Electronic Journal of Differential Equations, Vol. 2025 (2025), No. 65, pp. 1–20. ISSN: 1072-6691. URL: https://ejde.math.txstate.edu, https://ejde.math.unt.edu DOI: 10.58997/ejde.2025.65 THREE-POINT INTEGRAL BOUNDARY-VALUE PROBLEMS FOR PIECEWISE FRACTIONAL IMPULSIVE DIFFERENTIAL EQUATIONS WITH p-LAPLACIAN OPERATOR XIAO CHEN, WENXUE ZHOU Abstract. We study the existence and uniqueness of solutions for three-point integral boundary- value problems of piecewise fractional impulsive differential equations with p-Laplacian operator and delay. We prove the existence of solutions, by using the ψ-Hadamard fractional definition, the Leray-Schauder nonlinear alternative, and the Krasnosel’skiis fixed point theorem. Subse- quently, we prove uniqueness of the solution using the Banach fixed point theorem. To verify the feasibility of the main results, we give two examples. 1. Introduction Fractional calculus, as an extension of classical calculus, originated from the conjectures of Leibniz and Euler, and it has evolved to the present day. In recent decades, boundary value prob- lems (BVPs) associated with fractional ordinary differential equations (FODEs) have garnered significant attention, driving the rapid development and broad application of fractional calculus in various fields such as fractional physics, viscoelastic mechanics, stochastic processes, and reaction- diffusion equations. Additionally, as a crucial aspect of fractional calculus, fractional differential equations with boundary conditions have garnered significant attention and produced a wide range of research outcomes [8, 11, 19, 20, 30, 31]. In 1983, during studies on turbulence in porous me- dia, Leibenson introduced the p-Laplacian operator. Since then, fractional differential equations involving the p-Laplacian operator have attracted considerable attention from scholars. However, most of the resulting research has been conducted under the standard definitions of Caputo and Riemann-Liouville fractional derivatives [7, 17, 22, 25, 28, 32]. In many continuous gradual pro- cesses, the state of a system undergoes sudden changes at certain moments due to disturbances or external influences. This phenomenon is known as impulsive effects. Because of the significant role of impulsive differential equations in describing such abrupt changes in electronic technology and communication engineering, they have become an important subject of research in recent years. Dynamical systems with impulsive phenomena have extensive applications in fields such as physics, biology, economics, and engineering. Differential equations with impulsive conditions are often used to simulate processes that exhibit discontinuous jumps and abrupt changes. The significance of studying fractional impulsive differential equations lies in their extension of classical differential equation theory, their ability to reveal new characteristics of complex systems, their capacity to provide precise models for practical problems, and their promotion of innovation in related mathematical methodologies. These theoretical achievements and application values make them an important frontier topic in interdisciplinary research across modern mathematics, physics, engineering, biology, and economics [2, 3, 14, 23, 24, 26]. In perfect conditions, state variables change over time following a consistent motion law, but nearly all systems experience delays due to limited motion speeds. Different time intervals cause variations in the motion laws of state variables, which interact through time lags τ and time leads 2020 Mathematics Subject Classification. 34B15, 34B30, 34B37, 45J05. Key words and phrases. ψ-Hadamard fractional derivative; p-Laplacian operator; impulsive; delay. ©2025. This work is licensed under a CC BY 4.0 license. Submitted February 24, 2025. Published June 30, 2025. 1 2 X. CHEN, W. ZHOU EJDE-2025/65 −τ , thereby making the dynamics more realistic. In the real world, piecewise differential sys- tems serve as effective mathematical models, reflecting the differing motion laws of state variables across distinct time segments. Piecewise fractional differential systems can more accurately depict complex dynamical phenomena characterized by discontinuities or abrupt changes prevalent in reality. In application domains such as signal processing, image restoration, robotic path plan- ning, and power system stabilization control, piecewise fractional differential systems offer more precise models and algorithms. They better accommodate nonlinearity, time-varying behavior, and discontinuities inherent in real data, thereby enhancing system performance and robustness. Zhang et al. [34] studied the existence, uniqueness, and stability of solutions for piecewise continuous fractional impulsive differential systems with delay: Dα tk u(t) + gk ( t, u(t), u(t+ (−1) k τ) ) = 0, t ∈ (tk, tk+1), k = 0, 1, . . . ,m, ∆u(t)|t=tk = Ik(u(tk)), ∆u ′(t)|t=tk = Īk(u(tk)), k = 1, 2, . . . ,m, u(0) = a, u′(1) = b, where 1 < α < 2, Dα tk is conformable fractional derivative of order α starting from tk (k = 0, 1, 2, . . . ,m). Poovarasan et al. [27] investigated the existence and uniqueness of solutions to the three-point impulsive boundary value problem with ψ-Caputo fractional derivative CDζ,ψu(t) = g(t, u(t)), t ∈ J0 := (0, 1), t ̸= tk, k ∈ Nm, ∆u|t=tk = Ik(u(tk)), ∆u′|t=tk = Īk(x(tk)), k ∈ Nm, u(0) + u′(0) = u(1) + u′(ϑ) = 0, where 1 ≤ ζ ≤ 2, CDζ,ψ is the ψ-Caputo fractional derivative. Ali et al. [4] studied the existence and uniqueness of solutions to a class of nonlinear implicit impulsive fractional differential equations with three-point boundary-value problems cDq tk u(t) = f(t, u(t), cDq tk u(t)), t ∈ (tk, tk+1], k = 0, 1, . . . , b, 1 < q ≤ 2, u(t)|t=0 = 0, u(t)|t=1 = λu(t)|t=η, λ, η ∈ (0, 1), ∆u(t)|t=tk = Pk(u(tk)), ∆u′(t)|t=tk = Qk(u(tk)), k = 1, 2, . . . , b, where cDq tk is the Caputo fractional derivative of order q starting from tk. In 2024, Balachandran et al. [10] defined a new class of Hadamard fractional integrals and fractional derivatives for function ψ. From existing research results, it has been found that there are few achievements in piecewise fractional impulsive differential equation systems with delay and p-Laplacian operator, thus having great research prospects. Inspired by the aforementioned work and the [18, 12, 1, 6, 15], this paper uses Leray-Schauder nonlinear alternative, Krasnosel’skiis fixed point theorem, and Banach fixed point theorem to discuss the existence and uniqueness of solutions for a class of piecewise impulsive fractional differential equations with p-Laplacian operator and delay under the definition of ψ-Hadamard fractional derivatives ϕp( HDα,ψ tk x(t)) = λhk(t, x(t), x(t+ (−1) k τ)), ∆x(tk) = Pk(x(tk)), ∆x′(tk) = Qk(x(tk)), x(0) = x′(0), x′(1) = ∫ η 0 g(s, x(s))ds, (1.1) where 1 < α ≤ 2, HDα,ψ tk is the ψ-Hadamard fractional derivative of order α starting from tk (k = 0, 1, 2, . . . ,m), λ > 0 is a parameter, 1 < p < 2, η ∈ (ta, ta+1), a is a nonnegative integer, v = sup1≤i≤m+1{lnψ(ti)− lnψ(ti−1)}. 0 ≤ a ≤ m, J = [0, 1], 0 = t0 < t1 < t2 < · · · < tm < tm+1 = 1, hk ∈ C([0, 1]× R2,R)(k = 0, 1, 2, . . . ,m), g ∈ C(J × R,R), Pk, Qk ∈ C(R,R) (k = 1, 2, . . . ,m), 0 ≤ τ ≤ min1≤k≤m+1{tk − tk−1}, and ∆x(t)|t=tk = x(t+k )−x(t − k ), ∆x ′(t)|t=tk = x′(t+k )− x′(t−k ), x(t + k ), x ′(t+k ) and x(t − k ), x ′(t−k ) represent the right and left limits of x(t), x′(t) at t = tk(k = 1, 2, . . . ,m), respectively. EJDE-2025/65 THREE-POINT INTEGRAL BOUNDARY-VALUE PROBLEMS 3 2. Preliminaries This section includes crucial definitions and lemmas necessary for proving the primary results. Let ϖ0 = [0, t1], ϖ1 = (t1, t2], ϖ2 = (t2, t3], . . . , ϖm−1 = (tm−1, tm], ϖm = (tm, 1], ϖ ′ = ϖ\{t1, t2, . . . , tm}. We denote ℑ(ϖ,R) = {ϖ → R : x ∈ C(ϖ′,R), x(t+k ), x(t − k ) exist, x(t − k ) = x(t+k ), k = 1, 2, . . . ,m}, with norm ∥x∥ℑ = supt∈ϖ |x(t)|, then ℑ(ϖ,R) is a Banach space. p ≥ 1, p−1 + q−1 = 1, ϕp(s) = |s|p−2s, (ϕp) −1 = ϕq, λ > 0 is a parameter, let ψ(t) be an increasing differentiable function on ℑ(ϖ,R), then ψ′(t) > 0. Let ψ∗ = mint∈ϖ ψ ′(t). Definition 2.1 ([10]). Let f ∈ C[a, b] and φ ∈ Cn[a, b] be an increasing function, with φ′(x) ̸= 0, for all x ∈ [a, b]. The fractional integral of the left-sided Hadamard functional of order α > 0 is defined as HIα,φa+ f(x) = 1 Γ(α) ∫ x a φ′(t) φ(t) (lnφ(x)− lnφ(t)) α−1 f(t)dt, and the fractional integral of the right-sided Hadamard functional of order α is defined as HIα,φb− f(x) = 1 Γ(α) ∫ b x φ′(t) φ(t) (lnφ(t)− lnφ(x)) α−1 f(t)dt. Definition 2.2 ([10]). Let n−1 < α < n, n ∈ N, f ∈ Cn−1[a, b], and φ ∈ Cn[a, b] be an increasing function, with φ′(x) ̸= 0, for all x ∈ [a, b]. The left-sided Hadamard functional fractional derivative of f of order α is defined as HDα,φ a+ f(x) = ( φ(x) φ′(x) d dx )n HIn−α,φa+ f(x) = 1 Γ(n− α) ( φ(x) φ′(x) d dx )n ∫ x a φ′(t) φ(t) (lnφ(x)− lnφ(t))n−α−1f(t)dt, and the right-hand sided of Hadamard functional fractional derivative of f of order α is HDα,φ b− f(x) = 1 Γ(n− α) ( − φ(x) φ′(x) d dx )n ∫ b x φ′(t) φ(t) (lnφ(t)− lnφ(x))n−α−1f(t)dt. Obviously, if 0 < α < 1 , then we have HDα,φ a+ f(x) = ( φ(x) φ′(x) d dx )nHI1−α,φa+ f(x) = 1 Γ(1− α) ( φ(x) φ′(x) d dx )∫ x a φ′(t) φ(t) (lnφ(x)− lnφ(t))−αf(t)dt, and HDα,φ b− f(x) = 1 Γ(1− α) ( − φ(x) φ′(x) d dx )∫ b x φ′(t) φ(t) (lnφ(t)− lnφ(x))−αf(t)dt. Lemma 2.3 ([10]). Let n− 1 < α, β < n, f ∈ C[a, b] and φ ∈ Cn[a, b]. Then HIα,φa+ HIβ,φa+ f(x) = HIα+β,φa+ f(x) and HIα,φb− HIβ,φb− f(x) = HIα+β,φb− f(x). For functions f ∈ C[a, b], φ ∈ Cn[a, b] and n− 1 < α, β < n with n ∈ N, we obtain HDα,φ a+ HIα,φa+ f(x) = f(x) and HDα,φ b− HIα,φb− f(x) = f(x). Lemma 2.4 ([10]). For each a function f ∈ Cn−1[a, b] and n− 1 < α, β < n with n ∈ N, we have HIα,φa+ HDα,φ a+ f(x) = f(x)− n−1∑ k=0 [ ( φ(t) φ′(t) d dt ) k f(t) ] t=a ( (lnφ(x)− lnφ(a)) k k! ) , and HIα,φb− HDα,φ b− f(x) = f(x)− n−1∑ k=0 [ (− φ(t) φ′(t) d dt ) k f(t) ] t=b ( (lnφ(b)− lnφ(x)) k k! ) . 4 X. CHEN, W. ZHOU EJDE-2025/65 Lemma 2.5 ([29]). Let ϕp be the p-Laplacian operator, then we have (i) If 1 < p < 2, ζ1, ζ2 > 0, and |ζ1|, |ζ2| ≥ ρ, then |ϕp(ζ1)− ϕq(ζ2)| ≤ (p− 1)ρp−2|ζ1 − ζ2|; (ii) If p > 2 and |ζ1|, |ζ2| < ρ, then |ϕp(ζ1)− ϕq(ζ2)| ≤ (p− 1)ρp−2|ζ1 − ζ2|. Now we have the Leray-Schauder nonlinear alternative. Lemma 2.6 ([13]). Let ℑ(ϖ,R) be the Banach space, Br ⊂ ℑ(ϖ,R) is a convex closed set, BNx is an open set relative to Br and 0 ∈ BNx . If Λ : B̄Nx → ℑ(ϖ,R) be a completely continuous operator and Λ(Ū) is bounded, then it satisfies (i) Λ has a fixed point in B̄Nx ; or (ii) there exists x ∈ ∂BNx , and γ1 ∈ (0, 1), such that x = γ1Λx. Now we have the Banach fixed point theorem. Lemma 2.7 ([16]). Let ℑ(ϖ,R) be a Banach space, and mapping Λ : ℑ(ϖ,R) → ℑ(ϖ,R) be a contraction on ℑ(ϖ,R). Then there is a unique x∗ ∈ ℑ(ϖ,R) with Λx∗ = x∗. Now we have the Krasnosel’skiis fixed point theorem. Lemma 2.8 ([9]). Let BR be a bounded closed convex non-empty subset on Banach space ℑ(ϖ,R) , where operators Ψ, T satisfy (i) Ψx1 + Tx2 ∈ BR, where x1, x2 ∈ BR; (ii) operator Ψ is compact and continuous; (iii) operator T is a contraction mapping, Then there exists z ∈ BR , such that z = Tz +Ψz. Now we have the Arzela-Ascoli theorem. Lemma 2.9 ([21]). Let ℑ(ϖ,R) be the Banach space, If Br is a compact set of ℑ(ϖ,R), and sequence {xn} is uniformly bounded and equicontinuous in Br, then this sequence has uniformly continuous subsequences in Br. Lemma 2.10. Let 1 < α ≤ 2, yk ∈ C(ϖ,R) (k = 0, 1, 2, . . . ,m), pk, qk ∈ R (k = 1, 2, . . . ,m). Then the linear piecewise fractional impulsive differential equation ϕp ( HDα,ψ tk x(t) ) = λyk(t), ∆x(tk) = pk, ∆x′(tk) = qk, x(0) = x′(0), x′(1) = ∫ η 0 g(s, x(s))ds, (2.1) is equivalent to the integral equation x(t) = ψ(1) ψ′(1) [ψ′(0) ψ(0) + lnψ(t)− lnψ(0) ] ∫ η 0 g(s, x(s))ds+ k∑ i=1 pi − m∑ i=1 χ(t, ti)qi ψ(ti) ψ′(ti) + λ Γ(α) k∑ i=1 ∫ ti ti−1 ψ′(s) ψ(s) (lnψ(ti)− lnψ(s)) α−1 ϕq(yi−1(s))ds + λ Γ(α) ∫ t tk ψ′(s) ψ(s) (lnψ(t)− lnψ(s))α−1ϕq(yk(s))ds − λ Γ(α− 1) m+1∑ i=1 ∫ ti ti−1 χ(t, ti) ψ′(s) ψ(s) (lnψ(ti)− lnψ(s)) α−2 ϕq(yi−1(s))ds, (2.2) where χ(t, s) =  ψ′(0) ψ(0) + lnψ(s)− lnψ(0), 0 ≤ s ≤ t ≤ 1, ψ′(0) ψ(0) + lnψ(t)− lnψ(0), 0 ≤ t ≤ s ≤ 1. (2.3) EJDE-2025/65 THREE-POINT INTEGRAL BOUNDARY-VALUE PROBLEMS 5 Proof. Suppose x = x(t) represents the solution to BVP(2.1), HDα,ψ tk x(t) = λϕq(yk(t)), HIα,ψtk HDα,ψ tk x(t) = λ Γ(α) ∫ t tk ψ′(s) ψ(s) (lnψ(t)− lnψ(s)) α−1 ϕq(yk(s))ds. For each t ∈ ϖ0, there are constants c0,0 and c0,1 that belong to the set of real numbers, we have x(t) = c0,0 + c0,1(lnψ(t)− lnψ(0)) + λ Γ(α) ∫ t 0 ψ′(s) ψ(s) (lnψ(t)− lnψ(s))α−1ϕq(y0(s))ds, x′(t) = c0,1 ψ′(t) ψ(t) + λ Γ(α− 1) ∫ t 0 ψ′(s)ψ′(t) ψ(s)ψ(t) (lnψ(t)− lnψ(s))α−2ϕq(y0(s))ds, x(t−1 ) = c0,0 + c0,1(lnψ(t1)− lnψ(0)) + λ Γ(α) ∫ t1 0 ψ′(s) ψ(s) (lnψ(t1)− lnψ(s))α−1ϕq(y0(s))ds, x′(t−1 ) = c0,1 ψ′(t1) ψ(t1) + λ Γ(α− 1) ∫ t1 0 ψ′(s)ψ′(t1) ψ(s)ψ(t1) (lnψ(t1)− lnψ(s))α−2ϕq(y0(s))ds. For each t ∈ ϖ1, there are real constants c1,0 and c1,1 such that x(t) = c1,0 + c1,1(lnψ(t)− lnψ(t1)) + λ Γ(α) ∫ t t1 ψ′(s) ψ(s) (lnψ(t)− lnψ(s))α−1ϕq(y1(s))ds, x′(t) = c1,1 ψ′(t) ψ(t) + λ Γ(α− 1) ∫ t t1 ψ′(s)ψ′(t) ψ(s)ψ(t) (lnψ(t)− lnψ(s))α−2ϕq(y1(s))ds, x(t+1 ) = c1,0, x′(t+1 ) = c1,1 ψ′(t1) ψ(t1) , x(t−2 ) = c1,0 + c1,1(lnψ(t2)− lnψ(t1)) + λ Γ(α) ∫ t2 t1 ψ′(s) ψ(s) (lnψ(t2)− lnψ(s))α−1ϕq(y1(s))ds, x′(t−2 ) = c1,1 ψ′(t2) ψ(t2) + λ Γ(α− 1) ∫ t2 t1 ψ′(s)ψ′(t2) ψ(s)ψ(t2) (lnψ(t2)− lnψ(s))α−2ϕq(y1(s))ds. From the impulsive conditions ∆x(t1) = p1, ∆x ′(t1) = q1, we have p1 = ∆x(t1) = x(t+1 )− x(t−1 ) = c1,0 − c0,0 − c0,1(lnψ(t1)− lnψ(0)) − λ Γ(α) ∫ t1 0 ψ′(s) ψ(s) (lnψ(t1)− lnψ(s))α−1ϕq(y0(s))ds, q1 = ∆x′(t1) = x′(t+1 )− x′(t−1 ) = c1,1 ψ′(t1) ψ(t1) − c0,1 ψ′(t1) ψ(t1) − λ Γ(α− 1) ∫ t1 0 ψ′(s)ψ′(t1) ψ(s)ψ(t1) (lnψ(t1)− lnψ(s))α−2ϕq(y0(s))ds. Then c1,0 − c0,0 = p1 + c0,1(lnψ(t1)− lnψ(0)) + λ Γ(α) ∫ t1 0 ψ′(s) ψ(s) (lnψ(t1)− lnψ(s))α−1ϕq(y0(s))ds, (2.4) c1,1 − c0,1 = q1 ψ(t1) ψ′(t1) + λ Γ(α− 1) ∫ t1 0 ψ′(s) ψ(s) (lnψ(t1)− lnψ(s))α−2ϕq(y0(s))ds. (2.5) 6 X. CHEN, W. ZHOU EJDE-2025/65 For each t ∈ ϖ2, there are real constants c2,0 and c2,1 such that x(t) = c2,0 + c2,1(lnψ(t)− lnψ(t2)) + λ Γ(α) ∫ t t2 ψ′(s) ψ(s) (lnψ(t)− lnψ(s))α−1ϕq(y2(s))ds, x′(t) = c2,1 ψ′(t) ψ(t) + λ Γ(α− 1) ∫ t t2 ψ′(s)ψ′(t) ψ(s)ψ(t) (lnψ(t)− lnψ(s))α−2ϕq(y2(s))ds, x(t+2 ) = c2,0, x′(t+2 ) = c2,1 ψ′(t2) ψ(t2) . From the impulsive conditions ∆x(t2) = p2, ∆x ′(t2) = q2, we have p2 = ∆x(t2) = x(t+2 )− x(t−2 ) = c2,0 − c1,0 − c1,1(lnψ(t2)− lnψ(t1)) − λ Γ(α) ∫ t2 t1 ψ′(s) ψ(s) (lnψ(t2)− lnψ(s))α−1ϕq(y1(s))ds, q2 = ∆x′(t2) = x′(t+2 )− x′(t−2 ) = c2,1 ψ′(t2) ψ(t2) − c1,1 ψ′(t2) ψ(t2) − λ Γ(α− 1) ∫ t2 t1 ψ′(s)ψ′(t2) ψ(s)ψ(t2) (lnψ(t2)− lnψ(s))α−2ϕq(y1(s))ds. Then c2,0 − c1,0 = p2 + c1,1(lnψ(t2)− lnψ(t1)) + λ Γ(α) ∫ t2 t1 ψ′(s) ψ(s) (lnψ(t2)− lnψ(s))α−1ϕq(y1(s))ds, (2.6) c2,1 − c1,1 = q2 ψ(t2) ψ′(t2) + λ Γ(α− 1) ∫ t2 t1 ψ′(s) ψ(s) (lnψ(t2)− lnψ(s))α−2ϕq(y1(s))ds. (2.7) In the same manner, for each t ∈ ϖk, there are constants ck,0 and ck,1 that belong to the set of real numbers, we have x(t) = ck,0 + ck,1(lnψ(t)− lnψ(tk)) + λ Γ(α) ∫ t tk ψ′(s) ψ(s) (lnψ(t)− lnψ(s))α−1ϕq(yk(s))ds, x′(t) = ck,1 ψ′(t) ψ(t) + λ Γ(α− 1) ∫ t tk ψ′(s)ψ′(t) ψ(s)ψ(t) (lnψ(t)− lnψ(s))α−2ϕq(yk(s))ds, (2.8) ck,0 − ck−1,0 = pk + ck−1,1(lnψ(tk)− lnψ(tk−1)) + λ Γ(α) ∫ tk tk−1 ψ′(s) ψ(s) (lnψ(tk)− lnψ(s))α−1ϕq(yk−1(s))ds, ck,1 − ck−1,1 = qk ψ(tk) ψ′(tk) + λ Γ(α− 1) ∫ tk tk−1 ψ′(s) ψ(s) (lnψ(tk)− lnψ(s))α−2ϕq(yk−1(s))ds. (2.9) In particular, for each t ∈ ϖm, there are constants cm,0 and cm,1 that belong to the set of real numbers, we have x(t) = cm,0 + cm,1(lnψ(t)− lnψ(tm)) + λ Γ(α) ∫ t tm ψ′(s) ψ(s) (lnψ(t)− lnψ(s))α−1ϕq(ym(s))ds, x′(t) = cm,1 ψ′(t) ψ(t) + λ Γ(α− 1) ∫ t tm ψ′(s)ψ′(t) ψ(s)ψ(t) (lnψ(t)− lnψ(s))α−2ϕq(ym(s))ds, (2.10) EJDE-2025/65 THREE-POINT INTEGRAL BOUNDARY-VALUE PROBLEMS 7 cm,0 − cm−1,0 = pm + cm−1,1(lnψ(tm)− lnψ(tm−1)) + λ Γ(α) ∫ tm tm−1 ψ′(s) ψ(s) (lnψ(tm)− lnψ(s))α−1ϕq(ym−1(s))ds, cm,1 − cm−1,1 = qm ψ(tm) ψ′(tm) + λ Γ(α− 1) ∫ tm tm−1 ψ′(s) ψ(s) (lnψ(tm)− lnψ(s))α−2ϕq(ym−1(s))ds. (2.11) From (2.4)-(2.11), we have c1,0 − c0,0 = p1 + c0,1(lnψ(t1)− lnψ(0)) + λ Γ(α) ∫ t1 0 ψ′(s) ψ(s) (lnψ(t1)− lnψ(s))α−1ϕq(y0(s))ds, c2,0 − c1,0 = p2 + c1,1(lnψ(t2)− lnψ(t1)) + λ Γ(α) ∫ t2 t1 ψ′(s) ψ(s) (lnψ(t2)− lnψ(s))α−1ϕq(y1(s))ds, . . . ck,0 − ck−1,0 = pk + ck−1,1(lnψ(tk)− lnψ(tk−1)) + λ Γ(α) ∫ tk tk−1 ψ′(s) ψ(s) (lnψ(tk)− lnψ(s))α−1ϕq(yk−1(s))ds, . . . cm,0 − cm−1,0 = pm + cm−1,1(lnψ(tm)− lnψ(tm−1)) + λ Γ(α) ∫ tm tm−1 ψ′(s) ψ(s) (lnψ(tm)− lnψ(s))α−1ϕq(ym−1(s))ds, and c1,1 − c0,1 = q1 ψ(t1) ψ′(t1) + λ Γ(α− 1) ∫ t1 0 ψ′(s) ψ(s) (lnψ(t1)− lnψ(s))α−2ϕq(y0(s))ds, c2,1 − c1,1 = q2 ψ(t2) ψ′(t2) + λ Γ(α− 1) ∫ t2 t1 ψ′(s) ψ(s) (lnψ(t2)− lnψ(s))α−2ϕq(y1(s))ds, . . . ck,1 − ck−1,1 = qk ψ(tk) ψ′(tk) + λ Γ(α− 1) ∫ tk tk−1 ψ′(s) ψ(s) (lnψ(tk)− lnψ(s))α−2ϕq(yk−1(s))ds, . . . cm,1 − cm−1,1 = qm ψ(tm) ψ′(tm) + λ Γ(α− 1) ∫ tm tm−1 ψ′(s) ψ(s) (lnψ(tm)− lnψ(s))α−2ϕq(ym−1(s))ds. Then ck,0 − c0,0 = k∑ i=1 pi + k∑ i=1 ci−1,1(lnψ(ti)− lnψ(ti−1)) + λ Γ(α) k∑ i=1 ∫ ti ti−1 ψ′(s) ψ(s) (lnψ(ti)− lnψ(s)) α−1 ϕq(yi−1(s))ds, (2.12) cm,0 − c0,0 = m∑ i=1 pi + m∑ i=1 ci−1,1(lnψ(ti)− lnψ(ti−1)) + λ Γ(α) m∑ i=1 ∫ ti ti−1 ψ′(s) ψ(s) (lnψ(ti)− lnψ(s)) α−1 ϕq(yi−1(s))ds, (2.13) 8 X. CHEN, W. ZHOU EJDE-2025/65 and ck,1 − c0,1 = k∑ i=1 qi ψ(ti) ψ′(ti) + λ Γ(α− 1) k∑ i=1 ∫ ti ti−1 ψ′(s) ψ(s) (lnψ(ti)− lnψ(s)) α−2 ϕq(yi−1(s))ds, (2.14) cm,1 − c0,1 = m∑ i=1 qi ψ(ti) ψ′(ti) + λ Γ(α− 1) m∑ i=1 ∫ ti ti−1 ψ′(s) ψ(s) (lnψ(ti)− lnψ(s)) α−2 ϕq(yi−1(s))ds. (2.15) Applying the boundary conditions x(0) = x′(0), x′(1) = ∫ η 0 g(s, x(s))ds, we obtain x′(0) = c0,1 ψ′(0) ψ(0) , x(0) = c0,0, x′(1) = cm,1 ψ′(1) ψ(1) + λ Γ(α− 1) ∫ 1 tm ψ′(s)ψ′(1) ψ(s)ψ(1) (lnψ(1)− lnψ(s))α−2ϕq(ym(s))ds. Then c0,0 = c0,1 ψ′(0) ψ(0) , cm,1 = ψ(1) ψ′(1) ∫ η 0 g(s, x(s))ds− λ Γ(α− 1) ∫ 1 tm ψ′(s) ψ(s) (lnψ(1)− lnψ(s))α−2ϕq(ym(s))ds. From (2.15), we have c0,1 = cm,1 − m∑ i=1 qi ψ(ti) ψ′(ti) − λ Γ(α− 1) m∑ i=1 ∫ ti ti−1 ψ′(s) ψ(s) (lnψ(ti)− lnψ(s)) α−2 ϕq(yi−1(s))ds = ψ(1) ψ′(1) ∫ η 0 g(s, x(s))ds− λ Γ(α− 1) ∫ 1 tm ψ′(s) ψ(s) (lnψ(1)− lnψ(s))α−2ϕq(ym(s))ds − m∑ i=1 qi ψ(ti) ψ′(ti) − λ Γ(α− 1) m∑ i=1 ∫ ti ti−1 ψ′(s) ψ(s) (lnψ(ti)− lnψ(s)) α−2 ϕq(yi−1(s))ds, c0,0 = ψ(1) ψ′(1) ψ′(0) ψ(0) ∫ η 0 g(s, x(s))ds− ψ′(0) ψ(0) m∑ i=1 qi ψ(ti) ψ′(ti) − λ Γ(α− 1) ψ′(0) ψ(0) ∫ 1 tm ψ′(s) ψ(s) (lnψ(1)− lnψ(s))α−2ϕq(ym(s))ds − λ Γ(α− 1) ψ′(0) ψ(0) m∑ i=1 ∫ ti ti−1 ψ′(s) ψ(s) (lnψ(ti)− lnψ(s)) α−2 ϕq(yi−1(s))ds. From (2.12) and (2.14), we have ck,0 = ψ(1) ψ′(1) ψ′(0) ψ(0) ∫ η 0 g(s, x(s))ds− ψ′(0) ψ(0) m∑ i=1 qi ψ(ti) ψ′(ti) − λ Γ(α− 1) ψ′(0) ψ(0) m+1∑ i=1 ∫ ti ti−1 ψ′(s) ψ(s) (lnψ(ti)− lnψ(s)) α−2 ϕq(yi−1(s))ds + k∑ i=1 pi + k∑ i=1 ci−1,1 ( lnψ(ti)− lnψ(ti−1) ) + λ Γ(α) k∑ i=1 ∫ ti ti−1 ψ′(s) ψ(s) (lnψ(ti)− lnψ(s)) α−1 ϕq(yi−1(s))ds, EJDE-2025/65 THREE-POINT INTEGRAL BOUNDARY-VALUE PROBLEMS 9 ck,1 = ψ(1) ψ′(1) ∫ η 0 g(s, x(s))ds− m∑ i=1 qi ψ(ti) ψ′(ti) − λ Γ(α− 1) m+1∑ i=1 ∫ ti ti−1 ψ′(s) ψ(s) (lnψ(ti)− lnψ(s)) α−2 ϕq(yi−1(s))ds + k∑ i=1 qi ψ(ti) ψ′(ti) + λ Γ(α− 1) k∑ i=1 ∫ ti ti−1 ψ′(s) ψ(s) (lnψ(ti)− lnψ(s)) α−2 ϕq(yi−1(s))ds = ψ(1) ψ′(1) ∫ η 0 g(s, x(s))ds− m∑ i=k+1 qi ψ(ti) ψ′(ti) − λ Γ(α− 1) m+1∑ i=k+1 ∫ ti ti−1 ψ′(s) ψ(s) (lnψ(ti)− lnψ(s)) α−2 ϕq(yi−1(s))ds. So, for each t ∈ ϖk, x(t) = ψ(1) ψ′(1) ψ′(0) ψ(0) ∫ η 0 g(s, x(s))ds− ψ′(0) ψ(0) m∑ i=1 qi ψ(ti) ψ′(ti) − λ Γ(α− 1) ψ′(0) ψ(0) m+1∑ i=1 ∫ ti ti−1 ψ′(s) ψ(s) (lnψ(ti)− lnψ(s)) α−2 ϕq(yi−1(s))ds + k∑ i=1 pi + k∑ i=1 ci−1,1(lnψ(ti)− lnψ(ti−1)) + λ Γ(α) k∑ i=1 ∫ ti ti−1 ψ′(s) ψ(s) (lnψ(ti)− lnψ(s)) α−1 ϕq(yi−1(s))ds + ψ(1) ψ′(1) (lnψ(t)− lnψ(tk)) ∫ η 0 g(s, x(s))ds− (lnψ(t)− lnψ(tk)) m∑ i=k+1 qi ψ(ti) ψ′(ti) − λ Γ(α− 1) (lnψ(t)− lnψ(tk)) m+1∑ i=k+1 ∫ ti ti−1 ψ′(s) ψ(s) (lnψ(ti)− lnψ(s)) α−2 ϕq(yi−1(s))ds + λ Γ(α) ∫ t tk ψ′(s) ψ(s) (lnψ(t)− lnψ(s))α−1ϕq(yk(s))ds. Thus x(t) = ψ(1) ψ′(1) [ψ′(0) ψ(0) + lnψ(t)− lnψ(0) ] ∫ η 0 g(s, x(s))ds+ k∑ i=1 pi − m∑ i=1 χ(t, ti)qi ψ(ti) ψ′(ti) + λ Γ(α) k∑ i=1 ∫ ti ti−1 ψ′(s) ψ(s) (lnψ(ti)− lnψ(s)) α−1 ϕq(yi−1(s))ds + λ Γ(α) ∫ t tk ψ′(s) ψ(s) (lnψ(t)− lnψ(s))α−1ϕq(yk(s))ds − λ Γ(α− 1) m+1∑ i=1 ∫ ti ti−1 χ(t, ti) ψ′(s) ψ(s) (lnψ(ti)− lnψ(s)) α−2 ϕq(yi−1(s))ds. The proof is complete. □ From the definition of χ(t, s) we obtain |χ(t, s)| ≤ ∣∣ψ′(0) ψ(0) + lnψ(1)− lnψ(0) ∣∣. 10 X. CHEN, W. ZHOU EJDE-2025/65 We define operator Λ : ℑ(ϖ,R) → ℑ(ϖ,R) as (Λx)(t) = ψ(1) ψ′(1) [ ψ′(0) ψ(0) + lnψ(t)− lnψ(0)] ∫ η 0 g(s, x(s))ds+ k∑ i=1 Pi(x(ti)) − m∑ i=1 χ(t, ti)Qi(x(ti)) ψ(ti) ψ′(ti) + λ Γ(α) × k∑ i=1 ∫ ti ti−1 ψ′(s) ψ(s) (lnψ(ti)− lnψ(s)) α−1 ϕq ( hi−1 ( s, x(s), x(s+ (−1) i−1 τ) )) ds + λ Γ(α) ∫ t tk ψ′(s) ψ(s) (lnψ(t)− lnψ(s))α−1ϕq ( hk(s, x(s), x ( s+ (−1) k τ ) ) ) ds− λ Γ(α− 1) × m+1∑ i=1 ∫ ti ti−1 χ(t, ti) ψ′(s) ψ(s) (lnψ(ti)− lnψ(s)) α−2 ϕq ( hi−1 ( s, x(s), x(s+ (−1) i−1 τ) )) ds. Therefore, the existence of a solution to BVP (1.1) can be transformed into the existence of a fixed point of the operator Λ on ℑ(ϖ,R). For convenience in calculations, we define Θ1 = ψ′(0) ψ(0) + lnψ(1)− lnψ(0), Φ1 = (q − 1)(σ∗)q−2(M∗ h +N∗ h), Φ2 = λ(m+ 1)(vα +Θ1αv α−1) Γ(α+ 1) . 3. Existence and uniqueness of solutions We obtain the existence and uniqueness of solutions for BVP (1.1) by applying three fixed point theorems. First, we state some assumptions needed. (H1) There exist functions σk ∈ C(ϖk, [0,+∞)) (k = 0, 1, 2, . . . ,m), such that for all uk, vk ∈ R (k = 0, 1, 2, . . . ,m) and t ∈ ϖk, we have |hk(t, uk, vk)| ≤ σk(t); and we denote σ∗ = max { sup t∈ϖ0 σ0(t), sup t∈ϖ1 σ1(t), sup t∈ϖ2 σ2(t), . . . , sup t∈ϖm σm(t) } . (H2) There exist constants M1, N1,M2, N2 > 0, such that, for any u ∈ R, |Pk(u)| ≤M1|u|+N1, |Qk(u)| ≤M2|u|+N2, k = 1, 2, . . . ,m. (H3) There exist constants M∗ h , N ∗ h > 0, such that, for any uk, vk, ūk, v̄k ∈ R, t ∈ ϖk (k = 0, 1, 2, . . . ,m), we have |hk(t, uk, vk)− hk(t, ūk, v̄k)| ≤M∗ h |uk − vk|+N∗ h |ūk − v̄k|, k = 0, 1, 2, . . . ,m. (H4) There exist constants l, l∗ > 0, such that, for any u, ū ∈ R, |Pk(u)− Pk(ū)| ≤ l|u− ū|, |Qk(u)−Qk(ū)| ≤ l∗|u− ū|, k = 1, 2, . . . ,m. (H5) Function g ∈ C(ϖ × R,R), and there exists a function φ1(t) ∈ L1/2(ϖ,R+), such that, for all u, ū ∈ R, t ∈ ϖ, we have |g(t, u)− g(t, ū)| ≤ φ1(t)|u− ū|, where ∥φ1∥ = ( ∫ 1 0 φ1 2(s)ds)1/2. (H6) There exist constants φ∗ 2, φ ∗ 3 > 0, such that, for all u ∈ R, t ∈ ϖ, |g(t, u)| ≤ φ∗ 2 + φ∗ 3|u|. EJDE-2025/65 THREE-POINT INTEGRAL BOUNDARY-VALUE PROBLEMS 11 (H7) There exists a constant Nx > 0, such that Nx ψ(1) ψ′(1)Θ1(φ∗ 2 + φ∗ 3Nx)η +m(M1Nx +N1) + Θ1(M2Nx +N2) mψ(1) ψ∗ + [Φ1Nx + ϕq(K)]Φ2 > 1, where supt∈ϖk |hk(t, 0, 0)| = K <∞, (k = 0, 1, 2, . . . ,m). Theorem 3.1. Under assumptions (H1)–(H7), BVP (1.1) has at least one solution. Proof. When t ∈ ϖk (k = 0, 1, 2, . . . ,m), for any x1, x2 ∈ ℑ(ϖ,R), according to conditions (H1), (H3), and lemma 2.5, we have∣∣ϕq(hk(t, x1(t), x1(t+ (−1) k τ)))− ϕq(hk(t, x2(t), x2(t+ (−1) k τ))) ∣∣ ≤ (q − 1)(σ∗)q−2(M∗ h |x2(t)− x1(t)|+N∗ h |x1(t+ (−1) k τ)− x2(t+ (−1) k τ)|) ≤ (q − 1)(σ∗)q−2(M∗ h +N∗ h)∥x2 − x1∥ℑ, according to the definition of operator Λ and the continuity of hk (k = 0, 1, 2, . . . ,m), operator Λ is continuous. For any r > 0, denote Br = {x ∈ ℑ(ϖ,R) : ∥x∥ ≤ r}, it is easy to check that Br is a bounded closed ball in ℑ(ϖ,R). Firstly, prove that operator Λ maps a bounded set in ℑ(ϖ,R) to a bounded set. There exists a constant ρ1, for any x ∈ Br, we have ∥Λx∥ ≤ ρ1. Let supt∈ϖk |hk(t, 0, 0)| = K <∞. For any t ∈ ϖk(k = 0, 1, 2, . . . ,m), x ∈ Br, we have |(Λx)(t)| ≤ ψ(1) ψ′(1) [ψ′(0) ψ(0) + lnψ(t)− lnψ(0) ] ∫ η 0 |g(s, x(s))|ds+ k∑ i=1 |Pi(x(ti))| + m∑ i=1 |χ(t, ti)||Qi(x(ti))| ψ(ti) ψ′(ti) + λ Γ(α) × k∑ i=1 ∫ ti ti−1 ψ′(s) ψ(s) (lnψ(ti) − lnψ(s))α−1 ∣∣ϕq(hi−1(s, x(s), x(s+ (−1) i−1 τ)))− ϕq(hi−1(s, 0, 0)) ∣∣ds + λ Γ(α) k∑ i=1 ∫ ti ti−1 ψ′(s) ψ(s) (lnψ(ti)− lnψ(s)) α−1|ϕq(hi−1(s, 0, 0))|ds + λ Γ(α) ∫ t tk ψ′(s) ψ(s) (lnψ(t) − lnψ(s))α−1 ∣∣ϕq(hk(s, x(s), x(s+ (−1) k τ)))− ϕq(hk(s, 0, 0)) ∣∣ds + λ Γ(α) ∫ t tk ψ′(s) ψ(s) (lnψ(t)− lnψ(s))α−1|ϕq(hk(s, 0, 0))|ds+ λ Γ(α− 1) × m+1∑ i=1 ∫ ti ti−1 |χ(t, ti)| ψ′(s) ψ(s) (lnψ(ti) − lnψ(s)α−2 ∣∣ϕq(hi−1(s, x(s), x(s+ (−1) i−1 τ)))− ϕq(hi−1(s, 0, 0)) ∣∣ds + λ Γ(α− 1) m+1∑ i=1 ∫ ti ti−1 |χ(t, ti)| ψ′(s) ψ(s) (lnψ(ti)− lnψ(s)) α−2|ϕq(hi−1(s, 0, 0))|ds ≤ ψ(1) ψ′(1) (ψ′(0) ψ(0) + lnψ(1)− lnψ(0) ) (φ∗ 2 + φ∗ 3∥x∥)η +m(M1∥x∥+N1) + ( ψ′(0) ψ(0) + lnψ(1)− lnψ(0))(M2∥x∥+N2) mψ(1) ψ∗ 12 X. CHEN, W. ZHOU EJDE-2025/65 + (q − 1)(σ∗)q−2(M∗ h +N∗ h)∥x∥ λ(m+ 1)vα Γ(α+ 1) + λ(m+ 1)vα Γ(α+ 1) ϕq(K) + (ψ′(0) ψ(0) + lnψ(1)− lnψ(0) ) (q − 1)(σ∗)q−2(M∗ h +N∗ h)∥x∥ (m+ 1)λvα−1 Γ(α) + (ψ′(0) ψ(0) + lnψ(1)− lnψ(0) ) (m+ 1)λvα−1 Γ(α) ϕq(K). Then ∥Λx∥ ≤ ψ(1) ψ′(1) Θ1(φ ∗ 2 + φ∗ 3r)η +m(M1r +N1) + Θ1(M2r +N2) mψ(1) ψ∗ + λ(m+ 1)(q − 1)(σ∗)q−2(M∗ h +N∗ h)r vα +Θ1αv α−1 Γ(α+ 1) + λ(m+ 1)(vα + αΘ1v α−1) Γ(α+ 1) ϕq(K) = ρ1. From this we infer that Λ maps the bounded set in ℑ(ϖ,R) to a bounded set. Secondly, we show that Λ is equicontinuous. For any x ∈ Br, t1, t2 ∈ ϖk (k = 0, 1, 2, . . . ,m), where t1 < t2, we have |(Λx)(t2)− (Λx)(t1)| ≤ ψ(1) ψ′(1) [lnψ(t2)− lnψ(t1)] ∫ η 0 |g(s, x(s))|ds+ m∑ i=1 |χ(t2, ti)− χ(t1, ti)||Qi(x(ti))| ψ(ti) ψ′(ti) + λ Γ(α) ∫ t2 tk ψ′(s) ψ(s) (lnψ(t2)− lnψ(s))α−1 ∣∣∣ϕq(hk(s, x(s), x(s+ (−1) k τ) ))∣∣∣ds − λ Γ(α) ∫ t1 tk ψ′(s) ψ(s) (lnψ(t1)− lnψ(s))α−1 ∣∣∣ϕq(hk(s, x(s), x(s+ (−1) k τ) ))∣∣∣ds+ λ Γ(α− 1) × m+1∑ i=1 ∫ ti ti−1 |χ(t2, ti)− χ(t1, ti)| ψ′(s) ψ(s) (lnψ(ti) − lnψ(s))α−2 ∣∣∣ϕq(hi−1(s, x(s), x ( s+ (−1) i−1 τ ) ) )∣∣∣ds ≤ ψ(1) ψ′(1) [lnψ(t2)− lnψ(t1)] ∫ η 0 |g(s, x(s))|ds+ [lnψ(t2)− lnψ(t1)] m∑ i=1 |Qi(x(ti))| ψ(ti) ψ′(ti) + λ Γ(α) ∫ t2 tk ψ′(s) ψ(s) (lnψ(t2)− lnψ(s))α−1 ∣∣∣ϕq(hk(s, x(s), x(s+ (−1) k τ)) ) − ϕq ( hk(s, 0, 0) )∣∣∣ds + λ Γ(α) ∫ t2 tk ψ′(s) ψ(s) (lnψ(t2)− lnψ(s))α−1 ∣∣ϕq(hk(s, 0, 0))∣∣ds − λ Γ(α) ∫ t1 tk ψ′(s) ψ(s) (lnψ(t1)− lnψ(s))α−1 ∣∣ϕq(hk(s, x(s), x(s+ (−1) k τ)))− ϕq(hk(s, 0, 0)) ∣∣ds − λ Γ(α) ∫ t1 tk ψ′(s) ψ(s) (lnψ(t1)− lnψ(s))α−1 ∣∣ϕq(hk(s, 0, 0))∣∣ds+ λ Γ(α− 1) [lnψ(t2)− lnψ(t1)] × m+1∑ i=1 ∫ ti ti−1 ψ′(s) ψ(s) (lnψ(ti)− lnψ(s)) α−2|ϕq(hi−1(s, x(s), x(s+ (−1) i−1 τ)))− ϕq(hi−1(s, 0, 0))|ds + λ Γ(α− 1) [lnψ(t2)− lnψ(t1)] m+1∑ i=1 ∫ ti ti−1 ψ′(s) ψ(s) (lnψ(ti)− lnψ(s)) α−2|ϕq(hi−1(s, 0, 0))|ds ≤ ψ(1) ψ′(1) [lnψ(t2)− lnψ(t1)](φ ∗ 2 + φ∗ 3r)η + [lnψ(t2)− lnψ(t1)]m(M2r +N2) ψ(1) ψ∗ EJDE-2025/65 THREE-POINT INTEGRAL BOUNDARY-VALUE PROBLEMS 13 + (q − 1)(σ∗)q−2(M∗ h +N∗ h)r λ Γ(α+ 1) {[lnψ(t2)− lnψ(tk)] α − [lnψ(t1)− lnψ(tk)] α} + λ Γ(α+ 1) ϕq(K){[lnψ(t2)− lnψ(tk)] α − [lnψ(t1)− lnψ(tk)] α} + (m+ 1)[lnψ(t2)− lnψ(t1)](q − 1)(σ∗)q−2(M∗ h +N∗ h)r λvα−1 Γ(α) + λvα−1 Γ(α) (m+ 1)[lnψ(t2)− lnψ(t1)]ϕq(K). Therefore, |(Λx)(t2)− (Λx)(t1)| → 0, as t1 → t2. By Arzela-Ascoli theorem, Λ is compact, thus Λ : ℑ(ϖ,R) → ℑ(ϖ,R) is a completely continuous operator. If x is a solution to the BVP(1.1), for any t ∈ ϖk(k = 0, 1, 2, . . . ,m), similar to the previous proof method, we have ∥x∥ ≤ ψ(1) ψ′(1) Θ1(φ ∗ 2 + φ∗ 3∥x∥)η +m(M1∥x∥+N1) + Θ1(M2∥x∥+N2) mψ(1) ψ∗ + λ(m+ 1)Φ1∥x∥ vα +Θ1αv α−1 Γ(α+ 1) + λ(m+ 1)(vα + αΘ1v α−1) Γ(α+ 1) ϕq(K); thus, ∥x∥ ψ(1) ψ′(1)Θ1(φ∗ 2 + φ∗ 3∥x∥)η +m(M1∥x∥+N1) + Θ1(M2∥x∥+N2) mψ(1) ψ∗ + [Φ1∥x∥+ ϕq(K)]Φ2 ≤ 1, according to (H7), there exists a constant Nx, such that ∥x∥ ≠ Nx. Assume BNx = {x ∈ ℑ(ϖ,R) : ∥x∥ < Nx}, because Λ is a completely continuous operator, considering the choice of BNx , with respect to a particular γ1 ∈ (0, 1). There is no x ∈ B̄Nx , such that x = γ1Λx. From Lemma 2.6, it can be inferred that Λ has at least one fixed point x ∈ B̄Nx , which means that the BVP(1.1) has at least one solution. The verification has been accomplished. □ Theorem 3.2. Under assumptions (H3)–(H5), if M1 = ψ(1) ψ′(1) Θ1 √ η∥φ1∥+ml + ψ(1) ψ∗ Θ1ml ∗ + λαΘ1(m+ 1)vα−1 + λ(m+ 1)vα Γ(α+ 1) Φ1 < 1, then BVP (1.1) has a unique solution. Proof. For each x1, x2 ∈ ℑ(ϖ,R) and t ∈ ϖk(k = 0, 1, 2, . . . ,m), by (H5), we obtain∫ η 0 |g(s, x2(s))− g(s, x1(s))|ds ≤ ∥x2 − x1∥ ∫ η 0 φ1(s)ds ≤ ∥x2 − x1∥ (∫ η 0 12ds )1/2(∫ η 0 φ1 2(s)ds )1/2 ≤ ∥x2 − x1∥ √ η (∫ 1 0 φ1 2(s)ds )1/2 ≤ ∥x2 − x1∥ √ η∥φ1∥. (3.1) From this inequality we have |(Λx2)(t)− (Λx1)(t)| ≤ ψ(1) ψ′(1) Θ1 ∫ η 0 |g(s, x2(s))− g(s, x1(s))|ds+ m∑ i=1 |Pi(x2(ti))− Pi(x1(ti))| +Θ1 m∑ i=1 |Qi(x2(ti))−Qi(x1(ti))| ψ(ti) ψ′(ti) + λ Γ(α) m∑ i=1 ∫ ti ti−1 ψ′(s) ψ(s) (lnψ(ti) 14 X. CHEN, W. ZHOU EJDE-2025/65 − lnψ(s))α−1 ∣∣∣ϕq(hi−1 ( s, x2(s), x2(s+ (−1) i−1 τ) )) − ϕq ( hi−1 ( s, x1(s), x1(s+ (−1) i−1 τ) ))∣∣∣ds+ λ Γ(α) ∫ t tk ψ′(s) ψ(s) lnψ(t) − lnψ(s))α−1 ∣∣∣ϕq(hk(s, x2(s), x2(s+ (−1) k τ) )) − ϕq ( hk ( s, x1(s), x1(s+ (−1) k τ) ))∣∣∣ds + λΘ1 Γ(α− 1) m+1∑ i=1 ∫ ti ti−1 ψ′(s) ψ(s) (lnψ(ti) − lnψ(s))α−2 ∣∣∣ϕq(hi−1(s, x2(s), x2(s+ (−1) i−1 τ))) − ϕq ( hi−1 ( s, x1(s), x1(s+ (−1) i−1 τ) ))∣∣∣ds ≤ { ψ(1) ψ′(1) Θ1 √ η∥φ1∥+ml + ψ(1) ψ∗ Θ1ml ∗ + λαΘ1(m+ 1)vα−1 + λ(m+ 1)vα Γ(α+ 1) Φ1 } ∥x2 − x1∥. Then ∥(Λx2)(t)− (Λx1)(t)∥ ≤ { ψ(1) ψ′(1) Θ1 √ η∥φ1∥+ml + ψ(1) ψ∗ Θ1ml ∗ + λαΘ1(m+ 1)vα−1 + λ(m+ 1)vα Γ(α+ 1) Φ1}∥x2 − x1∥. Thus, ∥(Λx2)(t)− (Λx1)(t)∥ < M1∥x2 − x1∥. Therefore, by lemma 2.7, it can be inferred that Λ is compressed and has a unique fixed point on ℑ(ϖ,R), that is, the BVP (1.1) has a unique solution. □ For the next theorem we need the following assumptions; (H8) There exists a constant ς > 0, such that for any uk, vk ∈ R (k = 0, 1, 2, . . . ,m) and t ∈ ϖk, we have |ϕq(hk(t, uk, vk))| < ς, k = 0, 1, 2, . . . ,m; (H9) ψ(1) ψ′(1) Θ1∥φ1∥ √ η +ml +mΘ1l ∗ψ(1) ψ∗ < 1 . Theorem 3.3. Under assumptions (H2), (H4)–(H6), (H8), (H9), the BVP (1.1) has at least one solution. Proof. We denote BR = {x ∈ ℑ(ϖ,R) : ∥x∥ ≤ R}, where R > Φ2ς + ψ(1) ψ′(1)Θ1φ ∗ 2η +mN1 +Θ1N2 mψ(1) ψ∗ 1− ψ(1) ψ′(1)Θ1φ∗ 3η −mM1 −M2Θ1 mψ(1) ψ∗ . Then BR is a bounded closed convex non-empty subset on Banach space ℑ(ϖ,R). We define the two operators Ψ, T on BR as (Ψx)(t) = λ Γ(α) k∑ i=1 ∫ ti ti−1 ψ′(s) ψ(s) (lnψ(ti)− lnψ(s)) α−1 ϕq ( hi−1(s, x(s), x ( s+ (−1) i−1 τ ) ) ) ds + λ Γ(α) ∫ t tk ψ′(s) ψ(s) (lnψ(t)− lnψ(s))α−1ϕq ( hk ( s, x(s), x ( s+ (−1) k τ ))) ds− λ Γ(α− 1) × m+1∑ i=1 ∫ ti ti−1 χ(t, ti) ψ′(s) ψ(s) (lnψ(ti)− lnψ(s)) α−2 ϕq ( hi−1 ( s, x(s), x ( s+ (−1) i−1 τ ))) ds, EJDE-2025/65 THREE-POINT INTEGRAL BOUNDARY-VALUE PROBLEMS 15 and (Tx)(t) = ψ(1) ψ′(1) [ ψ′(0) ψ(0) + lnψ(t)− lnψ(0)] ∫ η 0 g(s, x(s))ds+ k∑ i=1 Pi(x(ti)) − m∑ i=1 χ(t, ti)Qi(x(ti)) ψ(ti) ψ′(ti) . Firstly, we verify that (Ψx1)(t) + (Tx2)(t) ∈ BR. When t ∈ ϖk (k = 0, 1, 2, . . . ,m), for any x1, x2 ∈ BR, we have |(Ψx1)(t) + (Tx2)(t)| ≤ λ Γ(α) k∑ i=1 ∫ ti ti−1 ψ′(s) ψ(s) (lnψ(ti)− lnψ(s)) α−1 ∣∣∣ϕq(hi−1 ( s, x1(s), x1(s+ (−1) i−1 τ) ))∣∣∣ds + λ Γ(α) ∫ t tk ψ′(s) ψ(s) (lnψ(t)− lnψ(s))α−1 ∣∣∣ϕq(hk(s, x1(s), x1(s+ (−1) k τ ) ) )∣∣∣ds+ λ Γ(α− 1) × m+1∑ i=1 ∫ ti ti−1 |χ(t, ti)| ψ′(s) ψ(s) (lnψ(ti)− lnψ(s)) α−2 ∣∣∣ϕq(hi−1 ( s, x1(s), x1(s+ (−1) i−1 τ) ))∣∣∣ds + ψ(1) ψ′(1) Θ1 ∫ η 0 |g(s, x2(s))|ds+ k∑ i=1 |Pi(x2(ti))|+ m∑ i=1 |χ(t, ti)||Qi(x2(ti))| ψ(ti) ψ′(ti) ≤ Φ2ς + ψ(1) ψ′(1) Θ1(φ ∗ 2 + φ∗ 3∥x∥)η +m(M1∥x∥+N1) + Θ1(M2∥x∥+N2) mψ(1) ψ∗ ≤ Φ2ς + ψ(1) ψ′(1) Θ1φ ∗ 2η +mN1 +Θ1N2 mψ(1) ψ∗ + ( ψ(1) ψ′(1) Θ1φ ∗ 3η +mM1 +M2Θ1 mψ(1) ψ∗ ) ∥x∥ ≤ Φ2ς + ψ(1) ψ′(1) Θ1φ ∗ 2η +mN1 +Θ1N2 mψ(1) ψ∗ + ( ψ(1) ψ′(1) Θ1φ ∗ 3η +mM1 +M2Θ1 mψ(1) ψ∗ ) R. Therefore, |(Ψx1)(t) + (Tx2)(t)| ≤ BR, and (Ψx1)(t) + (Tx2)(t) ∈ BR. Secondly, we prove that operator T is a contraction mapping within BR, when t ∈ ϖk, for any x1, x2 ∈ BR, from (3.1), we have |(Tx2)(t)− (Tx1)(t)| ≤ ψ(1) ψ′(1) Θ1 ∫ η 0 |g(s, x2(s))− g(s, x1(s))|ds+ k∑ i=1 |Pi(x2(ti))− Pi(x1(ti))| + m∑ i=1 |χ(t, ti)||Qi(x2(ti))−Qi(x1(ti))| ψ(ti) ψ′(ti) ≤ [ ψ(1) ψ′(1) Θ1∥φ1∥ √ η +ml +mΘ1l ∗ψ(1) ψ∗ ]∥x2 − x1∥. By (H9), operator T is a contraction mapping within BR. Finally, verify that operator Ψ is a completely continuous operator. From the definition of operator Ψ and the continuity of function h, it can be inferred that operator Ψ is continuous, thus, we only need to prove that operator Ψ is compact. The process is divided into the following two steps. Step 1. Ψ is uniformly bounded. When t ∈ ϖk (k = 0, 1, 2, . . . ,m), for any x ∈ BR, there exists a constant ξ, such that |(Ψx)t| ≤ ξ. According to (H8), for any t ∈ ϖk(k = 0, 1, 2, . . . ,m), x ∈ BR, we have |(Ψx)(t)| ≤ λ Γ(α) k∑ i=1 ∫ ti ti−1 ψ′(s) ψ(s) (lnψ(ti)− lnψ(s)) α−1 ∣∣∣ϕq(hi−1 ( s, x(s), x(s+ (−1) i−1 τ) ))∣∣∣ds 16 X. CHEN, W. ZHOU EJDE-2025/65 + λ Γ(α) ∫ t tk ψ′(s) ψ(s) (lnψ(t)− lnψ(s))α−1 ∣∣∣ϕq(hk(s, x(s), x(s+ (−1) k τ ) ) )∣∣∣ds+ λ Γ(α− 1) × m+1∑ i=1 ∫ ti ti−1 |χ(t, ti)| ψ′(s) ψ(s) (lnψ(ti)− lnψ(s)) α−2 ∣∣∣ϕq(hi−1 ( s, x(s), x ( s+ (−1) i−1 τ )))∣∣∣ds ≤ λ(m+ 1)ς(vα +Θ1αv α−1) Γ(α+ 1) := ξ. From this, we infer that operator Ψ is uniformly bounded. Step 2. Ψ is equicontinuous. For any x ∈ BR, t1, t2 ∈ ϖk(k = 0, 1, 2, . . . ,m), where t1 < t2, we obtain |(Ψx)(t2)− (Ψx)(t1)| ≤ λ Γ(α) ∫ t2 tk ψ′(s) ψ(s) (lnψ(t2)− lnψ(s))α−1 ∣∣∣ϕq(hk(s, x(s), x(s+ (−1) k τ )))∣∣∣ds − λ Γ(α) ∫ t1 tk ψ′(s) ψ(s) (lnψ(t1)− lnψ(s))α−1 ∣∣∣ϕq(hk(s, x(s), x(s+ (−1) k τ) ))∣∣∣ds + λ Γ(α− 1) m+1∑ i=1 ∫ ti ti−1 |χ(t2, ti)− χ(t1, ti) ∣∣∣ψ′(s) ψ(s) ( lnψ(ti) − lnψ(s) )α−2|ϕq ( hi−1 ( s, x(s), x(s+ (−1) i−1 τ) ))∣∣∣ds ≤ ςλ Γ(α+ 1) [(lnψ(t2)− lnψ(tk)) α − (lnψ(t1)− lnψ(tk)) α ] + λς(m+ 1)vα−1 Γ(α) [lnψ(t2)− lnψ(t1)]. Therefore, |(Ψx)(t2)− (Ψx)(t1)| → 0 as t1 → t2. Thus, the operator Ψ is equicontinuous on BR. In line with lemma 2.9, operator Ψ is a compact operator within BR, satisfying the conditions required by Lemma 2.8. In conclusion, there is at least one solution to BVP (1.1). □ 4. Examples In this section we verify the main results through two examples. Example 4.1. Consider the BVP ϕp ( HD 3/2,t2+t+1 0 x(t) ) = 1 10 ( t 100 + t sinx 150 + t sinx(t+ 1 3 ) 100 ) , t ∈ (0, 1 2 ) ϕp ( HD 3/2,t2+t+1 1/2 x(t) ) = 1 10 ( t 100 + t + t2 sinx 150 + t + t sinx(t− 1 3 ) 100 ) , t ∈ ( 1 2 , 1) ∆x ( t1/2 ) = |x(1/2)| 300 + |x(1/2)| , ∆x′(t1/2) = |x(1/2)| 200 + |x(1/2)| , x(0) = x′(0), x′(1) = 1 10 ∫ 1/2 0 essin1/2s 1 + 2s x(s) 1 + s ds, (4.1) where m = 1, α = 3/2, t0 = 0, t1 = 1/2, t2 = 1, τ = 1/3, λ = 1/10, η = 1/2, and p = 3/2. From p = 3/2 and q−1 + p−1 = 1, we obtain q = 3, ψ(t) = t2 + t+ 1 ψ′(t) = 2t+ 1, and ψ∗ = 1. Then ψ(t) = t2 + t+ 1, is an increasing function in t ∈ [0, 1], v = 0.81, h1(t, x(t), x(t+ τ)) = t 100 + t sinx 150 + t sinx(t+ 1 3 ) 100 ≤ 2t 75 , h2(t, x(t), x(t− τ)) = t 100 + t + t2 sinx 150 + t + t sinx(t− 1 3 ) 100 ≤ t2 + 3t 150 , EJDE-2025/65 THREE-POINT INTEGRAL BOUNDARY-VALUE PROBLEMS 17 where σ1(t) = 2t 75 , σ2(t) = t2+3t 150 , then σ∗ = max { sup t∈(0,1/2) ( 2t 75 ), sup t∈(1/2,1) ( t2 + 3t 150 ) } = 2 75 , P (x(1/2)) = |x(1/2)| 300 + |x(1/2)| , Q(x(1/2)) = |x(1/2)| 200 + |x(1/2)| , g(t, x(t)) = 1 10 (etsin1/2t 1 + 2t x(t) 1 + t ) . Obviously, hk(k = 1, 2) is a continuous function. For each u1, v1, ū1, v̄1, u2, v2, ū2, v̄2, u, ū ∈ R and t ∈ [0, 1], we have |h1(t, u1, v1)| ≤ 2t 75 , |h2(t, u2, v2)| ≤ t2 + 3t 150 , |P (u)| ≤ 1 300 |u|+ 1 10 , |Q(u)| ≤ 1 200 |u|+ 1 10 , |h1(t, u1, v1)− h1(t, ū1, v̄1)| ≤ 1 150 |u1 − ū1|+ 1 100 |v1 − v̄1|, |h2(t, u2, v2)− h2(t, ū2, v̄2)| ≤ 1 150 |u2 − ū2|+ 1 100 |v2 − v̄2|, |P (u)− P (ū)| ≤ 1 300 |u− ū|, |Q(u)−Q(ū)| ≤ 1 200 |u− ū|, |g(t, u)| ≤ 1 10 et|u|+ 1 100 , |g(t, u)− g(t, ū)| ≤ 1 10 etsin1/2t 1 + 2t |u− ū| ≤ 1 10 etsin1/2t|u− ū|. So that conditions (H1)–(H6) hold, where M1 = 1 300 , N1 = 1 10 , M2 = 1 200 , N2 = 1 10 , M∗ h = 1 150 , N∗ h = 1 100 , l = 1 300 , l∗ = 1 200 , φ∗ 2 = 1 100 , φ∗ 3 = e 10 . After calculations, it can be concluded that ∥φ1∥ = (∫ 1 0 ( 1 10 essin1/2s) 2 ds )1/2 ≈ 0.1265, Θ1 = ψ′(0) ψ(0) + lnψ(1)− lnψ(0) = 3.09, Φ1 = (q − 1)(σ∗)q−2(M∗ h +N∗ h) ≈ 0.00089, Φ2 = λ(m+ 1)(vα +Θ1αv α−1) Γ(α+ 1) ≈ 0.7283. From the definition of hk, k = 1, 2, it can be inferred that K = 1 100 , then ϕq(K) = 1 10000 . Then there exists a constant Nx = 10 > 0, such that Nx/ ( ψ(1) ψ′(1) Θ1(φ ∗ 2 + φ∗ 3Nx)η +m(M1Nx +N1) + Θ1(M2Nx +N2) mψ(1) ψ∗ + [Φ1Nx + ϕq(K)]Φ2 ) ≈ 1.5163 > 1. 18 X. CHEN, W. ZHOU EJDE-2025/65 Therefore, all the assumptions in Theorem 3.1 are satisfied, so there is at least one solution to the BVP (4.1). Because M1 = ψ(1) ψ′(1) Θ1 √ η∥φ1∥+ml + ψ(1) ψ∗ Θ1ml ∗ + λαΘ1(m+ 1)vα−1 + λ(m+ 1)vα Γ(α+ 1) Φ1 ≈ 0.3267 < 1, all the assumptions in Theorem 3.2 are satisfied, thus, BVP (4.1) has a unique solution. Example 4.2. Consider another BVP ϕp ( HD 3/2,t2+2t+1 0 x(t) ) = 1 20 ( t 100 + t sinx 150 + t sinx(t+ 1 3 ) 100 ), t ∈ (0, 1 3 ) ϕp ( HD 3/2,t2+2t+1 1/3 x(t) ) = 1 20 ( t 100 + t + t2 sinx 150 + t + t sinx(t− 1 3 ) 100 ), t ∈ ( 1 3 , 2 3 ) ϕp ( HD 3/2,t2+2t+1 2/3 x(t) ) = 1 20 ( t 100 + sinx 150(1 + t) + t sinx(t+ 1 3 ) 100 ) , t ∈ ( 2 3 , 1) ∆x ( t1/3 ) = |x(1/3)| 30 + |x(1/3)| , ∆x′(t1/3) = |x(1/3)| 40 + |x(1/3)| , x(0) = x′(0), x′(1) = 1 10 ∫ 1/3 0 e2ssin1/2s x(s) 1 + s ds, (4.2) where m = 1, α = 3/2, t0 = 0, t1 = 1/3, t2 = 2/3, t3 = 1, τ = 1/3, λ = 1/20, η = 1/3, p = 3/2, From p = 3/2 and q−1 + p−1 = 1, we obtain q = 3, ψ(t) = t2 + 2t + 1, ψ′(t) = 2t + 2, ψ∗ = 2. Then ψ(t) = t2 + 2t+ 1 is an increasing function in t ∈ [0, 1], h1(t, x(t), x(t+ τ)) = t 100 + t sinx 100 + t sinx(t+ 1 3 ) 150 ≤ 2t 75 , h2(t, x(t), x(t− τ)) = t 100 + t + t2 sinx 150 + t + t sinx(t− 1 3 ) 100 ≤ t2 + 3t 150 , h3(t, x(t), x(t+ τ)) = t 100 + sinx 150(1 + t) + t sinx(t+ 1 3 ) 100 ≤ 3t+ 1 150 . Obviously, hk (k = 1, 2, 3) is a continuous function, and |ϕq(hk(t, x(t), x(t+ (−1) k τ)))| < ( 2 75 )2 = 4 5625 , let ς = 4 5625 , then the inequality |ϕq(hk(t, x(t), x(t+ (−1) k τ)))| < ς holds, that is, condition (H8) is satisfied. P ( x(1/3) ) = |x(1/3)| 30 + |x(1/3)| , Q(x(1/3)) = |x(1/3)| 40 + |x(1/3)| , g(t, x(t)) = 1 10 e2tsin1/2t x(t) 1 + t . For each u, ū ∈ R, t ∈ [0, 1], we have |P (u)− P (ū)| ≤ 1 30 |u− ū|, |Q(u)−Q(ū)| ≤ 1 40 |u− ū|, |g(t, u)| ≤ 1 10 + e2 10 |u|, |g(t, u)− g(t, ū)| ≤ 1 10 e2tsin1/2t|u− ū|, EJDE-2025/65 THREE-POINT INTEGRAL BOUNDARY-VALUE PROBLEMS 19 therefore, conditions (H1)–(H7) are satisfied, where l = 1 30 , l∗ = 1 40 , φ∗ 2 = 1 10 , φ∗ 3 = e2 10 . After calculation, we concluded that ∥φ1∥ = (∫ 1 0 ( 1 10 e2ssin1/2s )2 ds )1/2 ≈ 0.2844 Θ1 = ψ′(0) ψ(0) + lnψ(1)− lnψ(0) = 3.39′ For assumption (H9), we have ψ(1) ψ′(1) Θ1∥φ1∥ √ η +ml +mΘ1l ∗ψ(1) ψ∗ ≈ 0.8442 < 1. 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Liu, et al.; The convergence analysis and error estimation for unique solution of a p-Laplacian fractional differential equation with singular decreasing nonlinearity, Boundary Value Problems, 2018(2018), 1-15. [33] W. Yao, H. Zhang; Multiple solutions for p-Laplacian fractional differential equations with Caputo derivative and impulsive effects, J. Appl. Anal. Comput., 15 (2025), 3480-3503. [34] L. Zhang, X. Liu, M. Jia, et al.; Piecewise conformable fractional impulsive differential system with delay: existence, uniqueness and Ulam stability, Journal of Applied Mathematics and Computing, 70(2024), 1543- 1570. Xiao Chen School of Mathematics and Physics, Lanzhou Jiaotong University, Lanzhou, Gansu 730070, China Email address: chxiao1005@126.com Wenxue Zhou (corresponding author) School of Mathematics and Physics, Lanzhou Jiaotong University, Lanzhou, Gansu 730070, China Email address: wxzhou2006@126.com 1. Introduction 2. Preliminaries 3. Existence and uniqueness of solutions 4. Examples Acknowledgments References