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This article was downloaded by: [Florida Atlantic University]On: 13 November 2014, At: 01:09Publisher: Taylor & FrancisInforma Ltd Registered in England and Wales Registered Number: 1072954Registered office: Mortimer House, 37-41 Mortimer Street, London W1T 3JH, UK
International Journal ofMathematical Education in Scienceand TechnologyPublication details, including instructions for authors andsubscription information:http://www.tandfonline.com/loi/tmes20
The hares and the foxesCarl W. David aa Department of Chemistry , University of Connecticut , 215Glenbrook Road, U‐60 Rm. 161, Storrs, Connecticut 06268,U.S.A.Published online: 09 Jul 2006.
To cite this article: Carl W. David (1989) The hares and the foxes, InternationalJournal of Mathematical Education in Science and Technology, 20:1, 21-28, DOI:10.1080/0020739890200103
To link to this article: http://dx.doi.org/10.1080/0020739890200103
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INT. J. MATH. EDUC. SCI. TECHNOL., 1989, VOL. 20, NO. 1, 21-28
The hares and the foxes
by CARL W. DAVIDDepartment of Chemistry, University of Connecticut, U-60 Rm. 161,
215 Glenbrook Road, Storrs, Connecticut 06268, U.S.A.
(Received 9 May 1986)
A wonderful problem in chemical kinetics is routinely ignored in physicalchemistry courses because it deals with population and ecology rather thanchemistry. In this paper, this Volterra-Lotke problem is discussed in detail, withenough mathematical details filled in to provide a reasonable basis for includingthis marvellous example in standard courses.
1. IntroductionA classic problem in ecology, which can be analysed in chemical kinetics terms, is
worth some study just for the fun of it. The mathematics needed for studying thisproblem is equivalent to that employed in standard physical chemistry courses.
2. The modelImagine a closed area (such as a forest), with a starting population of hares
(rabbits) and foxes. This problem, the Volterra—Lotka problem [1], starts with thepresumption that there is enough food for the hares to thrive, i.e. weeds, in the forest.Therefore, the hares, if left to their own devices, would grow, and we will hereassume that they would grow by first-order kinetics. Secondly, we assume that thecarnivorous foxes would die out if there were not enough hares about, i.e. they do noteat grass, and would starve if there was only grass to eat. We will assume that starvingfoxes die following first-order kinetics.
Now comes the killer. We assume that in any collision between a fox and a hare,there is a finite probability that the hare will die, and the fox will feed. The rate of lossof hares, and the rate of increase of foxes, will be proportional to the number of hare-fox collisions (under the assumption that feeding foxes mate and thrive). The finalrate law for the concentration of foxes, [F], and hares [H~\, is:
(1)
dt
d[F]
dt
We are making the assumption that the rate of loss of hares due to being eaten is equalto the rate of gain of foxes due to feeding. At this point the model is fixed, and we arefaced with the task of solving this set of differential equations for the dependentvariables [F] and [H].
0020-739X/89 83-00 © 1989 Taylor & Francis Ltd.
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(2)
= {kb-kc[F]}
(3)
22 C. W. David
3. Understanding the equationsMoving terms about we obtain:
^=[H]{kb-kc[F]}
d[F]dt
which can be re-written as:
d\n[H]Jt
dln[F~\~~dt
We have reached the end of the analytical trail here; we cannot explicitly solve thesetwo equations for [F] and [H~\ as functions of time.
There is one thing we can do with these equations: divide one by the other. Weobtain
dln[H]_ kh-kc[F]d\n[F] -kf + kc[H] ( '
which allows us to cross-multiply and obtain
Since
dln[H] = 1 d[H]
dt [H] dt ( '
we can integrate the equation (5) directly, and obtain:
- k{ In [H] + k,[H] = kH In [F] -k,[F] + a constant (7)
which relates the instantaneous concentration of Hares to the instantaneousconcentration of foxes.
It is not at all obvious why this equation predicts cyclic concentrations of Foxesand Hares as time progresses. Let us take the derivative of the concentration of Hareswith respect to the concentration of foxes, i.e. d[H]/d[F] and see where this is zero.We obtain
,7rFn(8)
which clearly can be zero only when [F] = khlkc, and can be 'infinite' when[H] = kflkc\ What is going on here? We expected two values of [F] where thederivative was zero, and instead we found.only one! Figure 1 shows a plot of thefunction
- k, In [H] + kc[H] - kH In [F] - kc[F]
which is a re-arranged form of equation (5), plotted against [H] and [F]. Figure 2 is acontour map of the surface in figure 1. The contours are closed loops! The
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The hares and the foxes 23
- 0 0 5 2
Figure 1. A plot of -kfln[H] + kc[H]-kH\n[F]-kc[F] versus [H] and [F]. fch = 001,k( = 002 and &c = 0/001 for this (and all subsequent) plots.
concentrations of hares and foxes are linked together, and as one goes about the loop,the two concentrations change.
Let us agree to use this single value of [F] (^kjk,.), and plug it into the right-hand side of equation (7). Now the entire right hand side of equation (7) is a constant(call it c'). We have
which we re-write as
(9)
(10)
where lnc' = t". Combining terms we get
\n{[H]k<exP(-kc[H]}=c" (11)
so that the argument of the logarithm must be yet another constant, call it c'". Thus
[Hf<exp(-kc[H]=c'" (12)
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24 C. W. David
- 0 . 0 5 1 2
Figure 2. A contour plot of the same function as shown in figure 1.
Plotting the left-hand side of this equation against [H] we see that there will be twovalues of [H] at which the left-hand side equals c'"\ Figure 3 shows how this ispossible.
Lastly, let us take the second derivative of In [H] versus t (equation (2)) and seethat there are both a maximum and a minimum value for In [H] depending on thesign of the derivative of [F] with time when [F]=kjk,..
d2\n[H] L d[F]dt dt
which equals
= -K{-k([F]+kcMisi0B[H][F]} (14)
Clearly, the sign of the second derivative can flip from plus to minus, depending onthe concentrations of hares and foxes. Therefore, there can be minima and maxima inthe concentration of hares as a function of time.
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The hares and the foxes 25
0.970
0.968
0.966
0.964
0.962
0.960
-
i i
i i
i 1
i
i i
i-
\
\
1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1
/
/
/
10 20
Hares
30 40
Figure 3. A plot of k(\n[H]+kc[H] versus [H] showing that there are two values of [H]which can give the same value of k(ln[H] + kc[H].
4. Brute force integrationIf one integrates the rate laws by brute force Euler integration, one can obtain the
plot (figure 4) of [F] and [H] on the same plot against time; we see that the extremumin [H] occurs when [F] reaches its assigned value, and depending on the slope of [F]versus t, the extremum is either a maximum or a minimum.
5. Lotka's assertionLotka [2], in his original paper, wrote that the period for the oscillation was given
by the expression:
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26 C. W. David
8
0 100 200 300 400 500 600 700 800 900 1000Time
Figure 4. A plot of [H] and [F] versus time. The method of integration (simplest possibleEuler integration) resulted in a slight error at each step, which can be seen in theincreasing amplitude of [H], The measured period (from measuring the [H]maldistance from the graph is about 480 (time units). This can be compared to the valuefrom Equation 15.
which he obtained through a Fourier Analysis. Let us see if we can also obtain such aresult. Lotka wrote the Fourier Series in their real form, i.e.
+ A2 cos 2cot + B2 sin 2cot
[F] = A'o + A\ cos cot + B\ sin cot
+ A'2 cos 2cot + B'2 sin loit
+ ... (16)
Lotka now tells us to evaluate n, {A}, {B}, {A1} and {B'} by substituting these lattertwo equations into the defining differential equations (equation (1)). The timederivative of [H] as a function of t, from the Ansatz, is
d[H]
dt= Al(o{ — sin cot) + B^Ci cos (Ot
— 2A2co sin 2cot + B2a> cos 2cot
(17)
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The hares and the foxes 27
and
d[F]—— = A\co{ — sin cot) + B\co sin cot
at
— 2A'2co sin lent + 2B'2(O cos 2cot
+ . . . (18)
Substituting [H], [F], d[H]/dt and d[F]jdt into equation (1) we obtain
— Anco sin (ncot) + Bnco cos (ncot) =I
2I
cos («»/) + Bn sin (ncot)] + ̂ h(^o + Bo)
- kc < J] [̂ 4; cos (MWO + B; sin
(19)
for just the equation for d[H]/dt one sees, immediately, that the last term in theequation, which involves the product of two infinite series, is going to cause animmense problem. Each term in the expansion of this product will contain a sine orcosine from [H] and another sine or cosine from [F]. A typical term will be
^17sin(17cot)x^74sin(74ct;0 (20)
and there will be cross terms in A' and B, A' and B', B and A' and B and A. What amess. This is getting nowhere. Is there another tactic we might employ?
Let us try adding the two sub-equations of equation (2). We obtain
which, when we substitute the Ansatzen (equation (16)) into this equation becomes
— (A!+A'Jco sin cot+ (Bi+B\)co cos cot+...
= khA0-k[A'0-kh(A1+A\)coscot-kt(B'1+B'1)smcot + ... (22)
We are forced to conclude that
khAo-ktA'o = 0 (23)
which leads us nowhere. Looking at the coefficient of coscot we have:
(B,+Bi)o) = Ah(^1+^'1) (24)
and the coefficient of sin cot we have
-(A1+A'1)to=-kt(Bl+B'1) (25)
which may be simultaneously solved to give us
co2 = k{kh (26)
'How sweet it is'.
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28 The hares and the foxes
6. DiscussionNo one claims that the Volterra-Lotka (prey-predator) model is the sine qua non
of ecological models. It is, however, of special interest to chemists. First, it is a simpleexample outside the chemist's normal experience, which uses the same mathematicsas used normally in chemical kinetics, in a language which is 'friendlier'. Second,there do exist oscillatory reactions, so-called autocatalytic [3] reactions, which areworthy of study for their own sake. The prey-predator model is a wonderfulintroduction to this intriguing area of a current subfield of chemical kinetics.
References[1] VOLTERRA, V., and CONSEIL, J., 1928, Permanent Intern. Exploration Mer III, 1;
LOTKE, A. J., 1910, J. chem. Phys., 14, 271; AOEL, N. S., MAITRA, S. C , andMONTROLL, E. W., 1971, Rev. Mod. Phys., 43, 231.
[2] LOTKA, A. J., 1920, J. Am. chem. Soc, 42, 1595.[3] DEGN, H., 1972, J. chem. Ed., 49,302; FELD, R. J., 1972, J. chem. Ed., 49,308; LEFELHOCZ,
J. F., 1972, J. chem. ed., 49, 312.
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