Existence and uniqueness of positive solution to a new class of nonlocal elliptic problem with parameter dependency

Abstract

In this paper, we prove that under weak assumptions on the reaction terms and diffusion coefficients, a positive solution exists for a one-dimensional case and a positive radial solution to a multidimensional case of a nonlocal elliptic problem. Additionally, we establish the uniqueness of the solution, with the fixed point theorem being the primary tool employed. Our results are new and generalize several existing results.

1 Introduction

We will establish the existence of both a positive solution and a positive radial solution to the following problems:

$$ \textstyle\begin{cases} \begin{aligned} - & \mathcal{A} \Big(t, \displaystyle{\int _{0}^{1}} g(u)dy \Big) \Delta u = \lambda f(t, u) \quad t \in (0,1), \\ &u'(0)= u(1) =0 , \end{aligned} \end{cases} $$

(1.1)

and

$$ \textstyle\begin{cases} \begin{aligned} - & \mathcal{A} \Big( |t|, \displaystyle{\int _{\varOmega}} g(u)dy \Big) \Delta u = \lambda f(|t|, u) \quad \text{in } \, \Omega , \\ u &=0 \quad \text{on}\, \partial \Omega , \end{aligned} \end{cases} $$

(1.2)

where $\Omega = \{t \in \mathbb{R}^{n} : a < |t| < b \}$ with $a < b$ and $n \geq 2$ and $\mathcal{A}: [a, b] \times [0, \infty ) \rightarrow (0, \infty )$ is a continuous function. Several researchers have studied the nonlocal elliptic problem and its applications in modeling various physical and biological systems. These include collections of particles in thermodynamic balance interacting through gravitational (Coulombic) potential (see [4, 5, 23, 24, 29, 36], thermal runaway in ohmic heating (see [8, 40]), one-dimensional fluid flow where the strain rate is directly related to the power of stress and influenced by a temperature-dependent function, and the population density of organisms (e.g., bacteria) inside a container (see [25]).

Many researchers are interested in studying and developing the following type of nonlocal elliptic problem with boundary conditions:

$$ - \mathcal{A} \left ( t, \int _{\Omega }g(u) \, dy \right ) \Delta u = \lambda f(t, u), \quad t \in \Omega . $$

In [1], the authors used dynamical methods to obtain a nontrivial solution, with the second term being $f(u) + f_{0}(t)$ on a smooth domain $\Omega \subset \mathbb{R}^{n}$, $n \geq 2$. In [35], where $g(u) = u^{p}$, bifurcation arguments were employed to prove the existence of positive solutions. The structure of the set of positive solutions heavily depends on the balance between the nonlocal and reaction terms. Furthermore, in the autonomous case of $\mathcal{A}$, as shown in [22], since g is a concave function, at least one positive solution was obtained by applying fixed point theory. Additionally, in [30], the existence and uniqueness of solutions were established for different values of λ with $g(u) = f(t,u) = e^{-u}$. The case $g(u) = u^{p}$ has received considerable attention, with the existence of a positive solution being proven using the Krasnoselskii fixed point theorem in [13]. In [16], fixed point theory was applied, and both papers introduced a sub-supersolution method, imposing various conditions on $\mathcal{A}$ and f. An analogous method with the fixed point index proved the uniqueness of a positive solution (see [37, 38]). In [10, 11], the authors applied the Schauder fixed point theorem to cases where f changes sign, and similar arguments in [7] proved the existence and multiplicity of positive solutions. In these papers [2, 14, 15, 18], researchers used the Bolzano theorem, bifurcation arguments, the Galerkin method, subcritical growth conditions, and variational techniques, respectively. Many authors have studied the following problem:

$$ \textstyle\begin{cases} - \Delta u = \lambda f(u) \quad \text{in}\, \Omega , \\ u=0\quad \text{on}\, \partial \Omega . \end{cases} $$

(1.3)

As detailed by Shivaji in [34], specific conditions were formulated to determine the uniqueness of positive solutions to (1.3). Dancer et al. also derived essential conditions for positive solutions of (1.3) in [17]. Additionally, Maya et al. utilized the sub-supersolution method to examine the existence of positive solutions [32]. Perera studied the quasilinear elliptic problem and employed variational arguments in [33]. In [9], the author focused on a broad bounded domain within $\mathbb{R}^{n}, n \geq 2$. Lin investigated the existence and multiplicity of positive radial solutions to (1.3) within the context of an annular domain in $\mathbb{R}^{N}$, where $N \geq 2$. They proved that there are at least two positive radial solutions for each λ in $(0, \lambda ^{*})$ and at least one solution when λ equals $\lambda ^{*}$ with a positive function f such that $\lim _{u \rightarrow \infty} \frac{f(u)}{u} = \infty$ and a specific value $\lambda ^{*} > 0$. Moreover, for $\lambda > \lambda ^{*}$, no solutions exist. Furthermore, if certain conditions are met, such as $f(0) = 0$, $\lim _{u \rightarrow 0} \frac{f(u)}{u} = 1$, and $uf'(u) > (1 + \varepsilon )f(u)$ for strictly positive u and ε, there is a variational solution to (1.3) in $(0, \lambda _{1})$, where $\lambda _{1}$ is the smallest eigenvalue of −Δ. Additionally, in the case of $f(0) = 0$, $\underset{u \rightarrow 0}{\lim} \frac{f(u)}{u} = 0$, and $\underset{u \rightarrow \infty}{\lim} \frac{f(u)}{u} = \infty$, there is at least one positive radial solution to (1.3) for $\lambda > 0$ (see [31]). Our results complete recent research on the existence of positive solutions with Dirichlet- or Neumann-type boundary conditions for local or nonlocal elliptic equations (see [3, 6, 12, 19, 20, 28, 39] and the references therein).

More precisely, we expand on the existence and uniqueness of solutions to the nonstationary problem previously established in [3, 13, 16]. To demonstrate the existence of a positive solution to the problem outlined in equation (1.1), a number of assumptions must be made. Let $g: [0, \infty ) \rightarrow [0, \infty )$ be a continuous function such that

$$ \alpha (u) \leq g(u) \leq \beta (u), $$

(1.4)

since $\alpha ,\beta : [0, \infty ) \to [0, \infty )$ are two nondecreasing continuous functions. The reaction term $f: [0,1] \times [0, \infty ) \rightarrow [0, \infty )$ is a continuous function,

$$ \mu (t) k(s) \leq f(t, s) \leq \eta (t) \varphi (s), $$

(1.5)

for some measurable functions $\mu , \eta : [0,1] \rightarrow [0, \infty )$, and increasing continuous functions $k,\varphi :[0, \infty )\rightarrow [0, \infty )$, $f(t, 0)=0$, and the diffusion coefficient $\mathcal{A}: [0,1] \times [0, \infty ) \rightarrow (0, \infty )$ is a continuous function,

$$ \xi (t) v(s) \leq \mathcal{A}(t, s) \leq \theta (t) w(s), $$

(1.6)

where $v,w: [0, \infty ) \to (0, \infty )$ are nondecreasing functions, $\xi ,\theta : [0,1] \to (0, \infty )$ are measurable functions, and $\lambda >0$ is a real number. For every $d \in (0,1)$, $s \in [0, \infty )$, we have

$$ \int _{0}^{1} \mu (s) \theta (s)^{-1} ds < \infty ,\, 0 < \int _{0}^{d}G(0,s) \mu (s)\theta (s)^{-1} ds< \infty , $$

(1.7)

denoting G as Green’s function,

$$ G(t,s)= \textstyle\begin{cases} \begin{aligned} & 1-t, \quad 0 \leq s \leq t \leq 1; \\ & 1-s, \quad 0 \leq t \leq s \leq 1. \end{aligned} \end{cases} $$

Assume that there exist two constants $C_{1}$, $C_{2}$ that verify

$$ \inf _{0 < s \leq C_{1}} \frac{\lambda k((1-d)s)}{sw(g(s))} \geq \left ( \int _{0}^{d} G(0, s) \frac{\mu (s)}{\theta (s)} ds \right )^{-1}, $$

(1.8)

and

$$ \sup _{s \geq C_{2}} \frac{\lambda \varphi (s)}{sv(dg((1-b)s))} < \left ( \int _{0}^{1} \frac{\eta (s)}{\xi (s)} ds \right )^{-1}. $$

(1.9)

The assumptions we will make to obtain the existence of radial positive solution to problem (1.2) are then stated. $\lambda > 0$, $g: [0, \infty ) \rightarrow [0, \infty )$ is a continuous function, the reaction term $f: [a, b] \times [0, \infty ) \rightarrow [0, \infty )$ is continuous with $f(\cdot , 0) = 0$, and the diffusion coefficient $\mathcal{A}: [a, b] \times [0, \infty ) \rightarrow (0, \infty )$ is continuous and satisfies

$$ \mu (x) v(s) \leq \mathcal{A}(x, s) \leq \theta (x) \eta (s), $$

(1.10)

where $v, \eta : [0, \infty ) \rightarrow (0, \infty )$ are nondecreasing measurable functions, and $\mu , \theta : [a, b] \rightarrow (0, \infty )$ are measurable functions. We set

$$ C = \frac{(ab)^{N-2}}{b^{N-2} - a^{N-2}}, \quad B = \frac{b^{N-2}}{b^{N-2} - a^{N-2}}. $$

(1.11)

Now, we are ready to state our main results

Theorem 1.1

Under assumptions (1.5)–(1.9), problem (1.1) has a positive solution.

Theorem 1.2

We get the uniqueness of a positive solution to the problem

$$ \textstyle\begin{cases} - \mathcal{A} \Big( \displaystyle{\int _{0}^{1}} g(u) d y \Big) u''= \lambda f(t,u) \quad t \in (0,1), \\ u'(0)=u(1)=0. \end{cases} $$

(1.12)

Under this assumption

$$\begin{aligned}& \exists \, M > 0, \forall s \in [0, \infty ):\quad \mathcal{A}(s) \geq M. \end{aligned}$$

(1.13)

$$\begin{aligned}& \Big( f(t, s_{1})- f(t, s_{2}) \Big)( s_{1} - s_{2}) \leq C | s_{1} - s_{2}|^{2}, \end{aligned}$$

(1.14)

and

$$ c_{1} c_{2}\| f(\cdot ,0) \|_{2} < ( M - \lambda C)^{2}, \quad \lambda C< M, $$

(1.15)

where $c_{1}$, $c_{2}$ are constants Lipschitz of $\mathcal{A}$, f respectively, and $\| f(\cdot ,0) \|_{2}= \Big( \int _{0}^{1} |f(t,0)|^{2}\Big)^{ \frac{1}{2}}$.

The paper is organized as follows. In Sect. 2, we prove Theorems 1.1 and 1.2. In Sect. 3, we introduce Theorems 3.1 and 3.2 that are dedicated to the radial case of problem (1.2), together with their proofs.

2 Proof of Theorems 1.1 and 1.2

Before proving the earlier theorems, we first need to establish the following theorem.

Theorem 2.1

(Guo-Krasnosel’skii)

[25] Let E be a Banach space, define a cone P on E, $\Omega _{1}$ and $\Omega _{2}$ two open bounded sets on E such that $0\in \Omega _{1}$, $\overline{\Omega _{1}} \subset \Omega _{2}$. Let $S:P \cap (\overline{\Omega _{2}} \setminus \Omega _{1} ) \longrightarrow P$ be a completely continuous operator, such that one of the following conditions is satisfied:

• 1. $\| Sx \| \leq \| x \|$ for all $x \in P \cap \partial \Omega _{1}$ and $\| Sx \| \geq \| x \|$ for all $x \in P \cap \partial \Omega _{2}$.

• 2. $\| Sx \| \geq \| x \|$ for all $x \in P \cap \partial \Omega _{1}$ and $\| Sx \| \leq \| x \|$ for all $x \in P \cap \partial \Omega _{2}$.

Then, S admits at least one fixed point in $P \cap ( \overline{\Omega _{2}} \setminus \Omega _{1} )$.

2.1 Proof of Theorem 1.1

Consider the following problem

$$ \textstyle\begin{cases} -u''(t)= \psi (t) \qquad t \in (0, 1), \\ u'(0)= u(1)=0, \end{cases} $$

(2.1)

which has a unique solution, given by

$$ \forall t \in [0, 1]:\quad u(t) = \int _{0}^{1} G(t, s) \psi (s) ds, $$

where $\psi : [0, 1] \to \mathbb{R}$ is a continuous function. Indeed, $u \in C([0, 1], \mathbb{R})$ is a solution to (2.1), which implies that u is a solution to (1.1):

$$ \forall t \in [0, 1]:\quad u(t) = \int _{0}^{1} G(t, s) \dfrac{\lambda f(s,u(s))}{ \mathcal{A} \Big(s, \displaystyle{\int _{0}^{1}} g(u) d y \Big)}ds. $$

(2.2)

By replacing ψ with the function

$$ s \mapsto \frac{\lambda f(s,u(s))}{ \mathcal{A}\Big(s, \int _{0}^{1} g(u) \, dy \Big)}, $$

if u satisfies (2.2), then $u \in C^{2}([0, 1],\mathbb{R})$ and

$$ \forall t \in [0, 1]:\quad u'(t)= - \lambda \int _{0}^{t} \frac{ f(s,u(s))}{ \mathcal{A}\Big(s, \int _{0}^{1} g(u) \, dy \Big)} \, ds, \quad u''(t)= -\lambda \frac{ f(s,u(s))}{ \mathcal{A}\Big(s, \int _{0}^{1} g(u) \, dy \Big)}. $$

Since $u'(0) = u(1) = 0$, it follows that u is a solution to (2.1).

Define the following map

$$ \begin{aligned} S : C([0, 1], \mathbb{R}) & \rightarrow C([0, 1], \mathbb{R}) \\ u & \mapsto S u, \end{aligned} $$

where

$$ \forall t \in [0,1]: \quad Su(t) = \int _{0}^{1} G(t, s) \frac{\lambda f(s,u(s))}{ \mathcal{A} \Big(s, \int _{0}^{1} g(u) \, dy \Big)} \, ds. $$

We seek a positive fixed point of the map S by applying Theorem 2.1. Let $E = C([0, 1], \mathbb{R})$ and $F = \{ u \in E \mid u(t) \geq 0 \}$. The map $S : (F, \| \cdot \|_{\infty}) \to (F, \| \cdot \|_{\infty})$ is completely continuous. We prove that S is continuous.

Let $(u_{j})_{j \in \mathbb{N}} \subset F$ be a sequence such that

$$ \lim _{j \rightarrow \infty} \|u_{j}-u\|_{\infty}=0. $$

(2.3)

By (2.3), there exists $c > 0$ such that

$$ \forall j \in \mathbb{N}: \quad u_{j}, u \in [0, c]. $$

(2.4)

Since $0 \leq G \leq 1$, we have

$$ 0 \leq \| Su_{j} - Su \|_{\infty }\leq \int _{0}^{1} \left | \frac{\lambda f(s,u_{j}(s))}{\mathcal{A} \Big(s, \int _{0}^{1} g(u_{j}) \, dy \Big)} - \frac{\lambda f(s,u(s))}{\mathcal{A} \Big(s, \int _{0}^{1} g(u) \, dy \Big)} \right | ds. $$

(2.5)

From the uniform continuity of g on $[0, c]$, (2.3), and (2.4), we obtain

$$ \lim _{j \rightarrow \infty} \int _{0}^{1} g(u_{j}) \, dy = \int _{0}^{1} g(u) \, dy. $$

(2.6)

According to (2.3)–(2.4), (2.6), and the continuity of f, $\mathcal{A}$, and g, we obtain

$$ \forall s \in [0,1]:\quad \lim _{j\rightarrow +\infty} \frac{\lambda f ( s,u_{j})}{ \mathcal{A} \left (s, \int _{0}^{1}g(u_{j}) dy \right ) } = \frac{\lambda f (s,u(s))}{ \mathcal{A}\left (s,\int _{0}^{1}g(u) dy \right ) }, $$

(2.7)

and

$$ \exists \, m > 0, \forall j \in \mathbb{N}, \forall s \in [0,1]: \quad \frac{\lambda f (s,u_{j}(s))}{ \mathcal{A}\left (s,\int _{0}^{1}g(u_{j}) dy \right ) } \leq m. $$

(2.8)

Using the Lebesgue dominated convergence theorem and (2.5)–(2.8), we deduce

$$ \lim _{j \rightarrow +\infty}\left \|Su_{j}\right \|_{\infty}= \|S u \|_{\infty}. $$

So, S is continuous. Next, we claim the compactness of S. Let $M \subset F$ be a bounded set, denote

$$ W=\{S u ; u \in M\},\qquad W(t)=\{S u(t) ; u \in M\}. $$

For all $t_{1}, t_{2} \in [0, 1]$ and $u \in M$, we have

$$ \left | S u\left (t_{1}\right )-S u \left (t_{2}\right )\right | \leq \int _{0}^{1} \left | G(t_{1}, s)-G\left (t_{2}, s\right ) \right | \frac{\lambda f ( s,u(s))}{\mathcal{A} \left (s,\int _{0}^{1}g(u)dy \right )} d s. $$

From the continuity of f, $\mathcal{A}$, and g, we get that

$$ s \mapsto \frac{\lambda f ( s,u(s))}{\mathcal{A} \left (s,\int _{0}^{1}g(u)dy \right )} $$

is bounded independently of u. The Green function G is uniformly continuous and bounded on $[0, 1] \times [0, 1]$, which implies that the set W is relatively compact according the Arzelà-Ascoli theorem. Therefore, S is compact.

Setting the cone

$$ P= \{ u \in E, \, \min _{t \in [0, d]} u(t) \geq (1-d)\|u \|_{\infty} \}, $$

see [25], then $S (P) \subset P$. We have,

$$ S u(0) = \lambda \int _{0}^{1} G(0, s) \frac{ f ( s,u(s))}{\mathcal{A} \left (s, \int _{0}^{1}g(u)dy\right )} d s . $$

Since,

$$ s \mapsto G(0, s) \frac{ f ( s,u(s))}{\mathcal{A} \left (s, \int _{0}^{1}g(u)dy\right )} $$

is nonnegative on $[0, 1]$ and $[0, d] \subset [0, 1]$, using (1.5) and (1.6) implies

$$ S u(0) \geq \frac{ \lambda }{ w \left ( \int _{0}^{1} g(u) dy\right )} \int _{0}^{d} G(0, s) \frac{ \mu (s) k(u(s))}{ \theta (s)} d s. $$

For $u \in P$, we get

$$ (1-d)\|u \|_{\infty} \leq u \, \, \text{in} \, [0, d],\quad \text{and} \quad u \leq \|u \|_{\infty} \quad \text{in} \, [0, 1]. $$

(2.9)

Since β, k and w are nondecreasing functions, it follows that

$$ Su(0) \geq \dfrac{\lambda k \left ((1-d)\|u\|_{\infty}\right )}{w( \beta ( \|u\|_{\infty} ))} \int _{0}^{d} G(0, s) \frac{\mu (s)}{ \theta (s)}ds. $$

If $\|u\|_{\infty} \leq C_{1}$, in view of the condition (1.8), we obtain

$$ \|S u\|_{\infty} \geq Su(0) \geq \|u\|_{\infty}. $$

Let us set

$$ \Omega _{1}= \{ u \in E; \quad \|u\|_{\infty} \leq C_{1} \}. $$

We get

$$ \forall u \in P\cap \partial \varOmega _{1}: \quad \|S u\|_{\infty} \geq \|u\|_{\infty}. $$

(2.10)

On the other hand, for $u \in P$ and $s \in [0,d]$, $g\geq 0$ is nondecreasing function. Then,

$$ \int _{0}^{d} g(u(s))ds \geq dg ((1-d)\|u\|_{\infty} ). $$

According to (1.5)-(1.7), and (2.9), φ, v and α are also increasing functions, and it follows that

$$ \begin{aligned} \forall t \in [0,1]: \quad S u(t) & \leq \dfrac{\lambda \varphi ( \|u\|_{\infty})}{ v(\int _{0}^{d} g(u) dy)} \int _{0}^{1} G(t, s) \frac{ \eta (s)}{ \xi (s)} d s \\ & \leq \dfrac{ \lambda \varphi ( \|u\|_{\infty})}{ v(d \alpha ((1-d)\|u\|_{\infty} )} \int _{0}^{1} \frac{ \eta (s)}{ \xi (s)} d s. \end{aligned} $$

If $\|u\|_{\infty} \geq C_{2}$ and (1.9), we obtain

$$ \forall t \in [0,1]: \quad Su(t)\leq \|u\|_{\infty}. $$

From this set

$$ \Omega _{2}= \{ u \in E; \, \|u\|_{\infty} \leq C_{2} \}, $$

we have,

$$ \forall u \in P\cap \partial \varOmega _{2}: \quad \|S u\|_{\infty} \leq \|u\|_{\infty}. $$

(2.11)

Using Theorem 2.1, we deduce that S possesses a fixed point u in $P \cap ( \overline{\Omega _{2}} \setminus \Omega _{1} )$. From choosing the suitable cone, we get that u is a positive solution to problem (1.1). Here,

$$ \forall t \in [0, 1]:\quad u(t) = \lambda \int _{0}^{1} G(t, s) \dfrac{ f ( s,u(s))}{ \mathcal{A} \Big(s, \displaystyle{\int _{0}^{1}} g(u) d y \Big)}ds. $$

This completes the proof.  □

2.2 Proof of Theorem 1.2

Consider u is a positive solution to

$$ - \mathcal{A} \Big( \int _{0}^{1} g(u) d y \Big) u''= \lambda f(t,u) \quad t \in (0,1). $$

Multiplying the above equation by u, and then integrating by parts over $[0,1]$ with $u'(0) = u(1) = 0$, we get

$$ \mathcal{A} \Big( \displaystyle{\int _{0}^{1}} g(u) d y \Big) \int _{0}^{1} u^{\prime \,2} dt = \lambda \int _{0}^{1} f(t,u)u dt. $$

(2.12)

Combining (1.13) and (1.14) and using the Cauchy-Schwarz inequality, $\|u\|_{2} \leq \|u'\|_{2}$ and (2.12), we obtain

$$ \begin{aligned} M \int _{0}^{1} u^{\prime \,2} dt &\leq \lambda C \int _{0}^{1} | u |^{2} dt + \lambda \int _{0}^{1}f(t,u)u dt \\ & \leq \lambda C \|u'\|_{2}^{2} + \lambda \|f(\cdot ,u)\|_{2} \|u'\|_{2}. \end{aligned} $$

This leads to

$$ \|u'\|_{2} \leq \frac{\| \lambda f(\cdot ,u)\|_{2}}{ M - \lambda C}. $$

(2.13)

Given that $M > \lambda C$, let us assume that there are two positive solutions $u_{1}$ and $u_{2}$ to (1.12), and then we have

$$\begin{aligned}& -\mathcal{A} \Big( \int _{0}^{1} g(u_{1}) d y \Big) u^{\prime \prime }_{1} = \lambda f(t,u_{1}), \end{aligned}$$

(2.14)

$$\begin{aligned}& - \mathcal{A} \Big( \int _{0}^{1} g(u_{2}) d y\Big) u^{\prime \prime }_{2} = \lambda f(t,u_{2}). \end{aligned}$$

(2.15)

From (2.14) and (2.15), note $u_{3} = u_{1} - u_{2}$, we derive that

$$ \begin{aligned} -\mathcal{A} \Big( \int _{0}^{1} g(u_{1}) d y \Big) u^{\prime \prime }_{3} & = \Bigl\{ \mathcal{A} \Big( \int _{0}^{1} g(u_{1}) d y \Big)- \mathcal{A} \Big( \int _{0}^{1} g(u_{2}) d y \Big) \Bigr\} u^{\prime \prime }_{2} \\ &+\lambda \int _{0}^{1} \Big( f(t,u_{1}) - f(t,u_{2})\Big)dt. \end{aligned} $$

(2.16)

Multiplying (2.16) by $u_{3}$ and then integrating by parts from 0 to 1 with $u'_{3}(0) = u_{3}(1) = 0$ gets

$$ \begin{aligned} \mathcal{A}\Big( \int _{0}^{1} g(u_{1}) d y \Big) \| u^{\prime }_{3}\|_{2}^{2} &= - \Biggl\{ \mathcal{A}\Big( \int _{0}^{1} g(u_{1}) d y \Big)- \mathcal{A} \Big( \int _{0}^{1} g(u_{2}) d y \Big)\Biggr\} \int _{0}^{1} u^{\prime }_{2} u^{\prime }_{3}dt \\ & + \lambda \int _{0}^{1} \Big( f(t,u_{1}) - f(t,u_{2}) \Big)u_{3} dt \\ &:= I+ J. \end{aligned} $$

(2.17)

Using (2.13) and (1.14), we obtain

$$ \begin{aligned} |I| & \leq c_{1} \Big| \int _{0}^{1} g(u_{1}) dy - \int _{0}^{1} g(u_{2}) dy \Big| \| u^{\prime }_{2}\|_{2}\| u^{\prime }_{3}\|_{2} \\ &\leq \frac{c_{1} c_{2}}{M - \lambda C} \| f( \cdot ,0)\|_{2} \| u_{3} \|_{2}^{2}, \end{aligned} $$

(2.18)

and

$$ J\leq \lambda C \int _{0}^{1} | u_{3}| dt =\lambda C \| u_{3}\|_{2}^{2} \leq \lambda C \| u^{\prime }_{3}\|_{2}^{2}. $$

(2.19)

According to (2.17), (2.19), and (1.13), we get

$$ M \| u^{\prime }_{3}\|_{2}^{2} \leq \Big( \frac{c_{1} c_{2}}{M - \lambda C} \| f(\cdot ,0) \|_{2} + \lambda C\Big) \| u^{\prime }_{3}\|_{2}^{2} . $$

This leads to

$$ \Big( M - \lambda C- \frac{c_{1} c_{2}}{M - \lambda C}\| f( \cdot ,0) \|_{2} \Big) \| u^{\prime }_{3}\|_{2}^{2} \leq 0. $$

Using (1.15), we can conclude that $\| u_{3}\|_{2} = 0$. Then, $u_{1} = u_{2}$. Now, we complete the proof. □

3 Existence of the radial positive solution to problem (1.2)

We obtained the existence of a positive radial solution to problem (1.2) under the following hypotheses. For $\lambda >0$ and $g: [0, \infty ) \rightarrow [0, \infty )$ is a continuous function, the reaction term $f: [a, b] \times [0, \infty ) \rightarrow [0, \infty )$ is a continuous function with $f(\cdot , 0)=0$, and the diffusion coefficient $\mathcal{A}: [a, b] \times [0, \infty ) \rightarrow (0, \infty )$ is a continuous function, which satisfies

$$ \mu (x) v(s)\leq \mathcal{A}(x, s) \leq \theta (x) \eta (s), $$

(3.1)

where $v,\eta : [0, \infty ) \rightarrow (0, \infty )$ are nondecreasing measurable functions, and $\mu ,\theta : [a, b] \rightarrow (0, \infty )$ are measurable functions.

Theorem 3.1

Assume that

$$ \textstyle\begin{cases} \begin{aligned} & \lim \limits _{u \rightarrow 0^{+}} \frac{f(b^{1-t}a^{t}, u)}{u} = 0 \quad \textit{uniformly for } 0 < t < 1, \quad \textit{if}\, N = 2, \\ & \lim \limits _{u \rightarrow 0^{+}} \frac{ f((\frac{C}{B-t})^{\frac{1}{N-2}},u)}{u}= 0 \quad \textit{uniformly for } 0 < t < 1, \quad \textit{if} \, N \geq 3, \end{aligned} \end{cases} $$

(3.2)

and

$$ \textstyle\begin{cases} \begin{aligned} & \lim \limits _{u \rightarrow 0^{+}} \frac{f(b^{1-t}a^{t}, u)}{u} = \infty \quad \textit{uniformly for } 0 < t < 1, \quad \textit{if}\, N = 2, \\ & \lim \limits _{u \rightarrow 0^{+}} \frac{ f((\frac{C}{B-t})^{\frac{1}{N-2}},u)}{u} =\infty \quad \textit{uniformly for } 0 < t < 1, \quad \textit{if} \, N \geq 3, \end{aligned} \end{cases} $$

(3.3)

where C, B are defined on (1.11). For $L> 0$, there exists a sufficiently large constant $\lambda _{L} > 0$ such that for $\lambda > \lambda _{L}$ and (3.2) is true, or $\lambda < \lambda _{L}$ and (3.3) is true, then the problem (1.2) admits a positive solution u such that $\underset{t \in [0,1]}{\sup } u(t) \leq L$.

Theorem 3.2

Assume that

$$ \textstyle\begin{cases} \begin{aligned} & \lim \limits _{u \rightarrow \infty} \frac{f(b^{1-t}a^{t}, u)}{u} = 0 \quad \textit{uniformly for } 0 < t < 1, \quad \textit{if}\, N = 2, \\ & \lim \limits _{u \rightarrow \infty} \frac{ f((\frac{C}{B-t})^{\frac{1}{N-2}},u)}{u}=0\quad \textit{uniformly for } 0 < t < 1, \quad \textit{if} \, N \geq 3, \end{aligned} \end{cases} $$

(3.4)

and

$$ \textstyle\begin{cases} \begin{aligned} & \lim \limits _{u \rightarrow \infty} \frac{f(b^{1-t}a^{t}, u)}{u} = \infty \quad \textit{uniformly for } 0 < t < 1, \quad \textit{if}\, N = 2, \\ & \lim \limits _{u \rightarrow \infty} \frac{ f((\frac{C}{B-t})^{\frac{1}{N-2}},u)}{u}=\infty \quad \textit{uniformly for } 0 < t < 1, \quad \textit{if} \, N \geq 3, \end{aligned} \end{cases} $$

(3.5)

where C, B are defined on (1.11). For any strictly positive number l, there exists a sufficiently large constant $\lambda _{l} > 0$. Since $\lambda > \lambda _{l}$ and (3.4) is true or $\lambda < \lambda _{l}$ and (3.5) is true, then the problem (1.2) admits a positive solution u such that $\underset{t \in [0,1]}{\min} u(t) \geq l$.

We give some basic results that will be used in the proof, and we will utilize updated versions of the Gustafson and Schmitt theorems, as outlined in references [21, 26].

Theorem 3.3

Let X be a Banach space, P is cone in X. Suppose there are two real numbers l, L where $0 < l < L$. Consider $T: D \rightarrow P$, where $D= \{ u \in P: \, l \leq \| u \| \leq L \}$ is a compact continuous operator:

• 1. $u \in D$, $\xi <1$ and $u= \xi T u$, then $\| u \| \neq L$.

• 2. $u \in D$, $\xi > 1$ and $u= \xi T u$, then $\| u \| \neq l$.

• 3. $\underset{\| u \| =l}{\inf} \| Tu \| > 0$.

Hence, there exists a fixed point for T in D.

Theorem 3.4

Let X be a Banach space, define a cone P on X. Suppose there are two real numbers l and L, where $0 < l < L$. Assume that $T: D \rightarrow P$, where $D= \{u \in P: \, l \leq \|u\| \leq L \}$ is a compact continuous operator:

• 1. $u \in D$, $\xi >1$ and $u= \xi T u$ then $\| u \| \neq L$.

• 2. $u \in D$, $\xi < 1$ and $u= \xi T u$ then $\| u \| \neq l$.

• 3. $\underset{\| u \| =L }{\inf} \| Tu \| > 0$.

Hence, there exists a fixed point for T in D.

Lemma 3.1

[21] Suppose that $\eta : [0,1] \rightarrow [ 0, + \infty )$ is a continuous function whose graph is concave down such that $\|\eta \| := \max \{ \eta (t): t \in [0,1] \}$. For every $t \in [\alpha , 1-\alpha ]$ with $0 < \alpha < \frac{1}{2}$, we get $\alpha \| \eta \| \leq \eta (t)$.

Lemma 3.2

[21] Assume that $\dfrac{h(t,u)}{\mathcal{A} \Big(t, \displaystyle{\int _{0}^{1}} j(u) dy \Big)} > 0$ for any $t \in (0,1)$ and $u > 0$. If $L> 0$, we have $0< \inf \{ \|T u \|: \, u \in P, \| u \| = L \}$.

3.1 Proof of Theorem 3.1

See [28], problem (1.2) can be transformed, and when $N = 2$, we suppose $r=b^{1-t}a^{t}$, $\quad U(y)=u(r)$, $\quad h(t, U)=f(b^{1-t}a^{t}, U)$, $\quad q_{2}(t)= \Big(b^{1-t}a^{t} \log (\frac{b}{a}) \Big)^{2}$ and

$$ \begin{aligned} \mathcal{A} \Big( t, \int _{0}^{1} j(U(y)) dy \Big)=& \mathcal{A} \Big(b^{1-t}a^{t}, \omega \int _{a}^{b}r g(u(r))dr \Big) \\ =& \mathcal{A} \Big( b^{1-t}a^{t}, \omega b^{2} \log (\frac{a}{b}) \int _{0}^{1} (\frac{a}{b})^{2s} g(U(s))ds \Big). \end{aligned} $$

For $N \geq 3$, we set

$$\begin{aligned}& t=B- Cr^{-N+2},\quad U(y)=u(r),\quad h(t, U)= f \Big( (\frac{C}{B-t})^{ \frac{1}{N-2}},U\Big),\\& q_{N}(t)= (N-2)^{-2} \frac{C^{\frac{2}{N-2}}}{(B-t)^{2 \frac{N-1}{N-2}}}, \end{aligned}$$

and

$$ \begin{aligned} \mathcal{A} \Big( t, \int _{0}^{1} j(U(y)) dy \Big)=& \mathcal{A} \Big( (\frac{C}{B-t})^{\frac{1}{N-2}},\, \omega \int _{a}^{b}r^{N-1}g(u(r))dr \Big) \\ =& \mathcal{A} \Big( (\frac{C}{B-t})^{\frac{1}{N-2}},\, \frac{\omega}{C(N-2)} \int _{0}^{1} (\frac{C}{B-s})^{2\frac{N-1}{N-2}} g(U(s))ds \Big). \end{aligned} $$

By above, this one-dimensional nonlocal elliptic problem is the transformation of problem (1.2)

$$ \textstyle\begin{cases} - \mathcal{A} \Big( t, \displaystyle{\int _{0}^{1}} j(U(\tau )) d \tau \Big) U''= \lambda q_{N}(t)h(t, U) \quad t \in (0,1), \\ U(0)=U(1)=0. \end{cases} $$

(3.6)

Remark 3.1

We can obtain the existence of positive radial solution to problem (3.6) by applying the Guo-Krasnosel’ski theorem. In addition, we can get the uniqueness by replacing $[ 0 ,1]$ to $[ a, b]$ in (1.15) with more modifications in (1.14).

Let us define the operator $T :P \rightarrow E$, $U \mapsto TU$, such that

$$ TU(t)= \lambda \int _{0}^{1}G(t,s)q_{N}(s) \frac{h(s, U(s)) }{\mathcal{A} \Big(s,\displaystyle{ \int _{0}^{1}} j(U) dy \Big)}ds, $$

(3.7)

where G is the Green function,

$$ G(t,s)= \textstyle\begin{cases} \begin{aligned} & s(1-t), \quad 0 \leq s \leq t \leq 1; \\ & t(1-s), \quad 0 \leq t \leq s \leq 1, \end{aligned} \end{cases} $$

which is proved in [27], satisfied

$$ \begin{aligned} G(t,s)& > 0 \quad \text{on} \quad (0,1)\times (0,1), \\ G(t,s)& \leq G(s,s)= s(1-s), \quad 0\leq s,\quad t \leq 1, \end{aligned} $$

and

$$ G(t,s) \geq \frac{1}{4} G(s,s)= \frac{1}{4} s(1-s), \quad \frac{1}{4} \leq t \leq \frac{3}{4},\quad 0 \leq s \leq 1. $$

Define a cone P and a set D on E,

$$ P= \{U\in E: U(t) \geq 0, U(0)=0 = U(1)\},\, \, D= \{ U \in P: U(t) \in [ 0, L] \},\, L> 0. $$

Using the Arzelà-Ascoli theorem, we get the compactness and continuity of mapping $T: D \rightarrow P$. To finalize the proof of theorem, we will use Theorem 3.4. For any $U \in D$ such that $U = \xi T U$, we have

$$ U(t)= \xi \lambda \int _{0}^{1}G(t,s)q_{N}(s) \frac{h(s, U(s)) }{\mathcal{A}\Big(s, \displaystyle{\int _{0}^{1}} j(U) dy \Big)}ds, \quad \xi > 1 . $$

(3.8)

Noting that $\tilde{U}(t)$ is solution to (3.6), the graph of Ũ is concave downward, and we can apply Lemma 3.1 with $\alpha = \frac{1}{4}$, which yields

$$ \frac{1}{4} \| \tilde{U} \| \leq \tilde{U}(t), \quad \forall t \in \Big[ \frac{1}{4}, \frac{3}{4} \Big]. $$

(3.9)

We define

$$ \min _{t \in \Big[ \frac{1}{4}, \frac{3}{4} \Big], \frac{L}{4} \leq U \leq L} \frac{h(t, U)}{U} = m_{L} . $$

We choose $\lambda \geq \lambda _{L}$ and ϵ such that

$$ \textstyle\begin{cases} \begin{aligned} & \lambda _{L} > \dfrac{16 \eta \Big( \omega b^{2} \log (\frac{a}{b}) \int _{0}^{1} (\frac{a}{b})^{2y} g(\tilde{U})dy \Big)}{m_{L} \int _{\frac{1}{4}}^{\frac{3}{4}} G(s,s) \Big(b^{1-s}a^{s} \log (\frac{b}{a}) \Big)^{2} \Big( \theta (b^{1-s}a^{s})\Big)^{-1} ds}, \quad \text{if N $=$ 2,} \\ & \lambda _{L} > \dfrac{16 \eta \Big( \frac{\omega}{C(N-2)} \int _{0}^{1} (\frac{C}{B-y})^{2\frac{N-1}{N-2}} g(\tilde{U})dy \Big)}{m_{L} \int _{\frac{1}{4}}^{\frac{3}{4}} G(s,s) C^{\frac{2}{N-2}}\Big( (N-2)^{2}(B-s)^{2 \frac{N-1}{N-2}} \theta \Big((\frac{C}{B-s})^{\frac{1}{N-2}} \Big) \Big)^{-1} ds}, \quad \text{if N} \geq 3, \end{aligned} \end{cases} $$

and

$$ \textstyle\begin{cases} \begin{aligned} & 0< \epsilon < \dfrac{\eta \Big( \omega b^{2} \log (\frac{a}{b}) \int _{0}^{1} (\frac{a}{b})^{2y} g(U)dy \Big) }{\lambda \int _{0}^{1} G(s,s) \frac{C^{\frac{2}{N-2}}}{(N-2)^{2}(B-s)^{2 \frac{N-1}{N-2}}} \Big( \theta ( b^{1-s}a^{s})\Big)^{-1} ds }, \quad \text{if N}=2, \\ &0< \epsilon < \dfrac{\eta \Big( \frac{\omega}{C(N-2)} \int _{0}^{1} (\frac{C}{B-y})^{2\frac{N-1}{N-2}} g(U)dy \Big)}{ \lambda \int _{0}^{1} G(s,s) \Big(b^{1-s}a^{s} \log (\frac{b}{a}) \Big)^{2} \Big( \theta \Big( (\frac{C}{B-s})^{\frac{1}{N-2}} \Big) \Big)^{-1} ds}, \quad \text{if N} \geq 3. \end{aligned} \end{cases} $$

Then, there exists $l>0$ such that $0 \leq U \leq l$. Using (3.2), the function $h(t, U)$ is bounded by ϵU. Define

$$ D_{1}= \{ U \in P: \quad U(t) \in [l, L], \, t \in (0,1) \}. $$

We show that there is no solution $U \in D_{1}$ satisfying (3.8) with $\|U\| = L$. By the absurd, if $N=2$, we have

$$ \begin{aligned} \| \tilde{U} \| & \geq \min _{t \in [\frac{1}{4} ,\frac{3}{4}] } \tilde{U}(t) \geq \xi \lambda \int _{0}^{1} \Big( \min _{t \in [ \frac{1}{4} ,\frac{3}{4}] } G(t,s)\Big) q_{N}(s) \frac{h(s, \tilde{U}(s)) }{\mathcal{A} \Big(s,\displaystyle{ \int _{0}^{1}} j(\tilde{U}) dy \Big)} ds \\ & \geq \frac{\xi \lambda m_{L}}{4 \eta \Big( \omega b^{2} \log (\frac{a}{b}) \int _{0}^{1} (\frac{a}{b})^{2y} g(\tilde{U}(y))dy \Big) } \int _{\frac{1}{4}}^{\frac{3}{4}} G(s,s) \Big(b^{1-s}a^{s} \log ( \frac{b}{a}) \Big)^{2} \frac{ \tilde{U}(s)}{\theta (b^{1-s}a^{s})} ds. \end{aligned} $$

(3.10)

From (3.9) and (3.10), we get

$$ \begin{aligned} L= \| \tilde{U}\| & \geq \frac{\lambda _{L} \xi m_{L}}{16 \eta \Big( \omega b^{2} \log (\frac{a}{b}) \int _{0}^{1} (\frac{a}{b})^{2y} g(\tilde{U})dy \Big)} \| \tilde{U} \| \int _{\frac{1}{4}}^{\frac{3}{4}} G(s,s) \frac{\Big(b^{1-s}a^{s} \log (\frac{b}{a}) \Big)^{2}}{\theta (b^{1-s}a^{s}) } ds \\ &> \xi \| \tilde{U} \| = \xi L> L. \end{aligned} $$

Since $N\geq 3$,

$$ \begin{aligned} \| \tilde{U} \| &\geq \min _{t \in [\frac{1}{4} ,\frac{3}{4}] } \tilde{U}(t) \geq \xi \lambda \int _{0}^{1} \Big( \min _{t \in [ \frac{1}{4} ,\frac{3}{4}] } G(t,s)\Big) q(s) \frac{h(s, \tilde{U}(s)) }{\mathcal{A} \Big(s, \displaystyle{\int _{0}^{1}} j(\tilde{U}) dy \Big)} ds \\ & \geq \frac{\xi \lambda m_{L}}{4 \eta \Big( \frac{\omega}{C(N-2)} \int _{0}^{1} (\frac{C}{B-y})^{2\frac{N-1}{N-2}} g(\tilde{u})dy \Big) } \int _{\frac{1}{4}}^{\frac{3}{4}} G(s,s) \frac{C^{\frac{2}{N-2}}}{(N-2)^{2}(B-s)^{2 \frac{N-1}{N-2}}} \frac{\tilde{U}(s)}{\theta \Big( (\frac{C}{B-s})^{\frac{1}{N-2}} \Big)} ds. \end{aligned} $$

(3.11)

Combining (3.9) and (3.11), we obtain

$$ \begin{aligned} L= \| \tilde{U}\|\geq{}& \frac{\lambda _{L} \xi m_{L}}{16 \eta \Big( \frac{\omega}{C(N-2)} \int _{0}^{1} (\frac{C}{B-y})^{2\frac{N-1}{N-2}} g(\tilde{U})dy \Big) } \| \tilde{U} \| \\ &{}\times \int _{\frac{1}{4}}^{\frac{3}{4}} G(s,s) \frac{C^{\frac{2}{N-2}}}{(N-2)^{2}(B-s)^{2 \frac{N-1}{N-2}} \theta \Big((\frac{C}{B-s})^{\frac{1}{N-2}} \Big)} ds \\ >{}& \xi \| \tilde{U} \| = \xi L> L. \end{aligned} $$

This leads to a contradiction. Then, $\| U \| \neq L$.

We will prove that $\| U\|\neq l$. By the absurd, we get

$$ \begin{aligned} l= \| \tilde{U} \| &\leq \xi \lambda \epsilon \| U \| \int _{0}^{1} G(s,s) \frac{q_{N}(s) }{\mathcal{A} \Big(s,\displaystyle{\int _{0}^{1}} j(U) dy \Big)} ds \\ &\leq \xi \lambda \epsilon l \int _{0}^{1} G(s,s) \frac{q_{N}(s) }{\mathcal{A} \Big(s,\displaystyle{\int _{0}^{1}} j(U) dy \Big)}ds \\ & < \xi l < l, \end{aligned} $$

which leads to a contradiction. Therefore, $\| U \| \neq l$.

Let us define

$$ \max _{t \in [\frac{1}{4}, \frac{3}{4}] \frac{L}{4} \leq U \leq L} \frac{h(t, U)}{U} = M_{L}. $$

By (3.3) and $\lambda <\lambda _{L}, \sigma$ such that

$$ \textstyle\begin{cases} \begin{aligned} & \lambda _{L} \leq \dfrac{ \eta \Big( \omega b^{2} \log (\frac{a}{b}) \int _{0}^{1} (\frac{a}{b})^{2y} g(U)dy \Big) }{M_{L} \int _{0}^{1} G(s,s) \Big(b^{1-s}a^{s} \log (\frac{b}{a}) \Big)^{2} \Big( \theta (b^{1-s}a^{s}) \Big)^{-1} ds }, \quad \text{if N $=$ 2} \\ & \lambda _{L} \leq \dfrac{ \eta \Big( \frac{\omega}{C(N-2)} \int _{0}^{1} (\frac{C}{B-y})^{2\frac{N-1}{N-2}} g(U)dy \Big)}{M_{L} \int _{0}^{1} G(s,s) C^{\frac{2}{N-2}}\Big( (N-2)^{2}(B-s)^{2 \frac{N-1}{N-2}} \theta \Big((\frac{C}{B-s})^{\frac{1}{N-2}} \Big) \Big)^{-1} ds} , \quad \text{if N} \geq 3, \end{aligned} \end{cases} $$

and

$$ \textstyle\begin{cases} \sigma > \dfrac{16 \eta \Big( \omega b^{2} \log (\frac{a}{b}) \int _{0}^{1} (\frac{a}{b})^{2y} g(U)dy \Big)}{\lambda} \Big( \int _{ \frac{1}{4} }^{\frac{3}{4}} G(s,s) \frac{\Big(b^{1-s}a^{s} \log (\frac{b}{a}) \Big)^{2} }{\theta (b^{1-s}a^{s})} ds\Big)^{-1},\quad \text{if}\, N=2 \\ \sigma > \dfrac{16 \Big( \frac{\omega}{C(N-2)} \int _{0}^{1} (\frac{C}{B-y})^{2\frac{N-1}{N-2}} g(U)dy \Big) }{\lambda}\Big( \int _{ \frac{1}{4} }^{\frac{3}{4}} G(s,s) \frac{ C^{\frac{2}{N-2}}}{(N-2)^{2}(B-s)^{2 \frac{N-1}{N-2}}\theta \Big( (\frac{C}{B-s})^{\frac{1}{N-2}} \Big)} \Big)^{-1}, \\ \quad \text{if}\, N\geq 3. \end{cases} $$

We find

$$ h(t,U) > \sigma U,\quad \forall U \in (0, l ], \quad t \in (0,1), l \in (0, L). $$

Assume that $U \in D$, where $U = \xi T U$, $\xi \in (0, 1)$. We claim $\|U\| \neq L$. So, by the absurd, for $N =2$

$$ \begin{aligned} U(t)& = \xi \lambda \int _{0}^{1} G(t,s) q(s) \frac{h(s, U(s)) }{\mathcal{A} \Big(s, \displaystyle{\int _{0}^{1}} j(U) dy \Big)} ds \\ & < \frac{\xi \lambda _{L} M_{L}}{\eta \Big( \omega b^{2} \log (\frac{a}{b}) \int _{0}^{1} (\frac{a}{b})^{2y} g(U)dy \Big) } \int _{0}^{1} G(s,s)U(s) \frac{\Big(b^{1-s}a^{s} \log (\frac{b}{a}) \Big)^{2} }{\theta (b^{1-s}a^{s}) } ds. \end{aligned} $$

Then,

$$ \begin{aligned} L= \| U\| &< \frac{\xi \lambda _{L} M_{L} \| U \|}{\eta \Big( \omega b^{2} \log (\frac{a}{b}) \int _{0}^{1} (\frac{a}{b})^{2y} g(U)dy \Big) } \int _{0}^{1} G(s,s) \frac{\Big(b^{1-s}a^{s} \log (\frac{b}{a}) \Big)^{2} }{\theta (b^{1-s}a^{s}) } ds \\ & < \xi L < L. \end{aligned} $$

If $N\geq 3$,

$$ \begin{aligned} U(t)={}& \xi \lambda \int _{0}^{1} G(t,s) q(s) \frac{h(s, U(s)) }{\mathcal{A}\Big(s, \displaystyle{\int _{0}^{1}} j(U) dy \Big)} ds \\ < {}& \frac{\xi \lambda _{L} M_{L}}{\eta \Big( \frac{\omega}{C(N-2)} \int _{0}^{1} (\frac{C}{B-y})^{2\frac{N-1}{N-2}} g(U)dy \Big)} \\ &{}\times \int _{0}^{1} G(s,s) U(s) \frac{ C^{\frac{2}{N-2}}}{(N-2)^{2}(B-s)^{2 \frac{N-1}{N-2}}\theta \Big( (\frac{C}{B-s})^{\frac{1}{N-2}} \Big)} ds, \end{aligned} $$

which implies

$$ \begin{aligned} L= \| U\| < {}& \frac{\xi \lambda _{L} M_{L} \| U \|}{\eta \Big( \frac{\omega}{C(N-2)} \int _{0}^{1} (\frac{C}{B-y})^{2\frac{N-1}{N-2}} g(U)dy \Big)} \\ &{}\times \int _{0}^{1} G(s,s) \frac{ C^{\frac{2}{N-2}}}{(N-2)^{2}(B-s)^{2 \frac{N-1}{N-2}} \theta \Big( (\frac{C}{B-s})^{\frac{1}{N-2}} \Big)} ds \\ < {}& \xi L < L. \end{aligned} $$

This result leads to a contradiction. Then, $\|U \| \neq L$.

On the other hand, will prove $\| U \|\neq l$. By the absurd, if $N = 2$,

$$ U(t) \geq \xi \lambda \frac{\sigma }{\eta \Big( \omega b^{2} \log (\frac{a}{b}) \int _{0}^{1} (\frac{a}{b})^{2y} g(U)dy \Big) } \int _{\frac{1}{4}}^{\frac{3}{4}} G(t,s)U(s) \frac{\Big(b^{1-s}a^{s} \log (\frac{b}{a}) \Big)^{2} }{\theta (b^{1-s}a^{s}) } ds, $$

which leads to

$$ \begin{aligned} \| U \| &\geq \min _{t \in [\frac{1}{4} ,\frac{3}{4}] } U(t) \\ & \geq \xi \lambda \frac{ \sigma}{4 \eta \Big( \omega b^{2} \log (\frac{a}{b}) \int _{0}^{1} (\frac{a}{b})^{2y} g(U)dy \Big)} \int _{ \frac{1}{4} }^{\frac{3}{4}} G(s,s) U(s) \frac{\Big(b^{1-s}a^{s} \log (\frac{b}{a}) \Big)^{2} }{\theta (b^{1-s}a^{s}) } ds. \end{aligned} $$

In view of (3.9), we get

$$ \begin{aligned} l= \| U \| & \geq \xi \lambda l \frac{ \sigma}{16 \eta \Big( \omega b^{2} \log (\frac{a}{b}) \int _{0}^{1} (\frac{a}{b})^{2y} g(U)dy \Big)} \int _{ \frac{1}{4} }^{\frac{3}{4}} G(s,s) \frac{\Big(b^{1-s}a^{s} \log (\frac{b}{a}) \Big)^{2} }{\theta (b^{1-s}a^{s}) } ds \\ &> \xi l > l. \end{aligned} $$

For $N\geq 3$,

$$\begin{aligned} U(t) \geq{}& \xi \lambda \frac{\sigma }{\eta \Big( \frac{\omega}{C(N-2)} \int _{0}^{1} (\frac{C}{B-y})^{2\frac{N-1}{N-2}} g(U)dy \Big)} \\ &{}\times \int _{\frac{1}{4}}^{\frac{3}{4}} G(t,s) U(s) \frac{ C^{\frac{2}{N-2}}}{(N-2)^{2}(B-s)^{2 \frac{N-1}{N-2}}\theta \Big( (\frac{C}{B-s})^{\frac{1}{N-2}} \Big)} ds. \end{aligned}$$

Then,

$$ \begin{aligned} \| U \| \geq{}& \min _{t \in [\frac{1}{4} ,\frac{3}{4}] } U(t) \\ \geq{}& \xi \lambda \frac{ \sigma}{4 \eta \Big( \frac{\omega}{C(N-2)} \int _{0}^{1} (\frac{C}{B-y})^{2\frac{N-1}{N-2}} g(U)dy \Big)} \\ &{}\times \int _{ \frac{1}{4} }^{\frac{3}{4}} G(s,s) U(s) \frac{ C^{\frac{2}{N-2}}}{(N-2)^{2}(B-s)^{2 \frac{N-1}{N-2}}\theta \Big( (\frac{C}{B-s})^{\frac{1}{N-2}} \Big)} ds. \end{aligned} $$

Using (3.9), we have

$$ \begin{aligned} l= \| U \| \geq{}& \xi \lambda l \frac{ \sigma}{16 \Big( \frac{\omega}{C(N-2)} \int _{0}^{1} (\frac{C}{B-y})^{2\frac{N-1}{N-2}} g(U)dy \Big)} \\ &{}\times \int _{ \frac{1}{4} }^{\frac{3}{4}} G(s,s) \frac{ C^{\frac{2}{N-2}}}{(N-2)^{2}(B-s)^{2 \frac{N-1}{N-2}}\theta \Big( (\frac{C}{B-s})^{\frac{1}{N-2}} \Big)} ds. \\ >{}& \xi l > l, \end{aligned} $$

which is a contradiction, confirming that $\|U\| \neq l$.

All in all, by Lemma 3.2 and Theorem 3.3, the problem (1.2) has a positive solution U, such that $l \leq U(t) \leq L$ for $t \in (0, 1)$. The proof is finished. □

3.2 Proof of Theorem 3.2

For $l>0$, we define

$$ \min _{t \in [\frac{1}{4}, \frac{3}{4}], \frac{l}{4} \leq U \leq l} \frac{ h(t, U)}{U}= m_{l} . $$

Using (3.4), $\lambda \geq \lambda _{l}, \epsilon$ such that

$$ \textstyle\begin{cases} \lambda _{l} > 4 \eta \Big( \omega b^{2} \log (\frac{a}{b}) \int _{0}^{1} (\frac{a}{b})^{2y} g(\tilde{U})dy \Big) \Big( m_{l} \int _{ \frac{1}{4}}^{\frac{3}{4}} G(s,s) \frac{\Big(b^{1-s}a^{s} \log (\frac{b}{a}) \Big)^{2} }{\theta (b^{1-s}a^{s}) } ds \Big)^{-1}, \quad \text{if}\, N=2 \\ \lambda _{l} >4 \eta \Big( \frac{\omega}{C(N-2)} \int _{0}^{1} ( \frac{C}{B-y})^{2\frac{N-1}{N-2}} g(\tilde{U})dy \Big) \Big( m_{l} \int _{ \frac{1}{4}}^{\frac{3}{4}} G(s,s) \frac{ C^{\frac{2}{N-2}}}{(N-2)^{2}(B-s)^{2} \frac{N-1}{N-2}\theta \Big( (\frac{C}{B-s})^{\frac{1}{N-2}} \Big)} ds \Big)^{-1}, \\ \quad \text{if}\, \, N\geq 3. \end{cases} $$

and

$$ \textstyle\begin{cases} \epsilon < \frac{\eta \Big( \omega b^{2} \log (\frac{a}{b}) \int _{0}^{1} (\frac{a}{b})^{2y} g(U)dy \Big) }{\lambda} \Big( \int _{0}^{1} G(s,s) \frac{\Big(b^{1-s}a^{s} \log (\frac{b}{a}) \Big)^{2} }{\theta (b^{1-s}a^{s}) } ds \Big)^{-1}, \quad \text{if}\, N=2 \\ \epsilon < \frac{\eta \Big( \frac{\omega}{C(N-2)} \int _{0}^{1} (\frac{C}{B-y})^{2\frac{N-1}{N-2}} g(U)dy \Big) }{\lambda} \Big( \int _{0}^{1} G(s,s) \frac{ C^{\frac{2}{N-2}}}{(N-2)^{2}(B-s)^{2} \frac{N-1}{N-2}\theta \Big( (\frac{C}{B-s})^{\frac{1}{N-2}} \Big)} ds \Big)^{-1},\quad \text{if}\, \, N\geq 3, \end{cases} $$

we get

$$ \exists \, L_{0}>l, U \geq L_{0}:\, \, h(t, U) < \epsilon U. $$

Our goal is to use Theorem 3.3. To achieve this, we aim to show that $U = \xi T U$, $\xi \in (0,1)$ has no solution to $\|L\| \geq L_{0}$. By the absurd, there exists a sequence $(L_{n})_{n\geq 1}$ approaching to infinity, where $L_{n} \geq L_{0}$, a sequence $(\xi _{n} )_{n\geq 1}$ of real numbers, $\xi _{n} \in (0,1)$, and a sequence of functions $(U_{n} )_{n\geq 1}$ such that $\| U_{n}\|= L_{n}$, we have

$$ U_{n} = \xi _{n} TU_{n}, \quad n \in \mathbb{N}^{\ast}. $$

(3.12)

Denoting $t_{n}$ is the unique point in $[0, 1]$ such that $U_{n}(t_{n}) = \|U_{n}\|$. Then, if $N=2$ and (3.12), we obtain

$$ \begin{aligned} L_{n} &= U(t_{n})= \xi _{n} \lambda \int _{0}^{1} G(t_{n},s) q(s) \frac{h(s, U_{n}(s)) }{\mathcal{A} \Big( s, \displaystyle{\int _{0}^{1}} j(U) dy \Big)} ds \\ & \leq \xi _{n} \lambda \frac{\epsilon \| U_{n}\| }{\eta \Big( \omega b^{2} \log (\frac{a}{b}) \int _{0}^{1} (\frac{a}{b})^{2y} g(U_{n})dy \Big) } \int _{0}^{1} G(s,s) \frac{\Big(b^{1-s}a^{s} \log (\frac{b}{a}) \Big)^{2} }{\theta (b^{1-s}a^{s}) } ds \\ & < \xi _{n} L_{n} < L_{n}. \end{aligned} $$

Since $N\geq 3$, we have

$$ \begin{aligned} L_{n} &= U(t_{n})=\xi _{n} \lambda \int _{0}^{1} G(t_{n},s) q(s) \frac{h(s, U_{n}(s)) }{\mathcal{A} \Big( s, \displaystyle{\int _{0}^{1}} j(U) dy \Big)} ds \\ & \leq \xi _{n} \lambda \frac{\epsilon \| U_{n}\| }{\eta \Big( \frac{\omega}{C(N-2)} \int _{0}^{1} (\frac{C}{B-y})^{2\frac{N-1}{N-2}} g(U_{n})dy \Big)} \int _{0}^{1} G(s,s) \frac{ C^{\frac{2}{N-2}}}{(N-2)^{2}(B-s)^{2 \frac{N-1}{N-2}}\theta \Big( (\frac{C}{B-s})^{\frac{1}{N-2}} \Big)} ds \\ & < \xi _{n} L_{n} < L_{n}, \end{aligned} $$

which implies a contradiction.

For $L > L_{0}$. Based on the preceding argument, $U = \xi TU$, $\xi \in (0,1)$ does not possess a solution if $\|U\| = L$. Assume that

$$ D = \{ U(t) \in P, t \in [0,1]: l \leq \|U\|\leq L\}. $$

In this selected value of L, we obtain that the first condition of Theorem 3.3 is satisfied. Let $U\in D$ such that $U = \xi T U$ for $\xi > 1$. We claim that $\|U\| \neq l$. By the absurd, as $\tilde{U} \in [0,l]$ and $\|\tilde{U}\| =l$, if $N=2$, it follows that

$$ \min _{t \in [\frac{1}{4} ,\frac{3}{4}] } \tilde{U}(t) > \frac{ \xi \lambda _{l}}{4 \eta \Big( \omega b^{2} \log (\frac{a}{b}) \int _{0}^{1} (\frac{a}{b})^{2y} g(\tilde{U})dy \Big)} m_{l} \int _{ \frac{1}{4}}^{\frac{3}{4}} G(s,s) \tilde{U}(s) \frac{\Big(b^{1-s}a^{s} \log (\frac{b}{a}) \Big)^{2} }{\theta (b^{1-s}a^{s}) } ds. $$

If $N \geq 3$,

$$\begin{aligned} \min _{t \in [\frac{1}{4} ,\frac{3}{4}] } \tilde{U}(t) >{}& \frac{ \xi \lambda _{l}}{4 \eta \Big( \frac{\omega}{C(N-2)} \int _{0}^{1} (\frac{C}{B-y})^{2\frac{N-1}{N-2}} g(U)dy \Big)} m_{l}\\ &{}\times \int _{ \frac{1}{4}}^{\frac{3}{4}} G(s,s)\tilde{U}(s) \frac{ C^{\frac{2}{N-2}}}{(N-2)^{2}(B-s)^{2 \frac{N-1}{N-2}}\theta \Big( (\frac{C}{B-s})^{\frac{1}{N-2}} \Big)} ds. \end{aligned}$$

For $\tilde{t} \in [ \frac{1}{4}, \frac{3}{4}]$, such that

$$ \tilde{U}( \tilde{t}) = \min _{t \in [\frac{1}{4}, \frac{3}{4}] } \tilde{U}(t). $$

So,

$$ \begin{aligned} \tilde{U}( \tilde{t}) & > \frac{\xi}{4 }\lambda _{l} \tilde{U}( \tilde{t} ) \int _{ \frac{1}{4}}^{\frac{3}{4}} G(s,s) \frac{q(s)}{\mathcal{A}\Big( s, \displaystyle{\int _{0}^{1}} j(\tilde{U}) dy \Big)} ds \\ & > \xi \tilde{U}( \tilde{t})> \tilde{U}( \tilde{t}). \end{aligned} $$

This also leads to a contradiction.

From Theorem 3.3 and Lemma 3.2, the problem (1.2) has a positive solution $U\in D$, such that $l < \| U \| < L$.

On the other hand, for $l>0$, we define

$$ \max _{t \in [\frac{1}{4}, \frac{3}{4}], \frac{l}{4} \leq u \leq l} \frac{h(t, U)}{U}= M_{l} . $$

By (3.5), $\lambda < \lambda _{l}, \sigma$ such that

$$ \begin{aligned} \textstyle\begin{cases} &0< \lambda _{l} < \dfrac{ \eta \Big( \omega b^{2} \log (\frac{a}{b}) \int _{0}^{1} (\frac{a}{b})^{2y} g(U)dy \Big)}{M_{l} \int _{0}^{1}G(s,s) \Big(b^{1-s}a^{s} \log (\frac{b}{a}) \Big)^{2} \Big( \theta (b^{1-s}a^{s})\Big)^{-1} ds }, \quad \text{if}\, N=2 \\ &0< \lambda _{l} < \dfrac{\eta \Big( \frac{\omega}{C(N-2)} \int _{0}^{1} (\frac{C}{B-y})^{2\frac{N-1}{N-2}} g(U_{n})dy \Big)}{M_{l} \int _{0}^{1}G(s,s) C^{\frac{2}{N-2}} \Big( (N-2)^{2}(B-s)^{2 \frac{N-1}{N-2}} \theta \Big((\frac{C}{B-s})^{\frac{1}{N-2}} \Big) \Big)^{-1} ds}, \quad \text{if}\, N\geq 3, \end{cases}\displaystyle \end{aligned} $$

and

$$ \begin{aligned} \textstyle\begin{cases} & \sigma > \dfrac{\eta \Big( \omega b^{2} \log (\frac{a}{b}) \int _{0}^{1} (\frac{a}{b})^{2y} g(U_{n})dy \Big)}{\lambda \int _{\frac{1}{4}}^{\frac{3}{4}} G(s,s) \Big(b^{1-s}a^{s} \log (\frac{b}{a}) \Big)^{2} \Big( \theta (b^{1-s}a^{s})\Big)^{-1} ds }, \quad \text{if}\, N=2 \\ & \sigma > \dfrac{\eta \Big( \frac{\omega}{C(N-2)} \int _{0}^{1} (\frac{C}{B-y})^{2\frac{N-1}{N-2}} g(U_{n})dy \Big)}{ \lambda \int _{\frac{1}{4}}^{\frac{3}{4}} G(s,s) C^{\frac{2}{N-2}}\Big( (N-2)^{2}(B-s)^{2 \frac{N-1}{N-2}} \theta \Big((\frac{C}{B-s})^{\frac{1}{N-2}} \Big)\Big)^{-1} ds} ,\quad \text{if}\, N\geq 3, \end{cases}\displaystyle \end{aligned} $$

it follows that

$$ h(t, U) \geq \sigma U, \quad \text{for}\, U \geq L_{0}, L_{0} > 0. $$

If $L > L_{0}$, the equation $U = \xi T U$, $\xi > 1$, does not have a solution to $\|U\| = L$. By the absurd, there is a sequence $(L_{n})_{n \geq 1 }$ that approaches to infinity, where $L_{n} > L_{0}$. Additionally, a sequence $(\xi _{n})_{n \geq 1}$ of real numbers with $\xi _{n} > 1$, and a sequence of function $( U_{n})_{n \geq 1}$ such that $\| U_{n}\| = L_{n}$ with $U_{n}$ satisfies (3.12). We get

$$ U_{n}(t) \geq \xi _{n} \lambda \sigma \int _{0}^{1} G(t,s)U_{n}(s) \frac{q_{N}(s)}{\mathcal{A} \Big(s, \displaystyle{\int _{0}^{1}} j(U_{n}) dy \Big)} ds. $$

Choosing $\tilde{t} \in [\frac{1}{4}, \frac{3}{4}]$,

$$ \min _{t \in [\frac{1}{4} ,\frac{3}{4}] } U_{n}(t) = U_{n}( \tilde{t}), \quad \forall n \in \mathbb{N}^{*}. $$

Since $N=2$,

$$ \begin{aligned} U_{n}( \tilde{t}) & \geq \xi _{n} \lambda \sigma \int _{0}^{1} \Big( \min _{t \in [\frac{1}{4} ,\frac{3}{4}] } G(t,s) \Big) U_{n}(s) \frac{q(s)}{\mathcal{A} \Big(s, \displaystyle{\int _{0}^{1}} j(U_{n}) dy \Big)} ds \\ & \geq \frac{\xi _{n} \lambda \sigma }{ 4\eta \Big( \omega b^{2} \log (\frac{a}{b}) \int _{0}^{1} (\frac{a}{b})^{2y} g(U_{n})dy \Big)} u_{n}(t^{*}) \int _{\frac{1}{4}}^{\frac{3}{4}} G(s,s) \frac{\Big(b^{1-s}a^{s} \log (\frac{b}{a}) \Big)^{2}}{\theta (b^{1-s}a^{s}) } ds \\ & > \xi _{n} U_{n}( \tilde{t}) > U_{n}( \tilde{t}). \end{aligned} $$

If $N \geq 3$, we have

$$ \begin{aligned} U_{n}( \tilde{t}) \geq{}& \xi _{n} \lambda \sigma \int _{0}^{1} \Big( \min _{t \in [\frac{1}{4} ,\frac{3}{4}] } G(t,s) \Big) U_{n}(s) \frac{q(s)}{\mathcal{A} \Big(s, \displaystyle{\int _{0}^{1}} j(U_{n}) dy \Big)} ds \\ \geq{}& \frac{\xi _{n} \lambda \sigma }{ 4 \eta \Big( \frac{\omega}{C(N-2)} \int _{0}^{1} (\frac{C}{B-y})^{2\frac{N-1}{N-2}} g(U_{n}(y))dy \Big)} U_{n}(t^{*})\\ &{}\times \int _{\frac{1}{4}}^{\frac{3}{4}} G(s,s) \frac{C^{\frac{2}{N-2}}}{(N-2)^{2}(B-s)^{2 \frac{N-1}{N-2}} \theta \Big((\frac{C}{B-s})^{\frac{1}{N-2}} \Big)} ds \\ >{}& \xi _{n} U_{n}(\tilde{t} ) > U_{n}( \tilde{t}). \end{aligned} $$

This contradiction confirms the validity of our claim.

Assume that $D= \{ U \in P: \, l \leq \| U \|\leq L \}$. Then, the first condition of Theorem 3.4 is satisfied. For $U \in D$ such that $U= \xi T U$, $0 < \xi < 1$, we will establish that $\|U\| \neq l$. By the absurd, suppose that $\|U\|= l$. If $N=2$, we have

$$ \begin{aligned} l= \| u \| & = \xi \lambda \int _{0}^{1}G(s,s) q_{N}(s) \frac{h(s, U(s))}{\mathcal{A} \Big(s, \displaystyle{\int _{0}^{1}} j(U) dy \Big) } ds \\ & \leq \frac{ \xi \lambda _{l} M_{l}\| U\|}{\eta \Big( \omega b^{2} \log (\frac{a}{b}) \int _{0}^{1} (\frac{a}{b})^{2y} g(U)dy \Big)} \int _{0}^{1}G(s,s) \frac{\Big(b^{1-s}a^{s} \log (\frac{b}{a}) \Big)^{2}}{\theta (b^{1-s}a^{s}) } ds \\ & < \xi \| U \| < \| U \| =l. \end{aligned} $$

Since $N \geq 3$,

$$\begin{aligned} l= \| u \| ={}& \xi \lambda \int _{0}^{1}G(s,s) q_{N}(s) \frac{h(s, u(s))}{\mathcal{A} \Big( s, \displaystyle{\int _{0}^{1}} j(u) dy \Big)} ds \\ \leq{}& \frac{ \xi \lambda _{l} M_{l}\| u\|}{\eta \Big( \frac{\omega}{C(N-2)} \int _{0}^{1} (\frac{C}{B-y})^{2\frac{N-1}{N-2}} g(u_{n})dy \Big)} \\ &{}\times \int _{0}^{1}G(s,s) \frac{C^{\frac{2}{N-2}}}{(N-2)^{2}(B-s)^{2 \frac{N-1}{N-2}} \theta \Big((\frac{C}{B-s})^{\frac{1}{N-2}} \Big)} ds \\ < {}& \xi \| u \| < \| u \| =l. \end{aligned}$$

This leads to a contradiction.

Then, problem (1.2) has a positive solution U, such that $\underset{t \in [0,1]}{\min} U(t) \geq l$. Using the previously mentioned steps, we complete the proof. □