2 Dynamics of single population Flashcards
LOGISTIC GROWTH model
Single population cannot exceed a maximum value, or “carrying capacity”, K
dN(t)/dt =
rN(t)* (1 - (N(t)/K))
r = b-d assumed r>0
[r] = T^-1
solve using sep of vars and partial fractions
N(t) =
[KN(0)exp(rt)]/ [K-N(0)+N(0)exp(rt)]
solution shape
N(t) → K as t → ∞
increases then flattens
for small initial contion 0.01K say more s like
Stability LOGISTIC GROWTH model
f(N)
dN/dt=0 gives
N=0 unstable
N=k stable
f’(0) >0
f’(K)<0
plotting f(N): upside down u shape
crosses at SS
max at k/2 from f’(N) and fastest growth here when N=k/2
If (N(0) chosen between 0 and K then f(N(0))>0 and N(t) moves right towards K
for a small initial population (N(0) ≈ 1), the rate of growth is slow at the beginning, increasing until it reaches a peak, and then decreasing again as N(t) approaches K .
dn/dt>0 pop grows until reaches K
if plotting N to f(N) means f(N) always positive (still upside down u)
usefulness
dN(t)/dt
= f(N(t))
plotting f(N)
Rate of change of N dep on N itself not on time for single pops sos et of possible trajectories can be classified
-N(t) always increases
-or always decreases
-or remains constant as a function of time
Shapes dep on function N and initial N(0)
represents pop so f(0) ≥ 0.
possible shapes of TRAJECTORIES
e.g exp decrease
exp increase, exp increase to carrying capacity (r shape)
cant have negative values
if f(N) > 0, N increases (moves to the right);
if f(N) < 0, N decreases (moves to the left).
impossible trajectories:
mountain shape above axis completely
decrease like right side of n
increase like right side of u(very similar to allowed one though for exp increase)
Stable vs unstable
If f’(N∗) > 0, SS at N∗
is unstable.
(if after a small perturbation, the system moves away )
If f’(N∗) < 0, SS at N∗
is stable.
UNSTABLE: If N(0) is slightly larger than N∗
then f(N(0)) > 0. The
population increases, taking it further away from N∗. If N(0) is slightly smaller than N∗ then f(N(0)) < 0. The population decreases, taking it further away from N∗ .
How does taylor series help to identify stability?
proven analytically by expanding f(N) in a Taylor series about N∗
:
f(n)=
f(N) + f’(N)(N-N) + 0.5f’‘(N)(N-N)^2 + ….
f(N) = 0
Define U(t) = N(t) -N*
dU/dt =dN/dt =f(N)
Du/dt = f’(N)U +0.5f’‘(N)U^2+….
for near SS we have small values of U
U(t) ∝ exp(f’(N∗)t).
That is, the difference between the population size and its steady-state value increases or decreases (depending on the sign of f’(N∗)), exponentially. The exponent is the value of the derivative of f evaluated at N∗
.
logistic growth model for large carrying capacity
dN/dt approx rN (exponential model)
Harvesting model
What does f(N) look like?
SS?
single population obeying logistic model to be harvested
dN(t)/dt =
rN(t)(1-[N(t)/K]) - αN(t)
= rN(t)(1-(α/r) -(N(t)/K
[α] =[r] =T^-1
harvesting, term is −αN
number of fish caught at any time is proportional to the number that are there
Two cases:
α < r : upside down U shape SS N=0 (unstable)and N = [(r- α)r]K (stable)
size of pop in long term reaches SS
α > r :
decreases from N=0 (stable) like rhs of n
f(N)<0 pop always decreases, overfishing causes extinction
Harvesting model
when α=0
K=N*
logistic model
Harvesting model steady state rate of harvesting
N=0 and N = [(r- α)r]*K
ss rate of harvesting
αN* = α[(r- α)r]K
Harvesting model: optimal value of α (if we want to
sustainably catch as many fish as possible?)
Define ss of harvesting function
h(α)
α[(r- α)r]K
plot against r
upside down u
h’(α) =K - (2K/r)*α
α_max=r/2
The way to sustainably harvest the maximum from the population (to catch the most fish you can without driving the population to extinction) is to choose α to be half of r. In that case, the steady state size of the fish population
will be half as big as if there were no fishing at all.
The Allee Effect (L5)
the model
dimensions
individuals cooperate to mutual benefit
dN/dt=
rN*[
(1-(N/K) - (η/(1+γN))]
term: - (η/(1+γN))
becomes less negative (increases) as N increases. Penalises small N term tends to - η for small N
[γ] = 1/[N] and η
is dimensionless.
The Allee Effect (L5)
f(N) and analysis
If η > 1 then f’(0) < 0.
The Allee Effect (L5)
scaled version
x(t) = γN(t)
we can’t plot f(n) easily
so insead scale using change of vars
dx/dt = rx*
[1-(x/M)-(η/(1 + x)) ]
steady states x*=0 and
x²+(1-M)x +M(1+η) =0
where M= γk
To analise…
The Allee Effect (L5)
scaled version
x(t) = γN(t) steady states x*=0 and
x²+(1-M)x +M(1+η) =0
where M= γk
To analise…
Define
η = (1/M)(M-x)(x+1) = g(x) assuming M>1 we plot
η against x giving n shape with max at x = (m-1)/2
Plot x against η
gives D shape. With ss dep on whether η intersects at 2 points, 1 point or 0
η_max = (M+1)²/ 4m
dep onM= γk
+2 ss if η >1
+1 ss if η <1
(if η =1 only N*=0 ss)
e.g plot f(N) against N for η =0.75 gives n shape 2 ss, η =2 gives 3 ss under over
