New existence and uniqueness results for high dimensional fractional differential systems

NEW EXISTENCE AND UNIQUENESS RESULTS FOR HIGH

DIMENSIONAL FRACTIONAL DIFFERENTIAL SYSTEMS

∗

Zoubir Dahmani and Amele Taeb

Abstract.In this paper, using the Caputo fractional derivative approach, we present new results on the existence and uniqueness of solutions to n−dimensional nonlinear coupled systems for differential equations. We also discuss some examples to illustrate the main results.

1. Introduction and Preliminaries

The physical laws of dynamics are not always described by ordinary order dif-ferential equations. In some cases, their behavior is governed by fractional or-der differential equations. For more details, we refer the reaor-der to the books of Hilfer in [14] and Podlubny in [28]. Other research works can be found in [2, 4, 7, 9, 20, 24, 25, 26, 27, 29, 31, 32]. Further, many authors have established the ex-istence and uniqueness of solutions for some fractional systems. The reader may re-fer to the following research papers [1, 3, 5, 6, 8, 10, 11, 12, 15, 16, 17, 18, 19, 21, 33, 34]. These papers treat problems with only one nonlinear term depending on two un-known functions. Other cases, where we have more than one nonlinearity de-pending on some unknown functions, are more complicated and have not been discussed in the above cited works.

In this paper, we are concerned with the following nonlinear system: (1.1)_⎧ ⎪⎪⎪⎪ ⎪⎪⎪⎪ ⎪⎪⎪⎪ ⎪⎪⎪⎨ ⎪⎪⎪⎪ ⎪⎪⎪⎪ ⎪⎪⎪⎪ ⎪⎪⎪⎩ Dα1_x 1(t)= m  i=1f 1 i  t, x1(t), x2(t), ..., xn(t), Dβ1x1(t), Dβ2x2(t), ..., Dβnxn(t)  , t ∈ J, Dα2_x2(t)= m  i=1f 2 i  t, x1(t), x2(t), ..., xn(t), Dβ1x1(t), Dβ2x2(t), ..., Dβnxn(t)  , t ∈ J, ... Dαnxn(t)= m  i=1f n i  t, x1(t), x2(t), ..., xn(t), Dβ1x1(t), Dβ2x2(t), ..., Dβnxn(t)  , t ∈ J, xk(0)= x  k(0)= ... = x (n−2) k (0)= x (n−1) k (1)= 0, k = 1, 2, ..., n, Received March 20, 2015

2010 Mathematics Subject Classification. 34A34, 34B10.

∗_{The authors were supported in part by . . .}

where n− 1 < αk< n, n − 2 < βk< n − 1, k = 1, 2, ..., n, n ∈ N∗− {1} , J := [0, 1] . The derivatives Dαk and Dβk, k = 1, 2, ..., n, are in the sense of Caputo. The functions 

fk i

k=1,2,...,n

i=1,...,m : J× R2n→ R will be specified later.

We present some basic definitions and properties which are used throughout this paper [13, 22, 28].

Definition 1.1. The Riemann-Liouville fractional integral operator of orderα ≥ 0 for a continuous function f on [0, ∞[ is defined as:

Jαf (t)= 1 Γ (α) t 0 (t− s)α−1f (s) ds, α > 0, t ≥ 0, (1.2) J0f (t)= f (t), t ≥ 0, whereΓ (α) :=₀∞e−xxα−1dx.

Definition 1.2. The Caputo derivative of order α for a function x : [0, ∞) → R, which is at least n−times differentiable can be defined as:

(1.3) Dαx(t)= 1 Γ (n − α) t 0 (t− s)n−α−1x(n)(s) ds= Jn−α_x(n)_(t), for n− 1 < α < n, n ∈ N∗.

We recall the following lemmas [11, 12, 23, 30]:

Lemma 1.1. Letα > 0. The solution to the differential equation Dαx(t)= 0, is given by

(1.4) x(t)= c0+ c1t+ c2t2+ · · · + cn−1tn−1,

such that cj∈ R, j = 0, ..., n − 1, n = [α] + 1.

Lemma 1.2. Letα > 0. Then

(1.5) JαDαx(t)= x(t) + c0+ c1t+ c2t2+ · · · + cn−1tn−1,

where cj∈ R, j = 0, 1, ..., n − 1, n = [α] + 1.

Lemma 1.3. Let q> p > 0, f ∈ L1_{([a, b]) . Then D}p_Jq_{f (t)= J}q−p_{f (t), t ∈ [a, b] .}

Lemma 1.4. Let E be a Banach space. Assume that T : E→ E is completely continuous operator and the set V := x∈ E, x = μTx, 0 < μ < 1 is bounded. Then T has a fixed point in E.

We prove the following auxiliary result:

Lemma 1.5. Suppose that Qk_ik=1,...,n

i=1,...,m∈ C ([0, 1] , R) , and consider the problem (1.6) Dαkxk(t)=

m

i=1

Qk_i(t), t ∈ J, n − 1 < αk< n; k = 1, 2, ..., n, m, n ∈ N∗,

associated with the conditions:

(1.7) xk(0)= x  k(0)= ... = x (n−2) k (0)= x (n−1) k (1)= 0, k = 1, 2, ..., n. Then, we have (1.8) xk(t)= m i=1 _t 0 (t− s)αk−1 Γ (αk) Qk i(s) ds − tn−1 (n−1)!Γ(αk−n+1) m  i=1 ₁ 0 (1− s) αk−n_Qk i(s) ds, k = 1, 2, ..., n.

Proof. We use Lemma 1.1, Lemma 1.2 and (1.6) . So, we can write

(1.9) xk(t)= m i=1 t 0 (t− s)αk−1 Γ (αk) Q k i(s) ds− c0− c1t− c2t2− ... − cn−1tn−1, where cj∈ R, j = 0, 1, 2, ..., n − 1 and n − 1 < αk< n, k ∈ N∗. For all k= 1, 2, ..., n, j = 0, 1, ..., n − 2, we have

x(_kj)(0)= −j!cj. Using Lemma 1.3 and the relation (1.7), we obtain:

(1.10) cj= ⎧ ⎪⎪⎪⎨ ⎪⎪⎪⎩ 0, j= 0, 1, ..., n − 2, m  i=1 ₁ 0 (1−s)αk−n (n−1)!Γ(αk−n+1)Qki(s) ds, j = n − 1.

Substituting the values of cjinto (1.9) , we obtain (1.8) . This ends the proof of the auxiliary result.

Now, let us introduce the Banach space:

S :=(x1, x2, ..., xn) : xk∈ C ([0, 1] , R) , Dβkxk∈ C ([0, 1] , R) , k = 1, 2, ..., n. 

equipped with the norm

(1.11) (x1, x2, ..., xn)S= max  x1 , x2 , ..., xn ,Dβ1x1 ,Dβ2x2 , ...,Dβnxn  , such that, xk = sup t∈J |xk(t)| ,Dβkxk = sup t∈J  Dβk_x k(t) , k = 1, 2, ..., n.

2. Main Results

We begin this section by introducing the following hypotheses: (H1) : There exist nonnegative constants

 λk

i 

j, i = 1, ..., m, k = 1, 2, ..., n, j = 1, 2, ..., 2n, such that for all t ∈ [0, 1] and all (x1, x2, ..., x2n),

 y1, y2, ..., y2n  ∈ R2n_, we have (2.1)  f_ik(t, x1, x2, ..., x2n)− f_ikt, y1, y2, ..., y2n ≤ 2n j=1  λk i  j  xj− yj .

(H2) : The functions f_ik: [0, 1] × R2n→ R are continuous for each i = 1, 2, ..., m, k =

1, 2, ..., n, m, n ∈ N∗_.

(H3) : There exist nonnegative constants



Lk i

k=1,2...,n

i=1,...,m , such that:

for each t∈ J and all (x1, x2, ..., x2n)∈ R2n

(2.2)  f_ik(t, x1, x2, ..., x2n) ≤ Lk_i, i = 1, ..., m, k = 1, 2, ..., n.

Then, we consider the following quantities:

Ok = _{Γ (α}Σk k+ 1), O ∗ k= Σk Γαk− βk+ 1 ; Σk = m i=1  λk i  1+  λk i  2+ ... +  λk i  2n  , k = 1, 2, ..., n, (2.3) Ak = 1 Γ (αk+ 1)+ 1 (n− 1)!Γ (αk+ 2 − n), k = 1, 2, ..., n, A∗_k = 1 Γαk− βk+ 1 + 1 Γn− βk  Γ (αk+ 2 − n), k = 1, 2, ..., n. (2.4)

Now, we are ready to prove the first main result:

Theorem 2.1. If the hypothesis (H1) and the inequality

(2.5) max 1≤k≤n  Ok, O∗_k  < 1

are satisfied, then (1.1) has a unique solution on J.

Proof. Let us define the nonlinear operator T : S→ S by

such that Tk(x1, x2, ..., xn) (t)= (2.6) m i=1 t 0 (t− s)αk−1 Γ (αk) ϕ k i(s) ds− tn−1 (n− 1)!Γ (αk− n + 1) m i=1 1 0 (1− s)αk−nϕk_i(s) ds, k = 1, 2, ..., n, where, ϕk i(s)= fik  s, x1(s), x2(s), ..., xn(s), Dβ1x1(s), Dβ2x2(s), ..., Dβnxn(s)  . We show that the operator T is contractive:

Let (x1, x2, ..., xn), 

y1, y2, ..., yn 

∈ S. Then, for each k = 1, 2, ..., n and t ∈ J, we have:  Tk(x1, x2, ..., xn) (t)− Tk  y1, y2, ..., yn  (t) ≤ (2.7) tαk Γ (αk+ 1) sup s∈J m i=1     fk i  s, x1(s), x2(s), ..., xn(s), Dβ1x1(s), Dβ2x2(s), ..., Dβnxn(s)  − fk i  s, y1(s), y2(s), ..., yn(s), Dβ1y1(s), Dβ2y2(s), ..., Dβnyn(s)  .     By (H1), we can write  Tk(x1, x2, ..., xn)− Tk  y1, y2, ..., yn ≤ 1 Γ (αk+ 1) m i=1  λk i  1+  λk i  2+ ... +  λk i  2n  (2.8) × maxx1− y1 ,x2− y2 , ...,xn− yn ,D β1x1− y1,Dβ2x2− y2,...,Dβnxn− yn. Therefore, for all k= 1, 2, ..., n,

 Tk(x1, x2, ..., xn)− Tky1, y2, ..., yn ≤ (2.9) Σk Γ (αk+ 1)  x1− y1, x2− y2, ..., xn− yn, Dβ1x1− y1, Dβ2x2− y2, ..., Dβnxn− yn S. On the other hand,

 Dβk_T k(x1, x2, ..., xn) (t)− DβkTky1, y2, ..., yn(t) ≤ (2.10) tαk−βk Γαk− βk+ 1sups∈J m i=1     f_iks, x1(s), x2(s), ..., xn(s), Dβ1x1(s), Dβ2x2(s), ..., Dβnxn(s)  − fk i  s, y1(s), y2(s), ..., yn(s), Dβ1y1(s), Dβ2y2(s), ..., Dβnyn(s)  _ , where k= 1, 2, ..., n.

Then, for k= 1, 2, ..., n, we have  Dβk_T k(x1, x2, ..., xn)− DβkTk  y1, y2, ..., yn ≤ (2.11) Σk Γαk− βk+ 1 x1− y1, x2− y2, ..., xn− yn, Dβ1x1− y1, Dβ2x2− y2, ..., Dβnxn− yn S. Using (2.9) and (2.11), we get

 Tk(x1, x2, ..., xn)− Tky1, y2, ..., yn_S≤ (2.12) max 1≤k≤n  Ok, O∗_k   x1− y1, ..., xn− yn, Dβ1x1− y1, ..., Dβnxn− yn S, k = 1, 2, ..., n. Thus, by (2.5) , we deduce that the operator T is contractive. Therefore, by the Banach fixed point theorem, T has a unique fixed point which is a solution of the system (1.1) .

We also prove the result:

Theorem 2.2. Assume that fk

i, i = 1, 2, ..., m, k = 1, 2, ..., n satisfy (H2) and (H3). Then

the nonlinear fractional system (1.1) has at least one solution on J. Proof. The proof will be given in two steps:

A : We show that T is completely continuous:

We begin by proving that T maps bounded sets into bounded sets in S : Let us con-sider the set Bδ:=(x1, x2, ..., xn)∈ S; (x1, x2, ..., xn)S≤ δ, δ > 0

and (x1, x2, ..., xn)∈

Bδ. Then, for each t ∈ J, k = 1, 2, ..., n, and using (H3), we can obtain

Tk(x1, x2, ..., xn) ≤ _{Γ (α}tαk k+ 1)× sups∈J m i=1  fk i  s, x1(s), x2(s), ..., xn(s), Dβ1x1(s), Dβ2x2(s), ..., Dβnxn(s) + 1 (n− 1)!Γ (αk+ 2 − n) sup s∈J m i=1  fk i  s, x1(s), x2(s), ..., xn(s), Dβ1x1(s), Dβ2x2(s), ..., Dβnxn(s) ≤  1 Γ (αk+ 1)+ 1 (n− 1)!Γ (αk+ 2 − n)  m i=1 Lk_i ≤ Ak m i=1 Lk_i (2.13)

and _ Dβk_T k(x1, x2, ..., xn) ≤ tαk−βk Γαk− βk+ 1  × sup s∈J m i=1  fk i  s, x1(s), x2(s), ..., xn(s), Dβ1x1(s), Dβ2x2(s), ..., Dβnxn(s) + tn−βk−1 Γn− βk  Γ (αk+ 2 − n) sup s∈J m i=1 f_iks, x1(s), x2(s), ..., xn(s), Dβ1x1(s), Dβ2x2(s), ..., Dβnxn(s)  ≤  1 Γαk− βk+ 1  + 1 Γn− βk  Γ (αk+ 2 − n)  m i=1 Lk i ≤ A∗ k m i=1 L(2.14)k_i. Hence, T (x1, x2, ..., xn)S≤ max  Ak, A∗k  m i=1 Lk_i < ∞.

This means that T maps bounded sets into bounded sets in S.

Thanks to (H2), the operator T is continuous on S. On the other hand, for any

0≤ t1< t2≤ 1 and (x1, x2, ..., xn)∈ Bδ, we have: Tk(x1, x2, ..., xn) (t2)− Tk(x1, x2, ..., xn) (t1) ≤ (2.15)_ 1 Γ (αk+ 1)  2 (t2− t1)αk+  tαk₂ − tαk₁ + 1 (n− 1)!Γ (αk+ 2 − n)  tn−1 2 − tn1−1  m i=1 Lk i and _ Dβk_T k(x1, x2, ..., xn) (t2)− DβkTk(x1, x2, ..., xn) (t1) ≤ (2.16)  1 Γαk− βk+ 1  2 (t2− t1)αk−βk+  tαk−βk₂ − tαk−βk₁ + 1 Γn− βkΓ (αk+ 2 − n)  tn₂−βk−1− tn₁−βk−1, where, k= 1, 2, ..., n.

The right-hand sides of (2.15) and (2.16) are independent of (x1, x2, ..., xn)∈ Bδand

tend to zero as t2− t1 → 0. Thus T is equi-continuous. Finally, we can see by the

above arguments that T is a completely continuous operator.

B : We consider the setΩ :=(x1, x2, ..., xn)∈ S, (x1, x2, ..., xn)= μT (x1, x2, ..., xn), 0 < μ < 1  and show that is bounded:

Let (x1, x2, ..., xn) ∈ Ω, then xk(t) = μTk(x1, x2, ..., xn) (t). Thus, for each t ∈ J and corresponding to (2.13) and (2.14), we have:

(2.17) xk ≤ μAk m i=1 Lk_i; Dβkxk ≤ A∗k m i=1 Lk_i, k = 1, 2, ..., n, which implies (x1, x2, ..., xn)S ≤ μ max ⎛ ⎜⎜⎜⎜ ⎜⎝A1 m i=1 L1_i, A2 m i=1 L2_i, ..., An m i=1 Ln_i, A∗₁ m i=1 L1_i, A∗₂ m i=1 L2_i, ..., A∗_n m i=1 Ln_i ⎞ ⎟⎟⎟⎟ ⎟⎠ . < ∞. (2.18) Therefore,Ω is bounded.

Consequently by the steps A, B and using lemma 2.4, we deduce that T has a fixed point which is a solution to (1.1) . Theorem 2.2 is thus proved.

3. Examples

We present two examples to illustrate our main results.

Example 3.1. We begin with the system:

(3.1)_⎧ ⎪⎪⎪⎪ ⎪⎪⎪⎪ ⎪⎪⎪⎪ ⎪⎪⎪⎪ ⎪⎪⎪⎪ ⎪⎪⎪⎪ ⎪⎪⎪⎪ ⎪⎪⎨ ⎪⎪⎪⎪ ⎪⎪⎪⎪ ⎪⎪⎪⎪ ⎪⎪⎪⎪ ⎪⎪⎪⎪ ⎪⎪⎪⎪ ⎪⎪⎪⎪ ⎪⎪⎩ D73x₁(t)=   x1(t)+x2(t)+x3(t)+D 4 3x1(t)+D 3 2x2(t)+D 5 3x3(t)    9π3  1+_x1(t)+x2(t)+x3(t)+D 4 3x1(t)+D 3 2x2(t)+D 5 3x3(t)     + 1 64π2e _sin(x 1(t))+sin(x2(t))+sin(x3(t)) et+1 + cos  D43_x1(t)  + cosD32_x2(t)  − sinD53_x3(t)  , t ∈ ]0, 1[ , D94x₂(t)= 1 24π3e2t+1  sinD43_x1(t)  + sinD32x₂(t)  + sinD53_x3(t)  + |x1(t)+x2(t)+x3(t)| π+|x1(t)+x2(t)+x3(t)|  + t2 16π2et2+1 ⎛ ⎜⎜⎜⎜

⎜⎜⎝ sin(x1(t))+cos(x2(t))+cos(x3(t))−cos

 D43x1(t)  +sin  D32x2(t)  +sin  D53x3(t) 

e+sin(x1(t))+cos(x2(t))+cos(x3(t))−cos

 D43x1(t)  +sin  D32x2(t)  +sin  D53x3(t)  ⎞ ⎟⎟⎟⎟ ⎟⎟⎠, t ∈ ]0,1[, D83x₃(t)= 1 32π ⎛ ⎜⎜⎜⎜

⎜⎜⎝cos(x1(t))+ sin (x2(t))+ sin (x3(t))+   D43x1(t)+D 3 2x2(t)+D 5 3x3(t)     1+_D43x1(t)+D 3 2x2(t)+D 5 3x3(t)     ⎞ ⎟⎟⎟⎟ ⎟⎟⎠ + 1 16π(t+e)2 ⎛ ⎜⎜⎜⎜ ⎜⎜⎝sinx1(t)+ sin D 4 3x₁(t)+ sin D32x₂(t)   x2(t)+x3(t)+D 5 3x3(t)    3πe  1+_x2(t)+x3(t)+D 5 3x3(t)     ⎞ ⎟⎟⎟⎟ ⎟⎟⎠, t ∈ ]0,1[ xk(0)= x  k(0)= x  k(1)= 0, k = 1, 2, 3.

For this example, we have:

n= 3, m = 2, α1=7₃, α2= 9₄, α3=8₃, β1= 4₃, β2=3₂, β3=5₃, J = [0, 1] .

On the other hand,

(3.2) f1 1 (t, x1, x2, x3, x4, x5, x6)= |x1+ x2+ x3+ x4+ x5+ x6| 9π3_{(1+ |x} 1+ x2+ x3+ x4+ x5+ x6|) , (3.3) f21(t, x1, x2, x3, x4, x5, x6)= 1 64π2_{e sin x} 1+ sin x2+ sin x3

et+1 + cos x4+ cos x5− sin x6

 ,

(3.4) f2

1 (t, x1, x2, x3, x4, x5, x6)= 1

24π3_e2t+1

sin x4+ sin x5+ sin x6+_{π + |x}|x1+ x2+ x3| 1+ x2+ x3|  , (3.5) f2 2(t, x1, x2, x3, x4, x5, x6)= t 2 16π2_et2_{+1  sin x}

1+ cos x2+ cos x3− cos x4+ sin x5+ sin x6

e+ sin x1+ cos x2+ cos x3− cos x4+ sin x5+ sin x6

 , (3.6) f3 1(t, x1, x2, x3, x4, x5, x6)= 1 32π 

cos x1+ sin x2+ sin x3+ |x4+ x5+ x6|

(1+ |x4+ x5+ x6|)  and (3.7) f3 2 (t, x1, x2, x3, x4, x5, x6)= 1 16π (t + e)2 

sin x1+ sin x4+ sin x5+

|x2+ x3+ x6|

3πe (1 + |x2+ x3+ x6|)

 . So, for t∈ [0, 1] and (x1, x2, x3, x4, x5, x6),

 y1, y2, y3, y4, y5, y6  ∈ R6_{, we have:}   f1 1(t, x1, x2, x3, x4, x5, x6)− f11  t, y1, y2, y3, y4, y5, y6 ≤ (3.8) _9π1₃x1− y1+_9π1₃ x2− y2+_9π1₃x3− y3+ _9π1₃ x4− y4+_9π1₃x 5− y5+_9π1₃ x6− y6 ,   f1 2(t, x1, x2, x3, x4, x5, x6)− f21  t, y1, y2, y3, y4, y5, y6 ≤ (3.9) 1 64π2_e2  x1− y1+ 1 64π2_e2  x2− y2+ 1 64π2_e2  x3− y3+ 1 64π2_e  x4− y4+ 1 64π2_e  x5− y5+ 1 64π2_e  x6− y6 ,   f2 1(t, x1, x2, x3, x4, x5, x6)− f12  t, y1, y2, y3, y4, y5, y6 ≤ (3.10) 1 24π2_ex1− y1+ 1 24π2_ex2− y2+ 1 24π2_ex3− y3+ 1 24π3_ex4− y4+ 1 24π3_ex5− y5+ 1 24π3_ex6− y6 ,   f2 2(t, x1, x2, x3, x4, x5, x6)− f22  t, y1, y2, y3, y4, y5, y6 ≤ (3.11) 1 16π2  x1− y1+_16π1₂x2− y2+_16π1₂x3− y3+_16π1₂x4− y4+_16π1₂x5− y5+_16π1₂x6− y6 ,   f3 1(t, x1, x2, x3, x4, x5, x6)− f 3 1  t, y1, y2, y3, y4, y5, y6 ≤ (3.12) 1 32πx1− y1 +_32π1 x2− y2 +_32π1 x3− y3 +_32π1 x4− y4 +_32π1 x5− y5 +_32π1 x6− y6 , and _  f3 2(t, x1, x2, x3, x4, x5, x6)− f23  t, y1, y2, y3, y4, y5, y6 ≤ (3.13) 1 16πe2  x1− y1+_48π1₂ e3  x2− y2+_48π1₂ e3  x3− y3+_16πe1 ₂x4− y4+_16πe1 ₂ x5− y5+_48π1₂ e3  x6− y6 . We can take: (3.14) λ1 1  1=  λ1 1  2=  λ1 1  3=  λ1 1  4=  λ1 1  5=  λ1 1  6= 1 9π3,

(3.15) λ1 2  1=  λ1 2  2=  λ1 2  3= 1 64π2_e2,  λ1 2  4=  λ1 2  5=  λ1 2  6= 1 64π2_e, (3.16) λ21  1=  λ2 1  2=  λ2 1  3= 1 24π2_e,  λ2 1  4=  λ2 1  5=  λ2 1  6= 1 24π3_e, (3.17) λ2 2  1=  λ2 2  2=  λ2 2  3=  λ2 2  4=  λ2 2  5=  λ2 2  6= 1 16π2, (3.18) λ3 1  1=  λ3 1  2=  λ3 1  3=  λ3 1  4=  λ3 1  5=  λ3 1  6= 1 32π, and (3.19) λ3 2  1=  λ3 2  4=  λ3 2  5= 1 16πe2,  λ3 2  2=  λ3 2  3=  λ3 2  6= 1 48π2_e3. It follows that: (3.20) Σ1= 0.023891, Σ2= 0.044138, Σ3= 0.068076. Since (3.21) Γ (α1+ 1) = 2.778062, Γ (α2+ 1) = 2.549257, Γ (α3+ 1) = 4.012356, (3.22) Γα1− β1+ 1  = 1, Γα2− β2+ 1  = 0.919062, Γα3− β3+ 1  = 1, then it yields that:

O1 = 0.008599, O2= 0.009372, O3= 0.005954, (3.23) O∗1 = 0.023891, O∗2= 0.048025, O∗3= 0.068076, maxO1, O2, O3, O∗1, O∗2, O∗3  < 1.

The condition (2.5) is satisfied. So by Theorem 2.1, we deduce that the system (3.1) has a unique solution on [0, 1].

Example 3.2. To illustrate the second main result, let us consider the system:

(3.24) ⎧ ⎪⎪⎪⎪ ⎪⎪⎪⎪ ⎪⎪⎪⎪ ⎪⎪⎪⎪ ⎪⎪⎪⎪ ⎪⎪⎪⎪ ⎪⎨ ⎪⎪⎪⎪ ⎪⎪⎪⎪ ⎪⎪⎪⎪ ⎪⎪⎪⎪ ⎪⎪⎪⎪ ⎪⎪⎪⎪ ⎪⎩ D94x₁(t)= π(t+1) sin  D32x1(t)+D 4 3x2(t)+D 5 4x3(t)  2−cos(x1(t)+x2(t)+x3(t)) + et(t2+1) 2π+cos  x2(t)+D 4 3x2(t)  cos  x3(t)+D 5 4x3(t)  +sin2  x₁(t)D32x₁(t) _{, t ∈ ]0, 1[ ,} D73x₂(t)= e 2_sin(x 1(t)+x2(t)+x3(t)) 2π+cos  D32x1(t)+D 4 3x2(t)+D 5 4x3(t)  + 3t2_cos(x 2(t)+x3(t)) et2+1_−cos  x1(t)−D 3 2x1(t)+D 4 3x2(t)+D 5 4x3(t) _{, t ∈ ]0, 1[ ,} D52x₃(t)= sin(x3(t)) 2e+cox  x1(t)+x2(t)+D 3 2x1(t)+D 4 3x2(t)+D 5 4x3(t)  + cos (x1(t)+ x2(t)+ x3(t)) sin  D32x₁(t)+ D43x₂(t)+ D54x₃(t)  , t ∈ ]0, 1[ , xk(0)= x  k(0)= x  k(1)= 0, k = 1, 2, 3.

We have: n= 3, m = 2, α1= 94, β1=32;α2= 37, β2=43, α3= 52, β3=54, J = [0, 1] . Since (3.25)  f1 1(t, x1, x2, x3, x4, x5, x6) =   π (t + 1) sin (x + x_{2− cos (x1+ x2+ x}5+ x₃₎6) ≤ 2π, (3.26)  f1 2 (t, x1, x2, x3, x4, x5, x6) =    e t_t2_{+ 1}

2π + cos (x2+ x5) cos (x3+ x6)+ sin2(x1x4)

   ≤πe, (3.27)  f2 1(t, x1, x2, x3, x4, x5, x6) =    e 2_{sin (x} 1+ x2+ x3) 2π + cos (x4+ x5+ x6)    ≤ e 2 2π − 1, (3.28)  f2 2(t, x1, x2, x3, x4, x5, x6) =    3t 2_{cos (x} 2+ x3) et2_{+1− cos (x} 1− x4+ x5+ x6)    ≤e− 13 ,   f3 1(t, x1, x2, x3, x4, x5, x6) =   _{2e+ cox (x1+ x}sin x_{2+ x}3 _{4+ x5+ x6)} ≤ _2e1_{− 1} and _  f3 2(t, x1, x2, x3, x4, x5, x6) = |cos (x1+ x2+ x3) sin (x4+ x5+ x6)| ≤ 1,

then, we can see that the functionsfk i

k=1,2,3

i=1,2 are continuous and bounded on [0, 1] × R 6_{. So,}

by Theorem 2.2, the system (3.24) has at least one solution on [0, 1].

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Zoubir DAHMANI Laboratory LPAM Faculty of SEI

UMAB, University of Mostaganem zzdahmani@yahoo.fr

Amele TA¨IEB Faculty of SEI

Department of Mathematics and Informatics UMAB, University of Mostaganem