About dynamical system modelling Illustration in micro-biology - - PowerPoint PPT Presentation

about dynamical system modelling
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About dynamical system modelling Illustration in micro-biology - - PowerPoint PPT Presentation

About dynamical system modelling Illustration in micro-biology Alain Rapaport alain.rapaport@montpellier.inra.fr Research School AgreenSkills Toulouse, 29-31 october 2014 Contents Modeling bacterial growth in batch and continuous cultures


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About dynamical system modelling

Illustration in micro-biology Alain Rapaport alain.rapaport@montpellier.inra.fr

Research School AgreenSkills

Toulouse, 29-31 october 2014

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Contents

◮ Modeling bacterial growth in batch and continuous cultures ◮ Simple representations of space ◮ Questions related to biodiversity ◮ Modeling and analysis of growth inhibition ◮ Facing models with experimental data

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Bacterial growth in batch

time number of cells per volume

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Bacterial growth in batch

time number of cells per volume

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The logistic growth (Verhulst, 1838)

→ x(t) = xmax 1 + e−rt

  • xmax

x0 − 1

  • Equivalent formulation: x(·) is solution of the differential equation

dx dt = rx

  • 1 −

x xmax

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Jacques Monod’s experiments (1930)

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Jacques Monod’s experiments (1930)

does not fit the logistic curves...

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Mathematical modeling

Hypothesis H0. There exists y s.t. x + ys = m = cste Hypothesis H1a. dx dt = µsx ⇒ dx dt = µm y

  • r

x

  • 1 − x

m

  • Hypothesis H1b. dx

dt = µ(s)x where µ(·) is not necessarily linear:

S µ

How to identify with accuracy this specific growth curve?

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The chemostat device

feed bootle culture vessel collection vessel pump pump

Monod 1950 – Novick & Szilard 1950

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Dynamical modelling

◮ Mechanics: 2nd motion law:

mass × acceleration =

  • forces

◮ Bio-reaction kinetics: mass balance (for biotic and abiotic):

mass variation = inputs − outputs + growth − death

  • for biotic only
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Dynamical modelling

◮ chemical kinetics:

  • Ex. : αA + βB −

→ P = AαBβ ⇒ v = dP dt = k[A]α[B]β

◮ microbial kinetics:

  • Ex. : B + kS −

→ B + B − → v = dB dt = ✘✘✘

✘ ❳❳❳ ❳

[B][S]k growth is not simply a matter of matching bacteria with molecules of substrate...

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The mathematical model (in concentrations)

   

dx dt ds dt

   

= µ(s)x −µ(s) y x + −Q V x Q V (sin − s) growth dilution

  • Remark. Q = 0 ⇒ x + ys = constant

Simplification and notations. y = 1 ˙= d dt D = Q V

  • ˙

s = −µ(s)x + D(sin − s) ˙ x = µ(s)x − Dx

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SLIDE 13

Determination of equilibria

˙ s = −µ(s)x + D(sin − s) ˙ x = µ(s)x − Dx ⇒ x⋆ = 0 s⋆ = sin

  • r

µ(s⋆) = D x⋆ = sin − s⋆ wash-out positive equilibrium

D Sin S µ S*

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Null-clines

˙ s = −µ(s)x + D(sin − s) ˙ x = µ(s)x − Dx

s x

˙ x = 0 ˙ s = 0

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Vector field

˙ s = −µ(s)x + D(sin − s) ˙ x = µ(s)x − Dx

s x

˙ x = 0 ˙ s = 0

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Phase portrait

˙ s = −µ(s)x + D(sin − s) ˙ x = µ(s)x − Dx

s x

˙ x = 0 ˙ s = 0

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Back to experiments

For each dilution rate D, one obtains a steady state s⋆ with µ(s⋆) = D:

6 2 1 3

s* s* s* s* D D D

2 3 4 5

D6 µ

1

D D s*s* s s

in 5 4

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The Monod’s law

S µ K µmax µ

max

2

µ(s) = µmaxs K + s :

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The chemostat model

˙ s = −µ(s)x + D(sin − s) ˙ x = µ(s)x − Dx Ecology of mountain lakes Industrial bioreactors

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For various increasing dilution rates

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For various increasing input concentrations

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About conversion yield at equilibrium

◮ The mathematical model of the chemostat predicts that the

substrate concentration at equilibrium is independant of the input concentration sin (provided that µ(sin) > D).

◮ Micro-biologists report that this property is not verified when

the tank is not homogeneous or in natural ecosystems such as soil ecosystems. Question: What is the influence of a spatial repartition on output substrate concentration at steady state?

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The paradigm of simple representations of space

  • input

input hot−spots microbial

  • utputs

roots main real soil input input

  • utput
  • utput

interconnected chemostats drainage

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Study of some simple spatial configurations

  • Q

Q Q Q Q Q Q = Q + Q 1 2 Q2 V 1 V 2 V 1 V 2 V 1 diffusion

OR

serial parallel

with V = V1 + V2

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Comparison of performances at steady-state

in in in

S =2.5 S =3 S =4 r Sout Sout S = 1.7

in

S = 1.3 S = 1.1

in in

S = 1.5

in

d

in

S =2 S = 2

in

S = 2.2

in in

S =1.5

in

S =1.75

serial parallel

0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1.0 0.6 0.8 1.0 1.2 1.4 1.6 1.8 2.0 0.0 0.2 0.4 0.6 0.8 1.0 1.2 1.4 1.6 1.8 2.0 0.80 0.85 0.90 0.95 1.00 1.05 1.10 1.15 1.20 1.25

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Roles of spatial structure and diffusion

Message 1. For any monotonic µ(·), there exists a threshold ¯ sin such that

◮ for sin > ¯

sin, the serial configuration is more efficient,

◮ for sin < ¯

sin, the parallel configuration is more efficient, Message 2. For the parallel configurations,

◮ for sin > ¯

sin, the map d → s⋆

  • ut(d) is decreasing,

◮ for sin < ¯

sin, the map d → s⋆

  • ut(d) admits an unique minimum

for d⋆ < +∞ Furthermore, there exists another threshold sin < ¯ sin s.t. d⋆ = 0 for sin < sin.

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Questions related to microbial biodiversity

◮ How to model microbial competition? ◮ Can a biodiversity be maintened? ◮ Does spatial structures impact biodiversity? ◮ Is biodiversity is in favor of bioconversion yielding?

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Having two species in the chemostat

˙ s = −µ1(s)x1 − µ2(s)x2 + D(sin − s) ˙ x1 = µ1(s)x1 − Dx1 ˙ x2 = µ2(s)x2 − Dx2 Equilibria: wash-out species 1 only species 2 only species coexistence

  

sin

     

s⋆

1

sin − s⋆

1

     

s⋆

2

sin − s⋆

2

  

would require µ1(s⋆) = µ2(s⋆) = D non generic condition!

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Species competition

D s species 1 species 2

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Species competition

species 1 species 2

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Species competition

D s

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Species competition

species 1 s p e c i e s 2

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Species competition

D s

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Species competition

species 1 s p e c i e s 2

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The Competitive Exclusion Principle

Assume that all µi(·) are increasing Let Λi(D) = {s ≥ 0 | µi(s) > D} and define the break-even concentrations λi(D) = inf Λi(D)

D s

Principle.

◮ Generically, there is at most one species at equilibrium. ◮ The equilibrium (if it exists) with the species that possesses

the smallest break-even concentration is the only attractive equilibrium.

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About niches and over-yielding

Consider two strains: and the spatial structure:

µ

1

µ2 D* S

Q Q S in Sin S1 S2 + S out

1 2

α (1−α) r (1−r) V = V V = V

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Bioconversion and over-yielding

Q Q S in Sin α S1 S2 + S out (1−α)Q Q S in Sin α S1 S2 + S out (1−α)Q Q S in Sin α S1 S2 + S out (1−α)Q Q S in Sin α S1 S2 + S out (1−α)

What is the most efficient configuration?

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Overyielding at equilibrium

For increasing µ(·), define the conditional break-even conc. ¯ λ(D) =

  • µ−1(D)

if D ≤ µ(sin) sin if D > µ(sin) and consider F(α, r) := α¯ λ

α

r D

  • + (1 − α)¯

λ

1 − α

1 − r D

  • Definition : There is over-yielding when

G(α, r) := α¯ λ1

α

r D

  • +(1−α)¯

λ2

1 − α

1 − r D

  • < min(F1(α, r), F2(α, r))
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Overyielding at equilibrium

Assume there exists D⋆ such that ¯ λ1(D⋆) = ¯ λ2(D⋆).

  • Proposition. If µ(·) is concave, then there exist configurations

(α, r) such that α r D < D⋆ < 1 − α 1 − r D < min(µ1(sin), µ2(sin)) that exhibit over-yielding.

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The high diversity case

DGGE

  • With tools of molecular biology (fingerprints), biologists observe

bioreactors with substrate and biomass concentrations at quasi steady state, while the diversity of the biomass is high and its composition is constantly evolving...

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The chemostat model with “many” species

˙ s = −

n

  • j=1

µj(s)xj + D(sin − s) ˙ xi = µi(s)xi − Dxi (i = 1 · · · n) Numerical simulations with n large (over hundred):

◮ random drawing of functions µi(·) among a family of Monod

functions

◮ random drawing of the initial composition of the biomass

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The chemostat model with “many” species

Growth curves S

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The chemostat model with “many” species

substrate total biomass 50 100 150 200 250 300 350 400 450 500 0.0 0.2 0.4 0.6 0.8 1.0 1.2

time

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The chemostat model with “many” species

Let b =

n

  • i=1

xi et pi = xi b ⇒

  • ˙

s = −˜ µ(s, p)b + D(sin − s) ˙ b = ˜ µ(s, p)b − Db with ˜ µ(s, p) =

n

  • i=1

piµi(s) and ˙ pi = (µi(s) − ˜ µ(s, p)) pi

substrate t

  • t

a l b i

  • m

a s s

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Splitting species in two groups

m efficient packet of

µ D λ λj

less efficient species species

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The fast dynamics

            

˙ s = −¯ µ(s)¯ x −

n

  • j=m+1

µj(s)xj + D(sin − s) ˙ ¯ x = ¯ µ(s)¯ x − D¯ x ˙ xj = µj(s)xj − Dxj (j = m + 1 · · · n) Competitive Exclusion Principle:

        

lim

t→+∞ s(t) = s⋆

lim

t→+∞ ¯

x(t) = x⋆ = sin − s⋆ lim

t→+∞ xj(t) = 0

(j = m + 1 · · · n)

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The slow dynamics

for i = 1 · · · m let write µi(s) = ¯ µ(s) + ενi(s) ⇒ ˙ pi = ε

 νi(s⋆) −

m

  • j=1

pjνj(s⋆)

  pi

Explicit solution: pi(t) = pi(0)eAiεt

m

  • j=1

pj(0)eAjεt with Aj = νj(s⋆)

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SLIDE 48

Properties of the slow dynamics

◮ Each proportion t → pi(t) is

  • either increasing up to a time Ti and then decreasing,
  • either increasing for any t,
  • either decreasing for any t.

◮ When ε → 0 then Ti → +∞ for all species excepted one. ◮ The “reactional entropy”

E(t) :=

  • j

µj(¯ s)pj(t) is increasing toward D.

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Concluding remark The Competitive Exclusion Principle provides only an asymptotic result.

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Substrate inhibition D Monod Haldane S

µ(S) = µmaxS K + S µ(S) = ¯ µS K + S + S2/Ki

e.g. J. F. Andrews, A mathematical model for the continuous culture of microorganisms utilizing inhibitory substrates, Biotech. Bioengrg., 10 (1968), pp. 707-723.

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The chemostat model with the Monod law

µ(S) = µmaxS K + S (S) D S* S

S S* B

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The chemostat model with the Haldane law

µ(S) = ¯ µS K + S + S2/Ki

D S*

2

S S*

1

S S*

1

S*

2

B

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Possible behaviors

Courbe de croissance S

D > max

s∈[0,Sin] µ(s)

1 equilibrium: wash-out

Courbe de croissance S

µ(sin) < D < max

s∈[0,Sin] µ(s)

3 equilibria : bi-stability

Courbe de croissance S

D < µ(sin) 2 equilibria : stability

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Operating real bioreactors

INRA LBE (Narbonne) “It happens that some additional species need to be added for the bioprocess to start, but these additional species are no longer present at steady state.”

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Generalization of the Exclusion Principle

Hypothesis: Ei(D) = {s > 0 | µi(s) > D} is an interval (λ−

i (D), λ+ i (D))

s D + λ − µ + λ λ − D µ 8 λ =+ s

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Generalization of the Exclusion Principle

Proposition. Let Q(D) =

  • i

Ei(D) =

  • i

(λ−

i (D), λ+ i (D)) ◮ There is generically competitive exclusion. ◮ There are as many species that could win the competition as

the number of connected components of the set Q(D).

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Adding another species

˙ s = −µ1(s)x1 − µ2(s)x2 + D(sin − s) ˙ x1 = µ1(s)x1 − Dx1

  • f Haldane type

˙ x2 = µ2(s)x2 − Dx2

  • f Monod type

S

case 1

S

case 2

S

case 3

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About the bio-augmentation

0.0 0.2 0.4 0.6 0.8 1.0 1.2 1.4 1.6 1.8 2.0 0.00 0.01 0.02 0.04 0.05 0.06

D Sin

max

[0,sin] µ < µ(sin) < D

⇒ the wash-out is attractive... What is the effect of adding strain “blue” or “green” (in small quantity)?

0.0 0.2 0.4 0.6 0.8 1.0 1.2 1.4 1.6 1.8 2.0 0.00 0.01 0.02 0.04 0.05 0.06

D Sin

0.0 0.2 0.4 0.6 0.8 1.0 1.2 1.4 1.6 1.8 2.0 0.00 0.01 0.02 0.04 0.05 0.06

D Sin

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SLIDE 59

Red species alone

100 200 300 400 500 600 700 800 900 1000 0.0 0.2 0.4 0.6 0.8 1.0 1.2 1.4 1.6 1.8 2.0

substrate time

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Adding green strain

100 200 300 400 500 600 700 800 900 1000 0.0 0.5 1.0 1.5

substrate time

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Phase portrait

s s* x alone red with green strain condition initial

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Adding blue strain

100 200 300 400 500 600 700 800 900 1000 0.0 0.2 0.4 0.6 0.8 1.0 1.2 1.4 1.6 1.8 2.0

substrate time

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Toward the “biological control”

Let E = (λ−, λ+) for the red species. Let λ be the break-even conc. for the additional species.

◮ λ ∈ E: stabilization of the red species ◮ λ /

∈ E: possibility of invasion

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Playing with interconnections

Consider D = Q V s.t. µ(sin) < D < max

s∈[0,Sin] µ(s).

  • Q

Q Q Q Q Q Q = Q + Q 1 2 Q2 V 1 V 2 V 1 V 2 V 1

OR

serial parallel

with V = V1 + V2

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Serial and parallel patterns

Q Q V 1 V 2

V1 < V ⇒ Q V1 > D , V2 < V ⇒ Q V2 > D The wash-out equilibrium is attractive in both tanks.

Q Q Q = Q + Q 1 2 Q2 V 1 V 2 1

Q1 V1 < µ(sin) ⇒ Q2 V2 > µ(sin) (and vice-versa) The wash-out equilibrium is attractive in at least one tank.

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The buffered pattern

V 1 V 2 Q1 Q2 Q2 Q Q = Q + Q 1 2 Q Q V

with V = V1 + V2

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The buffered chemostat

V 1 V 2 Q1 Q2 Q2 Q Q = Q + Q 1 2

Two parameters: (α, r) with Q2 V2 = αD et V1 = rV

◮ buffer tank: classical chemostat

⇒ unique positive equ. if αD < µ(sin)

◮ main tank: chemostat with double inputs:

µ(s⋆

1) = D

r − αD

  • 1 − 1

r

sin − s⋆

2

sin − s⋆

1

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A graphical characterization

1

S D S µ

  • r

1

S D S µ

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A matricial generalization

˙ S = −1 y R(S, X) + MS + DSin ˙ X = R(S, X) + MX + DX in where D is a diagonal matrix, and M is a compartmental matrix:

  • i. Mii ≤ 0 for any indice i,
  • ii. Mij ≥ 0 for any indices

i = j,

  • iii. for any indice i, one has
  • j

Mij ≤ 0

Q d Q d

ik ji ij ik

Qi Qi V

i

Vj

k

V

  • ut

in

that fulfills the Kirchoff law of mass conservation.

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About experimental identification

Principle: one fixes a dilution rate Di and “wait” for the corresponding equilibrium s⋆

i .

6 2 1 3

s* s* s* s* D D D

2 3 4 5

D6 µ

1

D D s*s* s s

in 5 4

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About experimental identification

Principle: one fixes a dilution rate Di and “wait” for the corresponding equilibrium s⋆

i .

6 5

D

2

s* s

in 6 5 4 3

D D D

2 3 4 1

D D µ µ

?

1

s s* s* s* s* s*

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Change of view point

One fixes a reference value s⋆ and look for stabilizing the chemostat at s = s⋆ adjusting the dilution rate D.

  • pen loop:

dilution chemostat model input

  • utput

concentration substrate rate

closed loop:

controller chemostat model

  • utput

input

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Adaptive controller

Fix ¯ D as a value of the dilution rate.

◮ Step 1: With the feedback law

D(¯ D, s) = ¯ D + G1(s⋆ − s) the closed-loop chemostat admits an equilibrium with s = ¯ s such that µ(¯ s) = D(¯ D,¯ s).

◮ Step 2: With the adaptation law for ¯

D: d ¯ D dt = G2(s⋆ − s) the closed-loop chemostat admits the equilibrium with s = s⋆ and ¯ D⋆ = µ(s⋆).

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Facing the real life...

controller input

  • utput

chemostat model controller input

  • utput
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The auxostat

see Gostomski, Muhlemann, Lin, Mormino & Bungay. Auxostats for continuous culture research. Journal of Biotechnology, 1994

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Simulations with noisy measurements

G1 = 1, G2 = 1 :

5 10 15 20 25 30 35 40 45 50 0.0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9

Concentration en substrat mesurée temps

5 10 15 20 25 30 35 40 45 50 0.0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8

Taux de dilution appliqué temps

G1 = 10, G2 = 10:

5 10 15 20 25 30 35 40 45 50 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9

Concentration en substrat mesurée temps

5 10 15 20 25 30 35 40 45 50 0.0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1.0

Taux de dilution appliqué temps

Dilemma accuracy/speed

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General comments

◮ A model is ONE representation of the reality. There exist

modelS.

◮ A model is generally built for a question of interest

⇒ different representations for the same object.

◮ A model is a representation ALONG WITH HYPOTHESES

⇒ a validity domain is associated to a model.

◮ Simple models can represent complex systems, depending on

the scale of representation.

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SLIDE 78

Inputs-outputs models

model inputs

  • utputs

inputs model

  • utputs

known known unknown prediction known unknown known identification unknown known known control

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SLIDE 79

Further reading

◮ J. Haefner. Modeling Biological Systems, Springer 2005. ◮ C. Taubes. Mathematical Differential Equations in Biology,

Cambridge 2008.

◮ L. Edelstein-Keshet. Mathematical Models in Biology,

SIAM 2005.

◮ N. Britton. Essential Mathematical Modelling, Springer

2005.