#### 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

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

- 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

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

- 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

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

- 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

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

- 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.

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.- 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

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

- 1 year 2 months ago

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

##### Samuel Kaelin

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

- 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:

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 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.

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.- 1 year 2 months ago

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

##### Cameron Demler

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

- 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

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

- 1 year 2 months ago

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

##### luke vanlaningham

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

- 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

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

- 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.

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.- 1 year 2 months ago

#### Clone of Logistic Growth

##### luke vanlaningham

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

- 1 year 2 months ago

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

##### Adheke Silas

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

- 1 year 2 months ago

#### Clone of Basic Model, Tyson Lynx and Hare

##### Thomas Browe

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.

- 1 year 2 months ago

#### Clone of Logistic Growth

##### Jon Ford

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

- 1 year 2 months ago

#### Clone of Basic Model, Tyson Lynx and Hare

##### Jacob Adkins

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.

- 1 year 2 months ago

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

##### Kate Quinn

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

- 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:

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 year 2 months ago

#### Clone of Basic Model, Tyson Lynx and Hare

##### Proctor Mercer

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.

- 1 year 2 months ago

#### Clone of Basic Model, Tyson Lynx and Hare

##### Jacob Adkins

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.

- 1 year 2 months ago

#### Clone of Basic Model, Tyson Lynx and Hare

##### Proctor Mercer

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.

- 1 year 2 months ago