SOME MATHEMATICAL MODELS AND MATHEMATICAL ANALYSIS ABOUTRAYLEIGH-TAYLOR INSTABILITY?

Boling Guo

(Institute of Applied Physics and Computational Math.,China Academy of Engineering Physics,100088,Beijing,PR China)

Binqiang Xie

(South China Research Center for Applied Math.and Interdisciplinary Studies,Guangzhou 510631,Guangdong,PR China)

Abstract In this paper we will review some mathematical models and mathematical analysis about Rayleigh-Taylor instability.

Keywords mathematical models;mathematical analysis;Rayleigh-Taylor instability

1 Introduction

The instability of the interface between two different densities of fluid under the action of gravity or inertial force,as early as 1950,was clearly pointed out by G.L.Taylor,and is often named after him,actually earlier than him,L.Rayleigh in 1900 and S.H.Lamb in 1932 also talked about this problem in some sense,people sometimes call Rayleigh-Taylor or Rayleigh-Lamb-Taylor instability.This interfacial instability phenomenon can be found not only in astrophysics,but also in laster fusion and high-speed collision.It is very important even for hydraulic machinery and various engines.The linear development stage of interface instability is relatively clear.However,there are still many problems in nonlinear development that need to be recognized.The relevant research has very important practical and theoretical value.

In the mathematical analysis theory,since 2003,there have been some breakthrough works on the RT instability of compressible fluids[1,2],free boundary prob-lems[3]and MHD fluids[4].Of course,these theoretical results are still far away form the actual physical mechanics.There is still a big gap in the problem.

2 Some RT Instability Mathematical Models

2.1 Double in finite fluid Taylor instability

We consider the two-layer in finite fluid shown in Figure 2.1,where each of densities ρ1and ρ2occupies a half plane of y>0 and y<0,and gravityis parallel to y axis,pointing to its negative direction,that is=?

Figure 2.1

The deviation of the initial position of the interface from y=0 is a small amount,that is y=εcoskx.

Assume that the two-layer fluid is in a static state at the initial moment,and thus is initially non-rotating.For an ideal incompressible fluid,the fluid remains free-curl in the subsequent movement.Thus,the velocity potential Φi(i=1,2)can be introduced corresponding to the upper layer and lower layer of fluid.Then under two-dimensional conditions,Φi(i=1,2)satisfies the Laplace equation

If we suppose that the interface position during the movement is y=η(x,t).And its derivative is the first-order small amount of ε,then

Therefore,

The normal vector of the interface is,and the normal velocity of the interface is D.The following subscripts(i=1,2)respectively denote the physical quantities in the flow on two sides of the interface,and the conditions of equal pressure and normal velocity on both sides of the fluid interface are continuous

where p is the pressure and q is the velocity.

On the interface F(r,t)=0,

In the case of potenial,the momentum equation can be integrated

where U(r)is the potential of mass power,f0(t)is an arbitrary function.In(2.6),we can choose ρ1f1(t)= ρ2f2(t),then under the condition of retaining a first-order small amount,on y=η(x,t),there are

Since η is a first-order small quantity,Φtis also a first-order small quantity,and the relevant quantity on y=η can be performed by Taylor expands,that is

That is,the relevant derivative of(2.7)can be expanded by the value of y=0,which implies(2.7)still holds on the unknown active boundary y=η,but it is also fixed on the fixed boundary y=0 after linearization.This greatly simplifies the solution of the problem.Φiobviously meets the in finity disappearing condition,that is

For the two-material interface,the initial conditions should also be given.We only discuss harmonics with zero initial velocity of k:

In general,for a variety of k harmonics,we can use the Fourier series expansion method,which can be solved by linear superposition method.

Since the equation is linear,the variable x appears only in the initial condition as coskx,so it is desirable to choose

We substitute this into(2.10)and get A(0)=ε,A0(0)=0.Also we insert them into Laplace equation,then obtain

Therefore the solution takes the following form: ?i=e±kyAi(t).By the in finity condition(2.9),we know ?1=A1(t)e?ky, ?2=A2(t)eky.Substituting these results into the boundary condition y=0 in(2.7),we have

Differentiate the second equation of the above system with respect to t,then substitute it into the first equation of the above system,and eliminate¨A1(t)and¨A2(t).Finally,we obtain

Now we will discuss it in two situations.

(1)ρ1> ρ2,that is,the heavy fluid is located above the light fluid.At this time n0is purely real,A(t)=ae?n0t+ben0t.Due to the existence of the en0tformal solution,A(t)grows exponentially over time,that is,the disturbance is unstable.This phenomenon is explained in Figure 2.2.When a fluid with a density of ρ1invades into a fluid with a density of ρ2,the gravity of the portion of the fluid is G v V ρ1g.

Figure 2.2

And its buoyancy is the weight of the invading fluid that is discharged in the same volume,that is,the buoyancy is F ～ V ρ2g.

When ρ1> ρ2,the gravity is greater than the buoyancy,and this part of the invading fluid continues to sink,causing the disturbance on the interface to develop further,which leads to instability.

Taking into account the initial conditions(2.10),we have

Here coshn0t=(en0t+e?n0t)/2 is the hyperbolic cosine.If the initial condition(2.10)is changed to η(x,0)=0, ηt(x,0)= εcoskx,then the form of the interface is y =sinhntcoskx.

In a more general case,it should be a linear superposition of two solutions.

After finding the expression of the interface shape η(x,t),it is not difficult to get the form of ?i(t).

(2)ρ1< ρ2,n0=im0,where n0is the pure imaginary number.At this time

Then the disturbance on the interface no longer develops,but it oscillates with time.The interface is stable corresponding to the gravity less than buoyancy in Figure 2.2,and the density ρ1of the fluid invading into the density ρ2of the fluid is retracted again,such that the interface disturbances are no longer considered.

Now we discuss the incompressible fluid with a density of ρ,a thickness of h under the gravitational field whose the direction is downward.In order for the fluid to be suspended and stationary,a pressure difference must be given to the lower surface.Since the ideal incompressible flow is initially at rest,it is curl-free.

As before,we can introduce the velocity potential.Let Φ0be the case corresponding to the undisturbed static,and ? be the additional velocity potential formed by the interface disturbance.The velocity potential is Φ = Φ(0)+?.In the case of ignoring high order quantities,the pressure can be expressed as p= ?ρ(+?t+gyf(t)).Firstly we take into account the zero-order situation.

On the upper surface y=0,p1= ?ρ(+f(t)).

On the lower surface y= ?1,p2= ?ρ(?gh+f(t)).

Now discuss the disturbance on the interface.On the upper surface,we have y= η1(x,t);on the lower surface,there is y= ?h+η2(x,t).Since the pressures of the upper and lower surfaces,p1and p2,satisfy the zero-order quantity,the disturbance pressure is

2.2 Finite thickness double interface fluid

In practice,we often encounter complex situations of fluids with limited thickness.

In addition,the second condition of(2.5)can be linearized to

The perturbation of the Fourier component with coskx is still discussed,so that ? satisfies the Laplace equation,by choosing ? =(Aeky+Be?ky)entcoskx,η1=a1entcoskx and η2=a2entcoskx.Substituting these into the boundary condition of y=0,we get

Substituting into the condition of y=?h,we have

Eliminating a1and a2from these four equations,we get

Thus,the conditions for the non-trivial solutions of A and B are n2=?kg.

Obviously the first term in square brackets represents a solution that vibrates over time,while the second term represents an unstable solution.Correspondingly,the two surface disturbances should also have the following form of solution:

Substituting these into the kinematic conditions of y=0,y=?h,we can get

Therefore,

where A2=A1e?kh,B2=B1ekh.Assuming that the upper and lower surface disturbances at the initial moment t=0 are η1= ε1coskx and η2= ε2coskx.Obviously we have A1+B1= ε1and A2+B2= ε2,which can be written as A1e?kh+B1ekh= ε2.Thus,we have

and

For ε2=0 and ε1=0,the expression of(2.16)is very interesting,so it can be seen as the correlation and perturbation between the upper and lower surfaces.

2.3 Fluid interface movement of finite thickness two-sided vacuum

As is shown in Figure 2.3,there is a finite thickness of d incompressible fluid with a density of ρ,where outside the horizontal in finite upper and lower surfaces is vacuum,and the gravity field force is

Figure 2.3

Set

which denote the equations of motion for the upper and lower interfaces respectively,where ηu(x,t)and ηl(x,t)respectively are perturbations of the upper and lower interfaces.Again,letting Φ0(t)be the initial undisturbed velocity potential,the velocity potential of the disturbing fluid is Φ(x,y,t)= Φ0(t)+?(x,y,t).Here ? is the additional velocity potential caused by the disturbance interface with?2?=0.According to the Bernoulli equation,the pressures on the upper and lower interfaces are

where f(t)is a function.For the undisturbed situations,

which satisfy the pressure balance condition pl(0)?pu(0)= ρgd.Therefore the disturbance pressure is zero on both interfaces,that is,

Substituting these into equations(2.17)and(2.18),we get

According to(2.22),(2.23),(2.26),(2.27),we expand ηu,ηl,? as the power of the small parameter ε and obtain

In the case of the first-order approximation o(ε),we can obtain a second-order coupled ordinary differential equation system

where γ2=gk is the square of the linear growth rate of classical RT1(AT=1),ξ=kd.

The initial conditions are

From these,we get

In the second order of o(ε2),the coupled ODE is

From these,we deduce

which are shown in Figure 2.4.

Figure 2.4:(a)kd=0.2, γt=4.2, ···,(b)kd=0.8, γt=4.9

2.4 Variable density fluid layer

Here we don’t discuss a two-layer fluid with a density discontinuity,but a singlelayer fluid with a density distribution in the y direction.This distribution is also unstable under certain conditions.The fluid is heated from the bottom and has different densities at different temperatures.Then it leads to convection.Without disturbance,the fluid is at rest.At this time

Therefore we get

In case of disturbance

And u,v are first-order small quantities.The continuity equation can be written as

We take

then,

In the momentum equation,for the disturbance term,we have

Assume that the disturbance has the form eikx+nt,that is

Substituting these into the above equation,we get

And by the second and fourth equations,there is

If you do not consider the density distribution in the y direction,,then

Substituting these into(2.38),we have

Therefore,we have

Here q1,q2is the two roots for the above equation,that is

Assuming that the layer of fluid is between the two solid walls of y=?h and y=0,by y=0,(y)=0,we get A2?A1=A,and(y)=A(eq1y?eq2y).Due to y=?h,(y)=0,we have e(q1?q2)h=1,which holds only when

Thus,

In accordance with(2.40),(2.41),we have

and

It can be seen from the above equation that β>0,that is,as the density gradient of the fluid is the same direction as g,the above density distribution of the fluid layer is unstable.For short waves,

That is,the index n is proportional to,the greater the density gradient,the more unstable the instability.

2.5 The Taylor instability of viscous fluids

The two-layer semi-in finite fluids occupy the y>0 and y<0 planes,respectively,with densities of ρ1and ρ2.The gravitational acceleration is perpendicular to the interface,along the negative direction of y.In the case of incompressibility,the continuous equation can be written as

The momentum equation is the following NS equation

On the material interface F(,t)=0,the interface normal velocity is continuous with

And the tangential velocity is continuous,the stress equation is

where σijk(i=1,2)is the viscous stress tensor of the two fluids with

piis the fluid pressure,and i=1,2 means two different fluids.

Set the form of the interface to be,whereis a small amount,then.And the corresponding u,p are first-order small quantities,then the momentum equation can be linearized to

For viscous fluids,the flow is not no-curl,since the potential function is insufficient to satisfy the momentum equation.We choose

Substituting this into the momentum equation,we have

According to the existing vector expression of,the equation of motion and stress conditions on the interface can be written as

Here,the surface tension is considered in the normal pressure difference,Σ is the surface tension coefficient,and the corresponding stress is

Representing ui,vias a function expression for Φiand Ψi,substituting them into the above expression and converting the condition on the interface y=0 to

we obtain

Assume that the initial perturbation of the interface is η(x,0)= εcoskx.In order for Φito satisfy the conditions of the Laplace equation and x= ±∞,we take

Set

Substituting these into(2.43)and considering the disappearance condition at y=±∞,we deduce

Substituting the above Φiand Ψiexpressions into the inner boundary conditions(2.44)and(2.45),eliminating a,we get

Here a=(ρ1? ρ2)kg?k3Σ.

The non-trivial solution of the above homogeneous equations of A,B,C and D exists under the condition that the corresponding coefficient determinant is zero,that is

Expanding this determinant and substituting m into this,we get

When the viscosityμiis large,n becomes smaller.We ignore the term with n in the root of(2.46),that is takeμik2? ρin,or approximately choose mik.Thus we get

Meanwhile we choose ν =(μ1+ μ2)/(ρ1+ ρ2),therefore

We focus on the unstable root where the coefficient before the square root is positive,so that in the case of no viscosity,the index n is n0

We discuss it two situations.

(1)n0νk2,that is,when k or viscosity is small,by(2.47)we obtain

From these we can see the viscosity reduces the RT instability.The larger k is,the more significant it is.

(2)n0νk2,by(2.47)we get

That is,n quickly decreases with the increase of k,especially as k→kc.By(2.48)we can obtain

This is the case where n tends to zero near kc.When Σ=0,regardless of surface tension,

That is,when k is large enough,n asymptotically approaches the k axis in the form of a curve.

2.6 Ablation instability

We consider a simplified model,discuss the ideal incompressible fluid,and assume that the e ff ect of heat conduction is mainly the motion that derives the phase change.It has a certain velocity or acceleration,which is ignored in the equation.Let the section be at an average speed of U0moves forward in the direction of y with an acceleration of g,the densities before and after the cross section are ρ1and ρ2respectively,taking the coordinate system moving along the section,here the speed of the“1”zone in the coordinate system is

along the opposite direction of y.

Conservation of mass momentum in a coordinate system with a constant crosssection can be written as

here n and τ represent the normal and tangential directions of the section,respectively.ρi,pi(i=1,2)are respectively the densities and pressures on both sides,qi(i=1,2)denote the velocities of the section on both sides that are moving in motion.Assume that the section is the following form:

where η is a small amount,then

Here Dnis the normal velocity of the section.In this way,the pressure and speed can be expanded to

Substituting the relevant quantities into(2.52)respectively we can obtain the relationship between the following zero-order and first-order quantities on both sides of the discontinuity:

The zero order quantity

and the first order quantity

Next we solve the first-order quantity.The front of the ablation section is originally static and therefore non-curl.However,the rear of the section,that is,the “2” area may be non-rotating,but may also be rotated.We discuss two cases below.

Case 1There is no swirling flow after the cross section,since it is non-rotating in both areas,we can choose

Under incompressible conditions,we have

Integrating the momentum equation and taking the first-order small quantity,the following pressure disturbance expression can be obtained:

Substituting this into(2.56),we can get the condition of the section between y=η(x,t).Since the zero-order quantity is uniform,the value of the first-order quantity on y=η(x,t)can be Taylor expanded.Ignoring the high order small quantity,we deduce the condition on the y=0 fixed boundary

For simplicity,we discuss the case where both layers are semi-in finite fluids,by assuming that a fluid with a density of ρ1occupies a semi-in finite plane of y>0,and a fluid with a density of ρ2occupies a semi-in finite plane of y<0.

Assume that the initial moment of disturbance is in the form of coskx.To guarantee that Φisatisfies the Laplace equation and the corresponding in finity disappearance condition,we choose

Substituting these into(2.60),we get

This is a homogeneous linear algebraic equation for a,A and B.The condition for this non-trivial solution is that the corresponding coefficient determinant is zero,thus we can obtain the following dispersion relation:

Considering that(2.55),we have

Next we discuss a few special cases:

(1)When ρ1ρ2,(2.60)can be written as

where g>0 corresponds to the RT instability case,but slows the development of instability due to the presence of ablation pressure.When?p(0)= ρ1g/k,the instability can be truncated.

(2)When ρ1～ ρ2,(2.63)can be written as

When g>0,the instability develops rapidly.When g<0,the section is quickly oscillated.

Now consider that the ablation product area is a swirling flow,and in the case of a spin in the“2”area,we choose

The “1” area is still non-rotating.There are four physical quantities to be determined,which are η, Φ1, Φ2,and Ψ2.But the boundary condition(2.56)contains only three equations.The problem is uncertain.Landau assumed the normal velocity is zero when discussing the combustion

There is a supplementary condition from(2.56)

To get the expression(2.59)for the perturbation pressurein the “2” area,we must take

Thus the momentum equation can be satisfied.If we take Ψ2～ ekt+m2ysinkx,then m2=?n/V2.

Note that V2<0,so m2>0.Therefore Φ2satisfies the disappearing condition as y→?∞.Set

Substituting these into the boundary condition of y=η(x,t)and expanding on y,we can get four constants which are determined relative to a,A,B and c.These four constants satisfy homogeneous linear algebraic equations,and the existence condition is that the corresponding coefficient determinant is zero.Thus,

Since V1<0,V2<0,when ρ1? ρ2,the second item in(2.65)square brackets can be omitted,we get

g=0 is the combustion instability case discussed by Landau,If ρ1? ρ2,

When the burning speed is vγ=?V1,there is

At this time,the instability is more serious.When k is a small constant,that is,for long waves,the term V1V2k2is ignored,there is

Storm used a similar formula and multiplied a correction factor to estimate the instability of the laser ablation interface.

2.7 RT instability of magnetic field in MHD

As is shown in Figure 2.5,the ideal magnetohydrodynamics equations can be expressed as:

Figure 2.5

Linearizing(2.67)-(2.70)and expressing as the “0” order and the “1” order small perturbation scale,there are

Substituting these into(2.67)-(2.70),we obtain the linearized equations

In additional,using the incompressible conditions

Use the following initial flow field distribution

and set u(1)=u(1)ex+v(1)ey,B(1)=B(1)ex+C(1)ey.Substituting the above initial fluid distribution and small perturbation into equations(2.71)-(2.73),there is

Setting

Substituting(2.81)into(2.75)-(2.80)gives u(0)=0 and

where n=γ?iw,γ represents the linear growth rate,and w=w(k)is the disturbance frequency.

The second-order eigenvalue equation for the perturbation velocitycan be obtained by simplifying the above algebraic equations

Integrating(2.88)from?∞to∞,we can obtain the following simple form:

Set

where Lpand LBare the density and the magnetic field filter layer widths,respectively.

where ΛT=(ρ1? ρ2)/(ρ1+ ρ2)is the Atwood number.

3 Some Mathematical Analysis Theory Results

In this section we mainly introduce some theoretical analysis results of Guo Yan and his collaborators.

3.1 Linear instability of compressible viscous fluids

As is shown in Figure 3.1,we consider a horizontal in finite flat area ? =R2×(?m,l)? R3.

Figure 3.1

Consider the compressible NS equation:

where

under the jump condition on the interface

where σ≥0,H is the mean curvature of the surface.

The boundary is

Set the density and the speed to be periodic on the space ?±(t),

In the Lagrange coordinates,we can rewritten the equations as another form.Thus we can solve this equations in the fixed domain.

Setting ??=R2×(?m,0),?+=R2×(0,l),there exists a mapping:

Define a flow mapping η±:

where ?±(t)= η±(?±,t),Σ(t)= η±{{x3=0},t}.

Defining the Lagrange unknowns

By introducing η±,v±,q±,AT=(Dη)?1.In Lagrange coordinates,the evolution equations for v,q,η are

where the viscous stress tensor T is

with Iijbeing an element of 3×3 matrix I.

To obtain the jump conditions,for a quantity f=f±,we define the interfacial jump as

The jump conditions across the interface are

where

for the unit normal to the surface Σ(t)= η(x3=0,t)and H for twice the mean curvature of Σ(t),

Finally,we require the no-slip boundary condition

In the following we seek a steady-state solution,by taking

We deduce the following ODE

subject to the jump condition

To solve this we introduce the enthalpy function defined by

To discuss the Rayleigh-Taylor instability,we must prove the fluid is denser above the interface,that is>,then

The solution is then given by P±(ρ)=K±ργ,K±>0,γ±≥ 1,

If γ+= γ?,for any>0,K?>K+is required.If γ+γ?,then K?,K+>0 can be arbitrary,but we must require that>0 must satisfy

We now linearize the nonlinear problem(3.8)around the steady-state solution v=0,η =Id,q= ρ0.The obtained linearized equations with η,v,q are

and

where ε0= ε0(ρ0),δ0= δ0(ρ0).

The jump conditions is linearized to

under the initial data η(0)= η0,v(0)=v0,q(0)=q0,that satisfy the jump and boundary conditions in addition to the assumption that[[η0]]=0,which implies that η(t)is continuous across x3=0,for any t≥ 0.

3.2 Growing mode ansatz

We will look for a growing normal mode solution to(3.22),(3.23)by first assuming an ansatz

for some λ>0.Plugging the ansatz into(3.22)-(3.23),we may solve the first and second equations for,in terms of v.By doing so and eliminating them from the third equation,we arrive at the time-invariant equation

This is coupled to the jump conditions[[w]]=0 and

with the boundary conditions w?(x1,x2,?m)=w+(x1,x2,l)=0.Notice that the first jump condition implies the assumptions on

Since the coefficients of the linear problem(3.26)depend only on the x3variable,we define

For each fixed ξ,we arrive at the following system of ODEs for ?(x3),θ(x3),ψ(x3)and λ,

The first jump condition yields

The second jump condition becomes

which implies that

and that

The boundary conditions

must also hold.If(?,θ,ψ)is a solution to(3.31)-(3.33)for ξ∈ R2and λ,then for any rotation operator R∈SO(2),=R(ψ,θ)is a solution to the same equations for=Rξ with ψ,λ unchanged.So,by choosing an appropriate rotation,we may assume without loss of generality that ξ2=0,and ξ1=|ξ|.In this setting θ solves

Multiplying this equation by θ,integrating over(?m,l),integrating by parts,and using the jump conditions yield

which implies that θ=0,since λ >0.This reduces to the pair of equations for ?,ψ:

along with the jump conditions

and the boundary conditions

3.3 Statement of main results

In the absence of viscosity,ε= δ=0 and for a fixed spatial frequency ξ 6=0,(3.41),(3.42)can be viewed as an eigenvalue problem with eigenvalue?λ2.Such a problem has a natural variational structure that allows for the construction of solutions via the direct methods and for a variational characterization of the eigenvalue according to

where

and

Unfortunately,when viscosity is present,the natural variational structure breaks down since λ2appears quadratically as a multiplier of ρ0and linearly as a multiplier of ε0and δ0.Since the equations imply a quadratic relationship between λ and various integrals of the solution,they can be solved for λ to determine the sign of Reλ.On the other hand,the appearance of λ both quadratically and linearly eliminates the capacity by using constrained minimization techniques to produce solutions to the equations.

In order to circumvent this problem and restore the ability to use variational methods,we artificially remove the linear dependence on λ.To this end,we define the modified viscosities=sε0,=sδ,where s>0 is an arbitrary parameter.We then introduce a family of modified problems given by

along with the jump conditions

and the boundary conditions

A solution to the modified problem with λ=s corresponds to a solution to the original problem.

Modifying the problem in this way restores the variational structure and allows us to apply a constrained minimization to the viscous analogue of the energy E defined above to find a solution to(3.49),(3.50)with λ = λ(|s|,s)>0,when s>0 is sufficiently small and precisely when

We then further exploit the variational structure to show that λ is a continuous function and is strictly increasing in s.Using this,we show in Theorem 3.6 that the parameter s can be uniquely chosen so that

which implies that there exists a solution to the original problem(3.41),(3.42).This choice of s allows us to consider λ = λ(|ξ|),which gives rise to a solution to system(3.32),(3.33)as well.

Theorem 3.1 For ξ∈ R2,so that 0<|ξ|20 to(3.32),(3.33)satisfying the appropriate jump and boundary conditions so that ψ = ψ(ξ,0)6=0.The solutions are smooth when restricted to(?m,0)or(0,l),and they are equivariant in ξ in the sense that if R∈SO(2),that is a rotation operator,then

Without surface tension(σ=0),it is possible to construct a solution to(3.55)with λ >0,for any ξ 6=0,but with surface tension(σ >0)there is a critical frequencyfor which there is no solution with λ >0 being available if|ξ|≥ |ξ|c.In the nonperiodic case,we capture a continuum|ξ|∈ (0,|ξ|c)of growing mode solutions,but in the 2πL periodic case we find only finitely many solutions.Indeed,if

then there is a positive but finite number of spatial frequencies ξ∈ (L?1Z)2with|ξ|<|ξ|c.On the other hand,if

then our method fails to construct any growing mode solutions at all.It is important to know the behavior of λ(|ξ|)as|ξ|varies within 0<|ξ|<|ξ|c.It is easy to show that λ(|ξ|)is continuous and satisfies

In the nonperiodic case,this implies that there is a largest growth rate

In the periodic case for L satisfying(3.57),the largest rate is always achieved and is given by

Note that in general ΛL< Λ.

The stabilizing e ff ects of viscosity and surface tension are evident in these results.Without viscosity or surface tension,there is λ(ξ) → ∞,|ξ|→ ∞.With viscosity but no surface tension,all spatial frequencies remain unstable,but the growth rate λ(ξ)is bounded.With viscosity and surface tension,only a critical interval of spatial frequencies is unstable,and λ(|ξ|)remains bounded.Finally,with viscosity,surface tension and the periodicity L satisfying(3.58),there do not exist any growing modes.

In the periodic case when L satisfies(3.57),the solutions to(3.32),(3.33)constructed in Theorem 3.1 immediately give rise to growing mode solutions to(3.22),(3.23).

Theorem 3.2 Suppose that L satisfies(3.57),and let ξ1,ξ2∈ (L?1E)2be lattice points such that ξ1= ?ξ2,λ(ξi)= ΛL,where ΛLis defined by(3.66).Define

where ?,θ,ψ are the solutions provided by Theorem 3.1.Writing x0=x1e1+x2e2,we define

and

Then η,v,q are real solutions to(3.22),(3.23)with the corresponding jump and boundary conditions.For every t>0,we have η(t),v(t),q(t)∈ Hk(?),and

Remark 3.1 In this theorem,the space Hk(?)is not the usual Sobolev space of order k,but what we call the piecewise Sobolev space of order k.In the nonperiodic case,although Λ = λ(|ξ|)may be achieved for some ξ∈ R2,|ξ|∈ (0,|ξ|c),no L2(?)solution to(3.22),(3.23)may be constructed.

where ?,θ,ψ are the solutions provided by Theorem 3.1.Writing x0=x1e1+x2e2we define

Then η,v,q are real-valued solutions to the linearized equations(3.22),(3.23)along with the corresponding jump and boundary conditions.The solutions are equivariant in the sense that if R∈SO(3)is a rotation that keeps the vector e3fixed,then

For every k∈N we have the estimate

To state the result,we first define the weighted L2norm and the viscosity seminorm by

Theorem 3.4 Let v,η,q be a solution to(3.22),(3.23)along with the corresponding jump and boundary conditions.Then in the nonperiodic case

Theorem 3.5 In the periodic case let L satisfy(3.58).For j≥1 define the constants Kj≥0 in terms of the initial data via

Then solutions to(3.22),(3.23)satisfy

and for j≥1

3.4 A family of modified variational problems

In order to understand λ in a variational framework we consider the two energies

and

which are both well defined on the space.Consider a set

We want to show that the in fimumsolves(3.49),(3.50)along with the corresponding jump and boundary conditions.Also notice that by employing the identity?2ab=(a?b)2?(a2+b2),and the constraint on J(?,ψ),we may rewrite

Proposition 3.1 E achieves its in fimum on A.

Proof First note that(3.82)shows that E is bounded below on A. Let(?n,ψn) ∈ A be a minimizing sequence.Then ?n,ψnare bounded in((?m,l))and ψn(0)is bounded in R,so we can choose a subsequence(?n,ψn)*(?,ψ)weakly in×,and(?n,ψn)→ (?,ψ)strongly in L2×L2.The compact embeddingimplies that ψn(0) → ψ(0)as well.Because of the quadratic structure of all the terms in the integrals defining E,the weak lower semicontinuity and strong L2convergence imply that

That(?,ψ)∈ A follows from the strong L2convergence.

We now show that the minimizer constructed in the previous result satisfies Euler-Langrange equations equivalent to(3.49),(3.50).

Proposition 3.2 Let(?,ψ)∈ A be the minimizers of E and μ =E(?,ψ),then(?,ψ)are smooth when restricted to(?m,0)or(0,l)and satisfy

along with the jump conditions

and the boundary conditions ?(?m)= ?(l)= ψ(?m)= ψ(l)=0.

Proof Fix(?0,ψ0)∈((?m,l))×((?m,l)).Define

and note that j(0,0)=1.Moreover,j is smooth,

So,by the inverse function theorem,we can solve τ= τ(t)in a neighborhood of 0 as a C1function of t so that τ(0)=0,j(t,z(t))=1.We differentiate the last equation to obtain

hence

Since(?,ψ)are minimizers over A,we make variations with respect to(?0,ψ0)to find that

which implies that

Rearranging and plugging in the value of τ0(0),we rewrite this equation as

where the Lagrange multiplier is μ =E(?,ψ).Since ?0,ψ0are independent,this gives rise to the pair of equations

and

By making variations with ?0,ψ0compactly supported in either(?m,0)or(0,l),we find that ?,ψ satisfy(3.86),(3.87)in a weak sense in(?m,0)or(0,l).The stand-ard bootstrapping arguments then show that(?,ψ)are in Hk(?m,0)(Hk(0,l))for any k≥0,and hence the functions are smooth when restricted to either interval.To show that the jump conditions are satisfied,integrating the terms in(3.97),we find that

Performing a similar integration by parts in(3.98)yields the jump condition

The conditions[[?]]=[[ψ]]=0, ?(?m)= ?(l)= ψ(?m)= ψ(l)=0 are satisfied trivially since

We now show that for s sufficiently small,the in fimum inf E over A is in fact negative.

Proposition 3.3 Suppose that 0<|ξ|20 depending on the quantities,p±,g, ε±, σ,m,l,|ξ|so that for s ≤ s0,there isμ(s)<0.

Proof Since both E and J are homogeneous of degree 2,it suffices to show that

but since J is positive definite,we may reduce to construct any pair(?,ψ)∈×such that E(?,ψ)<0.We assume that ? = ?ψ0/|ξ|so that the first integrand term in E(?,ψ)is vanished.We must then construct a ψ ∈so that

We employ the identity ψψ0=((ψ)2)0/2,an integration by parts,and the fact that ρ0solves

to write

Note that[[ρ0]]= ρ+?ρ?>0 so that the right-hand side is not positive definite.

For α >5 we define the test function ψα∈(?m,l)according to

Simple calculations then show that

where oα(1)is a quantity that vanishes as α → ∞,and that

for a constant C depending on α,m,l,|ξ|.Combining these,we find that

for a constant C depending on α,,p±,g,ε±,m,l,|ξ|.Since σ|ξ|20 depending on the various parameters so that for s≤ s0,we have(ψα)<0,thereby this proves the result.

Remark 3.2 We can proveμ(s)≤ ?C0+sC1by

ψ(0)6=0,|ξ|2

Lemma 3.1 Suppose that(?,ψ) ∈ A satisfies E(?,ψ)<0.Then ψ(0)6=0,|ξ|2

Proof A completion of the square allows us to write

Integrating by parts as in(3.102),we know that

Combining these equalities,we can rewrite E(?,ψ)as

From the nonnegativity of the integrals,if E(?,ψ)<0,then ψ(0)6=0,|ξ|2

Proposition 3.4 Letμ:(0,∞)→ R be given by(3.84),then the following conclusions hold:

2.There exists a positive constant C2=C2(,ρ±,g,ε±,σ,m,l)so that

3.μ(s)is strictly increasing.

Proof Fix a compact interval Q=[a,b]? C(0,∞)and any pair(?0,ψ0)∈ A.We may decompose E according to

for

and

The nonnegativity of E1implies that E is nondecreasing in s with a fixed(?,ψ)∈ A.

Now,by Proposition 3.1,for each s∈(0,+∞)we can find a pair(?,ψ)∈A so that

We deduce from the nonnegativity of E1,the minimality of(?s,ψs),and the equality(3.82)that

for all s ∈ Q.This implies that there exists a constant 0

Let s ∈ Q,i=1,2.Using the minimality of(?s1,ψs1)compared to(?s2,ψs2)we know that

but from the decomposition(3.110),we obtain

Together with these two inequalities and employing(3.115),we find that

Similarly we can obtainμ(s2)≤μ(s1)+K|s1?s2|.Thus

which proves the first assertion.

To prove(3.109)we note that equality(3.82)and the nonnegativity of E1imply that

We can easily verify that this in fimum,denoted by a constant C2,is positive.

Finally,to prove the third assertion,note that if 0

This shows thatμis nondecreasing in s.Now suppose by way of contradiction thatμ(s1)= μ(s2).Then the previous inequality implies that

which means that E1(?s2,ψs2)=0.This in turn forces ?s2= ψs2=0,which contradicts the fact that(?s2,ψs2)∈ A.Hence the equality cannot be achieved,andμis strictly increasing in s.

Now we know that when 0<|ξ|2

We can now state a result about the existence of solutions to(3.49),(3.50)for these values of|ξ|,s.To emphasize the dependence on the parameters,we write

Proposition 3.5 For each s ∈ S,0<|ξ|20 to problem(3.49),(3.50)along with the corresponding jump and boundary conditions.Solutions ψs(|ξ|,0)6=0 are smooth when restricted to either(?m,0)or(0,l).

Proof Let ?s(|ξ|,·),ψs(|ξ|,·) ∈ A be the solutions to(3.86),(3.87).Since s ∈ S,we may write μ = ?λ2,λ >0,which means that the pair ?s(|ξ|,s),ψs(|ξ|,s)solve problem(3.49),(3.50).The fact that ψs(|ξ|,0)6=0 follows from Lemma 3.1.

In order to obtain solutions to the original problem,we must be able to find an s ∈ S,so that s= λ(|ξ|,s).It turns out that the set S is sufficiently large to accomplish this.

Theorem 3.6 There exists a unique s ∈ S,so that λ(|ξ|,s)=>0 and

Proof According to Remark 3.2,we know thatμ(s)≤ ?C0+sC1.Moreover,the lower bound(3.109)implies thatμ(s)→+∞ as s→∞.Sinceμis continuous and strictly increasing,there exists an s ∈ (0,∞)so that S= μ?1((?∞,0))=(0,s?).Sinceμ >0 on S,we define λ =.Now define a function Φ :(0,s?) →(0,∞)according to Φ(s)=s/λ(|ξ|,s).It can be easily check that Φ is continuous and strictly increasing in s.Also,.Then by the intermediate value theorem,there exists an s ∈ (0,s?)so that Φ(s)=1,that is s= λ(|ξ|,s).This is unique since Φ is strictly increasing.

We now use Theorem 3.6 to study s=s(|ξ|).Since for each fixed 0<|ξ|2

Proof of Theorem 3.1 We may find a rotation operator R∈SO(2),so that Rξ=(|ξ|,0).For s=s(ξ)given by Theorem 3.6,define(?(ξ,x3),θ(ξ,x3))=R?1(?s(|ξ|,x3),0)and ψ(ξ,x3)= ψs(|ξ|,x3),where the functions ?s(|ξ|,x3),ψs(|ξ|,x3)are the solutions from Proposition 3.5.This gives a solution to(3.31)-(3.33).The equivariance in ξ follows from the definition.

3.5 Solutions to(3.22),(3.23)

In this section we will construct growing solutions to(3.22),(3.23)using the solutions to(3.31)-(3.33)constructed in Theorem 3.1.In the periodic case this can only be done when L satisfies(2.17),but the construction is essentially trivial since normal mode solutions are in L2(?).In the nonperiodic case,we must resort to a Fourier synthesis of the normal modes in order to produce L2solutions.

We begin by defining some terms.For a function f ∈ L2(?),we define the horizontal Fourier transform in the nonperiodic case via

for ξ∈R2.In the periodic case the integral over R2must be replaced with an integral over(2πLT)2for ξ∈ (L?1Z)2.In the nonperiodic case,by the Fubini and Parseval theorems,we havebf∈ L2(?)and

The period is replaced with 4πL2and the integral is replaced with a sum over(L?1Z)2on the right-hand side.

We now define the piecewise Sobolev spaces.For a function f defined on ? we write f+for the restriction to ?+and f?for the restriction to ??.For k ∈ N,define the piecewise Sobolev space of order k by

Lemma 3.2 Suppose 00 depending on the parameters a,b,,P±,g,ε±,δ±,m,l,

Also,there exists a B0>0 depending on the same parameters so that

Proof of Theorem 3.2 It is clear that η,v,q defined in this way are solutions to(3.22),(3.23).That they are real-valued follows from the equivariance in ξ stated in Theorem 3.1 The solutions are in Hk(?)at t=0 because of Lemma 3.2.The growth in time stated in(3.65)follows from the definition of η,v,q.

Proof of Theorem 3.3 For each fixed ξ∈ R2so that|ξ|∈ (0,|ξ|c),

then we obtain a solution to(3.22),(3.23).

Since supp(f) ? (0,|ξ|c),Lemma 3.2 implies that

This together with the definition of Λ,and the dominated convergence theorem implies that the Fourier synthesis of these solutions given by(3.67)-(3.69)is also a solution that is smooth when restricted to ?±.The Fourier synthesis is real-valued because f(|ξ|)is real-valued and radial because of the equivariance in ξ given in Theorem 3.1.This equivariance in ξ also implies the equivariance of η,v,q written in(3.70).

The bound(3.71)follows by applying Lemma 3.2 with arbitrary k≥0 and utilizing the fact that f is compactly supported.The compact support of f also implies that λ0(f)>0,so that λ0(f) ≤ λ(|ξ|) ≤ Λ,|ξ|∈ supp(f).This then yields the bounds(3.72).

3.6 Growth of solutions to the linearized problem

In this section we will prove estimates for the growth in time of arbitrary solutions to(3.22),(3.23)in terms of the largest growing mode:Λ in the nonperiodic case and ΛLin the periodic case,defined by(3.60)and(3.61)respectively.To this end,we suppose that η,v,q are real-valued solutions to(3.22),(3.23)along with the corresponding jump and boundary conditions.

It will be convenient to work with a second-order formulation of the equations.To arrive at this,we differentiate the third equation in time and eliminate the q,η terms using the other equations.This yields the equation

coupled to the jump conditions[[?tv]]=0,and

The function ?tv also satisfies?tv(x1,x2,?m,t)= ?tv(x1,x2,l,t)=0 at the upper and lower boundaries.The initial data for?tv(x1,x2,?m,t)= ?tv(x1,x2,l,t)=0 is given in terms of the initial data q(0),v(0)and η(0)via the third linear equation;that is,?tv(0)satisfies

The following lemma gives an energy and its evolution equation for solutions to the second-order problem.

Lemma 3.3 Let v solve(3.133)with the corresponding jump and boundary conditions.Then in the nonperiodic case,

In the periodic case,the same equation holds by replacing the integral over R2with an integral over(2πLT)2.

Proof We will prove the result in the nonperiodic case,Since the periodic case follows similarly.Take the dot product of(3.133)with vtand integrate over ?t.After integrating by parts,we get

Taking the derivatives of both sides of the above equation leads to

A similar result holds on ??=R2×(?m,0)with the opposite sign on the right-hand side.Adding these two equalities yields

Using the jump conditions,we find that

The result follows by substituting this into(3.138).

The following result is devoted to estimating the energy in terms of Λ.

Lemma 3.4 Let v∈ H1(?)be so that v(x1,x2,?m)=v(x1,x2,l)=0.In the nonperiodic case we have the inequality

ProofWe also only prove only the nonperiodic version.Take the horizontal Fourier transform and apply(3.126)to see that

Consider now the last integrand for fixed ξ 6=0,and letThat is,define

The expression for Z is invariant under simultaneous rotations of ξ and(?,θ),so without loss of generality we may assume that ξ=(|ξ|,0),|ξ|>0 and θ=0.If σ>0,then we assume that|ξ|<|ξ|c.Using(3.79)withwe may rewrite Z as

hence

For|ξ| ≥ ξc,the expression for Z is nonpositive,so the previous inequality holds obviously,and so we deduce that it holds for all|ξ|>0.

Together(3.141)with(3.143)for fixed ξ,we find

where B=Dv+DvT?2(divv)I/3.Integrating both sides of this inequality on R2and using(3.79),we prove the result.

When σ>0 and L is sufficiently small,a better result is available in the periodic case.

Lemma 3.5 Let v ∈ H1(?)be so that v(x1,x2,?m)=v(x1,x2,l)=0 and suppose in the periodic case that L satisfies(3.58).Then

Proof Apply the horizontal Fourier transform to see that

Because of(3.58),there exists a unique ξ∈ (L?1Z)2such that g[[ρ0]]? g|ξ|2≥ 0,that is ξ=0.Obviously,if ξ 6=0,there is

For this we expand the term in the integral and integrate by parts to get

which yields the desired inequality.

3.7 Proof of Theorems 3.4 and 3.5

Proof of Theorem 3.4 Again,we will only prove the nonperiodic case.Integrate(3.136)with t from 0 to t to find that

where

We may then apply Lemma 3.4 to get an inequality

Using the definitions of the norms k·k1and k·k2,we may compactly rewrite the previous inequality as

Integrati ng(3.152)with t and using Cauchy inequality,we obtain

On the other hand,note that

From(3.152)-(3.154),we derive the following differential inequality

To derive the corresponding bounds for,according(3.152),(3.153)and(3.156),we obtain

for a constant C>0 depending on,p±,Λ,ε±,δ±,σ,g,m,l.

Proof of Theorem 3.5 We again integrate(3.136)with t from 0 to t to find that

We may apply Lemma 3.5 to see that the sum of all of the integrals on the right side of the previous inequality are nonpositive,and hence

From this we deduce that

Then,using ?tη =v,we get

This implies(3.77)holds.Note that

Thus(3.78)holds.