ode biology

8/2/2019 ODE Biology

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CHAPTER 8First Order Ordinary

Differential Equations

8. 1 Exponential Growth and Growthin a Limited EnvironmentThe statement that a population grows exponentially is expressed mathe-ma tically in the differential equation

r/Ndt = k:V, [136]v.hich asserts that the growth rate dN jdt of sorne size N is proportional tothat size, where the proportionality constant /.; is positive; the reader willrecall the equation from Chapter 4; it \Vas used in Examples l and 5 inSection 4.1 and in Example 6 in Section 4.3. In Section 4.3 it was shown thatif equation [136] is valid, then the relation between size N and time t isN = N 0', where N 0 denotes the size of the po pulation at the initial timet = O. Equation [136] is called a mathematical model descript ie of ex-ponenCial growth of a population. In this section our purpose is to constructa different mathematical model, one based on more realistic assumptions.

Continu


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404 8 Fir st Orda Ordinary Di.fferentia! Equationsgrowth of a bacte ri a! or animal population cann ot coritinue indefinitely.F or example, consider a culture of Escherich ia co!i growing in an idealnutrient broth meditlln so that the generation period is 20 m inutes, as wedescribed in Exam ple 1 of Sec tion 4.1 ; the rela tion between the numbe rN of cells and time t m in is N = N 0 2'12 0 . In 24 hours, t = 24 60 =1,440 minutes, and a t the en d of that time, a single bacterium would ha ve

= 256. 18,446,744,073,709,551.616descendant cells. Suppose we let these descendant cells proliferate for a nother24 hours o f expo nen t ia! grow th ; th e co mbined mass of a l! o f the descendantsof th e single cell wo uld be severa! times greater th an the mass o f the earthitself.

Such fantastic growtb does not ha ppen fo r a t leas t two reasons. The firstis sim ply tha t the bac teria e\en tu a lly exhaust their own food supply. Atthe same time, howeve r, th e bacte ria excrete products tha t are poiso nous toth ems elves and these pollu te :heir environm ent. Deteriora tion of the en-

oFigure 8.1

1 Ea rth s mass- - - ~ - - - - - - - -

11

2Time 1, in


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8.1 Exponelllial Growlh aiJ(/ Grmrlh in a Limilul Environmml 405vironment eontinues at an increasing rate while the bacteria! populationincreases and results in a slowing down of bacteria! growth until the rateof growth is close to zero. From then on, the net number of cel!s remainsessentially constant and the bacteria! population is said to be in a stationaryplwsc. For a culture of Eschcrichiu coli in a nutrient broth medium, thestationary phase is reached \\hen the cell concentration is between 2 10 9and S 10 9 cells per milliliter.

Let us express these ideas about restricted growth in mathematical terms.Our purpose shall be to construct a mathematical mode1 for a populationthat is subject to restricted growth - that is, we wish to construct a modelfor a population that will nentually enter a stationary phase. Thus, Jet tdenote elapsed time and Jet N = N(t) denote population size at time t.Let us suppose that there exists a fixed upper limit, or maximal value, forthe population size. We denote the upper limit by K, a positive constant,so we have N (t) :::; K foral! r. In the bacteria! population discussed earlier,we recall that , as the pcpulation size N increases, the rate of growth dN jdtdecreases ; we should like to account for this behavior in our mathematicalmodel. A simple mathematical model that satisfies these requirements is theditTerential equation

dNdt k(K - N), [137]

where k is a positive constant. This equation states that the growth rate,/N irl r is pro po rtio na l to some population increase K - N yet to be realizedin order fo r thc population to reach the maximum size K. According to thiscq uat ion. as N increases to K . the differcnce K - N decreases to zero ; \hence tb e grovvth rate decreases to zero.

Th c cliffctcntial q u a t i [137] does not tell the \'. ho le story by itself. Inparticular, \Ve must express the total population size N explicitly as a functionoftime; thus we must salve the differential equation [137] for N as a functionof t.

Our first step in solving is to divide both sides of [137] by K - N toobtain the equation.1 dN- - - - - =k K- N dt '

to do this \re asswne that K - N > O. Now we integrate eacb sicle of thisequation with respect to t and we have

S1 dN--- -- dt = kt +e,K - N dt


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406 8 First Order Ordinary Dijferentia/ Equationswhere C is the arbitrary constant of integration on the right-hand side. Asfor the left-hand side of this equation, we have

I 1 dN dt = -ln(K - N)+ C',K - N d1where C' is the arbitrary co nsta nt of integration; hence it follows that

-ln(K - N) + C' = la + C,or

ln(K - N)= - k t + (C' - C).We take the exponential of each side of this equation to obtain the

equation

and solving for N, we have, finally,[138]

where e = e


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N

K - - - - - - - - - - - - - - - - - - - - - - - - - - -

:\ 'o

oFigure 8.2Th e g r ~ p h of the function N = K - e ~ _ , , with e = K - .\' 0 : this function is the uniquefunction that satisfics the diffcrential equationelN jdr = k( K - S 1an d the initial conditionN = N 0 at r = O. Th e constants k and K arepositive. As r increases without bound , the valuesof N = K - c e - ' ' approach K; tha t is . thegraph has the line .\ = K as a horizontalasymptote.

407

for all values of t and, as r approaches infinity, that is, as t increases withoutbound, N = K - ce-k' approaches K (because e-kr approaches zero). Thusthe graph of [ 137] les below the line N = K and has this lineas a horizontalasymptote as t approaches infini ty; tbe graph is shown in Figure 8.2. Thuswe ha ve constructecl a mutlw.mil tical model[or reslricted qro11 rh

Both the exponen tia! growth model [ l36] ancl the restricted growth m9clel[137] are special cases of the differential equation

a+ bN, [139]

where both a ancl b are constants. If, in [139], we Jet a = O ancl b = k, thenwe obtain [136]. If we Jet a = kK ancl b = - k, then we obtain [137]. Forthe sake of generality in [139], ' 'e replace N by y ancl t by x to obtain theequation

dydx a+ by. [140]


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408 8 First Order Ordinary Differentia/ EquationsOur purpose is to solve [140] for y as a function of x . The method is thesame as that for sol vin g [1 37] , except for a negative sign, as we sha!l see inthe following steps.

First, we divide both sides of [ 140] by ajb + y to obtain

then

thus,

---,-,--dy = baj b + y dx 'I ___ _ dy dx = Jbdx ;ajb + y dx

In la/b + Yl = bx + C",where C" is the combined constant of integration ; exponentiating, we have

if we Jet e = ec, then th e sign is accounted for and we ma y remove theab solute value sigu a nd write u,'IJ + y = cehx ; th us, fi na l! y,

(/)' = c ' - ; [ 141JIn order to verify th at [ 141 J satisfies th e differential equation [ 140], weditterentiate [141] and , usi ng [141] to obtam the second equahty, we obtain

=a+ by,which is the equation [140] , as required. Notice that the constante in [141]is determined by an initia l condition y = y0 at x = OWe shall write thisinitial cond ition as y (O) = Yo


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8.1 Exponenlial Grmvlh allil Grmnh in a Limired Enrironmenl 409One may obtain [138] from [141] by making the following substitutions

in [ 141]: let y = N, x = .t, b = - k , K =oo - ajb, and re place e by -c .These theoretical discussions are illustrated with the following examples.

Examplc l. The diiTerential equationdr___:__ = 2 + 3rdx

is of the type [140], with 2 in place of a and 3 in place of b. The general solutionofthis equation is. according to [141],

)r = ce 3" - =3

11 here the constan! e is determined by an initial condition _r(O) = Yo As a particularcase. suppose that IYe seek a solution that satisfies the initial condition _r(O) = 2.Then

2 = _r(O) = ce 3 0 232

= e- 3'and hence

8e= - .3Thus the only solution _r = _r(x) of the given differential equation thar satisfiesthe condition _r(O J = 2 is the fun ct ion

Q 2r = e3 x ,3 3its graph is sketched in Figure 8.3. We check that this function satisfies the givendifferential equation dy/dx = 2 + 3y as follows:

dy d (8 3x 2)---e - -dx dx 3 3= 3Gy)= 2 + 3y.


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410y

X

Fgur.e 8.3Graphs of various so lutions of th e ditTerentialequat ion dy j dx = 2 + 3y. Any solution is afunc tion of the for m y = ce 3x - 2/3 i ~ h theva lue of the constant e determined by the valueof y a t x = O. If e > Othen the graph is co nca ve upward an d is also above the line y= - 2/3.I f e < O, then the graph is concave dow m ta rdand is also below the !in e y = - 2/3. If e = Othen the solu tion is the line y = -2 /3.

has the gener al solution

dydx -9 + 4y

9Y = ce4x +-4'where e is a constan!. This solution is obtained from [141] with - 9 in place of aand 4 in place of b. If we wish to find the particular solution that satisfies the initialcondition y(O) = 3, th en we must equate

9 93 = y (O) = e 0 + 4 = e + 4,and we find that 3e = - .4


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'

Thus

Figure 8.4Graphs of three particular solutions of thedillere nt ia l equation cly jclx = - 9 + -l_r. Anysolution is a function of the form _r = ce""' + 9 4with the Yalue of th e cons tan! e determ ined bythe value . ~ " o of _r. at x = Oaccording to theequation e = _r 0 - 9 4. If y0 > 9:-1. then e > O.and the gra ph is concave upward. a b o ; ~ the liney ::o: 9 4. I y 0 = 9 '4 tHs shown), then t.' = O. Jn drhc graph is '-he linc : = 9 -l. lf _1 0 < 'l l . th


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412 8 First Order Ordinary Differential Equations

which is an equa tion that m ay be interpreted as a mathematical model of restrictedgrowth ofthe type [137]. Let us suppose th at it is such a mode l ; that is, supposethat this last equation describes the growth of a population whose size N mayapproach K = 107 bu t never excceds it. Wc also suppose tha t at time t = O thesize is N 0 = 500.From [138] we see that the function

expresses N as a function o f t; that is, in term s of [ 138] the constant e = K - N 0 =107 - 500. The graph is sketched in Figure 8.5: compare tha t gra ph with Figure 8.2 .N

Figure 8.5T he graph oft he pa rt icular solution N = 10 7 -(107 - SOO) e- '13 oft he di fferential equa tion dN jdt =(1/ 3)(107 - N) . Wh en t is greater th an 34.5, t-henthe val!...!e of 1f !s be t'.veen lCO aud lG7 .

A reasonable qu est ion is this : Wh en does the population become 100 individualssho rt of its upper limit 107 ? In o rder to answer thi s we must solve the equation

107 - 100 = 107 - (107 - 500)e - '13for t, as fo llows. Su bt rac t 107 from each side of this equat10n to obtain

-100 = - (107 - SOO)e - '13 ,or

t/3 107 - 500e = -- - -100


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8.1 E\ponenlial Grmrlh a/1(1 Grmrlh in a Limi1ct! Enrinmmcnl 413Taking natural logarithms of each sidc of thc last cquation, wc have

1 = 3 ln(105 - 5).Lct us use th e approxim a tions 1n( 10 5 - 5) ;::::: 1n 10 5 and In 1O ::::: 2.3 to compute

;::: 15. 2.3 = 34.5.Thus the po pulation becomes 100 individuals sh9rt of 10 7 wh en r is approx imatcly34.5.

Example 4. The differential equationdv = 8 - 2ydx

is


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414

Exercises

y

!ni tia! condition is y (O) = 13

Fi gure 8.6T he graphs o f threc parti cular sb lutions of the di tTerential cqua tiond_ridx = 8 - 2_-. Thc so lution l ' = 9e- 2" + 4 corresponds to theinitial condition y(OJ = 13 at x = O; the so lutionr = -3 e-2x + 4corresponds to the initial cond ition .r(Ol = 1 a t x = O. An y so lutionof th e differential equation is a function of the form )' = ce - 2-' + 4,where e denotes a const an ! determined by a choice of initial valuefor y at x = O.

X

In Exercises 1-5, solve the differential equation in cach for N, or for y, as a functionof t, or of x. Go through eaeh step in the derivation of the so lution.

dN1. dt = 3(8 - N).dN

2.- = 7 + 4N.dtdy3.- = 4 - y.dx

dy4. 3 - = 6- 9v.dx dN5. 5 dt = 50(198 - N).

In Exercises 6-12, find the solution, in each, of the d iffe rential equation that satisfiesthe initial condition at t = O. Sketch the graph of the solution as a function of t. Indicate the behavior of the solution as t ...... oo .


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8. 1 Exponentia/ Grmrth and Grmrth in a Limited Environment

dN6. - = 2(500 - N).dtdN

N(O) = 10 a t t = O.7. - = 3(8 - N), N(O) = 30 at t = O.dt

dr8. __ = 4 + 3v ;(O) = 1 a t t = O.dt . 't!N9. dt

12 (300 - /\' ). N(O ) = 20a t t = O.dr10. - = 3( - 1 + y), y(O) = 2 at r = O.drdr11. --=- = 3(9 - r), y(O) = 1 at t = O.dt .dN 1

12. - = - (20 - N). N(O) = 60 at t = O.dr 3

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13. Cons id e ra cu lture of Escht'rich iu coli. Suppost: that the cu lture is in an env iron ment in 11 hich the upper limit on the number of cells is 6 10 10 ce li s. Let N = N (t)deno te th e number of ce lls presen t at time t. an d suppose that the gro11 th rate ispr oportional to the differen ce 6 10 10 - N, wit h the constan t of propor tiona lit yequal to 4 1 0 (a) Express the growth rate of the ce ll population in terms of the number of cells.(b) Gi ve n that at r = O there ar e N (O) = 3 107 cells, ex press the number of cells

as a functio n of r. Ske tch th e graph o f this function. showing its be havio r asr __. :r.:.14. Let J\' = S(;) denote the size o f animr,l populat ion at time r. Suppose that the

birth r


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416 8 First Order Ordinary Differential Equations17. This exercise is based on data obtained in an actual agricultura! study. Let q denote

the dietary gross energy (in calories) consumed by an average 433 kg steer, andlet Q calories denote th e net energy recovered. Agricultura! scientists found thatthe theoretical maximum net energy obtainable per da y is 24,000 cal, and that therate of change, dQjdg, of the nel energy with respect to the gross energy is proportional to the ditference, 24,000 - Q, between the maximum net cnergy and theactualnet energy. The co nstant ofproport ionality was calculated to be numericallyequa l to 0.000025. Find the net energy recovered as a function of the dietary grossenergy consumed. The constan t of integration is determined by observing that , ifthere is no dietary gross enc rgy consurned, then the net energy recovered is zero.Sketch the graph of this function. [After S. Brody, ibid., page V9.]

8.2 The Logistic Equation: A Synthesis ofTwo Approaches to the Study of GrowthAs a means of introducing the logist istic equation, we shall need to havesorne understanding of rates of increase and decrease of populations. Supposewe Jet N = N(t) denote the size of a certain animal or cell population at atime t. The growth rate dN jdt of the population s the difference between therate of addition of individuals due to birth and immigration, and the rateof subtraction of indviduals dueto death and emigration. We assume thatthe population is closed to the outside in the sense that no individuals en terthe population from the outside (immigration = 0) , or leave the population(emigration = 0). Thus we obtain the equation

dN = Bi rth rate - D eath rate,dtwith which we shall work.

Let us consider the birth rate first. The quotientb Birth rateN

may be interpreted as an average fertility over al! of the indivduals in thepopub tion; the quotient is measured in units of births per unit time perindividual. In Example 6 in Section 4.3, we assume that the average fertilitybis constant with respect to time, which is equivalent to saying that the birthrate is proportional to N. Except perhaps for small populations, a morerealistic assumption is th at the average fertility b decreases as the population


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8.2 The Loyistic Equation: A Synthesis o( T11o Approaches 417sizc incrcascs. Thus, lct us assume that h is a dccreasing function of N. Thesimplest dccteasing function is a linear function of the form

where thc positivc constant b0 is thc value that b approachcs as the sizc Nof the population becomes very sma!L and \vhere k, is a positive constan!.Thus \1'1:' ussume that

' Birth rate = (h 0 - kbN)N.Now let us considcr thc quotient

m Death rateNwhich may be viewed asan average probability of dcath (mortality) through-out the entire population ofindividuals.lt is likely that, owing to such factorsas a reduction in the amount of food and space available to each individual,this average probability of death increases as the population increases.Accordingly, vve as sume that thc average probability misan increasing linearfunction of N, namely,

m= n1 0 + k,,N\Vhere the positive constant m0 is the value that m approaches as the size Nbecome s ~ r y smalL and ' 'he re k111 is a positive constant; th is is sho'-vn inFigur


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418b, or m

o

Figure 8.7

KPopulation size, N N

Th e average fertility b = Birth rate/N , and th e averagepr ob a b il ity o f deat h 111 = Death ra te/N, are assu med tobe linea r functions of the popu lation size N. Thepopulat ion size sta bilizes when h = 111: the individualbirth rate is eq ual lo the individual death ratc whenN = K = (h 0 - 111 0 ) '(k 1 + k.,.).

For convenience let

and K

then we have the differential eq uation

dNdt kN (K - N);

bo - mof


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8.2 The Logistic Equation: A S_nuhcsis of Two Approaches 4192. l f we compare the logistic equation [142] with our previous models,

one of wh ich is the exponen tia! growth model

dN = kNdt ,

where the rate of growth dN/dt is proportional to the size N, and theother of which is the restricted growth model

dN- = k (K - N).dtwhere the rate of growth dN jdt is proportional to th e difference K - N,we see that the logistic mod el [142] may be viewed as a combination ofth ese two previous models in the sense that the rate of growth dN/dt isno w proportional to the pro duct N(K - N) of the size N and thedifference K - N. Th is remark will be further clari fied after we so lve[ 142 ] for N as a fun ction of t.

We solve the logistic equation as follows. We di vide both sides of [142]by N(N - K) and have

dN-- -N(N- K) dt - k . [143]Now antid iffere ntiate eac h side of this equation with respect to t. For theieft-han sie of [1 43] \\e use lhe rnelhud u[ partial f1actions, discsscd inSection 7 3, in order to write

N(N - K) K(N - K) KN 'hence we have

f 1 dN dt = f 1 dN dt _ f _1_ dN dtN(N - K) dt K(N - K) dt KN dt1 1= K ln \N - K\ - K In N + C.


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420 8 First Order Ordinary Differential Equations

Notice that we do not need the absolute value sign on N because N cannotbe negative. Thus

f 1 dN - 1 . IN - KlN(N - K) ~ l t dt - K L1 N + e,which must be equal to an antiderivative of the right-hand side of [143] ,namely,

iJ-k ) dt = - la + C';here e and C' denote constants of integration.It follows that

~ l n IN- Kl +e= - l a+ C';K Nhence

I ~ Kl = - Kkt + K(C - e).NExponentiating, we have

hencewhereThusorhence, finally,

IN - Kl - Kkt K(C' - C)~ - - , - - - ' = e e N ,N - K = -ce-Kk'N,

N(1 + ce-Kkr) = K;

KN =- - - -+ ce-Kkr [144]is the solution of [142] that we seek. Function [144] is called the logisticfun ction. The value of the constant e is determined by the initial condition

N(O) = N 0


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8.2 The Logistic Equation: A Synthesis of Tlt'o Approachcs 421according to the equation

KC=-- 1,No [145]which is obtained from [144] by setting 1 = O and N = N 0 .How should we sketch the graph of the solution [144]? For the present1et us consider the most importan! case, 1111! cusi:' 11-lwri:' the initial si::.e N0is /i:'ss 1/wn K, which is true when the population size N is increasing. Thene= (K /N 0 ) - 1 is a positici:' constant and from [144] \Ve see that N(t)


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422 8 First Order Ordinary Dijferential Equations' dcrivatives, dNjdt and d2Njdt 2 . As N= N(t) is always less than K we seefrom the logistic equation [142] that dNjdt is positive for a l! values of t;hence N = N(t) is an increasing function of t. l t is tedious to differentiate[144] twice and we avoid this by differentiating [142] once, using the productrule for differentiation, as follows:

- = kN - - + k- (K - N)1 N ( dN) dNdt2 dt dtdN= k (K - 2N) - .dt

Notice that if N < K /2, then K - 2N > O; thus :Kif N< -2 '

d2 N dNthen - ,- = k(K - 2N) - > O,dr- drbeca use both k and dNjdt are positive. It follows that the graph of thelogistic function [144] is concave upwardjor N < J( j 2. Similarly, if N > K/2,then K - 2N < O; thus,

Kif N>-2 'd2 N dNthen - 2 = k(K - 2N)- < Odt dt

That is to say, the graph of [144] is con caue do\\"mvardfor N > K /2. Thusthere is an inflection point when N = K/2, which corresponds to the valuet = (In c)/K k, a value obtained from [144] by letting N = K/2 in [144] andthcn sol 'ing for t in the follovving steps:

and, finally,

K K---=-2 1 + ce--Kkr'

1 = ce-Kkt;

Kkt = In e;

In et = ---K.k .


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8.2 The Logistic Equation: A Synthcsis of T1ro Approaches 423In the majority of applications this value of t is po.sitive; hence In e is

po sitive, which means that we must have e > l. From the condition thate > 1 and [ 145], we obtain the ineq uality

hence

or

l > 1;

K-> 2No 'N 0 < K/2.

In Figure 8.8, N 0 < K/ 2. We see that K is the upper limit of the size of ourpopulation and hence our model is one of restricted growth.

Our resu lts lead us to the following desc ription of the growth of thepopulation in the case \Yhere the initial size N 0 < K/2: th e growth startsoff at a relatively slow rate when r = O. The growth then proceeds ata fasterand faster rate, and in this re spect it resemb les exponential growth. Duringthis early period the graph is con cave upward. In our logistic model, however,thc growth evcntually slows down. In fact, the growth bcgins to slow downwhen the time reache s t = {ln[(K/ N 0)- 1] }/Kk , which is at the point ofinflcction. At this time the population reaches the size Kj2, which is one-halfof its maximum sizc. From thi s point onward the growth ra tc decreases asthe popu l:nion size in creases and approaches its up per limit K: the graph isnow concave d\\.11\\ard. It is during this period that thc populat ion gro wthof our logistic model [ 142] rese mbles that of the restricted growth modeld.V jd1 = f.( K - N) ofthe previous section; compare Figure 8.8 with Figures8.9 and 8. 10. In our logistic model the growth is fas test in the neighborhoodof the point of inflection.

Before presenting some examples it is worth while mcntioning thc terminology th at has been used by population biologists to describe the logisticequa tion . This is well described in C. J. Krebs, Eeology: The ExperimentalAnalysis of Di stribittion and Abundanee, Harper & Row, New York, 1972.First of a ll, the constant K , which is the upper asymptotc or maximal valueof the size N, is called the earrying (apae ity of the environment. The quantity(K - N) j K is called the unutili::ed opportunity for population growth andis the remaining capacity of the population divided by th e maximum (orcarrying) capacity. The constant

r = kK


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424

Time, tFigure 8.9The growth curve for the exponen tia! growthmodel given by the differential equation dS dr =k N . The population size increases without limitat a fas ter and faster rate.

"2:;o.o

N

K - - ---------------- ------- - -

" Oo

Time,tFigure 8.10Th e growth curve for the restricted growthmodel dN /dt = k (K - N). The growth rate ismaximum at r = O, and decreases to zero as thepopulation size approaches its upper limit K.

is called the innate capacity for increase. Thus the logistic equation may bewritten

(G h ) ( Innate ) (p 1. ) (Unutilized )owt rate . opu atJon .f , . = capacJty x . x opportumty for ,o o Utatwn . srze .p p . _ for mcrease populatwn growth


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8.2 Tlw Logis tic Equation : A Synthesis of Two Approaches 425or

dNdr rN (K- N)K '

and the so lution may be writtenN K

.And now for severa! examples. These will be based on hypothetical data .Examples based on real data from actual experiments will be used in thenext section .

Example l. The average fertility of a certain populat ion is gi ven as a linearfu nc ti on of the popula tion size N by the equation

Birth rate 5 1--- - = - --- N.N 8 1,600T he average probability of death is given as a lin ear funct ion of the size by theequation

Death rate 1 1- -- - = - + - - N.N 8 4,800\V e wish to write down the diffe rent ial equation that describes the growth andregulation of our popu lation. assurn ing that the population is closed to the ou tside.from r.ivPn infnrmiltinn we see that the growth ra te is

dN- = Birth ra te - D eath ratedt

T hus the growth and regulation of the population is determ ined by th e differentialequation

dN 1- = - N(600 - N)dt 1,200 '


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426 8 First Order Ordinary Differential Equationswhich is an ecju ation of the form [142], with

1k= --1,200 and K= 600.Let us suppose that at the time t = O the si ze of the population is N(O) = 4. Thenaccording to our previous discussion, particularly our derivation of [144], thesolution of our differential equation is

where600N=---.,.+ce t/2 '

600e=- - 149.4The point of infiection is at

t = 2ln 149,which is where N = 300. At this po int the growth of the population is fastest. The

N

600

6001 + 149e-ti2

Figure 8.11Tbe growtb curve N = 600/ (1 + 149e - ''1 ).The average ferti lity of this population isb = (5/8) - (1/1600)N, and the averageprobability of death is m = (1 /8 ) + (1 /4800)N,from which we obtain the di fferential equa tiondN fdt = (1/1200)N(600- N ), which withthe initial condition N(O) = 4 , determinesthe growth curve. The growth rate of thepopulation is greatest at the point of inflection,at t = 2 ln 149. The maximum size of thcpopulation is 600.


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8.2 The Logislic Equa/ion: A Synlhesis of Two Approaches 427graph of our so lution, or growth law, is sketched in Figure 8. 11. We see that theupper limit on the size of the population is 600.

Example 2. In a hypothetical population,

Birth rate = G N) N.and

( 11 1 )Death rate = 4+ 200 N N.

We assume the population to be closed to the outs id e. What is required is that weexp ress the size N of the population as a function of the time t.So lution. We have

dN- = Birth rate - Death ratedt= - - 1- N) N + - 1- N) N2 800 4 200= ( 1 ~ 0 N) N

1= :-60 (4o- N)N :so we obtain an equation of the form [142] \1 ith

According to [144],

where

1k= -160 ilnrl

40

K = 40.

N = -- ----,.,.,-1 + ce '14 '40C=-- 1.No

Suppose that we are given that at t = O, N(O) = 2. Then e = 19; hence40N = - - - - - - : - : + 19e - rf4'

The graph of this growth law is sketched in Figure 8.12; the point of inflectionis at t = 4 In 19, which is when N = 20.


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428N

40 ~ ~ ~ ~ ~ - - - - - - - - - - - - - - -

401 + !9e - tf4

Time, tFigure 8.12The growth cur ve g iven by N = 40 '(1 + 19e- "l .which is the unique function th at satisfies thedifferent ial equation d1Y jdt = ( l / 160)N(40- S )an d th e initial cond ition N(O ) = 2.

Example 3. This and the next two examples deal with purely marhematicalaspects of our discussion. Suppose th at we are required lo solve the differcntialequation

dv_.:__ = 2y(3 - y)dx

for y as a function of x, subject to the condit ion that y(O) = 1. Furthermore, wemust do this direct ly, without ap pea ling to expression [ 144]. The method of soh ingth is equation is the same as that used for solving [142] , the logistic equation.Solution. We divide both sides of the equation by y(y - 3) and obtain theequation

1 dyy (y - 3) dx - 2.

Antidifferentiating eaeh side of this a ~ t equation with respec t to x, we have

f 1 dyy(y _ 3) dx dx = - 2x + C.Now, we use partial fractions,

1 A B A(y - 3) + By- - , - - ~ = - +-- = - -.,----'--:-,---- .:..._y(y - 3) y y - 3 y(y -- 3)


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8.2 The Logis tic Equation: A Sy1111iesis of Two Approaches 429from which we have, by equating numerators ,

1 = A(y - 3) + By.We choose y= O in thi s equation and obtain A = - 1/ 3 ; we choose y= 3 andobtain B = 1/ 3; hence

f 1 dy f - 1/3 dy f 1/ 3 dydx = --- dx + --- dxy( y - 3) dx y dx y - 3 dx1 1= -- IniYI + - In ly- 31 + e3 3

1 ly- 313 n - y - + C'.I t follows that

1 1)'- 31In -- + C' = - 2x + e,3 yor

ln lr 31 = - 6x + 3 ( e - C');exponentiating th is, we ha ve

Now we Jet - e = > 3


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430 8 First Order Ordinary Differential Equationswhich is defined for all values of x. This function, which we denote by y = y(x),satisfies the eq uation

and th e condition y (O) = l.dy- = 2y(3- y)dx

Now let 1JSgo through th e computations for sketching the graph ofthis function.Fi rst o f all, 1 + 2e- 6 x > 1 for all x; bence O < y(x ) < 3 for all x. Therefore,dyfdx = 2y(3 - y) > O for all valu es o f x; this means that the fun ction is increasing. Furthermore, from th e original differential eq uation, we obtain, by usingthe product rule for diffe rentia tion,

d2y ( dy ) dy- = 2y - - + 2 - (3 - y)dx 2 dx dxdy= 2 dx (3 - 2y),

which is positive if y < 3/2, nega tive if y > 3/2, and zero if y = 3/2. Consequently,there is a point of inflection at _r = 3j2; it corresponds to . that value of x which

11111111111O In 21- 6

Figure 8.13

X

Th e graph of the func tion y = 3/{1 + 2e- 6 ) ,defined fo r al! values of x . Th is func tion y(x ) isthe un ique so iut ion o f the diffe ren t ial equationdyjdx = 2y(3 - y) with t he initial conditiony(O) = l. As x inc reases witho ut bound (x-+ c:o ),y(x ) approaches 3; as x decreases withoutbound (x -+ - co ), y(x) approaches ze ro.Th ere is a point of inflect ion a t x = (in 2)/6.


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8.2 The Logis tic Equation: A Synthesis of Two Approachessatisfies

3 32 1 + 2e 6 '

From this equa tion we obtain

from which we obtain- 6x 1e = -2'

'and after exponentiat ing, we haveIn 2X=6.

The graph of the so lution is sketched in Figure 8.13.

Example 4. Consid er the differential equa tiondN- = N( l - N).de

431

We wish to so lve this eq uation for N as a function of c. Th is equat ion is of the forro[ 142] with k= K= l, and according to [144] , a solution is

1N= ---1 + ce- 'where e is so rne constan . Let us check that this fun ct ion is indeed a solution of

Solution. We have

hence

A .\T:::.:.:_ = N(1 - N).dt

dN - 1 d-- (1 +ce - ')dt - (1 + ce-')2 dt- 1(1 +ce ')2( - ce-')


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8.2 Tlze Logistic Equation: A Synrlzesis of Two Approaclzes 433y = N, x = t, 4 = k, and 18 = K. Thus our solution is of the form [144], fromwhich we obtain

where

Thus the required solution is

18V = - , - - - - - - - - - -= ; ; - . 1 + ce 71 ' '18 1e= 16- 1 = s

18Y = 1 + (1 /8)e 72 ' '

which is defined for al! values of x. The graph is sketched in Figure 8.14. Thereis a point of inflection at

ln(1/ 8) In 8X = 72= ---;:,which is negative, because y(O) = 16 > 18/ 2 = K/2.

In our discussion of the logistic equation [142] we considered onlypopulations in which the initial size N 0 is less than K. We now show thatifN 0 > K , then N(t) > Kfor al/ t >O and N(t) decreases to K as t approachesinfinity ; this is shown in the upper part of Figure 8.15.

To argue that this is so requires the juggling of sorne inequalities. First,let us state that ifthen

Ko


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434

Figure 8.15

N

KN = l+ ce -- Kk r= ~ ~ ~ ~ ~

N= K - - - - -

1/// /

o

11111111

/,-"'/

The sol id curve is the graph of the func t ion N = Kj ( l +ce-" ' ' ) in the case wherethe init ia! size N 0 at r = O is greater than K. This graph is ab ove the line N = K.As t approaches infin ity, ,\ ' = K /(1 + ce - K ') approaches K. This func tion sa t is fi esthe d ifferentia l equation dNjdr = kN (K - N ). Th e dashed curve is a lso a solutionof this differential equation , but it is a so!ution for which tbe initial size is less thanK; as t approaches infinity, this curve a lso approaches the line _V = K . T hus K iscalled the stab le equil ibrium size of the population whose growth is regulated bythe differential equation dN jr/1 = luV(K - N ).

hence, if we multiply each term of this last inequality by - 1, thus reversingthe inequality, we have

_ 1 < - e -Kk r . [147]Consequently, cornbining [1 46 ] and [147] , we have

- 1 < ce - Kkr < Ofrom which we obtain, by adding 1 to each terrn,

O < 1 + c e - Kkt < l.Frorn th is it fo llows th at

KN= >K1 + c e -K kt 'for all t > O, where e = (K/N 0 ) - l. Notice also that as t approachesinfinty, e-Kkr approaches ze ro, so N approaches K. In particular, frorn the


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8.2 The Logistic Equation: ASynthesis of Two Approaches 435logistic equation dNjdt = kN(K - N) we see that dN/dt is negative forall t > O because N > K for all t > O, so N = N(t) is decreasing. Finally,

d2 N dN- 1 ? = k - 1 (K - 2N)ct- eis positive because both dN jdt and K - 2N are negative; hence the graph ofN = N(t) is concave upward.

The condition that N 0 > K, which implies that N(t) > K for al! t > Osays, mathematically, that the population exceeds the capa K , the size N approaches thecarrying capacity K on a graph that is concave upward- that is , from above.I f N 0 < K , the size N approaches K from below. For these reasons the

carrying capacity K is ref erred to as the equilibrium size of the population. Inother words, any change of the population size from K affects the growthrate in such a way as to return the population to its equilibrium size K. I fthe initial size N 0 = K, then from [144] we deduce immediately that e = Oso that N (t) = K fo r all t. Thus we may say in conclusion that K is thestable equilibrium size ofthe population whose growth law N = N(t) satisfiesthe logis tic eq uation

dNdt = kN(K - N).

Example 6. Con sider the logistic equa tiond.V 1= --- - N(600 - N)dr 1,200

of Example 1, where the carrying ca pacity of the environment is 600. N ow Jet ussuppose that at time t = O the initia l size is N (O) = 1,000, which is greater thanthe capacity of the en vironment- unlike Example 1, where we assum ed that theinitial size was less th an the carrying capacity.

The growth law is600

1 + ce '12 'where

600 2C= - - -1=1,000 5That is, the growth law is

600N = 1 - (2/ S)e rz > 600


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436 8 First Order Ordinary DJji!rential Equar ionsbecause

O< 1 - (2/S)e-'12 < 1 for t > OThe graph is sketched in Figure 8.16.

600N = 1 - (2/S)e - tf l

600300 ---- --- --- N= 1 + 149e- '' 2

4oFigure 8.16Two so lutions ofthe differen tial equation dN jdt o= (1 1.200)N(600- N). Tb e solution N o= 600/ [ 1 - (2 /S)e-'12] sat isfi es the ini tia l condition N(O) = 1.000. Th esolu tio n N = 600/ (1 + 149 e- '12 ) sa tisfies the initial condi tio n S(O) = 4. As tapproaches infinity, each solution approaches 600, which is the stable cquilibriumsize of the populat ion.

ExercisesL Let N = N (t ) denote the size of an animal population at time t, and suppose thatboth the irnmigration rate and the emigration rateare equal to zero. Fu rthermore,

suppose that the quotien t of the bi rth rate by the popul ation size N is given by thefollowing linear fi.mct ion of N,

Birth ra te 3 1------ =- - -- N:N 2 1,000 .

and that the quotient of the death rate by the population size is given by the linearfunction of N,

Death rate 1 1- - -- =-+ - - N.N 2 3,000


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8.2 The Logis tic Equation: A Synthesis of Two Approaches 437Find the differential eq ua tion that describes the growth and regulation of thispopulation. Using this differential equation, find N explicitly as a function of thetime 1, gi ve n that at 1 = O, N(O) = 1OO. Go through ea eh step in the derivation ofthis function. Sketch the graph of this function, labe ling the concavities and theintlection point. Describe the behavior of the size N as r approaches infinity. Isth ere a time when the growth ra te of the population is at a maximum? Give adetaled exp lanation of your answer.

2. Proceed as directed in Exercise 1, given thatImmigraton = Emgration = O;

Birth Rate = (3 --1-N) N;1,000, Death Rate = (2 + , ~ 0 0 N)N;

N(O) = 10.3. Proceed as directed in Exercise l , given that

Immigration = Emigra tion = O;Bi rth Rate 3 1~ - - - N N 4 2,000 'Death Rate 1 1- - - -= - + --N N 4 3,000 '

N(O) = 200. P: ccced ::ts direcred in E x , ~ 1, giver; that

Jmm gration = Emigration = O;Birth Rate 1 1---- - ... - - --NN 2 1,000. '

Death Rate 1 1- ---=-+ - -N N 4 1,000 'N(O ) = 2.

5. Proceed as directed in Exercise 1, given thatImm igra tion = Emigration = O;

Birth Rate 5 1- - - N 6 5,000 'Death Rate 1 1

N = 3 + 7,500N;N(O) = 2,000.


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438 8 First Order Ordinary Dijferential EquationsIn Exercises 6-10, solve thedifferential equation for N as a function of t; subject to tbeinitial condition given. Describe the behavior of the solution as t -> oo. Make a roughsketch of the graph of the solution, labeling the concavities and the infl ection point.

dN 1 dN6. - = 3N(8 - N), N(O) = l. 9. -- = 7- N , N(O) = 3.dt N dtdN 6 dN7. - = N(13- N), N(O) = 3. 10. -- = 31 -N , N(O) = 10., dt N dt

dN 18. 5 - = 2N(1 -N) , N(O) = -2.! dt11. A biologist begins a laboratory study of a certain animal populat ion at a time

1 = O, when the population consists of 18 individuals. On the basis of the datacollected during the experiment the biologist finds that the differential equation

dN 1- = -- N(30- N)dt !,000describes the growth and regul a tion of the population . Solve this equation for thepopulation size N as a function of 1, and sketch the graph of this function . Is therean ins tant of time when the growth rate of th e population is a maximum Explainyour answer.

12. Proceed as directed in Exercises 11 , given that the differential equat ion

dN 1- = - - N(119 - N)dt 50describes the growth ancl regulation of an experimental animal popul ation andthat at the start of the study, when t = O, the population con sists of N (O) = 62individuals.

13. Tbe uiiTerent iai equationdN 1- = -- N(l ,SOO- N)dt 1,000

describes the growth and regulation of a fish population in an empty region of thesea. At time t = O tbere are N(O) = 2,000 fi sh. Find N as a function of t; sketchtbe graph of this function. De scribe the behavior of N at t approiches infinity.Is the population increasing or decreasing in size? Explain.

14. The differential equation104 dN = N(10 3 - N)dt

describes tbe growth and regulation of an an ima population, wbere N denotes thenum ber of individuals and where t denotes the time, with t- O. I f N(O) = 1,800


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8.3 The Logistic Equation: Some C/assica/ Examp/es 439at t = O, sketch the graph of N as a funct ion of r. Wi ll thc popula t ion cver exceed1,800 individuals? Why? Is it possib le that at some time there will be fewe r than1,000 ind ivid uals? Give a precise explanation for your answer.

15. A biologist is observing the growth of a co lon y of bacter ia. Let A de note the areaoccupicd, in square ccntimeters, by the bacteria! co lony, and lct t denote the timein days . Th e bio logist finds tha t the va lues for the occupied a rcas, calculated fromthc form ula

soA

are in extreme!y clase agreement with bis observations of the a reas occupied eachday. F ind the differential equat ion - that logistic equation- which describes thegrowth an d regulation of this bacteria! colony. Sketch the graph of A as a funct io nof t, showing concavit ies and inflect ion points.

16. A biologist is study ing th e growth ofa co lony offrui t flies . Let N denote the numberof fruit flies an d let t denote the time in days. The bio logist finds tha t the formula

1,066N = ------;;-=+ 8le o.z,is in close agreement with the corresponding census counts made each day. Findthe differential equa tion tha t describes the gr01vth and regula tion of th e colony.Sketch the graph of N as a funct ion of t, showing concavities an d inflection po in ts .

In Exercises 17 - 24, solve the differential eq uation for y as a function of x. Go throughea eh step in the dcrivation of the solution.

dr 1 dr17. - = 4y(l - y). 21. = 4- _r.dx 2y dxdv dy 3y2 .8. - - = 2y(5 - y) . 22. - = 6y -dx dxdy l . dr 8yz.9. - = 2y - 23 . __:__ = 4y -dx dx

20. dr - y+ l . 24. dy -2 y + 4l .dx dx

8.3 The Logis tic Equation : Sorne Classical ExamplesIn this section we consider the matter of how, in the course of studying thegrowth of a population, one passes from observations of experimental datato sorne logistic equation that describes that growth in a satisfactory way.We shall treat this consideration by relying on severa! classical examples.


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440 8 First Order Ordinary Di.fferential Equations

Examp le l. In 1913, Tor Carlson grew Saccharomyces, a common type of yeast,in laboratory cultures. He began his experiment by placing a few yeast cells ina suitable nutritive solution. Cell growth was rapid during the early stages of theexperiment, giving rise to an increasing population size.

Table 8.1The growth of Carlson's yeast population

Values of theAmount of yeast, N, logistic function,t

Hours, t as measured by Carlson* N= 665 /(l + e . lB96-0 .5J55r)

o 9.6 9.91 18.3 16.82 29 .0 28 .23 47.2 46.74 71.1 76.0S 119. 1 120. 16 174.6 181.97 257.3 260.38 35 0.7 348.29 441.0 433.910 513.3 506.9

11 559.7 562.312 594.8 600.813 629.4 625.814 640.8 641.515 651.1 65 1.016 655.9 656.717 659 .6 660 .1] 8 66 1.8 662 .1

..* This experiment is discussed by To r Carlson in bis original pa per,"Uber Geschwindigkeit und Grosse der Hefevermehrung in Wrze,"Biochemische Z eils chriji, 57 (19 13 ). These da ta a re plotted in Figure 8.18.

1 This logistic function was introduced, to fit the experimental data, byRaymond Pearl , in h is The Biology ofPopulation Gro\\'th, Alfred A.Knopf, Ne w York, 1925, p. 10 .

Instead of taking actual census counts of the yeast population, Carlson measuredthe quantity of yeas t at intervals in the growth indirectly, with a centrifuge. Thesemeasurements are given in Table 8.1, and they are numbers proportional to whatwould have resulted had actual census counts been taken. Indirect .methods ofmeasuring the size of a cell popul ation, such as with a centrifuge, are often usedin the laboratory beca use of the extremely small size of individual cells; the detailsof these methods need no t concern us he re.


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8.3 The Logis tic Equation: Some Classical Examp/es 441Rather, our purpose is to find a logistic function

[148]

that f1ts the data given in Table 8.1, where the positive constants K , k and e mustbe found from these data. We obtain estimates for these constants in the followingway. First, we express the quotient

K - NN

where N and K are given by [148], fn terms of t:

That is, we obtain

K- N 1-- = (K - N) -N NK ) 1 + ce - Kkr

1 + ce - Kkr K= ( - 1 . )(1 + ce - Kkr)1 + ce-Kkr= (1 + ce - Kkr - 1)

K -N= ce - KkrN ,

and if we take logarithms of both sides of this equation we haveK- N ,I n ~ ~ - = In e+ In e-KkrN

= ln e- (Kk)t.Thus we see that ln[(K - N)/N] is a linear function of t:

K - NIn - - = ln e - (Kk)t,Nwhose graph is a line with the negative slope - Kk, as shown in Figure 8.17.

[149]

Now, we can formulate our line of attack, as follows. The experimental datagive n in the second column of Table 8.1 determine the values of N(t) at certaintimes t. These data are plotted in Figure 8.18 and we estmate a value for thecarrying capacity, K, from this plot, by eye. A reasonable estmate for this upperasymptote is

K = 655.


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K-NIn --:-

o

Figure 8. 17

K-NIn--;:--- = In e - (Kk)t

The graph of [ 149], which is the so-called lin earform of the 1ogis tic function [1-iSJ.

NAsymptote = 665

665+ e4.187 - 0. 533 r

- Inflection at N= 332.5.In e 4. 187 _at t = Kk = 0_533 = 7 8)6 hours

o 2 4 6 8 1o 12 14 16 18 20Time t, in hours

Figure 8.18The growth of a popu lation of yeast cells. Tbe plottedpoints represent !he experimental da ta due to Carlson,as recorded in Table 8.1. The smooth curve is the graphofthe logistic fu nction N = 665/(l + e4 1B? - 0.5 3 3') ,and fits the experimental data ciosely.


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8.3 The Logistic Equation: Some Class ical Examples 443Now that we have chosen a va lue for K we may compute the va lues of(K - N)/Nat those times t listed in Table 8.1. These values of (K - N)/N are given in thethird column of Table 8.2. In the fourth column of Table 8.2 we find the corresponding values of ln[(K - N) jN]. We are now in a posit ion to plot thesevalues ofln[(K -N ) /N] as a function oft; this \VC do with Figure 8. 19. Accordingto [ 149] this plot should be a str aight !ine. But wc can see that the po ints plottedin Figure 8.19 do not all lie exacrly on a straight line; however, they do [alinear /yon a straight line and our problem is to choose a stra ight line that best fits theplotted po ints.

We choose thi s line by eye, with the aid of a little good luck. When we canno tbe so fortunate, we should prefer to have a more sophisticated technique forchoosing such lines, and such a technique, called the "method of linear leastsquares," is developed in Section 11.3 . Our choice is a line tha t passes throughl\VO of the plotted points of Figure 8.19. An obviously bad choice would be thatline passing through the points (12, - 2.137) and (13, - 2.865), for a ll ofthe otherplotted po ints lie al quite a distance from that line. A much better cho ice is theline that passes through the points (5, 1.522) an d (14, -3.276), as can be seen inFigure 8.19. These points correspond to times at which observations of the experiment were made , times of r = 5 hours and t = 14 hours, respectively. T he

Table 8.2Computations based on Ca rl son's da ta

Amount of yeast. N, K - N* K- N'In - - -1-!ours, t as measured by Carlson N .\"o 9.6 68.270 4.2241 18.3 35. 340 3.5652 29 .0 21.930 3.0883 47 .2 13.090 2.5724 71.1 8.353 2.123S 119.! 1!.584 1.51:26 174 .6 2.809 1.0337 257.3 1.585 0.4608 350.7 0.896 - 0 .1109 441.0 0.508 -0.67710 513 .3 0.296 -1.2 19

11 559.7 0.188 -1.67 112 594.8 0.118 - 2.13713 629.4 0.057 - 2.86514 640.8 0.038 -3.27615 651.1 0.021 -3.84716 655.9 0.014 - 4 .27817 659.6 0.008 - 4.82818 661.8 0.005 - 5 .298*K = 665.t These va lues are plotted in Figure 8.19.


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444K-NI n - : -

o 2 4 6 8 10 12 14 16 18Time t, in hoursFigure 8.19The growth of Carlson's yeast cells is plotted in linearform. The plotted points represen! the exper imentaldata from Table 8.2. The solid line passing throughthe data points (5, 1.522) and (14, - 3.276) is 2 goodfit to this da ta . The dashed line passing through theda ta points (12, -2 .137) and (13, -2.865) is anex tremely poor fit.

slope of the line through these points is- 3.276 - 1.522------14 - 5 4.7989 -0.533.

Now, as the equation ofthis line is givcn by [149], we see that we must haveKk = 0.533;

hence, in particular,0.533 0.533

k= K= 655 We conclude from [149] that the equation of our line is

655- Nln ~ - = In e - (0 .533)t,Nwhere the constant e is determincd by the condition that the line passes throughthe point (5, 1.522), as follows:

at t = S, 655- NIn---- = L522;


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8.3 Th e Logis tic Equation : Some C/assical Exampleshence, substituting these values into the equation of the line, we obtain

or,

1.522 = In e - (0.533) 5= In e - 2.665,In e= 4.187.

Exponentiating this last equation, we obtain

445

Now , co llet ting all of this inforrnation, we see th at our logistic fun ction [148]takes the forrn

665N= 1 + e4 . 187 e 0. 533t'or

655 [150]= + e4.1S7 0.533 r

The graph of this function appears in Figure 8.18. Our logistic function [150]differs on ly slightly frorn the function

665N = - - - - ~ ~ ~ ~ + e4.1896 0.5355 rthat was found by Raymond Pe ar l; sorne v


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446 8 First Order Ordinmy Differential Equations

Example 2. We now pass from yeast, a unicellular plant, toa multicellular animalthat is much higher than yeast on the biological scale. Drosophila melanogaster,sometimes called the fru it fly or vinegar fly, was studied by the biologist RaymondPearl in the early 1920s. D. melanoga ster is a good choice for Jaboratory experi-mentation, because it breeds rapidly and has a short life cycle so that an entiregeneration, from adult to adult offspring, on the average and under suitableconditions, passes in about 10 or 11 days. Thus Pearl was able to create a smalluniverse in a one-pint bottle containing sufficient food for a viable population of

Table 8.4

Table 8.3Raymond Pearl's population ofD. melanogasterTotal population,22 Aies

Male Female23 34 42 21

Age in da ys

12305595

Growth of Pearl 's wild-type Drosophila population in a pint bot tleVal ues of theDate Ob se rved po pulation logistic function.'of census Days, t as co unted by Pea rl* N = 1,035/(1 + e4 .27-0 .I 7

December 2 o 22 14.3December U 9 39 61.0December 14 12 105 96.7December 17 15 152 150.2December 20 18 225 226.0December 23 21 390 326.0December 27 25 499 488.4December 29 27 547 574.1December 31 29 618 656.8January 4 33 79 1 798.4January 7 36 877 877.1January 10 39 938 932.9* This expe riment is discussed in deta il by Pearl in his book , T he Bio!ogy of Population

Growth, Alfred Knop f, New York , 1925, Chap te r 2. These data are plotted in Figure 8.20.t This function is dueto Pearl. The variable 1 denotes thc time in days, start ing with t = Oon Decemb er 2.


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8.3 Tlze Logistic Equation: Some Classical Examples 447


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K-NIn - - : -

Table 8.5Computations based on Pearl's dataK- N*

Days, t Obse rved population , N No 22 46.0459 39 25.53812 105 8.85715 152 5.80918 225 3.60021 390 1.65425 499 1.07427 547 0.89229 618 0.67233 791 0.30836 877 0.18039 938 0.103* K = 1,035.1 The values a re pl otted in Figure 8.21.

Slope - 0.166

. / (1 8, 1.281)

Days, t, of durationof experiment ,

K- N 1In---N3.8303.2402.1811.7591.2810.5030.072- 0.114

- 0.397- 1.176-1.7 14- 2.269

from t ~ O ~ December 2to t 39 January 10 (36 , - 1.714)/

Figure 8.21The growth of a population of Dro sophila plotted in linear form. The do ts represent Pearl'sexperimental data, shown in Table 8.5. The straight lin e chosen to lit the dat a passes throughthe po ints (18, 1.281) and (36, - 1.714)_.


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8.3 The Logistic Equation: Some C/assica/ Examp/es 449points correspond to the observation times of t = 18 days (December 20) andt = 36 days (January 7), respective!y, and the slope of the line is

From [149] it follows thatand

-1 .714- 1.281---- - = - 0.166.3 6 - 18Kk = 0.166,

0.166k=--.1,035Thus the equa tion of our Jine for Figure 8.21 is

1.035 - Nln = ln e - (0.166)1,Nwhere the constant e is determined by the condition that the line passes throughthe point (18, 1.281 ), as follows:

at t = 18 , 1,035 - Nln---N 1.281;hence, substituting these values into the equation of the line for Figure 8.21 ,we have

1.281 = ln e - (0.166)18,or

ln e = 4.269;hence, exponentiating this, we obtain

Coilecting all of this information, we see that in this experiment our logistic function [148] takes the form

N= 1,035 [151]+ . 2 6 9 0.166,

\vhich is extremely close to the function found by Pearl, whose values are givenin the fourth column of Table 8.4. The logis tic funct ion [151 J atisfies the difieren tia]equation

dN 0.166- = - N(1 035 - N)dt 1 ~ 3 5 , ,


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450 8 First Order Ordinary Differential Equationsand the condition

N(18) = 225 .

The graph of [151J is sketched in Figure 8.20, where we see tha t it describes thegrowth ofthe experimental populat ion of D. melanogaster in a satisfactory manner.

Example 3. In this final example we consider the growth of the population ofSweden from 1750 to 1920. Sweden is a good choice for study because it has oneof the longest continuous records of census coun ts of any country, and exhibitsa growth tha t is well described by a logistic function. Th e size of the populationof Sweden a t its various census counts is shown in Table 8.6, an d these da ta areplotted in Figure 8.22. By a mathematical analysis similar to those . analysespresented in Examples 1 and 2, we find from the data ofTable 8.6 that the logistic

Table 8.6Th e population growth of Sweden

Values of theYear Time t, Population, lo!:!istic function .'of census in years in millions, * N= 1.535 + .336/ (1 + 7265e - 0 02 3 ' )1750 -50 1.763 1.8001760 -40 1.893 1.8641770 -30 2.030 1.9441780 -20 2 .1 18 2.0411790 - 10 2.158 2. 1601800 o 2.347 2.3021810 10 2.378 2.4711820 20 2.585 2.6691830 30 2.888 2.9001840 40 3.139 3.1621850 50 3.483 3.4551860 60 3.800 3.7761870 70 4.168 4.1191880 80 4.566 4.4771890 90 4.785 4.8411900 100 5.136 5.2021910 110 5.522 5.5491920 120 5.904 5.876* These data are from Pearl, Studies in Human Biology, Williams & Wilkins Co., Baltirnore,1924, Chapters 24 and 25 . A plot is given in Figure 8.22 .1 This function is dueto Pearl, in The Biology of Population GrO\Hh , Alfred A. Knopf,New York, 1925, p. 11. The graph is sketched in Figure 8.22.


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8.3 Th e Logistic Equation: Some C/assical Examp/es 45 1function

6.336N = 1.535 + 1 + 7.265e o.o231 [152]

describes the population growth of Sweden with extreme accuracy. In this equation N denotes the pop ulat ion in million s, and t denotes the time in years withe = Ocorresponding to the year 1800; thu s, for example, t = -30 corresponds to1770, and t = 20 corresponds to 1820. The graph of this function is given inFigure 8.22. Fo r the sake ofcomparison the values ofthe f u n c t i o ~ [152] a re givenin Table 8.6, where we see that they are extremely close to the correspondingcensus cou nts. Th e function [152] satisfie s the differential equation

N

o2 ) f - s 1: ' : 1< 4 - 1g 1"'o.o0...

Figure 8.22

dN 23- = - (N - 1.535)(7.871 - N)dt 633.6

Uppe r limit of growth cycle = 7.87 1

Lowe r limit of growth cycle = 1.535

Time t , of census , in years and in fifty-year periods

The population growth of Sweden. The dots represent actual census co unts. Th e smoo thcurve is the graph of the logis tic function N = 1.535 + 6.336/(1 + 7.265e- 0 023 ') , which fitsthe census counts closely.


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452 8 First Order Ordinary Dijferential Equationsand the condition

N(O) = 2.302.

Exercisesl. In 1922 H. G. Thornton observed the growth of a bacteria! colony of Bacil/us

dendroides. The arca occupied by the colony was measured each da y for six consecutive days; the measurements, in sq uare centimeters, are list ed in Table 8.7. Usingthese data , find a logistic function that expresses the area N , in cm2 , occupied by theco lcmy as a function of time t in days. Ca lcula te the area according to your logist icfunction at t = 2 days and at t = 4 days, and compare these va lues with thoseobserved in Table 8.6. Find the corresponding logistic equation- the appropriatedifferential equation-satisfied by your logistic fu nction. Finally, sketch the graphof your logistic function, showing also all of the observed data points of Table 8.7 .

Table 8.7Growth of Thorn ton 's bacteria! colony*Age of colonyin da ys, 1

o12345

Obse rved area, N,in square centimeters0.242.7813.5336.3047.5049.40

* See H. G. Thornton, "On the Development of a Standardised Agar Medium forCounting Soil Bacteria," Ann. Appl. Bio/. ,9 (!922): 265.

2. The biologists H. S. Reed and R. H. Holland stud ied the height to which sunflowerplants grow. The mean heights of the sunflower see'ct!ings they observed are listedin Table 8.8, in cm; the measurements were taken once every seven days. With thesedata find a logistic function that expresses tbe mean height, H, in cm, of the sunftower plants as a function of time t, in days. To simplify your calculatio ns use on lythe measurements taken on the following days: t = 14, 28, 49, 56, and 70. Calculate


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8.3 The Loistic Equation: Some C/assica/ Examp/es

Tablc 8.8Growth in height ofsunllower plants,acco rding to Reed and Holland*

Da ys, r714

2 1283542495663707784

O bsc rved mean heighl H.in een li mcters17 .9336.3667.7698 .1 0

131.00169 .50205.50228.30247.1025 0.50253.80254.50

* See H. S. Reed an d R. H. H o ll and ,"'Th e Growth Ra te o f a n An n ual Pla ntHelianrlllls."' Proc. Nar. Acad. Sci. U.S.A.,5 ( 1919 : I.JO.

453

the height s according to your logist ic function at t = 49 days and at t = 70 days,and compare these values with the o bserve d values listed in T ab le 8.8. Find thecorresponding log istic equa t ion sat isfied by your logistic function. Sketch th e graphof your 1og istic functi on, showing also a ll of the observed data points of Table 8.8.

3. T. B. Robertson used an S-shaped curve to describe the grO\vth of the pumpkinCucurbira pepo. The data he ga the red appear in Table 8.9 . Twenty-one observa tions11e re ma de on successive days, and weights were meas ured in grams. Using thesedata. fi nd a log ist ic function that expresses the weight W, in gm , as a function ofthe time 1, in days. For simplicity in your calcula ttO ns, use oniy the weights oi:Jservedon th e fo llO\ving days : t = 7, 10, 12 , 18, and 21. Calculate the weight accordingto your logistic function at t = 17 days and at t = 24 days, and compare theseva1u es 1vit h the weight s listed in Ta b le 8.9. Find the differe ntial eq uation that issa tisfied by your logistic function. Sketch th e graph of your logistic function,showing a ll of the ob serve d da ta points of Tab le 8.9.

4. Th e fruit fty experiment carried ou t by Pearl, discussed in Example 2 of thi s section,dealt with the normal wild-type Drosophila melanogaster. Pearl also studied anothertype of D. me/anogaster, a mutant type Quintuple, so named because five of itspheno typic characters are in mutant fo rm . In this experiment a pint bottle wasstarted with an initia1 population of 6 mutant ft ies. Th e census counts are given inTable 8.10. Using these data, find a logistic function that expresses the populationsize N as a function of the time t, in days. To simplify the calculations use only thecensus counts observed on the follo wing days : t = 3, 12, 18 , 24, and 32. Calculatethe popu lat ion size according to you r 1og istic function at t = 21 days and compare


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Ta ble 8.10

Table 8.9G rowth of a pumpkinCucurbita pepo*Observed weight W,Day s, r in grams

5 2676 4437 6588 9619 149810 220011 292012 336613 375814 409215 448816 472017 486418 498019 511420 517621 524222 529823 535224 536025 5366

* See Raymond Pearl, Th eBiology of Pomlarion Gro1rth,Alfred A. Knopf, New Yo rk, 1925,p. 215 .

Growth of Pearl's population of D. melanogaster(mutant type Quintuple) in a pint bottle.Date of census Days, r Observed population, NOc tober 6 o 6October 9 3 10October 13 7 21October 15 9October 18 12 67October 21 15 104October 24 18 163October 27 21 226O ctober 30 24 265November 3 28 282N ovember 7 32 319

See Raymond Pearl, The Eiology of Population Growth,Alfred A . Knopf, New York, 1925, p. 224.


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8.4 Separation of Variables 455it with thc count obscrvcd and rccordcd in Tablc 8.10. Find thc di!Tcrential equationthat is satisfied by your logistic function. Sketch its graph, showing also all of thedata points recorded in Table 8.10.

8.4 Separation of VariablesThe method we shall discuss in this section is cal!ed separation of variables;we shall see that it can be used to salve sorne differential equations bu t not.others. i

Let us begin with a familiar example. The rate of reaction by which theen zyme trypsin is formed from its enzyme precursor trypsinogen was discu ssed in Section 3.3 and Section 7.4. Let o and b denote the concentrations,in moles per liter, of trypsinogen and trypsin, respectively, present at thetime t = Ohours. Let y mol/liter denote the concentration of trypsin formedfrom trypsinogen wh en t hours have elapsed; then the total concentration oftr ysinogen present at t hours is a - y because ea ch molecule of trypsinogenyields o ne molecule of trypsin . Experiments show that the reaction ratedy fdt is proportional to the product (a - y)(b + y) of the concentrations oftrypsinogen and trypsin present ; that is to say, there is a positive constant ksuch that

dydt = k(a - y)(b + y). [153]We divide cach side u fthis differentia l equation by (a - y)(b +y) to obtainthe equat ion

,___ dy = k(a - _r) (b + y) dt [154]It is shown in Ex ample 1 of Section 7 4, that an indefini.te integral of theleft-hand side is

f 1 dy dt = In lb + yl + e(a - y)(b + y) dt a + b y - a 'where C is a constant of integra tion ; an integral of the right-hand side of[154] is kt; hence

1 lb+ yl- b In - - + e = la,a + y - aand we must solve this equation for y as a function of t.


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456 8 First Order Ordina1y Dijferential EquationsFirst muitipiy each side by -1 ; then appiy the fact that - In u = ln(1/u),

to write- - In - - -C= -kt .y- aa+ b b+ y

Now multiply each side by a + b, then add (a + b)C to obtain

\y - aIn b + y = - (a+ b)kt + (a + b)C.

Exponen tia te each side of this equation to obtain

If we let e = e


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8.4 Separar ion of Variables 457so the numerator a + be = O; hence e= -a j b is negative. Now substitutee = - aj b into [155] to obtain

a+ b(-a j b)e -


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458

-o""'E2.S : ~ - - - - - - - - - - - - - - - -t .+;::-=:::.4-


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8.4 Separation of Variables 459They are, respective]y, th e exponen tia] growth equatiori, the logistic equation,and the equation for th e rate of the reaction of trypsinogen -> trypsin. Eachequation is a relation between so rne unknown function, either N = N(() ory = y(t), and its de riva ti ve with respect to time. Be cause of the presence ofth e derivative, th ese are ca lled differential equa tions. In each, our mathemati-cal objective is to so lve th e equation for N or y as a fun c tion of . The techniquein each is to bring the function, N or y, and its derivative both to one sideof the eq uation, as was done, for instance, in going from [1 53] to [154], andthen to perform an integration with respect to t. In the remainder of thissection we studyo ther differential equations by using the same meth od, whichwe shall no w outline : it is the method of separation of variables.

Let us formu late our programas follows. Suppose that y and x are va ri ablessuch that y is a fun ction ofx, th at is , y = y(x). Suppose we know th at y = y(x)an d its deriva tive dy j dx must sa tisfy the equation

di'-1 = f(x)g(y)IXfor each x, where f(x) and g( y) are kno\\'11 functions of x and y respectively.We mu st sol ve th is eq uation for y explicitly as a function of x.

The firs t step in do ing this is to bring a ll terms with y in them to the sameside of the equa tion as the derivative dy j dx, and to bring a ll te rm s with x inthem together on the other side. This is achie ved by dividing both sides ofthe equa tion dy j dx = f(x )g(y) by g(y) to obt ain

_ 1_ dy = f(x).g(y) dxNow we in tegrate each side of th is equation with respect to x, and we ha ve

or f 1 dy f- dx = f(x) dx,g(y) dxf -1- dy = J(x) dx.g(y)

Assum ing th at th ese integrations can be performed, we obtain an equationinvolving y and x, but no derivatives, and we solve our la test equation for yin terms of x , whenever po ssible.


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460 8 First Order Ordinary Dif(erential Equations

Example l. Suppose that y is a function of x such that for each x, y = y(x)satisfies the equation

dy ?-= x-y.dxIn other words, for each x, the derivative dyjdx is equal to x 2 times y. Find yexplicitly as a function of x, given that at x = O, y(O) = l. In the notation usedabove, f(x) = x 2 and g( r ) = y. We divide both sides of the equation by g(y) = yand we have

1 dy ?-- = x- .y dxBy integrating both sides with respect to x, we obtain

or

After exponentiation,

S dy dx = Jx 2 dx ,y dx1ln y = - x 3 + C.3

where k = . We are given that at x = O, y(O) = y(O)i = 1; sok= ke0 = y(O)i = 1;in short,

as required. We drop the absolute value sign because r(O) = 1 > O.It is interesting to check directl y that the function y= cx '- 3 satisfies the given

differential equation dyjdx = x 2y:dy d x ' f J- =- - ( e )dx dx

holds for all values of x. l t is true that the function y = ex'13 is the only functionthat satisfies dyjdx = x 2 y and the initial condition y(O) = l.

Example 2. Suppose that y and x are variables such that the rate of change fy with respect to x satisfies the differential equation


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8.4 Separarion 4 Variables 461Suppose, further, it is given th at y(O) = 3 at x = O. Our task is then to express yexplicit ly as a function of x. Recall our sta ted method and think of f(x) = x andq(.l') = 1'-v in the differential equation we are working with . We divide each side ofthe differentia l eq ua tion by e- ) and we have

dve' _::_ = x.dxWe integrate with respect to x, to obtain

f dy f-dx = x dx,dxor

1 ,er =- x- + C.2T ak ing logarithms of each side, \Ve find

To determine the value for the constant C, we use the given condition _r(O) = 3,as follows:

3 = y(O) = In C;by ex ponentiat ion,

Thcrefore, the required fu nct ion is

( 1 , '\r = In :2 x- + eo )' J

Both of the examples given so far have used the method we discussed:separa tion of variables. The method may be used to solve those differentialequations where the derivative dy jdx is equal to the product of a functionf(x), of x alone, anda function g(y), of y alone. H owever, an ex ample wherethis method cannot be used is the equation dyjdx = x + y, because x + ycannot be expressed in the form f( x)g(y).

In the next three examples we consider biological applications in whichth e method can be used successfully.

Example 3. Let us consider a problem dealing wilh birth and im migration. Le tN = N( t) denote the size ofa population at time t ; we shall suppose that the birth


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462 8 First Order Ordinary Differential Equationsrate b minus the death (mortality) rate mis proportional to the size N:

b- m= kN ,where k is a positive constan . Fu rthermore, suppose that the rate of immigrationminus the rate of em igrat ion is a positive co nstan! , let us say A indi viduals per un ittime. Wha t we must do is express N explicitly as a function of time.So lution. First of a l!, the populatio n growth ra te dNjdt satisfies the equat ion

ordN- = b - m + Immigration rate - Em igrat ion rate,dtdN- = kN + A.dt

In arder to sa lve this Iast equation, we may separa te variab les, by dividing eachside by kN + A, to obtain

dN- - -k N + A dt l.Integrating each side '.Vi th respect to t, we obtain

orf l dNd t = f l d t ,kN + A dt

1- In [kN + A[ = t + C.kNow we multip ly each side by k and exponent iat e, to have

kN +A = e',where we have dropped the ab solute value because kN + A > O and wheree = eke_Solving fo r N,

e k t AN = - e - -k kLeting N 0 = N (O ) den ote the population size t the time t = Owe have

o r

Thus,

e 0 A e AN 0 = - e -- = - - -k k k k 'e A- =No + - .k k

( A) kt A= N 0 + - e - -k . k: ..


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8.4 Separation of Variables 463or

We see th at the growth o f the population dcpcnds upon two terms: first, thecxponcntial funclion N 0 ek', which dcpcnds upon/; alonc: ami sccond, (,1 h)(ek' - 1 ,which is also rapidly in creas ing bu t dc pencis on both k and A and is thc con-tribution to growth du e to net immigration.

Examplc 4. Allometry is the measure and st udy of the relative growth of a partof an organism with respec t to thc en ti re o rganism. In this example we cons ideranallometric law. Let t denote the time an d Jet x = x(t) denote the size of an organin a gi ven individual a t a time instant t. Th e size x ma y be th e length , vo lume, orweight of the organ . Th e gro11th rate of th e organ is given by the derivative dx jdt .Relatite groll'th rate is defined to be th e quotien t of the growth ra te dx fdt by thesize x: m a thematically .

1 c/xx c/r

Let _r = y(r) denote the size of anoth"er organ in the same individual at the sametime insta nt t. Emprica! ev idence lends plausibility to the asse rtion th a t the relativegrowth rates,

1 dry dr an d

1 dxx dr

are proportional. Th a t is to say. there is a positi ve constant k, depending only onthe two parts of the indi vidual undc:r consideration, such that the equatio n

1 d_r , 1 dx-=.1( --.r dt X dtis valid a t each time instant t . From this equation 1ve find a relation between y and x,as follo ws. Integra te eaeh side of the equation with respect to t to obtain

orS dy S 1 dx--dt = k - - dt)' dt . X dt '

In y = k In X + e,where y > Oan d x > O. By expo nentiation,

ory= ex\


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464 8 First Order Ordinary DifJerential Equationswhere e = e' and xk = ln ' . Thus y is a power function of x, where the exponentis k > Oand the coefficient is e = > ONotice in particular that the relationy = axk is independent of time, so this relation is not affected by the time rate ofgrowth of the individual.

Example 5. The Gompertz growth curve is the graph of a solution of the differential equation

dN-- = kN(A - In N)dt , [157]where A and k are positive constants. This model differs from the Jogistic modelof Section 8.2 in that the factor K - N, the unutilized opportunity for growth, inthe logistic model has been replaced here by A - ln N. As we shall see, this modelis also one oflimited growth. Let us solve the differen tial equation [157].

First, separate variables,dN

N(ln N- A) dt -k ;

then integra te each side ofthis equation to ob ta inlnjln N- Aj = - k t + C

where C is the constan! of integration. The integration

J-. 1- - dN = lnjln N-- Aj + CN(ln N- A)[158]

is carried cut with the substitution u = In N - A , as then du = (1 / N) d.V, so

J l dN = J~ d u = lnjuj + C' = 1n[ln 1\' - A[ + C.N(ln N- A) uExponen tia te both sides of [158] to obtain

Ifwe denote by -e , then we may remove the absolute value sign and write

then exponentiate: [159]The constant e is determined by the initial value N 0 of N at the time t = O, as

follows:


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8.4 Separa/ion 1 Variables 465now take logarithms, to have

In N 0 =A- eore= A- In N 0 .

Let us sketch the graph of the solution [ 159]. In order to have growth, of necessitydN jdt must be greate r th an zero; hence A - In N > O, as can be seen in the equation [157]. In particular, at t =O we must have e= A - In N 0 >O. Thu s, let usassume tha t In N 0 < A; then in [ 159] th e co nsta n! e = A - In N 0 is pos itive,which imp lies that

is pos itive. No w we shal l ca lculate th e de riva tive of each side of [157], using theproduct formula for differentiation o n the righ t hand s ide :

d2 N ( 1 dN) dN- 0 = J.:N - - - + k-(A - I n N)dr- N dt dtdN= J.: -(A - 1 - In N).dt

In th is expression for the seco nd derivative the factors J.: and dN jdt are positive,so the concavity of the gra ph of [ 159] is determined by the factor A - 1 - In N.In particular, the graph is concave upwa rd if A - 1 - In N > O, an d it is concavedown ward if A - 1 - In N < O. There is an infl ection po int a t th e va lue of Nthat sa tisfies the equa tion A- 1 - I n N= O, or In N = A- 1, or N= eA- 1;the correspond ing va!u e of t sa tisfies the equation

e.-1 ---1 -:= eAe - ce - kt'

obtained fro m [1 59]. D ividing ea c:h side of the last equation by eA - 1, we obtainthe equation

we take logarithms, andO= -ce-k' + l.

Thus e - kc = 1/c, so - kt = ln(l jcJ = - In c. It follows thatIn e ln(A- In N 0 )t = - = ---'---,-----=:.:.k k

is the locat ion of the on ly point of inflection. T he graph is sketched in Figure 8.24.The reader should verify that if t O,and the graph is concave upward; if t > (In c)jk, th en N> eA- 1 an d the graph isconcave downward. As t--> w, e -kc --> O; hence N approaches eA, as ca n be seenin [ 159].


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466NeA ------------------------

NoO (A - In No)

k

Figure 8.24Th e Gompertz growth curve is a solution of thedifferential equation dN jdr = kN(A - In N ),where k andA d e n o ~ positive constants. Thecurve describes a population in an environmemwherc there are limits to growth. The maximumsize is eA, whereas the maximum growth rate isat the size eA- 1.

A Gompertz growth curve was studied briefiy in Example 4 of Section 2.3. Thefom1 used there was

it is instructive to compare that earlier discussion with the present one. The mainpoint' in this section is that we began with the differenta l equation [157], whichexpresses how the growth rate dN jdt depends upon the size X; from this equationwe derived the explicit relation [159] between size and time. Notice that the upperlimit on the size N is eA and that the growth rate dNjdt begins to decrease whenthe size N is equal to eA- t = eA j e, which is less than one-half the maximum sizeeA because e > 2; the reader should compare the location of th is maximemgrowth rate with that of the logis tic model.

In each example discussed so far it has been possible to solvc for thedependent variable- either N or y-as a function of the independentvariable-either t or x. There are cases, however, where it is not possible towrite down the relation expressing the dependent variable explicitly as a

___ _[unctiqn of the incjepen_dent variab le, even though an equation that relates


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8.4 Separa/ion of Variables 467the two variables can be obtained- an equation free of derivatives. As anillustration, consider the following example.

Examplc 6. This example makes use of an equation, called the M ichaelis-Mentenequation, which describes the kinetic properties of many enzymes an d so is importan! in biochemistry; this equation is named after Leonor Michaelis and MaudMenten, who 11rst proposcd the model from which it is obtained. A substance S,called a substrate, is ac ted upon by an enzyme E to yield products P. Let y = y(t)denote the concentration of the substrate at the time t; the usual notation isy = [S] or y(r) = [S],. Th e velocity of the reaction is by definition the rate v =- dyjdt, which is positive because the concentration y = [S] of the substratedecreases as t increases , which is to say, dyfdt < O. lf v0 = - (dyjdt)0 denotes theinitial velocity an d y(O) = [SJ0 denotes the initial concentration at th e time t = O.then th e Michaelis-Menten equation is

V[S] 0t o = K + [S]o Vo.roK+ Yowhere V0 and K ar e positive constants that depend on the substrate and the enzyme.N ole : Fora deri vation of this eq uation , see Exercise 26 at the en d of this section.or the discussion given in R. C. Bohinski, Modem Conceprs in Biochemistry, Allynand Bacon, Boston, 1973, from which our derivation is taken .

Notice that the equation is 1101 a differential equation but is an equation expressing a relation at the initial time t = O; hence !he J'vfichaelis-JJemen equarionmar becolled un im ;m lse eqll(ttion. Tlic ini tial :clocity r 0 is the most accurate measur;:mentof activity iro. e n z y m e - c ~ 1 1 reactions because of severa! variabl es that changethe re;:ction rate w ith ri me. Let us assume, ho11 enT . that this cquation holds for atime arter thc: reaction has begun. Thu s, as r --" -- dr , ril . we obtain the difTerentialecu a tion

dydt

VoY- ---K+ y'and we solve this by separation o f variables.Solution. We have

K +y dyy dt

or+ 1) dy = - Vo,y dt

and integrating with respect to t, we obtainK In y + y = - V0 t + C


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468 8 First Order Ordinary Dijferential Equationswhere e is the constant of integration. Let Yo = y(O) at t = O; then

K ln Yo + Yo = e,so

K 1n y + y = - Y0 t + K ln Yo + YoN otice that we canno t solve this equation for the concentration y as a functionof the time t in any simple way. It is nevertheless possible to sketch the graph ofof y as a function of t . We simply sketch the graph of tasa function of y by solvingthe last equation fo r tasa function of y, thus:

K 1 K 1t = - - In y - - y+ - In Yo+ - Yo;Y0 Y0 Y0 Y0y

2.

Yo -o

o Yo yAmount of substrate, y Time, t

(a) (b )u

Time , t(e)

Figure 8.25At (a), th e grap h of th e func tion t = - (K/V0 )1n(y jy 0 ) - (1/V0 )(y - y0 ). At (b), thegraph of the solution y = y(t) of the di fferent ial eq uation dyjdt = - V0 yj(K + y)th a t satisfies the initial condition y(O) = Yo at t = O is obtained from (a) byreflection in the !ine t = y. At (e), the graph of the amount of product, ti = Yo - y,as a func tio n of t. The initial ve loci ty is v0 = (dujdt) 0 = - - (dyjdt) 0


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8.4 Separation of Variables

hence

anddtdy

K 1 l-----


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470 8 First Order Ordinary Differentia/ Equations7. Pepsinogen is converted to pepsin in the stomach during digestion in such a way

that one molecule of pcpsinogen yields on e molecule of pcpsin; this we discussed inExercise 1 in Section 3.3. Let y = y(t) mol es per liter denote the concentration ofpepsin for med at t hours after th e s tart of the reaction atO hours ; let a and b denote.respect ive]y, the initia l concentra t ions of pepsinogen an d pepsin such th a t 11 > h.E;{periments show that the reaction rate dyjdt. is pro portional to the product ofthe concentrations of pepsinogen and pe psin. Find y as a function of th e time t,given tha t y (O) = Oat t = O; sk etch the gra ph, show ing concavities and inflectionpoints. Find the time at which th e reaction rate is grea tes t. What is the beha vior ofthe concent ration y as t --> co?

8. H ydrogen and iodine combine chemically to form hyd rogen iodide in su ch a waythat one m olecule of hydrogen co m bines with on e molecule of iodine to fo rm twomolecules of hydroge n iodide , thus: H 2 + ! 2 --> 2HI. Let .r moles per liter denotethe concentration of hydrogen th at has und ergone cbemical change at th e time 1;let a moles per liter denote the in itial concentration of hyd rogen, a t t = O; and le t bdenote th e initial concentration of iodine . The reaction ra te dy/dt is proport ionalto the pr od uct of the remaining concentrations of hydrogen and iocline. Find .r as afunction of the time t, and sketch th e graph show ing co ncavit ies ond intlection points.Find the time at which the reac tion ra te is greatest. What is the beha vior of theconcentrat ion as t --> co ? Note : F or convenience, assume tha t u < b.

In Exerci ses 9 - 24, find th e function y = y(x) , N = !Y(t), o! .r = y(l) tha t sat isfies eachdifferential equ a tion and initial condition.

dy9. - = xy,dx y(O) = l.dy 2x10. - = - - y (O) = 2.dx 1 +y 'dN tN11 - = - - N (O) = 3.. dt 1 + t2 ,dy12. -:- = - e'v2 tl t

1y(O) = 2.dy l13. dx = 1 + x' y(O) = 5.dy14. - = (t 3 + t)e -y , y(O) = Odtdy15. - = y COS X,dxdy xy16. - = - - ,dx 1 + x

y(O) = 8.

y(O) = 7.

dy y17. - - = - - ,dt t + 1 r(O) = -2 .dy18. .riO) = - 2.dt ydy 219 - _r (O) = l.. dt y3 ,

20. dy = - = . . J :dx x -r 1 ' y(O) = l .dN ln( l + r)21. - - =-- - . N(O) = 2.dr NdN22. - = 3N (4 - In N ), N (O) = 2.dtdy23. dt = 2 + .J)', 1y(O) = -.2dy l24. dt = 5y - :3 y In y, y (O ) = 2.

25. Sketch th e graph of the solution y = y(t) o[ the equat iondy 2ydt 3 + y

sa tisfying the initial condition y (O) = 5.


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8.4 Separa/ion oj' Variables 47126. De ri ve the Michaelis-Menten equa tion, given in Example 6, from consideration of

thc fo llow ing four points.Poinr l. The enzyme E combines reversibly with the subst ra te S to fo rm an

int ermed iate enzyme-substrate complex ES; this is the fasr step of th e reacrionS --+ P ancl is denoted by E + S


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472 8 First Order Ordinary D(/Jerential Equationsattained when the concentration of the intermediatc complex at t = O is equal to[E] 0 + [ESJ 0 . Thus, explain why

v0 [ES] 0V0 [E]o + [ES]o.

Point 4. Use the last equation obtained in Point 1 and the last equation obtainedin Point 3 to deduce the Michael is-Menten eq uation

Vo[S]oVo =K+ [S] 0 'where K = (k 2 + k3 )/k 1.

27. Sketch the graph ofthe Michaelis-Menten equation; that is , sketch the graph of theinitial ve locity v0 as a function of the initial concentration [SJ 0 of substrate.

8.5 Geometric lriterpretations of Differential EquationsIn this section our purpose is to give a geometr ic interpretation of thegeneral diffe rential equation dyjdx = Q(x, y) and to consider sorne examplesin this connection.

Differential equations arise naturally in the mathematical analysis ofmany .systems found in nature; we have seen various examples ofthis already.The differential equations ~ v e ha ve studied so far are of the special type

dy-1 = f(x)g(y),where f(x) is a: function of .x al one and q( y) is a function of y alone. In par ticular, if g(y) = 1 for ali y, this differentiai equation reduces to dyjdx = f lx) ,which is equivalent to the "integral" equation

y = J.f(x)dx,which is solved by finding an indefi!1ite integral of f(x); and that is preciselywhat we did throughout Chapter 7. In general , the differential equationd.v/dx = f(x)g(y) is sol ved by the method of separation of variables.

An equation that contains a function y = y(x ) and its derivative dyjdx isa "differential" equation because thc equation contains not only the functionybut its derivative dyjdx. So far we have been studying first order ordinarydifferential equations; we say "first order" beca use only the first derivativeappears in the equation, and "ordinary" because the function y = y(x) is afui1ct ion of the one variable x.


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8.5 Geometric Interpretarions o( Dijferential Equations 473But now Jet us consider general first order ordinary d i f f e r n t i a l equations

written in the formdy = Q(x, y),{X

where Q(x, y) is a known function of the two variables x and y. A solution ofthc .forcgoing equation is, by dcjinition, a ji111ction y = y(x) defincd in someintcrwl ofx such that, 1\'hen y(x) is substitutcd.for y in thc cquation, !he rcsultingequotion in x afane, id :

y(x) = Q(x, y(x)),X [160]

holds for al/ x in that interval.For example, the function y = e' 2 is a solution of the differential equation

dy = 2xv,dx as \Ve can verify immediately by substituting ex 2 for y in the equation: onthe left-hand side we obtain

which is clearly eq ual to 2xy f .r == e ~ that is to say.

holds foral! values of x, and hence y = e" 2 is a solution of dy j dx = 2xy.As a first consideration of dy j dx = Q(x, y), notice that the derivative

dyjdx is the slope of th e line tangent to the graph of the function y = y(x)at the point (x, y(x) ). Therefore, for y = y(x) to be a solution ofthe differentialequation

dv~ ~ = Q(x, y),X

the value of Q(x, y(x)), for each x, must be the slope ofthe line tangent to thegraph of y = y(x) at the point (x, y(x) ), because equation [160] holds foreach x; see Figure 8.26. Thus we may interpret the function Q(x, y) as deter-mining a direction jield in the xy-plane, as follows: at each point (a, b) in


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y

Slope = Q(a, y( a))

aFigure 8 .26l f the function _1 ' = y(x) is a solution of thed ifferential equation dr j dx = Q(x, r) then, foreach point (u , r(a )) on the graph of 1' = y(x),the slope (dr/dx)" of the graph at (a, ,r(u)) isequal to Q(u, r (uJ); that is , (d r j d:\)a = Q(u, ,r(al)for eaeh point (u, r(a) ).

y1 1 1t y = 3 - - - - - - - - l - ~ ~ ~ P _ e _ = _ L _ , ______1 1 . :--j-_Y_:}___ - - - - - - - - ) f - s ~ ~ ~ - = - - ~ - - ~ - - - - - - 1 1 11 .!' y === 1 " 1 / Slone ::::. 1 1 /--;(-------- - - - - - - - j f - - - ~ - - - ? f - - - - - - -1 1 11 1 11 1 Slope = O 1

1 11 1- - - - - - ~ ________).________

Slope = --1 1'-- - - - - - - ~ - - - - - - - ~ - - - - - -__ - - r w , , o ~ : - _

1 Slope = - 3 :Figure 8.27Th e direction field of the differential equation dyjdx = y. Th edirection at each point (x, y) is given by the line through (x, y)with slop e equal to y.

X


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8.5 Geometric Jntcrpreratious o( Dijferential Equmions 475the xy-plane , consider the line through (a, b) with slope Q(a, b); th is linedetermines a direction at the point (a, b). In genera l, the direction at thepoint (.\ , y) is given by the line through (x, y) with slope Q(x, y). The totalityofall ofthese dir('ctions is cal/('(/ thedirection field o(the differentiul equation.For example , the direction field of the equation dyjdx = y is sketched inF igure 8.27. We imagine the direction field to give rise to a Huid tlow in theplane, as shown in Figure 8.28 for th e equation dyjdx = y. The reader shou ldcompare l "igure 8.27 \Vith Figure 8.28. The flow in the plane has stream lines,where each streamline )' = _r(x) is, by definition, the graph o{ a solution of t hediff"eren tiul equntion. Thus each streamline is tangent to th e direction fieldat ea eh point of the streamline; that is, dyjdx = Q(x , y(x)) at each po in t(x, y(x)) on the curve y = y(x). Thus all of the solutions of the equationdydx = Q(x, y) are curves. or streaml i

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