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A Projection Method for Compressible Generic Two-Fluid Model

Po-Yi Wu1
1Moldex3D (CoreTech System Co., Ltd.)
Abstract

A new projection method for a generic two-fluid model is presented in this work. To be specific, it is shown that the projection method for solving single-phase variable density incompressible flows or compressible flows can be extended to the case of viscous compressible two-fluid flows. The idea relies on the property that the single pressure pp can be uniquely determined by the products of volume fractions and densities ϕkρk\phi_{k}\rho_{k} of the two fluids, respectively. Moreover, a suitable assignment of the intermediate step variables is necessary to maintain the stability. The energy stability for the proposed numerical scheme is proved and the first order convergence in time is justified by three numerical tests.

1 Introduction

Multi-fluid flows are widely seen in both nature and industry. For instance, the oil/gas/water three-phase system is encountered in petroleum industry[1, 2, 3]; two-gas system is used in the semiconductor manufacture[4, 5, 6]; and gas/liquid system occurs in many chemical engineering processes[7, 8, 9, 10]. The simplest case is the two-fluid system. In this case, each fluid obey its own governing equations and interact with the other fluid through the free surface or some constitutive relations.

Several versions of the model for two-fluid flows have been proposed. They can be roughly divided into two classes: (i) Interface-capturing models [11, 12, 13, 14] and (ii) Averaged models [15, 16, 17, 18, 19]. The use of interface-capturing models is able to resolve the topology of phase distribution and extract the physical detail in the system. When the topology of the phase distribution is too complicated or the interface between phases involves large variety of length scales, the class of averaged models can be a good compromise for simulating the flow problem. Although the information of the interface is blurred by the use of averaged model, several advantages arise instead: (i) a cheaper computational cost; (ii) less sensitive to mesh than interfacial-capturing model; and (iii) simpler expression of the interfacial terms.

In this work, a generic two-fluid model with similar form as [15] is considered. The unknown variables of the system are two velocities, two densities, two volume fractions, and one pressure. It is worth noting that such model allows a disparity of motion between the two phases, which is quite common in gas-liquid system. As far as the author’s knowledge, there is no mathematically provable stable numerical scheme for solving this system. To begin with the development, some ideas are borrowed by the numerical method for single phase variable density incompressible Navier-Stokes equations[20] and for single phase compressible Navier-Stokes equations[21, 22]. Both of them are of the framework of projection method [23, 24, 25]. Several difficulties occur when extending their ideas to the two-fluid system. The main issues are the following two: (i) One will have to solve one pressure by the two intermediate velocities in the projection step. The asymmetric structure causes that the pressure cannot be obtained by means of the formulation proposed in [21] directly; (ii) A proper assignment of the intermediate volume fractions and densities to advection-diffusion step and renormalization step shall be taken care to maintain the stability.

A prediction-correction procedure in the framework of projection method for the two-fluid model being considered is proposed. To tackle the asymmetric structure in the projection step, we may observe that the pressure pp can be uniquely determined by the products of volume fraction and density (say αg\alpha_{g} and αl\alpha_{l}) of the two phases (see [15] or Section 2.2 in this paper). A symmetric formulation proposed in this paper for solving αg\alpha_{g} and αl\alpha_{l} together provides an available way to accomplish the projection step. For the other main issue, the choice of intermediate step of volume fractions and densities can be understood by the process of the stability analysis.

The method for analyzing the stability of the method is similar to [26, 20]. However, there are some difficulties for keeping the energy bounds of potential functions. To this end, auxiliary functions related to the equation of states are introduced to complete the proof of the energy estimates. Finally, it is shown that the proposed time-discrete scheme is unconditionally stable.

The paper is organized as follows. In Section 2, the governing equations and their reformulation being considered are presented. Some relations among the unknown variables are introduced. The a priori estimates for clarifying the motivation of the numerical scheme are exhibited. In Section 3, the time-discrete projection method for the considered two-fluid model is provided and the finite element implementation for numerical tests is introduced. In Section 4, the stability analysis for time-discrete scheme is conducted. In Section 5, three test problems implemented by finite element method with convergence test justify the feasibility of the numerical scheme.

2 Modeling equations

In the following, we assume that the fluid domain Ωd\Omega\subset\mathbb{R}^{d}, d=2,3d=2,3 is smooth, bounded and connected. Let T>0T>0 be the final time. We denote by Γ\Gamma its boundary. For convenience, we define ΩT:=Ω×(0,T)\Omega_{T}:=\Omega\times(0,T) and ΓT:=Γ×(0,T)\Gamma_{T}:=\Gamma\times(0,T).

2.1 Compressible generic two-fluid model

We denote by the subscript k=g,lk=g,l for phase gg and phase ll, respectively. Let ϕk\phi_{k} be the volume fractions, ρk\rho_{k} be the densities, 𝒖k\bm{u}_{k} be the velocities, pkp_{k} be the pressures. The governing equations are given by

ϕg+ϕl=1,inΩT,\phi_{g}+\phi_{l}=1,~{}~{}~{}\text{in}~{}\Omega_{T}, (2.1a)
t(ϕkρk)+(ϕkρk𝒖k)=0,inΩT,\partial_{t}(\phi_{k}\rho_{k})+\nabla\cdot(\phi_{k}\rho_{k}\bm{u}_{k})=0,~{}~{}~{}\text{in}~{}\Omega_{T}, (2.1b)
t(ϕkρk𝒖k)+(ϕkρk𝒖k𝒖k)(ϕkτk(𝒖k))+ϕkpk+FD,k=0,inΩT,\partial_{t}(\phi_{k}\rho_{k}\bm{u}_{k})+\nabla\cdot(\phi_{k}\rho_{k}\bm{u}_{k}\otimes\bm{u}_{k})-\nabla\cdot(\phi_{k}\tau_{k}(\bm{u}_{k}))+\phi_{k}\nabla p_{k}+F_{D,k}=0,~{}~{}~{}\text{in}~{}\Omega_{T}, (2.1c)
pg=pl=p,inΩT.p_{g}=p_{l}=p,~{}~{}~{}\text{in}~{}\Omega_{T}. (2.1d)
The constraint (2.1d) is to close the system so that the number of variables equals to the number of equations.

The stress tensor τk(𝒖k)\tau_{k}(\bm{u}_{k}) are of the form

τ(𝒖k)=2μkD(𝒖k)+λk(𝒖k)I,\tau(\bm{u}_{k})=2\mu_{k}D(\bm{u}_{k})+\lambda_{k}(\nabla\cdot\bm{u}_{k})I, (2.2)

where D(𝒖k)D(\bm{u}_{k}) denotes the symmetric part of the velocity gradient 𝒖k\nabla\bm{u}_{k}, μk\mu_{k} are the dynamic viscosities and λk\lambda_{k} are the second viscosities. The drag force FD,kF_{D,k} are assumed to be of the form:

FD,k=CDϕgϕl|𝒖g𝒖l|(𝒖k𝒖k~),k=g,l,k~=l,g,F_{D,k}=C_{D}\phi_{g}\phi_{l}|\bm{u}_{g}-\bm{u}_{l}|(\bm{u}_{k}-\bm{u}_{\widetilde{k}}),~{}~{}k=g,l,~{}~{}\widetilde{k}=l,g, (2.3)

where CDC_{D} is assumed to be a positive constant.

Equation of state

We assume the barotropic system such that

p=ζg(ρg)=ζl(ρl)p=\zeta_{g}(\rho_{g})=\zeta_{l}(\rho_{l}) (2.4)

with

ζg(z)=Agzγg,ζl(z)=Al(zγlρl,0γl)+p0,\zeta_{g}(z)=A_{g}z^{\gamma_{g}},~{}~{}~{}\zeta_{l}(z)=A_{l}(z^{\gamma_{l}}-\rho_{l,0}^{\gamma_{l}})+p_{0}, (2.5)

for some positive constants Ag,Al,γg,γl,ρl,0,p0A_{g},A_{l},\gamma_{g},\gamma_{l},\rho_{l,0},p_{0}. In this work, we further assume that γk>1\gamma_{k}>1 for k=g,lk=g,l.

Boundary conditions and Initial conditions

To complete the statement of the mathematical problem, we impose that

ρk|Γin=ρk,in,ϕk|Γin=ϕk,in,𝒖k|Γ=𝒃k,\displaystyle\rho_{k}|_{\Gamma_{in}}=\rho_{k,in},~{}~{}\phi_{k}|_{\Gamma_{in}}=\phi_{k,in},~{}~{}\bm{u}_{k}|_{\Gamma}=\bm{b}_{k}, (2.6)
ρk|t=0=ρk0,ϕk|t=0=ϕk0,𝒖k|t=0=𝒖k,0,\displaystyle\rho_{k}|_{t=0}=\rho_{k}^{0},~{}~{}\phi_{k}|_{t=0}=\phi_{k}^{0},~{}~{}\bm{u}_{k}|_{t=0}=\bm{u}_{k,0},

where Γin\Gamma_{in} is the inlet boundary such that

Γin:={xΓ|𝒃k(x)𝒏(x)<0},\Gamma_{in}:=\{x\in\Gamma|~{}\bm{b}_{k}(x)\cdot\bm{n}(x)<0\}, (2.7)

where 𝒏()\bm{n}(\cdot) is the outer normal on Γ\Gamma.

2.2 Reformulation of the governing equations

Let αk=ϕkρk\alpha_{k}=\phi_{k}\rho_{k}, (2.1) becomes

αgρg+αlρl=1,inΩT,\frac{\alpha_{g}}{\rho_{g}}+\frac{\alpha_{l}}{\rho_{l}}=1,~{}~{}~{}\text{in}~{}\Omega_{T}, (2.8a)
tαk+(αk𝒖k)=0,inΩT,\partial_{t}\alpha_{k}+\nabla\cdot(\alpha_{k}\bm{u}_{k})=0,~{}~{}~{}\text{in}~{}\Omega_{T}, (2.8b)
t(αk𝒖k)+(αk𝒖k𝒖k)(αkρkτk(𝒖k))\displaystyle\partial_{t}(\alpha_{k}\bm{u}_{k})+\nabla\cdot(\alpha_{k}\bm{u}_{k}\otimes\bm{u}_{k})-\nabla\cdot(\frac{\alpha_{k}}{\rho_{k}}\tau_{k}(\bm{u}_{k})) (2.8c)
+\displaystyle+ αkρkp+CDαgαlρgρl|𝒖g𝒖l|(𝒖k𝒖k~)=0,k=g,l,k~=l,g,inΩT.\displaystyle\frac{\alpha_{k}}{\rho_{k}}\nabla p+C_{D}\frac{\alpha_{g}\alpha_{l}}{\rho_{g}\rho_{l}}|\bm{u}_{g}-\bm{u}_{l}|(\bm{u}_{k}-\bm{u}_{\widetilde{k}})=0,~{}~{}k=g,l,~{}~{}\widetilde{k}=l,g,~{}~{}~{}\text{in}~{}\Omega_{T}.

In the following, we will show that with given αg\alpha_{g} and αl\alpha_{l}, the quantities ϕg\phi_{g}, ϕl\phi_{l}, ρg\rho_{g}, ρl\rho_{l}, and pp can be uniquely determined.

Pressure balance

By (2.1d), we have

sg2dρg=sl2dρl,sk=dζkdρk(ρk)s_{g}^{2}d\rho_{g}=s_{l}^{2}d\rho_{l},~{}~{}~{}s_{k}=\sqrt{\frac{d\zeta_{k}}{d\rho_{k}}(\rho_{k})} (2.9)

By αk=ϕkρk\alpha_{k}=\phi_{k}\rho_{k}, we have

dρg=1ϕg(dαgρgdϕg),dρl=1ϕl(dαlρldϕg)d\rho_{g}=\frac{1}{\phi_{g}}(d\alpha_{g}-\rho_{g}d\phi_{g}),~{}~{}~{}d\rho_{l}=\frac{1}{\phi_{l}}(d\alpha_{l}-\rho_{l}d\phi_{g})

Using (2.9), the differential of the gaseous phase volume fraction can be expressed as

dϕg=ϕlsg2ϕlρgsg2+ϕgρlsl2dαgϕlsl2ϕlρgsg2+ϕgρlsl2dαld\phi_{g}=\frac{\phi_{l}s_{g}^{2}}{\phi_{l}\rho_{g}s_{g}^{2}+\phi_{g}\rho_{l}s_{l}^{2}}d\alpha_{g}-\frac{\phi_{l}s_{l}^{2}}{\phi_{l}\rho_{g}s_{g}^{2}+\phi_{g}\rho_{l}s_{l}^{2}}d\alpha_{l}

Note that we always work with the case with ϕk>0\phi_{k}>0. The differential of the gaseous phase density can be expressed as

dρg=ρgρlsl2αlρg2sg2+αgρl2sl2(ρldαg+ρgdαl).d\rho_{g}=\frac{\rho_{g}\rho_{l}s_{l}^{2}}{\alpha_{l}\rho_{g}^{2}s_{g}^{2}+\alpha_{g}\rho_{l}^{2}s_{l}^{2}}(\rho_{l}d\alpha_{g}+\rho_{g}d\alpha_{l}).

Therefore, the differential of pressure can be expressed as

dp=C2(ρldαg+ρgdαl),dp=C^{2}(\rho_{l}d\alpha_{g}+\rho_{g}d\alpha_{l}), (2.10)

where

C2=sl2sg2ϕgρlsl2+ϕlρgsg2C^{2}=\frac{s_{l}^{2}s_{g}^{2}}{\phi_{g}\rho_{l}s_{l}^{2}+\phi_{l}\rho_{g}s_{g}^{2}} (2.11)

With (2.10) in hand, (2.8c) can be rewritten as

t(αk𝒖k)+(αk𝒖k𝒖k)+C2(ρlαg+ρgαl)(ϕkτk(𝒖k))+FD,k=0\partial_{t}(\alpha_{k}\bm{u}_{k})+\nabla\cdot(\alpha_{k}\bm{u}_{k}\otimes\bm{u}_{k})+C^{2}(\rho_{l}\nabla\alpha_{g}+\rho_{g}\nabla\alpha_{l})-\nabla\cdot(\phi_{k}\tau_{k}(\bm{u}_{k}))+F_{D,k}=0 (2.12)

Relations among ϕk\phi_{k}, ρk\rho_{k}, and αk\alpha_{k}

We recall (2.8a), which gives us

ρl=αlρgρgαg\rho_{l}=\frac{\alpha_{l}\rho_{g}}{\rho_{g}-\alpha_{g}} (2.13)

By the pressure equilibrium assumption (2.1d), we have

φ(ρg)=ζg(ρg)ζl(αlρgρgαg)=0\varphi(\rho_{g})=\zeta_{g}(\rho_{g})-\zeta_{l}\left(\frac{\alpha_{l}\rho_{g}}{\rho_{g}-\alpha_{g}}\right)=0 (2.14)

Differentiating φ\varphi with respect to ρg\rho_{g} yields

φ(ρg)=sg2+sl2αlαg(ρgαg)2\varphi^{\prime}(\rho_{g})=s_{g}^{2}+s_{l}^{2}\frac{\alpha_{l}\alpha_{g}}{(\rho_{g}-\alpha_{g})^{2}}

Therefore φ\varphi is a non-decreasing function of ρg\rho_{g}. For non-degenerate case ϕk<1\phi_{k}<1, we look for ρg(αg,+)\rho_{g}\in(\alpha_{g},+\infty). Since φ((αg,+))=(,+)\varphi((\alpha_{g},+\infty))=(-\infty,+\infty), ρg=ρg(αg,αl)\rho_{g}=\rho_{g}(\alpha_{g},\alpha_{l}) is uniquely determined by solving φ(ρg)=0\varphi(\rho_{g})=0 with given αg\alpha_{g} and αl\alpha_{l}. Finally, ρl\rho_{l} and ϕk\phi_{k}, k=g,lk=g,l are given by

ϕg(αg,αl)=αgρg(αg,αl),ϕl(αg,αl)=1αgρg(αg,αl),ρl(αg,αl)=αlρg(αg,αl)ρg(αg,αl)αg\phi_{g}(\alpha_{g},\alpha_{l})=\frac{\alpha_{g}}{\rho_{g}(\alpha_{g},\alpha_{l})},~{}~{}\phi_{l}(\alpha_{g},\alpha_{l})=1-\frac{\alpha_{g}}{\rho_{g}(\alpha_{g},\alpha_{l})},~{}~{}\rho_{l}(\alpha_{g},\alpha_{l})=\frac{\alpha_{l}\rho_{g}(\alpha_{g},\alpha_{l})}{\rho_{g}(\alpha_{g},\alpha_{l})-\alpha_{g}} (2.15)

2.3 A priori estimates to (2.8a)-(2.8c)

In this work, the stability issues are particularly payed attention and a discrete version associated to the numerical scheme is desired. Let us recall the a priori estimates associated to problem (2.8a)-(2.8c) with zero forcing term and zero velocities on Γ\Gamma for all time[15]. The following estimates present the positiveness of the mass, mass conservation, and the energy stability, respectively. For k=g,lk=g,l, we have

  • Positiveness of the mass

    αk(x,t)>0,(x,t)ΩT\alpha_{k}(x,t)>0,~{}~{}~{}\forall(x,t)\in\Omega_{T} (2.16)
  • Mass conservation

    Ωαk(x,t)𝑑x=Ωαk(x,0)𝑑x,t(0,T)\int_{\Omega}\alpha_{k}(x,t)dx=\int_{\Omega}\alpha_{k}(x,0)dx,~{}~{}~{}\forall t\in(0,T) (2.17)
  • Energy stability

    k=g,l[12ddtΩαk(x,t)𝒖k(x,t)2dx+ddtΩαk(x,t)ek(ρk(x,t))dx\displaystyle\sum_{k=g,l}\left[\frac{1}{2}\frac{d}{dt}\int_{\Omega}\alpha_{k}(x,t)\bm{u}_{k}(x,t)^{2}dx+\frac{d}{dt}\int_{\Omega}\alpha_{k}(x,t)e_{k}(\rho_{k}(x,t))dx\right. (2.18)
    +Ωϕk(x,t)τk(𝒖k(x,t)):𝒖k(x,t)dx]+ΩCDϕg(x,t)ϕl(x,t)|𝒖g(x,t)𝒖l(x,t)|3dx=0\displaystyle\left.+\int_{\Omega}\phi_{k}(x,t)\tau_{k}(\bm{u}_{k}(x,t)):\nabla\bm{u}_{k}(x,t)dx\right]+\int_{\Omega}C_{D}\phi_{g}(x,t)\phi_{l}(x,t)|\bm{u}_{g}(x,t)-\bm{u}_{l}(x,t)|^{3}dx=0

In the above, ek()e_{k}(\cdot) is the potential energy dervied from the equation of state such that

ek(z)=ζk(z)z2e^{\prime}_{k}(z)=\frac{\zeta_{k}(z)}{z^{2}} (2.19)

We may choose proper constants ρk,ref\rho_{k,ref} so that the following expression makes sense:

ek(z)=ρk,refzζk(s)s2𝑑se_{k}(z)=\int_{\rho_{k,ref}}^{z}\frac{\zeta_{k}(s)}{s^{2}}ds (2.20)

The estimate (2.18) is a little bit different from that proposed by [15] because of extra capillary terms in their work. Nevertheless, they can be obtained by a completely same way.

3 Numerical scheme

3.1 Time-discrete scheme

We proceed the following to get the time-discrete solution:

  • Step 1. Prediction of the mass fraction α~km+1\widetilde{\alpha}_{k}^{m+1}:

    α~km+1αkmδt+(α~km+1𝒖km)=0\frac{\widetilde{\alpha}_{k}^{m+1}-\alpha_{k}^{m}}{\delta t}+\nabla\cdot(\widetilde{\alpha}_{k}^{m+1}\bm{u}_{k}^{m})=0 (3.1)
  • Step 2. Solve the intermediate volume fraction ϕ~km+1\widetilde{\phi}_{k}^{m+1} and the density ρ~km+1\widetilde{\rho}_{k}^{m+1} by

    {ϕ~gm+1+ϕ~lm+1=1ϕ~km+1ρ~km+1=α~km+1,k=g,lζg(ρ~gm+1)=ζl(ρ~lm+1)\begin{cases}\widetilde{\phi}_{g}^{m+1}+\widetilde{\phi}_{l}^{m+1}=1\\ \widetilde{\phi}_{k}^{m+1}\widetilde{\rho}_{k}^{m+1}=\widetilde{\alpha}_{k}^{m+1},~{}~{}k=g,l\\ \zeta_{g}(\widetilde{\rho}_{g}^{m+1})=\zeta_{l}(\widetilde{\rho}_{l}^{m+1})\end{cases} (3.2)
  • Step 3. Renormalization of the intermediate pressure p~km+1\widetilde{p}_{k}^{m+1}:

    (ϕ~km+1ρ~km+1p~km+1)=(ϕ~km+1ϕ~kmρ~km+1ρ~kmpm)\nabla\cdot\left(\frac{\widetilde{\phi}_{k}^{m+1}}{\widetilde{\rho}_{k}^{m+1}}\nabla\widetilde{p}_{k}^{m+1}\right)=\nabla\cdot\left(\sqrt{\frac{\widetilde{\phi}_{k}^{m+1}\widetilde{\phi}_{k}^{m}}{\widetilde{\rho}_{k}^{m+1}\widetilde{\rho}_{k}^{m}}}\nabla p^{m}\right) (3.3)

    with the boundary constraint:

    ϕ~km+1ρ~km+1p~km+1n=ϕ~km+1ϕ~kmρ~km+1ρ~kmpmnonΓ\frac{\widetilde{\phi}_{k}^{m+1}}{\widetilde{\rho}_{k}^{m+1}}\frac{\partial\widetilde{p}_{k}^{m+1}}{\partial n}=\sqrt{\frac{\widetilde{\phi}_{k}^{m+1}\widetilde{\phi}_{k}^{m}}{\widetilde{\rho}_{k}^{m+1}\widetilde{\rho}_{k}^{m}}}\frac{\partial p^{m}}{\partial n}~{}~{}~{}\text{on}~{}\Gamma (3.4)
  • Step 4. Solve the intermediate velocities 𝒖~km+1\widetilde{\bm{u}}_{k}^{m+1}

    α~km+1𝒖~km+1αkm𝒖kmδt+(α~km+1𝒖km𝒖~km+1)+ϕ~km+1p~km+1\displaystyle\frac{\widetilde{\alpha}_{k}^{m+1}\widetilde{\bm{u}}_{k}^{m+1}-\alpha_{k}^{m}\bm{u}_{k}^{m}}{\delta t}+\nabla\cdot(\widetilde{\alpha}_{k}^{m+1}\bm{u}_{k}^{m}\otimes\widetilde{\bm{u}}_{k}^{m+1})+\widetilde{\phi}_{k}^{m+1}\nabla\widetilde{p}_{k}^{m+1} (3.5)
    \displaystyle- (ϕ~km+1τk(𝒖~km+1))+CDϕ~gm+1ϕ~lm+1|𝒖~gm+1𝒖~lm+1|(𝒖~km+1𝒖~k~m+1)=0\displaystyle\nabla\cdot(\widetilde{\phi}_{k}^{m+1}\tau_{k}(\widetilde{\bm{u}}_{k}^{m+1}))+C_{D}\widetilde{\phi}_{g}^{m+1}\widetilde{\phi}_{l}^{m+1}|\widetilde{\bm{u}}_{g}^{m+1}-\widetilde{\bm{u}}_{l}^{m+1}|(\widetilde{\bm{u}}_{k}^{m+1}-\widetilde{\bm{u}}_{\widetilde{k}}^{m+1})=0
  • Step 5. (Projection) Solve 𝒖¯km+1\overline{\bm{u}}_{k}^{m+1}, pm+1p^{m+1}, ϕkm+1\phi_{k}^{m+1}, ρkm+1\rho_{k}^{m+1}:

    {α~km+1𝒖¯km+1𝒖~km+1δt+ϕ~km+1(pm+1p~km+1)=0ϕkm+1ρkm+1ϕkmρkmδt+(ϕ~km+1ρkm+1𝒖¯km+1)=0ϕgm+1+ϕlm+1=1ρkm+1=ζk1(pm+1)\begin{cases}\displaystyle\widetilde{\alpha}_{k}^{m+1}\frac{\overline{\bm{u}}_{k}^{m+1}-\widetilde{\bm{u}}_{k}^{m+1}}{\delta t}+\widetilde{\phi}_{k}^{m+1}\nabla(p^{m+1}-\widetilde{p}_{k}^{m+1})=0\\ \displaystyle\frac{\phi_{k}^{m+1}\rho_{k}^{m+1}-\phi_{k}^{m}\rho_{k}^{m}}{\delta t}+\nabla\cdot(\widetilde{\phi}_{k}^{m+1}\rho_{k}^{m+1}\overline{\bm{u}}_{k}^{m+1})=0\\ \phi_{g}^{m+1}+\phi_{l}^{m+1}=1\\ \rho_{k}^{m+1}=\zeta_{k}^{-1}(p^{m+1})\end{cases} (3.6)

    with the boundary constraint

    pm+1n=p~km+1n\frac{\partial p^{m+1}}{\partial n}=\frac{\partial\widetilde{p}_{k}^{m+1}}{\partial n} (3.7)
  • Step 6. Renormalization of the end-of-step velocities 𝒖km+1\bm{u}_{k}^{m+1}:

    αkm+1𝒖km+1=α~km+1𝒖¯km+1,αkm+1=ϕkm+1ρkm+1,k=g,l\sqrt{\alpha_{k}^{m+1}}\bm{u}_{k}^{m+1}=\sqrt{\widetilde{\alpha}_{k}^{m+1}}\overline{\bm{u}}_{k}^{m+1},~{}~{}\alpha_{k}^{m+1}=\phi_{k}^{m+1}\rho_{k}^{m+1},~{}~{}k=g,l (3.8)

Equivalent problem of the projection step

To eliminate the variable 𝒖¯km+1\overline{\bm{u}}_{k}^{m+1} in the coupling problem (3.6), we may manipulate the first two equations in (3.6) to get

  • Step 5’. Solve pm+1p^{m+1}, ϕgm+1\phi_{g}^{m+1}, ϕlm+1\phi_{l}^{m+1}:

    {1δt2(ϕkm+1ζk1(pm+1)αkm)+1δt(ϕ~km+1ζk1(pm+1)𝒖~km+1)(ϕ~km+1ζk1(pm+1)ρ~km+1(pm+1p~km+1))=0ϕgm+1+ϕlm+1=1\begin{cases}\displaystyle\frac{1}{\delta t^{2}}(\phi_{k}^{m+1}\zeta_{k}^{-1}(p^{m+1})-\alpha_{k}^{m})+\frac{1}{\delta t}\nabla\cdot(\widetilde{\phi}_{k}^{m+1}\zeta_{k}^{-1}(p^{m+1})\widetilde{\bm{u}}_{k}^{m+1})\\ -\nabla\cdot\left(\frac{\widetilde{\phi}_{k}^{m+1}\zeta_{k}^{-1}(p^{m+1})}{\widetilde{\rho}_{k}^{m+1}}\nabla(p^{m+1}-\widetilde{p}_{k}^{m+1})\right)=0\\ \phi_{g}^{m+1}+\phi_{l}^{m+1}=1\end{cases} (3.9)

In view of the asymmetric structure of (3.9), it is not easy to solve the problem directly. To tackle the difficulty, the argument in Section 2.2 is applied so that we can express pm+1p^{m+1} in terms of αlm+1\alpha_{l}^{m+1} and αgm+1\alpha_{g}^{m+1}. A symmetric projection step can be expressed as

  • Step 5”. Solve αkm+1\alpha_{k}^{m+1}, pm+1p^{m+1}:

    {1δt2(αkm+1αkm)+1δt(ϕ~km+1ζk1(pm+1)𝒖~km+1)(ϕ~km+1ζk1(pm+1)ρ~km+1(pm+1p~km+1))=0pm+1=C(αgm+1,αlm+1)2(ρl(αgm+1,αlm+1)αgm+1+ρg(αgm+1,αlm+1)αlm+1)\begin{cases}\displaystyle\frac{1}{\delta t^{2}}(\alpha_{k}^{m+1}-\alpha_{k}^{m})+\frac{1}{\delta t}\nabla\cdot(\widetilde{\phi}_{k}^{m+1}\zeta_{k}^{-1}(p^{m+1})\widetilde{\bm{u}}_{k}^{m+1})-\nabla\cdot\left(\frac{\widetilde{\phi}_{k}^{m+1}\zeta_{k}^{-1}(p^{m+1})}{\widetilde{\rho}_{k}^{m+1}}\nabla(p^{m+1}-\widetilde{p}_{k}^{m+1})\right)=0\\ \nabla p^{m+1}=C(\alpha_{g}^{m+1},\alpha_{l}^{m+1})^{2}\left(\rho_{l}(\alpha_{g}^{m+1},\alpha_{l}^{m+1})\nabla\alpha_{g}^{m+1}+\rho_{g}(\alpha_{g}^{m+1},\alpha_{l}^{m+1})\nabla\alpha_{l}^{m+1}\right)\end{cases} (3.10)
Remark 1

Step 5” is solved iteratively in this way: Given αkm+1,l\alpha_{k}^{m+1,l}, ρkm+1,l\rho_{k}^{m+1,l} at the ll-th iteration step, the solutions for the (l+1)(l+1)-th iteration step are given by

{1δt2αkm+1,l+1+1δt(ϕ~km+1ρkm+1,l𝒖~km+1)(ϕ~km+1ρkm+1,lρ~km+1C(αgm+1,l,αlm+1,l)2(ρlm+1,lαgm+1,l+1+ρgm+1,lαlm+1,l+1))=(ϕ~km+1ρkm+1,lρ~km+1p~km+1)+1δt2αkmρkm+1,l+1=ρk(αgm+1,l+1,αlm+1,l+1)\begin{cases}\displaystyle\frac{1}{\delta t^{2}}\alpha_{k}^{m+1,l+1}+\frac{1}{\delta t}\nabla\cdot(\widetilde{\phi}_{k}^{m+1}\rho_{k}^{m+1,l}\widetilde{\bm{u}}_{k}^{m+1})\\ -\displaystyle\nabla\cdot\left(\frac{\widetilde{\phi}_{k}^{m+1}\rho_{k}^{m+1,l}}{\widetilde{\rho}_{k}^{m+1}}C(\alpha_{g}^{m+1,l},\alpha_{l}^{m+1,l})^{2}(\rho_{l}^{m+1,l}\nabla\alpha_{g}^{m+1,l+1}+\rho_{g}^{m+1,l}\nabla\alpha_{l}^{m+1,l+1})\right)\\ =\displaystyle-\nabla\cdot\left(\frac{\widetilde{\phi}_{k}^{m+1}\rho_{k}^{m+1,l}}{\widetilde{\rho}_{k}^{m+1}}\nabla\widetilde{p}_{k}^{m+1}\right)+\frac{1}{\delta t^{2}}\alpha_{k}^{m}\\ \rho_{k}^{m+1,l+1}=\rho_{k}(\alpha_{g}^{m+1,l+1},\alpha_{l}^{m+1,l+1})\end{cases} (3.11)

3.2 Finite element implementation

For simplicity, we consider the problem with 𝒖k=0\bm{u}_{k}=0 on Γ\Gamma, k=g,lk=g,l. The case with nonhomogeneous boundary conditions or outlet can be modified accordingly by a standard way.

Let us introduce finite element approximations YhH1Y_{h}\subset H_{1} for densities ρk,h\rho_{k,h} (also for intermediate step ρ~k,h\widetilde{\rho}_{k,h}), volume fractions ϕk,h\phi_{k,h} (also for intermediate step ϕ~k,h\widetilde{\phi}_{k,h}), their products αk,h\alpha_{k,h} (also for intermediate step α~k,h\widetilde{\alpha}_{k,h}), and for pressure php_{h} (also for intermediate step p~k,h\widetilde{p}_{k,h}); 𝑿0,h𝑯01\bm{X}_{0,h}\subset\bm{H}_{0}^{1} for intermediate step velocities 𝒖~k,h\widetilde{\bm{u}}_{k,h}.

To maintain the positiveness of αk\alpha_{k}, the weak formulation for Step 1 - (3.1) reads: For m0m\geq 0, find α~k,hm+1Yh\widetilde{\alpha}_{k,h}^{m+1}\in Y_{h} such that for α~k,hm+1|Γin=ϕk,inρk,in\widetilde{\alpha}_{k,h}^{m+1}|_{\Gamma_{in}}=\phi_{k,in}\rho_{k,in} and such that for all qhYhq_{h}\in Y_{h} with qh|Γin=0q_{h}|_{\Gamma_{in}}=0,

((1+12δt(𝒖k,hm)+12δt(𝒖k,hm)+)α~k,hm+1,qh)+δt(𝒖k,hmα~k,hm+1,qh)=((1+12δt(𝒖k,hm))αk,hm,qh)\left((1+\frac{1}{2}\delta t(\nabla\cdot\bm{u}_{k,h}^{m})+\frac{1}{2}\delta t(\nabla\cdot\bm{u}_{k,h}^{m})^{+})\widetilde{\alpha}_{k,h}^{m+1},q_{h}\right)+\delta t(\bm{u}_{k,h}^{m}\cdot\nabla\widetilde{\alpha}_{k,h}^{m+1},q_{h})=\left((1+\frac{1}{2}\delta t(\nabla\cdot\bm{u}_{k,h}^{m})^{-})\alpha_{k,h}^{m},q_{h}\right) (3.12)

The positiveness of the above scheme is shown in Remark 2.

At Step 2, the intermediate ϕ~k,hm+1\widetilde{\phi}_{k,h}^{m+1} and ρ~k,hm+1\widetilde{\rho}_{k,h}^{m+1} can be obtain pointwisely by the given α~k,hm+1\widetilde{\alpha}_{k,h}^{m+1} without error generation by spatial discretization since all of them belong to the same space YhY_{h}. One may conduct a root-finding technique (e.g. Ridder’s method[27]) to solve ϕ~k,hm+1\widetilde{\phi}_{k,h}^{m+1} and ρ~k,hm+1\widetilde{\rho}_{k,h}^{m+1} pointwisely.

The renormalization (Step 3) for intermediate pressure p~km+1\widetilde{p}_{k}^{m+1} can be proceeded by the following problem: For m0m\geq 0, find p~km+1Yh\widetilde{p}_{k}^{m+1}\in Y_{h} such that for all whYhw_{h}\in Y_{h},

(ϕ~k,hm+1ρ~k,hm+1p~k,hm+1,wh)=(ϕ~k,hm+1ϕ~k,hmρ~k,hm+1ρ~k,hmpk,hm,wh)\left(\frac{\widetilde{\phi}_{k,h}^{m+1}}{\widetilde{\rho}_{k,h}^{m+1}}\nabla\widetilde{p}_{k,h}^{m+1},\nabla w_{h}\right)=\left(\sqrt{\frac{\widetilde{\phi}_{k,h}^{m+1}\widetilde{\phi}_{k,h}^{m}}{\widetilde{\rho}_{k,h}^{m+1}\widetilde{\rho}_{k,h}^{m}}}\nabla p_{k,h}^{m},\nabla w_{h}\right) (3.13)

We note that the boundary terms are eliminated because of (3.4).

The weak formulation for Step 4 (advection-diffusion step) reads: For m0m\geq 0, find 𝒖~k,hm+1𝑿0,h\widetilde{\bm{u}}_{k,h}^{m+1}\in\bm{X}_{0,h} such that for all 𝒗h𝑿0,h\bm{v}_{h}\in\bm{X}_{0,h},

(α~k,hm+1𝒖~k,hm+1αk,hm𝒖k,hmδt,𝒗h)+((α~k,hm+1𝒖k,hm𝒖~k,hm+1),𝒗h)(p~k,hm+1,(ϕ~k,hm+1𝒗h))\displaystyle\left(\frac{\widetilde{\alpha}_{k,h}^{m+1}\widetilde{\bm{u}}_{k,h}^{m+1}-\alpha_{k,h}^{m}\bm{u}_{k,h}^{m}}{\delta t},\bm{v}_{h}\right)+\left(\nabla\cdot(\widetilde{\alpha}_{k,h}^{m+1}\bm{u}_{k,h}^{m}\otimes\widetilde{\bm{u}}_{k,h}^{m+1}),\bm{v}_{h}\right)-\left(\widetilde{p}_{k,h}^{m+1},\nabla\cdot(\widetilde{\phi}_{k,h}^{m+1}\bm{v}_{h})\right) (3.14)
+(ϕ~k,hm+1τk(𝒖~k,hm+1),𝒗h)+(CDϕ~g,hm+1ϕ~l,hm+1|𝒖~g,hm+1𝒖~l,hm+1|(𝒖~k,hm+1𝒖~k~,hm+1),𝒗h)=0.\displaystyle+(\widetilde{\phi}_{k,h}^{m+1}\tau_{k}(\widetilde{\bm{u}}_{k,h}^{m+1}),\nabla\bm{v}_{h})+\left(C_{D}\widetilde{\phi}_{g,h}^{m+1}\widetilde{\phi}_{l,h}^{m+1}|\widetilde{\bm{u}}_{g,h}^{m+1}-\widetilde{\bm{u}}_{l,h}^{m+1}|(\widetilde{\bm{u}}_{k,h}^{m+1}-\widetilde{\bm{u}}_{\widetilde{k},h}^{m+1}),\bm{v}_{h}\right)=0.

The above equation can be solved separately for gg and ll and the nonlinear term can be tackled by iteration (see for example [28]).

The projection step (Step 5) is handled by (3.10) and the according weak formulation reads: For m0m\geq 0, find αkm+1Yh\alpha_{k}^{m+1}\in Y_{h} such that for all qhYhq_{h}\in Y_{h},

(αk,hm+1αk,hm,qh)+δt((ϕ~k,hm+1ζk1(phm+1)𝒖~k,hm+1),qh)+δt2(ϕ~k,hm+1ζk,h1(phm+1)ρ~k,hm+1(phm+1p~k,hm+1),qh)=0\displaystyle(\alpha_{k,h}^{m+1}-\alpha_{k,h}^{m},q_{h})+\delta t\left(\nabla\cdot(\widetilde{\phi}_{k,h}^{m+1}\zeta_{k}^{-1}(p_{h}^{m+1})\widetilde{\bm{u}}_{k,h}^{m+1}),q_{h}\right)+\delta t^{2}\left(\frac{\widetilde{\phi}_{k,h}^{m+1}\zeta_{k,h}^{-1}(p_{h}^{m+1})}{\widetilde{\rho}_{k,h}^{m+1}}\nabla(p_{h}^{m+1}-\widetilde{p}_{k,h}^{m+1}),\nabla q_{h}\right)=0 (3.15)

with

phm+1=C(αg,hm+1,αl,hm+1)2(ρl(αg,hm+1,αl,hm+1)αg,hm+1+ρg(αg,hm+1,αl,hm+1)αl,hm+1)p_{h}^{m+1}=C(\alpha_{g,h}^{m+1},\alpha_{l,h}^{m+1})^{2}(\rho_{l}(\alpha_{g,h}^{m+1},\alpha_{l,h}^{m+1})\nabla\alpha_{g,h}^{m+1}+\rho_{g}(\alpha_{g,h}^{m+1},\alpha_{l,h}^{m+1})\nabla\alpha_{l,h}^{m+1}) (3.16)

The above system can be solved iteratively by the way provided in Remark 1. In (3.16), functions of αg,hm+1\alpha_{g,h}^{m+1} and αl,hm+1\alpha_{l,h}^{m+1}: C2C^{2}, ρl\rho_{l}, and ρg\rho_{g} are taken to be their projection on YhY_{h}. Similar to (3.13), the boundary terms of (3.15) are eliminated because of the boundary constraint (3.7).

4 Stability analysis

Let φ\varphi\in\mathbb{R} be any function. We denote by φ+:=max(f,0)\varphi^{+}:=\max(f,0) and by φ:=min(f,0)\varphi^{-}:=-\min(f,0). We denote by Lp:=Lp(Ω)\|\cdot\|_{L^{p}}:=\|\cdot\|_{L^{p}(\Omega)} the LpL^{p} norm on Ω\Omega, Wk,p:=Wk,p(Ω)\|\cdot\|_{W^{k,p}}:=\|\cdot\|_{W^{k,p}(\Omega)} the Wk,pW^{k,p} norm, and k=Hk=Wk,2\|\cdot\|_{k}=\|\cdot\|_{H^{k}}=\|\cdot\|_{W^{k,2}}, 0k+0\leq k\leq+\infty, 1p+1\leq p\leq+\infty. We denote by (,)(\cdot,\cdot) the inner product associated to L2(Ω)L^{2}(\Omega) space such that (u,v)=Ωu(x)v(x)𝑑x(u,v)=\int_{\Omega}u(x)v(x)dx for u,vu,v in L2(Ω)L^{2}(\Omega).

Mass conservation

Step 5 and Step 6 guarantee the mass conservation in the following sense:

Proposition 1

If 𝐮¯km+1\overline{\bm{u}}_{k}^{m+1}, k=g,lk=g,l satisfy the boundary conditions 𝐮¯km+1𝐧|Γ=0\overline{\bm{u}}_{k}^{m+1}\cdot\bm{n}|_{\Gamma}=0, then we have

Ωαkm+1𝑑x=Ωαkm𝑑x,k=g,l\int_{\Omega}\alpha_{k}^{m+1}dx=\int_{\Omega}\alpha_{k}^{m}dx,~{}~{}~{}k=g,l (4.1)

Proof. By (3.8) and (3.6), we have

αkm+1αkmδt+(ϕ~km+1ρkm+1𝒖¯km+1)=0\frac{\alpha_{k}^{m+1}-\alpha_{k}^{m}}{\delta t}+\nabla\cdot(\widetilde{\phi}_{k}^{m+1}\rho_{k}^{m+1}\overline{\bm{u}}_{k}^{m+1})=0

Taking the integration on both sides, the divergence theorem together with the assumption 𝐮¯km+1𝐧|Γ=0\overline{\bm{u}}_{k}^{m+1}\cdot\bm{n}|_{\Gamma}=0 lead to the conclusion. Q.E.D.

Energy estimates of (3.1)-(3.8)

We recall an important lemma which will be used several times:

Lemma 1

For sufficiently smooth φ\varphi and 𝐯\bm{v} with 𝐯𝐧|Γ=0\bm{v}\cdot\bm{n}|_{\Gamma}=0, we have

Ω(φ𝒗φ+12φ2𝒗)=0\int_{\Omega}\left(\varphi\bm{v}\cdot\nabla\varphi+\frac{1}{2}\varphi^{2}\nabla\cdot\bm{v}\right)=0

Let us start with the estimates for the intermediate pressures p~km+1\widetilde{p}_{k}^{m+1}. Taking the inner product of (3.3) with p~m+1\widetilde{p}^{m+1},

ϕ~km+1ρ~km+1p~km+102=(ϕ~km+1ρ~km+1p~km+1,ϕ~kmρ~km+1pm)ϕ~km+1ρ~km+1p~km+10ϕ~kmρ~kmpkm0\left\|\sqrt{\frac{\widetilde{\phi}_{k}^{m+1}}{\widetilde{\rho}_{k}^{m+1}}}\nabla\widetilde{p}_{k}^{m+1}\right\|_{0}^{2}=\left(\sqrt{\frac{\widetilde{\phi}_{k}^{m+1}}{\widetilde{\rho}_{k}^{m+1}}}\nabla\widetilde{p}_{k}^{m+1},\sqrt{\frac{\widetilde{\phi}_{k}^{m}}{\widetilde{\rho}_{k}^{m+1}}}\nabla p^{m}\right)\leq\left\|\sqrt{\frac{\widetilde{\phi}_{k}^{m+1}}{\widetilde{\rho}_{k}^{m+1}}}\nabla\widetilde{p}_{k}^{m+1}\right\|_{0}\left\|\sqrt{\frac{\widetilde{\phi}_{k}^{m}}{\widetilde{\rho}_{k}^{m}}}\nabla p_{k}^{m}\right\|_{0}

Therefore,

ϕ~km+1ρ~km+1p~km+10ϕ~kmρ~kmpkm0\left\|\sqrt{\frac{\widetilde{\phi}_{k}^{m+1}}{\widetilde{\rho}_{k}^{m+1}}}\nabla\widetilde{p}_{k}^{m+1}\right\|_{0}\leq\left\|\sqrt{\frac{\widetilde{\phi}_{k}^{m}}{\widetilde{\rho}_{k}^{m}}}\nabla p_{k}^{m}\right\|_{0} (4.2)

Summing the inner product of (3.5) with 𝒖~km+1\widetilde{\bm{u}}_{k}^{m+1} and the inner product of (3.1) with 12|𝒖~km+1|2-\frac{1}{2}|\widetilde{\bm{u}}_{k}^{m+1}|^{2}, we have

1δtα~km+1𝒖~km+1021δt(αkm𝒖km,𝒖~km+1)+((α~km+1𝒖km𝒖~km+1),𝒖~km+1)\displaystyle\frac{1}{\delta t}\|\sqrt{\widetilde{\alpha}_{k}^{m+1}}\widetilde{\bm{u}}_{k}^{m+1}\|_{0}^{2}-\frac{1}{\delta t}(\alpha_{k}^{m}\bm{u}_{k}^{m},\widetilde{\bm{u}}_{k}^{m+1})+(\nabla\cdot(\widetilde{\alpha}_{k}^{m+1}\bm{u}_{k}^{m}\otimes\widetilde{\bm{u}}_{k}^{m+1}),\widetilde{\bm{u}}_{k}^{m+1}) (4.3)
+(ϕ~km+1p~km+1,𝒖~km+1)+(ϕ~km+1τk(𝒖~km+1),𝒖~km+1)+(F~D,km+1,𝒖~km+1)\displaystyle+(\widetilde{\phi}_{k}^{m+1}\nabla\widetilde{p}_{k}^{m+1},\widetilde{\bm{u}}_{k}^{m+1})+(\widetilde{\phi}_{k}^{m+1}\tau_{k}(\widetilde{\bm{u}}_{k}^{m+1}),\nabla\widetilde{\bm{u}}_{k}^{m+1})+(\widetilde{F}_{D,k}^{m+1},\widetilde{\bm{u}}_{k}^{m+1})
12δtα~km+1𝒖~km+102+12δtαkm𝒖~km+10212((α~km+1𝒖km),|𝒖~km+1|2)=0,\displaystyle-\frac{1}{2\delta t}\|\sqrt{\widetilde{\alpha}_{k}^{m+1}}\widetilde{\bm{u}}_{k}^{m+1}\|_{0}^{2}+\frac{1}{2\delta t}\|\sqrt{\alpha_{k}^{m}}\widetilde{\bm{u}}_{k}^{m+1}\|_{0}^{2}-\frac{1}{2}(\nabla\cdot(\widetilde{\alpha}_{k}^{m+1}\bm{u}_{k}^{m}),|\widetilde{\bm{u}}_{k}^{m+1}|^{2})=0,

where F~D,km+1=CDϕ~gm+1ϕ~lm+1|𝒖~gm+1𝒖~lm+1|(𝒖~km+1𝒖~k~m+1)\widetilde{F}_{D,k}^{m+1}=C_{D}\widetilde{\phi}_{g}^{m+1}\widetilde{\phi}_{l}^{m+1}|\widetilde{\bm{u}}_{g}^{m+1}-\widetilde{\bm{u}}_{l}^{m+1}|(\widetilde{\bm{u}}_{k}^{m+1}-\widetilde{\bm{u}}_{\widetilde{k}}^{m+1}). Using the Cauchy’s inequality, Green’s theorem, and an analogue of Lemma 1, we have

12α~km+1𝒖~km+102+δt(ϕ~km+1p~km+1,𝒖~km+1)+δt(ϕ~km+1τk(𝒖~km+1),𝒖~km+1)+δt(F~D,km+1,𝒖~km+1)12αkm𝒖km02\frac{1}{2}\|\sqrt{\widetilde{\alpha}_{k}^{m+1}}\widetilde{\bm{u}}_{k}^{m+1}\|_{0}^{2}+\delta t(\widetilde{\phi}_{k}^{m+1}\nabla\widetilde{p}_{k}^{m+1},\widetilde{\bm{u}}_{k}^{m+1})+\delta t(\widetilde{\phi}_{k}^{m+1}\tau_{k}(\widetilde{\bm{u}}_{k}^{m+1}),\nabla\widetilde{\bm{u}}_{k}^{m+1})+\delta t(\widetilde{F}_{D,k}^{m+1},\widetilde{\bm{u}}_{k}^{m+1})\leq\frac{1}{2}\|\sqrt{\alpha_{k}^{m}}\bm{u}_{k}^{m}\|_{0}^{2} (4.4)

Taking the inner product of the first equation in (3.6) with δt𝒖¯km+1\delta t\overline{\bm{u}}_{k}^{m+1},

12α~km+1𝒖¯km+102+12α~km+1(𝒖¯km+1𝒖~km+1)0212α~km+1𝒖~km+102\displaystyle\frac{1}{2}\|\sqrt{\widetilde{\alpha}_{k}^{m+1}}\overline{\bm{u}}_{k}^{m+1}\|_{0}^{2}+\frac{1}{2}\|\sqrt{\widetilde{\alpha}_{k}^{m+1}}(\overline{\bm{u}}_{k}^{m+1}-\widetilde{\bm{u}}_{k}^{m+1})\|_{0}^{2}-\frac{1}{2}\|\sqrt{\widetilde{\alpha}_{k}^{m+1}}\widetilde{\bm{u}}_{k}^{m+1}\|_{0}^{2} (4.5)
+δt(ϕ~km+1pm+1,𝒖¯km+1)δt(ϕ~km+1p~km+1,𝒖¯km+1)=0\displaystyle+\delta t(\widetilde{\phi}_{k}^{m+1}\nabla p^{m+1},\overline{\bm{u}}_{k}^{m+1})-\delta t(\widetilde{\phi}_{k}^{m+1}\nabla\widetilde{p}_{k}^{m+1},\overline{\bm{u}}_{k}^{m+1})=0

Taking the inner product of the first equation in (3.6) again with δtp~km+1ρ~km+1\displaystyle\delta t\frac{\nabla\widetilde{p}_{k}^{m+1}}{\widetilde{\rho}_{k}^{m+1}},

(ϕ~km+1p~km+1,𝒖¯km+1)(ϕ~km+1p~km+1,𝒖~km+1)\displaystyle(\widetilde{\phi}_{k}^{m+1}\nabla\widetilde{p}_{k}^{m+1},\overline{\bm{u}}_{k}^{m+1})-(\widetilde{\phi}_{k}^{m+1}\nabla\widetilde{p}_{k}^{m+1},\widetilde{\bm{u}}_{k}^{m+1}) (4.6)
12δt(ϕ~km+1ρ~km+1p~km+102+ϕ~km+1ρ~km+1(p~km+1pm+1)02ϕ~km+1ρ~km+1pm+102)=0\displaystyle-\frac{1}{2}\delta t\left(\left\|\sqrt{\frac{\widetilde{\phi}_{k}^{m+1}}{\widetilde{\rho}_{k}^{m+1}}}\nabla\widetilde{p}_{k}^{m+1}\right\|_{0}^{2}+\left\|\sqrt{\frac{\widetilde{\phi}_{k}^{m+1}}{\widetilde{\rho}_{k}^{m+1}}}\nabla(\widetilde{p}_{k}^{m+1}-p^{m+1})\right\|_{0}^{2}-\left\|\sqrt{\frac{\widetilde{\phi}_{k}^{m+1}}{\widetilde{\rho}_{k}^{m+1}}}\nabla p^{m+1}\right\|_{0}^{2}\right)=0

Combining (4.2)-(4.6), we have

12α~km+1𝒖¯km+102+12α~km+1(𝒖¯km+1𝒖~km+1)02+δt(ϕ~km+1pm+1,𝒖¯km+1)\displaystyle\frac{1}{2}\|\sqrt{\widetilde{\alpha}_{k}^{m+1}}\overline{\bm{u}}_{k}^{m+1}\|_{0}^{2}+\frac{1}{2}\|\sqrt{\widetilde{\alpha}_{k}^{m+1}}(\overline{\bm{u}}_{k}^{m+1}-\widetilde{\bm{u}}_{k}^{m+1})\|_{0}^{2}+\delta t(\widetilde{\phi}_{k}^{m+1}\nabla p^{m+1},\overline{\bm{u}}_{k}^{m+1}) (4.7)
+δt(ϕ~km+1τk(𝒖~km+1),𝒖~km+1)+δt(F~D,km+1,𝒖~km+1)+12δt2ϕ~km+1ρ~km+1pm+102\displaystyle+\delta t(\widetilde{\phi}_{k}^{m+1}\tau_{k}(\widetilde{\bm{u}}_{k}^{m+1}),\nabla\widetilde{\bm{u}}_{k}^{m+1})+\delta t(\widetilde{F}_{D,k}^{m+1},\widetilde{\bm{u}}_{k}^{m+1})+\frac{1}{2}\delta t^{2}\left\|\sqrt{\frac{\widetilde{\phi}_{k}^{m+1}}{\widetilde{\rho}_{k}^{m+1}}}\nabla p^{m+1}\right\|_{0}^{2}
12αlm𝒖km02+12δt2ϕ~kmρ~kmpkm02+12δt2ϕ~km+1ρ~km+1(p~km+1pm+1)02\displaystyle\leq\frac{1}{2}\|\sqrt{\alpha_{l}^{m}}\bm{u}_{k}^{m}\|_{0}^{2}+\frac{1}{2}\delta t^{2}\left\|\sqrt{\frac{\widetilde{\phi}_{k}^{m}}{\widetilde{\rho}_{k}^{m}}}\nabla p_{k}^{m}\right\|_{0}^{2}+\frac{1}{2}\delta t^{2}\left\|\sqrt{\frac{\widetilde{\phi}_{k}^{m+1}}{\widetilde{\rho}_{k}^{m+1}}}\nabla(\widetilde{p}_{k}^{m+1}-p^{m+1})\right\|_{0}^{2}

Using the first equation in (3.6), we have

12α~km+1𝒖¯km+102+δt(ϕ~km+1pm+1,𝒖¯km+1)\displaystyle\frac{1}{2}\|\sqrt{\widetilde{\alpha}_{k}^{m+1}}\overline{\bm{u}}_{k}^{m+1}\|_{0}^{2}+\delta t(\widetilde{\phi}_{k}^{m+1}\nabla p^{m+1},\overline{\bm{u}}_{k}^{m+1}) (4.8)
+δt(ϕ~km+1τk(𝒖~km+1),𝒖~km+1)+δt(F~D,km+1,𝒖~km+1)+12δt2ϕ~km+1ρ~km+1pm+102\displaystyle+\delta t(\widetilde{\phi}_{k}^{m+1}\tau_{k}(\widetilde{\bm{u}}_{k}^{m+1}),\nabla\widetilde{\bm{u}}_{k}^{m+1})+\delta t(\widetilde{F}_{D,k}^{m+1},\widetilde{\bm{u}}_{k}^{m+1})+\frac{1}{2}\delta t^{2}\left\|\sqrt{\frac{\widetilde{\phi}_{k}^{m+1}}{\widetilde{\rho}_{k}^{m+1}}}\nabla p^{m+1}\right\|_{0}^{2}
12αlm𝒖km02+12δt2ϕ~kmρ~kmpkm02\displaystyle\leq\frac{1}{2}\|\sqrt{\alpha_{l}^{m}}\bm{u}_{k}^{m}\|_{0}^{2}+\frac{1}{2}\delta t^{2}\left\|\sqrt{\frac{\widetilde{\phi}_{k}^{m}}{\widetilde{\rho}_{k}^{m}}}\nabla p_{k}^{m}\right\|_{0}^{2}

To this stage, the problem is to estimate δt(ϕ~km+1pm+1,𝒖¯km+1)\delta t(\widetilde{\phi}_{k}^{m+1}\nabla p^{m+1},\overline{\bm{u}}_{k}^{m+1}). Let us introduce auxiliary functions fk(z)=zek(z)f_{k}(z)=ze_{k}(z). If γk>1\gamma_{k}>1, then fkf_{k} are C2C^{2}, strictly convex functions for z>0z>0. Indeed, we have the following by definition:

fk(z)=zρk,refzζk(s)s2𝑑s,z(0,+)f_{k}(z)=z\int_{\rho_{k,ref}}^{z}\frac{\zeta_{k}(s)}{s^{2}}ds,~{}~{}~{}z\in(0,+\infty)

Differentiating ff twice with respect to zz, we get

fk′′(z)=ζk(z)z2+ζk(z)zζk(z)z2=ζk(z)zf_{k}^{\prime\prime}(z)=\frac{\zeta_{k}(z)}{z^{2}}+\frac{\zeta_{k}^{\prime}(z)z-\zeta_{k}(z)}{z^{2}}=\frac{\zeta_{k}^{\prime}(z)}{z}

Therefore, we have

ζk(z)=zfk′′(z)\zeta_{k}^{\prime}(z)=zf_{k}^{\prime\prime}(z) (4.9)

We observe that

Ω(ϕ~km+1ρkm+1u¯km+1)fk(ρkm+1)=Ωϕ~km+1ρkm+1𝒖¯km+1[ddρ(ρek(ρ)]ρ=ρkm+1\displaystyle-\int_{\Omega}\nabla\cdot(\widetilde{\phi}_{k}^{m+1}\rho_{k}^{m+1}\overline{u}_{k}^{m+1})f_{k}^{\prime}(\rho_{k}^{m+1})=\int_{\Omega}\widetilde{\phi}_{k}^{m+1}\rho_{k}^{m+1}\overline{\bm{u}}_{k}^{m+1}\cdot\nabla\left[\frac{d}{d\rho}(\rho e_{k}(\rho)\right]_{\rho=\rho_{k}^{m+1}} (4.10)
=Ωϕ~km+1𝒖¯km+1ρkm+1(dζk(ρ)dρ)ρ=ρkm+1=Ωϕ~km+1pm+1u¯m+1\displaystyle=\int_{\Omega}\widetilde{\phi}_{k}^{m+1}\overline{\bm{u}}_{k}^{m+1}\cdot\nabla\rho_{k}^{m+1}\left(\frac{d\zeta_{k}(\rho)}{d\rho}\right)_{\rho=\rho_{k}^{m+1}}=\int_{\Omega}\widetilde{\phi}_{k}^{m+1}\nabla p^{m+1}\cdot\overline{u}^{m+1}

On the other hand

(αkm+1αkm)fk(ρkm+1)=αkm+1fk(ρkm+1)αkmfk(ρkm)+αkm(fk(ρkm)fk(ρkm+1))\displaystyle(\alpha_{k}^{m+1}-\alpha_{k}^{m})f_{k}^{\prime}(\rho_{k}^{m+1})=\alpha_{k}^{m+1}f_{k}^{\prime}(\rho_{k}^{m+1})-\alpha_{k}^{m}f_{k}^{\prime}(\rho_{k}^{m})+\alpha_{k}^{m}(f_{k}^{\prime}(\rho_{k}^{m})-f_{k}^{\prime}(\rho_{k}^{m+1})) (4.11)
=αkm+1ek(ρkm+1)αkmek(ρkm)+ϕkm+1ζk(ρkm)ϕkmζk(ρkm)+αkm(fk(ρkm)fk(ρkm+1))\displaystyle=\alpha_{k}^{m+1}e_{k}(\rho_{k}^{m+1})-\alpha_{k}^{m}e_{k}(\rho_{k}^{m})+\phi_{k}^{m+1}\zeta_{k}(\rho_{k}^{m})-\phi_{k}^{m}\zeta_{k}(\rho_{k}^{m})+\alpha_{k}^{m}(f_{k}^{\prime}(\rho_{k}^{m})-f_{k}^{\prime}(\rho_{k}^{m+1}))

Now,

ϕkm+1ζk(ρkm+1)ϕkmζk(ρkm)+αkm(fk(ρkm)fk(ρkm+1))\displaystyle\phi_{k}^{m+1}\zeta_{k}(\rho_{k}^{m+1})-\phi_{k}^{m}\zeta_{k}(\rho_{k}^{m})+\alpha_{k}^{m}(f_{k}^{\prime}(\rho_{k}^{m})-f_{k}^{\prime}(\rho_{k}^{m+1})) (4.12)
=ζk(ρkm+1)(ϕkm+1ϕkm)+ϕkm(ζk(ρkm+1)ζk(ρkm)+ρkm(fk(ρkm)fk(ρkm+1))\displaystyle=\zeta_{k}(\rho_{k}^{m+1})(\phi_{k}^{m+1}-\phi_{k}^{m})+\phi_{k}^{m}(\zeta_{k}(\rho_{k}^{m+1})-\zeta_{k}(\rho_{k}^{m})+\rho_{k}^{m}(f_{k}^{\prime}(\rho_{k}^{m})-f_{k}^{\prime}(\rho_{k}^{m+1}))

Since ϕkm0\phi_{k}^{m}\geq 0 and f′′(z)>0f^{\prime\prime}(z)>0, we have

ϕkm[ζk(ρkm+1)ζk(ρkm)+ρkm(fk(ρkm)fk(ρkm+1))]=ϕkm[ρkmρkm+1(zρkm)fk′′(z)𝑑z]0\phi_{k}^{m}\left[\zeta_{k}(\rho_{k}^{m+1})-\zeta_{k}(\rho_{k}^{m})+\rho_{k}^{m}(f_{k}^{\prime}(\rho_{k}^{m})-f_{k}^{\prime}(\rho_{k}^{m+1}))\right]=\phi_{k}^{m}\left[\int_{\rho_{k}^{m}}^{\rho_{k}^{m+1}}(z-\rho_{k}^{m})f_{k}^{\prime\prime}(z)dz\right]\geq 0 (4.13)

Now, we have

δtΩϕ~km+1pm+1𝒖¯km+1Ωαkm+1ek(ρkm+1)αkmek(ρkm)+pm+1(ϕkm+1ϕkm)\delta t\int_{\Omega}\widetilde{\phi}_{k}^{m+1}\nabla p^{m+1}\cdot\overline{\bm{u}}_{k}^{m+1}\geq\int_{\Omega}\alpha_{k}^{m+1}e_{k}(\rho_{k}^{m+1})-\alpha_{k}^{m}e_{k}(\rho_{k}^{m})+p^{m+1}(\phi_{k}^{m+1}-\phi_{k}^{m}) (4.14)

Taking the summation with respect to kk, we have

kδtΩϕ~km+1pm+1𝒖¯km+1kΩ(αkm+1ek(ρkm+1)αkmek(ρkm)+pm+1(ϕkm+1ϕkm))\displaystyle\sum_{k}\delta t\int_{\Omega}\widetilde{\phi}_{k}^{m+1}\nabla p^{m+1}\cdot\overline{\bm{u}}_{k}^{m+1}\geq\sum_{k}\int_{\Omega}\left(\alpha_{k}^{m+1}e_{k}(\rho_{k}^{m+1})-\alpha_{k}^{m}e_{k}(\rho_{k}^{m})+p^{m+1}(\phi_{k}^{m+1}-\phi_{k}^{m})\right) (4.15)
=kΩαkm+1ek(ρkm+1)αkmek(ρkm)\displaystyle=\sum_{k}\int_{\Omega}\alpha_{k}^{m+1}e_{k}(\rho_{k}^{m+1})-\alpha_{k}^{m}e_{k}(\rho_{k}^{m})

Using (4.8), (4.15), and (3.8), we have

k[12αkm+1𝒖km+102+Ωαkm+1ek(ρkm+1)+δt(ϕ~km+1τk(𝒖~km+1),𝒖~km+1)+12δt2ϕ~km+1ρ~km+1pm+102]\displaystyle\sum_{k}\left[\frac{1}{2}\|\sqrt{\alpha_{k}^{m+1}}\bm{u}_{k}^{m+1}\|_{0}^{2}+\int_{\Omega}\alpha_{k}^{m+1}e_{k}(\rho_{k}^{m+1})+\delta t(\widetilde{\phi}_{k}^{m+1}\tau_{k}(\widetilde{\bm{u}}_{k}^{m+1}),\nabla\widetilde{\bm{u}}_{k}^{m+1})+\frac{1}{2}\delta t^{2}\left\|\sqrt{\frac{\widetilde{\phi}_{k}^{m+1}}{\widetilde{\rho}_{k}^{m+1}}}\nabla p^{m+1}\right\|_{0}^{2}\right] (4.16)
+δtΩCDϕ~gm+1ϕ~lm+1|𝒖~gm+1𝒖~lm+1|3\displaystyle+\delta t\int_{\Omega}C_{D}\widetilde{\phi}_{g}^{m+1}\widetilde{\phi}_{l}^{m+1}|\widetilde{\bm{u}}_{g}^{m+1}-\widetilde{\bm{u}}_{l}^{m+1}|^{3}
k(12αkm𝒖km02+Ωαkmek(ρkm)+12δt2ϕ~kmρ~kmpm02)\displaystyle\leq\sum_{k}\left(\frac{1}{2}\|\sqrt{\alpha_{k}^{m}}\bm{u}_{k}^{m}\|_{0}^{2}+\int_{\Omega}\alpha_{k}^{m}e_{k}(\rho_{k}^{m})+\frac{1}{2}\delta t^{2}\left\|\sqrt{\frac{\widetilde{\phi}_{k}^{m}}{\widetilde{\rho}_{k}^{m}}}\nabla p^{m}\right\|_{0}^{2}\right)

Summing over mm, we arrive at the theorem:

Theorem 1

For any δt>0\delta t>0, the solution (ϕm,ρm,𝐮m,pm)(\phi^{m},\rho^{m},\bm{u}^{m},p^{m}), m=1,2,m=1,2,\ldots of the semi-discrete scheme (3.1)-(3.8) satisfies the stability estimate

k[12αkm+1𝒖km+102+Ωαkm+1ek(ρkm+1)+δtj=0m(ϕ~kj+1τk(𝒖~kj+1),𝒖~kj+1)+12δt2ϕ~km+1ρ~km+1pm+102]\displaystyle\sum_{k}\left[\frac{1}{2}\|\sqrt{\alpha_{k}^{m+1}}\bm{u}_{k}^{m+1}\|_{0}^{2}+\int_{\Omega}\alpha_{k}^{m+1}e_{k}(\rho_{k}^{m+1})+\delta t\sum_{j=0}^{m}(\widetilde{\phi}_{k}^{j+1}\tau_{k}(\widetilde{\bm{u}}_{k}^{j+1}),\nabla\widetilde{\bm{u}}_{k}^{j+1})+\frac{1}{2}\delta t^{2}\left\|\sqrt{\frac{\widetilde{\phi}_{k}^{m+1}}{\widetilde{\rho}_{k}^{m+1}}}\nabla p^{m+1}\right\|_{0}^{2}\right] (4.17)
+δtj=0mΩCDϕ~gj+1ϕ~lj+1|𝒖~gj+1𝒖~lj+1|3\displaystyle+\delta t\sum_{j=0}^{m}\int_{\Omega}C_{D}\widetilde{\phi}_{g}^{j+1}\widetilde{\phi}_{l}^{j+1}|\widetilde{\bm{u}}_{g}^{j+1}-\widetilde{\bm{u}}_{l}^{j+1}|^{3}
k(12αk0𝒖k002+Ωαk0ek(ρk0)+12δt2ϕ~k0ρ~k0p002)\displaystyle\leq\sum_{k}\left(\frac{1}{2}\|\sqrt{\alpha_{k}^{0}}\bm{u}_{k}^{0}\|_{0}^{2}+\int_{\Omega}\alpha_{k}^{0}e_{k}(\rho_{k}^{0})+\frac{1}{2}\delta t^{2}\left\|\sqrt{\frac{\widetilde{\phi}_{k}^{0}}{\widetilde{\rho}_{k}^{0}}}\nabla p^{0}\right\|_{0}^{2}\right)
Remark 2

The prediction step (3.1) does not guarantee the positiveness of α~km+1\widetilde{\alpha}_{k}^{m+1}. An O(δt)O(\delta t) modification is possible to force its positiveness:

(1+12δt(𝒖km)+12δt(𝒖km)+)α~km+1+δt𝒖kmα~km+1=(1+12δt(𝒖km))αkm(1+\frac{1}{2}\delta t(\nabla\cdot\bm{u}_{k}^{m})+\frac{1}{2}\delta t(\nabla\cdot\bm{u}_{k}^{m})^{+})\widetilde{\alpha}_{k}^{m+1}+\delta t\bm{u}_{k}^{m}\cdot\nabla\widetilde{\alpha}_{k}^{m+1}=(1+\frac{1}{2}\delta t(\nabla\cdot\bm{u}_{k}^{m})^{-})\alpha_{k}^{m} (4.18)

Indeed, taking the inner product of (4.18) with (α~km+1)(\widetilde{\alpha}_{k}^{m+1})^{-}, we have

Ω(1+12δt(𝒖km)+)|(α~km+1)|2δt(Ω12(𝒖km)|(α~km+1)|2+(α~km+1)𝒖km(α~km+1))\displaystyle-\int_{\Omega}(1+\frac{1}{2}\delta t(\nabla\cdot\bm{u}_{k}^{m})^{+})|(\widetilde{\alpha}_{k}^{m+1})^{-}|^{2}-\delta t\left(\int_{\Omega}\frac{1}{2}(\nabla\cdot\bm{u}_{k}^{m})|(\widetilde{\alpha}_{k}^{m+1})^{-}|^{2}+(\widetilde{\alpha}_{k}^{m+1})^{-}\bm{u}_{k}^{m}\cdot\nabla(\widetilde{\alpha}_{k}^{m+1})^{-}\right)
=Ω(1+12δt(𝒖km))αkm(α~km+1)\displaystyle=\int_{\Omega}(1+\frac{1}{2}\delta t(\nabla\cdot\bm{u}_{k}^{m})^{-})\alpha_{k}^{m}(\widetilde{\alpha}_{k}^{m+1})^{-}

Using Lemma 1, the second term on the left hand side can be eliminated. Therefore, we have (α~km+1)=0(\widetilde{\alpha}_{k}^{m+1})^{-}=0 almost everywhere provided αkm0\alpha_{k}^{m}\geq 0 almost everywhere. This implies the positiveness of α~km+1\widetilde{\alpha}_{k}^{m+1}.

5 Numerical test

Let LL be the reference length scale, UU be the reference velocity, and ρ0\rho_{0} be the reference density. The parameters and variables in use are scaled by the following:

μgμgρ0UL,μlμlρ0UL,ρl,0ρl,0ρ0,AgAgρ0γgρ0U2,AlAlρ0γlρ0U2,p0p0ρ0U2,CDCDρ0\displaystyle\mu_{g}\rightarrow\frac{\mu_{g}}{\rho_{0}UL},~{}~{}\mu_{l}\rightarrow\frac{\mu_{l}}{\rho_{0}UL},~{}~{}\rho_{l,0}\rightarrow\frac{\rho_{l,0}}{\rho_{0}},~{}~{}A_{g}\rightarrow\frac{A_{g}\rho_{0}^{\gamma_{g}}}{\rho_{0}U^{2}},~{}~{}A_{l}\rightarrow\frac{A_{l}\rho_{0}^{\gamma_{l}}}{\rho_{0}U^{2}},~{}~{}p_{0}\rightarrow\frac{p_{0}}{\rho_{0}U^{2}},~{}~{}C_{D}\rightarrow\frac{C_{D}}{\rho_{0}}
αgαgρ0,αlαlρ0,𝒖g𝒖gU,𝒖l𝒖lU,ppρ0U2\displaystyle\alpha_{g}\rightarrow\frac{\alpha_{g}}{\rho_{0}},~{}~{}\alpha_{l}\rightarrow\frac{\alpha_{l}}{\rho_{0}},~{}~{}\bm{u}_{g}\rightarrow\frac{\bm{u}_{g}}{U},~{}~{}\bm{u}_{l}\rightarrow\frac{\bm{u}_{l}}{U},~{}~{}p\rightarrow\frac{p}{\rho_{0}U^{2}}

In the following test, we assume that λk=0\lambda_{k}=0 for simplicity.

For Sections 5.1 and 5.2, we consider a square physical domain Ω\Omega of size 1m×1m1m\times 1m. The initial values are constructed by the following procedure:

  1. 1.

    Set the initial of volume fractions and densities: ϕk(0,x)=ϕk0(x)\phi_{k}(0,x)=\phi_{k}^{0}(x), ρk(0,x)=ρk0(x)\rho_{k}(0,x)=\rho_{k}^{0}(x).

  2. 2.

    Let αk(0,x)=αk0:=ϕk0ρk0\alpha_{k}(0,x)=\alpha_{k}^{0}:=\phi_{k}^{0}\rho_{k}^{0}.

  3. 3.

    Set the initial values of intermediate step: ϕ~k0=ϕk0\widetilde{\phi}_{k}^{0}=\phi_{k}^{0}, ρ~k0=ρk0\widetilde{\rho}_{k}^{0}=\rho_{k}^{0}.

  4. 4.

    Solve a steady state Stokes equation with admissible boundary conditions for initial velocities and pressure: Find (𝒖0,p0)(\bm{u}^{0},p^{0}) such that

    (μ𝒖0)+p0=𝒇,\displaystyle-\nabla\cdot(\mu\nabla\bm{u}^{0})+\nabla p^{0}=\bm{f},{}{}{} inΩ\displaystyle\text{in}~{}\Omega (5.1)
    𝒖0=0,\displaystyle\nabla\cdot\bm{u}^{0}=0,{}{}{} inΩ\displaystyle\text{in}~{}\Omega

    where μ=ϕg0μg+ϕl0μl\mu=\phi_{g}^{0}\mu_{g}+\phi_{l}^{0}\mu_{l}. We set 𝒖g(0,x)=𝒖l(0,x)=𝒖0\bm{u}_{g}(0,x)=\bm{u}_{l}(0,x)=\bm{u}^{0}, p(0,x)=p~g0=p~l0=p0p(0,x)=\widetilde{p}_{g}^{0}=\widetilde{p}_{l}^{0}=p^{0}.

In the following context, the P1-bubble element (see for instance [29]) is applied for velocities field 𝒖k\bm{u}_{k} and P1P_{1} element is applied for all other variables. The computer program for the implementation is written using Freefem++ [30].

5.1 Two-gas system

First, the case that the two fluids possess similar densities is taken into account. The parameters in use are listed in Table 1. The initial values for volume fractions and densities are

ϕg0(x,y)=0.2+0.2exp(30((x0.25)2+(y0.25)2)),ϕl0(x,y)=1ϕg0(x,y)\displaystyle\phi_{g}^{0}(x,y)=0.2+0.2\exp(-30((x-0.25)^{2}+(y-0.25)^{2})),~{}~{}\phi_{l}^{0}(x,y)=1-\phi_{g}^{0}(x,y) (5.2)
ρg0(x,y)=2(kg/m3),ρl0(x,y)=4(kg/m3)\displaystyle\rho_{g}^{0}(x,y)=2~{}(kg/m^{3}),~{}~{}\rho_{l}^{0}(x,y)=4~{}(kg/m^{3})

The initial velocities are obtained by solving (5.1) with 𝒖0|Γ=0\bm{u}^{0}|_{\Gamma}=0 and 𝒇=0.006(y,x)T(kg/m2s2)\bm{f}=0.006(y,-x)^{T}~{}(kg/m^{2}\cdot s^{2}).

For all test, a 40×4040\times 40 uniform triangular mesh is employed and we compare the test results with the numerical solution with δt=0.001\delta t=0.001 at the final time T=1.6T=1.6. The convergence test shows a first order accuracy in time for all variables (see Figure 1).

We note that in this case, the densities of both phases change only slightly (less than 1.0×1061.0\times 10^{-6} times of their original densities). Therefore, it is safe for us to discuss only the volume fractions ϕk\phi_{k} or their products with the densities αk\alpha_{k}.

the contours in Figure 2 show that the two phases tend to separate at the beginning. That is, the spinning velocity field (see Figure 3) reduces the volume fraction of the lighter phase gg in its region of lower volume fraction, and vice versa. Indeed, the difference between the velocities of the two fluids is observed: According to 3, the lower density material gg is accelerated. A possible explanation is the virtual force caused by the density difference. On the other hand, the material of higher density ll is decelerated in view of 4. This result is natural because of the dissipation by viscous force and the momentum transfer by the drag force from phase ll to phase gg.

Parameters Values Parameters Values
ρg0\rho_{g0} 2.0(kg/m3)2.0~{}(kg/m^{3}) ρl0\rho_{l0} 4.0(kg/m3)4.0~{}~{}(kg/m^{3})
μg\mu_{g} 3.0×104(kg/ms)3.0\times 10^{-4}~{}(kg/m\cdot s) μl\mu_{l} 1.86×104(kg/ms)1.86\times 10^{-4}~{}(kg/m\cdot s)
AgA_{g} 3.8395×104(m3.2/kg0.4s2)3.8395\times 10^{4}~{}(m^{3.2}/kg^{0.4}\cdot s^{2}) AlA_{l} 1.01325×105(m4.1/kg0.7s2)1.01325\times 10^{5}~{}(m^{4.1}/kg^{0.7}\cdot s^{2})
p0p_{0} 1.01325×105(kg/ms2)1.01325\times 10^{5}~{}(kg/m\cdot s^{2}) CdC_{d} 100.0(kg/m3)100.0~{}(kg/m^{3})
Table 1: Parameters in Section 5.1
Refer to caption
Figure 1: Convergence plot for the test of two-gas system in Section 5.1.
Refer to caption
(a) αg\alpha_{g} at t=0.4t=0.4
Refer to caption
(b) αg\alpha_{g} at t=0.8t=0.8
Refer to caption
(c) αg\alpha_{g} at t=1.2t=1.2
Refer to caption
(d) αg\alpha_{g} at t=1.6t=1.6
Figure 2: Results of αg\alpha_{g} by the test of two-gas system in Section 5.1.
Refer to caption
(a) 𝒖g\bm{u}_{g} at t=0.4t=0.4
Refer to caption
(b) 𝒖g\bm{u}_{g} at t=0.8t=0.8
Refer to caption
(c) 𝒖g\bm{u}_{g} at t=1.2t=1.2
Refer to caption
(d) 𝒖g\bm{u}_{g} at t=1.6t=1.6
Figure 3: Results of 𝒖g\bm{u}_{g} by the test of two-gas system in Section 5.1.
Refer to caption
(a) 𝒖l\bm{u}_{l} at t=0.4t=0.4
Refer to caption
(b) 𝒖l\bm{u}_{l} at t=0.8t=0.8
Refer to caption
(c) 𝒖l\bm{u}_{l} at t=1.2t=1.2
Refer to caption
(d) 𝒖l\bm{u}_{l} at t=1.6t=1.6
Figure 4: Results of 𝒖l\bm{u}_{l} by the test of two-gas system in Section 5.1.

5.2 Liquid-gas system

With the same domain as in Section 5.1, the parameters in use are listed in Table 2. We assume the same initial volume fractions as the test in Section 5.1 but different initial densities:

ϕg0(x,y)=0.2+0.2exp(30((x0.25)2+(y0.25)2)),ϕl0(x,y)=1ϕg0(x,y)\displaystyle\phi_{g}^{0}(x,y)=0.2+0.2\exp(-30((x-0.25)^{2}+(y-0.25)^{2})),~{}~{}\phi_{l}^{0}(x,y)=1-\phi_{g}^{0}(x,y) (5.3)
ρg0(x,y)=2(kg/m3),ρl0(x,y)=1000(kg/m3)\displaystyle\rho_{g}^{0}(x,y)=2~{}(kg/m^{3}),~{}~{}\rho_{l}^{0}(x,y)=1000~{}(kg/m^{3})

The initial velocities are obtained by solving (5.1) with 𝒖0|Γ=0\bm{u}^{0}|_{\Gamma}=0 and 𝒇=0.01(y,x)T(kg/m2s2)\bm{f}=0.01(y,-x)^{T}~{}(kg/m^{2}\cdot s^{2}).

For all test, a 40×4040\times 40 uniform triangular mesh is employed and we compare the test results with the numerical solution with δt=0.0001\delta t=0.0001 at the final time T=1.6T=1.6. The convergence in this case is worse than the test of two-gas system. Fortunately, a first order convergence is still kept (see Figure 5).

Similar to the test of two-gas system, the densities of both phases change only slightly (less than 1.0×1061.0\times 10^{-6} times of their original densities). According to Figure 6 a separation phenomenon can be observed in this case as well due to the difference between velocities of the two fluids. The acceleration for lower density phase and the deceleration for higher density phase can be seen from Figures 7 and 8. With finite element implementation for the case of large density difference, undesired oscillation tends to spread rapidly if the time step is not chosen sufficiently small. A smaller time step than that of the test for two-gas system is therefore adopted here.

Parameters Values Parameters Values
ρg0\rho_{g0} 2.0(kg/m3)2.0~{}(kg/m^{3}) ρl0\rho_{l0} 1000.0(kg/m3)1000.0~{}~{}(kg/m^{3})
μg\mu_{g} 1.86×104(kg/ms)1.86\times 10^{-4}~{}(kg/m\cdot s) μl\mu_{l} 2.3×103(kg/ms)2.3\times 10^{-3}~{}(kg/m\cdot s)
AgA_{g} 3.8395×104(m3.2/kg0.4s2)3.8395\times 10^{4}~{}(m^{3.2}/kg^{0.4}\cdot s^{2}) AlA_{l} 1.0×106(m5/kgs2)1.0\times 10^{6}~{}(m^{5}/kg\cdot s^{2})
p0p_{0} 1.01325×105(kg/ms2)1.01325\times 10^{5}~{}(kg/m\cdot s^{2}) CdC_{d} 100.0(kg/m3)100.0~{}(kg/m^{3})
Table 2: Parameters in Section 5.2
Refer to caption
Figure 5: Convergence plot for the test of liquid-gas system in Section 5.2.
Refer to caption
(a) αg\alpha_{g} at t=0.4t=0.4
Refer to caption
(b) αg\alpha_{g} at t=0.8t=0.8
Refer to caption
(c) αg\alpha_{g} at t=1.2t=1.2
Refer to caption
(d) αg\alpha_{g} at t=1.6t=1.6
Figure 6: Results of αg\alpha_{g} by the test of liquid-gas system in Section 5.2.
Refer to caption
(a) 𝒖g\bm{u}_{g} at t=0.4t=0.4
Refer to caption
(b) 𝒖g\bm{u}_{g} at t=0.8t=0.8
Refer to caption
(c) 𝒖g\bm{u}_{g} at t=1.2t=1.2
Refer to caption
(d) 𝒖g\bm{u}_{g} at t=1.6t=1.6
Figure 7: Results of 𝒍g\bm{l}_{g} by the test of liquid-gas system in Section 5.2.
Refer to caption
(a) 𝒖l\bm{u}_{l} at t=0.4t=0.4
Refer to caption
(b) 𝒖l\bm{u}_{l} at t=0.8t=0.8
Refer to caption
(c) 𝒖l\bm{u}_{l} at t=1.2t=1.2
Refer to caption
(d) 𝒖l\bm{u}_{l} at t=1.6t=1.6
Figure 8: Results of 𝒖l\bm{u}_{l} by the test of liquid-gas system in Section 5.2.

5.3 Liquid-gas system with initially large pressure gradient

In this case, we consider the same physical domain as what is used in Sections 5.1 and 5.2 and the parameters presented in Table 2. The initial values of the volume fractions are

ϕg0(x,y)=0.2+0.2exp(30((x0.5)2+(y0.5)2)),ϕl0(x,y)=1ϕg0(x,y)\phi_{g}^{0}(x,y)=0.2+0.2\exp(-30((x-0.5)^{2}+(y-0.5)^{2})),~{}~{}\phi_{l}^{0}(x,y)=1-\phi_{g}^{0}(x,y) (5.4)

The initial pressure distribution p0p^{0} is given by

p0(x,y)=p0+p0exp(30((x0.5)2+(y0.5)2))p^{0}(x,y)=p_{0}+p_{0}\exp(-30((x-0.5)^{2}+(y-0.5)^{2})) (5.5)

The initial densities are determined accordingly:

ρk0(x,y)=ζk1(p0(x,y))\rho^{0}_{k}(x,y)=\zeta_{k}^{-1}(p^{0}(x,y)) (5.6)

The initial velocities are given by

𝒖k0(x,y)=0\bm{u}_{k}^{0}(x,y)=0 (5.7)

and the homogeneous boundary conditions for velocities are imposed:

𝒖k=0onΓ.\bm{u}_{k}=0~{}~{}~{}\text{on}~{}\Gamma. (5.8)

For all test, a 40×4040\times 40 uniform triangular mesh is employed and we compare the test results with the numerical solution with δt=1.0×106\delta t=1.0\times 10^{-6} at the final time T=5.12×103T=5.12\times 10^{-3}. A much smaller time step is chosen in order to capture the rapid change of the density field of phase gg (see Figure 12). The convergence results given by Figure 9 present a first order accuracy in this case.

The magnitude contours of αg\alpha_{g} and ϕg\phi_{g} given by Figures 10 and 11 do not show a large change with time. However, their quotient ρg=αg/ϕg\rho_{g}=\alpha_{g}/\phi_{g} change rapidly so that a very small time step is needed to capture its wave-motion. This example implies that the wave velocity of the αk\alpha_{k} propagation may be of different scale with the wave velocity of the density change.

In the scenario of this test, the main driven force for the fluid velocities is the pressure. Since the pressure force for each phase is proportional to its volume fraction, the acceleration by the pressure force is reciprocal to the density. Indeed, Figures 13 and 14 present a large different between velocities of the two fluids.

Refer to caption
Figure 9: Convergence plot for the test of the liquid-gas system with initially large pressure gradient in Section 5.3.
Refer to caption
(a) αg\alpha_{g} at t=1.25×103t=1.25\times 10^{-3}
Refer to caption
(b) αg\alpha_{g} at t=2.5×103t=2.5\times 10^{-3}
Refer to caption
(c) αg\alpha_{g} at t=3.75×103t=3.75\times 10^{-3}
Refer to caption
(d) αg\alpha_{g} at t=5.0×103t=5.0\times 10^{-3}
Figure 10: Results of αg\alpha_{g} by the test of liquid-gas system with initially large pressure gradient in Section 5.3.
Refer to caption
(a) ϕg\phi_{g} at t=1.25×103t=1.25\times 10^{-3}
Refer to caption
(b) ϕg\phi_{g} at t=2.5×103t=2.5\times 10^{-3}
Refer to caption
(c) ϕg\phi_{g} at t=3.75×103t=3.75\times 10^{-3}
Refer to caption
(d) ϕg\phi_{g} at t=5.0×103t=5.0\times 10^{-3}
Figure 11: Results of ϕg\phi_{g} by the test of liquid-gas system with initially large pressure gradient in Section 5.3.
Refer to caption
(a) ρg\rho_{g} at t=1.25×103t=1.25\times 10^{-3}
Refer to caption
(b) ρg\rho_{g} at t=2.5×103t=2.5\times 10^{-3}
Refer to caption
(c) ρg\rho_{g} at t=3.75×103t=3.75\times 10^{-3}
Refer to caption
(d) ρg\rho_{g} at t=5.0×103t=5.0\times 10^{-3}
Figure 12: Results of ρg\rho_{g} by the test of liquid-gas system with initially large pressure gradient in Section 5.3.
Refer to caption
(a) 𝒖g\bm{u}_{g} at t=1.25×103t=1.25\times 10^{-3}
Refer to caption
(b) 𝒖g\bm{u}_{g} at t=2.5×103t=2.5\times 10^{-3}
Refer to caption
(c) 𝒖g\bm{u}_{g} at t=3.75×103t=3.75\times 10^{-3}
Refer to caption
(d) 𝒖g\bm{u}_{g} at t=5.0×103t=5.0\times 10^{-3}
Figure 13: Results of 𝒖g\bm{u}_{g} by the test of liquid-gas system with initially large pressure gradient in Section 5.3.
Refer to caption
(a) 𝒖l\bm{u}_{l} at t=1.25×103t=1.25\times 10^{-3}
Refer to caption
(b) 𝒖l\bm{u}_{l} at t=2.5×103t=2.5\times 10^{-3}
Refer to caption
(c) 𝒖l\bm{u}_{l} at t=3.75×103t=3.75\times 10^{-3}
Refer to caption
(d) 𝒖l\bm{u}_{l} at t=5.0×103t=5.0\times 10^{-3}
Figure 14: Results of 𝒖l\bm{u}_{l} by the test of liquid-gas system with initially large pressure gradient in Section 5.3.

6 Conclusion

In this work, a new projection method for solving a viscous two-fluid model is proposed. A symmetric formulation of the projection step enables the computation of the projected pressure. Moreover a suitable assignment of the intermediate densities and volume fractions maintain the stability of the numerical scheme, which is justify by the stability analysis for the time-discrete problem. Numerical tests show that the method can not only be recovered to treat the problem with fluids being slightly compressed but also be used for the case with heavily compressed fluids. A first order temporal accuracy of the scheme is justified by all numerical tests in Section 5.

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