4 Abstract and Applied Analysis
×
(
−
)
𝛼
𝑡
1
0
(
−
)
1−𝛼−1
(
)
𝑠=𝑡
𝑠=𝑡
1
+
1
Γ
(
)
Γ
(
1−
)
×
𝑡
𝑡
1
(
−
)
𝛼−1
𝑠
𝑡
1
(
−
)
1−𝛼−1
(
)
+
1
Γ
(
+1
)
Γ
(
1−
)
×
(
−
)
𝛼
𝑠
𝑡
1
(
−
)
1−𝛼−1
(
)
𝑠=𝑡
𝑠=𝑡
1
=
1
Γ
(
)
Γ
(
1−
)
×
𝑡
1
0
(
)
𝑡
1
𝜏
(
−
)
𝛼−1
(
−
)
1−𝛼−1
+
1
Γ
(
)
Γ
(
1−
)
×
𝑡
1
0
(
)
𝑡
𝑡
1
(
−
)
𝛼−1
(
−
)
1−𝛼−1
+
1
Γ
(
)
Γ
(
1−
)
×
𝑡
𝑡
1
(
)
𝑡
𝜏
(
−
)
𝛼−1
(
−
)
1−𝛼−1
−
1−𝛼
(
0
)
Γ
(
+1
)
𝛼
−
1−𝛼
1
Γ
(
+1
)
−
1
𝛼
=
𝑡
0
(
)
−
1−𝛼
(
0
)
Γ
(
+1
)
𝛼
−
1−𝛼
1
Γ
(
+1
)
−
1
𝛼
.
(14)
us, by (13)and(14), we get
𝛼
𝑡
𝛼
𝑡
(
)
=
(
)
−
1−𝛼
(
0
)
Γ
(
)
𝛼−1
−
1−𝛼
1
Γ
(
)
−
1
𝛼−1
.
(15)
Similarly, if ∈(
𝑘
,
𝑘+1
], =2,...,,wecanget
𝛼
𝑡
𝛼
𝑡
(
)
=
(
)
−
𝑚
𝑘=1
1−𝛼
𝑘
Γ
(
)
−
𝑘
𝛼−1
−
1−𝛼
(
)
𝑡=0
𝛼−1
Γ
(
)
.
(16)
e proof is completed.
e Laplace transform formula for the Riemann-Liouville
fractional integral is dened by
𝛼
𝑡
(
)
;=
1
𝛼
(
)
,
(17)
where
()is the Laplace of dened by
(
)
=
∞
0
−𝜆𝑡
(
)
,
>,
|
(
)
|
≤
𝜔𝑡
,is a constant.
(18)
Lemma 5. Let ∈(0,1]and ∈
𝑝
(,), >1/;if()∈
1−𝛼
(,),
1−𝛼
() ∈ (,),andis a solution of the
following problem:
𝛼
𝑡
(
)
=
(
)
+
(
)
,
∈
(
0,
]
,=
𝑘
, =1,2,...,,
1−𝛼
𝑡
(
)
𝑡=𝑡
𝑘
=
𝑘
−
𝑘
, =1,2,...,,
1−𝛼
𝑡
(
)
𝑡=0
=
0
∈,
(19)
then, satises the following equation:
(
)
=
𝛼−1
𝛼
(
)
0
+
𝑡
0
(
−
)
𝛼−1
𝛼
(
−
)
(
)
,
∈0,
1
,
𝛼−1
𝛼
(
)
0
+
𝑚
𝑘=1
𝛼
−
𝑘
−
𝑘
𝛼−1
1−𝛼
0
+
𝑘
+
𝑡
0
(
−
)
𝛼−1
𝛼
(
−
)
(
)
,
∈
𝑘
,
𝑘+1
,
(20)
where =1,2,...,,
𝛼
(
)
=
∞
0
𝛼
(
)
𝛼
,
𝛼
(
)
=
1
−1−(1/𝛼)
𝛼
−1/𝛼
,
𝛼
(
)
=
1
∞
𝑛=1
(
−1
)
𝑛−1
−𝑛𝛼−1
Γ
(
+1
)
!
sin
(
)
,
∈
(
0,∞
)
,
(21)
where
𝛼
is a probability density function dened on (0,∞);
that is,
𝛼
(
)
≥0, ∈
(
0,∞
)
,
∞
0
𝛼
(
)
=1.
(22)
Proof. We observe that (⋅)can be decomposed to (⋅)+(⋅),
where isthecontinuousmildsolutionfor
𝛼
𝑡
(
)
=
(
)
+
(
)
,∈
(
0,
]
,
1−𝛼
𝑡
(
)
𝑡=0
=
0
∈,
(23)
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