Dynamical Systems in Neuroscience:

Solutions to Exercises, Chap. 6 4256. The Jacobian matrix at the equilibrium (u, v) = (0, 0) has the form( )b −1L =.1 bIt has eigenvalues b±i. Therefore, the loss of stability occurs at b = 0, and the non-hyperbolicityand transversality conditions are satisfied. Since the model can be reduced to the polarcoordinatesystem (see Ex. 4), and a ≠ 0, the non-degeneracy condition is also satisfied, andthe system undergoes an Andronov-Hopf bifurcation.7. Since(cr + ar 3 ) ′ r = c + 3ar 2 = c + 3a|c/a| =the limit cycle is stable when a < 0{c + 3|c| when a > 0,c − 3|c| when a < 0,8. The sequence of bifurcations is similar to that of the RS neuron in Fig. 8.15. The restingstate is a globally asymptotically stable equilibrium for I < 5.64. At this value a stable(spiking) limit cycle appears via a big saddle homoclinic orbit bifurcation. At I = 5.8 asmall-amplitude unstable limit cycle is born via another saddle homoclinic orbit bifurcation.This cycle shrinks to the equilibrium and makes it lose stability via subcritical Andronov-Hopfbifurcation at I = 6.5. This unstable focus becomes an unstable node when I increases, andthen it coalesces with the saddle (at I = 7.3) and disappears. Notice that there is a saddle-nodebifurcation according to the I-V relation, but it corresponds to the disappearance of an unstableequilibrium.9. The Jacobian matrix of partial derivatives has the form( )−I′L = V (V, x) −I x(V, ′ x)x ′ ,∞(V )/τ(V ) −1/τ(V )so thatandThe characteristic equationhas two solutionstr L = −{I ′ V (V, x) + 1/τ(V )}det L = {I ′ V (V, x) + I ′ x(V, x)x ′ ∞(V )}/τ(V ) = I ′ ∞(V )/τ(V ) .which might be complex-conjugate.10. Let z = re ϕi , thenλ 2 − λ tr L + det L = 0c{ }} { { √ }} {(tr L)/2 ± {(tr L)/2}2 − det Lωr ′ = ar + r 3 − r 5 ,ϕ ′ = ω .Any limit cycle is an equilibrium of the amplitude equation, i.e.,a + r 2 − r 4 = 0 .The system undergoes fold limit cycle bifurcation when the amplitude equation undergoes asaddle-node bifurcation, i.e., whena + 3r 2 − 5r 4 = 0(check the non-degeneracy and transversality conditions). The two equations have the nontrivialsolution (a, r) = (−1/4, 1/ √ 2).