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global solutions to the one-dimensional navier-stokes-poission equation with large data

global solutions to the one-dimensional navier-stokes-poission equation with large data
global solutions to the one-dimensional navier-stokes-poission equation with large data

SIAM J.M ATH.A NAL .c 2013Society for Industrial and Applied Mathematics Vol.45,No.2,pp.547–571GLOBAL SOLUTIONS TO THE ONE-DIMENSIONAL COMPRESSIBLE NA VIER–STOKES–POISSON EQUATIONS WITH LARGE DATA ?ZHONG TAN ?,TONG YANG ?,HUIJIANG ZHAO §

,AND QINGYANG ZOU ?Abstract.In this paper,we study the global solutions with large data away from vacuum to the Cauchy problem of the one-dimensional compressible Navier–Stokes–Poisson system with density-dependent viscosity coe?cient and density-and temperature-dependent heat-conductivity coe?cient.The proof is based on some detailed analysis on the bounds on the density and temper-ature functions.Key words.Navier–Stokes–Poisson equations,global solutions with large data,viscosity and heat-conductivity coe?cients AMS subject classi?cations.35Q35,35D35,76D05DOI.10.1137/1208761741.Introduction.The compressible Navier–Stokes–Poisson (NSP)system con-sisting of the compressible Navier–Stokes equations coupled with the Poisson equation models the viscous ?uid under the in?uence of the self-induced electric force:?????????ρτ+?ξ·(ρu )=0,(ρu )τ+?ξ·(ρu ?u )+?ξp =ρ?ξΦ+?ξ·T ,(ρE )τ+?ξ·(ρu E +up )=ρu ·?ξΦ+?ξ·(u T )+?ξ·(κ(v,θ)?ξθ),ΔξΦ=ρ?ρ(ξ),lim |ξ|→+∞

Φ(τ,ξ)=0.Here,ρ>0,u =(u 1,u 2,u 3),θ>0,p =p (ρ,θ),e ,and Φdenote the density,ve-

locity,absolute temperature,pressure,internal energy,and the electrostatic potential function,respectively.Also the total energy E =12|u |2+e and the stress tensor T =μ(ρ,θ)(?ξu +(?ξu )t )+ν(ρ,θ)(?ξ·u )I with I being the identity matrix.The viscosity coe?cients μ(ρ,θ)>0and ν(ρ,θ)satisfy μ(ρ,θ)+23ν(ρ,θ)>0.The ther-modynamic variables p ,ρ,and e are related by Gibbs equation de =θds ?pdρ?1with s being the speci?c entropy.κ(ρ,θ)>0denotes the heat-conductivity coe?cient and ρ(ξ)>0is the background doping pro?le;see [30].To explain the purpose of this paper,we ?rst give the following remarks on the viscosity and heat-conductivity coe?cients:?Received by the editors May 7,2012;accepted for publication November 7,2012;published electronically March 14,2013.https://www.doczj.com/doc/366468998.html,/journals/sima/45-2/87617.html ?School of Mathematical Sciences,Xiamen University,Xiamen 361005,Fujian,China (tan85@https://www.doczj.com/doc/366468998.html,).The ?rst author was supported by the grant from the National Natural Science Foundation of China under contract 10976026.?Department of Mathematics,City University of Hong Kong,Kowloon,Hong Kong,China (matyang@https://www.doczj.com/doc/366468998.html,.hk).The second author was supported by the General Research Fund of Hong Kong,CityU 104310and the Croucher Foundation.§Corresponding author:School of Mathematics and Statistics,Wuhan University,Wuhan 430072,China (zgyheda@https://www.doczj.com/doc/366468998.html,).The third author was supported by two grants from the National Natural Science Foundation of China under contracts 10925103and 11271160.This work was also supported by the Fundamental Research Funds for the Central Universities.?School of Mathematics and Statistics,Wuhan University,Wuhan 430072,China (qy zou@https://www.doczj.com/doc/366468998.html,).This work was also supported by the Fundamental Research Funds for the Central Universities.547D o w n l o a d e d 09/30/13 t o 202.121.182.203. R e d i s t r i b u t i o n s u b j e c t t o S I A M l i c e n s e o r c o p y r i g h t ; s e e h t t p ://w w w .s i a m .o r g /j o u r n a l s /o j s a .p h p

548Z.TAN,T.YANG,H.ZHAO,AND Q.ZOU ?When the viscosity coe?cients μ(ρ,θ)>0,ν(ρ,θ),and the heat-conductivity coe?cient κ(ρ,θ)>0are constants,(1.1)is used in semiconductor theory to

model the transport of charged particles under the in?uence of self-induced electric ?eld;see [30].

?In the kinetic theory,the time evolution of the particle distribution func-tion for the charged particles in a dilute gas can be modeled by the Vlasov–Poisson–Boltzmann system;see [4],[3],[34].When we derive the NSP (1.1)

from the Vlasov–Poisson–Boltzmann system by using the Chapman–Enskog expansion,see [4],[12],[34],the viscosity coe?cients μ,ν,and the heat-conductivity coe?cient κdepend on the absolute temperature θand ν=?23μfor the monatomic gas.If the intermolecular potential is proportional to r ?αwith α>1,where r represents the intermolecular distance,then μ,ν,and κare proportional to the temperature to some power:μ,?ν,κ∝θα+42α.In particular,for the Maxwellian molecule (α=4),such dependence is lin-

ear,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 coe?cients μ,ν,and the heat-conductivity coe?cient κdepend on ρand θ.Unlike the small perturbation solutions,such dependence has a strong in?uence on the solution behavior and thus leads to di?culties in analysis but not for the case of constant coe?cients.In fact,for the one-dimensional compressible Navier–Stokes equations,there are a lot of recent papers on the construction of nonvacuum solutions to the one-dimensional compressible Navier–Stokes equations with density-and temperature-dependent transportation coe?cients in various forms;see [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 of this paper is the construction of globally smooth,non-vacuum solutions to the one-dimensional nonisentropic compressible NSP with density dependent viscous coe?cient and density-and temperature-dependent heat-conducti-vity coe?cient for arbitrarily large data.We hope that the analysis here can shed some light on the construction of global classical solutions to the ?uid model derived from the Vlasov–Poisson–Boltzmann system with large data.Let x be the Lagrangian space variable,t be the time variable,and v =1ρdenote the speci?c volume.Then the one-dimensional compressible NSP system (1.1)with viscous coe?cient μ(v )and heat-conductivity coe?cient κ(v,θ)becomes

?????????????????v t ?u x =0,u t +p (v,θ)x = μ(v )u x v x +Φx v ,e t +p (v,θ)u x =μ(v )u 2x v + κ(v,θ)θx v x , Φx v x =1?v,lim |x |→+∞Φ(t,x )=0.(1.1)Throughout this paper,we will concentrate on the ideal,polytropic gases:

(1.2)p (v,θ)=Rθv =Av ?γexp γ?1R s ,e =C v θ=Rθγ?1,D o w n l o a d e d 09/30/13 t o 202.121.182.203. R e d i s t r i b u t i o n s u b j e c t t o S I A M l i c e n s e o r c o p y r i g h t ; s e e h t t p ://w w w .s i a m .o r g /j o u r n a l s /o j s a .p h p

COMPRESSIBLE NAVIER–STOKES–POISSON EQUATIONS 549where the speci?c gas constant R and the speci?c heat at constant volume C v are

positive constants and γ>1is the adiabatic constant.Moreover,to simplify the presentation,we will only consider the case when the background doping pro?le ρis a positive constant which is normalized to 1as in (1.1)4.Take the initial data (v (0,x ),u (0,x ),θ(0,x ))=(v 0(x ),u 0(x ),θ0(x )),

(1.3)lim x →±∞(v 0(x ),u 0(x ),θ0(x ))=(v ±,u ±,θ±),satisfying v ?=v +,u ?=u +,θ?=θ+.Without loss of generality,we assume v ?=v +=1,u ?=u +=0,θ?=θ+=1.The ?rst result is concerned with the case (1.4)μ(v )=v ?a ,κ(v,θ)=θb ,

which is stated as follows.Theorem 1.1.Suppose

?(v 0(x )?1,u 0(x ),θ0(x )?1,Φ0x (x ))∈H 1(R ),and there exist positive constants V ,V ,Θ,Θsuch that (1.5)V ≤v 0(x )≤V ,Θ≤θ0(x )≤Θ;

?13

Then the Cauchy problem (1.1),(1.3)with μ(v )and κ(v,θ)given by (1.4)admits a

unique global solution (v (t,x ),u (t,x ),θ(t,x ))satisfying (1.6)(v (t,x )?1,u (t,x ),θ(t,x )?1)∈C 0 0,T ;H 1(R ) ,(u x (t,x ),θx (t,x ))∈L 2 0,T ;H 1(R ) ,Φx (t,x )∈C 0 0,T ;H 2(R ) ,

0

Here T >0is any given positive constant and V 0,Θ0are some positive constants which may depend on T .Note that the assumptions imposed on a and b in Theorem 1.1exclude the case

when the viscous coe?cient μand the heat-conductivity coe?cient κare positive constants.The next result will recover this in another setting.The main idea is to use the smallness of γ?1to deduce uniform lower and upper bounds on the absolute temperature.This can be achieved by showing that (v 0(x )?1,u 0(x ),s 0(x )?s )∈

H 1(R )are bounded in H 1(R )independent of γ?1so that θ0(x )?1 L ∞(R )can be

chosen to be small when γis close to 1.Here s =R γ?1ln R A is the far ?eld of the initial entropy s 0(x ),that is,lim |x |→+∞s 0(x )=lim |x |→+∞R γ?1ln Rθ0(x )v 0(x )γ?1A

=s.Taking (v,u,s )as the unknown functions,the second global existence theorem can

be stated as follows.D o w n l o a d e d 09/30/13 t o 202.121.182.203. R e d i s t r i b u t i o n s u b j e c t t o S I A M l i c e n s e o r c o p y r i g h t ; s e e h t t p ://w w w .s i a m .o r g /j o u r n a l s /o j s a .p h p

550Z.TAN,T.YANG,H.ZHAO,AND Q.ZOU Theorem 1.2.Suppose we ahve the following:? (v 0(x )?1,u 0(x ),s 0(x )?s,Φ0x (x )) H 1(R )is bounded by some positive con-

stant independent of γ?1and (1.5)holds for some (γ?1)-independent positive constants V ,V ,Θ,Θ.?(γ?1) s 0(x ) L ∞(R )is bounded by some constant independent of γ?1.?The smooth function μ(v )satis?es μ(v )>0for all v >0and

(1.7)lim v →0+Ψ(v )=?∞,lim v →+∞Ψ(v )=+∞.Here,(1.8)Ψ(v )= v 1√z ?ln z ?1z μ(z )dz.?For the heat-conductivity coe?cient,there are two cases.If κ(v,θ)=κ(θ)depends only on θ,we only assume κ(θ)>0for θ>0with some smoothness

condition.If it depends on both v and θ,then

in addition to κ(v,θ)>0for all v >0,θ>0,we also assume the following.Set κ1(v )=min Θ≤θ≤Θκ(v,θ)and assume (1.9)κθθ(v,θ)<0?v >0,θ>0and (1.10)lim v →0+μ(v )κ1(v )|Ψ(v )|2

=lim v →+∞μ(v )κ1(v )|Ψ(v )|2=0.?γ?1is su?ciently 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 (1.11)lim t →+∞sup x ∈R (v (t,x )?1,u (t,x ),θ(t,x )?1) =0.Remark 1.1.We give the following remarks on Theorems 1.1and 1.2.?From the proof of Theorem 1.2,one will notice that the assumption (1.10)can be replaced by the following weaker assumption:(1.12)lim v →0+μ(v )κ1(v )|Ψ(v )|2≤ε0,lim v →+∞μ(v )κ1

(v )|Ψ(v )|2≤ε0.Here ε0>0is a suitably chosen su?ciently small positive constant.?Under the assumptions in Theorem 1.2,when γ?1is su?ciently small,although θ0?1 H 1(R )is small, (v 0?1,u 0,s 0?s ) H 1(R )can be large.?When μ(v )satis?es 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 a similar result to Theorem 1.2also holds even when (v 0?1,u 0,s 0?s ) H 1(R ),V ,and V depend on 1γ?1with certain

growth conditions as γ→1+.?The same arguments for Theorems 1.1and 1.2can be applied directly to the compressible Navier–Stokes equations which generalize the previous re-sults [18]and [23],where the viscosity coe?cient is assumed to be a positive

constant.D o w n l o a d e d 09/30/13 t o 202.121.182.203. R e d i s t r i b u t i o n s u b j e c t t o S I A M l i c e n s e o r c o p y r i g h t ; s e e h t t p ://w w w .s i a m .o r g /j o u r n a l s /o j s a .p h p

COMPRESSIBLE NAVIER–STOKES–POISSON EQUATIONS 551?It is worth pointing out that since (1.13) μ(v )u x v x = μ(v )v t v x = μ(v )v x v 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 μdepends on θ.We now review some related results.First,there are some recent results on the construction of nonvacuum,large solutions to the one-dimensional compress-ible Navier–Stokes equations with constant viscosity coe?cient μand density-and temperature-dependent heat-conductivity coe?cient κ;see [18],[23].A key ingredi-ent in these works is the pointwise a priori estimates on the speci?c volume which guarantees that no vacuum or concentration of mass occurs.This together with the standard maximum principle deduce a lower-bound estimate on the absolute tempera-ture θ(t,x )and,consequently,the main e?ort in [18],[23]is to obtain the upper-bound

estimate on θ(t,x ).The strategy to prove Theoerem 1.1can be stated as follows.We note that,however,for the compressible NSP system (1.1),even when the viscosity coe?cient μ(v )is a positive constant,the argument in [25]does not give bounds on v (t,x )because of the nonlocal term Φx v .For this,to prove Theorem 1.1,we will ?rst apply the maximum principle for second-order parabolic equations to obtain a lower-bound estimate on 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 θ1?b L ∞([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 2.9then 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 ):(1.14)12Θ≤θ(t,x )≤2Θfor (t,x )∈[0,T ]×R .Then by some delicate energy-type estimates and by using the argument initiated in [21],we can deduce a uniform (with respect to the time variable t )lower and upper bound on v (t,x )and some uniform-energy estimates on v ?1,u,(θ?1)/√γ?1 (t ) H 1(R )in terms of v 0?1,u 0,(θ0?1)/√γ?1 H 1(R ),inf x ∈R v 0(x ),and sup x ∈R v 0(x ).At the end,to extend the solution globally in time,

we only need to close the a priori assumption (1.14)where we need the smallness of γ?1.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,the global existence of smooth small perturbative solutions away from vacuum with the optimal time-decay estimates was recently obtained in [26]for the isentropic ?ow,and in [37],[16]for the nonisentropic ?ow.There,it is observed that the electric ?eld a?ects the large time behavior of the solution so that the momentum decays at the rate (1+t )?14which is slower than the rate (1+t )?34for the compressible Navier–Stokes system,while the density tends to its asymptotic state at the rate (1+t )?34just like the compressible Navier–Stokes system.Moreover,the global existence of a strong solution in Besov-type space was obtained in [15].On the other hand,it is D o w n l o a d e d 09/30/13 t o 202.121.182.203. R e d i s t r i b u t i o n s u b j e c t t o S I A M l i c e n s e o r c o p y r i g h t ; s e e h t t p ://w w w .s i a m .o r g /j o u r n a l s /o j s a .p h p

552Z.TAN,T.YANG,H.ZHAO,AND Q.ZOU quite di?erent for the compressible Euler–Poisson system.In fact,it was shown in [14]

that the long time convergence rate of the global irrotational solution is enhanced by the dispersion e?ect due to the coupling of the electric ?eld,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 κare constants,it is straightforward to show that they hold when μ,ν,and κare smooth functions of density and temperature.

Finally,for the results with large initial data,the existence of renormalized solu-tions to the NSP system are obtained in [6],[33],[38].Note that for the compressible NSP system related to the dynamics of self-gravitating gases in 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 isen-tropic polytropic gas was studied with a special requirement on the range of adiabatic exponent,i.e.,γ>32with constant viscosity coe?cient.

For the nonisentropic 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”;see [11].The rest of the paper is organized as follows.The proofs of Theorems 1.1and 1.2will be given in sections 2and 3,respectively.Notations.O(1)or C i (i ∈N )stands for a generic positive constant which is independent of t and x ,while C i (·,···,·)(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. · s represents the norm in H s (R )with · = · 0and for 1≤p ≤+∞,L p (R )denotes the standard Lebesgue space.2.The proof of Theorem 1.1.To prove Theorem 1.1,we ?rst de?ne the following function space for the solution to the Cauchy problem (1.1),(1.3):X (0,T ;M 0,M 1;N 0,N 1)(2.1)=?????????????(v,u,θ,Φ)(t,x ) (v ?1,u,θ?1)(t,x )∈C 0 0,T ;H 1(R ) (u x ,θx )(t,x )∈L 2 0,T ;H 1(R ) Φx (t,x )∈C 0 0,T ;H 2(R ) M 0≤v (t,x )≤M 1,N 0≤θ(t,x )≤N 1?????????????.Here T >0,M 1≥M 0>0,N 1≥N 0>0are some positive constants.Under the assumptions given in either Theorems 1.1or 1.2,we can get the fol-lowing local existence result.Lemma 2.1(local existence ).Under the assumptions in either Theorems 1.1or 1.2,there exists a su?ciently small positive constant t 1,which depends only on V ,V ,Θ,Θ,and (v 0?1,u 0,θ0?1) 1,such that the Cauchy problem (1.1),(1.3)admits a unique smooth solution (v (t,x ),u (t,x ),θ(t,x ),Φ(t,x ))∈X 0,t 1;12V ,2V ;12Θ,2Θ and (v (t,x ),u (t,x ),θ(t,x ),Φ(t,x ))satisfy ???0

COMPRESSIBLE NAVIER–STOKES–POISSON EQUATIONS 553(2.4)lim |x |→∞(v (t,x )?1,u (t,x ),θ(t,x )?1,Φx (t,x ))=(0,0,0,0).Lemma 2.1can 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 t x =u x + Φx v x t =u x +(1?v )t =u x ?v t =0,we have the following lemma from (2.4).Lemma 2.2.Under the conditions in Lemma 2.1,we have

(2.5)u (t,x )=? Φx (t,x )v (t,x ) 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),and thus the Cauchy problem (1.1),(1.3)can be rewritten as ???????????????v t ?u x =0,u t +p (v,θ)x = u x v 1+a x +Φx v ,C v θt +p (v,θ)u x =u 2x v 1+a + θb θx v x , Φx v x =1?v,lim |x |→+∞Φ(t,x )=0,(2.6)(v (0,x ),u (0,x ),θ(0,x ))=(v 0(x ),u 0(x ),θ0(x )),lim |x |→+∞(v 0(x ),u 0(x ),θ0(x ))=(1,0,1).(2.7)Suppose that the local solution (v (t,x ),u (t,x ),θ(t,x ),Φ(t,x ))constructed in Lemma 2.1has been extended to t =T ≥t 1and satis?es the a priori assumption (H 1)V 0≤v (t,x )≤V 1,Θ0≤θ(t,x )≤Θ1

for all x ∈R ,0≤t ≤T,and some positive constants 0<Θ0≤Θ1,0

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 1but may depend on T .The ?rst one is concerned with the basic energy estimate.For this,note that η(v,u,θ)=Rφ(v )+u 22+Rφ(θ)γ?1with φ(x )=x ?ln x ?1is a convex entropy to (2.6)which satis?es (2.8)η(v,u,θ)t + Rθv ?R u x ? uu x v 1+a +(θ?1)θx vθ1?b x + u 2x v 1+a θ+θ2x vθ2?b =u Φx v .With (2.8),since u Φx v = u Φv +Φv Φx v t x ?12 Φx v 2 t +Φv x v 2 u + Φx v t ,D o w n l o a d e d 09/30/13 t o 202.121.182.203. R e d i s t r i b u t i o n s u b j e c t t o S I A M l i c e n s e o r c o p y r i g h t ; s e e h t t p ://w w w .s i a m .o r g /j o u r n a l s /o j s a .p h p

554Z.TAN,T.YANG,H.ZHAO,AND Q.ZOU we can deduce the following lemma by integrating (2.8)with respect to t and x over

[0,T ]×R and from (2.5).Lemma 2.3(basic energy estimates ).Let the conditions in Lemma 2.1hold and

suppose that the local solution (v (t,x ),u (t,x ),θ(t,x ),Φ(t,x ))constructed in Lemma 2.1satis?es the a priori assumption (H 1),then we have for 0≤t ≤T that R η(v,u,θ)+12 Φx v 2 (t,x )dx + t 0 R u 2x v 1+a θ+θ2x vθ2?b (τ,x )dxdτ= R η(v 0,u 0,θ0)+12 Φ0x v 0 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 <1that

(2.10)1θ(t,x )≤O (1)+O (1) 1v 1?a L ∞T,x ,x ∈R ,0≤t ≤T.Proof .First of all,(2.6)3implies C v 1θ t =?u 2x θ2v 1+a +Ru x vθ?2θ1+b v 1θ x 2+ θb v 1θ x x = θb v 1θ x x ? 2θ1+b v 1θ x 2+1v 1+a θ2 u x ?Rθv a 2 2 (2.11)+R 24v 1?a .Set h (t,x )=1θ?R 2t 4C v 1v 1?a L ∞T,x and we can deduce that h (t,x )satis?es ???C v h t ≤ θb v h x x ,x ∈R ,0≤t ≤T,h (0,x )=1θ0(x )≤1Θ,(2.12)and the standard maximum principle for parabolic equations implies that h (t,x )≤1Θholds for all (t,x )∈[0,T ]×R .That is,for x ∈R ,0≤t ≤T ,(2.13)1θ?R 2t 4C v 1v 1?a L ∞T,x ≤1Θ.This is (2.10)and the proof of Lemma 2.4is completed.To use Kanel’s method to deduce a lower bound and an upper bound on v (t,x ),we need to deduce an estimate on v x v 1+a ,which is the main concern of our next lemma.D o w n l o a d e d 09/30/13 t o 202.121.182.203. R e d i s t r i b u t i o n s u b j e c t t o S I A M l i c e n s e o r c o p y r i g h t ; s e e h t t p ://w w w .s i a m .o r g /j o u r n a l s /o j s a .p h p

COMPRESSIBLE NAVIER–STOKES–POISSON EQUATIONS 555It is worth pointing out that it is in this step that we ask the viscous coe?cient μ

depend only on v .Lemma 2.5.Under the assumptions in Lemma 2.3

,we have v x v 1+a 2+ t 0 R θv 2x v 3+a +g (v )(v ?1) dxds (2.14)≤ v 0x 2+ (v 0?1,u 0,θ0?1,Φ0x ) 2

+ t 0 R u 2x v 1+a dxds +O (1) t 0 R θ2x v 1+a θdxds and g (v )= v 1dz z 1+a =1?v ?a a .Proof .Notice that v x v 1+a t = v t v 1+a x = u x v 1+a x =u t +p (v,θ)x ?Φx v .We have by multiplying the above identity by v x v 1+a and integrating the resulting

equation with respect to t and x over [0,T ]×R that 12 v x v 1+a 2+ t 0 R Rθv 2x v 3+a dxds (2.15)≤O (1) v 0x 2+ t 0 R Rθx v x v 2+a dxds I 1+ t 0 R u t v x v 1+a dxds I 2? t 0 R v x v 1+a Φx v dxds I 3.Now we estimate I 1,I 2,and I 3term by term.First,we have from (2.6)4and the Cauchy–Schwarz inequality that I 3= t 0 R g (v )x Φx v dxds =? t 0 R g (v ) Φx v x dxds (2.16)=? t 0 R g (v )(1?v )dxds ≥0and (2.17)I 1≤12 t 0 R Rθv 2x v 3+a dxds +O (1) t 0 R θ2x v 1+a θdxds.As to I 2,we have from (2.9)that I 2= R uv x v 1+a dx t 0? t 0 R u v x v 1+a t dxds (2.18)≤ R uv x v 1+a dx +O (1) (u 0,v 0x ) 2? t 0 R u u x v 1+a x dxds ≤12 v x v 1+a 2+O (1) (v 0?1,v 0x ,u 0,θ0?1,Φ0x ) 2+ t 0 R u 2x v

1+a dxds.D o w n l o a d e d 09/30/13 t o 202.121.182.203. R e d i s t r i b u t i o n s u b j e c t t o S I A M l i c e n s e o r c o p y r i g h t ; s e e h t t p ://w w w .s i a m .o r g /j o u r n a l s /o j s a .p h p

556Z.TAN,T.YANG,H.ZHAO,AND Q.ZOU Inserting (2.16)–(2.18)into (2.15),we can deduce (2.14)immediately.This com-

pletes the proof of Lemma 2.5.To bound the two terms on the right-hand side of (2.14),we now estimate t 0 R u 2x v 1+a dxds in the following lemma.Lemma 2.6.Under the assumptions in Lemma 2.3,we have

(2.19) u (t ) 2+ t 0 R u 2x v 1+a dxds ≤O (1) (v 0?1,u 0,θ0?1,Φ0x ) 2+O (1) t 0 R (θ?1)2v 1?a dxds.Proof .Multiplying (2.6)2by u ,we have by integrating the resulting equation with respect to t and x over [0,T ]×R that 12 u (t ) 2+ t 0 R u 2x v 1+a dxds (2.20)≤O (1) u 0 2+ t 0 R R (θ?1)u x v dxds I 4+ t 0 R R 1?1v u x dxds I 5+ t 0 R u Φx v dxds I 6

.From the basic energy estimate (2.9)and the Cauchy–Schwarz inequality,we can bound I j (j =4,5,6)as follows:I 6≤ t 0 u (s ) Φx v (s ) ds ≤C (T ) (u 0,v 0?1,θ0?1,Φ0x ) 2,

I 5= t 0 R R 1?1v v t dxds =R R φ(v )dx t 0=R R φ(v )dx ? R φ(v 0)dx ≤O (1) (u 0,v 0?1,θ0?1,Φ0x ) 2,I 4≤12 t 0 R u 2x v 1+a dxds +O (1) t 0 R (θ?1)2v 1?a dxds.Substituting the above estimates into (2.20),we can deduce (2.19)and complete the proof of the lemma.To bound the terms appearing on the right-hand side of (2.19)and (2.14),we

need the following lemma.Lemma 2.7.Under the assumptions in Lemma 2.3,we have for b =0,?1that t 0max x ∈R |θ(s,x )|b ds ≤C (T ),

(2.21) t 0max x ∈R |θ(s,x )|b +1ds ≤C (T ) 1+ θ L ∞T,x ,(2.22)and (2.23) t 0max x ∈R |θ(s,x )|b +1ds ≤C (T ) 1+ v L ∞T,x .D o w n l o a d e d 09/30/13 t o 202.121.182.203. R e d i s t r i b u t i o n s u b j e c t t o S I A M l i c e n s e o r c o p y r i g h t ; s e e h t t p ://w w w .s i a m .o r g /j o u r n a l s /o j s a .p h p

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 for each i ∈Z ,there are positive constants A 0>0,A 1>0which are independent of i such that

(2.24)A 0≤ i +1i v (t,x )dx, i +1

i θ(t,x )dx ≤A 1?t ∈[0,T ].Hence,there exist a i (t )∈[i,i +1],b i (t )∈[i,i +1]such that

(2.25)A 0≤v (t,a i (t )),θ(t,b i (t ))≤A 1.

De?ne g 1(θ)=

θ1s b ?12ds =2b +1 θb +1

2?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 ≥b i (t ).Thus g 1(θ(t,x ))=g 1(θ(t,b i (t )))+

x

b i (t )

g 1(θ(t,y ))y dy ≤O (1)+ i +1i

θb ?12θx dx ≤O (1)+ R

θ2x vθ2?b dx 12

i +1i vθdx 12≤O (1)+ θ 12

L ∞T,x R θ2x vθ2?b dx 1

2.The above estimate and (2.9)give (2.22)and thus completes the proof of the lemma.As a direct corollary of (2.21)–(2.23),we have the following corollary.

Corollary 2.1.Under the conditions in Lemma 2.3,we have (2.26) t 0

R

(θ?1)2dxds ≤O (1) θ1?b L ∞T,x

.

Proof .In fact,(2.9)together with (2.21)imply that t 0 R (θ?1)2dxdτ≤O (1) t 0 R (θ+1)φ(θ)dxdτ≤O (1) t

0max x ∈R θ(τ,x )dτ+O (1)=O (1)

t

0max x ∈R

θ1?b θb dτ+O (1)≤O (1) θ1?b L ∞

T,x

t 0max x ∈R

θb (τ,x )dτ+O (1)≤O (1) θ1?b L ∞T,x

+O (1),and this completes the proof of the corollary.

D o w n l o a d e d 09/30/13 t o 202.121.182.203. R e d i s t r i b u t i o n s u b j e c t t o S I A M l i c e n s e o r c o p y r i g h t ; s e e h t t p ://w w w .s i a m .o r g /j o u r n a l s /o j s a .p h p

Having obtained (2.26),we can deduce that (2.27) t 0 R (θ?1)2v 1?a dxdτ≤ 1v 1?a L ∞T,x t 0 R (θ?1)2dxdτ≤O (1) θ1?b L ∞T,x 1v 1?a L ∞T,x .On the other hand,from (2.9),we have t 0 R θ2x θv 1+a dxdτ= t 0 R θ2

x vθ2?b 1

v a θb ?1

dxdτ≤ 1v a L ∞T,x

θ1?b L ∞

T,x t 0 R θ2

x

vθ2?b dxdτ≤O (1) (v 0?1,u 0,θ0?1,Φ0x ) 2 1v a L ∞T,x

θ1?b L ∞T,x .(2.28)Substituting (2.27)and (2.28)into (2.19)and (2.14),we have the following

corollary.

Corollary 2.2.Under the assumptions in Lemma 2.3,we have u (t ) 2+ t 0 R

u 2x v 1+a dxdτ(2.29)≤O (1) (v 0?1,u 0,θ0?1,Φ0x ) 2+O (1) 1v a L ∞T,x θ1?b L ∞T,x

, v x v 1+a 2+ t 0 R

θv 2

x

v 3+a +g (v )(v ?1) dxdτ(2.30)≤O (1) (v 0?1,u 0,θ0?1,Φ0x ) 2+O (1) 1v a L ∞T,x 1v 1?a L ∞

T,x

θ1?b L ∞T,x .Now we apply Kanel’s approach to deduce a lower bound and an upper bound on v (t,x )in terms of θ1?b L ∞T,x

.To this end,set

(2.31)Ψ(v )= v 1

φ(z )

z 1+a dz.Note that there exist positive constants A 2,A 3such that (2.32)|Ψ(v )|≥A 2 v ?a +v 12?a ?A 3.

Since

|Ψ(v )|= x ?∞

Ψ(v (t,y ))y dy ≤ R φ(v )v

1+a v x dx ≤

φ(v ) v x v 1+a (2.33)≤O (1) 1+ 1v a 2L ∞T,x +

1v 1?a 2L ∞T,x

θ1?b 1

2L ∞T,x ,D o w n l o a d e d 09/30/13 t o 202.121.182.203. R e d i s t r i b u t i o n s u b j e c t t o S I A M l i c e n s e o r c o p y r i g h t ; s e e h t t p ://w w w .s i a m .o r g /j o u r n a l s /o j s a .p h p

we have from (2.32)and (2.33)that (2.34) 1v a L ∞T,x + v 12?a L ∞T,x ≤O (1) 1+ 1v a 2L ∞T,x + 1v 1?a 2L ∞

T,x

θ1?b 12L ∞T,x .Thus if 13

2,we can deduce from (2.34)the following corollary.

Corollary 2.3.Under the conditions in Lemma 2.3,if we assume further that 13

(2.35)1v (t,x )≤O (1) 1+ θ1?b 13a ?1

L

∞T,x

and (2.36)v (t,x )≤O (1) 1+ θ1?b 2a (3a ?1)(1?2a )L ∞T,x hold for any (t,x )∈[0,T ]×R .

Consequently,(2.29)and (2.30)can be rewritten as u (t ) 2+ t 0 R u 2x v

1+a dxdτ≤O (1) 1+ θ1?b 2a 3a ?1L ∞T,x ,(2.37) v x v 1+a 2+ t 0 R

θv 2x v 3+a +g (v )(v ?1) dxdτ≤O (1) 1+ θ1?b 2a 3a

?1L ∞T,x .(2.38)To get an upper bound on θ(t,x ),we need also the estimate on u x (t ) 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 u x (t ) 2+ v (t ) 2+ t 0 R u 2xx

v 1+a

dxdτ(2.39)≤O (1) (v 0?1,u 0,θ0?1,Φ0x ) 2+O (1) θ2?b L ∞T,x 1+ θ1?b 2a 2(3a ?1)(1?2a )

L ∞

T,x +O (1) 1+ θ1?b 2(2a ?2a 2+1)(3

a ?1)(1?2a )

L ∞T,x

.

Proof .By di?erentiating (2.6)2with respect to x ,multiplying the resulting iden-tity by u x ,and integrating the result with respect to t and x over [0,T ]×R ,we have u x (t ) 2+ t 0 R u 2xx v 1+a

dxdτ+ v ?1 2

(2.40)≤O (1) u 0x 2+2 t 0 R u xx p (v,θ)x dxdτ I 7+2(1+a ) t 0

R

u x v x u xx v 2+a dxdτ I 8.D o w n l o a d e d 09/30/13 t o 202.121.182.203. R e d i s t r i b u t i o n s u b j e c t t o S I A M l i c e n s e o r c o p y r i g h t ; s e e h t t p ://w w w .s i a m .o r g /j o u r n a l s /o j s a .p h p

For I 7,we have from (2.9)that I 7=2R t 0

R

u xx θx v ?θv x

v 2 dxdτ(2.41)≤14 t 0 R u 2xx v 1+a dxdτ+O (1) t 0 R θ2x

v 1?a dxdτ+O (1) t 0 R θ2v 2

x v

3?a dxdτ≤14 t 0 R u 2xx v 1+a dxdτ+O (1) θ2?b L ∞T,x 1+ θ1?b 2a 2(3a ?1)(1?2a )

L ∞T,x +O (1) 1+ θ1?b 7

a ?4a 2?1(3a ?1)(1?2a )L ∞T,x .Here we have used the fact that

t 0 R θ2x v 1?a dxdτ= t 0 R θ2

x vθ2?b

v a θ2?b dxdτ≤O (1) v a L ∞

T,x θ2?b L ∞T,x

t 0

R θ2x vθ2?b dxdτ≤O (1) v a L ∞

T,x

θ2?b L ∞T,x ≤O (1) θ2?b L ∞T,x 1+ θ1?b 2a 2(3a ?1)(1?2a )

L ∞T,x

and

t 0 R θ2v 2x v 3?a dxdτ= t 0 R

v 2

x v 2+2a θ2

v 1?3a dxdτ≤ t 0 R max x ∈R θ2(s,x )ds v 3a ?1L ∞T,x R v 2x

v 2+2a dx dτ≤O (1) 1+ θ1?b 2a 3a ?1

L ∞T,x v 3a ?1L ∞T,x t 0 max x ∈R

θ2(s,x )ds

≤O (1) 1+ θ1?b 2a 3a ?1L ∞T,x v 3a ?1L ∞T,x t 0

max x ∈R θ1?b θ1+b

(s,x )ds

≤O (1) 1+ θ1?b 5a ?13a ?1

L ∞T,x v 3a ?1L ∞T,x t 0

max x ∈R θ1+b (s,x )ds ≤O (1) 1+ θ1?b 5

a ?13a ?1L ∞T,x

v 3a ?1L ∞T,x 1+ v L ∞

T,x ≤O (1) 1+ θ1?b 7

a ?4a 2?1(3a ?1)(1?2a )

L ∞T,x ,where (2.9),(2.21)–(2.23),and (2.38)are used.

As for I 8,since (2.36),(2.37),together with the Sobolev inequality,imply t 0 u x (τ) 2L ∞x dτ≤ t

u x (τ) u xx (τ) dτ(2.42)≤ t 0 u x (τ) 2dτ 12 t 0

u xx (τ) 2

dτ 12

D o w n l o a d e d 09/30/13 t o 202.121.182.203. R e d i s t r i b u t i o n s u b j e c t t o S I A M l i c e n s e o r c o p y r i g h t ; s e e h t t p ://w w w .s i a m .o r g /j o u r n a l s /o j s a .p h p

≤ v 1+a L ∞

T,x

t 0 u x v 1+a 2(τ) 2dτ 12 t 0 u xx v 1+a 2(τ) 2dτ 12≤O (1) 1+ θ1?b 3a (3a ?1)(1?2a )L ∞T,x t 0 u xx v 1+a 2(τ) 2dτ 12,we can deduce from (2.35)–(2.38)that

I 8≤14 t 0 R u 2xx v 1+a dxdτ+O (1) t 0 R u 2x v 2

x v 3+a dxdτ(2.43)

≤14 t 0 R u 2xx v 1+a dxdτ+O (1) t 0 u 2x v 1?a L ∞x

R v 2x v 2+2a dxdτ≤14 t 0 R u 2xx v 1+a dxdτ+O (1) 1v 1?a L ∞T,x 1+ θ1?b 2a 3a ?1L ∞T,x t 0 u x (τ) 2L ∞x dτ≤14 t 0 R u 2xx v 1+a dxdτ+O (1) 1+ θ1?b 2a ?2a 2+1(3a ?1)(1?2a )L ∞

T,x t 0 R u 2xx v

1+a dxdτ 12≤12 t 0 R u 2xx v 1+a dxdτ+O (1) 1+ θ1?b 2(2a ?2a 2+1)(3a ?1)(1?2a )L ∞T,x .Putting (2.40),(2.41),and (2.43)together and noticing that 2(2a ?2a 2+1)>7a ?4a 2?1imply (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 (2.44) θ L ∞T,x ≤O (1) 1+ t 0 u 2x v 1+a

L ∞x + u 2x v 2 L ∞x + θ 2L ∞x dτ .Proof .From (2.6)3,it is easy to see that for each p >1,C v (θ?1)2p t +2p (2p ?1)(θ?1)2(p ?1)θb θ2x v

(2.45)= 2p (θ?1)2p ?1θb θx v x

+2p (θ?1)2p ?1v 1+a u 2x ?2pRθ

v u x (θ?1)2p ?1.Integrating (2.45)with respect to x over R ,we have (2.46)C v θ?1 2p L 2p t ≤2p R u 2x (θ?1)2p ?1

v 1+a dx I 9?2pR R

θu x (θ?1)2p ?1v dx

I 10

.Since

I 9≤2pO (1) θ?1 2p ?1L 2p u 2x v 1+a L 2p

,I 10≤2pO (1) θ?1 2p ?1L 2p θu x v L 2p

hold for some positive constant O (1)independent of p ,we have D o w n l o a d e d 09/30/13 t o 202.121.182.203. R e d i s t r i b u t i o n s u b j e c t t o S I A M l i c e n s e o r c o p y r i g h t ; s e e h t t p ://w w w .s i a m .o r g /j o u r n a l s /o j s a .p h p

(2.47) θ?1 L 2p ≤O (1)+O (1) t 0 u 2x v 1+a

L 2p

+

θu x v L

2p dτ.Letting p →∞in (2.47)and exploiting the Cauchy inequality,we can deduce (2.44)immediately and the proof of Lemma 2.9is complete.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 ).First,we have from (2.42)and (2.39)that

t 0 u x (s ) 2L ∞x ds ≤O (1) 1+ θ1?b 3a (3a ?1)(1

?2a )L ∞T,x

(2.48)× θ2?b 12L ∞T,x 1+ θ1?b a 2(3a ?1)(1?2a )L ∞T,x +1+ θ1?b 2a ?2a 2+1(3a ?1)(1?2a )

L ∞T,x

≤O (1) θ2?b 12L ∞T,x 1+ θ1?b 3a +a 2(3a ?

1)(1?2a )L ∞T,x +O (1) θ1?b 5

a ?2a 2+1(3a ?1)(1?2a )

L ∞T,x

+O (1).Thus,we have from (2.35)–(2.36),(2.48)that

t 0 u 2x v 1+a L ∞x + u 2x v 2

L ∞x

dτ(2.49)

≤O (1) 1v 1+a L ∞T,x + 1v 2L ∞T,x

t

0 u x (τ) 2L ∞x dτ≤O (1) 1+ θ1?b 2

3a ?1L ∞T,x

t 0

u x (τ) 2L ∞x dτ≤O (1) θ2?b 12L ∞T,x 1+ θ1?b a

2?a +2(3a ?1)(1?2a )L ∞T,x +O (1) θ1?b 3+a ?2a 2(3a ?1)(1?2a )L ∞T,x

+O (1)and t 0max x ∈R θ2(s,x )ds ≤ t

0max x ∈R

θ1?b

(s,x

)θb +1(s,x ) ds (2.50)≤

θ1?b L ∞

T,x t 0max x ∈

R θ1+b (s,x )ds ≤O (1) θ1?b L ∞T,x

1+ v L ∞T,x ≤O (1) 1+ θ1?b 7a ?6a 2?1(3a ?1)(1?2a )L ∞T,x

.Inserting (2.49)and (2.50)into (2.44)yields

θ L ∞T,x

≤O (1)+O (1) θ2?b 12L ∞T,x 1+ θ1?b a

2?a +2(3a ?1)(1?2a )

L ∞T,x

(2.51)+O (1) θ1?b 3+a ?2a 2(3a ?1)(1?2a )L ∞T,x

+O (1) θ1?b 7a ?6a 2?1(3a ?1)(1?2a )

L ∞T,x D o w n l o a d e d 09/30/13 t o 202.121.182.203. R e d i s t r i b u t i o n s u b j e c t t o S I A M l i c e n s e o r c o p y r i g h t ; s e e h t t p ://w w w .s i a m .o r g /j o u r n a l s /o j s a .p h p

≤O (1)+O (1) θ2?b 12L ∞T,x 1+ θ1?b a 2?a +2(3a ?1)(1?2a )L ∞T,x +O (1) θ1?b 3+a ?2a 2(3a

?1)(1?2a )L ∞T,x

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

Corollary 2.4.Under the assumptions in Lemma 2.3,we further assume that 13

2and one of the following conditions holds:(i)1≤b <2a

1?a <2;

(ii)0

2+(a 2?a +2)(1?b )(3a ?1)(1?2a )<1,(1?b )(3+a ?2a 2)(3a ?1)(1?2a )<1.Then there exist positive constants V 1>0,Θ1>0,such that

(2.52) V ?1

1≤v (t,x )≤V 1,Θ?1

1≤θ(t,x )≤Θ1.Proof .We ?rst consider the case b ≥1.In this case,we have from (2.10),(2.35),and (2.36)that 1θ L ∞T,x ≤O (1)+O (1) θ1?b 1?a 3a ?1L ∞T,x

≤O (1)+O (1) 1θ

(1?a

)(b ?1)3a ?1L ∞T,x

which,together with the assumption b <2a

1?a ,implies that there exists a positive constant Θ1>0such that (2.53)θ(t,x )≥Θ?1

1>0?(t,x )∈[0,T ]×R .

Moreover,(2.35),(2.36),(2.53)together with the fact that b ≥1imply that there exists a positive constant V 1>0,which may depend on T ,such that

(2.54)V ?11≤v (t,x )≤V 1?(t,x )∈[0,T ]×R .

Thus to prove (2.52),we only need to deduce the upper bound on θ(t,x ).For this purpose,we have from the fact 1≤b <2a

1?a <2,(2.53),and (2.51)that (2.55) θ L ∞T,x ≤O (1)+O (1) θ 2?b 2L ∞

T,x 1+ 1θ (a 2?a +2)(b ?1)(3a ?1)(1?2a )L ∞T,x +O (1) 1θ (3+

a ?2a 2)(

b ?1)(3a

?1)(1?2a )L ∞T,x

≤O (1) 1+ θ 2?b

2L ∞

T,x

.From (2.55)and the fact that 0<2?b

2

<1,one can easily deduce an upper bound on θ(t,x ).This completes the proof of (2.52)for the case 1≤b <2a 1?a .

When b <1,we have from (2.51)that

(2.56) θ L ∞T,x ≤O (1)+O (1) θ 2?b 2+(a 2?a +2)(1?b )(3a ?1)(1?2a )L ∞T,x +O (1) θ (3+a ?2a 2)(1?b )(3a ?1)(1?2a )

L ∞T,x

.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 1.1follows from the standard continuation

argument.

D o w n l o a d e d 09/30/13 t o 202.121.182.203. R e d i s t r i b u t i o n s u b j e c t t o S I A M l i c e n s e o r c o p y r i g h t ; s e e h t t p ://w w w .s i a m .o r g /j o u r n a l s /o j s a .p h p

3.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,t 1;12V ,2V ;12Θ,2Θ with t 1depending

on V ,V ,Θ,Θ, (v 0?1,v 0,θ0?1,Φ0x ) 1can 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 ≥t 1.To apply the continuation argument for global existence,we ?rst set the following a priori estimate:

(H 2)

1

2Θ≤θ(t,x )≤2Θ,(t,x )∈[0,T ]×R .Here without loss of generality,we can assume that 0<Θ<1<Θ.

Note that the smallness of γ?1is needed to close the a priori estimate,the generic constants used later are independent of γ?1,and,the dependence on this factor will be clearly stated in the estimates when needed.

Similar to Lemma 2.3we have the following basic energy estimate.

Lemma 3.1.Under the conditions in Theorem 1.2,we have for 0≤t ≤T that

R Rφ(v )+u 22+R γ?1φ(θ)+Φ2x 2v 2 (t,x )dx + t 0 R

μ(v )u 2x vθ+κ(v,θ)θ2x

vθ2 dxdτ(3.1)

= R

Rφ(v 0)+u 202+R γ?1φ(θ0)+Φ2

0x

2v 20 (x )dx.Here,as in section 2,φ(x )=x ?ln x ?1.

Now,we deduce an estimate on μ(v )v x

v .For this,similar to Lemma 2.5,we can

deduce μ(v )v x v 2+ t 0 R μ(v )θv 2x v 3dxdτ+ t 0 R g (v )(1?v )dxdτ(3.2)≤O (1) v 0x 2+O (1)

t 0 R μ(v )u 2x v dxdτ J 1+O (1) t 0 R

μ(v )θ2

x vθdxdτ

J 2

.If the a priori estimate (H 2)holds,we have from (3.1)and the assumptions imposed on κ(v,θ)in Theorem 1.2that (3.3)J 1≤O (1) t 0 R μ(v )u 2x vθdxdτ≤O (1) v 0?1,u 0,θ0?1√γ?1,Φ0x 2and

J 2≤ t 0 R κ(v,θ)θ2x vθ2θμ(v )κ(v,θ)dxdτ(3.4)≤O (1) μ(v )κ1(v ) L

t 0 R κ(v,θ)θ2x

vθ2dxdτ≤O (1) v 0?1,u 0,θ0?1√γ?1,Φ0x 2

μ(v )κ1(v ) L ∞.Putting (3.2),(3.3),and (3.4)together,we obtain the following lemma.Lemma 3.2.Under the assumptions in Lemma 3.1and the a priori assumption

(H 2),we have

D o w n l o a d e d 09/30/13 t o 202.121.182.203. R e d i s t r i b u t i o n s u b j e c t t o S I A M l i c e n s e o r c o p y r i g h t ; s e e h t t p ://w w w .s i a m .o r g /j o u r n a l s /o j s a .p h p

μ(v )v x v 2+ t 0 R

μ(v )v 2x v 3dxdτ(3.5)≤O (1) v 0?1,u 0,θ0?1√γ?1,Φ0x ,v 0x 2

1+ μ(v )κ1(v ) L ∞T,x

.Having obtained (3.1)and (3.5),we can use Kanel’s argument,see [21],to deduce the lower and upper bounds on v (t,x )as follows.

Lemma 3.3.Under the assumptions in Theorem 1.2and Lemma 3.2,there exists a positive constant V 2≥1,which depends only on v 0?1,u 0,θ0?1√γ?1,Φ0x ,v 0x

,V ,V ,Θ,and Θ,but is independent of T ,such that (3.6)V ?12≤v (t,x )≤V 2,(t,x )∈[0,T ]×R .

Proof .De?ne

Ψ(v )= v 1

φ(z )

z μ(z )dz,φ(z )=z ?ln z ?1

and notice that

|Ψ(v )|= x ?∞Ψ(v (t,y ))y dy ≤ R

φ(v )μ(v )v x v dx ≤ φ(v ) 12L 1 μ(v )v x v ≤O (1) v 0?1,u 0,θ0?1√γ?1

,Φ0x ,v 0x 2 1+

μ(v )κ1(v ) L ∞T,x

12

.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 u x (t ) .

Lemma 3.4.Under the assumptions in Lemma 3.3,we have for each 0≤t ≤T that

u x (t ) 2+ v (t )?1 2+ t 0 R

u 2xx v 1+a dxdτ(3.7)≤O (1) v 0?1,v 0x ,u 0x ,u 0x ,θ0?1√γ?1

,Φ0x 6.Since v (t,x )satis?es (3.6)and θ(t,x )is assumed to satisfy the a priori estimate

(H 2),(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 (H 2),we need to deduce an estimate on θx (t ) .For the case when κ(v,θ)≡κ(θ),we have the following lemma.Lemma 3.5.Under the assumptions in Theorem 1.2and Lemma 3.3,we have

R |K (θ)x |2

γ?1dx + t 0 R vκ(θ) K (θ)x v x

2dxdτ(3.8)≤O (1) v 0?1,u 0,θ0?1√γ?1,Φ0x

101

.Here

(3.9)K (θ)= θ

1

κ(z )dz.D o w n l o a d e d 09/30/13 t o 202.121.182.203. R e d i s t r i b u t i o n s u b j e c t t o S I A M l i c e n s e o r c o p y r i g h t ; s e e h t t p ://w w w .s i a m .o r g /j o u r n a l s /o j s a .p h p

Proof .Multiplying (1.1)3by κ(θ)and di?erentiating the resulting equation with respect to x ,we get (3.10)C v K (θ)tx +(κ(θ)p (v,θ)u x )x = κ(θ)μ(v )u 2x v x + κ(θ) K (θ)x v x x

.

Multiplying (3.10)by K (θ)x and integrating with respect to t and x over [0,t ]×R give R C v 2|K (θ)x |2

dx + t 0

R vκ(θ) K (θ)x v x

2dxdτ(3.11)≤O (1) θ0x √γ?1 2+O (1) t 0 R

K (θ)x v x

K (θ)x v v x dxdτ

J 3

+ t 0 R K (θ)x κ(θ)μ(v )u 2x v x dxdτ J 4? t 0 R K

(θ)x (κ(θ)p (v,θ)u x )x dxdτ J 5

.Notice that

θx L ∞x ≤O (1) K (θ)x v L ∞x

≤O (1) K (θ)x v 12 K (θ)x v x 1

2(3.12)≤O (1) θx 12 K (θ)x v x 1

2;we have from (3.1),(3.5),(3.6),and the a priori estimate (H 2)that J 3≤112 t 0 R vκ(θ) K (θ)x v x 2dxdτ+O (1) t 0 R v 2x θ2x dxdτ(3.13)

≤112 t 0 R vκ(θ) K (θ)x v x

2dxdτ+O (1) t 0

v x (τ) 2 θx (τ) vκ(θ) K (θ)x v x dτ≤16 t

0 R vκ(θ) K (θ)x v x

2dxdτ+O (1) t 0 v x (τ) 4 θx (τ) 2dτ≤16 t 0

R vκ(θ) K (θ)x v x

2dxdτ+O (1) v 0?1,u 0,θ0?1√γ?1,Φ0x 61,J 4=? t 0 R κ(θ)μ(v )u 2x K (θ)x v x

dxdτ? t 0 R K (θ)x v x κ(θ)μ(v )u 2x v 2dxdτ(3.14)

≤112 t 0 R vκ(θ) K (θ)x v x 2dxdτ+O (1) t 0 R u 4x dxdτ+O (1) t 0 R θ2x v 2x dxdτ≤16 t 0

R vκ(θ) K (θ)x v x

2dxdτ+O (1) v 0?1,u 0,θ0?1√γ?1,Φ0x 101,D o w n l o a d e d 09/30/13 t o 202.121.182.203. R e d i s t r i b u t i o n s u b j e c t t o S I A M l i c e n s e o r c o p y r i g h t ; s e e h t t p ://w w w .s i a m .o r g /j o u r n a l s /o j s a .p h p

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