Mathematics 562
Homework Assignment 1
Due Thursday Feb 4, 2016
Nernst potential, Input resistance, Time-scale difference, and Numerical
simulations using XPPAUT
Question 1: At 10o C, Na+ concentration inside a cell is 80 mM while there is 100 mM outside of it. What is the reversal potential of Na+ across the membrane of this cell? (Absolute
out
temperature is defined as T=degree centigrade+273.16. At 20o C, E = 58
log [C]
mV .)
z
[C]in
Question 2: Consider a passive cable with electrotonic length l and a sealed end at ξ = l.
Suppose that V (0) = V0 . Show that the input resistance at ξ = 0 is
RIN P =
∞
RIN
P
,
tanh(l)
∞
where RIN
P is the input resistance of the same cable when it is semi-infinite and under the
same boundary condition at ξ = 0.
Question 3: Use the ode file provided with this assignment, study numerically the behaviour of the Hodgkin-Huxley equation.
(a) Start from the initial conditions provide in the file, integrate the system for long enough
time, find the steady state (Vs , ms , hs , ns ) accurate up to 5 digits after the decimal
point.
(b) Find all the eigenvalues of the linearized system near the steady state using XPP.
(c) Plot the solution curves using the following initial conditions: (−60, ms , hs , ns ) and
(−55, ms , hs , ns ). Generate a plot (V vs t plot) that contains solution curves for
both cases in the same plot.
(d) Find two integer values of V , say V1 and V2 such that the threshold IC for generating
an action potential spike is defined by (Vthresh , ms , hs , ns ) where V1 < Vthresh < V2 .
Question 4:
Consider the following system of linear differential equations
ẋ = −3x + y,
ẏ = 100(2x − y),
(1)
subject to initial condition x(0) = 1, y(0) = 0.
(a) Which variable is the fast variable and which is the slow variable?
(b) What is the quasi-steady state of the fast variable?
(c) Use the quasi-steady state approximation to eliminate the fast variable and show that
the solution of the reduced system is x(t) = e−t and y(t) = 2e−t .
(d) Show that the exact solution of the system is x(t) = 0.02e−102.02t + 0.98e−0.98t , y(t) =
−1.98e−102.02t + 1.98e−0.98t .
(e) Compare the approximate solution with the exact solution. Where do they agree and
where do they disagree? Why?
2