Dynamical Systems in Neuroscience:

Solutions to Exercises, Chap. 5 4216543g'stable focusf'=1unstable focusg'=(f'+1) 2unstable nodef'=g'210stable nodesaddle-1-2 -1 0 1 2 3 4f'Figure 10.24: Stability diagram ofthe Hindmarsh-Rose spiking neuronmodel; see Ex. 7.9. (I h -model) The steady-state I-V relation of the I h -model is monotone, hence it has a uniqueequilibrium denote here as ( ¯V , ¯h) ∈ R 2 , where ¯V < E h and ¯h = h ∞ ( ¯V ). The Jacobian at theequilibrium has the form(−(gL + ḡL =h¯h)/C −ḡh ( ¯V)− E h )/Ch ′ ∞( ¯V )/τ( ¯V ) −1/τ( ¯V ,)with the signs( )− +L =.− −Obviously, det L > 0 and tr L < 0, hence the equilibrium is always stable.10. (Bendixson’s criterion) The divergence of the vector field of the I K -model∂f(x,y)/∂x{ }} {(−g L − ḡ K m 4 )/C +∂g(x,y)/∂y{ }} {−1/τ(V )is always negative, hence the model cannot have a periodic orbit. Therefore, it cannot havesustained oscillations.11. The x-nullcline is y = a + x 2 and the y-nullcline is y = bx/c, as in Fig. 10.25. The equilibria(intersections of the nullclines) are¯x = b/c ± √ (b/c) 2 − 4a2, ȳ = b¯x/c ,provided that a < 1 4 (b/c)2 . The Jacobian matrix at (¯x, ȳ) has the formwith tr L = 2¯x − c and( )2¯x −1L =b −cdet L = −2¯xc + b = ∓ √ b 2 − 4ac 2 .Thus, the right equilibrium (i.e., (b/c + √ (b/c) 2 − 4a)/2) is always a saddle and the left equilibrium(i.e., (b/c − √ (b/c) 2 − 4a)/2) is always a focus or a node. It is always stable when itlies on the left branch of the parabola y = a + x 2 (i.e., when ¯x < 0), and can also be stable onthe right branch if it is not too far from the parabola knee (i.e., if ¯x < c/2); see Fig. 10.25.