Pattern formation Turing instability

Pattern formation Turing instability Tomáš Vejchodský Centre for Mathematical Biology Mathematical Institute Summer school, Prague, 6 8 August, 213

Outline Motivation Turing instability general conditions Schnakenberg system Turing conditions Schnakenberg system particular example

Motivation

Turing instability Reaction-diffusion system: du dt = d 1 u + f (u, v) dv dt = d 2 v + g(u, v) Homogeneous steady state: f (u, v ) = g(u, v ) = Linearisation at (u, v ): du dt = d 1 u + A 11 u + A 12 v dv dt = d 2 v + A 21 u + A 22 v Linearisation matrix: ( ) fu f A = v (u, v ) g u g v Turing instability occurs if solution (u, v ) of linearisation is: (a) stable with respect to spatially homogeneous disturbances (b) unstable to spatial disturbances

Turing conditions (a) (u, v ) is stable w.r.t. spatially homogeneous disturbances if (T1) tr A < (T2) det A > (b) (u, v ) is unstable to spatial disturbances if [ ] k > :: λ k = 1 2 b k + bk 2 4h k > h k < where b k = (d 1 + d 2 )µ k tr A h k = d 1 d 2 µ 2 k (d 2f u + d 1 g v )µ k + det A µ k... eigenvalue of Laplacian Remark 1: (b) (T3) d 2 f u + d 1 g v > (T4) 4d 1 d 2 det A < (d 2 f u + d 1 g v ) 2

Turing conditions (a) (u, v ) is stable w.r.t. spatially homogeneous disturbances if (T1) tr A < (T2) det A > (b) (u, v ) is unstable to spatial disturbances if [ ] k > :: λ k = 1 2 b k + bk 2 4h k > h k < where b k = (d 1 + d 2 )µ k tr A h k = d 1 d 2 µ 2 k (d 2f u + d 1 g v )µ k + det A µ k... eigenvalue of Laplacian Remark 2: w k (x, t) = e λ kt z k (x)u k is an unstable solution of linearisation. [A µ k D]U k = λ k U k

Schnakenberg model 2A + B k 1 3A k 2,k 3 A k 4 B Notation: u, v... number of molecules of A,B Reaction-diffusion system: Stationary state: u s = k 4 + k 2 k 3, v s = k 4 k 1 u 2 s du dt = d 1 u + k 1 u 2 v + k 2 k 3 u, dv dt = d 2 v k 1 u 2 v + k 4

Schnakenberg model shifting the system Shifted variables: u = u + u s, v = v + u s Shifted system: where α = k 3 k 4 k 2 du dt = d 1 u + αu + βv + f (u, v) dv dt = d 2 v γu βv f (u, v), β = k 1 k 4 + k 2 k3 2 f (u, v) = k 1 u 2 v + 2k 1 u s uv + k 1 v s u 2 (k 4 + k 2 ) 2, γ = 2k 3k 4 k 4 + k 2

Schnakenberg Turing conditions k 4 k 2 α = k 3, β = k 1 k 4 + k 2 k 2 3 (k 4 + k 2 ) 2, γ = 2k 3k 4 k 4 + k 2 Linearisation matrix: ( α β A = γ β ) Turing conditions: (T1) tr A = α β < (T2) det A = k 1 (k 4 + k 2 ) 2 /k 3 > (T3) d 2 α d 1 β > d 2 d 1 > β α > 1 (T4) 4d 1 d 2 det A < (d 2 f u + d 1 g v ) 2

Schnakenberg particular example Ω = (, 1) k 1 = 1 6, k 2 = 1, k 3 =.2, k 4 = 2 [sec 1 ] d 1 = 1 5, d 2 = 1 3 [mm 2 sec 1 ] Linearisation matrix: A = 1 ( 1 4 3 4 ) Turing conditions: (T1) tr A =.3 < (T2) det A = 1/125 > (T3) d 2 d 1 > 4 d 2 > 4d 1 1 3 > 4 1 5 (T4) d 2 > d 1 25 8d 1 1 3 > 1 3 4 1 5 25 8 1 5

Schnakenberg deterministic du dt = d 1u + αu + βv + f (u, v) in (, 1), u () = u (1) = dv dt = d 2v γu βv f (u, v) in (, 1), v () = v (1) = 5 3 number of B molecules 8 6 4 2.2.4.6.8 1.2.4.6.8 1

Schnakenberg stochastic (compartments) 2A + B k 1 3A k 2,k 3 A k 4 B 5 time=6 min stochastic steady state time=6 min stochastic steady state 3 number of B molecules 8 6 4 2.2.4.6.8 1.2.4.6.8 1

Schnakenberg stochastic vs. deterministic 5 3 time=6 min stochastic deterministic steady state number of B molecules 8 6 4 2 time=6 min stochastic deterministic steady state.2.4.6.8 1.2.4.6.8 1

Schnakenberg dependence on initial cond. 5 5 3 3.2.4.6.8 1 5 3.2.4.6.8 1.2.4.6.8 1 5 3.2.4.6.8 1

Schnakenberg dependence on initial cond. 5 3.2.4.6.8 1 3 5 3 5 5 3.2.4.6 3.8 1.2.4.6.8 1 5.2.4.6.8 1.2.4.6.8 1

Schnakenberg sharper conditions (b) (u, v ) is unstable to spatial disturbances if k > : [ ] λ k = 1 2 b k + bk 2 4h k >... dispersion relation h k = d 1 d 2 µ 2 k (d 2f u + d 1 g v )µ k + det A <... hyperbolas Solutions of linearization: where z k = µ kz k in (, 1) z k () = z k (1) = } ( ) u (x, t) = w v k (x, t) = e λkt z k (x)u k, { zk = cos(kπx), k =, 1, 2,... µ k = k 2 π 2 ( ) d1 [A µ k D]U k = λ k U k, D = d 2 General solution of linearization: w(x, t) = c k w k (x, t) k=

Schnakenberg Hyperbolas 1 1 k=9 k=4 5 x 1 3 k=4 Dispersion relation k=9 d2 [mm 2 sec 1 ] 1 2 1 3 λ k 5 1 1 4 1 6 1 5 1 4 1 3 1 2 d1 [mm 2 sec 1 ] 15 2 5 15 µ k d 1 = 1 5 d 2 = 1 3 [mm 2 sec 1 ] Point (d 1, d 2 ) lies inside hyperbolas for k = 4, 5,..., 9. λ 6 > λ 5 > λ 7 > λ 4 > λ 8 > λ 9

Schnakenberg Laplace eigenmodes cos(kπx) for k=6, (λ =3.44e 3) cos(kπx) for k=4, (λ =2.43e 3) 6 4 1 1 1 1.2.4.6.8 1.2.4.6.8 1 x x 1 cos(kπx) for k=5, (λ 5 =3.4e 3) 1 cos(kπx) for k=8, (λ 8 =1.9e 3) 1 1.2.4.6.8 1.2.4.6.8 1 x x 1 cos(kπx) for k=7, (λ 7 =2.88e 3) 1 cos(kπx) for k=9, (λ 9 =5.77e 4) 1 1.2.4.6.8 1.2.4.6.8 1 x x

Schnakenberg dependence on initial cond. k=6 time=9 min.5 1 k=5 time=9 min.5 1 number of B molecules number of B molecules 8 6 4 8 6 4 k=6 time=9 min.5 1 k=5 time=9 min.5 1

Schnakenberg 2D sharper conditions (b) (u, v ) is unstable to spatial disturbances if k, l > : [ ] λ k,l = 1 2 b k,l + bk,l 2 4h k,l >... dispersion relation h k,l = d 1 d 2 µ 2 k,l (d 2f u + d 1 g v )µ k,l + det A <... hyperbolas ( ) u Solutions of linearization: (x, t) = w v k,l (x, t) = e λk,lt z k,l (x)u k,l, where Ω = (, 1) 2, k, l =, 1, 2,... } { z k,l = µ k,l z k,l in Ω zk,l = cos(kπx) cos(lπx) z k,l / n = on Ω µ k,l = (k 2 + l 2 )π 2 ( ) d1 [A µ k,l D]U k,l = λ k,l U k,l, D = d 2 General solution of linearization: w(x, t) = c k,l w k,l (x, t) k,l=

Schnakenberg 2D Laplace eigenmode:

Schnakenberg 2D 1 1 Hyperbolas x 1 3 Dispersion relation d2 [mm 2 sec 1 ] 1 2 1 3 λ k 5 1 15 1 4 1 6 1 5 1 4 1 3 d1 [mm 2 sec 1 ] 2 3 µ k d 1 = 3.55 1 5 d 2 = 1.225 1 3 [mm 2 sec 1 ] Point (d 1, d 2 ) lies inside the hyperbola for k = 4, l =.

Schnakenberg 2D Hyperbolas x 1 3 Dispersion relation 1 2.89 d2 [mm 2 sec 1 ] 1 2.91 λ k 5 1 15 1 2.93 1 4.55 1 4.53 1 4.51 d1 [mm 2 sec 1 ] 2 3 µ k d 1 = 3.55 1 5 d 2 = 1.225 1 3 [mm 2 sec 1 ] Point (d 1, d 2 ) lies inside the hyperbola for k = 4, l =.

Schnakenberg 2D stationary state Number of A molecules Number of B molecules

Summary General conditions for Turing instability Schnakenberg system in 1D stochastically and deterministically Schnakenberg system in 2D

Acknowledgement I am thankful to Radek Erban and Philip K. Maini for their support and fruitful discussions about the topics presented during this summer school. Marie Curie Fellowship, StochDetBioModel The research leading to these results has received funding from the People Programme (Marie Curie Actions) of the European Union s Seventh Framework Programme (FP7/7-213) under REA grant agreement no. 3288.

Thank you for your attention Tomáš Vejchodský Centre for Mathematical Biology Mathematical Institute Summer school, Prague, 6 8 August, 213