DOI: 10.1515/auom-2015-0006
An. Şt. Univ. Ovidius Constanţa
Vol. 23(1),2015, 73–82
Some remarks on a fractional
integro-differential inclusion with boundary
conditions
Aurelian Cernea
Abstract
We study the existence of solutions for fractional integrodifferential
inclusions of order q ∈ (1, 2] with families of mixed, closed, strip and integral boundary conditions. We establish Filippov type existence results
in the case of nonconvex set-valued maps.
1
Introduction
Differential equations with fractional order have recently proved to be strong
tools in the modelling of many physical phenomena. As a consequence there
was an intensive development of the theory of differential equations and inclusions of fractional order ([13, 15, 16] etc.). Applied problems require definitions
of fractional derivative allowing the utilization of physically interpretable initial conditions. Caputo’s fractional derivative, originally introduced in [6] and
afterwards adopted in the theory of linear visco elasticity, satisfies this demand.
Very recently several qualitative results for fractional integro-differential equations were obtained in [1, 10, 12, 14, 17, 18] etc.
This paper is concerned with the following fractional integro-differential
inclusion
Dcq x(t) ∈ F (t, x(t), V (x)(t)) a.e. ([0, T ]),
(1)
Key Words: Differential inclusion, Fractional derivative, Boundary condition.
2010 Mathematics Subject Classification: Primary 34A60, 26A33; Secondary 26A42,
34B15.
Received: 20 April, 2014.
Revised: 10 June, 2014.
Accepted: 30 June, 2014.
73
A FRACTIONAL INTEGRO-DIFFERENTIAL INCLUSION
74
where q ∈ (1, 2], Dcq is the Caputo fractional derivative, F : [0, T ] × R × R →
P(R) is a set-valued map and V : C([0, T ], R) → C([0, T ], R) is a nonlinear
Rt
Volterra integral operator defined by V (x)(t) = 0 k(t, s, x(s))ds with k(., ., .) :
[0, T ] × R × R → R a given function.
We study (1) subject to four families of boundary conditions:
i) Mixed boundary conditions
T x0 (0) = −ax(0) − bx(T ),
T x0 (T ) = bx(0) + dx(T ).
(2)
T x0 (T ) = γx(0) + δT x0 (0),
(3)
ii) Closed boundary conditions
x(T ) = αx(0) + βT x0 (0),
where a, b, d, α, β, γ, δ ∈ R are given constants.
iii) Strip boundary conditions
Z β
Z
x(0) = σ
x(s)ds, x(1) = η
α
δ
x(s)ds,
(4)
γ
where σ, η ∈ R and 0 < α < β < γ < δ < 1.
iv) Nonlocal Riemann-Liouville type integral boundary conditions
x(0) = aI ω x(µ),
x(1) = bI ν x(θ),
(5)
q
where a, b ∈ R, ν, ω, µ, θ ∈ (0, 1) and I x(.) is the Riemann-Liouville fractional
integral of order q.
The aim of this note is to show that Filippov’s ideas ([7]) can be suitably adapted in order to obtain the existence of solutions for problems (1)-(2),
(1)-(3), (1)-(4) and (1)-(5). Recall that for a differential inclusion defined by
a lipschitzian set-valued map with nonconvex values, Filippov’s theorem ([7])
consists in proving the existence of a solution starting from a given ”quasi” solution. Moreover, the result provides an estimate between the ”quasi” solution
and the solution obtained.
We note that in the case when F does not depend on the last variable,
existence results for problems (1)-(2), (1)-(3), (1)-(4) and (1)-(5) may be found
in [7,8,9]. In fact, the results in the present paper extend the main results in
[7,8,9] to fractional integrodifferential inclusions.
The paper is organized as follows: in Section 2 we recall some preliminary
results that we need in the sequel and in Section 3 we prove our main results.
2
Preliminaries
Let (X, d) be a metric space. Recall that the Pompeiu-Hausdorff distance of
the closed subsets A, B ⊂ X is defined by
dH (A, B) = max{d∗ (A, B), d∗ (B, A)}, d∗ (A, B) = sup{d(a, B); a ∈ A},
A FRACTIONAL INTEGRO-DIFFERENTIAL INCLUSION
75
where d(x, B) = inf y∈B d(x, y).
Let I = [0, T ], we denote by C(I, R) the Banach space of all continuous
functions from I to R with the norm ||x(.)||C = supt∈I |x(t)| and L1 (I, R) is
the Banach space of integrable functions u(.) : I → R endowed with the norm
RT
||u(.)||1 = 0 |u(t)|dt.
Definition 2.1. a) The fractional integral of order α > 0 of a Lebesgue
integrable function f : (0, ∞) → R is defined by
α
Z
t
(t − s)α−1
f (s)ds,
Γ(α)
I f (t) =
0
provided the right-hand side is pointwise defined
R ∞ on (0, ∞) and Γ(.) is the
(Euler’s) Gamma function defined by Γ(α) = 0 tα−1 e−t dt.
b) The Caputo fractional derivative of order α > 0 of a function f :
[0, ∞) → R is defined by
Dα
c f (t) =
1
Γ(n − α)
Z
t
(t − s)−α+n−1 f (n) (s)ds,
0
where n = [α] + 1. It is assumed implicitly that f is n times differentiable
whose n-th derivative is absolutely continuous.
We recall (e.g., [13]) that if α > 0 and f ∈ C(I, R) or f ∈ L∞ (I, R) then
α α
(Dc I f )(t) ≡ f (t).
The next two technical results are proved in [2].
Lemma 2.2. The unique solution x(.) ∈ C(I, R) of problem
Dcq x(t) = f (t)
a.e. ([0, T ]),
(6)
RT
with boundary conditions (2) is given by x(t) = 0 G1 (t, s)f (s)ds, where
G1 (., .) is the Green function defined by

2
q−1
(t−s)q−1
1 [T (b+d)+(b −ad)t](T −s)

+
 Γ(q) − ∆1 (
T Γ(q)


 [(a+b)t−(1+b)T ](T −s)q−2
+
) if 0 ≤ s < t ≤ T,
Γ(q−1)
G1 (t, s) :=
2
q−1
[T
(b+d)+(b
−ad)t](T
−s)
](T −s)q−2

1

+ [(a+b)t−(1+b)T
)

T Γ(q)
Γ(q−1)
 − ∆1 (

if 0 ≤ t < s ≤ T,
with ∆1 = (1 + b)(b + d) − (a + b)(d − 1) 6= 0.
Note that
|G1 (t, s)| ≤
|b + d + b2 − ad| + (q − 1)|a − 1|
T q−1
· (1 +
) =: M1
Γ(q)
|∆1 |
∀t, s ∈ I.
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A FRACTIONAL INTEGRO-DIFFERENTIAL INCLUSION
Lemma 2.3. The unique solution x(.) ∈ C(I, R) of problem (6)-(3) is given
RT
by x(t) = 0 G2 (t, s)f (s)ds, where the Green function defined by
G2 (t, s) :=











−s)q−1
(t−s)q−1
− ∆12 ( [T (1−δ)+γt](T
+
Γ(q)
T Γ(q)
[(1−α)t−(1−β)T ](T −s)q−2
+
) if 0 ≤ s < t ≤ T,
Γ(q−1)
q−1
](T −s)q−2
1 [T (1−δ)+γt](T −s)
+ [(1−α)t−(1−β)T
)
∆2 (
T Γ(q)
Γ(q−1)
if
0 ≤ t < s ≤ T,
with ∆2 = γ(1 − β) + (1 − α)(1 − δ) 6= 0.
Note that
|G2 (t, s)| ≤
T q−1
|1 − δ + γ| + (q − 1)|α − β|
· (1 +
) =: M2
Γ(q)
|∆2 |
∀t, s ∈ I.
For simplicity, in the following results T = 1.
The next result is proved in [3].
Lemma 2.4. For a given function f (.) ∈ C(I, R) the unique solution of
problem (6)-(4) is given by
Rt
σ
(t − s)q−1 f (s)ds + ∆
[−( η2 (δ 2 − γ 2 ) − 1)+
R β R s (s−m)q−1
1 σ
f (m)dm)ds + ∆
[ 2 (β 2 − α2 )−
t(η(δ − γ) − 1)] α ( 0
Γ(q)
R δ R s (s−m)
R1
q−1
q−1
(σ(β − α) − 1)t][η γ ( 0
f (m)dm)ds − 0 (1−s)
Γ(q)
Γ(q) f (s)ds],
x(t) =
1
Γ(q)
0
where
η
σ
∆ = [ (δ 2 − γ 2 ) − 1)][σ(β − α) − 1] − [ (β 2 − α2 )][η(δ − γ) − 1] 6= 0.
2
2
q−1
η 2
σ
2
Denote A(t, s) = (t−s)
Γ(q) χ[0,t] (s), B(t, s) = ∆Γ(q) [−( 2 (δ − γ ) − 1) +
1
1
σ
t(η(δ−γ)−1)] q [(β−s)q χ[0,β] (s)−(α−s)q χ[0,α] (s)], C(t, s) = ∆Γ(q) [ 2 (β 2 −α2 )−
1
(σ(β − α) − 1)t] ηq [(δ − s)q χ[0,δ] (s) − (γ − s)q χ[0,γ] (s)], D(t, s) = − ∆Γ(q)
[ σ2 (β 2 −
α2 )−(σ(β −α)−1)t](1−s)q−1 and G3 (t, s) = A(t, s)+B(t, s)+C(t, s)+D(t, s),
where χS (.) is the characteristic function of the set S. Then the solution x(.)
R1
in Lemma 3 may be written as x(t) = 0 G3 (t, s)f (s)ds. Moreover, for any
t, s ∈ I we have
|G3 (t, s)| ≤
σ
η
β q + αq
1
+
[| (δ 2 − γ 2 ) − 1| + |η(δ − γ) − 1|]
+
Γ(q) |∆|Γ(q) 2
q
σ
η
1
[| (β 2 − α2 )| + |σ(β − α) − 1|][ (δ q + γ q ) + 1] =: M3 .
|∆|Γ(q) 2
q
77
A FRACTIONAL INTEGRO-DIFFERENTIAL INCLUSION
The proof of the following lemma may be found in [4].
Lemma 2.5. For a given f (.) ∈ C(I, R) the unique solution of the problem
(6)-(5) is given by
Rt
Rµ
q+ω−1
1
x(t) = Γ(q)
(t − s)q−1 f (s)ds + (c1 − tc4 ) 0 (µ−s)
Γ(q+ω) f (s)ds+
0
Rθ
R1
q+ν−1
1
q−1
(c2 + c3 t)[b 0 (θ−s)
f (s)ds],
Γ(q+ν) f (s)ds − Γ(q) 0 (1 − s)
where
c1 =
a
bθν+1
aµω+1
1
aµω
a
(1 −
), c2 =
, c3 = (1 −
), c4 = (1−
c
Γ(ν + 2)
cΓ(ω + 2)
c
Γ(ω + 1)
c
bθν
aµω
bθν+1
aµω+1
bθν
), c = (1 −
)(1 −
)+
(1 −
).
Γ(ν + 1)
Γ(ω + 1)
Γ(ν + 2)
Γ(ω + 2)
Γ(ν + 1)
It is implicitly assumed that c 6= 0.
q−1
(µ−s)q+ω−1
Denote A1 (t, s) = (t−s)
Γ(q) χ[0,t] (s), B1 (t, s) = (c1 − tc4 ) Γ(q+ω) χ[0,µ] (s),
−
q+ν−1
q−1
(1−s)
C1 (t, s) = b(c2 + c3 t) (θ−s)
and
Γ(q+ν) χ[0,θ] (s), D1 (t, s) = −(c2 + c3 t) Γ(q)
G4 (t, s) = A1 (t, s) + B1 (t, s) + C1 (t, s) + D1 (t, s), then the solution x(.) in
R1
Lemma 4 may be written as x(t) = 0 G4 (t, s)f (s)ds. Moreover, for any t, s ∈ I
we have
q+ω−1
q+ν−1
1
|G4 (t, s)| ≤ Γ(q)
+ (|c1 | + |c4 |) µΓ(q+ω) + |b|(|c2 | + |c3 |) θΓ(q+ν) +
1
=: M4
(|c2 | + |c3 |) Γ(q)
3
The main results
In order to prove our results we need the following hypotheses.
Hypothesis. i) F (., .) : I × R × R → P(R) has nonempty closed values and
is L(I) ⊗ B(R × R) measurable.
ii) There exists L(.) ∈ L1 (I, (0, ∞)) such that, for almost all t ∈ I, F (t, ., .)
is L(t)-Lipschitz in the sense that
dH (F (t, x1 , y1 ), F (t, x2 , y2 )) ≤ L(t)(|x1 − x2 | + |y1 − y2 |) ∀ x1 , x2 , y1 , y2 ∈ R.
iii) k(., ., .) : I × R × R → R is a function such that ∀x ∈ R, (t, s) →
k(t, s, x) is measurable.
iv) |k(t, s, x) − k(t, s, y)| ≤ L(t)|x − y| a.e. (t, s) ∈ I × I, ∀ x, y ∈ R.
We use next the following notations
Z t
M (t) := L(t)(1 +
L(u)du),
0
Z
t ∈ I,
M0 =
T
M (t)dt.
0
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A FRACTIONAL INTEGRO-DIFFERENTIAL INCLUSION
Theorem 3.1. Assume that Hypothesis is satisfied and M1 M0 < 1. Let
y(.) ∈ C(I, R) be such that T y 0 (0) = −ay(0) − by(T ), T y 0 (T ) = by(0) + dy(T )
and there exists p(.) ∈ L1 (I, R+ ) with d(Dcq y(t), F (t, y(t), V (y)(t))) ≤ p(t) a.e.
(I).
Then there exists x(.) ∈ C(I, R) a solution of problem (1)-(2) satisfying
for all t ∈ I
Z T
M1
p(t)dt.
(7)
|x(t) − y(t)| ≤
1 − M1 M0 0
Proof. The set-valued map t → F (t, y(t), V (y)(t)) is measurable with closed
values and
F (t, y(t), V (y)(t)) ∩ {Dcq y(t) + p(t)[−1, 1]} =
6 ∅
a.e. (I).
It follows (e.g., Theorem 1.14.1 in [5]) that there exists a measurable selection f1 (t) ∈ F (t, y(t), V (y)(t)) a.e. (I) such that
|f1 (t) − Dcq y(t)| ≤ p(t) a.e. (I)
Define x1 (t) =
RT
0
(8)
G1 (t, s)f1 (s)ds and one has
Z
T
|x1 (t) − y(t)| ≤ M1
p(t)dt.
0
We claim that it is enough to construct the sequences xn (.) ∈ C(I, R),
fn (.) ∈ L1 (I, R), n ≥ 1 with the following properties
T
Z
xn (t) =
G1 (t, s)fn (s)ds,
t ∈ I,
(9)
0
fn (t) ∈ F (t, xn−1 (t), V (xn−1 )(t)) a.e. (I),
(10)
Z t
|fn+1 (t)−fn (t)| ≤ L(t)(|xn (t)−xn−1 (t)|+
L(s)|xn (s)−xn−1 (s)|ds) a.e. (I)
0
(11)
If this construction is realized then from (8)-(11) we have for almost all
t∈I
Z T
n
|xn+1 (t) − xn (t)| ≤ M1 (M1 M0 )
p(t)dt ∀n ∈ N.
0
Indeed, assume that the last inequality is true for n − 1 and we prove it
for n. One has
Z T
|xn+1 (t) − xn (t)| ≤
|G1 (t, t1 )|.|fn+1 (t1 ) − fn (t1 )|dt1 ≤
0
79
A FRACTIONAL INTEGRO-DIFFERENTIAL INCLUSION
Z
T
Z
0
Z
t1
L(t1 )[|xn (t1 ) − xn−1 (t1 )| +
M1
L(s)|xn (s) − xn−1 (s)|ds]dt1 ≤ M1
0
T
Z
L(t1 )(1 +
0
t1
L(s)ds)dt1 .M1n M0n−1
0
Z
T
p(t)dt = M1 (M1 M0 )n
Z
T
p(t)dt
0
0
Therefore {xn (.)} is a Cauchy sequence in the Banach space C(I, R), hence
converging uniformly to some x(.) ∈ C(I, R). Therefore, by (11), for almost
all t ∈ I, the sequence {fn (t)} is Cauchy in R. Let f (.) be the pointwise limit
of fn (.).
Moreover, one has
Pn−1
|xn (t) − y(t)| ≤ |x1 (t) − y(t)| + i=1 |xi+1 (t) − xi (t)|
≤
R
RT
RT
(12)
Pn−1
M1 0T p(t)dt
i
M1 0 p(t)dt + i=1 (M1 0 p(t)dt)(M1 M0 ) = 1−M1 M0 .
On the other hand, from (8), (11) and (12) we obtain for almost all t ∈ I
Pn−1
|fn (t) −R Dcq y(t)| ≤ i=1 |fi+1 (t) − fi (t)| + |f1 (t) − Dcq y(t)| ≤
L(t)
M1 0T p(t)dt
1−M1 M0
+ p(t).
Hence the sequence fn (.) is integrably bounded and therefore f (.) ∈ L1 (I, R).
Using Lebesgue’s dominated convergence theorem and taking the limit in
(9), (10) we deduce that x(.) is a solution of (1)-(2). Finally, passing to the
limit in (12) we obtained the desired estimate on x(.).
It remains to construct the sequences xn (.), fn (.) with the properties in
(9)-(11). The construction will be done by induction.
Since the first step is already realized, assume that for some N ≥ 1 we
already constructed xn (.) ∈ C(I, R) and fn (.) ∈ L1 (I, R), n = 1, 2, ...N
satisfying (9), (11) for n = 1, 2, ...N and (10) for n = 1, 2, ...N − 1. The
set-valued map t → F (t, xN (t), V (xN )(t)) is measurable. Moreover, the map
Rt
t → L(t)(|xN (t) − xN −1 (t)| + 0 L(s)|xN (s) − xN −1 (s)|ds) is measurable. By
the lipschitzianity of F (t, .) we have that for almost all t ∈ I
F (t, xN (t)) ∩ {fN (t) + L(t)(|xN (t) − xN −1 (t)|+
Rt
L(s)|xN (s) − xN −1 (s)|ds)[−1, 1]} =
6 ∅.
0
Theorem 1.14.1 in [5] yields that there exist a measurable selection fN +1 (.) of
F (., xN (.), V (xN )(.)) such that for almost all t ∈ I
Z t
|fN +1 (t) − fN (t)| ≤ L(t)(|xN (t) − xN −1 (t)| +
L(s)|xN (s) − xN −1 (s)|ds).
0
We define xN +1 (.) as in (9) with n = N + 1. Thus fN +1 (.) satisfies (10)
and (11) and the proof is complete.
A FRACTIONAL INTEGRO-DIFFERENTIAL INCLUSION
80
The proofs of the next three theorems are similar to the proof of Theorem
3.1.
Theorem 3.2. Assume that Hypothesis is satisfied and M2 M0 < 1. Let
y(.) ∈ C(I, R) be such that y(T ) = αy(0) + βT y 0 (0), T y 0 (T ) = γy(0) + δT y 0 (0)
and there exists p(.) ∈ L1 (I, R) with d(Dcq y(t), F (t, y(t, V (y)(t)))) ≤ p(t) a.e.
(I).
Then there exists x(.) ∈ C(I, R) a solution of problem (1)-(3) satisfying
for all t ∈ I
Z T
M2
p(t)dt.
|x(t) − y(t)| ≤
1 − M2 M0 0
Theorem 3.3. Assume that Hypothesis is satisfied and M3 M0 < 1. Let
Rβ
Rδ
y(.) ∈ C(I, R) be such that y(0) = σ α y(s)ds, y(1) = η γ y(s)ds and there
exists p(.) ∈ L1 (I, R+ ) with d(Dcq y(t), F (t, y(t), V (y)(t))) ≤ p(t) a.e. (I).
Then there exists x(.) ∈ C(I, R) a solution of problem (1)-(4) satisfying
for all t ∈ I
Z 1
M3
|x(t) − y(t)| ≤
p(t)dt.
1 − M 3 M0 0
Theorem 3.4. Assume that Hypothesis is satisfied and M4 M0 < 1 Let y(.) ∈
C(I, R) be such that y(0) = aI ω y(µ), y(1) = bI ν y(θ) and there exists p(.) ∈
L1 (I, R+ ) with d(Dcq y(t), F (t, y(t), V (y)(t))) ≤ p(t) a.e. (I).
Then there exists x(.) ∈ C(I, R) a solution of problem (1)-(5) satisfying
for all t ∈ I
Z 1
M4
|x(t) − y(t)| ≤
p(t)dt.
1 − M 4 M0 0
Remark 3.5. If F (., ., .) does not depend on the last variable the fractional
integrodifferential inclusion (1) reduces to
Dcq x(t) ∈ F (t, x(t)) a.e. (I)
and Theorem 3.1 yields Theorem 3.3 in [7], Theorem 3.2 yields Theorem 3.4
in [7], Theorem 3.3 yields Theorem 3.4 in [8] and Theorem 3.4 yields Theorem
3.5 in [9].
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equations of mixed type with time varying generating operators and optimal control, Opuscula Math. 30 (2010), 361-381.
Aurelian CERNEA,
Faculty of Mathematics and Computer Science,
University of Bucharest,
Academiei 14, 010014 Bucharest, Romania.
Email: acernea@fmi.unibuc.ro