Global spherically symmetric flows for a viscous radiative and reactive gas in an exterior domain

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ScienceDirect J. Differential Equations 266 (2019) 6459–6506 www.elsevier.com/locate/jde

Global spherically symmetric flows for a viscous radiative and reactive gas in an exterior domain Yongkai Liao a,b , Tao Wang a,b,∗ , Huijiang Zhao a,b a School of Mathematics and Statistics, Wuhan University, Wuhan 430072, China b Hubei Key Laboratory of Computational Science, Wuhan University, Wuhan 430072, China

Received 9 May 2018 Available online 9 November 2018

Abstract We consider the equations for a viscous, compressible, radiative and reactive gas (pressure P = Rρθ + aθ 4 /3, internal energy e = cv θ + aθ 4 /ρ) over an unbounded exterior domain in Rn , where n ≥ 2 is the space dimension. The existence, uniqueness, and large-time behavior of global spherically symmetric solutions are established for large initial data. The key point in the analysis is to deduce certain uniform a priori estimates on the solutions, especially on lower and upper bounds of the specific volume and temperature. © 2018 Elsevier Inc. All rights reserved. Keywords: Viscous radiative and reactive gas; Spherically symmetric solutions; Large initial data; Global well-posedness; Large-time behavior

1. Introduction 1.1. The Eulerian description We are concerned with the existence, uniqueness, and large-time behavior of global spherically symmetric solutions to a model for the dynamic combustion of a viscous, compressible, radiative and reactive gas in the unbounded exterior domain  := {ξ ∈ Rn : |ξ | > 1}, where ξ is * Corresponding author.

E-mail addresses: [email protected] (Y. Liao), [email protected] (T. Wang), [email protected] (H. Zhao). https://doi.org/10.1016/j.jde.2018.11.008 0022-0396/© 2018 Elsevier Inc. All rights reserved.

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the spatial variable with space dimension n ≥ 2. The motion of the gas is described by the compressible Navier–Stokes system coupled to the reaction–diffusion equation, which can be written in the Eulerian coordinates as (cf. Williams [28, Chapter 1]) ⎧ ρt + ∇ · (ρu) = 0, ⎪ ⎪ ⎪ ⎪ ⎪ ⎨ (ρu)t + ∇ · (ρu ⊗ u) + ∇P = ∇ · S, ⎪ (ρe)t + ∇ · (ρeu) + P ∇ · u = ∇ · (κ∇θ ) + S : ∇u + λφρz, ⎪ ⎪ ⎪ ⎪ ⎩ (ρz)t + ∇ · (ρuz) = ∇ · (ρd∇z) − φρz.

(1.1)

Here t > 0 is the time variable, and the primary dependent variables are density ρ, velocity u, absolute temperature θ , and reactant mass fraction z. We treat the radiation as a continuous field and take both the wave and photonic effects into account. We also assume that the high-temperature radiation is at thermal equilibrium with the fluid. As a result, the pressure P and the internal energy e consist of a linear term in θ corresponding to the perfect polytropic contribution and a fourth-order radiative part due to the Stefan–Boltzmann law (see, e.g., Mihalas and Mihalas [20, p. 320]): a P = P (ρ, θ ) = Rρθ + θ 4 , 3

e = e(ρ, θ ) = cv θ + a

θ4 , ρ

(1.2)

where R, cv , and a are positive constants. In this paper we concentrate on Newtonian fluids for which the viscous stress tensor S reads as   S = μ ∇u + (∇u)T + ζ ∇ · u I, where (∇u)T is the transpose matrix of ∇u and I is the n × n identity matrix. The viscosity coefficients μ and ζ , the species diffusion d > 0, and the heat release λ > 0 are supposed to be constants and satisfy μ > 0,

nζ + 2μ > 0.

(1.3)

The heat conductivity κ is assumed to have the following form (cf. Ducomet [5]): κ(ρ, θ) = κ1 + κ2

θb , ρ

(1.4)

where κ1 , κ2 , and b are positive constants. The reaction function φ = φ (θ ) is defined by the first-order Arrhenius law: φ(θ) = θ β e−A/θ

(β ≥ 0),

(1.5)

where positive constant A stands for the activation energy. We shall consider the initial boundary value problem of system (1.1) in the region (0, ∞) ×  with the following initial and boundary conditions:

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(ρ, u, θ, z)|t=0 = (ρ0 , u0 , θ0 , z0 )(ξ ),

ξ ∈ ,

(u, ∂n θ, ∂n z)|∂ = 0,

t ≥ 0.

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(1.6)

Here n denotes the unit outer normal to ∂. We suppose that the boundary conditions are compatible with the initial data. The global well-posedness of large solutions to initial/initial–boundary value problems of system (1.1)–(1.2) has been studied by many authors. In the non-radiative (a = 0) case, the existence of global strong solutions was proved by Chen [1], Chen et al. [2,3], and Li [16], among others. For the radiative (a > 0) case, the well-posedness of global classical solutions in one-dimensional bounded domains was established by [5,12,18] for the Dirichlet–Neumann boundary conditions, and by [11,18,21,23,24] for the free and pure Neumann boundary conditions. Also see Ducomet and Zlotnik [6,7] for exponential stabilization of solutions by constructing new global Lyapunov functionals. The global existence and large-time behavior of solutions to the Cauchy problem have been treated very recently by Liao and Zhao [19] for the constant viscosity case and by He et al. [9] for the temperature-dependent viscosity case. In more than one dimension, the global existence of variational solutions was shown by Donatelli and Trivisa [4], Feireisl and Novotný [8] for bounded domains. As for the classical solutions to (1.12), the local existence was given in Jenssen et al. [10, Appendix] for sufficiently smooth initial data. Zhang [29] showed that spherically and cylindrically symmetric solutions exist globally-in-time for initial–boundary value problem of (1.1) in a bounded annular domain, which extended the results of [22,25] to large initial data. It is worth noting that the boundedness of domains is essential in the analysis of [22,25,29]. We are interested in global well-posedness of spherically symmetric large solutions to (1.1)–(1.6) in the unbounded domain . Supplemented with spherically symmetric initial data, the solution (ρ, u, θ ) to problem (1.1)–(1.6) is also spherically symmetric, i.e.

ξ (ρ(t, ξ ), u(t, ξ ), θ (t, ξ ), z(t, ξ )) = ρ(t, ˆ r), u(t, ˆ r), θˆ (t, r), zˆ (t, r) , r

r := |ξ | ≥ 1.

The system for (ρ, ˆ u, ˆ θˆ , zˆ ) then takes the form: ⎧ ⎪ ˆ r (r n−1 ρˆ u) ⎪ ⎪ = 0, ρˆt + ⎪ n−1 ⎪ r ⎪ ⎪ ⎪  ⎪ ⎪ ˆ r α(r n−1 u) ⎪ ⎪ ˆ ⎪ ρ( ˆ u ˆ + u ˆ u ˆ ) + P = , t r r ⎨ r n−1 r

⎪ ⎪ ˆ r Pˆ (r n−1 u) (κr ˆ n−1 θˆr )r ⎪ ⎪ = + Qˆ + λφρz, ρ( ˆ eˆt + uˆ eˆr ) + ⎪ ⎪ ⎪ r n−1 r n−1 ⎪ ⎪ ⎪ ⎪ ⎪ (ρdr ˆ n−1 zˆ r )r ⎪ ⎩ ρ(ˆ ˆ zt + uˆ ˆ zr ) = − φρz, r n−1 where α = 2μ + ζ and α[(r n−1 u) ˆ r ]2 2(n − 1)μ(r n−2 uˆ 2 )r Qˆ = − . r 2n−2 r n−1

(1.7)

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The initial and boundary conditions (1.6) are reduced to 

(ρ, ˆ u, ˆ θˆ , zˆ )|t=0 = (ρˆ0 , uˆ 0 , θˆ0 , zˆ 0 )(r),

r ≥ 1,

(u, ˆ ∂r θˆ , ∂r zˆ )|r=1 = 0,

t ≥ 0.

(1.8)

These boundary conditions are supposed to be compatible with the initial data. 1.2. Lagrangian coordinates and the main result In order to achieve the global existence, it is more convenient to transform initial boundary value problem (1.7)–(1.8) into that in the Lagrangian coordinates. We introduce the Lagrangian variables (t, x) and denote (ρ, u, θ, z)(t, x) := (ρ, ˆ u, ˆ θˆ , zˆ )(t, r), where

t r = r(t, x) = r0 (x) +

u(s, ˆ r(s, x))ds,

(1.9)

0

and −1

r0 (x) := h

r (x),

h(r) :=

y n−1 ρˆ0 (y)dy.

(1.10)

1

Then one can obtain the following identities: rt (t, x) = u(t, x),

rx (t, x) = r 1−n v(t, x),

(1.11)

where v := 1/ρ is the specific volume. In view of (1.11), system (1.7) can be reformulated to that for (v, u, θ, z)(t, x) as vt = (r n−1 u)x ,  α(r n−1 u)x ut = r n−1 −P , v x  2n−2  n−1 θx u)x κr α(r + et = − P (r n−1 u)x v v x −(2n − 2)μ(r n−2 u2 )x + λφz,  2n−2 zx dr zt = − φz, v2 x

(1.12a) (1.12b)

(1.12c) (1.12d)

where t > 0 and x ∈ R+ := (0, ∞). The corresponding initial and boundary conditions are

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(v, u, θ, z)|t=0 = (v0 , u0 , θ0 , z0 )

for x ≥ 0,

(1.13a)

(u, ∂x θ, ∂x z)|x=0 = 0

for t ≥ 0.

(1.13b)

The initial data are supposed to be compatible with boundary conditions (1.13b) and satisfy the far-field condition: lim (v0 , u0 , θ0 , z0 )(x) = (1, 0, 1, 0).

(1.14)

x→∞

We are now in a position to state the main result of the present paper, that is, the global wellposedness theorem for problem (1.12)–(1.13). Clearly, this result implies an equivalent statement for the corresponding problem (1.7)–(1.8) in the Eulerian coordinates. Theorem 1.1. Suppose that the transport coefficients μ, ζ , and κ satisfy (1.3)–(1.4), and b>

19 , 4

0 ≤ β < b + 9.

(1.15)

Let the initial data (v0 , u0 , θ0 , z0 ) be compatible with boundary conditions (1.13b) and satisfy (v0 − 1, u0 , θ0 − 1, z0 , ∂xx u0 ) ∈ L2 (R+ ),

inf {v0 (y), θ0 (y)} > 0,

y∈R+

(r n−1 ∂x v0 , r n−1 ∂x u0 , r n−1 ∂x θ0 , r n−1 ∂x z0 , r n−1 ∂xx v0 ) ∈ L2 (R+ ), z0 ∈ L1 (R+ ),

0 ≤ z0 (x) ≤ 1

for all x ∈ R+ .

(1.16)

Then there exist positive constants V , V , , , and C such that problem (1.12)–(1.13) admits a unique global solution (v, u, θ, z) satisfying V ≤ v(t, x) ≤ V ,

≤ θ (t, x) ≤ ,

0 ≤ z(t, x) ≤ 1,

(1.17)

and

(v − 1, u, θ − 1, z, uxx ) (t) 2L2 (R

+

 2  n−1  + r , u , θ , z , v (t) (v )   2 x x x x xx )

L (R+ )

   2   + z(t) L1 (R+ ) + r n−1 vx , ux , θx , zx , vxx , r n−1 uxx , θxx , zxx (s) 2

L (R+ )

R+

ds ≤ C, (1.18)

for all (t, x) ∈ R+ × R+ . Moreover, the global solution (v, u, θ, z)(t, x) converges to the nonvacuum equilibrium state (1, 0, 1, 0) as time goes to infinity: lim sup |(v − 1, u, θ − 1, z)(t, x)| = 0.

t→∞ x∈R

+

(1.19)

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1.3. Outline of the paper and our methodology As pointed out by Liao and Zhao [19], the crucial step to construct global solutions of initial– boundary value problem (1.12)–(1.13) with large initial data is to obtain the positive lower and upper bounds of the specific volume v(t, x) and temperature θ (t, x). In particular, in order to deduce the upper bound of θ (t, x), one has to derive certain estiamtes on some nonlinear terms, among which the most difficulty one is the higher-order term

t r 4n−4 u4x 0 I

for I := R+ . This term can be controlled in Zhang [29] where radius r(t, x) is assumed to satisfy r1 ≤ r(t, x) ≤ r2 with r1 and r2 being two positive constants so that domain I is bounded. However, the domain under our consideration is unbounded. To overcome such difficulty, we will deduce bounds on vx (t) L2 (R+ ) ,

t

t

r

n−1

r n−1 ux (s) 2L∞ (R+ ) ds,

uxx (s) 2L2 (R ) ds, +

0

r n−1 vx (t) L2 (R+ ) ,

0

t  and r n−1 ux (t) L2 (R+ ) in terms of θ L∞ ([0,T ]×R+ ) . Then the nonlinear term 0 R+ r 4n−4 u4x can be estimated as follows (cf. (4.59) along with the definitions of l2 and l3 in (3.12) and (3.16)):

t

t r 4n−4 u4x

0 R+



2l +l

2 3

r n−1 ux (s) 2L∞ (R+ ) r n−1 ux (s) 2 ds  θ L∞ ([0,T ]×R+ ) .

0

The above estimate turns out to play a central role in our analysis. The rest of this paper is organized as follows. In Section 2, we derive the uniform-in-time upper and lower bounds of the specific volume v(t, x). For this purpose, we first construct a normalized entropy η (cf. (2.4)) to system (1.12) and derive basic energy estimates for our problem. We emphasize that a different method from Zhang [29] needs to be applied because of the unboundedness of the domain under our consideration. Then we derive a local representation of v(t, x) (cf. (2.11)) and obtain the uniform pointwise bounds of v(t, x). This is motivated by the works of Jiang [13,14] on the one-dimensional, compressible Navier–Stokes system for a viscous and heat-conducting ideal polytropic gas. t Section 3 is devoted to the derivation of estimates on vx (t) L2 (R+ ) , 0 r n−1 uxx (s) 2L2 (R ) ds, +  t n−1 2 n−1 (v , u )(t)

∞ ([0,T ]×R ) , which will

r u (s)

ds, and

r in terms of

θ

2 ∞ x x x L L (R+ ) + L (R+ ) 0 be useful in getting the upper bound of θ (t, x). Motivated by Kawohl [15] and [19], we introduce in Section 4 the auxiliary functions X(t), Y (t), and Z(t) in order to deduce the upper bound of θ (t, x). We notice that our definition of Y (t) is different from that in Liao and Zhao [19, (2.51)] as a result of (4.16) and the fact that r(t, x) ≥ 1. Moreover, it is worth noting that the method used in Liang [17] does not work for our case owing to the fourth-order radiative parts in P (v, θ ) and e(v, θ ) (cf. (1.2)).

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In Section 5, we derive a local-in-time estimate on the lower bound of θ (t, x). Even though such a bound depends on time variable t , it is sufficient to extend the local solution to a global one by combining these a priori estimates with the continuation argument introduced in [27]. Then the proof of Theorem 1.1 is completed. Notations. To simplify the presentation, we employ C, c, and Ci (i ∈ N) to denote various positive constants, depending only on μ, λ1 , λ, A, d, R, cv , a, κ1 , κ2 , n and initial data (v0 , u0 , θ0 , z0 ). Hence C, c, and Ci are independent of t. Symbol A  B (or B  A) means that A ≤ CB holds uniformly for some uniform-in-time constant C. 2. Uniform bounds for the specific volume The main purpose of this section is to deduce the uniform-in-time positive bounds for the specific volume v(t, x) to initial–boundary value problem (1.12)–(1.13) in terms of initial data (v0 , u0 , θ0 , z0 ). To this end, we assume (v, u, θ, z) ∈ X(0, T ; M1 , M2 ; N1 , N2 ), where the set X(t1 , t2 ; M1 , M2 ; N1 , N2 ) is defined as:  X(t1 , t2 ; M1 , M2 ; N1 , N2 ) := (v, u, θ, z) : z ∈ C(t1 , t2 ; L2 (R+ ) ∩ L1 (R+ )), (v − 1, u, θ − 1, r n−1 vx , r n−1 ux , r n−1 θx , r n−1 zx , r n−1 vxx , uxx ) ∈ C(t1 , t2 ; L2 (R+ )), r n−1 (vx , ux , θx , zx , vxx , uxx , θxx , zxx ) ∈ L2 (t1 , t2 ; L2 (R+ )),

 ∀ (t, x) ∈ [t1 , t2 ] × R+ ,

M1 ≤ v(t, x) ≤ M2 , N1 ≤ θ (t, x) ≤ N2

for constants M, N , t1 , and t2 (t1 ≤ t2 ). For the sake of simplicity, we will use the following abbreviations:

· ∞ := · L∞ ([0,T ]×R+ ) ,

· := · L2 (R+ ) .

In the next lemma we have the basic energy estimates, which will play a fundamental role in deducing the desired positive lower and upper bounds of the specific volume v(t, x). Lemma 2.1. Assume that the conditions listed in Theorem 1.1 hold. Then for all t ∈ [0, T ],

t z(t, x)dx +

φz =

R+

0 R+



t

R+



z2 (t, x)dx + R+

0 R+

R+

dr 2n−2 zx2 2 + φz = z02 (x)dx, v2

(2.2)

R+

T η(t, x)dx +

sup

t∈[0,T ]

(2.1)

z0 (x)dx,

V (t)dt  1,

(2.3)

0

where the entropy η and the entropy dissipation rate functional V (t) are defined respectively by

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u2 av η(v, θ) := Rψ(v) + + cv ψ(θ ) + (θ − 1)2 (3θ 2 + 2θ + 1), ψ(y) := y − ln y − 1, (2.4) 2 3 

  n−1 2  n−1 2 (r κ r θx u)x vu2 r 2(n−1) u2x φz V (t) := + + 2 + + (t, x)dx. (2.5) vθ 2 vθ r θ vθ θ R+

Proof. Identities (2.1) and (2.2) follow directly from (1.12d) and integration by parts. According to Liao and Zhao [19], function η(v, θ ) is the normalized entropy around (v, u, θ ) = (1, 0, 1) for system (1.12) with constitutive relations (1.2). A straightforward calculation yields  2  2

α (r n−1 u)x  κ r n−1 θx λφz 1  n−2 2  ηt + + u + + 2μ(n − 1) 1 − r x vθ θ θ vθ 2 



 α(r n−1 u)x 1 κr 2n−2 θx Rθ a a  = r n−1 u + λφz. − + R + − θ 4 r n−1 u + 1 − v v 3 3 θ v x We integrate the above identity over (0, t) × R+ and use (2.1) to obtain

R+



t   n−1 2 κ r θx λφz η(t, x)dx + + θ vθ 2 0 R+

 n−2 2  

t   n−1 2 u x r α (r u)x − 2μ(n − 1)  1. + vθ θ

(2.6)

0 R+

By virtue of (1.11), we can get (cf. [26, (2.14)]) that  n−2 2  2  u x r α (r n−1 u)x  vu2 r 2(n−1) u2x − 2μ(n − 1) ≥ + . vθ θ Cr 2 θ Cvθ Then we plug the above estimate into (2.6) to derive (2.3).

2

Since the pointwise bounds of z(t, x) can be established by a standard maximum principle, we omit the proof for brevity (see, for instance, Chen [1]). Lemma 2.2. Assume that the conditions listed in Theorem 1.1 hold. Then 0 ≤ z(t, x) ≤ 1 for all (t, x) ∈ [0, T ] × R+ .

(2.7)

The next lemma follows from utilizing (2.3) and Jensen’s inequality. Lemma 2.3. Assume that the conditions listed in Theorem 1.1 hold. Then for all k ∈ N and t ∈ [0, T ], there exist ak (t), bk (t) ∈ k := [k, k + 1] such that

v(t, x)dx ∼ 1, θ (t, x)dx ∼ 1, v(t, ak (t)) ∼ 1, θ(t, bk (t)) ∼ 1. (2.8) k

k

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In the following lemma we deduce a rough estimate on θ (t, x) in terms of V (t) (cf. (2.5)). Lemma 2.4. Let 0 ≤ m ≤ b + 1 and x ∈ R+ . Then 1 − CV (t)  θ (t, x)m  1 + V (t),

for all 0 ≤ t ≤ T .

(2.9)

Proof. We assume x ∈ k without loss of generality. In light of (1.4), we have ⎤1 ⎡

⎡

  m m ⎢  θ (t, x) 2 − θ (t, bk (t)) 2   ⎣  k

⎡ ⎢ ⎣

k

2

vθ m

⎥ ⎢  dx ⎦ ⎣ 1 + vθ b r 2(n−1)

κ

2



r n−1 θ vθ 2

x

⎤1

2

⎥ dx ⎦

k

⎤1

2

vθ m 1 + vθ

1 1 ⎥ dx ⎦ V (t) 2  V (t) 2 . b

To derive the last inequality, we have used assumption m ≤ b + 1, boundedness of k , and estimate (2.8). 2 By defining the cut-off function ϕ ∈ W 1,∞ (R+ ) as follows: ⎧ ⎪ ⎨1, ϕ(x) = k + 2 − x, ⎪ ⎩ 0,

0 ≤ x ≤ k + 1, k + 1 ≤ x ≤ k + 2, x ≥ k + 2,

(2.10)

we can deduce a local representation of v(t, x) in the next lemma. Lemma 2.5. Assume that the conditions listed in Theorem 1.1 hold. Then 1 v(t, x) = B(t, x)Q(t, x) + α

t

B(t, x)Q(t, x)v(s, x)P (s, x) ds, B(s, x)Q(s, x)

(t, x) ∈ [0, T ] × k ,

0

(2.11) where B(t, x) := v0 (x) exp

⎧ ∞ ⎨1  ⎩α

x



⎫ ⎬

u0 r01−n − ur 1−n ϕ(y)dy , ⎭

⎧ ⎫ ⎪ ⎪

t ∞ ⎨ 1 t  ⎬ n−1 u)x n−1 α(r Q(t, x) := exp ϕu2 r −n . − −P + ⎪ ⎪ v α ⎩α ⎭ 0 k+1

0 x

With basic energy estimates (2.1)–(2.3) and local formulation (2.11) in hand, we are ready to derive uniform-in-time pointwise bounds of v(t, x).

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Lemma 2.6. Assume that the conditions listed in Theorem 1.1 hold. Then V ≤ v(t, x) ≤ V

for all (t, x) ∈ [0, T ] × R+ .

(2.12)

Proof. First we use the second identity of (1.11) to infer r(t, x) ≥ 1. Then it follows from (2.3) that C −1 ≤ B(t, x) ≤ C

for all x ∈ k .

(2.13)

By modifying the argument in [13, p. 186], we can get 

t −

inf θ (τ, x)dτ ≤

s

x∈k

0, −C(t − s),

0 ≤ t − s ≤ 1, t − s ≥ 1.

Consequently, we utilize Hölder’s and Jensen’s inequalities to obtain

t 



t Rθ α(r n−1 u)x − ≤C v v

s k+1

≤C−

vθ

R − 2

s k+1

R 2

t

R 2



t s

x∈k+1

k+1

⎡

θ v

s k+1

⎤−1

⎢ inf θ (τ, ·) ⎣

t

1 dxdτ v(τ, x)

inf θ (τ, ·)

s

≤C−

 n−1 2 (r u)x 

⎥ vdx ⎦

x∈k+1

dτ ≤ C − C(t − s),

0 ≤ s ≤ t ≤ T . (2.14)

k+1

From the definition of Q(t, x) and (2.14), we have for 0 ≤ s ≤ t ≤ T that ⎧ ⎫ ⎪ ⎪

t  ⎨ n−1 Q(t, x) u)x Rθ ⎬ α(r 1 ≤ exp − ≤ C exp{−C(t − s)}. ⎪ Q(s, x) v v ⎪ ⎩α ⎭ s k+1

Using (2.13)–(2.15) and Lemma 2.4, we have

t v(t, x)  Q(t, x) +

Q(t, x) v(s, x)P (s, x)ds Q(s, x)

0

t exp{−C(t − s)}(1 + v(s, x))(1 + θ 4 (s, x))ds

1+ 0

t exp{−C(t − s)}(1 + v(s, x))(1 + V (s))ds.

1+ 0

(2.15)

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Then applying Gronwall’s inequality yields the uniform upper bound of v: v(t, x) ≤ C.

(2.16)

Now we turn to deduce a positive lower bound of v(t, x). As in [17, Lemma 3.1], we can derive  t ∞    t   −n 2   ϕr u      0 x

0

   t   n−1 (r u)x      1.   v  0 k+1 

 u 2   ds  1,  n/2  ∞ r L (R+ )

As a consequence, one has ⎫ ⎧ ⎪ ⎪

t ⎬ ⎨ Q(t, x) 1 P (v, θ ) =: Q(s, t). ∼ exp − ⎪ ⎪ Q(s, x) ⎭ ⎩ α

(2.17)

s k+1

In view of (2.3) and (2.8), we have

vP dx  k

!



θ + v 1 + (θ − 1) 2

2

"

dx 

k

[θ + v + η] dx  1.

(2.18)

k

From (2.13), (2.17) and (2.18), we integrate (2.11) over k to conclude

t

1

Q(t, x) vP (s, x)dxds Q(s, x)

Q(t, x)dx + C 0 k

k

t  exp{−Ct} +

Q(s, t) 0

t

vP (s, x)dxds  exp{−Ct} +

Q(s, t)ds. 0

k

Plugging (2.9), (2.13), (2.15), (2.17), and (2.19) into (2.11) yields

t v(t, x) 

Q(t, x) θ (s, x)ds  Q(s, x)

0

t Q(s, t)ds − C

0

Q(t, x) (1 − CV (s))ds Q(s, x)

0

t 

t

Q(t, x) V (s)ds Q(s, x)

0

⎧ ⎫ ⎪ ⎨ t/2 t ⎪ ⎬  1 − C exp{−Ct} − C + exp{−C(t − s)}V (s)ds ⎪ ⎪ ⎩ ⎭ 0

t/2

(2.19)

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#

$

Ct  1 − C exp{−Ct} − C exp − 2

t V (s)ds  1,

−

(2.20)

t 2

for t ≥ T0 , where T0 is a positive constant independent of t . On the other hand, we use (2.13), (2.17), and integrate (2.11) over k to find

Q(0, t)

−1

t 1+

Q(0, s)−1

0

vP dxds,

k

which combined with (2.18) implies Q(0, t)  exp{−Ct}. Use again (2.11) to deduce that v(t, x)  exp{−Ct}. Then we combine with (2.20) to obtain the uniform lower bound of v and hence finish the proof of this lemma. 2 The next corollary follows directly from (1.9) and Lemma 2.6. Corollary 2.7. Assume that the conditions listed in Theorem 1.1 hold. Then 1 + C −1 x ≤ r n (t, x) ≤ 1 + Cx

for all (t, x) ∈ [0, T ] × R+ .

(2.21)

3. More energy estimates Lemma 3.1. Under the assumptions listed in Theorem 1.1, we have

t    u 2 ds  1  (s) ∞ r L (R+ )

for all t ∈ [0, T ].

(3.1)

0

Proof. Estimate (3.1) for n = 2 can be proved by a similar argument in [17, Lemma 3.1]. Now let n ≥ 3. Since ⎛ u =⎝

∞

2

⎞2 ux dx ⎠ 

x

R+

r 2n−2 u2x dx vθ

R+

vθ r 2n−2

dx,

and

R+

k+1

k+1 ∞ ∞ ∞ ) ) ) θ 1 1 dx = dr  θ dr   1, r 2n−2 r n−1 k n−1 k n−1

vθ

k=1 k

k=1

k

we use the fact that r ≥ 1 to complete the proof of our lemma. 2

k=1

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Lemma 3.2. Under the assumptions listed in Theorem 1.1, we have for any 0 ≤ t ≤ T that

t √ 1 ,

vx (t) + θ vx (s) 2 ds  1 + C θ l∞ 2

(3.2)

0

where l1 = max{1, (7 − b)+ }.

(3.3)

Proof. We first rewrite (1.12b) as r 1−n ut + Px = α

v  t

v

x

=α

v  x

v

.

(3.4)

t

Multiply the above equation by vx /v to get    vx Rθv 2 ! vt " α vx 2 − r 1−n u + 3 x + r 1−n u 2 v v t v v x u2 vx  1−n  vt Rθx vx 4a θ 3 vx θx = + u + (n − 1)r −n + r . 2 x v 3 v v v Integrating this last identity, we utilize (2.3), (2.12), and Cauchy’s inequality to deduce

t √

t ! " 2

vx (t) + θ vx (s) ds  1 + (θ −1 + θ 5 )θx2 + r −n u2 |vx | + |(r n−1 u)x vt | . (3.5) 2

0 R+

0

We now estimate the terms on the right-hand side of (3.5). First, it follows from (2.3) that

t (θ

−1

t +θ

0 R+

5

)θx2

κθx2 θ + θ 7 (7−b)  1 + θ ∞ + . 2 b θ 1+θ

 0 R+

(3.6)

By virtue of (2.3), we get

t r

t

−n 2

u |vx |  θ ∞

0 R+

0 R+

t  θ ∞ + 0

u2 + r 2θ

t  2 u

vx 2   ∞ r L (R+ ) 0

 u 2  

vx 2 .   ∞ r L (R+ )

Using (1.12a) and 



r 1−n u

x

= r 2(1−n) (r n−1 u)x + 2 (1 − n) r 1−2n vu,

(3.7)

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we have

t 

t |(r

n−1

u)x vt |  θ ∞

0 R+

  2 r 2(1−n)  r n−1 u x  vθ

0 R+

vu2 + 2 r θ

  θ ∞ .

Plug (3.6)–(3.8) into (3.5), apply Gronwall’s inequality and use (3.1) to obtain (3.2).

(3.8)

2

Lemma 3.3. Under the assumptions listed in Theorem 1.1, we have

T sup ux (t) +

2 ,

r n−1 uxx 2  1 + θ l∞

2

t∈[0,T ]

(3.9)

0

l2 /4 ,

u ∞  1 + C θ ∞

(3.10)

T 2 ,

r n−1 ux 2L∞ (R+ ) ds  1 + θ l∞

(3.11)

0

where + * l2 = max 2l1 + 1, (8 − b)+ .

(3.12)

Proof. Multiplying (1.12b) by uxx yields   αr 2(n−1) u2xx uv r 2(n−1) ux vx n−2 + + (n − 1) 2 − 2(n − 1)r ux − αuxx v v2 r t

4a θ vx θx = (ux ut )x + Ruxx r n−1 − 2 + uxx r n−1 · θ 3 θx . v 3 v



u2x 2



Integrating this last identity over [0, t] × R+ , we get from (2.3) and (3.2) that

t

ux (t) +

r n−1 uxx (s) 2 ds

2

0

t

 r 2(n−1) u2x vx2 +

1+ 0 R+

!

1  1 + 1 + θ l∞

"

u2 r 2(n+1)

+

u2x + θx2 + θ 2 vx2 + θ 6 θx2 r2

⎡

t  2  ⎣  r n−1 ux  ∞

L (R+ )

0



⎤ ds + θ ∞ ⎦ + θ ∞

(8−b)+

.

(3.13)

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Noticing 

r n−1 ux

 x

= (n − 1)r −1 vux + r n−1 uxx ,

we have

t    n−1 2 ux  ∞ r

L (R+ )

t         ds  r n−1 ux   r n−1 ux  ds x

0

0

t

r



n−1

C uxx + 

t

0

t

r

2

n−1

0

ux  

r n−1 uxx 2 +

2

C

θ ∞ , 

(3.14)

0

1 −1 for any positive . Plugging (3.14) with  := (1 + θ l∞ ) to (3.13) yields (3.9). Estimates (3.10) and (3.11) follow immediately from (2.3), (3.9), and (3.14). 2

Lemma 3.4. Under the assumptions listed in Theorem 1.1, we have for all 0 ≤ t ≤ T that

t

r

n−1

vx (t) + 2

√ 3 ,

r n−1 θ vx (s) 2 ds  1 + θ l∞

(3.15)

0

where l3 = max{1, l2 /2, (7 − b)+ }.

(3.16)

Proof. In view of (1.12a)–(1.12b), we derive



r n−1 vx α v

+ t

Rr n−1 θ vx Rr n−1 θx 4a n−2 vx = u − α(n − 1)r u + + r n−1 θ 3 θx . t v v 3 v2

Multiplying this identity by r n−1 vx /v, we utilize Cauchy’s inequality, (1.2), and (2.3) to deduce

t

r

n−1

vx (t) + 2

√

r n−1 θ vx (s) 2 ds

0

t  1+ 0 R+

 1 + θ

|(r n−1 u)x |2 κr 2(n−1) θx2 θ + θ 7 · + r 2n−3 |u|vx2 + r n−2 u2 |vx | + v vθ 2 1 + θb

(7−b)+

t

t  |(r n−1 u)x |2 n−2 2 + u |vx | + + r v 0 R+

0



 u 2  

r n−1 vx 2 .   ∞ r L (R+ ) (3.17)

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From (2.3) and (3.10), one has

t  |(r n−1 u)x |2 r n−2 u2 |vx | + v 0 R+

t 

t

2

t √ u v |(r n−1 u)x |2 C

r n−1 θ vx (s) 2 ds +

u 2L∞ (R+ ) +

θ

∞ 2  r θ vθ

0

0

t 

R+

0 R+

√ C l2 /2

r n−1 θ vx (s) 2 ds + (1 + θ ∞ ) + θ ∞ . 

0

Plug the last estimate into (3.17), employ Gronwall’s inequality, and use (3.1) to obtain (3.15). 2 Lemma 3.5. Under the assumptions listed in Theorem 1.1, we have for all 0 ≤ t ≤ T that

t

r

n−1

ux (t) + 2

2 +l3 .

r 2(n−1) uxx (s) 2 ds  1 + θ l∞

(3.18)

0

Proof. Multiplying (1.12b) by r 2n−2 uxx , we obtain



  αr 4n−4 u2xx − r 2n−2 ut ux + − (n − 1)r 2n−3 uu2x + 2(n − 1)vr n−2 ux ut x v  αr n−1 ux vx 3n−3 −1 −n−1 =r uxx Px − 2α(n − 1)r ux + α(n − 1)r uv + . (3.19) v2

∂t

r 2n−2 u2x 2

Integrating the above identity over [0, t) × R+ yields

t

r

n−1

ux (t) +

r 2(n−1) uxx (s) 2 ds  1 +

2

0

3 )

Hj ,

(3.20)

j =1

where Hj (j = 1, 2, 3) are defined and estimated below. In view of (2.3), (3.10) and (3.11), we have H1 :=

t !

r 2n−4 u2x + r −4 u2 + r 4(n−1) u2x vx2 + r 2n−3 |u|u2x

"

0 R+

t  θ ∞ +

t

r

0 2 +l3 .  1 + θ l∞

n−1

ux 2L∞ (R+ ) r n−1 vx 2

+ θ ∞ u ∞ 0 R+

r 2(n−1) u2x θ (3.21)

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6475

Then it follows from (2.3), (2.12), and (3.15) that

t H2 :=

r 2(n−1) Px2 0 R+

t

  r 2(n−1) θx2 + θ 2 vx2 + θ 6 θx2

 0 R+

t

κr 2(n−1) θx2 θ 2 + θ 8 · + θ ∞ vθ 2 1 + θb

 0 R+

t r 2(n−1) θ vx2 0 R+

(8−b) l3 +1  1 + θ ∞ + θ ∞ + .

(3.22)

For the last term    t      n−2 H3 :=  vr ux ut  ,    0 R+  we obtain from (1.11), (1.12b), (2.3) that

H3 

t !

" r n−4 |uux | + r 2(n−2) u2x + r 3n−4 |ux uxx | + r 3n−4 |vx |u2x + r 2n−3 |ux Px |

0 R+

t 

r

2(n−1)

t ! " uxx + C() θ ∞ + r 3n−4 |vx |u2x + r 2n−3 |ux Px | . 2

0 R+

0

,

-.

H3a

(3.23)

/

By virtue of (2.3), (3.11), and (3.15), we obtain

t H3a  θ ∞ +

t

r

n−1

ux 2L∞ (R+ ) r n−1 vx 2

+ 0 R+

0 2 +l3  1 + θ l∞

  r 2n−3 |ux | |θx | + θ |vx | + θ 3 |θx |

t

+ 0 R+

κr 2n−2 θx2 θ 2 + θ 8 + 1 + θb vθ 2

t

2 +l3 . r 2n−2 θ 2 vx2  1 + θ l∞

0 R+

Inserting estimates (3.21)–(3.24) into (3.20) yields (3.18).

2

(3.24)

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4. Upper bound of the temperature In this section we shall derive an estimate on the upper bound of θ . To this end, we set

t   X(t) : = 1 + θ b+3 (s, x) θt2 (s, x)dxds, 0 R+

Y (t) : = sup s∈(0,t)

(4.1)

  r 2n−2 1 + θ 2b (s, x) θx2 (s, x)dx,

(4.2)

u2xx (s, x)dx,

(4.3)

R+

Z(t) : = sup s∈(0,t)

R+

and then try to deduce certain estimates among them by employing the special structure of system (1.12). Lemma 4.1. Under the assumptions listed in Theorem 1.1, we have for all 0 ≤ t ≤ T that 1

θ (t) L∞ (R+ )  1 + Y (t) 2b+6 , 1

(4.4) 3

sup ux (s) 2  1 + Z(t) 2 ,

ux (t) L∞ (R+ )  1 + Z(t) 8 .

s∈(0,t)

(4.5)

Proof. We assume that x ∈ [k, k + 1] for some k ∈ N and x ≥ bk (t). Then

x (θ (t, x) − 1)

2b+6

= (θ (t, bk (t)) − 1)

2b+6

+

(2b + 6) (θ (t, y) − 1)2b+5 θx (t, y)dy

bk (t)

⎡ ⎢  1 + θ (t) − 1 b+3 L∞ (R+ ) ⎣

⎤1 ⎡ 2

⎥ ⎢ (θ − 1)4 dx ⎦ ⎣

R+

⎤1

2

⎥ (θ − 1)2b θx2 r 2n−2 dx ⎦

R+

1 2

 1 + C θ (t) − 1 b+3 L∞ (R+ ) Y (t), from which we can deduce (4.4) by using Cauchy’s inequality. Estimates (4.5) follow directly by applying Gagliardo–Nirenberg and Sobolev inequalities. 2 Our next result shows that X(t) and Y (t) can be controlled by Z(t). Lemma 4.2. Under the assumptions listed in Theorem 1.1, we have for 0 ≤ t ≤ T that X(t) + Y (t)  1 + Z(t)λ1 .

(4.6)

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6477

Here λ1 is given by # λ1 = max

$ b+3 3(b + 3) , . 4(b + 4) 2(3b + 9 − 2l3 )

(4.7)

Proof. As in Kawohl [15], we set

θ K (v, θ ) = 0

κ (v, ξ ) κ1 θ κ2 θ b+1 dξ = + . v v b+1

(4.8)

Thanks to (2.12), we have κθt Kt = Kv (v, θ )(r n−1 u)x + , v

κ  κθx + Kv (r n−1 u)xx + Kvv vx (r n−1 u)x + vx θt , Kxt = v t v v |Kv (v, θ )| + |Kvv (v, θ )|  θ.

(4.9) (4.10) (4.11)

We can rewrite (1.12c) as eθ θt + θ Pθ (r

n−1

2 

  α (r n−1 u)x  r 2n−2 κθx u)x − − 2μ(n − 1) r n−2 u2 + λφz. = x v v x (4.12)

Multiplying (4.12) by Kt and integrating the resulting identity over [0, t] × R+ yield  n−1 2  (r u)x  α Kt eθ (v, θ )θt + θ Pθ (v, θ )(r n−1 u)x − v

t  0 R+

t + 0 R+

r 2n−2 κ (v, θ ) θx Ktx + 2μ(n − 1) v

t  r

n−2 2



u

x

0 R+

t Kt =

λφzKt . (4.13) 0 R+

Combining (4.8)–(4.13), we have

t 0 R+

eθ (v, θ )κ (v, θ ) θt2 + v

t 0 R+

κ (v, θ ) θx v



κ (v, θ ) θx v

t

     t    t    2         1+ eθ (v, θ )θt Kv (v, θ )(r n−1 u)x  +  θpθ (v, θ ) (r n−1 u)x  Kv (v, θ )      0 R+    0 R+ -. / , -. / , I1

I2

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Y. Liao et al. / J. Differential Equations 266 (2019) 6459–6506

     t   t   n−1 2    α (r u)x  Kt (v, θ )  θpθ (v, θ )κ (v, θ ) (r n−1 u)x θt    + +      v v  0 R+   0 R+  , -. / , -. / I3

I4

    t     2n−2 r κ(v, θ )θx   n−1 n−1 Kvv (v, θ )vx (r + u)x + Kv (v, θ )(r u)xx    v   0 R+ , -. / I5

     t   t 

    2n−2 r κ(v, θ )θx κ(v, θ )     n−1 + vx θt  +  λφzKv (v, θ )(r u)x      v v v  0 R+   0 R+  , -. / , -. / I6

I7

     t   t

      λφzκ (v, θ ) θt   κ(v, θ )θt   Kv (v, θ )(r n−1 u)x + r n−2 u2 + +  . (4.14)     x v v  0 R+    0 R+ , -. / , -. / I8

I9

We now turn to control Ik (k = 1, . . . , 9) term by term. To do so, we first have

t

eθ (v, θ )κ (v, θ ) θt2  v

0 R+

t 

1 + θ3



 1 + θ b θt2  X(t),

(4.15)

0 R+

and

t 0 R+

1 = 2

r 2n−2 κ (v, θ ) θx v

r R+

2n−2

κθx v



κ (v, θ ) θx v

t

t

2 dx − C − (n − 1)

t  Y (t) − C − C 0 R+

0 R+

κθx v

κθx v

2 r 2n−3 u

2 r 2n−3 u.

Furthermore, we obtain    t  t

2

    κ r n−1 θx 2 θ 2 κθx  2n−3  b 1+θ r u  u L∞ (R+ )    v r v2θ 2  0  0 R+ R+

(4.16)

Y. Liao et al. / J. Differential Equations 266 (2019) 6459–6506





1 + θ b+2 ∞

 t

1 2

u ux

0



 1 + Y (t)

b+2 2b+6



1 + Z(t)

1 8



2  κ r n−1 θx v2θ 2

1 2

6479

R+

  b+3  C() 1 + Z(t) 4(b+4) + Y (t).

(4.17)

On the other hand, it follows from Cauchy’s inequality and (2.3) that

t

    (1 + θ )4 θt (r n−1 u)x 

I1  0 R+

≤ X(t) + C()



(6−b) 1 + θ ∞ +

0 R+

≤ X(t) + C() 1 + Y (t)

t

   t (r n−1 u)x 2 ·v vθ

(6−b)+ 2b+6



≤ (X(t) + Y (t)) + C(),

(4.18)

 1 + θ 6∞  C() + Y (t),

(4.19)

 2   (1 + θ )5 (r n−1 u)x 

I2  0 R+

t 

(1 + θ ) · 6

 n−1 2 (r u)x  vθ

0 R+

and

t I3 

    (1 + θ )b+4 (r n−1 u)x θt 

0 R+

≤ X(t) + C()



1 + θ b+6 ∞

   t (r n−1 u)x 2 vθ 0 R+

≤ X(t) + C()Y (t)

b+6 2b+6

≤  (X(t) + Y (t)) + C().

(4.20)

As for the term I4 , it is easy to see that    t  n−1 2

    α (r u)x κ (v, θ ) θt   I4 =  Kv (r n−1 u)x +    v v  0 R+ 

t 

t       n−1 3  n−1 2 a b u)x  θ + u)x  1 + θ b |θt | =: I13 + I13 .  (r (r 0 R+

0 R+

(4.21)

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We deduce from (3.1), (3.11), (3.18), and the fact 2l2 + l3 < b + 9 (since b > 19/4) that  a I  

t |(r

4

n−1

t    n−1 4 u)x | θ + u)x  (r 2 2

0 R+

0 R+

t  θ 3∞ +

  r −4 u4 + r 4n−4 u4x

0 R+

 1 + Y (t)

3 2b+6

t  2

t u 2 +  

u + r n−1 ux 2L∞ (R+ ) r n−1 ux 2 r L∞ (R+ ) 0

3

 1 + Y (t) 2b+6 + Y (t)

0

2l2 +l3 2b+6

≤ Y (t) + C()

(4.22)

and

t      b 1 + θ b−3 (r n−1 u)4x I4  ≤ X(t) + C() 0 R+

≤ X(t) + C()

t 

1+θ

b−3

0 R+





 u4



+ r 4n−4 u4x r4

⎛

⎝1 + ≤ X(t) + C() 1 + θ b−3 ∞

t

⎞

r n−1 ux 2L∞ (R+ ) r n−1 ux 2 ⎠

0





2 +l3 +b−3 ≤  (X(t) + Y (t)) + C(). ≤ X(t) + C() 1 + θ 2l ∞

(4.23)

Thus the combination of (4.21)–(4.23) and the assumption b > 19/4 yields I4 ≤  (X(t) + Y (t)) + C().

(4.24)

Now for the term I5 , we have from (4.11) that

t I5 

  "   !     r 2n−2 1 + θ b+1 θx vx (r n−1 u)x  + θx (r n−1 u)xx  =: I5a + I5b ,

0 R+

then we can obtain from (2.3), (3.1), (3.9), (3.11), (3.15) and the fact that (r n−1 u)xx = 2(n − 1)r −1 vux + (n − 1)r −1 vx u − (n − 1)r −1−n v 2 u + r n−1 uxx , that

(4.25)

Y. Liao et al. / J. Differential Equations 266 (2019) 6459–6506

t I5a

κr 2n−2 θx2 1 + θ 2b+4 · +C 1 + θb vθ 2

1 0 R+

t



 1 + C θ b+4 ∞ +C

r 2n−2 vx2 0 R+

 1 + CY (t)

b+4 2b+6

 1 + CY (t)

b+4 2b+6

t + 0

t

6481

 2   r 2n−2 vx2 (r n−1 u)x 

0 R+

u2 + r 2n−2 u2x r2



  2     u 2

r n−1 vx 2   + r n−1 ux  ∞ L (R+ ) r L∞ (R+ ) l2 +l3

+ CY (t) 2b+6 ≤ Y (t) + C(),

(4.26)

and

t I5b

κr 2n−2 θx2 1 + θ 2b+4 · + 1 + θb vθ 2

1 0 R+

 1 + θ b+4 ∞

t 0 R+

 2 r 2n−2 (r n−1 u)xx  v



t u2 2n−4 2 2n−4 2 2 4n−4 2 + ux + r u vx + 4 + r uxx r r 0 R+

 1 + Y (t)

b+4 2b+6

t + 0 R+

t + 0

vu2 θ r 2n−2 u2x vθ · 2 + 2 · 2 vθ r r θ vr



t  u 2   2n−2 2 r vx + r 4n−4 u2xx   ∞ r L (R+ ) R+

b+4

 1 + Y (t) 2b+6 + Y (t)

0 R+

l2 +l3 2b+6

≤ Y (t) + C().

(4.27)

Combining (4.25)–(4.27) together, we get I5 ≤ Y (t) + C().

(4.28)

As to the term I6 , it is easy to see that

t I6 

r

2n−2



1+θ

b



t |θx θt vx | ≤ X(t) + C()

0 R+

  r 4n−4 1 + θ b−3 θx2 vx2

0 R+

2  n−1   r κ(v, θ)θx   n−1 2   v r  x  ds .  ∞  v L (R+ ) 0 , -. /

t

≤ X(t) + C()

I6a

(4.29)

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We can deduce from (4.12) that

r n−1 κθx v

x

 2 α (r n−1 u)x  eθ θ t θ Pθ (r n−1 u)x = n−1 + − − (n − 1)r −1 κθx r r n−1 vr n−1   2μ(n − 1) r n−2 u2 x λφz + − n−1 . n−1 r r

(4.30)

Thus we conclude from (4.30) that

I6a





3 1 + θ l∞

 t  r n−1 κ(v, θ)θ   r n−1 κθ

x  x    v v 0 R+

x

   

⎛ ⎞1 ⎛ ⎞1 2  n−1

2 2

t  t r 2n−2 κθ 2  r κ(v, θ)θx  ⎟ ⎜ x⎟ ⎜ 3  1 + θ l∞ (1 + θ )b+2  ⎝ ⎠ ⎝  ⎠ 2 vθ v x 

0 R+

0 R+

⎛ ⎞1 2   t   l3 ⎜ b+2 2 2 2 2 2 4 2 2 ⎟  1 + θ ∞ ⎝ eθ θt + θ Pθ ux + ux + φ z ⎠ (1 + θ ) 0 R+



⎛ 2  2  n−1 4

t (r l3 θ 2 Pθ2 (r n−1 u)x  u)x  eθ2 θt2 ⎜ b+2 1 + Y (t) 2b+6 ⎝ + + (1 + θ ) 2(n−1) 2(n−1) 2(n−1) r r r 0 R+

+ r −2 κ 2 θx2 +

 n−2 2  2  r u  x

r 2(n−1)

+

φ 2 z2

33 12

r 2(n−1)

.

(4.31)

For the last terms on the right hand side of (4.31), one can deduce from (3.1) that

t 0 R+

(1 + θ )b+2 eθ2 θt2  r 2(n−1)

t 0 R+

 0 R+

0 R+

  5 (1 + θ )b+8 θt2  1 + Y (t) 2b+6 X(t),

(4.32)

0 R+

(1 + θ )b+2 θ 2 Pθ2 |(r n−1 u)x |2 r 2(n−1)

t

t

t

 2 (1 + θ )b+11 (r n−1 u)x  b+11  1 + Y (t) 2b+6 ,  1 + θ b+11 ∞ vθ

 4 t   (1 + θ )b+2 (r n−1 u)x  (1 + θ )b+2 r −4 u4 + r 4(n−1) u4x  r 2(n−1) r 2(n−1) 0 R+

(4.33)

Y. Liao et al. / J. Differential Equations 266 (2019) 6459–6506

6483

⎤ ⎡

2(n−1) 2   t  u 2 r ux ⎥ ⎢   1 + θ b+2

u 2 + θ ∞ ux 2L∞ (R+ ) ⎦ ⎣  ∞ ∞ r L (R+ ) vθ R+

0

   1 3  1 1 + Y (t) 2 1 + Z(t) 4 ,

(4.34)

and

t

2b+4

2b+4  1 + Y (t) 2b+6 . (1 + θ )b+2 r −2 κ 2 θx2  1 + θ ∞

(4.35)

0 R+

Moreover, from (2.2) and the fact that   r n−2 u2 = (n − 2)r n−3 rx u2 + 2r n−2 uux = (n − 2)r −2 vu2 + 2r n−2 uux ,

(4.36)

x

we have

t

  2 (1 + θ)b+2  r n−2 u2 x  r 2(n−1)

0 R+





1 + θ b+2 ∞

 1 + Y (t) 2

 0 R+

(1 + θ )b+2 u4 r 2(n−2) (1 + θ)b+2 u2 u2x + r 2(n−1)+4 r 2(n−1)



2  t u4   t u b+3 2 + 1 + θ ∞

ux L∞ (R+ ) r2 r 2θ

 1



t 

0 R+

 3

0

1 + Z(t) 4

(4.37)

and

t 0 R+

 (1 + θ )b+2 φ 2 z2  b+2+β  1 +

θ

∞ r 2(n−1)

t φz2  1 + Y (t)

b+2+β 2b+6

.

(4.38)

0 R+

Consequently, we obtain by combining the estimates (4.31)–(4.38) that 2

t  n−1 r κ(v, θ)θx      ∞ v 0



!    l3 5  n−1 2 2b+6 vx  ds  1 + Y (t) 1 + Y (t) 2b+6 X(t) r

L (R+ )

   "1 b+2+β 2 b+11 2b+4 1 3 + Y (t) 2b+6 + 1 + Y (t) 2 1 + Z(t) 4 + Y (t) 2b+6 + Y (t) 2b+6

3(b+3) 2(3b+9−2l3 ) ≤  (X(t) + Y (t)) + C() 1 + Z(t) .

(4.39)

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Y. Liao et al. / J. Differential Equations 266 (2019) 6459–6506

Combining (4.29)–(4.39), we can get the following estimate on I6 |I6 | ≤  (X(t) + Y (t)) + C() 1 + Z(t)

3(b+3) 2(3b+9−2l3 )

.

(4.40)

For I7 and I8 , we get from (2.2) and assumptions b > 19/4 and 0 ≤ β < b + 9 that    t      I7 =  λφzKv (v, θ )(r n−1 u)x     0 R+  ⎞1 ⎛ ⎞1 ⎛ 2

t

t  n−1 2 2 β+3 u)x ⎟ (r ⎜ ⎟ ⎜ φz2 ⎠ ⎝  1 + θ ∞2 ⎝ ⎠ vθ 0 R+

0 R+

β+3 4b+12

 1 + Y (t) ≤ Y (t) + C(),    t 

t    λφzκ (v, θ ) θt   1 + θ b+β−3 φz2 I8 =   ≤ X(t) + C()   v  0 R+  0 R+ ≤ X(t) + C()



b+β−3 1 + θ ∞

 t

φz2 ≤ (X(t) + Y (t)) + C().

(4.41)

(4.42)

0 R+

Finally, for the term I9 , one can infer from (4.36) that I9 

t 

r −2 u2 + r n−2 |uux |

   " !    θ (r n−1 u)x  + 1 + θ b |θt |

0 R+

t  

        r −2 u2 θ (r n−1 u)x  + r −2 u2 1 + θ b |θt | + r n−2 1 + θ b |uux θt |

0 R+

    1 2 3 4 + r n−2 θ uux (r n−1 u)x  := I18 + I18 + I18 + I18 .

(4.43)

j

We estimate I9 (j = 1, 2, 3, 4) term by term as follows. First, we have from (3.1) that I91

t  n−1 2

t 4 (r u)x u  · θ3 + vθ r4 0 R+

 θ 3∞

0 R+

t  2

3 u +   u2  1 + Y (t) 2b+6 ≤ Y (t) + C(), ∞ r L (R+ ) 0

R+

(4.44)

Y. Liao et al. / J. Differential Equations 266 (2019) 6459–6506

t I92

6485

  r −4 u4 1 + θ b−3

≤ X(t) + C() 0 R+

 ≤  (X(t) + Y (t)) + C(). ≤ X(t) + C() 1 + θ b−3 ∞ 

(4.45)

As for the term I93 , we have from (3.1) and (3.18) that

t I93

≤ X(t) + C() 0 R+

≤ X(t) + C()



  r 2(n−1) u2 u2x 1 + θ b−3 r2

1 + θ b−3 ∞

 t  u 2   r 2(n−1) u2x   ∞ r L (R+ ) 0



R+



2 +l3 ≤  (X(t) + Y (t)) + C(). ≤ X(t) + C() 1 + θ b−3+l ∞

(4.46)

For I94 , we can get from (2.3), (3.1) and the fact that l2 < 2b + 6 since b > 19/4 that

I94



t  n−1 2 (r u)x vθ

0 R+

 θ 3∞

t ·θ + 3

r 2n−4 u2 u2x 0 R+

t  2   + r n−1 ux  ∞

L (R+ ) R+

0 3

u2

l2

 1 + Y (t) 2b+6 + Y (t) 2b+6 ≤ Y (t) + C().

(4.47)

Thus it follows from (4.43)–(4.47) that I9 ≤  (X(t) + Y (t)) + C().

(4.48)

With the above estimates in hand, if we define λ1 as in (4.7), by combining all the above estimates and choosing  > 0 small enough, we can deduce estimate (4.6). Since b > 19/4, one gets from (3.3), (3.12), (3.16) and (4.7) that 0 < λ1 < 1. 2 Our next result in this section is to show that Z(t) can be bounded by X(t) and Y (t). Lemma 4.3. Under the assumptions listed in Theorem 1.1, we have for all 0 ≤ t ≤ T that Z(t)  1 + X(t) + Y (t) + Z(t)λ2 .

(4.49)

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Y. Liao et al. / J. Differential Equations 266 (2019) 6459–6506

Here λ2 is given by #

$ b+3 3(b + 3) λ2 = max , . 2(2b + 5) 2(2b + 6 − l1 ) It is easy to see that λ2 ∈ (0, 1) provided that b >

(4.50)

19 4 .

Proof. Differentiating (1.12b) with respect to t and multiplying it by ut , we obtain

u2t 2

= t

# 

$

(r n−1 u)x (r n−1 u)x r n−1 α ut − r n−1 α utx −P −P v v t t x 

 (r n−1 u) x − r n−1 ut . −P α x v t

Then integrating the above identity over [0, t] × R+ yields

ut (t) 2 + 2

t 0 R+

αr 2(n−1) u2tx + α(n − 1)2 v

t r

u |utx | −2α(n − 1)

n−3 2

0 R+

-.

,

,

0 R+

+C 0 R+

,

/

r 0 R+

,

/

r 0 R+

t r 0 R+

,

−2

Pt utx + (n − 1)

-.

/

r n−2 P uutx 0 R+

,

-.

|uut vt | +C

t /

r 0 R+

,

I19

/

I15

t

-.

P |ut vt | +C

/

t n−1

/

r n−3 |uux ut | 0 R+

,

I17

-.

-. I12

t ,

I16

+C

,

/

I14

|vu2 ut | +C r3

−1

0 R+

r 2n−3 |uux utx | v

I11

-.

t

ut utx +C

-.

r 2(n−1) ux utx vt + v2

-.

t n−2

0 R+

/,

I13

t

0 R+

r

I10

+α

vu2t r2

t

≤C +C

t

t

-.

/

I18 −2

t |vuP ut | +C

-.

/

vr −1 |ut Pt | .

0 R+

,

I20

-.

(4.51)

/

I21

Now we turn to estimate Ik (k = 10, 11, · · · , 21) term by term. To this end, we compute from (2.3) and (3.1) that

t I10 ≤  0 R+

αr 2(n−1) u2tx + C() v

t  2

t αr 2(n−1) u2tx u 2

u ≤  + C(), (4.52)   ∞ r L (R+ ) v 0

0 R+

Y. Liao et al. / J. Differential Equations 266 (2019) 6459–6506

6487

and

t 

t  n−2 2 r I11 = α(n − 1) ut = α(n − 1)(n − 2) r −2 vu2t . x

0 R+

(4.53)

0 R+

Since

t I12 ≤  0 R+

αr 2(n−1) u2tx + C() v

t r 2(n−1)−2 u2 u2x ,

(4.54)

0 R+

and

t

t r 2(n−1)−2 u2 u2x



0 R+

0 R+

t  θ ∞

u 2L∞ (R+ ) R+

0

r 2(n−1) u2x u2 θ · 2 vθ r

r 2(n−1) u2x  θ ∞ vθ



t

u

ux

0

R+

r 2(n−1) u2x vθ

     b+3 1 1  1 + Y 2b+6 (t) 1 + Z 4 (t)  1 + Y (t) + Z 2(2b+5) (t) ,

(4.55)

we can deduce that

t I12 ≤  0 R+

  b+3 αr 2(n−1) u2tx + C() 1 + Y (t) + Z 2(2b+5) (t) . v

(4.56)

For the term I13 , we get

t I13 ≤  0 R+

αr 2(n−1) u2tx + C() v

t 0 R+

r 2(n−1) u2x (r n−1 u)2x . v2

(4.57)

Since

t 0 R+

and

3  2 t 2 r 2(n−1) u2x (r n−1 u)x  r 2(n−1) u2x u2 4(n−1) 4  +r ux , v2 r2 0 R+

(4.58)

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Y. Liao et al. / J. Differential Equations 266 (2019) 6459–6506

t r 4(n−1) u4x

t  2    r n−1 ux  ∞

0 R+

L (R+ ) R+

0

2 +l3  1 + Y  1 + θ 2l ∞

2l2 +l3 2b+6

r 2(n−1) u2x

 1 + Y (t),

(4.59)

we combine (4.55), (4.58)–(4.59) to find  2 b+3 r 2(n−1) u2x (r n−1 u)x   1 + Y (t) + Z 2(2b+5) (t). v2

t 0 R+

(4.60)

Consequently, we can conclude from (4.57)–(4.60) that

t

  b+3 αr 2(n−1) u2tx + C() 1 + Y (t) + Z 2(2b+5) (t) . v

I13 ≤  0 R+

(4.61)

Finally for Ik (k = 14, 15, · · · , 21), it follows from (1.2) and (2.12) that

t I14 ≤  0 R+

t ≤

αr 2(n−1) u2tx + C() v

≤ 0 R+

1+θ

6



θt2

 2  n−1 + θ (r u)x  2

0 R+

⎛

αr 2(n−1) u2tx v

0 R+

t

t 

⎞

t  n−1 2 (r u)x ⎜ ⎟ + C() ⎝X(t) + · θ 3⎠ vθ 0 R+

αr 2(n−1) u2tx + C() (1 + X(t) + Y (t)) , v

t

I15 ≤  0 R+

t

≤ 0 R+

t I16 ≤  0 R+

αr 2(n−1) u2tx + C() v

t 0 R+

  u2 2 8 + θ · θ θ r 2θ

αr 2(n−1) u2tx + C() (1 + Y (t)) , v

vu2t + C() r2

t 0 R+

u4 ≤ r4

(4.62)

t 0 R+

vu2t + C(), r2

(4.63)

(4.64)

Y. Liao et al. / J. Differential Equations 266 (2019) 6459–6506

t I17 ≤  0 R+

u 2L∞ (R+ )

t

vu2t + C() θ ∞ r2

≤ 0 R+

u

ux

vθ

 n−1 2 (r u)x vθ

R+

0

t

 n−1 2 (r u)x

R+

0

t

  b+3 vu2t + C() 1 + Y (t) + Z 2(2b+5) (t) , 2 r

≤ 0 R+

t

vu2t + C() r2

I18 ≤  0 R+

t

t

0 R+

0 R+

t

u 2L∞ (R+ ) R+

0

r 2(n−1) u2x vθ

  b+3 vu2t 2(2b+5) (t) , + C() 1 + Y (t) + Z r2

≤ 0 R+

t I19 ≤  0 R+

t ≤ 0 R+

t ≤ 0 R+

vu2t + C() r2

(4.65)

r 2(n−1) u2x vθ u2 · 2 vθ r

vu2t + C() θ ∞ r2

≤

t

t

vu2t + C() θ ∞ r2

6489

(4.66)

t  n−1 2   (r u)x · θ θ2 + θ8 vθ 0 R+

  vu2t 9 + C() 1 +

θ

∞ r2 vu2t + C() (1 + Y (t)) , r2

t I20 ≤  0 R+

t ≤ 0 R+

vu2t + C() r2

t 0 R+

(4.67)

  u2 · θ θ2 + θ8 2 r θ

vu2t + C() (1 + Y (t)) , r2

(4.68)

and

t I21 ≤  0 R+

vu2t + C() (1 + X(t) + Y (t)) . r2

(4.69)

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Y. Liao et al. / J. Differential Equations 266 (2019) 6459–6506

Combining estimates (4.51)–(4.69) and choosing  > 0 small enough, we arrive at

t

ut (t) + 2

0 R+

r 2(n−1) u2tx + v

t 0 R+

b+3 vu2t  1 + X(t) + Y (t) + Z 2(2b+5) (t). 2 r

(4.70)

To yield an estimate on uxx (t) , (1.12)2 tells us that uxx =

ut v αr 2(n−1)

+

(n − 1)v 2 u 2(n − 1)vux Px v − + v −1 ux vx + n−1 . n 2n r r αr

(4.71)

Thus one gets from (4.71) that



uxx (t) 

u2t

2

r 4(n−1)

R+

1+



u2 u2x Px2 2 2 + 4n + 2n + ux vx + 2(n−1) dx r r r

 u2t + Px2 + u2x + u2x vx2 dx

R+

 1 + ut (t) 2 +





  1 + θ 6 θx2 dx + θ 2 + u2x vx2 dx + u2x dx

R+

 1 + X(t) + Y (t) + Z

R+ b+3 2(2b+5)

R+

(t) + θ 2∞ vx (t) 2 + ux 2∞ vx (t) 2 + ux 2 , (4.72)

while the last three terms in the right hand side of (4.72) can be bounded as follows: 2+l1

1 2b+6  1 + Y (t),

θ 2∞ vx (t) 2  1 + θ 2+l ∞  1 + Y (t)    3 1 1 + Z(t) 4

ux 2∞ vx (t) 2  1 + θ l∞

3(b+3) l1 l1 3 3 2b+6 2b+6 4 4 + Y (t) Z(t)  1 + Y (t) + Z(t) 2(2b+6−l1 ) ,  1 + Z(t) + Y (t) l2

2  1 + Y (t) 2b+6  1 + Y (t).

ux 2  1 + θ l∞

(4.73)

(4.74) (4.75)

Thus it follows from (4.72)–(4.75) that

uxx (t) 2  1 + X(t) + Y (t) + Z(t)λ2 ,

(4.76)

where λ2 is given by (4.50). Having obtained (4.76), we can get estimate (4.49) by using the definition of Z(t). 2 Combining Lemmas 4.1–4.3, we can deduce that Y (t)  1. Thus we can get the desired upper bounds on θ (t, x) from (4.4). In fact, we can deduce from Lemmas 2.1 and 4.3 that:

Y. Liao et al. / J. Differential Equations 266 (2019) 6459–6506

6491

Lemma 4.4. Under the assumptions listed in Theorem 1.1, there exists a positive constant , which depends only on the constants μ, λ1 , λ, K, A, d, R, cv , a, κ1 , κ2 , n and the initial data (v0 (x), u0 (x), θ0 (x), z0 (x)), such that θ (t, x) ≤ ∀ (t, x) ∈ [0, T ] × R+ .

(4.77)

Moreover, we have for 0 ≤ t ≤ T that  2  

(v − 1, u, θ − 1, z) (t) 2 + r n−1 (vx , ux , θx ) (t) + uxx (t) 2 + z(t) L1 (R+ )

t   2 √   +  r 2n−2 uxx , r n−1 θ vx , r n−1 utx , r n−1 ux , r n−1 θx , θt , r n−1 zx (s) ds  1

(4.78)

0

and

ux (t) L∞ (R+ )  1,

u(t) L∞ (R+ )  1,

t    n−1 2 ux  ∞ r

L (R+ )

ds  1.

(4.79)

0

Lemma 4.5. Under the assumptions listed in Theorem 1.1, we have for 0 ≤ t ≤ T that

t  2  

θx (t) + r n−1 θxx (s) ds  1, 2

(4.80)

0

t

 2  n−1  θx (s) ∞ r

L (R+ )

ds  1.

(4.81)

0

Proof. Multiplying (4.12) by θxx /eθ , one has 2 1  2 r 2n−2 κθxx θx − (θt θx )x + t 2 veθ

=

θ Pθ (r n−1 u)x θxx α(r n−1 u)2x θxx (2n − 2)r 2n−2 κθx θxx r 2n−2 κv vx θx θxx − − − eθ veθ eθ veθ  n−2 2  2n−2 2 2n−2 u x θxx 2μ(n − 1) r κθ θx θxx κvx θx θxx r r λφzθxx − + + − . (4.82) 2 veθ v eθ eθ eθ

Integrating the above identity with respect to x over R+ , one has 1 d

θx (t) 2 + 2 dt

R+

2 r 2n−2 κ(v, θ)θxx

veθ (v, θ )

         θ P (v, θ )(r n−1 u) θ   α (r n−1 u) 2 θ  x xx  θ x xx    ≤ +      eθ (v, θ ) veθ (v, θ ) R+  R+  -. / , -. / , I22

I23

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Y. Liao et al. / J. Differential Equations 266 (2019) 6459–6506

             r 2n−2 κ (v, θ )v θ θ   r 2n−2 κ (v, θ )θ 2 θ   r 2n−2 κ(v, θ)θ v θ  v x x xx   θ x x xx   x xx   +  + +       veθ (v, θ ) veθ (v, θ ) v 2 eθ (v, θ )  R+  R+  R+ -. / , -. / , -. / , I24

I25

I26

             n−2 2       2n−2 u x θxx  λφzθxx   (2n − 2)r κθx θxx   2μ(n − 1) r  + + + .  eθ (v, θ )      eθ (v, θ ) eθ (v, θ ) R+  R+  R+  -. / , -. / , -. / , I27

I28

(4.83)

I29

Now we turn to estimate the terms Ik (k = 22, . . . , 29). To begin with, we have

I22 ≤  R+

≤ R+

2 r 2n−2 κ(v, θ)θxx dx + C() veθ (v, θ )

 n−1 2 (r u)x vθ

R+

·

θ 3 Pθ2 (v, θ ) dx κ(v, θ)r 2n−2 eθ (v, θ )

2 r 2n−2 κ(v, θ)θxx dx + C()V (t). veθ (v, θ )

(4.84)

Estimates (4.77) and (4.79) imply

I23 ≤  R+

≤ R+

≤ R+

2 r 2n−2 κ(v, θ)θxx dx + C() veθ (v, θ )

 n−1 4 (r u)x 

R+

2 r 2n−2 κ(v, θ)θxx dx + C() veθ (v, θ )

R+

r 2n−2 κ(v, θ)eθ (v, θ )

dx

u4 + r 4(n−1) u4x dx r 2n−2 κ(v, θ)eθ (v, θ )

  2 r 2n−2 κ(v, θ)θxx  u 2 + V (t) . dx + C()   veθ (v, θ ) r L∞ (R+ )

(4.85)

By virtue of Sobolev’s inequality and Lemma 4.4, we infer

I24 ≤  R+

≤ 2 R+

≤ 2 R+

2 r 2n−2 κ(v, θ)θxx dx + C() veθ (v, θ )

r 2n−2 vx2 ·

R+

2 r 2n−2 κ(v, θ)θxx

veθ (v, θ )

dx + C() R+

κv2 (v, θ )θx2 dx eθ (v, θ )κ(v, θ )

r 2n−2 κ(v, θ)θx2 θ 2 eθ · dx vθ 2 r 4n−4 κ 2

2 r 2n−2 κ(v, θ)θxx dx + C()V (t), veθ (v, θ )

(4.86)

Y. Liao et al. / J. Differential Equations 266 (2019) 6459–6506

2 r 2n−2 κ(v, θ)θxx dx + C() veθ (v, θ )

I25 ≤  R+



r 2n−2 θx4 κθ2 (v, θ ) dx κ(v, θ)eθ (v, θ )

2 r 2n−2 κ(v, θ)θxx dx + C() θx (t) θxx (t)

veθ (v, θ )

≤ R+

6493

r 2n−2 θx2 dx

R+

2 r 2n−2 κ(v, θ)θxx dx + C()V (t), veθ (v, θ )

≤ 2 R+

(4.87)

and

2 r 2n−2 κ(v, θ)θxx dx + C() veθ (v, θ )

I26 ≤  R+

R+

2 r 2n−2 κ(v, θ)θxx

≤

veθ (v, θ )

R+

veθ (v, θ )

R+

dx + C() θx 2L∞ (R+ )

2 r 2n−2 κ(v, θ)θxx

≤ 2

r 2n−2 κ(v, θ)θx2 vx2 dx eθ (v, θ ) r 2n−2 vx2 dx

R+

dx + C()V (t).

(4.88)

It follows from (1.5), (4.77) and (4.78) that

I27 ≤  R+

2 r 2n−2 κ(v, θ)θxx dx + C() veθ (v, θ )

I28 ≤  R+

φz2 dx,

(4.89)

R+

2 r 2n−2 κ(v, θ)θxx dx + C()V (t), veθ (v, θ )

(4.90)

and

 I29  R+

≤ R+

 r −2 u2 + r n−2 |uux | θxx dx eθ (v, θ )

  2 r 2n−2 κ(v, θ)θxx  u 2 + V (t) . dx + C()   veθ (v, θ ) r L∞ (R+ )

(4.91)

Combining (4.83)–(4.91) and by choosing  > 0 small enough, we arrive at

 2  u 2 d  n−1    2 θxx (t)  V (t) +   + φ(t, x)z2 (t, x)dx.

θx (t) + r dt r L∞ (R+ )

(4.92)

R+

Integrating the inequality (4.92) over (0, t) and using estimates (2.2), (2.3) and (3.1) yield (4.80).

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In view of (2.3), Lemma 4.4, (4.80) and Sobolev’s inequality, we find

t  2   n−1 θx (s) ∞ r

L (R+ )

ds 

0

t       n−1   n−1  θx   r θx  ds r x

0

t 

t    2  n−1 2   θx  ds +  r n−1 θx  ds  r x

0

0

t 1+ 0 R+



t κr 2(n−1) θx2 2 θx2 2(n−1) 2 +r θxx  1 + · θ  1. 2 r vθ 2 0 R+

This completes the proof of this Lemma. 2 Lemma 4.6. Let the assumptions listed in Theorem 1.1 hold. Then for any 0 ≤ t ≤ T ,  2 t  n−1  zx (t) + zt (s) 2 ds  1 r

(4.93)

0

zx (t) 2 +

t  2   n−1 zxx (s) ds  1. r

(4.94)

0

Proof. To prove (4.93), we multiply equation (1.12d) by zt to get

dr 2n−2 zx2 2v 2



+ zt2 = d t

r 2n−2 zx zt v2

+ x

d(n − 1)r 2n−3 uzx2 dr 2n−2 (r n−1 u)x zx2 − − φzzt . v2 v3

Integrating this last identity with respect to t and x, we have

R+

dr 2n−2 zx2 dx + 2v 2

t + 0 R+

,

t

zt2

=

0 R+

R+

d(n − 1)r 2n−3 uzx2 − v2 -.

/

dr 2n−2 zx2 (0, x)dx 2v 2

t 0 R+

,

I30

dr 2n−2 (r n−1 u)x zx2 − v3 -. I31

/

t φzzt . 0 R+

,

-.

(4.95)

/

I32

To control Ik (k = 30, 31, 32), we can get from (2.2) and (4.79) that

t |I30 |  0 R+

r 2n−2 zx2  1, v2

(4.96)

Y. Liao et al. / J. Differential Equations 266 (2019) 6459–6506

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and

t |I31 |  0 R+

 r 2n−2 zx2  −1 n−1 |u| |u | r + r x v2

t  2    1 + r n−1 ux  ∞

L (R+ ) R+

0

r 2n−2 zx2 . v2

(4.97)

As for the term I32 , we get from (2.2) that 1 |I32 | ≤ 2

t

t zt2

+C

0 R+

1 φ z ≤ 2

t zt2 + C.

2 2

0 R+

(4.98)

0 R+

Substituting (4.96)–(4.98) into (4.95) and using Gronwall’s inequality yield (4.93). Multiply (1.12d) by zxx to infer

zx2 2

− (zt zx )x + t

2 dr 2n−2 zxx (2n − 2)dr n−2 zx zxx 2dr 2n−2 zx vx zxx = − + + φzzxx . v2 v2 v3

Integrating the above identity over [0, t) × R+ , we arrive at

t  2  

zx (t) + r n−1 zxx (s) ds 2

0

t

1+

t r

0 R+

,

n−2

-.

|zx zxx | + /

t r

0 R+

,

I33

2n−2

φz |zxx | .

|zx vx zxx | +

-.

/

I34

0 R+

,

-.

(4.99)

/

I35

The terms Ik (42 ≤ k ≤ 44) can be estimated by employing Cauchy’s and Sobolev’s inequalities, Lemmas 2.1 and 4.4 as follows: 1 |I33 | ≤ 8

t 

t

t 2 2 1   n−1  n−1   −2 2 zxx (s) ds + C r zx ≤ zxx (s) ds + C, r r 8 0 R+

0

1 |I34 | ≤ 8

0

t 

t 2  n−1  zxx (s) ds + C r 2n−2 vx2 zx2 r 0

1 ≤ 8

(4.100)

t 0

0 R+

t

t  2 2 1   n−1  n−1   zxx (s) ds + C zx zxx ds ≤ zxx (s) ds + C, (4.101) r r 4 0

0

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and 1 |I35 | ≤ 8

t 

t 2   n−1 zxx (s) ds + C φ 2 z2 r 0 R+

0

1 ≤ 8

t 

t

t 2 2 1     n−1  n−1 β 2 zxx (s) ds + C θ ∞ φz ds ≤ zxx (s) ds + C. (4.102) r r 8 0 R+

0

0

2

Then (4.94) follows from (4.99)–(4.102).

Lemma 4.7. Under the assumptions listed in Theorem 1.1, we get for any 0 ≤ t ≤ T that

R+

t

2 r 2n−2 vxx dx + v2

t

2 Rr 2n−2 θ vxx 1 v3

0 R+

2    n−1 vx (s) ∞ r

L (R+ )

(4.103)

ds  1.

(4.104)

0

Proof. Differentiate (3.4) with respect to x and multiply it by r n−1 vxx /v to obtain  2 2 α r 2n−2 vxx Rr 2n−2 θ vxx r n−1 utx vxx + = 2 v2 v3 v t



n−1 2 u)x α(n − 1)r 2n−3 uvxx αr 2n−2 vxx vx2 (r + + − α(n − 1)r n−2 vxx v v2 v v2 t x

4 Rθvx Rθx Rr 2n−2 vxx θxx + (n − 1)r n−2 vxx − 2 + aθ 3 θx + v 3 v v2 −

2Rr 2n−2 vxx vx θx 2Rr 2n−2 θ vxx vx2 4ar 2n−2 θ 2 vxx θx2 4aθ 3 r 2n−2 vxx θxx + + + . v 3v v3 v4

Integrating this last identity over [0, t] × R+ implies

R+

2 r 2n−2 vxx dx + 2 v

t 0 R+

2 Rr 2n−2 θ vxx 1+ 3 v

t 0 R+

,

r n−1 |utx vxx | v -.

/

I36

t +

r 0 R+

,



n−1  u)x (r vxx v

n−2 

-. I37

x

 t 2n−3 2  |u| vxx r +  v2

/

0 R+

,

-. I38

/

Y. Liao et al. / J. Differential Equations 266 (2019) 6459–6506

6497

 t 2n−2

t  2n−2 r |vxx θxx | vxx vx2  r  + +  v v2 t  v2 0 R+

-.

,

/

0 R+

,

-.

I39

t + 0 R+

/

I40



 t 2n−2   |vxx θx vx | r 4 Rθvx Rθx r n−2 vxx − 2 + aθ 3 θx  + v 3 v v3 -.

,

0 R+

/

,

-.

I41

t + 0 R+

r 2n−2 θ vx2 |vxx | + v4

,

-.

/

/

I42

t 0 R+

r 2n−2 θ 2 θx2 |vxx | + v

,

-.

I43

/

t

r 2n−2 θ 3 |vxx θxx | . (4.105) v

0 R+

,

-.

I44

/

I45

Now we turn to estimate terms Ij (j = 36, . . . , 45). For this purpose, we get first from (4.70), Lemma 4.4 and (2.9) with m = 1/4 that

t I36 ≤ 0 R+

t ≤ 0 R+

2 Rr 2(n−1) θ vxx +C 20v 3

2 Rr 2(n−1) θ vxx +C 20v 3

t

V (s)

0

R+

t

V (s) R+

0

2 r 2n−2 vxx +C v2

t 0 R+

r 2n−2 u2tx v

2 r 2n−2 vxx + C. v2

(4.106)

Next, due to the fact that

(r n−1 u)x v

= x

2(n − 1)ux r n−1 uxx (n − 1)uv r n−1 ux vx − + − , r v r n+1 v2

we can obtain from (2.3), (2.9), (3.2), (3.9), (3.11) and (4.77) that

t I37 ≤ C 0 R+

t ≤ 0 R+

v |uvxx | r 2n−3 |vxx vx ux | r 2n−3 |vxx uxx | + r n−3 |ux vxx | + + 3 v r v2

2 Rr 2(n−1) θ vxx +C 20v 3

t 0

t 2 +C 0 R+

V (s) R+

2 r 2n−2 vxx v2

vu2 r 2(n−1) u2x + + r 2n−2 u2x vx2 + r 2n−2 u2xx vθ r 2θ

3



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Y. Liao et al. / J. Differential Equations 266 (2019) 6459–6506

t ≤ 0 R+

t

≤ 0 R+

2 Rr 2(n−1) θ vxx +C +C 20v 3

t  V (s) r

n−1



 2   vxx + r n−1 ux  ∞ 2

L (R+ )

vx

2

0 2 Rr 2(n−1) θ vxx +C 3 20v

t V (s) r n−1 vxx 2 + C.

(4.107)

0

And

t I38 ≤ 0 R+

t ≤ 0 R+

2 Rr 2(n−1) θ vxx +C 3 20v



t V (s)

R+

0 2 Rr 2(n−1) θ vxx +C 20v 3

t

2 r 2n−2 vxx +C 2 v

2 r 2n−4 u2 vxx 2 v

0 R+

2n−2 2

t  u 2 r vxx   V (s) +   . 2 ∞ r L (R+ ) v

(4.108)

R+

0

For the term I39 , since

vx2 v2

= t

2vx vtx v − vx vt 2vx (r n−1 u)xx 2v 2 (r n−1 u)x · = − x 2 , 2 2 v v v v

then we have

t I39  0 R+

  t 2n−2   vx vxx (r n−1 u)xx  r 2n−2 vx2 vxx (r n−1 u)x  r + . v4 v3

,

-.

/

0 R+

,

-.

1 I39

(4.109)

/

2 I39

1 , we can deduce that As for the term I39

t 1 I39 ≤ 0 R+

t ≤ 0 R+

t ≤ 0 R+

2 Rr 2(n−1) θ vxx 80v 3

t +C 0

2 Rr 2(n−1) θ vxx +C 80v 3

⎡ ⎢ ⎣V (s)

R+

2 r 2n−2 vxx v2

+

!

"2

⎤

⎥ r 2n−2 vx4 (r n−1 u)x ⎦

R+

t   2   V (s) r n−1 vxx 2 + vx 2L∞ (R+ ) (r n−1 u)x  ∞



L (R+ )

0 2 Rr 2(n−1) θ vxx +C 3 40v

t ! " V (s) r n−1 vxx 2 + (r n−1 u)xx 2 . 0

(4.110)

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Here we have used (4.78) and the fact that

2   u  n−1 2 2n−2 2 u)x   +r ux dx  1. (r r2

(4.111)

R+

On the other hand, we can infer from (2.3), (3.1), (4.77) and (4.78) that

t

t

(r

n−1

u2

u)xx  2

r 2n+2

0 R+

0

t 1+ 0 R+

u2 ·θ + r 2θ

u2 v 2 u2 + 2 x + 2x + r 2n−2 u2xx r r



t  2

t 2n−2 2 r ux u 2 vx + θ  1.   ∞ r L (R+ ) vθ R+

0

(4.112)

0 R+

The combination of (4.110) and (4.112) shows that

t 1 I39

≤ 0 R+

2 Rr 2(n−1) θ vxx +C 40v 3

t V (s) r n−1 vxx 2 + C.

(4.113)

0

Moreover, it follows from (2.9), (3.1), (4.78), (4.79) that

t

2 r 2(n−1) θ vxx 80v 3

2 I39 ≤ 0 R+

t

t +C

⎡ ⎢ ⎣V (s)

R+

0

2 r 2(n−1) θ vxx +C 3 80v

≤ 0 R+

t

V (s) R+

0

t +C 0 R+

t 0 R+

2 r 2n−2 vxx v2

+

⎤ ! "2 ⎥ r 2n−2 vx2 (r n−1 u)xx ⎦

R+

2 r 2n−2 vxx 2 v

vx2 u2 r 2n−2 u2 vx4 r 2n−2 u2x vx2 + + + r 4n−4 vx2 u2xx r4 r2 r2

2 r 2(n−1) θ vxx 80v 3

≤



t +C

⎡ ⎢ ⎣V (s)

R+

0

2 r 2n−2 vxx v2



⎤  u 2 ⎥   + 

vx 2 ⎦ r L∞ (R+ )

t +C

vx 2L∞ (R+ ) (r n−1 vx , r n−1 ux , r 2n−2 uxx ) 2 0

1 ≤C+ 80

t

r 0

n−1

t ! " √ 2 θ vxx + C V (s) r n−1 vxx 2 + vx

vxx (1 + r 2n−2 uxx 2 ) 0

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1 ≤C+ 40

t

r

n−1

√

t θvxx + C

0

0

t +C

V (s) r n−1 vxx 2

2

t

vx 2 + C

0

r 2n−2 uxx 2 r n−1 vxx 2 ds.

(4.114)

0

Furthermore, (2.9) and (4.78) tell us that

t

t vx2



0 R+

t θ vx2

+C

0 R+

V (s)

vx2  1.

(4.115)

R+

0

Substituting (4.113)–(4.115) into (4.109) gives 1 I39 ≤ C + 40

t

2n−2 2 √ r vxx 2 θ vxx + C V (s) v2 t

r

n−1

0

0

R+

t +C

r 2n−2 uxx 2 r n−1 vxx 2 ds.

(4.116)

0

For I40 , by making use of (2.9) and (4.80), we find I40 ≤

1 20

1 ≤ 20

t 0 R+

t 0 R+

2 Rr 2(n−1) θ vxx +C v3

2 Rr 2(n−1) θ vxx +C v3

t

V (s)

0

R+

t

V (s) R+

0

2 r 2n−2 vxx +C v2

t 2 r 2n−2 θxx 0 R+

2 r 2n−2 vxx + C. v2

(4.117)

As for the term I41 , we have from (2.3), (4.77) and (4.78) that

t I41 ≤ C 0 R+

1 ≤ 20

≤

1 20

t 0 R+

t

0 R+

r n−2 |vxx θx | r n−2 θ |vxx vx | + θ 3 |vxx θx | + v v2

2 Rr 2(n−1) θ vxx +C v3

t  0 R+

2 Rr 2(n−1) θ vxx + C. 3 v



 r 2n−2 θx2  · θ + θ 7 + θ vx2 2 vθ



(4.118)

Y. Liao et al. / J. Differential Equations 266 (2019) 6459–6506

6501

In view of (4.78) and (4.115) and similar to that of (4.114), we conclude 1 I42 ≤ 40

≤

1 40

1 ≤ 20

t 0 R+

t

0 R+

t 0 R+

2 Rr 2(n−1) θ vxx +C v3

t

V (s) R+

0 2 Rr 2(n−1) θ vxx +C 3 v

2 Rr 2(n−1) θ vxx +C v3

t

V (s)

0

R+

t

V (s)

0

R+

2 r 2n−2 vxx +C v2

t

vx 2L∞ (R+ ) R+

0 2 r 2n−2 vxx +C 2 v

r 2n−2 θx2

t

vx

vxx ds 0

2 r 2n−2 vxx +C v2

(4.119)

and 1 I43 ≤ 40 1 ≤ 40 1 ≤ 20

t 0 R+

t 0 R+

t 0 R+

2 Rr 2(n−1) θ vxx +C 3 v

t

vx 2L∞ (R+ ) R+

0 2 Rr 2(n−1) θ vxx +C v3

r 2n−2 vx2

t

vx

vxx ds 0

2 Rr 2(n−1) θ vxx +C v3

t

V (s) R+

0

2 r 2n−2 vxx + C. v2

(4.120)

For the term I44 , one gets from (4.78) and (4.81) that 1 I44 ≤ 20

≤

1 20

1 ≤ 20

t 0 R+

t 0 R+

t 0 R+

2 Rr 2(n−1) θ vxx +C v3

2 Rr 2(n−1) θ vxx +C v3

t r 2n−2 θx4 0 R+

t  2   n−1 θx (s) ∞ r

L (R+ ) R+

0

θx2

2 Rr 2(n−1) θ vxx + C. v3

(4.121)

Furthermore, we conclude from (4.80) that 1 I45 ≤ 20

t 0 R+

2 Rr 2(n−1) θ vxx +C v3

t 2 r 2n−2 θxx 0 R+

1 ≤ 20

t 0 R+

2 Rr 2(n−1) θ vxx + C. v3

(4.122)

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Combining (4.105)–(4.122) and taking full advantage of (3.1), (4.78) and Gronwall’s inequality, we can obtain (4.103). 2 t  It remains to estimate the term 0 r n−1 vx (s)L∞ (R ) ds. To this end, noticing that +



r n−1 vx

 x

= (n − 1)r −1 vvx + r n−1 vxx ,

we have

t    n−1 2 vx  ∞ r

L (R+ )

t         ds  r n−1 vx   r n−1 vx  ds x

0



0

t 

2 r 2n−2 vx2 + vx2 + r 2n−2 vxx



0 R+

1+

t 

θ r 2n−2 vx2

2 + θ r 2n−2 vxx

0 R+



t V (s) (r n−1 vx , r n−1 vxx ) 2 ds  1. (4.123)

+ 0

Here we have used (2.9), (2.12), (4.78), (4.103) and (4.115).

2

5. Lower bound on temperature We shall derive a local-in-time lower bound on the temperature θ (t, x) in the following lemma. Lemma 5.1. Under the assumptions listed in Theorem 1.1, for all 0 ≤ s ≤ t ≤ T and x ∈ R+ , θ (t, x) ≥

minR+ θ (s, ·) . C + C(t − s + 1) minR+ θ (s, ·)

(5.1)

Proof. By setting h(t, x) := 1/θ (t, x), we multiply (4.12) by θ −2 to get

# 2n−2 2

κhx r 2n−2 κhx 2r h2 eθ ht = − + ((n − 1)δα − 2(n − 2)μ) v vh v x 2  2μ (nδα − 2(n − 1)μ) r 2n−2 u2x h2 uv (αδ − 2μ) r n−1 ux + + × r (n − 1)δα − 2(n − 2)μ v ((n − 1)δα − 2(n − 2)μ)  $ 2 vPθ2 α (1 − δ) h2 vPθ + + λφzh2 + (r n−1 u)x − , v 2α (1 − δ) h 4α (1 − δ)

(5.2)

where α = 2μ + ζ > 0 and 0≤

2(n − 2)μ 2(n − 1)μ < < δ < 1. (n − 1)α nα

(5.3)

Y. Liao et al. / J. Differential Equations 266 (2019) 6459–6506

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By virtue of (4.77), one can infer ht ≤

1 eθ



r 2n−2 κ (v, θ ) hx v



vPθ2 1 ≤ 4α (1 − δ) eθ eθ

+ x



r 2n−2 κ (v, θ ) hx v

+ C.

(5.4)

x

For g(t, x) := h(t, x) − 1, we have 1 gt ≤ eθ



r 2n−2 κ (v, θ ) gx v

+ C.

(5.5)

x

Multiply (5.5) by (g+ )2p−1 with g+ (t, x) := max{g(t, x), 0} to yield   2p−1

g+ L2p (R ) g+ L2p (R+ ) + t



(g+ )2p−1 r 2n−2 κ (v, θ ) gx  (g+ )2p−1 dx + dx. eθ v x R+

(5.6)

R+

And we also have (g+ )2p−1 eθ



r 2n−2 κ (v, θ ) gx v

2

= x

r 2n−2 κ (v, θ ) (g+ )2p−1 gx veθ

r 2n−2 κ (v, θ ) (g+ )2p−1 (g+ )x (eθ )x + − veθ2

3 x

(2p − 1) r 2n−2 κ (v, θ ) (g

2p−2 +)

4

(g+ )x

52 .

(5.7)

4 52  2 (2p − 1) r 2n−2 κ (v, θ ) (g+ )2p−2 (g+ )x C r 2n−2 κ (v, θ ) (g+ )2p (eθ )x ≤ + . veθ 2p − 1 veθ3

(5.8)

veθ

It is easy to see r 2n−2 κ (v, θ ) (g+ )2p−1 (g+ )x (eθ )x veθ2

Combining (5.6)–(5.8) and making use of Hölder’s inequality yield 

2p−1

g+ (t) L2p (R

+)

g+ (t) L2p (R+ )

 t

4p−3

1

≤ C g+ (t) 2 2p L

3

(R+ )

g+ (t) L2p2 (R

   r 2n−2 κ (v, θ ) g (e ) 2  C + θ x   2p−1

g+ (t) L2p (R )  +  +   2p − 1 veθ3

+)

.

L2p (R+ )

That is, 1



g+ (t) L2p (R+ )

g+ (t) 2 2p

 t

≤C

L

3

(R+ )

1

g+ (t) L2 2p (R

+)

  2   2n−2 κ (v, θ ) g C  + (eθ )x  r +    2p − 1  veθ3

L2p (R+ )

. (5.9)

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Y. Liao et al. / J. Differential Equations 266 (2019) 6459–6506

Integrating (5.9) with respect to t over (s, t), we arrive at 1

2

t g+ (τ ) 2p

L

g+ (t) L2p (R+ ) ≤ g+ (s) L2p (R+ ) + C

3

(R+ )

1 2

g+ (τ ) L2p (R

s

dτ

+)



t   r 2n−2 κ (v, θ ) g (e ) 2  C + θ x   +    2p − 1  veθ3

dτ.

(5.10)

L2p (R+ )

s

Moreover, one can conclude from (2.12) that 

(eθ )x

2

 2 = 4aθ 3 vx + 12avθ 2 θx  θ 6 vx2 + θ 4 θx2 .

(5.11)

If we define 1 (t) = {x ∈ R+ |0 < θ(t, x) < 1}, then it follows from (2.3), (2.12), (4.77), (4.81), (4.104) and (5.11) that 

t   r 2n−2 κ (v, θ ) g (e ) 2  + θ x       veθ3

L2p (R+ )

s



t   r 2n−2 κ (v, θ ) g (e ) 2  θ x   dτ =     veθ3



t   r 2n−2 1 + θ b  (1 − θ ) θ 6 v 2 + θ 4 θ 2    x x  ≤C     θ s



t      2n−2 (1 − θ ) vx2 + θx2  r

L2p (1 (τ ))

s

t  2   n−1 vx (τ ) ∞  r

L (R+ )

dτ

L2p (1 (τ ))

s

dτ L2p (1 (τ ))

dτ

 2   + r n−1 θx (τ ) ∞

L (R+ )

0

⎛ ⎜ ⎝

⎞

1 2p

⎟ (1 − θ (τ, x))2p dx ⎠

dτ  1.

R+

(5.12) Hence (5.10) and (5.12) give birth to 1

2

t g+ (τ ) 2p

L

g+ (t) L2p (R+ ) ≤ g+ (s) L2p (R+ ) + C s

3

(R+ )

1 2

g+ (τ ) L2p (R

dτ +

C . 2p − 1

(5.13)

+)

Then we can deduce from the fact g+ (t, x) ∈ L∞ (R+ ) ∩ L2 (R+ ) that lim g+ (t) Lp (R+ ) = g+ (t) L∞ (R+ ) .

p→+∞

(5.14)

Y. Liao et al. / J. Differential Equations 266 (2019) 6459–6506

6505

Letting p → +∞ in (5.13) and using (5.14), we have

g+ (t, ·) L∞ (R+ ) ≤ g+ (s, ·) L∞ (R+ ) + C (t − s) . Then (5.1) follows from (5.15) and the definition of g+ (t, x) immediately.

(5.15) 2

With Lemmas 2.1–5.1 in hand, we can deduce Theorem 1.1 by using the continuation argument introduced in Wang and Zhao [27] and we omit the details for brevity. Acknowledgment The research of Yongkai Liao was supported in part by National Postdoctoral Program for Innovative Talents of China under contract BX20180054. The research of Tao Wang was supported in part by the grants from National Natural Science Foundation of China under contracts 11601398 and 11731008. The research of Huijiang Zhao was supported in part by the grants from National Natural Science Foundation of China under contracts 11671309 and 11731008. References [1] G.Q. Chen, Global solutions to the compressible Navier–Stokes equations for a reacting mixture, SIAM J. Math. Anal. (ISSN 0036-1410) 23 (3) (1992) 609–634, https://doi.org/10.1137/0523031. [2] G.-Q. Chen, D. Hoff, K. Trivisa, On the Navier–Stokes equations for exothermically reacting compressible fluids, Acta Math. Appl. Sin. Engl. Ser. (ISSN 0168-9673) 18 (1) (2002) 15–36, https://doi.org/10.1007/s102550200002. [3] G.-Q. Chen, D. Hoff, K. Trivisa, Global solutions to a model for exothermically reacting, compressible flows with large discontinuous initial data, Arch. Ration. Mech. Anal. (ISSN 0003-9527) 166 (4) (2003) 321–358, https:// doi.org/10.1007/s00205-002-0233-6. [4] D. Donatelli, K. Trivisa, On the motion of a viscous compressible radiative–reacting gas, Comm. Math. Phys. (ISSN 0010-3616) 265 (2) (2006) 463–491, https://doi.org/10.1007/s00220-006-1534-7. [5] B. Ducomet, A model of thermal dissipation for a one-dimensional viscous reactive and radiative gas, Math. Methods Appl. Sci. (ISSN 0170-4214) 22 (15) (1999) 1323–1349, https://doi.org/10.1002/(SICI)1099-1476(199910)22: 15<1323::AID-MMA80>3.0.CO;2-8. [6] B. Ducomet, A. Zlotnik, Lyapunov functional method for 1D radiative and reactive viscous gas dynamics, Arch. Ration. Mech. Anal. (ISSN 0003-9527) 177 (2) (2005) 185–229, https://doi.org/10.1007/s00205-005-0363-8. [7] B. Ducomet, A. Zlotnik, On the large-time behavior of 1D radiative and reactive viscous flows for higher-order kinetics, Nonlinear Anal. (ISSN 0362-546X) 63 (8) (2005) 1011–1033, https://doi.org/10.1016/j.na.2005.03.064. [8] E. Feireisl, A. Novotný, Singular Limits in Thermodynamics of Viscous Fluids, Advances in Mathematical Fluid Mechanics, Birkhäuser Verlag, Basel, ISBN 978-3-7643-8842-3, 2009. [9] L. He, Y. Liao, T. Wang, H. Zhao, One-dimensional viscous radiative gas with temperature dependent viscosity, Acta Math. Sci. Ser. B (Engl. Ser.) (ISSN 0252-9602) 38 (5) (2018) 1515–1548, https://doi.org/10.1016/S02529602(18)30830-0. [10] H.K. Jenssen, G. Lyng, M. Williams, Equivalence of low-frequency stability conditions for multidimensional detonations in three models of combustion, Indiana Univ. Math. J. (ISSN 0022-2518) 54 (1) (2005) 1–64, https:// doi.org/10.1512/iumj.2005.54.2685. [11] J. Jiang, S. Zheng, Global solvability and asymptotic behavior of a free boundary problem for the one-dimensional viscous radiative and reactive gas, J. Math. Phys. (ISSN 0022-2488) 53 (12) (2012) 123704, https://doi.org/10.1063/ 1.4770049. [12] J. Jiang, S. Zheng, Global well-posedness and exponential stability of solutions for the viscous radiative and reactive gas, Z. Angew. Math. Phys. (ISSN 0044-2275) 65 (4) (2014) 645–686, https://doi.org/10.1007/s00033-013-0350-0. [13] S. Jiang, Large-time behavior of solutions to the equations of a one-dimensional viscous polytropic ideal gas in unbounded domains, Comm. Math. Phys. (ISSN 0010-3616) 200 (1) (1999) 181–193, https://doi.org/10.1007/ s002200050526.

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