Memristorbased circuit
A memristor is regarded as a 2terminal tool in which the magnetic flux \((\phi )\) between the terminals is considered as a function of the electric charge q that passes through the device [57]. Here, the memristor is of fluxcontrolled type specified by its incremental memductance function \(W(\phi )\) given as follows:
$$ W(\phi ) \triangleq \frac{\mathrm{d}q(\phi )}{\mathrm{d}\phi }. $$
(1)
From (1) we reach a relation between the current \(I_{M}(t)\) through the memristor and the voltage \(V_{1}(t)\) which is presented as
$$ I_{M}(t) \triangleq \frac{\mathrm{d}q}{\mathrm{d}t} = \frac{\mathrm{d}q}{\mathrm{d}\phi } \frac{\mathrm{d}\phi }{\mathrm{d}t} = W\bigl(\phi (t)\bigr)V_{1}(t). $$
(2)
By (2) and \(W(\phi (t)) = W(\int V_{1}(t))\), the integral operator on the memductance function indicates that the mentioned function remembers the history of the voltage values. On the other side, if
$$ W\bigl(\phi (t)\bigr) = W\biggl( \int V_{1}(t)\biggr) = \mathrm{const}, $$
then a memristor will be a resistor.
The memristor was derived from a circuit attributed to Chua by putting the Chua diode with the fluxcontrolled memristor [47, 57]. Chua’s electric circuit involves the resistor R, the inductor L, capacitors \(C_{j}\), \(j=1,2\), and a nonlinear resistor (NR). The dynamic equation for the memristorbased chaotic circuit is formulated by
$$\begin{aligned} &\frac{\mathrm{d}V_{1}(t)}{\mathrm{d}t}= \frac{1}{C_{1}} \biggl[ \frac{(V_{2}(t)V_{1}(t))}{R} + GV_{1}(t)  I_{M}(t) \biggr], \\ &\frac{\mathrm{d}V_{2}(t)}{\mathrm{d}t}= \frac{1}{C_{2}} \biggl[ \frac{(V_{2}(t)V_{1}(t))}{R} + I_{L}(t) \biggr], \\ &\frac{\mathrm{d}I_{L}(t)}{\mathrm{d}t}= \frac{1}{L} \bigl[ V_{2}(t)R_{L}I_{L}(t) \bigr], \\ &\frac{\mathrm{d}\phi }{\mathrm{d}t}= V_{1}(t), \end{aligned}$$
(3)
in which \(V_{1}\), \(V_{2}\), and \(R_{L}\) stand for the voltages over the capacitors \(C_{j}\), \(j=1,2\), \(I_{L}\) stands for the current through the existing inductance L, and \(I_{M}(t)\) is defined in (2).
Based on the motivation that smooth nonlinearity does give rise to chaos, let us choose a smooth continuous cubic monotone increasing nonlinearity presented by
$$ q(\phi ) = a_{0} + a_{1} \phi + a_{2} \phi ^{3},\quad a_{1} >0, a_{2} >0. $$
(4)
Consequently, the memductance function is formulated as
$$ W(\phi ) = a_{1} + 3a_{2}\phi ^{2}. $$
(5)
The system of differential equations (3) can be converted to a structure without the dimension given as follows:
$$ \textstyle\begin{cases} \frac{\mathrm{d}x}{\mathrm{d}t} = \beta _{1} ( W(\phi )x(t) + y(t)x(t)+ \varsigma x(t) ), \\\frac{\mathrm{d}y}{\mathrm{d}t} = x(t)y(t) + z(t), \\\frac{\mathrm{d}z}{\mathrm{d}t} =  \beta _{2} y(t)\beta _{3}z(t), \\\frac{\mathrm{d}\phi }{\mathrm{d}t} = x(t), \end{cases} $$
(6)
where
$$ \begin{gathered}x:=V_{1},\qquad y:=V_{2},\qquad z:=I_{L},\qquad C_{2}=1,\qquad R=1,\\ \beta _{1}:= \frac{1}{C_{1}},\qquad \varsigma :=G,\qquad \beta _{2} :=\frac{1}{L},\qquad \beta _{3} := \frac{R_{L}}{L}. \end{gathered} $$
(7)
Fractional representation of the memristorbased circuit
In this subsection, we recall the definitions and basic properties of Atangana–Baleanu derivative of Caputo type (ABC fractional derivative).
Definition 2.1
([35, 58, 59])
Let \(\phi \in C^{1}(a,b)\) and \(\alpha \in (0,1)\). The ABC \(\alpha ^{th}\)derivative of ϕ is defined as
$$ {}^{ABC}_{a} \mathcal{D}_{t}^{\alpha } \phi (t) = \frac{F(\alpha )}{1\alpha } \int _{a}^{t} \frac{\mathrm{d}\phi }{\mathrm{d}k} E_{\alpha } \biggl[  \frac{\alpha }{1\alpha } (tk)^{\alpha } \biggr] \,\mathrm{d}k, $$
(8)
where \(F(\alpha ) = 1\alpha  \frac{\alpha }{\Gamma (\alpha )}\), and \(E_{\alpha }[\cdot ]\) is the MittagLeffler function.
Definition 2.2
([35, 58, 59])
The Atangana–Baleanu (AB) \(\alpha ^{th}\)integral of \(\phi \in C^{1}(a,b)\) is given by
$$ {}^{AB}_{a} \mathcal{I}_{t}^{\alpha } \phi (t) = \frac{1\alpha }{F(\alpha )} \phi (t) + \frac{\alpha }{\Gamma (\alpha )F(\alpha )} \int _{a}^{t} \phi (k) (tk)^{ \alpha  1} \,\mathrm{d}k. $$
(9)
Lemma 2.3
([60])
The ABC \(\alpha ^{th}\)derivative and AB \(\alpha ^{th}\)integral of \(\phi \in C^{1}(a,b)\) fulfill the Newton–Leibniz relation
$$ {}^{AB}_{a} \mathcal{I}_{t}^{\alpha } \bigl( {}^{ABC}_{a} \mathcal{D}_{t}^{\alpha } \phi (t) \bigr) = \phi (t)\phi (a). $$
(10)
Lemma 2.4
([59])
For \(\phi _{1},\phi _{2}\in C^{1}(a,b)\), the AB \(\alpha ^{th}\)derivative in the Caputo sense satisfies the Lipschitz condition
$$ \bigl\vert {}^{ABC}_{a} \mathcal{D}_{t}^{\alpha }\phi _{1}(t)  {}^{ABC}_{a} \mathcal{D}_{t}^{\alpha }\phi _{2}(t) \bigr\vert \leq \Lambda \bigl\Vert \phi _{1}(t) \phi _{2}(t) \bigr\Vert . $$
(11)
Now we follow to formulate model (6) in terms of ABC fractional derivatives.
Lemma 2.5
([59])
Let \(G\in C^{1}(a,b)\). Then the following fractional differential equation
$$ {}^{ABC}_{0} \mathcal{D}_{t}^{\alpha }\phi (t) = G(t),\quad f(0)=G_{0} $$
admits a solution uniquely given by
$$ \phi (t) = G_{0} + \frac{1\alpha }{F(\alpha )} G(t) + \frac{\alpha }{\Gamma (\alpha )F(\alpha )} \int _{0}^{t} (tk)^{ \alpha 1} G(k)\,\mathrm{d}k. $$
We can describe the result in Lemma 2.5 in the form of the Banach fixed point theorem as follows. We begin by defining a Banach space
$$ \mathcal{X}=\bigl\{ x\in C(\mathbb{I},\mathbb{R}), \mathbb{I} = [0,1]\bigr\} , $$
with a norm defined as \(\Vert x\Vert _{\mathcal{X}} =\sup_{t\in \mathbb{I}}\vert x(t)\vert \).
Now define operators \(O_{1}, O_{2},O_{3}, O_{4}: \mathcal{X} \to \mathcal{X}\) by
$$\begin{aligned} O_{1}x(t) &= x_{0} + \frac{1\alpha }{F(\alpha )} \bigl( \beta _{1} \bigl(W( \phi )x(t) + y(t)+ (\varsigma 1)x(t)\bigr) \bigr) \\ &\quad{}+ \frac{\alpha }{\Gamma (\alpha )F(\alpha )} \int _{0}^{t} (tk)^{ \alpha 1} \bigl( \beta _{1} \bigl(y(k)+ (\varsigma 1)x(k)W(\phi )x(k)\bigr) \bigr) \,\mathrm{d}k, \end{aligned}$$
(12)
and
$$\begin{aligned} O_{2}y(t) &= y_{0} + \frac{1\alpha }{F(\alpha )} \bigl( x(t)y(t)+z(t) \bigr) \\ &\quad{}+ \frac{\alpha }{\Gamma (\alpha )F(\alpha )} \int _{0}^{t} (tk)^{ \alpha 1} \bigl( x(k)y(k)+z(k) \bigr) \,\mathrm{d}k, \end{aligned}$$
(13)
and
$$\begin{aligned} O_{3}z(t) &= z_{0} + \frac{1\alpha }{F(\alpha )} \bigl( \beta _{2} y(t)  \beta _{3} z(t) \bigr) \\ &\quad{}+ \frac{\alpha }{\Gamma (\alpha )F(\alpha )} \int _{0}^{t} (tk)^{ \alpha 1} \bigl( \beta _{2} y(k)  \beta _{3} z(k) \bigr) \,\mathrm{d}k, \end{aligned}$$
(14)
and
$$\begin{aligned} O_{4}\phi (t) = \phi _{0} + \frac{1\alpha }{F(\alpha )} x(t) + \frac{\alpha }{\Gamma (\alpha )F(\alpha )} \int _{0}^{t} (tk)^{ \alpha 1} x(k) \,\mathrm{d}k. \end{aligned}$$
(15)
We are now ready to describe the dynamic equation for the aforesaid memristorbased circuit given in (6) using the ABC fractional derivative
$$ \textstyle\begin{cases} {}^{ABC}_{0} \mathcal{D}_{t}^{\alpha }x(t) = \beta _{1} ( W(\phi )x(t) + y(t)x(t)+ \varsigma x(t) ), \\{}^{ABC}_{0} \mathcal{D}_{t}^{\alpha }y(t) = x(t)y(t) + z(t), \\{}^{ABC}_{0} \mathcal{D}_{t}^{\alpha }z(t) =  \beta _{2} y(t)\beta _{3}z(t), \\{}^{ABC}_{0} \mathcal{D}_{t}^{\alpha }\phi (t) = x(t), \end{cases} $$
(16)
in which \(x(0)=x_{0}\), \(y(0)=y_{0}\), \(z(0)=z_{0}\), and \(\phi (0)=\phi _{0}\) are initial conditions.
Results on the existence–uniqueness
Here, the existence–uniqueness of the solution for the ABC fractional model given in (16) are proved using Banach fixed point theorem (BFPT) for contraction mapping. The following two theorems are worth recalling before proceeding further.
Theorem 2.6
(BFPT, [61])
Let \(\Pi \neq \emptyset \) be any closed subset of a Banach space \(\mathcal{X}\). Then any contraction mapping \(O: \Pi \to \Pi \) admits a fixed point uniquely.
Theorem 2.7
Assume that x, y, z, ϕ are continuous mappings satisfying the following assumptions:

(D1)
\(\exists d_{1},d_{2},d_{3},d_{4}>0\) s.t. \(\vert x(t)\vert < d_{1}\), \(\vert y(t)\vert < d_{2}\), \(\vert z(t)\vert < d_{3}\), \(\vert \phi (t)\vert < d_{4}\);

(D2)
\(\beta _{1}((\varsigma 1)+ \vert a_{1}\vert + 3a_{2}d_{4}^{2})((1 \alpha )\Gamma (\alpha )+ 1) < \Gamma (\alpha )F(\alpha )\) and
$$ (1\alpha )\Gamma (\alpha ) + 1 < \Gamma (\alpha )F(\alpha ) \quad \textit{and} \quad \beta _{3} \bigl((1\alpha )\Gamma (\alpha ) + 1 \bigr) < \Gamma ( \alpha )F(\alpha ). $$
Then the ABC fractional derivative system given by (16) has a unique solution in the region \(\mathcal{X}\).
Proof
Let us show that the operator \(O_{1}\) defined in (12) is well defined in the sense that \(O_{1}x(t) \in \Im _{r}\) and \({}^{ABC}_{0} \mathcal{D}_{t}^{\alpha }(O_{1}x)(t) \in C(\mathbb{I}=[0,1], \mathbb{R})\), where
$$ \Im _{r} := \bigl\{ w\in \mathcal{X}, \Vert w \Vert _{\mathcal{X}} \leq r\bigr\} , $$
(17)
and
$$ r> \frac{ \vert x_{0} \vert F(\alpha )\Gamma (\alpha ) + ((1\alpha )\Gamma (\alpha ) +1) (\beta _{1}(d_{2}+d_{1}(\varsigma 1)+ \vert a_{1}+3a_{2}d_{4}^{2} \vert d_{1}) )}{F(\alpha )\Gamma (\alpha )} . $$
Now, for any \(x\in \Im _{r}\) and from (12), we have
$$\begin{aligned} & \bigl\Vert O_{1}x(t) \bigr\Vert _{\mathcal{X}} \\ &\quad \leq \vert x_{0} \vert + \frac{1\alpha }{F(\alpha )} \bigl( \beta _{1} \bigl(d_{2}+ (\varsigma 1)d_{1} + \bigl\vert a_{1}+3a_{2}d_{4}^{2} \bigr\vert d_{1}\bigr) \bigr) \\ &\qquad{}+ \sup_{t\in \mathbb{I}} \biggl( \frac{\alpha }{F(\alpha )\Gamma (\alpha )} \int _{0}^{t} (tk)^{ \alpha 1} \bigl\vert \beta _{1} \bigl(y(k)+ (\varsigma 1)x(k)W(\phi )x(k)\bigr) \bigr\vert \,\mathrm{d}k \biggr) \\ &\quad \leq \vert x_{0} \vert + \frac{1\alpha }{F(\alpha )} \bigl( \beta _{1} \bigl(d_{2}+ (\varsigma 1)d_{1} + \bigl\vert a_{1}+3a_{2}d_{4}^{2} \bigr\vert d_{1}\bigr) \bigr) \\ &\qquad{}+ \frac{\alpha }{F(\alpha )\Gamma (\alpha )} \bigl( \beta _{1} \bigl(d_{2}+ ( \varsigma 1)d_{1} + \bigl\vert a_{1}+3a_{2}d_{4}^{2} \bigr\vert d_{1}\bigr) \bigr) \sup_{t\in \mathbb{I}=[0,1]} \biggl( \int _{0}^{t} (tk)^{\alpha 1} \,\mathrm{d}k \biggr) \\ &\quad \leq \frac{ \vert x_{0} \vert F(\alpha )\Gamma (\alpha ) + ((1\alpha )\Gamma (\alpha ) + 1) (\beta _{1}(d_{2}+d_{1}(\varsigma 1)+ \vert a_{1}+3a_{2}d_{4}^{2} \vert d_{1}) )}{F(\alpha )\Gamma (\alpha )} < r. \end{aligned}$$
Consequently, we have \(O_{1}x(t) \in \Im _{r}\). To show continuous differentiability on \(\mathbb{I}\), we proceed from
$$\begin{aligned} {}^{ABC}_{0} \mathcal{D}_{t}^{\alpha }(O_{1}x) (t) &= {}^{ABC}_{0} \mathcal{D}_{t}^{\alpha }{}^{AB}_{0} \mathcal{I}_{t}^{\alpha } \bigl( \beta _{1} \bigl( W( \phi )x(t) + y(t)+(\varsigma 1) x(t) \bigr) \bigr) \\ &= \beta _{1} \bigl( W(\phi )x(t) + y(t) + (\varsigma 1) x(t) \bigr), \end{aligned}$$
which is continuous on \(\mathbb{I}\), and so we conclude that \({}^{ABC}_{0} \mathcal{D}_{t}^{\alpha }(O_{1}x)(t)\) is continuous on \(\mathbb{I}\). As a result \(O_{1}\Im _{r} \subset \Im _{r}\).
To show that the operator \(O_{1}\) has a fixed point, we apply Theorem 2.6. Based on this theorem, it is enough to show that \(O_{1}\) is a contraction mapping. Indeed, let \(x_{1},x_{2} \in \mathcal{X}\), \(t\in \mathbb{I}\). Hence,
$$\begin{aligned} & \Vert O_{1}x_{1} O_{1}x_{2} \Vert _{\mathcal{X}} \\ &\quad \leq \frac{1\alpha }{F(\alpha )} \beta _{1} \bigl( (\varsigma 1) \Vert x_{1} x_{2} \Vert _{\mathcal{X}} + \bigl\vert W(\phi ) \bigr\vert \Vert x_{1}x_{2} \Vert _{\mathcal{X}} \bigr) \\ &\qquad{}+ \frac{\alpha }{F(\alpha ) \Gamma (\alpha )} \beta _{1} \bigl( ( \varsigma 1) \Vert x_{1}x_{2} \Vert _{\mathcal{X}} + \bigl\vert W(\phi ) \bigr\vert \Vert x_{1}x_{2} \Vert _{\mathcal{X}} \bigr) \sup_{t\in \mathbb{I}} \int _{0}^{t} (tk)^{\alpha 1} \,\mathrm{d}k \\ &\quad \leq \biggl( \frac{\beta _{1}((\varsigma 1)+ \vert a_{1} \vert + 3a_{2}d_{4}^{2})(\Gamma (\alpha )(1\alpha )+ 1)}{F(\alpha )\Gamma (\alpha )} \biggr) \Vert x_{1}x_{2} \Vert _{\mathcal{X}} \\ &\quad = H \Vert x_{1}x_{2} \Vert _{\mathcal{X}}, \end{aligned}$$
where \(H= \frac{\beta _{1}((\varsigma 1)+ \vert a_{1}\vert + 3a_{2}d_{4}^{2})(\Gamma (\alpha )(1\alpha )+ 1)}{F(\alpha )\Gamma (\alpha )}\). Since \(H<1\), by the hypothesis of the theorem, we conclude that \(O_{1}\) is a contraction.
To show that the remaining operators \(O_{2}\), \(O_{3}\), \(O_{4}\) are contractions, we proceed in the same manner.
Let us verify that the operator \(O_{2}\) defined in (13) is well defined in the sense that \(O_{2}y(t) \in \Im '_{r'}\) and \({}^{ABC}_{0} \mathcal{D}_{t}^{\alpha }(O_{2}y)(t) \in C(\mathbb{I}=[0,1], \mathbb{R})\), where
$$ \Im '_{r'} := \bigl\{ w\in \mathcal{X}, \Vert w \Vert _{\mathcal{X}} \leq r' \bigr\} , $$
(18)
for
$$ r'> \frac{ \vert y_{0} \vert F(\alpha )\Gamma (\alpha ) + ((1\alpha )\Gamma (\alpha ) + 1) ( d_{1} + d_{2} + d_{3} )}{F(\alpha )\Gamma (\alpha )} . $$
Now, for any \(y\in \Im '_{r'}\) and from (13), we have
$$\begin{aligned} \bigl\Vert O_{2}y(t) \bigr\Vert _{\mathcal{X}} &\leq \vert y_{0} \vert + \frac{1\alpha }{F(\alpha )} ( d_{1} + d_{2} + d_{3} ) \\ &\quad{}+ \sup_{t\in \mathbb{I}} \biggl( \frac{\alpha }{F(\alpha )\Gamma (\alpha )} \int _{0}^{t} (tk)^{ \alpha 1} \bigl\vert x(k)y(k)+z(k) \bigr\vert \,\mathrm{d}k \biggr) \\ &\leq \vert y_{0} \vert + \frac{1\alpha }{F(\alpha )} ( d_{1} + d_{2} + d_{3} ) \\ &\quad{}+ \frac{\alpha }{F(\alpha )\Gamma (\alpha )} ( d_{1} + d_{2} + d_{3} ) \sup_{t\in \mathbb{I}=[0,1]} \biggl( \int _{0}^{t} (tk)^{ \alpha 1} \,\mathrm{d}k \biggr) \\ &\leq \frac{ \vert y_{0} \vert F(\alpha )\Gamma (\alpha ) + ((1\alpha )\Gamma (\alpha ) + 1) ( d_{1} + d_{2} + d_{3} )}{F(\alpha )\Gamma (\alpha )} < r'. \end{aligned}$$
Consequently, we have \(O_{2}y(t) \in \Im '_{r'}\). To show continuous differentiability on \(\mathbb{I}\), we proceed from
$$ {}^{ABC}_{0} \mathcal{D}_{t}^{\alpha }(O_{2}y) (t) = {}^{ABC}_{0} \mathcal{D}_{t}^{\alpha }{}^{AB}_{0} \mathcal{I}_{t}^{\alpha } \bigl( x(t)y(t)+z(t) \bigr) = x(t)y(t)+z(t), $$
which is continuous on \(\mathbb{I}\), and so we conclude that \({}^{ABC}_{0} \mathcal{D}_{t}^{\alpha }(O_{2}y)(t)\) is continuous on \(\mathbb{I}\). As a result, \(O_{2}\Im '_{r'} \subset \Im '_{r'}\).
To show that the operator \(O_{2}\) has a fixed point, we apply Theorem 2.6. Based on this theorem, it is enough to show that \(O_{2}\) is a contraction mapping. Indeed, let \(y_{1},y_{2} \in \mathcal{X}\), \(t\in \mathbb{I}\). Thus,
$$\begin{aligned} & \Vert O_{2}y_{1} O_{2}y_{2} \Vert _{\mathcal{X}} \\ &\quad \leq \frac{1\alpha }{F(\alpha )} \bigl( \Vert y_{1} y_{2} \Vert _{\mathcal{X}} \bigr) + \frac{\alpha }{F(\alpha ) \Gamma (\alpha )} \bigl( \Vert y_{1} y_{2} \Vert _{\mathcal{X}} \bigr) \sup _{t\in \mathbb{I}} \int _{0}^{t} (tk)^{\alpha 1} \,\mathrm{d}k \\ &\quad \leq \biggl( \frac{(1\alpha )\Gamma (\alpha ) + 1}{F(\alpha )\Gamma (\alpha )} \biggr) \Vert y_{1}y_{2} \Vert _{\mathcal{X}} \\ &\quad = H' \Vert y_{1}y_{2} \Vert _{\mathcal{X}}, \end{aligned}$$
where \(H'= \frac{(1\alpha )\Gamma (\alpha ) + 1}{F(\alpha )\Gamma (\alpha )}\). Since \(H'<1\), by the hypothesis of the theorem, we conclude that \(O_{2}\) is a contraction.
In the sequel, we verify that the operator \(O_{3}\) defined in (14) is well defined in the sense that \(O_{3}z(t) \in \Im ''_{r''}\) and \({}^{ABC}_{0} \mathcal{D}_{t}^{\alpha }(O_{3}z)(t) \in C(\mathbb{I}=[0,1], \mathbb{R})\), where
$$ \Im ''_{r''} := \bigl\{ w\in \mathcal{X}, \Vert w \Vert _{\mathcal{X}} \leq r'' \bigr\} $$
(19)
for
$$ r''> \frac{ \vert z_{0} \vert \Gamma (\alpha )F(\alpha ) + (\Gamma (\alpha )(1\alpha ) + 1) ( \beta _{2} d_{2} + \beta _{3} d_{3} )}{\Gamma (\alpha )F(\alpha )} . $$
Now, for any \(z\in \Im ''_{r''}\) and from (14), we have
$$\begin{aligned} \bigl\Vert O_{3}z(t) \bigr\Vert _{\mathcal{X}} &\leq \vert z_{0} \vert + \frac{1\alpha }{F(\alpha )} ( \beta _{2} d_{2} + \beta _{3} d_{3} ) \\ &\quad{}+ \sup_{t\in \mathbb{I}} \biggl( \frac{\alpha }{\Gamma (\alpha )F(\alpha )} \int _{0}^{t} (tk)^{ \alpha 1} \bigl\vert  \beta _{2} y(k) \beta _{3} z(k) \bigr\vert \,\mathrm{d}k \biggr) \\ &\leq \vert z_{0} \vert + \frac{1\alpha }{F(\alpha )} ( \beta _{2} d_{2} + \beta _{3} d_{3} ) \\ &\quad{}+ \frac{\alpha }{\Gamma (\alpha )F(\alpha )} ( \beta _{2} d_{2} + \beta _{3} d_{3} ) \sup_{t\in \mathbb{I}=[0,1]} \biggl( \int _{0}^{t} (tk)^{\alpha 1} \,\mathrm{d}k \biggr) \\ &\leq \frac{ \vert z_{0} \vert F(\alpha )\Gamma (\alpha ) + ((1\alpha )\Gamma (\alpha ) + 1) ( \beta _{2} d_{2} + \beta _{3} d_{3} )}{F(\alpha )\Gamma (\alpha )} < r''. \end{aligned}$$
Consequently, we have \(O_{3}z(t) \in \Im ''_{r''}\). To show continuous differentiability on \(\mathbb{I}\), we proceed from
$$ {}^{ABC}_{0} \mathcal{D}_{t}^{\alpha }(O_{3}z) (t) = {}^{ABC}_{0} \mathcal{D}_{t}^{\alpha }{}^{AB}_{0} \mathcal{I}_{t}^{\alpha } \bigl(  \beta _{2} y(t) \beta _{3} z(t) \bigr) = \beta _{2} y(t) \beta _{3} z(t), $$
which is continuous on \(\mathbb{I}\), and so we conclude that \({}^{ABC}_{0} \mathcal{D}_{t}^{\alpha }(O_{3}z)(t)\) is continuous on \(\mathbb{I}\). As a result, \(O_{3}\Im ''_{r''} \subset \Im ''_{r''}\).
To show that the operator \(O_{3}\) has a fixed point, we apply Theorem 2.6. Based on this theorem, it is enough to show that \(O_{3}\) is a contraction mapping. Indeed, let \(z_{1},z_{2} \in \mathcal{X}\), \(t\in \mathbb{I}\). Hence,
$$\begin{aligned} & \Vert O_{3}z_{1} O_{3}z_{2} \Vert _{\mathcal{X}} \\ &\quad \leq \frac{1\alpha }{F(\alpha )} \bigl( \beta _{3} \Vert z_{1} z_{2} \Vert _{\mathcal{X}} \bigr) + \frac{\alpha }{F(\alpha ) \Gamma (\alpha )} \bigl( \beta _{3} \Vert z_{1} z_{2} \Vert _{\mathcal{X}} \bigr) \sup_{t \in \mathbb{I}} \int _{0}^{t} (tk)^{\alpha 1} \,\mathrm{d}k \\ &\quad \leq \biggl( \frac{\beta _{3} ((1\alpha )\Gamma (\alpha ) + 1 ) }{F(\alpha )\Gamma (\alpha )} \biggr) \Vert z_{1}z_{2} \Vert _{\mathcal{X}} \\ &\quad = H'' \Vert z_{1}z_{2} \Vert _{\mathcal{X}}, \end{aligned}$$
where \(H''= \frac{\beta _{3} ((1\alpha )\Gamma (\alpha ) + 1 ) }{F(\alpha )\Gamma (\alpha )}\). Since \(H''<1\), by the hypothesis of the theorem, we conclude that \(O_{3}\) is a contraction.
Lastly, we verify that the operator \(O_{4}\) defined in (15) is well defined in the sense that \(O_{4}\phi (t) \in \Im '''_{r'''}\) and \({}^{ABC}_{0} \mathcal{D}_{t}^{\alpha }(O_{4}\phi )(t) \in C( \mathbb{I}=[0,1], \mathbb{R})\), where
$$ \Im '''_{r'''} := \bigl\{ w\in \mathcal{X}, \Vert w \Vert _{\mathcal{X}} \leq r'''\bigr\} $$
(20)
for
$$ r'''> \frac{ \vert \phi _{0} \vert \Gamma (\alpha )F(\alpha ) + (\Gamma (\alpha )(1\alpha ) + 1) d_{1}}{\Gamma (\alpha )F(\alpha )} . $$
Now, for any \(\phi \in \Im '''_{r'''}\) and from (15), we have
$$\begin{aligned} \bigl\Vert O_{4}\phi (t) \bigr\Vert _{\mathcal{X}} &\leq \vert \phi _{0} \vert + \frac{1\alpha }{F(\alpha )} d_{1} + \sup _{t\in \mathbb{I}} \biggl( \frac{\alpha }{F(\alpha )\Gamma (\alpha )} \int _{0}^{t} (tk)^{ \alpha 1} \bigl\vert x(k) \bigr\vert \,\mathrm{d}k \biggr) \\ &\leq \vert \phi _{0} \vert + \frac{1\alpha }{F(\alpha )} d_{1} + \frac{\alpha }{F(\alpha )\Gamma (\alpha )} d_{1} \sup_{t\in \mathbb{I}=[0,1]} \biggl( \int _{0}^{t} (tk)^{\alpha 1} \,\mathrm{d}k \biggr) \\ &\leq \frac{ \vert \phi _{0} \vert F(\alpha )\Gamma (\alpha ) + ((1\alpha )\Gamma (\alpha ) + 1) d_{1}}{F(\alpha )\Gamma (\alpha )} < r'''. \end{aligned}$$
Consequently, we have \(O_{4}\phi (t) \in \Im '''_{r'''}\). To show continuous differentiability on \(\mathbb{I}\), we proceed from
$$ {}^{ABC}_{0} \mathcal{D}_{t}^{\alpha }(O_{4} \phi ) (t) = {}^{ABC}_{0} \mathcal{D}_{t}^{\alpha }{}^{AB}_{0} \mathcal{I}_{t}^{\alpha }x(t) = x(t), $$
which is continuous on \(\mathbb{I}\), and so we conclude that \({}^{ABC}_{0} \mathcal{D}_{t}^{\alpha }(O_{4}\phi )(t)\) is continuous on \(\mathbb{I}\). As a result, \(O_{4}\Im '''_{r'''} \subset \Im '''_{r'''}\). Immediately, it follows that \(O_{4}\) is a contraction. Hence, by BFPT 2.6, system (16) admits a solution uniquely in \(\mathcal{X}\). □