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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..

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Exact solutions of a two-fluid model of two-phase compressible flows with gravity

Department of Mathematics, International University, Quarter 6, Linh Trung Ward, Thu Duc District, Ho Chi Minh City, Viet Nam

a r t i c l e i n f o

Article history:

Received 1 September 2010

Accepted 16 September 2011

Keywords:

Two-fluid

Two-phase flow

Conservation law

Source term

Discontinuity

Contact wave

Jump relation

a b s t r a c t

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

1 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 speedσcan be made stationary thanks to a classical Galilean change of variable(x,t) → (x− σt,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 6 governing equations is given by (see [1,2]):

∂t(αgρg) + ∂x(αgρg u g) =0,

∂t(αgρg u g) + ∂x(αg(ρg u2g+p)) =p∂xαg+ αgρg g,

∂t(αgρg E g) + ∂x(αg(ρg E g+p)u g) = −p∂tαg+ αgρg u g g,

∂t(αlρl) + ∂x(αlρl u l) =0,

∂t(αlρl u l) + ∂x(αl(ρl u2l +p)) =p∂xαl+ αlρl g,

∂t(αlρl E l) + ∂x(αl(ρl E l+p)u l) = −p∂tαl+ αlρl u l g,

(1.1)

whereαiis the volume fraction,ρi is the density, u i is the velocity, e iis the internal energy and

E i=e i+1

2u

2

i,

∗Tel.: +84 8 2211 6965; fax: +84 8 3724 4271.

E-mail addresses:mdthanh@hcmiu.edu.vn , hatothanh@yahoo.com

doi:10.1016/j.nonrwa.2011.09.009

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is the total energy, g is the gravity constant in the model with gravity, and g = 0 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=1.

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

and we use the stiffened gas equation of state for the liquid phase

p= γl−1

γl

ρl C lp T l−p∞,

e l= C pl

γl

T l+p∞

ρl .

(1.3)

For computing purposes, we will take

γg =1.4, C gp=1.012 kJ

kg K,

γl=1.9276, C lp=8.07673 kJ

kg K, p∞=11.5968.103atm.

(1.4)

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 = 0 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 eigenvaluesλ1=u g, λ2=u l 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 =u g,r1), there are five Riemann invariantsαg,p,u g,u l,S land that across the contact waves associated with the linearly degenerate field(λ2 =u l,r2), there are five Riemann invariantsαg,p,u g,u l,S g Consequently, only the entropy S gchanges across the

1-contact waves and only the entropy S lchanges 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 Section3, where we compute the eigenvalues, see also [6] Hence, the Riemann problem

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 fluid models of two-phase 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 Section2we present basic features and properties of the model(1.1)including alternative forms of the model(1.1), hyperbolicity and nonhyperbolicity, and Riemann invariants Section3provides 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

and finding a nonlinear equation for the pressure Finally, we discuss how to solve this equation and the equilibria resulted

by a stationary wave

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2 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(αgρg e g) + ∂x(αgρg e g u g) = −p(∂tαg + ∂x(αg u g)),

∂t(αlρl e l) + ∂x(αlρl e l u l) = −p(∂tαl+ ∂x(αl u l)).

(2.1)

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

∂t(αρu2/2) + ∂x(αρuu2/2+ αpu) = αρug+p∂x(αu),

or

∂t(αρu2/2) + ∂x(αρuu2/2) +p∂x(αu) + αu∂ −xp= αρug+p∂x(αu). (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

αρu∂t u/2+u/2∂t(αρu) + (u/2)∂x(αρu2/2) + αρu2∂x(u/2) + αu∂x p= αρug,

or

(u/2)(αρ∂t u+ ∂t(αρu) + ∂x(αρu2) + αρu∂x u+2α∂x p−2αρg) =0.

If u=0, then we are done Otherwise, canceling u/2̸=0 from the last equation, after re-arranging terms, we obtain

(∂t(αρu) + ∂x(α(ρu2+p)) − αρg−p∂xα) + α∂x p+ αρ∂t u+ αρu∂x u− αρg=0. (2.3) Using the equation of balance of momentum in(1.1), we get from(2.3)

Multiplying the equation of balance of mass in(1.1)by u and adding up the resulting equation with(2.4), we obtain

∂t(αρu) + ∂x(α(ρu2+p)) −p∂xα − αρg =0

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(αgρg S g) + ∂x(αgρg u g S g) =0,

∂ (α ρS) + ∂ (α ρu S) =0.

(2.5)

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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

∂t(αρ)e+ (αρ)∂t e+e∂x(αρu) + (αρu)∂x e+p(∂tα + ∂x(αu)) =0.

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

Using the thermodynamical identity

e=TdS−pdv, v = 1ρ ,

we obtain from(2.6)that

αρ(T∂t S−p∂tv +Tu∂x S−up∂xv) +p(∂tα + ∂x(αu)) =0.

Substitutingv =1/ρinto the last equation, after re-arranging terms, we get

Tαρ(∂t S+u∂x S) +p



∂tα + α

ρ ∂tρ + ∂x(αu) + αu

ρ ∂xρ



=0,

or

Tαρ(∂t S+u∂x S) + ρp(∂t(αρ) + ∂x(αρu)) =0.

Using the equation of conservation of mass, and canceling T ̸=0 from the last equation, we get

Multiplying the last equation byαρ, 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(αρS) + ∂x(αρuS) =0.

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

A straightforward calculation shows that the matrix A(U)in(2.8)is given by

A(V) =







a1 a2 a3 0 a4 0

b1 b2 b3 0 b4 0

ρg

u g 0 0 0

ρl







where

a1= ρg u gαl∂pρl+ ρl u lαg∂pρg

αl∂p(ρl)ρg+ αg∂p(ρg)ρl

, a2= αg∂p(ρg)αl∂p(ρl)(u g−u l)

,

a3= ρgαl∂p(ρl)αg

αl∂p(ρl)ρg + αg∂p(ρg)ρl

, a4= ρlαg∂p(ρg)αl

,

b1= ρlρg(u g−u l)

, b2= ρlαg u g∂pρg+ ρgαl u l∂pρl

,

b3= ρlρgαg

α ∂ (ρ )ρ + α ∂ (ρ )ρ , b4= ρlρgαl

α ∂ (ρ )ρ + α ∂ (ρ )ρ .

(2.10)

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Fig 1 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

where

Q(λ) = (a1− λ)(b2− λ)(u g− λ)(u l− λ) −a2b1(u g− λ)(u l− λ) −



b3

ρg

(u l− λ) +b4

ρl

(u g− λ)

 (a1− λ)

+a3b1

ρg (u l− λ) +a4b1

The polynomial P(λ)always admits the two obvious rootsλ1 = u g andλ2 = u l 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 SeeFig 1, where the polynomial Q(λ)has four roots, and seeFig 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 eigenvaluesλ1 = u g, λ2 = u l 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) ·r i(V) ≡0, i=1,2,

so that the first and the second characteristic fields are linearly degenerate

The 5 Riemann invariants associated with the first characteristic fields are

αg,p,u g,u l,S l,

the 5 Riemann invariants associated with the second characteristic fields are

αg,p,u g,u l,S g.

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)

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Fig 2 The polynomial Q(λ)defined by (2.12) has only two roots.

3 Equilibrium states and stationary solutions

3.1 Stationary smooth waves

As discussed earlier in Section2, 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,S i), leaving the remaining three thermodynamical variables pressure, internal energy, and absolute

temperature p,e i,T i to be dependent variables, i=g,l Therefore, the unknown function is

V(x,t) = (ρg,u g,S g, ρl,u l,S l)(x,t), x∈R, t>0.

Smooth stationary waves V =V(x,t), x∈R, t>0, of(2.5)are the ones that do not depend on time More precisely, they are solutions of the differential equations

d

dx(αiρi u i) =0,

d

dx(αi(ρi u2i +p)) =p d

dxαi+ αiρi g,

d

dx(αiρi S i u i) =0, i=l,g.

(3.1)

The last system can be written as

d

u i d

dx(αiρi u i) + αiρi u i d

dx u i+ αi

d

dx p+p

d

dxαi=p d

dxαi+ αiρi g,

S i d

dx(αiρi u i) + (αiρi u i)d

dx S i=0, i=l,g,

or

d

dx



u2i

2



+ 1

ρi

dp

dx −g =0,

u i d

dx S i =0, i=l,g.

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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

dx



u2i

2



+ 1

ρi

dp

dx−g=0,

d

dx(S i) =0, i=l,g.

(3.2)

In the sequel we will seek for stationary waves of(2.5)given by(3.2), which says that the entropies S l,S gare constants Let

us introduce the specific enthalpy

h i=e i+ p

ρi

which satisfies a thermodynamical identity

dh i =T i dS i+ 1

ρi

dp, i=g,l.

This implies

dh i(ρi,S i)

dρi

= 1

ρi

dp(ρi,S i)

dρi , i=g,l.

From the last equality, we can rewrite system(3.2)as

d

dx(αgρg u g) =0,

d

dx



u2

g

2 +h g−gx



=0,

d

dx(αlρl u l) =0,

d

dx



u2

l

2 +h l−gx



=0.

(3.4)

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 ≥ 0 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

[ αgρg u g] =0,



u2g

2 +h g



=0, [ αlρl u l] =0,

[

u2

l

2 +h l

]

=0,

where[ αiρi u i] := α+

i ρ+

i u+i − α−

i ρ−

i u−i , and so on, denotes the difference of the corresponding valueαiρi u i,i=g,l between

the right-hand and left-hand states of the stationary discontinuity

Set

U=







αgρg

αgρg u g

αgρg S g

αlρl

αlρl u l

αlρl S l





 , U0=







αg0ρg0

αg0ρg0 u g0

αg0ρg0 S g0

αl0ρl0

αl0ρl0 u l0

αl0ρl0 S l0













αg

p

u g

S g

u l

S l





 , V0=







αg0

p0

u g0

S g0

u l0

S l0







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 U or in the form V , and we look for the right-hand state either in the form U or in the form V , that can be

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connected to U0by a stationary wave In other words, we solve the following system of nonlinear equations

αgρg u g = αg0ρg0 u g0,

u2

g

2 +h g =

u2

g0

2 +h g0,

αlρl u l= αl0ρl0 u l0,

u2l

2 +h l=

u2l0

2 +h l0.

(3.6)

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 R g T g = (γg−1)ρg e g,

e g =C gvT = R g T g

where R g is the specific gas constant; R g = ηgR, whereηgis the mole-mass fraction, andRis the universal gas constant

Let C gv= γg R−g1be the specific heat at constant volume and C gp= R gγg

γg− 1be the specific heat at constant pressure Thus,

C gp= γg C gv.

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:

p=p g(ρg,S g) = (γg−1)ργg

g exp



S g−S g∗

C gv

 ,

h g =h g(ρg,S g) =G(S)ργg− 1

g , where G(S) = γgexp



S g−S g∗

C gv

for some constant S g∗ 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,S g) =



p

γg−1

1 /γg

exp



S g∗−S g

C gp

 ,

h g =h g(p,S g) = γg

(γg−1)(γg− 1 )/γg p(γg− 1 )/γgexp



S g−S g∗

C gp



(3.9)

Second, we consider the liquid phase where the equations of state are given by

p= (γl−1)ρl C lvT l−p∞, e l=C lvT l+p∞

where C lv = R l

γl− 1 is the specific heat at constant volume and C lp = R lγl

γl− 1 is the specific heat at constant pressure, so that

C lp= γl C lv, where R lis the specific gas constant

S l−S l∗

C lv

=log(T lvγl− 1

l ),

for some S l∗ This yields

p=p l(ρl,S l) = (γl−1)C lvexp



S l−S l∗

C lv



ργl

l −p∞,

h l=h l(ρl,S l) =L(S)ργl− 1

l , where L(S) =C lpexp



S l−S l∗

Cv



(3.11)

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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,S l) =



p+p∞

R l

1 /γl

exp



S l∗−S l

C lp

 ,

h l=h l(p,S l) =C lp



p+p∞

R l

(γl− 1 )/γl

exp



S l−S l∗

C lp



(3.12)

3.4 Solving the jump relations

Substituting the enthalpy h gfrom(3.8)and h lfrom(3.11)into the jump relations(3.6), observing that the entropies S g,S l

are constants, we can rewrite the jump relations as

αgρg u g = αg0ρg0 u g0,

u2g

2 +G(S)ργg− 1

g = u2g0

2 +h g0,

αlρl u l= αl0ρl0 u l0,

u2

l

2 +L(S)ργl− 1

l0

2 +h l0,

(3.13)

where G(S),L(S)defined by(3.8)and(3.11), respectively, are constants in this case Substituting

u g = αg0ρg0 u g0

from the first equation into the third equation, and substituting

u l= αl0ρl0 u l0

αlρl

(3.15) from the second equation into the fourth equation, after re-arranging terms, we get

α2

g



u2g0

2 +h g0



ρ2

g −G(S)ργg+ 1

g



= (αg0ρg0 u g0)2 2

α2

l



u2l0

2 +h l0



ρ2

l −L(S)ργl+ 1

l



= (αl0ρl0 u l0)2

This implies that the volume fractions can be resolved in terms of the pressures:

αg = αg0ρg0|u g0|

√

2



u2

g0

2 +h g0



ρ2

g−G(S)ργg+ 1

g

1 / 2

αl=1− αg = αl0ρl0|u l0|

√

2



u2

l0

2 +h l0



ρ2

l −G(S)ργl+ 1

l

1 / 2.

(3.16)

Adding up these equations, we have

αg0ρg0|u g0|

√

2



u2g0

2 +h g0



ρ2

g −G(S)ργg+ 1

g

1 / 2 + αl0ρl0|u l0|

√

2



u2

l0

2 +h l0



ρ2

l −L(S)ργl+ 1

l

1 / 2 =1,

or

√

2



u2g0

2 +h g0



ρ2

g −G(S)ργg+ 1

g

1 / 2

×



u2

l0

2 +h l0



ρ2

l −L(S)ργl+ 1

l

1 / 2

− αl0ρl0|u l0|



u2g0

2 +h g0



ρ2

g −G(S)ργg+ 1

g

1 / 2

− αg0ρg0|u g0|



u2

l0

2 +h l0



ρ2

l −L(S)ργl+ 1

l

1 / 2

=0. (3.17)

Trang 10

m g = αg0ρg0|u g0| , q g = u2g0

(γg−1)1 /γg exp



S g∗−S g

C gp

 ,

m l= αl0ρl0|u l0| , q l= u2l0

2 +h l0, ζl= 1

R1/γl l

exp



S l∗−S l

C lp



(3.18)

Observe that with the notations(3.18), we can writeρg, ρlfrom(3.9)and(3.12)in the form

Substituting these expressions for the densities into the Eq.(3.17), we obtain the following nonlinear equation

√

2



q gζ2

g p2/γg−G(S)ζγg+ 1

g p1+1/γg1 / 2

q lζ2

l (p+p∞)2 /γl−L(S)ζγl+ 1

l (p+p∞)1 + 1 /γl

1 / 2

−m l



q gζ2

g p2/γg −G(S)ζγg+ 1

g p1+1/γg

1 / 2

−m g



q lζ2

l (p+p∞)2 /γl−L(S)ζγl+ 1

l (p+p∞)1 + 1 /γl

1 / 2

Observe that from(3.8),(3.11)and(3.18)we deduce

G(S)ζγg+ 1

γg−1ζg, L(S)ζγl+ 1

γl−1ζl

so that we can write the Eq.(3.20)in the form

F(p) :=



2ζgζl



q gζg p2/γg− γg

γg−1p

1 + 1 /γg

1 / 2

q lζl(p+p∞)2 /γl− γl

γl−1(p+p∞)1 + 1 /γl

1 / 2

−m l



ζg



q gζg p2/γg − γg

γg−1p

1 + 1 /γg

1 / 2

−m g



ζl



q lζl(p+p∞)2 /γl− γl

γl−1(p+p∞)1 + 1 /γl

1 / 2

Let us discuss the finding roots of the nonlinear equation(3.21) First, the domain of the function F(p)is determined by

q gζg p2/γg− γg

γg−1p

1 + 1 /γg ≥0, q lζl(p+p∞)2 /γl− γl

γl−1(p+p∞)1 + 1 /γl≥0,

or

p≤

 γg−1

γg

q gζg

 γg

γg− 1 ,

p≤

 γl−1

γl

q lζl

 γl

γl− 1

−p∞.

Thus, providing that

 γl−1

γl

q lζl

 γl

γl− 1

−p∞>0,

the domain of F(p)is given by

0≤p≤ ¯p:=min



 γg−1

γg

q gζg

 γg

γg− 1 ,

 γl−1

γl

q lζl

 γl

γl− 1

−p∞



Furthermore, it is easy to see that

F(0) <0, F(¯p) <0, F(p0) =0.

Thus, if F(p) >0 for some p, then there are at least two roots of F(p) =0 Otherwise, F(p0)is the maximum value and the

equation F(p) =0 has a repeated root at p=p0 This establishes the following result (seeFig 3)

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