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
log [C]
mV .)
temperature is defined as T=degree centigrade+273.16. At 20o C, E = 58
z
[C]in
Solution:
Nernst potential is given by
EN a =
Thus, at 10o C,
RT [Na+ ]out
ln
zF
[Na+ ]in
RT
10 + 273.16
= 58
= 56
F
20 + 273.16
Therefore,
EN a = 56 log
100
[Na+ ]out
= 56 log
= 5.4 mV.
+
[Na ]in
80
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.
Solution:
Longitudinal current is expressed as
IL = −
1
rL
π(d/2)2
∂V
πd2 ∂V
=−
.
∂x
4rL ∂x
For semi-infinite cable,
V (ξ) = V (0)e−ξ = V0 e−ξ
⇒
∂V
∂V dξ
V0
=
= − e−ξ .
∂x
∂ξ dx
λ
Thus,
πd2 ∂V dξ
πd2
πd2 ∂V
|ξ=0 = −
|ξ=0 = −
I0 = −
4rL ∂x
4rL ∂ξ dx
4rL
V0 −ξ
πd2
− e
=
V0 .
λ
4rL λ
ξ=0
Therefore,
√
2 rM rL
V0
4rL λ
=
=
=
.
I0
πd2
πd3/2
For a cable of finite electrotonic length l with sealed end at ξ = l, the solution is
∞
RIN
P
V (ξ) = V0
cosh(l − ξ)
.
cosh(l)
Now, if a current I0 is injected longitudinally at ξ = 0, then
πd2
sinh(l − ξ)
πd2 V0 tanh(l)
πd2 ∂V dξ
πd2 ∂V
|ξ=0 = −
−V0
|ξ=0 =
|ξ=0 = −
I0 = −
4rL ∂x
4rL ∂ξ dx
4rL
λ cosh(l)
4rL λ
Therefore,
∞
RIN
P
√
2 rM rL
4rL λ
R∞
V0
= 2
= 3/2
= IN P .
=
I0
πd tanh(l)
πd tanh(l)
tanh(l)
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 .
Solution:
(a) (Vs , ms , hs , ns ) = (−64.99972, 0.05293, 0.59611, 0.31768).
(b) Eigenvalues:
-4.675830 + i
-0.202459 + i
-0.202459 + i
-0.120657 + i
0.000000
0.383560
-0.383560
0.000000
2
(c)
40
20
0
-20
-40
-60
-80
0
5
10
15
20
(d) V1 = −59 and V2 = −58, −59 < Vthresh < −58.
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 analytical solution of the system is approximated by x(t) = 0.02e−102.02t +
0.98e−0.98t , y(t) = −1.98e−102.02t + 1.98e−0.98t . (Use the approximate engenvalues: λ1,2 =
−0.98, −102.02 to find approximate eigenvectors.)
(e) Compare the approximate solution with the exact solution. Where do they agree and
where do they disagree? Why?
Solution:
(a) x is the slow variable; y is the fast one.
(b) y = 2x is the quasi-steady state.
(c) Substitute y = 2x into the slow equation, we obtain
ẋ = 3x + 2y
⇒
ẋ = −x
⇒
x(t) = x(0)e−t = e−t ,
3
y(t) = 2x(t) = 2e−t .
(d) This is a linear system for which T r = −103, Det = 300 − 200 = 100. Thus,
√
λ1,2 = −51.5 ± 51.52 − 100 ≈ −51.5 ± 50.52 = −0.98, −102.02.
Approximately, the eigenvectors are
1
~v1 =
,
2.02
~v2 =
1
−99.02
.
Thus,
~x(t) =
x(t)
y(t)
= c1~v1 e
−0.98t
+ c2~v2 e
= c1
−102.02t
1
2.02
e
−0.98t
+ c2
1
−99.02
e−102.02t .
Using the initial consition,
c1 = −99.02/(−101.04) ≈ 0.98,
c2 = −2.02/(−101.04) ≈ 0.02.
Therefore,
x(t)
1
1
0.98
0.02
−0.98t
−102.02t
−0.98t
= 0.98
e
+0.02
e
≈
e
+
e−102.02t .
y(t)
2.02
−99.02
1.98
−1.98
(e) At t = 0, the analytical solution gives
1
0.02
0.98
x(0)
.
=
+
=
0
−1.98
1.98
y(0)
However, based on quasi-steady state approximation, x(0) = 1, y(0) = 2x(0) = 2 in which
1
y(0) is completely wrong. However, for t > 34
, e−102.02t < 0.001, the solution becomes
x(t)
1
≈ 0.98
e−0.98t .
y(t)
2.02
This is a very good approximation of the analytical solution. This is because it takes a short
interval of time for the 2D system to jump from the initial condition onto the nullcline of the
fast variable. After that the two systems are basically identical. See the following figures.
y
2
y(t)
y=2x
2
1.8
1.6
1.4
1.5
1.2
1
1
0.8
0.6
0.5
0.4
0.2
x
0
0
0.2
0.4
0.6
0.8
1
1.2
1.4
1.6
1.8
0
2
4
0
1
2
3
4
5
t
6