# Maximum Norm Estimates of the Solution of the Navier-Stokes Equations in the Halfspace with Bounded Initial Data

Authors: Santosh Pathak
Journal: Abstract and Applied Analysis
 (2021)

---

## Abstract

In this paper, I consider the Cauchy problem for the incompressible Navier-Stokes equations inℝ+n for n≥3 with bounded initial data and derive a priori estimates of the maximum norm of all derivatives of the solution in terms of the maximum norm of the initial data. This paper is a continuation of my work in my previous papers, where the initial data are considered in Tn and ℝn respectively. In this paper, because of the nonempty boundary in our domain of interest, the details in obtaining the desired result are significantly different and more challenging than the work of my previous papers. This challenges arise due to the possible noncommutativity nature of the Leray projector with the derivatives in the direction of normal to the boundary of the domain of interest. Therefore, we only consider one derivative of the velocity field in that direction.


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## Body

## 1. Introduction

We consider the Cauchy problem of the incompressible Navier-Stokes equations inℝ+n,n≥3:
(1)ut+u·∇u+∇p=Δuforx∈ℝ+n,t>0,∇·u=0forx∈ℝ+n,t>0,ut=0=fforx∈ℝ+n,uxn=0=0fort>0,where u=ux,t=u1x,t,⋯,unx,t and p=px,t stand for the unknown velocity vector field of the fluid and its pressure, while f=fx=f1x,⋯fnx is the given initial velocity vector field, with ∇·f=0 and fxn=0=0. In what follows, we will use the same notations for the space of vector-valued and scalar functions for convenience in writing.There is a large literature on the existence and uniqueness of solution of the Navier-Stokes equations inℝn. For the given initial data, solutions of (1) have been constructed in various function spaces. For example, if f∈Lr for some r with 3≤r<∞, then it is well known that there is a unique classical solution in some maximum interval of time: 0≤t<Tf, where 0<Tf≤∞. But, for the uniqueness of the pressure, one requires ∣px,t∣⟶0 as ∣x∣⟶∞. See [1] and [2] for r=3 and [3] for 3<r<∞. The solution is C∞ for 0<Tf<∞.It is well known that forf∈L∞ℝn, there is a unique, smooth, and local-in-time solution u for the Navier-Stokes equations with
(2)p=∑i,jRiRjuiuj,where Ri=−Δ−1/2∂xi is the ith Riesz operator. It is known that in ℝ2, this solution can be extended globally in time. For f∈L∞ℝn, where n≥3, the existence of a regular solution follows from [4]. The solution is only unique if one puts some growth restrictions on the pressure as ∣x∣⟶∞. A simple example of nonuniqueness is demonstrated in [5], where the velocity u is bounded, but ∣px,t∣≤C∣x∣. In addition, an estimate ∣px,t∣≤C1+xσ with σ<1 (see [6]) implies uniqueness. Also, the assumption p∈Lloc10,T;BMO (see [7]) implies uniqueness.Forf∈L∞ℝ+n, where n≥3, the existence of a local mild solution is proved by Bae and Jin in [8]. In the same paper, it is also proved that such mild solution is indeed a strong solution of the Navier-Stokes equations (1). Before the result of Bae and Jin, the local-in-time existence of mild (strong) solution of the halfspace problem was provided in [9] by Solonikov for continuous bounded initial data in ℝ+n.In this paper, I am interested in obtaining estimates of the maximum norm of the derivatives ofu in terms of the maximum norm of the initial function f, assuming that the solution exists, and it is C∞ℝ+n for 0<t<Tf. The work of this paper is a continuation of the work of my papers [10] and [11] to the halfspace case for nondecaying initial data. Nonempty boundary in the domain in this paper makes this work different, in some aspects, and significantly more challenging in proving the key lemmas than the work in my previous works where the initial functions are in ℝn or Tn.We begin by transforming the momentum equations of (1) into the abstract ordinary differential equations:
(3)ut+Au=−ℙu·∇u,where A=−ℙΔ is the Stokes operator and ℙ is the Leray projector, which is given by
(4)ℙfx≡fx+∇x∫ℝ+n∇yGx,y·fydy,where fnxn=0=0. Note that
(5)Gx,y≡Nx−y+Nx−y∗,where y∗=y1,⋯yn−1,−yn,Nx=1/2−nωnx2−n, if n≥3, and ωn denotes the surface area of the unit sphere in ℝn which is given by ωn=2πn/2/Γn/2.The solution of (3) is formally expressed in the integral form:
(6)ut=e−Atf−∫0te−At−sℙu·∇usds.Solonikov [9] has expressed the solution operator of the Stokes equations in ℝ+n in the integral form
(7)e−Atf≡∫ℝ+nGx,y,t·fydy,whereG=Giji,j=1,⋯n is given by
(8)Gijx,y,t=δijΓx−y,t−Γx−y∗,t+41−δjn∂∂xj∫0xn∫ℝn−1∂∂xiNx−zΓz−y∗,tdz.The functionΓx,t is the n-dimensional Gaussian kernel defined by Γx,t≡Γtx≡1/4πtn/2e−x2/4t.A solution formula of the Stokes equations (3) in ℝ+n has also been provided by Ukai in [12]. Such solution formula has been used in the Lq setting, particularly for 1<q<∞ (see [13, 14]). For L1 and L∞ estimates of the Stokes flow or its gradient, see [15, 16]. The solution formula provided by Solonikov [9] has mainly been used for L∞ framework (see [14, 17]).To formulate the main result of this paper, we first introduce some notations as follows:(9)f∞=supx∣fx∣withfx2=∑ifi2x,and Dα=D1α1⋯Dnαn,∂xi=Di=∂/∂xiforamulti−indexα=α1,⋯,αn. In what follows, if ∣α∣=j, for any j=0,1,⋯, then we will denote Dα=D1α⋯Dnαn by Dj. We also set
(10)Djut∞≔Dju·,t∞=max∣α∣=jDαu·,t∞.Clearly,Djut∞ measures all space derivatives of order j in maximum norm. For later purposes, let us also introduce a few other notations:
(11)Γx,t=Γtx=Γ¯tx¯Γtnxn=14πtn−1/2e−x¯2/4t14πt1/2e−xn2/4t,Dβ=D1β1⋯,Dn−1βn−1,Dα=DβDn,α=β,1,∇·=div.Throughout this paper,Dj will be understood as the derivative of order j=∣α∣=∣β∣+1. In addition, 1ℝ+n denotes the characteristic function which is 1 on ℝ+n and 0 otherwise. τa is a translation operator defined by τafx≔fa−x.The goal of this paper is to prove the following theorem.Theorem 1.
Consider the Cauchy problem for the Navier-Stokes equations ((1)) wheref∈L∞ℝ+nand∇·f=0is understood in the sense of distribution. There is a constantc0>0, and for anyα=α1,⋯αn−1,1with∣α∣=jwherej=0,1,⋯there is a constantKjso that(12)tj/2Djut∞≤Kjf∞for0<t≤c0f∞2.
The constantsc0 and Kj are independent of t and f.One of the important tools in the proof of Theorem1 is the uniform estimates of the composite operator Dje−Atℙ∇div. But, obtaining such uniform estimates is complicated because of the possible noncommutativity nature of the Leray projector with the derivatives in the direction of normal to the boundary of the domain; hence, Dj and e−Atℙ may not be commutative.To overcome this difficulty, we will generalize the techniques of obtaining the uniform estimates on∇e−Atℙ∇div of the paper [8] by Bae and Jin to obtain our desired uniform estimates on Dje−Atℙ∇div. In their paper, they require the uniform estimates to prove the existence of the local solution of the Navier-Stokes equations in halfspace for bounded initial data.This paper is organized in the following ways. In “Some Auxiliary Results,” we introduce some auxiliary results which will be labelled as propositions. In “Estimate ofDje−AtP∇·g,” we derive an important estimate on the composite operator Dje−Atℙ∇div. In “Estimates for the Navier-Stokes Equations,” we establish some estimates on the solution of the Navier-Stokes equations. In “Estimates for the Navier-Stokes Equations,” a proof of Theorem 1 will be provided. Finally, Appendices A, B, and C contain proofs of the propositions which are introduced in “Some Auxiliary Results.”


## 2. Some Auxiliary Results

Let us consider the Stokes problem inℝ+n,n≥3:
(13)ut−Δu+∇p=−∇·guforx∈ℝ+n,t>0,∇·u=0forx∈ℝ+n,t>0,ut=0=fforx∈ℝ+n,uxn=0=0fort>0,where g=u⊗u=gij1≤i,j≤n. Here, we note that each gij is quadratic in components of u.Solonikov in [9] has obtained the solution of (13) which is given by
(14)ux,t=∫ℝ+nGx,y,t·fydy−∫0t∫ℝ+nGx,y,t−s·ℙ∇·gy,sdyds.Next, we state the following proposition.Proposition 2.
If
k
=
1
,
⋯
,
n
−
1
, then we have
(15)
∂
∂
x
k
G
i
j
x
,
y
,
t
=
−
∂
∂
y
k
G
i
j
x
,
y
,
t
and(16)∂∂xnGijx,y,t=∂∂ynGijx,y,t−2δij∂∂ynΓx−y,t+41−δjn∂∂xj∫ℝn−1∂∂xiNx¯−z¯,xnΓz¯−y¯,yn,tdz¯.Proof.
The proof is given in Appendix A.Proposition 3.
Let
x
∈
ℝ
+
n
and
f
be any Hölder continuous function with the exponent
0
<
α
<
1:
(17)fα=supx,z∣fx−fz∣x−zα<∞.
Then, fori,j≠n or i,j=n, we have
(18)∂xj∫0xn∫ℝn−1∂xiNx−zfzdz=−δij2nfx+∫0xn∫ℝn−1∂xixj2Nx−zfzdz+δjn∫ℝn−1∂xiNx¯−z¯fz¯,xndz¯.
Fori=n,j≠n or j=n,i≠n, we have
(19)∂xj∫0xn∫ℝn−1∂xiNx−zfzdz=∫0xn∫ℝn−1∂xixj2Nx−zfzdz.Proof.
The proof is given in Appendix B.Next, we define the Hardy spaceH1. Let Nhx≔sups>0∣h∗Γsx∣. Let H1ℝn be the space of functions h so that Nh∈L1ℝn with the norm hH1ℝn≔NhL1ℝn. Let H1ℝ+n be the space of functions h so that there is h~∈H1ℝn with h~ℝ+n=h with the norm hH1ℝ+n=infh~H1ℝn:h~ℝ+n=h.Next, we state a few well-known results related to the Hardy-norm estimates of the Gaussian kernelΓtProposition 4.
Fix
a
∈
ℝ
n
. Then
1
ℝ
+
n
D
j
τ
a
Γ
t
∈
H
1
ℝ
n
with
(20)
1
ℝ
+
n
D
j
τ
a
Γ
t
H
1
ℝ
n
≤
C
t
−
j
/
2and
(21)1ℝ+nDjτa∗Γt−τb∗ΓtH1ℝn≤Ct−j+α/2e−an2/4ta¯−b¯α
for0<α<1.We omit the proofs of well-known results of Proposition4.Proposition 5.
Let
j
=
1
⋯
,
n
−
1
and
i
=
1
,
⋯
n
. Then we have
(22)
∫
ℝ
+
n
∣
∫
0
x
n
∫
ℝ
n
−
1
∂
2
∂
z
i
∂
z
j
N
x
−
z
f
z
,
y
d
z
∣
d
y
≤
C
sup
x
∈
ℝ
+
n
∫
ℝ
+
n
∣
f
x
,
y
∣
d
y
+
C
x
n
α
sup
x
n
/
2
<
z
n
<
x
n
,
z
¯
∈
ℝ
n
−
1
∫
ℝ
+
n
∣
f
z
¯
,
z
n
,
y
−
f
x
¯
,
z
n
,
y
∣
d
y
x
¯
−
z
¯
α
.Proof.
The proof is given in Appendix C.


## 3. Estimate ofDje−Atℙ∇·g

Solonikov in [9] and Shimizu in [16] provide the following estimates:
(23)Dje−Atf∞≤Ct−j/2f∞,(24)e−Atℙ∇·g∞≤Ct−1/2g∞,where f∈L∞ℝ+n, g=gij,1≤i,j≤n, and gij∈L∞ℝ+n for each i,j. Also, f and g vanish on the boundary. In addition, in paper [8] by Bae and Jin, they prove
(25)∇e−Atℙ∇·g∞≤Ct−1/2∇g∞as a critical estimate to prove their desired result.With all the above estimates in hand, we begin to obtain the uniform estimate on the composite operatorDje−Atℙ∇·g. For that purpose, recall
(26)e−Atf=∫ℝ+nGx,y,t·fydy,where Gx,y,t is defined by (8). In the following, consider i≠n, and denote by H=Hijki,j,k=1n the kernel tensor of the operator e−Atℙ∇div. For simplicity in computational purpose, we consider gij as a Schwartz class function in ℝ+n vanishing on the boundary for each i,j. Thus, we begin by writing
(27)∂xie−AtℙDx¯β∇·gix=∑j,k=1n∫ℝ+n∂xiHijkx,y,tDy¯βgjkydy,where
(28)Hijkx,y,t=−∂ykGijx,y,t+∑l=1n∂yj∫ℝ+n∂ykGz,y∂zlGilx,z,tdz.With integration by parts, we obtain(29)∂xie−AtℙDx¯β∇·gix=∑j,k=1n−1∣β∣∫ℝ+n∂xiDy¯βHijkx,y,tgjkydy.UseGin=0 for i≠n to write the following:
(30)∂xiDy¯βHijkx,y,t=−∂xi∂ykDy¯βGijx,y,t+∑l=1n−1∂xi∂yjDy¯β∫ℝ+n∂ykGz,y∂zlGilx,z,tdz.Use∂xiGijx,y,t=−∂yiGijx,y,t for i≠n and Proposition 5 to justify the following expression:
(31)∂xiDy¯βHijkx,y,t=∂ykDy¯β+1Gijx,y,t+∑l=1n−1∂xi∂yj∫ℝ+n∂ykGz,y∂zlDz¯βGilx,z,tdz=∂ykDy¯β+1Gijx,y,t+∑l=1n−1∂yj∫ℝ+n∂ykGz,y∂xiDz¯β+1Gilx,z,tdz.Therefore, (29) can be rewritten as
(32)∂xie−AtℙDx¯β∇·gix=∑j,k=1n−1∣β∣∫ℝ+n∂ykDy¯β+1Gijx,y,t+∑l=1n−1∂yj∫ℝ+n∂ykGz,y∂xiDz¯β+1Gilx,z,tdzgjkydy=I1x,t+I2x,t,where(33)I1x,t≔−1∣β∣∑j,k=1n∫ℝ+n∂ykDy¯β+1Gijx,y,tgjkydy,I2x,t≔−1∣β∣∑j,k=1n∫ℝ+n∑l=1n−1∂yj∫ℝ+n∂ykGz,y∂xiDz¯β+1Gilx,z,tdzgjkydy.First, we estimateI1 for k≠n. For that purpose, recall ∂ykGijx,y,t=−∂xkGijx,y,t. Therefore,
(34)I1x,t=∑j,k=1,k≠nn∫ℝ+nDx¯β+2Gijx,y,tgjkydy.Clearly,I1x,t is the derivative of the ith component of the solution of the Stokes equations. So, using estimate (23), we get the desired estimate on I1 as below
(35)∣I1x,t∣≤Ct−j+1/2g∞fork≠n.Next, we estimateI1 for k=n. In [9], Gij is given as
(36)Gijx,y,t=δijΓx−y,t+Gij∗x,y,t,where
(37)Gij∗x,y,t=−δijΓx−y∗,t+41−δjn∂xj∫0xn∫ℝn−1∂xiNx−zΓz−y∗,tdz.Next, we use the estimate(38)∣DtsDxkDymGij∗x,y,t∣≤Ct−s−mn/2t+xn2−kn/2x−y∗2+t−n+∣k′∣+∣m′∣2e−Cyn2t,where m′=m1,m2,⋯mn−1,k′=k1,k2,⋯kn−1, provided in [18], to obtain the desired estimate on I1 for k=n.Therefore, we use modifiedGij for k=n and rewrite I1 as
(39)I1x,t=−1∣β∣∑j,k=1n∫ℝ+n∂ynDy¯β+1δijΓx−y,t+Gij∗x,y,tgjkydy=−12∣β∣+1∑j,k=1n∫ℝ+n∂ynDx¯β+1δijΓx−y,tgjkydy+−1∣β∣∑j,k=1n∫ℝ+n∂ynDy¯β+1Gij∗x,y,tgjkydy≔I1∗+I1∗∗.The estimate forI1∗ follows as
(40)∣I1∗x,t∣≤Cg∞∫ℝn∣1ℝ+n∂ynDx¯β+1δijΓx−y,t∣dy≤Cg∞1ℝ+nDx¯β+1∂xnΓx−,tH1ℝn.Applying Proposition4, we obtain
(41)∣I1∗x,t∣≤Cg∞t−j+1/2.To estimateI1∗∗, we use the estimate of (38) and obtain
(42)∣I1∗∗x,t∣≤Cg∞t−1/2∫ℝ+ne−Cyn2/tx−y∗2+tn+j/2dy≤Cg∞t−j+1/2∫ℝ+ne−Cyn2/tx−y∗2+tn/2dy≤Cg∞t−j+1/2t1/2t+xn2−1/2.Finally, we get(43)∣I1∗∗x,t∣≤Ct−j+1/2g∞.We obtain(44)I1∞≤I1∗∞+I1∗∗∞≤Ct−j+1/2g∞fork=n.Therefore, from (35) and (44), we obtain
(45)I1∣∞≤Ct−j+1/2g∞forallk.Next, we estimateI2: For that, let us begin by rewriting I2 after dropping the summation notations and negative signs for convenience in writing.
(46)I2x,t=∫ℝ+n∂yj∫ℝ+n∂ykGz,y∂xiDz¯β+1Gilx,z,tdzgjkydy.Equivalently, we write(47)I2x,t=∫ℝ+nTx,y,tgjkydy,where
(48)Tx,y,t≔∂yj∫ℝ+n∂ykGz,yDz¯β+2Gilx,z,tdz.Using expression forGij from (8) for i,l≠n, we obtain
(49)Tx,y,t=∂yj∫ℝ+n∂ykGz,yDz¯β+2δilΓx−z,t−Γx−z∗,t+4∂xl∫0xn∫ℝn−1∂xiNx−wΓw−z∗,tdwdz=T1+T2+T3,where
(50)T1x,y,t≔∂yj∫ℝ+n∂ykGz,yDz¯β+2δilΓx−z,tdz,T2x,y,t≔−∂yj∫ℝ+n∂ykGz,yDz¯β+2δilΓx−z∗,tdz,T3x,y,t≔∂yj∫ℝ+n∂ykGz,yDz¯β+24∂xl∫0xn∫ℝn−1∂xiNx−wΓw−z∗,tdwdz.To estimateT1, let us proceed by writing
(51)T1x,y,t=∂yj∫ℝ+n∂ykNz−y+Nz−y∗Dz¯β+2δilΓx−z,tdz=∂yj∫ℝn∂ykNz−y1zn>0Dz¯β+2δilΓx−z,tdz+∂yj∫ℝn∂ykNz−y1zn<0Dz¯β+2δilΓx−z∗,tdz.It is well known that1ℝ+nDz¯β+2Γx−z,t and 1ℝ−nDz¯β+2Γx−z∗,t are in Hardy space H1ℝn, for any fixed z∈ℝn. Since the Calderon-Zygmund type transforms are bounded in Hardy space, we obtain that
(52)∫ℝ+n∣T1x,y,t∣dy≤C1ℝ+nDz¯β+2Γtz−·H1ℝn+C1ℝ−nDz¯β+2Γtz−·H1ℝn.Using the estimates of Proposition4, we arrive at
(53)∫ℝ+n∣T1x,y,t∣dy≤Ct−j+1/2.With exactly the same argument as forT1, we also obtain
(54)∫ℝ+n∣T2x,y,t∣dy≤Ct−j+1/2.It remains to obtain an estimate forT3. We use Proposition 3 for i,l≠n by replacing f by Γw,. and also use Gln=0 to rewrite T3 as
(55)T3x,y,t=4∂yj∫ℝ+n∂ykGz,yDz¯β+2−δil2nΓx−z∗,t+∫0xn∫ℝn−1∂xixl2Nx−wΓw−z∗,tdwdz=−δil2n∂yj∫ℝ+n∂ykGz,yDz¯β+2Γx−z∗,tdz+∂yj∫ℝ+n∂ykGz,y∫0xn∫ℝn−1∂xixl2Nx−wDz¯β+2Γw−z∗,tdwdz≔T3∗+T3∗∗.By the same argument as forT1, we can obtain
(56)∫ℝ+n∣T3∗x,y,t∣dy≤Ct−j+1/2.Let us rewriteT3∗∗ as
(57)T3∗∗x,y,t=∫0xn∫ℝn−1∂xixl2Nx−w∂yj∫ℝ+n∂ykGz,yDz¯β+2Γw−z∗,tdzdw.Set(58)Tjkw,y,t≔∂yj∫ℝ+n∂ykGz,yDz¯β+2Γw−z∗,tdz.By Proposition5,
(59)∫ℝ+n∣T3∗∗x,y,t∣dy≤Csupw∈ℝ+n∫ℝ+n∣Tjkw,y,t∣dy+Cxnαsupxn/2<wn<xnsupw¯∈ℝn−1∫ℝ+n∣Tjkw¯,wn,y,t−Tjkx¯,wn,y,t∣dyx¯−w¯α.Notice that(60)Tjkw,y,t=∂yj∫ℝ+n∂ykNz−y+Nz−y∗Dz¯β+2Γw−z∗,tdz=∂yj∫ℝn∂ykNz−y1zn>0Dz¯β+2Γw−z∗,tdz+∂yj∫ℝn∂ykNz−y1zn<0Dz¯β+2Γw−z,tdz.We also recall that1ℝ+nDwβ+2Γw−z∗,t and 1ℝ−nDwβ+2Γw−z,t are in the Hardy space H1ℝn, for any fixed w∈ℝn. Since the Calderon-Zygmund type transforms are bounded in Hardy space, after using Dw¯β+2Γw−z,t=−1∣β∣+2Dz¯β+2Γw−z,t, we arrive at
(61)∫ℝ+n∣Tjkw,y,t∣dy≤C1ℝ+nDw¯β+2Γw−·,tH1ℝn+1ℝ−nDw¯β+2Γw−·,tH1ℝn≤Ct−j+1/2.Let us recall a result of Proposition4:
(62)1ℝ+nDwβ+2τa∗Γt−τb∗Γt∈H1ℝn,for any a,b∈ℝn, and is bounded by Ct−∣β∣+2+γ/2e−wn2/4tw¯−x¯γ for 0<γ<1. Hence, in similar way as for Pjk, we obtain
(63)∫ℝ+n∣Pjkw¯,wn,y,t−Pjkx¯,wn,y,t∣dy≤Ct−∣β∣+2+γ/2e−wn24tw¯−x¯γ.Therefore,(64)∫ℝ+n∣T3∗∗x,y,t∣dy≤Ct−∣β∣+2/2+Ct−∣β∣+2+γ/2xnγe−xn24t≤Ct−j+1/2.Using (56) and (64) leads us to obtain
(65)∫ℝ+n∣T3x,y,t∣dy≤Ct−j+1/2.SinceT=T1+T2+T3, with the use of (53), (54), and (65), we obtain
(66)∣I2x,t∣≤g∞∫ℝ+n∣Tx,y,t∣dy≤Ct−j+1/2g∞.Finally, using (45) and (66) with fact that e−At commutes with Dx¯β, we have proved the following important lemma.Lemma 6.
For any
g
=
g
i
j
,
1
≤
i
,
j
≤
n
with
g
i
j
∈
L
∞
ℝ
+
n
, and
g
i
j
x
¯
,
0
=
0
, there exists a constant
C
independent of
t
and
g
such that
(67)
D
j
e
−
A
t
ℙ
∇
·
g
∣
∞
≤
C
t
−
j
+
1
/
2
g
∞
,for 0<t<T, for some T>0.Corollary 7.
Let
g
be as in the previous lemma, then the solution of
(68)
u
t
+
A
u
=
ℙ
∇
·
g
,
u
t
=
0
=
0
,
u
x
n
=
0
=
0satisfies
(69)ut∣∞≤Ct1/2max0≤s≤tgs∞,0<t<T,for some T>0.Proof.
The solution of (38) is given by
(70)ut=∫0te−At−sℙ∇·gsds,0<t<T
and(71)ut∞≤∫0te−At−sℙ∇·gs∞ds.
Applying the estimate (24), we obtain
(72)ut∞≤max0≤s≤tgs∞∫0tt−s−1/2ds.
Hence, we obtain(73)ut∞≤Ct1/2max0≤s≤tgs∞.


## 4. Estimates for the Navier-Stokes Equations

Recall the transformed abstract ordinary differential equation (3):
(74)ut+Au=−ℙu·∇u.Solution of (74) with given initial and boundary condition as in (1) is given by
(75)ut=e−Atf−∫0te−At−sℙu·∇usds.Using the solution (75) along with the use of estimates (23), (24), and (25), we prove the following important lemma.Lemma 8.
Set
(76)
V
t
=
u
t
∞
+
t
1
/
2
D
u
t
∞
,
0
<
t
<
T
.
There is a constantC>0, independent of t and f, so that
(77)Vt≤Cf∞+Ct1/2max0≤s≤tV2t,0<t<T.Proof.
Using estimate (23) for the solution of the Stokes equations in (75), we obtain
(78)ut∞≤f∞+∫0te−At−sℙ∇·gsds∞.
From (74), after using estimate (24), with the fact that g is quadratic in u gives us
(79)ut∞≤f∞+C∫0tt−s−1/2us∞2ds=f∞+C∫0tt−s−1/2s−1/2s1/2us∞2ds≤f∞+Cmax0≤s≤ts1/2us∞2∫0tt−s−1/2s−1/2ds.
Since∫0tt−s−1/2s−1/2ds=C>0, which is independent of t, we arrive at the following estimate
(80)ut∞≤f∞+Cmax0≤s≤ts1/2∣us∞2,(81)ut∞≤f∞+Ct1/2max0≤s≤tV2s.
Next, applyDi to ut in the integral form to obtain and estimate for Diu≕v:
(82)vt=Die−Atf−Di∫0te−At−sℙ∇·gusds=Die−Atf−∫0tDie−At−sℙ∇·gusds.
Let us estimate the integral in the above expression as below.(83)∣∫0tDie−At−sℙ∇·gsds∣≤∫0t∣Die−At−sℙ∇·gs∣ds.
We use the estimate (25) again with the fact that g is quadratic in u to obtain
(84)∣∫0tDie−At−sℙ∇·gsds∣≤C∫0tt−s−1/2us∞Dus∞ds=C∫0tt−s−1/2s−1/2s1/2us∞Dus∞ds≤Cmax0≤s≤ts1/2us∞Dus∞∫0tt−s−1/2s−1/2ds≤Cmax0≤s≤tus∞2+sDus∞2.
Therefore, we have the following estimate forv=Diu(85)vt∣∞≤Ct−1/2f∞+Cmax0≤s≤tus∣∞2+sDus∞2,(86)t1/2Dut∞≤Cf∞+Ct1/2max0≤s≤tV2t.
The combination of (80) and (85), proves Lemma 8.Lemma 9.
Let
C
and
u
∈
L
∞
ℝ
+
n
×
0
,
T
be same as in Lemma 8for someT>0. Set(87)c0=116C4.
ThenT>c0/f∞2 and
(88)ut∞+t1/2Dut∞<2Cf∞for0≤t<c0f∞2.Proof.
We prove this lemma by contradiction after recalling the definition ofVt in (76). Suppose that (88) does not hold, then denote by t0 the smallest time with Vt0=2Cf∞. Use (77) to obtain
(89)2Cf∞=Vt0≤Cf∞+Ct01/24C2f∞2.
Thus(90)1≤4C2t01/2f∞2.
Therefore,t0≥c0/f∞2. This contradiction proves (88) and T>c0/f∞2.


## 5. Proof of Theorem1

Lemma9 proves Theorem 1 for j=0,1 for 0<t<c0/f∞2. Now, we apply induction on j to prove Theorem 1. Suppose j≥1 and assume
(91)tk/2Dkut∞≤Kkf∞,for0≤t≤c0f∞2,0≤k≤j−1.ApplyDβ≔Dx¯β to ut+Au=−ℙ∇·g with the fact that ℙ commutes with Dβ. Also let Dβu≔v to obtain
(92)vt+Av=−Dβℙ∇·gwithv∂ℝ+n=0,∇·v=0.The solution of above system can be written as(93)vt=Dβe−Atf−∫0te−At−sDβℙ∇·gusds.Since∇·v=0, we can write
(94)∂xnvt=−∑i=1n−1∂xivt.Using integral form ofvt from above, we can write
(95)∂xnvt=−∑i=1n−1∂xiDβe−Atf−∫0te−At−sDβℙ∇·gusds.Our goal is to prove∂xnvt∞≤Ct−j/2f∞. For that, let us start with the following where i≠n.
(96)∂xnvt∞≤CDβ+1e−Atf∞+C∂xi∫0te−At−sDβℙ∇·gusds∞.Using the estimate (23) in the first term of the above expression, we obtain
(97)∂xnvt∞≤Ct−j/2f∞+C∂xi∫0t/2e−At−sDβℙ∇·gusds∣∞+C∂xi∫t/2te−At−sDβℙ∇·gusds∞,(98)≤Ct−j/2f∞+I1+I2,where
(99)I1x,t≔C∂xi∫0t/2e−At−sDβℙ∇·gusds∞,I2x,t≔C∂xi∫t/2te−At−sDβℙ∇·gusds∞.To estimateI1 uniformly, we proceed as
(100)I1∣∞≤C∫0t/2∂xie−At−sDβℙ∇·gus∞ds.Using Lemma6, we obtain
(101)I1∞≤C∫0t/2t−s−j+1/2gus∞ds.We use simple integration, and the fact thatg is quadratic in u to arrive at
(102)I1∞≤Ct1−j/2f∞2.Next, we estimateI2. For that, we proceed in the following way:
(103)I2∣∞≤C∫t/2t∂xie−At−sDβℙ∇·gus∞ds.Since the order of the derivatives ofDβ∇·g∞ is ∣β∣+1, for convenience in writing, we use Djg∞ to estimate Dβ∇·g∞. Since gu is quadratic in u; therefore
(104)Djgu∞≤Cu∞Dju∞+∑k=1j−1Dku∞Dj−ku∞.By induction hypothesis (91) we obtain
(105)∑k=1j−1Dkus∞Dj−kus∞≤Cs−j/2f∞2.Apply estimate (25) to the integral (103) with the use of (104) to obtain
(106)I2t∞≤C∫t/2tt−s−1/2Cus∞Djus∞+∑k=1j−1Dkus∞Dj−kus∞ds=J1+J2,where
(107)J1x,t≔C∫t/2tt−s−1/2us∞Djus∞ds,J2x,t≔C∑k=1j−1∫t/2tt−s−1/2Dkus∞Dj−kus∞ds.Since∫t/2tt−s−1/2s−j/2ds=Ct1−j/2, where C is independent of t, and the using the estimate of (105), we obtain J2t∞≤Cf∞2t1−j/2. For J1, let us begin as below.
(108)J1t∞≤C∫t/2tt−s−1/2us∞Djus∞ds≤Cf∞∫t/2tt−s−1/2s−j/2sj/2Djus∞ds≤Cf∞t1−j/2max0≤s≤tsj/2Djus∞.Therefore(109)I2t∞≤J1t∞+J2t∞,I2t∞≤Ct1−j/2f∞2+Cf∞t1−j/2max0≤s≤tsj/2Djus∞.We use these bounds to bind the integral in (97). We have Dju=∂xnDβu. Then, maximizing the resulting estimate for tj/2Djut∞ over all derivatives Dj of order j and setting
(110)ϕt≔tj/2Djut∞,and from (98), we obtain the following estimate:
(111)ϕt≤Cf∞+Ct1/2f∞2+Cf∞t1/2max0≤s≤tϕsfor0≤t≤c0f∞2.Sincet1/2f∞≤c0, then Ct1/2f∞2≤Cc0f∞. Therefore
(112)ϕt≤Cjf∞+Cjf∞t1/2max0≤s≤tϕsfor0≤t≤c0/f∞2.Let us fixCj so that the above estimate holds and set
(113)cj=minc0,14Cj2.First, let us prove the following:(114)ϕt<2Cjf∞for0≤t<cjf∞2.Suppose there is a smallest timet0 such that 0<t0<cj/f∞2 with ϕt0=2Cjf∞. Then, using (88), we obtain
(115)2Cjf∞=ϕt0≤Cjf∞+2Cj2f∞2t01/2.Thus(116)1≤2Cjf∞t01/2givest0≥cj/f∞2,which contradicts the assertion. Therefore, we proved the estimate(117)tj/2Djut∞≤2Cjf∞for0≤t≤cj/f∞2.If(118)Tj≔cjf∞2<t≤c0f∞2≕T0,then we start the corresponding estimate at t−Tj. Using Lemma 9, we have ut−Tj∞≤2f∞ and obtain
(119)Tjj/2Djut∞≤4Cjf∞.Finally, for anyt satisfying (118)
(120)tj/2≤T0j/2=c0cjj/2Tjj/2,and (119) yield
(121)tj/2Djut∞≤4Cjc0cjj/2f∞.This completes the proof of Theorem1.In the following appendices, we provide proofs of the propositions that are introduced in “Some Auxiliary Results.” However, these proofs have also been provided in [8]. For the reader’s convenience, we provide them with more details in this paper as well.



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