Biomedical Engineering Reference
In-Depth Information
The first element of this result is the Right Hand Definition, but as mentioned above,
one must include a remainder term that is dependent on the value of the function at 0.
In classic linear ODEs, there are typically two forms:
u
(t)
=−
u(t)
+
q(t)
(A.33)
and
u
(t)
=−
u(t)
+
q(t)
(A.34)
Similarly, one may use the previously introduced definitions to obtain Linear
Fractional ODEs, which can be represented as follows:
u(t)
=
D
α
m
−
1
t
k
k
!
D
α
∗
u
(k)
(
0
)
u(t)
−
=−
u(t)
+
q(t)
(A.35)
k
=
0
Note the use of the Right Hand Definition in this definition. As was discussed above
in the properties of the Right Hand Definition and Left Hand Definition, the choice
to use this definition is based upon the ability to use integer order initial conditions
in the solution of problems of this kind. The most straightforward means of solving
(
A.35
) is by means of Laplace transform, and can be re-arranged as
m
−
1
m
−
1
s
α
−
k
−
1
s
α
s
α
s
α
−
k
−
1
u
(k)
(
0
)
u(s)
˜
=
=˜
u(s)
=
(A.36)
+
1
k
=
0
k
=
0
The terms inside the sum can be rewritten
s
α
−
k
−
1
s
α
s
α
1
=
L
J
k
E
α
−
t
α
1
s
k
1
=
(A.37)
s
α
+
+
and the terms
s
1
d
t
α
s
α
−
1
s
α
dt
E
α
−
1
1
=−
1
−
=
L
(A.38)
s
α
+
+
Finally, using both (
A.37
) and (
A.38
) to define the inverse Laplace transform, it is
possible to transform (
A.36
) into an expression for
u(t)
, and thus define the solution
to the fractional-order ODE:
m
−
1
J
k
E
α
−
t
α
u
(k)
(
0
)
−
q(t)
∗
E
α
−
t
α
u(t)
=
(A.39)
k
=
0
In this topic, the time-domain definitions have not been used; instead a simpler
form, that in the frequency domain, has been used. As the reader can see, the time-
domain definitions are of significant complexity and appealing to mathematicians
rather than pragmatic individuals. However, for the sake of completeness, these def-
initions have been included here, but a more comprehensive overview can be found
in [
105
,
118
,
126
].
