Solvability of Iterative Classes of Nonlinear Elliptic Equations on an Exterior Domain

Abstract

This work explores the possibility that iterative classes of elliptic equations have both single and coupled positive radial solutions. Our approach is based on using the well-known Guo–Krasnoselskii and Avery–Henderson fixed-point theorems in a Banach space. Furthermore, we utilize Rus’ theorem in a metric space, to prove the uniqueness of solutions for the problem. Examples are constructed for the sake of verification.

1. Introduction

The study of nonlinear elliptic systems has a strong motivation, and important research efforts have been made undertaken recently for these systems, aiming to apply the results of the existence and asymptotic behavior of positive solutions in applied fields (see [1,2,3,4,5]). The investigation of the following system of nonlinear elliptic equations in a bounded domain $\mho \subset {\mathbb{R}}^{\mathrm{N}}$,

$$▵{\mathfrak{z}}_{{}_{\mathsf{\beta }}}+\mathsf{\lambda }{\mathtt{F}}_{\mathsf{\beta }}\left({\mathfrak{z}}_{{}_{\mathsf{\beta }+1}}\right)=0,$$

(1)

where ${\mathfrak{z}}_{{}_{\mathsf{\beta }}}=0$ on $\partial \mho$ and ${\mathfrak{z}}_{{}_1}={\mathfrak{z}}_{{}_{\mathfrak{d}+1}},$ $\mathsf{\beta }\in \{1,2,3,\cdots ,\mathfrak{d}\},$ has an important application in science and technology [6,7]. In [8], Dalmasso discussed the existence of positive solutions to such systems for $\mathfrak{d}=2$ when the ${\mathtt{F}}_{}{\left(0\right)}^{\prime }s$ are non-negative with at least one ${\mathtt{F}}_{}\left(0\right)>0$ (positone problems). In [7], when $\mathfrak{d}=2,$ Ali–Ramaswamy–Shivaji discussed the existence of multiple positive solutions to such positone problems. In particular, in cases where one of $\frac{\mathfrak{z}}{{\mathtt{F}}_1\left(\mathfrak{z}\right)}$ or $\frac{\mathfrak{z}}{{\mathtt{F}}_2\left(\mathfrak{z}\right)}$ decreases for some range of $\mathfrak{z},$ they established conditions for the existence of at least three positive solutions for a certain range of $\mathsf{\lambda }.$ In [9], Hai–Shivaji discussed the existence of positive solutions for $\mathsf{\lambda }>>1$ for cases where no sign conditions are assumed on ${\mathtt{F}}_{}\left(0\right),$ $\mathsf{\beta }\in \{1,2\}$ (semipositone problems). In [10], again for $\mathfrak{d}=2,$ Ali–Shivaji discussed the existence of multiple positive solutions for $\mathsf{\lambda }>>1$ when ${\mathtt{F}}_{}\left(0\right)=0={\mathtt{F}}_{}^{\prime }\left(0\right)$ for $\mathsf{\beta }\in \{1,2\}.$ In addition, in [11,12,13,14,15,16,17,18,19,20], relevant references to the most recent works on (1) can be found. Next, we quote some recent works on elliptic equations.

In [21], Padhi et al. derived sufficient conditions to the following problem in an annular domain:

$$\begin{array}{cc}\hfill ▵\mathfrak{z}=\mathsf{\lambda }\mathtt{F}\left(\right|\mathsf{\nu }|,\mathfrak{z}),& \mathfrak{z}\in \mho =\{\mathsf{\nu }\in {\mathbb{R}}^{\mathrm{N}}:a_1<|\mathsf{\nu }|<a_2\},\hfill \\ & \mathfrak{z}=0,\mathfrak{z}\in \partial \mho ,\hfill \end{array}$$

for the existence of positive radial solutions, by utilizing Gustafson and Schmitt fixed-point theorems. In [22], Chrouda and Hassine established the uniqueness of positive radial solutions to the following Dirichlet boundary value problem for the semilinear elliptic equation in an annulus:

$$\begin{array}{cc}\hfill ▵\mathfrak{z}=\mathtt{F}\left(\mathfrak{z}\right),& \mathfrak{z}\in \mho =\{\mathsf{\nu }\in {\mathbb{R}}^{\mathrm{N}}:a_1<|\mathsf{\nu }|<a_2\},\hfill \\ & \mathfrak{z}=0,\mathfrak{z}\in \mathfrak{z}\in \partial \mho ,\hfill \end{array}$$

for any dimension $\mathrm{N}\ge 1.$ In [23], Dong and Wei established the existence of radial solutions for the following nonlinear elliptic equations with gradient terms in annular domains:

$$\begin{array}{cc}& ▵\mathfrak{z}+\mathtt{g}(\left|\mathsf{\nu }\right|,\mathfrak{z},\frac{\mathsf{\nu }}{\left|\mathsf{\nu }\right|}·\nabla \mathfrak{z})=0\mathrm{in}{\Omega }_a^b,\hfill \\ & \mathfrak{z}=0\mathrm{on}\partial {\Omega }_a^b,\hfill \end{array}$$

by using Schauder’s fixed-point theorem and contraction mapping theorem. In [24], R. Kajikiya and E. Ko established the existence of positive radial solutions for a semipositone elliptic equation of the form

$$\begin{array}{cc}& ▵\mathfrak{z}+\mathsf{\lambda }\mathtt{g}\left(\mathfrak{z}\right)=0\mathrm{in}\Omega ,\hfill \\ & \mathfrak{z}=0\mathrm{on}\partial \Omega ,\hfill \end{array}$$

where $\Omega$ is a ball or an annulus in ${\mathbb{R}}^{\mathbf{N}}.$ Recently, Son and Wang [25] considered the following system in an exterior ball ${\mho }_{\mathfrak{X}}$:

$$\begin{array}{cc}& ▵{\mathfrak{z}}_{{}_{\mathsf{\beta }}}+\mathsf{\lambda }{\mathtt{K}}_{\mathsf{\beta }}\left(\right|\mathsf{\nu }\left|\right){\mathtt{F}}_{\mathsf{\beta }}\left({\mathfrak{z}}_{{}_{\mathsf{\beta }+1}}\right)=0,\hfill \\ & {\mathfrak{z}}_{{}_{\mathsf{\beta }}}\to 0\mathrm{as}\left|\mathsf{\nu }\right|\to +\infty \hfill \\ & {\mathfrak{z}}_{{}_{\mathsf{\beta }}}=0\mathrm{on}\left|\mathsf{\nu }\right|=r_0,\hfill \end{array}$$

where $\mathsf{\beta }\in \{1,2,3,\cdots ,\mathfrak{d}\},$ ${\mathfrak{z}}_{{}_1}={\mathfrak{z}}_{{}_{\mathfrak{d}+1}},$ and derived sufficient conditions for the existence of positive radial solutions. The above-mentioned works motivated us to study the following iterative classes of nonlinear elliptic equations on an exterior domain:

$$\left.\begin{array}{cc}\hfill & ▵{\mathfrak{z}}_{{}_{\mathsf{\beta }}}-\frac{{(\mathrm{N}-2)}^2{r}_0^{2\mathrm{N}-2}}{{\left|\mathsf{\nu }\right|}^{2\mathrm{N}-2}}{\mathfrak{z}}_{{}_{\mathsf{\beta }}}+\varrho \left(\right|\mathsf{\nu }\left|\right){\mathtt{F}}_{\mathsf{\beta }}\left({\mathfrak{z}}_{{}_{\mathsf{\beta }+1}}\right)=0,\mathsf{\nu }\in \mho ,\hfill \\ \hfill & \underset{\left|\mathsf{\nu }\right|\to \infty }{lim}{\mathfrak{z}}_{{}_{\mathsf{\beta }}}\left(\mathsf{\nu }\right)=0,{\mathfrak{z}}_{{}_{\mathsf{\beta }}}{|}_{\partial \mho }=0,\hfill \end{array}\right\}$$

(2)

where $\mathsf{\beta }\in \{1,2,3,\cdots ,n\},$ ${\mathfrak{z}}_{{}_1}={\mathfrak{z}}_{n+1},$ $\Delta \mathfrak{z}=\mathtt{div}(\nabla \mathfrak{z}),$ $\mathrm{N}>2,$ $\mho =\{\mathfrak{z}\in {\mathbb{R}}^{\mathrm{N}}|\left|\mathfrak{z}\right|>r_0\},$ $\varrho ={\prod }_{i=1}^{\kappa }{\varrho }_i,$ each ${\varrho }_i\in \mathtt{C}((r_0,+\infty ),(0,+\infty )),$ $r^{\mathrm{N}-1}\varrho$ is integrable. The Guo–Krasnoselskii cone fixed-point theorem is a key tool for obtaining single positive radial solutions, whereas the Avery–Henderson cone fixed-point theorem is utilized to obtain the coupled solutions. We further study the uniqueness of solutions of the problem (2) via Rus’ theorem in a metric space.

The study of the positive solutions to the iterative classes of ordinary differential equations with two-point boundary conditions,

$$\left.\begin{array}{cc}\hfill & {\mathfrak{z}}_{{}_{\mathsf{\beta }}}^{″}\left(\widehat{\mathbf{r}}\right)-r_0^2{\mathfrak{z}}_{{}_{\mathsf{\beta }}}\left(\widehat{\mathbf{r}}\right)+\varrho \left(\widehat{\mathbf{r}}\right){\mathtt{F}}_{\mathsf{\beta }}\left({\mathfrak{z}}_{{}_{\mathsf{\beta }+1}}\left(\widehat{\mathbf{r}}\right)\right)=0,0<\widehat{\mathbf{r}}<1,\hfill \\ \hfill & {\mathfrak{z}}_{{}_{\mathsf{\beta }}}\left(0\right)=0,{\mathfrak{z}}_{{}_{\mathsf{\beta }}}\left(1\right)=0,\hfill \end{array}\right\}$$

(3)

where $\mathsf{\beta }\in \{1,2,3,\cdots ,\mathfrak{d}\},$ ${\mathfrak{z}}_{{}_1}={\mathfrak{z}}_{{}_{\mathfrak{d}+1}},$ $r_0>0$ and $\varrho \left(\widehat{\mathbf{r}}\right)=\frac{r_0^2}{{(\mathrm{N}-2)}^2}{\widehat{\mathbf{r}}}^{\frac{2(\mathrm{N}-1)}{2-\mathrm{N}}}{\prod }_{i=1}^{\kappa }{\varrho }_i\left(\widehat{\mathbf{r}}\right),$ ${\varrho }_i\left(\widehat{\mathbf{r}}\right)={\varrho }_i\left(r_0{\widehat{\mathbf{r}}}^{\frac{1}{2-\mathrm{N}}}\right)$ by a Kelvin-type transformation [26,27] through the change of variables $\mathfrak{m}=\left|\mathsf{\nu }\right|$ and $\widehat{\mathbf{r}}={\left(\frac{\mathfrak{m}}{r_0}\right)}^{2-\mathrm{N}},$ facilitates the investigation of the positive radial solutions of (2).

We impose the below-mentioned presumptions whenever necessary:

$\left({\mathcal{J}}_1\right)$

${\mathtt{F}}_{\mathsf{\beta }}:[0,+\infty )\to [0,+\infty )$ is continuous.

$\left({\mathcal{J}}_2\right)$

For $1\le \dot{\iota }\le \mathfrak{d},$ ${\varrho }_{\dot{\iota }}\in {\mathrm{L}}^{{\mathtt{p}}_{\dot{\iota }}}[0,1](1\le {\mathtt{p}}_{\dot{\iota }}\le +\infty )$ and $\exists {\varrho }_{\dot{\iota }}^{★}>0\ni {\varrho }_{\dot{\iota }}^{★}<{\varrho }_{\dot{\iota }}\left(\widehat{\mathbf{r}}\right)<\infty$ almost everywhere on the interval $[0,1].$

The remainder of the paper is structured as follows: The problem (3) is transformed into an analogous integral equation involving the kernel in Section 2. Additionally, we calculate the kernel boundaries that are crucial to our major findings. In Section 3, we employ Guo–Krasnoselskii’s cone fixed-point theorem, to provide a criterion for the single positive radial solution. In Section 4, the coupled solutions are established by the Avery–Henderson cone fixed-point theorem. The final portion deals with a unique solution. Meanwhile, some numerical examples are provided.

2. Preliminaries

The essential results are stated here, prior to proceeding to the main results in the subsequent sections.

Lemma 1. 

For every $\wp \in \mathtt{C}[0,1],$ the BVP

$$-{\mathfrak{z}}_{{}_1}^{″}\left(\widehat{\mathbf{r}}\right)+r_0^2{\mathfrak{z}}_{{}_1}\left(\widehat{\mathbf{r}}\right)=\wp \left(\widehat{\mathbf{r}}\right),0<\widehat{\mathbf{r}}<1,$$

$${\mathfrak{z}}_{{}_1}\left(0\right)={\mathfrak{z}}_{{}_1}\left(1\right)=0,$$

has a unique solution

$${\mathfrak{z}}_{{}_1}\left(\widehat{\mathbf{r}}\right)={\int }_0^1\mathtt{Q}(\widehat{\mathbf{r}},\mathsf{\zeta })\wp \left(\mathsf{\zeta }\right)\mathtt{d}\mathsf{\zeta },$$

where

$$\begin{array}{c}\hfill \mathtt{Q}(\widehat{\mathbf{r}},\mathsf{\zeta })=\frac{1}{r_{0}sinh\left(r_0\right)}\left\{\begin{array}{c}sinh\left(r_0\widehat{\mathbf{r}}\right)sinh\left(r_0(1-\mathsf{\zeta })\right),0\le \widehat{\mathbf{r}}\le \mathsf{\zeta }\le 1,\hfill \\ sinh\left(r_0\mathsf{\zeta }\right)sinh\left(r_0(1-\widehat{\mathbf{r}})\right),0\le \mathsf{\zeta }\le \widehat{\mathbf{r}}\le 1.\hfill \end{array}\right.\end{array}$$

Lemma 2. 

The kernel $\mathtt{Q}(\widehat{\mathbf{r}},\mathsf{\zeta })$ has the subsequent characteristics:

(i) 

$\mathtt{Q}(\widehat{\mathbf{r}},\mathsf{\zeta })\ge 0$ and continuous on $[0,1]\times [0,1];$

(ii) 

$\mathtt{Q}(\widehat{\mathbf{r}},\mathsf{\zeta })\le \mathtt{Q}(\mathsf{\zeta },\mathsf{\zeta }),$ $\widehat{\mathbf{r}},\mathsf{\zeta }\in [0,1];$

(iii) 

there exists $\mathsf{\xi }\in (0,\frac{1}{2})$ such that $\mathsf{\sigma }\left(\mathsf{\xi }\right)\mathtt{Q}(\mathsf{\zeta },\mathsf{\zeta })\le \mathtt{Q}(\widehat{\mathbf{r}},\mathsf{\zeta }),$ $(\widehat{\mathbf{r}},\mathsf{\zeta })\in [\mathsf{\xi },1-\mathsf{\xi }]\times [0,1],$ where $\mathsf{\sigma }\left(\mathsf{\xi }\right)=\frac{sinh\left(r_0\mathsf{\xi }\right)}{sinh\left(r_0\right)}.$

Proof. 

(i) is evident. The following proves (ii):

$$\begin{array}{cc}\hfill \frac{\mathtt{Q}(\widehat{\mathbf{r}},\mathsf{\zeta })}{\mathtt{Q}(\mathsf{\zeta },\mathsf{\zeta })}=& \left\{\begin{array}{c}{\displaystyle \frac{sinh\left(r_0\widehat{\mathbf{r}}\right)}{sinh\left(r_0\mathsf{\zeta }\right)},0\le \widehat{\mathbf{r}}\le \mathsf{\zeta }\le 1,}\hfill \\ {\displaystyle \frac{sinh\left(r_0(1-\widehat{\mathbf{r}})\right)}{sinh\left(r_0(1-\mathsf{\zeta })\right)},0\le \mathsf{\zeta }\le \widehat{\mathbf{r}}\le 1,}\hfill \end{array}\right.\hfill \\ \hfill \le & \left\{\begin{array}{c}{\displaystyle 1,0\le \widehat{\mathbf{r}}\le \mathsf{\zeta }\le 1,}\hfill \\ 1,0\le \mathsf{\zeta }\le \widehat{\mathbf{r}}\le 1,\hfill \end{array}\right.\hfill \end{array}$$

For (iii), we consider

$$\begin{array}{cc}\hfill \frac{\mathtt{Q}(\widehat{\mathbf{r}},\mathsf{\zeta })}{\mathtt{Q}(\mathsf{\zeta },\mathsf{\zeta })}=& \left\{\begin{array}{c}{\displaystyle \frac{sinh\left(r_0\widehat{\mathbf{r}}\right)}{sinh\left(r_0\mathsf{\zeta }\right)},0\le \widehat{\mathbf{r}}\le \mathsf{\zeta }\le 1,}\hfill \\ {\displaystyle \frac{sinh\left(r_0(1-\widehat{\mathbf{r}})\right)}{sinh\left(r_0(1-\mathsf{\zeta })\right)},0\le \mathsf{\zeta }\le \widehat{\mathbf{r}}\le 1,}\hfill \end{array}\right.\hfill \\ \hfill \ge & \left\{\begin{array}{c}{\displaystyle \frac{sinh\left(r_0\mathsf{\xi }\right)}{sinh\left(r_0\right)},0\le \widehat{\mathbf{r}}\le \mathsf{\zeta }\le 1,\mathsf{\xi }\le \widehat{\mathbf{r}}\le 1-\mathsf{\xi },}\hfill \\ {\displaystyle \frac{sinh\left(r_0\mathsf{\xi }\right)}{sinh\left(r_0\right)},0\le \mathsf{\zeta }\le \widehat{\mathbf{r}}\le 1,\mathsf{\xi }\le \widehat{\mathbf{r}}\le 1-\mathsf{\xi },}\hfill \end{array}\right.\hfill \\ \hfill =& \mathsf{\sigma }.\hfill \end{array}$$

The proof is now completed. □

We observe that a $\mathfrak{d}$-tuple $({\mathfrak{z}}_{{}_1},{\mathfrak{z}}_2,\cdots ,{\mathfrak{z}}_{\mathfrak{d}})$ is a solution of BVP (3) from Lemma 1 if and only if

$$\begin{array}{cc}\hfill {\mathfrak{z}}_{{}_1}\left(\widehat{\mathbf{r}}\right)=& {\int }_0^1\mathtt{Q}(\widehat{\mathbf{r}},{\mathsf{\zeta }}_1)\varrho \left({\mathsf{\zeta }}_1\right){\mathtt{F}}_1\left[{\int }_0^1\mathtt{Q}({\mathsf{\zeta }}_1,{\mathsf{\zeta }}_2)\varrho \left({\mathsf{\zeta }}_2\right){\mathtt{F}}_2\right[{\int }_0^1\mathtt{Q}({\mathsf{\zeta }}_2,{\mathsf{\zeta }}_3)\varrho \left({\mathsf{\zeta }}_3\right){\mathtt{F}}_4\cdots \hfill \\ & {\mathtt{F}}_{\mathfrak{d}-1}\left[{\int }_0^1\mathtt{Q}({\mathsf{\zeta }}_{\mathfrak{d}-1},{\mathsf{\zeta }}_{\mathfrak{d}})\varrho \left({\mathsf{\zeta }}_{\mathfrak{d}}\right){\mathtt{F}}_{\mathfrak{d}}\left({\mathfrak{z}}_{{}_1}\left({\mathsf{\zeta }}_{\mathfrak{d}}\right)\right)\mathtt{d}{\mathsf{\zeta }}_{\mathfrak{d}}\right]\cdots \left]\mathtt{d}{\mathsf{\zeta }}_3\right]\mathtt{d}{\mathsf{\zeta }}_2]\mathtt{d}{\mathsf{\zeta }}_1.\hfill \end{array}$$

In general,

$$\begin{array}{cc}\hfill {\mathfrak{z}}_{{}_{\mathsf{\beta }}}\left(\widehat{\mathbf{r}}\right)=& {\int }_0^1\mathtt{Q}(\widehat{\mathbf{r}},\mathsf{\zeta })\varrho \left(\mathsf{\zeta }\right){\mathtt{F}}_{\mathsf{\beta }}\left({\mathfrak{z}}_{{}_{\mathsf{\beta }+1}}\left(\mathsf{\zeta }\right)\right)\mathtt{d}\mathsf{\zeta },\mathsf{\beta }=1,2,3,\cdots ,\mathfrak{d},\hfill \\ \hfill {\mathfrak{z}}_{{}_1}\left(\widehat{\mathbf{r}}\right)=& {\mathfrak{z}}_{{}_{\mathfrak{d}+1}}\left(\widehat{\mathbf{r}}\right).\hfill \end{array}$$

Let $\aleph :=\mathtt{C}\left(\right(0,1),\mathbb{R})$ be a Banach space equipped with a norm ${\displaystyle \parallel \mathfrak{z}\parallel =\underset{\widehat{\mathbf{r}}\in [0,1]}{max}\left|\mathfrak{z}\left(\widehat{\mathbf{r}}\right)\right|}$, and

$${\mathfrak{X}}_{\mathsf{\xi }}=\left\{\mathfrak{z}\in \aleph :\mathfrak{z}\left(\widehat{\mathbf{r}}\right)\ge 0\mathrm{on}[0,1],\underset{\widehat{\mathbf{r}}\in [\mathsf{\xi },1-\mathsf{\xi }]}{min}\mathfrak{z}\left(\widehat{\mathbf{r}}\right)\ge \mathsf{\sigma }\left(\mathsf{\xi }\right)\parallel \mathfrak{z}\parallel \right\}$$

be a cone, for $\mathsf{\xi }\in (0,\frac{1}{2}).$ For any ${\mathfrak{z}}_{{}_1}\in \mathfrak{X},$ define an operator $\pounds :\mathfrak{X}\to \aleph$ by

$$\begin{array}{cc}\hfill \left(\pounds {\mathfrak{z}}_{{}_1}\right)\left(\widehat{\mathbf{r}}\right)=& {\int }_0^1\mathtt{Q}(\widehat{\mathbf{r}},{\mathsf{\zeta }}_1)\varrho \left({\mathsf{\zeta }}_1\right){\mathtt{F}}_1\left[{\int }_0^1\mathtt{Q}({\mathsf{\zeta }}_1,{\mathsf{\zeta }}_2)\varrho \left({\mathsf{\zeta }}_2\right){\mathtt{F}}_2\right[{\int }_0^1\mathtt{Q}({\mathsf{\zeta }}_2,{\mathsf{\zeta }}_3)\varrho \left({\mathsf{\zeta }}_3\right)\cdots \hfill \\ \hfill & {\mathtt{F}}_{\mathfrak{d}-1}\left[{\int }_0^1\mathtt{Q}({\mathsf{\zeta }}_{\mathfrak{d}-1},{\mathsf{\zeta }}_{\mathfrak{d}})\varrho \left({\mathsf{\zeta }}_{\mathfrak{d}}\right){\mathtt{F}}_{\mathfrak{d}}\left({\mathfrak{z}}_{{}_1}\left({\mathsf{\zeta }}_{\mathfrak{d}}\right)\right)\mathtt{d}{\mathsf{\zeta }}_{\mathfrak{d}}\right]\cdots \left]\mathtt{d}{\mathsf{\zeta }}_3\right]\mathtt{d}{\mathsf{\zeta }}_2]\mathtt{d}{\mathsf{\zeta }}_1.\hfill \end{array}$$

(4)

Lemma 3. 

£ is self–mapping on ${\mathfrak{X}}_{\mathsf{\xi }}$ and $\pounds :{\mathfrak{X}}_{\mathsf{\xi }}\to {\mathfrak{X}}_{\mathsf{\xi }}$ is completely continuous.

Proof. 

As ${\mathtt{F}}_{\mathsf{\beta }}\left({\mathfrak{z}}_{{}_{\mathsf{\beta }+1}}\left(\widehat{\mathbf{r}}\right)\right)\ge 0$ and $\mathtt{Q}(\widehat{\mathbf{r}},\mathsf{\zeta })\ge 0$ for $\widehat{\mathbf{r}},\mathsf{\zeta }\in [0,1],$ we have $\pounds \left({\mathfrak{z}}_{{}_1}\left(\widehat{\mathbf{r}}\right)\right)\ge 0$ for $\widehat{\mathbf{r}}\in [0,1],{\mathfrak{z}}_{{}_1}\in {\mathfrak{X}}_{\mathsf{\xi }}.$ Applying Lemmas 1 and 2, we obtain

$$\begin{array}{cc}\hfill \underset{\widehat{\mathbf{r}}\in [\mathsf{\xi },1-\mathsf{\xi }]}{min}\left(\pounds {\mathfrak{z}}_{{}_1}\right)\left(\widehat{\mathbf{r}}\right)& =\underset{\widehat{\mathbf{r}}\in [\mathsf{\xi },1-\mathsf{\xi }]}{min}\left\{{\int }_0^1\mathtt{Q}(\widehat{\mathbf{r}},{\mathsf{\zeta }}_1)\varrho \left({\mathsf{\zeta }}_1\right){\mathtt{F}}_1\right[{\int }_0^1\mathtt{Q}({\mathsf{\zeta }}_1,{\mathsf{\zeta }}_2)\varrho \left({\mathsf{\zeta }}_2\right){\mathtt{F}}_2[{\int }_0^1\mathtt{Q}({\mathsf{\zeta }}_2,{\mathsf{\zeta }}_3)\varrho \left({\mathsf{\zeta }}_3\right){\mathtt{F}}_4\cdots \hfill \\ & {\mathtt{F}}_{\mathfrak{d}-1}\left[{\int }_0^1\mathtt{Q}({\mathsf{\zeta }}_{\mathfrak{d}-1},{\mathsf{\zeta }}_{\mathfrak{d}})\varrho \left({\mathsf{\zeta }}_{\mathfrak{d}}\right){\mathtt{F}}_{\mathfrak{d}}\left({\mathfrak{z}}_{{}_1}\left({\mathsf{\zeta }}_{\mathfrak{d}}\right)\right)\mathtt{d}{\mathsf{\zeta }}_{\mathfrak{d}}\right]\cdots \left]\mathtt{d}{\mathsf{\zeta }}_3\right]\mathtt{d}{\mathsf{\zeta }}_2\left]\mathtt{d}{\mathsf{\zeta }}_1\right\}\hfill \\ & \ge \mathsf{\sigma }\left(\mathsf{\xi }\right)\left\{{\int }_0^1\mathtt{Q}({\mathsf{\zeta }}_1,{\mathsf{\zeta }}_1)\varrho \left({\mathsf{\zeta }}_1\right){\mathtt{F}}_1\right[{\int }_0^1\mathtt{Q}({\mathsf{\zeta }}_1,{\mathsf{\zeta }}_2)\varrho \left({\mathsf{\zeta }}_2\right){\mathtt{F}}_2[{\int }_0^1\mathtt{Q}({\mathsf{\zeta }}_2,{\mathsf{\zeta }}_3)\varrho \left({\mathsf{\zeta }}_3\right){\mathtt{F}}_4\cdots \hfill \\ & {\mathtt{F}}_{\mathfrak{d}-1}\left[{\int }_0^1\mathtt{Q}({\mathsf{\zeta }}_{\mathfrak{d}-1},{\mathsf{\zeta }}_{\mathfrak{d}})\varrho \left({\mathsf{\zeta }}_{\mathfrak{d}}\right){\mathtt{F}}_{\mathfrak{d}}\left({\mathfrak{z}}_{{}_1}\left({\mathsf{\zeta }}_{\mathfrak{d}}\right)\right)\mathtt{d}{\mathsf{\zeta }}_{\mathfrak{d}}\right]\cdots \left]\mathtt{d}{\mathsf{\zeta }}_3\right]\mathtt{d}{\mathsf{\zeta }}_2\left]\mathtt{d}{\mathsf{\zeta }}_1\right\}\hfill \\ & \ge \mathsf{\sigma }\left(\mathsf{\xi }\right)\left\{{\int }_0^1\mathtt{Q}(\widehat{\mathbf{r}},{\mathsf{\zeta }}_1)\varrho \left({\mathsf{\zeta }}_1\right){\mathtt{F}}_1\right[{\int }_0^1\mathtt{Q}({\mathsf{\zeta }}_1,{\mathsf{\zeta }}_2)\varrho \left({\mathsf{\zeta }}_2\right){\mathtt{F}}_2[{\int }_0^1\mathtt{Q}({\mathsf{\zeta }}_2,{\mathsf{\zeta }}_3)\varrho \left({\mathsf{\zeta }}_3\right){\mathtt{F}}_4\cdots \hfill \\ & {\mathtt{F}}_{\mathfrak{d}-1}\left[{\int }_0^1\mathtt{Q}({\mathsf{\zeta }}_{\mathfrak{d}-1},{\mathsf{\zeta }}_{\mathfrak{d}})\varrho \left({\mathsf{\zeta }}_{\mathfrak{d}}\right){\mathtt{F}}_{\mathfrak{d}}\left({\mathfrak{z}}_{{}_1}\left({\mathsf{\zeta }}_{\mathfrak{d}}\right)\right)\mathtt{d}{\mathsf{\zeta }}_{\mathfrak{d}}\right]\cdots \left]\mathtt{d}{\mathsf{\zeta }}_3\right]\mathtt{d}{\mathsf{\zeta }}_2\left]\mathtt{d}{\mathsf{\zeta }}_1\right\}\hfill \\ & \ge \mathsf{\sigma }\left(\mathsf{\xi }\right)\underset{\widehat{\mathbf{r}}\in [0,1]}{max}\left|\pounds {\mathfrak{z}}_{{}_1}\left(\widehat{\mathbf{r}}\right)\right|.\hfill \end{array}$$

Thus, $\pounds \left({\mathfrak{X}}_{\mathsf{\xi }}\right)\subset {\mathfrak{X}}_{\mathsf{\xi }}.$ In light of this, the operator £ is fully continuous according to the Arzela–Ascoli theorem. □

The following theorems are key tools for the existence of positive solutions:

Theorem 1 (Hölder’s [28]). 

For $\ell =1,2,\cdots ,\kappa ,$ and ${\mathtt{p}}_{\ell }>1,$ let $\hslash \in {\mathrm{L}}^{{\mathtt{p}}_{\ell }}[0,1]$ with ${\sum }_{\ell =1}^{\kappa }\frac{1}{{\mathtt{p}}_{\ell }}=1;$ then, ${\prod }_{\ell =1}^{\kappa }{\hslash }_{\ell }\in {\mathrm{L}}^1[0,1]$ and ${∥{\prod }_{\ell =1}^{\kappa }{\hslash }_{\ell }∥}_1\le {\prod }_{\ell =1}^{\kappa }{\parallel {\hslash }_{\ell }\parallel }_{{\mathtt{p}}_{\ell }}.$ Furthermore, if $\hslash \in {\mathrm{L}}^1[0,1]$ and $\overline{g}\in {\mathrm{L}}^{\infty }[0,1]$ then $\hslash \overline{g}\in {\mathrm{L}}^1[0,1]$ and $\parallel \hslash \overline{g}{\parallel }_1\le {\parallel \hslash \parallel }_1{\parallel \overline{g}\parallel }_{\infty }.$

Theorem 2 (Guo–Krasnoselskii [29]). 

Let $\mathfrak{G}$ be a Banach space, and let ${\mathfrak{N}}_1,{\mathfrak{N}}_2$ be bounded open subsets of $\mathfrak{G}$ with $0\in {\mathfrak{N}}_1\subset {\overline{\mathfrak{N}}}_1\subset {\mathfrak{N}}_2$ and $\aleph :\mathfrak{X}\cap ({\overline{\mathfrak{N}}}_2\backslash {\mathfrak{N}}_1)\to \mathfrak{X}$ $(\mathfrak{X}\subset \mathfrak{G}$ is a cone) as a completely continuous operator, such that

(i) 

$\parallel \aleph \mathfrak{z}\parallel \le \parallel \mathfrak{z}\parallel ,\mathfrak{z}\in \mathfrak{X}\cap \partial {\mathfrak{N}}_1,$ and $\parallel \aleph \mathfrak{z}\parallel \ge \parallel \mathfrak{z}\parallel ,\mathfrak{z}\in \mathfrak{X}\cap \partial {\mathfrak{N}}_2,$ or

(ii) 

$\parallel \aleph \mathfrak{z}\parallel \ge \parallel \mathfrak{z}\parallel ,\mathfrak{z}\in \mathfrak{X}\cap \partial {\mathfrak{N}}_1,$ and $\parallel \aleph \mathfrak{z}\parallel \le \parallel \mathfrak{z}\parallel ,\mathfrak{z}\in \mathfrak{X}\cap \partial {\mathfrak{N}}_2;$

then, ℵ has a fixed point in $\mathfrak{X}\cap ({\overline{\mathfrak{N}}}_2\backslash {\mathfrak{N}}_1).$

Let $\mathsf{\psi }\ge 0$ be a continuous functional on a cone $\mathfrak{X}$, and let $\mathfrak{f}>0$ and $\mathfrak{h}>0.$ Define $\mathfrak{X}(\mathsf{\psi },\mathfrak{h})=\{\mathfrak{z}\in \mathfrak{X}:\mathsf{\psi }(\mathfrak{z})<\mathfrak{h}\}$ and ${\mathfrak{X}}_{\mathfrak{f}}=\{\mathfrak{z}\in \mathfrak{X}:\parallel \mathfrak{z}\parallel <\mathfrak{f}\}.$

Theorem 3 (Avery–Henderson [30]). 

If ${\mathsf{\gamma }}_1\ge 0,$ ${\mathsf{\gamma }}_2\ge 0,$ ${\mathsf{\gamma }}_3\ge 0$ continuous and increasing functionals on $\mathfrak{X},$ ${\mathsf{\gamma }}_3\left(0\right)=0$, such that, for some positive numbers $\mathfrak{h}$ and $k,$ ${\mathsf{\gamma }}_2\left(\mathfrak{z}\right)\le {\mathsf{\gamma }}_3\left(\mathfrak{z}\right)\le {\mathsf{\gamma }}_1\left(\mathfrak{z}\right)$ and $\parallel \mathfrak{z}\parallel \le k{\mathsf{\gamma }}_2\left(\mathfrak{z}\right),$ for all $\mathfrak{z}\in \overline{\mathfrak{X}({\mathsf{\gamma }}_2,\mathfrak{h})},$ and there exist $\mathfrak{f}>0$ and $\mathfrak{g}>0$ with $\mathfrak{f}<\mathfrak{g}<\mathfrak{h}$, such that ${\mathsf{\gamma }}_3\left(\mathsf{\lambda }\mathfrak{z}\right)\le \mathsf{\lambda }{\mathsf{\gamma }}_3\left(\mathfrak{z}\right),$ for $0\le \mathsf{\lambda }\le 1$ and $\mathfrak{z}\in \partial \mathfrak{X}({\mathsf{\gamma }}_3,\mathfrak{g}).$ Furthermore, if $\pounds :\overline{\mathfrak{X}({\mathsf{\gamma }}_2,\mathfrak{h})}\to \mathfrak{X}$ is a completely continuous operator, such that

$\left(a\right)$

${\mathsf{\gamma }}_2\left(\pounds \mathfrak{z}\right)>\mathfrak{h},$ for all $\mathfrak{z}\in \partial \mathfrak{X}({\mathsf{\gamma }}_2,\mathfrak{h}),$

$\left(b\right)$

${\mathsf{\gamma }}_3\left(\pounds \mathfrak{z}\right)<\mathfrak{g},$ for all $\mathfrak{z}\in \partial \mathfrak{X}({\mathsf{\gamma }}_3,\mathfrak{g}),$

$\left(c\right)$

$\mathfrak{X}({\mathsf{\gamma }}_1,\mathfrak{f})\ne Ø$ and ${\mathsf{\gamma }}_1\left(\pounds \mathfrak{z}\right)>\mathfrak{f},$ for all $\partial \mathfrak{X}({\mathsf{\gamma }}_1,\mathfrak{f}),$

then £ has at least two fixed points ${}^1\mathfrak{z},{}^2\mathfrak{z}\in \mathtt{P}({\mathsf{\gamma }}_2,\mathfrak{h})$, such that $\mathfrak{f}<{\mathsf{\gamma }}_1\left({}^1\mathfrak{z}\right)$ with ${\mathsf{\gamma }}_3\left({}^1\mathfrak{z}\right)<\mathfrak{g}$ and $\mathfrak{g}<{\mathsf{\gamma }}_3\left({}^2\mathfrak{z}\right)$ with ${\mathsf{\gamma }}_2\left({}^2\mathfrak{z}\right)<\mathfrak{h}.$

Define the non-negative, increasing, continuous functional ${\mathsf{\gamma }}_2,{\mathsf{\gamma }}_3,$ and ${\mathsf{\gamma }}_1$ by

$$\begin{array}{c}\hfill {\mathsf{\gamma }}_2\left(\mathfrak{z}\right)=\underset{\widehat{\mathbf{r}}\in [\mathsf{\xi },1-\mathsf{\xi }]}{min}\mathfrak{z}\left(\widehat{\mathbf{r}}\right),{\mathsf{\gamma }}_3\left(\mathfrak{z}\right)=\underset{\widehat{\mathbf{r}}\in [0,1]}{max}\mathfrak{z}\left(\widehat{\mathbf{r}}\right),{\mathsf{\gamma }}_1\left(\mathfrak{z}\right)=\underset{\widehat{\mathbf{r}}\in [0,1]}{max}\mathfrak{z}\left(\widehat{\mathbf{r}}\right).\end{array}$$

It is obvious that for each $\mathfrak{z}\in \mathfrak{X},$ ${\mathsf{\gamma }}_2\left(\mathfrak{z}\right)\le {\mathsf{\gamma }}_3\left(\mathfrak{z}\right)={\mathsf{\gamma }}_1\left(\mathfrak{z}\right),$ and ${\mathsf{\gamma }}_2\left(\mathfrak{z}\right)\ge \mathsf{\sigma }\left(\mathsf{\xi }\right)\parallel \mathfrak{z}\parallel .$ Thus, $\parallel \mathfrak{z}\parallel \le \frac{1}{\mathsf{\sigma }\left(\mathsf{\xi }\right)}{\mathsf{\gamma }}_2\left(\mathfrak{z}\right)$ for all $\mathfrak{z}\in \mathfrak{X}.$ Furthermore, we observe that ${\mathsf{\gamma }}_3\left(\mathsf{\lambda }\mathfrak{z}\right)=\mathsf{\lambda }{\mathsf{\gamma }}_3\left(\mathfrak{z}\right),$ for $0\le \mathsf{\lambda }\le 1,$ $\mathfrak{z}\in \mathfrak{X}.$

3. Single Positive Radial Solution

In accordance with Guo–Krasnoselskii’s theorem, we demonstrate in this section that problem (3) has a single positive radial solution.

For ${\varrho }_i\in {\mathrm{L}}^{{\mathtt{p}}_i}[0,1],$ we have the following cases:

$$\sum _{i=1}^{\kappa }\frac{1}{{\mathtt{p}}_i}<1,\sum _{i=1}^{\kappa }\frac{1}{{\mathtt{p}}_i}=1,\sum _{i=1}^{\kappa }\frac{1}{{\mathtt{p}}_i}>1.$$

We discuss the positive radial solutions for ${\displaystyle \sum _{i=1}^{\kappa }\frac{1}{{\mathtt{p}}_i}<1,}$ in the following theorem:

Theorem 4. 

Suppose that $\left({\mathcal{J}}_1\right)$–$\left({\mathcal{J}}_2\right)$ hold, and there exist positive constants $a_2>a_1>0$, such that

$\left({\mathcal{J}}_3\right)$

${\mathtt{F}}_{\mathsf{\beta }}\left(\mathfrak{z}\left(\widehat{\mathbf{r}}\right)\right)\le {\mathfrak{R}}_2{a}_2$ for $0\le \widehat{\mathbf{r}}\le 1,0\le \mathfrak{z}\le a_2,$ where ${\displaystyle {\mathfrak{R}}_2={\left[\frac{r_0^2}{{(\mathrm{N}-2)}^2}\parallel \widehat{\mathtt{Q}}{\parallel }_{\mathtt{q}}\prod _{i=1}^{\kappa }{\parallel {\varrho }_i\parallel }_{{\mathtt{p}}_i}\right]}^{-1}}$ and $\widehat{\mathtt{Q}}\left(\mathsf{\zeta }\right)=\mathtt{Q}(\mathsf{\zeta },\mathsf{\zeta }){\mathsf{\zeta }}^{\frac{2(\mathrm{N}-1)}{2-\mathrm{N}}},$

$\left({\mathcal{J}}_4\right)$

${\mathtt{F}}_{\mathsf{\beta }}\left(\mathfrak{z}\left(\widehat{\mathbf{r}}\right)\right)\ge {\mathfrak{R}}_1{a}_1$ for $\mathsf{\xi }\le \widehat{\mathbf{r}}\le 1-\mathsf{\xi },\mathsf{\sigma }\left(\mathsf{\xi }\right)a_1\le \mathfrak{z}\le a_1,$ where

$${\displaystyle {\mathfrak{R}}_1={\left[\frac{\mathsf{\sigma }\left(\mathsf{\xi }\right)r_0^2}{{(\mathrm{N}-2)}^2}\prod _{i=1}^{\kappa }{\varrho }_i^{★}{\int }_{\mathsf{\xi }}^{1-\mathsf{\xi }}\mathtt{Q}(\mathsf{\zeta },\mathsf{\zeta }){\mathsf{\zeta }}^{\frac{2(\mathrm{N}-1)}{2-\mathrm{N}}}\mathtt{d}\mathsf{\zeta }\right]}^{-1},}$$

then the BVP (3) has a solution $({\mathfrak{z}}_{{}_1},{\mathfrak{z}}_2,\cdots ,{\mathfrak{z}}_{\mathfrak{d}})$, such that ${\mathfrak{z}}_{{}_{\mathsf{\beta }}}>0,a_1\le \parallel {\mathfrak{z}}_{{}_{\mathsf{\beta }}}\parallel \le a_2,\mathsf{\beta }=1,2,\cdots ,\mathfrak{d}.$

Proof. 

Let ${\mathfrak{N}}_1=\{\mathfrak{z}\in \aleph :\parallel \mathfrak{z}\parallel <a_1\}$ and ${\mathfrak{N}}_2=\{\mathfrak{z}\in \aleph :\parallel \mathfrak{z}\parallel <a_2\}.$ For ${\mathfrak{z}}_{{}_1}\in \partial {\mathfrak{N}}_2,$ $0\le {\mathfrak{z}}_1\le a_2$ for $\widehat{\mathbf{r}}\in [0,1].$ For ${\mathsf{\zeta }}_{\mathfrak{d}-1}\in [0,1],$ and from $\left({\mathcal{J}}_3\right),$ we obtain

$$\begin{array}{cc}\hfill {\int }_0^1\mathtt{Q}({\mathsf{\zeta }}_{\mathfrak{d}-1},{\mathsf{\zeta }}_{\mathfrak{d}})\varrho \left({\mathsf{\zeta }}_{\mathfrak{d}}\right){\mathtt{F}}_{\mathfrak{d}}\left({\mathfrak{z}}_{{}_1}\left({\mathsf{\zeta }}_{\mathfrak{d}}\right)\right)\mathtt{d}{\mathsf{\zeta }}_{\mathfrak{d}}& \le {\int }_0^1\mathtt{Q}({\mathsf{\zeta }}_{\mathfrak{d}},{\mathsf{\zeta }}_{\mathfrak{d}})\varrho \left({\mathsf{\zeta }}_{\mathfrak{d}}\right){\mathtt{F}}_{\mathfrak{d}}\left({\mathfrak{z}}_{{}_1}\left({\mathsf{\zeta }}_{\mathfrak{d}}\right)\right)\mathtt{d}{\mathsf{\zeta }}_{\mathfrak{d}}\hfill \\ & \le {\mathfrak{R}}_2{a}_2{\int }_0^1\mathtt{Q}({\mathsf{\zeta }}_{\mathfrak{d}},{\mathsf{\zeta }}_{\mathfrak{d}})\varrho \left({\mathsf{\zeta }}_{\mathfrak{d}}\right)\mathtt{d}{\mathsf{\zeta }}_{\mathfrak{d}}\hfill \\ & \le {\mathfrak{R}}_2{a}_2\frac{r_0^2}{{(\mathrm{N}-2)}^2}{\int }_0^1\mathtt{Q}({\mathsf{\zeta }}_{\mathfrak{d}},{\mathsf{\zeta }}_{\mathfrak{d}}){\mathsf{\zeta }}_{\mathfrak{d}}^{\frac{2(\mathrm{N}-1)}{2-\mathrm{N}}}\prod _{i=1}^{\kappa }{\varrho }_i\left({\mathsf{\zeta }}_{\mathfrak{d}}\right)\mathtt{d}{\mathsf{\zeta }}_{\mathfrak{d}}.\hfill \end{array}$$

Now, there exists $\mathtt{q}>1$, such that ${\displaystyle \sum _{i=1}^{\kappa }\frac{1}{{\mathtt{p}}_i}+\frac{1}{\mathtt{q}}=1.}$ From Theorem 1, we have

$$\begin{array}{cc}\hfill {\int }_0^1\mathtt{Q}({\mathsf{\zeta }}_{\mathfrak{d}-1},{\mathsf{\zeta }}_{\mathfrak{d}})\varrho \left({\mathsf{\zeta }}_{\mathfrak{d}}\right){\mathtt{F}}_{\mathfrak{d}}\left({\mathfrak{z}}_{{}_1}\left({\mathsf{\zeta }}_{\mathfrak{d}}\right)\right)\mathtt{d}{\mathsf{\zeta }}_{\mathfrak{d}}& \le {\mathfrak{R}}_2{a}_2\frac{r_0^2}{{(\mathrm{N}-2)}^2}\parallel \widehat{\mathtt{Q}}{\parallel }_{\mathtt{q}}\prod _{i=1}^{\kappa }{\parallel {\varrho }_i\parallel }_{{\mathtt{p}}_i}\hfill \\ & \le a_2.\hfill \end{array}$$

Similarly, for $0<{\mathsf{\zeta }}_{\mathfrak{d}-2}<1,$

$$\begin{array}{cc}\hfill {\int }_0^1\mathtt{Q}({\mathsf{\zeta }}_{\mathfrak{d}-2},{\mathsf{\zeta }}_{\mathfrak{d}-1})\varrho \left({\mathsf{\zeta }}_{\mathfrak{d}-1}\right){\mathtt{F}}_{\mathfrak{d}-1}& \left[{\int }_0^1\mathtt{Q}({\mathsf{\zeta }}_{\mathfrak{d}-1},{\mathsf{\zeta }}_{\mathfrak{d}})\varrho \left({\mathsf{\zeta }}_{\mathfrak{d}}\right){\mathtt{F}}_{\mathfrak{d}}\left({\mathfrak{z}}_{{}_1}\left({\mathsf{\zeta }}_{\mathfrak{d}}\right)\right)\mathtt{d}{\mathsf{\zeta }}_{\mathfrak{d}}\right]\mathtt{d}{\mathsf{\zeta }}_{\mathfrak{d}-1}\hfill \\ & \le {\int }_0^1\mathtt{Q}({\mathsf{\zeta }}_{\mathfrak{d}-1},{\mathsf{\zeta }}_{\mathfrak{d}-1})\varrho \left({\mathsf{\zeta }}_{\mathfrak{d}-1}\right){\mathtt{F}}_{\mathfrak{d}-1}\left(a_2\right)\mathtt{d}{\mathsf{\zeta }}_{\mathfrak{d}-1}\hfill \\ & \le {\mathfrak{R}}_2{a}_2{\int }_0^1\mathtt{Q}({\mathsf{\zeta }}_{\mathfrak{d}-1},{\mathsf{\zeta }}_{\mathfrak{d}-1})\varrho \left({\mathsf{\zeta }}_{\mathfrak{d}-1}\right)\mathtt{d}{\mathsf{\zeta }}_{\mathfrak{d}-1}\hfill \\ & \le {\mathfrak{R}}_2{a}_2\frac{r_0^2}{{(\mathrm{N}-2)}^2}\parallel \widehat{\mathtt{Q}}{\parallel }_{\mathtt{q}}\prod _{i=1}^{\kappa }{\parallel {\varrho }_i\parallel }_{{\mathtt{p}}_i}\hfill \\ & \le a_2.\hfill \end{array}$$

Following this bootstrapping reasoning, we arrive at

$$\begin{array}{cc}\hfill \left(\pounds {\mathfrak{z}}_{{}_1}\right)\left(t\right)=& {\int }_0^1\mathtt{Q}(\widehat{\mathbf{r}},{\mathsf{\zeta }}_1)\varrho \left({\mathsf{\zeta }}_1\right){\mathtt{F}}_1\left[{\int }_0^1\mathtt{Q}({\mathsf{\zeta }}_1,{\mathsf{\zeta }}_2)\varrho \left({\mathsf{\zeta }}_2\right){\mathtt{F}}_2\right[{\int }_0^1\mathtt{Q}({\mathsf{\zeta }}_2,{\mathsf{\zeta }}_3)\varrho \left({\mathsf{\zeta }}_3\right){\mathtt{F}}_4\cdots \hfill \\ & {\mathtt{F}}_{\mathfrak{d}-1}\left[{\int }_0^1\mathtt{Q}({\mathsf{\zeta }}_{\mathfrak{d}-1},{\mathsf{\zeta }}_{\mathfrak{d}})\varrho \left({\mathsf{\zeta }}_{\mathfrak{d}}\right){\mathtt{F}}_{\mathfrak{d}}\left({\mathfrak{z}}_{{}_1}\left({\mathsf{\zeta }}_{\mathfrak{d}}\right)\right)\mathtt{d}{\mathsf{\zeta }}_{\mathfrak{d}}\right]\cdots \left]\mathtt{d}{\mathsf{\zeta }}_3\right]\mathtt{d}{\mathsf{\zeta }}_2]\mathtt{d}{\mathsf{\zeta }}_1\hfill \\ \hfill \le & a_2.\hfill \end{array}$$

As ${\mathfrak{N}}_2=\parallel {\mathfrak{z}}_{{}_1}\parallel$ for ${\mathfrak{z}}_{{}_1}\in \mathfrak{X}\cap \partial {\mathfrak{N}}_2,$ we obtain

$$\parallel \pounds {\mathfrak{z}}_{{}_1}\parallel \le \parallel {\mathfrak{z}}_{{}_1}\parallel .$$

(5)

Let $\widehat{\mathbf{r}}\in [\mathsf{\xi },1-\mathsf{\xi }];$ then, ${\displaystyle a_1= \parallel {\mathfrak{z}}_{{}_1}\parallel \ge {\mathfrak{z}}_{{}_1}\left(\widehat{\mathbf{r}}\right)\ge \underset{\widehat{\mathbf{r}}\in [\mathsf{\xi },1-\mathsf{\xi }]}{min}{\mathfrak{z}}_{{}_1}\left(t\right)\ge \mathsf{\sigma }\left(\mathsf{\xi }\right)\parallel {\mathfrak{z}}_{{}_1}\parallel \ge \mathsf{\sigma }\left(\mathsf{\xi }\right)a_1.}$ By $\left({\mathcal{J}}_4\right)$ and for ${\mathsf{\zeta }}_{\mathfrak{d}-1}\in [\mathsf{\xi },1-\mathsf{\xi }],$ we have

$$\begin{array}{cc}\hfill {\int }_0^1\mathtt{Q}({\mathsf{\zeta }}_{\mathfrak{d}-1},{\mathsf{\zeta }}_{\mathfrak{d}})\varrho \left({\mathsf{\zeta }}_{\mathfrak{d}}\right){\mathtt{F}}_{\mathfrak{d}}\left({\mathfrak{z}}_{{}_1}\left({\mathsf{\zeta }}_{\mathfrak{d}}\right)\right)\mathtt{d}{\mathsf{\zeta }}_{\mathfrak{d}}& \ge {\int }_{\mathsf{\xi }}^{1-\mathsf{\xi }}\mathtt{Q}({\mathsf{\zeta }}_{\mathfrak{d}-1},{\mathsf{\zeta }}_{\mathfrak{d}})\varrho \left({\mathsf{\zeta }}_{\mathfrak{d}}\right){\mathtt{F}}_{\mathfrak{d}}\left({\mathfrak{z}}_{{}_1}\left({\mathsf{\zeta }}_{\mathfrak{d}}\right)\right)\mathtt{d}{\mathsf{\zeta }}_{\mathfrak{d}}\hfill \\ & \ge \mathsf{\sigma }\left(\mathsf{\xi }\right){\int }_{\mathsf{\xi }}^{1-\mathsf{\xi }}\mathtt{Q}({\mathsf{\zeta }}_{\mathfrak{d}},{\mathsf{\zeta }}_{\mathfrak{d}})\varrho \left({\mathsf{\zeta }}_{\mathfrak{d}}\right){\mathtt{F}}_{\mathfrak{d}}\left({\mathfrak{z}}_{{}_1}\left({\mathsf{\zeta }}_{\mathfrak{d}}\right)\right)\mathtt{d}{\mathsf{\zeta }}_{\mathfrak{d}}\hfill \\ & \ge \mathsf{\sigma }\left(\mathsf{\xi }\right){\mathfrak{R}}_1{a}_1{\int }_{\mathsf{\xi }}^{1-\mathsf{\xi }}\mathtt{Q}({\mathsf{\zeta }}_{\mathfrak{d}},{\mathsf{\zeta }}_{\mathfrak{d}})\varrho \left({\mathsf{\zeta }}_{\mathfrak{d}}\right)\mathtt{d}{\mathsf{\zeta }}_{\mathfrak{d}}\hfill \\ & \ge {\mathfrak{R}}_1{a}_1\frac{\mathsf{\sigma }\left(\mathsf{\xi }\right)r_0^2}{{(\mathrm{N}-2)}^2}{\int }_{\mathsf{\xi }}^{1-\mathsf{\xi }}\mathtt{Q}({\mathsf{\zeta }}_{\mathfrak{d}},{\mathsf{\zeta }}_{\mathfrak{d}}){\mathsf{\zeta }}_{\mathfrak{d}}^{\frac{2(\mathrm{N}-1)}{2-\mathfrak{d}}}\prod _{i=1}^{\kappa }{\varrho }_i\left({\mathsf{\zeta }}_{\mathfrak{d}}\right)\mathtt{d}{\mathsf{\zeta }}_{\mathfrak{d}}\hfill \\ & \ge {\mathfrak{R}}_1{a}_1\frac{\mathsf{\sigma }\left(\mathsf{\xi }\right)r_0^2}{{(\mathrm{N}-2)}^2}\prod _{i=1}^{\kappa }{\varrho }_i^{★}{\int }_{\mathsf{\xi }}^{1-\mathsf{\xi }}\mathtt{Q}({\mathsf{\zeta }}_{\mathfrak{d}},{\mathsf{\zeta }}_{\mathfrak{d}}){\mathsf{\zeta }}_{\mathfrak{d}}^{\frac{2(\mathrm{N}-1)}{2-\mathrm{N}}}\mathtt{d}{\mathsf{\zeta }}_{\mathfrak{d}}\hfill \\ & \ge a_1.\hfill \end{array}$$

Similarly, for $0<{\mathsf{\zeta }}_{\mathfrak{d}-2}<1,$

$$\begin{array}{cc}\hfill {\int }_0^1\mathtt{Q}({\mathsf{\zeta }}_{\mathfrak{d}-2},{\mathsf{\zeta }}_{\mathfrak{d}-1})\varrho \left({\mathsf{\zeta }}_{\mathfrak{d}-1}\right){\mathtt{F}}_{\mathfrak{d}-1}& \left[{\int }_0^1\mathtt{Q}({\mathsf{\zeta }}_{\mathfrak{d}-1},{\mathsf{\zeta }}_{\mathfrak{d}})\varrho \left({\mathsf{\zeta }}_{\mathfrak{d}}\right){\mathtt{F}}_{\mathfrak{d}}\left({\mathfrak{z}}_{{}_1}\left({\mathsf{\zeta }}_{\mathfrak{d}}\right)\right)\mathtt{d}{\mathsf{\zeta }}_{\mathfrak{d}}\right]\mathtt{d}{\mathsf{\zeta }}_{\mathfrak{d}-1}\hfill \\ & \ge {\int }_{\mathsf{\xi }}^{1-\mathsf{\xi }}\mathtt{Q}({\mathsf{\zeta }}_{\mathfrak{d}-2},{\mathsf{\zeta }}_{\mathfrak{d}-1})\varrho \left({\mathsf{\zeta }}_{\mathfrak{d}-1}\right){\mathtt{F}}_{\mathfrak{d}-1}\left(a_1\right)\mathtt{d}{\mathsf{\zeta }}_{\mathfrak{d}-1}\hfill \\ & \ge \mathsf{\sigma }\left(\mathsf{\xi }\right){\int }_{\mathsf{\xi }}^{1-\mathsf{\xi }}\mathtt{Q}({\mathsf{\zeta }}_{\mathfrak{d}-1},{\mathsf{\zeta }}_{\mathfrak{d}-1})\varrho \left({\mathsf{\zeta }}_{\mathfrak{d}-1}\right){\mathtt{F}}_{\mathfrak{d}-1}\left(a_1\right)\mathtt{d}{\mathsf{\zeta }}_{\mathfrak{d}-1}\hfill \\ & \ge \mathsf{\sigma }\left(\mathsf{\xi }\right){\mathfrak{R}}_1{a}_1{\int }_{\mathsf{\xi }}^{1-\mathsf{\xi }}\mathtt{Q}({\mathsf{\zeta }}_{\mathfrak{d}-1},{\mathsf{\zeta }}_{\mathfrak{d}-1})\varrho \left({\mathsf{\zeta }}_{\mathfrak{d}-1}\right)\mathtt{d}{\mathsf{\zeta }}_{\mathfrak{d}-1}\hfill \\ & \ge {\mathfrak{R}}_1{a}_1\frac{\mathsf{\sigma }\left(\mathsf{\xi }\right)r_0^2}{{(\mathrm{N}-2)}^2}{\int }_{\mathsf{\xi }}^{1-\mathsf{\xi }}\mathtt{Q}({\mathsf{\zeta }}_{\mathfrak{d}-1},{\mathsf{\zeta }}_{\mathfrak{d}-1}){\mathsf{\zeta }}_{\mathfrak{d}-1}^{\frac{2(\mathrm{N}-1)}{2-\mathfrak{d}}}\prod _{i=1}^{\kappa }{\varrho }_i\left({\mathsf{\zeta }}_{\mathfrak{d}-1}\right)\mathtt{d}{\mathsf{\zeta }}_{\mathfrak{d}-1}\hfill \\ & \ge {\mathfrak{R}}_1{a}_1\frac{\mathsf{\sigma }\left(\mathsf{\xi }\right)r_0^2}{{(\mathrm{N}-2)}^2}\prod _{i=1}^{\kappa }{\varrho }_i^{★}{\int }_0^1\mathtt{Q}({\mathsf{\zeta }}_{\mathfrak{d}-1},{\mathsf{\zeta }}_{\mathfrak{d}-1}){\mathsf{\zeta }}_{\mathfrak{d}-1}^{\frac{2(\mathrm{N}-1)}{2-\mathrm{N}}}\mathtt{d}{\mathsf{\zeta }}_{\mathfrak{d}-1}\hfill \\ & \ge a_1.\hfill \end{array}$$

It follows that

$$\begin{array}{cc}\hfill \left(\pounds {\mathfrak{z}}_{{}_1}\right)\left(\widehat{\mathbf{r}}\right)=& {\int }_0^1\mathtt{Q}(\widehat{\mathbf{r}},{\mathsf{\zeta }}_1)\varrho \left({\mathsf{\zeta }}_1\right){\mathtt{F}}_1\left[{\int }_0^1\mathtt{Q}({\mathsf{\zeta }}_1,{\mathsf{\zeta }}_2)\varrho \left({\mathsf{\zeta }}_2\right){\mathtt{F}}_2\right[{\int }_0^1\mathtt{Q}({\mathsf{\zeta }}_2,{\mathsf{\zeta }}_3)\varrho \left({\mathsf{\zeta }}_3\right){\mathtt{F}}_4\cdots \hfill \\ & {\mathtt{F}}_{\mathfrak{d}-1}\left[{\int }_0^1\mathtt{Q}({\mathsf{\zeta }}_{\mathfrak{d}-1},{\mathsf{\zeta }}_{\mathfrak{d}})\varrho \left({\mathsf{\zeta }}_{\mathfrak{d}}\right){\mathtt{F}}_{\mathfrak{d}}\left({\mathfrak{z}}_{{}_1}\left({\mathsf{\zeta }}_{\mathfrak{d}}\right)\right)\mathtt{d}{\mathsf{\zeta }}_{\mathfrak{d}}\right]\cdots \left]\mathtt{d}{\mathsf{\zeta }}_3\right]\mathtt{d}{\mathsf{\zeta }}_2]\mathtt{d}{\mathsf{\zeta }}_1\hfill \\ \hfill \ge & a_1.\hfill \end{array}$$

Thus, for ${\mathfrak{z}}_{{}_1}\in \mathfrak{X}\cap \partial {\mathfrak{N}}_1,$ we have

$$\parallel \pounds {\mathfrak{z}}_{{}_1}\parallel \ge \parallel {\mathfrak{z}}_{{}_1}\parallel .$$

(6)

It can be seen that $0\in {\mathfrak{N}}_1\subset {\overline{\mathfrak{N}}}_1\subset {\mathfrak{N}}_2$, and from (5), (6), and Theorem 2, the operator £ has a fixed point ${\mathfrak{z}}_{{}_1}\in \mathfrak{X}\cap \left({\overline{\mathfrak{N}}}_2\backslash {\mathfrak{N}}_1\right)$ and ${\mathfrak{z}}_{{}_1}\left(\widehat{\mathbf{r}}\right)\ge 0$ on $(0,1).$ Now, put ${\mathfrak{z}}_{{}_1}={\mathfrak{z}}_{{}_{\mathfrak{d}+1}},$ to obtain an infinite number of solutions:

$$\begin{array}{cc}\hfill {\mathfrak{z}}_{{}_{\mathsf{\beta }}}\left(\widehat{\mathbf{r}}\right)& ={\int }_0^1\mathtt{Q}(\widehat{\mathbf{r}},s)\varrho \left(s\right){\mathtt{F}}_{\mathsf{\beta }}\left({\mathfrak{z}}_{{}_{\mathsf{\beta }+1}}\left(s\right)\right)ds,\mathsf{\beta }=1,2,\cdots ,\mathfrak{d}-1,\mathfrak{d},\hfill \\ \hfill {\mathfrak{z}}_{{}_{\mathfrak{d}+1}}\left(\widehat{\mathbf{r}}\right)& ={\mathfrak{z}}_{{}_1}\left(\widehat{\mathbf{r}}\right),\widehat{\mathbf{r}}\in (0,1).\hfill \end{array}$$

□

For the cases ${\displaystyle \sum _{i=1}^{\kappa }\frac{1}{{\mathtt{p}}_i}=1}$ and ${\displaystyle \sum _{i=1}^{\kappa }\frac{1}{{\mathtt{p}}_i}>1,}$ we have the following theorems:

Theorem 5. 

Suppose $\left({\mathcal{J}}_1\right)$–$\left({\mathcal{J}}_2\right)$ hold, and there exist constants $b_2>b_1>0$ with ${\mathtt{F}}_{\mathsf{\beta }}(\mathsf{\beta }=1,2,\cdots ,\mathfrak{d})$ satisfies $\left({\mathcal{J}}_4\right)$ and

$\left({\mathcal{J}}_5\right)$

${\mathtt{F}}_{\mathsf{\beta }}\left(\mathfrak{z}\left(\widehat{\mathbf{r}}\right)\right)\le {\mathfrak{N}}_2{b}_2$ for $0\le \widehat{\mathbf{r}}\le 1,0\le \mathfrak{z}\le b_2,$ where ${\displaystyle {\mathfrak{N}}_2={\left[\frac{r_0^2}{{(\mathrm{N}-2)}^2}\parallel \widehat{\mathtt{Q}}{\parallel }_{\infty }\prod _{i=1}^{\kappa }{\parallel {\varrho }_i\parallel }_{{\mathtt{p}}_i}\right]}^{-1}}$ and $\widehat{\mathtt{Q}}\left(\mathsf{\zeta }\right)=\mathtt{Q}(\mathsf{\zeta },\mathsf{\zeta }){\mathsf{\zeta }}^{\frac{2(\mathrm{N}-1)}{2-\mathrm{N}}};$

then the BVP (3) has a solution $({\mathfrak{z}}_{{}_1},{\mathfrak{z}}_2,\cdots ,{\mathfrak{z}}_{\mathfrak{d}})$, such that ${\mathfrak{z}}_{{}_{\mathsf{\beta }}}>0,b_1\le \parallel {\mathfrak{z}}_{{}_{\mathsf{\beta }}}\parallel \le b_2,\mathsf{\beta }=1,2,\cdots ,\mathfrak{d}.$

Proof. 

The proof is similar to the proof of Theorem 4; therefore, we omit the details here. □

Theorem 6. 

Suppose $\left({\mathcal{J}}_1\right)$–$\left({\mathcal{J}}_2\right)$ hold, and there exist constants $c_2>c_1>0$ with ${\mathtt{F}}_{\mathsf{\beta }}(\mathsf{\beta }=1,2,\cdots ,\mathfrak{d})$ satisfying $\left({\mathcal{J}}_4\right)$ and

$\left({\mathcal{J}}_6\right)$

${\mathtt{F}}_{\mathsf{\beta }}\left(\mathfrak{z}\left(\widehat{\mathbf{r}}\right)\right)\le {\mathfrak{M}}_2{c}_2$ for all $0\le \widehat{\mathbf{r}}\le 1,0\le \mathfrak{z}\le c_2,$ where ${\displaystyle {\mathfrak{M}}_2={\left[\frac{r_0^2}{{(\mathrm{N}-2)}^2}\parallel \widehat{\mathtt{Q}}{\parallel }_{\infty }\prod _{i=1}^{\kappa }{\parallel {\varrho }_i\parallel }_1\right]}^{-1}}$ and $\widehat{\mathtt{Q}}\left(\mathsf{\zeta }\right)=\mathtt{Q}(\mathsf{\zeta },\mathsf{\zeta }){\mathsf{\zeta }}^{\frac{2(\mathrm{N}-1)}{2-\mathrm{N}}},$

then the BVP (3) has a solution $({\mathfrak{z}}_{{}_1},{\mathfrak{z}}_2,\cdots ,{\mathfrak{z}}_{\mathfrak{d}})$, such that ${\mathfrak{z}}_{{}_{\mathsf{\beta }}}>0,c_1\le \parallel {\mathfrak{z}}_{{}_{\mathsf{\beta }}}\parallel \le c_2,\mathsf{\beta }=1,2,\cdots ,\mathfrak{d}.$

Proof. 

The proof is similar to the proof of Theorem 4; therefore, we omit the details here. □

Example 1. 

Consider the problem

$$▵{\mathfrak{z}}_{{}_{\mathsf{\beta }}}-\frac{{(\mathrm{N}-2)}^2{r}_0^{2\mathrm{N}-2}}{{\left|\mathsf{\nu }\right|}^{2\mathrm{N}-2}}{\mathfrak{z}}_{{}_{\mathsf{\beta }}}+\varrho \left(\right|\mathsf{\nu }\left|\right){\mathtt{F}}_{\mathsf{\beta }}({\mathfrak{z}}_{{}_{\mathsf{\beta }+1}})=0,1<\left|\mathsf{\nu }\right|<3,$$

(7)

$${\mathfrak{z}}_{{}_{\mathsf{\beta }}}\left(0\right)=0,{\mathfrak{z}}_{{}_{\mathsf{\beta }}}\left(1\right)=0,$$

(8)

where $r_0=1,$ $\mathrm{N}=3,$ $\mathsf{\beta }\in \{1,2\},{\mathfrak{z}}_3={\mathfrak{z}}_{{}_1},$ $\varrho \left(\widehat{\mathbf{r}}\right)=\frac{1}{{\widehat{\mathbf{r}}}^4}{\prod }_{i=1}^2{\varrho }_i\left(\widehat{\mathbf{r}}\right),$ ${\varrho }_i\left(\widehat{\mathbf{r}}\right)={\varrho }_i\left(\frac{1}{\widehat{\mathbf{r}}}\right),$ in which ${\varrho }_1\left(t\right)=\frac{2}{t^2+1}$ and ${\varrho }_2\left(t\right)=\frac{1}{\sqrt{t+2}},$ then ${\varrho }_1,{\varrho }_2\in L^{\mathtt{p}}[0,1]$ and ${\prod }_{i=1}^2{\varrho }_i^{*}=\frac{1}{\sqrt{3}}.$ Let $\mathsf{\xi }=\frac{1}{3},$ ${\mathtt{F}}_1\left(\mathfrak{z}\right)={\mathtt{F}}_2\left(\mathfrak{z}\right)=1+\frac{1}{3}|sin(1+\mathfrak{z})|+\frac{1}{1+\mathfrak{z}}.$

$$\begin{array}{c}\hfill \mathtt{Q}(\widehat{\mathbf{r}},\mathsf{\zeta })=\frac{1}{sinh\left(1\right)}\left\{\begin{array}{c}sinh\left(\widehat{\mathbf{r}}\right)sinh(1-\mathsf{\zeta }),0\le \widehat{\mathbf{r}}\le \mathsf{\zeta }\le 1,\hfill \\ sinh\left(\mathsf{\zeta }\right)sinh(1-\widehat{\mathbf{r}}),0\le \mathsf{\zeta }\le \widehat{\mathbf{r}}\le 1,\hfill \end{array}\right.\end{array}$$

and $\mathsf{\sigma }\left(\mathsf{\xi }\right)=\frac{sinh\left(\mathsf{\xi }\right)}{sinh\left(1\right)}=\frac{sinh\left(\frac{1}{3}\right)}{sinh\left(1\right)}=0.2889212153.$ In addition,

$$\begin{array}{c}\hfill {\mathfrak{R}}_1={\left[\frac{\mathsf{\sigma }\left(\mathsf{\xi }\right)r_0^2}{{(\mathrm{N}-2)}^2}\prod _{i=1}^{\kappa }{\varrho }_i^{★}{\int }_{\mathsf{\xi }}^{1-\mathsf{\xi }}\mathtt{Q}(\mathsf{\zeta },\mathsf{\zeta }){\mathsf{\zeta }}^{\frac{2(\mathrm{N}-1)}{2-\mathrm{N}}}\mathtt{d}\mathsf{\zeta }\right]}^{-1}\approx 2.932844681.\end{array}$$

Let ${\mathtt{p}}_1=2,{\mathtt{p}}_2=3$ and $\mathtt{q}=6,$ then $\frac{1}{{\mathtt{p}}_1}+\frac{1}{{\mathtt{p}}_2}+\frac{1}{\mathtt{q}}=1$ and

$$\begin{array}{c}\hfill {\mathfrak{R}}_2={\left[\frac{r_0^2}{{(\mathrm{N}-2)}^2}\parallel \widehat{\mathtt{Q}}{\parallel }_{\mathtt{q}}\prod _{i=1}^{\kappa }{\parallel {\varrho }_i\parallel }_{{\mathtt{p}}_i}\right]}^{-1}\approx 4.284821634.\end{array}$$

Choose $a_1=\frac{1}{2}$ and $a_2=1.$ Then,

$$\begin{array}{cc}\hfill {\mathtt{F}}_1\left(\mathfrak{z}\right)={\mathtt{F}}_2\left(\mathfrak{z}\right)& =1+\frac{1}{3}|sin(1+\mathfrak{z})|+\frac{1}{1+\mathfrak{z}}\le 4.284821634={\mathfrak{R}}_2{a}_2,0\le \mathfrak{z}\le 1,\hfill \\ \hfill {\mathtt{F}}_1\left(\mathfrak{z}\right)={\mathtt{F}}_2\left(\mathfrak{z}\right)& =1+\frac{1}{3}|sin(1+\mathfrak{z})|+\frac{1}{1+\mathfrak{z}}\ge 1.466422340={\mathfrak{R}}_1{a}_1,0.1444606076\le \mathfrak{z}\le \frac{1}{2}.\hfill \end{array}$$

Thus, by Theorem 4, BVP (7) and (8) has at least one positive solution $({\mathfrak{z}}_{{}_1},{\mathfrak{z}}_2)$, such that $\frac{1}{2}\le \parallel {\mathfrak{z}}_{{}_{\mathsf{\beta }}}\parallel \le 1$ for $\mathsf{\beta }=1,2.$

4. Existence of Coupled Positive Radial Solutions

By utilizing the Avery–Henderson cone fixed-point theorem, we demonstrate in this section that there are coupled positive solutions for (3). Denote

$${\mathsf{\beta }}_1=\frac{\mathsf{\sigma }\left(\mathsf{\xi }\right)r_0^2}{{(\mathrm{N}-2)}^2}\prod _{i=1}^{\kappa }{\varrho }_i^{★}{\int }_0^1\mathtt{Q}(\mathsf{\zeta },\mathsf{\zeta }){\mathsf{\zeta }}^{\frac{2(\mathrm{N}-1)}{2-\mathrm{N}}}\mathtt{d}\mathsf{\zeta },$$

$${\mathsf{\beta }}_2=\frac{r_0^2}{{(\mathrm{N}-2)}^2}\parallel \widehat{\mathtt{Q}}{\parallel }_{\mathtt{q}}\prod _{i=1}^{\kappa }{\parallel {\varrho }_i\parallel }_{{\mathtt{p}}_i},$$

$${\mathsf{\beta }}_3=\frac{r_0^2}{{(\mathrm{N}-2)}^2}\parallel \widehat{\mathtt{Q}}{\parallel }_{\infty }\prod _{i=1}^{\kappa }{\parallel {\varrho }_i\parallel }_{{\mathtt{p}}_i},$$

$${\mathsf{\beta }}_4=\frac{r_0^2}{{(\mathrm{N}-2)}^2}\parallel \widehat{\mathtt{Q}}{\parallel }_{\infty }\prod _{i=1}^{\kappa }{\parallel {\varrho }_i\parallel }_1.$$

Theorem 7. 

Suppose that $\left({\mathcal{J}}_1\right)$–$\left({\mathcal{J}}_2\right)$ hold, and that there exist three positive real numbers $\mathfrak{f}<\mathfrak{g}<\mathfrak{h}$ with ${\mathtt{F}}_{\mathsf{\beta }}(\mathsf{\beta }=1,2,\cdots ,\mathfrak{d})$ satisfying

$\left({\mathcal{J}}_7\right)$

${\mathtt{F}}_{\mathsf{\beta }}\left(\mathfrak{z}\right)>\frac{\mathfrak{h}}{{\mathsf{\beta }}_1}$, $\mathfrak{h}\le \mathfrak{z}\le \frac{\mathfrak{h}}{\mathsf{\sigma }\left(\mathsf{\xi }\right)},$

$\left({\mathcal{J}}_8\right)$

${\mathtt{F}}_{\mathsf{\beta }}\left(\mathfrak{z}\right)<\frac{\mathfrak{g}}{{\mathsf{\beta }}_2}$, $0\le \mathfrak{z}\le \frac{\mathfrak{g}}{\mathsf{\sigma }\left(\mathsf{\xi }\right)},$

$\left({\mathcal{J}}_9\right)$

${\mathtt{F}}_{\mathsf{\beta }}\left(\mathfrak{z}\right)>\frac{\mathfrak{f}}{{\mathsf{\beta }}_1}$, $\mathfrak{f}\le \mathfrak{z}\le \frac{\mathfrak{f}}{\mathsf{\sigma }\left(\mathsf{\xi }\right)},$

then the BVP (3) has coupled positive solutions $\left\{({}^1{\mathfrak{z}}_{{}_1},{}^1{\mathfrak{z}}_2,\cdots ,{}^1{\mathfrak{z}}_{\mathfrak{d}})\right\}$ and $\left\{\right({}^2{\mathfrak{z}}_{{}_1},{}^2{\mathfrak{z}}_2,\cdots ,$ ${}^2{\mathfrak{z}}_{\mathfrak{d}}\left)\right\}$ satisfying

$$\mathfrak{f}<{\mathsf{\gamma }}_1\left({}^1{\mathfrak{z}}_{{}_{\mathsf{\beta }}}\right)\mathit{with}{\mathsf{\gamma }}_3\left({}^1{\mathfrak{z}}_{{}_{\mathsf{\beta }}}\right)<\mathfrak{g},\mathsf{\beta }=1,2,\cdots ,\mathfrak{d}$$

and

$$\mathfrak{g}<{\mathsf{\gamma }}_3\left({}^2{\mathfrak{z}}_{{}_{\mathsf{\beta }}}\right)\mathit{with}{\mathsf{\gamma }}_2\left({}^2{\mathfrak{z}}_{{}_{\mathsf{\beta }}}\right)<\mathfrak{h},\mathsf{\beta }=1,2,\cdots ,\mathfrak{d}.$$

Proof. 

It is easy to demonstrate that $\pounds :\overline{\mathfrak{X}({\mathsf{\gamma }}_2,\mathfrak{h})}\to \mathfrak{X}$ and £ are completely continuous from (4): first, we check that the condition $\left(a\right)$ of Theorem 3 holds; for this, we choose ${\mathfrak{z}}_{{}_1}\in \partial \mathfrak{X}({\mathsf{\gamma }}_2,\mathfrak{h});$ then, ${\mathsf{\gamma }}_2\left({\mathfrak{z}}_{{}_1}\right)={min}_{\widehat{\mathbf{r}}\in [\mathsf{\xi },1-\mathsf{\xi }]}{\mathfrak{z}}_{{}_1}\left(\widehat{\mathbf{r}}\right)=\mathfrak{h}$, so $\mathfrak{h}\le {\mathfrak{z}}_{{}_1}\left(\widehat{\mathbf{r}}\right)$ for $\widehat{\mathbf{r}}\in [\mathsf{\xi },1-\mathsf{\xi }].$ As $\parallel {\mathfrak{z}}_{{}_1}\parallel \le \frac{1}{\mathsf{\sigma }\left(\mathsf{\xi }\right)}{\mathsf{\gamma }}_2\left({\mathfrak{z}}_{{}_1}\right)=\frac{1}{\mathsf{\sigma }\left(\mathsf{\xi }\right)}\mathfrak{h},$ we have $\mathfrak{h}\le {\mathfrak{z}}_{{}_1}\left(\widehat{\mathbf{r}}\right)\le \frac{\mathfrak{h}}{\mathsf{\sigma }\left(\mathsf{\xi }\right)},\widehat{\mathbf{r}}\in [\mathsf{\xi },1-\mathsf{\xi }].$ Let ${\mathsf{\zeta }}_{\mathfrak{d}-1}\in [\mathsf{\xi },1-\mathsf{\xi }].$ Then, by $\left({\mathcal{J}}_7\right),$ we have

$$\begin{array}{cc}\hfill {\int }_0^1\mathtt{Q}({\mathsf{\zeta }}_{\mathfrak{d}-1},{\mathsf{\zeta }}_{\mathfrak{d}})\varrho \left({\mathsf{\zeta }}_{\mathfrak{d}}\right){\mathtt{F}}_{\mathfrak{d}}\left({\mathfrak{z}}_{{}_1}\left({\mathsf{\zeta }}_{\mathfrak{d}}\right)\right)\mathtt{d}{\mathsf{\zeta }}_{\mathfrak{d}}& \ge \mathsf{\sigma }\left(\mathsf{\xi }\right){\int }_{\mathsf{\xi }}^{1-\mathsf{\xi }}\mathtt{Q}({\mathsf{\zeta }}_{\mathfrak{d}},{\mathsf{\zeta }}_{\mathfrak{d}})\varrho \left({\mathsf{\zeta }}_{\mathfrak{d}}\right){\mathtt{F}}_{\mathfrak{d}}\left({\mathfrak{z}}_{{}_1}\left({\mathsf{\zeta }}_{\mathfrak{d}}\right)\right)\mathtt{d}{\mathsf{\zeta }}_{\mathfrak{d}}\hfill \\ & \ge \frac{\mathsf{\sigma }\left(\mathsf{\xi }\right)\mathfrak{h}}{{\mathsf{\beta }}_1}{\int }_{\mathsf{\xi }}^{1-\mathsf{\xi }}\mathtt{Q}({\mathsf{\zeta }}_{\mathfrak{d}},{\mathsf{\zeta }}_{\mathfrak{d}})\varrho \left({\mathsf{\zeta }}_{\mathfrak{d}}\right)\mathtt{d}{\mathsf{\zeta }}_{\mathfrak{d}}\hfill \\ & \ge \frac{\mathsf{\sigma }\left(\mathsf{\xi }\right)\mathfrak{h}{r}_0^2}{{(\mathrm{N}-2)}^2{\mathsf{\beta }}_1}{\int }_{\mathsf{\xi }}^{1-\mathsf{\xi }}\mathtt{Q}({\mathsf{\zeta }}_{\mathfrak{d}},{\mathsf{\zeta }}_{\mathfrak{d}}){\mathsf{\zeta }}_{\mathfrak{d}}^{\frac{2(\mathrm{N}-1)}{2-\mathrm{N}}}\prod _{i=1}^{\kappa }{\varrho }_i\left({\mathsf{\zeta }}_{\mathfrak{d}}\right)\mathtt{d}{\mathsf{\zeta }}_{\mathfrak{d}}\hfill \\ & \ge \frac{\mathsf{\sigma }\left(\mathsf{\xi }\right)\mathfrak{h}{r}_0^2}{{(\mathrm{N}-2)}^2{\mathsf{\beta }}_1}\prod _{i=1}^{\kappa }{\varrho }_i^{★}{\int }_{\mathsf{\xi }}^{1-\mathsf{\xi }}\mathtt{Q}({\mathsf{\zeta }}_{\mathfrak{d}},{\mathsf{\zeta }}_{\mathfrak{d}}){\mathsf{\zeta }}_{\mathfrak{d}}^{\frac{2(\mathrm{N}-1)}{2-\mathrm{N}}}\mathtt{d}{\mathsf{\zeta }}_{\mathfrak{d}}\hfill \\ & \ge \mathfrak{h}.\hfill \end{array}$$

Following this, we arrive at

$$\begin{array}{cc}\hfill {\mathsf{\gamma }}_2\left(\pounds {\mathfrak{z}}_{{}_1}\right)=& \underset{\widehat{\mathbf{r}}\in [0,1]}{min}{\int }_0^1\mathtt{Q}(\widehat{\mathbf{r}},{\mathsf{\zeta }}_1)\varrho \left({\mathsf{\zeta }}_1\right){\mathtt{F}}_1\left[{\int }_0^1\mathtt{Q}({\mathsf{\zeta }}_1,{\mathsf{\zeta }}_2)\varrho \left({\mathsf{\zeta }}_2\right){\mathtt{F}}_2\right[{\int }_0^1\mathtt{Q}({\mathsf{\zeta }}_2,{\mathsf{\zeta }}_3)\varrho \left({\mathsf{\zeta }}_3\right){\mathtt{F}}_4\cdots \hfill \\ & {\mathtt{F}}_{\mathfrak{d}-1}\left[{\int }_0^1\mathtt{Q}({\mathsf{\zeta }}_{\mathfrak{d}-1},{\mathsf{\zeta }}_{\mathfrak{d}})\varrho \left({\mathsf{\zeta }}_{\mathfrak{d}}\right){\mathtt{F}}_{\mathfrak{d}}\left({\mathfrak{z}}_{{}_1}\left({\mathsf{\zeta }}_{\mathfrak{d}}\right)\right)\mathtt{d}{\mathsf{\zeta }}_{\mathfrak{d}}\right]\cdots \left]\mathtt{d}{\mathsf{\zeta }}_3\right]\mathtt{d}{\mathsf{\zeta }}_2]\mathtt{d}{\mathsf{\zeta }}_1\hfill \\ \hfill \ge & \mathfrak{h}.\hfill \end{array}$$

Condition $\left(a\right)$ of Theorem 3 is proved. To prove $\left(b\right),$ choose ${\mathfrak{z}}_{{}_1}\in \partial \mathfrak{X}({\mathsf{\gamma }}_3,\mathfrak{g}).$ Then, ${\mathsf{\gamma }}_3\left({\mathfrak{z}}_{{}_1}\right)={max}_{\widehat{\mathbf{r}}\in [0,1]}{\mathfrak{z}}_{{}_1}\left(\widehat{\mathbf{r}}\right)=\mathfrak{g}$, so that $0\le {\mathfrak{z}}_{{}_1}\left(\widehat{\mathbf{r}}\right)\le \mathfrak{g}$ for $\widehat{\mathbf{r}}\in [0,1].$ As $\parallel {\mathfrak{z}}_{{}_1}\parallel \le \frac{1}{\mathsf{\sigma }\left(\mathsf{\xi }\right)}{\mathsf{\gamma }}_2\left({\mathfrak{z}}_{{}_1}\right)\le \frac{1}{\mathsf{\sigma }\left(\mathsf{\xi }\right)}{\mathsf{\gamma }}_3\left({\mathfrak{z}}_{{}_1}\right)=\frac{\mathfrak{g}}{\mathsf{\sigma }\left(\mathsf{\xi }\right)},$ we have $0\le {\mathfrak{z}}_{{}_1}\left(\widehat{\mathbf{r}}\right)\le \mathsf{\sigma }{\left(\mathsf{\xi }\right)}^2\mathfrak{g},\widehat{\mathbf{r}}\in [0,1].$ Let $0<{\mathsf{\zeta }}_{\mathfrak{d}-1}<1.$ Then, by $\left({\mathcal{J}}_8\right),$ we have

$$\begin{array}{cc}\hfill {\int }_0^1\mathtt{Q}({\mathsf{\zeta }}_{\mathfrak{d}-1},{\mathsf{\zeta }}_{\mathfrak{d}})\varrho \left({\mathsf{\zeta }}_{\mathfrak{d}}\right){\mathtt{F}}_{\mathfrak{d}}\left({\mathfrak{z}}_{{}_1}\left({\mathsf{\zeta }}_{\mathfrak{d}}\right)\right)\mathtt{d}{\mathsf{\zeta }}_{\mathfrak{d}}& \le \mathsf{\sigma }\left(\mathsf{\xi }\right){\int }_0^1\mathtt{Q}({\mathsf{\zeta }}_{\mathfrak{d}},{\mathsf{\zeta }}_{\mathfrak{d}})\varrho \left({\mathsf{\zeta }}_{\mathfrak{d}}\right){\mathtt{F}}_{\mathfrak{d}}\left({\mathfrak{z}}_{{}_1}\left({\mathsf{\zeta }}_{\mathfrak{d}}\right)\right)\mathtt{d}{\mathsf{\zeta }}_{\mathfrak{d}}\hfill \\ & \le \frac{\mathsf{\sigma }\left(\mathsf{\xi }\right)\mathfrak{g}}{{\mathsf{\beta }}_2}{\int }_0^1\mathtt{Q}({\mathsf{\zeta }}_{\mathfrak{d}},{\mathsf{\zeta }}_{\mathfrak{d}})\varrho \left({\mathsf{\zeta }}_{\mathfrak{d}}\right)\mathtt{d}{\mathsf{\zeta }}_{\mathfrak{d}}\hfill \\ & \le \frac{\mathsf{\sigma }\left(\mathsf{\xi }\right)\mathfrak{g}{r}_0^2}{{(\mathrm{N}-2)}^2{\mathsf{\beta }}_2}{\int }_0^1\mathtt{Q}({\mathsf{\zeta }}_{\mathfrak{d}},{\mathsf{\zeta }}_{\mathfrak{d}}){\mathsf{\zeta }}_{\mathfrak{d}}^{\frac{2(\mathrm{N}-1)}{2-\mathrm{N}}}\prod _{i=1}^{\kappa }{\varrho }_i\left({\mathsf{\zeta }}_{\mathfrak{d}}\right)\mathtt{d}{\mathsf{\zeta }}_{\mathfrak{d}}.\hfill \end{array}$$

For some $\mathtt{q}>1,$ we have ${\displaystyle \frac{1}{\mathtt{q}}+\sum _{i=1}^{\kappa }\frac{1}{{\mathtt{p}}_i}=1.}$ From Theorem 1, we have

$$\begin{array}{cc}\hfill {\int }_0^1\mathtt{Q}({\mathsf{\zeta }}_{\mathfrak{d}-1},{\mathsf{\zeta }}_{\mathfrak{d}})\varrho \left({\mathsf{\zeta }}_{\mathfrak{d}}\right){\mathtt{F}}_{\mathfrak{d}}\left({\mathfrak{z}}_{{}_1}\left({\mathsf{\zeta }}_{\mathfrak{d}}\right)\right)\mathtt{d}{\mathsf{\zeta }}_{\mathfrak{d}}& \le \frac{\mathsf{\sigma }\left(\mathsf{\xi }\right)\mathfrak{g}{r}_0^2}{{(\mathrm{N}-2)}^2{\mathsf{\beta }}_2}\parallel \widehat{\mathtt{Q}}{\parallel }_{\mathtt{q}}\prod _{i=1}^{\kappa }{\parallel {\varrho }_i\parallel }_{{\mathtt{p}}_i}\le \mathfrak{g}.\hfill \end{array}$$

It follows that

$$\begin{array}{cc}\hfill {\mathsf{\gamma }}_3\left(\pounds {\mathfrak{z}}_{{}_1}\right)=& \underset{\widehat{\mathbf{r}}\in [0,1]}{max}{\int }_0^1\mathtt{Q}(\widehat{\mathbf{r}},{\mathsf{\zeta }}_1)\varrho \left({\mathsf{\zeta }}_1\right){\mathtt{F}}_1\left[{\int }_0^1\mathtt{Q}({\mathsf{\zeta }}_1,{\mathsf{\zeta }}_2)\varrho \left({\mathsf{\zeta }}_2\right){\mathtt{F}}_2\right[{\int }_0^1\mathtt{Q}({\mathsf{\zeta }}_2,{\mathsf{\zeta }}_3)\varrho \left({\mathsf{\zeta }}_3\right){\mathtt{F}}_4\cdots \hfill \\ & {\mathtt{F}}_{\mathfrak{d}-1}\left[{\int }_0^1\mathtt{Q}({\mathsf{\zeta }}_{\mathfrak{d}-1},{\mathsf{\zeta }}_{\mathfrak{d}})\varrho \left({\mathsf{\zeta }}_{\mathfrak{d}}\right){\mathtt{F}}_{\mathfrak{d}}\left({\mathfrak{z}}_{{}_1}\left({\mathsf{\zeta }}_{\mathfrak{d}}\right)\right)\mathtt{d}{\mathsf{\zeta }}_{\mathfrak{d}}\right]\cdots \left]\mathtt{d}{\mathsf{\zeta }}_3\right]\mathtt{d}{\mathsf{\zeta }}_2]\mathtt{d}{\mathsf{\zeta }}_1\hfill \\ \hfill \le & \mathfrak{g}.\hfill \end{array}$$

Thus, $\left(b\right)$ holds. Finally, we also check that $\left(c\right)$ of Theorem 3 holds. Observe that ${\mathfrak{z}}_{{}_1}\left(\widehat{\mathbf{r}}\right)=\mathfrak{f}/4\subset \mathfrak{X}({\mathsf{\gamma }}_1,\mathfrak{f})$ and $\mathfrak{f}/4<\mathfrak{f},$ so that $\mathfrak{X}({\mathsf{\gamma }}_1,\mathfrak{f})\ne Ø.$ Next, if ${\mathfrak{z}}_{{}_1}\in \mathfrak{X}({\mathsf{\gamma }}_1,\mathfrak{f}),$ then $\mathfrak{f}={\mathsf{\gamma }}_1\left({\mathfrak{z}}_{{}_1}\right)={max}_{\widehat{\mathbf{r}}\in [0,1]}{\mathfrak{z}}_{{}_1}\left(\widehat{\mathbf{r}}\right)=\parallel {\mathfrak{z}}_{{}_1}\parallel =\frac{1}{\mathsf{\sigma }\left(\mathsf{\xi }\right)}{\mathsf{\gamma }}_2\left({\mathfrak{z}}_{{}_1}\right)\le \frac{1}{\mathsf{\sigma }\left(\mathsf{\xi }\right)}{\mathsf{\gamma }}_3\left({\mathfrak{z}}_{{}_1}\right)=\frac{1}{\mathsf{\sigma }\left(\mathsf{\xi }\right)}{\mathsf{\gamma }}_1\left({\mathfrak{z}}_{{}_1}\right)=\frac{\mathfrak{f}}{\mathsf{\sigma }\left(\mathsf{\xi }\right)},$ i.e., $\mathfrak{f}\le {\mathfrak{z}}_{{}_1}\left(\widehat{\mathbf{r}}\right)\le \frac{\mathfrak{f}}{\mathsf{\sigma }\left(\mathsf{\xi }\right)}$ for $\widehat{\mathbf{r}}\in [0,1].$ Let $0<{\mathsf{\zeta }}_{\mathfrak{d}-1}<1.$ Then, by $\left({\mathcal{J}}_9\right),$ we have

$$\begin{array}{cc}\hfill {\int }_0^1\mathtt{Q}({\mathsf{\zeta }}_{\mathfrak{d}-1},{\mathsf{\zeta }}_{\mathfrak{d}})\varrho \left({\mathsf{\zeta }}_{\mathfrak{d}}\right){\mathtt{F}}_{\mathfrak{d}}\left({\mathfrak{z}}_{{}_1}\left({\mathsf{\zeta }}_{\mathfrak{d}}\right)\right)\mathtt{d}{\mathsf{\zeta }}_{\mathfrak{d}}& \ge \mathsf{\sigma }\left(\mathsf{\xi }\right){\int }_{\mathsf{\xi }}^{1-\mathsf{\xi }}\mathtt{Q}({\mathsf{\zeta }}_{\mathfrak{d}},{\mathsf{\zeta }}_{\mathfrak{d}})\varrho \left({\mathsf{\zeta }}_{\mathfrak{d}}\right){\mathtt{F}}_{\mathfrak{d}}\left({\mathfrak{z}}_{{}_1}\left({\mathsf{\zeta }}_{\mathfrak{d}}\right)\right)\mathtt{d}{\mathsf{\zeta }}_{\mathfrak{d}}\hfill \\ & \ge \frac{\mathsf{\sigma }\left(\mathsf{\xi }\right)\mathfrak{f}}{{\mathsf{\beta }}_1}{\int }_{\mathsf{\xi }}^{1-\mathsf{\xi }}\mathtt{Q}({\mathsf{\zeta }}_{\mathfrak{d}},{\mathsf{\zeta }}_{\mathfrak{d}})\varrho \left({\mathsf{\zeta }}_{\mathfrak{d}}\right)\mathtt{d}{\mathsf{\zeta }}_{\mathfrak{d}}\hfill \\ & \ge \frac{\mathsf{\sigma }\left(\mathsf{\xi }\right)\mathfrak{f}{r}_0^2}{{(\mathrm{N}-2)}^2{\mathsf{\beta }}_1}{\int }_{\mathsf{\xi }}^{1-\mathsf{\xi }}\mathtt{Q}({\mathsf{\zeta }}_{\mathfrak{d}},{\mathsf{\zeta }}_{\mathfrak{d}}){\mathsf{\zeta }}_{\mathfrak{d}}^{\frac{2(\mathrm{N}-1)}{2-\mathrm{N}}}\prod _{i=1}^{\kappa }{\varrho }_i\left({\mathsf{\zeta }}_{\mathfrak{d}}\right)\mathtt{d}{\mathsf{\zeta }}_{\mathfrak{d}}\hfill \\ & \ge \frac{\mathsf{\sigma }\left(\mathsf{\xi }\right)\mathfrak{f}{r}_0^2}{{(\mathrm{N}-2)}^2{\mathsf{\beta }}_1}\prod _{i=1}^{\kappa }{\varrho }_i^{★}{\int }_{\mathsf{\xi }}^{1-\mathsf{\xi }}\mathtt{Q}({\mathsf{\zeta }}_{\mathfrak{d}},{\mathsf{\zeta }}_{\mathfrak{d}}){\mathsf{\zeta }}_{\mathfrak{d}}^{\frac{2(\mathrm{N}-1)}{2-\mathrm{N}}}\mathtt{d}{\mathsf{\zeta }}_{\mathfrak{d}}\hfill \\ & \ge \mathfrak{f}.\hfill \end{array}$$

Following this bootstrapping reasoning, we arrive at

$$\begin{array}{cc}\hfill {\mathsf{\gamma }}_1\left(\pounds {\mathfrak{z}}_{{}_1}\right)=& \underset{\widehat{\mathbf{r}}\in [0,1]}{max}{\int }_0^1\mathtt{Q}(\widehat{\mathbf{r}},{\mathsf{\zeta }}_1)\varrho \left({\mathsf{\zeta }}_1\right){\mathtt{F}}_1\left[{\int }_0^1\mathtt{Q}({\mathsf{\zeta }}_1,{\mathsf{\zeta }}_2)\varrho \left({\mathsf{\zeta }}_2\right){\mathtt{F}}_2\right[{\int }_0^1\mathtt{Q}({\mathsf{\zeta }}_2,{\mathsf{\zeta }}_3)\varrho \left({\mathsf{\zeta }}_3\right){\mathtt{F}}_4\cdots \hfill \\ & {\mathtt{F}}_{\mathfrak{d}-1}\left[{\int }_0^1\mathtt{Q}({\mathsf{\zeta }}_{\mathfrak{d}-1},{\mathsf{\zeta }}_{\mathfrak{d}})\varrho \left({\mathsf{\zeta }}_{\mathfrak{d}}\right){\mathtt{F}}_{\mathfrak{d}}\left({\mathfrak{z}}_{{}_1}\left({\mathsf{\zeta }}_{\mathfrak{d}}\right)\right)\mathtt{d}{\mathsf{\zeta }}_{\mathfrak{d}}\right]\cdots \left]\mathtt{d}{\mathsf{\zeta }}_3\right]\mathtt{d}{\mathsf{\zeta }}_2]\mathtt{d}{\mathsf{\zeta }}_1\hfill \\ \hfill \ge & \underset{\widehat{\mathbf{r}}\in [0,1]}{min}{\int }_0^1\mathtt{Q}(\widehat{\mathbf{r}},{\mathsf{\zeta }}_1)\varrho \left({\mathsf{\zeta }}_1\right){\mathtt{F}}_1\left[{\int }_0^1\mathtt{Q}({\mathsf{\zeta }}_1,{\mathsf{\zeta }}_2)\varrho \left({\mathsf{\zeta }}_2\right){\mathtt{F}}_2\right[{\int }_0^1\mathtt{Q}({\mathsf{\zeta }}_2,{\mathsf{\zeta }}_3)\varrho \left({\mathsf{\zeta }}_3\right){\mathtt{F}}_4\cdots \hfill \\ & {\mathtt{F}}_{\mathfrak{d}-1}\left[{\int }_0^1\mathtt{Q}({\mathsf{\zeta }}_{\mathfrak{d}-1},{\mathsf{\zeta }}_{\mathfrak{d}})\varrho \left({\mathsf{\zeta }}_{\mathfrak{d}}\right){\mathtt{F}}_{\mathfrak{d}}\left({\mathfrak{z}}_{{}_1}\left({\mathsf{\zeta }}_{\mathfrak{d}}\right)\right)\mathtt{d}{\mathsf{\zeta }}_{\mathfrak{d}}\right]\cdots \left]\mathtt{d}{\mathsf{\zeta }}_3\right]\mathtt{d}{\mathsf{\zeta }}_2]\mathtt{d}{\mathsf{\zeta }}_1\hfill \\ \hfill \ge & \mathfrak{f}.\hfill \end{array}$$

Thus, assumption $\left(c\right)$ of Theorem 3 holds. Hence, by Theorem 3, there exist coupled positive solutions as mentioned in the hypothesis. □

The following theorems are for the cases, ${\displaystyle \sum _{i=1}^{\kappa }\frac{1}{{\mathtt{p}}_i}=1}$ and ${\displaystyle \sum _{i=1}^{\kappa }\frac{1}{{\mathtt{p}}_i}>1,}$ respectively:

Theorem 8. 

Suppose that $\left({\mathcal{J}}_1\right)$–$\left({\mathcal{J}}_2\right)$ hold, and there exist three positive real numbers $\mathfrak{f}<\mathfrak{g}<\mathfrak{h}$ with ${\mathtt{F}}_{\mathsf{\beta }}(\mathsf{\beta }=1,2,\cdots ,\mathfrak{d})$ satisfying $\left({\mathcal{J}}_7\right),$ $\left({\mathcal{J}}_9\right)$, and

$\left({\mathcal{J}}_{10}\right)$

${\mathtt{F}}_{\mathsf{\beta }}\left(\mathfrak{z}\right)<\frac{\mathfrak{g}}{{\mathsf{\beta }}_3}$, $0\le \mathfrak{z}\le \frac{\mathfrak{g}}{\mathsf{\sigma }\left(\mathsf{\xi }\right)},$

then the BVP (3) has coupled positive solutions $\left\{({}^1{\mathfrak{z}}_{{}_1},{}^1{\mathfrak{z}}_2,\cdots ,{}^1{\mathfrak{z}}_{\mathfrak{d}})\right\}$ and $\left\{\right({}^2{\mathfrak{z}}_{{}_1},{}^2{\mathfrak{z}}_2,\cdots ,$ ${}^2{\mathfrak{z}}_{\mathfrak{d}}\left)\right\}$ satisfying

$$\mathfrak{f}<{\mathsf{\gamma }}_1\left({}^1{\mathfrak{z}}_{{}_{\mathsf{\beta }}}\right)\mathit{with}{\mathsf{\gamma }}_3\left({}^1{\mathfrak{z}}_{{}_{\mathsf{\beta }}}\right)<\mathfrak{g},\mathsf{\beta }=1,2,\cdots ,\mathfrak{d}$$

and

$$\mathfrak{g}<{\mathsf{\gamma }}_3\left({}^2{\mathfrak{z}}_{{}_{\mathsf{\beta }}}\right)\mathit{with}{\mathsf{\gamma }}_2\left({}^2{\mathfrak{z}}_{{}_{\mathsf{\beta }}}\right)<\mathfrak{h},\mathsf{\beta }=1,2,\cdots ,\mathfrak{d}.$$

Proof. 

The proof is similar to the proof of Theorem 7; therefore, we omit the details here. □

Theorem 9. 

Suppose that $\left({\mathcal{J}}_1\right)$–$\left({\mathcal{J}}_3\right)$ hold, and there exist three positive real numbers $0<\mathfrak{f}<\mathfrak{g}<\mathfrak{h}$ with ${\mathtt{F}}_{\mathsf{\beta }}(\mathsf{\beta }=1,2,\cdots ,\mathfrak{d})$ satisfying $\left({\mathcal{J}}_7\right),$ $\left({\mathcal{J}}_9\right)$ and

$\left({\mathcal{J}}_{11}\right)$

${\mathtt{F}}_{\mathsf{\beta }}\left(\mathfrak{z}\right)<\frac{\mathfrak{g}}{{\mathsf{\beta }}_4}$, $0\le \mathfrak{z}\le \frac{\mathfrak{g}}{\mathsf{\sigma }\left(\mathsf{\xi }\right)},$

then the BVP (3) has coupled positive solutions $\left\{({}^1{\mathfrak{z}}_{{}_1},{}^1{\mathfrak{z}}_2,\cdots ,{}^1{\mathfrak{z}}_{\mathfrak{d}})\right\}$ and $\left\{\right({}^2{\mathfrak{z}}_{{}_1},{}^2{\mathfrak{z}}_2,\cdots ,$ ${}^2{\mathfrak{z}}_{\mathfrak{d}}\left)\right\}$ satisfying

$$\mathfrak{f}<{\mathsf{\gamma }}_1\left({}^1{\mathfrak{z}}_{{}_{\mathsf{\beta }}}\right)\mathit{with}{\mathsf{\gamma }}_3\left({}^1{\mathfrak{z}}_{{}_{\mathsf{\beta }}}\right)<\mathfrak{g},\mathsf{\beta }=1,2,\cdots ,\mathfrak{d}$$

and

$$\mathfrak{g}<{\mathsf{\gamma }}_3\left({}^2{\mathfrak{z}}_{{}_{\mathsf{\beta }}}\right)\mathit{with}{\mathsf{\gamma }}_2\left({}^2{\mathfrak{z}}_{{}_{\mathsf{\beta }}}\right)<\mathfrak{h},\mathsf{\beta }=1,2,\cdots ,\mathfrak{d}.$$

Proof. 

The proof is similar to the proof of Theorem 7; therefore, we omit the details here. □

Example 2. 

Consider the problem

$$▵{\mathfrak{z}}_{{}_{\mathsf{\beta }}}-\frac{{(\mathrm{N}-2)}^2{r}_0^{2\mathrm{N}-2}}{{\left|\mathsf{\nu }\right|}^{2\mathrm{N}-2}}{\mathfrak{z}}_{{}_{\mathsf{\beta }}}+\varrho \left(\right|\mathsf{\nu }\left|\right){\mathtt{F}}_{\mathsf{\beta }}\left({\mathfrak{z}}_{{}_{\mathsf{\beta }+1}}\right)=0,1<\left|\mathsf{\nu }\right|<3,$$

(9)

$${\mathfrak{z}}_{{}_{\mathsf{\beta }}}\left(0\right)={\mathfrak{z}}_{{}_{\mathsf{\beta }}}\left(1\right)=0,$$

(10)

where $r_0=1,$ $\mathrm{N}=3,$ $\mathsf{\beta }\in \{1,2\},{\mathfrak{z}}_{{}_3}={\mathfrak{z}}_{{}_1},$ $\varrho \left(\widehat{\mathbf{r}}\right)=\frac{1}{{\widehat{\mathbf{r}}}^4}{\prod }_{i=1}^2{\varrho }_i\left(\widehat{\mathbf{r}}\right),$ ${\varrho }_i\left(\widehat{\mathbf{r}}\right)={\varrho }_i\left(\frac{1}{\widehat{\mathbf{r}}}\right),$ in which ${\varrho }_1\left(\widehat{\mathbf{r}}\right)=\frac{1}{\widehat{\mathbf{r}}+2}$ and ${\varrho }_2\left(\widehat{\mathbf{r}}\right)=\frac{3}{{\widehat{\mathbf{r}}}^2+1},$ then ${\varrho }_1,{\varrho }_2\in L^{\mathtt{p}}[0,1],{\prod }_{i=1}^2{\varrho }_i^{*}=\frac{1}{2},$ and $\mathsf{\sigma }\left(\mathsf{\xi }\right)=\frac{sinh\left(\mathsf{\xi }\right)}{sinh\left(1\right)}=\frac{sinh\left(\frac{1}{3}\right)}{sinh\left(1\right)}=0.2889212153.$ In addition,

$$\begin{array}{c}\hfill {\mathsf{\beta }}_1=\frac{\mathsf{\sigma }\left(\mathsf{\xi }\right)r_0^2}{{(\mathrm{N}-2)}^2}\prod _{i=1}^{\kappa }{\varrho }_i^{★}{\int }_{\mathsf{\xi }}^{1-\mathsf{\xi }}\mathtt{Q}(\mathsf{\zeta },\mathsf{\zeta }){\mathsf{\zeta }}^{\frac{2(\mathrm{N}-1)}{2-\mathrm{N}}}\mathtt{d}\mathsf{\zeta }\approx 0.1704829453.\end{array}$$

Let ${\mathtt{p}}_1=6,{\mathtt{p}}_2=3$ and $\mathtt{q}=2,$ then $\frac{1}{{\mathtt{p}}_1}+\frac{1}{{\mathtt{p}}_2}+\frac{1}{\mathtt{q}}=1$ and

$$\begin{array}{c}\hfill {\mathsf{\beta }}_2=\frac{r_0^2}{{(\mathrm{N}-2)}^2}\parallel \widehat{\mathtt{Q}}{\parallel }_{\mathtt{q}}\prod _{i=1}^{\kappa }{\parallel {\varrho }_i\parallel }_{{\mathtt{p}}_i}\approx 0.1255931381.\end{array}$$

Let

$$\begin{array}{c}\hfill {\mathtt{F}}_1\left(\mathfrak{z}\right)={\mathtt{F}}_2\left(\mathfrak{z}\right)=\left\{\begin{array}{c}3.9,\mathfrak{z}\le 1.8,\hfill \\ 3.9{(\mathfrak{z}-0.8)}^2+\mathfrak{z}-1.8,\mathfrak{z}>1.8.\hfill \end{array}\right.\end{array}$$

Choose $\mathfrak{f}=\frac{1}{3},\mathfrak{g}=\frac{1}{2}$ and $\mathfrak{h}=\frac{3}{5}.$ Then,

$$\begin{array}{cc}\hfill {\mathtt{F}}_1\left(\mathfrak{z}\right)={\mathtt{F}}_2\left(\mathfrak{z}\right)& \ge 3.519413622=\frac{\mathfrak{h}}{{\mathsf{\beta }}_1},\mathfrak{z}\in \left[\frac{3}{5},3.461\times \frac{3}{5}\right],\hfill \\ \hfill {\mathtt{F}}_1\left(\mathfrak{z}\right)={\mathtt{F}}_2\left(\mathfrak{z}\right)& \le 3.981109220=\frac{\mathfrak{g}}{{\mathsf{\beta }}_2},\mathfrak{z}\in \left[0,3.461\times \frac{1}{2}\right],\hfill \\ \hfill {\mathtt{F}}_1\left(\mathfrak{z}\right)={\mathtt{F}}_2\left(\mathfrak{z}\right)& \ge 1.955229790=\frac{\mathfrak{f}}{{\mathsf{\beta }}_1},\mathfrak{z}\in \left[\frac{1}{3},3.461\times \frac{1}{3}\right].\hfill \end{array}$$

Hence, by an application of Theorem 4, the BVP (9) and (10) has coupled positive solutions $({}^{\mathsf{\beta }}{\mathfrak{z}}_{{}_1},{}^{\mathsf{\beta }}{\mathfrak{z}}_2),\mathsf{\beta }=1,2$, such that

$$\frac{1}{3}<\underset{\widehat{\mathbf{r}}\in [0,1]}{max}{}^{\mathsf{\beta }}{\mathfrak{z}}_{{}_1}\left(\widehat{\mathbf{r}}\right)\mathit{with}\underset{\widehat{\mathbf{r}}\in [0,1]}{max}{}^{\mathsf{\beta }}{\mathfrak{z}}_{{}_1}\left(\widehat{\mathbf{r}}\right)<\frac{1}{2},\mathit{for}\mathsf{\beta }=1,2,$$

$$\frac{1}{2}<\underset{\widehat{\mathbf{r}}\in [0,1]}{max}{}^{\mathsf{\beta }}{\mathfrak{z}}_2\left(\widehat{\mathbf{r}}\right)\mathit{with}\underset{\widehat{\mathbf{r}}\in [0,1]}{min}{}^{\mathsf{\beta }}{\mathfrak{z}}_2\left(\widehat{\mathbf{r}}\right)<\frac{3}{5},\mathit{for}\mathsf{\beta }=1,2.$$

5. Uniqueness of Positive Radial Solution

We use two metrics, in accordance with Rus’ theorem [31,32], in this part, to test if there is a unique positive solution to the BVP (3). Consider the collection of continuous, real-valued functions defined on $[0,1]$: this space is symbolised by the letter $\mathtt{X}.$ Take into account the below metrics on $\mathtt{X},$ for functions $\mathfrak{y},\mathfrak{z}\in \mathtt{X}:$

$$\mathtt{d}(\mathfrak{y},\mathfrak{z})=\underset{\widehat{\mathbf{r}}\in [0,1]}{max}|\mathfrak{y}\left(\widehat{\mathbf{r}}\right)-\mathfrak{z}\left(\widehat{\mathbf{r}}\right)|;$$

(11)

$$\mathsf{\rho }(\mathfrak{y},\mathfrak{z})={\left[{\int }_0^1{|\mathfrak{y}\left(\widehat{\mathbf{r}}\right)-\mathfrak{z}\left(\widehat{\mathbf{r}}\right)|}^{\mathtt{p}}\mathtt{d}\widehat{\mathbf{r}}\right]}^{\frac{1}{\mathtt{p}}},\mathtt{p}>1.$$

(12)

The combination $(\mathtt{X},\mathtt{d})$ creates a complete metric space for $\mathtt{d}$ in (11). Then, $(\mathtt{X},\mathsf{\rho })$ constitutes a metric space for the value of $\mathsf{\rho }$ in (12). The equation expressing the connection between the two measures on $\mathtt{X}$ is

$$\mathsf{\rho }(\mathfrak{y},\mathfrak{z})\le \mathtt{d}(\mathfrak{y},\mathfrak{z})\mathrm{for}\mathrm{all}\mathfrak{y},\mathfrak{z}\in \mathtt{X}.$$

(13)

Theorem 10 (Rus [32]). 

Let $\mathtt{F}:\mathtt{X}\to \mathtt{X}$ be a continuous with respect to $\mathtt{d}$ on $\mathtt{X}$ and

$$\mathtt{d}(\mathtt{F}\mathfrak{y},\mathtt{F}\mathfrak{z})\le {\alpha }_1\mathsf{\rho }(\mathfrak{y},\mathfrak{z}),$$

(14)

for some ${\alpha }_1>0$ and for all $\mathfrak{y},\mathfrak{z}\in \mathtt{X},$

$$\mathsf{\rho }(\mathtt{F}\mathfrak{y},\mathtt{F}\mathfrak{z})\le {\alpha }_2\mathsf{\rho }(\mathfrak{y},\mathfrak{z}),$$

(15)

for some $0<{\alpha }_2<1$ for all $\mathfrak{y},\mathfrak{z}\in \mathtt{X},$ then there is a unique ${\mathfrak{y}}^{*}\in \mathtt{X}$ such that $\mathtt{F}{\mathfrak{y}}^{*}={\mathfrak{y}}^{*}.$

Denote $\mathsf{{\rm Y}}\left(\mathsf{\zeta }\right)=\mathtt{Q}(\mathsf{\zeta },\mathsf{\zeta }){\mathsf{\zeta }}^{\frac{2(\mathrm{N}-1)}{2-\mathrm{N}}}{\prod }_{i=1}^{\kappa }{\varrho }_i\left(\mathsf{\zeta }\right).$

Theorem 11. 

Suppose that $\left({\mathcal{J}}_1\right)$ and $\left({\mathcal{J}}_2\right)$ and the following

$\left({\mathcal{J}}_{12}\right)$

$|{\mathtt{F}}_{\mathsf{\beta }}\left(\mathfrak{z}\right)-{\mathtt{F}}_{\mathsf{\beta }}\left(\mathfrak{y}\right)|\le \mathtt{K}|\mathfrak{z}-\mathtt{y}|\mathit{for}\mathfrak{z},\mathfrak{y}\in \mathtt{X},$ for some $\mathtt{K}>0$

are satisfied. Furthermore, there are two real numbers $\mathtt{p}>1,\mathtt{q}>1$ satisfying $\frac{1}{\mathtt{p}}+\frac{1}{\mathtt{q}}=1$, and the following holds:

$${\left[\frac{\mathsf{\sigma }\left(\mathsf{\xi }\right)\mathtt{K}{r}_0^2}{{(\mathrm{N}-2)}^2}\right]}^{\mathfrak{d}+1}{\left[{\int }_0^1\left|\mathsf{{\rm Y}}\left(\mathsf{\zeta }\right)\right|\mathtt{d}\mathsf{\zeta }\right]}^{\mathfrak{d}}{\left[{\int }_0^1{\left|\mathsf{{\rm Y}}\left(\mathsf{\zeta }\right)\right|}^{\mathtt{q}}\mathtt{d}\mathsf{\zeta }\right]}^{\frac{1}{q}}<1;$$

(16)

then the BVP (3) has a unique positive solution in $\mathtt{X}.$

Proof. 

Let ${\mathfrak{z}}_{{}_1},{\mathfrak{y}}_1\in \mathtt{X}$ and ${\mathsf{\zeta }}_{n-1}\in [0,1].$ The Hölder’s inequality gives

$$\begin{array}{cc}\hfill |{\int }_0^1& \mathtt{Q}({\mathsf{\zeta }}_{\mathfrak{d}-1},{\mathsf{\zeta }}_{\mathfrak{d}})\varrho \left({\mathsf{\zeta }}_{\mathfrak{d}}\right){\mathtt{F}}_{\mathfrak{d}}\left({\mathfrak{z}}_{{}_1}\left({\mathsf{\zeta }}_{\mathfrak{d}}\right)\right)\mathtt{d}{\mathsf{\zeta }}_{\mathfrak{d}}-{\int }_0^1\mathtt{Q}({\mathsf{\zeta }}_{\mathfrak{d}-1},{\mathsf{\zeta }}_{\mathfrak{d}})\varrho \left({\mathsf{\zeta }}_{\mathfrak{d}}\right){\mathtt{F}}_{\mathfrak{d}}\left({\mathfrak{y}}_1\left({\mathsf{\zeta }}_{\mathfrak{d}}\right)\right)\mathtt{d}{\mathsf{\zeta }}_{\mathfrak{d}}|\hfill \\ & \le {\int }_0^1|\mathtt{Q}({\mathsf{\zeta }}_{\mathfrak{d}-1},{\mathsf{\zeta }}_{\mathfrak{d}})\varrho \left({\mathsf{\zeta }}_{\mathfrak{d}}\right)\left|\right|{\mathtt{F}}_{\mathfrak{d}}\left({\mathfrak{z}}_{{}_1}\left({\mathsf{\zeta }}_{\mathfrak{d}}\right)\right)-{\mathtt{F}}_{\mathfrak{d}}\left({\mathfrak{y}}_1\left({\mathsf{\zeta }}_{\mathfrak{d}}\right)\right)|\mathtt{d}{\mathsf{\zeta }}_{\mathfrak{d}}\hfill \\ & \le {\int }_0^1|\mathtt{Q}({\mathsf{\zeta }}_{\mathfrak{d}},{\mathsf{\zeta }}_{\mathfrak{d}})\varrho \left({\mathsf{\zeta }}_{\mathfrak{d}}\right)\left|\mathtt{K}\right|{\mathfrak{z}}_{{}_1}\left({\mathsf{\zeta }}_{\mathfrak{d}}\right)-{\mathfrak{y}}_1\left({\mathsf{\zeta }}_{\mathfrak{d}}\right)|\mathtt{d}{\mathsf{\zeta }}_{\mathfrak{d}}\le \frac{\mathtt{K}{r}_0^2}{{(\mathfrak{d}-2)}^2}{\int }_0^1|\mathsf{{\rm Y}}\left({\mathsf{\zeta }}_{\mathfrak{d}}\right)\left|\right|{\mathfrak{z}}_{{}_1}\left({\mathsf{\zeta }}_{\mathfrak{d}}\right)-{\mathfrak{y}}_1\left({\mathsf{\zeta }}_{\mathfrak{d}}\right)|\mathtt{d}{\mathsf{\zeta }}_{\mathfrak{d}}\hfill \\ & \le \frac{\mathtt{K}{r}_0^2}{{(\mathrm{N}-2)}^2}{\left[{\int }_0^1{\left|\mathsf{{\rm Y}}\left({\mathsf{\zeta }}_{\mathfrak{d}}\right)\right|}^{\mathtt{q}}\mathtt{d}{\mathsf{\zeta }}_{\mathfrak{d}}\right]}^{\frac{1}{\mathtt{q}}}{\left[{\int }_0^1{|{\mathfrak{z}}_{{}_1}\left({\mathsf{\zeta }}_{\mathfrak{d}}\right)-{\mathfrak{y}}_1\left({\mathsf{\zeta }}_{\mathfrak{d}}\right)|}^{\mathtt{p}}\mathtt{d}{\mathsf{\zeta }}_{\mathfrak{d}}\right]}^{\frac{1}{\mathtt{p}}}\hfill \\ & \le \frac{\mathtt{K}{r}_0^2}{{(\mathrm{N}-2)}^2}{\left[{\int }_0^1{\left|\mathsf{{\rm Y}}\left({\mathsf{\zeta }}_{\mathfrak{d}}\right)\right|}^{\mathtt{q}}\mathtt{d}{\mathsf{\zeta }}_{\mathfrak{d}}\right]}^{\frac{1}{\mathtt{q}}}\mathsf{\rho }({\mathfrak{z}}_{{}_1},{\mathfrak{y}}_1)\le {\alpha }_1^{★}\mathsf{\rho }({\mathfrak{z}}_{{}_1},{\mathfrak{y}}_1),\hfill \end{array}$$

where

$${\alpha }_1^{★}=\frac{\mathtt{K}{r}_0^2}{{(\mathrm{N}-2)}^2}{\left[{\int }_0^1{\left|\mathsf{{\rm Y}}\left({\mathsf{\zeta }}_{\mathfrak{d}}\right)\right|}^{\mathtt{q}}\mathtt{d}\mathsf{\zeta }\right]}^{\frac{1}{\mathtt{q}}}.$$

Similarly, for $0<{\mathsf{\zeta }}_{\mathfrak{d}-2}<1,$ we obtain

$$\begin{array}{cc}\hfill |{\int }_0^1& \mathtt{Q}({\mathsf{\zeta }}_{\mathfrak{d}-2},{\mathsf{\zeta }}_{\mathfrak{d}-1})\varrho \left({\mathsf{\zeta }}_{\mathfrak{d}-1}\right){\mathtt{F}}_{\mathfrak{d}-1}\left[{\int }_0^1\mathtt{Q}({\mathsf{\zeta }}_{\mathfrak{d}-1},{\mathsf{\zeta }}_{\mathfrak{d}})\varrho \left({\mathsf{\zeta }}_{\mathfrak{d}}\right){\mathtt{F}}_{\mathfrak{d}}\left({\mathfrak{z}}_{{}_1}\left({\mathsf{\zeta }}_{\mathfrak{d}}\right)\right)\mathtt{d}{\mathsf{\zeta }}_{\mathfrak{d}}\right]\mathtt{d}{\mathsf{\zeta }}_{\mathfrak{d}-1}\hfill \\ & -{\int }_0^1\mathtt{Q}({\mathsf{\zeta }}_{\mathfrak{d}-2},{\mathsf{\zeta }}_{\mathfrak{d}-1})\varrho \left({\mathsf{\zeta }}_{\mathfrak{d}-1}\right){\mathtt{F}}_{\mathfrak{d}-1}\left[{\int }_0^1\mathtt{Q}({\mathsf{\zeta }}_{\mathfrak{d}-1},{\mathsf{\zeta }}_{\mathfrak{d}})\varrho \left({\mathsf{\zeta }}_{\mathfrak{d}}\right){\mathtt{F}}_{\mathfrak{d}}\left({\mathfrak{y}}_1\left({\mathsf{\zeta }}_{\mathfrak{d}}\right)\right)\mathtt{d}{\mathsf{\zeta }}_{\mathfrak{d}}\right]\mathtt{d}{\mathsf{\zeta }}_{\mathfrak{d}-1}|\hfill \\ & \le \frac{\mathtt{K}{r}_0^2}{{(\mathrm{N}-2)}^2}{\int }_0^1\left|\mathsf{{\rm Y}}\left({\mathsf{\zeta }}_{\mathfrak{d}-1}\right)\right|{\alpha }_1\mathsf{\rho }({\mathfrak{z}}_{{}_1},{\mathfrak{y}}_1)\mathtt{d}{\mathsf{\zeta }}_{\mathfrak{d}-1}\le {\widehat{\alpha }}_1{\alpha }_1^{★}\mathsf{\rho }({\mathfrak{z}}_{{}_1},{\mathfrak{y}}_1),\hfill \end{array}$$

where

$${\widehat{\alpha }}_1=\frac{\mathtt{K}{r}_0^2}{{(\mathrm{N}-2)}^2}{\int }_0^1\left|\mathsf{{\rm Y}}\left(\mathsf{\zeta }\right)\right|\mathtt{d}\mathsf{\zeta }.$$

Thus, we have

$$\left|\mathtt{F}{\mathfrak{z}}_{{}_1}\left(\mathsf{\zeta }\right)-\mathtt{F}{\mathfrak{y}}_1\left(\mathsf{\zeta }\right)\right|\le {\widehat{\alpha }}_1^{\mathfrak{d}}{\alpha }_1^{★}\mathsf{\rho }({\mathfrak{z}}_{{}_1},{\mathfrak{y}}_1);$$

that is,

$$\mathtt{d}(\mathtt{F}{\mathfrak{z}}_{{}_1},\mathtt{F}{\mathfrak{y}}_1)\le {\alpha }_1\mathsf{\rho }({\mathfrak{z}}_{{}_1},{\mathfrak{y}}_1),$$

(17)

for some ${\alpha }_1={\widehat{\alpha }}_1^{\mathfrak{d}}{\alpha }_1^{★}>0$ for all ${\mathfrak{z}}_{{}_1},{\mathfrak{y}}_1\in \mathtt{X},$ this proves (14). Next, let ${\mathfrak{z}}_{{}_1},{\mathfrak{y}}_1\in \mathtt{X},$ and from (13) and (17), we obtain

$$\mathtt{d}(\mathtt{F}{\mathfrak{z}}_{{}_1},\mathtt{F}{\mathfrak{y}}_1)\le {\alpha }_1\mathsf{\rho }({\mathfrak{z}}_{{}_1},{\mathfrak{y}}_1)\le {\alpha }_1\mathtt{d}({\mathfrak{z}}_{{}_1},{\mathfrak{y}}_1).$$

Thus, for $\epsilon >0,$ select $\mathsf{\eta }=\epsilon /{\alpha }_1,$ we obtain $\mathtt{d}(\mathtt{F}{\mathfrak{z}}_{{}_1},\mathtt{F}{\mathfrak{y}}_1)<\epsilon ,$ whenever $\mathtt{d}({\mathfrak{z}}_{{}_1},{\mathfrak{y}}_1)<\mathsf{\eta },$ which shows that $\mathtt{F}$ is continuous on $\mathtt{X}$ with metric $\mathtt{d}.$ It remains to be shown that $\mathtt{F}$ is contractive on $\mathtt{X}$ with metric $\mathsf{\rho }.$ For each ${\mathfrak{z}}_{{}_1},{\mathfrak{y}}_1\in \mathtt{X},$ and from (17), we have

$$\begin{array}{cc}\hfill {\left[{\int }_0^1{|\left(\mathtt{F}{\mathfrak{z}}_{{}_1}\right)\left(\mathsf{\zeta }\right)-\left(\mathtt{F}{\mathfrak{y}}_1\right)\left(\mathsf{\zeta }\right)|}^{\mathtt{p}}\mathtt{d}\mathsf{\zeta }\right]}^{\frac{1}{\mathtt{p}}}& \le {\left[{\int }_0^1{\left|{\widehat{\alpha }}_1^{\mathfrak{d}}{\alpha }_1^{★}\mathsf{\rho }({\mathfrak{z}}_{{}_1},{\mathfrak{y}}_1)\right|}^{\mathtt{p}}\mathtt{d}\mathsf{\zeta }\right]}^{\frac{1}{\mathtt{p}}}\hfill \\ & \le {\left[\frac{\mathtt{K}{r}_0^2}{{(\mathrm{N}-2)}^2}\right]}^{\mathfrak{d}+1}{\left[{\int }_0^1\left|\mathsf{{\rm Y}}\left(\mathsf{\zeta }\right)\right|\mathtt{d}\mathsf{\zeta }\right]}^{\mathfrak{d}}{\left[{\int }_0^1{\left|\mathsf{{\rm Y}}\left(\mathsf{\zeta }\right)\right|}^{\mathtt{q}}\mathtt{d}\mathsf{\zeta }\right]}^{\frac{1}{q}}\mathsf{\rho }({\mathfrak{z}}_{{}_1},{\mathfrak{y}}_1);\hfill \end{array}$$

that is

$$\begin{array}{c}\hfill \mathsf{\rho }(\mathtt{F}{\mathfrak{z}}_{{}_1},\mathtt{F}{\mathfrak{y}}_1)\le {\left[\frac{\mathtt{K}{r}_0^2}{{(\mathrm{N}-2)}^2}\right]}^{\mathfrak{d}+1}{\left[{\int }_0^1\left|\mathsf{{\rm Y}}\left(\mathsf{\zeta }\right)\right|\mathtt{d}\mathsf{\zeta }\right]}^{\mathfrak{d}}{\left[{\int }_0^1{\left|\mathsf{{\rm Y}}\left(\mathsf{\zeta }\right)\right|}^{\mathtt{q}}\mathtt{d}\mathsf{\zeta }\right]}^{\frac{1}{q}}\mathsf{\rho }({\mathfrak{z}}_{{}_1},{\mathfrak{y}}_1).\end{array}$$

From assumption (16), we have

$$\mathsf{\rho }(\mathtt{F}{\mathfrak{z}}_{{}_1},\mathtt{F}{\mathfrak{y}}_1)\le {\alpha }_2\mathsf{\rho }({\mathfrak{z}}_{{}_1},{\mathfrak{y}}_1)$$

for some ${\alpha }_2<1$ and all ${\mathfrak{z}}_{{}_1},{\mathfrak{y}}_1\in \mathtt{X}.$ It follows from Theorem 10 that $\mathtt{F}$ has a unique fixed point in $\mathtt{X}.$ Moreover, from Lemma 3, $\mathtt{F}$ is positive. Hence, the BVP (1) has a unique positive solution. □

Example 3. 

Consider the problem,

$$▵{\mathfrak{z}}_{{}_{\mathsf{\beta }}}-\frac{{(\mathrm{N}-2)}^2{r}_0^{2\mathrm{N}-2}}{{\left|\mathsf{\nu }\right|}^{2\mathrm{N}-2}}{\mathfrak{z}}_{{}_{\mathsf{\beta }}}+\varrho \left(\right|\mathsf{\nu }\left|\right){\mathtt{F}}_{\mathsf{\beta }}({\mathfrak{z}}_{{}_{\mathsf{\beta }+1}})=0,1<\left|\mathsf{\nu }\right|<2,$$

(18)

$${\mathfrak{z}}_{{}_{\mathsf{\beta }}}\left(0\right)={\mathfrak{z}}_{{}_{\mathsf{\beta }}}\left(1\right)=0,$$

(19)

where $r_0=1,$ $\mathrm{N}=3,$ $\mathsf{\beta }\in \{1,2\},{\mathfrak{z}}_3={\mathfrak{z}}_{{}_1},$ $\varrho \left(\widehat{\mathbf{r}}\right)=\frac{1}{{\widehat{\mathbf{r}}}^4}{\prod }_{i=1}^2{\varrho }_i\left(\widehat{\mathbf{r}}\right),$ ${\varrho }_i\left(\widehat{\mathbf{r}}\right)={\varrho }_i\left(\frac{1}{\widehat{\mathbf{r}}}\right),$ in which ${\varrho }_1\left(\widehat{\mathbf{r}}\right)={\varrho }_2\left(\widehat{\mathbf{r}}\right)=\frac{{\widehat{\mathbf{r}}}^3}{\sqrt{\widehat{\mathbf{r}}+1}}$. Let ${\mathtt{F}}_1\left(\mathfrak{z}\right)=\frac{3}{2}sin\left(\mathfrak{z}\right)$ and ${\mathtt{F}}_2\left(\mathfrak{z}\right)=\frac{3}{2(\mathfrak{z}+1)};$ then,

$$|{\mathtt{F}}_1\left(\mathfrak{z}\right)-{\mathtt{F}}_1\left(\mathtt{y}\right)|=\frac{|sin(\mathfrak{z})-sin(\mathtt{y}\left)\right|}{{10}^3}\le \frac{3}{2}|\mathfrak{z}-\mathtt{y}|$$

and

$$|{\mathtt{F}}_2\left(\mathfrak{z}\right)-{\mathtt{F}}_2\left(\mathtt{y}\right)|=\frac{3}{2}\left|\frac{1}{\mathfrak{z}+1}-\frac{1}{\mathtt{y}+1}\right|\le \frac{3}{2}|\mathfrak{z}-\mathtt{y}|.$$

Thus, $\mathtt{K}=\frac{3}{2}.$ Let $\mathfrak{d}=2$ and $\mathtt{p}=\mathtt{q}=2;$ then,

$${\left[\frac{\mathtt{K}{r}_0^2}{{(\mathrm{N}-2)}^2}\right]}^{\mathfrak{d}+1}{\left[{\int }_0^1\left|\mathsf{{\rm Y}}\left(\mathsf{\zeta }\right)\right|\mathtt{d}\mathsf{\zeta }\right]}^{\mathfrak{d}}{\left[{\int }_0^1{\left|\mathsf{{\rm Y}}\left(\mathsf{\zeta }\right)\right|}^{\mathtt{q}}\mathtt{d}\mathsf{\zeta }\right]}^{\frac{1}{q}}\approx 0.1508078067<1.$$

Hence, as an application of Theorem 11, the BVP (18) and (19) has a unique positive radial solution.

6. Conclusions

In this paper, we developed a theory to study the existence of single and coupled positive radial solutions for a certain type of iterative system of nonlinear elliptic equations, by applying Krasnoselskii’s and Avery–Henderson’s fixed-point theorems in a Banach space. In the future, we will study the existence of positive radial solutions for an iterative system of elliptic equations with a logarithmic nonlinear term. In addition, we will study global existence and ground-state solutions to the addressed problem.