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This work is devoted to establishing the local-in-time well-posedness of strong solutions to the three-dimensional compressible primitive equations of atmospheric dynamics. It is shown that strong solutions exist, are unique, and depend continuously on the initial data, for a short time in two cases: with gravity but without vacuum, and with vacuum but without gravity. © 2021, The Author(s). Introduction The Compressible Primitive EquationsThe general hydrodynamic and thermodynamic equations (see, e.g., [36]) with Coriolis force and gravity are used to model the motion and state of the atmosphere, which is a specific compressible fluid. However, such equations are extremely complicated and prohibitively expensive computationally. However, since the vertical scale of the atmosphere is significantly smaller than the planetary horizontal scale, the authors in [17] take advantage, as it is commonly done in planetary scale geophysical models, of the smallness of this aspect ratio between these two orthogonal directions to formally derive the compressible primitive equations (CPE) from the compressible Navier-Stokes equations. Specifically, in the CPE the vertical component of the momentum in the compressible Navier-Stokes equations is replaced by the hydrostatic balance equation (1.1) 3 , below, which is also known as the quasi-static equilibrium equation. It turns out that the hydrostatic approximation equation is accurate enough for practical applications and has become a fundamental equation in atmospheric science. It is the starting point of many large scale models in the theoretical investigations and practical weather predictions (see, e.g., [35]). This has also been observed by meteorologists (see, e.g., [39,44]). In fact, such an approximation is reliable and useful in the sense that the balance of gravity and pressure dominates the dynamic in the vertical direction and that the vertical velocity is usually hard to observe in reality. In many simplified models, it is assumed that the atmosphere is under adiabatic process and therefore the entropy remains unchanged along the particle path. In particular, if the entropy is constant in the spatial variables initially, it remains so in later time. On the other hand, instead of the molecular viscosity, eddy viscosity is used to model the statistical effect of turbulent motion in the atmosphere. The observations above and more perceptions from the meteorological point of view can be found in [39,Chapter 4]. Therefore, under the above assumptions, one can write down the isentropic compressible primitive equations as in (1.1), below. Moreover, we also study the problem by further neglecting the gravity in (1.2), below. We remark here that, although it does not cause any additional difficulty, we have omitted the Coriolis force in this work for the convenience of presentation. That is, the local well-posedness theorems still work for systems (1.1) and (1.2) with the Coriolis force.The first mathematical treatment of the compressible primitive equations (CPE) can be tracked back to Lions, Temam and Wang [35]. Actually, the authors formulated the compressible primitive equations in the pressure coordinates ( p-coordinates) and show that in the new coordinate system, the equations are in the form of classical primitive equations (called primitive equations, or PE hereafter) with the incompressibility condition. In yet another work [34], the authors modeled the nearly incompressible ocean by the PE. It is formulated as the hydrostatic approximation of the Boussinesq equations. The authors show the existence of global weak solutions and therefore indirectly study the CPE (see, e.g., [32,33] for additional work by the authors). Notably, the PE have been the subject of intensive mathematical research. For instance, Guillén-González, Masmoudi and Rodríguez-Bellido in [21] study the local existence of strong solutions and global existence of strong solutions for small initial data to the PE. In [47] the authors address the global existence of strong solutions to PE in a domain with small depth for restricted large initial data depending on the depth. In [38], the authors study the Sobolev and Gevrey regularity of the solutions to PE. The first breakthrough concerning the global well-posedness of PE is obtained by Cao and Titi in [8], in which the authors show the existence of unique global strong solutions (see, also, [9,[22][23][24][25][27][28][29]46] and the references therein for related study). On the other hand, with partial anisotropic diffusion and viscosity, Cao, Li and Titi in [3][4][5][6][7]10] establish the global well-posedness of strong solutions to PE. For the inviscid primitive equations, or hydrostatic incompressible Euler equations, in [1,26,37], the authors show the short time existence of solutions in the analytic function space and in H s space. More recently, the authors in [2,45] construct finite-time blowup for the inviscid PE in the absence of rotation. Also, in [20], the authors establish the Gevrey regularity of hydrostatic Navier-Stokes equations with only vertical viscosity.Despite the fruitful study of the primitive equations, it still remains interesting to study the compressible equations. On the one hand, it is a more direct model to study the atmosphere and perform practical weather predictions. On the other hand, the former deviation of the PE from the CPE in the p-coordinates did not treat the corresponding derivation of the boundary conditions. In fact, due to the change of pressure on the boundary, the appropriate studying domain for the PE should be evolving together with the flows in order to recover the solutions to the CPE. Thus, even though the formulation of the PE significantly simplifies the equations of the CPE, the boundary conditions are more complicated than before in order to study the motion of the atmosphere. We believe that this might be one of the reasons that is responsible for the not-completely successful prediction of the weather by using the PE.Recently, Gatapov, Kazhikhov, Ersoy, Ngom construct a global weak solution to some variant of two-dimentional compressible primitive equations in [16,19]. Meanwhile, Ersoy, Ngom, Sy, Tang, Gao study the stability of weak solutions to the CPE in [17,41] in the sense that a sequence of weak solutions satisfying some entropy conditions contains a subsequence converging to another weak solution. In recent work, we show the existence of such weak solutions in [31]. See also [43].In this and subsequent works, we aim to address several problems concerning the compressible primitive equations. In this work, we start by studying the local well-posedness of strong solutions to the CPE. That is, we will establish the local strong solutions to (1.1) and (1.2), below, in the domain = h × (0, 1), withh = T 2 = [0, 1] 2 ⊂ R 2 beingthe fundamental periodic domain. In comparison with the compressible Navier-Stokes equations [18], the absence of evolutionary equations for the vertical velocity (vertical momentum) causes the main difficulty. This is the same difficulty as in the case of the PE. In fact, the procedure of recovering the vertical velocity is a classical one in the modeling of the atmosphere [39,Chapter 5]. This is done with the help of the hydrostatic equation, which causes the stratification of density profiles in the CPE. On the one hand, in (1.1), as one will see later, the hydrostatic equation implies that if there is vacuum in the physical domain , the sound speed will be at most 1/2-Hölder continuous. Thus the H 2 estimate of the density is not available in the presence of vacuum. However in (1.2), such an obstacle no longer exists. For this reason, the local well-posedness established in this work doesn't allow vacuum in the presence of gravity, but vacuum is allowed in the case without gravity. On the other hand, the hydrostatic equation does have some benefits. Indeed, such a relation yields that the density admits a stratified profile along the vertical direction. This fact will help us recover the vertical velocity from the continuity equation (see (1.8) and (1.13), below).In this work, we will first reformulate the compressible primitive equations (1.1), (1.2) by making use of the stratified density profile. Then we will study the local well-posedness of the reformulated systems under the assumption that there is no vacuum initially. This is done via a fixed point argument. Next, in order to obtain the existence of strong solutions to (1.2) with non-negative density, we establish some uniform estimates independent of the lower bound of the density. We point out that in comparison to the compressible Navier-Stokes equations (see, e.g., [12][13][14][15]), we will require H 2 estimate of ρ 1/2 in order to derive the above mentioned uniform estimates. Such estimates are not available in the case with gravity (1.1). To this end, continuity arguments are used to establish the solutions with vacuum. We also study the continuous dependence on the initial data and the uniqueness of the strong solutions.Through out this work, we will use x := (x, y, z) , x h := (x, y) to represent the coordinates in and h , respectively. In addition, we will use the following notations to denote the differential operators in the horizontal direction:∇ h := (∂ x , ∂ y ) , ∂ h ∈ {∂ x , ∂ y }, div h := ∇ h •, h := div h ∇ h .The isentropic compressible primitive equations with gravity are governed by the system⎧ ⎪ ⎪ ⎪ ⎨ ⎪ ⎪ ⎪ ⎩ ∂ t ρ + div h (ρv) + ∂ z (ρw) = 0 i n , ∂ t (ρv) + div h (ρv ⊗ v) + ∂ z (ρwv) + ∇ h P = μ h v + μ∂ zz v + (μ + λ)∇ h div h v in , ∂ z P − ρg = 0 i n , (1.1)with P := ρ γ . We will study in this work only the case when γ = 2 in (1.1) for the sake of simplifying our presentation. For general γ > 1, we refer to Remark 1, below.On the other hand, the isentropic compressible primitive equations without gravity are governed by the system⎧ ⎪ ⎪ ⎪ ⎨ ⎪ ⎪ ⎪ ⎩ ∂ t ρ + div h (ρv) + ∂ z (ρw) = 0 i n , ∂ t (ρv) + div h (ρv ⊗ v) + ∂ z (ρwv) + ∇ h P = μ h v + μ∂ zz v + (μ + λ)∇ h div h v in , ∂ z P = 0 i n , (1.2)with P := ρ γ and γ > 1.In the above systems, (1.1) and (1.2), the viscosity coefficients μ, λ are assumed to be strictly positive. Also, (1.1) and (1.2) are supplemented with the following boundary conditions:w = 0, ∂ z v = 0 on h × {0, 1}.(1.3)The rest of this paper will be organized as follows: in section 1.2, we present a reformulation of (1.1) and (1.2) by making use of the stratified density profiles. Also, we present the formula for recovering the vertical velocity and the main theorems of this work. After listing some useful inequalities and notations, we study in section 2 the existence theory. Next, in section 3 we show the continuous dependence on the initial data and the uniqueness of strong solutions. Reformulation, Analysis and Main TheoremsIn this section, we will reformulate (1.1) and (1.2) and point out how to recover the vertical velocity in terms of the density and the horizontal velocity.The Case with Gravity and γ = 2 We first consider (1.1). From (1.1) 3 , one hasρ γ −1 ( x, t) = γ − 1 γ gz + ρ γ −1 ( x h , 0, t).Denote by ξ = ξ( x h , t) := ρ γ −1 ( x h , 0, t). The continuity equation (1.1) 1 implies∂ t ξ + v • ∇ h ξ + (γ − 1) ξ + γ − 1 γ gz (div h v + ∂ z w) + γ − 1 γ gw = 0.(1.4)In particular, since γ = 2, we have ρ( x, t) = ξ( x h , t) + 1 2 gz and (1.1) can be written as⎧ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎨ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎩ ∂ t ξ + v • ∇ h ξ + ξ + 1 2 gz div h v + ∂ z ξw + 1 2 gzw = 0 in , ξ + 1 2 gz (∂ t v + v • ∇ h v + w∂ z v) + (2ξ + gz)∇ h ξ = μ h v + μ∂ zz v + (μ + λ)∇ h div h v in , ∂ z ξ = 0 i n . (1.5)Hereafter, we denote, for any f : → R,f := 1 0 f dz, f := f − f . (1.6)Then averaging over the vertical variable in (1.5) 1 yields, thanks to (1.3),∂ t ξ + v • ∇ h ξ + ξ div h v + 1 2 gzdiv h v = 0. (1.7)Then comparing (1.7) with (1.5) 1 implies∂ z (ρw) = ∂ z ξw + 1 2 gzw = − v • ∇ h ξ − ξ div h v − g 2 zdiv h v in .Therefore, the vertical velocity w is determined, thanks to the boundary condition (1.3), by the relationρw = ξ + 1 2 gz w = − z 0 div h (ξ v) + g 2 zdiv h v dz. (1.8)System (1.5) is complemented with the initial data(ξ, v)| t=0 = (ξ 0 , v 0 ),(1.9)with ξ 0 , v 0 ∈ H 2 ( ). Also the following compatible conditions are imposed:ρ 0 = ξ 0 + 1 2 gz ρ > 0 in , and ∂ z v 0 | z=0,1 = 0, μ h v 0 + μ∂ zz v 0 + (μ + λ)∇ h div h v 0 − (2ξ 0 + gz)∇ h ξ 0 − ρ 0 v 0 • ∇ h v 0 − ρ 0 w 0 ∂ z v 0 =: ρ 0 V 1 , with V 1 ∈ L 2 ( ),and ρ 0 w 0 = − z 0 div h (ξ 0 v 0 ) + g 2 zdiv h v 0 dz.(1.10) Also, we will denote the boundsξ 0 2 H 2 B g,1 , v 0 2 H 2 + V 1 2 L 2 B g,2 .(1.11)Theorem 1. Suppose the initial data (ρ 0 , v 0 ) = (ξ 0 + 1 2 gz, v 0 ) satisfy (1.11) and the compatible conditions (1.10). Then there is a unique strong solution (ρ, v) to system (1.1), with the boundary condition (1.3), in × (0, T ), for some positive constant T = T (B g,1 , B g,2 , ρ) > 0. Also, the solution satisfiesρ ∈ L ∞ (0, T ; H 2 ( )), ∂ t ρ ∈ L ∞ (0, T ; H 1 ( )), v ∈ L ∞ (0, T ; H 2 ( )) ∩ L 2 (0, T ; H 3 ( )), ∂ t v ∈ L ∞ (0, T ; L 2 ( )) ∩ L 2 (0, T ; H 1 ( )).Furthermore, for some positive constantC(B g,1 , B g,2 , ρ), inf ( x,t)∈ ×(0,T ) ρ( x, t) 1 2 ρ > 0, sup 0 t T ρ(t) 2 H 2 + ∂ t ρ(t) 2 H 1 + v(t) 2 H 2 + ∂ t v(t) 2 L 2 + T 0 v(t) 2 H 3 + ∂ t v(t) 2 H 1 dt C(B g,1 , B g,2 , ρ).Moreover, for any two solutions (ρ i , v i ), i = 1, 2 with initial data (ρ i,0 , v i,0 ), i = 1, 2 satisfying the conditions mentioned above, we have the following inequalityρ 1 − ρ 2 L ∞ (0,T ;L 2 ( )) + v 1 − v 2 L ∞ (0,T ;L 2 ( )) + ∇(v 1 − v 2 ) L 2 (0,T ;L 2 ( )) C μ,λ,B g,1 ,B g,2 ,ρ,T × ρ 1,0 − ρ 2,0 L 2 ( )) + v 1,0 − v 2,0 L 2 ( )) ,for some positive constant C μ,λ,B g,1 ,B g,2 ,ρ,T .Remark 1. For general γ > 1, after multiplying (1.4) withξ + γ −1 γ gz ξ 2−γ γ −1and averaging the resultant in the z-variable, one obtainsγ g (ξ + γ − 1 γ g) 1 γ −1 ξ γ −2 γ −1 − ξ ∂ t ξ + ξ γ −2 γ −1 ξ + γ − 1 γ gz 2−γ γ −1 v • ∇ h ξ + (γ − 1)ξ γ −2 γ −1 ξ + γ − 1 γ gz 1 γ −1 div h v = 0 in h .Consequently, by eliminating ∂ t ξ from the above equation and (1.4), it follows that(γ − 1)∂ z ξ + γ − 1 γ gz 1 γ −1 w = −(γ − 1) ξ + γ − 1 γ gz 1 γ −1 div h v − ξ + γ − 1 γ 2−γ γ −1 v • ∇ h ξ + gξ γ −2 γ −1 ξ + γ − 1 γ gz 2−γ γ −1 γ ξ + γ − 1 γ g 1 γ −1 ξ γ −2 γ −1 − ξ × ξ + γ − 1 γ gz 2−γ γ −1 v • ∇ h ξ + (γ − 1) ξ + γ − 1 γ gz 1 γ −1 div h v .Therefore, from above, and as in the case when γ = 2, the vertical velocity w can be represented in the formw = z 0 H (div h v, v, ∇ h ξ, ξ ) dz ,similarly to (1.8), for an explicit function H (•). Notably, the arguments and proofs, below, apply equally, and similar conclusion of Theorem 1 also holds for γ > 1.The Case Without Gravity and γ > 1 Concerning system (1.2), since (1.2) 3 already yields the independence of the density of the vertical variable, after taking the vertical average of (1.2) 1 , as before, one has∂ t ρ + div h (ρv) = 0.(1.12)Comparing (1.12) with (1.2) 1 yields, thanks to the boundary condition (1.3), that the vertical velocity w is determined by the relationρw = − z 0 div h (ρ v) dz. (1.13)In particular, by denoting σ := ρ 1/2 , from (1.12) and (1.13), one has either σ = 0 or∂ t σ + v • ∇ h σ + 1 2 σ div h v = 0, (1.14) σ w = − z 0 σ div h v + 2 v • ∇ h σ dz. (1.15)In fact, for (σ, v) regular enough, (1.14), (1.15) hold regardless of whether σ = 0 or not. See also the justification in the beginning of section 3.2. System (1.2) is complemented with the initial data(ρ, v)| t=0 = (ρ 0 , v 0 ), or equivalently (σ, v)| t=0 = (σ 0 , v 0 ),(1.16)withσ 0 = ρ 1/2 0 , v 0 ∈ H 2 (), and the initial total mass and physical energy satisfy0 < ρ 0 d x = σ 2 0 d x = M < ∞, 0 < ρ 0 v 0 2 d x + 1 γ − 1 ρ γ 0 d x = σ 2 0 v 0 2 d x + 1 γ − 1 σ 2γ 0 d x = E 0 < ∞.(1.17) Also the following compatible conditions are imposed:ρ 0 0, ∂ z v 0 | z=0,1 = 0, μ h v 0 + μ∂ zz v 0 + (μ + λ)∇ h div h v 0 − ∇ h ρ γ 0 − ρ 0 v 0 • ∇ h v 0 − ρ 0 w 0 ∂ z v 0 =: ρ 1/2 0 h 1 , with h 1 ∈ L 2 ( ),and ρ 0 w 0 = − z 0 div h (ρ 0 v 0 ) dz. (1.18)Also, we will denote the boundsσ 0 2 H 2 = ρ 1/2 0 2 H 2 B 1 , v 0 2 H 2 + h 1 2 L 2 B 2 . (1.19) Moreover, if ρ = σ 2 > 0, (1.2) can be written as ⎧ ⎪ ⎪ ⎪ ⎪ ⎨ ⎪ ⎪ ⎪ ⎪ ⎩ ∂ t σ + v • ∇ h σ + w∂ z σ + 1 2 σ (div h v + ∂ z w) = 0 in , σ 2 (∂ t v + v • ∇ h v + w∂ z v) + ∇ h σ 2γ = μ h v + μ∂ zz v + (μ + λ)∇ h div h v in , ∂ z σ = 0 i n .(1.20)Theorem 2. Suppose the initial data (ρ 0 , v 0 ) = (σ 2 0 , v 0 ) satisfy (1.17), (1.19) and the compatible conditions (1.18). Then there is a unique strong solution (ρ, v) to system (1.2), with the boundary condition (1.3), in × (0, T * ), for some positive constant T * = T * (B 1 , B 2 ) > 0. Also, the solution satisfiesρ 1/2 ∈ L ∞ (0, T * ; H 2 ( )), ∂ t ρ 1/2 ∈ L ∞ (0, T * ; H 1 ( )), v ∈ L ∞ (0, T * ; H 2 ( )) ∩ L 2 (0, T * ; H 3 ( )), ∂ t v ∈ L 2 (0, T * ; H 1 ( )) ρ 1/2 ∂ t v ∈ L ∞ (0, T * ; L 2 ( )).Furthermore, for some positive constantC(B 1 , B 2 ), inf ( x,t)∈ ×(0,T * ) ρ( x, t) 0, sup 0 t T * ρ 1/2 (t) 2 H 2 + ∂ t ρ 1/2 (t) 2 H 1 + v(t) 2 H 2 + (ρ 1/2 v t )(t) 2 L 2 + T * 0 v(t) 2 H 3 + v t (t) 2 H 1 dt C(B 1 , B 2 ).Moreover, for any two strong solutions (ρ i , v i ), i = 1, 2, with initial data (ρ i,0 , v i,0 ), i = 1, 2, satisfying the conditions mentioned above, we have the inequalityρ 1/2 1 − ρ 1/2 2 L ∞ (0,T * ;L 2 ( )) + ρ 1/2 1 (v 1 − v 2 ) L ∞ (0,T * ;L 2 ( )) + ρ 1/2 2 (v 1 − v 2 ) L ∞ (0,T * ;L 2 ( )) + v 1 − v 2 L 2 (0,T * ;L 2 ( )) + ∇(v 1 − v 2 ) L 2 (0,T * ;L 2 ( )) C μ,λ,B 1 ,B 2 ,T * ρ 1/2 1,0 − ρ 1/2 2,0 L 2 ( )) + v 1,0 − v 2,0 L 2 ( ))for some positive constant C μ,λ,B 1 ,B 2 ,T * . PreliminariesWe will use • , • to denote norms in h ⊂ R 2 and ⊂ R 3 , respectively. After applying Ladyzhenskaya's and Agmon's inequalities in h and , directly we havef L 4 C f 1/2 L 2 f 1/2 H 1 , f L ∞ C f 1/2 L 2 f 1/2 H 2 , f L 3 C f 1/2 L 2 f 1/2 H 1 (1.21)for any function f with bounded right-hand sides. Also, f L p , f L p C f L p , for every p 1. Considering any quantities A, B, we use the notation A B to denote A C B for some generic positive constant C, which may be different from line to line. In what follows δ, ω > 0 are arbitrary constants which will be chosen later in the relevant paragraphs to be adequately small. C q represents a positive constant depending on the quantity q. We will also need the following classical inequality:Lemma 1. Let 2 p 6, and ρ 0 such that 0 < ρ d x = M < ∞, and ρ γ d x E 0 , for some γ ∈ (1, ∞). Then one has f L p C ∇ f L 2 + C ρ 1/2 f L 2 (1.22)for some constant C = C(M, E 0 ), provided the right-hand side is finite.Proof. This is standard. See, e.g., [18, Lemma 3.2]. Associated Linear Systems and Existence TheoryIn this section, we will establish the local existence theory of (1.1) and (1.2). To do this, we will first study the local existence of solutions to (1.5) and (1.20) via the Schauder-Tchonoff fixed point theorem capitalizing on some a priori estimates. In fact, under the assumption thatρ 0 = ⎧ ⎨ ⎩ ξ 0 + 1 2gz in the case with gravity (σ 0 ) 2 in the case without gravity> ρ > 0,(2.1)we will first introduce linear systems and the function spaces Y associated with (1.5) and (1.20) with some given input states (ξ o , v o ) and (σ o , v o ), respectively, in section 2.1 and 2.2. Here, Y are compactly embedded in some corresponding spaces V. Also, we will show that the maps T : X → X, for some convex bounded subsets X of Y, given by(ξ o , v o ) (ξ, v)in the case with gravity, and(σ o , v o ) (σ, v)in the case without gravity, are well-defined; observing that X are convex subsets of Y and hence compact in V. We will use the same notations X, Y, V, T to denote the convex bounded sets, the compact function spaces, the embedded function spaces and the constructed maps in both cases. We summarize the relevant regularity estimates in section 2.3 and show that the Schauder-Tchonoff fixed point theorem will yield the existence of solutions to (1.5) and (1.20) in the corresponding set. Recall that the Schauder-Tchonoff fixed point theorem states that for a Banach space V with a convex compact subset X ⊂ V , if F : X → X is continuous, then F has at least one fixed point in X . In our case, we will take X = X andV = V := {(ξ, v)|ξ, v ∈ L ∞ (0, T ; L 2 ( )), ∇v ∈ L 2 (0, T ; L 2 ( ))} in the case with gravity, or V = V := {(σ, v)|σ, v ∈ L ∞ (0, T ; L 2 ( )), ∇v ∈ L 2 (0, T ; L 2 ( ))} in the case without gravity, with the corresponding norms. We will only sketch the key steps in this paper. For more detailed calculation, we refer to our preprint [30]. Y = Y T :={(ξ, v)|ξ ∈ L ∞ (0, T ; H 2 ( )), ∂ t ξ ∈ L ∞ (0, T ; H 1 ( )), v ∈ L ∞ (0, T ; H 2 ( )) ∩ L 2 (0, T ; H 3 ( )), ∂ t v ∈ L ∞ (0, T ; L 2 ( )) ∩ L 2 (0, T ; H 1 ( ))}, (2.2)with the norm(ξ, v) Y := ξ L ∞ (0,T ;H 2 ( )) + ∂ t ξ L ∞ (0,T ;H 1 ( )) + v L ∞ (0,T ;H 2 ( )) + v L 2 (0,T ;H 3 ( )) + ∂ t v L ∞ (0,T ;L 2 ( )) + ∂ t v L 2 (0,T ;H 1 ( )) .Notice that, thanks to the Aubin compactness theorem (see, e.g., [42, Theorem 2.1] and [11,40]), every bounded subset of Y is a compact subset of the spaceV = V T := {(ξ, v)|ξ, v ∈ L ∞ (0, T ; L 2 ( )), ∇v ∈ L 2 (0, T ; L 2 ( ))}, (2.3) with the norm (ξ, v) V := ξ L ∞ (0,T ;L 2 ( )) + v L ∞ (0,T ;L 2 ( )) + v L 2 (0,T ;H 1 ( )) .(2.4)Let X = X T be a bounded subset of Y defined byX =X T := (ξ, v) ∈ Y|(ξ, v)| t=0 = (ξ 0 , v 0 ), ∂ z v| z=0,1 = 0, ∂ z ξ = 0, ξ + 1 2 gz 1 2 ρ > 0, sup 0 t T ξ(t) 2 H 2 2M 0 , sup 0 t T ∂ t ξ(t) 2 H 1 C 2 , sup 0 t T { v(t) 2 H 2 + ∂ t v(t) 2 L 2 } + T 0 v(t) 2 H 3 + ∂ t v(t) 2 H 1 dt C 1 M 1 ,(2.5)where M 0 , M 1 are the bounds of initial data in (2.9) andC 1 = C 1 (M 0 , μ, λ, ρ), C 2 = C 2 (M 0 , C 1 M 1 )are given below in (2.34), (2.17), respectively. Notice, for (ξ, v) ∈ X,1 0 div h (ξ v) + g 2 zdiv h v dz = 0. Let (ξ o , v o ) ∈ X.The following inhomogeneous linear system is inferred from (1.5) using (ξ o , v o ) as an input:⎧ ⎪ ⎪ ⎪ ⎪ ⎪ ⎨ ⎪ ⎪ ⎪ ⎪ ⎪ ⎩ ∂ t ξ + v o • ∇ h ξ + ξ div h v o + g 2 zdiv h v o = 0 i n , (ξ o + 1 2 gz)(∂ t v + v o • ∇ h v o + w o ∂ z v o ) + (2ξ o + gz)∇ h ξ o = μ h v + μ∂ zz v + (μ + λ)∇ h div h v in , ∂ z ξ = 0 i n . (2.6)Here w o is given by (1.8) with (ξ o , v o ) instead of (ξ, v), i.e., ρ o w o = (ξ o + 1 2 gz)w o := − z 0 div h (ξ o v o ) + g 2 zdiv h v o dz. (2(ξ, v)| t=0 = (ξ 0 , v 0 ), ∂ z v| z=0,1 = 0.(2.8)The compatible conditions in (1.10) are still imposed and we requireξ 0 2 H 2 M 0 , v 0 2 H 2 + V 1 2 L 2 M 1 . (2.9) Recall that V 1 is given in (1.10), essentially, V 1 = v t | t=0 .Then the map T , in this case, is defined asT : (ξ o , v o ) (ξ, v), (2.10)where (ξ, v) is the unique solution to the linear system (2.6) with (ξ o , v o ) ∈ X. We claim that T is a well defined map from X to X, which is the consequence of the following two propositions:Proposition 1. For given (ξ o , v o ) ∈ C ∞ ( ×[0, T ])∩X T , there is a unique strong solution (ξ, v) ∈ Y T of system (2.6) with the initial and boundary conditions (2.8). Suppose, in addition, that ξ 0 , v 0 ∈ H 3 ( ). One will have the following regularity of the unique solution (ξ, v) of system (2.6):ξ ∈ L ∞ (0, T ; H 3 ( )), ∂ t ξ ∈ L ∞ (0, T ; H 2 ( )), v ∈ L ∞ (0, T ; H 3 ( )) ∩ L 2 (0, T ; H 4 ( )), ∂ t v ∈ L ∞ (0, T ; H 1 ( )) ∩ L 2 (0, T ; H 2 ( )).(2.11) Proposition 2. Consider the initial data with the bounds M0 , M 1 in (2.9) and (ξ o , v o ) ∈ X = X T .There is a T g = T g (M 0 , M 1 , μ, λ, ρ) > 0 sufficiently small such that for any T ∈ (0, T g ), there exists a unique solution to (2.6). Moreover, the solution belongs to X = X T . Therefore, for any such T , the map T in (2.10) is a well defined map from X into X.We omit the proof of Proposition 1, and refer to [30] for the details. The existence of solutions in Proposition 2 follows from Proposition 1 and a standard approximating argument. We only show the required a priori estimates in the rest of this subsection, which are sufficient to establish Proposition 2. See Propositions 3 and 4, below. A Priori Estimates for the Inhomogeneous Linear SystemHereafter, we assume that the solution (ξ, v) to the linear system (2.6) is smooth enough so that the following estimates are rigorous.We start by establishing some estimates for the solutions of (2.6) 1 . In particular, we will establish the following: Proposition 3. There exists a T = T (M 0 , C 1 M 1 , ρ) > 0 sufficiently small such that for any T ∈ (0, T ], the solution ξ to (2.6) 1 satisfies thatξ + 1 2 gz 1 2 ρ; sup 0 t T ξ(t) 2 H 2 2M 0 ; sup 0 t T ∂ t ξ(t) 2 H 1 C 2 ,where M 0 is as in (2.9). The lower bound for ξIn order to derive the lower bound of ξ , we employ the following Stampaccialike argument. Let M = M(t) > 0 be a nonnegative integrable function to be determined later. Consider η = η(x, y, t) := ξ − ρ + t 0 M(s) ds. Then according to (2.6) 1 , η satisfies the equation∂ t η + v o • ∇ h η + ηdiv h v o = −(ρ − t 0 M(s) ds)div h v o − g 2 zdiv h v o + M(t).Let 1 {η<0} = 1 whenever {η < 0}, 0 otherwise, and denote by η − := −η1 {η<0} 0. Observe that since ξ ∈ H 1 ( × [0, T ]), so is η − . Thus, one hasd dt h η − d x h = {η<0} (ρ − t 0 M(s) ds)div h v o + g 2 zdiv h v o − M(t) d x h . Now, let 0 < M(t) := C max{ div h v o L ∞ , zdiv h v o L ∞ } C v o H 3 < ∞, a.e.,, for some constant C > 0. Then the integrand on the right-hand side of the above equation satisfiesρ − t 0 M(s) ds div h v o + g 2 zdiv h v o − M(t) 1 C ρ + C T 0 v o H 3 (s) ds + g 2 M(t) − M(t) < 0,provided C is large enough and T is small enough such that1 C ρ + 1 + g 2 < 1, andC T 0 v o H 3 (s) ds CT 1/2 T 0 v o 2 H 3 (s) ds 1/2 < 1.Therefore, we haved dt h η − d x h 0 a.e.,which, after integrating over [0, t 0 ] for any t 0 ∈ [0, T ], thanks to the fact η − (0) ≡ 0, yieldsh η − (t 0 ) d x h 0.(2.12)Hence η − = 0 in h × [0, T ]. That is, η(t) = ξ(t) − ρ + t 0 M(s) ds 0 and ξ(t) + 1 2 gz > ξ(t) ρ − C T 0 v o H 3 (s) ds ρ − CC 1/2 1 M 1/2 1 T 1/2 1 2 ρ (2.13) for t ∈ [0, T ], T T 1 , with T 1 = T 1 (C 1 M 1 , ρ) sufficiently small. The H 2 ( ) norm for ξApplying the standard H 2 estimate of linear transport equations to (2.6) 1 yields, since ξH 2 = ξ H 2 , d dt ξ 2 H 2 C v o H 3 ξ 2 H 2 + C v o H 3 ξ H 2 .Thanks to the Grönwall and the Hölder inequalities, one hassup 0 t T ξ(t) 2 H 2 1 4 sup 0 t T ξ 2 H 2 + C 2 C 1 M 1 e 2CC 1/2 1 M 1/2 1 T 1/2 T + e CC 1/2 1 M 1/2 1 T 1/2 M 0 .(2.14) Then for T ∈ (0, T 2 ] with T 2 (M 0 , C 1 M 1 )∂ t ξ 2 H 1 C v o 2 H 2 ξ 2 H 2 + C v o 2 H 2 C 2 , (2.16)whereC 2 = C 2 (M 0 , C 1 M 1 ) is given by C(1 + 2M 0 )C 1 M 1 =: C 2 .(2.17)Proof of Proposition 3. By choosing T = min{T 1 , T 2 }, the proof of Proposition 3 follows from (2.13), (2.15) and (2.16). Next, we showProposition 4. There exists a T = T (M 0 , M 1 , C 1 , C 2 , ρ) ∈ (0, ∞), sufficiently small, such that for every T ∈ (0, T ], the solution v to (2.6) 2 satisfiessup 0 t T ( v(t) 2 H 2 + v t (t) 2 L 2 ) + T 0 v(t) 2 H 3 + v t (t) 2 H 1 dt C 1 M 1 . Horizontal spatial derivative estimates for vApplying ∂ hh to (2.6) 2 will yieldρ o ∂ t ∂ hh v − μ h ∂ hh v − μ∂ zz ∂ hh v − (μ + λ)∇ h div h ∂ hh v = −2∂ h ρ o ∂ t ∂ h v − ∂ hh ρ o ∂ t v − ∂ hh (ρ o v o • ∇ h v o ) − ∂ hh (ρ o w o ∂ z v o ) − ∂ hh ((2ξ o + gz)∇ h ξ o ),(2.18)where ρ o = ξ o + 1 2 gz. After taking the inner product of (2.18) with ∂ hh v and integrating by parts, one hasd dt 1 2 ρ o ∂ hh v 2 d x + μ ∇ h ∂ hh v 2 + μ ∂ hhz v 2 + (μ + λ) × div h ∂ hh v 2 d x = − 2∂ h ρ o ∂ t ∂ h v • ∂ hh v + ∂ hh ρ o ∂ t v • ∂ hh v d x + 1 2 ∂ t ρ o ∂ hh v o 2 d x + ∂ h (ρ o v o • ∇ h v o ) • ∂ hhh v d x + ∂ h (ρ o w o ∂ z v o ) • ∂ hhh v d x + ∂ h ((2ξ o + gz)∇ h ξ o ) • ∂ hhh v d x =: 5 i=1 I i . (2.19)While we only will omit the detailed estimates, which are standard, we list below the estimates for the I i terms (see [30] for details). We will use the factρ o 2 H 2 C ξ o 2 H 2 + Cg 2 C M 0 + C, ∂ t ρ o 2 L 2 = ∂ t ξ o 2 L 2 C 2 .Also, hereafter the estimates hold for every δ, ω > 0 which will be chosen later to be adequately small. Correspondingly, C δ , C ω , C δ,ω are some positive constants depending on δ, ω:I 1 δ ∂ hh v 2 H 1 + ω v t 2 H 1 + C δ,ω (M 2 0 + 1)( v 2 H 2 + ∂ t v 2 L 2 ). I 2 ω v o 2 H 3 + C ω C 2 2 C 1 M 1 . I 3 δ ∂ hh v 2 H 1 + C δ (M 0 + 1)C 2 1 M 2 1 . I 5 δ ∂ hh v 2 H 1 + C δ (M 2 0 + 1).In order to estimate I 4 , we shall plug in (2.7). One hasI 4 = ∂ h (ρ o w o )∂ z v o • ∂ hhh v d x + ρ o w o ∂ hz v o • ∂ hhh v d x = − 1 0 h z 0 ∂ h div h (ξ o v o ) + g 2 zdiv h ∂ h v o dz ∂ z v o • ∂ hhh v d x h dz − 1 0 h z 0 div h (ξ o v o ) + g 2 zdiv h v o dz ∂ hz v o • ∂ hhh v d x h dz =: I 4 + I 4 .Then applying the Minkowski and the Sobolev embedding inequalities yieldsI 4 = − 1 0 z 0 h div h (ξ o v o ) + g 2 zdiv h v o ( x h , z , t) × ∂ hz v o • ∂ hhh v ( x h , z, t) d x h dz dz 1 0 ∇ h ξ o L 4 v o L ∞ + ξ o L ∞ ∇ h v o L 4 + ∇ h v o L 4 dz × 1 0 ∂ hz v o L 4 ∂ hhh v L 2 dz ( ξ o H 2 + 1) v o 3/2 H 2 v o 1/2 H 3 × ∂ hh v H 1 δ ∂ hh v 2 H 1 + ω v o 2 H 3 + C δ,ω (M 2 0 + 1)C 3 1 M 3 1 ,and, similarly,I 4 δ ∂ hh v 2 H 1 + ω v o 2 H 3 + C δ,ω (M 2 0 + 1)C 3 1 M 3 1 ,where we have employed (1.21). Summing up the above inequalities, with δ small enough, yields the following estimated dt ρ o ∂ hh v 2 L 2 + c μ,λ ∂ hh v 2 H 1 ω( ∂ t v 2 H 1 + v o 2 H 3 ) + C ω H(M 0 , C 1 M 1 , C 2 )( v 2 H 2 + ∂ t v 2 L 2 + 1).Hereafter, H will be used to denote a polynomial quantity of its arguments (i.e., the norms of the initial data and ξ o , v o ) which may be different from line to line. Also c μ,λ , C ω denote positive constants depending on μ, λ and ω, respectively. Similar arguments also hold for the lower order derivatives. Then after suitable choice of ω, one hasd dt ρ o v 2 L 2 + ρ o ∇ h v 2 L 2 + ρ o ∇ 2 h v 2 L 2 + c μ,λ v 2 H 1 + ∇ h v 2 H 1 + ∇ 2 h v 2 H 1 ω ∂ t v 2 H 1 + v o 2 H 3 + C ω H(M 0 , C 1 M 1 , C 2 ) v 2 H 2 + ∂ t v 2 L 2 + 1 .(2.20) Time derivative estimates for vApplying ∂ t to (2.6) 2 yieldsρ o ∂ t v t − μ h v t − μ∂ zz v t − (μ + λ)∇ h div h v t = −∂ t ρ o ∂ t v − ∂ t (ρ o v o • ∇ h v o ) − ∂ t (ρ o w o ∂ z v o ) − ∂ t ((2ξ o + gz)∇ h ξ o ). (2.21)Consequently, one hasd dt 1 2 ρ o ∂ t v 2 d x + μ ∇ h v t 2 + μ ∂ z v t 2 + (μ + λ) div h v t 2 d x = − 1 2 ∂ t ρ o v t 2 d x − ∂ t (ρ o v o • ∇ h v o ) • ∂ t v d x − ∂ t (ρ o w o ∂ z v o ) • ∂ t v d x − ∂ t ((2ξ o + gz)∇ h ξ o ) • ∂ t v d x =: 9 i=6 I i . (2.22)Then applying similar estimates as before to the right-hand side of (2.22) (see [30] for details) and making use of the fact that w o is given by (2.7), one hasd dt ρ o v t 2 L 2 + c μ,λ v t 2 H 1 ω v o 2 H 3 + ∂ t v o 2 H 1 + C ω H(M 0 , C 1 M 1 , C 2 ) ∂ t v 2 L 2 + 1 .(2.23) Vertical derivative estimates for vTaking the inner product of (2.6) 2 with v t and applying similar estimates as before to the resultant impliesd dt μ ∇ h v 2 L 2 + μ ∂ z v 2 L 2 + (μ + λ) div h v 2 L 2 + ρ o v t 2 L 2 H(M 0 , C 1 M 1 , C 2 , ρ).(2.24)On the other hand, (2.6) 2 can be written asμ∂ zz v − ρ o ∂ t v = −μ h v − (μ + λ)∇ h div h v + ρ o v o • ∇ h v o + ρ o w o ∂ z v o + (2ξ o + gz)∇ h ξ o . (2.25)Taking the inner product of (2.25) with ∂ t ∂ zz v and applying similar estimates as before will yield,d dt (μ ∇ h ∂ z v 2 L 2 + μ ∂ zz v 2 L 2 + (μ + λ) div h ∂ z v 2 L 2 ) + c μ,λ ρ o ∂ z v t 2 L 2 ω v o 2 H 3 + C ω H(M 0 , C 1 M 1 , C 2 , ρ)( v t 2 L 2 + 1). (2.26) Next, applying ∂ ∈ {∂ x , ∂ y , ∂ z } to (2.25) yields μ∂∂ zz v − ρ o ∂v t = ∂ρ o ∂ t v − μ h ∂v − (μ + λ)∇ h div h ∂v + ∂(ρ o v o • ∇ h v o ) + ∂(ρ o w o ∂ z v o ) + ∂((2ξ o + gz)∇ h ξ o ). (2.27) This implies μ ∂∂ zz v L 2 ∇ 2 h v H 1 + (M 1/2 0 + 1)( v t H 1 + M 1/2 0 + C 1 M 1 ) + ∂(ρ o w o )∂ z v o L 2 + ρ o w o ∂∂ z v o L 2 .Also, by employing the Minkowski's inequality, one obtains∂ h (ρ o w o )∂ z v o 2 L 2 1 0 ∂ h (ρ o w o ) 2 L 2 ∂ z v o 2 L ∞ dz 1 0 ∇ 2 h (ξ o v o ) L 2 + ∇ 2 h v o L 2 dz 2 × 1 0 ∂ z v o H 1 ∂ z v o H 2 dz ( ξ o 2 H 2 + 1) v o 3 H 2 v o H 3 ω v o 2 H 3 + C ω (M 2 0 + 1)C 3 1 M 3 1 ,and, similarly,ρ o w o ∂∂ z v o 2 L 2 ω v o 2 H 3 + C ω (M 2 0 + 1)C 3 1 M 3 1 , ∂ z (ρ o w o )∂ z v o 2 L 2 (M 0 + 1)C 2 1 M 2 1 .Therefore, we have ∂ zz v 2 H 1 C ∇ 2 h v 2 H 1 + C(M 0 + 1) v t 2 H 1 + ω v o 2 H 3 + C ω H(M 0 , C 1 M 1 , C 2 ). (2(t) + c μ,λ,ρ v 2 H 1 + ∇ h v 2 H 1 + ∇ 2 h v 2 H 1 + v t 2 H 1 ω v t 2 H 1 + ω v o 2 H 3 + v o t 2 H 1 + C ω H(M 0 , C 1 M 1 , C 2 , ρ)(E g (t) + 1),(2.29)whereE g (t) := ρ o v 2 L 2 + ρ o ∇ h v 2 L 2 + ρ o ∇ 2 h v 2 L 2 + ρ o v t 2 L 2 + μ ∇v 2 L 2 + (μ + λ) div h v 2 L 2 + μ ∇∂ z v 2 L 2 + (μ + λ) div h ∂ z v 2 L 2 .(2.30)Notice that for some positive constants C i,μ,λ,ρ,M 0 , i = 1, 2, depending on μ, λ, ρ, M 0 , we haveC 1,μ,λ,ρ,M 0 ( v 2 H 2 + v t 2 L 2 ) E g (t) C 2,μ,λ,ρ,M 0 ( v 2 H 2 + v t 2 L 2 ). (2.31) For 0 < ω c μ,λ,ρ 2 , one infers from (2.29), that d dt E g (t) ω( v o 2 H 3 + v o t 2 H 1 ) + C ω H(M 0 , C 1 M 1 , C 2 , ρ)(E g (t) + 1).Therefore, applying the Grönwall's inequality yields sup 0 t T E g (t) e C ω H(M 0 ,C 1 M 1 ,C 2 ,ρ)T E g (0) + ω T 0 ( v o 2 H 3 + v o t 2 H 1 ) dt + T 0 C ω H(M 0 , C 1 M 1 , C 2 , ρ) dt e C ω H(M 0 ,C 1 M 1 ,C 2 ,ρ)T C 2,μ,λ,ρ,M 0 M 1 + ωC 1 M 1 + C ω H(M 0 , C 1 M 1 , C 2 , ρ)T ,2 T 0 v 2 H 1 + ∇ h v 2 H 1 + ∇ 2 h v 2 H 1 + v t 2 H 1 dt E g (0) + E g (t) + ω T 0 v o 2 H 3 + v o t 2 H 1 dt + C ω H(M 0 , C 1 M 1 , C 2 , ρ) T 0 E g (t) + 1 dt 2 + T C ω H(M 0 , C 1 M 1 , C 2 , ρ) e C ω H(M 0 ,C 1 M 1 ,C 2 ,ρ)T × C 2,μ,λ,ρ,M 0 M 1 + ωC 1 M 1 + C ω H(M 0 , C 1 M 1 , C 2 , ρ)T .Additionally, from (2.28), we haveT 0 ∂ zz v 2 H 1 dt C T 0 ∇ 2 h v 2 H 1 + (M 0 + 1) v t 2 H 1 dt + ωC 1 M 1 + C ω H(M 0 , C 1 M 1 , C 2 )T.Therefore, we conclude that sup0 t T ( v(t) 2 H 2 + v t (t) 2 L 2 ) + T 0 v 2 H 3 + v t 2 H 1 dt C −1 1,μλ,ρ,M 0 sup 0 t T E g (t) + T 0 v 2 H 1 + ∇ h v 2 H 1 + ∇ 2 h v 2 H 1 + v t 2 H 1 dt + T 0 ∂ zz v 2 H 1 (M 0 + 1) C 3,μ,λ,ρ,M 0 + C ω H(M 0 , C 1 M 1 , C 2 , ρ)T × e C ω H(M 0 ,C 1 M 1 ,C 2 ,ρ)T C 2,μ,λ,ρ,M 0 M 1 + ωC 1 M 1 + C ω H(M 0 , C 1 M 1 , C 2 , ρ)T + ωC 1 M 1 + C ω H(M 0 , C 1 M 1 , C 2 , ρ)T, (2.32)for some positive constant C 3,μ,λ,ρ,M 0 , depending on μ, λ, ρ, M 0 . Now fix ω =1 2 min{ c μ,λ,ρ 2 , 1 C 1 } and let T ∈ (0, T ], where T = T (M 0 , M 1 , C 1 , C 2 , ρ) is small enough and satisfying C ω H(M 0 , C 1 M 1 , C 2 , ρ)T min{1, M 1 }. Then (2.32) yields sup 0 t T ( v(t) 2 H 2 + v t (t) 2 L 2 ) + T 0 v 2 H 3 + v t 2 H 1 dt C 1 M 1 , (2.33)where C 1 is given byC 1 := (M 0 + 1)(C 3,μ,λ,ρ,M 0 e + e)(C 2,μ,λ,ρ,M 0 + 2) + 2. (2.34)This concludes the proof. The Case Without Gravity and γ > 1Consider a finite positive time T , which will be determined later. Let Y = Y T be the function space defined byY = Y T :={(σ, v)|σ ∈ L ∞ (0, T ; H 2 ( )), ∂ t σ ∈ L ∞ (0, T ; H 1 ( )), v ∈ L ∞ (0, T ; H 2 ( )) ∩ L 2 (0, T ; H 3 ( )), ∂ t v ∈ L ∞ (0, T ; L 2 ( ))L 2 (0, T ; H 1 ( ))}. (2.35)Notice that, thanks to Aubin compactness theorem (see [42,Theorem 2.1] and [11,40]), every bounded subset of Y is a compact subset of the spaceV = V T := {(σ, v)|σ, v ∈ L ∞ (0, T ; L 2 ( )), ∇v ∈ L 2 (0, T ; L 2 ( ))}.(2.36) Let X = X T be a bounded subset of Y defined byX =X T := (σ, v) ∈ Y|(σ, v)| t=0 = (σ 0 , v 0 ), ∂ z v| z=0,1 = 0, ∂ z σ = 0, σ 2 1 2 ρ > 0, sup 0 t T σ (t) 2 H 2 2M 0 , sup 0 t T ∂ t σ (t) 2 H 1 C 2 , sup 0 t T v(t) 2 H 2 + v t (t) 2 L 2 + T 0 v 2 H 3 + ∂ t v 2 H 1 dt C 1 M 1 , (2.37)where ρ is the positive lower bound of initial density profile as in (2.1), for some positive constantsC 1 = C 1 (M 0 , μ, λ, ρ), C 2 = C 2 (M 0 , C 1 M 1 ). Notice, for (σ, v) ∈ X, 1 0 div h (σ 2 v) dz = 0. Let (σ o , v o ) ∈ X.The linear system inspired by (1.20) with (σ o , v o ) as an input is given by⎧ ⎪ ⎪ ⎪ ⎪ ⎨ ⎪ ⎪ ⎪ ⎪ ⎩ ∂ t σ + v o • ∇ h σ + 1 2 σ div h v o = 0 i n , ρ o ∂ t v + ρ o v o • ∇ h v o + σ o σ o w o ∂ z v o + 2γ (σ o ) 2γ −1 ∇ h σ o = μ h v + μ∂ zz v + (μ + λ)∇ h div h v in , ∂ z σ = 0 i n , (2.38)where ρ o = (σ o ) 2 and w o is determined, as in (1.15), byσ o w o := − z 0 σ o div h v o + 2 v o • ∇ h σ o dz. (2.39)The initial and boundary conditions for system (2.38) are given by(σ, v)| t=0 = (σ 0 , v 0 ) = (ρ 1/2 0 , v 0 ), ∂ z v| z=0,1 = 0. (2.40)Here, in addition to the compatible conditions in (1.18), we require ρ 0 ρ > 0 , for some positive constant ρ as in (2.1). Also, we denote byV 1 := h 1 /ρ 1/2 0 . Recall that h 1 is given in (1.18). Then V 1 ∈ L 2 ( ) and we require σ 0 2 H 2 M 0 , v 0 2 H 2 + V 1 2 L 2 M 1 . Essentially V 1 = v t | t=0 .Then the map T , in the case without gravity, is defined asT : (σ o , v o ) (σ, v), (2.41)where (σ, v) is the unique solution to (2.38) for given(σ o , v o ) ∈ X. Proposition 5. There is a T v = T v (M 0 , M 1 , μ, λ, ρ) > 0,sufficiently small, such that for every T ∈ (0, T v ], there is a unique solution (σ, v) to (2.38) in the set X = X T . Therefore, for such T , the map T defined in (2.41) is a well defined map from X into X.The proof is similar as Proposition 1 and Proposition 2 in section 2.1 and therefore is omitted. Existence TheoryIn this subsection, we will establish the existence of solutions for (1.1) and (1.2) for given corresponding initial data and boundary conditions.1 − ρ o 2 v o 2 • ∇ h v o 12 − (ρ o 1 w o 1 − ρ o 2 w o 2 )∂ z v o 1 − ρ o 2 w o 2 ∂ z v o 12 .(2.42)Now we perform standard L 2 estimates for (2.42). Take the L 2 -inner product of (2.42) 1 with 2ξ 12 and take the L 2 -inner product of (2.42) 2 with 2v 12 . Then applying similar estimates as before yields, together with the Grönwall inequality, that sup0 t T ( ξ 12 (t) 2 L 2 + v 12 (t) 2 L 2 ) + T 0 ∇v 12 (t) 2 L 2 dt C M 0 ,C 1 M 1 ,C 2 ,ρ sup 0<t<T ( ξ o 12 2 L 2 + v o 12 2 L 2 ) + T 0 ∇v o 12 2 L 2 dt . (2.43)This implies the continuity of T in V. Therefore, after applying the fixed point theorem mentioned before, we have the following:Proposition 6. Consider (ρ 0 , v 0 ) = (ξ 0 + 1 2 gz, v 0 ),given in (1.9) satisfying (1.10) and (1.11). There is a positive constant T depending on the initial data such that there is a strong solution (ρ, v) = (ξ + 1 2 gz, v) to (1.1) (or equivalently (1.5)) with the boundary conditions (1.3) and with (ξ, v) ∈ X T . The Case With Vacuum and γ > 1, but Without GravityWhen ρ 0 ρ > 0, the existence of strong solutions to (1.2) follows from the estimates in section 2.2 and similar arguments to those in section 2.3.1. In fact, taking M 0 = B 1 , M 1 = B 2 + ρ −1 B 2 , we have the following: Proposition 7. Suppose that (1.17), (1.18), (1.19) hold for the given initial data (1.16) with ρ 0 ρ > 0. Then there is a positive constant T , depending on the initial data and ρ, such that there exists a strong solution (ρ, v) = (σ 2 , v) to (1.2) (or equivalently (1.20)) satisfying the boundary conditions (1.3) and that (σ, v) ∈ X T .In the following, we shall present some estimates independent of ρ and show that for a given non-negative initial density ρ 0 0, there are strong solutions to equations (1.2). We will use here the notation σ 2 = ρ and the alternative form of equations (1.20), as well as (1.2). Meanwhile, let us assume thatσ 0 H 2 = ρ 1/2 0 H 2 K 1 , v 0 H 2 + h 1 L 2 K 2 , (2.44)for given 18). Also, taking inner product of (1.2) 2 with v yields, the conservation of physical energyK 1 , K 2 > 0. Recall essentially h 1 = (σ v t )| t=0 from (1.1 2 ρ v 2 d x + 1 γ − 1 ρ γ d x + T 0 μ ∇v 2 + (μ + λ) div h v 2 d x = 1 2 ρ 0 v 0 2 d x + 1 γ − 1 ρ γ 0 d x < ∞,(2.45)where (1.2) 1 is also applied. Also, integrating (1.2) 1 over × (0, T ) yields the conservation of total mass0 < ρ d x = ρ 0 d x = M < ∞.(2.46)These facts are important when applying (1.22) in what follows. A priori assumptionsLet (σ, v) be the solution to (1.20) given in Proposition 7. We assume first, for some constants C d K 2 2 , T d (may depend on ρ),sup 0 t T d v(t) 2 H 2 + σ v t (t) 2 L 2 + T d 0 v 2 H 3 + v t 2 H 1 dt < C d .(In what follows, we will derive some a priori estimates independent of ρ. Also, we set T ∈ (0, T d ] to be determined later. We emphasize that the smallness of T in what follows is independent of ρ. ρ-independent lower bound: non-negativity of ρ We will use the same Stampaccia-like argument as before to derive the lower bound of ρ. Considerη = η(x, y, t) := ρ inf x h ∈ h ρ 0 ( x h ) − 1 + t 0 2 div h v(s) L ∞ ds, for t ∈ [0, T d ].Due to (1.12), η satisfies∂ t η + v • ∇ h η + ηdiv h v = t 0 2 div h v(s) L ∞ ds − 1 × div h v + 2 div h v(s) L ∞ −2 div h v + 2 div h v L ∞ 0,for every t ∈ [0, T ] with T ∈ (0, T 1 ], and T 1 sufficiently small such that2 t 0 div h v(s) L ∞ ds 2C t 0 v(s) H 3 ds 2CT 1/2 t 0 v(s) 2 H 3 ds 1/2 2CC 1/2 d T 1/2 1 1 2 .Denote by η − := −η1 {η<0} 0. Then multiplying the above equation with −1 {η<0} and integrating the resultant in the spatial variable yieldd dt h η − d x h 0.Hence, η − = 0 in h × (0, T ], since η − (0) ≡ 0. Therefore, η 0 andρ = inf x h ∈ h ρ 0 ( x h ) × η + 1 − t 0 2 div h v(s) L ∞ ds inf x h ∈ h ρ 0 ( x h ) × 0 + 1 − 1 2 = 1 2 inf x h ∈ h ρ 0 ( x h ).(2.48)ρ-independent estimate: H 2 ( ) for σ = ρ 1/2After performing standard H 2 estimate of (1.14) and applying the Grönwall inequality to the result, one hassup 0 t T σ (t) 2 H 2 e C T 0 v H 3 dt σ 0 2 H 2 e CC 1/2 d T 1/2 K 2 1 < 2K 2 1 ,(2.49)for all T ∈ (0, T 2 ], provided T 2 is sufficiently small. ρ-independent estimate: H 1 ( ) for σ t It directly follows from (1.14) that∂ t σ (t) H 1 C v(t) H 2 σ (t) H 2 √ 2CC 1/2 d K 1 =: K 3 , (2.50) for all t ∈ [0, T ]. ρ-independent estimate: L 2 ( ) for v tTaking the time derivative of (1.20) 2 yieldsσ 2 ∂ t v t − μ h v t − μ∂ zz v t − (μ + λ)∇ h div h v t = −2σ ∂ t σ ∂ t v − ∂ t (σ 2 v • ∇ h v) − ∂ t (σ 2 w∂ z v) − ∂ t ∇ h σ 2γ . (2.51)Taking the L 2 -inner product of (2.51) with∂ t v gives 1 2 d dt σ v t 2 L 2 + μ ∇ h v t 2 L 2 + μ ∂ z v t 2 L 2 + (μ + λ) div h v t 2 L 2 = − σ ∂ t σ v t 2 d x − ∂ t (σ 2 v • ∇ h v) • v t d x − ∂ t (σ 2 w∂ z v) • v t d x + ∂ t σ 2γ div h v t d x =: 4 i=1 L i . (2.52)Then one has the following estimates to the terms in the right-hand side of (2.52) (see [30] for details):L 1 ∂ t σ L 2 σ v t L 3 v t L 6 ∂ t σ L 2 σ v t 1/2 L 2 σ 1/2 L ∞ v t 3/2 L 6 δ ∇v t 2 L 2 + C δ C d (K 2 1 K 4 3 + 1). L 2 ∂ t σ L 2 v L ∞ ∇ h v L 6 σ v t L 3 + σ v t L 3 ∇ h v L 6 σ v t L 2 + σ L ∞ v L ∞ ∇ h v t L 2 σ v t L 2 δ ∇v t 2 L 2 + C δ (C d + 1)(K 2 1 C d + K 1 K 3 C 2 d + 1). L 4 σ 2γ −1 L ∞ ∂ t σ L 2 ∇v t L 2 δ ∇v t 2 L 2 + C δ K 4γ −2 1 K 2 3 .We have applied above the Hölder inequality and (1.22). Notice that σ = ρ 1/2 and that the conservations of energy and mass (2.45), (2.46) hold. In order to estimate L 3 term, we substitute (1.13) and integrate by parts. ThenL 3 = − ∂ t (ρw)∂ z v • v t d x − ρw∂ z v t • v t d x = − 1 0 h z 0 (σ 2 v) t dz • ∇ h (∂ z v • v t ) d x h dz + 1 0 h z 0 div h (σ 2 v) dz × ∂ z v t • v t d x h dz =: L 3 + L 3 .Now we use (1.21), the Minkowski and Hölder inequalities,L 3 = − 1 0 h z 0 σ v t + 2σ t v dz • (∇ h ∂ z v • σ v t + σ ∇ h v t • ∂ z v) d x h dz 1 0 σ v t L 2 + σ t L 2 v L ∞ dz × 1 0 σ L ∞ ∇ h ∂ z v L 4 v t L 4 + σ L ∞ ∇ h v t L 2 ∂ z v L ∞ dz 1 0 σ v t L 2 + σ t L 2 v H 2 dz × 1 0 σ H 2 ∂ z v 1/2 H 1 ∂ z v 1/2 H 2 v t H 1 dz σ H 2 ( σ v t L 2 + σ t L 2 v H 2 ) v 1/2 H 2 v 1/2 H 3 ( σ v t L 2 + ∇v t L 2 ) δ ∇v t 2 L 2 + ω v 2 H 3 + C δ,ω C d (K 4 1 (K 4 3 + 1)C 2 d + 1), similarly, L 3 δ ∇v t 2 L 2 + C δ (K 10 1 C 4 d + K 10/3 1 C 2 d). After summing the above inequalities, (2.52) impliesd dt σ v t 2 L 2 + c μ,λ ∇v t 2 L 2 ω v 2 H 3 + C ω H(K 1 , K 3 , C d ),(2.53)where, as before, H denotes a polynomial quantity of its arguments. ρ-independent estimate: spatial derivatives of v Now we are able to derive the estimates on the spatial derivatives of v. Standard L 2 estimate of (1.20) 2 yields thatd dt σ v 2 L 2 + c μ,λ ∇v 2 L 2 C σ 2γ 2 L 2CH(K 1 ).(2.54) Furthermore, taking the L 2 -inner product of (1.20) 2 with v t yields, after applying similar estimates as before,d dt (μ ∇ h v 2 L 2 + μ ∂ z v 2 L 2 + (μ + λ) div h v 2 L 2 ) + σ v t 2 L 2 H(K 1 , C d ).(2.55)Next we estimate the second order spatial derivatives. Taking the L 2 -inner product of (1.20) 2 with ∂ zz v t yields 1 2 d dt (μ ∇ h ∂ z v 2 L 2 + μ ∂ zz v 2 L 2 + (μ + λ) div h ∂ z v 2 L 2 ) + σ ∂ z v t 2 L 2 = − ∂ z (σ 2 v • ∇ h v) • ∂ z v t d x − ∂ z (σ 2 w∂ z v) • ∂ z v t d x =: L 8 + L 9 .(2.56)At the same time, taking the L 2 -inner product of (1.20) 2 with h v t yields1 2 d dt (μ ∇ 2 h v 2 L 2 + μ ∇ h ∂ z v 2 L 2 + (μ + λ) ∇ h div h v 2 L 2 ) + σ ∇ h v t 2 L 2 = −2 σ ∇ h σ • ∇ h v t • v t d x − ∇ h (σ 2 v • ∇ h v) : ∇ h v t d x − ∇ h (σ 2 w∂ z v) : ∇ h v t d x − ∇ 2 h σ 2γ : ∇ h v t d x =: 13 i=10 L i .(2.57)Applying similar estimates as before to the right-hand sides of (2.56) and (2.57) yields, after summing up the results,d dt (μ ∇ 2 h v 2 L 2 + 2μ ∇ h ∂ z v 2 L 2 + μ ∂ zz v 2 L 2 + (μ + λ) ∇div h v 2 L 2 ) + σ ∇v t 2 L 2 ω( ∇v t 2 L 2 + v 2 H 3 ) + C ω H(K 1 , C d ).(2.58)Finally, we provide estimates for the spatial derivative of v.Applying ∂ ∈ {∂ x , ∂ y , ∂ z } to (1.20) 2 yields μ h ∂v + μ∂ zz ∂v + (μ + λ)∇ h div h ∂v = ∂(σ 2 v t ) + ∂(σ 2 v • ∇ h v) + ∂(σ 2 w∂ z v) + ∂∇ h σ 2γ . (2.59)Taking the L 2 -inner product of (2.59) with h ∂v, for ∂ ∈ {∂ h , ∂ z }, and integrating by parts imply, after applying the Cauchy-Schwarz inequality,∇ 3 v 2 L 2 ∇(σ 2 v t ) 2 L 2 + ∇(σ 2 v • ∇ h v) 2 L 2 + ∇(σ 2 w∂ z v) 2 L 2 + ∇ 2 h σ 2γ 2 L 2 .(2.60)Then, similarly to (2.28), noticing the fact that w are given by (1.15), (2.60) yields v 2 H 3 ∇ 3 v 2 L 2 + C d C K 4 1 ∇v t 2 L 2 + H(K 1 , C d ), (2ρ( x, t) 1 2 inf x∈ ρ 0 > 0, sup 0 t T σ (t) H 2 2K 1 , sup 0 t T ∂ t σ (t) H 1 K 3 , sup 0 t T v(t) 2 H 2 + (σ v t )(t) 2 L 2 + T 0 v(t) 2 H 3 + v t (t) 2 H 1 dt K 2 4 ,where K 4 = 2C μ,λ K 2 , K 3 = C K 1 K 4 are given in (2.63) and (2.64). Notably, the bounds in these estimates depend only on the initial bounds K 1 , K 2 and do not depend on the lower bound of density. Also, the smallness of T * does not depend on ρ, even though T d may depend on ρ.Proof. Denote byE(t) := σ v t 2 L 2 + σ v 2 L 2 + μ ∇v 2 L 2 + (μ + λ) div h v 2 L 2 + μ ∇ 2 h v 2 L 2 + 2μ ∇ h ∂ z v 2 L 2 + μ ∂ zz v 2 L 2 + (μ + λ) ∇div h v 2 L 2 .(2.62)Then from (2.53), (2.54), (2.55) and (2.58), we haved dt E(t) + c μ,λ ∇v t 2 L 2 + ∇v 2 L 2 + σ v t 2 L 2 + σ ∇v t 2 L 2 ω v 2 H 3 + ∇v t 2 L 2 + C ω H(K 1 , K 3 , C d ).Then integrating the above inequality yields, for T (0, T d ], where T d is as in (2.47),sup 0 t T E(t) + c μ,λ T 0 ∇v t 2 L 2 + ∇v 2 L 2 + σ v t 2 L 2 + σ ∇v t 2 L 2 dt E(0) + ωC d + C ω T H(K 1 , K 3 , C d ).Then together with (2.61), we have, after choosing ω small enough and then T sufficiently small, sup0 t T v(t) 2 H 2 + σ v t (t) 2 L 2 + T 0 v 2 H 3 + v t 2 H 1 dt C μ,λ K 2 2 + ωC d C μ,λ (K 4 1 + 1) + C ω T H(K 1 , K 3 , C d ) 2C μ,λ K 2 2 =: K 2 4 ,(2.63)where we have employed inequality (1.22) and the fact for some positive constant C μ,λ > 0 we haveC −1 μ,λ ( v 2 H 2 + σ v t 2 L 2 ) E(t) C μ,λ ( v 2 H 2 + σ v t 2 L 2). Then plugging in (2.63) back into (2.50) implies∂ t σ H 1 C K 4 K 1 =: K 3 .(2.64)Thus the conclusion is drawn from (2.48), (2.49), (2.63) and (2.64).Existence of strong solutions with vacuum but no gravity and γ > 1 Now we are in the place to remove the strict positivity (of the initial density profile) assumption in Proposition 7. In order to do so, we introduce a sequence of approximating initial data (ρ 0,n , v 0,n ) satisfying, in addition to (1.17), (1.18), (1.19),ρ 0,n 1 n > 0, such that ρ 1/2 0,n → ρ 1/2 0 , v 0,n → v 0 in H 2 ( ), as n → ∞, where (ρ 0 , v 0 ) (or equivalently (σ 0 , v 0 )) is given in (1.16)satisfying (1.18) and (1.19). We require that the initial physical energy and total mass given in (1.17) with ρ 0 , v 0 replaced by ρ 0,n , v 0,n satisfy0 < ρ 0,n d x = M < ∞, 0 < ρ 0,n v 0,n 2 d x + 1 γ − 1 ρ γ 0,n d x E 0 + 1 < ∞,uniformly in n, so that when we apply inequality the constant in the inequality is independent of n. Now we apply Proposition 7 with the initial data (ρ 0,n , v 0,n ). Indeed, considerM 0 = B 1 , M 1 = B 2 + n B 2 . Then Proposition 7 guarantees that there is a T 1 = T 1 (n, B 1 , B 2 ) such that (1.2) admits a strong solution (ρ n , v n ) = (σ 2 n , v n ) satisfying sup 0 t T 1 σ n (t) 2 H 2 2B 1 , sup 0 t T 1 ∂ t σ n (t) 2 H 1 C 1 (B 1 , B 2 , n), sup 0 t T 1 v n (t) 2 H 2 + ∂ t v n (t) 2 L 2 + T 1 0 v n (t) 2 H 3 + ∂ t v n (t) 2 H 1 dt C 2 (B 1 , B 2 , n), and ρ n = σ 2 n 1 2n.Sequently, we apply Proposition 8 withK 1 = B 1/2 1 , K 2 = 2B 1/2 2 , C d = (1 + 2n)C 2 (B 1 , B 2 , n) and T d = T 1 . It yields that there is a T 2 = T 2 (B 1 , B 2 , n) T 1 such that the following bounds are satisfied sup 0 t T 2 σ n (t) H 2 2B 1/2 1 , sup 0 t T 2 ∂ t σ n (t) H 1 C 3 (B 1 , B 2 ), sup 0 t T 2 v n (t) 2 H 2 + (σ n v n,t )(t) 2 L 2 + T 2 0 v n (t) 2 H 3 + v n,t (t) 2 H 1 dt C 4 (B 1 , B 2 ),and inf( x,t)∈ ×[0,T 2 ]ρ n 1 2n .( (B 1 , B 2 ). That is, for some δT = δT (B 1 , B 2 , n) > 0, inf ( x,t)∈ ×(T 2 ,T 2 +δT ) ρ n 1 4n , sup T 2 <t<T 2 +δT σ n (t) H 2 4B 1/2 1 , sup T 2 <t<T 2 +δT ∂ t σ n (t) H 1 C 3 (4B 1 , C 4 (B 1 , B 2 )), sup T 2 <t<T 2 +δT v n (t) 2 H 2 + σ n v n,t (t) 2 L 2 + T 2 +δT T 2 v n (t) 2 H 3 + v n,t (t) 2 H 1 dt C 4 (4B 1 , C 4 (B 1 , B 2 )). Now we apply Proposition 8 withT d = T 2 + δT and C d = C 4 (4B 1 , C 4 (B 1 , B 2 ))in the time interval (0, T d ). This will yield that there is a T * = T * (B 1 , B 2 ) and T 3 := min{T * , T 2 + δT }, the bounds in (2.65) hold with T 2 replaced by T 3 .If T 3 = T * , we have got an existence time independent of n and this finishes the job. Otherwise, let (σ n , v n )| t=T 3 as a new initial data and repeat the arguments to get the bounds in (2.65) with T 2 replaced by T 4 min{T * , T 3 + δT } = min{T * , T 2 + 2δT }. Keep repeating this process, one will eventually get that there is a m ∈ Z + sufficiently large that T m := min{T * , T 2 + (m − 2)δT } = T * . Therefore, we have got a sequence of approximating solutions (ρ n , v n ) = (σ 2 n , v n ) with a uniform existence time T * independent of n for the approximating initial data (ρ 0,n , v 0,n ) constructed above. In particular, (σ n , v n ) satisfies the bounds in (2.65) with T 2 replaced by T * . Thus by taking n → ∞, it is straightforward to check that we have got a strong solution (ρ, v) = (σ 2 , v) to (1.2). In fact, we have the following: Proposition 9. Consider the initial data (ρ 0 , v 0 ) (or equivalently (σ 0 , v 0 )) given in (1.16) satisfying (1.17), (1.18) and (1.19). There is a constant T * > 0 such that there exists a solution(ρ, v) = (σ 2 , v) to equation (1.2) satisfying σ ∈ L ∞ (0, T * ; H 2 ( )), ∂ t σ ∈ L ∞ (0, T * ; H 1 ( )), v ∈ L ∞ (0, T * ; H 2 ( )) ∩ L 2 (0, T * ; H 3 ( )), ∂ t v ∈ L 2 (0, T * ; H 1 ( )) σ ∂ t v ∈ L ∞ (0, T * ; L 2 ( )), (2.66) and sup 0 t T * σ (t) H 2 2B 1/2 1 , sup 0 t T * ∂ t σ (t) H 1 C 3 (B 1 , B 2 ), sup 0 t T * v(t) 2 H 2 + (σ v t )(t) 2 L 2 + T * 0 v(t) 2 H 3 + v t (t) 2 H 1 dt C 4 (B 1 , B 2 ), and inf ( x,t)∈ ×[0,T * ] ρ 0,(2.67)for some constant C 3 = C 3 (B 1 , B 2 ), C 4 = C 4 (B 1 , B 2 ). Continuous Dependence on Initial Data and UniquenessIn this section, we will show the continuous dependence of the solutions of (1.1) and (1.2) on the initial data. This will also imply the uniqueness of strong solutions constructed in Proposition 6 and Proposition 9. The Case With Gravity and γ = 2, but Without VacuumConsider two sets of initial data (ρ i,0 , v i,0 ) = (ξ i,0 + 1 2 gz, v i,0 ), i = 1, 2, in (1.9) for (1.1) satisfying (1.10), (1.11). Denote (ρ i , v i ) = (ξ i + 1 2 gz, v i ), i = 1, 2, as the corresponding strong solutions constructed in Proposition 6 in the interval [0, T ] for some T > 0. Then we have (ξ i , v i ) ∈ X T , i = 1, 2. Throughout this section we will denote the constant C > 0 which may be different from line to line and depends on μ, B g,1 , B g,2 , ρ, T . Also, we will use the notationsξ 12 := ξ 1 − ξ 2 , v 12 := v 1 − v 2 , ξ 12,0 := ξ 1,0 − ξ 2,0 , v 12,0 := v 1,0 − v 2,0 .Taking the difference of the equations satisfied by (ξ i , v i ), i = 1, 2, as in (2.42), then (ξ 12 , v 12 ) satisfies⎧ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎨ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎩ ∂ t ξ 12 + v 1 • ∇ h ξ 12 + ξ 12 div h v 1 + v 12 • ∇ h ξ 2 + ξ 2 div h v 12 + g 2 zdiv h v 12 = 0, ρ 1 ∂ t v 12 − μ h v 12 − μ∂ zz v 12 − (μ + λ)∇ h div h v 12 = −ξ 12 ∂ t v 2 − ∇ h (ξ 12 (ρ 1 + ρ 2 )) − ξ 12 v 1 • ∇ h v 1 − ρ 2 v 12 • ∇ h v 1 − ρ 2 v 2 • ∇ h v 12 − (ρ 1 w 1 − ρ 2 w 2 )∂ z v 1 − ρ 2 w 2 ∂ z v 12 .Then after applying standard L 2 estimates to the above system and applying the Grönwall's inequality to the resultant, one can show the following:Proposition 10. Given two sets of initial data (ρ i,0 , v i,0 ) = (ξ i,0 + 1 2 gz, v i,0 ), i = 1, 2, satisfying (1.10) and (1.11), the corresponding strong solutions (ρ i , v i ) = (ξ i + 1 2 gz, v i ), i = 1, 2, of (1.1) constructed in Proposition 6 in the interval [0, T ], for some T > 0, satisfyρ 1 − ρ 2 L ∞ (0,T ;L 2 ( )) + v 1 − v 2 L ∞ (0,T ;L 2 ( ))+ ∇(v 1 − v 2 ) L 2 (0,T ;L 2 ( )) C μ,λ,B g,1 ,B g,2 ,ρ,T × ( ρ 1,0 − ρ 2,0 L 2 ( )) + v 1,0 − v 2,0 L 2 ( )) ).In particular, if ρ 1,0 = ρ 2,0 , v 1,0 = v 2,0 , we have ρ 1 = ρ 2 , v 1 = v 2 in [0, T ]. The Case With Vacuum and γ > 1, but Without GravityFirst, we claim that any solution (ρ, v) = (σ 2 , v) to (1.2) satisfying (2.66) with the bounds in (2.67) will also satisfy the following equations:⎧ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎨ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎩ ∂ t σ + v • ∇ h σ + 1 2 σ div h v = 0 i n , σ w = − z 0 σ div h v + 2 v • ∇ h σ dz in , σ 2 ∂ t v + σ 2 v • ∇ h v + σ σ w∂ z v + ∇ h σ 2γ = μ h v + μ∂ zz v + (μ + λ)∇ h div h v in , ∂ z σ = 0 i n . (1.2')To show this claim, we first consider the non-degenerate variable ρ+ε = σ 2 +ε, for some constant ε > 0. From (1.12), one has∂ t (ρ + ε) + v • ∇ h (ρ + ε) + (ρ + ε)div h v − εdiv h v = 0.Then after dividing (ρ + ε) 1/2 , one has2∂ t (ρ + ε) 1/2 + 2v • ∇ h (ρ + ε) 1/2 + (ρ + ε) div h v − ε (ρ + ε) 1/2 div h v = 0. (3.1)Now it is easy to verify that (3.1) will converge to (1.2') 1 in the sense of distribution, as ε → 0. On the other hand, from (1.13), one hasσ 2 w = −σ z 0 σ div h v + 2 v • ∇ h σ dz.We defineσ w σ := − z 0 σ div h v + 2 v • ∇ h σ dz.Then σ σ w σ = ρw and we will use hereafter the notation σ w = σ w σ . As before it is easy to verify that (1.2') 3 is equivalent to (1.2') 2 in the sense of distribution. Summing up the facts above, we have shown that the solutions to (1.2) satisfying the (2.66) regularity with the bounds in (2.67) are also solutions to (1.2').Consider two sets of initial data (ρ i,0 , v i,0 ) = (σ 2 i,0 , v i,0 ), i = 1, 2, in (1.16) for (1.2) satisfying (1.18) and (1.19). Denote (ρ i , v i ) = (σ 2 i , v i ), i = 1, 2, as the corresponding strong solutions constructed in Proposition 9 in the interval [0, T * ], for some T * > 0. Then we have (σ i , v i ), i = 1, 2, satisfying the bounds in (2.67). Also (σ i , v i ), i = 1, 2, are solutions to (1.2'). Throughout this section, we will denote the constant C > 0 which may be different from line to line and depends on μ, λ, B 1 , B 2 , T * . Also, we will use the notationsσ 12 := σ 1 − σ 2 , v 12 := v 1 − v 2 ,σ 12,0 := σ 1,0 − σ 2,0 , v 12,0 := v 1,0 − v 2,0 .Taking the difference of the equations satisfied by (σ i , v i ), i = 1, 2, we have Therefore, after employing (1.22) and noticing the fact that we can interchange (σ 1 , v 1 ), (σ 2 , v 2 ) in the previous arguments, we will have the following: Proposition 11. Given two sets of initial data (ρ i,0 , v i,0 ) = (σ 2 i,0 , v i,0 ), i = 1, 2, for (1.2) satisfying (1.17), (1.18) and (1.19), the corresponding strong solutions (ρ i , v i ) = (σ 2 i , v i ), i = 1, 2, constructed in Proposition 9 in the interval [0, T * ], for some T * > 0, satisfy⎧ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎨ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎩ ∂ t σ 12 + v 1 • ∇ h σ 12 + 1 2 σ 12 div h v 1 + v 12 • ∇ h σ 2 + 1 2 σ 2 div h v 12 = 0, σ 2 1 ∂ t v 12 − μ h v 12 − μ∂ zz v 12 − (μ + λ)∇ h div h v 12 = −σ 12 (σ 1 + σ 2 )∂ t v 2 − ∇ h σ 12 σ 2γ 1 − σ 2γ 2 σ 1 − σ 2 − σ 12 (σ 1 + σ 2 )v 2 • ∇ h v 2 − σ 2 1 v 12 • ∇ h v 2 − σ 2 1 v 1 • ∇ h v 12 − σ 12 σ 2 w 2 ∂ z v 2 − σ 1 (σ 1 w 1 − σ 2 w 2 )∂ z v 2 − σ 1 σ 1 w 1 ∂ z v 12 , σ i w i = − z 0 σ i div h v i + 2 v i • ∇ h σ i dz, i = 1, 2.σ 1 − σ 2 L ∞ (0,T * ;L 2 ( )) + σ 1 (v 1 − v 2 ) L ∞ (0,T * ;L 2 ( )) + σ 2 (v 1 − v 2 ) L ∞ (0,T * ;L 2 ( )) + v 1 − v 2 L 2 (0,T * ;L 2 ( )) + ∇(v 1 − v 2 ) L 2 (0,T * ;L 2 ( )) C μ,λ,B 1 ,B 2 ,T * σ 1,0 − σ 2,0 L 2 ( )) + v 1,0 − v 2,0 L 2 ( )) .In particular, if ρ 1,0 = ρ 2,0 , v 1,0 = v 2,0 , we have ρ 1 = ρ 2 , v 1 = v 2 in [0, T * ]. Proofs of the Main TheoremsTheorem 1 follows from Proposition 6 and Proposition 10. Theorem 2 follows from Proposition 9 and Proposition 11. 2. 1 . The Case with Gravity and γ = 2 2.1.1. Associated Linear Inhomogeneous System Consider a finite positive time T , which will be determined later. Let Y = Y T be the function space defined by 2 L 2 + 2 L 2 + 2 C σ 12 ,0 2 L 2 + v 12 C μ,λ (σ 1 v 12 )(t) C μ,λ σ 1,0 v 12,0 2 L The H 1 ( ) norm for ∂ t ξ Using equation (2.6) 1 , ∂ t ξ can be represented in terms of the spatial derivatives of ξ, v o . Then applying the Hölder and the Sobolev embedding inequalities implies sufficiently small, (2.14) yields sup 0 t T ξ(t) 2 H 2 2M 0 . (2.15) .28) Proof of Proposition 4. From (2.20), (2.23), (2.24) and (2.26), there is a constant c μ,λ,ρ such that d dt E g whereωis as above. Now, we integrate with respect to the time variable inequality (2.29). It follows, since 0 < ω < c μ,λ,ρ 2 , that c μ,λ,ρ .61) where C d is as in(2.47) We summarize the estimates obtained, so far, in this section in the following:Consider the solution (σ, v) = (ρ 1/2 , v) to (1.2) with the bound (2.47) and initial data satisfying (1.17), (2.44). There is a positive constant T * = T * (C d , K 1 , K 2 ), sufficiently small, such that (σ, v) admits the following bounds, for T = min{T * , T d }, Proposition 8. inf ( x,t)∈ ×[0,T ] .65) Next, let (σ n , v n )| t=T 2 as a new initial data for (1.2). The same arguments as above yield the bound (2.65) with lower bound of ρ n replaced by 1 4n , B 1 replaced by 4B 1 and B 2 replaced by C 4 Acknowledgements. The authors would like to thank the referee for the useful remarks that led to the improvement of the paper. They are also thankful to the École Polytechnique for its kind hospitality, where this work was completed, and the École Polytechnique Foundation for its partial financial support through the 2017-2018 "Gaspard Monge Visiting Professor" Program. This work is supported in part by the NSF grant number DMS-1516866 and by the ONR grant N00014-15-1-2333. The work of E.S.T. was also supported in part by the Einstein Stiftung/Foundation-Berlin, through the Einstein Visiting Fellow Program. The Case WithGravity and γ = 2, but Without Vacuum We will apply the Schauder-Tchonoff fixed point theorem to establish the existence of strong solutions to (1.1). With Proposition 2, it is sufficient to verify that T , defined by (2.10), is continuous in V = V T given in (2.3), where the norm is given by(2.4)In order to show this, let M 0 = B g,1 , M 1 = B g,2 in X T and T ∈ (0, T g ], with T g given in Proposition 2. Here B g,1 , B g,2 are given in (1.11). We denoteDenote byThen (ξ 12 , v 12 )| t=0 = 0. 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