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 va
Logistic Mat375 Non-dimensionalize Math Modeling
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=

Clone of Non-dimensionalized Logistic Growth
Kimmy
The basic model of Modelling the Canada lynx and snowshoe hare population cycle: The role of specialist predators ( Tyson, et al.) demonstrates logistic growth in prey, and in predator (with prey dependence for carrying capacity). But interestingly, one possibility is limit cycles, which mimic the
Mat375 Math Modeling
The basic model of Modelling the Canada lynx and snowshoe hare population cycle: The role of specialist predators (Tyson, et al.) demonstrates logistic growth in prey, and in predator (with prey dependence for carrying capacity). But interestingly, one possibility is limit cycles, which mimic the cycling of the populations in nature.

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

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

where K is the logistic carrying capacity of the prey (hare), in the absence of predation; the second term is a "generalist predation" term;  and the third term is the "specialist predation" (in the limit as the prey gets big, this becomes simply proportional to y (the lynx population)).

The differential equations for the population of lynx (y) is

y'(t) = sy(1- qy/x) = sy - sqy^2/x

for the predator (lynx), which is essentially logistic growth. Its growth term suggests exponential growth, but there is a loss term of the form sqy^2/x -- loss is proportional to population (crowding), and inversely proportional to prey density. As the hare population goes to zero, so shall the lynx....

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

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

So what we're really looking at here is the system

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

where w(t)=qy(t).

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

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

Author: Andy Long, Northern Kentucky University (2020)

Reference: Tyson, Rebecca, Sheena Haines,  Karen Hodges. Modelling the Canada lynx and snowshoe hare population cycle: The role of specialist predators. Theoretical Ecology. 3, 97–111 (2010). https://doi.org/10.1007/s12080-009-0057-1

Resource: A comparable Mathematica model can be found at  http://ceadserv1.nku.edu/longa//classes/mat375/days/Mathematica/BasicModel.nb,
which allows one to experiment a little more easily than one can with this InsightMaker model.
Clone of Basic Model, Tyson Lynx and Hare
Adheke Silas
This is an example I thought of after reading Olinick's book An Introduction to Mathematical Models in the Social and Life Sciences . ​ It's an SIR-type model, but one where the equilibrium (ws,wi,wr) is always the same, even as the weights in the transition matrix change. Actually it might be be
Mat375 Markov Chain
This is an example I thought of after reading Olinick's book An Introduction to Mathematical Models in the Social and Life Sciences. ​

It's an SIR-type model, but one where the equilibrium (ws,wi,wr) is always the same, even as the weights in the transition matrix change.

Actually it might be better to think of this as a poisoning model: the rate of infection is constant, and independent of the existence of an infected population. That's more like disease due to an environmental effect (e.g. lead-poisoning from smelters, or mercury poisoning from the burning of coal). So infected would mean "effected", and "recovered" might be "treated" -- and ultimately released, to be exposed again.

This shows that the equilibrium does not determine the transition probabilities: two different transition matrices can have the same ultimate equilibrium.

There is a constraint on the infection rate that I haven't figured out how to build in:

InfectionRate < Min[1,wi/ws, wr/ws]

I can allow InfectionRate to vary up to 1 if I take
ws < wi
and
ws < wr
However if you violate that, you'll get interesting solutions with negative values of populations. The dynamics are pretty interesting in that case, however! If you want to see them, you'll have to remove the constraints that I put on the parameters in the Recover and LossOfImmunity parameters.

Thanks Mike! Interesting examples, as always....
Andy Long

Clone of SIR (poisoning would be better) Markov Model
Christopher Milesky
This simple model demonstrates logistic growth.The differential equation looks like y'(t)=by(t)(1-y(t)/K) where K is the carrying capacity of the quantity y. Alternatively, y'(t)=by(t) - b/K*y(t)^2 so the growth term suggests exponential growth, but there is a loss term is of the form b/K y(t
Mat375 Math Modeling
This simple model demonstrates logistic growth.The differential equation looks like

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

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

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

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

A comparable Mathematica file is available at
http://www.nku.edu/~longa/classes/2018spring/mat375/mathematica/LogisticGrowth-and-DecayModel.nb
Clone of Logistic Growth
Jacob Adkins
This simple model demonstrates logistic growth.The differential equation looks like y'(t)=by(t)(1-y(t)/K) where K is the carrying capacity of the quantity y. Alternatively, y'(t)=by(t) - b/K*y(t)^2 so the growth term suggests exponential growth, but there is a loss term is of the form b/K y(t
Mat375 Math Modeling
This simple model demonstrates logistic growth.The differential equation looks like

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

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

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

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

A comparable Mathematica file is available at
http://www.nku.edu/~longa/classes/2018spring/mat375/mathematica/LogisticGrowth-and-DecayModel.nb
Clone of Logistic Growth
Thomas Browe
This is a first example of a simple SIR (Susceptible, Infected, Recovered) model. There are three pools of individuals: those who are infected (without them, no disease!), the pool of those who are at risk (susceptible), and the recovered -- who may lose their immunity and become susceptible again
Mat375 SIR Math Modeling
This is a first example of a simple SIR (Susceptible, Infected, Recovered) model.

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

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

Clone of A Simple SIR (Susceptible, Infected, Recovered) Example
Jake Moore
The basic model of Modelling the Canada lynx and snowshoe hare population cycle: The role of specialist predators ( Tyson, et al.) demonstrates logistic growth in prey, and in predator (with prey dependence for carrying capacity). But interestingly, one possibility is limit cycles, which mimic the
Math Modeling Mat375
The basic model of Modelling the Canada lynx and snowshoe hare population cycle: The role of specialist predators (Tyson, et al.) demonstrates logistic growth in prey, and in predator (with prey dependence for carrying capacity). But interestingly, one possibility is limit cycles, which mimic the cycling of the populations in nature.

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

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

where K is the logistic carrying capacity of the prey (hare), in the absence of predation; the second term is a "generalist predation" term;  and the third term is the "specialist predation" (in the limit as the prey gets big, this becomes simply proportional to y (the lynx population)).

The differential equations for the population of lynx (y) is

y'(t) = sy(1- qy/x) = sy - sqy^2/x

for the predator (lynx), which is essentially logistic growth. Its growth term suggests exponential growth, but there is a loss term of the form sqy^2/x -- loss is proportional to population (crowding), and inversely proportional to prey density. As the hare population goes to zero, so shall the lynx....

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

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

So what we're really looking at here is the system

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

where w(t)=qy(t).

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

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

Author: Andy Long, Northern Kentucky University (2020)

Reference: Tyson, Rebecca, Sheena Haines,  Karen Hodges. Modelling the Canada lynx and snowshoe hare population cycle: The role of specialist predators. Theoretical Ecology. 3, 97–111 (2010). https://doi.org/10.1007/s12080-009-0057-1

Resource: A comparable Mathematica model can be found at  http://ceadserv1.nku.edu/longa//classes/mat375/days/Mathematica/BasicModel.nb,
which allows one to experiment a little more easily than one can with this InsightMaker model.
Clone of Basic Model, Tyson Lynx and Hare
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
Math Modeling Mat375
The basic model of Modelling the Canada lynx and snowshoe hare population cycle: The role of specialist predators (Tyson, et al.) demonstrates logistic growth in prey, and in predator (with prey dependence for carrying capacity). But interestingly, one possibility is limit cycles, which mimic the cycling of the populations in nature.

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

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

where K is the logistic carrying capacity of the prey (hare), in the absence of predation; the second term is a "generalist predation" term;  and the third term is the "specialist predation" (in the limit as the prey gets big, this becomes simply proportional to y (the lynx population)).

The differential equations for the population of lynx (y) is

y'(t) = sy(1- qy/x) = sy - sqy^2/x

for the predator (lynx), which is essentially logistic growth. Its growth term suggests exponential growth, but there is a loss term of the form sqy^2/x -- loss is proportional to population (crowding), and inversely proportional to prey density. As the hare population goes to zero, so shall the lynx....

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

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

So what we're really looking at here is the system

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

where w(t)=qy(t).

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

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

Author: Andy Long, Northern Kentucky University (2020)

Reference: Tyson, Rebecca, Sheena Haines,  Karen Hodges. Modelling the Canada lynx and snowshoe hare population cycle: The role of specialist predators. Theoretical Ecology. 3, 97–111 (2010). https://doi.org/10.1007/s12080-009-0057-1

Resource: A comparable Mathematica model can be found at  http://ceadserv1.nku.edu/longa//classes/mat375/days/Mathematica/BasicModel.nb,
which allows one to experiment a little more easily than one can with this InsightMaker model.
Clone of Basic Model, Tyson Lynx and Hare
Thomas Browe
The basic model of Modelling the Canada lynx and snowshoe hare population cycle: The role of specialist predators ( Tyson, et al.) demonstrates logistic growth in prey, and in predator (with prey dependence for carrying capacity). But interestingly, one possibility is limit cycles, which mimic the
Mat375 Math Modeling
The basic model of Modelling the Canada lynx and snowshoe hare population cycle: The role of specialist predators (Tyson, et al.) demonstrates logistic growth in prey, and in predator (with prey dependence for carrying capacity). But interestingly, one possibility is limit cycles, which mimic the cycling of the populations in nature.

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

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

where K is the logistic carrying capacity of the prey (hare), in the absence of predation; the second term is a "generalist predation" term;  and the third term is the "specialist predation" (in the limit as the prey gets big, this becomes simply proportional to y (the lynx population)).

The differential equations for the population of lynx (y) is

y'(t) = sy(1- qy/x) = sy - sqy^2/x

for the predator (lynx), which is essentially logistic growth. Its growth term suggests exponential growth, but there is a loss term of the form sqy^2/x -- loss is proportional to population (crowding), and inversely proportional to prey density. As the hare population goes to zero, so shall the lynx....

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

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

So what we're really looking at here is the system

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

where w(t)=qy(t).

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

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

Author: Andy Long, Northern Kentucky University (2020)

Reference: Tyson, Rebecca, Sheena Haines,  Karen Hodges. Modelling the Canada lynx and snowshoe hare population cycle: The role of specialist predators. Theoretical Ecology. 3, 97–111 (2010). https://doi.org/10.1007/s12080-009-0057-1

Resource: A comparable Mathematica model can be found at  http://ceadserv1.nku.edu/longa//classes/mat375/days/Mathematica/BasicModel.nb,
which allows one to experiment a little more easily than one can with this InsightMaker model.
Clone of 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
Math Modeling Mat375
The basic model of Modelling the Canada lynx and snowshoe hare population cycle: The role of specialist predators (Tyson, et al.) demonstrates logistic growth in prey, and in predator (with prey dependence for carrying capacity). But interestingly, one possibility is limit cycles, which mimic the cycling of the populations in nature.

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

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

where K is the logistic carrying capacity of the prey (hare), in the absence of predation; the second term is a "generalist predation" term;  and the third term is the "specialist predation" (in the limit as the prey gets big, this becomes simply proportional to y (the lynx population)).

The differential equations for the population of lynx (y) is

y'(t) = sy(1- qy/x) = sy - sqy^2/x

for the predator (lynx), which is essentially logistic growth. Its growth term suggests exponential growth, but there is a loss term of the form sqy^2/x -- loss is proportional to population (crowding), and inversely proportional to prey density. As the hare population goes to zero, so shall the lynx....

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

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

So what we're really looking at here is the system

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

where w(t)=qy(t).

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

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

Author: Andy Long, Northern Kentucky University (2020)

Reference: Tyson, Rebecca, Sheena Haines,  Karen Hodges. Modelling the Canada lynx and snowshoe hare population cycle: The role of specialist predators. Theoretical Ecology. 3, 97–111 (2010). https://doi.org/10.1007/s12080-009-0057-1

Resource: A comparable Mathematica model can be found at  http://ceadserv1.nku.edu/longa//classes/mat375/days/Mathematica/BasicModel.nb,
which allows one to experiment a little more easily than one can with this InsightMaker model.
Clone of Basic Model, Tyson Lynx and Hare
Cameron Demler
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
Math Modeling Mat375
This simple model demonstrates logistic growth.The differential equation looks like

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

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

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

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

A comparable Mathematica file is available at
http://www.nku.edu/~longa/classes/2018spring/mat375/mathematica/LogisticGrowth-and-DecayModel.nb
Clone of Logistic Growth
Proctor Mercer
This model illustrates predator prey interactions using real-life data of wolf and moose populations on the Isle Royale. We incorporate logistic growth into the moose dynamics, and we replace the death flow of the moose with a kill rate modeled from the kill rate data found on the Isle Royale websi
Mat375 Midterm Environment Populations Ecology
This model illustrates predator prey interactions using real-life data of wolf and moose populations on the Isle Royale.

We incorporate logistic growth into the moose dynamics, and we replace the death flow of the moose with a kill rate modeled from the kill rate data found on the Isle Royale website.

Thanks to Jacob Englert for the model if-then-else structure.

I start with these parameters:
Wolf Death Rate = 0.15
Wolf Birth Rate = 0.0187963
Moose Birth Rate = 0.4
Carrying Capacity = 2000
Initial Moose: 563
Initial Wolves: 20

I used RK-4 with step-size 0.1, from 1959 for 60 years.

The moose birth flow is logistic, MBR*M*(1-M/K)
Moose death flow is Kill Rate (in Moose/Year)
Wolf birth flow is WBR*Kill Rate (in Wolves/Year)
Wolf death flow is WDR*W

MAT 375 Midterm file: Model of Isle Royale: Predator Prey Interactions
Andrew E Long
3
This simple model demonstrates exponential growth or decay in a population. A comparable Mathematica file is at http://www.nku.edu/~longa/classes/2018spring/mat375/mathematica/ExponentialGrowth-and-DecayModel.nb
Mat375 Math Modeling
This simple model demonstrates exponential growth or decay in a population.

A comparable Mathematica file is at
http://www.nku.edu/~longa/classes/2018spring/mat375/mathematica/ExponentialGrowth-and-DecayModel.nb
Clone of Exponential Growth
Travis Rhoades
The basic model of Modelling the Canada lynx and snowshoe hare population cycle: The role of specialist predators ( Tyson, et al.) demonstrates logistic growth in prey, and in predator (with prey dependence for carrying capacity). But interestingly, one possibility is limit cycles, which mimic the
Mat375 Math Modeling
The basic model of Modelling the Canada lynx and snowshoe hare population cycle: The role of specialist predators (Tyson, et al.) demonstrates logistic growth in prey, and in predator (with prey dependence for carrying capacity). But interestingly, one possibility is limit cycles, which mimic the cycling of the populations in nature.

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

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

where K is the logistic carrying capacity of the prey (hare), in the absence of predation; the second term is a "generalist predation" term;  and the third term is the "specialist predation" (in the limit as the prey gets big, this becomes simply proportional to y (the lynx population)).

The differential equations for the population of lynx (y) is

y'(t) = sy(1- qy/x) = sy - sqy^2/x

for the predator (lynx), which is essentially logistic growth. Its growth term suggests exponential growth, but there is a loss term of the form sqy^2/x -- loss is proportional to population (crowding), and inversely proportional to prey density. As the hare population goes to zero, so shall the lynx....

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

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

So what we're really looking at here is the system

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

where w(t)=qy(t).

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

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

Author: Andy Long, Northern Kentucky University (2020)

Reference: Tyson, Rebecca, Sheena Haines,  Karen Hodges. Modelling the Canada lynx and snowshoe hare population cycle: The role of specialist predators. Theoretical Ecology. 3, 97–111 (2010). https://doi.org/10.1007/s12080-009-0057-1

Resource: A comparable Mathematica model can be found at  http://ceadserv1.nku.edu/longa//classes/mat375/days/Mathematica/BasicModel.nb,
which allows one to experiment a little more easily than one can with this InsightMaker model.
Clone of Basic Model, Tyson Lynx and Hare
Thomas Browe
This simple model demonstrates logistic growth.The differential equation looks like y'(t)=by(t)(1-y(t)/K) where K is the carrying capacity of the quantity y. Alternatively, y'(t)=by(t) - b/K*y(t)^2 so the growth term suggests exponential growth, but there is a loss term is of the form b/K y(t
Math Modeling Mat375
This simple model demonstrates logistic growth.The differential equation looks like

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

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

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

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

A comparable Mathematica file is available at
http://www.nku.edu/~longa/classes/2018spring/mat375/mathematica/LogisticGrowth-and-DecayModel.nb
Clone of Clone of Logistic Growth
Adheke Silas
This is a first example of a simple SIR (Susceptible, Infected, Recovered) model. There are three pools of individuals: those who are infected (without them, no disease!), the pool of those who are at risk (susceptible), and the recovered -- who may lose their immunity and become susceptible again
SIR Math Modeling Mat375
This is a first example of a simple SIR (Susceptible, Infected, Recovered) model.

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

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

Clone of A Simple SIR (Susceptible, Infected, Recovered) Example
Jon Ford
This simple model demonstrates logistic growth.The differential equation looks like y'(t)=by(t)(1-y(t)/K) where K is the carrying capacity of the quantity y. Alternatively, y'(t)=by(t) - b/K*y(t)^2 so the growth term suggests exponential growth, but there is a loss term is of the form b/K y(t
Mat375 Math Modeling
This simple model demonstrates logistic growth.The differential equation looks like

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

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

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

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

A comparable Mathematica file is available at
http://www.nku.edu/~longa/classes/2018spring/mat375/mathematica/LogisticGrowth-and-DecayModel.nb
Clone of Clone of Logistic Growth
Adheke Silas
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 va
Math Modeling Non-dimensionalize Logistic Mat375
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=

Clone of Non-dimensionalized Logistic Growth
luke vanlaningham
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 va
Non-dimensionalize Logistic Math Modeling Mat375
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=

Clone of Non-dimensionalized Logistic Growth
Proctor Mercer
This is an example I thought of after reading Olinick's book An Introduction to Mathematical Models in the Social and Life Sciences . ​ It's an SIR-type model, but one where the equilibrium (ws,wi,wr) is always the same, even as the weights in the transition matrix change. Actually it might be be
Mat375 Markov Chain
This is an example I thought of after reading Olinick's book An Introduction to Mathematical Models in the Social and Life Sciences. ​

It's an SIR-type model, but one where the equilibrium (ws,wi,wr) is always the same, even as the weights in the transition matrix change.

Actually it might be better to think of this as a poisoning model: the rate of infection is constant, and independent of the existence of an infected population. That's more like disease due to an environmental effect (e.g. lead-poisoning from smelters, or mercury poisoning from the burning of coal). So infected would mean "effected", and "recovered" might be "treated" -- and ultimately released, to be exposed again.

This shows that the equilibrium does not determine the transition probabilities: two different transition matrices can have the same ultimate equilibrium.

There is a constraint on the infection rate that I haven't figured out how to build in:

InfectionRate < Min[1,wi/ws, wr/ws]

I can allow InfectionRate to vary up to 1 if I take
ws < wi
and
ws < wr
However if you violate that, you'll get interesting solutions with negative values of populations. The dynamics are pretty interesting in that case, however! If you want to see them, you'll have to remove the constraints that I put on the parameters in the Recover and LossOfImmunity parameters.

Thanks Mike! Interesting examples, as always....
Andy Long

Driehaus SIR (poisoning would be better) Markov Model
Rachel Driehaus
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
Math Modeling Mat375
This simple model demonstrates logistic growth.The differential equation looks like

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

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

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

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

A comparable Mathematica file is available at
http://www.nku.edu/~longa/classes/2018spring/mat375/mathematica/LogisticGrowth-and-DecayModel.nb
Clone of Logistic Growth
Samuel Kaelin
This simple model demonstrates logistic growth.The differential equation looks like y'(t)=by(t)(1-y(t)/K) where K is the carrying capacity of the quantity y. Alternatively, y'(t)=by(t) - b/K*y(t)^2 so the growth term suggests exponential growth, but there is a loss term is of the form b/K y(t
Mat375 Math Modeling
This simple model demonstrates logistic growth.The differential equation looks like

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

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

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

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

A comparable Mathematica file is available at
http://www.nku.edu/~longa/classes/2018spring/mat375/mathematica/LogisticGrowth-and-DecayModel.nb
Clone of Logistic Growth
Adheke Silas
The basic model of Modelling the Canada lynx and snowshoe hare population cycle: The role of specialist predators ( Tyson, et al.) demonstrates logistic growth in prey, and in predator (with prey dependence for carrying capacity). But interestingly, one possibility is limit cycles, which mimic the
Mat375 Math Modeling
The basic model of Modelling the Canada lynx and snowshoe hare population cycle: The role of specialist predators (Tyson, et al.) demonstrates logistic growth in prey, and in predator (with prey dependence for carrying capacity). But interestingly, one possibility is limit cycles, which mimic the cycling of the populations in nature.

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

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

where K is the logistic carrying capacity of the prey (hare), in the absence of predation; the second term is a "generalist predation" term;  and the third term is the "specialist predation" (in the limit as the prey gets big, this becomes simply proportional to y (the lynx population)).

The differential equations for the population of lynx (y) is

y'(t) = sy(1- qy/x) = sy - sqy^2/x

for the predator (lynx), which is essentially logistic growth. Its growth term suggests exponential growth, but there is a loss term of the form sqy^2/x -- loss is proportional to population (crowding), and inversely proportional to prey density. As the hare population goes to zero, so shall the lynx....

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

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

So what we're really looking at here is the system

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

where w(t)=qy(t).

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

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

Author: Andy Long, Northern Kentucky University (2020)

Reference: Tyson, Rebecca, Sheena Haines,  Karen Hodges. Modelling the Canada lynx and snowshoe hare population cycle: The role of specialist predators. Theoretical Ecology. 3, 97–111 (2010). https://doi.org/10.1007/s12080-009-0057-1

Resource: A comparable Mathematica model can be found at  http://ceadserv1.nku.edu/longa//classes/mat375/days/Mathematica/BasicModel.nb,
which allows one to experiment a little more easily than one can with this InsightMaker model.
Clone of Clone of Basic Model, Tyson Lynx and Hare
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
SIR Mat375 Math Modeling
This is a first example of a simple SIR (Susceptible, Infected, Recovered) model.

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

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

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