modeling and analysis of anaerobic digestion in a bioreactor

Introduction Systems to Study Mathematical Approach to Solving Systems Results Concluding Thoughts

Modeling and Analysis of Anaerobic Digestion in aBioreactor

Christina Berti

REU 2013

Christina Berti Modeling and Analysis of Anaerobic Digestion in a Bioreactor


Advantages of Wastewater Treatment

Recycle water

Produce biogas

Minimal pollution



Degradation Process

Biodegradable Organic Matter Inert Material,Biproducts

Acidogenesis

Hydrolysis

Acidogenesis

Acetogenesis

46%

70%

20%

40%

13%

100%

21%

30%

34%

21%5%

Methanogenesis

S2

S2

X2

X1

S1

Amino Acids, Simple Sugars Fatty Acids

Proteins Carbohydrates Lipids

Propionate, Butyrate, etc.

Hydrogen, Carbon DioxideAcetate

Biogas

Complex Organic MatterRaw Material: sludge, slurries, lump compound

Methane, Carbon Dioxide

Figure : Detailed Flowchart of Model for Biogas Production



Condensed 4-Dimensional (two-step reaction process)System

Definition

4-Dimensional System:

S ′1 = −X1k1(µ1(S1)) + D(S1in − S1)X ′1 = X1(µ1(S1) − Dα)S ′2 = D(S2in − S2) + X1k2(µ1(S1)) − X2k3(µ2(S2))X ′2 = X2(µ2(S2) − Dα)

(1)



Parameter Value Ranges and Definitions

S1in and S2in : Input substrate concentrations.

k1, k2, k3 : Pseudo-stoichiometric coefficients based on natureof bioreactions.

α : Fraction of biomass not retained in the digester (accountsfor decoupling of Hydraulic Retention Time from SolidRetention Time).

D: Dilution factor for incoming and outgoing substrate andbacteria.

µ1(S1) and µ2(S2) are functions used to demonstrate thegrowth of bacteria 1 and 2, respectively.



Simplified Two-Step Reaction Process

Definition

Two Steps:

Acidogenesis : k1S1µ1(S1)X1−→ X1 + k2S2

Methanogenesis : k3S2µ2(S2)X2−→ X2 + k4CH4

(2)

Acidogenesis: Organic substrate (S1) is broken down into volatilefatty acids (S2) by acidogenic bacteria (X1).

Methanogenesis: Volatile fatty acids (S2) are degraded to produceCH4 and CO2 by methanogenic bacteria (X2).



Concerns Regarding Approach to Attaining Steady-State

Two Main Concerns:

Growth of Bacteria

Substrate Degredation and Product Formation

To attain a steady-state: The substrate flow and gas productionmust remain constant and continuous. The growth requirementsfor bacteria must remain constant over time.

Definition

4-Dimensional System:

0 = −X1k1(µ1(S1)) + D(S1in − S1)0 = X1(µ1(S1) − Dα)0 = D(S2in − S2) + X1k2(µ1(S1)) − X2k3(µ2(S2))0 = X2(µ2(S2) − Dα)

(3)



Steady-State

Substrate Balance: dSdt = DS0 − DS + dS

dt

Bacteria Balance: dXdt = DX0 − DX + µ(S)X + kdX

Equilibrium point: dXdt = 0 dS

dt = 0 as t −→∞

dSdt and dX

dt : Accumulation

DS0 and DX0: Diluted Input DS and DX : Diluted Output



Different Approaches

Each differential system under study was characterized by itsunique combination of two of the numerous hypothesized bacterialgrowth functions; our study included application of the Monod andHaldane functions of bacteria growth [d ],

Monod: µ1(S1) = m1S1K1+S1

Haldane: µ2(S2) = m2S2

K2+S2+S22

KI

m1 and m2: Define the maximum attainable speeds of X1 andX2 growth, respectively.

K1 and K2 : Substrate Concentrations at 50 percent ofmaximum specific growth rate(see graph).

KI : Substrate concentration where bacteria growth is reducedto 50 percent of it’s maximum growth rate due to substrateinhibition (see graph).



Bacteria Growth Kinetics

0 4 80

m

S (Substrate Concentration)

Monod Model of Bacteria Growth Kineticsu(S) = (mS)/(K+S)

u(S

)

Approaching Maximum Growth Rate (m)

50% Maximum Growth Rate (m/2)

Substrate Concentration (K) at m/2

Figure : Monod Model for Bacteria Growth Kinetics



Bacteria Growth Kinetics

0 10 20 30 40 50 600

0.1

0.2

0.3

0.4

0.5

0.6


Haldane Model of Bacteria Growth Kinetics u(S) = (mS)/(K+S+(S2)/KI)

u(S

)

50% Maximum GrowthRate of Bacteria

Substrate Concentration (KI) at 50% ofMaximum Growth Rate of Bacteria

Substrate Concentration(K1) at 50% of MaximumGrowth Rate of Bacteria

Figure : Haldane Model for Bacteria Growth Kinetics



Theory Behind Four-Dimensional Systems A and B

Two Hypotheses:

4-Dimensional System A: The growth rates of X1 and X2 areboth increasing functions of added substrate (S1 and S2).

µ1(S1) = m1S1K1+S1

µ2(S2) = m2S2K2+S2

4-Dimensional System B: The growth rate of X1 is anincreasing function of substrate (S1) and the growth rate ofX2 approaches a maximum at a medium substrateconcentration (KI = medium S2 concentration).

µ1(S1) = m1S1K1+S1

µ2(S2) = m2S2

K2+S2+S22

KI



Introduction of Foreign Toxin

S ′1 = −X1k1e−yµ(µ1(S1)) + D(S1in − S1)

X ′1 = X1(e−yµµ1(S1) − Dα)S ′2 = D(S2in − S2) + X1k2e

−yµ(µ1(S1)) − X2k3(µ2(S2))X ′2 = X2(µ2(S2) − Dα)y ′ = D(yin − y)− X2k4µ3(y)

(4)

where y represents the toxin.



Inhibition Caused by Foreign Species

Types of Inhibition

Competitive Inhibition: A foreign species similar in structureto the substrate binds to the enzymes, inhibiting reactionspots.

Noncompetitive Inhibition: A foreign species not necessarilysimilar in structure to the substrate binds to the enzymesand/or enzyme-substrate complexes, preventing completion ofthe reaction.

Uncompetitive Inhibition: A foreign species not necessarilysimilar in structure to the substrate binds to theenzyme-substrate complexes, preventing completion of thereaction.



Effect of a Toxin

0


Inhibition Caused by Foreign Species

Gro

wth

Rat

e o

f B

acte

ria

Maximum Attainable Growth Rateof Bacteria in Presence of

Noncompetitively orUncompetitively Behaving

ForeignSpecies

MaximumAttainable

Growth Rate ofBacteria

Maximum AttainableGrowth Rate of Bacteria in

Presence of Competitively BehavingForeign Species

Figure : Illustrations of Competition by Non-Substrate Species



Five-Dimensional Systems

The four systems differed amongst each other in terms of themodels used to represent X2 growth and toxin consumption.

Four Hypotheses:

5-Dimensional System A: The Monod model is used torepresent the growth rate of X1, the Haldane model is used torepresent the growth rate of X2, and the Monod model is usedto represent the consumption rate of the toxin.

µ(S1) = m1S1K1+S1

µ(S2) = m2S2

K2+S2+S22

KI1

µ(y) = m4yK4+y

5-Dimensional System B: The Monod model is used torepresent the growth rate of X1, the Haldane model is used torepresent both the growth rate of X2 and the consumptionrate of the toxin.

µ(S1) = m1S1K1+S1

µ(S2) = m2S2

K2+S2+S22

KI1

µ(y) = m4y

K4+y+ y2

KI2



Five-Dimensional Systems

The four systems differed amongst each other in terms of themodels used to represent X2 growth and toxin consumption.

Four Hypotheses:

5-Dimensional System C: The Monod model is used torepresent the growth rate of X1, the growth rate of X2, andthe consumption rate of the toxin.

µ(S1) = m1S1K1+S1

µ(S2) = m2S2K2+S2

µ(y) = m4yK4+y

5-Dimensional System D: The Monod model is used torepresent the growth rate of Bacteria 1 and 2, and theHaldane model is used to represent the consumption rate ofthe toxin.

µ(S1) = m1S1K1+S1

µ(S2) = m2S2K2+S2

µ(y) = m4y

K4+y+ y2

KI2



Six-Dimensional System

S ′1 = −X1k1(µ1(S1)) + D(S1(1 + (rv)) − S1)X ′1 = X1(P(µ1(S1) − Dα))S ′2 = D(0− S2) + X1k2(µ1(S1)) − X2k3(µ2(S2))X ′2 = X2(µ2(S2) − Dα)u′ = u(1 − u2 − v2) − 2πv

v ′ = v(1 − u) − v2) + 2πu

(5)

where S1(1 + (rv)) = S1(t) = S1(1 + (r sin(2πt)))and 0 ≤ r ≤ 1.



Solving the 4-Dimensional Systems

Algebraically Determine all Attainable Equilibria

x0 = (S1, X1, S2, X2)

Linearize nonlinear system by formulating Jacobian matricesand solving for respective eigenvalue functions.

A = Df (x0) det(A− Iλ) = 0

Determine the number of potential equilibrium points andexpected behavior of each one according to the calculatedeigenvalues.

Verify Algebraically Determined Discoveries with Illustrationsof Behavior



Possible Results

Hyperbolic Equilibria

All eigenvalues are nonzero values, nor are any ofthem purely imaginary.

Saddle

Stable NodeFocus

Nonhyperbolic Equilibria

One eigenvalue is equal to zero or is purelyimaginary. The system is susceptible to a bifurcationwith small changes in parameter values.

Center



Possible Results



SaddleStable Node

Focus



Center



Possible Results



SaddleStable NodeFocus



Center



Types of Bifurcations

Bifurcation Value = Parameter Value at which Bifurcation Occurs(Dilution Factor, D, acted as the variable parameter value)

Transcritical BifurcationNo change in the number of equilibrium points.Switch in stability of equilibria at bifurcation value.

Fold/Saddle-Node Bifurcation

Change in number of equilibrium points into stableand unstable points.

Pitchfork BifurcationChange in number of equilibrium points from one tothree.

Hopf Bifurcation

Periodic orbits arise from an equilibrium point as itchanges stability at bifurcation value.



Solving 5 and 6 Dimensional Systems

Precise solutions for equilibria were not found algebraically.

Investigation carried out by analysis of the systems usingXPPAUT.

Bifurcations identified and verified algebraically usingSotomayor’s Theorem, then further analyzed using MATLABR2012b when necessary.

Wolfram Mathematica 9.0 used to find any equilibria in themodels, behavior determination of equilibria using MATLABR2012b.



Solving the 5 and 6 Dimensional Systems

When linearizing a six-dimensional system around a periodicorbit of period τ , a total of six Floquet multipliers are solvedfor from a 6 x 6 Monodromy matrix:

λi , 1 < i < 6

Using XPPAUT, find the number of stable Floquet multiplierscorresponding to each periodic orbit of interest.

x ′ = f (x , λ) → x ′ = A(t)x M = A(τ)

Solution: x(t) = x(t + τ), for all t ∈ IR.



Solving the 5 and 6 Dimensional Systems

Hyperbolic periodic orbit:Exactly one Floquet multiplier must be equal to one.

Stable Hyperbolic: Remaining Floquet multipliersare less than one.Unstable Hyperbolic: At least one remaining Floquetmultiplier is greater than one.

Nonhyperbolic periodic orbit:More than one Floquet multiplier is located on the unit circle.



Results from 4-Dimensional System A

Four equilibria were found:

Equilibrium 1: (S1in, 0, S2in, 0)

Always Exists





Equilibrium 2: (S1in, 0, S∗2 (D), X ∗2 (D))

S∗2 (D) = DK2αm2−Dα X ∗2 (D) = 1

k3α(S2in − DK2α

m2−Dα)

Conditions that must hold for point to exist:

m2 > Dα

S2in ≥ ( DK2αm2−Dα)





Equilibrium 2: (S1in, 0, S∗2 (D), X ∗2 (D))

S∗2 (D) = DK2αm2−Dα X ∗2 (D) = 1

k3α(S2in − DK2α

m2−Dα)


m2 > DαS2in ≥ ( DK2α

m2−Dα)





Equilibrium 3: (S∗1 (D), X ∗1 (D), S∗2in(D), 0)

S∗1 (D) = DK1αm1−Dα X ∗1 (D) = 1

k1α(S1in − DK1α

m1−Dα)

S∗2in(D) = S2in + (k2k1 )(S1in − DK1αm1−Dα)


m1 > Dα


S2in ≥ (k2k1 )(S1in − DK1αm1−Dα)






S∗1 (D) = DK1αm1−Dα X ∗1 (D) = 1

k1α(S1in − DK1α

m1−Dα)

S∗2in(D) = S2in + (k2k1 )(S1in − DK1αm1−Dα)



m1−Dα)

S2in ≥ (k2k1 )(S1in − DK1αm1−Dα)





Equilibrium 4: (S∗1 (D), X ∗1 (D), S∗2 (D),X ∗2in(D))

S∗1 (D) = DK1αm1−Dα X ∗1 (D) = 1

k1α(S1in− DK1α

m1−Dα) S∗2 (D) = DK2αm2−Dα

X ∗2in(D) = 1k3α

[S2in − DK2αm2−Dα + (k2k1 )(S1in − DK1α

m1−Dα)]




Conditions that must hold for point to exist (Equilibrium Point 4):

m1 > Dα m2 > Dα S1in ≥ ( DK1αm1−Dα)

S2in + (k2k1 )(S1in − DK1αm1−Dα) ≥ ( DK2α

m2−Dα)




Equilibrium Point 1

0 5 106.5

7

7.5

8

8.5

0 5 100.09

0.095

0.1

0.105

0 5 1040

60

80

100

0 5 10−0.14

−0.12

−0.1

−0.08

0 5 10 15 207.9

8

8.1

8.2

0 5 10 15 20−10

−8

−6

−4x 10

−3

0 5 10 15 2035

40

45

50

0 5 10 15 200

0.02

0.04

0.06

0.08

0.1

Figure : Case 8: Saddle (left) and Case 9: Stable Node (right). In Case8, solutions are moving away from (8, 0, 50, 0) as t −→∞. In Case 9,solutions are moving towards (8, 0, 50, 0) as t −→∞.




Equilibrium Point 2

0 20 40 600

2

4

6

8

10

0 20 40 600.1

0.2

0.3

0.4

0.5

0 20 40 600

20

40

60

80

0 20 40 600.2

0.4

0.6

0.8

1

0 5 10

0.35

0.4

0.45

0.5

0 5 100

0.02

0.04

0.06

0.08

0.1

0 5 1015.5

16

16.5

17

17.5

0 5 100.26

0.27

0.28

0.29

0.3

Figure : Case 7: Saddle (left) and Case 11: Stable Node (right). In Case7, solutions are moving away from (9.0811, 0, 12.9454, 0.28312) ast −→∞. In Case 11, solutions are moving towards (0.4285, 0, 17.0765,0.2638) as t −→∞.




Equilibrium Point 3

0 5 10 150.054

0.056

0.058

0.06

0 5 10 150.96

0.965

0.97

0.975

0.98

0 5 10 150

100

200

300

0 5 10 150

0.5

1

1.5

2

0 5 100.02

0.021

0.022

0.023

0 5 102.5

2.55

2.6

0 5 1066.5

67

67.5

68

0 5 109.6

9.8

10

10.2x 10

−9

Figure : Case 7: Saddle (left) and Case 11: Stable Node (right). In Case7, solutions are moving away from (0.0554, 0.9742, 221.273, 0) ast −→∞. In Case 11, solutions are moving towards 0.02167, 2.577,67.603, 0) as t −→∞.




Equilibrium Point 4

0 2000 4000 60007

7.2

7.4

7.6

7.8

8x 10

−3

0 2000 4000 60000.065

0.07

0.075

0.08

0 2000 4000 6000

1.4

1.6

1.8

2

0 2000 4000 60000.006

0.008

0.01

0.012

0.014

0 50 100 1500.46

0.465

0.47

0.475

0.48

0 50 100 1500.49

0.5

0.51

0.52

0 50 100 15010

20

30

40

0 50 100 1500

0.1

0.2

0.3

0.4

Figure : Case 4: Stable Node (left) and Case 5: Stable Node (right). InCase 4, solutions are moving towards (0.0076, 0.0743, 1.9314, 0.00735)as t −→∞. In Case 5, solutions are moving towards (0.4623, 0.5108,37.5477, 0.00667) as t −→∞.



Set of Parameter Values used to Solve 4-DimensionalSystems

m1 m2 k1 k2 k3 α1.2 1.1 25 250 268 0.5

K1 K2 KI S1in S2in D2 10 40 8 50 variable

Table : Parameter values used to solve 4-Dimensional Systems



Bifurcations found for 4-Dimensional System A

Equilibrium Point 1 (S1in, 0, S2in, 0):

Proved Transcritical bifurcation exists when

D = D∗1 = ( 1α)( m2S2in

K2+S2in) or when

D = D∗2 = ( 1α)( m1S1in

K1+S1in)

When given parameter values are substituted into system, D∗1 = 116

and D∗2 = 1.92.




Equilibrium Point 2: (S1in, 0, S∗2 (D), X ∗2 (D)):


D = D∗ = ( 1α)( m1S1in

K1+S1in)

When the following parameter values are substituted into system,D∗ = 0.1173.

Randomly generated parameter values using MATLAB:

m1 = 1.3038, k2 = 168.9636, K1 = 2.3323, m2 = 0.0253,α = 0.1739, K2 = 19.892, S1in = 9.7551, S2in = 83.0481,k1 = 33.5826




Equilibrium Point 3 (S∗1 (D), X ∗1 (D), S∗2in(D), 0):


D = D∗ = ( 1α)( m2q1

K2+q1) such that

q1 = S2in + (k2k1 )(S1in − DK1αm1−Dα)

When given parameter values are substituted into system,D∗ = 1.877.




Equilibrium Point 4 (S∗1 (D), X ∗1 (D), S∗2 (D), X ∗2in(D)):


D = D∗ = ( 1K2α

)(S2in + (k2k1 )(S1in − K1Dαm1−Dα))(m2 − Dα)

such that

S2in + (k2k1 )(S1in − K1Dαm1−Dα) = K2Dα

m2−Dα

A Hopf Bifurcation test was performed algebraically and usingMATLAB, but no results were generated.




7

8

9

10

11

12

13

S1

1.7 1.8 1.9 2 2.1 2.2D

Figure : Illustration of Transcritical Bifurcations found for Monod Modelusing given parameter values: D∗ = 1.833, D∗ = 1.92, D∗ = 1.877.



Results from 4-Dimensional System B

Up to six equilibria were found:

Equilibrium 1: (S1in, 0, S2in, 0)

Always Exists






S∗1 (D) = DK1αm1−Dα X ∗1 (D) = 1

k1α(S1in − DK1α

m1−Dα)

S∗2in(D) = S2in + X ∗1 (D)k2α


m1 > Dα


S2in + (k2k1 )(S1in − DK1αm1−Dα) ≥ 0






S∗1 (D) = DK1αm1−Dα X ∗1 (D) = 1

k1α(S1in − DK1α

m1−Dα)

S∗2in(D) = S2in + X ∗1 (D)k2α



m1−Dα)

S2in + (k2k1 )(S1in − DK1αm1−Dα) ≥ 0





Equilibrium 3: (S1in, 0, S1∗2 (D), X 1

2 (D))

S1∗2 (D) = (KI

2y )[(1− y) + ((1− y)2 − (4K2KI

)(y2))1/2]

X 12 (D) = 1

k3α(S2in − S1∗

2 (D))

y = Dαm2

Conditions that must hold for point to exist:Amputate my foot...





Equilibrium 4: (S1in, 0, S2∗2 (D), X 2

2 (D))

S2∗2 (D) = (KI

2y )[(1− y)− ((1− y)2 − (4K2KI

)(y2))1/2]

X 22 (D) = 1

k3α(S2in − S2∗

2 (D))

y = Dαm2

Conditions that must hold for point to exist:End world hunger...





Equilibrium 5: (S∗1 (D), X ∗1 (D), S1∗2 (D), X 1∗

2 (D))

S∗1 (D) = DK1αm1−Dα X ∗1 (D) = 1

k1α(S1in − DK1α

m1−Dα)

S1∗2 (D) = (KI

2y )[(1− y) + ((1− y)2 − (4K2KI

)(y2))1/2]

X 1∗2 (D) = ( 1

k3α)(S∗2in(D)− S1∗

2 (D))

y = Dαm2

Conditions that must hold for point to exist:Give up first born son...





Equilibrium 6: (S∗1 (D), X ∗1 (D), S2∗2 (D), X 2∗

2 (D))

S∗1 (D) = DK1αm1−Dα X ∗1 (D) = 1

k1α(S1in − DK1α

m1−Dα)

S2∗2 (D) = (KI

2y )[(1− y)− ((1− y)2 − (4K2KI

)(y2))1/2]

X 2∗2 (D) = ( 1

k3α)(S∗2in(D)− S2∗

2 (D))

y = Dαm2

Conditions that must hold for point to exist:Amputate my other foot...




Equilibrium Point 1

0 5 102

4

6

8

0 5 100.1

0.2

0.3

0.4

0.5

0 5 1040

60

80

100

120

0 5 100

0.02

0.04

0.06

0.08

0.1

0 10 20 30 406.5

7

7.5

8

0 10 20 30 400

0.02

0.04

0.06

0.08

0.1

0 10 20 30 4045

50

55

60

0 10 20 30 400

0.02

0.04

0.06

0.08

0.1

Figure : Case 5: Saddle (left) and Case 8: Stable Node (right). In Case5, solutions are moving away from (8, 0, 50, 0) as t −→∞. In Case 8,solutions are moving towards (8, 0, 50, 0) as t −→∞.




Equilibrium Point 2

0 50 100 150 2000.16

0.165

0.17

0.175

0.18

0 50 100 150 2000.95

1

1.05

1.1

0 50 100 150 200102

104

106

108

110

0 50 100 150 2000

0.02

0.04

0.06

0.08

0.1

0 5 10 15 200.02

0.022

0.024

0.026

0.028

0.03

0 5 10 15 200.069

0.07

0.071

0.072

0.073

0 5 10 15 200

20

40

60

0 5 10 15 200.1

0.2

0.3

0.4

0.5

Figure : Case 2: Stable Node (left) and Case 8: Saddle (right). In Case2, solutions are approaching (0.1672, 1.059, 110.474, 0) as t −→∞. InCase 8, solutions are moving away from (0.0272, 0.07557, 42.014, 0) ast −→∞.




Equilibrium Points 3 and 4

0 5 10 15 200

2

4

6

8

10

0 5 10 15 200

0.5

1

1.5

0 5 10 15 200

20

40

60

0 5 10 15 200

0.01

0.02

0.03

0.04

0 100 200 300 4002.2

2.4

2.6

2.8

3

0 100 200 300 4000

0.02

0.04

0.06

0.08

0.1

0 100 200 300 4000.4

0.5

0.6

0.7

0 100 200 300 4000.18

0.2

0.22

0.24

Figure : Equilibrium Point 3 Case A6: Saddle (left) and Equilibrium Point4 Case C4: Stable Node (right). In EqPt 3 Case A6, solutions are movingaway from (8.6266, 0, 7.08424, 0.03633) as t −→∞. In EqPt 4 Case C4,solutions are moving towards (3.0224, 0, 0.4495, 0.1948) as t −→∞.




Equilibrium Point 6

0 50 100 150 2000.4

0.41

0.42

0.43

0 50 100 150 2000.0125

0.013

0.0135

0.014

0 50 100 150 2000.3

0.305

0.31

0.315

0.32

0.325

0 50 100 150 2000.42

0.43

0.44

0.45

Figure : Equilibrium Point 6 Case B5: Stable Node. In EqPt 6 Case B5,solutions are moving towards (0.4116, 0.014, 0.3016, 0.4401) ast −→∞. No random parameters were generated that agreed with thepredicted results from any cases of Equilibrium Point 5.



Bifurcations found from 4-Dimensional System B

Equilibrium Point 1 (S1in, 0, S2in, 0):


D = D∗1 = ( 1α)( m1S1in

K1+S1in) or when

D = D∗2 = ( 1α)( m2S2in

K2+S2in+S22inKI

)

When given parameter values are substituted into system,D∗1 = 1.92 and D∗2 = 0.898.

A Pitchfork Bifurcation test was performed algebraically and

using MATLAB in the case where K2 =S22inKI

, but no resultswere generated.




Equilibrium Point 2: (S∗1 (D), X ∗1 (D), S∗2in(D), 0):


D = D∗ = ( 1α)( m2x

K2+x+ x2

KI

) such that

x = S2in + (k2k1 )(S1in − K1Dαm1−Dα)

When the following parameter values are substituted intosystem, D∗ = 0.4305.Randomly generated parameter values using MATLAB:

m1 = 1.6362, k2 = 250.9017, K1 = 1.419,m2 = 1.109, α = 0.2349, K2 = 3.7185,S1in = 10.7751, S2in = 66.2991, k1 = 15.396,KI = 24.1864, k3 = 268.

A Pitchfork Bifurcation test was performed algebraically and usingMATLAB in the case where K2 = x2

KI, but no results were

generated.Christina Berti Modeling and Analysis of Anaerobic Digestion in a Bioreactor



Equilibrium Point 3: (S1in, 0, S1∗2 (D), X 1

2 (D)):


D = D∗ = ( 1α)( m1S1in

K1+S1in)

When the following parameter values are substituted into system,D∗ = 0.4993.Randomly generated parameter values using MATLAB:

m1 = 0.6323, k2 = 101.7029, K1 = 1.6611, m2 = 1.824,α = 0.9722, K2 = 8.0747, S1in = 5.4899, S2in = 77.3912,k1 = 5.0144, k3 = 268, KI = 25.164.




Equilibrium Point 4: (S1in, 0, S2∗2 (D), X 2

2 (D)):


D = D∗ = ( 1α)( m1S1in

K1+S1in)

When the following parameter values are substituted into system,D∗ = 0.2567.Randomly generated parameter values using MATLAB:

m1 = 0.2811, k2 = 51.6183, K1 = 1.0457, m2 = 1.3676,α = 0.9074, K2 = 13.2164, S1in = 5.0548, S2in = 6.2729,k1 = 8.4958, k3 = 268, KI = 49.88.




Equilibrium Point 5: (S∗1 (D), X ∗1 (D), S1∗2 (D), X 1∗

2 (D))

Equilibrium Point 6: (S∗1 (D), X ∗1 (D), S2∗2 (D), X 2∗

2 (D))

Hopf Bifurcation tests were performed for both equilibria, but nodata was generated.




m1 m2 k1 k2 k3 α1 3.2 30 215 100 0.3

K1 K2 KI1 KI2 S1in S2in D7 200 500 400 40 5 variable

m4 k4 K4 yin µ1 5 1 5 0.06

Table : Parameter values used for solving 5-Dimensional Systems



5-Dimensional System A

Combination of functions applied:

Monod model for the growth of Bacteria 1

µ(S1) = m1S1K1+S1

Haldane model for the growth of Bacteria 2

µ(S2) = m2S2

K2+S2+S22

KI1

Monod model for the consumptions of Toxin

µ(y) = m4yK4+y



Bifurcations found when Solving 5-Dimensional System A

30

32

34

36

38

40

42

S1

1.7 1.8 1.9 2 2.1 2.2 2.3 2.4D

Figure : Illustration of Bifurcations found for 5-Dimensional System A,Saddle-Node Bifurcation point: D∗ = 2.30067.



Stable Periodic Orbit of 5-Dimensional System A

33

33.5

34

34.5

35

35.5

0.5

0.6

0.7

0.8

0.930

40

50

60

Figure : Presence of Stable Periodic Orbit in 5D System A. Though aHopf bifurcation exists in this system from which stable periodic orbitsare generated, the nonhyperbolic equilibrium point corresponding to thedetected Hopf bifurcation displayed a negative solution for X2. Unstableequilibria found: (40, 0, 5, 0, 5), (40, 0, 48.24, -1.44, -0.61). Wegraphed only 3-D projections of the solutions: S1, X1, and S2.






µ(S1) = m1S1K1+S1


µ(S2) = m2S2

K2+S2+S22

KI1

Haldane model for the consumptions of Toxin

µ(y) = m4y

K4+y+ y2

KI2



Bifurcations found when Solving 5-Dimensional System B

0

10

20

30

40

50

60

S1

-0.5 0 0.5 1 1.5 2 2.5D1

Figure : Illustration of Bifurcations found for 5-Dimensional System B,Saddle-Node Bifurcation point: D∗ = 2.03477.



Stable Periodic Orbit of 5-Dimensional System B

3333.5

3434.5

3535.5

0.5

0.6

0.7

0.8

0.935

40

45

50

55

3234

3638

4042

0.5

0.6

0.7

0.8

0.90

10

20

30

40

50

60

Figure : Proof of Stable Periodic Orbit in 5D System B, D∗ = 2.035.Though a Hopf bifurcation exists in this system from which stableperiodic orbits are generated, the nonhyperbolic equilibrium pointcorresponding to the detected Hopf bifurcation displayed a negativesolution for X2. Unstable equilibria found using Wolfram Mathematica9.0: (40, 0, 5, 0, 5) and (34.22, 0.64, 48.24, -0.06, 5.12) whenD∗ = 2.03477.



Results from 5-Dimensional System C



µ(S1) = m1S1K1+S1


µ(S2) = m2S2K2+S2

Monod model for the consumptions of Toxin

µ(y) = m4yK4+y



Bifurcations found when Solving 5-Dimensional System C

0

10

20

30

40

50

60

S1

-0.5 0 0.5 1 1.5 2 2.5D1

Figure : Illustration of Bifurcations found for 5-Dimensional System C,D∗ = 2.30567 at saddle-node bifurcation point.



Results from 5-Dimensional System D



µ(S1) = m1S1K1+S1


µ(S2) = m2S2K2+S2

Haldane model for the consumptions of Toxin

µ(y) = m4y

K4+y+ y2

KI2



Bifurcations found when Solving 5-Dimensional System D

0

10

20

30

40

50

60

S1

-0.5 0 0.5 1 1.5 2 2.5D1

Figure : Illustration of Bifurcations found for 5-Dimensional System D,D∗ = 2.305139 at saddle-node bifurcation point.



Parameter values used when solving 6-Dimensional Systems

m1 = 3, m2 = 0.75

k1 = 10, k2 = 5.2254, k3 = 40

K1 = 0.5, K2 = 0.15, KI = 1

S1 = 6, S2 = 0, α = 1, p = 1, r = 0.5.



6-Dimensional System

Functions applied:


µ(S2) = m2S2

K2+S2+S22

KI

Periodic inflow of S1in of amplitude ”r”, in the hopes ofgenerating oscillating solutions.




Bifurcation Results of 6-Dimensional System

0

1

2

3

4

5

6

7

S2

0 0.05 0.1 0.15 0.2 0.25 0.3 0.35 0.4 0.45 0.5D

0.064

0.065

0.066

0.067

0.068

0.069

0.07

X2

0 0.2 0.4 0.6 0.8r

Figure : Saddle-Node Bifurcation of Periodic Orbits, D∗ = 0.4196 atsaddle-node bifurcation point.

Surprising Results: Saddle-Node Bifurcation of Periodic Orbits





00.05

0.10.15

0.2

0.570.580.590.60.610.620

0.2

0.4

0.6

0.8

1

1.2

1.4

1.6

1.8

0

0.1

0.2 0.580.590.60.610.620.63

0

0.1

0.2

0.3

0.4

0.5

0.6

0.7

0.8

Figure : Periodic Orbits of Opposite Stability converging at Saddle-NodeBifurcation Point. Green: stable, Blue: unstable.





0.02

0.025

0.03

0.035

0.04

0.045

0.05

0.055

0.06

0.065

0.07

0.580.585

0.590.595

0.60.605

0.610.615

0

0.05

0.1

0.15

0.2

Figure : Stability statuses of Periodic Orbits 6 and 14. Left: Orbit 6,D∗ = 0.2313 (stable), Right: Orbit 14, D∗ = 0.3937 (unstable).



Equilibria Results of 6-Dimensional System

0.0417

0.0417

0.0418

0.0418

0.59580.5958

0.59580.5958

0

0.5

1

1.5

2

2.5

Figure : Unstable Equilibria found for system using specific set ofparameter values. (0.0417725, 0.595823, 0.0690168, 0.0761099, 0, 0)and (0.0417725, 0.595823, 2.17338, 0.0235007, 0, 0) when D∗ = 0.2313.



Further Research to be Done

6-Dimensional System: Find heteroclinic orbits connectingtwo unstable periodic orbits rather than solely orbits ofopposite stability.

5-Dimensional Systems:

Toxin has a profound influence on the growth of X2.Toxin acts immediately on S2 accumulation ratherthan indirectly through S1 reaction by X1.

Analysis of environmental conditions to find the most idealsetting for digestion. Provide flexibility for maximum bacteriagrowth velocity parameters: m1 , m2, and m4.



References

[1] B. Benyahia, T. Sari, B. Cherki, J. Harmand, “Bifurcation andStability Analysis of a Two Step Model for Monitoring AnaerobicDigestion Process”. Journal of Process Control 22 (2012).

[2] O. Bernard and J. Hess, “Design and Study of a RiskManagement Criterion for an Unstable Anaerobic WastewaterTreatment Process”. French Research Institute of ComputerScience and Automatic Control. (2007).

[3] R. Cooke, “Wastewater Treatment Methods and Disposal”@ONLINE. July 2013. URL =http://water.me.vccs.edu/courses/ENV149/methods.htm.

[4] K. Cornely and C. Pratt, “Essential Biochemistry”. 2ndedition New York: Wiley. (2011).



References

[5] B. Ermentrout, “Simulating, Analyzing, and AnimatingDynamical Systems: A Guide to XPPAUT for Researchers andStudents”. Philadelphia: Society for Industrial and AppliedMathematics. (2002).

[6] M. Gerber, R. Span, “An Analysis of Available MathematicalModels for Anaerobic Digestion of Organic Substances forProduction of Biogas”. International Gas Union ResearchConference. (2008).

[7] Y. Li and J.S. Muldowney, “On Bendixson’s Criterion*”.Journal of Differential Equations 106, 27-39 (1993).



References

[8] J. Rebaza, “A First Course in Applied Mathematics”. NewJersey: Wiley. (2012).

[9] J. Senisterra, “Dynamical Analysis of the Anaerobic DigestionModel as Proposed by Hess and Bernard”. Print.

[10] M. Weederman, “Analysis of a Model for the Effects of anExternal Toxin on Anaerobic Digestion”. Mathematical Biosciencesand Engineering 9 (2012).


modeling and analysis of anaerobic digestion in a bioreactor

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