Global Solutions to the One-dimensional Compressible Navier-Stokes ...

Global Solutions to the One-dimensional Compressible Navier-Stokes-Poisson Equations with Large Data Zhong Tan School of Mathematical Sciences, Xiamen University Xiamen 361005, Fujian, China Tong Yang Department of Mathematics, City University of Hong Kong Tat Chee Avenue, Kowloon, Hong Kong, China Huijiang Zhao School of Mathematics and Statistics Wuhan University, Wuhan 430072, China Qingyang Zou School of Mathematics and Statistics Wuhan University, Wuhan 430072, China

Abstract This paper is concerned with the construction of global smooth solutions away from vacuum to the Cauchy problem of the one-dimensional compressible Navier-Stokes-Poisson system with large data and density dependent viscosity coefficient and density and temperature dependent heat conductivity coefficient. The proof is based on some detailed analysis on the bounds on the density and temperature functions. Key Words and Phrases: Navier-Stokes-Poisson equations, global solutions with large data, viscosity and heat conductivity coefficients. A.M.S. Mathematics Subject Classification: 35Q35, 35D35, 76D05

1

Introduction

The compressible Navier-Stokes-Poisson (denoted by NSP) system consisting of the compressible Navier-Stokes equations coupled with Poisson equation models the viscous fluid under the influence of the self-induced electric force:   ρτ + ∇ξ · (ρu) = 0,     (ρu)τ + ∇ξ · (ρu ⊗ u) + ∇ξ p = ρ∇ξ Φ + ∇ξ · T, (ρE)τ + ∇ξ · (ρuE + up) = ρu · ∇ξ Φ + ∇ξ · (uT) + ∇ξ · (κ(v, θ)∇ξ θ) ,     ∆ξ Φ = ρ − ρ(ξ), lim Φ(τ, ξ) = 0.  |ξ|→+∞

1

2

Zhong Tan, Tong Yang, Huijiang Zhao, and Qingyang Zou

Here, ρ > 0, u = (u1 , u2 , u3 ), θ > 0, p = p(ρ, θ), e and Φ denote the density, velocity, absolute temperature, pressure, internal energy and the electrostatic potential function, respectively. And
1 2 t E = 2 |u| + e is the specific total energy, T = µ(ρ, θ) ∇ξ u + (∇ξ u) + ν(ρ, θ)(∇ξ · u)I is the stress tensor with I being the identity matrix. The viscosity coefficients µ(ρ, θ) > 0 and ν(ρ, θ) satisfy µ(ρ, θ) + 32 ν(ρ, θ) > 0. The thermodynamic variables p, ρ, and e are related by Gibbs equation de = θds − pdρ−1 with s being the specific entropy. κ(ρ, θ) > 0 denotes the heat conductivity coefficient, and ρ(ξ) > 0 is the background doping profile, cf. [30]. To explain the purpose of this paper, we firstly give the following remarks on the viscosity and heat conductivity coefficients: • When the viscosity coefficients µ(ρ, θ) > 0, ν(ρ, θ) and the heat-conductivity coefficient κ(ρ, θ) > 0 are constants, (1.1) is used in semiconductor theory to model the transport of charged particles under the influence of self-induced electric field, cf. [30]. • In the kinetic theory, the time evolution of the particle distribution function for the charged particles in a dilute gas can be modelled by the Vlasov-Poisson-Boltzmann system, cf. [4], [3], [34]. When we derive the NSP (1.1) from the Vlasov-Poisson-Boltzmann system by using the Chapman-Enskog expansion, cf. [4], [12], [34], the viscosity coefficients µ, ν and the heat-conductivity coefficient κ depend on the absolute temperature θ and ν = − 23 µ for the monatomic gas. If the inter-molecular potential is proportional to r−α with α > 1, where r represents the intermolecular distance, then µ, ν and κ are proportional to the temperature to some power: µ, −ν, κ ∝ θ

α+4 2α

.

In particular, for the Maxwellian molecule (α = 4), such dependence is linear, while for the √ hard sphere model and also the case when α → +∞, the dependence is in the form of θ. This paper is concerned with the global existence of large data solutions when the viscosity coefficients µ, ν and the heat conductivity coefficient κ depend on ρ and θ. Unlike the small perturbation solutions, such dependence has strong influence on the solution behavior and thus leads to difficulties in analysis not for the case of constant coefficients. In fact, for the one-dimensional compressible Navier-Stokes equations, recently there are a lot of works on the construction of non-vacuum solutions to the one-dimensional compressible Navier-Stokes equations with density and temperature dependent transportation coefficients in various forms, cf. [1] [5], [18], [19], [21], [22], [23], [24], [25] and the references therein. However, there is a gap between the physical models and the satisfactory existence theory. The main purpose in this paper is devoted to the construction of globally smooth, nonvacuum solutions to the one-dimensional non-isentropic compressible NSP with degenerate density dependent viscous coefficient and degenerate density and temperature dependent heat conductivity coefficient for arbitrarily large data. We hope that the analysis here can shed some light on the construction of global classical solutions to the fluid model derived from the VlasovPoisson-Boltzmann system with large data. Let x be the Lagrangian space variable, t be the time variable, and v = ρ1 denote the specific volume. Then the one-dimensional compressible NSP system (1.1) with viscous coefficient µ(v)

One-dimensional Compressible Navier-Stokes-Poisson Equation with Large Data

3

and heat conductivity coefficient κ(v, θ) becomes   vt − ux = 0,        µ(v)ux   + Φvx , u + p(v, θ) = x  t v x  κ(v,θ)θx µ(v)u2x  + , e + p(v, θ)u = t x  v v  x       Φx  = 1 − v, lim Φ(t, x) = 0.  v x

(1.1)

|x|→+∞

Throughout this paper, we will concentrate on the ideal, polytropic gases: Rθ γ−1 = Av −γ exp s , v R 

p(v, θ) =



e = Cv θ =

Rθ , γ−1

(1.2)

where the specific gas constant R and the specific heat at constant volume Cv are positive constants and γ > 1 is the adiabatic constant. Moreover, to simplify the presentation, we will only consider the case when the background doping profile ρ is a positive constant which is normalized to 1 as in (1.1)4 . Take the initial data (v(0, x), u(0, x), θ(0, x)) = (v0 (x), u0 (x), θ0 (x)),

lim (v0 (x), u0 (x), θ0 (x)) = (v± , u± , θ± )

x→±∞

(1.3) satisfying v− = v+ , u− = u+ , θ− = θ+ . Without loss of generality, we assume v− = v+ = 1, u− = u+ = 0, θ− = θ+ = 1. The first result is concerned with the case µ(v) = v −a ,

κ(v, θ) = θb ,

(1.4)

which is stated as follows. Theorem 1.1 Suppose • (v0 (x)−1, u0 (x), θ0 (x)−1, Φ0x (x)) ∈ H 1 (R), and there exist positive constants V , V , Θ, Θ such that V ≤ v0 (x) ≤ V , Θ ≤ θ0 (x) ≤ Θ; (1.5) •

1 3

< a < 12 ;

• b satisfies one of the following conditions (i) 1 ≤ b < (ii) 0 < b
0 is any given positive constant and V0 , Θ0 are some positive constants which may depend on T .

4

Zhong Tan, Tong Yang, Huijiang Zhao, and Qingyang Zou

Note that the assumptions imposed on a and b in Theorem 1.1 exclude the case when the viscous coefficient µ and the heat conductivity coefficient κ are positive constants. The next result will recover this in another setting. The main idea is to use the smallness of γ − 1 to deduce uniform lower and upper bounds on the absolute temperature. This can be achieved by showing that (v0 (x) − 1, u0 (x), s0 (x) − s) ∈ H 1 (R) are bounded in H 1 (R) independent of γ − 1 R so that kθ0 (x) − 1kL∞ (R) can be chosen to be small when γ is close to 1. Here s = γ−1 ln R A is the far field of the initial entropy s0 (x), that is, lim s0 (x) =

|x|→+∞

Rθ0 (x)v0 (x)γ−1 R ln = s. A |x|→+∞ γ − 1 lim

Take (v, u, s) as the unknown function, the second global existence theorem can be stated as follows. Theorem 1.2 Suppose • k(v0 (x)−1, u0 (x), s0 (x)−s, Φ0x (x))kH 1 (R) is bounded by some positive constant independent of γ − 1 and (1.5) holds for some γ − 1−independent positive constants V , V , Θ, Θ; • (γ − 1)ks0 (x)kL∞ (R) is bounded by some constant independent of γ − 1; • The smooth function µ(v) satisfies µ(v) > 0 for all v > 0 and lim Ψ(v) = −∞,

v→0+

Here Ψ(v) =

lim Ψ(v) = +∞.

Z v√ z − ln z − 1

z

1

(1.7)

v→+∞

µ(z)dz;

(1.8)

• For the heat conductivity coefficient, there are two cases. If κ(v, θ) = κ(θ) depends only on θ, we only assume κ(θ) > 0 for θ > 0 with some smoothness condition. If it depends on both v and θ, then in addition to κ(v, θ) > 0 for all v > 0, θ > 0, we also assume the following. Set κ1 (v) = min κ(v, θ), assume Θ≤θ≤Θ

κθθ (v, θ) < 0,

∀v > 0, θ > 0,

µ(v) κ1 (v)

µ(v) κ1 (v)

and lim

v→0+ |Ψ(v)|2

= lim

v→+∞ |Ψ(v)|2

(1.9)

= 0;

(1.10)

• γ − 1 is sufficiently small. Then the Cauchy problem (1.1), (1.3) admits a unique global solution (v(t, x), u(t, x), θ(t, x)) satisfying (1.6) and lim sup (v(t, x) − 1, u(t, x), θ(t, x) − 1) = 0. (1.11) t→+∞ x∈R

Although in Theorem 1.2, the case µ and κ are positive constants can be covered, it does ask that γ − 1 to be sufficiently small, our final result in this paper shows that for the case when µ is a positive constant, similar result hold provided that κ(v, θ) satisfies κ(v, θ) > 0

∀v > 0, θ > 0,

min e >0, θ≥Θ e >0 v≥V





e > 0. κ(v, θ) ≥ κ Ve , Θ

(1.12)

One-dimensional Compressible Navier-Stokes-Poisson Equation with Large Data

5

Theorem 1.3 Suppose • (v0 (x)−1, u0 (x), θ0 (x)−1, Φ0x (x)) ∈ H 1 (R), and there exist positive constants V , V , Θ, Θ satisfying (1.5); • µ is a positive constant and κ(v, θ) satisfies (1.12). Then the Cauchy problem (1.1), (1.3) admits a unique global solution (v(t, x), u(t, x), θ(t, x)) satisfying (1.6). Remark 1.1 We give the following remarks on Theorem 1.1-Theorem 1.3. • From the proof of Theorem 1.2, one will notice that the assumption (1.10) can be replaced by the following weaker assumption lim

µ(v) κ1 (v)

v→0+ |Ψ(v)|2

≤ ε0 ,

µ(v) κ1 (v)

lim

v→+∞ |Ψ(v)|2

≤ ε0 .

(1.13)

Here ε0 > 0 is a suitably chosen sufficiently small positive constant. • Under the assumptions in Theorem 1.2, when γ − 1 is sufficiently small, although kθ0 − 1kH 1 (R) is small, k(v0 − 1, u0 , s0 − s)kH 1 (R) can be large. • When µ(v) satisfies certain growth conditions when v → 0+ and v → +∞, for example, µ(v) ∼ v a as v → 0+ and µ(v) ∼ v b as v → +∞ with a < 0, b > − 12 , then similar result 1 to Theorem 1.2 also holds even when (v0 − 1, u0 , s0 − s)kH 1 (R) , V , and V depend on γ−1 with certain growth condition as γ → 1+ . • The same arguments for Theorem 1.1-Theorem 1.3 can be applied directly the compressible Navier-Stokes equations which generalize the previous results [18] and [23] where the viscosity coefficient is assumed to be a positive constant and/or is bounded from below and above by some positive constants, which means that viscosity coefficient µ(v) is non-degenerate. • It is worth to pointing out that since the fact that 

µ(v)ux v





= x

µ(v)vt v





= x

µ(v)vx v



,

(1.14)

t

plays an important role in the following analysis, we can only treat the case when µ(v) is a smooth function of v. Hence, it is interesting to study the case when µ depends on θ. We now review some related results. Firstly, recently there are some results on the construction of non-vacuum, large solutions to the one-dimensional compressible Navier-Stokes equations with constant viscosity coefficient µ and density and temperature dependent heat conductivity coefficient κ, cf. [18], [23]. A key ingredient in these works is the pointwise a priori estimates on the specific volume which guarantees that no vacuum nor concentration of mass occur. It is worth pointing out that it was in deducing the above mentioned upper and lower bounds on the specific volume that the viscosity coefficient µ(v) is assumed to be non-degenerate, for example µ(v) is assumed to satisfy 0 < µ0 ≤ µ(v) ≤ µ1 in [23] and µ(v) ≡ µ0 > 0 in [18], . The strategy to prove Theoerem 1.1 can be stated as follows. We will firstly apply the maximum principle for second order parabolic equation to obtain a lower bound estimate on

6

Zhong Tan, Tong Yang, Huijiang Zhao, and Qingyang Zou

θ(t, x) in terms of the lower bound on v(t, x) in Lemma 2.4. And then by combining the arguments used in [21] and [25], we can deduce a lower bound and an upper bound on v(t, x) in terms of kθ1−b kL∞ ([0,T ]×R) , that is, the estimates (2.35) and (2.36). These two estimates together with the L∞ ([0, T ] × R)−estimate on θ(t, x) given in Lemma 4.9 then yield the desired lower and upper bound on v(t, x) and θ(t, x) provided that the parameters a and b satisfy certain conditions. To prove Theorem 1.2, the main idea is to assume the following a priori assumption on the absolute temperature θ(t, x) 1 Θ ≤ θ(t, x) ≤ 2Θ, (1.15) 2 for (t, x) ∈ [0, T ] × R. Then by some delicate energy type estimates and by using the argument initiated in [21], we can deduce an uniform (with respect to the time variable

lower and upper  t)

θ−1 in terms of bound on v(t, x) and some uniform energy estimates on v − 1, u, √γ−1 (t) 1

H (R)

 

θ0 −1 , inf v (x), and sup v (x). At the end, to extend the solution globally

v0 − 1, u0 , √

0 0 γ−1 H 1 (R) x∈R x∈R

in time, we only need to close the a priori assumption (1.15) where we need the smallness of γ − 1. For Theorem 1.3, since the viscosity coefficient µ is assumed to be a positive constant, even though the heat conductivity coefficient κ(v, θ) may depend on v and θ and there is a nonlocal term Φvx in the momentum equation, the argument in [25] can be adopted to deduce an explicit formula for the specific volume v(t, x). Based on this formula, we can deduce a lower bound on v(t, x) and from which and the maximum principle for the absolute temperature θ(t, x), we can deduce a lower bound on θ(t, x). Having obtained the lower bound estimates on both v(t, x) and θ(t, x), we can deduce the desired upper bound on v(t, x) by employing the explicit formula for v(t, x) again provided that κ(v, θ) satisfies (1.12). With these estimates in hand, the upper bound on θ(t, x) can be obtained by following the argument used in the proof of Theorem 1.1 by considering the case κ(v, θ) is uniformly bounded for 0 < V0−1 ≤ v ≤ V0 , θ ≥ Θ−1 0 (such a bound can depends on V0 and Θ0 ) and the case lim κ(v, θ) = +∞ for 0 < V0−1 ≤ v ≤ V0 , θ ≥ Θ−1 0 θ→+∞

respectively. Before concluding the introduction, we point out that there are many results on the construction of global solutions to the NSP system (1.1). In particular, recently, the global existence of smooth small perturbative solutions away from vacuum with the optimal time decay estimates was obtained in [26] for the isentropic flow, and in [37], [16] for the non-isentropic flow. There, it is observed that the electric field affects the large time behavior of the solution so 1 3 that the momentum decays at the rate (1 + t)− 4 which is slower than the rate (1 + t)− 4 for the compressible Navier-Stokes system, while the density tends to its asymptotic state at the 3 rate (1 + t)− 4 just like the compressible Navier-Stokes system. Moreover, the global existence of strong solution in Besov type space was obtained in [15]. On the other hand, it is quite different for the compressible Euler-Poisson (EP) system. In fact, it was shown in [14] that the long time convergence rate of global irrotational solution is enhanced by the dispersion effect due to the coupling of electric field, namely, both density and velocity tend to the equilibrium constant state at the rate (1 + t)−p for any p ∈ (1, 32 ). Note that even though most of the results for the small perturbative solutions are considered for the case when µ, ν, and κ are constants, it is straightforward to show that they hold when µ, ν, and κ are smooth functions of density and temperature. Finally, for the results with large initial data, the existence of re-normalized solutions to

One-dimensional Compressible Navier-Stokes-Poisson Equation with Large Data

7

the NSP system are obtained in [6], [33], [38]. Note that for the compressible NSP system related to the dynamics of selfgravitating gases stars, some existence results on the weak solution (renormalized solution) were given in [8], [9], [38]. Since the analysis in these works is based on the weak convergence argument, only isentropic polytropic gas was studied with a special requirement on the range of adiabatic exponent, i.e. γ > 23 with constant viscosity coefficient. For the non-isentropic case, even for the compressible Navier-Stokes system, the only available global existence theory for large data is the construction of the so called “variational solution”, cf. [11]. The rest of the paper is organized as follows. The proofs of Theorem 1.1 and Theorem 1.2 will be given in Section 2 and Section 3 respectively. Notations: O(1) or Ci (i ∈ N) stands for a generic positive constant which is independent of t and x, while Ci (·, · · · , ·) (i ∈ N) is used to denote some positive constant depending on the arguments listed in the parenthesis. Note that all these constants may vary from line to line. k · ks represents the norm in H s (R) with k · k = k · k0 and for 1 ≤ p ≤ +∞, Lp (R) denotes the standard Lebesgue space.

2

The proof of Theorem 1.1

To prove Theorem 1.1, we first define the following function space for the solution to the Cauchy problem (1.1), (1.3)

X(0, T ; M0 , M1 ; N0 , N1 ) =

              

   v − 1, u, θ − 1 (t, x) ∈ C 0 0, T ; H 1 (R)
      
  2 2  u , θ (t, x) ∈ L 0, T ; H (R) x x . v, u, θ, Φ (t, x)
 0 0, T ; H 3 (R)  Φ(t, x) ∈ C     M ≤ v(t, x) ≤ M , N ≤ θ(t, x) ≤ N  0

1

0

1

(2.1) Here T > 0, M1 ≥ M0 > 0, N1 ≥ N0 > 0 are some positive constants. Under the assumptions given in either Theorems 1.1 or 1.2, we can get the following local existence result. Lemma 2.1 (Local existence) Under the assumptions in either Theorems 1.1 or 1.2, there exists a sufficiently small positive constant t1 , which depends only on V , V , Θ, Θ and k(v0 − 1, u0 , θ0 − 1)k1 , such that the Cauchy  problem (1.1), (1.3)  admits a unique smooth solution 1 1 (v(t, x), u(t, x), θ(t, x), Φ(t, x)) ∈ X 0, t1 ; 2 V , 2V ; 2 Θ, 2Θ and (v(t, x), u(t, x), θ(t, x), Φ(t, x)) satisfies   0 < V2 ≤ v(t, x) ≤ 2V , (2.2)  0 < Θ2 ≤ θ(t, x) ≤ 2Θ, 

sup k(v − 1, u, θ − 1, Φx )(t)k1



≤ 2k(v0 − 1, u0 , θ0 − 1, Φ0 )k1 ,

(2.3)

[0,t1 ]

and lim

|x|→∞





v(t, x) − 1, u(t, x), θ(t, x) − 1, Φx (t, x) = (0, 0, 0, 0).

(2.4)

8

Zhong Tan, Tong Yang, Huijiang Zhao, and Qingyang Zou

Lemma 2.1 can be proved by the standard iteration argument as in [32] for the one-dimensional compressible Navier-Stokes system, we thus omit the details for brevity. Now we give some properties on the local solution (v(t, x), u(t, x), θ(t, x), Φ(t, x)) constructed above. Noticing that 



u+

Φx v





= ux + t x

Φx v

 

= ux + (1 − v)t = ux − vt = 0, x t

we have the following lemma from (2.4). Lemma 2.2 Under the conditions in Lemma 2.1, we have Φx (t, x) u(t, x) = − v(t, x) 



.

(2.5)

t

Now we turn to prove Theorem 1.1. Recall that µ(v) = v −a , κ(v, θ) = θb , and the constitutive equations (1.2), the Cauchy problem (1.1), (1.3) can be rewritten as          

vt − ux = 0, ut + p(v, θ)x =



ux



+ Φvx , v 1+a x  b  u2x θ θx , + 1+a v v x

 Cv θt + p(v, θ)ux =         Φx  = 1 − v,  v x

(v(0, x), u(0, x), θ(0, x)) = (v0 (x), u0 (x), θ0 (x)),

(2.6)

lim Φ(t, x) = 0,

|x|→+∞

lim (v0 (x), u0 (x), θ0 (x)) = (1, 0, 1). (2.7)

|x|→+∞

Suppose that the local solution (v(t, x), u(t, x), θ(t, x), Φ(t, x)) constructed in Lemma 2.1 has been extended to t = T ≥ t1 and satisfies the a priori assumption V 0 ≤ v(t, x) ≤ V 1 ,

(H1 )

Θ0 ≤ θ(t, x) ≤ Θ1

for all x ∈ R, 0 ≤ t ≤ T, and some positive constants 0 < Θ0 ≤ Θ1 , 0 < V 0 ≤ V 1 , we now deduce certain a priori estimates on (v(t, x), u(t, x), θ(t, x), Φ(t, x)) which are independent of Θ0 , Θ1 , V 0 , V 1 but may depend on T . The first one is concerned with the basic energy estimate. For this, note that η(v, u, θ) = Rφ(v) +

u2 Rφ(θ) + , 2 γ−1

with φ(x) = x − ln x − 1,

is a convex entropy to (2.6) which satisfies 

η(v, u, θ)t +

Rθ −R u v

 



− x

uux (θ − 1)θx + 1+a v vθ1−b

(



+ x

u2x

θx2 + v 1+a θ vθ2−b

)

=

uΦx . v

(2.8)

With (2.8), since uΦx = v



uΦ Φ + v v



Φx v

1 − 2 x

 t

"

Φx v

2 #

Φvx + 2 u+ v t 



Φx v



, t

we can deduce the following lemma by integrating (2.8) with respect to t and x over [0, T ] × R and from (2.5).

One-dimensional Compressible Navier-Stokes-Poisson Equation with Large Data

9

Lemma 2.3 (Basic energy estimates) Let the conditions in Lemma 2.1 hold and suppose that the local solution (v(t, x), u(t, x), θ(t, x), Φ(t, x)) constructed in Lemma 2.1 satisfies the a priori assumption (H1 ), then we have for 0 ≤ t ≤ T that 1 η(v, u, θ) + 2

Z R



Φx v

1 η(v0 , u0 , θ0 ) + 2

Z

= R



2 !

u2x

θx2 + v 1+a θ vθ2−b

Z tZ

(t, x)dx + 0

Φ0x v0

R

!

(τ, x)dxdτ

2 !

(x)dx.

(2.9)

The next estimate is concerned with a lower bound estimate on θ(t, x) in terms of the lower bound on v(t, x). Lemma 2.4 Under the assumptions in Lemma 2.3, we have for a < 1 that

1 1−a 1

≤ O(1) + O(1)

v ∞ , θ(t, x) LT,x



x ∈ R, 0 ≤ t ≤ T.

(2.10)

Proof: First of all, (2.6)3 implies 1 θ

 

Cv

t

u2x Rux 2θ1+b = − 2 1+a + − θ v vθ v "

= +

# ! b  

θ v

1 θ

(

−

"

  2

1 θ

+ x

  2 2θ1+b 1

x x

v

θ

+

x

θb v

!  #

1 v 1+a θ2

1 θ



x x

Rθv a ux − 2

2 )

(2.11)

R2 . 4v 1−a

Set h(t, x) =

1 R2 t − θ 4Cv

1−a

1



v ∞ , LT,x

we can deduce that h(t, x) satisfies  b    Cv ht ≤ θv hx , x



h(0, x) =

1 θ0 (x)

≤

x ∈ R, 0 ≤ t ≤ T, (2.12)

1 Θ,

and the standard maximum principle for parabolic equation implies that h(t, x) ≤ all (t, x) ∈ [0, T ] × R. That is, for x ∈ R, 0 ≤ t ≤ T 1 R2 t − θ 4Cv

1−a

1 1



v ∞ ≤ Θ. L

1 Θ

holds for

(2.13)

T,x

This is (2.10) and the proof of Lemma 2.4 is completed. To use Y. Kanel’s method

to deduce a lower bound and an upper bound on v(t, x), we need

vx to deduce an estimate on v1+a

, which is the main concern of our next lemma. It is worth to pointing out that it is in this step that we ask the viscous coefficient µ depends only on v.

10

Zhong Tan, Tong Yang, Huijiang Zhao, and Qingyang Zou

Lemma 2.5 Under the assumptions in Lemma 2.3, we have !

Z tZ

vx 2

+

v 1+a 0

R

θvx2 + g(v)(v − 1) dxds v 3+a 



≤ kv0x k2 + k(v0 − 1, u0 , θ0 − 1, Φ0x )k2 +

u2x dxds + O(1) v 1+a

Z tZ 0

R

θx2

Z tZ 0

R

v 1+a θ

dxds, (2.14)

and

Z v dz

g(v) =

z 1+a

1

1 − v −a . a

=

Proof: Notice that vx





v 1+a

vt



=



v 1+a

t



= x

we have by multiplying the above identity by respect to t and x over [0, T ] × R that

2 1

vx + 2 v 1+a

ux v 1+a vx v 1+a



= ut + p(v, θ)x − x

Φx , v

and integrating the resulting equation with

Rθvx2 dxds 3+a 0 R v Z tZ Z tZ Z tZ Rθx vx ut vx vx Φx ≤ O(1)kv0x k2 + dxds dxds dxds . + − 2+a 1+a 1+a v 0 R v 0 R v 0 R v



Z tZ

|

{z

}

I1

|

{z

}

I2

|

{z

(2.15)

}

I3

Now we estimate I1 , I2 and I3 term by term. First, we have from (2.6)4 and the CauchySchwarz inequality that Z tZ



g(v)x

I3 = 0

R

Φx dxds = − v 

Z tZ



g(v) 0

R

Φx v



dxds = −

Z tZ 0

x

g(v)(1 − v)dxds ≥ 0,

R

(2.16) and I1 ≤

1 2

Z tZ 0

R

Rθvx2 dxds + O(1) v 3+a

Z tZ 0

R

θx2 v 1+a θ

dxds.

(2.17)

As to I2 , we have from (2.9) that t uvx t vx I2 = dx − u 1+a dxds 1+a 0 v R v 0 R t   Z tZ Z uvx ux 2 ≤ dx + O(1) k(u , v )k − u dxds 0 0x 1+a v 1+a x R v 0 R

2 Z tZ 1 vx u2x 2

≤ 1+a + O(1) k(v0 − 1, v0x , u0 , θ0 − 1, Φ0x )k + dxds. 1+a 2 v 0 R v

Z

Z Z





(2.18)

Inserting (2.16)-(2.18) into (2.15), we can deduce (2.14) immediately. This completes the proof of Lemma 2.5. To bound the two terms on the right hand side of (2.14), we now estimate the following lemma.

u2x 0 R v 1+a dxds

RtR

in

One-dimensional Compressible Navier-Stokes-Poisson Equation with Large Data

11

Lemma 2.6 Under the assumptions in Lemma 2.3, we have ku(t)k2 +

Z tZ 0

R

u2x dxds ≤ O(1)k(v0 − 1, u0 , θ0 − 1, Φ0x )k2 + O(1) v 1+a

Z tZ 0

R

(θ − 1)2 dxds. (2.19) v 1−a

Proof: Multiplying (2.6)2 by u, we have by integrating the resulting equation with respect to t and x over [0, T ] × R that u2x dxds 1+a 0 R v   Z tZ Z tZ Z tZ R(θ − 1)ux uΦx 1 2 ≤ O(1)ku0 k + R 1− ux dxds + dxds + dxds(2.20) . v v 0 R 0 R v 0 R 1 ku(t)k2 + 2

Z tZ

|

{z

}

I4

|

{z

}

I5

|

{z I6

}

From the basic energy estimate (2.9) and the Cauchy-Schwarz inequality, we can bound Ij (j = 4, 5, 6) as follows: I6 ≤

Z t 0

 

φx

ku(s)k

v

Z tZ



1 vt dxds = R v 

R 1−

I5 = 0



2 (s)

ds ≤ C(T )k(u0 , v0 − 1, θ0 − 1, Φ0x )k ,

ZR

φ(v)dx −

=R

Z

 R

Z

t

φ(v)dx

0

φ(v0 )dx R

R

≤ O(1)k(u0 , v0 − 1, θ0 − 1, Φ0x )k2 , t (θ − 1)2 u2x 1 t dxds + O(1) dxds. 1−a 2 0 R v 1+a 0 R v Substituting the above estimates into (2.20), we can deduce (2.19) and complete the proof of the lemma.

Z Z

Z Z

I4 ≤

To bound the terms appearing on the right hand side of (2.19) and (2.14), we need the following Lemma 2.7 Under the assumptions in Lemma 2.3, we have for b 6= 0, −1 that Z t

max |θ(s, x)|b ds ≤ C(T ),

(2.21)

0 x∈R

Z t





(2.22)





(2.23)

max |θ(s, x)|b+1 ds ≤ C(T ) 1 + kθkL∞ , T,x

0 x∈R

and

Z t

max |θ(s, x)|b+1 ds ≤ C(T ) 1 + kvkL∞ . T,x

0 x∈R

Proof: We only prove (2.22) because (2.21)and (2.23) can be proved similarly. From the argument used in [25], we have from the basic energy estimate (2.9), the Jenssen inequality that from each i ∈ Z, there are positive constants A0 > 0, A1 > 0 such that A0 ≤

Z i+1

Z i+1

v(t, x)dx, i

i

θ(t, x)dx ≤ A1 ,

∀t ∈ [0, T ].

(2.24)

12

Zhong Tan, Tong Yang, Huijiang Zhao, and Qingyang Zou

Hence, there exist ai (t) ∈ [i, i + 1], bi (t) ∈ [i, i + 1] such that A0 ≤ v(t, ai (t)), θ(t, bi (t)) ≤ A1 . Define

Z θ

s

g1 (θ) =

b−1 2

(2.25)

 2  b+1 θ 2 −1 , b+1

ds =

1

for each x ∈ R, there exists an integer i ∈ Z such that x ∈ [i, i + 1] and we can assume without loss of generality that x ≥ bi (t). Thus Z x

g1 (θ(t, y))y dy

g1 (θ(t, x)) = g1 (θ(t, bi (t))) + bi (t)

Z i+1 b−1 ≤ O(1) + θ 2 θx dx i

≤ O(1) +

θx2 dx vθ2−b

Z R 1 2

Z

≤ O(1) + kθkL∞

T,x

R

! 1 Z 2 i+1

 21

vθdx i

θx2 dx vθ2−b

!1

2

.

The above estimate and (2.9) give (2.22) and then complete the proof of the lemma. As a direct corollary of (2.21)-(2.23), we have Corollary 2.1 Under the conditions in Lemma 2.3, we have Z tZ 0

R





(θ − 1)2 dxds ≤ O(1) θ1−b

L∞ T,x

.

(2.26)

Proof: In fact (2.9) together with (2.21) imply that Z tZ 0

2

(θ − 1) dxdτ ≤ O(1)

R

Z tZ

(θ + 1)φ(θ)dxdτ 0

≤ O(1) = O(1)

R

Z t

max θ(τ, x)dτ + O(1)

0 x∈R Z t





max θ1−b θb dτ + O(1)

0 x∈R

Z t



≤ O(1) θ1−b ∞ max θb (τ, x)dτ + O(1) LT,x 0 x∈R



≤ O(1) θ1−b ∞ + O(1), LT,x

and this completes the proof of corollary. Having obtained (2.26), we can deduce that Z tZ 0

R

1 1−a (θ − 1)2

dxdτ ≤

v ∞ v 1−a LT,x



Z tZ 0

R





(θ − 1)2 dxdτ ≤ O(1) θ1−b

L∞ T,x

1−a

1



v ∞ . LT,x

(2.27)

One-dimensional Compressible Navier-Stokes-Poisson Equation with Large Data

13

On the other hand, from (2.9), we have Z tZ 0

R

θx2 dxdτ = θv 1+a

Z tZ 0

R

θx2 1 dxdτ 2−b a vθ v θb−1

a Z tZ

1

1−b

≤

θ

∞ L v ∞ LT,x

T,x

0

θx2 dxdτ vθ2−b

R

a

2 1





≤ O(1)k(v0 − 1, u0 , θ0 − 1, Φ0x )k

θ 1−b ∞ . LT,x v L∞ T,x

(2.28)

Substituting (2.27) and (2.28) into (2.19) and (2.14), we have Corollary 2.2 Under the assumptions in Lemma 2.3, we have 2

ku(t)k +

Z tZ 0

R



1 a u2x

1−b 2

θ dxdτ ≤ O(1) k(v − 1, u , θ − 1, Φ )k +O(1)

∞ , (2.29) 0 0 0 0x

v ∞ LT,x v 1+a LT,x



!

Z tZ

vx 2

+

v 1+a 0

R

θvx2 + g(v)(v − 1) dxdτ v 3+a

a

1−a !



1 1

1−b

≤ O(1) k(v0 − 1, u0 , θ0 − 1, Φ0x )k2 + O(1) + θ

∞ .

v ∞

v ∞ LT,x L L T,x

(2.30)

T,x

Now we apply Y. Kanel’s approach to deduce a lower bound and an upper bound on v(t, x)

in terms of θ1−b ∞ . To this end, set LT,x

Ψ(v) =

Z vp φ(z)

z 1+a

1

dz.

(2.31)

Note that there exist positive constants A2 , A3 such that 1





|Ψ(v)| ≥ A2 v −a + v 2 −a − A3 .

(2.32)

Since Z x

|Ψ(v)| =



Ψ(v(t, y))y dy

−∞ Z p φ(v) ≤ 1+a vx dx R v



q

vx



≤ φ(v) 1+a

v !

a

1−a !

1

1 2

1 2

2 1−b

≤ O(1) 1 +

∞ ,

v ∞ + v ∞ θ LT,x L L T,x

(2.33)

T,x

we have from (2.32) and (2.33) that

a 1

1 −a 2

≤ O(1) 1 +

v ∞ + kvkL∞ T,x LT,x

Thus if

1 3

!

a

1−a !

1

1 2

1 2

2 1−b





∞ .

v ∞ + v ∞ θ LT,x LT,x LT,x

< a < 12 , we can deduce from (2.34)

(2.34)

14

Zhong Tan, Tong Yang, Huijiang Zhao, and Qingyang Zou

Corollary 2.3 Under the conditions in Lemma 2.3, if we assume further that we have !

1 1

1−b 3a−1 ≤ O(1) 1 + θ ∞ , LT,x v(t, x) and 2a



v(t, x) ≤ O(1) 1 + θ1−b (3a−1)(1−2a) ∞

1 3

< a < 21 , then (2.35)

!

(2.36)

LT,x

hold for any (t, x) ∈ [0, T ] × R. Consequently, (2.29) and (2.30) can be rewritten as 2

ku(t)k +

Z tZ 0

,

(2.37)

!

Z tZ

vx 2

+

v 1+a 0

R

!

2a u2x

1−b 3a−1 dxdτ ≤ O(1) 1 + θ

∞ LT,x v 1+a

2a θvx2

1−b 3a−1 + g(v)(v − 1) dxdτ ≤ O(1) 1 + θ

∞ LT,x v 3+a

R

!

.

(2.38)

To get an upper bound on θ(t, x), we need also the estimate on kux (t)k which is given in the following lemma. Lemma 2.8 Under the conditions listed in Lemma 2.3, we have for 0 ≤ t ≤ T , that u2xx dxdτ 1+a 0 R v ≤ O(1) k(v0 − 1, u0 , θ0 − 1, Φ0x )k2 kux (t)k2 + kv(t)k2 +

2a2



1 + θ1−b (3a−1)(1−2a) ∞



+O(1) θ2−b

L∞ T,x

!

LT,x

2(2a−2a2 +1) (3a−1)(1−2a)





Z tZ

+O(1) 1 + θ1−b

L∞ T,x

!

.

(2.39)

Proof: By differentiating (2.6)2 with respect to x, multiplying the resulting identity by ux , and integrating the result with respect to t and x over [0, T ] × R, we have 2

kux (t)k +

u2xx dxdτ + kv − 1k2 v 1+a

Z tZ 0

2

R

≤ O(1)ku0x k + 2

Z tZ

Z tZ

uxx p(v, θ)x dxdτ + 2(1 + a)

| 0

R

0

{z

}

I7

R

|

ux vx uxx dxdτ . v 2+a

{z

(2.40)

}

I8

For I7 , we have from (2.9) that Z tZ



I7 = 2R 1 4 1 ≤ 4 ≤

θx θvx − 2 dxdτ v v 

uxx 0 R Z tZ u2xx dxdτ 1+a 0 R v Z tZ u2xx dxdτ 1+a 0 R v

θx2

Z tZ

+ O(1) 0

v

R



+ O(1) θ2−b

L∞ T,x



+O(1) 1 + θ1−b

7a−4a2 −1 (3a−1)(1−2a)

L∞ T,x

Z tZ

dxdτ + O(1) 1−a

0

R

θ2 vx2 dxdτ v 3−a

2a2

1−b (3a−1)(1−2a) 1 + θ ∞

!

LT,x

!

.

(2.41)

One-dimensional Compressible Navier-Stokes-Poisson Equation with Large Data

15

Here we have used the fact that θx2

Z tZ 0

R

v

Z tZ

dxdτ = 1−a

0

R

θx2 a 2−b v θ dxdτ vθ2−b

Z tZ



≤ O(1)kvkaL∞ θ2−b T,x ≤

O(1)kvkaL∞ T,x

L∞ T,x

0

R

θx2 dxdτ vθ2−b



2−b

θ

L∞ T,x

2a2



1 + θ1−b (3a−1)(1−2a) ∞



≤ O(1) θ2−b

L∞ T,x

!

LT,x

,

and Z tZ 0

R

θ2 vx2 dxdτ = v 3−a ≤

vx2

Z tZ 0

R

θ2

v 2+2a v 1−3a

Z tZ 

dxdτ 

2

max θ (s, x)ds 0

x∈R

R

2a

≤ O(1) 1 + θ1−b 3a−1 ∞

!

2a

≤ O(1) 1 + θ1−b 3a−1 ∞

!

5a−1

≤ O(1) 1 + θ1−b 3a−1 ∞

!

5a−1

!

3a−1 kvkL ∞

LT,x

3a−1 kvkL ∞ T,x

LT,x

3a−1 kvkL ∞

LT,x



≤ O(1) 1 + θ1−b 3a−1 ∞ LT,x





v 2+2a

R

Z t

T,x

!

vx2

Z

3a−1 kvkL ∞ T,x

max θ2 (s, x)ds



x∈R

0

Z t

max θ

1−b 1+b

x∈R

0

Z t

T,x

dx dτ

θ



(s, x)ds

max θ1+b (s, x)ds



x∈R

0



3a−1 1 + kvkL∞ kvkL ∞ T,x



T,x

7a−4a2 −1

!

≤ O(1) 1 + θ1−b (3a−1)(1−2a) ∞ LT,x

,

where (2.9), (2.21)-(2.23), and (2.38) are used. As for I8 , since (2.36), (2.37) together with the Sobolev inequality imply Z t 0

kux (τ )k2L∞ dτ ≤ x

Z t

kux (τ )kkuxx (τ )kdτ

0

Z t

≤

2

kux (τ )k dτ

 1 Z t 2

0

≤

kvk1+a L∞ T,x

2

1 2

kuxx (τ )k dτ

0

2 ! 21 Z t

ux



1+a (τ ) dτ

v

0

2

2 ! 21 Z t

uxx



1+a (τ ) dτ 0

v

2

! Z

2 ! 21 3a

t u

xx

1−b (3a−1)(1−2a)

≤ O(1) 1 + θ ∞ , 1+a (τ ) dτ

L T,x

we can deduce from (2.35)-(2.38) that 1 I8 ≤ 4

Z tZ 0

R

u2xx dxdτ + O(1) v 1+a

Z tZ 0

R

u2x vx2 dxdτ v 3+a

0

v

2

(2.42)

16

Zhong Tan, Tong Yang, Huijiang Zhao, and Qingyang Zou 1 ≤ 4

Z tZ

1 ≤ 4

Z tZ

1 ≤ 4

Z tZ

1 2

Z tZ

≤

0

R

0

R

0

R

0

R

u2xx dxdτ + O(1) v 1+a

Z t

u2

x

1−a

0 v

vx2

Z R

L∞ x

v 2+2a

dxdτ

2a

1 1−a u2xx

1−b 3a−1

1 + θ dxdτ + O(1)

∞

v ∞ LT,x v 1+a LT,x



2

2a−2a +1

u2xx

1−b (3a−1)(1−2a) dxdτ + O(1) 1 + θ

∞

LT,x v 1+a 2(2a−2a2 +1) (3a−1)(1−2a)

u2xx

1−b dxdτ + O(1) 1 + θ

∞ LT,x v 1+a

!Z

t

0

dτ kux (τ )k2L∞ x

! Z Z t 0

R

u2xx dxdτ v 1+a

!1 2

!

.

(2.43)

Putting (2.40), (2.41), and (2.43) together and noticing that 2(2a − 2a2 + 1) > 7a − 4a2 − 1 imply (2.39), and this completes the proof of Lemma 2.8. Now we turn to deduce the upper bound on θ(t, x). Lemma 2.9 Under the conditions in Lemma 2.3, we have kθkL∞ T,x

  



u2 

u2

x   dτ . + 2x + kθk2L∞ ≤ O(1) 1 +

1+a

x 

v

∞ v ∞  0 L L  

Z t

x

(2.44)

x

Proof: From (2.6)3 , it is easy to see that for each p > 1, h

Cv (θ − 1)2p (

=

2p(θ −

i

+ 2p(2p − 1)(θ − 1)2(p−1)

t 2p−1 1) θb θ

) x

v

+ x

θb θx2 v

2p(θ − 1)2p−1 2 2pRθ ux − ux (θ − 1)2p−1 . v 1+a v

(2.45)

Integrating (2.45) with respect to x over R, we have 

Cv kθ − 1k2p L2p

 t

≤ 2p

Z R

|

u2x (θ − 1)2p−1 dx − 2pR v 1+a {z

}

I9

Z R

θux (θ − 1)2p−1 dx . v

|

{z

(2.46)

}

I10

Since



I9 ≤ 2pO(1)kθ −

2 2p−1 ux 1kL 2p 1+a

v

2p L

I10 ≤ 2pO(1)kθ −

2p−1

θux 1kL 2p v L2p

,

hold for some positive constant O(1) independent of p, we have kθ − 1kL2p ≤ O(1) + O(1)

Z t 0



u2

x

1+a

v

L2p



θux

+ v

!

dτ.

(2.47)

L2p

Letting p → ∞ in (2.47) and by exploiting the Cauchy inequality, we can deduce (2.44) immediately and the proof of Lemma 2.9 is complete.

One-dimensional Compressible Navier-Stokes-Poisson Equation with Large Data

17

We are now ready to use (2.35), (2.36), and (2.44) to deduce a lower bound and an upper bound on θ(t, x). Firstly, we have from (2.42) and (2.39) that Z t 0

kux (s)k2L∞ ds x

3a



≤ O(1) 1 + θ1−b (3a−1)(1−2a) ∞

!

LT,x

"

1

× θ2−b 2 ∞

LT,x

a2

1−b (3a−1)(1−2a) 1 + θ ∞

2a−2a2 +1

1−b (3a−1)(1−2a) + 1 + θ ∞

!

LT,x

1

≤ O(1) θ2−b 2 ∞



1 + θ1−b

LT,x

3a+a2 (3a−1)(1−2a)

!

L∞ T,x

LT,x

#

5a−2a2 +1

1−b (3a−1)(1−2a) + O(1) θ ∞ + O(1). LT,x

(2.48)

Thus, we have from (2.35)-(2.36), (2.48) that 



u2

u2

x  + 2x  dτ

1+a

v ∞

v 0 L∞ Lx x

1+a

2 ! Z t

1

1

≤ O(1) kux (τ )k2L∞ dτ

v ∞ + v ∞ x Z t

LT,x

0

LT,x

2

≤ O(1) 1 + θ1−b 3a−1 ∞

!Z

LT,x

0

t

kux (τ )k2L∞ dτ x

a2 −a+2

1−b (3a−1)(1−2a) 1 + θ ∞

1

≤ O(1) θ2−b 2 ∞

LT,x

!

LT,x

3+a−2a2

1−b (3a−1)(1−2a) + O(1) θ ∞ + O(1), LT,x

(2.49)

and Z t

Z t

2

max θ (s, x)ds ≤





max θ1−b (s, x)θb+1 (s, x) ds

0 x∈R

0 x∈R

Z t

1−b max θ1+b (s, x)ds ≤ θ ∞ LT,x 0 x∈R

 

≤ O(1) θ1−b ∞ 1 + kvkL∞ T,x LT,x

7a−6a2 −1

1−b (3a−1)(1−2a) ≤ O(1) 1 + θ ∞

!

LT,x

.

(2.50)

Inserting (2.49) and (2.50) into (2.44) yields kθkL∞ T,x

1

≤ O(1) + O(1) θ2−b 2 ∞

LT,x





3+a−2a2

a2 −a+2

1−b (3a−1)(1−2a) 1 + θ ∞

!

LT,x





7a−6a2 −1

+O(1) θ1−b (3a−1)(1−2a) + O(1) θ1−b (3a−1)(1−2a) ∞ ∞ LT,x

LT,x

1

≤ O(1) + O(1) θ2−b 2 ∞

LT,x





a2 −a+2

1−b (3a−1)(1−2a) 1 + θ ∞ LT,x

3+a−2a2

+O(1) θ1−b (3a−1)(1−2a) . ∞ LT,x

Based on the estimate (2.10), (2.35), (2.36) and (2.51), we have

!

(2.51)

18

Zhong Tan, Tong Yang, Huijiang Zhao, and Qingyang Zou

Corollary 2.4 Under the assumptions in Lemma 2.3, we further assume that one of the following conditions holds (i). 1 ≤ b
0 such that θ(t, x) ≥ Θ−1 (2.53) 1 > 0, ∀(t, x) ∈ [0, T ] × R.

Moreover, (2.35), (2.36), (2.53) together with the fact that b ≥ 1 imply that there exists a positive constant V1 > 0, which may depends on T , such that V1−1 ≤ v(t, x) ≤ V1 , ∀(t, x) ∈ [0, T ] × R.

(2.54)

Thus to prove (2.52), we only need to deduce the upper bound on θ(t, x). For this purpose, 2a we have from the fact 1 ≤ b < 1−a < 2, (2.53), and (2.51) that kθkL∞ ≤ O(1) + O(1) kθk T,x 

≤ O(1) 1 + kθk

(a2 −a+2)(b−1)

1 + 1θ (3a−1)(1−2a) ∞

2−b 2 L∞ T,x

2−b 2 L∞ T,x

LT,x

!

(3+a−2a2 )(b−1)

(3a−1)(1−2a) + O(1) 1θ ∞ LT,x



(2.55)

.

From (2.55) and the fact that 0 < 2−b 2 < 1, one can easily deduce an upper bound on θ(t, x). 2a This completes the proof of (2.52) for the case 1 ≤ b < 1−a . When b < 1, we have from (2.51) that 2−b

kθkL∞ ≤ O(1) + O(1)kθkL2∞ T,x

+

T,x

(a2 −a+2)(1−b) (3a−1)(1−2a)

(3+a−2a2 )(1−b) (3a−1)(1−2a) + O(1)kθkL∞ . T,x

(2.56)

From (2.56) and the assumption (ii) of Corollary 2.4, we can deduce an upper bound on θ(t, x). With this, the lower and upper bound on v(t, x) can be deduced from (2.35) and (2.36). And then (2.10) implies the lower bound on θ(t.x). This completes the proof of the corollary. With Corollary 2.4, Theorem 1.1 follows from the standard continuation argument.

One-dimensional Compressible Navier-Stokes-Poisson Equation with Large Data

3

19

The proof of Theorem 1.2

First of all, the local solvability of the Cauchy problem (1.1),(1.3) in the function space X (0, t1 ;  1 1 2 V , 2V ; 2 Θ, 2Θ with t1 depending on V , V , Θ, Θ, k(v0 − 1, v0 , θ0 − 1, Φ0x )k1 can be proved as in Lemma 3.1. Suppose this solution (v(t, x), u(t, x), θ(t, x), Φ(t, x)) is extended to t = T ≥ t1 . To apply the continuation argument for global existence, we first set the following a priori estimate: 1 Θ ≤ θ(t, x) ≤ 2Θ, (t, x) ∈ [0, T ] × R. 2

(H2 )

Here without loss of generality, we can assume that 0 < Θ < 1 < Θ. Note that the smallness of γ −1 is needed to close the a priori estimate, the generic constants used later are independent of γ − 1 and whenever the dependence on this factor will be clearly stated in the estimates. Similar to Lemma 2.3 we have the following basic energy estimate. Lemma 3.1 Under the conditions in Theorem 1.2, we have for 0 ≤ t ≤ T that Z R

(

u2 R Φ2 Rφ(v) + + φ(θ) + x2 2 γ−1 2v Z

= R

)

µ(v)u2x κ(v, θ)θx2 + vθ vθ2

Z tZ

(t, x)dx + 0

u2 R Φ2 Rφ(v0 ) + 0 + φ(θ0 ) + 0x2 2 γ−1 2v0

R

!

dxdτ

!

(x)dx.

(3.1)

Here, as in Section 2, φ(x) = x − ln x − 1.



x Now, we turn to deduce an estimate on µ(v)v v . For this, similar to Lemma 2.5, we can deduce

Z tZ

µ(v)vx 2

+

v

t µ(v)θvx2 g(v)(1 − v)dxdτ dxdτ + v3 0 R 0 R Z tZ Z tZ µ(v)u2x µ(v)θx2 ≤ O(1)kv0x k2 + O(1) dxdτ +O(1) dxdτ . v vθ 0 R 0 R

Z Z

|

{z

}

J1

|

{z

(3.2)

}

J2

If the a priori estimate (H2 ) holds, we have from (3.1) and the assumptions imposed on κ(v, θ) in Theorem 1.2 that J1 ≤ O(1)

Z tZ 0

R

2 θ0 − 1 µ(v)u2x

√ dxdτ ≤ O(1) v − 1, u , , Φ 0 0x ,

0 vθ γ−1





(3.3)

and κ(v, θ)θx2 θµ(v) dxdτ vθ2 κ(v, θ) 0 R

Z tZ

µ(v) κ(v, θ)θx2

dxdτ ≤ O(1)

κ1 (v) L∞ 0 R vθ2

J2 ≤

Z tZ

2 µ(v) θ0 − 1



√ ≤ O(1) v − 1, u , , Φ . 0 0x

0 γ−1 κ1 (v) L∞



Putting (3.2), (3.3) and (3.4) together, we obtain



(3.4)

20

Zhong Tan, Tong Yang, Huijiang Zhao, and Qingyang Zou

Lemma 3.2 Under the assumptions in Lemma 3.1 and the a priori assumption (H2 ), we have

Z tZ

µ(v)vx 2

+

v 0



R





2

µ(v)vx2 θ0 − 1

1 + µ(v)

v0 − 1, u0 , √ . dxdτ ≤ O(1) , Φ , v 0x 0x

v3 γ−1 κ1 (v) L∞ 



T,x

(3.5) Having obtained (3.1) and (3.5), we can use Y. Kanel’s argument, cf. [21], to deduce the lower and upper bounds on v(t, x) as follows. Lemma 3.3 Under the assumptions in Theorem 1.2 and Lemma 3.2,

  there exists a positive

θ −1 0 constant V2 ≥ 1, which depends only on v0 − 1, u0 , √γ−1 , Φ0x , v0x , V , V , Θ, and Θ, but is independent of T , such that V2−1 ≤ v(t, x) ≤ V2 , (t, x) ∈ [0, T ] × R. Proof: Define Ψ(v) =

Z vp φ(z) 1

z

µ(z)dz,

(3.6)

φ(z) = z − ln z − 1,

and notice that Z x Ψ(v(t, y))y dy |Ψ(v)| = −∞ Z q φ(v) µ(v)vx dx ≤ v R

1 µ(v)v x ≤ kφ(v)kL2 1

v  1



2  2

µ(v) θ − 1 0



  ≤ O(1) v0 − 1, u0 , √ , Φ0x , v0x 1 + . γ−1 κ1 (v) L∞ T,x

It is straightforward to deduce (3.6) from the assumptions in Theorem 1.2. This completes the proof of the lemma. The next lemma is about the estimate on kux (t)k. Lemma 3.4 Under the assumptions in Lemma 3.3, we have for each 0 ≤ t ≤ T that kux (t)k2 + kv(t) − 1k2 +

Z tZ 0

R

6 u2xx θ0 − 1

v0 − 1, v0x , ux , u0x , √

. (3.7) dxdτ ≤ O(1) , Φ 0x

v 1+a γ−1





Since v(t, x) satisfies (3.6) and θ(t, x) is assumed to satisfy the a priori estimate (H2 ), (3.7) can be proved by applying the argument used in the proof of Lemma 2.8. Thus, we omit the detail for brevity. To close the a priori estimate (H2 ), we need to deduce an estimate on kθx (t)k. For the case when κ(v, θ) ≡ κ(θ), we have Lemma 3.5 Under the assumptions in Theorem 1.2 and Lemma 3.3, we have Z R

|K(θ)x |2 dx + γ−1

Z tZ 0

R

  K(θ)x vκ(θ) v

2

  10

θ0 − 1 dxdτ ≤ O(1) v0 − 1, u0 , √ , Φ0x

. (3.8) γ−1 x 1

Here

Z θ

K(θ) =

κ(z)dz. 1

(3.9)

One-dimensional Compressible Navier-Stokes-Poisson Equation with Large Data

21

Proof: Multiplying (1.1)3 by κ(θ) and differentiating the resulting equation with respect to x, we get !     κ(θ)µ(v)u2x K(θ)x Cv K(θ)tx + (κ(θ)p(v, θ)ux )x = + κ(θ) . (3.10) v v x x x Multiplying (3.10) by K(θ)x and integrating with respect to t and x over [0, t] × R give t K(θ)x 2 Cv 2 dxdτ vκ(θ) |K(θ)x | dx + v R 2 0 R x

 Z t Z 

θ0x 2 K(θ)x K(θ)x

+ O(1) √ ≤ O(1) v x dxdτ

γ − 1 v v 0 R x

Z



Z Z



|

K(θ)x 0

R

|

}

J3

κ(θ)µ(v)u2x v

Z tZ

+

{z !

dxdτ −

|0

x

{z

}

J4

Z tZ R

K(θ)x (κ(θ)p(v, θ)ux )x dxdτ . {z

(3.11)

}

J5

Notice that kθx kL∞ x

1

K(θ)x 2

≤ O(1) v



K(θ)x

≤ O(1) v

L∞ x

 

K(θ)x

v

x

1

2 1

≤ O(1)kθx k 2

 

K(θ)x

v

x

1

2

,

(3.12) we have from (3.1), (3.5), (3.6), and the a priori estimate (H2 ) that   Z tZ K(θ)x 2 vκ(θ) vx2 θx2 dxdτ dxdτ + O(1) v 0 R 0 R x

q   2   Z t Z tZ

1 K(θ)x K(θ)x 2

dτ kv (τ )k kθ (τ )k ≤ vκ(θ) dxdτ + O(1) vκ(θ) x x



12 0 R v v 0 x x   Z t Z Z K(θ)x 2 1 t dxdτ + O(1) ≤ kvx (τ )k4 kθx (τ )k2 dτ vκ(θ) 6 0 R v 0 x

    6 Z Z

K(θ)x 2

θ0 − 1 1 t

vκ(θ) dxdτ + O(1) v0 − 1, u0 , √ , Φ0x (3.13) ≤

, 6 v γ−1

1 J3 ≤ 12

Z tZ

0

J4 = −

R

x

Z tZ 0

R

κ(θ)µ(v)u2x



1

K(θ)x v



dxdτ − x

Z tZ 0

R

K(θ)x vx κ(θ)µ(v)u2x dxdτ v2

  Z tZ Z tZ K(θ)x 2 1 4 ≤ vκ(θ) ux dxdτ + O(1) θx2 vx2 dxdτ dxdτ + O(1) 12 0 R v 0 R 0 R x 

   10 Z Z K(θ)x 2

1 t θ0 − 1

√ ≤ vκ(θ) dxdτ + O(1) v0 − 1, u0 , , Φ0x (3.14)

, 6 v γ−1 Z tZ

0

R

x

1

and Z tZ 

J5 = 0

R

K(θ)x v

Z tZ



vκ(θ)p(v, θ)ux dxdτ + 0

x

R

K(θ)x vx κ(θ)p(v, θ)ux dxdτ v

  Z tZ   K(θ)x 2 2 2 2 dxdτ + O(1) vκ(θ) u + θ v x x x dxdτ v 0 R 0 R x 

   6 Z Z K(θ)x 2

θ0 − 1 1 t

v0 − 1, u0 , √

vκ(θ) dxdτ + O(1) Φ ≤ 0x .

6 v γ−1

1 ≤ 12

Z tZ

0

R

x

1

(3.15)

22

Zhong Tan, Tong Yang, Huijiang Zhao, and Qingyang Zou

Here we have used the fact that Z tZ 0

R

u4x dxdτ

≤ O(1)

Z t 0

≤ O(1)

Z t

dτ kux (τ )k2 kux (τ )k2L∞ x kux (τ )k3 kuxx (τ )kdτ

0

  10

θ0 − 1

√ v − 1, u , ≤ O(1) Φ 0 0x .

0 γ−1 1

(3.16)

Inserting (3.13)-(3.15) into (3.11), we deduce (3.8) and complete the proof of the lemma. Now we turn to the case when κ(v, θ) depends on both v and θ. For this , we have Lemma 3.6 Under the assumptions in Lemma 3.5, if κθθ (v, θ) < 0 holds for v > 0, θ > 0, then we have

Z t Z tZ

θx (t) 2 κ(v, θ) 2 κθθ (v, θ) 4

√

+ θ dxdτ − θx dxdτ xx

γ − 1 v v 0 0 R

  6

θ0 − 1

≤ O(1) v0 − 1, u0 , √ , Φ0x

. γ−1

(3.17)

1

Proof: Differentiating (1.1)3 with respect to x and multiplying the resulting equation by θx , we have by integrating it over [0, t] × R that t κ(v, θ) 2 Cv kθx (t)k2 + θxx dxdτ 2 v 0 R     Z tZ Z tZ Cv κ(v, θ) µ(v) 2 2 = θx θx kθ0x k + u dxdτ − θxx dxdτ 2 v x x v 0 R 0 R x

Z Z

|

{z J6

}

|

{z J7

}

Z tZ

θxx p(v, θ)ux dxdτ .

+ 0

(3.18)

R

|

{z J8

}

For J6 , J7 and J8 , we have from Lemma 3.1-Lemma 3.4 and the a priori estimate (H2 ) that Z tZ

µ(v) 2 ux θxx dxdτ 0 R v Z tZ Z tZ 1 κ(v, θ) 2 ≤ θxx dxdτ + O(1) u4x dxdτ 6 0 R v 0 R Z Z Z tZ 1 t κ(v, θ) 2 ≤ θxx dxdτ + O(1) kux (τ )k3 kuxx (τ )kdxdτ 6 0 R v 0 R

  10 Z Z

1 t κ(v, θ) 2 θ0 − 1

√ ≤ θxx dxdτ + O(1) v − 1, u , , Φ 0 0 0x ,

6 0 R v γ−1 1

J6 = −

t κ(v, θ) 2 1 t J8 ≤ θxx dxdτ + O(1) u2x dxdτ 6 0 R v 0 R

  2 Z Z

1 t κ(v, θ) 2 θ0 − 1

√ ≤ θxx dxdτ + O(1) v − 1, u , , Φ 0 0 0x ,

6 0 R v γ−1 1

Z Z

(3.19)

Z Z

(3.20)

One-dimensional Compressible Navier-Stokes-Poisson Equation with Large Data

23

and Z tZ

t κ(v, θ) κ(v, θ) θx v x J7 = − θxx dxdτ − θxx dxdτ v v 0 R 0 R θ v     Z tZ Z tZ Z tZ κ(v, θ) 1 κθ (v, θ) 1 4 κθθ (v, θ) 3 θx v x = θ θ vx dxdτ − θxx dxdτ dxdτ + 3 0 R x v 3 0 R x v v 0 R v v Z Z Z Z Z tZ 1 t κθθ (v, θ) κ(v, θ) 2 1 t ≤ θx4 θx2 vx2 dxdτ dxdτ + θxx dxdτ + O(1) 6 0 R v 12 0 R v 0 R Z Z Z Z 1 t κθθ (v, θ) 1 t κ(v, θ) 2 θx4 dxdτ + θxx dxdτ ≤ 6 0 R v 6 0 R v

  6

θ0 − 1

(3.21) , Φ0x +O(1) v0 − 1, u0 , √

. γ−1 1

θx2





Z Z





Here we have used the fact that Z tZ 0

R

θx2 vx2 dxdτ

  2 Z t

θ0 − 1

≤ ≤ O(1) v0 − 1, u0 , √ , Φ0x kθx (τ )k2L∞ dτ

x γ−1 0 1 0

  2 Z t

θ0 − 1

√ ≤ O(1) v0 − 1, u0 , , Φ0x kθx (τ )kkθxx (τ )kdτ

γ−1 1 0

  4 Z t Z Z Z

θ0 − 1 1 t κ(v, θ) 2

√ v − 1, u , θx2 dxdτ. ≤ θxx dxdτ + O(1) , Φ 0 0 0x

12 v γ−1 Z t

kθx (τ )k2L∞ kvx (τ )k2 dτ x

0

1 0

R

R

Inserting (3.19)-(3.21) into (3.18), we obtain Cv 1 kθx (t)k2 + 2 2

Z tZ

κ(v, θ) 2 1 θxx dxdτ − v 6 0 R

  10

θ − 1 0

≤ O(1)

v0 − 1, u0 , √γ − 1 , Φ0x . 1

Z tZ 0

R

κθθ (v, θ) 4 θx dxdτ v

(3.22)

This is (3.17) and the proof of Lemma 3.6 is completed. Lemma 3.1-Lemma 3.6 imply that under the a priori estimate (H2 ), there exist two   positive

θ −1 0 constants V2 ≥ 1 and C1 ≥ 1 with V2 depending only on v0 − 1, u0 , √γ−1 , Φ0x , v0x , V , V , Θ, and Θ but independent of T and γ − 1, and C1 depending only on V2 but independent of T > 0 and γ − 1, such that V2−1 ≤ v(t, x) ≤ V2 ,

(t, x) ∈ [0, T ] × R,



  2 Z t Z   2 



θ0 − 1 2 2

v − 1, u, √θ − 1 , Φx (t) +

√ u + θ (τ, x)dxdτ ≤ C v − 1, u , , Φ 1 0 0 0x , x x

γ−1 γ−1 0 R

  2 Z tZ

θ0 − 1

√ kvx (t)k2 + vx2 (τ, x)dxdτ ≤ C1 v − 1, u , , Φ 0x , 0

0 γ−1 0 R 1

  6 Z tZ

θ − 1 0

kux (t)k2 + u2xx (τ, x)dxdτ ≤ C1

v0 − 1, u0 , √γ − 1 , Φ0x , 0 R 1



 10 Z tZ

θx (τ ) 2

θ − 1 0 2

√

θxx (τ, x)dxdτ ≤ C1 (3.23)

v0 − 1, u0 , √γ − 1 , Φ0x

γ − 1 + 0

R

1

24

Zhong Tan, Tong Yang, Huijiang Zhao, and Qingyang Zou

hold for 0 ≤ t ≤ T . To obtain the global existence of solutions, we only need to close the a priori estimate (H2 ). For this, we need the smallness of γ − 1 > 0. In fact, we have from (3.23)2 , (3.23)5 and Sobolev’s inequality that kθ(t) − 1k

L∞ T,x

  4

θ0 − 1

√ ≤ kθ(t) − 1k kθx (t)k ≤ C1 (γ − 1) v0 − 1, u0 , , Φ0x

. γ−1 1 1 2

On the other hand, since θ =

A 1−γ Rv

1 2

exp

1 2



γ−1 R s



, if we set s =

R γ−1

(3.24)

ln R A , we have

A 1−γ γ−1 v exp s −1 R R     γ−1 A γ−1 A s s − exp = v 1−γ exp R R R R         A  1−γ γ−1 γ−1 A γ−1 = v − 1 exp exp s . s + s − exp R R R R R 



θ−1=

Consequently,

A(γ − 1) γ−1 1

−γ (3.25) kθ0 − 1k ≤ O(1) exp ks0 kL∞ v ks0 (x) − sk ,

0 L∞ kv0 − 1k + x R R R x    1−γ A(γ − 1) γ−1 1  −γ kθ0x k ≤ O(1) ks k . exp ks0 kL∞ inf v (x) (inf v (x)) kv k + 0x 0 0 0x x R R R x 





∞ Since kv0 (x)kL∞ , inf v0 (x), γ−1 x A ks0 (x)kLx are assumed to be independent of γ − 1, we have from (3.24) and (3.25) that 1

kθ(t) − 1kL∞ ≤ C2 (γ − 1) 2 k(v0 − 1, u0 , Φ0x )k31 + C3 (γ − 1)2 k(v0 − 1, s0 − s)k31 x

(3.26)

holds for 0 ≤ t ≤ T . Thus if γ − 1 > 0 is chosen to be sufficiently small such that 1

n

o

C2 (γ − 1) 2 k(v0 − 1, u0 , Φ0x )k31 + C3 (γ − 1)2 k(v0 − 1, u0 , s0 − s)k31 ≤ min Θ − 1, 1 − Θ , (3.27) we have from (3.26) and (3.27) that for any 0 ≤ t ≤ T, x ∈ R, o

n

θ(t, x) ≤ kθ(t, x) − 1kL∞ + 1 ≤ 1 + min Θ − 1, 1 − Θ ≤ Θ, T,x

(3.28)

n

(3.29)

and o

θ(t, x) ≥ 1 − kθ(t, x) − 1kL∞ ≥ 1 − min Θ − 1, 1 − Θ ≥ 1 − (1 − Θ) = Θ. T,x That is Θ ≤ θ(t, x) ≤ Θ,

x ∈ R, 0 ≤ t ≤ T.

(3.30)

This closes the a priori estimate (H2 ) and then Theorem 1.2 follows from the standard continuation argument.

One-dimensional Compressible Navier-Stokes-Poisson Equation with Large Data

4

25

The proof of Theorem 1.3

When µ is a positive constant, the Cauchy Problem (1.1), (1.3) can be rewritten as   vt − ux = 0,        ux
Φx Rθ    ut + v x = µ v x + v ,   µu2x κ(v,θ)θx Rθ  u = + , C θ +  v t v x v v  x       Φx  = 1 − v, lim Φ(t, x) = 0  v x

(4.1)

|x|→+∞

with prescribed initial data (v(0, x), u(0, x), θ(0, x)) = (v0 (x), u0 (x), θ0 (x)),

lim (v0 (x), u0 (x), θ0 (x)) = (1, 0, 1).

x→±∞

(4.2)

Let (v(t, x), u(t, x), θ(t, x)) ∈ X(0, T ; M0 , M1 ; N0 , N1 ) be a solution of the Cauchy Problem (1.1), (1.3) which is defined in the time strip [0, T ] for some T > 0, to extend such a solution globally, as pointed out in the proofs of Theorems 1.1 and 1.2, we only need to deduce positive lower and upper bounds on v(t, x) and θ(t, x) which are independent of M0 , M1 , N0 and N1 but may depend on T . First, similar to the proof of Theorem 1.2, we have the following basic energy estimate Lemma 4.1 Under the conditions in Theorem 1.3, we have for 0 ≤ t ≤ T that Z R

(

R Φ2 u2 + φ(θ) + x2 Rφ(v) + 2 γ−1 2v Z

= R

)

µu2x κ(v, θ)θx2 + vθ vθ2

Z tZ

(t, x)dx + 0

u2 R Φ2 Rφ(v0 ) + 0 + φ(θ0 ) + 0x2 2 γ−1 2v0

R

!

dxdτ

!

(x)dx.

(4.3)

Here, as in Section 2, φ(x) = x − ln x − 1. Based on the estimate (4.3), we now turn to deduce the desired lower and upper bounds on v(t, x) and θ(t, x). To this end, for each x ∈ R, we can find some i ∈ Z such that x ∈ [i, i + 1]. Recall ai (t), bi (t) defined in (2.24) and (2.25) and notice that in Theorem 1.3, since µ is a positive constant, we have by employing the argument developed in [25] that 1+

R µ

v(t, x) =

Z t

θ(τ, x)Bi (τ, x)Yi (τ )Ai (τ, x)dτ 0

.

Bi (t, x)Yi (t)Ai (t, x)

(4.4)

Here Ai (t, x) = exp −

 Z t Z ai (t)  Φx 0

x

v

v0 (ai (t)) 1 exp Bi (t, x) = v0 (x)v(t, ai (t)) µ 

Yi (t) = exp

R µ

Z t  0

!

(τ, y)dydτ

Z ai (t)

,

(4.5) !

(u(t, y) − u0 (y))dy ,

(4.6)

x

θ (τ, ai (t))dτ . v 

(4.7)

26

Zhong Tan, Tong Yang, Huijiang Zhao, and Qingyang Zou (2.5) implies that 

Φx (t, x) = − v 

Z t



u(τ, x)dx + 0

Φx (0, x), v 

(4.8)

we have from (4.8) and (4.3) that there exist positive constants A, A, B, B such that 0 < A ≤ Ai (t, x) ≤ A,

0 < B ≤ Bi (t, x) ≤ B,

Yi (t) ≥ 1.

(4.9)

On the other hand, notice that (4.4) can be rewritten as 1+

R µ

Z t

θ(τ, x)Bi (τ, x)Yi (τ )Ai (τ, x)dτ 0

v(t, x)Yi (t) =

.

Bi (t, x)Ai (t, x)

(4.10)

Integrating (4.10) with respect to x over [i, i + 1], we can get from (2.24) and (4.9) that 

Yi (t) ≤ O(1) 1 +

Z t Z i+1





θ(τ, y)dy Yi (τ )dτ

(4.11)

i

0

≤ O(1) 1 +



Z t



Yi (τ )dτ . 0

From which and the Gronwall inequality, we can deduce that there exists a positive constant Y which is independent of M0 , M1 , N0 , N1 and i but may depend on T such that 1 ≤ Yi (t) ≤ Y ,

∀i ∈ Z,

∀t ∈ [0, T ].

(4.12)

From (4.4), (4.9), and (4.12), one easily deduce that there exists a positive constant V3 > 0 which is independent of M0 , M1 , N0 , N1 but may depend on T such that v(t, x) ≥ V3−1 > 0,

∀(t, x) ∈ [0, T ] × R.

(4.13)

(4.13) together with Lemma 2.4 imply that there exists a positive constant Θ−1 3 > 0 which is independent of M0 , M1 , N0 , N1 but may depend on T such that θ(t, x) ≥ Θ−1 3 > 0,

∀(t, x) ∈ [0, T ] × R.

(4.14)

Now we turn to deduce an upper bound for v(t, x). For this purpose, notice from (2.24), (2.25), (4.9)-(4.14), (4.12) and the fact min

v≥V3−1 , θ≥Θ−1 3





κ(v, θ) ≥ κ V3−1 , Θ−1 >0 3

that q 2 q θ(t, x) ≤ 2 θ(t, x) − θ(t, bi (t)) + 4θ(t, bi (t)) Z 2 bi (t)  |θ |  x √ = (t, y)dy + 4θ(t, bi (t)) x θ Z i+1 Z i+1 2

≤ O(1) 1 +

i

κ(v, θ)θx dx · vθ2

≤ O(1) 1 + kv(t)kL∞ x

i

Z i+1 κ(v, θ)θx2 i

vθ2

vθ dx κ(v, θ) !

dx .

!

(4.15)

One-dimensional Compressible Navier-Stokes-Poisson Equation with Large Data

27

(4.4) together with (4.3), (2.24), (2.25), (4.9)-(4.14), (4.12), and (4.15) imply Z t



≤ O(1) 1 + kv(t)kL∞ x



max θ(τ, x)dτ

0 x∈R

Z t

≤ O(1) 1 +

0

kv(τ )kL∞ x

Z R

κ(v, θ)θx2 vθ2

!

!

(τ, x)dxdτ

.

Based on the above inequality, we can conclude from (4.3), (4.13), (4.14) and the Gronwall inequality that Lemma 4.2 Under the conditions listed in Lemma 4.1, there exist positive constants V3 and Θ3 such that 0 < V3−1 ≤ v(t, x) ≤ V3 , θ(t, x) ≥ Θ−1 (4.16) 3 >0 hold for (t, x) ∈ [0, T ] × R. Here the constants V3 and Θ3 are independent of M0 , M1 , N0 , N1 but may depend on T . To complete the proof of Theorem 1.3, we only need to deduce an upper bound on θ(t, x). To do so, as a direct consequence of (4.15), (4.16), we can deduce from (4.3) and the fact (θ − 1)2 ≤ O(1)(1 + |θ − 1|)φ(θ) that Corollary 4.1 Under the conditions listed in Lemma 4.1, we have Z t

max θ(τ, x)dτ ≤ O(1)

(4.17)

0 x∈R

and

Z tZ 0

(θ(τ, x) − 1)2 dxdτ ≤ O(1).

(4.18)

R

Here and throughout this section, O(1) is used to denote some positive constant independent of M0 , M1 , N0 , N1 but may depend on T . With (4.17) and (4.18) in hand, we have from (4.16) and the proof of Lemma 2.6 that Lemma 4.3 Under the conditions listed in Lemma 4.1, we have 2

ku(t)k +

Z t

kux (s)k2 ds ≤ O(1).

(4.19)

0

As to the estimate on kvx (t)k, if we set κ1 (θ) =

min

V3−1 ≤v≤V3

κ(v, θ),

(4.20)

we have from the proof of Lemma 2.5 and the fact Z tZ 0

that

R

θ θx2

dxdτ ≤

κ (θ) ∞ θ 1 L



t,x

Z tZ 0

R

θ κ(v, θ)θx2

dxdτ ≤ O(1)

κ (θ) ∞ 2 θ 1 L



t,x

28

Zhong Tan, Tong Yang, Huijiang Zhao, and Qingyang Zou

Lemma 4.4 Under the conditions listed in Lemma 4.1, we have 



2 Z t q



θ(s)vx (s) ds ≤ O(1) 1 + θ kvx (t)k2 +

κ (θ)

0

L∞ t,x

1

.

(4.21)

To deduce an estimate on kux (t)k, noticing that Z tZ 0

R

u2x vx2 dxds ≤ ε

Z t

kuxx (s)k2 ds + O(1)

Z t

kux (s)k2 kvx (s)k4 ds,

0

0

we have from the proof of Lemma 2.8, (4.3), (4.19), and (4.21) that Lemma 4.5 Under the conditions listed in Lemma 4.1, we have kux (t)k2 +

Z t

kuxx (s)k2 ds

0

≤ O(1) + O(1)

Z tZ  0

R



θx2 + θ2 vx2 + u2x vx2 dxds





θ 2

θ

∞ ≤ O(1) 1 + + kθk Lt,x

κ (θ) ∞ κ (θ) 1

L∞ t,x

1

Lt,x

2 

θ

. + p

κ1 (θ) ∞

(4.22)

Lt,x

Now we turn to deduce an upper bound on θ(t, x) for the case lim κ1 (θ) = +∞ based on θ→+∞

the above estimates. In fact (2.44) together with (4.17) and the Gronwall inequality imply 

kθ(t)k

L∞ x

≤ O(1) 1 +

Z t 0



≤ O(1) 1 +

Z t

kux (s)k2L∞ x

+ kθ −

1k2L∞ ds x

kux (s)kkuxx (s)k + kθ −

0



1k2L∞ ds x



(4.23)





1

1

θ 2

θ 2 θ

2

p ≤ O(1) 1 + + kθk +

∞

κ (θ) ∞

∞ Lt,x

κ (θ) κ (θ) 1 1 1 L L t,x

Z t

+

L∞ t,x

t,x

0



kθ − 1k2L∞ ds . x

Here we have used (4.19) and (4.22). From (4.23) and the assumption that lim κ1 (θ) = +∞, one easily deduce an upper bound θ→+∞

on θ(t, x). For the case κ(v, θ) is bounded from above, the above argument does not apply and we have to use another method to deduce an upper bound on θ(t, x). For this purpose, as in [1], set 1 w(t, x) = u2 (t, x) + Cv (θ(t, x) − 1), 2

(4.24)

we can easily deduce that 

wt = µ

wx v





+ (κ(v, θ) − Cv µ) x

θx v





− x

Rθu v



+ x

uΦx . v

(4.25)

Multiplying (4.25) by w and integrating the result with respect to t and x over [0, t] × R, we have by some integrations by parts that 1 kw(t)k2 + 2

Z tZ 0

R

µwx2 dxds ≤ O(1) − v

Z tZ 0

Z tZ 

+ |0

R

(κ(v, θ) − Cv µ)

R

wx θ x dxds v

Rθuwx wuΦx + dxds . v v 

{z

K1

}

(4.26)

One-dimensional Compressible Navier-Stokes-Poisson Equation with Large Data

29

Since µwx2 µu2 u2x 2µCv uux θx Cv2 µθx2 = + + , v v v v (κ(v, θ) − Cv µ)θx wx (Cv µ − κ(v, θ))Cv θx2 (cv µ − κ(v, θ))θx uux − = + , v v v Z t

K1 ≤ O(1) + ε

kwx (s)k2 ds

0

Z tZ

2 2

≤ O(1) + ε

R

Z t

kwx (s)k2 ds + O(1)

0

|uux θx |dxds ≤ ε

Z t

0

kθx (s)k2 ds + O(1)

Z tZ 0

0

R

R

Φx v

Z tZ 

0

Z tZ



(θ − 1) u + w u +

+O(1) 0

2 2

2 !

dxds 

(θ − 1)2 u2 + w2 u2 dxds,

R

u2 u2x dxds,

we have by substituting the above estimates into (4.26) that kw(t)k2 +

Z tZ 0

≤ O(1) + O(1)

(Cv κ(v, θ) − ε)θx2 dxds

R tZ

Z 0

R



(4.27) 

(θ − 1)2 u2 + w2 u2 + u2 u2x dxds.

Here and in the above analysis, we have used the fact that κ(v, θ) is uniformly bounded for 0 < V3−1 ≤ v ≤ VR 3 ,R θ ≥ Θ−1 3 . t 2 2 To estimate 0 R u ux dxds, we multiply (4.1) by u3 and integrate the result with respect to t and x over [0, t] × R to yield ku(t)k4L4 +

Z tZ 0

R

µu2 u2x dxds ≤ O(1) + O(1) v |

Z t Z 2 Rθu ux dxds v 0 R {z }

(4.28)

K

2 Z t Z 3 Φ u x + O(1) dxds . 0 R v | {z } K3

Due to Z tZ Z tZ Φx 2 u6 dxds K3 ≤ O(1) v dxds + O(1) 0 R 0 R Z t

≤ O(1) + O(1)

0

≤ O(1) + ε K2 ≤ ε

Z tZ 0

R

Z t

2

ku(s)ux (s)k ds + O(1)

0 µu2 u2x

v

ku(s)k4L∞ ds x

Z t 0

Z tZ

dxds + O(1) 0

ku(s)k4L4 ds,

(θ − 1)2 u2 dxds + O(1),

R

we have by inserting the above estimates into (4.28) and by employing the Gronwall inequality that Z t Z tZ 4 2 ku(t)kL4 + ku(s)ux (s)k ds ≤ O(1) + O(1) (θ − 1)2 u2 dxds. (4.29) 0

0

R

30

Zhong Tan, Tong Yang, Huijiang Zhao, and Qingyang Zou A suitable linear combination of (4.29) and (4.27) yields kθ(t) − 1k2 + ku(t)k4L4 + ≤ O(1) + O(1)

Z tZ  0

Z t



kθx (s)k2 + ku(s)ux (s)k2 ds

(4.30)

0



θ − 1)2 u2 + u2 w2 dxds.

R

Here again we have used the fact that κ(v, θ) is bounded both from below and above for 0 < V3−1 ≤ v ≤ V3 , θ ≥ Θ−1 3 > 0. From (4.19), we have Z tZ 0

(θ − 1)2 u2 dxds ≤ O(1)

Z t 0

R

Z tZ 0

u2 w2 dxds ≤ O(1)

0

R

≤

Z t 0

Z t

Z t

kθ(s) −

0

Z t

1k2L∞ ds x

ku(s)k4L∞ ds x

0

≤ε

Z t

ds, kθ(s) − 1k2L∞ x ds kw(s)k2L∞ x 

ds, + ku(s)k4L∞ kθ(s)k2L∞ x x kθx (s)k2 ds + O(1)

Z t

0

≤ε

kθ(s) − 1k2 ds,

0

Z t

2

ku(s)ux (s)k ds + O(1)

Z t 0

0

ku(s)k4L4 ds.

Putting the above estimates into (4.30), we have by the Gronwall inequality that 2

kθ(t) − 1k +

ku(t)k4L4

+

Z t



kθx (s)k2 + ku(s)ux (s)k2 ds ≤ O(1).

(4.31)

0

A direct consequence of (4.31) is Z t 0

kθ(s)k2L∞ ds ≤ O(1) x

(4.32)

and the estimate (4.21) obtained in Lemma 4.4 can be improved as

2 Z t q



kvx (t)k +

θ(s)vx (s) ds ≤ O(1). 2

0

(4.19), (4.32) together with (4.33) imply Z tZ 0

R

θ2 vx2 dxds ≤ O(1) ≤ O(1)

Z t 0

Z t 0

Z tZ 0

R

u2x vx2 dxds ≤ O(1) ≤ O(1)

Z t 0

Z t 0

≤ε

Z t

kθ(s)k2L∞ kvx (s)k2 ds x kθ(s)k2L∞ ds ≤ O(1), x kux (s)k2L∞ kvx (s)k2 ds x kux (s)k2L∞ ds x

kuxx (s)k2 ds + O(1)

Z t

0

≤ε

Z t 0

0 2

kuxx (s)k ds + O(1).

kux (s)k2 ds

(4.33)

One-dimensional Compressible Navier-Stokes-Poisson Equation with Large Data

31

Combining the above estimates with the first inequality of (4.22) yields 2

kux (t)k +

Z t

kuxx (s)k2 ds ≤ O(1).

(4.34)

0

(4.34) together with (4.19) imply Z t 0

kux (s)k2L∞ ds ≤ O(1). x

(4.35)

Having obtained (4.35), the upper bound on θ(t, x) can be obtained immediately from (4.23) for the case when κ(v, θ) is uniformly bounded for 0 < V3−1 ≤ v ≤ V3 , θ ≥ Θ−1 3 > 0. This completes the proof of Theorem 1.3.

Acknowledgment The research of Zhong Tan was supported by the grant from the National Natural Science Foundation of China under contract 10976026, the research of the second author was supported by the General Research Fund of Hong Kong, CityU No.104310, and the Croucher Foundation. And research of the third author was supported by the grant from the National Natural Science Foundation of China under contract 10925103. This work is also supported by “the Fundamental Research Funds for the Central Universities”.

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