Non-Trivial Solutions of Non-Autonomous Nabla Fractional Difference Boundary Value Problems

symmetry S S Article Non-Trivial Solutions of Non-Autonomous Nabla Fractional Difference Boundary Value Problems Alberto Cabada 1,*,† , Nikolay D. Dimitrov 2,† and Jagan Mohan Jonnalagadda 3,†   Citation: Cabada, A.; Dimitrov, N.D.; Jonnalagadda, J.M. Non-Trivial Solutions of Non-Autonomous Nabla Fractional Difference Boundary Value Problems. Symmetry 2021,13, 1101. https://doi.org/10.3390/sym13061101 Academic Editor: Juan Luis García Guirao Received: 5 May 2021 Accepted: 16 June 2021 Published: 21 June 2021 Publisher’s Note: MDPI stays neutral with regard to jurisdictional claims in published maps and institutional affiliations. Copyright: © 2021 by the authors. Licensee MDPI, Basel, Switzerland. This article is an open access article distributed under the terms and conditions of the Creative Commons Attribution (CC BY) license (https:// creativecommons.org/licenses/by/ 4.0/). 1Departamento de Estatística, Análise Matemática e Optimización, Instituto de Matemáticas, Facultade de Matemáticas, Universidade de Santiago de Compostela, 15782 Santiago de Compostela, Spain 2Department of Mathematics, University of Ruse, 7017 Ruse, Bulgaria; [email protected] 3Department of Mathematics, Birla Institute of Technology and Science Pilani, Hyderabad 500078, India; [email protected] or [email protected] *Correspondence: [email protected] or [email protected] † These authors contributed equally to this work. Abstract: In this article, we present a two-point boundary value problem with separated boundary conditions for a finite nabla fractional difference equation. First, we construct an associated Green’s function as a series of functions with the help of spectral theory, and obtain some of its properties. Under suitable conditions on the nonlinear part of the nabla fractional difference equation, we deduce two existence results of the considered nonlinear problem by means of two Leray– Schauder fixed point theorems. We provide a couple of examples to illustrate the applicability of the established results. Keywords: nabla fractional difference; boundary value problem; separated boundary conditions; Green’s function; existence of solutions 1. Introduction Denote the set of all real numbers and positive real numbers by R and R+ , respectively. Define by Na={a , a+ 1, a+ 2, . . .} and Nb a={a , a+ 1, a+ 2, . . . , b} for any a , b∈R such that b−a∈N1. In this article, we consider the following nabla fractional difference equation associated with separated boundary conditions: −∇ν−1 a∇u(t) + g(t)u(t) = f(t,u(t)),t∈Nb a+2, αu(a+1)−β∇u(a+1) = 0, γu(b) + δ∇u(b) = 0. (1) Here a , b∈R with b−a∈N1 ; 1 <ν< 2; g:Nb a→R ; f:Nb a×R→R ; ∇ν−1 a denotes the (ν− 1 ) -th order Riemann–Liouville backward (nabla) difference operator; ∇ denotes the first order backward (nabla) difference operator; α , β , γ , δ∈R such that α2+β2> 0 and γ2+δ2>0. Gray and Zhang [ 1 ], Atici and Eloe [ 2 ] and Anastassiou [ 3 ] initiated the study of nabla fractional sums and differences. The combined efforts of a number of researchers has resulted in a fairly strong foundation to the basic theory of nabla fractional calculus during the past decade. For a detailed discussion on the evolution of nabla fractional calculus, we refer to the recent monograph [4] and the references therein. We point out that problem (1) is a discrete version of the second order ordinary differential Hill’s equation, which has a lot of applications in engineering and physics. We can find, among others, several problems in astronomy, circuits, electric conductivity of metals and cyclotrons. Hill’s equation is named after the pioneering work of the mathematical astronomer George William Hill (1838–1914), see [ 5 ]. There is a long literature Symmetry 2021,13, 1101. https://doi.org/10.3390/sym13061101 https://www.mdpi.com/journal/symmetry Symmetry 2021,13, 1101 2 of 15 in the study of the oscillation of the solutions of such an equation and the constant sign solutions. The reader can consult the monographs [ 6 , 7 ] and references therein. We note that the boundary conditions cover the Sturm–Liouville conditions, which include, as particular cases, the Dirichlet, Neumann and Mixed ones. Recently, there has been a surge of interest in the development of the theory of nabla fractional boundary value problems. Brackins [ 8 ] initiated the study of boundary value problems for linear and nonlinear nabla fractional difference equations. Following this work, several authors have studied nabla fractional boundary value problems extensively. We refer to [9–18] and the references therein to name a few. Brackins [8] showed that for all (t,s)∈Nb a×Nb a+1(see Figure 1) G0(t,s) = (v1(t,s),t∈Nρ(s) a, v2(t,s),t∈Nb s (2) is the Green’s function related to the following boundary value problem:        −∇ν−1 a∇u(t) = 0, t∈Nb a+2, αu(a+1)−β∇u(a+1) = 0, γu(b) + δ∇u(b) = 0. (3) Here, v1(t,s) = 1 ξhαγHν−1(t,a)Hν−1(b,ρ(s)) + αδHν−1(t,a)Hν−2(b,ρ(s)) + (β−α)γHν−1(b,ρ(s)) + (β−α)δHν−2(b,ρ(s))i, v2(t,s) = v1(t,s)−Hν−1(t,ρ(s)), ξ= (β−α)γ+αγHν−1(b,a) + αδHν−2(b,a)6=0. 10 15 20 25 30 35 40 0.5 1.0 1.5 2.0 2.5 3.0 Figure 1. Graphic of G0(t ,20 ) for α=β=γ= 1, δ= 0 (Dirichlet case), µ= 3 / 2, a= 5 and b= 40. This result was obtained by expressing the general solution of the nabla fractional difference equation in (3) , using the method of variation of constants. Notice that, for a non-constant function g the expression of the general solution does not exist and, as a consequence, the method used in [ 8 ] is not applicable for the following boundary value problem: Symmetry 2021,13, 1101 3 of 15        −∇ν−1 a∇u(t) + g(t)u(t) = 0, t∈Nb a+2, αu(a+1)−β∇u(a+1) = 0, γu(b) + δ∇u(b) = 0. (4) Due to this reason, Graef et al. [ 19 ] and Cabada et al. [ 20 ] followed a different approach. Graef et al. [19] studied the following Dirichlet problem: (−Dµ 0u(t) + g(t)u(t) = w(t)f(t,u(t)), 0 <t<1, u(0) = u(1) = 0, where 1 <µ< 2; g:[ 0,1 ]→R , w:[ 0,1 ]→R+∪ { 0 } , f:[ 0,1 ]×R→R are continuous functions, and Dµ 0 denotes the µth -th order Riemann–Liouville fractional derivative. Cabada et al. [20] studied the following Dirichlet problem: −∆µu(t) + g(t+µ−1)u(t+µ−1) = w(t)f(t+µ−1, u(t+µ−1)), u(µ−2) = u(µ+b+1) = 0, where t∈Nb+1 0 , b∈N5 ; 1 <µ< 2; g , w:Nb+1 0→R with w6≡ 0 on Nb+1 0 ; f:Nµ+b µ−1×R→ R is a continuous function, and ∆µ denotes the µ -th order Riemann–Liouville forward (delta) difference operator. Similar to these works, we obtained the Green’s function related to (4) as a series of functions by using the spectral theory. Then, under suitable conditions on g , w and f , we proved the existence of at least one solution of the boundary value problem (1) . This work provides a new approach for constructing Green’s functions for nabla fractional boundary value problems. This article is organized as follows: In Section 2, we recall some definitions and preliminary results. In Section 3, we obtain the Green’s function related to (4) , and deduce some of its important properties. In Section 4, we establish a couple of existence results for the boundary value problem (1) , using two different Leray–Schauder fixed point theorems and under different assumptions on the data of the problem. Finally, we give some examples to demonstrate the applicability of these results. 2. Preliminaries In this section, we recall some elementary definitions and fundamental facts of nabla fractional calculus, which will be used throughout the article. Denote by Na={a , a+ 1, a+ 2, . . .} and Nb a={a , a+ 1, a+ 2, . . . , b} for any a , b∈R such that b−a∈N1 . The backward jump operator ρ:Na+1→Nais defined by ρ(t) = max {a,t−1},t∈Na. The Euler gamma function is defined by Γ(z) = Z∞ 0tz−1e−tdt,<(z)>0. Using its reduction formula, the Euler gamma function can also be extended to the halfplane <(z)≤ 0 except for z∈ {. . . , − 2, − 1,0 } . For t∈R\ {. . . , − 2, − 1,0 } and r∈R such that (t+r)∈R\ {. . . , − 2, − 1,0 } , the generalized rising function is defined by the following: tr=Γ(t+r) Γ(t). Symmetry 2021,13, 1101 4 of 15 If t∈ {. . . , − 2, − 1,0 } and r∈R such that (t+r)∈R\ {. . . , − 2, − 1,0 } , then we find that tr=0. Let µ∈R\ {. . . , − 2, − 1 } , define the µ -th order nabla fractional Taylor monomial by the following: Hµ(t,a) = (t−a)µ Γ(µ+1), provided that the right-hand side exists. Observe that Hµ(a , a) = 0 and Hµ(t , a) = 0 for all µ∈ {. . . , −2, −1}and t∈Na. Let u:Na→R and N∈N1 . The first order backward (nabla) difference of u is defined by the following: ∇u(t) = u(t)−u(t−1),t∈Na+1, and the N-th order nabla difference of uis defined recursively by ∇Nu(t) = ∇∇N−1u(t),t∈Na+N. Let u:Na+1→R and N∈N1 . The N -th order nabla sum of u based at a is given by the following: ∇−N au(t) = t ∑ s=a+1 HN−1(t,ρ(s))u(s),t∈Na, where, by convention, ∇−N au(a) = 0. We define ∇−0 au(t) = u(t)for all t∈Na+1. Definition 1. Let u:Na+1→R and ν> 0. The ν -th order nabla sum of u based at a is given by the following [4]: ∇−ν au(t) = t ∑ s=a+1 Hν−1(t,ρ(s))u(s),t∈Na, where, by convention, ∇−ν au(a) = 0. Definition 2. Let u:Na+1→R , ν> 0and choose N∈N1 such that N− 1 <ν≤N . The ν -th order Riemann–Liouville nabla difference of u is given by the following [4]: ∇ν au(t) = ∇N∇−(N−ν) au(t),t∈Na+N. In [ 21 , 22 ], Jonnalagadda obtained the following properties of the Green’s function G0(t , s) . Theorem 1. Assume that the following condition holds [22]: (A0) α,β,γ,δ≥0,α2+β2>0,γ2+δ2>0and β≥α. Then, 1. G0(t,s)≥0for all (t,s)∈Nb a×Nb a+1; 2. max t∈Nb a G0(t,s) = G0(ρ(s),s)for all s ∈Nb a+1; 3. G0(ρ(s),s)<Λ, where Λ=1 ξhαγHν−1(b,a)Hν−1(b,a) + αδHν−1(b,a) + (β−α)γHν−1(b,a) + (β−α)δi. Symmetry 2021,13, 1101 5 of 15 Theorem 2. Assume that the condition (A0) holds [21]. Then, b ∑ s=a+1 G0(t,s)≤Ω, for all (t,s)∈Nb a×Nb a+1, where Ω=1 ξhαγH2ν−1(b,a+1) + αδH2ν−2(b,a+1) + (β−α)γHν(b,a) + (β−α)δHν−1(b,a)i. We mention the following classical result that will be used in the next section. Lemma 1. Let X be a Banach space, A:X→X be a linear operator with the operator norm kAk[23] (page 795). Then, if kAk<1, we have that (I−A)−1exists and (I−A)−1= ∞ ∑ n=0 An. Here, I is the identity operator. 3. Green’s Function and Its Properties In this section, we construct the Green’s function related to problem (4) , and deduce some significant properties. We denote by X the set of all maps from Nb a into R . Clearly, X is a Banach space endowed with the maximum norm k · k . We assume the following condition throughout the paper. (A1) There exists ¯ g>0 such that |g(t)|≤¯ g<1 Ω,t∈Nb a. We define G:Nb a×Nb a+1→Rby the following: G(t,s) = ∞ ∑ n=0 (−1)nGn(t,s), (5) where G0(t,s)is given by (2), and set (see Figures 2–4). Gn(t,s) = b ∑ τ=a+1 G0(t,τ)Gn−1(τ,s)g(τ),n∈N1. (6) Then, we have the following result. Theorem 3. Assume that conditions (A 0 ) and (A 1 ) are fulfilled, then function G(t , s) , defined in (5) as a series of functions, is convergent for (t , s)∈Nb a×Nb a+1 . Moreover, G(t , s) is the Green’s function for the boundary value problem (4). Symmetry 2021,13, 1101 6 of 15 10 15 20 25 30 35 40 0.2 0.4 0.6 0.8 Figure 2. Graphic of G1(t ,20 ) for α=β=γ= 1, δ= 0 (Dirichlet case), µ= 3 / 2, a= 5, b= 40 and g≡1/100. 10 15 20 25 30 35 40 0.05 0.10 0.15 0.20 0.25 0.30 0.35 Figure 3. Graphic of G2(t ,20 ) for α=β=γ= 1, δ= 0 (Dirichlet case), µ= 3 / 2, a= 5, b= 40 and g≡1/100. 10 15 20 25 30 35 40 0.5 1.0 1.5 2.0 2.5 Figure 4. Graphic of the first three iterates of G(t ,20 ) for α=β=γ= 1, δ= 0 (Dirichlet case), µ=3/2, a=5, b=40 and g≡1/100. Proof. For any h∈Xand t∈Nb a, consider the following linear boundary value problem: −∇ν−1 a∇u(t) + g(t)u(t) = h(t),t∈Nb a+2, αu(a+1)−β∇u(a+1) = 0, γu(b) + δ∇u(b) = 0. (7) Symmetry 2021,13, 1101 7 of 15 By definition of the Green’s function G0 , the solutions u of this problem satisfy the following identity: u(t) = b ∑ s=a+1 G0(t,s)[h(s)−g(s)u(s)], which is the same to u(t) + b ∑ s=a+1 G0(t,s)g(s)u(s) = b ∑ s=a+1 G0(t,s)h(s). (8) Now, define the operators T1:X→Xand T2:X→Xby the following: (T1h)(t) = b ∑ s=a+1 G0(t,s)h(s),t∈Nb a, (T2u)(t) = b ∑ s=a+1 G0(t,s)g(s)u(s),t∈Nb a. Then, (8) can be expressed as the following: (I+T2)u=T1h. Using condition (A1) and Theorem 1the following is true: kT2k=max kuk=1kT2uk=max kuk=1"max t∈Nb a |(T2u)(t)|# =max kuk=1"max t∈Nb a b ∑ s=a+1 G0(t,s)g(s)u(s)# ≤max kuk=1"max t∈Nb a b ∑ s=a+1 G0(t,s)|g(s)||u(s)|# ≤max kuk=1"¯ gkukmax t∈Nb a b ∑ s=a+1 G0(t,s)# <max kuk=1[¯ gkukΩ]=¯ gΩ<1. Then, by Lemma 1, we have the following: u= (I+T2)−1T1h= ∞ ∑ n=0 (−T2)nT1h. (9) Arguing in a similar manner than in [20], we can deduce the following: (−T2)nT1h(t) = b ∑ s=a+1 (−1)nGn(t,s)h(s),t∈Nb a,n=0, 1,2, . . . (10) Let us see now that the following inequality is fulfilled: |(−1)nGn(t,s)|<Λ(¯ gΩ)n,n=0, 1,2, . . . (11) From Theorem 1, we have that (11) holds for n= 0. Assume now that (11) is true for some n=k. Then, the following is true: Symmetry 2021,13, 1101 8 of 15 (−1)k+1Gk+1(t,s)= (−1)k+1b ∑ τ=a+1 G0(t,τ)Gk(τ,s)g(τ) = − b ∑ τ=a+1 G0(t,τ)(−1)kGk(τ,s)g(τ) ≤ b ∑ τ=a+1 G0(t,τ)(−1)kGk(τ,s)|g(τ)| <Λ(¯ gΩ)k¯ g b ∑ τ=a+1 G0(t,τ) <Λ(¯ gΩ)k¯ gΩ=Λ(¯ gΩ)k+1. Thus, (11) holds for n=k+ 1. By mathematical induction, (11) holds for any n= 0,1,2, . . . . As a direct consequence of previous inequality and condition (A1), we deduce that for all (t,s)∈Nb a×Nb a+1the following property is fulfilled: |G(t,s)|= ∞ ∑ n=0 (−1)nGn(t,s) ≤ ∞ ∑ n=0 |(−1)nGn(t,s)| <Λ ∞ ∑ n=0 (¯ gΩ)n=Λ 1−¯ gΩ<∞, and, a a consequence, G(t,s)converges on Nb a×Nb a+1. Finally, expressions (5) , (9) and (10) imply that for all t∈Nb a the following equality is fulfilled: u(t) = ∞ ∑ n=0"b ∑ s=a+1 (−1)nGn(t,s)h(s)#= b ∑ s=a+1"∞ ∑ n=0 (−1)nGn(t,s)#h(s) = b ∑ s=a+1 G(t,s)h(s). (12) It is not difficult to verify that any function defined by (12) is a solution of the boundary value problem (7) . So we conclude that problem (7) has a unique solution and, as a consequence, Gis its related Green’s function. Lemma 2. Assume conditions (A0)and (A1). Let G be defined by (5)and the following: ¯ G(s) = G0(ρ(s),s) 1−¯ gΩ,s∈Nb a+1. (13) Then, |G(t,s)|≤¯ G(s),(t,s)∈Nb a×Nb a+1. Proof. First, we prove the following: |(−1)nGn(t,s)|<G(s−1, s)(¯ gΩ)n,s∈Nb a+1,n=0,1, 2,. . . (14) Theorem 1implies that inequality (14) is true for n=0. Assume now that (14) holds for some n=k . We will show that (14) holds for n=k+1. Consider the following: Symmetry 2021,13, 1101 9 of 15 (−1)k+1Gk+1(t,s)= (−1)k+1b ∑ τ=a+1 G0(t,τ)Gk(τ,s)g(τ) = − b ∑ τ=a+1 G0(t,τ)(−1)kGk(τ,s)g(τ) ≤ b ∑ τ=a+1 G0(t,τ)(−1)kGk(τ,s)|g(τ)| <G(s−1, s)(¯ gΩ)k¯ g b ∑ τ=a+1 G0(t,τ) <G(s−1, s)(¯ gΩ)k¯ gΩ=G(s−1, s)(¯ gΩ)k+1. Thus, (14) holds for n=k+ 1 and the inequalities are deduced from mathematical induction. Now, from (5), (13) and (14), for s∈Nb a+1, we obtain the following: |G(t,s)|= ∞ ∑ n=0 (−1)nGn(t,s) ≤ ∞ ∑ n=0 |(−1)nGn(t,s)| <G(s−1, s) ∞ ∑ n=0 (¯ gΩ)n =G0(ρ(s),s) 1−¯ gΩ=¯ G(s), and the proof is complete. From the previous result, we deduce the following consequence for g≤0. Corollary 1. Assume that condition (A) is fulfilled and −¯ g<g(t)≤0, t∈Nb a. Then, G(t,s)≥0for each (t,s)∈Nb a×Nb a+1. Proof. From Theorem 1, we know that G0(t , s)≥ 0 for each (t , s)∈Nb a×Nb a+1 . The result follows immediately from (5) and (6). 4. Existence of Solutions In this section, we derive two existence results for the nonlinear problem (1) . Define the operator T:X→X(Xdefined in previous section) by the following: Tu(t) = b ∑ s=a+1 G(t,s)f(s,u(s)),t∈Nb a. (15) In view of (12) , it is clear that u is a fixed point of T if and only if u is a solution of (1) . For any R>0 given, we define the following set: KR={u∈X:kuk<R}. Clearly, KRis a non-empty open subset of X, 0 ∈ KRand T:KR→X. Now, denoting by max t∈Nb a |f(t,0)|=Mand b ∑ s=a+1 ¯ G(s) = K(>0),