Positive solutions to a fractional thermostat model in Banach spaces via fixed point results

Abstract

The motive behind this manuscript is to set up the existence and uniqueness of positive solutions to a fractional thermostat model for certain values of the parameter $\lambda >0$. We accomplish sufficient conditions for the existence of positive solutions to the model, and afterwards formulate a couple of non-trivial examples to authenticate the grounds of our obtained results. Our findings are based on certain fixed point results for contractions depending on a couple of altering distance functions $\phi$ and $\psi$ in the setting of Banach spaces.

1 Introduction and preliminaries

Metric fixed point theory is extensively employed in different mathematical branches as well as in real-world problems originating in applied sciences. The results on fixed points of contractive maps considered on different underlying spaces are mostly applied on the validation of the existence and uniqueness of solutions of functional, differential or integral equations. The plurality of these types of problems elicits the probe to more and better techniques, which is a salient feature of the recent research works in this literature.

The dawning of fixed point theory on a complete metric space is integrated with the Banach contraction principle due to Banach [6].

Theorem 1.1

Let (X, d) be a complete metric space and T be a self-mapping on X satisfying

$$\begin{aligned} d(Tx,Ty)\le kd(x,y) \end{aligned}$$

for all $x,y \in X$ and $k\in [0,1)$. Then T has a unique fixed point $z\in X$, and for any $x\in X$, the sequence of iterates $\{T^nx\}$ converges to z.

Because of its inferences and huge usability in mathematical theory, Banach contraction principle has been improved and generalized in metric spaces, partially ordered metric spaces, Banach spaces and many other spaces, see [1, 3, 4, 7, 11,12,13,14, 17, 21, 24].

In 1962, Rakotch [23] proved that the Theorem 1.1 still holds if the constant k is replaced by a contraction monotone decreasing function. He proved the following theorem as a corollary.

Theorem 1.2

Let (X, d) be a complete metric space and $T:X\rightarrow X$ be a mapping such that

$$\begin{aligned} d(Tx,Ty) \le \alpha (x,y)d(x,y) \end{aligned}$$

for all $x,y \in X$, where $\alpha$ is a function defined on $[0,\infty )$ satisfying the following conditions:

• (i) $\alpha (x,y)=\alpha (d(x,y)),~~ i.e.,~~ \alpha$ is dependent on the distance of x and y only;

• (ii) $0\le \alpha (\tau ) <1$ for all $\tau >0$;

• (iii) $\alpha (\tau )$ is monotonically decreasing function of $\tau .$

Then T has a unique fixed point.

In his research article, Jaggi [19] used the continuity and some different contractive conditions on the mapping to attain the succeeding result.

Theorem 1.3

Let f be a continuous self-map defined on a complete metric space (X, d). Further let f satisfy the following condition:

$$\begin{aligned} d(f(x),f(y)) \le \frac{\alpha d(x,f(x))d(y,f(y))}{d(x,y)} + \beta d(x,y) \end{aligned}$$

for all $x,y \in X,$ with $x\ne y$ and for some $\alpha , \beta \in [0,1)$ with $\alpha + \beta <1$. Then f has a unique fixed point in X.

In this connection, the readers are referred to the pertinent papers [25, 26] for more interesting results.

Thereafter, Khan et al. [20] extended and generalized the Banach principle using a control function, known as altering distance function.

Definition 1.4

A function $\varphi :[0,\infty )\rightarrow [0,\infty )$ is called an altering distance function if it satisfies the following conditions:

• (i) $\varphi$ is monotone increasing and continuous;

• (ii) $\varphi (t)=0$ if and only if $t=0$.

In [20], the authors also proved the following fixed point theorem by means of the newly originated concept of control functions.

Theorem 1.5

Let (X, d) be a complete metric space and $\psi :[0,\infty )\rightarrow [0,\infty )$. Also suppose that $f:X\rightarrow X$ is a mapping satisfying

$$\begin{aligned} \psi (d(fx,fy)) \le a\psi (d(x,y)) \end{aligned}$$

for all $x,y \in X$ and for some $0\le a<1$. Then f has a unique fixed point.

Alber and Guerre-Delabriere [2] introduced the notion of weak contractions in a Hilbert space.

Definition 1.6

[2]. Let (X, d) be a metric space. A mapping $T:X \rightarrow X$ is called weakly contractive if and only if

$$\begin{aligned} d(Tx,Ty) \le d(x,y) - \phi (d(x,y)) \end{aligned}$$

for all $x,y \in X$, where $\phi$ is an altering distance function.

In 2015, Salazar and Reich [28] proved that a self-mapping defined on a bounded set is of Rakotch type contractive map iff it is a weak contraction in the sense of Guerre-Delabriere. Rhoades [27] generalized the weak contraction condition in metric spaces and proved the following fixed point result in complete metric spaces.

Theorem 1.7

Let (X, d) be a complete metric space. If $T:X \rightarrow X$ is a weakly contractive map, then T has a unique fixed point.

In their research paper, Dutta and Choudhury [16] generalized Theorems 1.5 and 1.7 to obtain the following theorem.

Theorem 1.8

Let (X, d) be a complete metric space and $T:X\rightarrow X$ be a mapping satisfying

$$\begin{aligned} \psi (d(Tx,Ty)) \le \psi (d(x,y)) - \phi (d(x,y)) \end{aligned}$$

for all $x,y \in X$, where $\psi ~~ and ~~ \phi$ are two altering distance functions. Then T has a unique fixed point.

Fractional calculus has been explored for many decades mostly as a pure analytic mathematical branch. Though in recent times, many authors are showing a lot of interest in its applications for solving ordinary differential equations. Fractional differential equations appear in different engineering and scientific branches as the mathematical modeling of systems and techniques in the domains of physics, chemistry, aerodynamics, robotics and many more. For a few recent articles in this direction, see [5, 8, 9, 15, 18, 22, 29] and the references in that respect.

Considering exclusively positive solutions are effective for several applications, inspired by the aforementioned works, in our draft, we set up an existence and uniqueness theorem to find positive solutions to a fractional thermostat model with a positive parameter. With a view to inspect the solutions, we enquire into some new fixed point results in a Banach space by considering a pair of altering distance functions in a more adequate appearance. We also extend our results in a Banach space which is equipped with an arbitrary binary relation and keeps the order-preserving property of the mappings. Finally, some suitable constructive examples are furnished to substantiate the effectiveness of our results.

2 Fixed point results

This section deals with the results on the existence and uniqueness of fixed points of maps satisfying a contractive condition with a pair of control functions in a Banach space and also their proofs. Moreover, we formulate an example to elucidate our attained results.

Theorem 2.1

Let $(X,\Vert .\Vert )$ be a Banach space and C be a closed subset of X. Let $T : C \rightarrow C$ be a mapping. Assume that there exist two altering distance functions $\phi , \psi :[0,\infty )\rightarrow [0,\infty )$ such that

$$\begin{aligned} \phi (\Vert Tx-Ty\Vert )&\le \phi (\Vert Tx-y\Vert )-\psi (\Vert x-y\Vert ) \end{aligned}$$

(2.1)

for all $x,y \in C$. Then T has a unique fixed point in C.

Proof

Let $x_0\in C$ be arbitrary but fixed. Consider, the iterated sequence $\{x_n\}$ where $x_n=T^nx_0$ for each natural number n.

Therefore, by the given condition we have

$$\begin{aligned}&\phi (\Vert Tx_{n-1}-Tx_{m-1}\Vert )\le \phi (\Vert Tx_{n-1}-x_{m-1}\Vert )-\psi (\Vert x_{n-1}-x_{m-1}\Vert ) \nonumber \\&\qquad \Rightarrow \phi (\Vert x_n-x_m\Vert )\le \phi (\Vert x_n-x_{m-1}\Vert )-\psi (\Vert x_{n-1}-x_{m-1}\Vert ), \end{aligned}$$

(2.2)

which implies that

$$\begin{aligned} \phi (\Vert x_n-x_m\Vert )\le \phi (\Vert x_n-x_{m-1}\Vert ) \end{aligned}$$

for all $n,m\in {\mathbb {N}}$. Since $\phi$ is monotone increasing, we have

$$\begin{aligned} \Vert x_n-x_m\Vert \le \Vert x_n-x_{m-1}\Vert \end{aligned}$$

for all $n,m\in {\mathbb {N}}$.

Interchanging the role of $x_n$ and $x_m$ in the above equation, we get

$$\begin{aligned} \Vert x_n-x_m\Vert \le \Vert x_{n-1}-x_m\Vert \end{aligned}$$

(2.3)

for all $n,m\in {\mathbb {N}}$.

Thus, for each fixed $n\in {\mathbb {N}}$, we can conclude that the sequence $\{s^{(n)}_m\}_{m\in {\mathbb {N}}}$ of non-negative real numbers is monotone decreasing, where $s^{(n)}_m=\Vert x_n-x_m\Vert$ for each $m\in {\mathbb {N}}$. So, $\{s^{(n)}_m\}_{m\in {\mathbb {N}}}$ is convergent for each $n\in {\mathbb {N}}$.

Let

$$\begin{aligned} \lim _{m\rightarrow \infty }s^{(n)}_m=a^{(n)} \end{aligned}$$

for each $n\in {\mathbb {N}}.$

Now from Eq. (2.2), we have

$$\begin{aligned} \phi (\Vert x_n-x_m\Vert )+\psi (\Vert x_{n-1}-x_{m-1}\Vert )\le \phi (\Vert x_n-x_{m-1}\Vert ). \end{aligned}$$

Keeping n fixed, taking limit as $m\rightarrow \infty$ on both sides of the above equation and using the continuity of $\phi , \psi$ on $[0,\infty )$, we get

$$\begin{aligned}&\lim _{m\rightarrow \infty }\phi (\Vert x_n-x_m\Vert )+ \lim _{m\rightarrow \infty }\psi (\Vert x_{n-1}-x_{m-1}\Vert ) \le \lim _{m\rightarrow \infty }\phi (\Vert x_n-x_{m-1}\Vert ) \\&\quad \Rightarrow \phi ( \lim _{m\rightarrow \infty }\Vert x_n-x_m\Vert )+\psi ( \lim _{m\rightarrow \infty }\Vert x_{n-1}-x_{m-1}\Vert ) \le \phi ( \lim _{m\rightarrow \infty }\Vert x_n-x_{m-1}\Vert )\\&\quad \Rightarrow \phi (a^{(n)})+\psi (a^{(n-1)}) \le \phi (a^{(n)})\\&\quad \Rightarrow \psi (a^{(n-1)}) \le 0\\&\quad \Rightarrow \psi (a^{(n-1)}) = 0\\&\quad \Rightarrow a^{(n-1)}=0 \hbox { [since, }\psi (t)=0 \, \hbox {if and only if} \, t=0 ]. \end{aligned}$$

Therefore, $a^{(n)}=0$ for all $n\in {\mathbb {N}}$, i.e., $\lim _{m\rightarrow \infty }\Vert x_n-x_m\Vert =0$ for all $n\in {\mathbb {N}}$.

Now, we consider the sequence of functions $\{f_m\}$ defined on C by

$$\begin{aligned}f_m(x)= \left\{ \begin{array}{ll} \Vert x_n-x_m\Vert , &{}\quad \text{ if } x =x_n \text{ for } \text{ some } n\in {\mathbb {N}};\\ 0, &{}\quad \text{ otherwise }. \end{array} \right. \end{aligned}$$

Therefore, $\lim _{m\rightarrow \infty }f_m(x)=0$ for all $x\in C.$ Thus, the limit function f of the sequence of functions $\{f_m\}$ is given by

$$\begin{aligned} f(x)=0\quad \text{ for } \text{ all } x\in C. \end{aligned}$$

Now, let

$$\begin{aligned} M_m= \sup _{x\in C}|f_m(x)-f(x)|. \end{aligned}$$

Therefore,

$$\begin{aligned} M_m&= \sup _{x\in C}|f_m(x)| \text{[since }, \quad f(x)=0 \quad \hbox {for all}\quad { x\in C]}\\&= \sup _{n}|f_m(x_n)|\\&= \sup _{n}\Vert x_n-x_m\Vert . \end{aligned}$$

But, we know from (2.3) that

$$\begin{aligned} \Vert x_n-x_m\Vert \le \Vert x_{n-1}-x_m\Vert \le \Vert x_{n-2}-x_m\Vert \le \cdots \le \Vert x_1-x_m\Vert , \end{aligned}$$

which implies that

$$\begin{aligned}&\sup _{n}\Vert x_n-x_m\Vert \le \Vert x_1-x_m\Vert \\&\quad \Rightarrow M_m \le \Vert x_1-x_m\Vert \\&\quad \Rightarrow \lim _{m\rightarrow \infty } M_m \le \lim _{m\rightarrow \infty }\Vert x_1-x_m\Vert =0\\&\quad \Rightarrow \lim _{m\rightarrow \infty } M_m=0. \end{aligned}$$

Let $\epsilon >0$ be arbitrary. Since, $\lim _{m\rightarrow \infty } M_m=0$, there exists a natural number N such that

$$\begin{aligned}&|M_m|< \epsilon \text{ for } \text{ all } m\ge N\\&\quad \Rightarrow \sup _{x\in C} |f_m(x)-f(x)|<\epsilon \text{ for } \text{ all } m\ge N\\&\quad \Rightarrow |f_m(x)-f(x)|< \epsilon \text{ for } \text{ all } m\ge N \text{ and } \text{ for } \text{ all } x\in C\\&\quad \Rightarrow |f_m(x)|< \epsilon \text{ for } \text{ all } m\ge N \text{ and } \text{ for } \text{ all } x\in C. \end{aligned}$$

In particular, we have

$$\begin{aligned} |f_m(x_n)|< \epsilon \text{ for } \text{ all } m\ge N \text{ and } \text{ for } \text{ all } n\in {\mathbb {N}}. \end{aligned}$$

Therefore, we can write

(2.4)

for all $n,m \ge N$.

Next, we consider the double sequence $\{s_{nm}\}_{n,m \in {\mathbb {N}}}$ of real numbers, where

$$\begin{aligned} s_{nm}=\Vert x_n-x_m\Vert \end{aligned}$$

for all $n,m \in {\mathbb {N}}.$ Here using (2.4), we have

$$\begin{aligned} |s_{nm}-0|< \epsilon \quad \text{ for } \text{ all } n,m \ge N. \end{aligned}$$

This implies the double sequence $\{s_{nm}\}_{n,m \in {\mathbb {N}}}$ converges to 0, i.e,

$$\begin{aligned} \lim _{n,m\rightarrow \infty }\Vert x_n-x_m\Vert =0. \end{aligned}$$

Thus, $\{x_n\}$ is a Cauchy sequence in C. C being complete, $\{x_n\}$ must converge to some $z\in C.$

Now from (2.1), we have

$$\begin{aligned}&\phi (\Vert x_{n+1}-Tz\Vert ) \le \phi (\Vert x_{n+1}-z\Vert )-\psi (\Vert x_n-z\Vert )\\&\quad \Rightarrow \phi (\Vert x_{n+1}-Tz\Vert ) \le \phi (\Vert x_{n+1}-z\Vert )\\&\quad \Rightarrow \lim _{n\rightarrow \infty }\phi (\Vert x_{n+1}-Tz\Vert ) \le \lim _{n\rightarrow \infty }\phi (\Vert x_{n+1}-z\Vert )\\&\quad \Rightarrow \phi \left( \lim _{n\rightarrow \infty }\Vert x_{n+1}-Tz\Vert \right) \le \phi \left( \lim _{n\rightarrow \infty }\Vert x_{n+1}-z\Vert \right) \\&\quad \Rightarrow \phi \left( \lim _{n\rightarrow \infty }\Vert x_{n+1}-Tz\Vert \right) \le \phi (0)=0\\&\quad \Rightarrow \phi \left( \lim _{n\rightarrow \infty }\Vert x_{n+1}-Tz\Vert \right) =0\\&\quad \Rightarrow \lim _{n\rightarrow \infty }\Vert x_{n+1}-Tz\Vert =0. \end{aligned}$$

The above equation shows that the sequence $\{x_n\}$ converges to Tz. Thus, $Tz=z$ and z is a fixed point of T.

Finally, we check the uniqueness of the fixed point z. To check this, let $z_1$ be another fixed point of T, i.e., $Tz_1=z_1$.

From (2.1) and using Definition 1.4, we have

$$\begin{aligned}&\phi (\Vert Tz-Tz_1\Vert )\le \phi (\Vert Tz-z_1\Vert )-\psi (\Vert z-z_1\Vert )\\&\quad \Rightarrow \phi (\Vert z-z_1\Vert )+\psi (\Vert z-z_1\Vert )\le \phi (\Vert z-z_1\Vert )\\&\quad \Rightarrow \psi (\Vert z-z_1\Vert )\le 0\\&\quad \Rightarrow \psi (\Vert z-z_1\Vert )=0\\&\quad \Rightarrow z=z_1. \end{aligned}$$

Therefore, z is the only fixed point of T. $\square$

Now, we generalize Theorem 2.1 in a Banach space which is equipped with an arbitrary binary relation and state the subsequent theorem.

Theorem 2.2

Let $(X,\Vert .\Vert )$ be a Banach space and ${\mathscr {R}}$ be an equivalence relation on X. Assume that X has the property that if $\{x_n\}$ be any sequence in X converging to $z\in X$ and $x_n {\mathscr {R}}x_m$ for each pair of natural numbers n and m, then $x_n {\mathscr {R}} z$ for each natural number n. Let C be a closed subset of X and $T : C \rightarrow C$ be a mapping such that T satisfies the following conditions:

• (i) T is order-preserving with respect to ${\mathscr {R}}$,

• (ii) $\phi (\Vert Tx-Ty\Vert )\le \phi (\Vert Tx-y\Vert )-\psi (\Vert x-y\Vert )$ for all $x,y\in C$ such that $x{\mathscr {R}}y$,

where $\phi , \psi :[0,\infty )\rightarrow [0,\infty )$ are two altering distance functions. Then T has a unique fixed point in C if there exists $x_0 \in X$ such that $x_0 {\mathscr {R}} Tx_0$.

Proof

The proof of this theorem is analogous to the previous one and so omitted. $\square$

In the next portion of this section, we present a result which not only gives the guarantee of existence of fixed point but also properly points out the fixed point.

Theorem 2.3

Let $(X,\Vert .\Vert )$ be a Banach space and C be a closed subspace of X. Let $T : C \rightarrow C$ be a mapping. Also assume that there exist two altering distance functions $\phi , \psi :[0,\infty )\rightarrow [0,\infty )$ such that T satisfies the following conditions:

• (i) $\phi (\Vert Tx-Ty\Vert ) \le \phi (\Vert x-y\Vert )-\psi (\Vert x-y\Vert )$,

• (ii) $\phi (\Vert Tx-y\Vert ) \le \phi (\Vert x-y\Vert )-\psi (\Vert x-y\Vert )$

for all $x,y \in C$. Then the null vector of X is the only fixed point of T.

Proof

Let $x_0 \in C$ be arbitrary but fixed and consider the iterated sequence $\{x_n\}$ where $x_n=T^nx_0$ for all $n \in {\mathbb {N}}$.

Let $s_n=\Vert x_n-x_{n+1}\Vert$ for all $n \in {\mathbb {N}}$. Now, by condition (i) we get

$$\begin{aligned}&\phi (\Vert Tx_n-Tx_{n+1}\Vert ) \le \phi (\Vert x_n-x_{n+1}\Vert )-\psi (\Vert x_n-x_{n+1}\Vert )\\&\quad \Rightarrow \phi (\Vert x_{n+1}-x_{n+2}\Vert ) \le \phi (\Vert x_n-x_{n+1}\Vert )\\&\quad \Rightarrow \phi (s_{n+1}) \le \phi (s_n)\\&\quad \Rightarrow s_{n+1} \le s_n. \end{aligned}$$

This is true for all natural numbers n, which implies that $\{s_n\}$ is a decreasing sequence of non-negative reals and hence this sequence must converge. Let

$$\begin{aligned} \lim _{n \rightarrow \infty }s_n=a. \end{aligned}$$

Again, from (i) we have

$$\begin{aligned}&\phi (s_{n+1}) \le \phi (s_n)-\psi (s_n)\\&\quad \Rightarrow \lim _{n \rightarrow \infty }\phi (s_{n+1}) \le \lim _{n \rightarrow \infty }\phi (s_n)- \lim _{n \rightarrow \infty }\psi (s_n)\\&\quad \Rightarrow \phi (a) \le \phi (a)-\psi (a)\\&\quad \Rightarrow \psi (a) \le 0\\&\quad \Rightarrow \psi (a)=0\\&\quad \Rightarrow a=0\\&\quad \Rightarrow \lim _{n \rightarrow \infty }s_n=0. \end{aligned}$$

Therefore,

$$\begin{aligned} \lim _{n \rightarrow \infty }{\Vert x_n-x_{n+1}-\theta \Vert }=0. \end{aligned}$$

This shows that the sequence $\{u_n\}$ in C converges strongly to $\theta$, where $\theta$ is the null vector in X and $u_n=x_n-x_{n+1}$ for all natural numbers n. Now,

$$\begin{aligned}&\phi (\Vert Tu_n-T\theta \Vert ) \le \phi (\Vert u_n-\theta \Vert )-\psi (\Vert u_n-\theta \Vert )\\&\quad \Rightarrow \lim _{n \rightarrow \infty }\phi (\Vert Tu_n-T\theta \Vert ) \le \lim _{n \rightarrow \infty }\phi (\Vert u_n-\theta \Vert )- \lim _{n \rightarrow \infty }\psi (\Vert u_n-\theta \Vert )\\&\quad \Rightarrow \lim _{n \rightarrow \infty }\phi (\Vert Tu_n-T\theta \Vert ) \le 0\\&\quad \Rightarrow \lim _{n \rightarrow \infty }\phi (\Vert Tu_n-T\theta \Vert )=0\\&\quad \Rightarrow \phi ( \lim _{n \rightarrow \infty }\Vert Tu_n-T\theta \Vert )=0\\&\quad \Rightarrow \lim _{n \rightarrow \infty }\Vert Tu_n-T\theta \Vert =0. \end{aligned}$$

Again, by condition (ii) we get

$$\begin{aligned}&\phi (\Vert Tu_n-\theta \Vert ) \le \phi (\Vert u_n-\theta \Vert )-\psi (\Vert u_n-\theta \Vert )\\&\quad \Rightarrow \lim _{n \rightarrow \infty }\phi (\Vert Tu_n-\theta \Vert ) \le \lim _{n \rightarrow \infty }\phi (\Vert u_n-\theta \Vert )- \lim _{n \rightarrow \infty }\psi (\Vert u_n-\theta \Vert )\\&\quad \Rightarrow \lim _{n \rightarrow \infty }\phi (\Vert Tu_n-\theta \Vert ) \le 0\\&\quad \Rightarrow \lim _{n \rightarrow \infty }\phi (\Vert Tu_n-\theta \Vert )=0\\&\quad \Rightarrow \phi \left( \lim _{n \rightarrow \infty }\Vert Tu_n-\theta \Vert \right) =0\\&\quad \Rightarrow \lim _{n \rightarrow \infty }\Vert Tu_n-\theta \Vert =0. \end{aligned}$$

Therefore, by the uniqueness of limit, we obtain

$$\begin{aligned} T\theta =\theta , \end{aligned}$$

i.e., $\theta$ is a fixed point of T.

Finally, suppose z be another fixed point of T. Therefore,

$$\begin{aligned}&\phi (\Vert Tz-T\theta \Vert ) \le \phi (\Vert z-\theta \Vert )-\psi (\Vert z-\theta \Vert )\\&\quad \Rightarrow \phi (\Vert z-\theta \Vert ) \le \phi (\Vert z-\theta \Vert )-\psi (\Vert z-\theta \Vert )\\&\quad \Rightarrow \psi (\Vert z-\theta \Vert )\le 0\\&\quad \Rightarrow \psi (\Vert z-\theta \Vert )=0\\&\quad \Rightarrow z=\theta . \end{aligned}$$

Therefore, $\theta$ is the only fixed point of T in C. $\square$

Example 2.4

Consider the Banach space ${\mathbb {R}}$ endowed with the usual norm and define a relation ${\mathscr {R}}$ on ${\mathbb {R}}$ by: for $x,y \in {\mathbb {R}}$ $x {\mathscr {R}} y$ if and only if either $x,y \in [-(n+1),-n]$ or $x,y \in [n,n+1]$ for some $n\in {\mathbb {N}}$ or $x=y$. Then clearly ${\mathscr {R}}$ is an equivalence relation on ${\mathbb {R}}$.

Now, let $C=C_1 \cup C_2 \cup C_3$, where $C_1=[-2,-1],~~ C_2=[1,2], ~~ C_3=\{0\}$. Then C is a closed subset of ${\mathbb {R}}$.

Define a mapping $T:C \rightarrow C$ by

$$\begin{aligned} Tx = \left\{ \begin{array}{ll} -x, &{}\quad \text {if } x\in C_1;\\ 0, &{}\quad \text {if } x\in C_2 \cup C_3. \end{array}\right. \end{aligned}$$

Therefore,

$$\begin{aligned} \Vert Tx-Ty\Vert = \left\{ \begin{array}{ll} |x-y|, &{}\quad \text {if } x,y\in C_1;\\ |x|, &{}\quad \text {if } x\in C_1 \text { and } y\in C_2 \cup C_3;\\ 0, &{}\quad \text {if } x,y \in C_2 \cup C_3, \end{array}\right. \end{aligned}$$

and

$$\begin{aligned} \Vert Tx-y\Vert = \left\{ \begin{array}{ll} |x+y|, &{}\quad \text {if } x\in C_1;\\ |y|, &{}\quad \text {if } x\notin C_1. \end{array}\right. \end{aligned}$$

Consider the functions $\phi , \psi :[0,\infty )\rightarrow [0,\infty )$ defined by

$$\begin{aligned}&\phi (t)=t^2 \quad ~~\text{ and }\\&\psi (t)=\frac{t^2}{100000} \end{aligned}$$

for all $t\in [0,\infty )$.

Then, clearly $\phi , \psi$ are two altering distance functions. Let $x,y \in C$ be arbitrary such that $x{\mathscr {R}} y$. Then the following cases arise.

Case 1 Let $x,y \in C_1$. Then

$$\begin{aligned}&\phi (\Vert Tx-Ty\Vert )+\psi (\Vert x-y\Vert )-\phi (\Vert Tx-y\Vert )\\&\quad = |x-y|^2+\frac{|x-y|^2}{100000}-|x+y|^2\\&\quad = -4xy + \frac{(x-y)^2}{100000}\\&\quad \le 0\\ \Rightarrow \phi (\Vert Tx-Ty\Vert )&\le \phi (\Vert Tx-y\Vert )-\psi (\Vert x-y\Vert ). \end{aligned}$$

Case 2 Let $x,y \in C_2$. Then

$$\begin{aligned}&\phi (\Vert Tx-Ty\Vert )+\psi (\Vert x-y\Vert )-\phi (\Vert Tx-y\Vert )\\&\quad = 0+ \frac{(x-y)^2}{100000} - y^2\\&\quad \le 0\\&\quad \Rightarrow \phi (\Vert Tx-Ty\Vert ) \le \phi (\Vert Tx-y\Vert )-\psi (\Vert x-y\Vert ). \end{aligned}$$

Case 3 Let $x,y \in C_3$. Then clearly the equality holds.

Thus,

$$\begin{aligned} \phi (\Vert Tx-Ty\Vert ) \le \phi (\Vert Tx-y\Vert )-\psi (\Vert x-y\Vert ) \end{aligned}$$

for all $x,y \in C$ with $x {\mathscr {R}} y$.

Also it is easily seen that T is order-preserving and by Theorem 2.2, 0 is the only fixed point of T.

3 Application to fractional thermostat model

The motivation of this section is to provide an application of the results discussed in this manuscript. For this purpose, we consider the following fractional thermostat model:

$$\begin{aligned} ^CD^{\alpha }u(t) + \lambda f(t,u(t))=0,\quad t\in [0,1], \end{aligned}$$

(3.1)

subject to the boundary conditions:

$$\begin{aligned} u'(0)=0, \beta ^CD^{\alpha - 1}u(1) + u(\eta )=0, \end{aligned}$$

(3.2)

where $^CD^{\alpha }$ stands for Caputo fractional derivative of order $\alpha$, $\lambda$ is a positive constant and $1 <\alpha \le 2,~~ 0 \le \eta \le 1,~~ \beta >0$ such that the following conditions hold:

• (1) $\beta \Gamma (\alpha )-(1-\eta )^{(\alpha -1)}>0$;

• (2) $f: [0,1] \times {\mathbb {R}} \rightarrow {\mathbb {R}}^+$ is a continuous function;

• (3) $u:[0,1] \rightarrow {\mathbb {R}}$ is continuous.

Our aim is to derive some sufficient conditions under which the problem (3.1) with the boundary conditions (3.2) possesses a unique positive solution for certain values of the parameter $\lambda$. To proceed further, we first recall the following lemmas.

Lemma 3.1

[22]. Assume $f\in C[0,1]$. A function $u\in C[0,1]$ is a solution of the boundary value problem

$$\begin{aligned} ^CD^{\alpha }u(t) + \lambda f(t,u(t))= & {} 0, t\in [0,1], \end{aligned}$$

(3.3)

$$\begin{aligned} u'(0)= & {} 0, \beta ^CD^{\alpha - 1}u(1) + u(\eta )=0, \end{aligned}$$

(3.4)

if and only if it satisfies the integral equation

$$\begin{aligned} u(t)= \int _{0}^{1}G(t,s)f(s)\mathrm{d}s, \end{aligned}$$

(3.5)

where G(t, s) is the Green’s function (depending on $\alpha )$ given by

$$\begin{aligned} G(t,s)= \beta + H_{\eta }(s) - H_t(s) \end{aligned}$$

and for $r \in [0,1],$ $H_r(s):[0,1] \rightarrow {\mathbb {R}}$ is defined as $H_r(s)= \frac{(r-s)^{\alpha - 1}}{\Gamma (\alpha )} \text{ for } s \le r$ and $H_r(s)=0 \text{ for } s> r$, i.e.,

$$\begin{aligned}G(t,s) = \left\{ \begin{array}{ll} \beta - \frac{(t-s)^{\alpha - 1}}{\Gamma (\alpha )} + \frac{(\eta -s)^{\alpha - 1}}{\Gamma (\alpha )}, &{}\quad \text{ if } 0\le s \le \eta ,~~ s\le t;\\ \beta + \frac{(\eta -s)^{\alpha - 1}}{\Gamma (\alpha )}, &{}\quad \text{ if } 0\le s \le \eta ,~~ s\ge t;\\ \beta - \frac{(t-s)^{\alpha - 1}}{\Gamma (\alpha )}, &{}\quad \text{ if } \eta \le s \le 1,~~ s\le t;\\ \beta , &{}\quad \text{ if } \eta \le s \le 1,~~ s\ge t. \end{array} \right. \end{aligned}$$

Lemma 3.2

[30]. The function G(t, s) arising in Lemma 3.1 satisfies the following conditions:

• (i) G(t, s) is a continuous map defined on $[0,1]\times [0,1]$;

• (ii) for $t,s \in (0,1)$, we have $G(t,s)>0$.

Now we prove the following lemma.

Lemma 3.3

The Green’s function G(t, s) derived in Lemma 3.1 satisfies

$$\begin{aligned} \sup _{t \in [0,1]}\int _{0}^{1}G(t,s)\mathrm{d}s=\beta + \frac{ \eta ^{\alpha } }{\Gamma (\alpha + 1)} \end{aligned}$$

and

$$\begin{aligned} \inf _{t \in [0,1]}\int _{0}^{1}G(t,s)\mathrm{d}s=\beta + \frac{ \eta ^{\alpha } - 1}{\Gamma (\alpha + 1)}. \end{aligned}$$

Proof

Let us consider the function $\varphi$ defined on [0, 1] by

$$\begin{aligned} \varphi (t)=\int _{0}^{1}G(t,s)\mathrm{d}s \end{aligned}$$

for all $t \in [0,1]$.

Now, for $t \in [0,1]$ and $t \le \eta , ~~s\ge \eta ,$ we have $t \le s$ and thus,

$$\begin{aligned} \varphi (t)&= \int _{0}^{1}G(t,s)\mathrm{d}s\\&= \int _{0}^{\eta }G(t,s)\mathrm{d}s + \int _{\eta }^{1}G(t,s)\mathrm{d}s\\&= \int _{0}^{t}G(t,s)\mathrm{d}s + \int _{t}^{\eta }G(t,s)\mathrm{d}s + \int _{\eta }^{1}G(t,s)\mathrm{d}s\\&= \int _{0}^{t}\left\{ \beta - \frac{(t-s)^{\alpha - 1}}{\Gamma (\alpha )} + \frac{(\eta -s)^{\alpha - 1}}{\Gamma (\alpha )}\right\} \mathrm{d}s \\&\quad + \int _{t}^{\eta }\left\{ \beta + \frac{(\eta -s)^{\alpha - 1}}{\Gamma (\alpha )}\right\} \mathrm{d}s + \int _{\eta }^{1}\beta \mathrm{d}s\\&= \beta + \frac{ \eta ^{\alpha } - t^{\alpha }}{\Gamma (\alpha + 1)}. \end{aligned}$$

Again, for $t \in [0,1]$ and $t \ge \eta , ~~s\le \eta ,$ we have $t \ge s$ and so

$$\begin{aligned} \varphi (t)&= \int _{0}^{1}G(t,s)\mathrm{d}s\\&= \int _{0}^{\eta }G(t,s)\mathrm{d}s + \int _{\eta }^{1}G(t,s)\mathrm{d}s\\&= \int _{0}^{\eta }G(t,s)\mathrm{d}s + \int _{\eta }^{t}G(t,s)\mathrm{d}s + \int _{t}^{1}G(t,s)\mathrm{d}s\\&= \int _{0}^{\eta }\left\{ \beta - \frac{(t-s)^{\alpha - 1}}{\Gamma (\alpha )} + \frac{(\eta -s)^{\alpha - 1}}{\Gamma (\alpha )}\right\} \mathrm{d}s \\&\quad + \int _{\eta }^{t}\left\{ \beta - \frac{(t-s)^{\alpha - 1}}{\Gamma (\alpha )}\right\} \mathrm{d}s + \int _{t}^{1}\beta \mathrm{d}s\\&= \beta + \frac{ \eta ^{\alpha } - t^{\alpha }}{\Gamma (\alpha + 1)}. \end{aligned}$$

Thus, from the above calculations we get,

$$\begin{aligned} \varphi (t)=\beta + \frac{ \eta ^{\alpha } - t^{\alpha }}{\Gamma (\alpha + 1)} \end{aligned}$$

for all $t \in [0,1]$.

Therefore,

$$\begin{aligned} \varphi '(t)=\frac{-\alpha t^{\alpha -1}}{\Gamma (\alpha + 1)}<0 \end{aligned}$$

for all $t\in [0,1].$

This implies that the function $\varphi$ is a decreasing function on [0, 1]. So,

$$\begin{aligned} \sup _{t \in [0,1]}\int _{0}^{1}G(t,s)\mathrm{d}s&= \sup _{t\in [0,1]} \varphi (t)\\&= \varphi (0)\\&= \beta + \frac{ \eta ^{\alpha } }{\Gamma (\alpha + 1)}, \end{aligned}$$

and

$$\begin{aligned} \inf _{t \in [0,1]}\int _{0}^{1}G(t,s)\mathrm{d}s&= \inf _{t\in [0,1]} \varphi (t)\\&= \varphi (1)\\&= \beta + \frac{ \eta ^{\alpha } - 1}{\Gamma (\alpha + 1)}. \end{aligned}$$

This completes the proof of the lemma. $\square$

As a special case of Proposition 1 of [10], we have the following lemma.

Lemma 3.4

For the Green’s function G(t, s) derived in Lemma 3.1,

$$\begin{aligned} \beta - \frac{(1-\eta )^{\alpha - 1}}{\Gamma (\alpha )} \le G(t,s) \le \beta + \frac{\eta ^{\alpha - 1}}{\Gamma (\alpha )} \end{aligned}$$

for all $t,s \in [0,1]$ holds.

Now we prove the following theorems concerning the existence and uniqueness of a positive solution to the fractional thermostat model given by Eqs. (3.1) and (3.2).

Theorem 3.5

Let us consider the fractional thermostat model with parameter $\lambda >0$ given by Eqs. (3.1) and (3.2). Assume that the following conditions hold:

• (i) $\beta \Gamma (\alpha + 1)+ \eta ^{\alpha }>1;$

• (ii) for all $s\in [0,1]$,

  $$\begin{aligned} \lambda |f(s,u(s))-f(s,v(s))| \le \lambda |f(s,u(s))| - \lambda \sup _{t\in [0,1]} |v(t)| -\psi (\sup _{t\in [0,1]} |u(t) - v(t)|) \end{aligned}$$

  for some altering distance function $\psi$ and for all real-valued continuous functions u(s), v(s) defined on [0, 1]; 

• (iii) f is non-decreasing with respect to the second argument and there exists $t_0 \in (0,1)$ such that $f(t_0,0)>0$.

Then the fractional thermostat model with parameter $\lambda$ given by Eqs. (3.1) and (3.2) has a unique positive solution for $\lambda \ge \frac{1}{k}$, where $k= \beta + \frac{ \eta ^{\alpha } - 1}{\Gamma (\alpha + 1)}.$

Proof

Consider the Banach space C[0, 1] of all real-valued continuous functions defined on [0, 1] equipped with the sup norm.

Define a mapping $T:C[0,1] \rightarrow C[0,1]$ by

$$\begin{aligned} Tu(t)=\lambda \int _{0}^{1}G(t,s)f(s,u(s))\mathrm{d}s \end{aligned}$$

for all $u \in C[0,1]$, where G(t, s) is defined as in Lemma 3.1.

From Lemma 3.1, it is obvious that the thermostat model given by Eqs. (3.1) and (3.2) has u(t) as a solution if and only if u(t) is a fixed point of T.

Now, by condition (ii), we have

$$\begin{aligned} \lambda |f(s,u(s))-f(s,v(s))|&\le \lambda |f(s,u(s))| - \lambda \sup _{t\in [0,1]} |v(t)| \\&\quad -\psi ( \sup _{t\in [0,1]} |u(t) - v(t)|)\\&= \lambda |f(s,u(s))| - \lambda \Vert v\Vert -\psi (\Vert u-v\Vert ). \end{aligned}$$

Multiplying both sides by |G(t, s)|, we get

$$\begin{aligned} \lambda |f(s,u(s))-f(s,v(s))||G(t,s)|&\le \lambda |f(s,u(s))| |G(t,s)| \\&\quad - \lambda \Vert v\Vert |G(t,s)|\\&\quad -\,\psi (\Vert u-v\Vert )|G(t,s)|\\ \Rightarrow \lambda \int _{0}^{1} |f(s,u(s))-f(s,v(s))| |G(t,s)|\mathrm{d}s&\le \lambda \int _{0}^{1} |f(s,u(s))| |G(t,s)|\mathrm{d}s \\&\quad -\, \lambda \int _{0}^{1}\Vert v\Vert |G(t,s)| \mathrm{d}s\\&\quad -\,\int _{0}^{1}\psi (\Vert u-v\Vert )|G(t,s)|\mathrm{d}s\\&= \lambda \int _{0}^{1} G(t,s)f(s,u(s)) \mathrm{d}s \\&\quad -\, \lambda \Vert v\Vert \int _{0}^{1} G(t,s) \mathrm{d}s \\&\quad -\, \psi (\Vert u-v\Vert )\int _{0}^{1}G(t,s)\mathrm{d}s\\&\le \lambda \int _{0}^{1} G(t,s)f(s,u(s)) \mathrm{d}s \\&\quad -\, \lambda \Vert v\Vert \inf _{t\in [0,1]}\int _{0}^{1} G(t,s) \mathrm{d}s \\&\quad -\, \psi (\Vert u-v\Vert ) \inf _{t\in [0,1]}\int _{0}^{1}G(t,s)\mathrm{d}s\\&\le \lambda \int _{0}^{1} G(t,s)f(s,u(s)) \mathrm{d}s \\&\quad - \lambda k\Vert v\Vert - k\psi (\Vert u-v\Vert ). \end{aligned}$$

Now if $\lambda k \ge 1$, then from the above equation we obtain,

$$\begin{aligned} \lambda \int _{0}^{1} |f(s,u(s))-f(s,v(s))| |G(t,s)|\mathrm{d}s&\le \lambda \int _{0}^{1} G(t,s)f(s,u(s)) \mathrm{d}s - \Vert v\Vert \nonumber \\&\quad -\,k\psi (\Vert u-v\Vert )\nonumber \\&\le \lambda \int _{0}^{1} G(t,s)f(s,u(s)) \mathrm{d}s - |v(t)|\nonumber \\&\quad -\,k\psi (\Vert u-v\Vert )\nonumber \\&\le \left| \lambda \int _{0}^{1} G(t,s)f(s,u(s)) \mathrm{d}s - v(t) \right| \nonumber \\&\quad -\, k\psi (\Vert u-v\Vert ). \end{aligned}$$

(3.6)

Therefore, using Eq. (3.6) we get

$$\begin{aligned} |Tu(t) - Tv(t)|&= \left| \lambda \int _{0}^{1}G(t,s)f(s,u(s))\mathrm{d}s - \lambda \int _{0}^{1}G(t,s)f(s,v(s))\mathrm{d}s \right| \\&= \left| \lambda \int _{0}^{1}G(t,s)(f(s,u(s))- f(s,v(s)))\mathrm{d}s \right| \\&\le \left| \lambda \int _{0}^{1} G(t,s)f(s,u(s)) \mathrm{d}s - v(t) \right| - k\psi (\Vert u-v\Vert ).\\ \end{aligned}$$

The above inequality holds for all $t\in [0,1]$ and so we have

$$\begin{aligned} \sup _{t\in [0,1]}|Tu(t) - Tv(t)|&\le \sup _{t\in [0,1]}\lambda \left| \int _{0}^{1} G(t,s)f(s,u(s)) \mathrm{d}s - v(t) \right| - k\psi (\Vert u-v\Vert ) \nonumber \\ \Rightarrow \Vert Tu - Tv\Vert&\le \Vert Tu - v\Vert - k\psi (\Vert u-v\Vert ). \end{aligned}$$

(3.7)

It is easily perceived by condition (i) that $k>0.$

Define two functions $\phi , \psi _1: [0,\infty )\rightarrow [0,\infty )$ by

$$\begin{aligned} \phi (t)=t ~~ and \\ \psi _1(t)=k\psi (t) \end{aligned}$$

for all $t \in [0,\infty ).$ Then one can easily verify that $\phi , \psi _1$ are two altering distance functions and also from Eq. (3.7) we get

$$\begin{aligned} \phi (\Vert Tu - Tv\Vert ) \le \phi (\Vert Tu - v\Vert ) - \psi _1(\Vert u-v\Vert ). \end{aligned}$$

(3.8)

The above inequality holds for all $u,v \in C[0,1]$ and so by Theorem 2.1, T has a unique fixed point u(t), say, in C[0, 1].

Note that Eq. (3.8) holds if $\lambda k \ge 1$. So, T has u(t) as a fixed point if $\lambda k \ge 1$, i.e., u(t) is a solution of the thermostat model (3.1) and (3.2) if $\lambda k \ge 1$, i.e., $\lambda \ge \frac{1}{k}$.

Now we have $\lambda>0,~~ G(t,s) >0$ and $f(s,u(s)) \ge 0$ for all $t,s \in [0,1]$. Therefore, it is clear that

$$\begin{aligned} \lambda \int _{0}^{1} G(t,s)f(s,u(s))\mathrm{d}s \ge 0 \end{aligned}$$

for all $t\in [0,1]$. This means that $Tu(t) \ge 0$ for all $t\in [0,1]$ and which leads us to the fact that $u(t) \ge 0$ for all $t\in [0,1].$

Finally, we show that the unique solution u(t) is always positive. To show this, first we show that the zero function 0 is not a fixed point of T.

Suppose to the contrary that the zero function 0 is a fixed point of T. Then, we have

$$\begin{aligned} 0=\lambda \int _{0}^{1}G(t,s)f(s,0)\mathrm{d}s, \end{aligned}$$

for all $t \in [0,1].$ Since $G(t,s)f(s,0) \ge 0$ for all $t\in [0,1]$ and for all $s\in [0,1]$, we have

$$\begin{aligned} G(t,s)f(s,0) = 0, \end{aligned}$$

for all $t\in [0,1]$ and for almost all $s\in [0,1]$. This fact leads us to

$$\begin{aligned} f(s,0)=0\quad \text{ for } \text{ almost } \text{ all } s\in [0,1]. \end{aligned}$$

(3.9)

By condition (iii), there exists $t_0 \in (0,1)$ such that $f(t_0,0)>0$. Again, since f is continuous at $(t_0,0)$, there exists a subset A of [0, 1] of positive Lebesgue measure such that $f(s,0)>0$ for all $s\in A$. This is a contradiction to Eq. (3.9). So the zero function 0 is not a fixed point of T.

Now, let $u(t_1)=0$ for some $t_1\in (0,1)$. Therefore, we have

$$\begin{aligned} \int _{0}^{1}G(t_1,s)f(s,u(s))\mathrm{d}s&= 0. \end{aligned}$$

(3.10)

But $u(s) \ge 0$ for all $s\in [0,1]$ and f is non-decreasing with respect to the second argument. Hence,

$$\begin{aligned} 0 \ge \int _{0}^{1}G(t_1,s)f(s,u(s))\mathrm{d}s&\ge \int _{0}^{1}G(t_1,s)f(s,0)\mathrm{d}s \ge 0. \end{aligned}$$

(3.11)

Therefore, from (3.10) and (3.11), we obtain

$$\begin{aligned} \int _{0}^{1}G(t_1,s)f(s,0)\mathrm{d}s = 0. \end{aligned}$$

As $G(t_1,s)f(s,0) \ge 0$, it follows that $G(t_1,s)f(s,0) = 0$ for almost all $s\in [0,1]$. This implies that $f(s,0) = 0$ for almost all $s\in [0,1]$, which is a contradiction.

Hence, it follows that $u(t)>0$ for all $t\in (0,1).$ Again, since u is continuous on [0, 1], we have $u(t)>0$ for all $t\in [0,1].$ Thus, the fractional thermostat model, given by Eqs. (3.1) and (3.2), has a unique positive solution for $\lambda \ge \frac{1}{k}$, where $k= \beta + \frac{ \eta ^{\alpha } - 1}{\Gamma (\alpha + 1)}.$ $\square$

Theorem 3.6

Let us consider the fractional thermostat model with parameter $\lambda$ given by Eqs. (3.1) and (3.2). Assume that the following conditions hold:

• (i) $\beta \Gamma (\alpha + 1)+ \eta ^{\alpha }>1;$

• (ii) for all $s\in [0,1]$,

  $$\begin{aligned} \lambda |f(s,u(s))-f(s,v(s))| \le \lambda |f(s,u(s))| - \lambda \sup _{t\in [0,1]} |v(t)| -\psi ( \sup _{t\in [0,1]} |u(t) - v(t)|), \end{aligned}$$

  for some bounded altering distance function $\psi$ and for all u(s), v(s) in the set $C= \{u(s)\in C[0,1]: R_1 \le u(s) \le R,$ for all $s\in [0,1]$, where $R, R_1$ are constants with ${R>1, R_1>0}\}$;

• (iii) $k_1=\beta + \frac{\eta ^{\alpha - 1}}{\Gamma (\alpha )}$, $k_2= \beta - \frac{(1-\eta )^{\alpha - 1}}{\Gamma (\alpha )} >0$, $k=\beta + \frac{ \eta ^{\alpha } - 1}{\Gamma (\alpha + 1)}<1$;

• (iv) $\int _{0}^{1} f(s,R) \mathrm{d}s \le \frac{6R}{\lambda k_1}$ and $\int _{0}^{1} f(s,0) \mathrm{d}s > \frac{R_1}{5\lambda k_2}$;

• (v) f is non-decreasing with respect to the second argument and there exists $t_0 \in (0,1)$ such that $f(t_0,0)>0$.

Then the fractional thermostat model with parameter $\lambda$ given by Eqs. (3.1) and (3.2) has a unique positive solution in C for $\lambda \ge \frac{1}{k}.$

Proof

Let us take the Banach space C[0, 1] endowed with the sup norm. We consider the set $C'$ defined as

$$\begin{aligned}C'=\left\{ u(s)\in C[0,1]: \frac{R_1}{5}\le u(s) \le 6R \text{ for } \text{ all } ~~s\in [0,1] ~~\text{ or } u(s)=0 \text{ for } \text{ all } s \in [0,1]\right\} .\end{aligned}$$

Then it is easily noticeable that $C'$ is a closed subset of C[0, 1]. The fact that f is non-decreasing with respect to the second argument gives us

$$\begin{aligned} \int _{0}^{1} f(s,u(s)) \mathrm{d}s&\le \int _{0}^{1} f(s,R) \mathrm{d}s \\&\le \frac{6R}{\lambda k_1}, \end{aligned}$$

and

$$\begin{aligned} \int _{0}^{1} f(s,u(s)) \mathrm{d}s&\ge \int _{0}^{1} f(s,R_1) \mathrm{d}s \\&\ge \int _{0}^{1} f(s,0) \mathrm{d}s\\&> \frac{R_1}{5\lambda k_2}\\ \end{aligned}$$

for all $u \in C$. Therefore,

$$\begin{aligned} \lambda \int _{0}^{1} G(t,s) f(s,u(s))\mathrm{d}s&\le \lambda \left( \beta + \frac{\eta ^{\alpha - 1}}{\Gamma (\alpha )} \right) \int _{0}^{1} f(s,u(s)) \mathrm{d}s \\&\le \lambda k_1 \frac{6R}{\lambda k_1}\\&= 6R, \end{aligned}$$

i.e.,

$$\begin{aligned} \lambda \int _{0}^{1} G(t,s) f(s,u(s)) \mathrm{d}s \le 6R, \end{aligned}$$

(3.12)

and

$$\begin{aligned} \lambda \int _{0}^{1} G(t,s) f(s,u(s))\mathrm{d}s&\ge \lambda \left( \beta - \frac{(1-\eta )^{\alpha - 1}}{\Gamma (\alpha )} \right) \int _{0}^{1} f(s,u(s)) \mathrm{d}s \\&\ge \lambda k_2\frac{R_1}{5\lambda k_2}\\&= \frac{R_1}{5}, \end{aligned}$$

i.e.,

$$\begin{aligned} \lambda \int _{0}^{1} G(t,s) f(s,u(s)) \mathrm{d}s \ge \frac{R_1}{5}. \end{aligned}$$

(3.13)

Next, we define a mapping $T:C' \rightarrow C'$ by

$$\begin{aligned}Tu(t)= \left\{ \begin{array}{ll} \lambda \int _{0}^{1}G(t,s)f(s,u(s))\mathrm{d}s, &{}\quad \text{ for } \text{ all } u \in C\cup \{0\};\\ 0, &{}\quad \text{ elsewhere }.\end{array} \right. \end{aligned}$$

From Eqs. (3.12) and (3.13) one can easily check that T is well defined on $C'$.

We now define a relation ${\mathscr {R}}$ on C[0, 1] by the following:

for $u,v \in C[0,1]$, $u {\mathscr {R}} v$ if and only if

• (1) either $u,v \in C' \backslash \{0\}$ and both $Tu, Tv \ne 0;$

• (2) or $u,v \in C' \backslash \{0\}$ and both $Tu, Tv =0;$

• (3) or $u=v=0.$

Then it is clear that ${\mathscr {R}}$ is an equivalence relation on C[0, 1]. We claim that T is order-preserving on $C'$ with respect to ${\mathscr {R}}$.

Let $u,v \in C'$ be arbitrary with $u {\mathscr {R}} v$. Then the following cases may arise.

Case 1 When $u,v \in C' \backslash \{0\}$ and both $Tu, Tv \ne 0.$

Then

$$\begin{aligned} Tu&\ge \frac{R_1}{5}>0\\ \Rightarrow T(Tu)&> 0. \end{aligned}$$

Similarly, $T(Tv)>0.$ So, $T(Tu),T(Tv)\ne 0$, and $Tu {\mathscr {R}} Tv.$

Case 2 When $u,v \in C' \backslash \{0\}$ and both $Tu, Tv =0.$

Since ${\mathscr {R}}$ is an equivalence relation, $Tu {\mathscr {R}} Tv.$

Case 3 When $u=v=0$.

Then we have $Tu=Tv$ and so $Tu {\mathscr {R}} Tv$, since ${\mathscr {R}}$ is an equivalence relation. Therefore, T is order-preserving on $C'$ with respect to ${\mathscr {R}}$.

Since $\psi$ is bounded, so there exists a constant $M>0$ such that $\psi (t)\le M$ for all $t \in [0,\infty )$. Without loss of generality, we may assume that $\frac{R_1}{5}\ge M$. Now we define two altering distance functions by $\phi (t)=(2M+1)kt$ and $\psi _1(t)=(2M+1)k^2\psi (t)$ for all $t \in [0,\infty )$.

Let $u,v \in C'$ be arbitrary with $u {\mathscr {R}} v$. Then the succeeding cases may arise.

Case 1 When $u,v \in C' \backslash \{0\}$ and both $Tu, Tv \ne 0.$

So $u,v \in C$ and

$$\begin{aligned}Tu=\lambda \int _{0}^{1}G(t,s)f(s,u(s))\mathrm{d}s\end{aligned}$$

and

$$\begin{aligned}Tv=\lambda \int _{0}^{1}G(t,s)f(s,v(s))\mathrm{d}s.\end{aligned}$$

Then proceeding as in Theorem 3.5, we get

$$\begin{aligned}&\Vert Tu - Tv\Vert \le \Vert Tu - v\Vert - k \psi (\Vert u-v\Vert ), ~~ \text{ if } \lambda \ge \frac{1}{k}\\&\Rightarrow (2M+1)k \Vert Tu - Tv\Vert \le (2M+1)k \Vert Tu - v\Vert - (2M+1)k^2 \psi (\Vert u-v\Vert )\\&\Rightarrow \phi (\Vert Tu - Tv\Vert ) \le \phi (\Vert Tu - v\Vert )- \psi _1(\Vert u-v\Vert ), ~~ \text{ if } \lambda \ge \frac{1}{k}. \end{aligned}$$

Case 2 When $u,v \in C' \backslash \{0\}$ and both $Tu, Tv =0.$

Therefore, $u,v \notin C$ and $\phi (\Vert Tu - Tv\Vert )=\phi (0)=0.$ Also, $\Vert v\Vert > \frac{R_1}{5}$. Therefore,

$$\begin{aligned} \phi (\Vert v\Vert )&>\phi \left( \frac{R_1}{5} \right) \\&= (2M+1)k \frac{R_1}{5}\\&\ge (2M+1)k^2 \frac{R_1}{5}\\&\ge (2M+1)k^2 M\\&\ge (2M+1)k^2 \psi (t)\\&= \psi _1(t)\\ \Rightarrow \phi (\Vert Tu - v\Vert )&\ge \psi _1(t)\\ \Rightarrow \phi (\Vert Tu - v\Vert )&\ge \psi _1(\Vert u - v\Vert )\\ \Rightarrow \phi (\Vert Tu - v\Vert )-\psi _1(\Vert u - v\Vert )&\ge \phi (\Vert Tu - Tv\Vert ). \end{aligned}$$

Case 3 When $u=v=0$.

In this case, clearly we have

$$\begin{aligned}\phi (\Vert Tu - Tv\Vert ) \le \phi (\Vert Tu - v\Vert )- \psi (\Vert u-v\Vert ).\end{aligned}$$

Thus, we see that

$$\begin{aligned}\phi (\Vert Tu - Tv\Vert ) \le \phi (\Vert Tu - v\Vert )- \psi (\Vert u-v\Vert )\end{aligned}$$

holds for all $u,v \in C'$ with $u {\mathscr {R}} v$ if $\lambda \ge \frac{1}{k}$. So by Theorem 2.2, T has a unique fixed point in $C'$ if $\lambda \ge \frac{1}{k}$. But by the definition of T, it can have the fixed point only in C. So T has a unique fixed point in C.

Thus, u(t) is the unique solution of the fractional thermostat model given by Eqs. (3.1) and (3.2), which follows by Lemma 3.1 and the definition of T. Also, since $u \in C$, u is positive. Hence, the fractional thermostat model given by Eqs. (3.1) and (3.2) satisfying the hypotheses of Theorem 3.6 has a unique positive solution in C for $\lambda \ge \frac{1}{k}.$ $\square$

Now, we demonstrate an example which validates the effectiveness of the aforementioned result.

Example 3.7

Let us consider the fractional thermostat model

$$\begin{aligned} ^CD^{\alpha }u(t) + \lambda f(t,u(t))=0, t\in (0,1), \end{aligned}$$

(3.14)

$$\begin{aligned} u'(0)=0, \beta ^CD^{\alpha - 1}u'(1) + u(\eta )=0. \end{aligned}$$

(3.15)

We choose

$$\begin{aligned} \alpha = \frac{3}{2},~~ \beta = \frac{4}{5}, ~~ \eta =\frac{1}{2} \end{aligned}$$

and

$$\begin{aligned} f(t,u)= \left\{ \begin{array}{ll} \ln (3^{20} + t^2) + t^3 + \frac{1}{24 - u},&{} \text{ if } u\le 20;\\ \ln (3^{20} + t^2) + t^3 + \frac{1}{4}, &{} \text{ elsewhere }. \end{array}\right. \end{aligned}$$

Then, $\beta \Gamma (\alpha )-(1-\eta )^{(\alpha -1)} = \frac{4}{5}.\frac{1}{2}. \sqrt{\pi } -(\frac{1}{2})^\frac{1}{2} >0.$

Clearly, $f:[0,1] \times {\mathbb {R}} \rightarrow {\mathbb {R}}^+$ is a continuous function and also f is non-decreasing with respect to the second argument and there exists $\frac{1}{2} \in (0,1)$ such that $f(\frac{1}{2},0) >0$.

We take

$$\begin{aligned} C=\left\{ u(s) \in C[0,1] : \frac{1}{10} \le u(s) \le 20, \text{ for } \text{ all } s\in [0,1]\right\} , \end{aligned}$$

i.e., here $R_1=\frac{1}{10},~~ R=20$.

Now,

$$\begin{aligned} \beta \Gamma (\alpha + 1) + \eta ^{\alpha } = \frac{4}{5}. \frac{3}{2}. \frac{1}{2}. \sqrt{\pi } + \left( \frac{1}{2}\right) ^\frac{3}{2} \approx 1.4165 >1, \end{aligned}$$

and

$$\begin{aligned} k&= \beta + \frac{\eta ^{\alpha } - 1}{\Gamma (\alpha + 1)}\\&= \frac{4}{5} +\frac{(\frac{1}{2})^\frac{3}{2} -1}{\frac{3}{2}. \frac{1}{2}. \sqrt{\pi }}\\&\approx 0.3135 < 1\\ \Rightarrow \frac{1}{k}&\approx 3.1897. \end{aligned}$$

Also we have

$$\begin{aligned} k_1&= \beta + \frac{\eta ^{\alpha - 1}}{\Gamma (\alpha )}\\&= \frac{4}{5} +\frac{(\frac{1}{2})^\frac{1}{2} }{ \frac{1}{2}. \sqrt{\pi }}\\&\approx 1.5981\,\,\mathrm{and}\\ k_2&= \beta - \frac{(1-\eta )^{\alpha - 1}}{\Gamma (\alpha )}\\&= \frac{4}{5} - \frac{(\frac{1}{2})^\frac{1}{2} }{ \frac{1}{2}. \sqrt{\pi }}\\&\approx 0.0019. \end{aligned}$$

We choose $\lambda = 3.2$ and clearly $\lambda \ge \frac{1}{k}.$

Now,

$$\begin{aligned} \int _{0}^{1} f(s,R)\mathrm{d}s&\le \ln {(1 + 3^{20})} + \frac{1}{4} + \frac{1}{4}\\&\approx 22.4722\\&\le \frac{6R}{\lambda k_1} \end{aligned}$$

as

$$\begin{aligned} \frac{6R}{\lambda k_1}&\approx \frac{6 \times 20}{3.2 \times 1.5981}\\&\approx 23.4653. \end{aligned}$$

Also,

$$\begin{aligned} \int _{0}^{1} f(s,0)\mathrm{d}s&= \ln {( 3^{20})} + \frac{1}{4} + \frac{1}{24}\\&\approx 22.2639\\&> \frac{R_1}{5\lambda k_2} \end{aligned}$$

as

$$\begin{aligned} \frac{R_1}{5\lambda k_2}&\approx \frac{6 \times 20}{3.2 \times 1.5981}\\&\approx 3.2894. \end{aligned}$$

Next, we define a mapping $\psi :[0,\infty ) \rightarrow [0,\infty )$ by

$$\begin{aligned}\psi (t)= \left\{ \begin{array}{ll} t^2, &{}\quad \text{ if } 0 \le t <1 ;\\ 1, &{}\quad \text{ if } t \ge 1 . \end{array} \right. \end{aligned}$$

Then it is an easy task to note that $\psi$ is a bounded altering distance function.

Finally, for any $u(s), v(s) \in C$, we have

$$\begin{aligned} \lambda |f(s,u(s)) - f(s,v(s))|&= 3.2 \left| \frac{1}{24 - u(s)} - \frac{1}{24 - v(s)} \right| \\&\le 3.2 \left( \frac{1}{4} + \frac{1}{4} \right) \\&= 1.6. \end{aligned}$$

But,

$$\begin{aligned} \lambda |f(s,u(s))| - \lambda \sup _{t\in [0,1]} |u(t)| - \psi ( \sup _{t\in [0,1]} |u(t) - v(t)|)&\ge 3.2\times 21 - 3.2 \times 20 -\psi (40)\\&= 2.20. \end{aligned}$$

Therefore,

$$\begin{aligned} \lambda |f(s,u(s)) - f(s,v(s))| \le \lambda |f(s,u(s))| - \lambda \sup _{t\in [0,1]} |u(t)| - \psi \left( \sup _{t\in [0,1]} |u(t) - v(t)|\right) \end{aligned}$$

for all $u(s), v(s) \in C$. So, by Theorem 3.6, the thermostat model given by Eqs. (3.14) and (3.15) has a unique positive solution in C for $\lambda = 3.2$.

Remark 3.8

It is quite interesting to note that if we take $\lambda < \frac{1}{k}$ in Theorem 3.6, then the thermostat model given by Eqs. (3.1) and (3.2) may not have a positive solution in C. We present the following example in support of our claim.

Example 3.9

Let us consider the fractional thermostat model

$$\begin{aligned} ^CD^{\alpha }u(t) + \lambda f(t,u(t))= & {} 0, t\in (0,1), \end{aligned}$$

(3.16)

$$\begin{aligned} u'(0)= & {} 0, \beta ^CD^{\alpha - 1}u'(1) + u(\eta )=0. \end{aligned}$$

(3.17)

We take

$$\begin{aligned} \alpha= & {} \frac{3}{2},~~ \beta = \frac{6}{5}, ~~ \eta =\frac{1}{2}, \\ f(t,u)= & {} 6 +5t^2 \end{aligned}$$

and

$$\begin{aligned} C=\left\{ u(s) \in C[0,1] : \frac{1}{15} \le u(s) \le 5, \text{ for } \text{ all } s\in [0,1]\right\} , \end{aligned}$$

i.e., here $R_1=\frac{1}{15},~~ R=5$.

Then, clearly $f:[0,1] \times {\mathbb {R}} \rightarrow {\mathbb {R}}^+$ is a continuous function and also f is non-decreasing with respect to the second argument and there exists $\frac{1}{2} \in (0,1)$ such that $f(\frac{1}{2},0) >0$. Also, $\beta \Gamma (\alpha )-(1-\eta )^{(\alpha -1)} = \frac{6}{5}.\frac{1}{2}. \sqrt{\pi } -(\frac{1}{2})^\frac{1}{2} >0.$ Again we have $\beta \Gamma (\alpha + 1) + \eta ^{\alpha } \approx 2.3019 >1$, $k= \beta + \frac{\eta ^{\alpha } - 1}{\Gamma (\alpha + 1)}=0.7136 <1,$ $\frac{1}{k} \approx 1.1014$, $k_1 = \beta + \frac{\eta ^{\alpha - 1}}{\Gamma (\alpha )} \approx 1.9981,$ $k_2 = \beta - \frac{(1-\eta )^{\alpha - 1}}{\Gamma (\alpha )} \approx 0.4019.$ We choose $\lambda = \frac{1}{230}$, then clearly $\lambda < \frac{1}{k}$. Now,

$$\begin{aligned} \int _{0}^{1} f(s,R)\mathrm{d}s&\approx 7.66\\&\le \frac{6R}{\lambda k_1}\\&\approx 3303.13 \end{aligned}$$

and

$$\begin{aligned} \int _{0}^{1} f(s,0)\mathrm{d}s&\approx 7.66\\&> \frac{R_1}{5\lambda k_2}\\&= 7.6304. \end{aligned}$$

Now, we consider the bounded altering distance function $\psi :[0,\infty ) \rightarrow [0,\infty )$ defined by

$$\begin{aligned} \psi (t)= \left\{ \begin{array}{ll} \frac{t}{230},&{}\quad \text{ if } t \le 1;\\ \frac{1}{230}, &{}\text{ if } t > 1. \end{array}\right. \end{aligned}$$

Therefore, for any $u(s),~~ v(s) \in C$ we have

$$\begin{aligned} \lambda |f(s,u(s))| - \lambda \sup _{t\in [0,1]} |u(t)| - \psi \left( \sup _{t\in [0,1]} |u(t) - v(t)|\right)&\ge \frac{1}{230} \times 6 - \frac{1}{230} \times 5 -\psi (10)\\&= 0 \\&= \lambda |f(s,u(s)) - f(s,v(s))|. \end{aligned}$$

Thus, all the conditions of Theorem 3.6 are satisfied but the thermostat model does not have a positive solution in C, because if this thermostat model has a positive solution u(t), say in C, then we have

$$\begin{aligned} T(u(t)) = u(t)&= \lambda \int _{0}^{1} G(t,s) f(s,u(s)) \mathrm{d}s\\&\le \frac{1}{230} \times 1.9981 \times \left( 6 +\frac{5}{3} \right) \\&< \frac{1}{15} \end{aligned}$$

which contradicts the fact that $u(t) \in C$.