Nonlinear Analysis: Real World Applications 13 (2012) 987–998 Contents lists available at SciVerse ScienceDirect Nonlinear Analysis: Real World Applications journal homepage: www.elsevier.com/locate/nonrwa Exact solutions of a two-fluid model of two-phase compressible flows with gravity Mai Duc Thanh ∗ Department of Mathematics, International University, Quarter 6, Linh Trung Ward, Thu Duc District, Ho Chi Minh City, Viet Nam article info Article history: Received September 2010 Accepted 16 September 2011 Keywords: Two-fluid Two-phase flow Conservation law Source term Discontinuity Contact wave Jump relation abstract We aim at determining and computing a class of exact solutions of a two-fluid model of two-phase flows with/without gravity The model is described by a non-hyperbolic system of balance laws whose characteristic fields may not be given explicitly, making it perhaps impossible to solve the Riemann problem First, we investigate Riemann invariants in the linearly degenerate characteristic fields and obtain a surprising result on the corresponding contact waves of the model without gravity Second, even when gravity is allowed, we show that smooth stationary solutions can be governed by a system of differential equations in divergence form, which determines jump relations for any stationary discontinuity wave Using these relations, we establish a nonlinear equation for the pressure and propose a method to compute the pressure and then the equilibria resulted by a stationary wave © 2011 Elsevier Ltd All rights reserved Introduction In general it is always interesting to build exact solutions whenever they are available Furthermore, exact solutions of a hyperbolic problem, if they can be computed, are very useful for justifying a numerical scheme This in particular serves for the case of two-fluid models, since the Euler part of the flow equations may yield complex-conjugate eigenvalues and consequently make the initial value problem ill-posed On the other hand, any given unsteady discontinuity of balance laws propagating with constant speed σ can be made stationary thanks to a classical Galilean change of variable (x, t ) → (x − σ t , t ) We therefore develop in this paper a way for computing stationary waves of a one-pressure two-fluid model of two-phase flows with/without gravity The two phases are labeled as ‘‘gas phase’’ and ‘‘liquid phase’’ The model consisting of governing equations is given by (see [1,2]): ∂t (αg ρg ) + ∂x (αg ρg ug ) = 0, ∂t (αg ρg ug ) + ∂x (αg (ρg u2g + p)) = p∂x αg + αg ρg g , ∂t (αg ρg Eg ) + ∂x (αg (ρg Eg + p)ug ) = −p∂t αg + αg ρg ug g , ∂t (αl ρl ) + ∂x (αl ρl ul ) = 0, ∂t (αl ρl ul ) + ∂x (αl (ρl u2l + p)) = p∂x αl + αl ρl g , ∂t (αl ρl El ) + ∂x (αl (ρl El + p)ul ) = −p∂t αl + αl ρl ul g , where αi is the volume fraction, ρi is the density, ui is the velocity, ei is the internal energy and Ei = ei + ∗ u2i , Tel.: +84 2211 6965; fax: +84 3724 4271 E-mail addresses: mdthanh@hcmiu.edu.vn, hatothanh@yahoo.com 1468-1218/$ – see front matter © 2011 Elsevier Ltd All rights reserved doi:10.1016/j.nonrwa.2011.09.009 (1.1) 988 M.D Thanh / Nonlinear Analysis: Real World Applications 13 (2012) 987–998 is the total energy, g is the gravity constant in the model with gravity, and g = in the model without gravity, and the subscript ‘‘i’’ can be ‘‘g’’ or ‘‘l’’, representing the gas or liquid phase of fluids respectively The volume fractions of the fluid satisfy αg + αl = The non-conservative terms on the right-hand side (1.1) are the interphase interaction terms, indicating the momentum and energy exchanged between phases The system of equations is closed by supplementing the equations of state of gas and liquid phases In this paper, we assume that the gas phase has the equation of state of the perfect gas for the air p = (γg − 1)ρg eg , (1.2) and we use the stiffened gas equation of state for the liquid phase p= el = γl − ρl Clp Tl − p∞ , γl Cpl γl Tl + p∞ ρl (1.3) For computing purposes, we will take γg = 1.4, γl = 1.9276, Cgp = 1.012 kJ kg K , Clp = 8.07673 kJ kg K (1.4) , p∞ = 11.5968.103 atm To understand more about the system (1.1), we consider its Euler part, i.e., the system without gravity Formally, this can be done by setting g = in (1.1) The system without gravity also represents a one-pressure model of two-phase flows, and has been studied by many authors, see for example [3–5] Investigating the system shows that: first, the system can yield complex eigenvalues; and second, the system admits two linearly degenerate characteristic fields associated with the eigenvalues λ1 = ug , λ2 = ul We go on further by studying the Riemann invariants for these two linearly degenerate fields This gives us a surprising result that across the contact waves associated with the linearly degenerate field (λ1 = ug , r1 ), there are five Riemann invariants αg , p, ug , ul , Sl and that across the contact waves associated with the linearly degenerate field (λ2 = ul , r2 ), there are five Riemann invariants αg , p, ug , ul , Sg Consequently, only the entropy Sg changes across the 1-contact waves and only the entropy Sl changes across the 2-contact waves Compressible multi-fluid flow models such as (1.1) have been widely used to model multi-phase flows One of the major concerns for this kind of model is that the system is not hyperbolic, and the analytical form of the characteristic fields are not available This will be clear in Section 3, where we compute the eigenvalues, see also [6] Hence, the Riemann problem for (1.1) cannot be solved in the standard way and therefore non-trivial exact solutions of (1.1) in general are not available Numerically, it is hard to use the conventional Roe or Godunov-type of schemes to calculate the numerical fluxes Thus, there are challenging problems for this kind of model in both theoretical study and numerical treatments Finding exact solutions, even stationary, therefore is important, since they would be the unique kind of exact solutions that can be obtained These solutions can be used as the references for testing advanced new numerical schemes for (1.1) In this paper we aim at obtaining a key nonlinear equation in the pressure that can serve as a decisive step for solving the jump relations for stationary solutions First, we derive a system of differential equations in divergence form for the stationary smooth solutions of (1.1) This gives us the jump relations for stationary jumps between equilibrium states of the flow By rewriting the equations of states both in terms of the pressure, we establish a nonlinear equation of the pressure We then provide discussions solving this equation and the equilibria resulted by a stationary wave That the stationary jump can be obtained as the limit of smooth solutions is quite related to viscous traveling waves, see for example [7–10], and the references therein We observe that the general theory for nonconservative systems of balance laws was introduced by [11] The Riemann problem for simpler systems of balance laws was considered in [12–14], etc Properties of several two-fluid models of twophase flows were established in [15,6,16] Well-balanced schemes that are capable of capturing stationary waves were presented in [17,18] for the model of fluid flows in a nozzle with variable cross-section, in [19,20] for shallow water equations Well-balanced schemes for multi-phase flows were studied in [21–26], etc The paper is organized as follows In Section we present basic features and properties of the model (1.1) including alternative forms of the model (1.1), hyperbolicity and nonhyperbolicity, and Riemann invariants Section provides us with the main results of this paper Here, we first establish a system of ordinary differential equations in divergence form for stationary smooth solutions of (1.1) Arguing that the stationary waves of (1.1) are merely the limit of smooth solutions of (1.1), we derive jump relations for stationary waves We then transform the equations of state in terms of the pressure and finding a nonlinear equation for the pressure Finally, we discuss how to solve this equation and the equilibria resulted by a stationary wave M.D Thanh / Nonlinear Analysis: Real World Applications 13 (2012) 987–998 989 Preliminaries 2.1 Equivalent forms of the two-fluid model As seen below, several equivalent forms of the two-fluid model (1.1) for smooth solutions are also available We note that similar work for the case without gravity was done by Stewart and Wendroff [4] Each of the equivalent forms differ from each other by the equation of balance of energy There are three equivalent models for smooth solutions where the equation for the balance of energy in each phase is written in terms of the total energy as in (1.1), in terms of the internal energy, or in terms of the specific entropy Model involving equation for internal energy The first alternative form of the two-fluid model (1.1) has the equations of balance of energy expressed in terms of internal energy: ∂t (αg ρg ) + ∂x (αg ρg ug ) = 0, ∂t (αg ρg ug ) + ∂x (αg (ρg u2g + p)) = p∂x αg + αg ρg g , ∂t (αg ρg eg ) + ∂x (αg ρg eg ug ) = −p(∂t αg + ∂x (αg ug )), ∂t (αl ρl ) + ∂x (αl ρl ul ) = 0, (2.1) ∂t (αl ρl ul ) + ∂x (αl (ρl u2l + p)) = p∂x αl + αl ρl g , ∂t (αl ρl el ) + ∂x (αl ρl el ul ) = −p(∂t αl + ∂x (αl ul )) Since the governing equations in the two phases are similar, we need only to treat one phase only, say, the gas phase And thus for simplicity, we remove the index Subtract side-by-side the equations of balance of energy in (2.1) from the ones in (1.1) we get ∂t (αρ u2 /2) + ∂x (αρ uu2 /2 + α pu) = αρ ug + p∂x (α u), or ∂t (αρ u2 /2) + ∂x (αρ uu2 /2) + p∂x (α u) + α u∂ − xp = αρ ug + p∂x (α u) (2.2) We will show that (2.2) is always true Indeed, applying the chain rule and canceling the terms, we obtain from the last equation αρ u∂t u/2 + u/2∂t (αρ u) + (u/2)∂x (αρ u2 /2) + αρ u2 ∂x (u/2) + α u∂x p = αρ ug , or (u/2)(αρ∂t u + ∂t (αρ u) + ∂x (αρ u2 ) + αρ u∂x u + 2α∂x p − 2αρ g ) = If u = 0, then we are done Otherwise, canceling u/2 ̸= from the last equation, after re-arranging terms, we obtain (∂t (αρ u) + ∂x (α(ρ u2 + p)) − αρ g − p∂x α) + α∂x p + αρ∂t u + αρ u∂x u − αρ g = (2.3) Using the equation of balance of momentum in (1.1), we get from (2.3) α∂x p + αρ∂t u + αρ u∂x u − αρ g = (2.4) Multiplying the equation of balance of mass in (1.1) by u and adding up the resulting equation with (2.4), we obtain ∂t (αρ u) + ∂x (α(ρ u2 + p)) − p∂x α − αρ g = which is the equation of balance of momentum in (1.1) This shows that the system (1.1) and the system (2.1) are equivalent for smooth solutions Model involving equation for the specific entropy The second alternative form of the two-fluid model (1.1) has the equations of balance of energy expressed in terms of the specific entropy: ∂t (αg ρg ) + ∂x (αg ρg ug ) = 0, ∂t (αg ρg ug ) + ∂x (αg (ρg u2g + p)) = p∂x αg + αg ρg g , ∂t (αg ρg Sg ) + ∂x (αg ρg ug Sg ) = 0, ∂t (αl ρl ) + ∂x (αl ρl ul ) = 0, ∂t (αl ρl ul ) + ∂x (αl (ρl u2l + p)) = p∂x αl + αl ρl g , ∂t (αl ρl Sl ) + ∂x (αl ρl ul Sl ) = (2.5) 990 M.D Thanh / Nonlinear Analysis: Real World Applications 13 (2012) 987–998 As in the first form (2.1), we need only to treat one phase only, and we remove the index for simplicity We will derive the equation of balance of energy in (2.5) from the one in (2.1) as follows By the chain rule, the equation of balance of energy in (2.5) can be expressed as ∂t (αρ)e + (αρ)∂t e + e∂x (αρ u) + (αρ u)∂x e + p(∂t α + ∂x (α u)) = The second and the third terms of the last equation cancel each other by the equation of conservation of mass in (1.1) Thus, we obtain the result as αρ(∂t e + u∂x e) + p(∂t α + ∂x (α u)) = (2.6) Using the thermodynamical identity e = TdS − pdv, v= ρ , we obtain from (2.6) that αρ(T ∂t S − p∂t v + Tu∂x S − up∂x v) + p(∂t α + ∂x (α u)) = Substituting v = 1/ρ into the last equation, after re-arranging terms, we get T αρ(∂t S + u∂x S ) + p ∂t α + α αu ∂t ρ + ∂x (α u) + ∂x ρ ρ ρ = 0, or T αρ(∂t S + u∂x S ) + p ρ (∂t (αρ) + ∂x (αρ u)) = Using the equation of conservation of mass, and canceling T ̸= from the last equation, we get ∂t S + u∂x S = (2.7) Multiplying the last equation by αρ , multiplying the equation of conservation of mass in (1.1), and then summing up these two resulting equations, we obtain the equation of balance of energy in (2.5): ∂t (αρ S ) + ∂x (αρ uS ) = This indicates that the system (1.1) and the system (2.5) are equivalent for smooth solutions 2.2 Non-hyperbolicity In this subsection we will investigate the hyperbolicity and nonhyperbolicity of the model (1.1) As above, we show that for smooth solutions, the systems (1.1) and (2.5) are equivalent Therefore, we can consider the model in the form (2.5) and when investigating the hyperbolicity We will re-write the system (1.1) without the gravity in the matrix form ∂t V + A(V )∂x V = 0, V = (αg , p, ug , Sg , ul , Sl )T (2.8) A straightforward calculation shows that the matrix A(U ) in (2.8) is given by a a2 b2 a3 b3 0 a4 b4 0 0 0 0 b1 0 A(V ) = 0 0 ρg ug ug 0 0 ul 0 0 0 ul ρl , (2.9) where ρg ug αl ∂p ρl + ρl ul αg ∂p ρg , αl ∂p (ρl )ρg + αg ∂p (ρg )ρl ρg αl ∂p (ρl )αg a3 = , αl ∂p (ρl )ρg + αg ∂p (ρg )ρl ρl ρg (ug − ul ) b1 = , αl ∂p (ρl )ρg + αg ∂p (ρg )ρl ρl ρg αg , b3 = αl ∂p (ρl )ρg + αg ∂p (ρg )ρl a1 = αg ∂p (ρg )αl ∂p (ρl )(ug − ul ) , αl ∂p (ρl )ρg + αg ∂p (ρg )ρl ρl αg ∂p (ρg )αl a4 = , αl ∂p (ρl )ρg + αg ∂p (ρg )ρl ρl αg ug ∂p ρg + ρg αl ul ∂p ρl b2 = , αl ∂p (ρl )ρg + αg ∂p (ρg )ρl ρl ρg αl b4 = αl ∂p (ρl )ρg + αg ∂p (ρg )ρl a2 = (2.10) M.D Thanh / Nonlinear Analysis: Real World Applications 13 (2012) 987–998 991 Fig The polynomial Q (λ) defined by (2.12) has four roots The characteristic polynomial det(A(V ) − λI ) of the matrix A(U ) in (2.9) is found to be P (λ) = (ug − λ)(ul − λ)Q (λ), (2.11) where Q (λ) = (a1 − λ)(b2 − λ)(ug − λ)(ul − λ) − a2 b1 (ug − λ)(ul − λ) − + a3 b1 ρg (ul − λ) + a4 b1 ρl (ug − λ) b3 ρg (ul − λ) + b4 ρl (ug − λ) (a1 − λ) (2.12) The polynomial P (λ) always admits the two obvious roots λ1 = ug and λ2 = ul These two roots of course can coincide Furthermore, the polynomial Q (λ) may or may not have four roots This demonstrates that the matrix A(U ) may or may not admit a complete set of six eigenvalues We illustrate these situations numerically by plotting the graph of the polynomial Q for different values of the parameters See Fig 1, where the polynomial Q (λ) has four roots, and see Fig 2, where Q (λ) has only two roots Consequently, there are regions where the system is hyperbolic and there are regions where the system is not hyperbolic 2.3 The linearly degenerate characteristic fields and Riemann invariants Now, consider the two eigenvalues λ1 = ug , λ2 = ul of the matrix A(U ) in (2.9) It is not difficult to verify that the associated eigenvectors can be chosen as r1 (V ) = (0, 0, 0, 1, 0, 0)T , r2 (V ) = (0, 0, 0, 0, 0, 1)T , respectively It is easy to see that Dλi (V ) · ri (V ) ≡ 0, i = 1, 2, so that the first and the second characteristic fields are linearly degenerate The Riemann invariants associated with the first characteristic fields are αg , p, ug , ul , Sl , the Riemann invariants associated with the second characteristic fields are αg , p, ug , ul , Sg Accordingly, the volume fractions, the pressure, the velocities in both phases, the entropy in the liquid phase are constant across the 1-contact waves Similarly, the volume fractions, the pressure, the velocities in both phases and the entropy in the gas phase are constant across the 2-contact waves This in particular shows that the system (1.1) can be reduced to the divergence form for the 1- and 2-contact waves Consequently, the contacts waves verify the system (1.1) in the usual sense of weak solutions (in the sense of distributions) 992 M.D Thanh / Nonlinear Analysis: Real World Applications 13 (2012) 987–998 Fig The polynomial Q (λ) defined by (2.12) has only two roots Equilibrium states and stationary solutions 3.1 Stationary smooth waves As discussed earlier in Section 2, smooth solutions of (1.1) are the ones of (2.5) and vice-versa To study stationary solutions, it is convenient to deal with (2.5) Let us choose the two thermodynamical independent variables to be the density and the entropy (ρi , Si ), leaving the remaining three thermodynamical variables pressure, internal energy, and absolute temperature p, ei , Ti to be dependent variables, i = g , l Therefore, the unknown function is V (x, t ) = (ρg , ug , Sg , ρl , ul , Sl )(x, t ), x ∈ R, t > Smooth stationary waves V = V (x, t ), x ∈ R, t > 0, of (2.5) are the ones that not depend on time More precisely, they are solutions of the differential equations d dx d dx d dx (αi ρi ui ) = 0, (αi (ρi u2i + p)) = p (αi ρi Si ui ) = 0, d dx αi + αi ρi g , (3.1) i = l, g The last system can be written as d dx ui Si (αi ρi ui ) = 0, d dx d dx (αi ρi ui ) + αi ρi ui d ui + αi dx d (αi ρi ui ) + (αi ρi ui ) dx dx ui d dx + Si = 0, ρi dx − g = 0, i = l, g dx Si = 0, or d (αi ρi ui ) = 0, dx 2 d ui dp d p+p d dx αi = p i = l, g , d dx αi + αi ρi g , M.D Thanh / Nonlinear Analysis: Real World Applications 13 (2012) 987–998 993 Thus, any solution of the following system is also a solution of (3.1) and therefore a stationary wave of the system (2.5): d (αi ρi ui ) = 0, dx 2 ui dp d dx + ρi dx − g = 0, (3.2) d (Si ) = 0, i = l, g dx In the sequel we will seek for stationary waves of (2.5) given by (3.2), which says that the entropies Sl , Sg are constants Let us introduce the specific enthalpy hi = e i + p ρi , i = g , l, (3.3) which satisfies a thermodynamical identity dhi = Ti dSi + ρi dp, i = g , l This implies dhi (ρi , Si ) dρi = dp(ρi , Si ) ρi dρi , i = g , l From the last equality, we can rewrite system (3.2) as d dx (αg ρg ug ) = 0, u2g d dx + hg − gx = 0, (3.4) d (αl ρl ul ) = 0, dx d ul + hl − gx = dx 3.2 The jump relations and equilibria We look for stationary discontinuity waves of (2.5) to be the limit of smooth stationary solutions of (3.4) The discontinuity happens along the ray t ≥ and the solution is continuous elsewhere Values of the solution on both sides of the discontinuity constitute equilibrium states which can be determined as follows Since the system (3.4) is under a perfect divergence form, it is easy to see that stationary waves satisfy the jump relations [αg ρg ug ] = 0, u2g + hg = 0, [αl ρl ul ] = 0, [ ] ul + hl = , − − − where [αi ρi ui ] := αi+ ρi+ u+ i −αi ρi ui , and so on, denotes the difference of the corresponding value αi ρi ui , i = g , l between the right-hand and left-hand states of the stationary discontinuity Set α ρ g g αg ρg ug α g ρg Sg U = αl ρl , αl ρl ul αl ρl Sl α ρ g0 g0 αg0 ρg0 ug0 αg0 ρg0 Sg0 U0 = αl0 ρl0 , αl0 ρl0 ul0 αl0 ρl0 Sl0 α g p ug V = Sg , ul Sl α g0 p0 ug0 V0 = Sg0 (3.5) ul0 Sl0 To find the equilibrium states on both sides of a stationary wave, we fix a state on one side, say the left-hand state, either in the form U0 or in the form V0 , and we look for the right-hand state either in the form U or in the form V , that can be 994 M.D Thanh / Nonlinear Analysis: Real World Applications 13 (2012) 987–998 connected to U0 by a stationary wave In other words, we solve the following system of nonlinear equations αg ρg ug = αg0 ρg0 ug0 , u2g + hg = u2g0 + hg0 , (3.6) αl ρl ul = αl0 ρl0 ul0 , u2l + hl = u2l0 + hl0 We are finding the way to solve the system of algebraic equations 3.3 Equations of state A fluid is characterized by its equations of state In the following, we will present alternative forms of the equations of state with different choices of the thermodynamical independent variables First, assume that the fluid in the gas phase has the equation of state of a polytropic ideal gas as p = ρg Rg Tg = (γg − 1)ρg eg , eg = Cg v T = Rg Tg γg − , (3.7) hg = Cpg Tg where Rg is the specific gas constant; Rg = ηg R, where ηg is the mole-mass fraction, and R is the universal gas constant R γ R g Let Cg v = γ − be the specific heat at constant volume and Cgp = γ g−g1 be the specific heat at constant pressure Thus, g g Cgp = γg Cg v It is a basic fact to verify that the pressure and the specific enthalpy in the gas phase can be written as a function of the density and the specific entropy: γg p = pg (ρg , Sg ) = (γg − 1)ρg exp γ g −1 hg = hg (ρg , Sg ) = G(S )ρg , Sg − Sg ∗ Cg v , where G(S ) = γg exp Sg − Sg ∗ (3.8) Cg v for some constant Sg ∗ It follows from (3.8) that the density and the specific enthalpy can be written as a function of the pressure and the specific entropy as ρg = ρg (p, Sg ) = p 1/γg Sg ∗ − Sg , γg − Cgp γg Sg − Sg ∗ (γg −1)/γg hg = hg (p, Sg ) = p exp Cgp (γg − 1)(γg −1)/γg exp (3.9) Second, we consider the liquid phase where the equations of state are given by p = (γl − 1)ρl Clv Tl − p∞ , e l = C lv T l + p∞ ρl , (3.10) Rγ R where Clv = γ −l is the specific heat at constant volume and Clp = γ l−l1 is the specific heat at constant pressure, so that l l Clp = γl Clv , where Rl is the specific gas constant From (3.10), using the thermodynamical identity del = Tl dSl − pdvl , vl = 1/ρl , we obtain Sl − Sl∗ Clv γ −1 = log(Tl vl l ), for some Sl∗ This yields p = pl (ρl , Sl ) = (γl − 1)Clv exp γ −1 hl = hl (ρl , Sl ) = L(S )ρl l , Sl − Sl∗ γ ρl l − p∞ , Sl − Sl∗ where L(S ) = Clp exp C lv C lv (3.11) M.D Thanh / Nonlinear Analysis: Real World Applications 13 (2012) 987–998 995 From (3.11), we can express the density and the specific enthalpy in the solid phase as a function of the pressure and the specific entropy: ρl = ρl (p, Sl ) = p + p∞ 1/γl Rl hl = hl (p, Sl ) = Clp exp p + p∞ Sl∗ − Sl Clp (γl −1)/γl exp Rl , Sl − Sl∗ Clp (3.12) 3.4 Solving the jump relations Substituting the enthalpy hg from (3.8) and hl from (3.11) into the jump relations (3.6), observing that the entropies Sg , Sl are constants, we can rewrite the jump relations as αg ρg ug = αg0 ρg0 ug0 , u2g γ −1 + G(S )ρg g u2g0 = + hg0 , (3.13) αl ρl ul = αl0 ρl0 ul0 , u2l γ −1 + L(S )ρl l = u2l0 + hl0 , where G(S ), L(S ) defined by (3.8) and (3.11), respectively, are constants in this case Substituting ug = αg0 ρg0 ug0 , αg ρg (3.14) from the first equation into the third equation, and substituting ul = αl0 ρl0 ul0 αl ρl (3.15) from the second equation into the fourth equation, after re-arranging terms, we get α g u2g0 αl2 u2l0 + hg0 ρ − G(S )ρg γg +1 g γ +1 + hl0 ρl2 − L(S )ρl l = = (αg0 ρg0 ug0 )2 (αl0 ρl0 ul0 )2 This implies that the volume fractions can be resolved in terms of the pressures: αg = √ u2g0 αl = − αg = αg0 ρg0 |ug0 | 1/2 γ g +1 + hg0 ρg − G(S )ρg √ u2l0 (3.16) αl0 ρl0 |ul0 | 1/2 γl +1 + hl0 ρl − G(S )ρl Adding up these equations, we have √ u2g0 αg0 ρg0 |ug0 | αl0 ρl0 |ul0 | 1/2 + 1/2 = 1, √ u2l0 γg +1 γ l +1 2 + hg0 ρg − G(S )ρg + hl0 ρl − L(S )ρl or √ u2g0 γ g +1 + hg0 ρ − G(S )ρg g − αl0 ρl0 |ul0 | u2g0 1/2 × γ +1 + hg0 ρg2 − G(S )ρg g u2l0 1/2 γ +1 + hl0 ρl2 − L(S )ρl l 1/2 − αg0 ρg0 |ug0 | u2l0 1/2 γ +1 = + hl0 ρl2 − L(S )ρl l (3.17) 996 M.D Thanh / Nonlinear Analysis: Real World Applications 13 (2012) 987–998 Denote mg = αg0 ρg0 |ug0 |, ml = αl0 ρl0 |ul0 |, u2g0 qg = ql = u2l0 + hg0 , ζg = + hl0 , ζl = exp (γg − 1)1/γg exp 1/γ Rl l Sl∗ − Sl Clp Sg ∗ − Sg Cgp , (3.18) Observe that with the notations (3.18), we can write ρg , ρl from (3.9) and (3.12) in the form ρg = ζg p1/γg , ρl = ζl (p + p∞ )1/γl (3.19) Substituting these expressions for the densities into the Eq (3.17), we obtain the following nonlinear equation √ γg +1 1+1/γg 1/2 qg ζg2 p2/γg − G(S )ζg p γ l +1 ql ζl2 (p + p∞ )2/γl − L(S )ζl (p + p∞ )1+1/γl 1/2 1/2 γ +1 − ml qg ζg2 p2/γg − G(S )ζg g p1+1/γg 1/2 γ +1 − mg ql ζl2 (p + p∞ )2/γl − L(S )ζl l (p + p∞ )1+1/γl = (3.20) Observe that from (3.8), (3.11) and (3.18) we deduce γg +1 G(S )ζg = γg ζg , γg − γ l +1 L(S )ζl = γl ζl γl − so that we can write the Eq (3.20) in the form F (p) := 2ζg ζl qg ζg p 2/γg γg p1+1/γg − γg − − ml ζg qg ζg p2/γg − 1/2 γg p1+1/γg γg − − mg ζl ql ζl (p + p∞ )2/γl − 2/γl ql ζl (p + p∞ ) γl − (p + p∞ )1+1/γl γl − 1/2 1/2 γl (p + p∞ )1+1/γl γl − 1/2 = (3.21) Let us discuss the finding roots of the nonlinear equation (3.21) First, the domain of the function F (p) is determined by qg ζg p2/γg − γg p1+1/γg ≥ 0, γg − ql ζl (p + p∞ )2/γl − γl (p + p∞ )1+1/γl ≥ 0, γl − or γ γ−g g γg − p≤ qg ζg , γg γ γ−l l γl − p≤ ql ζl − p∞ γl Thus, providing that γl − ql ζl γl γ γ−l l − p∞ > , the domain of F (p) is given by ≤ p ≤ p¯ := γg − qg ζg γg γ γ−g γ γ−l g l γl − , q l ζl − p∞ γl Furthermore, it is easy to see that F (0) < 0, F (¯p) < 0, F (p0 ) = Thus, if F (p) > for some p, then there are at least two roots of F (p) = Otherwise, F (p0 ) is the maximum value and the equation F (p) = has a repeated root at p = p0 This establishes the following result (see Fig 3) M.D Thanh / Nonlinear Analysis: Real World Applications 13 (2012) 987–998 997 Fig Graph of the function F (p) Lemma 3.1 Given a left-hand state U0 = (αg0 , p0 , ug0 , Sg0 , ul0 , Sl0 ) The corresponding pressure p of the right-hand state given by U = (αg , p, ug , Sg , ul , Sl ) of the stationary wave of (2.5) satisfies the nonlinear equation (3.21), which always admits at least two roots: the trivial root p = p0 , and p = p1 for some p1 in the domain of F (p) These two roots coincide if F (p0 ) is the absolute maximum value We can now summarize the above arguments as follows: to find the equilibrium right-hand (left-hand) state U = (αg , p, ug , Sg , ul , Sl ) resulted by a stationary wave from a given left-hand (right-hand) state U0 = (αg0 , p0 , ug0 , Sg0 , ul0 , Sl0 ), we follow the strategy stated in the following theorem Theorem 3.2 (Equilibrium States) Given a left-hand (right-hand) state either in the form U0 or V0 defined by (3.5), the corresponding right-hand (left-hand) state U that can be connected to U0 by a stationary wave satisfying (3.13) can be determined as follows: (i) First, we evaluate the pressure p from the nonlinear equation (3.21) This can be done, for example, by using the Regula–Falsi method (ii) Second, we compute the densities ρg , ρl from (3.19), having observed that the entropies are constants across the stationary wave: Sg = Sg0 , Sl = Sl0 (iii) Third, the volume fractions αg , αl can be computed using (3.16) (iv) Finally, we compute the velocities ug , ul from (3.14) and (3.15), respectively 3.5 Continuous values of the stationary discontinuity wave As discussed earlier, a stationary discontinuity wave is discontinuous only along the ray t ≥ in the (x, t )-plane Apart from this ray, the wave is continuous The values of the solution satisfy the ordinary differential equation in the perfect divergence form (3.4) and thus satisfy the nonlinear algebraic equations αi (x)ρi (x)ui (x) − gx = αgk ρgk ugk , ui (x)2 + hi (x) − gx = u2ik + hik , i = g , l, (3.22) where k = for x < and k = for x > and the index k represents the values on both sides of the discontinuity: k = for the left-hand state, and k = for the right-hand state The Eqs (3.22) determine implicit continuous functions U = Uk (x), k = 1, on each ray x ≤ and x ≥ 0, respectively These two functions correspond to the left and right parts of the stationary discontinuity wave Solving (3.22) can follow the strategy of finding equilibria above, or any manner evaluating (continuous) implicit functions References [1] H Staedtke, et al., Advanced three-dimensional two-phase flow simulation tools for application to reactor safety (ASTAR), Nucl Eng Des 235 (2005) 379–400 [2] J.R García-Cascales, H Paillère, Application of ausm schemes to multi-dimensional compressible two-phase flow problems, Nucl Eng Des 236 (2006) 1225–1239 [3] M Ishii, Thermo-Fluid Dynamic Theory of Two-Phase Flow, Eyrolles, Paris, 1975 998 M.D Thanh / Nonlinear Analysis: Real World Applications 13 (2012) 987–998 [4] H.B Stewart, B Wendroff, Two-phase flow: models and methods, J Comput Phys 56 (1984) 363–409 [5] C.H Chang, M.S Liou, A conservative compressible multifluid model for multiphase flow: shock-interface interaction problems, in: 17th AIAA Computational Fluid Dynamics Conference, June 6-9, 2005, Toronto, Ontario, Canada, American Institute of Aeronautics and Astronautics, 2005, Paper 2005-5344 [6] B.L Keyfitz, R Sander, M Sever, Lack of hyperbolicity in the two-fluid model for two-phase incompressible flow, Discrete Contin Dyn Syst Ser B (2003) 541–563 [7] J Bona, M.E Schonbek, Traveling-wave solutions to the Korteweg–de Vries–Burgers equation, Proc Roy Soc Edinburgh Sect A 101 (1985) 207–226 [8] M.D Thanh, Global existence of traveling wave for general flux functions, Nonlinear Anal TMA 72 (2010) 231–239 [9] M.D Thanh, Attractor and traveling waves of a fluid with nonlinear diffusion and dispersion, Nonlinear Anal TMA 72 (2010) 3136–3149 [10] M.D Thanh, Existence of traveling waves in elastodynamics with variable viscosity and capillarity, Nonlinear Anal RWA 12 (2011) 236–245 [11] G Dal Maso, P.G LeFloch, F Murat, Definition and weak stability of nonconservative products, J Math Pures Appl 74 (1995) 483–548 [12] P.G LeFloch, M.D Thanh, The Riemann problem for fluid flows in a nozzle with discontinuous cross-section, Commun Math Sci (2003) 763–797 [13] P.G LeFloch, M.D Thanh, The Riemann problem for shallow water equations with discontinuous topography, Commun Math Sci (2007) 865–885 [14] M.D Thanh, The Riemann problem for a non-isentropic fluid in a nozzle with discontinuous cross-sectional area, SIAM J Appl Math 69 (2009) 1501–1519 [15] B.L Keyfitz, Change of Type in Simple Models for Two-Phase Flow Viscous Profiles and Numerical Methods for Shock Waves, Raleigh, NC, 1990, SIAM, Philadelphia, PA, 1991, pp 84–104 [16] B.L Keyfitz, M Sever, F Zhang, Viscous singular shock structure for a nonhyperbolic two-fluid model, Nonlinearity 17 (2004) 1731–1747 [17] D Kröner, M.D Thanh, Numerical solutions to compressible flows in a nozzle with variable cross-section, SIAM J Numer Anal 43 (2005) 796–824 [18] D Kröner, P.G LeFloch, M.D Thanh, The minimum entropy principle for fluid flows in a nozzle with discontinuous cross-section, Math Model Numer Anal 42 (2008) 425–442 [19] M.D Thanh, K Md Fazlul, A.I.Md Ismail, Well-balanced scheme for shallow water equations with arbitrary topography, Int J Dyn Syst Differ Equ (2008) 196–204 [20] S Jin, X Wen, An efficient method for computing hyperbolic systems with geometrical source terms having concentrations, J Comput Math 22 (2004) 230–249 [21] F Bouchut, Nonlinear Stability of Finite Volume Methods for Hyperbolic Conservation Laws, and Well-Balanced Schemes for Sources, in: Frontiers in Mathematics Series, Birkhäuser, 2004 [22] M.-H Lallemand, R Saurel, Pressure relaxation procedures for multiphase compressible flows, INRIA Report No 4038, 2000 [23] R Saurel, R Abgrall, A multi-phase Godunov method for compressible multifluid and multiphase flows, J Comput Phys 150 (1999) 425–467 [24] A Ambroso, C Chalons, F Coquel, T Galié, Relaxation and numerical approximation of a two-fluid two-pressure diphasic model, Math Model Numer Anal 43 (2009) 1063–1097 [25] M.D Thanh, A.Izani.Md Ismail, Well-balanced scheme for a one-pressure model of two-phase flows, Phys Scr 79 (2009) 065401 pp [26] M.D Thanh, D Kröner, N.T Nam, Numerical approximation for a Baer–Nunziato model of two-phase flows, Appl Numer Math 61 (2011) 702–721 ... the equations of state with different choices of the thermodynamical independent variables First, assume that the fluid in the gas phase has the equation of state of a polytropic ideal gas as p... We are finding the way to solve the system of algebraic equations 3.3 Equations of state A fluid is characterized by its equations of state In the following, we will present alternative forms of. .. equation and the equilibria resulted by a stationary wave That the stationary jump can be obtained as the limit of smooth solutions is quite related to viscous traveling waves, see for example