Positive Solutions of a Fractional Thermostat Model with a Parameter

Abstract

We study the existence, multiplicity, and uniqueness results of positive solutions for a fractional thermostat model. Our approach depends on the fixed point index theory, iterative method, and nonsymmetry property of the Green function. The properties of positive solutions depending on a parameter are also discussed.

1. Introduction

In this paper, we investigate a fractional nonlocal boundary value problem (BVP)

$$\left\{\begin{array}{l}{}^{c}D_{0+}^{\alpha }x\left(t\right)+\lambda g\left(t\right)f\left(x\left(t\right)\right)=0,t\in (0,1),\\ x^{\prime }\left(0\right)=0,{\beta }^c{D}_{0+}^{\alpha -1}x\left(1\right)+x\left(\eta \right)=0,\end{array}\right.$$

(1)

where $1<\alpha \le 2,\beta >0,0\le \eta \le 1,\beta \mathsf{\Gamma }\left(\alpha \right)>{(1-\eta )}^{\alpha -1},{}^c{D}_{0+}^{\alpha }$ is the Gerasimov–Caputo fractional derivative of order $\alpha$, $\lambda >0$ is a parameter, $f\in C\left(\right[0,+\infty ),[0,+\infty \left)\right),g\in C\left(\right(0,1),[0,+\infty \left)\right)$, and $0<{\int }_0^{1}g\left(t\right)dt<+\infty$.

One motivation is that the thermostat model

$$\left\{\begin{array}{l}{x}^{″}\left(t\right)+g\left(t\right)f(t,x\left(t\right))=0,t\in (0,1),\\ x^{\prime }\left(0\right)=0,\beta x^{\prime }\left(1\right)+x\left(\eta \right)=0,\end{array}\right.$$

(2)

which is a special case with $\alpha =2$ and $\lambda =1$, has been discussed by Infante and Webb [1,2]. They established multiplicity results of BVP (2). These types of problems have been investigated by various scholars, see References [3,4,5,6,7,8,9,10,11,12,13,14,15,16,17].

Recently, the thermostat model was extended to the fractional case

$$\left\{\begin{array}{l}{}^{c}D_{0+}^{\alpha }x\left(t\right)+f(t,x\left(t\right))=0,t\in (0,1),\alpha \in (1,2],\\ x^{\prime }\left(0\right)=0,{\beta }^c{D}_{0+}^{\alpha -1}x\left(1\right)+x\left(\eta \right)=0,\end{array}\right.$$

(3)

where $\beta >0,0\le \eta \le 1,f\in C\left(\right[0,1]\times [0,+\infty ),[0,+\infty \left)\right)$. Nieto and Pimentel [18] proved the existence of positive solutions based on the Krasnosel’skii fixed point theorem. Cabada and Infante [19] discussed the multiplicity results of positive solutions for BVP (3).

In Reference [20], Shen, Zhou, and Yang studied a fractional thermostat model

$$\left\{\begin{array}{l}{}^{c}D_{0+}^{\alpha }x\left(t\right)+\lambda f(t,x\left(t\right))=0,t\in (0,1),1<\alpha \le 2,\\ x^{\prime }\left(0\right)=0,{\beta }^c{D}_{0+}^{\alpha -1}x\left(1\right)+x\left(\eta \right)=0,\end{array}\right.$$

where $\beta >0,0\le \eta \le 1,\beta \mathsf{\Gamma }\left(\alpha \right)>{(1-\eta )}^{\alpha -1},\lambda >0$, $f:[0,1]\times [0,+\infty )\to [0,+\infty )$ is continuous. The authors obtained intervals of parameter $\lambda$ that correspond to at least one and no positive solutions. Similar fractional thermostat problems have been studied in References [21,22,23,24].

In this paper, we deal with positive solutions for the fractional thermostat model (1). The existence, multiplicity, and uniqueness results are established by the fixed point index theory and iterative method. The properties of positive solutions depending on a parameter are also discussed. Some of the ideas in this paper are from References [25,26]. Let us remark that the definition of the Gerasimov–Caputo derivative was first introduced and applied by Gerasimov in 1947 and then by Caputo in 1967, see for example, the overview by Novozhenova in Reference [27]. For details on the theory and applications of the fractional derivatives and integrals and fractional differential equations, see References [28,29,30,31].

2. Preliminaries

Lemma 1 ([20]).

Given $u\left(t\right)\in C(0,1)\cap L^1(0,1)$, the solution of the problem

$$\left\{\begin{array}{l}{}^{c}D_{0+}^{\alpha }x\left(t\right)+u\left(t\right)=0,t\in (0,1),\\ x^{\prime }\left(0\right)=0,{\beta }^c{D}_{0+}^{\alpha -1}x\left(1\right)+x\left(\eta \right)=0\end{array}\right.$$

is

$$x\left(t\right)={\int }_0^{1}G(t,s)u\left(s\right)ds,t\in [0,1],$$

where

$$G(t,s)=\left\{\begin{array}{l}\beta -\frac{{(t-s)}^{\alpha -1}}{\mathsf{\Gamma }\left(\alpha \right)}+\frac{{(\eta -s)}^{\alpha -1}}{\mathsf{\Gamma }\left(\alpha \right)},0\le s\le \eta ,s\le t,\\ \beta +\frac{{(\eta -s)}^{\alpha -1}}{\mathsf{\Gamma }\left(\alpha \right)},0\le s\le \eta ,s\ge t,\\ \beta -\frac{{(t-s)}^{\alpha -1}}{\mathsf{\Gamma }\left(\alpha \right)},\eta \le s\le 1,s\le t,\\ \beta ,\eta \le s\le 1,s\ge t,\end{array}\right.$$

and $G(t,s)$ satisfies:

(i) 

$G(t,s):[0,1]\times [0,1]\to (0,+\infty )$ is continuous;

(ii) 

$\frac{\partial }{\partial t}G(t,s)\le 0,t,s\in [0,1]$;

(iii) 

$\gamma \overline{G}=\underline{G}\le G(1,s)\le G(t,s)\le G(0,s)\le \overline{G},t,s\in [0,1],$

where

$$\gamma =\frac{\beta \mathsf{\Gamma }\left(\alpha \right)-{(1-\eta )}^{\alpha -1}}{\beta \mathsf{\Gamma }\left(\alpha \right)+{\eta }^{\alpha -1}},\underline{G}=\frac{\beta \mathsf{\Gamma }\left(\alpha \right)-{(1-\eta )}^{\alpha -1}}{\mathsf{\Gamma }\left(\alpha \right)},\overline{G}=\frac{\beta \mathsf{\Gamma }\left(\alpha \right)+{\eta }^{\alpha -1}}{\mathsf{\Gamma }\left(\alpha \right)}.$$

Denote $E=C[0,1]$ and $\parallel x\parallel ={sup}_{t\in [0,1]}\left|x\left(t\right)\right|$. We define the cone

$$P=\{x\in E:x\left(t\right)\ge 0,\underset{t\in [0,1]}{inf}x\left(t\right)\ge \gamma \parallel x\parallel \}.$$

For any $0<r<+\infty$, let $P_r=\{x\in P:\parallel x\parallel <r\}$. We define $T:(0,+\infty )\times E\to E$ as

$$T(\lambda ,x)\left(t\right)=\lambda {\int }_0^{1}G(t,s)g\left(s\right)f\left(x\left(s\right)\right)ds,t\in [0,1].$$

It is obvious from Lemma 1 that if $x\in P$ is a fixed point of operator T, then x is a positive solution of Problem (1). By regularity arguments, we can show that T is completely continuous and $T\left(P\right)\subset P$.

Define the linear operator $L:E\to E$ by

$$Lx\left(t\right)={\int }_0^{1}G(t,s)g\left(s\right)x\left(s\right)ds,t\in [0,1].$$

By the Krein–Rutman theorem, we see that the spectral radius $r\left(L\right)$ of the operator L is positive, and L has positive eigenfunction ${\phi }_1$ corresponding to its first eigenvalue ${\mu }_1={\left(r\left(L\right)\right)}^{-1}$.

Lemma 2 ([32]).

Let P be a cone in Banach space E. Suppose that $T:P\to P$ is a completely continuous operator. (i) If $Tu\ne \mu u$ for any $u\in \partial P_r$ and $\mu \ge 1$, then $i(T,P_r,P)=1$. (ii) If $Tu\ne u$ and $\parallel Tu\parallel \ge \parallel u\parallel$ for any $u\in \partial P_r$, then $i(T,P_r,P)=0$.

Denote

$$f_0=\underset{s\to 0}{lim}\frac{f\left(s\right)}{s},f_{\infty }=\underset{s\to \infty }{lim}\frac{f\left(s\right)}{s},A={\int }_0^{1}G(0,s)g\left(s\right)ds,l=\underset{s\in (0,\infty )}{min}\frac{f\left(s\right)}{s}.$$

We assume that:

(H1)

f is nondecreasing on $[0,+\infty )$;

(H2)

there exists a function $\varphi :(0,1]\to [0,1]$ continuous nondecreasing, such that $f\left(\kappa x\right)\ge \varphi \left(\kappa \right)f\left(x\right)$ for $0<\kappa <1,x>0$, and $F\left(\kappa \right):=\frac{\kappa }{\varphi \left(\kappa \right)}$ is strictly increasing on $(0,1]$ and $F\left(1\right)=1$.

Lemma 3.

Suppose that $\left(H_1\right)$ holds, $f_0=\infty$ and $l>0$. If $0<{\lambda }_1<{\lambda }_2<\frac{1}{lA}$, then there exist $x_1,x_2\in P\setminus \left\{\theta \right\},x_1\le x_2$, such that $T({\lambda }_1,x_1)\left(t\right)=x_1\left(t\right)$ and $T({\lambda }_2,x_2)\left(t\right)=x_2\left(t\right)$.

Proof. 

Assume $s_0\in (0,\infty )$ such that $f\left(s_0\right)=l{s}_0$. Since $0<{\lambda }_1<{\lambda }_2<\frac{1}{lA}$, we have $l<\frac{1}{{\lambda }_{2}A}<\frac{1}{{\lambda }_{1}A}.$ We define

$$x_0\left(t\right)=\frac{s_0}{A}{\int }_0^{1}G(t,s)g\left(s\right)ds,t\in [0,1],$$

then

$$\parallel x_0\parallel =x_0\left(0\right)=s_0,x_0\left(t\right)\ge \frac{s_0}{A}{\int }_0^1\gamma G(0,s)g\left(s\right)ds=\gamma \parallel x_0\parallel ,t\in [0,1].$$

Therefore, $x_0\in P$ and $\parallel x_0\parallel =s_0$. Direct computations yield

$$\begin{array}{cc}\hfill T({\lambda }_1,x_0)\left(t\right)& ={\lambda }_1{\int }_0^{1}G(t,s)g\left(s\right)f\left(x_0\left(s\right)\right)ds\le {\lambda }_1{\int }_0^{1}G(t,s)g\left(s\right)f(\parallel x_0\parallel )ds\hfill \\ \hfill & ={\lambda }_{1}l{s}_0{\int }_0^{1}G(t,s)g\left(s\right)ds<\frac{s_0}{A}{\int }_0^{1}G(t,s)g\left(s\right)ds=x_0\left(t\right),t\in [0,1].\hfill \end{array}$$

Define

$$x_1^1\left(t\right)=T({\lambda }_1,x_0)\left(t\right),x_1^j\left(t\right)=T({\lambda }_1,x_1^{j-1})\left(t\right)=T^j({\lambda }_1,x_0)\left(t\right),j=2,3,\cdots ,t\in [0,1].$$

Direct calculations show that $x_0>x_1^1>x_1^2>\cdots >x_1^j>x_1^{j+1}>\cdots \ge \theta .$ Hence, sequence ${\left\{x_1^j\right\}}_{j=1}^{\infty }$ is decreasing and bounded from below, ${lim}_{j\to \infty }{x}_1^j\left(t\right)$ exists and convergence is uniform for $t\in [0,1]$. Assume that ${lim}_{j\to \infty }{x}_1^j=x_1$, we claim that $x_1\left(t\right)>0$. Otherwise, since $x_1\in P$, $x_1\left(t\right)=0$, i.e., ${lim}_{j\to \infty }{x}_1^j\left(t\right)=0,t\in [0,1]$, and hence from $x_1^j\in P$, we deduce $\parallel x_1^j\parallel \to 0$. Since $f_0=\infty$, for any $H>\frac{1}{{\lambda }_1\gamma A}$, there is integral $Z>0$ such that for $j>Z$, we have $f(x_1^j\left(t\right))>H{x}_1^j\left(t\right)$, and hence

$$\begin{array}{cc}\hfill x_1^{j+1}\left(0\right)& ={\lambda }_1{\int }_0^{1}G(0,s)g\left(s\right)f\left(x_1^j\left(s\right)\right)ds\hfill \\ \hfill & >{\lambda }_{1}H\gamma {\int }_0^{1}G(0,s)g\left(s\right)\parallel x_1^j\parallel ds\hfill \\ \hfill & \ge x_1^j\left(0\right){\lambda }_{1}H\gamma A>x_1^j\left(0\right).\hfill \end{array}$$

The contradiction shows that $x_1\in P\setminus \left\{\theta \right\}$ and $x_1=T({\lambda }_1,x_1)$.

Similarly, from $x_2^1\left(t\right)=T({\lambda }_2,x_0)\left(t\right)$ and $x_2^j\left(t\right)=T({\lambda }_2,x_2^{j-1})\left(t\right),j=2,3,\cdots ,$ we deduce

$$x_0>x_2^1>x_2^2>\cdots >x_2^j>x_2^{j+1}>\cdots \ge \theta ,$$

${lim}_{j\to \infty }{x}_2^j=x_2\in P\setminus \left\{\theta \right\}$, and $x_2=T({\lambda }_2,x_2)$. It follows from $x_1^1=T({\lambda }_1,x_0)<T({\lambda }_2,x_0)=x_2^1$ and the monotonicity of f that $x_1^j\le x_2^j,j=2,3,\cdots$. Therefore, $x_1\le x_2$. ☐

Lemma 4.

If $f_{\infty }=\infty$, then for any $\mu >0$, the set $S_{\mu }=\{x\in P:T(\lambda ,x)=x,\lambda \in [\mu ,\infty )\}$ is bounded.

Proof. 

Otherwise, there exists $x_n\in S_{\mu }$ corresponding to ${\lambda }_n\in [\mu ,\infty )$ such that

$$T({\lambda }_n,x_n)=x_n,\underset{n\to \infty }{lim}\parallel x_n\parallel =\infty .$$

Because $f_{\infty }=\infty$, there is $X>0$ such that $f\left(s\right)>Hs$ for $s>X,$ where $H>\frac{1}{\mu \gamma A}$. Since ${lim}_{n\to \infty }\parallel x_n\parallel =\infty$, there exists $N_0>0$ such that $\parallel x_n\parallel >\frac{X}{\gamma }$ for $n>N_0$, and $x_n\left(t\right)\ge \gamma \parallel x_n\parallel >X,t\in [0,1]$. Then, for any $n>N_0$, we obtain

$$\parallel x_n\parallel >{\lambda }_n{\int }_0^{1}G(0,s)g\left(s\right)H{x}_n\left(s\right)ds>\mu H\gamma \parallel x_n\parallel A>\parallel x_n\parallel ,$$

which is absurd, and hence $S_{\mu }$ is bounded.  ☐

Lemma 5.

Assume that $\left(H_1\right)$ holds, and that $f_0=f_{\infty }=\infty$. Then, T admits a fixed point for $\lambda =\frac{1}{lA}$.

Proof. 

Choosing a sequence $0<{\lambda }_1<{\lambda }_2<\cdots <{\lambda }_n<{\lambda }_{n+1}<\cdots <\frac{1}{lA}$ such that ${lim}_{n\to \infty }{\lambda }_n=\frac{1}{lA}$. By Lemma 3, there exists a nondecreasing sequence ${\left\{x_n\right\}}_{n=1}^{\infty }\subset P\setminus \left\{\theta \right\}$ such that $x_n=T({\lambda }_n,x_n)$. By Lemma 4, we know that ${\left\{x_n\right\}}_{n=1}^{\infty }$ is uniformly bounded and equicontinuous. By using the Arzela–Ascoli theorem, we can prove that there exists ${\left\{x_{n_k}\right\}}_{k=1}^{\infty }\subset {\left\{x_n\right\}}_{n=1}^{\infty }$ such that $x_{n_k}\to \tilde{x}\in E$ uniformly on $[0,1]$. Therefore, $x_{n_k}$ satisfies

$$x_{n_k}\left(t\right)=T({\lambda }_{n_k},x_{n_k})\left(t\right)={\lambda }_{n_k}{\int }_0^{1}G(t,s)g\left(s\right)f\left(x_{n_k}\left(s\right)\right)ds,t\in [0,1].$$

Passing to the limit as $k\to \infty$, we obtain

$$\tilde{x}\left(t\right)=\frac{1}{lA}{\int }_0^{1}G(t,s)g\left(s\right)f\left(\tilde{x}\left(s\right)\right)ds,t\in [0,1].$$

Hence, $\tilde{x}=T\left(\frac{1}{lA},\tilde{x}\right)$. ☐

Lemma 6.

Assume that $\left(H_1\right)$ holds, and that $f\left(0\right)>0$. Then, for any $x\in P$, there exist $U_x\ge V>0$ such that

$$V{K}_{\lambda }\le T(\lambda ,x)\left(t\right)\le U_x{K}_{\lambda },t\in [0,1],$$

where

$$K_{\lambda }=\lambda {\int }_0^{1}g\left(t\right)dt.$$

Proof. 

By $\left(H_1\right)$, for any $x\in P$ and $t\in [0,1]$, we have

$$T(\lambda ,x)\left(t\right)\ge \underline{G}f\left(0\right)\lambda {\int }_0^{1}g\left(t\right)dt:=V{K}_{\lambda },$$

and

$$T(\lambda ,x)\left(t\right)\le \overline{G}f(\parallel x\parallel )\lambda {\int }_0^{1}g\left(t\right)dt:=U_x{K}_{\lambda }.$$

 ☐

3. Main Results

Theorem 1.

Assume that $f_{\infty }=\infty$ and $0<f_0<\infty$. Then, for any $0<\lambda <\frac{{\mu }_1}{f_0}$, BVP (1) admits a positive solution.

Proof. 

Since $0<\lambda <\frac{{\mu }_1}{f_0}$, there exist $\epsilon >0$ small enough and $r>0$ such that $\lambda (f_0+\epsilon )<{\mu }_1$, and $\frac{f\left(s\right)}{s}<f_0+\epsilon$ for $s\in (0,r]$. We claim that

$$T(\lambda ,x)\ne \mu x,x\in \partial P_r,\mu \ge 1.$$

Otherwise, there exist $x_0\in \partial P_r$ and ${\mu }_0\ge 1$ such that $T(\lambda ,x_0)={\mu }_0{x}_0$. Since $0<\gamma r\le x_0\left(t\right)\le \parallel x_0\parallel =r$, we have

$${\mu }_0{x}_0\left(t\right)\le \lambda (f_0+\epsilon ){\int }_0^{1}G(t,s)g\left(s\right)x_0\left(s\right)ds=\lambda (f_0+\epsilon )L{x}_0\left(t\right),$$

then $L{x}_0\left(t\right)\ge \frac{{\mu }_0}{\lambda (f_0+\epsilon )}{x}_0\left(t\right)$. Thus, $r\left(L\right)\ge \frac{{\mu }_0}{\lambda (f_0+\epsilon )}\ge \frac{1}{\lambda (f_0+\epsilon )}$. It follows that ${\mu }_1\le \lambda (f_0+\epsilon )$, which is a contradiction. Then, $i(T,P_r,P)=1.$

Next, we prove that $i(T,P_R,P)=0$ for some $R>r$. In fact, $f_{\infty }=\infty$ implies that $f\left(s\right)>Ms$ for some large $R_1>0$ and $s\ge R_1,$ where $M>{\left(\lambda \gamma A\right)}^{-1}.$ Let $R>max\{r,\frac{R_1}{\gamma }\}$. For $x\in \partial P_R$, we have $x\left(t\right)\ge \gamma \parallel x\parallel =\gamma R>R_1,t\in [0,1]$, then

$$\parallel T(\lambda ,x)\parallel \ge \lambda M{\int }_0^{1}G(0,s)g\left(s\right)x\left(s\right)ds\ge \lambda M\gamma \parallel x\parallel A>\parallel x\parallel .$$

Hence, $i(T,P_R,P)=0$, and $i(T,P_R\setminus {\overline{P}}_r,P)=-1.$ Therefore, T admits a fixed point $x^{*}\in P_R\setminus {\overline{P}}_r$. ☐

Theorem 2.

Assume that $\left(H_1\right)$ holds, and that $f_0=f_{\infty }=\infty$. Then, BVP (1) has at least one and two positive solutions for $\lambda =\frac{1}{lA}$ and $\lambda \in (0,\frac{1}{lA})$, respectively.

Proof. 

By Lemma 5, BVP (1) admits a positive solution for $\lambda =\frac{1}{lA}$. For $\lambda \in (0,\frac{1}{lA})$, by Lemmas 3 and 5, there exist $\tilde{x},x_{\lambda }\in P\setminus \left\{\theta \right\},x_{\lambda }\le \tilde{x}$ such that

$$T\left(\frac{1}{lA},\tilde{x}\right)\left(t\right)=\tilde{x}\left(t\right),T(\lambda ,x_{\lambda })\left(t\right)=x_{\lambda }\left(t\right),t\in [0,1].$$

If $x_{\lambda }=\tilde{x}$, we have

$$T(\lambda ,x_{\lambda })=x_{\lambda }=\tilde{x}=T\left(\frac{1}{lA},\tilde{x}\right)=T\left(\frac{1}{lA},x_{\lambda }\right).$$

This contradiction shows that $x_{\lambda }<\tilde{x}$.

Define ${\mathsf{\Omega }}_1=\{x\in E:-r<x\left(t\right)<\tilde{x}\left(t\right),t\in [0,1]\},$ where $r>0$ is the same as in the first part of Theorem 1. For any $x\in P\cap \partial {\mathsf{\Omega }}_1$, we obtain $\parallel x\parallel =\parallel \tilde{x}\parallel$, and

$$\parallel T(\lambda ,x)\parallel <\frac{1}{lA}{\int }_0^{1}G(0,s)g\left(s\right)f\left(\tilde{x}\left(s\right)\right)ds=\tilde{x}\left(0\right)=\parallel \tilde{x}\parallel .$$

Therefore,

$$\parallel T(\lambda ,x)\parallel <\parallel x\parallel ,x\in P\cap \partial {\mathsf{\Omega }}_1.$$

As in the proof in Theorem 1, there is $R>0$ large enough such that

$$\parallel T(\lambda ,x)\parallel >\parallel x\parallel ,x\in P\cap \partial {\mathsf{\Omega }}_2,$$

where ${\mathsf{\Omega }}_2=\{x\in E:\parallel x\parallel <R\}$. By compression expansion fixed point theorem, we see that T has a fixed point ${\overline{x}}_{\lambda }\in P\cap ({\mathsf{\Omega }}_2\setminus {\overline{\mathsf{\Omega }}}_1)$. Since $x_{\lambda }\in {\mathsf{\Omega }}_1,x_{\lambda }\ne {\overline{x}}_{\lambda }$, problem (1) has a second positive solution. ☐

Theorem 3.

Assume that $\left(H_1\right)$ and $\left(H_2\right)$ hold, and that $f\left(0\right)>0$. Then, for any $\lambda \in (0,\infty )$, BVP (1) admits a unique positive solution ${\dot{x}}_{\lambda }\left(t\right)$, and ${\dot{x}}_{\lambda }\left(t\right)$ satisfies:

(i) 

${\dot{x}}_{\lambda }\left(t\right)$ is nondecreasing with respect to λ;

(ii) 

${lim}_{\lambda \to 0^{+}}\parallel {\dot{x}}_{\lambda }\parallel =0,{lim}_{\lambda \to \infty }\parallel {\dot{x}}_{\lambda }\parallel =\infty$;

(iii) 

$\parallel {\dot{x}}_{\lambda }-{\dot{x}}_{{\lambda }_0}\parallel \to 0$ as $\lambda \to {\lambda }_0$.

Proof. 

Since T is nondecreasing, for $u\in P$, we have

$$T(\lambda ,\kappa x)\left(t\right)\ge \varphi \left(\kappa \right)\lambda {\int }_0^{1}G(t,s)g\left(s\right)f\left(x\left(s\right)\right)ds=\varphi \left(\kappa \right)T(\lambda ,x)\left(t\right),t\in [0,1].$$

(4)

Define $\widehat{x}\left(t\right)=K_{\lambda }$, where $K_{\lambda }$ is given by Lemma 6, then $\widehat{x}\in P$ and $V{K}_{\lambda }\le T(\lambda ,\widehat{x})\left(t\right)\le U_x{K}_{\lambda }.$ Denote

$$\overline{V}=sup\{\mu :\mu K_{\lambda }\le T(\lambda ,\widehat{x})\left(t\right)\},\overline{U}=inf\{\mu :\mu K_{\lambda }\ge T(\lambda ,\widehat{x})\left(t\right)\},$$

then $\overline{V}\ge V$ and $\overline{U}\le U_x$. Select $\tilde{V}$ and $\tilde{U}$ so that

$$0<\tilde{V}<min\{1,F^{-1}\left(\overline{V}\right)\},0<\frac{1}{\tilde{U}}<min\left\{1,F^{-1}\left(\frac{1}{\overline{U}}\right)\right\}.$$

We define

$$x_1\left(t\right)=\tilde{V}{K}_{\lambda },x_{k+1}\left(t\right)=T(\lambda ,x_k)\left(t\right),t\in [0,1],k=1,2,\cdots ,$$

$$y_1\left(t\right)=\tilde{U}{K}_{\lambda },y_{k+1}\left(t\right)=T(\lambda ,y_k)\left(t\right),t\in [0,1],k=1,2,\cdots .$$

Combining the properties of T and (4), we get

$$\tilde{V}{K}_{\lambda }=x_1\left(t\right)\le x_2\left(t\right)\le \cdots \le x_k\left(t\right)\le \cdots \le y_k\left(t\right)\le \cdots \le y_2\left(t\right)\le y_1\left(t\right)=\tilde{U}{K}_{\lambda }.$$

(5)

Let $d=\frac{\tilde{V}}{\tilde{U}}$, obviously $0<d<1$. We claim that

$$x_k\left(t\right)\ge {\varphi }_{k-1}\left(d\right)y_k\left(t\right),t\in [0,1],k=1,2,\cdots ,$$

(6)

where ${\varphi }_0\left(d\right)=d,{\varphi }_k\left(d\right)=\varphi \left({\varphi }_{k-1}\left(d\right)\right),k=1,2,\cdots .$ In fact, $x_1\left(t\right)=d{y}_1\left(t\right)={\varphi }_0\left(d\right)y_1\left(t\right),t\in [0,1]$. Suppose $x_n\left(t\right)\ge {\varphi }_{n-1}\left(d\right)y_n\left(t\right)$ for $t\in [0,1]$, then

$$x_{n+1}\left(t\right)\ge T(\lambda ,{\varphi }_{n-1}\left(d\right)y_n)\left(t\right)\ge \varphi \left({\varphi }_{n-1}\left(d\right)\right)T(\lambda ,y_n)\left(t\right)={\varphi }_n\left(d\right)y_{n+1}\left(t\right).$$

Hence, it follows by induction that (6) is true. According to (5) and (6), one has

$$0\le x_{n+m}\left(t\right)-x_n\left(t\right)\le y_n\left(t\right)-x_n\left(t\right)\le (1-{\varphi }_{n-1}\left(d\right))y_1\left(t\right)=(1-{\varphi }_{n-1}\left(d\right))\tilde{U}{K}_{\lambda },$$

where $m\ge 0$ is an integer. Thus,

$$\parallel x_{n+m}-x_n\parallel \le \parallel y_n-x_n\parallel \le (1-{\varphi }_{n-1}\left(d\right))\tilde{U}{K}_{\lambda }.$$

(7)

We claim that ${lim}_{n\to \infty }{\varphi }_n\left(d\right)=1$. From $\left(H_2\right)$ and $0<d<1$, we see that $\varphi \left(d\right)\in (d,1)$ and $d={\varphi }_0\left(d\right)<{\varphi }_1\left(d\right)<\cdots <{\varphi }_n\left(d\right)<\cdots <1$. Sequence ${\left\{{\varphi }_n\left(d\right)\right\}}_{n=1}^{\infty }$ is increasing and bounded, there is $p\in [d,1]$ such that ${lim}_{n\to \infty }{\varphi }_n\left(d\right)=p$. By the continuity of $\varphi$ and ${\varphi }_n\left(d\right)=\varphi \left({\varphi }_{n-1}\left(d\right)\right)$, we conclude that $p=\varphi \left(p\right)$, i.e., $F\left(p\right)=1.$ It follows that $p=1.$ Inequality (7) implies that there exists $\overline{x}\in P$ such that ${lim}_{n\to \infty }{x}_n\left(t\right)={lim}_{n\to \infty }{y}_n\left(t\right)=\overline{x}\left(t\right)$ for $t\in [0,1]$. Clearly, $\overline{x}\left(t\right)$ is a positive solution of problem (1).

Suppose that ${\overline{x}}_1\left(t\right)$ and ${\overline{x}}_2\left(t\right)$ are positive solutions of problem (1), then $T(\lambda ,{\overline{x}}_1)\left(t\right)={\overline{x}}_1\left(t\right)$ and $T(\lambda ,{\overline{x}}_2)\left(t\right)={\overline{x}}_2\left(t\right),t\in [0,1]$. Define $\tilde{\delta }=sup\{\delta :{\overline{x}}_1\left(t\right)\ge \delta {\overline{x}}_2\left(t\right)\},$ then ${\overline{x}}_1\left(t\right)\ge \tilde{\delta }{\overline{x}}_2\left(t\right)$. We claim that $\tilde{\delta }\ge 1$. Otherwise, $\tilde{\delta }<1$. Assumption $\left(H_2\right)$ implies that $f(\tilde{\delta }{\overline{x}}_2\left(t\right))>\phi (\tilde{\delta })f\left({\overline{x}}_2\left(t\right)\right),t\in [0,1]$. Since f is nondecreasing,

$${\overline{x}}_1\left(t\right)=T(\lambda ,{\overline{x}}_1)\left(t\right)\ge T(\lambda ,\tilde{\delta }{\overline{x}}_2)\left(t\right)>\varphi (\tilde{\delta })T(\lambda ,{\overline{x}}_2)\left(t\right)>\tilde{\delta }{\overline{x}}_2\left(t\right),t\in [0,1],$$

a contradiction. Then, ${\overline{x}}_1\left(t\right)\ge {\overline{x}}_2\left(t\right)$ for $t\in [0,1]$. Similarly, ${\overline{x}}_2\left(t\right)\ge {\overline{x}}_1\left(t\right)$. Therefore, ${\overline{x}}_1\left(t\right)={\overline{x}}_2\left(t\right),t\in [0,1]$. This proves the uniqueness result.

Next, we show that $\left(i\right)-\left(iii\right)$ hold. Let

$$\left(Hx\right)\left(t\right)={\int }_0^{1}G(t,s)g\left(s\right)f\left(x\left(s\right)\right)ds,t\in [0,1],$$

then $T(\lambda ,x)=\lambda Hx$. Since $P^o=\{x\in P:x\left(t\right)>0,t\in [0,1]\}$ is nonempty, the operator $H:P^o\to P^o$ is increasing, and $H\left(\kappa x\right)\ge \varphi \left(\kappa \right)Hx$ for $0<\kappa <1$. Let $\omega =\frac{1}{\lambda }$. We now write $H{x}_{\omega }=\omega x_{\omega }$ instead of $\lambda H{x}_{\lambda }=x_{\lambda }.$ Assume $0<{\omega }_1<{\omega }_2$, then $x_{{\omega }_1}\ge x_{{\omega }_2}$. Indeed, denote $\overline{\omega }=sup\{t>0:x_{{\omega }_1}\ge t{x}_{{\omega }_2}\}$, then $\overline{\omega }\ge 1.$ Otherwise $0<\overline{\omega }<1$. Direct computations yield ${\omega }_1{x}_{{\omega }_1}=H{x}_{{\omega }_1}\ge H\left(\overline{\omega }{x}_{{\omega }_2}\right)\ge \varphi \left(\overline{\omega }\right)H{x}_{{\omega }_2}=\varphi \left(\overline{\omega }\right){\omega }_2{x}_{{\omega }_2}$, then $x_{{\omega }_1}\ge \frac{{\omega }_2}{{\omega }_1}\varphi \left(\overline{\omega }\right)x_{{\omega }_2}>\overline{\omega }{x}_{{\omega }_2}$. This is a contradiction to the definition of $\overline{\omega }$. Thus, $\overline{\omega }\ge 1,x_{{\omega }_1}\ge x_{{\omega }_2}$, and further

$$x_{{\omega }_1}=\frac{1}{{\omega }_1}H{x}_{{\omega }_1}\ge \frac{1}{{\omega }_1}H{x}_{{\omega }_2}=\frac{{\omega }_2}{{\omega }_1}{x}_{{\omega }_2}\gg x_{{\omega }_2},0<{\omega }_1<{\omega }_2.$$

(8)

Then, $x_{\omega }\left(t\right)$ is strong decreasing in $\omega$, that is, $x_{\lambda }\left(t\right)$ is strong increasing in $\lambda$. Let ${\omega }_2=\omega$ and fix ${\omega }_1$ in (8), for $\omega >{\omega }_1$, we have $x_{{\omega }_1}\ge \frac{\omega }{{\omega }_1}{x}_{\omega }$, and

$$\parallel x_{\omega }\parallel \le \frac{N{\omega }_1}{\omega }\parallel x_{{\omega }_1}\parallel ,$$

where $N>0$ is a normal constant of cone P. Because $\omega =\frac{1}{\lambda }$, then ${lim}_{\lambda \to 0^{+}}\parallel x_{\lambda }\parallel =0$. Let ${\omega }_1=\omega$ and fix ${\omega }_2$ in (8), we obtain ${lim}_{\lambda \to +\infty }\parallel x_{\lambda }\parallel =+\infty .$

Finally, for given ${\omega }_0$, by (8), we have

$$x_{\omega }\ll x_{{\omega }_0},\omega >{\omega }_0.$$

(9)

Let $t_{\omega }=sup\{t>0:x_{\omega }\ge t{x}_{{\omega }_0},\omega >{\omega }_0\}$, then $0<t_{\omega }<1$ and $x_{\omega }\ge t_{\omega }{x}_{{\omega }_0}$. Direct computations yield $\omega x_{\omega }=H{x}_{\omega }\ge H\left(t_{\omega }{x}_{{\omega }_0}\right)\ge \varphi \left(t_{\omega }\right)H{x}_{{\omega }_0}=\varphi \left(t_{\omega }\right){\omega }_0{x}_{{\omega }_0}.$ By the definition of $t_{\omega }$, we have $\frac{{\omega }_0}{\omega }\varphi \left(t_{\omega }\right)\le t_{\omega }$, and

$$t_{\omega }\ge F^{-1}\left(\frac{{\omega }_0}{\omega }\right),\forall \omega >{\omega }_0.$$

(10)

Combining (9) with (10), one has that

$$\parallel x_{{\omega }_0}-x_{\omega }\parallel \le N\left[1-F^{-1}\left(\frac{{\omega }_0}{\omega }\right)\right]\parallel x_{{\omega }_0}\parallel \to 0,\omega \to {\omega }_0+0.$$

Similarly, $\parallel x_{{\omega }_0}-x_{\omega }\parallel \to 0,\omega \to {\omega }_0-0$. Hence, $\parallel x_{{\omega }_0}-x_{\omega }\parallel \to 0$ as $\omega \to {\omega }_0$.  ☐