23 moving populations and travelling waves

Question 1

23 moving pops and travelling waves we consider:

Answer 1

consider another example of reaction-diffusion equation,
nonlinear: Fisher’s equation for the spread of an
advantageous allele (1937) (still have diffusion, in 1D)

study the dynamics using phase-plane techniques(as non-linear we dont find explicit solution)

look for a travelling wave solution

find a critical minimal speed for a travelling wave
solution to exist (necessary condition)

Question 2

WAVE OF ADVANCE OF ADVANTAGEOUS GENES

By R. A. FISHER, Sc.D., F.R.S.

I. THE PROBLEM OF GENE DISPERSION

CONSIDER a population distributed in a linear habitat, such as a shore line, which it occupies with uniform density. If at any point of the habitat a mutation occurs, which happens to be in some degree, however slight, advantageous to survival, in the totality of its efferts, we may expect the mutant gene to increase at the expense of the allelomorph or allelomorphs previously occupying the same locus. This process will be first completed in the neighbour-hood of the occurrence of the mutation, and later, as the advantageous gene is diffused into the surrounding population, in the adjacent portions of its range. Supposing the range to be long compared with the distances separating the sites of offspring from those of their parents, there will be, advancing from the origin, a wave of increase in the gene frequency-

Answer 2

WAVE OF ADVANCE OF ADVANTAGEOUS GENES

By R. A. FISHER, Sc.D., F.R.S.

I. THE PROBLEM OF GENE DISPERSION

CONSIDER a population distributed in a linear habitat, such as a shore line, which it occupies with uniform density. If at any point of the habitat a mutation occurs, which happens to be in some degree, however slight, advantageous to survival, in the totality of its efferts, we may expect the mutant gene to increase at the expense of the allelomorph or allelomorphs previously occupying the same locus. This process will be first completed in the neighbour-hood of the occurrence of the mutation, and later, as the advantageous gene is diffused into the surrounding population, in the adjacent portions of its range. Supposing the range to be long compared with the distances separating the sites of offspring from those of their parents,** there will be, advancing from the origin, a wave of increase in the gene frequency**-

Question 3

Fisher equation assumptions

Answer 3

Assume that :
individuals move randomly in a one-dimensional domain, with diffusion coefficient D

allele A and B only; A is advantageous over B

replacement of allele B by A is proportional to the product of their respective frequencies (σ

Let:
F(x, t) = fraction of allele A, advantageous over allele B. Hence:
F(x, t) ∈ [0, 1]

1 − F(x, t) is the fraction of B
If F=1 pop is only allele A
If F=0 pop is only allele B

σ = selection intensity, so that replacement of B by A happens at rate σF(1 − F) assumed positive

we assume the advantageous gene is introduced from the left, diffuses through the population

What is the equation that governs the dynamics of F?

Question 4

Fisher equation

Answer 4

F satisfies the equation
∂F(x, t)/∂t =
D∂²F(x, t)/∂x + σF(x, t)(1 − F(x, t))
x ∈ IR
σ = selection intensity,

∂F(x, t)/∂t =diffusion + replacement of B by A

Hence, the reaction term is a logistic growth
R(F) = σF(1 − F)

Question 5

recap:
d(F)/dt = σF(1 − F)

Logistic growth without diffusion

Answer 5

d(F)/dt = σF(1 − F)
w/o diffusion solution F increases, bounded above by 1 for different initial conditions
0<F<1

has two steady states:
F~ = 0: only gene B present in the population
F~ = 1: only gene A present in the population

σ > 0: F~ = 1 is asymptotically stable

Without movement (no diffusion), each point in space where the frequency of gene A is initially positive, will tend towards replacement of all B’s by A. Every point with B only, will remain with B only.

F(x,t) against x

starts at 0 remains at 0 for all x
starts at 1 remains at 1 for all x
if some points of x only formed of allele B F=0 and some other points only A F=1 then no movements, wont be mixing the system

Question 6

logistic growth with diffusion

Answer 6

if we perturb the system movement and gene A will increase due to advantage finally replacing B and equilibrium

If we have diffusion then part of pop with gene A will move randomly and end the regions where only B was present, fraction of A will grow and replace the whole pop

With diffusion, the constant solutions 0 and 1 (homogeneous in space) are still steady states, but what happen when we have
only A on one side, and only B on the other?
Diffusion mixes the genes at the boundary, and the advantageous gene A overtakes the region with gene B

Question 7

logisitc growth with diffusion:
Travelling waves

Answer 7

We look for a travelling wave solution, i.e.,
F(x, t) = U(x − ct)
where c is the constant speed of the wave,
U is the shape of the wave, with special condition (wave profile which moves at a constant speed, respectively if we move at that speed its stationary imagine fixing a t then moving that image for another time t )
(invasion from left to right so c >0)

(F(x,t) plotted against space x, shifts to the right as time increases)

U(x)=F(x,0) profile at t=0 which moves

y=x-ct

U(y) → 0 as y → +∞ (far right in space x to infinity, past time)
U(y) → 1 as y → −∞ (far left in spacex to -infinity, future time)

and U′(y) → 0 as y → ±∞

Question 8

logistic growth with diffusion:
Travelling waves
Does this exist, and under what conditions?
discussion

Answer 8

;like SS,
if the initial conditions are exactly this travelling wave then the solution is exact
after infinite time will tend to this

if the initial condition is not exactly the travelling wave but is a pertubation of it then as time goes on it will smooth out and tend to this travellign wave as time goes on, it is a limiting constant

Question 9

logistic growth with diffusion:
Travelling waves
Does this exist, and under what conditions?

Fisher eq
∂F(x, t)/∂t =
D∂²F(x, t)/∂x + σF(x, t)(1 − F(x, t))
unbounded 1d domain x in R

Answer 9

Let F(x, t) = U(x − ct), then:

∂F(x, t)/∂t = −cU′(x − ct)
from dU/dt x ∂y/∂t

∂²F(x, t)/∂x²= U′′(x − ct)

Y=x-ct

Hence U must satisfy the second-order ODE
−cU′ = DU′′ − σU(1 − U)
Scalar 2nd order non linear ODE
We can’t find sol so we do stability analysis in the long term
To do this we transfer it into a system of first order equations

Question 10

travelling waves visually

ecological situation

invasion of species
random movement spreads as time goes on

Answer 10

imagine you have a population in 1D space with gene B
Gene A individuals enter through the left side boundary, as this is the advantageous their introduction diffuses through this boundary and replaces gene B from the left
for a fixed time
their distribution would decrease the further right you go, but over time this pop enters diffuses and replaces more and more until constant F=1

Question 11

Using fisher we look for travelling wave sol

Hence U must satisfy the second-order ODE
−cU′ = DU′′ − σU(1 − U)
Scalar 2nd order non linear ODE
We can’t find sol so we do stability analysis in the long term
To do this we transfer it into a system of first order equations

Answer 11

Introduce new var v
U’ = V
Now
-cV = DV’ - σU(1-U)

V’= -cV/D + σU(1-U)/D
Standard way to transform into first order eq we have a SYSTEM OF 2

And we can look at stability

Question 12

Fisher eq looked for travelling wave sim and transformed ode
SYSTEM OF EQ
U’ = V
V’= -cV/D - σU(1-U)/D
SS

Answer 12

SS V’= 0 and U’= 0
V= 0

-cV/D - σU(1-U)/D = 0

(U, V)

= (0, 0 )
U = 0 means F = 0 no gene A only gene B RHS of travelling wave
In the past

and (1, 0)
U=1 F=1
No gene B in the population only gene A LHS of the travelling wave
Future state

Question 13

Fisher eq looked for travelling wave sim and transformed ode
SYSTEM OF EQ
U’ = V
V’= -cV/D - σU(1-U)/D

with U(y)=F(x,t)
y=x-ct
Stability

= (0, 0 )

= (0, 1 )

Answer 13

Consider phase plane for U-V
SS marked
We are looking for a sol that starts in gene b only and ends in gene A only
But we consider the limits for y for past and future previously defined

Join a U shape under the U axis
From (0, 1 ) to (0, 0 ).
It is below because we want derivative U’(y) to be negative V(y)<0 so negative V axis part

We want (0, 1 ) to be unstable and (0, 0 ) to be stable as y tends to infinity

Jacobian
A*=
[0 1]
[{-σ +2σU}/D. -c/D]

Compute at SS

= (0, 1 )

[0 1]
[{-σ /D. -c/D]

Consider eq to find the eigenvalies
We have one root which is real and positive one real and negative
The SS is unstable

= (0, 0 )
We want this to be stable and this have eigen values with negative real part
[0 1]
[-σ/D. -c/D]

Trace -c/D <0
Det-c/D + σ/D
Positive if c> σ
Negative if c <σ
Eigen values could my complex and give a spiral
But for the sol to be biologically feasible we want
U(y) = F(x,t) > 0
Thus we need to have real eigenvalues

Question 14

fishers look for travelling wave sol

Answer 14

We look for a travelling wave solution, i.e.,
F(x, t) = U(x − ct)
where c is the constant speed of the wave, U is the shape of the
wave, with special conditions (invasion from left to right)
U(y) → 0 as y → +∞ (i.e., x → +∞ or t → −∞)

U(y) → 1 as y → −∞ (i.e., x → −∞ or t → +∞)
and U′(y) → 0 as y → ±∞

from
F→ 1 as x → −∞
F→ 0 as x → +∞

Question 15

Travelling wave
Plot of F(x, t) = U(x − ct) as time increases
descrive

Answer 15

F(x,t) against x

frequency 1 LHS x=-100
decreasesto RHS
curve
RHS F(x,t)=0 x=100

Question 16

Travelling wave
Plot of F(x, t) = U(x − ct) as time increases
describe

Answer 16

F(x,t) against x
similarly but slightly to the right as the t value is increased

Question 17

Travelling wave
F(x,t)=U(x-ct)

U’ = V
V’= -cV/D - σU(1-U)/D

with U(y)=F(x,t)
y=x-ct
Once we have (U,V)= (0,0) (1,0) we need to look for a solution st

Answer 17

For our travelling wave, we look for a solution with:
(U, V ) → (1, 0) as y → −∞
(U, V ) → (0, 0) as y → +∞
(y=x-ct so as y → −∞ we consider LHS or a past time, we want F to be 1)

Hence:
**
(1, 0) should be unstable (limit point as y → −∞)
(0, 0) should be stable (limit point as y → ∞)
biologically feasible solution U ∈ [0, 1], hence (0, 0) should
be a stable node (real eigenvalues of the jacobian)** and no spirals

⇒ Necessary condition: c larger than the minimal wave speed
c ≥ c_min = 2√{σD}

Question 18

condition for speed of a travelling wave

Answer 18

⇒ Necessary condition: c larger than the minimal wave speed
c ≥ c_min = 2√{σD}

Question 19

Exercise
If given a reaction-diffusion system and we are looking for a travelling wave what do we do?

e.g.
∂F/∂t = D∂²F/∂x²+ F(α + F)(1 − F),

α ∈ [0, 1]
What’s the minimal wave speed for a species invading left to right?

Answer 19

We are looking for a solution of the form prev:

1) Make the substitution of variable
F(x,t)= U(x-ct)= U(y)
2)Find the ODE system of 2 ODEs from V=U’
3) look at equilibria (0,0) and (1,0)
F=-α is not biologically feasible
4) (1,0) should be unstable (from the left )
(0,0) should be stable
phase portrait curve under the axis
from (1,0) and both should be nodes, real eigenvalues , otherwise spiral is not biologically feasible

Question 20

Spatial epidemics modelling:
Spatial spread of rabies in a fox population

Answer 20

SI-type model: individuals can be susceptible or become
infected, but cannot recover

infected foxes are disoriented: move randomly (diffusion D)

susceptible healthy foxes are territorial, do not diffuse

∂S(x, t)/∂t = −rS(x, t)I(x, t)
(no diffusion)
∂I(x, t)/∂t = rS(x, t)I(x, t)− aI(x, t)+ D∂²I(x, t)∂x²

new infections-death+diffusion
#new infections proportional to #I and #S

, x ∈ IR where:
a is the death rate (no recovery)

r is the infection rate per unit time (contact rate of one infected (accounts for how often get in contact and probability of transmission)

individual with one susceptible individual, times probability of transmission
(we still keep track of susceptible location as how many susceptibles dep on how many infected)

{z }

Question 21

Spatial epidemics modelling:
Spatial spread of rabies in a fox population

Transformation of parameters
to new time and space vars
linearly
which are fractions over the initial pop density
∂S(x, t)/∂t = −rS(x, t)I(x, t)
(no diffusion)
∂I(x, t)/∂t = rS(x, t)I(x, t)− aI(x, t)+ D∂²I(x, t)∂x²
t → τ = rS₀t
x → x˜ = x sqrt(rS₀/D)

Answer 21

Let S0 be the initial (fully susceptible) population density. We
can rescale the system for dimensionless variables (exercise):
t → τ = rS₀t
x → x˜ = x sqrt(rS₀/D)

S(x, t) −→ S˜(˜x, τ ) = S(x, t)/S₀
∈ [0, 1]

I(x, t) −→ ˜I(˜x, τ ) = I(x, t)/S₀
∈ [0, 1]

and further we define
m :=a/rS₀

(note that 1/m = R₀ is the classical basic reproductive number without diffusion, see Chapter 7- exp #new infections one individual is expected to produce)

The dimensionless system reads
∂S˜/∂τ = −S˜˜I
∂ ˜I/∂τ = S˜˜I − m˜I + ∂ ²˜I/∂x˜²

, x˜ ∈ IR
with S˜ = S˜(˜x, τ ) and ˜I = ˜I(˜x, τ )

Question 22

SI model:
foxes transformed system assumptions

Answer 22

the population is initially completely susceptible,
S˜ = S/S₀ = 1

infection is seeded to the left of the domain and spreads
towards the right as time increases

when the infection is introduced in a susceptible population, the epidemic grows until depletion of
susceptibles occurs, the n decreases leaving the susceptibles

at a lower post-epidemic level S˜ = S˜_f < 1

diagram:
susceptibles against x: from thr LHS =above to 0, increases and then hits 1 constant
infected against x: at 0 from LHS, increases then decreases back to 0

for later times this is the same but translated to the right

Question 23

si model: foxes
epidemic persists if…

if we are in the case where epidemic takes over?

Answer 23

We show that:
the epidemic persists only if m < 1 (i.e., R₀ > 1), otherwise
it dies out

if m < 1, the final size is S˜_f = 1- (S_f /S₀) satisfies S˜_f < m and can be found as the solution of the final size equation

S˜_f = 1 +(a/rS₀)ln S˜_f
(compare with final size in the standard SIR model without
diffusion from Chapter 7)

We are looking for a travelling wave that moves at constant speed from L to R with speed above:

if a travelling wave exists, it must satisfy
c ≥ 2√(1 − m) = 2sqrt(1 −a/(rS₀) =: cmin

where cmin is the minimal wave speed

speed must be larger than this value

(no parameter D, epidemological parameters are the only ones playing a part in epidemic number final number on the c_min)

Question 24

What kind of travelling wave sol do we look for?
RECIPE
SI model with change of vars

How do limits in x~ and tau translate into limits in new var y=x-ct

Answer 24

We are looking for a travelling wave that moves at constant speed from L to R

1) introduce 2 functs U AND V
**
We look for a travelling wave solution of the form
S˜(˜x, τ ) = U(˜x − cτ ), (profile of susceptibles)
˜I(˜x, τ ) = V (˜x − cτ ) (profile of infected)**

2)Plot against x: Susceptibles S~ to the right of the domain are at 1 (initially) and the after they decrease on the LHS of the domain towards S_f

V is 0 before and after the epidemic LHS and RHS so I
**
so that, as x˜ → +∞ and τ → −∞ we have
S˜ → 1,˜I → 0 (before epidemic wave)**
**
and, as x˜ → −∞ and τ → +∞ we have
S˜ → S˜_f ,˜I → 0 (after epidemic wave)**

We are looking for a travelling wave that moves at constant speed from L to R

3) For S˜(˜x, τ ) = U(˜x − cτ ), (profile of susceptible) we want that
as y˜ → −∞ U→ S˜_f , (final susceptibles left)
as x → −∞ S~→ S˜_f ,
(for fixed time as y tends to -infinity so does x~)
(for a fixed x~ and look at tau as y tends to -infinity we have tau to infinity
:future time so U and S~ tend to S~_f final #susceptibles)
for past time tau tends to neg infinity so y tends to +infinity or x tends to +infinity,

Question 25

What kind of
What SS are we looking for travelling wave sol do we look for?
RECIPE continued
SI model
equilibria

Answer 25

Hence we want the solutions (U, V ) = (S, ˜ ˜I) as a function of y = ˜x − cτ to go from
(U, V ) = (S˜_f , 0) to (U, V ) = (1, 0),
i.e.,

(U, V ) → (S˜_f , 0) as y → −∞ (⇔ x˜ → −∞)

(U, V ) → (1, 0) as y → +∞ (⇔ x˜ → +∞)

plotting U-V plane
we want a solution from (S~_f,0) to (1,0) but this time in the sol is in the positive plane for V because it now represents the infected population

V starts at 0 becomes positive and then decreases to 0
cant have negative U
or U above 1

so we want:
1)(S˜_f , 0) to be unstable
2) (1, 0) to be asymptotically stable
3) viable solution (U, V ≥ 0), hence both should be nodes:
real eigenvalues of the jacobian matrix

Question 26

For a general ODE
x’=f(x,y)
y’=g(x,y)

phase plane

Answer 26

x-y plane
x and y are time dependent
x(t) y(t)
phase plane doesnt plot time but we look at trajectories as time goes on,
so we might have for example (0,0) and (10,A) trajectories move away from origin to (1,1) say

plotting x(t) against time means for negative time we have x(t) close to 0, increasing with time until x(10)=A) and bounded above by this value A

Question 27

SI model:
But first: are these steady states and what is S˜_f ?

Hence we want the solutions (U, V ) = (S, ˜ ˜I) as a function of y = ˜x − cτ to go from
(U, V ) = (S˜_f , 0) to (U, V ) = (1, 0),
i.e.,

(U, V ) → (S˜_f , 0) as y → −∞ (⇔ x˜ → −∞)

(U, V ) → (1, 0) as y → +∞ (⇔ x˜ → +∞)

Answer 27

(U, V ) = (S, ˜ ˜I) satisfies (check by exercise!)
(∂S~/∂𝜏) = dU/dy ∂y/∂𝜏 = -cU’
(∂S~/∂x~) = dU/dy ∂y/∂x~ = U’
as ∂y/∂x~=1
(∂²S~/∂x~²) = U’’
plug into diffusion eq and for I~

we get
cU′ = UV
−cV ′ = UV− mV+V’’
(UV = cU’ and mV = mcU’/U)

(different from prev, V is different interpretation and now second order)

Question 28

SI model:
What do we do with this system?

cU′ = UV (1)
−cV ′ = UV− mV+V’’ (2)

transform into
U’= (1/c)UV
V’=mcln(U) - c(U+V) +C

Answer 28

(UV = cU’ and mV = mcU’/U)
(2) rewritten V as a function of U’ and U
−cV ′ = cU’− mcU’/U+V’’
-cV’=

Hence, using U′/U =d/dy[lnU], the second equation becomes:
d/dy[V′ + cV + cU − mc lnU]= 0
and hence as total deriv =0 we have a constant

V′ + c(U + V ) − mc lnU = constant = c

where the last equality comes from
(U, V ) → (1, 0) as y → +∞
( mc lnU → 0, V′ → 0 , U → 1 , V →0)
we can replace this last equation into the system!

we get **
U’= (1/c)UV
V’=mcln(U) - c(U+V) +C**

Question 29

U’= (1/c)UV
V’=mcln(U) - c(U+V) +C

SS of this system?

Answer 29

(U,V)
NOT (0,) we have a logarithm of U so U must be positive and non 0, ln (0) undefined
V*=0 mcln(U) - cU+C = 0

f(U)= -m ln(U) +U-1
f is 0 when U =1 ln(1)=0
plotting f
( as u→0 ln(U) →-infinity, f→+infinity, as u →+infinity ln(u)→+infinity,

f’u)=m/u +1 >0 iff u>m and m<1
so plot kinda U shape with first U interscept below m, S~
next at U=1

SS satisfy F(U)=0
a sol exists 0<S~_f<1 exists if m <1
SS are
(S~_f,0) and (1,0)

phase plane UV
(S~_f,0) unstable in y
(1,0) stable

then look at Jacobian
A=
[1/c U/c ]
[(mc/U)-c -c ]

(S~_f,0) :
[0 S~_f/c]
[mc/SD~_f - c]

Eigenvalues: stable
trace <0 and det>0 required
we know S~_f<m

(1,0)
[0 1/c]
[c(m-1) -c]

tr=-c<0
det=-m+1>0
stable

find the eqigenvalues specifically….

Answer 30

pgf of RV
PGF of a stachastic process (a family of random variables)
use the properties of PGF

branching process in discrete time:
main characteristic, individuals behaving IID taking positions at discrete time steps or im observing the process at discrete time steps

each cell is dividing at different times not discrete but if i observe i want to observe in each generation it is discrete, branchin happening in continuous time but i dont care when divisions happen only at discrete steps

mean size of a pop/N
#cells in duration n
each cell #offspring RV and pgf of Z
dynamics of offspring depend on PGF
formula for mean value

Extinction of pop for branching: prob of extinction u_n at gen n
formula given U_n phi _n-1
what happens as U goes to infinity: probability of extinction in long term eq in long term

pgf of branching proicess

continuous time branching process: pure death, pure birth, birth and death, linear as Lambda_n =nlambda and mu_n=nmu
probability transition p_n (t)=p(Xsometimes possible sometimes not
but we can analise the mean
probability of extinction, extinction time,
density function of extinction

Question 31

t