THE differential equations of fractional order arise in

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1 IAENG International Journal of Applied Mathematics 45:4 IJAM_45_4_4 Applications of Fixed Point Theorems for Coupled Systems of Fractional Integro-Differential Equations Involving Convergent Series Mohamed Amin Abdellaoui Zoubir Dahmani and Nabil Bedjaoui Abstract The existence and uniqueness of solutions in an appropriate Banach space is established for a coupled system of integro-differential equations involving convergent series with derivatives of non-integer order on the unnown functions and integrals of Riemann-Liouville in the nonlinearity as well as in the initial conditions Some illustrative examples are also presented to demonstrate our main results Index Terms Caputo derivative convergent series differential system fixed point existence and uniqueness Riemann- Liouville integral where c D γ + denotes the Caputo fractional derivative for γ > < α β 2 < p q t J = [ T ] f f 2 C [ T ] R+ 2 R 2 R + Recently in [24] the authors discussed the following coupled system with integral boundary conditions: I INTRODUCTION THE differential equations of fractional order arise in many scientific disciplines such as physics chemistry control theory signal processing and biophysics For more details we refer the reader to [5] [9] [6] [7] [8] and the references therein Recently there has been a significant progress in the investigation of this theory see [] [2] [3] [4] [6] [9] [2] [23] Moreover the study of coupled systems of fractional differential equations is also of a great importance Such systems occur in various problems of applied science For some recent results on the fractional systems we refer the reader to [7] [8] [] [4] [2] [22] [24] Let us now present some results that have inspired our wor We begin by [] where the authors studied the following problem: D α x t + f t y t D δ y t = t J D β y t + f 2 t x t D σ x t = t J x = x y = y x + x + y + y = x = J p x η y = J q y ξ where α β ]3 4] δ α σ β η ξ [ ] J p J q are the Riemann-Liouville fractional integrals D α D β D δ D σ are the Caputo fractional derivatives J = [ ] x y are real constants Then M Li and Y Liu [4] studied the following fractional differential problem: c D α +u t = f t u t v t c D p + u t c D q + v t = c D β + v t = f 2 t u t v t c D p + u t c D q + v t = u + λ u = u + λ c 2D p + u = v + λ v = v + λ c 2D q + v = 2 Manuscript received January 7 25; revised April 3 25 M A Abdellaoui and Z Dahmani corresponding author zzdahmani@yahoofr are with the Department LPAM Faculty SEI UMAB of Mostaganem Algeria N Bedjaoui is with the LAMFA laboratory of mathematics UPJV Rue St Leu Amiens 8 France c D α +u t = f t v t c D p + v t c D β + v t = g t u t c D q + u t = < t < au + u η = φ s v s ds u η 2 + bu = ψ s v s ds cv + v ξ = ϕ s u s ds v ξ 2 + dv = ρ s u s ds 3 where < α β < 2 < p q < α p β q η < η 2 φ ψ ϕ ρ L [ ] and f g C R 2 R c D γ + are in the sense of Caputo Very recently A Taieb and Z Dahmani [9] established some results on the existence and uniqueness of solutions for the problem: D α x t = m fi t x t x 2 t x n t t J D α2 x 2 t = m fi 2 t x t x 2 t x n t t J D αn x n t = m fi n t x t x 2 t x n t t J = 2 x 2 = 3 x 3 = n x n = x = = x = = x 2 = x = = = x 2 = x = 4 where < α < = 2 n J := [ ] and the derivatives D α are in the sense of Caputo In this paper we discuss the existence and uniqueness of solutions for the following coupled system of fractional Advance online publication: 4 November 25

2 IAENG International Journal of Applied Mathematics 45:4 IJAM_45_4_4 integro-differential equations: D α u t = f t u t v t t s α i Γα i ϕ i s g i s u s v s ds t J D β v t = f 2 t u t v t t s β i Γβ i φ i s h i s u s v s ds t J n 2 = u + v = u n = γi p u η η ] [ v n = δi q v ζ ζ ] [ 5 where D α D β denote the Caputo fractional derivatives with α β ]n n[ n N α i β i p q R+ and J = [ ] The functions f and f 2 as well as the general terms ϕ i φ i g i and h i ; i N will be specified later To the best of our nowledge the case of differential systems involving series no contribution exists So the main aim of this wor is to fill the gap in the relevant literature The rest of the paper is organized as follows: In section we present some preliminaries and lemmas Section 2 is devoted to existence of solutions of problem 5 In the last section some examples are presented to illustrate our results The following preliminaries will be used in the proof of our main results [2] [3] [5] Definition : A real function f t ; t > is said to be in space C µ µ R if there exists a real number p > µ such that f t = t p f t where f t C + and it is said to be in the space Cµ n if f n C µ n N Definition 2: The Riemann-Liouville fractional integral operator of order α for a function f C µ µ is defined as: I α f t = t Γα t τα f τ dτ α > t f t ifα = 6 Definition 3: The Caputo fractional derivative of order α > n < α n n N of a function f C n is defined as: D α f t = t Γn α t τn α f n τ dτ t > f n t ifα = n 7 The following lemma gives some properties of Riemann- Liouville fractional integral and Caputo derivative [2] [3] Lemma : Let f C Then for all t > we have: I r I s ft = I r+s ft; r > s > D s I s ft = ft; s > D r I s ft = I s r ft; s > r > To study the coupled system 5 we need the following two lemmas [2]: Lemma 2: Let n < α < n n N The general solution of D α x t = is given by x t = c + c t + c 2 t c n t n 8 where c j R j = 2 n Lemma 3: Let n < α < n n N Then for a given function f C n we have I α D α f t = f t + c + c t + c 2 t c n t n 9 for some c j R j = 2 n n = [α] + The following result gives an integral representation for the problem 5: Lemma 4: Suppose that x C n n < α < n α i If F G i Φ i C [ ] R such that Φ i < and G i is uniformly bounded then the integral solution of the problem D α t s αi x t = F t + Φ i s G i s ds Γ α i associated with the conditions n 2 x = x n = γi p x η = is given by: x t = t sα F s ds Γα γγp+nt n t s α i +α Γα i+α Φ i s G i s ds + ΓnΓp+n γη p+n η η where γ Γp+n η p+n η s α+p Γα+p η s α i +α+p Γα+α i+p Φ i s G i s ds Proof Applying Lemma 3 to we can write x t = t Γα t sα F s ds t s α i +α Γα i+α Φ i s G i s ds Then c c t c n t n F s ds 2 3 x =! c = n 4 Since x = x = = x n 2 = it follows then that c = c = c 2 = = c n 2 = Thans to Lemma it yields that I p x t = t t sα+p F s ds+ Γα+p t s α i +α+p Γα i+α+p Φ i s G i s ds c n Γnt p+n Γp+n 5 Advance online publication: 4 November 25

3 IAENG International Journal of Applied Mathematics 45:4 IJAM_45_4_4 Using we get γγp+n c n = ΓnΓp+n γη p+n I α+p F η I α+αi+p Φ i η G i η 6 Substituting c c c 2 c n in 3 we obtain 2 Lemma 4 is thus proved II MAIN RESULTS In this paragraph we introduce the following assumptions: H : There exist non negative real numbers m j m j n j n j j = 2 such that for all t [ ] and u v u 2 v 2 R 2 we have: f t u 2 v 2 f t u v m u 2 u +m 2 v 2 v f 2 t u 2 v 2 f 2 t u v n u 2 u +n 2 v 2 v g i t u 2 v 2 g i t u v m u 2 u +m 2 v 2 v g i t u 2 v 2 g i t u v m u 2 u +m 2 v 2 v where i N H2 : i : Suppose that ϕ i φ i C [ ] R and g i and h i are uniformly bounded ie there exist nonnegative real numbers L L 2 such that for all t [ ] and u v R 2 we have: g i t u v L h i t u v L 2 i = 2 3 ii : Assume that + ϕ i < + + φ i < + H3 : The functions f f 2 g i and h i : [ ] R 2 R; i N are continuous and there exist nonnegative real numbers L L 2 such that for all t [ ] and u v R 2 we have: f t u v L f 2 t u v L 2 We also need to introduce the quantities: + + ω := ω 2 := M := γγ p + n Γ n Γ p + n γη p+n δγ q + n Γ n Γ q + n δζ q+n Γ α + + ω Γ α + p + ϕ i Γ α + α i + + ω ϕ i Γ α + α i + p + M 2 := Γ β + + ω 2 Γ β + q + φ i Γ β + β i + + ω 2 ϕ i Γ β + β i + q + L = max { m j n j m j n } j j=2 Now we are ready to state and prove our main results We begin by the following theorem: Theorem 2: Suppose that γ Γp+n η δ Γq+n p+n ζ and q+n assume that H and H2 hold If L M + M 2 < 7 then the fractional system 5 has exactly one solution in X X Proof We apply the Banach contraction principle To this end we introduce the Banach space: X := C n [ ] R equipped with the norm X = So the product space X X u v X X is also a Banach space with norm u v X X = u X + v X We also define the operator Ψ : X X X X by: where Ψ u v t = Ψ u v t Ψ 2 u v t 8 Ψ u v t = +ω t n η +ω t n + and t s α Γα f s u s v s ds t s α+α i Γα+α i ϕ i s g i s u s v s ds η Ψ 2 u v t = +ω 2 t n + +ω 2 t n ζ ζ η s α+p Γα+p f s u s v s ds+ η s α+α i +p Γα+α i+p ϕ i s g i s u s v s ds 9 t s β Γβ f 2 s u s v s ds t s β+β i Γβ+β i φ i s h i s u s v s ds ζ s β+q Γβ+q f 2 s u s v s ds ζ s β+β i +q Γβ+β i+q φ i s h i s u s v s ds 2 We shall prove that Ψ is contractive: Let u v u 2 v 2 X X Then for each t [ ] we have Ψ u 2 v 2 t Ψ u v t t s α Γα ds + ω η η s α+p Γα+p ds sup s f s u 2 s v 2 s f s u s v s + sup s g i s u 2 s v 2 s g i s u s v s + sup s ϕ i s t s α+α i Γα+α i ds + ω η η s α+α i +p Γα+α i+p ds 2 For all t [ ] we can write Ψ u 2 v 2 t Ψ u v t Γα+ + ω Γα+p+ sup s f s u 2 s v 2 s f s u s v s + m ϕi Γα+α + ω ϕi i+ Γα+α i+p+ sup s g i s u 2 s v 2 s g i s u s v s 22 Using H yields the following inequality Advance online publication: 4 November 25

4 IAENG International Journal of Applied Mathematics 45:4 IJAM_45_4_4 Ψ u 2 v 2 t Ψ u v t Γα+ + ω Γα+p+ L sup t u 2 t u t + sup t v 2 t v t ϕi Γα+α + ω ϕi i+ Γα+α i+p+ L sup t u 2 t u t + sup t v 2 t v t 23 By H2 we obtain So that Ψ u 2 v 2 t Ψ u v t M L u 2 u X + v 2 v X Ψ u 2 v 2 Ψ u v X L M u 2 u v 2 v X X With the same arguments as before we have Ψ 2 u 2 v 2 Ψ 2 u v X LM 2 u 2 u v 2 v X X Using 25 and 26 we deduce that Ψ u 2 v 2 Ψ u v X X L M + M 2 u 2 u v 2 v X X Thans to 7 we conclude that Ψ is a contraction mapping Hence by Banach fixed point theorem there exists a unique fixed point which is a solution of 5 The second main result is given by the following theorem: Theorem 22: Assume that H2 and H3 are satisfied Then the system 5 has at least one solution in X X Proof We use Schaefer fixed point theorem to prove this result: First of all we show that the operator Ψ is completely continuous The continuity of Ψ on X X is trivial and hence it is omitted By the continuity of g i and h i imposed in H2 we can state that the series in 5 are uniformly convergent On the other thans to ii of H2 for any t [ ] we can write: t s α+αi ϕ i s g i s u s v s ds Γ α + α i X L ϕi X Γα+α i+ 28 The conditions imposed on α and α i allow us to obtain: L ϕ i X Γ α + α i + L ϕ i < + 29 With the same arguments we can write: L 2 φ i X Γ β + β i + L 2 φ i < + 3 Step : Let us tae r > C > and define B r := { u v X X u v X X r } ; C := L M + M 2 L := max { } L j L j B r and t [ ] we have: Ψ u v t t α Γα+ + ηα+p Γα+p+ ϕi t α+α i Γα+α i+ By 29 and H3 yields Similarly for Ψ 2 we have j=2 Then for u v sup t f t u t v t + ϕi ηα+αi+p Γα+α i+p+ sup t g i t u t v t 3 Ψ u v X LM < + 32 Ψ 2 u v X LM 2 < + 33 Thans to 32 and 33 we can write Hence Ψ B r B r Ψ u v X X C r 34 Step 2: Let t t 2 [ ] t < t 2 and u v B r We have: Ψ u v t 2 Ψ u v t t 2 s α Γα f s u s v s ds t s α Γα f s u s v s ds t 2 s α+α i Γα+α i ϕ i s g i s u s v s ds t s α+α i Γα+α i ϕ i s g i s u s v s ds + ω t n 2 t n η s α+p Γα+p f s u s v s ds η + ω t n 2 t n + η η s α+α i +p Γα+α i+p ϕ i s g i s u s v s ds 35 Using H2 and H3 we can write Ψ u v t 2 Ψ u v t Ltα 2 tα +t2 tα Γα+ L ϕi t α+α i 2 t α+α i +t 2 t α+α i Γα+α i+ + ω Ltn 2 t n Γα+p+ + ω + L ϕi t n 2 t n Γα+α i+p+ With the same arguments as before we have Ψ 2 u v t 2 Ψ 2 u v t L2tβ 2 tβ +t2 tβ Γβ+ L 2 φ i t β+β i 2 t β+β i +t 2 t β+β i + ω2 L2tn 2 t n Γβ+p+ Γβ+β i+ + ω 2 + L 2 φi t n 2 t n Γβ+β i+p Advance online publication: 4 November 25

5 IAENG International Journal of Applied Mathematics 45:4 IJAM_45_4_4 Thans to 36 and 37 we get Ψ u v t 2 Ψ u v t L t α 2 tα +t2 tα Γα+ + ω Ltn 2 t n Γα+p+ L ϕi t α+α i 2 t α+α i +t 2 t α+α i Γα+α i+ + ω + L ϕi t n 2 t n Γα+α i+p+ + L2tβ 2 tβ +t2 tβ Γβ+ + ω2 L2tn 2 t n Γβ+q+ L 2 ϕi t β+β i 2 t β+β i +t 2 t β+β i Γβ+β i+ + ω 2 + L 2 φi t n 2 t n Γβ+β i+q+ 38 As t 2 t the right-hand side of 38 tends to zero Then as a consequence of Steps 2 and by Arzela-Ascoli theorem we conclude that Ψ is completely continuous Next we show that the set Ω := {u v X X/ u v = λψ u v < λ < } 39 is bounded: Let u v Ω then u v = λψ u v for some < λ < Hence for t [ ] we have: Thus u t = λψ u v t v t = λψ 2 u v t 4 By 34 we obtain u v X X = λ Ψu v X X 4 u v X X λc 42 Consequently Ω is bounded As a conclusion of Schaefer fixed point theorem we deduce that Ψ has at least one fixed point which is a solution of 5 Corollary 2: Under the assumptions of Theorem 2 if α i = β i = then the fractional system 5 has exactly one solution in X X Corollary 22: Under the assumptions of Theorem 22 if α i = β i = then the system 5 has at least one solution in X X III ILLUSTRATIVE EXAMPLES Example 3: We begin by the following system D 2 u t = sinut+vt 64t+ + 2 t exp s sinus+vs 2 66!s 2 + ds t [ ] = D sin ut+sin vt 2 v t = 64t exp s 2 sin us+sin vs ds t [ ] = 2 66! s exps 2 + u = 2 πi 2 u 4 v = 2 πi 2 u 4 43 where α = β = 2 α = β = = 2 3 η = ζ = 4 p = q = 2 γ = δ = 2 π and f t u v = sinu+v 64t+ + 2 sin u+sin v f 2 t u v = 64t sin u+sin v 3 g t u v = 66!t 2 + sin u+sin v h t u v = 66! ϕ t expt 2 + t = exp t and 2 φ t = exp s2 For all u 2 v u 2 v 2 R 2 t [ ] we have f t u 2 v 2 f t u v 64 u 2 u + v 2 v f 2 t u 2 v 2 f 2 t u v 64 u 2 u + v 2 v g t u 2 v 2 g t u v 66 u 2 u + v 2 v h t u 2 v 2 h t u v 66 u 2 u + v 2 v So M = M 2 = 823 where + exp t = + 2 exp t2 = + 2 = = = π 2 6 and L = max m j m j = j=2 64 Hence L M + M 2 = 25 < 2 = The conditions of the Theorem 2 hold Therefore the problem 43 has a unique solution To illustrate the second main result we give the following example: Example 32: Let us tae D 7 4 u t = cosut+vt = t s 3 Γ4 π+t 2 + h D 2 5 v t = sinut+vt 5+t 2 = 3 j= t s 3 Γ4 exp s e s cosus vs ds t [ ] exp s 2 2 u j + v j = e s sinus vs ds t [ ] u 4 = 2I 2 u 3 v 4 = 5I 23 v 5 44 For this example we have α = 7 4 β = 2 5 α = β = 4; = 2 For t [ ] we have ϕ t = exp t φ 4 t = exp s2 2 and for each u v R 2 f t u v = cosu+v π+t 2 f 2 t u v = sinu+v 5+t 2 g t u v = e t cosu v 2 +2 h t u v = e t sinu v For all = 2 and t [ ] we have = = ϕ t = + = φ t = + = 4 < 2 < Therefore by Theorem 22 the problem 44 has at least one solution Advance online publication: 4 November 25

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