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www.i-csrs.org

Zakaria Bekkouche, Zoubir Dahmani, Guo Zhang

School of Mathematics and Statistics, Central China Normal University, Wuhan 430079, China

e-mail:bezack@mails.ccnu.edu.cn

Laboratory LPAM, Faculty SEI, UMAB Mostaganem, Algeria e-mail:zzdahmani@yahoo.fr

School of Mathematics and Statistics, Central China Normal University, Wuhan 430079, China

e-mail:guozhang625@hotmail.com

Received 10 December 2017; Accepted 13 March 2018

Abstract

In this work, we are concerned with a two dimension frac-tional Lane Emden differential system with right hand side depending on an unknown vector function. Using Banach con-traction principle on an appropriate product Banach space, we establish some results on the existence and uniqueness of solu-tions. The existence of at least one solution of the considered problem is also studied. Some notions of Ulam type stabili-ties are presented and illustrated. At the end, an example is discussed.

Keywords: Caputo derivative, fixed point, existence, Lane-Emden system of two dimension, uniqueness.

2010 Mathematics Subject Classification: 03F55, 46S40.

1

Introduction

Recently fractional calculus started to attract serious attention in a lot of scientific areas, such as mathematics, biology and engineering. For getting a better understanding of the theory we suggest the reader to address the

Solutions and Stabilities for a 2D−non Homogeneous

Lane-Emden Fractional System

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following papers [12, 14, 15] and the reference therein. It is also important to mention how relevant it is to do research on fractional deferential equations. Nowadays many branches of science and technology are making use of this theory (see [4, 13, 18]). The existence and uniqueness of solutions for nonlinear fractional differential equations was studied by many scholars. For getting further information the reader can address the following papers [6, 9, 20]. Along with that it is needed to mention that the Ulam type stabilities for fractional differential problems are useful for solving practical problems in biology, economics and mechanics. The examples of the application of this theory can be found in [1, 3, 11].

Now we would like to bring to the attention the Lane-Emden model, which serves as the basis for our research.

It is generally known that the Lame-Emden equations are found in a few models of mathematical physics and astrophysics, such as aspects of stellar structure, isothermal gas spheres and thermionic currents [5]. The Lane-Emden equation 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, g are continuous real valued functions. This equation and the problems related to it has occupied the minds of a number of researchers. For getting further information the reader is recommended to turn to [8, 16, 19].

In [10] the authors studied coupled Lane-Emden equations arising in catalytic diffusion reaction by reproducing kernel Hilbert space method while giving consideration to the following problem:

   u00(x) + K1 x u 0(x) = f 1(u, v), 0 < x ≤ 1 v00(x) + K2 x v 0(x) = f 2(u, v), 0 < x ≤ 1,

subject to the initial conditions

u0(0) = 0, u(1) = α1,

v0(0) = 0, v(1) = α2,

where K1 and K2 are constants, f1(u, v) and f2(u, v) are analytic functions in

u and v. Such boundary value problems arise in catalytic diffusion reaction. A. Akgl, M. Inc, E. Karatas, D. Baleanu applied the reproducing kernel method in [2] and suggested a numerical study for the following Lane-Emden problem:

 Dαy (t) + k tα−βD

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

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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]) .

Most recently, in [7] Z. Dahmani and M.Z. Sarikaya studied the following generalized Lane Emden system:

   Dβ1(Dα1 + b 1g1(t)) x1(t) + f1(t, x1(t) , x2(t)) = h1(t) , 0 < t < 1, Dβ2(Dα2 + b 2g2(t)) x2(t) + f2(t, x1(t) , x2(t)) = h2(t) , 0 < t < 1, xk(0) = 0, Dαxk(1) + bkgk(1)xk(1) = 0,

where 0 < βk < 1, 0 < αk< 1, bk ≥ 0, k = 1, 2 and the derivatives Dβk and

Dαk are in the sense of Caputo.

Motivated by the above work, this paper considers a more general system of Lane Emden type by injecting the unknown functions (solutions) not only on the left hand side of the system, but on right hand side of the problem too. This injection makes the problem very difficult to study, since basically the problem is singular. So let us consider the following problem:

   Dβ1(Dα1 + b 1g1(t)) x1(t) + f1(t, x1(t) , x2(t)) = ω1S1(t, x1(t) , x2(t)) , 0 < t < 1, Dβ2(Dα2 + b 2g2(t)) x2(t) + f2(t, x1(t) , x2(t)) = ω2S2(t, x1(t) , x2(t)) , 0 < t < 1, xk(0) = 0, Dαxk(1) + bkgk(1)xk(1) = 0, (1) where 0 < βk < 1, 0 < αk < 1, bk ≥ 0, 0 < ωk < ∞, k = 1, 2 and the

derivatives Dβk and Dαk are in the sense of Caputo. The functions f

k : [0, 1]×

are continuous, gk : ]0, 1] → [0, +∞) is continuous and singular at t = 0.

2

Preliminaries

Definition 2.1 The Riemann-Liouville integral operator [14]: Jαf (t) = 1 Γ (α) Z t 0 (t − s)α−1f (s) ds, α > 0, t ≥ 0, (2) where Γ (α) :=R∞ 0 e −xxα−1dx,

and the Caputo fractional derivative Dα

Dαf (t) = 1 Γ (n − α)

Z t

0

(t − s)n−α−1f(n)(s) ds, n − 1 < α < n. (3) We need the following auxiliary results [12]:

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Lemma 2.2 For α > 0, the general solution of the fractional differential equation Dαx(t) = 0 is given by x(t) = n−1 X j=0 cjtj, (4) where cj ∈ R, j = 0, ..., n − 1, n = [α] + 1.

Lemma 2.3 Let α > 0. We have

JαDαx(t) = x(t) + n−1 X j=0 cjtj, (5) where cj ∈ R, j = 0, 1, ..., n − 1, n = [α] + 1.

Lemma 2.4 Let q > p > 0, g ∈ L1([a, b]) . Then DpJqg (t) = Jq−pg (t) , t ∈ [a, b] .

Lemma 2.5 Let E be a Banach space and let’s assume that T : E → E is a completely continuous operator. If the set V := {x ∈ E : x = µT x, 0 < µ < 1} is bounded, then T has a fixed point in E.

To give the integral representation of (1) , we need to prove the following auxiliary result:

Lemma 2.6 Let H1, H2 ∈ C ([0, 1], R) . Then, the problem

   Dβ1(Dα1 + b 1g1(t)) x1(t) = H1(t) , t ∈ [0, 1], Dβ2(Dα2 + b 2g2(t)) x2(t) = H2(t) , t ∈ [0, 1] (6)

associated with the conditions

xk(0) = 0, Dαxk(1) + bkgk(1)xk(1) = 0, k = 1, 2 (7)

has a unique solution (x1, x2) given by:

xk(t) = Jαk+βkHk(t) − bkJαkgk(t) xk(t) − JβkHk(1)

tαk

Γ(αk+ 1)

. (8) Proof. We use Lemma 2.3 to obtain:

x1(t) = t Z 0 (t − τ )α1−1 Γ (α1)   τ Z 0 (τ − s)β1−1 Γ (β1) H1(s) ds − b1g1(t) x1(τ )  dτ − c1Jα1(1) − c2. (9)

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Then, by (7) we obtain:

c2 = 0, c1 = Jβ1H1(1).

With the same arguments we obtain the component x2(t).

Lemma 2.6 is thus proved.

Let us now introduce the Banach space X × X, k(u, v)kX×X , with k(u, v)kX×X = max{kukX, kvkX} and X := C([0, 1], R), ||.||X = ||.||∞.

3

Main results

3.1

Existence and Uniqueness

We prove the following theorem.

Theorem 3.1 Let g1, g2 : ]0, 1] → [0, +∞) be continuous, lim

t→0g1(t) = limt→0g2(t) =

∞. Suppose that there exist 0 < λ1, λ2 < 1, t 7−→ (tλ1g1(t) , tλ2g2(t)) are

con-tinuous on [0, 1] . If |f1(t, x1, x2) − f1(t, y1, y2)| ≤ 2 X j=1 Li|xj − yj| , |f2(t, x1, x2) − f2(t, y1, y2)| ≤ 2 P j=1 L0i|xj − yj| , (10) |S1(t, x1, x2) − S1(t, y1, y2)| ≤ 2 X j=1 Ri|xj − yj| , |S2(t, x1, x2) − S2(t, y1, y2)| ≤ 2 X j=1 R0i|xj − yj| , for all t ∈ [0, 1] , (x1, x2) , (y1, y2) ∈ R2,

then the problem (1) has a unique solution on [0, 1] provided that

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where D1 = b1M1 β(α1, 1 − λ1) Γ(α1) D2 = b2M2 β(α2, 1 − λ2) Γ(α2) K1 = 1 Γ(α1+ β1+ 1) + 1 Γ(α1+ 1)Γ(β1+ 1) K2 = 1 Γ(α2+ β2+ 1) + 1 Γ(α2+ 1)Γ(β2+ 1) and L := max{ L1, L2, L01, L 0 2}, R := max{ R1, R2, R01, R 0 2}, Mk := M axt∈[0,1]|tλkgk(t)| , k = 1, 2.

Proof. Let us consider the operator T : X × X → X × X defined by T (x1, x2) :=  T1(x1, x2), T2(x1, x2)  , (12) where Tk(x1, x2) (t) := Jαk+βk(ωkSk(x1, x2)(t) − fk(x1, x2)(t)) − bkJαkgk(t) xk(t) −Jβk kSk(x1, x2)(1) − fk(x1, x2)(1)) t αk Γ(αk+1), k = 1, 2. (13) We need to prove that T is contractive.

Let (x1, x2), (y1, y2) ∈ X × X. We have T1(x1, x2) (t) − T1(y1, y2) (t) = Jα1+β1(ω1S1(x1, x2)(t) − f1(x1, x2)(t)) − b1Jα1(g1(t) x1(t)) −Jβ1 1S1(x1, x2)(1) − f1(x1, x2)(1)) t α1 Γ(α1+1) −  Jα1+β1 1S1(x1, x2)(t) − f1(y1, y2)(t)) −b1Jα1(g1(t) y1(t)) − Jβ1(ω1S1(x1, x2)(1) − f1(y1, y2)(1)) t α1 Γ(α1+1)  . (14) Some easy techniques allow us to write

|T1(x1, x2) (t) − T1(y1, y2) (t)| ≤ |ω1Jα1+β1(S1(y1, y2)(t) − S1(x1, x2)(t))| + Γ(αω11+1)|Jβ1(S1(y1, y2)(1) − S1(x1, x2)(1))| +|Jα1+β1(f 1(y1, y2)(t) − f1(x1, x2)(t))| + Γ(α11+1)|Jβ1(f1(y1, y2)(1) − f1(x1, x2)(1))| +b1M1|x1− y1|(t)Jα1t−λ1. (15)

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Thanks to the conditions on f1, S1 and tλ1g1, we obtain ||T1(x1, x2) − T1(y1, y2) ||X ≤ ω1  1 Γ(α1+β1+1) + 1 Γ(α1+1)Γ(β1+1)  R1||x1 − y1|| + R2||x2− y2||   1 Γ(α1+β1+1)+ 1 Γ(α1+1)Γ(β1+1)  L1||x1− y1|| + L2||x2− y2||  +b1M1||x1 − y1|| β(α1,1−λ1) Γ(α1) . (16) Therefore, ||T1(x1, x2) − T1(y1, y2) ||X ≤b1M1β(αΓ(α1,1−λ1) 1) + 2(L + ω1R)( 1 Γ(α1+β1+1) + 1 Γ(α1+1)Γ(β1+1))  ||(x1− y1, x2− y2)||X×X. (17) With the same arguments as before we can write

||T2(x1, x2) − T2(y1, y2) ||X ≤b2M2β(αΓ(α2,1−λ2) 2) + 2(L + ω2R)( 1 Γ(α2+β2+1) + 1 Γ(α2+1)Γ(β2+1))  ||(x1− y1, x2− y2)||X×X. (18) Using these two inequalities we get

||T (x1, x2) − T (y1, y2) ||X×X

≤D1+ D2+ 2L(K1+ K2) + 2R(ω1K1+ ω2K2)



||(x1− y1, x2− y2)||X×X.

(19) Since D1+ D2+ 2L(K1+ K2) + 2R(ω1K1+ ω2K2) < 1, we can state that the

operator T is contractive. Thus the theorem is proved. Thus the theorem is proved.

3.2

Existence

In the case where 0 < λk ≤ αk < 1, we present the following theorem:

Theorem 3.2 For k = 1, 2, suppose that gk : ]0, 1] → [0, +∞) are continuous,

lim

t→0gk(t) = ∞, and there exist λk; 0 < λk ≤ αk < 1, t 7−→ t λkg

k(t) are

con-tinuous on [0, 1] . Assume that fk : [0, 1] × R2 → R and Sk : [0, 1] × R2 → R

are bounded respectively by Ik and Qk. Then the problem (1) has at least one

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Proof. We will prove the theorem through the following steps:

The continuity of the functions fk, Sk, tλkgk, k = 1, 2 implies that T is

con-tinuous on X × X.

Step 2: The operator T is completely continuous:

We define the set Ωr :=(x1, x2) ∈ X × X, k(x1, x2)kX×X ≤ r , where r > 0.

For (x1, x2) ∈ Ωr, we obtain ||Tk(x1, x2) ||X ≤ Ik+ ωkQk Γ(αk+ βk+ 1) + rbkMkβ(αk, 1 − λk) Γ(αk) + Ik+ ωkQk Γ(βk+ 1)Γ(αk+ 1) . (20) This is to say that

||T (x1, x2) ||X×X ≤ maxk=1,2  Ik+ωkQk Γ(αk+βk+1) + rbkMkβ(αk,1−λk) Γ(αk) + Ik+ωkQk Γ(βk+1)Γ(αk+1)  . (21)

Hence, the operator T maps bounded sets into bounded sets in X × X. Step 3: Equi-continuity of T (Ωr) :

For t1, t2 ∈ [0, 1] ; t1 < t2, and (x1, x2) ∈ Ωr, we have:

||Tk(x1, x2) (t2) − Tk(x1, x2) (t1)||X ≤ (Ik+ωkQk)(tαk+βk2 −tαk+βk2 ) Γ(αk+βk+1) + rbkMkΓ(1−λk)(tαk−λk2 −tαk−λk2 ) Γ(αk−λk+1)Γ(αk) + (Ik+ωkQk)(tαk2 −tαk1 ) Γ(βk+1)Γ(αk+1) := Ck. (22) In these inequalities the right hand sides are independent of x1, x2 and tend to

zero as t1 tends to t2.

In view of the results obtained in steps 2, 3 and according to Arzela-Ascoli theorem, it is seen that T is completely continuous.

Step 4: The set

Ω := {(x1, x2) ∈ X × X; (x1, x2) = λT (x1, x2) , 0 < λ < 1} (23)

is bounded:

Let (x1, x2) ∈ Ω, then (x1, x2) = λT (x1, x2) , for some 0 < λ < 1. Hence, for

t ∈ [0, 1] , we have:

x1(t) = λT1(x1, x2) (t) , x2(t) = λT2(x1, x2) (t) . (24)

Thus,

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Since the functions fk and Sk are bounded, then by (22) we obtain

k(x1, x2)kX×X ≤ λ(C1+ C2). (26)

Consequently, Ω is bounded.

As a conclusion of Schaefer fixed point theorem, we deduce that T has at least one fixed point, which is a solution of (1).

3.3

∆−Ulam Stabilities

In this section, we will focus our attention on the ∆−Ulam-Hyers and general-ized ∆−Ulam-Hyers stabilities for the problem (1) . We start with the following definitions:

Definition 3.3 The problem (1) is ∆−Ulam-Hyers stable, if there exists a real number R > 0, such that for each k> 0, k = 1, 2 and for for each solution

(x1, x2) ∈ X × X of the inequalities Dβk(Dαk+ b kgk(t)) xk(t) + fk(t, x1(t) , x2(t)) − ωkSk(t, x1(t) , x2(t)) < k, t ∈ [0, 1], (27) there exists a solution (y1, y2) ∈ X × X of (1) ,

such that

k(x1, x2) − (y1, y2)kX×X < ∆ + (1+ 2)R. (28)

Definition 3.4 The problem (1) is ∆−generalized Ulam-Hyers stable, if there exists an increasing function Z ∈ C (R+, R+) , Z(0) = ∆, such that for all

k> 0, and for each solution (x1, x2) ∈ X × X of (27), there exists a solution

(y1, y2) ∈ X × X of (1) (with the same conditions as in (1)), such that

k(x1, x2) − (y1, y2)kX×X < Z(1+ 2). (29)

Let us consider the equation (1) and the inequalities (27). We prove the following stability result:

Theorem 3.5 Let the assumptions of Theorem 3.1 hold. If the inequality 1 − 2[L + R(ω1+ ω2)]  Jα1+β1(1) + Jα2+β2(1)  −β(α1, 1 − λ1)b1M1+ β(α2, 1 − λ2)b2M2  > 0 (30)

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Proof. By Theorem 3.1, the problem (1) has a unique solution (y1, y2) ∈

X × X. Let (x1, x2) be a solution of (28) . By definition, we can state that

there exist lk (depending on (x1, x2)) that satisfy |lk(t)| ≤ k, such that

xk(t) = t Z 0 (t − τ )αk−1 Γ (αk)   τ Z 0 (τ − s)βk−1 Γ (βk) (ωkSk− fk+ lk)(s)ds − bkgk(t) xk(τ )  dτ −ckJαk(1) − dk, ck, dk ∈ R. (31) So, we have |xk(t) − yk(t)| ≤ |Jαk+βk(ωkSk(x1, x2)(t) − fk(x1, x2)(t) + lk(t)) − bkJαkgk(t) xk(t) − ckJαk − dk −Jαk+βk kSk(y1, y2)(t) − fk(y1, y2)(t)) −bkJαkgk(t) yk(t) − Jβk(ωkSk(y1, y2)(1) − fk(y1, y2)(1)) t αk Γ(αk+1)  |. (32) Therefore, |xk(t) − yk(t)| ≤ ωk|Jαk+βk(Sk(y1, y2)(t) − Sk(x1, x2)(t)) + ||Jαk+βk(fk(y1, y2)(t) − fk(x1, x2)(t))| +Γ(α1 k+1)|J βk(f k(y1, y2)(1) + ωkJβkSk(y1, y2)(1))| +bkMk|xk(t) − yk(t)|Jαkt−λk+ |ck|Jαk(1) + |dk| + kJαk+βk(1). (33) Consequently, ||xk− yk||X ≤ 2[L + Rωk]Jαk+βk(1)||xk− yk|| + Γ(α1k+1)  |Jβkf k(y1, y2)(1)| + ωk|JβkSk(y1, y2)(1)|  +β(αk, 1 − λk)bkMk||xk− yk|| + |ck|Jαk(1) + |dk| + kJαk+βk(1). (34)

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Adding these two inequalities (for k = 1, 2,) we obtain ||(x1, x2) − (y1, y2)||X×X ≤ 2Rω1Jα1+β1(1) + ω2Jα2+β2(1)  ||(x1, x2) − (y1, y2)||X×X +2L  Jα1+β1(1) + Jα2+β2(1)  ||(x1, x2) − (y1, y2)||X×X +Γ(α1 1+1)  |Jβ1f 1(y1, y2)(1)| + |ω1Jβ1S1(y1, y2)(1)|  +Γ(α1 2+1)  |Jβ2f 2(y1, y2)(1)| + ω2|Jβ2S2(y1, y2)(1)|  +β(α1, 1 − λ1)b1M1+ β(α2, 1 − λ2)b2M2  ||(x1, x2) − (y1, y2)||X×X +|c1|Jα1(1) + |d1| + |c2|Jα2(1) + |d2| +   Jα1+β1(1) + Jα2+β2(1)  . (35)

Consequently, we deduce that

h 1 − 2Rω1Jα1+β1(1) + ω2Jα2+β2(1)  − 2LJα1+β1(1) + Jα1+β1(1)  β(α1, 1 − λ1)b1M1+ β(α2, 1 − λ2)b2M2 i ×||(x1, x2) − (y1, y2)||X×X ≤ 1 Γ(α1+1)  |Jβ1f 1(y1, y2)(1)| + |ω1Jβ1S1(y1, y2)(1)|  +Γ(α1 2+1)  |Jβ2f 2(y1, y2)(1)| + ω2Jβ1S2(y1, y2)(1)|  +|c1|Jα1(1) + |d1| + |c2|Jα2(1) + |d2| +   Jα1+β1(1) + Jα2+β2(1)  . (36)

Thanks to (36), we deduce that (1) is ∆−Ulam-Hyers stable. Hence, the problem (1) is ∆−generalized Ulam-Hyers stable.

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3.4

Illustrations

We consider the following system        D34  D34 + g1(t)  x1(t) + f1(t, x1(t) , x2(t)) = 12S1(t, x1(t) , x2(t)) , 0 < t < 1, D34  D34 + g2(t)  x2(t) + f2(t, x1(t) , x2(t)) = 14S2(t, x1(t) , x2(t)) , 0 < t < 1, xk(0) = 0, Dαxk(1) + bkgk(1)xk(1) = 0, (37) where,

f1(t, u, v) = t2(cos u.v) , f2(t, u, v) = t2(sin u.v),

S1(t, u, v) = sin(u) sin(v) + t, S2(t, u, v) = cos(tv) sin(tv),

g1(t) = 1 2 √ t, g2(t) = 1 3 √ t

and λ1 = 12 λ2 = 13. Then the problem (1) has at least one solution on [0, 1].

4

Open Problem

We end this paper by proposing the following open questions:

What will happen if the function g admits an arbitrary singularity on the whole t−positive real line? What about the stability of the associated Lane-Emden system in this case?

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