# The Asymptotic Stability of the Generalized 3D Navier-Stokes Equations

Authors: Wen-Juan Wang; Yan Jia
Journal: Journal of Applied Mathematics
 (2013)

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## Abstract

We study the stability issue of the generalized 3D Navier-Stokes equations. It is shown that if the weak solutionu of the Navier-Stokes equations lies in the regular class ∇u∈Lp(0,∞;Bq,∞0(ℝ3)), (2α/p)+(3/q)=2α, 2<q<∞, 0<α<1, then every weak solution v(x,t) of the perturbed system converges asymptotically to u(x,t) as vt-utL2→0, t→∞.


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## Body

## 1. Introduction and Main Result

In this study, we consider the Cauchy problem of the generalized 3D Navier-stokes equations:(1)ut+(-Δ)αu+(u·∇)u+∇π=f,(x,t)∈ℝ3×(0,∞),∇·u=0,u(x,0)=u0.
Here, 0<α<1, and u and π denote unknown velocity and pressure, respectively. f is the external force and u0 is a given initial velocity.It is well known that whenα=1, system (1) becomes the classic Navier-Stokes equations. For the Navier-Stokes equations, it is proved that it has a global weak solution
(2)u(x,t)∈L∞(0,T;L2)∩L2(0,T;H1),∀T>0
for given u0∈L2(ℝ3) with ∇·u0=0 [1]. However, the regularity of Leray weak solutions is still an open problem in mathematical fluid mechanics even if much effort has been made [2–4]. It is an interesting problem to investigate the stability properties of the Navier-Stokes equations and related fluid models [5–11]. As regard to the above system (1), the asymptotic stability of weak solution of the generalized 3D Navier-Stokes equation is described as follows. If u is perturbed initially by ω0 without any smallness assumption, then the perturbed system v is governed by the following equations:
(3)vt+(-Δ)αv+(v·∇)v+∇π=f,∇·v=0,v(x,0)=u0+ω0,
where ω0 is the initial perturbation. There is large literature on the stability issue of the classic Navier-Stokes equations and related fluid models [12–17]. The aim of this paper is to show the stability of weak solution in the framework of the homogeneous Besov space. More precisely, with the use of the Littlewood-Paley decomposition and the classic Fourier splitting technique, we can show that when the initial perturbation ω0∈L2(ℝ3), then every weak solution v(t) of the perturbed system (2) converges asymptotically to u(t) as ∥v(t)-u(t)∥L2→0,t→∞.Now our result reads as follows.Theorem 1.
Letf∈L2(0,T;H-α(ℝ3)), ω0∈L2(ℝ3); Suppose that u(x,t) is a weak solution of (1) and that v(x,t) is a weak solution of the perturbed problem (2), respectively. Moreover, if ∇u also lies in the following regular class:
(4)∇u∈Lp(0,∞;Bq,∞0(ℝ3)),2αp+3q=2α,2<q<∞,
then ∥v(t)-u(t)∥L2→0(t→∞).The remainder of this paper is organized as follows. In the Section2, we first recall the Littlewood-Paley decomposition and the Bony decomposition; then we give three key lemmas. And we prove asymptotic stability of the weak solution in the Section 3.


## 2. Some Auxiliary Lemmas

We recall some basic facts about the Littlewood-Paley decomposition (refer to [18]). Let 𝒮(ℝ3) be Schwartz class of rapidly decreasing functions; supposing f∈𝒮(ℝ3), the Fourier transformation ℱ is defined by
(5)ℱf(ξ)=∫ℝ3‍e-ix·ξf(x)dx.
Choose two nonnegative radial functions χ,φ∈𝒮(ℝ3), supported in ℬ={ξ∈ℝ3,|ξ|≤4/3} and 𝒞={ξ∈ℝ3,3/4≤|ξ|≤8/3}, respectively, such that
(6)χ(ξ)+∑j≥0‍φ(2-jξ)=1,ξ∈ℝ3.Leth=ℱ-1φ and h~=ℱ-1χ, we define the dyadic blocks as follows:
(7)Δjf=φ(2-jD)f=23j∫ℝ3‍h(2jy)f(x-y)dy,forj≥0,Sjf=χ(2-jD)f=∑-1≤k≤j-1‍Δkf=23j∫ℝ3‍h~(2jy)f(x-y)dy,Δ-1f=S0f,Δjf=0forj≤-2.
We can easily verify that
(8)ΔjΔkf=φ(2-jξ)φ(2-kξ)f^=0,if|j-k|≥2,Δj(Sk-1fΔkf)=φ(2-jξ)χ(2-(k-1)ξ)f^×φ(2-kξ)f^=0,if|j-k|≥5.
Especially for any f∈L2(ℝ3), we have the Littlewood-Paley decomposition:
(9)f=S0(f)+∑j≥0‍Δjf,f∈𝒮'(ℝ3).Now we give the definition of the Besov space. Lets∈ℝ and p,q∈[1,∞]; the inhomogeneous Besov space Bp,qs(ℝ3) (see [18]) is defined by the full-dyadic decomposition, such as
(10)Bp,qs(ℝ3)={f∈𝒮'(ℝ3):∥f∥Bp,qs<∞},
where
(11)∥f∥Bp,qs={(∑j=-1∞‍2jsq∥Δjf∥Lpq)1/q,1≤q<∞,supj≥-12js∥Δjf∥Lp,q=∞,
and 𝒮'(ℝ3) is a dual space of 𝒮(ℝ3).The Bony decomposition (see [19]) will be frequently used; it is followed by
(12)uv=Tuv+Tvu+R(u,v),
where
(13)Tuv=∑j‍Sj-1uΔjv,R(u,v)=∑|j′-j|≤1‍ΔjuΔj′v.The following Bernstein inequality (see [18]) will be used in the next section.Lemma 2.
Assume thatk,j∈Z and 1≤p≤q≤∞, for f∈𝒮(ℝ3), one has
(14)sup|α|=k∥∂αΔjf∥Lq(ℝ3)≤C2jk+3j((1/p)-(1/q))∥Δjf∥Lp(ℝ3),
and the constant C is independent of j and k.In the following, we will introduce two lemmas, which will be employed in the proof of our theorem.Lemma 3.
Suppose thatu,w∈L∞(0,T;L2)∩L2(0,T;Hα), for all T>0, ∇v∈Lp(0,∞;Bq,∞0), (2α/p)+(3/q)=2α, 2<q<∞.
Then the trilinear form(15)F(u,v,w)=∫0T‍∫ℝ3‍(u·∇v)wdxdt
is continuous and
(16)|F(u,v,w)|≤C∥u∥L∞(0,T;L2)1/p∥u∥L2(0,T;Hα)1-(1/p)∥w∥L∞(0,T;L2)1/p×∥w∥L2(0,T;Hα)1-(1/p)∥∇v∥Lp(0,T;Bq,∞0).
In particular, if u=w, then
(17)|F(w,v,w)|≤12∫0T‍∥Λαw∥L22dt+C∫0T‍∥w∥L22∥∇v∥Bq,∞0pdt.Proof of Lemma3.
We borrow the idea of [20] to prove this lemma. By using of the Littlewood-Paley decomposition and the Bony decomposition, we obtain
(18)F(u,v,w)=∫0T‍∫ℝ3‍(uiw)∂ivdxdt=∫0T‍∫ℝ3‍(Tuiw+Twui+R(ui,w))ffffffff×(∑j‍Δj∂iv)dxdt=∑|k-j|≤4‍∫0T‍∫ℝ3‍Sk-1uiΔkwΔj∂ivdxdt+∑|k-j|≤4‍∫0T‍∫ℝ3‍ΔkuiSk-1wΔj∂ivdxdt+∑|k-k′|≤1‍∑k,k′≥j-3‍∫0T‍∫ℝ3‍ΔkuiΔk′wΔj∂ivdxdt=I1+I2+I3.
Then we estimateI1, I2, and I3 one by one. Applying the Hölder inequality and the Bernstein inequality (40), we derive
(19)|I1|≤C∑|k-j|≤4‍∑k′≤k-2‍∫0T‍∥Δk′ui∥L2q/(q-2)∥Δkw∥L2∥Δj∂iv∥Lqdt≤C∑|k-j|≤4‍∑k′≤k-2‍∫0T‍2(3/q)k′∥Δk′ui∥L2∥Δkw∥L2∥Δj∂iv∥Lqdt≤C∑|k-j|≤4‍∑k′≤k-2‍∫0T‍(2(α/p′)k′∥Δk′u∥L2)ffffffffffff×(2(α/p′)k∥Δkw∥L2)ffffffffffff×∥Δj∇v∥Lq2((3/q)-(α/p′))k′-(α/p′)kdt,
where (1/p)+(1/p′)=1.
Since|k-j|≤4,k′<k and (2α/p)+(3/q)=2α with 2<q<∞, then
(20)2((3/q)-(α/p′))k′-(α/p′)k=2((3/q)-α+(α/p))k′-(α-(α/p))k=2(3/2q)(k′-k)≤C.
Thanks to the Sobolev embeddingB2,∞α/p′(ℝ3)↪B2,2α/p′(ℝ3)=Hα/p′(ℝ3), we have the following estimate:
(21)|I1|≤C∫0T‍∥u∥Hα/p′∥w∥Hα/p′∥∇v∥Bq,∞0dt.
Similarly, forI2, we also have
(22)|I2|≤C∫0T‍∥u∥Hα/p′∥w∥Hα/p′∥∇v∥Bq,∞0dt.
To estimate the last termI3, by using the Hölder inequality and the Bernstein inequality we obtain
(23)|I3|≤C∑|k-k′|≤1‍∑k,k′≥j-3‍∫0T‍∥Δkui∥L2∥Δk′w∥L2∥Δj∂iv∥L∞dt≤C∑|k-k′|≤1‍∑k,k′≥j-3‍∫0T‍∥Δkui∥L2∥Δk′w∥L2∑|k-k′|≤1‍fff∑k,k′≥j-3‍∫0T‍×(2(3/q)j∥Δj∂iv∥Lq)dt≤C∑|k-k′|≤1‍∑k,k′≥j-3‍∫0T‍(2(α/p′)k∥Δku∥L2)fffffffffffffff×(2(α/p′)k′∥Δk′w∥L2)fffffffffffffff×∥Δj∇v∥Lq2-(3/q)j-(α/p′)(k+k′)dt.
Since |k-k′|≤1,k,k′≥j-3 and (2α/p)+(3/q)=2α,2<q<∞, we have
(24)2-(3/q)j-(α/p′)(k+k′)=2-(3/q)j-(3/2)(k+k′)(1/q)≤29/q≤C,|I3|≤C∫0T‍∥u∥Hα/p′∥w∥Hα/p′∥∇v∥Bq,∞0dt.
So, we can derive(25)|F(u,v,w)|≤C∫0T‍∥u∥Hα/p′∥w∥Hα/p′∥∇v∥Bq,∞0dt≤C(∫0T‍∥u∥Hα/p′2p′dt)1/2p′(∫0T‍∥w∥Hα/p′2p′dt)1/2p′×(∫0T‍∥∇v∥Bq,∞0pdt)1/p≤C∥u∥L2p′(0,T;Hα/p′)∥w∥L2p′(0,T;Hα/p′)∥∇v∥Lp(0,T;Bq,∞0).
Applying the interpolation inequality, we have(26)∥u∥L2p′(0,T;Hα/p′)≤C∥u∥L∞(0,T;L2)1-(1/p′)·∥u∥L2(0,T;Hα)1/p′≤C∥u∥L∞(0,T;L2)1/p·∥u∥L2(0,T;Hα)1-(1/p).
Then(27)|F(u,v,w)|≤C∥u∥L∞(0,T;L2)1/p∥u∥L2(0,T;Hα)1-(1/p)∥w∥L∞(0,T;L2)1/p×∥w∥L2(0,T;Hα)1-(1/p)∥∇v∥Lp(0,T;Bq,∞0).
Especially ifu=w, by using the interpolation inequality, we get
(28)|F(u,v,w)|≤C∫0T‍∥w∥Hα/p′2∥∇v∥Bq,∞0dt≤C∫0T‍∥w∥L22(1-(1/p′))∥Λαw∥L22/p′∥∇v∥Bq,∞0dt≤12∫0T‍∥Λαw∥L22dt+C∫0T‍∥w∥L22∥∇v∥Bq,∞0pdt.
Hence, the proof of the lemma is complete.Letw(x,t)=v(x,t)-u(x,t) denote the difference of v(x,t) and u(x,t), where u(x,t) is a weak solution of (1) and v(x,t) is a weak solution of the perturbed problem (2). Thus w(x,t) satisfies the following equations:(29)wt+(-Δ)αw+(v·∇)w+(w·∇)u+∇π=0,f(x,t)∈ℝ3×(0,∞),∇·w=0,w(x,0)=w0.Lemma 4.
Letw(x,t) be the solution of the above problem. Then
(30)|w^(ξ,t)|≤e-|ξ|2αt|w^0(ξ)|+C|ξ|t.Proof of Lemma4.
Taking the Fourier transformation of the first equation of (38), we get
(31)w^t+|ξ|2αw^=F[-(v·∇)w-(w·∇)u-∇π]=:G(ξ,t).
We can easily obtain(32)|F[-(v·∇)w]|≤∑i,j‍∫ℝ3‍|viwj||ξj|dx≤|ξ|∥v∥L2∥w∥L2,|F[-(w·∇)u]|≤∑i,j‍∫ℝ3‍|wiuj||ξj|dx≤|ξ|∥w∥L2∥u∥L2.
Applying the operator ∇div to the first equation of (38), we have
(33)Δπ=∑i,j‍∂2∂xi∂xj(-viwj-wiuj),
and taking the Fourier transformation, we get
(34)|ξ|2F[π]=∑i,j‍ξiξjF[-viwj-wiuj];
thus
(35)|F[∇π]|≤|ξ||F[π]|≤|ξ|∥w∥L2(∥u∥L2+∥v∥L2).
Then we have(36)|G(ξ,t)|≤|ξ|∥w∥L2(∥u∥L2+∥v∥L2).
Thus solving the ordinary differential equation (31) and using (36) gives
(37)|w^(ξ,t)|=|w^0(ξ)e-|ξ|2αt+∫0t‍e-|ξ|2α(t-s)G(ξ,s)ds|≤|w^0(ξ)|e-|ξ|2αt+C|ξ|∫0t‍∥w∥L2(∥u∥L2+∥v∥L2)ds≤e-|ξ|2αt|w^0(ξ)|+C|ξ|t,
which is the desired assertion of Lemma 4.


## 3. Proof of Theorem1

The following argument is follows the classic Fourier splitting methods which is first used by Schonbek [21] (see also [22]).Taking the inner product of the first equation in (38) with w together with the divergence-free condition of v,w we have
(38)12ddt∥w∥L22+∫ℝ3‍|Λαw|2dx=-∫ℝ3‍(w·∇)u·wdx.Applying Plancherel’s theorem to (38) yields
(39)12ddt∫ℝ3‍|w^(ξ,t)|2dξ+∫ℝ3‍|ξ|2α|w^(ξ,t)|2dξ=-∫ℝ3‍(w·∇)u·wdx.Letf(t) be a continuous function of t with f(0)=1, f(t)>0 and f′(t)>0, we can derive the following:
(40)ddt(f(t)∫ℝ3‍|w^(ξ,t)|2dξ)+2f(t)∫ℝ3‍|ξ|2α|w^(ξ,t)|2dξ=-2f(t)∫ℝ3‍(w·∇)u·wdx+f′(t)∫ℝ3‍|w^(ξ,t)|2dξ.By integrating in time from0 to t for (40), we have
(41)f(t)∫ℝ3‍|w^(ξ,t)|2dξ+2∫0t‍f(s)∫ℝ3‍|ξ|2α|w^(ξ,s)|2dξds=∫ℝ3‍|w^0|2dξ-2∫0t‍f(s)∫ℝ3‍(w·∇)u·wdxds+∫0t‍f′(s)∫ℝ3‍|w^(ξ,s)|2dξds.Noting thatf(t) is a scalar function and applying Lemma 3, we get
(42)|∫0t‍f(s)∫ℝ3‍(w·∇)u·wdxds|≤12∫0t‍f(s)∥Λαw∥L22ds+C∫0t‍f(s)∥w∥L22∥∇u∥Bq,∞0pds≤12∫0t‍f(s)∫ℝ3‍|ξ|2α|w^(ξ,s)|2dξdt+C∫0t‍f(s)∥w∥L22∥∇u∥Bq,∞0pds.Then,(43)f(t)∫ℝ3‍|w^(ξ,t)|2dξ+∫0t‍f(s)∫ℝ3‍|ξ|2α|w^(ξ,s)|2dξds≤∫ℝ3‍|w^0|2dξ+∫0t‍f′(s)∫ℝ3‍|w^(ξ,s)|2dξds+C∫0t‍f(s)∥w∥L22∥∇u∥Bq,∞0pds.LetB(t)={ξ∈ℝ3:f(t)|ξ|2α<f′(t)}, we have
(44)f(s)∫ℝ3‍|ξ|2α|w^(ξ,s)|2dξ≥f′(s)∫ℝ3‍|w^(ξ,s)|2dξ-f′(s)∫B(s)‍|w^(ξ,s)|2dξ.Then,(45)f(t)∫ℝ3‍|w^(ξ,t)|2dξ≤∫ℝ3‍|w^0(ξ)|2dξ+C∫0t‍f(s)∥w∥L22∥∇u∥Bq,∞0pds+∫0t‍f′(s)∫B(s)‍|w^(ξ,s)|2dξds.In addition,(46)∫0t‍f′(s)∫B(s)‍|w^(ξ,s)|2dξds≤C∫0t‍f′(s)∫B(s)‍(e-2|ξ|2αs|w^0(ξ)|2+|ξ|2s2)dξds≤C∫0t‍f′(s)(∫ℝ3‍e-2|ξ|2αs|w^0(ξ)|2dξ)ds+C∫0t‍f′(s)s2(f′(s)f(s))5/2αds.Choosef(t)=(1+t)2, then
(47)(1+t)2∫ℝ3‍|w^(ξ,t)|2dξ≤C+C∫0t‍(1+s)2∥w∥L22∥∇u∥Bq,∞0pds+C∫0t‍(1+s)∫ℝ3‍e-2|ξ|2αs|w^0(ξ,s)|2dξds+C(1+t)4-(5/2α),(1+t)2∥w∥L22≤C∫0t‍(1+s)∫ℝ3‍e-2|ξ|2αs|w^0(ξ)|2dξds+C∫0t‍(1+s)2∥w∥L22∥∇u∥Bq,∞0pds+C(1+t)4-(5/2α).By using the Gronwall inequality, it follows that(48)(1+t)2∥w∥L22≤{C∫0t‍(1+s)∫ℝ3‍e-2|ξ|2αs|w^0(ξ)|2dξds+C(1+t)4-(5/2α)}×exp(∫0t‍∥∇u∥Bq,∞0pds).Since(49)∫ℝ3‍e-2|ξ|2αt|w^0(ξ)|2dξ≤C(1+t)-3/2α⟶0,t⟶∞,
we derive
(50)∥w∥L2≤C(1+t)-2∫0t‍(1+s)∫ℝ3‍e-2|ξ|2αs|w^0(ξ)|2dξds+C(1+t)2-(5/2α)⟶0,t⟶∞,
which completes the proof of Theorem 1.



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