Dynamical Systems in Neuroscience:

518 Solutions to Exercises, Chap. 1019. Let χ = ϕ 2 − ϕ 1 ; then we have˙χ = ω − c sin χ .If |ω/c| ≤ 1, then there are two synchronized states, χ = arcsin (ω/c) and χ = π − arcsin (ω/c),one stable and the other unstable.20. From the theorem by Hoppensteadt and Izhikevich (1997) presented in Sect. 10.3.3 it followsthat Kuramoto’s model is a gradient system when c ij = c ji and ψ ij = −ψ ji . From Ermentrout’stheorem presented in the same section, it follows that the synchronized state ϕ i = ϕ j is stableif, e.g., all ψ ij = 0 and c ij > 0.21. Since the probability density function g(ω) is symmetric, the averaged frequency deviation ofthe network is zero, and rotating the coordinate system, we can make the cluster phase ψ = 0.The network is split into two populations: One oscillatinf with the cluster (|ω| < Kr), therebyforming the cluster, and one drifting in and out of the cluster. The latter does not contributeto the Kuramoto synchronization index, because contributions from different oscillators canceleach other on average. In the limit n → ∞, the sum (10.21) becomes the integral∫∫r = e iϕ(ω) g(ω)d ω ≈ e iϕ(ω) g(ω)d ω .|ω| 0 .(x)25. (Brown et al. 2004) The solution of x ′ = λx with x(0) = x 0 is x(t) = x 0 e λt . The period T =log(∆/x 0 )/λ is found from the condition x(T ) = ∆. Hence, Q(ϑ) = 1/(λx(ϑ)) = 1/(λx 0 e λϑ ) =e λ(T −ϑ) /(∆λ).26. Let us first consider the SNIC case ẋ = 1 + x 2 . The solution staring with x(0) = x reset hasthe form (check by differentiating) x(t) = tan(t + t 0 ), where t 0 = atan x reset . The periodshould be found from the condition tan(T + t 0 ) = +∞, and it is T = π/2 − t 0 . Hence, x(t) =tan(t+π/2−T ) = − cot(t−T ). Now, Q(ϑ) = 1/(1+x(ϑ) 2 ) = 1/(1+cot 2 (ϑ−T )) = sin 2 (ϑ−T ).The homoclinic case ẋ = −1 + x 2 is quite similar. The solution staring with x(0) = x reset > 1has the form (check by differentiating) x(t) = − coth(t + t 0 ), where t 0 = acoth(−x reset ) = −acoth(x reset ). The period is found from the condition − coth(T +t 0 ) = +∞ resulting in T = −t 0 .Hence, x(t) = − coth(t − T ). Finally, Q(ϑ) = 1/(−1 + x(ϑ) 2 ) = 1/(1 + coth 2 (ϑ − T )) =sinh 2 (ϑ − T ).