Biomedical Engineering Reference
In-Depth Information
EXAMPLE PROBLEM 7.17
Consider the unilateral five-compartment model with no loss of solute to the environment from
any compartments and an input for compartment 3 only. Additionally, all transfer rates equal
2 and
f
3
(
t
)
¼
5d(
t
). Assume that the initial conditions are zero. Solve for the quantity in compart-
ment 3.
Solution
As before, we transform the input,
f
3
(
t
)
¼
5d(
t
), into a change in initial condition for compart-
ment 3 to
q
3
(0)
¼
5 and no input. The conservation of mass for each compartment yields
q
1
¼
2
q
1
þ
2
q
5
q
2
¼
2
q
1
2
q
2
q
3
¼
2
q
2
q
2
3
q
4
¼
2
q
3
2
q
4
q
5
¼
2
q
4
2
q
5
Using MATLAB, we have
>>
syms D
>>
¼
A
[-2 0 0 0 2;2 -2 0 0 0;0 2 -2 0 0;0 0 2 -2 0;0 0 0 2 -2];
>>
det(D*eye(5)-A)
ans
¼
^
þ
^
þ
^
þ
^
þ
D
5
10*D
4
40*D
3
80*D
2
80*D
and
5
4
3
2
d
q
3
dt
10
d
q
3
dt
40
d
q
3
dt
80
d
q
3
dt
80
dq
3
þ
þ
þ
þ
dt
¼
0
5
4
3
2
The roots from the characteristic equation (i.e., eig(
A
)) are 0,
3.6180
j
1.1756, and
1.3820
j
1.9021. The complete solution for
q
3
is the natural solution, since the forced response is zero, and is
given by
q
3
¼
B
1
þ
e
3:618
t
B
2
cos 1
Þ þ
e
1:382
t
B
4
cos 1
ð
:
1756
t
þ
B
3
sin 1
:
1756
t
ð
:
9021
t
þ
B
5
sin 1
:
9021
t
Þ
The initial conditions for
q
3
are found using the conservation of mass equations and successive
derivatives, giving
q
3
ð
0
Þ¼
5,
q
3
ð
0
Þ¼
10,
q
3
ð
0
Þ¼
20,
__
q
3
ð
0
Þ¼
40, and
____
q
3
ð
0
Þ¼
80
:
Solving for
the unknown coefficients using the initial conditions, we have
q
3
ð
0
Þ¼
5
¼
B
1
þ
B
2
þ
B
4
q
3
ð
0
Þ¼
10
¼
3
:
61
B
2
þ
1
:
1756
B
3
1
:
382
B
4
þ
1
:
9021
B
5
q
3
ð
0
Þ¼
20
¼
11
:
72
B
2
8
:
51
B
3
1
:
71
B
4
5
:
26
B
5
__
q
3
ð
0
Þ¼
40
¼
32
:
4
B
2
þ
44
:
58
B
3
þ
7
:
64
B
4
þ
4
:
02
B
5
____
q
ð
0
Þ¼
80
¼
64
:
79
B
2
199
:
4
B
18
:
18
B
4
þ
9
:
02
B
5
3
3
and using MATLAB yields
B
1
¼
1.0,
B
2
¼
2.0,
B
3
¼
0.0,
B
4
¼
2.00, and
B
5
¼
0.0. Thus, for t
0, we
have
e
3:618
t
cos 1
e
1:382
t
cos 1
q
3
¼
1
þ
2
:
1756
t
þ
2
:
00
:
9021
t
