# Lr-Lp Stability of the Incompressible Flows with Nonzero Far-Field Velocity

Authors: Jaiok Roh
Journal: Abstract and Applied Analysis
 (2011)

---

## Abstract

We consider the stability of stationary solutionsw for the exterior Navier-Stokes flows with a nonzero constant velocity u∞ at infinity. For u∞=0 with nonzero stationary solution w, Chen (1993), Kozono and Ogawa (1994), and Borchers and Miyakawa (1995) have studied the temporal stability in Lp spaces for 1<p and obtained good stability decay rates. For the spatial direction, we recently obtained some results. For u∞≠0, Heywood (1970, 1972) and Masuda (1975) have studied the temporal stability in L2 space. Shibata (1999) and Enomoto and Shibata (2005) have studied the temporal stability in Lp spaces for p≥3. Then, Bae and Roh recently improved Enomoto and Shibata's results in some sense. In this paper, we improve Bae and Roh's result in the spaces Lp for p>1 and obtain Lr-Lp stability as Kozono and Ogawa and Borchers and Miyakawa obtained for u∞=0.


---

## Body

## 1. Introduction

The motion of nonstationary flow of an incompressible viscous fluid past an isolated rigid body is formulated by the following initial boundary value problem of the Navier-Stokes equations:(1.1)∂∂tu-Δu+(u⋅∇)u+∇p=f,∇⋅u=0inΩ×(0,∞),u|t=0=u0,u|∂Ω=0,lim⁡|x|→∞u(x,t)=u∞,
where Ω is an exterior domain in Rn with a smooth boundary ∂Ω, and u∞ denotes a given constant vector describing the velocity of the fluid at infinity. In this paper, we consider a nonzero constant u∞. The physical model of the exterior Navier-Stokes equations with a nonzero constant u∞ can be considered as the motion of water in the sea when a boat is moving with the speed -u∞, while the one with zero constant u∞ can be considered when a boat is stopped. There are few known results for the case u∞≠0, while, with u∞=0, many results were obtained for the temporal decay and weighted estimates of solutions of (1.1) (refer [1–12]).Now, we setu=u∞+v in (1.1) and have(1.2)∂∂tv-Δv+(u∞⋅∇)v+(v⋅∇)v+∇p1=f,∇⋅v=0inΩ×(0,∞),v|t=0=u0-u∞,v|∂Ω=-u∞,lim⁡|x|→∞v(x,t)=0.Consider the following linear problem:(1.3)∂∂tu-Δu+(u∞⋅∇)u+∇p=0,∇⋅u=0inΩ×(0,∞),u|t=0=u0,u|∂Ω=0,lim⁡|x|→∞u(x,t)=0,
which is referred to as the Oseen equations; see [13].In order to formulate the problem (1.3), Enomoto and Shibata [14] used the Helmholtz decomposition: (1.4)Lp(Ω)n=Jp(Ω)⊕Gp(Ω),
where 1<p<∞, (1.5)Lp(Ω)n={u=(u1,…,un):uj∈Lp(Ω),j=1,…,n},C0,σ∞={u=(u1,…,un)∈C0∞(Ω)n:∇⋅u=0inΩ},Jp(Ω)=thecompletionofC0,σ∞(Ω),inLp(Ω)n,Gp(Ω)={∇π∈Lp(Ω)n:π∈Lp,loc(Ω¯)}.
The Helmholtz decomposition of Lp(Ω)n was proved by Fujiwara and Morimoto [15], Miyakawa [16], and Simader and Sohr [17]. Let P be a continuous projection from Lp(Ω)n onto Jp(Ω)n.By applyingP into (1.3) and setting 𝒪u∞ = P(-Δ+u∞·∇), one has (1.6)ut+Ou∞u=0,fort>0,u(0)=u0,
where the domain of 𝒪u∞ is given by (1.7)Dp(Ou∞)={u∈Jp(Ω)∩Wp2(Ω)n:u|∂Ω=0}.
Then, Enomoto and Shibata [14] proved that 𝒪u∞ generates an analytic semigroup {T(t)}t≥0 which is called the Oseen semigroup (one can also refer to [16, 18]) and obtained the following properties.Proposition 1.1.
Letσ0>0 and assume that |u∞|≤σ0. Let 1≤r≤q≤∞, then
(1.8)‖T(t)a‖Lq(Ω)≤Cr,q,σ0t-(3/2)(1/r-1/q)‖a‖Lr(Ω),t>0,
where (r,q)≠(1,1) and (∞,∞),
(1.9)‖∇T(t)a‖Lq(Ω)≤Cr,q,σ0t-(3/2)(1/r-1/q)-1/2‖a‖Lr(Ω),t>0,
where 1≤r≤q≤3 and (r,q)≠(1,1).The main purpose of this paper is to discuss the temporal stability of stationary solutionw of the nonlinear Navier-Stokes equation (1.2). One can note that w satisfies the following equations:(1.10)-Δw+(u∞⋅∇)w+(w⋅∇)w+∇p2=f,∇⋅w=0,w|∂Ω=-u∞,lim⁡|x|→∞w(x)=0.For suitablef, Shibata [19] proved that, for any given 0<δ<1/4, there exists ϵ such that if 0<|u∞|≤ϵ, then one has(1.11)‖w‖p,2+|‖w‖|δ+‖p2‖p,1≤|u∞|β,
where (1.12)‖u‖p,m=‖∂mu‖Lp(Ω),|‖u‖|δ=sup⁡x∈Ω(1+|x|)(1+su∞(x))δ|u(x)|+sup⁡x∈Ω(1+|x|)3/2(1+su∞(x))1/2+δ|∇u(x)|,su∞(x)=|x|-xT⋅u∞|u∞|δ<β<1-δ.
Throughout this paper, we assume that f satisfies the assumption in Shibata [19]. Now, we consider the polar coordinate system (1.13)y1=rcos⁡θ,y2=rsin⁡θcos⁡ϕ,y3=rsin⁡θsin⁡ϕ,
for 0≤θ≤π, 0≤ϕ≤2π, and 0≤r<∞. Let S be an orthogonal matrix such that Su∞ = |u∞|(1,0,0)T and put s(y) = |y|-y1. By a change of variable y=Sx, (1.14)|x|=|y|=r,su∞(x)=s(y)=r(1-cos⁡θ).
See Shibata [19] for the detail. Now, by using the above change of variable, we can see easily that w satisfies(1.15)‖w‖L3/(1+δ1)(Ω)+‖w‖L3/(1-δ2)(Ω)+‖∇w‖L3/(2+δ1)(Ω)+‖∇w‖L3/(2-δ2)(Ω)≤C|u∞|1/2,
for small δ1, δ2, where C is independent on u∞.One can also refer to [20] for more general cases of the existence and regularity of stationary Navier-Stokes equations.For the stability of stationary solutionsw, by setting u=v-w and p=p1-p2 for v, p1, w, p2 in (1.2) and (1.10), we have the following equations in Ω:(1.16)∂∂tu-Δu+(u∞⋅∇)u+(u⋅∇)w+(w⋅∇)u+(u⋅∇)u+∇p=0,∇⋅u=0,u(x,0)=u0(x)forx∈Ω,u(x,t)=0forx∈∂Ω,lim⁡|x|→∞u(x,t)=0.
Here, in fact, the initial data should be u0-u∞-w, but for our convenience, we denote by u0 for u0-u∞-w if there is no confusion.First, Heywood [21, 22] and Masuda [23] have studied the temporal stability in L2 space. Shibata [19] proved that there exists small ϵ such that if 0<|u∞|≤ϵ and ∥u0∥3≤ϵ, then a unique solution u(x,t) of (1.16) has the following properties: for any 3<p<∞, (1.17)[u]3,0,t+[u]p,μ(p),t+[∇u]3,1/2,t≤ϵ,lim⁡t→0+[‖u(t)-u0‖3+[u]p,μ(p),t+[∇u]3,1/2,t]=0,
where (1.18)[z]p,ρ,t=sup⁡0<s<tsρ‖z(s,⋅)‖p,μ(p)=12-32p.After that, Enomoto and Shibata [14] considered the stability for arbitrary u∞ by deleting the smallness condition of |u∞|. But in this case, all constants in their results depend on σ0 when |u∞|≤σ0. Also, they assumed the existence of stationary solution w with (1.19)‖w‖L3/(1+δ1)(Ω)+‖w‖L3/(1-δ2)(Ω)+‖∇w‖L3/(2+δ1)(Ω)+‖∇w‖L3/(2-δ2)(Ω)≤α,
for small δ1, δ2 and α. Then, as a result, they proved (1.16) has a unique solution u(x,t) with (1.20)lim⁡t→0+{‖u(t)-u0‖3+t1/2(‖u(t)‖L∞+‖∇u(t)‖L3)}=0,‖u(t)‖u(t)=o(t-((1/2)-(3/2p))),forany3≤p≤∞,‖∇u(t)‖3=o(t-1/2)
as t→∞ when u0 is small enough in the space L3(Ω).Also, Bae and Roh [24] improved Enomoto-Shibata's result in some sense. But their result is limited in the space Lp for 3/2<p, while we consider all 1<p. Moreover, their result depends on s and r, while ours only depends on r, where w∈Ls and u0∈Lr. Also, their optimal decay rate is 2/3+δ, while ours is 3/2+δ.Now, in the next main Theorem, we settle the temporal stability of stationary solutions for the Navier-Stokes equations with a nonzero constant vector at infinity. The idea of the proof is initiated by Kato [25] for w=0 and a very well-known method. Also, for w≠0 with u∞=0, Kozono and Ogawa [12] also used similar method.Theorem 1.2.
There exists smallϵ(p,q,r) such that if0<|u∞|≤ϵ and ∥u0∥L3(Ω)<ϵ, then a unique solution u(x,t) of (1.16) has the following properties:
(1.21)‖u(t)‖Lp(Ω)≤Cϵt-3/2(1/r-1/p)‖u0‖rfor1<r<p≤∞,t>0,‖∇u(t)‖Lq(Ω)≤Cϵt-3/2(1/r-1/q)-1/2‖u0‖rfor1<r<q≤3,t>0,
where u0∈L3(Ω)∩Lr(Ω).


## 2. Proof of Main Theorem

First, we consider the following linear problem:(2.1)∂∂tu-Δu+(u∞⋅∇)u+(w⋅∇)u+(u⋅∇)w+∇p=0,∇⋅u=0,u|t=0=u0,u|∂Ω=0,lim⁡|x|→∞u(x,t)=0.By applying Helmholtz-Leray projectionP and setting (2.2)Lu=P[-Δu+(u∞⋅∇)u+(w⋅∇)u+(u⋅∇)w]=Ou∞u+P[(w⋅∇)u+(u⋅∇)w],
we have (2.3)ut+Lu=0,fort>0,u(0)=u0.
And we note that the domain of ℒ is (2.4)Dp(L)=Dp(Ou∞)={u∈Jp(Ω)∩Wp2(Ω)n|u|∂Ω=0}.LetS(t) be a semigroup generated by the linear operator ℒ, then, by Duharmel's Principle, a solution u(x,t) of (2.1) can be written as in the following integral form,(2.5)u(x,t)=S(t)u0=T(t)u0+∫0tT(t-τ)P[(w⋅∇)u+(u⋅∇)w]dτ,
where T(t) is an analytic semigroup generated by the Oseen operator 𝒪u∞.Lemma 2.1.
Letu0∈L3(Ω)∩Lr(Ω) for 1<r<3, then there exists a small ϵ(p,q,r) such that if |u∞|≤ϵ and ∥u0∥L3(Ω)<ϵ, then a solution u(x,t) represented by (2.5) satisfies 1<p≤∞ with 1/r-1/p<2/3,
(2.6)‖u(t)‖Lp(Ω)=‖S(t)u0‖Lp(Ω)≤Cϵt-3/2(1/r-1/p)‖u0‖Lr(Ω),t>0,
and for 1<q≤3 with 1/r-1/q<1/3,
(2.7)‖∇u(t)‖Lq(Ω)=‖∇S(t)u0‖Lq(Ω)≤Cϵt-3/2(1/r-1/q)-1/2‖u0‖Lr(Ω),t>0.Proof.
Before we prove Lemma2.1 note from (1.15) that we have
(2.8)‖w‖L3/(1+δ1)(Ω)+‖w‖L3/(1-δ2)(Ω)+‖∇w‖L3/(2+δ1)(Ω)+‖∇w‖L3/(2-δ2)(Ω)≤C|u∞|1/2,
for small δ1,δ2>0. In fact, by straight calculations, we can choose any δ1,δ2≤3/16.
Step 1.
Let3<p≤∞ with 1/3≤1/r-1/p<2/3 and 3/2<q≤3 with 1/r-1/q<1/3. We consider the following iteration method to obtain our estimates:
(2.9)uk+1(t)=T(t)u0+∫0tT(t-τ)P[(w⋅∇)uk+(uk⋅∇)w]dτ.
We let 1/q-1/p=1/3 and
(2.10)Mpk=sup⁡t∈[0,∞)tn/2(1/r-1/p)‖uk(t)‖p,Nqk=sup⁡t∈(0,∞)tn/2(1/r-1/q)+1/2‖∇uk(t)‖q.
If t≥2, then by Proposition 1.1, for small δ1,δ2>0, we have
(2.11)∫0t‖T(t-τ)P[(w⋅∇)uk+(uk⋅∇)w]‖pdτ≤C[∫0t-1(t-τ)-n/2(1/r1-1/p)‖(w⋅∇)uk‖r1dτ+∫t-1t(t-τ)-n/2(1/r2-1/p)‖(w⋅∇)uk‖r2dτ+∫0t-1(t-τ)-n/2(1/r1-1/p)‖(uk⋅∇)w‖r1dτ+∫t-1t(t-τ)-n/2(1/r2-1/p)‖(uk⋅∇)w‖r2dτ]≤C|u∞|Nqk[∫0t-1(t-τ)-1-δ1/2τ-3/2(1/r-1/q)-1/2dτ+∫t-1t(t-τ)-1+δ2/2τ-3/2(1/r-1/q)-1/2dτ]+C|u∞|Mpk[∫0t-1(t-τ)-1-δ1/2τ-3/2(1/r-1/p)dτ+∫t-1t(t-τ)-1+δ2/2τ-3/2(1/r-1/p)dτ]≤C|u∞|(Mpk+Nqk)t-3/2(1/r-1/p),
where 1/r1=1/p+2/3+δ1/3 and 1/r2=1/p+2/3-δ2/3. If 0<t<2, then we have
(2.12)∫0t‖T(t-τ)P[(w⋅∇)uk+(uk⋅∇)w]‖pdτ≤C[∫0t(t-τ)-n/2(1/r3-1/p)‖(w⋅∇)uk‖r3dτ+∫0t(t-τ)-n/2(1/r3-1/p)‖(uk⋅∇)w‖r3dτ]≤C|u∞|(Mpk+Nqk)t-3/2(1/r-1/p),
where 1/r3=1/p+2/3-δ2/3. So, we obtain
(2.13)‖uk+1(t)‖p≤Ct-n/2(1/r-1/p)‖u0‖r+C|u∞|t-3/2(1/r-1/p)[Mpk+Nqk],∀t>0,
which implies
(2.14)Mpk+1≤C‖u0‖r+C|u∞|(Mpk+Nqk).
Similarly, we obtain for t≥2,
(2.15)‖∇uk+1(t)‖q≤‖∇T(t)u0‖q+∫0t‖∇T(t-τ)P[(w⋅∇)uk+(uk⋅∇)w]‖qdτ≤Ct-n/2(1/r-1/q)-1/2‖u0‖r+C|u∞|Nqk∫0t-1(t-τ)-1-δ1/2τ-3/2(1/r-1/q)-1/2dτ+C|u∞|Nqk∫t-1t(t-τ)-1+δ2/3τ-3/2(1/r-1/q)-1/2dτ+C|u∞|Mpk∫0t(t-τ)-n/2(1/r4-1/q)-1/2τ-3/2(1/r-1/p)dτ≤Ct-n/2(1/r-1/q)-1/2‖u0‖r+C|u∞|t-3/2(1/r-1/q)-1/2[Mpk+Nqk],
where 1/r4=2/3+1/p=1/3+1/q. Also, for 0<t<2, we have
(2.16)∫0t‖∇T(t-τ)P[(w⋅∇)uk+(uk⋅∇)w]‖qdτ≤C|u∞|(Mpk+Nqk)t-3/2(1/r-1/q)-1/2+δ2/2≤C|u∞|(Mpk+Nqk)t-3/2(1/r-1/q)-1/2.
Therefore, we get(2.17)Mpk+1+Nqk+1≤C‖u0‖r+C|u∞|(Mpk+Nqk).
So if C|u∞|<1 (the constant C is bounded as |u∞| goes to zero, so we can make C|u∞|<1 by choosing small u∞), then we have some K such that
(2.18)Mpk+1+Nqk+1<K,
for all k. Hence, by taking the limit, we complete the proof.
Step 2.
Now, we want to prove1<r<p≤3. For this case, we choose 3/2<q≤3 and p1>3 such that
(2.19)1r-1q<13,1r-1p1<23.
Then, we have
(2.20)‖u(t)‖p≤‖T(t)u0‖p+∫0t‖T(t-τ)P[(w⋅∇)u+(u⋅∇)w]‖pdτ≤Ct-3/2(1/r-1/p)‖u0‖r+C∫0t(t-τ)-3/2(1/r1-1/p)‖w‖3‖∇u‖qdτ+C∫0t(t-τ)-3/2(1/r2-1/p)‖u‖p1‖∇w‖3/2dτ≤Cϵt-3/2(1/r-1/p)‖u0‖r,
where 1/r1=1/3+1/q and 1/r2=1/p1+2/3. One can note that 1/r1-1/p<2/3 and 1/r2-1/p<2/3.
Step 3.
Now, we want to prove1<r<q≤3/2. For this case, we choose 3/2<q1≤3 and p>3 such that
(2.21)1r-1q1<13,1r-1p<23.
Similar to Step 2, we have
(2.22)‖∇u(t)‖q≤‖∇T(t)u0‖q+∫0t‖∇T(t-τ)P[(w⋅∇)u+(u⋅∇)w]‖qdτ≤Ct-3/2(1/r-1/q)-1/2‖u0‖r+C∫0t(t-τ)-3/2(1/r1-1/q)-1/2‖w‖3‖∇u‖q1dτ+C∫0t(t-τ)-3/2(1/r2-1/q)-1/2‖u‖p‖∇w‖3/2dτ≤Ct-3/2(1/r-1/q)-1/2‖u0‖r,
where 1/r1=1/3+1/q1 and 1/r2=1/p+2/3. One can note that 1/r1-1/q<1/3 and 1/r2-1/q<1/3.
Step 4.
At last, we want to prove3<p<∞ with 1/r-1/p<1/3. In this case, we can do easily, by interpolation inequality, Steps 1 and 2.
Therefore, we complete the proof by Steps1–4.Now, by applying the Helmholtz-Leray projectionP into (1.16), we can obtain(2.23)ut+Lu+P[(u⋅∇)u]=0,fort>0,u(0)=u0,
where (2.24)Lu=P[-Δu+(u∞⋅∇)u+(w⋅∇)u+(u⋅∇)w]=Ou∞u+P[(w⋅∇)u+(u⋅∇)w],Dp(L)=Dp(Ou∞)={u∈Jp(Ω)∩Wp2(Ω)n|u|∂Ω=0}.One can note from of [14, Lemma 2.6] that for 1<p<∞ and u∈𝒟p(ℒ)=𝒟p(𝒪u∞), (2.25)‖u‖W2,p(Ω)≤Cp(‖Ou∞u‖p+‖u‖p).
Also, from (1.11), we have (2.26)‖(w⋅∇)u+(u⋅∇)w‖p≤(‖w‖∞+‖∇w‖∞)‖u‖W2,p(Ω)≤|u∞|‖u‖W2,p(Ω)≤Cp|u∞|(‖Ou∞u‖p+‖u‖p).
Since the linear operator 𝒪u∞ generates an analytic semigroup T(t) (refer to [14, 19]), we obtain an analytic semigroup S(t) generated by the linear operator ℒ if |u∞| is small enough. The proof is from perturbation theory of analytic semigroup (refer to [26, Theorem 2.4, page 499]).Remark 2.2.
In Lemma2.1, by the property of a semigroup, we can remove the conditions 1/r-1/p<2/3 for ∥u(t)∥Lp(Ω) and 1/r-1/p<1/3 for ∥∇u(t)∥Lp(Ω), because we have u(x,t) = S(t)u0 = S(t/2)S(t/2)u0.Now, we are in the position to prove Theorem1.2. For the proof, we consider a solution u(x,t) (1.16) as the limit of the following usual iteration method:(2.27)uk+1(t)=S(t)u0-∫0tS(t-τ)P[(uk⋅∇)uk]dτ.Here, we will prove by a similar method with the proof of Lemma2.1. One can note that we will prove without Remark 2.2.Step 1.
We prove that, for anyp>3, we have
(2.28)‖∇u(t)‖3<Ct-1/2,‖u(t)‖p<Ct-1/2+3/2p,∀t>0.
Let(2.29)Mpk=sup⁡t∈[0,∞)t1/2-3/2p‖uk(t)‖p,forp>3,N3k=sup⁡t∈(0,∞)t1/2‖∇uk(t)‖3.
By Lemma 2.1 and (2.27), we obtain
(2.30)‖uk+1(t)‖p≤Ct-1/2+3/2p‖u0‖3+C∫0t(t-τ)-1/2‖uk(t)‖p‖∇uk(t)‖3dτ≤Ct-1/2+3/2p‖u0‖3+CMpkN3k∫0t(t-τ)-1/2τ-1/2+3/2pτ-1/2dτ≤t-1/2+3/2p[C‖u0‖3+CMpkN3k],
which implies
(2.31)Mpk+1≤C‖u0‖3+CMpkN3k.
Similarly, we have
(2.32)‖∇uk+1(t)‖3≤Ct-1/2‖u0‖3+C∫0t(t-τ)-3/2p-1/2‖uk(t)‖p‖∇uk(t)‖3dτ≤t-1/2[C‖u0‖3+CMpkN3k],
which implies
(2.33)N3k+1≤C‖u0‖3+CMpkN3k.
Hence, we have(2.34)Mpk+1+N3k+1<C‖u0‖3+C(Mpk+N3k)2.
Now, we have a sequence of the form(2.35)xk+1≤α+βxk2,
and we know that such sequence satisfies
(2.36)xk≤M≡1-(1-4αβ)1/22β<12β,ifα<14β.
Therefore, by recurrence estimates, smallness of∥u0∥3 implies
(2.37)Mpk+1+N3k+1<K,
for some constant K. Finally, we obtain
(2.38)‖∇u(t)‖3<Ct-1/2,‖u(t)‖p<Ct-1/2+3/2p,∀t>0.Step 2.
We prove that if3/2<p with 1/r-1/p<1/3 and u0∈Lr(Ω)∩L3(Ω), then we have
(2.39)‖u(t)‖p≤Ct-3/2(1/r-1/p),∀t>0.
Let
(2.40)Mp=sup⁡t∈(0,∞)t3/2(1/r-1/p)‖u(t)‖p.
From estimates of Step1, one can note that we have
(2.41)‖∇u(t)‖3≤Ct-1/2‖u0‖3,∀t>0.
So, we have(2.42)‖u(t)‖p≤Ct-3/2(1/r-1/p)‖u0‖r+C∫0t(t-τ)-n/2(1/r8-1/p)‖u(t)‖p‖∇u(t)‖3dτ≤Ct-n/2(1/r-1/p)‖u0‖r+C‖u0‖3∫0t(t-τ)-1/2τ-n/2(1/r-1/p)τ-1/2dτ≤t-n/2(1/r-1/p)[C‖u0‖r+C‖u0‖3Mp],
which implies
(2.43)Mp<C‖u0‖r+C‖u0‖3Mp,
where 1/r8=1/3+1/p.
Hence, we complete the proof withC∥u0∥3<1.Step 3.
We prove that if3/2<q≤3 with 1/r-1/q<1/3 and u0∈Lr(Ω)∩L3(Ω), then we have
(2.44)‖∇u(t)‖q≤Ct-3/2(1/r-1/q)-1/2,∀t>0.
Let(2.45)Nq=sup⁡t∈(0,∞)tn/2(1/r-1/q)+1/2‖∇u(t)‖q.
We choose some p1≈3 with p1>3 such that
(2.46)∥∇u∥q≤Ct-n/2(1/r-1/q)-1/2‖u0‖r+C∫0t(t-τ)-n/2(1/r7-1/q)-1/2‖u‖p1‖∇u‖qdτ≤Ct-n/2(1/r-1/q)-1/2‖u0‖r+C‖u0‖3Nq∫0t(t-τ)-1/2-3/2pτ-1/2+3/2p1τ-n/2(1/r-1/q)-1/2dτ≤t-n/2(1/r-1/q)-1/2[C‖u0‖r+C‖u0‖3Nq].
So we complete the proof with C∥u0∥3<1.Step 4.
We prove that if1<r<p<∞, 1<r<3, and u0∈Lr(Ω)∩L3(Ω), then we have
(2.47)‖u(t)‖p≤Ct-3/2(1/r-1/p),∀t>0.
Case 1 (letp>3/2).
Since we proved for1/r-1/p<1/3 in Step 2, we can assume that 1/3≤1/r-1/p. One notes that we can rewrite solutions u(t) in the form
(2.48)u(t)=S(t2)u(t2)-∫t/2tS(t-τ)P[(u⋅∇)u]dτ.
For any r>1, we choose l>3/2 such that 1/r-1/l<1/3 and 1/l-1/p<2/3. Also, for any 1<r<p≤∞ with 1<r<3, we choose s1>3 and 3/2<s2<3 such that
(2.49)1r-1s2<13,1s1+1s2-1p<23.
Then, by Steps 1–3, we have
(2.50)‖u(t)‖p≤Ct-3/2(1/l-1/p)‖u(t2)‖l+C∫t/2t(t-τ)-3/2(1/s-1/p)‖(u⋅∇)u‖sdτ≤Ct-3/2(1/r-1/p)‖u0‖r+C‖u0‖r∫t/2t(t-τ)-3/2(1/s-1/p)τ-1/2-3/2(1/r-1/s2)τ-1/2+3/2s1dτ≤Ct-3/2(1/r-1/p)‖u0‖r,∀t>0.
Case 2 (let1<p≤3/2).
By Step1–3, we have
(2.51)‖u(t)‖p≤Ct-3/2(1/r-1/p)‖u0‖r+C∫0t(t-τ)-3/2(1/s-1/p)‖(u⋅∇)u‖sdτ≤Ct-3/2(1/r-1/p)‖u0‖r+C‖u0‖r∫0t(t-τ)-3/2(1/s-1/p)τ-3/2(1/r-1/s1)τ-1/2dτ≤Ct-3/2(1/r-1/p)‖u0‖r,∀t>0,
where s1>3/2, 1/r-1/s1<1/3, 1/s=1/s1+1/3.Step 5.
We prove that if1<r<q≤3 and u0∈Lr(Ω)∩L3(Ω), then
(2.52)‖∇u(t)‖q≤Ct-3/2(1/r-1/q)-1/2.
Case 1 (let3/2<q≤3).
Since we proved1/r-1/q<1/3 in Step 3, we can assume that 1/3≤1/r-1/q. Now, we choose l>3/2 such that 1/r-1/l<1/3 and 1/l-1/q<1/3. We also can have s1>3 and 3/2<s2<3 with
(2.53)1s=1s1+1s2,1r-1s2<13,1s-1q<13.
So, by Step 1–4, we obtain
(2.54)‖∇u(t)‖q≤Ct-3/2(1/l-1/q)-1/2‖u(t2)‖l+C∫t/2t(t-τ)-3/2(1/s-1/q)-1/2‖u(t)‖s1‖∇u(t)‖s2dτ≤Ct-3/2(1/r-1/q)-1/2‖u0‖r+C‖u0‖r×∫t/2t(t-τ)-3/2(1/s-1/q)-1/2τ-1/2+3/2s1τ-3/2(1/r-1/s2)-1/2dτ≤Ct-3/2(1/r-1/q)-1/2‖u0‖r.
Case 2 (let1<q≤3/2).
By Step1-Step 3, we have
(2.55)‖∇u(t)‖q≤Ct-3/2(1/r-1/q)-1/2‖u0‖r+C∫0t(t-τ)-3/2(1/s-1/q)-1/2‖(u⋅∇)u‖sdτ≤Ct-3/2(1/r-1/q)-1/2‖u0‖r+C‖u0‖r∫0t(t-τ)-3/2(1/s-1/q)-1/2τ-3/2(1/r-1/s1)τ-1/2dτ≤Ct-3/2(1/r-1/q)-1/2‖u0‖r,∀t>0,
where s1>3/2, 1/r-1/s1<1/3, 1/s=1/s1+1/3, and 1/s-1/q<1/3.
Therefore, by Step1–5, we complete the proof of Theorem 1.2.



---

*Source: 856896-2011-11-03.xml*