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== INTRODUCTION ==
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{{Trabajo|Predator-Prey Lotka-Volterra model|[[:Categoría:Ecuaciones Diferenciales|Ecuaciones Diferenciales]]|[[:Categoría:Trabajos 2012-13|2012-13]]}}
 +
== Introduction ==
  
 
The Lotka – Volterra equations, also known by the name of predator-prey equations, are a pair of first order and non linear differential equations.
 
The Lotka – Volterra equations, also known by the name of predator-prey equations, are a pair of first order and non linear differential equations.
Línea 5: Línea 6:
  
  
\begin{matrix}\frac{\mathrm{d} R}{\mathrm{d} t}=aR-bRF\\\frac{\mathrm{d} F}{\mathrm{d} t}=-cF+dRF\\R(t_{0})=R_{0}, F(t_{0})=F_{0}\end{matrix}
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<math>
 +
\left\{\begin{matrix}\frac{\mathrm{d} R}{\mathrm{d} t}=aR-bRF\\\frac{\mathrm{d} F}{\mathrm{d} t}=-cF+dRF\\R(t_{0})=R_{0}, F(t_{0})=F_{0}\end{matrix}\right.
  
== INTERPRETATION ==
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</math>
 +
 
 +
== Interpretation ==
  
 
First of all,we are going to describe the physical meanings term by term:
 
First of all,we are going to describe the physical meanings term by term:
-        R is the number of prey, in this case, Rabbits.
+
 
-        F is the number of predators, for this example, theFoxes.
+
-        '''R''' is the number of preys, in this case, Rabbits.
-        dR/dt is the growth of the population of Rabbits.
+
 
-        dR/dt is the growth of the population of Foxes.
+
-        '''F''' is the number of predators, for this example, theFoxes.
-        a, b, c and d are parameters describing the interaction of the species.
+
 
Now we aregoing to analyze how they are related with themselves.
+
-        '''dR/dt''' is the growth of the population of Rabbits.
-                First we assume that thepreys have an unlimited food supply. The increase of rabbits, in absence offoxes (F=0) is proportional to the initial number of rabbits. The growth of therabbits is affected by the interaction between the rabbits and foxes (FR). The“b” parameter indicates this process. So we can conclude that the change inrabbitpopulation is given by its growth minus the number of eaten by the foxes.
+
 
dR/dt= aR– bRF
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-        '''dF/dt''' is the growth of the population of Foxes.
-        We have to assume that thefood supply for the foxes depends entirely on the rabbit population. The“-c”parameter tell us about the decreasing rhythm of the foxes, affected by theirown death, emigration, etc. In absence of prey (R=0), it leads to anexponential decay.  
+
 
 +
-        '''a''', '''b''', '''c''' and '''d''' are parameters describing the interaction of the species.
 +
 
 +
Now we are going to analyze how they are related with themselves.
 +
 
 +
-                First we assume that the preys have an unlimited food supply. The increase of rabbits, in absence of foxes (F=0) is proportional to the initial number of rabbits. The growth of the rabbits is affected by the interaction between the rabbits and foxes (FR). The“b” parameter indicates this process. So we can conclude that the change in rabbit's population is given by its growth minus the number of eaten by the foxes.
 +
 
 +
<math>
 +
\frac{\mathrm{d} R}{\mathrm{d} t}=aR-bRF
 +
</math>
 +
 
 +
 
 +
-        We have to assume that the food supply for the foxes depends entirely on the rabbit population. The “-c” parameter tell us about the decreasing rhythm of the foxes, affected by their own death, emigration, etc. In absence of preys (R=0), it leads to an exponential decay.  
 
   
 
   
dF/dt= −cF+ dRF
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<math>
== NUMERICAL RESOLUTION ==
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\frac{\mathrm{d} F}{\mathrm{d} t}=-cF+dRF
 +
</math>
 +
 
 +
== Numerical Resolution ==
 +
 
 +
We have to apply iterative methods to solve the equations, because we are in the case that the equations are non linear.
  
We have to apply iterative methods to solve the equations, because weare in the case that the equations are non linear.
 
 
NOTE: at allthe exercises we use the scale 1:5000.  
 
NOTE: at allthe exercises we use the scale 1:5000.  
=== EULER METHOD ===
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=== Euler Method ===
  
For an initial population of R=3000 and F=1000, using an interval of time(0,100) and using a=0.4, b=0.37, c=0.3, d=0.05. The OCTAVE code for 5000intervals is:
+
For an initial population of '''R=3000''' and '''F=1000''', using an interval of time(0,100) and using '''a=0.4''', '''b=0.37''', '''c=0.3''', '''d=0.05'''. The OCTAVE code for 5000 intervals is:
 
<gallery>
 
<gallery>
 
Archivo:311.png
 
Archivo:311.png
Línea 40: Línea 60:
 
</gallery>
 
</gallery>
  
We have taken 5000iterations trying to obtain the realest solution (this is because, according tothe Euler Method the higher the number of N is, the shorter will be thedistance h and closer to the real solution it will be. The system solution is cyclical.
+
We have taken 5000 iterations trying to obtain the realest solution (this is because, according to the Euler Method the higher the number of N is, the shorter will be the distance h and closer to the real solution it will be. The system solution is cyclical.
We can affirm that theecosystem is not stable because, as we can see clearly in the graphics, as weadvance in time, there are more differences between foxes and rabbits(sometimes even the population of foxes is bigger than the rabbit’s), so thesystem is less stable.
+
We can affirm that the ecosystem is not stable because, as we can see clearly in the graphics, as we advance in time, there are more differences between foxes and rabbits(sometimes even the population of foxes is bigger than the rabbit’s), so the system is less stable.
 
   
 
   
Now we are going to suppose three cases with three different values of rabbit’spopulation.
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Now we are going to suppose three cases with three different values of rabbit’s population.
 
R = 1500<gallery>
 
R = 1500<gallery>
 
Archivo:314.png
 
Archivo:314.png
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</gallery>
 
</gallery>
  
As we can see,in all the graphics the number of rabbits is growing up among time. Logically,the less rabbits there are, more foxes will die at the beginning, and this willcause that the number of rabbits will reach more maximums.
+
As we can see,in all the graphics the number of rabbits is growing up among time. Logically,the less rabbits there are, more foxes will die at the beginning, and this will cause that the number of rabbits will reach more maximums.
The number offoxes will reach the lowest minimums when the initial population of rabbits isthe lowest, and as a consequence of this, when the number of rabbits reach itsmaximum, the population of foxes will grow up again to reach its maximum.
+
The number of foxes will reach the lowest minimums when the initial population of rabbits is the lowest, and as a consequence of this, when the number of rabbits reach its maximum, the population of foxes will grow up again to reach its maximum.
Talking aboutthe proximity to the limit of foxes before extinction (50), in the last case italmost reach this point, getting very close and getting into a dangerous pointto these animal.
+
Talking about the proximity to the limit of foxes before extinction (50), in the last case it almost reaches this point, getting very close and getting into a dangerous point for these animals.
=== RUNGE-KUTTA METHOD ===
+
=== Runge-Kutta Method ===
  
Now we are going to solve the system by the 4rd order Runge-Kuttamethod, with the same initial populations and values for the constancies as inthe Euler method. The OCTAVE code is:
+
Now we are going to solve the system by the 4rd order Runge-Kutta method, with the same initial populations and values for the constancies as in the Euler method. The OCTAVE code is:
 
<gallery>
 
<gallery>
 
Archivo:321.png
 
Archivo:321.png
Línea 76: Línea 96:
 
</gallery>
 
</gallery>
  
As we have explainedbefore, the Euler Method gets closer to the real solution when N is bigger. Abetter approximation to that real solution is obtained by the Runge KuttaMethod. The solution using the Euler Method with the highest N seems to be likethe solutions given by the Runge-Kutta Method explained now.  
+
As we have explained before, the Euler Method gets closer to the real solution when N is bigger. A better approximation to that real solution is obtained by the Runge KuttaMethod. The solution using the Euler Method with the highest N seems to be like the solutions given by the Runge-Kutta Method explained now.  
The solutions given bythis system are much more precise than the Euler’s. This is because the EulerMethod the approximation is given by the tangent line to the curve and becauseof this, in cyclic solutions like the ones in our cases, it does not give usgood solutions.
+
The solutions given by this system are much more precise than the Euler’s. This is because in the Euler Method the approximation is given by the tangent line to the curve and becauseof this, in cyclic solutions like the ones in our cases, it does not give us good solutions.
=== TRAPEZOIDAL METHOD ===
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=== Trapezoidal Method ===
  
This one is an implicit method that requires solving our non-linearequation and solves the following value of (R, F).It´s too difficult to solveit by using this method.
+
This one is an implicit method that requires solving our non-linear equation and solves the following value of (R, F).It´s too difficult to solve it by using this method.
=== BACKWARD SOLUTION ===
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=== Backward Solution ===
  
Assuming that now at time t = 100 thepopulation of rabbits and foxes is 2200 and 320 respectively. The approximationof the population at time t = 0 using the OCTAVE code is:
+
Assuming that now at time t = 100 thepopulation of rabbits and foxes is 2200 and 320 respectively. The approximation of the population at time t = 0 using the OCTAVE code is:
 
<gallery>
 
<gallery>
 
Archivo:341.png
 
Archivo:341.png
 
</gallery>
 
</gallery>
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[[Categoría:Grado en Ingeniería Civil y Territorial]]
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[[Categoría:Ecuaciones Diferenciales]]
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[[Categoría:Articles in English]]
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[[Categoría:Trabajos 2012-13]]

Revisión actual del 00:29 3 jun 2013

1 Introduction

The Lotka – Volterra equations, also known by the name of predator-prey equations, are a pair of first order and non linear differential equations. They are commonly used to describe the model in which two species (predator and prey) interact one with the other, their interactions and competitions.


[math] \left\{\begin{matrix}\frac{\mathrm{d} R}{\mathrm{d} t}=aR-bRF\\\frac{\mathrm{d} F}{\mathrm{d} t}=-cF+dRF\\R(t_{0})=R_{0}, F(t_{0})=F_{0}\end{matrix}\right. [/math]

2 Interpretation

First of all,we are going to describe the physical meanings term by term:

- R is the number of preys, in this case, Rabbits.

- F is the number of predators, for this example, theFoxes.

- dR/dt is the growth of the population of Rabbits.

- dF/dt is the growth of the population of Foxes.

- a, b, c and d are parameters describing the interaction of the species.

Now we are going to analyze how they are related with themselves.

- First we assume that the preys have an unlimited food supply. The increase of rabbits, in absence of foxes (F=0) is proportional to the initial number of rabbits. The growth of the rabbits is affected by the interaction between the rabbits and foxes (FR). The“b” parameter indicates this process. So we can conclude that the change in rabbit's population is given by its growth minus the number of eaten by the foxes.

[math] \frac{\mathrm{d} R}{\mathrm{d} t}=aR-bRF [/math]


- We have to assume that the food supply for the foxes depends entirely on the rabbit population. The “-c” parameter tell us about the decreasing rhythm of the foxes, affected by their own death, emigration, etc. In absence of preys (R=0), it leads to an exponential decay.

[math] \frac{\mathrm{d} F}{\mathrm{d} t}=-cF+dRF [/math]

3 Numerical Resolution

We have to apply iterative methods to solve the equations, because we are in the case that the equations are non linear.

NOTE: at allthe exercises we use the scale 1:5000.

3.1 Euler Method

For an initial population of R=3000 and F=1000, using an interval of time(0,100) and using a=0.4, b=0.37, c=0.3, d=0.05. The OCTAVE code for 5000 intervals is:

The graphic solutions are:

We have taken 5000 iterations trying to obtain the realest solution (this is because, according to the Euler Method the higher the number of N is, the shorter will be the distance h and closer to the real solution it will be. The system solution is cyclical. We can affirm that the ecosystem is not stable because, as we can see clearly in the graphics, as we advance in time, there are more differences between foxes and rabbits(sometimes even the population of foxes is bigger than the rabbit’s), so the system is less stable.

Now we are going to suppose three cases with three different values of rabbit’s population.

R = 1500 R = 1000 R = 250

As we can see,in all the graphics the number of rabbits is growing up among time. Logically,the less rabbits there are, more foxes will die at the beginning, and this will cause that the number of rabbits will reach more maximums. The number of foxes will reach the lowest minimums when the initial population of rabbits is the lowest, and as a consequence of this, when the number of rabbits reach its maximum, the population of foxes will grow up again to reach its maximum. Talking about the proximity to the limit of foxes before extinction (50), in the last case it almost reaches this point, getting very close and getting into a dangerous point for these animals.

3.2 Runge-Kutta Method

Now we are going to solve the system by the 4rd order Runge-Kutta method, with the same initial populations and values for the constancies as in the Euler method. The OCTAVE code is:

As we have explained before, the Euler Method gets closer to the real solution when N is bigger. A better approximation to that real solution is obtained by the Runge KuttaMethod. The solution using the Euler Method with the highest N seems to be like the solutions given by the Runge-Kutta Method explained now. The solutions given by this system are much more precise than the Euler’s. This is because in the Euler Method the approximation is given by the tangent line to the curve and becauseof this, in cyclic solutions like the ones in our cases, it does not give us good solutions.

3.3 Trapezoidal Method

This one is an implicit method that requires solving our non-linear equation and solves the following value of (R, F).It´s too difficult to solve it by using this method.

3.4 Backward Solution

Assuming that now at time t = 100 thepopulation of rabbits and foxes is 2200 and 320 respectively. The approximation of the population at time t = 0 using the OCTAVE code is: