#### 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.

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

##### Sean Field

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 Basic Model, Tyson Lynx and Hare

##### 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 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

##### 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) -- 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 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

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 Logistic Growth

##### Adheke Silas

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

##### 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.

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

##### Jon Ford

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

##### Jon Ford

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

##### Jon Ford

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

##### Jon Ford

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

##### Jon Ford

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 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

##### Jon Ford

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

##### 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.

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 Basic Model, Tyson Lynx and Hare

##### Jon Ford

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

##### Jon Ford

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

##### Jon Ford

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 Sleek, non-dimensionalized SIR (Susceptible, Infected, Recovered) model

##### Jiangkun Wang

This is an example of an SIR (Susceptible, Infected, Recovered) model that has been re-parameterized down to the bare minimum, to illustrated the dynamics possible with the fewest number of parameters.

We're rescaled this SIR model, so that time is given in infection rate-appropriate time units, "rates" are now ratios of rates (with infectivity rate in the denominator), and populations are considered proportions (unfortunately InsightMaker doesn't function properly if I give them all values from 0 to 1, which sum to 1 -- so, at the moment, I give them values that sum to 100, and consider the results percentages).

The new display includes the asymptotics: the three sub-populations will tend to fixed values as time goes to infinity; the infected population goes to zero if the recovery rate is greater than the infectivity rate -- i.e., the disease dies out.

Note the use of a "ghost" stock (for Total Population), which I think is a pretty cool idea. It cuts down on the number of arcs in the model graph.

A comparable model in Mathematica is available at

http://www.nku.edu/~longa/classes/2018spring/mat375/mathematica/SIRModel-rescaled.nb

We're rescaled this SIR model, so that time is given in infection rate-appropriate time units, "rates" are now ratios of rates (with infectivity rate in the denominator), and populations are considered proportions (unfortunately InsightMaker doesn't function properly if I give them all values from 0 to 1, which sum to 1 -- so, at the moment, I give them values that sum to 100, and consider the results percentages).

The new display includes the asymptotics: the three sub-populations will tend to fixed values as time goes to infinity; the infected population goes to zero if the recovery rate is greater than the infectivity rate -- i.e., the disease dies out.

Note the use of a "ghost" stock (for Total Population), which I think is a pretty cool idea. It cuts down on the number of arcs in the model graph.

A comparable model in Mathematica is available at

http://www.nku.edu/~longa/classes/2018spring/mat375/mathematica/SIRModel-rescaled.nb

- 10 months 1 week ago

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

##### Jon Ford

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

##### Jiangkun Wang

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:

- 10 months 2 weeks ago

#### Clone of Non-dimensionalized Logistic Growth

##### Proctor Mercer

This (simplest!) model demonstrates logistic growth.The original differential equation looks like

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

where K is the carrying capacity of the quantity y.

But if we divide each side of the equation by K, we obtain

d(y/K)/dt = b (y/K) (1-y/K)

Defining a new variable w, the population relative to its carrying capacity, we obtain

dw/dt = b w (1 - w)

Finally we divide both sides by b, to write

dw/d(bt) = w (1 - w)

So if we work in dimensionless time units of bt, we have

w' = w (1 - w)

where the derivative is with respect to the variable bt=τ. .

τ=τ

This This equation, as simple as possible, contains all the dynamics (all the ways the population can behave), while masking the "trivialities"; but it kind of hides the physical aspects of the problem. So it's easy to study, but harder to interpret: alas, you can't have it all!:)

τ=1 when t=1b: so if b=.5/year, then τ=1 when t=2.

So the larger b (the greater the birthrate), the shorter the real time t to give τ=1. τ=τ=

τ=

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

where K is the carrying capacity of the quantity y.

But if we divide each side of the equation by K, we obtain

d(y/K)/dt = b (y/K) (1-y/K)

Defining a new variable w, the population relative to its carrying capacity, we obtain

dw/dt = b w (1 - w)

Finally we divide both sides by b, to write

dw/d(bt) = w (1 - w)

So if we work in dimensionless time units of bt, we have

w' = w (1 - w)

where the derivative is with respect to the variable bt=τ. .

τ=τ

This This equation, as simple as possible, contains all the dynamics (all the ways the population can behave), while masking the "trivialities"; but it kind of hides the physical aspects of the problem. So it's easy to study, but harder to interpret: alas, you can't have it all!:)

τ=1 when t=1b: so if b=.5/year, then τ=1 when t=2.

So the larger b (the greater the birthrate), the shorter the real time t to give τ=1. τ=τ=

τ=

- 1 year 2 months ago