Dynamical Systems in Neuroscience:

424 Solutions to Exercises, Chap. 62. (a) The systemẋ = bx 2 , b ≠ 0cannot exhibit saddle-node bifurcation: It has one equilibrium for any non-zero b, or aninfinite number of equilibria when b = 0. The equilibrium x = 0 is non-hyperbolic and thenon-degeneracy condition is satisfied (a = b ≠ 0). However, the transversality conditionis not satisfied at the equilibrium x = 0. Another example is ẋ = b 2 + x 2 .(b) The systemẋ = b − x 3has a single stable equilibrium for any b. However, the point x = 0 is non-hyperbolic whenb = 0, and the transversality condition is also satisfied. The non-degeneracy condition isviolated, though.3. It is easy to check (by differentiating) that√c(b − bsn )V (t) = √ tan( √ ac(b − b sn )t)ais a solution to the system. Since tan(−π/2) = −∞ and tan(+π/2) = +∞, it takesfor the solution to go from −∞ to +∞.T =π√ac(b − bsn )4. The first system can be transformed into the second one if we use complex coordinates z = u+iv.To obtain the third system, we use polar coordinatesso thatre iϕ = z = u + iv ∈ C ,ż(c(b)+iω(b))z+(a+id)z|z| 2{ }} { { }} {ṙe iϕ + re iϕ i ˙ϕ = (c(b) + iω(b))re iϕ + (a + id)r 3 e iϕ .Next, we divide both sides of this equation by e iϕ and separate the real and imaginary partsto obtain{ṙ − c(b)r − ar 3 } + ir{ ˙ϕ − ω(b) − dr 2 } = 0 ,which we can write in the polar-coordinates form.5. (a) The equilibrium r = 0 of the systemṙ = br 3 ,˙ϕ = 1 ,has a pair of complex-conjugate eigenvalues ±i for any b, and the non-degeneracy conditionis satisfied for any b ≠ 0. However, the transversality condition is violated, andthe system does not exhibit Andronov-Hopf bifurcation (no limit cycle exists near theequilibrium).(b) The equilibrium r = 0 for b = 0ṙ = br ,˙ϕ = 1 ,has a pair of complex-conjugate eigenvalues ±i and the transversality condition is satisfied.However, the bifurcation is not of the Andronov-Hopf type because no limit cycleexists near the equilibrium for any b.