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 Lp−Lq estimates developed by Borchers and Miyakawa [
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 Lp−Lq estimates developed by Borchers and Miyakawa [
In this paper, we study the non-Newtonian fluids associated with Maxwell equations:
{ut−∇⋅S(Du)+(u⋅∇)u+∇π−(b⋅∇)b=0,bt−△b+(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(∂ui∂xj+∂uj∂xi), 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|p−2)Du, | (1.2) |
where μ0≥0 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)x∈R3+, | (1.3) |
together with the boundary conditions defined as follows:
u=0andb⋅n=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|q−2)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 u0∈H2(Ω) and b0∈H1(Ω) with the following boundary conditions (1.4) for a bounded domain, it was shown in [6] that if 52<q≤6, a generalized solution exists, and moreover, it satisfies
u∈L∞(0,T;L2(Ω)∩H1(Ω)),∇u∈L∞(0,T;Lq(Ω)), |
b∈L∞(0,T;L2(Ω)∩H1(Ω)),∇b∈L∞(0,T;L2(Ω)),b∈L2(0,T;H2(Ω)). |
ut∈L2(Ω×(0,T)),bt∈L2(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∈(W1,2∩W1,p) and b0∈W1,2, 52≤p 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)‖L2≤C(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 b≡0 in (1.1) enjoys the optimal algebraic decay estimates
{‖u(t)‖2L2≤c(1+t)−32(1r−12),if1≤r<∞,‖(u,b)(t)‖2L2≤c(1+t)−32(1r−12)−12,∫R3+|x3u0(x)|rdx<∞,if1≤r<2. | (1.7) |
by using the Lp−Lq 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 p≥5/2. Then
(A) limt→∞‖(u,b)(t)‖L2=0, whenever u0,b0∈L2σ(R3+),
(B) ‖(u,b)(t)‖L2≤c(1+t)−32(1r−12), whenever u0,b0∈(L2σ∩Lr)(R3+) (1≤r<2).
Theorem 1.2. Suppose that (u,b) is a strong solution of (1.1)–(1.4) with p≥5/2. under u0,b0∈(L2σ∩Lr)(R3+) for 1<r≤2
∫R3+|x3u0(x)|r+|x3b0(x)|rdx<∞. | (1.8) |
Then we have
‖(u,b)(t)‖L2≤c(1+t)−32(1r−12)−12. |
Corollary 1.3. Suppose that u is a strong solution of Naiver-Stokes equations of non-Newtonian type with p≥11/5, namely b=0 in (1.1)–(1.4). Then
(A) limt→∞‖u(t)‖L2=0, whenever u0∈L2σ,
(B) ‖u(t)‖L2≤c(1+t)−32(1r−12), whenever u0∈L2σ∩Lr (1≤r<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,u∈D(A):=W2,q(R3+)∩W1,q0,σ(R3+), |
and
Ab=∇×(∇×b), b∈D(Bq)={b∈W2,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|q−2Du), |
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<r≤q<∞ or 1≤r<q≤∞ and v∈L2σ∩Lr. Then we have the Lr–Lq estimate
‖∇ke−Atv‖Lq≤ct−k2(1r−1q)‖v‖Lr,k≥0, |
and
‖e−Atv‖Lq≤ct−32(1r−1q)−12(∫R3+|xnu(x)|r)1/r, |
where e−At 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(‖u‖2L2+‖b‖2L2+‖∇u‖p−1Lp−1)λ54, |
and
‖E(λ)B2(u,b)‖≤c(‖u‖2L2+‖b‖2L2)λ54,∀ λ>0. |
Proof. Following [2, Theorem 3.6], we note that for 1<r<∞, α>0 and 0<1q:=1r−2α3≤1
3∑i,j=1‖∂xi∂xjv‖Lr≤C‖Asv‖Lr,‖∇v‖Lr≤C‖A1/2sv‖Lr,‖v‖Lq≤C‖A1/2sv‖Lr. | (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⋅∇)u−P(b⋅∇)b−P∇⋅|e(u)|p−2e(u)),v⟩|=|⟨(P(u⋅∇)u−P(b⋅∇)b−P∇⋅|e(u)|p−2e(u)),E(λ)v⟩|≤|⟨u,(u⋅∇)E(λ)v⟩|+|⟨b,(b⋅∇)E(λ)v⟩|+|⟨|e(u)|p−2e(u),∇E(λ)v⟩|≤c‖u‖2L2‖∇E(λ)v‖L∞+c‖b‖2L2‖∇E(λ)v‖L∞+c‖∇u‖p−1Lp−1‖∇E(λ)v‖L∞≤c(‖u‖2L2+‖b‖2L2+‖∇u‖p−1Lp−1)‖∇E(λ)v‖1/2L6‖∇2E(λ)v‖1/2L6≤c(‖u‖2L2+‖b‖2L2+‖∇u‖p−1Lp−1)‖A1/2sE(λ)v‖1/2L6‖AsE(λ)v‖1/2L6≤c(‖u‖2L2+‖b‖2L2+‖∇u‖p−1Lp−1)‖A12+24sE(λ)v‖1/2L2‖A1+24sE(λ)v‖1/2L2=c(‖u‖2L2+‖b‖2L2+‖∇u‖p−1Lp−1)λ54‖v‖L2 | (2.2) |
Similarly, we have
|⟨E(λ)B2(u,b),v⟩|=|⟨E(λ)((u⋅∇)b−(b⋅∇)u),v⟩|≤c(‖u‖2L2+‖b‖2L2)λ54‖v‖L2, |
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+)+2‖A1/2su‖2L2(R3+)+‖B1/2b‖2L2(R3+)≤0. | (2.3) |
To obtain a lower bound of the second term in (2.3), we get for ρ>0,
‖A1/2sz‖2=∫∞0λd‖E(λ)z(t)‖2≥∫∞ρλd‖E(λ)z(t)‖2≥ρ∫∞ρλd‖E(λ)z(t)‖2≥ρ2(‖z(t)‖2−‖E(ρ)z(t)‖2), |
and
‖B1/2z‖2≥ρ2(‖z(t)‖2−‖E(ρ)z(t)‖2). |
And thus, we have
ddt‖(u,b)(t)‖2L2+ρ‖(u,b)(t)‖2≤ρ‖E(ρ)u(t)‖2L2+ρ‖E(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)=e−tAsu0+∫t0e−(t−s)AsB1(u,b)ds, |
and
b(t)=e−tAsu0+∫t0e−(t−s)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)e−tAsu0+∫t0∫ρ0e−λ(t−s)d(E(λ)B1(u,b))ds |
=E(ρ)e−tAsu0+∫t0e−ρ(t−s)(E(ρ)B(u))ds+∫t0(t−s)(e−λ(t−s)(E(λ)B1(u,b))dλ)ds. |
This together with Lemma 2.1 implies
‖E(ρ)u(t)−E(ρ)e−tAsu0‖≤cρ54∫t0e−ρ(t−s)(‖u‖2+‖∇u‖p−1p−1)+cρ54∫t0(t−s)(e−λ(t−s)((‖u‖2+‖∇u‖p−1p−1))dλ)ds |
≤cρ54∫t0e−ρ(t−s)(‖u‖2+‖∇u‖p−1p−1). |
where we use Korn's inequality and the following inequality in the last inequality: For 115≤p<3,
∫t0‖∇z(s)‖p−1Lp−1ds≤∫t0‖z(s)‖7−p4L2‖∇2z(s)‖5p−114L2ds≤C(∫t0‖z(s)‖14−2p19−5pds)19−5p8(∫∞0‖∇2z(t)‖2L2dt)5q−118≤C(∫t0‖z(s)‖14−2p19−5pL2ds)19−5p8, | (2.5) |
and for p≥3
∫t0‖∇z(s)‖p−1Lq−1ds≤C∫t0‖∇z(s)‖2q−2L2‖∇z‖p(p−3)p−2Lpds≤C||∇z||2q−2L2((0,t);L2)||∇z||q(q−3)q−2Lq((0,t);Lq)<∞. | (2.6) |
Hence, we have
‖E(ρ)u(t)‖L2≤‖e−tAsu0‖L2+cρ54∫t0‖u(t)‖2L2ds+Cρ54(∫t0‖z(s)‖14−2p19−5pL2ds)19−5p8+cρ54. | (2.7) |
In the same way, we get
‖E(ρ)b(t)‖L2≤‖e−tBb0‖L2+cρ54∫t0‖(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)‖2L2≤‖e−tAsu0‖2L2+‖e−tBb0‖2L2 |
+cρ52(∫t0‖u(t)‖2L2ds)2+cρ54[(∫t0‖(u,b)(s)‖14−2p19−5pL2ds)19−5p8]2+cρ52 |
≤‖e−tAsu0‖2L2+‖e−tBb0‖2L2+cρ52t2+cρ52,∀ρ>0. |
Here, we use the following estimate
[(∫t0‖z(s)‖14−2p19−5pL2ds)19−5p8]2≤(∫t0‖z(s)‖2(14−2p)19−5pL2dst)19−5p8 |
≤∫t0‖z(s)‖2(14−2p)19−5pL2dst+C≤∫t0‖z(s)‖2L2dst+C≤t2+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(α‖e−tAsu0‖2L2+α‖e−tBb0‖2L2+(1+t)1−32+(1+t)−52) |
≤c(1+t)2‖e−tAsu0‖2L2+c(1+t)2‖e−tBb0‖2L2+(1+t)32+(1+t)−12. |
And thus, we get
‖(u,b)(t)‖2L2≤c+c(1+t)−32+c(1+t)−72. | (2.9) |
Since ‖e−tAsu0‖2L2→0 and ‖e−tBb0‖2L2→0 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)‖2L2≤c(t−(3r−32)‖(u0,b0)‖2Lr+(1+t)−32+(1+t)−72). | (2.10) |
Since 3r−32<52, the desired assertion is obtained if 3r−32≤32. It remains to consider the case 3r−32>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)‖2L2≤ρt−(3r−32)‖(u0,b0)‖2Lr+cρ72t+cρ72. |
Let ρ=3t−1 and then multiply the both sides of this equation by t3 to get
ddt(t3‖(u,b)(t)‖|2L2)≤ρt2−3r−32+cρ12+cρ−12. |
Since 1≤r<2 implies 3r−32≤32, we have
‖(u,b)(t)‖2L2≤c(1+t)−(3r−32)fort≥1, |
and complete the proof of Theorem 3.1.
Proof of Theorem 2.
ddt‖(u,b)(t)‖2L2+ρ‖(u,b)(t)‖2L2≤‖e−tAsb0‖2L2+‖e−tBu0‖2L2 |
+cρ72[(∫t0‖(u,b)(t)‖2ds)2+1]+cρ72[(∫t0‖u(s)‖14−2p19−5pL2ds)19−5p8]2. |
Let ρ=3t−1 and multiply both sides above by t3 to obtain
ddt(t3‖(u,b)(t)‖2L2)≤ct3(t−(3r−32)−2+t−72(∫t0‖(u,b)(t)‖2L2ds)2+t−72[(∫t0‖u(s)‖14−2p19−5pL2ds)19−5p8]2+t−72) |
≤c(t−(3r−32)+1+t−12(∫t0‖(u,b)(t)‖2L2ds)2+t−12[(∫t0‖z(s)‖14−2p19−5pL2ds)19−5p8]2+t−12), |
where we have used Lemma 1.5 and the assumption (1.8). Integrating with respect to t, we get
‖(u,b)(t)‖2L2≤ct−(3r−32)−1+t−52(∫t0‖u(t)‖2ds)2+t−52[(∫t0‖z(s)‖14−2p19−5pL2ds)19−5p8]2+t−52 |
≤ct−(3r−32)−1+t−52(∫t0‖u(t)‖2L2ds)2+t−32(∫t0‖u(s)‖2L2ds)+t−52. |
From Theorem 3.1(ii), we know
‖(u,b)(t)‖2L2≤c(1+t)−32(1r−12). | (2.12) |
A. If 32(1r−12)>12, we have
∫t0‖(u,b)(s)‖2L2<∞. | (2.13) |
Substituting this equation into (2.13) yields
‖(u,b)(t)‖2L2≤ct−(3r−32)−1+t−52+t−32≤ct−(3r−32)−1. |
B. if 32(1r−12)=12, then
‖(u,b)(t)‖2L2≤c(1+t)−(3r−32)−1+(1+t)−52(ln(1+t)2) |
+(1+t)−32ln(1+t)+t−52≤c(1+t)−(3r−32)−1. |
C. if 32(1r−12)<12, we have
‖(u,b)(t)‖2L2≤c(1+t)−(3r−32)−1+c(1+t)(−6r+32)+c(1+t)−3r+c(1+t)−52≤c(1+t)−(3r−32)−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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