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Global well-posedness and viscosity vanishing limit of a new initial-boundary value problem on two/three-dimensional incompressible Navier-Stokes equations and/or Boussinesq equations

  • The global well-posedness theory and viscosity vanishing limit of the initial-boundary value problem on two/three-dimensional (2D/3D) incompressible Navier-Stokes (NS) equations and/or Boussinesq equations with nonlinear boundary conditions are studied. The global existence of weak solution to the initial boundary value problem for 2D/3D incompressible NS equation with one kind of boundary of pressure-velocity's relation and the global existence and uniqueness of the smooth solution to the corresponding problem in 2D case for large smooth initial data are proven. The viscosity vanishing limit of the corresponding initial-boundary value problem for 2D/3D incompressible NS equations in the bounded domain is also established. And the corresponding results are extended to the 2D/3D incompressible Boussinesq equations.

    Citation: Shu Wang. Global well-posedness and viscosity vanishing limit of a new initial-boundary value problem on two/three-dimensional incompressible Navier-Stokes equations and/or Boussinesq equations[J]. Communications in Analysis and Mechanics, 2025, 17(2): 582-605. doi: 10.3934/cam.2025023

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  • The global well-posedness theory and viscosity vanishing limit of the initial-boundary value problem on two/three-dimensional (2D/3D) incompressible Navier-Stokes (NS) equations and/or Boussinesq equations with nonlinear boundary conditions are studied. The global existence of weak solution to the initial boundary value problem for 2D/3D incompressible NS equation with one kind of boundary of pressure-velocity's relation and the global existence and uniqueness of the smooth solution to the corresponding problem in 2D case for large smooth initial data are proven. The viscosity vanishing limit of the corresponding initial-boundary value problem for 2D/3D incompressible NS equations in the bounded domain is also established. And the corresponding results are extended to the 2D/3D incompressible Boussinesq equations.



    We study the global well-posedness theory and viscosity vanishing limit of the following problem for two/three-dimensional (2D/3D) incompressible Navier-Stokes (NS) equations in the bounded domain

    {vt+vv+p=εΔv+f0,xΩ,t>0, div v=0,xΩ,t>0,εvnpn12|v|2n=k0v+g0,xΩ,t>0,v(0,x)=v0(x),xΩ (1.1)

    and the following problem for 2D/3D incompressible Boussinesq equations in the bounded domain

    {vt+vv+p=εΔv+ϱe+f0,xΩ,t>0, div v=0,xΩ,t>0,ϱt+vϱ=κΔϱ+f1,xΩ,t>0,εvnpn12|v|2n=k0v+g0,xΩ,t>0,κϱn12ϱvn=k1ϱ+g1,xΩ,t>0,(v,ϱ)(0,x)=(v0,ϱ0)(x),xΩ. (1.2)

    Here the functions v=v(t)=v(t,x) and p=p(t)=p(t,x) are the velocity vector and scalar pressure of the fluid respectively, and the function ϱ=ϱ(t)=ϱ(t,x) is the temperature or density, Ω is the smooth bounded domain of Rd,d=2,3 with the boundary Γ=Ω, n is the unit outer normal vector to Γ=Ω, x=(x,xd)Ω with x=x1 when d=2 and x=(x1,x2) when d=3, the vector e denotes the unit one of xd-direct, the constants ε>0 and κ>0 are the viscosity and diffusion coefficients, and k0,k1 are fixed given constants. Also, the functions f0=f0(t)=f0(t,x),f1=f1(t)=f1(t,x),g0=g0(t)=g0(t,x),g1=g1(t)=g1(t,x) are the given forces satisfying  div f0(t)=0 and v0(x),ϱ0(x) are the given initial data satisfying and  div v0=0.

    It is well known that the 3D incompressible NS equations (with homogeneous boundary conditions in the case of the bounded domain) have at least one globally-in-time weak solution having the finite energy [1,2,3,4]. But, the issue of the global regularity and uniqueness for the weak solution in 3D case is still one open problem in the field of mathematical theory of fluid mechanics [5,6,7,8,9,10,11]. For global wellposedness theory on the smooth solution to some special cases for 3D incompressible NS equation, e.g., in the case of the axi-symmetric flow, see, for example, [6,8,12], in the case of the helical flow, see, for example, [13], and in the case of the absence of simple hyperbolic blow-up regimes for the three-dimensional incompressible Euler and quasi-geostrophic equations, see, for example, [14]. In the 2D case, global wellposedness and regularity theory of incompressible NS equation in the whole space and in the bounded domain with the classical boundary conditions, such as the Dirichlet boundary condition and Navier boundary condition are well-known, see [4,6,11]. Recently, in [15], the globally dynamical stabilizing effects of the geometry of the domain at which the flow locates and of the geometry structure of the solution to 3D incompressible NS equations are investigated, and the existence and uniqueness of the global and smooth solution to the Cauchy problem for 3D incompressible NS and Euler equations for a class of the large and smooth initial data in orthogonal curvilinear coordinate systems are also established.

    Also, the two/three-dimensional incompressible Boussinesq equation is an important model in the applied sciences such as atmospheric and oceanographic, see [12,16,17,18]. There are many important progresses on global wellposedness on incompressible Boussinesq equations. For example, for global smooth solution results of two-dimensional incompressible Boussinesq equations with the viscosity or diffusive coefficients in the smooth large initial data, see, for example, [8,19,20,21,22], for global well-posedness in the 3D no-swirl case of axi-symmetric Boussinesq system with partial viscosity or partial thermal diffusivity, see [23,24,25]. Of course, global regularity theory for three-dimensional incompressible Boussinesq equation in the case of general smooth initial data is also still open.

    Physically the boundary condition (1.1)3 represents that the normal stress of the fluid at the boundary is a linear function of the velocity v if we use p+12|v|2 as one new pressure function since we can re-write the equation (1.1)1 into the form tv+curl v×v+div (ϵv+(p+12|v|2)I)=f0. Moreover, the boundary condition (1.1)3 can be equivalently re-written into the following one

    {(εvnk0vg0)×n=0,xΓ,t>0,p=εvnn12|v|2k0vng0n,xΓ,t>0.

    Since, for the known study on wellposedness of initial-boundary value problem for incompressible NS/Euler or Boussinesq equations in one bounded domain with the boundary, the boundary conditions involved there are all given zero velocity Dirichlet/Navier boundary condition etc., so far we do not see any study progress on the problems (1.1) and (1.2) for incompressible fluid. In this paper we establish the global well-poseness theory and viscosity vanishing limit of a class of the initial-boundary value problem for the incompressible fluid with non-homogeneous unknown nonlinear boundary conditions of pressure-velocity relation's type or pressure-velocity coupled with density-velocity relation's type. Our main purpose is that we obtain the existence of the global weak solution to the problems (1.1) and (1.2) in 2D/3D cases and the global existence and uniqueness of the strong and smooth solution for the problems (1.1) and (1.2) in 2D case for any smooth and large initial data. Also, we want to establish the viscosity vanishing limit of the corresponding initial boundary value problem under the suitable assumption on the solution to the initial-boundary value problem for 2D/3D incompressible Euler equation with a class of nonlinear boundary condition.

    Let us introduce basic difficulties and some key points of our success of this paper. Different from the traditional Cauchy problem and initial boundary value problem with homogeneous boundary condition for incompressible fluid, the main difficulty involved here is caused by the more complex nonlinear boundary condition (1.1)3 or (1.2)4,5, which yields to that the vorticity equation can not be used here since the boundary condition of the vorticity is very complex if one use the vorticity equation. More interesting, the boundary condition (1.2)4,5 in the problem for incompressible Boussinesq equation shows that the non-homogeneous boundary condition (1.2)4 for the velocity field shall transfer to the density to get the coupled non-homogeneous density-velocity relation's boundary condition (1.2)5 due to the convection of the fluid. To deal with these complex boundary conditions, motivated by [4], where the global existence and uniqueness of the weak solution for incompressible NS fluid-structure interaction problem are established, we first introduce one new kind of definition of a class of global weak solution to the problem (1.1) or the problem (1.2), which is transformed into one equivalent integration system with nonlinear boundary integration term caused by the non-homogeneous nonlinear boundary condition (1.1)3 or (1.2)4,5. In this definition, the appearance of boundary integration term brings some difficulties when one obtain the more high regularities for the incompressible velocity of the involved approximating system by the Galerkin method. For example, we need to establish one new trace theorem (see Proposition 1, below, in this section) for the incompressible velocity field so as to obtain the uniform estimate of tvl(0) on l=1,2,, see (3.27) in section 3.1. Secondly, we construct the approximating solution by the Galerkin method and establish the uniformly a priori estimates for the approximating solution so as to obtain global weak solution by the compactness argument. In this steps, the key point of our success is to sufficiently use the advantage of the positive viscosity coefficient and the positive diffusion coefficient so as to perform the limit on nonlinear convection term and the boundary integration term in the present definition of global weak solution. Then, we obtain the more better regularity estimates in the boundary and also in the time direction so as to obtain the globally strong and smooth solution to the problem (1.1) and the problem (1.2) in 2D case for smooth and large initial data, where some important interpolation inequality and the trace inequality in two-dimensional case are used and the regularity theory for the boundary value problem for the Stokes equation with Dirichlet boundary condition can be used to conclude our desired results on the strong and smooth solution for smooth initial data satisfying suitable compatibility conditions.

    We also point out that the nonlinear boundary condition (1.1)3 or (1.2)4,5 involved in this paper is in fact related to the free boundary value problem (see, for example, [26,27,28,29,30,31]) or the fluid structure interaction problems (see, for example, [4,32,33,34]) for the incompressible fluid, where the boundary integration terms does not appear in the corresponding definition of the global weak solution because of Lagrangian transform or the cancellation between the fluid and the structure at the interface.

    Now we recall some preliminary knowledge and prove one theorem for the trace of incompressible v velocity field, which will be used in our regularity estimates for the strong and smooth solution for incompressible fluid.

    We recall some standard Hilbert and Sobolev spaces as follows ([6,35,36,37]). The space L2(Ω) is the Hilbert one with the inner product (v,w)Ω and L2(Ω) norm =L2(Ω). The space H1(Ω) and Hs(Ω) are the standard Sobolev one, and Hs(Γ) is the standard fractional order Sobolev's space with Hs(Γ)=(Hs0(Γ)) for s>0. The spaces Ws,q(Ω) and Ws,q(Γ)Lq(Γ) are the Sobolev one. Denote H={vL2(Ω)| div v=0}, V1={wH1(Ω)| div w=0 on Ω}. Also, Y denotes the dual space of the space Y, v=(v1,,vd)Y denotes viX,i=1,,d or vYd=Y××Y, C>0 denotes a positive constant depending only upon the domain but independent of the given time T and C(T)>0 denotes a positive constant depending upon T and the domain.

    Some basic embedding results and basic trace inequalities are needed as follows [6,36,37].

    Lemma 1. If ΩRd is one given smooth bounded domain satisfying Γ=ΩRd1, or Ω=Rd+={xRd|y=(y1,,yd1)Rd1,yd>0} is one half space with one boundary Γ=Ω={xRd|y=(y1,,yd1)Rd1,yd=0}, then the embedding

    Hs1(Ω)Lq1(Ω),1q1=12s1d,0s1<d2,d=1,2,3, (1.3)
    Hs2(Γ)Lq2(Γ),1q2=12s2d1,0s2<d12,d=2,3 (1.4)

    and

    Hs(Ω)Hs12(Γ),s>12,d=2,3 (1.5)

    are continuous. Moreover, there exists one linear continuous right inverse lifting mapping R:wH12(Γ)RwH1(Ω) satisfying

    Rw|Γ=w,RwH1(Ω)CwH12(Γ) (1.6)

    for some positive constant C.

    Also, the following trace inequality is basic:

    Γ|u|2dΓδΩ|u|2dx+CΩ|u|2dx,vH1(Ω) (1.7)

    for any δ>0.

    We also recall one regularity result on the Stokes problem, see [6,11,38].

    Lemma 2. Assume that ΩRd,d=2,3, is one smooth bounded domain with the boundary Γ=ΩCs, s=max{l,2}. Assume that f1Wl2,q(Ω),g1Wl1,q(Ω),Φ1Wl1q,q(Γ), l1,1<q<, and Ωg1dx=ΓΦ1ndΓ, where n is the unit outer normal vector on Γ. Then the solution (u,p) of the Stokes problem

    {Δup=f1,xΩ,divu=g1,xΩ,u=Φ1,xΓ (1.8)

    satisfies the estimate

    uWl,q(Ω)+pWl2,q(Ω)C(f1Wl2,q(Ω)+g1Wl1,q(Ω)+Φ1Wl1q,q(Γ)). (1.9)

    To establish our regularity results on the strong and smooth solution for incompressible fluid, we state one theorem for the trace of incompressible velocity field vH, see, e.g., [39], and outline its proof for completeness, based on the lifting operator technique in Lemma 1. We point out that this result for general function vL2(Ω) does not hold, i.e., the general trace operator T0:L2(Ω)H12(Γ) is not continuous, see [36].

    Proposition 1. The trace operator T:vH(vn)|ΓH12(Γ) is continuous, i.e., there exists a positive constant C=C(Ω) such that, for vH, it holds

    (vn)|ΓH12(Γ)Cv. (1.10)

    Proof. Here, for completeness, we outline the key points of the proof. For fixed vH and for any wH12(Γ)=H120(Γ), we define one functional

    Zv(w)=Ωv(Rw)dx, (1.11)

    where R:H12(Γ)H1(Ω) is any linear and continuous right inverse operator satisfying (1.6).

    We want to verify that, for any given vH, Zv(w) is one bounded linear functional defined on H12(Γ). Firstly, the operator Zv(w) is independent of the operator R. In fact, for any two operators R1,R2 satisfying the property (1.6), we have (R1wR2w)|Γ=ww=0 on Γ, and, hence

    Ωv(R1wR2w)dx=Γ(vn)(R1wR2w)dΓΩ div v(R1wR2w)dx=0 (1.12)

    due to  div v=0. Thus, we get from (1.12) that the functional Zv(w) is independent of the operator R and is determined uniquely by functions vH and wH12(Γ).

    Secondly, the operator Zv(w) is bounded on H12(Γ). In fact, by the definition (1.11), for any wH12(Γ),

    |Zv(w)|=|Ωv(Rw)dx|v(Rw)vRwH1(Ω)CvwH12(Γ). (1.13)

    Also, it is obvious that Zv(w) is linear with respect to wH12(Γ).

    Lastly, by Riesz-Frechet representation theorem, there exists γ0vH12(Γ) such that

    Zv(w)=(γ0v,w)H12(Γ),H12(Γ),γ0vH12(Γ)ZvH12(Γ)Cv (1.14)

    by using (1.13). On the other hand, for vC1(¯Ω), div v=0, and wC1(¯Ω), we have

    Zv(w)=ΩvR(w|Γ)dx=Ωvwdx=ΓvnwdΓ=(vn,w)H12(Γ),H12(Γ). (1.15)

    Combining (1.14) and (1.15), we get γ0v=(vn)|Γ and (1.10).

    This completes the proof of Proposition 1.

    Finally, we recall one result on the functional analysis ([40]).

    Lemma 3. Let the space Y be a separable Hilbert space, then the space Y has a complete orthogonal system consisting of an at most countable number of elements.

    The rest of this paper is organized as follows. In section 2 we state the main results of this paper, and section 3 gives the proofs of our main results.

    Introduce

    aΩ(v,w)=(v,w)Ω,bΩ(v,w,z)=(vw,z)Ω.

    Definition 1. (The definition of the global weak solution to the problem (1.1)) (v,p) is called to be one global weak solution to the problem (1.1) in time, if for any given positive T, there exists (v,p), defined in the interval [0,T], satisfying vL(0,T;L2(Ω))L2(0,T;V1), tvLs(0,T;V1) for some constant s>1,

    (tv(t),w)Ω+εaΩ(v(t),w)+bΩ(v(t),v(t),w)12Γ|v(t)|2nwdΓ=Γ(k0v(t)+g0(t))wdΓ+(f0(t),w)Ω,wV1,0tT,v(0)=v0(x) in V1. (2.1)

    Here dΓ is the arc differential for d=2 and the area differential for d=3.

    Remark 1. We point out that if v is one global weak solution of the system (2.1) in the definition 1, then there exists a function p such that (v,p) satisfies the system (1.1) almost everywhere for (x,t) in some sense. In fact, setting wC0(Ω) with  div w=0 in (2.1) and denoting

    tvϵΔv+vvf0=S,

    we have S((D(Ω×(0,T))))d and (S,w)=0 in the sense of (D((0,T))) for any wC0(Ω) with  div w=0, which gives S=p and the equation (1.1)1 for some p((D(Ω×(0,T))))d. Also, (1.1)2 is obvious because vV1. Thus, the rest one is to verify the nonlinear boundary condition (1.1)3. Multiplying (1.1)1 by wV1 and using integration by parts, we get

    (tv,w)Ω+εaΩ(v,w)+bΩ(v,v,w)+Γ(pnεvn)wdΓ=(f0,w)Ω,wV1,0tT,

    which, together with (2.1), gives

    Γ(εvnpn12|v|2n(k0v+g0))wdΓ=0,vV1,0tT. (2.2)

    Then (1.1)3 is obtained from (2.2) and ΓnwdΓ=0 because wV1 is arbitrary.

    Now we state the main results of this paper on the problem (1.1) for 2D/3D incompressible NS equations.

    Theorem 1. Assume that the domain ΩRd,d=2,3, is smooth and bounded. Also, assume that Γ=ΩC1, v0H, f0L2(0,;L2(Ω)) and g0L2(0,;H1(Ω)). Then the problem (1.1) has one global weak solution in time satisfying the the following energy inequality

    v(t)2+2εt0v(t)2dtv02+2t0{Γ(k0v(t)+g0(t))v(t)dΓ+Ωf0(t)v(t))dx}dt (2.3)

    for any 0tT and any given T>0. Moreover, the global weak solution to the problem (1.1) is unique when d=2.

    Theorem 2. Set d=2 and let Ω=T×[1,1], where T=R2πZ is a torus. Assume that f0H1(0,;L2(Ω))L2(0,;H1(Ω)) and g0H1(0,;H1(Ω))L2(0,;H2(Ω)). Also, assume that v0V1H2(Ω) satisfies the following zero order compatibility condition

    εv0nk0v0g0(0)=((εv0nk0v0g0(0))n)n,xΩ=T×{1,1}, (2.4)

    where n is the unit outer normal vector of the boundary Γ=Ω. Then the problem (1.1) has one unique and global strong solution in time satisfying

    tvL(0,T;L2(Ω))L2(0,T;V1)vL(0,T;V1)L2(0,T;H2(Ω)). (2.5)

    Furthermore, if f0,g0,v0 are smooth and satisfy suitable higher order compatibility conditions at the boundary, then the global-in-time weak solution to the problem (1.1) is smooth.

    Theorem 3. (Viscosity vanishing limit ε0) Let f0(t),g00(t),v00 be the given smooth functions and the assumptions in Theorem 1 hlod. Let (v,p)=(v,p)(t) be one solution to the problem (1.1) given by Theorem 1, and let (v0,p0)=(v0,p0)(t) defined on the interval [0,T] be one smooth solution to the initial boundary value problem for 2D/3D incompressible Euler equations

    {v0t+v0v0+p0=f0,xΩ,0tT,divv0=0,xΩ,0tT,p0=12|v0|2g00,xΩ,0tT,v(0,x)=v00(x),xΩ. (2.6)

    Also, assume that g0(t)=k0v0(t)+g00(t)n and v0v000 as ε0. Then there exists one positive constant K=K(T) such that, for any k0K, it holds that v(t)v0(t)0 for 0tT when ε0.

    Remark 2. If we replace the boundary condition (1.1)3 by the boundary condition

    εvnpn12(vn)v=k0v+g0,xΩ,0tT,

    then similar results to Theorem 1 and Theorem 2 hold. The similar results to Theorem 2 in the case of the general bounded smooth domain is also true, which will be discussed in another paper.

    Remark 3. If f0=0,g00=0, and the smooth function v000 satisfies div v00=0,curl v00=0, (v00n)|Γ0, and (v00×n)|Γ0, then (v0,p0)(t,x)=(v00,12|v00|2g00)(x) is one smooth solution to the system (2.6). But, the well-posedness locally in time for the incompressible Euler system (2.6) with general initial data is open and will be remained to the future.

    Definition 2. (The definition of the global weak solution to the problem (1.2)) (v,p,ϱ) is called to be one global-in-time weak solution to the problem (1.2), if for any T>0, there exists the function pairs (v,p,ϱ), defined in the time interval [0,T], satisfying vL(0,T;L2(Ω))L2(0,T;V1) and ϱL(0,T;L2(Ω))L2(0,T;H1(Ω)) with tvLs(0,T;V1) and tϱLs(0,T;(H1(Ω))) for some s>1 satisfying

    (tv(t),w)Ω+εaΩ(v(t),w)+bΩ(v(t),v(t),w)12Γ|v(t)|2nwdΓ=Γ(k0v(t)+g0(t))wdΓ+(ϱ(t)e+f0(t),w)Ω,wV1,0tT,(tϱ(t),ψ)+κaΩ(ϱ(t),ψ)+bΩ(v(t),ϱ(t),ψ)12Γ(ϱ(t)v(t)n)ψdΓ=Γ(k1ϱ(t)+g1(t))ψdΓ+(f1(t),ψ)Ω,ψH1(Ω),0tT,(v(0),ϱ(0))=(v0(x),ϱ0(x)) in V1×(H1(Ω)). (2.7)

    Here dΓ is the arc differential for d=2 and the area differential for d=3.

    For the incompressible Boussinesq system, we state our results as follows.

    Theorem 4. Assume that the domain ΩRd,d=2,3, is bounded and smooth. Also, assume that the boundary Γ=ΩC1, v0H, ϱ0L2(Ω), f0,f1L2(0,;L2(Ω)) and g0,g1L2(0,;H1(Ω)). Then the problem (1.2) has one globally-in-time weak solution satisfying the following energy inequality

    (ϱ(t),v(t))2+2εt0v(t)2dt+2κt0ϱ(t)2dt2t0{Γ(k0v(t)+g0(t))v(t)dΓ+Ω(ϱ(t)e+f0(t))v(t)dx}dt+2t0{Γ(k1ϱ(t)+g1(t))ϱ(t)dΓ+Ωf1(t)ϱ(t))dx}dt,0tT (2.8)

    for any given positive T. Moreover, the global-in-time weak solution to the problem (1.2) is unique when d=2.

    Theorem 5. Let Ω=T×[1,1], where T=R2πZ is the torus. Assume that f0,f1H1(0,;L2(Ω))L2(0,;H1(Ω)) and g0,g1H1(0,;H1(Ω))L2(0,;H2(Ω)). Also, assume that v0V1H2(Ω), ϱ0H2(Ω) satisfies the following zero order compatibility condition

    εv0nk0v0g0(0)=((εv0nk0v0g0(0))n)n,xΩ=T×{1,1},κϱ0n12ϱ0v0n=k1ϱ0+g1(0),xΩ=T×{1,1}, (2.9)

    where n is the unit outer normal vector of Γ=Ω. Then the problem (1.2) has one unique and globally-in-time strong solution satisfying

    tvL(0,T;L2(Ω))L2(0,T;V1),tϱL(0,T;L2(Ω))L2(0,T;H1(Ω)),vL(0,T;V1)L2(0,T;H2(Ω)),ϱL(0,T;H1(Ω))L2(0,T;H2(Ω)). (2.10)

    Furthermore, if the functions f0,f1,g0,g1,ϱ0,v0 are smooth and satisfy suitable higher order compatibility conditions at the boundary, then the globally-in-time weak solution to the problem (1.2) is smooth.

    Remark 4. We can not extend the present results in Theorems 3 and 4 to the case of κ=0 and ε>0 or the case of κ>0 and ε=0 because the corresponding initial boundary value problem for the inviscid or/and non-diffusive case for incompressible Boussinesq fluid in the bounded domain is different from (1.2). The discussion of these cases will be considered in the future.

    The proof of Theorem 1.

    We prove Theorem 1 by constructing the approximating solution based on the Galerkin method and by establishing the a priori estimates. Let T>0 be an arbitrarily given positive constant. By using Lemma 3, there exists the orthogonal basis {el(x)}l=1 of the space V1L2(Ω) with el=1. For any given l1, we take the following l order approximating solution vl of the solution v to the problem (1.1):

    vl(t)=vl(t,x)=mj=1glj(t)ej(x),l=1,2,, (3.1)

    where glj(t),j=1,,l, satisfies the system

    (tvl(t),ej)Ω+εaΩ(vl(t),ej)+bΩ(vl(t),vl(t),ej)12Γ|vl(t)|2nejdΓ=Γ(k0vl(t)+g0(t))ejdΓ+(f0(t),ej)Ω,j=1,2,,l,0tT,vl(0)=vl0(x)=lk=1vk0ek(x),v0(x)=l=1vl0el(x)V1, (3.2)

    which is one ordinary differential equations(ODE) for glj(t),j=1,,l. By standard existence and uniqueness theory for the ODE, there exist some tl>0 and one unique solution {glj(t)}lj=1 of the system (3.2), defined in the interval [0,tl] and satisfying glj(t)H1(0,tl) and limttl|glj(t)|= for some j:1jl. Now we want to prove tl=T.

    Step 1: The a priori L(0,T;L2(Ω)) estimate for vl(t,x).

    Multiplying (3.2) by glj(t) and summing the resulting one from j=1 to l, by using the fact that

    bΩ(vl,vl,vl)12Γ|vl|2nvldΓ=0

    according to  div vl=0, we have

    12ddtvl(t)2+εvl(t)2=Γ(k0vl(t)+g0)vl(t)dΓ+(f0(t),vl(t))Ω,0tT. (3.3)

    With the help of the Young's inequality and the trace inequality (1.7), it follows from (3.3) that

    12ddtvl(t)2+ε2vl(t)2Cvl(t)2+Γ|g0(t)|2dΓ+f0(t)2,0tT. (3.4)

    By the Gronwall's inequality, we get from (3.4) that

    vl(t)2+εt0vl(t)2dteCt(v02+t0Γ|g0(t)|2dΓdt+t0f0(t)2dt)C(T)<,0tT, (3.5)

    which implies that there exists a constant C(T)>0 such that |glj(t)|C(T)< for any l1, j=1,2,,l, and 0ttl, and, hence, tl=T.

    Step 2: The a priori Lq(0,T;V1) estimate for tvl(t,x).

    For j=1,2,,l, noting the fact that vl|Γ0 and ej|Γ0, and by integration by parts, with the help of the Holder inequality and the trace inequalities (1.7), from (3.2) we get

    |(tvl(t),ej)Ω|=|εaΩ(vl(t),ej)+bΩ(vl(t),ej,vl(t))Γ(vl(t)n)(vl(t)ej)dΓ12Γ|vl(t)|2nejdΓ+Γ(k0vl(t)+g0(t))ejdΓ+(f0(t),ej)Ω|εvl(t)ej+|bΩ(vl(t),ej,vl(t))|+|Γ(vl(t)n)(vl(t)ej)dΓ12Γ|vl(t)|2nejdΓ|+C(vl(t)H1(Ω)+g0(t)H1(Ω))ejH1(Ω)+f0(t)ej. (3.6)

    To estimate the remaining integral terms in (3.6), we need the following basic inequality.

    For any wH1(Ω), one have

    wL4(Ω)Cwd4H1(Ω)w1d4,d=2,3 (3.7)

    and

    w|ΓL3(Γ)Cwd+26H1(Ω)w1d+26,d=2,3. (3.8)

    In fact, on one hand, by taking q1=4 in (1.3) of Lemma 1 and using 14=12s1d, we get Hd4(Ω)L4(Ω), which, together with the interpolation inequality ([36,41]), yields

    wL4(Ω)CwHd4(Ω)Cwd4H1(Ω)w1d4,d=2,3,

    which gives (3.7).

    On the other hand, by taking q2=3 in (1.4) of Lemma 1 and using 13=12s2d1, we get Hd16(Γ)L3(Γ), which, together with the trace inequality (1.5) and the interpolation inequality ([36,41]), yields to

    w|ΓL3(Γ)Cw|ΓHd16(Γ)CwHd16+12(Ω)Cwd+26H1(Ω)w1d+26,d=2,3,

    which gives (3.8).

    By the Holder's inequality, using the fact that nLC< thanking to the smoothness of the domain, (3.7), (3.8), and the estimate (3.5), we get, for j=1,2,,l,

    |bΩ(vl(t),ej,vl(t))|+|Γ(vl(t)n)(vl(t)ej)dΓ12Γ|vl(t)|2nejdΓ|vl(t)2L4(Ω)ej+Cvl(t)|Γ2L3(Γ)ejL3(Γ)C(T)(vl(t)d2H1(Ω)+vl(t)d+23H1(Ω))ejH1(Ω) (3.9)

    Combining (3.6) and (3.9), using the Young's inequality and the fact that 1d2d+23<2 for d=2,3, we get, 0tT,

    |(tvl(t),ej)Ω|C(T)(vl(t)H1(Ω)+vl(t)d2H1(Ω)+vl(t)d+23H1(Ω)+g0(t)H1(Ω)+f0(t))ejH1(Ω)C(T)(vl(t)d+23H1(Ω)+g0(t)H1(Ω)+f0(t)+C)ejH1(Ω),

    which gives

    tvl(t)V1C(T)(vl(t)d+23H1(Ω)+g0(t)H1(Ω)+f0(t)+C),t[0,T], (3.10)

    which implies that there exists a constant C(T)>0 such that

    tvl(t)Lq(0,T;V1)C(T)<,1q=6d+22,d=2,3. (3.11)

    Combining the estimates (3.5) and (3.11), and using Lions-Aubin lemma, we have that there exists a subsequence, denoted still by vl(t), satisfying, when l,

    tvl(t)tv(t) weakly in Lq(0,T;V1), (3.12)
    vl(t)v(t) weakly* in L(0,T;L2(Ω)), (3.13)
    vl(t)v(t) weakly in L2(0,T;H1(Ω)), (3.14)
    vl(t)v(t) strongly in L2(0,T;H1δ(Ω)),δ(0,12). (3.15)

    Furthermore, according to the fact that the embedding H1δ(Ω)L2(Γ) is compact when δ<12, we have

    vl(t)|Γv(t)|Γ strongly in L2(0,T;L2(Γ)). (3.16)

    Then, it follows from (3.15) and (3.16) that, for any wV1, when l,

    bΩ(vl(t),vl(t),w)bΩ(v(t),v(t),w) strongly in L1(0,T), (3.17)

    and

    12Γ|vl(t)|2nwdΓ12Γ|v(t)|2nwdΓ strongly in L1(0,T). (3.18)

    Now, performing one limit as l in the equation (3.2) and using (3.12)–(3.18), we know that v satisfies the equation (2.1), and, then, integrating the equation (3.3) over [0,T] with respect to the time t, setting l in the resulting one and using the convergence (3.15) and (3.16), we obtain the desired energy inequality (2.3) and the estimate

    v(t)2+t0v(t)2dtC(T)<,0tT. (3.19)

    Step 3: Uniqueness of the solution when d=2

    Let v,v be any two solution to the problem (1.1). Then ω=vv satisfies

    (tω,w)Ω+εaΩ(ω,w)+bΩ(v,ω,w)+bΩ(ω,v,w)bΩ(ω,ω,w)+12Γ|ω|2nwdΓΓ(ωw)(nw)dΓ=Γk0ωwdΓ,wV1,0tT,ω(0)=0. (3.20)

    Taking w=ω(t) in the system (3.20), one have

    12ddtω(t)2+εω(t)2=bΩ(ω(t),v(t),ω(t))12Γ|ω(t)|2nv(t)dΓ+Γ(ω(t)v(t))(nω(t))dΓ+k0Γ|ω(t)|2dΓ. (3.21)

    By the inequalities (3.7) and (3.8) with d=2, and using (3.19), the Holder inequality, the Young inequality, and the trace inequality (1.7), we get

    bΩ(ω(t),v(t),ω(t))|Ω(ω(t))v(t)ω(t)dx|v(t)ω(t)2L4(Ω)Cv(t)ω(t)H1(Ω)ω(t)ε8ω(t)2+Cω(t)2+Cv(t)2ω(t)2, (3.22)
    12Γ|ω(t)|2nv(t)dΓ+Γ(ω(t)v(t))(nω(t))dΓCΓ|ω(t)|2|v(t)|dΓCv(t)|ΓL3(Γ)ω(t)|Γ2L3(Γ)Cv(t)23H1(Ω)v(t)13ω(t)43H1(Ω)ω(t)23ε8ω(t)2+Cω(t)2+Cv(t)2H1(Ω)v(t)ω(t)2ε8ω(t)2+C(T)ω(t)2+C(T)v(t)2ω(t)2 (3.23)

    and

    k0Γ|ω(t)|2dΓε8ω(t)2+Cω(t)2. (3.24)

    Combining (3.21) with (3.22)–(3.24), we get

    ddtω(t)2+εω(t)2Cω(t)2+Cv(t)2ω(t)2. (3.25)

    Using Gronwall's inequality to (3.25), ω(0)=0, and the inequality T0v(t)2dtC(T)< which is given by (3.19), we obtain ω=0 for 0tT. This gives the uniqueness result on the global weak solution in the case of d=2.

    The proof of Theorem 1 is complete.

    The proof of Theorem 2

    Based on the proof of Theorem 1, so as to get Theorem 2, we need establish the more regularity estimates on the approximating solution vl(t) given by the system (3.1)-(3.2) when d=2. To do this, we choose the functions ej(x),j=1,2,, to be the orthogonal basic function sequence of V1H2(Ω) and choose initial data vl0(x)=vl(0)=vl(t=0,x) satisfying

    vl0v0 strongly in V1H2(Ω)

    and

    εvl0nk0vl0g0(0)=((εvl0nk0vl0g0(0))n)n,xΩ=T×{1,1}. (3.26)

    Step 1: The estimate for the initial data tvl(0) uniformly on l.

    We obtain one estimate for tvl(0) uniformly on l by using the trace theorem given by Proposition 1. By (3.2), using integration by parts, (3.26), the Holder's inequality, and the trace property (1.10) in Proposition 1 thanking to the fact that  div tvl(0)=0 due to (3.1) and ejV1, we get

    tvl(0)2=εΩΔvl(0)tvl(0)dxεΓvl(0)ntvl(0)dΓbΩ(vl(0),vl(0),tvl(0))+12Γ|vl(0)|2ntvl(0)dΓ+Γ(k0vl(0)+g0(0))tvl(0)dΓ+(f0(0),tvl(0))Ω=εΩΔvl(0)tvl(0)dxbΩ(vl(0),vl(0),tvl(0))Γ((εvl(0)nk0vl(0)g0(0))n)(ntvl(0))dΓ+12Γ|vl(0)|2ntvl(0)dΓ+(f0(0),tvl(0))ΩεΔvl(0)tvl(0)+vl(0)vl(0)tvl(0)+C(vl(0)n,vl(0),g0(0))|ΓH12(Γ)(ntvl(0))|ΓH12(Γ)+C|vl(0)|Γ|2H12(Γ)(ntvl(0))|ΓH12(Γ)+f0(0)tvl(0)CΔvl(0)tvl(0)+Cvl(0)2H2(Ω)tvl(0)+C(vl(0)nH1(Ω)+vl(0)H1(Ω)+g0(0))H1(Ω))tvl(0)+Cvl(0)2H2(Ω)tvl(0)+f0(0)tvl(0)C(v0H2(Ω)+v02H2(Ω)+g0(0)H1(Ω)+f0(0))tvl(0), (3.27)

    which gives

    tvl(t)|t=0C(v0H2(Ω)+v02H2(Ω)+g0(0)H1(Ω)+f0(0))C<, (3.28)

    which means that, by using (1.1)1 at t=0,

    tv|t=0=εΔv0v0v0p(t=0)+f0(0)C<,

    where the pressure function p(t=0)=p0=p(0,x) is solved by

    {Δp0= div (v0v0f0(0)),xΩ,p0=εv0nn12|v0|2k0v0ng0(0)n,xΓ=Ω. (3.29)

    Thus, by comparing the condition (3.29)2 and the boundary condition (1.1)3, we know that the compatibility condition (2.4) is necessary.

    Step 2: L(0,T;L2(Ω)) estimate for tvl(t) and vl(t).

    Differentiating the system (3.2) with respect to t, multiplying the resulting one by tglj(t) and summing from j=1 to l, by noticing that

    bΩ(vl,tvl,tvl)=12Γ|tvl|2nvl(t)dΓ

    due to  div vl=0, we get

    12ddttvl(t)2+εtvl(t)2=bΩ(tvl,vl,tvl)Γ[12|tvl|2nvl(t)(vltvl(t))(ntvl(t))]dΓ+Γ(k0tvl(t)+tg0(t))tvl(t)dΓ+(tf0(t),tvl(t))Ω,0tT. (3.30)

    Now we estimate each term in the right hand side of (3.30).

    Using the inequality (3.7) with d=2, with the help of the Holder's inequality and the Young's inequality, we get

    |bΩ(tvl,vl,tvl)|vl(t)tvl(t)2L4(Ω)Cvl(t)tvl(t)H1(Ω)tvl(t)ε8tvl(t)2+Ctvl(t)2+Cvl(t)2tvl(t)2. (3.31)

    Using the fact that n is one constant unit vector, the inequality (3.8) with d=2 and the estimate (3.5), with the help of the Holder's inequality and the Young's inequality, we get

    Γ[12|tvl|2|nvl(t)|+|(vltvl(t))(ntvl(t))|]dΓC(Γ|tvl|3dΓ)23(Γ|vl(t)|3dΓ)13Ctvl(t)43H1(Ω)tvl(t)23vl(t)23H1(Ω)vl(t)13ε8tvl(t)2+ε8tvl(t)2+Ctvl(t)2vl(t)2H1(Ω)vl(t)ε8tvl(t)2+C(T)tvl(t)2+C(T)vl(t)2tvl(t)2. (3.32)

    Also, by the trace inequality (1.7), with the help of the Holder's inequality and the Young's inequality, we get

    Γ(k0tvl(t)+tg0(t))tvl(t)dΓ+(tf0(t),tvl(t))Ωε8tvl(t)2+Ctvl(t)2+Ctg0(t)2H1(Ω)+Ctf0(t)2. (3.33)

    Combining (3.30) and (3.31)–(3.33), we get

    ddttvl(t)2+εtvl(t)2C(T)tvl(t)2+C(T)vl(t)2tvl(t)2+Ctg0(t)2H1(Ω)+Ctf0(t)2,0tT. (3.34)

    Using the estimate (3.5), with the help of Gronwall's inequality, from (3.34), we get

    tvl(t)2eC(T)t0(1+vl(t)2)dt(tvl(0)2+Ct0(tg0(t)2H1(Ω)+tf0(t)2)dt)C(T)<,0tT. (3.35)

    Combining (3.34) and (3.35), we have

    tvl(t)2+εt0tvl(t)2dtC+Ct0tg0(t)2H1(Ω)dt+t0tf0(t)2dtC(T)<,0tT. (3.36)

    Hence, by (3.4), using (3.5), (3.36), the assumptions on f0(t) and g0(t), and the Holder's inequality, we get

    vl(t)C(T)<,0tT. (3.37)

    Step 3: One L(0,T;L2(Ω)) estimate for x1vl(t) and the boundary regularity estimate for vl(t) at x2=0

    In this case, we deal with the special domain Ω=T×[1,1], where T=R2πZ. Thus, Γ={x=(x1,x2)R2|x1T,x2=1, and x2=1}, dΓ=ds=dx1 on xΓ, n={0,1} in the line x2=1 and n={0,1} in the line x2=1.

    Denote Dhvl(t)=Dhvl(t,x)=vl(t,x1+h,x2)vl(t,x1,x2)h and

    DhDhvl(t,x)=vl(t,x1+h,x2)2vl(t,x1,x2)+vl(t,x1h,x2)h2.

    Using (w,DhDhw)=(Dhw,Dhw), Dh(fg)=hDhfDhg+gDhf+fDhg and noting that Dhn=0 for one constant vector n, we get

    bΩ(vl,vl,DhDhvl)12Γ|vl|2nDhDhvmdΓ=hbΩ(Dhvl,Dhvl,Dhvl)+bΩ(Dhvl,vl,Dhvl)+bΩ(vl,Dhvl,Dhvl)12Γ(h|Dhvl|2+2vlDhvl)nDhvldΓ=bΩ(Dhvl,vl,Dhvl)+12Γ|Dhvl|2(nvl)dΓΓ(vlDhvl)(nDhvl)dΓ. (3.38)

    Taking w=DhDhvl(t,x) in the equation (3.2) and using (w,DhDhw(t,x))=(Dhw,Dhw) and (3.38), we get

    12ddtDhvl(t)2+εDhvl(t)2=bΩ(Dhvl(t),vl(t),Dhvl(t))12Γ|Dvul(t)|2(nvl(t))dΓ+Γ(vl(t)Dhvl(t))(nDhvl(t))dΓ+Γ(k0Dhvl(t)+Dhg0(t))Dhvl(t)dΓ+(Dhf0(t),Dhvl(t))Ω,0tT,Dhvl(0)x1v0(x)C<, as h0. (3.39)

    Because each term in the right hand side of (3.39) is completely similar to that of (3.30), as one deal with each term in the right hand side of (3.30) in Step 2, by (3.39), with the help of (3.37) and the fact that DhvlCx1vl for small h, we can get

    ddtDhvl(t)2+εDhvl(t)2C(T)Dhvl(t)2+C(T)vl(t)2Dhvl(t)2+CDhg0(t)2H1(Ω)+CDhf0(t)2,0tT. (3.40)

    By (3.40) and using the estimate (3.19), the fact that DhwCx1w, and the assumptions on f0(t) and g0(t), with the help of Gronwall's inequality, we have

    Dhvl(t)2+εt0Dhvl(t)2dtC(T)<,0tT,

    which gives

    x1vl(t)2+εt0x1vl(t)2dtC(T)<,t[0,T]. (3.41)

    This means that x1vl(t,x)L2(0,T;H1(Ω)). Using the trace's theorem, we have x1vl(t)|Γ=x1vl(t,x1,x2=±1)L2(0,T;H12(Γ)), i.e., vl(t)|ΓL2(0,T;H32(Γ)) and

    vl(t)|ΓL2(0,T;H32(Γ))C(T)<. (3.42)

    Also, it is easy to verify that vl(t) satisfies the following incompressible NS system in the weak sense

    {εΔvl+pl=f0tvlvlvl,xΩ,t[0,T], div vl=0,xΩ,t[0,T],vl=vl,xΓ,t[0,T]. (3.43)

    Applying Lemma 2 to (3.43), and using the Holder inequality and the Sobolev's embedding inequality H1(Ω)L4(Ω), with the help of the Gagliardo-Nirenberg's inequality DvlL4(Ω)C1vl14D2vl34+C2vl for some constants C1>0 and C2>0 (see [41]), we have

    vl(t)H2(Ω)+pl(t)L2(Ω)C((f0tvlvlvl)+vl(t)|ΓH32(Γ))C(f0(t)+tvl(t)+vl(t)L4(Ω)vl(t)L4(Ω)+vl(t)|ΓH32(Γ))C(f0(t)+tvl(t)+vl(t)H1(Ω)vl(t)14vl(t)34H2(Ω)+vl(t)|ΓH32(Γ)). (3.44)

    Using f0(t)L2(0,T;L2(Ω)) and the estimates (3.5), (3.36), (3.37), and (3.42), with the help of the Young's inequality, from (3.44) we obtain

    t0vl(t)2H2(Ω)dtCt0(f(t)2+tvl(t)2+vl(t)8H1(Ω)vl(t)2)dt+Ct0vl(t)|Γ2H32(Γ)dtC(T)<,0tT. (3.45)

    Now, performing the limit l in (3.36), (3.37), and (3.45), we obtain the estimate (2.5).

    Similarly, if the initial data is smooth and satisfies the more high compatibility conditions, the more high regularity stated in Theorem 2 can also be established.

    The proof of Theorem 2 is complete.

    We use so-called modified energy method. Firstly, it is easy to know that the strong solution to the system (2.6) satisfies

    (tv0(t),w)Ω+bΩ(v0(t),v0(t),w)12Γ|v0(t)|2nwdΓ=Γ(g00(t)nwdΓ+(f0(t),w)Ω,wV1,0tT,v0(0)=v00(x). (3.46)

    Now take w=v(t)v0(t)V1 in (3.46), we have

    (tv0(t),v(t)v0(t))Ω+bΩ(v0(t),v0(t),v(t)v0(t))12Γ|v0(t)|2n(v(t)v0(t))dΓ=Γ(g00(t)n(v(t)v0(t))dΓ+(f0(t),v(t)v0(t))Ω,0tT. (3.47)

    Then, combining (2.1) with v=v(t)v0(t)V1 and (3.47), by the direct computation, and noting that g0(t)=k0v0(t)+g00(t)n and the smoothness of v0(t), we get

    12ddtΩ|v(t)v0(t)|2dx=Ω(tv(t)tv0(t))(v(t)v0(t))dx=εΩ1v(t)(v(t)v0(t))dxΩ1(v(t)v(t)v0(t)v0(t))(v(t)v0(t))dx+12Γ1(|v(t)|2|v0(t)|2)n(v(t)v0(t))dΓ1+Γ1[k0(v(t)v0(t))(v(t)v0(t))dΓ1=εΩ|(v(t)v0(t))|2dxεΩv0(t)(v(t)v0(t))dxΩ((v(t)v0(t)))v0(t)(v(t)v0(t))dx+k0Γ|v(t)v0(t)|2dΓ12Γ|v(t)v0(t)|2(v0(t)n)dΓ+Γ((v(t)v0(t))v0(t))((v(t)v0(t))n)dΓεΩ|(v(t)v0(t))|2dxεΩv0(t)(v(t)v0(t))dxΩ((v(t)v0(t)))v0(t)(v(t)v0(t))dx+(k0+C(T))Γ|v(t)v0(t)|2dΓ (3.48)

    for some C(T)>0. Thus, there exists a constant K=K(T)>0 such that k0+C(T)0 for any k0K and it holds, by using (3.48) and the Young's inequality, that

    12ddtΩ|v(t)v0(t)|2dxε2Ω|(v(t)v0(t))|2dx+v0(t,x)L([0,T]×Ω)Ω|v(t)v0(t)|2dx+ε2Ω|v0(t)|2dx

    which yields to the desired convergence estimate

    Ω|v(t)v0(t)|2dxC(T)Ω|v0v00|2dx+C(T)ε0 as ε0.

    Here C(T)>0 is one constant depending upon v0(t).

    The proof of Theorem 3 is complete.

    The proof of Theorem 4 are similar to that of Theorem 1 by constructing the following approximating solution (vl,ϱl) of (v,ϱ) to the system (1.2):

    vl(t)=vl(t,x)=mi=1gli(t)ei(x),l=1,2,,ϱl(t)=ϱl(t,x)=li=1ϱli(t)ϕi(x),l=1,2,,

    where (gli,ϱli)(t),i=1,,l, satisfies the ODEs

    (tvl(t),ei)Ω+εaΩ(vl(t),ei)+bΩ(vl(t),vl(t),ei)12Γ|vl(t)|2neidΓ=Γ(k0vl(t)+g0(t))eidΓ+(ϱl(t)e+f0(t),ei)Ω,i=1,2,,l,0tT,(tϱl(t),ϕi)Ω+κaΩ(ϱl(t),ϕi)+bΩ(vl(t),ϱl(t),ϕi)12Γϱl(t)vl(t)nϕidΓ=Γ(k1ϱl(t)+g1(t))ϕidΓ+(f0(t),ϕi)Ω,i=1,2,,l,0tT,vl(0)=vl0(x)=lk=0vk0ek(x),v0(x)=k=0vk0ek(x)V1,ϱl(0)=ϱl0(x)=lk=0ϱk0ϕk(x)),ϱ0(x)=k=0ϱk0ϕk(x)H1(Ω),

    and

    κϱl(0)n+12ϱl(0)vl(0)n+(k1ϱl(0)+g1(0))=0.

    Here {ei}i=1 is chosen to be same as in the proof of Theorem 1 and {ϕi}i=1 is taken as an orthogonal basic function sequence of the Hilbert space H1(Ω) with ϕi=1.

    Also the proof of Theorem 5 are similar to that of Theorem 2 by using the following fact that

    tϱl(0)=κaΩ(ϱl(0),tϱl(0))bΩ(vl(0),ϱl(0),tϱl(0))+12Γϱl(0)vl(0)ntϱl(0)dΓ+Γ(k1ϱl(0)+g1(0))tϱl(0)dΓ=κΩΔϱl(0)tϱl(0)dxbΩ(vl(0),ϱl(0),tϱl(0))+Γ[κϱl(0)n+12ϱl(0)vl(0)n+(k1ϱl(0)+g1(0))]tϱl(0)dΓ=κΩΔϱl(0)tϱl(0)dxbΩ(vl(0),ϱl(0),tϱl(0))C(ϱl(0)H2(Ω)+vl(0)H1(Ω)ϱl(0)H2(Ω))tϱl(0)

    due to another compatibility condition for ϱ0(x) on Ω.

    The details of the proofs of Theorem 4 and Theorem 5 are omitted.

    The authors declare they have not used Artificial Intelligence (AI) tools in the creation of this article.

    The work is supported by the National Science Foundation of China (No. 12171111, 12426610, 12426614) of China and the Beijing Science Foundation (KZ202110005011) of China.

    The authors declare there is no conflict of interest.



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