# Global Well-Posedness for thed-Dimensional Magnetic Bénard Problem without Thermal Diffusion

Authors: Yinhong Cao
Journal: Complexity
 (2021)

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## Abstract

This paper focuses on the global existence of strong solutions to the magnetic Bénard problem with fractional dissipation and without thermal diffusion inℝd with d≥3. By using the energy method and the regularization of generalized heat operators, we obtain the global regularity for this model under minimal amount dissipation.


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## Body

## 1. Introduction

Consider the global well-posedness problem to thed-dimensional (dD) magnetic Bénard problem(1)∂tu+u⋅∇u+μΛ2αu=−∇p+b⋅∇b+θed,x∈ℝd,t>0,∂tb+u⋅∇b+νΛ2βb=b⋅∇u,x∈ℝd,t>0,∂tθ+u⋅∇θ=u⋅ed,x∈ℝd,t>0,∇⋅u=0,∇⋅b=0,x∈ℝd,t>0,ux,0=u0x,bx,0=b0x,θx,0=θ0x,x∈ℝd,where ux,t, bx,t, θx,t, and px,t denote the velocity field, the magnetic field, the temperature, and the pressure, respectively. μ≥0 is kinematic viscosity and ν≥0 is magnetic diffusion. θed and u⋅ed model the acting of the buoyancy force on fluid motion and the Rayleigh–Bénard convection in a heated inviscid fluid, respectively. The parameters α and β are nonnegative, and Λrf with r≥0 is defined via Λrf^ξ=ξrf^ξ. The magnetic Bénard problem can be used to model the behavior of the thermal instability under the influence of the magnetic field. One can refer to [1–7] for more physical background.The global regularity problem of the magnetic Bénard problem (1) has caught much attention. For the 2D case, Zhou et al. in [8] obtained the global regularity for the case α=β=1. When α=2 and ν=0, the global existence and uniqueness of strong solutions were established by Yamazaki in [9]. Recently, Shang in [10] showed the global well-posedness of solutions for the case α=1/2 and β=1. Compared to the magnitude results on the 2D case, it appears that there are only few global regularity results on dD (d≥3) magnetic Bénard problem (1).This paper focuses its attention on the global regularity problem of (1) with minimal amount dissipation. More precisely, we are able to establish the following global regularity result.Theorem 1.
Consider (1) with μ>0 and ν>0. Suppose that u0,b0,θ0∈Hsℝd with s>1+d/2, ∇⋅u0=0, and ∇⋅b0=0. Assume also that(2)α≥12+d4,α+β≥1+d2.
Then, (1) has a unique global strong solution u,b,θ satisfying, for any T>0,(3)u,b,θ∈L∞0,T;Hsℝd,u∈L20,T;Hs+αℝd,w∈L20,T;Hs+βℝd.In the subsequent section, we prove Theorem1. Moreover, the definitions of the Besov spaces are provided in the appendix. We shall separately write Lp=Lpℝd, Hs=Hsℝd, and ∫=∫ℝd for notational convenience.


## 2. Proof of Theorem1

This section proves Theorem1. The key step is to establish global a prioriHs-bound for u,b,θ. More specifically, we shall establish the following result.Proposition 1.
Consider (1) with μ>0 and ν>0. Suppose that u0,b0,θ0∈Hsℝd with s>1+d/2 and ∇⋅u0=∇⋅b0=0, and(4)α≥12+d4,α+β≥1+d2.
Then, the corresponding solution of (1) is globally bounded in Hsℝd.We only prove Proposition1 for the case α+β=1+d/2. In fact, the case α+β>1+d/2 is even simpler.


### 2.1. Preparations

To prove the main theorem, as preparations we give three lemmas in this section. The first contains two calculus inequalities.Lemma 1.
(see [11, 12]). Let s>0. Let 1<r<∞ and 1/r=1/p1+1/q1=1/p2+1/q2 with q1,p2∈1,∞ and p1,q2∈1,∞. Then,(5)Λs,fgLr≤C∇fLp1Λs−1gLq1+ΛsfLp2gLq2,with Λs,fg=Λsfg−fΛsg and(6)ΛsfgLr≤CΛsfLp1gLq1+fLp2ΛsgLq2,where C′s are constants.The second is the property of the generalized heat operator.Lemma 2.
(see, e.g., [13, 14]). Let A=ξ∈ℝd:r1≤ξ≤r2 with 0<r1<r2 being constants. Then, there exist two positive constants c and C such that, for(7)f∈Lpℝd,suppf^⊂λA,p∈1,∞,and we have, for α>0, t>0, and λ>0,(8)e−−ΔαtfLpℝd≤Ce−ctλ2αfLpℝd.The last is the following logarithmic type interpolation inequality.Lemma 3.
(see [14, 15]). Let s>1+d/2 and q∈1,∞. Then, there exists C>0 such that(9)∇fL∞≤C1+fL2+Λd/q∇fLqloge+ΛsfL2.


### 2.2. GlobalHs Bound

This section gives the proof of Proposition1. We first prove the following global L2-bound.Lemma 4.
For anyt>0, the solution u,b,θ of (1) obeys(10)uL22+bL22+θL22+∫0tΛαuτL22dτ+∫0tΛβbτL22dτ≤C.Proof.
Taking theL2-inner product of (1) with u,b,θ, together with the Young inequality, we obtain(11)12ddtuL22+bL22+θL22+μΛαuL22+νΛβbL22=∫θed⋅u+∫u⋅edθ≤uL22+θL22.
Then, (10) follows from this and Gronwall’s inequality.
The second one is a better regularity forb.Lemma 5.
Letu,b,θ be the solution of (1). Then, for all t>0 and 0<σ<α−1,(12)ΛρbtL22+∫0tΛβ+σbτL22dτ≤C.Proof.
ApplyingΛσ to 1.12 and dotting the result with Λσb, we have(13)12ddtΛσbL22+νΛβ+σbL22=∫Λσb⋅∇u⋅Λσb−∫Λσu⋅∇b⋅Λσb≔I1+I2.
Applying Lemma1, we arrive at(14)I1≤CΛσbL2d/d−2β∇uLd/β+Λσ∇uL2d/d+2+2σ−2αbLd/α−1−σΛσbL2≤CΛβ+σbL2ΛαuL2ΛσbL2≤ν4Λβ+σbL22+CΛαuL22ΛσbL22.
Note that(15)I2=−∫Λσu⋅∇b−u⋅∇Λσb⋅Λσb.
Then, by again applying Lemma1, we find that I2 obeys the same bound as I1. Combing the above estimates up, together with (10), Gronwall’s inequality yields (12).
The last preparation is stated as follows.Lemma 6.
Letu,b,θ be the solution of (1). Then, for all t>0 and d/2α−2<q<∞,(16)∫0tΛd/qωτLqdτ<∞,where ω=∇×u.
In addition, for all2≤q<∞,(17)θtLq<∞.Proof.
Taking curl on the both sides of1.11, we have(18)∂tω+μ−Δαω=∇×b⋅∇b−∇×u⋅∇u+∇×θed,with ω=∇×u. Write (18) into the integral form(19)ωt=e−μ−Δαtω0+∫0te−μ−Δαt−τ∇×∇⋅b⊗bτdτ−∫0te−μ−Δαt−τ∇×∇⋅u⊗uτdτ+∫0te−μ−Δαt−τ∇×θedτdτ.
We further localize it by applyingΔj with j≥−1:(20)Δjωt=Δje−μ−Δαtω0++∫0tΔje−μ−Δαt−τ∇×∇⋅b⊗bτdτ−∫0tΔje−μ−Δαt−τ∇×∇⋅u⊗uτdτ+∫0tΔje−μ−Δαt−τ∇×θedτdτ.
Forq∈1,∞, taking the Lq-norm to this equation, we derive that(21)ΔjωtLq≤Ce−c0μt22αjΔjω0Lq+C∫0t22je−c0μt−τ22αjΔjb⊗bτLqdτ+C∫0t22je−c0μt−τ22αjΔju⊗uτLqdτ+C∫0t2je−c0μt−τ22αjΔjθτLqdτ.
Chooseσ such that(22)0<σ<α−1,σ+β=d21−1q.
By Sobolev’s inequality,(23)Δjb⊗bτLq≤b⊗bτLq≤bL2q2≤CΛσ+βbL22.
Similarly, we have(24)Δju⊗uτLq≤CuL22+Λ2α−1uL22=CuH2α−12.
We multiply the second equation in (1) by θθq−2 and integrate the result in space domain to obtain(25)1qddtθLqq≤u⋅edθθq−2≤uLqθLqq−1,which yields(26)θLq≤θ0Lq+∫0tuτLqdτ≤C+C∫0tuτH2α−1dτ.
Note thatΔjθLq≤θLq; then, inserting (23)–(26) into (21), we derive that(27)ΔjωtLq≤Ce−c0μt22αjΔjω0Lq+C∫0t22je−c0μt−τ22αjΛσ+βbτL22dτ+C∫0t22je−c0μt−τ22αjuτH2α−12dτ+C∫0t2je−c0μt−τ2αjuτH2α−1dτ.
Integrating this in time, together with (12), we obtain(28)22α−2j∫0tΔjωτLqdτ≤C2−2jΔjω0Lq+C∫0tΛσ+βbτL22dτ+C∫0tuτH2α−12+uτH2α−1dτ≤C+C∫0tuτH2α−12+uτH2α−1dτ.
Next, we applyΛα−1 to the first equation of (1) to obtain(29)12ddtΛα−1uL22+μΛ2α−1uL22=∫Λα−1b⋅∇b⋅Λα−1u−∫Λα−1u⋅∇u⋅Λα−1u+∫Λα−1θed⋅Λα−1u≔J1+J2+J3.
Using∇⋅b=0,(30)J1≤CΛ−1∇⋅b⊗bL2Λ2α−1uL2≤CbL4Λ2α−1uL2≤μ8Λ2α−1uL22+CΛσ0+βbL22,where we have chosen σ0 satisfying(31)0<σ0<α−1,σ0+β=d4.
Similarly, we have(32)J2≤CΛ−1∇⋅u⊗uL2Λ2α−1uL2≤CuL4Λ2α−1uL2≤μ8Λ2α−1uL22+CΛαuH12,(33)J3≤CθedL2Λ2α−2uL2≤CθL2uL21/2α−1Λ2α−1uL22α−2/2α−1≤μ8Λ2α−1uL22+CuL22+θL22.
Inserting (30)–(33) into (29),(34)ddtΛα−1uL22+μΛ2α−1uL22≤CΛσ0+βbL22+ΛαuH12+uL22+θL22.
Then, Gronwall’s inequality and (10)–(12) yields(35)Λα−1uL22+∫0tΛ2α−1uτL22dτ≤C.
This together with (26) yields (17). Substituting (35) into (28), we obtain(36)supj≥−122α−2j∫0tΔjωLqdτ<∞.
Thus, forq>d/2α−2,(37)Λd/qωτLqdτ≤C∫0tωτBq,1d/qdτ=C∫0t∑j≥−12d/qjΔjωτLqdτ≤C∫0t∑j≥−12d/q−2α−2j22α−2jΔjωτLqdτ=C∑j≥−12d/q−2α−2j22α−2j∫0tΔjωτLqdτ≤Csupj≥−122α−2j∫0tΔjωτLqdτ∑j≥−12d/q−2α−2j=Csupj≥−122α−2j∫0tΔjωτLqdτ<∞.
This is (16).
With Lemmas4–6 at our disposal, we are ready to prove Proposition 1.Proof.
of Proposition1. Applying Λs to (1) and taking the L2-inner product with Λsu,Λsb,Λsθ yields(38)12ddtΛsuL22+ΛsbL22+ΛsθL22+μΛs+αuL22+νΛs+βbL22=K1+K2+K3+K4+K5+K6+K7,where(39)K1=−∫Λs,u⋅∇u⋅Λsu,K2=∫Λsθed⋅Λsu,K3=∫Λsb⋅∇b⋅Λsu,K4=−∫Λs,u⋅∇b⋅Λsb,K5=∫Λs,b⋅∇u⋅Λsb,K6=−∫Λs,u⋅∇θ⋅Λsθ,K7=∫Λsu⋅edΛsθ.
Taking advantage of Lemma1,(40)K1≤C∇uL∞ΛsuL22.
By Young’s inequality,(41)K2+K7≤ΛsuL22+ΛsθL22.
Using∇⋅b=0, we have(42)K3≤CΛsbL2bLdα−1Λs+1uL2d/d+2−2α≤CΛsbL2ΛβbL2Λs+αuL2≤μ8Λs+αuL22+CΛβbL22ΛsbL22.
Similarly, by Lemma1, K4 and K5 are bounded by(43)K4+K5≤C∇uL∞ΛsbL22+CΛsbL2bLdα−1Λs+1uL2d/d+2−2α≤μ8Λs+αuL22+C∇uL∞ΛsbL22+CΛβbL22ΛsbL22.
By Lemma1,(44)K6≤C∇uL∞ΛsθL22+CΛsθL2Λ1+suL2d/d+2−2αθLd/α−1≤C∇uL∞ΛsθL22+CΛsθL2Λs+αuL2θL2/α−1≤μ8Λs+αuL22+C∇uL∞ΛsθL22+CθLd/α−12ΛsθL22.
Combining the above bounds with (38), we infer that(45)ddtΛsuL22+ΛsbL22+ΛsθL22+μΛs+αuL22+νΛs+βbL22≤C1+∇uL∞+ΛβbL22+θLdα−12ΛsuL22+ΛsbL22+ΛsθL22.
We bound‖∇u‖L∞ by Lemma 3. By Lemma 6, (10), and (17), we obtain(46)∫0tΛd/qωτLqdτ<∞,∫0tΛβbL22dτ<∞,∫0tθL2/α−12dτ<∞.
Then, the desired globalHs-bound follows from these bounds and Osgood’s inequality. This completes the proof of Proposition 1.


## 2.1. Preparations

To prove the main theorem, as preparations we give three lemmas in this section. The first contains two calculus inequalities.Lemma 1.
(see [11, 12]). Let s>0. Let 1<r<∞ and 1/r=1/p1+1/q1=1/p2+1/q2 with q1,p2∈1,∞ and p1,q2∈1,∞. Then,(5)Λs,fgLr≤C∇fLp1Λs−1gLq1+ΛsfLp2gLq2,with Λs,fg=Λsfg−fΛsg and(6)ΛsfgLr≤CΛsfLp1gLq1+fLp2ΛsgLq2,where C′s are constants.The second is the property of the generalized heat operator.Lemma 2.
(see, e.g., [13, 14]). Let A=ξ∈ℝd:r1≤ξ≤r2 with 0<r1<r2 being constants. Then, there exist two positive constants c and C such that, for(7)f∈Lpℝd,suppf^⊂λA,p∈1,∞,and we have, for α>0, t>0, and λ>0,(8)e−−ΔαtfLpℝd≤Ce−ctλ2αfLpℝd.The last is the following logarithmic type interpolation inequality.Lemma 3.
(see [14, 15]). Let s>1+d/2 and q∈1,∞. Then, there exists C>0 such that(9)∇fL∞≤C1+fL2+Λd/q∇fLqloge+ΛsfL2.


## 2.2. GlobalHs Bound

This section gives the proof of Proposition1. We first prove the following global L2-bound.Lemma 4.
For anyt>0, the solution u,b,θ of (1) obeys(10)uL22+bL22+θL22+∫0tΛαuτL22dτ+∫0tΛβbτL22dτ≤C.Proof.
Taking theL2-inner product of (1) with u,b,θ, together with the Young inequality, we obtain(11)12ddtuL22+bL22+θL22+μΛαuL22+νΛβbL22=∫θed⋅u+∫u⋅edθ≤uL22+θL22.
Then, (10) follows from this and Gronwall’s inequality.
The second one is a better regularity forb.Lemma 5.
Letu,b,θ be the solution of (1). Then, for all t>0 and 0<σ<α−1,(12)ΛρbtL22+∫0tΛβ+σbτL22dτ≤C.Proof.
ApplyingΛσ to 1.12 and dotting the result with Λσb, we have(13)12ddtΛσbL22+νΛβ+σbL22=∫Λσb⋅∇u⋅Λσb−∫Λσu⋅∇b⋅Λσb≔I1+I2.
Applying Lemma1, we arrive at(14)I1≤CΛσbL2d/d−2β∇uLd/β+Λσ∇uL2d/d+2+2σ−2αbLd/α−1−σΛσbL2≤CΛβ+σbL2ΛαuL2ΛσbL2≤ν4Λβ+σbL22+CΛαuL22ΛσbL22.
Note that(15)I2=−∫Λσu⋅∇b−u⋅∇Λσb⋅Λσb.
Then, by again applying Lemma1, we find that I2 obeys the same bound as I1. Combing the above estimates up, together with (10), Gronwall’s inequality yields (12).
The last preparation is stated as follows.Lemma 6.
Letu,b,θ be the solution of (1). Then, for all t>0 and d/2α−2<q<∞,(16)∫0tΛd/qωτLqdτ<∞,where ω=∇×u.
In addition, for all2≤q<∞,(17)θtLq<∞.Proof.
Taking curl on the both sides of1.11, we have(18)∂tω+μ−Δαω=∇×b⋅∇b−∇×u⋅∇u+∇×θed,with ω=∇×u. Write (18) into the integral form(19)ωt=e−μ−Δαtω0+∫0te−μ−Δαt−τ∇×∇⋅b⊗bτdτ−∫0te−μ−Δαt−τ∇×∇⋅u⊗uτdτ+∫0te−μ−Δαt−τ∇×θedτdτ.
We further localize it by applyingΔj with j≥−1:(20)Δjωt=Δje−μ−Δαtω0++∫0tΔje−μ−Δαt−τ∇×∇⋅b⊗bτdτ−∫0tΔje−μ−Δαt−τ∇×∇⋅u⊗uτdτ+∫0tΔje−μ−Δαt−τ∇×θedτdτ.
Forq∈1,∞, taking the Lq-norm to this equation, we derive that(21)ΔjωtLq≤Ce−c0μt22αjΔjω0Lq+C∫0t22je−c0μt−τ22αjΔjb⊗bτLqdτ+C∫0t22je−c0μt−τ22αjΔju⊗uτLqdτ+C∫0t2je−c0μt−τ22αjΔjθτLqdτ.
Chooseσ such that(22)0<σ<α−1,σ+β=d21−1q.
By Sobolev’s inequality,(23)Δjb⊗bτLq≤b⊗bτLq≤bL2q2≤CΛσ+βbL22.
Similarly, we have(24)Δju⊗uτLq≤CuL22+Λ2α−1uL22=CuH2α−12.
We multiply the second equation in (1) by θθq−2 and integrate the result in space domain to obtain(25)1qddtθLqq≤u⋅edθθq−2≤uLqθLqq−1,which yields(26)θLq≤θ0Lq+∫0tuτLqdτ≤C+C∫0tuτH2α−1dτ.
Note thatΔjθLq≤θLq; then, inserting (23)–(26) into (21), we derive that(27)ΔjωtLq≤Ce−c0μt22αjΔjω0Lq+C∫0t22je−c0μt−τ22αjΛσ+βbτL22dτ+C∫0t22je−c0μt−τ22αjuτH2α−12dτ+C∫0t2je−c0μt−τ2αjuτH2α−1dτ.
Integrating this in time, together with (12), we obtain(28)22α−2j∫0tΔjωτLqdτ≤C2−2jΔjω0Lq+C∫0tΛσ+βbτL22dτ+C∫0tuτH2α−12+uτH2α−1dτ≤C+C∫0tuτH2α−12+uτH2α−1dτ.
Next, we applyΛα−1 to the first equation of (1) to obtain(29)12ddtΛα−1uL22+μΛ2α−1uL22=∫Λα−1b⋅∇b⋅Λα−1u−∫Λα−1u⋅∇u⋅Λα−1u+∫Λα−1θed⋅Λα−1u≔J1+J2+J3.
Using∇⋅b=0,(30)J1≤CΛ−1∇⋅b⊗bL2Λ2α−1uL2≤CbL4Λ2α−1uL2≤μ8Λ2α−1uL22+CΛσ0+βbL22,where we have chosen σ0 satisfying(31)0<σ0<α−1,σ0+β=d4.
Similarly, we have(32)J2≤CΛ−1∇⋅u⊗uL2Λ2α−1uL2≤CuL4Λ2α−1uL2≤μ8Λ2α−1uL22+CΛαuH12,(33)J3≤CθedL2Λ2α−2uL2≤CθL2uL21/2α−1Λ2α−1uL22α−2/2α−1≤μ8Λ2α−1uL22+CuL22+θL22.
Inserting (30)–(33) into (29),(34)ddtΛα−1uL22+μΛ2α−1uL22≤CΛσ0+βbL22+ΛαuH12+uL22+θL22.
Then, Gronwall’s inequality and (10)–(12) yields(35)Λα−1uL22+∫0tΛ2α−1uτL22dτ≤C.
This together with (26) yields (17). Substituting (35) into (28), we obtain(36)supj≥−122α−2j∫0tΔjωLqdτ<∞.
Thus, forq>d/2α−2,(37)Λd/qωτLqdτ≤C∫0tωτBq,1d/qdτ=C∫0t∑j≥−12d/qjΔjωτLqdτ≤C∫0t∑j≥−12d/q−2α−2j22α−2jΔjωτLqdτ=C∑j≥−12d/q−2α−2j22α−2j∫0tΔjωτLqdτ≤Csupj≥−122α−2j∫0tΔjωτLqdτ∑j≥−12d/q−2α−2j=Csupj≥−122α−2j∫0tΔjωτLqdτ<∞.
This is (16).
With Lemmas4–6 at our disposal, we are ready to prove Proposition 1.Proof.
of Proposition1. Applying Λs to (1) and taking the L2-inner product with Λsu,Λsb,Λsθ yields(38)12ddtΛsuL22+ΛsbL22+ΛsθL22+μΛs+αuL22+νΛs+βbL22=K1+K2+K3+K4+K5+K6+K7,where(39)K1=−∫Λs,u⋅∇u⋅Λsu,K2=∫Λsθed⋅Λsu,K3=∫Λsb⋅∇b⋅Λsu,K4=−∫Λs,u⋅∇b⋅Λsb,K5=∫Λs,b⋅∇u⋅Λsb,K6=−∫Λs,u⋅∇θ⋅Λsθ,K7=∫Λsu⋅edΛsθ.
Taking advantage of Lemma1,(40)K1≤C∇uL∞ΛsuL22.
By Young’s inequality,(41)K2+K7≤ΛsuL22+ΛsθL22.
Using∇⋅b=0, we have(42)K3≤CΛsbL2bLdα−1Λs+1uL2d/d+2−2α≤CΛsbL2ΛβbL2Λs+αuL2≤μ8Λs+αuL22+CΛβbL22ΛsbL22.
Similarly, by Lemma1, K4 and K5 are bounded by(43)K4+K5≤C∇uL∞ΛsbL22+CΛsbL2bLdα−1Λs+1uL2d/d+2−2α≤μ8Λs+αuL22+C∇uL∞ΛsbL22+CΛβbL22ΛsbL22.
By Lemma1,(44)K6≤C∇uL∞ΛsθL22+CΛsθL2Λ1+suL2d/d+2−2αθLd/α−1≤C∇uL∞ΛsθL22+CΛsθL2Λs+αuL2θL2/α−1≤μ8Λs+αuL22+C∇uL∞ΛsθL22+CθLd/α−12ΛsθL22.
Combining the above bounds with (38), we infer that(45)ddtΛsuL22+ΛsbL22+ΛsθL22+μΛs+αuL22+νΛs+βbL22≤C1+∇uL∞+ΛβbL22+θLdα−12ΛsuL22+ΛsbL22+ΛsθL22.
We bound‖∇u‖L∞ by Lemma 3. By Lemma 6, (10), and (17), we obtain(46)∫0tΛd/qωτLqdτ<∞,∫0tΛβbL22dτ<∞,∫0tθL2/α−12dτ<∞.
Then, the desired globalHs-bound follows from these bounds and Osgood’s inequality. This completes the proof of Proposition 1.



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*Source: 9943271-2021-06-04.xml*