Electronic Journal of Differential Equations, Vol. 2025 (2025), No. 66, pp. 1–15. ISSN: 1072-6691. URL: https://ejde.math.txstate.edu, https://ejde.math.unt.edu DOI: 10.58997/ejde.2025.66 SOLUTIONS TO NONLINEAR ELLIPTIC PROBLEMS WITH NONHOMOGENEOUS OPERATORS AND MIXED NONLOCAL BOUNDARY CONDITIONS EUN KYOUNG LEE, INBO SIM, BYUNGJAE SON Abstract. We investigate the existence, multiplicity and nonexistence of positive solutions to nonlinear (singular) elliptic problems involving nonhomogeneous operators and mixed nonlocal boundary conditions based on the behaviors of the nonlinear term near 0 and ∞. In particular, we discuss the existence of at least three positive solutions to the mixed nonlocal boundary problems, which is new finding even for the problems involving homogeneous operators. The novelty of this study lies in constructing completely continuous operators related to nonlinear elliptic problems involving complicated boundary conditions. We emphasize that only one fixed point theorem is used to obtain the existence and multiplicity results, despite generalizing and extending most of the problems in previous literature. 1. Introduction and main results We consider the nonlinear (singular) elliptic problems with nonhomogeneous operators and mixed nonlocal boundary conditions: −(w(t)ϕ(u′))′ = λh(t)f(u), t ∈ (0, 1), u(0)− au′(0) = ∫ 1 0 g0(s)k 0 0(s, u)ds+ m∑ i=1 αik 0 i (ζi, u(ζi)), u(1) + bu′(1) = ∫ 1 0 g1(s)k 1 0(s, u)ds+ n∑ j=1 βjk 1 j (ξj , u(ξj)), (1.1) where λ > 0, a ≥ 0, b ≥ 0, 0 ≤ αi < 1, 0 ≤ βj < 1, m ∈ N, n ∈ N, and {ζi}mi=1 and {ξj}nj=1 are increasing sequences in (0, 1). Here ϕ, f , w, h, g0, g1, k 0 i and k1j satisfy the following conditions: (H1) ϕ ∈ C(R,R) is an odd increasing homeomorphism such that there exists ψ ∈ C((0,∞), (0,∞)) such that ψ(0) = 0 and ϕ(rs) ≤ ψ(r)ϕ(s) for r > 0 and s > 0, (H2) f ∈ C((0,∞), (0,∞)) with lim infs→∞ f(s) > 0 and there exist c > 0 and γ ≥ 0 such that f(s) < c sγ for 0 < s < 1, (H3) w ∈ C([0, 1], (0,∞)) and h ∈ C((0, 1), (0,∞)) with ∫ 1 0 h(r) d(r)γ dr < ∞, where d(r) := min{r, 1− r}, (H4) g0, g1 ∈ C((0, 1), [0,∞)) with G := max{∥g0∥1 + ∑m i=1 αi, ∥g1∥1 + ∑n j=1 βj} < 1, (H5) k0i , k 1 j ∈ C((0, 1) × [0,∞), [0,∞)) are such that k0i (s, r) ≤ r and k1j (s, r) ≤ r, where i ∈ {0, 1, 2, . . . ,m} and j ∈ {0, 1, 2, . . . , n}. Nonlocal boundary value problems of ordinary differential equations arise in various areas of applied mathematics and physics. In particular, multipoint boundary value problems arise in a fluid flow problem [11] and the theory of elastic stability [20], and integral boundary value problems 2020 Mathematics Subject Classification. 34B10, 34B16, 34B18. Key words and phrases. Singular elliptic problem; nonhomogeneous operator; integral boundary condition; multipoint boundary condition; positive solution. ©2025. This work is licensed under a CC BY 4.0 license. Submitted February 20, 2025. Published June 30, 2025. 1 2 E. K. LEE, I. SIM, B. SON EJDE-2025/66 arise in blood flow problems [17, 24] and thermal conduction problems [5, 13]. Recently, many studies have been conducted on these two nonlocal boundary value problems. One can find several works on multipoint and integral boundary value problems in a series of papers [6, 7, 15, 16, 21] and [1, 3, 4, 12, 14, 22, 23], respectively. In [16], Ma studied the existence of positive solutions of (1.1) with the homogeneous operator u 7→ (w(t)u′)′ (i.e., ϕ(s) = s), a nonsingular nonlinear term f and the following m-point boundary condition au(0)− bw(0)u′(0) = m−2∑ i=1 αiu(ζi) and cu(1) + dw(1)u′(1) = m−2∑ i=1 βiu(ζi) with ac + ad + bc > 0. They used the Guo-Krasnoselskii fixed point theorem in a cone to get the result. Webb and Infante [22, 23] provided a unified method of establishing the existence and multiplicity of positive solutions of (1.1) with the homogeneous operator u 7→ u′′ (i.e., w(t) ≡ 1 and ϕ(s) = s), a nonsingular nonlinear term f and various nonlocal boundary conditions involving Stieltjes integrals (thus allowing for m-point and integral boundary conditions). But one can see that our mixed nonlocal boundary conditions contain their boundary conditions. Recently, Hai and Wang [10] addressed the problem (1.1) with the homogeneous operator u 7→ (|u′|p−2u′)′ (i.e., w(t) ≡ 1 and ϕ(s) = |s|p−2s), a singular nonlinear term f and the following boundary conditions: au(0)− bu′(0) = ∫ 1 0 g(s)u(s)ds and u′(1) = 0, or au(0)− bu′(0) = ∫ 1 0 g(s)u(s)ds and u(1) = 0, with a > 0 and b ≥ 0. Assuming that the nonlinear term f(s) could have negative values near 0, they showed the existence of positive solutions for the cases lims→∞ f(s) ϕ(s) = 0 and lims→∞ f(s) ϕ(s) = ∞ by applying the Krasnoselskii fixed point theorem in a Banach space. We note that they did not discuss multiplicity results. The problem involving both multipoint and integral boundary conditions simultaneously was initially introduced in [2] as follows: −u′′(t) = f(t, u), t ∈ (T1, T2), α1u(T1) + α2u(T2) = α3 ∫ ζ T1 u(s)ds+ m∑ i=1 γiu(νi), β1u ′(T1) + β2u ′(T2) = β3 ∫ ζ T1 u′(s)ds+ m∑ i=1 ρiu ′(νi), where 0 < T1 < ζ ≤ νi ≤ T2, αi, βi ∈ R, and γi, ρi ∈ (0,∞). The existence of solutions (which may not be positive solutions) was discussed via three different fixed point theorems: Schaefer, Krasnoselskii and Leray-Schauder. Motivated by the aforementioned studies, we extend these results to the more general case (1.1), which has a nonhomogeneous operator, a (non)singular nonlinear term, and mixed multipoint and integral boundary conditions. Our objective is to study the existence, multiplicity and nonexis- tence of positive solutions of (1.1) in C1[0, 1] according to the behaviors of f near 0 and ∞, that is, the values of f0 := lims→0 f(s) ϕ(s) and f∞ := lims→∞ f(s) ϕ(s) . In particular, to discuss the existence of three positive solutions, we assume (H6) f(s) := fγ(s) sγ , where fγ is continuous and nondecreasing, (H7) there exist η > 0 and θ > 0 such that f(θ) ϕ(θ) / f(η) ϕ(η) > 4γw∗∥hγ∥1ψ( 16δ )ψ( 1+a 1−G ) w∗h∗δ2γ , and one or two of the following: (H8a) η < 4θ δ , (H8b) η > 4θ δ , EJDE-2025/66 SOLUTIONS TO NONLINEAR ELLIPTIC PROBLEMS 3 (H9a) f(θ) ϕ(θ) > 21+2γw∗∥h∥1ψ( 16 δ )ψ( 1+a 1−G )min{f0,f∞} w∗h∗δγ , (H9b) f(η) ϕ(η) < w∗h∗δ γ max{f0,f∞} 2w∗∥hγ∥1ψ( 16 δ )ψ( 1+a 1−G ) , (H10a) max{f0,f∞} min{f0,f∞} > 4w∗∥h∥1ψ( 16 δ )ψ( 1+a 1−G ) w∗h∗ (> 1), (H10b) f0 > f∞, (H8a), (H9a) and (H9b), (H10c) f0 < f∞, (H8b), (H9a) and (H9b), where w∗ := mint∈[0,1] w(t), w ∗ := maxt∈[0,1] w(t), ∥hγ∥1 := ∫ 1 0 h(r) d(r)γ dr, h∗ := min {∫ 1 2 1 4 h(r)dr, ∫ 3 4 1 2 h(r)dr } , and δ is the largest constant such that maxs∈[0,δ] ψ(s) ≤ w∗ 2w∗ . We note that ψ(1) ≥ 1 since ϕ(1) ≤ ψ(1)ϕ(1) by (H1). This implies δ < 1. Noting that f0 = ∞ implies f(s) could be singular at 0, we state theorems according to the following value of f0: f0 = ∞, f0 = 0 and f0 ∈ (0,∞). 1. Case f0 = ∞. Let λ∗ := 4γw∗ψ( 16 δ )ϕ(θ) h∗δγf(θ) , λ∗ := w∗δ γϕ(η) ∥hγ∥1ψ( 1+a 1−G )f(η) and λ∞ := w∗ 2∥h∥1ψ( 1+a 1−G )f∞ . We establish the following results. Theorem 1.1. Assume (H1)–(H5), f0 = ∞ and f∞ = ∞. Then (1.1) has no positive solution for λ≫ 1 and has two positive solutions u1 and u2 for λ ≈ 0 such that ∥u1∥∞ → 0 and ∥u2∥∞ → ∞ as λ→ 0. Theorem 1.2. Assume (H1)–(H5), f0 = ∞ and f∞ = 0. Then (1.1) has a positive solution u for λ > 0 such that ∥u∥∞ → 0 as λ → 0 and ∥u∥∞ → ∞ as λ → ∞. In addition, if (H6), (H7), (H8a) are satisfied, then (1.1) has three positive solutions u1, u2 and u3 for λ ∈ (λ∗, λ ∗) such that ∥u1∥∞ < η < ∥u2∥∞ < 4θ δ < ∥u3∥∞. Theorem 1.3. Assume (H1)–(H5), f0 = ∞ and f∞ ∈ (0,∞). Then (1.1) has no positive solution for λ ≫ 1 and has a positive solution u for λ < λ∞ such that ∥u∥∞ → 0 as λ → 0. In addition, if (H6), (H7), (H8a), (H9a) are satisfied, then (1.1) has three positive solutions u1, u2 and u3 for λ ∈ (λ∗,min{λ∗, λ∞}) such that ∥u1∥∞ < η < ∥u2∥∞ < 4θ δ < ∥u3∥∞. 2. Case f0 = 0. Let λ∞ := 2w∗ψ( 16 δ ) h∗f∞ . We establish the following results. Theorem 1.4. Assume (H1)–(H5), f0 = 0 and f∞ = ∞. Then (1.1) has a positive solution u for λ > 0 such that ∥u∥∞ → ∞ as λ → 0 and ∥u∥∞ → 0 as λ → ∞. In addition, if (H6), (H7), (H8b) are satisfied, then (1.1) has three positive solutions u1, u2 and u3 for λ ∈ (λ∗, λ ∗) such that ∥u1∥∞ < 4θ δ < ∥u2∥∞ < η < ∥u3∥∞. Theorem 1.5. Assume (H1)–(H5), f0 = 0 and f∞ = 0. Then (1.1) has no positive solution for λ ≈ 0 and has two positive solutions u1 and u2 for λ≫ 1 such that ∥u1∥∞ → 0 and ∥u2∥∞ → ∞ as λ→ ∞. Theorem 1.6. Assume (H1)–(H5), f0 = 0 and f∞ ∈ (0,∞). Then (1.1) has no positive solution for λ ≈ 0 and has a positive solution u for λ > λ∞ such that ∥u∥∞ → 0 as λ → ∞. In addition, if (H6), (H7), (H8b), (H9b) are satisfied, then (1.1) has three positive solutions u1, u2 and u3 for λ ∈ (max{λ∗, λ∞}, λ∗) such that ∥u1∥∞ < 4θ δ < ∥u2∥∞ < η < ∥u3∥∞. 3. Case f0 ∈ (0,∞). Let λ0 := w∗ 2∥h∥1ψ( 1+a 1−G )f0 and λ0 := 2w∗ψ( 16 δ ) h∗f0 . We establish the following results. Theorem 1.7. Assume (H1)–(H5), f0 ∈ (0,∞) and f∞ = ∞. Then (1.1) has no positive solution for λ ≫ 1 a nd has a positive solution u for λ < λ0 such that ∥u∥∞ → ∞ as λ → 0. In addition, if (H6), (H7), (H8b), (H9a) are satisfied, then (1.1) has three positive solutions u1, u2 and u3 for λ ∈ (λ∗,min{λ∗, λ0}) such that ∥u1∥∞ < 4θ δ < ∥u2∥∞ < η < ∥u3∥∞. 4 E. K. LEE, I. SIM, B. SON EJDE-2025/66 Theorem 1.8. Assume (H1)–(H5), f0 ∈ (0,∞) and f∞ = 0. Then (1.1) has no positive solution for λ ≈ 0 and has a positive solution u for λ > λ0 such that ∥u∥∞ → ∞ as λ → ∞. In addition, if (H6), (H7), (H8a), (H9b) are satisfied, then (1.1) has three positive solutions u1, u2 and u3 for λ ∈ (max{λ∗, λ0}, λ∗) such that ∥u1∥∞ < η < ∥u2∥∞ < 4θ δ < ∥u3∥∞. Theorem 1.9. Assume (H1)–(H5), f0 ∈ (0,∞) and f∞ ∈ (0,∞). Then (1.1) has no pos- itive solution for λ ≈ 0 and λ ≫ 1. If (H10a) is satisfied, then (1.1) has a positive solu- tion for λ ∈ (min{λ0, λ∞},max{λ0, λ∞}). If (H6), (H7), (H10a), (H10b) are satisfied, then (1.1) has three positive solutions u1, u2 and u3 for λ ∈ (max{λ∗, λ0},min{λ∗, λ∞}). If (H6), (H7), (H10a), (H10c) are satisfied, then (1.1) has three positive solutions u1, u2 and u3 for λ ∈ (max{λ∗, λ∞},min{λ∗, λ0}). We use the following Krasnoselskii-type fixed point theorem to get the existence and multiplicity results (see [9, Lemma A]). Proposition 1.10. Let X be a Banach space and I : X → X be a completely continuous operator. Suppose that there exist a nonzero element z ∈ X and positive constants r and R with r ̸= R such that (a) if y ∈ X satisfies y = σIy for σ ∈ (0, 1], then ∥y∥X ̸= r, (b) if y ∈ X satisfies y = Iy + τz for τ ≥ 0, then ∥y∥X ̸= R. Then I has a fixed point y ∈ X with min{r,R} < ∥y∥X < max{r,R}. The main challenges of this study are constructing the completely continuous operator (Tλ in Section 2) for the modified problem of (1.1) that reflects the mixed nonlocal boundary conditions and finding lower estimates of functions obtained through the operator (Tλy, where y ∈ C[0, 1]). In general, due to the boundary conditions, completely continuous operators related to nonlocal boundary value problems are more complicated than those related to local boundary value prob- lems, and functions obtained through the operators related to nonlocal boundary value problems could have maximums either in (0, 1) or at a boundary of (0, 1). To overcome the difficulty of con- structing the operator Tλ, we modify the completely continuous operators for the local (Dirichlet and nonlinear) boundary value problems in [18, 19] by adding some functions representing the boundary conditions (A(y), B(y) and C(s) in Section 2). To find necessary lower estimates of Tλy, we use different representations of Tλy depending on where it has a maximum. It is also noteworthy that this study complements the existing results by dealing with the nonlocal boundary value problem in which it has a nonhomogeneous operator and a (non)singular nonlinear term. In particular, we discuss the existence of at least three positive solutions that has not been treated much in previous studies. In Section 2, we construct the completely continuous operator Tλ to find fixed points of the modified problem of (1.1) and show that the fixed points are eventually positive solutions of (1.1). We prove Theorems 1.1 - 1.3, 1.4 - 1.6, and 1.7 - 1.9 in Sections 3, 4, and 5, respectively. Section 6 provides an example of (1.1) with a nonhomogeneous operator and a singular nonlinear term. 2. Preliminaries In this section, we construct the completely continuous operator Tλ for the modified problem of (1.1) and show that fixed points of Tλ are positive solutions of (1.1). We first construct the completely continuous operator Tλ. To do this, we extend the ideas in [18, 19], which are the local boundary value problems. For y ∈ C[0, 1], we define the following functions related to the boundary conditions: A(y) := ∫ 1 0 g0(s)k 0 0(s, |y(s)|)ds+ m∑ i=1 αik 0 i (ζi, |y(ζi)|), B(y) := ∫ 1 0 g1(s)k 1 0(s, |y(s)|)ds+ n∑ j=1 βjk 1 j (ξj , |y(ξj)|), C(s) := ϕ−1 (w(1)ϕ(s) w(0) + λ w(0) ∫ 1 0 h(r)f∗(y)dr ) for s ∈ R, EJDE-2025/66 SOLUTIONS TO NONLINEAR ELLIPTIC PROBLEMS 5 where f∗(y(t)) := f(max{y(t), δρλd(t)}) and ρλ > 0 is to be determined in each section. Then we define the operator Tλ : C[0, 1] → C[0, 1] by Tλy(t) := A(y) + aC(my) + ∫ t 0 ϕ−1 (w(1)ϕ(my) w(s) + λ w(s) ∫ 1 s h(r)f∗(y)dr ) ds, where y ∈ C[0, 1] and my ∈ R is the constant such that B(y)− bmy = A(y) + aC(my) + ∫ 1 0 ϕ−1 (w(1)ϕ(my) w(s) + λ w(s) ∫ 1 s h(r)f∗(y)dr ) ds. It can be shown that Tλy ∈ C1[0, 1], my = (Tλy) ′(1), C(my) = (Tλy) ′(0), and my is continuous for y. Further, Tλ : C[0, 1] → C[0, 1] is completely continuous, Tλy(t) is the solution to the boundary value problem −(w(t)ϕ(x′))′ = λh(t)f∗(y), t ∈ (0, 1), x(0)− ax′(0) = A(y), x(1) + bx′(1) = B(y), and Tλy satisfies the following property. Lemma 2.1. Assume (H1)–(H5). Then Tλy(t) ≥ δ∥Tλy∥∞d(t). Proof. Let x(t) := Tλy(t). We first show x(0) ≥ 0. If a = 0, then it is clear because x(0) = A(y) ≥ 0. Let a > 0. Assume to the contrary that x(0) < 0. Then x′(0) < 0 by the boundary condition at 0. If there exists tx ∈ (0, 1] such that x′(tx) = 0 and x′(t) < 0 for t ∈ (0, tx), then we have x′(t) = ϕ−1 ( λ w(t) ∫ tx t h(s)f∗(y)ds ) ≥ 0, which is a contradiction. Thus x′(t) < 0 for t ∈ (0, 1]. This implies x′(1) < 0 and x(1) < 0. However, this is a contradiction since x(1) = B(y)− bx′(1) ≥ 0. Hence x(0) ≥ 0. We can show x(1) ≥ 0 by similar arguments. Then we obtain x(t) ≥ δ∥x∥∞d(t) by Lemma 2.1 in [8]. □ Next we find a condition for fixed points of Tλ to be positive solutions of (1.1). Lemma 2.2. Assume (H1)–(H5). If Tλy = y for some y ∈ C[0, 1] with ∥y∥∞ ≥ ρλ, then y is a positive solution of (1.1). Proof. It is clear that y ∈ C1[0, 1] since Tλy ∈ C1[0, 1]. Further, we have y(t) ≥ δ∥y∥∞d(t) ≥ δρλd(t) ≥ 0 by Lemma 2.1. Thus y satisfies −(w(t)ϕ(y′))′ = λh(t)f∗(y) = λh(t)f(y), t ∈ (0, 1), y(0)− ay′(0) = A(y) = ∫ 1 0 g0(s)k 0 0(s, y(s))ds+ m∑ i=1 αik 0 i (ζi, y(ζi)), y(1) + by′(1) = B(y) = ∫ 1 0 g1(s)k 1 0(s, y(s))ds+ n∑ j=1 βjk 1 j (ξj , y(ξj)). Hence y is a positive solution of (1.1). □ Now we introduce different representations of Tλy to be used to find lower estimates. If ∥Tλy∥∞ = Tλy(tm) for some tm ∈ (0, 1), then (Tλy) ′(t) ≥ 0 for t ∈ (0, tm), (Tλy) ′(t) ≤ 0 for t ∈ (tm, 1) and Tλy can be written as Tλy(t) = A(y) + a(Tλy) ′(0) + ∫ t 0 ϕ−1 ( λ w(s) ∫ tm s h(r)f∗(y)dr ) ds, 0 ≤ t ≤ tm, B(y)− b(Tλy) ′(1) + ∫ 1 t ϕ−1 ( λ w(s) ∫ s tm h(r)f∗(y)dr ) ds, tm ≤ t ≤ 1. If ∥Tλy∥∞ = Tλy(0), then (Tλy) ′(t) ≤ 0 for t ∈ (0, 1) and Tλy can be written as Tλy(t) = B(y)− b(Tλy) ′(1) + ∫ 1 t ϕ−1 ( − w(0)ϕ((Tλy) ′(0)) w(s) + λ w(s) ∫ s 0 h(r)f∗(y)dr ) ds. 6 E. K. LEE, I. SIM, B. SON EJDE-2025/66 If ∥Tλy∥∞ = Tλy(1), then (Tλy) ′(t) ≥ 0 for t ∈ (0, 1) and Tλy can be written as Tλy(t) = A(y) + a(Tλy) ′(0) + ∫ t 0 ϕ−1 (w(1)ϕ((Tλy)′(1)) w(s) + λ w(s) ∫ 1 s h(r)f∗(y)dr ) ds. 3. Proofs of Theorems 1.1–1.3 We use Proposition 1.10 with I = Tλ, X = C[0, 1] and z ≡ 1 to show the existence and multiplicity results in Theorems 1.1–1.3. Let fm(s) := infr∈(s,∞) f(r). Noting that lims→0 fm(s) ϕ(s) = ∞ since f0 = ∞, we can choose rλ ∈ (0, 1) such that fm(s) ϕ(s) > 2w∗ψ( 16 δ ) λh∗ for s ≤ rλ. Then we define ρλ = rλ in Tλ. Additionally, we assume rλ < min{η, 4θδ } to prove the existence of three positive solutions in Theorems 1.2 - 1.3. Proof of Theorem 1.1. We first show the multiplicity result for λ ≈ 0. Let σ ∈ (0, 1] and u ∈ C[0, 1] be a solution of u = σTλu and ∥u∥∞ ≥ rλ. Then u(t) = σTλu(t) ≥ 0 by Lemma 2.1. If ∥u∥∞ = u(0), then u′(0) = σ(Tλu) ′(0) ≤ 0 and u satisfies ∥u∥∞ = u(0) ≤ Tλu(0) ≤ A(u) = ∫ 1 0 g0(s)k 0 0(s, |u(s)|)ds+ m∑ i=1 αik 0 i (ζi, |u(ζi)|) ≤ G∥u∥∞. However, this is a contradiction since G < 1. If ∥u∥∞ = u(1), then u′(1) = σ(Tλu) ′(1) ≥ 0 and u satisfies ∥u∥∞ = u(1) ≤ Tλu(1) ≤ B(u) = ∫ 1 0 g1(s)k 1 0(s, |u(s)|)ds+ n∑ j=1 βjk 1 j (ξj , |u(ξj)|) ≤ G∥u∥∞. However, this is a contradiction. Hence there exists tm ∈ (0, 1) such that ∥u∥∞ = u(tm). Then u′(1) = σ(Tλu) ′(1) ≤ 0. We note that δ < 1, A(u) ≤ G∥u∥∞, (Tλu) ′(0) = C((Tλu) ′(1)), and u(t) = σTλu(t) ≥ σδ∥Tλu∥∞d(t) = δ∥u∥∞d(t) by Lemma 2.1. Thus u satisfies ∥u∥∞ ≤ A(u) + a(Tλu) ′(0) + ∫ tm 0 ϕ−1 ( λ w(s) ∫ tm s h(r)f∗(u)dr ) ds = A(u) + aϕ−1 (w(1)ϕ((Tλu)′(1)) w(0) + λ w(0) ∫ 1 0 h(r)f∗(u)dr ) + ∫ tm 0 ϕ−1 ( λ w(s) ∫ tm s h(r)f∗(u)dr ) ds ≤ A(u) + (1 + a)ϕ−1 ( λ w∗ ∫ 1 0 h(r)f∗(u)dr ) ≤ A(u) + (1 + a)ϕ−1 ( λ w∗ ∫ 1 0 h(r) ( c max{u, δrλd(r)}γ + fM (max{u, δrλd(r)}) ) dr ) ≤ A(u) + (1 + a)ϕ−1 ( λ w∗ ∫ 1 0 h(r) ( c (δ∥u∥∞d(r))γ + fM (∥u∥∞) ) dr ) ≤ G∥u∥∞ + (1 + a)ϕ−1 ( λ w∗ ( c∥hγ∥1 δγ∥u∥γ∞ + ∥h∥1fM (∥u∥∞) )) , where fM ∈ C([0,∞), [0,∞)) is such that fM (0) = 0, fM (s) is nondecreasing for s ≤ 1 and fM (s) := maxr∈[1,s] f(r) for s > 1. Then we obtain ϕ(∥u∥∞) ≤ ψ ( 1 + a 1−G ) ϕ ( (1−G)∥u∥∞ 1 + a ) ≤ λ w∗ ψ ( 1 + a 1−G )( c∥hγ∥1 δγ∥u∥γ∞ + ∥h∥1fM (∥u∥∞) ) . EJDE-2025/66 SOLUTIONS TO NONLINEAR ELLIPTIC PROBLEMS 7 This implies 1 ≤ λ w∗ ψ ( 1 + a 1−G )( c∥hγ∥1 δγ∥u∥γ∞ϕ(∥u∥∞) + ∥h∥1fM (∥u∥∞) ϕ(∥u∥∞) ) . (3.1) If ∥u∥∞ = 1, then we have 1 ≤ λ w∗ ψ( 1+a 1−G )( c∥hγ∥1 δγϕ(1) + ∥h∥1fM (1) ϕ(1) ). However, this is a contradiction for λ ≈ 0. Hence ∥u∥∞ ̸= 1 for λ ≈ 0. Now we show that there exist two constants (one is greater than 1 and one is ess than 1) satisfying (b) in Proposition 1.10. Let τ ≥ 0 and u ∈ C[0, 1] be a solution of u = Tλu + τ . Then there are three cases: (i) ∥u∥∞ = u(tm) for some tm ∈ (0, 1), (ii) ∥u∥∞ = u(0), and (iii) ∥u∥∞ = u(1). We first consider the case (i). Then (Tλu) ′(tm) = 0. We note that u(t) = Tλu(t) + τ ≥ δ∥Tλu∥∞d(t) + τ ≥ δ(∥Tλu∥∞ + τ)d(t) = δ∥u∥∞d(t) ≥ δ∥u∥∞ 4 (3.2) for t ∈ [ 14 , 3 4 ] by δ < 1 and Lemma 2.1. If tm ≥ 1 2 , then (Tλu) ′(0) ≥ 0 and u satisfies ∥u∥∞ ≥ A(u) + a(Tλu) ′(0) + ∫ tm 0 ϕ−1 ( λ w(s) ∫ tm s h(r)f∗(u)dr ) ds ≥ ∫ 1/4 0 ϕ−1 ( λ w(s) ∫ tm s h(r)f∗(u)dr ) ds ≥ ∫ 1/4 0 ϕ−1 ( λ w∗ ∫ 1 2 1 4 h(r)fm(u)dr ) ds ≥ 1 4 ϕ−1 (λh∗ w∗ fm( δ∥u∥∞ 4 ) ) . By similar arguments, we can show that if tm < 1 2 then ∥u∥∞ ≥ 1 4ϕ −1(λh∗ w∗ fm( δ∥u∥∞ 4 )). For case (ii), (Tλu) ′(0) ≤ 0, (Tλu) ′(1) ≤ 0 and u satisfies ∥u∥∞ ≥ B(u)− b(Tλu) ′(1) + ∫ 1 3/4 ϕ−1 ( − w(0)ϕ((Tλu) ′(0)) w(s) + λ w(s) ∫ s 0 h(r)f∗(u)dr ) ds ≥ ∫ 1 3/4 ϕ−1 ( λ w(s) ∫ s 0 h(r)f∗(u)dr ) ds ≥ ∫ 1 3/4 ϕ−1 ( λ w∗ ∫ 3 4 1 2 h(r)fm(u)dr ) ds ≥ 1 4 ϕ−1 (λh∗ w∗ fm( δ∥u∥∞ 4 ) ) . For case (iii), (Tλu) ′(0) ≥ 0, (Tλu) ′(1) ≥ 0 and u satisfies ∥u∥∞ ≥ A(u) + a(Tλu) ′(0) + ∫ 1/4 0 ϕ−1 (w(1)ϕ((Tλu)′(1)) w(s) + λ w(s) ∫ 1 s h(r)f∗(u)dr ) ds ≥ ∫ 1/4 0 ϕ−1 ( λ w(s) ∫ 1 s h(r)f∗(u)dr ) ds ≥ ∫ 1/4 0 ϕ−1 ( λ w∗ ∫ 1 2 1 4 h(r)fm(u)dr ) ds ≥ 1 4 ϕ−1 (λh∗ w∗ fm( δ∥u∥∞ 4 ) ) . Hence we obtain ∥u∥∞ ≥ 1 4ϕ −1(λh∗ w∗ fm( δ∥u∥∞ 4 )) for all cases. Then we have λh∗ w∗ fm (δ∥u∥∞ 4 ) ≤ ϕ(4∥u∥∞) ≤ ψ (16 δ ) ϕ (δ∥u∥∞ 4 ) . 8 E. K. LEE, I. SIM, B. SON EJDE-2025/66 This implies fm( δ∥u∥∞ 4 ) ϕ( δ∥u∥∞ 4 ) ≤ w∗ψ( 16δ ) λh∗ . (3.3) By the definition of rλ, we obtain rλ < δ∥u∥∞ 4 . Thus ∥u∥∞ ̸= rλ for λ > 0. Noting that lims→∞ fm(s) ϕ(s) = ∞ since f∞ = ∞, we can also find Rλ ≫ 1 such that Rλ > 1 and fm( δs4 ) ϕ( δs4 ) > w∗ψ( 16δ ) λh∗ for s ≥ Rλ. Thus ∥u∥∞ ̸= Rλ for λ > 0. By Proposition 1.10, Tλ has two fixed points v1, v2 ∈ C[0, 1] for λ ≈ 0 such that rλ < ∥v1∥∞ < 1 < ∥v2∥∞ < Rλ. Hence v1 and v2 are positive solutions of (1.1) by Lemma 2.2. Further, we obtain ∥v1∥∞ → 0 and ∥v2∥∞ → ∞ as λ→ 0 from (3.1). Next we show the nonexistence result for λ ≫ 1. Assume that (1.1) has a positive solution u. Then u satisfies (3.3). Since lims→0 fm(s) ϕ(s) = ∞ = lims→∞ fm(s) ϕ(s) , we have 0 < inf s∈(0,∞) fm(s) ϕ(s) ≤ fm( δ∥u∥∞ 4 ) ϕ( δ∥u∥∞ 4 ) ≤ w∗ψ( 16δ ) λh∗ . (3.4) However, this is a contradiction for λ≫ 1. Hence (1.1) has no positive solution for λ≫ 1. □ Proof of Theorem 1.2. We first show the existence result for λ > 0. If u ∈ C[0, 1] is a solution of u = Tλu+ τ with τ ≥ 0, then u satisfies (3.3). This implies ∥u∥∞ ̸= rλ for λ > 0. If u ∈ C[0, 1] is a solution of u = σTλu with σ ∈ (0, 1] and ∥u∥∞ ≥ rλ, then u satisfies (3.1). Noting that lims→∞ fM (s) ϕ(s) = 0 since f∞ = 0, we can find R̂λ ≫ 1 such that R̂λ > rλ and 1 > λ w∗ ψ ( 1 + a 1−G )( c∥hγ∥1 δγsγϕ(s) + ∥h∥1fM (s) ϕ(s) ) for s ≥ R̂λ. Thus we obtain ∥u∥∞ ̸= R̂λ for λ > 0 from (3.1). By Proposition 1.10, Tλ has a fixed point v for λ > 0 such that rλ < ∥v∥∞ < R̂λ. By Lemma 2.2, v is a positive solution of (1.1). Further, we obtain ∥v∥∞ → 0 as λ→ 0 from (3.1) and ∥v∥∞ → ∞ as λ→ ∞ from (3.3). Next we show the multiplicity result for λ ∈ (λ∗, λ ∗). Let σ ∈ (0, 1] and u ∈ C[0, 1] be a solution of u = σTλu. We note that u(t) = σTλu(t) ≥ σδ∥Tλu∥∞d(t) = δ∥u∥∞d(t) by Lemma 2.1. By (H6), if ∥u∥∞ = η (≥ rλ) then u satisfies ∥u∥∞ ≤ A(u) + a(Tλu) ′(0) + ∫ tm 0 ϕ−1 ( λ w(s) ∫ tm s h(r)f∗(u)dr ) ds ≤ A(u) + (1 + a)ϕ−1 ( λ w∗ ∫ 1 0 h(r)f∗(u)dr ) = A(u) + (1 + a)ϕ−1 ( λ w∗ ∫ 1 0 h(r) fγ(max{u, δrλd(r)}) max{u, δrλd(r)}γ dr ) ds ≤ A(u) + (1 + a)ϕ−1 ( λ w∗ ∫ 1 0 h(r) fγ(∥u∥∞) (δ∥u∥∞d(r))γ dr ) ≤ G∥u∥∞ + (1 + a)ϕ−1 (λ∥hγ∥1f(∥u∥∞) w∗δγ ) . Therefore, ϕ(∥u∥∞) ≤ ψ ( 1 + a 1−G ) ϕ ( (1−G)∥u∥∞ 1 + a ) ≤ λ∥hγ∥1f(∥u∥∞) w∗δγ ψ ( 1 + a 1−G ) . EJDE-2025/66 SOLUTIONS TO NONLINEAR ELLIPTIC PROBLEMS 9 This implies w∗δ γ λ∥hγ∥1ψ( 1+a 1−G ) ≤ f(∥u∥∞) ϕ(∥u∥∞) = f(η) ϕ(η) . However, this is a contradiction for λ < λ∗. Hence ∥u∥∞ ̸= η for λ < λ∗. Let τ ≥ 0 and u ∈ C[0, 1] be a solution of u = Tλu+ τ . Assume ∥u∥∞ = 4θ δ . Then three cases can occur: (i) ∥u∥∞ = u(tm) for some tm ∈ (0, 1), (ii) ∥u∥∞ = u(0) and (iii) ∥u∥∞ = u(1). We first consider case (i). Noting that u(t) ≥ δ∥u∥∞ 4 for t ∈ [ 14 , 3 4 ] from (3.2), if tm ≥ 1 2 , then ∥u∥∞ ≥ A(u) + a(Tλu) ′(0) + ∫ tm 0 ϕ−1 ( λ w(s) ∫ tm s h(r)f∗(u)dr ) ds ≥ ∫ 1/4 0 ϕ−1 ( λ w(s) ∫ tm s h(r) fγ(max{u, δrλd(r)}) max{u, δrλd(r)}γ dr ) ds ≥ ∫ 1/4 0 ϕ−1 ( λ w∗ ∫ 1 2 1 4 h(r) fγ ( δ∥u∥∞ 4 ) ∥u∥γ∞ dr ) ds ≥ 1 4 ϕ−1 (λh∗δγ 4γw∗ f (δ∥u∥∞ 4 )) . By similar arguments, we can show that if tm < 1 2 then ∥u∥∞ ≥ 1 4ϕ −1(λh∗δ γ 4γw∗ f( δ∥u∥∞ 4 )). For case (ii), we have ∥u∥∞ ≥ B(u)− b(Tλu) ′(1) + ∫ 1 3/4 ϕ−1 ( − w(0)ϕ((Tλu) ′(0)) w(s) + λ w(s) ∫ s 0 h(r)f∗(u)dr ) ds ≥ ∫ 1 3/4 ϕ−1 ( λ w(s) ∫ s 0 h(r) fγ(max{u, δrλd(r)}) max{u, δrλd(r)}γ dr ) ds ≥ ∫ 1 3/4 ϕ−1 ( λ w∗ ∫ 3 4 1 2 h(r) fγ ( δ∥u∥∞ 4 ) ∥u∥γ∞ dr ) ds ≥ 1 4 ϕ−1 (λh∗δγ 4γw∗ f (δ∥u∥∞ 4 )) . For case (iii), we have ∥u∥∞ ≥ A(u) + a(Tλu) ′(0) + ∫ 1/4 0 ϕ−1 (w(1)ϕ((Tλu)′(1)) w(s) + λ w(s) ∫ 1 s h(r)f∗(u)dr ) ds ≥ ∫ 1/4 0 ϕ−1 ( λ w(s) ∫ 1 s h(r) fγ(max{u, δrλd(r)}) max{u, δrλd(r)}γ dr ) ds ≥ ∫ 1/4 0 ϕ−1 ( λ w∗ ∫ 1 2 1 4 h(r) fγ ( δ∥u∥∞ 4 ) ∥u∥γ∞ dr ) ds ≥ 1 4 ϕ−1 (λh∗δγ 4γw∗ f (δ∥u∥∞ 4 )) . Hence we obtain ∥u∥∞ ≥ 1 4ϕ −1(λh∗δ γ 4γw∗ f( δ∥u∥∞ 4 )) for all cases. Since ∥u∥∞ = 4θ δ , we have λh∗δ γf(θ) 4γw∗ = λh∗δ γ 4γw∗ f (δ∥u∥∞ 4 ) ≤ ϕ ( 4∥u∥∞ ) ≤ ψ (16 δ ) ϕ (δ∥u∥∞ 4 ) = ψ (16 δ ) ϕ(θ). However, this is a contradiction for λ > λ∗. Hence ∥u∥∞ ̸= 4θ δ for λ > λ∗. We can choose R̂λ ≫ 1 such that R̂λ > 4θ δ . Further, rλ < η < 4θ δ and (λ∗, λ ∗) is nonempty by (H8a) and (H7), respectively. Thus (1.1) has three positive solutions v1, v2 and v3 for λ ∈ (λ∗, λ ∗) such that rλ < ∥v1∥∞ < η < ∥v2∥∞ < 4θ δ < ∥v3∥∞ < R̂λ by Proposition 1.10 and Lemma 2.2. □ Proof of Theorem 1.3. We first show the existence result for λ < λ∞. If u ∈ C[0, 1] is a solution of u = Tλu+ τ with τ ≥ 0, then u satisfies (3.3). Thus ∥u∥∞ ̸= rλ for λ > 0. 10 E. K. LEE, I. SIM, B. SON EJDE-2025/66 Let σ ∈ (0, 1] and u ∈ C[0, 1] be a solution of u = σTλu. Assume ∥u∥∞ ≥ rλ. Then u satisfies (3.1). Since lims→∞ fM (s) ϕ(s) = f∞ ∈ (0,∞), there exists Rλ ≫ 1 such that Rλ > rλ and λ w∗ ψ ( 1 + a 1−G )( c∥hγ∥1 δγsγϕ(s) + ∥h∥1fM (s) ϕ(s) ) < 2λ∥h∥1f∞ w∗ ψ ( 1 + a 1−G ) for s ≥ Rλ. If ∥u∥∞ = Rλ, then 1 < 2λ∥h∥1f∞ w∗ ψ( 1+a 1−G ) from (3.1). However, this is a contradiction for λ < λ∞. Hence ∥u∥∞ ̸= Rλ for λ < λ∞. By Proposition 1.10 and Lemma 2.2, (1.1) has a positive solution v for λ < λ∞ such that rλ < ∥v∥∞ < Rλ. Further, we obtain ∥v∥∞ → 0 as λ→ 0 from (3.1). Next we show the multiplicity result for λ ∈ (λ∗,min{λ∗, λ∞}). The following were proven in the proof of Theorem 1.2: (i) if u ∈ C[0, 1] is a solution of u = σTλu with σ ∈ (0, 1], then ∥u∥∞ ̸= η for λ < λ∗, and (ii) if u ∈ C[0, 1] is a solution of u = Tλu + τ with τ ≥ 0, then ∥u∥∞ ̸= 4θ δ for λ > λ∗. Further, we can choose Rλ ≫ 1 such that Rλ > 4θ δ . Since rλ < η < 4θ δ and (λ∗,min{λ∗, λ∞}) is nonempty by (H7), (H8a), and (H9a), (1.1) has three positive solutions v1, v2 and v3 for λ ∈ (λ∗,min{λ∗, λ∞}) such that rλ < ∥v1∥∞ < η < ∥v2∥∞ < 4θ δ < ∥v3∥∞ < Rλ. Now we show the nonexistence result for λ≫ 1. Assume to the contrary that (1.1) has a positive solution u for λ≫ 1. Then u satisfies (3.4) since lims→0 fm(s) ϕ(s) = ∞ and lims→∞ fm(s) ϕ(s) = f∞ > 0. However, this is a contradiction for λ≫ 1. Hence (1.1) has no positive solution for λ≫ 1. □ 4. Proofs of Theorems 1.4–1.6 In this section, we consider the case f0 = 0. Since f0 = 0 implies lims→0 f(s) = 0, we can define f∗M (s) := maxr∈[0,s] f(r). Noting that lims→0 f∗ M (s) ϕ(s) = 0 since f0 = 0, there exists r∗λ ∈ (0, 1) such that f∗ M (s) ϕ(s) < w∗ λ∥h∥1ψ( 1+a 1−G ) for s ≤ r∗λ. Then we define ρλ = r∗λ in Tλ. Additionally, r∗λ < min{η, 4θδ } is assumed to show the multiplicity results in Theorem 1.4 and Theorem 1.6. Proof of Theorem 1.4. We first show the existence result for λ > 0. Let σ ∈ (0, 1] and u ∈ C[0, 1] be a solution of u = σTλu. Assume ∥u∥∞ ≥ r∗λ. Following the arguments in the proof of Theorem 1.1, we obtain ∥u∥∞ ≤ A(u) + (1 + a)ϕ−1 ( λ w∗ ∫ 1 0 h(r)f∗(u)dr ) ≤ A(u) + (1 + a)ϕ−1 ( λ w∗ ∫ 1 0 h(r)f∗M (max{u, δr∗λd(r)})dr ) ≤ G∥u∥∞ + (1 + a)ϕ−1 (λ∥h∥1f∗M (∥u∥∞) w∗ ) . Then u satisfies ϕ(∥u∥∞) ≤ ψ ( 1 + a 1−G ) ϕ ( (1−G)∥u∥∞ 1 + a ) ≤ λ∥h∥1f∗M (∥u∥∞) w∗ ψ ( 1 + a 1−G ) . This implies w∗ λ∥h∥1ψ( 1+a 1−G ) ≤ f∗M (∥u∥∞) ϕ(∥u∥∞) . (4.1) Thus ∥u∥∞ ̸= r∗λ for λ > 0 by the definition of r∗λ. If u ∈ C[0, 1] is a solution of u = Tλu+ τ with τ ≥ 0, then u satisfies (3.3). Then ∥u∥∞ ̸= Rλ (> 1) for λ > 0, where Rλ is the constant found in the proof of Theorem 1.1. By Proposition 1.10 and Lemma 2.2, (1.1) has a positive solution v for λ > 0 such that r∗λ < ∥v∥∞ < Rλ. Further, we obtain ∥v∥∞ → 0 as λ→ ∞ from (3.3) and ∥v∥∞ → ∞ as λ→ 0 from (4.1). Next we show the multiplicity result for λ ∈ (λ∗, λ ∗). Following the arguments in the proof of Theorem 1.2, we can show that (i) if u ∈ C[0, 1] is a solution of u = σTλu with σ ∈ (0, 1], then ∥u∥∞ ̸= η for λ < λ∗, and (ii) if u ∈ C[0, 1] is a solution of u = Tλu+τ with τ ≥ 0, then ∥u∥∞ ̸= 4θ δ for λ > λ∗. Further, we can choose Rλ ≫ 1 such that Rλ > η. We note that r∗λ < 4θ δ < η and EJDE-2025/66 SOLUTIONS TO NONLINEAR ELLIPTIC PROBLEMS 11 (λ∗, λ ∗) is nonempty by (H8b) and (H7), respectively. Hence (1.1) has three positive solutions v1, v2 and v3 for λ ∈ (λ∗, λ ∗) such that r∗λ < ∥v1∥∞ < 4θ δ < ∥v2∥∞ < η < ∥v3∥∞ < Rλ. □ Proof of Theorem 1.5. We first show the multiplicity result for λ ≫ 1. Let τ ≥ 0 and u ∈ C[0, 1] be a solution of u = Tλu + τ . Then u satisfies (3.3). If ∥u∥∞ = 1, we have fm( δ 4 ) ϕ( δ 4 ) ≤ w∗ψ( 16 δ ) λh∗ . However, this is a contradiction for λ≫ 1. Hence ∥u∥∞ ̸= 1 for λ≫ 1. Let σ ∈ (0, 1] and u ∈ C[0, 1] be a solution of u = σTλu and ∥u∥∞ ≥ r∗λ. Then u satisfies (4.1). This implies ∥u∥∞ ̸= r∗λ for λ > 0. Further, since lims→∞ f∗ M (s) ϕ(s) = 0, we can find R̃λ ≫ 1 such that R̃λ > 1 and f∗M (s) ϕ(s) < w∗ λ∥h∥1ψ( 1+a 1−G ) for s ≥ R̃λ. Thus we obtain ∥u∥∞ ̸= R̃λ for λ > 0 from (4.1). By Proposition 1.10 and Lemma 2.2, there exist positive solutions v1 and v2 for λ ≫ 1 such that r∗λ < ∥v1∥∞ < 1 < ∥v2∥∞ < R̃λ. Further, we obtain ∥v1∥∞ → 0 and ∥v2∥∞ → ∞ as λ→ ∞ from (3.3). Next we show the nonexistence result for λ ≈ 0. If u is a positive solution of (1.1), then u satisfies (4.1). Thus we have w∗ λ∥h∥1ψ( 1+a 1−G ) ≤ sup s∈(0,∞) f∗M (s) ϕ(s) <∞. (4.2) However, this is a contradiction for λ ≈ 0. Hence (1.1) has no positive solution for λ ≈ 0. □ Proof of Theorem 1.6. We first show the existence result for λ > λ∞. If u ∈ C[0, 1] is a solution of u = σTλu with σ ∈ (0, 1] and ∥u∥∞ ≥ r∗λ, then u satisfies (4.1). This implies ∥u∥∞ ̸= r∗λ for λ > 0. Let τ ≥ 0 and u ∈ C[0, 1] be a solution of u = Tλu+τ . Then u satisfies (3.3). We note that there exists R∗ λ ≫ 1 such that R∗ λ > 1 and fm(s) ϕ(s) > f∞ 2 for s ≥ R∗ λ since lims→∞ fm(s) ϕ(s) = f∞ ∈ (0,∞). If ∥u∥∞ = 4R∗ λ δ , then we have f∞ 2 ≤ fm(R∗ λ) ϕ(R∗ λ) ≤ w∗ψ( 16δ ) λh∗ (4.3) from (3.3). However, this is a contradiction for λ > λ∞. Hence ∥u∥∞ ̸= 4R∗ λ δ for λ > λ∞. By Proposition 1.10 and Lemma 2.2, (1.1) has a positive solution v for λ > λ∞ such that r∗λ < ∥v∥∞ < 4R∗ λ δ . Further, we obtain ∥v∥∞ → 0 as λ→ ∞ from (3.3). Next we show the multiplicity result for λ ∈ (max{λ∗, λ∞}, λ∗). The following were proven in the proof of Theorem 1.2: (i) if u ∈ C[0, 1] is a solution of u = σTλu with σ ∈ (0, 1], then ∥u∥∞ ̸= η for λ < λ∗ and (ii) if u ∈ C[0, 1] is a solution of u = Tλu + τ with τ ≥ 0, then ∥u∥∞ ̸= 4θ δ for λ > λ∗. Further, r∗λ < 4θ δ < η and (max{λ∗, λ∞}, λ∗) is nonempty by (H7), (H8b), and (H9b), and we can choose R∗ λ ≫ 1 such that 4R∗ λ δ > η. Hence (1.1) has three positive solutions v1, v2 and v3 for λ ∈ (max{λ∗, λ∞}, λ∗) such that r∗λ < ∥v1∥∞ < 4θ δ < ∥v2∥∞ < η < ∥v3∥∞ < 4R∗ λ δ . Now we show the nonexistence result for λ ≈ 0. If u is a positive solution of (1.1), then we can show that u satisfies (4.2) following the arguments in Theorem 1.5. However, this is a contradiction for λ ≈ 0. Hence (1.1) has no positive solution for λ ≈ 0. □ 5. Proofs of Theorems 1.7–1.9 Proof of Theorem 1.7. We choose ρλ = rλ in Tλ, where rλ ∈ (0, 1) is such that f∗ M (s) ϕ(s) < 2f0 for s ≤ rλ. To show the multiplicity result, we also assume rλ < min{η, 4θδ }. We first show the existence result for λ < λ0. Let σ ∈ (0, 1] and u ∈ C[0, 1] be a solution of u = σTλu. If ∥u∥∞ = rλ, then u satisfies (4.1). Thus we have w∗ λ∥h∥1ψ( 1+a 1−G ) ≤ f∗M (rλ) ϕ(rλ) < 2f0. (5.1) However, this is a contradiction for λ < λ0. Hence ∥u∥∞ ̸= rλ for λ < λ0. 12 E. K. LEE, I. SIM, B. SON EJDE-2025/66 If u ∈ C[0, 1] is a solution of u = Tλu+ τ with τ ≥ 0, then u satisfies (3.3). Then ∥u∥∞ ̸= Rλ (> 1) for λ > 0, where Rλ is the constant found in the proof of Theorem 1.1. By Proposition 1.10 and Lemma 2.2, (1.1) has a positive solution v for λ < λ0 such that rλ < ∥v∥∞ < Rλ. Further, we obtain ∥v∥∞ → ∞ as λ→ 0 from (4.1). Next we show the multiplicity result for λ ∈ (λ∗,min{λ∗, λ0}). Following the arguments in the proof of Theorem 1.2, we can show that (i) if u ∈ C[0, 1] is a solution of u = σTλu with σ ∈ (0, 1], then ∥u∥∞ ̸= η for λ < λ∗, and (ii) if u ∈ C[0, 1] is a solution of u = Tλu + τ with τ ≥ 0, then ∥u∥∞ ̸= 4θ δ for λ > λ∗. Further, rλ < 4θ δ < η and (λ∗,min{λ∗, λ0}) is nonempty by (H7), (H8b) and (H9a), and we can choose Rλ ≫ 1 such that Rλ > η. Hence (1.1) has three positive solutions v1, v2 and v3 for λ ∈ (λ∗,min{λ∗, λ0}) such that rλ < ∥v1∥∞ < 4θ δ < ∥v2∥∞ < η < ∥v3∥∞ < Rλ. Now we show the nonexistence result for λ ≫ 1. If u is a positive solution of (1.1), then u satisfies (3.4). However, this is a contradiction for λ≫ 1. Hence (1.1) has no positive solution for λ≫ 1. □ Proof of Theorem 1.8. We choose ρλ = r̃λ in Tλ, where r̃λ ∈ (0, 1) is such that fm(s) ϕ(s) > f0 2 for s ≤ r̃λ. To show the multiplicity result, we also assume r̃λ < min{η, 4θδ }. We first show the existence result for λ > λ0. Let τ ≥ 0 and u ∈ C[0, 1] be a solution of u = Tλu+ τ . If ∥u∥∞ = r̃λ, then we have f0 2 < fm( δr̃λ4 ) ϕ( δr̃λ4 ) ≤ w∗ψ( 16δ ) λh∗ (5.2) from (3.3). However, this is a contradiction for λ > λ0. Hence ∥u∥∞ ̸= r̃λ for λ > λ0. Let σ ∈ (0, 1] and u ∈ C[0, 1] be a solution of u = σTλu with ∥u∥∞ ≥ r̃λ. Then u satisfies (4.1). Since lims→∞ f∗ M (s) ϕ(s) = 0, we obtain ∥u∥∞ ̸= R̃λ (> r̃λ) for λ > 0, where R̃λ is the constant found in the proof of Theorem 1.5. By Proposition 1.10 and Lemma 2.2, (1.1) has a positive solution v for λ > λ0 such that r̃λ < ∥v∥∞ < R̃λ. Further, we obtain ∥v∥∞ → ∞ as λ→ ∞ from (3.3). Next we show the multiplicity result for λ ∈ (max{λ∗, λ0}, λ∗). Following the arguments in the proof of Theorem 1.2, we can show that (i) if u ∈ C[0, 1] is a solution of u = σTλu with σ ∈ (0, 1], then ∥u∥∞ ̸= η for λ < λ∗ and (ii) if u ∈ C[0, 1] is a solution of u = Tλu + τ with τ ≥ 0, then ∥u∥∞ ̸= 4θ δ for λ > λ∗. Further, r̃λ < η < 4θ δ and (max{λ∗, λ0}, λ∗) ̸= ∅ by (H7), (H8a) and (H9b), and we can choose R̃λ ≫ 1 such that R̃λ > 4θ δ . Hence (1.1) has three positive solutions v1, v2 and v3 for λ ∈ (max{λ∗, λ0}, λ∗) such that r̃λ < ∥v1∥∞ < η < ∥v2∥∞ < 4θ δ < ∥v3∥∞ < R̃λ. Now we show the nonexistence result for λ ≈ 0. If u is a positive solution of (1.1), then u satisfies (4.2). However, this is a contradiction for λ ≈ 0. Hence (1.1) has no positive solution for λ ≈ 0. □ Proof of Theorem 1.9. We first consider the case f0 > f∞. Then λ0 = min{λ0, λ∞} and λ∞ = max{λ0, λ∞}. We choose ρλ = r̃λ, which is the constant in the proof of Theorem 1.8. We show the existence result for λ ∈ (λ0, λ ∞) = (min{λ0, λ∞},max{λ0, λ∞}). Let τ ≥ 0 and u ∈ C[0, 1] be a solution of u = Tλu + τ . If ∥u∥∞ = r̃λ, then (5.2) is satisfied. This implies ∥u∥∞ ̸= r̃λ for λ > λ0. Let σ ∈ (0, 1] and u ∈ C[0, 1] be a solution of u = σTλu. Since lims→∞ f∗ M (s) ϕ(s) = f∞, there exists R⋄ λ ≫ 1 such that R⋄ λ > r̃λ and f∗ M (s) ϕ(s) < 2f∞ for s ≥ R⋄ λ. If ∥u∥∞ = R⋄ λ, then w∗ λ∥h∥1ψ( 1+a 1−G ) ≤ f∗ M (R⋄ λ) ϕ(R⋄ λ) < 2f∞ from (4.1). However, this is a contradiction for λ < λ∞. Thus ∥u∥∞ ̸= R⋄ λ for λ < λ∞. Since (λ0, λ ∞) is nonempty by (H10a), (1.1) has a positive solution for λ ∈ (λ0, λ ∞). Now we show the multiplicity results. We have (i) if u ∈ C[0, 1] is a solution of u = σTλu with σ ∈ (0, 1], then ∥u∥∞ ̸= η for λ < λ∗ and (ii) if u ∈ C[0, 1] is a solution of u = Tλu + τ with τ ≥ 0, then ∥u∥∞ ̸= 4θ δ for λ > λ∗. Further, r̃λ < min{η, 4θδ } and we can choose R⋄ λ ≫ 1 such that R⋄ λ > max{η, 4θδ }. Since (max{λ∗, λ0},min{λ∗, λ∞}) is nonempty by (H7), (H10a) and (H10b), (1.1) has three positive solutions v1, v2 and v3 for λ ∈ (max{λ∗, λ0},min{λ∗, λ∞}) such that r̃λ < ∥v1∥∞ < η < ∥v2∥∞ < 4θ δ < ∥v3∥∞ < R⋄ λ. EJDE-2025/66 SOLUTIONS TO NONLINEAR ELLIPTIC PROBLEMS 13 Next we consider the case f0 < f∞. Then λ∞ = min{λ0, λ∞} and λ0 = max{λ0, λ∞}. We choose ρλ = rλ, which is the constant in the proof of Theorem 1.7. We show the existence result for λ ∈ (λ∞, λ 0) = (min{λ0, λ∞},max{λ0, λ∞}). Let σ ∈ (0, 1] and u ∈ C[0, 1] be a solution of u = σTλu. If ∥u∥∞ = rλ, then (5.1) is satisfied. This implies ∥u∥∞ ̸= rλ for λ < λ0. Let τ ≥ 0 and u ∈ C[0, 1] be a solution of u = Tλu + τ . If ∥u∥∞ = 4R∗ λ δ , then (4.3) is satisfied. This implies ∥u∥∞ ̸= 4R∗ λ δ for λ > λ∞. Since (λ∞, λ 0) is nonempty by (H10a), (1.1) has a positive solution for λ ∈ (λ∞, λ 0). Since (max{λ∗, λ∞},min{λ∗, λ0}) is nonempty by (H7), (H10a) and (H10c), we can also show that (1.1) has three positive solutions v1, v2 and v3 for λ ∈ (max{λ∗, λ∞},min{λ∗, λ0}) such that rλ < ∥v1∥∞ < 4θ δ < ∥v2∥∞ < η < ∥v3∥∞ < 4R∗ λ δ following the above arguments. The proofs of the nonexistence results for λ ≈ 0 and for λ ≫ 1 follow the arguments in the proofs of Theorem 1.5 and Theorem 1.1, respectively. □ 6. Example In this section, we discuss an example of a mixed nonlocal boundary value problem involving a nonhomogeneous operator and a singular nonlinear term. We consider the (p, q)-Laplacian problem −(w(t)(|u′|p−2u′ + |u′|q−2u′))′ = λh(t) ( A uγ1 +Buγ2 + Ce γ3u γ3+u +D ) , t ∈ (0, 1), u(0)− au′(0) = ∫ 1 0 g0(s)k 0 0(s, u)ds+ m∑ i=1 αik 0 i (ζi, u(ζi)), u(1) + bu′(1) = ∫ 1 0 g1(s)k 1 0(s, u)ds+ n∑ j=1 βjk 1 j (ξj , u(ξj)), (6.1) where 1 < p < q < ∞, γ1 > 0, γ2 > 0, γ3 > 0, A > 0, B ≥ 0, C ≥ 0, D ∈ R and C +D ≥ 0. We assume a ≥ 0, b ≥ 0, 0 ≤ αi < 1, 0 ≤ βj < 1, w and h satisfy (H3), g0 and g1 satisfy (H4), and k00, k 1 0, k 0 i , and k 1 j satisfy (H5), where i ∈ {1, 2, . . . ,m} and j ∈ {1, 2, . . . , n}. The operator of this problem is u 7→ (|u′|p−2u′ + |u′|q−2u′)′. It is a nonhomogeneous operator and ϕ(s) = |s|p−2s + |s|q−2s satisfies (H1) with ψ(s) = max{sp−1, sq−1}. The nonlinear term is f(s) = A sγ1 +Bsγ2 +Ce γ3s γ3+s +D. It is singular at 0 and f0 = ∞ since A > 0. Further, it satisfies (H2) and (H6) with γ = γ1. Hence, (6.1) satisfies (H1)–(H6), f0 = ∞ and γ = γ1 under the above conditions. 1. If B > 0 and γ2 > q− 1, then f∞ = ∞. Thus (6.1) has no positive solution for λ≫ 1 and has two positive solutions u1 and u2 for λ ≈ 0 such that ∥u1∥∞ → 0 and ∥u2∥∞ → ∞ as λ→ 0. 2. If γ2 < q − 1, then f∞ = 0. Thus (6.1) has a positive solution u for λ > 0 such that ∥u∥∞ → 0 as λ → 0 and ∥u∥∞ → ∞ as λ → ∞. In addition, if C > 0, η = 1 and θ = γ3, then we have f(θ) ϕ(θ) / f(η) ϕ(η) = A γ γ1 3 +Bγγ23 + Ce γ3 2 +D γp−1 3 + γq−1 3 2 A+B + Ce γ3 γ3+1 +D ≥ Ce γ3 2 +D 2γq−1 3 2 A+B + Ce+D ≥ Ce γ3 2 +D γq−1 3 (A+B + Ce+D) ≫ 1 (6.2) 14 E. K. LEE, I. SIM, B. SON EJDE-2025/66 for γ3 ≫ 1. Thus (H7) and (H8a) are satisfied for γ3 ≫ 1. Further, we have λ∗ = 4γ1w∗( 16δ ) q−1(γp−1 3 + γq−1 3 ) h∗δγ1( A γ γ1 3 +Bγγ23 + Ce γ3 2 +D) , λ∗ = 2w∗δ γ1 ∥hγ1∥1( 1+a 1−G ) q−1(A+B + Ce γ3 γ3+1 +D) . (6.3) Hence if γ3 ≫ 1, then (6.1) has three positive solutions for λ ∈ (λ∗, λ ∗). 3. If B > 0 and γ2 = q − 1, then f∞ = B ∈ (0,∞) and λ∞ = w∗ 2B∥h∥1( 1+a 1−G )q−1 . Thus (6.1) has no positive solution for λ ≫ 1 and has a positive solution u for λ < λ∞ such that ∥u∥∞ → 0 as λ→ 0. In addition, if C > 0, η = 1 and θ = γ3, then we have f(θ) ϕ(θ) = A γ γ1 3 +Bγq−1 3 + Ce γ3 2 +D γp−1 3 + γq−1 3 ≥ Ce γ3 2 +D 2γq−1 3 ≫ 1, f(θ) ϕ(θ) / f(η) ϕ(η) ≥ Ce γ3 2 +D γq−1 3 (A+B + Ce+D) ≫ 1 for γ3 ≫ 1 from (6.2). Thus (H7), (H8a) and (H9a) are satisfied for γ3 ≫ 1. Noting that λ∗ and λ∗ are the same as those in (6.3), if γ3 ≫ 1 then (6.1) has three positive solutions for λ ∈ (λ∗,min{λ∗, λ∞}). Acknowledgments. E. K. Lee was supported by the National Research Foundation of Korea (NRF) grant funded by the Korea Government (NRF-2020R1F1A1A01048442). I. Sim was sup- ported by the National Research Foundation of Korea (NRF) grant funded by the Korea Gov- ernment (NRF-2021R1I1A3A0403627013). B. Son was supported by a Summer Research Grant funded by Ohio Northern University. References [1] B. Ahmad, A. Alsaedi; Existence of approximate solutions of the forced Duffing equation with discontinuous type integral boundary conditions, Nonlinear Anal. Real World Appl. 10 (2009), no. 1, 358–367. [2] A. Alsaedi, M. Alsulami, R. P. Agarwal, B. Ahmad; Some new nonlinear second-order boundary value problems on an arbitrary domain, Adv. Difference Equ. (2018), no. 227, 18 pp. [3] A. Boucherif; Second-order boundary value problems with integral boundary conditions, Nonlinear Anal. 70 (2009), no. 1, 364–371. [4] M. Boukrouche, D. A. Tarzia; A family of singular ordinary differential equations of the third order with an integral boundary condition, Bound. Value Probl. (2018), no. 32, 11 pp. [5] R. Yu. Chegis; Numerical solution of a heat conduction problem with an integral condition, Litovsk. Mat. Sb. 24 (1984), no. 4, 209–215 (Russian). [6] C. P. Gupta; Solvability of a three-point nonlinear boundary value problem for a second order ordinary differ- ential equation, J. Math. Anal. Appl. 168 (1992), no. 2, 540–551. [7] C. P. Gupta; A generalized multi-point boundary value problem for second order ordinary differential equations, Appl. Math. Comput. 89 (1998), no. 1-3, 133–146. [8] D. D. Hai, R. Shivaji; On radial solutions for singular combined superlinear elliptic systems on annular domains, J. Math. Anal. Appl. 446 (2017), no. 1, 335–344. [9] D. D. Hai, R. Shivaji; Positive radial solutions for a class of singular superlinear problems on the exterior of a ball with nonlinear boundary conditions, J. Math. Anal. Appl. 456 (2017), no. 2, 872–881. [10] D. D. Hai, X. Wang; On singular p-Laplacian boundary value problems involving integral boundary conditions, Electron. J. Qual. Theory Differ. Equ. (2019), no. 90, 13 pp. [11] M. A. Hajji; Multi-point special boundary-value problems and applications to fluid flow through porous media, Proceedings of International Multi-Conference of Engineers and Computer Scientists (IMECS 2009), Hong Kong. Vol. 31. 2009. [12] J. Henderson; Smoothness of solutions with respect to multi-strip integral boundary conditions for nth order ordinary differential equations, Nonlinear Anal. Model. Control 19 (2014), no. 3, 396–412. [13] N. I. Ionkin; The solution of a certain boundary value problem of the theory of heat conduction with a non- classical boundary condition, Differ. Uravn. 13 (1977), no. 2, 294–304 (Russian). [14] I. Y. Karaca, F. T. Fen; Positive solutions of nth-order boundary value problems with integral boundary conditions, Math. Model. Anal. 20 (2015), 188–204. [15] R. Ma; Positive solutions of a nonlinear three-point boundary-value problem, Electron. J. Differential Equations (1999), no. 34, 8 pp. EJDE-2025/66 SOLUTIONS TO NONLINEAR ELLIPTIC PROBLEMS 15 [16] R. Ma; Existence of positive solutions for superlinear semipositone m-point boundary-value problems, Proc. Edinb. Math. Soc. (2) 46 (2003), no. 2, 279–292. [17] F. Nicoud, T. Schfönfeld; Integral boundary conditions for unsteady biomedical CFD applications, Int. J. Numer. Methods Fluids 40 (2002), 457–465. [18] I. Sim, B. Son; Positive radial solutions to singular nonlinear elliptic problems involving nonhomogeneous operators, Appl. Math. Lett. 125 (2022), no. 107757, 7 pp. [19] B. Son, P. Wang; Analysis of positive radial solutions for singular superlinear p-Laplacian systems on the exterior of a ball, Nonlinear Anal. 192 (2020), no. 111657, 15 pp. [20] S. Timoshenko; Theory of elastic stability (McGraw-Hill, New York, 1961). [21] J. R. L. Webb; Positive solutions of some three point boundary value problems via fixed point index theory, Nonlinear Anal. 47 (2001), no. 7, 4319–4332. [22] J. R. L. Webb, G. Infante; Positive solutions of nonlocal boundary value problems: a unified approach, J. London Math. Soc. (2) 74 (2006), no. 3, 673–693. [23] J. R. L. Webb, G. Infante; Positive solutions of nonlocal boundary value problems involving integral conditions, NoDEA Nonlinear Differential Equations Appl. 15 (2008), no. 1-2, 45–67. [24] J. R. Womersley; Method for the calculation of velocity, rate of flow and viscous drag in arteries when the pressure gradient is known, J. Physiol. 127 (1955), 553–563. Eun Kyoung Lee Department of Mathematics Education, Pusan National University, Busan 46241, Republic of Korea Email address: eklee@pusan.ac.kr Inbo Sim Department of Mathematics, University of Ulsan, Ulsan 44610, Republic of Korea Email address: ibsim@ulsan.ac.kr Byungjae Son School of Science, Technology, and Mathematics, Ohio Northern University, OH 45810, USA Email address: b-son@onu.edu 1. Introduction and main results 2. Preliminaries 3. Proofs of Theorems 1.1–1.3 4. Proofs of Theorems 1.4–1.6 5. Proofs of Theorems 1.7–1.9 6. Example Acknowledgments References