22 Models of population growth and diffusion
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why we consider Reaction–diffusion models
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
Reaction–diffusion models are of the form
∂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
Reaction–diffusion models are of the form variations
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
Skellam work from 1951
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
Kierstead and Slobodkin work from 1953
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?
The KISS (Kierstead–Slobodkin–Skellam) model assumptions and requirements
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]
The KISS (Kierstead–Slobodkin–Skellam) model model
∂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)
∂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?
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
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)
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
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?
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)
summary model of pop growth and diffusion on [0,L] and absorbing dirichlet homogenous boundary conditions
“a population in circulation for a finite region can support itself against diffusion only if reproductive rate exceeds leakage”
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?
(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
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),
(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
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)
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
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),
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
dirac delta δ(x-x₀)=
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
if we are interested in the specific case of N₀(x)= δ(x-x₀)
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
we consider: Exponential growth and diffusion without boundaries in IR2
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
Diffusion without boundaries in IR2 : diffusion and growth Radial symmetry: polar coordinates in IR2 model PDE:
∂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
Diffusion without boundaries in IR2 :diffusion and growth Radial symmetry: polar coordinates in IR2 model PDE: Fundamental solution in the unbounded plane
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
Diffusion without boundaries in IR2 :diffusion and growth what does it look like
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
Diffusion without boundaries in IR2 :diffusion and growth N𝒸 how does limit of r𝒸 change as t →∞?
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)
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
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
general method for solving: ∂N(r, t)/∂t = D∇²N(r, t) + αN(r, t) N(r,0)=N₀(r) =δ(r)
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