25 turing instability Flashcards
morphogenesis
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)
Diffusion-driven instability
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
The reaction-diffusion equations
Consider two populations, N₁ and N₂ , in a one-dimensional
domain Ω ⊂ IR satisfying
model
what sols we look at?
∂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
The reaction-diffusion equations
Consider two populations, N₁ and N₂ , in a one-dimensional
domain Ω ⊂ IR satisfying:
∂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
what sols we look at?
uniform coexistence steady state
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)
Reminder: steady states and stability in ODEs
eigen values stability?
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!
Diffusion-driven instability
ss requirments>
we want a uniform equilibrium (homogenous) that is stable without diffusion
we want the same equilibrium to be unstable with diffusion
Diffusion-driven instability
Stability of the uniform steady state without 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]
Diffusion-driven instability
Stability of the uniform steady state without diffusion
eigenvalues
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 ]
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 }
Diffusion-driven instability
uniform steady state
what do we assume?
What happens to the stability equilibrium with diffusion?
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
Diffusion-driven instability
Stability of the uniform steady state with diffusion
∂N₁/∂t
=D₁∂²N₁/∂x² + f(N₁, N₂)
∂N₂/∂t
=D₂∂²N₂/∂x² + g (N₁, N₂)
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]
explaining:
Diffusion-driven instability
Stability of the uniform steady state with diffusion
∂N₁/∂t
=D₁∂²N₁/∂x² + f(N₁, N₂)
∂N₂/∂t
=D₂∂²N₂/∂x² + g (N₁, N₂)
linear PDE started with
already linear for D term
reaction terms f and g need to be linearised
∂N₁/∂t
=D₁∂²N₁/∂x² + f(N₁, N₂)
∂N₂/∂t
=D₂∂²N₂/∂x² + g (N₁, N₂)
so;utions form
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
det[σI − J(k²]= 0
THIS IS THE NEW JACOBIAN OF THE DIFFUSION EQ GIVEN DIFFUSION IS HAPPENING
∂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]
given diffusion is happening what requirements on new eigenvalue
σ
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)
comparing the situations with diffusion and now diffusion (D_1, D_2 =0)
jacobian (linearise around same homogenous equilibrium)
we look for
conditions
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)
