Electronic Journal of Differential Equations, Vol. 2014 (2014), No. 98, pp. 1–18.
ISSN: 1072-6691. URL: http://ejde.math.txstate.edu or http://ejde.math.unt.edu
ftp ejde.math.txstate.edu
COMPONENT REDUCTION FOR REGULARITY CRITERIA OF
THE THREE-DIMENSIONAL MAGNETOHYDRODYNAMICS
SYSTEMS
KAZUO YAMAZAKI
Abstract. We study the regularity of the three-dimensional magnetohydrodynamics system, and obtain criteria in terms of one velocity field component
and two magnetic field components. In contrast to the previous results such as
[22], we have eliminated the condition on the third component of the magnetic
field completely while preserving the same upper bound on the integrability
condition.
1. Introduction and statement of results
We study the three-dimensional magnetohydrodynamics (MHD) system
∂u
+ (u · ∇)u − (b · ∇)b + ∇π = ν∆u,
∂t
∂b
+ (u · ∇)b − (b · ∇)u = η∆b,
∂t
∇ · u = ∇ · b = 0, (u, b)(x, 0) = (u0 , b0 )(x), t ∈ R+ ∪ {0},
(1.1)
where u : R3 × R+ → R3 represents the velocity field, b : R3 × R+ → R3 the
magnetic field, π : R3 × R+ → R the pressure field, ν, η > 0 the viscosity and
diffusivity constants respectively. Hereafter let us assume ν = η = 1 and write
∂
∂
∂t = ∂t and ∂xi = ∂i and the components of u and b by
u = (u1 , u2 , u3 ),
b = (b1 , b2 , b3 ),
bh := (b1 , b2 , 0).
Due to the works of [22], we know that (1.1) possesses at least one global L2 weak solution pair for any initial data pair (u0 , b0 ) ∈ L2 . However, whether the
local solution pair remains smooth for all time remains open as in the case of the
Navier-Stokes equations (NSE), the system (1.1) at b ≡ 0.
To show that the weak solution pair is actually strong, there has been a large
amount of research conducted by many mathematicians to obtain a sufficient condition on (u, b) so that imposing such conditions lead to the H 1 -norm bound on
(u, b). We discuss some of them in particular.
2000 Mathematics Subject Classification. 35B65, 35Q35, 35Q86.
Key words and phrases. Magnetohydrodynamics system; Navier-Stokes equations;
regularity criteria.
c
2014
Texas State University - San Marcos.
Submitted December 9, 2013. Published April 11, 2014.
1
2
K. YAMAZAKI
EJDE-2014/98
Following the pioneering work by Serrin [23], Beir̃ao da Veiga [2] obtained regularity criteria on ∇u. Similar results followed for the MHD system; in particular,
Zhou [34] showed that it suffices to bound only u dropping the conditions on b
completely. For example, the following regularity criteria was obtained by He and
Xin [11].
Z
T
3 2
+ ≤ 1, 3 < p.
p r
0
In an accompanying paper [29], the author reduced this criteria to any two components of u. For the regularity criteria in terms of other quantities for the NSE such
as vorticity, π, we refer readers to [1, 3, 7, 24, 33, 35].
Results related on the reduction of components appeared for example in Kukavica
and Ziane [16]:
Z T
5
24
3 2
+ ≤ ,
≤ p,
ku3 krLp dτ < ∞,
p r
8
5
0
(see also [17, 31, 32]). A few of the most recent results are the following:
Z T
2(p + 1)
7
3 2
+ <
,
< p,
ku3 krLp dτ < ∞,
p r
3p
2
0
for the NSE see Cao and Titi [4] (also [5, 14, 20, 21, 36] followed by many in the
case of the MHD system (e.g. [6, 13, 18, 25]). In particular, Jia and Zhou [12]
showed that if
Z T
3 2
3
1
10
ku3 krLp + kbkrLp dτ < ∞,
+ ≤ + ,
< p,
(1.2)
p r
4 2p
3
0
then the solution pair (u, b) remains smooth (cf. [37] for the case of the NSE).
More variations of (1.2) were also obtained in [12, 19]; however, in any case, if
condition is given only on u3 and no other component of u, then without a new
idea, it seems we need to impose some integrability condition on every component
of b. This is due to the difficulty in decomposing the four non-linear terms in the
k∇h uk2L2 + k∇h bk2L2 -estimates so that every term has either have u3 or bh , where
∇h = (∂1 , ∂2 , 0) (see the Appendix for details). Now we present our results.
kukrLp dτ < ∞,
Theorem 1.1. Suppose (u, b) solves (1.1) in time interval [0, T ] and satisfies
Z T
8/3
ku3 kL∞ + kbh krLp dτ < ∞, bh = (b1 , b2 , 0),
(1.3)
0
where 10/3 < p < ∞ and 8 ≤ r satisfy
3
1
10
3 2
+ ≤ + ,
< p ≤ 5,
p r
4 2p
3
Then there is no singularity up to time T .
r = 8,
5 ≤ p < ∞.
(1.4)
The boarder-line case of p = 10/3 may be obtained as well, via a slight modification of the proof for Theorem 1.1.
Theorem 1.2. Suppose (u, b) solves (1.1) in time interval [0, T ] and
Z T
8/3
ku3 kL∞ dτ + sup kbh (τ )kL10/3 < ∞, bh = (b1 , b2 , 0).
0
τ ∈[0,T ]
Then there is no singularity up to time T .
EJDE-2014/98
COMPONENT REDUCTION
3
Theorem 1.3. Suppose (u, b) solves (1.1) in time interval [0, T ] and satisfies
Z T
8/3
8/3
ku2 kL∞ + ku3 kL∞ + kb1 krLp dτ < ∞,
0
where 10/3 < p < ∞, 8 ≤ r satisfy (1.4). Then there is no singularity up to time
T.
Theorem 1.4. Suppose (u, b) solves (1.1) in time interval [0, T ] and
Z T
8/3
8/3
ku2 kL∞ + ku3 kL∞ dτ + sup kb1 (τ )kL10/3 < ∞.
τ ∈[0,T ]
0
Then there is no singularity up to time T .
Remark 1.5. (1) We may replace the role of u3 with any other component of u
as long as the two components of b will be the different two components.
(2) We emphasize that in particular in Theorem 1.1, we have eliminated the
condition on b3 completely while preserving the integrability condition on bh and
p = ∞, r = 8/3 also satisfies (1.2). Thus, it is clear that (1.4) is an improvement of
the special case of (1.2). We were also able to obtain results in case when p 6= ∞ for
u3 in (1.3); however, the integrability conditions became worse; thus, for simplicity,
we chose not to present those results. We also remark that in contrast to results
from [12], Theorem 1.2 is not a smallness result.
(3) We also wish to emphasize that previously when the regularity criteria for
the three-dimensional MHD system was obtained in terms of three terms, they have
always been all from the velocity vector field; e.g. ∂3 u1 , ∂3 u2 , ∂3 u3 from [6] and [13],
any three partial derivatives of u1 , u2 , u3 from [15, 18, 25].
(4) The new idea in our proof is to make use of the structure of the magnetic
vector field equation and estimate kb3 kLp and obtain its bound in terms of bh
and u3 . Our proof was inspired by the others including [27], in particular [8, 9]
concerning the [26, Theorems 1.3-1.4] and [28, Propositions 3.1-3.2]. Modification of
Propositions 3.1-3.2 are possible indicating that in the future to obtain a regularity
criteria of the MHD system in terms of one component of the velocity vector field,
which has been done for the NSE but not for the MHD system, it suffices to discover
a decomposition of the four non-linear terms that separate u3 and b3 , not necessarily
just u3 .
(5) After this manuscript was completed, the author discovered in [30] a new
decomposition of the four non-linear terms of (4.1) which led to a regularity criteria
of (1.1) in terms of u3 and j3 where j3 is the third component of the current density
j := ∇ × b.
In the Preliminary section, we set notation. Thereafter, we prove two crucial
propositions and then prove Theorems 1.1–1.4.
2. Preliminary
Let us denote a constant that depends on a, b by c(a, b) and A . B when there
exists
R
Ra constant c ≥ 0 of no significance such that A ≤ cB. We shall also denote
f = R3 f (x)dx,
2
2
∆h = ∂11
+ ∂22
,
X(t) := k∇u(t)k2L2 + k∇b(t)k2L2 ,
Y (t) := k∇h u(t)k2L2 + k∇h b(t)k2L2 ,
Z(t) := k∆u(t)k2L2 + k∆b(t)k2L2 ,
∇h = (∂1 , ∂2 , 0),
4
K. YAMAZAKI
T
Z
M1 :=
0
Z
N1 :=
0
8/3
ku3 kL∞ dτ,
T
EJDE-2014/98
M2 := kb3 (0)kL10/3 + sup kbh (t)kL10/3 ,
t∈[0,T ]
8/3
ku2,3 kL∞ dτ,
N2 := kb2,3 (0)kL10/3 + sup kb1 (t)kL10/3 ,
t∈[0,T ]
where e.g. b2,3 is a two dimensional vector of two entries b2 and b3 .
We have the following special case of Troisi inequality (cf. [6, 10])
1/3
1/3
1/3
kf kL6 ≤ ck∂1 f kL2 k∂2 f kL2 k∂3 f kL2 .
(2.1)
Finally, we obtain the basic energy inequality by taking L2 -inner products of
(1.1) with (u, b) respectively, integrating by parts and using the incompressibility
of u and b to deduce after integrating in time
Z T
2
2
sup (ku(t)kL2 + kb(t)kL2 ) + 2
X(τ )dτ . 1.
(2.2)
t∈[0,T ]
0
3. Two propositions
Proposition 3.1. Suppose (u, b) is the solution to (1.1) in time interval [0, T ].
Then for any p ∈ (2, ∞), the following inequality holds: for any distinct choices of
j1 , j2 , j3 ∈ {1, 2, 3}
sup kbj1 (t)k2Lp ≤ kbj1 (0)k2Lp e(p−1)
RT
0
kuj1 k2L∞ dλ
t∈[0,T ]
Z
+ 2(p − 1)
(3.1)
T
e
(p−1)
RT
τ
kuj1 k2L∞ dλ
kuj1 k2L∞ kbj2 ,j3 k2Lp dτ,
0
where bj2 ,j3 is a two dimensional vector of two entries bj2 and bj3 .
Remark 3.2. We remark that we cannot obtain an analogous bound for u3 due to
the ∇π-term in the first equation of (1.1). Moreover, we were able to obtain various
modifications of this inequality; however, we emphasize that (3.1) in particular
implies that if we have a sufficient bound on uj1 , then we may bound the Lp -norm
of bj1 by the same Lp -norm of bj2 ,j3 .
Proof of Proposition 3.1. From the second equation of (1.1), we have the equation
that governs the growth of bj1 in time
∂t bj1 + (u · ∇)bj1 − (b · ∇)uj1 = ∆bj1 .
(3.2)
We multiply by |bj1 |p−2 bj1 and integrate in space to obtain
Z
Z
Z
1
p
p−2
p−2
∂t kbj1 kLp − ∆bj1 |bj1 | bj1 = − (u · ∇)bj1 |bj1 | bj1 + (b · ∇)uj1 |bj1 |p−2 bj1 .
p
By the incompressibility condition, we see that the first term on the right hand
side after integrating by parts equals zero. We compute the diffusive term after
integrating by parts as follows:
Z
3 Z
3 Z
X
X
2
p−2
2
p−2
|∂k bj ||bj | p−2
2 .
− ∆bj1 |bj1 | bj1 = −
(∂kk bj1 )|bj1 | bj1 = (p − 1)
1
1
k=1
k=1
EJDE-2014/98
COMPONENT REDUCTION
5
Therefore, we obtain by integrating by parts and using the incompressibility condition of b,
3 Z
X
2
1
|∂k bj ||bj | p−2
2 ∂t kbj1 kpLp + (p − 1)
1
1
p
k=1
Z
3
X
=−
∂k bk uj1 |bj1 |p−2 bj1 + bk uj1 ∂k (|bj1 |p−2 bj1 )
k=1
=−
3 Z
X
(p − 1)1/2 bk uj1 |bj1 |
p−2
2
(p − 1)1/2 |bj1 |
p−2
2
∂k bj1
k=1
3
p − 1 X
≤
2
3
Z
2
2
|bk | |uj1 | |bj1 |
p−2
k=1
(p − 1) X
+
2
Z
||∂k bj1 |bj1 |
p−2
2
|2
k=1
by Young’s inequality. Absorbing the diffusive term, we have
3 Z
3 Z
p−2
(p − 1) X
p − 1 X
1
p
2
2
∂t kbj1 kLp +
|(∂k bj1 )|bj1 |
| ≤
|bk |2 |uj1 |2 |bj1 |p−2 .
p
2
2
k=1
k=1
Therefore, Hölder’s inequalities and then dividing by
∂t kbj1 k2Lp
− (p −
1)kuj1 k2L∞ kbj1 k2Lp
≤ 2(p −
p−2
1
2 kbj1 kLp
lead to
1)kuj1 k2L∞ kbj2 ,j3 k2Lp .
This leads to (3.1) completing the proof of Proposition 3.1.
The next proposition may be obtained by an identical procedure.
Proposition 3.3. Suppose (u, b) is the solution pair to (1.1) in time interval [0, T ].
Then for any p ∈ (2, ∞), the following inequality holds: for any distinct choices of
j1 , j2 , j3 ∈ {1, 2, 3}
sup kbj1 ,j2 (t)k2Lp ≤ kbj1 ,j2 (0)k2Lp e2(p−1)
RT
0
kuj1 ,j2 k2L∞ dλ
t∈[0,T ]
Z
+ (p − 1)
T
e2(p−1)
RT
τ
kuj1 ,j2 k2L∞ dλ
kuj1 ,j2 k2L∞ kbj3 k2Lp dτ.
0
(3.3)
4. Proof of Theorem 1.1
k∇h ukL2 +k∇h bk2L2 -estimate.
We now fix p and r that satisfy (1.4), take L2 -inner
products of the first equation of (1.1) with −∆h u and the second with −∆h b to
estimate
1
∂t Y (t) + k∇∇h uk2L2 + k∇∇h bk2L2
2 Z
(4.1)
= (u · ∇)u · ∆ u − (b · ∇)b · ∆ u + (u · ∇)b · ∆ b − (b · ∇)u · ∆ b
h
h
h
h
:= I1 + I2 + I3 + I4 ,
where Y (t) := k∇h u(t)k2L2 +k∇h b(t)k2L2 . The following decomposition was obtained
in [12]; we provide details in the Appendix for convenience of readers:
Z
I1 . |u3 ||∇u||∇∇h u|, I2 + I3 + I4 . |b||∇b||∇∇h u| + |b||∇u||∇∇h b|. (4.2)
6
K. YAMAZAKI
EJDE-2014/98
Now by Hölder’s and Young’s inequalities we immediately obtain
I1 . ku3 kL∞ k∇ukL2 k∇∇h ukL2 ≤
1
k∇∇h uk2L2 + cku3 k2L∞ k∇uk2L2 .
4
(4.3)
Next, by Hölder’s inequalities and interpolation inequalities we estimate
I2 + I3 + I4
. kbkLp k∇bk
2p
L p−2
k∇∇h ukL2 + kbkLp k∇uk
2p
L p−2
p−3
k∇∇h bkL2
p−3
3/p
3/p
. kbkLp k∇bkL2p k∇bkL6 k∇∇h ukL2 + kbkLp k∇ukL2p k∇ukL6 k∇∇h bkL2 .
By (2.1) and Young’s inequalities we have
p−3
2/p
1/p
I2 + I3 + I4 . kbkLp k∇bkL2p k∇∇h bkL2 k∆bkL2 k∇∇h ukL2
p−3
2/p
1/p
+ kbkLp k∇ukL2p k∇∇h ukL2 k∆ukL2 k∇∇h bkL2
(4.4)
2p
2
2( p−3
1
1
p−2 )
≤ k∇∇h uk2L2 + k∇∇h bk2L2 + c(kbkLp−2
k∆bkLp−2
p k∇bkL2
2
4
2
2( p−3 )
2p
2
p−2
k∆ukLp−2
+ kbkLp−2
p k∇ukL2
2 ).
Thus, with (4.3) and (4.4) applied to (4.2), absorbing the dissipative and diffusive
terms, integrating in time we obtain
Z t
sup Y (τ ) +
k∇∇h uk2L2 + k∇∇h bk2L2 dτ
τ ∈[0,t]
0
(4.5)
Z
t
.1+
0
2p
p−3
1
p−2 (τ )Z p−2 (τ )dτ.
ku3 k2L∞ k∇uk2L2 + kbkLp−2
p X
k∇uk2L2 + k∇bk2L2 -estimate. For both equations in (1.1), we take the L2 -inner
products with −∆u and −∆b respectively and integrate by parts to obtain
1
∂t X(t) + Z(t)
2 Z
=
:=
(u · ∇)u · ∆u − (b · ∇) · ∆u + (u · ∇)b · ∆b − (b · ∇)u · ∆b
4
X
(4.6)
IIi .
i=1
For II2 and II4 we have the estimate
II2 + II4 ≤ kbkLp k∇bk
2p
L p−2
≤ c(kbk2Lp k∇bk2
k∆ukL2 + kbkLp k∇uk
2p
L p−2
2p
L p−2
+ kbk2Lp k∇uk2
2p
L p−2
k∆bkL2
1
) + Z(t),
8
by Hölder’s and Young’s inequalities. Now we use a Gagliardo-Nirenberg inequality
p−3
kf k
2p
L p−2
3/p
. kf kL2p k∇f kL2
(4.7)
EJDE-2014/98
COMPONENT REDUCTION
7
and Young’s inequalities to obtain
2( p−3
p )
II2 + II4 ≤ c(kbk2Lp k∇bkL2
2( p−3
p )
6
k∆bkLp 2 + kbk2Lp k∇ukL2
6
1
k∆ukLp 2 ) + Z(t)
8
2p
1
≤ ckbkLp−3
Z(t).
p X +
4
(4.8)
For II3 , we integrate by parts twice to deduce
Z
3
X
2
II3 = −
∂k ui ∂i bj ∂k bj + ui ∂ik
bj ∂k bj
i,j,k=1
=−
Z
3
X
i,j,k=1
Z
3
X
=
1
∂k ui ∂i bj ∂k bj − ∂i ui (∂k bj )2
2
2
∂k ui bj ∂ik
bj
Z
.
|∇u||b||∇∇b|
i,j,k=1
so that similarly as before, Hölder’s and Young’s inequalities, (4.7) and another
Young’s inequality lead to
II3 . kbkLp k∇uk
2p
L p−2
k∆bkL2
2( p−3
p )
≤ ckbk2Lp k∇ukL2
2p
p−3
Lp
≤ ckbk
6
1
k∆ukLp 2 + k∆bk2L2
4
(4.9)
1
X(t) + Z(t).
4
Finally, on II1 , we write
II1 =
2 Z
X
i=1
1
ui ∂i u · ∆u + u3 ∂3 u · ∆h u + u3 ∂3 (∂3 u)2
2
and then integrate by parts on each to obtain
2 X
3 Z
X
2
II1 = −
∂k ui ∂i u · ∂k u + ui ∂ik
u · ∂k u
i=1 k=1
2
X
1
2
∂k u3 ∂3 u · ∂k u + u3 ∂3k
u · ∂k u + ∂3 u3 (∂3 u)2
2
k=1
2 X
3 Z
2
X
X
1
∂k u3 ∂3 u · ∂k u
=−
∂k ui ∂i u · ∂k u − ∂i ui (∂k u)2 +
2
i=1
+
k=1
k=1
1
1
− ∂3 u3 (∂k u)2 − (∂1 u1 + ∂2 u2 )(∂3 u)2
2
2
Z
. |∇h u||∇u|2 .
Now Hölder’s, interpolation inequalities, and (2.1) lead to
II1 . k∇h ukL2 k∇uk2L4
1/2
3/2
1/2
1/2
. k∇h ukL2 k∇ukL2 k∇ukL6 . k∇h ukL2 k∇ukL2 k∇∇h ukL2 k∆ukL2 .
(4.10)
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K. YAMAZAKI
EJDE-2014/98
We apply the bounds of (4.8)–(4.10) into (4.6) to obtain, after absorbing the dissipative and diffusive terms,
2p
1/2
1/2
∂t X(t) + Z(t) . kbkLp−3
p X(t) + k∇h ukL2 k∇ukL2 k∇∇h ukL2 k∆ukL2 .
(4.11)
Integrating in time, we obtain
Z t
X(t) +
Z(τ )dτ
0
Z t
Z t
2p
1/2
1/2
kbkLp−3
k∇h ukL2 k∇ukL2 k∇∇h ukL2 k∆ukL2 dτ
≤ X(0) + c
p X(τ )dτ + c
0
0
Z t
2p
p−3
kbkLp X(τ )dτ
.1+
0
Z t
1/2 Z t
1/4 Z t
1/4
+ sup k∇h u(τ )kL2
k∇∇h uk2L2 dτ
X(τ )dτ
Z(τ )dτ
τ ∈[0,t]
0
0
0
by Hölder’s inequality. By (4.5) and (2.2) we obtain
Z t
X(t) +
Z(τ )dτ
0
Z t
2p
.1+
kbkLp−3
p X(τ )dτ
0
Z
t
Z t
1/4
2p
p−3
1
p−2
2
2
p−2
p−2
+ 1+
ku3 kL∞ k∇ukL2 + kbkLp X
(τ )Z
(τ )dτ
Z(τ )dτ
0
≤ c0 +
4
X
0
IIIi ,
i=1
(4.12)
where
Z t
1/4
2p
III2 = c0
Z(τ )dτ
,
kbkLp−3
p X(τ )dτ,
0
0
Z t
Z t
1/4
III3 = c0
ku3 k2L∞ k∇uk2L2 dτ
Z(τ )dτ
,
0
0
Z t
Z t
1/4
2p
p−3
1
p−2 (τ )Z p−2 (τ )dτ
X
III4 = c0
kbkLp−2
Z(τ )dτ
,
p
Z
t
III1 = c0
0
0
and c0 does not depend on t. By (3.1) with j1 = 3, j2 = 1, j3 = 2, we have
Z T
RT
2
sup kb3 (t)k2Lp ≤ c(p)ec 0 ku3 (λ)kL∞ dλ 1 +
ku3 (λ)k2L∞ kbh (λ)k2Lp dλ .
t∈[0,T ]
0
Using the elementary inequality
(a + b)p ≤ 2p (ap + bp ) p ≥ 0,
a, b ≥ 0,
(4.13)
we obtain
2p
2p
2p kb(τ )kLp−3
≤ c(p) kbh (τ )kLp−3
+ kb3 (τ )kLp−3
p
p
p
2p
p−3
Lp
≤ c(p)kbh k
+e
c(p)
RT
0
ku3 k2L∞ dλ
1+
Z
T
ku3 k2L∞ kbh k2Lp dλ
p p−3
.
0
(4.14)
EJDE-2014/98
COMPONENT REDUCTION
9
Thus, by (2.2),
Z
III1 ≤ c(p)
0
+ ec(p)
t
2p
kbh (τ )kLp−3
p X(τ )dτ
RT
0
ku3 (λ)k2L∞ dλ
1+
T
Z
ku3 (λ)k2L∞ kbh (λ)k2Lp dλ
p p−3
(4.15)
.
0
The estimate on III2 is immediate by Young’s inequality
Z
Z t
1/4
1 t
III2 = c0
Z(τ )dτ
Z(τ )dτ.
≤c+
8 0
0
Next, by Young’s and Hölder’s inequalities and (2.2),
Z
Z t
1 t
8/3
2
Z(τ )dτ.
III3 ≤ c
ku3 (τ )kL∞ k∇u(τ )kL2 dτ +
8 0
0
(4.16)
(4.17)
Finally, by successive applications of Hölder’s and Young’s inequalities,
p+2
Z t
p−3
Z t
4(p−2)
2p
p−2
Z(τ )dτ
III4 .
kbkLp−3
p X(τ )dτ
0
0
t
( 4(p−3)
3p−10 )
Z
1 t
≤c
kbk
X(τ )dτ
+
Z(τ )dτ
8 0
0
Z
Z
1 t
t
8p
X(τ )dτ +
Z(τ )dτ,
≤c
kbkL3p−10
p
8 0
0
Z
2p
p−3
Lp
(4.18)
where using (4.14), we may obtain
8p
2p 4(p−3)
3p−10
kb(τ )kL3p−10
= kb(τ )kLp−3
p
p
8p
RT
2
+ ec(p) 0 ku3 kL∞ dλ
≤ c(p)kbh kL3p−10
p
Z
4p
T
3p−10
× 1+
ku3 k2L∞ kbh k2Lp dλ
0
and therefore,
Z t
Z t
8p
8p
RT
2
X(τ )dτ ≤ c(p)
kbh kL3p−10
Xdτ + ec(p) 0 ku3 kL∞ dλ
kb(τ )kL3p−10
p
p
0
0
Z
× 1+
T
ku3 k2L∞ kbh k2Lp dλ
4p
3p−10
(4.19)
0
due to (2.2). We apply (4.19) into (4.18) and along with (4.15)-(4.17) applied to
(4.12), obtain after absorbing dissipative and diffusive terms
Z t
X(t) +
Z(τ )dτ
0
t
p Z T
p−3
2p
R
c(p) 0T ku3 k2L∞ dλ
2
2
.1+
kbh kLp−3
Xdτ
+
e
1
+
ku
k
kb
k
dλ
∞
p
p
3 L
h L
0
0
Z t
Z t
8p
8/3
kbh kL3p−10
+
ku3 kL∞ k∇uk2L2 dτ +
Xdτ
p
Z
0
+ ec(p)
RT
ku3 k2L∞ dλ
0
Z
1+
0
T
0
ku3 k2L∞ kbh k2Lp dλ
4p
3p−10
10
K. YAMAZAKI
t
Z
(
0
c(p)
2p
(1 + kbh kLp−3
p
.1+
RT
ku3 k2L∞ dλ
)(
4(p−3)
3p−10 )
Z
EJDE-2014/98
)Xdτ
T
+e
1+
ku3 k2L∞ kbh k2Lp dλ
0
Z t
8/3
+
ku3 kL∞ k∇uk2L2 dτ
0
Z t
8p
8/3
+ ku3 kL∞ )Xdτ
.1+
(kbh kL3p−10
p
0
4(p−3)
p
( p−3
)( 3p−10 ) 0
4p
Z T
3p−10
RT
8/3
3/4
1/4
+ ec(p)T ( 0 ku3 kL∞ dλ)
1+
ku3 k2L∞ kbh k2Lp dλ
0
Z t
8p
RT
8/3
3/4
1/4
8/3
.1+
(kbh kL3p−10
+ ku3 kL∞ )Xdτ + ec(p)T ( 0 ku3 kL∞ dλ)
p
0
T
Z
× 1+
8/3
ku3 kL∞ dλ
0
Z
≈
t
(kbh k
0
8p
3p−10
Lp
+
3/4 Z
T
kbh k8Lp dλ
4p
1/4 3p−10
0
8/3
ku3 kL∞ )Xdτ
+ c(p, M1 ) 1 +
Z
T
kbh k8Lp dλ
p
3p−10
.
0
By Gronwall’s inequality, the proof of Theorem 1.1 is complete if
Z
T
8p
8/3
+ ku3 (τ )kL∞ + kbh (τ )k8Lp dτ < ∞.
kbh (τ )kL3p−10
p
0
For p ∈ (10/3, 5), we use Hölder’s inequality to obtain
T
Z
kbh (τ )k8Lp dτ ≤ T
10−2p
p
Z
0
T
8p
dτ
kbh (τ )kL3p−10
p
0
3p−10
p
<∞
by (1.4) whereas if p ∈ (5, ∞), Hölder’s inequality again by (1.4) implies
T
Z
0
8p
2p−10
dτ ≤ T 3p−10
kbh (τ )kL3p−10
p
Z
T
kbh (τ )k8Lp dτ
p
3p−10
< ∞.
0
5. Proof of Theorem 1.2
k∇h uk2L2 +k∇h bk2L2 -estimate. For fixed T > 0, firstly, from (3.1) with j1 = 3, j2 =
1, j2 = 2, by Hölder’s inequalities we have
sup kb3 (t)k2L10/3
t∈[0,T ]
7
≤ kb3 (0)k2L10/3 e 3 T
+
14 37 T 1/4
e
3
7
≤ M22 e 3 T
1/4
3/4
M1
1/4
RT
0
RT
0
8/3
ku3 (λ)kL∞ dτ
8/3
ku3 (λ)kL∞ dτ
3/4
3/4
sup kbh (t)k2L10/3 T 1/4
t∈[0,T ]
14 7 T 1/4 M13/4 2 1/4 3/4
+
e3
M2 T M1 .
3
Z
0
T
8/3
ku3 (τ )kL∞ dτ
3/4
EJDE-2014/98
COMPONENT REDUCTION
11
Thus, using (4.13), we compute
sup kb(t)k2L10/3 ≤ 28/5 sup kbh (t)k2L10/3 + sup kb3 (t)k2L10/3
t∈[0,T ]
t∈[0,T ]
t∈[0,T ]
14 7 T 1/4 M13/4 2 1/4 3/4 M2 T M1
.
e3
3
(5.1)
Next, we choose t1 ∈ [0, T ] to be specified subsequently and as before, we may
obtain by (4.1)-(4.4) which only required p > 3 and integrating in time over [0, t1 ]
as in (4.5)
Z t1
k∇∇h uk2L2 + k∇∇h bk2L2 dτ
sup Y (t) +
7
≤ 28/5 M22 + M22 e 3 T
t∈[0,t1 ]
1/4
3/4
M1
+
0
t1
Z
.1+
0
ku3 k2L∞ k∇uk2L2 + kbk5L10/3 X 1/4 (τ )Z 3/4 (τ )dτ.
k∇uk2L2 + k∇bk2L2 -estimate. By (4.8)–(4.10), all of which only required p > 3,
applied to (4.6) we have
1/2
1/2
∂t X(t) + Z(t) . kbk20
L10/3 X(t) + k∇h ukL2 k∇ukL2 k∇∇h ukL2 k∆ukL2
as in (4.11). Integrating in time and by Hölder’s inequality as before in (4.12), we
have
Z t1
sup X(t) +
Z(τ )dτ
t∈[0,t1 ]
0
Z
≤ X(0) + c
0
×
t1
kbk20
L10/3 X(τ )dτ
t1
Z
X(τ )dτ
. 1 + sup kb(t)k20
L10/3
Z
t∈[0,t1 ]
Z
+ 1+
0
.1+
4
X
t∈[0,t1 ]
t1
1/4 Z
0
+ sup k∇h u(τ )kL2
k∆uk2L2 dτ
t1
Z
k∇∇h uk2L2 dτ
1/2
0
1/4
0
t1
X(τ )dτ
0
t1
ku3 k2L∞ k∇uk2L2 + kbk5L10/3 X 1/4 (τ )Z 3/4 (τ )dτ
t1
Z
Z(τ )dτ
1/4
0
IIIIi ,
i=1
(5.2)
where
IIII1 = c1 sup
t∈[0,t1 ]
kb(t)k20
L10/3 ,
IIII2 = c1
t1
Z(τ )dτ
1/4
,
0
t1
1/4
IIII3 = c1
Z(τ )dτ
,
0
0
Z
Z
t1
t1
1/4
IIII4 = c1
kbk5L10/3 X 1/4 (τ )Z 3/4 (τ )dτ
Z(τ )dτ
,
Z
0
t1
Z
ku3 k2L∞ k∇uk2L2 dτ
Z
0
12
K. YAMAZAKI
EJDE-2014/98
for c1 ≥ 0 independent of time t1 . Now due to (2.2), we can choose t1 ∈ [0, T ] so
that
3/4
7 1/4
14 7 T 1/4 M13/4 2 1/4 3/4 5/2
c1 28/5 M22 + M22 e 3 T M1 +
M 2 T M1
e3
3
Z
t1
1/4
(5.3)
×
X(τ )dτ
0
1
≤ .
8
Then, by (5.1),
3/4
7 1/4
14 7 T 1/4 M13/4 2 1/4 3/4 10
IIII1 ≤ c1 28/5 M22 + M22 e 3 T M1 +
e3
M 2 T M1
. 1.
3
(5.4)
The estimate of IIII2 is same as before in (4.16) and the estimate of IIII3 is also
same as (4.17). Finally,
IIII4
Z
≤ c1
0
t1
kbk20
L10/3 X(τ )dτ
≤ c1 sup kb(t)k5L10/3
Z
t∈[0,t1 ]
1/4 Z
0
t1
X(τ )dτ
0
≤ c1 28/5 M22 + M22 e
Z t1
×
Z(τ )dτ
0
Z
1 t1
Z(τ )dτ,
≤
8 0
t1
3/4
7 1/4
M1
3T
Z(τ )dτ
1/4 Z t1
Z(τ )dτ
0
14 7 T 1/4 M13/4 2 1/4 3/4 5/2
e3
+
M2 T M1
3
(5.5)
by Hölder’s inequality, (5.1) and (5.3). Using (5.4) and (5.5) in (5.2), absorbing the
dissipative and diffusive terms, we have
Z
Z t1
1 t1
8/3
Z(τ )dτ . 1 +
sup X(t) +
ku3 kL∞ k∇uk2L2 dτ.
2
t∈[0,t1 ]
0
0
By Gronwall’s inequality, we have the bound
Z
1 t1
Z(τ )dτ.
sup X(t) +
2 0
t∈[0,t1 ]
We restart on time interval [t1 , 2t1 ] and after finite number of repetitions obtain
the same bound on [0, T ]. This completes the proof of Theorem 1.2.
6. Proof of Theorem 1.3
This proof is similar to that of Theorem 1.1. We sketch it for completeness.
k∇h uk2L2 + k∇h bk2L2 - estimate. As before, from (4.1)–(4.4) which only required
p > 3 leading to (4.5), we have
Z t
sup Y (τ ) +
k∇∇h uk2L2 + k∇∇h bk2L2 dτ
τ ∈[0,t]
0
EJDE-2014/98
COMPONENT REDUCTION
Z
.1+
0
k∇uk2L2
+
(4.10) and
t
2p
p−3
13
1
p−2 (τ )Z p−2 (τ )dτ.
ku3 k2L∞ k∇uk2L2 + kbkLp−2
p X
k∇bk2L2 -estimate. As in the proof
using k∇h uk2L2 + k∇h bk2L2 -estimate
of Theorem 1.1, from (4.6), (4.8)–
leading to (4.12), we have
4
X
t
Z
Z(τ )dτ ≤ c0 +
X(t) +
0
IIIi .
(6.1)
i=1
By (3.3) with j1 = 2, j2 = 3, j3 = 1, we have
Z
RT
2
sup kb2,3 (t)k2Lp ≤ ec(p) 0 ku2,3 (λ)kL∞ dλ 1 +
t∈[0,T ]
T
ku2,3 (λ)k2L∞ kb1 (λ)k2Lp dλ .
0
so that by (4.13), similarly to (4.14),
2p
p−3
Lp
kb(τ )k
2p
p−3
Lp
≤ c(p)kb1 (τ )k
+e
c(p)
RT
0
ku2,3 k2L∞ dλ
1+
Z
T
ku2,3 k2L∞ kb1 k2Lp dλ
p p−3
0
(6.2)
and hence
t
Z
III1 ≤ c(p)
0
+e
c(p)
2p
kb1 (τ )kLp−3
p X(τ )dτ
RT
0
ku2,3 (λ)k2L∞ dλ
Z
1+
T
ku2,3 (λ)k2L∞ kb1 (λ)k2Lp dλ
p p−3
(6.3)
0
by (6.2) and (2.2). We take the identically same estimates on III2 and III3 as
before from (4.16) and (4.17). On III4 , from (4.18), we have
Z t
1Z t
8p
III4 ≤ c
kbkL3p−10
X(τ )dτ +
Z(τ )dτ,
(6.4)
p
8 0
0
where due to (6.2) and (4.13) we have
8p
kb(τ )kL3p−10
p
8p
≤ c(p)kb1 (τ )kL3p−10
+ ec(p)
p
RT
0
ku2,3 k2L∞ dλ
1+
Z
T
ku2,3 k2L∞ kb1 k2Lp dλ
4p
3p−10
.
0
Thus, by (2.2) similarly to (4.19),
Z t
Z t
8p
8p
RT
2
3p−10
kb(τ )kLp X(τ )dτ ≤ c(p)
kb1 kL3p−10
Xdτ + ec(p) 0 ku2,3 kL∞ dλ
p
0
0
× 1+
Z
T
ku2,3 k2L∞ kb1 k2Lp dλ
4p
3p−10
(6.5)
.
0
With (6.5) applied to (6.4), along with (4.16), (4.17) and (6.3) applied to (6.1),
absorbing dissipative and diffusive terms, we have by Hölder’s inequalities
Z t
X(t) +
Z(τ )dτ
Z
0
t
2p
p−3
Lp
c(p)
RT
ku2,3 k2L∞ dλ
.1+
kb1 k
Xdτ + e
1+
0
Z t
Z t
8p
8/3
ku3 kL∞ k∇uk2L2 dτ +
kb1 kL3p−10
+
Xdτ
p
0
0
0
Z
0
T
ku2,3 k2L∞ kb1 k2Lp dλ
p p−3
14
K. YAMAZAKI
EJDE-2014/98
4p
3p−10
Z T
RT
2
ku2,3 k2L∞ kb1 k2Lp dλ
+ ec(p) 0 ku2,3 kL∞ dλ 1 +
0
Z t
8p
8/3
3p−10
kb1 kLp
.1+
+ ku3 kL∞ Xdτ
0
+e
R
8/3
c(p)T 1/4 ( 0T ku2,3 kL∞ dλ)3/4
1+
Z
0
×
T
Z
kb1 k8Lp dλ
T
8/3
ku2,3 kL∞ dλ
3/4
4p
1/4 3p−10
0
.1+
Z t
Z
8p
8/3
Xdτ
+
c(p,
N
)
1
+
+
ku
k
kb1 kL3p−10
p
1
3 L∞
T
kb1 k8Lp dλ
p
3p−10
.
0
0
Thus, the proof of Theorem 1.3 is complete because by Hölder’s inequalities as in
the proof of Theorem 1.1, we have
Z T
8p
8/3
kb1 (τ )kL3p−10
+ ku3 (τ )kL∞ + kb1 (τ )k8Lp dλ < ∞.
p
0
7. Proof of Theorem 1.4
The proof is similar to that of Theorem 1.2; we sketch it for completeness. For
fixed T > 0, firstly, from (3.3) with j1 = 2, j2 = 3, j3 = 1, we obtain by Hölder’s
inequality
3/4
14 1/4
7 14 T 1/4 N13/4 2 1/4 3/4
N2 T N1
e3
sup kb2,3 (t)k2L10/3 ≤ N22 e 3 T N1 +
3
t∈[0,T ]
so that by (4.13),
sup kb(t)k2L10/3 ≤ sup 28/5 kb1 k2L10/3 + kb2,3 k2L10/3
t∈[0,T ]
t∈[0,T ]
3/4
14 1/4
7 14 T 1/4 N13/4 2 1/4 3/4 N2 T N1
e3
≤ 28/5 N22 + N22 e 3 T N1 +
3
(7.1)
similarly to (5.1). Now as in the proof of Theorem 1.2, we choose t1 ∈ [0, T ] to be
specified subsequently and use the previous estimate of
Z t1
4
X
sup X(t) +
Z(τ )dτ . 1 +
IIIIi
(7.2)
t∈[0,t1 ]
0
i=1
from (5.2). By (2.2), we can choose t1 ∈ [0, T ] so that
3/4
14 1/4
7 14 T 1/4 N13/4 2 1/4 3/4 5/2
c1 28/5 N22 + N22 e 3 T N1 +
e3
N2 T N1
3
Z
t1
1/4
×
X(τ )dτ
(7.3)
0
1
.
8
Firstly, by (7.1),
3/4
14 1/4
7 7 T 1/4 N13/4 2 1/4 3/4 10
IIII1 ≤ c1 28/5 N22 + N22 e 3 T N1 +
e3
N2 T N1
.
3
≤
(7.4)
EJDE-2014/98
COMPONENT REDUCTION
15
We use the same estimates of (4.16) and (4.17) for IIII2 and IIII3 as before.
Finally, from (2.2), (7.1) and (7.3) and Hölder’s inequality,
Z t1
1/4 Z t1
IIII4 ≤ c1
kbk20
X(τ
)dτ
Z(τ
)dτ
10/3
L
0
0
5/2 Z t1
1/4 Z t1
≤ c1 sup kb(t)k2L10/3
X(τ )dτ
Z(τ )dτ
t∈[0,T ]
0
14
≤ c1 28/5 N22 + N22 e 3 T
Z t1
×
Z(τ )dτ
0
Z
1 t1
Z(τ )dτ.
≤
8 0
1/4
3/4
N1
0
7 14 T 1/4 N13/4 2 1/4 3/4 5/2 (7.5)
e3
+
N2 T N1
3
After absorbing dissipative and diffusive terms, due to (4.16), (4.17), (7.4) and (7.5)
applied to (7.2), Gronwall’s inequality gives the bound on
Z
1 t1
sup X(t) +
Z(τ )dτ.
2 0
t∈[0,t1 ]
Reiterating on [t, 2t1 ], after finite times we obtain the bound on the whole interval
[0, T ] completing the proof of Theorem 1.4.
8. Appendix
In this section we give details of the decomposition in the k∇h uk2L2 + k∇h bk2L2 estimate, namely (4.2) (cf. [5, 12]). The following lemma is useful:
Lemma 8.1 ([17]). Assume that u ∈ H 2 (R3 ) is smooth and ∇ · u = 0. Then
Z
Z
Z
2 Z
2
X
X
1
ui ∂i uj ∆h uj =
∂i uj ∂i uj ∂3 u3 − ∂1 u1 ∂2 u2 ∂3 u3 + ∂1 u2 ∂2 u1 ∂3 u3
2
i,j=1
i,j=1
Firstly, for I1 applying this Lemma and integrating by parts, we have
Z
2 Z
X
ui ∂i uj ∆h uj . |u3 ||∇h u||∇∇h u|.
i,j=1
Thus,
I1 =
2 Z
X
ui ∂i uj ∆h uj +
i,j=1
2 Z
X
u3 ∂3 uj ∆h uj +
j=1
3
X
ui ∂i u3 ∆h u3
i=1
Z
.
|u3 ||∇u||∇∇h u|.
Next, we decompose I2 : integrating by parts
Z
2
2 Z
3 X
2 Z
X
X
X
2
2
2
I2 = −
bi ∂i bj ∂kk
uj −
bi ∂i b3 ∂kk
u3 −
b3 ∂3 bj ∂kk
uj
i,j,k=1
=−
2
X
i,j,k=1
j=1 k=1
i,k=1
Z
2
bi ∂i bj ∂kk
uj −
2
X
i,k=1
Z
2
bi ∂i b3 ∂kk
u3
16
K. YAMAZAKI
+
3 X
2 Z
X
EJDE-2014/98
2
∂k b3 ∂3 bj ∂k uj + b3 ∂k3
bj ∂ k u j
j=1 k=1
Z
2
X
=−
2
bi ∂i bj ∂kk
uj −
i,j,k=1
+
2 Z
X
2
bi ∂i b3 ∂kk
u3
i,k=1
3 X
2 Z
X
2
2
2
−∂3k
b3 bj ∂k uj − ∂k b3 bj ∂3k
uj + b3 ∂k3
bj ∂k uj
j=1 k=1
Z
|b||∇h b||∇∇h u| + |b||∇h u||∇∇h b|.
.
Next, we write
Z
2
X
I3 =
2
ui ∂i bj ∂kk
bj +
i,j,k=1
2 Z
X
2
ui ∂i b3 ∂kk
b3 +
i,k=1
3 X
2
X
2
u3 ∂3 bj ∂kk
bj
j=1 k=1
= I31 + I32 + I33 .
Integrating by parts,
Z
2
X
2
∂k ui ∂i bj ∂k bj + ui ∂ik
bj ∂k bj
I31 = −
i,j,k=1
=
Z
2
X
1
2
2
∂kk
ui ∂i bj bj + ∂k ui ∂ik
bj bj + ∂i ui ∂k bj ∂k bj
2
i,j,k=1
=
Z
2
X
1 2
2
2
2
∂kk
ui ∂i bj bj + ∂k ui ∂ik
bj bj − (∂ik
ui ∂k bj bj + ∂i ui ∂kk
bj bj )
2
i,j,k=1
Z
|b||∇h b||∇∇h u| + |b||∇h u||∇∇h b|,
.
I32 = −
2 Z
X
i,k=1
=
2 Z
X
1
2
2
∂kk
ui ∂i b3 b3 + ∂k ui ∂ik
b3 b3 + ∂ i u i ∂ k b3 ∂ k b3
2
i,k=1
=
1
∂k ui ∂i b3 ∂k b3 + ui ∂i (∂k b3 )2
2
2 Z
X
1 2
2
2
2
∂kk
ui ∂i b3 b3 + ∂k ui ∂ik
b3 b3 − (∂ik
ui ∂k b3 b3 + ∂i ui ∂kk
b3 b3 )
2
i,k=1
Z
.
I33 = −
|b||∇h b||∇∇h u| + |b||∇h u||∇∇h b|,
3 X
2 Z
X
j=1 k=1
=
3 X
2 Z
X
j=1 k=1
1
∂k u3 ∂3 bj ∂k bj + u3 ∂3 (∂k bj )2
2
1
2
2
∂3k
u3 bj ∂k bj + ∂k u3 bj ∂3k
bj + ∂3 u3 ∂k bj ∂k bj
2
EJDE-2014/98
=
COMPONENT REDUCTION
3 X
2 Z
X
2
2
2
∂3k
u3 bj ∂k bj + ∂k u3 bj ∂3k
bj −
1X
∂i ui ∂k bj ∂k bj
2 i=1
2
2
∂3k
u3 bj ∂k bj + ∂k u3 bj ∂3k
bj +
1X 2
2
(∂ ui ∂k bj bj + ∂i ui ∂kk
bj bj )
2 i=1 ik
j=1 k=1
=
3 X
2 Z
X
17
2
j=1 k=1
Z
|b||∇h b||∇∇h u| + |b||∇h u||∇∇h b|.
.
Therefore,
Z
I3 .
|b||∇h b||∇∇h u| + |b||∇h u||∇∇h b|.
Finally,
I4 = −
Z
2
X
i,j,k=1
2
bi ∂i uj ∂kk
bj −
2
X
2
bi ∂i u3 ∂kk
b3 −
i,k=1
3 X
2
X
2
b3 ∂3 uj ∂kk
bj
j=1 k=1
Z
.
|b||∇h b||∇∇h u| + |b||∇u||∇∇h b|.
Hence,
I2 + I3 + I4 . |b||∇b||∇∇h u| + |b||∇u||∇∇h b|.
This completes the decomposition claimed.
Acknowledgments. The author wants to expresses his gratitude to Professors
Jiahong Wu and David Ullrich for their teachings, and to the referees for their
valuable comments.
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Kazuo Yamazaki
Department of Mathematics, Oklahoma State University, 401 Mathematical Sciences,
Stillwater, OK 74078, USA
E-mail address: kyamazaki@math.okstate.edu