Biomedical Engineering Reference
In-Depth Information
• W
ð
0
Þ¼
0;
• The increments, W
ð
t
k
þ
1
Þ
W
ð
t
k
Þ
and W
ð
t
k
Þ
W
ð
t
k
1
Þ
are independent for any
0
t
k
1
\t
k
\t
k
þ
1
;
• For 0
s
t
;
the increment W
ð
t
Þ
W
ð
s
Þ
has the Gaussian distribution with
mean 0 and variance t
s
;
i.e. W
ð
t
Þ
W
ð
s
Þ
N
ð
0
;
t
s
Þ:
Further, W
ð
t
Þ
is 'stochastically continuous' (lim
t
!
s
P
ðj
W
ð
t
Þ
W
ð
s
Þj
[
Þ¼
0),
where P stands for the probability. The formal analytic solution,
X
ð
t
Þ¼
X
0
þ
rW
ð
t
Þ;
Y
ð
t
Þ¼
Y
0
þ
rW
ð
t
Þ;
Z
ð
t
Þ¼
Z
0
þ
rW
ð
t
Þ;
ð
2
Þ
for t [ 0
;
can be given, however the differential form is more useful in this study from a
practical point of view. The equations are classically numerically solved using the
Euler-Maruyama Method, given by
X
i
þ
1
¼
X
i
þ
rDW
i
þ
1
;
X
0
¼ð
X
0
;
Y
0
;
Z
0
Þ:
ð
3
Þ
Here each component of DW is a normally distributed stochastic parameter with
zero mean and variance Dt
;
denoted by N
ð
0
;
Dt
Þ;
and it can be proved that [
24
]
each component m satisfies DW
i
þ
1
¼
W
i
þ
1
W
i
N
ð
0
;
1
Þ
p
;
in other words, a
Gaussian distribution with zero mean and a variance of Dt
:
We show a run of the
solution of the stochastic differential equations with one bacterium initially located
at
ð
0
;
0
;
0
Þ
with mobility r
¼
2
:
6833
10
5
m/
Dt
p
:
This value was chosen from
[
25
] and corresponds to the classical bacillum in Fig.
1
. Figure
1
shows the tra-
jectory of the bacterium over time in three dimensions. Since Fig.
1
only gives one
specific run, the trajectory itself is a stochastic parameter and hence for many
purposes the probability density function is of more importance. To this extent,
since dW
ð
t
Þ
N
ð
0
;
1
Þ
p
and W
ð
t
Þ
N
ð
0
;
1
Þ
t
p
;
the probability density for the
position of the bacterium at time t for each coordinate direction satisfies
dt
exp
ð
ð
m
m
0
Þ
2
2r
2
t
1
2pr
2
t
p
f
m
ð
t
;
m
Þ¼
Þ;
m
2ð
X
;
Y
;
Z
Þ:
ð
4
Þ
Since the Brownian motion in each coordinate direction is an independent
stochastic event, the multi-variate probability density is given by
ð
2pr
2
t
Þ
2
exp
ð
ð
x
X
0
Þ
2
1
f
ð
x
;
y
;
z
Þ¼
Þ;
ð
5
Þ
2r
2
t
3
which solves the initial value problem in R
ot
r
2
of
2
Df
¼
0
;
f
ð
0
; ð
x
;
y
;
z
ÞÞ ¼
d
ð
x
X
0
Þ:
ð
6
Þ
Here d
ð
x
Þ
represents the Dirac Delta Distribution in three dimensions, with
characteristics
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