Dynamical Systems in Neuroscience:

420 Solutions to Exercises, Chap. 4a=-1.2a=-0.8y10.50-0.5-2 -1 0 1 2 -2 -1 0 1 2xxFigure 10.23: Nullclines and phase portrait of the Bonhoeffer–van der Pol oscillator (b = 0.05 andc = 0).-1shown in Fig. 10.23. They intersect at the point x = a, y = a − a 3 /3. The Jacobian matrix atthe equilibrium (a, a − a 3 /3) has the form( )1 − a2−1L =.b 0Since tr L = 1 − a 2 and det L = b > 0, the equilibrium is a stable (unstable) focus when |a| > 1(|a| < 1), as we illustrate in Fig. 10.23.7. (Hindmarsh-Rose spiking neuron) The Jacobian matrix at the equilibrium (¯x, ȳ) is( )f′−1L =g ′ ,−1thereforetr L = f ′ − 1 and det L = −f ′ + g ′ .The equilibrium is a saddle (det L < 0) when g ′ < f ′ , i.e., in the region below the diagonal inFig. 10.24. When g ′ > f ′ , the equilibrium is stable (tr L < 0) when f ′ < 1, which is the lefthalf-plane in Fig. 10.24. Using the classification in Fig. 4.15, we conclude that it is focus when(f ′ − 1) 2 − 4(g ′ − f ′ ) < 0, i.e., wheng ′ > 1 4 (f ′ + 1) 2which is the upper part of the parabola in Fig. 10.24.8. (I K -model) The steady-state I-V relation of the I K -model is monotone, hence it has a uniqueequilibrium, which we denote here as ( ¯V , ¯m) ∈ R 2 , where ¯V > E K and ¯m = m ∞ ( ¯V ). TheJacobian at the equilibrium has the form(−(gL + ḡL =K ¯m 4 )/C −4ḡ K ¯m 3 ( ¯V)− E K )/Cm ′ ∞( ¯V )/τ( ¯V ) −1/τ( ¯V ,)with the signsL =(− −+ −Obviously, det L > 0 and tr L < 0, hence the equilibrium (focus or node) is always stable.).