Remarks on analytical solutions to compressible Navier–Stokes equations with free boundaries

Jianwei Dong; Manwai Yuen

doi:10.1515/ans-2023-0146

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Remarks on analytical solutions to compressible Navier–Stokes equations with free boundaries

Jianwei Dong and Manwai Yuen

Published/Copyright: July 2, 2024

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From the journal Advanced Nonlinear Studies Volume 24 Issue 4

Abstract

In this paper, we consider the free boundary problem of the radially symmetric compressible Navier–Stokes equations with viscosity coefficients of the form μ(ρ) = ρ ^θ, λ(ρ) = (θ − 1)ρ ^θ in R N . Under the continuous density boundary condition, we correct some errors in (Z. H. Guo and Z. P. Xin, “Analytical solutions to the compressible Navier–Stokes equations with density-dependent viscosity coefficients and free boundaries,” J. Differ. Equ., vol. 253, no. 1, pp. 1–19, 2012) for N = 3, θ = γ > 1 and improve the spreading rate of the free boundary, where γ is the adiabatic exponent. Moreover, we construct an analytical solution for θ = 2 3 , N = 3 and γ > 1, and we prove that the free boundary grows linearly in time by using some new techniques. When θ = 1, under the stress free boundary condition, we construct some analytical solutions for N = 2, γ = 2 and N = 3, γ = 5 3 , respectively.

Keywords: compressible Navier–Stokes equations; density-dependent viscosity; analytical solution; free boundary

Mathematics Subject Classification: 35Q30; 76D09

1 Introduction

In this paper, we consider the following compressible isentropic Navier–Stokes equations (CNS) with density-dependent viscosity coefficients in R N (N = 2, 3):

(1.1) ρ t + d i v ( ρ U ) = 0 ,

(1.2) ( ρ U ) t + d i v ( ρ U ⊗ U ) − d i v ( μ ( ρ ) D ( U ) ) − ∇ ( λ ( ρ ) d i v U ) + ∇ P ( ρ ) = 0 ,

where the fluid density ρ = ρ(x, t) and the velocity field U = U(x, t) are the unknown variables, P(ρ) = ρ ^γ (γ > 1) denotes the pressure, D ( U ) = ∇ U + ∇ U T 2 represents the strain tensor, μ(ρ) and λ(ρ) stand for the Lamé viscosity coefficients satisfying

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

There are many similar models to system (1.1) and (1.2) in geophysical flows [1]. Many researchers have been devoted to studying the well-posedness of the solutions of CNS (1.1) and (1.2). When N = 2, μ(ρ) = ρ, λ(ρ) = 0 and γ = 2, model (1.1) and (1.2) corresponds to the viscous Saint-Venant system for shallow water, whose local smooth solutions or global smooth solutions for data close to equilibrium were obtained by Sundbye [2]. When μ(ρ) = ρ ^α, λ(ρ) = 0 ( α ∈ ( 0 , 3 2 ) ) , many authors studied the well-posedness of the solutions to the free boundary problem of the one-dimensional compressible Navier–Stokes equations, see Refs. [3], [4], [5], [6], [7], [8], [9], [10], [11], [12]. For the significant progress in the existence of weak solutions to the multi-dimensional compressible Navier–Stokes equations with density-dependent viscosity coefficients, we can refer to Refs. [13], [14], [15], [16], [17], [18] and the references therein.

There are also some works on the radially symmetric compressible Navier–Stokes equations with density-dependent viscosity coefficients which have been published in Refs. [19], [20], [21], [22], [23], [24], [25] and so on. In Ref. [24], for system (1.1) and (1.2) with μ(ρ) = 0, λ(ρ) = ρ ^θ in R N , Yeung and Yuen constructed a family of self-similar analytical solutions of the form:

(1.4) ρ ( r , t ) = y r a ( t ) a N ( t ) , u ( r , t ) = a ′ ( t ) a ( t ) r ,

where r = |x|, and a(t), y(z) ∈ C ¹ are two functions satisfying some ordinary differential equations. However, such ordinary differential equations were not solved in Ref. [24]. For system (1.1) and (1.2) with μ(ρ) = ρ ^θ, λ(ρ) = (θ − 1)ρ ^θ, θ = γ > 1 in R N , Liang and Shi [23] constructed a class of analytical solutions of the form (1.4) and investigated the blowup and expanding properties by studying the ordinary differential equation for a(t). Their approach is based on the phase plane method. For system (1.1) and (1.2) with μ(ρ) = ρ, λ(ρ) = 0 and large, discontinuous and spherically symmetric initial data in a fixed ball Ω = { x ∈ R 3 ‖ x | ≤ R } , Guo, Jiu and Xin [19] obtained a global weak solution for 1 < γ < 3 under the boundary conditions ρu(0, t) = 0, ρu(R, t) = 0. Later, Guo, Li and Xin [20] extended the existence results of Ref. [19] to the free boundary value problem with stress free boundary condition and investigated the Lagrange structure and dynamics for the solutions, such as the existence and uniqueness of particle paths, transportation of initial regularities, formation or non-formation of vacuum, etc. Note that the flow density is required to be positive up to the free interface in Ref. [20]. Recently, Li and Zhang [22] obtained the existence and large time behavior of global radially symmetric strong solution to the free boundary value problem for (1.1) and (1.2) in R 2 with μ(ρ) = 2μ, λ(ρ) = ρ ^β, β > 1 under stress free boundary condition even if the flow density vanishes at the free boundary. As stressed in Ref. [21], it seems difficult to apply the method in Ref. [20] for continuous density boundary condition due to the high degeneracy in vacuum. In Ref. [21], Guo and Xin constructed some analytical solutions of the form (1.4) to the free boundary value problem for the radially symmetric CNS (1.1) and (1.2) in R 3 with continuous density boundary condition and stress free boundary condition, respectively. Interesting new results such as the formation of vacuum at the center of the symmetry as time tends to infinity and explicit regularities and large time decay estimates of the velocity field were exhibited in Ref. [21]. For a further extension of the results of Ref. [21], we can refer to Ref. [26].

Letting

(1.5) ρ ( x , t ) = ρ ( r , t ) , U ( x , t ) = u ( r , t ) x r

with r = |x|, system (1.1) and (1.2) with μ(ρ) = ρ ^θ and λ(ρ) = (θ − 1)ρ ^θ becomes

(1.6) ρ t + ( ρ u ) r + N − 1 r ρ u = 0 ,

(1.7) ( ρ u ) t + ( ρ u 2 ) r + ( ρ γ ) r + N − 1 r ρ u 2 − θ ρ θ u r + N − 1 r u r + N − 1 r ( ρ θ ) r u = 0 .

In addition, we impose the following initial and boundary conditions:

(1.8) ( ρ , u ) | t = 0 = ( ρ 0 ( r ) , u 0 ( r ) ) , 0 ≤ r ≤ a 0 ,

(1.9) ρ ( a ( t ) , t ) = 0 ,

(1.10) ρ γ = θ ρ θ u r + N − 1 r u o n r = a ( t ) ,

where a ₀ > 0 is a constant, and a(t) is the free boundary satisfying

(1.11) a ′ ( t ) = u ( a ( t ) , t ) , a ( 0 ) = a 0 , ∀ t ≥ 0 .

Condition (1.9) is called the continuous density boundary condition and condition (1.10) is called the stress free boundary condition, see Ref. [21].

For the problem (1.6)–(1.9) with N = 3 and θ = γ > 1, Guo and Xin [21] constructed an analytical solution and proved that the free boundary a(t) tends to +∞ as t → +∞ with the following rates:

(1.12) C 1 ( 1 + t ) 2 − α 3 ( θ − 1 ) ≤ a ( t ) ≤ C 2 ( 1 + t ) 9 θ − 8 3 θ − 3 , ∀ t ≥ 0 ,

where C ₁ and C ₂ are two constants and

(1.13) α = 1 , i f θ > 4 3 , 1 + σ , σ > 0 s m a l l , i f θ = 4 3 , 5 − 3 θ , i f 1 < θ < 4 3 .

However, there are some typos in Ref. [21] and the upper bound of a(t) in (1.12) is not accuracy. So we will correct the typos of Ref. [21] and improve the upper bound of a(t). Moreover, we will construct an analytical solution for θ = 2 3 , N = 3 and γ > 1, and prove that the free boundary grows linearly in time by using some new techniques. On the other hand, for the problem (1.6)–(1.8) and (1.10) with N = 3, γ = 2 θ − 1 3 and γ > 5 3 , Guo and Xin [21] constructed an analytical solution. Unfortunately, this excludes the very interesting case θ = 1 and γ = 5 3 . So we will give some analytical solutions to the problem (1.6)–(1.8) and (1.10) with θ = 1 for N = 2, γ = 2 and N = 3, γ = 5 3 , respectively. We remark that when θ = 1, N = 2 and γ = 2, model (1.1) and (1.2) corresponds to the viscous Saint–Venant system for shallow water.

The rest of this paper is organized as follows. In Section 2, we deal with the continuous density boundary problem (1.6)–(1.9). In Section 3, we treat the stress free boundary problem (1.6)–(1.8) and (1.10).

2 The continuous density boundary problem

We first point out some typos in Ref. [21]. The Equations (23) (or (41)), (38)–(40) and (42) in Ref. [21] should be changed to

(2.1) ρ ( r , t ) = 1 2 ( θ − 1 ) 1 − r 2 a 2 ( t ) 1 θ − 1 a 3 ( t ) , u ( r , t ) = a ′ ( t ) a ( t ) r ,

(2.2) − z = f θ − 2 ( z ) f ′ ( z ) ,

(2.3) f ( z ) = f θ − 1 ( 1 ) + 1 2 ( θ − 1 ) ( 1 − z 2 ) 1 θ − 1 ,

(2.4) a ″ ( t ) − θ a 2 − 3 θ ( t ) − ( 2 − 3 θ ) θ a ′ ( t ) a 1 − 3 θ ( t ) = 0 ,

and

(2.5) a ′ ( t ) = a 1 − θ a 0 2 − 3 θ + θ a 2 − 3 θ ( t ) + θ ∫ 0 t a 2 − 3 θ ( s ) d s , a ( 0 ) = a 0 , a ′ ( 0 ) = a 1 .

This is because that the density ρ(r, t) is nonnegative when 0 ≤ r ≤ a(t).

We multiply (2.4) by a′(t) and integrate it over [0, t] to have

1 2 a ′ ( t ) 2 + θ 3 θ − 3 a 3 − 3 θ ( t ) = 1 2 a 1 2 + θ 3 θ − 3 a 0 3 − 3 θ − ( 3 θ − 2 ) θ ∫ 0 t a ′ ( s ) 2 a 1 − 3 θ ( s ) d s

(2.6) ≤ 1 2 a 1 2 + θ 3 θ − 3 a 0 3 − 3 θ ,

which implies that a′(t) ≤ C for a constant C > 0, then we get

(2.7) a ( t ) ≤ C ( 1 + t ) .

Compared with C 2 ( 1 + t ) 9 θ − 8 3 θ − 3 in (1.12), where 9 θ − 8 3 θ − 3 > 3 for θ > 1, (2.7) is more accurate, so we improved the spreading rate of the free boundary in Ref. [21].

In order to investigate the spreading rate of the free boundary for the case of θ = 2 3 , N = 3 and γ > 1, we introduce the following averaged quantities:

(2.8) m ( t ) = ∫ 0 a ( t ) ρ r 2 d r ,

(2.9) I ( t ) = ∫ 0 a ( t ) ρ r 4 d r ,

(2.10) F ( t ) = ∫ 0 a ( t ) ρ u r 3 d r .

Theorem 2.1.

Let m(0), I(0), F(0) ∈ (0, +∞). When θ = 2 3 , N = 3 and γ > 1, the problem (1.6)–(1.9) has an analytical solution:

(2.11) ρ ( r , t ) = 1 2 ( γ − 1 ) 1 − r 2 a 2 ( t ) 1 γ − 1 a 3 ( t ) , u ( r , t ) = a ′ ( t ) a ( t ) r ,

where the free boundary a(t) satisfies the following ordinary differential equation:

(2.12) a ″ ( t ) − γ a 2 − 3 γ ( t ) = 0 , a ( 0 ) = a 0 > 0 , a ′ ( 0 ) = a 1 .

Furthermore, the free boundary a(t) satisfies that

(2.13) B ( 1 + t ) ≤ a ( t ) ≤ C ( 1 + t )

with B, C > 0 being two constants.

Proof.

When θ = 2 3 , N = 3 and γ > 1, the fact that the problem (1.6)–(1.9) has an analytical solution (2.11) with a(t) satisfying (2.12) is a special case of Ref. [21], so we omit the proof. In the following, we prove (2.13). By (2.8), (1.6) and (1.9), we know that

(2.14) m ′ ( t ) = ∫ 0 a ( t ) ρ t r 2 d r = − ∫ 0 a ( t ) ( ρ u ) r r 2 d r − 2 ∫ 0 a ( t ) ρ u r d r = 0 ,

which implies that

(2.15) m ( t ) = m ( 0 ) .

In view of (1.6), (1.9), (2.9), and (2.10), one has

(2.16) I ′ ( t ) = ∫ 0 a ( t ) ρ t r 4 d r = − ∫ 0 a ( t ) ( ρ u ) r r 4 d r − 2 ∫ 0 a ( t ) ρ u r 3 d r = 2 ∫ 0 a ( t ) ρ u r 3 d r = 2 F ( t ) .

It follows from (1.7), (1.9), and (2.10) that

(2.17) F ′ ( t ) = ∫ 0 a ( t ) ( ρ u ) t r 3 d r = − ∫ 0 a ( t ) ( ρ u 2 ) r r 3 d r − 2 ∫ 0 a ( t ) ρ u 2 r 2 d r − ∫ 0 a ( t ) ( ρ γ ) r r 3 d r + θ ∫ 0 a ( t ) ρ θ u r + 2 u r r r 3 d r − 2 ∫ 0 a ( t ) ( ρ θ ) r u r 2 d r = ∫ 0 a ( t ) ρ u 2 r 2 d r + 3 ∫ 0 a ( t ) ρ γ r 2 d r + ( 2 − 3 θ ) ∫ 0 a ( t ) ρ θ u r r 2 d r + ( 4 − 6 θ ) ∫ 0 a ( t ) ρ θ u r d r = ∫ 0 a ( t ) ρ u 2 r 2 d r + 3 ∫ 0 a ( t ) ρ γ r 2 d r ≥ ∫ 0 a ( t ) ρ u 2 r 2 d r , i f θ = 2 3 ,

which together with F(0) > 0 implies that F(t) > 0, and this leads to I(t) > 0 because of I(0) > 0 and (2.16).

By the Schwartz inequality, we have

(2.18) F ( t ) 2 = ∫ 0 a ( t ) ρ u r 3 d r 2 ≤ ∫ 0 a ( t ) ρ u 2 r 2 d r ⋅ ∫ 0 a ( t ) ρ r 4 d r = I ( t ) ⋅ ∫ 0 a ( t ) ρ u 2 r 2 d r ,

which together with (2.17) leads to

(2.19) F ( t ) 2 ≤ I ( t ) ⋅ F ′ ( t ) .

By (2.19) and (2.16), we know that

(2.20) 2 I ″ ( t ) I ′ ( t ) ≥ I ′ ( t ) I ( t ) .

We integrate (2.20) over [0, t] and use (2.16) to get

(2.21) ln I ′ ( t ) 2 I ( t ) = 2 ⁡ ln I ′ ( t ) − ln ⁡ I ( t ) ≥ 2 ⁡ ln I ′ ( 0 ) − ln ⁡ I ( 0 ) = ln 4 F ( 0 ) 2 I ( 0 ) ,

which implies that

(2.22) I ′ ( t ) 2 I ( t ) ≥ F ( 0 ) I ( 0 ) .

We can solve out (2.22) as

(2.23) I ( t ) ≥ [ F ( 0 ) t + I ( 0 ) ] 2 I ( 0 ) .

On the other hand, by (2.9), (2.8) and (2.15), we obtain

(2.24) I ( t ) ≤ a ( t ) 2 ∫ 0 a ( t ) ρ r 2 d r = m ( t ) a ( t ) 2 = m ( 0 ) a ( t ) 2 ,

which together with (2.23) leads to a(t) ≥ B(1 + t) with B > 0 being a constant.

Next, we integrate (2.12) over [0, t] to have:

(2.25) a ′ ( t ) = a 1 + γ ∫ 0 t a ( s ) 2 − 3 γ d s ,

which together with a(t) ≥ B(1 + t) implies that

(2.26) a ′ ( t ) ≤ a 1 + γ B 2 − 3 γ 3 − 3 γ [ ( 1 + t ) 3 − 3 γ − 1 ] ≤ C

with C > 0 being a constant. Thus, we obtain a(t) ≤ C(1 + t). □

3 The stress free boundary problem

For the problem (1.6)–(1.8) and (1.10), we have the following two theorems.

Theorem 3.1.

Assume that θ = 1, N = 2, γ = 2. Then the free boundary value problem (1.6)–(1.8) and (1.10) has a solution with the free boundary a(t) given by

(3.1) a ( t ) = 2 t + a 0 2

and

(3.2) ρ ( r , t ) = f r a ( t ) a 2 ( t ) , u ( r , t ) = a ′ ( t ) a ( t ) r = r 2 t + a 0 2 ,

where f(z) ∈ C ¹([0, 1]) satisfies the following ordinary differential equation:

(3.3) z f ( z ) + f ′ ( z ) − 2 f ( z ) f ′ ( z ) = 0 , z ∈ [ 0,1 ] , f ( 1 ) = 2 .

Moreover, f(0) = A ∈ (1, 2) and f′(z) > 0 for all z ∈ (0, 1].

Theorem 3.2.

Assume that θ = 1, N = 3, γ = 5 3 . Then the free boundary value problem (1.6)–(1.8) and (1.10) has a solution with the free boundary a(t) given by (3.1), and

(3.4) ρ ( r , t ) = f r a ( t ) a 3 ( t ) , u ( r , t ) = a ′ ( t ) a ( t ) r = r 2 t + a 0 2 ,

where f(z) ∈ C ¹([0, 1]) satisfies the following ordinary differential equation:

(3.5) z f ( z ) + f ′ ( z ) − 5 3 f 2 3 ( z ) f ′ ( z ) = 0 , z ∈ [ 0,1 ] , f ( 1 ) = 3 3 .

Moreover, f ( 0 ) = B ∈ ( 1,3 3 ) and f′(z) > 0 for all z ∈ (0, 1].

Remark 3.1.

The solutions we constructed in Theorems 3.1 and 3.2 have the properties: For all 0 ≤ r ≤ a(t), we have

(3.6) ρ ( r , t ) → 0 a s t → + ∞ ,

(3.7) u ( r , t ) = r 2 t + a 0 2 → 0 a s t → + ∞ .

Theorems 3.1 and 3.2 imply that the free boundary expands at an algebraic rate in time, and the fluid density tends to zero almost everywhere away from the symmetry centre as the time grows up. These phenomena have been seen in Ref. [20] for weak solutions to the stress free boundary value problem (1.6)–(1.8) and (1.10), but here we provide such an analytical solution to this problem.

Remark 3.2.

Different from Theorem 2.2 of Ref. [21], the functions f(z) in Theorems 3.1 and 3.2 of this paper belong to C ¹[0, 1], whereas f(z) ∈ C[0, 1] ∩ C ¹(0, 1] in Theorem 2.2 of Ref. [21].

To prove Theorems 3.1 and 3.2, we recall a lemma from Ref. [25].

Lemma 3.1.

(Lemma 3 of Ref. [25]) For the equation of conservation of mass in radial symmetry (1.6), there exists solutions

(3.8) ρ ( r , t ) = f r a ( t ) a N ( t ) , u ( r , t ) = a ′ ( t ) a ( t ) r ,

with the form f ≥ 0 ∈ C ¹ and a(t) > 0 ∈ C ¹.

For N = 2 and γ = 2, system (1.6) and (1.7) becomes

(3.9) ρ t + ( ρ u ) r + ρ u r = 0 ,

(3.10) ρ ( u t + u u r ) + ( ρ 2 ) r − ρ u r + u r r + ρ r u r = 0 .

By (3.8), we plug ρ ( r , t ) = f ( r a ( t ) ) a 2 ( t ) , u ( r , t ) = a ′ ( t ) a ( t ) r into (3.10) to obtain

(3.11) f ( z ) a ″ ( t ) a 3 ( t ) r − a ′ ( t ) a 4 ( t ) f ′ ( z ) + 2 a 5 ( t ) f ( z ) f ′ ( z ) = 0 ,

where z = r a ( t ) . In view of a(t) > 0, (3.11) implies that

(3.12) z f ( z ) a ″ ( t ) a ( t ) − a ′ ( t ) a ( t ) f ′ ( z ) + 2 a 2 ( t ) f ( z ) f ′ ( z ) = 0 .

In order to find an analytical solution, we let

(3.13) a ″ ( t ) a ( t ) = − a ′ ( t ) a ( t ) = − 1 a 2 ( t ) ,

which can be solved as

(3.14) a ( t ) = 2 t + a 0 2 ,

where a ₀ > 0 is a constant. Then, from (3.12) and (3.13), we obtain the following ordinary differential equation:

(3.15) z f ( z ) + f ′ ( z ) − 2 f ( z ) f ′ ( z ) = 0 , z ∈ [ 0,1 ] .

We choose a(t) as a free boundary which satisfies the stress free boundary condition (1.10) for θ = 1, N = 2 and γ = 2. Then, we get

(3.16) f ( 1 ) = 2 .

In the following, we will prove that the problem (3.15) and (3.16) can be solved on [0,1]. For this purpose, we first show the following lemma.

Lemma 3.2.

Let f(z) > 0, z ∈ [0, 1] be a solution to the problem (3.15) and (3.16) in C ¹([0, 1]). Then f(z) increases strictly in [0,1] and

(3.17) A = f ( 0 ) < f ( z ) ≤ 2 , f ′ ( z ) > 0 , ∀ z ∈ ( 0,1 ] , f ′ ( 0 ) = 0 ,

where A ∈ (1, 2) is the solution of the following equation

(3.18) g ( x ) = 2 x − ln ⁡ x + ln ⁡ 2 − 7 2 = 0 .

Furthermore, such a solution is unique.

Proof.

Since f(z) > 0 is a solution to the problem (3.15) and (3.16) in C ¹([0, 1]), we divide (3.15) by f(z) and integrate it from 1 to z, and use (3.16) to obtain

(3.19) 1 2 z 2 + ln ⁡ f ( z ) − ln ⁡ 2 − 2 f ( z ) + 7 2 = 0 .

Let f(0) = A and let z → 0 in (3.19), this leads to the fact that A ∈ (1, 2) is the solution of the Equation (3.18). Here, we claim that the Equation (3.18) has a unique solution x = A ∈ (1, 2). This claim is due to the fact that g ( 1 ) = ln ⁡ 2 − 3 2 < 0 , g ( 2 ) = 1 2 > 0 and g ′ ( x ) = 2 − 1 x > 0 for x ∈ [1, 2]. We plug f(0) = A ∈ (1, 2) into (3.15) to obtain f′(0) = 0.

We rewrite (3.15) as

(3.20) f ′ ( z ) [ 1 − 2 f ( z ) ] = − z f ( z ) , z ∈ [ 0,1 ] .

To prove f′(z) > 0, ∀z ∈ (0, 1], we first claim that f ( z ) > 1 2 , ∀ z ∈ [ 0,1 ] . Otherwise, if there exists z ₀ ∈ [0, 1] such that f ( z 0 ) = 1 2 , then by (3.20) we get z ₀ = 0, which contradicts the fact that f(0) = A ∈ (1, 2); if there exists z ₀ ∈ [0, 1] such that f ( z 0 ) < 1 2 , then by using f(0) = A ∈ (1, 2) and the continuity of the function f(z), we obtain that there exists z ₁ ∈ (0, z ₀) such that f ( z 1 ) = 1 2 , which together with (3.20) implies that z ₁ = 0, which also contradicts the fact that f(0) = A ∈ (1, 2). By (3.20) and f ( z ) > 1 2 , ∀ z ∈ [ 0,1 ] , we have f′(z) > 0, ∀z ∈ (0, 1], which together with f(0) = A, f(1) = 2 implies that A = f(0) < f(z) ≤ 2, ∀z ∈ (0, 1].

Now, we turn to prove the uniqueness result. Let f ̄ ( z ) ∈ C 1 ( [ 0,1 ] ) be another solution to the problem (3.15) and (3.16) with A = f ̄ ( 0 ) < f ̄ ( z ) ≤ 2 and f ̄ ′ ( z ) > 0 for all z ∈ (0, 1]. Then one has

(3.21) z f ̄ ( z ) + f ̄ ′ ( z ) − 2 f ̄ ( z ) f ̄ ′ ( z ) = 0 , z ∈ [ 0,1 ] ,

(3.22) f ̄ ( 1 ) = 2 .

We multiply (3.15) and (3.21) by f ̄ ( z ) and f(z), respectively, then the difference of the two resultant equations is

(3.23) f ̄ ( z ) f ′ ( z ) − 2 f ( z ) f ′ ( z ) f ̄ ( z ) − f ̄ ′ ( z ) f ( z ) + 2 f ̄ ( z ) f ̄ ′ ( z ) f ( z ) = 0 , z ∈ [ 0,1 ] .

Define w ( z ) = f ( z ) − f ̄ ( z ) . Then by (3.16), (3.22) and (3.23), w(z) is a solution of the following problem:

(3.24) f ̄ ( z ) w ′ ( z ) − f ̄ ′ ( z ) w ( z ) − 2 f ̄ ( z ) w ( z ) w ′ ( z ) − 2 f ̄ ( z ) 2 w ′ ( z ) = 0 , z ∈ [ 0,1 ] , w ( 0 ) = w ( 1 ) = 0 .

Dividing (3.24) ₁ by f ̄ ( z ) 2 and integrating it over [0,1], we have

(3.25) ∫ 0 1 w ( z ) f ̄ ( z ) ′ d z − ∫ 0 1 2 w ( z ) w ′ ( z ) f ̄ ( z ) d z − 2 ∫ 0 1 w ′ ( z ) d z = 0 ,

which together with (3.24) ₂ implies that

(3.26) ∫ 0 1 w ( z ) 2 ⋅ f ̄ ′ ( z ) f ̄ ( z ) 2 d z = 0 .

Thus w(z) ≡ 0 for z ∈ [0, 1] since f ̄ ′ ( z ) > 0 for all z ∈ (0, 1]. So f ( z ) = f ̄ ( z ) for all z ∈ [0, 1]. □

Based on Lemma 3.2, we can give an existence result to the problem (3.15) and (3.16), whose proof is simpler than the one given in Lemma 3.5 of Ref. [21].

Lemma 3.3.

The problem (3.15) and (3.16) has a solution f(z) > 0 in C ¹([0, 1]).

Proof.

We first rewrite (3.15) and (3.16) as

(3.27) f ′ ( z ) = F ( f ( z ) , z ) = − z f ( z ) 1 − 2 f ( z ) , z ∈ [ 0,1 ] , f ( 1 ) = 2 .

We next seek for a solution to (3.27) such that

(3.28) f ( z ) ∈ C 1 ( [ 0,1 ] ) , A ≤ f ( z ) ≤ 2 , ∀ z ∈ [ 0,1 ] , A ∈ ( 1,2 ) .

Set

(3.29) S = { ( z , f ( z ) ) | z ∈ [ 0,1 ] , f ( z ) ∈ [ A , 2 ] , A ∈ ( 1,2 ) } .

Note that F is continuous on S and Lipschitz continuous with respect to f(z). Thus, we can obtain a solution f(z) > 0 in C ¹([0, 1]) for the problem (3.27) by the standard theory of ordinary differential equations. □

In view of the above analysis, we have proved the global existence f(z) ∈ C ¹([0, 1]) to the problem (3.15) and (3.16) and the proof of Theorem 3.1 is completed.

Next, we turn to prove Theorem 3.2. For N = 3 and γ = 5 3 , system (1.6) and (1.7) becomes

(3.30) ρ t + ( ρ u ) r + 2 ρ u r = 0 ,

(3.31) ρ ( u t + u u r ) + ρ 5 3 r − ρ u r + 2 u r r + 2 ρ r u r = 0 .

By (3.8), we plug ρ ( r , t ) = f ( r a ( t ) ) a 3 ( t ) , u ( r , t ) = a ′ ( t ) a ( t ) r into (3.31) to obtain

(3.32) f ( z ) a ″ ( t ) a 4 ( t ) r + 5 3 f 2 3 ( z ) f ′ ( z ) ⋅ 1 a 6 ( t ) − f ′ ( z ) a ′ ( t ) a 5 ( t ) = 0 ,

where z = r a ( t ) . In view of a(t) > 0, (3.32) implies that

(3.33) z f ( z ) a ″ ( t ) a ( t ) − a ′ ( t ) a ( t ) f ′ ( z ) + 5 3 a 2 ( t ) f 2 3 ( z ) f ′ ( z ) = 0 .

For the case of N = 2, γ = 2, suppose that (3.13) holds, then we deduce that a(t) satisfies (3.14) and f(z) satisfies the following ordinary differential equation:

(3.34) z f ( z ) − 5 3 f 2 3 ( z ) f ′ ( z ) + f ′ ( z ) = 0 , z ∈ [ 0,1 ] .

We choose a(t) as a free boundary which satisfies the stress free boundary condition (1.10) for θ = 1, N = 3 and γ = 5 3 . Thus, we get

(3.35) f ( 1 ) = 3 3 .

Similar to Lemmas 3.2 and 3.3, for the problem (3.34) and (3.35), we have the following two lemmas.

Lemma 3.4.

Let f(z) > 0 be a solution to the problem (3.34) and (3.35) in C ¹([0, 1]). Then, f(z) increases strictly in [0,1] and

(3.36) B = f ( 0 ) < f ( z ) ≤ 3 3 , f ′ ( z ) > 0 , ∀ z ∈ ( 0,1 ] , f ′ ( 0 ) = 0 ,

where B ∈ ( 1,3 3 ) is the solution of the following equation

(3.37) h ( x ) = 5 2 x 2 3 − ln ⁡ x − 7 + ln ⁡ 3 3 = 0 .

Furthermore, such a solution is unique.

Proof.

Since f(z) > 0 is a solution to the problem (3.34) and (3.35) in C ¹([0, 1]), we divide (3.34) by f(z), then, integrate it from 1 to z and use (3.35) to obtain

(3.38) 1 2 z 2 − 5 2 f 2 3 ( z ) + ln ⁡ f ( z ) − ln ⁡ 3 3 + 7 = 0 .

Let f(0) = B and z → 0 in (3.38). Then B ∈ ( 1,3 3 ) is the solution of the Equation (3.37). Here, we claim that the Equation (3.37) has a unique solution x = B ∈ ( 1,3 3 ) . This claim is due to that h ( 1 ) = ln ⁡ 3 3 − 9 2 < 0 , h ( 3 3 ) = 1 2 > 0 and h ′ ( x ) = 5 3 x − 1 3 − 1 x > 0 for x ∈ [ 1,3 3 ] . We plug f ( 0 ) = B ∈ ( 1,3 3 ) into (3.34) to obtain f′(0) = 0.

We rewrite (3.34) as

(3.39) f ′ ( z ) 1 − 5 3 f 2 3 ( z ) = − z f ( z ) , z ∈ [ 0,1 ] .

To prove f′(z) > 0, ∀z ∈ (0, 1], we first claim that f ( z ) > ( 3 5 ) 3 2 , ∀ z ∈ [ 0,1 ] . Otherwise, if there exists z ₀ ∈ [0, 1] such that f ( z 0 ) = ( 3 5 ) 3 2 , then by (3.39) we get z ₀ = 0, which contradicts the fact that f ( 0 ) = B ∈ ( 1,3 3 ) ; if there exists z ₀ ∈ [0, 1] such that f ( z 0 ) < ( 3 5 ) 3 2 , then by using f ( 0 ) = B ∈ ( 1,3 3 ) and the continuity of the function f(z), we obtain that there exists z ₁ ∈ (0, z ₀) such that f ( z 1 ) = ( 3 5 ) 3 2 , which together with (3.39) implies that z ₁ = 0, which also contradicts the fact that f ( 0 ) = B ∈ ( 1,3 3 ) . By (3.39) and f ( z ) > ( 3 5 ) 3 2 , ∀ z ∈ [ 0,1 ] , we have f′(z) > 0, ∀z ∈ (0, 1], which together with f(0) = B, f ( 1 ) = 3 3 implies that B = f ( 0 ) < f ( z ) ≤ 3 3 , ∀ z ∈ ( 0,1 ] .

The proof of the uniqueness result is similar to the one of Lemma 3.2. For the sake of simplicity, we omit the proof. □

Lemma 3.5.

The problem (3.34) and (3.35) has a solution f(z) > 0 in C ¹([0, 1]).

Proof.

The proof is similar to that of Lemma 3.3. □

Using Lemmas 3.4 and 3.5, we complete the proof of Theorem 3.2.

Corresponding author: Manwai Yuen, Department of Mathematics and Information Technology, The Education University of Hong Kong, 10 Lo Ping Road, Tai Po, New Territories, Hong Kong, E-mail: nevetsyuen@hotmail.com

Acknowledgments

We would like to express our sincere thanks to the referee for the valuable and helpful suggestions and comments, which have improved the quality of this paper.

Research ethics: Not applicable.
Author contributions: The authors have accepted responsibility for the entire content of this manuscript and approved its submission.
Competing interests: Authors state no conflict of interest.
Research funding: The first author was partially supported by the Henan Natural Science Foundation (Nos. 242300421397 and 222300420579), the Vital Science Research Foundation of Henan Province Education Department (Nos. 22A110024 and 22A110026), the Project of Youth Backbone Teachers of Colleges and Universities in Henan Province (No. 2019GGJS176) and the Scientific Research Team Plan of Zhengzhou University of Aeronautics (No. 23ZHTD01003). The second was partially supported by the Departmental Research Grant 2023-24 (MIT/DRG04-24) from the Education University of Hong Kong.
Data availability: Not applicable.

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Received: 2023-07-11

Accepted: 2024-05-24

Published Online: 2024-07-02

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Keywords for this article

compressible Navier–Stokes equations; density-dependent viscosity; analytical solution; free boundary

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