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{SECT 0 {EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 40 "restart: with (plo
ts): with(DEtools):" }}}{EXCHG {PARA 0 "" 0 "" {TEXT 256 16 "competit
ion.mws " }{TEXT -1 20 " Competition Models" }}{PARA 0 "" 0 "" {TEXT
-1 48 "BIOL 763 and SCCC 411 B - Mathematical Biology" }}{PARA 0 ""
0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 110 "This discussion is \+
adapted from Robert H. MacArthur, 1972. Geographical Ecology. Harper
& Row. pp. 33 - 58." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0
"" {TEXT -1 578 "Volterra developed the basic theory of competition, u
sing the logistic equation as his framework. His premise was that the
`effective carrying capacity' of the environment is reduced by the pr
esence of competitors for resources. Therefore we start from the logi
stic equation. First, however, we tell Maple that when we make assump
tions about unknowns (e.g., that parameter values are positive), we do
n't want the unknowns to be flagged with Maple's ~ mark (this would be
the default, 1), nor do we want the assumptions to be completely hidd
en (this would be selected with 0)." }}}{EXCHG {PARA 0 "> " 0 ""
{MPLTEXT 1 0 27 "interface(showassumed = 2);" }}}{EXCHG {PARA 0 "> "
0 "" {MPLTEXT 1 0 49 "logist:= diff(X1(t),t) = r1* X1 * (K1 - X1) / K1
;" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1
137 "and introduce a term to describe the reduction in carrying capaci
ty caused by the presence of members of a second species in the habita
t:" }}{PARA 0 "> " 0 "" {MPLTEXT 1 0 58 "pop1:= diff(X1(t),t) = r1* X1
* (K1 - X1 - alpha*X2) / K1;" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 0 "
" }}{PARA 0 "" 0 "" {TEXT -1 371 "X2 represents the population size of
the competitors and alpha represents the per capita reduction in carr
ying capacity of species 1 caused by competition with species 2. Alph
a is called the 'competition coefficient' of species 2 acting on speci
es 1. Sometimes it is written with subscripts, i.e. alpha21, indicat
ing the effect of species 2 on the growth of species 1." }}{PARA 0 ""
0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 122 "A similar equation \+
can be written for the second species, whose carrying capacity is redu
ced by the presence of the first:" }}{PARA 0 "> " 0 "" {MPLTEXT 1 0
58 "pop2:= diff(X2(t),t) = r2 * X2 * (K2 - X2 - beta * X1)/K2;" }}}
{EXCHG {PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 271 "B
eta represents the per capita reduction in carrying capacity of specie
s2 caused by competition with species 1. For similarity with equation
1, it is sometimes written as alpha with subscripts, i.e. alpha12, in
dicating the effect of species 1 on the growth of species 2." }}{PARA
0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 369 "
These two equations when examined simultaneously, describe the dynamic
s of two species in competition with each other. There are no explici
t functions that serve as solutions to these simultaneous equations. \+
Hence a variety of methods have been used to examine their predictions
. We can use variations on 'equilibrium analysis' to examine the beha
vior of the system." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "
" {TEXT -1 820 "A basic question we can ask is whether the two species
can coexist or whether one drives the other to extinction. To examin
e this question, we first need to think about the biological condition
s for coexistence. The real world is a mosaic of patches of habitat w
hich are colonized and vacated regularly. New habitat opens up after \+
environmental disturbances caused by winter cold, summer heat, hurrica
nes, fires, glaciers, volcanos, locust swarms, herds of goats, humans \+
and other destructive forces. It is rare that two or more species wil
l colonize an empty patch simultaneously. Instead, usually one specie
s will enter a patch, establish itself, and then later be subject to i
nvasion by other species. Therefore for COEXISTENCE to be possible, i
nvaders must be able to enter a fully occupied patch of habitat. " }}
{PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}{EXCHG {PARA 0 "" 0 "" {TEXT
-1 393 "Mathematically, this means that the invader (for instance spec
ies 2) must have a positive growth rate (dX2/dt > 0) during the first \+
stages of invasion when it is rare (X2 is a small number). Since it i
s invading a habitat fully occupied by another species (for instance s
pecies 1), we need for dX2/dt > 0 even when the population X1 is at o
r near the carrying capacity K1. We write this as:" }}{PARA 0 "> " 0
"" {MPLTEXT 1 0 27 "invade2:= rhs( pop2 ) > 0;" }}}{EXCHG {PARA 0 ""
0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 313 "To manipulate this \+
inequality Maple has to be told some things that you know in your slee
p: for instance, that the populations, the growth rates, and the carry
ing capacities are positive quantities. WARNING! Maple puts an annoy
ing little flag (~) on all variables that come with assumptions; try t
o ignore it. " }}{PARA 0 "> " 0 "" {MPLTEXT 1 0 55 "assume( X1 > 0 );
assume( K1 > 0 ); assume( r1 > 0 );" }}{PARA 0 "> " 0 "" {MPLTEXT
1 0 55 "assume( X2 > 0 ); assume( K2 > 0 ); assume( r2 > 0 );" }}}
{EXCHG {PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 154 "W
hen species 1 is invading, species 2 is at, or near, its carrying capa
city and species 1 (the invader) is rare. We seek to have dX1/dt > 0 \+
in this case." }}{PARA 0 "> " 0 "" {MPLTEXT 1 0 26 "invade1:= rhs( po
p1) > 0;" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 ""
{TEXT -1 366 "Since the world presumably consists of patches of specie
s 1, patches of species 2, and empty space, coexistence (or patches o
f both) can only occur if BOTH SPECIES CAN INVADE THE OTHER. So we ar
e interested in values of alpha and beta that make both invasion condi
tions true. Let us rearrange to put alpha and beta on one side of t
heir respective inequalities." }}{PARA 0 "> " 0 "" {MPLTEXT 1 0 79 "c
ondition2:= solve( \{invade2\}, beta); condition1:= solve( \{invade1
\}, alpha); " }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "
" {TEXT -1 223 "The following command gets rid of the ugly braces (tec
hnically we are telling Maple to select the first operand of the expre
ssion, which just so happens to be a set that contains exactly one ope
rand, namely the inequality)." }}{PARA 0 "> " 0 "" {MPLTEXT 1 0 34 "co
ndition2:= op( 1, condition2); " }}}{EXCHG {PARA 0 "> " 0 ""
{MPLTEXT 1 0 32 "condition1:= op( 1, condition1);" }}}{EXCHG {PARA 0 "
" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 250 "Recall that these
conditions were supposed to hold under the most extreme conditions: o
ne population (essentially) zero, and the other (essentially) at carry
ing capacity. Strictly speaking, we should compute limits here, but s
ubstituting is easier. " }}{PARA 0 "> " 0 "" {MPLTEXT 1 0 53 "conditio
n2:= subs( \{X2 = 0, X1 = K1\}, condition2 );" }}}{EXCHG {PARA 0 "> \+
" 0 "" {MPLTEXT 1 0 54 "condition1:= subs( \{X1 = 0, X2 = K2\}, con
dition1 );" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 ""
{TEXT -1 180 "Note that the right hand sides of the two inequalities a
re reciprocals of one another. Let's call K1/K2 by a new name, R, and
rewrite the condtions that allow for mutual invasion." }}{PARA 0 "> \+
" 0 "" {MPLTEXT 1 0 46 "condition1:= subs( K1 = R * K2 , condition1);
" }}{PARA 0 "> " 0 "" {MPLTEXT 1 0 48 "condition2:= subs( K1 = R * K
2 , condition2 );" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "
" 0 "" {TEXT -1 72 "Let's think about the consequences of this. What \+
if alpha = beta = 0.9?" }}{PARA 0 "> " 0 "" {MPLTEXT 1 0 71 "restricti
on:= subs( \{alpha=0.9, beta=0.9\}, \{condition1, condition2\} );" }}
}{EXCHG {PARA 0 "" 0 "" {TEXT -1 626 "We can tell Maple to solve the s
econd of these inequalities and put it together with the first one, bu
t this turns out to be harder than you might think. Anyhow, at some p
oint it really makes more sense to do the work by hand. It's pretty e
asy to see that the two species can coexist only if the ratio R of the
ir carrying capacities lies between 0.9 and 1.1. In the real world, c
arrying capacities fluctuate dramatically, sometimes by orders of magn
itude, because of weather, predators, etc. Therefore it is very unlik
ely that two species could remain that close in ratio of K's for very \+
long. What if alpha = beta = 0.1?" }}{PARA 0 "> " 0 "" {MPLTEXT 1 0
73 "restriction:= subs( \{alpha=0.1, beta=0.1\}, \{condition1, conditi
on2\} ); " }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "
" {TEXT -1 409 "How about this set of conditions for coexistence? Giv
en fluctuations in K, is coexistence more or less likely than the last
example? Try some other values for alpha and beta, not necessarily w
ith alpha = beta. What general pattern do you see about the likelihoo
d of coexistence, and the magnitudes of alpha and beta? How did we or
iginally define alpha and beta? How does this relate to your observat
ions?" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 265
"Let's now look at the range of conditions under which coexistence can
occur. From the above inequalities, if alpha = beta, coexistence can
happen if the ratio of carrying capacities is between alpha and 1/alp
ha. We can plot this over the range 0.1 < alpha < 1.0:" }}{PARA 0 ">
" 0 "" {MPLTEXT 1 0 0 "" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 63
"plot( \{alpha, 1 / alpha\}, alpha = 0.1 .. 1.0, thickness = 2 );" }}
}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}{EXCHG {PARA 0 "" 0 ""
{TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 161 "An alternative analysis \+
of the competition equations uses equilibrium methods directly. In th
is case we set the population growth rate to zero for both species." }
}{PARA 0 "> " 0 "" {MPLTEXT 1 0 11 "pop1; pop2;" }}}{EXCHG {PARA 0 ""
0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 104 "At equilibrium both
of these rates are zero, so we can simplify by multiplying each equat
ion by K/(r*X):" }}{PARA 0 "> " 0 "" {MPLTEXT 1 0 53 "p1:=rhs(pop1)*K1
/(r1*X1); p2:=rhs(pop2)*K2/(r2*X2);" }}}{EXCHG {PARA 0 "" 0 ""
{TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 344 "We examine a case where \+
the K1 = 900, K2 = 800, and both alpha and beta are 0.5. To see on th
e phase plane where these equilibria lie, we need to substitute these \+
values in to the appropriate equilibrium expressions. Notice that R =
K1 / K2 = 1.125, which is safely between 0.5 = alpha and 2.0 = 1 / al
pha. For species 1, the equilibrium is:" }}{PARA 0 "> " 0 ""
{MPLTEXT 1 0 36 "pp1:= subs(\{K1=900, alpha=0.5\}, p1);" }}}{EXCHG
{PARA 0 "" 0 "" {TEXT -1 152 "In order to plot this on the phase plane
(X2 vs. X1) we need to rearrange this so it is X2 as a function of X1
, so we solve the above expression for X2:" }}{PARA 0 "> " 0 ""
{MPLTEXT 1 0 28 "nullcline1:= solve(pp1, X2);" }}}{EXCHG {PARA 0 "" 0
"" {TEXT -1 63 "We look at the equilibrium for species 2 in terms of s
pecies 1:" }}{PARA 0 "> " 0 "" {MPLTEXT 1 0 34 "pp2:=subs(\{K2=800, be
ta=0.5\}, p2);" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 28 "nullcline
2:= solve(pp2, X2);" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 172 "Now we ca
n plot both of the nullclines. When the populations lie above and to \+
the right of these lines, they grow, and when they lie below and to th
e left, they decline. " }}{PARA 0 "> " 0 "" {MPLTEXT 1 0 137 "regions
:= plot([nullcline1, nullcline2], X1=0 .. 1400, X2= 0 .. 1400, color=[
brown, orange], thickness = 2): # save this graph for later!" }}}
{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 31 "regions; # and look at it n
ow!" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}{EXCHG {PARA 0 "
" 0 "" {TEXT -1 141 "\nNow let's do a numerical solution of the equati
ons, using the same values for K1, K2, alpha, beta. We will pick r1=r
2=0.1 for this attempt:" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 147
"rate_equations:= subs( \{K1=900, alpha=0.5, r1=0.1, K2=800, beta=0.5
, r2=0.1, X1(t)= x1(t), X1 = x1(t), X2(t) = x2(t), X2 = x2(t)\}, [pop1
, pop2]) ;" }}{PARA 0 "> " 0 "" {MPLTEXT 1 0 21 "vars:=[x1(t), x2(t)];
" }}{PARA 0 "> " 0 "" {MPLTEXT 1 0 18 "domain:= 0 .. 120;" }}}{EXCHG
{PARA 0 "> " 0 "" {MPLTEXT 1 0 124 "init1:=[x1(0)=20, x2(0)=150]; init
2:=[x1(0)=1400, x2(0)=100]; init3:=[x1(0)=100, x2(0)=1400]; init4:= [x
1(0)=200, x2(0)=20];" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 162 "Ph
asePortrait:= DEplot(rate_equations, vars, domain, [init1, init2, init
3, init4], stepsize=1, scene=[x1, x2], linecolor = [black, blue, green
, grey] ): # save it" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 80 "Pha
sePortrait; # see it (sometimes the color order gets scrambled unfortu
nately)" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}{EXCHG
{PARA 0 "" 0 "" {TEXT -1 168 "\nHave a look at the trajectories as t -
> infinity. Where do they lie relative to the zero growth lines on th
e previous graph? The following command might be revealing." }}{PARA
0 "> " 0 "" {MPLTEXT 1 0 108 "display( \{ PhasePortrait, regions\}); #
overlay the phase portrait and the regions defined by the nullclines
" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1
368 "Try some other values for K1, K2, alpha, and beta and see if you \+
can demonstrate that the invasion criterion we derived earlier is corr
ect. Note that you will have to start with your invader being RARE, N
OT ABSENT. When we did the invasion analysis we set the invader to ze
ro; obviously you will need to start with a few individuals or you wil
l never get any growth." }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "
" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1
533 "Next we try to account for alpha and beta in a more mechanistic. \+
less magical, way. What one species does to the other to affect its g
rowth rate may involve several different aspects. There may be overt c
ompetition, physical exclusion from prime feeding or nesting sites, fo
r example. Or the competition may be covert, say in consumption of a \+
common food resource. Let us consider a pair of consumer populations \+
X1 and X2 which eat resources R1 and R2, depriving each other in the p
rocess. We begin with MacArthur's equation (3):" }}}{EXCHG {PARA 0 ">
" 0 "" {MPLTEXT 1 0 147 "pop1:= diff(X1(t), t)/X1 = C[1] * (a[11]*w[1
]*R1 + a[12]*w[2]*R2 - T[1]);\npop2:= diff(X2(t), t)/X2 = C[2] * (a[21
]*w[1]*R1 + a[22]*w[2]*R2 - T[2]);" }}}{EXCHG {PARA 0 "" 0 "" {TEXT
-1 61 " X1 and X2 are the population sizes of the consumers"
}}{PARA 0 "" 0 "" {TEXT -1 66 " R1 and R2 are the population \+
sizes of the two resources." }}{PARA 0 "" 0 "" {TEXT -1 80 " \+
w1 and w2 are the weights of individuals of each of the two resources.
" }}{PARA 0 "" 0 "" {TEXT -1 110 " aij is the probability tha
t a consumer of species i encounters and eats an individual of resourc
e j." }}{PARA 0 "" 0 "" {TEXT -1 81 " T1 and T2 are the maint
enance requirements of consumer species 1 and 2." }}{PARA 0 "" 0 ""
{TEXT -1 85 " C1 and C2 represent the conversion of grams of \+
prey into population growth." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA
0 "" 0 "" {TEXT -1 134 "In this model, the per capita rate of populati
on growth is determined by the amount of resource that exceeds mainten
ance requirements." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 ""
{TEXT -1 281 "We now need to consider the population dynamics of the r
esources. We assume that each grows logistically when alone, but has \+
its growth reduced by its encounters with consumers. Thus growth rate
= logistic - loss to predator 1 - loss to predator 2; see MacArthur'
s equation (4):" }}{PARA 0 "> " 0 "" {MPLTEXT 1 0 159 "res1:= diff(R1(
t), t)/R1 = (r[1]/K[1]) * (K[1] - R1) - a[11] * X1 - a[21] * X2;\nres2
:= diff(R2(t), t)/R2 = (r[2]/K[2]) * (K[2] - R2) - a[12] * X1 - a[22] \+
* X2;" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 54 "\nDetermine equilibria e
quations for the two resources:" }}{PARA 0 "> " 0 "" {MPLTEXT 1 0 53 "
equil_R1:= rhs(res1) = 0; equil_R2:= rhs(res2) = 0;" }}}{EXCHG
{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 111 "Solve fo
r the equilibrium population sizes of the two resources (this is equa
tion (5) of MacArthur on p. 38): " }}{PARA 0 "> " 0 "" {MPLTEXT 1 0
56 "Req1:= solve(equil_R1, R1); Req2:= solve(equil_R2, R2);" }}}
{EXCHG {PARA 0 "" 0 "" {TEXT -1 143 "\nDetermine the equilibria equati
ons for the consumers. Since C1 and C2 cannot be 0, we can divide thr
ough by these and simplify our equations." }}{PARA 0 "> " 0 ""
{MPLTEXT 1 0 67 "equil_X1:= rhs(pop1) / C[1] = 0; equil_X2:= rhs(pop
2) / C[2] = 0;" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 ""
0 "" {TEXT -1 368 "Now we substitute the equilibrium values for the re
source populations into the equations for the consumers. In Maple we \+
use the subs command to replace all occurrences of R2 and R1 in the eq
uil_X1 and equil_X2 equations with the values we obtained by solving f
or R2 and R1 in the resource equations. This gives us the equations n
ear the bottom of p. 38 in MacArthur." }}{PARA 0 "> " 0 "" {MPLTEXT 1
0 86 "eX1:= subs(\{R2=Req2, R1=Req1\}, equil_X1); eX2:= subs(\{R2=Re
q2, R1=Req1\}, equil_X2);\n" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 0 "" }
}{PARA 0 "" 0 "" {TEXT -1 368 "We collect the constant terms, and the \+
terms including X1 and X2, as done at the bottom of p. 38 and the top \+
of p. 39. To understand the Maple commands convert and parfrac go to \+
the help browser and use the keyword search. Or enter the commands ?c
onvert, ?parfrac. Since we are doing nothing more than rearraging the
equations, we might as well keep the same names." }}{PARA 0 "> " 0 "
" {MPLTEXT 1 0 82 "eX1:= convert(eX1, parfrac, \{X1, X2\}); eX2:= \+
convert(eX2, parfrac, \{X2, X1\});\n" }}}{EXCHG {PARA 0 "" 0 "" {TEXT
-1 100 "Note that what we got in both cases was: (-mess_1) * X1 + (-
mess_2) * X2 + (mess_3) = 0," }}{PARA 0 "" 0 "" {TEXT -1
117 "which looks a lot like (mess_3)
- (mess_1) * X1 - (mess_2) * X2 = 0," }}{PARA 0 "" 0 "
" {TEXT -1 106 "which could be converted to (mes
s_3 / mess_1) - X1 - (mess_2 / mess_1) * X2 = 0," }}{PARA 0 "" 0 "
" {TEXT -1 127 "which plainly has the form \+
K1 - X1 - alpha * X2 =
0," }}{PARA 0 "" 0 "" {TEXT -1 61 "if (mess_3 / mess_1) = K and (
mess_2 / mess_1) = alpha. " }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA
0 "" 0 "" {TEXT -1 121 "So let's try to extract the 'mess_1' by asking
Maple to find the coefficient of X1 on the left hand side of the equa
tion:" }}{PARA 0 "> " 0 "" {MPLTEXT 1 0 26 "b1:= coeff( lhs(eX1), X1);
" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 70 "Let's divide our original equ
ation by this constant and collect terms:" }}{PARA 0 "> " 0 ""
{MPLTEXT 1 0 45 "Eq6a:= convert( eX1 / b1, parfrac, \{X1, X2\});" }}}
{EXCHG {PARA 0 "" 0 "" {TEXT -1 123 "Let's do it again for the other s
pecies. Get the coefficient of X2, divide through by this coefficient
, and collect terms." }}{PARA 0 "> " 0 "" {MPLTEXT 1 0 26 "b2:= coeff(
lhs(eX2), X2);" }}{PARA 0 "> " 0 "" {MPLTEXT 1 0 45 "Eq6b:= convert( \+
eX2 / b2, parfrac, \{X1, X2\});" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 0
"" }}{PARA 0 "" 0 "" {TEXT -1 150 "Now we have a definition for alpha \+
and beta, which are the coefficients in X2 in the first equation and t
he coefficients in X1 in the second equation:" }}{PARA 0 "> " 0 ""
{MPLTEXT 1 0 64 "alpha:= coeff(lhs(Eq6a), X2); beta:= coeff(lhs(Eq6
b), X1);\n " }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "
" {TEXT -1 273 "Let's examine these coefficients to see what they may \+
signify. Remember that the parameters a11, a12 are the probabilities \+
of consumption of resources 1 and 2 by consumer 1. Parameters a21 and
a22 are the probabilities of consumption of resources 1 and 2 by cons
umer 2. " }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1
136 "First consider the case where consumer 1 only eats resource 1 and
consumer 2 only eats resource 2. This means that a12 = 0 and a21 = 0
:" }}{PARA 0 "> " 0 "" {MPLTEXT 1 0 73 "A:= subs(\{a[12]=0, a[21]=0\},
alpha); B:= subs(\{a[12]=0, a[21]=0\}, beta);" }}}{EXCHG {PARA 0 ""
0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 92 "What does this mean \+
about the relationship between consumption of resources and competitio
n?" }}{PARA 0 "" 0 "" {TEXT -1 96 "\nNow let's consider the case where
each consumer has 50% of its diet comprised of each resource:" }}
{PARA 0 "> " 0 "" {MPLTEXT 1 0 124 "A:= subs(\{a[12]=0.5, a[21]=0.5, a
[11]=0.5, a[22]=0.5\}, alpha);\nB:= subs(\{a[12]=0.5, a[21]=0.5, a[11]
=0.5, a[22]=0.5\}, beta);" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 219 "\nD
o you still believe what you thought about the relation between diet a
nd competition? Consider consumer 1 eating 75% of resource 1 and 25% \+
of resource 2, while consumer 2 eats 25% of resource 1 and 75% of reso
urce 2:" }}{PARA 0 "> " 0 "" {MPLTEXT 1 0 132 "A:= subs(\{a[11]=0.75, \+
a[12]=0.25, a[21]=0.25, a[22]=0.75\}, alpha);\nB:= subs(\{a[11]=0.75, \+
a[12]=0.25, a[21]=0.25, a[22]=0.75\}, beta);" }}}{EXCHG {PARA 0 "" 0 "
" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 199 "Notice that under thes
e conditions we do not have enough information to determine alpha and \+
beta. If we assume the carrying capacities, growth rates and body siz
es of the resources are similar then:" }}{PARA 0 "> " 0 "" {MPLTEXT 1
0 198 "A:= subs(\{a[11]=0.75, a[12]=0.25, a[21]=0.25, a[22]=0.75, K[1]
=K[2], w[1]=w[2], r[1]=r[2]\}, alpha);\nB:= subs(\{a[11]=0.75, a[12]=0
.25, a[21]=0.25, a[22]=0.75, K[1]=K[2], w[1]=w[2], r[1]=r[2]\}, beta);
" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1
225 "On the other hand if the carrying capacities and growth rates of \+
the resources are different, then the competition coefficients are dif
ferent, even with equivalent consumption. Here we assume K1 = 2*K2 \+
and r1 = 0.5 * r2." }}{PARA 0 "> " 0 "" {MPLTEXT 1 0 210 "A:= subs(
\{a[11]=0.75, a[12]=0.25, a[21]=0.25, a[22]=0.75, K[1]=2*K[2], w[1]=w[
2], r[1]=0.5*r[2]\}, alpha);\nB:= subs(\{a[11]=0.75, a[12]=0.25, a[21]
=0.25, a[22]=0.75, K[1]=2*K[2], w[1]=w[2], r[1]=0.5*r[2]\}, beta);" }}
}{EXCHG {PARA 0 "" 0 "" {TEXT -1 134 "What generalization can you make
from what you have seen about the relationship between diet overlap a
nd the intensity of competition?" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}
{PARA 0 "" 0 "" {TEXT -1 29 "Look at alpha and beta again:" }}{PARA 0
"> " 0 "" {MPLTEXT 1 0 28 "alpha:= alpha; beta:= beta;" }}}{EXCHG
{PARA 0 "" 0 "" {TEXT -1 170 "\nSome artful substitutions can make the
se complicated expressions look a lot simpler, and will suggest a natu
ral generalization from 2 resources to an arbitrary number N." }}
{PARA 0 "> " 0 "" {MPLTEXT 1 0 184 "substitute_set1:= \{ a[11]^2 * w[1
] * K[1] / r[1] = u[11]^2, a[12]^2 * w[2] * K[2] / r[2] = u[12]^2, a
[21]^2 * w[1] * K[1] / r[1] = u[21]^2, a[22]^2 * w[2] * K[2] / r[2] =
u[22]^2 \};" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 75 "A:= subs( s
ubstitute_set1, alpha) ; B:= subs(substitute_set1, beta ) ; " }}}
{EXCHG {PARA 0 "" 0 "" {TEXT -1 197 "\nIf you take the square root on \+
each side of each of the above substitution equations, you will notice
that each term in the numerators of alpha and beta can also be writee
n in terms of the u_ij's." }}{PARA 0 "> " 0 "" {MPLTEXT 1 0 126 "subst
itute_set2:= \{a[11] * a[21] * w[1] * K[1] / r[1] = u[11] * u [21] , a
[22] * a[12] * w[2] * K[2] / r[2] = u[12] * u[22] \};" }}}{EXCHG
{PARA 0 "> " 0 "" {MPLTEXT 1 0 86 "A:=simplify( subs( substitute_set2,
A ) ); B:= simplify( subs( substitute_set2, B) );" }}}{EXCHG {PARA
0 "" 0 "" {TEXT -1 304 "\nPretty clearly then, if there are 2 consumer
s and N resources (potentially) in common, we get competition coeffici
ents ALPHA and BETA in terms of numbers U_ij that reflect the amou
nt of resource sharing between the consumers and among themselves. Ca
n you interpret what each U_ij term is measuring?" }}{PARA 0 "> " 0 "
" {MPLTEXT 1 0 140 "ALPHA:= Sum(U[1][j]*U[2][j], j = 1 .. N) / Sum(U[1
][j]^2, j= 1 .. N);\nBETA := Sum(U[2][j]*U[1][j], j = 1 .. N) / Sum(U
[2][j]^2, j= 1 .. N);" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }
}}}{MARK "0 0 0" 40 }{VIEWOPTS 1 1 0 1 1 1803 1 1 1 1 }{PAGENUMBERS 0
1 2 33 1 1 }