# A New Regularity Criterion for the Three-Dimensional Incompressible Magnetohydrodynamic Equations in the Besov Spaces

Authors: TianLi LI; Wen Wang; Lei Liu
Journal: Journal of Function Spaces
 (2021)

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## Abstract

Regularity criteria of the weak solutions to the three-dimensional (3D) incompressible magnetohydrodynamic (MHD) equations are discussed. Our results imply that the scalar pressure fieldπ plays an important role in the regularity problem of MHD equations. We derive that the weak solution u,b is regular on 0,T, which is provided for the scalar pressure field π in the Besov spaces.


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## Body

## 1. Introduction

In this article, we consider the global regularity problem concerning the 3D incompressible MHD equations(1)∂tu+u·∇u+∇π=b·∇b+Δu,∂tb+u·∇b=b·∇u+Δb,∇·u=0,∇·b=0,that satisfy the initial condition
(2)ut=0=u0,bt=0=b0.Here,u=u1,u2,u3,b=b1,b2,b3, and π=πx,t represent the velocity field, the magnetic field, and the pressure, respectively; u0x,b0x are the corresponding initial data which satisfied ∇·u0=0,∇·b0=0 in the sense of distribution.MHD equations govern the dynamics of the velocity and magnetic fields of electrically conducting fluids such as plasmas, liquid metals, and salt water. Besides their important physical applications, the MHD equations also have important mathematical significance. It is well known [1] that problem (1) is locally well-posed for any given initial datum u0,b0∈HsR3. However, whether a local strong solution can exist globally, or equivalently, whether global weak solutions are smooth is still a challenging open problem. Nevertheless, there exist plenty of results in the literature showing that the answer to this problem is positive if some additional conditions are imposed on the weak solutions [2–12]. Some of them are motivated by the works on the Navier-Stokes equations (b=0)
(3)∂tu+u·∇u+∇π=Δu,∇·u=0,ut=0=u0.Among these results, Zhou [7] and He and Xin [13] obtained some Ladyzhenskaya-Prodi-Serrin-type regularity criteria for the 3D MHD equations in terms of velocity and the gradient of velocity, independently. They proved that the velocity u satisfies
(4)∫0Tu·,tLrsdt<+∞,2s+3r≤1,r>3,then the weak solution u,b is regular on 0,T.However, we are interested in regularity criteria involving only in terms of the pressureπ (for details, refer to [14–17]). For regularity criteria results for the 3D MHD equations, Jia and Zhou [14] obtained the following:
(5)∂3π∈Lp0,T;LqR3,2p+3q=2,3≤q<∞.In [15], Liu established the new regularity criteria in terms of the pressure as follows:
(6)∂3π∈Lp0,T;Lq1Rx1,x22;Lq2Rx3,2p+2q1+1q2=s∈2,3,3s≤q2≤q1≤1k−2.Later, Gala and Ragusa [16] extended the regularity criteria to the BMO space and Besov space. If the pressure π or the pressure gradient ∇π satisfies
(7)π∈L20,T;Ḃ∞,∞−1R3,or
(8)∇π∈L2/30,T;BMOR3,then the weak solution u,b is regular on 0,T.Very recently, Tong and Wang [17] showed the following regularity criterion for the 3D MHD:
(9)∫0TπḂ∞,∞−11+lne+πḂ∞,∞−1dt<∞,or
(10)∫0T∇uḂ∞,∞−12+∇uḂ∞,∞−121+lne+∇uḂ∞,∞−1+∇uḂ∞,∞−1dt<∞.It is the aim of the paper to give the complete description on the regularity criteria of the weak solutions of the 3D MHD equations on the pressure field. Let us introduce the assignment of this paper; we first recall some preliminaries on functional settings and state the main results in Section2 and prove the main results in Section 3.


## 2. Preliminaries and Main Result

Throughout this text,C stands for a generic positive constant which may differ in value from one line to another. We use ·p to denote the norm of the Lebesgue space Lp1≤p≤∞, and the norm as follows:
(11)fp=∫R3fxpdx1/p,1≤p<∞,esssupx∈R3fx,p=∞.In order to define the Besov space, let us first recall the Littlewood-Paley theory (see Ref. [18]). For a given function f∈S, its Fourier transformation f^ of F is defined by
(12)Ffξ=f^ξ=∫R3e−ix·ξfxdx,where S is the Schwartz class of rapidly decreasing functions defined on R3. Choose a nonnegative radial function ψ∈SR3 supported in B≜ξ∈R3;ξ≤2 such that for ξ∈R3 and ξ≤1. Setting the radial function
(13)ϕξ=ψξ2−ψξ,ξ∈R3,for the integer set Z, we have
(14)∑j∈Zϕ2−jξ=1,ξ∈R3\0.Leth=F−1ϕ and define the dyadic blocks as follows:
(15)Δjfx=23j∫R3h2jsfx−sds,x∈R3.We thus have the following Littlewood-Paley decomposition:(16)f=∑j∈ZΔjfx.Forα∈R,1≤p,q≤∞, we can now define the homogeneous Besov space Ḃp,qαR3 as
(17)Ḃp,qαR3=f∈S′R3PR3;fḂp,qα<∞,where
(18)fḂp,qα=∑j=−∞∞2jαqΔjfxpq1/q,1≤q<∞,supj∈ZΔjfxp,q=∞.PR3 is the set of all scalar polynomials defined on R3. S′R3 is the space of all tempered distributions on R3.Definition 1.
Assumingu0,b0∈L2ℝ3, and ∇·u0=∇·b0=0,T>0, the measurable function u,b defined on 0,T×R3 is called the weak solution of Equation (1), if
(1)
u,b∈L∞0,T;L2R3∩L20,T;H1R3.(2)
∇·u=∇·b=0 and ∀φ∈C0∞0,T×R3 have(19)∫0T∫R3u,b·∇φdxdt=0,Equation (1) holds in the sense of distributions.
For∀φ∈C0∞0,T×R3,∇·φ=0, we have
(20)∫0T∫R3∂tφ+u·∇φ·udxdt+∫R3u0·φx,0dx=∫0T∫R3∇u:∇φdxdt,∫0T∫R3∂tφ+b·∇φ·bdxdt+∫R3b0·φx,0dx=∫0T∫R3∇b:∇φdxdt,where A:B=∑i,j=13aijbij,A=aij,B=bij.
(3)
The strong energy inequality, that is,(21)ut,btL22+2∫0t∇u,bτL22dτ≤u0,b0L22,∀0≤t<T.In this paper, we establish the following theorem.Theorem 2.
Letu0,b0∈L2ℝ3∩L4ℝ3, with ∇·u0=0,∇·b0=0,T>0 in the sense of distribution. Let u,b be a weak solution of the MHD Equations (1) on 0,T which satisfies the strong energy inequality (21). If the corresponding pressure π satisfies
(22)π∈L2/α+10,T;Ḃ∞,∞αR3,0<α<1,then the weak solution u,b is regular on 0,T.When the time critical index2/α+1=1, we can derive the following Theorem 3. And if π∈L4/30,T;Ḃ∞,∞0, we prove that the weak solution u,b is regular on 0,T in Theorem 4.Theorem 3.
Under the same assumption in Theorem2, if the corresponding pressure p satisfies
(23)π∈L10,T;Ḃ∞,∞αR3,12≤α<1,then the weak solution u,b is regular on 0,T.Theorem 4.
Under all the assumptions in Theorem2, the weak solution u,b is regular on 0,T. If the corresponding pressure p satisfies the following condition:
(24)π∈L4/30,T;Ḃ∞,∞0R3.Remark 5.
In article [17], the regularity condition of ∫0TπḂ∞,∞−1/1+lne+∥π∥Ḃ∞,∞−1dt<∞ is better than the regularity condition of Theorems 2–4. However, in article [17], the initial condition is required to satisfy u0,b0∈H3R3, while Theorems 2–4 only need to satisfy u0,b0∈L2R3∩L4R3.


## 3. Pressure Regularity Criteria

In order to avoid the difficulties caused by the magnetic field when we do energy estimates, by adding and subtracting (1)1 with (1)2, we convert the 3D MHD Equations (1) into a mathematically symmetric form as follows:
(25)∂tw++w−·∇w++∇π=Δw+,∂tw−+w+·∇w−+∇π=Δw−,∇·w+=0,∇·w−=0,w+x,0=w0+x,w−x,0=w0−x,with w±≜u±b.Let us introduce some auxiliary results. To establish some new regularity criteria in terms of pressure, an effective method is to find a “bridge” between the desired results and the known criteria. The following lemma plays such a role in the proof of Theorems2–4.Lemma 6 (Bernstein inequality [18]).
For1≤p≤q≤∞ and an integer k, the following estimate is true:
(26)∇kΔjfq≤C2jk+3j1/p−1/qΔjfp.Lemma 7.
Letu0,b0∈L2R3∩L4R3 and ∇·u0=∇·b0=0. Let u,b be the corresponding solution to the 3D MHD Equations (1). If w+ and w− satisfy
(27)sup0≤t≤Tw+·,t4+w+·,t4<+∞.
Then, the weak solutionu,b is regular on 0,T.Proof.
By the definition ofw+ and w−, one can deduce that
(28)u·,t4+b·,t4≤w+·,t4+w−·,t4.
By (4), we can derive that the weak solution u,b is regular on 0,T.
Now, we are ready to prove the theorem. Taking the inner product of the first equation of (21) with w3+2w3+ and the second equation of (21) with w3−2w3− in L2R3, adding them together, and noticing that ∇·w±=0, we have
(29)14ddtw+44+w−44+12∫R3∇w+22+∇w−22dx+∫R3∇w+2w+2+∇w−2w−2dx=−∫R3∇π·w+w+2dx−∫R3∇π·w−w−2dx≜I+II.
Here, we have used the following identities due to the fact that the divergence free condition∇w+=∇w−=0:
(30)∫R3w+·∇w−·w−w−2dx=∫R3w−·∇w+·w+w+2dx=0.
By the Littlewood-Paley decomposition,p can be written as follows:
(31)π=∑j∈ZΔjπ=∑j≤NΔjπ+∑j≥N+1Δjπ.
Hence, by (29), we obtain
(32)I=∑j≤N∫R3Δj∇π·w+w+2dx+∑j≥N+1∫R3Δj∇π·w+w+2dx≜I1+I2.☐


### 3.1. The Proof of Theorem2

We first consider that the velocityπ satisfies the growth condition (22):
(33)π∈L2/α+10,T;Ḃ∞,∞α,0<α<1.For0<α<1, by the Hölder inequality, we estimate Iii=1,2 of the right-hand side of (32) one by one.
(34)I1≤∑j≤NΔj∇π∞w+2w+22≤∑j≤N2j−α+12jα−1Δj∇π∞w+2w+42≤C2N−α+1∇πḂ∞,∞α−1w+2w+42.Similarly, forI2, integration by parts gives
(35)I2≤∑j≥1+NΔjπ∞w+2w+∇w+2≤∑j≥1+N2−jα2jαΔjπ∞w+2w+∇w+2≤C2−αNπḂ∞,∞αw+2w+∇w+2.Hence, choosing the integerN such that (34) is equal to (35),
(36)2N−α+1∇πḂ∞,∞α−1w+2w+42=2−αNπḂ∞,∞αw+2w+∇w+2,then
(37)log2w+∇w+2w+42≤N≤log2w+∇w+2w+42+1.Plugging the estimates (34) and (35) into (32), and using Young inequality, we have
(38)I≤C2−αlog2w+∇w+2/w+42πḂ∞,∞αw+2w+∇w+2≤Cw+∇w+2w+42−απḂ∞,∞αw+∇w+2≤Cw+∇w+21−αw+42απḂ∞,∞α≤Cw+42απḂ∞,∞α2/α+1+14w+∇w+21−α2/1−α≤Cw+44α/α+1πḂ∞,∞α2/α+1+14w+∇w+22.By letting0<α<1, such that 4α/α+1≤4, and by using the Young inequality, it follows that
(39)w+44α/α+1≤Cw+44+1.Hence, we obtain(40)I≤CπḂ∞,∞α2/α+1w+44+1+14w+∇w+22.Using the similar way with the estimateII, we see that
(41)II≤CπḂ∞,∞α2/α+1w−44+1+14w−∇w−22.Combination of (40), (41), and (29) implies that
(42)14ddtw+44+w−44+1+12∫R3∇w+22+∇w−22dx+∫R3∇w+2w+2+∇w−2w−2dx=−∫R3∇π·w+w+2dx−∫R3∇π·w−w−2dx=Cw+L44+w−44+1πḂ∞,∞α2/α+1.Using the Gronwall inequality, we have(43)sup0<t<Tw+44+w−44+1≤Cw0+44+w0−44+1expC∫0tπḂ∞,∞α2/α+1.By assuming Theorem2 and the definition of W0±, we have
(44)w+04+w−04≤C.By using Lemma6 and (22), we have
(45)sup0<t<Tw+44+w−44+1≤C.This completes the proof of Theorem2.


### 3.2. The Proof of Theorem3

Next, we are ready to prove Theorem3. Let us begin with the case of growth condition (23):
(46)π∈L10,T;Ḃ∞,∞α,12≤α<1.For1/2≤α<1, by the Hölder inequality, we estimate Iii=1,2 one by one.
(47)I1≤∑j≤NΔj∇π∞w+2w+22≤∑j≤N2j−α+12jα−1Δj∇π∞w+2w+22≤2N−α+1∇πḂ∞,∞α−1w+2w+22≤C2N−α+1πḂ∞,∞αw+42.Similarly, forI2 of (32), by (26) and integration by parts, gives
(48)I2≤∑j≥1+NΔjπ4w+4w+∇w+2≤∑j≥1+N23j/4Δjπ2w+4w+∇w+2≤∑j≥1+N2−j/21/2∑j≤1+N22jΔjπ221/2w+4w+∇w+2≤C2−N/4πḂ2,21w+4w+∇w+2≤C2−N/4w+·∇w−2w+4w+∇w+2.Hence, choosing the integerN such that (47) is equal to (48)
(49)2N−α+1πḂ∞,∞αw+42=2−N/4w+·∇w−2w+4w+∇w+2,then
(50)log2w+·∇w−2w+∇w+2πḂ∞,∞αw+44/5−4α≤N≤log2w+·∇w−2w+∇w+2πḂ∞,∞αw+44/5−4α+1.Plugging the estimates (47) and (48) into (32), we have
(51)I≤C2log2w+·∇w−2w+∇w+2/πḂ∞,∞αw+441−α/5−4απḂ∞,∞αw+42≤CπḂ∞,∞α1/5−4αw+46−4α/5−4αw+·∇w−241−α/5−4αw+∇w+241−α/5−4α≤CπḂ∞,∞α1/5−4αw+46−4α/5−4α5−4α+14w+·∇w−241−α/5−4αw+∇w+241−α/5−4α5−4α/41−α≤CπḂ∞,∞αw+46−4α+14w+·∇w−2w+∇w+2≤CπḂ∞,∞αw+46−4α+18w+·∇w−22+18w+∇w+22.Using the similar way with the estimateII, we see that
(52)II≤CπḂ∞,∞αw−46−4α+18w−·∇w+22+18w−∇w−22.Combination of (51), (52), and (29) implies that
(53)14ddtw+L44+w−L44+14∫R3∇w+22+∇w−22dx+∫R3∇w+2w+2+∇w−2w−2dx=CπḂ∞,∞αw+46−4α+w−46−4α+14w−·∇w+22+w−·∇w+22.By letting1/2≤α<1, such that 6−4α≤4, and by using the Young inequality, it follows that
(54)w±46−4α≤Cw±44+1.Using the triangle inequality(55)ur=12w++w−r=12w+r+w−r,br=12w+−w−r=12w+r+w−r,w+r=u+br≤ur+br,w−r=u−bLr≤ur+br,we obtain
(56)14ddtu44+b44+1+14∫R3∇u22+∇b22dx+∫R3∇u2u2+∇u2b2+∇b2b2+∇b2u2dx≤CπḂ∞,∞αu44+b44+1+12u·∇u22+u·∇b22+b·∇b22+b·∇u22.Hence,(57)14ddtu44+b44+1+14∫R3∇u22+∇b22dx+12u·∇u22+u·∇b22+b·∇b22+b·∇u22≤CπḂ∞,∞αu44+b44+1.Then, the bounds for theL4-norms of u and b follow from the standard Gronwall inequality. Thanks to (4), this completes the proof of Theorem 3.


### 3.3. The Proof of Theorem4

For the final assumptions (24), that is,
(58)π∈L4/30,T;Ḃ∞,∞0R3.After applying the Hölder inequality, we estimateIii=1,2one by one.
(59)I1≤∑j≤NΔj∇π∞w+2w+22≤∑j≤N2j2−jΔj∇π∞w+2w+42≤2N∇πḂ∞,∞−1w+2w+42≤C2NπḂ∞,∞0w+42.Similarly, forI2 of (32), by (27), the Hölder inequality, and integration by parts, we have
(60)I2≤∑j≥1+NΔjπ3w+6w+∇w+2≤∑j≥1+N2j/2Δjπ2w+41/2w+121/2w+∇w+2≤∑j>1+N2−j1/2∑j>1+N2jΔjπ221/2w+41/2w+261/4w+∇w+2≤C2−N/2πḂ2,21w+41/2w+∇w+25/4≤C2−N/2w+·∇w−2w+41/2w+∇w+25/4.Hence, choosing the integerN such that (59) is equal to (60),
(61)2NπḂ∞,∞0w+42=2−N/2w+·∇w−2w+41/2w+∇w+25/4,then
(62)log2w+·∇w−22/3w+∇w+25/6πḂ∞,∞02/3w+4≤N≤log2w+·∇w−22/3w+∇w+25/6πḂ∞,∞02/3w+4+1.Plugging the estimates (59) and (60) into (32), we have
(63)I≤C2log2w+·∇w−22/3w+∇w+25/6/πḂ∞,∞02/3w+4πḂ∞,∞0w+42≤CπḂ∞,∞01/3w+4w+·∇w−22/3w+∇w+25/6≤CπḂ∞,∞01/3w+44+14w+·∇w−22/3w+∇w+25/64/3≤CπḂ∞,∞04/3w+44+14w+·∇w−28/9w+∇w+210/9≤CπḂ∞,∞04/3w+44+18w+·∇w−22+18w+∇w+22.Using the similar way with the estimateII, we see that
(64)II≤CπḂ∞,∞04/3w−44+18w−·∇w+22+18w−∇w−22.Combination of (63), (64), and (29) implies that
(65)14ddtw+44+∥w−∥44+14∫R3∣∇∣w+22+∣∇∣w−22dx+∫R3∇w+2w+2+∇w−2w−2dx=CπḂ∞,∞04/3w+44+w−44+14w−·∇w+22+w−·∇w+22.Using the same triangle inequality(66)ur=12w++w−r≤12w+4+w−r,br=12w+−w−r≤12w+4+w−r,w+r=u+br≤ur+br,w−r=u−br≤ur+br,we obtain
(67)14ddtu44+b44+14∫R3∇u22+∇b22dx+∫R3∇u2u2+∇b2b2+∇u2b2+∇b2u2dx≤CπḂ∞,∞04/3u44+b44+12u·∇u22+u·∇b22+b·∇b22+b·∇u22.Hence,(68)14ddtu44+b44≤CπḂ∞,∞04/3u44+b44.By Gronwall’s inequality and (4), Theorem 4 is proven.


## 3.1. The Proof of Theorem2

We first consider that the velocityπ satisfies the growth condition (22):
(33)π∈L2/α+10,T;Ḃ∞,∞α,0<α<1.For0<α<1, by the Hölder inequality, we estimate Iii=1,2 of the right-hand side of (32) one by one.
(34)I1≤∑j≤NΔj∇π∞w+2w+22≤∑j≤N2j−α+12jα−1Δj∇π∞w+2w+42≤C2N−α+1∇πḂ∞,∞α−1w+2w+42.Similarly, forI2, integration by parts gives
(35)I2≤∑j≥1+NΔjπ∞w+2w+∇w+2≤∑j≥1+N2−jα2jαΔjπ∞w+2w+∇w+2≤C2−αNπḂ∞,∞αw+2w+∇w+2.Hence, choosing the integerN such that (34) is equal to (35),
(36)2N−α+1∇πḂ∞,∞α−1w+2w+42=2−αNπḂ∞,∞αw+2w+∇w+2,then
(37)log2w+∇w+2w+42≤N≤log2w+∇w+2w+42+1.Plugging the estimates (34) and (35) into (32), and using Young inequality, we have
(38)I≤C2−αlog2w+∇w+2/w+42πḂ∞,∞αw+2w+∇w+2≤Cw+∇w+2w+42−απḂ∞,∞αw+∇w+2≤Cw+∇w+21−αw+42απḂ∞,∞α≤Cw+42απḂ∞,∞α2/α+1+14w+∇w+21−α2/1−α≤Cw+44α/α+1πḂ∞,∞α2/α+1+14w+∇w+22.By letting0<α<1, such that 4α/α+1≤4, and by using the Young inequality, it follows that
(39)w+44α/α+1≤Cw+44+1.Hence, we obtain(40)I≤CπḂ∞,∞α2/α+1w+44+1+14w+∇w+22.Using the similar way with the estimateII, we see that
(41)II≤CπḂ∞,∞α2/α+1w−44+1+14w−∇w−22.Combination of (40), (41), and (29) implies that
(42)14ddtw+44+w−44+1+12∫R3∇w+22+∇w−22dx+∫R3∇w+2w+2+∇w−2w−2dx=−∫R3∇π·w+w+2dx−∫R3∇π·w−w−2dx=Cw+L44+w−44+1πḂ∞,∞α2/α+1.Using the Gronwall inequality, we have(43)sup0<t<Tw+44+w−44+1≤Cw0+44+w0−44+1expC∫0tπḂ∞,∞α2/α+1.By assuming Theorem2 and the definition of W0±, we have
(44)w+04+w−04≤C.By using Lemma6 and (22), we have
(45)sup0<t<Tw+44+w−44+1≤C.This completes the proof of Theorem2.


## 3.2. The Proof of Theorem3

Next, we are ready to prove Theorem3. Let us begin with the case of growth condition (23):
(46)π∈L10,T;Ḃ∞,∞α,12≤α<1.For1/2≤α<1, by the Hölder inequality, we estimate Iii=1,2 one by one.
(47)I1≤∑j≤NΔj∇π∞w+2w+22≤∑j≤N2j−α+12jα−1Δj∇π∞w+2w+22≤2N−α+1∇πḂ∞,∞α−1w+2w+22≤C2N−α+1πḂ∞,∞αw+42.Similarly, forI2 of (32), by (26) and integration by parts, gives
(48)I2≤∑j≥1+NΔjπ4w+4w+∇w+2≤∑j≥1+N23j/4Δjπ2w+4w+∇w+2≤∑j≥1+N2−j/21/2∑j≤1+N22jΔjπ221/2w+4w+∇w+2≤C2−N/4πḂ2,21w+4w+∇w+2≤C2−N/4w+·∇w−2w+4w+∇w+2.Hence, choosing the integerN such that (47) is equal to (48)
(49)2N−α+1πḂ∞,∞αw+42=2−N/4w+·∇w−2w+4w+∇w+2,then
(50)log2w+·∇w−2w+∇w+2πḂ∞,∞αw+44/5−4α≤N≤log2w+·∇w−2w+∇w+2πḂ∞,∞αw+44/5−4α+1.Plugging the estimates (47) and (48) into (32), we have
(51)I≤C2log2w+·∇w−2w+∇w+2/πḂ∞,∞αw+441−α/5−4απḂ∞,∞αw+42≤CπḂ∞,∞α1/5−4αw+46−4α/5−4αw+·∇w−241−α/5−4αw+∇w+241−α/5−4α≤CπḂ∞,∞α1/5−4αw+46−4α/5−4α5−4α+14w+·∇w−241−α/5−4αw+∇w+241−α/5−4α5−4α/41−α≤CπḂ∞,∞αw+46−4α+14w+·∇w−2w+∇w+2≤CπḂ∞,∞αw+46−4α+18w+·∇w−22+18w+∇w+22.Using the similar way with the estimateII, we see that
(52)II≤CπḂ∞,∞αw−46−4α+18w−·∇w+22+18w−∇w−22.Combination of (51), (52), and (29) implies that
(53)14ddtw+L44+w−L44+14∫R3∇w+22+∇w−22dx+∫R3∇w+2w+2+∇w−2w−2dx=CπḂ∞,∞αw+46−4α+w−46−4α+14w−·∇w+22+w−·∇w+22.By letting1/2≤α<1, such that 6−4α≤4, and by using the Young inequality, it follows that
(54)w±46−4α≤Cw±44+1.Using the triangle inequality(55)ur=12w++w−r=12w+r+w−r,br=12w+−w−r=12w+r+w−r,w+r=u+br≤ur+br,w−r=u−bLr≤ur+br,we obtain
(56)14ddtu44+b44+1+14∫R3∇u22+∇b22dx+∫R3∇u2u2+∇u2b2+∇b2b2+∇b2u2dx≤CπḂ∞,∞αu44+b44+1+12u·∇u22+u·∇b22+b·∇b22+b·∇u22.Hence,(57)14ddtu44+b44+1+14∫R3∇u22+∇b22dx+12u·∇u22+u·∇b22+b·∇b22+b·∇u22≤CπḂ∞,∞αu44+b44+1.Then, the bounds for theL4-norms of u and b follow from the standard Gronwall inequality. Thanks to (4), this completes the proof of Theorem 3.


## 3.3. The Proof of Theorem4

For the final assumptions (24), that is,
(58)π∈L4/30,T;Ḃ∞,∞0R3.After applying the Hölder inequality, we estimateIii=1,2one by one.
(59)I1≤∑j≤NΔj∇π∞w+2w+22≤∑j≤N2j2−jΔj∇π∞w+2w+42≤2N∇πḂ∞,∞−1w+2w+42≤C2NπḂ∞,∞0w+42.Similarly, forI2 of (32), by (27), the Hölder inequality, and integration by parts, we have
(60)I2≤∑j≥1+NΔjπ3w+6w+∇w+2≤∑j≥1+N2j/2Δjπ2w+41/2w+121/2w+∇w+2≤∑j>1+N2−j1/2∑j>1+N2jΔjπ221/2w+41/2w+261/4w+∇w+2≤C2−N/2πḂ2,21w+41/2w+∇w+25/4≤C2−N/2w+·∇w−2w+41/2w+∇w+25/4.Hence, choosing the integerN such that (59) is equal to (60),
(61)2NπḂ∞,∞0w+42=2−N/2w+·∇w−2w+41/2w+∇w+25/4,then
(62)log2w+·∇w−22/3w+∇w+25/6πḂ∞,∞02/3w+4≤N≤log2w+·∇w−22/3w+∇w+25/6πḂ∞,∞02/3w+4+1.Plugging the estimates (59) and (60) into (32), we have
(63)I≤C2log2w+·∇w−22/3w+∇w+25/6/πḂ∞,∞02/3w+4πḂ∞,∞0w+42≤CπḂ∞,∞01/3w+4w+·∇w−22/3w+∇w+25/6≤CπḂ∞,∞01/3w+44+14w+·∇w−22/3w+∇w+25/64/3≤CπḂ∞,∞04/3w+44+14w+·∇w−28/9w+∇w+210/9≤CπḂ∞,∞04/3w+44+18w+·∇w−22+18w+∇w+22.Using the similar way with the estimateII, we see that
(64)II≤CπḂ∞,∞04/3w−44+18w−·∇w+22+18w−∇w−22.Combination of (63), (64), and (29) implies that
(65)14ddtw+44+∥w−∥44+14∫R3∣∇∣w+22+∣∇∣w−22dx+∫R3∇w+2w+2+∇w−2w−2dx=CπḂ∞,∞04/3w+44+w−44+14w−·∇w+22+w−·∇w+22.Using the same triangle inequality(66)ur=12w++w−r≤12w+4+w−r,br=12w+−w−r≤12w+4+w−r,w+r=u+br≤ur+br,w−r=u−br≤ur+br,we obtain
(67)14ddtu44+b44+14∫R3∇u22+∇b22dx+∫R3∇u2u2+∇b2b2+∇u2b2+∇b2u2dx≤CπḂ∞,∞04/3u44+b44+12u·∇u22+u·∇b22+b·∇b22+b·∇u22.Hence,(68)14ddtu44+b44≤CπḂ∞,∞04/3u44+b44.By Gronwall’s inequality and (4), Theorem 4 is proven.



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*Source: 4227796-2021-08-29.xml*