FREE BOUNDARY VALUE PROBLEM OF ONE DIMENSIONAL TWO-PHASE LIQUID-GAS MODEL

Available online at www.sciencedirect.comMalhemiscHtientiaScienceDirect数学物理学报Acta Mathematica Scientia 2012,32B(1):413 432http:// actams. wipm.ac.cnFREE BOUNDARY VALUE PROBLEM OF ONEDIMENSIONAL TWO-PHASE LIQUID-GAS MODEL*Dedicated to Professor Constantine M. Dafermos on the occasion of his 70th birthdayWang Zhen (王振)Zhang Hui (张卉)Wuhan Institute of Physics and Mathermatics, The Chinese Academy of Sciences, Wuhan 430071, ChinaE-mail: zhenwang@wipm. ac. cn; huizhang1987@126.comAbstract In this paper, we study a free boundary value problem for two phase liquid-gas model with mass- dependent viscosity coefficient when both the initial liquid and gas ;masses connect to vacuum continuously. The gas is assumed to be polytropic whereas theliquid is treated as an incompressible fAuid. We give the proof of the global existence anduniqueness of weak solutions when β∈(0, 1), which have improved the result of Evje andKarlsen, and we obtain the regularity of the solutions by energy method.Key words two phase flow; weak solutions; Lagrangian coordinates; free boundary prob-lem; vacuum; uniqueness2000 MR Subject Classifcation 35B40; 35L721 IntroductionIn this paper, we consider a one dimensional viscous two phase liquid and gas model com-posed of two separate mass conservation equations corresponding to each of the two phases andone mixture momentum conservation equation in following form, cf [1- 3, 8].8-(agPg) + de(agPgug)=0, .8+(atp) + a:(aupru) = 0,(1.1)8,(arpru +agPgug) + a(apu2 + CogPou吗 + P)= -9+ a(e4mix),where Umix= agUg+Qu，P,ε ≥0.The unknown variables ag, au denote volume fractions satisfying the fundamental relationag+a=1.(1.2)Furthermore, the other unknown variables Pl, Pg,中国煤化工nsity, gas den-sity, velocities of liquid and gas respectively, P is comnMYHC N M H Ges, q representsexternal forces like gravity and friction and ε denotes viscosity coefficient.* Received January 4, 2012. Supported by the National Natural Science Foundation of China (11171340).ACTA MATHEMATICA SCIENTIAVol.32 Ser.BJust as in [11], in order to avoid some unsolved dificulties, we consider a simplified modelobtained as follows:(1) We assune that fuid velocites are equal ug = u = u and neglect the external forces,i.e., q= 0.(2) Due to the fact that the liquid phase density is much higher than the gas phasedensity, typically to the order pl/Pg = 0(103), we can neglect the gas phase efets in the .mixture momentum conservation equation.Consequently, the model (1.1) can be simplifed into the fllowing model of the form1 &.(agPg) + 0<(QgPgu) =0,8-(anp) + 8<(arpru)= 0,(1.3)( 8-(arpru) + ae(arpru2 + P) = e(e8 1 andC> 0. Let .n= QgPg, m= Q{PI.(1.4)By (1.2)-(1.3), we have( 8,n+ 8a(nu)= 0,8rm + ac(mu)= 0,(1.5)( 8-(mu) + a-(mu2 + P) = Be(e0eu),andP(n,m)=cp(="m)”.(1.6)In this paper, we consider (1.5) in a free boundary value problem setting where the masses .n and m initially occupy only a finite interval [a,b] C IR. That isn(,0) = no(E)>0, m(6,0) = mo(E)>0, u(5,0)= uo(), ξ∈[a,b),(1.7)and no(E) = mo(E) = 0 outside [a, b].The boundary conditions are given asn(a(+),r) = n(b(T),r)= 0,(1.8)m(a(),r)= m(b(r),T)= 0,where a(r) and 6() are the vacuurn boundary, i.e., the particle paths separating the gas andthe vacuum, satisfying:( da(r)= u(a(r),r),dT( a(0)=a.中国煤化工db(r)= u(b(+),),MYHCNMHGd(b(0)= b..No.1z. Wang & H. Zhang: FREE BOUNDARY VALUE PROBLEM415The viscosity coffcient ε is in general assumed to be a function of the masses n and m,we will considernβε=ε(n,m)= .(1.9)0q- m)B+I'where β∈(0, 1).To solve the free boundary value problem (1.5)- (1.8), it is convenient to convert the free .boundaries to fixed boundaries by using Lagrangian coordinates. Letx=m(z, r)dz,t= T.(1.10)Then the free boundariesξ = a(T) and ξ = b(T) becomex=0，and工=jN、,m(z, r)dz=mo(z)dz,(1.11)by the conservation of mass, where so mo(z)dz is the total liquid mass itially. We normalize .S。mo(z)dz= 1.Furthermore, in the Lagrangian coordinates, the free boundary value problem (1.5)-(1.8)becomes the following fixed boundary value problem:Ern + (nm)8xu= 0,8m + m2xu= 0,(1.12). Bru + 8xP(n, m) = 8x(e(n, m)m08_u), .with initial datan(x,0)= no(x), m(x,0) = mo(x)>0, u(x,0) = uo(x), x∈[0, 1],(1.13)and the boundary conditionsn(0,t)= n(1,t)= 0,(1.14)m(0,t)= m(1,t)= 0.For simplicity, we set Cp? = 1, then (1.6) can be simplified asP(n,m)=(_ n(1.15)(pi-m)， γ>1,andε=ε(n,m) =β∈(0,1).(1.16)(pr- m)8+1'LetC=n中国煤化工Then by (1.12)MYHCNMHGCt =-nt-m.__nm」nm--Ux= 0,mnt - m2mt=mm2416ACTA MATHEMATICA SCIENTIAVol.32 Ser.Band the initial boundary value problem (1.5)- -(1.8) can be rewritten in terms of variables (C, m, u)in the form8;c=0,8m+ m28xu= 0,(1.17)( Oru + 8x P(c,m) = Ax(E(c, m)axu), .with initial datac(x,0)= co(x), m(x,0) = mo(x)>0, u(x,0)= uo(x), x∈[0,1],(1.18)and the boundary conditionsc(0,t) = c(1,t)= 0,(1.19)m(0,t) = m(1,t)= 0,wheremcP(c,m) = .\pl- m)γ> 1,andE(c,m) = e(c, m)m=CmB+1(0pi-m)9HI，β∈(0,1).In particular, (1.17)1 gives thatnc(x,t)= co(x) :=mo(x)， for x∈[0,1], t≥0.Precisely, introduce the variablesmQ(m)=ρl- m'and observe thatQ(m)t= (pl-m)t~ \ρu-mt (ρr-m)互}mm2(p-m)m= -PI((PI- m)j) = -pQ(m)9ux.Consequently, we can rewrite the initial boundary value problem (1.17) in the form( 0rc= 0,atQ(m) + pnQ(m)28xu= 0,(1.20)( Oru + 8_P(c, Q(m)) = a(E(C, Q(m))8xu),c(x,0)=co(x)，Q(x,0)= Qo(x)>0, u(r cI∩11(1.21)中国煤化工fYHCNMHGc(0,t)=c(1,t)= 0,(1.22)Q(0,t)= Q(1,t)=0,No.1z. Wang & H. Zhang: FREE BOUNDARY VALUE PROBLEM17whereS P(c,m) = (cQ(m))"， γ>1,(1.23)E(c,Q(m)) = e(C, m)m = &Q(m)+1，β∈ (0,1),andQo(x) = Q(mo()) =mo(x)ρl - mo(x)For related free boundary problems for one dimensional two phase liquid -gas flow, whenviscosity is constant, S. Evje and T. Flatten in [10] obtained the global existence of weak solu-tion. When the viscosity depends on the liquid mass, like μ= CTor myprt, and the assumptionthat both the initial liquid and gas masses connect to vacuum with a discontinuity, S. Evjeand T. Flatten in [11] obtained the global existence of weak solutions when0< β<素. LaterL.Yao and C.J. Zhu in [12] obtained the existence and uniqueness of global weak solution when0< β≤1. When the viscosity depends on the liquid and gas mass, like μ = CTor mijsr,and the assumption that both the initial liquid and gas masses connect to vacuum continu-ously, S. Evje, T. Flatten, H.A. Fris in [13] obtained the global existence of weak solution when0<β<系In this paper, we establish the existence of global weak solution to two phase liquid-gas fow with the viscosity depends on the liquid and gas mass μ= Cton p,0<β< 1 andthe assumption that both the initial liquid and gas masses connect to vacuum continuously.The rest of this paper is organized as follows. In Section 2 we give the main result to themodel (1.12) obtained from (1.5). In Section 3 we describe a priori estimates for an auxiliarymodel obtained from (1.12) by employing an appropriate variable transformation. In Section4 we proof the uniqueness of the weak solution. Finally, in Section 5 we consider a family ofapproximate solutions obtained by deining a semi- discrete approximation to (1 20), which inturn imply compactness and convergence to a global weak solution, as stated in Theorem 2.1.2 The Main ResultsBefore we state the main result for the model (1.12), we describe the notation we applythroughout the paper. H'(I) represents the usual Sobolev space defined over I = (0, 1) withnorm II . |H1(I) Moreover, LP(K, B) with norm |I. |Lp(K,B)denotes the space of all stronglymeasurable, pth-power integrable functions ftom K to B where K typically is a subset of R andB is a Banach space. In addition, let a∈(0, 1), Ca[0, 1] denote the Banach space of functionson [0,1] which are uniformly Hoder continuous with exponent a and Ca,号 (Dr), the Banachspace of functions on Dr = [0, 1] X [0, T] which are uniformly Hoder continuous with exponentainx and号in t.In what follows, we always use C or C(T) to denote a generic positive constant dependingonly on the initial data (and the given time T).Main Assumptions(A1) no, mo be continuous function on [0, 1], van中国煤化工re are psitivieconstants Ci and C2 such thatYHCNMHGmo~ C1φ(x)°，co(x):= 心~ C2中(x)"，0<θ 1,元 0n∈L°([0,1} x (0,T])nC'(0,T]; L(I0, 1)),m∈L∞([0,1] x [0,T])∩C"([0,T]; L2([0, 1)),u∈L∞([0,1]x [0,T])nC"(0,T]; L2([0, 1)), .e(n, m)mux∈L∞([0,1] x [0,T)nC*([0,T]; L(0, 1)),and_lim. n(x,t)= lim n(x,t)=0，lim. m(x,t)= lim m(x,t)= 0.工- +0(2) Furthermore, the following equations holdnt+mnux=0,mt+m2ux=0,(n, m)(x,0) = (no(x),mo(x)), for a.e. x∈(0,1) and anyt≥0,[upr + (P(c,Q) - E(c,Q)ux)Q]Jdxdt +uo(x)φ(x,0)dx= 0,for any test function中(x,t)∈C*(D), where D:= {(x,t)|0≤x≤1,t≥0}.Remark 2.1 It is noticed that the set of initial data (no, mo, uo) satisfying all the as-sumptions is not empty. For example, if we choose no = A(x(1 - x))* with the exponent 8satisfying云≤8, then it satisfies all assumptions on density. Notice also that when initial datais given in the form of A(x(1 - x))", the condition (A3) implies that 兹28.Basic Estimates中国煤化工Lemma 3.1 (Some Useful Identities) we have foMHCNMHG.u(,td=-J .u(y,t)dy, .(3.1)dt JodNo.1z. Wang & H. Zhang: FREE BOUNDARY VALUE PROBLEM419(CQ+()u)(x,t) = (0.(3.4)Moreover,cQ(m)(x,t)≤C，V(x,t)∈ [0,1]x [0,T],(3.5)and for any positive integer lu21-2c@Q(m)+(ux)2dxds≤C(T).(3.6)Corollary 3.1 We have the estimatesc(x)Q(m)≤C(T)$(x)9+a，and Q(m) ≤C(T)$(x)°.The above lemmas can be found in [13] and the references therein.Lemma 3.3 We have for sufficiently large integer k(3.7)for a suitable constant C(T).Proof From (1.20), we have[(cQ)]t = -Bp(u + P(c, Q)x).(3.8)Integrating (3.8) with respect to t over [0, t], and then multiplying (3.8) by(x(1 - ()-1(]Q)2-1-and integrating in x over [0, 1] give(x(1 - x)k-1(cQ)£]2^x= [ (1 - )1C<)-1-(10)]x、中国煤化工+βpl[(ar )_HCNMHG-βpr(3.9)420ACTA MATHEMATICA SCIENTIAVol.32 Ser.BBy Young's inequalityf" (x(1- x)k +(Q)2r≤C.' (x(1- ()1-(01]*x+0号[' (x(l ()-1-(0)*2+of[((x(1- 1)2k-1u(x)*dx +cf (1 ()-1-(x)2*dx2o‘lkP(c,Q)xdsdx.(3.10)o、Consequently, applying (3.6) and (A1)- (A2), it fllows that2k、≤C(T) +Cf[(c( -)-(。P(c, Q)xdsdx/o≤C(T)+ C(T)max(cQ)2k(+-A) | (x(1 - x))2k-1[(cQ)出]2*dxds.(3.11)JIn view of (3.5), we can easily obtain (3.7) by Gronwall's inequality.CLemma 3.4 Let k been as in (A1), for any k1 >乖, we haveφk1dx≤C(T).(3.12)Jo Q(m(x,t))The proof of this lemma can be found in [13]. Specially takek1 =一in Lemma 3.4, wehave the following corollary.Corollary 3.2 The following estimate holds,φ的。Q(m(x, t))(3.13)Lemma 3.5 For any given β∈(0, 1) and suficiently large integer k,letk2= 1-78-店>B(1+0)+店、0, we haveQ(m(,))≥C(T)(x)]+k2.(3.14)Proof Note that c(x) = co(x) ~ C2o(x)9, by Sobolev's embedding theorem,φ1+k2Q,。"(*0)Jp1+ke2c(x)≤C(T)+cQ4)。|dx-dx+ Cφ1+ka|calarc { φ1+k2c(){(c9)a|Ld中国煤化工”From (3.7), (3.12) and (3.13), we haveYHCNMHG.≤C(T)+C「' φ+(x()(cQ)图lx(CQ)1+βNo.1Z. Wang & H. Zhang: FREE BOUNDARY VALUE PROBLEM421≤C(T)+C(\去()((kz+乖-830)z与验元(9)(+)码φ一乖+β≤C(T) +C(T)max(6n(工, t))咖)曲)(φ1+k2\兹+≤C(T) + C(T) max，(3.15) .sinceβ+永< 1, we can conclude that( φ1+k2(营):≤C(T).CLemma 3.6 The following estimate holdsQ(m(x, t))≥C(T)中(x)",(3.16)for a suitable constant C(T).Proof We firstly show that Q(m(x,t))≥C(T)$(x)a for all x∈(0,8)∪(1 - 8, 1), whenδ > 0 is small enough.When x∈(0,8), from (3.3), Lemma 3.2 and conditions (A1)-(A3), we get(cQ)(x,t) = (cQ)(x,0) - βρnus.(y,s)dy + P(c,Q) )ds= (coQo)P -βpl I (u(y,t)- uo)dy-Bpn[" IP(c, Q)dyds≥((x)a+9)9 - C(T)(x)多- C(T)(x)(a+)y≥C(T)()a+0),(3.17)which implies (3.16) for x∈(0, 8). Similarly, the inequality holds for x∈(1- δ,1).Next, we prove (3.16) whenx∈(6,1- 8).Since中(x)> δ for allx∈(8,1- 8). Form Lemma 3.5, we haveQ(m(,))≥C(T)(x)2+kz≥C(T)(;)+kz-*$(x)^,and Lemma 3.6 follows.Corollary 3.3 We have the following estimates|8xmldx≤C(T)，and ||8xn|dx < C(T).(3.18)Proof In view of Q'(m)= &Q(m)2 and m =中国煤化工8x[(cQ(m))门] = C0[(Q(m))月] + βc9-1eun; Ox:YHCNMHG=(cQ(m)9-1[1 + Q(m)28xm + Bec-1Q(m)9O2c,(3.19)Pl^422ACTA MATHEMATICA SCIENTIAVol.32 Ser.Bwe have[cQ(m)1-[Q(m)]1-βμ + Q(m)28xm=-8x[(cQ(m)) - !-Bc9-1Q(m)Oxc[Q(m)]1-B-o[(cQ(m)] - pCQ(m28xc.(3.20)cConsequently, from Corollary 3.1 and (A1), we get|8.mldx≤C1 [cQ(m)1-3 ;-82[(cQ(m))9]dx +C[ Q(m)B2eldx。≤C(T)+ C(T))。"些cQca|dx + C(T)Q(x)c2中(x)≤C(T)+C(T)「2(凹)-dx s C(T).(3.21)J。中(工)Clearly,|Ixn|dx≤|m|l8cldx +c|8xm)dx≤C(T).(3.22)JCorollary 3.3 is proved.Lemma 3.7 The following estimate holds:(u)2dx +. [ f@C@m+(u)PardscC).(3.23)Proof Diferentiating the (1 20)3 with respect to t, multiplying the resulting equation by2ut, and then integrating over [0, 1] x [0, t], we obtain(u)2(x,)dx +2,['cc(Q(m)tu4dxds: [ ()(0)x+2。' f(m0)+uxudrds.3.24)Using the equation (1.20)3 at t = 0 and the assumption (A2), it follows thatf" ()(,0)≤C(T),(3.25)Moreover, by (1.20)2, we have(@Q(m)+lu)turdxrds= (B+ 1)pland中国煤化工CNMHG.= YPI(3.27)No.1Z. Wang & H. Zhang: FREE BOUNDARY VALUE PROBLEM423inserting this in (3.24), we get(u()(x,t)dx +2.^&Q(m)+(ux)Pdrds≤C + 2(β+ 1)puf" f" cQ(m2(u)utdrds-2γprc"Q(m)+1uztU4drds.(3.28)For the last two terms on the right-hand side, we estimate as follows.2(β+ 1)pn≤1]° ," (Qm()(u)Pards +2(0+192。[" (Qn+2a)urds, (3.2)andc"Q(m)+1uatuzdxds≤今f°。" (Qmjiaude+22.F s" f" 2-Q(m)2) -(audud. (3.30)Inserting this in (3.28), we obtain。" ()C,0)+.: f [。cQ(m+(ux)Pdrds≤C+2(9+ 12pIrs +272pI, (3.3)where .Is=I4=SetV(8)=.' CQ(m)+(u)dx,thenIs≤max(Q2(ux)2)V(s)ds.(3.32)From (3.2), (3.5) and Lemma 3.6,Q2(ux)2 = (c@Q(m)B+1ux)(Q(m))-29pa= (cQ(m)-23(uydx + P(c,Q(m)) )≤C(cQ(m)-28(x) [(4)2dx + C(cQ)2(r-B)Jo中国煤化工≤C(x)-20(a+0)I。(rMHCNMHG≤C(T)/our)?dx +C.(3.33)424ACTA MATHEMATICA SCIENTIAVol.32 Ser.BConsequently,Is≤C(T)+ C(T)(u)drds.(3.34)Moreover, we have from (3.5) thatI4≤[ max(Q)2(-B)V()ds ≤C(T) [ V(<)ds.(3.35)Inserting (3.34)- (3.35) in (3.31), we get、 (u)(,t)dx +J[。&Q(m)*+l(uxt)2drds≤C(T)o(u)2drds + C(T),(3.36)which yields by (3.4) and Gronwall's inequality[" ()(,t)dx≤C(T)exp(C(T) [ V()d)≤C(T).The proof of Lemma 3.7 is complete.Lemma 3.8 We have|ue(x,t)|dx≤C(T)，l(,()(r) ≤C(T),(3.37)where Dr= {(x,t)|x∈[0,1],t∈[0,T]}.Proof From the momentum equation of (1.20)3, we getu = cr-BQ(m)7-9-1 +c~'Q(m)-9-1udy.Then[ lu]lda≤ [ (-(m)-d-+- [ c-8Q(m)-9-1( (°I|ut}dy dx\≤[cC~-(Q(m)-9--dx +C[" c-8Q(m)-1()dx(|u{dx(3.38)By the assumption (A1), Corollary 3.1 and Lemma 3.6, we obtaine-m)---s (dx≤C(T),(3.39)and。c-9Q(m)-8- ~中(x)dx≤I中(x)(cQ)-日dx≤中国煤化工MYHCNMHG.≤C(T),(3.40)the first inequality of (3.37) is proved.No.1Z. Wang & H. Zhang: FREE BOUNDARY VALUE PROBLEM425Consequently, the second inequality in (3.37) follows from Sobolev's embedding theoremW1,1([0, 1])→L∞(0, 1]) and Lemma 3.2.This completes the proof of the Lemma 3.8.CFrom Lemma 3.3, Lemma 3.7 and (3.5), we get the next result.Lemma 3.9 The following estimate holds,Ilc(x)*Q(m)+u(0,t)lx(Dr)≤C(T), Dr= [0,1]x [0,T],(3.41)。l(c(x)fQ(m)+(1+(x,)(x,t)dx ≤C(T).(3.42)Now, we give the L2-continuity in time for (m, n, u).Lemma 3.10 For0 0,q:1+8qz2Q2dx+ |w崆dx =0.which implies Q= 0，Wx=0, i.e., Q1 =Q2，U1 = u2.5Proof of Existence ResultIn order to construct weak solutions to the initial-boundary problem (IBVP) (1.12), weapply the line method which can be described as follows [13]. For any given positive integer N,let h=六. Discretizing the derivatives with respect to x in (1.12), a system of ODEs of thefollowing form is obtained会()+ (2()m会()Dxu公(t)=0， i=1,-,N-1,m2(t) + [m2()]2 Dxu名(t) =0,i=1.,.,N- 1,ut'中国煤化工二会-1(t)+ DxP(n2;-1(t), m21())MYHCNMHG= ((()()()D。u名- E(n会2(),mai- 2())Dxu2p 2)， i=1,.N， (5.1)No.1z. Wang & H. Zhang: FREE BOUNDARY VALUE PROBLEM429with the boundary conditionsn哈(t)=n哈v(t)=0，and m(t)= m2N(t)=0，t≥0,(5.2)and initial datan2:(0) = no(20哈)> 0,ρl> m2(0) = mo(哈)>0，i-=.,..N-1哈-(0) = u0((2-1号),i= 1,.,N,(5.3)where we have used the assumption (A1) for the initial data no and mo. Here Dxai= (ai+1 -ai- -1)/h and Dxai-1 = (ai- ai- -2)/h. Moreover, for i = 1,N we set u1 = u哈N+1(t) = 0when we apply the third equation of (5.1), without loss of generality since E(n名(t), m(t)) =E(n名N(), mav()) = 0The theory for ordinary diferential equations ensures that the Cauchy problem (5.1) admitsa temporarily local solution in the domainR3N-1= {(n2i(t), m2(t),2- ()i=1,,N-1,j-=1..N}in the regularity class C(0,T) x C(0,T) x C(0,T), satisfyingn哈(t)>0，pi> m2(t)>0，for0 0 and Q&(t) > 0 are well deined in (5.4).Moreover, by applying (5.1) we observe that the quantities (C%(), Q路(),吃1 (t)) are describedby a system of ODEs of the formd会(t)=0， i=1,..,N-1,Q2&(t) + pl[Q&(t)]2Dxu&()=0， i=1,-,N-1,哈a(t)+ D.P(% -(),Q%, -())()()公) - E(c4 2(),Q% 2())Dxa _2), i=1,.,N, (5.6)where P(c, Q) and E(C, Q) are given by (1.23). Boundary conditions are given byc(t)=c2Nv(t)=0，and Q路(t)=Q验v(t)=0， t≥0,(5.7)中国煤化工Q2(0)= Qo0HCNMHG_(0)=u0(2i-)冷)，1-1...(5.8)430ACTA MATHEMATICA SCIENTIAVol.32 Ser.BIn the following we neglect the index h and use the notation (2i(t), Qzi(),u2i-1(t). First ofall, we obtain the following basic energy estimate by mimicking the arguments used in Lemma3.2 similarly to the references [13-17].Lemma 5.1 Let (2i(t), Q2i(),u2i- 1()) be the solution to (5.6) with boundary andinitial data given by (5.7) and (5.8), respectively. Then we haveQ&f()(2())2hds宫(品710+万h, Vt∈[0,T]. .(5.9)p(r- 1)As a consequence, by application of the variable transformation (5.5), the existence anduniqueness of global solutions (n名。(t), m2;(t), u2 1(t)) to (5.1)-(5.3) follows as well, where (5.4)holds for any T <∞. In the following we list a number of estimates that hold for the approx-imate solutions (n2;(t), m2;(t), u2i- 1(t)) that are uniform with respect to the discretizationparameter h.Lemma 5.2 Let (n名(t), m,(),哈1()) be the solution to (5.1)- (5.3). Then we haven2i(t)ρl - m2i(t): = C2;Q2i(t)≤C(T)中(ih)9+a， Qzi()≤C(T)(ih)*,(5.10)and(ih(1 - ih))2k- -1 D(cz;Qz(t)])2kh≤C(T).(5.11)Lemma 5.3 Let (n名(t), m名(t),吃一1(t)) be the solution to (5.1)- (5.3). Then we havefor any positive integer kNN φ(ih)=r-h≤C(T),(5.12)2-Qzi(t)i1三啮()n+k(2k- 1))。"EruZ-(j)2Q2()(Dx2()2hds s C(T),二1hds≤C(T).(5.13)Finally, we haveQ2z() > C(T)Qo(ih)，i= 0..,N.(5.14) .Lemma 5.4 Let (0名(t), m&(t),吃名()) be the solution to (5.1)-(5.3). Then we havem2() - ma(t)1≤C(T)，2I(5.15)中国煤化工i=1TYHCNMHG2n|u2i+1(t)- U2i-1(t)|≤C(T)， |u42i+1(t)|≤C(T),(5.16)No.1z. Wang & H. Zhang: FREE BOUNDARY VALUE PROBLEM431andN(E(n2;(t), ma()Dxu2;(t) - E(n2i- 2(), m2i- 2())Dxru2i. _2(t)I≤C(T),(5.17)|E(n2:(t), m()Dxuz;()|≤C(T).(5.18)Moreover, we have the time continuity estimates|m2() - mz;()2h≤C(T)t-s|， 2I|n2i(t) - n2i(s)|2h≤C(T)|t -s|,(5.19)台i=12 lu2i+1(t) - u2i-()Ph≤C(T)t-s|,(5.20)isi2(2(b(n1()()Dxv2() - E(n2(s), m()ruzx()2h≤C(T)t -s|.(5.21)Following along the lines of previous works, see for [13-17], approximate solutions (n2(t),m2;(t),u2 . n(t)) for (x,t)∈Dr= [0,1]x [0,T] are defined as fllws,nn(x,t) = n2i(),mh(x,t) = m2z(t),un(x,t)=方([ - (i- 1/)2+() + (i + 1/2)h-x)u2r -(),forx∈((i- 1/2)h, (i + 1/2)h),i.=0,1,..,N.In particular, we havea2xu(x,t)= (4z+() -U2- 1(t) = Dxuz;()-By the estimates in the above lemmas, Helly's theorem and arguments in one of thereferences [10-13], we complete the proof of the existence and uniqueness of weak solution tothe model (1.12), i.e., the limit function (n, m, u) obtained from (nn, mh, un) is weak solutiondefined in Theorem 2.1.References[] Brennen C E. Fundamentals of Multiphase Flow. New York: Cambridge Univ Press, 2005[2] Ishi M. Thermo Fluid Dynamic Theory of Two Phase Flow. Paris: Eyrolles, 19753] Prosperetti A, Tryggvason G, eds. Computational Methods for Multiphase Flow. New York: CambridgeUniv Press, 2007中国煤化工[4] Evje S, Fjelde K -K. Hybrid fux spliting schemes for a two4YC N M H G.,2002, 175(2):674- 701.[5] Evje s, Fjelde K -K. Relaxation schemes for the calculation of two-phase flow in pipes. Math CompModelling, 2002, 36: 535- 567432ACTA MATHEMATICA SCIENTIAVol.32 Ser.B[6] Evje S, Fjelde K -K. On a rough AUSM scheme for a one dimensional two phase model. Comp Fluids,2002, 32(10): 1497 -1530[7] Evje S, Flatten T. On the wave structure of two -phase flow models. SIAM J Appl Math, 2007, 67(2):487- 511[8] Fille S, Heintze E. A rough finite volume scherme for modeling two -phase flow in a pipeline. Comp Fluids,1999, 28: 213-241[9] Fjelde K -K, Karlsen K -H. High-resolution hybrid primitive -conservative upwind schemes for the driftfAux model. Comp Fluids, 2002, 31: 335- -367[10] Evje S, Karlsen K H. Global existence of weak solutions for a viscous two-phase model. J Difer Equ, 2008,245: 2660- 2703[11] Evje s, Karlsen K H. Global weak solutions for a viscous liquid-gas model with singular pressure law.http://www.irisresearch.no/docsent/emp.nsf/wvAnsatte/SEV.[12] Yao L, Zhu C J. Free boundary value problem for a viscous two phase model with mass-dependent viscosity.J Difer Equ, 2009, 247: 2705- 2739[13] Evje S, Flatten T, Fris H A. Global weak solutions for a viscous liquid-gas model with transition tosingle- phase gas flow and vacuum. Nonlinear Anal, 2009, 70: 3864 -3886[14] Fang D Y, Zhang T. Compressible Navier -Stokes equations with vacuum state in one dimension. CommPure Appl Anal, 2004, 3(4): 675- 694[15] Yang T, Zhu C J. Compressible Navier-Stokes equations with degenerate viscosity coeffcient and vacuum.Comm Math Phys, 2002, 230: 329 -363[16] Guo Z H, Zhu C J. Remarks on one dimensional compressible Navier Stokes equations with Density-Dependent viscosity and vacuum. Acta Mathematica Sinica, English Series, 2010, 26(10): 2615- -2030[17] Okada M, Matusu-Necasova s, Makino T. Free boundary problem for the equation of one- dimensionalmotion of compressible gas with densitydependent viscosity. Ann Univ Ferrara Sez VII (N S), 2002, 48:1-2018] Yang T, Yao Z -A, Zhu C -J. Compressible Navier- Stokes equations with densitydependent viscosity andvacuum. Comm Partial Difer Equ, 2001, 26: 965 -981[19] Yang T, Zhao H -J. A vacuum problem for the one-dimensional compressible Navier-Stokes equations withdensity-dependent viscosity. J Difer Equ, 2002, 184: 163- -184[20] VongS W, Yang T, Zhu C J. Compressible Navier Stokes equations with degenerate viscosity coefficientand vacuum II. J Differ Equ, 2003, 192: 475- 501中国煤化工MYHCNMHG