Qualitative Aspects of a Fractional-Order Integro-Differential Equation with a Quadratic Functional Integro-Differential Constraint

Abstract

This manuscript investigates a constrained problem of an arbitrary (fractional) order quadratic functional integro-differential equation with a quadratic functional integro-differential constraint. We demonstrate that there is at least one solution $x\in C[0,T]$ to the problem. Moreover, we outline the necessary demands for the solution’s uniqueness. In addition, the continuous dependence of the solution and the Hyers–Ulam stability of the problem are analyzed. In order to illustrate our results, we provide some particular cases and instances.

1. Introduction

Fractional-order differential and integral equations have a wide range of applications across various fields with examples in physics, engineering, and biomedical engineering. The nonlocal conditions are often encountered in mathematical and physical problems, where the behavior of a system depends on different factors or parameters; see [1,2,3,4,5,6,7,8,9,10].

In recent years, several scholars have concentrated their efforts on constrained integral equations. Their findings about functional integral equations have been expanded to include a particular set of constrained integral equations on a bounded interval (see [11,12,13]) and unbounded intervals (see [14]). Constrained problems are essential in the mathematical depiction of real-world situations, where such problems are transformed into mathematical models [15,16,17]. The relevance of handling constraints or control variables arises from the unanticipated elements that persistently disrupt biological systems in the real world; biological traits like survival rates might change as a result. The question of whether an ecosystem can survive those erratic, disruptive occurrences that happen for a short while is of practical significance to ecology. The disturbance functions are what we refer to as control variables in the context of control variables. Numerous papers address this type of problem; for instance, in [18], the authors discussed a nonlinear constrained problem involving a nonlinear functional integral equation. They also examined the appropriate conditions for the solution’s uniqueness and its continuous dependence on certain parameters. The authors applied Schauder’s fixed-point theorem to prove the existence of solutions. In [14], the authors studied the solvability of a constrained problem involving a nonlinear-delay functional equation subject to a quadratic functional integral constraint. By applying the De Blasi measure of noncompactness, they studied nondecreasing solutions in the bounded interval $L_1[0,T]$ and nonincreasing solutions in the unbounded interval $L_1(R+)$.

Problems with a feedback control or control variable have great importance in numerous fields due to unforeseen factors that disrupt ecosystems in the real world. It could lead to changes in biological characteristics like survival rates; see [19,20,21,22]. Furthermore, ecology faces a practical challenge in determining whether an ecosystem can withstand unpredictable, disruptive events; see [15,23,24,25]. In addition, feedback control problems are crucial to establishing the solutions to delay population models; see [26,27,28,29]. In [23], the authors investigated the effect of feedback control on chemostat models; they studied a sufficient condition for the existence of a positive periodic solution tp the model. In [30], the author discussed a positive periodic solution to a nonlinear neutral delay population equation with feedback control. In [12], the authors studied fractional-order models of thermostats; they proved the existence of a solution and the continuous dependence of the unique solution on the control variable. In [13], the author investigated the solvability and the asymptotic stability of a class of nonlinear functional-integral equations with feedback control. For further relevant works, see [12,31,32,33,34,35].

Fixed-point theorems are a great tool for discussing the solvability of differential equation problems that have been studied in a number of monographs and publications; see [6,31,36,37,38,39,40].

Inspired by the above, we consider the constrained problem

$$\frac{dx}{dt}=f\left(t,g_1(t,D^{\zeta }x(t))·{\int }_0^{\vartheta (t)}{g}_2(s,D^{\gamma }x(s))ds\right),\zeta ,\gamma \in (0,1),t\in (0,T]$$

(1)

with the quadratic functional integro-differential constrained

$$x(\tau )=x_0+{\int }_0^{T-\tau }h\left(s,x(s)·D^{\eta }x(s)\right)ds,\eta \in (0,1),\tau \in [0,T].$$

(2)

Our aim in this paper is to examine the existence of a solution $x\in C(0,T]$ to the constrained problems (1) and (2). A sufficient hypothesis for the solution’s uniqueness will be given. Furthermore, we prove the Hyers–Ulam stability of the problem. The continuous dependence of the solution on the fractional-order derivative $D^{\zeta }x(t)$, the parameter $x_0$, and the function h will be studied. To highlight our results, we present several examples and special cases. This study establishes conditions for the existence and uniqueness of the solution according to Schauder’s fixed-point theorem.

2. Main Result

2.1. Formulation of the Problem

Consider the constrained problem (1) and (2) under the next hypothesis. Let $I=[0,T]$.

(i)

$\vartheta :I\to I$ is continuous function such that $\vartheta (t)\le t$.

(ii)

$f,h$ and $g_i,i=1,2$: $I\times R\to R$ are Caratheodry functions [41]. There exist bounded measurable functions [42] a and $a_i:I\to R$ and a positive constants b and $b_i$ such that

$$\begin{array}{ccc}\hfill |f(t,x)|& \le & \left|a(t)\right|+b\left|x\right|\le a^{*}+b\left|x\right|,a^{*}=\underset{t\in I}{sup}\left|a(t)\right|.\hfill \\ \hfill |g_i(t,x)|& \le & |a_i(t)|+b_i\left|x\right|\le a_i^{*}+b_i\left|x\right|,a_i^{*}=\underset{t\in I}{sup}\left|a_i(t)\right|,i=1,2.\hfill \\ \hfill |h(t,x)|& \le & |a_3(t)|+b_3\left|x\right|,\underset{s\in I}{sup}{\int }_0^{T-\tau }\left|a_3(s)\right|ds\le N.\hfill \end{array}$$

(iii)

The following algebraic equation has a real positive root $r_1$.

$$\begin{array}{c}\hfill b{b}_1{b}_2{T}^{2-\gamma }{r}_1^2+\left(a^{*}{b}_2{T}^{2-\gamma }+b{a}_1^{*}{b}_2{T}^{2-\gamma }+b{b}_1{a}_2^{*}{T}^{2-\zeta }-1\right)r_1+a^{*}{a}_2^{*}{T}^{2-\zeta }+b{a}_1^{*}{a}_2^{*}{T}^{2-\zeta }=0.\end{array}$$

(iv)

$r_1{b}_3{T}^{\zeta -\eta +1}<1.$

The next lemma demonstrates the equivalence between the constrained problem (1) and (2) and its corresponding integral equations.

Lemma 1.

If the solution to (1) and (2) exists, then it can be expressed by

$$x(t)=x_0+{\int }_0^{T-\tau }\left(h(s,x(s)·I^{\zeta -\eta }y(s))\right)ds-I^{\zeta }y(\tau )+I^{\zeta }y(t)$$

(3)

and

$$\begin{array}{c}\hfill y(t)=I^{1-\zeta }f\left(t,g_1(t,y(t))·{\int }_0^{\vartheta (t)}{g}_2(s,I^{\zeta -\gamma }y(s))ds\right).\end{array}$$

(4)

Proof. 

Let x be the solution to (1) and (2). Operating by $I^{1-\zeta }$ in on both sides of (1), we obtain

$$\begin{array}{ccc}& & D^{\zeta }x(t)=I^{1-\zeta }\frac{dx}{dt}=I^{1-\zeta }\left(f(t,g_1(t,D^{\zeta }x(t)))·{\int }_0^{\vartheta (t)}{g}_2(s,D^{\gamma }x(s))ds\right).\hfill \end{array}$$

Taking $D^{\zeta }x(t)=y(t)$, then,

$$x(t)=x(0)+I^{\zeta }y(t).$$

(5)

And we can deduce that

$$\begin{array}{ccc}\hfill I^{\zeta -\gamma }y(t)& =& I^{\zeta -\gamma }{D}^{\zeta }x(t)=I^{\zeta -\gamma }{I}^{1-\zeta }\frac{dx}{dt}\hfill \\ & =& I^{1-\gamma }\frac{dx}{dt}=D^{\gamma }x(t),\hfill \end{array}$$

(6)

and similarly,

$$\begin{array}{ccc}\hfill I^{\zeta -\eta }y(t)& =& I^{\zeta -\eta }{D}^{\zeta }x(t)=I^{\zeta -\eta }{I}^{1-\zeta }\frac{dx}{dt}\hfill \\ & =& I^{1-\eta }\frac{dx}{dt}=D^{\eta }x(t).\hfill \end{array}$$

(7)

Substituting from (5)–(7) in (1) and (2), we obtain (4) and (3). Conversely, let x be a solution to (3). Differentiating (3), we obtain

$$\begin{array}{ccc}\hfill \frac{dx}{dt}& =& \frac{d}{dt}[x_0+{\int }_0^{T-\tau }\left(h(s,x(s)·I^{\zeta -\eta }y(s))\right)ds-I^{\zeta }y(\tau )+I^{\zeta }y(t)].\hfill \\ & =& \frac{d}{dt}{I}^{\zeta }y=\frac{d}{dt}{I}^{\zeta }{I}^{1-\zeta }f(t,g_1(t,D^{\zeta }x(t))·{\int }_0^{\vartheta (t)}{g}_2(s,D^{\gamma }x(s))ds\hfill \\ & =& f(t,g_1(t,D^{\zeta }x(t)))·{\int }_0^{\vartheta (t)}{g}_2(s,D^{\gamma }x(s))ds\hfill \end{array}$$

This proves the equivalence between the two systems (1) and (2) and (3) to (4). □

2.2. Existence of the Solution

Here, we prove the existence of the continuous solution $x\in C(I)$ of (1) and (2). For this purpose, we present the next theorem.

Theorem 1.

Assume that the hypotheses (i)–(iv) are satisfied; then, the solution $x\in C(I)$ of (1) and (2) exists.

Proof. 

Define the closed sphere $Q_{r_1}$ and the operator $F_1$ with

$$\begin{array}{c}\hfill Q_{r_1}=\{y\in C(I):\parallel y\parallel \le r_1\}.\end{array}$$

and

$$\begin{array}{c}\hfill F_{1}y(t)=I^{1-\zeta }f\left(t,g_1(t,y(s))·{\int }_0^{\vartheta (t)}{g}_2(s,I^{\zeta -\gamma }{y}_2(s))ds\right).\end{array}$$

Let $y\in Q_{r_1}$; then, for $t\in [0,T]$, and assumptions (i)–(ii), we obtain

$$\begin{array}{ccc}\hfill |F_{1}y(t)|& =& \left|I^{1-\zeta }f\left(t,g_1(t,y(s))·{\int }_0^{\vartheta (t)}{g}_2(s,I^{\zeta -\gamma }{y}_2(s))ds\right)\right|\hfill \\ & \le & I^{1-\zeta }\left(a^{*}+b(a_1^{*}+b_1\left|y(s)\right|)·{\int }_0^t(a_2^{*}+b_2{I}^{\zeta -\gamma }\left|y(s)\right|)ds\right)\hfill \\ \\ & \le & (a^{*}+b(a_1^{*}+b_1{r}_1))(I^{2-\zeta }{a}_2^{*}+I^{2-\gamma }{b}_2{r}_1)\hfill \\ \\ & \le & (a^{*}+b(a_1^{*}+b_1{r}_1))\left(\frac{a_2^{*}{t}^{2-\zeta }}{\Gamma (3-\zeta )}+\frac{b_2{r}_1{t}^{2-\gamma }}{\Gamma (3-\gamma )}\right)\hfill \\ \\ & \le & (a^{*}+b(a_1^{*}+b_1{r}_1))(a_2^{*}{T}^{2-\zeta }+b_2{r}_1{T}^{2-\gamma })=r_1.\hfill \end{array}$$

From assumption (iii), we obtain

$$\begin{array}{ccc}\hfill \parallel F_{1}y\parallel & \le & (a^{*}+b(a_1^{*}+b_1{r}_1))(a_2^{*}{T}^{2-\zeta }+b_2{r}_1{T}^{2-\gamma })=r_1.\hfill \end{array}$$

This proves that $\left\{F_{{}_1}y\right\}$ is uniformly bounded on $Q_{r_1}$. Let $y\in Q_{r_1}$, $t_1,t_2\in I$ such that $t_2>t_1$ and $\mid t_1-t_2\mid \le \delta$. By using assumption (ii), then,

$$\begin{array}{ccc}& & |F_{1}y(t_2)-F_{1}y(t_1)|=\hfill \\ \\ & & |{\int }_0^{t_2}\frac{{(t_2-s)}^{-\zeta }}{\Gamma (1-\zeta )}\left(f(s,g_1(s,y(s)))·{\int }_0^{\vartheta (s)}{g}_2(\theta ,I^{\zeta -\gamma }y(\theta ))d\theta \right)ds\hfill \\ & -& {\int }_0^{t_1}\frac{{(t_1-s)}^{-\zeta }}{\Gamma (1-\zeta )}\left(f(s,g_1(s,y(s)))·{\int }_0^{\vartheta (s)}{g}_2(\theta ,I^{\zeta -\gamma }y(\theta ))d\theta \right)ds|\hfill \\ & \le & |{\int }_0^{t_1}\frac{{(t_2-s)}^{-\zeta }}{\Gamma (1-\zeta )}\left(f(s,g_1(s,y(s)))·{\int }_0^{\vartheta (s)}{g}_2(\theta ,I^{\zeta -\gamma }y(\theta ))d\theta \right)ds\hfill \\ & +& {\int }_{t_1}^{t_2}\frac{{(t_2-s)}^{-\zeta }}{\Gamma (1-\zeta )}\left(f(s,g_1(s,y(s)))·{\int }_0^{\vartheta (s)}{g}_2(\theta ,I^{\zeta -\gamma }y(\theta ))d\theta \right)ds\hfill \\ & -& {\int }_0^{t_1}\frac{{(t_1-s)}^{-\zeta }}{\Gamma (1-\zeta )}\left(f(s,g_1(s,y(s)))·{\int }_0^{\vartheta (s)}{g}_2(\theta ,I^{\zeta -\gamma }y(\theta ))d\theta \right)ds|\hfill \\ & \le & |{\int }_0^{t_1}\frac{{(t_2-s)}^{-\zeta }}{\Gamma (1-\zeta )}-\frac{{(t_1-s)}^{-\zeta }}{\Gamma (1-\zeta )}((a^{*}+b(a_1^{*}+b_1{r}_1))(a_2^{*}{T}^{2-\zeta }+b_2{r}_1{T}^{2-\gamma }))ds\hfill \\ & +& {\int }_{t_1}^{t_2}\frac{1}{\Gamma (1-\zeta ){(t_2-s)}^{\zeta }}((a^{*}+b(a_1^{*}+b_1{r}_1))(a_2^{*}{T}^{2-\zeta }+b_2{r}_1{T}^{2-\gamma }))ds{\displaystyle |}\hfill \\ & \le & {\int }_0^{t_1}|\frac{{(t_2-s)}^{\zeta }-{(t_1-s)}^{\zeta }}{\Gamma (1-\zeta ){(t_1-s)}^{\zeta }{(t_2-s)}^{\zeta }}|((a^{*}+b(a_1^{*}+b_1{r}_1))(a_2^{*}{T}^{2-\zeta }+b_2{r}_1{T}^{2-\gamma }))ds\hfill \\ & +& {\int }_{t_1}^{t_2}\frac{1}{\Gamma (1-\zeta ){(t_2-s)}^{\zeta }}((a^{*}+b(a_1^{*}+b_1{r}_1))(a_2^{*}{T}^{2-\zeta }+b_2{r}_1{T}^{2-\gamma }))ds.\hfill \end{array}$$

This proves that $F_1:Q_{r_1}\to Q_{r_1}$ and that $\left\{F_{1}y\right\}$ is equi-continuous on $Q_{r_1}$. From [41], $\left\{F_{1}y\right\}$ is relatively compact. Hence, the operator $F_1$ is compact.

Let $\left\{y_n\right\}\subset Q_{r_1}$ be such that $y_n\to y$; then,

$$\begin{array}{ccc}\hfill F_1{y}_n(t)& =& I^{1-\zeta }f\left(t,g_1(t,y_n(t))·{\int }_0^{\vartheta (t)}{g}_2(s,I^{\zeta -\gamma }{y}_n(s))ds\right),\hfill \end{array}$$

Thus, by taking the limits for both sides and in view of Lebesgues dominated convergence Theorem [41] and assumption (ii), we obtain

$$\begin{array}{ccc}\hfill \underset{n\to \infty }{lim}{F}_1{y}_n(t)& =& \underset{n\to \infty }{lim}{I}^{1-\zeta }f\left(t,g_1(t,y_n(t))·{\int }_0^{\vartheta (t)}{g}_2(s,I^{\zeta -\gamma }{y}_n(s))ds\right)\hfill \\ & =& I^{1-\zeta }f\left(t,g_1(t,\underset{n\to \infty }{lim}{y}_n(t))·{\int }_0^{\vartheta (t)}{g}_2(s,I^{\zeta -\gamma }\underset{n\to \infty }{lim}{y}_n(s))ds\right)\hfill \\ & =& I^{1-\zeta }f\left(t,g_1(t,y(t))·{\int }_0^{\vartheta (t)}{g}_2(s,I^{\zeta -\gamma }y(s))ds\right)\hfill \\ & =& F_{1}y(t),\hfill \end{array}$$

Hence, $F_1$ is continuous and the solution to (4) exists.

Now, for the validity of solutions $x\in C(I)$ of (3), let the assumptions (i)–(iv) be satisfied. Define $Q_{r_2}$ as the closed sphere

$$\begin{array}{c}\hfill Q_{r_2}=\{x\in C(I):\parallel x\parallel \le r_2\},r_2=\frac{|x_0|+N+2r_1{T}^{\zeta }}{1-b_3{r}_1{T}^{\zeta -\eta +1}}\end{array}$$

and define the operator $F_2$ as

$$\begin{array}{c}\hfill F_{2}x(t)=x_0+{\int }_0^{T-\tau }\left(h(s,x(s)·I^{\zeta -\eta }y(s))\right)ds-I^{\zeta }y(\tau )+I^{\zeta }y(t).\end{array}$$

Let $x\in Q_{r_2}$; then, by using assumption (ii), we obtain

$$\begin{array}{ccc}\hfill |F_{2}x(t)|& =& |x_0+{\int }_0^{T-\tau }h\left(s,x(s)·I^{\zeta -\eta }y(s)\right)ds-I^{\zeta }y(\tau )+I^{\zeta }y(t)|\hfill \\ & \le & |x_0|+{\int }_0^{T-\tau }|h(s,x(s)·I^{\zeta -\eta }y(s))|ds+I^{\zeta }\left|y(\tau )\right|+I^{\zeta }\left|y(t)\right|\hfill \\ & \le & |x_0|+{\int }_0^{T-\tau }\left(|a_3(s)|+b_3(|x(s)I^{\zeta -\eta }y(s)|)\right)ds+2r_1{I}^{\zeta }\hfill \\ & \le & |x_0|+{\int }_0^{T-\tau }\left(a_3+b_3{r}_1{r}_2\frac{T^{\zeta -\eta }}{\Gamma (\zeta -\eta +1)}\right)ds+\frac{2r_1{T}^{\zeta }}{\Gamma (\zeta +1)}\hfill \\ & \le & |x_0|+N+\frac{r_1{r}_2{b}_3{T}^{\zeta -\eta }}{\Gamma (\zeta -\eta +1)}+\frac{2r_1{T}^{\zeta }}{\Gamma (\zeta +1)}\hfill \end{array}$$

and from assumption (iv), we obtain

$$\begin{array}{c}\hfill \parallel F_{2}x\parallel \le |x_0|+N+r_1{r}_2{b}_3{T}^{\zeta -\eta }+2r_1{T}^{\zeta }=r_2.\end{array}$$

This shows that $\left\{F_{{}_2}x\right\}$ is uniformly bounded on $Q_{r_2}$. Now, for $x\in Q_{r_2}$ and $t_1,t_2\in I$, where $t_2>t_1$ and $\mid t_1-t_2\mid \le \delta$, we obtain

$$\begin{array}{ccc}\hfill |F_{2}x(t_2)-F_{2}x(t_1)|& =& |x_0+{\int }_0^{T-\tau }h\left(s,x(s)·I^{\zeta -\eta }y(s)\right)ds-I^{\zeta }y(\tau )+I^{\zeta }y(t_2)\hfill \\ & -& x_0+{\int }_0^{T-\tau }h\left(s,x(s)·I^{\zeta -\eta }y(s)\right)ds-I^{\zeta }y(\tau )+I^{\zeta }y(t_1)|\hfill \\ & \le & {\int }_0^{t_2}\left|f\left(s,g_1(s,y(s))·{\int }_0^{\vartheta (t)}{g}_2(s,I^{\zeta -\gamma }y(s))ds\right)\right|ds\hfill \\ & -& {\int }_0^{t_1}\left|f\left(s,g_1(s,y(s))·{\int }_0^{\vartheta (t)}{g}_2(s,I^{\zeta -\gamma }y(s))ds\right)\right|ds\hfill \\ & \le & {\int }_{t_1}^{t_2}\left|f\left(s,g_1(s,y(s))·{\int }_0^{\vartheta (t)}{g}_2(s,I^{\zeta -\gamma }y(s))ds\right)\right|ds.\hfill \end{array}$$

This means that $F_2:Q_{r_2}\to Q_{r_2}$ and that $\left\{F_{2}x\right\}$ is equi-continuous on $Q_{r_2}$. From [41], $\left\{F_{2}x\right\}$ is relatively compact. Hence, $F_2$ is compact.

Assuming that $\left\{x_n\right\}\subset Q_{r_2}$, where $x_n\to x$, then,

$$\begin{array}{ccc}\hfill F_2{x}_n(t)& =& x_0+{\int }_0^{T-\tau }h\left(s,x_n(s)·I^{\zeta -\eta }y(s)\right)ds-I^{\zeta }y(\tau )+I^{\zeta }y(t)\hfill \end{array}$$

and by passing the limit, we have

$$\begin{array}{ccc}\hfill \underset{n\to \infty }{lim}{F}_2{x}_n(t)& =& \underset{n\to \infty }{lim}\left(x_0+{\int }_0^{T-\tau }h\left(s,x_n(s)·I^{\zeta -\eta }y(s)\right)ds-I^{\zeta }y(\tau )+I^{\zeta }y(t)\right)\hfill \end{array}$$

Applying the Lebesgue dominated convergence Theorem [41], then,

$$\begin{array}{ccc}\hfill \underset{n\to \infty }{lim}{F}_2{x}_n(t)& =& x_0+{\int }_0^{T-\tau }h\left(s,\underset{n\to \infty }{lim}x(s)·I^{\zeta -\eta }y(s)\right)ds-I^{\zeta }y(\tau )+I^{\zeta }y(t)\hfill \\ & =& x_0+{\int }_0^{T-\tau }h\left(s,x(s)·I^{\zeta -\eta }y(s)\right)ds-I^{\zeta }y(\tau )+I^{\zeta }y(t)=F_{2}x(t).\hfill \end{array}$$

This means that $F_2{x}_n(t)\to F_{2}x(t)$. Therefore, $F_2$ is continuous. From [41], the solution $x\in C(I)$ of (3) exists. As a result, the solution $x\in C[0,T]$ to Problem (1) and (2) exists. □

3. Uniqueness of the Solution

Consider the next additional hypothesis:

${(i)}^{*}$

$f,h$ and $g_i$: $I\times R\to R$ are measurable in $t\in I,\forall x\in R$ and satisfy the Lipschitz condition [43]

$$\begin{array}{ccc}\hfill |f(t,x)-f(t,y)|& \le & b|x-y|,\hfill \\ \hfill |g_i(t,x)-g_i(t,y)|& \le & b_i|x-y|\hfill \\ \hfill |h(t,x)-h(t,y)|& \le & b_3|x-y|\hfill \end{array}$$

with Lipschitz constants $b,b_i,b_3>0$ and $t\in I,x,y\in R,i=1,2$.

Remark 1.

From assumption ${(i)}^{*}$, we deduce assumption (ii) as follows:

$$|f(t,x)|\le |f(t,0)|+b\left|x\right|,$$

$$\begin{array}{c}\hfill \left|f(t,x)\right|\le a+b\left|x\right|,wherea=\underset{t\in I}{sup}\left|f(t,0)\right|.\end{array}$$

Also,

$$|g_i(t,x)|\le |g_i(t,0)|+b_i\left|x\right|,$$

$$\begin{array}{c}\hfill |g_i(t,x)|\le a_i+b_i\left|x\right|,where{a}_i=\underset{t\in I}{sup}\left|g_i(t,0)\right|,i=1,2.\end{array}$$

and

$$\left|h(t,x)\right|\le \left|h(t,0)\right|+b_3\left|x\right|,$$

$$\begin{array}{c}\hfill \left|h(t,x)\right|\le a_3+b_3\left|x\right|,where{a}_3=\underset{t\in I}{sup}\left|h(t,0)\right|.\end{array}$$

Theorem 2.

Let the hypotheses (i)–(iv) and (i*) be valid. If

$$\begin{array}{c}\hfill (a^{*}{b}_2+b{b}_2(a_1^{*}+b_1{r}_1))T^{2-\gamma }+b{b}_1(a_2^{*}{T}^{2-\zeta }+r_1{b}_2{T}^{2-\gamma })<1,\end{array}$$

Hence, the solution to (1) and (2) is unique.

Proof. 

It is clear that all hypotheses of Theorem 1 are valid, and thus, the solution to (4) exists. Now, assume that $y_1,y_2$ are two solutions of (4); then,

$$\begin{array}{ccc}& & |y_2(t)-y_1(t)|\hfill \\ & =& |I^{1-\zeta }f\left(t,g_1(t,y_2(s))·{\int }_0^{\vartheta (t)}{g}_2(s,I^{\zeta -\gamma }{y}_2(s))ds\right)\hfill \\ & -& I^{1-\zeta }f\left(t,g_1(t,y_1(s))·{\int }_0^{\vartheta (t)}{g}_2(s,I^{\zeta -\gamma }{y}_1(s))ds\right)|\hfill \\ & =& |I^{1-\zeta }f\left(t,g_1(t,y_2(s))·{\int }_0^{\vartheta (t)}{g}_2(s,I^{\zeta -\gamma }{y}_2(s))ds\right)\hfill \\ & -& I^{1-\zeta }f\left(t,g_1(t,y_2(s))·{\int }_0^{\vartheta (t)}{g}_2(s,I^{\zeta -\gamma }{y}_1(s))ds\right)\hfill \\ & +& I^{1-\zeta }f\left(t,g_1(t,y_2(s))·{\int }_0^{\vartheta (t)}{g}_2(s,I^{\zeta -\gamma }{y}_1(s))ds\right)\hfill \\ & -& I^{1-\zeta }f\left(t,g_1(t,y_1(s))·{\int }_0^{\vartheta (t)}{g}_2(s,I^{\zeta -\gamma }{y}_1(s))ds\right)|\hfill \\ & \le & I^{1-\zeta }|f(t,g_1(t,y_2(t))|·{\int }_0^{\vartheta (t)}|g_2(s,I^{\zeta -\gamma }{y}_2(s))-g_2(s,I^{\zeta -\gamma }{y}_1(s))|ds\hfill \\ & +& I^{1-\zeta }(|f(t,g_1(t,y_2(t))-f(t,g_1(t,y_1(t))|)·{\int }_0^{\vartheta (t)}\left|g_2(s,I^{\zeta -\gamma }{y}_1(s))\right|ds\hfill \\ & \le & I^{1-\zeta }(a^{*}+b(a_1^{*}+b_1{r}_1))·b_2{\int }_0^t{I}^{\zeta -\gamma }|y_2(s)-y_1(s)|ds\hfill \\ & +& I^{1-\zeta }\left(b{b}_1|y_2(s)-y_1(s)|{\int }_0^t(a_2^{*}+b_2{I}^{\zeta -\gamma }\left|y(s)\right|)ds\right)\hfill \\ & \le & (a^{*}+b(a_1^{*}+b_1{r}_1))·b_2{I}^{2-\gamma }\parallel y_2-y_1\parallel +b{b}_1(a_2^{*}{T}^{2-\zeta }+r_1{b}_2{T}^{2-\gamma })\parallel y_2-y_1\parallel .\hfill \end{array}$$

Hence,

$$\begin{array}{c}\hfill \parallel y_2-y_1\parallel \left(1-\left[(a^{*}{b}_2+b{b}_2(a_1^{*}+b_1{r}_1))T^{2-\gamma }+b{b}_1(a_2^{*}{T}^{2-\zeta }+r_1{b}_2{T}^{2-\gamma })\right]\right)\le 0.\end{array}$$

Since

$$(a^{*}{b}_2+b{b}_2(a_1^{*}+b_1{r}_1))T^{2-\gamma }+b{b}_1(a_2^{*}{T}^{2-\zeta }+r_1{b}_2{T}^{2-\gamma })<1.$$

Then, the solution of (4) is unique.

Now, for every solution $y\in C(I)$ to (4), there exists a unique solution $x\in C(I)$ of (3).

Let $y\in C(I)$ be a solution to (4), and let $x_1,x_2$ be two solutions to Equation (3); then,

$$\begin{array}{ccc}\hfill |x_2(t)-x_1(t)|& =& |x_0+{\int }_0^{T-\tau }h\left(s,x_2(s)·I^{\zeta -\eta }y(s)\right)ds-I^{\zeta }y(\tau )+I^{\zeta }y(t)\hfill \\ & -& x_0-{\int }_0^{T-\tau }h\left(s,x_1(s)·I^{\zeta -\eta }y(s)\right)ds+I^{\zeta }y(\tau )-I^{\zeta }y(t)|.\hfill \\ & \le & {\int }_0^{T-\tau }|h(s,x_2(s)·I^{\zeta -\eta }y(s))-h(s,x_1(s)·I^{\zeta -\eta }y(s))|ds\hfill \\ & \le & r_1{b}_3{\int }_0^{T-\tau }|x_2(s)-x_1(s)|I^{\zeta -\eta }ds,\hfill \\ & \le & r_1{b}_3\parallel x_2-x_1\parallel \frac{T^{\zeta -\eta +1}}{\Gamma (1+\zeta -\eta )},\hfill \end{array}$$

from assumption (iv), we obtain

$$\begin{array}{c}\hfill \parallel x_2-x_1\parallel (1-(\frac{r_1{b}_3{T}^{\zeta -\eta +1}}{\Gamma (1+\zeta -\eta )}))\le 0.\end{array}$$

Thus, there is only one solution to (3). As a result, there is only one solution to (1) and (2). □

4. Hyers–Ulam Stability

Definition 1.

[44] Let the solution to (1) and (2) exist. The constrained problem (1) and (2) is Hyers–Ulam-stable if $\forall ϵ>0$, $\exists \delta (ϵ)>0$ such that, for any solution $x_s\in C[0,T]$ of (1) and (2) satisfying

$$\begin{array}{c}\hfill |\frac{d{x}_s}{dt}-f(t,g_1(t,D^{\zeta }{x}_s(t))·{\int }_0^{\vartheta (t)}{g}_2(s,D^{\gamma }{x}_s(s)))|\le \delta .\end{array}$$

(8)

Then

$$\begin{array}{c}\hfill \parallel x-x_s{\parallel }_c\le ϵ.\end{array}$$

Theorem 3.

Assume that the hypothesis of Theorem 2 is satisfied; then, problem (1) and (2) is Hyers–Ulam-stable.

Proof. 

Let the condition of Equation (8) be satisfied; then, we have

$$\begin{array}{ccc}& -\delta & \le \frac{d{x}_s(t)}{dt}-f(t,g_1(t,D^{\zeta }{x}_s(t))·{\int }_0^{\vartheta (t)}{g}_2(s,D^{\gamma }{x}_s(s))ds)\le \delta ,\hfill \\ & -\frac{T^{1-\zeta }\delta }{\Gamma (2-\zeta )}& \le I^{1-\zeta }\frac{d{x}_s(t)}{dt}-I^{1-\zeta }f(t,g_1(t,D^{\zeta }{x}_s(t))·{\int }_0^{\vartheta (t)}{g}_2(s,D^{\gamma }{x}_s(s))ds)\le \frac{T^{1-\zeta }\delta }{\Gamma (2-\zeta )},\hfill \\ & -{\delta }_1& \le y_s(t)-I^{1-\zeta }f(t,g_1(t,y_s(t))·{\int }_0^{\vartheta (t)}{g}_2(s,I^{\zeta -\gamma }{y}_s(s))ds)\le {\delta }_1.\hfill \end{array}$$

Now,

$$\begin{array}{ccc}& & |y(t)-y_s(t)|=\hfill \\ \\ & & |I^{1-\zeta }f(t,g_1(t,y(t))·{\int }_0^{\vartheta (t)}{g}_2(s,I^{\zeta -\gamma }y(s))ds)-y_s(t)\hfill \\ & -& I^{1-\zeta }f(t,g_1(t,y_s(t))·{\int }_0^{\vartheta (t)}{g}_2(s,I^{\zeta -\gamma }{y}_s(s))ds)\hfill \\ & +& I^{1-\zeta }f(t,g_1(t,y_s(t))·{\int }_0^{\vartheta (t)}{g}_2(s,I^{\zeta -\gamma }{y}_s(s))ds)|\hfill \\ & \le & |I^{1-\zeta }f(t,g_1(t,y(t))·{\int }_0^{\vartheta (t)}{g}_2(s,I^{\zeta -\gamma }y(s))ds)\hfill \\ & -& I^{1-\zeta }f(t,g_1(t,y_s(t))·{\int }_0^{\vartheta (t)}{g}_2(s,I^{\zeta -\gamma }{y}_s(s))ds)|\hfill \\ & +& |I^{1-\zeta }f(t,g_1(t,y_s(t))·{\int }_0^{\vartheta (t)}{g}_2(s,I^{\zeta -\gamma }{y}_s(s))ds)-y_s(t)|\hfill \\ & \le & I^{1-\zeta }|f(t,g_1(t,y(t))|·{\int }_0^{\vartheta (t)}|g_2(s,I^{\zeta -\gamma }y(s))-g_2(s,I^{\zeta -\gamma }{y}_s(s)))|ds\hfill \\ & +& I^{1-\zeta }|f(t,g_1(t,y(t))-f(t,g_1(t,y_s(t))|·{\int }_0^{\vartheta (t)}|g_2(s,I^{\zeta -\gamma }{y}_s(s)))|ds+{\delta }_1\hfill \\ & \le & I^{1-\zeta }(a^{*}+b(a_1^{*}+b_1{r}_1))·b_2{\int }_0^t{I}^{\zeta -\gamma }|y(s)-y_s(s)|ds.\hfill \\ & +& I^{1-\zeta }b{b}_1|y(s)-y_s(s)|·{\int }_0^t(a_2^{*}+b_2{I}^{\zeta -\gamma }|y_s|)ds+{\delta }_1\hfill \\ & \le & (a^{*}+b(a_1^{*}+b_1{r}_1))·b_2{T}^{2-\gamma }\parallel y-y_s\parallel +b_{1}b(a_2^{*}{T}^{2-\zeta }+r_1.b_2{T}^{2-\gamma })\parallel y-y_s\parallel +{\delta }_1.\hfill \end{array}$$

Hence,

$$\begin{array}{c}\hfill \parallel y-y_s\parallel \left(1-\left[(a^{*}+b(a_1^{*}+b_1{r}_1))·b_2{T}^{2-\gamma }+b_{1}b(a_2^{*}{T}^{2-\zeta }+r_1·b_2{T}^{2-\gamma })\right]\right)\le {\delta }_1\end{array}$$

and

$$\begin{array}{c}\hfill \parallel y-y_s\parallel \le \frac{{\delta }_1}{1-\left[(a^{*}+b(a_1^{*}+b_1{r}_1))·b_2{T}^{2-\gamma }+b_{1}b(a_2^{*}{T}^{2-\zeta }+r_1·b_2{T}^{2-\gamma })\right]}.\end{array}$$

Since

$$\begin{array}{c}\hfill (a^{*}+b(a_1^{*}+b_1{r}_1))·b_2{T}^{2-\gamma }+b_{1}b(a_2^{*}{T}^{2-\zeta }+r_1·b_2{T}^{2-\gamma })<1,\end{array}$$

then

$$\begin{array}{c}\hfill \parallel y-y_s\parallel <ϵ.\end{array}$$

Also, using assumption (iv), we obtain

$$\begin{array}{ccc}\hfill |x(t)-x_s(t)|& =& |x_0+{\int }_0^{T-\tau }h(s,x(s)·I^{\zeta -\eta }y(s))ds-I^{\zeta }y(\tau )+I^{\zeta }y(t)\hfill \\ & -& x_0-{\int }_0^{T-\tau }h(s,x_s(s)·I^{\zeta -\eta }{y}_s(s))ds+I^{\zeta }{y}_s(\tau )-I^{\zeta }{y}_s(t)|\hfill \\ & \le & {\int }_0^{T-\tau }|h(s,x(s)·I^{\zeta -\eta }y(s))-h(s,x_s(s)·I^{\zeta -\eta }{y}_s(s))|ds+2I^{\zeta }\parallel y-y_s\parallel \hfill \\ & \le & b_3{\int }_0^{T-\tau }(|x(s)-x_s(s)|I^{\zeta -\eta }\left|y(s)\right|+I^{\zeta -\eta }|y(s)-y_s(s)\left|\right|x_s|)ds+\frac{2T^{\zeta }}{\Gamma (\zeta +1)}ϵ\hfill \\ & \le & \frac{r_1{b}_3{T}^{\zeta -\eta +1}}{\Gamma (\zeta -\eta +1)}\parallel x-x_s\parallel +\frac{r_2{b}_3{T}^{\zeta -\eta +1}}{\Gamma (\zeta -\eta +1)}ϵ+\frac{2T^{\zeta }}{\Gamma (\zeta +1)}ϵ,\hfill \end{array}$$

$$\begin{array}{c}\hfill \parallel x-x^{*}\parallel \le \frac{(\frac{r_2{b}_3{T}^{\zeta -\eta +1}}{\Gamma (\zeta -\eta +1)}+\frac{2T^{\zeta }}{\Gamma (\zeta +1)})ϵ}{1-(\frac{r_1{b}_3{T}^{\zeta -\eta +1}}{\Gamma (\zeta -\eta +1)})}.\end{array}$$

Since

$$\begin{array}{c}\hfill \frac{r_1{b}_3{T}^{\zeta -\eta +1}}{\Gamma (\zeta -\eta +1)}<1,\end{array}$$

thus,

$$\begin{array}{c}\hfill \parallel x-x^{*}\parallel \le ϵ.\end{array}$$

Then, the problem (1) and (2) is Hyers–Ulam-stable. □

5. Continuous Dependence

Definition 2.

The solution to (1) and (2) depends continuously on $y=D^{\zeta }x,h$ and $x_0$, and if $\forall ϵ>0$, $\exists \delta (ϵ)>0$ such that

$$max\{\parallel y-\check{y}\parallel ,\parallel h-\check{h}\parallel ,|x_0-\check{x_0}|\le \delta \}\Rightarrow \parallel x-\check{x}\parallel \le ϵ,$$

where $\check{x}$ and $\check{y}$ are the solutions to

$$\begin{array}{ccc}\hfill \check{x}(t)& =& \check{x_0}+{\int }_0^{T-\tau }\check{h}(s,\check{x}(s)·I^{\zeta -\eta }\check{y}(s))ds-I^{\zeta }\check{y}(\tau )+I^{\zeta }\check{y}(t),\hfill \end{array}$$

(9)

$$\begin{array}{ccc}\hfill \check{y}(t)& =& I^{1-\zeta }f\left(t,g_1(t,\check{y}(t))·{\int }_0^{\vartheta (t)}{g}_2(s,I^{\zeta -\gamma }\check{y}(s))ds\right).\hfill \end{array}$$

(10)

Theorem 4.

Suppose that the hypotheses of Theorem 2 are satisfied; then, the solution to (1) and (2) depends continuously on $y,h,and{x}_0$.

Proof. 

If $x(t)$ and $\check{x}(t)$ are the solutions to (3) and (9), respectively, using assumption ${(i)}^{*}$, we obtain

$$\begin{array}{ccc}& & |x(t)-\check{x}(t)|\hfill \\ & =& |x_0+{\int }_0^{T-\tau }h(s,x(s)·I^{\zeta -\eta }y(s))ds-I^{\zeta }y(\tau )+I^{\zeta }y(t)\hfill \\ & -& \check{x_0}-{\int }_0^{T-\tau }\check{h}(s,\check{x}(s)I^{\zeta -\eta }\check{y}(s))ds+I^{\zeta }\check{y}(\tau )-I^{\zeta }\check{y}(t)|\hfill \\ & \le & |x-\check{x_0}|+|{\int }_0^{T-\tau }\left(h(s,x(s)·I^{\zeta -\eta }y(s))-\check{h}(s,\check{x}(s)·I^{\zeta -\eta }\check{y}(s))ds\right)ds\hfill \\ & +& I^{\zeta }(y(\tau )-\check{y}(\tau ))+I^{\zeta }(y(t)-\check{y}(t))|\hfill \\ & \le & |x-\check{x_0}|+{\int }_0^{T-\tau }|h(s,x(s)·I^{\zeta -\eta }y(s))-h(s,\check{x}(s)·I^{\zeta -\eta }\check{y}(s))\hfill \\ & +& h(s,\check{x}(s)·I^{\zeta -\eta }\check{y}(s))-\check{h}(s,\check{x}(s)·I^{\zeta -\eta }\check{y}(s))|ds+2I^{\zeta }\parallel y-\check{y}\parallel \hfill \\ & \le & |x-\check{x_0}|+b_3{\int }_0^{T-\tau }|x(s)I^{\zeta -\eta }y(s)-\check{x}(s)I^{\zeta -\eta }\check{y}(s)|ds\hfill \\ & +& b_3{\int }_0^{T-\tau }\parallel h-\check{h}\parallel ds+2I^{\zeta }\parallel y-\check{y}\parallel \hfill \\ & \le & |x-\check{x_0}|+b_3{\int }_0^{T-\tau }|x(s)I^{\zeta -\eta }y(s)-x(s)I^{\zeta -\eta }\check{y}(s)\hfill \\ & +& x(s)I^{\zeta -\eta }\check{y}(s)-\check{x}(s)I^{\zeta -\eta }\check{y}(s)|ds+b_3\parallel h-\check{h}\parallel T+\parallel y-\check{y}\parallel \frac{2T^{\zeta }}{\Gamma (\zeta +1)}\hfill \\ & \le & \delta +b_3{\int }_0^{T-\tau }\left|x(s)\right|I^{\zeta -\eta }|y(s)-\check{y}(s)|ds\hfill \\ & +& b_3{\int }_0^{T-\tau }|x(s)-\check{x}(s)|I^{\zeta -\eta }\left|\check{y}(s)\right|ds+b_{3}T\delta +\frac{2T^{\zeta }\delta }{\Gamma (\zeta +1)}\hfill \\ & \le & \delta +b_3{r}_2\parallel y-\check{y}\parallel \frac{T^{\zeta -\eta +1}}{\Gamma (\zeta -\eta +1)}+b_3{r}_1\parallel x-\check{x}\parallel \frac{T^{\zeta -\eta +1}}{\Gamma (\zeta -\eta +1)}+b_{3}T\delta +\frac{2\delta T^{\zeta }}{\Gamma (\zeta +1)}\hfill \\ & \le & \delta +b_3{r}_2\delta \frac{T^{\zeta -\eta +1}}{\Gamma (\zeta -\eta +1)}+b_3{r}_1\parallel x-\check{x}\parallel \frac{T^{\zeta -\eta +1}}{\Gamma (\zeta -\eta +1)}+b_{3}T\delta +\frac{2\delta T^{\zeta }}{\Gamma (\zeta +1)}.\hfill \end{array}$$

Hence,

$$\begin{array}{c}\hfill \parallel x-\check{x}\parallel (1-\frac{r_1{b}_3{T}^{\zeta -\eta +1}}{\Gamma (\zeta -\eta +1)})\le \left(1+\frac{r_2{b}_3{T}^{\zeta -\eta +1}}{\Gamma (\zeta -\eta +1)}+\frac{2T^{\zeta }}{\Gamma (\zeta +1)}+b_{3}T\right)\delta \end{array}$$

and

$$\begin{array}{c}\hfill \parallel x-\check{x}\parallel =\frac{\left(1+\frac{r_2{b}_3{T}^{\zeta -\eta +1}}{\Gamma (\zeta -\eta +1)}+\frac{2T^{\zeta }}{\Gamma (\zeta +1)}+b_{3}T\right)\delta }{1-\frac{r_1{b}_3{T}^{\zeta -\eta +1}}{\Gamma (\zeta -\eta +1)}}=ϵ.\end{array}$$

Since $\frac{r_1{b}_3{T}^{\zeta -\eta +1}}{\Gamma (\zeta -\eta +1)}<1,$ therefore, the solution to (3) depends continuously on $y,h,x_0$. Consequently, the solution $x\in C[0,T]$ of (1) and (2) depends continuously on $y,h,x_0$. □

6. Special Cases and Examples

Corollary 1.

Let the hypothesis of Theorem 1 be valid; if we put $\tau =T$ in (2), then the backward problem

$$\begin{array}{c}\hfill \frac{dx}{dt}=f(t,g_1(t,D^{\zeta }x(t)){\int }_0^{\vartheta (t)}{g}_2(s,D^{\gamma }x(s)ds)),\zeta ,\gamma \in (0,1),t\in (0,T],\end{array}$$

$$x(T)=x_0,$$

has a solution $x\in C[0,T]$. Consequently, if the hypotheses of Theorem 2 are valid, it has a unique solution $x\in C[0,T]$.

Corollary 2.

Let the hypothesis of Corollary 1 be valid. If $\tau =T,\gamma =1-\zeta$, then the backward problem

$$\begin{array}{c}\hfill \frac{dx}{dt}=f(t,g_1(t,D^{\zeta }x(t)){\int }_0^{\vartheta (t)}{g}_2(s,D^{1-\zeta }x(s)ds)),\zeta \in (\frac{1}{2},1),t\in (0,T],\end{array}$$

$$x(T)=x_0.$$

has a solution $x\in C[0,T]$. Consequently, if the hypotheses of Theorem 2 are valid, it has a unique solution $x\in C[0,T]$.

Example 1.

Consider the next problem,

$$\begin{array}{ccc}\hfill \frac{dx}{dt}& =& \frac{1}{2}(\frac{e^{-t}}{1+e^{-t}})+\frac{1}{8}(\frac{t^2}{2}+\frac{1}{6}{D}^{\frac{1}{2}}x(t))·{\int }_0^{\rho t}(\frac{s^3}{4}+\frac{1}{3}{D}^{\frac{1}{2}}x(s))ds,t\in (0,1],\hfill \end{array}$$

(11)

$$\begin{array}{ccc}\hfill x(\tau )& =& \frac{1}{4}+{\int }_0^{1-\tau }(\frac{sins}{6}+\frac{1}{2}x(s)·D^{\frac{1}{2}}x(s))ds,\hfill \end{array}$$

(12)

where

$$\begin{array}{c}\hfill \zeta =\eta =\gamma =\frac{1}{2},\rho \in (0,1),x(0)=\frac{1}{4}.\end{array}$$

Then

$$\begin{array}{ccc}& & f\left(t,g_1(t,D^{\zeta }x(t))·{\int }_0^{\vartheta (t)}{g}_2(s,D^{\gamma }x(s)ds)\right)\hfill \\ & =& \frac{1}{2}(\frac{e^{-t}}{1+e^{-t}})+\frac{1}{8}\left(\frac{t^2}{2}+\frac{1}{6}{D}^{\frac{1}{2}}x(t)\right)·{\int }_0^{\rho t}(\frac{s^3}{4}+\frac{1}{3}{D}^{\frac{1}{2}}x(s))ds.\hfill \end{array}$$

Set

$$\begin{array}{ccc}\hfill g_1(t,D^{\zeta }x(t))& =& \frac{t^2}{2}+\frac{1}{6}{D}^{\frac{1}{2}}x(t)\hfill \\ \hfill g_2(t,D^{\gamma }x(t))& =& \frac{t^3}{4}+\frac{1}{3}{D}^{\frac{1}{2}}x(t)\hfill \\ \hfill h(t,x(t)·D^{\eta }x(t))& =& \frac{sint}{6}+\frac{1}{2}x(t)·D^{\frac{1}{2}}x(t).\hfill \end{array}$$

Here, we have

$$\begin{array}{ccc}& & a^{*}=a_1^{*}=\frac{1}{2},a_2^{*}=\frac{1}{4},b=\frac{1}{8},b_1=\frac{1}{6},b_2=\frac{1}{3},b_3=\frac{1}{2}\hfill \\ & & N=\frac{1}{6},r_1\approx 0.2,and{r}_2\approx 0.9.\hfill \end{array}$$

It is obvious that all the hypotheses of Theorem 1 are valid. Hence there exists at least one solution $x\in [0,1]$ of (12)–(15). Moreover, we have

$$\begin{array}{c}\hfill a^{*}{b}_2+b{b}_2(a_1^{*}+b_1{r}_1)T^{2-\gamma }+b{b}_1(a_2^{*}{T}^{2-\zeta }+r_1{b}_2{T}^{2-\gamma })=0.1950<1.\end{array}$$

Thus, all the hypotheses of Theorem 2 are valid, so the solution of Problem (12)–(15) is unique.

Example 2.

Consider the problem

$$\begin{array}{c}\hfill \frac{dx}{dt}=\frac{1}{4}{e}^{-t^2}cos(2t)+\frac{1}{3}\left(\frac{1}{3}(\frac{1}{5-t}+D^{\frac{1}{3}}x(t))·{\int }_0^{\frac{1}{2}t}\frac{1}{5}(\frac{e^{-s}}{s+2}+D^{\frac{1}{4}}x(s))ds\right)\\ \hfill t\in [0,1],\end{array}$$

(13)

$$\begin{array}{ccc}\hfill x(\tau )& =& \frac{1}{5}+{\int }_0^{1-\tau }\left(\frac{1}{18}{s}^2(sin(2s+1))+\frac{1}{6}x(s)D^{\frac{1}{5}}x(s)\right)ds,\hfill \end{array}$$

(14)

where

$$\begin{array}{c}\hfill \zeta =\frac{1}{3},\eta =\frac{1}{5},\gamma =\frac{1}{4},t\in (0,1],x(0)=\frac{1}{5}.\end{array}$$

Then,

$$\begin{array}{ccc}& & f(t,g_1(t,D^{\zeta }x(t)){\int }_0^{\vartheta (t)}{g}_2(s,D^{\gamma }x(s)ds))\hfill \\ & =& \frac{1}{4}{e}^{-t^2}cos(2t)+\frac{1}{3}(\frac{1}{3}\frac{1}{5-t}+D^{\frac{1}{3}}x(t))·{\int }_0^{\frac{1}{2}t}\frac{1}{5}(\frac{e^{-s}}{s+2}+D^{\frac{1}{4}}x(s))ds.\hfill \end{array}$$

Set

$$\begin{array}{ccc}\hfill g_1(t,D^{\zeta }x(t))& =& \frac{1}{3}(\frac{1}{5-t}+D^{\frac{1}{3}}x(t))\hfill \\ \hfill g_2(t,D^{\gamma }x(t))& =& \frac{1}{5}(\frac{e^{-t}}{t+2}+D^{\frac{1}{4}}x(t))\hfill \\ \hfill h(t,x(t)·D^{\eta }x(t))& =& \frac{1}{6}(\frac{1}{3}{t}^2(sin(2t+1))+x(t)D^{\frac{1}{5}}x(t)).\hfill \end{array}$$

Here, we have

$$\begin{array}{ccc}& & a^{*}=\frac{1}{4},a_1^{*}=\frac{1}{12},a_2^{*}=\frac{1}{10},b=b_1=\frac{1}{3},b_2=\frac{1}{5},b_3=\frac{1}{6},\hfill \\ & & N=\frac{1}{18},r_1=0.3,r_2=0.05.\hfill \end{array}$$

It is obvious that all the hypotheses of Theorem 1 are valid. Hence the solution $x\in [0,T]$ of (13) and (14) exists. Moreover, we have

$$\begin{array}{c}\hfill a^{*}{b}_2+b{b}_2(a^{*}+b_1{r}_1)t^{2-\gamma }+b{b}_1(a_2^{*}{t}^{2-\gamma }+r_1{b}_2{t}^{2-\gamma })=0.2314<1,\end{array}$$

Thus, all the hypotheses of Theorem 2 are valid, and then the solution to (13)–(14) is unique.

Example 3.

Consider the next problem

$$\begin{array}{c}\hfill \frac{dx}{dt}=\frac{1}{4}(\frac{t}{t^3+1})+\frac{1}{3}(\frac{7+3t}{16}+\frac{ln(1+|D^{\frac{1}{5}}x(t)|)}{5t+7})·{\int }_0^{t^4}(\frac{1}{9-s}+\frac{{(D^{\frac{1}{7}}x(s))}^2}{6(1+|D^{\frac{1}{7}}x(s)|)})ds,\\ \hfill t\in (0,\frac{1}{3}],\end{array}$$

(15)

$$\begin{array}{ccc}\hfill x(\tau )& =& \frac{1}{4}+{\int }_0^{\frac{1}{3}-\tau }(\frac{s^2}{s^2+1}+\frac{ln(1+|x(s)·D^{\frac{1}{4}}x(s)|)}{8+s^2})ds.\hfill \end{array}$$

(16)

Here, we have

$$\begin{array}{c}\hfill x_0=\frac{1}{4},\zeta =\frac{1}{5},\eta =\frac{1}{7},\gamma =\frac{1}{4},\vartheta (t)=t^4,\end{array}$$

and

$$\begin{array}{ccc}& & f\left(t,g_1(t,D^{\zeta }x(t))·{\int }_0^{\vartheta (t)}{g}_2(s,D^{\eta }x(s)ds)\right)\hfill \\ & =& \frac{1}{4}(\frac{t}{t^3+1})+\frac{1}{3}(\frac{7+3t}{16}+\frac{ln(1+|D^{\frac{1}{5}}x(t)|)}{5t+7})·{\int }_0^{t^4}(\frac{1}{9-s}+\frac{{(D^{\frac{1}{7}}x(s))}^2}{6(1+|D^{\frac{1}{7}}x(s)|)})ds.\hfill \end{array}$$

Set

$$\begin{array}{c}\hfill \vartheta (t)=t^4,\\ \hfill g_1(t,D^{\zeta }x(t))& =& \frac{7+3t}{16}+\frac{ln(1+|D^{\frac{1}{5}}x(t)|)}{5t+7},\hfill \\ \hfill g_2(t,D^{\gamma }x(t))& =& \frac{1}{9-s}+\frac{{(D^{\frac{1}{7}}x(s))}^2}{6(1+|D^{\frac{1}{7}}x(s)|)},\hfill \\ \hfill h(t,x(t)·D^{\eta }x(t))& =& \frac{t^2}{t^2+1}+\frac{ln(1+|x(t)·D^{\frac{1}{4}}x(t)|)}{8+t^2}.\hfill \end{array}$$

Thus, we obtain

$$\begin{array}{ccc}& & a^{*}=\frac{1}{12},a_1^{*}=\frac{1}{2},a_2^{*}=\frac{3}{26},N=\frac{1}{9},b=\frac{1}{3},b_1=\frac{1}{7},b_2=\frac{1}{6},b_3=\frac{1}{8},\hfill \\ & & r_1\approx 0.03,and{r}_2\approx 0.42.\hfill \end{array}$$

It is clear that all assumptions of Theorem 1 are satisfied. Hence, there exists at least one solution $x\in [0,T]$ of (12)–(15). Moreover, we have

$$\begin{array}{c}\hfill a^{*}{b}_2+b{b}_2(a_1^{*}+b_1{r}_1)t^{2-\gamma }+b{b}_1(a_2^{*}{t}^{2-\gamma }+r_1{b}_2{t}^{2-\gamma })=0.043<1.\end{array}$$

Thus, all assumptions of Theorem 2 are satisfied, and then the solution of the problem (12)–(15) is unique.

7. Conclusions

In this manuscript, we considered the constrained problem of the fractional-order integro-differential equation (1) under the quadratic functional integro-differential constraint (2). We proved the existence of solutions to the problem (1) and (2). The sufficient conditions for the uniqueness of the solution have been presented. The Hyers–Ulam stability of the problem (1) and (2) has been analyzed. The continuous dependence of the unique solution on its fractional-order derivative $D^{\zeta }x(t)$, the parameter $x_0$, and the function h has been studied. We introduced several examples and special cases.