Infinitely many positive solutions for an iterative system of singular multipoint boundary value problems on time scales

Abstract

In this paper, we consider an iterative system of singular multipoint boundary value problems on time scales. The sufficient conditions are derived for the existence of infinitely many positive solutions by applying Krasnoselskii’s cone fixed point theorem in a Banach space.

1 Introduction

Differential equations with state-dependent delays have attracted a great deal of interest to the researchers since they widely arise from application models, such as population models [4], mechanical models [19], infection disease transmission [28], the dynamics of economical systems [5], position control [9], two-body problem of classical electrodynamics [15], etc. As special type of state-dependent delay-differential equations, iterative differential equations have distinctive characteristics and have been investigated in recent years, e.g. equivariance [30], analyticity [31], convexity [27], monotonicity [16], smoothness [12]. Recently [17], Feckan, Wang and Zhao established the maximal and minimal nondecreasing bounded solutions of the following iterative functional differential equations

$$\begin{aligned} \mathtt {x}'(t)=\mathtt {g}\left( t,\mathtt {x}^{(1)}(t), \mathtt {x}^{(2)}(t),\ldots ,\mathtt {x}^{(n)}(t)\right) , \end{aligned}$$

where $\mathtt {x}^{(i)}(t):=x(\mathtt {x}^{(i-1)})(t)$ indicates the i-th iterate of $\mathtt {x}$, where $i=1,2,\ldots ,n,$ by the method of lower and upper solutions.

On the other hand, the theory of time scales was created to unify continuous and discrete analysis. Difference and differential equations can be studied simultaneously by studying dynamic equations on time scales. A time scale is any closed and nonempty subset of the real numbers. So, by this theory, we can extend the continuous and discrete theories to cases ”in between.” These types of time scales play an important role for applications, since most of the phenomena in the environment are neither only discrete nor only continuous, but they possess both behaviours. Research in this area of mathematics has exceeded by far a thousand publications, and numerous applications to literally all branches of science such as statistics, biology, economics, finance, engineering, physics, and operations research have been given. Moreover, basic results on this issue have been well documented in the articles [1, 2] and monographs of Bohner and Peterson [7, 8]. There is a great deal of research activity devoted to positive solutions of dynamic equations on time scales, see for example [14, 20, 21, 24,25,26] and references therein.

In [22], Liang and Zhang studied countably many positive solutions for nonlinear singular $m-$point boundary value problems on time scales,

$$\begin{aligned} \begin{aligned}&\big (\varphi (\mathtt {x}^{\Delta }(t))\big )^{\nabla }+a(t)f\big (\mathtt {x}(t)\big )=0,~t\in [0, a]_{{\mathbb {T}}}\\&\mathtt {x}(0)=\sum _{i=1}^{m-2}a_i\mathtt {x}(\xi _i),~\mathtt {x}^{\Delta }(a)=0, \end{aligned} \end{aligned}$$

by using the fixed-point index theory and a new fixed-point theorem in cones.

In [13], Dogan considered second order m–point boundary value problem on time scales,

$$\begin{aligned} \begin{aligned}&\big (\upphi _p(\mathtt {x}^{\Delta }(t))\big )^{\nabla }+\upomega (t)f\big (t, \mathtt {x}(t)\big )=0,~t\in [0,T]_{{\mathbb {T}}}\\&\mathtt {x}(0)=\sum _{i=1}^{m-2}a_i\mathtt {x}(\xi _i),~ \upphi _p(\mathtt {x}^{\Delta }(T))=\sum _{i=1}^{m-2}b_i\upphi _p(\mathtt {x}^{\Delta }(\xi _i)), \end{aligned} \end{aligned}$$

and established existence of multiple positive solutions by applying fixed-point index theory.

Many researchers have concentrated on studying first order iterative differential equations by different approaches such as fixed point theory, Picard’s successive approximation and the technique of nonexpansive operators. But the literature related to the equations of higher order is limited since the presence of the iterates increases the difficulty of studying them. This motivates us to investigate the following second order dynamical iterative system of boundary value problems with singularities on time scales,

$$\begin{aligned}&\left. \begin{aligned}&\mathtt {x}_\ell ^{\Delta \nabla }(t)+\uplambda (t)\mathtt {g}_\ell \big (\mathtt {x}_{\ell +1}(t)\big )=0,~1\le \ell \le n,~t\in (0,\upsigma (a)]_{{\mathbb {T}}}\\&\quad \mathtt {x}_{n+1}(t)=\mathtt {x}_1(t),~t\in (0,\upsigma (a)]_{{\mathbb {T}}}, \end{aligned}\right\} \end{aligned}$$

(1)

$$\begin{aligned}&\mathtt {x}_\ell ^{\Delta }(0)=0,~\mathtt {x}_\ell (\upsigma (a))=\sum _{k=1}^{n-2}c_k\mathtt {x}_\ell (\upzeta _k),~1\le \ell \le n, \end{aligned}$$

(2)

where $n\in {\mathbb {N}}$, $c_k\in {\mathbb {R}}^+:=[0,+\infty )$ with $\sum _{k=1}^{n-2}c_k<1,$ $0<\upzeta _k<\upsigma (a)/2,$ $k\in \{1,2,\ldots ,n-2,\},$ $\uplambda (t)=\prod _{i=1}^{m}\uplambda _i(t)$ and each $\uplambda _i(t)\in L^{p_i}_{\nabla }((0, \upsigma (a)]_{\mathbb {T}}) (p_i\ge 1)$ has a singularity in the interval $(0, \upsigma (a)/2]_{\mathbb {T}}.$ By applying Hölder’s inequality and Krasnoselskii’s cone fixed point theorem in a Banach space, we establish the existence of infinitely many positive solutions for the system (1). Equation (1) in real continuous time scales describes diffusion phenomena with a source or a reaction term. For instance, in thermal conduction, it can be interpreted as the one-dimensional heat conduction equation which models the steady-states of a heated bar of length a with a controller at $\mathtt {x}=a$ that adds or removes heat according to a sensor, while the left endpoint is maintained at ${0}^\circ$C and $\mathtt {g}$ is the distributed temperature source function depending on delayed temperatures. We refer the interested reader to [10, 11] and the references therein for more details.

We assume the following conditions are true throughout the paper:

$({\mathcal {H}}_1)$:

$\mathtt {g}_\ell :[0, +\infty )\rightarrow [0, +\infty )$ is continuous.

$({\mathcal {H}}_2)$:

there exists a sequence $\{t_r\}_{r=1}^{\infty }$ such that $0<t_{r+1}<t_r<\upsigma (a)/2,$

$$\begin{aligned} \lim _{r\rightarrow \infty } t_r =t^*<\upsigma (a)/2,\, \lim _{t\rightarrow t_r}\uplambda _i(t) = +\infty ,\,i=1,2,\ldots ,m. \end{aligned}$$

Further, for each $i\in \{1,2,\ldots ,m\},$ there exist $\updelta _i>0$ such that $\uplambda _i(t)>\updelta _i.$

2 Preliminaries

In this section, we introduce some basic definitions and lemmas which are useful for our later discussions.

Definition 2.1

[7] A time scale ${{\mathbb {T}}}$ is a nonempty closed subset of the real numbers ${{\mathbb {R}}}.$ ${{\mathbb {T}}}$ has the topology that it inherits from the real numbers with the standard topology. It follows that the jump operators $\upsigma , \rho :{{\mathbb {T}}}\rightarrow {{\mathbb {T}}},$ and the graininess $\mu :{{\mathbb {T}}}\rightarrow [0,+\infty )$ are defined by $\upsigma (t)=\inf \{\uptau \in {{\mathbb {T}}}:\uptau > t\},$ $\rho (t)=\sup \{\uptau \in {{\mathbb {T}}}:\uptau < t\},$ and $\mu (t)=\upsigma (t)-t,$ respectively.

• The point $t \in {{\mathbb {T}}}$ is left-dense, left-scattered, right-dense, right-scattered if $\rho (t)=t,$ $\rho (t)<t,$ $\upsigma (t)=t,$ $\upsigma (t)>t,$ respectively.

• If ${{\mathbb {T}}}$ has a right-scattered minimum m, then ${{\mathbb {T}}}_\kappa ={{\mathbb {T}}} \backslash \{m\}$; otherwise ${{\mathbb {T}}}_\kappa ={{\mathbb {T}}}.$

• If ${{\mathbb {T}}}$ has a left-scattered maximum m, then ${{\mathbb {T}}}^\kappa ={{\mathbb {T}}} \backslash \{m\}$; otherwise ${{\mathbb {T}}}^\kappa ={{\mathbb {T}}}.$

• A function $f:{{\mathbb {T}}}\rightarrow {{\mathbb {R}}}$ is called rd-continuous provided it is continuous at right-dense points in ${{\mathbb {T}}}$ and its left-sided limits exist (finite) at left-dense points in ${{\mathbb {T}}}.$ The set of all rd-continuous functions $f:{\mathbb {T}}\rightarrow {\mathbb {R}}$ is denoted by $C_{rd}=C_{rd}({\mathbb {T}})=C_{rd}({\mathbb {T}},{\mathbb {R}}).$

• A function $f:{{\mathbb {T}}}\rightarrow {{\mathbb {R}}}$ is called ld-continuous provided it is continuous at left-dense points in ${{\mathbb {T}}}$ and its right-sided limits exist (finite) at right-dense points in ${{\mathbb {T}}}.$ The set of all ld-continuous functions $f:{\mathbb {T}}\rightarrow {\mathbb {R}}$ is denoted by $C_{ld}=C_{ld}({\mathbb {T}})=C_{ld}({\mathbb {T}},{\mathbb {R}}).$

• By an interval time scale, we mean the intersection of a real interval with a given time scale. i.e., $[a, b]_{{\mathbb {T}}}=[a, b]\cap {\mathbb {T}}.$ Other intervals can be defined similarly.

Definition 2.2

[6] Let $\mu _\Delta$ and $\mu _\nabla$ be the Lebesgue $\Delta -$ measure and the Lebesgue $\nabla -$measure on ${\mathbb {T}},$ respectively. If $A \subset {\mathbb {T}}$ satisfies $\mu _\Delta (A)=\mu _\nabla (A),$ then we call A is measurable on ${\mathbb {T}},$ denoted $\mu (A)$ and this value is called the Lebesgue measure of A. Let P denote a proposition with respect to $t\in {\mathbb {T}}.$

• (i) If there exists $\Gamma _1\subset A$ with $\mu _\Delta (\Gamma _1)=0$ such that P holds on $A\backslash \Gamma _1,$ then P is said to hold $\Delta$–a.e. on A.

• (ii) If there exists $\Gamma _2\subset A$ with $\mu _\nabla (\Gamma _2)=0$ such that P holds on $A\backslash \Gamma _2,$ then P is said to hold $\nabla$–a.e. on A.

Definition 2.3

[3, 6] Let $E\subset {\mathbb {T}}$ be a $\Delta$–measurable set and $p\in \bar{{\mathbb {R}}}\equiv {\mathbb {R}}\cup \{-\infty , +\infty \}$ be such that $p\ge 1$ and let $f:E\rightarrow \bar{{\mathbb {R}}}$ be $\Delta$–measurable function. We say that f belongs to $L^p_\Delta (E)$ provided that either

$$\begin{aligned} \int _E \vert f \vert ^p (s)\Delta s < \infty \quad \text {if} \quad p\in [1, +\infty ), \end{aligned}$$

or there exists a constant $M\in {\mathbb {R}}$ such that

$$\begin{aligned} \vert f \vert \le M, ~~ \Delta -a.e.~on~E~~ \text {if}~~ p=+\infty . \end{aligned}$$

Lemma 2.4

[29] Let $E\subset {\mathbb {T}}$ be a $\Delta$–measurable set. If $f:{\mathbb {T}}\rightarrow {\mathbb {R}}$ is $\Delta$–integrable on E,  then

$$\begin{aligned} \int _E f(s)\Delta s = \int _E f(s)ds+\sum _{i\in I_E}\big (\upsigma (t_i)-t_i\big ) f(t_i)+r(f,E), \end{aligned}$$

where

$$\begin{aligned} r(f,E)=\left\{ \begin{array}{ll} \mu _\mathtt {N}(E)f(M),~\text {if}~~ \mathtt {N}\in {\mathbb {T}},\\ 0,~\text {if}~~\mathtt {N}\notin {\mathbb {T}}, \end{array}\right. \end{aligned}$$

$I_E:=\{i\in I: t_i \in E\}$ and $\{t_i\}_{i \in I}, I \subset {\mathbb {N}},$ is the set of all right-scattered points of ${\mathbb {T}}.$

Definition 2.5

[29] Let $E\subset {\mathbb {T}}$ be a $\nabla$–measurable set and $p\in \bar{{\mathbb {R}}}\equiv {\mathbb {R}}\cup \{-\infty , +\infty \}$ be such that $p\ge 1$ and let $f:E\rightarrow \bar{{\mathbb {R}}}$ be $\nabla$–measurable function. Say that f belongs to $L^p_\nabla (E)$ provided that either

$$\begin{aligned} \int _E \vert f \vert ^p (s)\nabla s < \infty \quad \text {if} \quad p\in {\mathbb {R}}, \end{aligned}$$

or there exists a constant $C \in {\mathbb {R}}$ such that

$$\begin{aligned} \vert f \vert \le C, ~~ \nabla -a.e.~on~E~~ \text {if}~~ p=+\infty . \end{aligned}$$

Lemma 2.6

[29] Let $E\subset {\mathbb {T}}$ be a $\nabla$–measurable set. If $f:{\mathbb {T}}\rightarrow {\mathbb {R}}$ is a $\nabla$–integrable on E,  then

$$\begin{aligned} \int _E f(s)\nabla s = \int _E f(s)ds+\sum _{i\in I_E}\big (t_i-\rho (t_i)\big ) f(t_i), \end{aligned}$$

where $I_E:=\{i\in I: t_i \in E\}$ and $\{t_i\}_{i \in I}, I \subset {\mathbb {N}},$ is the set of all left-scattered points of ${\mathbb {T}}.$

Lemma 2.7

For any $\mathtt {y}(t)\in {\mathcal {C}}_{ld}((0, \upsigma (a)]_{\mathbb {T}}),$ the boundary value problem,

$$\begin{aligned} \mathtt {x}_1^{\Delta \nabla }(t)+\mathtt {y}(t)=0,~ t\in (0,\upsigma (a)]_{\mathbb {T}}, \end{aligned}$$

(3)

$$\begin{aligned} \mathtt {x}_1^{\Delta }(0)=0,~\mathtt {x}_1(\upsigma (a))=\sum _{k=1}^{n-2}c_k\mathtt {x}_1(\upzeta _k) \end{aligned}$$

(4)

has a unique solution

$$\begin{aligned} \mathtt {x}_1(t)=\int _0^{\upsigma (a)} \aleph (t, \uptau )\mathtt {y}(\uptau )\nabla \uptau +\frac{1}{1-\sum _{k=1}^{n-2}c_k}\sum _{k=1}^{n-2}c_k\int _0^{\upsigma (a)} \aleph (\upzeta _k, \uptau )\mathtt {y}(\uptau )\nabla \uptau , \end{aligned}$$

(5)

where

$$\begin{aligned} \aleph (t, \uptau )=\left\{ \begin{array}{ll} \upsigma (a)-t,~~\text{if}~~\, 0\le \uptau \le t\le \upsigma (a),\\ \upsigma (a)-\uptau ,~~\text{if}~~ 0\le t\le \uptau \le \upsigma (a). \end{array} \right. \end{aligned}$$

(6)

Proof

Suppose $\mathtt {x}_1$ is a solution of (3), then

$$\begin{aligned} \begin{aligned} \mathtt {x}_1(t)&=-\int _0^t\int _0^\uptau \mathtt {y}(\uptau _1)\nabla \uptau _1\Delta \uptau +\mathtt {A}t+\mathtt {B}\\&=-\int _0^t (t-\uptau )\mathtt {y}(\uptau )\nabla \uptau +\mathtt {A}t+\mathtt {B}, \end{aligned} \end{aligned}$$

where $\mathtt {A}=\mathtt {x}_1^\Delta (0)$ and $\mathtt {X}=\mathtt {x}_1(0)$. Using conditions (4), we get $\mathtt {A}=0$ and

$$\begin{aligned} \mathtt {B}=\int _0^{\upsigma (a)}(\upsigma (a)-\uptau )\mathtt {y}(\uptau )\nabla \uptau +\sum _{k=1}^{n-2}c_k\mathtt {x}_1(\upzeta _k). \end{aligned}$$

So, we have

$$\begin{aligned} \mathtt {x}_1(t)=&-\int _0^t (t-\uptau )\mathtt {y}(\uptau )\nabla \uptau +\int _0^{\upsigma (a)}(\upsigma (a)-\uptau )\mathtt {y}(\uptau )\nabla \uptau +\sum _{k=1}^{n-2}c_k\mathtt {x}_1(\upzeta _k)\nonumber \\ =&\int _0^{\upsigma (a)}\aleph (t,\uptau )\mathtt {y}(\uptau )\nabla \uptau +\sum _{k=1}^{n-2}c_k\mathtt {x}_1(\upzeta _k). \end{aligned}$$

(7)

Plugging $t=\upzeta _k$ and multiplying with $c_k$ then summing from 1 to $n-2$ in the above equation (7), we obtain

$$\begin{aligned} \mathtt {x}_1(\upzeta _k)=\frac{1}{1-\sum _{k=1}^{n-2}c_k}\sum _{k=1}^{n-2}c_k\int _0^{\upsigma (a)}\aleph (\upzeta _k,\uptau )\mathtt {y}(\uptau )\nabla \uptau . \end{aligned}$$

(8)

Substituting (8) into (7), we get required solution (5). This completes the proof. $\square$

Lemma 2.8

Suppose $({\mathcal {H}}_1)$–$({\mathcal {H}}_2)$ hold. Let $\upeta \in (0, \upsigma (a)/2)_{{\mathbb {T}}}$ with $\upzeta _k\in [\upeta , \upsigma (a)-\upeta ]_{\mathbb {T}},$ $k\in \{1,2,\cdots ,n-2\},$ the kernel $\aleph (t, \uptau )$ have the following properties:

• (i) $0\le \aleph (t, \uptau )\le \aleph (\uptau , \uptau )$ for all $\,t, \uptau \in [0, \upsigma (a)]_{\mathbb {T}},$

• (ii) $\frac{\upeta }{\upsigma (a)} \aleph (\uptau , \uptau )\le \aleph (t, \uptau )$ for all $t \in [\upeta , \upsigma (a)-\upeta ]_{{\mathbb {T}}}$ and $\uptau \in [0, \upsigma (a)]_{\mathbb {T}}.$

Proof

(i) is evident. To prove (ii), let $t \in [\upeta , \upsigma (a)-\upeta ]_{{\mathbb {T}}}$ and $\uptau \le t.$ Then

$$\begin{aligned} \begin{aligned} \frac{\aleph (t, \uptau )}{\aleph (\uptau , \uptau )}=\frac{ \upsigma (a)-t}{\upsigma (a)-\uptau }\ge \frac{\upeta }{\upsigma (a)}. \end{aligned} \end{aligned}$$

For $t\le \uptau ,$

$$\begin{aligned} \begin{aligned} \frac{\aleph (t, \uptau )}{\aleph (\uptau , \uptau )}=\frac{ \upsigma (a)-\uptau }{\upsigma (a)-\uptau }=1\ge \frac{\upeta }{\upsigma (a)}. \end{aligned} \end{aligned}$$

This completes the proof. $\square$

Notice that an $n-$tuple $(\mathtt {x}_1(t), \mathtt {x}_2(t), \mathtt {x}_3(t),\ldots ,\mathtt {x}_n(t))$ is a solution of the iterative boundary value problem (1)–(2) if and only if

$$\begin{aligned} \begin{aligned}&\mathtt {x}_\ell (t)=\int _0^{\upsigma (a)}\aleph (t,\uptau )\uplambda (\uptau ) \mathtt {g}_\ell (\mathtt {x}_{\ell +1}(\uptau ))\nabla \uptau \\&\quad +\frac{1}{1-\sum _{k=1}^{n-2}c_k}\sum _{k=1}^{n-2}c_k\int _0^{\upsigma (a)} \aleph (\upzeta _k, \uptau )\uplambda (\uptau ) \mathtt {g}_\ell (\mathtt {x}_{\ell +1}(\uptau ))\nabla \uptau \end{aligned} \end{aligned}$$

and

$$\begin{aligned} \mathtt {x}_{\ell +1}(t)=\mathtt {x}_1(t),~t\in (0,a]_{{\mathbb {T}}},~1\le \ell \le n. \end{aligned}$$

That is

$$\begin{aligned} \begin{aligned} \mathtt {x}_1(t)&=\int _0^{\upsigma (a)}\aleph (t,\uptau _1)\uplambda (\uptau _1) \mathtt {g}_1\bigg [\int _0^{\upsigma (a)}\aleph (\uptau _1,\uptau _2)\uplambda (\uptau _2)\mathtt {g}_2\bigg [\int _0^{\upsigma (a)}\aleph (\uptau _2,\uptau _3)\cdots \\&\quad \times \mathtt {g}_{n-1}\bigg [\int _0^{\upsigma (a)}\aleph (\uptau _{n-1},\uptau _n) \uplambda (\uptau _n)\mathtt {g}_n(\mathtt {x}_1(\uptau _n))\Delta \uptau _n\bigg ]\cdot \cdot \cdot \Delta \uptau _3\bigg ]\Delta \uptau _2\bigg ]\Delta \uptau _1\\&\quad +\frac{1}{1-\sum _{k=1}^{n-2}c_k}\sum _{k=1}^{n-2}c_k\int _0^{\upsigma (a)} \aleph (\upzeta _k, \uptau _1)\uplambda (\uptau _1) \mathtt {g}_1\bigg [\int _0^{\upsigma (a)}\aleph (\uptau _1,\uptau _2)\uplambda (\uptau _2)\cdots \\&\quad \times \mathtt {g}_{n-1}\bigg [\int _0^{\upsigma (a)}\aleph (\uptau _{n-1},\uptau _n) \uplambda (\uptau _n)\mathtt {g}_n(\mathtt {x}_1(\uptau _n))\Delta \uptau _n\bigg ]\cdot \cdot \cdot \Delta \uptau _3\bigg ]\Delta \uptau _2\bigg ]\Delta \uptau _1. \end{aligned} \end{aligned}$$

Let $\mathtt {X}$ be the Banach space $C_{ld}((0,\upsigma (a)]_{\mathbb {T}},{\mathbb {R}})$ with the norm $\Vert \mathtt {x}\Vert =\displaystyle \max _{t\in (0,\upsigma (a)]_{\mathbb {T}}}\vert \mathtt {x}(t)\vert .$ For $\upeta \in (0, \upsigma (a)/2)_{{\mathbb {T}}},$ we define the cone $\mathtt {P}_\upeta \subset \mathtt {X}$ as

$$\begin{aligned} \mathtt {P}_\upeta =\left\{ \mathtt {x}\in \mathtt {X} : \mathtt {x}(t)\text {~is nonnegative~and~} \min _{t\in {[\upeta ,\,\upsigma (a)-\upeta ]_{{\mathbb {T}}}}}\mathtt {x}(t)\ge \frac{\upeta }{\upsigma (a)}\Vert \mathtt {x}(t)\Vert \right\} , \end{aligned}$$

For any $\mathtt {x}_1\in \mathtt {P}_\upeta ,$ define an operator ${\mathscr {L}}:\mathtt {P}_\upeta \rightarrow \mathtt {X}$ by

$$\begin{aligned} \begin{aligned} (&{\mathscr {L}}\mathtt {x}_1)(t)=\int _0^{\upsigma (a)}\aleph (t,\uptau _1)\uplambda (\uptau _1) \mathtt {g}_1\bigg [\int _0^{\upsigma (a)}\aleph (\uptau _1,\uptau _2)\uplambda (\uptau _2)\mathtt {g}_2\bigg [\int _0^{\upsigma (a)}\aleph (\uptau _2,\uptau _3)\cdots \\&\times \mathtt {g}_{n-1}\bigg [\int _0^{\upsigma (a)}\aleph (\uptau _{n-1},\uptau _n) \uplambda (\uptau _n)\mathtt {g}_n(\mathtt {x}_1(\uptau _n))\Delta \uptau _n\bigg ]\cdot \cdot \cdot \Delta \uptau _3\bigg ]\Delta \uptau _2\bigg ]\Delta \uptau _1\\&+\frac{1}{1-\sum _{k=1}^{n-2}c_k}\sum _{k=1}^{n-2}c_k\int _0^{\upsigma (a)} \aleph (\upzeta _k, \uptau _1)\uplambda (\uptau _1) \mathtt {g}_1\bigg [\int _0^{\upsigma (a)}\aleph (\uptau _1,\uptau _2)\uplambda (\uptau _2)\cdots \\&\times \mathtt {g}_{n-1}\bigg [\int _0^{\upsigma (a)}\aleph (\uptau _{n-1},\uptau _n) \uplambda (\uptau _n)\mathtt {g}_n(\mathtt {x}_1(\uptau _n))\Delta \uptau _n\bigg ]\cdot \cdot \cdot \Delta \uptau _3\bigg ]\Delta \uptau _2\bigg ]\Delta \uptau _1. \end{aligned} \end{aligned}$$

Lemma 2.9

Assume that $({\mathcal {H}}_1)$–$({\mathcal {H}}_2)$ hold. Then for each $\upeta \in (0, \upsigma (a)/2)_{{\mathbb {T}}},$ ${\mathscr {L}}(\mathtt {P}_\upeta )\subset \mathtt {P}_\upeta$ and ${\mathscr {L}}:\mathtt {P}_\upeta \rightarrow \mathtt {P}_\upeta$ are completely continuous.

Proof

From Lemma 2.8, $\aleph (t,\uptau )\ge 0$ for all $t, \uptau \in (0,\upsigma (a)]_{\mathbb {T}}.$ So, $({\mathscr {L}}\mathtt {x}_1)(t)\ge 0.$ Also, for $\mathtt {x}_1\in \mathtt {P}_\upeta ,$ we have

$$\begin{aligned} \begin{aligned} \Vert {\mathscr {L}}\mathtt {x}_1\Vert =&\max _{t\in (0,\upsigma (a)]_{\mathbb {T}}}\int _0^{\upsigma (a)}\aleph (t,\uptau _1)\uplambda (\uptau _1) \mathtt {g}_1\bigg [\int _0^{\upsigma (a)}\aleph (\uptau _1,\uptau _2)\uplambda (\uptau _2)\cdots \\&\times \mathtt {g}_{n-1}\bigg [\int _0^{\upsigma (a)}\aleph (\uptau _{n-1},\uptau _n) \uplambda (\uptau _n)\mathtt {g}_n(\mathtt {x}_1(\uptau _n))\Delta \uptau _n\bigg ]\cdot \cdot \cdot \Delta \uptau _3\bigg ]\Delta \uptau _2\bigg ]\Delta \uptau _1\\&+\frac{1}{1-\sum _{k=1}^{n-2}c_k}\sum _{k=1}^{n-2}c_k\int _0^{\upsigma (a)} \aleph (\upzeta _k, \uptau _1)\uplambda (\uptau _1) \mathtt {g}_1\bigg [\int _0^{\upsigma (a)}\aleph (\uptau _1,\uptau _2)\uplambda (\uptau _2)\cdots \\&\times \mathtt {g}_{n-1}\bigg [\int _0^{\upsigma (a)}\aleph (\uptau _{n-1},\uptau _n) \uplambda (\uptau _n)\mathtt {g}_n(\mathtt {x}_1(\uptau _n))\Delta \uptau _n\bigg ]\cdot \cdot \cdot \Delta \uptau _3\bigg ]\Delta \uptau _2\bigg ]\Delta \uptau _1\\&\le \,\int _0^{\upsigma (a)}\aleph (\uptau _1,\uptau _1)\uplambda (\uptau _1) \mathtt {g}_1\bigg [\int _0^{\upsigma (a)}\aleph (\uptau _1,\uptau _2)\uplambda (\uptau _2)\cdots \\&\times \mathtt {g}_{n-1}\bigg [\int _0^{\upsigma (a)}\aleph (\uptau _{n-1},\uptau _n) \uplambda (\uptau _n)\mathtt {g}_n(\mathtt {x}_1(\uptau _n))\Delta \uptau _n\bigg ]\cdot \cdot \cdot \Delta \uptau _3\bigg ]\Delta \uptau _2\bigg ]\Delta \uptau _1\\&+\frac{1}{1-\sum _{k=1}^{n-2}c_k}\sum _{k=1}^{n-2}c_k\int _0^{\upsigma (a)} \aleph (\uptau _1, \uptau _1)\uplambda (\uptau _1) \mathtt {g}_1\bigg [\int _0^{\upsigma (a)}\aleph (\uptau _1,\uptau _2)\uplambda (\uptau _2)\cdots \\&\times \mathtt {g}_{n-1}\bigg [\int _0^{\upsigma (a)}\aleph (\uptau _{n-1},\uptau _n) \uplambda (\uptau _n)\mathtt {g}_n(\mathtt {x}_1(\uptau _n))\Delta \uptau _n\bigg ]\cdot \cdot \cdot \Delta \uptau _3\bigg ]\Delta \uptau _2\bigg ]\Delta \uptau _1. \end{aligned} \end{aligned}$$

Again from Lemma 2.8, we get

$$\begin{aligned} \begin{aligned}&\min _{t\in [\upeta ,a-\upeta ]_{\mathbb {T}}}\big \{({\mathscr {L}}\mathtt {x}_1)(t)\big \}\ge \frac{\upeta }{\upsigma (a)}\bigg [\int _0^{\upsigma (a)}\aleph (\uptau _1,\uptau _1)\uplambda (\uptau _1) \mathtt {g}_1\bigg [\int _0^{\upsigma (a)}\aleph (\uptau _1,\uptau _2)\uplambda (\uptau _2)\cdots \\&\times \mathtt {g}_{n-1}\bigg [\int _0^{\upsigma (a)}\aleph (\uptau _{n-1},\uptau _n) \uplambda (\uptau _n)\mathtt {g}_n(\mathtt {x}_1(\uptau _n))\Delta \uptau _n\bigg ]\cdot \cdot \cdot \Delta \uptau _3\bigg ]\Delta \uptau _2\bigg ]\Delta \uptau _1\\&+\frac{1}{1-\sum _{k=1}^{n-2}c_k}\sum _{k=1}^{n-2}c_k\int _0^{\upsigma (a)} \aleph (\uptau _1, \uptau _1)\uplambda (\uptau _1) \mathtt {g}_1\bigg [\int _0^{\upsigma (a)}\aleph (\uptau _1,\uptau _2)\uplambda (\uptau _2)\cdots \\&\times \mathtt {g}_{n-1}\bigg [\int _0^{\upsigma (a)}\aleph (\uptau _{n-1},\uptau _n) \uplambda (\uptau _n)\mathtt {g}_n(\mathtt {x}_1(\uptau _n))\Delta \uptau _n\bigg ]\cdot \cdot \cdot \Delta \uptau _3\bigg ]\Delta \uptau _2\bigg ]\Delta \uptau _1\bigg ]. \end{aligned} \end{aligned}$$

It follows from the above two inequalities that

$$\begin{aligned} \min _{t\in [\upeta ,a-\upeta ]_{\mathbb {T}}}\big \{({\mathscr {L}}\mathtt {x}_1)(t)\big \}\ge \frac{\upeta }{\upsigma (a)}\Vert {\mathscr {L}}\mathtt {x}_1\Vert . \end{aligned}$$

So, ${\mathscr {L}}\mathtt {x}_1\in \mathtt {P}_\upeta$ and thus ${\mathscr {L}}(\mathtt {P}_\upeta )\subset \mathtt {P}_\upeta .$ Next, by standard methods and Arzela-Ascoli theorem, it can be proved easily that the operator ${\mathscr {L}}$ is completely continuous. The proof is complete. $\square$

3 Infinitely many positive solutions

For the the existence of infinitely many positive solutions for iterative system of boundary value problem (1)–(2). We apply following theorems.

Theorem 3.1

(Krasnoselskii’s [18]) Let ${\mathcal {B}}$ be a cone in a Banach space ${\mathcal {E}}$ and $\mathtt {Q}_1,\, \mathtt {Q}_2$ are open sets with $0\in \mathtt {Q}_1, \overline{\mathtt {Q}}_1\subset \mathtt {Q}_2.$ Let ${\mathcal {K}}:{\mathcal {B}}\cap (\overline{\mathtt {Q}}_2\backslash \mathtt {Q}_1)\rightarrow {\mathcal {B}}$ be a completely continuous operator such that

(a):

$\Vert {\mathcal {K}}v\Vert \le \Vert v\Vert ,\, v\in {\mathcal {B}}\cap \partial \mathtt {Q}_1,$ and $\Vert {\mathcal {K}}v\Vert \ge \Vert v\Vert ,\, v\in {\mathcal {B}}\cap \partial \mathtt {Q}_2,$ or

(b):

$\Vert {\mathcal {K}}v\Vert \ge \Vert v\Vert ,\,v\in {\mathcal {B}}\cap \partial \mathtt {Q}_1,$ and $\Vert {\mathcal {K}}v\Vert \le \Vert v\Vert ,\, v\in {\mathcal {B}}\cap \partial \mathtt {Q}_2.$

Then ${\mathcal {K}}$ has a fixed point in ${\mathcal {B}}\cap (\overline{\mathtt {Q}}_2\backslash \mathtt {Q}_1).$

Theorem 3.2

(Hölder’s Inequality [3, 23]) Let $f\in L_\nabla ^p(I)$ with $p>1,\, g\in L_\nabla ^q(I)$ with $q>1,$ and $\frac{1}{p}+\frac{1}{q}=1.$ Then $fg\in L_\nabla ^1(I)$ and $\Vert fg\Vert _{L_\nabla ^1}\le \Vert f\Vert _{L_\nabla ^p}\Vert g\Vert _{L_\nabla ^q}.$ where

$$\begin{aligned} \Vert f\Vert _{L_\nabla ^p}:=\left\{ \begin{array}{ll}\displaystyle {\Big [\int _I\vert f\vert ^p(s)\nabla s\Big ]^{\frac{1}{p}}}, p\in {\mathbb {R}},\\ \displaystyle {inf\Big \{K\in {\mathbb {R}} \, / \, \vert f \vert \le K \,\, \nabla -a.e., \,on\, I\Big \}}, ~~ p=\infty , \end{array}\right. \end{aligned}$$

and $I=[a, b]_{{\mathbb {T}}}.$ Moreover, if $f\in L_\nabla ^1(I)$ and $g\in L_\nabla ^\infty (I).$ Then $fg \in L_\nabla ^1(I)$ and $\Vert fg\Vert _{L_\nabla ^1}\le \Vert f\Vert _{L_\nabla ^1}\Vert g\Vert _{L_\nabla ^\infty }.$

Consider the following three possible cases for $\uplambda _i\in L^{p_i}_{\Delta }(0,\upsigma (a)]_{{\mathbb {T}}}:$

$$\begin{aligned} \displaystyle \sum _{i=1}^{m}\frac{1}{p_i}<1, ~\sum _{i=1}^{m}\frac{1}{p_i}=1,~ \sum _{i=1}^{m}\frac{1}{p_i}>1. \end{aligned}$$

Firstly, we seek infinitely many positive solutions for the case $\displaystyle \sum _{i=1}^{m}\frac{1}{p_i}<1.$

Theorem 3.3

Suppose $({\mathcal {H}}_1)$–$({\mathcal {H}}_2)$ hold, let $\{\upeta _r\}_{r=1}^\infty$ be a sequence with $t_{r+1}<\upeta _r<t_r.\,$ Let $\{\Gamma _r\}_{r=1}^\infty$ and $\{\Lambda _r\}_{r=1}^\infty$ be such that

$$\begin{aligned} \Gamma _{r+1}<\frac{\upeta _r}{\upsigma (a)} \Lambda _r<\Lambda _r<\uptheta \Lambda _r<\Gamma _r~~\text {and}~~\frac{\upeta _r}{\upsigma (a)}<\frac{1}{2},~r\in {\mathbb {N}}, \end{aligned}$$

where

$$\begin{aligned} \begin{aligned} \uptheta&=\max \Bigg \{ \bigg [\frac{\upeta _1}{\upsigma (a)}\prod _{i=1}^{m}\updelta _i\int _{\upeta _1}^{\upsigma (a)-\upeta _1}\aleph (\uptau ,\uptau )\Delta \uptau \bigg ]^{-1},\\&\left[ \frac{\sum _{k=1}^{n-2}c_k}{1-\sum _{k=1}^{n-2}c_k}\frac{\upeta _1}{\upsigma (a)}\prod _{i=1}^{m}\updelta _i\int _{\upeta _1}^{\upsigma (a)-\upeta _1}\aleph (\uptau ,\uptau )\nabla \uptau \right] ^{-1}\Bigg \}. \end{aligned} \end{aligned}$$

Assume that $\mathtt {g}_\ell$ satisfies

$(\mathtt {J}_1)$:

$\displaystyle \mathtt {g}_\ell (\mathtt {x})\le \frac{{\mathfrak {N}}_1\Gamma _r}{2}$ $\forall$ $t\in (0,\upsigma (a)]_{\mathbb {T}},\, 0\le \mathtt {x} \le \Gamma _r,$ where

$$\begin{aligned} {\mathfrak {N}}_1<\min \left\{ \left[ \Vert \aleph \Vert _{L_{\nabla }^q}\prod _{i=1}^{m}\left\| \uplambda _i\right\| _{L_{\nabla }^{p_i}}\right] ^{-1},\,\left[ \frac{\sum _{k=1}^{n-2}c_k}{1-\sum _{k=1}^{n-2}c_k}\Vert \aleph \Vert _{L_{\nabla }^q}\prod _{i=1}^{m}\left\| \uplambda _i\right\| _{L_{\nabla }^{p_i}}\right] ^{-1}\right\} , \end{aligned}$$

$(\mathtt {J}_2)$:

$\displaystyle \mathtt {g}_\ell (\mathtt {x})\ge \frac{\uptheta \Lambda _r}{2}$ $\forall$ $t\in [\upeta _r, \upsigma (a)-\upeta _r]_{\mathbb {T}},~\displaystyle \frac{\upeta _r}{\upsigma (a)}\Lambda _r\le \mathtt {x} \le \Lambda _r.$

Then the iterative boundary value problem (1)–(2) has infinitely many solutions $\{(\mathtt {x}_1^{[r]}, \mathtt {x}_2^{[r]},\cdot \cdot \cdot ,\mathtt {x}_n^{[r]})\}_{r=1}^\infty$ such that $\mathtt {x}_\ell ^{[r]}(t)\ge 0$ on $(0,\upsigma (a)]_{\mathbb {T}},$ $\ell =1,2,\cdot \cdot \cdot ,n$ and $r\in {\mathbb {N}}.$

Proof

Let

$$\begin{aligned} \mathtt {Q}_{1, r}=\{ \mathtt {x}\in \mathtt {X} : \Vert \mathtt {x} \Vert< \Gamma _r\},~\mathtt {Q}_{2, r}=\{ \mathtt {x}\in \mathtt {X} : \Vert \mathtt {x} \Vert < \Lambda _r\} \end{aligned}$$

be open subsets of $\mathtt {X}.\,$ Let $\{\upeta _r\}_{r=1}^\infty$ be given in the hypothesis and we note that

$$\begin{aligned} t^*<t_{r+1}<\upeta _r<t_r<\frac{\upsigma (a)}{2}, \end{aligned}$$

for all $r\in {\mathbb {N}}$. For each $r\in {\mathbb {N}},$ we define the cone $\mathtt {P}_{\upeta _r}$ by

$$\begin{aligned} \mathtt {P}_{\upeta _r}=\Big \{ \mathtt {x}\in \mathtt {X} : \mathtt {x}(t)\ge 0,\, \min _{t\in {[\upeta _r,\upsigma (a)-\upeta _r]_{\mathbb {T}}}}\mathtt {x}(t)\ge \frac{\upeta _r}{\upsigma (a)}\Vert \mathtt {x}(t)\Vert \Big \}. \end{aligned}$$

Let $\mathtt {x}_1\in \mathtt {P}_{\upeta _r}\cap \partial {\mathtt {Q}_{1, r}}.$ Then, $\mathtt {x}_1(\uptau )\le \Gamma _r=\Vert \mathtt {x}_1 \Vert$ for all $\uptau \in (0,\upsigma (a)]_{\mathbb {T}}.$ By $(\mathtt {J}_1)$ and for $\uptau _{m-1}\in (0,\upsigma (a)]_{{\mathbb {T}}},$ we have

$$\begin{aligned} \begin{aligned} \int _0^{\upsigma (a)}\aleph (\uptau _{n-1},\uptau _n) \uplambda (\uptau _n)\mathtt {g}_n(\mathtt {x}_1(\uptau _n))\nabla \uptau _n&\le \int _0^{\upsigma (a)}\aleph (\uptau _n,\uptau _n) \uplambda (\uptau _n)\mathtt {g}_n(\mathtt {x}_1(\uptau _n))\nabla \uptau _n\\&\le \frac{{\mathfrak {N}}_1\Gamma _r}{2}\int _0^{\upsigma (a)}\aleph (\uptau _n,\uptau _n) \prod _{i=1}^{m}\uplambda _i(\uptau _n)\nabla \uptau _n. \end{aligned} \end{aligned}$$

There exists a $q>1$ such that $\displaystyle \frac{1}{q}+\sum _{i=1}^{n}\frac{1}{p_i}=1.$ So,

$$\begin{aligned} \begin{aligned} \int _0^{\upsigma (a)}\aleph (\uptau _{n-1},\uptau _n) \uplambda (\uptau _n)\mathtt {g}_n(\mathtt {x}_1(\uptau _n))\Delta \uptau _{n}&\le \frac{{\mathfrak {N}}_1\Gamma _r}{2}\big \Vert \aleph \big \Vert _{L_{\nabla }^q}\left\| \prod _{i=1}^{m}\uplambda _i\right\| _{L_{\nabla }^{p_i}}\\&\le \frac{{\mathfrak {N}}_1\Gamma _r}{2}\Vert \aleph \Vert _{L_{\nabla }^q}\prod _{i=1}^{m}\left\| \uplambda _i\right\| _{L_{\nabla }^{p_i}}\le \frac{\Gamma _r}{2}<\Gamma _r. \end{aligned} \end{aligned}$$

It follows in similar manner (for $\uptau _{n-2}\in (0,\upsigma (a)]_{{\mathbb {T}}},$ ) that

$$\begin{aligned} \begin{aligned}&\int _0^{\upsigma (a)}\aleph (\uptau _{n-2},\uptau _{n-1}) \uplambda (\uptau _{n-1})\mathtt {g}_{n-1}\bigg [\int _0^{\upsigma (a)}\aleph (\uptau _{n-1},\uptau _n)\uplambda (\uptau _n)\mathtt {g}_n(\mathtt {x}_1(\uptau _n))\nabla \uptau _{n}\bigg ]\nabla \uptau _{n-1}\\&\quad \le \int _0^{\upsigma (a)}\aleph (\uptau _{n-2},\uptau _{n-1}) \uplambda (\uptau _{n-1})\mathtt {g}_{n-1}(\Gamma _r)\nabla \uptau _{n-1}\\&\quad \le \int _0^{\upsigma (a)}\aleph (\uptau _{n-1},\uptau _{n-1}) \uplambda (\uptau _{n-1})\mathtt {g}_{n-1}(\Gamma _r)\nabla \uptau _{n-1}\\&\quad \le \frac{{\mathfrak {N}}_1\Gamma _r}{2}\int _0^{\upsigma (a)}\aleph (\uptau _{n-1},\uptau _{n-1}) \prod _{i=1}^{m}\uplambda _i(\uptau _{n-1})\nabla \uptau _{n-1}\\&\quad \le \frac{{\mathfrak {N}}_1\Gamma _r}{2}\Vert \aleph \Vert _{L_{\nabla }^q}\prod _{i=1}^{m}\left\| \uplambda _i\right\| _{L_{\nabla }^{p_i}}\le \frac{\Gamma _r}{2}< \Gamma _r. \end{aligned} \end{aligned}$$

Continuing with this bootstrapping argument, we get

$$\begin{aligned} \begin{aligned}&\int _0^{\upsigma (a)}\aleph (t,\uptau _1)\uplambda (\uptau _1) \mathtt {g}_1\bigg [\int _0^{\upsigma (a)}\aleph (\uptau _1,\uptau _2)\uplambda (\uptau _2)\mathtt {g}_2\bigg [\int _0^{\upsigma (a)}\aleph (\uptau _2,\uptau _3)\cdots \\&\quad \times \mathtt {g}_{n-1}\bigg [\int _0^{\upsigma (a)}\aleph (\uptau _{n-1},\uptau _n) \uplambda (\uptau _n)\mathtt {g}_n(\mathtt {x}_1(\uptau _n))\nabla \uptau _n\bigg ]\cdot \cdot \cdot \nabla \uptau _3\bigg ]\nabla \uptau _2\bigg ]\nabla \uptau _1 \le \frac{\Gamma _r}{2}. \end{aligned} \end{aligned}$$

Also, we note that

$$\begin{aligned} \begin{aligned}&\frac{1}{1-\sum _{k=1}^{n-2}c_k}\sum _{k=1}^{n-2}c_k\int _0^{\upsigma (a)} \aleph (\upzeta _k, \uptau _1)\uplambda (\uptau _1) \mathtt {g}_1\bigg [\int _0^{\upsigma (a)}\aleph (\uptau _1,\uptau _2)\uplambda (\uptau _2)\cdots \\&\quad \times \mathtt {g}_{n-1}\bigg [\int _0^{\upsigma (a)}\aleph (\uptau _{n-1},\uptau _n) \uplambda (\uptau _n)\mathtt {g}_n(\mathtt {x}_1(\uptau _n))\nabla \uptau _n\bigg ]\cdot \cdot \cdot \nabla \uptau _2\bigg ]\nabla \uptau _1\\&\quad \le \frac{1}{1-\sum _{k=1}^{n-2}c_k}\sum _{k=1}^{n-2}c_k\int _0^{\upsigma (a)} \aleph (\uptau _1, \uptau _1)\uplambda (\uptau _1) \mathtt {g}_1(\Gamma _r)\nabla \uptau _1\\&\quad \le \frac{1}{1-\sum _{k=1}^{n-2}c_k}\sum _{k=1}^{n-2}c_k\frac{{\mathfrak {N}}_1\Gamma _r}{2}\Vert \aleph \Vert _{L_{\nabla }^q}\prod _{i=1}^{m}\left\| \uplambda _i\right\| _{L_{\nabla }^{p_i}}\le \frac{\Gamma _r}{2}. \end{aligned} \end{aligned}$$

Thus, $({\mathscr {L}}\mathtt {x}_1)(t)\le \frac{\Gamma _r}{2}+\frac{\Gamma _r}{2}=\Gamma _r.$ Since $\Gamma _r=\Vert \mathtt {x}_1\Vert$ for $\mathtt {x}_1\in \mathtt {P}_{\upeta _r}\cap \partial {\mathtt {Q}_{1, r}},$ we get

$$\begin{aligned} \Vert {\mathscr {L}}\mathtt {x}_1\Vert \le \Vert \mathtt {x}_1\Vert . \end{aligned}$$

(9)

Next, let $t\in [\upeta _r, \upsigma (a)-\upeta _r]_{\mathbb {T}}.$ Then,

$$\begin{aligned} \Lambda _r=\Vert \mathtt {x}_1\Vert \ge \mathtt {x}_1(t)\ge \,\min _{t\in [\upeta _r, a-\upeta _r]_{\mathbb {T}}}\,\mathtt {x}_1(t)\ge \,\frac{\upeta _r}{\upsigma (a)}\,\Vert \mathtt {x}_1 \Vert \ge \frac{\upeta _r}{\upsigma (a)}\Lambda _r. \end{aligned}$$

By $(\mathtt {J}_2)$ and for $\uptau _{n-1}\in [\upeta _r, \upsigma (a)-\upeta _r]_{\mathbb {T}},$ we have

$$\begin{aligned} \begin{aligned}&\int _0^{\upsigma (a)}\aleph (\uptau _{n-1},\uptau _n) \uplambda (\uptau _n)\mathtt {g}_n(\mathtt {x}_1(\uptau _n))\big ]\nabla \uptau _{n}\\&\quad \ge \int _{\upeta _r}^{\upsigma (a)-\upeta _r}\aleph (\uptau _{n-1},\uptau _n) \uplambda (\uptau _n)\mathtt {g}_n(\mathtt {x}_1(\uptau _n))\nabla \uptau _n\\&\quad \ge \frac{\upeta _r}{\upsigma (a)}\frac{\uptheta \Lambda _r}{2}\int _{\upeta _r}^{\upsigma (a)-\upeta _r}\aleph (\uptau _n,\uptau _n)\uplambda (\uptau _n))\nabla \uptau _n\\&\quad \ge \frac{\upeta _r}{\upsigma (a)}\frac{\uptheta \Lambda _r}{2}\int _{\upeta _r}^{\upsigma (a)-\upeta _r}\aleph (\uptau _n,\uptau _n)\prod _{i=1}^{m}\uplambda _i(\uptau _n))\nabla \uptau _n\\&\quad \ge \frac{\upeta _1}{\upsigma (a)}\frac{\uptheta \Lambda _r}{2}\prod _{i=1}^{m}\updelta _i\int _{\upeta _1}^{\upsigma (a)-\upeta _1}\aleph (\uptau _n,\uptau _n)\nabla \uptau _n\\&\quad \ge \frac{\Lambda _r}{2}. \end{aligned} \end{aligned}$$

and

$$\begin{aligned} \begin{aligned}&\frac{1}{1-\sum _{k=1}^{n-2}c_k}\sum _{k=1}^{n-2}c_k\int _0^{\upsigma (a)} \aleph (\upzeta _k, \uptau _1)\uplambda (\uptau _1) \mathtt {g}_1\bigg [\int _0^{\upsigma (a)}\aleph (\uptau _1,\uptau _2)\uplambda (\uptau _2)\cdots \\&\quad \times \mathtt {g}_{n-1}\bigg [\int _0^{\upsigma (a)}\aleph (\uptau _{n-1},\uptau _n) \uplambda (\uptau _n)\mathtt {g}_n(\mathtt {x}_1(\uptau _n))\nabla \uptau _n\bigg ]\cdot \cdot \cdot \nabla \uptau _2\bigg ]\nabla \uptau _1\\&\quad \ge \frac{1}{1-\sum _{k=1}^{n-2}c_k}\sum _{k=1}^{n-2}c_k\frac{\upeta _1}{\upsigma (a)}\int _0^{\upsigma (a)} \aleph (\uptau _1, \uptau _1)\uplambda (\uptau _1) \mathtt {g}_1(\Gamma _r)\nabla \uptau _1\\&\quad \ge \frac{1}{1-\sum _{k=1}^{n-2}c_k}\sum _{k=1}^{n-2}c_k\frac{\upeta _1}{\upsigma (a)}\frac{\uptheta \Lambda _r}{2}\prod _{i=1}^{m}\updelta _i\int _{\upeta _1}^{\upsigma (a)-\upeta _1}\aleph (\uptau _1,\uptau _1)\nabla \uptau _1 \end{aligned} \end{aligned}$$

Continuing with bootstrapping argument, we get $({\mathscr {L}}\mathtt {x}_1)(t)\ge \frac{\Lambda _r}{2}+\frac{\Lambda _r}{2}=\Lambda _r.$ Thus, if $\mathtt {x}_1\in \mathtt {P}_{\upeta _r}\cap \partial \mathtt {P}_{2, r},$ then

$$\begin{aligned} \Vert {\mathscr {L}}\mathtt {x}_1\Vert \ge \Vert \mathtt {x}_1\Vert . \end{aligned}$$

(10)

It is evident that $0\in \mathtt {Q}_{2, k}\subset \overline{\mathtt {Q}}_{2, k}\subset \mathtt {Q}_{1, k}.$ From (9),(10), it follows from Theorem 3.1 that the operator ${\mathscr {L}}$ has a fixed point $\mathtt {x}_1^{[r]}\in \mathtt {P}_{\upeta _r}\cap \big (\overline{\mathtt {Q}}_{1, r}\backslash \mathtt {Q}_{2, r}\big )$ such that $\mathtt {x}_1^{[r]}(t)\ge 0$ on $(0,a]_{\mathbb {T}},$ and $r\in {\mathbb {N}}.$ Next setting $\mathtt {x}_{m+1}=\mathtt {x}_1,$ we obtain infinitely many positive solutions $\{(\mathtt {x}_1^{[r]}, \mathtt {x}_2^{[r]},\ldots ,\mathtt {x}_m^{[r]})\}_{r=1}^\infty$ of (1)–(2) given iteratively by

$$\begin{aligned} \begin{aligned} \mathtt {x}_\ell (t)=\int _0^{\upsigma (a)}\aleph (t,\uptau )\uplambda (\uptau ) \mathtt {g}_\ell (\mathtt {x}_{\ell +1}(\uptau ))\nabla \uptau ,~t\in (0,\upsigma (a)]_{{\mathbb {T}}},~\ell =n,n-1,\ldots ,1. \end{aligned} \end{aligned}$$

The proof is completed. $\square$

For $\displaystyle \sum _{i=1}^{m}\frac{1}{p_i}=1,$ we have the following theorem.

Theorem 3.4

Suppose $({\mathcal {H}}_1)$–$({\mathcal {H}}_2)$ hold, let $\{\upeta _r\}_{r=1}^\infty$ be a sequence with $t_{r+1}<\upeta _r<t_r.\,$ Let $\{\Gamma _r\}_{r=1}^\infty$ and $\{\Lambda _r\}_{r=1}^\infty$ be such that

$$\begin{aligned} \Gamma _{r+1}<\frac{\upeta _r}{\upsigma (a)} \Lambda _r<\Lambda _r<\uptheta \Lambda _r<\Gamma _r~~~\text {and}~~~\frac{\upeta _r}{\upsigma (a)}<\frac{1}{2},~r\in {\mathbb {N}}. \end{aligned}$$

Assume that $\mathtt {g}_\ell$ satisfies $(\mathtt {J}_2)$ and

$(\mathtt {J}_3)$:

$\displaystyle \mathtt {g}_\ell (\mathtt {x})\le \frac{{\mathfrak {N}}_2\Gamma _r}{2}$ $\forall$ $t\in (0,\upsigma (a)]_{\mathbb {T}},\, 0\le \mathtt {x} \le \Gamma _r,$ where

$$\begin{aligned} {\mathfrak {N}}_2<\min \left\{ \left[ \Vert \aleph \Vert _{L_{\nabla }^\infty }\prod _{i=1}^{m}\left\| \uplambda _i\right\| _{L_{\nabla }^{p_i}}\right] ^{-1},\,\left[ \frac{\sum _{k=1}^{n-2}c_k}{1-\sum _{k=1}^{n-2}c_k}\Vert \aleph \Vert _{L_{\nabla }^\infty }\prod _{i=1}^{m}\left\| \uplambda _i\right\| _{L_{\nabla }^{p_i}}\right] ^{-1}\right\} . \end{aligned}$$

Then the iterative boundary value problem (1)–(2) has infinitely many solutions $\left\{ \left( \mathtt {x}_1^{[r]}, \mathtt {x}_2^{[r]},\ldots ,\mathtt {x}_n^{[r]}\right) \right\} _{r=1}^\infty$ such that $\mathtt {x}_\ell ^{[r]}(t)\ge 0$ on $(0,\upsigma (a)]_{\mathbb {T}},$ $\ell =1,2,\ldots ,n$ and $r\in {\mathbb {N}}.$

Proof

For a fixed r,  let $\mathtt {Q}_{1, r}$ be as in the proof of Theorem 3.3 and let $\mathtt {x}_1\in \mathtt {P}_{\upeta _r}\cap \partial \mathtt {Q}_{2, r}.$ Again

$$\begin{aligned} \mathtt {x}_1(\uptau )\le \Gamma _r=\Vert \mathtt {x}_1 \Vert , \end{aligned}$$

for all $\uptau \in (0,\upsigma (a)]_{\mathbb {T}}.$ By $(\mathtt {J}_3)$ and for $\uptau _{\ell -1}\in (0,\upsigma (a)]_{{\mathbb {T}}},$ we have

$$\begin{aligned} \begin{aligned} \int _0^{\upsigma (a)}\aleph (\uptau _{n-1},\uptau _n) \uplambda (\uptau _n)\mathtt {g}_n(\mathtt {x}_1(\uptau _n))\nabla \uptau _n&\le \int _0^{\upsigma (a)}\aleph (\uptau _n,\uptau _n) \uplambda (\uptau _n)\mathtt {g}_n(\mathtt {x}_1(\uptau _n))\nabla \uptau _n\\&\le \frac{{\mathfrak {N}}_1\Gamma _r}{2}\int _0^{\upsigma (a)}\aleph (\uptau _n,\uptau _n) \prod _{i=1}^{m}\uplambda _i(\uptau _n)\nabla \uptau _n\\&\le \frac{{\mathfrak {N}}_1\Gamma _r}{2}\big \Vert \aleph \big \Vert _{L_{\nabla }^\infty }\left\| \prod _{i=1}^{m}\uplambda _i\right\| _{L_{\nabla }^{p_i}}\\&\le \frac{{\mathfrak {N}}_1\Gamma _r}{2}\Vert \aleph \Vert _{L_{\nabla }^\infty }\prod _{i=1}^{m}\left\| \uplambda _i\right\| _{L_{\nabla }^{p_i}}\le \frac{\Gamma _r}{2}<\Gamma _r. \end{aligned} \end{aligned}$$

It follows in similar manner (for $\uptau _{n-2}\in (0,\upsigma (a)]_{{\mathbb {T}}},$ ) that

$$\begin{aligned} \begin{aligned} \int _0^{\upsigma (a)}\aleph (\uptau _{n-2},\uptau _{n-1})&\uplambda (\uptau _{n-1})\mathtt {g}_{n-1}\bigg [\int _0^{\upsigma (a)}\aleph (\uptau _{n-1},\uptau _n)\uplambda (\uptau _n)\mathtt {g}_n(\mathtt {x}_1(\uptau _n))\nabla \uptau _{n}\bigg ]\nabla \uptau _{n-1}\\&\le \int _0^{\upsigma (a)}\aleph (\uptau _{n-2},\uptau _{n-1}) \uplambda (\uptau _{n-1})\mathtt {g}_{n-1}(\Gamma _r)\nabla \uptau _{n-1}\\&\le \int _0^{\upsigma (a)}\aleph (\uptau _{n-1},\uptau _{n-1}) \uplambda (\uptau _{n-1})\mathtt {g}_{n-1}(\Gamma _r)\nabla \uptau _{n-1}\\&\le \frac{{\mathfrak {N}}_1\Gamma _r}{2}\int _0^{\upsigma (a)}\aleph (\uptau _{n-1},\uptau _{n-1}) \prod _{i=1}^{m}\uplambda _i(\uptau _{n-1})\nabla \uptau _{n-1}\\&\le \frac{{\mathfrak {N}}_1\Gamma _r}{2}\Vert \aleph \Vert _{L_{\nabla }^\infty }\prod _{i=1}^{m}\left\| \uplambda _i\right\| _{L_{\nabla }^{p_i}}\le \frac{\Gamma _r}{2}< \Gamma _r. \end{aligned} \end{aligned}$$

Continuing with this bootstrapping argument, we get

$$\begin{aligned} \begin{aligned} \int _0^{\upsigma (a)}&\aleph (t,\uptau _1)\uplambda (\uptau _1) \mathtt {g}_1\bigg [\int _0^{\upsigma (a)}\aleph (\uptau _1,\uptau _2)\uplambda (\uptau _2)\mathtt {g}_2\bigg [\int _0^{\upsigma (a)}\aleph (\uptau _2,\uptau _3)\cdots \\&\times \mathtt {g}_{n-1}\bigg [\int _0^{\upsigma (a)}\aleph (\uptau _{n-1},\uptau _n) \uplambda (\uptau _n)\mathtt {g}_n(\mathtt {x}_1(\uptau _n))\nabla \uptau _n\bigg ]\cdot \cdot \cdot \nabla \uptau _3\bigg ]\nabla \uptau _2\bigg ]\nabla \uptau _1 \le \frac{\Gamma _r}{2}. \end{aligned} \end{aligned}$$

Also, we note that

$$\begin{aligned} \begin{aligned} \frac{1}{1-\sum _{k=1}^{n-2}c_k}&\sum _{k=1}^{n-2}c_k\int _0^{\upsigma (a)} \aleph (\upzeta _k, \uptau _1)\uplambda (\uptau _1) \mathtt {g}_1\bigg [\int _0^{\upsigma (a)}\aleph (\uptau _1,\uptau _2)\uplambda (\uptau _2)\cdots \\&\times \mathtt {g}_{n-1}\bigg [\int _0^{\upsigma (a)}\aleph (\uptau _{n-1},\uptau _n) \uplambda (\uptau _n)\mathtt {g}_n(\mathtt {x}_1(\uptau _n))\nabla \uptau _n\bigg ]\cdot \cdot \cdot \nabla \uptau _2\bigg ]\nabla \uptau _1\\&\le \frac{1}{1-\sum _{k=1}^{n-2}c_k}\sum _{k=1}^{n-2}c_k\int _0^{\upsigma (a)} \aleph (\uptau _1, \uptau _1)\uplambda (\uptau _1) \mathtt {g}_1(\Gamma _r)\nabla \uptau _1\\&\le \frac{1}{1-\sum _{k=1}^{n-2}c_k}\sum _{k=1}^{n-2}c_k\frac{{\mathfrak {N}}_1\Gamma _r}{2}\Vert \aleph \Vert _{L_{\nabla }^\infty }\prod _{i=1}^{m}\left\| \uplambda _i\right\| _{L_{\nabla }^{p_i}}\le \frac{\Gamma _r}{2}. \end{aligned} \end{aligned}$$

Thus, $({\mathscr {L}}\mathtt {x}_1)(t)\le \frac{\Gamma _r}{2}+\frac{\Gamma _r}{2}=\Gamma _r.$ Since $\Gamma _r=\Vert \mathtt {x}_1\Vert$ for $\mathtt {x}_1\in \mathtt {P}_{\upeta _r}\cap \partial {\mathtt {Q}_{1, r}},$ we get

$$\begin{aligned} \Vert {\mathscr {L}}\mathtt {x}_1\Vert \le \Vert \mathtt {x}_1\Vert . \end{aligned}$$

(11)

Now define $\mathtt {Q}_{2, r}=\{\mathtt {x}_1\in \mathtt {X}:\Vert \mathtt {x}_1 \Vert <\Lambda _r\}.$ Let $\mathtt {x}_1\in \mathtt {P}_{\upeta _r}\cap \partial \mathtt {Q}_{2, r}$ and let $\uptau \in [\upeta _r,\, \upsigma (a)-\upeta _r]_{\mathbb {T}}.$ Then, the argument leading to (11) can be done to the present case. Hence, the theorem. $\square$

Lastly, the case $\displaystyle \sum _{i=1}^{m}\frac{1}{p_i}>1.$

Theorem 3.5

Suppose $({\mathcal {H}}_1)$–$({\mathcal {H}}_2)$ hold, let $\{\upeta _r\}_{r=1}^\infty$ be a sequence with $t_{r+1}<\upeta _r<t_r.\,$ Let $\{\Gamma _r\}_{r=1}^\infty$ and $\{\Lambda _r\}_{r=1}^\infty$ be such that

$$\begin{aligned} \Gamma _{r+1}<\frac{\upeta _r}{\upsigma (a)} \Lambda _r<\Lambda _r<\uptheta \Lambda _r<\Gamma _r~~~\text {and}~~~\frac{\upeta _r}{\upsigma (a)}<\frac{1}{2},~r\in {\mathbb {N}}. \end{aligned}$$

Assume that $\mathtt {g}_\ell$ satisfies $(\mathtt {J}_2)$ and

$(\mathtt {J}_4)$:

$\displaystyle \mathtt {g}_\ell (\mathtt {x})\le \frac{{\mathfrak {N}}_2\Gamma _r}{2}$ $\forall$ $t\in (0,\upsigma (a)]_{\mathbb {T}},\, 0\le \mathtt {x} \le \Gamma _r,$ where

$$\begin{aligned} {\mathfrak {N}}_2<\min \left\{ \left[ \Vert \aleph \Vert _{L_{\nabla }^\infty }\prod _{i=1}^{m}\left\| \uplambda _i\right\| _{L_{\nabla }^{1}}\right] ^{-1},\,\left[ \frac{\sum _{k=1}^{n-2}c_k}{1-\sum _{k=1}^{n-2}c_k}\Vert \aleph \Vert _{L_{\nabla }^\infty }\prod _{i=1}^{m}\left\| \uplambda _i\right\| _{L_{\nabla }^{1}}\right] ^{-1}\right\} . \end{aligned}$$

Then the iterative boundary value problem (1)–(2) has infinitely many solutions $\{(\mathtt {x}_1^{[r]}, \mathtt {x}_2^{[r]},\ldots ,\mathtt {x}_n^{[r]})\}_{r=1}^\infty$ such that $\mathtt {x}_\ell ^{[r]}(t)\ge 0$ on $(0,\upsigma (a)]_{\mathbb {T}},$ $\ell =1,2,\ldots ,n$ and $r\in {\mathbb {N}}.$

Proof

The proof is similar to the proof of Theorem 3.1. So, we omit the details here. $\square$

4 Example

In this section, we provide two examples to check validity of our main results.

Example 4.1

Consider the following boundary value problem on ${\mathbb {T}}=[0, 1].$

$$\begin{aligned} \left. \begin{aligned}&\mathtt {x}_\ell ''(t)+\uplambda (t) \mathtt {g}_\ell (\mathtt {x}_{\ell +1}(t))=0, t\in (0,\upsigma (1)]_{{\mathbb {T}}},\,\ell =1,2,3,4,\\&\mathtt {x}_5(t)=\mathtt {x}_1(t),t\in (0,\upsigma (1)]_{{\mathbb {T}}}, \end{aligned}\right\} \end{aligned}$$

(12)

$$\begin{aligned} \mathtt {x}_\ell '(0)=0,~~\mathtt {x}_\ell (1)=\frac{1}{2}\mathtt {x}_\ell \left( \frac{1}{3}\right) +\frac{1}{3}\mathtt {x}_\ell \left( \frac{1}{4}\right) , \end{aligned}$$

(13)

where we take $n=4,m=2,$ $c_1=\frac{1}{2},$ $c_2=\frac{1}{3},$ $\upzeta _1=\frac{1}{3},$ $\upzeta _2=\frac{1}{4}$ and $\uplambda (t)=\uplambda _1(t)\uplambda _2(t)$ in which

$$\begin{aligned} \uplambda _1(t)=\frac{1}{\vert t-\frac{1}{4}\vert ^{\frac{1}{2}}}~~\text {~~and~~}~~\uplambda _2(t)=\frac{1}{\vert t-\frac{3}{4}\vert ^{\frac{1}{2}}}. \end{aligned}$$

Then $\sum _{k=1}^{n-2}c_k=\frac{5}{6}<1$ and $\updelta _1=\updelta _2=\left( 4/3\right) ^{1/2}.$ For $\ell =1,2,3,4,$ let

$$ \begin{aligned}\mathtt {g}_\ell (\mathtt {x})=\left\{ \begin{array}{llllll} 0.05\times 10^{-4},\mathtt {x}\in (10^{-4}, +\infty ), \\ \frac{62\times 10^{-(4r+3)}-0.05\times 10^{-4r}}{10^{-(4r+3)}-10^{-4r}}(\mathtt {x}-10^{-4r})+0.05\times 10^{-8r},\\ ~\mathtt {x}\in \bigg [10^{-(4r+3)}, 10^{-4r}\bigg ],\\ 62\times 10^{-(4r+3)},\mathtt {x}\in \bigg (\frac{1}{5}\times 10^{-(4r+3)}, 10^{-(4r+3)}\bigg ),\\ \frac{62\times 10^{-(4r+3)}-0.05\times 10^{-8r}}{0.05\times 10^{-(4r+3)}-10^{-(4r+4)}}(\mathtt {x}-10^{-(4r+4)})+0.05\times 10^{-8r},\\ ~\mathtt {x}\in \bigg (10^{-(4r+4)}, \frac{1}{5}\times 10^{-(4r+3)}\bigg ],\\ 0, ~\mathtt {x}=0, \end{array} \right. \end{aligned}$$

for all $r\in {\mathbb {N}}.$ Let

$$\begin{aligned} t_r=\frac{31}{64}-\sum _{k=1}^{r}\frac{1}{4(k+1)^4}~~\text {and}~~\upeta _r=\frac{1}{2}(t_r+t_{r+1}),~r\in {\mathbb {N}}, \end{aligned}$$

then

$$\begin{aligned} \upeta _1=\frac{15}{32}-\frac{1}{648}<\frac{15}{32} \end{aligned}$$

and

$$\begin{aligned} t_{r+1}<\upeta _r<t_r,\, \upeta _r>\frac{1}{5}. \end{aligned}$$

Therefore,

$$\begin{aligned} \frac{\upeta _r}{a}=\frac{\upeta _r}{1}>\frac{1}{5},\,r\in {\mathbb {N}}. \end{aligned}$$

It is clear that

$$\begin{aligned} t_1=\frac{15}{32}<\frac{1}{2},~ t_r-t_{r+1}=\frac{1}{4(r+2)^4},~ r\in {\mathbb {N}}. \end{aligned}$$

Since $\displaystyle \sum _{j=1}^{\infty }\frac{1}{j^4}=\frac{\uppi ^4}{90}$ and $\displaystyle \sum _{j=1}^{\infty }\frac{1}{j^2}=\frac{\uppi ^2}{6},$ it follows that

$$\begin{aligned} t^*=\lim _{r\rightarrow \infty }t_r=\frac{31}{64}-\sum _{k=1}^{\infty }\frac{1}{4(r+1)^4}=\frac{47}{64}-\frac{\uppi ^4}{360}=0.46. \end{aligned}$$

Also, we have

$$\begin{aligned} \begin{aligned} \int _{\upeta _1}^{\upsigma (a)-\upeta _1}\aleph (\uptau ,\uptau )\Delta \uptau =\int _{\frac{15}{32}-\frac{1}{648}}^{1-\frac{15}{32}+\frac{1}{648}}(1-\uptau )d\uptau =0.03. \end{aligned} \end{aligned}$$

Thus, we get

$$\begin{aligned} \begin{aligned} \uptheta =\max \bigg \{\frac{1}{0.0163}, \frac{1}{5\times 0.0163}\bigg \}=61.35. \end{aligned} \end{aligned}$$

Next, let $0<{\mathfrak {a}}<1$ be fixed. Then $\uplambda _1,\uplambda _2\in L^{1+{\mathfrak {a}}}[0,1].$ A simple calculations shows that

$$\begin{aligned} \int _0^{\upsigma (1)}\uplambda _1(t)\uplambda _2(t)dt=\uppi -\ln (7-4\sqrt{3}). \end{aligned}$$

So, let $p_i=1$ for $i=1,2.$ Then

$$\begin{aligned} \prod _{i=1}^{m}\left\| \uplambda _i\right\| _{L_\nabla ^{p_i}}=\uppi -\ln (7-4\sqrt{3})\approx 5.78, \end{aligned}$$

and also $\Vert \aleph \Vert _{L_\nabla ^{\infty }}=1.$ Therefore,

$$\begin{aligned} {\mathfrak {N}}_1<\left[ \Vert \aleph \Vert _{\infty }\prod _{i=1}^{m}\left\| \uplambda _i\right\| _{L_\nabla ^{p_i}}\right] ^{-1}\approx 0.173. \end{aligned}$$

Taking ${\mathfrak {N}}_1=\frac{1}{10}.$ In addition if we take

$$\begin{aligned} \Gamma _r=10^{-4r},\,\Lambda _r=10^{-(4r+3)}, \end{aligned}$$

then

$$\begin{aligned} \begin{aligned} \Gamma _{r+1}&=10^{-(4r+4)}<\frac{1}{5}\times 10^{-(4r+3)}<\frac{\upeta _r}{a}\Lambda _r\\&<\Lambda _r=10^{-(4r+3)}<\Gamma _r=10^{-4r}, \end{aligned} \end{aligned}$$

$\uptheta \Lambda _r=61.35\times 10^{-(4r+3)}<\frac{1}{10}\times 10^{-4r}={\mathfrak {N}}_1\Gamma _r,\,r\in {\mathbb {N}}$ and $\mathtt {g}_\ell (\ell =1,2,3,4)$ satisfies the following growth conditions:

$$\begin{aligned} \begin{aligned} \mathtt {g}_\ell (\mathtt {x})\le \,&{\mathfrak {N}}_1\Gamma _r=\frac{1}{10}\times 10^{-4r},~~\mathtt {x}\in \bigg [0, 10^{-4r}\bigg ],\\ \mathtt {g}_\ell (\mathtt {x})\ge \,&\uptheta \Lambda _r=61.35\times 10^{-(4r+3)},~~\mathtt {x}\in \bigg [\frac{1}{5}\times 10^{-(4r+3)}, 10^{-(4r+3)}\bigg ], \end{aligned} \end{aligned}$$

for $r\in {\mathbb {N}}.$ Then all the conditions of Theorem 3.3 are satisfied. Therefore, by Theorem 3.3, the iterative boundary value problem (1) has infinitely many solutions $\{(\mathtt {x}_1^{[r]}, \mathtt {x}_2^{[r]},\mathtt {x}_3^{[r]},\mathtt {x}_4^{[r]})\}_{r=1}^\infty$ such that $\mathtt {x}_\ell ^{[r]}(t)\ge 0$ on [0, 1],  $\ell =1,2,3,4$ and $r\in {\mathbb {N}}.$

Example 4.2

Let $\displaystyle {\mathbb {T}}=\{0\}\cup [1/2, 1]\cup \left\{ \frac{1}{2^{k+1}}:k\in {\mathbb {N}}\right\} .$ Consider the boundary value problem

$$\begin{aligned} \left. \begin{aligned}&\mathtt {x}_\ell ^{\Delta \nabla }(t)+\uplambda (t) \mathtt {g}_\ell (\mathtt {x}_{\ell +1}(t))=0,\,t\in (0,\upsigma (1)]_{{\mathbb {T}}},\,\ell =1,2,3,\\&\mathtt {x}_4(t)=\mathtt {x}_1(t),\,t\in (0,\upsigma (1)]_{{\mathbb {T}}}, \end{aligned}\right\} \end{aligned}$$

(14)

$$\begin{aligned} \mathtt {x}_\ell '(0)=0,~~\mathtt {x}_\ell (1)=\frac{1}{5}\mathtt {x}_\ell \left( \frac{1}{4}\right) , \end{aligned}$$

(15)

where we take $n=3,$ $m=2,$ $c_1=\frac{1}{5},$ $\upzeta _1=\frac{1}{4}$ and $\uplambda (t)=\uplambda _1(t)\uplambda _2(t)$ in which

$$\begin{aligned} \uplambda _1(t)=\frac{1}{\vert t-\frac{2}{5}\vert ^{1/4}}~~~~\text {~~and~~}~~~~\uplambda _2(t)=\frac{1}{\vert t-\frac{3}{4}\vert ^{1/4}}. \end{aligned}$$

Then $\sum _{k=1}^{n-2}c_k=\frac{1}{5}<1$ and $\updelta _1=\updelta _2=\left( 4/3\right) ^{1/4}.$ For $\ell =1,2,3,$ let

$$\begin{aligned} \begin{aligned}\mathtt {g}_\ell (\mathtt {x})=\left\{ \begin{array}{llllll} \frac{1}{5}\times 10^{-9},\mathtt {x}\in (10^{-9}, +\infty ), \\ \frac{62\times 10^{-(8r+3)}-\frac{1}{5}\times 10^{-(8r+1)}}{10^{-(8r+3)}-10^{-(8r+1)}}(\mathtt {x}-10^{-(8r+1)})+\frac{1}{5}\times 10^{-(8r+1)},\\ ~\mathtt {x}\in \bigg [10^{-(8r+3)}, 10^{-(8r+1)}\bigg ],\\ 62\times 10^{-(8r+3)},\mathtt {x}\in \bigg (\frac{1}{5}\times 10^{-(8r+3)}, 10^{-(8r+3)}\bigg ),\\ \frac{62\times 10^{-(8r+3)}-\frac{1}{5}\times 10^{-(8r+4)}}{\frac{1}{5}\times 10^{-(8r+3)}-10^{-(8r+4)}}(\mathtt {x}-10^{-(8r+4)})+\frac{1}{5}\times 10^{-(8r+4)},\\ ~\mathtt {x}\in \bigg (10^{-(8r+4)}, \frac{1}{5}\times 10^{-(8r+3)}\bigg ],\\ 0, ~\mathtt {x}=0, \end{array} \right. \end{aligned} \end{aligned}$$

for all $r\in {\mathbb {N}}.$

Let $t_r,\upeta _r$ be the same as in example 4.1. Then $\upeta _1=\frac{15}{32}-\frac{1}{648}<\frac{15}{32},$ $t_{r+1}<\upeta _r<t_r,\, \upeta _r>\frac{1}{5}$ and $t_1=\frac{15}{32}<\frac{1}{2},~ t_r-t_{r+1}=\frac{1}{4(r+2)^4},~ r\in {\mathbb {N}}.$ Also, $t^*=\lim _{r\rightarrow \infty }t_r=\frac{31}{64}-\sum _{i=1}^{\infty }\frac{1}{4(i+1)^4}=\frac{47}{64}-\frac{\uppi ^4}{360}=0.46.$ Also, we have

$$\begin{aligned} \begin{aligned} \int _{\upeta _1}^{\upsigma (a)-\upeta _1}\aleph (\uptau ,\uptau )\Delta \uptau =\int _{\frac{15}{32}-\frac{1}{648}}^{1-\frac{15}{32}+\frac{1}{648}}(1-\uptau )d\uptau =0.03. \end{aligned} \end{aligned}$$

Thus, we get

$$\begin{aligned} \begin{aligned} \uptheta =\max \bigg \{\frac{1}{0.0161845}, \frac{1}{4\times 0.0161845}\bigg \}=61.79. \end{aligned} \end{aligned}$$

By Lemma 2.4, we obtain

$$\begin{aligned} \begin{aligned} \int _0^{\upsigma (1)}\uplambda _1(t)\uplambda _2(t)dt&=\int _\frac{1}{2}^1\uplambda _1(t)\uplambda _2(t)dt+\sum _{k=1}^{\infty }\left[ \upsigma \left( \frac{1}{2^k}\right) -\frac{1}{2^k}\right] \uplambda _1\left( \frac{1}{2^k}\right) \uplambda _2\left( \frac{1}{2^k}\right) \\&\approx 2.311909422 \end{aligned} \end{aligned}$$

So, let $p_i=1$ for $i=1,2.$ Then

$$\begin{aligned} \prod _{i=1}^{m}\left\| \uplambda _i\right\| _{L_\nabla ^{p_i}}\approx 2.311909422, \end{aligned}$$

and also $\Vert \aleph \Vert _{L_\nabla ^{\infty }}=1.$ Therefore,

$$\begin{aligned} {\mathfrak {N}}_1<\left[ \Vert \aleph \Vert _{\infty }\prod _{i=1}^{m}\left\| \uplambda _i\right\| _{L_\nabla ^{p_i}}\right] ^{-1}\approx 0.4325428974. \end{aligned}$$

Taking ${\mathfrak {N}}_1=\frac{1}{3}.$ In addition, if we take

$$\begin{aligned} \Gamma _r=10^{-8r}~~\text {and}~~\Lambda _r=10^{-(8r+3)}, \end{aligned}$$

then

$$\begin{aligned} \begin{aligned} \Gamma _{r+1}=10^{-(8r+8)}<\frac{1}{5}\times 10^{-(8r+3)}<\frac{\upeta _r}{a}\Lambda _r<\Lambda _r=10^{-(8r+3)}<\Gamma _r=10^{-8r}, \end{aligned} \\ \uptheta \Lambda _r=61.79\times 10^{-(8r+3)}<\frac{1}{3}\times 10^{-8r}={\mathfrak {N}}_1\Gamma _r,\,r\in {\mathbb {N}} \end{aligned}$$

and $\mathtt {g}_\ell (\ell =1,2,3)$ satisfies the following growth conditions:

$$\begin{aligned} \begin{aligned} \mathtt {g}_\ell (\mathtt {x})\le \,&{\mathfrak {N}}_1\Gamma _r=\frac{1}{3}\times 10^{-8r},~~\mathtt {x}\in \bigg [0, 10^{-8r}\bigg ],\\ \mathtt {g}_\ell (\mathtt {x})\ge \,&\uptheta \Lambda _r=61.79\times 10^{-(8r+3)},~~\mathtt {x}\in \bigg [\frac{1}{5}\times 10^{-(8r+3)}, 10^{-(8r+3)}\bigg ], \end{aligned} \end{aligned}$$

for $r\in {\mathbb {N}}.$ Then all the conditions of Theorem 3.3 are satisfied. Therefore, by Theorem 3.3, the iterative boundary value problem (1) has infinitely many solutions $\{(\mathtt {x}_1^{[r]}, \mathtt {x}_2^{[r]},\mathtt {x}_3^{[r]})\}_{r=1}^\infty$ such that $\mathtt {x}_\ell ^{[r]}(t)\ge 0$ on [0, 1],  $\ell =1,2,3$ and $r\in {\mathbb {N}}.$