2.1 The basic problem
In this paper, we study the existence and uniqueness of solutions for the multi-term nonlinear fractional integro-differential equation
(1)
with boundary values and , where , , , , is continuous, and for the mappings
with the property and , the maps ϕ and ψ are defined by and . In this way, we need the following result, which has been proved in [2].
Lemma 2.1 Let and . Then
where are some real numbers.
The proof of the following result by using Lemma 2.1 is straightforward.
Lemma 2.2 Let , and . Then the problem with boundary values and has the unique solution
2.2 Some results on solving the problem
Let be the space of all continuous real-valued functions on and
endowed with the norm . It is known that is a Banach space.
Theorem 2.3
Assume that there exist
and
such that
for all and . Then problem (1) has a unique solution whenever
where , , .
Proof Define the mapping by
For each and , by using the Hölder inequality, we have
Also, we have
Since , we obtain
for all . Hence, we get
Since , F is a contraction mapping, therefore, by using the Banach contraction principle, F has a unique fixed point, which is the unique solution of problem (1) by using Lemma 2.2. □
Corollary 2.4
Assume that there exists
such that
for all and . Then problem (1) has a unique solution whenever
where , .
Now, we restate the Schauder’s fixed point theorem, which is needed to prove next result (see Theorem 1.10.16 in [23]).
Theorem 2.5 Let E be a closed, convex and bounded subset of a Banach space X, and let be a continuous mapping such that is a relatively compact subset of X. Then F has a fixed point in E.
Theorem 2.6 Let be a continuous function such that there exists a constant and a real-valued function such that
(*)
where and for , or
where and for . Then problem (1) has a solution.
Proof First, suppose that f satisfy condition (∗). Define , where
and . Note that is a closed, bounded and convex subset of the Banach space X. For each , we have
Also, we have
Since and, on the other hand, , we conclude that
for all . Thus,
Hence, F maps into . Now, suppose that f satisfy the second condition. In this case, choose
By using similar arguments, one can show that , and so F maps into . Since f is continuous, it is easy to get that F is also continuous. Now, we show that F is completely continuous operator on . For each , put
For each with , we have
On the other hand, for each , we have