#### Clone of Newton's Method

##### Sean Field

This insight implements Newton's method as an InsightMaker model.

It is important to use Euler's method, with step-size of 1. That's what allows us to get away with this!:)

Fun to try a couple of different cases, so I have built four choices into this example. You can choose the function ("Function Choice" of 0, 1, 2, or 3) using the slider.

Andy LongSpring, 2020

- 1 month 1 week ago

#### Clone of MAT375 Clone of Bio103 Predator-Prey Model ("Lotka'Volterra")

##### Kate Quinn

**Clone of Bio103 Predator-Prey Model ("Lotka'Volterra")**

Tags: Education, Chaos, Ecology, Biology, PopulationThanks to Insight Author: John Petersen

Edits by Andy Long

Everything that follows the dashes was created by John Petersen (or at least came from his Insight model). I just wanted to make a few comments.

We are looking at Hare and Lynx, of course. Clone this insight, and change the names.

Then read the text below, to get acquainted with one of the most important and well-known examples of a simple system of differential equations in all of mathematics.

http://www.nku.edu/~longa/classes/mat375/mathematica/Lotka-Volterra.nb------------------------------------------------------------

Dynamic simulation modelers are particularly interested in understanding and being able to distinguish between the behavior of stocks and flows that result from internal interactions and those that result from external forces acting on a system.

For some time modelers have been particularly interested in internal interactions that result in stable oscillations in the absence of any external forces acting on a system.

The model in this last scenario was independently developed by Alfred Lotka (1924) and Vito Volterra (1926). Lotka was interested in understanding internal dynamics that might explain oscillations in moth and butterfly populations and the parasitoids that attack them. Volterra was interested in explaining an increase in coastal populations of predatory fish and a decrease in their prey that was observed during World War I when human fishing pressures on the predator species declined.

Both discovered that a relatively simple model is capable of producing the cyclical behaviors they observed.

Since that time, several researchers have been able to reproduce the modeling dynamics in simple experimental systems consisting of only predators and prey. It is now generally recognized that the model world that Lotka and Volterra produced is too simple to explain the complexity of most predator-prey dynamics in nature. And yet, the model significantly advanced our understanding of the critical role of feedback in predator-prey interactions and in feeding relationships that result in community dynamics.

The Lotka–Volterra model makes a number of assumptions about the environment and evolution of the predator and prey populations:

1. The prey population finds ample food at all times.2. The food supply of the predator population depends entirely on the size of the prey population.3. The rate of change of population is proportional to its size.4. During the process, the environment does not change in favour of one species and genetic adaptation is inconsequential.5. Predators have limitless appetite.

As differential equations are used, the solution is deterministic and continuous. This, in turn, implies that the generations of both the predator and prey are continually overlapping.[23]

**Prey**

When multiplied out, the prey equation becomesdx/dt = αx - βxy The prey are assumed to have an unlimited food supply, and to reproduce exponentially unless subject to predation; this exponential growth is represented in the equation above by the term αx. The rate of predation upon the prey is assumed to be proportional to the rate at which the predators and the prey meet; this is represented above by βxy. If either x or y is zero then there can be no predation.

With these two terms the equation above can be interpreted as: the change in the prey's numbers is given by its own growth minus the rate at which it is preyed upon.

PredatorsThe predator equation becomes

dy/dt = -

In this equation, {\displaystyle \displaystyle \delta xy} represents the growth of the predator population. (Note the similarity to the predation rate; however, a different constant is used as the rate at which the predator population grows is not necessarily equal to the rate at which it consumes the prey). {\displaystyle \displaystyle \gamma y} represents the loss rate of the predators due to either natural death or emigration; it leads to an exponential decay in the absence of prey.

Hence the equation expresses the change in the predator population as growth fueled by the food supply, minus natural death.

Education Chaos Ecology Biology Population Mat375 Lotka Volterra

- 2 months 18 hours ago

#### Clone of Clone of Logistic Growth

##### Adheke Silas

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

- 2 months 5 days 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 month 3 weeks 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 month 3 weeks ago

#### Clone of Logistic Growth

##### Andrew E Long

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 month 2 weeks 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 month 3 weeks 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 month 3 weeks ago

#### Clone of Clone of Logistic Growth

##### Adheke Silas

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

- 2 months 5 days 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 month 2 weeks ago

#### Clone of Newton's Method

##### Jared Slavey

This insight implements Newton's method as an InsightMaker model.

It is important to use Euler's method, with step-size of 1. That's what allows us to get away with this!:)

Fun to try a couple of different cases, so I have built four choices into this example. You can choose the function ("Function Choice" of 0, 1, 2, or 3) using the slider.

Andy LongSpring, 2020

- 1 month 1 week 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

- 2 months 5 days ago

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

##### Sean Field

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 month 3 weeks ago

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

##### Sean Field

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 month 2 weeks ago

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

##### Proctor Mercer

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 month 3 weeks 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 month 3 weeks 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 month 3 weeks ago

#### Clone of Logistic Growth

##### Jacob Adkins

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

- 2 months 3 weeks ago

#### Clone of Newton's Method

##### Jake Moore

This insight implements Newton's method as an InsightMaker model.

It is important to use Euler's method, with step-size of 1. That's what allows us to get away with this!:)

Fun to try a couple of different cases, so I have built four choices into this example. You can choose the function ("Function Choice" of 0, 1, 2, or 3) using the slider.

Andy LongSpring, 2020

- 1 month 1 week 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 month 3 weeks ago

#### Clone of Clone of Logistic Growth

##### Jacob Adkins

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

- 2 months 5 days ago

#### Clone of 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 month 3 weeks ago

#### Clone of Logistic Growth

##### Thomas Browe

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

- 2 months 3 weeks ago

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

##### luke vanlaningham

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 month 3 weeks ago