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
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
This model illustrates predator prey interactions using real-life data of wolf and moose populations on the Isle Royale. It was "cloned" from a model that InsightMaker provides to its users, at  https://insightmaker.com/insight/2068/Isle-Royale-Predator-Prey-Interactions  Thanks Scott Fortmann-Roe.
This model illustrates predator prey interactions using real-life data of wolf and moose populations on the Isle Royale. It was "cloned" from a model that InsightMaker provides to its users, at
https://insightmaker.com/insight/2068/Isle-Royale-Predator-Prey-Interactions
Thanks Scott Fortmann-Roe.

I've created a Mathematica file that replicates the model, at
http://www.nku.edu/~longa/classes/2018spring/mat375/mathematica/Moose-n-Wolf-InsightMaker.nb

It allows one to experiment with adjusting the initial number of moose and wolves on the island.

I used steepest descent in Mathematica to optimize the parameters, with my objective data being the ratio of wolves to moose. You can try my (admittedly) kludgy code, at
http://www.nku.edu/~longa/classes/2018spring/mat375/mathematica/Moose-n-Wolf-InsightMaker-BestFit.nb

{WolfBirthRateFactorStart,
WolfDeathRateStart,
MooseBirthRateStart,
MooseDeathRateFactorStart,
moStart,
woStart} =
{0.000267409,
0.239821,
0.269755,
0.0113679,
591,
23.};

 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
Spring, 2020: in the midst of on-line courses, due to the pandemic of Covid-19.

With the onset of the Covid-19 coronavirus crisis, we focus on SIRD models, which might realistically model the course of the disease.

We start with an SIR model, such as that featured in the MAA model featured in

Without mortality, with time measured in days, with infection rate 1/2, recovery rate 1/3, and initial infectious population I_0=1.27x10-4, we reproduce their figure

With a death rate of .005 (one two-hundredth of the infected per day), an infectivity rate of 0.5, and a recovery rate of .145 or so (takes about a week to recover), we get some pretty significant losses -- about 3.2% of the total population.

Resources:
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
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

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
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
 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
Spring, 2020: in the midst of on-line courses, due to the pandemic of Covid-19.

With the onset of the Covid-19 coronavirus crisis, we focus on SIRD models, which might realistically model the course of the disease.

We start with an SIR model, such as that featured in the MAA model featured in

Without mortality, with time measured in days, with infection rate 1/2, recovery rate 1/3, and initial infectious population I_0=1.27x10-4, we reproduce their figure

With a death rate of .005 (one two-hundredth of the infected per day), an infectivity rate of 0.5, and a recovery rate of .145 or so (takes about a week to recover), we get some pretty significant losses -- about 3.2% of the total population.

Resources:
This is an example from Cushing's book  An Introduction to Structured Population Dynamics . ​  The parameters initially included reproduce the bifurcation results on p. 39 of Cushing's manuscript.  The tuning parameter is b, the birthrate.   p. 37: The LPA flour beetle model.  The bifurcation diagra
This is an example from Cushing's book An Introduction to Structured Population Dynamics. ​

The parameters initially included reproduce the bifurcation results on p. 39 of Cushing's manuscript.
The tuning parameter is b, the birthrate.

p. 37: The LPA flour beetle model.

The bifurcation diagram for parameter b is on page 39;
The bifurcation diagram for mu adult is on p. 59;
The bifurcation diagram for C pa is on p. 60.

Andy Long

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
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
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
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
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
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
 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
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.
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,
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

 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
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.
 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
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.
This model illustrates predator prey interactions using real-life data of wolf and moose populations on the Isle Royale. It was "cloned" from a model that InsightMaker provides to its users, at  https://insightmaker.com/insight/2068/Isle-Royale-Predator-Prey-Interactions  Thanks Scott Fortmann-Roe.
This model illustrates predator prey interactions using real-life data of wolf and moose populations on the Isle Royale. It was "cloned" from a model that InsightMaker provides to its users, at
https://insightmaker.com/insight/2068/Isle-Royale-Predator-Prey-Interactions
Thanks Scott Fortmann-Roe.

I've created a Mathematica file that replicates the model, at
http://www.nku.edu/~longa/classes/2018spring/mat375/mathematica/Moose-n-Wolf-InsightMaker.nb

It allows one to experiment with adjusting the initial number of moose and wolves on the island.

I used steepest descent in Mathematica to optimize the parameters, with my objective data being the ratio of wolves to moose. You can try my (admittedly) kludgy code, at
http://www.nku.edu/~longa/classes/2018spring/mat375/mathematica/Moose-n-Wolf-InsightMaker-BestFit.nb

{WolfBirthRateFactorStart,
WolfDeathRateStart,
MooseBirthRateStart,
MooseDeathRateFactorStart,
moStart,
woStart} =
{0.000267409,
0.239821,
0.269755,
0.0113679,
591,
23.};

This is an example from Cushing's book  An Introduction to Structured Population Dynamics . ​  The parameters initially included reproduce the bifurcation results on p. 39 of Cushing's manuscript.  The tuning parameter is b, the birthrate.   p. 37: The LPA flour beetle model.  The bifurcation diagra
This is an example from Cushing's book An Introduction to Structured Population Dynamics. ​

The parameters initially included reproduce the bifurcation results on p. 39 of Cushing's manuscript.
The tuning parameter is b, the birthrate.

p. 37: The LPA flour beetle model.

The bifurcation diagram for parameter b is on page 39;
The bifurcation diagram for mu adult is on p. 59;
The bifurcation diagram for C pa is on p. 60.

Andy Long

This is a first example of a simple SIR (Susceptible, Infected, Recovered) model.   There are three pools of individuals: those who are infected (without them, no disease!), the pool of those who are at risk (susceptible), and the recovered -- who may lose their immunity and become susceptible again
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

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
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

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
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
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
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
 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
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.
This is an introductory example from Olinick's book  An Introduction to Mathematical Models in the Social and Life Sciences . ​  Next up: and SIR, and his interesting model of female birth weights.
This is an introductory example from Olinick's book An Introduction to Mathematical Models in the Social and Life Sciences. ​

Next up: and SIR, and his interesting model of female birth weights.
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
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
This model illustrates predator prey interactions using real-life data of wolf and moose populations on the Isle Royale. It was "cloned" from a model that InsightMaker provides to its users, at  https://insightmaker.com/insight/2068/Isle-Royale-Predator-Prey-Interactions  Thanks Scott Fortmann-Roe.
This model illustrates predator prey interactions using real-life data of wolf and moose populations on the Isle Royale. It was "cloned" from a model that InsightMaker provides to its users, at
https://insightmaker.com/insight/2068/Isle-Royale-Predator-Prey-Interactions
Thanks Scott Fortmann-Roe.

I've created a Mathematica file that replicates the model, at
http://www.nku.edu/~longa/classes/2018spring/mat375/mathematica/Moose-n-Wolf-InsightMaker.nb

It allows one to experiment with adjusting the initial number of moose and wolves on the island.

I used steepest descent in Mathematica to optimize the parameters, with my objective data being the ratio of wolves to moose. You can try my (admittedly) kludgy code, at
http://www.nku.edu/~longa/classes/2018spring/mat375/mathematica/Moose-n-Wolf-InsightMaker-BestFit.nb

{WolfBirthRateFactorStart,
WolfDeathRateStart,
MooseBirthRateStart,
MooseDeathRateFactorStart,
moStart,
woStart} =
{0.000267409,
0.239821,
0.269755,
0.0113679,
591,
23.};