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Morphology of Experimental and Simulated Turing Patterns

Morphology of Experimental and Simulated Turing Patterns

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withχ u = 2γ u sin 2 k x∆2+ 2γ u sin 2 k y∆2(3.26)<strong>and</strong>χ v = 2γ v sin 2 k x∆2 + 2γ v sin 2 k y∆2 . (3.27)Eq. (3.25) can be written as a matrix equation:( ) ( )ξ − αu −∆t f v δu0· = 0 (3.28)−∆t g u ξ − α v δv 0withα u = 1 − 2χ u + ∆t f u <strong>and</strong> α v = 1 − 2χ v + ∆t g v , (3.29)which has nontrivial solutions only if the determinant is zero, i.e.(ξ − α u ) (ξ − α v ) − (∆t) 2 f v g u = 0. (3.30)So the condition |ξ| < 1 has to be fullfilled for the solutionsξ ± = 1 [](α u + α v ) ±√(α u + α v ) 2 + 4(∆t) 2 f v g u . (3.31)2Obviously the stability depends on ∆t, ∆ <strong>and</strong> the linearized reaction terms f u ,f v<strong>and</strong> g u ,g v . Consequently the time <strong>and</strong> spatial discretization have to be adjusted forthe considered part <strong>of</strong> the parameter space, when a system is solved numerically.For the numerical solutions in chapter 4 the fixed discretization parameters where∆ = 1 or ∆ = 0.5 with ∆t = 0.01 or ∆t = 0.001 respectively. This ensures |ξ ± | < 1for the analyzed parameter space.3.3.2 Semi-implicit Crank-Nicolson schemeWe linearize the difference equation Eq. (3.11) <strong>and</strong> Eq. (3.12) about the homogeneoussteady state using Eq. (3.20) to obtain a linearized difference equation for δu(1 + 2γ u ) δu n+1j,l− γ ()uδu n+1j,l + δu n+12j−1,l + δun+1 j,l+1 + δun+1 j,l−1= (1 − 2γ u ) δu n j,l + γ u (δun2 j ,l + δu n j−1,l + δun j,l+1 + ) ( δun j,l−1 + ∆t fu δu n j,l + f vδvj,l)n(3.32)<strong>and</strong> similar for δv :(1 + 2γ v ) δv n+1j,l− γ v2= (1 − 2γ v ) δv n j,l + γ v2)δvj n+1 ,l + δv n+1j−1,l + δvn+1 j,l+1 + δvn+1 j,l−1(δvnj ,l + δvj−1,l n + δvn j,l+1 + ) ( δvn j,l−1 + ∆t gv δvj,l n + g uδu n j,l().(3.33)33

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