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Math Modeling

Clone of A Simple SIR (Susceptible, Infected, Recovered) Example

Proctor Mercer
This is a first example of a simple SIR (Susceptible, Infected, Recovered) model.

There are three pools of individuals: those who are infected (without them, no disease!), the pool of those who are at risk (susceptible), and the recovered -- who may lose their immunity and become susceptible again.

A comparable model in Mathematica is available at
http://www.nku.edu/~longa/classes/2018spring/mat375/mathematica/SIRModel.nb

SIR Math Modeling Mat375

  • 1 year 2 months ago

Clone of A Simple SIR (Susceptible, Infected, Recovered) Example

Jared Slavey
This is a first example of a simple SIR (Susceptible, Infected, Recovered) model.

There are three pools of individuals: those who are infected (without them, no disease!), the pool of those who are at risk (susceptible), and the recovered -- who may lose their immunity and become susceptible again.

A comparable model in Mathematica is available at
http://www.nku.edu/~longa/classes/2018spring/mat375/mathematica/SIRModel.nb

SIR Math Modeling Mat375

  • 1 year 2 months ago

Clone of Clone of A Simple Infection-only SIR (Susceptible, Infected, Recovered) Example

Jiangkun Wang
This is a simple example of (part of a) simple SIR (Susceptible, Infected, Recovered) model, suggested by De Vries, et al. in A Course in Mathematical Biology.

They wanted to illustrate the comparative behavior of differential equations and discrete difference equations. We know that differential equations are generally solved numerically by discretizing them, so that the comparison is a little bit rigged....

A comparable model in Mathematica is available at
http://www.nku.edu/~longa/classes/2018spring/mat375/mathematica/SIRModel-w-discrete-version.nb

SIR Math Modeling Mat375

  • 8 months 3 weeks ago

Clone of Logistic Growth

Jared Slavey
This simple model demonstrates logistic growth.The differential equation looks like

y'(t)=by(t)(1-y(t)/K)

where K is the carrying capacity of the quantity y. Alternatively,

y'(t)=by(t) - b/K*y(t)^2

so the growth term suggests exponential growth, but there is a loss term is of the form b/K y(t) -- loss is proportional to population (crowding).

A comparable Mathematica file is available at
http://www.nku.edu/~longa/classes/2018spring/mat375/mathematica/LogisticGrowth-and-DecayModel.nb

Math Modeling Mat375

  • 1 year 3 months ago

Clone of Basic Model, Tyson Lynx and Hare

Cameron Demler
The basic model of Modelling the Canada lynx and snowshoe hare population cycle: The role of specialist predators (Tyson, et al.) demonstrates logistic growth in prey, and in predator (with prey dependence for carrying capacity). But interestingly, one possibility is limit cycles, which mimic the cycling of the populations in nature.

The differential equations for the population of hare (x) is

x'(t) = rx(1-x/K)            - gamma x^2/(x^2+eta^2)            - alpha y x/(x+mu)
where K is the logistic carrying capacity of the prey (hare), in the absence of predation; the second term is a "generalist predation" term;  and the third term is the "specialist predation" (in the limit as the prey gets big, this becomes simply proportional to y (the lynx population)).

The differential equations for the population of lynx (y) is
y'(t) = sy(1- qy/x) = sy - sqy^2/x
for the predator (lynx), which is essentially logistic growth. Its growth term suggests exponential growth, but there is a loss term of the form sqy^2/x -- loss is proportional to population (crowding), and inversely proportional to prey density. As the hare population goes to zero, so shall the lynx....

As one can see, the prey density won't change if y=x/q. If the prey density were not changing at the same time, the system would be at equilibrium.

In this InsightMaker model, I scaled the second equation by multiplying by q, then replace y by w=qy throughout both equations. This requires a slight change in the prey equation -- alpha replaced by the ratio of alpha/q.  (I used my favorite mathematical trick, of multiplying by the appropriate form of 1!)
So what we're really looking at here is the system

x'(t) = rx(1-x/K)            - gamma x^2/(x^2+eta^2)            - alpha/q w x/(x+mu)w'(t) = sw(1- w/x)

where w(t)=qy(t).

Tyson, et al. took q to be about 212 for hare and lynx -- so that it requires about 212 hare to allow for one lynx to survive at "equilibrium".

However, when alpha -- the hares/lynx/year -- gets sufficiently large (e.g. 1867 -- and that does seem like a lot of hares per lynx per year...:), limit cycles develop (rather than a stable equilibrium). This means that the populations oscillate about the equilibrium values, rather than stabilize at those values.

Author: Andy Long, Northern Kentucky University (2020)

Reference: Tyson, Rebecca, Sheena Haines,  Karen Hodges. Modelling the Canada lynx and snowshoe hare population cycle: The role of specialist predators. Theoretical Ecology. 3, 97–111 (2010). https://doi.org/10.1007/s12080-009-0057-1
Resource: A comparable Mathematica model can be found at  http://ceadserv1.nku.edu/longa//classes/mat375/days/Mathematica/BasicModel.nb, which allows one to experiment a little more easily than one can with this InsightMaker model.

Math Modeling Mat375

  • 1 year 2 months ago

Clone of A Simple SIR (Susceptible, Infected, Recovered) Example

Cameron Demler
This is a first example of a simple SIR (Susceptible, Infected, Recovered) model.

There are three pools of individuals: those who are infected (without them, no disease!), the pool of those who are at risk (susceptible), and the recovered -- who may lose their immunity and become susceptible again.

A comparable model in Mathematica is available at
http://www.nku.edu/~longa/classes/2018spring/mat375/mathematica/SIRModel.nb

SIR Math Modeling Mat375

  • 1 year 2 months ago

Clone of A Simple SIR (Susceptible, Infected, Recovered) Example

Samuel Kaelin
This is a first example of a simple SIR (Susceptible, Infected, Recovered) model.

There are three pools of individuals: those who are infected (without them, no disease!), the pool of those who are at risk (susceptible), and the recovered -- who may lose their immunity and become susceptible again.

A comparable model in Mathematica is available at
http://www.nku.edu/~longa/classes/2018spring/mat375/mathematica/SIRModel.nb

SIR Math Modeling Mat375

  • 1 year 2 months ago

Clone of Coronavirus: A Simple SIR (Susceptible, Infected, Recovered) with death

Jon Ford
Spring, 2020: in the midst of on-line courses, due to the pandemic of Covid-19.

With the onset of the Covid-19 coronavirus crisis, we focus on SIRD models, which might realistically model the course of the disease.
We start with an SIR model, such as that featured in the MAA model featured inhttps://www.maa.org/press/periodicals/loci/joma/the-sir-model-for-spread-of-disease-the-differential-equation-model
Without mortality, with time measured in days, with infection rate 1/2, recovery rate 1/3, and initial infectious population I_0=1.27x10-4, we reproduce their figure
With a death rate of .005 (one two-hundredth of the infected per day), an infectivity rate of 0.5, and a recovery rate of .145 or so (takes about a week to recover), we get some pretty significant losses -- about 3.2% of the total population.
Resources:
  1. http://www.nku.edu/~longa/classes/2020spring/mat375/mathematica/SIRModel-MAA.nb
  2. https://www.maa.org/press/periodicals/loci/joma/the-sir-model-for-spread-of-disease-the-differential-equation-model

SIR Math Modeling Mat375 COVID-19 Coronavirus SIRD

  • 1 year 1 month ago

Clone of Basic Model, Tyson Lynx and Hare

Jordan
The basic model of Modelling the Canada lynx and snowshoe hare population cycle: The role of specialist predators (Tyson, et al.) demonstrates logistic growth in prey, and in predator (with prey dependence for carrying capacity). But interestingly, one possibility is limit cycles, which mimic the cycling of the populations in nature.

The differential equations for the population of hare (x) is

x'(t) = rx(1-x/K)            - gamma x^2/(x^2+eta^2)            - alpha y x/(x+mu)
where K is the logistic carrying capacity of the prey (hare), in the absence of predation; the second term is a "generalist predation" term;  and the third term is the "specialist predation" (in the limit as the prey gets big, this becomes simply proportional to y (the lynx population)).

The differential equations for the population of lynx (y) is
y'(t) = sy(1- qy/x) = sy - sqy^2/x
for the predator (lynx), which is essentially logistic growth. Its growth term suggests exponential growth, but there is a loss term of the form sqy^2/x -- loss is proportional to population (crowding), and inversely proportional to prey density. As the hare population goes to zero, so shall the lynx....

As one can see, the prey density won't change if y=x/q. If the prey density were not changing at the same time, the system would be at equilibrium.

In this InsightMaker model, I scaled the second equation by multiplying by q, then replace y by w=qy throughout both equations. This requires a slight change in the prey equation -- alpha replaced by the ratio of alpha/q.  (I used my favorite mathematical trick, of multiplying by the appropriate form of 1!)
So what we're really looking at here is the system

x'(t) = rx(1-x/K)            - gamma x^2/(x^2+eta^2)            - alpha/q w x/(x+mu)w'(t) = sw(1- w/x)

where w(t)=qy(t).

Tyson, et al. took q to be about 212 for hare and lynx -- so that it requires about 212 hare to allow for one lynx to survive at "equilibrium".

However, when alpha -- the hares/lynx/year -- gets sufficiently large (e.g. 1867 -- and that does seem like a lot of hares per lynx per year...:), limit cycles develop (rather than a stable equilibrium). This means that the populations oscillate about the equilibrium values, rather than stabilize at those values.

Author: Andy Long, Northern Kentucky University (2020)

Reference: Tyson, Rebecca, Sheena Haines,  Karen Hodges. Modelling the Canada lynx and snowshoe hare population cycle: The role of specialist predators. Theoretical Ecology. 3, 97–111 (2010). https://doi.org/10.1007/s12080-009-0057-1
Resource: A comparable Mathematica model can be found at  http://ceadserv1.nku.edu/longa//classes/mat375/days/Mathematica/BasicModel.nb, which allows one to experiment a little more easily than one can with this InsightMaker model.

Math Modeling Mat375

  • 1 year 2 months ago

Clone of A Simple SIR (Susceptible, Infected, Recovered) Example

Cameron Demler
This is a first example of a simple SIR (Susceptible, Infected, Recovered) model.

There are three pools of individuals: those who are infected (without them, no disease!), the pool of those who are at risk (susceptible), and the recovered -- who may lose their immunity and become susceptible again.

A comparable model in Mathematica is available at
http://www.nku.edu/~longa/classes/2018spring/mat375/mathematica/SIRModel.nb

SIR Math Modeling Mat375

  • 1 year 2 months ago

Clone of Logistic Growth

Proctor Mercer
This simple model demonstrates logistic growth.The differential equation looks like

y'(t)=by(t)(1-y(t)/K)

where K is the carrying capacity of the quantity y. Alternatively,

y'(t)=by(t) - b/K*y(t)^2

so the growth term suggests exponential growth, but there is a loss term is of the form b/K y(t) -- loss is proportional to population (crowding).

A comparable Mathematica file is available at
http://www.nku.edu/~longa/classes/2018spring/mat375/mathematica/LogisticGrowth-and-DecayModel.nb

Math Modeling Mat375

  • 1 year 2 months ago

Clone of A Simple SIR (Susceptible, Infected, Recovered) Example

luke vanlaningham
This is a first example of a simple SIR (Susceptible, Infected, Recovered) model.

There are three pools of individuals: those who are infected (without them, no disease!), the pool of those who are at risk (susceptible), and the recovered -- who may lose their immunity and become susceptible again.

A comparable model in Mathematica is available at
http://www.nku.edu/~longa/classes/2018spring/mat375/mathematica/SIRModel.nb

SIR Math Modeling Mat375

  • 1 year 2 months ago

Clone of Logistic Growth

Thomas Browe
This simple model demonstrates logistic growth.The differential equation looks like

y'(t)=by(t)(1-y(t)/K)

where K is the carrying capacity of the quantity y. Alternatively,

y'(t)=by(t) - b/K*y(t)^2

so the growth term suggests exponential growth, but there is a loss term is of the form b/K y(t) -- loss is proportional to population (crowding).

A comparable Mathematica file is available at
http://www.nku.edu/~longa/classes/2018spring/mat375/mathematica/LogisticGrowth-and-DecayModel.nb

Math Modeling Mat375

  • 1 year 2 months ago

Clone of Basic Model, Tyson Lynx and Hare

Cameron Demler
The basic model of Modelling the Canada lynx and snowshoe hare population cycle: The role of specialist predators (Tyson, et al.) demonstrates logistic growth in prey, and in predator (with prey dependence for carrying capacity). But interestingly, one possibility is limit cycles, which mimic the cycling of the populations in nature.

The differential equations for the population of hare (x) is

x'(t) = rx(1-x/K)            - gamma x^2/(x^2+eta^2)            - alpha y x/(x+mu)
where K is the logistic carrying capacity of the prey (hare), in the absence of predation; the second term is a "generalist predation" term;  and the third term is the "specialist predation" (in the limit as the prey gets big, this becomes simply proportional to y (the lynx population)).

The differential equations for the population of lynx (y) is
y'(t) = sy(1- qy/x) = sy - sqy^2/x
for the predator (lynx), which is essentially logistic growth. Its growth term suggests exponential growth, but there is a loss term of the form sqy^2/x -- loss is proportional to population (crowding), and inversely proportional to prey density. As the hare population goes to zero, so shall the lynx....

As one can see, the prey density won't change if y=x/q. If the prey density were not changing at the same time, the system would be at equilibrium.

In this InsightMaker model, I scaled the second equation by multiplying by q, then replace y by w=qy throughout both equations. This requires a slight change in the prey equation -- alpha replaced by the ratio of alpha/q.  (I used my favorite mathematical trick, of multiplying by the appropriate form of 1!)
So what we're really looking at here is the system

x'(t) = rx(1-x/K)            - gamma x^2/(x^2+eta^2)            - alpha/q w x/(x+mu)w'(t) = sw(1- w/x)

where w(t)=qy(t).

Tyson, et al. took q to be about 212 for hare and lynx -- so that it requires about 212 hare to allow for one lynx to survive at "equilibrium".

However, when alpha -- the hares/lynx/year -- gets sufficiently large (e.g. 1867 -- and that does seem like a lot of hares per lynx per year...:), limit cycles develop (rather than a stable equilibrium). This means that the populations oscillate about the equilibrium values, rather than stabilize at those values.

Author: Andy Long, Northern Kentucky University (2020)

Reference: Tyson, Rebecca, Sheena Haines,  Karen Hodges. Modelling the Canada lynx and snowshoe hare population cycle: The role of specialist predators. Theoretical Ecology. 3, 97–111 (2010). https://doi.org/10.1007/s12080-009-0057-1
Resource: A comparable Mathematica model can be found at  http://ceadserv1.nku.edu/longa//classes/mat375/days/Mathematica/BasicModel.nb, which allows one to experiment a little more easily than one can with this InsightMaker model.

Math Modeling Mat375

  • 1 year 2 months ago

Clone of Logistic Growth

luke vanlaningham
This simple model demonstrates logistic growth.The differential equation looks like

y'(t)=by(t)(1-y(t)/K)

where K is the carrying capacity of the quantity y. Alternatively,

y'(t)=by(t) - b/K*y(t)^2

so the growth term suggests exponential growth, but there is a loss term is of the form b/K y(t) -- loss is proportional to population (crowding).

A comparable Mathematica file is available at
http://www.nku.edu/~longa/classes/2018spring/mat375/mathematica/LogisticGrowth-and-DecayModel.nb

Math Modeling Mat375

  • 1 year 2 months ago

Clone of A Simple SIR (Susceptible, Infected, Recovered) Example

Adheke Silas
This is a first example of a simple SIR (Susceptible, Infected, Recovered) model.

There are three pools of individuals: those who are infected (without them, no disease!), the pool of those who are at risk (susceptible), and the recovered -- who may lose their immunity and become susceptible again.

A comparable model in Mathematica is available at
http://www.nku.edu/~longa/classes/2018spring/mat375/mathematica/SIRModel.nb

SIR Math Modeling Mat375

  • 1 year 2 months ago

Clone of Basic Model, Tyson Lynx and Hare

Thomas Browe
The basic model of Modelling the Canada lynx and snowshoe hare population cycle: The role of specialist predators (Tyson, et al.) demonstrates logistic growth in prey, and in predator (with prey dependence for carrying capacity). But interestingly, one possibility is limit cycles, which mimic the cycling of the populations in nature.

The differential equations for the population of hare (x) is

x'(t) = rx(1-x/K)            - gamma x^2/(x^2+eta^2)            - alpha y x/(x+mu)
where K is the logistic carrying capacity of the prey (hare), in the absence of predation; the second term is a "generalist predation" term;  and the third term is the "specialist predation" (in the limit as the prey gets big, this becomes simply proportional to y (the lynx population)).

The differential equations for the population of lynx (y) is
y'(t) = sy(1- qy/x) = sy - sqy^2/x
for the predator (lynx), which is essentially logistic growth. Its growth term suggests exponential growth, but there is a loss term of the form sqy^2/x -- loss is proportional to population (crowding), and inversely proportional to prey density. As the hare population goes to zero, so shall the lynx....

As one can see, the prey density won't change if y=x/q. If the prey density were not changing at the same time, the system would be at equilibrium.

In this InsightMaker model, I scaled the second equation by multiplying by q, then replace y by w=qy throughout both equations. This requires a slight change in the prey equation -- alpha replaced by the ratio of alpha/q.  (I used my favorite mathematical trick, of multiplying by the appropriate form of 1!)
So what we're really looking at here is the system

x'(t) = rx(1-x/K)            - gamma x^2/(x^2+eta^2)            - alpha/q w x/(x+mu)w'(t) = sw(1- w/x)

where w(t)=qy(t).

Tyson, et al. took q to be about 212 for hare and lynx -- so that it requires about 212 hare to allow for one lynx to survive at "equilibrium".

However, when alpha -- the hares/lynx/year -- gets sufficiently large (e.g. 1867 -- and that does seem like a lot of hares per lynx per year...:), limit cycles develop (rather than a stable equilibrium). This means that the populations oscillate about the equilibrium values, rather than stabilize at those values.

Author: Andy Long, Northern Kentucky University (2020)

Reference: Tyson, Rebecca, Sheena Haines,  Karen Hodges. Modelling the Canada lynx and snowshoe hare population cycle: The role of specialist predators. Theoretical Ecology. 3, 97–111 (2010). https://doi.org/10.1007/s12080-009-0057-1
Resource: A comparable Mathematica model can be found at  http://ceadserv1.nku.edu/longa//classes/mat375/days/Mathematica/BasicModel.nb, which allows one to experiment a little more easily than one can with this InsightMaker model.

Math Modeling Mat375

  • 1 year 2 months ago

Clone of Logistic Growth

Jon Ford
This simple model demonstrates logistic growth.The differential equation looks like

y'(t)=by(t)(1-y(t)/K)

where K is the carrying capacity of the quantity y. Alternatively,

y'(t)=by(t) - b/K*y(t)^2

so the growth term suggests exponential growth, but there is a loss term is of the form b/K y(t) -- loss is proportional to population (crowding).

A comparable Mathematica file is available at
http://www.nku.edu/~longa/classes/2018spring/mat375/mathematica/LogisticGrowth-and-DecayModel.nb

Math Modeling Mat375

  • 1 year 2 months ago

Clone of Basic Model, Tyson Lynx and Hare

Jacob Adkins
The basic model of Modelling the Canada lynx and snowshoe hare population cycle: The role of specialist predators (Tyson, et al.) demonstrates logistic growth in prey, and in predator (with prey dependence for carrying capacity). But interestingly, one possibility is limit cycles, which mimic the cycling of the populations in nature.

The differential equations for the population of hare (x) is

x'(t) = rx(1-x/K)            - gamma x^2/(x^2+eta^2)            - alpha y x/(x+mu)
where K is the logistic carrying capacity of the prey (hare), in the absence of predation; the second term is a "generalist predation" term;  and the third term is the "specialist predation" (in the limit as the prey gets big, this becomes simply proportional to y (the lynx population)).

The differential equations for the population of lynx (y) is
y'(t) = sy(1- qy/x) = sy - sqy^2/x
for the predator (lynx), which is essentially logistic growth. Its growth term suggests exponential growth, but there is a loss term of the form sqy^2/x -- loss is proportional to population (crowding), and inversely proportional to prey density. As the hare population goes to zero, so shall the lynx....

As one can see, the prey density won't change if y=x/q. If the prey density were not changing at the same time, the system would be at equilibrium.

In this InsightMaker model, I scaled the second equation by multiplying by q, then replace y by w=qy throughout both equations. This requires a slight change in the prey equation -- alpha replaced by the ratio of alpha/q.  (I used my favorite mathematical trick, of multiplying by the appropriate form of 1!)
So what we're really looking at here is the system

x'(t) = rx(1-x/K)            - gamma x^2/(x^2+eta^2)            - alpha/q w x/(x+mu)w'(t) = sw(1- w/x)

where w(t)=qy(t).

Tyson, et al. took q to be about 212 for hare and lynx -- so that it requires about 212 hare to allow for one lynx to survive at "equilibrium".

However, when alpha -- the hares/lynx/year -- gets sufficiently large (e.g. 1867 -- and that does seem like a lot of hares per lynx per year...:), limit cycles develop (rather than a stable equilibrium). This means that the populations oscillate about the equilibrium values, rather than stabilize at those values.

Author: Andy Long, Northern Kentucky University (2020)

Reference: Tyson, Rebecca, Sheena Haines,  Karen Hodges. Modelling the Canada lynx and snowshoe hare population cycle: The role of specialist predators. Theoretical Ecology. 3, 97–111 (2010). https://doi.org/10.1007/s12080-009-0057-1
Resource: A comparable Mathematica model can be found at  http://ceadserv1.nku.edu/longa//classes/mat375/days/Mathematica/BasicModel.nb, which allows one to experiment a little more easily than one can with this InsightMaker model.

Math Modeling Mat375

  • 1 year 2 months ago

Clone of A Simple SIR (Susceptible, Infected, Recovered) Example

Kate Quinn
This is a first example of a simple SIR (Susceptible, Infected, Recovered) model.

There are three pools of individuals: those who are infected (without them, no disease!), the pool of those who are at risk (susceptible), and the recovered -- who may lose their immunity and become susceptible again.

A comparable model in Mathematica is available at
http://www.nku.edu/~longa/classes/2018spring/mat375/mathematica/SIRModel.nb

SIR Math Modeling Mat375

  • 1 year 2 months ago

Clone of Coronavirus: A Simple SIR (Susceptible, Infected, Recovered) with death

Jon Ford
Spring, 2020: in the midst of on-line courses, due to the pandemic of Covid-19.

With the onset of the Covid-19 coronavirus crisis, we focus on SIRD models, which might realistically model the course of the disease.
We start with an SIR model, such as that featured in the MAA model featured inhttps://www.maa.org/press/periodicals/loci/joma/the-sir-model-for-spread-of-disease-the-differential-equation-model
Without mortality, with time measured in days, with infection rate 1/2, recovery rate 1/3, and initial infectious population I_0=1.27x10-4, we reproduce their figure
With a death rate of .005 (one two-hundredth of the infected per day), an infectivity rate of 0.5, and a recovery rate of .145 or so (takes about a week to recover), we get some pretty significant losses -- about 3.2% of the total population.
Resources:
  1. http://www.nku.edu/~longa/classes/2020spring/mat375/mathematica/SIRModel-MAA.nb
  2. https://www.maa.org/press/periodicals/loci/joma/the-sir-model-for-spread-of-disease-the-differential-equation-model

SIR Math Modeling Mat375 COVID-19 Coronavirus SIRD

  • 1 year 2 months ago

Clone of Basic Model, Tyson Lynx and Hare

Proctor Mercer
The basic model of Modelling the Canada lynx and snowshoe hare population cycle: The role of specialist predators (Tyson, et al.) demonstrates logistic growth in prey, and in predator (with prey dependence for carrying capacity). But interestingly, one possibility is limit cycles, which mimic the cycling of the populations in nature.

The differential equations for the population of hare (x) is

x'(t) = rx(1-x/K)            - gamma x^2/(x^2+eta^2)            - alpha y x/(x+mu)
where K is the logistic carrying capacity of the prey (hare), in the absence of predation; the second term is a "generalist predation" term;  and the third term is the "specialist predation" (in the limit as the prey gets big, this becomes simply proportional to y (the lynx population)).

The differential equations for the population of lynx (y) is
y'(t) = sy(1- qy/x) = sy - sqy^2/x
for the predator (lynx), which is essentially logistic growth. Its growth term suggests exponential growth, but there is a loss term of the form sqy^2/x -- loss is proportional to population (crowding), and inversely proportional to prey density. As the hare population goes to zero, so shall the lynx....

As one can see, the prey density won't change if y=x/q. If the prey density were not changing at the same time, the system would be at equilibrium.

In this InsightMaker model, I scaled the second equation by multiplying by q, then replace y by w=qy throughout both equations. This requires a slight change in the prey equation -- alpha replaced by the ratio of alpha/q.  (I used my favorite mathematical trick, of multiplying by the appropriate form of 1!)
So what we're really looking at here is the system

x'(t) = rx(1-x/K)            - gamma x^2/(x^2+eta^2)            - alpha/q w x/(x+mu)w'(t) = sw(1- w/x)

where w(t)=qy(t).

Tyson, et al. took q to be about 212 for hare and lynx -- so that it requires about 212 hare to allow for one lynx to survive at "equilibrium".

However, when alpha -- the hares/lynx/year -- gets sufficiently large (e.g. 1867 -- and that does seem like a lot of hares per lynx per year...:), limit cycles develop (rather than a stable equilibrium). This means that the populations oscillate about the equilibrium values, rather than stabilize at those values.

Author: Andy Long, Northern Kentucky University (2020)

Reference: Tyson, Rebecca, Sheena Haines,  Karen Hodges. Modelling the Canada lynx and snowshoe hare population cycle: The role of specialist predators. Theoretical Ecology. 3, 97–111 (2010). https://doi.org/10.1007/s12080-009-0057-1
Resource: A comparable Mathematica model can be found at  http://ceadserv1.nku.edu/longa//classes/mat375/days/Mathematica/BasicModel.nb, which allows one to experiment a little more easily than one can with this InsightMaker model.

Math Modeling Mat375

  • 1 year 2 months ago

Clone of Basic Model, Tyson Lynx and Hare

Jacob Adkins
The basic model of Modelling the Canada lynx and snowshoe hare population cycle: The role of specialist predators (Tyson, et al.) demonstrates logistic growth in prey, and in predator (with prey dependence for carrying capacity). But interestingly, one possibility is limit cycles, which mimic the cycling of the populations in nature.

The differential equations for the population of hare (x) is

x'(t) = rx(1-x/K)            - gamma x^2/(x^2+eta^2)            - alpha y x/(x+mu)
where K is the logistic carrying capacity of the prey (hare), in the absence of predation; the second term is a "generalist predation" term;  and the third term is the "specialist predation" (in the limit as the prey gets big, this becomes simply proportional to y (the lynx population)).

The differential equations for the population of lynx (y) is
y'(t) = sy(1- qy/x) = sy - sqy^2/x
for the predator (lynx), which is essentially logistic growth. Its growth term suggests exponential growth, but there is a loss term of the form sqy^2/x -- loss is proportional to population (crowding), and inversely proportional to prey density. As the hare population goes to zero, so shall the lynx....

As one can see, the prey density won't change if y=x/q. If the prey density were not changing at the same time, the system would be at equilibrium.

In this InsightMaker model, I scaled the second equation by multiplying by q, then replace y by w=qy throughout both equations. This requires a slight change in the prey equation -- alpha replaced by the ratio of alpha/q.  (I used my favorite mathematical trick, of multiplying by the appropriate form of 1!)
So what we're really looking at here is the system

x'(t) = rx(1-x/K)            - gamma x^2/(x^2+eta^2)            - alpha/q w x/(x+mu)w'(t) = sw(1- w/x)

where w(t)=qy(t).

Tyson, et al. took q to be about 212 for hare and lynx -- so that it requires about 212 hare to allow for one lynx to survive at "equilibrium".

However, when alpha -- the hares/lynx/year -- gets sufficiently large (e.g. 1867 -- and that does seem like a lot of hares per lynx per year...:), limit cycles develop (rather than a stable equilibrium). This means that the populations oscillate about the equilibrium values, rather than stabilize at those values.

Author: Andy Long, Northern Kentucky University (2020)

Reference: Tyson, Rebecca, Sheena Haines,  Karen Hodges. Modelling the Canada lynx and snowshoe hare population cycle: The role of specialist predators. Theoretical Ecology. 3, 97–111 (2010). https://doi.org/10.1007/s12080-009-0057-1
Resource: A comparable Mathematica model can be found at  http://ceadserv1.nku.edu/longa//classes/mat375/days/Mathematica/BasicModel.nb, which allows one to experiment a little more easily than one can with this InsightMaker model.

Math Modeling Mat375

  • 1 year 2 months ago

Clone of Basic Model, Tyson Lynx and Hare

Proctor Mercer
The basic model of Modelling the Canada lynx and snowshoe hare population cycle: The role of specialist predators (Tyson, et al.) demonstrates logistic growth in prey, and in predator (with prey dependence for carrying capacity). But interestingly, one possibility is limit cycles, which mimic the cycling of the populations in nature.

The differential equations for the population of hare (x) is

x'(t) = rx(1-x/K)            - gamma x^2/(x^2+eta^2)            - alpha y x/(x+mu)
where K is the logistic carrying capacity of the prey (hare), in the absence of predation; the second term is a "generalist predation" term;  and the third term is the "specialist predation" (in the limit as the prey gets big, this becomes simply proportional to y (the lynx population)).

The differential equations for the population of lynx (y) is
y'(t) = sy(1- qy/x) = sy - sqy^2/x
for the predator (lynx), which is essentially logistic growth. Its growth term suggests exponential growth, but there is a loss term of the form sqy^2/x -- loss is proportional to population (crowding), and inversely proportional to prey density. As the hare population goes to zero, so shall the lynx....

As one can see, the prey density won't change if y=x/q. If the prey density were not changing at the same time, the system would be at equilibrium.

In this InsightMaker model, I scaled the second equation by multiplying by q, then replace y by w=qy throughout both equations. This requires a slight change in the prey equation -- alpha replaced by the ratio of alpha/q.  (I used my favorite mathematical trick, of multiplying by the appropriate form of 1!)
So what we're really looking at here is the system

x'(t) = rx(1-x/K)            - gamma x^2/(x^2+eta^2)            - alpha/q w x/(x+mu)w'(t) = sw(1- w/x)

where w(t)=qy(t).

Tyson, et al. took q to be about 212 for hare and lynx -- so that it requires about 212 hare to allow for one lynx to survive at "equilibrium".

However, when alpha -- the hares/lynx/year -- gets sufficiently large (e.g. 1867 -- and that does seem like a lot of hares per lynx per year...:), limit cycles develop (rather than a stable equilibrium). This means that the populations oscillate about the equilibrium values, rather than stabilize at those values.

Author: Andy Long, Northern Kentucky University (2020)

Reference: Tyson, Rebecca, Sheena Haines,  Karen Hodges. Modelling the Canada lynx and snowshoe hare population cycle: The role of specialist predators. Theoretical Ecology. 3, 97–111 (2010). https://doi.org/10.1007/s12080-009-0057-1
Resource: A comparable Mathematica model can be found at  http://ceadserv1.nku.edu/longa//classes/mat375/days/Mathematica/BasicModel.nb, which allows one to experiment a little more easily than one can with this InsightMaker model.

Math Modeling Mat375

  • 1 year 2 months ago

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