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Faculty of Computing, Engineering & Technology
Staffordshire
UN E IV RSIT Y
Assignment for
Mathematical Modeling
Module Code Number: Semester: Assessment Tutor: Written By:
CE63032-3 Two Dr Martin Paisley gv004727
A REPORT ON I Will Survive
By Yijun Gu
Abstract
This paper deals with a classical predator-prey model which in the absence of predators, the prey population increases without bound. Then using the given constants to figure out the change of values of e will bring what effect to the population. The equilibrium point of this predator-prey system will be identified. Some necessary implementation by Maple will be developed. At last, the applicability of this model in ecological reality will be discussed.
CONTENTES
Page 1. INTRODUCTION 2. MAIN SECTIONS
2.1 Assumptions
2.2 2.3 2.4 2.5 2.6 2.7 2.8 Equation of model Equilibrium points The Classification MAPLE Implementation Applicability of the model Conclusions Recommendations 1
1
1~2 2~3 3~4 4~6 7 7 8
3. Appendix Reference 4. Appendix
8 8
1. INTRODUCTION
This report should include analysis and implementation of a classical predator-prey model by be given values of constants and equation. At first, the meaning of each of the constants will be explained. After that, the indentifying of the equilibrium point of the system and the effect of parameter e will be included. Then the investigation of finding and classifying the equilibrium point of the system, and MAPLE worksheet will illustrate the range of the behaviors expected with graphics. At the final part of this paper, the applicability of models to actual ecological populations will be discussed.
2. MAIN SECTIONS
2.1 Assumptions
To keep this model simple, before justify the meaning of constants, some assumptions that would be unrealistic in most of these predator-prey situations should be declared. Specifically, these assume as follows: ? ? ? ? The predator species is totally dependent on a single prey species as its only food supply. The prey species has an unlimited food supply and sufficient space to grow indefinitely. There is no threat to the prey other than the specific predator. The prey will have competition to avoid eaten by predator in their own species.
2.2 Equation of model The given model as follow:
dx = ax ? bxy ? ex 2 dt dy = ?cy + dxy dt
Where a, b, c and d are positive constants.
There are two species interacting: a prey species x and a predator species y . The meaning of each of the constants: ? ? In the absence of predators, the population of prey grows exponentially. In particular a . As the assumptions, there is sufficient food and space to grow indefinitely. In the absence of prey, the number of predators decreases exponentially. In particular c . Thus there is no enough food to sustain the population of predator with the prey??s population decreased. The number of prey is reduced at a rate proportional to the number of possible interactions between prey and predators. In particular b . Implicit predator will eat prey when they interacting. There is a corresponding increase in predators at a rate proportional to the number of interactions. In particular d . The same reason like constant b .
There is competition in the prey species, the rate of competition is value of parameter e . The given values of constants are: a=0.11; b=0.00009; c=0.24; d=0.00021;
2.3 Equilibrium points
Equilibrium occurs when
dx dy = 0 and = 0. This gives the following: dt dt
dx =0 dt
ax ? bxy ? ex 2 =0.
x = 0 or a ? by ? ex = 0 ?Ù
Hence x( a ? by ? ex) = 0
dy =0 dt
cy + dxy = 0 .
Hence y ( ?c + dx ) = 0 , so that y=0 or ? c + dx = 0 Use y=0 get into ?Ù
x=
c . d
x=
a . e y= a? b ec d . a? b ec d ), the further analysis is
Use x =
c get into ?Ù d
a c Hence there are three equilibrium points: (0, 0), ( , 0) and ( , e d
required to find the character of each point, and hence the behavior of the solution nearby. The MAPLE solution of equilibrium points as follow:
> equil:= solve( {rhs(e1), rhs(e2)}, {x , y});
dx dy = ax ? bxy ? ex 2 ??, e2 represent = ?cy + dxy . dt dt Before explain what should happen to their locations as the value of e increases. The range of ec a? d > 0 and e > 0 , so e should be classified. By reason of these species can??t extinct, hence b ad that 0 < e < , by the given value of constant a, d and c 0 < e < 0.00009625 c
Where e1 represent the first equation ??
a If the value of e increase, then and e
a? b
ec d will decrease. So that two equilibrium points
will tend to 0, which means these two creatures?? population will decrease and extinct finally.
2.4 The Classification The predator-prey system is
dx = ax ? bxy ? ex 2 dt dy = ?cy + dxy dt
Hence the partial derivatives required for the Jacobian Matrix are:
f x = a ? by ? 2ex , g x = dy ,
At the equilibrium point (0, 0) the Jacobian is This has eigenvalues given by or?Ë
f y = ?bx
g y = ?c + dx
a
?Ë 0 = a ?Ë c ?Ë 0 c ?Ë c. These are of opposite signs, so (0, 0) is a saddle point.
a
0
0 . c
0. Hence ?Ë
a
At the equilibrium point (
a a , 0) the Jacobian is e 0 ?a
0 ?Ë c ?Ë c
c
. c 0. Hence
This has eigenvalues given by
a
?Ë
?Ë
?Ë
?Ë
a >0 or?Ë
c
, because e
0, so (
a , 0) is a saddle point. e
c At the equilibrium point ( , d
a? b
ec d ) the Jacobian is
0 ?Ë 0 ?Ë
. 0.
?Ë This has eigenvalues given by d?Ë ? e c ?Ù If ? ?Ë ec ec?Ë 4d acd 2d 4cd ad ec 0 then 4cd ad ec acd ec ec 0 ec
0
?Ë
e c , because e>0
0
0.00008719818612
0.00009625
?Ú If ?
0 then 4cd ad 0, so (
ec
e c , because e
0.00008719818612
e
If e is ?Ù then ?
c , d
a? b
ec d ) is a stable spiral. c , d a? b ec d ) is a stable node
If e is ?Ú then ?
0 and ?Ë
0, so (
The MAPLE implementation as follow:
> a:=0.11; b:=0.00009; c:=0.24; d:=0.00021; > solve(4*c*d*(a*d-e*c)>e^2*c^2);
> solve(4*c*d*(a*d-e*c)<=e^2*c^2); >
2.5 MAPLE Implementation
After the equilibrium and classification parts, the change of parameter e will effect of both the equilibrium points and classification. If 0 0.00008719818612, then the last equilibrium points
can get a stable spiral, otherwise it only gets a steady state point. So I assume e=0.00003 as a normal case of this model.
The MAPLE implementation as follow:
> restart: a:=0.11; b:=0.00009; c:=0.24; d:=0.00021; e:=0.00003;
> e1 := diff(x(t), t) = a*x(t)-b*x(t)*y(t)-e*x(t)^2; e2:= diff(y(t),
t) = -c*y(t)+d*x(t)*y(t);
> vars:= [x(t), y(t)]; > init1:=[x(0)=4000,y(0)=1000]; init2:=[x(0)=4000,y(0)=2000]; init3:=[x(0)=4000,y(0)=4000]; domain:= 0 .. 400;
> with(DEtools): with(plots): > L:= DEplot([e1, e2], vars, domain,{init1 },stepsize=0.5, scene=[t, y],arrows=NONE): H:= DEplot([e1, e2], vars, domain, {init1 }, stepsize=0.5, scene=[t, x],arrows=NONE): > display( {L,H}, title = `Prey and 100 * Predator vs. time` );
> DEplot([e1, e2], vars, domain,{init2, init3 }, stepsize=0.5, scene=[x,y], title = `Prey vs. 100 * Predaor for t = 0 .. 400`, arrows= SLIM ) ;
*carefully look at the point (11000/3,0) where is second Equilibrium point > equil:= solve( {rhs(e1), rhs(e2)}, {x , y});
As you can see, when ??0 0.00008721130148?? these two species will have a steady population at the second equilibrium point. If ??e?? becomes large than 0.00008721130148, and maintain other conditions, then we can get some graphics as follow (Where e=0.00009):
These two species will decrease gradually and extinct finally.
2.6 Applicability of the model
Due to this model was built based on some assumption which are purely imaginary, there is no such species which only eat just one other creature. So there are lots of other elements need to add in this model to change the equations if we need apply the model to complicated real ecological populations.
2.7 Conclusion
Above all calculation and analysis show the parameter e is the key issue of the destiny of these two species. If e large than 0.00008719818612, they will extinct eventually, the decreasing of e will be the condition of the increase of steady population. But as far as I concerned why there is no competition in the species of predator. If there is another ??e?? like ??g?? to represent the competition rate in the predators, the model will have more applicability in the actual ecological population.
2.8 Recommendations If there is any possible to do some further research in predator-prey system model, in my opinion, some constant will be necessary which like separate the predator to two groups, one is immature predator and another is mature predator. The mature predator can catch more preys and immature predator cannot procreate. And also can import the disease factor to influence the population of prey and predator. Taking one with another, predator and prey system is a magnificent model which included tremendous factors. There is a long way to figure out a prefect model.
3. REFERENCE
Fundamentals of Ecology, Second Edition, by Eugene P. Odum and Howard T. Odum, copyright@1959 by Saunders College Publishing and renewed 1987 by P. Odum and Howard T. Odum. Reproduced by Permissin of the Publisher Bak, P., 1996, How Nature Works: The Science of Self-organized Criticality, Corpernicus (New York). Murray, J. D., 1993, Mathematical Biology, Springer (Berlin). Strogatz, S. H., 1988, Love Affairs and Differential Equations. Mathematics Magazine, 61, 35. J. Maynard Smith, Models in Ecology, Cambridge University Press, Cambridge, 1974.
4. Appendix
Some MAPLE calculation: Diff:
> f(x,y):=a*x-b*x*y-e*x^2; > g(x,y):=-c*y+d*x*y; >
diff(f(x,y),x); > diff(f(x,y),y); > diff(g(x,y),x); > diff(g(x,y),y);
Ad/c
> a:=0.11; b:=0.00009; c:=0.24; d:=0.00021; > a*d/c;
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