24 Age-structured populations (MATH5566M extra topic) Flashcards
consider the pop growth equation, what are the limitations?
e.g.: the Malthusian population growth equation:
N˙(t) = βN(t) − µN(t)
(constant birth and death rate)
But:
Mortality is often NOT constant
Individuals go through juvenile / fertile / sterile periods- we all age
Ageing is not a random walk!
consider structure of a population
age structured
What is a ‘structure’? An individual property that uniquely characterises the vital properties of an individual.
E.g. age, size of individuals, maturation of a cell. . .
If we assume a population is age-structured, we prescribe: µ(a)
(mortality), β(a) (fertility), . . . all dependent on age
drawback of this means that at every time t we need to keep track of how the pop is distrubted over the ages e.g bell curve
If we want to know #individuals we can integrate over the denisty
density over the continuous ages
p(a, t) = density of individuals of age a ∈ [0, ∞] at time t
density over the continuous ages
finding the number number of individuals with age in the interval
p(a, t)Δa = number of individuals with age in the interval
[a, a + Δa] at time t
density over the continuous ages
#individuals with age in the interval
[a1, a2] at time t
integral_[a_1,a_2] of
p(a, t)da = number of individuals with age in the interval
[a1, a2] at time t
considering ages intro to distribution
time against age of pop.
At time t we know # individuals with age a
consider a small interval Δt
these individuals will have age a +Δt
How can I find probabilities?
p(t+Δt,a) for age structured populations
p(t+Δt,a)
= p(t,a-Δt) -μp(t,a)Δt
(ageing)
(#individuals who died in this time)
[p(t+Δt,a)-p(t,a)+ p(t,a)- p(t,a-Δt)]/Δt
=-[μp(t,a)Δt]/Δt
gives limit as Δt tends to 0:
∂p(t,a)/∂t +∂p(t,a)/∂a =-μp(t,a)
holds for all a >0
differs from the diffusion equation and makes it a transport equation instead
as age is not random, you get older as time goes on older unless you die
what happens for a=0 newborns
How can I find probabilities?
p(t+Δt,a) for age structured populations
what happens for a=0 newborns
density?
count how many newborns:
#in the time interval:
p(t,0)Δt =
Δt∫_[-∞,∞]
β(a)p(t,a).da
each individual of age a produces β(a) newborns
p(t,a) look at distribution of all ages
The McKendrick–von Foerster partial differential equation
intro
This approach has been generalized to a population
consisting of individuals that can give birth and others that cannot pop of just females
McKendrick–von Foerster PDE:
well defined?
(∂/∂t)p(a, t) =
−(∂/∂a)p(a, t) − µ(a)p(a, t).
to be well defined we need BC is at boundary of age =0
and initial condition at t=0
McKendrick–von Foerster PDE:
p(a,0)=
p(a,0)=p₀(a)
means
given initial distribution)
McKendrick–von Foerster PDE:
p(0, t)
p(0, t) = B(t) = ∫_[0,∞] β(a)p(a, t)da
(population birth rate)
the pop density determines the BC which makes this more complictes
McKendrick–von Foerster PDE:
summary
first-order linear PDE of transport (hyperbolic) type
ageing happens with velocity 1 with respect to time
could have a maximal age amax if e.g. µ → ∞ as a → amax could be that every individual dies at some point or can be infinity
difficulty in that the BC depends on the variable p itself (pop density determines boundary condition which makes it complicated)
(not symmetric r deterministic like the diffusion eq)
transport PDE
McKendrick–von Foerster PDE
The transport equation can be seen as the generalization of the conservation equation:
Convection of a scalar along a flow field moving cyclicly, from left to right.
whereas Diffusion of a scalar, becoming less concentrated at the original injection place, diffusing into the surrounding fluid.
Integration on characteristics
dP(t(η),a(η))/dη
method will be used to solve McKendrick–von Foerster PDE (implicit solution):
(∂/∂t)p(a, t) =
−(∂/∂a)p(a, t) − µ(a)p(a, t).
one For now we “forget” the BC is given by how many in the population and just assume the pop birth rate is given and we know it for now
p(0, t) = B(t) = ∫_[0,∞] β(a)p(a, t)da
assumed
p(0,t)=B(t) assumed given
two the derivative is a directional derivative in the direction 1,1
time against age. So we are taking the derivative on a line of slope 1
so if individuals age 1 year as 1 year goes on the slope is 1. So actually we are just checking if these individuals who are ageing are dying or not dying
boundary of age =0 line (time axis) on this boundary we have new-borns which we have assumed as B(t)
The initial condition at time t=0 might be we have certain individuals distributed over the ages, could be a function p₀(x)
three we could simplify by thinking as if we know your age at this time we have the year you were born. This would allow us to assume we know the birth rate at negative times. We project on the diagram for these parallel lines of slope 1. Each line can be described as lines t=a+c for each c we would have the point at which the time axis was intercepted
rewrite
t=t(η)=η+c
a=a(η)=η
I have one single particle η that describes the position of this line
p(a(η),t(η))rewritten and
dp/dη written :
dp(a(η),t(η))/dη
=∂P/∂t dt/dη + ∂P/∂a da/dη
by lines dt/dη =1 and da/dη = 1
so
dp(a(η),t(η))/dη
=∂P/∂t + ∂P/∂a
− µ(a(η)) p(a(η),t(η))
gives a new ODE for the dummy variable η (family )
which is actually the LHS of our original PDE so equated this
four
to be clear call this new var p~ and we know that dp~/dη =-µ~(η) p~(η)
which gives an exponential solution
p~(η)=p~(0) exp( - ∫_[0,η] µ(σ ) dσ )
(given initial condition p(0)
If we replace p~ with the original we would have:
p(a,t) = p~
η=a =age thus we know the exponential
now what is p~(0)?
p~(0) = p(a(0), t(0))
=p(0, 0+c) = p(0,c)=p(0/t-a)
=B(t-a)
where c is the intercept of a specific linec=t-a
so this is the birth rate at time t-a as it is the pop at this time, of newborns at this time
five putting this all together
p(a,t) = B(t-a) exp(- ∫_[0,a] µ(σ ) dσ)
B(t-a)- birth rate at time t-a based on # newborns
exp term gives the probability that these newborns survive until age a ie that they are still alive now.
So this method assumes we know the birth rate at all years
we might instead say we have the initial condition for age which would give another solution
method will be used to solve McKendrick–von Foerster PDE (implicit solution):
summary of solutions
The (implicit) solution reads
p(a, t) =
{p₀(a − t) exp(− ∫_[a−t,a] µ(s)ds
if a > t
{B(t − a) exp(−∫_[0,a] µ(s)ds) if a ≤ t
Can be found via integration
along characteristics
Has biological interpretation on
its own (check how many born, check how many suvrvive a years later, exp of cumulative mortality rate descibes the prob of survival up to age a)
S(a) = exp(−∫_[0,a] µ(s)ds)
is the probability to survive up to age
we look at cases η>0, η=0, η<0 etc
does our solution
p(0,t-a)
give a pdf?
B(t − a) is a pop birth rate birth rate for a small interval
so if we multiply in a small time interval this rate by a small time interval it will give #newborns in this small time interval,
looking at the sol
we have a rate (#/T) from the exp integral
and then a density over time
method will be used to solve McKendrick–von Foerster PDE (implicit solution):
we look at cases η>0, η=0, η<0 etc
η>0
lines of slope 1
which are above the diagonal 1,1 they intercept the birth rates B(t) on the time axis (y-axis) (positive) and we can describe as derived
η<0 those lines below this, we assume the initial condition for age is given instead then derive other part of sol
method will be used to solve McKendrick–von Foerster PDE (implicit solution):
why implicit?
not explicit as i have assumed i know the birth rate B
but actually this depends on the ODE
method will be used to solve McKendrick–von Foerster PDE (implicit solution):
looking at B(t)
looking at B(t)
=rate of newborns
=∫_[0,∞ ]β(a) p(a,t ) .da
integral of the (per capita fertility rate x pop density)
= ∫[0,∞ ]β(a) **B(t − a) exp(−∫[0,a] µ(σ)dσ) . da
now we have found an expression for the pop density using the survival probability
integral equation
The pop birth rate now dep on the pop birth rate in the past
fully equivalent alternative solution
McKendrick–von Foerster PDE:
Asymptotic behaviour at t → ∞
cant really solve but check what happens in the long term, pop explode, extinct?
We try a solution of the form:
p(a, t) = q(a)e^(λt)
We find that λ must satisfy the Lotka–Euler equation
this sol is seperable in age and time with an exponential in time
(exp in time is typical as the model is linear)
∂p/∂t= λq(a)e^(λt)
∂p/∂a= q’(a)e^(λt)
plug into PDE
λq(a)e^(λt) + q’(a)e^(λt)
= − µ(a)q(a)e^(λt)
simplifies to
λq(a) + q’(a)
= − µ(a)q(a) (q’ satisfies an ODE)
q’(a)= − (λq(a)+µ(a)q(a))
gives sol
q(a) = q(0)exp(-∫_[0,a] (λ+µ( σ)).d σ)
and BC at age 0 : β x pop
q(0)e^(λt) =∫_[0,∞] β(a)q(a)e^(λt)
simplifies to
q(0) =∫_[0,∞] β(a)q(a).da
to be used
q(0)=∫_[0,∞] β(a)q(0) e^(-λa) exp(-∫_[0,a] µ( σ)).d σ).da
simplifies q(0) assumed not 0
1=∫[0,∞] β(a)e^(-λa) exp(-∫[0,a] µ( σ)).d σ).da
We find that λ must satisfy the Lotka–Euler equation
Lotka–Euler equation
what is it
1
=∫_[0,∞] β(a)S(a) e^(−λa) da
=: F(λ)
for S(a) =
exp(− ∫_[0,a] µ(s)ds)
survival probability
important in being able to find if pops go extinct or explode
Lotka–Euler equation
properties
1
=∫_[0,∞] β(a)S(a) e^(−λa) da
=: F(λ)
for S(a) =
exp(− ∫_[0,a] µ(s)ds)
survival probability
1=F(lambda)
if we fix, beta, survival prob then as all coeffs are positive
F is monotonically decreasing with lambda (lambda is real)
thinking about e^(λt):
If Re(λ)>0 grow
Re(λ)<0 extinction
as t goes to infinity
*has unique real root λ∗ which has the largest real part
**F(λ) is decreasing in λ and crosses the vertical axis at
F(0) = ∫_[0,aₘₐₓ]β(a)S(a) da =: R₀
(net reproductive rate or basic reproduction
number)
*the sign of λ∗ determines the asymptotic behaviour:
if R₀ > 1 then λ∗ > 0 and the population grows as t → ∞
if R₀ < 1 then λ∗ < 0 and the population decays as t → ∞
if R₀ = 1 then λ∗ = 0 and the population is stationary
*q(a) gives the stationary distribution of the population
over the ages (snapshot)
The Lotka–Euler equation
graphs
F against lambda
decrease exponentially?
when lambda* >0 crosses yaxis at R_0 above 1
F is monotonically decreasing then as lambda tends to infinity F tends to 0
we will always have F(lambda)=1
lamdba will be positive if the intersection at y-axis F(0) is above 1 growth
if lambda* <0 then R_0 below 1 and curve shifted left, F(0) will be below 1 and this lambda* is negative
F(lambda*)=1 here too
extinction
F(0)?
*F(λ) is decreasing in λ and crosses the vertical axis at
F(0) = ∫_[0,aₘₐₓ]β(a)S(a) da =: R₀
(net reproductive rate or basic reproduction
number)
F(0) is always the net reproductive rate/ basic reproductive number of the population
for one single individual how many individuals they will produce in their lifetime
and integrating over the whole lifetime of the individual counts will give you how many offspring if they survive- average number of offspring
so if on avg produce more than one offspring pop grows, on avrg less than they pop decay
please check example sheet 5
a few qs sheet 5
past exams
Jan 2022
Jan 2020
on age structured populations
check if everything is clear!
