Research article

Time decay rates for the coupled modified Navier-Stokes and Maxwell equations on a half space

  • Received: 07 June 2021 Accepted: 10 September 2021 Published: 17 September 2021
  • MSC : 35Q30, 76A05, 35B35

  • This paper is concerned with time decay rates of the strong solutions of an incompressible the coupled modified Navier-Stokes and Maxwell equations in a half space R3+. With the use of the spectral decomposition of the Stokes operator and LpLq estimates developed by Borchers and Miyakawa [2], we study the L2-decay rate of strong solutions.

    Citation: Jae-Myoung Kim. Time decay rates for the coupled modified Navier-Stokes and Maxwell equations on a half space[J]. AIMS Mathematics, 2021, 6(12): 13423-13431. doi: 10.3934/math.2021777

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  • This paper is concerned with time decay rates of the strong solutions of an incompressible the coupled modified Navier-Stokes and Maxwell equations in a half space R3+. With the use of the spectral decomposition of the Stokes operator and LpLq estimates developed by Borchers and Miyakawa [2], we study the L2-decay rate of strong solutions.



    In this paper, we study the non-Newtonian fluids associated with Maxwell equations:

    {utS(Du)+(u)u+π(b)b=0,btb+(u)b(b)u=0.div u=0 and  div b=0, in QT:=R3+×(0,T), (1.1)

    Here u:QT×(0,T)R3 is the flow velocity vector, b:QT×(0,T)R3 is the magnetic vector, π:QT×(0,T)R is the total pressure and Du is the symmetric part of the velocity gradient, i.e.

    Du=Diju:=12(uixj+ujxi), i,j=1,2,3.

    To motivate the conditions on the stress tensor S, we recall the following examples of constitutive laws

    S(Du)=(μ0+μ1|Du|p2)Du, (1.2)

    where μ00 and μ1>0 are constants (see e.g. [1,16]). We consider the initial-boundary value problem of (1.1), which requires initial conditions

    u(x,0)=u0(x)andb(x,0)=b0(x)xR3+, (1.3)

    together with the boundary conditions defined as follows:

    u=0andbn=0, (×b)×n=0, (1.4)

    where n is the outward unit normal vector along boundary R3+.

    The fluids such as coal-water, faint, soaps, etc., is not followed linear relationship between rate of strain and shear stress. In languages of mathematics, commonly, the standard Navier-Stokes equations refers to the equations of motion of an incompressible fluid type of S=μ0Du, where μ0>0 is constant. On the other hand, one class of non-Newtonian fluids is defined by S=μ(|Du|)Du (μ() a positive nonlinear function). For example, we note that

    S(Du)=(μ0+μ1|Du|q2)Du,1<q<,μ0>0, μ1>0. (1.5)

    For the existence of solutions to the model with the stress tensor (1.5), we make a brief comment. After the pioneering study done by Ladyzhenskaya [10], the global-in-time existence of strong solutions is proved for q>115 in [11]. On the other hand, they also established the small data global-in-time existence of strong solution for 53<q< in three dimensional space (see also [3] for weak solutions)

    The Eqs (1.1), which are the generalized incompressible magnetohydrodynamics equations is regarded as one of the simplest model describing the dynamics of electrically conducting liquid with involved rheological structure in a magnetic field. For the model (1.1), Gunzburger et al. in [6] considered (1.1)–(1.4) for the case of bounded or periodic domains, and they established unique solvability of the initial-boundary value problem. More specifically, assuming that u0H2(Ω) and b0H1(Ω) with the following boundary conditions (1.4) for a bounded domain, it was shown in [6] that if 52<q6, a generalized solution exists, and moreover, it satisfies

    uL(0,T;L2(Ω)H1(Ω)),uL(0,T;Lq(Ω)),
    bL(0,T;L2(Ω)H1(Ω)),bL(0,T;L2(Ω)),bL2(0,T;H2(Ω)).
    utL2(Ω×(0,T)),btL2(Ω×(0,T)). (1.6)

    Furthermore, they shown the uniqueness of solutions. Here strong solutions means that solutions satisfy (1.1) pointwise a.e. and the energy equality holds. Recently, the authors in [8] establish global unique solvability to (1.1)–(1.4) for u0(W1,2W1,p) and b0W1,2, 52p in the same class above (see [12] for weak solutions). For a half space, the proof in [6] is also held. And thus, we will not comment further on the existence and uniqueness of strong solutions to (1.1) and (1.2).

    For the asymptotic behavior of strong solutions to (1.1) and (1.2), the author in [9] recently examined the L2-algebraic decay in the whole space R3 with respect to the monopolar shear thickening fluids using Fourier splitting method in [13]. Precisely, he shown that

    (u,b)(t)L2C(1+t)34,t>0.

    We also refer to [5,7] for Navier-Stokes equations of non-Newtonian type.

    On the other hand, in the case of a half space Rn+, Dong and Chen [4] obtain that the weak solution u(t) of the Naiver-Stokes equations of non-Newtonian type, that is b0 in (1.1) enjoys the optimal algebraic decay estimates

    {u(t)2L2c(1+t)32(1r12),if1r<,(u,b)(t)2L2c(1+t)32(1r12)12,R3+|x3u0(x)|rdx<,if1r<2. (1.7)

    by using the LpLq estimates based on the explicit solution formula to the Stokes equation given by Ukai [15] and the spectral decomposition method and fractional powers of the Stokes operator derived by Borchers and Miyakawa [2].

    For a domain, we briefly some comments. For a whole space, we obtain the same result like as Theorem 1.1 by the fourier splitting method (see [8]). However, in R3+ case with boundary conditions, we needs to be handled the solution form due to the boundary effects. To effectively deal with boundary effect, we use the well-known spectral method, we would like to obtain temporal rate of the strong solutions. So far, there are few known results on the time decay problem to (1.1) in R3+. In this direction, our main results are as follows:

    Theorem 1.1. Suppose that (u,b) is a strong solution of (1.1)(1.4) with p5/2. Then

    (A) limt(u,b)(t)L2=0, whenever u0,b0L2σ(R3+),

    (B) (u,b)(t)L2c(1+t)32(1r12), whenever u0,b0(L2σLr)(R3+) (1r<2).

    Theorem 1.2. Suppose that (u,b) is a strong solution of (1.1)(1.4) with p5/2. under u0,b0(L2σLr)(R3+) for 1<r2

    R3+|x3u0(x)|r+|x3b0(x)|rdx<. (1.8)

    Then we have

    (u,b)(t)L2c(1+t)32(1r12)12.

    Corollary 1.3. Suppose that u is a strong solution of Naiver-Stokes equations of non-Newtonian type with p11/5, namely b=0 in (1.1)(1.4). Then

    (A) limtu(t)L2=0, whenever u0L2σ,

    (B) u(t)L2c(1+t)32(1r12), whenever u0L2σLr (1r<2).

    Remark 1.4. Comparing to [4], since the result of Corollary 1 is about the time decay rate for strong solution to Naiver-Stokes equations of non-Newtonian type, the restriction to the range of p can be slightly relaxed.

    Let us rewrite the abstract formulation of (1.1),

    {ut+Asu+B1(u,b)=0,u(0)=u0,bt+Au+B2(u,b)=0,b(0)=b0, (1.9)

    where the Stokes operators As and Laplacian operator A to be specially considered the boundary conditions (1.4) are defined as follows:

    Asu=PΔu,uD(A):=W2,q(R3+)W1,q0,σ(R3+),

    and

    Ab=×(×b), bD(Bq)={bW2,q(R3+)W1,20,σ(R3+) | (×b)×n=0 on R3+}.

    (see [14, page 262]). And also the bilinear operator B1 and B2 is defined as follows:

    B1(u,b):=(u)u(b)b(|Du|q2Du),

    and

    B2(u,b):=(u)b(b)u.

    Here P is the orthogonal projection mapping Lq(R3+) onto Lqσ(R3+) (see, e.g. Ukai [15]).

    Lemma 1.5. ([2,Theorem 3.6]). Let either 1<rq< or 1r<q and vL2σLr. Then we have the LrLq estimate

    keAtvLqctk2(1r1q)vLr,k0,

    and

    eAtvLqct32(1r1q)12(R3+|xnu(x)|r)1/r,

    where eAt denotes the analytic semigroup generated by the Stokes operator A. Here P is the orthogonal projection mapping Lq(R3+) onto Lqσ(R3+) (see, for example, Ukai [12]).

    Lemma 2.1. Suppose that (u,b) is a strong solution of (1.1)(1.4). Then

    E(λ)B1(u,b)c(u2L2+b2L2+up1Lp1)λ54,

    and

    E(λ)B2(u,b)c(u2L2+b2L2)λ54, λ>0.

    Proof. Following [2, Theorem 3.6], we note that for 1<r<, α>0 and 0<1q:=1r2α31

    3i,j=1xixjvLrCAsvLr,vLrCA1/2svLr,vLqCA1/2svLr. (2.1)

    The proof is almost same to that in [4, Lemma 2.3]. For the convenience of readers, we give a proof. Indeed, a proof is based on (2.1) and Gagliardo-Nirenberg inequality as follows:

    |E(λ)B1(u,b),v|=|E(λ)(P(u)uP(b)bP|e(u)|p2e(u)),v|=|(P(u)uP(b)bP|e(u)|p2e(u)),E(λ)v||u,(u)E(λ)v|+|b,(b)E(λ)v|+||e(u)|p2e(u),E(λ)v|cu2L2E(λ)vL+cb2L2E(λ)vL+cup1Lp1E(λ)vLc(u2L2+b2L2+up1Lp1)E(λ)v1/2L62E(λ)v1/2L6c(u2L2+b2L2+up1Lp1)A1/2sE(λ)v1/2L6AsE(λ)v1/2L6c(u2L2+b2L2+up1Lp1)A12+24sE(λ)v1/2L2A1+24sE(λ)v1/2L2=c(u2L2+b2L2+up1Lp1)λ54vL2 (2.2)

    Similarly, we have

    |E(λ)B2(u,b),v|=|E(λ)((u)b(b)u),v|c(u2L2+b2L2)λ54vL2,

    which is complete of the proof.

    Proof of Theorem 1. Following the argument in [4], we will prove Theorem.

    From the energy inequality, we know

    ddt(u,b)(t)2L2(R3+)+2A1/2su2L2(R3+)+B1/2b2L2(R3+)0. (2.3)

    To obtain a lower bound of the second term in (2.3), we get for ρ>0,

    A1/2sz2=0λdE(λ)z(t)2ρλdE(λ)z(t)2ρρλdE(λ)z(t)2ρ2(z(t)2E(ρ)z(t)2),

    and

    B1/2z2ρ2(z(t)2E(ρ)z(t)2).

    And thus, we have

    ddt(u,b)(t)2L2+ρ(u,b)(t)2ρE(ρ)u(t)2L2+ρE(ρ)b(t)2L2 (2.4)

    To estimate the right-hand side of (2.4), we consider the integral form of (1.9),

    u(t)=etAsu0+t0e(ts)AsB1(u,b)ds,

    and

    b(t)=etAsu0+t0e(ts)BB2(u,b)ds.

    Applying the operator E(ρ) to the both sides of this integral equation and integrating by parts, we obtain

    E(ρ)u(t)=E(ρ)etAsu0+t0ρ0eλ(ts)d(E(λ)B1(u,b))ds
    =E(ρ)etAsu0+t0eρ(ts)(E(ρ)B(u))ds+t0(ts)(eλ(ts)(E(λ)B1(u,b))dλ)ds.

    This together with Lemma 2.1 implies

    E(ρ)u(t)E(ρ)etAsu0cρ54t0eρ(ts)(u2+up1p1)+cρ54t0(ts)(eλ(ts)((u2+up1p1))dλ)ds
    cρ54t0eρ(ts)(u2+up1p1).

    where we use Korn's inequality and the following inequality in the last inequality: For 115p<3,

    t0z(s)p1Lp1dst0z(s)7p4L22z(s)5p114L2dsC(t0z(s)142p195pds)195p8(02z(t)2L2dt)5q118C(t0z(s)142p195pL2ds)195p8, (2.5)

    and for p3

    t0z(s)p1Lq1dsCt0z(s)2q2L2zp(p3)p2LpdsC||z||2q2L2((0,t);L2)||z||q(q3)q2Lq((0,t);Lq)<. (2.6)

    Hence, we have

    E(ρ)u(t)L2etAsu0L2+cρ54t0u(t)2L2ds+Cρ54(t0z(s)142p195pL2ds)195p8+cρ54. (2.7)

    In the same way, we get

    E(ρ)b(t)L2etBb0L2+cρ54t0(u,b)(t)2L2ds+cρ54. (2.8)

    Using (2.7), (2.8) and the energy inequality, it implies

    ddt(u,b)(t)2L2+ρ(u,b)(t)2L2etAsu02L2+etBb02L2
    +cρ52(t0u(t)2L2ds)2+cρ54[(t0(u,b)(s)142p195pL2ds)195p8]2+cρ52
    etAsu02L2+etBb02L2+cρ52t2+cρ52,ρ>0.

    Here, we use the following estimate

    [(t0z(s)142p195pL2ds)195p8]2(t0z(s)2(142p)195pL2dst)195p8
    t0z(s)2(142p)195pL2dst+Ct0z(s)2L2dst+Ct2+C.

    Now let ρ=3(1+t)1 and multiply both sides above by (1+t)3 to obtain

    ddt((1+t)3(u,b)(t)2L2)c(1+t)2(αetAsu02L2+αetBb02L2+(1+t)132+(1+t)52)
    c(1+t)2etAsu02L2+c(1+t)2etBb02L2+(1+t)32+(1+t)12.

    And thus, we get

    (u,b)(t)2L2c+c(1+t)32+c(1+t)72. (2.9)

    Since etAsu02L20 and etBb02L20 as t, we conclude that u(t)0 and b(t)0 as t and thus it complete the proof of first part in Theorem 1.1.

    For second part, it follows from (2.9) and Lemma 2.2 that

    (u,b)(t)2L2c(t(3r32)(u0,b0)2Lr+(1+t)32+(1+t)72). (2.10)

    Since 3r32<52, the desired assertion is obtained if 3r3232. It remains to consider the case 3r32>32 Hence (2.10) implies that

    |(u,b)(t)2L2(1+t)3/2. (2.11)

    Using (2.11), through the same method as before, we have

    ddt(u,b)(t)2L2+ρ(u,b)(t)2L2ρt(3r32)(u0,b0)2Lr+cρ72t+cρ72.

    Let ρ=3t1 and then multiply the both sides of this equation by t3 to get

    ddt(t3(u,b)(t)|2L2)ρt23r32+cρ12+cρ12.

    Since 1r<2 implies 3r3232, we have

    (u,b)(t)2L2c(1+t)(3r32)fort1,

    and complete the proof of Theorem 3.1.

    Proof of Theorem 2.

    ddt(u,b)(t)2L2+ρ(u,b)(t)2L2etAsb02L2+etBu02L2
    +cρ72[(t0(u,b)(t)2ds)2+1]+cρ72[(t0u(s)142p195pL2ds)195p8]2.

    Let ρ=3t1 and multiply both sides above by t3 to obtain

    ddt(t3(u,b)(t)2L2)ct3(t(3r32)2+t72(t0(u,b)(t)2L2ds)2+t72[(t0u(s)142p195pL2ds)195p8]2+t72)
    c(t(3r32)+1+t12(t0(u,b)(t)2L2ds)2+t12[(t0z(s)142p195pL2ds)195p8]2+t12),

    where we have used Lemma 1.5 and the assumption (1.8). Integrating with respect to t, we get

    (u,b)(t)2L2ct(3r32)1+t52(t0u(t)2ds)2+t52[(t0z(s)142p195pL2ds)195p8]2+t52
    ct(3r32)1+t52(t0u(t)2L2ds)2+t32(t0u(s)2L2ds)+t52.

    From Theorem 3.1(ii), we know

    (u,b)(t)2L2c(1+t)32(1r12). (2.12)

    A. If 32(1r12)>12, we have

    t0(u,b)(s)2L2<. (2.13)

    Substituting this equation into (2.13) yields

    (u,b)(t)2L2ct(3r32)1+t52+t32ct(3r32)1.

    B. if 32(1r12)=12, then

    (u,b)(t)2L2c(1+t)(3r32)1+(1+t)52(ln(1+t)2)
    +(1+t)32ln(1+t)+t52c(1+t)(3r32)1.

    C. if 32(1r12)<12, we have

    (u,b)(t)2L2c(1+t)(3r32)1+c(1+t)(6r+32)+c(1+t)3r+c(1+t)52c(1+t)(3r32)1.

    The proof is complete.

    The author thank the knowledgeable referee for his/her valuable comments and helpful suggestions. Jae-Myoung Kim was supported by National Research Foundation of Korea Grant funded by the Korean Government (NRF-2020R1C1C1A01006521).

    The authors declare that they have no competing interests.



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