Existence of solutions of discrete fractional problem coupled to mixed fractional boundary conditions
Abstract
In this paper, we introduce a two-point nonlinear boundary value problem for a finite fractional difference equation. An associated Green’s function is constructed as a series of functions and some of its properties are obtained. Some existence results are deduced from fixed point theory and lower and upper solutions
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Rev. Real Acad. Cienc. Exactas Fis. Nat. Ser. A-Mat. (2022) 116: 175 https://doi.org/10.1007/s13398-022-01321-6 ORIGINAL PAPER Existence of solutions of discrete fractional problem coupled to mixed fractional boundary conditions Rim Bourguiba1·Alberto Cabada2,3 ·Wanassi Om Kalthoum1 Received: 8 January 2022 / Accepted: 26 August 2022 / Published online: 6 September 2022 © The Author(s) 2022 Abstract In this paper, we introduce a two-point nonlinear boundary value problem for a finite fractional difference equation. An associated Green’s function is constructed as a series of functions and some of its properties are obtained. Some existence results are deduced from fixed point theory and lower and upper solutions. Keywords Fractional equations ·Green’s functions ·Nonlinear boundary value problems Mathematics Subject Classification 34A08 ·34B27 ·34K37 1 Introduction In this paper, we consider the following discrete fractional problem coupled to mixed fractional boundary conditions, −αy(t)+a(t+α−1)y(t+α−1)=f(t+α−1,y(t+α−1)), t∈NT+1 0, y(α −2)=βy(α +T+1−β) =0,(1) where αis the standard Riemman–Liouville type discrete fractional difference operator, t∈NT+1 0,1<α≤2,0≤β≤1, Tis a positive integer and function f:Nα+T α−1×R−→ R is continuous. Here we denote NK+r r={r,...,K+r}for any r∈Rand Ka positive integer. So, we will look for solutions y:NT+α+1 α−2−→ R. BAlberto Cabada [email protected] Rim Bourguiba [email protected] Wanassi Om Kalthoum [email protected] 1Department of Mathematics, University of Monastir, Monastir, Tunisia 2CITMAga, 15782 Santiago de Compostela, Galicia, Spain 3Departamento de Estatística, Análise Matemática e Optimización , Universidade de Santiago de Compostela, Santiago de Compostela, Santiago, Spain 123
175 Page 2 of 15 R. Bourguiba et al. The continuous fractional calculus has a long history, but the discrete fractional calculus has been investigated recently. The development of this theory starts with the paper of Díaz and Osler [5] where is defined a fractional difference as an infinite series and a generalization of the binomial formula for the n-th order difference nfoperator. This study continues with the work of Miller and Ross [10] which deals with the linear υ-th order fractional differential equation as an analogue of the linear n-th order ordinary differential equation. Since them, a great progress has been made in the study of boundary value problems for fractional difference equations (see [1,2,6–8] and references therein). We recall some classical definitions from discrete fractional calculus theory and preliminary results. Definition 1 We define t(ν) =(t+1)/ (t+1−ν),foranytand νfor which the right hand side is well defined. We also appeal to the convention that if t+1−νis a pole of the Gamma function and t+1 is not a pole, then t(ν) =0. Definition 2 The ν-th fractional sum of a function f, for ν>0andt∈Nr+ν,isdefinedas −νf(t)(≡−νf(t,r)) =1 (ν) t−ν s=r (t−s−1)(ν−1)f(s). We define the ν-th fractional difference for ν>0,by νf(t):= Nν−Nf(t), where t∈Nr+νand N∈Nis chosen so that 0 ≤N−1≤ν≤N. Lemma 3 [9]Let 0≤N−1≤ν≤N. Then −ννy(t)=y(t)+c1t(ν−1)+c2t(ν−2)+ ···+cNt(ν−N)for some ci∈R, with 1≤i≤N. The paper is scheduled as follows: In next section we deduce some properties of the Green’s function related to the linear problem −αy(t)+a(t+α−1)y(t+α−1)=h(t+α−1), t∈NT+1 0, y(α −2)=βy(α +T+1−β) =0,(2) for the particular case of a(t+α−1)=0forallt∈NT+1 0. The case 0 <α−β<1 has been treated in [9], we will continue this study, by improving some of the obtained results in [9] and by considering the case 1 ≤α−β≤2. Section 3 is devoted to deduce the expression of the Green’s function related to the problem (2)fora nontrivial function a(t), with small enough bounded absolute value. Moreover, some a priori bounds of the Green’s function are obtained. The arguments are in the line of the ones given in the paper [3]. In last section we show the applicability of the given results by obtaining some existence and uniqueness results of the nonlinear problem (1). 2 Properties of function G0 In this section we will extend previous results related to the Green’s function G0of problem (2) with aidentically zero. Some of these results have been given in [9] for the case 0 <α−β<1, we improve some of them and study also the case 1 ≤α−β≤2. Lemma 4 [9]The unique solution of the linear fractional mixed boundary value problem −αy(t)=h(t+α−1), t∈NT+1 0, y(α −2)=βy(α +T+1−β) =0, 123
Existence of solutions of discrete fractional... Page 3 of 15 175 has the form y(t)= T+1 s=0 G0(t,s)h(s+α−1), where G0:NT+α+1 α−2×NT+1 0−→ Ris given by the expression G0(t,s)=1 (α) ⎧ ⎪ ⎪ ⎨ ⎪ ⎪ ⎩ t(α−1)(α +T−β−s)(α−β−1) (α +T−β+1)(α−β−1)−(t−s−1)(α−1),0≤s<t−α+1≤T+2, t(α−1)(α +T−β−s)(α−β−1) (α +T−β+1)(α−β−1),0≤t−α+1≤s≤T+2. (3) In the following Lemma, we will show the properties of the Green’s function, which extends the one given in [9, Lemma 3.3]. Lemma 5 The Green’s function G0(t,s)given by (3)has the following properties: (i) Suppose that 0<α−β≤2,thenG 0(t,s)>0for t ∈Nα+T+1 α−1and s ∈NT+1 0. (ii) Suppose that 0<α−β≤2,then max t∈Nα+T+1 α−1 G0(t,s)=G0(s+α−1,s), for each s ∈NT+1 1. Proof The case 0 <α−β<1 has been proved in [9, Lemma 3.3 (i)], so, we only need to prove the case 1 ≤α−β≤2. (i) For 0 ≤t−α+1≤s≤T+1, it is clear that (α)G0(t,s)=(T+3)t(α−1) (α +T−β+2)(α +T−β−s)(α−β−1) =(T+3)(t+1)(α +T−β−s+1) (α +T−β+2)(t−α+2)(T−s+2) >0. Now, for 0 ≤s<t−α+1≤T+1, we have (α)G0(t,s)=(t−s−1)(α−1)t(α−1)(α +T−β−s)(α−β−1) (t−s−1)(α−1)(α +T−β+1)(α−β−1)−1.(4) Let F(t,s,α,β) := t(α−1)(α +T−β−s)(α−β−1) (t−s−1)(α−1)(α +T−β+1)(α−β−1) =t(α−1)(T+3) (t−s−1)(α−1)(T−s+2)(α +T−β−s)(−s−1). Since ∂ ∂β F(t,s,α,β) =(t+1)(T+3)(−s+t−α+1)(−s+T+α−β+1) (t−s)(−s+T+2)(t−α+2)(T+α−β+2) ×(ψ(0)(T+α−β+2)−ψ(0)(−s+T+α−β+1)) > 0, 123
175 Page 4 of 15 R. Bourguiba et al. where by ψ(0)it is denoted the PolyGamma special function [11]. This implies that F(t,s,α,β)> F(t,s,α,0) =t(α−1)(α +T−s)(α−1) (t−s−1)(α−1)(α +T+1)(α−1) =t(t−1)···(t−s)(T+2)(T+1)···(T−s+2) (t−α+1)(t−α)···(t−α+1−s)(α +T+1)(α +T)···(α +T+1−s) = s i=0 P(t,α,i), where, for every i∈{0,...,s}, we denote P(t,α,i):= (t−i)(T+2−i) (t−α+1−i)(α +T+1−i). Since ∂P ∂i(t,α,i)=(α −1)(2i−t−T−2)(−α+t−T−1) (α +i−t−1)2(α −i+T+1)2>0,0≤i≤s, we have that P(t,α,i)>P(t,α,0)=t(T+2) (−α+t+1)(α+T+1). Now, using that ∂P ∂t(t,α,0)=(1−α)(T+2) (−α+t+1)2(α+T+1)<0,we deduce that P(t,α,i)> P(T+1+α, α, 0)=1,for all i∈{1,...,s}and, from (4), the positiveness of the Green’s function is shown. (ii) The case 0 <α−β<1 has been proved in [9, Lemma 3.3 (ii)]. Now, let 1 ≤α−β≤2. If 0 ≤t−α+1≤s≤T+1. Then tG0(t,s)=(α −1)(T+3)(t+1)(α +T−β−s+1) (α)(α +T−β+2)(t−α+3)(T−s+2)>0. Hence, we deduce that tG0(t,s)>0for0≤t−α+1≤s≤T+1. And so, G0(t,s)is increasing with respect to tfor 0 ≤t−α+1≤s≤T+1, which implies that G0(t,s)≤G0(s+α−1,s). If 0 ≤s<t−α+1≤T+1. Then (α)tG0(t,s)=(α −1)t(α−2)(α +T−β−s)(α−β−1) (α +T−β+1)(α−β−1)−(α −1)(t−s−1)(α−2) =(α −1)(t−s−1)(α−2) ×t(α−2)(α +T−β−s)(α−β−1) (α +T−β+1)(α−β−1)(t−s−1)(α−2)−1. Thus, to show that (α)tG0(t,s)<0, we must prove that H(t,s,α,β):= t(α−2)(α +T−β−s)(α−β−1) (α +T−β+1)(α−β−1)(t−s−1)(α−2)<1.(5) 123
Existence of solutions of discrete fractional... Page 5 of 15 175 Now, since tH(t,s,α,β) =(2−α)(s+1)(t+1)(T+3)(−s+t−α+2)(−s+T+α−β+1) (−s+t+1)(−s+T+2)(t−α+4)(T+α−β+2)>0, we have that H(t,s,α,β) ≤H(T+1+α, s,α,β) =(−s+T+2)(T+α+2)(−s+T+α−β+1) (T+3)(−s+T+α+1)(T+α−β+2). So, by using that sH(T+1+α, s,α,β) =(−α+2β−βs+s+(β −1)T)(T+α+2)(−s+T+α−β) (T+3)(−s+T+α+1)(T+α−β+2), since −α+2β−βs+s+(β −1)T≤0foralls∈{0,...,T}if and only if α−β≥β, which is true whenever α−β≥1, we have that sH(T+1+α, s,α,β)≤0foralls∈{0,...,T}. As a direct consequence, we deduce that H(T+1+α, s,α,β) ≤H(T+1+α, 0,α,β) =(T+2)(α +T+1) (T+3)(α −β+T+1)for all s∈NT+1 0. Now, from the fact that ∂ ∂β H(T+1+α, 0,α,β)=(T+2)(α +T+1) (T+3)(α −β+T+1)2>0, we conclude that H(t,s,α,β)≤H(T+1+α, 0,α,α−1)=α+T+1 T+3<1, for all t∈{α−1,...,T+1+α},0≤t−α+1≤s≤T+1and0<β≤α−1<1, which implies, from (5), that G0(t,s)≤G0(s+α−1,s). As a result, we get that G0(t,s)is increasing with respect to tfor 0 ≤t−α+1≤s≤T+1 and decreasing with respect to tfor 0 ≤s<t−α+1≤T+1. Hence, it follows that max t∈Nα+T+1 α−1 G0(t,s)=G0(s+α−1,s) and the proof is concluded. In the following, we obtain a bound from above of the Green’s function. Theorem 6 Let G0(t,s)be the Green’s function defined by (3). Then, for any (t,s)∈ Nα+T+1 α−1×NT+1 0, the following inequalities are fulfilled: 123
175 Page 6 of 15 R. Bourguiba et al. (i) If 0<α−β≤1,then G0(t,s)≤C1(T,α,β):= (T+2)(α −β)(T+α+1) (α)(T+α−β+2). (ii) If α−β>1,then G0(t,s)≤C2(T,α,β) := (T+3) α+−αT+T−β −2α+β+2α−β+−αT+βT+T+β −2α+β+2+1 (α)(T+α−β+2) −(T+2)(α−1) −2α+β+2−αT+βT+T+β −2α+β+2+2. Proof From previous result we know that 0<G0(t,s)≤G0(s+α−1,s)=(T+3)(s+α)(−s+T+α−β+1) (α)(s+1)(−s+T+2)(T+α−β+2). So, the bounds come from the ones of the previous inequality. Since, sG0(s+α−1,s) =(T+3)(s+α)(β +s(−2α+β+2)+(α −1)T)(−s+T+α−β) (α)(s+2)(−s+T+2)(T+α−β+2),(6) we have that sG0(s+α−1,s)|s=0=(T+2)(β +(α −1)T) (α −β+T)(α −β+T+1)>0. Moreover, it is clear that (6) vanishes if and only if s=s0(α, β) := β+(α −1)T 2α−β−2. It is immediate to verify that s0(α, β) ∈{0,...,T+1}if and only if α−β>1. (i) Consider the first case: 0 <α−β≤1. In this situation we have that G0(s+α−1,s)is increasing in s∈NT+1 0. Thus G0(s+α−1,s)≤G0(T+α, T+1)=C1(T,α,β). (ii) Assume now that 1 <α−β. Now the maximum is attained at s0(α, β), in consequence: G0(s+α−1,s)≤G0(s0(α, β) +α−1,s0(α, β)) =C2(T,α,β) and the proof is concluded. 3 Green’s function of problem (2) In this section we deduce the existence of the Green’s function Grelated to Problem (2). The expression of the function Gis given as a functional series and we prove its convergence for suitable values of the function a(t). The arguments are in the line to the ones given in [3]. 123
Existence of solutions of discrete fractional... Page 7 of 15 175 Let G:NT+α+1 α−2×NT+1 0−→ Rby G(t,s)=∞ n=0 (−1)nGn(t,s)(7) where G0(t,s)is given by (3)and Gn:NT+α+1 α−2×NT+1 0−→ R is defined as Gn(t,s)= T+1 τ=0 a(τ +α−1)G0(t,τ)Gn−1(τ +α−1,s), n≥1.(8) In order to express the Green’s function associated with the linear problem (2) we shall use the spectral theory in Banach space given by the following Lemma: Lemma 7 [12, Theorem 1.B] Let X be a Banach space and A:X−→ X be a linear operator with the operator norm A.ThenifA<1,then(I−A)−1exists and (I−A)−1= ∞ n=0An,whereIstands for the identity operator. Let X:= {y:NT+α+1 α−2−→ R:y(α −2)=βy(α +T+1−β) =0} be the Banach space with norm y=max{|y(t)|; t∈NT+α+1 α−2}. Consider the following assumption: (H)There exists a>0 such that (i) If 0 <α−β≤1, then |a(t+α−1)|≤a<1 (T+2)C1(T,α,β) for all t∈NT+1 0. (ii) If α−β>1, then |a(t+α−1)|≤a<1 (T+2)C2(T,α,β) for all t∈NT+1 0, where C1(T,α,β)and C2(T,α,β)are introduced in Theorem 6. In next result we prove the existence and uniqueness of solution of problem (2) by means of the construction of its related Green’s function. Theorem 8 If condition (H)is fulfilled, then the function G(t,s)defined in (7)as a series of functions is convergent for all (t,s)∈NT+α+1 α−2×NT+1 0. Moreover, G(t,s)is the Green’s function related to problem (2). Proof The solution yof (2) satisfies, for all t∈NT+α+1 α−2, the following equality: y(t)= T+1 s=0 G0(t,s)h(s+α−1)−a(s+α−1)y(s+α−1), where G0is defined by (3). 123
175 Page 8 of 15 R. Bourguiba et al. This expression can be reformulated as y(t)+ T+1 s=0 G0(t,s)a(s+α−1)y(s+α−1)= T+1 s=0 G0(t,s)h(s+α−1). (9) Now, denote operators Aand Bby Ah(t)= T+1 s=0 G0(t,s)h(s+α−1)(10) By(t)= T+1 s=0 G0(t,s)a(s+α−1)y(s+α−1). (11) Then, Eq. (9) becomes (I+B)y=Ah. First, let us show that B<1. For any y∈Xwith y=1andt∈NT+α+1 α−2,by(11), we have (i) If 0 <α−β≤1, then, using Theorem 6(i) and condition (H),wehavethatforall t∈NT+α+1 α−2the following inequalities are fulfilled: |By(t)|≤ T+1 s=0 aG0(t,s)≤ T+1 s=0 aC 1(T,α,β)=(T+2)aC 1(T,α,β)<1. (ii) If α−β>1, then using Theorem 6(ii) and condition (H),weget |By(t)|≤ T+1 s=0 aG0(t,s)≤ T+1 s=0 aC 2(T,α,β)=(T+2)aC 2(T,α,β)<1. Therefore, by Lemma 7, we deduce that y=I+B−1Ah=∞ n=0 (−B)nAh.(12) By using analogous arguments to [3], we show that for n∈N0={0,1,2,...}the following identity holds: ((−B)nAh)(t)= T+1 s=0 (−1)nGn(t,s)h(s+α−1). (13) In the sequel we obtain the following a priori bounds for function Gnfor all n∈N0: (i) If 0 <α−β≤1, then |(−1)nGn(t,s)|≤an((T+2)C1(T,α,β) )n+1.(14) (ii) If α−β>1, then |(−1)nGn(t,s)|≤an((T+2)C2(T,α,β) )n+1,(15) where ais defined in (H). 123
Existence of solutions of discrete fractional... Page 9 of 15 175 In the first case, for n=0, (14) holds. Assume (14) holds for n=m≥0. Then, for any (t,s)∈NT+α+1 α−2×NT+1 0,wehave |(−1)m+1Gm+1(t,s)|≤ T+1 τ=0|a(τ +α−1)|G0(t,τ)Gm(τ +α−1,s) ≤ T+1 τ=0 aC 1(T,α,β)Gm(τ +α−1,s) ≤ T+1 τ=0 am+1(T+2)m+1(C1(T,α,β) )m+2 =am+1(T+2)m+2(C1(T,α,β) )m+2, hence (14) holds for n=m+1. By induction, (14)iffulfilledforanyn∈N0. Arguing as at the first case, we conclude that (15)istrueforalln∈N0. Finally, by (H)and (7), for all (t,s)∈NT+α+1 α−2×NT+1 0, we deduce the following inequalities: (i) If 0 <α−β≤1, then |G(t,s)|= ∞ n=0 (−1)nGn(t,s)≤∞ n=0 an((T+2)C1(t,α,β) )n+1<∞. (ii) If α−β>1, then |G(t,s)|= ∞ n=0 (−1)nGn(t,s)≤∞ n=0 an((T+2)C2(t,α,β) )n+1<∞. Therefore, G(t,s)is uniformly convergent on NT+α+1 α−2×NT+1 0. Moreover, from (7), (12)and(13), we obtain y(t)=∞ n=0 T+1 s=0 (−1)nGn(t,s)h(s+α−1)= T+1 s=0 G(t,s)h(s+α−1). (16) On the other hand, let ybe defined by (16). Using (7), (10)and(11) together (8)and(13), we conclude that ysatisfies (12). Then, yis the unique solution of problem (2)andGis the Green’s function related to problem (2). As a direct consequence of Theorem 8and Eqs. (7)and(8), we deduce the following result Corollary 9 If a(t)∈[−¯a,0]for all t ∈Nα+T+1 α−1,thenG(t,s)>0for all t ∈Nα+T+1 α−1and s∈NT+1 0. In next result, we obtain an upper bound for function |G(t,s)|for any sfixed. Lemma 10 We introduce the following function G:NT+1 0→(0,∞): (i) For 0<α−β≤1and for all (t,s)∈Nα+T+1 α−2×NT+1 0 G(s):= G0(s+α−1,s)T+2 1−a(T+2)C1(t,α,β).(17) 123