Now, we are ready to state and prove our main results. First, we give the following one.
Lemma 3.1 Let , , , and be continuous functions. The unique solution of the fractional differential problem
via the boundary value conditions (2) is given by
Proof It is known that the general solution of (5) is
that is
(6)
where , , are real arbitrary constants (see [10, 13] and [14]). Thus,
and . Hence,
and
By using the boundary conditions, we obtain
and
This is a linear system of equations of triangular form, having , , and as unknowns. We solve by back substitution and find
and
Now, we replace , , and in (6) and find the solution as we stated. This completes the proof. □
Let endowed with the norm . Then is a Banach space. For , define
For the study of problem (1) and (2), we shall consider the following conditions.
(H1) is an integrable bounded multifunction such that is measurable for all ;
(H2) be continuous functions, a nondecreasing upper semi-continuous map such that and for all ;
(H3) There exist such that
and for all , and , where
and , and finally
(H4) is given by
where
Theorem 3.1 Assume that (H1)-(H4) are satisfied. If the multifunction N has the approximate endpoint property, then the boundary value inclusion problem (1) and (2) has a solution.
Proof We show that the multifunction has a endpoint which is a solution of the problem (1) and (2).
Note that the multivalued map is measurable and has closed values for all . Hence, it has measurable selection and so is nonempty for all . First, we show that is closed subset of X for all .
Let and be a sequence in with . For each , choose such that
for all .
Since F has compact values, has a subsequence which converges to some . We denote this subsequence again by .
It is easy to check that and
for all . This implies that and so N has closed values.
Since F is a compact multivalued map, it is easy to check that is a bounded set for all .
Now, we show that .
Let and . Choose such that
for almost all .
Since
for all , there exists such that
for all .
Consider the multivalued map defined by
Since and are measurable, the multifunction is measurable.
Choose such that
for all .
Now, consider the element , which is defined by
for all . Thus,
and
Hence,
Thus, it is easy to get for all .
Since the multifunction N has approximate endpoint property, by using Lemma 2.1 there exists such that . Hence by using Lemma 3.1, is a solution of the problem (1) and (2). □
Now, we investigate the existence of solution for the fractional differential inclusion problem
via integral boundary value conditions
where , , , , for all , , and is a multifunction.
Lemma 3.2 Let , , , and for . The unique solution of the fractional differential problem via the boundary value conditions
with , is
where
is the Green function given by
whenever ,
whenever ,
whenever ,
whenever ,
whenever
and
whenever , where and .
Proof It is known that the general solution of the equation is
where are arbitrary constants (see [10, 13] and [14]). Thus,
and . Hence,
and
By using the boundary conditions, we obtain
and
Thus,
Hence,
This completes the proof. □
Suppose that endowed with the norm . Then is a Banach space [15]. For , define
Now, put
and for all .
Theorem 3.2 Let a nondecreasing upper semi-continuous map such that and for all , a multifunction such that is measurable and integrable bounded for all . Assume that there exists such that
Define
by
If the multifunction Ω has the approximate endpoint property, then the boundary value inclusion problem (3) and (4) has a solution.
Proof We show that the multifunction has a endpoint which is a solution of the problem (3) and (4).
First, we show that is closed subset of X for all .
Let and be a sequence in with . For each , choose such that for all . Since F has compact values, has a subsequence which converges to some . We denote this subsequence again by .
It is easy to check that and for all . This implies that and so Ω has closed values. Since F is a compact multivalued map, it is easy to check that is a bounded set for all .
Now, we show that for all , .
Let and . Choose such that for almost all . Since
for all , there exists such that
for all . Consider the multivalued map defined by the rule
Since and
are measurable, the multifunction
is measurable.
Choose such that
for all . Now, consider the element which is defined by for all .
Thus,