CUBO, A Mathematical Journal Vol. 24, no. 03, pp. 413–437, December 2022 DOI: 10.56754/0719-0646.2403.0413 Existence of positive solutions for a nonlinear semipositone boundary value problems on a time scale Saroj Panigrahi1, B Sandip Rout1 1 School of Mathematics and Statistics, University of Hyderabad, Hyderabad, 500 046, India. panigrahi2008@gmail.com B sandiprout7@gmail.com ABSTRACT In this paper, we are concerned with the existence of positive solution of the following semipositone boundary value problem on time scales: (ψ(t)y ∆ (t)) ∇ +λ1g(t, y(t))+λ2h(t, y(t)) = 0, t ∈ [ρ(c), σ(d)]T, with mixed boundary conditions αy(ρ(c)) − βψ(ρ(c))y∆(ρ(c)) = 0, γy(σ(d)) + δψ(d)y ∆ (d) = 0, where ψ : C[ρ(c), σ(d)]T, ψ(t) > 0 for all t ∈ [ρ(c), σ(d)]T; both g and h : [ρ(c), σ(d)]T × [0, ∞) → R are continuous and semipositone. We have established the existence of at least one positive solution or multiple positive solutions of the above boundary value problem by using fixed point theorem on a cone in a Banach space, when g and h are both superlinear or sublinear or one is superlinear and the other is sublinear for λi > 0; i = 1, 2 are sufficiently small. Accepted: 28 September, 2022 Received: 12 April, 2022 ©2022 S. Panigrahi 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/10.56754/0719-0646.2403.0413 https://orcid.org/0000-0003-4704-5102 https://orcid.org/0000-0002-6575-9910 mailto:panigrahi2008@gmail.com mailto:sandiprout7@gmail.com CUBO, A Mathematical Journal Vol. 24, no. 03, pp. 413–437, December 2022 DOI: 10.56754/0719-0646.2403.0413 RESUMEN En este art́ıculo estudiamos la existencia de soluciones posi- tivas del siguiente problema de valor de frontera semipositón en escalas de tiempo: (ψ(t)y ∆ (t)) ∇ +λ1g(t, y(t))+λ2h(t, y(t)) = 0, t ∈ [ρ(c), σ(d)]T, con condiciones de frontera mixtas αy(ρ(c)) − βψ(ρ(c))y∆(ρ(c)) = 0, γy(σ(d)) + δψ(d)y ∆ (d) = 0, donde ψ : C[ρ(c), σ(d)]T, ψ(t) > 0 para todo t ∈ [ρ(c), σ(d)]T; ambas g y h : [ρ(c), σ(d)]T × [0, ∞) → R son continuas y semipositón. Hemos establecido la existencia de al menos una solución positiva o múltiples soluciones positivas del problema de valor en la frontera anterior usando un teorema de punto fijo en un cono en un espacio de Banach, cuando g y h son ambas superlineales o sublineales o una es superlineal y la otra es sublineal para λi > 0; i = 1, 2 suficientemente pequeños. Keywords and Phrases: Positive solutions, boundary value problems, fixed point theorem, cone, time scales. 2020 AMS Mathematics Subject Classification: 34B15, 34B16, 34B18, 34N05, 39A10, 39A13. Accepted: 28 September, 2022 Received: 12 April, 2022 ©2022 S. Panigrahi 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/10.56754/0719-0646.2403.0413 CUBO 24, 3 (2022) Existence of positive solutions for a nonlinear semipositone... 415 1 Introduction The study of dynamic equations on time scales goes to the seminal work of Stefan Hilger [11] and has received a lot of attention in recent years. Time scales were created to unify the study of continuous and discrete mathematics and particularly used in differential and difference equations. We are interested to prove the results for a dynamic equation where the domain of the unknown function is a time scale T, which is a non-empty closed subset of real numbers R. We consider the second order semipositone boundary value problem on time scales: (ψ(t)y∆(t))∇ + λ1g(t, y(t)) + λ2h(t, y(t)) = 0, t ∈ [ρ(c), σ(d)]T, (1.1) with mixed boundary conditions αy(ρ(c)) − βψ(ρ(c))y∆(ρ(c)) = 0, γy(σ(d)) + δψ(d)y∆(d) = 0, (1.2) where λ1 and λ2 are positive and (H1) ψ : C[ρ(c), σ(d)]T, ψ(t) > 0 for all t ∈ [ρ(c), σ(d)]T; (H2) α, β, γ, δ, ≥ 0 and αδ + βγ + αγ > 0; (H3) g and h : [ρ(c), σ(d)]T × [0, ∞) → R are continuous satisfying with both g and h are semipositone. D. R. Anderson and P. Y. Wong [1], have established the existence result for the SL-BVP (1.1) and (1.2) where g is superlinear such that g(t, y) ≥ −M for some constant M > 0 and λ is in some interval of R with h(t, y) = 0. They did not establish any results concerning the existence of positive solutions for the boundary value problem (1.1) and (1.2), when g is sublinear. Many findings have also been obtained for the existence of positive solution of the boundary value problem (1.1) and (1.2), when h(t, y) = 0, but only a few results have been established for the existence of positive solutions when h(t, y) ̸= 0. Motivated by the work of [1] and the references cited therein, we would like to establish the sufficient conditions for the existence of positive solution of the boundary value problem (1.1) and (1.2), when g and h are both superlinear or sublinear or one is superlinear and the other is sublinear for λi > 0; i = 1, 2 are sufficiently small. It is worthy of mention that results of this paper not only apply to the set of real numbers or the set of integers but also to more general time scales such as T = N20 = {t2 : t ∈ N0}, T = { √ n : n ∈ N0}, etc. For basic notations and concepts on time scale calculus, we refer the readers to monographs [5, 6] and references cited therein. The study of nonlinear, semipositone boundary value problem has considerable importance even in differential equations. In recent years, several researchers studied 416 S. Panigrahi & S. Rout. CUBO 24, 3 (2022) semipositone boundary value problem on time scales [1, 2, 4, 7, 10, 16, 17]. Semipositone problems arise in many physical and chemical processes such as in chemical reactor theory, astrophysics, gas dynamics and fluidmechanics, relativistic mechanics, nuclear physics, design of suspension bridges, bulking of mechanical systems, combustion and management of natural resources (see [3, 9, 12, 15]). Let a and b such that 0 ≤ ρ(a) ≤ a < b ≤ σ(b) < ∞ and (ρ(a), σ(b))T has at least two points. The plan of the paper is as follows. In Section 2, we provided some preliminary results concerning the Green’s function for the homogeneous boundary value problem and some important Lemmas. These results allow us in Section 3 to discuss the existence of at least one or multiple positive solutions. Finally, in Section 4, we illustrate few examples to justify the results obtained in the previous section. 2 Preliminaries In this section, we have obtained some basic results related to Green’s function for the homogeneous boundary value problem and some important Lemmas. Now let us consider the homogenoeous dynamic boundary value problem (ψ(t)y∆(t))∇ = 0, t ∈ [ρ(c), σ(d)]T, (2.1) with boundary conditions (1.2). Green’s function G(t, s) (see [7]) for the boundary value problem (2.1) and with the boundary conditions (1.2) is given by G(t, s) = 1 φ   ( β + α ∫ t ρ(c) ∇τ ψ(τ) )( δ + γ ∫σ(d) s ∇τ ψ(τ) ) , ρ(c) ≤ t ≤ s ≤ σ(d), ( β + α ∫ s ρ(c) ∇τ ψ(τ) )( δ + γ ∫σ(d) t ∇τ ψ(τ) ) , ρ(a) ≤ s ≤ t ≤ σ(b), (2.2) where φ = αδ + βγ + αγ ∫ σ(d) ρ(c) ∇τ ψ(τ) > 0. Lemma 2.1 ([17]). Assume (H1) and (H2) hold. Then the Green function G(t, s) satisfies (ψ(t)y∆(t))∇ + Q(t) = 0, t ∈ (ρ(c), σ(d))T, (2.3) with mixed boundary conditions (1.2), where Q ∈ Crd[ρ(c), σ(d)]T, Q(t) ≥ 0; then y(t) ≥ q(t)∥y∥, t ∈ [ρ(c), σ(d)]T, s ∈ [a, b]T, (2.4) CUBO 24, 3 (2022) Existence of positive solutions for a nonlinear semipositone... 417 where q(t) is given by q(t) = min   β + α ∫ t ρ(c) ∇τ ψ(τ) β + α ∫σ(d) ρ(c) ∇τ ψ(τ) , δ + γ ∫σ(d) t ∇τ ψ(τ) δ + γ ∫σ(d) ρ(c) ∇τ ψ(τ)   . Lemma 2.2 ([1]). For all t ∈ [ρ(c), σ(d)]T and s ∈ [c, d]T, then q(t)G(s, s) ≤ G(t, s) ≤ G(s, s), (2.5) where G(t, s) is given in (2.2) and q(t) is defined as in Lemma 2.1. Lemma 2.3 ([1]). Let (H1) and (H2) hold and let y1 be the solution of (ψ(t)y∆(t))∇ + 1 = 0, t ∈ (ρ(c), σ(d))T, (2.6) with mixed boundary conditions (1.2), then there exists a positive constant C such that y1(t) ≤ C q(t), t ∈ [ρ(c), σ(d)]T, (2.7) where C = 1 φ (σ(d) − ρ(c)) ( β + α ∫ σ(d) ρ(c) ∇τ ψ(τ) )( δ + γ ∫ σ(d) ρ(c) ∇τ ψ(τ) ) . Lemma 2.4 ([8]). Let lim y→∞ g(t, y) y = ∞ and define G : [0, ∞) → [0, ∞) by G = max ρ(c)≤t≤σ(d), 0≤y≤r g(t, y). (2.8) Then (I) G is non-decreasing; (II) lim r→∞ G(r) r = ∞; (III) there exists r∗ > 0 such that G(r) > 0 for r ≥ r∗. Lemma 2.5 ([8]). Let limy→∞ g(t, y) y = 0 holds. Then G defined by (2.8) is a nondecreasing function, such that lim r→∞ G(r) r = 0. Define a function for y ∈ C[ρ(c), σ(d)]T, g(t, y) =   g(t, y), y ≥ 0, g(t, 0), y < 0. 418 S. Panigrahi & S. Rout. CUBO 24, 3 (2022) and h(t, y) =   h(t, y), y ≥ 0, h(t, 0), y < 0. Let us consider the nonlinear boundary value problem: (ψ(t)y∆)∇ = −[λ1g(t, y − x) + λ2h(t, y − x) + M], (2.9) with boundary conditions (1.2). Lemma 2.6. Assume that x(t) = My1(t), where y1(t) is a unique solution of the boundary value problem (2.6) and (1.2). Then y(t) is a solution of the boundary value problem (1.1) and (1.2) if and only if y(t) = y(t) + x(t) is a positive solution of the boundary value problem (2.9) and (1.2) with y(t) > x(t) for t ∈ [ρ(c), σ(d)]T. Proof. Let us assume that y(t) is a solution of the boundary value problem (2.9) and (1.2) such that y(t) ≥ x(t) for any t ∈ [ρ(c), σ(d)]T. Let y(t) = y(t) −x(t) > 0 on [ρ(c), σ(d)]T as y(t) ≥ x(t). Now, for any t ∈ [ρ(c), σ(d)]T, we have (ψ(t)y∆(t))∇ + [λ1g(t, (y(t) − x(t)) + λ2h(t, (y(t) − x(t))) + M] = 0, that is, (ψ(t)y∆(t))∇ + (ψ(t)x∆(t))∇ + [λ1g(t, (y(t) − x(t)) + λ2h(t, (y(t) − x(t))) + M] = 0. By using the definition of y together with the definition of x, we have (ψ(t)y∆(t))∇ + [λ1g(t, y(t)) + λ2h(t, y(t)) + M] + M(ψ(t)y ∆ 1 (t)) ∇(t) = 0. Thus, (ψ(t)y∆(t))∇ + λ1g(t, y(t)) + λ2h(t, y(t)) = 0. On the other hand, αy(ρ(c)) − βψ(ρ(c))y(ρ(c)) = (αy(ρ(c)) − βψ(ρ(c))y∆(ρ(c))) − (αx(ρ(c)) − βψ(ρ(c))x∆(ρ(c))) = (αy(ρ(c)) − βψ(ρ(c))y∆(ρ(c))) − M(αy1(ρ(c)) − βψ(ρ(c))y∆1 (ρ(c))) = 0, and γy(σ(d)) + δψ(d)y∆(d) = γy(σ(d)) + δψ(d)y∆(d) − (γx(σ(d)) + δψ(d)x∆(d)) CUBO 24, 3 (2022) Existence of positive solutions for a nonlinear semipositone... 419 = γy(σ(d)) + δψ(d)y∆(d) − M(γy1(σ(d)) + δψ(d)y∆1 (d)) = 0. Hence, y(t) is a solution of the boundary value problem (1.1) and (1.2). Hence this completes the proof of the lemma. Let us define a Banach space E = {y : C[ρ(c), σ(d)]T → R} endowed with the norm ∥y∥ = max{|y(t)|, t ∈ [ρ(c), σ(d)]T}. Define a cone K on E by K = {y ∈ C[ρ(c), σ(d)]T : y(t) ≥ q(t)∥y∥, t ∈ [ρ(c), σ(d)]T}, where q(t) is defined as in Lemma 2.1. Let us define an operator Tλ on K by Tλy(t) = ∫ d ρ(c) G(t, s)[λ1g(s, y(s) − x(s)) + λ2h(s, y(s) − x(s)) + M]∇s. (2.10) Lemma 2.7. Assume that (H1)–(H3) hold. Then Tλ(K) ⊂ K and Tλ : K → K is a completely continuous operator. Proof. First we show that Tλ(K) ⊂ K. Let y ∈ K and t ∈ [ρ(a), σ(b)]T. Note that (Tλy)(t) = ∫ d ρ(c) G(t, s)[λ1g(s, y(s) − x(s)) + λ2h(s, y(s) − x(s)) + M]∇s, that is, (Tλy)(t) ≤ ∫ d ρ(c) G(s, s)[λ1g(s, y(s) − x(s)) + λ2h(s, y(s) − x(s)) + M]∇s. Hence, ∥Tλy∥ ≤ ∫ d ρ(c) G(s, s)[λ1g(s, y(s) − x(s)) + λ2h(s, y(s) − x(s)) + M]∇s. By use of the Lemma (2.2), we obtain (Tλy)(t) ≥ q(t) ∫ d ρ(c) G(s, s)[λ1g(s, y(s) − x(s)) + λ2h(s, y(s) − x(s)) + M]∇s, which implies (Tλy)(t) ≥ q(t)∥Tλy∥. Thus, Tλ(K) ⊂ K. Since f and g are continuous, it shows that Tλ is continuous and by the 420 S. Panigrahi & S. Rout. CUBO 24, 3 (2022) Arzelà-Ascoli Theorem [14], it is easy to verify that Tλ is a completely continuous operator. Hence this completes the proof of the lemma Lemma 2.8 ([13]). Let E be a real Banach space, and let K ⊂ E be a cone. Let Ω1, Ω2 be two bounded open subsets of E with 0 ∈ Ω1, Ω1 ⊂ Ω2. Assume that T : K ∩ (Ω2 \ Ω1) → K be a completely continuous operator such that either ∥Ty∥ ≤ ∥y∥ for all y ∈ K ∩ ∂Ω1 and and ∥Ty∥ ≥ ∥y∥ for all y ∈ K ∩ ∂Ω2, or ∥Ty∥ ≥ ∥y∥ for all y ∈ K ∩ ∂Ω1 and ∥Ty∥ ≤ ∥y∥ for all y ∈ K ∩ ∂Ω2, then T has at least one fixed point in K ∩ (Ω2 \ Ω1). Let us define the following: (L1) lim y→∞ g(t, y) y = ∞; (L2) lim y→∞ g(t, y) y = 0; (L3) lim y→0 g(t, y) y = 0; (L4) lim y→0 g(t, y) y = ∞; (L5) lim y→∞ h(t, y) y = ∞; (L6) lim y→∞ h(t, y) y = 0; (L7) lim y→0 h(t, y) y = 0; (L8) lim y→0 h(t, y) y = ∞. Note that the limits (Li), i ∈ N81, are assumed to be inform with respect t. We would like to establish the existence of solutions for the boundary value problem (1.1) and (1.2) under the following cases: (I) L1 and L5; (II) L1 and L6; (III) L1 and L7; (IV ) L2 and L5; (V ) L2 and L6; (V I) L2 and L8; (V II) L3 and L5; (V III) L3 and L7; (IX) L3 and L8; (X) L4 and L6; (XI) L4 and L7; (XII) L4 and L8. Remark 2.9. We fails to apply the Lemma 2.8 for the pairs such as (XIII) L1 and L8, (XIV ) L2 and L7, (XV ) L3 and L6 & (XV I) L4 and L5. CUBO 24, 3 (2022) Existence of positive solutions for a nonlinear semipositone... 421 3 Main Results Theorem 3.1. Let (H1)–(H3), (L1) and (L5) hold. Then the boundary value problem (1.1) and (1.2) has a positive solution for λi, i = 1, 2 are sufficiently small. Proof. Let λ1 and λ2 satisfy 0 < λ1 + λ2 < 1 max ρ(c)≤t≤σ(d) 0≤y≤r1 g(t, y) + max ρ(c)≤t≤σ(d) 0≤y≤r1 h(t, y) , (3.1) where r1 = max{(M + 1)∥y1∥, r∗, CM}, C and r∗ are defined as in Lemma 2.3 and Lemma 2.4, respectively and y1 be the solution of (1.2) and (2.6). Define Ωr1 = {y ∈ C[ρ(c), σ(d)]T : ∥y∥ < r1}. For y ∈ K ∩ ∂Ωr1, we have (Tλy)(t) = ∫ d ρ(c) G(t, s)[λ1g(s, y − x) + λ2h(s, y − x) + M]∇s ≤ (λ1 + λ2)   max ρ(c)≤t≤σ(d) 0≤y≤r1 g(t, y) + max ρ(c)≤t≤σ(d) 0≤y≤r1 h(t, y)  ∫ d ρ(c) G(t, s)∇s + ∫ d ρ(c) G(t, s)M∇s =  (λ1 + λ2)   max ρ(c)≤t≤σ(d) 0≤y≤r1 g(t, y) + max ρ(c)≤t≤σ(d) 0≤y≤r1 h(t, y)   + M  y1(t) ≤ (1 + M)y1(t) ≤ r1 = ∥y∥. Thus, ∥Tλy∥ ≤ ∥y∥ for y ∈ K ∩ ∂Ωr1. (3.2) Let us choose a constant M > 0 such that 1 2 M(λ1 + λ2)µ ( min t1≤t≤t2 ∫ t2 t1 G(t, s)∇s ) ≥ 1, (3.3) where µ = min t1≤s≤t2 q(s). (3.4) From (L1) and (L5), we have for same M > 0 there exists a constant l > 0 such that g(t, y) ≥ My for y ∈ [l, ∞), h(t, y) ≥ My for y ∈ [l, ∞). Now set r2 = max { 2r1, 2CM, 2l1 µ } . Define Ωr2 = {y ∈ C[ρ(c), σ(d)]T : ∥y∥ < r2}. For y ∈ 422 S. Panigrahi & S. Rout. CUBO 24, 3 (2022) K ∩ ∂Ωr2, we have y(s) − x(s) = y(s) − My1(s) ≥ y(s) − MCq(s) ≥ y(s) − CM ∥y∥ y(s) ≥ y(s) − CM r2 y(s) ≥ 1 2 y(s), and min t1≤s≤t2 (y(s) − x(s)) ≥ min t1≤s≤t2 y(s) 2 ≥ min t1≤s≤t2 ∥y∥ 2 q(s) = r2µ 2 ≥ l. For y ∈ K ∩ ∂Ωr2, we have min t∈[t1, t2] (Tλy)(t) = min t∈[t1, t2] ∫ d ρ(c) G(t, s)[λ1g(s, y − x) + λ2h(s, y − x) + M]∇s ≥ min t∈[t1, t2] ∫ t2 t1 G(t, s)[λ1g(s, y − x) + λ2h(s, y − x) + M]∇s ≥ min t∈[t1, t2] ∫ t2 t1 G(t, s)(λ1 + λ2)M(y(s) − x(s))∇s ≥ min t∈[t1, t2] ∫ t2 t1 G(t, s)(λ1 + λ2)M y(s) 2 ∇s ≥ 1 2 (λ1 + λ2)Mµ min t∈[t1, t2] ∫ t2 t1 G(t, s)∥y∥∇s ≥ ∥y∥. Thus, ∥Tλy∥ ≥ ∥y∥ for y ∈ K ∩ ∂Ωr2. (3.5) By Lemma 2.8, Tλ has a fixed point y with r1 ≤ ∥y∥ ≤ r2. By use of the Lemma 2.3, it follows CUBO 24, 3 (2022) Existence of positive solutions for a nonlinear semipositone... 423 that y(t) ≥ r1q(t) ≥ r1 y1(t) C ≥ My1(t) = x(t). Hence, y = y − x is a positive solution of the boundary value problem (1.1) and (1.2). This completes the proof of the theorem. Theorem 3.2. Let (H1)–(H3), (L4) and (L8) hold. Then the boundary value problem (1.1) and (1.2) has a positive solution for λi, i = 1, 2 are sufficiently small. Proof. The proof of Theorem 3.2 is similar to that of Theorem 3.1, hence it is omitted. Theorem 3.3. Assume that (H1), (H2), (L2) and (L6) hold. Let there exist two constant D > 0 and η > 0 such that g(t, y) ≥ η for t ∈ [ρ(c), σ(d)], y ∈ [D, ∞), h(t, y) ≥ η for t ∈ [ρ(c), σ(d)], y ∈ [D, ∞), then the boundary value problem (1.1) and (1.2) has a positive solution for λi, i = 1, 2 are suffi- ciently small. Proof. Set r1 = max { 2D µ , 2MC } , (3.6) and A = 2r1 ( min t1≤t≤t2 ∫ t2 t1 G(t, s)(λ1 + λ2)η∇s )−1 , (3.7) where µ = min t1≤s≤t2 q(s). Our claim is that for λi ∈ [A, ∞), i = 1, 2, the boundary value problem (1.1) and (1.2) has a positive solution. Define Ωr1 = {y ∈ C[ρ(c), σ(d)]T : ∥y∥ < r1}. For y ∈ K ∩ ∂Ωr1, we have y(s) − x(s) = y(s) − My1(s) ≥ y(s) − MCq(s) ≥ y(s) − CM r3 y(s) ≥ 1 2 y(s), 424 S. Panigrahi & S. Rout. CUBO 24, 3 (2022) and min t1≤s≤t2 (y(s) − x(s)) ≥ min t1≤s≤t2 y(s) 2 ≥ min t1≤s≤t2 ∥y∥ 2 q(s) = r1µ 2 ≥ D. For y ∈ K ∩ ∂Ωr1, we have min t∈[t1, t2] (Tλy)(t) = min t∈[t1, t2] ∫ d ρ(c) G(t, s)[λ1g(s, y − x) + λ2h(s, y − x) + M]∇s ≥ min t∈[t1, t2] ∫ t2 t1 G(t, s)[λ1g(s, y − x) + λ2h(s, y − x) + M]∇s ≥ min t∈[t1, t2] ∫ t2 t1 G(t, s)(λ1 + λ2)η∇s = r1 = ∥y∥. Thus, ∥Tλy∥ ≥ ∥y∥ for y ∈ K ∩ ∂Ωr1. (3.8) From (L2) and (L6), we have g(t, y) ≤ ϵy for t ∈ [ρ(c), σ(d)], y ≥ l, h(t, y) ≤ ϵy for t ∈ [ρ(c), σ(d)], y ≥ l, On the other hand, by use of the Lemma 2.4, there exists a R > 0 such that R > max { 2r1, max ρ(c)≤t≤σ(d) ∫ d ρ(c) [G(t, s)M + 1]∇s } . and ϵ satisfies max ρ(c)≤t≤σ(d) ∫ d ρ(c) G(t, s)[ϵλ1R + ϵλ2R + M]∇s ≤ R. Let ΩR = {y ∈ C[ρ(c), σ(d)]T : ∥y∥ < R}. CUBO 24, 3 (2022) Existence of positive solutions for a nonlinear semipositone... 425 For y ∈ K ∩ ∂ΩR, we have Tλy(t) = ∫ d ρ(c) G(t, s)[λ1g(s, y − x) + λ2h(s, y − x) + M]∇s ≤ ∫ d ρ(c) G(t, s)[λ1ϵR + ϵλ2R + M]∇s ≤ R = ∥y∥. Thus, ∥Tλy∥ ≤ ∥y∥ for y ∈ K ∩ ∂ΩR. (3.9) By Lemma 2.8, Tλ has a fixed point y with r1 ≤ ∥y∥ ≤ R. It follows that y(t) ≥ r1q(t) ≥ r1 y1(t) C ≥ 2My1(t) ≥ x(t). Hence, y = y − x is a positive solution of the boundary value problem (1.1) and (1.2). This completes the proof of the theorem. Theorem 3.4. Assume that (H1)–(H3), (L3) and (L7) hold. Let there exist two constant D > 0 and η > 0 such that g(t, y) ≥ η for t ∈ [ρ(c), σ(d)]y ∈ [D, l], h(t, y) ≥ η for t ∈ [ρ(c), σ(d)]y ∈ [D, l], then the boundary value problem (1.1) and (1.2) has a positive solution for λi, i = 1, 2 are suffi- ciently small. Proof. The proof of the Theorem 3.4 is similar to that of Theorem 3.3, hence it is omitted. Theorem 3.5. Let (H1)–(H3), (L1) and (L6) hold. Then the boundary value problem (1.1) and (1.2) has at least two positive solutions for λi, 1 = 1, 2 are sufficiently small. Proof. If (L6) holds, then by the Lemma 2.5, there exists a constant r1 > 0 such that G(r1) ≤ Nr1. 426 S. Panigrahi & S. Rout. CUBO 24, 3 (2022) Since λ1 and λ2 are sufficiently small, we have λ1 max ρ(c)≤t≤σ(d) 0≤y≤r1 g(t, y) + λ2 max ρ(c)≤t≤σ(d) 0≤y≤r1 h(t, y) + M  ∫ d ρ(c) G(s, s)∇s ≤ r1. Let Ωr1 = {y ∈ C[ρ(c), σ(d)]T : ∥y∥ < r1}. For y ∈ ∂Ωr1, we have (Tλy)(t) = ∫ d ρ(c) G(t, s) [ λ1g(s, y − x) + λ2h(s, y − x) + M ] ∇s ≤  λ1 max ρ(c)≤t≤σ(d) 0≤y≤r1 g(t, y) + λ2 max ρ(c)≤t≤σ(d) 0≤y≤r1 h(t, y)  ∫ d ρ(c) G(s, s)∇s + ∫ d ρ(c) G(s, s)M∇s ≤ r1 = ∥y∥. Thus, ∥Tλy∥ ≤ ∥y∥ for all y ∈ K ∩ ∂Ωr1. (3.10) From (L1), we have g(t, y) > N1y for all y ≤ l. Let r2 = max { 2CM, 2l µ ,2r1 } and Ωr2 = {y ∈ C[ρ(c), σ(d)]T : ∥y∥ < r2}. For y ∈ ∂K ∩ Ωr2, we have y(s) − x(s) = y(s) − My1(s) ≥ y(s) − MCq(s) ≥ y(s) − CM r5 y(s) ≥ 1 2 y(s), and min t1≤s≤t2 (y(s) − x(s)) ≥ min t1≤s≤t2 y(s) 2 ≥ min t1≤s≤t2 ∥y∥ 2 q(s) = r2µ 2 ≥ l. CUBO 24, 3 (2022) Existence of positive solutions for a nonlinear semipositone... 427 For y ∈ K ∩ ∂Ωr2, we have min t∈[t1, t2] (Tλy)(t) = min t∈[t1, t2] ∫ d ρ(c) G(t, s)[λ1g(s, y − x) + λ2h(s, y − x) + M]∇s ≥ min t∈[t1, t2] ∫ t2 t1 G(t, s)[λ1g(s, y − x) + λ2h(s, y − x) + M]∇s ≥ min t∈[t1, t2] ∫ t2 t1 G(t, s)λ1N1(y(s) − x(s))∇s = r2 = ∥y∥. Thus, ∥Tλy∥ ≥ ∥y∥ for all y ∈ K ∩ ∂Ωr2. (3.11) Let R = max    λ1 max ρ(c)≤t≤σ(d) 0≤y≤R g(t, y) + λ2 max ρ(c)≤t≤σ(d) 0≤y≤R h(t, y) + M  (∫ d ρ(c) G(s, s)∇s ) ,2r2   , then r1 < r2 < R. Let ΩR = {y ∈ C[ρ(c), σ(d)]T : ∥y∥ < R}. For y ∈ K ∩ ΩR, t ∈ [ρ(c), σ(d)]T, we have (Tλy)(t) = ∫ d ρ(c) G(t, s) [ λ1g(s, y − x) + λ2h(s, y − x) + M ] ∇s ≤  λ1 max ρ(c)≤t≤σ(d) 0≤y≤R g(t, y) + λ2 max ρ(c)≤t≤σ(d) 0≤y≤R h(t, y) + M  ∫ d ρ(c) G(s, s)∇s ≤ R = ∥y∥. Thus, ∥Tλy∥ ≤ ∥y∥ for all y ∈ K ∩ ∂ΩR. (3.12) Thus by the Lemma 2.8, Tλ has at least two fixed points. Hence, the boundary value problem (1.1) and (1.2) has at least two positive solutions. Theorem 3.6. Let (H1)–(H3), (L2) and (L5) hold. Then the boundary value problem (1.1) and (1.2) has at least two positive solutions for λi, i = 1, 2 are sufficiently small. Proof. The proof of the Theorem 3.6 is similar to that of Theorem 3.5. Theorem 3.7. Let (H1)–(H3), (L4) and (L7) hold. Then the boundary value problem (1.1) and (1.2) has at least two positive solutions for λi, i = 1, 2 are sufficiently small. 428 S. Panigrahi & S. Rout. CUBO 24, 3 (2022) Proof. From (L7), we have lim y→0 h(t, y) y = 0. For ϵ > 0, there exists a r1 > 0 such that h(t, y) ≤ ϵy for y ∈ [0, r1). Since λ1 and λ2 are sufficiently small, we have (λ1 + λ2)   max ρ(c)≤t≤σ(d) 0≤y≤r1 g(t, y) + max ρ(c)≤t≤σ(d) 0≤y≤r1 h(t, y)   + M  ∫ d ρ(c) G(s, s)∇s ≤ r1. Let Ωr1 = {y ∈ C[ρ(c), σ(d)]T : ∥y∥ < r1}. For y ∈ ∂Ωr1, we have (Tλy)(t) = ∫ d ρ(c) G(t, s) [ λ1g(s, y − x) + λ2h(s, y − x) + M ] ∇s ≤ ∫ d ρ(c) G(s, s) ( λ1g(s, y − x) + λ2h(s, y − x) ) ∇s + ∫ d ρ(c) G(s, s)M∇s ≤  (λ1 + λ2)   max ρ(c)≤t≤σ(d) 0≤y≤r1 g(t, y) + max ρ(c)≤t≤σ(d) 0≤y≤r1 h(t, y)   + M  ∫ d ρ(c) G(s, s)∇s ≤ r1 = ∥y∥ Thus, ∥Tλy∥ ≤ ∥y∥ for all y ∈ K ∩ ∂Ωr1. (3.13) From (L4), we have g(t, y) > N1y for all y ≤ l. Let r2 = max { 2CM, 2l µ ,2r1 } and Ωr2 = {y ∈ C[ρ(c), σ(d)]T : ∥y∥ < r2}. For y ∈ K ∩ ∂Ωr2, we have y(s) − x(s) = y(s) − My1(s) ≥ y(s) − MCq(s) ≥ y(s) − CM r5 y(s) ≥ 1 2 y(s), CUBO 24, 3 (2022) Existence of positive solutions for a nonlinear semipositone... 429 and min t1≤s≤t2 (y(s) − x(s)) ≥ min t1≤s≤t2 y(s) 2 ≥ min t1≤s≤t2 ∥y∥ 2 q(s) = r2µ 2 ≥ l. For y ∈ K ∩ ∂Ωr2, we have min t∈[t1, t2] (Tλy)(t) = min t∈[t1, t2] ∫ d ρ(c) G(t, s) [ λ1g(s, y − x) + λ2h(s, y − x) + M ] ∇s ≥ min t∈[t1, t2] ∫ t2 t1 G(t, s) [ λ1g(s, y − x) + λ2h(s, y − x) + M ] ∇s ≥ min t∈[t1, t2] ∫ t2 t1 G(t, s)λ1N1(y(s) − x(s))∇s = r2 = ∥y∥. Thus, ∥Tλy∥ ≥ ∥y∥ for all y ∈ K ∩ ∂Ωr2. (3.14) Let R = max    λ1 max ρ(c)≤t≤σ(d) 0≤y≤R g(t, y) + λ2 max ρ(c)≤t≤σ(d) 0≤y≤R h(t, y) + M  (∫ d ρ(c) G(s, s)∇s ) ,2r2   , then r1 < r2 < R. Let ΩR = {y ∈ C[ρ(c), σ(d)]T : ∥y∥ < R}. For y ∈ K ∩ ΩR, t ∈ [ρ(c), σ(d)]T, we have (Tλy)(t) = ∫ d ρ(c) G(t, s) [ λ1g(s, y − x) + λ2h(s, y − x) + M ] ∇s ≤  (λ1 + λ2)   max ρ(c)≤t≤σ(d) 0≤y≤R g(t, y) + max ρ(c)≤t≤σ(d) 0≤y≤R h(t, y)   + M  ∫ d ρ(c) G(s, s)∇s ≤ R = ∥y∥ Thus, ∥Tλy∥ ≤ ∥y∥ for all y ∈ K ∩ ∂ΩR. (3.15) Thus by the Lemma 2.8, Tλ has at least two fixed points. Hence, the boundary value problem (1.1) and (1.2) has at least two positive solutions. 430 S. Panigrahi & S. Rout. CUBO 24, 3 (2022) Theorem 3.8. Let (H1)–(H3), (L3) and (L8) hold. Then the boundary value problem (1.1) and (1.2) has at least two positive solutions for λi, i = 1, 2 are sufficiently small. Proof. The proof of the Theorem 3.8 is similar to that of Theorem 3.5. Theorem 3.9. Let (H1)–(H3), (L1) and (L7) hold. Then the boundary value problem (1.1) and (1.2) has at least one positive solution for λi, i = 1, 2 are sufficiently small. Proof. From (L1), we have lim y→∞ g(t, y) y = ∞. For k > 0, there exists a r1 > 0 such that g(t, y) ≥ ky for y > r1. Let Ωr1 = {y ∈ C[ρ(c), σ(d)]T : ∥y∥ < r1} and let k satisfy kµ 2 λ1 min t∈[t1, t2] ∫ t2 t1 G(t, s)∇s ≥ 1. For y ∈ K ∩ ∂Ωr1, we have min t∈[t1, t2] (Tλy)(t) = min t∈[t1, t2] ∫ d ρ(c) G(t, s) [ λ1g(s, y − x) + λ2h(s, y − x) + M ] ∇s ≥ min t∈[t1, t2] ∫ t2 t1 G(t, s)λ1k(y − x)∇s ≥ k 2 λ1 min t∈[t1, t2] ∫ t2 t1 G(t, s)∥y∥q(s)∇s ≥ ∥y∥. Thus, ∥Tλy∥ ≥ ∥y∥ for y ∈ K ∩ ∂Ωr1. (3.16) From (L7), we have lim y→0 h(t, y) y = 0. For ϵ > 0, there exists a r2 > 0 such that h(t, y) ≤ ϵy for y ∈ [0, ∞). CUBO 24, 3 (2022) Existence of positive solutions for a nonlinear semipositone... 431 Since λ1 and λ2 are sufficiently small, let (λ1 + λ2)   max ρ(c)≤t≤σ(d) 0≤y≤r2 g(t, y) + max ρ(c)≤t≤σ(d) 0≤y≤r2 h(t, y) + M    ∫ d ρ(c) G(s, s)∇s ≤ r2. Let Ωr2 = {y ∈ C[ρ(c), σ(d)]T : ∥y∥ < r2}. Now for any y ∈ K ∩ ∂Ωr2, we have (Tλy)(t) = ∫ d ρ(c) G(t, s) [ λ1g(t, y − x) + λ2h(s, y − x) + M ] ∇s ≤ ∫ d ρ(c) G(s, s) [ λ1g(t, y − x) + λ2h(s, y − x) + M ] ∇s ≤  (λ1 + λ2)   max ρ(c)≤t≤σ(d) 0≤y≤r2 g(t, y) + max ρ(c)≤t≤σ(d) 0≤y≤r2 h(t, y)   + M  ∫ d ρ(c) G(s, s)∇s ≤ r2 = ∥y∥. Thus, ∥Tλy∥ ≤ ∥y∥ for y ∈ K ∩ ∂Ωr2. (3.17) Hence, by the Lemma 2.8, Tλ has a fixed point y with r1 < ∥y∥ < r2. By the Lemma 2.6, the boundary value problem (1.1) and (1.2) has at least one positive solution. Theorem 3.10. Let (H1)–(H3), (L3) and (L5) hold. Then the boundary value problem (1.1) and (1.2) has at least one positive solution for λi, i = 1, 2 are sufficiently small. Proof. The proof of the Theorem 3.10 is similar to that of Theorem 3.9. Theorem 3.11. Let (H1)–(H3), (L2) and (L8) hold. Then the boundary value problem (1.1) and (1.2) has at least one positive solution for λi, i = 1, 2 are sufficiently small. Proof. From (L2), we have lim y→∞ g(t, y) y = 0. By Lemma 2.5, there exist r1 > 0 and k1 > 0 such that G(r1) ≤ k1r1. Let Ωr1 = {y ∈ C[ρ(c), σ(d)]T : ∥y∥ < r1}. Since λ1 and λ2 are sufficiently small, we have (λ1 + λ2)   max ρ(c)≤t≤σ(d) 0≤y≤r1 g(t, y) + max ρ(c)≤t≤σ(d) 0≤y≤r1 h(t, y)   + M  ∫ d ρ(c) G(s, s)∇s ≤ r1. 432 S. Panigrahi & S. Rout. CUBO 24, 3 (2022) For any y ∈ K ∩ ∂Ωr1, we obtain (Tλy)(t) = ∫ d ρ(c) G(t, s) [ λ1g(t, y − x) + λ2h(s, y − x) + M ] ∇s ≤ ∫ d ρ(c) G(s, s) [ λ1g(t, y − x) + λ2h(s, y − x) + M ] ∇s ≤  (λ1 + λ2)   max ρ(c)≤t≤σ(d) 0≤y≤r2 g(t, y) + max ρ(c)≤t≤σ(d) 0≤y≤r2 h(t, y)   + M  ∫ d ρ(c) G(s, s)∇s ≤ r1 = ∥y∥. Thus, ∥Tλy∥ ≤ ∥y∥ for y ∈ K ∩ ∂Ωr1. (3.18) From (L8), we have lim y→0 h(t, y) y = ∞. For k > 0, there exists a l > 0 such that h(t, y) ≥ ky for y ∈ [0, l]. Let r2 = { 2cm, 2l µ ,2r1 } and Ωr2 = {y ∈ C[ρ(c), σ(d)]T : ∥y∥ < r2}. For any y ∈ K ∩ ∂Ωr2, we have min t∈[t1, t2] (Tλy)(t) = min t∈[t1, t2] ∫ d ρ(c) G(t, s) [ λ1g(t, y − x) + λ2h(s, y − x) + M ] ∇s ≥ min t∈[t1, t2] ∫ t2 t1 G(t, s)λ2k(y(s) − x(s))∇s ≥ k 2 λ2µ min t∈[t1, t2] ∫ t2 t1 G(t, s)∥y∥∇s ≥ r2 = ∥y∥. Thus, ∥Tλy∥ ≥ ∥y∥ for y ∈ K ∩ ∂Ωr2. (3.19) Hence, by the Lemma 2.8, Tλ has a fixed point y with r1 < ∥y∥ < r2. By the Lemma 2.6, the boundary value problem (1.1) and (1.2) has at least one positive solution. Theorem 3.12. Let (H1)–(H3), (L4) and (L6) hold. Then the boundary value problem (1.1) and (1.2) has at least one positive solution for λi, , i = 1, 2 are sufficiently small. Proof. The proof of the Theorem 3.12 is similar to that of Theorem 3.11. CUBO 24, 3 (2022) Existence of positive solutions for a nonlinear semipositone... 433 4 Examples We shall illustrate few examples in different time scales to justify the results obtained in the preceding section. Example 4.1. Let us consider the following boundary value problem on time scale T = R, ((1 + t2)y′)′ + 1 2 1 + y2 52 + 1 4 y2 sin2 y 35 = 0, t ∈ [0, 1], (4.1) with boundary conditions y(0) − y′(0) = 0, y(1) + 2y′(1) = 0, (4.2) where ψ(t) = 1+t2, M = 1, α, β, γ, δ ≥ 0, g(t, y) = 1+y 2 52 and h(t, y) = y 2 sin2 y 35 . Green’s function for the boundary value problem (4.1) and (4.2) is given by G(t, s) = 1 2 + π 4   ( 1 + tan−1 t )( 1 + π 4 − tan−1 s ) , t ≤ s,( 1 + tan−1 s )( 1 + π 4 − tan−1 t ) , s ≤ t. All the conditions (H1)–(H3), (L1) and (L5) are satishfied for (t, y) ∈ [0, 1]×[0, 100]. By Theorem 3.1, boundary value problem (4.1) and (4.2) has at least one positive solution for λ1 = 1 2 and λ2 = 1 4 . Example 4.2. Let us consider the following boundary value problem on time scale T = Z, ∇((1 + t)−1y∆) + λ1 sin2 y + λ2 √ y cos y = 0, t ∈ [0, 3], (4.3) with boundary conditions y(0) − ∆y(0) = 0, y(3) + 1 3 ∆y(2) = 0, (4.4) where ψ(t) = (1 + t)−1, M = 1, α, β, γ, δ ≥ 0, g(t, y) = sin2 y and h(t, y) = √ y cos y. Green’s function for the boundary value problem (4.3) and (4.4) is given by G(t, s) = 1 11   ( 1 + t 2+3t 2 )( 1 + (3−s)(s+6) 2 ) , t ≤ s,( 1 + s 2+3s 2 )( 1 + (3−t)(t+6) 2 ) , s ≤ t. All the conditions (H1)–(H3), (L2) and (L6) are satishfied for (t, y) ∈ [0, 3] × [0, 100]. Let D = 1 434 S. Panigrahi & S. Rout. CUBO 24, 3 (2022) and η = 1 2 such that g(t, y) ≥ 1 2 and h(t, y) ≥ 1 2 for t ∈ [0, 3], y ∈ [1, ∞). By Theorem 3.3, boundary value problem (4.3) and (4.4) has at least one positive solution for λi; i = 1, 2 are sufficiently small. Example 4.3. Consider the boundary value problem on time scale T = qZ = {2k : k ∈ Z} ∪ {0}, where q = 2 > 1, Dq ( (1 + t)−1Dqy(t) ) + λ1 y2 sin y + λ2 ln(y) = 0, t ∈ [0, 2], (4.5) with boundary conditions y(0) − Dqy(0) = 0, y(2) + 1 2 Dqy(1) = 0, (4.6) where ψ(t) = (1+t)−1, M = 1, α, β, γ, δ ≥ 0, g(t, y) = y 2 sin y and h(t, y) = ln(y). Green’s function for the boundary value problem (4.5) and (4.6) is given by G(t, s) = 3 20   ( 2t2+3t+3 3 )( 17−3s−2s2 3 ) , t ≤ s,( 2s2+3s+3 3 )( 17−3t−2t2 3 ) , s ≤ t. The conditions (H1)–(H3), (L1) and (L6) are satishfied for (t, y) ∈ [0, 2] × [0, 100]. By Theorem 3.5, boundary value problem (4.5) and (4.6) has at least two positive solutions for λi; i = 1, 2 are sufficiently small. Example 4.4. Let us consider the time scale T = Pa,b = ∞⋃ k=0 [k(a + b),k(a + b) + a] = P1, 1 = ∞⋃ k=0 [2k,2k + 1], where a = b = 1. Consider the following boundary value problem:   y∆∇ + λ1 √ y + λ2 y ln(1 + y), t ∈ (0, 2), y(0) = 0, y(2) = 0, (4.7) where ψ(t) = 1, M = 1, α, β, γ, δ ≥ 0, g(t, y) = √ y and h(t, y) = y ln(y). Green’s function for the boundary value problem (4.7) is given by G(t, s) = 1 2   t(1 − s), t ≤ s, (1 + s)(2 − t), s ≤ t. The conditions (H1)–(H3), (L4) and (L7) are satishfied for (t, y) ∈ [0, 2] × [0, 100]. By Theorem 3.7, boundary value problem (4.7) has at least two positive solutions for λi; i = 1, 2 are sufficiently small. CUBO 24, 3 (2022) Existence of positive solutions for a nonlinear semipositone... 435 Example 4.5. Consider the boundary value problem on time scale T = {n 2 : t ∈ N0}: y∆∇(t) + λ1y ln(1 + y) + λ2 √ y siny 6 = 0, t ∈ [ 0, 3 2 ] , (4.8) with boundary conditions y(0) − y∆(0) = 0, y ( 3 2 ) + y∆(1) = 0, (4.9) where ψ(t) = 1, M = 1, α, β, γ, δ ≥ 0, g(t, y) = y ln(1+y) and h(t, y) = √ y sin y 6 . Green’s function for the boundary value problem (4.8) and (4.9) is given by G(t, s) = 2 7   (1 + s) ( 5 2 − t ) , t ≤ s, (1 + t) ( 5 2 − s ) , s ≤ t. The conditions (H1)–(H3), (L1) and (L7) are satisfied for (t, y) ∈ [0, 32] × [0, 100]. By Theorem 3.9, boundary value problem (4.8) and (4.9) has at least one positive solutions for λi; i = 1, 2 are sufficiently small. Example 4.6. Consider the following boundary value problem in time scale T = hZ = {hk : k ∈ Z}, where h = 1 2 > 0, ( (1 + t)−1y∆ )∇ + λ1 √ y siny + λ2 = 0 for t ∈ [0, 2], (4.10) with boundary conditions y(0) − y∆(0) = 0, y(2) + 2 5 y∆ ( 3 2 ) = 0, (4.11) where ψ(t) = (1+t)−1, M = 1, α, β, γ, δ ≥ 0, g(t, y) = √ y siny and h(t, y) = 1. Green’s function for the boundary value problem (4.10) and (4.11) is given by G(t, s) = 2 13   ( 1 + s(2s+5) 4 )( 1 + (2−t)(2t+9) 4 ) , s ≤ t,( 1 + t(2t+5) 4 )( 1 + (2−s)(2s+9) 4 ) , t ≤ s. The conditions (H1)–(H3), (L2) and (L8) are satishfied for (t, y) ∈ [0, 2] × [0, 100]. 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Modelling, vol. 52, no. 3–4, pp. 481–489, Introduction Preliminaries Main Results Examples