Local existence for the free boundary problem for the non-relativistic and relativistic compressible Euler equations with a vacuum boundary condition
aa r X i v : . [ m a t h . A P ] F e b Local existence for the free boundary problemfor the non-relativistic and relativistic compressibleEuler equations with a vacuum boundary condition
Yuri Trakhinin
Sobolev Institute of Mathematics, Koptyug av. 4, 630090 Novosibirsk, Russiae-mail: [email protected]
Abstract
We study the free boundary problem for the equations of compressible Euler equations with a vacuumboundary condition. Our main goal is to recover in Eulerian coordinates the earlier well-posedness resultobtained by Lindblad [11] for the isentropic Euler equations and extend it to the case of full gas dynamics.For technical simplicity we consider the case of an unbounded domain whose boundary has the form of agraph and make short comments about the case of a bounded domain. We prove the local-in-time existence inSobolev spaces by the technique applied earlier to weakly stable shock waves and characteristic discontinuities[5, 21]. It contains, in particular, the reduction to a fixed domain, using the “good unknown” of Alinhac[1], and a suitable Nash-Moser-type iteration scheme. A certain modification of such an approach is causedby the fact that the symbol associated to the free surface is not elliptic. This approach is still directlyapplicable to the relativistic version of our problem in the setting of special relativity and we briefly discussits extension to general relativity.
Consider the compressible Euler equations with the gravitational field
G ∈ R : ∂ t ρ + div ( ρv ) = 0 , (1) ∂ t ( ρv ) + div ( ρv ⊗ v ) + ∇ p = ρ G , (2) ∂ t (cid:0) ρ (cid:0) e + | v | (cid:1)(cid:1) + div (cid:0)(cid:0) ρ (cid:0) e + | v | (cid:1) + p (cid:1) v (cid:1) = 0 , (3)where ρ denotes density, v ∈ R fluid velocity, p = p ( ρ, S ) pressure, S entropy, and e = e ( ρ, S ) internalenergy. With a state equation of gas, p = p ( ρ, S ), and the first principle of thermodynamics, (1)–(3) is aclosed system. As the unknown we can fix, for example, the vector U = U ( t, x ) = ( p, v, S ).We can easily symmetrize system (1)–(3) by rewriting it in the nonconservative form1 ρc d p d t + div v = 0 , ρ d v d t + ∇ p = ρ G , d S d t = 0 , (4)where c = p ρ ( ρ, S ) is the square of the sound velocity and d / d t = ∂ t + ( v, ∇ ) (by ( , ) we denote the scalarproduct). Equations (4) read as the symmetric quasilinear system A ( U ) ∂ t U + X j =1 A j ( U ) ∂ j U + Q ( U ) = 0 , (5)1here Q ( U ) = (0 , − ρ G , A = ρc ρ ρ ρ
00 0 0 0 1 , A = v ρc ρv ρv ρv
00 0 0 0 v ,A = v ρc ρv ρv ρv
00 0 0 0 v , A = v ρc ρv ρv ρv
00 0 0 0 v . System (5) is symmetric hyperbolic if the the hyperbolicity condition A > ρ > , p ρ > . (6)One can alternatively consider the isentropic Euler equations, i.e., system (1), (2) for the same variablesexcept for the entropy S . Then, the state equation of gas is p = p ( ρ ) and the second inequality in (6) isunderstood in the sense that p ′ ( ρ ) > t ) which boundary Σ( t ) = { F ( t, x ) = 0 } is to be determined and moves with the velocity of the gas particles at the boundary:d F d t = 0 , p = 0 on Σ( t ) (7)(for all t ∈ [0 , T ]). This free boundary problem can be used for modeling the motion of the ocean or a star.Most results for such kind of problems were earlier obtained for incompressible fluids and the history ofmathematical studies of incompressible versions of problem (1)–(3), (7) can be found, for example, in [11].The first result for compressible fluids was obtained by Makino [13] (see also [14]) who proved thelocal-in-time existence of solutions to problem (1)–(3), (7) for the case of a polytropic gas and when theboundary condition p = 0 in (7) is replaced by ρ = 0. This was done by using a special symmetrizationof the gas dynamics system that supports vacuum regions. That is, the corresponding symmetric systemfor a new unknown U (see [13, 14]) is always hyperbolic without assumptions (6). However, employing thissymmetrization leads to certain non-physical restrictions on the initial data. Therefore, Makino’s resultdoes not cover the general case. On the other hand, from the physical point of view, the vacuum boundarycondition ρ | Σ = 0 is, of course, more natural than p | Σ = 0. In particular, (6) and (7) does not formally allowthe equation of state of a polytropic gas p = aρ γ exp( S/c V ). In this connection, as was recommended in [11],for the case of boundary condition p | Σ = 0 one can alternatively think of the pressure as a small constanton the boundary (see also Remark 2.1 below).The local-in-time existence for the general case of initial data was recently proved by Lindblad [11] forthe free boundary problem with non-vanishing density on the boundary for the isentropic Euler equations.Namely, the local-in-time existence of smooth solutions of problem (1), (2), (7) (with G = 0) was shown in[11] under the natural physical assumption ∂p∂N ≤ − ǫ < , (8)2here ∂/∂N = ( ∇ F, ∇ ), together with the hyperbolicity condition (6), provided that the initial domain Ω(0)is diffeomorfic to a ball. The main tool in [11] is the passage to the Lagrangian coordinates for reducingthe original problem to that in a fixed domain. Such a technique seems most natural for free boundaryproblems with boundary conditions like (7). At the same time, for compressible fluids it is connected witha lot of technical difficulties and it is not quite clear how to extend the results to similar problems for morecomplicated fluid dynamics models like, for example, relativistic gas dynamics or magnetohydrodynamics.Even the extension of the existence theorem in [11] to full gas dynamics does not seem to be just a technicalmatter. Remark 1.1
If the domain Ω( t ) is unbounded, we should additionally assume that the velocity vanishes atinfinity (as | x | → ∞ ). As follows from the second vector equation in (4), in the absence of gravity ( G = 0)this contradicts condition (8). That is, in the case of an unbounded domain, the presence of gravity is absolutely necessary . However, if the domain is bounded, without loss of generality and as was done in [11],the gravity can be neglected as a lower order term (it plays no role in the proof of well-posedness).In this paper we propose another approach to studying the well-posedness of problem (1)–(3), (7) (or(1), (2), (7)) and similar free boundary problems for other systems of hyperbolic conservation laws. Thisapproach could be probably called “hyperbolic” or ”shock waves” approach because it was first applied byBlokhin (see [3] and references therein) and Majda [12] to prove the short-time persistence of discontinuousshock front solutions to hyperbolic conservation laws. The “hyperbolic” approach to free boundary problemsdoes not propose to pass to the Lagrangian coordinates (the more so as this is impossible for shock waves).Instead of this we work in the Eulerian coordinates and reduce our free boundary problem to that in a fixeddomain. More precisely, such a procedure is indeed quite simple if our domain Ω( t ) is unbounded and itsboundary has the form of a graph. In this case we reduce our problem to that in a half-space by simplestraightening of the unknown free surface (for example, a shock front). Otherwise, the technique of reductionto a fixed domain is more technically involved (see [12]), but the resulting problem in a fixed domain has noprincipal differences from that for the case of unbounded domains. We can then follow standard argumentsand reduce the corresponding linearized problem to a linear problem in a half-space by using a fixed partitionof unity flattering the boundary. Therefore, without loss of generality we can restrict ourself to an unboundedinitial domain and we do so in this paper. On the other hand, the possibility to treat unbounded domainsis already a certain advantage of the “hyperbolic” approach.Regarding the free boundary problem (1)–(3), (7), it should be noted that its linearized version is well-posed only in a weak sense. It means that the corresponding linear problem satisfies the Kreiss–Lopatinskicondition but violates the uniform Kreiss–Lopatinski condition [9, 12, 15]. This yields losses of derivatives ina priori estimates for the linearized problem. Therefore, we are not able to use such estimates to prove theexistence of solutions to the original nonlinear problem by the fixed-point argument as was done by Blokhinor Majda (see also [15]) for uniformly stable shock waves (the uniform Kreiss–Lopatinski condition holds forsuch shocks). Thus, we have to modify the “hyperbolic” approach to apply it to free boundary problemswhose linearized versions are weakly well-posed. In some sense, this was already done in previous works. Weshould first mention Alinhac’s study [1] of rarefaction waves for hyperbolic conservation laws.It is well-known that the Nash-Moser method can sometimes compensate the loss of derivatives phe-nomenon and to use it we should perform a genuine linearization of our nonlinear problem, i.e., to keep allthe lower-order terms while linearizing. One of these terms is a first-order term for the perturbation of thefree surface in the linearized interior equations. To neutralize such a bad term Alinhac proposed to pass to a3ew unkwnown (so-called “good unknown”) and we use this idea for problem (1)–(3), (7). Such a techniquewas recently applied to other hyperbolic free boundary value problems. We mean the results of Coulombeland Secchi [5] for 2D supersonic vortex sheets and weakly stable shock waves in isentropic gas dynamicsand author’s result for compressible current-vortex sheets [20, 21]. The local-in-time existence of the listedweakly stable discontinuities was shown in [5, 21] by a suitable Nash-Moser-type iteration scheme.At last, we should note that problem (1)–(3), (7) is not a quite standard “weakly stable” hyperbolic freeboundary problem like those studied in [1, 5, 21]. Actually, regardless of the fact that the constant (“frozen”)coefficients linearized problem for (1)–(3), (7) always satisfies the weak Kreiss–Lopatinski condition, thecorresponding variable coefficients problem is not unconditionally well-posed and (8) is an extra conditionwhich is necessary for well-posedness (though, the question on its necessity is a separate and non-trivialproblem). This unusual feature is a consequence of the fact that the symbol associated with the free surfaceis not elliptic (see Remark 2.4) that leads to a loss of “control on the boundary.” Therefore, we have tomodify somewhat the energy method which we use for deriving a priori estimates for the linearized problem.Having in hand a good a priori estimate (so-called tame estimate [1]) for the linearized problem, we provethe local existence (and uniqueness) theorem for our nonlinear problem (see Theorem 2.1 below) by theNash-Moser method.Such a modified “hyperbolic” approach outlined above allows one to prove a counterpart of Theorem2.1 for the relativistic version of problem (1)–(3), (7) in the setting of special relativity without furthermodifications. Actually, the proof is absolutely the same as for the non-relativistic case and we may dropit. Since in the framework of our “hyperbolic” approach we use the energy method (but not the Kreisssymmetrizer technique [9, 12, 15]), the only important point is that the system of relativistic Euler equations ∇ α ( ρu α ) = 0 , ∇ α T αβ = 0 (9)can be symmetrized (we write down its symmetric form in the last section of the paper). Here ∇ α is thecovariant derivative with respect to the metric g with the components g αβ ; ρ is the particle number densityin the rest frame (for convenience we use the notations that are consistent with the non-relativistic case); T αβ = ρhu α u β + pg αβ ; h = 1 + e + ( p/ρ ) is the specific enthalpy, p is the pressure, e = e ( ρ, S ) is the specific internal energy perparticle, S is the entropy per particle, u α are components of the four-velocity. The metric g should satisfythe Einstein equations. Following [18] (see also [6]), in the last section of the paper we write down them inso-called harmonic coordinates. In the case of special relativity g = diag ( − , , ,
1) and equations (9) (inthe presence of gravity) take the form ∂ t ( ρ Γ) + div ( ρu ) = 0 , (10) ∂ t ( ρh Γ u ) + div ( ρhu ⊗ u ) + ∇ p = ρ G , (11) ∂ t ( ρh Γ − p ) + div ( ρh Γ u ) = 0 , (12)where t := x , div := div x , x = ( x , x , x ) , u = ( u , u , u ) , v = ( v , v , v ) = u/ Γ , Γ = 1 + | u | ;Γ = u = (1 − | v | ) − / is the Lorentz factor, and the speed of the light is equal to unity.4egarding the free boundary problem for relativistic fluids with a vacuum boundary condition, its local-in-time existence was proved by Rendall [18] for the boundary condition ρ | Σ = 0 and a special class of initialdata by generalizing Makino’s symmetrization [13, 14] to the relativistic case. This result was obtained forthe setting of general relativity and under the simplifying assumption that the relativistic fluid is isentropic.Actually, in the framework of Makino’s approach this assumption was just a technical simplification. Thatis, our main goal in this paper is to cover the general case of initial data but for the boundary condition p | Σ = 0.As was already noted above, we do not almost need to make efforts for extending Theorem 2.1 to therelativistic Euler equations in the setting of special relativity. Concerning the case of general relativity,the proof of the existence theorem is based on using harmonic coordinates and the facts that the Einsteinequations for the metric g can be written in the form of a symmetric hyperbolic system [18] and the metricshould be smooth on the fluid-vacuum boundary Σ. More precisely, for the relativistic Euler equationswe easily obtain a counterpart of Theorem 2.1 for any fixed metric, but not only for g = diag ( − , , , H s with s ≥
3. In Section 4, we first specify compatibility conditions for the initial data and, by constructingan approximate solution, reduce our problem to that with zero initial data. Then, we solve the reducedproblem by a suitable Nash-Moser-type iteration scheme. At last, in Section 5 we describe extensions of theresult of Theorem 2.1 to special and general relativity.
For technical simplicity (see Remark 2.2 below), we assume that the space-time domain Ω( t ) is unboundedand lies from one side of its free boundary Σ( t ) which has the form of a graph, x = ϕ ( t, x ′ ), x ′ = ( x , x ).That is, Ω( t ) = { x > ϕ ( t, x ′ ) } (13)and the function ϕ ( t, x ′ ) is to be determined. As for shock waves, using Majda’s arguments [12] , we cangeneralize the technique below to the case of an arbitrary compact free surface Σ. The mapping of Ω( t ) toa fixed domain is just more technically involved when Ω( t ) is bounded (see Remark 2.2).For domain (13) the boundary conditions (7) take the form ∂ t ϕ = v N , p = 0 on Σ( t ) , (14)5nd the gravitational field G = ( G, , , where v N = ( v, N ), N = (1 , − ∂ ϕ, − ∂ ϕ ), and G denotes Newton’s gravitational constant. Our final goal isto find conditions on the initial data U (0 , x ) = U ( x ) , x ∈ Ω(0) , ϕ (0 , x ′ ) = ϕ ( x ′ ) , x ′ ∈ R , (15)providing the existence of a smooth solution ( U, ϕ ) of the free boundary value problem (5), (14), (15) in Ω( t )for all t ∈ [0 , T ], where the time T is small enough.To reduce the free boundary value problem (5), (14), (15) to that in a fixed domain we straighten,as usual, the unknown free surface Σ. That is, the unknown U being smooth in Ω( t ) is replaced by thevector-function e U ( t, x ) := U ( t, Φ( t, x ) , x ′ ) , that is smooth in the fixed domain R = { x > , x ′ ∈ R } , where Φ( t, , x ′ ) = ϕ ( t, x ′ ) and ∂ Φ >
0. Asin [21], to avoid assumptions about compact support of the initial data in the nonlinear existence theoremand work globally in R we use the choice of Φ( t, x ) similar to that suggested by M´etivier [15]:Φ( t, x ) := x + Ψ( t, x ) , Ψ( t, x ) := χ ( x ) ϕ ( t, x ′ ) , where χ ∈ C ∞ ( R ) equals to 1 on [0 , k χ ′ k L ∞ ( R ) < /
2. Then, the fulfillment of the requirement ∂ Φ > k ϕ k L ∞ ([0 ,T ] × R ) ≤
1. The last is fulfilled if, withoutloss of generality, we consider the initial data satisfying k ϕ k L ∞ ( R ) ≤ /
2, and the time T in our existencetheorem is sufficiently small.Dropping for convenience tildes in e U , we reduce (5), (14), (15) to the initial boundary value problem L ( U, Ψ) = 0 in [0 , T ] × R , (16) B ( U, ϕ ) = 0 on [0 , T ] × { x = 0 } × R , (17) U | t =0 = U in R , ϕ | t =0 = ϕ in R , (18)where L ( U, Ψ) = L ( U, Ψ) U + Q ( U ), L ( U, Ψ) = A ( U ) ∂ t + e A ( U, Ψ) ∂ + A ( U ) ∂ + A ( U ) ∂ , e A ( U, Ψ) = 1 ∂ Φ (cid:16) A ( U ) − A ( U ) ∂ t Ψ − X k =2 A k ( U ) ∂ k Ψ (cid:17) ( ∂ Φ = 1 + ∂ Ψ), and (17) is the compact form of the boundary conditions ∂ t ϕ − v N = 0 , p = 0 on [0 , T ] × { x = 0 } × R . We are now in a position to state the local-in-time existence theorem for problem (16)–(18). Clearly, thistheorem implies a corresponding theorem for the original problem (5), (14), (15).
Theorem 2.1
Let m ∈ N and m ≥ . Suppose the initial data (16), with ( U − ˇ U , ϕ ) ∈ H m +7 ( R ) × H m +7 ( R ) and ρ ( p , S ) − ǫ ∈ H m +7 ( R ) , atisfy the hyperbolicity condition (6) for all x ∈ R and are compatible up to order m + 7 in the sense ofDefinition 4.1. Here ˇ U = (2 ǫx , , , , , ǫ = 2 ǫ/G ( ǫ = const > . Let also the initial data satisfy the physical condition ∂ p ≥ ǫ > at x = 0 (19) for all x ′ ∈ R . Then, there exists a sufficiently short time T > such that problem (16)–(18) has a uniquesolution ( U, ϕ ) ∈ (cid:8) ˇ U + H m ([0 , T ] × R ) (cid:9) × H m ([0 , T ] × R ) . Moreover, ρ − ǫ ∈ H m ([0 , T ] × R ) . Remark 2.1
The hyperbolicity condition (6) which should be satisfied for all x ∈ R implies that thefunction ρ ( x ) = ρ ( p , S )( x ) cannot vanish at infinity. Indeed, in Theorem 2.1 we assume that ρ − ǫ ∈ H m +7 ( R ). On the other hand, (6) together with the boundary condition p | Σ = 0 do not formally allowthe equation of state of a polytropic gas (or a γ –law gas for isentropic gas dynamics). However, as wasnoted in [11], from a physical point of view we can alternatively think of the pressure as a small positiveconstant ε on the boundary. One can easily generalize the result of Theorem 2.1 to the case of the boundarycondition p | x =0 = ε . More precisely, we now assume that U − ˇ U − C ∈ H m +7 ( R ) and prove that U − ˇ U − C ∈ H m ([0 , T ] × R ), where C = ( ε, , , , p ′ = p − ε and omitting the primes, we obtain problem (16)–(18) with the matrices A α ( U + C ). The further argumentsare almost the same as in the proof of Theorem 2.1 (see below). Remark 2.2
Inequality (19) is a counterpart of the physical condition (8) for the unbounded domain (13).If the domain is bounded and its initial boundary Σ(0) is a compact co-dimension one surface in R , we canfollow Majda’s arguments [12] (see also [2, sect. 12.4.2]). More precisely, we can make ( locally in time ) achange of variables that sends all boundary locations Σ( t ) to the initial surface Σ(0). We refer the reader to[12, 2] for details of such a change of variables. In particular, it requires the application of the Weingartenmap while writing down boundary conditions on Σ(0). The resulting initial boundary value problem is aproblem in the fixed domain Ω(0). Its principal difference from problem (16)–(18) is that we have to dealwith a problem in a fixed compact domain instead of a half-space. For this problem the proof of a counterpartof Theorem 2.1 is more technical, but the ideas are basically the same as for Theorem 2.1. For instance, weshould reduce the corresponding linearized problem to that in a half-space by using a fixed partition of unityflattering the boundary. The resulting linearized problem in a half-space will not have principal differencesfrom the linearized problem for (16)–(18). Only its coefficients will be more technically complicated thanthose for the linearization of (16)–(18). Therefore, as is usually done for shock waves or other types or strongdiscontinuities (see, e.g., [2, 3, 15]), in this paper we restrict ourself to the case of an unbounded domainwhose boundary has a form of a graph.The existence of solutions in Theorem 2.1 will be proved by Nash-Moser iterations. The main tool forproving the convergence of the Nash-Moser iteration scheme is a so-called tame estimate [1, 5, 21] for thelinearized problem. In this section, we derive a basic a priori L –estimate for the linearized problem bythe energy method. This estimate is a basis for deriving the tame estimate in Sobolev spaces (see the nextsection) and implies uniqueness of a solution to the nonlinear problem (16)–(18) that can be proved bystandard argument. 7et us first pass to the new unknown U ′ = ( p ′ , v, S ) = U − ˇ U . For U ′ system (16) is rewritten as L ′ ( U ′ , Ψ) := L ( U ′ + ˇ U , Ψ) U ′ + A ν ( U ′ + ˇ U , Ψ) ∂ ˇ U + Q ( U ′ + ˇ U ) = 0 , where ∂ ˇ U = (2 ǫ, , , , ρ ′ ( p ′ , S ) := ρ ( p, S ), A ′ α ( U ′ ) := A α ( U ), Q ′ ( U ′ ) := Q ( U ), and U ′ := U − ˇ U .Then, omitting the primes, for the new unknown we get the system L ( U, Ψ) := L ( U, Ψ) U + A ν ( U, Ψ) ∂ ˇ U + Q ( U ) = 0 in [0 , T ] × R (20)with the boundary conditions (17) and the initial data (18). From now on we will work with problem(20), (17), (18). We should now prove the existence of its solution, U ∈ H m ([0 , T ] × R ), assuming that U ∈ H m +7 ( R ). For the initial data for the new unknown we assume that ∂ p | x =0 > − ǫ ∀ x ′ ∈ R . (21)This guarantees the fulfillment of assumption (19) for the original unknown. Remark 2.3
We easily compute the boundary matrix: e A ( U, Ψ) = 1 ∂ Φ f ρc − ∂ Ψ − ∂ Ψ 01 ρ f − ∂ Ψ 0 ρ f − ∂ Ψ 0 0 ρ f
00 0 0 0 f , where f = v − v ∂ Ψ − v ∂ Ψ − ∂ t Ψ. The vector-function e A ( U, Ψ a ) ∂ ˇ U cannot belong to a Sobolev spaceon R because its second component is 2 ǫ/ ( ∂ Φ). However, if problem (20), (17), (18) has a solution from aSobolev space and Theorem 2.1 takes place, then the sum e A ( U, Ψ) ∂ ˇ U + Q ( U ) already belongs to a Sobolevspace because 2 ǫ∂ Φ − Gρ = − G ( ρ − ǫ ) − ǫ ∂ Ψ ∂ Φ ∈ H m ([0 , T ] × R ) . Thus, for our case of an unbounded domain the presence of gravity is of great importance (see also Remark1.1).We now formulate the linearized problem. ConsiderΩ T := ( −∞ , T ] × R , ∂ Ω T := ( −∞ , T ] × { x = 0 } × R . Let ( b U ( t, x ) , ˆ ϕ ( t, x ′ )) ∈ W ∞ (Ω T ) × W ∞ ( ∂ Ω T ) (22)be a given sufficiently smooth vector-function, with b U = (ˆ p, ˆ v, b S ), and k b U k W ∞ (Ω T ) + k ˆ ϕ k W ∞ ( ∂ Ω T ) ≤ K, (23)where K > k ˆ ϕ k L ∞ ( ∂ Ω T ) <
1. Thisimplies ∂ b Φ ≥ /
2, with b Φ( t, x ) := x + b Ψ( t, x ), b Ψ( t, x ) := χ ( x ) ˆ ϕ ( t, x ′ ). We also assume that the basic state(22) about which we shall linearize problem (20), (17) satisfies the hyperbolicity condition (6) in Ω T , ρ (ˆ p, b S ) > , ρ p (ˆ p, b S ) > , (24)8he first boundary condition in (17), ∂ t ˆ ϕ − ˆ v N | x =0 = 0 , (25)and the assumption (21), ∂ ˆ p | x =0 > − ǫ, (26)where ˆ v N = ˆ v − ˆ v ∂ ˆ ϕ − ˆ v ∂ ˆ ϕ .The linearized equations for (20) and (17) for determining small perturbations ( δU, δϕ ) read (below wedrop δ ): L ′ ( b U , b Ψ)( U, Ψ) := L ( b U , b Ψ) U + C ( b U , b Ψ) U − (cid:8) L ( b U , b Ψ)Ψ (cid:9) ∂ ( b U + ˇ U ) ∂ b Φ = f, B ′ ( b U , ˆ ϕ )( U, ϕ ) := ∂ t ϕ + ˆ v ∂ ϕ + ˆ v ∂ ϕ − v N p = g, where v N = v − v ∂ ˆ ϕ − v ∂ ˆ ϕ , and the matrix C ( b U , b Ψ) is determined as follows: C ( b U , b Ψ) U = ( U, ∇ u A ( b U )) ∂ t b U + ( U, ∇ u A ν ( b U , b Ψ)) ∂ b U + X k =2 ( U, ∇ u A k ( b U )) ∂ k b U + − gρ p (ˆ p, b S ) p − gρ S (ˆ p, b S ) S . ( Y, ∇ y A ( b U )) := X i =1 y i (cid:18) ∂A ( Y ) ∂y i (cid:12)(cid:12)(cid:12)(cid:12) Y = b U (cid:19) , Y = ( y , . . . , y ) . Here, as usual, we introduce the source terms f = ( f , . . . , f ) and g = ( g , g ) to make the interior equationsand the boundary conditions inhomogeneous.The differential operator L ′ ( b U , b Ψ) is a first order operator in Ψ = χ ( x ) ϕ ( t, x ′ ). Following Alinhac [1]and introducing the “good unknown” ˙ U := U − Ψ ∂ b Φ ∂ ( b U + ˇ U ) , (27)we simplify the linearized interior equations: L ( b U , b Ψ) ˙ U + C ( b U , b Ψ) ˙ U − Ψ ∂ b Φ ∂ (cid:8) L ( b U , b Ψ) (cid:9) = f. (28)As in [1, 5, 20, 21], we drop the zero-order term in Ψ in (28) and consider the effective linear operators L ′ e ( b U , b Ψ) ˙ U := L ( b U , b Ψ) ˙ U + C ( b U , b Ψ) ˙ U = A ( b U ) ∂ t ˙ U + e A ( b U , b Ψ) ∂ ˙ U + A ( b U ) ∂ ˙ U + A ( b U ) ∂ ˙ U + C ( b U , b Ψ) ˙ U (29)In the subsequent nonlinear analysis the dropped term in (28) will be considered as an error term at eachNash-Moser iteration step.Regarding the boundary differential operator B ′ , in terms of unknown (27) it reads: B ′ e ( b U , ˆ ϕ )( ˙ U , ϕ ) := B ′ ( b U , ˆ ϕ )( U, ϕ ) = ∂ t ϕ + ˆ v ∂ ϕ + ˆ v ∂ ϕ − ˙ v N − ϕ ∂ ˆ v N ˙ p + ϕ (2 ǫ + ∂ ˆ p ) , (30)9here ˙ v N = ˙ v − ˙ v ∂ ˆ ϕ − ˙ v ∂ ˆ ϕ . Thus, the linear problem for ( ˙ U , ϕ ) has the form L ′ e ( b U , b Ψ) ˙ U = f in Ω T , (31) B ′ e ( b U , ˆ ϕ )( ˙ U , ϕ ) = g on ∂ Ω T , (32)( ˙ U , ϕ ) = 0 for t < , (33)where f and g vanish in the past. We consider the case of zero initial data, that is usual assumption, andpostpone the case of nonzero initial data to the nonlinear analysis (construction of a so-called approximatesolution).On the basic state the boundary matrix e A has the form e A ( b U , b Ψ) = 1 ∂ b Φ ˆ f ˆ ρ ˆ c − ∂ b Ψ − ∂ b Ψ 01 ˆ ρ ˆ f − ∂ b Ψ 0 ˆ ρ ˆ f − ∂ b Ψ 0 0 ˆ ρ ˆ f
00 0 0 0 ˆ f , where ˆ ρ = ρ (ˆ p, b S ) , ˆ c = ρ p (ˆ p, b S ) , ˆ f = ˆ v − ˆ v ∂ b Ψ − ˆ v ∂ b Ψ − ∂ t b Ψ . In view of (25), ˆ f | x =0 = ˆ v N | x =0 − ∂ t ˆ ϕ = 0 . We see that the boundary matrix e A ( b U , b Ψ) is singular on the boundary x = 0 (it is of constant rank 2 at x = 0). That is, (31)–(33) is a hyperbolic problem with characteristic boundary of constant multiplicity.It is convenient to separate “characteristic” and “noncharacteristic” unknowns. For this purpose weintroduce the new unknown V = ( ˙ p, ˙ v n , ˙ v , ˙ v , ˙ S ) , where ˙ v n = ˙ v − ˙ v ∂ b Ψ − ˙ v ∂ b Ψ ( ˙ v n | x =0 = ˙ v N | x =0 ). We have ˙ U = JV , with J = ∂ b Ψ ∂ b Ψ 00 0 1 0 00 0 0 1 00 0 0 0 1 , Then, system (31) is equivalently rewritten as A ( b U , b Ψ) ∂ t V + X k =1 A k ( b U , b Ψ) ∂ k V + A ( b U , b Ψ) V = F ( b U , b Ψ) , (34)where A α = J T A α J ( α = 0 , , , A = J T e A J, F = J T f . The boundary matrix A in system (34)has the form A = 1 ∂ b Φ A (1) + A (0) , A (1) = , A (0) | x =0 = 0 , (35)10.e., V n = ( ˙ p, ˙ v n ) is the “noncharacteristic” part of the vector V . The explicit form of A (0) is of no interest,and it is only important that, in view (25), A (0) | x =0 = 0. The boundary matrix A on the boundary x = 0has one positive (“outgoing”) eigenvalue. Since one of the boundary conditions is needed for determiningthe function ϕ , the correct number of boundary conditions is two (that is the case in (32)). Hence, thehyperbolic problem (31)–(33) has the property of maximality [16].By standard argument we get for system (31) the energy inequality I ( t ) − Z ∂ Ω t ˙ p ˙ v N | x =0 d x ′ d s ≤ C ( K ) (cid:18) k f k L (Ω T ) + Z t I ( s ) d s (cid:19) , (36)where I ( t ) = R R ( A V, V ) d x and C = C ( K ) > K (see (23)). In view of theboundary conditions (32), one has − p ˙ v N | x =0 = 2( ϕ ˆ a − g )( ∂ t ϕ + ˆ v ∂ ϕ + ˆ v ∂ ϕ − ϕ ∂ ˆ v N − g ) | x =0 = ∂ t (cid:8) ˆ a | x =0 ϕ − g ϕ (cid:9) − (cid:8) ∂ t ˆ a + ∂ (ˆ v ˆ a ) + ∂ (ˆ v ˆ a ) − a∂ ˆ v N (cid:9) | x =0 ϕ +2 { ∂ t g + ∂ (ˆ v g ) + ∂ (ˆ v g ) + g ∂ ˆ v N − g ˆ a } | x =0 ϕ + 2 g g + ∂ (cid:8) ˆ v ˆ aϕ − v g ϕ (cid:9) + ∂ (cid:8) ˆ v ˆ aϕ − v g ϕ (cid:9) , where ˆ a = 2 ǫ + ∂ ˆ p . Then, using the Young inequality, from (36) we obtain I ( t ) + Z R (2 ǫ + ∂ ˆ p | x =0 ) ϕ d x ′ ≤ C ( K ) n k f k L (Ω T ) + k g k H ( ∂ Ω T ) + Z t (cid:16) I ( s ) + k ϕ ( s ) k L ( R ) (cid:17) d s o . Taking into account assumption (26) and applying Gronwall’s lemma, we finally deduce the basic a priori L –estimate k ˙ U k L (Ω T ) + k ϕ k L ( ∂ Ω T ) ≤ C ( K ) (cid:8) k f k L (Ω T ) + k g k H ( ∂ Ω T ) (cid:9) . (37) Remark 2.4
In the a priori estimate (37) we have a loss of one derivative from the source term g to thesolution (more precisely, we loose one derivative only from g but not from g ). This is quite naturalbecause one can check that the constant coefficients linearized problem, i.e., problem (31)–(33) with frozencoefficients satisfies the Kreiss–Lopatinski condition but violates the uniform Kreiss–Lopatinski condition[9, 15]. Although the weak Kreiss–Lopatinski condition holds we had to assume the fulfillment of the extracondition (26) while deriving the a priori estimate (37). This is very unusual for hyperbolic initial boundaryvalue problems because, as a rule (see, e.g., [5, 21]), the fulfillment of the Kreiss–Lopatinski condition isenough for obtaining a priori estimates. Actually, in our case the appearance of an extra condition on thelevel of variable coefficients linear analysis is caused by the fact that the symbol associated to the free surfaceis not elliptic , i.e., we are not able to resolve our boundary conditions (32) for the gradient ( ∂ t ϕ, ∂ ϕ, ∂ ϕ ).Therefore, it is also natural that in estimate (37) we “lose one derivative from the front”, i.e., we do nothave the H –norm of ϕ in the left-hand side of (37).Since in estimate (37) we do not lose derivatives from the source term f to the solution, the existence ofsolutions to problem (31)–(33) can be proved by the classical argument of Lax and Phillips [10]. Indeed, wefirst reduce our problem to one with homogeneous boundary conditions by subtracting from the solution amore regular function (see, e.g., [17]). Namely, there exists e U = (˜ p, ˜ v, e S ) ∈ H s +1 (Ω T ) vanishing in the pastsuch that − ˜ v N = g , ˜ p = g on ∂ Ω T , v N = ˜ v − ˜ v ∂ ˆ ϕ − ˜ v ∂ ˆ ϕ . If ˙ U = U ♮ + e U , then U ♮ satisfies (31)–(33) with g = 0 and f = f ♮ , where f ♮ = f − L ′ e ( b U , b Ψ) e U . That is, it is enough to prove the existence of a solution ( ˙ U , ϕ ) to problem (31)–(33)with g = 0. For this problem we have the estimate k ˙ U k L (Ω T ) + k ϕ k L ( ∂ Ω T ) ≤ C ( K ) k f k L (Ω T ) . (38)Having in hand estimate (38) with no loss of derivatives we may use the classical argument in [10]. Inparticular, we define a dual problem for (31)–(33) as follows: L ′ e ( b U , b Ψ) ∗ ¯ U = ¯ f in Ω T , (39) ∂ t ¯ p + ∂ ( ˆ v ¯ p ) + ∂ ( ˆ v ¯ p ) + ¯ p∂ ˆ v N + ¯ v N ˆ a = 0 on ∂ Ω T , (40)¯ U = 0 for t < , (41)where ¯ U = (¯ p, ¯ v, ¯ S ), ¯ v N = ¯ v − ¯ v ∂ ˆ ϕ − ¯ v ∂ ˆ ϕ , and L ′ e ∗ = − L ′ e + C + C T − ∂ t A − ∂ e A − ∂ A − ∂ A . Problem (39)–(41) is indeed a dual problem for (31)–(33) because for all ˙ U ∈ H (Ω T ) and ¯ U ∈ H (Ω T ),with ¯ U | t = T = 0, satisfying the homogeneous boundary conditions (32) (with g = 0) and (40) respectively,one has ( L ′ e ˙ U , ¯ U ) L (Ω T ) − ( ˙ U , L ′ e ∗ ¯ U ) L (Ω T ) = − ( e A ˙ U , ¯ U ) L ( ∂ Ω T ) = − ( A V, ¯ V ) L ( ∂ Ω T ) = 0 , where ¯ V = J − ¯ U . For the dual problem (39)–(41) we can easily get the inequality¯ I ( t ) + Z R ǫ + ∂ ˆ p | x =0 ¯ p | x =0 d x ′ ≤ C ( K ) (cid:26) k ¯ f k L (Ω T ) + Z t (cid:16) ¯ I ( s ) + k ¯ p | x =0 ( s ) k L ( R ) (cid:17) d s (cid:27) ( ¯ I ( t ) = R R ( A ¯ V , ¯ V ) d x ) which, in view of condition (26), implies the L –estimate k ¯ U k L (Ω T ) ≤ C ( K ) k ¯ f k L (Ω T ) . We omit further arguments which are really classical and refer to [10] (see also, e.g., [4, 15]). Thus, we havethe following well-posedness theorem for the linearized problem (31)–(33).
Theorem 2.2
Let assumptions (23) – (26) are fulfilled for the basic state (22). Then for all ( f, g ) ∈ L (Ω T ) × H ( ∂ Ω T ) that vanish in the past problem (31)–(33) has a unique solution ( ˙ U , ϕ ) ∈ L (Ω T ) × L ( ∂ Ω T ) . Thissolution obeys the a priori estimate (37) . Remark 2.5
Strictly speaking, the uniqueness of the solution to problem (31)–(33) follows from estimate(37), provided that our solution belongs to H (Ω T ) × H ( ∂ Ω T ). We omit here a formal proof of theexistence of solutions having an arbitrary degree of smoothness, and we shall suppose that the existenceresult of Theorem 2.2 is also valid for the function spaces H s (Ω T ) × H s ( ∂ Ω T ), with s ≥
1. In this case exactassumptions about the regularity of the basic state will be made in Sect. 3, where we prove a tame a prioriestimate in H s (Ω T ) × H s ( ∂ Ω T ) with s large enough. We are going to derive a tame a priori estimate in H s for problem (31)–(33), with s large enough. This tameestimate (see Theorem 3.1 below) being, roughly speaking, linear in high norms (that are multiplied by low12orms) is with no loss of derivatives from f , with the loss of one derivative from g , and with a fixed lossof derivatives with respect to the coefficients, i.e., with respect to the basic state (22). Although problem(31)–(33) is a hyperbolic problem with characteristic boundary that implies a natural loss of control onderivatives in the normal direction we manage to compensate this loss and derive higher order estimates inusual Sobolev spaces. This is achieved by using the same idea as in [19, 5] and estimating missing normalderivatives through a vorticity-type linearized system. Theorem 3.1
Let
T > and s ∈ N , with s ≥ . Assume that the basic state ( b U , ˆ ϕ ) ∈ H s +3 (Ω T ) × H s +3 ( ∂ Ω T ) satisfies assumptions (23)–(26) and k b U k H (Ω T ) + k ˆ ϕ k H ( ∂ Ω T ) ≤ b K, (42) where b K > is a constant. Let also the data ( f, g ) ∈ H s (Ω T ) × H s +1 ( ∂ Ω T ) vanish in the past. Then thereexists a positive constant K that does not depend on s and T and there exists a constant C ( K ) > suchthat, if b K ≤ K , then there exists a unique solution ( ˙ U , ϕ ) ∈ H s (Ω T ) × H s ( ∂ Ω T ) to problem (31)–(33) thatobeys the a priori tame estimate k ˙ U k H s (Ω T ) + k ϕ k H s ( ∂ Ω T ) ≤ C ( K ) n k f k H s (Ω T ) + k g k H s +1 ( ∂ Ω T ) + (cid:0) k f k H (Ω T ) + k g k H ( ∂ Ω T ) (cid:1)(cid:0) k b U k H s +3 (Ω T ) + k ˆ ϕ k H s +3 ( ∂ Ω T ) (cid:1)o (43) for a sufficiently short time T . Proof . Since arguments below are quite standard we somewhere will drop detailed calculations. By applyingto system (34) the operator ∂ α tan = ∂ α t ∂ α ∂ α , with | α | = | ( α , α , α ) | ≤ s , one gets Z R ( A ∂ α tan V, ∂ α tan V )d x − Z ∂ Ω t ∂ α tan ˙ p ∂ α tan ˙ v N | x =0 d x ′ d s = R , (44)where R = Z Ω t (cid:16)(cid:8) div A ∂ α tan V − X j =0 [ ∂ α tan , A j ] ∂ j V − ∂ α tan ( A V ) + 2 ∂ α tan F (cid:9) , ∂ α tan V (cid:17) d x d s, div A = P j =0 ∂ j A j ( ∂ := ∂ t ), and we use the notation of commutator: [ a, b ] c := a ( bc ) − b ( ac ). Using theMoser-type calculus inequalities k uv k H s (Ω T ) ≤ C (cid:0) k u k H s (Ω T ) k v k L ∞ (Ω T ) + k u k L ∞ (Ω T ) k v k H s (Ω T ) (cid:1) , (45) k F ( u ) k H s (Ω T ) ≤ C ( M ) (cid:0) k u k H s (Ω T ) (cid:1) , (46)where the function F is a C ∞ function of u , and M is such a positive constant that k u k L ∞ (Ω T ) ≤ M , weestimate the right-hand side in (44): R ≤ C ( K ) n k V k H s (Ω t ) + k f k H s (Ω T ) + (cid:16) k ˙ U k W ∞ (Ω T ) + k f k L ∞ (Ω T ) (cid:17) (cid:0) k coeff k s +1 (cid:1)o , (47)with k coeff k m := k b U k H m (Ω T ) + k ˆ ϕ k H m ( ∂ Ω T ) .Taking into account the boundary conditions, we have: − ∂ α tan ˙ p ∂ α tan ˙ v N | x =0 = 2 ∂ α tan ( ϕ ˆ a − g ) ∂ α tan ( ∂ t ϕ + ˆ v ∂ ϕ + ˆ v ∂ ϕ − ϕ ∂ ˆ v N − g ) | x =0 = ∂ t (cid:8) ˆ a | x =0 ( ∂ α tan ϕ ) − ∂ α tan g ∂ α tan ϕ (cid:9) + . . . + ∂ { ˆ v | x =0 [ ∂ α tan , ∂ ˆ p | x =0 ] ϕ } ∂ α tan ϕ + . . . , k ∂ { ˆ v | x =0 [ ∂ α tan , ∂ ˆ p | x =0 ] ϕ } k L ( R ) ≤ C ( K ) n k ϕ k H s ( ∂ Ω t ) + k ϕ k L ∞ ( ∂ Ω T ) (cid:16) k b U | x =0 k H s +2 ( ∂ Ω T ) (cid:17)o ≤ C ( K ) n k ϕ k H s ( ∂ Ω t ) + k ϕ k L ∞ ( ∂ Ω T ) (cid:16) k b U k H s +3 (Ω T ) (cid:17)o . Omitting detailed calculations, from (44) and (47) we obtain ||| V ( t ) ||| ,s + ||| ϕ ( t ) k H s ( R ) ≤ C ( K ) M ( t ) , (48)where M ( t ) = N ( T ) + Z t I ( s ) d s, I ( t ) = ||| V ( t ) ||| H s ( R ) + ||| ϕ ( t ) ||| H s ( R ) , N ( T ) = k f k H s (Ω T ) + k g k H s +1 ( ∂ Ω T ) + (cid:16) k ˙ U k W ∞ (Ω T ) + k ϕ k W ∞ ( ∂ Ω T ) + k f k L ∞ (Ω T ) (cid:17) (cid:0) k coeff k s +3 (cid:1) , ||| u ( t ) ||| ,m := X | α |≤ m k ∂ α tan u ( t ) k L ( R ) , ||| u ( t ) ||| H m ( D ) := m X j =0 k ∂ jt u ( t ) k H m ( D ) ( D = R or D = R ). Since only the biggest loss of derivatives from the coefficients will play the role forobtaining the final tame estimate, we have roughened inequality (48) by choosing the biggest loss.It follows from (34) and (35) that( ∂ V n , , ,
0) = (cid:0) ∂ b Φ (cid:1) A (1) (cid:16) F − A ∂ t V − X k =2 A k ∂ k V − A V − A (0) ∂ V (cid:17) . (49)Applying to (49) the operator ∂ β tan , with | β | ≤ s −
1, using decompositions like ∂ β tan ( B∂ i V ) = B∂ β tan ∂ i V + [ ∂ β tan , B ] ∂ i V, taking into account the fact that A (0) | x =0 = 0, and employing counterparts of the calculus inequalities (45)and (46) for the “layerwise” norms ||| ( · )( t ) ||| (see [19]), one gets k ∂ ∂ β tan V n ( t ) k L ( R ) ≤ C ( K ) n ||| V ( t ) ||| ,s + k σ∂ ∂ β tan V ( t ) k L ( R ) + ||| V ( t ) ||| H s − ( R ) + ||| f ( t ) ||| H s − ( R ) + (cid:16) k ˙ U k W ∞ (Ω T ) + k f k L ∞ (Ω T ) (cid:17) (cid:0) ||| coeff(t) ||| s +1 (cid:1)o , (50)where σ = σ ( x ) ∈ C ∞ ( R + ) is a monotone increasing function such that σ ( x ) = x in a neighborhoodof the origin and σ ( x ) = 1 for x large enough. Since σ | x =0 = 0 we do not need to use the boundaryconditions to estimate σ∂ j ∂ γ tan V , with j + | γ | ≤ s , and we easily get the inequality k σ∂ j ∂ γ tan V ( t ) k L ( R ) ≤ C ( K ) n k V k H s (Ω t ) + k f k H s (Ω T ) + (cid:16) k ˙ U k W ∞ (Ω T ) + k f k L ∞ (Ω T ) (cid:17) (cid:0) k coeff k s +1 (cid:1)o . (51)Taking into account Sobolev’s embedding in one space dimension, ||| u ( t ) ||| H m − ( D ) ≤ k u k L ∞ ([0 ,t ] ,H m − ( D )) ≤ C k u k H m ([0 ,t ] × D ) , and combining (48), (50), and (51) for j = 1, we obtain ||| V ( t ) ||| ,s + ||| ϕ ( t ) k H s ( R ) + k X i =1 X | α |≤ s − i k ∂ i ∂ α tan V n ( t ) k L ( R ) ≤ C ( K ) M ( t ) , (52)14ith k = 1.Estimate (52) for k = s is easily proved by finite induction and equivalently rewritten as ||| V ( t ) ||| ,s + k V n ( t ) k H s ( R ) + ||| ϕ ( t ) k H s ( R ) ≤ C ( K ) M ( t ) . (53)Missing normal derivatives in (53) for the “characteristic” part ( ˙ v , ˙ v , ˙ S ) of the unknown V can be estimatedfrom the last equation in (31), ∂ t ˙ S + 1 ∂ b Φ n ( ˆ w, ∇ ) ˙ S + ( ˙ u, ∇ ) b S o = f , (54)and a system for the linearized vorticity ξ = ∇ × ˜ v , where˜ v = ( ˙ v , ˙ v τ , ˙ v τ ) , ˙ v τ k = ( ˙ v, τ k ) , τ = ( ∂ ˆ ϕ, , , τ = ( ∂ ˆ ϕ, , , ˆ w = (ˆ v n − ∂ t b Ψ , ˆ v ∂ b Φ , ˆ v ∂ b Φ) , ˙ u = ( ˙ v n , ˙ v ∂ b Φ , ˙ v ∂ b Φ) . This system is obtained by applying the curl operator to the equation for ˜ v following from (31), ∂ t ˜ v + 1 ∂ b Φ ( ( ˆ w, ∇ )˜ v + 1 ρ (ˆ p, b S ) ∇ ˙ p ) + l.o.t = ˜ f v ( ˜ f v = ( f , f τ , f τ ) , f τ k = ( f v , τ k ) , f v = ( f , f , f )), and has the form ξ t + 1 ∂ b Φ ( ˆ w, ∇ ) ξ + l.o.t = ∇ × ˜ f v , (55)where l.o.t. are lower-order terms which exact form has no meaning.Both equations (54) and (55) do not need boundary conditions because, in view of (25), the first compo-nent of the vector ˆ w is zero on the boundary x = 0. Therefore, omitting detailed calculations and combiningcorresponding estimates for the normal derivatives of the “characteristic” unknown ( ˙ v , ˙ v , ˙ S ) with (53), wededuce the inequality I ( t ) ≤ C ( K ) (cid:26) N ( T ) + Z t I ( s ) d s (cid:27) . Applying then Gronwall’s lemma, one gets I ( t ) ≤ C ( K ) e C ( K ) T N ( T )( I (0) = 0, see (33)). Integrating the last inequality over the interval [0 , T ], we come to the estimate k V k H s (Ω T ) + k ϕ k H s ( ∂ Ω T ) ≤ C ( K ) T e C ( K ) T N ( T ) . (56)Recall that ˙ U = JV . Taking into account the decomposition J ( ˆ ϕ ) = I + J ( ˆ ϕ ) and J (0) = 0, using (45)together with the improved calculus inequality (46) for the case F (0) = 0, k F ( u ) k H s (Ω T ) ≤ C ( M ) k u k H s (Ω T ) , and applying Sobolev’s embedding in one space dimension, we obtain k ˙ U k H s (Ω T ) = k V + J V k H s (Ω T ) ≤ C ( K ) (cid:0) k V k H s (Ω T ) + k ˙ U k L ∞ (Ω T ) k coeff k s (cid:1) ≤ C ( K ) k V k H s (Ω T ) + T C ( K ) k ˙ U k L ∞ (Ω T ) k coeff k s +1 . (57)15nequalities (56) and (57) imply k ˙ U k H s (Ω T ) + k ϕ k H s ( ∂ Ω T ) ≤ C ( K ) T e C ( K ) T N ( T ) . (58)Taking into account Theorem 2.2 and Remark 2.5, we have the well-posedness of problem (31)–(33) in H s (Ω T ) × H s ( ∂ Ω T ). Applying Sobolev’s embeddings, from (58) with s ≥ k ˙ U k H s (Ω T ) + k ϕ k H s ( ∂ Ω T ) ≤ C ( K ) T / e C ( K ) T n k f k H s (Ω T ) + k g k H s +1 ( ∂ Ω T ) + (cid:16) k ˙ U k H (Ω T ) + k ϕ k H ( ∂ Ω T ) + k f k H (Ω T ) (cid:17) (cid:0) k b U k H s +3 (Ω T ) + k ˆ ϕ k H s +3 ( ∂ Ω T ) (cid:1)o , (59)where we have absorbed some norms k ˙ U k H (Ω T ) and k ϕ k H ( ∂ Ω T ) in the left-hand side by choosing T smallenough. Considering (59) for s = 3 and using (42), we obtain for T small enough that k ˙ U k H (Ω T ) + k ϕ k H ( ∂ Ω T ) ≤ C ( K ) (cid:8) k f k H (Ω T ) + k g k H ( ∂ Ω T ) (cid:9) . (60)It is natural to assume that T < C ( K ) does not depend on T . Inequalities (59) and (60) imply (43). (cid:3) To use the tame estimate (43) for the proof of convergence of the Nash-Moser iteration, we should reduceour nonlinear problem (20), (17), (18) on [0 , T ] × R to that on Ω T which solutions vanish in the past.This is achieved by the classical argument suggesting to absorb the initial data into the interior equationsby constructing a so-called approximate solution . Before constructing the approximate solution we have todefine compatibility conditions for the initial data (18),( U , ϕ ) = ( p , v , , v , , v , , S , ϕ ) . Assuming that the hyperbolicity condition (6) is satisfied, we rewrite system (20) in the form ∂ t U = − ( A ( U )) − (cid:16) e A ( U, Ψ) ∂ U + A ( U ) ∂ U + A ( U ) ∂ U + A ν ( U, Ψ) ∂ ˇ U + Q ( U ) (cid:17) . (61)The traces U j = ( p j , v ,j , v ,j , v ,j , S j ) = ∂ jt U | t =0 and ϕ j = ∂ jt ϕ | t =0 , with j ≥
1, are recursively defined by the formal application of the differential operator ∂ j − t to the boundarycondition ∂ t ϕ = ( v − v ∂ ϕ − v ∂ ϕ ) | x =0 (62)and (61) and evaluating ∂ jt ϕ and ∂ jt U at t = 0. Moreover, Ψ j = ∂ jt Ψ | t =0 = χ ( x ) ϕ j .We naturally define the zero-order compatibility condition as p | x =0 = 0. Note that, unlike the casewhen the symbol associated with the free surface is elliptic [5, 15, 21], this condition does not contain thefunction ϕ . Evaluating (62) at t = 0, we get ϕ = ( v , − v , ∂ ϕ − v , ∂ ϕ ) | x =0 , (63)and then, with ∂ t Φ | t =0 := Φ = χ ( x ) ϕ , from (61) evaluated at t = 0 we define U . The first-ordercompatibility condition p | x =0 = 0 will implicitly depend on ϕ and ϕ . Knowing ϕ and U we can thenfind ϕ , U , etc. The following lemma is the analogue of Lemma 4.2.1 in [15], Lemma 2 in [5], and Lemma5 in [21]. 16 emma 4.1 Let µ ∈ N , µ ≥ , U ∈ H µ ( R ) , and ϕ ∈ H µ ( R ) . Then, the procedure described abovedetermines U j ∈ H µ − j ( R ) and ϕ j ∈ H µ − j ( R ) for j = 1 , . . . , µ . Moreover, µ X j =1 (cid:16) k U j k H µ − j ( R ) + k ϕ j k H µ − j ( R ) (cid:17) ≤ CM , (64) where M = k U k H µ ( R ) + k ϕ k H µ ( R ) , (65) the constant C > depends only on µ and the norms k U k W ∞ ( R ) and k ϕ k W ∞ ( R ) . The proof is almost evident and based on the multiplicative properties of Sobolev spaces (Remark 2.3should be also taken into account).
Definition 4.1
Let µ ∈ N , µ ≥ . The initial data ( U , ϕ ) ∈ H µ ( R ) × H µ ( R ) are said to be compatibleup to order µ when ( U j , ϕ j ) satisfy p j | x =0 = 0 (66) for j = 0 , . . . , µ . We are now ready to construct the approximate solution.
Lemma 4.2
Suppose the initial data (18) are compatible up to order µ and satisfy the assumptions ofTheorem 2.1 (i.e., (6) for all x ∈ R and (21)). Then there exists a vector-function ( U a , ϕ a ) ∈ H µ +1 (Ω T ) × H µ +1 ( ∂ Ω T ) , that is further called the approximate solution to problem (20) , (17) , (18) , such that ∂ jt L ( U a , Ψ a ) | t =0 = 0 for j = 0 , . . . , µ − , (67) and it satisfies the boundary conditions (17), where Ψ a = χ ( x ) ϕ a . Moreover, the approximate solutionobeys the estimate k U a k H µ +1 (Ω T ) + k ϕ a k H µ +1 ( ∂ Ω T ) ≤ C ( M ) (68) and satisfies the hyperbolicity condition (6) on Ω T as well as condition (21) on ∂ Ω T , where C = C ( M ) > is a constant depending on M (see (65)). Moreover, ρ a − ǫ = ρ ( p a , S a ) − ǫ ∈ H µ +1 (Ω T ) . Proof . Consider functions U a ∈ H µ +1 ( R × R ) and ϕ a ∈ H µ +1 ( R ) such that ∂ jt U a | t =0 = U j ∈ H µ − j ( R ) , ∂ jt ϕ a | t =0 = ϕ j ∈ H µ − j ( R ) for j = 0 , . . . , µ, where U j and ϕ j are given by Lemma 4.1. Thanks to (63) and (66) we can choose U a and ϕ a that satisfy theboundary conditions (17). By using a cut-off C ∞ function we can suppose that ( U a , ϕ a ) vanishes outside ofthe interval [ − T, T ], i.e., ( U a , ϕ a ) ∈ H µ +1 (Ω T ) × H µ +1 ( ∂ Ω T ). Applying Sobolev’s embeddings, we rewriteestimate (64) as µ X j =1 (cid:16) k U j k H µ − j ( R ) + k ϕ j k H µ − j ( R ) (cid:17) ≤ C ( M ) , (69)where C = C ( M ) > M . The estimate (68) follows from (69) and thecontinuity of the lifting operators from the hyperplane t = 0 to R × R . Conditions (67) hold thanks to theproperties of ( U j , ϕ j ) given by Lemma 4.1. At last, since ( U a , ϕ a ) satisfies the hyperbolicity condition (6)17nd condition (21) at t = 0, in the above procedure we can choose ( U a , ϕ a ) that it satisfies (6) and (21) forall times t ∈ [ − T, T ]. The condition ρ a − ǫ ∈ H µ +1 (Ω T ) is just an assumption on the state equation. (cid:3) Without loss of generality we can suppose that k U k H µ ( R ) + k ϕ k H µ ( R ) ≤ , k ϕ k H µ ( R ) ≤ / . (70)Then for a sufficiently short time interval [0 , T ] the smooth solution which existence we are going to provesatisfies k ϕ k L ∞ ([0 ,T ] × R ) ≤ ∂ Φ ≥ / k χ ′ k L ∞ ( R ) < /
2, see Section 2). Let µ isan integer number that will appear in the regularity assumption for the initial data in the existence theoremfor problem (20), (17), (18). Running ahead, we take µ = m + 7, with m ≥ k U a k H m +8 (Ω T ) + k ϕ a k H m +8 ( ∂ Ω T ) ≤ C ∗ , (71)where C ∗ = C (1).Let us introduce f a := ( − L ( U a , Ψ a ) for t > , t < . (72)Since ( U a , ϕ a ) ∈ H m +8 (Ω T ) × H m +8 ( ∂ Ω T ), using (67), we get f a ∈ H m +7 (Ω T ) and k f a k H m +7 (Ω T ) ≤ δ ( T ) , (73)where the constant δ ( T ) → T →
0. The crucial role in the proof of the fact that f a belongs to a Sobolevspace is played by the presence of gravity (see Remark 2.3). To prove estimate (73) we use the Moser-typeand embedding inequalities and the fact that f a vanishes in the past. Then, given the approximate solutiondefined in Lemma 4.2, ( U, ϕ ) = ( U a , ϕ a ) + ( e U , ˜ ϕ ) is a solution of the original problem (20), (17), (18) on[0 , T ] × R if ( e U , ˜ ϕ ) satisfies the following problem on Ω T (tildes are dropped): L ( U, Ψ) = f a in Ω T , (74) B ( U, ϕ ) = 0 on ∂ Ω T , (75)( U, ϕ ) = 0 for t < , (76)where L ( U, Ψ) := L ( U a + U, Ψ a + Ψ) − L ( U a , Ψ a ) , B ( U, ϕ ) := B ( U a + U, ϕ a + ϕ ). From now on weconcentrate on the proof of the existence of solutions to problem (74)–(76).We solve problem (74)–(76) by a suitable Nash-Moser-type iteration scheme. In short, this schemeis a modified Newton’s scheme and at each Nash-Moser iteration step we smooth the coefficient u n of acorresponding linear problem for δu n = u n +1 − u n . Errors of a classical Nash-Moser iteration are the“quadratic” error of Newton’s scheme and the “substitution” error caused by the application of smoothingoperators S θ (see, e.g., [7] and references therein). As in [5, 21], in our case the Nash-Moser procedure isnot completely standard and we have the additional error caused by the introduction of an intermediate (ormodified) state u n +1 / satisfying some nonlinear constraints. In our case, the main constraint is condition(25) that was required to be fulfilled for the basic state (22). Also the additional error is caused by droppingthe zero-order term in Ψ in the linearized interior equations written in terms of the “good unknown” (see(27)–(29)). We first list the important properties of smoothing operators [1, 5, 7].18 roposition 4.1 There exists such a family { S θ } θ ≥ of smoothing operators in H s (Ω T ) acting on the classof functions vanishing in the past that k S θ u k H β (Ω T ) ≤ Cθ ( β − α ) + k u k H α (Ω T ) , α, β ≥ , (77) k S θ u − u k H β (Ω T ) ≤ Cθ β − α k u k H α (Ω T ) , ≤ β ≤ α, (78) (cid:13)(cid:13) ddθ S θ u (cid:13)(cid:13) H β (Ω T ) ≤ Cθ β − α − k u k H α (Ω T ) , α, β ≥ , (79) where C > is a constant, and ( β − α ) + := max(0 , β − α ) . Moreover, there is another family of smoothingoperators (still denoted S θ ) acting on functions defined on the boundary ∂ Ω T and meeting properties (77)–(79), with the norms k · k H α ( ∂ Ω T ) . Now, following [5, 21], we describe the iteration scheme for problem (74)–(76). We choose U = 0 , ϕ = 0and assume that ( U k , ϕ k ) are already given for k = 0 , . . . , n . Moreover, let ( U k , ϕ k ) vanish in the past, i.e.,they satisfy (76). We define U n +1 = U n + δU n , ϕ n +1 = ϕ n + δϕ n , where the differences δU n and δϕ n solve the linear problem L ′ e ( U a + U n +1 / , Ψ a + Ψ n +1 / ) δ ˙ U n = f n in Ω T , B ′ n +1 / ( δ ˙ U n , δϕ n ) = g n on ∂ Ω T , ( δ ˙ U n , δϕ n ) = 0 for t < . (80)Here δ ˙ U n := δU n − δ Ψ n ∂ (Φ a + Ψ n +1 / ) ∂ ( ˇ U + U a + U n +1 / ) (81)is the “good unknown” (cf. (27)), B ′ n +1 / := B ′ e (( U a + U n +1 / ) | x =0 , ϕ a + ϕ n +1 / ) , the operators L ′ e and B ′ e are defined in (29), (30), and ( U n +1 / , ϕ n +1 / ) is a smooth modified state suchthat ( U a + U n +1 / , ϕ a + ϕ n +1 / ) satisfies constraints (24)–(26) (Ψ n , Ψ n +1 / , and δ Ψ n are associated to ϕ n , ϕ n +1 / , and δϕ n like Ψ is associated to ϕ ). The right-hand sides f n and g n are defined through theaccumulated errors at the step n .The errors of the iteration scheme are defined from the following chains of decompositions: L ( U n +1 , Ψ n +1 ) − L ( U n , Ψ n )= L ′ ( U a + U n , Ψ a + Ψ n )( δU n , δ Ψ n ) + e ′ n = L ′ ( U a + S θ n U n , Ψ a + S θ n Ψ n )( δU n , δ Ψ n ) + e ′ n + e ′′ n = L ′ ( U a + U n +1 / , Ψ a + Ψ n +1 / )( δU n , δ Ψ n ) + e ′ n + e ′′ n + e ′′′ n = L ′ e ( U a + U n +1 / , Ψ a + Ψ n +1 / ) δ ˙ U n + e ′ n + e ′′ n + e ′′′ n + D n +1 / δ Ψ n and B ( U n +1 | x =0 , ϕ n +1 ) − B ( U n | x =0 , ϕ n )= B ′ (( U a + U n ) | x =0 , ϕ a + ϕ n )( δU n | x =0 , δϕ n ) + ˜ e ′ n = B ′ (( U a + S θ n U n ) | x =0 , ϕ a + S θ n ϕ n )( δU n | x =0 , δϕ n ) + ˜ e ′ n + ˜ e ′′ n = B ′ n +1 / ( δ ˙ U n , δϕ n ) + ˜ e ′ n + ˜ e ′′ n + ˜ e ′′′ n , S θ n are smoothing operators enjoying the properties of Proposition 4.1, with the sequence ( θ n ) definedby θ ≥ , θ n = p θ + n, and we use the notation D n +1 / := 1 ∂ (Φ a + Ψ n +1 / ) ∂ (cid:8) L ( U a + U n +1 / , Ψ a + Ψ n +1 / ) (cid:9) . The errors e ′ n and ˜ e ′ n are the usual quadratic errors of Newton’s method, and e ′′ n , ˜ e ′′ n and e ′′′ n , ˜ e ′′′ n are the firstand the second substitution errors respectively.Let e n := e ′ n + e ′′ n + e ′′′ n + D n +1 / δ Ψ n , ˜ e n := ˜ e ′ n + ˜ e ′′ n + ˜ e ′′′ n , (82)then the accumulated errors at the step n ≥ E n = n − X k =0 e k , e E n = n − X k =0 ˜ e k , (83)with E := 0 and e E := 0. The right-hand sides f n and g n are recursively computed from the equations n X k =0 f k + S θ n E n = S θ n f a , n X k =0 g k + S θ n e E n = 0 , (84)where f := S θ f a and g := 0. Since S θ N → I as N → ∞ , one can show that we formally obtain thesolution to problem (74)–(76) from L ( U N , Ψ N ) → f a and B ( U N | x =0 , ϕ N ) →
0, provided that ( e N , ˜ e N ) → Remark 4.1
In general, the realization of the Nash-Moser procedure for problem (74)–(76) below is muchsimpler as in [21] for current-vortex sheets. As in [5] and unlike [21], we work in usual Sobolev spaces H s (in [21] one works in the anisotropic weighted Sobolev spaces H s ∗ ). More precisely, in [5] the exponentiallyweighted Sobolev spaces H sγ := e γt H s were used, but for H sγ , Sobolev’s embeddings, Moser-type inequalities,etc. are internally the same as for the usual Sobolev spaces H s . Therefore, in some places below ourcalculations are almost the same as in [5]. However, for convenience of the reader we prefer to present allthe calculations (at least, in brief). Moreover, since, unlike [5], we do not assume that our initial data areclose to a constant solution and in our tame estimate (43) we lose, as [21], “one derivative from the front”,somewhere we have to modify arguments of [5].Below we closely follow the plan of [5] and [21]. Let us first formulate an inductive hypothesis. As in [21]and unlike [5], we do not require more regularity for δϕ k in our inductive hypothesis. Inductive hypothesis.
Given a small number δ >
0, the integer α := m + 1, and an integer ˜ α , ourinductive hypothesis reads:( H n − ) a ) ∀ k = 0 , . . . , n − , ∀ s ∈ [3 , ˜ α ] ∩ N , k δU k k H s (Ω T ) + k δϕ k k H s ( ∂ Ω T ) ≤ δθ s − α − k ∆ k ,b ) ∀ k = 0 , . . . , n − , ∀ s ∈ [3 , ˜ α − ∩ N , kL ( U k , Ψ k ) − f a k H s (Ω T ) ≤ δθ s − α − k ,c ) ∀ k = 0 , . . . , n − , ∀ s ∈ [4 , α ] ∩ N , kB ( U k | x =0 , ϕ k ) k H s ( ∂ Ω T ) ≤ δθ s − α − k , k = θ k +1 − θ k . Note that the sequence (∆ n ) is decreasing and tends to zero, and ∀ n ∈ N , θ n ≤ ∆ n = p θ n + 1 − θ n ≤ θ n . Recall that ( U k , ϕ k ) for k = 0 , . . . , n are also assumed to satisfy (76). Running a few steps forward, weobserve that we will need to use inequalities (71) and (73) with m = ˜ α −
4. That is, we now choose˜ α = m + 4. Our goal is to prove that ( H n − ) implies ( H n ) for a suitable choice of parameters θ ≥ δ >
0, and for a sufficiently short time
T >
0. After that we shall prove ( H ). From now on we assume that( H n − ) holds. As in [5], we have the following consequences of ( H n − ). Lemma 4.3 If θ is big enough, then for every k = 0 , . . . , n and for every integer s ∈ [3 , ˜ α ] we have k U k k H s (Ω T ) + k ϕ k k H s ( ∂ Ω T ) ≤ δθ ( s − α ) + k , α = s, (85) k U k k H α (Ω T ) + k ϕ k k H α ( ∂ Ω T ) ≤ δ log θ k , (86) k ( I − S θ k ) U k k H s (Ω T ) + k (1 − S θ k ) ϕ k k H s ( ∂ Ω T ) ≤ Cδθ s − αk . (87) For every k = 0 , . . . , n and for every integer s ∈ [3 , ˜ α + 4] we have k S θ k U k k H s (Ω T ) + k S θ k ϕ k k H s ( ∂ Ω T ) ≤ Cδθ ( s − α ) + k , α = s, (88) k S θ k U k k H α (Ω T ) + k S θ k ϕ k k H α ( ∂ Ω T ) ≤ Cδ log θ k . (89)Estimates (87)–(89) follow from (85), (86), and Proposition 4.1. Moreover, (87) and (88) hold actuallyfor every integer s ≥ s ∈ [3 , ˜ α ] and s ∈ [3 , ˜ α + 4] respectively. Estimate of the quadratic errors.
The quadratic errors e ′ k = L ( U k +1 , Ψ k +1 ) − L ( U k , Ψ k ) − L ′ ( U k , Ψ k )( δU k , δ Ψ k ) , ˜ e ′ k = (cid:0) B ( U k +1 , ϕ k +1 ) − B ( U k , ϕ k ) − B ′ ( U k , ϕ k )( δU k , δϕ k ) (cid:1) | x =0 can be rewritten as e ′ k = Z (1 − τ ) L ′′ ( U a + U k + τ δU k , Ψ a + Ψ k + τ δ Ψ k ) (cid:0) ( δU k , δ Ψ k ) , ( δU k , δ Ψ k ) (cid:1) dτ, (90)˜ e ′ k = 12 B ′′ (cid:0) ( δU k | x =0 , δϕ k ) , ( δU k | x =0 , δϕ k ) (cid:1) (91)by using the second derivatives of the operators L and B : L ′′ ( b U , b Ψ)(( U ′ , Ψ ′ ) , ( U ′′ , Ψ ′′ )) := dd ε L ′ ( U ε , Ψ ε )( U ′ , Ψ ′ ) | ε =0 ( L ′ ( b U , b Ψ)( U ′′ , Ψ ′′ ) := dd ε L ( U ε , Ψ ε )) , B ′′ (( W ′ , ϕ ′ ) , ( W ′′ , ϕ ′′ )) := dd ε B ′ ( W ε , ϕ ε )( W ′ , ϕ ′ ) | ε =0 ( B ′ ( b U | x =0 , ˆ ϕ )( W ′′ , ϕ ′′ ) = dd ε B ( W ε , ϕ ε )) , where U ε = b U + εU ′′ , W ε = b U | x =0 + εW ′′ , ϕ ε = ˆ ϕ + εϕ ′′ , and Ψ ′ and Ψ ′′ are associated to ϕ ′ and ϕ ′′ respectively like Ψ is associated to ϕ . We easily compute the explicit form of B ′′ , that do not depend on thestate ( b U , ˆ ϕ ): B ′′ (( W ′ , ϕ ′ ) , ( W ′′ , ϕ ′′ )) = v ′ ∂ ϕ ′′ + v ′ ∂ ϕ ′′ + v ′′ ∂ ϕ ′ + v ′′ ∂ ϕ ′ ! . To estimate the quadratic errors by utilizing representations (90) and (91) we need estimates for L ′′ and B ′′ . They can easily be obtained from the explicit forms of L ′′ and B ′′ by applying the Moser-type andembedding inequalities. Omitting detailed calculations, we get the following result.21 roposition 4.2 Let
T > and s ∈ N , with s ≥ . Assume that ( b U , ˆ ϕ ) ∈ H s +1 (Ω T ) × H s +1 ( ∂ Ω T ) and k b U k H ((Ω T ) + k ˆ f k H ( ∂ Ω T ) ≤ e K. Then there exists a positive constant e K , that does not depend on s and T , and there exists a constant C ( e K ) > such that, if e K ≤ e K and ( U ′ , ϕ ′ ) , ( U ′′ , ϕ ′′ ) ∈ H s +1 (Ω T ) × H s +1 ( ∂ Ω T ) , then k L ′′ ( b U , b Ψ)(( U ′ , Ψ ′ ) , ( U ′′ , Ψ ′′ )) k H s (Ω T ) ≤ C ( e K ) n hh ( b U , ˆ f ) ii s +1 hh ( U ′ , ϕ ′ ) ii hh ( U ′′ , ϕ ′′ ) ii + hh ( U ′ , ϕ ′ ) ii s +1 hh ( U ′′ , ϕ ′′ ) ii + hh ( U ′′ , ϕ ′′ ) ii s +1 hh ( U ′ , ϕ ′ ) ii o , where hh ( U, ϕ ) ii ℓ := k U k H ℓ (Ω T ) + k ϕ k H ℓ ( ∂ Ω T ) . If ( W ′ , ϕ ′ ) , ( W ′′ , ϕ ′′ ) ∈ H s ( ∂ Ω T ) × H s +1 ( ∂ Ω T ) , then k B ′′ (( W ′ , ϕ ′ ) , ( W ′′ , ϕ ′′ )) k H s ( ∂ Ω T ) ≤ C ( e K ) n k W ′ k H s ( ∂ Ω T ) k ϕ ′′ k H ( ∂ Ω T ) + k W ′ k H ( ∂ Ω T ) k ϕ ′′ k H s +1 ( ∂ Ω T ) + k W ′′ k H s ( ∂ Ω T ) k ϕ ′ k H ( ∂ Ω T ) + k W ′′ k H ( ∂ Ω T ) k ϕ ′ k H s +1 ( ∂ Ω T ) + k W ′ k H s ( ∂ Ω T ) k W ′′ k H ( ∂ Ω T ) + k W ′ k H ( ∂ Ω T ) k W ′′ k H s ( ∂ Ω T ) o . Without loss of generality we assume that the constant e K = 2 C ∗ , where C ∗ is the constant from (71).By using (90), (91), and Proposition 4.2, we obtain the following result. Lemma 4.4
Let α ≥ . There exist δ > sufficiently small, and θ ≥ sufficiently large, such that for all k = 0 , . . . n − , and for all integer s ∈ [3 , e α − , we have the estimates k e ′ k k H s (Ω T ) ≤ Cδ θ L ( s ) − k ∆ k , (92) k ˜ e ′ k k H s ( ∂ Ω T ) ≤ Cδ θ L ( s ) − k ∆ k , (93) where L ( s ) = max { ( s + 1 − α ) + + 4 − α, s + 2 − α } . Proof . In view of (71) (recall that m = ˜ α − H n − ), and (85), we estimate the “coefficient” of L ′′ in (90)as follows:sup τ ∈ [0 , hh ( U a + U k + τ δU k , ϕ a + ϕ k + τ δϕ k ) ii ≤ C ∗ + δθ (3 − α ) + k + δθ − αk ∆ k ≤ C ∗ + Cδ ≤ C ∗ for δ sufficiently small. Therefore, we may apply Proposition 4.2: k e ′ k k H s (Ω T ) ≤ C (cid:16) δ θ − αk ∆ k (cid:0) C ∗ + hh ( U k , ϕ k ) ii s +1 + hh ( δU k , δϕ k ) ii s +1 (cid:1) + δ θ s +2 − αk ∆ k (cid:17) for s ∈ [3 , e α − s + 1 = α , it follows from (85) that k e ′ k k H s (Ω T ) ≤ Cδ ∆ k n θ ( s +2 − α ) + +12 − αk + θ s +7 − αk o ≤ Cδ θ L ( s ) − k ∆ k (here we have used the inequality θ k ∆ k ≤ / s + 1 = α and α ≥ k e ′ k k H s (Ω T ) ≤ Cδ ∆ k (cid:8) ( C ∗ + δ log θ k + δθ − k ∆ k ) θ − αk + θ − αk (cid:9) ≤ Cδ ∆ k θ − αk ≤ Cδ θ L ( α − − k ∆ k . Analogously, by using (91), Proposition 4.2, and the trace theorem, we get (93). (cid:3) stimate of the first substitution errors. The first substitution errors can be rewritten as follows: e ′′ k = L ′ ( U k , Ψ k )( δU k , δ Ψ k ) − L ′ ( S θ k U k , S θ k Ψ k )( δU k , δ Ψ k )= Z L ′′ (cid:0) U a + S θ k U k + τ ( I − S θ k ) U k , Ψ a + S θ k Ψ k + τ ( I − S θ k )Ψ k (cid:1)(cid:0) ( δU k , δ Ψ k ) , (( I − S θ k ) U k , ( I − S θ k )Ψ k ) (cid:1) dτ, (94)˜ e ′′ k = (cid:0) B ′ ( U k , ϕ k )( δU k , δϕ k ) − B ′ ( S θ k U k , S θ k ϕ k )( δU k , δϕ k ) (cid:1) | x =0 = B ′′ (cid:0) ( δU k | x =0 , δϕ k ) , (( U k − S θ k U k ) | x =0 , ϕ k − S θ k ϕ k ) (cid:1) . (95) Lemma 4.5
Let α ≥ . There exist δ > sufficiently small, and θ ≥ sufficiently large, such that for all k = 0 , . . . n − , and for all integer s ∈ [6 , e α − , one has k e ′′ k k H s (Ω T ) ≤ Cδ θ L ( s ) − k ∆ k , (96) k ˜ e ′′ k k H s ( ∂ Ω T ) ≤ Cδ θ L ( s ) − k ∆ k , (97) where L ( s ) = max { ( s + 1 − α ) + + 6 − α, s + 5 − α } . Proof . It follows from (71), ( H n − ), (87), and (88) thatsup τ ∈ [0 , hh ( U a + S θ k U k + τ ( I − S θ k ) U k , ϕ a + S θ k ϕ k + τ ( I − S θ k ) ϕ k ) ii ≤ C ∗ for δ sufficiently small, i.e., we may apply Proposition 4.2 for estimating L ′′ in (94). Using again (71),( H n − ), (87), and (88), for s + 1 = α and s + 1 ≤ ˜ α we get k e ′′ k k H s (Ω T ) ≤ C n δ θ − αk ∆ k (cid:0) C ∗ + δθ ( s +1 − α ) + k + δθ s +1 − αk (cid:1) + δ θ s +3 − αk ∆ k o ≤ Cδ θ L ( s ) − k ∆ k . Similarly, but exploiting (89) instead of (88), for the case s + 1 = α we obtain k e ′′ k k H s (Ω T ) ≤ C n δ θ − αk ∆ k ( C ∗ + δ log θ k + δ ) + δ θ − αk ∆ k o ≤ Cδ ∆ k (cid:8) θ − αk + θ − αk (cid:9) ≤ Cδ θ L ( α − − k ∆ k for α ≥ k ˜ e ′′ k k H s (Ω T ) ≤ C n [ δU k ] s +1 , ∗ ,T k (1 − S θ k ) ϕ k k H ( ∂ Ω T ) + k δU k k H (Ω T ) k (1 − S θ k ) ϕ k k H s +1 ( ∂ Ω T ) + k ( I − S θ k ) U k ] H s +1 (Ω T ) k δϕ k k H ( ∂ Ω T ) + k ( I − S θ k ) U k k H (Ω T ) k δϕ k k H s +1 ( ∂ Ω T ) + k δU k k H s +1 (Ω T ) k ( I − S θ k ) U k k H (Ω T ) + k δU k k H (Ω T ) k ( I − S θ k ) U k k H s +1 (Ω T ) o . Then, ( H n − ) and (87) imply (97). (cid:3) Construction and estimate of the modified state.
Since the approximate solution satisfies the strictinequalities (6) (for all x ∈ Ω T ) and (21) (see Lemma 4.2) and since we shall require that the smooth modifiedstate vanishes in the past, the state ( U a + U n +1 / , ϕ a + ϕ n +1 / ) will satisfy (6) and (21) for a sufficientlyshort time T >
0. Therefore, while constructing the modified state we may focus only on constraint (25),i.e., the first boundary condition in (17). 23 roposition 4.3
Let α ≥ . The exist some functions U n +1 / and ϕ n +1 / , that vanish in the past, and suchthat ( U a + U n +1 / , ϕ a + ϕ n +1 / ) satisfies (25), and inequalities (6) and (21) for a sufficiently short time T .Moreover, these functions satisfy ϕ n +1 / = S θ n ϕ n , p n +1 / = S θ n p n , v j,n +1 / = S θ n v j,n ( j = 2 , , S n +1 / = S θ n S n , (98) and k U n +1 / − S θ n U n k H s (Ω T ) ≤ Cδθ s +1 − αn for s ∈ [3 , ˜ α + 3] . (99) for sufficiently small δ > and T > , and a sufficiently large θ ≥ . Proof . Actually, estimate (99) which we are going to prove hold for every s ≥ s ∈ [3 , ˜ α + 3]. Let ϕ n +1 / , the pressure p n +1 / , the entropy S n +1 / , and the tangential componentsof the velocity v n +1 / are defined by (98). We define v ,n +1 / as in [21]: v ,n +1 / := S θ n v ,n + R T G , where G = ∂ t ϕ n +1 / − ( S θ n v ,n ) | x =0 + X j =2 (cid:0) ( v aj + v j,n +1 / ) ∂ j ϕ n +1 / + v j,n +1 / ∂ j ϕ a (cid:1)(cid:12)(cid:12) x =0 , and R T : H s ( ∂ Ω T ) −→ H s +1 (Ω T ) is the lifting operator from the boundary to the interior. To get theestimate of v ,n +1 / − S θ n v ,n we use the following decompositions: G = S θ n B ( U n | x =0 , ϕ n ) − ∂ t (1 − S θ n ) ϕ n + (1 − S θ n ) ∂ t ϕ n + X j =2 (cid:0) ( v aj + S θ n v j,n ) ∂ j S θ n ϕ n − S θ n (( v aj + v j,n ) ∂ j ϕ n ) + ( S θ n v j,n ) ∂ j ϕ a − S θ n ( v j,n ∂ j ϕ a ) (cid:1)(cid:12)(cid:12) x =0 and B ( U n | x =0 , ϕ n ) = B v ( U n − | x =0 , ϕ n − ) + ∂ t ( δϕ n − )+ X j =2 (cid:0) ( v aj + v j,n − ) ∂ j ( δϕ n − ) + δv j,n − ∂ j ( ϕ a + ϕ n ) − δv ,n − (cid:1)(cid:12)(cid:12) x =0 , where B denotes the first row of the boundary operator B in (75).Exploiting point c ) of ( H n − ), one has kR T ( S θ n B ( U n − | x =0 , ϕ n − )) k H s (Ω T ) ≤ C k S θ n B ( U n − | x =0 , ϕ n − ) k H s ( ∂ Ω T ) ≤ Cθ s − αn kB ( U n − | x =0 , ϕ n − ) k H α ( ∂ Ω T ) for s ∈ [ α, ˜ α + 3] ,C kB ( U n − | x =0 , ϕ n − ) k H s +1 ( ∂ Ω T ) for s ∈ [3 , α − ≤ Cδθ s − αn for s ∈ [3 , ˜ α + 3] . Using (77) and point a ) of ( H n − ), we get kR T ( S θ n ∂ t ( δϕ n − )) k H s (Ω T ) ≤ C k S θ n ∂ t ( δϕ n − ) k H s ( ∂ Ω T ) ≤ Cθ s − n k δϕ n − k H ( ∂ Ω T ) ≤ Cθ s − n δθ − αn − θ − n − ≤ Cδθ s − α − n for s ∈ [3 , ˜ α + 3]. We also obtain kR T (cid:0) S θ n (( v aj + v j,n − ) | x =0 ∂ j ( δϕ n − )) (cid:1) k H s (Ω T ) ≤ Cθ s − n k ( v aj + v j,n − ) | x =0 ∂ j ( δϕ n − ) k H ( ∂ Ω T ) ≤ Cθ s − n n k δϕ n − k H ( ∂ Ω T ) k U a + U n − k H (Ω T ) + k δϕ n − k H ( ∂ Ω T ) k U a + U n − k H (Ω T ) o ≤ Cθ s − n δθ − αn C ∗ ≤ Cδθ s − α − n j = 2 , s ∈ [3 , ˜ α +3]. Estimating similarly the remaining terms containing in R T ( S θ n B ( U n | x =0 , ϕ n )),we finally obtain kR T ( S θ n B ( U n | x =0 , ϕ n )) k H s (Ω T ) ≤ Cδθ s − αn , s ∈ [3 , ˜ α + 3] . We now need to derive estimates for the remaining terms containing in R T G . For s ∈ [ α, ˜ α + 3] one has kR T ( − ∂ t (1 − S θ n ) ϕ n + (1 − S θ n ) ∂ t ϕ n ) k H s (Ω T ) ≤ C (cid:8) k ∂ t ( S θ n ϕ n ) k H s ( ∂ Ω T ) + k S θ n ( ∂ t ϕ n ) k H s ( ∂ Ω T ) (cid:9) ≤ C (cid:8) k S θ n ϕ n k H s +1 ( ∂ Ω T ) + θ s − αn k ϕ n k H α +1 ( ∂ Ω T ) (cid:9) ≤ Cδθ s +1 − αn , while for s ∈ [3 , ˜ α −
1] we obtain (recall that ˜ α = α + 3) kR T ( ∂ t (1 − S θ n ) ϕ n ) (cid:13)(cid:13) H s (Ω T ) ≤ Cδθ s +1 − αn , kR T ((1 − S θ n ) ∂ t ϕ n ) k H s (Ω T ) ≤ Cθ s − αn k ϕ n k H α +1 ( ∂ Ω T ) ≤ Cδθ s +1 − αn . Here we have, in particular, used Lemma 4.3. We do not get estimates for all the remaining terms containingin R T G and leave corresponding calculations to the reader. Collecting these estimates and the estimatesabove, we finally have k v ,n +1 / − S θ n v ,n k H s (Ω T ) ≤ Cδθ s +1 − αn , s ∈ [3 , ˜ α + 3] , that is equivalent to (99). (cid:3) Estimate of the second substitution errors.
The second substitution errors e ′′′ k = L ′ ( S θ k U k , S θ k Ψ k )( δU k , δ Ψ k ) − L ′ ( U k +1 / , Ψ k +1 / )( δU k , δ Ψ k )and ˜ e ′′′ k = (cid:0) B ′ ( S θ k U k , S θ k ϕ k )( δU k , δϕ k ) − B ′ ( U k +1 / , ϕ k +1 / )( δU k , δϕ k ) (cid:1) | x =0 can be written as e ′′′ k = Z L ′′ (cid:0) U a + U k +1 / + τ ( S θ k U k − U k +1 / ) , Ψ a + S θ k Ψ k ) (cid:0) ( δU k , δ Ψ k ) , ( S θ k U k − U k +1 / , (cid:1) dτ, (100)˜ e ′′′ k = B ′′ (cid:0) ( δU k | x =0 , δϕ k ) , (( S θ k U k − U k +1 / ) | x =0 , (cid:1) . (101)Employing (100) and (101), we get the following result. Lemma 4.6
Let α ≥ . There exist δ > , T > sufficiently small, and θ ≥ sufficiently large, such thatfor all k = 0 , . . . n − , and for all integer s ∈ [3 , e α − , one has k e ′′′ k k H s (Ω T ) ≤ Cδ θ L ( s ) − k ∆ k (102) and ˜ e ′′′ k = 0 , where L ( s ) = max { ( s + 1 − α ) + + 8 − α, s + 5 − α } . Proof . Using Lemma 4.3 and Proposition 4.3, we obtain the estimatesup τ ∈ [0 , hh ( U a + U k +1 / + τ ( S θ k U k − U k +1 / ) , ϕ a + S θ k ϕ k ) ii ≤ C ∗ δ sufficiently small, i.e., we may apply Proposition 4.2. Similarly, one gets hh ( U a + U k +1 / + τ ( S θ k U k − U k +1 / ) , ϕ a + S θ k ϕ k ) ii s +1 ≤ C (cid:8) C ∗ + δθ s +2 − αk + δθ ( s +1 − α ) + +1 k (cid:9) ≤ Cδθ ( s +1 − α ) + +1 n . Applying Proposition 4.2, we obtain (102): k e ′′′ k k H s (Ω T ) ≤ C n δθ ( s +1 − α ) + +1 k δθ − αk ∆ k δθ − αk + δθ s + − αk ∆ k δθ − αk + δθ − αk ∆ k δθ s +2 − αk o ≤ Cδ θ L ( s ) − k ∆ k . Using the explicit form of B ′′ , we easily get ˜ e ′′′ k = 0. (cid:3) Estimate of the last error term.
We now estimate the last error term D k +1 / δ Ψ k = δ Ψ k ∂ (Φ a + Ψ n +1 / ) R k , where R k := ∂ (cid:8) L ( U a + U k +1 / , Ψ a + Ψ k +1 / ) (cid:9) . Note that | ∂ (Φ a + Ψ n +1 / ) | = | ∂ (Ψ a + Ψ n +1 / ) | ≥ / , provided that T and δ are small enough. Lemma 4.7
Let α ≥ . There exist δ > , T > sufficiently small, and θ ≥ sufficiently large, such thatfor all k = 0 , . . . n − , and for all integer s ∈ [3 , e α − , one has k D k +1 / δ Ψ k k H s (Ω T ) ≤ Cδ θ L ( s ) − k ∆ k , (103) where L ( s ) = max { ( s + 2 − α ) + + 8 − α, ( s + 1 − α ) + + 9 − α, s + 6 − α } . Proof . The proof follows from the arguments as in [1, 5] (see also [21]). Using the Moser-type and embeddinginequalities, we obtain k [ D k +1 / δ Ψ k k H s (Ω T ) ≤ C n k δϕ k k H s ( ∂ Ω T ) k R k k H (Ω T ) + k δϕ k k H ( ∂ Ω T ) (cid:0) k R k k H s (Ω T ) + k R k k H (Ω T ) k ϕ a + ϕ k +1 / k H s ( ∂ Ω T ) (cid:1)o (104)(note that k ∂ (Ψ a +Ψ n +1 / ) k H s (Ω T ) ≤ C k ϕ a + ϕ k +1 / k H s ( ∂ Ω T ) ). To estimate R k we utilize the decomposition L ( U a + U k +1 / , Ψ a + Ψ k +1 / ) = L ( U k , Ψ k ) − f a + L ( U a + U k +1 / , Ψ a + Ψ k +1 / ) − L ( U a + U k , Ψ a + Ψ k ) = L ( U k , Ψ k ) − f a + Z L ′ (cid:0) U a + U k + τ ( U k +1 / − U k ) , Ψ a + Ψ k + τ (Ψ k +1 / − Ψ k ) (cid:1) ( U k +1 / − U k , Ψ k +1 / − Ψ k ) dτ. Clearly, k R k H s (Ω T ) ≤ kL ( U k , Ψ k ) − f a k H s (Ω T ) + sup τ ∈ [0 , k L ′ ( . . . )( . . . ) k H s +1 (Ω T ) (105)(for short we drop the arguments of L ′ ). It follows from point b ) of ( H n − ) that kL ( U k , Ψ k ) − f a k H s +1 (Ω T ) ≤ δθ s − αk (106)26or s ∈ [3 , ˜ α − L ′ similarly to L ′′ (see Proposition 4.2). One hassup τ ∈ [0 , hh ( U a + U k + τ ( U k +1 / − U k ) , ϕ a + ϕ k + τ ( ϕ k +1 / − ϕ k )) ii ≤ C ∗ for δ small enough. Then, omitting detailed calculations, we get the estimate k L ′ ( . . . )( . . . ) k H s +1 (Ω T ) ≤ Cδ ( θ s +3 − αk + θ ( s +2 − α ) + +5 − αk )for s ∈ [3 , ˜ α − k R k H s (Ω T ) ≤ Cδ ( θ s +3 − αk + θ ( s +2 − α ) + +5 − αk ) (107)for s ∈ [3 , ˜ α − s = ˜ α − k R k H s (Ω T ) ≤ k L ( U a + U k +1 / , Ψ a + Ψ k +1 / ) k H s +1 (Ω T ) ≤ C hh ( U a + ( U k +1 / − S θ n U k ) + S θ n U k , ϕ a + S θ n ϕ k ) ii s +2 ≤ Cδθ s +3 − αk . That is, we get estimate (107) for s ∈ [3 , ˜ α − α ≥ (cid:3) Convergence of the iteration scheme.
Lemmas 4.4–4.7 yield the estimate of e n and ˜ e n defined in (82)as the sum of all the errors of the k th step. Lemma 4.8
Let α ≥ . There exist δ > , T > sufficiently small, and θ ≥ sufficiently large, such thatfor all k = 0 , . . . n − , and for all integer s ∈ [3 , e α − , one has k e k k H s (Ω T ) + k ˜ e k k H s ( ∂ Ω T ) ≤ Cδ θ L ( s ) − k ∆ k , (108) where L ( s ) is defined in Lemma 4.7. Remark 4.2
In principle, we could try to use the advantage of the fact that in the tame estimate (43) wedo not lose derivatives from the source term f to the solution. To this end, in Lemma 4.8 we could estimateerrors e n and ˜ e n separately. However, this does not reduce the number of derivatives lost from the initialdata to the solution in the existence Theorem 2.1. In fact, we can even use a roughened version of estimate(43) in which we lose one derivative from f to the solution.Lemma 4.8 gives the estimate of the accumulated errors E n and e E n . Lemma 4.9
Let α ≥ . There exist δ > , T > sufficiently small, and θ ≥ sufficiently large, such that k E n k H α +2 (Ω T ) + k e E n k H α +2 ( ∂ Ω T ) ≤ Cδ θ n , (109) where L ( s ) is defined in Lemma 4.7. Proof . One can check that L ( α + 2) ≤ α ≥
7. It follows from (108) that hh ( E n , e E n ) ii α +2 ≤ n − X k =0 hh ( e k , e e k ) ii α +2 ≤ n − X k =0 Cδ ∆ k ≤ Cδ θ n for α ≥ α + 2 ∈ [3 , ˜ α − α ≥ α + 4. The minimal possible ˜ α is α + 4, i.e., our choice ˜ α = α + 4is suitable. (cid:3) We now derive the estimates of the source terms f n and g n defined in (84).27 emma 4.10 Let α ≥ . There exist δ > , T > sufficiently small, and θ ≥ sufficiently large, suchthat for all integer s ∈ [3 , e α + 1] , one has k f n k H s (Ω T ) ≤ C ∆ n (cid:8) θ s − α − n (cid:0) k f a k H α +1 (Ω T ) + δ (cid:1) + δ θ L ( s ) − n (cid:9) , (110) k g n k H s ( ∂ Ω T ) ≤ Cδ ∆ n (cid:0) θ L ( s ) − n + θ s − α − n (cid:1) . (111) Proof . It follows from (84) that f n = ( S θ n − S θ n − ) f a − ( S θ n − S θ n − ) E n − − S θ n e n − . Using (77), (79), (108), and (109), we obtain the estimates k ( S θ n − S θ n − ) f a k H s (Ω T ) ≤ Cθ s − α − n − k f a k H α +1 (Ω T ) ∆ n − , k ( S θ n − S θ n − ) E n − k| H s (Ω T ) ≤ Cθ s − α − n − k E n − k H α +2 (Ω T ) ∆ n − ≤ Cδ θ s − α − n − ∆ n − , k S θ n e n − k H s (Ω T ) ≤ Cδ θ L ( s ) − n ∆ n − . Using the inequalities θ n − ≤ θ n ≤ √ θ n − , θ n − ≤ θ n , and ∆ n − ≤ n , from the above estimates wededuce (110). Similarly, we get (111). (cid:3) We are now in a position to obtain the estimate of the solution to problem (80) by employing the tameestimate (43). Then the estimate of ( δU n , δϕ n ) follows from formula (81). Lemma 4.11
Let α ≥ . There exist δ > , T > sufficiently small, and θ ≥ sufficiently large, suchthat for all integer s ∈ [3 , e α ] , one has k δU n k H s (Ω T ) + k δϕ n k H s ( ∂ Ω T ) ≤ δθ s − α − n ∆ n . (112) Proof . Without loss of generality we can take the constant K appearing in estimate (43) that K = 2 C ∗ ,where C ∗ is the constant from (71). In order to apply Theorem 3.1, by using (88) and (99), we check that k U a + U n +1 / k H (Ω T ) + k ϕ a + S θ n ϕ n k H ( ∂ Ω T ) ≤ C ∗ for α ≥ δ small enough. That is, assumption (42) is satisfied for the coefficients of problem (80). Byapplying the tame estimate (43), for T small enough one has k δ ˙ U n k H s (Ω T ) + k δϕ n k H s ( ∂ Ω T ) ≤ C n k f n k H s (Ω T ) + k g n k H s +1 ( ∂ Ω T ) + (cid:0) k f n k H (Ω T ) + k g n k H ( ∂ Ω T ) (cid:1)(cid:0) k U a + U n +1 / k H s +3 (Ω T ) + k ϕ a + S θ n ϕ n k H s +3 ( ∂ Ω T ) (cid:1)o . (113)Using Moser-type inequalities, from formula (81) we obtain k δU n k H s (Ω T ) ≤ k δ ˙ U n k H s (Ω T ) + C (cid:8) k δϕ n k H s ( ∂ Ω T ) + k δϕ n k H ( ∂ Ω T ) k ϕ a + S θ n ϕ n k H s ( ∂ Ω T ) (cid:9) . Then (113) yields k δU n k H s (Ω T ) + k δϕ n k H s ( ∂ Ω T ) ≤ C n k f n k H s (Ω T ) + k g n k H s +1 ( ∂ Ω T ) + (cid:0) k f n k H (Ω T ) + k g n k H ( ∂ Ω T ) (cid:1)(cid:0) k U a + U n +1 / k H s +3 (Ω T ) + k ϕ a + S θ n ϕ n k H s +3 ( ∂ Ω T ) (cid:1)o (114)28or all integer s ∈ [6 , e α ]. Below we can actually use a roughened version of (114) (see Remark 4.2). ApplyingLemma 4.11, (88), and Proposition 4.3, from (114) we derive the estimate k δU n k H s (Ω T ) + k δϕ n k H s ( ∂ Ω T ) ≤ C (cid:8) θ s − α − n (cid:0) k f a k H α +1 (Ω T ) + δ (cid:1) + δ θ L ( s +1) − n (cid:9) ∆ n + Cδ ∆ n (cid:8) θ − αn (cid:0) k f a k H α +1 (Ω T ) + δ (cid:1) + δ θ − αn (cid:9)(cid:8) C ∗ + θ ( s +3 − α ) + n + θ s +4 − αn (cid:9) . (115)Exactly as in [5], we can check that the inequalities L ( s + 1) ≤ s − α, ( s + 3 − α ) + + 2 − α ≤ s − α − , ( s + 3 − α ) + + 9 − α ≤ s − α − ,s + 6 − α ≤ s − α − , s + 13 − α ≤ s − α − α ≥ s ∈ [3 , ˜ α ]. Thus, (115) and (73) yield k δU n k H s (Ω T ) + k δϕ n k H s ( ∂ Ω T ) ≤ C (cid:0) δ ( T ) + δ (cid:1) θ s − α − n ∆ n ≤ δθ s − α − n ∆ n for δ and T small enough. (cid:3) Remark 4.3
As we can see, Lemma 4.11 with ˜ α = α + 4 is absolutely analogous to Lemma 16 in [5]. In thissense, the “gain of one derivative for the front” in the tame estimate gives no advantage in the realization ofthe Nash-Moser method. This is caused by the fact that even if in point a) of ( H n − ) we had the H s +1 –normof δϕ k we could never use this advantage before the proof of Lemma 4.11.Inequality (112) is point a ) of ( H n ). It remains to prove points b ) and c ) of ( H n ). Lemma 4.12
Let α ≥ . There exist δ > , T > sufficiently small, and θ ≥ sufficiently large, suchthat for all integer s ∈ [3 , e α − kL ( U n , Ψ n ) − f a k H s (Ω T ) ≤ δθ s − α − n . (117) Moreover, for all integer s ∈ [4 , α ] one has kB ( U n | x =0 , ϕ n ) k H s ( ∂ Ω T ) ≤ δθ s − α − n . (118) Proof . One can show that L ( U n , Ψ n ) − f a = ( S θ n − − I ) f a + ( I − S θ n − ) E n − + e n − . (119)For s ∈ [ α + 1 , ˜ α − k ( I − S θ n − ) f a k H s (Ω T ) ≤ θ s − α − n ( C k f a k H α +1 (Ω T ) + k f a k H s (Ω T ) ) ≤ Cδ ( T ) θ s − α − n , while for s ∈ [3 , α + 1], applying (78), we get k ( I − S θ n − ) f a k H s (Ω T ) ≤ Cθ s − α − n − k f a k H α +1 (Ω T ) ≤ Cδ ( T ) θ s − α − n . Lemma 4.9 and (78) imply k ( I − S θ n − ) E n − k H s (Ω T ) ≤ Cθ s − α − n − k E n − k H α +2 (Ω T ) ≤ Cδ θ s − α − n ≤ s ≤ α + 2 = ˜ α − k e n − k H s (Ω T ) ≤ Cδ θ L ( s ) − n − ∆ n − ≤ Cδ θ L ( s ) − n ≤ Cδ θ s − α − n . From the above estimates and decomposition (119), by choosing
T > δ > B ( U n | x =0 , ϕ n ) = ( I − S θ n − ) e E n − + ˜ e n − , we can prove estimate (118). (cid:3) As follows from Lemmas 4.11 and 4.12, we have proved that ( H n − ) implies ( H n ), provided that α ≥ α = α + 4, the constant θ ≥ T > δ > α , δ , and θ , we prove ( H ). Lemma 4.13
If the time
T > is sufficiently small, then ( H ) is true. Proof . We recall that ( U , f ) = 0. Then, by the definition of the approximate solution in Lemma 4.2 thestate ( U a + U , ϕ a + ϕ ) = 0 satisfies already (6), (17), and (21). That is, it follows from the construction ofProposition 4.3 that ( U / , ϕ / ) = 0. Consequently, ( δ ˙ U , δϕ ) solves the linear problem (31)–(33) with thecoefficients ( b U , ˆ ϕ ) = ( U a , ϕ a ) and the source terms f = S θ f a and g = 0. Thanks to (71) the assumption(42) is satisfied (recall that K = 2 C ∗ ). Applying (43), we get the estimate k δ ˙ U k H s (Ω T ) + k δϕ k H s ( ∂ Ω T ) ≤ C k S θ f a k H s +1 (Ω T ) . Together with (74) and formula (81) this estimate yields k δU k H s (Ω T ) + k δϕ k H s ( ∂ Ω T ) ≤ C k S θ f a k H s +1 (Ω T ) ≤ Cθ ( s − α ) + δ ( T ) ≤ δθ s − α − ∆ for all integer s ∈ [3 , ˜ α ], provided that T is sufficiently small. Likewise, points b ) and c ) of ( H ) can beshown to be satisfied for a sufficiently short time T > (cid:3) (cid:3) The proof of Theorem 2.1.
We consider initial data ( U , ϕ ) ∈ H m +7 ( R ) × H m +7 ( R ) satisfyingall the assumptions of Theorem 2.1. In particular, they satisfy the compatibility conditions up to order µ = m + 7 (see Definition 4.1). Then, thanks to Lemmas 4.1 and 4.2 we can construct an approximatesolution ( U a , ϕ a ) ∈ H m +8 (Ω T ) × H m +8 ( ∂ Ω T ) that satisfies (71). As follows from Lemmas 4.11–4.13, ( H n )holds for all integer n ≥
0, provided that α ≥
7, ˜ α = α + 4, the constant θ ≥ T > δ > H n ) implies ∞ X n =0 (cid:8) k δU n k H m (Ω T ) + k δϕ n k H m ( ∂ Ω T ) (cid:9) ≤ ∞ . Hence, the sequence ( U n , ϕ n ) converges in H m (Ω T ) × H m ( ∂ Ω T ) to some limit ( U, ϕ ). Recall that m = α − ≥
6. Passing to the limit in (117) and (118) with s = m , we obtain (74)–(76). Consequently, U := U + U a , ϕ := ϕ + ϕ a is a solution of problem (20), (17), (18). This completes the proof of Theorem 2.1.30 Free boundary problem in relativistic gas dynamics: special andgeneral relativity
Let us first write down a suitable symmetric form of the relativistic Euler equations. First of all, we notethat for the set of covariant laws (8) we have the supplementary covariant law ∇ α ( ρSu α ) = 0 (120)that arises as a consequence of (8) and the first principle of thermodynamics. In the setting of specialrelativity (120) becomes the entropy conservation law ∂ t ( ρ Γ S ) + div ( ρSu ) = 0 . (121)In principle, taking into account (121) and using Godunov’s symmetrization method, we can rewrite system(10)–(12) for the unknown U = ( p, u, S ) as a symmetric system for a new (canonical) unknown Q and thenreturn to the original unknown U keeping the symmetry property: A ( U ) ∂ t U + A j ( U ) ∂ j U + Q ( U ) = 0 , (122)where A α = ( A α ) T , ∂ j = ∂/∂x j , and Q ( U ) = − (0 , − ρ G , A α were written for the special case u = u = 0. Such a procedure is absolutelyalgorithmic and always works, but it is however connected with very long calculations. Therefore, here weprefer to symmetrize the conservation laws (10)–(12) by rewriting them in a suitable nonconservative form.Equations (10) and (121) imply d S d t = 0 , (123)where d / d t = ∂ t + ( v, ∇ ) is the material derivative as for the non-relativistic case (4). Using (123), we firstrewrite (10) in a nonconservative form. Combining then (11) and (12) and employing again (123), we finallyget the relativistic counterpart of system (4):Γ ρc d p d t + ( v, ∂ t u ) + div u = 0 , ( ρh Γ) (cid:18) d u d t − v (cid:18) v, d u d t (cid:19)(cid:19) + ( ∂ t p ) v + ∇ p = ρ G , d S d t = 0 , (124)where c = ( p ρ ( ρ, S )) / . System (124) being written in the quasilinear form (122) is already symmetric with A = Γ ρc v T v ρh Γ B
00 0 1 , A j = u j ρc e T j e j ρhu j B
00 0 v j , (125)where B = ( b ij ) , b ij = δ ij − v i v j , e j = ( δ j , δ j , δ j ), and a T is the vector-row for a correspondingvector-column a (recall also that u j = Γ v j ). The matrix A > relativistic causality condition0 < c s < , (126)31here c s is the relativistic speed of sound, c s = c /h . Of course, (126) will be an additional restriction onthe initial data in a counterpart of Theorem 2.1.Now, for system (122), (125) in the domain (13) endowed with the boundary conditions (14) we canliterally repeat arguments of Sections 2–4. The only important point is that the boundary matrix A on theboundary x = 0 for system (34) written now for matrices (125) and V = ( ˙ p, ˙ u n , ˙ u , ˙ u , ˙ S ) coincides withthe matrix A | x =0 in (35), where˙ u n := b Γ ˙ v n , ˙ v n = ˙ v − ˙ v ∂ b Ψ − ˙ v ∂ b Ψ , b Γ = (1 + | ˆ u | ) / , ˆ v = ˆ u/ b Γ , (127)˙ U = ( ˙ p, ˙ u, ˙ S ) is the “good unknown”, b U = (ˆ p, ˆ u, b S ) is the basic state, and ˙ v = ( ˙ v , ˙ v , ˙ v ) is defined from theformula ˙ u = b Γ ˙ v + b Γˆ u (ˆ u, ˙ v ) (128)suggested by the relation between the perturbations δu and δv .Indeed, we easily compute: e A ( b U , b Ψ) = 1 ∂ b Φ b Γˆ f ˆ ρ ˆ c ˆ a T a ˆ ρ ˆ h b Γˆ f b B
00 0 ˆ f , where ˆ a = (1 − ˆ v ∂ t b Ψ , − ∂ b Ψ − ˆ v ∂ t b Ψ , − ∂ b Ψ − ˆ v ∂ t b Ψ) , ˆ f = ˆ v − ˆ v ∂ b Ψ − ˆ v ∂ b Ψ − ∂ t b Ψ , and b B is the matrix B calculated for the basic state. Taking into account (25), (127), and (128), we haveˆ f | x =0 = 0 and ( ∂ b Φ) ( e A ( b U , b Ψ) ˙
U , ˙ U ) | x =0 = 2 ˙ p | x =0 (cid:16) ˙ u − ˙ u ∂ b Ψ − ˙ u ∂ b Ψ − (ˆ v, ˙ u ) ∂ t ˆ ϕ (cid:17)(cid:12)(cid:12)(cid:12) x =0 = 2 ˙ p | x =0 (cid:16)b Γ ˙ v n + (ˆ u, ˙ v ) ∂ t ˆ ϕ ( b Γ − − | ˆ u | ) (cid:17)(cid:12)(cid:12)(cid:12) x =0 = 2( ˙ p ˙ u n ) | x =0 = ( A (1) V | x =0 , V | x =0 )(the matrix A (1) was defined in (35)). Then( e A ( b U , b Ψ) ˙
U , ˙ U ) | x =0 = ( e A ( b U , b Ψ) JV, JV ) | x =0 = ( J T e A ( b U , b Ψ) JV, V ) | x =0 = ( A V, V ) | x =0 , where the matrix A | x =0 is the same as in Section 2 and the transition matrix J can be easily writtendown. Thus, we obtain the local-in-time existence (and uniqueness) theorem for the relativistic version ofproblem (16)–(18) (in the framework of special relativity) in the form of Theorem 2.1. Clearly, we shouldalso supplement conditions (6) with (126) while writing assumptions on the initial data. It means that theinitial data should satisfy inf x ∈ R (cid:8) ρ ( p , S ) , ρ p ( p , S ) , c s ( p , S ) , − c s ( p , S ) (cid:9) > , where c s ( p , S ) = 1 ρ p ( p , S ) h ( p , S ) , h ( p , S ) = 1 + e ( ρ ( p , S ) , S ) + p ρ ( p , S ) . Let us now briefly discuss the case of general relativity. The metric g appearing in the relativistic Eulerequations (9) should satisfy the Einstein equations G αβ = κT αβ . As in [6], following Rendall [18] andintroducing g αβγ := ∂ γ g αβ ,
32e write the Einstein equations in harmonic coordinates as − g ∂ t g αβ − g i ∂ i g αβ − g ij ∂ i g αβj + 2 H αβ ( g γδ , g γδσ ) = κ (2 T αβ − T γγ g αβ ) ,g ij ∂ t g αβi − g ij ∂ i g αβi = 0 ,∂ t g αβ − g αβ = 0 . (129)System (129) written in the compact form B ( W ) ∂ t W + B j ( W ) ∂ j W + Q ( W, U ) = 0 (130)is symmetric for the vector W whose components are g αβ and g αβγ . Recall that U = ( p, u, S ). The symmetricsystem (130) is hyperbolic if g < g ij ) > g . This was done by Rendall [18] for isentropic fluids. In the general case we can however just repeatarguments from [18] by taking into account the entropy law (121) which has form (123) for constant metrics.Roughly speaking, the calculations in [18] are just a “tensor” variant of our simple calculations towardsobtaining the nonconservative form (124). With reference to [18], we write equations (9) for a fixed constantmetric g in the symmetric form (122), (125) with B = ( b ij ) , b ij = g ij + g i v j + g j v i + g v i v j = g ij + g i u j u + g j u i u + g u i u j ( u ) . For a non-fixed metric g the balance laws (9) are written as the symmetric system A ( U ) ∂ t U + A j ( U ) ∂ j U + B ( U, W ) = 0 . (131)It is worth noting that for system (131) for any fixed (and not necessarily constant) metric we can prove acounterpart of Theorem 2.1 under suitable assumptions on W .Now we consider the free boundary problem for the symmetric hyperbolic system (131), (130) with theboundary conditions (14). However, in the setting of general relativity it is actually an interface problembecause we should consider system (130) for the metric variables not only in the domain Ω( t ) but also inthe vacuum region R \ Ω( t ) = { x < ϕ ( t, x , x ) } . As was shown in [8], the jump conditions on an interfaceΣ( t ) written for the Einstein tensor are satisfied if the metric g is smooth on this interface, i.e,[ W ] = W + − W − = 0 on Σ( t ) . (132)In our case W + and W − are the metric variables in the fluid domain Ω( t ) and the vacuum region R \ Ω( t )respectively. Constraints on the initial data under which condition (132) is not only sufficient but alsonecessary for the fulfillment of the jump conditions for the Einstein tensor are discussed in [6] and connectedwith the notion of so-called natural coordinates [8]. That is, as for shock waves in general relativity studiedin [6], we will treat our problem in harmonic natural coordinates .Thus, we have the symmetric hyperbolic systems A ( U ) ∂ t U + A j ( U ) ∂ j U + B ( U, W + ) = 0 in Ω( t ) , (133) B ( W + ) ∂ t W + + B j ( W + ) ∂ j W + + Q ( W + , U ) = 0 in Ω( t ) , (134) B ( W − ) ∂ t W − + B j ( W + ) ∂ j W − + Q ( W − ,
0) = 0 in R \ Ω( t ) (135)endowed with the boundary conditions (14) and (132) on a time-like hypersurface Σ( t ) = { x = ϕ ( t, x , x ) } .Here (135) is the symmetric form of the vacuum Einstein equations. We reduce problem (133)–(135), (14),33132) to the fixed domain R by straightening the free surface Σ: e U ( t, x ) := U ( t, Φ + ( t, x ) , x ′ ) , f W ± := W ± ( t, Φ ± ( t, x ) , x ′ )Φ ± ( t, x ) := ± x + Ψ ± ( t, x ) , Ψ ± ( t, x ) := χ ( ± x ) ϕ ( t, x ′ ) , x ′ = ( x , x )(the cut-off function χ ( x ) was described in the beginning of Section 2).Regarding further arguments towards the proof of the local-in-time existence theorem for the reducedproblem in the domain R , we give here only a rough scheme or even an idea of this proof and postponedetailed arguments to a future work. The main idea is the following. The existence of solutions of problem(133), (14) reduced to the fixed domain R is proved by Nash-Moser iterations for any fixed metric g . Theboundary conditions (132) are linear and, therefore, we do not need introduce source terms for them in thelinearized problem. Moreover, for the linearized problem these boundary conditions are dissipative . Though,they are not strictly dissipative, but the crucial point is that they are homogeneous. Hence, we can prove theexistence of solutions to the reduced problem for (134), (135), (132) in [0 , T ] × R by the classical fixed-pointargument for any fixed fluid unknown U . Then, the existence of solutions to the whole problem (133)–(135),(14), (132) reduced to the fixed domain R is proved by Nash-Moser iterations for the “fluid” part of theproblem whereas at each Nash-Moser iteration step the metric g is found as a solution of the problem whoselinear version has maximally dissipative boundary conditions. More presicely, at each ( n + 1)th iterationstep before solving the linear problem for δ ˙ U n with W + = W + n we find W ± n as a unique solution of thecorresponding problem for W ± with U = U n and ϕ = ϕ n taken from the n th iteration step. At last, wenote that the constraints [6] on the initial data connected with the introduction of natural coordinates arenot needed to be satisfied at each Nash-Moser iteration step and we may therefore not care about them. Acknowledgements
The principal part of this work was done during the short stay of the author at thethe Department of Mathematics and Statistics of the University of Konstanz. The author gratefully thanksHeinrich Freist¨uhler for his kind hospitality and many helpful discussions during this visit.
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