Theory article

Blowup for regular solutions and C1 solutions of the two-phase model in RN with a free boundary

  • Received: 16 February 2022 Revised: 29 April 2022 Accepted: 12 May 2022 Published: 17 June 2022
  • MSC : 35B44, 35R35, 76T10

  • In this paper, under the assumption of an initial bounded region Ω(0), we establish the blowup phenomenon of the regular solutions and C1 solutions to the two-phase model in RN. If the total energy E and the total mass M>0 satisfy

    maxx0Ω(0)Ni=1u2i(0,x0)<min{2,N(Γ1),N(γ1)}EM,

    where E=Ω(0)(12n|u|2+12ρ|u|2+1Γ1nΓ+1γ1ργ)dV and M=Ω(0)(n+ρ)dV>0, then the blowup of the solutions to the two-phase model will be formed in finite time in RN. Furthermore, under the assumptions that the radially symmetric initial data and initial density contain vacuum states, the blowup of the smooth solutions to the two-phase model will be formed in finite time in RN(N2).

    Citation: Jingjie Wang, Xiaoyong Wen, Manwai Yuen. Blowup for regular solutions and C1 solutions of the two-phase model in RN with a free boundary[J]. AIMS Mathematics, 2022, 7(8): 15313-15330. doi: 10.3934/math.2022839

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  • In this paper, under the assumption of an initial bounded region Ω(0), we establish the blowup phenomenon of the regular solutions and C1 solutions to the two-phase model in RN. If the total energy E and the total mass M>0 satisfy

    maxx0Ω(0)Ni=1u2i(0,x0)<min{2,N(Γ1),N(γ1)}EM,

    where E=Ω(0)(12n|u|2+12ρ|u|2+1Γ1nΓ+1γ1ργ)dV and M=Ω(0)(n+ρ)dV>0, then the blowup of the solutions to the two-phase model will be formed in finite time in RN. Furthermore, under the assumptions that the radially symmetric initial data and initial density contain vacuum states, the blowup of the smooth solutions to the two-phase model will be formed in finite time in RN(N2).



    The compressible inviscid liquid-gas two-phase model in RN that will be considered is as follows:

    {nt+div(nu)=0,ρt+div(ρu)=0,[(ρ+n)u]t+div[(ρ+n)uu]+P(n,ρ)=0, (1.1)

    where u=u(t,x), P=P(n,ρ)=nΓ+ργ are the velocity and pressure while n=n(t,x)0, ρ=ρ(t,x)0 are the densities of two phases, where γ, Γ>1.

    Euler equations have been used as one of the basic models for studying fluids, plasmas, atmospheric dynamics, and condensed matter in [2,5,6,7,11,30], and the blowup analysis for these equations is carried out in [3,4,12,17,21,25,26,28,34,36]. In 1985, Sideris [25] constructed the functional

    F(0)=R3xρudx (1.2)

    to prove that the C1 solutions of the three-dimensional compressible Euler equations will blow up in a finite time when the initial functional F(0) is sufficiently large. Yuen [35] used the energy method to consider the blowup results of the C1 solutions and the weakened regular solutions of the Euler equations in RN. In [18], the authors rewrote the system in the form of a quasilinear wave equation about the density ρ to study the blowup of solutions to Euler equations. Liu, Wang and Yuen studied the blowup results of solutions to the compressible Euler equations with time-dependent damping with vacuum and C1 solutions of the irrotational compressible Euler equations with time-dependent damping in [19]. In [24], the author studied the three-dimensional Euler equations with a free boundary subjected to tension.

    In the two-phase fluid, because of its wide application in aerospace, micro-technology, chemical engineering and other fields, it has aroused many researchers' interest. Zuber studied the two-phase model firstly in [39]. The motion of liquid and gas mixture is studied by the two-phase model in [16]. In [23], the authors considered a hyperbolic two-phase model. For existence, asymptotic and uniqueness of global weak solutions to the two-phase flow model with vacuum, Yao, Zhang and Zhu used the line method and a priori estimate to obtain relevant results in [31,32,33], and there are also studies of these issues in [1,8,13,14,15,27]. In [37], under the assumption of H2norm of the initial perturbation with a constant state is sufficiently small and L1norm is bounded, Zhang and Zhu studied the global existence of Cauchy problem to viscous liquid-gas two-phase flow in three dimensions. Furthermore, for the Cauchy problem of 3D inviscid liquid-gas two-phase flow, Zhang considered the optimal LPL2 (1p<65) time decay rates of the solutions with the damping on the qualitative behaviors in [38]. Wen and Zhu [29] considered global existence of weak solutions to two-fluid about the Dirichlet problem in one dimension. In [10], Dong et al. considered the energy integration method to prove the singularity of the smooth solutions to the Cauchy problem for the viscous two-phase model in arbitrary dimensions. Furthermore, Dong and Yuen provided the blowup phenomena of self-similar solutions for the inviscid liquis-gas two-phase flow [9]. By introducing the definition of regular solutions, Makino and Perthame studied the blowup phenomena of radical symmetric solutions to the Euler-Poisson equations with compact support and with a repulsive force and an attractive force in [20,22].

    In this paper, we study the blowup results of regular solutions and C1 solutions of the two-phase model in RN with a free boundary. Our method depends on the energy integration method and a quasi-linear wave equation about ρ, the singularity of the two-phase model will form in finite time.

    In this section, we give the blowup results of regular solutions and C1 solutions of the two-phase model in RN with a free boundary. Firstly, we introduce the definition of regular solutions to the two-phase (1.1).

    Definiton 3.1. (Weakened regular solution) A solution (n,ρ,u) to the system (1.1) which is in RN and in the non-vacuum region is regular for 1<Γ<2 or 1<γ<2, if

    (nΓ1η,ργ1η,u)C1, (3.1)

    with a fixed constant η>1.

    In the following, we state the blowup results of the two-phase model with a free boundary in RN.

    Theorem 3.1. Assume that the fluid enters a bounded open region Ω(t)RN, with the contacting vacuum boundaryΩ(t). Suppose that the (n,ρ,u) is a weakened regular solution on [0,T)×Ω(t) of the two-phase model (1.1) in RN. If

    maxx0Ω(0)Ni=1u2i(0,x0)<min{2,N(Γ1),N(γ1)}EM, (3.2)

    where the total energy E=Ω(0)(12n|u|2+12ρ|u|2+1Γ1nΓ+1γ1ργ)dV and thetotal mass M=Ω(0)(n+ρ)dV>0, then the solutions of the two-phase model will blow up in finite time T.

    Furthermore, we consider the two-phase model (1.1) with the pressure P=P(n,ρ)=nγ+ργ. In this paper, we will rewrite the two-phase model (1.1) by forming a quasi-linear wave equation about the density ρ and studying the blowup results for solutions of the two-phase model in RN(N2) with the initial data

    t=0:n=n0(x),  ρ=ρ0(x),  u=u0(x). (3.3)

    Introducing radial symmetry, the initial data become

    n0(x)=n0(r),    ρ0(x)=ρ0(r),    u0(x)=xrV0(r), (3.4)

    where r=(Ni=1x2i)12.

    In the following, we give the blowup results of the two-phase model in RN(N3).

    Theorem 3.2. Assume that the initial density n00 and ρ00 in (1.2), and (n0,ρ0,u0)H3(RN). Consider the solutions (n,ρ,u) of the two-phase model (1.1) in RN(N3). If the initial conditions

    n0(0)=0,    ρ0(0)=0, (3.5)
    RN(n0+ρ0)(r)dx>0, (3.6)

    and

    RN(1+r)(n0+ρ0)V0r2erdx12γ1(1+γ)[Γ(1+N2)(N2)!πN2]γ1(RN(n0+ρ0)rerdx)γ+12 (3.7)

    are satisfied, where πN2Γ(1+N2) is the volume ofthe unit ball in RN, then the solutions will blow up on or before the finite time 2H(0)γ12(γ1)C0, where H(0)=RN(n0+ρ0)rerdx and C0=12γ1(1+γ)[Γ(1+N2)(N2)!πN2]γ1.

    Then, under the appropriate assumptions, we will give the blowup results of the two-phase model in the two dimensional cases.

    Theorem 3.3. Assume that the initial density n00,ρ00 in (1.2), and (n0,ρ0,u0)H3(R2). Let K0(r) be the modified Bessel function

    K0(r)=0ercoshtdt. (3.8)

    Consider the solution (n,ρ,u) of the two-phase model (1.1) in R2. If the initial conditions

    n0(0)=0,    ρ0(0)=0, (3.9)
    R2(n0+ρ0)(r)dx>0, (3.10)

    and

    R2(n0+ρ0)V0K0(r)dx12γ1(1+γ)(R2(n0+ρ0)(r)K0(r)dx)γ+12(R2K0(r)dx)γ12 (3.11)

    are satisfied, then the solutions will blow up on or before the finite time 2G(0)γ12(γ1)C1, where G(0)=R2(n0+ρ0)K0(r)dx and C1=12γ1(γ+1)(R2K0(r)dx)γ12.

    Remark 3.1. For the two-phase model in RN with pressure P=P(n,ρ)=(n+ρ)γ, we can also obtain the same blowup results.

    In this section, we will show the proof of the blowup result of solutions to the two-phase model in RN. Firstly, we give some lemmas for the conserved energy and the properties of the local second inertia function to the two-phase model.

    Lemma 3.1. Assume the density n and ρ have compact support in the region Ω(t) for non-trivial C1 solutions of the two-phase model (1.1) in RN, where Ω(t) is a moving region, and the region Ω(0) is bounded by the contacting vacuum boundary, then

    M(t)=M(0), (3.12)

    and

    0<E=E(t)=Ω(t)(12n|u|2+12ρ|u|2+1Γ1nΓ+1γ1ργ)dV<, (3.13)

    that is, the total energy energy E(t) is conserved, where M(t)=Ω(t)(n+ρ)dV represents the total mass.

    Proof. By (1.1)1 and (1.1)2, we obtain

    ddtM(t)=Ω(t)(nt+ρt)dV=Ω(t)div(nu)+div(ρu)dV=0. (3.14)

    Therefore, M(t)=M(0).

    Multiplying u on both sides of the Eq (1.1)3 and integrating over Ω(t), we obtain

    Ω(t)(nu)tudV+Ω(t)div(nuu)udx+Ω(t)(ρu)tudV+Ω(t)div(ρuu)udV+Ω(t)P(n,ρ)udV=0. (3.15)

    On the one hand, from (1.1)1, we have

    12nt|u|212|u|2div(nu)=0. (3.16)

    Due to

    (12n|u|2)t=(nu)tu12nt|u|2, (3.17)

    we obtain

    Ω(t)(nu)tudx+Ω(t)div(nuu)udV=Ω(t)(12n|u|2)tdV+Ω(t)12nt|u|2dV+Ω(t)u[(nuu)]dV=Ω(t)(12n|u|2)tdVΩ(t)12|u|2(nu)dV+Ω(t)u[(nuu)]dV. (3.18)

    We note that

    u[(nuu)]=Ni,j=1ui[j(ρuj)ui+nujjui]=|u|2Ni=1i(ρui)+Ni,j=1nuiujjui. (3.19)

    Thus, we get

    Ω(t)[12|u|2(nu)+u[(nuu)]]dV=Ω(t)[12|u|2Ni=1i(nui)+|u|2Ni=1i(nui)Ni,j=1nuiujjui]dV=Ω(t)[12|u|2Ni=1i(nui)+Ni,j=1nuiujjui]dV=Ω(t)[Ni=1uiunui+Ni,j=1nuiujjui]dV=Ω(t)[Ni,j=1nuiujiuj+Ni,j=1nuiujjui]dV=0. (3.20)

    Therefore, we obtain

    Ω(t)(nu)tudV+Ω(t)div(nuu)udV=Ω(t)(12n|u|2)tdV. (3.21)

    Similarly, we get

    Ω(t)(ρu)tudV+Ω(t)div(ρuu)udV=Ω(t)(12ρ|u|2)tdV. (3.22)

    On the other hand,

    Pt=Γ nΓ1tn+γργ1tρ=Γ nΓ1[Ni=1i(nui)]+γργ1[Ni=1i(ρui)]=Ni=1i(nΓ)uiNi=1ΓnΓiuiNi=1i(ργ)uiNi=1γργiui. (3.23)

    Integrating over the region Ω(t) and applying the integration by parts, we obtain

    Ω(t)PtdV=Ω(t)Ni=1i(nΓ)uidVΩ(t)Ni=1ΓnΓiuidVΩ(t)Ni=1i(ργ)uidVΩ(t)Ni=1γργiuidV=Ω(t)Ni=1i(nΓ)uidV+Ω(t)Ni=1Γi(nΓ)uidVΩ(t)Ni=1i(ργ)uidVΩ(t)Ni=1γi(ργ)uidV=Ω(t)Ni=1(Γ1)i(nΓ)uidV+Ω(t)Ni=1(γ1)i(ργ)uidV. (3.24)

    Therefore, we have

    Ω(t)uP(n,ρ)dV=1Γ1Ω(t)t(nΓ)dV+1γ1Ω(t)t(ργ)dV. (3.25)

    Thus, by (3.15), we obtain

    ddtE(t)=ddt(Ω(t)12n|u|2+12ρ|u|2+1Γ1nΓ+1γ1ργdV)=Ω(t)(nu)tudV+Ω(t)div(nuu)udx+Ω(t)(ρu)tudV+Ω(t)div(ρuu)udV+Ω(t)P(n,ρ)udV=0. (3.26)

    Therefore, for non-trivial C1 solutions, we obtain

    E(t)=E(0)<. (3.27)

    The proof is complete.

    Next, we show the second derivative of the local second inertia function for the two-phase model (1.1) in RN for the solutions on [0,T)×Ω(t).

    Before stating the following lemma, we first give some physical quantities as follows:

    F(t)=Ω(t)nuxdV+Ω(t)ρuxdV=Fn(t)+Fρ(t), (3.28)
    H(t)=12Ω(t)n|x|2dV+12Ω(t)ρ|x|2dV=Hn(t)+Hρ(t), (3.29)

    and

    E(t)=Ω(t)12n|u|2dV+Ω(t)12ρ|u|2dV+1Γ1Ω(t)nΓdV+1γ1Ω(t)ργdV=Ekn(t)+Ekρ(t)+Ein(t)+Eiρ(t), (3.30)

    where Fn(t) and Fρ(t), Hn(t) and Hρ(t), Ekn(t) and Ekρ(t), Ein(t) and Eiρ(t) represent the momentum weight, the local second inertia, the kinetic energy and the internal energy for the two-phase fluid, respectively.

    Lemma 3.2. For the two-phase model (1.1) in RN, we have

    H(t)=F(t), (3.31)

    and

    H(t)=2Ekn(t)+2Ekρ(t)+N(Γ1)Ein(t)+N(γ1)Eiρ(t). (3.32)

    Proof. Differentiating H(t) with respect to t, we obtain

    H(t)=12Ω(t)nt|x|2dV+12Ω(t)ρt|x|2dV=Hn(t)+Hρ(t). (3.33)

    Applying the integration by parts to (3.33) and using (1.1)1, we get

    Hn(t)=12Ω(t)div(nu)|x|2dV=12Ω(t)Ni,ji(nui)x2jdV=12Ω(t)Ni,j2nuixjdV=Ω(t)NinuixidV=Ω(t)xnudV=Fn(t). (3.34)

    Similarly, we have

    Hρ(t)=Fρ(t). (3.35)

    Therefore, we obtain

    H(t)=F(t). (3.36)

    Next, we calculate H(t) as follows:

    H(t)=F(t)=Ω(t)(nu)txdV+Ω(t)(ρu)txdV=Ω(t)div(nuu)xdVΩ(t)div(ρuu)xdVΩ(t)P(n,ρ)xdV. (3.37)

    By using the integration by parts, we get

    Ω(t)div(nuu)xdV=Ω(t)Ni,ji(nuiuj)xjdV=Ω(t)Ni,jnuiujixjdV=Ω(t)nuudV=Ω(t)n|u|2dV. (3.38)

    Similarly, we have

    Ω(t)div(ρuu)xdV=Ω(t)ρ|u|2dV, (3.39)

    and

    Ω(t)P(n,ρ)xdV=NΩ(t)nΓdVNΩ(t)ργdV. (3.40)

    From (3.38) to (3.40) and Gauss formula, we obtain

    H(t)=F(t)=Ω(t)n|u|2dV+Ω(t)ρ|u|2dV+NΩ(t)nΓdV+NΩ(t)ργdV=2Ekn(t)+2Ekρ(t)+N(Γ1)Ein(t)+N(γ1)Eiρ(t). (3.41)

    The proof is complete.

    Using the above lemmas, we will give the proof of the blowup results of the solutions to the two-phase model with a free boundary in RN.

    Proof of Theorem 3.1. Introducing

    φ=nΓ1η,ω=ργ1η, (3.42)

    for n(t,x)0 and ρ(t,x)0, we transform the momentum equations (1.1) into

    (φηΓ1+ωηγ1)(uit+Nl=1uluixl)+xi(φηΓΓ1+ωηγγ1)=0. (3.43)

    Simplifying the above equation, we have

    (φηΓ1+ωηγ1)(uit+Nl=1uluixl)+ηΓΓ1φηΓΓ11xiφ+ηγγ1ωηγγ11xiω=0, (3.44)

    where η>1 is an arbitrary constant, (η1)Γ+1η>0 and (η1)γ+1η>0.

    We study the solutions near the contacting vacuum boundary point x0(t) in the region Ω(t). Firstly, we consider the local second inertial function

    H(t)=12Ω(t)n|x|2dV+12Ω(t)ρ|x|2dV. (3.45)

    By Lemma 3.2, we obtain

    H(t)=2Ekn(t)+2Ekρ(t)+N(Γ1)Ein(t)+N(γ1)Eiρ(t)min{2,N(Γ1),N(γ1)}E(t)=C1E, (3.46)

    with the conserved total energy E=Ω(0)(12n|u|2+12ρ|u|2+1Γ1nΓ+1γ1ργ)dV by Lemma 3.1, where C1=min{2,N(Γ1),N(γ1)}.

    Therefore, we obtain

    H(t)H(0)+˙H(0)t+C1E2t2. (3.47)

    For an arbitrary point x0(t)Ω(t) with the contacting vacuum, by applying Lemma 3.1 and ρ(0,x0)=0, we obtain

    uit+Nl=1uluixl=0. (3.48)

    We consider the governing differential dynamic system at the contacting vacuum point x0(t):

    {D2x0,i(t)Dt2=0,   fori=1,2,,N,x0,i(0,x0)=x0,i,   ˙x0,i(0,x0)=ui(0,x0), (3.49)

    where dx0,i(t)dt=ui(t,x0(t)).

    The solutions of the above system are as follows:

    x0,i(t)=x0,i+ui(0,x0)t. (3.50)

    Then, we get

    ui(t,x0(t))=dx0,i(t)dt=ui(0,x0). (3.51)

    By applying the Euclidean norm, we obtain

    maxxΩ(t)(x21+x22++x2N)12maxxΩ(t)(x21+x22++x2N)120. (3.52)

    Therefore, we have

    maxx0Ω(t)|x0|2maxxΩ(t)|x|2. (3.53)

    Furthermore, we obtain

    12maxx0Ω(t)Ni=1x20,i(t)Ω(t)ndV12Ω(t)n|x|2dV. (3.54)

    Similarly, we obtain

    12maxx0Ω(t)Ni=1x20,i(t)Ω(t)ρdV12Ω(t)ρ|x|2dV. (3.55)

    Therefore,

    12maxx0Ω(t)Ni=1x20,i(t)(Ω(t)ndV+Ω(t)ρdV)12(Ω(t)n|x|2dV+Ω(t)ρ|x|2dV)=H(t)H(0)+˙H(0)t+C1E2t2, (3.56)

    that is

    maxx0Ω(t)Ni=1x20,i(t)2M(H(0)+˙H(0)t+C1E2t2), (3.57)

    with the conserved total mass M=Ω(0)(n+ρ)dV>0 in Lemma 3.1.

    By (3.50), we have

    2M(H(0)+˙H(0)t+C1E2t2)maxx0Ω(0)Ni=1(x20,i+2x0,iui(0,x0)t+u2i(0,x0)t2). (3.58)

    When t+, we obtain

    limt+2H(0)M+2˙H(0)Mt+C1EMt2t2limt+maxx0Ω(0)Ni=1(x20,i+2x0,iui(0,x0)t+u2i(0,x0)t2)t2. (3.59)

    Therefore, we have

    C1EMmaxx0Ω(0)u2i(0,x0), (3.60)

    that is

    min(2,N(Γ1),N(γ1))EMmaxx0Ω(0)u2i(0,x0), (3.61)

    which contradicts with inequality (3.2) in Theorem 3.1. Therefore, we conclude that the solutions for the two-phase model (1.1) in RN will blow up in finite time.

    This completes the proof.

    For proofing the blowup results of solutions to the two-phase model in RN(N2), we first introduce the properties of the two-phase model.

    Lemma 3.3. (Theorem 2.1 in [18])Assume that n00, ρ00 and (n0,ρ0,u0)H3(RN), then there exists a unique solution (n,ρ,u) to the two-phase model on some time interval [0,T), which satisfies

    n,ρC([0,T)×RN), (3.62)

    and

    uC([0,T),H3(RN))C1([0,T),H2(RN))C2([0,T),H1(RN)). (3.63)

    Lemma 3.4. (Lemma 4 in [35])From the mass equations (1.1)1 and (1.1)2, we obtain

    n(t,x(t))=n(0,x0)exp(t0uds) (3.64)

    and

    ρ(t,x(t))=ρ(0,x0)exp(t0uds). (3.65)

    Next, we give the lemma to show the property of the modified Bessel function K0(r).

    Lemma 3.5. (Lemma 3.1 in [18])The modified Bessel function K0(r)=0ercoshtdt satisfies

    {K0(r)3r,|K0(r)|1r2,    0<r<12,K0(r)Ckrk,|K0(r)|Ckrk,    r>1, (3.66)

    for some constants Ck depending only on k>1.

    Then, we give the proof of the blowup results of solutions to the two-phase model in RN(N2).

    Proof of Theorem 3.2. We consider the solution (n,ρ,u) that satisfies the conditions in Theorem 3.2. From the two-phase system (1.1), we obtain

    ntt+ρtt=((nu)t+(ρu)t)=ΔP+[((n+ρ)uu)]. (3.67)

    Then, multiplying Eq (3.67) by 1rer and integrating over RN to obtain

    d2dt2RNn+ρrerdx=RNΔP1rerdx+RN[((n+ρ)uu)]1rerdx. (3.68)

    By the divergence theorem, we obtain

    RNΔP1rerdx=RNPΔ1rerdxlimε0r=ε(Pr1rerr(1rer)P)ds. (3.69)

    By applying the assumption (3.5) and Lemma 3.4, we obtain n(t,0)0,ρ(t,0)0. Using the continuity of P, we get that for any sufficiently small δ>0, there is 0<ε<1, such that P(t,x)<δ while r=ε, then,

    r=ε(1rer)rPds=r=εrer+err2e2rPdsδr=εrer+err2e2rds=δr=ε(ε+1)eεε2e2εds=δNα(N)εN1(ε+1)eεε2e2ε=δNα(N)εN3(ε+1)eεe2ε, (3.70)

    where α(N)=N(N2)πN2Γ(1+N2) and πN2Γ(1+N2) is the volume of the unit ball in RN.

    Therefore, for N3, we obtain

    limε0r=ε(1rer)rPds=0. (3.71)

    Similarly, we get

    limε0r=εPr1rerds=0. (3.72)

    Thus, we obtain

    RNΔP1rerdx=RNP1rerdx=RNnγrerdx+RNργrerdx. (3.73)

    Using H¨older inequality, we get

    (RNnrerdx)γ(RNnγrerdx)(RN1rerdx)γ1, (3.74)

    and

    (RNρrerdx)γ(RNργrerdx)(RN1rerdx)γ1. (3.75)

    Therefore, (3.73) becomes

    RNΔP1rerdx1(RN1rerdx)γ1[(RNnrerdx)γ+(RNρrerdx)γ]12γ1(RN1rerdx)γ1(RNn+ρrerdx)γ=12γ1[Γ(1+N2)(N2)!πN2]γ1(RNn+ρrerdx)γ. (3.76)

    Applying the integration by parts, we obtain

    RN[((n+ρ)uu)]1rerdx=RN1rerni,j=12((n+ρ)uiuj)xixjdx=RN(n+ρ)V2(1rer)dx=RN(n+ρ)V2[1r+2r2+2r3]erdx>0. (3.77)

    Therefore, we obtain

    d2dt2H(t)12γ1[Γ(1+N2)(N2)!πN2]γ1Hγ(t), (3.78)

    where H(t)=RN(n+ρ)(t,r)rerdx. Using integration by parts and (3.7), we obtain

    H(0)=ddtRN(n+ρ)rerdx|t=0=RN[(n+ρ)u]rerdx|t=0=RNNi=1(n+ρ)xirV1+rr2erxirdx|t=0=RN(1+r)(n0+ρ0)V0r2erdx>0. (3.79)

    Therefore,

    H(t)=H(0)+t0d2ds2H(s)dsH(0)>0. (3.80)

    Multiplying the both sides of (3.78) by 2H(t), we obtain

    (H(t)2)12γ1(1+γ)[Γ(1+N2)(N2)!πN2]γ1(Hγ+1(t)). (3.81)

    Integrate over [0,t], we obtain

    H(t)212γ1(1+γ)[Γ(1+N2)(N2)!πN2]γ1Hγ+1(t)+(ddtRN(n+ρ)rerdx)2|t=012γ1(1+γ)[Γ(1+N2)(N2)!πN2]γ1Hγ+1(0). (3.82)

    Applying the assumption (3.7), we obtain

    H(t)2C20Hγ+1(t)+(ddtRN(n+ρ)rerdx)2|t=0C20(RN(n0+ρ0)rerdx)γ+1C20Hγ+1(t), (3.83)

    where C0=12γ1(1+γ)[Γ(1+N2)(N2)!πN2]γ1.

    Therefore,

    H(t)C0Hγ+12(t). (3.84)

    By integrating over [0,t], we obtain

    t0dH(s)Hγ+12(s)C0t. (3.85)

    Therefore,

    2γ1(Hγ12(t)Hγ12(0))C0t, (3.86)

    which means that

    H(t)(Hγ12(0)γ12C0t)2γ1. (3.87)

    From (3.6), we have

    RNn0+ρ0rerdx>0. (3.88)

    Applying the mass conservation, we obtain

    H(t)=RNn+ρrerdxBr0n+ρrerdx+1r0RN(n0+ρ0)dx, (3.89)

    where Br0 is the N-dimensional ball centered at the origin with any given radius r0.

    When t2H(0)γ12(γ1)C0, H(t) cannot be bounded. Therefore, we obtain that rr0(n+ρ)(t,r)rdr cannot be bounded as t2H(0)γ12(γ1)C0, hence the solutions will blow up on or before the finite time t2H(0)γ12(γ1)C0.

    The proof is finished.

    In the following, we will prove Theorem 3.3.

    Proof of Theorem 3.3. We consider the solution (n,ρ,u) satisfies the conditions in Theorem 3.3. From (3.67), we obtain

    ntt+ρtt=ΔP+[((n+ρ)uu)]. (3.90)

    Then, multiplying Eq (3.90) by K0(r) and taking the integration over R2 to obtain

    d2dt2R2(n+ρ)K0(r)dx=R2ΔPK0(r)dx+R2[((n+ρ)uu)]K0(r)dx. (3.91)

    Using the same method as that we obtain from (3.69) to (3.73), we obtain from Lemma 3.5 that

    R2ΔPK0(r)dx=R2(nγ+ργ)K0(r)dx12γ1(R2K0(r)dx)γ1(R2(n+ρ)K0(r)dx)γ. (3.92)

    Using the integration by parts, we obtain

    R2[((n+ρ)uu)]K0(r)dx=R22i,j=12((n+ρ)uiuj)xixjK0(r)dx=R2(n+ρ)V2K0(r)dx>0. (3.93)

    Therefore,

    d2dt2G(t)12γ1(R2K0(r)dx)γ1Gγ(t), (3.94)

    where G(t)=R2(n+ρ)(t,r)K0(r)dx.

    Using integration by parts, (1.1)1, (1.1)2 and (3.11), we obtain

    G(0)=ddtR2(n+ρ)K0(r)dx|t=0=R2[(n+ρ)u]K0(r)dx|t=0=R22i=1(n+ρ)uiK0(r)xirdx|t=0=R22i=1(n+ρ)xirVK0(r)xirdx|t=0=R2(n+ρ)VK0(r)dx|t=0=R2(n0+ρ0)V0K0(r)dx>0. (3.95)

    Similarly, we obtain G(t)0.

    Multiplying both sides of (3.94) by 2G(t), we obtain

    (G(t)2)12γ1(1+γ)(R2K0(r)dx)γ1(Gγ+1(t)). (3.96)

    Integrate over [0,t] and using the assumption (3.11), we obtain

    G(t)2C21Gγ+1(t)+(ddtR2(n+ρ)K0(r)dx)2|t=0C21Gγ+1(0)C21Gγ+1(t), (3.97)

    where C1=12γ1(1+γ)(R2K0(r)dx)γ12.

    Therefore,

    G(t)C1Gγ+12(t). (3.98)

    By integrating over [0,t], we obtain

    t0dG(s)G1+γ2(s)C1t. (3.99)

    Therefore,

    2γ1(Gγ12(t)Gγ12(0))C1t, (3.100)

    which means that

    G(t)(Gγ12(0)γ12C1t)2γ1. (3.101)

    From (3.10), we have

    R2(n0+ρ0)K0(r)dx>0. (3.102)

    Applying the mass conservation, we obtain

    G(t)=R2(n+ρ)K0(r)dxBr0(n+ρ)K0(r)dx+maxrr0K0(r)R2(n0+ρ0)dx, (3.103)

    where Br0 is the 2-dimensional ball centered at the origin with any given radius r0.

    When t2G(0)γ12(γ1)C1, G(t) cannot be bounded. Thus, the solutions will blow up on or before the finite time t2G(0)γ12(γ1)C1. Therefore, we conclude that the solutions of the two-phase model in R2 will blow up.

    In this paper, we study the blowup results of solutions to the two-phase model in RN. Our method depends on the energy integration method and a quasi-linear wave equation about ρ, the singularity of the two-phase model will form in finite time.

    The first author is very grateful to the second and corresponding authors for their support of Beijing Natural Science Foundation under Grant No. 1202006 and the Small Grant for Academic Staff (MIT/SGA08/20-21), Department of Mathematics and Information Technology, the Education University of Hong Kong.

    On behalf of all authors, the corresponding author states that there is no conflict of interest.



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