Boundary Value Problems for Differential Equations with Fractional Order and Nonlinear

Mouffak Benchohra ^∗ , Samira Hamani

Boundary Value Problems for Differential Equations with Fractional Order and Nonlinear

Integral Conditions

Abstract. In this paper, we shall establish sufficient conditions for the existence of solutions for a class of boundary value problem for fractional differential equations involving the Caputo fractional derivative and nonlinear integral conditions.

2000 Mathematics Subject Classification: 26A33, 34B15.

Key words and phrases: boundary value problem, Caputo fractional derivative, frac- tional integral, existence, uniqueness, fixed point, integral conditions.

1. Introduction. This paper deals with the existence and uniqueness of solu- tions for the boundary value problems (BVP for short), for fractional order diffe- rential equations and nonlinear integral conditions of the form

(1) ^c D ^α y(t) = f (t, y), for each t ∈ J = [0, T ], 1 < α ¬ 2,

(2) y(0) =

Z T

0 g(s, y)ds,

(3) y(T ) =

Z T

0 h(s, y)ds,

where ^c D ^α is the Caputo fractional derivative, f, g, h : J × ℝ → ℝ are given continuous functions.

Differential equations of fractional order have recently proved to be valuable to- ols in the modeling of many phenomena in various fields of science and engineering.

∗

Corresponding author

Indeed, we can find numerous applications in viscoelasticity, electrochemistry, con- trol, porous media, electromagnetic, ect. (see [10, 16, 17, 19, 26, 27, 31]). There has been a significant development in fractional differential and partial differential equations in recent years; see the monographs of Kilbas et al [22], Miller and Ross [28], Samko et al [35] and the papers of Delbosco and Rodino [9], Diethelm et al [10, 11, 12], El-Sayed [13, 14, 15], Kaufmann and Mboumi [20], Kilbas and Ma- rzan [21], Mainardi [26], Momani and Hadid [29], Momani et al [30], Podlubny et al [34], Yu and Gao [36] and Zhang [37] and the references therein. Very recently some basic theory for the initial value problems of fractional differential equations involving Riemann-Liouville differential operator of order 0 < α ¬ 1 has been di- scussed by Lakshmikantham and Vatsala [23, 24, 25]. In [4, 7] the authors studied the existence and uniqueness of solutions of classes of initial value problems for functional differential equations with infinite delay and fractional order, and in [3] a class of perturbed functional differential equations involving the Caputo fractional derivative has been considered. Related problems to (1)-(3) have been considered by means of different methods by Belarbi et al. [2] and Benchohra et al. in [5, 6] in the case of α = 2.

Applied problems require definitions of fractional derivatives allowing the utili- zation of physically interpretable initial conditions, which contain y(0), y ⁰ (0), etc.

the same requirements of boundary conditions. Caputo’s fractional derivative satis- fies these demands. For more details on the geometric and physical interpretation for fractional derivatives of both the Riemann-Liouville and Caputo types see [33].

Boundary value problems with integral boundary conditions constitute a very interesting and important class of problems. They include two, three, multipoint and nonlocal boundary value problems as special cases. Integral boundary conditions appear in population dynamic [8], and Cellular systems [1].

In this paper, we present existence results for the problem (1)-(3). We give three results, one based on Banach fixed point theorem (Theorem 2.10) and another one based on Schaefer’s fixed point theorem (Theorem 3.6) and the third on the nonlinear alternative of Leray-Schauder type (Theorem 3.7). These results can be considered as a contribution to this emerging field.

2. Preliminaries. In this section, we introduce notations, definitions, and preliminary facts which are used throughout this paper. By C(J, ℝ) we denote the Banach space of all continuous functions from J into ℝ with the norm

kyk ∞ := sup{|y(t)| : t ∈ J}.

Definition 2.1 ([22, 32]). The fractional (arbitrary) order integral of the function h ∈ L ¹ ([a, b], ℝ + ) of order α ∈ ℝ + is defined by

I _a ^α h(t) = Z t

a

(t − s) ^{α −1} Γ(α) h(s)ds,

where Γ is the gamma function. When a = 0, we write I ^α h(t) = [h ∗ ϕ ^α ](t), where ϕ α (t) = t ^{α −1}

Γ(α) for t > 0, and ϕ α (t) = 0 for t ¬ 0, and ϕ ^α → δ(t) as α → 0, where δ

is the delta function.

Definition 2.2 ([22, 32]). For a function h given on the interval [a, b], the αth Riemann-Liouville fractional-order derivative of h, is defined by

(D ^α _a+ h)(t) = 1 Γ(n − α)

 d dt

 ⁿ Z t

a (t − s) ^{n −α−1} h(s)ds.

Here n = [α] + 1 and [α] denotes the integer part of α.

Definition 2.3 ([21]). For a function h given on the interval [a, b], the Caputo fractional-order derivative of h, is defined by

( ^c D _a+ ^α h)(t) = 1 Γ(n − α)

Z t

a (t − s) ^{n −α−1} h ⁽ⁿ⁾ (s)ds.

Here n = [α] + 1.

3. Existence of Solutions. Let us start by defining what we mean by a solution of the problem (1)–(3).

Definition 3.1 A function y ∈ C ² (J, ℝ) is said to be a solution of (1)–(3) if y satisfies equations (1)–(3).

For the existence of solutions for the problem (1)–(3), we need the following auxiliary lemma:

Lemma 3.2 ([22]) Let α > 0, then the differential equation

c D ^α h(t) = 0

has solutions h(t) = c 0 +c 1 t+c 2 t ² +. . .+c _n−1 t ^{n −1} , c i ∈ ℝ, i = 0, 1, 2, . . . , n−1, n = [α] + 1.

Lemma 3.3 [22] Let α > 0, then

I ^αc D ^α h(t) = c 0 + c 1 t + c 2 t ² + . . . + c _n−1 t ^{n −1} for some c i ∈ ℝ, i = 0, 1, 2, . . . , n − 1, n = [α] + 1.

As a consequence of Lemma 3.2 and Lemma 3.3 we have the following result which is useful in what follows.

Lemma 3.4 Let 1 < α ¬ 2 and let σ, ρ ¹ , ρ 2 : J → ℝ be continuous. A function y is a solution of the fractional integral equation

(4)

y(t) = 1

Γ(α) Z t

0 (t − s) ^{α −1} σ(s)ds

− t

T Γ(α) Z T

0 (T − s) ^{α −1} σ(s)ds

−

 t T − 1

 Z T

0 ρ 1 (s)ds + t T

Z T

0 ρ 2 (s)ds.

if and only if y is a solution of the fractional BVP

(5) ^c D ^α y(t) = σ(t), t ∈ J,

(6) y(0) =

Z T 0

ρ 1 (s)ds

(7) y(T ) =

Z T

0 ρ 2 (s)ds.

Proof Assume y satisfies (5), then Lemma 3.3 implies that

y(t) = c 0 + c 1 t + 1 Γ(α)

Z t

0 (t − s) ^α−1 h(s)ds.

From (6) and (7), we get

c 0 = Z T

0 ρ 1 (s)ds, and

y(T ) = Z T

0 ρ 1 (s)ds + c 1 T + 1 Γ(α)

Z T

0 (T − s) ^{α −1} h(s)ds

= Z T 0

ρ 2 (s)ds.

Hence we obtain equation (4). Inversely, it is clear that if y satisfies equation (4),

then equations (5)-(7) hold. _■

Our first result is based on Banach fixed point theorem.

Theorem 3.5 Assume that:

(H1) There exists a constant k > 0 such that

|f(t, u) − f(t, u)| ¬ k|u − u|, for each t ∈ J, and all u, u ∈ ℝ.

(H2) There exists a constant k ^∗ > 0 such that

|g(t, u) − g(t, u)| ¬ k ^∗ |u − u|, for each t ∈ J, and all u, u ∈ ℝ.

(H3) There exists a constant k ^∗∗ > 0 such that

|h(t, u) − h(t, u)| ¬ k ^∗∗ |u − u|, for each t ∈ J, and all u, u ∈ ℝ.

If

(8) 2kT ^α

Γ(α + 1) + T (k ^∗ + k ^∗∗ ) < 1

then the BVP (1)-(3) has at least one solution on J.

Proof Transform the problem (1)–(3) into a fixed point problem. Consider the operator

F : C(J, ℝ) → C(J, ℝ) defined by

F (y)(t) = 1 Γ(α)

Z t

0 (t − s) ^{α −1} f (s, y(s))ds

− t

T Γ(α) Z T

0 (T − s) ^α−1 f (s, y(s))ds

−

 t T − 1

 Z T

0 g(s, y(s))ds + t T

Z T

0 h(s, y(s))ds.

Clearly, the fixed points of the operator F are solution of the problem (1)-(3). We shall use the Banach contraction principle to prove that F has a fixed point. We shall show that F is a contraction.

Let x, y ∈ C(J, ℝ). Then, for each t ∈ J we have

|F (x)(t) − F (y)(t)| ¬ 1 Γ(α)

Z t

0 (t − s) ^{α −1} |f(s, x(s)) − f(s, y(s))|ds

+ 1

Γ(α) Z T

0 (T − s) ^{α −1} |f(s, x(s)) − f(s, y(s))|ds + Z T

0 |g(s, x(s)) − g(s, y(s))|ds + Z T

0 |h(s, x(s)) − h(s, y(s))|ds

¬ k kx − yk ∞

Γ(α) Z t

0 (t − s) ^α−1 ds + k kx − yk ∞

Γ(α) Z T

0 (T − s) ^α−1 ds + T k ^∗ kx − yk ∞ + T k ^∗∗ kx − yk ∞

¬ 2kT ^α

Γ(α + 1) k x − yk ∞ + T (k ^∗ + k ^∗∗ )kx − yk ∞ . Thus

kF (x) − F (y)k ∞ ¬

 2kT ^α

Γ(α + 1) + T (k ^∗ + k ^∗∗ )



kx − yk ∞ .

Consequently by (8) F is a contraction. As a consequence of Banach fixed point theorem, we deduce that F has a fixed point which is a solution of the problem

(1) − (3). ■

The second result is based on Schaefer’s fixed point theorem [18].

Theorem 3.6 Assume that:

(H4) The function f : J × ℝ → ℝ is continuous.

(H5) There exists a constant M > 0 such that

|f(t, u)| ¬ M for each t ∈ J and all u ∈ ℝ.

(H6) There exists a constant M 1 > 0 such that

|g(t, u)| ¬ M ¹ for all t ∈ J and all u ∈ ℝ.

(H7) There exists a constant M 2 > 0 such that

|h(t, u)| ¬ M ² for all t ∈ J and all u ∈ ℝ.

Then the BVP (1)-(3) has at least one solution on J.

Proof We shall use Schaefer’s fixed point theorem to prove that F has a fixed point. The proof will be given in several steps.

Step 1: F is continuous.

Let {y ⁿ } be a sequence such that y ⁿ → y in C(J, ℝ). Then for each t ∈ J

|F (y ⁿ )(t) − F (y)(t)| ¬ 1 Γ(α)

Z t

0 (t − s) ^{α −1} |f(s, y ⁿ (s)) − f(s, y(s))|ds

+ 1

Γ(α) Z T

0 (T − s) ^{α −1} |f(s, y ⁿ (s)) − f(s, y(s))|ds + Z T

0 |g(s, y ⁿ (s)) − g(s, y(s))|ds + Z T

0 |h(s, y ⁿ (s)) − h(s, y(s))|ds.

Since f, g and h are continuous functions, then we have kF (y ⁿ ) − F (y)k ∞ → 0 as n → ∞.

Step 2: F maps bounded sets into bounded sets in C(J, ℝ).

Indeed, it is enough to show that for any η ^∗ > 0, there exists a positive constant

` such that for each y ∈ B ^η

= {y ∈ C(J, ℝ) : kyk ∞ ¬ η ^∗ }, we have kF (y)k ∞ ¬ `.

By (H5)-(H7) we have for each t ∈ J,

|F (y)(t)| ¬ 1 Γ(α)

Z t

0 (t − s) ^{α −1} |f(s, y(s))|ds

+ 1

Γ(α) Z T

0 (T − s) ^α−1 |f(s, y(s))|ds +

Z T

0 |g(s, y(s))|ds + Z T

0 |h(s, y(s))|ds

¬ M

Γ(α) Z t

0 (t − s) ^α−1 ds + M Γ(α)

Z T

0 (T − s) ^α−1 ds + T M 1 + T M 2

¬ M

αΓ(α) T ^α + M

αΓ(α) T ^α + T M ₁ + T M ₂ . Thus

kF (y)k ∞ ¬ 2MT ^α

Γ(α + 1) + T M 1 + T M 2 := `.

Step 3: F maps bounded sets into equicontinuous sets of C(J, ℝ).

Let t 1 , t 2 ∈ J, t ¹ < t 2 , B η

be a bounded set of C(J, ℝ) as in Step 2, and let y ∈ B ^η

. Then

|F (y)(t ² ) − F (y)(t ¹ )| =    1 Γ(α)

Z t

0

[(t 2 − s) ^α−1 − (t ¹ − s) ^α−1 ]f(s, y(s))ds

+ 1

Γ(α) Z t

t

(t ₂ − s) ^{α −1} f (s, y(s))ds    + (t 2 − t ¹ )

T Z T

0 |g(s, y(s))|ds + (t 2 − t ¹ ) T

Z T

0 |h(s, y(s))|ds

¬ M

Γ(α) Z t

0

[(t ₁ − s) ^{α −1} − (t ² − s) ^{α −1} ]ds

+ M

Γ(α) Z t

t

(t 2 − s) ^{α −1} ds + (t ₂ − t 1 )

T M 1 + (t ₂ − t 1 )

T M 2

¬ M

Γ(α + 1) [(t 2 − t ¹ ) ^α + t ^α ₁ − t ^α 2 ] + M

Γ(α + 1) (t 2 − t ¹ ) ^α + (t ₂ − t 1 )

T M 1 + (t ₂ − t 1 ) T M 2 .

As t 1 −→ t ² , the right-hand side of the above inequality tends to zero. As a con- sequence of Steps 1 to 3 together with the Arzel´a-Ascoli theorem, we can conclude that F : C(J, ℝ) −→ C(J, ℝ) is continuous and completely continuous.

Step 4: A priori bounds.

Now it remains to show that the set

E = {y ∈ C(J, ℝ) : y = λF (y) for some 0 < λ < 1}

is bounded.

Let y ∈ E, then y = λF (y) for some 0 < λ < 1. Thus, for each t ∈ J we have

y(t) = λ

Γ(α) Z t

0 (t − s) ^{α −1} f (s, y(s))ds

− λt

T Γ(α) Z T

0 (T − s) ^{α −1} f (s, y(s))ds

− λ

 t T − 1

 Z T

0 g(s, y(s))ds + λ t T

Z T

0 h(s, y(s))ds.

This implies by (H5)-(H7) that for each t ∈ J we have

|y(t)| ¬ M Γ(α)

Z t

0 (t − s) ^α−1 ds

+ M

Γ(α) Z T

0 (T − s) ^{α −1} ds + T M 1 + T M 2

¬ M

αΓ(α) T ^α + M

αΓ(α) T ^α + T (M 1 + M 2 ).

Thus for every t ∈ J, we have

kyk ∞ ¬ 2MT ^α

Γ(α + 1) + T (M 1 + M 2 ) := R.

This shows that the set E is bounded. As a consequence of Schaefer’s fixed point theorem, we deduce that F has a fixed point which is a solution of the problem

(1) − (3). ■

In the following theorem we shall give an existence result for the problem (1)−(3) by means of an application of the nonlinear alternative of Leray-Schauder type [18], and which the conditions (H5)-(H7) are weakened.

Theorem 3.7 Assume that (H4) and

(H8) There exists φ f ∈ L ¹ (J, ℝ ⁺ ) and ψ : [0, ∞) → (0, ∞) continuous and nonde- creasing such that

|f(t, u)| ¬ φ ^f (t)ψ(|u|) for each t ∈ J and all u ∈ ℝ.

(H9) There exists φ g ∈ L ¹ (J, ℝ ⁺ ) and ψ ^∗ : [0, ∞) → (0, ∞) continuous and nonde- creasing such that

|g(t, u)| ¬ φ ^g (t)ψ ^∗ (|u|) for each t ∈ J and all u ∈ ℝ.

(H10) There exists φ h ∈ L ¹ (J, ℝ ⁺ ) and ψ : [0, ∞) → (0, ∞) continuous and nonde- creasing such that

|h(t, u)| ¬ φ ^h (t)ψ(|u|) for each t ∈ J and all u ∈ ℝ.

(H11) There exists an number M > 0 such that

(9) M

kI ^α φ f k L

ψ(M ) + (I ^α φ f )(T )ψ(M) + aψ ^∗ (M) + bψ(M) > 1, where

a = Z T

0 φ g (s)ds, b = Z T

0 φ h (s)ds.

Then the BVP (1)-(3) has at least one solution on J.

Proof Consider the operator F defined in Theorems 3.5 and 3.6. It can be easily shown that F is continuous and completely continuous. For λ ∈ [0, 1], let y be such that for each t ∈ J we have y(t) = λ(F y)(t). Then from (H8)-(H10) we have for each t ∈ J

|y(t)| ¬ 1 Γ(α)

Z t

0 (t − s) ^{α −1} φ f (s)ψ(|y(s)|)ds

+ 1

Γ(α) Z T

0 (T − s) ^{α −1} φ f (s)ψ(|y(s)|)ds + Z T

0 φ g (s)ψ ^∗ (|u(s)|)ds + Z T

0 φ h (s)ψ(|u(s)|)ds

¬ ψ(kyk ∞ ) 1 Γ(α)

Z t

0 (t − s) ^{α −1} φ f (s)ds + ψ(kyk ∞ ) 1

Γ(α) Z T

0 (T − s) ^{α −1} φ f (s)ds + ψ ^∗ (kyk ∞ ) Z T

0 φ g (s)ds + ψ(kyk ∞ ) Z T

0 φ h (s)ds.

Thus

kyk ∞

ψ( kyk ∞ )kI ^α φ f k ^L

+ (I ^α φ f )(T )ψ(kyk ∞ ) + aψ ^∗ (kyk ∞ ) + bψ(kyk ∞ ) ¬ 1.

Then by condition (8), there exists M such that kyk ∞ 6= M.

Let

U = {y ∈ C(J, ℝ) : kyk ∞ < M }.

The operator F : U → C(J, ℝ) is continuous and completely continuous. From the choice of U, there is no y ∈ ∂U such that y = λF (y) for some λ ∈ (0, 1). As a consequence of the nonlinear alternative of Leray-Schauder type we deduce that F has a fixed point y in U which is a solution of the problem (1)–(3). This completes

the proof. _■

4. An Example. In this section we give an example to illustrate the usefulness of our main results. Let us consider the following fractional boundary value problem, (10) ^c D ^α y(t) = e ^−t |y(t)|

(9 + e ^t )(1 + |y(t)|) , t ∈ J := [0, 1], 1 < α ¬ 2,

(11) y(0) =

X ∞ i=0

c i y(t i ),

(12) y(1) =

X ∞ j=0

d j y(˜ t j ),

where 0 < t 0 < t 1 < t 2 < . . . < 1, 0 < ˜ t 0 < ˜ t 1 < ˜ t 2 < . . . < 1, c i , d j , i, j = 0, . . . , are given positive constants with

X ∞ i=0

c i < ∞, X ∞ j=0

d i < ∞,

and X ∞

i=0

c i + X ∞ j=0

d i = 4 5 . Set

f (t, x) = e ^−t x

(9 + e ^t )(1 + x) , (t, x) ∈ J × [0, ∞).

Let x, y ∈ [0, ∞) and t ∈ J. Then we have

|f(t, x) − f(t, y)| = e ^−t (9 + e ^t )

   x

1 + x − y 1 + y

= e ^−t |x − y|

(9 + e ^t )(1 + x)(1 + y)

¬ e ^−t

(9 + e ^t ) | x − y|

¬ 1

10 | x − y|.

Hence the condition (H1) holds with k = 1

10 . We shall check that condition (8) is satisfied with T = 1, k ^∗ =

X ∞ i=0

c i and k ^∗∗ = X ∞ i=0

d i . Indeed

(13) 2kT ^α

Γ(α + 1) + T (k ^∗ + k ^∗∗ ) = 1 5Γ(α + 1) +

X ∞ i=0

c i + X ∞ j=0

d j < 1 ⇔ Γ(α + 1) > 1,

which is satisfied for each α ∈ (1, 2]. Then by Theorem 3.5 the problem (10)-(12)

has a unique solution on [0, 1].

Remark 4.1 We can choose, for instance, the constants c i and d j as

c i = 2 5

 1 3

 i

, d j = 2 15

 1 3

 j

. In this case we have

X ∞ i=0

c i = 3 5 ,

X ∞ j=0

d j = 1 5 .

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Mouffak Benchohra

Laboratoire de Math´ematiques, Universit´e de Sidi Bel-Abb`es, B.P. 89, 22000, Sidi Bel-Abb`es, Alg´erie E-mail: benchohra@univ-sba.dz

Samira Hamani

Laboratoire de Math´ematiques, Universit´e de Sidi Bel-Abb`es, B.P. 89, 22000, Sidi Bel-Abb`es, Alg´erie E-mail: hamani samira@yahoo.fr

(Received: 22.01.2008)