Homework 8.1 Solutions
Math 5110/6830
1. (a) Current:
IX
= ḡX ω(V − VX )
(b) It will be hyperpolarized (more negative) since ions are flowing into the cell.
(c) VX is −100mV < −65mV.
(d) Current-balance equation:
dV
dt
C
= −ḡX ω(V − VX ) + Iapp
(e) Equilibria point:
V∗
=
Iapp
+ VX
ḡX
For stability, let
f (V )
=
1
[−ḡX ω(V − VX ) + Iapp ]
C
then
0
f (V )
=
−
ḡX ω
C
So, V ∗ is stable.
(f) No, the parameters are all positive & the derivative is always negative.
(g) Equation:
C
dV
dt
= − [ḡX ωX (V − VX ) + ḡY ωY (V − VY )] + Iapp
Steady state voltage:
V∗
=
ḡX ωX VX + ḡY ωY VY
ḡX ωX + ḡY ωY
Homework 8.2 Solutions
Math 5110/6830
1. Recall that Iss (V ∗ ) = Iion (V ∗ , ω ∗ )
− 1c (ḡca m∞ + ḡk ω ∗ + ḡL ) − 1c (ḡk V ∗ − ḡk Vk )
J(V , ω ) =
φ ωτ∞
−φ τ1ω
ω
1
det(J) = φ [ḡca m∞ + ḡk ω∞ + ḡL + ḡK ω∞ (V ∗ − VK )]
τω
∗
∗
ss
If det(J)=0, then ḡca m∞ + ḡk ω∞ + ḡL + ḡK ω∞ (V ∗ − VK ) = 0. Note that dI
ḡL . The
dV = ḡca m∞ + ḡk ω∞ +
dI
dIss
∗
only way to satisfy this is to have V = VK and dV = 0. In addition, if Iss is monotone, then dVss > 0
for all V and det(J) 6= 0.
2. (a) Plot of Iss (V ):
1800
1600
1400
1200
Iss
1000
800
600
400
200
0
−200
−100
−80
−60
−40
−20
0
V
20
40
60
80
100
(b) See graph in part (c).
(c) Nullclines with Iapp = 0:
0.5
0.4
0.3
w
0.2
0.1
0
−0.1
V−nullcline, Iapp=0
V−nullcline, Iapp=−50
w−nullcline
−0.2
0
20
40
60
80
V
100
120
140
(d) If Iapp is held at −50 for sufficiently long, then the solution will approach the new steady state (shown
in the figure below); however, once Iapp is turned off then it returns back to the original steady state.
3. (a) Steady states for different values of Iapp :
(b) There is clearly a bifurcation happening since we lose/gain equilibria points. As you can see in the
following figure, we go from having one equilibria point (Iapp = 50), to having two (not shown here),
to having three (Iapp = 0), to again having two (the not shown), and finally one (Iapp = −50).
0.8
0.7
0.6
0.5
0.4
w
0.3
0.2
0.1
0
−0.1
Iapp=0
Iapp=−50
Iapp=50
w−nullcline
−0.2
0
20
40
60
80
V
100
120
140