Vietnam J. Math. DOI 10.1007/s10013-016-0218-7

Positive Solutions for Higher-Order Nonlinear Fractional Differential Equations Abdourazek Souahi1 Amara Hitta1

· Assia Guezane-Lakoud2 ·

Received: 18 November 2015 / Accepted: 27 April 2016 © Vietnam Academy of Science and Technology (VAST) and Springer Science+Business Media Singapore 2016

Abstract This paper investigates the existence of positive solutions for a class of higherorder nonlinear fractional differential equations with initial conditions given for ordinary as well as fractional derivatives of the unknown function. We assume that the nonlinear term f involves also derivatives of fractional order. The results are established by converting the problem into an equivalent integral equation and applying Guo–Krasnoselskii’s fixed-point theorem in cones. Keywords Boundary value problem · Fractional derivative · Fixed point theorem Guo–Krasnoselskii · Positive solution · Fractional differential equation Mathematics Subject Classification (2010) 34B18 · 34A08 · 26A33

1 Introduction Let us consider the following higher order fractional boundary value problem with Riemann–Liouville type derivative ⎧ q ⎨ D u(t) + g(t)f (u(t), u (t), u (t), . . . , u(m) (t), D α u(t)) = 0, u(0) = 0, u(i) (0) = 0, i = 1, . . . , n − 2, ⎩ β D u(1) = 0, 2 ≤ β ≤ n − 2, 1 ≤ m ≤ α ≤ β − 1,

n − 1 < q ≤ n,

 Abdourazek Souahi

[email protected] 1

Laboratory of Applied Mathematics and Modeling, University of 8 May 1945 Guelma, P.O. Box 401, 24000 Guelma, Algeria

2

Laboratory of Advanced Materials, University Badji Mokhtar-Annaba, P.O. Box 12, 23000 Annaba, Algeria

(1)

A. Souahi et al.

where n > 3 and (H1) (H2)

g ∈ L1 ([0, 1], R∗+ ), f ∈ C(Rm+2 + , R+ ).

The research on fractional differential boundary value problems plays a very important role in both theory and applications. It is widely used in modeling many phenomena in sciences, such the behavior of viscoelastic and viscoplastic materials under external influences [3] and the continuum and statistical mechanics for visco-elasticity problems [8] and so on, consequently, a great number of papers and books on the theory and applications of fractional calculus have appeared, we refer to [1, 7, 9, 11, 12]. Boundary value problems for fractional differential equations have been studied extensively in the last decades and by different methods as fixed point theorems, upper and lower solution method, coincidence degree theory of Mawhin. For some recent contributions in fractional boundary value problems (FBVP), one can see the papers of Zhang et al. [13–18], Bai [2], Goodrich [4], Sotiris et al. [10] and Guezane-Lakoud et al. [5] and the references therein. Bai [2] considered the existence of positive solutions of the fractional boundary value problem: D0α u(t) + f (t, u(t)) = 0, u(0) = 0,

0 < t < 1, 1 < α ≤ 2

βu(η) = u(1),

where D0α denotes the Riemann–Liouville fractional derivative. Goodrich [4] studied a similar problem −D0ν u(t) = f (t, u(t)), 0 < t < 1,  α  D u(t) t=1 = 0, 0 ≤ i ≤ n − 2, 1 ≤ α ≤ n − 2, ν > 3. u(i) (0) = 0, The author established the existence of a positive solution using cone-theoretic techniques. Sotiris et al. [10] extended his problem to a nonlinear fractional order differential equation with advanced arguments D0α u(t) + a(t)f (u(θ (t))), 0 < t < 1, n − 1 < α ≤ n  β  D u(t) t=1 = 0, 0 ≤ i ≤ n − 2, 1 ≤ β ≤ n − 2, n > 3. u(i) (0) = 0, Guezane-Lakoud et al. [5] studied the following problem c

D0α u(t) + a(t)f (t, u(t), D σ (t)), 





0 < t < 1, 2 < α < 3

u(0) = u (0) = u (0) = 0, u (1) = D σ u(t), c

where c D0α represents the Caputo fractional derivative of order α. Motivated by the previously mentioned works, we consider problem (1) where the nonlinear term does not depend solely on the unknown solution but on its fractional derivative of high order too, and we will introduce an appropriate Banach space to use the well known Guo–Krasnoselskii fixed-point theorem. The rest of the paper is organised as follows. In Section 2, we present the necessary definitions and properties from the fractional calculus theory and some auxiliary results which will be used in the sequel. The last section is devoted to the main result.

Positive Solutions for Higher Order Nonlinear Fractional Differential...

2 Preliminaries Now, let us recall basic definitions about fractional calculus and fixed point theorems. Throughout the paper,  is the Gamma function and [α] is the integer part of α and we denote by {α} = α − [α] the fractional part of α, m and n are integer numbers. We set   Cu0 = I q (1 − t)−β g(t)f (u(t), u (t), u (t), . . . , u(m) (t), D α u(t)) , t=1

Cui

(q) C0 , = (q − i)

Cuα

(q) C0 = (q − α)

and Ru (t) = f (u(t), u (t), u (t), . . . , u(m) (t), D α u(t)). Definition 1 ([7]) The fractional integral of the function h : (0, ∞) → R of order α ∈ R+ is defined by t 1 α (t − s)α−1 h(s)ds, I h(t) = (α) 0 provided the right side is pointwise defined on (0, ∞). Definition 2 ([7]) For a function h ∈ C((0, ∞), R),the Riemann–Liouville fractional derivative of h is defined by

n t d 1 (t − s)n−α−1 h(s)ds, D α h(t) = (n − α) dt 0 where n ∈ N∗ and n = [α] + 1, provided that the right side is pointwise defined on (0, ∞). The following lemmas give some properties of Riemann–Liouville fractional integrals and derivatives. Lemma 1 ([7]) Let p, q ≥ 0 and f ∈ L1 ([0, 1]), then p

q

q

p

p+q

I0+ I0+ h(t) = I0+ I0+ h(t) = I0+ h(t) and q

D q I0+ h(t) = h(t). Lemma 2 ([7]) Let p > q > 0, and h ∈ L1 ([a, b]), then for all t ∈ [a, b] we have p

p−q

D q I0+ h(t) = I0+ h(t). Lemma 3 The general solution to D q y(t) = 0, where 0 ≤ n − 1 < q ≤ n, is the function y(t) = nj=1 cj t (q−j ) , where cj ∈ R for each j . In this paper, we will use the following lemmas: Lemma 4 ([4]) Assume that y ∈ C([0, 1]), then the following problem ⎧ q ⎨ D u(t) + y(t) = 0, n − 1 < q ≤ n, u(0) = 0, u(i) (0) = 0, i = 1, . . . , n − 2, ⎩ β D u(1) = 0, 1 ≤ β ≤ n − 2,

A. Souahi et al.

has the unique solution



1

u(t) =

G(t, s)y(s)ds, 0

where

⎧ q−1 t (1 − s)q−β−1 − (t − s)q−1 ⎪ ⎪ , 0 ≤ s ≤ t ≤ 1, ⎨ (q) G(t, s) = q−1 q−β−1 (1 − s) ⎪ ⎪t ⎩ , 0 ≤ t ≤ s ≤ 1. (q)

(2)

Lemma 5 Let G(t, s) be as given in (2). Then for every i = 1, . . . , m there exist constants γ , γi , γα given by ⎫ ⎧  q−β−1 ⎪

q−1 ⎪ ⎬ ⎨ 12 1 , , γ = min β ⎪ ⎪ 2 ⎭ ⎩ 2 −1 ⎫ ⎧  q−β−1 ⎪

q−1−i ⎪ ⎬ ⎨ 12 1 , , γi = min β−i − 1 ⎪ ⎪ 2 ⎭ ⎩ 2 ⎫ ⎧  q−β−1 ⎪

q−1−α ⎪ ⎬ ⎨ 12 1 , , γα = min β−α − 1 ⎪ ⎪ 2 ⎭ ⎩2 such that min   G(t, s) ≥ γ max G(t, s) = γ G(1, s), t∈

1 2 ,1

t∈[0,1]

min   Gi (t, s) ≥ γi max Gi (t, s) = γi Gi (1, s), t∈

1 2 ,1

t∈[0,1]

min   Gα (t, s) ≥ γα max Gα (t, s) = γα Gα (1, s), t∈

1 2 ,1

t∈[0,1]

where

⎧ q−1−i t (1 − s)q−β−1 − (t − s)q−1−i ⎪ ⎪ , 0 ≤ s ≤ t ≤ 1, ⎨ i d (q − i) Gi (t, s) = i G(t, s) = q−1−i q−β−1 t (1 − s) ⎪ dt ⎪ ⎩ , 0 ≤ t ≤ s ≤ 1, (q − i) ⎧ q−1−α t (1 − s)q−β−1 − (t − s)q−1−α ⎪ ⎪ , 0 ≤ s ≤ t ≤ 1, ⎨ (q − α) Gα (t, s) = D α G(t, s) = q−1−α q−β−1 (1 − s) t ⎪ ⎪ ⎩ , 0 ≤ t ≤ s ≤ 1. (q − α) Moreover

  m  γ G(1, s) + Gα (1, s) + Gi (1, s) . (3) min   G(t, s) ≥ γ G(1, s) ≥ (m + 2)(q) t∈ 1 ,1 2

i=1

Positive Solutions for Higher Order Nonlinear Fractional Differential...

Proof Using the same process as in [4], we get for every i = 1, . . . , m the desired constants γ , γi and γα . Moreover, using the fact that 1 ≤ i ≤ m ≤ α ≤ β − 1, it follows for every 0 ≤ s ≤ 1 that (q − i) 1 Gi (1, s) ≥ Gi (1, s), (q) (q) 1 (q − α) G(1, s) ≥ Gα (1, s) ≥ Gα (1, s). (q) (q) G(1, s) ≥

Noting that (·) is increasing in (2, ∞), we obtain the desired inequality (3). Theorem 1 ([1, 6]) Let E be a Banach space, and let P ⊂ E be a cone. Assume that 1 and 2 are open bounded subsets contained in E with 0 ∈ 1 , 1 ⊂ 2 . Assume further, that T :

P ∩ ( 2 \ 1 ) → P

is a completely continuous operator. If either 1. 2.

T u ≤ u, u ∈ P ∩ ∂ 1 and T u ≥ u, u ∈ P ∩ ∂ 2 or T u ≤ u, u ∈ P ∩ ∂ 2 and T u ≥ u, u ∈ P ∩ ∂ 1 .

Then T has at least one fixed point in P ∩ ( 2 \ 1 ). Definition 3 Let f : Rm+2 → R and put A0 =

lim

m+2 + i=1 |vi |→0

f (v1 , . . . , vm+2 ) , m+2 i=1 |vi |

A∞ =

lim

m+2 i=1 |vi |→+∞

f (v1 , . . . , vm+2 ) . m+2 i=1 |vi |

The case where A0 = 0 and A∞ = ∞ is called the superlinear case and the case where A0 = ∞ and A∞ = 0 is called the sublinear case. Definition 4 A closed subset P of a Banach space X is called a cone if: 1. ∀x, y ∈ P : x + y ∈ P , 2. ∀x ∈ P , ∀α > 0 : αx ∈ P , 3. P ∩ (−P ) = {0}. A cone P defines a partial order by x ≤ y ⇔ y − x ∈ P . We define the space X as follows X = {u|u ∈ C m [0, 1], D α u ∈ C([0, 1]) and u(0) = 0, u(i) (0) = 0, i = 1, . . . , m}, (i) α endowed with the norm uX = m i=0 {max0≤t≤1 |u (t)|} + max0≤t≤1 |D u(t)|, where (0) u = u and m ≤ α ≤ β. Also, define the cone P ⊂ X by ⎧ ⎫ ⎨ ⎬ γ uX . P = u ∈ X : u(t) > 0 and min   u(t) ≥ ⎩ ⎭ (m + 2)(q) t∈ 1 ,1 2

Lemma 6 (X,  · X ) is a Banach space.

A. Souahi et al. ∞ Proof Let {uk }∞ k=1 be a Cauchy sequence in the space (X,  · X ), obviously {uk }k=1 (i) ∞ and {D α uk }∞ k=1 are Cauchy sequences in the space C([0, 1]). Therefore, {uk }k=1 and ∞ α (i) (i) {D uk }k=1 converge to some v and w in C([0, 1]) uniformly, v , w ∈ C([0, 1]) and v(0) = 0 and w(0) = 0. It suffices to show that w = D α v. We have t 1 α α α |I D uk (t) − I w(t)| ≤ (t − s)q−2 |D α uk (s) − w(s)|ds (q − 1) 0 1 ≤ max |D α uk (t) − w(t)|. (q) t∈[0,1] (i)

α α α Since {D α uk }∞ k=1 converges uniformly, then limk→∞ I D uk (t) = I w(t) uniformly α α on [0, 1]. From Lemma 3, we get I D uk (t) = uk (t). Hence, we have v(t) = I α w(t). From Lemma 2, we obtain w = D α v. Thus, {uk }∞ k=1 converges in X. This completes the proof.

We define T : X → X by 1 G(t, s)g(s)f (u(s), u (s), u (s), . . . , u(m) (s), D α u(s))ds, T u(t) = 0

where g and f satisfy (H1)–(H2). Clearly from Lemma 4, a function u is a solution of the BVP (1) if and only if it is a fixed point of T , u(t) = T u(t)

∀t ∈ [0, 1].

3 Existence Result In this section we deduce the existence of a positive solution to problem (1). To accomplish this we first introduce the following lemma which is crucial to the proof of our result based on the Guo–Krasnoselskii fixed-point theorem. Lemma 7 Assume that the conditions (H1)–(H2) are satisfied, then the operator T : P → P is completely continuous. The proof of this lemma is standard, thus we omit it. Now, we establish the existence result of a positive solution to problem (1). Theorem 2 Suppose that f satisfies (H1)–(H2), then problem (1) has at least one non trivial non-negative solution in the cone P in either the sublinear case or the superlinear case. Proof From Lemma 7, T is completely continuous. (i) Sublinear case: since A0 = +∞, there exists r1 > 0, for any 0 < m+2 i=1 |vi | ≤ r1 , m+2 then f (v1 , . . . , vm+2 ) ≥ M( i=1 |vi |), where M > 0 satisfies 1   γ G 12 , s g(s)ds ≥ 1. M (m + 2)(q) 12

Positive Solutions for Higher Order Nonlinear Fractional Differential...

Let 1 = {u ∈ E, uX < r1 }. Take u ∈ P ∩ ∂ 1 , then T uX ≥ T u∞ ≥ T u

1 2

= ≥ ≥ ≥

 G

0



1

G

1 2



1

G

1 2



1 1 2

G

≥ uX M

1 2,s

1 2,s 1 2,s 1 2,s



 

  1 2





1

=



1 2,1

G 0



g(s)Ru (s)ds +

g(s)Ru (s)ds 

1 1 2

G

1 2,s

 g(s)Ru (s)ds

g(s)f (u(s), u (s), u (s), . . . , u(m) (s), D α u(s))ds   g(s)M |u(s)| + |u (s)| + |u (s)| + · · · + |u(m) (s)| + |D α u(s)| ds

 g(s)M|u(s)|ds

γ (m + 2)(q)





1 1 2

G

1 2,s



g(s)ds ≥ uX .

Thus, T uX ≥ uX ∀u ∈ P ∩ ∂ 1 . Next, since A∞ = 0, there exists R such that f (v1 , . . . , vm+2 ) ≤ m+2 i=1 |vi | for m+2 |v | > r, where satisfies i i=1 1 (m + 2)(q) G(1, s)g(s)ds ≤ 1. 0

Let 2 = {u ∈ E : u < r2 , r2 > max{r1 , R}}, it is easy to see that 1 ⊂ 2 . Now we may choose u ∈ P ∩ ∂ 2 , then 

1

T uX = max

t∈[0,1]

G(t, s)g(s)Ru (s)ds +

0

 1 + Gα (t, s)g(s)Ru (s)ds

m  i=1

1

Gi (t, s)g(s)Ru (s)ds

0

0

1 ≤ (m + 2)(q) G(1, s)g(s)f (u(s), u (s), u (s), . . . , u(m) (s), D α u(s))ds 0

1 ≤ (m + 2)(q) G(1, s)g(s)uX ds ≤ uX . 0

Hence, by the second part of Guo–Krasnoselskii fixed-point Theorem 1, we can conclude that problem (1) has at least one positive solution. case: since A0 = 0, there exists r1 such that f (v1 , . . . , vm+2 ) ≤ m+2 (ii) Superlinear δ m+2 i=1 |vi | for i=1 |vi | ≤ r1 , where δ satisfies 1 δ(m + 2)(q) G(1, s)g(s)ds ≤ 1. 0

A. Souahi et al.

Let 1 = {u ∈ E, uX < r1 }. Take u ∈ P ∩ ∂ 1 , then  m 1 1  T uX = max G(t, s)g(s)Ru (s)ds + Gi (t, s)g(s)Ru (s)ds t∈[0,1]

0

1 + Gα (t, s)g(s)Ru (s)ds

i=1



0

0

1 ≤ (m + 2)(q) G(1, s)g(s)f (u(s), u (s), u (s), . . . , u(m) (s), D α u(s))ds 0

1 ≤ δ(m + 2)(q) G(1, s)g(s)uX ds ≤ uX . 0

there exists a constant r > r1 , such that f (v1 , . . . , vm+2 ) ≥ Next, since A∞ = +∞, m+2 λ( m+2 v ) for any |v i i | ≥ r, where λ > 0 satisfies i=1 i=1 1   γ λ G 12 , s g(s)ds ≥ 1. (m + 2)(q) 12 }, it is easy to see that 1 ⊂ 2 . Let 2 = {u ∈ E : uX < r2 , r2 > (q)(m+2)r γ γ Now we may choose u ∈ P ∩ ∂ 2 . Then (m+2)(q) uX > r. Hence,

 1   1 = G 12 , 1 g(s)Ru (s)ds T uX ≥ T u∞ ≥ T u 2 0 1   1 G 2 , s g(s)f (u(s), u (s), u (s), . . . , u(m) (s), D α u(s))ds ≥ ≥ ≥

1 2



1

G

1 2



1 1 2

G

≥ uX λ

1 2, s 1 2, s



g(s)λ(|u(s)| + |u (s)| + |u (s)| + · · · + |u(m) (s)| + |D α u(s)|)ds

 g(s)λ|u(s)|ds

γ (m + 2)(q)





1 1 2

G

1 2, s



g(s)ds ≥ uX .

Thus, T uX ≥ uX ∀u ∈ P ∩ ∂ 2 . Therefore, by the second part of Theorem 1, we conclude that problem (1) has at least one positive solution in P ∩ ( 2 \ 1 ). Example 1 Let us consider the following BVP: ⎧ 22 2  2  4 ⎪ D 5 u(t) + u2 (t) + u (t) + D 3 u(t) = 0, t ∈ [0, 1]\Q, ⎪ ⎪ ⎨

2    2   2  4 22 5 u(t) + t 2 + 1 3 u(t) D = 0, t ∈ [0, 1] ∩ Q, u (t) + u (t) + D ⎪ ⎪ ⎪ ⎩ 5 u(0) = 0, u (0) = 0, u(2) (0) = 0, u(3) (0) = 0, D 2 u(1) = 0, where n = 5, q =

22 5 ,

β = 52 , α = 43 ,  1 + t 2, g(t) = 1,

t ∈ [0, 1] ∩ Q, t ∈ [0, 1]\Q,

(4)

Positive Solutions for Higher Order Nonlinear Fractional Differential...

and f (x, y, z) = x 2 + y 2 + z2 . Thus g ∈ L1 ((0, 1), R+ ), and f ∈ C(Rm+2 + , R+ ). Since f is superlinear, by Theorem 2 we deduce that problem (4) has at least one positive solution on [0, 1]. Example 2 Let us consider the following BVP: ⎛ ⎞ ⎧ ⎪ ⎪ ⎪ 16 ⎜ 1 ⎟ ⎪ ⎪ D 3 u(t) + ⎝ u2 (t)+1 +  1 2 +  1 2 + 5 1 2 ⎠ = 0, t ∈ [0, 1]\Q, ⎪ ⎪ (u (t)) +1 (u (t)) +1 ⎪ 2 ⎪ D u(t) +1 ⎪ ⎪ ⎞ ⎛ ⎪ ⎨ 2 ⎜ 1 16 ⎟ ⎪ D 3 u(t) + t + 1 ⎝ u2 (t)+1 +  1 2 +  1 2 + 5 1 2 ⎠ = 0, ⎪ ⎪ (u (t)) +1 (u (t)) +1 ⎪ 2 ⎪ D u(t) +1 ⎪ ⎪ ⎪ ⎪ ⎪ t ∈ [0, 1] ∩ Q, ⎪ ⎪ 7 ⎩  (2) (3) u(0) = 0, u (0) = 0, u (0) = 0, u (0) = 0, D 2 u(1) = 0, (5) where n = 6, q =

16 3 ,

β = 72 , α = 52 ,  1 + t 2, g(t) = 1,

t ∈ [0, 1] ∩ Q, t ∈ [0, 1]\Q,

and f (x, y, z, w) =

x2

1 1 1 1 + 2 + 2 + 2 . +1 y +1 z +1 w +1

Thus g ∈ L1 ((0, 1), R+ ), and f ∈ C(Rm+2 + , R+ ). Since f is sublinear, by Theorem 2 we deduce that problem (5) has at least one positive solution on [0, 1]. Acknowledgments We would like to express our gratitude to the anonymous referees and the editor for their valuable comments and suggestions.

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