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Cours : Dynamique Non-Lin eaire Laurette TUCKERMAN laurette@pmmh.espci.fr VII. Reaction-Diffusion Equations: 1. Excitability 2. Turing patterns 3. Lyapunov functionals 4. Spatial analysis and fronts Reaction-Diffusion Systems t u i =


  1. Cours : Dynamique Non-Lin´ eaire Laurette TUCKERMAN laurette@pmmh.espci.fr VII. Reaction-Diffusion Equations: 1. Excitability 2. Turing patterns 3. Lyapunov functionals 4. Spatial analysis and fronts

  2. Reaction-Diffusion Systems ∂ t u i = f i ( u 1 , u 2 , . . . ) + D i ∆ u i � �� � � �� � diffusion reaction Reactions f i couple different species u i at same location Diffusivity D i couples same species u i at different locations Describe oscillating chemical reactions, such as famous Belousov-Zhabotinskii reaction, discovered by two Soviet scientists in 1950s-1960s. Also describe phenomena in –biology (population biology, epidemiology, neurosciences) –social sciences (economics, demography) –physics

  3. Two species Spatially homogeneous ∂ t u = f ( u, v ) + D u ∆ u ∂ t u = f ( u, v ) ∂ t v = g ( u, v ) + D v ∆ v ∂ t v = g ( u, v ) FitzHugh-Nagumo model Barkley model � � f ( u, v ) = 1 u − v + b f ( u, v ) = u − u 3 / 3 − v + I ǫ u (1 − u ) a g ( u, v ) = 0 . 08 ( u + 0 . 7 − 0 . 8 v ) g ( u, v ) = u − v u -nullclines f ( u, v ) = 0 , v -nullclines g ( u, v ) = 0 , • steady states � f u f v � stable if eigenvalues of have negative real parts g u g v

  4. Excitability � � f ( u, v ) = 1 u − v + b ǫ u (1 − u ) g ( u, v ) = u − v a O ( ǫ − 1 ) ∂ t u = f = 0 separates ← − and − → ∂ t v = g = 0 separates ↑ and ↓ O (1) u = 1 excited phase u = 0 v ∼ 1 refractory phase u = 0 v ≪ 1 excitable phase u = ( v + b ) /a excitation threshold

  5. Waves in Excitable Medium Spatial variation + diffusion + excitability = ⇒ propagating waves Excitable media in physiology: –neurons –cardiac tissue (the heart) Pacemaker periodically emits electrical signals, propagated to rest of heart

  6. Simulations from Barkley model , Scholarpedia Spiral waves in 2D Spiral waves in 3D

  7. Turing patterns Instability of homogeneous solutions (¯ u, ¯ v ) to reaction-diffusion systems � 0 = f (¯ � 0 = f (¯ � � u, ¯ v ) u, ¯ v ) + D u ∆¯ u = ⇒ 0 = g (¯ u, ¯ v ) 0 = g (¯ u, ¯ v ) + D v ∆¯ v What about stability? Does diffusion damp spatial variations? Linear stability analysis: � u ( x, t ) = ¯ � σ ˜ � � ue σt + i k · x v − D u k 2 ˜ u + ˜ u = f u ˜ u + f v ˜ u = ⇒ ve σt + i k · x v − D v k 2 ˜ v ( x, t ) = ¯ v + ˜ σ ˜ v = g u ˜ u + g v ˜ v � f u − D u k 2 f v � f u f v � D u 0 � � � − k 2 M k ≡ = g v − D v k 2 g u g u g v 0 D v If D u = D v ≡ D , then σ k ± = σ 0 ± − k 2 D ≤ σ 0 ± (¯ u, ¯ v ) stable to homogeneous perturbations = ⇒ (¯ u, ¯ v ) stable to inhomogeneous perturbations. Diffusion is stabilizing.

  8. Alan Turing (famous WW II UK cryptologist, founder of computer science) 1952: homogeneous state can be unstable if D u � = D v For instability, need Tr k > 0 or Det k < 0

  9. For instability, need Tr k > 0 or Det k < 0 � Tr 0 = f u + g v < 0 � and Homogeneous stability ⇐ ⇒ Det 0 = f u g v − f v g u > 0 Tr k = f u + g v − ( D u + D v ) k 2 = Tr 0 − ( D u + D v ) k 2 < Tr 0 < 0 So for instability, need Det k < 0 Det k = f u g v − f v g u + D u D v k 4 − ( D v f u + D u g v ) k 2 D u D v k 4 − ( D v f u + D u g v ) k 2 = Det 0 + � �� � � �� � > 0 > 0 , dominates for k ≫ 1 Find negative minimum for intermediate k 2 : � d Det k � = 2 D u D v k 2 0 = ∗ − ( D v f u + D u g v ) � d k 2 � k ∗ ∗ = D v f u + D u g v k 2 = ⇒ need D v f u + D u g v > 0 2 D u D v

  10. Need Det k < 0 at k 2 ∗ = ( D v f u + D u g v ) / (2 D u D v ) : 0 > Det k | k ∗ = Det 0 + D u D v k 4 ∗ − ( D v f u + D u g v ) k 2 ∗ = Det 0 + ( D v f u + D u g v ) 2 − 2( D v f u + D u g v ) 2 4 D u D v 4 D u D v = Det 0 − ( D v f u + D u g v ) 2 4 D u D v 0 > 4 D u D v ( f u g v − f v g u ) − ( D v f u + D u g v ) 2 Collecting the four conditions: Tr 0 = f u + g v < 0 Det 0 = f u g v − f v g u > 0 2 D u D v k 2 ∗ = D v f u + D u g v > 0 4 D u D v Det k | k ∗ = 4 D u D v ( f u g v − f v g u ) − ( D v f u + D u g v ) 2 < 0

  11. Turing patterns were first produced experimentally: –in 1990 by de Kepper et al. at Univ. of Bordeaux –in 1992 by Swinney et al. at Univ. of Texas at Austin Turing pattern in a chlorite- iodide-malonic acid chemical laboratory experiment. From R.D. Vigil, Q. Ouyang & H.L. Swinney, Turing patterns in a simple gel reactor , Physica A 188, 17 (1992) Might be mechanism for: –differentiation within embryos –formation of patterns on animal coats, e.g. zebras and leopards

  12. Lyapunov functionals 1D systems: no limit cycles, usually just convergence to fixed point Generalize to multidimensional variational, potential, or gradient flows: d u du i dt = − ∂ Φ dt = −∇ Φ ⇐ ⇒ ∂u i For gradient flow, Jacobian is Hessian matrix:   ∂ 2 Φ / ( ∂u 1 ∂u 1 ) ∂ 2 Φ / ( ∂u 1 ∂u 2 ) . . . ∂ 2 Φ / ( ∂u 2 ∂u 1 ) ∂ 2 Φ / ( ∂u 2 ∂u 2 ) . . .   H = −   . . . . . . . . . H symmetric = ⇒ no complex eigenvalues = ⇒ no Hopf bifurcations d Φ ∂ Φ du i ∂ Φ ∂ Φ � � = −|∇ Φ | 2 dt = dt = − ∂u i ∂u i ∂u i i i Φ decreases monotonically, either to −∞ or to point where d u /dt = −∇ Φ = 0 = ⇒ no limit cycles

  13. Generalize to reaction-diffusion systems involving potential Φ( u ) : ∂t = −∇ Φ + ∂ 2 u ∂ u on x lo ≤ x ≤ x hi ∂x 2 Boundary conditions: u( x lo ) = u lo u( x hi ) = u hi Dirichlet ∂ u ∂ u or Neumann (homogeneous) ∂x ( x lo ) = 0 ∂x ( x hi ) = 0 Define free energy or Lyapunov functional: � x hi � � � 2 � + 1 ∂ u( x, t )) � � F (u) ≡ dx Φ(u( x, t )) � � 2 � ∂x � � �� � x lo potential energy � �� � kinetic energy Seek quantity analogous to gradient: F (x + dx) = F(x) + ∇ F(x) · dx + O( | dx | ) 2 for all dx The functional derivative δ F /δ u is defined to be such that � x hi dx δ F δ u · δ u + O ( δ u) 2 for every δ u F (u + δ u) = F (u) + x lo

  14. Expand: � 2 � � x hi � � Φ(u + δ u) + 1 ∂ (u + δ u) � � F (u + δ u) = dx � � 2 � ∂x � x lo � 2 � � x hi � � Φ(u) + ∇ Φ(u) · δ u + . . . + 1 ∂x + ∂δ u ∂ u � � = dx ∂x + . . . � � 2 � � x lo � 2 � � x hi � � Φ(u) + 1 ∂ u � � = dx � � 2 � ∂x � x lo � x hi � � ∇ Φ(u) · δ u + ∂ u ∂x · ∂δ u + O ( δ u) 2 + dx ∂x x lo Integrate by parts: � x hi � x hi � ∂ u � x hi dx∂ 2 u dx ∂ u ∂x · ∂δ u ∂x = ∂x · δ u − ∂x 2 · δ u x lo x lo x lo � ∂ u ∂x ( x lo ) = ∂ u ∂x ( x hi ) = 0 for Neumann BCs Surface term vanishes since δ u( x lo ) = δ u( x hi ) = 0 for Dirichlet BCs

  15. � 2 � � x hi � x hi � � � � ∇ Φ(u) − ∂ 2 u ∂ u � � · δ u+ O ( δ u) 2 F (u+ δ u)= dx Φ(u) + + dx � � ∂x 2 � ∂x � x lo x lo The functional derivative δ F /δ u is defined to be such that � x hi dx δ F δ u · δ u + O ( δ u) 2 for every δ u = F (u + δ u) = F (u) + ⇒ x lo � x hi � x hi � � ∇ Φ(u) − ∂ 2 u dx δ F δ u · δ u = dx · δ u ∂x 2 x lo x lo Choosing δ u to be delta function centered on any x and pointing in any vector direction leads to pointwise equality: δ u = ∇ Φ(u) − ∂ 2 u δ F ∂x 2 = − ∂ u ∂t

  16. d F 1 = lim ∆ t [ F ( t + δt ) − F ( t )] dt ∆ t → 0 1 = lim ∆ t [ F (u( t + ∆ t )) − F (u( t ))] ∆ t → 0 � � � � 1 u( t ) + ∂ u = lim F ∂t ∆ t + . . . − F (u( t )) ∆ t ∆ t → 0 � x hi � � 1 dxδ F δ u · ∂ u = lim F (u( t )) + ∂t ∆ t + . . . − F (u( t )) ∆ t ∆ t → 0 x lo �� x hi � 1 dxδ F δ u · ∂ u = lim ∂t ∆ t + . . . ∆ t ∆ t → 0 x lo � x hi � x hi � � dxδ F δ u · ∂ u − ∂ u · ∂ u = ∂t = dx ∂t ∂t x lo x lo � x hi � � 2 ∂ u � � = − dx ≤ 0 � � � ∂t � x lo F decreases so limit cycles cannot occur. Can be applied in higher spatial dimensions via volume integration and Gauss’s Divergence Theorem.

  17. Spatial Analysis and Fronts du + ∂ 2 u ∂u ∂t = − d Φ ∂x 2 Travelling wave solutions: u ( x, t ) = U ( x − ct ) with c = 0 for steady states ξ ≡ x − ct ∂u ∂t ( x, t ) = − c dU dξ ( ξ ) ∂ 2 u ∂x 2 ( x, t ) = d 2 U dξ 2 ( ξ ) Equation obeyed by steady states and travelling waves becomes du + d 2 u d 2 u − c du dξ = − d Φ dξ 2 = d Φ du − c du = ⇒ dξ 2 dξ Analogy between space and time = ⇒ x must be 1D

  18. Spatial analysis or Mechanical analogy d 2 u d ( − Φ) − c du = − dξ 2 du dξ � �� � ���� � �� � “potential gradient” “acceleration” “friction” u ξ position time � � 2 du E ( ξ ) ≡ − Φ + 1 du velocity − Φ potential energy dξ 2 dξ d 2 u − c du acceleration friction dξ 2 dξ � � 2 � � du E = dE d − Φ + 1 ˙ = dξ dξ 2 dξ d 2 u = − d Φ du dξ + du dξ 2 du dξ  � du < 0 if c > 0 � � du � 2 du + d 2 u − d Φ  = dξ = − c = 0 if c = 0 dξ 2 dξ  > 0 if c < 0 � “Increase in energy” � c < 0 ⇐ ⇒ ⇐ ⇒ just leftwards motion “Negative friction”

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