Information Technology Reference
In-Depth Information
An alternative representation of the relaxation equation can be expressed in terms of
the anti-derivative operator by
D
−
1
t
Q
(
t
)
−
Q
(
0
)
=−
λ
[
Q
(
t
)
]
.
(5.41)
This equation suggests replacing the anti-derivative operator to describe a more complex
network with the RL fractional integral operator
)
=
0
D
−
α
[−
λ
α
Q
(
)
−
(
(
)
]
Q
t
Q
0
t
(5.42)
t
and, by inverting the fractional integral operator, replacing the relaxation equation with
the fractional relaxation equation
t
−
α
0
D
t
[
)
]+
λ
α
Q
Q
(
t
(
t
)
=
Q
(
0
).
(5.43)
(
1
−
α)
Here the initial value is a time-dependent inhomogeneous term and the relaxation time
is raised to a power to match the index of the fractional derivative in order to maintain a
consistent system of units. This rather intimidating-looking equation has a remarkable
solution with application to a wide variety of complex phenomena. But it should be
emphasized that we are incorporating the history of the dynamical web into its solution;
what happens at a given time depends on what happened at earlier times, not just on the
initial value.
Equations of the form (
5.43
) are mathematically well defined, and strategies for solv-
ing them have been developed by a number of investigators, particularly by Miller and
Ross [
18
], whose topic is devoted almost exclusively to solving such equations when
the fractional index
is rational. Here we allow the index to be irrational and posit the
solution to the dynamical equation through Laplace transforms. Consider the Laplace
transform of (
5.43
),
α
u
α
λ
α
+
Q
(
0
)
Q
(
u
)
=
u
α
,
(5.44)
u
whose inverse is the solution to the fractional relaxation equation. Inverting Laplace
transforms with irrational powers, such as (
5.44
), can be non-trivial and special tech-
niques for this purpose have been developed, but we do not review those techniques
here. The inverse Laplace transform of (
5.44
) yields the solution to the fractional
relaxation equation in terms of the Mittag-Leffler function (MLF)
∞
k
(
−
1
)
)
α
)
=
k
α
E
α
(
−
(λ
t
α)
(λ
t
)
(5.45)
(
1
+
k
k
=
0
given by
)
α
).
Q
(
t
)
=
Q
(
0
)
E
α
(
−
(λ
t
(5.46)
To verify that the series (
5.46
) and (
5.44
) are Laplace-transform pairs, consider the
Laplace transform of the MLF
∞
∞
∞
k
k
α
(
−
1
)
λ
e
−
ut
E
)
α
)
e
−
ut
t
k
α
dt
α
(
−
(λ
t
dt
=
,
(
1
+
k
α)
0
0
k
=
0
