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THE HIGHT ORDER LANE-EMDEN FRACTIONAL DIFFERENTIAL SYSTEM: EXISTENCE, UNIQUENESS AND

ULAM TYPE STABILITIES

AMELE TAIEB2AND ZOUBIR DAHMANI1

Abstract. In this paper, by considering a more general Lane-Emden system of high order fractional differential equations with two arbitrary orders in each equation, we obtain some results on the existence and uniqueness of solutions using some fixed point theorems. Furthermore, we define and study some types of Ulam stability. Some examples are presented to illustrate the main results.

1. Introduction and Preliminaries

In recent years, fractional calculus has attracted great attention. It provides an excellent tool for the description of hereditary properties of various materials and processes. Moreover, the fractional differential equations theory arises in many en-gineering and scientific disciplines such as mechanics, physics, chemistry, biology, economics, control theory and signal processing, (see [22, 23, 25]). Many authors inves-tigated the existence and uniqueness of solutions for nonlinear fractional differential equations. We refer the reader to [1–5, 10–17, 21, 26, 33] for more information and applications.

On the other hand, the Ulam type stabilities for fractional differential problems are quite significant in realistic problems, numerical analysis, biology and economics. Some results concerning these fractional stabilities have been obtained in [8, 19, 20, 34]. Let us now introduce some other important research papers related to the Lane-Emden model which has inspired our work: we know that modeling of several physical phenomena, such as pattern formation, population evolution and chemical reactions

Key words and phrases. Caputo derivative, fixed point, differential equation, existence, uniqueness, Ulam-Hyers stability, generalized Ulam-Hyers stability.

2010 Mathematics Subject Classification. Primary: 30C45. Secondary: 39B72, 39B82. Received: May 23, 2015.

Accepted: December 26, 2015.

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gives rise to the Lane-Emden differential problem [9]. Ever after, the Lane-Emden systems and other related systems have exhausted the attention of many authors [6, 7, 18, 27, 29, 30, 35, 36].

Lane-Emden model has the following form:

x00(t) + a tx

0

(t) + f (t, x (t)) = g (t) , t ∈ [0, 1] ,

with the initial conditions:

x (0) = A, x0(0) = B,

where A and B are constants, f is a continuous real valued function and g ∈ C ([0, 1]), (see the paper of J. Serrin and H. Zou [32]).

Recently, S. M. Mechee and N. Senu proposed a numerical study of the Lane-Emden differential problem of fractional order. The imposed Lane-Emden model has a more importance in applied mathematics, mathematical physics and astrophysics. The order appeared in two different fractional order as follows [28]:

Dαy (t) + k tα−βD

β

y (t) + f (t, y (t)) = g (t) ,

where t ∈ [0, 1], k ≥ 0, 1 < α ≤ 2, and 0 < β ≤ 1, with the initial conditions y (0) = A, y0(0) = B, where A and B are constants, f is a continuous real valued function and g ∈ C ([0, 1]).

Very recently, R. W. Ibrahim [19] imposed the Ulam-Hyers stability for the following singular variable Lane-Emden equation:

DβDα+a t



u (t) + f (t, u (t)) = g (t) ,

where u (0) = µ, u (1) = ν, 0 < α, β ≤ 1, 0 ≤ t ≤ 1, a ≥ 0, and where Dγ denotes

the Caputo fractional derivative for γ > 0, f is continuous real valued function and g ∈ C ([0, 1]).

In this paper, we consider a more general and high dimensional Lane-Emden coupled system of fractional differential equations. Then, we discuss the existence, uniqueness and some types of Ulam stabilities for the proposed coupled nonlinear fractional

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system. So, let us consider:                                              Dβ1  Dα1 + a1 t  x1(t) + f1(t, x1(t) , x2(t) , . . . , xn(t)) = g1(t) , t ∈ J, Dβ2  Dα2 + a2 t  x2(t) + f2(t, x1(t) , x2(t) , . . . , xn(t)) = g2(t) , t ∈ J, .. . Dβn  Dαn+ an t  xn(t) + fn(t, x1(t) , x2(t) , . . . , xn(t)) = gn(t) , t ∈ J, n X k=1 |xk(0)| = n X k=1 |x0k(0)| = · · · = n X k=1 x (l−1) k (0) = 0, n X k=1 |Dαkx k(0)| = n X k=1 |Dαk+1x k(0)| = · · · = n X k=1 |Dαk+l−2x k(0)| = 0, Dαk+l−1x k(1) = 0, k = 1, 2, . . . , n, (1.1) where l − 1 < βk < l, l − 1 < αk < l, ak ≥ 0, l ∈ N∗− {1}, k = 1, 2, . . . , n, n ∈ N∗

and J := [0, 1]. The derivatives Dβk and Dαk, k = 1, 2, . . . , n, are in the sense of Caputo. For each k = 1, 2, . . . , n, the functions fk : J × Rn→ R and gk: J → R will

be specified later.

To the best of our knowledge, there are no papers that have developed the Lane-Emden system in multi-variables, with arbitrary orders in each equation.

Definition 1.1 ([22, 31]). The Riemann-Liouville fractional integral operator of order α > 0, for a continuous function f on [0, ∞[ is defined as:

Jαf (t) =      1 Γ(α) Rt 0 (t − s) α−1 f (s) ds, α > 0, f (t), α = 0, where t ≥ 0, and Γ (α) :=R0∞e−xxα−1dx.

Definition 1.2 ([22,31]). The Caputo derivative of order α for a function x : [0, ∞) → R, which is at least l-times differentiable can be defined as:

Dαx(t) = 1 Γ (l − α) Z t 0 (t − s)l−α−1x(l)(s) ds = Jl−αx(l)(t), for l − 1 < α < l, l ∈ N∗.

Lemma 1.1 ([22]). For α > 0, the general solution of the fractional differential equation Dαx(t) = 0, is given by x(t) = l−1 X j=0 cjtj, where cj ∈ R, j = 0, . . . , l − 1, l = [α] + 1.

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Lemma 1.2 ([22]). Let α > 0. Then JαDαx(t) = x(t) + l−1 X j=0 cjtj, where cj ∈ R, j = 0, 1, . . . , l − 1, l = [α] + 1.

Lemma 1.3 ([22]). Let q > p > 0, g ∈ L1([a, b]). Then DpJqg (t) = Jq−pg (t), t ∈ [a, b].

Lemma 1.4 ([24], Krasnoselskii). Let M be a closed convex and nonempty subset of a Banach space X. Let A and B be the operators such that

(i) Ax + By ∈ M , whenever x, y ∈ M , (ii) A is a compact and continuous, (iii) B is a contraction mapping.

Then there exists z ∈ M such that z = Az + Bz.

The following auxiliary result is important to give the integral solution of (1.1). Lemma 1.5. Suppose that (Gk)nk=1 ∈ C (J, R), J = [0, 1] and consider the nonlinear

system            Dβ1 Dα1 + a1 t x1(t) = G1(t) , t ∈ J, Dβ2 Dα2 + a2 t x2(t) = G2(t) , t ∈ J, .. . Dβn Dαn+an t  xn(t) = Gn(t) , t ∈ J, (1.2)

where l − 1 < βk, αk < l, ak ≥ 0, k = 1, 2, . . . , n, l ∈ N∗− {1}, with the conditions: n X k=1 |xk(0)| = n X k=1 |x0k(0)| = · · · = n X k=1 x (l−1) k (0) = 0, (1.3) n X k=1 |Dαkx k(0)| = n X k=1 |Dαk+1x k(0)| = · · · = n X k=1 |Dαk+l−2x k(0)| = 0, where Dαk+l−1x

k(1) = 0, k = 1, 2, . . . , n, l ∈ N∗− {1}. Then, (1.2)–(1.3) has a unique

solution given by (x1, x2, . . . , xn) (t), where

xk(t) = t Z 0 (t − τ )αk−1 Γ (αk)   τ Z 0 (τ − s)βk−1 Γ (βk) Gk(s) ds − ak τ xk(τ )  dτ − t αk+l−1 Γ (αk+ l) 1 Z 0 (1 − τ )αk−1 Γ (αk)   τ Z 0 (τ − s)βk−1 Γ (βk) Gk(s) ds − ak τ xk(τ )  dτ, (1.4) and k = 1, 2, . . . , n.

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Proof. By Applying Lemma 1.2 to the problem (1.2), we get:  Dαk+ ak τ  xk(τ ) = τ Z 0 (τ − s)βk−1 Γ (βk) Gk(s) ds − ck0 − c k 1τ − · · · − c k l−1τ l−1, where k = 1, 2, . . . , n, ckjj=0,...,l−1 ∈ R, and l − 1 < βk< l, l ∈ N∗− {1}.

Now, applying Lemma 1.2 to the last assertion, yields xk(t) = t Z 0 (t − τ )αk−1 Γ (αk)   τ Z 0 (τ − s)βk−1 Γ (βk) Gk(s) ds − ak τ xk(τ )  dτ − c k 0tαk Γ (αk+ 1) − c k 1tαk+1 Γ (αk+ 2) − · · · − (l − 1)!c k l−1tαk+l−1 Γ (αk+ l) (1.5) − c00k− c01kt − · · · − c0l−1k tl−1, where k = 1, 2, . . . , n and c0jkj=0,...,l−1∈ R, l − 1 < αk < l, l ∈ N∗− {1}.

Using Lemma 1.3 and applying the conditions (1.3), we obtain the values of ckj and c0jk. Substituting the last condition in (1.5), we obtain (1.4). The proof of Lemma 1.5

is thus completed. 

Now, let us introduce the Banach space:

S := {(x1, x2, . . . , xn) : xk ∈ C (J, R) , k = 1, 2, . . . , n} ,

endowed with the norm:

k(x1, x2, . . . , xn)kS = max (kx1k , kx2k , . . . , kxnk) , kxkk = sup t∈J |xk(t)| , where k = 1, 2, . . . , n. 2. Main Results

In this section, we will formulate and establish sufficient conditions for the existence and uniqueness of solutions to (1.1). Then, we continue our study by imposing some types of Ulam stability: Ulam-Hyers stability, generalized Ulam-Hyers stability and Ulam-Heyers-Rassias stability for the problem (1.1).

We begin by list the following hypotheses:

(H1): There exist nonnegative constants (µk)j, j, k = 1, 2, . . . , n, such that for all

t ∈ [0, 1] and all (x1, x2, . . . , xn), (y1, y2, . . . , yn) ∈ S, we have

|fk(t, x1, x2, . . . , xn) − fk(t, y1, y2, . . . , yn)| ≤ n

X

j=1

(µk)j|xj− yj| .

(H2): The functions fk : [0, 1] × Rn → R and gk : [0, 1] → R are continuous for each

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(H3): For all k = 1, 2, . . . , n, n ∈ N∗, the function fk maps bounded subsets of

J × Rn into relatively compact subsets of R.

(H4): There exist nonnegative constants (Lk)k=1,2...,n, such that, for each t ∈ J and

all (x1, x2, . . . , xn) ∈ Rn,

|fk(t, x1, x2, . . . , xn)| ≤ Lk, k = 1, 2, . . . , n.

(H5): There exist nonnegative constants (Mk)k=1,2...,n,, such that, for each t ∈ J ,

|gk(t)| ≤ Mk, k = 1, 2, . . . , n.

Then, we set the following quantities:

(2.1) zk= Γ (αk+ l) + 1 Γ (αk+ l) Γ (αk+ βk+ 1) , Λk = Γ (αk+ l) + 1 (αk+ l − 1) Γ (2αk+ l − 1) , and Σk= n X j=1 (µk)j, k = 1, 2, . . . , n.

2.1. Existence and Uniqueness of Solutions. The first result is based on Banach contraction principle. We have:

Theorem 2.1. Assume that hypotheses (H1), (H4) and (H5) are satisfied. Then, the

system (1.1) has a unique solution on J provided that

(2.2) λk =: Σkzk+ akΛk < 1, k = 1, 2, . . . , n.

Proof. Define the nonlinear operator T : S → S by T (x1, x2, . . . , xn) (t) := (T1(x1, x2, . . . , xn) (t) ,

T2(x1, x2, . . . , xn) (t) , . . . , Tn(x1, x2, . . . , xn) (t)) , t ∈ J,

where, for all k = 1, 2, . . . , n, Tk(x1, . . . , xn) (t) = t Z 0 (t − τ )αk−1 Γ (αk) τ Z 0 (τ − s)βk−1 Γ (βk) × (gk(s) − fk(s, x1(s) , . . . , xn(s))) ds − ak τ xk(τ ) ! dτ − t αk+l−1 Γ (αk+ l) 1 Z 0 (1 − τ )αk−1 Γ (αk) τ Z 0 (τ − s)βk−1 Γ (βk) × (gk(s) − fk(s, x1(s) , . . . , xn(s))) ds − ak τ xk(τ ) ! dτ. For (2.3) r ≥ (Mk+ Lk) zk 1 − σ , λk≤ σ < 1,

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we consider the set Br := {(x1, x2, . . . , xn) ∈ S; k(x1, x2, . . . , xn)kS ≤ r} and we will

show that T Br ⊂ Br.

For (x1, x2, . . . , xn) ∈ Br and each k = 1, 2, . . . , n, we have

kTk(x1, x2, . . . , xn)k ≤ sup t∈J t Z 0 (t − τ )αk−1 Γ (αk) τ Z 0 (τ − s)βk−1 Γ (βk) × |gk(s) − fk(s, x1(s) , . . . , xn(s)) − fk(s, 0, . . . , 0) + fk(s, 0, . . . , 0)| ds + ak τ xk(τ ) ! dτ + sup t∈J tαk+l−1 Γ (αk+ l) 1 Z 0 (1 − τ )αk−1 Γ (αk) τ Z 0 (τ − s)βk−1 Γ (βk) × |gk(s) − fk(s, x1(s) , . . . , xn(s)) − fk(s, 0, . . . , 0) + fk(s, 0, . . . , 0)| ds + ak τ xk(τ ) ! dτ.

The hypotheses (H4) and (H5) allow us to write

kTk(x1, x2, . . . , xn)k ≤ sup t∈J t Z 0 (t − τ )αk−1 Γ (αk)   τ Z 0 (τ − s)βk−1 Γ (βk) ds  dτ (Mk+ Lk+ Σkk(x1, x2, . . . , xn)kS) + sup t∈J t Z 0 (t − τ )αk−1 Γ (αk) 1 τdτ akk(x1, x2, . . . , xn)kS + sup t∈J tαk+l−1 Γ (αk+ l) 1 Z 0 (1 − τ )αk−1 Γ (αk)   τ Z 0 (τ − s)βk−1 Γ (βk) ds  dτ × (Mk+ Lk+ Σkk(x1, x2, . . . , xn)kS) + sup t∈J tαk+l−1 Γ (αk+ l) 1 Z 0 (1 − τ )αk−1 Γ (αk) 1 τdτ akk(x1, x2, . . . , xn)kS. Therefore, by (H1)

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kTk(x1, x2, . . . , xn)k ≤ (Mk+ Lk+ Σkk(x1, x2, . . . , xn)kS) ×  1 + 1 Γ (αk+ l) Z1 0 (1 − τ )αk−1 Γ (αk)   τ Z 0 (τ − s)βk−1 Γ (βk) ds  dτ + ak  1 + 1 Γ (αk+ l)  k(x1, x2, . . . , xn)kS 1 Z 0 (1 − τ )αk−1 Γ (αk) ταk+l−2dτ. Hence, kTk(x1, x2, . . . , xn)k ≤ (Mk+ Lk+ Σkr)  1 Γ (αk) Γ (βk+ 1) + 1 Γ (αk) Γ (βk+ 1) Γ (αk+ l)  × 1 Z 0 (1 − τ )αk−1τβk + akr  1 Γ (αk) + 1 Γ (αk) Γ (αk+ l) Z1 0 (1 − τ )αk−1 ταk+l−2dτ.

Using the Beta function, we get kTk(x1, x2, . . . , xn)k ≤ (Mk+ Lk+ Σkr)  1 Γ (αk) Γ (βk+ 1) + 1 Γ (αk+ l) Γ (αk) Γ (βk+ 1)  × β (αk, βk+ 1) + akr  1 Γ (αk) + 1 Γ (αk) Γ (αk+ l)  β (αk, αk+ l − 1) . By (2.3), we obtain kTk(x1, x2, . . . , xn)k ≤ (1 − σ) r + λkr = (1 − σ + λk) r and kTk(x1, x2, . . . , xn)k ≤ r, where k = 1, 2, . . . , n. Thus, kT (x1, x2, . . . , xn)kS ≤ r.

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For (x1, x2, . . . , xn), (y1, y2, . . . , yn) ∈ S and for all t ∈ J , we have: kTk(x1, x2, . . . , xn) − Tk(y1, y2, . . . , yn) k ≤ sup t∈J t Z 0 (t − τ )αk−1 Γ (αk) τ Z 0 (τ − s)βk−1 Γ (βk) × |fk(s, x1(s) , . . . , xn(s)) − fk(s, y1(s) , . . . , yn(s))| ds + ak τ |xk(τ ) − yk(τ )| ! dτ + sup t∈J tαk+l−1 Γ (αk+ l) 1 Z 0 (1 − τ )αk−1 Γ (αk) τ Z 0 (τ − s)βk−1 Γ (βk) ×|fk(s, x1(s) , . . . , xn(s)) − fk(s, y1(s) , . . . , yn(s)) |ds + ak τ |xk(τ ) − yk(τ )| ! dτ.

Thanks to (H1), we can write

k Tk(x1, x2, . . . , xn) − Tk(y1, y2, . . . , yn)k ≤ sup t∈J t Z 0 (t − τ )αk−1 Γ (αk)   τ Z 0 (τ − s)βk−1 Γ (βk) ds  dτ Σkk(x1− y1, x2− y2, . . . , xn− yn)kS + sup t∈J t Z 0 (t − τ )αk−1 Γ (αk) 1 τdτ akk(x1− y1, x2− y2, . . . , xn− yn)kS+ supt∈J tαk+l−1 Γ (αk+ l) × 1 Z 0 (1 − τ )αk−1 Γ (αk)   τ Z 0 (τ − s)βk−1 Γ (βk) ds  dτ Σkk(x1− y1, x2− y2, . . . , xn− yn)kS + sup t∈J tαk+l−1 Γ (αk+ l) 1 Z 0 (1 − τ )αk−1 Γ (αk) 1 τdτ akk(x1− y1, x2− y2, . . . , xn− yn)kS. Consequently, kTk(x1, x2, . . . , xn) − Tk(y1, y2, . . . , yn)k ≤ Σk Γ (αk+ l) + 1 Γ (αk+ l) k(x1− y1, x2− y2, . . . , xn− yn)kS

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× 1 Z 0 (1 − τ )αk−1 Γ (αk)   τ Z 0 (τ − s)βk−1 Γ (βk) ds  dτ + ak Γ (αk+ l) + 1 Γ (αk+ l) k(x1− y1, x2− y2, . . . , xn− yn)kS (2.4) × 1 Z 0 (1 − τ )αk−1 Γ (αk) ταk+l−2dτ ≤ λkk(x1− y1, x2− y2, . . . , xn− yn)kS,

which implies that

kT (x1, x2, . . . , xn) − T (y1, y2, . . . , yn)kS

≤ max (λ1, λ2, . . . , λn) k(x1− y1, x2− y2, . . . , xn− yn)kS,

where, k = 1, 2, . . . , n.

Then by (2.2), T is contractive. 

Theorem 2.2. Assume that the hypotheses (Hi)i=1,2,...,5 and the inequalities

(2.5) Ck :=

λk

Γ (αk+ l) + 1

< 1, k = 1, 2, . . . , n, are satisfied. Then system (1.1) has at least one solution on J . Proof. On BR, such that

R ≥ (Mk+ Lk) zk 1 − akΛk

, akΛk6= 1,

we define the operators P and Q as follows: P (x1, x2, . . . , xn) (t) := (P1(x1, x2, . . . , xn) (t), P2(x1, x2, . . . , xn) (t), . . . , Pn(x1, x2, . . . , xn) (t)) , Q(x1, x2, . . . , xn) (t) := (Q1(x1, x2, . . . , xn) (t), Q2(x1, x2, . . . , xn) (t), . . . , Qn(x1, x2, . . . , xn) (t)) . For each k = 1, 2, . . . , n, Pk(x1, x2, . . . , xn)(t) := t Z 0 (t − τ )αk−1 Γ (αk)   τ Z 0 (τ − s)βk−1 Γ (βk) (gk(s) − fk(s, x1(s) , . . . , xn(s))) ds − ak τ xk(τ )  dτ, and

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Qk(x1, x2, . . . , xn) (t) := − tαk+l−1 Γ (αk+ l) 1 Z 0 (1 − τ )αk−1 Γ (αk) τ Z 0 (τ − s)βk−1 Γ (βk) × (gk(s) − fk(s, x1(s) , . . . , xn(s))) ds − ak τ xk(τ ) ! dτ.

For (x1, x2, . . . , xn) , (y1, y2, . . . , yn) ∈ BR and for each k = 1, 2, . . . , n, we can write

kPk(x1, x2, . . . , xn) + Qk(y1, y2, . . . , yn)k ≤ sup t∈J t Z 0 (t − τ )αk−1 Γ (αk) τ Z 0 (τ − s)βk−1 Γ (βk) × (|gk(s)| + |fk(s, x1(s) , x2(s) , . . . , xn(s))|) ds + ak τ xk(τ ) ! dτ + sup t∈J tαk+l−1 Γ (αk+ l) 1 Z 0 (1 − τ )αk−1 Γ (αk) τ Z 0 (τ − s)βk−1 Γ (βk) × (|gk(s)| + |fk(s, y1(s) , y2(s) , . . . , yn(s))|) ds + ak τ yk(τ ) ! dτ.

Using (H4) and (H5), we obtain

kPk(x1, x2, . . . , xn) + Qk(y1, y2, . . . , yn)k ≤ sup t∈J t Z 0 (t − τ )αk−1 Γ (αk)   τ Z 0 (τ − s)βk−1 Γ (βk) ds  dτ (Mk+ Lk) + sup t∈J t Z 0 (t − τ )αk−1 Γ (αk) 1 τdτ akk(x1, x2, . . . , xn)kS + sup t∈J tαk+l−1 Γ (αk+ l) 1 Z 0 (1 − τ )αk−1 Γ (αk)   τ Z 0 (τ − s)βk−1 Γ (βk) ds  dτ (Mk+ Lk) + sup t∈J tαk+l−1 Γ (αk+ l) 1 Z 0 (1 − τ )αk−1 Γ (αk) 1 τdτ akk(y1, y2, . . . , yn)kS. And then,

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kPk(x1, x2, . . . , xn) + Qk(y1, y2, . . . , yn) k ≤ (Mk+ Lk)  1 + 1 Γ (αk+ l) Z1 0 (1 − τ )αk−1 Γ (αk)   τ Z 0 (τ − s)βk−1 Γ (βk) ds  dτ + akR  1 + 1 Γ (αk+ l) Z1 0 (1 − τ )αk−1 Γ (αk) ταk+l−2dτ ≤  Mk+ Lk Γ (αk) Γ (βk+ 1) + Mk+ Lk Γ (αk) Γ (βk+ 1) Γ (αk+ l) Z1 0 (1 − τ )αk−1τβk + akR  1 Γ (αk) + 1 Γ (αk) Γ (αk+ l) Z1 0 (1 − τ )αk−1 ταk+l−2dτ.

Such that R ≥ (Mk+ Lk)Fk/(1 − αkΛk), αkΛk6= 1, we have

kPk(x1, x2, . . . , xn) + Qk(y1, y2, . . . , yn)k ≤ (Mk+ Lk) Γ (αk+ l) + 1 Γ (αk+ l) Γ (αk+ βk+ 1) + akR (Γ (αk+ l) + 1) (αk+ l − 1) Γ (2αk+ l − 1) = (Mk+ Lk) zk+ akΛkR ≤ R. Consequently, kP (x1, x2, . . . , xn) + Q (y1, y2, . . . , yn)kS ≤ R. Thus, P (x1, x2, . . . , xn) + Q (y1, y2, . . . , yn) ∈ BR.

We proceed to prove that Q is a contraction mapping in BR. For

(x1, x2, . . . , xn) , (y1, y2, . . . , yn) ∈ BR and for each k = 1, 2, . . . , n, we obtain:

kQk(x1, x2, . . . , xn) − Qk(y1, y2, . . . , yn) k ≤ sup t∈J tαk+l−1 Γ (αk+ l) 1 Z 0 (1 − τ )αk−1 Γ (αk) τ Z 0 (τ − s)βk−1 Γ (βk) × |fk(s, x1(s) , . . . , xn(s)) − fk(s, y1(s) , . . . , yn(s))| ds +ak τ |xk(τ ) − yk(τ )| Using (H1), we get kQk(x1, x2, . . . , xn) − Qk(y1, y2, . . . , yn) k ≤ Σk Γ (αk+ l) k(x1 − y1, x2− y2, . . . , xn− yn)kS

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× 1 Z 0 (1 − τ )αk−1 Γ (αk)   τ Z 0 (τ − s)βk−1 Γ (βk) ds  dτ + ak Γ (αk+ l) k(x1− y1, x2− y2, . . . , xn− yn)kS 1 Z 0 (1 − τ )αk−1 Γ (αk) ταk+l−2 ≤ Σkk(x1− y1, x2− y2, . . . , xn− yn)kS Γ (αk+ l) Γ (αk) Γ (βk+ 1) 1 Z 0 (1 − τ )αk−1 τβk +akk(x1− y1, x2− y2, . . . , xn− yn)kS Γ (αk+ l) Γ (αk) 1 Z 0 (1 − τ )αk−1ταk+l−2 ≤ Σkk(x1− y1, x2− y2, . . . , xn− yn)kS Γ (αk+ l) Γ (αk) Γ (βk+ 1) β (αk, βk+ 1) +akk(x1− y1, x2− y2, . . . , xn− yn)kS Γ (αk+ l) Γ (αk) β (αk, αk+ l − 1) . Then, kQk(x1, x2, . . . , xn) − Qk(y1, y2, . . . , yn) k ≤ Σkk(x1− y1, x2− y2, . . . , xn− yn)kS Γ (αk+ l) Γ (αk+ βk+ 1) + akΓ (αk+ l − 1) k(x1− y1, x2− y2, . . . , xn− yn)kS Γ (αk+ l) Γ (2αk+ l − 1) ≤  Σk Γ (αk+ l) Γ (αk+ βk+ 1) + ak (αk+ l − 1) Γ (2αk+ l − 1)  × k(x1− y1, x2− y2, . . . , xn− yn)kS. Thus, kQ (x1, x2, . . . , xn) − Q (y1, y2, . . . , yn) kS ≤ max 1≤k≤nCkk(x1− y1, x2− y2, . . . , xn− yn)kS.

Thanks to (2.2) and (2.5), we conclude that Q is a contractive.

The hypothesis (H2) implies that P is continuous. Then, for all (x1, x2, . . . , xn) ∈

BR and each t ∈ J , we obtain

kPk(x1, x2, . . . , xn) k ≤ sup t∈J t Z 0 (t − τ )αk−1 Γ (αk) τ Z 0 (τ − s)βk−1 Γ (βk)

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× (|gk(s)| + |fk(s, x1(s) , x2(s) , . . . , xn(s))|) ds + ak τ xk(τ ) ! dτ ≤ sup t∈J t Z 0 (t − τ )αk−1 Γ (αk)   τ Z 0 (τ − s)βk−1 Γ (βk) ds  dτ (Mk+ Lk) (2.6) + sup t∈J t Z 0 (t − τ )αk−1 Γ (αk) ταk−l−2dτ a kk(x1, x2, . . . , xn)kS ≤ Mk+ Lk Γ (αk) Γ (βk+ 1) 1 Z 0 (1 − τ )αk−1τβkdτ + a kR 1 Γ (αk) 1 Z 0 (1 − τ )αk−1ταk−l−2 ≤ Mk+ Lk Γ (αk+ βk+ 1) + Γ (αk+ l − 1) Γ (2αk+ l − 1) akR. And by (2.6), we get kP (x1, x2, . . . , xn)kS ≤ R.

Therefore, P is uniformly bounded on BR. Furthermore, we show that P is a compact

operator in BR.

Let 0 ≤ t1 < t2 ≤ 1 and (x1, x2, . . . , xn) ∈ BR, then

kPk(x1, x2, . . . , xn) (t2) − Pk(x1, x2, . . . , xn) (t1) k ≤ sup t∈J t2 Z 0 (t1− τ ) αk−1τβkdτ − t1 Z 0 (t2− τ ) αk−1τβk Mk+ Lk Γ (αk) Γ (βk+ 1) + akR Γ (αk) sup t∈J t2 Z 0 (t2− τ ) αk−1 ταk−l−2dτ − t1 Z 0 (t1− τ ) αk−1 ταk−l−2 = Mk+ Lk Γ (αk+ βk+ 1)  tαk+βk 2 − t αk+βk 1  +akRΓ (αk+ l − 1) Γ (2αk+ l − 1)  t2αk+l−2 2 − t 2αk+l−2 1  . Thus, kP (x1, x2, . . . , xn) (t2) − P (x1, x2, . . . , xn) (t1) kS ≤ max 1≤k≤n Mk+ Lk Γ (αk+ βk+ 1)  tαk+βk 2 − t αk+βk 1  + max 1≤k≤n akRΓ (αk+ l − 1) Γ (2αk+ l − 1)  t2αk+l−2 2 − t 2αk+l−2 1  . (2.7)

The right-hand side of the above inequalities (2.7) is independent of (x1, x2, . . . , xn)

and tend to zero as t2 → t1. Therefore, P is equicontinuous. Furthermore, by the

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compact operator on BR. Then thanks to Lmma 1.6, we can state that (1.1) has at

least one solution on J . Theorem 2.2 is thus proved. 

2.2. Ulam Stabilities. We introduce the following definitions.

Definition 2.1. The Lane-Emden fractional system (1.1) has the Ulam-Hyers stability if there exists a positive constant K with the following property:

For every  > 0, for any t ∈ J , and (x1, x2, . . . , xn) ∈ S of (1.1), with

D βk  Dαk+ ak t  xk(t) + fk(t, x1(t) , x2(t) , . . . , xn(t)) − gk(t) < , and k = 1, . . . , n, then, there exists a solution (y1, y2, . . . , yn) ∈ S that satisfies

(2.8) Dβk  Dαk +ak t  yk(t) + fk(t, y1(t) , y2(t) , . . . , yn(t)) = gk(t) , k = 1, . . . , n, and n X k=1 |yk(0)| = n X k=1 |yk0 (0)| = · · · = n X k=1 y (l−1) k (0) = n X k=1 |Dαky k(0)| = n X k=1 |Dαk+1y k(0)| = · · · = n X k=1 |Dαk+l−2y k(0)| = 0, where Dαk+l−1y k(1) = 0, such that, k(x1− y1, x2 − y2, . . . , xn− yn)kS < K.

Definition 2.2. The system (1.1) is Ulam-Hyers stable in the generalized sense if there exists ϕ ∈ C (R+, R+), such that ϕ (0) = 0 and for each  > 0, and for each solution

(x1, x2, . . . , xn) ∈ S of (1.1), with |Dβk(Dαk+αtk)yk(t) + fk(t, x1(t), x2(t), . . . , xn(t)) −

gk(t)| < , and k = 1, . . . , n, there exists (y1, y2, . . . , yn) ∈ S of (2.8), with

k(x1− y1, x2− y2, . . . , xn− yn)kS < ϕ () .

Definition 2.3. The system (1.1) is stable in the sense of Ulam-Hyers-Rassias if there exist Ψ ∈ C (J, R+) and some positive ρ

i, such that for each i > 0, i = 1, . . . n, and

for all solution (x1, x2, . . . , xn) ∈ S of the inequalities

                   D β1  Dα1+a1 t  x1(t) + f1(t, x1(t) , x2(t) , . . . , xn(t)) − g1(t) ≤ 1Ψ (t) , D β2  Dα2 +a2 t  x2(t) + f2(t, x1(t) , x2(t) , . . . , xn(t)) − g2(t) ≤ 2Ψ (t) , .. . D βn  Dαn+ an t  xn(t) + fn(t, x1(t) , x2(t) , . . . , xn(t)) − gn(t) ≤ nΨ (t) , (2.9)

there exists (y1, y2, . . . , yn) ∈ S of (2.8) that satisfies

k(x1− y1, x2− y2, . . . , xn− yn)kS ≤ ρΨ (t) , ρ := max1≤k≤nρk,  := max 1≤k≤nk.

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Remark 2.1. An element (x1, x2, . . . , xn) of S is a solution of (1.1) if and only if there

exists hk : [0, 1] → R, k = 1, 2, . . . , n, such that

(i) |hk(t)| < k, t ∈ [0, 1], (k> 0, and is sufficient small), and

(ii) Dβk Dαk+ ak

t  xk(t) + fk(t, x1(t) , x2(t) , . . . , xn(t)) = gk(t) + hk(t), t ∈ [0, 1].

Theorem 2.3. Under the assumptions of Theorem 2.1, if the inequalities (2.10) sup t∈J D βk  Dαk +ak t  xk(t) ≥ (Mk+ Lk) zk+ (Σkzk+ akΛk) r, Σk< 1,

are satisfied, then the Lane-Emden problem (1.1) is Ulam-Hyers stable in S.

Proof. Thanks to Theorem 2.1, we can state that the problem (1.1) has a solution (y1, y2, . . . , yn) ∈ S that satisfies (2.8).

Now, suppose (x1, x2, . . . , xn) ∈ S is a solution of (1.1), where

(2.11) D βk  Dαk+ ak t  xk(t) + fk(t, x1(t) , x2(t) , . . . , xn(t)) − gk(t) < k.

According to assumptions of Theorem 2.1, we have

(2.12) |xk(t)| ≤ (Mk+ Lk) zk+ (Σkzk+ akΛk) r, k = 1, 2, . . . , n,

where k> 0.

Then, by (2.10) and (2.12), we obtain

(2.13) sup t∈J |xk(t)| ≤ sup t∈J D βk  Dαk + ak t  xk(t) . Using (2.13), we get sup t∈J |xk(t) − yk(t) | ≤ sup t∈J D βk  Dαk +ak t  (xk(t) − yk(t)) ≤ sup t∈J Dβk  Dαk +ak t  xk(t) − gk(t) + fk(t, x1(t) , x2(t) , . . . , xn(t)) −Dβk  Dαk +ak t  yk(t) − gk(t) + fk(t, y1(t) , y2(t) , . . . , yn(t))  − fk(t, x1(t) , x2(t) , . . . , xn(t)) + fk(t, y1(t) , y2(t) , . . . , yn(t))

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sup t∈J |xk(t) − yk(t)| ≤ k+ (µk)1|x1(t) − y1(t)| + · · · + (µk)n|xn(t) − yn(t)| ≤ k+ n X j=1 (µk)j1≤j≤nmax |xj(t) − yj(t)| ≤ k+ n X j=1 (µk)j1≤k≤nmax |xk(t) − yk(t)| . Hence, max 1≤k≤nkxk− ykk∞ ≤ max1≤k≤nωkk, ωk := 1 1 − Σk . Therefore, k(x1 − y1, x2− y2, . . . , xn− yn)kS ≤ max 1≤k≤nωkk:= K,  = max1≤k≤nk.

Since Σk < 1, we get ωk> 0, that is K is a positive constant.

From (1.4), we see that

n X k=1 |yk(0)| = n X k=1 |y0k(0)| = · · · = n X k=1 y (l−1) k (0) = n X k=1 |Dαky k(0)| = n X k=1 |Dαk+1y k(0)| = · · · = n X k=1 |Dαk+l−2y k(0)| = Dαk+l−1yk(1) = 0. Then, k(x1− y1, x2− y2, . . . , xn− yn)kS ≤ K.

Therefore, the Lane-Emden fractional system (1.1) is Ulam-Hyers stable.  Remark 2.2. Taking ϕ () = K, we conclude that the problem (1.1) is generalized Ulam-Hyers stable.

Theorem 2.4. Assume that:

(i) The assumptions and the condition (2.2) of Theorem 2.1 are satisfied. (ii) There exists a function Ψ ∈ C ([0, 1] , R+) that satisfies (2.9).

Then, the fractional system (1.1) is Ulam-Hyers-Rassias stable in S.

Proof. Let (x1, x2, . . . , xn) ∈ S be a solution of (1.1). In virtue of Remark 2.1 and by

some easy calculations, we get: max 1≤k≤nkxk− ykk ≤ max1≤k≤n kΨ (t) 1 − λk := max 1≤k≤nρkkΨ (t) . Thus, k(x1− y1, x2− y2, . . . , xn− yn)kS = max 1≤k≤nkxk− ykk ≤ ρΨ (t) , where, ρ = max 1≤k≤nρk,  = max1≤k≤nk.

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The Condition (2.2) implies that ρk= 1/(1 − λk) > 0, k = 1, 2, . . . , n. Hence, (1.1) is

Ulam-Hyers-Rassias stable. 

3. Applications

In this section, we present some examples to illustrate some applications of the main results.

Example 3.1. Consider the following system:                                                                                                        D94  D52 +5 t  x1(t) + |x1(t) + x2(t) + x3(t) + x4(t)| 9π2et(1 + |x 1(t) + x2(t) + x3(t) + x4(t)|) = t 2, t ∈ [0, 1] , D125  D83 + 10.5 t  x2(t) + 1 32π (t + 1) × sin (x1(t)) + sin (x2(t)) + cos (x3(t))

2et+1 + sin (x4(t))  = t, t ∈ [0, 1] , D52  D114 + 15.2 t  x3(t) + 1 t2+ 1

×cos (x1(t)) + cos (x2(t)) + cos (x3(t)) + cos (x4(t))

4πe2 = 3t 2, t ∈ [0, 1] , D73  D115 + 3 × 10 8 t  x4(t) + t2 4π2et2+1 ×  sin x1(t) + |x2(t) + x3(t) + x4(t)| 2π3(1 + |x 2(t) + x3(t) + x4(t)|)  = t 4, t ∈ [0, 1] , |x1(0)| + |x2(0)| + |x3(0)| + |x4(0)| = 0, |x01(0)| + |x02(0)| + x 0 3(0) = |x 0 4(0)| = 0, x 00 1(0) + x 00 2(0) + x 00 3(0) + x 00 4(0) = 0, D 5 2x1(0) + D 8 3x2(0) + D 11 4 x3(0) + D 11 5 x4(0) = 0, D 7 2x1(0) + D 11 3x2(0) + D 15 4 x3(0) + D 16 5 x4(0) = 0, D92x1(1) = D 14 3x2(1) = D 19 4 x3(1) = D 21 5x4(1) = 0, (3.1)

In this example, we have: n = 4, l = 3, β1 = 9/4, β2 = 12/5, β3 = 5/2, β4 = 7/3,

α1 = 5/2, α2 = 8/3, α3 = 11/4, α4 = 11/5, a1 = 5, a2 = 10.5, a3 = 15.2, a4 = 3 × 108,

J = [0, 1].

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|f1(t, x1, x2, x3, x4) − f1(t, y1, y2, y3, y4) | ≤ 1 9π2 |x1− y1| + 1 9π2 |x2− y2| + 1 9π2 |x3− y3| + 1 9π2 |x4− y4| , |f2(t, x1, x2, x3, x4) − f2(t, y1, y2, y3, y4) | ≤ 1 64πe|x1− y1| + 1 64πe|x2− y2| + 1 64πe|x3− y3| + 1 32π|x4− y4| , |f3(t, x1, x2, x3, x4) − f3(t, y1, y2, y3, y4) | ≤ 1 4πe2|x1 − y1| + 1 4πe2 |x2− y2| + 1 4πe2 |x3− y3| + 1 4πe2 |x4− y4| , |f4(t, x1, x2, x3, x4) − f4(t, y1, y2, y3, y4) | ≤ 1 4π2e|x1 − y1| + 1 8π5e|x2− y2| + 1 8π5e|x3− y3| + 1 8π5e|x4− y4| , and sup t∈J |g1(t)| = 1 2, supt∈J |g2(t)| = 1, sup t∈J |g3(t)| = 3 2, supt∈J |g4(t)| = 1 4, sup t∈J |f1(t, x1, x2, x3, x4)| = 1 9π2, sup t∈J |f2(t, x1, x2, x3, x4)| = 2e + 1 64πe , sup t∈J |f3(t, x1, x2, x3, x4)| = 1 4πe2, sup t∈J |f4(t, x1, x2, x3, x4)| = 2π3+ 1 8π5e . We can take (µ1)1 = (µ1)2 = (µ1)3 = (µ1)4 = 1 9π2, (µ2)1 = (µ2)2 = (µ2)3 = 1 64πe, (µ2)4 = 1 32π, (µ3)1 = (µ3)2 = (µ3)3 = (µ3)4 = 1 4πe2, (µ4)1 = 1 4π2e = 2 (µ4)i, where i = 2, 3, 4. In addition Σ1 = 0.045032, Σ2 = 0.015436, Σ3 = 0.043079, Σ4 = 0.009769, z1 = 0.012936, z2 = 0.007538, z3 = 0.005478, z4 = 0.018669, Λ1 = 0.016461, Λ2 = 0.011025, Λ3 = 0.008978, Λ4 = 2.03516 × 10−12. Furthermore, we have: λ1 = 0.082888 < 1, λ2 = 0.115879 < 1, λ3 = 0.136702 < 1, λ4 = 0.000793 < 1.

Then, by Theorem 2.1, the fractional coupled system (3.1) has a unique solution on [0, 1].

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To illustrate the second main result, we consider the following example: Example 3.2.                                                  D103  D72 +10 2 t  x1(t) + |x1(t) + x2(t)| e6(t+1)(1 + |x 1(t) + x2(t)|) = e t 2 + t, t ∈ [0, 1] , D227  D154 +5 × 10 3 t  x2(t) + 1 12π2(t2+ 1)(cos (x1(t)) + cos (x2(t))) = t 2 , t ∈ [0, 1] , |x1(0)| + |x2(0)| = |x01(0)| + |x 0 2(0)| = 0, x 00 1(0) + x 00 2(0) = x 000 1 (0) + x 000 2 (0) = 0, D 7 2x1(0) + D 15 4x2(0) = D 9 2x1(0) + D 19 4 x2(0) = 0, D 11 2x1(0) + D 23 4 x2(0) = 0, D132x1(1) = D 27 4 x2(1) = 0. (3.2)

For this example, we have: n = 2, l = 4, β1 = 10/3, β2 = 22/7, α1 = 7/2, α2 = 15/4,

a1 = 102, a2 = 5 × 103, J = [0, 1].

We see that, for all t ∈ [0, 1] and (x1, x2), (y1, y2) ∈ R2, we get:

|f1(t, x1, x2) − f1(t, y1, y2)| ≤ 1 e6 |x1− y1| + 1 e6 |x2− y2| , |f2(t, x1, x2) − f2(t, y1, y2)| ≤ 1 12π2 |x1− y1| + 1 12π2 |x2− y2| , and sup t∈J |g1(t)| = e + 2 2 , supt∈J |g2(t)| = 1, sup t∈J |f1(t)| = 1 e6, sup t∈J |f2(t)| = 1 12π2. We can take: (µ1)1 = (µ1)2 = 1 e6, (µ2)1 = (µ2)2 = 1 12π2. In addition: Σ1 = 0.004958, Σ2 = 0.016887. Moreover, we obtain: C1 = 0.000042 < 1, C2 = 0.000654 < 1.

The functions fk : [0, 1] × R2 → R and gk : [0, 1] → R are also continuous. So, by

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2

LPAM, Faculty SEI,

UMAB, Mostaganem, Algeria E-mail address: taieb5555@yahoo.fr 1

LPAM, Faculty SEI,

UMAB Mostaganem, Algeria E-mail address: zzdahmani@yahoo.fr

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