Existence of Common Fixed Points Through Auxiliary Contractions and Applications

Abstract

In this paper, we introduce a new type of contraction, an M-auxiliary contraction, by modifying existing concepts involving auxiliary functions. We establish existence and uniqueness results for common fixed points of the proposed contraction mapping under suitable conditions. Applications to fractional differential equations and ordinary differential equations are provided to demonstrate the effectiveness of the main theorem.

1. Introduction

Fixed point theorems are fundamental tools for establishing the existence of solutions to complex equations that arise in science and engineering. In particular, they play a crucial role in solving fractional differential equations [1,2], which are used to model systems characterized by memory and distributed effects. Fixed-point theorems are also instrumental in the study of ordinary differential equations that describe phenomena such as economic growth [3], urban development, and agricultural assets [4]. As challenges in areas such as smart cities, the economy, and agriculture become increasingly intricate, fixed-point theory provides the necessary mathematical framework to ensure that models accurately reflect real-world behavior and that solutions are guaranteed to exist. An important advancement in this field is the extension of the contraction concept, as demonstrated in [5,6,7].

Geraghty introduced a new type of contraction, known as a Geraghty contraction, in [8]. Since then, the notion of Geraghty contraction has been further developed to address a broader range of mathematical problems [9,10,11]. Furthermore, the work of Karapinar et al. [12] highlighted the effectiveness of auxiliary functions in the context of fixed-point theorems for fractional differential equations. These contributions illustrate the important role of contractions with auxiliary functions and inspired us to introduce a new type of contraction called an “M-auxiliary contraction”.

Traditionally, common fixed-point results for two mappings have been obtained by extending contraction principles originally designed for a single mapping, such as Banach or Geraghty contractions. The motivation for introducing the M-auxiliary contraction arises from the limitations of these results, which often impose strong assumptions. For the existence of common fixed points, various earlier contraction conditions typically require additional assumptions, including conditions on the images of the mappings, continuity, compatibility or commutativity, and specific contractive inequalities. To overcome these limitations, our work develops a contraction type defined directly for pairs of mappings, aiming to simplify assumptions and broaden applicability. Specifically, our results demonstrate the existence of common fixed points using just one condition, the M-auxiliary contraction, that generalizes previous contraction approaches.

The structure of the paper is as follows. Section 2 introduces the concept of M-auxiliary contraction and presents results on the existence and uniqueness of common fixed points under suitable conditions. An example is provided to demonstrate the validity of the main theorems. In Section 3, we apply the developed framework to both fractional and ordinary differential equations, highlighting its practical relevance and broad applicability.

2. Main Results

This section begins with a discussion of fundamental concepts related to M-auxiliary contractions. The structure $(X,d,f,g)$ is adopted throughout this study as follows:

(1)

X is a nonempty set;

(2)

$(X,d)$ is a metric space;

(3)

f and g are self-mappings on X.

We recall the class $\mathcal{A}\left(X\right)$, which consists of all auxiliary functions $h:X\times X\to [0,1)$ that satisfy the condition

$$\mathrm{if}\underset{n\to \infty }{\lim }h(x_n,y_n)=1,\mathrm{then}\underset{n\to \infty }{\lim }d(x_n,y_n)=0,$$

for any sequences $\left\{x_n\right\}$ and $\left\{y_n\right\}$ in X, where the sequence $\left\{d(x_n,y_n)\right\}$ is decreasing, as outlined in [12]. Building on this, we define a new class of auxiliary functions, denoted ${\mathcal{A}}^{*}\left(X\right)$, which consists of all functions $\alpha :X\times X\times X\times X\to [0,1)$ that satisfy the condition

$$\mathrm{if}\underset{n\to \infty }{\lim }\alpha (x_n,y_n,u_n,v_n)=1,\mathrm{then}\underset{n\to \infty }{\lim }d(x_n,y_n)=0\mathrm{or}\underset{n\to \infty }{\lim }d(u_n,v_n)=0,$$

for any sequences $\left\{x_n\right\}$, $\left\{y_n\right\}$, $\left\{u_n\right\}$, and $\left\{v_n\right\}$ in X, where the sequences $\left\{d(x_n,y_n)\right\}$ and $\left\{d(u_n,v_n)\right\}$ are decreasing. Notably, if $h\in \mathcal{A}\left(X\right)$, then h can be defined as an element of ${\mathcal{A}}^{*}\left(X\right)$.

Example 1. 

The following are examples of functions $\alpha \in {\mathcal{A}}^{*}\left(X\right)$:

(1) 

$\alpha (x,y,u,v)=\left\{\begin{array}{cc}{\displaystyle 0}\hfill & \mathit{if}x=y,\hfill \\ {\displaystyle \frac{\ln (1+d(x,y)+d(u,v\left)\right)}{1+d(x,y)+d(u,v)}}\hfill & \mathit{if}x\ne y,\hfill \end{array}\right.$

(2) 

$\alpha (q,r,s,t)=\left\{\begin{array}{cc}{\displaystyle \frac{1}{3}}\hfill & \mathit{if}q=r,\hfill \\ \\ {\displaystyle \frac{1}{1+d(q,r)+d(s,t)}}\hfill & \mathit{if}q\ne r.\hfill \end{array}\right.$

Consequently, we are prepared to introduce a new type of contraction, defined by combining a new class of auxiliary functions with the maximum of certain expressions involving distances between points and their images under the mappings f and g, as described below.

Definition 1. 

On $(X,d,f,g)$, the pair $(f,g)$ is called an M-auxiliary contraction if there exists an auxiliary function $\alpha \in {\mathcal{A}}^{*}\left(X\right)$ such that for all $x,y\in X$, we have

$$d(fx,gy)\le \alpha (x,y,fx,gy)M(x,y),$$

where $M:X\times X\to [0,\infty )$ is a function such that for any $x,y\in X$,

$$\begin{array}{cc}\hfill M(x,y)=\max \{& d(x,y)+\left|d\right(x,fx)-d(y,gy\left)\right|,\hfill \\ \hfill & d(x,fx)+\left|d\right(x,y)-d(y,gy\left)\right|,\hfill \\ \hfill & d(y,gy)+\left|d\right(x,y)-d(x,fx\left)\right|\}.\hfill \end{array}$$

(1)

Lemma 1. 

On $(X,d,f,g)$, for all $x\in Fix\left(f\right)=\{x\in X:fx=x\}$ and $y\in Fix\left(g\right)=\{y\in X:gy=y\}$, if $(f,g)$ is an M-auxiliary contraction or $x=y$, then $M(x,y)=0$.

Proof. 

Assume that $(f,g)$ is an M-auxiliary contraction and let $x,y\in X$ such that

$$\begin{array}{c}\hfill d(fx,gy)\le \alpha (x,y,fx,gy)M(x,y).\end{array}$$

From (1), we have $M(x,y)=d(x,y).$ Assume that $M(x,y)\ne 0$, implying $d(x,y)>0$. We then obtain

$$\begin{array}{c}\hfill d(x,y)=d(fx,gy)\le \alpha (x,y,fx,gy)M(x,y)=\alpha (x,y,fx,gy)d(x,y)<d(x,y),\end{array}$$

which leads to a contradiction. Therefore, $M(x,y)=0$. Furthermore, assume that $x=y$. Since $x\in Fix\left(f\right)$ and $y\in Fix\left(g\right)$, we have $fx=x=y=gy$. Then

$$d(x,y)=d(x,fx)=d(y,gy)=0.$$

Hence,

$$\begin{array}{cc}\hfill d(x,y)+\left|d\right(x,fx)-d(y,gy\left)\right|& =d(x,fx)+\left|d\right(x,y)-d(y,gy\left)\right|\hfill \\ \hfill & =d(y,gy)+\left|d\right(x,y)-d(x,fx\left)\right|=0.\hfill \end{array}$$

Applying the definition of maximum, we conclude that $M(x,y)=0$. □

Next, we define the concept of common fixed points and present the associated lemma.

Definition 2. 

On $(X,d,f,g)$, let $Cm(f,g)$ denote the set of all common fixed points of f and g, defined as

$$Cm(f,g)=\{x\in X:fx=gx=x\}.$$

Lemma 2. 

On $(X,d,f,g)$, for all $x,y\in X$,

$$M(x,y)=0\mathit{if}\mathit{and}\mathit{only}\mathit{if}x=y\in Cm(f,g).$$

Proof. 

Assume that $M(x,y)=0$. By the definition of maximum, we have

$$\begin{array}{c}\hfill 0=M(x,y)\ge d(x,y)+\left|d\right(x,fx)-d(y,gy\left)\right|\ge d(x,y)\ge 0,\\ \hfill 0=M(x,y)\ge d(x,fx)+\left|d\right(x,y)-d(y,gy\left)\right|\ge d(x,fx)\ge 0,\end{array}$$

and

$$\begin{array}{c}\hfill 0=M(x,y)\ge d(y,gy)+\left|d\right(x,y)-d(x,fx\left)\right|\ge d(y,gy)\ge 0.\end{array}$$

It follows that $d(x,y)=d(x,fx)=d(y,gy)=0$, which implies $fx=x=y=gy$, that is, $x=y\in Cm(f,g)$. For the converse, suppose $x=y\in Cm(f,g)$. Then, $x\in Fix\left(f\right)$ and $y\in Fix\left(g\right)$, which implies that $M(x,y)=0$ by Lemma 1. □

To understand how the maps f and g interact, we define a sequence that applies them alternately.

Definition 3. 

On $(X,d,f,g)$, a sequence $\left\{c_n\right\}$ is called an even–odd split sequence if it satisfies

$$c_{2n+1}=g{c}_{2n}\mathit{and}{c}_{2n+2}=f{c}_{2n+1}$$

for every n in the set of natural numbers $\mathbb{N}$ including 0.

Lemma 3. 

On $(X,d,f,g)$, if $(f,g)$ is an M-auxiliary contraction and there exists an even–odd split sequence $\left\{c_n\right\}$ with $c_{n_0}=c_{n_0+1}$ for some $n_0\in \mathbb{N}\cup \left\{0\right\}$, then

$$Cm(f,g)\ne \varnothing .$$

Proof. 

Let $\left\{c_n\right\}$ be an even–odd split sequence in X such that

$$c_{2n+1}=g{c}_{2n}\mathrm{and}{c}_{2n+2}=f{c}_{2n+1},\mathrm{for}\mathrm{all}n\in \mathbb{N}\cup \left\{0\right\}.$$

Suppose there exists $n_0\in \mathbb{N}\cup \left\{0\right\}$ such that $c_{n_0}=c_{n_0+1}.$ We now consider two cases for $n_0$. In the first case, assume that $n_0$ is odd, that is, $n_0=2k-1$ for some $k\in \mathbb{N}$. Then, we have

$$c_{2k-1}=c_{2k}=f{c}_{2k-1},$$

which implies that $c_{2k-1}$ is a fixed point of f. Next, we aim to show that $f{c}_{2k-1}=g{c}_{2k}$. Suppose, for contradiction, that $f{c}_{2k-1}\ne g{c}_{2k}$, meaning $d(f{c}_{2k-1},g{c}_{2k})>0$. Since

$$\begin{array}{cc}\hfill M(c_{2k-1},c_{2k})=\max \{& d(c_{2k-1},c_{2k})+|d(c_{2k-1},f{c}_{2k-1})-d(c_{2k},g{c}_{2k})|,\hfill \\ \hfill & d(c_{2k-1},f{c}_{2k-1})+|d(c_{2k-1},c_{2k})-d(c_{2k},g{c}_{2k})|,\hfill \\ \hfill & d(c_{2k},g{c}_{2k})+|d(c_{2k-1},c_{2k})-d(c_{2k-1},f{c}_{2k-1})|\},\hfill \end{array}$$

we obtain

$$M(c_{2k-1},c_{2k})=d(c_{2k},g{c}_{2k})=d(c_{2k},c_{2k+1}).$$

Given that $(f,g)$ is an M-auxiliary contraction, we have

$$d(c_{2k},c_{2k+1})=d(f{c}_{2k-1},g{c}_{2k})\le \alpha (c_{2k-1},c_{2k},f{c}_{2k-1},g{c}_{2k})M(c_{2k-1},c_{2k})<d(c_{2k},c_{2k+1}),$$

which leads to a contradiction. Hence,

$$c_{2k-1}=c_{2k}=f{c}_{2k-1}=g{c}_{2k},$$

so $c_{2k-1}=c_{2k}\in Cm(f,g)$. Additionally, the case where $n_0$ is even can be obtained by applying the procedure from the first case. □

Lemma 4. 

Let $\left\{c_n\right\}$ be a sequence in a metric space $(X,d)$ such that

$$\underset{n\to \infty }{\lim }d(c_n,c_{n+1})=0.$$

If $\left\{c_{2n}\right\}$ contains two subsequences $\left\{c_{2m\left(k\right)}\right\}$ and $\left\{c_{2n\left(k\right)}\right\}$ along with $\epsilon >0$, such that for each $k\in \mathbb{N}$, $n\left(k\right)>m\left(k\right)\ge k$, where $n\left(k\right)$ is the smallest possible number, as well as

$$d(c_{2n\left(k\right)},c_{2m\left(k\right)})\ge \epsilon \mathit{and}d(c_{2n\left(k\right)-2},c_{2m\left(k\right)})<\epsilon ,$$

then

$${\displaystyle \epsilon =\underset{k\to \infty }{\lim }d(c_{2n\left(k\right)},c_{2m\left(k\right)})=\underset{k\to \infty }{\lim }d(c_{2n\left(k\right)-1},c_{2m\left(k\right)}).}$$

Proof. 

Assume that $\left\{c_{2n}\right\}$ contains two subsequences $\left\{c_{2m\left(k\right)}\right\}$ and $\left\{c_{2n\left(k\right)}\right\}$ along with $\epsilon >0$, such that for each $k\in \mathbb{N}$, $n\left(k\right)>m\left(k\right)\ge k$, where $n\left(k\right)$ is the smallest possible number, as well as

$$d(c_{2n\left(k\right)},c_{2m\left(k\right)})\ge \epsilon \mathrm{and}d(c_{2n\left(k\right)-2},c_{2m\left(k\right)})<\epsilon .$$

Then, we have

$$\begin{array}{cc}\hfill \epsilon & \le d\left(c_{2n\left(k\right)},c_{2m\left(k\right)}\right)\hfill \\ \hfill & \le d\left(c_{2n\left(k\right)},c_{2n\left(k\right)-2}\right)+d\left(c_{2n\left(k\right)-2},c_{2m\left(k\right)}\right)\hfill \\ \hfill & <d\left(c_{2n\left(k\right)},c_{2n\left(k\right)-1}\right)+d\left(c_{2n\left(k\right)-1},c_{2n\left(k\right)-2}\right)+\epsilon .\hfill \end{array}$$

It follows from ${\displaystyle \underset{n\to \infty }{\lim }d(c_n,c_{n+1})=0}$ that

$${\displaystyle \underset{k\to \infty }{\lim }d\left(c_{2n\left(k\right)},c_{2m\left(k\right)}\right)=\epsilon .}$$

Moreover, by applying the triangle inequality,

$$\begin{array}{cc}\hfill d\left(c_{2n\left(k\right)},c_{2m\left(k\right)}\right)& \le d\left(c_{2n\left(k\right)},c_{2n\left(k\right)-1}\right)+d\left(c_{2n\left(k\right)-1},c_{2m\left(k\right)}\right)\hfill \\ \hfill & <d\left(c_{2n\left(k\right)},c_{2n\left(k\right)-1}\right)+d\left(c_{2n\left(k\right)-1},c_{2n\left(k\right)-2}\right)+\epsilon .\hfill \end{array}$$

Taking $k\to \infty$, we obtain ${\displaystyle \underset{k\to \infty }{\lim }d\left(c_{2n\left(k\right)-1},c_{2m\left(k\right)}\right)=\epsilon .}$ □

We now proceed to prove the main theorem concerning the existence and uniqueness of common fixed points. In particular, we establish these fundamental properties by relying on a single condition, the M-auxiliary contraction, which generalizes and extends previous contraction frameworks.

Theorem 1. 

On $(X,d,f,g)$, if $(f,g)$ is an M-auxiliary contraction, then $Cm(f,g)$ contains exactly one element.

Proof. 

Let $c_0\in X$, and define a sequence $\left\{c_n\right\}$ in X by

$$c_{2n+1}=g{c}_{2n}\mathrm{and}{c}_{2n+2}=f{c}_{2n+1}$$

for all $n\in \mathbb{N}\cup \left\{0\right\}$. If $c_{n_0}=c_{n_0+1}$ for some $n_0\in \mathbb{N}\cup \left\{0\right\}$, then Lemma 3 implies that $Cm(f,g)\ne \varnothing$. Otherwise, we continue under the assumption that $c_n\ne c_{n+1}$ for each $n\in \mathbb{N}\cup \left\{0\right\}$. Let $n\in \mathbb{N}\cup \left\{0\right\}$, and set $C_n=d(c_{n+1},c_n)$. We aim to show that the sequence $\left\{C_n\right\}$ is non-increasing, that is,

$$C_{2n+1}\le C_{2n}\le C_{2n-1}$$

for all $n\in \mathbb{N}\cup \left\{0\right\}$. To show this, we consider two cases:

Case I: Suppose $C_{2a}<C_{2a+1}$ for some $a\in \mathbb{N}\cup \left\{0\right\}$. Then,

$$M(c_{2a+1},c_{2a})=\max \{C_{2a}+|C_{2a+1}-C_{2a}|,C_{2a+1},C_{2a}+|C_{2a}-C_{2a+1}\left|\right\}=C_{2a+1}.$$

Since $\alpha \in {\mathcal{A}}^{*}\left(X\right)$ and $(f,g)$ is an M-auxiliary contraction, it follows that

$$\begin{array}{c}0<C_{2a+1}=d(f{c}_{2a+1},g{c}_{2a})\le \alpha (c_{2a+1},c_{2a},f{c}_{2a+1},g{c}_{2a})M(c_{2a+1},c_{2a})<C_{2a+1},\end{array}$$

which leads to a contradiction. Therefore,

$$C_{2n}\ge C_{2n+1}\mathrm{for}\mathrm{all}n\in \mathbb{N}\cup \left\{0\right\}.$$

Case II: Suppose $C_{2b}>C_{2b-1}$ for some $b\in \mathbb{N}\cup \left\{0\right\}$. In this case,

$$M(c_{2b-1},c_{2b})=\max \{C_{2b-1}+|C_{2b-1}-C_{2b}|,C_{2b-1}+|C_{2b-1}-C_{2b}|,C_{2b}\}=C_{2b}.$$

Then,

$$\begin{array}{c}\hfill 0<C_{2b}=d(f{c}_{2b-1},g{c}_{2b})\le \alpha (c_{2b-1},c_{2b},f{c}_{2b-1},g{c}_{2b})M(c_{2b-1},c_{2b})<C_{2b},\end{array}$$

which leads to a contradiction. Therefore,

$$C_{2n}\le C_{2n-1},\mathrm{for}\mathrm{all}n\in \mathbb{N}\cup \left\{0\right\}.$$

Combining both cases, we conclude that

$$C_{2n+1}\le C_{2n}\le C_{2n-1}\mathrm{for}\mathrm{all}n\in \mathbb{N}\cup \left\{0\right\},$$

so the sequence $\left\{C_n\right\}$ is non-increasing. Since it is monotone and bounded, there exists $C\ge 0$ such that

$${\displaystyle \underset{n\to \infty }{\lim }{C}_n=C.}$$

Assume that $C>0$. Since $C_{2n+1}\le C_{2n}$, we have

$$\begin{array}{cc}\hfill M(c_{2n+1},c_{2n})& =\max \{C_{2n}+|C_{2n+1}-C_{2n}|,C_{2n+1},C_{2n}+|C_{2n}-C_{2n+1}\left|\right\}\hfill \\ \hfill & =\max \{2C_{2n}-C_{2n+1},C_{2n+1}\}.\hfill \end{array}$$

As $n\to \infty$, this yields

$$\begin{array}{c}\hfill \underset{n\to \infty }{\lim }M(c_{2n+1},c_{2n})=\underset{n\to \infty }{\lim }\max \{2C_{2n}-C_{2n+1},C_{2n+1}\}=C.\end{array}$$

Since $(f,g)$ is an M-axiliary contraction, we obtain

$$\begin{array}{c}\hfill C_{2n+1}=d(f{c}_{2n+1},g{c}_{2n})\le \alpha (c_{2n+1},c_{2n},f{c}_{2n+1},g{c}_{2n})M(c_{2n+1},c_{2n})<M(c_{2n+1},c_{2n}).\end{array}$$

Letting $n\to \infty$, this inequality forces

$$d(c_{2n+1},c_{2n})\to 0\mathrm{or}d(f{c}_{2n+1},g{c}_{2n})=d(c_{2n+2},c_{2n+1})\to 0,$$

because $\alpha (c_{2n+1},c_{2n},f{c}_{2n+1},g{c}_{2n})\to 1$. This contradicts the assumption that $C>0$. Therefore, $C=0$, that is,

$$\underset{n\to \infty }{\lim }{C}_n=\underset{n\to \infty }{\lim }d(c_{n+1},c_n)=0.$$

(2)

Next, we show that $\left\{c_{2n}\right\}$ is Cauchy. Suppose, for contradiction, that $\left\{c_{2n}\right\}$ is not Cauchy. This means there exists $\epsilon >0$ and two sequences of positive integers ${\left\{n\left(k\right)\right\}}_{k=1}^{\infty }$ and ${\left\{m\left(k\right)\right\}}_{k=1}^{\infty }$ such that $n\left(k\right)>m\left(k\right)>k$ for all $k\in \mathbb{N}$, and

$$\begin{array}{c}\hfill d\left(c_{2n\left(k\right)},c_{2m\left(k\right)}\right)\ge \epsilon ,\mathrm{while}d\left(c_{2n\left(k\right)-2},c_{2m\left(k\right)}\right)<\epsilon .\end{array}$$

By Lemma 4, we obtain

$$\underset{k\to \infty }{\lim }d\left(c_{2n\left(k\right)},c_{2m\left(k\right)}\right)=\epsilon =\underset{k\to \infty }{\lim }d\left(c_{2n\left(k\right)-1},c_{2m\left(k\right)}\right).$$

Since

$$\begin{array}{cc}\hfill M\left(c_{2n\left(k\right)-1},c_{2m\left(k\right)}\right)=\max & \{d\left(c_{2n\left(k\right)-1},c_{2m\left(k\right)}\right)+|C_{2n\left(k\right)-1}-C_{2m\left(k\right)}|,\hfill \\ \hfill & C_{2n\left(k\right)-1}+|d\left(c_{2n\left(k\right)-1},c_{2m\left(k\right)}\right)-C_{2m\left(k\right)}|,\hfill \\ \hfill & C_{2m\left(k\right)}+|d\left(c_{2n\left(k\right)-1},c_{2m\left(k\right)}\right)-C_{2n\left(k\right)-1}|\},\hfill \end{array}$$

and by (2), it follows that

$$\begin{array}{c}\hfill \underset{k\to \infty }{\lim }M\left(c_{2n\left(k\right)-1},c_{2m\left(k\right)}\right)=\epsilon >0.\end{array}$$

Given that $\alpha \in {\mathcal{A}}^{*}\left(X\right)$ and $(f,g)$ is an M-auxiliary contraction, we obtain

$$\begin{array}{cc}\hfill \epsilon & =\underset{k\to \infty }{\lim }d\left(c_{2n\left(k\right)},c_{2m\left(k\right)+1}\right)=\underset{k\to \infty }{\lim }d\left(f{c}_{2n\left(k\right)-1},g{c}_{2m\left(k\right)}\right)\hfill \\ \hfill & \le \underset{k\to \infty }{\lim }\alpha \left(c_{2n\left(k\right)-1},c_{2m\left(k\right)},f{c}_{2n\left(k\right)-1},g{c}_{2m\left(k\right)}\right)\underset{k\to \infty }{\lim }M\left(c_{2n\left(k\right)-1},c_{2m\left(k\right)}\right)\hfill \\ \hfill & \le \underset{k\to \infty }{\lim }M\left(c_{2n\left(k\right)-1},c_{2m\left(k\right)}\right)=\epsilon .\hfill \end{array}$$

This implies that ${\displaystyle \underset{k\to \infty }{\lim }\alpha \left(c_{2n\left(k\right)-1},c_{2m\left(k\right)},f{c}_{2n\left(k\right)-1},g{c}_{2m\left(k\right)}\right)=1}$, so

$${\displaystyle \underset{k\to \infty }{\lim }d\left(c_{2n\left(k\right)-1},c_{2m\left(k\right)}\right)=0\mathrm{or}\underset{k\to \infty }{\lim }d\left(c_{2n\left(k\right)},c_{2m\left(k\right)+1}\right)=0}$$

which leads to a contradiction. Therefore, $\left\{c_{2n}\right\}$ is Cauchy. Applying the triangle inequality yields

$$d(c_{2n+1},c_{2m})\le d(c_{2n+1},c_{2n})+d(c_{2n},c_{2m}).$$

Since $\left\{c_{2n}\right\}$ is a Cauchy sequence and, by (2), the first term on the right tends to zero as $n\to \infty$, it follows that $\left\{c_n\right\}$ is Cauchy. As $(X,d)$ is complete, there exists $\tilde{c}\in X$ such that

$${\displaystyle \underset{n\to \infty }{\lim }{c}_n=\tilde{c}.}$$

In particular, we obtain

$$\underset{n\to \infty }{\lim }{c}_{2n}=\underset{n\to \infty }{\lim }{c}_{2n+1}=\tilde{c}.$$

We now show that $\tilde{c}\in Cm(f,g)$ by proving that $\tilde{c}=f\tilde{c}$ and $\tilde{c}=g\tilde{c}$. Assume, for contradiction, that $\tilde{c}\ne f\tilde{c}$. Then, $d(\tilde{c},f\tilde{c})>0$. Consider

$$\begin{array}{cc}\hfill M(\tilde{c},c_{2n})=\max & \left\{d(\tilde{c},c_{2n})+|d(\tilde{c},f\tilde{c})-d(c_{2n},c_{2n+1})|,d(\tilde{c},f\tilde{c})+|d(\tilde{c},c_{2n})-d(c_{2n},c_{2n+1})|,\right.\hfill \\ \hfill & \left.d(c_{2n},c_{2n+1})+|d(\tilde{c},c_{2n})-d(\tilde{c},f\tilde{c})|\right\}.\hfill \end{array}$$

Taking the limit as $n\to \infty$, we obtain

$$\begin{array}{c}\hfill \underset{n\to \infty }{\lim }M(\tilde{c},c_{2n})=d(\tilde{c},f\tilde{c}).\end{array}$$

Since $\alpha \in {\mathcal{A}}^{*}\left(X\right)$ and $(f,g)$ is an M-auxiliary contraction, we have

$$\begin{array}{cc}\hfill d(f\tilde{c},\tilde{c})& =\underset{n\to \infty }{\lim }d(f\tilde{c},c_{2n+1})=\underset{n\to \infty }{\lim }d(f\tilde{c},g{c}_{2n})\hfill \\ \hfill & \le \underset{n\to \infty }{\lim }\alpha (\tilde{c},c_{2n},f\tilde{c},g{c}_{2n})\underset{n\to \infty }{\lim }M(\tilde{c},c_{2n})\hfill \\ \hfill & \le \underset{n\to \infty }{\lim }M(\tilde{c},c_{2n})=d(\tilde{c},f\tilde{c}).\hfill \end{array}$$

Dividing both sides by $d(\tilde{c},f\tilde{c})$, we find that ${\displaystyle \underset{n\to \infty }{\lim }\alpha (\tilde{c},c_{2n},f\tilde{c},g{c}_{2n})=1.}$ As $\alpha \in {\mathcal{A}}^{*}\left(X\right)$, it follows that

$${\displaystyle \underset{n\to \infty }{\lim }d(\tilde{c},f\tilde{c})=d(\tilde{c},f\tilde{c})=0\mathrm{or}\underset{n\to \infty }{\lim }d(f\tilde{c},g{c}_{2n})=\underset{n\to \infty }{\lim }d(f\tilde{c},c_{2n+1})=d(f\tilde{c},\tilde{c})=0,}$$

contradicting the assumption. Therefore, $\tilde{c}=f\tilde{c}$. Similarly, to show that $\tilde{c}=g\tilde{c}$, assume for contradiction that $\tilde{c}\ne g\tilde{c}$, so $d(\tilde{c},g\tilde{c})>0$. Since

$$\begin{array}{cc}\hfill M(c_{2n-1},\tilde{c})=\max & \left\{d(c_{2n-1},\tilde{c})+|d(c_{2n-1},c_{2n})-d(\tilde{c},g\tilde{c})|,\right.\hfill \\ \hfill & d(c_{2n-1},c_{2n})+|d(c_{2n-1},\tilde{c})-d(\tilde{c},g\tilde{c})|,\hfill \\ \hfill & \left.d(\tilde{c},g\tilde{c})+|d(c_{2n-1},\tilde{c})-d(c_{2n-1},c_{2n})|\right\},\hfill \end{array}$$

it follows that

$$\begin{array}{c}\hfill \underset{n\to \infty }{\lim }M(c_{2n-1},\tilde{c})=d(\tilde{c},g\tilde{c}).\end{array}$$

Since $\alpha \in {\mathcal{A}}^{*}\left(X\right)$ and $(f,g)$ is an M-auxiliary contraction, we have

$$\begin{array}{cc}\hfill d(\tilde{c},g\tilde{c})& =\underset{n\to \infty }{\lim }d(c_{2n},g\tilde{c})=\underset{n\to \infty }{\lim }d(f{c}_{2n-1},g\tilde{c})\hfill \\ \hfill & \le \underset{n\to \infty }{\lim }\alpha (c_{2n-1},\tilde{c},f{c}_{2n-1},g\tilde{c})\underset{n\to \infty }{\lim }M(c_{2n-1},\tilde{c})\hfill \\ \hfill & \le \underset{n\to \infty }{\lim }M(c_{2n-1},\tilde{c})=d(\tilde{c},g\tilde{c}).\hfill \end{array}$$

As in the earlier case, dividing both sides by $d(\tilde{c},g\tilde{c})$, we obtain

$${\displaystyle \underset{n\to \infty }{\lim }d(\tilde{c},g\tilde{c})=d(\tilde{c},g\tilde{c})=0\mathrm{or}\underset{n\to \infty }{\lim }d(f{c}_{2n-1},g\tilde{c})=\underset{n\to \infty }{\lim }d(c_{2n},g\tilde{c})=d(\tilde{c},g\tilde{c})=0,}$$

which contradicts the assumption. Hence, $\tilde{c}=g\tilde{c}$, and thus $\tilde{c}\in Cm(f,g)$. To conclude, we show that the common fixed point is unique. Suppose there exists another point $p\in Cm(f,g)$. Since $\tilde{c}\in Fix\left(f\right)$, $p\in Fix\left(g\right)$, and $(f,g)$ is an M-auxiliary contraction, Lemma 1 implies that

$$d(\tilde{c},p)=M(\tilde{c},p)=0,$$

which gives $p=\tilde{c}$. Therefore, the set $Cm(f,g)$ contains exactly one element. □

In order to further illustrate our results, we provide the following examples.

Example 2. 

Let $X=[0,1]\cup \left\{2\right\}$ equipped with the metric $d(q,r)=|q-r|$ for all $q,r\in X$. Define the mappings $f,g:X\to X$ by

$$fq=\left\{\begin{array}{cc}1\hfill & \mathit{if}q\in [0,1],\hfill \\ \\ {\displaystyle \frac{2}{3}}\hfill & \mathit{if}q=2,\hfill \end{array}\right.\mathit{and}gq=\left\{\begin{array}{cc}1\hfill & \mathit{if}q\in [0,1],\hfill \\ \\ {\displaystyle \frac{5}{6}}\hfill & \mathit{if}q=2,\hfill \end{array}\right.$$

for all $q\in X$. Also, define a function $\alpha :X\times X\times X\times X\to [0,1)$ as

$$\alpha (q,r,s,t)=\left\{\begin{array}{cc}{\displaystyle \frac{1}{3}}\hfill & \mathit{if}q=r,\hfill \\ \\ {\displaystyle \frac{1}{1+d(q,r)+d(s,t)}}\hfill & \mathit{if}q\ne r,\hfill \end{array}\right.$$

for all $q,r,s,t\in X$. Then, $\alpha (q,r,s,t)\in {\mathcal{A}}^{*}\left(X\right)$.We proceed to verify that $(f,g)$ is an M-auxiliary contraction in all possible cases.

Case I: Let $q,r\in [0,1]$ and $s,t\in X$. Then $fq=gq=1$, so

$$d(fq,gr)=0\le \alpha (q,r,s,t)M(q,r).$$

Case II: Let $q=2$, $r\in [0,1]$, and $s,t\in X$. Then, we compute

$$d(fq,gr)={\displaystyle \frac{1}{3}},d(q,r)=2-r,d(q,fq)={\displaystyle \frac{4}{3}},\mathit{and}d(r,gr)=1-r.$$

It follows that

$$d(q,r)+|d(q,fq)-d(r,gr)|={\displaystyle \frac{7}{3}},d(q,fq)+|d(q,r)-d(r,gr)|={\displaystyle \frac{7}{3}},$$

and

$$d(r,gr)+|d(q,r)-d(q,fq)|\le {\displaystyle \frac{5}{3}}.$$

Hence,

$$M(q,r)={\displaystyle \frac{7}{3}},\mathit{and}\alpha (q,r,s,t)={\displaystyle \frac{1}{1+d(q,r)+d(s,t)}}={\displaystyle \frac{1}{5-r}}\ge {\displaystyle \frac{1}{5}}.$$

Therefore,

$$d(fq,gr)={\displaystyle \frac{1}{3}}<{\displaystyle \frac{7}{15}}={\displaystyle \frac{1}{5}}·{\displaystyle \frac{7}{3}}\le \alpha (q,r,s,t)M(q,r).$$

Case III: Let $q\in [0,1]$,$r=2$, and $s,t\in X$. Then, we compute

$$d(fq,gr)={\displaystyle \frac{1}{6}},d(q,r)=2-q,d(q,fq)=1-q,\mathit{and}d(r,gr)={\displaystyle \frac{7}{6}}.$$

It follows that

$$d(q,r)+|d(q,fq)-d(r,gr)|={\displaystyle \frac{13}{6}},d(q,fq)+|d(q,r)-d(r,gr)|\le {\displaystyle \frac{11}{6}},$$

and

$$d(r,gr)+|d(q,r)-d(q,fq)|={\displaystyle \frac{13}{6}}.$$

Hence,

$$M(q,r)={\displaystyle \frac{13}{6}},\mathit{and}\alpha (q,r,s,t)={\displaystyle \frac{1}{1+d(q,r)+d(s,t)}}={\displaystyle \frac{1}{5-q}}\ge {\displaystyle \frac{1}{5}}.$$

Therefore,

$$d(fq,gr)={\displaystyle \frac{1}{6}}<{\displaystyle \frac{13}{30}}={\displaystyle \frac{1}{5}}·{\displaystyle \frac{13}{6}}\le \alpha (q,r,s,t)M(q,r).$$

Case IV: Let $q=r=2$ and $s,t\in X$. Then, we compute

$$d(fq,gr)={\displaystyle \frac{1}{6}},d(q,r)=0,d(q,fq)={\displaystyle \frac{4}{3}},\mathit{and}d(r,gr)={\displaystyle \frac{7}{6}}.$$

It follows that

$$d(q,r)+|d(q,fq)-d(r,gr)|={\displaystyle \frac{1}{6}},d(q,fq)+|d(q,r)-d(r,gr)|={\displaystyle \frac{5}{2}},$$

and

$$d(r,gr)+|d(q,r)-d(q,fq)|={\displaystyle \frac{5}{2}}.$$

Hence,

$$M(q,r)={\displaystyle \frac{5}{2}},\mathit{and}\alpha (q,r,s,t)={\displaystyle \frac{1}{3}}.$$

Therefore,

$$d(fq,gr)={\displaystyle \frac{1}{6}}<{\displaystyle \frac{5}{6}}={\displaystyle \frac{1}{3}}·{\displaystyle \frac{5}{2}}=\alpha (q,r,s,t)M(q,r).$$

In all cases, the condition

$$d(fq,gr)\le \alpha (q,r,fq,gr)M(q,r)$$

is satisfied. As a result, we conclude that $(f,g)$ is an M-auxiliary contraction.

Example 3. 

Let $X=[0,1]$ equipped with the metric $d(q,r)=|q-r|$ for all $q,r\in X$. Define the mappings $f,g:X\to X$ by

$$fq=\ln (1+q)\mathit{and}gq=\ln (1+{\displaystyle \frac{q}{2}}),$$

for all $q\in X$. Also, define a function $\alpha :X\times X\times X\times X\to [0,1)$ as

$$\alpha (q,r,s,t)=\left\{\begin{array}{cc}{\displaystyle \frac{1}{3}}\hfill & \mathit{if}q=r=0,\hfill \\ \\ {\displaystyle \frac{\ln (1+\mathfrak{M}(q,r,s,t\left)\right)}{\mathfrak{M}(q,r,s,t)}}\hfill & \mathit{otherwise},\hfill \end{array}\right.$$

where

$$\begin{array}{cc}\hfill \mathfrak{M}(q,r,s,t)=\max \{& d(q,r)+\left|d\right(q,s)-d(r,t\left)\right|,\hfill \\ \hfill & d(q,s)+\left|d\right(q,r)-d(r,t\left)\right|,\hfill \\ \hfill & d(r,t)+\left|d\right(q,r)-d(q,s\left)\right|\},\hfill \end{array}$$

for all $q,r,s,t\in X$. Then, $\alpha (q,r,s,t)\in {\mathcal{A}}^{*}\left(X\right)$ and $\mathfrak{M}(q,r,fq,gr)=M(q,r)$. We now verify that $(f,g)$ satisfies the conditions of an M-auxiliary contraction.

If $q=r=0$, then $d(fq,gr)=0$. Consequently, $d(fq,gr)\le \alpha (q,r,fq,gr)M(q,r)$. In all other cases, we have either $q\ne r$, or $q\ne 0$, or $r\ne 0$, that is, $d(q,r)\ne 0$, $d(q,fq)\ne 0$, or $d(r,gr)\ne 0$, which implies that $M(q,r)\ne 0$. Since

$$d\left(fq,gr\right)=\left|\ln \right(1+q)-\ln \left(1+{\displaystyle \frac{r}{2}}\right)|\le \ln (1+{\displaystyle \frac{|q-{\displaystyle \frac{r}{2}}|}{1+{\displaystyle \frac{r}{2}}}})$$

and

$$|q-{\displaystyle \frac{1}{2}}r|=|q-r+r-{\displaystyle \frac{1}{2}}r|\le |q-r|+{\displaystyle \frac{r}{2}},$$

it follows that

$$\begin{array}{cc}\hfill {\displaystyle d(fq,gr)}& \le \ln (1+{\displaystyle \frac{|q-r|+{\displaystyle \frac{r}{2}}}{1+{\displaystyle \frac{r}{2}}}}).\hfill \end{array}$$

As $q,r\in [0,1]$, the following inequality holds:

$$1+{\displaystyle \frac{|q-r|+{\displaystyle \frac{r}{2}}}{1+{\displaystyle \frac{r}{2}}}}\le 1+\mathfrak{M}(q,r,fq,gr)=1+M(q,r),$$

which implies that

$$d(fq,gr)\le \ln (1+\mathfrak{M}(q,r,fq,gr))={\displaystyle \frac{\ln (1+M(q,r\left)\right)}{M(q,r)}}M(q,r)=\alpha (q,r,fq,gr)M(q,r).$$

Therefore, we conclude that $(f,g)$ is an M-auxiliary contraction, and hence 0 is a common fixed point of the mappings f and g.

3. Applications to Fractional and Ordinary Differential Equations

This section presents applications involving both fractional and ordinary differential equations to demonstrate that the proposed contraction framework not only extends existing results but also remains effective in classical settings. We begin with Caputo fractional differential equations, which provide a powerful tool for developing more accurate and insightful economic growth models by capturing memory effects and enhancing model adaptability. Motivated by this, we now turn to the study of Caputo fractional boundary value problems of order $\delta \in (n-1,n]$, where $n\ge 2$, in the following form:

$${}^{c}D^{\delta }x\left(t\right)=f(t,x\left(t\right)),{}^{c}D^{\delta }y\left(t\right)=g(t,y\left(t\right)),t\in [0,1],n-1<\delta \le n,$$

(3)

with the boundary conditions

$$\begin{array}{c}\hfill x\left(0\right)=x^{\prime }\left(0\right)=\cdots =x^{(n-2)}\left(0\right)=0,x\left(1\right)={\int }_0^{\eta }x\left(s\right)ds,\\ \hfill y\left(0\right)=y^{\prime }\left(0\right)=\cdots =y^{(n-2)}\left(0\right)=0,y\left(1\right)={\int }_0^{\eta }y\left(s\right)ds,\end{array}$$

(4)

where $\eta \in (0,1)$, and $f,g:[0,1]\times \mathbb{R}\to \mathbb{R}$. The solution of the problem (3)–(4) is expressed as

$$\begin{array}{cc}\hfill x\left(t\right)=& {\displaystyle \frac{n{t}^n}{(n-{\eta }^n)\mathsf{\Gamma }(n-\delta )}}{\int }_0^{\eta }\widehat{I}(s,r,f,x)ds\hfill \\ \hfill & -{\displaystyle \frac{n{t}^n}{(n-{\eta }^n)\mathsf{\Gamma }(n-\delta )}}\widehat{I}(1,s,f,x)\hfill \\ \hfill & +{\displaystyle \frac{1}{\mathsf{\Gamma }(n-\delta )}}\widehat{I}(t,s,f,x),\hfill \end{array}$$

and

$$\begin{array}{cc}\hfill y\left(t\right)=& {\displaystyle \frac{n{t}^n}{(n-{\eta }^n)\mathsf{\Gamma }(n-\delta )}}{\int }_0^{\eta }\widehat{I}(s,r,g,y)ds\hfill \\ \hfill & -{\displaystyle \frac{n{t}^n}{(n-{\eta }^n)\mathsf{\Gamma }(n-\delta )}}\widehat{I}(1,s,g,y)\hfill \\ \hfill & +{\displaystyle \frac{1}{\mathsf{\Gamma }(n-\delta )}}\widehat{I}(t,s,g,y),\hfill \end{array}$$

where the Gamma function is defined by $\mathsf{\Gamma }\left(\delta \right)={\int }_0^{\infty }{s}^{\delta -1}{e}^{-s}ds$, and the integral operator $\widehat{I}$ is accordingly defined as $\widehat{I}(u,v,f,h)={\int }_0^u{(u-v)}^{n-\delta -1}f(v,h\left(v\right))dv$. To facilitate the analysis, we employ integral operators to construct an equivalent fixed-point problem, where F and G are defined as operators from $C[0,1]$ into $C[0,1]$:

$$\begin{array}{cc}\hfill F\left(x\right(t\left)\right)& {\displaystyle \frac{n{t}^n}{(n-{\eta }^n)\mathsf{\Gamma }(n-\delta )}}{\int }_0^{\eta }\widehat{I}(s,r,f,x)ds\hfill \\ \hfill & -{\displaystyle \frac{n{t}^n}{(n-{\eta }^n)\mathsf{\Gamma }(n-\delta )}}\widehat{I}(1,s,f,x)\hfill \\ \hfill & +{\displaystyle \frac{1}{\mathsf{\Gamma }(n-\delta )}}\widehat{I}(t,s,f,x),\hfill \end{array}$$

(5)

and

$$\begin{array}{cc}\hfill G\left(y\right(t\left)\right)=& {\displaystyle \frac{n{t}^n}{(n-{\eta }^n)\mathsf{\Gamma }(n-\delta )}}{\int }_0^{\eta }\widehat{I}(s,r,g,y)ds\hfill \\ \hfill & -{\displaystyle \frac{n{t}^n}{(n-{\eta }^n)\mathsf{\Gamma }(n-\delta )}}\widehat{I}(1,s,g,y)\hfill \\ \hfill & +{\displaystyle \frac{1}{\mathsf{\Gamma }(n-\delta )}}\widehat{I}(t,s,g,y),\hfill \end{array}$$

(6)

Therefore, the existence of a common solution to the problem (3)–(4) can be viewed as the existence of a common fixed point for the operators F and G.

Let $X=C[0,1]$, equipped with the uniform metric d, given by

$$d(x,y)=∥x-y∥=\sup \left\{\right|x\left(t\right)-y\left(t\right)|:t\in [0,1]\},$$

which makes $(X,d)$ a complete metric space. We will use this metric space throughout the remainder of this section.

Theorem 2. 

Let $t\in [0,1]$, $x,y\in C[0,1]$, and let $F,G\in C[0,1]$ be defined as in (5) and (6). If

$$\left|f\right(t,x\left(t\right))-g(t,y\left(t\right)\left)\right|\le K\arctan \left(\right|x-y\left|\right),$$

where the constant K is defined by

$$K={\displaystyle \frac{1}{{\sup }_{t\in (0,1)}\frac{(n-{\eta }^n)}{n\mathsf{\Gamma }(n-\delta )}\left({\int }_0^{\eta }{\int }_0^s{|s-r|}^{n-\delta -1}drds+{\int }_0^1{|1-s|}^{n-\delta -1}ds+{\displaystyle \frac{n}{n-{\eta }^n}}{\int }_0^t{|t-s|}^{n-\delta -1}ds\right)}},$$

then the problem (3)–(4) has a common solution.

Proof. 

Let $F,G\in C[0,1]$ be defined as in (5) and (6), then we have

$$\begin{array}{cc}\hfill \left|F\right(x\left(t\right))-G(y\left(t\right)\left)\right|& =|{\displaystyle \frac{n{t}^n}{(n-{\eta }^n)\mathsf{\Gamma }(n-\delta )}}{\int }_0^{\eta }{\int }_0^s{(s-r)}^{n-\delta -1}\left(f(r,x(r\left)\right)-g(r,y(r\left)\right)\right)drds\hfill \\ \hfill & -{\displaystyle \frac{n{t}^n}{(n-{\eta }^n)\mathsf{\Gamma }(n-\delta )}}{\int }_0^1{(1-s)}^{n-\delta -1}\left(f(r,x(s\left)\right)-g(r,y(s\left)\right)\right)ds\hfill \\ \hfill & +{\displaystyle \frac{1}{\mathsf{\Gamma }(n-\delta )}}{\int }_0^t{(t-s)}^{n-\delta -1}\left(f(s,x(s\left)\right)-g(s,y(s\left)\right)\right)ds|\hfill \\ \hfill & \le {\displaystyle \frac{n{t}^n}{(n-{\eta }^n)\mathsf{\Gamma }(n-\delta )}}{\int }_0^{\eta }{\int }_0^s{|s-r|}^{n-\delta -1}\left|f(r,x(r\left)\right)-g(r,y(r\left)\right)\right|drds\hfill \\ \hfill & +{\displaystyle \frac{n{t}^n}{(n-{\eta }^n)\mathsf{\Gamma }(n-\delta )}}{\int }_0^1{|1-s|}^{n-\delta -1}\left|f(s,x(s\left)\right)-g(s,y(s\left)\right)\right|ds\hfill \\ \hfill & +{\displaystyle \frac{1}{\mathsf{\Gamma }(n-\delta )}}{\int }_0^t{|t-s|}^{n-\delta -1}\left|f(s,x(s\left)\right)-g(s,y(s\left)\right)\right|ds\hfill \\ \hfill & \le \arctan \left(\right|x-y\left|\right){\displaystyle \frac{nK{t}^n}{(n-{\eta }^n)\mathsf{\Gamma }(n-\delta )}}{\int }_0^{\eta }{\int }_0^s{|s-r|}^{n-\delta -1}drds\hfill \\ \hfill & +\arctan \left(\right|x-y\left|\right){\displaystyle \frac{nK{t}^n}{(n-{\eta }^n)\mathsf{\Gamma }(n-\delta )}}{\int }_0^1{|1-s|}^{n-\delta -1}ds\hfill \\ \hfill & +\arctan \left(\right|x-y\left|\right){\displaystyle \frac{K}{\mathsf{\Gamma }(n-\delta )}}{\int }_0^t{|t-s|}^{n-\delta -1}ds.\hfill \end{array}$$

That

$$\begin{array}{c}\hfill \left|F\right(x\left(t\right))-G(y\left(t\right)\left)\right|\le mK\arctan \left(\right|x-y\left|\right),\end{array}$$

(7)

where

$$m=\underset{t\in (0,1)}{\sup }{\displaystyle \frac{(n-{\eta }^n)}{n\mathsf{\Gamma }(n-\delta )}}\left({\int }_0^{\eta }{\int }_0^s{|s-r|}^{n-\delta -1}drds+{\int }_0^1{|1-s|}^{n-\delta -1}ds+{\displaystyle \frac{n}{n-{\eta }^n}}{\int }_0^t{|t-s|}^{n-\delta -1}ds\right).$$

To proceed, define the function $\alpha :C[0,1]\times C[0,1]\times C[0,1]\times C[0,1]\to [0,1]$ by

$$\alpha (x,y,u,v)=\left\{\begin{array}{c}{\displaystyle \frac{\arctan \left(d\right(x,y\left)\right)}{d(x,y)}}\\ \\ 0\end{array}\begin{array}{c}\\ \end{array}\begin{array}{c}\mathrm{if}x\ne y,\\ \\ \mathrm{if}x=y.\end{array}\right.$$

Since $K={\displaystyle \frac{1}{m}}$, the inequality (7) implies that

$$\left|F\right(x\left(t\right))-G(y\left(t\right)\left)\right|\le \alpha (x,y,Fx,Gy)M(x,y),$$

where

$$\begin{array}{cc}\hfill M(x,y)=\max \{& d(x,y)+\left|d\right(x,Fx)-d(y,Gy\left)\right|,\hfill \\ \hfill & d(x,Fx)+\left|d\right(x,y)-d(y,Gy\left)\right|,\hfill \\ \hfill & d(y,Gy)+\left|d\right(x,y)-d(x,Fx\left)\right|\}.\hfill \end{array}$$

It follows that $(F,G)$ is an M-auxiliary contraction. Therefore, by Theorem 1, there exists a common solution to the problem (3)–(4). □

Finally, we demonstrate the applicability of our results to ordinary differential equations through the following second-order boundary value problem:

$$\left\{\begin{array}{cc}-x^{″}-f(t,x\left(t\right))=0,& t\in [0,1],\\ \\ x\left(0\right)=x\left(1\right)=0,& \end{array}\right.$$

(8)

and

$$\left\{\begin{array}{cc}-y^{″}-g(t,y\left(t\right))=0,& t\in [0,1],\\ \\ y\left(0\right)=y\left(1\right)=0,& \end{array}\right.$$

(9)

where $f,g:[0,1]\times \mathbb{R}\to \mathbb{R}$ are continuous functions. It is worth noting that $x,y\in C[0,1]$ are solutions to the differential Equations (8) and (9) if and only if they satisfy the associated integral equations:

$$x\left(t\right)={\int }_0^{1}k(t,s)f(s,x\left(s\right))ds,$$

and

$$y\left(t\right)={\int }_0^{1}k(t,s)g(s,y\left(s\right))ds,$$

where Green’s function $k(t,s)$ is given by

$$k(t,s)=\left\{\begin{array}{cc}t(1-s)& \mathrm{if}0\le t\le s\le 1,\\ \\ s(1-t)& \mathrm{if}0\le s\le t\le 1.\end{array}\right.$$

It is clear that

$${\int }_0^{1}k(t,s)ds={\displaystyle \frac{t(1-t)}{2}},\mathrm{and}\underset{t\in [0,1]}{\sup }{\int }_0^{1}k(t,s)ds={\displaystyle \frac{1}{8}}.$$

Define operators $F,G:C[0,1]\to C[0,1]$ by

$$F\left(x\left(t\right)\right)={\int }_0^{1}k(t,s)f(s,x\left(s\right))ds,\mathrm{and}G\left(y\left(t\right)\right)={\int }_0^{1}k(t,s)g(s,y\left(s\right))ds.$$

(10)

Then, the existence of a common solution to the problem (8)–(9) is equivalent to the existence of a common fixed point of the operators F and G.

Theorem 3. 

Let $t\in [0,1]$, $x,y\in C[0,1]$, and let $F,G\in C[0,1]$ be defined as in (10). If

$$|f(t,x\left(t\right))-g(t,y\left(t\right))|\le {\displaystyle \frac{8M(x,y)}{1+M(x,y)}},$$

(11)

where

$$\begin{array}{cc}\hfill M(x,y)=\max \{& \parallel x-y\parallel +|\parallel x-Fx\parallel -\parallel y-Gy\parallel |,\hfill \\ \hfill & \parallel x-Fx\parallel +|\parallel x-y\parallel -\parallel y-Gy\parallel |,\hfill \\ \hfill & \parallel y-Gy\parallel +|\parallel x-y\parallel -\parallel x-Fx\parallel |\},\hfill \end{array}$$

then the problem (8)–(9) has a common solution.

Proof. 

Let $F,G\in C[0,1]$ be defined as in (10). Using the inequality (11), we obtain

$$\begin{array}{cc}\hfill \left|F\right(x\left(t\right))-G(y\left(t\right)\left)\right|& =\left|{\int }_0^{1}k(t,s)(f(s,x\left(s\right))-g(s,y\left(s\right)))ds\right|\hfill \\ \hfill & \le {\int }_0^{1}k(t,s)\left|(f(s,x\left(s\right))-g(s,y\left(s\right)))\right|ds\hfill \\ \hfill & \le {\int }_0^{1}k(t,s)\left({\displaystyle \frac{8M(x,y)}{1+M(x,y)}}\right)ds\hfill \\ \hfill & \le \left({\displaystyle \frac{8M(x,y)}{1+M(x,y)}}\right)\underset{t\in [0,1]}{\sup }{\int }_0^{1}k(t,s)ds\hfill \\ \hfill & ={\displaystyle \frac{M(x,y)}{1+M(x,y)}}.\hfill \end{array}$$

Now define the function $\alpha :C[0,1]\times C[0,1]\times C[0,1]\times C[0,1]\to [0,1]$ by

$$\alpha (x,y,u,v)=\left\{\begin{array}{cc}{\displaystyle \frac{1}{1+\parallel x-y\parallel }}& \mathrm{if}x\ne y,\\ \\ 0& \mathrm{if}x=y.\end{array}\right.$$

Then we have

$$\begin{array}{c}d(Fx,Gy)\le {\displaystyle \frac{M(x,y)}{1+M(x,y)}}\le {\displaystyle \frac{M(x,y)}{1+\parallel x-y\parallel }}=\alpha (x,y,Fx,Gy)M(x,y).\end{array}$$

Therefore, $(F,G)$ is an M-auxiliary contraction. By Theorem 1, there exists a common solution to the problem (8)–(9). □

4. Conclusions

This paper presents a new definition of M-auxiliary contractions and develops a theoretical framework for fixed-point theorems involving auxiliary functions. It demonstrates their effectiveness in solving Caputo fractional differential equations, as well as ordinary differential equations. The results contribute to the advancement of fixed-point theorems and establish a strong foundation for future research, such as extending common fixed-point results to more generalized metric spaces, broadening the class of contractions, and weakening some assumptions to enhance applicability to a wider range of problems.