CUBO, A Mathematical Journal Vol. 24, no. 03, pp. 467-484, December 2022 DOI: 10.56754/0719-0646.2403.0467 Positive solutions of nabla fractional boundary value problem N. S. Gopal1 J. M. Jonnalagadda1, B 1 Department of Mathematics, Birla Institute of Technology and Science Pilani, Hyderabad - 500078, Telangana, India. nsgopal94@gmail.com j.jaganmohan@hotmail.com B ABSTRACT In this article, we consider the following two-point discrete fractional boundary value problem with constant coefficient associated with Dirichlet boundary conditions. − ( ∇νρ(a)u ) (t) + λu(t) = f(t, u(t)), t ∈ Nba+2, u(a) = u(b) = 0, where 1 < ν < 2, a, b ∈ R with b−a ∈ N3, Nba+2 = {a+2, a+ 3, . . . , b}, |λ| < 1, ∇νρ(a)u denotes the ν th-order Riemann– Liouville nabla difference of u based at ρ(a) = a − 1, and f : Nba+2 × R → R+. We make use of Guo–Krasnosels’kǐı and Leggett–Williams fixed-point theorems on suitable cones and under appropri- ate conditions on the non-linear part of the difference equa- tion. We establish sufficient requirements for at least one, at least two, and at least three positive solutions of the consid- ered boundary value problem. We also provide an example to demonstrate the applicability of established results. Accepted: 10 November, 2022 Received: 25 April, 2022 ©2022 N. S. Gopal et al. This open access article is licensed under a Creative Commons Attribution-NonCommercial 4.0 International License. http://cubo.ufro.cl/ http://dx.doi.org/10.56754/0719-0646.2403.0467 https://orcid.org/0000-0002-1166-3446 https://orcid.org/0000-0002-1310-8323 mailto:nsgopal94@gmail.com mailto:j.jaganmohan@hotmail.com CUBO, A Mathematical Journal Vol. 24, no. 03, pp. 467–485, December 2022 DOI: 10.56754/0719-0646.2403.0467 RESUMEN En este art́ıculo consideramos el siguiente problema de valor en la frontera de dos puntos discreto fraccional con coefi- cientes constantes asociado a condiciones de frontera de tipo Dirichlet − ( ∇νρ(a)u ) (t) + λu(t) = f(t, u(t)), t ∈ Nba+2, u(a) = u(b) = 0, donde 1 < ν < 2, a, b ∈ R con b − a ∈ N3, Nba+2 = {a + 2, a + 3, . . . , b}, |λ| < 1, ∇νρ(a)u denota la nabla diferencia de Riemann–Liouville de u de orden ν basada en ρ(a) = a − 1, y f : Nba+2 × R → R+. Usamos los teoremas de punto fijo de Guo–Krasnosels’kĭı y Leggett–Williams en conos adecuados y bajo condiciones apropiadas en la parte nolineal de la ecuación en diferen- cias. Establecemos requerimientos suficientes para al menos una, al menos dos, y al menos tres soluciones positivas del problema de valor en la frontera considerado. También en- tregamos un ejemplo para mostrar la aplicabilidad de los resultados. Keywords and Phrases: Nabla fractional difference, boundary value problem, Dirichlet boundary conditions, positive solution, existence, fixed-point. 2020 AMS Mathematics Subject Classification: 39A12. Accepted: 10 November, 2022 Received: 25 April, 2022 ©2022 N. S. Gopal et al. This open access article is licensed under a Creative Commons Attribution-NonCommercial 4.0 International License. http://cubo.ufro.cl/ http://dx.doi.org/10.56754/0719-0646.2403.0467 CUBO 24, 3 (2022) Positive solutions of nabla fractional boundary value problem 469 1 Introduction Nabla fractional calculus is a branch of mathematics that deals with arbitrary order differences and sums in the backward sense. The theory of nabla fractional calculus is still in its early stages, with the most important contributions appearing in the last two decades. Gray & Zhang [15] and Miller & Ross in [34] first introduced the concept of nabla fractional difference and sum. Atici & Eloe [2] developed the Riemann–Liouville type nabla fractional difference operator. They also studied the nabla fractional initial value problem, and established the exponential law, product rule, and nabla Laplace transform in this line. Several mathematicians [2, 3, 4, 5, 6, 7, 8, 16, 17, 21, 22] have contributed to the development of the theory of discrete fractional calculus in line with the theory of continuous fractional calculus. For historical references on continuous fractional calculus, see [28, 31, 32]. As a result of their works, today discrete fractional calculus has turned into a fruitful field of research in science and engineering. We refer here to recent monographs [9, 12, 29] and the references therein, which are important resources pertaining to this field of work. The study of boundary value problems (BVPs) has a long past and can be followed back to the work of Euler and Taylor on vibrating strings. On the discrete fractional side, there is a sudden growth in interest for the development of nabla fractional BVPs. Many authors have studied nabla fractional BVPs recently. To name a few, Goar [11] and Ikram [18] worked with self-adjoint Caputo nabla BVPs. Gholami et al. [10] obtained the Green’s function for a non-homogeneous Riemann– Liouville nabla BVP with Dirichlet boundary conditions. Jonnalagadda [19, 20, 23] analysed some qualitative properties of two-point non-linear Riemann–Liouville nabla fractional BVPs associated with a variety of boundary conditions. As pointed out earlier, many authors have studied the discrete fractional two-point boundary value problem like in [4, 19] and recently authors in [23] have worked with general nabla fractional difference equation with constant coefficients coupled with Dirichlet conditions, which resulted in for the first time Green’s function in terms of discrete Mittag–Leffler function along with a few properties of the same. Compared to discrete Taylor monomial, discrete Mittag–Leffler function is an infinite series because of which it poses a challenge while proving positivity of Green’s function. In the article, [23] the authors have overcome this challenge of proving positivity of Green’s function. In the present article, we use the positivity of Green’s function and prove an important lemma which helps us deal with conical mappings by proving that a ratio of infinite series is increasing or decreasing with respect to the ratio of its coefficient. To the best of our knowledge, no work has been done with Leggett–Williams fixed-point theorem in the nabla setting. We consider the following boundary value problem  − ( ∇ν ρ(a) u ) (t) + λu(t) = f(t, u(t)), t ∈ Nba+2, u(a) = u(b) = 0, (1.1) 470 N. S. Gopal & J. M. Jonnalagadda CUBO 24, 3 (2022) where 1 < ν < 2, a, b ∈ R with b−a ∈ N3, Nba+2 = {a+2, a+3, . . . , b}, |λ| < 1, ∇νρ(a)u denotes the νth-order Riemann–Liouville nabla difference of u based at ρ(a) = a − 1, and f : Nba+2 × R → R+. The present paper is organized as follows: Section 2 contains preliminaries on nabla fractional calculus. In Section 3, we establish some properties of the Green’s function associated with the nabla fractional boundary value problem (1.1) and construct the existence of at least one, at least two and at least three positive solutions with the help of Guo–Krasnosel’skĭı and Leggett–Williams fixed-point theorems on suitable cones and under appropriate conditions on the non-linear part of the difference equation. Finally, we conclude this article with an example to demonstrate the applicability of our results. 2 Preliminaries Denote the set of all real numbers and positive integers by R and Z+, respectively. We use the following notations, definitions and known results of nabla fractional calculus [12]. Assume empty sums and products are 0 and 1, respectively. Definition 2.1. For a ∈ R, the sets Na and Nba, where b − a ∈ Z+, are defined by Na = {a, a + 1, a + 2, . . .}, Nba = {a, a + 1, a + 2, . . . , b}. Let u : Na → R and N ∈ N1. The first order backward (nabla) difference of u is defined by( ∇u ) (t) = u(t) − u(t − 1), for t ∈ Na+1, and the Nth-order nabla difference of u is defined recursively by ( ∇Nu ) (t) = ( ∇ ( ∇N−1u )) (t), for t ∈ Na+N. Definition 2.2 ([12]). For t ∈ R\{. . . , −2, −1, 0} and r ∈ R such that (t+r) ∈ R\{. . . , −2, −1, 0}, the generalized rising function (many authors employ the Pochhammer symbol [33] to denote the same) is defined by tr = Γ(t + r) Γ(t) . Here Γ(·) denotes the Euler gamma function. Also, if t ∈ {. . . , −2, −1, 0} and r ∈ R such that (t + r) ∈ R \ {. . . , −2, −1, 0}, then we use the convention that tr = 0. Definition 2.3 ([12]). Let t, a ∈ R and µ ∈ R \ {. . . , −2, −1}. The µth-order nabla fractional Taylor monomial is given by Hµ(t, a) = (t − a)µ Γ(µ + 1) , provided the right-hand side exists. We observe the following properties of the nabla fractional Taylor monomials. CUBO 24, 3 (2022) Positive solutions of nabla fractional boundary value problem 471 Lemma 2.4 ([18, 19]). Let µ > −1 and s ∈ Na. Then the following hold: (1) If t ∈ Nρ(s), then Hµ(t, ρ(s)) ≥ 0 and if t ∈ Ns, then Hµ(t, ρ(s)) > 0. (2) If t ∈ Ns and −1 < µ < 0, then Hµ(t, ρ(s)) is an increasing function of s. (3) If t ∈ Ns+1 and −1 < µ < 0, then Hµ(t, ρ(s)) is a decreasing function of t. (4) If t ∈ Nρ(s) and µ > 0, then Hµ(t, ρ(s)) is a decreasing function of s. (5) If t ∈ Nρ(s) and µ ≥ 0, then Hµ(t, ρ(s)) is a non-decreasing function of t. (6) If t ∈ Ns and µ > 0, then Hµ(t, ρ(s)) is an increasing function of t. (7) If 0 < v ≤ µ, then Hv(t, a) ≤ Hµ(t, a), for each fixed t ∈ Na. Definition 2.5 ([12]). Let u : Na+1 → R and ν > 0. The νth-order nabla sum of u is given by ( ∇−νa u ) (t) = t∑ s=a+1 Hν−1(t, ρ(s))u(s), t ∈ Na+1. Definition 2.6 ([12]). Let u : Na+1 → R, ν > 0 and choose N ∈ N1 such that N − 1 < ν ≤ N. The νth-order Riemann–Liouville nabla difference of u is given by ( ∇νau ) (t) = ( ∇N ( ∇−(N−ν)a u )) (t), t ∈ Na+N. Lemma 2.7 ([13]). Let a, b be two real numbers such that 0 < a ≤ b and 1 < α < 2. Then (a−s)α−1 (b−s)α−1 is a decreasing function of s for s ∈ Na−10 . Lemma 2.8 ([12]). Assume the successive fractional nabla Taylor monomials are well defined. (1) Let ν > 0 and α ∈ R. Then, ∇−νa Hα(t, a) = Hα+ν(t, a), for t ∈ Na. (2) Let ν, α ∈ R and n ∈ N1 such that n − 1 < ν ≤ n. Then, ∇νaHα(t, a) = Hα−ν(t, a), for t ∈ Na+n. Finally, we present the definition of the nabla Mittag–Leffler function which is the nabla analogue of classical Mittag-Leffler function [14, 30]. Definition 2.9 ([12]). Let α, β, λ ∈ R such that α > 0 and |λ| < 1. The nabla Mittag–Leffler function is defined by Eλ,α,β(t, a) = ∞∑ n=0 λnHαn+β(t, a), for t ∈ Na. 472 N. S. Gopal & J. M. Jonnalagadda CUBO 24, 3 (2022) Theorem 2.10 ([23]). Assume 1 < ν < 2, −1 < λ < 1 and h : Na+2 → R. The unique solution of the nabla fractional boundary value problem  − ( ∇ν ρ(a) u ) (t) + λu(t) = h(t), t ∈ Nba+2, u(a) = u(b) = 0, (2.1) is given by u(t) = b∑ s=a+2 G(t, s)h(s), t ∈ Nba, (2.2) where G(t, s) =   G1(t, s) = Eλ,ν,ν−1(t, a) Eλ,ν,ν−1(b, a) Eλ,ν,ν−1(b, ρ(s)), s ∈ Nbt+1, G2(t, s) = Eλ,ν,ν−1(t, a) Eλ,ν,ν−1(b, a) Eλ,ν,ν−1(b, ρ(s)) − Eλ,ν,ν−1(t, ρ(s)), s ∈ Nta+2. (2.3) Now, we state some positive properties of the Green’s function (2.3). Lemma 2.11 ([23]). Assume 1 < ν < 2 and t ∈ Na+2. For each 0 ≤ λ < 1, denote by g(λ) = ∞∑ n=0 λnHνn+ν−3(t, ρ(a)) (2.4) = ∞∑ n=0 λn Γ(t − a + νn + ν − 2) Γ(t − a + 1)Γ(νn + ν − 2) . (2.5) Then there exists a unique λ̄ = λ̄(t) ∈ (0, 1) such that g(λ̄) = 0. (2.6) Take λ∗ = min t∈Nb a+2 λ̄(t). Then, 0 < λ∗ < 1. We observe the following properties of the nabla Mittag-Leffler function Lemma 2.12 ([23]). Assume 1 < ν < 2 and 0 ≤ λ < 1. Then, (1) 0 < Hν−1(t, ρ(a)) ≤ Eλ,ν,ν−1(t, ρ(a)) for t ∈ Na; (2) Eλ,ν,ν−1(t, ρ(a)) is an increasing function with respect to t for t ∈ Na; (3) 0 < Hν−2(t, ρ(a)) ≤ ∇Eλ,ν,ν−1(t, ρ(a)) for t ∈ Na+1; (4) ∇Eλ,ν,ν−1(t, ρ(a)) is a decreasing function with respect to t for t ∈ Na+1 and λ ∈ (0, λ∗]; (5) Eλ,ν,ν−1(t, ρ(s)) ≤ Eλ,ν,ν−1(t, a) for t ∈ Ns and s ∈ Na+1; (6) ∇Eλ,ν,ν−1(t, ρ(s)) ≥ ∇Eλ,ν,ν−1(t, a) for t ∈ Ns, s ∈ Na+1 and λ ∈ (0, λ∗]. CUBO 24, 3 (2022) Positive solutions of nabla fractional boundary value problem 473 Lemma 2.13 ([27]). Let (an) and (bn) (n = 0, 1, 2, . . . ) be real numbers and let the power series A(x) = ∞∑ n=0 anx n and B(x) = ∞∑ n=0 bnx n be convergent for |x| < r. If bn > 0, n = 0, 1, 2, . . . and the sequence ( an bn ) n≥0 is (strictly) increasing (decreasing), then the function A(x) B(x) is also (strictly) increasing (decreasing) on [0, r). Theorem 2.14 ([23]). Assume 1 < ν < 2 and 0 ≤ λ < 1 such that λ ∈ (0, λ∗]. The Green’s function G(t, s) defined in (2.3) satisfies G(t, s) ≥ 0 for each (t, s) ∈ Nba × Nba+2. In particular, G(a, s) = G(b, s) = 0 and G(t, s) > 0 for each (t, s) ∈ Nb−1a+1 × N b a+2. 3 Multiple Positive Solutions In this section, we establish sufficient conditions on existence of at least one, at least two and at least three positive solutions of (1.1) using Guo–Krasnosel’skĭı and Leggett–Williams fixed-point theorems on conical shells. Definition 3.1. Let B be a Banach space over R. A closed nonempty convex set K ⊂ B is called a cone provided, (i) λ1u ∈ K, for all u ∈ K and λ1 ≥ 0. (ii) u ∈ K and −u ∈ K implies u = 0. Definition 3.2. A functional α2 is said to be a non-negative continuous concave functional on a cone K of a real Banach space β, if α2 : K → [0, ∞) is continuous and α2(tx + (1 − t)y) ≥ tα2(x) + (1 − t)α2(y), for all x, y ∈ K and t ∈ [0, 1]. Definition 3.3. An operator is called completely continuous, if it is continuous and maps bounded sets into precompact sets. Theorem 3.4 (Guo–Krasnosel’skĭı fixed-point theorem, [24]). Let B be a Banach space and K ⊆ B be a cone. Assume that Ω1 and Ω2 are open sets contained in B such that 0 ∈ Ω1 and Ω1 ⊆ Ω2. Assume further that T : K ∩ (Ω2 \ Ω1) −→ K is a completely continuous operator. If, either (1) ∥Tu∥ ≤ ∥u∥ for u ∈ K ∩ ∂Ω1 and ∥Tu∥ ≥ ∥u∥ for u ∈ K ∩ ∂Ω2; or (2) ∥Tu∥ ≥ ∥u∥ for u ∈ K ∩ ∂Ω1 and ∥Tu∥ ≤ ∥u∥ for u ∈ K ∩ ∂Ω2; holds, then T has at least one fixed-point in K ∩ (Ω2 \ Ω1). The following results are useful for the main results of this section. 474 N. S. Gopal & J. M. Jonnalagadda CUBO 24, 3 (2022) Lemma 3.5. Let a, b be two real numbers such that 0 < a ≤ b and 1 < ν < 2. Then Eλ,ν,ν−1(a, ρ(s)) Eλ,ν,ν−1(b, ρ(s)) is a decreasing function of s for s ∈ Na−10 . Proof. For each s ∈ Na−10 , denote by an = Hνn+ν−1(a, ρ(s)) and bn = Hνn+ν−1(b, ρ(s)), n ∈ N0. Clearly, an and bn for n ∈ N0 are real numbers. Further, denote by A(λ) = Eλ,ν,ν−1(a, ρ(s)) and B(λ) = Eλ,ν,ν−1(b, ρ(s)). We know that the power series A(λ) and B(λ) are convergent for |λ| < 1. Also, bn > 0, n ∈ N0 and the sequence ( an bn ) n≥0 = ( Hνn+ν−1(a, ρ(s)) Hνn+ν−1(b, ρ(s)) ) n≥0 is strictly decreasing, by Lemma 2.7. Then, by Lemma 2.13, the function A(λ) B(λ) = Eλ,ν,ν−1(a, ρ(s)) Eλ,ν,ν−1(b, ρ(s)) is also strictly decreasing on [0, 1) for each s ∈ Na−10 . The proof is complete. Theorem 3.6. There exists a number γ ∈ (0, 1), such that min t∈Ndc G(t, s) ≥ γ max t∈Nba G(t, s) = γG(s − 1, s), (3.1) for λ ∈ (0, λ∗] and c, d ∈ Nb−1a+1 such that c = a + ⌈ b − a + 1 4 ⌉ and d = a + 3 ⌊b − a + 1 4 ⌋ . Proof. It follows from the proof of Theorem 2.14 in [23] that for each λ ∈ (0, λ∗], G(t, s) is an increasing function of t for ∈ Ns−1a and is a decreasing function of t for ∈ Nbs. Thus, we have max t∈Nba G(t, s) = G(s − 1, s) for s ∈ Nba+2. Consider G(t, s) G(s − 1, s) =   Eλ,ν,ν−1(t, a) Eλ,ν,ν−1(s − 1, a) , s ∈ Nbt+1, Eλ,ν,ν−1(t, a) Eλ,ν,ν−1(s − 1, a) − Eλ,ν,ν−1(t, ρ(s))Eλ,ν,ν−1(b, a) Eλ,ν,ν−1(b, ρ(s))Eλ,ν,ν−1(s − 1, a) , s ∈ Nta+2. Now, for s > t and c ≤ t ≤ d, G1(t, s) is an increasing function with respect to t. Then, we have min t∈Ndc G1(t, s) = G1(c, s) = Eλ,ν,ν−1(c, a) Eλ,ν,ν−1(b, a) Eλ,ν,ν−1(b, ρ(s)), s ∈ Nbt+1. For t > s and c ≤ t ≤ d, G2(t, s) is a decreasing function with respect to t. Then, we have min t∈Ndc G2(t, s) = G2(d, s) = Eλ,ν,ν−1(d, a) Eλ,ν,ν−1(b, a) Eλ,ν,ν−1(b, ρ(s)) − Eλ,ν,ν−1(d, ρ(s)), s ∈ Nta+2. CUBO 24, 3 (2022) Positive solutions of nabla fractional boundary value problem 475 Thus, min t∈Ndc G(t, s) =   G1(c, s), for s ∈ Nbd, min{G2(d, s), G1(c, s)}, for s ∈ Nd−1c+1, G2(d, s), for s ∈ Nca+2, =   G2(d, s), for s ∈ Nra+2, G1(c, s), for s ∈ Nbr, where c < r < d. Consider mint∈Ndc G(t, s) G(s − 1, s) =   Eλ,ν,ν−1(c, a) Eλ,ν,ν−1(s − 1, a) , s ∈ Nbr, Eλ,ν,ν−1(d, a) Eλ,ν,ν−1(s − 1, a) − Eλ,ν,ν−1(d, ρ(s))Eλ,ν,ν−1(b, a) Eλ,ν,ν−1(b, ρ(s))Eλ,ν,ν−1(s − 1, a) , s ∈ Nra+2. Hence, min t∈Ndc G(t, s) ≥ γ(s) max t∈Nba G(t, s), (3.2) where γ(s) = min [ Eλ,ν,ν−1(c, a) Eλ,ν,ν−1(s − 1, a) , Eλ,ν,ν−1(d, a) Eλ,ν,ν−1(s − 1, a) − Eλ,ν,ν−1(d, ρ(s))Eλ,ν,ν−1(b, a) Eλ,ν,ν−1(b, ρ(s))Eλ,ν,ν−1(s − 1, a) ] . For s ∈ Nbr, denote by γ1(s) = Eλ,ν,ν−1(c, a) Eλ,ν,ν−1(s − 1, a) ≥ Eλ,ν,ν−1(c, a) Eλ,ν,ν−1(b − 1, a) . Similarly, for s ∈ Nra+2, we take γ2(s) = Eλ,ν,ν−1(d, a) Eλ,ν,ν−1(s − 1, a) − Eλ,ν,ν−1(d, ρ(s))Eλ,ν,ν−1(b, a) Eλ,ν,ν−1(b, ρ(s))Eλ,ν,ν−1(s − 1, a) . By Lemma 3.5, we see that Eλ,ν,ν−1(d, ρ(s)) Eλ,ν,ν−1(b, ρ(s)) is a decreasing function for s ∈ Nra+2. Then, γ2(s) ≥ 1 Eλ,ν,ν−1(s − 1, a) [ Eλ,ν,ν−1(d, a) − Eλ,ν,ν−1(d, a + 1)Eλ,ν,ν−1(b, a) Eλ,ν,ν−1(b, a + 1) ] > 1 Eλ,ν,ν−1(d, a) [ Eλ,ν,ν−1(d, a) − Eλ,ν,ν−1(d, a + 1)Eλ,ν,ν−1(b, a) Eλ,ν,ν−1(b, a + 1) ] . Thus, min t∈Ndc G(t, s) ≥ γ max t∈Nba G(t, s), (3.3) where γ = min [ Eλ,ν,ν−1(c, a) Eλ,ν,ν−1(b − 1, a) , 1 − Eλ,ν,ν−1(d, a + 1)Eλ,ν,ν−1(b, a) Eλ,ν,ν−1(b, a + 1)Eλ,ν,ν−1(d, a) ] . 476 N. S. Gopal & J. M. Jonnalagadda CUBO 24, 3 (2022) Since G1(c, s) > 0 and G2(d, s) > 0, we have γ(s) > 0 for all s ∈ Nba+2, implying that γ > 0. It would be suffice to prove that one of the terms Eλ,ν,ν−1(c, a) Eλ,ν,ν−1(b − 1, a) , 1− Eλ,ν,ν−1(d, a + 1)Eλ,ν,ν−1(b, a) Eλ,ν,ν−1(b, a + 1)Eλ,ν,ν−1(d, a) is less than 1. It follows from Lemma 2.12 that Eλ,ν,ν−1(c, a) Eλ,ν,ν−1(b − 1, a) < 1. Therefore, we conclude that γ ∈ (0, 1). The proof is complete. By Theorem 2.10, we observe that u is a solution of (1.1) if and only if u is a solution of the summation equation u(t) = b∑ s=a+2 G(t, s)f(s, u(s)), t ∈ Nba. (3.4) Note that any solution u : Nba → R of (1.1) can be viewed as a real (b − a + 1)-tuple vector. Consequently, u ∈ Rb−a+1. Define the operator T : Rb−a+1 → Rb−a+1 by ( Tu ) (t) = b∑ s=a+2 G(t, s)f(s, u(s)), t ∈ Nba. (3.5) Clearly, u is a fixed-point of T if and only if u is a solution of (1.1). We use the fact that Rb−a+1 is a Banach space equipped with the maximum norm ∥u∥ = maxt∈Nba |u(t)|, for any u ∈ R b−a+1. Denote by B = {u : Nba → R | u(a) = u(b) = 0} ⊆ R b−a+1. (3.6) Clearly B is a Banach space equipped with the maximum norm i.e. ∥u∥ = max t∈Nba |u(t)|. Since T is defined on a discrete finite domain, it is trivially completely continuous. Define the cone K = {u ∈ B : u(t) ≥ 0 for t ∈ Nba, and min t∈Ndc u(t) ≥ γ∥u∥}. (3.7) Lemma 3.7. For λ ∈ (0, λ∗] the operator T maps K into itself. Proof. Let u ∈ K. Clearly, (Tu) (t) ≥ 0, whenever u ∈ K. Consider min t∈Ndc (Tu) (t) = min t∈Ndc b∑ s=a+2 G(t, s)f(s, u(s)) ≥ b∑ s=a+2 min t∈Ndc [G(t, s)] f(s, u(s)) ≥ b∑ s=a+2 γ max t∈Nba [G(t, s)] f(s, u(s)) ≥ γ max t∈Nba b∑ s=a+2 G(t, s)f(s, u(s)) = γ max t∈Nba ∣∣∣∣∣ b∑ s=a+2 G(t, s)f(s, u(s)) ∣∣∣∣∣ = γ∥Tu∥. Thus, we have T : K → K and it is completely continuous. The proof is complete. CUBO 24, 3 (2022) Positive solutions of nabla fractional boundary value problem 477 Take η = 1 b∑ s=a+2 G(s − 1, s) . Theorem 3.8. Assume f(t, u(t)) satisfies the following conditions for 0 < r1 < r2 (i) There exists a number r1 > 0 such that f(t, u(t)) ≤ ηr1, whenever 0 ≤ u ≤ r1. (ii) There exists a number r2 > 0 such that f(t, u(t)) ≥ ηr2γ , whenever γr2 ≤ u ≤ r2. Then, for λ ∈ (0, λ∗] the BVP (1.1) has at least one positive solution. Proof. We know that T : K → K is completely continuous. Define the set Ω1 = {u ∈ K : ∥u∥ < r1}. Clearly, Ω1 ⊆ β is an open set with 0 ∈ Ω1. Since ∥u∥ = r1 for u ∈ ∂Ω1, condition (i) holds for all u ∈ ∂Ω1. So, it follows that ∥Tu∥ = max t∈Nba b∑ s=a+2 G(t, s)f(s, u(s)) ≤ b∑ s=a+2 max t∈Nba G(t, s)f(s, u(s)) ≤ ηr1 b∑ s=a+2 G(s − 1, s) = r1 = ∥u∥. implying that ∥Tu∥ ≤ ∥u∥ whenever u ∈ K ∩ ∂Ω1. On the other hand, define the set Ω2 = {u ∈ K : ∥u∥ < r2}. Clearly, Ω2 ⊆ β is an open set and Ω1 ⊆ Ω2. Since ∥u∥ = r2 for u ∈ ∂Ω2, condition (ii) holds for all u ∈ ∂Ω2. Thus, we have ∥Tu∥ ≥ min t∈Ndc b∑ s=a+2 G(t, s)f(s, u(s)) ≥ b∑ s=a+2 min t∈Ndc G(t, s)f(s, u(s)) ≥ γ b∑ s=a+2 G(s − 1, s)f(s, u(s)) ≥ ηr2 b∑ s=a+2 G(s − 1, s) = r2 = ∥u∥ implying that ∥Tu∥ ≥ ∥u∥ whenever u ∈ K ∩ ∂Ω2. Hence by part 1 of Theorem 3.4, T has at least one fixed-point in K ∩ (Ω1\Ω1), say u0 satisfying r1 < ∥u0∥ < r2 478 N. S. Gopal & J. M. Jonnalagadda CUBO 24, 3 (2022) Theorem 3.9. Assume f(t, u(t)) satisfies the following conditions (i) There exists a number r2 > 0 such that f(t, u(t)) ≤ ηr2, whenever 0 ≤ u ≤ r2. (ii) lim u→0+ min t∈Nba f(t, u(t)) u = ∞, lim u→∞ min t∈Nba f(t, u(t)) u = ∞. Then, for λ ∈ (0, λ∗] the BVP (1.1) has at least two positive solution. Proof. Let us choose a number N > 0 such that Nγ η > 1, by condition (ii) there exists a number r∗ > 0 such that r∗ < r1 < r2 and f(t, u(t)) ≥ Nu for u ∈ [0, r∗] and t ∈ Nba. Define the set Ωr∗ = {u ∈ K : ∥u∥ < r∗}. It can easily be shown that ∥Tu∥ > ∥u∥, for u ∈ ∂Ωr∗ ∩ K. Next for the same N, we can find a number R1 > 0 such that f(t, u) ≥ Nu for u ≥ R1 and t ∈ Nba. Choose R such that R = max { r2, R1 γ } . Define the set ΩR = {u ∈ K : ∥u∥ < R}. We can show that ∥Tu∥ > ∥u∥, for u ∈ ∂ΩR ∩ K. Finally define the set Ω2 = {u ∈ K : ∥u∥ < r2}. Since ∥u∥ = r2 condition (i) holds for all u ∈ ∂Ω2. Then, we have ∥Tu∥ = max t∈Na b b∑ s=a+2 G(t, s)f(s, u(s)) ≤ b∑ s=a+2 max t∈Na b [G(t, s)] f(s, u(s) ≤ r2η b∑ s=a+2 G(s − 1, s) = r2. Implying ∥Tu∥ ≤ ∥u∥, for u ∈ ∂Ωr2 ∩ K. Hence, we conclude that T has at least two fixed-points say u1 ∈ Ω2\Ω̂r∗ and u2 ∈ ΩR\Ω̂2, where Ω̂ denoted the interior of the set Ω. In particular (1.1) has at least two positive solutions, say u1 and u2 satisfying 0 < ∥u1∥ < r2 < ∥u2∥. The proof is complete. We state here the Leggett–Williams fixed-point theorem as follows. The proof can be found in [26] and also, we would like to refer here a paper by Kwong [25] on the same. Denote Kc ={u ∈ K : ∥u∥ < c}, Kα2(a, b) ={u ∈ K : a ≤ α2(u), ∥u∥ ≤ b}, where α2 is defined as in Definition 3.2. CUBO 24, 3 (2022) Positive solutions of nabla fractional boundary value problem 479 Theorem 3.10 ([1]). Let T : K̄c → K̄c be completely continuous and α2 be a non-negative continuous concave functional on K, such that α2(u) ≤ ∥u∥, for all u ∈ K̄c. Suppose there exists 0 < d < a < b ≤ c, such that (1) {u ∈ Kα2(a, b) : α2(u) > a} ≠ ∅ and α2(Tu) > a, for u ∈ Kα2(a, b); (2) ∥Tu∥ < d, for ∥u∥ ≤ d; (3) α2(Tu) > a, for u ∈ Kα2(a, c) with ∥Tu∥ > b. Then, T has at least three fixed-points u1, u2, u3 satisfying ∥u1∥ < d, a < α2(u2), and ∥u3∥ > d and α2(u3) < a. We introduce here growth conditions on the non-linear function f in line with [1]. Theorem 3.11. Suppose there exists numbers a′, b′, d′ ∈ R+, where 0 < d′ < a′ < γb′ < b′, such that f satisfies the following (1) f(t, u(t)) > a′η γ , if u ∈ [a′, b′]; (2) f(t, u(t)) < d′η, if u ∈ [0, d′]; (3) There exists c′ such that c′ > b′ and if u ∈ [0, c′] then f(t, u(t)) < c′η; Then, the boundary value problem (1.1) for λ ∈ (0, λ∗] has at least three positive solutions. Proof. Define a non-negative continuous concave functional α2 : K → [0, ∞) with α2(u) ≤ ∥u∥, for all u ∈ K, by α2(u) = min t∈Ndc u(t). Claim 1: If there exists a positive number r such that u ∈ [0, r] implies f(u) < rη, then T : K̄r → Kr. Suppose that u ∈ K̄r. Then, ∥Tu∥ = max t∈Nba [ b∑ s=a+2 G(t, s)f(s, u(s)) ] ≤ b∑ s=a+2 max t∈Nba [G(t, s)] f(s, u(s)) = b∑ s=a+2 G(s − 1, s)f(s, u(s)) < rη b∑ s=a+2 G(s − 1, s) = r. 480 N. S. Gopal & J. M. Jonnalagadda CUBO 24, 3 (2022) Thus, T : K̄r → Kr. Hence, we have that if condition (3) holds, then there exists a number c′ such that c′ > b′ and T : K̄c′ → Kc′. Note that with r = d′ and using condition (2), we get that T : K̄d′ → Kd′. Claim 2: {u ∈ Kα2(a′, b′) : α2(u) > a′} ≠ ∅ and α2(Tu) > a′ for u ∈ Kα2(a′, b′). Since u = a ′+b′ 2 ∈ {u ∈ Kα2(a′, b′) : α2(u) > a′} ≠ ∅. Let u ∈ Kα2(a′, b′). By using condition (1), we have α2(Tu) = min t∈Ndc [ b∑ s=a+2 G(t, s)f(s, u(s)) ] ≥ b∑ s=a+2 min t∈Ndc [G(t, s)] f(s, u(s)) ≥ γ b∑ s=a+2 G(s − 1, s)f(s, u(s)) > a′ Thus, if u ∈ Kα2(a′, b′), then α2(Tu) > a′. Claim 3: If u ∈ Kα2(a′, c′) and ∥Tu∥ > b′ then α2(Tu) > a′. Suppose u ∈ Kα2(a′, c′) and ∥Tu∥ > b′. Then, α2(Tu) = min t∈Ndc [ b∑ s=a+2 G(t, s)f(s, u(s)) ] ≥ b∑ s=a+2 min t∈Ndc [G(t, s)] f(s, u(s)) ≥ γ b∑ s=a+2 max t∈Nba [G(t, s)] f(s, u(s)) ≥ γ max t∈Ndc [ b∑ s=a+2 G(t, s)f(s, u(s)) ] = γ∥Tu∥ > γb′ > a′. Thus, α2(Tx) > a ′. Hence all the hypothesis of the Theorem 3.10 are satisfied. Therefore, the boundary value problem (1.1) has at least three positive solutions u1, u2 and u3 satisfying ∥u1∥ < d′, a′ < α2(u2), and ∥u3∥ > d′ and α2(u3) < a′. The proof is complete. CUBO 24, 3 (2022) Positive solutions of nabla fractional boundary value problem 481 Example In this section, we have constructed a suitable example to illustrate the applicability of the estab- lished results. Example 3.12. Take ν = 1.5, a = 0, b = 5, and f(t, u(t)) = 1 20 (√ u + u2 ) . Then, (1.1) becomes  − ( ∇1.5 ρ(0) u ) (t) + λu(t) = 1 20 (√ u + u2 ) , t ∈ N52, u(0) = 0 = u(5). (3.8) Choose λ∗ = 0.007. Then, we get η = 1 5∑ s=2 G(s − 1, s) = Eλ,1.5,0.5(5, 0) 5∑ s=2 Eλ,1.5,0.5(s − 1, 0)Eλ,1.5,0.5(5, s − 1) = 0.2473. By taking r2 = 2, we have f(t, u) = 1 20 (√ u + u2 ) ≤ 1 20 (√ r2 + r 2 2 ) = 0.270 < ηr2 = 0.4946, implying that f(t, u) satisfies conditions (i) and (ii) of Theorem 3.9. Thus, all conditions of Theorem 3.9 are satisfied. Hence, (3.8) has at least two positive solutions u1 and u2 such that 0 < ∥u1∥ < 2 < ∥u2∥. Acknowledgement Authors acknowledge the review and editorial board for their comments and valuable suggestions. Author N. S. 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