Dynamical Systems in Neuroscience:

Solutions to Exercises, Chap. 10 51716. Winfree approach: Using results of Ex. 4,at A = 0.Z(ϑ) =∂∂A acos 1 + A cos ϑ√ = − sin ϑ1 + 2A cos ϑ + A2Kuramoto approach: Since grad ϑ(x) is orthogonal to the contour line of the function ϑ(x) atx, i.e., the isochron of x, and the results of Ex. 1 that isochrons are radial, we get grad (ϑ) =(− sin ϑ, cos ϑ) using purely geometrical considerations. Since p(t) is real, we need to keep onlythe first component.Malkin approach: Let us work in the complex domain. On the circle z(t) = e it we get Df = i.Since Df ⊤ is equivalent to complex-conjugation in the complex domain, we get ˙Q = i · Q,which has the solution Q(t) = Ce it . The free constant C = i is found from the normalizationcondition Q(0) ∗ i = 1, where ∗ means complex-conjugate.Alternatively, on the circle z(t) = e it , we have f(z(t)) = f(e it ) = ie it . From the normalizationcondition Q(t) ∗ f(z(t)) = 1 we find Q(t) = ie it = − sin ϑ + i cos ϑ.17. Rescaling the state variable z = √ bu and the time, τ = ɛt, we obtain the reduced systemu ′ = (1 + i)u + (−1 + di)u|u| 2 + εp(t) .We can apply the theory only when ε is small. That is, the theory is guaranteed to work in avery weak limit ɛ ≪ b √ b ≪ 1. As it is often the case, numerical simulations suggest that thetheory works well outside the guaranteed interval. Substituting u = re iθ into this equation,r ′ e iθ + re iθ iθ ′ = (1 + i)re iθ + (−1 + di)r 3 e iθ + εp(t) ,dividing by e iθ and separating real and imaginary terms, we represent the oscillator in polarcoordinatesr ′ = r − r 3 + ε Re p(t)e −iθθ ′ = 1 + dr 2 + ε Im r −1 p(t)e −iθ .When ε = 0, this system has a limit cycle attractor r(t) = 1 and θ(t) = (1 + d)t, provided thatd ≠ −1. On the attractor, the solution to Malkin’s adjoint equation (10.10),(Q ′ −2 0= −2d 0) ⊤ (0Q with Q(t) ·1 + d)= 1 ,is Q(t) = (d, 1)/(1+d). Indeed, the normalization condition results in Q 2 (t) = 1/(1+d). Hence,unique periodic solution of the first equation, Q ′ 1 = 2Q 1 − 2d/(1 + d), is Q 1 (t) = d/(1 + d). Onecan also use Kuramoto’s approach and the results of Ex. 2. The corresponding phase model,ϑ ′ = 1 + ε{d Re p(t)e −i(1+d)ϑ + Im p(t)e −i(1+d)ϑ }/(d + 1) ,can be simplified via θ = (1 + d)ϑ (notice the font difference) to get the result.18. (Delayed coupling) Let ϑ(t) = t + ϕ(τ), where τ = εt is the slow time. Since ϑ(t − d) =t − d + ϕ(τ − εd) = t − d + ϕ(τ) + O(ε), we have x j (ϑ i (t − d ij )) = x j (t − d ij + ϕ(τ)) so thatwe can proceed as in Sect. 10.2.5 except that there is an extra term, −d ij , in (10.16). See alsoIzhikevich (1998).