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Global strong solution of compressible flow with spherically symmetric data and density-dependent viscosities

  • Zhenhua Guo EMAIL logo , Lei Xu and Xueyao Zhang
Published/Copyright: August 13, 2025
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Advances in Nonlinear Analysis
From the journal Advances in Nonlinear Analysis Volume 14 Issue 1

Abstract

In this article, the Cauchy problem of a compressible Navier-Stokes system with density-dependent viscosities when the initial data are spherically symmetric is considered. Firstly, we construct the classical solution for the system in B a ( t ) = { r : 0 ≤ r ≤ a ( t ) } with the stress-free boundary, where a ( t ) is the “particle path,” which arising from a ( 0 ) = a 0 > 0 . Next, the strong solution for the system with the stress-free condition on the boundary a ( t ) in the exterior domain R 3 \ B a ( t ) is proved by the entropy estimate and the subtle analysis on the upper and lower bounds of the density. Finally, by continuously splicing the two parts of the solution via the stress-free boundary, we establish the global strong solution of the Cauchy problem. In particular, our analysis gives a positive example that it does not exhibit vacuum states provided that no vacuum states are present initially for multi-dimensional compressible flow with density-dependent viscosities, which is consistent with Hoff and Smoller [Hoff and Smoller, Non-formation of vacuum states for compressible Navier-Stokes equations, Comm. Math. Phys.  216 (2001), no. 2, 255–276], where the viscosity coefficients are assumed to be positive constants for one-dimensional case.

Keywords: compressible Navier-Stokes; strong solutions; density-dependent viscosities; spherically symmetric; Lagrangian coordinates
MSC 2010: 35Q30; 35B65; 76N10

1 Introduction

The system of isentropic compressible Navier-Stokes equations with density-dependent viscosity coefficients in R N can be read as follows:

(1) ρ t + div ( ρ U ) = 0 , ( ρ U ) t + div ( ρ U ⊗ U ) − div ( μ ( ρ ) D ( U ) ) − ∇ ( λ ( ρ ) div U ) + ∇ P ( ρ ) = 0 ,

where ρ ( x , t ) , U ( x , t ) , and P ( ρ ) = ρ γ ( γ ≥ 1 ) are the fluid density, velocity, and pressure, respectively, and D ( U ) = ∇ U + ∇ U T 2 . The shear viscosity coefficient μ ( ρ ) and the bulk viscosity coefficient λ ( ρ ) satisfy

(2) μ ( ρ ) ≥ 0 , μ ( ρ ) + N λ ( ρ ) ≥ 0 .

Throughout the process of studying the compressible Navier-Stokes equations with the viscosity coefficients as positive constants, a critical problem that one has to face is the possible appearance of vacuum, which can lead to a loss of regularity in the solution. In 1991, Hoff and Serre [13] show that physical solutions of the Navier-Stokes equations for one-dimensional, compressible flow need not depend continuously on their initial data, at least when vacuum states are allowed, and then can construct an nonphysical solutions of the Navier-Stokes equations for the compressible flow [14]. Later, Hoff and Smoller [15] considered the properties of vacuum states to the one-dimensional problem and proved that there are no vacuum states that develop in finite time, provided that there are no initial vacuum states. Xin and Yuan [25] extended this result to the multidimensional spherically symmetric case in a region far away from the center of symmetry, providing a sufficient condition on the regularity of the velocity to ensure no formation of vacuum. Moreover, for the one-dimensional initial-boundary value problem where the discontinuous initial data are connected by a vacuum state, Guo et al. [9] proved that the vacuum region is compressed at an algebraic rate and disappears in a finite time.

However, in general, for multidimensional, vacuum states may appear in the weak solutions of (1), even if the initial data are far from vacuum. For the spherically symmetric flow, an important problem is whether the initial nonvacuum states are retained over time, especially near the symmetry center at r = 0 . The difficulties here lie in the lack of estimates for the density and velocity near the symmetry center, which could induce the concentration of singularities. For instance, as initial data away from vacuum states, for γ = 1 , Hoff [16] first constructed the global weak solutions of (1) with constant viscosities and proved that such solutions may develop vacuum regions near the symmetry center, where the boundaries of these vacuum regions are Hölder continuous in space-time. Hoff and Jenssen [12] generalized these results and proved the global existence of weak solutions for the nonbarotropic flow. As for γ > 1 , Guo and Li [8] and Guo et al. [10] further obtained the global existence of weak solutions, and similarly, vacuum may appear near the center.

On the other hand, the case that both the shear and bulk viscosities depend on the density has also received a lot of attention recently, see [1,2,6,7,22,23] and the references therein. By some physical considerations, Liu et al. in [22] introduced the modified compressible Navier-Stokes equations with density-dependent viscosity coefficients for isentropic fluids. In fact, as presented in [21], while deriving the compressible Navier-Stokes equations from the Boltzmann equations by the Chapman-Enskog expansions, the viscosity depends on the temperature and correspondingly depends on the density for isentropic cases. Moreover, in geophysical flow, the viscous Saint-Venant system for the shallow water corresponds exactly to a kind of compressible Navier-Stokes equation with density-dependent viscosities, see details in [2]. In particular, if the viscosity coefficients satisfy the Bresch-Desjardins (BD) relation, λ ( ρ ) = 2 ( ρ μ ′ ( ρ ) − μ ( ρ ) ) , a new mathematical entropy estimate of the bounds on the derivative of the density was obtained by Bresch et al. [1,2]. Later, in 2007, by obtaining a new a priori estimate on smooth approximate solutions, Mellet and Vasseur [23] studied the stability of (1). Building on these results, the global existence of weak solutions to the initial-boundary-value problem was first proved by Guo et al. [6,7] in the case of spherically symmetric initial data. Recently, in 2015, for general initial data, Li and Xin [20] obtained the global weak solutions of the two- and three-dimensional periodic cases and the Cauchy problems, and Bresch et al. [3,24] obtained similar results for some special cases with a different approach. In particular, under the smallness assumptions on initial data, Xin and Zhu [26] proved the global well-posedness of regular solutions to the three-dimensional Cauchy problem with vacuum for a class of smooth initial data that are of small density but possibly large velocities.

For the issue that the possible appearance of vacuum, there are only a few one dimensional results in the case of density-dependent viscosities [4,11,27,28]. For one-dimensional large initial data, it is shown that for any global entropy weak solution, any (possibly existing) vacuum state must vanish within finite time [19]. Later, Lian et al. [18] proved that for the initial boundary value problem when the initial data contain discontinuously a piece of continuous vacuum and is regular away from the vacuum, the unique weak solution exists, and the vacuum state is compressed at an algebraic rate and vanishes within the finite time.

While for the high-dimensional case, the situation becomes more complex. Accordingly, when the viscosity coefficients depend on the density, some new information (BD-entropy) provides hope to study the properties of vacuum states for the way forward. For example, it is well known that away from the symmetry center, the strong solutions of the system (1) can be obtained easily, as similar as in the 1D case with density-dependent viscosities. In 2012, an analytical solution of (1) with density-dependent viscosities was constructed by Guo and Xin [5], which solutions with stress-free boundary conditions mean no vacuum state appears near the center. This discovery inspires us to construct a strong solution by dividing the Cauchy problem (1) into two free boundary problems that are continuously connected by the free boundary a ( t ) and splicing the solutions continuously through the stress-free conditions, and then gives positive examples that it does not exhibit vacuum states provided that no vacuum states are present initially for multi-dimensional compressible flow.

In this article, we consider the Cauchy problem of the compressible Navier-Stokes system with spherically symmetric initial data, where the viscosities satisfy the BD relation. To overcome the difficulties induced by the symmetry center, we first deal with a free boundary problem in B a ( t ) with the stress-free boundary by means of a constructive proof and then solve the system in the exterior domain for r ∈ [ a ( t ) , + ∞ ) , where a ( t ) is a special “particle path” defined by

(3) a ′ ( t ) = u ( a ( t ) , t ) , a ( 0 ) = a 0 > 0 ,

and the system satisfies the stress-free boundary on the curve of a ( t ) :

(4) ρ γ ( a ( t ) , t ) = α ρ α u r + 2 u r ( a ( t ) , t ) .

It is worth noticing that the stress-free boundary condition is natural, which is compatible with the conservation of momentum in each part.

  • For the free boundary problem of r ∈ [ 0 , a ( t ) ] , a class of analytical solutions has been obtained in the previous work [5]. Here, we relax the range of the adiabatic exponent γ and further discuss the boundedness of the analytical solution.

  • For the free boundary problem of r ∈ [ a ( t ) , + ∞ ) , based on the exploitation of the one-dimensional feature of the symmetric solution, we obtain higher-integrability estimates. Specifically, for any γ > 1 and n ∈ N + that satisfies (55), we have:

    ‖ ρ u 2 n r 2 ‖ L ∞ ( 0 , T ; L 1 ) + ρ α − 1 + 1 2 n r 2 n r 2 L ∞ ( 0 , T ; L 1 ) ≤ C ( n , T ) .

    Furthermore, we can obtain upper and lower bounds on the density, which allows us to establish the global existence of strong solutions for r ∈ [ a ( t ) , + ∞ ) . This generalizes the previous one-dimensional results to the multi-dimensional case.

Finally, by using the stress-free boundary condition, we can continuously concatenate the two parts of the solutions and obtain the global solution of the Cauchy problem.

2 Main result

Let

(5) μ ( ρ ) = ρ α , λ ( ρ ) = ( α − 1 ) ρ α , P ( ρ ) = ρ γ

in (1) and consider the spherically symmetric solutions in a ball in R 3 , such that

(6) ∣ x ∣ = r , ρ ( x , t ) = ρ ( r , t ) , U ( x , t ) = u ( r , t ) x r .

Then, (1) becomes

(7) ρ t + ( ρ u ) r + 2 r ρ u = 0 , ( ρ u ) t + ( ρ u 2 ) r + ( ρ γ ) r + 2 r ρ u 2 − α r 2 ρ α ( r 2 u ) r r + 2 r ( ρ α ) r u = 0 .

The initial value and boundary conditions are given by

(8) ( ρ , ρ u ) ∣ t = 0 = ( ρ 0 , m 0 ) , r ∈ [ 0 , + ∞ )

and

(9) u ( 0 , t ) = 0 ,

(10) ρ ( r , t ) → ρ ˜ > 0 , u ( r , t ) → 0 , as r → + ∞ .

Denote the potential energy by

(11) ψ ( ρ , ρ ˜ ) ≜ 1 γ − 1 ( ρ γ − ρ ρ ˜ γ − 1 ) + ρ ˜ γ ( 1 − ρ ρ ˜ − 1 ) .

In the sequel, f ∈ L p ( [ 0 , + ∞ ) ; r 2 d r ) ( p ≥ 1 ) means that ∫ 0 + ∞ ∣ f ∣ p r 2 d r 1 p < + ∞ .

The main result of this article reads:

Theorem 1

Let

α = 3 γ + 1 6 , γ 0 < γ ≤ 5 3 .

Assume that the initial data ( ρ 0 , u 0 ) ( r ) satisfy, for some given a 0 > 0 , ρ ˜ > 0 ,

( I D ) ρ 0 ( r ) = f r a 0 a 0 3 , u 0 ( r ) = r ρ 0 γ − α ( a 0 ) 3 α , 0 ≤ r ≤ a 0 , inf r ∈ [ a 0 , + ∞ ) ρ 0 ( r ) > 0 , ψ ( ρ 0 , ρ ˜ ) ∈ L 1 ( [ a 0 , + ∞ ) ; r 2 d r ) , ρ 0 ∈ L ∞ ( [ a 0 , + ∞ ) , r 2 d r ) , ( ρ 0 ) r ∈ L 2 ∩ L ∞ ( [ a 0 , + ∞ ) ; r 2 d r ) , u 0 ∈ L ∞ ( [ a 0 , + ∞ ) ; r 2 d r ) ∩ H 1 ( [ a 0 , + ∞ ) ; r 2 d r ) .

Here, γ 0 = 4 − 73 3 and f ( z ) ∈ C ( [ 0 , 1 ] ) ∩ C 1 ( ( 0 , 1 ] ) , ∀ z ∈ [ 0 , 1 ] with

f ( 0 ) = γ − 1 2 γ 6 3 γ − 1 a 0 3 ρ 0 ( a 0 ) > 0 , f ( 1 ) = a 0 3 ρ 0 ( a 0 ) > 0

is the solution of

(12) 6 ( γ − 1 ) ( 3 γ + 1 ) 2 z − γ f 1 3 − γ ( 1 ) f γ − 2 ( z ) f ′ ( z ) + γ − 1 2 f 1 6 − γ 2 ( 1 ) f 3 γ − 11 6 ( z ) f ′ ( z ) = 0 .

For any given T > 0 , there exists a global solution ( ρ , u ) ( r , t ) to the problems (7)–(10) satisfying that (7) holds for a.e. ( r , t ) ∈ [ 0 , + ∞ ) × [ 0 , T ] . Moreover, for any n ∈ N + satisfying (55) that,

(13) C − 1 ( T ) ≤ ρ ( r , t ) ≤ C ( T ) ,

(14) ψ ( ρ , ρ ˜ ) ∈ L ∞ ( [ 0 , T ] ; L 1 ( [ 0 , + ∞ ) ; r 2 d r ) ) , ρ r ∈ L ∞ ( [ 0 , T ] ; L 2 n ( ( 0 , + ∞ ) ; r 2 d r ) ) ,

(15) u ∈ L ∞ ( [ 0 , T ] ; H 1 ∩ L ∞ ( [ 0 , + ∞ ) ; r 2 d r ) ) ∩ L 2 ( [ 0 , T ] ; H 2 ( [ 0 , + ∞ ) ; r 2 d r ) ) .

Remark 1

Theorem 1 constructs a special strong solution that does not develop vacuum states in any finite time, provided the initial density be satisfied with some conditions away from vacuum states. This gives a positive example to the results in [25], which only show such phenomenon in a region far away from the center of symmetry.

Remark 2

For the weak solutions of (7)–(10) in a bounded domain obtained in [6], Theorem 1 means that for some special initial data without small conditions, the weak solution is strong ones. For the general initial data, the stability of the solution will be studied in the future, which gives the information of the existence of solutions in some sense.

Remark 3

The findings from our Theorem 1 encompass several cases of adiabatic indices that hold significant physical relevance by virtue of the facts that γ ∈ ( γ 0 , 5 3 ] , γ 0 = 4 − 73 3 ≅ 1.152 . For instance, triatomic gases such as carbon dioxide exhibit an adiabatic index of γ = 4 3 , diatomic gases like oxygen, nitrogen, and air correspond to γ = 1.4 , while monatomic gases, including noble gases, demonstrate an adiabatic index of γ = 5 3 .

Remark 4

Our method seems not effective for the case of the compressible Navier-Stokes equations with the positive constant viscosity coefficients because of the special condition on α , γ of constructing a strong solution on [ 0 , a ( t ) ] . Anyway, the structure of the solution near the symmetry center will bring hope to reveal the possible appearance of a vacuum in the classical compressible Navier-Stokes equations.

Remark 5

Noticing that our solution obtained in Theorem 1 is also a finite energy solution.

In fact, combining the Proposition 2, Lemma 3 and Lemma 9, it holds that

E ( t ) = E [ 0 , a ( t ) ] ( t ) + E [ a ( t ) , + ∞ ) ( t ) = ∫ 0 a ( t ) 1 2 ρ u 2 + ψ ( ρ , ρ ˜ ) r 2 d r + ∫ a ( t ) ∞ 1 2 ρ u 2 + ψ ( ρ , ρ ˜ ) r 2 d r ≤ C sup 0 ≤ t ≤ T ( a ′ ( t ) ) 2 a ( t ) − 5 ∫ 0 a ( t ) r 4 d r + C ( ρ ˜ ) sup 0 ≤ t ≤ T ( a ( t ) − 3 γ + a ( t ) − 3 ) ∫ 0 a ( t ) r 2 d r + C ≤ C sup 0 ≤ t ≤ T ( a ′ ( t ) ) 2 + C sup 0 ≤ t ≤ T ( a ( t ) − 3 γ + 3 + 1 ) ≤ C sup 0 ≤ t ≤ T ρ 0 1 − 3 γ 6 ( a 0 ) + 3 γ − 1 3 γ + 1 t 6 ( 1 − γ ) 3 γ − 1 + C ≤ C ,

where C is independent of T for all γ > 1 .

3 The proof of main result

The proof of the problem can be divided three steps. For the “particle path” r = a ( t ) arising from a 0 , which satisfies

u ( a ( t ) , t ) = a ′ ( t ) , a ( 0 ) = a 0 ,

we first construct the classical solution in B a ( t ) with the stress-free boundary

ρ γ ( a ( t ) , t ) = α ρ α u r + 2 u r ( a ( t ) , t ) .

Then, we use the curve of a ( t ) as the boundary of the system through the stress-free boundary to establish the well-posedness of strong solutions in R 3 \ B a ( t ) for the continuous initial data.

Thus, the solutions of two regions are both continuous and are connected by a ( t ) continuously due to the stress-free condition, which mean that the solution of the system in R 3 is continuous, see Figure 1.

Figure 1 The strong solution of the system.
Figure 1

The strong solution of the system.

Exactly as pointed as the aforementioned analysis, we can look for a solution of the system (7) with

( ρ , u ) ( r , t ) = ( ρ I , u I ) ( r , t ) , 0 ≤ r ≤ a ( t ) , ( ρ I I , u I I ) ( r , t ) , a ( t ) ≤ r < + ∞ ,

where ( ρ I , u I ) and ( ρ I I , u I I ) satisfy the following two combinations of systems:

(16) ( I ) ρ t + ( ρ u ) r + 2 r ρ u = 0,0 ≤ r ≤ a ( t ) , t ≥ 0 , ( ρ u ) t + ( ρ u 2 ) r + ( ρ γ ) r + 2 r ρ u 2 − α r 2 ρ α ( r 2 u ) r r + 2 r ( ρ α ) r u = 0 , ρ γ ( a ( t ) , t ) = α ρ α u r + 2 u r ( a ( t ) , t ) , u ( a ( t ) , t ) = a ′ ( t ) , u ( 0 , t ) = 0 , a ( 0 ) = a 0 , ( ρ , u ) ( r , 0 ) = ( ρ 0 , u 0 ) ( r ) , r ∈ [ 0 , a 0 ] ,

(17) ( I I ) ρ t + ( ρ u ) r + 2 r ρ u = 0 , a ( t ) ≤ r < + ∞ , t ≥ 0 , ( ρ u ) t + ( ρ u 2 ) r + ( ρ γ ) r + 2 r ρ u 2 − α r 2 ρ α ( r 2 u ) r r + 2 r ( ρ α ) r u = 0 , ρ γ ( a ( t ) , t ) = α ρ α u r + 2 u r ( a ( t ) , t ) , u ( a ( t ) , t ) = a ′ ( t ) , ρ ( r , t ) → ρ ˜ > 0 , u ( r , t ) → 0 , a s r → + ∞ , a ( 0 ) = a 0 , ( ρ , u ) ( r , 0 ) = ( ρ 0 , u 0 ) ( r ) , r ∈ [ a 0 , + ∞ ) .

3.1 The exact solution of system (I)

Firstly, we give the global existence of the exact solution to system (I), which can be derived similar as in [5,29].

Proposition 2

Under the assumptions of Theorem 1, for any γ > 1 , equation (12) exists a unique positive solution f ( z ) ∈ C ( [ 0 , 1 ] ) ∩ C 1 ( ( 0 , 1 ] ) , such that the system (16) has an exact solution of the form

(18) ρ ( r , t ) = f r a ( t ) a 3 ( t ) , u ( r , t ) = a ′ ( t ) a ( t ) r , 0 ≤ r ≤ a ( t ) , t ∈ [ 0 , T ] ,

with

(19) ρ ( a ( t ) , t ) = ρ 0 α − γ ( a 0 ) + γ − α α t 1 α − γ ,

(20) a ( t ) = f 1 3 ( 1 ) ρ 0 α − γ ( a 0 ) + γ − α α t 1 3 ( γ − α ) , f ( 1 ) = a 0 3 ρ 0 ( a 0 ) > 0 .

Let α = 3 γ + 1 6 , putting (18), (19), and (20) into (16)2, we have

(21) 6 ( γ − 1 ) ( 3 γ + 1 ) 2 z − γ f 1 3 − γ ( 1 ) f γ − 2 ( z ) f ′ ( z ) + γ − 1 2 f 1 6 − γ 2 ( 1 ) f 3 γ − 11 6 ( z ) f ′ ( z ) = 0 .

In what follows, we will discuss the properties of f ( z ) and then obtain Proposition 2 by virtue of Lemmas 1 and 2 together with Lemma 3.1 in [5].

First, we can obtain the low and upper bounds for the solutions of (21).

Lemma 1

For any γ > 1 , let f ( z ) be a solution to (21) in C ( [ 0 , 1 ] ) ∩ C 1 ( ( 0 , 1 ] ) . Then

(22) f ( 0 ) ≤ f ( z ) ≤ f ( 1 )

for z ∈ [ 0 , 1 ] , where f ( 0 ) = γ − 1 2 γ 6 3 γ − 1 a 0 3 ρ 0 ( a 0 ) > 0 , f ( 1 ) = a 0 3 ρ 0 ( a 0 ) > 0 . Furthermore, such a solution is unique.

Proof

As γ > 1 ( ≠ 5 3 ) , denoting g ( z ) ≜ ( f ( z ) f ( 1 ) ) 3 γ − 5 6 for any z ∈ [ 0 , 1 ] , then (21) becomes

(23) g ′ ( z ) g 3 γ − 1 3 γ − 5 ( z ) − γ − 1 2 γ = f 2 3 ( 1 ) ( γ − 1 ) ( 3 γ − 5 ) z γ ( 3 γ + 1 ) 2 , g ( 1 ) = 1 .

If g 3 γ − 1 3 γ − 5 ( z ) − γ − 1 2 γ = 0 , then (23) implies that z must be zero, i.e., g 3 γ − 1 3 γ − 5 ( 0 ) = γ − 1 2 γ . Namely, if z ≠ 0 , then one has g 3 γ − 1 3 γ − 5 ( z ) ≠ γ − 1 2 γ .

If γ > 5 3 , and it is assumed that g 3 γ − 1 3 γ − 5 ( z ) belongs to [ 0 , γ − 1 2 γ ) for any z ∈ ( 0 , 1 ] , then (23) implies that g ′ ( z ) ≤ 0 and 1 ≤ g ( z ) < γ − 1 2 γ 3 γ − 5 3 γ − 1 < 1 , which is a contradiction. Thus, we can deduce that g 3 γ − 1 3 γ − 5 ( z ) > γ − 1 2 γ for all z ∈ ( 0 , 1 ] , together with (23) to obtain g ′ ( z ) ≥ 0 , and consequently,

(24) g ( 0 ) = γ − 1 2 γ 3 γ − 5 3 γ − 1 < g ( z ) ≤ 1 , ∀ z ∈ ( 0 , 1 ] .

If 1 < γ < 5 3 , and it is assumed that g 3 γ − 1 3 γ − 5 ( z ) belongs to [ 0 , γ − 1 2 γ ) for any z ∈ ( 0 , 1 ] , then (23) implies that g ′ ( z ) ≥ 0 and 1 ≤ γ − 1 2 γ 3 γ − 5 3 γ − 1 < g ( z ) ≤ 1 , which is a contradiction. Thus, we can deduce that g 3 γ − 1 3 γ − 5 ( z ) > γ − 1 2 γ for all z ∈ ( 0 , 1 ] , together with (23) to obtain g ′ ( z ) ≤ 0 , and consequently,

(25) 1 ≤ g ( z ) < γ − 1 2 γ 3 γ − 5 3 γ − 1 = g ( 0 ) , ∀ z ∈ ( 0 , 1 ] .

When γ = 5 3 , denoting h ( z ) ≜ ( f ( z ) f ( 1 ) ) 2 3 , then (21) becomes

(26) f − 2 3 ( 1 ) h ′ ( z ) [ 5 2 − 1 2 h − 1 ( z ) ] = 1 9 z , h ( 1 ) = 1 .

Similarly, if 5 2 − 1 2 h − 1 ( z ) = 0 , then (26) implies that z must be zero, i.e., h ( 0 ) = 1 5 . Namely, if z ≠ 0 , then one has h ( z ) ≠ 1 5 . Supposed that h − 1 ( z ) > 5 , then (26) implies h ′ ( z ) ≤ 0 , and thus, 1 ≤ h ( z ) < 1 5 , which is a contradiction. Thus, we can deduce that, for all z ∈ ( 0 , 1 ] ,

(27) h ( 0 ) = 1 5 < h ( z ) ≤ 1 .

By combining (24), (25), and (27), we can prove that (22) holds.

Finally, the uniqueness can be proved by a similar way as in [5], and here, we give the uniqueness as γ = 5 3 for completeness.

To this end, let h ¯ ( z ) ∈ C ( [ 0 , 1 ] ) ∩ C 1 ( ( 0 , 1 ] ) be another solution to (26) with h ¯ ( 1 ) = 1 . Define w ( z ) = h ( z ) − h ¯ ( z ) . Then w ( z ) solves the following problem:

(28) d d z w ( z ) − 1 5 { ln [ w ( z ) + h ¯ ( z ) ] − ln [ h ¯ ( z ) ] } = 0 , w ( 1 ) = 0 .

Set

I = { z ∈ [ 0 , 1 ] ∣ w ( ξ ) ≡ 0 , z ≤ ξ ≤ 1 } .

Here, I ≠ ∅ because of 1 ∈ I . Define z 0 = inf I and then z 0 ∈ [ 0 , 1 ] . Obviously, the uniqueness of solutions to the system (26) will be showed by proving that z 0 = 0 and continuity argument.

We will prove the statement by contradiction. Suppose the statement is false. Then, z 0 ∈ ( 0 , 1 ] , and w ( z 0 ) = 0 . For any z ∈ ( 0 , z 0 ) , (27) tells us that

(29) 1 5 < h ¯ ( z ) ≤ 1 , ∀ z ∈ ( 0 , z 0 ) .

Integrating (28) over [ z , z 0 ] shows

(30) w ( z ) − 1 5 { ln [ w ( z ) + h ¯ ( z ) ] − ln [ h ¯ ( z ) ] } = 0 .

Taylor expansion gives

(31) ln [ w ( z ) + h ¯ ( z ) ] − ln [ h ¯ ( z ) ] = 1 h ¯ ( z ) w ( z ) + O ( 1 ) w 2 ( z )

for sufficiently small w ( z ) .

Putting (31) into (30) and using the fact w ( z 0 ) = 0 , one has

(32) 1 − 1 5 1 h ¯ ( z ) w ( z ) + O ( 1 ) w 2 ( z ) = 0

for z close to z 0 .

Observe that by virtue of (29), we have

1 − 1 5 1 h ¯ ( z ) > 0 , ∀ z ∈ ( 0 , z 0 ) .

From this, we can easily deduce that w ( z ) ≡ 0 , ∀ z ∈ ( z 0 − δ , z 0 ) , for some small δ > 0 . However, this contradicts the fact that z 0 = inf I . Therefore, we must have z 0 = 0 , and the proof of Lemma 1 is complete.□

Now we are ready to give an existence result to (12).

Lemma 2

For any γ > 1 , there is a positive function y = f ( z ) in C ( [ 0 , 1 ] ) ∩ C 1 ( ( 0 , 1 ] ) satisfying (12).

Proof

Note that the existence can be proved by a similar way as in [5], and here, we give the existence as γ = 5 3 for completeness.

As γ = 5 3 , we can rewrite (26) as follows:

(33) h ′ ( z ) = H ( h ( z ) , z ) ≜ f 2 3 ( 1 ) z 9 [ 5 2 − 1 2 h − 1 ( z ) ] , h ( 1 ) = 1 .

We seek a solution to (33) such that

(34) h ( z ) ∈ C ( [ 0 , 1 ] ) ∩ C 1 ( ( 0 , 1 ] ) , 1 5 ≤ h ( z ) ≤ 1 , ∀ z ∈ [ 0 , 1 ] .

Let

R = { ( z , h ( z ) ) ∣ 0 ≤ 1 − z ≤ a , 0 ≤ 1 − h ( z ) ≤ b }

for small a ∈ ( 0 , 1 ) and b ∈ ( 0 , 4 5 ) . Then we can easily deduce that

∣ H ( h ( z ) , z ) ∣ ≤ M , ∀ ( z , h ( z ) ) ∈ R ,

where M is a positive constant depends only on γ , a , and b . Since H ( h ( z ) , z ) is continuous in R , by choosing η = min { a , b M } and one can show that the solution to the initial value problem (32) exists in the neighborhood 0 ≤ 1 − z ≤ η by the Peano theorem. Then we can extend this solution from the left of the neighborhood 0 ≤ 1 − z ≤ η step by step until ( 0 , 1 ] .

As γ > 1 ( ≠ 5 3 ) , we can rewrite (23) as follows:

(35) g ′ ( z ) = f 2 3 ( 1 ) ( γ − 1 ) ( 3 γ − 5 ) z γ ( 3 γ + 1 ) 2 ( g 3 γ − 1 3 γ − 5 ( z ) − γ − 1 2 γ ) , g ( 1 ) = 1 .

Similarly as in [5], one can obtain the existence of g ( z ) with (24) and (25) when γ > 1 ( ≠ 5 3 ) .□

Remark 6

Lemmas 1 and 2 relax the range of γ in [5] from γ > 5 3 to γ > 1 .

At this point, we are in a position to complete the proof of Proposition 2.

The Proof of Proposition 2

First, for any C 1 function f ( z ) ≥ 0 and a ( t ) > 0 , define

(36) ρ ( r , t ) = f r a ( t ) a 3 ( t ) , u ( r , t ) = a ′ ( t ) a ( t ) r .

We can easily check that ( ρ ( r , t ) , u ( r , t ) ) is a self-similar solution to the continuity equation.

Then it follows from the free boundary (16) 3 and the continuity equation (16) 1 that

d d t ρ α = − α ρ α u r + 2 u r = − ρ γ , ρ ( r , 0 ) = ρ 0 ( a 0 ) , on  r = a ( t ) ,

which yields

ρ ( a ( t ) , t ) = ρ 0 α − γ ( a 0 ) + γ − α α t 1 α − γ .

Using the ansatz in (36) shows that

a ( t ) = f 1 3 ( 1 ) ρ 0 α − γ ( a 0 ) + γ − α α t 1 3 ( γ − α ) , f ( 1 ) = a 0 3 ρ 0 ( a 0 ) > 0 .

Finally, Lemmas 1 and 2 imply that there exists a unique positive function y = f ( z ) ∈ C ( [ 0 , 1 ] ) ∩ C 1 ( ( 0 , 1 ] ) such that (12) holds. In other words, ( ρ ( r , t ) , u ( r , t ) ) in (36) with f = y solves (16) 2 . Thus, the system (16) has an exact solution ( ρ ( r , t ) , u ( r , t ) ) of the form (18)–(20). The proof of this Proposition is complete.

3.2 The global strong solution of (II)

At this stage, we will establish the following proposition, which primarily focuses on the study of the global solution to subproblem (II).

Proposition 3

Under the assumptions of Theorem 1, there exists a unique global strong solution to the system (17) with (20) in [ a ( t ) , + ∞ ) × [ 0 , T ] satisfying

C − 1 ( T ) ≤ ρ ( r , t ) ≤ C ( T )

and

ψ ( ρ , ρ ˜ ) ∈ L ∞ ( [ 0 , T ] ; L 1 ( [ a ( t ) , + ∞ ) ; r 2 d r ) ) , ρ ∈ L ∞ ( [ 0 , T ] ; L ∞ ( [ a ( t ) , + ∞ ) ; r 2 d r ) ) , ρ r ∈ L ∞ ( [ 0 , T ] ; L 2 n ( [ a ( t ) , + ∞ ) ; r 2 d r ) ) , ρ t ∈ L 2 ( [ 0 , T ] ; L 2 ( [ a ( t ) , + ∞ ) ; r 2 d r ) ) , u ∈ L ∞ ( [ 0 , T ] ; H 1 ( [ a ( t ) , + ∞ ) ; r 2 d r ) ) ∩ L 2 ( [ 0 , T ] ; H 2 ( [ a ( t ) , + ∞ ) ; r 2 d r ) ) , u ∈ L ∞ ( [ 0 , T ] ; L ∞ ( [ a ( t ) , + ∞ ) ; r 2 d r ) ) , u t ∈ L 2 ( [ 0 , T ] ; L 2 ( [ a ( t ) , + ∞ ) ; r 2 d r ) ) ,

where n ∈ N + satisfies (55).

It is convenient to deal with the problem (17) in the Lagrangian coordinates. Define the coordinates transformation

(37) y ( r , t ) = ∫ a ( t ) r ρ r 2 d r , τ = t ,

which translates the domain [ a ( t ) , + ∞ ) × [ 0 , T ] into [ 0 , + ∞ ) × [ 0 , T ] and satisfies

(38) ∂ y ∂ r = ρ r 2 , ∂ y ∂ t = − ρ u r 2 , ∂ τ ∂ r = 0 , ∂ τ ∂ t = 1 , ∂ r ∂ τ = u .

Then, (17) is changed to

(39) ρ τ + ρ 2 ( r 2 u ) y = 0 , u τ + r 2 ( ρ γ ) y − α r 2 ( ρ 1 + α ( r 2 u ) y ) y + 2 r u ( ρ α ) y = 0 ,

and the initial and boundary conditions become

(40) ( ρ , u ) ∣ τ = 0 = ( ρ 0 , u 0 ) ,

(41) ρ γ ( 0 , τ ) = ( α ρ 1 + α ( r 2 u ) y ) ( 0 , τ ) ,

(42) ρ ( y , τ ) → ρ ˜ , u ( y , τ ) → 0 , as y → + ∞ ,

where the fixed boundary y = 0 corresponds to the free boundary a ( τ ) = r ( 0 , τ ) determined by

(43) ρ ( 0 , τ ) = ρ 0 1 − 3 γ 6 ( a 0 ) + 3 γ − 1 3 γ + 1 τ 6 1 − 3 γ ,

(44) u ( 0 , τ ) = d d τ a ( τ ) = 2 3 γ + 1 a 0 ρ 0 1 3 ( a 0 ) ρ 0 1 − 3 γ 6 ( a 0 ) + 3 γ − 1 3 γ + 1 τ 3 − 3 γ 3 γ − 1 ,

where for any γ > 1 , γ = 2 α − 1 3 , i.e., α = 3 γ + 1 6 , and a ( τ ) = a 0 ρ 0 1 3 ( a 0 ) ρ 0 1 − 3 γ 6 ( a 0 ) + 3 γ − 1 3 γ + 1 τ 2 3 γ − 1 , a ( 0 ) = a 0 > 0 .

Now, we establish a priori estimates for ( ρ , u ) to (39)–(42) under the assumptions as α = 3 γ + 1 6 , γ > 1 .

First, we give the following energy estimate and the BD-entropy estimate [1] in the Lagrangian coordinates.

Lemma 3

Under the assumption of Proposition 3, for any γ > 1 , it holds that

(45) ∫ 0 + ∞ ( u 2 + 1 ρ ψ ( ρ , ρ ˜ ) ) d y + ∫ 0 τ ∫ 0 + ∞ ρ γ 2 − 5 6 u 2 r 2 + ρ γ 2 + 7 6 u y 2 r 4 d y d τ ≤ C ,

(46) ∫ 0 + ∞ ( u + r 2 ρ γ 2 + 1 6 y ) 2 d y + ∫ 0 τ ∫ 0 + ∞ ( ρ 3 4 γ + 1 12 ) y 2 r 4 d y d τ ≤ C ,

where C is independent of T.

Proof

Multiplying (39)2 by u and integrating the resulting equation over ( 0 , + ∞ ) . Then, by (39)1 and the boundary conditions, we have

(47) d d τ ∫ 0 + ∞ 1 2 u 2 d y + ∫ 0 + ∞ 3 α − 2 α ρ γ 2 − 5 6 u 2 r 2 + 3 α − 2 2 ( 2 α − 1 ) ρ γ 2 + 7 6 u y 2 r 4 d y + α 2 ∫ 0 + ∞ 2 α − 1 2 α 2 r ρ γ 4 − 5 12 u + α − 1 α 2 α 2 α − 1 ρ γ 4 + 7 12 u y r 2 2 d y + α 2 ∫ 0 + ∞ α − 1 α 2 r ρ γ 4 − 5 12 u + ρ γ 4 + 7 12 u y r 2 2 d y = ( 2 ρ γ 2 + 1 6 u 2 r ) ( 0 , τ ) − ( ρ ˜ γ u r 2 ) ( 0 , τ ) + ∫ 0 + ∞ ( ρ γ − ρ ˜ γ ) ( r 2 u ) y d y = ( 2 ρ γ 2 + 1 6 u 2 r ) ( 0 , τ ) − ( ρ ˜ γ u r 2 ) ( 0 , τ ) + ∫ 0 + ∞ ( ρ γ − 2 − ρ − 2 ρ ˜ γ ) ( − ρ τ ) d y ≤ − d d τ ∫ 0 + ∞ [ 1 γ − 1 ( ρ γ − 1 − ρ ˜ γ − 1 ) + ρ ˜ γ ( ρ − 1 − ρ ˜ − 1 ) ] d y + ( 2 ρ γ 2 + 1 6 u 2 r − ρ ˜ γ u r 2 ) ( 0 , τ ) ≤ − d d τ ∫ 0 + ∞ 1 ρ ψ ( ρ , ρ ˜ ) d y + ( 2 ρ γ 2 + 1 6 u 2 r ) ( 0 , τ ) , ≤ − d d τ ∫ 0 + ∞ 1 ρ ψ ( ρ , ρ ˜ ) d y + C ρ 0 1 − 3 γ 6 ( a 0 ) + 3 γ − 1 3 γ + 1 τ 7 − 9 γ 3 γ − 1 ,

which leads to (45) after integration with τ and combining γ > 1 , ρ ˜ > 0 , inf [ a 0 , + ∞ ) ρ 0 > 0 , (43) and (44), u 0 ∈ L 2 ( [ a 0 , + ∞ ) ; r 2 d r ) , ρ 0 ∈ L ∞ ( [ a 0 , + ∞ ) ; r 2 d r ) , and ψ ( ρ 0 , ρ ˜ ) ∈ L 1 ( [ a 0 , + ∞ ) ; r 2 d r ) , where C is independent of T .

Combining (39)1 and (39)2 shows

(48) u + r 2 ( ρ γ 2 + 1 6 ) y τ + ( ρ γ ) y r 2 = 0 .

Multiplying (48) by u + r 2 ρ γ 2 + 1 6 y and integrating the resulting equation over ( 0 , + ∞ ) × ( 0 , τ ) , we have

∫ 0 + ∞ ( u + r 2 ( ρ γ 2 + 1 6 ) y ) 2 + 1 ρ ψ ( ρ , ρ ˜ ) d y + ∫ 0 τ ∫ 0 + ∞ ( ρ γ ) y ρ γ 2 + 1 6 y r 4 d y d τ ≤ C ∫ 0 + ∞ ( u 0 + ( r 2 ( ρ γ 2 + 1 6 ) y ) ∣ τ = 0 ) 2 d y + C ∫ 0 τ ( ( ρ γ − ρ ˜ γ ) u r 2 ) ( 0 , τ ) d τ ≤ C ,

where we have used the fact that ( ρ 0 ) r ∈ L 2 ( [ a 0 , + ∞ ) ; r 2 d r ) . Consequently, the proof of this lemma is finished.□

With the aforementioned estimates at hand, we are ready to obtain the upper bound of ρ .

Lemma 4

Under the assumption of Proposition 3, for any γ > 1 , there is a positive constant V = V ( T ) such that

(49) ρ ( y , τ ) ≤ 2 ρ ˜ , ∀ ( y , τ ) ∈ [ V , + ∞ ) × [ 0 , T ] .

Proof

Define the spatial domain Ω τ = { y ∈ [ 0 , + ∞ ) ∣ ρ ≥ 2 ρ ˜ } for any fixed 0 < τ ≤ T . Then, it holds that

(50) ∣ Ω τ ∣ = meas { y ∈ [ 0 , + ∞ ) ∣ ρ ≥ 2 ρ ˜ } = ∫ ρ ≥ 2 ρ ˜ 1 d y ≤ ∫ ρ ≥ 2 ρ ˜ ρ ψ ( ρ , ρ ˜ ) ρ ψ ( ρ , ρ ˜ ) d y ≤ C ∫ 0 + ∞ ψ ( ρ , ρ ˜ ) ρ d y ≤ C .

Thus, there exists a suitably large positive constant V , such that Ω τ ⊂ [ 0 , V ] , then

ρ ( y , τ ) ≤ 2 ρ ˜ , for any ( y , τ ) ∈ [ V , + ∞ ) × [ 0 , T ] ,

which completes the proof of Lemma 7.□

Lemma 5

Under the assumption of Proposition 3, for any γ > 1 , it holds that

(51) ρ ( y , τ ) ≤ C ( T ) , ∀ ( y , τ ) ∈ [ 0 , V ] × [ 0 , T ] .

Proof

Given ( y , τ ) ∈ [ 0 , M ] × [ 0 , T ] with M > V , integrating (48) over the domain ( 0 , y ) × ( 0 , τ ) , we obtain

(52) r 2 ρ γ 2 + 1 6 ( y , τ ) = ( r 2 ρ γ 2 + 1 6 ) ( 0 , τ ) + ∫ 0 y r 2 ρ γ 2 + 1 6 y τ = 0 d y + ∫ 0 y 2 r ρ γ 2 − 5 6 d y − ∫ 0 y ( u − u 0 ) d y − ∫ 0 τ ∫ 0 y ( ρ γ ) y r 2 d y d τ ≤ ∫ 0 M 2 r ρ γ 2 − 5 6 d y + ∫ 0 M ( u 2 + u 0 2 ) d y + C ( M , T ) + C max ( y , τ ) ∈ [ 0 , M ] × [ 0 , T ] ( ρ γ 2 + 1 6 r 2 ) 1 − 2 3 γ + 1 ∫ 0 τ ∫ 0 M ρ γ 2 + 1 6 y 2 r 4 + r 4 − 12 γ 3 γ + 1 d y d τ ≤ C ( M , T ) + ∫ 0 M 2 r ρ γ 2 − 5 6 d y + C ( M , T ) max ( y , τ ) ∈ [ 0 , M ] × [ 0 , T ] ( ρ γ 2 + 1 6 r 2 ) 1 − 2 3 γ + 1 ,

where we have used the bound of ρ ( 0 , τ ) on [ 0 , M ] × [ 0 , T ] due to (43). Furthermore, by (39)1 and (42), if we choose M suitable large, we have

∫ 0 y 2 r ρ γ 2 − 5 6 d y ≤ C ∫ 0 M ρ γ 2 − 5 6 d y ≤ C ∫ 0 M [ ρ 0 γ 2 − 5 6 + 5 6 − γ 2 ∫ 0 τ ρ γ 2 + 1 6 ( r 2 u ) y d τ ] d y ≤ C ∫ 0 M ρ 0 γ 2 − 5 6 d y + C ∫ 0 τ ∫ 0 M ∣ ρ γ 2 + 1 6 y r 2 u ∣ d y d τ + C ∫ 0 τ ( r 2 ∣ u ∣ ρ γ 2 + 1 6 ) ( M , τ ) + ( r 2 ∣ u ∣ ρ γ 2 + 1 6 ) ( 0 , τ ) d τ ≤ C ( M , T ) + C ∫ 0 τ ( ∫ 0 M ρ γ 2 + 1 6 y 2 r 4 d y ) 1 2 ( ∫ 0 M u 2 d y ) 1 2 d τ ≤ C ( T ) ,

where the term

∫ 0 τ ( r 2 ∣ u ∣ ρ γ 2 + 1 6 ) ( M , τ ) + ( r 2 ∣ u ∣ ρ γ 2 + 1 6 ) ( 0 , τ ) d τ ≤ C ( M , T )

is due to (42)–(44) and the choice of the M . Thus, we deduce from (52) that

( r 2 ρ γ 2 + 1 6 ) ( y , τ ) ≤ C ( T ) + C ( T ) max ( y , τ ) ∈ [ 0 , M ] × [ 0 , T ] ( ρ γ 2 + 1 6 r 2 ) 1 − 2 3 γ + 1 ≤ 1 2 max ( y , τ ) ∈ [ 0 , M ] × [ 0 , T ] ( ρ γ 2 + 1 6 r 2 ) + C ( T ) .

Then, due to r ≥ a ( τ ) ,

max ( y , τ ) ∈ [ 0 , M ] × [ 0 , T ] ρ γ 2 + 1 6 ≤ a − 2 ( T ) C ( T ) ≤ C ( T ) .

The proof of Lemma 5 is finished.□

By combining Lemmas 4 and 5, we can obtain the upper bound of the density in the entire time-space domain ( y , τ ) ∈ [ 0 , + ∞ ) × [ 0 , T ] .

Lemma 6

Under the assumption of Proposition 3, for any γ > 1 , it holds that

(53) ρ ( y , τ ) ≤ C ( T ) , ∀ ( y , τ ) ∈ [ 0 , + ∞ ) × [ 0 , T ] .

Lemma 7

Under the assumption of Proposition 3. As for any γ > 1 , the following inequality holds:

(54) ∫ 0 + ∞ u 2 n d y + ∫ 0 τ ∫ 0 + ∞ ρ γ 2 + 7 6 u y 2 u 2 n − 2 r 4 d y d s ≤ C ( n , T ) ,

where n ∈ N + satisfying with

(55) n = n γ , 1 ≤ n γ < n 2 ( γ ) , as 1 < γ < 5 3 , arbitrary p o s i t i v e i n t e g e r , a s γ ≥ 5 3 ,

with n 2 ( γ ) = 1 1 − Λ ( γ ) , Λ ( γ ) = 3 3 ( γ − 1 ) ( 3 γ + 1 ) ( 3 γ − 2 ) ∈ ( 0 , 1 ) , 1 < γ < 5 3 .

Proof

Multiplying (39) 2 by u 2 n − 1 for any 1 ≤ n ∈ N + , after integrating by parts with respect to y over ( 0 , + ∞ ) , it deduces that

(56) 1 2 n d d τ ∫ 0 + ∞ u 2 n d y + ∫ 0 + ∞ γ 2 + 1 6 ρ γ 2 + 7 6 ( r 2 u ) y 2 u 2 n − 2 + γ 2 + 1 6 ρ γ 2 + 7 6 ( r 2 u ) y ( r 2 u ) ( u 2 n − 2 ) y d y − ∫ 0 + ∞ 2 ρ γ 2 + 1 6 ( r u 2 n ) y d y = ∫ 0 + ∞ ρ γ ( r 2 u 2 n − 1 ) y d y + 2 r ρ γ 2 + 1 6 u 2 n ( 0 , τ ) .

First, we can deal with the case of 1 < γ < 5 3 .

By (43) and (44) and Young’s inequality, with any ε , η > 0 , we have from (56) that

1 2 n d d τ ∫ 0 + ∞ u 2 n d y + ∫ 0 + ∞ 2 γ − 4 3 ρ γ 2 − 5 6 u 2 n r 2 + ( 2 n − 1 ) γ 2 + 1 6 ρ γ 2 + 7 6 u y 2 u 2 n − 2 r 4 d y = ∫ 0 + ∞ ( 2 n − 1 ) ρ γ u 2 n − 2 u y r 2 + 2 ρ γ − 1 u 2 n − 1 r + 4 n 5 6 − γ 2 ρ γ 2 + 1 6 u 2 n − 1 u y r d y + 2 r ρ γ 2 + 1 6 u 2 n ( 0 , τ ) ≤ η ∫ 0 + ∞ ρ γ 2 + 7 6 u y 2 u 2 n − 2 r 4 d y + 2 ∫ 0 + ∞ ρ γ − 1 u 2 n − 1 r d y + C ( n , η ) ∫ 0 + ∞ ρ 3 2 γ − 7 6 u 2 n − 2 d y + ∫ 0 + ∞ ε 2 5 6 − γ 2 ρ γ 2 − 5 6 u 2 n r 2 + 8 n 2 ε 5 6 − γ 2 ρ γ 2 + 7 6 u y 2 u 2 n − 2 r 4 d y + C ( T ) .

Thus, by Lemma 6,

(57) 1 2 n d d τ ∫ 0 + ∞ u 2 n d y + ∫ 0 + ∞ 2 γ − 4 3 − ε 2 5 6 − γ 2 ρ γ 2 − 5 6 u 2 n r 2 + ( 2 n − 1 ) γ 2 + 1 6 − η − 8 n 2 ε 5 6 − γ 2 ρ γ 2 + 7 6 u y 2 u 2 n − 2 r 4 d y ≤ C ( n , η ) ∫ 0 + ∞ ( ρ γ − 1 u 2 n − 1 r + ρ 3 2 γ − 7 6 u 2 n − 2 ) d y + C ( T ) ≤ C ( n , η ) ∫ 0 + ∞ u 2 n d y + ∫ 0 + ∞ u 2 n − 2 d y + C ( T ) ,

due to 1 < γ < 5 3 .

Taking ε = 4 γ − 8 3 5 6 − γ 2 = 24 γ − 16 5 − 3 γ , we have from (57) that

(58) 1 2 n d d τ ∫ 0 + ∞ u 2 n d y + ( 2 n − 1 ) γ 2 + 1 6 − η − n 2 5 − 3 γ 3 γ − 2 5 6 − γ 2 ∫ 0 + ∞ ρ γ 2 + 7 6 u y 2 u 2 n − 2 r 4 d y ≤ C ( n , η ) ∫ 0 + ∞ u 2 n d y + ∫ 0 + ∞ u 2 n − 2 d y + C ( T ) .

Note that, if we choose

( 2 n − 1 ) γ 2 + 1 6 − n 2 5 − 3 γ 3 γ − 2 5 6 − γ 2 = − 1 6 ( 3 γ − 2 ) ( ( 5 − 3 γ ) 2 n 2 − 2 ( 3 γ + 1 ) ( 3 γ − 2 ) n + ( 3 γ + 1 ) ( 3 γ − 2 ) ) = − ( 5 − 3 γ ) 2 6 ( 3 γ − 2 ) n − ( 3 γ + 1 ) ( 3 γ − 2 ) ( 5 − 3 γ ) 2 2 − ( 3 γ + 1 ) ( 3 γ − 2 ) ( 27 γ − 27 ) ( 5 − 3 γ ) 4 > 0 ,

which is equivalent to

n − ( 3 γ + 1 ) ( 3 γ − 2 ) ( 5 − 3 γ ) 2 2 − ( 3 γ + 1 ) ( 3 γ − 2 ) ( 27 γ − 27 ) ( 5 − 3 γ ) 4 < 0 ,

i.e.

n 1 ( γ ) < n < n 2 ( γ ) ,

where

n 1 ( γ ) = 1 ( 5 − 3 γ ) 2 ( ( 3 γ + 1 ) ( 3 γ − 2 ) − 3 3 ( 3 γ + 1 ) ( 3 γ − 2 ) ( γ − 1 ) ) = ( 3 γ + 1 ) ( 3 γ − 2 ) ( 3 γ + 1 ) ( 3 γ − 2 ) + 3 3 ( γ − 1 ) = 1 1 + Λ ( γ ) < 1 , n 2 ( γ ) = 1 ( 5 − 3 γ ) 2 ( ( 3 γ + 1 ) ( 3 γ − 2 ) + 3 3 ( 3 γ + 1 ) ( 3 γ − 2 ) ( γ − 1 ) ) = ( 3 γ + 1 ) ( 3 γ − 2 ) ( 3 γ + 1 ) ( 3 γ − 2 ) − 3 3 ( γ − 1 ) = 1 1 − Λ ( γ ) > 1 ,

with Λ ( γ ) = 3 3 ( γ − 1 ) ( 3 γ + 1 ) ( 3 γ − 2 ) and 0 < Λ ( γ ) < 1 , Λ ( 1 ) = 0 , Λ 5 3 = 1 .

Furthermore, n 1 ( γ ) and n 2 ( γ ) have the following properties:

0 < n 1 ( γ ) < 1 < n 2 ( γ ) , ∀ 1 < γ < 5 3 , n 1 ( 1 ) = 1 = n 2 ( 1 ) , lim γ → 5 3 − n 2 ( γ ) = + ∞ .

Thus, for any fixed 1 < γ < 5 3 , together with the fact of Lemma 3 that

∫ 0 + ∞ u 2 d y ≤ C ,

an application of Grönwall’s inequality to (58) gives

(59) ∫ 0 + ∞ u 2 n d y + ∫ 0 τ ∫ 0 + ∞ ρ γ 2 + 7 6 u y 2 u 2 n − 2 r 4 d y d s ≤ C ( n , T ) ( ∫ 0 + ∞ u 0 2 n d y + ∫ 0 T ∫ 0 + ∞ u 2 n − 2 d y d τ ) ≤ C ( n , T ) ( 1 + ∫ 0 T ∫ 0 + ∞ u 2 n − 2 d y d τ ) .

Then, by (59) and (45) and the iteration method, we can see that (54) holds for any 1 < γ < 5 3 and n satisfying (55).

On the other hand, when γ ≥ 5 3 , we return to (56) and then

1 2 n d d τ ∫ 0 + ∞ u 2 n d y + ∫ 0 + ∞ 2 γ − 4 3 ρ γ 2 − 5 6 u 2 n r 2 + ( 2 n − 1 ) γ 2 + 1 6 ρ γ 2 + 7 6 u y 2 u 2 n − 2 r 4 d y = ∫ 0 + ∞ ( 2 n − 1 ) ρ γ u 2 n − 2 u y r 2 + 2 ρ γ − 1 u 2 n − 1 r + 4 n 5 6 − γ 2 ρ γ 2 + 1 6 u 2 n − 1 u y r d y + ( 2 r ρ γ 2 + 1 6 u 2 n ) ( 0 , τ ) ≤ η ∫ 0 + ∞ ρ γ 2 + 7 6 u y 2 u 2 n − 2 r 4 d y + 2 ∫ 0 + ∞ ρ γ − 1 u 2 n − 1 r d y + C ( n , η ) ∫ 0 + ∞ ρ 3 2 γ − 7 6 u 2 n − 2 d y + C ( n , η ) ∫ 0 + ∞ ρ γ 2 − 5 6 u 2 n r 2 d y + C ( T ) .

Choosing η = 1 2 ( 2 n − 1 ) γ 2 + 1 6 and then using Lemma 6, for any γ ≥ 5 3 , one has

(60) 1 2 n d d τ ∫ 0 + ∞ u 2 n d y + 1 2 ( 2 n − 1 ) γ 2 + 1 6 ∫ 0 ∞ ρ γ 2 + 7 6 u y 2 u 2 n − 2 r 4 d y ≤ C ( n , T ) ∫ 0 + ∞ u 2 n d y + ∫ 0 + ∞ u 2 n − 2 d y + C ( T ) .

By employing a same process as (58)–(60), we can conclude that for any γ ≥ 5 3 and n ≥ 1 , (54) holds. The proof of Lemma 7 is completed.□

Remark 7

Note that

n 2 ( γ ) = 1 1 − Λ ( γ ) , Λ ( γ ) = 3 3 ( γ − 1 ) ( 3 γ + 1 ) ( 3 γ − 2 ) .

Since

d d γ Λ ( γ ) = 27 ( 5 − 3 γ ) ( 3 γ − 1 ) 6 ( 3 γ − 3 ) 1 2 ( 3 γ + 1 ) 3 2 ( 3 γ − 2 ) 3 2 > 0 , ∀ 1 < γ < 5 3 ,

and then

d d γ Λ ( γ ) ∣ γ = 5 3 = 0 , Λ ( 1 ) = 0 , Λ 5 3 = 1 ,

which means that Λ ( γ ) ∈ ( 0 , 1 ) is a monotonically increasing function as γ ∈ ( 1 , 5 3 ) and sup γ ∈ ( 1 , 5 3 ) Λ ( γ ) = Λ 5 3 .

Therefore, as for any γ ∈ ( 1 , 5 3 ) and Λ ( γ ) ∈ ( 0 , 1 ) ,

1 ≤ n < n 2 ( γ ) ⇔ 1 − Λ ( γ ) < 1 n ⇔ 1 − 1 n < Λ ( γ ) ,

and the aforementioned inequality is apparently right for some suitable positive integer n = n γ .

Lemma 8

Under the assumptions of Proposition 3, for any γ > 1 and n ∈ N + that satisfies (55), it holds that

(61) ∫ 0 + ∞ r 2 ρ γ 2 + 1 6 y 2 n d y ≤ C ( n , T ) .

Proof

By integrating (48) with τ over ( 0 , τ ) , then multiplying the resulting equation by r 2 ρ γ 2 + 1 6 y 2 n − 1 for any 2 ≤ n ∈ N + , we have

∫ 0 + ∞ r 2 ρ γ 2 + 1 6 y 2 n d y = ∫ 0 + ∞ r 2 ρ γ 2 + 1 6 y 2 n − 1 ( r 2 ρ γ 2 + 1 6 y ) τ = 0 d y + ∫ 0 + ∞ r 2 ρ γ 2 + 1 6 y 2 n − 1 ( u 0 − u ) d y + ∫ 0 + ∞ r 2 ρ γ 2 + 1 6 y 2 n − 1 ∫ 0 τ ( ρ γ ) y r 2 d τ d y .

According to Young’s inequality and Lemmas 6 and 7, we compute

∫ 0 + ∞ r 2 ρ γ 2 + 1 6 y 2 n d y ≤ 1 2 ∫ 0 + ∞ r 2 ρ γ 2 + 1 6 y 2 n d y + C ∫ 0 + ∞ [ ( r 2 ρ γ 2 + 1 6 y ) τ = 0 2 n + u 2 n + u 0 2 n ] d y + C ( T ) ∫ 0 τ max y ∈ [ 0 , + ∞ ) ρ n ( γ − 1 3 ) ( ∫ 0 + ∞ r 2 ρ γ 2 + 1 6 y 2 n d y ) d τ ≤ 1 2 ∫ 0 + ∞ r 2 ρ γ 2 + 1 6 y 2 n d y + C ( n , T ) + C ( T ) ∫ 0 τ max y ∈ [ 0 , + ∞ ) ρ n ( γ − 1 3 ) ( ∫ 0 + ∞ r 2 ρ γ 2 + 1 6 y 2 n d y ) d τ ≤ 1 2 ∫ 0 + ∞ r 2 ρ γ 2 + 1 6 y 2 n d y + C ( n , T ) + C ( n , T ) ∫ 0 τ ∫ 0 + ∞ r 2 ρ γ 2 + 1 6 y 2 n d y d τ .

Applying Grönwall’s inequality to last inequality gives

∫ 0 + ∞ r 2 ρ γ 2 + 1 6 y 2 n d y ≤ C ( n , T ) .

This completes the proof of Lemma 8.□

From now until the end of this section, we return to the Eulerian coordinates. In what follows, we derive the lower bound of the density.

First, we translate the estimates in Lemma 3 and Lemmas 7 and 8 back to the Eulerian coordinates.

Lemma 9

Under the assumptions of Proposition 3, for any γ > 1 and n ∈ N + that satisfies (55), it holds that

sup 0 ≤ t ≤ T ∫ a ( t ) + ∞ ( ρ u 2 + ψ ( ρ , ρ ˜ ) ) r 2 d r ≤ C , sup 0 ≤ t ≤ T ∫ a ( t ) + ∞ ρ u 2 n r 2 d r + ∫ 0 t ∫ a ( t ) + ∞ ρ γ 2 + 1 6 u r 2 u 2 n − 2 r 2 d r d s ≤ C ( n , T ) , sup 0 ≤ t ≤ T ∫ a ( t ) + ∞ ρ n ( γ − 11 3 ) + 1 ( ρ r ) 2 n r 2 d r ≤ C ( n , T ) .

With the aforementioned estimates at hands, we are ready to obtain the lower bound of ρ .

Lemma 10

Under the assumptions of Proposition 3, for any γ > 1 , there exists a positive constant R = R ( T ) > a ( T ) such that

(62) ρ ( r , t ) ≥ ρ ˜ 2 , as ( r , t ) ∈ [ R , + ∞ ) × [ 0 , T ] .

Proof

First, for any fixed 0 < τ ≤ T , we define the spatial domain Ω t that satisfies

Ω t = { r ∈ ( a ( t ) , + ∞ ) ∣ 0 ≤ ρ < ρ ˜ 2 } .

Then, it holds that

∣ Ω t ∣ ≤ ∫ 0 ≤ ρ < ρ ˜ 2 ( r 2 + 1 [ 0 , 1 ] ) d r ≤ ∫ 0 ≤ ρ < ρ ˜ 2 ψ ( ρ , ρ ˜ ) ψ ( ρ , ρ ˜ ) r 2 d r + 1 ≤ C ∫ a ( t ) + ∞ ψ ( ρ , ρ ˜ ) r 2 d r + 1 ≤ C .

Therefore, there exists a suitably large positive constant R = R ( T ) , such that Ω t ⊂ B R , then

ρ ( r , t ) ≥ ρ ˜ 2 , ∀ ( r , t ) ∈ [ R , + ∞ ) × [ 0 , T ] .

The proof of Lemma 10 is completed.□

Next, we only need to prove the lower bound of the density in the annular domain B R \ B a ( t ) . For this purpose, we give some useful integrability estimates of ρ − 1 first.

Lemma 11

Under the assumptions of Proposition 3, for any γ 0 < γ < 5 3 , it holds that

(63) sup 0 ≤ t ≤ T ∫ a ( t ) R ρ − β d r ≤ C ( T ) ,

where β satisfies

(64) β = β 0 ∈ ( 0 , ( 3 γ − 3 ) ( 3 γ + 1 ) ( 3 γ − 2 ) − ( 19 − 15 γ ) 10 − 6 γ ] , if γ 0 < γ ≤ 19 15 , 15 γ − 19 10 − 6 γ , if 19 15 < γ < 5 3 ,

and γ 0 = 4 − 73 3 is the unique positive solution of the following equation:

( 3 γ − 3 ) ( 3 γ + 1 ) ( 3 γ − 2 ) − ( 19 − 15 γ ) = 0 .

Proof

Multiplying the equation (17)1 by − β ρ − β − 1 r 2 yields that

(65) ( ρ − β r 2 ) t − β ( β + 1 ) ρ − β − 1 ρ r u r 2 − β ( ρ − β u r 2 ) r = 0 ,

for β > 0 to be determined later.

By (10) and Lemma 9, integrating by parts to (65) with respect to ( r , t ) over ( a ( t ) , R ) × ( 0 , t ) gives

(66) ∫ a ( t ) R ρ − β r 2 d r ≤ ∫ a 0 R ρ 0 − β r 2 d r + ∫ 0 T ∫ a ( t ) R β ( β + 1 ) ρ − β − 1 ∣ ρ r u ∣ r 2 d r d t + ∫ 0 T ( ( β ρ − β ∣ u ∣ r 2 ) ( a ( t ) , t ) + ( β ρ − β ∣ u ∣ r 2 ) ( R , t ) ) d t ≤ C ( R , T ) + C ∫ 0 T ∫ a ( t ) R ρ − β − 1 ∣ ρ r ∣ ∣ u ∣ r 2 d r d t ≤ C ( R , T ) + C ∫ 0 T ∫ a ( t ) R ρ n ( γ − 11 3 ) + 1 ∣ ρ r ∣ 2 n r 2 d r d t 1 2 n ∫ 0 T ∫ a ( t ) R ρ − 2 ( β + 1 ) n 2 n − 1 − n ( γ − 11 3 ) + 1 2 n − 1 ∣ u ∣ 2 n 2 n − 1 r 2 d r d t 2 n − 1 2 n ≤ C ( R , T ) + C ( T ) ∫ 0 T ∫ a ( t ) R ρ − n ( γ + 2 β − 5 3 ) + 1 2 n − 1 ∣ u ∣ 2 n 2 n − 1 r 2 d r d t 2 n − 1 2 n ≤ C ( R , T ) + C ( T ) ∫ 0 T ∫ a ( t ) R ρ ∣ u ∣ 2 n r 2 d r d t 1 2 n ∫ 0 T ∫ a ( t ) R ρ − n ( γ + 2 β − 5 3 ) + 1 2 n − 1 + 1 2 n − 1 n ( 2 n − 1 ) n ( 2 n − 1 ) − n r 2 d r d t n − 1 n ≤ C ( R , T ) + C ( T ) ∫ 0 T ∫ a ( t ) R ρ ( 5 − 3 γ ) n − 6 ( β + 1 ) 6 n − 6 ρ − β r 2 d r d t ,

for any n ∈ N + that satisfies (55) and 1 < γ < 5 3 . In the second inequality of (66), we have used (18)–(20) to handle the term ∫ 0 T ( β ρ − β ∣ u ∣ r 2 ) ( a ( t ) , t ) d t and for the term ∫ 0 T ( β ρ − β ∣ u ∣ r 2 ) ( R , t ) d t , we have, for M = R + 1 , together with ρ ( R , t ) ≥ ρ ˜ 2 ,

(67) ∫ 0 T ( β ρ − β ∣ u ∣ r 2 ) ( R , t ) d t ≤ C ( ρ ˜ , R ) ∫ 0 T ‖ ρ u ‖ L ∞ ( a ( t ) , M ) d t ≤ C ∫ 0 T ‖ ρ u ‖ L 1 ( a ( t ) , M ) + ‖ ρ u r ‖ L 1 ( a ( t ) , M ) d s + ‖ ρ r u ‖ L 1 ( a ( t ) , M ) ≤ C ( n , T ) + C sup 0 ≤ t ≤ T ‖ ρ 1 2 u ‖ L 2 ( a ( t ) , M ; r 2 d r ) + C ‖ ρ γ 4 + 1 12 u r ‖ L 2 ( 0 , T ; L 2 ( a ( t ) , M ; r 2 d r ) ) + C sup 0 ≤ t ≤ T ‖ ρ 1 2 u ‖ L 2 ( a ( t ) , M ; r 2 d r ) ‖ ρ ( γ 2 − 11 6 ) + 1 2 ρ r ‖ L 2 ( a ( t ) , M ; r 2 d r ) ≤ C .

To control the last term in (66), we need:

( 5 − 3 γ ) n − 6 ( β + 1 ) 6 n − 6 ≥ 0 ,

i.e.,

n ≥ 6 ( β + 1 ) 5 − 3 γ for any 1 < γ < 5 3 .

Therefore, in combination with condition (55), for any 1 < γ < 5 3 , we need to find a positive integer n γ satisfying the following inequality

(68) 6 ( β + 1 ) 5 − 3 γ ≤ n γ < n 2 ( γ ) = ( 3 γ + 1 ) ( 3 γ − 2 ) ( 3 γ + 1 ) ( 3 γ − 2 ) − 3 3 γ − 3 ,

where 6 ( β + 1 ) 5 − 3 γ > 1 , ∀ 1 < γ < 5 3 , β > 0 .

So, it is only necessary to prove that

(69) 6 ( β + 1 ) 5 − 3 γ + k ≤ ( 3 γ + 1 ) ( 3 γ − 2 ) ( 3 γ + 1 ) ( 3 γ − 2 ) − 3 3 γ − 3 ⇔ ( k − 1 ) ≤ 3 3 γ − 3 ( 3 γ + 1 ) ( 3 γ − 2 ) − 3 3 γ − 3 − 6 ( β + 1 ) 5 − 3 γ ⇔ 1 3 ( k − 1 ) ≤ ( 3 γ − 3 ) ( 3 γ + 1 ) ( 3 γ − 2 ) − [ ( 10 β + 19 ) − ( 6 β + 15 ) γ ] ( 5 − 3 γ ) 2 ,

for any 1 < γ < 5 3 and some positive integer k ≥ 1 .

If we choose k = 1,0 < β = 15 γ − 19 10 − 6 γ , i.e., γ = 10 β + 19 6 β + 15 ,

( 10 β + 19 ) − ( 6 β + 15 ) γ = 0 ,

then (69) holds consequently for any 19 15 < γ < 5 3 . In particular, when γ = 19 15 , (69) also holds for any 0 < β ≤ 30 5 .

As 1 < γ < 19 15 , (69) can be rewritten as following:

1 3 ( k − 1 ) ( 5 − 3 γ ) 2 + ( 10 − 6 γ ) β ≤ ( 3 γ − 3 ) ( 3 γ + 1 ) ( 3 γ − 2 ) − ( 19 − 15 γ ) .

Let γ 0 be a solution of G ( γ ) ≐ ( 3 γ − 3 ) ( 3 γ + 1 ) ( 3 γ − 2 ) − ( 19 − 15 γ ) = 0 and choose 0 < β ≤ ( 3 γ − 3 ) ( 3 γ + 1 ) ( 3 γ − 2 ) − ( 19 − 15 γ ) 10 − 6 γ , then for any γ 0 < γ < 19 15 , (69) remains true.

Notices that

G ( γ ) ≐ ( 3 γ − 3 ) ( 3 γ + 1 ) ( 3 γ − 2 ) − ( 19 − 15 γ ) = 0 ⇔ ( 3 γ − 3 ) ( 3 γ + 1 ) ( 3 γ − 2 ) = ( 19 − 15 γ ) 2 ⇔ 27 γ 3 − 261 γ 2 + 573 γ − 355 = 0 ⇔ ( 3 γ − 5 ) ( 3 γ − 12 − 73 ) ( 3 γ − 12 + 73 ) = 0 .

Since

G ′ ( γ ) = ( 81 γ − 72 ) γ + 3 2 ( 3 γ − 3 ) ( 3 γ + 1 ) ( 3 γ − 2 ) + 15 > 0 , ∀ 1 < γ < 19 15 , G ( 1 ) = − 4 < 0 , G ( γ 0 ) = 0 , G ( 19 15 ) = 12 30 25 > 0 ,

and then γ 0 ∈ ( 1 , 19 15 ) , γ 0 = 4 − 73 3 consequently.

That is, we can find a positive integer n γ such that (68) holds for any β satisfying (64).

Then, by Lemma 6, it follows from (66) that

(70) ∫ a ( t ) R ρ − β r 2 d r ≤ C ( R , T ) + C ( T ) ∫ 0 T ∫ a ( t ) R ρ − β r 2 d r d t .

An application of Grönwall’s inequality to (70) leads to

∫ a ( t ) R ρ − β d r ≤ C ( T ) ∫ a ( t ) R ρ − β r 2 d r ≤ C ( R , T ) ≤ C ( T ) .

This proves assertion (63).□

Lemma 12

Under the assumptions of Lemma 11, for any γ 0 < γ < 5 3 , it holds that

(71) sup 0 ≤ t ≤ T ∫ a ( t ) R ∣ ( ρ − β ) r ∣ d r ≤ C ( T ) ,

for β > 0 satisfying (64).

Proof

By Lemmas 9 and 11, we have

(72) ∫ a ( t ) R ∣ ( ρ − β ) r ∣ d r = ∫ a ( t ) R β ρ − β − 1 ∣ ρ r ∣ d r ≤ C ∫ a ( t ) R ρ n γ ( γ − 11 3 ) + 1 ∣ ρ r ∣ 2 n γ r 2 d r 1 2 n γ ∫ a ( t ) R ρ − 2 ( β + 1 ) n γ 2 n γ − 1 − n γ ( γ − 11 3 ) + 1 2 n γ − 1 r − 2 2 n γ − 1 d r 2 n γ − 1 2 n γ ≤ C ( T ) + C ( T ) ∫ a ( t ) R ρ − 3 γ n γ + 5 n γ − 3 ( β + 1 ) 6 n γ − 3 ρ − β r 2 d r .

Then, in view of (68), we have

− 3 γ n γ + 5 n γ − 3 ( β + 1 ) 6 n γ − 3 ≥ − 3 n γ 5 3 − 2 ( β + 1 ) n γ + 5 n γ − 3 ( β + 1 ) 6 n γ − 3 = β + 1 2 n γ − 1 > 0 ,

from which it follows that

∫ a ( t ) R ∣ ( ρ − β ) r ∣ d r ≤ C ( T ) + C ( T ) ∫ a ( t ) R ρ − β r 2 d r ≤ C ( T ) .

This ends the proof of Lemma 12.□

Lemma 13

Under the assumptions of Proposition 3, for any γ 0 < γ < 5 3 , there exists a positive constant C ( T ) such that

(73) ρ ( r , t ) ≥ C − 1 ( T ) , ∀ ( r , t ) ∈ [ a ( t ) , R ] × [ 0 , T ] .

Proof

By Sobolev’s inequality and Lemmas 11 and 12, for any β > 0 satisfying (64), one has

sup 0 ≤ t ≤ T ‖ ρ − β ‖ L ∞ ( a ( t ) , R ) ≤ C sup 0 ≤ t ≤ T ∫ a ( t ) R ρ − β d r + C sup 0 ≤ t ≤ T ∫ a ( t ) R ∣ ( ρ − β ) r ∣ d r ≤ C ( T ) .

That is,

ρ ≥ C − 1 ( T )

for any ( r , t ) ∈ [ a ( t ) , R ] × [ 0 , T ] . Thus, we finish the proof of Lemma 13.□

By combining Lemmas 10 and 13, we can obtain the lower bound of the density in the entire time-space domain ( r , t ) ∈ [ a ( t ) , + ∞ ) × [ 0 , T ] when γ 0 < γ < 5 3 as follows.

Lemma 14

Under the assumption of Proposition 3 with γ 0 < γ < 5 3 , it holds that

(74) ρ ( r , t ) ≥ C − 1 ( T ) , ∀ ( r , t ) ∈ [ a ( t ) , + ∞ ) × [ 0 , T ] .

For the critical case of γ = 5 3 , the aforementioned argument is not valid anymore. Fortunately, the lower bound of the density can be derived by the similar way in [7, Lemmas 3.8–3.10], which can be described as in the following lemma.

Lemma 15

Under the assumptions of Proposition 3 with γ = 5 3 , there exists a positive constant C ( T ) such that

(75) ρ ( r , t ) ≥ C − 1 ( T ) , ∀ r ∈ [ a ( t ) , + ∞ ) × [ 0 , T ] .

Proof

This lemma can be proved by a similar argument as in [7]. For completeness, we give a brief proof below.

Similar to the proof of Lemma 10, we only need to prove the lower bound of the density in the domain [ a ( t ) , R ] × [ 0 , T ] when γ = 5 3 .

First, using the Sobolev imbedding theorem, by virtue of Lemmas 3, 6, and 9, we obtain for given s ∈ ( 0 , T ) , ∀ 0 < N < ∞ and n ∈ N + ,

‖ ρ 4 n 3 u 2 n ‖ L ∞ ( a ( s ) , N ) ≤ C ( ∫ a ( s ) N ρ 4 n 3 u 2 n d r + ∫ a ( s ) N ∣ ( ρ 4 n 3 u 2 n ) r ∣ d r ) ≤ C ( n , T ) + C ∫ a ( s ) N ρ u 2 n − 2 u y 2 r 4 d r ,

which, after integration over [ 0 , t ] and together with Lemma 9, yields

(76) ∫ 0 t ‖ ρ 4 n 3 u 2 n ‖ L ∞ ( 0 , N ) d s ≤ C ( n , T ) , 0 < t < T .

Next, integrating (48) over [ 0 , t ] to obtain

(77) ( ln ρ ) r ( r , t ) = ( ln ρ 0 ) r − u ( r , t ) + u 0 ( r ) − 5 2 ∫ 0 t ( ρ 2 3 ) r d s .

Then it follows form Lemma 6 and 9, and (77) that ∀ r ∈ ( a ( t ) , R ) ,

∫ 0 t ( ρ 2 3 ) r 2 n d s = 2 3 2 n ∫ 0 t ρ 4 n 3 ( ln ρ 0 ) r − 5 2 ∫ 0 s ( ρ 2 3 ) r d s ′ − u ( r , t ) + u 0 ( r ) 2 n d s ≤ C ∫ 0 t ρ 4 n 3 [ ( ρ 0 ) r 2 n + u 2 n + u 0 2 n ] d s + C ∫ 0 t ρ 4 n 3 ∫ 0 s ( ρ 2 3 ) r 2 n d s ′ d s ≤ C ( n , T ) + C ( T ) ∫ 0 t ∫ 0 s ( ρ 2 3 ) r 2 n d s ′ d s ,

from which and Grönwall’s lemma, we have

(78) ∫ 0 t ( ρ 2 3 ) r 2 n d s ≤ C ( n , T ) , r ∈ [ a ( t ) , R ] , 0 < t < T .

Now set

(79) v ( r , t ) = 1 r 2 ρ ( r , t ) , V ¯ ( t ) = max [ a ( t ) , R ] × [ 0 , t ] v ( r , s ) .

Then form (17) 1 , it is easy to verify that for any δ > 1 ,

(80) ( v δ ) t + ( v δ u ) r − ( δ + 1 ) ( v δ u ) r + ( δ + 1 ) δ v δ − 1 v r u = 0 .

Integrating (80) over [ a ( t ) , R ] × [ 0 , t ] and using (77) show

∫ a ( t ) R v δ d r + δ ( δ + 1 ) ∫ 0 t ∫ a ( t ) R v δ u 2 d r d s = ∫ a 0 R v 0 δ d r + δ ( δ + 1 ) ∫ 0 t ∫ a ( t ) R v δ u ( u 0 + ( ln ρ 0 ) r ) d r d s − 5 2 δ ( δ + 1 ) ∫ 0 t ∫ a ( t ) R v δ u ∫ 0 s ( ρ 2 3 ) r d s ′ d r d s + 2 δ ( δ + 1 ) ∫ 0 t ∫ a ( t ) R u v δ r d r d s + ( δ + 1 ) ∫ 0 t ( v δ u ) ( R , s ) − ( v δ u ) ( a ( t ) , s ) d s ≤ C ( R , T ) + δ ( δ + 1 ) 2 ∫ 0 t ∫ a ( t ) R v δ u 2 d r d s + C ∫ 0 t ∫ a ( t ) R v δ ( u 0 + ( ln ρ 0 ) r ) 2 d r d s + C ( n , T ) ∫ 0 t ∫ a ( t ) R v δ ∫ 0 s [ ( ρ 2 3 ) r ] 2 n d s ′ 1 2 n d r d s ,

where we have used (18)–(20) to handle the term ∫ 0 t ( v δ u ) ( a ( t ) , s ) d s and the similar process as (67) for ∫ 0 t ( v δ u ) ( R , s ) d s .

Due to inf [ a 0 , + ∞ ) ρ 0 > 0 , u 0 ∈ L 2 ( [ a 0 , + ∞ ) ; r 2 d r ) , ( ρ 0 ) r ∈ L ∞ ( [ a 0 , + ∞ ) ; r 2 d r ) and (78), we can obtain

(81) ∫ a ( t ) R v δ d r ≤ C ( R , T ) + C ∫ 0 t ∫ a ( t ) R v δ d r d s ,

which, together with Grönwall’s inequality, gives for any δ > 2 ,

(82) ∫ a ( t ) R v δ d r ≤ C ( R , T ) , t ∈ [ 0 , T ] .

Thus, one can deduce from Lemma 6, (82) that for δ > 2 ,

(83) V ¯ ( t ) δ = max [ a ( t ) , R ] × [ 0 , T ] v δ ( r , s ) ≤ C max t ∈ [ 0 , t ] ( ∫ a ( t ) R v δ d r + ∫ a ( t ) R ∣ ( v δ ) r ∣ d r ) ≤ C max t ∈ [ 0 , t ] ∫ a ( t ) R v δ d r + δ ( ∫ a ( t ) R v 2 δ + 2 d r ) 1 2 + C ∫ a ( t ) R v δ + 1 d r ≤ C 1 + V ¯ ( t ) 1 + δ 2 ,

which shows that

(84) V ¯ ( t ) ≤ C ( T ) , 0 ≤ t ≤ T .

This, together with r ∈ [ a ( t ) , R ] , completes the proof of this lemma.□

Finally, with all the estimates at hand, we are ready to prove Proposition 3.

The Proof of Proposition 3

With the upper and lower bound of the density, we can obtain the higher-order regularity estimates. Under the assumptions of Theorem 1, the global existence of the strong solution to the system (17) with (20) in [ a ( t ) , + ∞ ) × [ 0 , T ] can be proved in a way similar to that in [17]. We omit it for brevity.

3.3 The proof of Theorem 1

Combining subsections 3.1 and 3.2, the Cauchy problem (7)–(10) has a global solution of the form, where α = 3 γ + 1 6 and γ 0 < γ ≤ 5 3 ,

( ρ , u ) ( r , t ) = f r a ( t ) a 3 ( t ) , 2 r ( 3 γ + 1 ) ρ 0 1 6 − γ 2 ( a 0 ) + ( 3 γ − 1 ) t ( r , t ) , 0 ≤ r ≤ a ( t ) , ( ρ I I , u I I ) ( r , t ) , a ( t ) ≤ r < + ∞ .

From (22), we can derive the following bounds:

(85) ρ ( r , t ) ≥ γ − 1 2 γ 6 3 γ − 1 ρ 0 1 6 − γ 2 ( a 0 ) + 3 γ − 1 3 γ + 1 t − 6 3 γ − 1 > 0 ,

(86) ρ ( r , t ) ≤ ρ 0 1 6 − γ 2 ( a 0 ) + 3 γ − 1 3 γ + 1 t − 6 3 γ − 1 ≤ C

for any 0 ≤ r ≤ a ( t ) . Moreover, Proposition 3 and Lemmas 6, 14, and 15 indicate that ( ρ I I , u I I ) ( r , t ) is a global strong solution to the system (17) with (20) satisfying

(87) C − 1 ( T ) ≤ ρ I I ( r , t ) ≤ C ( T )

for any a ( t ) ≤ r < + ∞ and some positive constant C ( T ) depending on the initial data. Therefore, through the stress-free condition on the “particle path” a ( t ) , we can obtain the global solution of the Cauchy problem of (7)–(10) which satisfies (13)–(15). Consequently, the proof of Theorem 1 is complete.

Acknowledgments

The authors would like to thank the anonymous referees for several valuable suggestions and comments that help to improve the article.

  1. Funding information: Z. H. Guo has received funding from the National Natural Science Foundation of China under contract No. 11931013 and the Natural Science Foundation of Guangxi Province under contract No. 2022GXNSFDA035078. X. Y. Zhang has received funding from the National Natural Science Foundation of Yulin University under contract No. 2025GK11.

  2. Author contributions: All authors have accepted responsibility for the entire content of this manuscript and consented to its submission to the journal, reviewed all the results and approved the final version of the manuscript. All authors participate in writing, review, and editing. We declare that all authors contribute equally to this work.

  3. Conflict of interest: The authors have no competing interests to declare that are relevant to the content of this article.

  4. Data availability statement: No data were used for the research described in the article.

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Received: 2025-03-13
Revised: 2025-06-09
Accepted: 2025-06-24
Published Online: 2025-08-13

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  28. Existence of positive radial solutions of general quasilinear elliptic systems
  29. Low Mach number and non-resistive limit of magnetohydrodynamic equations with large temperature variations in general bounded domains
  30. Sharp viscous shock waves for relaxation model with degeneracy
  31. Bifurcation and multiplicity results for critical problems involving the p-Grushin operator
  32. Asymptotic behavior of solutions of a free boundary model with seasonal succession and impulsive harvesting
  33. Blowing-up solutions concentrated along minimal submanifolds for some supercritical Hamiltonian systems on Riemannian manifolds
  34. Stability of rarefaction wave for relaxed compressible Navier-Stokes equations with density-dependent viscosity
  35. Singularity for the macroscopic production model with Chaplygin gas
  36. Global strong solution of compressible flow with spherically symmetric data and density-dependent viscosities
  37. Global dynamics of population-toxicant models with nonlocal dispersals
  38. α-Mean curvature flow of non-compact complete convex hypersurfaces and the evolution of level sets
  39. High-energy solutions for coupled Schrödinger systems with critical growth and lack of compactness
  40. On the structure and lifespan of smooth solutions for the two-dimensional hyperbolic geometric flow equation
  41. Well-posedness for physical vacuum free boundary problem of compressible Euler equations with time-dependent damping
  42. On the existence of solutions of infinite systems of Volterra-Hammerstein-Stieltjes integral equations
  43. Remark on the analyticity of the fractional Fokker-Planck equation
  44. Continuous dependence on initial data for damped fourth-order wave equation with strain term
  45. Unilateral problems for quasilinear operators with fractional Riesz gradients
  46. Boundedness of solutions to quasilinear elliptic systems
  47. Existence of positive solutions for critical p-Laplacian equation with critical Neumann boundary condition
  48. Non-local diffusion and pulse intervention in a faecal-oral model with moving infected fronts
  49. Nonsmooth analysis of doubly nonlinear second-order evolution equations with nonconvex energy functionals
  50. Qualitative properties of solutions to the viscoelastic beam equation with damping and logarithmic nonlinear source terms
  51. Shape of extremal functions for weighted Sobolev-type inequalities
  52. One-dimensional boundary blow up problem with a nonlocal term
  53. Doubling measure and regularity to K-quasiminimizers of double-phase energy
  54. General solutions of weakly delayed discrete systems in 3D
  55. Global well-posedness of a nonlinear Boussinesq-fluid-structure interaction system with large initial data
  56. Optimal large time behavior of the 3D rate type viscoelastic fluids
  57. Local well-posedness for the two-component Benjamin-Ono equation
  58. Self-similar blow-up solutions of the four-dimensional Schrödinger-Wave system
  59. Existence and stability of traveling waves in a Keller-Segel system with nonlinear stimulation
  60. Existence and multiplicity of solutions for a class of superlinear p-Laplacian equations
  61. On the global large regular solutions of the 1D degenerate compressible Navier-Stokes equations
  62. Normal forms of piecewise-smooth monodromic systems
  63. Fractional Dirichlet problems with singular and non-locally convective reaction
  64. Sharp forced waves of degenerate diffusion equations in shifting environments
  65. Global boundedness and stability of a quasilinear two-species chemotaxis-competition model with nonlinear sensitivities
  66. Non-existence, radial symmetry, monotonicity, and Liouville theorem of master equations with fractional p-Laplacian
  67. Global existence, asymptotic behavior, and finite time blow up of solutions for a class of generalized thermoelastic system with p-Laplacian
  68. Formation of singularities for a linearly degenerate hyperbolic system arising in magnetohydrodynamics
  69. Linear stability and bifurcation analysis for a free boundary problem arising in a double-layered tumor model
  70. Hopf's lemma, asymptotic radial symmetry, and monotonicity of solutions to the logarithmic Laplacian parabolic system
  71. Generalized quasi-linear fractional Wentzell problems
  72. Existence, symmetry, and regularity of ground states of a nonlinear Choquard equation in the hyperbolic space
  73. Normalized solutions for NLS equations with general nonlinearity on compact metric graphs
  74. Boundedness and global stability in a predator-prey chemotaxis system with indirect pursuit-evasion interaction and nonlocal kinetics
  75. Review Article
  76. Existence and stability of contact discontinuities to piston problems
https://doi.org/10.1515/anona-2025-0100
Keywords for this article
compressible Navier-Stokes; strong solutions; density-dependent viscosities; spherically symmetric; Lagrangian coordinates
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  24. Qualitative properties of solutions for dual fractional parabolic equations involving nonlocal Monge-Ampère operator
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  26. Uniform boundedness and compactness for the commutator of an extension of Riesz transform on stratified Lie groups
  27. Normalized solutions to nonlinear Schrödinger equations with mixed fractional Laplacians
  28. Existence of positive radial solutions of general quasilinear elliptic systems
  29. Low Mach number and non-resistive limit of magnetohydrodynamic equations with large temperature variations in general bounded domains
  30. Sharp viscous shock waves for relaxation model with degeneracy
  31. Bifurcation and multiplicity results for critical problems involving the p-Grushin operator
  32. Asymptotic behavior of solutions of a free boundary model with seasonal succession and impulsive harvesting
  33. Blowing-up solutions concentrated along minimal submanifolds for some supercritical Hamiltonian systems on Riemannian manifolds
  34. Stability of rarefaction wave for relaxed compressible Navier-Stokes equations with density-dependent viscosity
  35. Singularity for the macroscopic production model with Chaplygin gas
  36. Global strong solution of compressible flow with spherically symmetric data and density-dependent viscosities
  37. Global dynamics of population-toxicant models with nonlocal dispersals
  38. α-Mean curvature flow of non-compact complete convex hypersurfaces and the evolution of level sets
  39. High-energy solutions for coupled Schrödinger systems with critical growth and lack of compactness
  40. On the structure and lifespan of smooth solutions for the two-dimensional hyperbolic geometric flow equation
  41. Well-posedness for physical vacuum free boundary problem of compressible Euler equations with time-dependent damping
  42. On the existence of solutions of infinite systems of Volterra-Hammerstein-Stieltjes integral equations
  43. Remark on the analyticity of the fractional Fokker-Planck equation
  44. Continuous dependence on initial data for damped fourth-order wave equation with strain term
  45. Unilateral problems for quasilinear operators with fractional Riesz gradients
  46. Boundedness of solutions to quasilinear elliptic systems
  47. Existence of positive solutions for critical p-Laplacian equation with critical Neumann boundary condition
  48. Non-local diffusion and pulse intervention in a faecal-oral model with moving infected fronts
  49. Nonsmooth analysis of doubly nonlinear second-order evolution equations with nonconvex energy functionals
  50. Qualitative properties of solutions to the viscoelastic beam equation with damping and logarithmic nonlinear source terms
  51. Shape of extremal functions for weighted Sobolev-type inequalities
  52. One-dimensional boundary blow up problem with a nonlocal term
  53. Doubling measure and regularity to K-quasiminimizers of double-phase energy
  54. General solutions of weakly delayed discrete systems in 3D
  55. Global well-posedness of a nonlinear Boussinesq-fluid-structure interaction system with large initial data
  56. Optimal large time behavior of the 3D rate type viscoelastic fluids
  57. Local well-posedness for the two-component Benjamin-Ono equation
  58. Self-similar blow-up solutions of the four-dimensional Schrödinger-Wave system
  59. Existence and stability of traveling waves in a Keller-Segel system with nonlinear stimulation
  60. Existence and multiplicity of solutions for a class of superlinear p-Laplacian equations
  61. On the global large regular solutions of the 1D degenerate compressible Navier-Stokes equations
  62. Normal forms of piecewise-smooth monodromic systems
  63. Fractional Dirichlet problems with singular and non-locally convective reaction
  64. Sharp forced waves of degenerate diffusion equations in shifting environments
  65. Global boundedness and stability of a quasilinear two-species chemotaxis-competition model with nonlinear sensitivities
  66. Non-existence, radial symmetry, monotonicity, and Liouville theorem of master equations with fractional p-Laplacian
  67. Global existence, asymptotic behavior, and finite time blow up of solutions for a class of generalized thermoelastic system with p-Laplacian
  68. Formation of singularities for a linearly degenerate hyperbolic system arising in magnetohydrodynamics
  69. Linear stability and bifurcation analysis for a free boundary problem arising in a double-layered tumor model
  70. Hopf's lemma, asymptotic radial symmetry, and monotonicity of solutions to the logarithmic Laplacian parabolic system
  71. Generalized quasi-linear fractional Wentzell problems
  72. Existence, symmetry, and regularity of ground states of a nonlinear Choquard equation in the hyperbolic space
  73. Normalized solutions for NLS equations with general nonlinearity on compact metric graphs
  74. Boundedness and global stability in a predator-prey chemotaxis system with indirect pursuit-evasion interaction and nonlocal kinetics
  75. Review Article
  76. Existence and stability of contact discontinuities to piston problems
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