The existence and space-time decay rates of strong solutions to Navier-Stokes Equations in weighed $L^\infty(|x|^\gamma{\rm dx})\cap L^\infty(|x|^\beta{\rm dx})$ spaces

0.17 MB·

by  D. Q. Khai

The existence and space-time decay rates of strong solutions to
Navier-Stokes Equations in weighed L∞( x γdx) L∞( x βdx) spaces
| | ∩ | |
D. Q. Khai, N. M. Tri
Institute of Mathematics, VAST
18 Hoang Quoc Viet, 10307 Cau Giay, Hanoi, Vietnam
6
1
0
2 Abstract: In this paper, we prove some results on the existence and

n space-time decay rates of global strong solutions of the Cauchy problem
a for the Navier-Stokes equations in weighed L∞(Rd, x γdx) L∞(Rd, x βdx)
J
7 spaces. | | ∩ | |

]
P §1. Introduction
A
h. This paper studies the Cauchy problem of the incompressible Navier-
t Stokes equations (NSE) in the whole space Rd for d 2,
a
≥
m
[ ∂ u = ∆u .(u u) p,
t
−∇ ⊗ −∇
 .u = 0,
1
 ∇
v u(0,x) = u ,
0
3
2 
which is a condensed writing for
7
1
0 1 k d, ∂ u = ∆u d ∂ (u u ) ∂ p,
. ≤ ≤ t k k − l=1 l l k − k
1  d ∂ u = 0, P
0  l=1 l l
6 P1 k d, uk(0,x) = u0k.
≤ ≤
1

:
v The unknown quantities are the velocity u(t,x) = (u (t,x),...,u (t,x)) of
1 d
i
X the fluid element at time t and position x and the pressure p(t,x).
r There is an extensive literature on the existence and decay rate of strong so-
a
lutionsoftheCauchyproblemforNSE.MariaE.Schonbek[1]established the
decay of the homogeneous Hm norms for solutions to NSE in two dimensions.
She showed that if u is a solution to NSE with an arbitrary u Hm L1(R2)
0
∈ ∩
with m 3 then
≥
kDαuk22 ≤ Cα(t+1)−(|α|+1) andkDαuk∞ ≤ Cα(t+1)−(|α|+21) for allt ≥ 1,α ≤ m.
2Keywords: Navier-Stokes equations; space-time decay rate
3e-mail address: [email protected],[email protected]
1
Zhi-Min Chen [2] showed that if u L1(Rd) Lp(Rd),(d p < ) and
0
∈ ∩ ≤ ∞
u + u is small enough then there is a unique solution
0 1 0 p
k k k k
u BC([0, );L1 Lp), which satisfies decay property
∈ ∞ ∩
suptd2 u ∞ +t12 Du ∞ +t12 D2u ∞ < .
k k k k k k ∞
t>0 (cid:0) (cid:1)
Kato[3] studiedstrong solutionsinthespaces Lq(Rd)by applying theLq Lp
−
estimates for the semigroup generated by the Stokes operator. He showed
that there is T > 0 and a unique solution u, which satisfies
t21(1−dq)u BC([0,T);Lq), for d q ,
∈ ≤ ≤ ∞
t21(2−dq) u BC([0,T);Lq), for d q ,
∇ ∈ ≤ ≤ ∞
as u Ld(Rd). He showed that T = if u is small enough.
0 ∈ ∞ 0 Ld(Rd)
(cid:13) (cid:13)
In 2002, Cheng He and Ling Hsiao [4] extended the results of Kato, they
(cid:13) (cid:13)
estimated on decay rates of higher order derivatives in time variable and
space variables for the strong solution to NSE with initial data in Ld(Rd).
They showed thatif u issmall enoughthenthereisaunique solution
0 Ld(Rd)
u, which satisfies (cid:13) (cid:13)
(cid:13) (cid:13)
t12(1+|α|+2α0−dq)DαDα0u BC([0, );Lq), for q d,
x t ∈ ∞ ≥
t12(2+|α|−dq)Dαp BC([0, );Lq), for q d,
x ∈ ∞ ≥
where α = (α ,α ,...,α ), α = α +α +...+α and α N. Dα denotes
1 2 d | | 1 2 d 0 ∈ x
∂x|α| = ∂|α|/(∂xα11∂xα22...∂xαdd), ∂tα0 = ∂α0/∂tα0.
In 2005, Okihiro Sawada [5] obtained the decay rate of solution to NSE with
initial data in H˙ d2−1(Rd). He showed that every mild solution in the class
u BC([0,T);H˙ d2−1) and t12(d2−dp)u BC([0,T);H˙pd2−1),
∈ ∈
for some T > 0 and p (2, ] satisfies
∈ ∞
d d
(cid:13)u(t)(cid:13)H˙qd2−1+α ≤ K1(K2α˜)α˜t−α2˜ for α > 0,q ≥ 2, and α˜ := α+ 2 − q
(cid:13) (cid:13)
where constants K and K depend only on d,p,M , and M with
1 2 1 2
M1 = 0<sut<pT(cid:13)u(t)(cid:13)H˙ d2−1 and M2 = 0<sut<pTtd2(12−p1)(cid:13)u(t)(cid:13)H˙pd2−1.
The time-de(cid:13)cay p(cid:13)roperties are therefore well un(cid:13)derst(cid:13)ood. However, there are
few results on the spatial decay properties. Farwing and Sohr [7] showed
a class of weighted x α weak solutions with second derivatives in space
| |
2
variables and one order derivatives in time variable in Ls([0,+ );Lq) for
∞
1 < q < 3/2,1 < s < 2 and 0 3/q+2/s 4 α < min 1/2,3 3/q in the
≤ − ≤ { − }
case of exterior domains. In [10], they also showed that there exists a class
of weak solutions satisfying
C(u ,f,α) if 0 α < 1,
t 0 ≤ 2
|x|α2u 22 +Z |x|α2∇u 22dt ≤  C(u0,f,α′,α)tα2′−1/4 if 21 ≤ α < α′ < 1,
(cid:13) (cid:13) 0 (cid:13) (cid:13) C(u ,f)(t1/4 +t1/2) if α = 1.
(cid:13) (cid:13) (cid:13) (cid:13) 0

While in [11], a class of weak solutions
(1+ x 2)1/4u L∞([0,+ );Lp(R3))
| | ∈ ∞
was constructed for 6/5 p < 3/2, which satisfies
≤
t
|x|21u 22 +Z |x|12∇u 22dt ≤ C(u0,f)(t1/4 +t1/2).
(cid:13) (cid:13) 0 (cid:13) (cid:13)
(cid:13) (cid:13) (cid:13) (cid:13)
In 2002 Takahashi [9] studied the existence and space-time decay rates of
globalstrongsolutionsoftheCauchyproblemfortheNavier-Stokesequations
in the weighted L∞(Rd,(1 + x )βdx) spaces. Takahashi showed that if u
0
| |
satisfies
(et∆u0)(x) < δ(1+ x )−β, (et∆u0)(x) < δ(1+t)−β2, (1)
| | | | | |
with sufficiently small δ, then NSE has a global mild solution u such that
u(x,t) C(1+ x )−β, u(x,t) C(1+t)−β2,
| | ≤ | | | | ≤
where β is restricted by the condition 1 β d+1.
≤ ≤
Takahashi also showed that if
u (x) c(1+ x )−β for some 0 < β d,
0
| | ≤ | | ≤
then
(et∆u0)(x) c(1+ x )−β, (et∆u0)(x) c(1+t)−2β.
| | ≤ | | | | ≤
In this paper, we discuss the existence and space-time decay rates of global
strongsolutionsoftheCauchy problemfortheNavier-Stokes equations inthe
weightedL∞(Rd, x γdx) L∞(Rd, x βdx)spaces. ThespacesL∞(Rd, x γdx)
| | ∩ | | | | ∩
L∞(Rd, x βdx) are more general than the spaces L∞(Rd,(1 + x )βdx).
| | | |
In particular, L∞(Rd, x γdx) L∞(Rd, x βdx) = L∞(Rd,(1+ x )βdx) when
| | ∩ | | | |
γ = 0, and so this result improves the previous one.
The content of this paper is as follows: in Section 2, we state our main
3
theorems after introducing some notations. In Section 3, we first prove the
some estimates concerning the heat semigroup with the Helmholtz-Leray
projection and some auxiliary lemmas. Finally, in Section 4, we will give the
proof of the main theorems.
§2. Statement of the results
Now, for T > 0, we say that u is a mild solution of NSE on [0,T] cor-
responding to a divergence-free initial datum u when u solves the integral
0
equation
t
u = et∆u e(t−τ)∆P . u(τ,.) u(τ,.) dτ.
0 −Z ∇ ⊗
0 (cid:0) (cid:1)
Above we have used the following notation: for a tensor F = (F ) we define
ij
thevector .F by( .F) = d ∂ F andfortwovectorsuandv,wedefine
∇ ∇ i j=1 j ij
their tensor product (u v) P= u v . The operator P is the Helmholtz-Leray
ij i j
⊗
projection onto the divergence-free fields
(Pf) = f + R R f ,
j j j k k
X
1≤k≤d
where R is the Riesz transforms defined as
j
∂ iξ
j j
R = i.e. R g(ξ) = gˆ(ξ).
j j
√ ∆ ξ
− d | |
The heat kernel et∆ is defined as
et∆u(x) = ((4πt)−d/2e−|.|2/4t u)(x).
∗
For a space of functions defined on Rd, say E(Rd), we will abbreviate it
as E and we do not distinguish between the vector-valued and scalar-value
spaces of functions. Throughout the paper, we sometimes use the notation
A . B as an equivalent to A CB with a uniform constant C. The
≤
notation A B means that A . B and B . A. Let β 0, we define the
≃ ≥
space L∞( x βdx) := L∞(Rd, x βdx) which is made up by the measurable
| | | |
functions u such that
u := esssup x β u(x) < + .
L∞(|x|βdx) | | | | ∞
(cid:13) (cid:13) x∈Rd
(cid:13) (cid:13)
Now we can state our result
4
Theorem 1. Assume that d 1, and 0 γ 1 β < d. Then for all
≥ ≤ ≤ ≤
f L∞( x γdx) L∞( x βdx) we have
∈ | | ∩ | |
sup x γ˜t21(γ−γ˜) + x αt12(1−α) + x β˜t21(β−β˜) et∆f
| | | | | | | |
x∈Rd,t>0(cid:0) (cid:1)
. f + f
L∞(|x|γdx) L∞(|x|βdx)
(cid:13) (cid:13) (cid:13) (cid:13)
(cid:13) (cid:13) (cid:13) (cid:13)
for 0 γ˜ γ,0 α 1, and 0 β˜ β.
≤ ≤ ≤ ≤ ≤ ≤
Theorem 2. Let 0 γ 1 β < d be fixed, then for all γ˜,α, and β˜
≤ ≤ ≤
satisfying
˜ ˜ ˜ ˜
0 γ˜ γ,β 0,β 2 < β β,0 < α < 1, and β β 1 < α < d β,
≤ ≤ ≥ − ≤ − − −
there exists a positive constant δ such that for all u L∞( x γdx)
γ,γ˜,α,β,β˜,d 0 ∈ | | ∩
L∞( x βdx) with div(u ) = 0 satisfying
0
| |
sup |x|γ˜t21(γ−γ˜) +|x|αt21(1−α) +|x|β˜t12(β−β˜) |et∆u0| ≤ δγ,γ˜,α,β,β˜,d, (2)
x∈Rd,t>0(cid:0) (cid:1)
NSE has a global mild solution u on (0, ) Rd such that
∞ ×
sup x γ +tγ2 + x β +tβ2 u(x,t) < + . (3)
| | | | | | ∞
x∈Rd,t>0(cid:0) (cid:1)
Remark 1. Our result improves the previous result for L∞(Rd,(1+ x )βdx).
| |
This space, studied in [9], is a particular case of the space L∞( x γdx)
| | ∩
L∞( x βdx) when γ = 0. Furthermore, we prove that Takahashi’s result
| |
holds true under a much weaker condition on the initial data. Indeed,
from Lemma 4 and Theorem 1, it is easily seen that the condition (2) of
Theorem 2 is weaker than the condition (1).
Remark 2. We invoke Theorem 1 to deduce that if u L∞( x γdx)
0
∈ | | ∩
L∞( x βdx) and u + u is small enough then the
| | 0 L∞(|x|γdx) 0 L∞(|x|βdx)
(cid:13) (cid:13) (cid:13) (cid:13)
condition (2) of Theorem 2 is valid.
(cid:13) (cid:13) (cid:13) (cid:13)
Theorem 3. Let 1 β < d be fixed, then for all α satisfying 0 < α < 1, there
≤
exists a positive constant δ such that for all u L∞( x dx) L∞( x βdx)
α,d 0
∈ | | ∩ | |
with div(u ) = 0 satisfying
0
sup x αt21(1−α) et∆u0 δα,d, (4)
| | | | ≤
x∈Rd,t>0
NSE has a global mild solution u on (0, ) Rd such that
∞ ×
1
sup x +t2 u(x,t) < +
| | | | ∞
x∈Rd,t>0(cid:0) (cid:1)
5
and
sup x β u(x,t) < + , for all T (0, ).
| | | | ∞ ∈ ∞
x∈Rd,0<t<T
Remark 3. We invoke Theorem 1 to deduce that if u L∞( x dx) and
0
∈ | |
u is small enough then the condition (4) of Theorem 3 is valid.
0 L∞(|x|dx)
(cid:13) (cid:13)
(cid:13) (cid:13)
§3. Some auxiliary results
In this section we establish some auxiliary lemmas. We first prove a
version of Young’s inequality type for convolutions in L∞( x βdx) spaces.
| |
Lemma 1. Assume that d 1,0 < α < d,0 < β < d and α+β > d. Then
≥
for all f L∞( x αdx) and for all g L∞( x βdx) we have
∈ | | ∈ | |
f g . f g .
∗ L∞(|x|α+β−ddx) L∞(|x|αdx) L∞(|x|βdx)
(cid:13) (cid:13) (cid:13) (cid:13) (cid:13) (cid:13)
(cid:13) (cid:13) (cid:13) (cid:13) (cid:13) (cid:13)
Proof. Since f g isbilinear onL∞( x αdx) L∞( x βdx), wemay assume
∗ | | × | |
f = g = 1. We have
L∞(|x|αdx) L∞(|x|βdx)
(cid:13) (cid:13) (cid:13) (cid:13)
(cid:13) (cid:13) (cid:13) (cid:13)
(f g)(x) = f(x y)g(y)dy = + + = I +I +I .
∗ ZRd − Z|y|<|x| Z|x|<|y|<3|x| Z|y|>3|x| 1 2 3
2 2 2 2
From
f(x) x −α, and g(x) x −β,
| | ≤ | | | | ≤ | |
we get
dy 2α dy 1
I .
| 1| ≤ Z|y|<|x2| |x−y|α|y|β ≤ |x|α Z|y|<|x2| |y|β ≃ |x|α+β−d
dy 2β dy 1
I .
| 2| ≤ Z|x2|<|y|<3|2x| |x−y|α|y|β ≤ |x|β Z|y|<5|2x| |y|α ≃ |x|α+β−d
dy dy 1
I 3α .
| 3| ≤ Z|y|>3|2x| |x−y|α|y|β ≤ Z|y|>3|2x| |y|α|y|β ≃ |x|α+β−d
We thus obtain
1
(f g)(x) . .
| ∗ | x α+β−d
| |
The proof Lemma 1 is complete.
WenowdeducetheL∞( x γdx) L∞( x βdx)estimatefortheheatsemigroup.
| | − | |
6
Lemma 2. Assume that d 1 and 0 γ β < d. Then for all
≥ ≤ ≤
f L∞( x βdx) we have
∈ | |
et∆f . t−21(β−γ) f , for t > 0. (5)
L∞(|x|γdx) L∞(|x|βdx)
(cid:13) (cid:13) (cid:13) (cid:13)
(cid:13) (cid:13) (cid:13) (cid:13)
Proof. We have
(et∆f)(x) = 1 E x−y f(y)dy, where E(x) = (4π)−d2e−|x4|2.
ZRd td/2 (cid:0) √t (cid:1)
Recall the simate
t−d2e−|x4|t2 . x −αt−12(d−α), for 0 α d. (6)
| | ≤ ≤
We first consider the case 0 < γ < β. From the inequality (6) and Lemma
1, we have
f
(et∆f)(x) . (cid:13) (cid:13)L∞(|x|βdx) dy . t−12(β−γ) x −γ f .
| | ZRd t12(β−(cid:13)γ) x(cid:13) y γ+d−β y β | | (cid:13) (cid:13)L∞(|x|βdx)
| − | | | (cid:13) (cid:13)
This proves (5).
We consider the case 0 = γ < β. Applying Proposition 2.4 (b) in ([6], pp.
20) and note that x −β Lβd,∞
| | ∈
.
|et∆f(x)| . t−d2(cid:13)E(cid:0)√t(cid:1)(cid:13)Ld−dβ,1(cid:13)f(cid:13)Lβd,∞ . t−β2(cid:13)E(cid:13)Ld−dβ,1(cid:13)f(cid:13)L∞(|x|βdx).
(cid:13) (cid:13) (cid:13) (cid:13) (cid:13) (cid:13) (cid:13) (cid:13)
This proves (5).
Suppose finally that 0 γ = β. We have
≤
1 x y
E − f(y)dy = + = I +I .
ZRd td/2 (cid:0) √t (cid:1) Z|y|<|x| Z|y|>|x| 1 2
2 2
From the inequality (6), we have
I . f x y −d y −βdy
| 1| (cid:13) (cid:13)L∞(|x|βdx)Z|y|<|x| | − | | | ≤
(cid:13) (cid:13) 2
x
f | | −d y −βdy f x −β.
(cid:13) (cid:13)L∞(|x|βdx)(cid:0) 2 (cid:1) Z|y|<|x| | | ≃ (cid:13) (cid:13)L∞(|x|βdx)| |
(cid:13) (cid:13) 2 (cid:13) (cid:13)
1 x y
I f E − y −βdy
| 2| ≤ (cid:13) (cid:13)L∞(|x|βdx)Z|y|>|x| td/2 (cid:0) √t (cid:1)| | ≤
(cid:13) (cid:13) 2
x 1 y
f | | −β E dy = C f x −β,
(cid:13) (cid:13)L∞(|x|βdx)(cid:0) 2 (cid:1) Zy∈Rd td/2 (cid:0)√t(cid:1) (cid:13) (cid:13)L∞(|x|βdx)| |
(cid:13) (cid:13) (cid:13) (cid:13)
7
where
C = 2β E(y)dy < + .
Z ∞
y∈Rd
Therefore,
et∆f(x) . f x −β.
| | L∞(|x|βdx)| |
(cid:13) (cid:13)
The proof of Lemma 2 is complete.(cid:13) (cid:13)
We now deduce the L∞( x γdx) L∞( x βdx) estimate for the operator
| | − | |
et∆P . Asshownin[6], thekernel functionF ofet∆P satisfiesthefollowing
t
∇ ∇
inequalities
x 1
Ft(x) = t−d+21F , F(x) . , (7)
√t | | (1+ x )d+1
(cid:0) (cid:1) | |
Ft(x) . x −αt−21(d+1−α), for 0 α d+1. (8)
| | | | ≤ ≤
By using the inequalities (7) and (8) and arguing as in the proof of
Lemma 2, we can easily prove the following lemma.
Lemma 3. Assume that d 1 and 0 γ β < d. Then for all
≥ ≤ ≤
f L∞( x βdx) we have
∈ | |
et∆P .f . t−21(β+1−γ) f , for t > 0.
∇ L∞(|x|γdx) L∞(|x|βdx)
(cid:13) (cid:13) (cid:13) (cid:13)
(cid:13) (cid:13) (cid:13) (cid:13)
Lemma 4. Let 0 γ < β d. Assume that f ′(Rd) and satisfies the
≤ ≤ ∈ S
following inequality
sup x γ + x β (et∆f)(x) = C < + , (9)
| | | | | | ∞
x∈Rd,t>0(cid:0) (cid:1)
then
f L∞( x γdx) L∞( x βdx)
∈ | | ∩ | |
and
esssup x γ + x β f(x) C. (10)
| | | | | | ≤
x∈Rd (cid:0) (cid:1)
Proof. Since 1 Lβd,∞ Lγd,∞ and Lβd,∞ Lγd,∞ Lq for all q
|x|γ+|x|β ∈ ∩ ∩ ⊂
satisfying d < q < d, it follows that et∆f L∞(0, ;Lq) for all q d, d ,
β γ ∈ ∞ ∈ β γ
by a compactness theorem in Banach space, there exists a sequence t (cid:0)whic(cid:1)h
k
converges to 0 such that etk∆f converges weakly to f′ in Lq with f′ Lq.
∈
Since et∆ is a continuous semigroup on ′(Rd), it follows that f = f′ Lq.
S ∈
Since et∆ is a continuous semigroup on Lq(Rd),(1 q < ), we get
≤ ∞
d d
lim etk∆f f = 0, for q , .
k→∞(cid:13) − (cid:13)Lq ∈ (cid:0)β γ(cid:1)
(cid:13) (cid:13)
8
Therefore, there exists a subsequence t of the sequence t such that
kj k
lim(etkj∆f)(x) = f(x) for almost everywhere x Rd. (11)
j→∞ ∈
The inequality (10) is deduced from equalities (9) and (11).
Remark 4. We invoke Lemma 4 for γ = 0 and Lemma 2 for γ = β to deduce
that the condition (1) of Takahashi on the initial data is equivalent to the
condition
u δ.
0 L∞((1+|x|)βdx) ≤
(cid:13) (cid:13)
Lemma 5. Let γ,θ R an(cid:13)d t(cid:13)> 0, then
∈
(a) If θ < 1 then
t 1
2 2
(t τ)−γτ−θdτ = Ct1−γ−θ, where C = (1 τ)−γτ−θdτ < .
Z − Z − ∞
0 0
(b) If γ < 1 then
t 1
(t τ)−γτ−θdτ = Ct1−γ−θ, where C = (1 τ)−γτ−θdτ < .
Zt − Z1 − ∞
2 2
(c) If γ < 1 and θ < 1 then
t 1
(t τ)−γτ−θdτ = Ct1−γ−θ, where C = (1 τ)−γτ−θdτ < .
Z − Z − ∞
0 0
The proof of this lemma is elementary and may be omitted.
Let us recall the following result on solutions of a quadratic equation in
Banach spaces (Theorem 22.4 in [6], p. 227).
Theorem 4. Let E be a Banach space, and B : E E E be a continuous
× →
bilinear map such that there exists η > 0 so that
B(x,y) η x y ,
k k ≤ k kk k
for all x and y in E. Then for any fixed y E such that y 1 , the
∈ k k ≤ 4η
equation x = y B(x,x) has a unique solution x E satisfying x 1 .
− ∈ k k ≤ 2η
§4. Proofs of Theorems 1, 2, and 3
In this section we will give the proofs of Theorems 1, 2, and 3. We now
need eight more lemmas. In order to proceed, we define an auxiliary space
9
Kβ . Let α,β, and T be such that 0 α β < d,0 < T + , we define
α,T ≤ ≤ ≤ ∞
theauxiliaryspaceKβ whichismadeupbythemeasurablefunctionsu(t,x)
α,T
such that
esssup x αt21(β−α) u(x,t) < + .
| | | | ∞
x∈Rd,0<t<T
The auxiliary space Kβ is equipped with the norm
α,T
u := esssup x αt12(β−α) u(x,t) .
Kβ | | | |
(cid:13) (cid:13) α,T x∈Rd,0<t<T
(cid:13) (cid:13)
We rewrite Lemma 2 as follows
Lemma 6. Assume that d 1 and 0 α β < d. Then for all
≥ ≤ ≤
f L∞( x βdx) we have et∆f Kβ and et∆f C f ,
∈ | | ∈ α,T Kβ ≤ L∞(|x|βdx)
(cid:13) (cid:13) α,T (cid:13) (cid:13)
where C is a positive constant independent of T.
(cid:13) (cid:13) (cid:13) (cid:13)
Lemma 7. Assume that d 1 and 0 α β < d. Then
≥ ≤ ≤
Kβ Kβ Kβ .
α,T ⊂ β,T ∩ 0,T
The proof of this lemma is elementary and may be omitted.
Lemma 8. Assume that d 1,T < + , and 0 α β β˜ < d. Then
≥ ∞ ≤ ≤ ≤
Kβ Kβ˜ .
α,T ⊂ α,T
The proof of this lemma is elementary and may be omitted.
In the following lemmas a particular attention will be devoted to the study
of the bilinear operator B(u,v)(t) defined by
t
B(u,v)(t) = e(t−τ)∆P . u(τ) v(τ) dτ. (12)
Z ∇ ⊗
0 (cid:0) (cid:1)
Lemma 9. Let β,β˜,βˆ, and α be such that
˜ ˜ ˜ ˜
0 β < d,β > β 2,0 β β,0 < α < 1,β β 1 < α < d β,
≤ − ≤ ≤ − − −
ˆ ˜ ˆ ˜
0 β β, and α+β 1 < β α+β.
≤ ≤ − ≤
Then the bilinear operator B is continuous from K1 Kβ into Kβ and
α,T × β˜,T βˆ,T
the following inequality holds
B(u,v) C u v , (13)
(cid:13) (cid:13)Kββˆ,T ≤ (cid:13) (cid:13)Kα1,T(cid:13) (cid:13)Kββ˜,T
(cid:13) (cid:13) (cid:13) (cid:13) (cid:13) (cid:13)
where C is a positive constant independent of T.
10