KBS Course, lecture 02

Density-dependence

Outline
-Single species, density-dependent model (Logistic)
	Derive, compare graphs, examine in Mathematica
-Alternative continuous-time models (Smith; Theta-logistic)
-Non-dimensionalized models
-Discrete-time models (Discrete Logistic; Ricker; Beverton-Holt)
-Stability; cob-web plots; overcompensatory
-[Problem set: negative density-dependence]

Density Dependent Growth - Logistic

Derive model

(dN/dt)=b N-m N

(dN/dt)=(b-m) N

Make functions of density
☺ What should be the slope of b’ vs N? What about m’ vs N?

b'=b-a N

(dN/dt)=(b-a N) N-m N

(dN/dt)=b N-m N -a (N^2)

(dN/dt)=(b-m )N-a (N^2)

(dN/dt)=(b-m)(N-(a/(b-m)) (N^2))

(dN/dt)=(b-m)(1-(a/(b-m)) N)N

(dN/dt)=(b-m)(1- (N/((b-m/a)))) N

r=b-m

k=((b-m)/a)

(dN/dt)=r(1- (N/k)) N

We call these the growth rate (really instantaneous growth rate at small densities) and carrying capacity
What are the units?

(density/time)=(1/time)(1-(density/density))density

So t is (1/time) and k is density

First, think about it as per capita growth.

Plot[2.2` (1-(n/10)),{n,0,10},Frame->True,FrameLabel->{"n","(dn/dt)(1/n)"}]

Label k and r on graph.

Walk through what happens on above graph given a starting values of 2.

Find Equilibrium

(dN/dt)=0
r(1- ((SuperStar[N])/k)) (SuperStar[N])=0
 (SuperStar[N])=0 (* one *)
(1- ((SuperStar[N])/k)) =0
 (SuperStar[N])=k (* two *)

Plot Total Pop Growth versus N
☺Draw axes, what does this plot look like?

Plot[2.2` (1-(n/10)) n,{n,0,11},Frame->True,FrameLabel->{"n","dn/dt"}]

Analyze Stability

Easier to do if we approximate the change (i.e. the dn/dt equation) as linear around each equilibrium. Positive slopes mean deviations will increase, while a negative slope means deviations will decrease. Note: work through the intuition on this once you have slopes.

Take the derivative AND evaluate it at the equilibrium. This gives the equivalent linear line at that point.

(dN/dt)=r(1- (N/k)) N
f(N)=r(1- (N/k)) N=r N- (r/k)(N^2)
(df/dN)=r-2(r/k)N
(Subscript[((df/dN)), N=(SuperStar[N])])
(* at (SuperStar[N])=0 *) 
r-2(r/k)0=r
(* unstable if r > 0 *)

(* at (SuperStar[N])=k *)

r-2(r/k)k=r-2r=-r
(* stable if r > 0 *)

Typical model includes the expectation that all parameters are defined as positive. In that case, 0 is always unstable and k is always stable.

Note, in more complex models, we’ll call these values eigenvalues because they describe the amount of change and qualitative direction of that change, i.e. increase or decrease, in a linear approximation.

Show[Plot[2.2` (1-(n/10)) n,{n,0,11},Frame->True,FrameLabel->{"n","dn/dt"}],Graphics[{EdgeForm[Thick],White,Disk[{0,0},0.3]}],Graphics[{EdgeForm[Thick],Black,Disk[{10,0},0.3]}]]

Find Solution

(dN/dt)=r(1- (N/k)) N=r N(1 -(N/k))
(dN/(N(1 -(N/k))))=r dt

The next steps are only for the math geeks, because Mathematica will do all of this for you. [Skip ahead, could have as bonus question in Problem Set for math geeks].

Partial fractions (algebra trick)

(1/(N(1 -(N/k))))=(A/N)+(B/(1 -(N/k)))
1=A(1 -(N/k))+B N
1=A +(B-(A/k))N

Because there is no N on left-hand side

(B-(A/k))=0
A=1

Which means

(B-(1/k))=0
B=(1/k)

Putting it A and B back in:

(1/N)+((1/k)/((1-(N/k))))

Which means the master equation is

(1/N)dN+((1/k)/((1-(N/k))))dN=r dt

Math magic

∫(1/N)\[DifferentialD]N=ln(N)+c

Use the substitution rule (which includes the chain rule )

u=1-(N/k)
du=-(1/k)
-du=(1/k)

∫((1/k)/((1-(N/k))))\[DifferentialD]N=∫-(1/u)\[DifferentialD]u=-ln(u)+c=-ln(1-(N/k))+c

Applying integration to each term gives

(1/N)dN+((1/k)/((1-(N/k))))dN=r dt
ln(N)-ln(1-(N/k))=rt+c

And do some algebra

ln((N/(1-(N/k))))=rt+c
(N/(1-(N/k)))=(e^(rt+c))=(e^rt)c

When t=0

((Subscript[N, 0])/(1-((Subscript[N, 0])/k)))=c=((k (Subscript[N, 0]))/(k-(Subscript[N, 0])))

Solve for N

(N/(1-(N/k)))=c (e^rt)
N=c (e^rt)(1-(N/k))=c (e^rt)-(N/k)c (e^rt)
N+(N/k)c (e^rt)=c (e^rt)
N(1+((c (e^rt))/k))=c (e^rt)
N=((c (e^rt))/(1+((c (e^rt))/k)))=((((k (Subscript[N, 0]))/(k-(Subscript[N, 0]))) (e^rt))/(1+((k (Subscript[N, 0]))/(k-(Subscript[N, 0])))((e^rt)/k)))=((((k (Subscript[N, 0]))/(k-(Subscript[N, 0]))) (e^rt))/(1+((Subscript[N, 0])/(k-(Subscript[N, 0])))(e^rt)))=((k (Subscript[N, 0]) (e^rt))/(k-(Subscript[N, 0])+(Subscript[N, 0])(e^rt)))=((k (Subscript[N, 0]) (e^rt))/(k+(Subscript[N, 0])((e^rt)-1)))

[Pick up here]
Mathematica does this in one step

n[t]/.({{DSolve[{n'[t]==r (1-((n[t])/k))n[t],n[0]==n0},n,t]}, {\[Placeholder]}})

Plot out Population versus Time

Plot[(((E^(r t)) k n0)/(k-n0+(E^(r t)) n0))/.{r->2,k->10,n0->0.2},{t,0,5},PlotRange->{0,All}]

Plot[(((E^(r t)) k n0)/(k-n0+(E^(r t)) n0))/.{r->2,k->10,n0->15},{t,0,5},PlotRange->{0,All}]

Nondimensionalize

(dN/dt)=r(1- (N/k)) N

Create dimensionless form:
	1) Create new parameters which are combinations of existing parameters
	2) Retains the “form” of the dynamics, which can now be analyzed by exploring fewer parameters
	3)Furthermore, those parameter relationships themselves can sometimes be informative about dynamics
-Consider units when proposing combinations
-Typically reduce by at least a parameter for each variable, i.e. make N and t dimensionless
-Often more than one possible dimensionless form, some with more utility

Systematic approach

Separate each variable into scalar (SuperStar[N]) and unit (OverHat[N]). Don’t do this for parameters, just variables.

N=(SuperStar[N]) (OverHat[N])
t=(SuperStar[t]) (OverHat[t])

(dN/dt)=r(1- (N/k)) N
((d(SuperStar[N]) (OverHat[N]))/((SuperStar[dt]) (OverHat[t])))=r(1- (((SuperStar[N]) (OverHat[N]))/k)) (SuperStar[N]) (OverHat[N])

Rearrange

((d(SuperStar[N]) (OverHat[N]))/((SuperStar[dt]) ))=r(1- (((SuperStar[N]) (OverHat[N]))/k)) (SuperStar[N]) (OverHat[N]) (OverHat[t])
((d(SuperStar[N]))/((SuperStar[dt]) ))=r(1- (((SuperStar[N]) (OverHat[N]))/k)) (SuperStar[N]) (OverHat[t])
((d(SuperStar[N]))/((SuperStar[dt]) ))=r(OverHat[t])(1- ((OverHat[N])/k)(SuperStar[N])) (SuperStar[N])

Proposed units for variables. Note, in more complex models there can be lots of options

(OverHat[N])=k
(OverHat[t])=(1/r)

units
density = density
time = (1/(1/time))= time

In this case, all parameters go away, but in more complex models combined dimensionless parameters are retained

And these can be communicated as

N=(SuperStar[N]) (OverHat[N])=(SuperStar[N]) k
t=(SuperStar[t]) (OverHat[t])=(SuperStar[t]) (1/r)

(SuperStar[N]) =(N/k)
(SuperStar[t]) =t r

For Z, the units are (density/density), so that Z is unitless.
For s, the units are (1/time) x time, so that t is unitless.

((d(SuperStar[N]))/((SuperStar[dt]) ))=(1- (SuperStar[N])) (SuperStar[N]) 
(dN/(dt ))=(1- N) N

Stars are dropped for simplicity of presentation

Review by working through summary graphs in PowerPoint

Smith

Smith 1963

(dN/dt)=r N ((1-(N/k))/(1+c (N/k)))

☺Ask students to walk through together finding equilibrium, analyze stability, plotting, etc. Can use Mathematica as wanted.

Per capita growth

Plot[{r ((1-(n/k))/(1+c (n/k)))}/.{r->2,c->4,k->10},{n,0,12}]

Equilibrium

0=r N ((1-(N/k))/(1+c (N/k)))
(SuperStar[N])=0
(SuperStar[N])=k

Stability

d(r N ((1-(N/k))/(1+c (N/k))))/dN

r N ((1-(N/k))/(1+c (N/k)))

D[r n ((1-(n/k))/(1+c (n/k))),n]

Can do the derivative by hand using product and division, but easy to get Mathematica output

-((c n (1-(n/k)) r)/(k (((1+((c n)/k)))^2)))-((n r)/(k (1+((c n)/k))))+(((1-(n/k)) r)/(1+((c n)/k)))/.n->0

Therefore “trivial” equilibrium is always unstable.

-((c n (1-(n/k)) r)/(k (((1+((c n)/k)))^2)))-((n r)/(k (1+((c n)/k))))+(((1-(n/k)) r)/(1+((c n)/k)))/.n->k

Therefore k equilibrium is always stable.

Time solution

Cannot be solve analytically, only numerically

Plot[Evaluate[n[t]/.NDSolve[{n'[t]==r n[t] ((1-((n[t])/k))/(1+c ((n[t])/k))),n[0]==n0}/.{r->2,c->1,k->10,n0->0.2},n,{t,0,20}]],{t,0,10}]

Nondimensionalize

((dN (OverHat[N]))/(dt (OverHat[t])))=r N (OverHat[N])((1-((N(OverHat[N]))/k))/(1+c ((N(OverHat[N]))/k)))

((dN )/dt)=r N ((1-((N(OverHat[N]))/k))/(1+c ((N(OverHat[N]))/k)))(OverHat[t])

((dN )/dt)=r (OverHat[t]) N ((1-((OverHat[N])/k)N)/(1+c ((OverHat[N])/k)N))
(OverHat[N])=k
(OverHat[t])=(1/r)
((dN )/dt)=N ((1-N)/(1+c N))

Theta logistic, aka Richards

Gilpin and Ayala 1973

(dN/dt)=r N(((1- (N/k)))^θ)

Plot[{r (((1- (n/k)))^θ)/.{r->2.2,k->10,θ->3},r (((1- (n/k)))^θ)/.{r->2.2,k->10,θ->1},r (((1- (n/k)))^θ)/.{r->2.2,k->10,θ->0.3}},{n,0,11},Frame->True,FrameLabel->{"n","(dn/dt)(1/n)"}]

Early-acting and late-acting density-dependence

Discrete Time

Logistic

(Subscript[N, t+1])=r (Subscript[N, t])(1-((Subscript[N, t])/k))

map={
r n(1-(n/k))
};
vars={n};
start={0.1};
paramset={r->2,k->2};
tmax=10;

next[x_]:=map/.Thread[vars->x]/.paramset

next[3]

addT=Module[{i},Table[vars[[i]]->Map[vars[[i]],{t}][[1]],{i,Length[vars]}]]

ListLinePlot[NestList[next,0.2,3]]

Ricker

Originally derived as spawning fish, cannibalism of young
Can be derived from scramble competition with neighbors.

(Subscript[N, t+1])=(Subscript[N, t])(e^(r(1-((Subscript[N, t])/k))))

Beverton-Holt

Can be derived as consequence of context competition with neighbors (and some clustering). 
Brannstrom and Sumpter 2005. http://rspb.royalsocietypublishing.org/content/royprsb/272/1576/2065.full.pdf

(Subscript[N, t+1])=(((Subscript[R, 0]) k (Subscript[N, t]))/(k+((Subscript[R, 0])-1)(Subscript[N, t])))

Nonmechanistic

Single-species models can be criticized for being non-mechanistic.
What is “carrying capacity”. Can only be measured by looking at the result.

However, often a useful building block of more complex food webs.

Many of the models can be derived from consumer-resource models

(dC/dt)=f(C,R)
(dR/dt)=g(C,R)

Biotic resource and type I functional response (which we’ll talk about later) leads to Logistic Model if the resources approach equilibrium quickly, i.e. quasi-equilibrium. (Reynolds and Brassil 2013)

Abiotic resource and type I functional response leads to Smith model is quasi-equilibrium

Problem Set

1) Biology

In the classical logistic equation, per capita growth rates always decline with increasing density of individuals.  In some species, per capita growth rates increase when the density of individuals increase from very low densities (in addition to the decrease at high densities illustrated above).  In terms of biology, what kinds of conditions or scenarios could result in this scenario? 

Draw 1 or 2 graphs (per capita growth rate versus N) illustrating this concept. (Be sure to include a decrease at high densities as well.)

2) Graphs

Draw 1 or 2 graphs of per capita growth rate versus N illustrating this concept. (Be sure to include a decrease at high densities as well.)

3) Model

Create a model that has these properties

4) Time solution

Draw a single graph showing Population Size versus Time.  Draw a line showing what would happen on this graph for the classical logistic model.  Now draw a line (from the same starting density at time 0) illustrating what happens with your new model.