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ARTICLE IN PRESSNonlinear Analysis ( ) –

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Nonlinear Analysis

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Stability of plane Couette flows with respect to small periodicperturbationsHorst Heck a, Hyunseok Kim b,∗, Hideo Kozono ca Institut fuer Mathematik, TU Darmstadt, Schlossgartenstrasse 7, 64289 Darmstadt, Germanyb Department of Mathematics, Sogang University, Seoul, 121-742, Republic of KoreacMathematical Institute, Tohoku University, Sendai, 980-8578, Japan

a r t i c l e i n f o

Article history:Received 5 August 2008Accepted 6 February 2009

MSC:35Q3076E05

Keywords:The Navier–Stokes equationsPlane Couette flowStability

a b s t r a c t

We consider the plane Couette flow v0 = (xn, 0, . . . , 0) in the infinite layer domainΩ = Rn−1 × (−1, 1), where n ≥ 2 is an integer. The exponential stability of v0 in Ln isshown under the condition that the initial perturbation is periodic in (x1, . . . , xn−1) andsufficiently small in the Ln-norm.

© 2009 Elsevier Ltd. All rights reserved.

1. Introduction

The motion of a viscous incompressible fluid occupying a domain Ω in Rn, n ≥ 2, is governed by the Navier–Stokesequations

∂tv − ν∆v +∇p = −(v · ∇)v inΩ × (0,∞)div v = 0 inΩ × (0,∞), (1)

where v = (v1, . . . , vn) and p denote the unknown velocity vector and pressure scalar fields of the fluid, respectively, whilea fixed positive number ν stands for the viscosity constant.It has been an important problem inmathematical fluidmechanics to prove the stability or instability of a given stationary

solution of (1). In this paper, we study the stability of the plane Couette flow

v0 = (xn, 0, . . . , 0)

defined in the infinite layer domain

Ω =x = (x′, xn) ∈ Rn : x′ = (x1, . . . , xn−1) ∈ Rn−1, −1 < xn < 1

.

The stability of the plane Couette flow v0 has been well-known for the physically relevant case n = 3 since the classicalpaper [1] by Romanov in 1973. See [2–4] for recent progress. Following a traditional approach (see [5,6] and [4]), Romanovconsidered the Orr–Sommerfeld boundary value problem to investigate the linear stability of v0. The main achievement

∗ Corresponding author.E-mail addresses: [email protected] (H. Heck), [email protected] (H. Kim), [email protected] (H. Kozono).

0362-546X/$ – see front matter© 2009 Elsevier Ltd. All rights reserved.doi:10.1016/j.na.2009.02.034

Please cite this article in press as: H. Heck, et al., Stability of plane Couette flows with respect to small periodic perturbations, Nonlinear Analysis (2009),doi:10.1016/j.na.2009.02.034

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http://dx.doi.org/10.1016/j.na.2009.02.034

ARTICLE IN PRESS2 H. Heck et al. / Nonlinear Analysis ( ) –

of [1] is to show, with the help of a numerical calculation, that if λ is an eigenvalue of the Orr–Sommerfeld problem,then Re λ ≤ −δ for some constant δ > 0. Then, from this result, Romanov concluded that the plane Couette flow v0 isexponentially stable in H10,σ (Ω): if the initial perturbation a = (a1(x), . . . , an(x)) belongs to H10,σ (Ω) and if ‖a‖H1(Ω) issufficiently small, then the initial-boundary value problem for the perturbation u = v − v0

∂tu− ν∆u+ (v0 · ∇)u+ (u · ∇)v0 +∇p = −(u · ∇)u inΩ × (0,∞)div u = 0 inΩ × (0,∞)u = 0 on ∂Ω × (0,∞), u(0) = a inΩ

(2)

has a unique global solution u ∈ C([0,∞);H10,σ (Ω)) and u(t) decays exponentially in H1(Ω) as t → ∞. Here H10,σ (Ω)

denotes the closure of the space C∞0,σ (Ω) of all divergence-free test functions inΩ with respect to H1(Ω)-norm.

However Romanov’s argument has not been made completely rigorous yet. Let us denote by σ(−L) the spectrum of theperturbed Stokes operator −L defined by Lu = P (−ν∆u+ (v0 · ∇)u+ (u · ∇)v0), where P is the Helmholtz projection.Then it is quite easy to deduce from Romanov’s estimate on the Orr–Sommerfeld eigenvalue problem that the real part ofeach eigenvalue of−L is bounded above by−δ; see the proof of Lemma 2.3 below. But this is not sufficient to conclude theexponential stability of v0. For the stability of v0, one has to prove the following stronger result:

Re λ ≤ −δ for all λ ∈ σ(−L), (3)

whichwas indeedutilizedbyRomanovwithout a detailedproof. Note thatσ(−L)maycontain points other than eigenvaluesof−L because the underlying domainΩ is unbounded. Hence Romanov’s proof of the stability of v0 is incomplete.On the other hand, a proof of the exponential stability in Lnσ (Ω) of v0, where L

nσ (Ω) is the closure of C

∞

0,σ (Ω) in Ln(Ω),

was obtained by Abe and Shibata [7] in cases when the viscosity ν is sufficiently large. But this result is not so surprisingfrom the viewpoint of the hydrodynamical stability theory because the operator−Lwith large viscosity ν can be regardedas a small perturbation of the Stokes operator which has been known to satisfy analogues of (3); see [8,7,9] for instance.Moreover, even though Ω is an unbounded domain, Poincaré inequalities hold for all vector fields in H10,σ (Ω) and thus astandard energy argument (see [4] for instance) allows us to deduce the exponential decay of u(t) in L2σ (Ω) as t → ∞ forany initial perturbation a ∈ L2σ (Ω) provided that ν is sufficiently large. Hence the major problem is to prove the stability ofv0 for the case of small viscosity ν.The purpose of this paper is to show that the plane Couette flow v0 is exponentially stable for any viscosity constant

ν provided that the initial perturbation a = a(x′, xn) is periodic in the variable x′ and sufficiently small in the Ln-norm. Tostate our result precisely, let us first introduce the Sobolev spaces consisting of periodic functions in x′. Given a fixed numberl > 0, letD be the space of all complex-valued functions f onΩ which can be written as

f (x′, xn) =∑k∈J

fk(xn) eiω〈k,x′〉 (4)

for some finite subset J of Zn−1 and some fk ∈ C∞([−1, 1]), where ω = πl . Note that the set e

iω<k,·>: k ∈ Zn−1 is

orthogonal in L2(T) with T = [−l, l]n−1 being the (n − 1)-dimensional torus. Hence if f ∈ D can be expanded into a finiteseries of the form (4), then its coefficients are determined uniquely by the partial Fourier series of f :

fk(xn) =1

(2l)n−1

∫Tf (x′, xn)e−iω〈k,x

′〉 dx′ (k ∈ Zn−1).

It is also necessary to introduce the spaceD0,σ of divergence-free test functions defined by

D0,σ =f ∈ [D]n : div f = 0 inΩ and f = 0 on ∂Ω

.

Ifm ≥ 0 is an integer and 1 < q <∞, we define the norms ‖ · ‖q and ‖ · ‖m,q by

‖f ‖qq =∫|f |q dx =

∫ 1

−1

∫T|f (x′, xn)|q dx′dxn

and

‖f ‖m,q =

[∑|α|≤m

‖Dα f ‖qq

] 1q

for each f ∈ D or f ∈ [D]n. Then our basic Sobolev spaces are defined as the closures ofD orD0,σ with respect to the norm‖ · ‖m,q as follows:

Hm,q = [D]n‖·‖m,q

(orD‖·‖m,q), Lq = H0,q,

Lqσ = D0,σ‖·‖q and H1,q0,σ = D0,σ

‖·‖1,q

form ≥ 0 and 1 < q <∞.We can now state the main result of this paper.


ARTICLE IN PRESSH. Heck et al. / Nonlinear Analysis ( ) – 3

Theorem 1.1. Let n ≥ 2 and l > 0 be given. Then there is a positive small number ε such that for any a ∈ Lnσ with ‖a‖n ≤ ε,the perturbation problem (2) has a unique solution (u, p) with the strong regularity

u ∈ C([0,∞); Lnσ ) ∩ C((0,∞);H1,n0,σ ∩ H

2,n),

ut ∈ C((0,∞); Lnσ ), p ∈ C((0,∞);H1,n) and∫p dx = 0.

Furthermore there are positive constants δ and C such that

‖u(t)‖n + t12 ‖∇u(t)‖n ≤ Ce−δt‖a‖n for all t > 0.

Here the constants ε, δ and C depend only on n, ν and l.

The exponential stability in Lnσ of the plane Couette flow v0 follows immediately from Theorem 1.1. One striking featureof our result should be emphasized: the stability of v0 is proved for any viscosity constant ν. The case of large viscosity wasalready verified by Abe and Shibata [7] observing that−L is then a small perturbation of the Stokes operator in Lnσ (Ω). Butsuch a perturbation argument is not applicable to the general case. Instead, we will make crucial use of Romanov’s result toprove (3) for the perturbed Stokes operator−L = −Lq in Lqσ with the domain D(−Lq) = H

1,q0,σ ∩ H

2,q. A rigorous proof ofa similar result remains mathematically challenging if the domain of −L is given by H1,q0,σ (Ω) ∩ H

2,q(Ω). However in ourperiodic setting, (3) can be deduced from the result in [1] because the spectrum of−Lq consists only of isolated eigenvaluesand has no accumulation points except infinity, which follows from the compact embedding of H1,q0,σ into L

qσ .

A detailed proof of Theorem 1.1 will be provided in the remaining sections. In Section 2, we consider the Stokes operator−Aq which is an unbounded operator in Lqσ with the domain H

1,q0,σ ∩ H

2,q. We prove the Helmholtz decomposition of Lqand derive some resolvent estimates which implies in particular that −Aq generates an analytic C0-semigroup in Lqσ . Thenan analogue of (3) for the perturbed Stokes operator −Lq is proved by making crucial use of Romanov’s estimate. TheLq−Lr -estimates for the semigroup e−tLqt≥0 are also derived. These estimates enable us to prove Theorem 1.1 in Section 3by applying the Banach fixed point theorem. The final section, Section 4, is devoted to proving two lemmas necessary forcomplete proofs of the Helmholtz decomposition and resolvent estimates in Section 2.

2. The operators −Aq and −Lq

In this section, we study the Stokes operator −Aq and its perturbation −Lq that are unbounded operators in Lqσ withdomain H1,q0,σ ∩ H

2,q.

2.1. The Stokes operator−Aq

To define the Stokes operator in Lqσ , we need to introduce the Helmholtz projection from Lq onto Lqσ . The Helmholtz

projection in the infinite layerΩ has been studied by Miyakawa [10], Farwig [11] and Abels andWiegner [12]. Applying thepartial Fourier transform, they have shown the existence and Lq-boundedness of the Helmholtz projection from Lq(Ω) ontoLqσ (Ω). Adapting their arguments with the partial Fourier series applied instead, we can prove the following result.

Proposition 2.1. Let 1 < q <∞. Then for each vector field u ∈ Lq, there exists a unique vector field v ∈ Lqσ such that

u = v +∇p for some scalar p ∈ H1,q.

Moreover we have

‖v‖q + ‖∇p‖q ≤ C‖u‖q

for some constant C = C(n, q) > 0. Hence the mapping

u ∈ Lq 7→ v = Pqu ∈ Lqσ

defines a bounded linear operator Pq (called the Helmholtz projection) from Lq onto Lqσ .

For 1 < q <∞, we define the Stokes operator −Aq in Lqσ by

Aqu = Pq(−ν∆u) for u ∈ D(Aq) = H1,q0,σ ∩ H

2,q.

Note that−Aq is an unbounded operator in Lqσ with a dense domain. The Stokes operator in Lqσ (Ω) has been studied by Abe

and Shibata [8,7] and Abels and Wiegner [12], independently, applying the partial Fourier transform. Their arguments canbe adapted to prove the following resolvent result for our Stokes operator−Aq.



Proposition 2.2. Let 1 < q <∞, 0 < ε < π2 and λ ∈ Σπ−ε . Then for each f ∈ L

qσ , there exists a unique solution u ∈ D(Aq) of

the resolvent equation

(λ+ Aq)u = f . (5)

Moreover we have

|λ|‖u‖q + ν‖u‖2,q ≤ Cε‖f ‖q (6)

for some constant Cε = C(ε, n, q) > 0.

HereΣδ denotes the sector of angle δ > 0, that is,Σδ = z ∈ C \ 0 : |arg z| < δ. Proposition 2.2 implies in particularthat the Stokes operator −Aq generates an analytic C0-semigroup e−tAqt≥0 in Lqσ and the set C \ (−∞, 0] is contained inthe resolvent set ρ(−Aq) of−Aq. Moreover, since the estimate (6) is uniform in λ ∈ (0,∞), a standard argument allows usto deduce from Proposition 2.2 that

0 ∈ ρ(−Aq) and ν‖(−Aq)−1f ‖2,q ≤ C‖f ‖q

for some constant C = C(n, q) > 0.Propositions 2.1 and 2.2 will be proved in Section 4 from solvability results in Lq for the Poisson equation and the

Stokes resolvent equations complemented with somemixed boundary conditions–periodic condition in x′ and Neumann orDirichlet condition in xn. Although these kinds of mixed boundary value problems have not been considered in the Lq settingyet (to the best of our knowledge), the corresponding problems in the infinite layer have been solved byMiyakawa [10], Abeand Shibata [8,7] and Abels and Wiegner [12] making use of the partial Fourier transform and Fourier multiplier theorems.Hence the partial Fourier series should be useful to prove the solvability in Lq for our mixed boundary value problems.In Section 4, we show that the approach of Abels and Wiegner [12] can be refined to give reasonably simple proofs ofPropositions 2.1 and 2.2.We finish this subsection with some interpolation inequalities. First of all, from the one-dimensional Poincaré inequality∫ 1

−1|ϕ|q dxn ≤

∫ 1

−1|∂xnϕ|

q dxn for all ϕ ∈ H1,q0 (−1, 1), (7)

we deduce that

2−1‖u‖q1,q ≤ ‖∇u‖qq ≡

∑|α|=1

‖Dαu‖qq ≤ ‖u‖q1,q (8)

for all u ∈ H1,q0,σ . Moreover using the elementary interpolation inequality (see [13, Example III.2.2] for instance)∫ l

−l|∂x1ϕ|

q dx1 ≤ η∫ l

−l|∂2x1ϕ|

q dx1 +9η

∫ l

−l|ϕ|q dx1

for any η > 0, we have

‖∇u‖qq ≤ η‖∆u‖qq +

9η‖u‖qq for all u ∈ H1,q0,σ ∩ H

2,q. (9)

Combining (8) and (9), we also obtain

C−1q ‖u‖q2,q ≤ ‖∇

2u‖qq ≡∑|α|=2

‖Dαu‖qq ≤ ‖u‖q2,q (10)

for all u ∈ H1,q0,σ ∩ H2,q.

2.2. The perturbed Stokes operator−Lq

For 1 < q <∞, we define unbounded operators Bq andLq in Lqσ by

Bqu = Pq ((v0 · ∇)u+ (u · ∇)v0) for u ∈ D(Bq) = H1,q0,σ

and

Lqu =(Aq + Bq

)u for u ∈ D(Lq) = D(Aq),

where v0 = (xn, 0, . . . , 0) is the plane Couette flow. We will show that−Lq generates an analytic semigroup e−tLqt≥0 onLqσ , which is uniformly exponentially stable. To show this, we first recall fromProposition 2.2 that the operator−Aq generates



a C0-semigroup e−tAqt≥0 on Lqσ , which is analytic and bounded in every sector Σ π2 −ε, 0 < ε < π

2 . On the other hand, byvirtue of the interpolation inequality (9) and Proposition 2.1, we have

‖ − Bqg‖q ≤ C‖g‖1,q ≤ η‖g‖2,q +Cη‖g‖q (11)

for all g ∈ D(Aq) and η > 0, which implies that −Bq is an (−Aq)-bounded operator with the (−Aq)-bound equal tozero. Hence it follows from a standard perturbation result (see [13] for instance) that −Lq is the generator of an analyticsemigroup e−tLqt≥0 on Lqσ and for each ε ∈ (0, π) there is a constant r > 0 such that the setΣπ−ε ∩ λ ∈ C : |λ| > r iscontained in the resolvent set ρ(−Lq) of−Lq. Let λ be any point in ρ(−Lq). Then λ+Lq : D(Lq)→ Lqσ is bijective. Notethat D(Lq) = H

1,q0,σ ∩ H

2,q and λ+ Lq is obviously bounded on H1,q0,σ ∩ H

2,q. Hence by virtue of the open mapping theorem,there is a constant C > 0 such that

‖(λ+Lq)−1f ‖2,q ≤ C‖f ‖q for all f ∈ Lqσ (12)

and so (λ+ Lq)−1 is a compact operator on Lqσ because the embedding H

1,q0,σ ∩ H

2,q → Lqσ is compact. Therefore, it followsfrom a classical result on operators with compact resolvents (see [14, Section III.6] for instance) that the spectrum σ(−Lq)of −Lq consists entirely of isolated eigenvalues with finite multiplicities and has no accumulation points except infinity.Furthermore we can show that σ(−Lq) lies in the left half plane by making crucial use of Romanov’s result in [1] on theeigenvalues of the Orr–Sommerfeld problem.

Lemma 2.3. There exists a positive constant δ = δ(n, l, ν) such that

Reλ ≤ −δ for all λ ∈ σ(−Lq).

Proof. Recall that for a fixed ε ∈ (0, π2 ) there is a constant r > 0 such that

Σπ−ε ∩ λ ∈ C : |λ| ≥ r ⊂ ρ(−L2).

But σ(−L2) has no accumulation points in λ ∈ C : |λ| ≤ r. Hence to prove the lemma, it suffices to show that ifλ ∈ σ(−Lq), then λ ∈ σ(−L2) and Reλ < 0.To show this, suppose that λ ∈ σ(−Lq). Then since λ is an eigenvalue of −Lq, there exists a nontrivial u ∈ D(Lq) =

H1,q0,σ ∩ H2,q such that

(λ+Lq)u = Pq (λu−∆u+ (v0 · ∇)u+ (u · ∇)v0) = 0.

Then it follows from Proposition 2.1 that

λu−∆u+ (v0 · ∇)u+ (u · ∇)v0 = −∇p inΩ

for some p ∈ H1,q. Hence the pair (u, p) is a nontrivial strong solution of the following resolvent problemλu− ν∆u+ (v0 · ∇)u+ (u · ∇)v0 +∇p = 0 inΩdiv u = 0 inΩu = 0 on ∂Ω.

(13)

By virtue of a standard regularity theory on the Stokes equations (see [15] for instance), we deduce that (u, p) ∈ C∞(Ω)and in particular λ ∈ σ(−L2). The equations in (13) can be thus rewritten componentwise as

λu1 − ν∆u1 + y∂x1u1+ w + ∂x1p = 0

λuj − ν∆uj + y∂x1uj+ ∂xjp = 0 (2 ≤ j ≤ n− 1)

λw − ν∆w + y∂x1w + ∂xnp = 0∂x1u

1+ · · · + ∂xn−1u

n−1+ ∂yw = 0

(14)

inΩ , where u = (u′, un) = (u1, . . . , un−1, un) andw = un. Using the Fourier series, we have

u(x′, xn) =∑k∈Zn−1

uk(xn)eiω〈k,x′〉

and

p(x′, xn) =∑k∈Zn−1

pk(xn)eiω〈k,x′〉.



Note that all the Fourier coefficients uk, pk are smooth on [−1, 1]. To derive equations for these coefficients, we multiply(14) by e−iω〈k,x

′〉 and integrate over T. Then since u and p are T-periodic in x′, we can use the integration by parts to deduce

that λu1k − ν

(∂2y − ω

2|k|2

)u1k + iωk1y u

1k + wk + iωk1pk = 0

λujk − ν(∂2y − ω

2|k|2

)ujk + iωk1y u

jk + iωkjpk = 0 (2 ≤ j ≤ n− 1)

λwk − ν(∂2y − ω

2|k|2

)wk + iωk1y wk + ∂ypk = 0

iωk1u1k + · · · + iωkn−1un−1k + ∂ywk = 0

(15)

for−1 ≤ y ≤ 1 and k ∈ Zn−1. Moreover, since u(x′,±1) = 0 for x′ ∈ Rn−1, it follows that

u1k(±1) = · · · = un−1k (±1) = wk(±1) = 0 for all k ∈ Zn−1. (16)

Recall that u was assumed to be nontrivial inΩ . Hence there is a multi-index k ∈ Zn−1 such that uk 6≡ 0 in (−1, 1). Fixingthis k and dropping the subscript k in (15) and (16) for simplicity, we deduce that

λu1 − ν(∂2y − ω

2|k|2

)u1 + iωk1y u1 + w + iωk1p = 0

λuj − ν(∂2y − ω

2|k|2

)uj + iωk1y uj + iωkjp = 0 (2 ≤ j ≤ n− 1)

λw − ν(∂2y − ω

2|k|2

)w + iωk1y w + ∂yp = 0

iωk1u1 + · · · + iωkn−1un−1 + ∂yw = 0

(17)

for−1 ≤ y ≤ 1 and

u1(±1) = · · · = un−1(±1) = w(±1) = 0. (18)

Suppose now that

k1u1 + · · · + kn−1un−1 = 0 identically in (−1, 1). (19)

Then from (17) and (18), we readily deduce that

w ≡ 0 in (−1, 1)

and

λuj − ν(∂2y − ω

2|k|2

)uj + iωk1y uj + iωkjp = 0 (1 ≤ j ≤ n− 1). (20)

Hence multiplying (20) by uj, integrating over (−1, 1) and summing up, we haven−1∑j=1

[(λ+ νω2|k|2

) ∫ 1

−1|uj|2 dy+ ν

∫ 1

−1|∂yuj|2 dy+ iωk1

∫ 1

−1y|uj|2 dy

]= 0

and taking the real part, we also haven−1∑j=1

[(Re λ+ νω2|k|2

) ∫ 1

−1|uj|2 dy+ ν

∫ 1

−1|∂yuj|2 dy

]= 0.

Therefore, in view of the Poincaré inequality (7), we conclude that

Re λ ≤ −ν.

Next we consider the case when (19) doesn’t hold. In this case, it follows from (17) and (18) that

w 6≡ 0 in (−1, 1) and w(±1) = ∂yw(±1) = 0.

Moreover, deleting p in (17) by a simple algebra, we derive that

λ(∂2y − ω

2|k|2

)w − ν

(∂2y − ω

2|k|2

)2w + iωk1y

(∂2y − ω

2|k|2

)w = 0

for−1 < y < 1. Hence λ is an eigenvalue of the Orr–Sommerfeld boundary value problemν(∂2y − ω

2|k|2

)2ϕ = (iωk1y+ λ)

(∂2y − ω

2|k|2

)ϕ − 1 < y < 1,

ϕ(±1) = ∂yϕ(±1) = 0.

Therefore, in view of Romanov’s result in [1], we conclude that

Re λ < 0.

This completes the proof of Lemma 2.3.



As a consequence of Lemma 2.3, we obtain the following Lq− Lr -estimates, which can be proved by a standard argumentbased on the analyticity of e−tLqt≥0 and Gagliardo–Nirenberg interpolation inequalities.

Lemma 2.4. Suppose that 1 < q ≤ r <∞. Then

‖Dαe−tLqa‖r ≤ Cq,r t−n2

(1q−

1r

)−|α|2 e−δt‖a‖q

for all a ∈ Lqσ , |α| ≤ 1 and t ∈ (0,∞). Here δ is a positive constant depending only on n, l and ν while Cq,r depends on q, r aswell.

3. The proof of Theorem 1.1

Theorem 1.1 can be deduced from Lemma 2.4 by several standard methods, for instance Kato’s iteration method [16].But for the sake of completeness, we provide a proof based on the Banach fixed point theorem for contraction mappings.See [7,9,1,17] for other approaches.Let us consider the nonlinear abstract Cauchy problem

u′ +Lnu = F(u), t > 0u(0) = a, (21)

where F(u) = Pn (−(u · ∇)u) and a ∈ Lnσ . We will prove the existence of a mild solution to (21) in the Banach space X of allvector fields v ∈ C([0,∞); Lnσ ) such that

t1−n2q∇v ∈ C([0,∞); Lq), lim

t→0t1−

n2q ‖∇v(t)‖q = 0 for n ≤ q ≤

32n

and

‖v‖X = sup0≤t<∞

eδt(‖v(t)‖n + t

12 ‖∇v(t)‖n + t

23 ‖∇v(t)‖ 3

2 n

)<∞,

where δ > 0 is the constant in Lemma 2.4. It follows easily from Lemma 2.4 that if v(t) = e−tLna for t ≥ 0, then v ∈ X and‖v‖X ≤ C‖a‖n.To solve (21) in X , we consider the operator G on X defined by

v ∈ X 7→ G(v); G(v)(t) =∫ t

0e−(t−s)LnF(v)(s) ds (t ≥ 0).

Lemma 3.1. The operator G maps X into X continuously. Moreover, there exists a constant C∗ > 0 such that

‖G(v1)− G(v2)‖X ≤ C∗ (‖v1‖X + ‖v2‖X ) ‖v1 − v2‖X (22)

for all v1, v2 ∈ X.

Proof. Let v ∈ X be fixed. We will show that G(v) ∈ X . First, using the Gagliardo and Nirenberg inequality, we deduce thatif n ≤ r <∞, then

‖v(t)‖r ≤ Cr‖v(t)‖nrn ‖∇v(t)‖

1− nrn ≤ Cre−δt t

12 (nr −1)‖v‖X

for t > 0. Hence for each qwith 35n ≤ q <32n, we have

‖F(v)(t)‖q = ‖Pn(−(v(t) · ∇)v(t))‖q≤ ‖v(t)‖ 3nq

3n−2q‖∇v(t)‖ 3

2 n

≤

(Ce−δt t

n2q−

32 ‖v‖X

) (t23 ‖∇v(t)‖ 3

2 n

)(23)

for t > 0. Combining (23) with Lemma 2.4, we can easily estimate ‖G(v)(t)‖n: for any fixed qwith 35n < q < n, we have

‖G(v)(t)‖n ≤∫ t

0‖e−(t−s)LqF(v)(s)‖n ds

≤

∫ t

0Ce−δ(t−s)(t − s)−

12

(nq−1

)‖F(v)(s)‖q ds

≤ Ce−δt(∫ t

0(t − s)−

12

(nq−1

)s−12

(3− nq

)ds)‖v‖X sup

0<s<ts23 ‖∇v(s)‖ 3

2 n

≤ Ce−δt(sup0<s<t

s23 ‖∇v(s)‖ 3

2 n

)‖v‖X



for each t > 0. Similar calculations yield

‖∇G(v)(t)‖n ≤ Ce−δt t−12

(sup0<s<t

s23 ‖∇v(s)‖ 3

2 n

)|v||X

and

‖∇G(v)(t)‖ 32 n≤ Ce−δt t−

23

(sup0<s<t

s23 ‖∇v(s)‖ 3

2 n

)‖v‖X

for t > 0. These a priori estimates enable us to conclude that

G(v)(t) ∈ Lnσ ∩ H1,n∩ H1,

32 n for each t > 0, ‖G(v)‖X ≤ C‖v‖2X (24)

and

limt→0‖G(v)(t)‖n = lim

t→0t12 ‖∇G(v)(t)‖n = lim

t→0t23 ‖∇G(v)(t)‖ 3

2 n= 0. (25)

Hence to show that G(v) ∈ X , it suffices to show that

G(v) ∈ C((0,∞); Lnσ ∩ H1,n∩ H1,

32 n).

Suppose that 0 < t0 <∞. Then for each t > t0, we write

G(v)(t) =∫ t0

0g(s, t) ds+ R(t),

where

g(s, t) = e−(t−s)LnF(v)(s) and R(t) =∫ t

t0g(s, t) ds.

Adapting the proof of (25), we can show that

R(t)→ 0 in Lnσ ∩ H1,n∩ H1,

32 n as t0 < t → t0;

for instance, the convergence of ∇R(t) in L32 n follows from

‖∇R(t)‖ 32 n≤ C

∫ t

t0(t − s)−

56 ‖F(v)(s)‖ 3

4 nds ≤ Ct

−56

0 (t − t0)16 ‖v‖2X .

On the other hand, it follows from the analyticity of the semi-group e−tLnt≥0 that

g(s, t)→ g(s, t0) in Lnσ ∩ H2,n as t0 < t → t0

for each s ∈ (0, t0). Hence the dominated convergence theorem allows us to show that∫ t0

0g(s, t) ds→

∫ t0

0g(s, t0) ds = G(v)(t0) in Lnσ ∩ H

2,n

as t0 < t → t0. This implies that G(v) is right-continuous in (0,∞). The left-continuity of G(v) in (0,∞) can be provedsimilarly by writing

G(v)(t) =∫ t0

0g(s, t) ds,

wheret02< t < t0 and g(s, t) =

0, t ≤ s < t0g(s, t), 0 < s < t.

We have shown that Gmaps X into X .Next, let v1, v2 ∈ X be given. Then since

F(v1)− F(v2) = −Pn (((v1 − v2) · ∇)v1 + (v2 · ∇)(v1 − v2)) ,

we can adapt the proof of (23) to show that

‖F(v1)(t)− F(v2)(t)‖q ≤ Ce−δt tn2q−

32 (‖v1‖X + ‖v2‖X ) ‖v1 − v2‖X

for 35n ≤ q <32n and t > 0. Hence using the same argument as in the derivation of (24), we prove (22). This completes the

proof of Lemma 3.1.



To complete the proof of Theorem 1.1, let us define the operator T on X by

v ∈ X 7→ T (v); T (v)(t) = e−tLna+ G(v)(t) (t ≥ 0).

Then in view of Lemma 3.1, there is a constant C∗ > 1 such that

‖T (v)‖X ≤ C∗(‖a‖n + ‖v‖2X

)for v ∈ X and

‖T (v1)− T (v2)‖X ≤ C∗ (‖v1‖X + ‖v2‖X ) ‖v1 − v2‖Xfor all v1, v2 ∈ X . Suppose now that

‖a‖n ≤ ε ≡18C2∗

.

Then we can show that T has a fixed point in X . In fact, if a = 0, then v = 0 is obviously a fixed point of T in X . Otherwiseit is easy to show that T is a contraction on the closed ball in X of radius 2C∗‖a‖n centered at 0. Hence T has a fixed point vin X such that ‖T (v)‖X ≤ 2C∗‖a‖n. This proves the existence of a mild solution u to (21) in X satisfying the desired stabilityestimate. The strong regularity of u and its associated pressure p can be proved by following the same arguments as in [18,9].Finally it is easy to prove the uniqueness of solutions to (21) with the regularity stated in Theorem 1.1. This completes theproof of Theorem 1.1.

4. Proofs of Propositions 2.1 and 2.2

Our proof of Proposition 2.1 is based on the following result on the decomposition of vector fields in D0 = w ∈ D :w|xn=±1 = 0 in T.

Lemma 4.1. Let 1 < q <∞. Then for each vector field u ∈ D0, there exist a vector field v ∈ D0,σ and a scalar function p ∈ Dsuch that

u = v +∇p and ‖v‖q + ‖∇p‖q ≤ C‖u‖q

for some constant C = C(n, q) > 0.

Proof of Proposition 2.1 from Lemma 4.1. We first prove the uniqueness part of Proposition 2.1 by a simple dualityargument. Suppose that 0 = v + ∇p for some v ∈ Lqσ and p ∈ H

1,q. To show that v = −∇p = 0, we have only toshow that∫

v · u dx = 0 for all u ∈ Lq′

. (26)

Let u be any vector field inD0. Then by virtue of Lemma 4.1, there existw ∈ D0,σ and ϕ ∈ D such that u = w +∇ϕ. Thensince v = −∇p ∈ Lqσ ,w ∈ D0,σ and Lqσ is the closure ofD0,σ in L

q, it follows that∫v · u dx = −

∫∇p · w dx+

∫v · ∇ϕ dx = 0.

On the other hand, it is easy to show that D0 is dense in Lq′

. This proves (26) and thus the proof of the uniqueness part iscompleted.The above argument also shows that the decomposition in Lemma 4.1 is unique. Hence, recalling again thatD0 is dense

in Lq, we can easily deduce the existence part and Lq-estimate in Proposition 2.1 from Lemma 4.1 by a standard densityargument. We have completed the proof of Proposition 2.1.

Similarly, Proposition 2.2 can be deduced from the following result.

Lemma 4.2. Let 1 < q < ∞, 0 < ε < π2 and λ ∈ Σπ−ε . Then for each f ∈ D0,σ , there exist a vector field u ∈ D0,σ and a

scalar field p ∈ D such that

(λ− ν∆)u+∇p = f inΩ (27)

and

|λ|‖u‖q + ν‖∇2u‖q ≤ Cε‖f ‖q (28)


Hence to complete the proofs of Propositions 2.1 and 2.2, it remains to prove Lemmas 4.1 and 4.2, which is the main taskof this final section. We will give reasonably simple proofs of the two lemmas, refining the approach of Abels and Wiegnerin [12].



4.1. Fourier multiplier theorems

The Lq estimates for the Helmholtz projection and the Stokes operator will be derived from a Fourier multiplier theoremof Mikhlin–Lizorkin type. To state the theorem precisely, we first introduce the definition of Fourier multipliers in a discretesetting.

Definition 4.3. A sequence a = (ak)k∈Zd in the complex field C is said to be a Fourier Multiplier on Lq((−π, π)d) if there is a

constant C > 0 such that∣∣∣∣∣∣∣∣∣∣∑k∈Zdakckei〈k,·〉

∣∣∣∣∣∣∣∣∣∣Lq((−π,π)d)

≤ C

∣∣∣∣∣∣∣∣∣∣∑k∈Zdckei〈k,·〉

∣∣∣∣∣∣∣∣∣∣Lq((−π,π)d)

, (29)

for any sequence c = (ck)k∈Zd in Cwith ck 6= 0 for only finitely many k ∈ Zd.

Let a = (ak)k∈Zd be a Fourier multiplier on Lq((−π, π)d). Then the mapping∑

k∈Zdckei〈k,·〉 7→

∑k∈Zdakckei〈k,·〉

extends uniquely to a bounded operator Ta on Lq((−π, π)d) with the norm ‖Ta‖ being the greatest lower bound of the setof all constants C such that (29) holds for any finite sequence c = (ck)k∈Zd in C. Moreover, an obvious change of variablesenables us to show that a = (ak)k∈Zd is in fact a Fourier multiplier on L

q((−l1, l1)× · · · × (−ld, ld)) for every li > 0 and itscorresponding operator T (l1,···,ld)a has the same norm as Ta.A classical result due to Marcinkiewicz [19] gives a sufficient condition for a sequence a to be a Fourier multiplier on

Lq((−π, π)d) for every q ∈ (1,∞). To state the Marcinkiewicz multiplier theorem, we need to introduce some notations.

(1) For any α, β ∈ Zd with α ≤ β (componentwise), the interval [α, β] in Zd is defined by

[α, β] = k ∈ Zd : α ≤ k ≤ β.

(2) For each ν = (ν1, . . . , νd) =∑dj=1 νjej ∈ Nd0, the difference operator∆

ν on a sequence a = (ak)k∈Zd is defined by

(∆νjej a)k =ak − ak−νjej , νj 6= 0ak, νj = 0

(k ∈ Zd)

and

∆νa = ∆ν1e1 · · ·∆νded a.

Then we define the variation of a = (ak)k∈Zd on an interval [α, β] in Zd by

var[α,β]

a =∑α≤k≤β

|(∆ν(k)a)k|,

where

ν(k) = (ν1(k), . . . , νd(k)) and νj(k) =1, kj 6= αj,0, kj = αj.

(3) We define the dyadic decomposition Ijj∈N0 of Z by

I0 = 0 and Ij = m ∈ Z : 2j−1 ≤ |m| ≤ 2j for j ∈ N

and extend it naturally to the dyadic decomposition Dνν∈Nd0 of Zd by defining

Dν = Iν1 × · · · × Iνd for ν = (ν1, . . . , νd) ∈ Nd0.

Note that each Dν can be written uniquely as the disjoint union of at most 2d dyadic intervals

Dν =2s⋃i=1

[αi, β i],

where s = s(ν) ∈ [0, d] is the number of nonzero components of ν ∈ Nd0. Hence the variation of a sequence a = (ak)k∈Zdon such Dν is defined by

varDνa =

2s∑i=1

var[αi,β i]

a.



The Marcinkiewicz multiplier theorem reads now as follows.

Proposition 4.4. Let (ak)k∈Zd be a sequence in C. Suppose that

supν∈Nd0

varDνa <∞.

Then for any q ∈ (1,∞), the sequence a = (ak)k∈Zd is a Fourier multiplier on Lq((−π, π)d) and there is a constant

C = C(q, d) > 0 such that

‖Ta‖Lq→Lq ≤ C supν∈Nd0

varDνa.

For a proof of Proposition 4.4, see [19], [20] or [21].From Proposition 4.4, we can deduce the following multiplier theorem.

Proposition 4.5. Let a = (ak)k∈Zd be a sequence in C such that

ak = m(k) (k ∈ Zd \ 0)

for some m ∈ Cd(Rd \ 0). Suppose that

[m] := supγ∈0,1d

supξ 6=0|ξ γDγm(ξ)| <∞.

Then for any q ∈ (1,∞), the sequence a = (ak)k∈Zd is a Fourier multiplier on Lq((−π, π)d) and there is a constant

C = C(q, d) > 0 such that

‖Ta‖Lq→Lq ≤ C max [m], |a0| .

Remark 4.6. The norm [·] of continuous multipliers is invariant under the scalings

m(ξ , . . . , ξd) 7→ m(r1ξ, . . . , rdξd)

for all (r1, . . . , rd) ∈ Rd.

Proof of Proposition 4.5. By virtue of Proposition 4.4, we have only to show that

var[α,β]

a ≤ 2dmax( [m], |a0|) (30)

for any dyadic interval [α, β]. To show this, we first observe that

var[α,β]

a =∑

γ∈0,1d

∑k:ν(k)=γ

|(∆γ a)k|. (31)

It is obvious that

|(∆0a)k| = |ak| ≤ max[m], |a0| for all k ∈ Zd. (32)

In particular this proves (30) in cases when α = β .Assume now that α < β . Let γ ∈ 0, 1d with γ 6= 0. Then applying the mean value theorem to m, we deduce that if

k ∈ Zd and ν(k) = γ , then

(∆γ a)k = Dγm(k)

for some k ∈ Rd \ 0 such that kj = kj if γj = 0 and kj ∈ (kj − 1, kj) if γj = 1. Moreover, since there are exactly (β − α)γdifferent multi-indices kwith ν(k) = γ , it follows that∑

k:ν(k)=γ

|(∆γ a)k| ≤ (β − α)γ supα≤ξ≤β

|Dγm(ξ)|. (33)

On the other hand, it is easy to show that if [α, β] is a dyadic interval and α ≤ ξ ≤ β , then

βj − αj ≤ |ξj| for all j.

Combining this fact with (31)–(33), we derive (30). This completes the proof of Proposition 4.5.

Finally, we provide three lemmas which are convenient technical tools to verify the hypothesis of Proposition 4.5. Proofsof the first two lemmas are quite elementary; see the paper [12] by Abels and Wiegner for instance.



Lemma 4.7. Let m,m1,m2 ∈ Cd(Rd \ 0).

(a) [m1m2] ≤ C[m1][m2].(b) If |m(ξ)| ≥ δ > 0 for ξ 6= 0, then [m−1] ≤ Cδ(1+ [m])d.(c) If m(ξ) = m(|ξ |) for some m ∈ Cd((0,∞)), then

[m] ≤ C sup0<r<∞

d∑i=0

r i|m(i)(r)|.

Lemma 4.8. Suppose that f , g ∈ Cd((0,∞)) satisfy

Re f (r) ≤ −CA r, sup1≤i≤d

r i−1|f (i)(r)| ≤ CA (r ≥ 0)

and

sup0≤i≤d

r i|g(i)(r)| ≤ Crp (r ≥ 0)

for some constants C, A > 0 and p ∈ N0. Then there exists a constant C > 0 independent of A such that

[ g(|ξ |) exp(f (|ξ |)) ] ≤ CA−p.

Lemma 4.9. Let 0 < θ < π and λ ∈ Σθ . For each r ≥ 0, we define µ = µ(r) by the unique complex number µ ∈ Σθ/2 suchthat µ2 = λ+ r2. Then there exists a constant Cθ > 0 such that[

|ξ |

µ(|ξ |)

]+

[µ(|ξ |)

|ξ | + µ(|ξ |)

]≤ Cθ , (34)[

|ξ |p exp(−A|ξ | − Bµ(|ξ |))]≤ Cθ (A+ B)−p (35)

and

[µ(|ξ |) exp(−Aµ(|ξ |))] ≤ CθA−1 (36)

for A, B ≥ 0 and p ∈ N0.

Proof. By a simple computation, we first derive

Reµ(r) ≥ |µ(r)| cos(θ

2

)and |µ(r)| ≥ max

|λ|

12 , r

sin

12 θ. (37)

On the other hand, it is easy to show that µ′(r) = rµ(r) and for each i ∈ N, there is a polynomial Pi such that µ(i)(r) =

r1−iPi( rµ(r) ). Hence it follows from (37) and result (c) in Lemma 4.7 that[|ξ |

µ(|ξ |)

]=[µ′(|ξ |)

]≤ C sup

0≤r<∞

d∑i=0

r i|µ(i+1)(r)| ≤ Cθ . (38)

Moreover, observing that

µ(|ξ |)

|ξ | + µ(|ξ |)=

(1+

|ξ |

µ(|ξ |)

)−1and Re

(1+

|ξ |

µ(|ξ |)

)≥ 1 > 0,

we deduce from (38) and result (b) in Lemma 4.7 that[µ(|ξ |)

|ξ | + µ(|ξ |)

]≤ C

(2+

[|ξ |

µ(|ξ |)

])d≤ Cθ .

This proves (34). (35) follows immediately from Lemma 4.8 because if we take g(r) = rp and f (r) = −Ar − Bµ(r), then thehypotheses of Lemma 4.8 are satisfied due to (37) and (38). Finally, to prove (36), we observe that

|exp(−Aµ(r))| = exp(−AReµ(r)) < (AReµ(r))−1 ≤ Cθ (A|µ(r)|)−1 ;

the last inequality follows from (37). Using this observation together with (37) and (38), we can easily show that if we define

m(r) = µ(r) exp(−Aµ(r)) and m(ξ) = m(|ξ |),



then

[m] ≤ C sup0<r<∞

d∑j=0

r j|m(j)(r)| ≤ CθA−1.

This completes the proof of Lemma 4.9.

Remark 4.10. It follows from Lemmas 4.7 and 4.9 that[ξj

|ξ |

]<∞ and

[ξj

µ(|ξ |)

]≤

[|ξ |

µ(|ξ |)

] [ξj

|ξ |

]≤ Cθ

for j = 1, . . . , d. In particular, the Riesz multipliers

rj = (rj)kk∈Zd; (rj)k6=0 =kj|k|

are Fourier multipliers on Lq((−π, π)d) for every q ∈ (1,∞).

4.2. Resolvent estimates for the Laplace operator

To prove Lemmas 4.1 and 4.2, we will follow the approach of Abels and Wiegner in [12]. Hence we first consider theresolvent problem for the Laplacian with periodic-Dirichlet boundary condition:

(λ−∆)u = f in Rn−1 × (−1, 1), (39)

where λ ∈ C \ (−∞, 0), u ∈ D(−∆q) = H1,q0 ∩ H

2,q and f ∈ Lq. Here we denote by H1,q0 the closure of D0 = w ∈ D :

w|xn=±1 = 0 in Rn−1 in H1,q.

Proposition 4.11. Let 1 < q < ∞, 0 < ε < π2 and λ ∈ Σπ−ε ∪ 0. Then for any f ∈ L

q, there exists a unique solutionu ∈ D(−∆q) of the resolvent equation (39). Furthermore we have

|λ|‖u‖q + ‖u‖2,q ≤ Cε‖f ‖q


Proposition 4.11 can be easily deduced from the following existence result inD .

Lemma 4.12. Let 1 < q <∞, 0 < ε < π2 and λ ∈ Σπ−ε . Then for any f ∈ D , there exists a solution u ∈ D0 of (39) satisfying

the estimate

|λ|‖u‖q + ‖∇2u‖q ≤ Cε‖f ‖q


Proof. Let f ∈ D be given by

f (x′, xn) =∑k∈Zn−1

fk(xn) eiω〈k,x′〉

for some fk ∈ C∞([−1, 1])with fk 6= 0 for only finitely many k ∈ Zn−1. Then the function u ∈ D , given by

u(x′, xn) =∑k∈Zn−1

uk(xn) eiω〈k,x′〉 (40)

for some uk ∈ C∞([−1, 1]), is a solution inD0 of (39) if and only if each Fourier coefficient uk satisfies(µ2 − ∂2xn

)uk = fk − 1 < xn < 1

uk(±1) = 0,(41)

where µ = µ(|ωk|) is the unique µ ∈ Σ(π−ε)/2 such that µ2 = λ+ |ωk|2. Since µ2 ∈ Σπ−ε ⊂ C \ (−∞, 0], it follows fromthe theory of ordinary differential equations that the boundary value problem (41) has a unique solution uk ∈ C∞([−1, 1]),which can be represented by

uk(xn) =∫ 1

−1K(µ(|ωk|), xn, yn)fk(yn) dyn (42)



with the kernel K defined by

K(µ, xn, yn) =e−µ(2+xn+yn) + e−µ(2−xn−yn) − e−µ|xn−yn| − e−µ(4−|xn−yn|)

2µ(1− e−4µ). (43)

Hence, if we define u by (40), (42) and (43), then u is a (actually unique) solution inD0 of (39). Therefore, to complete theproof, it remains to derive the following Lq-estimate:

‖Dαu‖Lq(T×(−1,1)) ≤ Cε‖f ‖Lq(T×(−1,1)) (44)

for all α ∈ Nn0 with |α| = 2. The idea of the proof of (44) is to find nice decompositions of the kernel K and the solution u byutilizing an identity due to Abels and Wiegner [12]:

e−µ|xn−yn| + e−µ(4−|xn−yn|)

2µ(1− e−4µ)=e−µ(4−xn+yn) + e−µ(4+xn−yn)

2µ(1− e−4µ)+e−µ|xn−yn|

2µ.

Substituting this into (43), we obtain

K(µ, xn, yn) =5∑m=1

Km(µ, xn, yn),

where

K 1 =e−µ(2+xn+yn)

2µ(1− e−4µ), . . . , K 4 = −

e−µ(4+xn−yn)

2µ(1− e−4µ), K 5 = −

e−µ|xn−yn|

2µ.

Then to each kernel Km, we associate um ∈ D defined by

um(x′, xn) =∑k∈Zn−1

umk (xn) eiω〈k,x′〉

and

umk (xn) =∫ 1

−1Km(µ(|ωk|), xn, yn)fk(yn) dyn.

To prove (44), we have to show that

‖Dαum‖Lq(T×(−1,1)) ≤ Cε‖f ‖Lq(T×(−1,1)) (m = 1, . . . , 5) (45)

for all α = (α′, αn)with α′ ∈ Nn−10 , αn ∈ N0 and |α| = |α′| + αn = 2.We first consider the case that 1 ≤ m ≤ 4. Then since Km = Km(µ, xn, yn) is a smooth function of xn, the k-th Fourier

coefficient of Dαum is given by

(iωk)α′

∂αnxn umk (xn) =

∫ 1

−1(iωk)α

′

∂αnxn Km(µ(|ωk|), xn, yn)fk(yn) dyn.

Hence by virtue of Proposition 4.5, we have

‖Dαum(·, xn)‖Lq(T) ≤ C∫ 1

−1

[(ωξ)α

′

∂αnxn Km(µ(|ωξ |), xn, yn)

]‖f (·, yn)‖Lq(T) dyn (46)

for some C = C(n, q) > 0 independent of l. Making use of the results of Section 4.1, we estimate the kernel in (46). In fact,using Remark 4.6, Lemmas 4.7 and 4.9, we easily obtain[

(ωξ)α′

∂αnxn K1(µ(|ωξ |), xn, yn)

]=

[ξα′

µ(|ξ |)αne−µ(|ξ |)(2+xn+yn)

2µ(|ξ |)(1− e−4µ(|ξ |))

]

≤ C

[ξα′

µ(|ξ |)αn

µ(|ξ |)2

] [µ(|ξ |)e−µ(|ξ |)(2+xn+yn)

] [(1− e−4µ(|ξ |))−1

]≤ Cε

12+ xn + yn

,

where the last expression is essentially the kernel of the Hilbert transform. Therefore, by virtue of the Lq-boundedness ofthe Hilbert transform, we easily deduce (45) with m = 1. Exactly the same argument also proves (45) with m = 2, 3 or 4because the corresponding kernel is bounded from above by



Cε1

2− xn − yn, Cε

14− xn + yn

or Cε1

4+ xn − yn.

The remaining kernel K 5 is treated in a different way; instead of estimating its multiplier norm in Rn−1, we apply theFourier transform with respect to the n-th variable and show that its Fourier coefficients form a Fourier multiplier onLq(T× [−2, 2]). For this, let us denote by g the T× [−2, 2]-periodic extension to Rn of f such that

g(x′, xn) =f (x′, xn) for xn ∈ (−1, 1)0 for xn ∈ [−2, 2] \ (−1, 1)

and by F the T× [−2, 2]-periodic function on Rn whose partial Fourier series Fk, k ∈ Zn−1, is given by

Fk(xn) = (iωk)α′

∂αnxn

∫ 2

−2Km(µ, xn, yn)gk(yn) dyn for xn ∈ [−2, 2],

where µ = µ(|ωk|). Note that each Fk is actually the convolution of two [−2, 2]-periodic functions, that is,

Fk = ∂αnxn[((iωk)α

′

Γ

)∗ gk

]with Γ = Γ (µ, ·) being the [−2, 2]-periodic function in R such that Γ (µ, xn) = e−µ|xn |

2µ for xn ∈ [−2, 2]. Hence applyingthe Fourier transform with respect to xn, we get

F(k,kn) = (iωk)α′(iπ

2kn)αn

Γkn g(k,kn) for (k, kn) ∈ Zn.

On the other hand, by direct computations, we easily obtain

Γkn =

∫ 2

−2e−i

π2 knxn

e−µ|xn|

2µdxn =

1µ2 + (π2 kn)

2

(1− e−(2µ+iπkn)

)and [

(iωξ)α′ (iπ2 ξn

)αnµ(|ωξ |)2 + (π2 ξn)

2

(1− e−(2µ(|ωξ |)+iπξn)

)]Rn≤ Cε.

Therefore, noting that Dαu5 = F in T× (−1, 1), we deduce from Proposition 4.5 that

‖Dαu5‖Lq(T×(−1,1)) ≤ ‖F‖Lq(T×(−2,2))≤ Cε‖g‖Lq(T×(−2,2)) = Cε‖f ‖Lq(T×(−1,1)),

which completes the proof of (45) and thus (44). This completes the proof of Lemma 4.12.

Remark 4.13. It should be emphasized that the various constants Cε are independent ofω = πl due to the scaling-invariance

of the multiplier norm [·].

4.3. Proof of Lemma 4.1

To prove Lemma 4.1, it suffices to show that for each vector field u ∈ D0, there exists a scalar function p ∈ D such that

−∆p = div u inΩ, ∂xnp = 0 on ∂Ω (47)

and

‖∇p‖q ≤ C‖u‖q (48)

for some C = C(n, q) > 0. Let u ∈ D0 be given by

u(x′, xn) =∑k∈Zn−1

uk(xn) eiω〈k,x′〉

for some uk ∈ C∞0 (−1, 1) = w ∈ C∞([−1, 1]) : w(±1) = 0 with uk 6= 0 for only finitely many k ∈ Zn−1. Then a scalar

function p ∈ D is a solution to (47) if and only if each partial Fourier coefficient pk of p satisfies(µ2 − ∂2xn

)pk = iωk · u′k + ∂xn u

nk, −1 < xn < 1

∂xn pk(±1) = 0,



where µ = |ωk|. We can easily solve this boundary value problem to derive an explicit solution:

p0(xn) =∫ xn

−1un0(yn) dyn

and

pk(xn) =∫ 1

−1G(|ωk|, xn, yn)

(iωk · u′(k, yn)+ ∂yn u

nk(yn)

)dyn

for k 6= 0, where

G(µ, xn, yn) =e−µ(2+xn+yn) + e−µ(2−xn−yn) + e−µ|xn−yn| + e−µ(4−|xn−yn|)

2µ(1− e−4µ).

Let us now define the function p by

p(x′, xn) =∑k∈Zn−1

pk(xn) eiω〈k,x′〉.

Then p is obviously a solution inD to the Neumann problem (47). To prove (48), we first decompose G and p by using thesame notation as in the proof of Lemma 4.12 as follows:

G = K 1 + K 2 − K 3 − K 4 − K 5 =5∑m=1

Gm, p =5∑m=1

(pm,1 + pm,2

),

p1,10 = · · · = p5,10 = p

2,20 = · · · = p

4,20 = 0, p5,20 = p0,

pm,1k (xn) =∫ 1

−1Gm(|ωk|, xn, yn)iωk · u′k(yn) dyn

and

pm,2k (xn) =∫ 1

−1Gm(|ωk|, xn, yn)∂yn u

nk(yn) dyn

for k 6= 0. Let α = (α′, αn) be a fixed multi-index such that α′ ∈ Nn−10 , αn ∈ N0 and |α| = |α′| + αn = 1. Then for eachk ∈ Zn−1, the k-th Fourier coefficient of Dαpm,1 is given by

(iωk)α′

∂αnxn pm,1k (xn) = (±1)∂αnxn

∫ 1

−1(iωk)α

′

Km(|ωk|, xn, yn)iωk · u′k(yn) dyn.

Ifm = 5, then the proof of Lemma 4.12 can be easily adapted to deduce that

‖∇p5,1‖Lq(T×(−1,1)) ≤ Cε‖u‖Lq(T×(−1,1)).

However, if 1 ≤ m ≤ 4, then we cannot adapt the previous argument directly, since the function ξ 7→ 1 − e−4|ωξ | is nolonger bounded from below by a positive constant. To overcome this difficulty, we introduce a non-decreasing function µin C∞(R) such that µ(r) = 1

2 for all r ≤12 and µ(r) = r for all r ≥ 1. Noting that 1− e

−4µ(|ωξ |)≥ 1− e−2 > 0 for ξ 6= 0,

we deduce as in the proof of Lemma 4.12 that[(ωξ)α

′

(ωξj)∂αnxn K

1(|ωξ |, xn, yn)]=

[ξα′

ξj|ξ |αne−|ξ |(2+xn+yn)

2|ξ |(1− e−4µ(|ξ |))

]

≤ C

[ξα′

ξj|ξ |αn

|ξ |2

] [|ξ |e−µ(|ξ |)(2+xn+yn)

] [(1− e−4µ(|ξ |))−1

]≤ Cε

12+ xn + yn

,

which implies that

‖∇p1,1‖Lq(T×(−1,1)) ≤ Cε‖u‖Lq(T×(−1,1)). (49)

The casem = 2, 3 or 4 can be treated similarly. Therefore, to prove (48), it remains to show that

‖∇pm,2‖Lq(T×(−1,1)) ≤ Cε‖u‖Lq(T×(−1,1)) (50)



for 1 ≤ m ≤ 5. In the case 1 ≤ m ≤ 4, we use the integration by parts to obtain

pm,2k (xn) = −∫ 1

−1∂ynG

m(|ωk|, xn, yn)unk(yn) dyn

= (±1)∫ 1

−1|ωk|Km(|ωk|, xn, yn)unk(yn) dyn

because Gm is smooth and unk(±1) = 0. Hence the proof of (50) is exactly the same as that of (49). We finally consider thecasem = 5. Let v be the T× [−2, 2]-periodic extension to Rn of u such that

v(x′, xn) =u(x′, xn) for xn ∈ (−1, 1)0 for xn ∈ [−2, 2] \ (−1, 1).

Then since unk(±1) = 0, it follows that vnk ∈ W

1,2loc (R) and

p5,2k (xn) = −∫ 2

−2Γ (|ωk|, xn − yn)∂yn v

nk (yn) dyn

= −∂xn(Γ (|ωk|, ·) ∗ vnk

)(xn).

Hence the proof of Lemma 4.12 can be adapted again to prove (50) form = 5. This completes the proof.

4.4. Proof of Lemma 4.2

Let f ∈ D0,σ be fixed. First of all, by the simple scaling (λ, u, p) 7→ (ν−1λ, νu, p), we may assume without loss ofgenerality that ν = 1. We next observe that if (u, p) ∈ D0,σ ×D satisfies (27), then p is harmonic inΩ , that is,−∆p = 0inΩ because div u = div f = 0 inΩ .Let p ∈ D be a harmonic scalar field in Ω . Then by virtue of Proposition 4.11 and Lemma 4.12, there exists a unique

solution u = u(p) ∈ D0 of the vectorial Laplace resolvent equation

(λ−∆)u = f −∇p inΩ, (51)

which satisfies the following Lq-estimate

|λ|‖u‖q + ‖∇2u‖q ≤ Cε(‖f ‖q + ‖∇p‖q

).

Taking the divergence operator in (51), we deduce that U = div u satisfies

(λ−∆)U = 0 inΩ.

Moreover, since u ∈ D0, it follows that U ∈ D and U(x′,±1) = ∂xnun(x′,±1) for x′ ∈ Rn−1. Hence it follows from

Proposition 4.11 that

u ∈ D0,σ ⇔ U ∈ D0 ⇔ ∂xnun(x′,±1) = 0 for x′ ∈ Rn−1.

Therefore, to prove Lemma 4.2, we only have to find a function p ∈ D such that

−∆p = 0 inΩ, (52)‖∇p‖q ≤ Cε‖f ‖q (53)

and

∂xnun|xn=±1 = 0 for a unique solution u = u(p) ∈ D0 of (51). (54)

Such a scalar p can be determined uniquely up to additive constants by means of the partial Fourier transform. In fact, forthe Fourier coefficients for k ∈ Zn−1 we deduce from (51), (52) and (54) that(

µ2 − ∂2xn

)unk + ∂xn pk = f

nk in (−1, 1) (55)(

|ωk|2 − ∂2xn)pk = 0 in (−1, 1) (56)

unk(±1) = ∂xn unk(±1) = 0, (57)

where µ = µ(|ωk|) =(λ+ |ωk|2

) 12 ∈ Σ(π−ε)/2.

If k = 0, then we can easily solve (55)–(57). Recalling again that div f = 0 inΩ , we find that ∂xn fn0 = 0 in (−1, 1), that

is, f n0 is constant and thus the solutions of (55)–(57) are given by

un0(xn) = 0 and p0(xn) = f n0 xn + (constants).



Suppose next that k 6= 0. Then the general solution of (56) is given by

pk(xn) = c1e|ωk|xn + c2e−|ωk|xn (58)

for some c1 and c2 independent of xn. Given this pk, let unk be the unique solution of (55)with unk(±1) = 0. Then the integration

by parts gives

∂xn unk(xn)e

±µxn |1−1 =

∫ 1

−1

(∂2xn u

nk(xn)− µ

2unk(xn))e±µxn dxn

=

∫ 1

−1

(∂xn pk(xn)− f

nk (xn)

)e±µxn dxn.

Since µ 6= 0, it follows immediately that

∂xn unk(±1) = 0 ⇔

∫ 1

−1∂xn pk(xn)e

±µxn dxn =∫ 1

−1f nk (xn)e

±µxn dxn.

Combining this and (58), we deduce that

∂xn unk(±1) = 0 ⇔

m+c1 −m−c2 =

∫ 1

−1f nk (xn)e

µxn dxn

m−c1 −m+c2 =∫ 1

−1f nk (xn)e

−µxn dxn,(59)

wherem± = m±(|ωk|) is given by

m±(|ωk|) =∫ 1

−1|ωk|e(|ωk|±µ)xn dxn =

|ωk||ωk| ± µ

(e(|ωk|±µ) − e−(|ωk|±µ)

).

It is easy to check thatm0 = m2+ −m2−6= 0. Hence solving the linear equations in (59), we obtain

c1 =1m0

(m+

∫ 1

−1f nk (yn)e

µyn dyn −m−

∫ 1

−1f nk (yn)e

−µyn dyn

)=

∫ 1

−1f nk (yn)m(|ωk|, yn) dyn

and

c2 =1m0

(m+

∫ 1

−1f nk (yn)e

−µyn dyn −m−

∫ 1

−1f nk (yn)e

µyn dyn

)=

∫ 1

−1f nk (yn)m(|ωk|,−yn) dyn,

where

m(|ωk|, yn) =1

m0(|ωk|)

(m+(|ωk|)eµ(|ωk|)yn −m−(|ωk|)e−µ(|ωk|)yn

).

We have shown that if pk is a solution of (56), then (55) and (57) are satisfied by some unk , uniquely determined by pk, if andonly if pk is defined by

pk =∫ 1

−1f nk (yn)

(m(|ωk|, yn)e|ωk|xn +m(|ωk|,−yn)e−|ωk|xn

)dyn.

Combining all the above arguments, we conclude that the solution p ∈ D of (52) and (54) is unique up to additiveconstants and given via the partial Fourier transform by

∂xn p0 = fn0 and pk = p+k + p−k for k 6= 0,

where

p±k(xn) =∫ 1

−1f nk (yn)m(|ωk|,±yn)e

±|ωk|xn dyn (p±0 = 0).

Hence, to complete the proof of the lemma, it remains to derive the Lq-estimate (53) or equivalently

‖∇p±‖Lq(T×(−1,1)) ≤ Cε‖f ‖Lq(T×(−1,1)). (60)



We first estimate ∂xnp±. Since

∂xn p±k(xn) =∫ 1

−1

(±|ωk|e±|ωk|xnm(|ωk|,±yn)

)f nk (yn) dyn,

it follows from Proposition 4.5 and Remark 4.6 that

‖∂xnp±(xn)‖Lq(T) ≤ C

∫ 1

−1

[|ξ ′|e±|ξ

′|xnm(|ξ ′|,±yn)

]‖f n(·, yn)‖Lq(T) dyn.

By virtue of Lemmas 4.7 and 4.9, we estimate[|ξ ′|e±|ξ

′|xnm(|ξ ′|,±yn)

]≤

[|ξ ′|e±|ξ

′|xn m+(|ξ

′|)

m0(|ξ ′|)eµ(|ξ

′|)yn

]+

[|ξ ′|e±|ξ

′|xn m−(|ξ

′|)

m0(|ξ ′|)e−µ(|ξ

′|)yn

]=

[|ξ ′|e−|ξ

′|(1±xn)−µ(|ξ ′|)(1−yn) ·

m+(|ξ ′|)m0(|ξ ′|)

e|ξ′|+µ(|ξ ′|)

]+

[|ξ ′|e−|ξ

′|(1±xn)−µ(|ξ ′|)(1+yn) ·

m−(|ξ ′|)m0(|ξ ′|)

e|ξ′|+µ(|ξ ′|)

]≤

Cε2± xn − yn

·

[m+(|ξ ′|)m0(|ξ ′|)

e|ξ′|+µ(|ξ ′|)

]+

Cε2± xn + yn

·

[m−(|ξ ′|)m0(|ξ ′|)

e|ξ′|+µ(|ξ ′|)

]and [

m+(|ξ ′|)±m−(|ξ ′|)m0(|ξ ′|)

e|ξ′|+µ(|ξ ′|)

]≤ C

(1+

[(m+(|ξ ′|)∓m−(|ξ ′|)

)e−(|ξ

′|+µ(|ξ ′|))

])n−1.

Using Lemmas 4.8 and 4.9, we easily obtain[m+(|ξ ′|)e−(|ξ

′|+µ(|ξ ′|))

]=

[|ξ ′|

|ξ ′| + µ(|ξ ′|)

(1− e−2(|ξ

′|+µ(|ξ ′|))

)]≤ Cε.

On the other hand since

m−(|ξ ′|) =∫ 1

−1|ξ ′|e(|ξ

′|−µ(|ξ ′|))t dt

it follows from Lemma 4.9 that[m−(|ξ ′|)e−(|ξ

′|+µ(|ξ ′|))

]≤

∫ 1

−1

[|ξ ′|e−((1+t)|ξ

′|+(1−t)µ(|ξ ′|))

]dt ≤ Cε.

Combining all the estimates, we conclude that[|ξ ′|e±|ξ

′|xnm(|ξ ′|,±yn)

]≤

Cε2± xn − yn

+Cε

2± xn + yn, (61)

which is essentially the kernel of the Hilbert transform. If 1 ≤ j ≤ n− 1, then

∂xjp±k(xn) =∫ 1

−1

(iωkj e±|ωk|xnm(|ωk|,±yn)

)f nk (yn) dyn

and

‖∂xjp±(xn)‖Lq(T) ≤ C

∫ 1

−1

[ξje±|ξ

′|xnm(|ξ ′|,±yn)

]‖f n(·, yn)‖Lq(T) dyn.

But since[ξj/|ξ

′|]≤ C for each j, it follows immediately from (61) that[

ξje±|ξ′|xnm(|ξ ′|,±yn)

]≤

Cε2± xn − yn

+Cε

2± xn + yn.

Therefore, (60) is now an immediate consequence of the Lq-boundedness of the Hilbert transform.

Acknowledgments

The authors would like to thank the reviewers very much for their helpful comments on the paper.The first author was supported by Japan Society for the Promotion of Science. The second author was supported by the

Korea Research FoundationGrant funded by theKoreanGovernment (MOEHRD, Basic Research Promotion Fund) (KRF-2006-331-C00023).



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