Solving Fractional Differential Equations via New Relation-Theoretic Fuzzy Fixed Point Theorems

Abstract

In this paper, we present the notion of fuzzy $\mathcal{R}-F-$contractive mappings and provide some fuzzy fixed point results in the setting of fuzzy metric spaces, which are endowed with binary relations. Furthermore, we apply our newly established fuzzy fixed point results to solve certain boundary value problems for nonlinear fractional differential equations involving the Caputo fractional derivatives. Also, we provide some examples to show the utility of our new results.

1. Introduction and Preliminaries

The concept of fuzzy sets was introduced by Zadeh [1] in 1965. He defined a fuzzy set M on a non-empty set X as a function that assigns to each element in X a value between 0 and 1, representing its degree of membership in the set X. After that, Kramosil and Michalek [2] formulated the notion of fuzzy metric spaces. The concept of fuzzy metric spaces was refined by George and Veeramani [3], who modified it to ensure the Hausdorff property (see also [4]).Subsequently, the concepts of fuzzy sets and fuzzy metric spaces have played a vital role in various scientific fields. They have found wide applications in several areas, such as artificial intelligence, control systems, decision-making, pattern recognition, image processing, medical diagnosis, and optimization problems, where uncertainty and imprecision are inherent.

Metric fixed point theory was first introduced by Banach [5] through his famous Banach contraction principle in 1922. This theory has proven to be a powerful and versatile tool for addressing problems in a wide range of disciplines, including mathematics, computer science, economics, and engineering. It has been extensively developed and generalized across various types of metric spaces including fuzzy metric spaces. The study of fixed point theory in fuzzy metric spaces began with the work of Grabiec [6], who established a fuzzy version of the Banach contraction principle. Thereafter, several researchers have proposed different types of contractive conditions and investigated the existence of fixed points for these mappings in fuzzy metric spaces (see [7,8]). In 2012, Wardowski [9], initiated the idea of F-contraction with a view to consider a new class of nonlinear contractions which generalizes the Banach contraction principle. Thereafter, many authors generalized and improved F-contraction in different ways (see [10,11] and references cited therein). To provide a fuzzy version for F-contractions, Huang et al. [12] presented fuzzy F-contractions, which can be viewed as a generalization of F-contractions with sharpener conditions in the setting of fuzzy metric spaces. Recently, several researchers have studied fuzzy F-contractions (e.g., [13,14] and references therein). Turinici [15] initiated the idea of relation-theoretic fixed point theory, which became a very active area after the great results obtained by Ran and Reurings [16] and Nieto and Lopez [17,18], who they presented a new version of the Banach contraction principle equipping the contractive condition with an ordered binary relation. Thereafter, a lot of fixed point theorems have been provided in which various definitions of binary relations are equipped (e.g., [19,20,21] and several others). Alfaqih et al. [22] provided a relation theoretic version for the Banach contraction principle in the setting of fuzzy metric spaces.

In this paper, motivated by the works of Alfaqih et al. [22] and Huang et al. [12], we introduce the notion of fuzzy $\mathcal{R}-F-$contractions in the setting of fuzzy metric spaces equipped with binary relations. Our findings generalize and unify the main results of [12,22]. We utilize fuzzy $\mathcal{R}-F-$contractions to solve some Caputo fractional differential equations. Also, we provide some examples to demonstrate the utility of our new results. The main contributions of this paper are organized as follows. In Section 2, we present our relation-theoretic fuzzy fixed point results, along with some illustrative examples. In Section 3, we consider the boundary value problem for the fractional order differential Equation (18). It is well known in the literature that the existence of a lower solution $\alpha$ and an upper solution $\gamma$ with $\alpha \le \gamma$ implies the existence of a solution to (18).Theorem 3 presents some suitable conditions ensuring the existence of a lower solution only, and guarantees the existence and uniqueness of a solution to (18).

Before proceeding to our results, we recall some basic definitions, notions, and results, which will be used throughout the rest of our work.

Definition 1

([2]). A continuous t-norm $\mathcal{T}$ is a continuous binary operation $\mathcal{T}$: $[0,1]\times [0,1]\to [0,1]$ which is commutative and associative and satisfies:

(i)

$\mathcal{T}(t,1)=t$$\forall t\in [0,1]$;

(ii)

$\mathcal{T}(t,s)\le T(u,v)$ whenever $t\le u$ and $s\le v$ $\forall t,s,u,v\in [0,1].$

The following are some well-known examples of continuous t-norm: $\mathcal{T}(t,s)=min\{t,s\}$, $\mathcal{T}(t,s)=ts$ and $\mathcal{T}(t,s)=max\{t+s-1,0\}$, $\forall t,s\in [0,1]$.

Definition 2

([3,23]). Let M be a fuzzy set on $X^2\times (0,\infty )$ and $\mathcal{T}$ a continuous t-norm. Assume that ($\forall y,z,w\in X$ and $t,s>0$):

(FMS-i) $M(y,z,t)>0$;

(FMS-ii) $M(y,z,t)=1$ iff $y=z$;

(FMS-iii) $M(y,z,t)=M(z,y,t)$;

(FMS-iv) $\mathcal{T}\left(M(y,z,t),M(z,w,s)\right)\le M(y,w,s+t))$;

(FMS-v) $M(y,z,.):(0,\infty )\to [0,1]$ is continuous.

Then, $(X,M,\mathcal{T})$ is called a fuzzy metric space. Moreover, if condition FMS-iv is replaced by the following one:

(FMS-iv*) $\mathcal{T}\left(M(y,z,t),M(z,w,t)\right)\le M(y,w,t))$, for all $y,z,w\in X$ and all $t>0$.

Then, $(X,M,\mathcal{T})$ is called a strong fuzzy metric space.

Definition 3

([3,23]). Let $(X,M,\mathcal{T})$ be a fuzzy metric space. A sequence $\left\{y_n\right\}\subseteq X$ is said to be

(i)

Convergent to $y\in X$ (written $\underset{n\to \infty }{lim}{y}_n=y$) if

$$\underset{n\to \infty }{lim}M(y_n,y,t)=1,\mathit{for}\mathit{all}t>0.$$

(ii)

Cauchy if $\forall ϵ>0$ and all $t>0$, $\exists N\in \mathbb{N}$ satisfying

$$M(y_n,y_{n+p},t)>1-ϵ\forall n\ge N\mathit{and}p\in \mathbb{N}.$$

(iii)

$(X,M,\mathcal{T})$ is complete if every Cauchy sequence in X is convergent.

Lemma 1

([3]). If $(X,M,\mathcal{T})$ is a fuzzy metric space, then M is a continuous function on $X^2\times (0,\infty )$.

Definition 4

([24]). A subset $\mathcal{R}$ of $X^2$ is called a binary relation on X. If $(y,z)\in \mathcal{R}$ (we may write $y\mathcal{R}z$ instead of $(y,z)\in \mathcal{R}$), then we say that “y is related to z under $\mathcal{R}$". If $y\mathcal{R}z$ and $y\ne z$, then we write $y{\mathcal{R}}^{\ne }z$.

Definition 5

([25,26]). A binary relation $\mathcal{R}$ on a non-empty set X is said to be the following:

(i)

Reflexive if $y\mathcal{R}y$$\forall y\in X$;

(ii)

Transitive if $y\mathcal{R}z$ and $z\mathcal{R}w$ imply $y\mathcal{R}w$$\forall y,z,w\in X$;

(iii)

Antisymmetric if $y\mathcal{R}z$ and $z\mathcal{R}y$ imply $y=z$$\forall y,z\in X$;

(iv)

Partial order if it is reflexive, antisymmetric, and transitive;

(v)

h-closed if $y\mathcal{R}z\Rightarrow hy\in \mathcal{R}hz\forall y,z\in X$, where h is a self mapping on X.

Definition 6

([22]). A binary relation $\mathcal{R}$ on X is said to be an M-self-closed if given any convergent sequence $\left\{y_n\right\}\subseteq X$ with $y_n\mathcal{R}{y}_{n+1}$ (for all n) which converges to some $y\in X$, $\exists \left\{y_{n_k}\right\}\subseteq \left\{y_n\right\}$ with $y_{n_k}\mathcal{R}y$.

Definition 7

([22]). A Cauchy sequence $\left\{y_n\right\}\subseteq X$ is called an $\mathcal{R}$-Cauchy if $y_n\mathcal{R}{y}_{n+1}$, for all $n\in {\mathbb{N}}_0$.

Definition 8

([22]). A fuzzy metric space $(X,M,\mathcal{T})$ which is endowed with a binary relation $\mathcal{R}$ is said to be $\mathcal{R}$-complete if every $\mathcal{R}$-Cauchy sequence is convergent in X.

Remark 1

([22]). Given any binary relation $\mathcal{R}$ on a non empty set X, then we have the following:

(i)

Every Cauchy sequence is an $\mathcal{R}$-Cauchy sequence. Indeed, $\mathcal{R}$-Cauchyness coincides with Cauchyness if $\mathcal{R}$ is taken to be the universal relation.

(ii)

Every complete fuzzy metric space is an $\mathcal{R}$-complete fuzzy metric space. Indeed, $\mathcal{R}$-completeness coincides with completeness if $\mathcal{R}$ is taken to be the universal relation.

Lemma 2

([12]). Let $(X,M,\mathcal{T})$ be a fuzzy metric space and $\left\{y_n\right\}$ be a sequence in X such that for any $n\in \mathbb{N}$,

$$\underset{t\to 0^{+}}{lim}M(y_n,y_{n+1},t)>0$$

(1)

and for any $t>0$,

$$\underset{n\to \infty }{lim}M(y_n,y_{n+1},t)=1.$$

(2)

If $\left\{y_n\right\}$ is not Cauchy in X, then there exist $ϵ\in (0,1)$, $t_0>0$ and two subsequences $\left\{y_{n\left(k\right)}\right\}$ and $\left\{y_{m\left(k\right)}\right\}$ of $\left\{y_n\right\}$ with $k<m\left(k\right)<m\left(k\right)$, $k\in \mathbb{N}$, such that the sequences

$$\left\{M(y_{m\left(k\right)},y_{n\left(k\right)},t_0)\right\},\left\{M(y_{m\left(k\right)},y_{n\left(k\right)+1},t_0)\right\},\left\{M(y_{m\left(k\right)-1},y_{n\left(k\right)},t_0)\right\},$$

$$\left\{M(y_{m\left(k\right)-1},y_{n\left(k\right)+1},t_0)\right\}\mathit{and}\left\{M(y_{m\left(k\right)+1},y_{n\left(k\right)+1},t_0)\right\}$$

tend to $1-ϵ$ as $k\to \infty$.

2. Fuzzy Fixed Point Results

Let $\mathcal{F}$ denotes the family of all strictly increasing functions F: $[0,1]\to \mathbb{R}$. Now, we introduce the notion of fuzzy $\mathcal{R}-F-$contractive mappings.

Definition 9.

Let $(\mathcal{Y},\mathcal{M},\mathcal{T})$ be a fuzzy metric space endowed with a binary relation $\mathcal{R}$ on $\mathcal{Y}$ and h: $\mathcal{Y}\to \mathcal{Y}$. We say that h is a fuzzy $\mathcal{R}-F-$contractive mapping if there exists $\tau \in (0,1)$ and $F\in \mathcal{F}$ such that (for all $y,z\in \mathcal{Y}$ and all $t>0$)

$$y{\mathcal{R}}^{\ne }z\Rightarrow \tau .F\left(\mathcal{M}\left(h\right(y),h(z),t)\right)\ge F\left(\mathcal{M}(y,z,t)\right).$$

(3)

The following is an example for a fuzzy $\mathcal{R}-F-$contractive mapping.

Example 1.

Let $\mathcal{Y}=[0,2)$ and $\mathcal{T}$ be the product continuous t-norm. Define $\mathcal{M}:{\mathcal{Y}}^2\times (0,\infty )\to [0,1]$ as:

$$\mathcal{M}(y,z,t)=e^{{\displaystyle \frac{-|y-z|}{t}}},\forall y,z\in \mathcal{Y}.$$

Define a binary relation $\mathcal{R}$ on $\mathcal{Y}$ as follows:

$$y\mathcal{R}z\iff y,z\in [0,1].$$

Consider the mapping $h:\mathcal{Y}\to \mathcal{Y}$ given by the following:

$$h\left(y\right)=\left\{\begin{array}{cc}{\displaystyle \frac{1}{9}}{y}^3,\hfill & \mathit{if}0\le y\le 1;\hfill \\ {\displaystyle \frac{1}{9}}y+1,\hfill & \mathit{if}1<y<2.\hfill \end{array}\right.$$

Let $F(0,1]\to \mathbb{R}$ be given by: $F\left(s\right)=lns,$ for all $s\in (0,1]$. We prove that h is a fuzzy $\mathcal{R}-F-$contractive mapping with $\tau ={\displaystyle \frac{1}{4}}$. Notice that

$$\begin{array}{ccc}\hfill {\displaystyle \frac{1}{4}}F\left(\mathcal{M}\left(h\right(y),h(z),t)\right)& \ge & F\left(\mathcal{M}(y,z,t)\right)\hfill \\ \hfill \Rightarrow {\displaystyle \frac{1}{4}}F\left(\mathcal{M}({\displaystyle \frac{y^3}{9}},{\displaystyle \frac{z^3}{9}},t)\right)& \ge & F\left(\mathcal{M}(y,z,t)\right)\hfill \\ \hfill \Rightarrow {\displaystyle \frac{1}{4}}F\left(e^{{\displaystyle \frac{-|y^3-z^3|}{9t}}}\right)& \ge & F\left(e^{{\displaystyle \frac{-|y-z|}{9t}}}\right)\hfill \\ \hfill \Rightarrow {\displaystyle \frac{1}{4}}{\displaystyle \frac{(-|y^3-z^3\left|\right)}{9t}}& \ge & {\displaystyle \frac{-|y-z|}{9t}}\hfill \\ \hfill \Rightarrow {\displaystyle \frac{1}{4}}{\displaystyle \frac{(-|y-z\left|\right|y^2+yz+z^2\left|\right)}{9t}}& \ge & {\displaystyle \frac{-|y-z|}{9t}}\hfill \\ \hfill \Rightarrow (y^2+yz+z^2)& \le & 4,\hfill \end{array}$$

which is true for all $y,z\in [0,1]$. Therefore, h is a fuzzy $\mathcal{R}-F-$contractive mapping.

Next, we provide our first main result in this section as follows.

Theorem 1.

Let $(\mathcal{Y},\mathcal{M},\mathcal{T})$ be a fuzzy metric space which equipped with a binary relation $\mathcal{R}$ on $\mathcal{Y}$ and $h:\mathcal{Y}\to \mathcal{Y}$. Assume that $\mathcal{Y}$ is an $\mathcal{R}-$complete and h is a fuzzy $\mathcal{R}-F-$contractive mapping such that the following hold:

(i)

$y_0\mathcal{R}h{y}_0$ for some $y_0\in \mathcal{Y}$;

(ii)

$\mathcal{R}$ is transitive and $h-$closed;

(iii)

one of the following holds:

(a)

h is continuous;

(b)

$\mathcal{R}$ is $\mathcal{M}-$self-closed.

Then, h has a fixed point in $X.$

Proof. 

Let $y_0\in \mathcal{Y}$ be such that $y_0\mathcal{R}h{y}_0$ (due to $\left(i\right)$). Define $y_{n+1}=h{y}_n,$ for all $n\in {\mathbf{N}}_0.$ If $y_n=y_{n+1},$ for some $n\in {\mathbf{N}}_0,$ then $y_n$ is a fixed point of h and the result is established. Assume that $y_n\ne y_{n+1},$ for all $n\in {\mathbf{N}}_0.$ We use the h-closedness of $\mathcal{R}$ to show that $\left\{y_n\right\}$ is monotone under ${\mathcal{R}}^{\ne }$ (i.e., $y_n{\mathcal{R}}^{\ne }{y}_{n+1}$ for all n). Since $y_1=h{y}_0$, we have $y_0\mathcal{R}{y}_1$ (in view of $\left(i\right)$). Now, as $\mathcal{R}$ is h-closed and $y_0\mathcal{R}{y}_1$, we get $h{y}_0\mathcal{R}h{y}_1$ or $y_1\mathcal{R}{y}_2$. Inductively, we have $y_n\mathcal{R}{y}_{n+1}$ for all n. Since $y_n\ne y_{n+1},$$y_n\mathcal{R}{y}_{n+1}$ for all n, we have $y_n{\mathcal{R}}^{\ne }{y}_{n+1}$ for all n. Now, h is a fuzzy $\mathcal{R}-F-$contractive mapping and $y_n{\mathcal{R}}^{\ne }{y}_{n+1}$ for all n, so there exist $\tau \in (0,1)$ and $F\in \mathcal{F}$ such that ($\forall n\in {\mathbb{N}}_0$ and all $t>0$)

$$F\left(\mathcal{M}(h\left(y_n\right),h\left(y_{n+1}\right),t)\right)>\tau .F\left(\mathcal{M}(h\left(y_n\right),h\left(y_{n+1}\right),t)\right)\ge F\left(\mathcal{M}(y_n,y_{n+1},t)\right).$$

(4)

As F is strictly increasing, we obtain ($\forall n\in {\mathbb{N}}_0$ and all $t>0$)

$$\mathcal{M}(h{y}_n,h{y}_{n+1},t)>\mathcal{M}(y_n,y_{n+1},t),$$

or

$$\mathcal{M}(y_{n+1},y_{n+2},t)>\mathcal{M}(y_n,y_{n+1},t),$$

which implies that $\left\{\mathcal{M}(y_n,y_{n+1},t)\right\}$ is a strictly increasing sequence in $(0,1)$, so it is convergent. That is, for any $t>0$, there must be $\delta \left(t\right)$ in $(0,1]$ such that

$$\underset{n\to \infty }{lim}\mathcal{M}(y_n,y_{n+1},t)=\delta \left(t\right).$$

(5)

Claim: $\delta \left(t\right)=1$, for all $t>0.$ To accomplish our claim, let us assume the contrary. That is, we assume that there exists $t_0>0$, such that $\delta \left(t_0\right)<1$. Notice that, for all $n\in \mathbb{N}$, we have

$$\mathcal{M}(y_n,y_{n+1},t_0))<\delta \left(t_0\right).$$

(6)

Making use of (5) and (6), we obtain

$$\underset{n\to \infty }{lim}F\left(\mathcal{M}(y_n,y_{n+1},t_0)\right)=F(\delta \left(t_0\right)-0).$$

(7)

Due to (4), we have

$$F\left(\mathcal{M}(y_{n+1},y_{n+2},t_0)\right)>\tau .F\left(\mathcal{M}(y_{n+1},y_{n+2},t_0)\right)\ge F\left(\mathcal{M}(y_n,y_{n+1},t_0)\right).$$

(8)

Taking the limit on both sides of (8) and using (7), we obtain

$$F(\delta \left(t_0\right)-0)>\tau .F(\delta \left(t_0\right)-0)\ge F(\delta \left(t_0\right)-0),$$

which provides $F(\delta \left(t_0\right)-0)=0$. This contradicts the fact that $F(\delta \left(t_0\right)-0)>0$. Therefore, we must have

$$\underset{n\to \infty }{lim}\mathcal{M}(y_n,y_{n+1},t)=1,\forall t>0.$$

(9)

Next, we need to show that $\left\{y_n\right\}$ is an $\mathcal{R}$-Cauchy sequence in $(\mathcal{Y},\mathcal{M},\mathcal{T}).$ Assume, on the contrary, that $\left\{y_n\right\}$ is not a Cauchy. In view of (9) and Lemma 2, there exist $\epsilon \in (0,1)$ and some $t_0>0$ such that for all $k\in {\mathbf{N}}_0$ there exist $m\left(k\right),n\left(k\right)\in {\mathbf{N}}_0$ with $k\le n\left(k\right)\le m\left(k\right)$ satisfying

$$\underset{k\to \infty }{lim}\mathcal{M}(y_{m\left(k\right)},y_{n\left(k\right)},t_0)=1-\epsilon .$$

(10)

Making use of (4), we obtain

$$F\left(\mathcal{M}(h{y}_{m\left(k\right)},h{y}_{n\left(k\right)},t_0)\right)>\tau .F\left(\mathcal{M}(h{y}_{m\left(k\right)},h{y}_{n\left(k\right)},t_0)\right)\ge F(\mathcal{M}(y_{m\left(k\right)},y_{n\left(k\right)},t_0).$$

Letting $k\to \infty$ and using (10), we obtain

$$F\left(\right(1-ϵ)-0)\ge \tau .F\left(\right(1-ϵ)-0)\ge F\left(\right(1-ϵ)-0),$$

this implies that $F\left(\right(1-ϵ)-0)=0$, a contradiction (as $F\left(\right(1-ϵ)-0)>0$). Hence, $\{y_n\}$ must be a Cauchy sequence. As $y_n\mathcal{R}{y}_{n+1}$ for all n, we have $\{y_n\}$ is an $\mathcal{R}-$Cauchy sequence. Since $(\mathcal{X},\mathcal{M},\mathcal{T})$ is $\mathcal{R}-$complete, there exists $y\in \mathcal{Y}$ such that $y_n\to y.$

Now, in view of condition $\left(iii\right)$, we distinguish the following two cases. Case1, h, is a continuous mapping. Taking the limit as $n\to \infty$ on both sides of $y_{n+1}=h{y}_n,$ $n\in {\mathbf{N}}_0,$ we obtain $y=hy$, and hence y is a fixed point of h. The proof is completed in this case. In Case2, $\mathcal{R}$ is $M-$self-closed; then, there exists a subsequence $\left\{y_{n\left(k\right)}\right\}\subseteq \left\{y_n\right\}$ such that $y_{n\left(k\right)}{\mathcal{R}}^{\ne }y$, for all $k\in {\mathbf{N}}_0.$ We need to prove that $y=hy.$ On the contrary, assume that $y\ne hy$. Since $\underset{k\to \infty }{lim}{y}_{n\left(k\right)}=y$, we have $\underset{k\to \infty }{lim}\mathcal{M}(y_{n\left(k\right)},y,t)=1$, for all $t>0.$ Then, $\mathcal{M}(y_{n\left(k\right)},y,t)>0$ when k is large enough for all $t>0$ and as $y_{n\left(k\right)}{\mathcal{R}}^{\ne }y,$ from (3), we deduce that

$$F\left(\mathcal{M}(h{y}_{n\left(k\right)},hy,t)\right)>\tau .F\left(\mathcal{M}(h{y}_{n\left(k\right)},hy,t)\right)\ge F(\mathcal{M}(y_{n\left(k\right)},y,t),\forall t>0.$$

Since, F is strictly increasing, we obtain

$$1\ge \mathcal{M}(y_{n\left(k\right)+1},hy,t_0))>\mathcal{M}(y_{n\left(k\right)},y,t_0).$$

Letting $k\to \infty$, we obtain

$$1\ge \underset{k\to \infty }{lim}\mathcal{M}(y_{n\left(k\right)+1},hy,t_0))\ge \underset{k\to \infty }{lim}\mathcal{M}(y_{n\left(k\right)},y,t_0)=1.$$

Thus, $\underset{k\to \infty }{lim}\mathcal{M}(y_{n\left(k\right)+1},hy,t_0))=1$ or $\underset{k\to \infty }{lim}{y}_{n\left(k\right)+1}=hy$. The uniqueness of the limit guarantees that $hy=y.$ This ends the proof. □

Now, we present the following example, which exhibits the utility of Theorem 1.

Example 2.

Let $\mathcal{Y}=[0,2)$ and $\mathcal{T}$ be the product’s continuous t-norm. Let $\mathcal{M}$ and $\mathcal{R}$ be defined as in Example 1. Then, $\mathcal{R}$ is a transitive binary relation on $\mathcal{Y}$. Also, $(\mathcal{Y},\mathcal{M},\mathcal{T})$ forms an $\mathcal{R}$-complete fuzzy metric space. Consider the fuzzy $\mathcal{R}-F-$contractive mapping $h:\mathcal{Y}\to \mathcal{Y}$ defined in Example 1. Let $y,z\in \mathcal{Y}$ be such that $y\mathcal{R}z$. Since $y\mathcal{R}z$, we have $y,z\in [0,1]$. Now, $y,z\in [0,1]$ implies that $hy,hz\in [0,1]$ (due to the definition of h). Hence, $hy\mathcal{R}hz$. Therefore, $\mathcal{R}$ is h-closed. Also, $0\in \mathcal{Y}$ and $0\mathcal{R}h0$. Now, consider the sequence $\left\{y_n\right\}={\left\{{\displaystyle \frac{1}{n}}\right\}}_{n\ge 1}\subseteq \mathcal{Y}$. Then, $y_n{\mathcal{R}}^{\ne }{y}_{n+1}$ for all $n\ge 1$ and $y_n\to 0$. Thus, $\mathcal{R}$ is $\mathcal{M}-$self-closed. Hence, all the requirements of Theorem 1 are fulfilled and h has a fixed point (namely, $y=0$).

Next, we provide the following uniqueness theorem.

Theorem 2.

In addition to the hypotheses of Theorem 1, if the following condition holds:

(iv)

for all $y,z\in Fix\left(h\right)$ with $y\ne z$, there exists $w\in \mathcal{Y}$ such that $y{\mathcal{R}}^{\ne }w$ and $w{\mathcal{R}}^{\ne }z$.

Then, $Fix\left(h\right)$ is a singleton set.

Proof. 

In view of Theorem 1, $Fix\left(h\right)$ is not empty. On the contrary, let us assume that $Fix\left(h\right)$ is not a singleton set. Let $y,z\in Fix\left(h\right)$ be such that $y\ne z$. Due to condition $\left(iv\right)$, there exists $w\in \mathcal{Y}$ such that $y{\mathcal{R}}^{\ne }w$ and $w{\mathcal{R}}^{\ne }z$. Define $w_0=w$ and $w_{n+1}=h{w}_n$, $\forall n\ge 0.$ Notice that $y{\mathcal{R}}^{\ne }{w}_0$. Therefore, (4) ensures the existence of $\tau \in (0,1)$ such that (for all $t>0$)

$$F\left(\mathcal{M}(y,w_1,t)\right)>\tau .F\left(\mathcal{M}(hy,h{w}_0,t)\right)\ge F(\mathcal{M}(y,w_0,t).$$

As F is strictly increasing, we obtain $\mathcal{M}(y,w_1,t)>\mathcal{M}(y,w_0,t).$ Similarly, one can show that $\mathcal{M}(y,w_2,t)>\mathcal{M}(y,w_1,t)$, for all $t>0$. Inductively, we obtain $\mathcal{M}(y,w_{n+1},t)>\mathcal{M}(y,w_n,t)$, for all $n\in {\mathbf{N}}_0$ and all $t>0$. Hence, for any $t>0$, $\left\{\mathcal{M}(y,w_n,t)\right\}$ is a strictly increasing sequence in $(0,1)$. So, there exists $0<\delta \left(t\right)\le 1$ for all $t>0$ such that

$$\underset{n\to \infty }{lim}\mathcal{M}(y,w_n,t)=\delta \left(t\right).$$

(11)

Claim: $\delta \left(t\right)=1$, for all $t>0.$ To prove our claim, let us assume the contrary. That is, we assume that there exists $t_0>0$ such that $\delta \left(t_0\right)<1$. Notice that, for all $n\in \mathbb{N}$, we have

$$\mathcal{M}(y,w_n,t_0))<\delta \left(t_0\right).$$

(12)

Making use of (11) and (12), we obtain

$$\underset{n\to \infty }{lim}F\left(\mathcal{M}(y,w_n,t_0)\right)=F(\delta \left(t_0\right)-0).$$

(13)

Now, as $y\mathcal{R}{w}_0,$ and $\mathcal{R}$ is h-closed, we can see that $y\mathcal{R}{w}_n,$ for all $n\in {\mathbb{N}}_0$. If $y=w_n$, for some $n\in \mathbb{N}$, then $\left\{w_n\right\}$ converges to y. Assume that $y\ne w_n$ for all $n\in \mathbb{N}$. Then, condition (4) guarantees the existence of $\tau >0$ such that (for all $n\in {\mathbb{N}}_0$)

$$F\left(\mathcal{M}(y,w_{n+1},t_0)\right)>\tau .F\left(\mathcal{M}(hy,h{w}_n,t_0)\right)\ge F(\mathcal{M}(y,w_n,t_0).$$

(14)

Taking limit on both sides of (14) and using (13), we obtain

$$F(\delta \left(t_0\right)-0)\ge \tau .F(\delta \left(t_0\right)-0)\ge F(\delta \left(t_0\right)-0),$$

which provides $F(\delta \left(t_0\right)-0)=0$. This contradicts the fact that $F(\delta \left(t_0\right)-0)>0$. Therefore, we must have

$$\underset{n\to \infty }{lim}\mathcal{M}(y,w_n,t)=1,\mathrm{for}\mathrm{all}t>0.$$

(15)

Similarly, one can show that

$$\underset{n\to \infty }{lim}\mathcal{M}(z,w_n,t)=1,\mathrm{for}\mathrm{all}t>0.$$

(16)

Observe that Equation (15) provides $\underset{n\to \infty }{lim}{w}_n=y$ and Equation (16) provides $\underset{n\to \infty }{lim}{w}_n=z$. The uniqueness of the limit implies that $y=z$, a contradiction. Therefore, $Fix\left(h\right)$ must be a singleton set, as required. □

Now, we present the following example, which supports Theorem 2.

Example 3.

Consider the fuzzy $\mathcal{R}-F-$contractive mapping h provided in Example 1. Observe that zero is the only fixed point of h in $\mathcal{Y}$. Example 2 shows that all the requirements of Theorem 2 are fulfilled, and hence the fixed point of h is unique.

Remark 2.

(1)

Under the universal relation, Theorem 2 reduces to the main result of Huang et al. [12].

(2)

Through setting $F\left(s\right)=lns$ in Theorems 1 and 2, we obtain the main results of Alfaqih et al. [22].

3. Nonlinear Fractional Differential Equations

In this section, we apply our fixed point results to study the existence of solutions of boundary value problems for fractional differential equations involving the Caputo fractional derivative.

Let $X=C\left(\right[0,1],R)$ be the Banach space of all continuous functions from $[0,1]$ into R with the norm

$${\parallel y\parallel }_{\infty }=\underset{t\in [0,1]}{sup}\left|y\left(t\right)\right|.$$

Define $\mathcal{M}:{\mathcal{Y}}^2\times (0,\infty )\to [0,1]$ using the following:

$$\mathcal{M}(y,z,t)=e^{{\displaystyle \frac{-\parallel y-z{\parallel }_{\infty }}{t}}},\forall y,z\in X,t\in (0,\infty ).$$

Then, $(X,M,T)$ is a complete fuzzy metric space with the product continuous t-norm (see [27]). Define a binary relation $\mathcal{R}$ on X using the following:

$$y\mathcal{R}z\iff y\left(t\right)\le z\left(t\right)\mathrm{for}\mathrm{all}y,z\in X,t\in [0,1].$$

Then, $(X,M,T)$ is an $\mathcal{R}-$complete fuzzy metric space the product continuous t-norm. Also, observe that $\mathcal{R}$ is a transitive binary relation on X.

Now, let us recall the following basic notions, which will be needed in the sequel.

Definition 10

([28]). The Caputo fractional derivative of a function $\sqcap :[a,b]\to [a,b]$ of order $\beta >0$ is defined by

$$\left({}^c\mathcal{D}_{0^{+}}^{\beta }\right)\sqcap \left(t\right)={\displaystyle \frac{1}{\Gamma (n-\beta )}}\underset{a}{\overset{t}{\int }}{(t-s)}^{n-\beta -1}{\sqcap }^{\left(n\right)}\left(s\right)ds,(n-1\le \beta <n,n=\left[\beta \right]+1),$$

(17)

where Γ is the gamma function and $\left[\beta \right]$ is the integer part of β.

Consider the boundary value problem for fractional order differential equation given by

$$\left\{\begin{array}{cc}{}^c\mathcal{D}_{0^{+}}^{\beta }\left(y\left(t\right)\right)=h(t,y\left(t\right)),\hfill & (t\in [0,1],2<\beta \le 3);\hfill \\ y\left(0\right)=c_0,y^{\prime }\left(0\right)=c_0^{*},y^{″}\left(1\right)=c_1,\hfill \end{array}\right.$$

(18)

where ${}^c\mathcal{D}_{0^{+}}^{\beta }$ denotes the Caputo fractional derivative of order $\beta ,$$h:[0,1]\to R$ is a continuous function and $c_0,c_0^{*},c_1\in R$.

Definition 11

([29]). A function $y\in C^3([0,1],R),$ which has β-derivative on $[0,1]$ is said to be a solution of (18) if ${}^c\mathcal{D}_{0^{+}}^{\beta }\left(y\left(t\right)\right)=h(t,y\left(t\right))$ on $[0,1]$ and the conditions $y\left(0\right)=c_0,$$y^{\prime }\left(0\right)=c_0^{*},$$y^{″}\left(1\right)=c_1$ are satisfied.

The following lemma is needed in what follows.

Lemma 3

([29]). Let $2<\beta \le 3$ and $\sqcap :[0,1]\to R$ be a continuous function. A function y is a solution of the fractional integral equation

$$\begin{array}{ccc}\hfill y\left(t\right)& =& {\displaystyle \frac{1}{\Gamma \left(\beta \right)}}\underset{0}{\overset{t}{\int }}{(t-s)}^{\beta -1}\sqcap \left(s\right)ds-{\displaystyle \frac{t^2}{2\Gamma (\beta -2)}}\underset{0}{\overset{1}{\int }}{(1-s)}^{\beta -3}\sqcap \left(s\right)ds+c_0+c_0^{*}t+{\displaystyle \frac{c_1}{2}}{t}^2\hfill \end{array}$$

if and only if y is a solution of the fractional boundary value problems

$$\begin{array}{c}{}^c\mathcal{D}_{0^{+}}^{\beta }\left(y\left(t\right)\right)=\sqcap \left(t\right),\hfill \\ y\left(0\right)=c_0,y^{\prime }\left(0\right)=c_0^{*},y^{″}\left(1\right)=c_1,\hfill \end{array}$$

where

$$y^{″}\left(1\right)=2c_2+{\displaystyle \frac{1}{\Gamma (\beta -2)}}{\int }_0^1{(1-s)}^{\beta -3}\sqcap \left(s\right)ds=c_1,c_i,c_0^{*}\in R,i=0,1,2.$$

Now, we state and prove our main result in this section.

Theorem 3.

Suppose that

(i)

there exists $\lambda >0$ such that (for all $y,z\in X,$$y\le z,$$t\in [0,1]$)

$$\begin{array}{c}\hfill \left|h\right(t,y\left(t\right))-h(t,z\left(t\right)\left)\right|\le \lambda \left|y\right(t)-z(t\left)\right|,\mathit{where}\end{array}$$

$$0<\tau =\lambda \left({\displaystyle \frac{1}{\Gamma (\beta +1)}}+{\displaystyle \frac{1}{2\Gamma (\beta -1)}}\right)<1,$$

(19)

(ii)

there exists a lower solution $y_0\in X$ of (18), i.e., $\exists y_0\in X$ such that

$$\begin{array}{ccc}\hfill y_0\left(t\right)& \le & {\displaystyle \frac{1}{\Gamma \left(\beta \right)}}\underset{0}{\overset{t}{\int }}{(t-s)}^{\beta -1}h(s,y_0\left(s\right))ds\hfill \\ & & -{\displaystyle \frac{t^2}{2\Gamma (\beta -2)}}\underset{0}{\overset{1}{\int }}{(1-s)}^{\beta -3}h(s,y_0\left(s\right))ds+c_0+c_0^{*}t+{\displaystyle \frac{c_1}{2}}{t}^2,\hfill \end{array}$$

(iii)

h is nondecreasing in the second variable.

Then, the Equation (18) has a unique solution in $X.$

Proof. 

Define $\mathcal{H}:X\to X$ by

$$\begin{array}{ccc}\hfill \mathcal{H}y\left(t\right)& =& {\displaystyle \frac{1}{\Gamma \left(\beta \right)}}\underset{0}{\overset{t}{\int }}{(t-s)}^{\beta -1}h(s,y\left(s\right))ds\hfill \\ & & -{\displaystyle \frac{t^2}{2\Gamma (\beta -2)}}\underset{0}{\overset{1}{\int }}{(1-s)}^{\beta -3}h(s,y\left(s\right))ds+c_0+c_0^{*}t+{\displaystyle \frac{c_1}{2}}{t}^2,\hfill \end{array}$$

where

$$c_1=2c_2+{\displaystyle \frac{1}{\Gamma (\alpha -2)}}{\int }_0^1{(1-s)}^{\beta -3}h(s,y\left(s\right))ds,c_i,c_0^{*}\in R,(i=0,1,2)\mathrm{are}\mathrm{constant}.$$

Firstly, we prove that $\mathcal{H}$ is continuous. Let $\left\{y_n\right\}$ be a sequence such that $\underset{n\to \infty }{lim}{y}_n=y$ in $X.$ Then, for each $t\in [0,1]$, we have

$$\begin{array}{cc}\hfill \left|\mathcal{H}{y}_n\left(t\right)-\mathcal{H}y\left(t\right)\right|\le & {\displaystyle \frac{1}{\Gamma \left(\beta \right)}}{\int }_0^t{(t-s)}^{\beta -1}\left|h\left(s,y_n\left(s\right)\right)-h(s,y\left(s\right))\right|ds\hfill \\ \hfill & +{\displaystyle \frac{1}{2\Gamma (\beta -2)}}{\int }_0^1{(1-s)}^{\beta -3}\left|h\left(s,y_n\left(s\right)\right)-h(s,y\left(s\right))\right|ds.\hfill \end{array}$$

Since h is a continuous function, we have

$$\begin{array}{ccc}& & \underset{n\to \infty }{lim}{\parallel h\left(s,y_n\left(s\right)\right)-h(s,y\left(s\right))\parallel }_{\infty }=0\hfill \\ & \iff & \underset{n\to \infty }{lim}{∥\mathcal{H}{y}_n-\mathcal{H}y∥}_{\infty }=0\hfill \\ & \iff & \underset{n\to \infty }{lim}{e}^{{\displaystyle \frac{-\parallel \mathcal{H}{y}_n{-\mathcal{H}y\parallel }_{\infty }}{\tau }}}=1\hfill \\ & \iff & \underset{n\to \infty }{lim}\mathcal{M}(\mathcal{H}{y}_n,\mathcal{H}y,\tau )=1\hfill \\ & \iff & \underset{n\to \infty }{lim}\mathcal{H}{y}_n=\mathcal{H}y.\hfill \end{array}$$

Hence, $\mathcal{H}$ is a continuous function.

Clearly, the fixed points of the operator $\mathcal{H}$ are solutions of the Equation (18). Now, we use Theorem 2 to prove that the operator $\mathcal{H}$ has a fixed point.

Let $y,z\in X,$ be such that $y{\mathcal{R}}^{\ne }z$. Then, $y\left(t\right)\le z\left(t\right)$, for all $t\in [0,1].$ Observe that

$$\begin{array}{ccc}\hfill \left|\mathcal{H}y\right(t)-\mathcal{H}z(t\left)\right|& \le & {\displaystyle \frac{1}{\Gamma \left(\beta \right)}}\underset{0}{\overset{t}{\int }}{(t-s)}^{\beta -1}|h(s,y\left(s\right))-h(s,z\left(s\right))|ds\hfill \\ & & +{\displaystyle \frac{t^2}{2\Gamma (\beta -2)}}\underset{0}{\overset{1}{\int }}{(1-s)}^{\beta -3}|h(s,y\left(s\right))-h(s,z\left(s\right))|ds\hfill \\ & \le & {\displaystyle \frac{1}{\Gamma \left(\beta \right)}}\underset{0}{\overset{t}{\int }}{(t-s)}^{\beta -1}|h(s,y\left(s\right))-h(s,z\left(s\right))|ds\hfill \\ & & +{\displaystyle \frac{1}{2\Gamma (\beta -2)}}\underset{0}{\overset{1}{\int }}{(1-s)}^{\beta -3}|h(s,y\left(s\right))-h(s,z\left(s\right))|ds\hfill \\ & \le & {\displaystyle \frac{1}{\Gamma \left(\beta \right)}}\underset{0}{\overset{t}{\int }}{(t-s)}^{\beta -1}\lambda |y\left(s\right)-z\left(s\right)|ds\hfill \\ & & +{\displaystyle \frac{1}{2\Gamma (\beta -2)}}\underset{0}{\overset{1}{\int }}{(1-s)}^{\beta -3}\lambda |y\left(s\right)-z\left(s\right)|ds\hfill \\ & \le & {\displaystyle \frac{1}{\Gamma \left(\beta \right)}}\underset{0}{\overset{t}{\int }}{(t-s)}^{\beta -1}\lambda \parallel y-z{\parallel }_{\infty }ds+{\displaystyle \frac{1}{2\Gamma (\beta -2)}}\underset{0}{\overset{1}{\int }}{(1-s)}^{\beta -3}\lambda \parallel y-z{\parallel }_{\infty }ds\hfill \\ & \le & {\displaystyle \frac{\lambda \parallel y-z{\parallel }_{\infty }}{\Gamma \left(\beta \right)}}\underset{0}{\overset{t}{\int }}{(t-s)}^{\beta -1}ds+{\displaystyle \frac{\lambda \parallel y-z{\parallel }_{\infty }}{2\Gamma (\beta -2)}}\underset{0}{\overset{1}{\int }}{(1-s)}^{\beta -3}ds\hfill \\ & \le & \lambda \left({\displaystyle \frac{1}{\Gamma (\beta +1)}}+{\displaystyle \frac{1}{2\Gamma (\beta -1)}}\right)\parallel y-z{\parallel }_{\infty }\hfill \\ & =& {\displaystyle \frac{\parallel y-z{\parallel }_{\infty }}{\tau }}.\hfill \end{array}$$

Hence,

$$\parallel \mathcal{H}y-\mathcal{H}z{\parallel }_{\infty }\le {\displaystyle \frac{\parallel y-z{\parallel }_{\infty }}{\tau }}.$$

This yields

$${\displaystyle \frac{\parallel \mathcal{H}y-\mathcal{H}z{\parallel }_{\infty }}{t}}\le {\displaystyle \frac{\parallel y-z{\parallel }_{\infty }}{t.\tau }},\forall t>0.$$

Which yields that

$$ln{e}^{\tau {\displaystyle \frac{-\parallel \mathcal{H}y-\mathcal{H}z{\parallel }_{\infty }}{t}}}\ge ln{e}^{{\displaystyle \frac{-\parallel y-z{\parallel }_{\infty }}{t}}},\forall t>0,$$

or

$$\tau ln{e}^{{\displaystyle \frac{-\parallel \mathcal{H}y-\mathcal{H}z{\parallel }_{\infty }}{t}}}\ge ln{e}^{{\displaystyle \frac{-\parallel y-z{\parallel }_{\infty }}{t}}},\forall t>0.$$

Therefore,

$$\tau F\left(\mathcal{M}\right(\mathcal{H}y,\mathcal{H}z,t\left)\right)\ge F\left(\mathcal{M}\right(y,z,t\left)\right),$$

with $F\left(t\right)=lnt$, for all $t>0.$ This shows that $\mathcal{H}$ is a fuzzy $\mathcal{R}-F-$contractive mapping. From $\left(ii\right)$, we conclude that $y_0\left(t\right)\mathcal{R}H{y}_0\left(t\right),$ for all $t\in [0,1]$; then, $y_0\mathcal{R}\mathcal{H}{y}_0$; that is, the condition $\left(i\right)$ of Theorem 1 is satisfied. Let $y,z\in X,$$y\left(t\right)\le z\left(t\right)$ for all $t\in [0,1],$ from $\left(iii\right)$; since h is nondecreasing in the second variable, we have

$$\begin{array}{ccc}\hfill \mathcal{H}y\left(t\right)& =& {\displaystyle \frac{1}{\Gamma \left(\beta \right)}}\underset{0}{\overset{t}{\int }}{(t-s)}^{\beta -1}h(s,y\left(s\right))ds+c_0+c_0^{*}t+c_2{t}^2\hfill \\ & \le & {\displaystyle \frac{1}{\Gamma \left(\beta \right)}}\underset{0}{\overset{t}{\int }}{(t-s)}^{\beta -1}h(s,z\left(s\right))ds+c_0+c_0^{*}t+c_2{t}^2\hfill \\ & =& \mathcal{H}z\left(t\right).\hfill \end{array}$$

we conclude that as $\mathcal{H}y\left(t\right)\le \mathcal{H}z\left(t\right)$ for all $t\in [0,1],$ then $\mathcal{H}y\le \mathcal{H}z$. Hence, $\mathcal{R}$ is $\mathcal{H}$-closed. Therefore, all the assumptions of Theorem 1 are fulfilled. Thus, $\mathcal{H}$ has a fixed point in X which provides a solution for Equation (18). Finally, observe that if $y,z\in \mathcal{Y}$ are two distinct fixed points of $\mathcal{H}$ in $\mathcal{Y}$, then $y\le max\{y,z\}$, $z\le max\{y,z\}$ and $w=max\{y,z\}\in \mathcal{Y}$. So, Theorem 2 ensures the uniqueness of the fixed point of $\mathcal{H}$, and thus the solution of (18) is also unique in $X.$ This ends the proof. □

Finally, we present the following example, which supports Theorem 3.

Example 4.

Consider the boundary value problem of fractional differential equation

$$\begin{array}{c}{D}_{0^{+}}^{{\displaystyle \frac{5}{2}}}y\left(t\right)={\displaystyle \frac{x\left(t\right)}{8(1+y(t\left)\right)}},t\in [0,1],\\ y\left(0\right)=0,y^{\prime }\left(0\right)=0,y^{″}\left(1\right)=1.\end{array}$$

(20)

Take

$$h(t,y\left(t\right))={\displaystyle \frac{y\left(t\right)}{8(1+y(t\left)\right)}},(t,y\left(t\right))\in [0,1]\times [0,\infty )$$

Let $y\left(t\right),z\left(t\right)\in [0,\infty )$ and $t\in [0,1].$ Then,

$$\begin{array}{cc}\hfill \left|h\right(t,y\left(t\right))-h(t,z\left(t\right)\left)\right|& ={\displaystyle \frac{1}{8}}|{\displaystyle \frac{y\left(t\right)}{1+y\left(t\right)}}-{\displaystyle \frac{z\left(t\right)}{1+z\left(t\right)}}|\hfill \\ \hfill & ={\displaystyle \frac{1}{8}}\left|{\displaystyle \frac{y\left(t\right)-z\left(t\right)}{(1+y(t\left)\right)(1+z(t\left)\right)}}\right|\hfill \\ \hfill & \le {\displaystyle \frac{1}{8}}|y\left(t\right)-z\left(t\right)|.\hfill \end{array}$$

Hence, condition $\left(i\right)$ of Theorem 3 holds, with $\lambda ={\displaystyle \frac{1}{8}}.$ Next, we investigate the following:

$$\tau =\lambda \left[{\displaystyle \frac{1}{\Gamma (\beta +1)}}+{\displaystyle \frac{1}{2\Gamma (\beta -1)}}\right]<1.$$

Observe that

$$0<\tau ={\displaystyle \frac{1}{8}}\left[{\displaystyle \frac{1}{\Gamma \left(\frac{7}{2}\right)}}+{\displaystyle \frac{1}{2\Gamma \left(\frac{3}{2}\right)}}\right]={\displaystyle \frac{23}{120\sqrt{\pi }}}<1.$$

Thus, (19) holds. Let $y_0=0$; then,

$$\begin{array}{c}\hfill 0\le {\displaystyle \frac{1}{\Gamma \left(\frac{5}{2}\right)}}\underset{0}{\overset{t}{\int }}{(t-s)}^{{\displaystyle \frac{3}{2}}}h(s,0)ds-{\displaystyle \frac{t^2}{2\Gamma \left(\frac{1}{2}\right)}}\underset{0}{\overset{1}{\int }}{(1-s)}^{{\displaystyle \frac{-1}{2}}}h(s,0)ds+{\displaystyle \frac{t^2}{2}}={\displaystyle \frac{t^2}{2}},t\in [0,1].\end{array}$$

Hence, the assumption $\left(ii\right)$ of Theorem 3 is satisfied. Also, if $y\le z$, we conclude $hy\le hz.$ Thus, condition $\left(iii\right)$ of Theorem 3 is satisfied. Therefore, Equation (20) has a unique solution on $[0,1]$.

Remark 3.

Although Example 4 uses $\beta ={\displaystyle \frac{5}{2}}$, the same analysis applies for any $\beta \in (2,3]$. The nonlinearity, bounds, and Green’s function estimates remain valid across this range and Theorem 3 guarantees the existence and uniqueness of a solution for all such β.

4. Conclusions

In this work, we introduced fuzzy $\mathcal{R}-F-$contractive mappings and presented key results concerning the existence and uniqueness of fixed points for such mappings in the setting of relational fuzzy metric spaces. By employing our newly established fixed point theorems, we demonstrated the existence and uniqueness of solutions for Caputo fractional differential equations. To highlight the usefulness of our main findings, we included some illustrative examples. As shown in Remark 2, our fixed point results extend and generalize the main results of Alfaqih et al. [22] and Huang et al. [12]. The novelty of this work lies in establishing relation-theoretic fuzzy fixed point results using F-contractions, which allow for a broader class of mappings and relax certain classical assumptions, such as continuity and completeness. Our findings facilitate the analysis of boundary value problems for fractional differential equations involving the Caputo fractional derivative. In particular, the incorporation of fuzziness and relational dependencies offers a more flexible framework for studying nonlinear boundary value problems. This framework enable the investigation of solutions to fractional differential equations in relational fuzzy metric spaces when addressing the existence of either lower or upper solutions only. As a direction for future research, this framework opens up new avenues for investigation and applications using several well-known contractive mappings.