We study a free boundary problem for compressible Navier-Stokes equations with density-dependent viscosity and a non-autonomous external force, which can be written in Eulerian coordinates as:

ρ τ + ( ρ u ) ξ = 0 , τ > 0

( ρ u ) τ + ( ρ u 2 + P ( ρ ) ) ξ = ( μ u ξ ) ξ + ρ f , a ( τ ) < ξ < b ( τ )

with initial data

( ρ , u ) ( ξ , 0 ) = ( ρ 0 , u 0 ) ( ξ ) , a = a ( 0 ) ≤ ξ ≤ b ( 0 ) = b ,

where ρ = ρ (ξ,τ), u = u(ξ,τ), P = P(ρ) and f = f(ξ,t) denote the density, velocity, pressure and a given external force, respectively, μ = μ(ρ) is the viscosity coefficient. a(τ) and b(τ) are the free boundaries with the following property:

d d τ a ( τ ) = u ( a ( τ ) , τ ) , d d τ b ( τ ) = u ( b ( τ ) , τ ) ,

( - P ( ρ ) + μ ( ρ ) u ξ ) ( ξ , τ ) = 0 , ξ = a ( τ ) , b ( τ ) .

The investigation in 1 showed that the continuous dependence on the initial data of the solutions to the compressible Navier-Stokes equations with vacuum failed. The main reason for the failure at the vacuum is because of kinematic viscosity coefficient being independent of the density. On the other hand, we know that the Navier-Stokes equations can be derived from the Boltzmann equation through Chapman-Enskog expansion to the second order, and the viscosity coefficient is a function of temperature. For the hard sphere model, it is proportional to the square-root of the temperature. If we consider the isentropic gas flow, this dependence is reduced to the dependence on the density function by using the second law of thermal dynamics.

For simplicity of presentation, we consider only the polytropic gas, i.e. P(ρ) = Aρ ^γ with A > 0 being constants. Our main assumption is that the viscosity coefficient μ is assumed to be a functional of the density ρ, i.e. μ = cρ ^θ , where c and θ are positive constants. Without loss of generality, we assume A = 1 and c = 1.

Since the boundaries x = a(τ) and x = b(τ) are unknown in Euler coordinates, we will convert them to fixed boundaries by using Lagrangian coordinates. We introduce the following coordinate transformation

x = ∫ a ( τ ) ξ ρ ( y , τ ) d y , t = τ ,

then the free boundaries ξ = a(τ) and ξ = b(τ) become

x = 0 , x = ∫ a ( τ ) b ( τ ) ρ ( z , τ ) d z = ∫ a b ρ 0 ( z ) d z

where ∫ a b ρ 0 ( z ) d z is the total initial mass, and without loss of generality, we can normalize it to 1. So in terms of Lagrangian coordinates, the free boundaries become fixed. Under the coordinate transformation, Eqs. (1.1)-(1.2) are now transformed into

ρ t + ρ 2 u x = 0 , t > 0 ,

u t + P ( ρ ) x = ( ρ μ ( ρ ) u x ) x + f ( r , t ) , 0 < x < 1

where r = ∫ 0 x ρ - 1 ( y , t ) d y . The boundary conditions (1.4)-(1.5) become

( - ρ γ + ρ 1 + θ u x ) ( d , t ) = 0 , d = 0 , 1 ,

and the initial data (1.3) become

( ρ , u ) ( x , 0 ) = ( ρ 0 , u 0 ) ( x ) , x ∈ [ 0 , 1 ] .

Now let us first recall some previous works in this direction. When the external force f ≡ 0, there have been many works (see, e.g., 2 3 4 5 6 7 8 9 ) on the existence and uniqueness of global weak solutions, based on the assumption that the gas connects to vacuum with jump discontinuities, and the density of the gas has compact support. Among them, Liu et al. 4 established the local well-posedness of weak solutions to the Navier-Stokes equations; Okada et al. 5 obtained the global existence of weak solutions when 0 < θ < 1/3 with the same property. This result was later generalized to the case when 0 < θ < 1/2 and 0 < θ < 1 by Yang et al. 7 and Jiang et al. 3 , respectively. Later on, Qin et al. 8 9 proved the regularity of weak solutions and existence of classical solution. Fang and Zhang 2 proved the global existence of weak solutions to the compressible Navier-Stokes equations when the initial density is a piece-wise smooth function, having only a finite number of jump discontinuities.

For the related degenerated density function and viscosity coefficient at free boundaries, see Yang and Zhao 10 , Yang and Zhu 11 , Vong et al. 12 , Fang and Zhang 13 14 , Qin et al. 15 , authors studied the global existence and uniqueness under some assumptions on initial data.

When f ≠ 0, Qin and Zhao 16 proved the global existence and asymptotic behavior for γ = 1 and μ = const with boundary conditions u(0,t) = u(1,t) = 0; Zhang and Fang 17 established the global behavior of the Equations (1.1)-(1.2) with boundary conditions u(0,t) = ρ(1,t) = 0. In this paper, we obtain the global existence of the weak solutions and regularity with boundary conditions (1.4)-(1.5). In order to obtain existence and higher regularity of global solutions, there are many complicated estimates on external force and higher derivations of solution to be involved, this is our difficulty. To overcome this difficulty, we should use some proper embedding theorems, the interpolation techniques as well as many delicate estimates. This is the novelty of the paper.

The notation in this paper will be as follows:

L p , 1 ≤ p ≤ + ∞ , W m , p , m ∈ N , H 1 = W 1 , 2 , H 0 1 = W 0 1 , 2 denote the usual (Sobolev) spaces on [0,1]. In addition, || · || _B denotes the norm in the space B; we also put | | ⋅ | | = | | ⋅ | | L 2 ( [ 0 , 1 ] ) .

The rest of this paper is organized as follows. In Section 2, we shall prove the global existence in H ¹. In Section 3, we shall establish the global existence in H ². In Section 4, we give the detailed proof of Theorem 4.1.

In this section, we shall establish the global existence of solutions in H ¹.

Theorem 2.1 Let 0 < θ < 1, γ > 1, and assume that the initial data (ρ ₀,u ₀) satisfies inf [ 0 , 1 ] ρ 0 > 0 , ρ 0 ∈ W 1 , 2 n , u 0 ∈ H 1 and external force f satisfies f(r(x,·),·) ∈ L ²ⁿ ([0,T], L ²ⁿ [0,1]) for some n ∈ N satisfying n(2n - 1)/(2n ² + 2n - 1) > θ, then there exists a unique global solution (ρ (x,t),u(x,t)) to problem (1.8)-(1.11), such that for any T > 0,

0 < C 1 - 1 ( T ) ≤ ρ ( x , t ) ≤ C 1 ( T ) , ρ ∈ L ∞ ( [ 0 , T ] , H 1 [ 0 , 1 ] ) , u ∈ L ∞ ( [ 0 , T ] , H 1 [ 0 , 1 ] ) ∩ ∈ L 2 ( [ 0 , T ] , H 2 [ 0 , 1 ] ) , u t ∈ L 2 ( [ 0 , T ] , L 2 [ 0 , 1 ] ) .

The proof of Theorem 2.1 can be done by a series of lemmas as follows.

Lemma 2.1 Under conditions of Theorem 2.1, the following estimates hold

∫ 0 1 1 2 u 2 + 1 γ - 1 ρ γ - 1 d x + ∫ 0 t ∫ 0 1 ρ 1 + θ u x 2 ( x , s ) d x d s ≤ C 1 ( T ) ,

ρ ( x , t ) ≤ C 1 ( T ) , ( x , t ) ∈ [ 0 , 1 ] × [ 0 , T ] ,

∫ 0 1 u 2 n d x + n ( 2 n - 1 ) ∫ 0 t ∫ 0 1 ρ 1 + θ u 2 n - 2 u x 2 ( x , s ) d x d s ≤ C 1 ( T )

where C ₁(T) denotes generic positive constant depending only on | | ρ 0 | | W 1 , 2 n [ 0 , 1 ] , | | u 0 | | H 1 [ 0 , 1 ] , time T and | | f | | L 2 n ( [ 0 , T ] , L 2 n [ 0 , 1 ] ) .

Proof Multiplying (1.8) and (1.9) by ρ ^γ-2 and u, respectively, using integration by parts, and considering the boundary conditions (1.10), we have

d d t ∫ 0 1 1 2 u 2 + 1 γ - 1 ρ γ - 1 d x + ∫ 0 1 ρ 1 + θ u x 2 d x = ∫ 0 1 f u d x

Integrating (2.4) with respect to t over [0,t], using Young's inequality, we have

∫ 0 1 1 2 u 2 + 1 γ - 1 ρ γ - 1 d x + ∫ 0 t ∫ 0 1 ρ 1 + θ u x 2 d x d s ≤ C 1 ( T ) + 1 2 ∫ 0 t ∫ 0 1 u 2 d x + C 1 ∫ 0 t ∫ 0 1 f 2 d x d s ≤ 1 2 ∫ 0 t ∫ 0 1 u 2 d x + C 1 ( T )

which, by virtue of Gronwall's inequality and assumption f(r(x,·),·) ∈ L ²ⁿ ([0,T], L ²ⁿ [0,1]), gives (2.1).

We derive from (1.8) that

( ρ θ ) t = - θ ρ 1 + θ u x

Integrating (2.5) with respect to t over [0,t] yields

ρ θ ( x , t ) = ρ 0 θ - θ ∫ 0 t ρ 1 + θ u x ( x , s ) d s .

Integrating (1.9) with respect to x, applying the boundary conditions (1.10), we obtain

ρ 1 + θ u x = ∫ 0 x u t d y + ρ γ - ∫ 0 x f ( r ( y , t ) , t ) d y

Inserting (2.7) into (2.6) gives

ρ θ + θ ∫ 0 t ρ γ d s = ρ 0 θ + θ ∫ 0 t ∫ 0 x f ( r , ( y , s ) , s ) d y d s - θ ∫ 0 x ( u - u 0 ) d y

Thus, the Hölder inequality and (2.1) imply

∫ 0 x u ( y , t ) d y ≤ C 1

and (2.2) follows from (2.8) and (2.9).

Multiplying (1.9) by 2nu ^2n-1 and integrating over x and t, applying the boundary conditions (1.10), we have

∫ 0 1 u 2 n d x + 2 n ( 2 n - 1 ) ∫ 0 t ∫ 0 1 u 2 n - 2 ρ 1 + θ u x 2 d x d s = ∫ 0 1 u 0 2 n d x + 2 n ( 2 n - 1 ) ∫ 0 t ∫ 0 1 u 2 n - 2 ρ γ u x d x d s + 2 n ∫ 0 t ∫ 0 1 f u 2 n - 1 d x d s .

Applying the Young inequality and condition f(r(x, ·), ·) ∈ L ²ⁿ ([0,T],L ²ⁿ [0,1]) to the last two terms in (2.10) yields

∫ 0 1 u 2 n d x + n ( 2 n - 1 ) ∫ 0 t ∫ 0 1 u 2 n - 2 ρ 1 + θ u x 2 d x d s ≤ C 1 + ∫ 0 t ∫ 0 1 f 2 n d x d s + ( 2 n - 1 ) ∫ 0 t ∫ 0 1 u 2 n d x d s + n ( 2 n - 1 ) ∫ 0 t ∫ 0 1 u 2 n - 2 ρ 2 γ - 1 - θ d x d s ≤ C 1 ( T ) + n ( 2 n - 1 ) ∫ 0 t ∫ 0 1 1 n ρ ( 2 γ - 1 - θ ) n + n - 1 n u 2 n d x d s + ( 2 n - 1 ) ∫ 0 t ∫ 0 1 u 2 n d x d s ≤ C 1 ( T ) + n ( 2 n - 1 ) ∫ 0 t ∫ 0 1 u 2 n d x d s .

Applying Gronwall's inequality, we conclude

∫ 0 1 u 2 n d x ≤ C 1 ( T )

, which, along with (2.11), yields (2.3). The proof of Lemma 2.1 is complete.

Lemma 2.2 Under conditions of Theorem 2.1, the following estimates hold

∫ 0 1 ( ρ θ ) x 2 n d x ≤ C 1 ( T ) ,

ρ ( x , t ) ≥ C 1 - 1 ( T ) > 0 .

Proof We derive from (2.5) and (1.9) that

( ρ θ ) x t = - θ ( u t + ( ρ γ ) x - f ) .

Integrating it with respect to t over [0,t], we obtain

( ρ θ ) x = ( ρ 0 θ ) x - θ ( u - u 0 ) - θ ∫ 0 t ( ρ γ ) x d s + θ ∫ 0 t f d s .

Multiplying (2.16) by [(ρ ^θ ) _x ]^2n-1, and integrating the resultant with respect to x to get

∫ 0 1 ( ρ θ ) x 2 n d x = ∫ 0 1 ( ρ θ ) x 2 n - 1 ( ρ 0 θ ) x d x - θ ∫ 0 1 ( u - u 0 ) + ∫ 0 t ( ρ γ ) x d s - ∫ 0 t f d s ( ρ θ ) x 2 n - 1 d x ≤ C ∫ 0 1 ( ρ θ ) x 2 n d x 2 n - 1 2 n ∫ 0 1 ( ρ 0 θ ) x 2 n d x 1 2 n + u - u 0 L 2 n + ∫ 0 1 ∫ 0 t ( ρ γ ) x d s 2 n d x 1 2 n + ∫ 0 1 ∫ 0 t f d s 2 n d x 1 2 n ≤ C ∫ 0 1 ( ρ θ ) x 2 n d x 2 n - 1 2 n ∫ 0 1 ( ρ 0 θ ) x 2 n d x 1 2 n + u - u 0 L 2 n + ∫ 0 t ∫ 0 1 ( ρ γ ) x 2 n d s 1 2 n d x + ∫ 0 t ∫ 0 1 f 2 n d x 1 2 n d s

here, we use the inequality ∫ g ( ⋅ , s ) L p ≤ ∫ g ( ⋅ , s ) L p d s . Using Young's inequality and assumptions of external of f, we get from (2.17) that

∫ 0 1 ( ρ θ ) x 2 n d x ≤ 1 2 ∫ 0 1 ( ρ θ ) 2 n x d x + C ∫ 0 t ∫ 0 1 ( ρ γ ) x 2 n d x d s + C ∫ 0 t ∫ 0 1 f 2 n d x d s + C 1 ( T ) ≤ 1 2 ∫ 0 1 ( ρ θ ) x 2 n d x + C 1 ( T ) ∫ 0 t ∫ 0 1 ( ρ γ ) x 2 n d x d s + C 1 ( T ) .

Hence,

∫ 0 1 ( ρ θ ) x 2 n d x ≤ C 1 ( T ) + C 1 ( T ) ∫ 0 t ∫ 0 1 ( ρ γ ) x 2 n d x d s

Using the Gronwall inequality to (2.18), we obtain (2.13).

The proof of (2.14) can be found in 3 , please refer to Lemma 2.3 in 3 for detail.

Lemma 2.3 Under the assumptions in Theorem 2.1, for any 0 ≤ t ≤ T, we have the following estimate

u x ( t ) 2 + ∫ 0 t u t ( s ) 2 d s ≤ C 1 ( T ) .

Proof Multiplying (1.9) by u _t , then integrating over [0,1] × [0,t], we obtain

∫ 0 t ∫ 0 1 u t 2 d x d s = ∫ 0 t ∫ 0 1 u t ( ρ 1 + θ u x - ρ γ ) x d x d s + ∫ 0 t ∫ 0 1 u t f d x d s .

Using integration by parts, (1.8) and the boundary conditions (1.10), we have

∫ 0 t ∫ 0 1 u 1 ρ 1 + θ u x - ρ γ x d x d s = ∫ 0 t ∫ 0 1 u t x ( ρ γ - ρ 1 + θ u x ) d x d s = ∫ 0 1 u x ρ γ - 1 2 ρ 1 + θ u x - u 0 x ρ 0 γ - 1 2 ρ 0 1 + θ u 0 x d x + ∫ 0 t ∫ 0 1 γ u x 2 ρ γ + 1 - 1 + θ 2 u x 3 ρ 2 + θ d x d s .

Thus,

∫ 0 t ∫ 0 1 u t 2 d x d s + 1 2 ∫ 0 1 ρ 1 + θ u x 2 d x = ∫ 0 1 u x ρ γ - u 0 x ρ 0 γ - 1 2 ρ 0 1 + θ u 0 x d x + ∫ 0 t ∫ 0 1 γ u x 2 ρ γ + 1 - 1 + θ 2 u x 3 ρ 2 + θ d x d s + ∫ 0 t ∫ 0 1 u t f d x d s ≤ C 1 ( T ) + ∫ 0 1 1 4 ρ 1 + θ u x 2 + ρ 2 γ - 1 - θ d x + C 1 ( T ) ∫ 0 t sup [ 0 , 1 ] ρ γ - θ ∫ 0 1 ρ 1 + θ u x 2 d x d s + C 1 ( T ) ∫ 0 t ∫ 0 1 ρ 1 + θ | u x | 3 d x d s + 1 4 ∫ 0 t ∫ 0 1 u t 2 d x d s + C 1 ( T ) ∫ 0 t ∫ 0 1 f 2 d x d s .

Using Lemmas 2.1-2.2, we derive

∫ 0 1 u x 2 d x + ∫ 0 t ∫ 0 1 u t 2 d x d s ≤ C 1 ( T ) + C 1 ( T ) ∫ 0 t ∫ 0 1 ρ 1 + θ | u x | 3 d x d s

The last term on the right-hand side of (2.21) can be estimated as follows, using (1.8), conditions (1.10) and Lemmas 2.1-2.2,

C 1 ( T ) ∫ 0 t ∫ 0 1 ρ 1 + θ | u x | 3 d x d s ≤ C 1 ( T ) ∫ 0 t max [ 0 , 1 ] | ρ 1 + θ u x | u x 2 d x d s ≤ C 1 ( T ) ∫ 0 t max [ 0 , 1 ] | ρ 1 + θ u x - ρ γ | ∫ 0 1 u x 2 d x d s + C 1 ( T ) ∫ 0 t ∫ 0 1 u x 2 d x d s ≤ C 1 ( T ) + C 1 ( T ) ∫ 0 t ∫ 0 1 | ( ρ 1 + θ u x - ρ γ ) x d s ∫ 0 1 u x 2 d x d s ≤ C 1 ( T ) + C 1 ( T ) ∫ 0 t ∫ 0 1 | u t —d s ∫ 0 1 u x 2 d x d s + C 1 ( T ) ∫ 0 t ∫ 0 1 | f | d s ∫ 0 1 u x 2 d x d s ≤ C 1 ( T ) + 1 4 ∫ 0 t ∫ 0 1 u t 2 d x d s + C 1 ( T ) ∫ 0 t ∫ 0 1 f 2 d x d s + C 1 ( T ) ∫ 0 t ∫ 0 1 u x 2 d x 2 d s ≤ C 1 ( T ) + 1 4 ∫ 0 t ∫ 0 1 u t 2 d x d s + C 1 ( T ) ∫ 0 t ∫ 0 1 u x 2 d x 2 d s .

Inserting the above estimate into (2.21),

∫ 0 1 u x 2 d x + ∫ 0 t ∫ 0 1 u t 2 d x d s ≤ C 1 ( T ) + C ∫ 0 t u x 2 ∫ 0 1 u x 2 d x d s .

which, by virtue of Gronwall's inequality, (2.1) and (2.14), gives (2.19).

Proof of Theorem 2.1 By Lemmas 2.1-2.3, we complete the proof of Theorem 2.1.

For external force f(r, t), we suppose

f ( r , t ) ∈ L ∞ ( [ 0 , T ] , L 2 [ 0 , 1 ] ) , f r ( r , t ) ∈ L 2 ( [ 0 , T ] , L 2 [ 0 , 1 ] ) , f t ( r , t ) ∈ L 2 ( [ 0 , T ] , L 2 [ 0 , 1 ] )

Constant C ₂(T) denotes generic positive constant depending only on the H ²-norm of initial data ( ρ 0 , u 0 ) , ‖ f ‖ L ∞ ( [ 0 , T ] ) , L 2 [ 0 , 1 ] ) ,   ‖ f r ‖ L 2 ( [ 0 , T ] , L 2 [ 0 , 1 ] ) ,   ‖ f t ‖ L 2 ( [ 0 , T ] , L 2 [ 0 , 1 ] ) , time T and constant C ₁(T).

Remark 3.1 By (3.1), we easily know that assumptions (3.1) is equivalent to the following conditions

f ( r ( x , t ) , t ) ∈ L ∞ ( [ 0 , T ] , L 2 [ 0 , 1 ] ) ,

f r ( r ( x , t ) , t ) ∈ L 2 ( [ 0 , T ] , L 2 [ 0 , 1 ] ) , f t ( r ( x , t ) , t ) ∈ L 2 ( [ 0 , T ] , L 2 [ 0 , 1 ] ) .

Therefore, the generic constant C ₂(T) depending only on the norm of initial data (ρ ₀,u ₀) in H ² , the norms of f in the class of functions in (3.2)-(3.3) and time T.

Theorem 3.1 Let 0 < θ < 1, γ > 1, and assume that the initial data satisfies (ρ ₀,u ₀) ∈ H ² and external force f satisfies conditions (3.1), then there exists a unique global solution (ρ (x,t),u(x,t)) to problem (1.8)-(1.11), such that for any T > 0,

ρ ∈ L ∞ ( [ 0 , T ] , H 2 [ 0 , 1 ] ) ,   u ∈ L ∞ ( [ 0 , T ] , H 2 [ 0 , 1 ] ∩ ∈ L 2 ( [ 0 , T ] , H 3 [ 0 , 1 ] ) ,

u t ∈ L ∞ ( [ 0 , T ] , L 2 [ 0 , 1 ] ) ∩ ∈ L 2 ( [ 0 , T ] , H 1 [ 0 , 1 ] ) .

The proof of Theorem 3.1 can be divided into the following several lemmas.

Lemma 3.2 Under the assumptions in Theorem 3.1, for any 0 ≤ t ≤ T, we have the following estimates

u t ( t ) 2 + ∫ 0 t ∫ 0 1 u t x 2 ( x , s ) d x d s ≤ C 2 ( T ) ,

u x ( t ) L 2 ∞ + u x x ( t ) 2 d x ≤ C 2 ( T ) .

Proof Differentiating (1.9) with respect to t, multiplying the resulting equation by u _t in L ²[0,1], performing an integration by parts, and using Lemma 2.1, we have

1 2 d d t u t 2 + ∫ 0 1 ρ 1 + θ u t x 2 d x = ∫ 0 1 ( θ + 1 ) ρ θ + 2 u x 2 - γ ρ γ + 1 u x + ∂ f ∂ t u t x d x ≤ 1 2 ∫ 0 1 ρ 1 + θ u t x 2 d x + C 1 ( T ) ∫ 0 1 ρ 2 θ + 3 u x 4 + ρ 2 γ + 1 - θ u x 2 d x + C 1 ( T ) ∫ 0 1 ( f r r t ) 2 + f t 2 d x .

Integrating (3.8) with respect to t, applying the interpolation inequality, we conclude

u t ( t ) 2 + ∫ 0 t ∫ 0 1 ρ 1 + θ u t x 2 d x d s ≤ u t ( x , 0 ) + C 1 ( T ) ∫ 0 t ∫ 0 1 u x 4 + u x 2 + f r 2 u 2 + f t 2 d x d s ≤ u t ( x , 0 ) + C 1 ( T ) ∫ 0 t u x 2 + u x x 1 4 u x 3 4 + u x 4 ( s ) d s + ∫ 0 t u L ∞ 2 ∫ 0 1 f r 2 d x d s + C 1 ( t ) ∫ 0 t ∫ 0 1 f t 2 d x d s .

On the other hand, by (1.9), we get

u 0 t = - γ ρ 0 γ - 1 ρ 0 x + ρ 0 θ + 1 u 0 x x + ( θ + 1 ) ρ 0 θ ρ 0 x u 0 x + f ( r 0 , 0 ) .

We derive from assumption (3.1) and (3.10) that

∫ 0 1 u 0 t 2 ( x ) d x ≤ C 2 ( T ) .

Inserting (3.11) into (3.9), by virtue of Lemmas 2.1-2.3 and assumption (3.1), we obtain (3.6). We infer from (1.9),

u t = - γ ρ γ - 1 ρ x + ρ θ + 1 u x x + ( θ + 1 ) ρ θ ρ x u x + f ( r , t ) .

Multiplying (3.12) by u _xx in L ²[0,1], we deduce

∫ 0 1 ρ θ + 1 u x x 2 d x = ∫ 0 1 u x x u t + γ ρ γ - 1 ρ x - ( θ + 1 ) ρ θ ρ x u x - f ( r , t ) d x .

Using Young's inequality and Sobolev's embedding theorem W ^1,1 ↪ W ^∞, Lemma 2.1 and (3.6), we deduce from (3.13) that

∫ 0 1 u x x 2 d x ≤ C 1 ( T ) ∫ 0 1 u t 2 + ρ x 2 + ρ x 2 u x 2 + f 2 d x + 1 4 ∫ 0 1 u x x 2 d x ≤ C 2 ( T ) + C 1 ( T ) u x L ∞ 2 ∫ 0 1 ρ x 2 d x + 1 4 ∫ 0 1 u x x 2 d x ≤ C 2 ( T ) + 1 2 ∫ 0 1 u x x 2 d x

whence

∫ 0 1 u x x 2 d x ≤ C 2 ( T ) .

Applying embedding theorem, we derive from (3.14) that

u x L ∞ 2 ≤ C 1 ( T ) u x 2 + u x x 2 ≤ C 2 ( T )

which, along with (3.14), gives (3.7). The proof is complete.

Lemma 3.3 Under the assumptions in Theorem 3.1, for any 0 ≤ t ≤ T, we have the following estimates

ρ x x ( t ) 2 + ∫ 0 t ρ x x ( s ) 2 d s ≤ C 2 ( T ) ,

∫ 0 t u x x x ( s ) 2 d x ≤ C 2 ( T ) .

Proof Differentiating (1.9) with respect to x, exploiting (1.8), we have

u t x = - ρ γ + ρ 1 + θ u x x x + d f d x = - γ ( γ - 1 ) ρ γ - 2 ρ x 2 - γ ρ γ - 1 ρ x x + ( θ + 1 ) θ ρ θ - 1 ρ x 2 u x + ( θ + 1 ) ρ θ ρ x x u x + 2 ( θ + 1 ) ρ θ ρ x u x x + ρ θ + 1 u x x x + 1 ρ f r