Biomedical Engineering Reference
In-Depth Information
λ
2
d
2
V
m
dx
2
−
V
m
=
0
.
(4.19)
The solution to Eq. (4.19) is the Helmholtz equation:
Ae
−|
x
|
/λ
Be
|
x
|
/λ
.
V
m
(x)
=
+
(4.20)
The second term does not make physical sense so
Ae
−|
x
|
/λ
.
=
V
m
(x)
(4.21)
To find
A
we will integrate Eq. (4.18) around the stimulus (
x
=
0):
λ
2
0
+
0
−
0
+
V
m
dx
=−
r
m
i
stim
0
+
0
−
d
2
V
m
dx
2
dx
−
δ(x) dx .
(4.22)
0
−
r
m
·
i
stim
2
λ
Evaluation shows that
A
=
so the steady-state solution is
r
m
i
stim
2
λ
e
−|
x
|
/λ
.
V
m
(x)
=
(4.23)
Note that the value of the constant
A
will be dependent upon the nature of the stimulus. For example,
the solution would change if we applied the stimulus to one end of the cable rather than at the middle.
4.1.6 FindingThe Length Constant
If a cable with unknown membrane properties is encountered, Eq. (4.23) is a way to find
λ
and
r
m
.Ifa
known stimulus is applied for a long time, the membrane will eventually reach steady-state. At the point
of the stimulus,
x
0 so the exponential term becomes 1.Therefore, the voltage at the stimulus is
r
m
·
i
stim
2
λ
=
.
Furthermore, this voltage level will fall off in space (in both directions because of
) at an exponential
rate governed by
λ
. Therefore,
λ
can be found as the rate of fall off in a similar way to finding
τ
m
in Ch. 2.
Once
λ
is known,
r
m
can be found from
r
m
·
i
stim
|
x
|
.
2
λ
4.1.7 Time and Space Dependent Solution
Although we will not show the solution here, it is possible to solve Eq. (4.17) for
V
m
as a function of
both time and space:
e
−|
x
|
/λ
erfc
|
x
|
2
λ
τ
m
t
−
t
τ
m
r
m
·
i
stim
4
λ
V
m
(x, t)
=
−
e
|
x
|
/λ
erfc
|
x
|
2
λ
τ
m
t
+
t
τ
m
(4.24)
