Biomedical Engineering Reference
In-Depth Information
4 Solution to the Diffusion Equation
The oxygen distribution in the tumor and its immediate surroundings during the
growth process is governed by the diffusion equation
@
c
ð
r
;
t
Þ
2
c r
t
¼Dr
ðÞ
;
t
k r
ðÞ
(1)
@
10
5
cm
2
s
1
[
14
], is the coefficient of diffusion; c(r,t) is the
magnitude of oxygen concentration at the CA element at location r at time t; k(r)is
the rate of oxygen consumption by the CA element at r and is dependent on the type
of cell. In two dimensions, the equation becomes
where
D
¼
1
2
c
2
c
@
c
ð
x
;
t
¼D
@
y
;
t
Þ
ð
x
;
y
;
t
Þ
þ
@
ð
x
;
y
;
t
Þ
k x
ð :
;
y
(2)
@
@
x
2
@
y
2
Initially, we will be interested in finding the distribution of oxygen in the lattice
before it is consumed by cells. Therefore, the initial distribution prior to consump-
tion is given by the equation
2
c
2
c
@
c
ð
x
;
t
¼D
@
y
;
t
Þ
ð
x
;
y
;
t
Þ
þ
@
ð
x
;
y
;
t
Þ
:
(3)
x
2
y
2
@
@
@
On an average, the time taken by a cell to undergo cell division is about 20 h
[
22
,
23
]. Therefore, once proliferation has taken place, a significantly high amount
of time has to elapse before another proliferation step takes place. Therefore, the
left hand side of (
3
) tends to zero leading to the following equation
2
c
2
c
@
ð
x
;
y
Þ
þ
@
ð
x
;
y
Þ
t
¼
0
:
(4)
@
x
2
@
y
2
The diffusion equation finally reduces to an equilibrium equation suggesting that
the concentration levels of oxygen reaches steady state during the time period
between two subsequent cell divisions or proliferation steps. Equation (
4
)is
equivalent to the following equation in two dimensions
2
c
r
ð
r
Þ
t
¼
0
:
(5)
We solve this initial-boundary value problem using the central difference
method to determine the oxygen concentration level at each node in the lattice.
We use the Dirichlet boundary condition with the oxygen concentration at the
boundary being constant and equal at all times to the oxygen concentration level
of a healthy tissue at 1
10
4
gcm
3
[
14
,
16
,
21
] i.e.,
C
t
C
t
0
ðÞ¼
x
;
0
ðÞ¼
;
y
C
t
N
C
t
0
10
4
gcm
3
ð
;
0
Þ¼
ð
;
N
Þ¼
1
at all times. To save computation time,
