122 6 Initialized Differintegrals and Generalized Calculus
6.7.2 Side Charging
The case for side charging is less definitive. Criteria for backward compatibility is the same as the terminal charging case. Relative to zero property the condition ψ( f,−p,a,c,t)= Γ1( p)
c a
(t−τ)p−1f (τ)dτ =0=ψ(h,m,a,c,t) is required for side charging sinceψis arbitrary. When these conditions are not met, the zero-order operation on f (t) will return f (t)+g(t), i.e. the original function with extra time function ((g(t)), the effect of initialization. Relative to linearity, the side charging demands additional requirements about initialization.
These are not so much of an issue as it appears for practical applications. In the solution of fractional differential equations,ψ(t) will be chosen in the much the same manner as initialization are currently chosen for ordinary differential equations in integer order. This will imply the nature of f (t) from a to c. The new aspect is that to achieve a particular initialization for a given composition now requires attention to the initialization of the composing elements.
This equation can be rearranged to obtain the system transfer function:
X (s)
U (s) =G(s)= b
sq+a (6.29)
This is the transfer function of the fundamental linear fractional order differential equation. As such, it contains the fundamental “fractional pole” and is the funda- mental building block for more complicated fractional order systems. As the con- stant b in (6.29) is a constant multiplier, it can be assumed without loss of generality to be unity. Typically, transfer functions are used to study various properties of a particular system. Specifically, they can be inverse Laplace transformed to obtain the system impulse response, which can then be used with convolution approaches to the problem. Generally, if U (s) is given, then the product G(s)U (s) can be expanded using partial fractions, and forced response obtained by inverse transforming each separately. To accomplish these tasks, it is necessary to obtain the inverse of (6.29), which is the impulse response, or generalized exponential function, of the funda- mental fractional order system.
6.8.1 The Generalized Impulse Response Function
Although the Laplace transform of (6.29) is not contained in standard Laplace trans- form table, the following transform pair is available:
1 sq =L
%tq−1 Γ(q)
&
, q >0 (6.30)
If we expand the right-hand side of (6.29) in describing powers of s, we can then inverse transform the series term by term and obtain the generalized impulse response. Then expanding (6.29) about s= ∞, we get
G(s)= 1
sq+a = 1 sq − a
s2q + a2
s3q −. . .= 1 sq
∞ n=0
(−a)n
snq (6.31)
This series can be inverse transformed term by term using (6.30). The result is L−1{G(s)} =L−1
%1 sq − a
s2q + a2 s3q −. . .
&
= tq−1
Γ(q)−at2q−1
Γ(2q) +a2t3q−1 Γ(3q) −. . .
(6.32) The right side of (6.32) can now be collected into summation and used as definition of the generalized impulse response function:
Fq[−a,t]≡tq−1 ∞ n=0
(−a)ntnq
Γ({n+1}q), q >0 (6.33)
124 6 Initialized Differintegrals and Generalized Calculus We thus have important Laplace identity as L'
Fq[a,t](
↔ sq1−a, q >0. Also the F function is generalization of exponential function for q =1,
F1[−a,t]=∞
n=0
(−at)n
Γ(n+1) =e−at
This generalization is the basis for the solution of most linear fractional order dif- ferential equations.
Here we have established the F function as the impulse response of the funda- mental linear differential equations. This function is important because it will allow the creation of concise theory for fractional order systems, which is a generalization of that of integer order systems, and where the F function generalizes and replaces the usual exponential function.
Several other variants of this F function is possible for the solution of the fun- damental equation (6.27), like Miller–Ross function, R function, and G function, listed in Chap. 2.
From (6.32) and (6.33), we obtained impulse response as:
g(t)=L−1{G(s)} =tq−1 ∞ n=0
(−a)ntnq
Γ(nq+q)≡Fq[−a,t], q >0 (6.34)
The function Fq[a,t] is closely related to the Mittag-Leffler function Eq[atq], where one-parameter Mittag-Leffler function in series form is defined as:
Eq[x ]≡ ∞ n=0
xn
Γ(nq+1), q>0 (6.35)
Letting x= −atq, this becomes
Eq
−atq
≡∞
n=0
(−a)ntnq
Γ(nq+1), q >0 (6.36)
which is similar to Fq[−a,t] expressed in (6.34), but not same as (6.34).
The Laplace transform of this Mittag-Leffler function (6.36) can also be obtained via term by term transform of the series expansion as:
L'
Eq[−atq](
=L 0 1
Γ(1) −Γ(1+q)atq +Γa(1+2q)2t2q +. . .1
=1s −sq+1a +s2q+1a2 +. . . (6.37)
or compactly:
L' Eq
−atq(
= 1 s
1− 1 sq + a2
s2q +. . .
=1 s
∞ n=0
−a sq
n
= 1 s
∞ n=0
(−a)n snq
(6.38) It should be noted that the summation expression (6.38) is similar to (6.31). Using (6.31), (6.38) can be rewritten as:
L' Eq
−atq(
=1 s
sq sq+a
(6.39) or equivalently:
L' Eq
−atq(
= 1 s
sqL'
Fq[−a,t](
(6.40) Thus a general result can be expressed as:
L' Eq
±atq(
= sq−1
sq∓a, q >0 (6.41)
From (6.40), the relation between Mittag-Leffler (Eq) function and Robotnov–
Hartley function (Fq) is
0dtq−1Fq[a,t]=Eq
atq
(6.42) This demonstration was to show a method of obtaining solution of the “fundamental fractional order (linear) differential equation” that is (6.27) by use of Robotnov–
Hartley function. This F function was utilized by Robotnov to study hereditary inte- grals in solid mechanics. Solution to (6.27) may be obtained in terms of Miller–Ross function and its Laplace transform. Miller–Ross function is the fractional derivative of the exponential function defined as
Et(v,a)≡0dt−vexp(at),whose Laplace transform is L{Et(v,a)} = ss−−va. Also recent developments to study diffusion and fractional kinetic equations use more complicated Fox functions in solving of fractional order differential equations.
Extending this developed technique, we obtain the solution of (6.27) for a unit step input excitation. This can be obtained via Laplace transforms by transforming the input function u(t), its Laplace is 1/s, which must be multiplied by the transfer function G(s), (6.29), where b=1 is taken. We get
X (s)=1 s
1 sq+a
(6.43) Manipulating (6.43), we obtain
126 6 Initialized Differintegrals and Generalized Calculus X (s)=1/a
s a
sq+a
= 1/a s
1− sq sq+a
= 1/a
s − sq/a
s(sq+a) (6.44) Using expression (6.40), (6.44) is inverse Laplace transformed; the result is the step response of the system:
x (t)=L−1
% 1 s(sq+a)
&
= 1 a
1−Eq
−atq
= 1 a
h(t)−Eq
−atq (6.45)
Heaviside step is h(t)=
%1 t ≥0 0 t <0
Taking integer derivative of (6.45) gives Fq[−a,t], the impulse response:
Fq[−a,t]= 1 a
d dt
h(t)−Eq
−atq
(6.46) Referring to (6.40) and multiplying the Laplace transforms, thereby s−qgives
s−qL' Eq
−atq(
=s−1L'
Fq[−a,t](
= 1 s(sq+a) ,
which is (6.43). Inverse transforming this equation using expression (6.45) shows that the step response of (6.27) with b=1 is also the qth fractional integral of the Mittag-Leffler function, that is:
x (t)=L−1
% 1 s(sq+a)
&
= 1 a
h(t)−Eq(−atq)
=0dt−qEq
−atq
(6.47) Few more interesting Laplace pairs can be obtained by taking the qth derivative of the F function. Taking the un-initialized fractional derivative (0dtq), and in Laplace domain multiplying by sqgives
L−1
% sq sq+a
&
=0dtqFq[−a,t] (6.48)
This expression (6.48) is also the integer derivative of the Mittag-Leffler function:
L−1
% sq sq +a
&
=0dtqFq[−a,t]=0dtqEq
−atq
(6.49) The expression (6.49) can also be rewritten as:
L−1
% sq sq+a
&
=L−1
%
1− a
sq+a
&
=δ(t)−a Fq[−a,t] (6.50)
Observing the expressions (6.48) and (6.50), we can write the following:
0dtqFq[−a,t]=δ(t)−a Fq[−a,t] (6.51) This Fq[a.t] function is generalization of the exponential function of the integer order calculus where it demonstrates the “eigenfunction property”; i.e., returning the same function upon the qth fractional differentiation. Also (6.51) shows that x (t) = Fq[−a,t], Robotnov–Hartley function is impulse response of the funda- mental fractional order differential equation given by (6.27), for u(t) = δ(t) and b=1.