Chapter 2: Growth & Decay Introduction

Introduction & BasicsGrowth and Decay

Linear & Non-Linear Interaction Models

Introduction & Simple ModelsLogistic Growth ModelsLogistic Growth with Constant CullingLogistic Growth Revisited: The Chemostat

Chapter 2:

Growth & Decay

Dr. Martin Crane CA659Mathematical Methods/Computational Science




Introduction

In this chapter, model biological systems whose population isso large or where growth is so fine-grained that continuity canbe assumed.

These continuous models give rise to differential equations.

As a consequence, there is no natural time delay in the modelsand hence none of the over-compensation of differenceequations.

Extend the logistic growth equation model above & add aculling term.

Develop the ideas from this to a non-linear model for thechemostat.





Simple Models

For cell pop’n whose env’t doesn’t alter with time (i.e. ∞nutrients/ space), can assume that rate of increase of cells ∝number of cells.

Mathematically, can put this as:

dN

dt= rN (2.1)

for some constant of proportionality, r .

r measures average growth rate per unit time per individual(i.e. “2.4 children per person per life” number).

This equation (a.k.a.the law of organic growth) was namedafter Malthus1 a gloomy philosopher called who predicted (ca1798) that world pop’n would soon outgrow its resources.

1T.R. Malthus (1766-1834)Dr. Martin Crane CA659Mathematical Methods/Computational Science




Simple Models cont’d

In real life experiments there is always a finite starting value ofN at t = 0, N0. Can show that:

N(t) = N0ert (2.2)

i.e. (given ∞ space/ food) growth of cells is exponential.

Note that the value of the constant r can be -ive or +ive,corresponding to decay or growth respectively.

For a human population with α births per year & β deaths,obviously r = α− β & population viability depends on thisbeing +ive.

Can be shown from Eqn.(2.2) that doubling time for pop’ngiven by τ2 = ln 2/r so (if r is a percent) world’s populationdoubles itself every 70/r years.





Restricted Growth: the Verhulst or logistic model

Pop’ns do not grow in above manner; restrictions (due tofood, space, predation or any other factors) apply on allgrowth rates.

These put a limit on max sustainable pop’n size. This limit iscalled the carrying capacity of environment, K .

Instead of constant r given by Eqn.(2.1), have a timedependent growth rate 1

NdNdt .

Eqn.(2.1) implies a constant growth rate given by r .





Restricted Growth cont’d

In logistic growth model, growth rate is assumed to start at rwhen N = 0 & reach zero when N = K ,

Growth Rate ∝(

1− N

K

)i.e. the growth rate decreases linearly with increasingpopulation.

Mathematically logistic growth model is given by:

1

N

dN

dt= r

(1− N

K

)(2.3)

This differential equation involves parameters r , K :

r is relevant to initial growth phase of a pop’n (beforerestrictions have an effect)K imposes an upper limit on population growth.





Applications of the Model: r/K selection theory

In ecology, so-called r/K selection theory, relates to selectionof combinations of traits in a species that trade off btwquantity or quality of offspring.

r ,K refer to low- & high-density conditions, respectively.

Terminology came about from the work of the ecologistsMacArthur & Wilson based on their work on insulatedecosystems.

Species are often called either r-strategist or K-strategistdepending upon the selective processes shaping theirreproductive strategies.

Theory contends that adaptation to high or low densityenvironments involve different characteristics.





Applications of the Model: r/K selection theory cont’d

r-selection

In unstable/unpredictable environments, r-selection isdominant mechanism because ability to reproduce quickly iscrucial.

Adaptations that permit successful competition with otherorganisms are unnecessary, due to environmental volatility.

Characteristic traits of r-selection are thought to include:

short generation time,high fecundity & ability to disperse offspring widely,early maturity onset with small body size.

Examples range from bacteria, through insects (e.g.mosquitos) & weeds, to various mammals, especially smallrodents (e.g. rats).





Applications of the Model: r/K selection theory cont’d

K-selection

In stable/predictable environments, K-selection is dominantmechanism because ability to compete successfully for limitedresources is crucial.

Pop’n sizes of K-selected organisms typically very stable &approach max that the environment can bear.

Differ from r-selected populations, where population numbersare more volatile.Characteristic traits of K-selection are thought to include:

long life expectancy,mate choice,production of fewer offspring requiring extensive parental careuntil maturity & large body size.

Examples large organisms such as elephants, trees, humans &whales, but also some smaller organisms.





Further Applications of the Model

As well as above, Verhulst Model has been applied to other areas:

1 Spread of a Disease.

Initial no. of susceptible individuals to an infectious disease inpop’n is (a const) K.So y & (K − y) are no. of infectives & susceptibles after t.Chance encounters spread disease at a rate proportional to(product of) infectives & susceptibles.So model is y = ry(K − y)Spread of rumors is an example of an identical model to this.





Further Applications of the Model cont’d

2 Explosion/Extinction.

No. y(t) of crocs in a swamp satisfies y = ry wheregrowth-decay constant r ∝ (y − M) & M is a thresholdpopulation.The logistic model y = k(y − M)y gives extinction for initialpopulations smaller than M & a doomsday populationexplosion y(t) → ∞ for y(0) > M.This model ignores harvesting/culling.






3 Logistic Growth with Culling.

No. y(t) of fish in a lake satisfies a logistic modely = (a − by)y − c, provided fish are harvested at a rate c > 0.This rate c can be either constant, variable (i.e. c = hy) orinvolve periodic harvesting & restocking c = hsin(ωt).These models are dealt with in more detail below






4 The Chemostat Model.

A chemostat (from Chemical environment is static) is abioreactor taking in fresh medium continuously, while cultureliquid is continuously removed to keep the culture volumeconstant.By changing rate with which medium is added to bioreactor,growth rate of microorganism can be easily controlled. Thismodel is examined below.






Solving the general form of logistic growth equationEqn.(2.3), (left as an exercise) to get:

N(t) =K(

KN0− 1)

e−rt + 1(2.4)

for some inital population N0.

Can be seen from Figs 2.1(a) & (b) that logistic growth ofpop’n depends qualitatively on the initial population:

For N0 > K/2, second derivative N is always -tive,For N0 < K/2 there is a period of growth which is exp’l innature (N > 0) which then levels off (N = 0) before headingasymptotically to K (N < 0) as t → ∞






0.00 0.

25 0.50 0.

75 1.00 1.

25 1.50 1.

75 2.00 2.

25 2.50

14

710

1316

1922

2528

3134

0

1

2

3

4

5

6

7

8

Population Size, N

Time, tGrowth Rate, k

(a) N0 > K/2

0.00 0.

25 0.50 0.

75 1.00 1.

25 1.50 1.

75 2.00 2.

25 2.50

14

710

1316

1922

2528

3134

0

1

2

3

4

5

6

7

8

Population Size, N

Time, tGrowth Rate, k

(b) N0 < K/2

Figure 2.1: Logistic Growth for Different N0






Taking Eqn.(2.3) & adding a culling/ harvesting term H:

dN

dt= rN

(1− N

K

)− H (2.5)

which may be non-dimensionalised (removal of dimensions, bydividing through by carrying capacity, K ), as follows:

dy

dt= ry (1− y)− h (2.6)

where y = N/K & h = H/K are non-dimensional pop’n &culling rates resp.

Can get steady states of this eqn (all y for which dy/dt = 0):

dy

dt= 0 hence − r

(y2 − y +

h

r

)= 0 (2.7)






Provided they meet the harvesting condition 4h < r , bothroots of Eqn 2.7:

y = p± =1

2

(1±√

1− 4h

r

)(2.8)

are steady states since both have the property y = N/K < 1.

Note: condition 4h < r gives us population harvestingcapacity (i.e. how much may be harvested).

In order to find stable states of Eqn.(2.6), it is necessary tograph the derivative of y against y itself.

This allows us to examine the behaviour of the derivative forvarious (especially initial) values of y .






Examining the graph (Fig. 2.2), see that points p− and p+

from Eqn.(2.8) demarcate areas of stability.

For y(0) < p−, pop’n will never grow, as y < 0 for all t andthus extinction will occur.

For y(0) > p−, y > 0 & pop’n will eventually settle down to asteady state y = p+ (this follows from working out a solutionto Eqn.(2.6) and examining behaviour of y(t) as t →∞).

Can be seen from Fig 2.2, that the time of optimum culling isthe point where y is maximal. Before this pop’n is notgrowing at an optimal rate. After this growth rate isdecreasing to the steady state.






dydt

y = p− y = p+

y(t)

Figure 2.2: y V y for Logistic Growth with Culling





Aside: Equilibrium & Stability

A constant solution of a differential equation is called anequilibrium solution or steady state.

Locating/ classifying steady states helps determine family ofall solutions of DE equation.

Soln to a DE of form N(t) = constant, because dN/dt = 0 iscalled an equilibrium point.





Aside: Equilibrium & Stability cont’d

Classify steady states according to behavior of other solutionsthat start nearby:

Steady states N = c is called stable if any solution N(t) thatstarts near N = c stays near it.Equilibrium N = c is called asymptotically stable if anysolution N(t) that starts near N = c actually converges to it,that is limt→∞ = c.If an equilibrium state is not stable, it is called unstable. Thismeans there is at least one solution that starts near equilibrium& runs away from it.Example of this is first state N = 0 of the Logistic Growthmodel with no culling (similar to Fig. 2.2, only shifted to theleft to have roots at N = 0 and N = K ) Eqn.(2.3), this is anunstable state. This is because, if N = +ε (for small ε), thepopulation will increase; if N = −ε, the population willdecrease further.





The Chemostat

Logistic Growth model can be used for Chemostat, (used tostudy growth of μorganisms under various lab conditions).

For the growth models so far, since the nutrients are notrenewed, exp’l growth is limited to a few generations.

Bacterial cultures can be maintained in a state of exp’l growthfor long time using a system of continuous culture, designed torelieve the conditions that stop exp’l growth in batch cultures.

Such continuous culture, in the chemostat, is a situation thatreflects bacterial growth in natural environments.

In chemostat, growth chamber is connected to a reservoir ofsterile medium.

Once growth is initiated, fresh medium is continuouslysupplied from reservoir.





The Chemostat cont’d

Fluid volume in growth chamber is maintained at constantlevel by an overflow drain.

Fresh medium enters growth chamber at a rate that limitsbacterial growth.

Bacteria grow at same rate that overflow removes bacterialcells & spent medium.

Fresh medium addition rate determines growth rate cos freshmedium contains limiting amount of essential nutrient.

Thus, chemostat relieves lack of nutrients, toxic substanceaccumulation, & accumulation of excess cells in culture, whichinitiate growth cycle’s stationary phase.

Bacterial culture can be grown and maintained at relativelyconstant conditions, depending on nutrient flow rate.






For Chemostat, model is more complicated than for purelogistic growth with a limited volume of nutrient Eqn.(2.3); ascan be seen in Fig. 2.3(b), there is an inflow rate into thecentral culture chamber and also an outflow rate from thechamber.

These latter two terms are both F to conserve mass inchamber.

To start, will derive logistic growth model again, formicroorganism pop’n N, taking into account nutrient concC (t).






Can define the equations to be:

dN

dt= K (C )N = κCN (2.9)

i.e. bacteria reproducing at a rate ∝ C , & change in nutrientconc in culture chamber:

dC

dt= −α

dN

dt= −ακCN (2.10)

i.e. α units of nutrient giving one unit of pop’n growth.

integrating w.r.t. t Eqn.(2.10) get C (t) = −αN(t) + C0,which, on subst’n in Eqn.(2.9), gives:

dN

dt= κN(C0 − αN) (2.11)

≡ logistic growth, Eqn.(2.3) for r = κC0, K = C0/α.






(a)

��

�

�

�

�

��

��

��

��

(b)

Figure 2.3: Chemostat: Schematic Representations





Chemostat Parameters

Quantity Symbol units

Nutrient Concentration C mass/volumeReservoir Nutrient Concentration C0 mass/volumeBacterial population N mass/volumeVolume of Growth Chamber V VolumeInflow/ Effluent Flow Rate F Volume/time

Table 2.1: Chemostat Parameters

Note from Table 2.1 that by introducing a concentration terminto Eqn.(2.9), forced dimensionally to consider N to be themass of bacteria per unit volume (i.e. a pop’n density).






Modify above chemostat eqns (2.10, 2.11) to account forinflux/eflux to/from culture chamber:

dN

dt= K (C )N − FN

V(2.12)

where 2nd term on RHS is removal rate of N, &

dC

dt= −αK (C )N − FC

V+

FC0

V(2.13)

where 2nd & 3rd terms on RHS are decrease in conc’n ratedue to removal & increase due to inflow, resp.

Volume term V makes eqns dimensionally correct.





The Chemostat cont’d: Michaelis-Menten Kinetics

Turning to growth rate term K (C ), will be obvious that K (C )is some limiting function of nutrient availability C 2.

Mechanism that is commonly adopted in literature is that ofMichaelis-Menten Kinetics:

K (C ) =KmaxC

Kn + C(2.14)

Will be seen from fig.2.4 that value of Kn corresponds tonutrient conc’n required to achieve half the maximum growthrate Kmax .

2i.e. growth rate increases with nutrient availability only to a limiting value(determined by capacity of bacteria to reproduce themselves only at a certainrate)






Kmax

Kmax2

K (C )

C = Kn C

Figure 2.4: Michaelis-Menten Kinetics, Eqn.(2.14)






Inserting model from Eqn.(2.14) into chemostat equationsEqn.s(2.12, 2.13), to get:

dN

dt=

(KmaxC

Kn + C

)N − FN

V(2.15)

anddC

dt= −α

(KmaxC

Kn + C

)N − FC

V+

FC0

V(2.16)

which include no fewer than two unknowns (N, C ) & sixparameters (Kn, Kmax ,F ,V ,C0,α).






It can be shown (Using Dimensional Analysis) that aboveequations reduce to:

dn

dτ= f (n, c) = α1

(nc

1 + c

)− n (2.17)

and

dc

dτ= g(n, c) = −

(nc

1 + c

)− c + α2 (2.18)

which now only contain the (dimensionless) parameters:

α1 =VKmax

Fand α2 =

C0

Kn

& dimensionless variables:

τ =tF

V, n =

NαVKmax

FKn, c =

C

Kn






Dimensional Analysis amounts looks at an equation & reducesit to its simplest form without dimensions.For example, in 1st of dimensionless variables above:

τ =tF

V

Now from Table 2.1 in dimensional form, this amounts to:

units of τ =[��time]

[��Volume��time

][��Volume]

(2.19)

rhs of which is dimensionless.“dimensionless” ≡ “the game is half-over”, i.e. not specifyingany units (e.g. minutes) for time gone but, expressing it interms of total game time, a more ‘natural’ time unit.DA also has benefit of removing physical dimensions (& units)






Eqn.s(2.17) & (2.18) are nonlinear due to term nc/(1 + c) &thus do not have exact solutions for n(τ) and c(τ).

Can, however, examine their steady-state solutions as did forlogistic equation, eqn.(2.7).

Setting the derivatives dndτ , dc

dτ to zero in Eqn.s(2.17) & (2.18),get:

α1

(nc

1 + c

)− n = 0 (2.20)

and

−(

nc

1 + c

)− c + α2 = 0 (2.21)

where n, c denote the values at the steady states.






Solving for n, c, get two solutions:

(n1, c1) =

[α1

(α2 − 1

α1 − 1

),

1

α1 − 1

](2.22)

and(n2, c2) = (0, α2) (2.23)

n2, c2 is a trivial solution: no bacteria left and nutrient hassame conc’n as reservoir.For n1, c1, eqn.(2.22), clearly need α1 > 1 (no -ive conc’s) &α2 > 1/(α1 − 1) (no -ive pop’n densities).

What is the stability of these steady states?





Aside: Stability at Steady States

Close to steady states (n, c), solution can be taken to behavelinearly. Writing the system of equations (2.17) & (2.18) as:

dX

dτ= F (X) (2.24)

where X is the vector [n(τ), c(τ)].

Considering close to the steady states X = X− X (i.e. smallX), can say that:

dX

dτ=

dX

dτ= F (X + X) (2.25)

(since dX/dτ = 0) which, by Taylor’s theorem, simplifies to:

dX

dτ= ��0

F (X) + F ′(X)X +��0O(X2) (2.26)





Aside: Stability at Steady States cont’d

First & last terms on rhs go to zero (first is value of derivativeat steady state, & last is small). So:

dX

dτ=

dX

dτ= F ′(X)X ≈ AX (2.27)

where A is known as the Jacobian of F (effectively thederivative of a vector function) and is given by:

A =

⎛⎝ ∂f

∂n∂f∂c

∂g∂n

∂g∂c

⎞⎠ (2.28)

Have a linear system eqn.(2.27) to solve. For convenience,rewrite eqn.(2.27) as

dX

dτ= AX for X ≈ X (2.29)






Solution of eqn system (2.29) can be written as a sum:

X(τ) =n∑

i=1

cieλiτvi (2.30)

i.e. a linear combination of eigenpairs (vi , λi ) & someunknown coefficients ci .

For stability of equilibria (steady states), need to know, willtrajectories approach an equilibrium as t →∞?

Since X measures proximity to our steady states, thiseffectively means, does X(t)→ 0 for solutions of eqn.(2.27)?(Recall that A is Jacobian of F close to steady states).






Know that cieλiτvi → 0, provided that eλiτ → 0. But for

what values of λ does eλiτ → 0?

Taking magnitudes of bothsides, get∣∣∣eλiτ∣∣∣→ 0. (2.31)

Substituting λ = a + ib (i =√−1), eqn.(2.31) becomes:

eaτ√

(cos bτ)2 + (sin bτ)2 → 0. (2.32)

Since term inside√

is just 1, stability of solutions aroundsteady states simply depends on real parts of all eigenvaluesbeing less than zero.






Now for a matrix A given by

A =

(a11 a12

a21 a22

)(2.33)

eigenvalues are given by solutions of characteristic polynomial,

λ2 − pλ + q = 0 (2.34)

where p = trace(A) and q = det(A).

Roots of eqn.(2.34) are λ =p±√

p2−4q2 .

Two parameters p, q thus determine stability of solution ofsystem under consideration.






Figure 2.5: Stability points on p V q






From Fig. 1.2, observe a number of cases:

1 λ1, λ2 real and distinct:

1 If λ1, λ2 have same sign, steady state called a node. ifλ2 < λ1 < 0, soln for X = (x y)t → 0 in Eqn.(2.30) & it is astable node. If λ1 > λ2 > 0, it is a unstable node.

2 If λ1, λ2 have different signs, one of the parts of soln inEqn.(2.30) will always go to 0 and other will go to ∞. This isa saddle point singularity & is always unstable.

2 E-values are complex: λ1, λ2 = α ± iβ, solns to Eqn.(2.30) areoscillatory (product of eατ & e iβτ ); there are two cases:

1 If α �= 0 this is a focus. α < 0 gives a stable focus and α > 0gives an unstable focus.

2 If α = 0 this is a centre. In order to strictly determine stabilitywe must examine higher order terms.






In terms of the trace and determinant p, q, these results aresummarized in Table 2.2

Type Eigenvalues Trace p, Det q

Saddle Point different signs q < 0Stable Node λ2 < λ1 < 0 p < 0 & q > 0Unstable Node λ1 > λ2 > 0 p > 0 & q > 0

Stable Focus (λ) < 0 p2 < 4q & p < 0Unstable Focus (λ) > 0 p2 < 4q & p > 0

Centre (λ) = 0 p2 < 4q & p = 0

Table 2.2: Stability & Matrix Properties






Eqn.(2.34) roots are λ =p±√

p2−4q2 , so:

√p2 − 4q < p, i.e.

p < 0 (trace(A)< 0) & q > 0 (det(A)> 0) for stability.

These latter two conditions ensure that soln’s exp’l part → 0as t →∞, i.e. conditions for stability. Hence for Stability

dX

dτ= F (X) =

(f (n, c)g(n, c)

)(2.35)

Routh-Hurwitz Conditions at X = (n(τ), c(τ)) must hold:

∂f

∂n

∣∣∣∣X=(n,c)

+∂g

∂c

∣∣∣∣X=(n,c)

< 0 (2.36)

and

∂f

∂n

∣∣∣∣(n,c)

∂g

∂c

∣∣∣∣(n,c)

− ∂f

∂c

∣∣∣∣(n,c)

∂g

∂n

∣∣∣∣(n,c)

> 0 (2.37)





The Chemostat: Stability at Steady States

So, getting back to the chemostat problem, if F = (f , g) aregiven by eqn.s(2.17, 2.18):

dn

dτ= f (n, c) = α1

(nc

1 + c

)− n

anddc

dτ= g(n, c) = −

(nc

1 + c

)n − c + α2

the Jacobian A of F (from eqn.(2.28)) is given by

A =

(α1

c1+c − 1 α1n

1+c2

− c1+c − n

1+c2 − 1

)(2.38)






So at the steady state:

c =1

α1 − 1and n =

α1μ

α1 − 1,

(where μ = α2(α1 − 1)− 1) given by eqn.(2.22), the Jacobianis given by:

A(n, c) =

[0 μ(α1 − 1)

− 1α1

−μ(α1−1)+α1

α1

](2.39)

Now, for stability, as seen, trace of A must be -ive, which istrue if μ(α1 − 1) + α1 > 0 which it is since α1 > 1 (no -iveconc’ns) & α2 > 1/(α1 − 1) (no -ive pop’n densities)






Thus a chemostat has a steady state solution (given byeqn.(2.22)) with bacteria in the growth chamber.

This equilibrium will only produce biologically meaningfulresults subject to:

α1 > 1 and α2 >1

α1 − 1,

i.e. no -ive conc’ns and no -ive pop’n densities resp.






So, in dimensional form, α1 > 1 corresponds to 1Kmax

< VF .

As Kmax is max bacterial repro rate with unlimited nutrientsdN/dt = KmaxN, ln 2

Kmaxgives doubling time τ2 of bacterial

pop’n.V /F = time to refill whole growth chamber volume with freshnutrient.So if emptying time (given by V /F ) × ln 2 is greater than thedoubling time (τ2 = ln 2

Kmax), bacteria will be washed out quicker

than they can reproduce.

α2 > 1/(α1 − 1) corresponds to C0/Kn > c so max(non-dimensional) conc’n c will never be more than(non-dimensional) conc’n in stock solution.


Documents

Chapter 2: Growth & Decay Introduction