25 turing instability

the biological process that causes a cell, tissue or organism to develop its shape How can different types of tissues in (body, animals) differentiate when they grow how can an embryo divide and create different organs? How can the patterns in animals be formed? Diffusion can be the mechanism behind morphogenesis interactions of two different (cells, chemicals, species to activate the production of proteins)

Consider two types of particles (activator / inhibitor), diffusing in the same domain and interacting Turing (diffusion-driven) instability happens when: 1 without diffusion, the system admits a stable spatially-uniform coexistence steady-state (homogenous, coexist in SS, stable in time, as its PDE we can have a stable SS constant in time but differences in space) stable and uniform without diffusion when we add diffusion same SS unstable 2 with diffusion, the uniform steady state becomes unstable, and spatial patterns may appear We will see that diffusivities must be different (+ more specific condition) (we need diffusion for patterns) emergence of space dep solutions, with a periodic patterns emerging we will see that for this to happen the two species must have different diffusion coefficients, behaving differently in space

∂N₁/∂t = D₁∂²N₁∂x² + f(N₁, N₂), ∂N₂/∂t = D₂∂²N₂/∂x²+ g(N₁, N₂) x ∈ Ω ⊂ IR, t ≥ 0 with D₁, D₂ > 0 diffusion coefficients of the two species. different! o/w no chance we can find We look at stationary solutions, i.e., constant in time. we consider 1d a line: R f and g describe how the species interact could be pred prey, cooperation etc REACTION TERMS

We look at stationary solutions, i.e., constant in time. uniform coexistence steady state is a solution (N∗₁, N∗₂), with N∗₁, N∗₂ > 0 **independent of time and space ** (i.e., uniform for x ∈ Ω), such that f(N*₁, N∗₂) = 0 and g(N∗₁, N∗₂) = 0. independent of space so PDE of space =0 time deriv=0 (PDE if we only looked at constant in time so function varies with x, but instead also constant in space so constant function for x)

ordinary differential equation (ODE) we: linearise the system around the equilibrium look at the spectrum of the Jacobian matrix if all eigenvalues have negative real part, the equilibrium is locally asymptotically stable; if one eigenvalue has positive real part, the equilibrum is unstable (and solution diverge exponentially near the equillibrium) Same steps work for PDEs!

we want a uniform equilibrium (homogenous) that is stable without diffusion we want the same equilibrium to be unstable with diffusion

Without diffusion: D1 = D2 = 0, the system becomes dN₁/dt = f(N₁, N₂), dN₂/dt = g(N₁, N₂) equilibrium (N∗₁, N∗₂), satsifies f(N*₁, N∗₂) = 0 and g(N∗₁, N∗₂) = 0. linearisations at this equilibrium obtained by looking at the perturbations N₁(x,t)= N₁+ u₁(x,t) N₂(x,t)= N₂+ u₂(x,t) linearised eq du₁(x,t)/dt = (∂ f(N₁,* N₂)/ ∂N₁ ) u₁(t)+ (∂ f(N₁, N₂*)// ∂N₂ ) u₂(t) du₂(x,t)/dt= (∂ g(N₁,* N₂)/ ∂N₁ ) u₁(t)+ (∂ g(N₁, N₂*)// ∂N₂ ) u₂(t) in matrix form d/dt ((u₁) (u₂)) =A*((u₁) (u₂)) A* is the jacobian at equilibrium = [∂ f(N₁,* N₂)/ ∂N₁ ∂ f(N₁, N₂)/ ∂N₂ ]] [∂ g(N₁, N₂)/ ∂N₁ ∂ g(N₁, N₂*)/ ∂N₂ ] (Jacobian equilibrium without diffusion]

The steady state is asymptotically stable without diffusion if all eigenvalues of A∗ have negative real part The eigenvalues λ must satisfy λ² − (a∗11 + a∗22)λ + a∗11a∗22 − a∗12a∗21= 0, λ² − traceλ + det= 0, And the roots of this polynomial have negative real parts if and only if: trace(A∗) < 0 and det(A∗) > 0 {z }

In the following, we assume we have a spatially uniform equilibrium f(N₁,* N₂), i.e., f(f(N₁, N₂)=0 g(f(N₁, N₂*)= 0 that is stable in the system without diffusion, i.e., trace(A∗) < 0 and det(A∗) > 0 and with diffusion the same equilibrium is: unstable

With diffusion: D1, D2 ≠ 0, linearisations at this equilibrium obtained by looking at the perturbations N₁(x,t)= N₁+ v₁(x,t) N₂(x,t)= N₂+ v₂(x,t) substitute into PDE linearised eq (∂N₁*/∂t =0) ∂v₁/∂t =[∂ f(N₁,* N₂)/ ∂N₁ ](v₁)+ [∂ f(N₁, N₂)/ ∂N₂]v₂ + D₁ ∂²v₁/∂x² - ∂v₂/∂t=∂g(N₁,* N₂*)/ ∂N₁ + [∂g(N₁, N₂*)/ ∂N₂]v₂ + D₁ ∂²v₁/∂x² in matrix form ∂ /∂t ((v₁) (v₂)) =A*((v₁) + [D₁ 0] ∂² v (v₂)) [0 D₂] ∂x² A* is the jacobian at equilibrium = [∂ f(N₁,* N₂)/ ∂N₁ ∂ f(N₁, N₂)/ ∂N₂ ]] [∂ g(N₁, N₂)/ ∂N₁ ∂ g(N₁, N₂*)/ ∂N₂ ] (Jacobian equilibrium without diffusion]

linear PDE started with already linear for D term reaction terms f and g need to be linearised

when we have a linear system we look for a linear sols that are seperable and exp in time: (v₁(x,t)) (v₂(x,t)) = (w₁(x)) exp(σt) (w₂(x)) for some σ ∈ C (temporal eigenvalue) such that the x-function(w₁, w₂) satisfies the additional condition (w₁’‘(x)) (w₂’‘(x))= -k² (w₁(x)) (w₂(x)) (we want to find an unstable sol: σ >0 for real part) because the Laplacian very conveniently disappears we will use the following : ∂ /∂t ((v₁) (v₂)) =A*((v₁) + [D₁ 0] ∂² v (v₂)) [0 D₂] ∂x² becomes by replacing space derivative and k² denotes a positive eigenvalue σv = A*v + D -k² v rearranging giving [σ [10] - A* -k² [D₁ 0] ] [01] [0 D₂] • ((v₁) =0 (v₂)) where v is non trivial sol of the system which is true if the matrix σI − J(k²) is non-invertible, ie det=0 J(k²) is the matrix in the [] i.e., σ is an eigenvalue of J(k²): det[σI − J(k²]= 0 THIS IS THE NEW JACOBIAN OF THE DIFFUSION EQ GIVEN DIFFUSION IS HAPPENING This determines σ as a function of k²

∂v₁/∂t= D₁(∂²v₁/∂x²)+ (∂f/∂N₁)v₁ + (∂f/∂N₂)v₂ ∂v₂/∂t= D₂(∂²v₂/∂x²)+ (∂g/∂N₁)v₁ + (∂g/∂N₂)v₂ linear eq linearization now with the assumption that the second derivative (∂²/ ∂x²)(v₁) = -k² (v₁) (v₂) (v₂) rewrite: above as [-k² [D₁ 0] ] +A*] (v₁) [0 D₂] ] (v₂) J(k²)= [σ [10] - A* -k² [D₁ 0] ] [01] [0 D₂]

For instability to occur, we want a growing time exponential, i.e., at least one eigenvalue σ with Re σ > 0 Earlier we saw the conditions for both having negative real part: traceJ(k²) < 0 and det J(k²) > 0 Now, one of the conditions must be broken now. (we need something unstable in time; meaning temporal eigenvalue with pos real part so exponential is growing)

homogenous SS constant in time and space NO diffusion A* Jacobian @ equilibrium stability (eigenvalues with Re < 0) trace < 0 det > 0 with diffusion J(k²) = A* -k² [D₁ 0] ] [0 D₂] extra term linearisation changes coming from diffusion assuming spacial term is an eigenfunction of the laplacian giving a family of jacobians for each k instability (at least one eig with Re > 0) trace <0 det<0 (trace negative with and without diffusion stays the same so only det changes det condition broken)

two populations: Linear R domain Consider two types of particles (activator / inhibitor), diffusing in the same domain and interacting Turing (diffusion-driven) instability happens when: 1 without diffusion, the system admits a stable spatially-uniform coexistence steady-state as soon as add diffusion: 2 with diffusion, the uniform steady state becomes unstable, and spatial patterns may appear

Gone through