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
SIRD Math Modeling COVID-19 Coronavirus Mat375 SIR
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 in

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:
Clone of Coronavirus: A Simple SIR (Susceptible, Infected, Recovered) with death
Xuexiao Zhang
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
COVID-19 SIR Mat375 SIRD Coronavirus Math Modeling
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 in

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:
Clone of Coronavirus: A Simple SIR (Susceptible, Infected, Recovered) with death
Proctor Mercer
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
SIR Mat375 SIRD COVID-19 Coronavirus Math Modeling
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 in

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:
Clone of Coronavirus: A Simple SIR (Susceptible, Infected, Recovered) with death
Jon Ford
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
Mat375 Math Modeling SIR
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

LabSIR Key of A Simple SIR (Susceptible, Infected, Recovered) Example
Andrew E Long
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
SIR Mat375 Math Modeling
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

Clone of A Simple SIR (Susceptible, Infected, Recovered) Example
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
Mat375 Math Modeling
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
Clone of Logistic Growth
Adheke Silas
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
Mat375 Math Modeling
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.
Clone of Basic Model, Tyson Lynx and Hare
Adheke Silas
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
Math Modeling Mat375
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.
Clone of Basic Model, Tyson Lynx and Hare
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
Mat375 Math Modeling
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
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
Mat375 Math Modeling
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
Clone of Logistic Growth
Adheke Silas
5
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
Mat375 Math Modeling
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.
Clone of Basic Model, Tyson Lynx and Hare
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
Mat375 Math Modeling SIR
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

Clone of A Simple SIR (Susceptible, Infected, Recovered) Example
Sean Field
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
Math Modeling Mat375
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.
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
Mat375 Math Modeling
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.
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
Math Modeling Mat375
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.
Clone of Basic Model, Tyson Lynx and Hare
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
Mat375 Math Modeling
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
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
Mat375 Math Modeling
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
Clone of Logistic Growth
Jon Ford
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
Mat375 Math Modeling
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.
Clone of Basic Model, Tyson Lynx and Hare
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
Mat375 Math Modeling
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
Clone of Logistic Growth
Sean Field
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
Math Modeling Mat375
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.
Clone of Basic Model, Tyson Lynx and Hare
Thomas Browe
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
SIR Math Modeling Mat375
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

Clone of A Simple SIR (Susceptible, Infected, Recovered) Example
Jon Ford
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
Math Modeling Mat375
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.
Clone of Basic Model, Tyson Lynx and Hare
Jacob Adkins
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
Mat375 Math Modeling
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
Clone of Logistic Growth
Cameron Demler
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 different
SIR Mat375 Math Modeling
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

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