Fractional and operational calculus with generalized fractional ...

804 Ž. Tomovski et al.
Proof Our demonstration of Theorem 5 is based upon the Laplace transform method. Indeed,
if we make use of the notational convention depicted in (1.7) and the Laplace transform formula
(1.6), by applying the operator L to both the sides of (3.7), it is easily seen that
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Furthermore, since
and
s β 1(α 1 −1)
s β 2(α 2 −1)
F (s)
Y (s) = ac 1 + bc
as α 1 + bs
α 2 +
.
2 + c as α 1 + bs
α 2 + c as α 1 + bs
α 2 + c
s β i(α i −1)
= 1 ( s
β i (α i −1)
) ⎛ ⎝
1
as α 1 + bs
α 2 + c b s α 2 +
c
b 1 + a b
[ 1
∞∑
= L
b
F (s)
as α 1 + bs
α 2 + c
= 1 b
= L
m=0
∞∑
m=0
[
1
b
(
s α 1
s α 2 + c b
⎞
) ⎠ = 1 b
∞∑
m=0
(
(−1) m a
) m
x
(α 2 −α 1 )m+α 2 +β i (1−α i )−1
b
· E m+1
α 2 ,(α 2 −α 1 )m+α 2 +β i (1−α i )
(
− a b
) m
(
s α 1m
(
s
α 2 +
c
b
(
− a ) m s α 1m+β i α i −β i
(
b s
α 2 +
c
b
(
− c b xα 2) ] (s) (i = 1, 2)
) m+1
F (p)
)
[ 1
= L
b
(
(
· x (α 2−α 1 )m+α 2 −1 E m+1
α 2 ,(α 2 −α 1 )m+α 2
− c b xα 2
∞∑
m=0
∞∑
m=0
(
− a ) m
b
)
∗ f (x)
) m+1
)]
(s)
(
(−1) m a
) m (
)(
E m+1
α
b
2 ,(α 2 −α 1 )m+α 2 ,− c ;0+f − c 2) ]
b b xα (s)
in terms of the Laplace convolution, by applying the inverse Laplace transform, we get the solution
(3.9) asserted by Theorem 5.
3.2. An Application of Theorem 5
The next problem is to solve the fractional differential equation (3.7) in the space of Lebesgue
integrable functions y ∈ L(0, ∞) when
α 1 = α 2 = α and β 1 ̸= β 2 ,
that is, the following fractional differential equation:
( ) ( )
a D α,β 1
0+ y (x) + b D α,β 2
0+ y (x) + cy (x) = f (x) (3.10)
under the initial conditions given by
(
)
I (1−β i)(1−α)
0+ y (0+) = c i (i = 1, 2). (3.11)
We now assume, without loss of generality, that β 2 ≦ β 1 . If c 1 < ∞, then
c 2 = 0 unless β 1 = β 2 .