Introduction to the Modeling and Analysis of Complex Systems

290 CHAPTER 14. CONTINUOUS FIELD MODELS II: ANALYSISLet’s test how well this prediction applies to actual dynamics of Turing models. In theprevious chapter, we used (a, b, c, d) = (1, −1, 2, −1.5) and (D u , D v ) = (10 −4 , 6 × 10 −4 )to generate the simulation result shown in Fig. 13.17. With these parameter settings,det(A) = −1.5 − (−2) = 0.5 > 0 and Tr(A) = −0.5 < 0, so the system would be stable ifthere were no diffusion terms. However,aD v + dD u = 6 × 10 −4 − 1.5 × 10 −4 = 4.5 × 10 −4 , (14.108)2 √ D u D v det(A) = 2 √ 10 −4 × 6 × 10 −4 × 0.5 = 2 × 10 −4√ 3 ≈ 3.464 × 10 −4 , (14.109)therefore inequality (14.107) holds. This indicates that the homogeneous equilibrium statemust be unstable and non-homogeneous spatial patterns should arise, which you canactually see in Fig. 13.17. As briefly mentioned in Section 13.6, this is called the diffusioninducedinstability. It is quite a counter-intuitive phenomenon, because diffusion is usuallyconsidered a process where a non-homogeneous structure is being destroyed by randommotion. But here, the system is stable at its homogeneous state without diffusion, butit can spontaneously create non-homogeneous structures with diffusion. This is a reallynice example of how mind-boggling the behavior of complex systems can be sometimes.Exercise 14.6 Conduct a linear stability analysis of the spatially extendedpredator-prey model, around its non-zero homogeneous equilibrium state, and discussthe results:∂r∂t = ar − brf + D r∇ 2 r (14.110)∂f∂t = −cf − drf + D f∇ 2 f (14.111)Assume that all model parameters are positive.Exercise 14.7 Conduct a linear stability analysis of the Gray-Scott model, aroundits homogeneous equilibrium state (u eq , v eq ) = (1, 0), and discuss the results:∂u∂t = F (1 − u) − uv2 + D u ∇ 2 u (14.112)∂v∂t = −(F + k)v + uv2 + D v ∇ 2 v (14.113)Again, assume that all model parameters are positive.