22 Models of population growth and diffusion

Question 1

why we consider Reaction–diffusion models

Answer 1

What if you not only have random movement in time, but also vital processes (e.g. birth, death, competition for resources,predations. . . ) are taking place while individuals move in space?

other reactivons besides diffusion
diffusion + a function describing the reaction

Question 2

Reaction–diffusion models are of the form

Answer 2

∂N(x, t)/∂t=
= D∇²N(x, t) + f(N(x, t))
diffusion + reaction

x ∈ Ω, t ≥ 0

where f(N) accounts for population vital processes that are not related to diffusion in space

N(x,t)~pop density at x ∈ Ω
∇² ~ d dimensional Laplace operator
R(N)~ reaction term models how the pop changes besides spreading diffusively and eq reduces to diffusion when R=0

firstly we look at R(N) is a linear function of N and hence describes how the population grows

Question 3

Reaction–diffusion models are of the form variations

Answer 3

systems of equations if more than one species
could be interacting together
x could be 1d,3d

we focus on exponential growth: f(N)= αN,
where α is the Malthusian parameter (assumed α > 0)
linear function f
(if α<0 pop would go extinct)

we consider cases:
* finite one-dimensional domain [0, L] in R
*plane R^2 , unbounded growth

Question 4

Skellam work from 1951

Answer 4

Considers questions related to the dispersal of:

oak trees
‘small animals animals such as earthworms and snails’
in one dimension (‘linear habitat’):
costal zones, bank of a river. . .
and in two dimensions with radial symmetry

Question 5

Kierstead and Slobodkin work from 1953

Answer 5

considers
a growing phytoplankton population
in a mass of (favourable) water (constrained)
surrounded by water that is unsuitable for the survival of
the population

how can they persist in the limited space?

Question 6

The KISS (Kierstead–Slobodkin–Skellam) model

assumptions and requirements

Answer 6

We want to model a population that:
lives in a one-dimensional habitat (river, coast,. . . ),
x ∈ [0, L] (boundary domain)

diffuses in the habitat with diffusion coefficient D > 0

the habitat is favourable: the population has a net growth
at per capita rate α > 0 [T⁻¹] (if we balance reproduction and death we have a positive growth)

is surrounded by extremely unfavourable habitat:
immediate death out of [0, L]

Question 7

The KISS (Kierstead–Slobodkin–Skellam) model

model

Answer 7

∂N(x, t)/∂t=
= D ∂²N(x, t)/∂x ² +αN(x,t)
diffusion + reproduces at rate α (per capita rate)
x ∈ [0,L] , t ≥ 0

(∂²N(x, t)/∂x ² Laplacian in 1D!)

(immediate death at the boundary:
N(x, t)=0 for x not in (0,L)
favourable habitat:)

unfavourable gives absorbing Dirichlet homogeneous boundary conditions
N(L,t)=0
N(0,t)=0

and initial condition N(x, 0) = N₀(x), where:
D is the diffusion coefficient
α > 0 is a constant growth rate ([α] = 1/time)

Question 8

∂N(x, t)/∂t=
= D ∂²N(x, t)/∂x ² +αN(x,t)

x ∈ [0,L] , t ≥ 0

absorbing Dirichlet homogeneous boundary conditions
N(L,t)=0
N(0,t)=0

and initial condition N(x, 0) = N₀(x), where:
D is the diffusion coefficient
α > 0 is a constant growth rate ([α] = 1/time)

model for pop growth and diffusion in 1D : what do we know about the sol from the ODE?

Answer 8

we have studied how +αN(x,t) term know it will be an exponentially growing pop

we have studied how diffusion eq on [0,L] with absorbing boundary conditions is doomed to be extinct for any initial condition

so …
While α > 0 implies individuals have a positive net production,
the absorbing boundaries ‘remove’ individuals/mass from the domain.

they oppose
How much do individuals need to reproduce for the population
to persist? The critical size given the rate of growth

Question 9

SOLUTION STANDARD FORM

∂N(x, t)/∂t=
= D ∂²N(x, t)/∂x ² +αN(x,t)

x ∈ [0,L] , t ≥ 0

absorbing Dirichlet homogeneous boundary conditions
N(L,t)=0
N(0,t)=0

and initial condition N(x, 0) = N₀(x), where:
D is the diffusion coefficient
α > 0 is a constant growth rate ([α] = 1/time)

Answer 9

N(x, t) = Σ[n=1,∞]
a ₙ sin(nπx/L)*
exp (α − [[D(nπ)² ]/L² ]]t)

infinite sum,time dep only in exp

pop will explode if positive exponent , if all exponents are neg then doomed to die out

as n becomes larger the neg term begins to dominate so we consider the smallest [[D(nπ)² ]/L² ]]t

Question 10

standard form sol for models of growth and diffusion in 1D with absorbing Dirichlet homogenous conditions
N(x, t) = Σ[n=1,∞]

a ₙ sin(nπx/L)*
exp (α − [[D(nπ)² ]/L² ]]t)

when does the pop grow?

Answer 10

largest exp of time dep function
if grows only if exp is positive and largest is obtained for n=1 with value
α − (Dπ²/L²)

THUS the population grows if and only if

α − (Dπ²/L²)>0

ie L> π* sqrt(D/α)
KISS size
(critical length)

Growth is promoted by:
larger α (more growth)
larger favourable domain L (better growth before hitting)
slower diffusion D (sends to unfavourable habitat)

Question 11

summary model of pop growth and diffusion on [0,L] and absorbing dirichlet homogenous boundary conditions

Answer 11

“a population in circulation for a finite region can support itself against diffusion only if reproductive rate exceeds leakage”

Question 12

Imagine you have mixed absorbing/reflecting boundaries
(unfavourable environment at one end, perfectly reflecting at the other end).

Can you guess the condition for population growth?

Answer 12

(Exercise: example sheet 5, Q3)

good practice for the
refresher for how to solve the diffusion eq (alpha =0)

You could have an exam q about “this is the diffusion eq show that.. or find the solution based on BCs etc”

For absorbing conditions we had L> π* sqrt(D/α)
L_crit= π* sqrt(D/α)

Now if we have mixed boundary conditions one reflecting( aborbing at L unfavourable and reflecting at 0):
∂N(0,t)/∂x=0
N(L,t)=0
it is actually like we have the interval twice so L_crit = π* sqrt(D/α)/2 showing this by assuming seperability we get again…
N(x,t)=
Cexp((α -λₙ)Dt * F(x)=acos(√{λ/D}x)+bsin(√{λ/D}x)

but now different conditions:
F’(0)=0
a√{λ} = 0 a=0
(if we had swapped the conditions the solution would just be cosines)
F(L)=0

√{λ} L + (π/2) +nπ
= π(0.5+n)
meaning √{λ} =π(0.5+n)/L
because of the mixed boundaries

family of sols dep on n and we sum from n=0 as the 0 term is not 0 as no longer just sines

N(x,t)= Σₙ bₙcos((n+0.5)π/L)x)* exp((α -D((n+0.5)π)²/L²)t)
???check D
from n=0 to∞

if we want to check when this pop grows we look at largest time exp which occurs including when n=0 which we get an exp((α -D((0.5)π)²/L²)t)

inverting this condition gives us
L_crit= [π* sqrt(D/α)] /2
b_n would be determined by initial condition by integration against cosines method used

Question 13

Solution to the KISS model

SOLUTION STANDARD FORM DERIVATION

∂N(x, t)/∂t=
= D ∂²N(x, t)/∂x ² +αN(x,t)

x ∈ [0,L] , t ≥ 0

Boundary condition N(L,t)=N(0,t)=0

initial condition N(x, 0) = N₀(x),

Answer 13

(solved similar to case where α=0)
1:assume seperable function
N(x,t)=F(x)G(t)
∂N(x, t)/∂t= F(x)G(t)
∂²N(x, t)/∂x ² = G(t)F’‘(x)
2 into the eq for PDE
F(x)G(t)= DG(t)F’‘(x) +α F(x)G(t)
(we look for a nontrivial sol we can assume both functs are not 0 so we can divide by)

G(t)/G(t) = DF’‘(x)/F(x) +α
(1/D)G(t)/G(t) - α = F’‘(x)/F(x)
= -λ

λ>0
as LHS dep on t only, RHS on x (space) only then must be constant
The constant must be negative as o/w we can’t satisfy BCs

(1/D)G(t)/G(t) - α =-λ**
G(t)/G(t) = α -λD (linear ODE)
(D check)
G(t) = Cexp((α -λD)t) constant C

Gₙ(t)= Cexp((α -λₙD)t
(CHECK D IS CORRECT?)

F’‘(x)/F(x)
= -λ
(2nd order ODE so look at the characteristic terms)

characteristic eq roots
DX² +λ =0 imaginary roots
solution is a form of sines and cosines
F(x)=asin(√{λ/D}x)+bcos(√{λ/D}x)
three use BCs
absorbing conditions

N(L,t)=N(0,t)=0
G(t)[b]=0 for all t
means b=0
G(t)[asin(√{λ}L))=0 for all t
means (as a non zero)
√{λ}L= (nπ)
√{λ}= (nπ)/L for integer n
Fₙ(x) = aₙsin(nπ/L))

Fₙ(x) Gₙ(t)
= Cexp((α -D(nπ)²/L²)t)[ aₙsin(nπ/L)x)]

four superposition
superposition principle, because this is a linear equation a general solution is given by a linear combination of all the particular solutions thus

N(x,t)= Σₙ aₙsin(nπ/L)x)* exp((α -(nπ)²/L²)Dt)

from n=1 to∞

with aₙ determined by the IC:
N(x, 0) = N₀(x),

initial condition N(x, 0) = N₀(x),
N(x,0)= Σₙ aₙsin(nπ/L)x)

now if we integrate:
∫₀ᴸ sin(nπx/L)*sin(mπx/L).dx
this is non zero only if n=m
=
{2/L if n=m
{0 if n≠m

aₙ= (2/L) ∫₀ᴸ N₀(x) sin(nπx/L).dx

Question 14

explaining why
(2nd order ODE so look at the characteristic terms)
DF’‘(x)/F(x)
= -λ

characteristic eq roots
X² +λ =0 imaginary roots
solution is a form of sines and cosines
[asin(√{λ}x)+bcos(√{λ}x)

Answer 14

we have BC [0,L] abosrbing dirichlet homogeneous

as we expect for solutions to be a family of functions depending on sines with 0 as sin(0) and sin(function of L)=0

Question 15

explaining the steps for:
used the principle of superposition
N(x,t)= Σₙ aₙsin(nπ/L)x)* exp((α -(nπ)²/L²)Dt)

from n=1 to∞

with aₙ determined by the IC:
N(x, 0) = N₀(x),

Answer 15

this is non zero only if n=m

N(x,0)= Σₙ aₙsin(nπ/L)x)
The family of functs of sines satisfy an orthogonality condition: in [0,L] at all frequencies

now if we integrate the product of sines:
∫₀ᴸ sin(nπx/L)*sin(mπx/L).dx

{2/L if n=m
{0 if n≠m

dep on length of domain.
So because of this we can compute the integral of out general sol (the simplest case of it; when t=0)
N(x,0)= Σₙ aₙsin(nπ/L)x)
then as linear
∫₀ᴸN₀(x)* [sin(mπ/L)x)] .dx =
∫₀ᴸ Σₙ aₙsin(nπ/L)x) * [sin(mπ/L)x)] .dx
=Σₙ ∫₀ᴸ aₙsin(nπ/L)x) * [sin(mπ/L)x)] .dx
=2aₘ/L
then compute using n instead and rearrange
aₙ= (L/2) ∫₀ᴸN₀(x)* [sin(nπ/L)x)] .dx

Question 16

dirac delta
δ(x-x₀)=

Answer 16

a measure not a function
centred at 0:δ(x)=
{0 everywhere
{infinite in 0
measures 0

defined using the integral:
∫-∞,∞ δ(x)f(x)= f(0)
any function f

centred at x₀:
δ(x-x₀)=
infinite at (x-x₀)
0 everywhere else
defined using the integral:
∫-∞,∞ δ(x-x₀)f(x)= f( x-x₀)
any function f

links to gaussian distribution limit as all mass concentrated at one point and 0 elsewhere

Question 17

if we are interested in the specific case of N₀(x)= δ(x-x₀)

Answer 17

aₙ= (L/2) ∫₀ᴸ N₀(x) sin(nπx/L).dx
= (L/2) ∫₀ᴸ δ(x-x₀) sin(nπx/L).dx
=(L/2) sin(nπx₀/L)

EXPLICIT EXPRESSION now for coeffs
finally GS given by infinite sum where we have explicitly computed aₙ

Question 18

we consider:
Exponential growth and diffusion without boundaries in IR2

Answer 18

Let us now consider the case of the diffusion equation (diffusivity D) with a linear growth
term (grow rate α) in the unbounded two-dimensional space IR^2 (no geography just space)

IR2 darker more towards centre-Exponential growth and diffusion in the unbounded two-dimensional space IR2

Initially, the population is at the origin; the density N(r, t) is a radial function and varies only
with the distance r from the origin: Density is high in dark regions and low in light regions

pop only varies with distance r from the centre- no angular dependence
N=N(r,t) radial function of the space and of time

if growth rate negative at first for sure pop goes extinct so assume α>0

Question 19

Diffusion without boundaries in IR2
: diffusion and growth

Radial symmetry: polar coordinates in IR2
model PDE:

Answer 19

∂N(r, t)/∂t = D∇²N(r, t) + αN(r, t)

= D(1/r)∂/∂r (r (∂N(r, t)/∂r)+ αN(r, t)

we let the population free to expand on the plane IR², so
r ∈ (0, ∞) (no boundary),

but we assume that N(r, t) is bounded everywhere, and in particular N(r, t) → 0 as r → ∞;further away tends to 0 assumed

Question 20

Diffusion without boundaries in IR2
:diffusion and growth

Radial symmetry: polar coordinates in IR2
model PDE:

Fundamental solution in the unbounded plane

Answer 20

N(r, t) = N~₀ /{4πDt}*exp( αt −r²/4Dt)

where N₀~ is the total initial population density so that
N(x, 0) = N~₀ δ(r)

Fundamental sol as assumed:
(where δ is the Dirac delta in r = 0;
as we assumed initially the population was concentrated at a specific point the centre, r=origin at t=0)

N(r,0)=N₀(r) =N~₀ δ(r)=0
pop initially concentrated at origin

as δ is a distribution integrates to 1 :
∫∫_R² N~₀ δ(r) dr dθ = N~₀
total number of individuals in space

Question 21

Diffusion without boundaries in IR2
:diffusion and growth

what does it look like

Answer 21

pop initially concentrated at origin

immediately diffusion happens, gaussian like density: positive probability to find the species anywhere on the plane, including infinity. diffusion means you find the population at any location r > 0 at any time t > 0.

But we imagine we have a “detectable level of pop” under which we cannot detect, limit of N as r tends to infinity is thus 0

So now we look for N( r𝒸,t) = N𝒸

Question 22

Diffusion without boundaries in IR2
:diffusion and growth

N𝒸

how does limit of r𝒸 change as t →∞?

Answer 22

a detectable level
What is the edge of the detectable population then? r𝒸, for which for x> r𝒸 we have N(x,t)=0, radius for which we have a detectable level

N( r𝒸,t) = N𝒸

r𝒸 =
2t √[
(αD)[1 − (1/αt) * ln((4π/N~₀)DtN𝒸)]
]

we consider the speed at which this detectable level moves
For large t the velocity =distance/time of the edge of the pop is

lim_t→∞ r𝒸/t = 2√[αD]
this is constant

dispersal is faster (velocity is larger) if:
larger α (larger growth)
or/& larger diffusion D (individuals move away faster)

Question 23

Find the solution:
∂N(r, t)/∂t = D∇²N(r, t) + αN(r, t)

= D(1/r)∂/∂r (r (∂N(r, t)/∂r)+ αN(r, t)

we let the population free to expand on the plane IR², so
r ∈ (0, ∞) (no boundary),

but we assume that N(r, t) is bounded everywhere, and in particular N(r, t) → 0 as r → ∞;further away tends to 0 assumed

Answer 23

N(r, t) = N~₀ /{4πDt}*exp( αt −r²/4Dt)

where N₀~ is the total initial population density so that
N(x, 0) = N~₀ δ(r)

METHOD: general used

Question 24

general method for solving:
∂N(r, t)/∂t = D∇²N(r, t) + αN(r, t)
N(r,0)=N₀(r) =δ(r)

Answer 24

U(r,t)=N(r,t) exp(-αt)
(expansion in time factored out)
Laplacian:
∇²U(r, t) = ∂²U(r, t)/∂r²
= ∂²N(r, t)/∂r² * exp(-αt)
=∇²N(r, t)*exp(-αt)

∂U(r, t)/∂t=
∂N(r, t)/∂t exp(-αt) -αN(r,t)exp(-αt)

thus
D∇²U(r, t)exp(αt) + αU(r,t)exp(αt)
=[∂U(r, t)/∂t * exp(αt)+ α*U(r,t)exp(αt)]

just standard diffusion in R^2:
D∇²U(r, t)=[∂U(r, t)/∂t]

then assume separability, equate to constant, solve for family of solutions, superposition principle for infinite sum from =0,…
guassian distrbution in R^2
fundamental solution uses:

N(r,0)=N₀(r) =δ(r)
U(r,0)=N(r,0) =δ(r)
U(r,t)=
{N₀~/4πDt}* exp({− r²/4Dt} }
(in R²)

N(r,t)=exp(αt) U(r,t)=
{N₀~/4πDt}* exp({− r²/4Dt} + αt}

This is the distribution of the pop over space, guassian in R²

Question 25

solution looking at closer:
N(r,t)=exp(αt) U(r,t)=
{N₀~/4πDt}* exp({− r²/4Dt} + αt}

This is the distribution of the pop over space, guassian in R²

What happens to the total population:?

Answer 25

Looking at R², pop ceded at origin,
guassian becomes flatter and flatter over time

To look at the total population we look at the integral:
∫ ∫_R² N(r,t) .dr.dθ
= ∫ ∫_R² [ N₀~exp{αt}{{1/4πDt}* exp({− r²/4Dt} )
]
.dr.dθ

N₀~exp{αt} is constant in space, we take it out of the intergal as integrating over _R²
**
{1/4πDt}* exp({− r²/4Dt} is the normal distribution, thus integrates to 1 on the plane**

=N₀~exp{αt}
So if we integrate over the whole plane we get the
if exponential growth population that we are used to if we forgot about space and just looked at the numbers growing

Question 26

Recap: Modelling of stationary tumours
Reaction-diffusion equation for nutrient concentration c(r, t)
with constant nutrient uptake:
a simple reaction-diffusion eq
model and bcs
how did we solve?
critical condition for no necrotic core?

Answer 26

we modelled everything thats happening in the system as a function to the centre. For tumor growth w assume a constant uptake of nutrients, -k

∂c(r, t)/∂t=
D∇²c(r, t) − k
=D(1/r²)d/dr(r² d/dr(c(R))) -k

diffusion in R^3 radial symmetry with boundary conditions

c(r₂, t) = c₂ (concentration in healthy tissue)

∂c(0, t) /∂r = 0 (no nutrients exchange at the core)

we solved for a stationary tumour (independent of time)

found a critical condition for no necrotic core
(c ≥ c₁):
r₂ < srt[(6D/k) (c₂ − c₁)] if r bigger the tumor now has a growing necrotic core and eq changes

Question 27

Recap: The KISS model

application to ecology

a population diffusing in a favourable environment

diffusion reaction model example

Answer 27

The KISS model for a population in a favourable environment surrounded by unfavourable environment: positive growth rate for favourable,
∂N(x, t)/∂t =
D∂²N(x, t)/∂x² + αN(x, t), x ∈ [0, L], t ≥ 0

with absorbing Dirichlet homogeneous boundary conditions:
N(0, t) = N(L, t) = 0

where D is the diffusion coefficient, α > 0 the growth rate.

we solved the solution on the bounded domain
found a critical domain length for population growth:
L > πsqrt(D/α)
allows a pop to grow, we want this favourable environment to be large enough