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Axioms
, Volume 11 (1) – Dec 23, 2021

/lp/multidisciplinary-digital-publishing-institute/multiplicity-of-positive-solutions-to-nonlocal-boundary-value-problems-q6hHMCBD0l

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axioms Article Multiplicity of Positive Solutions to Nonlocal Boundary Value Problems with Strong Singularity Chan-Gyun Kim Department of Mathematics Education, Chinju National University of Education, Jinju 52673, Korea; cgkim75@cue.ac.kr Abstract: In this paper, we consider generalized Laplacian problems with nonlocal boundary condi- tions and a singular weight, which may not be integrable. The existence of two positive solutions to the given problem for parameter l belonging to some open interval is shown. Our approach is based on the ﬁxed point index theory. Keywords: generalized Laplacian problems; multiplicity of positive solutions; singular weight function; nonlocal boundary conditions 1. Introduction Consider the following singular j-Laplacian problem: 0 0 (q(t)j(u (t))) + lh(t) f (u(t)) = 0, t 2 (0, 1), (1) Z Z 1 1 u(0) = u(r)da (r), u(1) = u(r)da (r), (2) 0 0 where j : R ! R is an odd increasing homeomorphism, q 2 C([0, 1], (0, ¥)), l 2 R := [0, ¥) is a parameter, f 2 C(R ,R ), h 2 C((0, 1),R ), and the integra- + + + + Citation: Kim, C.-G. Multiplicity of tor functions a (i = 1, 2) are nondecreasing on [0, 1]. Positive Solutions to Nonlocal All integrals in (2) are meant in the sense of Riemann–Stieltjes. Throughout this paper, Boundary Value Problems with we assume the following hypotheses: Strong Singularity. Axioms 2022, 11, 7. https://doi.org/10.3390/ (F ) There exist increasing homeomorphisms y , y : [0, ¥) ! [0, ¥) such that: 1 1 2 axioms11010007 j(x)y (y) j(yx) j(x)y (y) for all x, y 2 [0, ¥). (3) 1 2 Academic Editor: Chiming Chen (F ) For i = 1, 2, a := a (1) a (0) 2 [0, 1). 2 i i i Received: 18 November 2021 Accepted: 21 December 2021 Let x : [0, ¥) ! [0, ¥) be an increasing homeomorphism. Then, we denote by H Published: 23 December 2021 the set: ( ! ) Z Z 1 Publisher’s Note: MDPI stays neutral g 2 C((0, 1),R ) : x g(t)dt ds < ¥ . 0 s with regard to jurisdictional claims in published maps and institutional afﬁl- It is well known that if (F ) is assumed, then: iations. 1 1 1 1 1 j (x)y (y) j (xy) j (x)y (y) for all x, y 2 R (4) 2 1 and Copyright: © 2021 by the authors. 1 L (0, 1)\ C(0, 1) ( H H H y j y 1 2 Licensee MDPI, Basel, Switzerland. This article is an open access article (see, e.g., ([1], Remark 1)). distributed under the terms and It is not hard to see that any function of the form conditions of the Creative Commons Attribution (CC BY) license (https:// p 2 j(s) = jsj s creativecommons.org/licenses/by/ k=1 4.0/). Axioms 2022, 11, 7. https://doi.org/10.3390/axioms11010007 https://www.mdpi.com/journal/axioms Axioms 2022, 11, 7 2 of 9 p 1 p 1 p 1 p 1 n 1 n 1 satisfies the assumption (F ) with y (s) = minfs , s g and y (s) = maxfs , s g 1 1 2 for s 2 R (see, e.g., [1,2]). Here, n 2 N, p 2 (1, ¥) for 1 k n and p p for k i j p2 1 i j n. If n = 1, it follows that j(s) = jsj s for some p 2 (1, ¥), that is, Equation (1) becomes the classical p-Laplacian one. The study of problems with nonlocal boundary conditions is motivated by a variety of applications such as beam deﬂection [3], chemical reactor theory [4], and thermostatics [5]. For this reason, the existence of positive solutions for nonlocal boundary value problems has been extensively studied. For example, Liu [6] studied the multi-point boundary value problem, which is a special case of problem (1)–(2) with l = 1. Under various assumptions of the nonlinearity f , the existence of positive solutions was shown. Bachouche, Djebali and Moussaoui [7] proved, under suitable assumptions of the nonlinearity f = f (t, u, u ) satisfying the L -Carathe ´ odory condition, several existence results for positive solutions to j-Laplacian boundary value problems involving linear bounded operators in the boundary conditions. Yang [8], by using the Avery– Peterson ﬁxed point theorem, obtained the exis- tence of at least three positive solutions to the p-Laplacian equation with integral boundary conditions. Goodrich [9] studied perturbed Volterra integral operator equations and, as an application, established the existence of at least one positive solution to the p-Laplacian differential equation with nonlocal boundary conditions. Jeong and Kim [10] obtained sufﬁcient conditions on the nonlinearity f for the existence of multiple positive solutions to problem (1)–(2) with l = 1. For the nonlinearity f = f (t, s) satisfying f (t, 0) 6 0, Kim [11] showed the existence, nonexistence and multiplicity of positive solutions to problem (1)–(2) by investigating the shape of the unbounded solution continuum. For the historical de- velopment of the theory of the problems with nonlocal boundary conditions, we refer the reader to the survey papers [12–15]. In this paper, we show the existence of two positive solutions to nonlocal boundary value problems (1)–(2) for l belonging to some open interval in the case when either f = f = ¥ or f = f = 0. Here, 0 ¥ 0 ¥ f (s) f (s) f := lim and f := lim . s!¥ s!0 j(s) j(s) For problems with zero Dirichlet boundary conditions, that is, a ˆ = a ˆ = 0, there 1 2 have been several works for problems with such assumptions on the nonlinearity f . For p2 example, when j(s) = jsj s for some p 2 (1, ¥), q 1 and h 2 H , Agarwal, Lü and O’Regan [16] investigated the existence of two positive solutions to problem (1)–(2). After that, Wang [17] obtained the same multiplicity results in [16] for generalized j-Laplacian problems with the assumptions that j satisﬁes (F ) and h 2 C[0, 1]. Recently, Lee and Xu [18] extended the result of [17] to the singularly weighed j-Laplacian problem under the assumptions that q 1 and h 2 H , that is, h may be singular at t = 0 and/or t = 1. The aim of this paper is to generalize the results for the previous papers [16–18]. The main result is stated as follows: Theorem 1. Assume that (F ), (F ) and h 2 H nf0g hold. 1 2 y (1) If f = f = ¥, then there exist l 2 (0, ¥) and m 2 (0, ¥) such that problem (1) has 0 ¥ two positive solutions u (l) and u (l) for any l 2 (0, l ). Moreover, u (l) and u (l) can 2 2 1 1 be chosen with the property that: 0 < ku (l)k < m < ku (l)k , lim ku (l)k = 0 and lim ku (l)k = ¥. ¥ 2 ¥ ¥ 2 ¥ 1 1 l!0 l!0 (2) If f = f = 0, then there exist l 2 (0, ¥) and m 2 (0, ¥) such that (1) has two positive 0 ¥ solutions u (l) and u (l) for any l 2 (l , ¥). Moreover, u (l) and u (l) can be chosen 1 2 1 2 with the property that: 0 < ku (l)k < m < ku (l)k , lim ku (l)k = 0 and lim ku (l)k = ¥. 1 ¥ 2 ¥ 1 ¥ 2 ¥ l!¥ l!¥ Axioms 2022, 11, 7 3 of 9 The rest of this paper is organized as follows. In Section 2, preliminary results which are essential for proving Theorem 1 are provided. In Section 3, the proof of Theorem 1 is given. Finally, the summary of this paper is provided in Section 4. 2. Preliminaries Throughout this section, we assume that (F ), (F ) and h 2 H n f0g hold. For 1 2 j convenience, we use some notations which were used by Jeong and Kim ([10]). The usual maximum norm in a Banach space C[0, 1] is denoted by: kuk := max ju(t)j for u 2 C[0, 1], t2[0,1] and let a := inffx 2 (0, 1) : h(x) > 0g, b := supfx 2 (0, 1) : h(x) > 0g, h h a ¯ := supfx 2 (0, 1) : h(y) > 0 for all y 2 (a , x)g, h h b := inffx 2 (0, 1) : h(y) > 0 for all y 2 (x, b )g, h h 1 1 1 2 g := (3a + a ¯ ) and g := (b + 3b ). h h h h h h 4 4 Then, since h 2 C((0, 1),R )nf0g, we have two cases, either: (i) 0 a < a ¯ b < b 1 h h h h or (ii) 0 a = b < b 1 and 0 a < a ¯ = b 1. h h h h h h Consequently, 1 2 h(t) > 0 for t 2 (a , a ¯ )[ (b , b ), and 0 a < g < g < b 1. (5) h h h h h h h h 1 2 Let r := r minfg , 1 g g 2 (0, 1), where h 1 h h 1 1 1 1 q := min q(t) > 0 and r := y y 2 (0, 1]. 0 1 2 1 kqk q t2[0,1] ¥ 0 Then 1 2 K := fu 2 C([0, 1],R ) : u(t) r kuk for t 2 [g , g ]g + ¥ h h is a cone in C[0, 1]. For r > 0, let: K := fu 2 K : kuk < rg, ¶K := fu 2 K : kuk = rg r ¥ r ¥ and K := K [ ¶K . Let r r r ( ) Z Z Z Z g g g s h h 1 h 1 1 1 C := y min y h(t)dt ds, y h(t)dt ds ; 2 2 2 kqk ¥ g s g g h h Z Z Z Z g g 1 s h h 1 1 1 C := y max A y h(t)dt ds, A y h(t)dt ds . 2 1 2 1 1 1 0 s g g h h 1 2 g + g h h 1 Here, g := and A := (1 a ˆ ) 1 for i = 1, 2. Clearly, by (5), h i i C > 0 and C > 0. 1 2 Deﬁne continuous functions f , f : R ! R by, for m 2 R , + + + f (m) := minf f (y) : r m y mg and f (m) := maxf f (y) : 0 y mg. h Axioms 2022, 11, 7 4 of 9 Deﬁne R , R : (0, ¥) ! (0, ¥) by: 1 2 1 m 1 m R (m) := j and R (m) := j for m 2 (0, ¥). 1 2 f (m) C f (m) C 1 2 1 1 1 By (4) and (F ), y (y) y (y) for all y 2 R and A = (1 a ˆ ) 1 for i = 1, 2. 2 i i 2 1 Consequently, 0 < C < C and 1 2 0 < R (m) < R (m) for all m 2 (0, ¥). (6) 2 1 L(m) Remark 2. (1) For any L 2 C(R ,R ), let L := lim for c 2 f0, ¥g. Then it is easy to + + c m!c j(m) prove that: ( f ) = ( f ) = 0 if f = 0 , and ( f ) = ( f ) = ¥ if f = ¥. (7) c c c c c c For the reader’s convenience, we give the proof for the case ( f ) = ( f ) = 0 if f = 0. ¥ ¥ ¥ The proofs for other cases are similar. Indeed, let e > 0 be given and let f = 0 be assumed. Then, there exists M > 0 such that: f (s) < e for all s M, (8) j(s) and f (s) f (M) + f (x ) for s M. M,s Here x is the point in [M, s] satisfying M,s f (x ) = maxf f (x) : M x sg. M,s By (8), for s M, f (x ) f (s) f (s) f (M) f (M) M,s 0 + + e, j(s) j(s) j(s) j(x ) j(s) M,s which implies f (s) f (s) 0 lim sup lim sup e. (9) j(s) j(s) s!¥ s!¥ Consequently, ( f ) = ( f ) = 0, since (9) is true for all e > 0. ¥ ¥ (2) By (3) and (7), for i 2 f1, 2g, lim R (m) = 0 if f = ¥, and lim R (m) = 0 if f = ¥; (10) 0 ¥ i i m!¥ m!0 lim R (m) = ¥ if f = 0, and lim R (m) = ¥ if f = 0. (11) i 0 i ¥ m!¥ m!0 For g 2 H , consider the following problem: 0 0 (q(t)j(u (t))) + g(t) = 0, t 2 (0, 1), R R (12) 1 1 u(0) = u(r)da (r), u(1) = u(r)da (r). 1 2 0 0 Deﬁne a function T : H ! C[0, 1] by T(0) = 0 and, for g 2 H nf0g, j j R R R 1 r t A I (s, s)dsda (r) + I (s, s)ds, if 0 t s, g g 1 1 0 0 0 T(g)(t) = R R R (13) 1 1 1 A I (s, s)dsda (r) I (s, s)ds, if s t 1, 2 g 2 g 0 r t where I (s, x) := j g(t)dt for s, x 2 (0, 1) q(s) and s = s(g) is a constant satisfying: Axioms 2022, 11, 7 5 of 9 Z Z Z Z Z Z 1 r s 1 1 1 A I (s, s)dsda (r) + I (s, s)ds = A I (s, s)dsda (r) I (s, s)ds. (14) 1 g 1 g 2 g 2 g 0 0 0 0 r s For any g 2 H and any s satisfying (14), T(g) is monotone increasing on [0, s) and monotone decreasing on (s, 1]. We notice that s = s(g) is not necessarily unique, but T(g) is independent of the choice of s satisfying (14) (see [10], [Remark 2]). Lemma 3. ([10], [Lemma 2]) Assume that (F ), (F ) and g 2 H hold. Then T(g) is a unique 1 2 solution to problem (12), satisfying the following properties: (i) T(g)(t) minfT(g)(0), T(g)(1)g 0 for t 2 [0, 1]; (ii) for any g 6 0, maxfT(g)(0), T(g)(1)g < kT(g)k ; (iii) s is a constant satisfying (14) if and only if T(g)(s) = kT(g)k ; (iv) T(g)(t) r minft, 1 tgkT(g)k for t 2 [0, 1] and T(g) 2 K. Deﬁne a function F : R K ! C(0, 1) by F(l, u)(t) := lh(t) f (u(t)) for (l, u) 2 R K and t 2 (0, 1). Clearly, F(l, u) 2 H for any (l, u) 2 R K, since h 2 H . Let us deﬁne an operator j + j H : R K ! K by H(l, u) := T(F(l, u)) for (l, u) 2 R K. By Lemma 3 (iv), H(R K) K, and consequently H is well deﬁned. Moreover, u is a solution to BVP (1)–(2) if and only if H(l, u) = u for some (l, u) 2 R K. Lemma 4. ([11], [Lemma 4]) Assume that (F ), (F ) and h 2 H nf0g hold. Then, the operator 1 2 H : R K ! K is completely continuous. Finally, we recall a well-known theorem of the ﬁxed point index theory. Theorem 5. ([19,20]) Assume that, for some m > 0, H : K ! K is completely continuous. Then the following assertions are true: (i) i(H,K ,K) = 1 if kH(u)k < kuk for u 2 ¶K ; m ¥ ¥ m (ii) i(H,K ,K) = 0 if kH(u)k > kuk for u 2 ¶K . m ¥ ¥ m 3. Proof of Theorem 1 In this section, we give the proof of Theorem 1. Proof of Theorem 1. (1) Since f = f = ¥, from (10), it follows that, for i = 1, 2, 0 ¥ lim R (m) = lim R (m) = 0. (15) i i m!0 m!¥ We can choose l > 0 and m > 0 satisfying: l = maxfR (m) : m 2 R g and R (m ) = l . 2 2 Let l 2 (0, l ) be ﬁxed. By (6), there exist m = m (l), m = m (l), 1 1 2 2 M = M (l), M = M (l) such that: 1 1 2 2 m < m < m < M < M 1 2 2 1 and maxfR (m ), R (M )g < l < minfR (m ), R (M )g. 1 1 1 1 2 2 2 2 Axioms 2022, 11, 7 6 of 9 Since l < R (m ), 2 2 l m m 2 2 0 l f (v(t)) l f (m ) = j < j for t 2 [0, 1]. (16) R (m ) C C 2 2 2 2 Let u 2 ¶K be given and let s be a number satisfying H(l, u)(s) = kH(l, u)k . We m ¥ have two cases: either (i) s 2 (0, g ) or (ii) s 2 [g , 1). We only give the proof for the case h h (i), since the case (ii) can be proved in a similar manner. First, we show that: kH(l, u)k A I (s, s) for s 2 [0, s]. (17) ¥ 1 F(l,u) Since I (s, x) 0 for x s and I (s, x) 0 for x s, F(l,u) F(l,u) Z Z 1 r I (s, s)dsda (r) F(l,u) 0 s Z Z Z Z s s 1 r = I (s, s)dsda (r) + I (s, s)dsda (r) 0. 1 1 F(l,u) F(l,u) 0 r s s Consequently, Z Z Z 1 r s H(l, u)(s) = A I (s, s)dsda (r) + I (s, s)ds 1 F(l,u) 1 F(l,u) 0 0 0 Z Z Z Z 1 r 1 s = A I (s, s)dsda (r) + 1 da (r) I (s, s)ds 1 1 1 F(l,u) F(l,u) 0 0 0 0 Z Z Z 1 r s = A I (s, s)dsda (r) + I (s, s)ds 1 F(l,u) 1 F(l,u) 0 s 0 A I (s, s)ds. F(l,u) From (4), (16), (17) and the deﬁnition of C , it follows that: Z Z s s kH(l, u)k A j lh(t) f (u(t))dt ds ¥ 1 q(s) 0 s Z Z g g h h 1 m < A j h(t)dt j ds q C 0 s 0 2 Z Z g g h h 1 m 1 1 A y h(t)dt dsj j q C 0 s 0 2 Z Z g g h h 1 m 1 1 A y h(t)dt dsy m = kuk . 1 2 ¥ 1 1 q C 0 s 0 2 By Theorem 5 (i), i(H(l,),K ,K) = 1. (18) 1 2 Let v 2 ¶K be given. Since l > R (m ) and r m v(t) m for t 2 [g , g ], and 1 1 h 1 1 1 h h l m m 1 1 1 2 l f (v(t)) l f (m ) = j > j for t 2 [g , g ]. (19) h h R (m ) C C 1 1 1 1 Let s be a constant satisfying H(l, v)(s) = kH(l, v)k . Then we have two cases: either (i) s 2 [g , 1) or (ii) s 2 (0, g ). We only give the proof for the case (i), since the case h h (ii) can be proved in a similar manner. By Lemma 3 (i), H(l, v)(0) 0, and it follows from (4), (19) and the deﬁnition of C that: 1 Axioms 2022, 11, 7 7 of 9 Z Z s s kH(l, v)k = H(l, v)(0) + j lh(t) f (v(t))dt ds q(s) 0 s Z Z g g h h 1 m > j h(t)dt j ds kqk C g s ¥ Z Z g g h h 1 m 1 1 y h(t)dt dsj j kqk C g s ¥ 1 Z Z g g h h 1 m 1 1 1 y h(t)dt dsy m = kvk . 1 ¥ 2 2 kqk C g s ¥ 1 By Theorem 5 (ii), i(H(l,),K ,K) = 0. (20) From (18), (20) and the additivity property, i(H(l,),K nK ,K) = 1. m m 2 1 1 1 1 Then there exists u 2 K nK such that H(l, u ) = u by the solution property. m m l 1 l l 1 1 Consequently, problem (1)–(2) has a positive solution u satisfying ku k 2 (m , m ). ¥ 1 2 l l By the similar argument above, one can show the existence of another positive solu- 2 2 tion u to problem (1) satisfying ku k 2 (M , M ). Moreover, by (15), we may choose ¥ 2 1 l l m (l), M (l) satisfying m (l) ! 0 and M (l) ! ¥ as l ! 0 , and thus (1) has two 2 2 2 2 1 2 1 2 positive solutions u , u for any l 2 (0, l ) satisfying ku k ! 0 and ku k ! ¥ as ¥ ¥ l l l l l ! 0 . (2) Since f = f = 0, from (11), it follows that, for i = 1, 2, 0 ¥ lim R (m) = lim R (m) = ¥. (21) i i m!¥ m!0 We can choose l > 0 and m > 0 satisfying l = minfR (m) : m 2 R g and R (m ) = l . 1 + 1 Let l 2 (l , ¥) be ﬁxed. By (6), there exist m = m (l), m = m (l), 1 1 2 2 M = M (l), M = M (l) such that 1 1 2 2 m < m < m < M < M 2 1 1 2 and maxfR (m ), R (M )g < l < minfR (m ), R (M )g. 1 1 1 1 2 2 2 2 By the argument similar to those in the proof of (1), i(H(l,),K nK ,K) = i(H(l,),K nK ,K) = 1. m m M M 1 2 2 1 1 2 Thus, problem (1)–(2) has two positive solutions u , u for any l 2 (l , ¥) satis- l l 1 2 fying ku k 2 (m , m ) and ku k 2 (M , M ). Moreover, by (21), we may choose ¥ 2 1 ¥ 1 2 l l m (l), M (l) satisfying m (l) ! 0 and M (l) ! ¥ as l ! ¥, and thus (1)–(2) has two 1 1 1 1 1 2 1 2 positive solutions u , u for any l 2 (l , ¥) satisfying ku k ! 0 and ku k ! ¥ as ¥ ¥ l l l l l ! ¥. 4. Conclusions In this paper, we establish the existence of two positive solutions to nonlocal boundary value problems (1)–(2) for l belonging to some open interval in the case when either f = f = ¥ or f = f = 0. 0 ¥ 0 ¥ Let j be an odd function satisfying j(x) = x + x for x 2 R . Then, j satisﬁes (F ) + 1 2 2 with y (y) = minfy, y g and y (y) = maxfy, y g. Deﬁne h : (0, 1) ! R by: 1 2 + 1 1 c 1 h(t) = 0 for t 2 [0, ] and h(t) = (t )(1 t) for t 2 ( , 1). 4 4 4 Axioms 2022, 11, 7 8 of 9 Then, since y (s) = s for all s 1, h 2 H n L (0, 1) for any c 2 [1, 2). We give some examples for nonlinearity f to illustrate the main result (Theorem 1). Let s , for s 2 [0, 1]; f (s) = and f (s) = s for s 2 R . 1 2 + s , for s 2 (1, ¥) Then, ( f ) = ( f ) = ¥ and ( f ) = ( f ) = 0. 1 0 1 ¥ 2 0 2 ¥ Consequently, by Theorem 1, problem (1)–(2) with f = f has two positive solutions for all small l > 0, and problem (1)–(2) with f = f has two positive solutions for all large l > 0. As shown in the examples of nonlinearity f = f (s) above, f (0) may be 0. What this means is that nonnegative solutions may be trivial ones. The existence of an unbounded solution component to problem (1)–(2) can be obtained as in the paper [11], where the nonlinearity f = f (t, s) satisﬁes f (t, 0) 6 0, but we cannot get any information about positive solutions from the solution component. Thus, the ﬁxed point index theory was used in order to show the existence of two positive solutions to problem (1)–(2). Funding: Not applicable. Conﬂicts of Interest: The author declares no conﬂict of interest. References 1. Jeong, J.; Kim, C.G. Existence of Positive Solutions to Singular Boundary Value Problems Involving j-Laplacian. Mathematics 2019, 7, 654. [CrossRef] 2. Karakostas, G.L. Positive solutions for the F-Laplacian when F is a sup-multiplicative-like function. Electron. J. Differ. Equ. 2004, 68, 1–12. 3. Infante, G.; Pietramala, P. A cantilever equation with nonlinear boundary conditions. Electron. J. Qual. Theory Differ. Equ. 2009, 15, 1–14. [CrossRef] 4. Infante, G.; Pietramala, P.; Tenuta, M. 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Axioms – Multidisciplinary Digital Publishing Institute

**Published: ** Dec 23, 2021

**Keywords: **generalized Laplacian problems; multiplicity of positive solutions; singular weight function; nonlocal boundary conditions

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