Dynamical Systems in Neuroscience:

Solutions to Exercises, Chap. 6 423-30membrane voltage, V (mV)-40-50-60-70(-24, -41)(0.45, -63)-80-30 -20 -10 0 10injected dc-current, II Na,t -modelmembrane voltage, V (mV)0I A -model-20-40(10.75, -19)(12.7, -42)-600 5 10 15injected dc-current, IFigure 10.27: The saddle-node bifurcation diagrams of the I Na,t - and I A -minimal models.3. The curvesandare depicted in Fig. 10.27.I = g L (V − E L ) + ḡ Na m 3 ∞(V )h ∞ (V )(V − E Na )I = g L (V − E L ) + ḡ A m ∞ (V ) h ∞ (V )(V − E K )4. g is not an absolute conductance, but is taken relative to the conductance at resting state.Negative values occur because the initial holding voltage value in the voltage-clamp experimentdescribed in Fig. 5.22a corresponds to the resting potential, at which the K + conductance ispartially activated. Indeed, in the I Na,p +I K -model the K + gating variable n ≈ 0.04, hencethe K + conductance is approximately 0.4 (because ḡ K = 10). According to the procedure,this value corresponds to g = 0. Any small decrease in conductance would result in negativevalues of g. If the initial holding voltage were very negative, say below −100 mV, then the slowconductance g would have non-negative values in the relevant voltage range (above −100 mV).5. The curve I slow (V ) defines slow changes of the membrane voltage. The curve I −I fast (V ) definesfast changes. Its middle part, which has positive slope, is unstable. If the I-V curves intersectin the middle part, the equilibrium is unstable, and the system exhibits periodic spiking: Thevoltage slides down slowly along the left branch of the fast I-V curve toward the slow I-V curveuntil it reaches the left knee, and then it jumps quickly to the right branch. After the jump,the voltage slides up slowly along the right branch until it reaches the right knee, and then itquickly jumps to the left branch along the straight line that connects the knee and the point(E K , 0) (see also previous exercise). Notice that the direction of the jump is not horizontal,as in relaxation oscillators, but along a sloped line. On that line the slow conductance g isconstant, but the slow current I slow (V ) = g(V − E K ) changes fast because the driving forceV − E K changes fast. When the I-V curves intersect at the stable point (negative slope ofI − I fast (V )), the voltage variable may produce a single action potential, and then slides slowlytoward the intersection, which is a stable equilibrium.Solutions to Chapter 61. There are two equilibria: x = 0 and x = b. The stability is determined by the sign of thederivativeλ = (x(b − x)) ′ x = b − 2xat the equilibrium. Since λ = b when x = 0, this equilibrium is stable (unstable) when b < 0(b > 0). Since λ = −b when x = b, this equilibrium is unstable (stable) when b < 0 (b > 0).