In this paper, we study space-time decay rates of a nonconservative compressible two-phase flow model with common pressure in the whole space R3. Based on previous temporal decay results, we establish the space-time decay rate of the strong solution. The main analytical techniques involve delicate weighted energy estimates and interpolation.
Citation: Linyan Fan, Yinghui Zhang. Space-time decay rates of a nonconservative compressible two-phase flow model with common pressure[J]. Electronic Research Archive, 2025, 33(2): 667-696. doi: 10.3934/era.2025031
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In this paper, we study space-time decay rates of a nonconservative compressible two-phase flow model with common pressure in the whole space R3. Based on previous temporal decay results, we establish the space-time decay rate of the strong solution. The main analytical techniques involve delicate weighted energy estimates and interpolation.
As is well-known, multi-fluids are very common in nature as well as in various industry applications such as nuclear power, chemical processing, oil and gas manufacturing. The classic approach to simplify the complexity of multi-phase flows and satisfy the engineer's need of some modeling tools is the well-known volume-averaging method. This approach leads to so-called averaged multi-phase models, see [1,2,3] for details. As a result of such a procedure, one can obtain the following generic compressible two-phase fluid model:
{∂t(α±ρ±)+div(α±ρ±u±)=0,∂t(α±ρ±u±)+div(α±ρ±u±⊗u±)+α±∇P=div(α±τ±)+σ±α±ρ±∇Δ(α±ρ±),P=P±(ρ±)=A±(ρ±)ˉγ±, | (1.1) |
where 0≤α±≤1 are the volume fractions of the fluid + and fluid −, satisfying α++α−=1; ρ±(x,t)≥0,u±(x,t) and P±(ρ±)=A±(ρ±)ˉγ± denote the densities, velocities of each phase, and the two pressure functions, respectively; σ±≥0 denote the capillary coefficients of each phase; ˉγ±≥1,A±>0 are positive constants. In what follows, we set A+=A−=1 without loss of any generality. Moreover, τ± are the viscous stress tensors
τ±:=μ±(∇u±+∇tu±)+λ±divu±Id, | (1.2) |
where the constants μ± and λ± are shear and bulk viscosity coefficients satisfying the physical condition: μ±>0 and 2μ±+3λ±≥0, which implies that μ±+λ±>0. For more information about this model, we refer to [1,2,3,4,5] and references therein. In particular, a nice summary of the model was given in the introduction of [6]. However, it is well-known that as far as the mathematical analysis of two-fluid is concerned, there are many technical challenges. Some of them involve, for example:
● The corresponding linear system of the model has multiple eigenvalues, which makes the mathematical analysis (well-posedness and stability) of the model quite difficult and complicated;
● Transition to single-phase regions, i.e., regions where the mass α+ρ+ or α−ρ− becomes zero, may occur when the volume fractions α± or the densities ρ± become zero;
● The system is non-conservative, since the non-conservative terms α±∇P± are involved in the momentum equations. This brings various mathematical difficulties for us to employ methods used for single phase models to the two-fluid model.
Bresch et al. [6] investigated the generic two-fluid model (1.1) with the following density-dependent viscosities:
μ±(ρ±)=μ±ρ±,λ±(ρ±)=0. | (1.3) |
Under the assumption that 1<¯γ±<6, they obtained the global weak solutions for the 3D periodic domain problem. Later, Bresch et al. [7] established the global existence of weak solutions in one space dimension without capillary effects (i.e., σ±=0) when ¯γ±>1 by taking advantage of the one space dimension. Under the assumption that
μ±(ρ±)=νρ±,λ±(ρ±)=0,σ+=σ−=σ, | (1.4) |
Cui et al. [8] proved the time-decay rates of global small strong solutions for model (1.1). Recently, Li et al. [9] extended this result to the general constant viscosities as in (1.2). Very recently, Wu et al. [10] proved global existence and large time behavior of global classic solutions for the model (1.1) without capillary effects (i.e., σ±=0). We also mention the seminal work by Evje et al. [11], who considered the two-fluid model (1.1) with unequal pressures. More precisely, they made the following assumptions on pressures:
P+(ρ+)−P−(ρ−)=(ρ+)ˉγ+−(ρ−)ˉγ−=f(α−ρ−), | (1.5) |
where f is so-called capillary pressure which belongs to C3([0,∞)), and is a strictly decreasing function near the equilibrium, i.e., f′(1)<0. When the initial data is sufficiently small, they established global existence and large time behavior of global strong solutions. After that, this model has been studied by several authors. We refer to [12,13,14] and references therein.
The space-time decay rate of strong solution has attracted more and more attention. In the following, we will state the progress on the topic about the space-time decay in weighted Sobolev space HNγ. Takahashi first established the space-time decay of strong solutions to the Navier-Stokes equations in [15]. In [16,17], Kukavica et al. used the parabolic interpolation inequality to obtain sharp decay rates of the higher-order derivatives for the solutions in weighted Lebesgue space L2γ. In [18,19], Kukavica et al. also established the strong solution's space-time decay rate in Lpγ(2≤p≤∞) and extended the result to n(n≥2) dimensions.
However, to the best of our knowledge, up to now, there is no result on the space-time decay rate of the nonconservative compressible two-phase flow model (1.1). The main motivation of this paper is to give a definite answer to this issue. More precisely, we establish space-time decay rate of the k(0≤k≤N)–order derivative of strong solution to the Cauchy problem of the model (1.1) in weighted Lebesgue space L2γ.
In this subsection, we devote ourselves to reformulating the system (1.1) and stating the main results. The relations between the pressures of (1.1)3 imply
dP=s2+dρ+=s2−dρ−,wheres±:=√dPdρ±(ρ±). | (1.6) |
Here s± represent the sound speed of each phase respectively. As in [6], we introduce the fraction densities
R±=α±ρ±, | (1.7) |
which together with the relation α++α−=1 leads to
dρ+=1α+(dR+−ρ+dα+),dρ−=1α−(dR−+ρ−dα+). | (1.8) |
From (1.6)–(1.7), we finally get
dα+=α−s2+α−ρ+s2++α+ρ−s2−dR+−α+s2−α−ρ+s2++α+ρ−s2−dR−. |
Substituting the above equality into (1.8) yields
dρ+=s2−α−ρ+s2++α+ρ−s2−(ρ−dR++ρ+dR−), |
and
dρ−=s2+α−ρ+s2++α+ρ−s2−(ρ−dR++ρ+dR−), |
which together with (1.6) imply for the pressure differential dP
dP=C2(ρ−dR++ρ+dR−), | (1.9) |
where
C2:=s2+s2−α−ρ+s2++α+ρ−s2−,ands2±=dP(ρ±)dρ±=˜γ±P(ρ±)ρ±. |
Next, by using the relation: α++α−=1 again, we can get
R+ρ++R−ρ−=1,and thereforeρ−=R−ρ+ρ+−R+. | (1.10) |
By virtue of (1.1)3, we have
φ(ρ+,R+,R−):=P(ρ+)−P(R−ρ+ρ+−R+)=0. |
Consequently, for any given two positive constants ˜R+, ˜R−, there exists ˜ρ+>˜R+ such that
φ(˜ρ+,˜R+,˜R−)=0. |
Differentiating φ with respect to ρ+, we get
∂φ∂ρ+(ρ+,R+,R−)=s2++s2−R−R+(ρ+−R+)2, |
which implies
∂φ∂ρ+(˜ρ+,˜R+,˜R−)>0. |
Thus, this together with implicit function theorem implies that the unknowns ρ±, α± and C can be given by
ρ±=ϱ±(R+,R−),α±=α±(R+,R−),and thereforeC=C(R+,R−). |
We refer to [6] for the details.
Therefore, we can rewrite system (1.1) into the following equivalent form:
{∂tR±+div(R±u±)=0,∂t(R+u+)+div(R+u+⊗u+)+α+C2[ρ−∇R++ρ+∇R−]=div{α+[μ+(∇u++(∇u+)T)+λ+divu+Id]}+σ+R+∇ΔR+,∂t(R−u−)+div(R−u−⊗u−)+α−C2[ρ−∇R++ρ+∇R−]=div{α−[μ−(∇u−+(∇u−)T)+λ−divu−Id]}+σ−R−∇ΔR−. | (1.11) |
In the present paper, we consider the Cauchy problem of (1.11) subject to the initial condition
(R+,u+,R−,u−)(x,0)=(R+0,u+0,R+0,u−0)(x)→(ˉR+,→0,ˉR−,→0)as|x|→∞∈R3, | (1.12) |
where two positive constants ˉR+ and ˉR− denote the background doping profile, and in the present paper are taken as 1 for simplicity.
Before presenting our results, let us provide a brief explanation of the notation used in this paper.
We use Lp and Hℓ to denote the usual Lebesgue space Lp(R3) and Sobolev spaces Hℓ(R3)=Wℓ,2(R3) with norms ‖⋅‖Lp and ‖⋅‖Hℓ respectively. We denote ‖(f,g)‖X:=‖f‖X+‖g‖X for simplicity. The notation f≲g means that f≤Cg for a generic positive constant C>0 that only depends on the parameters coming from the problem.
We often drop x-dependence of differential operators, that is ∇f=∇xf=(∂x1f,∂x2f,∂x3f) and ∇k denotes any partial derivative ∂α with multi-index α,|α|=k. Furthermore, ∇∇αg=∇∂αg=(∂x1(∂αg),∂x2(∂αg),∂x3(∂αg)).
For any γ∈R, we denote the weighted Lebesgue space by Lpγ(R3)(2≤p<+∞) with respect to the spatial variables:
Lpγ(R3):={f(x):R3→R,‖f‖pLpγ(R3):=∫R3|x|pγ|f(x)|pdx<+∞}. |
Then, we can define the weighted Sobolev space:
Hsγ(R3)≜{f∈L2γ(R3)∣‖f‖2Hsγ(R3):=∑k≤s‖∇kf‖2L2γ(R3)<+∞}. |
Let Λs be the pseudo differential operator defined by
Λsf=F−1(|ξ|sˆf), for s∈R, |
where ˆf and F(f) are the Fourier transform of f. The homogenous Sobolev space ˙Hs(R3) with norm given by
‖f‖˙Hs≜‖Λsf‖L2. |
Schwartz class S consists of function f, which is infinitely differentiable and all of its derivative decrease rapidly at infinity, such that
supx|xαDβf(x)|<∞, |
for all α,β∈N3.
Theorem 1.1. Let (R+,u+,R−,u−) be the strong solution to the Cauchy problem (1.11)–(1.12) with initial data (R+0−1,u+0,R−0−1,u−0) belonging to the Schwartz class S. In addition, assume that (R+0−1,u+0,R−0−1,u0)∈HN(R3)⋂HNγ(R3)⋂L1(R3) for an integer N≥2. Then if there is a small constant δ0>0 such that
‖(R+0−1,R−0−1)‖HN+1∩L1+‖(u+0,u−0)‖HN∩L1≤δ0, | (1.13) |
then there exists a large enough T and for any 0≤k≤N such that
‖(∇ku+,∇ku−,∇∇kR+,∇∇kR−)‖L2γ≲tγ−34−k2,‖(∇k(R+−1),∇k(R−−1),β1β2∇k(R+−1)+√β4β3∇k(R−−1))‖L2γ≲tγ−14−k2, | (1.14) |
for all t>T.
Remark 1.2. Applying the Gagliardo-Nirenberg-Sobolev inequality, we can obtain the space-time decay rates of smooth solution in weighted normed linear space (Banach space) Lpγ as follows. For any f∈L2(R3)∩˙H2(R3), we have ‖f‖L∞(R3)≤‖f‖14L2(R3)‖f‖34H2(R3). So we can obtain the estimate ‖|x|γ∇k(R+−1,u+,R−−1,u−)(t)‖L∞(k∈[0,N−2]) from the estimates ‖|x|γ∇k(R+−1,u+,R−−1,u−)(t)‖L2 and ‖|x|γ∇k(R+−1,u+,R−−1,u−)(t)‖˙H2. Using the interpolation inequality, we can show that there exists a large enough T such that
‖∇k(R+−1,u+,R−−1,u−)(t)‖Lpγ≤Ct−32(1−1p)−k2+γ, |
for t>T, 2≤p≤∞ and 0≤k≤N−2, where C is a positive constant independent of t.
Now, let's briefly describe the proof process of the main results and the difficulties encountered during this process. For the proof of Theorem 1.1, we employ sophisticated weighted energy estimates, interpolation inequality, and inductive strategies. The proof mainly involves the following three steps.
Firstly, using several lemmas in Section 2 and energy methods, we obtain
ddtEk(t)≲Ek(t)1−12γ(1+t)−1γ(34+k2)+Ek(t)1−1γ(1+t)−2γ(34+k2)+Ek(t)12+Ek(t)12−12γ(1+t)−1γ(34+k2)k∑l=0(El(t))12(1+t)−(32+k−l2)+(Ek(t))12k∑l=0(El(t))12(1+t)−(1+k−l2), | (1.15) |
for t large enough, and according to the different values of k, the value of γ also varies,
Ek(t):=‖(∇kn+,∇ku+,∇kn−,∇ku−,√β1β2∇kn++√β4β3∇kn−)‖2L2γ and the range of values for k is from 0 to N.
Secondly, for the case of k=0, the fourth and fifth terms on the left side of inequality (1.15) can be directly written as E0(t)1−12γ(1+t)−32−1γ34 and E0(t)(1+t)−1. If we want to use Lemma 2.9 to obtain the result for the case k=0, the main difficulty is to handle the term E0(t)(1+t)−1. We will multiply (1+t)−1 on both sides of (1.15) simultaneously, and then apply Lemma 2.9 to control E0(t)(1+t)−1.
Thirdly, using the similar method as k=0 and the decay estimate already obtained by E0(t), we can show that the Theorem 1.1 holds for k=1, and according to the strategy of induction, we prove that Theorem 1.1 holds for 2≤k≤N. The main difficulties come from those terms like
⟨∇kF2,|x|2γ1β2∇ku+⟩,⟨∇kF4,|x|2γ1β3∇ku−⟩∫R3|x|2γ∇ku+∇∇kn−dx,and∫R3|x|2γ∇ku−∇∇kn+dx |
which involve three main difficulties. In section 3, we will provide detailed proofs and explain the methods and processes for dealing with these difficulties.
In this subsection, we reformulate the Cauchy problem (1.11)–(1.12). Setting n±=R±−1, we have
∂tR±+div(R±u±)=∂tn±+div(n±u±)+div(u±)=0 | (2.1) |
From this, we can directly obtain F1=−div(n+u+),F3=−div(n−u−).
According to the left of (1.11)2, we have
∂t(R+u+)+div(R+u+⊗u+)+α+C2(ρ−∇R++ρ+∇R−)=R+∂tu+−(R+divu++∇R+⋅u+)u++R+u+∇u++(R+divu+)u++(∇R+⋅u+)u++R+(C2ρ−ρ+∇R++C2∇R−)=R+∂tu++R+u+∇u++R+(C2ρ−ρ+∇R++C2∇R−)=R+∂tu++R+u+∇u++R+(C2ρ−ρ+∇n++C2∇n−) | (2.2) |
Before discussing the right of (1.11)2, make a transformation to α+ first:
∇α+=α−s2+α−ρ+s2++α+ρ−s2−∇R+−α+s2−α−ρ+s2++α+ρ−s2−∇R−=α−C2s2−∇R+−α+C2s2+∇R−=α−C2s2−∇n+−α+C2s2+∇n− | (2.3) |
And then, according to the right of (1.11)2 and (2.3), we have
div(α+(μ+(∇u++(∇u+)T)+(λ+divu+)I3))+σ+R+∇ΔR+=α+μ+Δu++α+(μ++λ+)∇divu++μ+(∇α+∇u+)+μ+(∇α+(∇u+)T)+λ+(∇α+(divu+)I3)+σ+R+∇ΔR+=α+μ+Δu++α+(μ++λ+)∇divu++μ+α−C2s2−∇n+∇u+−μ+α+C2s2+∇n−∇u++μ+α−C2s2−∇n+(∇u+)T−μ+α+C2s2+∇n−(∇u+)T+λ+α−C2s2−divu+∇n+−λ+α+C2s2+divu+∇n−+σ+R+∇Δn+ | (2.4) |
Divide both sides by R+, we can get
∂tu++u+∇u++(C2ρ−ρ+∇n++C2∇n−)=μ+ρ+Δu++μ+ρ+(μ++λ+)∇divu++μ+α−C2(n++1)s2−∇n+∇u+−μ+C2ρ+s2+∇n−∇u++μ+α−C2(n++1)s2−∇n+(∇u+)T−μ+C2ρ+s2+∇n−(∇u+)T+λ+α−C2(n++1)s2−divu+∇n+−λ+C2ρ+s2+divu+∇n−+σ+∇Δn+ | (2.5) |
Finally, by adding some initial terms of ρ on both sides of the equation and shifting the terms, we can obtain F2, F4.
After the above operations, the Cauchy problem (1.11)–(1.12) can be rewritten as
{∂tn++divu+=F1,∂tu++β1∇n++β2∇n–ν+1Δu+−ν+2∇divu+−σ+∇Δn+=F2,∂tn−+divu−=F3,∂tu−+β3∇n++β4∇n–ν−1Δu–ν−2∇divu–σ−∇Δn−=F4,(n+,u+,n−,u−)(x,0)=(n+0,u+0,n−0,u−0)(x)→(0,→0,0,→0),as|x|→+∞, | (2.6) |
where ν±1=μ±ˉρ±, ν±2=μ±+λ±ˉρ±>0, β1=C2(1,1)ˉρ−ˉρ+, β2=β3=C2(1,1), β4=C2(1,1)ˉρ+ˉρ− (which imply β1β4=β2β3=β22=β23), and the nonlinear terms are given by
F1=−div(n+u+), | (2.7) |
Fi2=−g+(n+,n−)∂in+−ˉg+(n+,n−)∂in−−(u+⋅∇)u+i+μ+h+(n+,n−)∂jn+∂ju+i+μ+k+(n+,n−)∂jn−∂ju+i+μ+h+(n+,n−)∂jn+∂iu+j+μ+k+(n+,n−)∂jn−∂iu+j | (2.8) |
+λ+h+(n+,n−)∂in+∂ju+j+λ+k+(n+,n−)∂in−∂ju+j+μ+l+(n+,n−)∂2ju+i+(μ++λ+)l+(n+,n−)∂i∂ju+j,F3=−div(n−u−), | (2.9) |
Fi4=−g−(n+,n−)∂in−−ˉg−(n+,n−)∂in+−(u−⋅∇)u−i+μ−h−(n+,n−)∂jn+∂ju−i+μ−k−(n+,n−)∂jn−∂ju−i+μ−h−(n+,n−)∂jn+∂iu−j+μ−k−(n+,n−)∂jn−∂iu−j+λ−h−(n+,n−)∂in+∂ju−j+λ−k−(n+,n−)∂in−∂ju−j+μ−l−(n+,n−)∂2ju−i+(μ−+λ−)l−(n+,n−)∂i∂ju−j, | (2.10) |
where
{g+(n+,n−)=(C2ρ−)(n++1,n−+1)ρ+(n++1,n−+1)−(C2ρ−)(1,1)ρ+(1,1),g−(n+,n−)=(C2ρ+)(n++1,n−+1)ρ−(n++1,n−+1)−(C2ρ+)(1,1)ρ−(1,1), | (2.11) |
{ˉg+(n+,n−)=C2(n++1,n−+1)−C2(1,1)ˉg−(n+,n−)=C2(n++1,n−+1)−C2(1,1), | (2.12) |
{h+(n+,n−)=(C2α−)(n++1,n−+1)(n++1)s2−(n++1,n−+1),h−(n+,n−)=−(C2)(n++1,n−+1)(ρ−s2−)(n++1,n−+1), | (2.13) |
{k+(n+,n−)=−C2(n++1,n−+1)(n++1)(s2+ρ+)(n++1,n−+1),k−(n+,n−)=−(α+C2)(n++1,n−+1)(n−+1)s2+(n++1,n−+1), | (2.14) |
l±(n+,n−)=1ρ±(n++1,n−+1)−1ρ±(1,1). | (2.15) |
In the following, we recall several useful tools, which will be frequently used throughout this paper.
Lemma 3.1. (Gagliardo-Nirenberg inequality) Let 1≤q≤+∞, j and m be non-negative integers such that j< m. Let 1≤r≤+∞, p≥1and θ∈[0,1] then
‖∇jf‖Lp≤C‖∇mf‖θLr‖f‖1−θLq(C=C(j,m,n,q,r,θ)), |
where θ satisfies
1p−jn=(1r−mn)θ+1q(1−θ). |
It is worth noting that there are additional requirements when taking some special values for the coefficients.
1) If j=0,q=+∞ and rm<n, then an additional assumption is needed either u→0(|x|→+∞) or u∈Lsfor some finite of s.
2) If r>1 and m−j−nris a non-negative integer, then jm≤θ<1 is needed.
3) Notice that p usually assumed to be finite. However, there are sharper formulations in which p=+∞ is considered, but other values maybe excluded j=0.
4) Setting f=∇lu, we have
‖∇Ju‖Lp≤C‖∇Mu‖θLr‖∇lu‖1−θLq(C=C(j,m,n,q,r,θ)), |
where θ satisfies
1p−Jn=(1r−Mn)θ+(1r−ln)(1−θ)(J=j+l,M=m+l). |
Proof. This is a special case of [15].
Lemma 3.2. Let f and g be smooth functions belonging to Hk∩L∞ for any integer k≥1, then
‖∇k(fg)‖L2≲‖f‖L∞‖∇kg‖L2+‖g‖L∞‖∇kf‖L2, |
‖∇k(fg)‖L1≲‖f‖L2‖∇kg‖L2+‖g‖L2‖∇kf‖L2, |
‖∇k−1(fg)‖L32≲‖f‖L2‖∇k−1g‖L6+‖g‖L6‖∇k−1f‖L2. |
Proof. The proof can be found in [[20], Lemma 3.1].
Lemma 3.3. Let f and g be smooth functions belonging to Hk∩L∞ for any integer k≥1 and define commutator [∇k,f]g=∇k(fg)−f∇kg, then
‖[∇k,f]g‖L2≲‖∇1f‖L∞‖∇k−1g‖L2+‖∇kf‖L2‖g‖L∞. |
Proof. The proof can be found in [[20], Lemma 3.1].
Lemma 3.4. Let F(f) be a smooth function of f with bounded derivatives of any order and f belong to Hk for any integer k≥3, then
‖∇k(F(f))‖L2≲sup0≤i≤k‖F(i)(f)‖L∞(k∑m=2‖f‖m−1−n(m−1)2kL2‖∇kf‖1+n(m−1)2kL2+‖∇kf‖L2). | (3.1) |
Proof. By direct calculation, we have
∇k(F(f))=k∑i=1(F(i)(f)⋅∑n∑j=1sj=i,0≤sj≤i∇k−i((∂1f)s1(∂2f)s2⋯(∂nf)sn)))≤sup0≤i≤k‖F(i)(f)‖L2∑m∑d=1hd=kC(s1,s2⋯sn)∇h1f∇h2f⋯∇hmf. | (3.2) |
Using Gagliardo-Nirenberg inequality, we have
‖∇h1f∇h2f⋯∇hmf‖L2≤‖∇h1f‖L2‖∇h2f‖L∞⋯‖∇hmf‖L∞≤‖∇kf‖1−a1L2‖f‖a1L2‖∇kf‖1−a2L2‖f‖a2L2⋯‖∇kf‖1−amL2‖f‖amL2, | (3.3) |
where
12−h1n=12a1+(12−kn)(1−a1),−h2n=12a2+(12−kn)(1−a2),⋯,−hmn=12am+(12−kn)(1−am),a1+a2+⋯+am=m−1−n(m−1)2k,(1−a1)+(1−a2)+⋯+(1−am)=1+n(m−1)2k. | (3.4) |
‖∇k(F(f))‖L2≲‖∇kf‖L2. |
Lemma 3.5. Let F(f) be a smooth function of vector function f=(f1,f2,⋯,fN) with bounded partial derivatives of any order and fi=fi(x1,x2,⋯,xn)(1≤i≤N) belonging to Hk for any integer k≥3, then
‖∇kx(F(f))‖L2≲sup0≤l≤k‖∇lyF‖L∞‖f‖N−1−n(N−1)2kL2‖∇kf‖1+n(N−1)2kL2. |
Proof. By direct calculation, we have
|∇kx(F(f))|=|∇k−1(∇yF⋅∇1f)|≲sup1≤l≤k‖∇lyF‖L∞∑∑Nd=1hd=k|∇h1f1∇h2f2⋯∇hNfN|. | (3.5) |
Using Gagliardo-Nirenberg inequality, we have
‖∇h1f1∇h2f2⋯∇hNfN‖L2≤‖∇h1f1‖L2‖∇h2f2‖L∞⋯‖∇hNfN‖L∞≤‖∇h1f‖L2‖∇h2f‖L∞⋯‖∇hNf‖L∞≤‖∇kf‖1−a1L2‖f‖a1L2‖∇kf‖1−a2L2‖f‖a2L2⋯‖∇kf‖1−aNL2‖f‖aNL2. | (3.6) |
where
12−h1n=12a1+(12−kn)(1−a1),−h2n=12a2+(12−kn)(1−a2),⋯,−hNn=12aN+(12−kn)(1−aN),a1+a2+⋯+aN=N−1−n(N−1)2k,(1−a1)+(1−a2)+⋯+(1−aN)=1+n(N−1)2k. | (3.7) |
Moreover, according to fi∈Hk(1≤i≤N,‖∇pf‖L2≤Mp(0≤p≤k)), we have
‖∇k(F(f))‖L2≲‖∇kf‖L2≲N∑i=1‖∇kfi‖L2. |
Lemma 3.6. For any vector function f∈C∞0(R3) and bounded scalar function g, it holds that
|∫R3(∇|x|2γ)⋅fgdx|≲‖g‖L2γ‖f‖L2γ−1. | (3.8) |
Proof. The left side of the above inequality can be rewritten as
|2γ∫R3|x|2γ−2xj∂ixjgfidx|. |
Using Hölder's inequality, we have
|∫R3(∇|x|2γ)⋅fgdx|≲‖g‖L2γ‖f‖L2γ−1. |
Lemma 3.7. (Interpolation inequality with weights) If p,r⩾1,s+n/r,α+n/p,β+n/q>0, and 0⩽θ⩽1, then
‖f‖Lrs≤‖f‖θLpα‖f‖1−θLqβ, |
for f∈C∞0(Rn), where
1r=θp+1−θq, |
and
s=θα+(1−θ)β. |
Especially, when s=p=q=2,θ=γ−1γ,s=γ−1,α=γ,β=0, we have
‖f‖L2γ−1≤‖f‖γ−1γL2γ‖f‖1γL2. | (3.9) |
Proof. We compute
∫U|x|sr|f|rdx=∫U|x|αθr|f|θr|x|β(1−θ)r|f|(1−θ)rdx≤(∫U(|x|αθr|f|θr)pθrdx)θrp(∫U(|x|β(1−θ)r|f|(1−θ)r)q(1−θ)rdx)(1−θ)rq. |
Thus, we complete the proof of Lemma 3.7.
Lemma 3.8. (Gronwall-type Lemma) Let α0>1,α1<1,α2<1, and β1<1,β2<2. Assume that a continuously differential function F:[1,∞)→[0,∞) satisfies
ddtF(t)≤C0t−α0F(t)+C1t−α1F(t)β1+C2t−α2F(t)β2+C3tγ1−1,t≥1F(1)≤K0, | (3.10) |
where C0,C1,C2,C3,K0≥0 and γi=1−αi1−βi>0 for i=1,2. Assume that γ1≥γ2, then there exists a constant C∗ depending on α0,α1,β1,α2,β2,K0,Ci,i=1,2,3, such that
F(t)≤C∗tγ1, |
for all t≥1.
Proof. This is Lemma 2.1 of [21].
We will generalize this lemma into the following one.
Lemma 3.9. Let α0>1,0<αi,βi<1(i=1,2,⋯,n). Assume that a continuously differential function F:[1,∞)→[0,∞) satisfies
ddtF(t)≤C0t−α0F(t)+n∑i=1Cit−αiF(t)βi+n∑i=1¯Citγi−1,t≥1,F(1)≤K0, | (3.11) |
where C0,Ci,¯Ci,K0≥0 and γi=1−αi1−βi>0 for i=1,2,⋯,n. Assume that γ1≥γi(2≤i≤n), then there exists a constant C∗ depending on α0,αi,βi,K0,Ci,¯Ci(i=1,2,⋯,n), such that
F(t)≤C∗tγ1, |
for all t≥1.
Proof. For any t≥1, according to the conditions provided by lemma, it can be concluded that
ddtF(t)≤C0t−α0F(t)+n∑i=1Cit−αiF(t)βi+n∑i=1¯Citγi−1≤C0F(t)+n∑i=1Ci((1−βi)t−αi1−βi+βiF(t))+n∑i=1¯Citγi−1≤CF(t)+n∑i=1Ci(1−βi)t−αi1−βi+n∑i=1¯Citγi−1(C=C(C0,Ci,βi)). |
Multiplying both sides of the equation by eC(t−1) and integrating the resulting equation from 1 to t, we have
F(t)≤eC(t−1)(F(1)+∫t1eC(t−1)n∑i=1(Ci(1−βi)sαi1−βi+¯Cisγi−1)ds)≤eC(t−1)(K0+∫t1n∑i=1(Ci(1−βi)sαi1−βi+¯Cisγi−1)ds)≤eC(t−1)(K0+(1−t)n∑i=1Ci(1−βi)+n∑i=1¯Ciγi(tγi−1−1)). |
Setting t0=(γ12C)−1α0−1, then we have
F(t0)≤eC(t0−1)(K0+(1−t0)n∑i=1Ci(1−βi)+n∑i=1¯Ciγi(tγi−10−1))=K1. |
Choosing
K≥max1≤i≤n{(nCi2βi+2γ−11)11−βi,4n¯Ciγ1,K1}, |
and considering the set R={t≥t0|F(t)≤2Ktγ1}, we clearly have F(t0)≤K1≤K. It's easy for us to know t0∈R. Therefore, R is not empty. Since F(t)−2Ktγ1 is continuous function, if there exists maximal interval [t0,b)⊂R (if the maximal interval does not exist, the proof is completed), then F(b)=2Kbγ1, (F(t)−2Ktγ1)′|t=b≥0. Through the above discussion, we can conclude that
2Kγ1bγ1−1≤F′(b)≤C0b−α0F(b)+n∑i=1Cib−αiF(b)βi+n∑i=1¯Cibγi−1=C02Kbγ1−α0+n∑i=1Cib−αi(2Kbγ1)βi+n∑i=1¯Cibγi−1≤Kγ1bγ1−1(2C0γ−11b1−α0+γ−11n∑i=1Ci2βiKβi−1+(Kγ1)−1n∑i=1¯Ci)≤Kγ1bγ1−1(1+n14n+n14n)=32Kγ1bγ1−1.(contradiction!) |
Lemma 3.10. (Gronwall's inequality of differential form). Let η(⋅) be a nonnegative, absolutely continuous function on [0,T], which satisfies for a.e. t the differential inequality
η′(t)≤ϕ(t)η(t)+ψ(t), |
where ϕ(t) and ψ(t) are nonnegative and summable functions on [0,T]. Then
η(t)≤e∫t0ϕ(s)ds[η(0)+∫t0e−∫s0ϕ(τ)dτψ(s)ds], | (3.12) |
for all 0≤t≤T.
Proof. The proof can be found in [22].
Based on the time decay results of [12] and our hypothesis in Theorem 1.1, it is clear that there exists a large enough T such that for any 0≤k≤N and t>T,
‖∇ku±‖L2≲(1+t)−34−k2‖∇kn±‖L2≲(1+t)−14−k2 | (4.1) |
In the following, we will prove Theorem 1.1. The proof mainly involves four steps.
Step 1: k-order the energy estimates.
By multiplying ∇k(2.6)1, ∇k(2.6)2, ∇k(2.6)3 and ∇k((2.6)4 by |x|2γβ1β2∇kn+, |x|2γ1β2∇ku+, |x|2γβ4β3∇kn− and |x|2γ1β3∇ku− respectively, summing up and then integrating the resultant equation over R3 by parts, we have
12ddt(12‖√β1β2∇kn++√β4β3∇kn−‖2L2γ+σ+β2‖∇∇kn+‖2L2γ+σ−β3‖∇∇kn−‖2L2γ+β14β2‖∇kn+‖2L2γ+β34β4‖∇kn−‖2L2γ+1β2‖∇ku+‖2L2γ+1β3‖∇ku−‖2L2γ)+1β2(ν+1‖∇∇ku+‖2L2γ+ν+2‖∇kdivu+‖2L2γ)+1β3(ν−1‖∇∇ku−‖2L2γ+ν−2‖∇kdivu−‖2L2γ)=⟨∇kF1,|x|2γβ1β2∇kn+⟩+⟨∇kF2,|x|2γ1β2∇ku+⟩+⟨∇kF3,|x|2γβ4β3∇kn−⟩+⟨∇kF4,|x|2γ1β3∇ku−⟩−⟨σ+β1∇kF1,|x|2γ∇kΔn+⟩−⟨σ−β3∇kF3,|x|2γ∇kΔn−⟩−⟨σ−β3∇kF1,∇(|x|2γ)∇∇kn+⟩−⟨σ+β1∇kF3,∇(|x|2γ)∇∇kn−⟩+⟨12∇kF1,|x|2γ∇kn−⟩+⟨12∇kF3,|x|2γ∇kn+⟩+β1β2∫R3∇kn+∇(|x|2γ)∇ku+dx+β4β3∫R3∇kn−∇(|x|2γ)⋅∇ku−dx−ν+1β2∫R3∇(|x|2γ)∇∇ku+∇ku+dx−ν−1β3∫R3∇(|x|2γ)∇∇ku−∇ku−dx−ν+2β2∫R3∇kdivu+∇(|x|2γ)∇ku+dx−ν−2β3∫R3∇kdivu−∇(|x|2γ)∇ku−dx−σ+β2∫R3∇kΔn+∇(|x|2γ)∇ku+dx−σ−β3∫R3∇kΔn−∇(|x|2γ)∇ku−dx+σ+β2∫R3∇kdivu+∇(|x|2γ)∇∇kn+dx+σ−β3∫R3∇kdivu−∇(|x|2γ)∇∇kn−dx+12∫R3∇(|x|2γ)∇ku+∇kn−dx+12∫R3∇(|x|2γ)∇ku−∇kn+dx−12∫R3|x|2γ∇ku+∇∇kn−dx−12∫R3|x|2γ∇ku−∇∇kn+dx:=24∑i=1Ji. | (4.2) |
We set
Ek(t)=12‖√β1β2∇kn++√β4β3∇kn−‖2L2γ+σ+β2‖∇∇kn+‖2L2γ+σ−β2‖∇∇kn−‖2L2γ+β14β2‖∇kn+‖2L2γ+β34β4‖∇kn−‖2L2γ+1β2‖∇ku+‖2L2γ+1β3‖∇ku−‖2L2γ. | (4.3) |
Next, we will discuss the items on the right separately.
Applying Hölder's inequality, we have
|J11|+|J21|≲‖∇kn+∇(|x|2γ)∇ku+‖L1+‖∇kn−∇(|x|2γ)∇ku+‖L1≲‖|x|2γ−1|∇ku+||∇kn+|‖L1+‖|x|2γ−1|∇ku+||∇kn−|‖L1≲(‖∇kn+‖L2γ+‖∇kn−‖L2γ)‖∇ku+‖L2γ−1≲(Ek(t))12‖∇ku+‖L2γ−1≲(Ek(t))12‖∇ku+‖γ−1γL2γ‖∇ku+‖1γL2≲(Ek(t))2γ−12γ(1+t)−1γ(34+k2). | (4.4) |
Similarly, we can obtain
|J12|+|J22|≲(Ek(t))2γ−12γ(1+t)−1γ(34+k2). | (4.5) |
Applying Hölder's inequality and mean value theorem, we have
|J13|+|J14|≤ν+1β2‖∇(|x|2γ)∇∇ku+∇ku+‖L1+ν−1β3‖∇(|x|2γ)∇∇ku−∇ku−‖L1≤2γν+1β2‖|x|2γ−1|∇ku+||∇∇ku+|‖L1+2γν−1β3‖|x|2γ−1|∇ku−||∇∇ku−|‖L1≤2γν+1β2‖∇∇ku+‖L2γ‖∇ku+‖L2γ−1+2γν−1β2‖∇∇ku−‖L2γ‖∇ku−‖L2γ−1≤2γν+1β2‖∇∇ku+‖L2γ‖∇ku+‖γ−1γL2γ‖∇ku+‖1γL2+2γν−1β2‖∇∇ku−‖L2γ‖∇ku−‖γ−1γL2γ‖∇ku−‖1γL2≤ε((2γν+2β2)2+(2γν−2β3)2)(‖∇∇ku+‖2L2γ+‖∇∇ku−‖2L2γ)+1ε(‖∇ku+‖2(γ−1γ)L2γ‖∇ku+‖2γL2+‖∇ku−‖2(γ−1γ)L2γ‖∇ku−‖2γL2)≤ε((2γν+1β2)2+(2γν−1β3)2)(‖∇∇ku+‖2L2γ+‖∇∇ku−‖2L2γ)+1ε(βγ−1γ2+βγ−1γ3)C1(Ek(t))γ−1γ(1+t)−2γ(34+k2). | (4.6) |
Employing similar methods used in estimating J15 and J16, we can get
|J15|+|J16|≤ν+2β2‖∇kdivu+∇(|x|2γ)∇ku+‖L1+ν−2β3‖∇kdivu−∇(|x|2γ)∇ku−‖L1≤ε((2γν+2β2)2+(2γν−2β3)2)(‖div∇ku+‖2L2γ+‖div∇ku−‖2L2γ)+1ε(‖∇ku+‖2(γ−1γ)L2γ‖∇ku+‖2γL2+‖∇ku−‖2(γ−1γ)L2γ‖∇ku−‖2γL2)≤ε((2γν+2β2)2+(2γν−2β3)2)(‖∇kdivu+‖2L2γ+‖∇kdivu−‖2L2γ)+1ε(βγ−1γ2+βγ−1γ3)C2(Ek(t))γ−1γ(1+t)−2γ(34+k2). | (4.7) |
As for J19 and J20, we have
|J19|+|J20|≤σ+β2‖∇kdivu+∇(|x|2γ)∇∇kn+‖L1+σ−β3‖∇kdivu−∇(|x|2γ)∇∇kn−‖L1≤2γσ+β2‖|∇kdivu+||x|2γ−1|∇∇kn+|‖L1+2γσ−β3‖|∇kdivu−||x|2γ−1|∇∇kn−|‖L1≤2γσ+β2‖∇kdivu+‖L2γ‖∇∇kn+‖L2γ−1+2γσ−β3‖∇kdivu−‖L2γ‖∇∇kn−‖L2γ−1≤2γσ+β2‖∇kdivu+‖L2γ‖∇∇kn+‖L2γ‖∇∇kn+‖1γL2+2γσ−β3‖∇kdivu−‖L2γ‖∇∇kn−‖γ−1γL2γ‖∇∇kn−‖1γL2≤ε((2γσ+β2)2+(2γσ−β3)2)(‖∇kdivu+‖2L2γ+‖∇kdivu−‖2L2γ)+1ε(Ek(t))γ−1γ(‖∇∇kn+‖1γL2+‖∇∇kn−‖1γL2)≤ε((2γσ+β2)2+(2γσ−β3)2)(‖∇kdivu+‖2L2γ+‖∇kdivu−‖2L2γ)+3ε((β2σ+)γ−1γ+(β3σ−)γ−1γ)C3(Ek(t))γ−1γ(1+t)−2γ(14+k+12). | (4.8) |
The terms J17 and J18 are more complicated. To begin with, we use integration by parts to get
|J17|+|J18|=σ+β2|∫R3div(∇k∇n+(∇(|x|2γ)∇ku+))dx−∫R3∇k∇n+∇(∇(|x|2γ)∇ku+)dx|+σ−β3|∫R3div(∇k∇n−(∇(|x|2γ)∇ku−))dx−∫R3∇k∇n−∇(∇(|x|2γ)∇ku−)dx|=σ+β2|∫R3∇k∇n+∇(∇(|x|2γ)∇ku+)dx|+σ−β3|∫R3∇k∇n−∇(∇(|x|2γ)∇ku−)dx|≤σ+β2∫R3|∇k∇n+||∇(∇(|x|2γ)∇ku+)|dx+σ−β3∫R3|∇k∇n−||∇(∇(|x|2γ)∇ku−)|dx≤σ+β2∫R3|∇k∇n+||3∑i=1[∇∂i(|x|2γ)∇ku+i+∂i(|x|2γ)∇∇ku+i]|dx+σ−β3∫R3|∇k∇n−||3∑i=1[∇∂i(|x|2γ)∇ku−i+∂i(|x|2γ)∇∇ku−i]|dx≤2γσ+β2∫R3|∇k∇n+|3∑i=1[|∇(|x|2γ−2xi)∇ku+i|+||x|2γ−2xi∇∇ku+i|]dx+2γσ−β3∫R3|∇k∇n+|3∑i=1[|∇(|x|2γ−2xi)∇ku−i|+||x|2γ−2xi∇∇ku−i|]dx≤2γσ+β2∫R3|∇k∇n+|3∑i=1[2√3|x|2γ−2|∇ku+i|+|x|2γ−1|∇∇ku+i|]dx+2γσ−β3∫R3|∇k∇n+|3∑i=1[2√3|x|2γ−2|∇ku−i|+|x|2γ−1|∇∇ku−i|]dx≤12√3γσ+β2‖∇∇kn+‖L2γ‖∇ku+‖L2γ−2+6γσ+β2‖∇∇kn+‖L2γ−1‖∇∇ku+‖L2γ+12√3γσ−β3‖∇∇kn−‖L2γ‖∇ku−‖L2γ−2+6γσ−β3‖∇∇kn−‖L2γ−1‖∇∇ku−‖L2γ≤12√3γσ+β2‖∇∇kn+‖L2γ‖∇ku+‖γ−2γL2γ‖∇ku+‖2γL2+6γσ+β2‖∇∇kn+‖γ−1γL2γ‖∇∇kn+‖1γL2‖∇∇ku+‖L2γ+12√3γσ−β3‖∇∇kn−‖L2γ‖∇ku−‖γ−2γL2γ‖∇ku−‖2γL2+6γσ−β3‖∇∇kn−‖γ−1γL2γ‖∇∇kn−‖1γL2‖∇∇ku−‖L2γ≤12√3γ(√σ+(1β2)1γ+√σ−(1β3)1γ)C4(Ek(t))γ−1γ(1+t)−2γ(34+k2)+1ε((β2σ+)γ−1γ+(β2σ−)γ−1γ)C5(Ek(t))γ−1γ(1+t)−2γ(14+k+12)+ε36γ2((σ+β2)2+(σ−β3)2)(‖∇∇ku+‖2L2γ+‖∇∇ku−‖2L2γ). | (4.9) |
It can be inferred from (1.15) that
|J23|+|J24|≲‖∇ku+‖L2γ‖∇∇kn−‖L2γ+‖∇ku−‖L2γ‖∇∇kn+‖L2γ≲(Ek(t))12. | (4.10) |
By choosing ε small enough, we can obtain
ddtEk(t)+C′(‖∇∇ku+‖2L2γ+‖∇kdivu+‖2L2γ+‖∇∇ku−‖2L2γ+‖∇kdivu−‖2L2γ)≲(Ek(t))2γ−12γ(1+t)−1γ(34+k2)+(Ek(t))γ−1γ(1+t)−2γ(34+k2)+(Ek(t))12+|⟨∇kF2,|x|2γ1β2∇ku+⟩|+|⟨∇kF4,|x|2γ1β3∇ku−⟩|+|⟨σ+β1∇kF1,|x|2γ∇kΔn+⟩|+|⟨σ−β3∇kF3,|x|2γ∇kΔn−⟩|+|⟨σ+β1∇kF1,∇(|x|2γ)∇∇kn+⟩|+|⟨σ−β3∇kF3,∇(|x|2γ)∇∇kn−⟩|+|⟨∇kF1,|x|2γ∇kn+⟩|+|⟨∇kF1,|x|2γ∇kn−⟩|+|⟨∇kF3,|x|2γ∇kn−⟩|+|⟨∇kF3,|x|2γ∇kn+⟩|≲(Ek(t))2γ−12γ(1+t)−1γ(34+k2)+(Ek(t))γ−1γ(1+t)−2γ(34+k2)+(Ek(t))12+8∑i=2,i≠3|Ji|+|J1|+|J9|+|J3|+|J10|. | (4.11) |
From direct observation, it can be inferred that due to good symmetry, we only need to calculate and .
Let's first consider and . Applying Lemma 2.1 and (3.1), we have
(4.12) |
(4.13) |
(4.14) |
Combining the above relations, we can conclude that
(4.15) |
Next, we will calculate . Applying the mean value theorem of binary functions, we have
(4.16) |
where .
Next, we will discuss the above items separately.
(4.17) |
Similarly, we have
(4.18) |
(4.19) |
Applying Lemma 2.1, Lemma 2.2 and Lemma 2.5, we have
(4.20) |
Similarly, we have
(4.21) |
For and , we only need to make simple transformations, and then follow the same process as above to get
(4.22) |
(4.23) |
For the last four items, we use the previous techniques to deal with.
(4.24) |
Like (4.24), by using the same operation, we can obtain
(4.25) |
Combining the relations (4.18)–(4.25), we have
(4.26) |
By relying on good symmetry and combining (4.11)–(4.14) and (4.26), for , we finally conclude that
(4.27) |
Step 2: Proof of Theorem 1.1 with
When , we have
(4.28) |
Multiplying ( is the coefficient of ) on both sides simultaneously, and noticing that is large enough (), we have
(4.29) |
If , then we can apply Lemma 2.8 with any , is the largest of them. An obvious fact is that when is large enough, Thus, we have
(4.30) |
which directly implies (1.14) with .
Step 3: Proof of Theorem 1.1 with
When , we have
(4.31) |
Employing similar arguments used in estimating , we have
(4.32) |
Set ; ; ; ; , and then , , , , . Applying Lemma 2.9, we have
(4.33) |
which directly implies (1.14) with .
Step 4: Proof of Theorem 1.1 with
When , we have
(4.34) |
Similar to the estimates of and , we have
(4.35) |
which together with Lemma 2.9, directly implies (1.14) with .
Therefore, combing the above results in Steps 2–4, we complete the proof of Theorem 1.1.
The authors declare they have not used Artificial Intelligence (AI) tools in the creation of this article.
This work is founded by National Natural Science Foundation of China (12271114), Guangxi Natural Science Foundation (2024GXNSFDA010071, 2019JJG110003), Science and Technology Project of Guangxi (GuikeAD21220114), the Innovation Project of Guangxi Graduate Education (JGY2023061), Center for Applied Mathematics of Guangxi (Guangxi Normal University) and the Key Laboratory of Mathematical Model and Application (Guangxi Normal University), Education Department of Guangxi Zhuang Autonomous Region (G2023KY05102).
The authors declare there is no conflicts of interest.
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