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the hydrolysis of particulate matter

R. Fekih-Salem, Nahla Abdellatif, T. Sari, Jérôme Harmand

To cite this version:

R. Fekih-Salem, Nahla Abdellatif, T. Sari, Jérôme Harmand. On a three step model of anaerobic

digestion including the hydrolysis of particulate matter. MATHMOD 2012 - 7th Vienna International

Conference on Mathematical Modelling, Jan 2012, Vienne, Austria. 6 p. �hal-00777559�

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On a Three Step Model of Anaerobic Digestion Including the Hydrolysis of

Particulate Matter

R. Fekih-Salem∗,∗∗ N. Abdellatif∗∗,∗∗∗ T. Sari∗,∗∗∗∗

J. Harmand∗,†

EPI Modemic Inra-Inria, UMR Mistea, SupAgro, 34060 Montpellier, France.

∗∗Universit´e de Tunis El Manar, Ecole Nationale d’Ing´enieurs de Tunis, LAMSIN, 1002 Tunis Le Belv´ed`ere, Tunisie (e-mail:

fekih@supagro.inra.fr).

∗∗∗Ecole Nationale des Sciences de l’Informatique, Campus Universitaire de Manouba, 2010 Manouba, Tunisie (e-mail:

nahla.abdellatif@ensi.rnu.tn).

∗∗∗∗Irstea, UMR Itap, 34196 Montpellier Cedex, France (e-mail:

tewfik.sari@irstea.fr).

Inra, Laboratoire de Biotechnologie de l’Environnement, 11100 Narbonne, France (e-mail: harmand@supagro.inra.fr).

Abstract:In this work, we focus on the mathematical analysis of the model of chemostat with enzymatic degradation of a substrate (organic matter) that can partly be under a solid form Simeonov and Stoyanov (2003). The study of this 3-step model is derived from a smaller order sub-model since some variables can be decoupled from the others. We study the existence and the stability of equilibrium points of the sub-model considering both Monod or Haldane growth rates and distinct dilution rates. In the classical chemostat model with monotonic kinetics, it is well known that only one equilibrium point attracts all solutions and that bistability never occurs Smith and Waltman (1995). In the present study, although (i) only monotonic growth rates are considered and (ii) the concentrations of input substrate concentration is less than the break-even concentration, it is shown that the considered sub-model may exhibit bistability. Hence, the importance of hydrolysis in the appearance of positive equilibrium points and the bistability is pointed out. If a non monotonic growth rate is considered, depending on the input substrate concentration, it is shown that at most four positive equilibrium points exist. Furthermore, for any positive initial condition, the solution converges towards one of the positive equilibrium points for which the washout is unstable. Finally, we study the case where the growth rate is density-dependent, such as the Contois kinetics, which may be of interest if we consider that we work in a non homogeneous environment Lobry and Harmand (2006).

Depending on the input substrate concentration, we show that the system can exhibit either a bistability or the global stability of the positive equilibrium point or of the washout.

Keywords:Enzymatic degradation, chemostat, models, growth rate, equilibrium, bistability.

1. INTRODUCTION

Anaerobic digestion is a biological process in which organic matter is transformed into methane and carbon dioxide (biogas) by microorganisms in the absence of oxygen. The search for models simple enough to be used for control design is of prior importance today to optimize fermen- tation processes and solve important problems such as the development of renewable energy from waste. Within the studies of microbiology, biochemistry and technology, the anaerobic digestion is generally considered as a three step process: hydrolysis and liquefaction of the large, in- soluble organic molecules by extracellular enzymes, acid production by an acidogenic microbial consortium and a methane production stage realized by a methanogenic

ecosystem. Several mathematical models describing these phenomena have been proposed in the literature. How- ever, they are usually too complex to be used for control synthesis Simeonov and Stoyanov (2003); B. Benyahia and Harmand (2010a,b); Bastin and Dochain (1991). The chemical reactions of anaerobic digestion which converts the substrate into biomass is :

X0 −−−−−−→r00X0 k0S1

k1S1 −−−−−−→r11X1 X1+k2S2+CO2

k3S2 −−−−−−→r22X2 X2+CO2+CH4

where ri = μiXi, i = 0· · ·2, denotes the reaction rate, respectively, μ0 is the specific growth rate of X1 on X0

andμi is the specific growth rate ofXi onSi fori= 1,2.

(3)

ki, i= 0. . .3, denote the pseudo-stochiometric coefficients associated to the chemical reactions,

Motor Biogas

CO2, CH4

S1in, S2in, X0in

Q1

V

S1, S2

Q2

S1, S2, X0, X1, X2

Q1Q2

Fig. 1. Chemostat.

We consider a continuous culture, i.e the input flow rate Q1is equal to the output flow rate. For low concentrations of substrate, the biomass residence time is greater than the substrate one, then the output flow rate of biomass and substrate in the form macromolecules isQ1−Q2. The three step model Simeonov and Stoyanov (2003) is :

⎧⎪

⎪⎪

⎪⎪

⎪⎨

⎪⎪

⎪⎪

⎪⎪

X˙0 = DX0in−αDX0−μ0(X0)X1,

S˙1 = D(S1in−S1) +k0μ0(X0)X1−k1μ1(S1)X1, X˙1 =

μ1(S1)−αD X1,

S˙2 = D(S2in−S2) +k2μ1(S1)X1−k3μ2(S2)X2, X˙2 =

μ2(S2)−αD

X2,

(1)

where

D= Q1

V and Q1−Q2

V =αD.

D denotes the dilution rate of the chemostat and α [0,1] represents the fraction of the biomass leaving the reactor.V denotes the volume of the bioreactor,X0(t) the concentration of the substrate in the form macromolecules at time t, with X0in the concentration of the input nutriment.Sj(t) denote the concentration of the substrates in the effluent, j = 1,2, at time t; with Sjin the input substrate concentrationsj.Xi(t) denote the concentration of the ith population of microorganisms,i= 1,2, at time t.

According to the principle of conservation of matter within the reaction scheme we have t2

t1

μ0(X0)X1V dτ t2

t1

k0μ0(X0)X1V dτ i.e 1k0, which means that, the quantity ofX0 degraded is greater than or equal to the quantity of S1 produced. Similarly, we have

k11 +k2 and k31

which means that, the quantity of S1 degraded is greater than or equal to the quantity ofX1andS2produced. The quantity of S2 degraded is greater than or equal to the quantity ofX2 produced.

In the following, we focus on the study of the sub-model given by the first three equations of system (1), the last two equations being decoupled since the first three equations are independent of variables X2 and S2. Thus, we study the existence and stability of equilibrium points of the following sub-model :

⎧⎪

⎪⎩

X˙0=D(X0in−αX0)−μ0(X0)X1,

S˙1=D(S1in−S1) +k0μ0(X0)X1−k1μ1(S1)X1, X˙1= μ1(S1)−αD

X1.

(2) First, we establish the following result :

Proposition 1.1.

(1) For any non-negative initial condition, the solutions of system (2) stay positive at any time and are bounded whent→+.

(2) The set

Ω ={(X0, S1, X1)R3+ :Z = k0X0+S1+k1X1

max

Z(0),SαDin } is positively invariant and attractor of all solutions of (2), with Sin=D(k0X0in+S1in).

2. STUDY OF THE SUB-MODEL The washout equilibrium E0 = X

0inα , S1in,0

, always exists. To look for positive equilibria, we consider the function

ξ(X0) = D(X0in−αX0) μ0(X0) . We assume that

H0: The function μ0(·) is increasing, μ0(0) = 0 and μ0(X0)0 for allX0

0,X0inα .

Lemma 2.1. Under assumptionH0, the functionξ(·) van- ishes on Xα0in, is decreasing and convex.

In the case where the functionμ0(·) is linear or of Monod type, the assumptionH0is satisfied.

2.1 Study of the sub-model with monotonic growth rate μ1(·)

In this section, we study the existence of equilibrium points of system (2) under the following assumption

H1: μ1(0) = 0 andμ1(S1)>0 for all S10.

H2: The equation μ1(S1) = αD has a finite solution λ1=μ−11 (αD).

Let Δ the line of equation : X1=δ(X0) = 1

k1α (S1in−λ1) +k0(X0in−αX0)

. Lemma 2.2. The equation ξ(X0) = kk01 has a unique solution ¯X0

0,Xα0in

if and only if ξ

X0in α

>−k0 k1. Moreover,

ξ X0in

α

>−k0

k1 ⇐⇒ k0μ0

X0in

α

> k1αD.

Ifξ

X0in

α

>−kk01, the intersection of the line Δ with the curve of the function ξ, has at most two points. Let us denote by E1 = (X0, λ1, X1) and E1∗∗ = (X0∗∗, λ1, X1∗∗) (see Fig. 2). By Lemma 2.2, there exists a unique solution X¯0

0,Xα0in

of equationξ(X0) =kk01. Thus, there is a

(4)

limit value X1Min for which the curveξ is tangent to the line Δ and who satisfied

X1Min= ¯X1+k0

k1

X¯0

with ¯E1= ( ¯X0, λ1,X¯1) an equilibrium of (2) (see case 4 of the Fig. 3). At this limite valueX1Min, we associateS1inMin which satisfied

X1Min=δ(0) = 1

k1α (SMin1in −λ1) +k0X0in

. In the generic case where S1inMin >0, we have shown the following result :

Proposition 2.1.

If λ1 S1in, there exists a unique positive equilib- riumE1 = (X0, λ1, X1).

If S1Min < S1in < λ1, there exist two positive equilibriaE1 etE1∗∗.

If S1in = S1Min, there exists a unique positive equi- librium ¯E1= ( ¯X0, λ1,X¯1).

IfS1in< S1Min, there is no positive equilibrium.

X1

Cξ

E1

Δ E0 X0in

α

X0

X1 Cξ

E1

Δ E0 X0in

α

X0

X1 Cξ

E1

Δ E∗∗1

E0

X0in

α

X0

Fig. 2. Case 1: λ1 < S1in = 5.5, case 2: S1in= 4.5 =λ1, case 3:SMin1 < S1in= 2< λ1.

X1 Cξ

XMin1

Δ E¯1= ( ¯X0, λ1,X¯1)

E0 X0in α

X0

X1 Cξ

Δ

E0 X0in

α

X0

Fig. 3. Case 4: S1in= 0.75 =SMin1 , case 5: S1in= 0.2<

S1Min.

In the following, we show the asymptotic behavior of equilibrium points. We choose the red color for Locally Asymptotically Stable (LAS) equilibrium, the green color for saddle node equilibrium and blue color for unstable equilibrium.

Proposition 2.2.

E0 is LAS if and only ifμ1(S1in)< αD.

IfE1 exists, then it is LAS (stable node).

IfE1∗∗ exists, then it is unstable (saddle point).

Proof. At washout E0 = (Xα0in, S1in,0), the Jacobian matrix of (2) is

J0=

−αD 0 −μ0(X0inα ) 0 −D k0μ0(Xα0in)−k1μ1(S1in) 0 0 μ1(S1in)−αD

⎦ The eigenvalues are −αD,−D andμ1(S1in)−αD. Then, the equilibriumE0 is LAS if and only ifμ1(S1in)< αD.

At a positive equilibriumE1= (X0, λ1, X1), the Jacobian matrix is

J1=

−m11 0 −m13

m21 −m22 θ

0 m32 0

where

m11=αD+μ0(X0)X1, m13=μ0(X0), m21=k0μ0(X0)X1, m22=D+k1μ1(λ1)X1,

θ=k0μ0(X0)−k1αD, m32=μ1(λ1)X1, with m11, m13, m21, m22 and m32 are positive. The char- acteristic polynomial ofJ1 is given by

PJ1(λ) =a0λ3+a1λ2+a2λ+a3

with

a0=1, a1=(m11+m22), a2=−m11m22+θm32, a3=−m32(m21m13−θm11).

According to the Routh-Hurwitz criterion, E1 is LAS if and only if

ai<0, i= 0· · ·3 a1a2−a0a3>0.

We have

a2 = k0μ0(X0)−k1αD−k1μ0(X0)X1

μ1(λ1)X1

0(X0)X1+αD

D+k1μ1(λ1)X1

. Since

ξ(X0) +k0

k1 = −k1αD−k1μ0(X0)X1+k0μ0(X0) k1μ0(X0)

then we deduce

a2 = ξ(X0) +kk0

1

k1μ0(X0)μ1(λ1)X1

0(X0)X1+αD

D+k1μ1(λ1)X1 therefore, ifξ(X0)<−kk01, thena2<0. Moreover

a3 = m32αD[k0μ0(X0)−k1αD−k1μ0(X0)X1]

= m32αD ξ(X0) +kk0

1

k1μ0(X0) is negative if and only ifξ(X0)<−kk01. Finally

a1a2−a0a3 = ξ(X0) +kk0

1

k1μ0(X0)μ0(X0)μ1(λ1)X1∗2 + P

where P =Dm211+

(αD)2+αDμ0(X0)X1

k1μ1(λ1)X1−m22a2

is positive if ξ(X0) < kk01. Since E1 satisfies ξ(X0) <

kk01, then it is LAS andE1∗∗ satisfiesξ(X0∗∗)>−kk01, then it is unstable.

The simulations shown in Fig. 4 where obtained for the following Monod functions

μ0(X0) = 2.5X0

1.5 +X0 and μ1(S) = 2S 1.5 +S and the following values of the parameters

X0in= 3, D= 1, α= 0.75, k0= 1 and k1= 1.2 The value of the break-even concentration isλ1= 0.9. We illustrate the case of bistability forS1in= 0.7 such as

ξ X0in

α

=−0.412>−0.833 =−k0

k1.

(5)

In this case, the equation ξ(X0) = δ(X0) admits two solutions (see Fig. 4 on the left) and the system (2) has a washout equilibrium

E0= (4,0.5,0) and two positive equilibria

E1= (1.201,0.9,1.887), E1∗∗= (2.808,0.9,0.548). The Fig. 4 in the middle shows the convergence to the positive equilibriumE1 for the initial condition

X0(0) = 4.5, S1(0) = 2 and X1(0) = 0.368 and the washout for the initial condition (see Fig. 4 on the right)

X0(0) = 4.5, S1(0) = 2 and X1(0) = 0.367

0 1 2 3 4

0.0 0.5 1.0 1.5 2.0 2.5 3.0 3.5 4.0 4.5

0 40 80 120 160 200

0.0 0.5 1.0 1.5 2.0 2.5 3.0 3.5 4.0 4.5

0 40 80 120 160 200

0.0 0.5 1.0 1.5 2.0 2.5 3.0 3.5 4.0 4.5

X1

E1

E1∗∗

E0

X0

X1

X0

S1

T ime

X0

S1

X1

T ime

Fig. 4. Existence of two positive equilibria and bistability forS1in< λ1.

2.2 Study of the sub-model with non-monotonic growth rate μ1(·)

In the following, we study the existence of equilibrium points of system (2) under the following assumption H3: The functionμ1(·) is non monotonic and is such that

the equationμ1(S1) =αDadmits two solutionsλ1and λ2 withλ1< λ2.

Let Δithe line of equation : X1=δi(X0) = 1

k1α (S1in−λi)+k0(X0in−αX0)

, i= 1,2. The positive equilibriumEi= (X0i, λi, X1i) of system (2) is a solution of the equation

X1i =δi(X0i) =ξ(X0i), i= 1,2.

When the intersection of the curve ξ and the line Δi is formed by two points, let us denote them by Ei = (X0i, λi, X1i) if ξ(X0i) < kk01 and by Ei∗∗ = (X0i∗∗, λi, X1i∗∗), otherwise (see Fig. 5 and Fig. 6). In the case ξX0in

α

> kk01, there exists a unique solution X¯0

0,Xα0in

of the equation ξ(X0) =kk01. As the two lines Δ1and Δ2 are parallel, then there exists fori= 1,2 a limit value X1Min for which the curve of ξis tangent to the line Δi and which satisfies

X1Min= ¯X1+k0

k1

X¯0

with ¯Ei = ( ¯X0, λi,X¯1) an equilibrium of (2) for i = 1,2.

At this limit valueX1Min, we associateS1iMin such that X1Min =δ(0) = 1

k1α[S1iMin−λi+k0X0in]. In the generic case where S1iMin>0 for i= 1,2, the cases S12Min< λ1andS12Min> λ1have to be distinguished. When S12Min< λ1, we have shown the following result :

Proposition 2.3.

IfS1inλ2> λ1, there exist two positive equilibrium Ei,i= 1,2.

If λ1S1in< λ2, there exist three positive equilib- riumEi andE2∗∗.

If SMin12 < S1in < λ1, there exist four positive equilibrium EiandEi∗∗.

IfS12Min=S1in, there exist three positive equilibrium E1,E1∗∗ and ¯E2.

If S11Min < S1in < S12Min, there exist two positive equilibrium E1andE1∗∗.

If S11Min = S1in, there exists a unique positive equi- librium ¯E1.

IfS1in< S11Min, there is no positive equilibrium.

In the case whereS12Min> λ1, we can prove, similarly, that if S1in SMin11 , we have one, two, three or four positive equilibria, depending on the position ofS1in.

X1

Cξ

Δ1 Δ2E1

E2

E0

X0in α

X0

X1 Cξ

E1

E2

E2∗∗E0 X0in

α

X0

X1 Cξ

E1

E2

E∗∗2 E1∗∗

E0

X0in α

X0

Fig. 5. Case 1: S1in = 8 > λ2 = 7, case 2: λ1 = 4.5 <

S1in= 5.75< λ2, case 3:SMin12 < S1in= 3.7< λ1.

X1

Cξ

E1

E¯2

E∗∗1

E0 X0in α

X0

X1 Cξ

E1

E∗∗1

E0 X0in α

X0

X1 Cξ

E0 X0in

α

X0

Fig. 6. Case 4: S1in = 3.125 = S12Min, case 5: S11Min = 0.62< S1in= 1.8< S12Min, case 6:S1in= 0.1< S11Min. We study now the local stability properties of the equilib- ria. For E0, E1 and E∗∗1 , we have the same results as in Proposition 2.2. Moreover, we can prove :

Proposition 2.4.

IfE2 exists, then it is unstable.

If E2∗∗ exists and D+k1μ1(λ2)X12∗∗ > 0, then it is LAS.

Proof.The Jacobian of the system (2) atE2 is J2=

−m11 0 −m13

m21 β θ

0 −m32 0

where

m11=αD+μ0(X02)X12 , m13=μ0(X02 ), m21=k0μ0(X02)X12 , β=

D+k1μ1(λ2)X12 , θ=k0μ0(X02)−k1αD, m32=−μ1(λ2)X12 , withm11, m13, m21andm32are positive. The characteris- tic polynomial ofJ2 is given by

PJ2(λ) =a0λ3+a1λ2+a2λ+a3

(6)

with

a0=−1, a1=β−m11, a2=m11β−θm32, a3=m32(m21m13−θm11).

We have

a2 = ξ(X02 ) +kk0

1

k1μ0(X02 )μ1(λ2)X12

0(X02 )X12 +αD

D+k1μ1(λ2)X12

. IfD+k1μ1(λ2)X12 >0 andξ(X02)>−kk01, thena2<0.

It is easy to check that

a3=−m32αD ξ(X02) +k0

k1

k1μ0(X02)

who is negative if and only ξ(X02 ) > kk01. Thus, E2 is unstable. One can readily check that

a1a2−a0a3=m11β2−m211β+m32(m21m13−θβ) and

θX12∗∗ =k0μ0(X02∗∗)X12∗∗−k1μ1(λ2)X12∗∗=−D(S1in−λ2). IfE2∗∗ exists thenS1in< λ2, thusθ >0. We conclude that according to the Routh-Hurwitz criterion, ifE2∗∗exists and

D+k1μ1(λ2)X12∗∗>0, then it is LAS.

3. STUDY OF CONTOIS MODEL

In this section we study the case where the growth rate μ0(.) depends on X0 and also on X1. Models with such growth rates may be of interest if we consider that we work in a non homogeneous environment Lobry and Harmand (2006). The Contois function is an exemple of these growth rates. We consider the model

⎧⎪

⎪⎩

X˙0=D(X0in−αX0)−μ0(X0, X1)X1,

S˙1=D(S1in−S1) +k0μ0(X0, X1)X1−k1μ1(S1)X1, X˙1= μ1(S1)−αD

X1.

(3) Let us denote by

f(X0, X1) =μ0(X0, X1)X1−D(X0in−αX0). We assume that

H4: μ0(0, X1) = 0 and μ0(0, X1)>0 for all X0 >0 and allX10.

H5: ∂μ0

∂X0 >0 and ∂μ0

∂X1 <0 for allX0>0 and allX10.

H6: μ1(0) = 0, μ1(S1)>0 for allS10 and the equation μ1(S1) =αD has a unique solutionλ1.

H7: ∂f

∂X1(X0, X1) = μ0(X0, X1) + ∂μ0

∂X1X1 > 0 for all X0>0 and all X10.

H8: There exista∈]0,Xα0in[ such that

Xlim0→a X1→+∞

f(X0, X1) = 0.

We consider, now, the existence of positive equilibria. We first prove :

Lemma 3.1. Assume that H4-H8 hold. The equation f(X0, X1) = 0 defines a decreasing function

F :

a,X0inα

−→ R+

X0 −→ F(X0) =X1

where 0< a < Xα0in and such that

Xlim0→aF(X0) = +∞, FX0in

α

= 0 and F(X0)<0. Then we state the following result :

Proposition 3.1.

For S1in> λ1, ifF(X0)>0 for allX0 a,Xα0in

, then there exists a unique positive equilibrium. If F(X0) changes sign forX0

a,Xα0in

, then there exists at least one positive equilibrium. Generically one has an odd number of positive equilibria (see Fig.

7).

For S1in< λ1, ifF(X0)>0 for allX0 a,Xα0in

, then there exist at most two positive equilibria. If F(X0) changes sign for X0

a,X0inα

, then the system has generically no positive equilibria or an even number of positive equilibria (see Fig. 8).

X1

X1=F(X0)

X1=δ(X0) E1

E0 a X0in

α

X0

X1

X1=F(X0) X1=δ(X0) E1

E1∗∗ E∗∗∗1

E0 a X0in

α

X0

Fig. 7. Null-clines X1 = F(X0) and X1 = δ(X0) for S1in > λ1. On the left, F(·) concave. On the right, F(·) changes sign.

X1

X1=F(X0)

X1=δ(X0) E1

E1∗∗

E0

a X0in

α

X0

X1

X1=F(X0) X1=δ(X0) E1

E1∗∗

E0

a X0in

α

X0

Fig. 8. Null-clines X1 = F(X0) and X1 = δ(X0) for S1in < λ1. On the left, F(·) concave. On the right, F(·) changes sign.

ExampleWe consider the following function of Contois μ0(X0, X1) = m0X0

X0+a0X1

wherem0>0 anda0>0. We have

∂f

∂X1(X0, X1) = m0X02

(X0+a0X1)2 >0 thus AssumptionH7is satisfied. Moreover,

X1lim→+∞f(X0, X1) = 0

= X0= DX0in m0

a0 +αD =a

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The presentation deals with the Swiss model of the press, the crisis of the Swiss press, the strategy of concentration as well as the strategy of digital diversification..

When one or several chemical species (solid, liquid or gaseous) dissolve in a liquid, we obtain an homogeneous mixture called solution.. The minority species are called