On Incompressible Vibrations of the Stratified Atmosphere on the Flat Earth
aa r X i v : . [ m a t h . A P ] F e b On Incompressible Vibrations of the StratifiedAtmosphere on the Flat Earth
Tetu Makino ∗ February 25, 2020
Abstract
We consider the vibrations and waves in the atmosphere under thegravitation on the flat earth. The stratified density distribution of theback ground equilibrium is supposed to touch the vacuum at the finiteheight of the stratosphere. We show that incompressible motions arepossible to create vibrations or progressive waves with countably manytime frequencies of the vibrations or speeds of the wave propagation whichaccumulate to 0, say, of countably infinitely many slow and slow modes.
Key Words and Phrases.
Euler equations, Atmospheric Waves, Vac-uum boundary, Gravity modes. Eigenvalue problem of Strum-Liouvilletype.
We investigate the motion of the atmosphere on the flat earth under the constantgravitational force. We suppose that the motion is governed by the compressibleEuler equations.The pioneering mathematical investigation can be found as the paper ‘Onthe Vibrations of an Atmosphere’ by Lord Rayleigh, 1890, [8]. He wroteIn order to introduce greater precision into our ideas respecting thebehavior of the Earth’s Atmosphere, it seems advisable to solve anyproblems that may present themselves, even though the search forsimplicity may lead us to stray rather far from the actual question.It is supposed here to consider the case of an atmosphere composedof gas which obeys Boyle’s law, viz. such that the pressure is alwaysproportional to the density. And in the first instance we shall neglectthe curvature and rotation of the Earth, supposing that the strata ∗ Professor Emeritus at Yamaguchi University, Japan; e-mail: [email protected];Supported by JSPS KAKENHI Grant Number JP18K03371
1f equal density are parallel planes perpendicular to the direction inwhich gravity acts.Our investigation in this article will be done under the same spirit as thatof Lord Rayleigh quoted above, except for his starting point that the background state is supposed to be that of the isothermal stratified gas, say, ρ = ρ exp( − g z/A ), where ρ is the density and z is the height, ρ , g , A being pos-itive constants. Instead we are interested in a back ground equilibrium like ρ = C ( z + − z ) ν , where z + is the height of the stratosphere, C, ν ( >
1) beingpositive constants, which touches the vacuum ρ = 0 on z > z + at the height z = z + . We shall analyze the waves of the small amplitude around the background stratified equilibrium of this form assuming that the motion is done inthe incompressible manner. Therefore it might be better to call the problem asthat of waves in a heterogeneous liquid, as [5, Chapter IX, Section 235], ratherthan the problem of waves in an atmosphere. In contrast with the existingdiscussions, we should introduce the Lagrangian co-ordinate to treat the vac-uum boundary as a free boundary. Anyway the linearized analysis leads us tothe investigation of the eigenvalue problem described by the so called Taylor-Goldstein equations. This equation is formally well known, e. g., as [5, p.378,(12)], but, usually it was discussed in a bounded strap with both fixed bound-aries at which the back ground density distribution behaves regularly and theusual Dirichlet boundary conditions are posed. If we consider it on the background which touches the vacuum at a finite height, it requires a careful analysisof the singular boundary at the vacuum boundary in order to justify that theeigenvalue problem can be considered as that of the Strum-Liouville type. Asthe result we have vibrations with time frequency √ λ n and progressive waveswith speed √ λ n /l , 2 π/l being the wave length in the horizontal direction, while λ > λ > · · · > λ n > · · · →
0, in the linearized approximation level. Thisanalysis cannot be found in the existing literatures and is newly exhibited inthis article.Let us describe the problems more precisely.We consider the motions of the atmosphere on the flat earth governed bythe Euler equations ∂ρ∂t + X k =1 ∂∂x k ( ρv k ) = 0 , (1.1a) ρ (cid:16) ∂v j ∂t + X k =1 v k ∂v j ∂x k (cid:17) + ∂P∂x j + ρ ∂ Φ ∂x j = 0 , j = 1 , , , (1.1b)Φ = g x . (1.1c)Here t ≥ , x = ( x , x , x ) ∈ Ω := { x ∈ R | x > } . The unknowns ρ ≥ , P are density and pressure, and v = v ∂∂x + v ∂∂x + v ∂∂x is the velocity field.2 is a positive constant. The boundary condition is ρv = 0 on x = 0 , (1.2)and the initial condition is ρ = ◦ ρ ( x ) , v = ◦ v ( x ) = ( ◦ v ( x ) , ◦ v ( x ) , ◦ v ( x )) at t = 0 . (1.3)Moreover we suppose that the motion of the atmosphere is incompressible,that is, it holds that DρDt := ∂ρ∂t + X k =1 v k ∂ρ∂x k = 0 , (1.4)which means that the density is constant along each stream line. Under theequation of continuity (1.1a), this is equivalent to ρ X k =1 ∂v k ∂x k = 0 . (1.5)Suppose that there is fixed a flat equilibrium ρ = ¯ ρ, P = ¯ P , v = whichsatisfy (1.1a)(1.1b) with (1.1c), where ¯ ρ, ¯ P are functions of z = x only. Namelywe suppose that d ¯ Pdz + g ¯ ρ = 0 for z > . (1.6)We assume that ¯ ρ ∈ C ([0 , + ∞ [) and { ¯ ρ > } = { ≤ z < z + } , where 0 < z + ≤ + ∞ . We suppose d ¯ ρdz < ≤ z < z + . (1.7)We consider the perturbation ξ j = δx j , j = 1 , , , δρ, δP at this fixed equi-librium. Notation 1
We use the Lagrangian coordinate which will be denoted by thediversion of the letter x of the Eulerian coordinate. Instead, hereafter, we shalluse the symbols x = ( x , x , x ) to denote the original Eulerian co-ordinates sothat x = x + ξ ( t, x ) , ξ being ( ξ , ξ , ξ ) . Of course we put ξ = ξ = ξ = 0 at t = 0 (1.8)The boundary condition is ξ = 0 on x = 0 , (1.9)3nd the initial condition is ∂ξ j ∂t (cid:12)(cid:12)(cid:12) t =0 = ◦ v j ( x ) , j = 1 , , . (1.10)Here let us recall the definition of the Euler perturbation δQ and the La-grange perturbation ∆ Q of a quantity Q :∆ Q ( t, x ) = Q ( t, ϕ ( t, x )) − ¯ Q ( x ) ,δQ ( t, x ) = Q ( t, ϕ ( t, x )) − ¯ Q ( ϕ ( t, x )) , where x = ϕ ( t, x ) = x + ξ ( t, x ) is the steam line given by ∂∂t ϕ ( t, x ) = v ( t, ϕ ( t, x )) , ϕ (0 , x ) = x . We consider that the Lagrangian co-ordinates variables x = ( x , x , x ) runson the fixed domain Π := { x ∈ R | ◦ ρ ( x ) > } . (1.11)Note that ◦ ρ ( x ) = ¯ ρ ( x ) + ∆ ρ | t =0 ( x ) = ¯ ρ ( x ) + δρ | t =0 ( x ) . (1.12)Then the equation (1.1b) reads ∂ ξ ∂t + g δρ ¯ ρ + δρ e + 1¯ ρ + δρ J − ∇ δP = 0 , (1.13)where ξ = X k =1 ξ k ∂∂x k , ∇ δP = X k =1 ∂δP∂x k ∂∂x k , e = ∂∂x , and J − = (( J − ) kj ) k,j = ( ∂x k /∂x j ) k,j is the inverse matrix of J = ( J jk ) j,k = (cid:16) ∂x j ∂x k (cid:17) j,k = (cid:16) δ jk + ∂ξ j ∂x k (cid:17) j,k . (1.14)Now the assumption (1.5)0 = X k ∂v k ∂x k = X k,j ( J − ) jk ∂∂t ∂ξ k ∂x j = X j,k ( J − ) jk ∂∂t J kj says that tr (cid:16) ∂J − ∂t J (cid:17) = 0 , and, equivalently, that det J − = 1det J = 1 (1.15)4olds for t ≥ x ∈ Π, since J | t =0 = I , I = ( δ jk ) j,k being the unit matrix.Note that we consider the equation (1.13) on the fixed domain [0 , T [ × Π andthe dominator ¯ ρ + δρ in the terms of the equation (1.13) should read¯ ρ + δρ = ¯ ρ ( x + ξ ( t, x )) + δρ ( t, x ) . (1.16)Moreover we see that the Lagrangian perturbation ∆ ρ is independent of t thanksto (1.4). Therefore by the definition we have δρ ( t, x ) = − (¯ ρ ( x + ξ ( t, x )) − ¯ ρ ( x )) + ∆ ρ | t =0 ( x )= − (¯ ρ ( x + ξ ( t, x )) − ¯ ρ ( x )) + δρ | t =0 ( x ) . (1.17)Accordingly we should read (1.16) as¯ ρ + δρ = ¯ ρ ( x ) + δρ | t =0 ( x ) = ◦ ρ ( x ) , (1.18)and the dominator ¯ ρ + δρ in the terms of the equation (1.13) does not containunknown functions and is nothing but the initial density distribution.If we insert P = A ¯ ρ γ with 1 < γ < dP/dz = − g ¯ ρ , then we get¯ ρ ( z ) = C (( z + − z ) ∨ ν = ( C ( z + − z ) ν (0 ≤ z < z + )0 ( z + ≤ z )Here z + is an arbitrary finite positive number, ν := 1 γ − C = (cid:16) ( γ − g Aγ (cid:17) ν . Keeping in mind this case, we consider an equilibrium ρ = ¯ ρ ( z ) with P = ¯ P ( z )which satisfies the following conditions:1) { ¯ ρ > } = { ≤ z < z + } with a finite positive z + , and¯ ρ ∈ C ([0 , + ∞ [) ∩ C ∞ ([0 , z + [); (1.19)2) It holds d ¯ ρdz < ≤ z < z + ; (1.20)3) ¯ ρ is analytic at z = 0 and¯ ρ = C ¯ ρ ( z + − z ) ν (1 + Λ( z + − z )) (1.21)as z → z + −
0, where C ¯ ρ is a positive constant and Λ is an analytic functionnear 0 such that Λ(0) = 0.Hereafter we shall use the following notation so that Λ( z + − z ) = [ z + − z ] :5 otation 2 Let K be a non-negative integer. Then the symbol [ X ] K stands forvarious convergent power series of the form P k ≥ K a k X k . Note that d ¯ Pdz = − g ¯ ρ = g C ¯ ρ ( z + − z ) ν (1 + [ z + − z ] ) ,d ¯ ρdz = − νC ¯ ρ ( z + − z ) ν − (1 + [ z + − z ] ) , and therefore d ¯ Pd ¯ ρ = g ν ( z + − z )(1 + [ z + − z ] )so that −∞ < ddz (cid:16) d ¯ Pd ¯ ρ (cid:17)(cid:12)(cid:12)(cid:12) z = z + − = − g ν < , that is, the boundary z = z + is a so called ‘physical vacuum boundary’. Seethe review [4]. Here d ¯ Pd ¯ ρ means (cid:16) d ¯ ρdz (cid:17) − (cid:16) d ¯ Pdz (cid:17) = g H [¯ ρ ] , H [¯ ρ ] = (cid:16) − ddz log ¯ ρ (cid:17) − being the so called ‘density scale height’.In this article we fix such an equilibrium ¯ ρ with ¯ P and investigate smallperturbations around this fixed equilibrium. Let us derive the linearized approximation of the equations.The linearized approximation of (1.17) reads δρ = − ( ∇| ¯ ρ ξ ) + δρ | t =0 . (2.1)Here we have used the fact that (1.5) implies( ∇| ξ ) = 0 (2.2)in the linearized approximation. The linearized approximation of the equation(1.13) is ∂ ξ ∂t + g δρ ¯ ρ e + 1¯ ρ ∇ δP = 0 . (2.3) Let us consider particular solutions of (2.1)(2.2)(2.3) of the following two typesof the form: 6 ype (1): ξ = u ( z ) cos lx w ( z ) sin lx sin √ λt, (2.4a) δP = δ ˇ P ( z ) sin lx sin √ λt. (2.4b)Here x = x , z = x , λ is a positive constant and l is a non-negative constant.In this type, taking δρ | t =0 = 0, that is, ◦ ρ = ¯ ρ , we put δρ = − ( ∇| ¯ ρ ξ ) . (2.5) Type (2): ξ = u ( z )(cos( lx − √ λt ) − cos lx )0 w ( z )(sin( lx − √ λt ) − sin lx ) , (2.6a) δP = δ ˇ P ( z )(sin( lx − √ λt ) − sin lx ) (2.6b)instead of (2.4a)(2.4b). In this type we should specify the initial perturbationof the density ∆ ρ | t =0 = δρ | t =0 by δρ | t =0 = h l ¯ ρu − ddz (¯ ρw ) i sin lx. (2.7)Then we have δρ = − ( ∇| ¯ ρ ξ ) + δρ | t =0 = h l ¯ ρu − ddz (¯ ρw ) i sin( lx − √ λt ) , (2.8)Then, for each type, the equations (2.2) and (2.3) are reduced to − lu + dwdz = 0 , (2.9)and − λu + l ¯ ρ δ ˇ P = 0 , (2.10a) − λw − g ρ d ¯ ρdz w + 1¯ ρ ddz δ ˇ P = 0 . (2.10b)7ote that Π = { ≤ z < z + } for the Type (1), butΠ = { ¯ ρ ( z ) − d ¯ ρdz w ( z ) sin lx > } = { ≤ z < z + and 1 − ρ d ¯ ρdz w ( z ) sin lx > } for the Type (2), since (2.9) implies that (2.8) reads δρ = − d ¯ ρdz w ( z ) sin( lx − √ λt ) . (2.11)The set of equations (2.10a)(2.9) is equivalent to δ ˇ P = λl ¯ ρu = λl ¯ ρ dwdz , (2.12)and therefore (2.10b) turns out to be ddz (cid:16) ¯ ρ dwdz (cid:17) + l λ (cid:16) − g d ¯ ρdz − ¯ ρλ (cid:17) w = 0 . (2.13)Putting N := s − g ¯ ρ d ¯ ρdz , (2.14)we can write (2.13) as ddz (cid:16) ¯ ρ dwdz (cid:17) + l λ ¯ ρ ( N − λ ) w = 0 , (2.15)which is called the ‘Taylor-Goldstein equation’. The boundary condition is w = 0 at z = 0 (2.16)The equation (2.15) can be written as − ddz (cid:16) ¯ ρ dwdz (cid:17) + l ¯ ρw = 1 λ µ ( z ) w, (2.17)where µ ( z ) := − g l d ¯ ρdz (2.18)8ote that µ ( z ) > ≤ z < z + . Let us perform the Liouville transformationon the eigenvalue problem (2.17), where the eigenvalue is 1 λ . Namely, putting ζ = Z z r µ ¯ ρ ( z ′ ) dz ′ , (2.19a) υ = (¯ ρµ ) w, (2.19b) q = ¯ ρµ h l + 14 d dz log (cid:16) − ¯ ρ d ¯ ρdz (cid:17) + − (cid:16) ddz log (cid:16) − ¯ ρ d ¯ ρdz (cid:17)(cid:17) + 14 (cid:16) ddz log ¯ ρ (cid:17)(cid:16) ddz log (cid:16) − ¯ ρ d ¯ ρdz (cid:17)(cid:17)i , (2.19c)we transform (2.17) to the standard form − d υdζ + qυ = 1 λ υ. (2.20)By the property 3) of ¯ ρ we have, as z → z + − µ ¯ ρ = − ν g l ( z + − z ) − (1 + [ z + − z ] ) , (2.21)therefore ζ + := Z z + r µ ¯ ρ ( z ) dz (2.22)is finite and ζ + − ζ = 2 p ν g l p z + − z (1 + [ z + − z ] ) . (2.23)Thus the z -interval [0 , z + ] is mapped onto the ζ -interval [0 , ζ + ].On the other hand, be a tedious calculation, we have q = (2 ν − ν − ν g l ( z + − z ) − (1 + [ z + − z ] )if ν = 3 / q = [ z + − z ] if ν = 3 /
2, or , thanks to (2.23), q = (2 ν − ν − ζ + − ζ ) − (1 + [( ζ + − ζ ) ] ) (2.24)if ν = 3 / q = [( ζ + − ζ ) ] (2.25)if ν = 3 /
2. Note that (2 ν − ν − > −
14 for ν >
1. Since q ∈ C ([0 , ζ + [),we can claim the following 9 roposition 1 There are constants K > −∞ , K > − such that q ( ζ ) ≥ K + K ( ζ + − ζ ) on ≤ ζ < ζ + . (2.26)Consequently we can claim the following Theorem 1
The operator − d dζ + q defined on C ∞ (]0 , ζ + [) [( or The opera-tor − ddz ¯ ρ ddz + l ¯ ρ defined on C ∞ (]0 , z + [) )] admits the Friedrichs extension,which is a self-adjoint operator, in the Hilbert space L ([0 , ζ + ]) [( in the Hilbertsapce L ([0 , z + ] , g l (cid:16) − d ¯ ρdz (cid:17) dz ) , respectively )] . The spectrum of the self-adjointoperator consists of eigenvalues < λ < λ < · · · < λ n < · · · → + ∞ . For a proof see, e.g., [2, Kapiter VII]. Let us fix an eigenvalue λ n and anassociated eigenfunction υ = υ n ( ζ ). We are going to prove the following Proposition 2
There is a constant a such that a = 0 and υ n ( ζ ) = a ( ζ + − ζ ) ν − (1 + [( ζ + − ζ ) ] ) (2.27)In order to prove the Proposition 2, we shall use the following Lemma 1
Let us consider the equation − d ydx + V ( x ) y = 0 , (2.28) where V ( x ) = 1 x ( K + [ x β ] ) . (2.29) with constants K, β such that K + 1 > , β > . Put α ± := 12 (1 ± √ K + 1) . (2.30) Then there is a fundamental system of solutions y = ϕ ( x ) , y = ϕ ( x ) of theequation (2.28) such that ϕ ( x ) = x α + (1 + [ x β ] ) , (2.31a) ϕ ( x ) = x α − (1 + [ x β ] ) + h (log x ) ϕ ( x ) . (2.31b) Here h is a constant, which may = 0 or = 0 generally, but h = 0 if α + − α − isnot an integer. § β = 1,we can reduce the proof to that of the case with β = 1, say, the case of the reg-ular singular point, by the change of variables x ❀ x ♮ := x β , y ❀ y ♮ := x β − y .Proof of Proposition 2. Let us suppose first that ν = 3 /
2. When K =(2 ν − ν − /
4, then α ± of Lemma 1 turn out to be α + = 12 (2 ν − , α − = 12 ( − ν + 3) . So, by the Lemma 1 we have a fundamental system of solutions υ (1) , υ (2) of(2.20) such that υ (1) ( ζ ) = ( ζ + − ζ ) ν − (1 + [( ζ + − ζ ) ] ) ,υ (2) ( ζ ) = ( ζ + − ζ ) − ν +32 (1 + [( ζ + − ζ ) ] )++ h (log( ζ + − ζ )) υ (1) ( ζ ) . But υ (2) does not belong to the domain of the Friedrichs self-adjoint extension of − d dζ + q ↾ C ∞ in L , since dυ (2) /dζ ∼ ν − ( ζ + − ζ ) − ν +12 L . Thus υ n = aυ (1) with a = 0. Next we suppose ν = 3 /
2. Then the boundary point z = z + is aregular boundary point, if we reduce the equation (2.20) in the variable ( ζ + − ζ ) .Therefore there is a fundamental system of solutions υ (1) , υ (2) such that υ (1) ( ζ ) = ( ζ + − ζ )(1 + [( ζ + − ζ ) ] ,υ (2) ( ζ ) = 1 + [( ζ + − ζ ) ] . Since a function υ belongs to the domain of the Firedrichs extension if andonly if it satisfies the Dirichret boundary conditions υ = 0 both at ζ = 0 and ζ = ζ + , it should hold that υ n = aυ (1) with a = 0. This completes the proof ofProposition 2 (cid:3) By the argument in the proof of Proposition 2 we can claim the following
Corollary 1
The eigenvalues /λ n of Theorem 1 are simple. Note that, if and only if the continuation of the solution υ (1) to the left hits0 at the regular boundary point ζ = 0, 1 /λ is an eigenvalue.Let us fix λ = λ n and an eigenfunction υ n ( ζ ) such that υ n ( ζ ) = ( ζ + − ζ ) ν − (1 + [( ζ + − ζ ) ] )and put w n ( z ) = κ (¯ ρµ ) − υ n ( ζ ) with κ = 2 ν − ( ν g l ) − ν − C ¯ ρ . (2.32)11hen the solution w n of the equation (2.15) satisfies w n ( z ) = 1 + [ z + − z ] as z → z + − . (2.33)Of course w n satisfies the usual Dirichlet boundary condition (2.16) at the reg-ular boundary z = 0.We have the associated particular solution (2.4a)(2.4b) of the Type (1) or(2.6a)(2.6b) of the Type (2) of the linearized problem by putting u ( z ) = − l ddz w n ( z ) , (2.34a) w ( z ) = w n ( z ) , (2.34b) δ ˇ P ( z ) = − λ ¯ ρl ddz w n ( z ) , (2.34c) λ = λ n . (2.34d)Let us fix this particular solution of the linearized problem and denote u = u L ( l,n ) , w = w L ( l,n ) , δ ˇ P = δ ˇ P L ( l,n ) . Accordingly for the Type (1) we shall denote ξ L ( l,n ) = u L ( l,n ) ( z ) cos lx w L ( l,n ) ( z ) sin lx sin √ λt, (2.35a) δP L ( l,n ) = δ ˇ P L ( l,n ) ( z ) sin lx sin √ λt. (2.35b)and for the Type (2) we shall denote ξ L ( l,,n ) = u L ( l,n ) ( z )(cos( lx − √ λt ) − cos lx )0 w L ( l,n ) ( z )(sin( lx − √ λt ) − sin lx ) , (2.36a) δP L ( l,,n ) = δ ˇ P L ( l,n ) ( z )(sin( lx − √ λt ) − sin lx ) (2.36b) Let us observe the movement of the vacuum boundary surface Γ, which is theboundary of the domainΠ = { ρ > } = { ¯ ρ + ∆ ρ > } = { ¯ ρ + δρ | t =0 > } = { ◦ ρ > } , described in the Eulerian co-ordinates x = ( x , x , x ) = ( x, y, z ).Type (1): We consider the solution ξ = ε ξ L ( l,n ) , δP = εδP L ( l,n ) with smallparameter ε, | ε | ≪
1. Then the vacuum boundary Γ turns out to be { z = z + } δρ | t =0 = 0, and it is expressed in the Eulerian co-ordinates as x = x + εu L ( l,n ) ( z + ) cos lx sin p λ n t, (2.37a) y = y, (2.37b) z = z + + εw L ( l,n ) ( z + ) sin lx sin p λ n t = z + + ε sin lx sin p λ n t (2.37c)Since ∂x∂x ( x, y, z + , t ) = 1 − εlu L ( l,n ) ( z + ) sin lx sin p λ n t = 1 + O ( ε )uniformly, we can assume that ∂x/∂x > x = ϕ ( x, t ) = x + O ( ε ) , (2.38)provided that | ε | is sufficiently small. Then Γ can be expressed as z = Z ( x, y, t ):= z + + ε sin lϕ ( x, t ) sin p λ n t = z + + ε (sin lx + O ( ε )) sin p λ n t. (2.39)This means that the vacuum boundary surface vibrates around the backgroundstratospheric hight z = z + as z = z + + ε sin lx sin p λ n t approximately modulo O ( ε ).Type (2). We consider the solution ξ = ε ξ L ( l,,n ) , δP = εδP L ( l,,n ) with | ε | ≪
1. The vacuum boundary Γ is the boundary of Π = { ¯ ρ ( x )+ δρ | t =0 ( x ) > } .Since ¯ ρ ( x ) + δρ | t =0 ( x ) = ¯ ρ ( z ) − ε d ¯ ρdz w L ( l,n ) ( z ) sin lx, we see that Π = { z + − z > − ε ρ d ¯ ρdz w L ( l,n ) ( z ) sin lx > } . But recall − ρ d ¯ ρdz = ν z + − z (1 + [ z + − z ] ) ,w L ( l,n ) ( z ) = 1 + [ z + − z ] . X ) = 1 + [ X ] such that, for z + − z > − ε ρ d ¯ ρdz w L ( l,n ) ( z ) sin lx > ⇔ z < z + + Υ( εν sin lx ) , provided that | ε | ν ≪
1. Put Z ( x ) := z + + min { Υ( εν sin lx ) , } . (2.40)Then Π turns out to be { z < Z ( x ) } so that Γ is { z = Z ( x ) } .Now the vacuum boundary Γ can expressed in the Eulerian co-ordinates as x = x + εu L ( l,n ) ( Z ( x ))(cos( lx − p λ n t ) − cos lx ) (2.41a) y = y, (2.41b) z = Z ( x )++ εw L ( l,n ) ( Z ( x ))(sin( lx − p λ n t ) − sin lx ) . (2.41c)We see that (2.41a) can be solved monotonically as x = ϕ ( x, t ) = x + O ( ε ) . (2.42)So Γ can be expressed as z = Z ( x, y, t ):= Z ( ϕ ( x, t ))++ εw L ( l,n ) ( Z ( ϕ ( x, t )) h sin( lϕ ( x, t ) − p λ n t ) − sin lϕ ( x, t ) i . (2.43)This means that the vacuum boundary surface is a traveling wave of the form z = z + + ( εν sin lx ) ∧ ε [sin( lx − p λ n t ) − sin lx ]approximately modulo O ( ε ). We should justify the linearized approximation discussed in the preceding Sec-tion in the following sense: