On Magnetic Inhibition Theory in Non-resistive Magnetohydrodynamic Fluids
aa r X i v : . [ m a t h - ph ] M a y On Magnetic Inhibition Theory in Non-resistiveMagnetohydrodynamic Fluids
Fei Jiang a , Song Jiang b a College of Mathematics and Computer Science, Fuzhou University, Fuzhou, 350108, China. b Institute of Applied Physics and Computational Mathematics, P.O. Box 8009, Beijing, 100088, China.
Abstract
We investigate why the non-slip boundary condition for the velocity, imposed in the directionof impressed magnetic fields, can contribute to the magnetic inhibition effect based on the mag-netic Rayleigh–Taylor (abbr. NMRT) problem in nonhomogeneous incompressible non-resistivemagnetohydrodynamic (abbr. MHD) fluids. Exploiting an infinitesimal method in Lagrangiancoordinates, the idea of (equivalent) magnetic tension, and the differential version of magneticflux conservation, we give an explanation of physical mechanism for the magnetic inhibitionphenomenon in a non-resistive MHD fluid. Moreover, we find that the magnetic energy in thenon-resistive MHD fluid depends on the displacement of fluid particles, and thus can be re-garded as elastic potential energy. Motivated by this observation, we further use the well-knownminimum potential energy principle to explain the physical meaning of the stability/instabilitycriteria in the NMRT problem. As a result of the analysis, we further extend the results onthe NMRT problem to the stratified MHD fluid case. We point out that our magnetic inhibi-tion theory can be used to explain the inhibition phenomenon of other flow instabilities, suchas thermal instability, magnetic buoyancy instability, and so on, by impressed magnetic fields innon-resistive MHD fluids.
Keywords:
Magnetic inhibition phenomenon; incompressible magnetohydrodynamic fluids;Rayleigh–Taylor instability; thermal instability; stabilizing effect.
1. Introduction
The study of the inhibition of flow instability by (impressed) magnetic fields goes back tothe theoretical work of Chandrasekhar, who first discovered the inhibiting effect of a sufficientlylarge (impressed) vertical magnetic field on the thermal (or convective) instability based on thelinearized magnetic Boussinesq equations of magnetohydrodynamic (abbr. MHD) fluids in ahorizontal layer domain in 1952 [4, 5]. Then Nakagawa experimentally verified Chandrasekhar’slinear magnetic inhibition theory in 1955 [26, 27]. Later, many authors tried to provide a rigorousmathematical proof of the magnetic inhibiting theory for the nonlinear case. In 1985, Gladi firstsuccessfully showed the theory for the nonlinear magnetic Boussinesq equations with resistivity byusing a so-called generalized energy method [10]. Now, it still remains open to mathematicallyshow Chandrasekhar’s assertion that thermal instability could be also inhibited by a vertical(magnetic) field in non-resistive MHD fluids, see [6, page 160]. An alternative question ariseswhether one can mathematically verify the inhibition of other instabilities by a magnetic field innon-resistive MHD fluids. The answer is positive, for example, the authors of this paper recently
Email addresses: [email protected] (Fei Jiang), [email protected] (Song Jiang)
Preprint submitted to Elsevier May 22, 2018 ave mathematically verified the magnetic inhibition phenomenon (or stability result) in thenonhomogeneous magnetic Rayleigh–Taylor (abbr. NMRT) problem [19]. Such a result in theNMRT problem supports Chandrasekhar’s assertion in the certain sense.In [19], the authors also showed the inhibition effect of a horizonal field for the case that theNMRT problem is considered in a vertical layer domain. Moreover, the authors further pointedout that the nonslip boundary condition of the velocity at the two parallel fixed slabs, imposedin the direction of the magnetic field, can contribute to the magnetic inhibition effect. This alsogives a reason why the horizonal field does not have inhibition effect as the vertical magnetic inthe RT instability in a horizontal layer domain. Recently, the inhibition effect of a horizontalfield is also found by the authors in the study of the magnetic buoyancy instability (the Parkerinstability) problem [17].As mentioned above, progress has been made on the mathematical analysis of the magneticinhibition phenomenon in non-resistive MHD fluids, however, to our best knowledge, there isstill no physical interpretation why the nonslip boundary condition, imposed in the direction ofthe (impressed) magnetic field, can contribute to the magnetic inhibition effect. In this article,we investigate the physical mechanism of the magnetic inhibition theory based on the NMRTproblem. Next, we briefly introduce our main results.First, we exploit an infinitesimal method in Lagrangian coordinates to give a compellingphysical mechanism of the inhibition effect of magnetic fields on the RT instability for the knownmathematical results in the NMRT problem. The physical mechanism can be described as follows.In the NMRT problem, we can think that the non-resistive MHD fluid under equilibrium is madeup of infinite (fluid) element lines which are parallel to the impressive field. Once we disturbthe rest state, the element lines will be bend. By using a differential version of magnetic fluxconservation in Lagrange coordinates, we can compute out that the direction of the magneticfield at each point of element lines is just tangent to the element lines in motion. Combining withthe idea that the Lorentz force (a body force) can be equivalent to a surface force, i.e., a so-calledmagnetic tension, we can find that each element line can be regarded as an elastic string, and themagnetic tension intensity is proportional to the impressed field intensity. Thus, the magnetictension will resist gravity and straighten the all bent element lines, when the impressed fieldintensity is sufficiently large. Noting that the endpoints of all element lines are fixed due to thenon-slip boundary condition for the velocity, thus all bent element lines will try to restore totheir initial locations, and can vibrate around their initial location under the magnetic tension.In particular, due to the viscosity, all bent element lines will asymptotically converge to theirinitial location. The corresponding details will be presented in Section 2. Moreover, we obtain aso-called equivalence theorem of magnetic flux conservation in the analysis process, see Theorem2.1. We mention that our magnetic inhibition mechanism with nonslip boundary condition ofthe velocity can be also used to explain the inhibition phenomenon of thermal instability andmagnetic buoyancy instability [33] by magnetic fields.Second, we further give the physical meaning of stability and instability criteria in the NMRTproblem. By the physical mechanism of the magnetic inhibition effect and the mathematicalrepresentation of magnetic energy, we find that the magnetic energy in non-resistive MHD fluidscan be regarded as the elastic potential energy. This means that the well-known minimumpotential energy principle can be applied to the NMRT problem. More precisely, if the totalpotential energy in the magnetic energy and the gravitational potential energy in the rest stateis minimal, then the NMRT problem is stable. Otherwise, the NMRT problem is unstable.Motivated by the minimum potential energy principle, and using some mathematical techniques,we can indeed prove that under the stability criterion, the total potential energy in the rest state2eaches its minimum, while under the instability criterion, the total potential energy in the reststate is not minimal, see Theorem 3.1. We shall also extend the results on the NMRT problemto the stratified magnetic RT (abbr. SMRT) problem (see Theorem 3.2), and the correspondinganalysis details will be presented in Section 3 and the rigorous proof in Section 4.Finally, in Section 5, we extend the mathematical result of the magnetic inhibition in theNMRT problem to the magnetic Boussinesq problem without heat conduction, and shall see thatthe obtained result supports Chandrasekhar’s assertion in the absence of heat conduction.We end this section by listing some notations which will be used throughout this article.(1) Basic notations:The superscript T denotes the transposition. I always denotes the 3 × e i stands for the unit vector, in which the i -th component is 1. det A denotes the determinant ofthe matrix A . For x := ( x , x , x ) ∈ R , we define x h := ( x , x ) and x v := ( x , x ). T := R / Z is the usual 1-torus. Let f := ( f , f , f ) T be a vector function defined in a three-dimensionaldomain, we define f h := ( f , f ) T , ∇ h f h := ( ∂ j f i ) × , ∂ ~n f := ~n · ∇ f , and ∂ ~n f := ( ~n · ∇ ) f , where ~n is a constant vector. d y h := d y d y is the infinitesimal on a plane. ϕ denotes the initial dataof a scalar, or a vector, or a matrix function ϕ ( x, t ), where t is the time variable.(2) Definitions of domains and boundaries: T = (2 πL T ) × (2 πL T ) , Ω + := { x := ( x , x v ) ∈ R | x v ∈ T , < x < h } , Ω ba := { x := ( x h , x ) ∈ R | x h ∈ T , a < x < b } , Ω + := Ω h , Ω − := Ω − l , Ω := Ω + ∪ Ω − , Ω ba − := R × ( a, b ) , Σ a := T × { x = a } , Σ + := Σ h , Σ := Σ , Σ − := Σ − l , where L , L , h , l , a , b >
0, and a > b . Let D be a domain, then D denotes the closure of D ,and ∂D denotes the boundary of Ω , if Ω is a bounded domain; { x = 0 } ∪ { x = h } , if D = Ω + ; { x = a } ∪ { x = b } , if D = Ω ba . We define f ( P ) := { x ∈ R | x = f ( y ) for y ∈ P } , where the vector function f is defined on theset P . It is should be noted that if a function is defined on Ω + , then the function is horizontallyperiodic, i.e., f ( x , x , x ) = f (2 mπL + x , nπL + x , x ) for any integer m and n. Similarly, if a function is defined on Ω + , then the function is vertically periodic function.(3) Notations of function spaces and simplified norms: H i ( D ) := W i, ( D ) denotes a Soblev space defined in D,H σ ( D ) := { η ∈ H ( D ) | div η = 0 , η | ∂D = 0 in the sense of trace } ,C ∞ σ (Ω ba ) := { w ∈ C ∞ (Ω ba ) | w = 0 on Ω ba \ Ω dc for some Ω dc satisfying a < c < d < b, div w = 0 } ,H k, , ∗ (Ω + ) := { ̟ ∈ H (Ω + ) ∩ H k (Ω + ) | ̟ ( y ) + y : Ω h − 7→ Ω h − is a homeomorphism mapping , det( ∇ ( ̟ ( y ) + y )) = 1 } ,H k, , ∗ (Ω) := { ̟ ∈ H (Ω h − l ) ∩ H k (Ω) | ̟ ( y ) + y : Ω h − l − 7→ Ω h − l − is a homeomorphism mapping , det( ∇ ( ̟ ( y ) + y )) = 1 } , k · k i,D := k · k H i ( D ) , | · | i := k · k H i ( T ) , i >
0, and k > W ,p ( D ) := L p ( D ) isthe usual Lebesgue space. In addition, if a norm is defined in a periodic domain or a horizon-tally/vertically periodic domain, then the norm is equal to the one defined in a periodic cell.For examples, k · k H k (Ω ba ) = k · k H k ((0 , πL ) × (0 , πL ) × ( a,b )) , and k · k H k ( T ) = k · k H k ((0 , πL ) × (0 , πL )) .Similarly, we have R Ω ba = R (0 , πL ) × (0 , πL ) × ( a,b ) and R T = R (0 , πL ) × (0 , πL ) .
2. Magnetic inhibition mechanism
This section is devoted to providing the physical mechanism of magnetic inhibition phe-nomenon. We first recall the known mathematical results on the NMRT problem in Subsection2.1 and define the direction of a (impressed) magnetic field in a non-resistive MHD fluid. We thenreformulate the momentum equations of the perturbed MHD fluid in Lagrangian coordinates inSubsection 2.2. Finally, we use the differential version of magnetic flux conservation to definethe direction of the magnetic field, and give thus the reason why the no-slip boundary conditionof the velocity, imposed in the direction of the impressed field, can contribute to the magneticinhibition effect in Subsection 2.3.
Let us first recall the stability and instability results of the NMRT problem. To start with,we introduce the three-dimensional homogeneous incompressible viscous MHD equations withzero resistivity in the presence of a uniform gravitational field in a domain D ⊂ R , ρ t + v · ∇ ρ = 0 ,ρv t + ρv · ∇ v + ∇ p − µ ∆ v = λ ( ∇ × M ) × M − ρge ,M t = M · ∇ v − v · ∇ M, div v = div M = 0 . (2.1)Here the unknowns ρ := ρ ( x, t ), v := v ( x, t ), M := M ( x, t ) and p := p ( x, t ) denote the density,velocity, magnetic field and the kinetic pressure of a MHD fluid, respectively; the positive con-stants λ , µ and g stand for the permeability of vacuum divided by 4 π , the coefficient of shearviscosity and the gravitational constant, respectively. For the system (2.1), we impose the initialand boundary conditions: ( ρ, v, N ) | t =0 = ( ρ , v , N ) in D, (2.2) v ( x, t ) | ∂D = 0 for any t > . (2.3)We call (2.1)–(2.3) the NMHD model. We mention that the well-posedeness problem of viscousMHD equations with zero resistivity have been extensively investigated, see [15, 22, 23, 30–32, 35, 37] for examples.Now we consider a uniform magnetic field (i.e., every component of the magnetic field isconstant, and at least one component is non-zero) ¯ M = ( ¯ M , ¯ M , ¯ M ) and a smooth RT densityprofile ¯ ρ ∈ C ( D ), which is independent of ( x , x ) and satisfiesinf x ∈ D ¯ ρ > , ¯ ρ ′ | x = x > x ∈ D x := { x | ( x h , x ) ∈ D } , (2.4)where we have denoted ¯ ρ ′ := d ¯ ρ/ d x . Then, r N := ( ¯ ρ, , ¯ M ) is a rest state (or equilibrium)solution of the NMHD model with an associated pressure ¯ p defined by the following relation ∇ ¯ p = − ¯ ρge . (2.5)4e often call ¯ M the impressed magnetic field. The problem whether the rest state r N is stableor unstable to the NMHD model is called the NMRT problem.The second condition in (2.4) is called the RT condition and assures that there is at least aregion in which the RT density profile has larger density with increasing height x , thus leading tothe RT instability under small perturbation for a sufficiently small ¯ M . However, for a sufficientlylarge | ¯ M | , the RT instability may be inhibited. Recently, the authors have shown the inhibitionof the RT instability by a magnetic field [19]. Before recalling the results in [19], we introducesome notations.Denote Π := ( ¯ M , , T , Π := (0 , , ¯ M ) T , Ω = Ω + and Ω = Ω + . If D takes Ω + , we shallfurther assume that the density profile ¯ ρ is a vertically horizontal function, i.e.,¯ ρ | x =2 πnL = ¯ ρ | x =2 πmL for any integer n, m. We define V Dg ¯ ρ ′ ( w ) := g Z D ¯ ρ ′ w d x and V D~n ( w ) := λ k ∂ ~n w k ,D , where ~n is a constant vector. We further define m D N ,j := s sup w ∈ H σ ( D ) V Dg ¯ ρ ′ ( w ) V De j ( w ) . (2.6)We remark here that if j = 3 and D = Ω + , we rewrite m Ω + N , by m N for the sake of simplicity.Then the authors have established the following stability and instability results for the NMRTproblem [19]:(1) Stability criterion: if | ¯ M j | > m Ω j N ,j (called asymptotic stability condition) , then the rest state r N with ¯ M = Π j is asymptotically stable to the NMRT model definedon D = Ω j with proper initial conditions under small perturbation for j = 1 and 3.(2) Instability criterion: if | ¯ M j | < m Ω j N ,j (called instability condition),then the rest state r N with ¯ M = Π j is unstable to the NMRT model defined on D = Ω j inthe Hadamard sense for j = 1 and 3. In addition, the rest state r N with ¯ M = Π k is alwaysunstable to the NMRT model defined on D = Ω j in the Hadamard sense, if k = j .Since ¯ ρ satisfies the RT condition and the velocity is non-slip on the boundary of Ω j , we canverify that m Ω j N ,j ∈ (0 , ∞ ). Thus the above positive constant m Ω j N ,j is called a strength-threshold ofΠ j for stability and instability of the NMRT problem. Moreover, for ¯ ρ ′ being a positive constant,we can compute out that m Ω j N ,j = 2 h p g ¯ ρ ′ /σπ. (2.7)Considering the case D = Ω + , we find that the rest state r N with ¯ M = ( ¯ M , , T is alwaysunstable to the NMRT model defined on Ω . This means that a horizontal field can not inhibitthe RT instability in a horizontally periodic domain. However, in view of the stability criterionfor j = 1, we find that a horizontal magnetic field can inhibit the RT instability in a vertically5eriodic domain with finite width. Thus, one observes that the non-slip boundary condition,imposed in the direction of the magnetic field, can contribute to the magnetic inhibition effect.Of course, we can use the threshold (2.6) to explain the above fact. Considering the case D = Ω + and ¯ M = (0 , , ¯ M ), for any w ∈ H σ (Ω + ), since w | ∂ Ω + = 0 and Ω + is horizontallyperiodic with finite height, then one has k w k , Ω + c k ∂ w k , Ω + for some constant c dependingon the domain, which implies m N ∈ (0 , ∞ ). However, for any positive constat j , we can alwaysconstruct a function w ∈ H σ (Ω + ) satisfying k w k , Ω + > j k ∂ w k , Ω + , which implies m Ω + N , = ∞ .Thus, the strength of any horizontal field ¯ M is always less than m Ω + N , , and this shows why therest state r N with ¯ M = ( ¯ M , ,
0) is always unstable to the NMRT model defined Ω + .Unfortunately, from the above mentioned stability/instability results we can not see any phys-ical mechanism to explain why the non-slip boundary condition can contribute to the magneticinhibition effect. To reveal the physical mechanism of the magnetic inhibition phenomenon, weshall carry out analysis of forces based the equations of non-resistive MHD fluids in Lagrangiancoordinates. From now on, we always assume that ¯ M is a general non-zero uniform magnetic field. How-ever, to avoid a discussion of the geometry structure of a general domain D in the analysis of(fluid) element lines, we only consider the simplest domain, i.e., D = Ω + . Of course, the magneticinhibition mechanism for D = Ω + can be easily generalized to a general domain that is boundedin the direction of ¯ M .We assume that ( ρ, v, M, q ) be a classical solution to the NMHD model defined on Q T + :=Ω + × [0 , T ] with T >
0, in which v satisfies the following regularity v ∈ C ( Q T + ) and ∇ v ∈ C ( Q T + ) . (2.8)It is well-known that the (generalized) Lorentz force on the right-hand side of (2.1) can bewritten as the divergence of the magnetic part of the electromagnetic stress, i.e., λ ( ∇ × M ) × M = λ div( M ⊗ M − | M | I/ . (2.9)We consider a bounded domain V with smooth surface in Ω + , then the Lorentz force acting onthe MHD fluid in V is given by the formula λ Z V ( ∇ × M ) × M d x = λ Z S F ν d S, (2.10)where F ν = λM · νM − λ | M | ν/ ν denotes the unit outer normal vector of V . The first term λM · νM in F ν is called the(equivalent) magnetic tension, the direction of which is along that of the magnetic field, andthe strength of which is λ | M | . The second term − λ | M | ν/ as follows: ρv t + ρv · ∇ v + ∇ p ∗ − µ ∆ v = λ div( M ⊗ M ) − ρge , (2.11)where p ∗ denotes the sum of the kinetic pressure p and the magnetic pressure λ | M | /
2. Hence,one can think that the momentum equation (2.11) describes the motion of a fluid driven by the6agnetic tension λM · νM and the gravity force − ρge . We denote the equations (2.1) with(2.11) in place of (2.1) by the system (2.1) ∗ .In view of the system (2.1) ∗ , it is very important to analyze how the magnetic tension affectsthe motion of the MHD fluid. To this end, we shall rewrite (2.1) ∗ in Lagrange coordinates. Wefirst label the particles (or element points) of the non-resistive fluid by the relation y = x , where x denote the Eulerian coordinates of particles at t = 0, and we call y the Lagrange coordinates(or Lagrangian particle markers). Then we define a location function (or particle-trajectorymapping) ζ of the fluid particles y as the solution to (cid:26) ζ t ( y, t ) = v ( ζ ( y, t ) , t ) ζ ( y,
0) = y, where y ∈ Ω + . Obviously, η := ζ − y represents the displacement function of particles. By theregularity (2.8) and the classical ODE theory [40], the solution ζ enjoys the following regularity: ζ ∈ C ( Q T + ) and ζ ∈ C ( Q T + ) . Note that the non-slip boundary condition (2.3) with Ω + in place of D is essential here, since itguarantees that for each t ∈ [0 , T ], ζ ( · , t ) : Ω + → Ω + is a homeomorphism mapping (referring toLemma 4.2).Before giving the motion equations (2.1) in Lagrangian coordinates, we temporarily introducesome notations involving ζ . Define A := ( A ij ) × via A T = ( ∇ ζ ) − := ( ∂ j ζ i ) − × , (2.12) J := det( ∇ ζ ), and the differential operators ∇ A and div A as follows. ∇ A w := ( ∇ A w , ∇ A w , ∇ A w ) T , ∇ A w i := ( A k ∂ k w i , A k ∂ k w i , A k ∂ k w i ) T , div A ( f , f , f ) T = (div A f , div A f , div A f ) T , div A f i := A lk ∂ k f il , ∆ A w := (∆ A w , ∆ A w , ∆ A w ) T and ∆ A w i := div A ∇ A w i for vector functions w := ( w , w , w ) T and f i := ( f i , f i , f i ) T , where we have used the Einsteinconvention of summation over repeated indices, and ∂ k denotes the partial derivative with respectto the k -th component of y . Moreover, ζ enjoys the following properties:(1) Let δ ij be the Kronecker delta, then ∂ i ζ k A kj = A ik ∂ k ζ j = δ ij . (2.13)(2) Since div v = 0, one has J = J , where J is the initial value of J . In particular, if J = 1, then J = 1 . (2.14)(3) In view of the definition of A and (2.14), we can see that A = ( A ∗ ij ) × , where A ∗ ij is thealgebraic complement minor of the ( i, j )-th entry ∂ j ζ i . Moreover, we can compute outthat ∂ k A ∗ ik = 0, which implies div A u = ∂ l ( A ∗ kl u k ) = 0 . (2.15)7ow, we define the Lagrangian unknowns by( ̺, u, q ∗ , B )( y, t ) = ( ρ, v, p ∗ , M )( ζ ( y, t ) , t ) for ( y, t ) ∈ Ω + × R + . Thus in Lagrangian coordinates, the evolution equations for ̺ , u , B and q ∗ read as ζ t = u,̺ t = 0 ,̺u t + ∇ A q ∗ − µ ∆ A u = λB · ∇ A B − ̺ge ,B t − B · ∇ A u = 0 , div A u = div A B = 0 . (2.16)Since we slightly disturb the rest state only in the velocity at the initial time t = 0, we have( ζ , ̺, u, B ) | t =0 = ( y, ¯ ρ, v , ¯ M ) . (2.17)We mention that div A B = 0 automatically holds since the initial value of div A B is just zero.In addition, in view of the non-slip boundary condition of v , we obtain the following boundarycondition: ( ζ , u ) | ∂ Ω + = ( y, . (2.18) The most important advantage of using Lagrange coordinates lies in that the magnetic fieldcan be expressed by the location function ζ , so that we can determine the direction of themagnetic tension. Next, we derive this fact.Consider the equations ζ t = u,̺ t = − ̺ div A u,B t − B · ∇ A u = − B div A u (2.19)with initial condition ( ζ , ̺, B ) | t =0 = ( ζ , ̺ , B ). We should note here that div u may not be 0 in(2.19). It is well-known that the system (2.19) can imply the mass conservation J ̺ = J ̺ andmagnetic flux conservation J A T B = J A T0 B , (2.20)which can be rewritten as a so-called Cauchy’s integral of the magnetic field [34]: B = J ∇ ζ A T0 B /J. (2.21)Here we call (2.20) the magnetic flux conservation, since we can directly verify that the conser-vation relation (2.20) is equivalent to the well-known theorem of magnetic flux conservation innon-resistive MHD fluids, see Theorem 2.1 in the next subsection.For an incompressible non-resistive MHD fluid, since div A u = 0, one obtains J = J inLagrangian coordinates. Hence, (2.21) reduces to B = ∇ ζ A T B . (2.22)The above formula can be also directly deduced from the equations (2.16) and (2.16) , and wealso call (2.22) the vorticity-transport formula in the theory of the vorticity equation, see [24,Propotition 1.8]. Since B | t =0 = ¯ M and A T0 = I , we further deduce from (2.22) that B = ∂ ¯ M ζ . (2.23)8he above expression has the prominent physical meaning that the larger the stretching ∇ ζ ofa non-resistive MHD fluid along the direction of ¯ M is, the stronger the magnetic field becomes.Next, we use (2.23) to determine the direction of the magnetic tension of each particle in motion.Let us think that the fluid is made up of infinite element lines, which are parallel to ¯ M . Weslightly disturb the rest state in the velocity by a perturbation v at t = 0. The motion equationsin Lagrangian coordinates after perturbation are described by (2.16) with initial-boundary valueconditions (2.17) and (2.18). Now, consider a straight element line denoted by l at t = 0. Wedenote by the set l (it is a segment for M = 0, and a line without endpoints for M = 0) theinitial location occupied by l . Under a slight perturbation, the element line l may be bent, andmay move to a new location at time s , which is occupied by l and denoted by the set l s (it is acurve, if l is bent). We further choose a particle Y on l at t = 0, and denote the coordinates of Y at t = 0 and t = s by y and x s , respectively. Hence, x s = ζ ( y , s ).Noting that the curve l s can be defined by the location function ζ ( y, s ) defined on l , onecan easily verify that ∂ ¯ M ζ ( y s , s ) is a tangential direction at point x s of the curve l s . In otherwords, the curve l s is tangent to the direction of the magnetic field at each point. Recallingthe definition of the magnetic line (i.e., a line which is tangent to the direction of the magneticfield at each point, and the direction of a magnetic line can be defined by that of the magneticfield), we find that l s is just a magnetic line. This is just the well-known (magnetic field) lineconservation theorem (or loosely called the frozen-in magnetic field lines in some literature [21]),i.e., any two particles in a non-resistive MHD fluid that are at an instant on a common magneticfield line will keep on the common magnetic field line at any other instant [9, 34].Since l s is a magnetic field line, by the mathematical expression of the magnetic tension, wesee that the element line l at any time s can be always regarded as an elastic string due to themagnetic tension along the curve l s . Thus, if the element line is bent at time s , then the magnetictension will drive the bend part of the element line back to a straight line, thus playing a roleof restoring force and representing the stabilizing effect in the motion of the non-resistive MHDfluid. We take the following figure to illustrate this fact by infinitesimal method. • y ✻ F y • y ✲ F y O (cid:0)(cid:0)(cid:0)✒ F r ❅❅❅❅❅ The element curve y Oy in the above figure is a bend part, denoted by l p , of element line l attime s , and the element segment y y is the initial location of l p . We assume that the directionof ¯ M is parallel to the vector −−→ y y . Now, we denote by F Y i the magnetic tension acting on theelement endpoint Y i of l p at time s for i = 1 and 2. Noting that y Oy is a magnetic field line,in view of the expression of the magnetic tension, we see that the total magnetic tension actingon the element curve y Oy without two element endpoints Y and Y is zero, and the magnetictension acting on the element endpoint Y i is given by the formula F y i = λ∂ ¯ M ζ ( y i , s ) · ν y i ∂ ¯ M ζ ( y i , s ) , where ν y i denotes the unit outer normal vector at element endpoint y i for i = 1 and 2. Conse-quently, the magnetic tension acting on the bend element curve y Oy is just the resultant force9 r of F y and F y from the above figure. Obviously, the resultant force F r drives the elementcurve y Oy to recover to a straight element line. From the above figure and (2.23), we canintuitively see that the more bent the magnetic line y Oy is, the stronger the resultant force F r (or called the magnetic restoring force) becomes.By the above analysis on the physical mechanism of stabilizing effect of magnetic fields,we can easily explain the stability/instability phenomenon in the NMRT problem under smallperturbation. First, we consider the case ¯ M = 0. Since the fluid is incompressible and v is zeroon the boundary, there always exist some bent element lines after disturbing the rest state. Oncesome element lines are bent, the destabilizing factor (i.e. gravity) may promote the developmentof the RT instability, which will further make the element lines more bent. Though the magnetictension resists the bend of the element lines, it can not prevent the heavier fluid from sinkingunder gravity, if it is too small. Hence, we shall increase the strength of ¯ M so that a properly largemagnetic tension can resist gravity . Noting that the endpoints of all element lines are fixed, onesees that all bent element lines will recover to their initial location, and may vibrate around theirinitial location under the magnetic tension with sufficiently large ¯ M , and the magnetic tensionrepresents the inhibition effect. Consequently, under the effect of viscosity, all bent elements willasymptotically converge to their initial location.Now, we roughly explain why the small height h of the domain also contributes to stability,see (2.7). To clearly see the reason, we consider a particle Y m which just lies at the midpointon the element line l at t = 0. The fixed two endpoints of l are labeled by y i for i = 1 and 2.We assume that there exists a force F pulling Y m along the perpendicular bisector of l . So, Y m moves to a new location y s at time s . Obviously, the magnetic line y y s y is more bent if h is smaller. As aforementioned, the more bent the magnetic line is, the stronger the magneticrestoring force becomes. Thus, if h is getting smaller, the magnetic line y y s y will result in astronger magnetic restoring force, which will resist the pulling force F . This means that whenthe height h is getting smaller, it is more difficult for the RT instability in a non-resistive MHDfluid to occur. In addition, for the two-dimensional case, by the magnetic inhibition mechanismwe can easily guess that a sufficiently large ( ¯ M ,
0) also has magnetic inhibition effect in thedomain T × (0 , h ). More precisely, if one disturbs the rest state, i.e., ( ρ, M, v ) = ( ¯ ρ ( x ) , ¯ M ,
0) inthe velocity, then the density, magnetic field and velocity will finally converge to ( ¯ ρ ( x ) , ¯ M , ζ ( x ) , x ), where ζ ( x ) onlydepends on x , and ζ ( x ) | x =0 ,h = 0.Now we consider the case ¯ M = 0. Since the domain is horizontally periodic, points onthe element lines can move freely. Thus, there may exist some element lines that are far awayfrom their initial location and parallel to ¯ M . Since such element lines are parallel to ¯ M , themagnetic tension can not pull the element lines back to their initial location. Therefore, ahorizonal magnetic field has no inhibition effect upon the RT instability in the horizontallyperiodic domain Ω + . Of course, if Ω + is a vertically horizontal domain with finite width, then ahorizonal magnetic field also has inhibition effect.From the above analysis on both ¯ M = 0 and ¯ M = 0 cases, we have seen why the non-slip In Theorem 3.1 in the next section, we shall see that only the increment | ¯ M | of ¯ M contributes to the inhibitioneffect. Here we provide an explanation for this. Assume that y and y in the above figure are fixed on the upperand lower boundaries of Ω + , respectively. For the slightly bent case in the above figure, one can intuitively seethat the stronger the magnetic restoring force acting on the element curve y Oy is, the bigger the intensity of¯ M is. However, we can also easily observe that the bigger the value | ¯ M i | ( i = 1 or 2) is, the longer the length of y Oy is. This means that one can not improve the intensity of the magnetic restoring force acting on the unitlength of y Oy by increasing the value of ¯ M i . This explains why the threshold is independent of | M i | in Ω + . M is sufficiently large, although theauthors have no idea to prove such a conclusion mathematically.In the analysis of magnetic inhibition for ¯ M = (0 , , ¯ M ) T , we consider the non-slip boundarycondition u | ∂ Ω + = 0. An interesting question is that what will happen for the case that particleson ∂ Ω + can freely slip on the boundary of ∂ Ω + , i.e., the condition u | ∂ Ω + = 0 is kept only. Weguess that, if the element lines are slightly bent and ¯ M is sufficiently large, then, due to themagnetic tension and viscosity, it could be possible that the bent element lines finally becomestraight again. Unfortunately, we can not verify this mathematically. We further discuss the conservation relation (2.20). The conservation relation is not onlyvery useful in the investigation of the dynamic behavior of a magnetic field [34], but also inthe study of some mathematical problems from MHD fluids, such as the global well-posdnessproblem [1, 14, 35], and the inhibition effect duo to magnetic fields upon flow instabilities, see[17–20, 38, 39]. As aforementioned, we can consider the conservation relation as a differentialversion of magnetic flux conservation due to their equivalence. Next, we establish the equivalencetheorem of magnetic flux conservation in Eulerian coordinates and (2.20).We begin with recalling the magnetic flux conservation in Eulerian coordinates. We choosea smooth surface S in a MHD fluid in the rest state. When the MHD fluid flows, the particleson the surface S ⊂ Ω + will form a new surface denoted by S ( t ) at time t . The equation of S ( t )can be given by S ( t ) = { x ∈ R | x = ζ ( y, t ) for y ∈ S } . (2.24)In particular, S (0) = S . Then the magnetic flux passing through the surface S ( t ) at t can beformally given by the following formula: Z S ( t ) M ( x, t ) · ν ( x, t )d S, (2.25)and the magnetic flux conservation in Eulerian coordinates formally reads as follows [3]. Z S ( t ) M ( x, t ) · ν ( x, t )d S = Z S ¯ M · ν ( x, S, where ν denotes the unit normal vector. However, each point on S ( t ) has two unit normal vectors,this results in a non-unique representation of the magnetic flux defined by (2.25). Therefore, weshall use the theory of surfaces to refine the definition (2.25), so that ν can be definitely computedout. To this end, we next introduce the definition of a regular surface.A surface S in Ω is called a regular surface, if S enjoys the following two properties: • The surface S can be parameterized by y = f ( α, β ) for ( α, β ) ∈ D S , where D S is abounded and closed domain. • The function y = f S ( α, β ): D S → S is one to one, and belongs to C ( D S ). Moreover,the corresponding Jacobi matrix, i.e., J ∂f S ∂α , ∂f S ∂β ! = ∂f S ∂α , ∂f S ∂α , ∂f S ∂α∂f S ∂β ∂f S ∂β ∂f S ∂β T ,
11s full rank at every point of D S .Now we define ω := ζ ( f S ( α, β ) , t ). Then the Jacobi matrix of ω can be given by J (cid:18) ∂ω∂α , ∂ω∂β (cid:19) = ∇ ζ J ∂f S ∂α , ∂f S ∂β ! . Keeping in mind that ζ satisfies ζ ( y,
0) = y in Ω + , ζ ( · , t ) : Ω + Ω + is one to one and belongs to C (Ω + ) , (2.26)det( ∇ ζ ( · , t )) = 0 on Ω + for any given t ∈ [0 , T ) , (2.27)we immediately see that, for any given t , S ( t ) is also a regular surface with the parameterizedequation f S ( t ) := ζ ( f S ( α, β ) , t ) defined on D S , and ν defined by the following formula is anormal vector of each point x on S ( t ): ν ( x, t ) := ∂ α ζ ( f S ( α, β ) , t ) × ∂ β ζ ( f S ( α, β ) , t )) / | ∂ α ζ ( f S ( α, β ) , t ) × ∂ β ζ ( f S ( α, β ) , t )) | (2.28)for t ∈ [0 , T ), where α and β satisfies x = f S ( α, β ).From now on, we stipulate that once we choose a parameterized equation f S of the surface S , then the normal vector of each point x on S ( t ) is defined by (2.28). Then, we can show thefollowing general equivalence theorem. Theorem 2.1.
Let D be a domain, ζ satisfy (2.26) – (2.27) with D in place of Ω + , and let J = det A with A being defined by (2.12) . Assume that M ( · , t ) ∈ C ( D ) for any t ∈ (0 , T ) , M ( · ,
0) = ¯ M , B := M ( ζ , t ) and the initial value ζ ( y,
0) = y .(1) For any given sub-domain D ⊆ D , if J A T B = ¯ M in D for some t ∈ (0 , T ) (2.29) then, for any regular surface S ⊂ D with a parameterized equation f S ( α, β ) defined on D S , we have Z S ( t ) M ( x, t ) · ν ( x, t )d S = Z S ¯ M · ν ( x, S, (2.30) where S ( t ) is defined by (2.24) , and ν ( x, t ) is defined by (2.28) .(2) Let D ⊆ D , if, for any bounded plane domain S ⊂ D , it holds that Z S ( t ) M ( x, t ) · ν ( x, t )d S = Z S ¯ M · ν ( x, S for some t ∈ (0 , T ) , (2.31) where S ( t ) is defined by (2.24) , then J A T B = ¯ M in D at time t . Remark 2.1.
Theorem 2.1 also holds for the case that D is a closed domain, or a horizontallyperiodic domain. 12 roof. (1) We prove the first assertion. Since ν is defined by (2.28), exploiting the integralformula on surfaces, we know that Z S ( t ) M ( x, t ) · ν ( x, t )d S = Z D S M ( ζ ( f S ( α, β ) , t ) , t ) · ( ∂ α ζ ( f S ( α, β ) , t ) × ∂ β ζ ( f S ( α, β ) , t ))d S. (2.32)Due to ∂ α ζ ( f S ( α, β ) , t ) × ∂ β ζ ( f S ( α, β ) , t )= ( ∂ ζ × ∂ ζ , ∂ ζ × ∂ ζ , ∂ ζ × ∂ ζ ) | y = f S ( α,β ) ( ∂ α f S × ∂ β f S )= ( J A ) | y = f S ( α,β ) ( ∂ α f S × ∂ β f S ) , we insert the above relation into (2.32) to deduce Z S ( t ) M ( x, t ) · ν ( x, t )d S = Z D S ( J M ( ζ ( y, t ) , t )) | y = f S ( α,β ) · ( A| y = f S ( α,β ) ( ∂ α f S × ∂ β f S ))d S = Z D S ( J A T M ( ζ ( y, t ) , t )) | y = f S ( α,β ) · ( ∂ α f S × ∂ β f S )d S. (2.33)Noting that S ⊂ D , we make use of (2.29) to get( J A T M ( ζ ( y, t ) , t )) y = f S ( α,β ) = ¯ M in D S . Recalling ζ ( y,
0) = y , one has ν ( x,
0) = ∂ α f S × ∂ β f S / | ∂ α f S × ∂ β f S | . Thus, one further findsthat Z S ( t ) M ( x, t ) · ν ( x, t )d S = Z D S ¯ M · ( ∂ α f S × ∂ β f S )d S = Z S ¯ M · ν ( x, ~S. Hence, (2.30) holds.(2) Now we verify the second assertion. Let S ⊂ D be a bounded plan domain that isperpendicular to x -axis. Then there is a constant a , such that S ⊂ { x ∈ R | x = a } . Thenwe can choose a parameterized equation f S of S as follows. f S := f S ( α, β ) = ( α, β, a ) , where ( α, β ) ∈ S x = a := { ( x , x ) | ( x , x , a ) ∈ S } . Noting that ∂ α f S × ∂ β f S = e , we canutilize (2.31) and (2.33) to further have Z S J A T B · e d S = Z D S ( J A T M ( ζ ( y, t ) , t )) y = f S ( α,β ) · e d S = Z S ¯ M · e d S. In view of arbitrariness of S ⊂ D , we obtain immediately( J A T B ) e = ¯ M e in D at time t. Similarly, we can also show that for i = 1 and 2,( J A T B ) e i = ¯ M e i in D at time t. Consequently, we have J A T B = ¯ M in D at time t . This completes the proof of the desiredconclusion. (cid:3) . Physical interpretation of the stability and instability criteria In this section, we give a physical interpretation for the stability and instability criteria inthe NMRT problem, and then extend the obtained results for the NMRT problem to the SMRTproblem. We recall here again that ¯ M is a general uniform magnetic field in this section. We proceed to discuss the physical interpretation for the stability and instability criteria inthe NMRT problem. We start with the discussion of the physical meaning of V Ω + ¯ M ( η ( t )) and V Ω + g ¯ ρ ′ ( η ( t )).It is well-known that the total magnetic energy E Ω + M ( t ) (defined on a periodic cell) of a MHDfluid in domain Ω + at the time t is given by the formula: E Ω + M ( t ) = λ Z Ω + | M | ( x, t )d x. Making use the transform of Lagrange coordinates, (2.14) and the magnetic field expression(2.23), we see that E Ω + M ( t ) = λ Z Ω + | B ( x, t ) | d y = λ Z Ω + | ∂ ¯ M ζ ( x, t ) | d y. Denote by δ ¯ M ( t ) the variation of total magnetic energy from 0 to t moment. Then, we canuse an integration by parts and the boundary condition η | ∂ Ω + = 0 to deduce δ ¯ M ( t ) := E Ω + M ( t ) − E Ω + M (0) = λ Z Ω + ( | ∂ ¯ M ζ | − | ¯ M | )d y = λ Z Ω + ∂ ¯ M η · ¯ M d y + λ Z Ω + | ∂ ¯ M η | d y = V Ω + ¯ M ( η ( t )) / , (3.1)from which we immediately have the following physical conclusion. Conclusion 3.1. V Ω + ¯ M ( η ( t )) / represents the variation of total magnetic energy of all the parti-cles of the non-resistive MHD fluid (defined on a periodic cell) deviating from their initial locationat time t . From the relation (3.1) and the estimate k η k , Ω + h V Ω + ~n ( η ) /π (see Remark 4.2 for aderivation), we find that the total magnetic energy increases, once the element lines are bent.On the other hand, the bent element lines will be straightened by the magnetic tension from thefigure on page 9. This dynamic behavior is very similar to that of the bent elastic string. Basedon this dynamic behavior, we can think that the non-resistive MHD fluid is made up of infiniteelastic strings, and the magnetic energy is the elastic potential energy.Next, we turn to the analysis of the physical meaning of V Ω + g ¯ ρ ′ ( η ( t )). During the developmentof RT instability, the potential energy will be released by the interchange of heavier and lighterparts of the fluid. Hence, motivated by the physical meaning of V Ω + ¯ M ( η ( t )), we naturally guessthat V Ω + g ¯ ρ ′ ( η ( t )) may be related to the variation of gravitational potential energy from time 0 to t . We denote by δ g ¯ ρ ′ ( t ) the variation of gravity potential energy. Recalling that initially onlythe initial velocity is perturbed, by the mass equation (2.16) we have ̺ = ̺ = ρ ( ζ ) = ¯ ρ ( y ) , (3.2)14hence, δ g ¯ ρ ′ ( t ) = g Z Ω + ( ρ − ¯ ρ ) x d x = g Z Ω + ¯ ρζ d y − g Z Ω + ¯ ρx d x = g Z Ω + ¯ ρη d x, where we have used the transform of Lagrange coordinates in the second equality.To analyze the relation between V g ¯ ρ ′ ( η ( t )) and δ g ¯ ρ ( t ), we next recall the mathematical deriva-tion of V g ¯ ρ ′ ( η ( t )).The relation (2.5) in Lagrange coordinates reads as ∇ A ¯ p ( ζ ) = − ¯ ρ ( ζ ) ge , (3.3)while using (2.13) and (2.23), we calculate that B · ∇ A B = ( ¯ M · ∇ ) η. (3.4)Exploiting (3.2)–(3.4), the momentum equation in Lagrange coordinates can be rewritten as¯ ρu t − µ ∆ A u + ∇ A q = λ ( ¯ M · ∇ ) η + G g ¯ ρ e , (3.5)where q := q ∗ − ¯ p ( ζ ) and G g ¯ ρ := G g ¯ ρ ( η , y ) := g ( ¯ ρ ( y + η ( y, t )) − ¯ ρ ( y )) e . By the equation(3.5), we can regard G g ¯ ρ as a force, which drives the growth of the RT instability in the fluid.Let N g ¯ ρ ′ ( τ, y ) := g Z τ ( τ − z ) d d z ¯ ρ ( y + z )d z, then G g ¯ ρ = g ¯ ρ ′ η + N g ¯ ρ ′ ( η , y ) . (3.6)Multiplying (3.5) by u in L (Ω + ), and making use integration by parts, the condition div A u = 0,(2.16) , the boundary condition (2.18), and the relation (3.6), we can deduce the followingevolution law for the variation of total energy of the non-resistive MHD fluid.d δ NE d t + µ k∇ A u k , Ω + = 0 , (3.7)where δ NE ( t ) := 12 Z Ω + ¯ ρ | u ( t ) | d y + 12 V Ω + ¯ M ( η ( t )) − V ∗ g ¯ ρ ′ ( η ( t )) , V ∗ g ¯ ρ ′ ( η ( t )) = 12 V Ω + g ¯ ρ ′ ( η ( t )) + N g ¯ ρ ( η ( t )) , N g ¯ ρ ( η ( t )) := Z Ω + Z η ( t )0 N g ¯ ρ ′ ( τ, y )d τ d y. Noting that the first two integrals in δ NE represent the variations of kinetic and magneticenergies respectively, one could guess that − V ∗ g ¯ ρ ( η ( t )) may represent the variation of potentialenergy from time 0 to t , i.e., − V ∗ g ¯ ρ ′ ( η ( t )) = δ g ¯ ρ ′ ( t ) . (3.8)In particular, if ¯ ρ ′ is a constant, (3.8) reduces to − V Ω + g ¯ ρ ′ ( η ( t )) = δ g ¯ ρ ′ ( t ) . Next we verify (3.8). 15et W g ¯ ρ ′ ( t ) := Z Ω + Z t G g ¯ ρ ( η ( y, τ ) , y ) u ( y, τ )d τ d y. The physical meaning of W g ¯ ρ ′ ( t ) will be discussed at the end of this section. Recalling the relation(3.6), we have dd t W g ¯ ρ ′ ( t ) = Z Ω + G g ¯ ρ ( η ( y, t ) , y ) u ( y, t )d y = dd t V ∗ g ¯ ρ ′ ( t ) , which, together with the fact that W g ¯ ρ ′ | t =0 = 0 and η | t =0 = 0, implies W g ¯ ρ ′ ( t ) = V ∗ g ¯ ρ ′ ( t ) . Thus, to get (3.8), we only need to show − W g ¯ ρ ′ ( t ) = δ g ¯ ρ ′ ( t ) . (3.9)Recalling the definition of G g ¯ ρ ′ and the fact η | t =0 = 0, we find that W g ¯ ρ ′ ( t ) = g Z Ω + Z t ( ¯ ρ ( y + η ( y, τ )) − ¯ ρ ( y )) u ( y, τ )d τ d y = g Z Ω + Z t ¯ ρ ( y + η ( y, τ )) u ( y, τ )d τ d y − g Z Ω + ¯ ρη d y. (3.10)In addition, we havedd t Z Ω + Z t ¯ ρ ( y + η ( y, τ )) u ( y, τ )d τ d y = Z Ω + ¯ ρ ( ζ ) u d y = Z Ω + ¯ ρv d x = Z h ¯ ρ Z Σ x v d x h d x = Z h ¯ ρ Z Ω x div v d x d x = 0 , where we have used the transform of Lagrangian coordinates and (2.14) in the second equality,integration by parts and the non-slip boundary condition for the velocity in the fourth equality,and the divergence-free condition in the last equality. Therefore, one concludes that Z Ω + Z t ¯ ρ ( y + η ( y, τ )) u ( y, τ )d τ d y = 0 . Substitution of the above identity into (3.10) yields (3.9). Consequently, one sees that − V Ω + g ¯ ρ ′ ( η ( t )) / t except for ¯ ρ ′ being a constant.However, − V Ω + g ¯ ρ ′ ( η ( t )) / w ∈ L ∞ (Ω + ) satisfying w + y ∈ Ω + a.e. in Ω + , one can estimate that | N g ¯ ρ ( w ) | c Z Ω + | w | d y c k w k L ∞ (Ω + ) k w k , Ω + , (3.11)16here the condition ¯ ρ ∈ C [0 , h ] has been used, and the constant c only depends on g and ¯ ρ ′ . Inparticular, if inf x ∈ (0 ,h ) { ¯ ρ ′ ( x ) } > , (3.12)one further has | N g ¯ ρ ( w ) | c k w k L ∞ (Ω + ) V Ω + g ¯ ρ ′ ( w ) . By virtue of the above estimate, for any given t , V Ω + g ¯ ρ ′ ( η ( t )) is an equivalent infinitesimal of V ∗ g ¯ ρ ′ ( η ( t )) as k η ( t ) k L ∞ (Ω + ) → Conclusion 3.2. − V ∗ g ¯ ρ ′ ( η ( t )) just represents the variation of potential energy of the MHD fluid(defined on a periodic cell) from time to t . Under the conditions ¯ ρ ∈ C [0 , h ] and (3.12) , V Ω + g ¯ ρ ′ ( η ( t )) is approximately equal to V ∗ g ¯ ρ ′ ( η ( t )) for any t in the following sense: V ∗ g ¯ ρ ′ ( η ( t )) V Ω + g ¯ ρ ′ ( η ( t )) = 1 + O ( k η ( t ) k L ∞ (Ω + ) ) , where O ( k η ( t ) k L ∞ (Ω + ) ) → as k η ( t ) k L ∞ (Ω + ) → . In particular, if ¯ ρ ′ is a constant, then − V g ¯ ρ ′ ( η ( t )) / just represents the variation of potential energy from time to t . With Conclusions 3.1–3.2 in hand, we are in a position to analyze the physical meaning ofthe stability and instability criteria for the NMRT problem.It is well-known that the RT instability can be explained by the minimum potential energyprinciple, which is a fundamental concept used in physics, chemistry, biology, and engineering,etc. It dictates that a structure or body shall deform or displace to a position that (locally)minimizes the total potential energy, with the lost potential energy being converted into kineticenergy (specifically heat). As aforementioned, the magnetic energy can be regarded as the elasticpotential energy, we thus call δ NE P ( η ( t )) := 12 V Ω + ¯ M ( η ) − V ∗ g ¯ ρ ′ ( η )the variation of total potential energy, which depends on the displacement function of particlesin the non-resistive MHD fluid.Now we further denote the total potential energy in the rest state by E N , rP , then the totalpotential energy of the non-resistive MHD fluid, denoted by E NP ( η ( t )) at time t , can be given by E NP ( η ( t )) = E N , rP + δ N E P ( η ( t )) . If one inserts the above relation into the evolution law of the total energy variation (3.7), onefinds the following evolution law for the total energy of the non-resistive MHD fluid:dd t (cid:18) Z Ω + ¯ ρ | u | d y + E NP ( η ) (cid:19) + µ k∇ A u k , Ω + = 0 . (3.13) If ¯ ρ is a constant, g = 0, µ = 0 and ¯ M := (0 , , ¯ M ) in (3.13), then one has Z T dd t E l ( y h , t )d y h = 0 ,
17n view of the above evolution law, we naturally believe that the minimum energy principle canbe also used to explain the stability and instability results in the NMRT problem. In other words,whether the potential energy in the rest state is minimal determines whether the NMRT problemis stable. Hence, we guess that whether the potential energy in the rest state is minimal shouldbe determined by the stability/instability conditions. The following theorem indeed supportsthis conjecture.
Theorem 3.1.
Let ¯ ρ ∈ C [0 , h ] satisfy the RT condition: ¯ ρ ′ | x = x > for some x ∈ (0 , h ) , and ¯ M be a non-zero constant vector. Then, the following assertions hold.(1) Under the instability condition | ¯ M | < m N , (3.14) we have the non-minimal condition of total potential energy in the following sense:For any given k > , there are a function w ∈ C ∞ σ (Ω + ) and a constant ε ∈ (0 , (dependingon w and k ), such that for any ε ∈ (0 , ε ) , there exist functions ̟ ε, r (depending on ε )satisfying k ̟ ε, r k k, Ω + c , and ̟ := εw + ε ̟ ε, r ∈ H k, , ∗ (Ω + ) , and δ N E P ( ̟ ) < (i.e. E NP ( ̟ ) < E N , rP ) , (3.15) where c depends on k , k ̟ k k, Ω + and Ω + .(2) Under the stability condition | ¯ M | > m N , (3.16) we have the minimal condition of total potential energy, i.e., there is a constant ε > suchthat, for any non-zero ̟ ∈ H (Ω + ) satisfying k ̟ k L ∞ (Ω + ) ε , δ N E P ( ̟ ) > (i.e. E NP ( ̟ ) > E N , rP ) . (3.17) Proof.
The proof of Theorem 3.1 needs some new preliminary mathematical results, so we willprovide the detailed proof in Sections 4.2 and 4.3. (cid:3)
By Theorem 3.1, we indeed have the following physical conclusion.
Conclusion 3.3.
Under the stability condition (3.16) for j = 3 , the total potential energy in therest state (defined on a periodic cell) is minimal in the sense of (3.17) . Under the instabilitycondition (3.14) for j = 3 , the potential energy in the rest state is not minimal in the sense of (3.15) . where E l ( y h , t ) := 12 Z h | ∂ t η | d y + a Z h | ∂ η | d y ! and a = s λ ¯ ρ | ¯ M | . We easily see that E l looks very similar to the energy of the one-dimensional (linear) wave equation (or thechord vibration equation) with fixed boundary. By virtue of the figure on page 10, it is not surpring to see thissimilarity, since the bent element line will vibrate around its initial location. Hence E l should repreasent the totalenergy of each bent element line, and the wave caused by this vibratation will propagate along the bent elementline. This means that a stands for the propagatation speed, and such a wave is called the Alfv´en wave. Therigirous derivation of Alfv´en wave based on the linearized MHD equation can be found in the classical textbookon MHD flows.
18e have given the physical explanation of the stability/instability criteria for the NMRTproblem. However, we can not judge whether the NMRT problem is stable or not for the criticalcase | ¯ M | = m N . In addition, under the instability condition (3.14), we may expect that theperturbed MHD flow should tend to another rest (equilibrium) state, the potential energy ofwhich is minimal. Unfortunately, we can not show this mathematically.Finally, we discuss the physical meaning of W g ¯ ρ ′ ( t ). Since W g ¯ ρ ′ ( t ) is closely related to theforce G g ¯ ρ , we first discuss the physical meaning of the force G g ¯ ρ . To this end, we take a particle Y labeled by y in the MHD fluid to analyze the action of G g ¯ ρ on the motion of the fluid withoutconsidering other forces. For the sake of simplicity, we assume that ¯ ρ ′ >
0. Then, from therelation G g ¯ ρ = g Z y + η ( y,t ) y ¯ ρ ′ ( s )d s we easily observe the following dynamic phenomena in the movement process of Y from time 0to t : • if η ( y, t ) < G g ¯ ρ drives the element point Y sinking.This implies that η ( y, t ) further decreases, and meanwhile Y further goes down away fromits initial location. • if η ( y, t ) > G g ¯ ρ drives the element point Y up. Thisshows that η ( y, t ) further increases, and meanwhile y further goes up away from its initiallocation.The above analysis is consistent with the early growth stage of the RT instability underthe force G g ¯ ρ . Moreover, the above dynamic phenomena are caused by the pressure differencebetween the heavier and lighter fluids. Hence, we easily think that G g ¯ ρ represents the pressuredifference. Moreover, W g ¯ ρ ′ ( t ) shall represent the total work done by the pressure difference G g ¯ ρ to all particles of the fluid from time 0 to t . We verify this fact below.Consider a particle Y labeled by y deviating from its initial location to a new location, thenthe pressure difference will do work to Y in the motion process. To evaluate the work, we denotethe location and the volume element of Y at time t by ζ ( y, t ) and d y , respectively. Since the pathΓ of the particle Y from its initial location to the new location ζ is continuously differentiablewith respect to ( x, t ), and the equation of Γ can be given byΓ : x = ζ ( y, s ) , s t. Thus, using the line integral of the second type and the relation d ζ = u d t , we find that the workcan be given by Z Γ d yG g ¯ ρ ( x − y , y ) e · d x = Z t G g ¯ ρ ( η ( y, τ ) , y ) u ( y, τ )d τ d y =: W Yg ¯ ρ ′ ( y, t ) . We integrate W Yg ¯ ρ ′ ( y, t ) with respect to all particles of the fluid to see that W g ¯ ρ ′ ( t ) indeed representsthe total work done by the pressure difference G g ¯ ρ to the all particles of the MHD fluid fromtime 0 to t . In addition, the relation (3.9) tells us that the work done by the pressure differencecomes from the release of potential energy. 19 .2. Case of the SMRT problem In this subsection we further extend the physical conclusions on the NMRT problem in theprevious section to the SMRT problem. We begin with introducing the mathematical model forthe SMRT problem. Consider two distinct, immiscible, incompressible MHD fluids evolving in amoving domain Ω( t ) = Ω + ( t ) ∪ Ω − ( t ) for time t >
0, where the upper fluid fills the upper domainΩ + ( t ) := { x := ( x h , x ) ∈ R | x h ∈ T , d ( x h , t ) < x < h } , and the lower fluid fills the lower domainΩ − ( t ) := { x ∈ R | x h ∈ T , − l < x < d ( x h , t ) } . We assume that h and l are given constants satisfying h > − l , but the internal surface function d := d ( x h , t ) is free and unknown. The internal surfaceΣ( t ) := { x ∈ R | x h ∈ T , x = d ( x h , t ) } moves between the two MHD fluids, and { x = − l } and { x = h } are the fixed lower and upperboundaries of Ω( t ), respectively.Now, we use the equations (2.1) , (2.1) and (2.11) with constant density to describe themotion of stratified (uniform) incompressible MHD fluids (without resistivity), and add thesubscript +, resp. − to the notations of the known physical parameters, and other unknownfunctions in (2.1) , (2.1) and (2.11) for the upper, resp. lower fluids. Thus, the motion equationsof stratified incompressible MHD fluids driven by the uniform gravitational field read as follows. ρ ± ∂ t v ± + ρ ± v ± · ∇ v ± + div S g ± = 0 in Ω ± ( t ) ,∂ t M ± = M ± · ∇ v ± − v ± · ∇ M ± in Ω ± ( t ) , div M ± = 0 , in Ω ± ( t ) , (3.18)where ρ ± are constants satisfying the RT jump condition ρ + > ρ − , and S g ± := S g ( v ± , M ± , p g ± ) := p g ± I − µ ± D ( v ± ) − λM ± ⊗ M ± , p g ± := p ∗± + ρ ± gx and D ( v ± ) = ∇ v ± + ∇ v T ± .For two viscous MHD fluids meeting at a free boundary, the standard assumptions are thatthe velocity is continuous across the interface and that the jump in the normal stress is zerounder ignoring the internal surface tension. This requires us to enforce the jump conditions J v K = 0 on Σ( t ) , (3.19)and J S g K ν = G g J ρ K := g J ρ K dν on Σ( t ) , (3.20)where ν represents the unit out normal vector of Ω − ( t ), J f K := f + | Σ( t ) − f − | Σ( t ) denotes theinterfacial jump, and f ± | Σ( t ) are the traces of the functions f ± on Σ( t ). We will also enforce thecondition that the fluid velocity vanishes at the upper and lower fixed boundaries, i.e., v = 0 on ∂ Ω h − l . (3.21)We also call G g J ρ K the pressure difference caused by gravity, since, similarly to G g ¯ ρ ′ , it drivesthe growth of the RT instability by acting on the particles on the interface. At the end of thissection, we will further discuss the behavior of G g J ρ K .20o simplify the formulation of (3.18), we introduce the indicator function χ and denote µ := µ + χ Ω + ( t ) + µ − χ Ω − ( t ) , ρ := ρ + χ Ω + ( t ) + ρ − χ Ω − ( t ) , v := v + χ Ω + ( t ) + v − χ Ω − ( t ) ,M := M + χ Ω + ( t ) + M − χ Ω − ( t ) , p ∗ := p ∗ + χ Ω + ( t ) + p ∗− χ Ω − ( t ) , S g := S g + χ Ω + ( t ) + S g − χ Ω − ( t ) to arrive at ρv t + ρv · ∇ v + div S g = 0 in Ω( t ) ,M t = M · ∇ v − v · ∇ M in Ω( t ) , div M = 0 in Ω( t ) . (3.22)Moreover, by virtue of the boundary condition (3.19), the internal surface function is defined by v , i.e., ∂ t d + v ∂ d + v ∂ d = v on Σ( t ) . (3.23)Finally, we impose the initial condition for ( v, M, d ):( v, M ) | t =0 := ( v , M ) in Ω h − l \ Σ(0) and d | t =0 = d on T , (3.24)where Σ(0) = { x ∈ R | x h ∈ T , x = d } .Then (3.19)–(3.24) constitute an initial boundary value problem of incompressible stratifiedMHD fluids with a free interface, which we call the SMF model for simplicity.Now, let us construct a rest state of the SMF model to be studied. Without loss of generality,assume the interface in the rest state is { x = 0 } . Let ¯ M be a uniform magnetic field in Ω, and¯ p ∗± satisfy (cid:26) ∇ ¯ p ∗± = 0 in Ω ± , J ¯ p ∗ K = 0 on Σ . Then r S := ( u = 0 , ¯ M , d = 0 , ¯ p ∗ ) is called the rest state of the SMF model. The problem whetherthe rest state r S is stable or unstable is called the SMRT problem.In [20], Jiang, et.al. showed that the value p gπ J ρ K /σ is a strength-threshold of the impressedfield ¯ M = (0 , , ¯ M ) T for stability/instability of the linearized SMRT problem with h = 1 and l = 1. Then, Wang [39] further established the existence of an asymptotically stable solution forthe (nonlinear) SMRT problem defined on Ω ′ := R × ( − l, m ) under the (asymptotical) stabilitycondition | ¯ M | > m ′ S with m ′ S := s sup w ∈ H σ (Ω ′ ) V ′ g J ρ K ( w ) V Ω ′ e ( w ) , V ′ g J ρ K ( w ) := g J ρ K k w ( · , k L ( R ) . Moreover, Wang further gave that m ′ S = 2 s gπ J ρ K σ ( h − + l − ) . Of course, Wang’s stability result also holds for the domain Ω h − l .The above stability result tells us that under the stability condition, the RT instability canbe inhibited by a magnetic field. Of course, the magnetic inhibition mechanism in the NMRTproblem can be also used to explain the above stability result. Moreover, we believe that thestability condition has physical meaning as in the case of the NMRT problem.21n the following, we give the physical meaning of the stability condition | ¯ M | > m S , (3.25)and the instability condition | ¯ M | < m S (3.26)in Lagrangian coordinates with proper regularity assumption on the solutions, where we havedefined that m S := s sup w ∈ H σ (Ω h − l ) V g J ρ K ( w ) V Ω e ( w ) , V g J ρ K ( w ) := g J ρ K | w ( · , | . In particular, we pay attention to the detailed derivation of the physical meaning of − V g J ρ K ( η ( t )),where some mathematical techniques are required. We mention that our analysis results also holdfor the domain Ω ′ .Now, let us disturb the rest state r S in the velocity by v , and assume that the SMHD modeldescribing the motion of the MHD fluid after perturbation defines a classical solution ( v, M, d, p ∗ )defined on Q t,T + ∪ Q t,T − , where Q t,T ± := { ( x, t ) | t ∈ [0 , T ] , x ∈ Ω ± ( t ) } for some T >
0. To makesome integrals involving v and p ∗ sense, we assume that v ± ∈ C ( Q t,T ± ) and p ∗± ∈ C ( Q t,T ± ) . We further assume that ζ ± is the solution of (cid:26) ∂ t ζ ± ( y, t ) = v ± ( ζ ± ( y, t ) , t ) , y ∈ Ω ± ,ζ ± ( y,
0) = y, y ∈ Ω ± with the regularity ζ ± ( · , t ) ∈ C ( Q T ± ) and ζ ± ∈ C ( Q T ± ), where Q T ± := Ω ± × [0 , T ]. We furtherassume that for each t ∈ [0 , T ], ζ ± ( t ) : Ω ± → Ω ± ( t ) are reversible , (3.27)and Σ( t ) = ζ ± (Σ , t ). Moreover, by virtue of the non-slip boundary condition (3.21) and thecontinuity of the velocity across Σ, ζ satisfies y = ζ ± ( y, t ) on Σ ± and J ζ K = 0 on Σ . From now on, we denote ζ = ζ + χ Ω + + ζ − χ Ω − , ζ = ζ χ Ω + + ζ − χ Ω − and η = ζ − y. (3.28)Obviously, ζ still enjoys the properties (2.13)–(2.15). Denoting( u, B, q )( y, t ) = ( v, M, p ∗ )( ζ ( y, t ) , t ) , we obtain B = ∂ ¯ M ζ and div A ( B ⊗ B ) = ∂ M η, where A is defined by (2.12) with ζ given by (3.28). Moreover, B ⊗ B A e = ∂ ¯ M ζ · ( A e ) ∂ ¯ M ζ = ¯ M ∂ ¯ M ζ . η, u, B, q ) satisfies the following initial-boundary value problem with an internal interface: ζ t = u in Ω ,ρu t + div A ( qI − µ D A ( u )) = λ∂ M η in Ω , J qI − µ D A ( u ) K A e − λ ¯ M J ∂ ¯ M η K = g J ρ K η A e on Σ , J η K = 0 on Σ , ( u, η ) = 0 on Σ − l ∪ Σ h , ( u, η ) = ( u ,
0) at t = 0 , (3.29)where D A ( u ) := ∇ A u + ∇ A u T .Since ζ ( · , t ) is continuous across Σ, one has J A e K = 0 on Σ . (3.30)Now, multiplying (3.29) by u in L (Ω), and employing integration by parts, and the boundaryconditions (3.29) –(3.29) and (3.30), we inferd δ S E ( t )d t + µ k D A ( u ( t )) k , Ω = 0 , (3.31)where δ S E ( t ) := 12 k√ ρu ( t ) k , Ω + δ S E P ( t ) , δ S E P ( t ) := 12 V Ω¯ M ( η ( t )) − V ∗ g J ρ K ( t ) , V ∗ g J ρ K ( t ) := 12 V g J ρ K ( η ( t )) + N g J ρ K ( t ) , N g J ρ K ( t ) = g J ρ K Z Σ Z t η ( y, τ ) ˜ A ( y, τ ) e · u ( y, τ )d τ d y h . Then, inspired by Conclusions 3.1 and 3.2, we have the following physical conclusions.
Conclusion 3.4. (1) V Ω¯ M ( η ( t )) / represents the variation of magnetic energy of the stratifiedMHD fluid without resistivity (defined on a periodic cell) deviating from its rest state r S attime t ;(2) − V ∗ g J ρ K ( t ) represents the variation of gravity potential energy of the stratified MHD fluiddeviating from its rest state r S at time t .(3) For any given t , V g J ρ K ( η ( t )) / is approximately equal to V ∗ g J ρ K ( t ) in the following sense: V ∗ g J ρ K ( t ) V g J ρ K ( η ( t )) = 1 + O ( f ( t )) , where f ( t ) := k∇ h η h ( t ) k L ∞ (Σ) (1 + k∇ h η h ( t ) k L ∞ (Σ) ) , and O ( f ( t )) → as f ( t ) → . D ERIVATION . (1) The first conclusion obviously holds by following the arguments in the deriva-tion of (3.1). 232) The derivation of the second conclusion requires more mathematical techniques. Let δ g J ρ K ( t ) denote the variation of gravity potential energy from time 0 to t . Then, we evaluate that δ g J ρ K ( t ) = g (cid:18)Z Ω( t ) ρx d x − Z Ω ρx d x (cid:19) = g (cid:18)Z Ω ρζ d y − Z Ω ρx d x (cid:19) = g Z Ω ρη d y = g Z h − l ρ Z Ω τ − l div η d y d τ =: ˜ δ g J ρ K ( t ) . Thus, to establish the second conclusion, it suffices to show˜ δ g J ρ K ( t ) = − V ∗ g J ρ K ( t ) . (3.32)Recalling J = 1, we have thatdiv η = ∂ η ∂ η + ∂ η ∂ η + ∂ η ∂ η − ∂ η ∂ η − ∂ η ∂ η − ∂ η ∂ η + ∂ η ( ∂ η ∂ η − ∂ η ∂ η )+ ∂ η ( ∂ η ∂ η − ∂ η ∂ η ) + ∂ η ( ∂ η ∂ η − ∂ η ∂ η ) . (3.33)Integrating by parts, we further get Z Ω τ − l div η d y = g Z Σ τ ( η ( ∂ η ∂ η − ∂ η ∂ η ) − η ∂ η − η ∂ η )d y h . So, we obtain ˜ δ g J ρ K ( t ) = g Z Ω ρ ( η ( ∂ η ∂ η − ∂ η ∂ η ) − η ∂ η − η ∂ η )d y = g Z Ω ρ∂ η η d y − I ( t ) = − V g J ρ K ( η ) − I ( t ) , (3.34)where I ( t ) := g Z Ω ρ (div ηη − η ( ∂ η ∂ η − ∂ η ∂ η ))d y. Obviously, to show (3.32), we have to establish I ( t ) = N g J ρ K ( t ) . Since η | t =0 = 0 and N g J ρ K (0) = 0, to get the above relation, it suffices to establish thatd I ( t )d t = d N g J ρ K ( t )d t . (3.35)Employing the second identity in (2.15) and partial integrations, we arrive atd I ( t )d t = g Z Ω ρ (div ηu − ∂ t ( η ( ∂ η ∂ η − ∂ η ∂ η )))d y + g Z Ω ρ div uη d y = g Z Ω ρ (div ηu − ∂ t ( η ( ∂ η ∂ η − ∂ η ∂ η )))d y − g Z Ω ρ div( u T ˜ A ) η d y = I ( t ) + d N g J ρ K ( t )d t , I ( t ) := g Z Ω ρ (cid:0) div ηu + ( ˜ A∇ η ) · u − ∂ t ( η ( ∂ η ∂ η − ∂ η ∂ η )) (cid:1) d y. Next, we shall verify the following identity in order to get (3.35): I ( t ) = 0 . (3.36)By a straightforward computation, we find that Z Ω ρ ˜ A∇ η · u d y = Z Ω ρ (( A ∗ ij ) × − I ) ∇ η · u d y = Z Ω ρ ( u ( ∂ η ∂ η − ∂ η ∂ η ) + u ( ∂ η ∂ η − ∂ η ∂ η )+ u (( ∂ η ∂ η − ∂ η ∂ η − ∂ η ) ∂ η + ( ∂ η ∂ η − ∂ η ∂ η − ∂ η ) ∂ η + ( ∂ η ∂ η − ∂ η ∂ η + ∂ η + ∂ η ) ∂ η ))d y, where A ∗ ij is the algebraic complement minor of the ( i, j )-th entry ∂ j ζ i . Thus, we employ (3.33)and integration by parts to deduce Z Ω ρ (div ηu + ( ˜ A∇ η ) · u )d y = Z Ω ρ ( u ( ∂ η ∂ η − ∂ η ∂ η ) + u ( ∂ η ∂ η − ∂ η ∂ η ) + u ( ∂ η ∂ η − ∂ η ∂ η ))d y = Z Ω ρ∂ t ( η ( ∂ η ∂ η − ∂ η ∂ η ))d y. Hence, (3.36) holds, and one gets the desired conclusion: − V ∗ g J ρ K ( t ) = δ g J ρ K ( t ) . (3.37)(3) Let I = g J ρ K Z Σ η ( ∂ η ∂ η + ∂ η + ∂ η − ∂ η ∂ η )d y h , then one can deduce thatd N g J ρ K ( t )d t = g J ρ K Z Σ η (( ∂ η ∂ η − ∂ η ∂ ζ ) u + ( ∂ η ∂ η − ∂ ζ ∂ η ) u + ( ∂ ζ ∂ ζ − ∂ η ∂ η − u )d y h = g J ρ K Z Σ (cid:18) η ( ∂ η ∂ η + ∂ η + ∂ η − ∂ η ∂ η ) u + 12 η ( ∂ u ∂ η + ∂ η ∂ u + ∂ u + ∂ u − ∂ u ∂ η − ∂ η ∂ u ) (cid:19) d y h = dd t I ( t ) , which gives I ( t ) = N g J ρ K ( t ) . (3.38)25ecalling the definition of I ( t ), we infer that (cid:12)(cid:12) N g J ρ K ( t ) (cid:12)(cid:12) k∇ h η h ( t ) k L ∞ (Σ) (1 + k∇ h η h ( t ) k L ∞ (Σ) ) V g J ρ K ( η ( t )) . (3.39)Thus, the desired conclusion follows immediately. (cid:3) In view of the relation (3.38), from now on, we renew to define N g J ρ K ( t ) as follows. N g J ρ K ( t ) := N g J ρ K ( η ( t )):= J ρ K Z Σ η ( t )( ∂ η ( t ) ∂ η ( t ) + ∂ η ( t ) + ∂ η ( t ) − ∂ η ( t ) ∂ η ( t ))d y h . (3.40)Similarly to Theorem 3.1, we have Theorem 3.2.
Let N g J ρ K ( t ) ∈ δ S E P ( ̟ ) be defined by (3.40) , then the following assertions hold.(1) Under the instability condition (3.26) , we have the non-minimal condition of total potentialenergy in the following sense:For any k > , there are a function w ∈ C ∞ σ (Ω) and a constant ε ∈ (0 , (depending on w and k ), such that for any ε ∈ (0 , ε ) , there exist functions ̟ ε, r (depending on ε ) satisfying ̟ := εw + ε ̟ ε, r ∈ H k, , ∗ (Ω) , k ̟ ε, r k k, Ω c and δ S E P ( ̟ ) < , (3.41) where c depends on k , k w k k, Ω and Ω .(2) Under the stability condition (3.25) , we have the minimal condition of total potential en-ergy, i.e., there is a constant ε > , such that for any non-zero ̟ ∈ H (Ω) satisfying k∇ h ̟ h k C ( T ) ε , δ S E P ( ̟ ) > . (3.42) Proof.
We shall mention how to prove Theorem 3.2 in Section 4.4. (cid:3)
By Theorem 3.2, we immediately have the following physical explanation:
Conclusion 3.5.
Under the stability condition (3.25) , the total potential energy in the rest state r S is minimal in the sense of (3.42) . Under the instability condition (3.26) , the total potentialenergy in the rest state r S is not minimal in the sense of (3.41) . Next, we explain how to find the relation (3.38) from the physical point of view. First, δ g J ρ K ( t )can be expressed by d , i.e., δ g J ρ K ( t ) = g (cid:18)Z Ω( t ) ρx d x − Z Ω ρx d x (cid:19) = g (cid:18) ρ + Z T Z hd x d x d x h + ρ − Z T Z d − l x d x h (cid:19) − π L L (cid:0) ρ + h − ρ − l (cid:1) = − g J ρ K Z T d d x h . By (3.27), the function ζ ( · , , t ) : T Σ( t ) is reversible for any t ∈ [0 , T ]. Then, for anygiven ( x , x , d ( x h , t )) on Σ( t ), there is a point ( y h , x = ζ ( y h , , t ) , x = ζ ( y h , , t ) and d ( x h , t ) = ζ ( y h , , t ) , d ( ζ ( y h , , t ) , ζ ( y h , , t ) , t ) = ζ ( y h , , t ) . Thus, if η is suitably small, then we can use the above change of variables to formally have δ g J ρ K ( t ) := − J ρ K Z Σ ζ ( ∂ ζ ∂ ζ − ∂ ζ ∂ ζ )d y h = − V g J ρ K ( η ( t )) / − I ( t ) . Recalling the definition of V ∗ g J ρ K ( t ) and the relation (3.37), we easily obtain (3.38).Finally we derive some additional physical results involving the pressure difference G g J ρ K . Weconsider some part of the upper heavier fluid at time t , which sinks below the interface Σ( t ).Assume that the domain D s occupied by the sinking part is just a bounded connected domainthat is bounded by the plane { x = 0 } and the surface Σ( t ). Then, the total pressure difference G D s acting on the lower surface D s can be given by the following integral formula: G D s = Z D s ∩ Σ( t ) g J ρ K x ν d S, where ν is the unit inner normal on the boundary of D s . Thus, a partial integration results in G D s = (0 , , − g J ρ K | D s | ) T , where | D s | denotes the volume of D s . The above formula reveals that the value of total pressuredifference acting on the sinking part in the domain D s is just equal to the difference betweenthe weight of sinking part of heaver fluid and that of the lighter fluid, which is displaced bythe sinking part. Moreover, the direction of total pressure difference is just along the negativedirection of x -axis. Obviously, the total pressure G D s further drives the heavier fluid below theinterface Σ( t ) to sink due to the increasing volume of the sinking part of the heavier fluid.Similarly, if we consider some part of the lower lighter fluid at time t which rises above theinterface Σ( t ), and assume that the domain D r occupied by the rising part is just a boundedconnected domain, which is bounded by the plane Σ and the surface Σ( t ). Then the totalpressure difference G D r acting on the upper surface of D r can be represented by the followingintegral formula: G D r = (0 , , g J ρ K | D r | ) T . This means that the total pressure G D r further drives the lighter fluid above the interface Σ( t )to rise up. The previous analysis is consistent with the early growth of the RT instability underthe force G g J ρ K .In addition, if we denote the total pressure difference acting on Σ( t ) by P d ( t ), then P d ( t ) canbe given by P d ( t ) = Z Σ( t ) G g J ρ K d S. We evaluate that P d ( t ) = 0. In fact, we have after a straightforward calculation thatdd t Z Ω − ( t ) d x = Z T d t d x h = Z T v · ( − ∂ d , − ∂ d, T d x h = Z Σ( t ) v · ν d x h = Z Ω − ( t ) div v d x = 0 , where ν denotes the unit outer normal vector of ∂ Ω − ( t ). Therefore, Z Ω − ( t ) d x = 4 π L L , P d ( t ) = Z Σ( t ) g J ρ K x ν d S = g J ρ K (cid:18) , , π L L l − Z Ω( t ) d x (cid:19) e = 0 . Finally, we discuss the work done by the pressure difference G g J ρ K acting on all particles onthe interface from time 0 to t . Denote the work done by W g J ρ K ( t ). Then, similarly to (3.9), weshould have − W g J ρ K ( t ) = δ g J ρ K ( t ) . (3.43)We can derive the above formula by an infinitesimal method. In fact, considering a particle Y labeled by ( y h , T on the interface deviating from its initial location to a new location, then thepressure difference will do work to Y in the motion process. To evaluate W g J ρ K ( t ), we denote thelocation of Y at time t and the area element of ( y h , T by ζ ( y h , , t ) and d y h , respectively. Whenthe particle slightly moves to the location ζ ( y h , , t ), the area element of Y will become |A e | d y h .Thus, the work done by the pressure difference on Y from time τ to τ + d τ in Lagrangiancoordinates can be given by g J ρ K ζ ( y h , , t ) A| y =( y h , e · u ( y h , , τ )d τ d y h . Integrating the above identity over (0 , t ), we obtain the work done by the pressure difference on Y from time 0 to t . Denote this work by W y h g J ρ K ( t ) for the sake of simplicity. Finally, an integrationof W y h g J ρ K ( t ) over T immediately yields W g J ρ K ( t ) = V ∗ g J ρ K ( t ). Therefore, (3.43) follows from (3.37).
4. Proof of Theorems 3.1–3.2
In this section we rigorously prove Theorems 3.1–3.2 stated in Section 3. We first give somepreliminary results in Section 4.1, then prove two assertions of Theorem 3.1 in Sections 4.2–4.3.Finally we mention how to show Theorem 3.1 in Section 4.4.
Next, we want to establish Lemma 4.2 below, which shows that any η ∈ H σ (Ω ba ) ∩ H k +2 (Ω ba )can be modified to a new function that belongs to H k, , ∗ (Ω ba ) and is close to η . For this purpose,we first recall the following lemma on the global existence of inverse functions. Lemma 4.1.
Let N > , D ⊆ R N be an open set, ζ : D → R N belongs to C ( D ) , and ∇ ζ ( x ) beinvertible for all x ∈ D . Suppose that K is a connected compact subset of D , and ζ : ∂K → R N is injective, Then ζ : K → R N is injective. We are now able to use the above lemma and the classical theory for the Stokes problem toestablish the following desired lemma:
Lemma 4.2.
Let a < b , k > , If η ∈ H σ (Ω ba ) ∩ H k +2 (Ω ba ) , then there exists a constant ε (depending on k and η ), such that for any ε ∈ (0 , ε ) , there is an η r ∈ H (Ω ba ) ∩ H k +2 (Ω ba ) satisfying det ∇ ζ = 1 and k η r k k +2 , Ω ba c, (4.1) where ζ := y + εη + ε η r , the constant c > depends on k , the norm k η k k +2 , Ω ba and the domain Ω ba , but not on ε . Moreover, ζ : Ω ba − → Ω ba − is a homeomorphism mapping . (4.2)28 roof. Throughout the proof, the letter c denotes a general positive constant, which maydepend on k , the norm k η k k +2 , Ω ba and the domain Ω ba , but independent of ε .(1) We begin with the construction of η r . So, we consider the following Stokes problem forgiven ξ ∈ H (Ω ba ): (cid:26) − ∆ w + ∇ ̟ = 0 , div w = O ( ξ ) in Ω ba ,w = 0 on ∂ Ω ba , (4.3)where O ( ξ ) = (1 + ε div ξ − det ∇ ( y + εη + ε ξ )) /ε . By virtue of the product estimate k f g k k +1 , Ω ba c k f k k +1 , Ω ba k g k k +1 , Ω ba with some constant c depending on k and Ω ba only, it is easy to see that k O ( ξ ) k k +1 , Ω ba c (1 + ε k ξ k k +2 , Ω ba + ε k ξ k k +2 , Ω ba + ε k ξ k k +2 , Ω ba ) c (1 + ε k ξ k k +2 , Ω ba ) . By the classical existence and regularity theory on the Stokes problem, there exists a solution( w, ̟ ) of (4.3). Moreover, for ε <
1, it holds that k w k k +2 , Ω ba + k ̟ k k +1 , Ω ba k O ( ξ ) k k +1 , Ω ba c (cid:16) ε k ξ k k +2 , Ω ba (cid:17) , (4.4)where the letter c denotes a fixed constant depending on k , k η k k +2 , Ω ba and Ω ba , but not on ε .Therefore, one can construct an approximate function sequence { η n } ∞ n =1 , such that for any n > (cid:26) − ∆ η n + ∇ ̟ n = 0 , div η n = O ( η n − ) in Ω ba ,η n = 0 on ∂ Ω ba , (4.5)where k η k k +2 , Ω ba c . Moreover, from (4.4) one gets k η n k k +2 , Ω ba + k ̟ n k k +1 , Ω ba c (1 + ε k η n − k k +2 , Ω ba ) for any n > , which implies k η n k k +2 , Ω ba + k ̟ n k k +1 , Ω ba c (4.6)for any n >
2, and any ε / c .Next we want to show that { η n } ∞ n =1 is a Cauchy sequence in H (Ω ba ). Noting that (cid:26) − ∆( η n +1 − η n ) + ∇ ( ̟ n +1 − ̟ n ) = 0 , div( η n +1 − η n ) = O ( η n ) − O ( η n − ) in Ω ba ,η n +1 − η n = 0 on ∂ Ω ba , we obtain k η n +1 − η n k k +2 , Ω ba + k ( ̟ n +1 − ̟ n ) k k +1 , Ω ba c k O ( η n ) − O ( η n − ) k k +1 , Ω ba . (4.7)On the other hand, using (4.6) and the product estimates, we arrive at k O ( η n ) − O ( η n − ) k k +1 , Ω ba cε k η n − η n − k k +2 , Ω ba . Substituting the above inequality into (4.7), one sees by taking ε appropriately small that { ( η n , ̟ n ) } ∞ n =1 is a Cauchy sequence in H k +2 (Ω ba ) × H k +1 (Ω ba ). Consequently, we can take tothe limit in (4.5) as n → ∞ to see that the limit function ( η r , ̟ ) solves (cid:26) − ∆ η r + ∇ ̟ = 0 , div η r = O ( η r ) in Ω ba ,η r = 0 on ∂ Ω ba . (4.8)29urthermore, k η r k k +2 , Ω ba c by (4.6).(2) We proceed to the proof ζ : Ω ba − → Ω ba − is injective . (4.9)Since k > ζ − y ) ∈ H (Ω ba ) ∩ H k +2 (Ω ba ), by the Sobolev embedding theorem in [28,Section 1.3.5.8], one sees that ( ζ − y ) ∈ L ∞ (Ω ba − ) ∩ C (Ω ba − ) and ζ | ∂ Ω ba = 0. Let S := [0 , πL ] × [0 , πL ] × [ a, b ]. We choose a function χ ∈ C ∞ ( R ) such that χ = 1 in S , and 0 χ R . Then there exists a ball B R (0) of radius R and center 0, such that χ = 0 in R \ B R (0) and S ⊂ B R (0). Let ζ χ := ( ζ − y ) χ/ε , where ζ := y + εη + ε η r . Then ζ χ ∈ H k +2 (Ω ba − ) ∩ C (Ω ba − ).Since Ω ba − is locally Lipschitz (see [2, Section 4.9] for the definition), by virtue of the well-knownStein extension theorem (see [2, Section 5.24]), there is an extension operator for Ω ba − , such that E ( ζ χ ) = ζ χ a.e. in Ω ba − , k E ( ζ χ ) k k +2 , R ˜ c k ζ χ k k +2 , Ω ba − , (4.10)where the constant ˜ c depends on k and Ω ba − .Define ζ m := y + εχE ( ζ χ ), then ζ m = ζ in S. (4.11)From (4.1), (4.10) and the periodicity of η and η r , it follows that k E ( ζ χ ) k k +2 , R ˜ c k χ ( η + εη r ) k k +2 , Ω ba − c, where the constant c > k and the norms of η and η r , but not on ε . Thus, in termsof the Sobolev embedding theorem, ∇ ζ m is invertible in R for sufficiently small ε .Let K := Ω ba − ∩ B R (0), then K is a connected compact set. Since ζ m ( y ) = y on ∂K , then ζ m : ∂K → R N is injective. Thus, ζ m : K → R N is injective by Lemma 4.1. Noting that S ⊂ K ,we see by (4.11) that ζ : S → R N is also injective, which implies that ζ : Ω ba − → R N is injective.Now, we further show that ζ : Ω ba − → Ω ba − . To this end, it suffices to prove that, for any x h ∈ [0 , πL ] × [0 , πL ], ζ : l x h → Ω ba − , where l x h := { ( x h , x ) | x ∈ ( a, b ) } . We prove this by contradiction.Assume that there exists a point x ∈ l x h , such that ζ ( x ) ∈ / Ω ba − . Without loss of generalization,we assume that ζ ( x ) is above the closed domain Ω ba − . Recalling ζ = y on ∂ Ω ba − , one has x ∈ ( a, b ),where x denotes the third component of x . Obviously, the curve function ζ defined on l := { x ∈ l x h | x ∈ ( a, x ) } must go through the upper boundary of { x = b } . We denote by z theintersection of the curve and the upper boundary. This means that there is a point y ∈ l , suchthat y ∈ ( a, b ) and ζ ( y ) = z . Noting that ζ ( z ) = z , we have ζ ( y ) = ζ ( z ) where y, z ∈ S and y = z . This contradicts with injectivity of ζ on S . Hence (4.9) holds.(3) Now we turn to prove that ζ : Ω ba − → Ω ba − is surjective . (4.12)We define that K := { x ∈ Ω ba − | x = ζ ( y ) for any y ∈ Ω ba − } . By (4.9), K ⊂ Ω ba − . Obviously, to get (4.12), it suffices to prove that K = ∅ . (4.13)Next we verify (4.13) by contradiction. 30e assume that K = ∅ . Noting that K ⊂ Ω ba − by (4.9), then ∅ 6 = ∂ K ⊆ Ω ba − . Moreover thereexists a point x ∈ ∂ K ∩ Ω ba − . (4.14)Then, there exists a ball B δ ( x ) ⊂ Ω ba − , such that, for any B δ ( x ) ⊂ B δ ( x ), B δ ( x ) ∩ K = ∅ (4.15)and B δ ( x ) ∩ (Ω ba − \ K ) = ∅ . (4.16)By (4.16), we can choose sequences { x n } ∞ n = m ⊂ (Ω ba − \ K ) and { y n } ∞ n = m ⊂ Ω ba − for some m > /δ ,such that x n ∈ B /n ( x ) ⊂ B δ ( x ), x n → x and x n = ζ ( y n ). Since ( ζ − y ) ∈ L ∞ (Ω ba − ) ∩ C (Ω ba − ),then { y n } ∞ n = m is a bounded sequence. Therefore, there exists a subsequence (still labeled by y n )such that y n → y for some y ∈ Ω ba − . By the continuity of ζ , x n = ζ ( y n ) → x = ζ ( y ). Since x ∈ Ω ba − , then y ∈ Ω ba − . Noting that ∇ ζ is invertible for y , thus, by the well-known inversefunction theorem [29, Theorem 9.24], there exist two open sets U and V ⊂ Ω ba − such that y ∈ U , x ∈ V , and ζ ( U ) = V , which imply that there exists a ball B δ ( x ) ⊂ (Ω ba − \ K ) for δ < δ , i.e. B δ ( x ) ∩ K = ∅ , which contradicts with to (4.15). Hence (4.13) holds.(4) Finally, since ζ : Ω ba − → Ω ba − is bijective, we can consider the inverse mapping ζ − definedon Ω ba − by ζ − ( ζ ( y )) = y for y ∈ Ω ba − . (4.17)Obviously, ζ − : Ω ba − → Ω ba − is bijective. Next we verify that ζ − is a continuous mapping of Ω ba − onto Ω ba − (4.18)by contradiction.We assume that ζ − is not continuous for some x ∈ Ω ba − . Then, there exists a constant ε > ι >
0, there exists a point x ι ∈ Ω ba − satisfying | x ι − x | < ι and | ζ − ( x ι ) − ζ − ( x ) | > ε. (4.19)Let ι = 1 /n , we denote y n = ζ − ( x /n ). Since ( ζ − y ) ∈ L ∞ (Ω ba − ) ∩ C (Ω ba − ), and { x /n } ∞ n =1 ⊂ B ( x ) ∩ Ω ba − , then { y n } ∞ n =1 is a bounded sequence. Thus there exists a subsequence (still labelledby y n ) such that y n → y for some y ∈ Ω ba − . By the continuity of ζ , x /n = ζ ( y n ) → ζ ( y ) = x .Thus ζ − ( x /n ) → ζ − ( x ), which contracts with (4.19). Hence (4.18) holds. Consequently, weobtain (4.2) from (4.9), (4.12) and (4.18). This completes the proof of (4.2). (cid:3) In addition, we can slightly modify the proof of Lemma 4.2 to get the following conclusionfor the stratified case:
Lemma 4.3.
Let − l < τ , k > , If η ∈ H σ (Ω h − l ) ∩ H k +2 (Ω) , then there exists a constant ε (depending on k and η ), such that for any ε ∈ (0 , ε ) , there is an η r ∈ H (Ω h − l ) ∩ H k +2 (Ω) satisfying det ∇ ζ = 1 in Ω and k η r k k +2 , Ω c, where ζ := y + εη + ε η r , the constant c > depends on k , the norm k η k k +2 , Ω and the domain Ω , but not on ε . Moreover, ζ : Ω h − l − → Ω h − l − is a homeomorphism mapping . V Ω e j ( w ) in the definition of m N can be replaced by V Ω ~n ( w ), where the third component of ~n is 1.To get the desired conclusion, we shall first derive an auxiliary result. Lemma 4.4.
Let a < b , and ¯ ρ ∈ C [ a, b ] satisfy ¯ ρ ′ | s = s > for some s ∈ [ a, b ] , (4.20) Then there exists a function ψ ∈ H ( a, b ) , such that ¯ ρ ′ ψ = 0 , ψ ′ = 0 , k ψ ′ k L ( a,b ) = 1 , and Z ba ¯ ρ ′ ψ d s = sup ψ ∈ H ( a,b ) R ba ¯ ρ ′ ψ d s k ψ ′ k L ( a,b ) =: γ. Remark 4.1.
When ¯ ρ ′ is a positive constant, γ reduces to (see [19])sup ψ ∈ H ( a,b ) k√ ¯ ρ ′ ψ k L ( a,b ) k ψ ′ k L ( a,b ) = ( b − a ) ¯ ρ ′ π . (4.21) Proof.
In view of the RT condition (4.20) and the definition of γ , we see that γ >
0. Thusthere is a function sequence { ψ m } ∞ m =1 satisfying 0 = ψ m ∈ H ( a, b ) and R ba ¯ ρ ′ ψ m d s k ψ ′ m k L ( a,h ) → γ as m → ∞ . Let ˜ ψ m := ψ m / k ψ ′ m k , then k ˜ ψ ′ m k = 1 and k ˜ ψ m k c , where c is independent of m . Thus, thereare a subsequence of { ˜ ψ m } ∞ m =1 (still denoted by ˜ ψ m ) and a function ψ = 0, such that˜ ψ ′ m → ψ ′ weakly in H ( a, b ) and Z ba ¯ ρ ′ ˜ ψ m d s → Z ba ¯ ρ ′ ψ d s, which gives R ha ¯ ρ ′ ψ d s = γ . Hence, ¯ ρ ′ ψ = 0 and ψ ′ = 0. In addition, by the weakly lowersemi-continuity, k ψ ′ k L ( a,b ) lim inf m →∞ k ˜ ψ ′ m k L ( a,b ) = 1 , which, by recalling the definition of γ , implies that γ = Z ba ¯ ρ ′ ψ d s R ba ¯ ρ ′ ψ d s k ψ ′ k L ( a,b ) γ. Hence k ψ ′ k L ( a,b ) = 1. This completes the proof of Lemma 4.4. (cid:3) Next we further use the above result to establish the desired conclusion.
Lemma 4.5.
Let a < b , ¯ ρ ∈ C [ a, b ] satisfy (4.20) , and the third component of the constantvector ~n be , then we have b N := sup w ∈ H σ (Ω ba ) V Ω ba g ¯ ρ ′ ( w ) V Ω ba ~n ( w ) = sup w ∈ H (Ω ba ) V Ω ba g ¯ ρ ′ ( w ) V Ω ba ~n ( w ) = sup ψ ∈ H ( a,b ) g R ba ¯ ρ ′ ψ d y λ k ψ ′ k L ( a,b ) =: c N . (4.22)32 emark 4.2. In view of (4.21) and (4.22), we know that for any w ∈ H (Ω + ), k ̟ k , Ω + h λπ V Ω + ~n ( ̟ ) . (4.23) Proof.
We prove Lemma 4.5 by four steps.(1) First of all, we claim thatsup w ∈ H (Ω ba ) V Ω ba g ¯ ρ ′ ( w ) V Ω ba ~n ( w ) sup w ∈ H (Ω ba ) V Ω ba g ¯ ρ ′ ( w ) V Ω ba e ( w ) =: a N . (4.24)In fact, for any given w := w ( y , y , y ) ∈ H (Ω ba ), one can verify that˜ w ( y h , y ) := w ( y h + y ~n h , y ) ∈ H (Ω ba ) , V Ω ba g ¯ ρ ′ ( ˜ w ) = V Ω ba g ¯ ρ ′ ( w ) , V Ω ba e ( ˜ w ) = k ~n · ∇ w ( y h + y ~n h , y ) k , Ω ba = V Ω ba ~n ( w ) . Thus, (4.24) holds obviously.(2) Then we prove that a N c N . (4.25)Let ˆ w ( ξ, x ) be the horizontal Fourier transform of w ( x ) ∈ H (Ω ba ), i.e.,ˆ w ( ξ, y ) = Z T w ( y h , y ) e − i y h · ξ d y h , ξ = ( ξ , ξ ) , then [ ∂ w = ∂ b w . Denote ψ ( ξ, y ) := ψ ( ξ, y ) + i ψ ( ξ, y ) =: ˆ w ( ξ, y ) where ψ and ψ are realfunctions. By the Fubini and Parseval theorems (see [13, Proposition 3.1.16]), we infer that V Ω ba g ¯ ρ ′ ( w ) = g π L L X ξ ∈ L − Z × L − Z Z ba ¯ ρ ′ ( ψ ( ξ, y ) + ψ ( ξ, y ))d y (4.26)and V Ω ba e ( w ) > V Ω ba e ( w ) = λ π L L X ξ ∈ L − Z × L − Z Z ba ( | ∂ ψ ( ξ, y ) | + | ∂ ψ ( ξ, y ) | )d y . (4.27)Keeping in mind that ψ i ∈ H ( a, b ) if w ∈ H (Ω ba ) by the definition of c N , we get c N λ Z ba | ψ ′ ( ξ ) | d y > g Z ba ¯ ρ ′ ( y ) | ψ ( ξ, y ) | d y . Thus, using (4.26)–(4.27), we deduce that for any 0 = w ∈ H (Ω ba ), c N > g P ξ ∈ L − Z × L − Z R ba ¯ ρ ′ | ψ ( ξ, y ) | d y λ P ξ ∈ L − Z × L − Z R ba | ∂ ψ ( ξ, y ) | d y = V Ω ba g ¯ ρ ′ ( w ) λ k ∂ w k , Ω ba > V Ω ba g ¯ ρ ′ ( w ) V Ω ba e ( w ) , from which (4.25) follows. 333) Finally, we show that c N b N . (4.28)By virtue of Lemma 4.4, there is a function ψ ∈ H ( a, b ), such that g R ba ¯ ρ ′ | ψ ( y ) | d y k ψ ′ k L ( a,b ) = c N . (4.29)Moreover, there is a function sequence { ψ j } ∞ j =1 ⊂ C ∞ ( a, b ) satisfying ψ j = 0 and ψ j → ψ in H ( a, b ) as j → ∞ . (4.30)We only consider the cases ~n = 0 and | ~n h | = 0, because the other case ~n = 0 can be dealtwith similarly. When ~n = 0 and | ~n h | = 0, for any given integer i , there always exists a realnumber sequence l i = (cid:26) − ~n L i/~n L , if ~n = 00 , if | ~n h | = 0 , such that r i := ~n L − l i + ~n L − i = 0 . Now we define v i,j ( y ) = (0 , ψ ′ j cos z, L − iψ j sin z ) , (4.31)where z := L − l i y + L − iy . Then v i,j ( y ) ∈ C ∞ σ (Ω ba ). Moreover, we have after a direct calculationthat V Ω ba g ¯ ρ ′ ( v i,j ) = gL − i Z ba ¯ ρ ′ | ψ j | d y k sin z k L ( T ) , (4.32) k ~n · ∇ v i,j ( z ) k = k ψ ′′ j k L ( a,b ) k cos z k L ( T ) + L − i k ψ ′ j k L ( a,b ) k sin z k L ( T ) . (4.33)Noting that k sin z k L ( T ) = 2 π L L − Z πL − πL Z πL − πL cos 2 z d y d y and k cos z k L ( T ) = 2 π L L + 12 Z πL − πL Z πL − πL cos 2 z d y d y , we apply the Riemann lemma (see [7, Theorem 16.2.1])lim p →∞ Z ba ψ ( s ) sin ps d s = lim p →∞ Z ba ψ ( s ) cos ps d s = 0 for any ψ ∈ C [ a, b ] , to infer that k sin z k L ( T ) , k cos z k L ( T ) → π L L as i → ∞ . (4.34)In view of (4.30), and (4.32)–(4.34), one has b N > lim j →∞ lim i →∞ V Ω ba g ¯ ρ ′ ( v i,j ) V Ω ba ~n ( v i,j ) = lim j →∞ g R ba ¯ ρ ′ | ψ j ( y ) | d y k ψ ′ j k L ( a,b ) = c N , (4.35)which gives (4.28). 344) We are now in a position to show (4.22). From (4.24), (4.25) and (4.28), we getsup w ∈ H (Ω ba ) V Ω ba g ¯ ρ ′ ( w ) V Ω ba ~n ( w ) a N c N b N . On the other hand, since H σ (Ω ba ) ⊂ H (Ω ba ), we see that b N sup w ∈ H (Ω ba ) V Ω ba g ¯ ρ ′ ( w ) V Ω ba ~n ( w ) . Consequently, we obtain the desired conclusion (4.22).Similarly to Lemma 4.5, we also an equivalence lemma of threshold for the stratified case.
Lemma 4.6.
Let h , l , g J ρ K > , and the third component of the constant vector ~n be , then wehave sup w ∈ H σ (Ω h − l ) V g J ρ K ( w ) V Ω h − l ~n ( w ) = sup w ∈ H (Ω h − l ) V g J ρ K ( w ) V Ω h − l e ( w ) = sup ψ ∈ H ( − l,h ) g J ρ K | ψ (0) | λ k ψ ′ k L ( − l,h ) . Remark 4.3.
In [20] Wang gave the following formulasup ψ ∈ H ( − l,h ) | ψ (0) | k ψ ′ k L ( − l,h ) = ( h − + l − ) − , where the supremum can be achieved by choosing ψ ( s ) = (cid:26) s/l, s ∈ ( − l, − s/h, s ∈ (0 , h ) . Proof.
We omit the trivial derivation here, since Lemma 4.6 can be easily shown by slightlymodifying the arguments in the proof of Lemma 4.5. (cid:3)
We now prove the first assertion in Theorem 3.1. Without loss of generality, we assume that¯ M = 0. By Lemmas 4.4 and 4.5, there exists a ψ ∈ H (0 , h ) satisfying m N = vuut g R h ¯ ρ ′ ψ d sλ k ψ ′ k L (0 ,h ) . Since ψ ∈ H (0 , h ) and m N >
0, there is a function sequence { ψ j } ∞ j =1 ⊂ C ∞ (0 , h ) satisfying¯ ρ ′ ψ j = 0 , ψ j = 0 and ψ j → ψ in H (0 , h ) as j → ∞ . (4.36)Let i be a positive integer to be determined later. Let v i,j be defined by (4.31) with ψ j con-structed as above. Then v i,j ∈ C ∞ σ (Ω + ). Moreover, we have after a straightforward computationthat V Ω + g ¯ ρ ′ ( v i,j ) = gL − i Z h ¯ ρ ′ | ψ j | d s k sin z k L ( T ) = 0 , (4.37) V Ω + ¯ M ( v i,j ) = ¯ M (cid:16) k ψ ′′ j k L (0 ,h ) k cos z k L ( T ) + L − i k ψ ′ j k L (0 ,h ) k sin z k L ( T ) (cid:17) . (4.38)35n what follows, the letters c ( i, j ), c k ( i, j ) and ε ( i, j ) denote positive constants for k = 2 and 3,which may depend on v i,j , but not on ε ∈ (0 , c ( i, j ) can vary from line to line.Since v i,j may not belong to H k, , ∗ (Ω + ), we have to further modify it. For v i,j ∈ C ∞ σ (Ω + ), byvirtue of Lemma 4.2, there is an ε ( i, j ) ∈ (0 , ε ∈ (0 , ε ( i, j )), there existsan ̟ i,jε, r ∈ H k (Ω + ) satisfying k ̟ i,jε, r k k, Ω + c ( i, j ) ,̟ i,jε := εv i,j + ε ̟ i,jε, r ∈ H k, , ∗ (Ω + ) , where the constant ε ( i, j ) > v i,j , but not on ε . It is easy to see that k ( v i,j , ̟ i,jε, r ) k C (Ω + ) c ( i, j ) . (4.39)We further define R i,jε := ε (cid:18) g Z Ω + ¯ ρ ′ v i,j ( ̟ i,jε, r ) d x − λ Z Ω + ∂ ~n v i,j · ∂ ~n ̟ i,jε, r d y (cid:19) + ε V Ω + g ¯ ρ ′ (( ̟ i,jε, r ) ) − V Ω + ~n ( ̟ i,jε, r )) + N g ¯ ρ ′ ( ̟ i,jε ) , where ( ̟ i,jε, r ) denotes the third component of ̟ i,jε, r ,¯ M ∗ := (cid:26) ¯ M / ¯ M , if ¯ M = 0 , ¯ M , if ¯ M = 0 and ~n := (cid:26) ¯ M , if ¯ M = 0 , ¯ M ∗ , if ¯ M = 0 . Utilizing (3.11) and (4.39), we infer that (cid:13)(cid:13) R i,jε (cid:13)(cid:13) L ∞ (Ω + ) c ( i, j ) ε . (4.40)Now we consider the case ¯ M = 0, and choose a positive integer i = i and a function ψ j .Thanks to (4.37), one has V Ω + g ¯ ρ ′ ( εv i ,j ) > c ( i , j ) ε , where c is independent of ε . Noting V Ω + ¯ M ( v i,j ) = 0, we obtain12 V Ω + ¯ M ( ̟ i ,j ε ) + δ g ¯ ρ ′ ( ̟ i ,j ε ) = − V Ω + g ¯ ρ ′ ( εv i ,j ) − R i ,j ε ε ( c ( i , j ) ε − c ( i , j )) < − ε c ( i , j ) / ε < ε := min { ε ( i , j ) , c ( i , j ) / c ( i , j ) } , which yields δ NE P ( ̟ i ,j ε ) <
0. Therefore,the first assertion in Theorem 3.1 holds for the case ¯ M = 0.For the case ¯ M = 0, we make use of (4.35) and (4.36) to find that m > lim j →∞ lim i →∞ V Ω + g ¯ ρ ′ ( v i,j ) V Ω + ¯ M ∗ ( v i,j ) = lim j →∞ g R h ¯ ρ ′ ψ j d sλ k ψ ′ j k L (0 ,h ) = m . (4.41)Then, for any δ >
0, there are i δ and j δ (depending on δ ), such that0 m − V Ω + g ¯ ρ ′ ( v i δ ,j δ ) V Ω + ¯ M ∗ ( v i δ ,j δ ) < δ. δ := ( m − ¯ M ) /
2. Due to ¯ M < m N , one sees V Ω + g ¯ ρ ′ ( v i δ ,j δ ) > ( ¯ M + ( m − ¯ M ) / V Ω + ¯ M ∗ ( v i δ ,j δ ) . (4.42)From (4.38) one gets 0 < c ( i δ , j δ ) ε ( m − ¯ M ) V Ω + ¯ M ∗ ( εv i δ ,j δ ) . Hence, by virtue of (4.40), there is a sufficiently small constant ε ( i δ , j δ ) < ε ( i δ , j δ ), such thatfor any ε ∈ (0 , ε ( i δ , j δ )), m − ¯ M V Ω + ¯ M ∗ ( εv i δ ,j δ ) + R i δ ,j δ ε > . (4.43)If we make use of (4.42) and (4.43), we derive that − δ g ¯ ρ ′ ( ̟ i δ ,j δ ε ) = ε V Ω + g ¯ ρ ′ ( v i δ ,j δ ) + gε Z Ω + ¯ ρ ′ v i δ ,j δ ( ̟ i δ ,j δ ε, r ) d x + ε V Ω + g ¯ ρ ′ ( v i δ ,j δ ε, r ) + N g ¯ ρ ′ ( ̟ i δ ,j δ ε ) > ε M + ( m − ¯ M ) / V Ω + ¯ M ∗ ( v i δ ,j δ )+ gε Z Ω + ¯ ρ ′ v i δ ,j δ ( ̟ i δ ,j δ ε, r ) d x + ε V Ω + g ¯ ρ ′ (( ̟ i δ ,j δ ε, r ) ) + N g ¯ ρ ′ ( ̟ i δ ,j δ ε )= ¯ M V Ω + ¯ M ∗ ( ̟ i δ ,j δ ε ) + m − ¯ M V Ω + ¯ M ∗ ( εv i δ ,j δ ) + R i δ ,j δ ε > V Ω + ¯ M ( ̟ i δ ,j δ ε ) , which yields δ NE P ( ̟ i δ ,j δ ε ) <
0. Therefore, the first assertion in Theorem 3.1 also holds for ¯ M = 0.This completes the proof. Since ¯ M = 0, we denote ~n := ¯ M / ¯ M . By (3.11), (3.16) and (4.23), we see that there isa sufficiently small constant ε , such that for any non-zero function ̟ ∈ H (Ω + ) satisfying k ̟ k L ∞ (Ω + ) ε , it holds that N g ¯ ρ ′ ( ̟ ) c k ̟ k L (Ω + ) c k ̟ k L ∞ (Ω + ) V Ω + ~n ( ̟ ) < ¯ M − m V Ω + ~n ( ̟ ) . On the other hand, from the definition m N one gets m V Ω + ~n ( ̟ ) > V g ¯ ρ ′ ( ̟ ) . Thus, for any non-zero function ̟ ∈ H (Ω + ), δ g ¯ ρ ′ ( ̟ ) m V Ω + ~n ( ̟ ) + N g ¯ ρ ′ ( ̟ ) < m V Ω + ~n ( ̟ ) + ¯ M − m V Ω + ~n ( ̟ ) = 12 V Ω + ¯ M ( ̟ ) . Hence, the second assertion in Theorem 3.1 holds.
With the help of (3.39), Lemmas 4.3 and 4.6, and Remark 4.3, we can easily establish Theorem3.2 by following the arguments in the proof of Theorem 3.1, and we omit the details here.37 . Extension to the magnetic B´enard problem
In this section we further extend the results on the magnetic inhibition in the NMRT problemto the magnetic B´enard problem without heat conduction. We begin with a brief introduction ofthe thermal instability. Thermal instability often arises when a fluid is heated from below. Theclassic example of this is a horizontal layer of fluid with its lower side hotter than its upper. Thebasic state is then one of rest states with light and hot fluid below heavy and cool fluid. Whenthe temperature difference across the layer is great enough, the stabilizing effects of viscosity andthermal conductivity are overcome by the destabilizing (thermal) buoyancy, and an overturninginstability ensues as thermal convection: hotter part of fluid is lighter and tends to rise as colderpart tends to sink according to the action of the gravity force [8].The effect of an impressed magnetic field on the onset of thermal instability in MHD fluids isfirst considered by Thompson [36] in 1951. Then Chandrasekhar theoretically further discoveredthe inhibiting effect of the magnetic field on the thermal instability in 1952 [4, 5], Later, thenonlinear magnetic inhibition theory with resistivity was given by Galdi [10], also see [11, 12, 25].However, by now it is still an open problem to show the inhibition of thermal instability bymagnetic fields based on the following 3D magnetic Boussinesq equations without resistivity inΩ + (see [6, Chapter IV] for a derivation): βv t + βv · ∇ v + ∇ ( p + λ | M | /
2) = gβ ( α (Θ − Θ ) − e + µ ∆ v + λM · ∇ M, Θ t + v · ∇ Θ = κ Θ ,M t + v · ∇ M = M · ∇ v, div v = div M = 0 . (5.1)We shall complementally introduce the new mathematical notations appearing in the equations(5.1). The unknown function Θ := Θ( x, t ) represents the temperature of the incompressibleMHD fluid. The new known physical parameters β , α and κ denote the density constant atsome properly chosen temperature parameter Θ , the coefficient of volume expansion and thecoefficient of heat conductivity, respectively.However, if we omit the thermometric conductivity (i.e. κ =0), we can easily get the inhibitionof thermal instability by a magnetic field. In fact, when κ =0, the system (5.1) reduces to βv t + βv · ∇ v + ∇P = gαβ Θ e + µ ∆ v + λM · ∇ M, Θ t + v · ∇ Θ = 0 ,M t + v · ∇ M = M · ∇ v, div v = div M = 0 , (5.2)where P := p + λ | M | / gβ ( α Θ + 1) x . For the well-posedness of (5.2), we need that following initial and boundary conditions:( v, θ, M ) | t =0 = ( v , θ , N ) in D, (5.3) v ( x, t ) = 0 on ∂D. (5.4)For simplicity, we call the model (5.2)–(5.4) the magnetic B´enard model without heat conduction.The rest state of the above magnetic B´enard model can be given by r B := (0 , ¯Θ , ¯ M ) with anassociated equilibrium pressure ¯ p , where ¯Θ and ¯ p are smooth functions defined on D , depend on x only, and satisfy the equilibrium relation ∇ ¯ p = gβ ( α ( ¯Θ − Θ ) − e ′ | x = x < x ∈ D x := { x | ( x h , x ) ∈ D } . (5.5)If D takes Ω + , we shall further assume that the density profile ¯Θ is a vertically horizontal function,i.e., ¯Θ | x =2 πnL = ¯Θ | x =2 πmL for any integer n, m. The convection condition (5.5) assures that there is at least a region in which the temperatureprofile ¯Θ has lower temperature with increasing height x , and thus may lead to the classicalconvective instability [6]. The problem whether r B is stable or unstable to the magnetic B´enardmodel without heat conduction is called the magnetic B´enard problem without heat conduction.We easily see that the magnetic B´enard problems is extremely similar to the NMRT problem.This is not surprising from the physical point of view: the thermal instability is caused by theinterchange of the lower hotter fluid and the upper cooler fluid driven by buoyancy, like theRT instability induced by the interchange of the upper heavier fluid and the lower lighter fluiddriven by gravity. Mathematically, both instabilities correspond to two terms gα Θ e and − gρe respectively. Hence, the thermal and RT instabilities belong to the type of interchange instability.In view of this similarity, we easily observe that all mathematically results of stability/instabilityfor the NMRT problem can be directly extended to the magnetic B´enard problem without heatconduction. In fact, if we define V Dgαβ ¯Θ ′ ( w ) := − Z D gαβ ¯Θ ′ | w | d x and m D B ,j := vuut sup w ∈ H σ ( D ) V Dgαβ ¯Θ ′ ( w ) V De j ( w ) , then, similarly to the NMRT problem, we have the following stability/instability results on themagnetic B´enard problem without heat conduction.(1) Stability criterion: if | ¯ M j | > m Ω j B ,j , then the rest state r B with ¯ M = Π j is asymptoticallystable to the magnetic B´enard model defined on D = Ω j with proper initial conditionunder small perturbation for j = 1 and 3.(2) Instability criterion: if | ¯ M j | < m Ω j B ,j , then the rest state r B with ¯ M = Π j is unstable to themagnetic B´enard model defined on D = Ω j in the Hadamard sense for j = 1 and 3. Inaddition, the rest state r B with ¯ M = Π k is always unstable to the magnetic B´enard modeldefined on D = Ω j in the Hadamard sense, when k = j .The above stability result shows that the thermal instability can be inhibited by (impressed)magnetic fields in a non-resistive MHD fluid with heat conduction. We mention that a sufficientlylarge µ in the system (5.1) can inhibit the thermal instability. However, µ in the system (5.2)can never inhibit the thermal instability due to the absence of κ . Of course, we can rigourouslyprove that µ can slow down the development of the linear thermal instability in (5.2) in somesense, see [16] on the effect of viscosity on the linear RT instability. Acknowledgements.
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