2d3x
2 t | | − 20.33 |∇ | ≤
Z Z
3 0.331·66·(4π)83( ω¯(x¯)2d3x)3+ε+√24π ω 3 . (5)
≤ 4π ε232+133εν3+2ε Z | | | |2!
Note that ε can be chosen arbitrarily small. Below we assume that ε = 1 .
100
Since ω¯(x¯)2d3x < for t [t ,t ], there exists N > 0 such that
1 2
| | ∞ ∈
ω¯(x¯)2d3x(t) N for t [t ,t ]. Denote
| | R ≤ ∈ 1 2
R t2 t
C(t ,t ) = dt ω¯(x¯)2d3x,C(t)= ds ω¯(x¯)2d3x(s).
1 2
Zt1 Z | | Z0 Z | |
Then we obtain
1 1 1 t2
ω¯(x¯)2d3x(t ) ω¯(x¯)2d3x(t )+ν(1 ) dt ¯ω¯(x¯)2d3x
2 Z | | 2 − 2 Z | | 1 − 20.33 Zt1 Z |∇ |
2.74 106 2.35 1021
≤ ε232+133·εν3+2εN2+εC(t1,t2)+2.1N12C(t1,t2) ≤ ν3·.02 N2+εC(t1,t2)+2.1N21C(t1,t2).
Therefore
Zt1t2dtZ d3x|∇¯ω¯(x¯)|2 ≤ 1.0ν754+·21ε022N2+εC(t1,t2)+ 9ν.6N21C(t1,t2)+ 4.ν57N. (6)
Part 2: Take the inner product of the Navier-Stokes equation for ω¯(x¯)
with ω(x¯) 32εω¯(x¯), integrate over space and integrate the viscosity term by
| |
parts to obtain
1
2+ 2ε∂t |ω|2+32ε(x¯)d3x+ν ∇¯(|ω|23ε(x¯)ω¯(x¯))(∇¯ω¯(x¯))d3x (7)
3 Z Z
3 d3xd3y
= (ω 23ε(x¯)ω¯(x¯) yˆ)(ω¯(x¯) ω¯(x¯+y¯) yˆ) .
4π | | · × · y 3
Z Z | |
The absolute value of the integral on the right hand side of (7) can be
estimated from above as
d3xd3y
(ω 23ε(x¯)ω¯(x¯) yˆ)(ω¯(x¯) ω¯(x¯+y¯) yˆ) (8)
(cid:12) | | · × · y 3 (cid:12)≤
(cid:12)Z Z | | (cid:12)
(cid:12) (cid:12)
(cid:12) (cid:12)
(cid:12) 4 (cid:12)
d3xd3y
(ω 23ε(x¯)ω¯(x¯) yˆ)(ω¯(x¯) (ω¯(x¯+y¯) ω¯(x¯)) yˆ) +
≤ (cid:12)(cid:12)Z Z|y|<1 | | · × − · |y|3 (cid:12)(cid:12)
+ (cid:12)(cid:12)(cid:12) (ω 23ε(x¯)ω¯(x¯) yˆ)(ω¯(x¯) ω¯(x¯+y¯) yˆ)d3xd3y (cid:12)(cid:12)(cid:12)
(cid:12)(cid:12)Z Z|y|≥1 | | · × · |y|3 (cid:12)(cid:12) ≤
(cid:12)(cid:12)(cid:12)3√6 1ds d3xω 2+23ε(x¯) |∇¯ω¯(x¯+sy¯)|d3y+ (cid:12)(cid:12)(cid:12)
≤ Z0 Z | | Z|y|<1 |y|2
d3xd3y
+3√2 ω 2+23ε(x¯)ω (x¯+y¯)
Z Z|y|≥1| | | | |y|3 ≤
3√6 1ds d3x′ ¯ω¯(x¯′) |ω|2+23ε(x¯′−sy¯)d3y+
≤ Z0 Z |∇ |Z|y|<1 |y|2
ω¯(x¯+y¯)
+3√2 ω 2+23ε(x¯)d3x | |d3y
Z | | Z|y|≥1 |y|3 ≤
3√6 1ds d3x′ |ω|2+32ε(x¯′−sy¯)d3y |ω|2+32ε(x¯′−sz¯)d3z ¯ω¯
≤ Z0 vuuZ Z|y|<1 |y|2 Z|z|<1 |z|2 |∇ |2
t +√24π ω 2+23ε ω
| |2+2ε| |2
3
3√6 1ds d3x′ |ω|2+32ε(x¯′−sy¯)d3y |ω|2+23ε(x¯′−sz¯)d3z ¯ω¯(x¯)2d3x
≤ Z0 vuuZ Z|y|≤1 |y|3−ε Z|z|≤1 |z|1+ε sZ |∇ |
t +√24π ω 2+23ε ω
| |2+2ε| |2
3
≤ 3√6Z01dsvuuZ |ω|2+23ε(x¯′′)d3x′′Z|y|≤1 |yd|33y−ε Z|z|≤1 |ω|2+23ε(x¯|z′′|1−+εs(z¯−y¯))d3z |∇¯ω¯|2
t +√24π ω 2+23ε ω
| |2+2ε| |2 ≤
3
≤ 3√6Z01dsvuuZ d3x′′|ω|2+32ε(x′′)Z|y|≤1 |yd|33y−ε Z|z′|≤s |ω|2+23ε(|zx′′|′1++εsy−z′)sd23−zε′|∇¯ω¯|2
t +√24π ω 2+23ε ω
| |2+2ε| |2 ≤
3
≤ 3√6Z01 s1d−s2εvuuZ d3x|ω|2+23ε(x¯)Z|y|≤1 |yd|33y−ε Z|z′|≤1 |ω|2+23ε|(zx′|1++εsy−z′)d3z′|∇¯ω¯|2+
t +√24π ω 2+23ε ω
| |2+2ε| |2 ≤
3
5
4π 1
3√6( )6(1−ε)
≤ 3ε2
Z01 s1d−s2εsZ d3x|ω|2+23ε(x¯)Z|y|≤1 |yd|33y−ε(Z |ω|(2+23ε)32−−33εε(x+sy−z′)d3z′)32−−33εε|∇¯ω¯|2
+√24π ω 2+23ε ω
| |2+2ε| |2
3
≤ 3√6(4επ)21(34επ2)6(11−ε)|∇¯ω¯|2
1 ds 1+(1−73ε+ε32)(1−23ε) (1+3ε)(1−32ε)
εv( ω¯(x¯)2+23εd3x) 2(1−35ε+ε42)(1−ε)( ¯ω¯(x¯)2d3x)2(1−35ε+ε42)(1−ε) +
Z0 s1−2uu Z | | Z |∇ |
t +√24π ω 2+23ε ω
| |2+2ε| |2
3
6√6(4π)21(4π )6(11−ε)( ω 2+23ε(x¯)d3x)21+(41(−1−735ε3+ε+ε32ε42)()1(−1−32εε))( ¯ω¯(x¯)2d3x)12+4(1(−1+353εε+)(ε142−)32(1ε−)ε)
≤ ε ε 3ε2 | | |∇ |
Z Z
+√24π ω 2+23ε ω
| |2+2ε| |2
3
α
≤ 3.0618·105 |∇¯ω¯(x¯)|2d3x+1 |ω¯(x¯)|2+32εd3x +√24π|ω|22++232εε|ω|2,
(cid:18)Z (cid:19)(cid:18)Z (cid:19) 3
where α = 1+ 1(1−73ε+ε32)(1−32ε) 0.747. Putting the estimates together we
2 4 (1−5ε+ε2)(1−ε) ≈
3 4
obtain
1
2+ 2ε∂t |ω¯(x¯)|2+23εd3x+ν ∇¯(|ω¯(x¯)|23εω¯(x¯))(∇¯ω¯(x¯))d3x≤ (9)
3 Z Z
3 3.0618 105 ω¯(x¯)2+23εd3x α ¯ω¯(x¯)2d3x+1 + 3√24π ω 2+23εN21
≤ 4π · | | |∇ | 4π | |2+2ε
(cid:18)Z (cid:19) (cid:18)Z (cid:19) 3
α
0.7305 105 ω 2+23εd3x ¯ω¯(x¯)2d3x+1 +2.1ω 2+23εN21.
≤ · | | |∇ | | |2+2ε
(cid:18)Z (cid:19) (cid:18)Z (cid:19) 3
Since we can rewrite the viscosity term as shown in the (10) we obtain that
the viscosity term is positive definite
∇¯(|ω|23ε(x¯)ω¯(x¯))(∇¯ω¯)d3x= (1+23εε)2 (∇¯|ω|1+ε3(x¯))2d3x+ |ω|32ε(x¯)(∇¯ω¯(x¯))2d3x (10)
Z 3 Z Z
From (9) we obtain
ω¯(x¯)2+23εd3x(t2) (11)
| | ≤
Z
6
1
e4.22 N12(t2−t1) 0.371 105 t2dt ¯ω¯ 2(x¯)d3x+1 + ω¯(x¯)2+32εd3x(t1) 1−α 1−α
≤ · Zt1 (cid:18)Z |∇ | (cid:19) (cid:18)Z | | (cid:19) !
1
e4.22N2(t2−t1)
≤
3.99 1026 3.562 105 N
ν4+·2ε N2+εC(t1,t2)+ ν· N21C(t1,t2)+1.7·105 ν +0.371·105(t2−t1) +
1
(1−α)(2+2ε) 1−α
+1+ ω 3 (t )
| |2+2ε 1
3 (cid:19)
Therefore we obtain that
ν t2dt ¯(ω¯(x¯) 23εω¯(x¯))(¯ω¯(x¯))d3x (12)
Zt1 Z ∇ | | ∇ ≤
≤ 0.7305·105maxt∈[t1,t2](cid:18)Z |ω¯(x¯)|2+32εd3x(t)(cid:19)αZt1t2dt(cid:18)Z |∇¯ω¯(x¯)|2d3x+1(cid:19)+
1
+2+ 2ε |ω¯(x¯)|2+23εd3x(t1)+ |ω¯(x¯)|2+32εd3x(t2) +
3 (cid:18)Z Z (cid:19)
+2.1maxt∈[t1,t2] |ω¯(x¯)|2+23εd3xN21(t2−t1)
Z
≤ 1.97e4.22N12(t2−t1)· 1+ 3.9ν94+·12ε026N2+εC(t1,t2)+ 3.562ν·105N12C(t1,t2)+
+1.7·105Nν +0.371·105(t2−t1)+1.066N12(t2−t1)+|ω|(21+−2αε)(2+23ε)(t1) 5
3 (cid:19)
Therefore
(1+23ε3ε)2 Zt1t2dtZ (∇¯|ω¯(x¯)|1+3ε)2d3x+Zt1t2dtZ |ω¯(x¯)|32ε(∇¯ω¯(x¯))2d3x≤ (13)
≤ 1.ν97e4.22N21(t2−t1)· 1+ 3.9ν94+·12ε026N2+εC(t1,t2)+ 3.562ν·105N12C(t1,t2)+
+1.7·105Nν +0.371·105(t2−t1)+1.066N21(t2−t1)+|ω|(21+−2αε)(2+23ε) 5.
3 (cid:19)
Part 3: Differentiate with respect to x the Navier-Stokes equation for
ω¯.
∂ ∂ ω¯(x¯) ν∂ ∆ω¯(x¯)= ∂ (ω¯(x¯) ¯)u¯(x¯) ∂ (u¯(x¯) ¯)ω¯(x¯) . (14)
t x x x x
− ·∇ − ·∇
Take inner product with ∂(cid:0)ω¯ (cid:1) (cid:0) (cid:1)
x
1
∂ (∂ ω¯(x¯))2 ν∂ ω¯(x¯) ∂ ∆ω¯(x¯)= ∂ ω¯(x¯) ∂ (ω¯(x¯) ¯)u¯(x¯) ∂ ω¯(x¯) ∂ (u¯(x¯) ¯)ω¯(x¯)
t x x x x x x x
2 − · · ·∇ − · ·∇
(cid:0) (cid:1) (cid:0) (cid:1)
7
Integrate over space and by parts.
1
∂ (∂ ω¯(x¯))2d3x+ν (∂ ω¯(x¯))2d3x+ν (∂ ω¯(x¯))2d3x+ν (∂ ω¯(x¯))2d3x =
t x xx xy xz
2
Z Z Z Z
= ∂ ω¯(x¯) (ω¯(x¯) ¯)u¯(x¯)d3x+ ∂ ω¯ (u¯(x¯) ¯)ω¯(x¯)d3x. (15)
xx xx
− · ·∇ · ·∇
Z Z
Add the equations obtained similarly but with differentiation with respect
to y or z instead of x.
1
∂ (∂ ω¯(x¯))2d3x+ν (∂ ω¯(x¯))2d3x+ν (∂ ω¯(x¯))2d3x+ν (∂ ω¯(x¯))2d3x=
t y yy xy yz
2
Z Z Z Z
= ∂ ω¯(x¯) (ω¯(x¯) ¯)u¯(x¯)d3x+ ∂ ω¯(x¯) (u¯(x¯) ¯)ω¯(x¯)d3x. (16)
yy yy
− · ·∇ · ·∇
Z Z
and
1
∂ (∂ ω¯(x¯))2d3x+ν (∂ ω¯(x¯))2d3x+ν (∂ ω¯(x¯))2d3x+ν (∂ ω¯(x¯))2d3x=
t z zz xz yz
2
Z Z Z Z
= ∂ ω¯(x¯) (ω¯(x¯) ¯)u¯(x¯)d3x+ ∂ ω¯(x¯) (u¯(x¯) ¯)ω¯(x¯)d3x. (17)
zz zz
− · ·∇ · ·∇
Z Z
to get
1
∂ ω 2(x¯)d3x+ν ¯ ¯ω 2(x¯)d3x= (18)
t
2 |∇ | |∇∇ |
Z Z
= ∆ω¯(x¯) (ω¯(x¯) ¯)u¯(x¯)d3x+ ∆ω¯(x¯) (u¯(x¯) ¯)ω¯(x¯)d3x.
− · ·∇ · ·∇
Z Z
Plugging the expression for u¯ in terms of ω¯
1 ω¯(y¯) (x¯ y¯) 1 ω¯(x¯+y¯) yˆ
u¯(x¯) = × − d3y = × d3y
4π x y 3 4π y 2
Z | − | Z | |
on the right hand side, we obtain
1 1 ω¯(x¯+y¯) yˆ
∂ ¯ω¯(x¯)2d3x+ν ¯ ¯ω¯(x¯)2d3x= ∆ω¯(x¯) (ω¯(x¯) ¯) × d3xd3y
t
2 |∇ | |∇∇ | −4π · ·∇ y 2
Z Z Z Z | |
1 ((ω¯(x¯+y¯) yˆ) ¯)ω¯(x¯)
+ ∆ω¯(x¯) × ·∇ d3xd3y. (19)
4π · y 2
Z Z | |
We can rewrite the right hand side of (19) as
1 ω¯(x¯+y¯) yˆ
∆ω¯(x¯) (ω¯(x¯) ¯) × d3xd3y (20)
−4π Z Z|y|≤1 · ·∇ |y|2 −
8
1 ω¯(x¯+y¯) yˆ
∆ω¯(x¯) (ω¯(x¯) ¯) × d3xd3y+
−4π Z Z|y|>1 · ·∇ |y|2
1 ((ω¯(x¯+y¯) yˆ) ¯)ω¯(x¯)
+ ∆ω¯(x¯) × ·∇ d3xd3y+
4π Z Z|y|≤1 · |y|2
1 ((ω¯(x¯+y¯) yˆ) ¯)ω¯(x¯)
+ ∆ω¯(x¯) × ·∇ d3xd3y.
4π Z Z|y|>1 · |y|2
Theabsolute value of the integrals over y 1 can beestimated from above
| | ≤
by
3 ¯ω¯(x¯+y¯) 3 ¯ω¯(x¯)
∆ω¯(x¯) ω¯(x¯) |∇ |d3xd3y+ ∆ω¯(x¯) ω¯(x¯+y¯) |∇ |d3xd3y
2π Z Z|y|≤1| || | |y|2 2π Z Z|y|≤1| || | |y|2
3 ¯ω¯(x¯)
= ∆ω¯(x¯ y¯) ω¯(x¯ y¯) |∇ |d3xd3y
2π Z Z|y|≤1| − || − | |y|2
3 ¯ω¯(x¯)
+ ∆ω¯(x¯) ω¯(x¯+y¯) |∇ |d3xd3y
2π Z Z|y|≤1| || | |y|2
3 d3y
= ∆ω¯(x¯ y¯) ω¯(x¯ y¯) ¯ω¯(x¯)d3x+
2π Z|y|≤1 |y|2 Z | − || − ||∇ |
3 ω¯(x¯+y¯)
+ ∆ω¯(x¯) ¯ω¯(x¯) | |d3y
2π Z | ||∇ |Z|y|≤1 |y|2
3 d3y ν 5
( ∆ω¯(x¯ y¯)2+ ω¯(x¯ y¯)2 ¯ω¯(x¯)2)d3x+
≤ 4π Z|y|≤1 |y|2 Z 5| − | ν| − | |∇ |
3 ν 5 ω¯(x¯+y¯) ω¯(x¯+z¯)
+ ( ∆ω¯(x¯)2+ ¯ω¯(x¯)2 | |d3y | |d3z)d3x
4π Z 5| | ν|∇ | Z|y|≤1 |y|2 Z|z|≤1 |z|2
3 ν d3y 3 5 ω¯(x¯+y¯)2 d3z
= ∆ω¯(x¯)2(1+ )+ ¯ω¯(x¯)2d3x | | d3y( +1)
4π 5 Z | | Z|y|≤1 |y|2 4πν Z |∇ | Z|y|≤1 |y|2 Z |z|2
3 ν d3y
∆ω¯(x¯)2d3x(1+ )+
≤ 4π 5 Z | | Z|y|≤1 |y|2
+ 3 5 ¯ω¯(x¯)2d3x( ω 6+2ε(x¯)d3x)3+1ε(4π+1)(4π)23++εε(2+ε)23++εε
4πν |∇ | | | ε
Z Z
3(1+4π)ν ∆ω(x¯)2d3x+ 30(4π +1) ¯ω¯(x¯)2d3x( ¯ ω¯ 1+3ε(x¯)2d3x)3+3ε.
≤ 4π 5 | | νε |∇ | |∇| | |
Z Z Z
The absolute value of the integral over y > 1 is bounded by
| |
1 ω¯(x¯+y¯) yˆ
∆ω¯(x¯) (ω¯(x¯) ¯) × d3xd3y+ (21)
4π Z Z|y|>1(cid:12) · ·∇ |y|2 (cid:12)
(cid:12) (cid:12)
(cid:12) 1 (cid:12) ((ω¯(x¯+y¯) yˆ) ¯)ω¯(x¯)
(cid:12) + ∆ω¯(x¯)(cid:12) × ·∇ d3xd3y
4π Z Z|y|>1(cid:12)(cid:12) · |y|2 (cid:12)(cid:12)
(cid:12) (cid:12)
(cid:12) (cid:12)
9 (cid:12) (cid:12)
3 d3y
∆ω¯(x¯) ω¯(x¯) ¯ω¯(x¯+y¯)d3x +
≤ 2π Z Z|y|>1| || ||∇ | |y|2
3 ω¯(x¯+y¯)
+ ∆ω¯(x¯) ¯ω¯(x¯) | |d3yd3x
2π Z | ||∇ |Z|y|≥1 |y|2
3 ν 15
∆ω¯(x¯)2d3x+ ω 2(x¯)d3x( ¯ω¯ 2(x¯+y¯)d3y)+
≤ 4π 5 | | ν | | |∇ |
Z Z Z
3 ν 15
+ ∆ω¯(x¯)2d3x+ ¯ω¯(x¯)2( ω¯(x¯+y¯)2d3y)d3x.
4π 5 | | ν |∇ | | |
Z Z Z
Therefore we obtain from (19) that
1 (8π 9)ν
∂ ¯ω¯(x¯)2d3x+ − ¯ ¯ω¯(x¯)2d3x (22)
t
2 |∇ | 20π |∇∇ | ≤
Z Z
30(4π +1) ¯ω¯(x¯)2d3x( (¯ ω¯(x¯)1+3ε)2d3x)3+3ε + 30 ¯ω¯(x¯)2d3x ω¯(x¯)2d3x.
≤ εν |∇ | ∇| | ν |∇ | | |
Z Z Z Z
We derive that
∂tln ¯ω¯(x¯)2d3x 816( ¯ ω¯(x¯)1+3ε 2d3x+1)+ 60 ω¯(x¯)2d3x. (23)
|∇ | ≤ εν |∇| | | ν | |
Z Z Z
Therefore
ln ¯ω¯(x¯)2d3x(t ) ln ¯ω¯(x¯)2d3x(t )
2 1
|∇ | − |∇ | ≤
Z Z
8.16·104 t2dt ¯ ω¯(x¯)1+3ε 2d3x(t)+(t2 t1) + 60C(t1,t2)
≤ ν (cid:18)Zt1 Z |∇| | | − (cid:19) ν
8.16 104 60 2.432 107 1
· (t t )+ C(t ,t )+ · e4.22N2(t2−t1) (24)
≤ ν 2− 1 ν 1 2 ν2 ×
3.99 1026 3.562 105 N
1+ ν4·.02 N2.01C(t1,t2)+ ν· N12C(t1,t2)+1.7·105 ν +0.371·105(t2−t1)
1 (2+2ε)(1−α) 5
+1.066N2(t2−t1)+|ω¯|2+23ε
3 (cid:19)
This implies
8.16 104 60
¯ω¯(x¯)2d3x(t ) ¯ω¯(x¯)2d3x(t )exp · (t t )+ C(t ,t )+ (25)
2 1 2 1 1 2
|∇ | ≤ |∇ | ν − ν
Z Z
+2.43ν22·107e4.22N21(t2−t1) 1+ 3.9ν94·.012026N2.01C(t1,t2)+ 3.562ν·105N12C(t1,t2)+

+1.7·105Nν +0.371·105(t2−t1)+1.066N12(t2−t1)+|ω|(22++3223εε)(1−α)(t1)(cid:19)5!
10