Pontryagin Maximum Principle for Fractional Delay Differential Equations and Controlled Weakly Singular Volterra Delay Integral Equations

Abstract

This article explores two distinct issues. To begin with, we analyze the Pontriagin maximum principle concerning fractional delay differential equations. Furthermore, we investigate the most effective method for resolving the control problem associated with Eq. (1.1) and its corresponding payoff function (1.2). Subsequently, we explore the Pontryagin Maximum principle within the framework of Volterra delay integral equations (1.3). We enhance the outcomes of our investigation by presenting illustrative examples towards the conclusion of the article.

1 Introduction

The Pontryagin Maximum Principle is a powerful tool in optimal control theory that is used to derive necessary conditions for finding optimal control trajectories. This principle was first introduced by Lev S. Pontryagin in 1956, and it is widely used in various mathematical and applied fields.

In recent years, the Pontryagin Maximum Principle has been extended to fractional delay differential equations (FDDEs) and delayed Volterra integral equations (DVIEs). These equations are widely used in various areas of science and engineering, including control systems, ecology, finance, biology, physics, and chemistry.

The Caputo FDDE is a type of fractional differential equation that contains a time delay term. It describes the dynamics of systems that exhibit memory effects, and it has been used to model various phenomena such as heat conduction, viscoelasticity, and fractional order control systems. The delayed Volterra integral equation, on the other hand, is a type of integral equation that contains a time delay term. It describes the behavior of systems that exhibit history-dependent dynamics, and it has been used to model various phenomena such as population dynamics, chemical reaction kinetics, and electrical networks.

The Pontryagin Maximum Principle for Caputo FDDEs and DVIEs is a set of necessary conditions that must be satisfied by optimal control trajectories for these equations. It provides a method to solve optimization problems involving these equations, such as finding the control inputs that minimize the energy consumption or maximize the system performance.

The necessary conditions derived from the Pontryagin Maximum Principle for Caputo FDDEs and DVIEs involve the optimal control trajectory, the corresponding adjoint function, and other variables that depend on the specific problem being considered. These conditions are often expressed in terms of differential or integral equations, and they can be used to derive optimal control strategies for a wide range of systems.

In recent times, there has been a growing interest among researchers in the study of fractional differential equations and their control processes. Noteworthy contributions in this area include works such as [1,2,3,4,5,6,7,8,9,10,11,12,13]. Additionally, there has been significant attention on Volterra-type integral equations, as evidenced by publications like [14,15,16,17,18,19,20,21,22,23,24,25,26,27,28,29,30]. Furthermore, numerous articles have explored topics related to the Pontriagin maximum principle, as seen in [31,32,33,34,35,36].

So that, the existence of theorems about optimal control problems governed by a nonlinear Volterra-type integral equation was articulated and demonstrated. Consideration was given to growth conditions, which was essential for ensuring compactness in the set of trajectories and establishing the existence of standard optimal solutions for the specified problem in [15].

Additionally, Carlson DA contributed an elementary proof of the maximum principle for optimal control problems with states governed by Volterra integral equations, and his proof relied solely on fundamental results from analysis and mathematical programming, reducing the optimal control problem to a mathematical programming problem through Pontryagin-type perturbations of controls in [19].

Burnap C and Kazemi M A investigated an optimal control problem where the states were determined by a nonlinear Volterra integral equation incorporating a delay. The problem encompassed both control constraints and equality/inequality restrictions on terminal states. Initially, employing Pontryagin-type control variations, they established a maximum principle under a rather general set of assumptions, and demonstrated that those findings promptly led to the classical Pontryagin maximum principle for a control process governed by an ordinary differential equation, regardless of the presence of delay in [18].

The maximum principle for optimal terminal time control problems, wherein the state was governed by a Volterra integral equation and constraints depended on both the terminal time and the state, was proven by Vega CD. Pontryagin-type perturbations were utilized to successfully reduce the problem to a well-established result in optimization theory in [20].

A new method for solving optimal control problems for systems governed by Volterra integral equations was formulated and analyzed in the paper. The original Volterra-controlled system was discretized, and a novel type of dynamic programming was employed, where the Hamilton-Jacobi function was parametrized by the control function, in contrast to ordinary dynamic programming where it is typically parametrized by the state. Estimates for the computational cost of this method were also derived in [16].

A novel method for solving optimal control problems for Volterra integral equations was presented and analyzed. This method was based on the approximation of the controlled Volterra integral equations by a sequence of systems of controlled ordinary differential equations. The resulting approximating problems could then be solved using dynamic programming methods tailored for systems governed by ordinary differential equations. Additionally, the connection between our version of dynamic programming and the Hamiltonian equations for Volterra-controlled systems was derived, as detailed in [17].

Lin P., and Yong J. investigated a class of controlled singular Volterra integral equations, providing a framework applicable to problems involving memory. Notably, well-known fractional-order ordinary differential equations of the Riemann–Liouville or Caputo types are special cases within the broader scope studied there. The paper establishes well-posedness and regularity results in suitable function spaces for these integral equations. For an associated optimal control problem, a novel approach is developed using a Liapunov-type theorem and the spike variation technique. This leads to a Pontryagin-type maximum principle for optimal controls, distinct from existing literature on fractional differential equations. Importantly, this method handles problems without assuming regularity conditions on controls, convexity on the control domain, or additional unnecessary constraints on the nonlinear terms of the integral equation and cost functional in [14].

While the Pontryagin Maximum Principle is typically formulated for systems without delays, it can be extended to handle delay systems as well. Delay systems involve equations where the state at a given time depends not only on the current state and control input but also on the state and/or control input at previous times. This introduces a history dependence that makes the analysis more complex. The extension of the Pontryagin Maximum Principle to delay systems is not straightforward and may involve additional considerations. One approach is to use state augmentation, where additional state variables are introduced to represent the delayed states. Another approach involves using advanced mathematical tools such as functional analysis and the calculus of variations. Research in the field of optimal control of delay systems has been ongoing, and there are various techniques and methodologies developed to address the challenges posed by delay dynamics. However, the application of the Pontryagin Maximum Principle to delay systems may not be as widespread or straightforward as in the case of systems without delays. While the Pontryagin Maximum Principle provides powerful tools for analyzing and solving optimal control problems, its direct application to delay systems requires additional considerations and may involve more advanced mathematical techniques.

In this research, motivated by the results in the paper [14], we explore a controlled process with time delays.

The article addresses two main issues. First, it examines a controlled Caputo-type fractional delayed differential equation represented as:

$$\begin{aligned} {\left\{ \begin{array}{ll} {^{C}}D^{\alpha }_{t}y(t)= f(t,y(t),y(t-h),u(t)), \quad a.e.\quad t\in [0,T],\\ y(t)=0,\quad -h\le t\le 0,\quad h>0. \end{array}\right. } \end{aligned}$$

(1.1)

Here, $f(\cdot ,\cdot ,\cdot ,\cdot )$ denotes the free term and the generator of the state equation, respectively. $y(\cdot )$ represents the state trajectory in Euclidean space $R^{n}$, and $u(\cdot )$ is the control map in a separable metric space U. To assess the control map’s effectiveness, a payoff functional is introduced with a running cost term on the right-hand side:

$$\begin{aligned} J(u(\cdot ))=\int _{0}^{T} g(t,y(t),y(t-h),u(t))dt. \end{aligned}$$

(1.2)

Secondly, the article investigates an optimal control problem governed by Volterra delay integral equations with a weakly singular kernel, as an alternative to the Caputo-type fractional delay differential equations. The Volterra delay integral equations are given by:

$$\begin{aligned} {\left\{ \begin{array}{ll} y(t)=\eta (t)+\int _{0}^{t}\frac{f(t,s,y(s),y(s-h),u(s))}{(t-s)^{1-\alpha }}ds, \quad a.e.\quad t\in [0,T],\\ y(t)=0,\quad -h\le t\le 0,\quad h>0. \end{array}\right. } \end{aligned}$$

(1.3)

where $\eta (\cdot )$ and $f(\cdot , \cdot , \cdot , \cdot ,\cdot )$ are given functions, representing the free term and the generator of the state equation, respectively. $y(\cdot )$ denotes the state trajectory in Euclidean space $R^{n}$, and $u(\cdot )$ is the control map in a separable metric space U.

2 Preliminaries

In the following passage, we will share initial discoveries that will be helpful for future investigations. Firstly, we will focus on a specific period of time labeled as $T>0$. Then, we will establish certain spaces:

$$\begin{aligned} L^{p}(0,T;R^{n})&=\bigg \lbrace \phi :[0,T]\rightarrow R^{n} \quad | \quad \phi (\cdot ) \quad is \quad measurable, \\&\qquad \Vert \phi (\cdot )\Vert _{p} \equiv \bigg (\int _{0}^{T} \vert \phi (t)\vert ^{p}dt\bigg )^{1/p}<\infty \bigg \rbrace , 1\le p<\infty , \end{aligned}$$

$$\begin{aligned} L^{\infty }(0,T;R^{n})&=\bigg \lbrace \phi :[0,T]\rightarrow R^{n} \quad | \quad \phi (\cdot ) \quad is \quad measurable, \\&\qquad \Vert \phi (\cdot )\Vert _{\infty }\equiv {{\,\mathrm{ess\,sup}\,}}_{t\in [0,T]}\vert \phi (t)\vert <\infty \bigg \rbrace . \end{aligned}$$

Also, we define

$$\begin{aligned} L^{p+}(0,T;R^{n})&=\bigcup _{r>p} L^{r}(0,T;R^{n}), \quad 1\le p<\infty ,\\ L^{p-}(0,T;R^{n})&=\bigcap _{r<p} L^{r}(0,T;R^{n}), \quad 1<p<\infty . \end{aligned}$$

In the subsequent analysis, we utilize the notation $\Delta =\lbrace (t,s)\in [0,T]^{2} | 0\le s<t\le T\rbrace$. It is important to note that the "diagonal line" represented by $\lbrace (t,t)| t\in [0,T]\rbrace$ does not belong to $\Delta$. Consequently, if we consider a continuous mapping $\phi :\Delta \rightarrow R^{n}$ where $(t,s)\mapsto \phi (t,s)$, the function $\phi (\cdot ,\cdot )$ may become unbounded as the difference $\vert t-s\vert \rightarrow 0$.

Throughout this paper, we adopt the notation $t_{1} \vee t_{2}=\max \lbrace t_{1},t_{2}\rbrace$ and $t_{1} \wedge t_{2}=\min \lbrace t_{1},t_{2}\rbrace$, for any $t_{1},t_{2}\in R$. Notably, $t^{+}=t\vee 0$. The characteristic function of any set E is denoted by $1_{E(\cdot )}$.

We call a continuous and strictly increasing function $\omega (\cdot ): R_{+}\rightarrow R_{+}$ a modulus of continuity if $\omega (0)=0$.

Definition 2.1

[8] The fractional integral of order $\alpha >0$ for a function $\phi :[0,T]\rightarrow R$ is defined by

$$\begin{aligned} (I^{\alpha }_{0+}\phi )(t)=\frac{1}{\Gamma (\alpha )}\int _{0}^{t}(t-\tau )^{\alpha -1}\phi (\tau )d\tau ,\quad t>0. \end{aligned}$$

where $\Gamma : (0,\infty )\rightarrow R$ is the well-known Euler gamma function defined as

$$\begin{aligned} \Gamma (\alpha )=\int _{0}^{\infty }\tau ^{\alpha -1}e^{-\tau }d\tau , \quad \alpha >0. \end{aligned}$$

Definition 2.2

[8] The Riemann–Liouville fractional derivative of the order $0<\alpha \le 1$ for a function $\phi (\cdot )\in L^{\infty }[0,T]$ is defined by

$$\begin{aligned} ({^{RL}} D^{\alpha }_{0+}\phi )(t)=\frac{1}{\Gamma (1-\alpha )}\frac{d}{dt}\int _{0}^{t}(t-\tau )^{-\alpha }\phi (\tau )d\tau ,\quad t>0. \end{aligned}$$

Definition 2.3

[8] The Caputo fractional derivative of the order $0<\alpha \le 1$ for a function $\phi (\cdot )\in L^{\infty }[0,T]$ is defined by

$$\begin{aligned} ({^{C}} D^{\alpha }_{0+}\phi )(t)=\frac{1}{\Gamma (1-\alpha )}\int _{0}^{t}(t-\tau )^{-\alpha }\phi ^{\prime }(\tau )d\tau ,\quad t>0. \end{aligned}$$

The relationship Caputo and Riemann–Liouville fractional differentiation operators for the function $\phi (\cdot )\in L^{\infty }[0,T]$ is as follows:

$$\begin{aligned} {^{C}} D^{\alpha }_{0+}\phi (t)={^{RL}} D^{\alpha }_{0+}(\phi (t)-\phi (0)),\quad \alpha \in (0,1]. \end{aligned}$$

Gronwall’s inequality is a fundamental tool in the analysis of ordinary differential equations and integral inequalities. It provides a powerful means to establish bounds on functions given certain conditions. While Gronwall’s inequality is widely used for ordinary differential equations, there are extensions and variations for delay differential equations as well. Applications of Gronwall-type inequalities in delay equations are widespread, especially in stability analysis of systems with delays, such as those arising in biology, control theory, and engineering. The subsequent lemma will play a pivotal role in analyzing the convergence of the primary results.

Lemma 2.1

[26] Let $\alpha \in (0,1)$ and $q>\frac{1}{\alpha }$. Let $L(\cdot ), a(\cdot ),y(\cdot )$ be nonnegative functions with $L(\cdot )\in L^{q}(0,T)$ and $a(\cdot ),y(\cdot )\in L^{\frac{q}{q-1}}(0,T)$. Assume

$$\begin{aligned} y(t)\le a(t)+\int _{0}^{t}\frac{L(s)y(s)}{(t-s)^{1-\alpha }}ds+\int _{0}^{t}\frac{L(s)y(s-h)}{(t-s)^{1-\alpha }}ds, \quad a.e. \quad t\in [0,T].\nonumber \\ \end{aligned}$$

(2.1)

Then there is a constant $K>0$ such that

$$\begin{aligned} y(t)\le K A(t)\exp \bigg [\int _{0}^{t}L(s)ds+\int _{0}^{t}\gamma (s)ds\bigg ], \quad 0\le t<T, \end{aligned}$$

(2.2)

where

$$\begin{aligned} A(t)&= a(t)+K_{0}\int _{0}^{t}\frac{L(s)a(s)}{(t-s)^{1-\alpha }}ds+K_{0}\int _{0}^{t}\frac{L(s)a(s-h)}{(t-s)^{1-\alpha }}ds\\&\quad + K_{1}\int _{0}^{t-h}\frac{L(s)a(s)}{(t-h-s)^{1-\alpha }}ds + K_{1}\int _{0}^{t-h}\frac{L(s)a(s-h)}{(t-h-s)^{1-\alpha }}ds\\&\quad + K_{2}\int _{0}^{t-2h}\frac{L(s)a(s)}{(t-2h-s)^{1-\alpha }}ds + K_{2}\int _{0}^{t-2h}\frac{L(s)a(s-h)}{(t-2h-s)^{1-\alpha }}ds\\&\quad +\cdots +K_{n-1}\int _{0}^{t-(n-1)h}\frac{L(s)a(s)}{(t-(n-1)h-s)^{1-\alpha }}ds\\&\quad + K_{n-1}\int _{0}^{t-(n-1)h}\frac{L(s)a(s-h)}{(t-(n-1)h-s)^{1-\alpha }}ds.\\ \end{aligned}$$

Let for any $\delta >0$,

$$\begin{aligned} {\mathscr {E}}_{\delta }=\lbrace E\subseteq [0,T] | \vert E\vert =\delta T\rbrace , \end{aligned}$$

where $\vert E\vert$ stands for the Lebesgue measure of E.

Lemma 2.2

[14] Let $\varphi : \Delta \rightarrow R^{n}$ be measurable such that

$$\begin{aligned} {\left\{ \begin{array}{ll} \vert \varphi (0,s)\vert \le {\bar{\varphi }}(s),\quad 0<s\le T,\\ \vert \varphi (t,s)-\varphi (t^{\prime },s)\vert \le \omega (\vert t-t^{\prime }\vert ) {\bar{\varphi }}(s), \quad (t,s), (t^{\prime },s)\in \Delta . \end{array}\right. } \end{aligned}$$

(2.3)

for some $\varphi (\cdot )\in L^{q}(0,T)$ with $q>\frac{1}{\alpha }, \alpha \in (0,1)$, and some modulus of continuity $\omega :R_{+}\rightarrow R_{+}$. Then

$$\begin{aligned} \inf _{E\in {\mathscr {E}}_{\delta }} \sup _{t\in [0,T]} \bigg \vert \int _{0}^{t}\bigg (\frac{1}{\delta }1_{E}(s)-1\bigg )\frac{\varphi (t,s)}{(t-s)^{1-\alpha }}ds\bigg \vert =0. \end{aligned}$$

(2.4)

Lemma 2.3

[8] Let $\alpha \in (0,1,)\quad p\ge 1, q\ge 1$ and $\frac{1}{p}+\frac{1}{q}\le 1+\alpha .$ (If $\frac{1}{p}+\frac{1}{q}=\alpha +1,$ then $p\not =1$ and $q\not =1).$If $f(\cdot )\in I^{\alpha }_{b-}(L^{p})$ and $g(\cdot )\in I^{\alpha }_{a+}(L^{q}),$ then

$$\begin{aligned} \int _{a}^{b}f(t)(D^{\alpha }_{a+}g)(t)dt=\int _{a}^{b}g(t)(D^{\alpha }_{b-}f)(t)dt. \end{aligned}$$

3 Main results

Suppose that U, a separable metric space with the metric $\rho$. U can be either a nonempty bounded or unbounded set in $R^{m}$ with the metric induced by the usual Euclidean norm. We can view U as a measurable space by considering the Borel $\sigma$-field. Let $u_{0}$ be a fixed element in U. For any $p \ge 1$, we define

$$\begin{aligned} U^{p}[0,T]=\lbrace u:[0,T]\rightarrow U | u(\cdot ) \quad is\quad measurable\quad \rho (u(\cdot ),u_{0})\in L^{p}(0,T;R)\rbrace . \end{aligned}$$

3.1 Pontryagin’s maximum principle for the fractional differential equation

Let us consider the following problem

$$\begin{aligned}{} & {} {\left\{ \begin{array}{ll} {^{C}}D^{\alpha }_{t}y(t)= f(t,y(t),y(t-h),u(t)), \quad a.e.\quad t\in [0,T],\\ y(t)=0,\quad -h\le t\le 0,\quad h>0. \end{array}\right. } \end{aligned}$$

(3.1)

$$\begin{aligned}{} & {} \quad J(u(\cdot ))=\int _{0}^{T} g(t,y(t),y(t-h),u(t))dt. \end{aligned}$$

(3.2)

In this paragraph, we negotiate the optimal control problem for (3.1) with payoff functional (3.2) Let us acquaint the following hypothesises. The assumed conditions are more than sufficient. Without loss generality, we state to utilize these conditions.

(H1): Let $f: [0,T]\times R^{n}\times R^{n}\times U\rightarrow R^{n}$ be a transformation with $t\mapsto f(t,y,y_{h},u)$ being measurable, $(y,y_{h})\mapsto f(t,y,y_{h},u)$ being continuously differentiable, $(y,y_{h},u)\mapsto f(t,y,y_{h},u)$, $(y,y_{h},u)\mapsto f_{y}(t,y,y_{h},u)$ and $(y,y_{h},u)\mapsto f_{y_{h}}(t,y,y_{h},u)$being continuous. There exist non-negative functions. $L_{0}(\cdot ), L(\cdot )$ with

$$\begin{aligned}&L_{0}(\cdot )\in L^{\frac{1}{\alpha }{+}}(0,T),\quad L(\cdot )\in L^{\frac{p}{p\alpha -1}{+}}(0,T) \end{aligned}$$

(3.3)

for some $p>\frac{1}{\alpha }$ and $\alpha \in (0,1), u_{0}\in U$.

$$\begin{aligned}&\vert f(t,0,0,u_{0})\vert \le L_{0}(t), \quad t\in [0,T], \end{aligned}$$

(3.4)

$$\begin{aligned}&\vert f(t,y,y_{h},u)-f(t,y^{\prime },y^{\prime }_{h},u^{\prime })\vert \le L(t)[\vert y-y^{\prime }\vert +\vert y_{h}-y^{\prime }_{h}\vert +\rho (u,u^{\prime })],\quad t\in [0,T], \nonumber \\&\quad y,y^{\prime }, y_{h},y^{\prime }_{h}\in R^{n}, \quad u,u^{\prime }\in U. \end{aligned}$$

(3.5)

We point out (3.4)–(3.5) declare

$$\begin{aligned} \vert f(t,y,y_{h},u)\vert&\le L_{0}(t)+L(t)[\vert y\vert +\vert y_{h}\vert +\rho (u,u_{0})],\nonumber \\&\quad (t,y,y_{h},u)\in [0,T]\times R^{n}\times R^{n} \times U. \end{aligned}$$

(3.6)

$$\begin{aligned} \vert f(t,y,y_{h},u)-f(t^{\prime },y,y_{h},u)\vert&\le K\omega (\vert t-t^{\prime }\vert )(1+\vert y\vert +\vert y_{h}\vert ), \nonumber \\&\quad t,t^{\prime }\in [0,T],\quad y,y_{h}\in R^{n}, \quad u\in U, \end{aligned}$$

(3.7)

for some modulus of continuity $\omega (\cdot )$. Moreover, it is evident that L is included in a smaller space, compared to the space to which $L_{0}$ belongs.

(H2): Let $g: [0,T]\times R^{n}\times R^{n}\times U\rightarrow R$ be a transformation with $t\mapsto g(t,y,y_{h},u)$ being measurable, $(y,y_{h})\mapsto g(t,y,y_{h},u)$ being continuously differentiable, and $(y,y_{h},u)\mapsto (g(t,y,y_{h},u),g_{y}(t,y,y_{h},u),g_{y_{h}}(t,y,y_{h},u))$ being continuous. There is a constant $L>0$ such that

$$\begin{aligned}&\vert g(t,0,0,u)\vert + \vert g_{y}(t,0,0,u)\vert + \vert g_{y_{h}}(t,0,0,u)\vert \le L,\\&\quad (t,y,y_{h},u)\in [0,T]\times R^{n}\times R^{n}\times U. \end{aligned}$$

Obviously, the payoff functional (3.2) is well-defined under (H1) and (H2). Thus, we are able to consider the following optimal control problem (OCP).

Problem (P) Find a $u^{*}(\cdot )\in U^{p}[0,T]$ such that

$$\begin{aligned} J(u^{*}(\cdot ))=\inf _{u(\cdot )\in U^{p}[0,T]}J(u(\cdot )). \end{aligned}$$

(3.8)

Arbitrary $u^{*}(\cdot )$ satisfying (3.8) is called an optimal control of Problem (P), the appropriate state $y^{*}(\cdot )$ is called an optimal state, and $(y^{*},u^{*})$ is called optimal pair.

Now, Pontryagin’s maximum principle can be given for the Problem (P).

Theorem 3.1

Let (H1) and (H2) satisfy. Assume $(y^{*}(\cdot ),u^{*}(\cdot ))$ is an optimal pair of Problem(P). Then there is a solution $\varphi \in L^{\frac{p}{p-1}}(0,T;R^{n})$ of the following adjoint equation

$$\begin{aligned} {\left\{ \begin{array}{ll} {^{C}}D^{\alpha }_{T}\psi (t)=-g_{y}(t,y^{*}(t),y^{*}(t-h),u^{*}(t))-\chi _{(0,T-h)}(t) g_{y_{h}}(t+h,y^{*}(t+h),\\ y^{*}(t),u^{*}(t+h))+f_{y}(t,y^{*}(t),y^{*}(t-h),u^{*}(t))\psi (t)\\ \quad \quad +\chi _{(0,T-h)}(t)f_{y_{h}}(t+h,y^{*}(t+h),\\ y^{*}(t),u^{*}(t+h))\psi (t+h),\psi (t)=0,\quad T\le t\le T+h,\quad h>0, \end{array}\right. } \end{aligned}$$

(3.9)

such that the following maximum condition satisfies:

$$\begin{aligned}&\psi ^{\top }(t)f(t,y^{*}(t),y^{*}(t-h),u^{*}(t))-g(t,y^{*}(t),y^{*}(t-h),u^{*}(t))\nonumber \\&\quad = \max _{u\in U}\big [\psi ^{\top }(t)f(t,y^{*}(t),y^{*}(t-h),u(t))-g(t,y^{*}(t),y^{*}(t-h),u(t))\big ],\nonumber \\&\qquad \qquad \forall u\in U, \quad a.e. \quad t\in [0,T]. \end{aligned}$$

(3.10)

Proof

Proof of theorem consists of two steps.

Step 1. A variational inequality. Let $(y^{*}(\cdot ),u^{*}(\cdot ))$ be an optimal pair of Problem(P). Fix an arbitrary $u(\cdot )\in U^{p}[0,T]$. Insert

$$\begin{aligned} {\left\{ \begin{array}{ll} {\widehat{f}}(s){=}f(s,y^{*}(s),y^{*}(s-h),u(s))-f(s,y^{*}(s),y^{*}(s-h),u^{*}(s))\\ {\widehat{g}}(s){=}g(s,y^{*}(s),y^{*}(s-h),u(s))-g(s,y^{*}(s),y^{*}(s-h),u^{*}(s)), \quad s \in [0,T]. \end{array}\right. } \end{aligned}$$

(3.11)

Taking

$$\begin{aligned} \varphi (s)={\widehat{f}}(s), \quad h(s)={\widehat{g}}(s), \quad \quad s \in [0,T]. \end{aligned}$$

and by using Lemma 2.2, accordingly, for arbitrary $\delta >0$, there exists an $E_{\delta }\in \mathcal {E_{\delta }}$ such that

$$\begin{aligned} {\left\{ \begin{array}{ll} \vert \int _{0}^{t}\bigg (\frac{1}{\delta } 1_{E_{\delta }}(s)-1\bigg )\frac{{\widehat{f}}(s)}{(t-s)^{1-\alpha }}ds\vert \le \delta , \quad t\in [0,T]\\ \vert \int _{0}^{T}\bigg (\frac{1}{\delta } 1_{E_{\delta }}(s)-1\bigg ){\widehat{g}}(s)ds\vert \le \delta . \end{array}\right. } \end{aligned}$$

(3.12)

Signify

$$\begin{aligned} u^{\delta }(t)= {\left\{ \begin{array}{ll} u^{*}(t),\quad t\in [0,T]\setminus E_{\delta },\\ u(t),\quad t\in E_{\delta }. \end{array}\right. } \end{aligned}$$

(3.13)

Obviously, $u^{\delta }(\cdot )\in U^{p}[0,T]$. Set $y^{\delta }(\cdot )=y(\cdot ;y_{0}, u^{\delta }(\cdot ))$ be the appropriate state. Then

$$\begin{aligned} y^{\delta }(t)-y^{*}(t)&= \frac{1}{\Gamma (\alpha )} \int _{0}^{t} \frac{f(s,y^{\delta }(s),y^{\delta }(s-h),u^{\delta }(s))-f(s,y^{*}(s),y^{*}(s-h),u^{*}(s))}{(t-s)^{1-\alpha }}ds\\&=\frac{1}{\Gamma (\alpha )} \int _{0}^{t} \frac{f(s,y^{\delta }(s),y^{\delta }(s-h),u^{\delta }(s))-f(s,y^{*}(s),y^{*}(s-h),u^{\delta }(s))}{(t-s)^{1-\alpha }}ds\\&\quad +\frac{1}{\Gamma (\alpha )} \int _{0}^{t} \frac{f(s,y^{*}(s),y^{*}(s-h),u^{\delta }(s))-f(s,y^{*}(s),y^{*}(s-h),u^{*}(s))}{(t-s)^{1-\alpha }}ds\\&=\frac{1}{\Gamma (\alpha )}\int _{0}^{t} \Bigg (\frac{f^{\delta }_{y}(s)}{(t-s)^{1-\alpha }}[y^{\delta }(s)-y^{*}(s)]+\frac{f^{\delta }_{y_{h}}(s)}{(t-s)^{1-\alpha }}[y^{\delta }(s-h)-y^{*}(s-h)]\\&\quad +1_{E_{\delta }}(s)\frac{{\widehat{f}}(s)}{(t-s)^{1-\alpha }}\Bigg )ds, \quad t\in [0,T], \end{aligned}$$

where

$$\begin{aligned} f^{\delta }_{y}(s)&=\int _{0}^{1} f_{y}(s,y^{*}(s)+\tau (y^{\delta }(s)-y^{*}(s),y^{*}(s-h)\\&\quad +\tau (y^{\delta }(s-h)-y^{*}(s-h)),u^{\delta }(s))d\tau \\ f^{\delta }_{y_{h}}(s)&=\int _{0}^{1} f_{y_{h}}(s,y^{*}(s)+\tau (y^{\delta }(s)-y^{*}(s),y^{*}(s-h)\\&\quad +\tau (y^{\delta }(s-h)-y^{*}(s-h)),u^{\delta }(s))d\tau \end{aligned}$$

By using (H1) and (3.11), we obtain

$$\begin{aligned} \vert {\widehat{f}}(s)\vert&=\vert f(s,y^{*}(s),y^{*}(s-h),u(s))-f(s,y^{*}(s),y^{*}(s-h),u^{*}(s))\vert \\&\le 2 L_{0}(s)+L(s)[2\vert y^{*}(s)\vert +2\vert y^{*}(s-h)\vert \\&\quad +\rho (u(s),u_{0})+\rho (u^{*}(s),u_{0})]\equiv \varphi (s), \quad s\in [0,T],\\ \vert f^{\delta }_{y}(s)\vert&\le L(s), \quad \vert f^{\delta }_{y_{h}}(s)\vert \le L(s), \quad s\in [0,T]. \end{aligned}$$

Obviously, $\varphi (\cdot )\in L^{q}(0,T)$ for some $q\in (\frac{1}{\alpha },p)$. We get

$$\begin{aligned} \vert y^{\delta }(t)-y^{*}(t)\vert&\le \frac{1}{\Gamma (\alpha )}\int _{0}^{t} \frac{L(s)}{(t-s)^{1-\alpha }}\vert y^{\delta }(s)-y^{*}(s)\vert ds\\&\quad +\frac{1}{\Gamma (\alpha )}\int _{0}^{t}\frac{L(s)}{(t-s)^{1-\alpha }}\vert y^{\delta }(s-h)-y^{*}(s-h)\vert ds\\&\quad +\frac{1}{\Gamma (\alpha )}\int _{0}^{t}1_{E_{\delta }}(s)\frac{\varphi (s)}{(t-s)^{1-\alpha }}ds, \quad t\in [0,T]. \end{aligned}$$

By applying the Gronwall’s inequality (Lemma 2.1), selecting $q^{\prime }\in (\frac{1}{\alpha },q)$, and using Hölder inequality,

we have

$$\begin{aligned}&\vert y^{\delta }(t)-y^{*}(t)\vert \\&\quad \le \int _{0}^{t}1_{E_{\delta }}(s)\frac{\varphi (s)}{(t-s)^{1-\alpha }}ds+K\int _{0}^{t} \frac{L(s)}{(t-s)^{1-\alpha }}\int _{0}^{s}1_{E_{\delta }}(\tau )\frac{\varphi (\tau )}{(s-\tau )^{1-\alpha }}d\tau ds\\&\qquad + K\int _{0}^{t} \frac{L(s)}{(t-s)^{1-\alpha }}\int _{0}^{s-h}1_{E_{\delta }}(\tau )\frac{\varphi (\tau )}{(s-h-\tau )^{1-\alpha }}d\tau ds\\&\qquad +K\int _{0}^{t-h} \frac{L(s)}{(t-h-s)^{1-\alpha }}\int _{0}^{s}1_{E_{\delta }}(\tau )\frac{\varphi (\tau )}{(s-\tau )^{1-\alpha }}d\tau ds\\&\qquad +K\int _{0}^{t-h} \frac{L(s)}{(t-h-s)^{1-\alpha }}\int _{0}^{s-h}1_{E_{\delta }}(\tau )\frac{\varphi (\tau )}{(s-h-\tau )^{1-\alpha }}d\tau ds\\&\qquad +\cdots +K\int _{0}^{t-(n-1)h} \frac{L(s)}{(t-(n-1)h-s)^{1-\alpha }}\int _{0}^{s}1_{E_{\delta }}(\tau )\frac{\varphi (\tau )}{(s-\tau )^{1-\alpha }}d\tau ds\\&\qquad +K\int _{0}^{t-(n-1)h} \frac{L(s)}{(t-(n-1)h-s)^{1-\alpha }}\int _{0}^{s-h}1_{E_{\delta }}(\tau )\frac{\varphi (\tau )}{(s-h-\tau )^{1-\alpha }}d\tau ds\\&\quad \le \bigg (\int _{0}^{t}1_{E_{\delta }}(s)^{q^{\prime }} \varphi (s)^{q^{\prime }}ds\bigg )^{\frac{1}{q^{\prime }}}+K\int _{0}^{t}\frac{L(s)}{(t-s)^{1-\alpha }}\bigg (\int _{0}^{s}1_{E_{\delta }}(\tau )^{q^{\prime }} \varphi (\tau )^{q^{\prime }}d\tau \bigg )^{\frac{1}{q^{\prime }}}ds\\&\qquad +K\int _{0}^{t}\frac{L(s)}{(t-s)^{1-\alpha }}\bigg (\int _{0}^{s-h}1_{E_{\delta }}(\tau )^{q^{\prime }} \varphi (\tau )^{q^{\prime }}d\tau \bigg )^{\frac{1}{q^{\prime }}}ds\\&\qquad +K\int _{0}^{t-h}\frac{L(s)}{(t-h-s)^{1-\alpha }}\bigg (\int _{0}^{s}1_{E_{\delta }}(\tau )^{q^{\prime }} \varphi (\tau )^{q^{\prime }}d\tau \bigg )^{\frac{1}{q^{\prime }}}ds\\&\qquad +K\int _{0}^{t-h}\frac{L(s)}{(t-h-s)^{1-\alpha }}\bigg (\int _{0}^{s-h}1_{E_{\delta }}(\tau )^{q^{\prime }} \varphi (\tau )^{q^{\prime }}d\tau \bigg )^{\frac{1}{q^{\prime }}}ds\\&\qquad +\cdots +K\int _{0}^{t-(n-1)h}\frac{L(s)}{(t-(n-1)h-s)^{1-\alpha }}\bigg (\int _{0}^{s}1_{E_{\delta }}(\tau )^{q^{\prime }} \varphi (\tau )^{q^{\prime }}d\tau \bigg )^{\frac{1}{q^{\prime }}}ds\\&\qquad +K\int _{0}^{t-(n-1)h}\frac{L(s)}{(t-(n-1)h-s)^{1-\alpha }}\bigg (\int _{0}^{s-h}1_{E_{\delta }}(\tau )^{q^{\prime }} \varphi (\tau )^{q^{\prime }}d\tau \bigg )^{\frac{1}{q^{\prime }}}ds\\&\quad \le K^{\prime }\vert E_{\delta }\vert ^{\frac{q-q^{\prime }}{q^{\prime }q}}\le K^{\prime }\delta ^{\frac{q-q^{\prime }}{q^{\prime }q}}\rightarrow 0 \end{aligned}$$

as $\delta \rightarrow 0$, uniformly in $t\in [0,T]$. Denote

$$\begin{aligned} Y^{\delta }(t)=\frac{y^{\delta }(t)-y^{*}(t)}{\delta }, \quad t\in [0,T]. \end{aligned}$$

It follows $Y^{\delta }(\cdot )$ be a solution of the following:

$$\begin{aligned} Y^{\delta }(t)=&\frac{1}{\Gamma (\alpha )}\int _{0}^{t}\bigg [ \frac{f^{\delta }_{y}(s)}{(t-s)^{1-\alpha }}Y^{\delta }(s)+\frac{f^{\delta }_{y_{h}}(s)}{(t-s)^{1-\alpha }}Y^{\delta }(s-h)\\&+\frac{1_{E_{\delta }}(s)}{\delta }\frac{{\widehat{f}}(s)}{(t-s)^{1-\alpha }}\bigg ]ds, \quad t\in [0,T]. \end{aligned}$$

Let be a solution of the following variational equation

$$\begin{aligned} Y(t)&=\frac{1}{\Gamma (\alpha )}\int _{0}^{t}\bigg [ \frac{f_{y}(s,y^{*}(s),y^{*}(s-h),u^{*}(s))}{(t-s)^{1-\alpha }}Y(s)\nonumber \\ {}&\quad +\frac{f_{y_{h}}(s,y^{*}(s),y^{*}(s-h),u^{*}(s))}{(t-s)^{1-\alpha }}Y(s-h) +\frac{{\widehat{f}}(s)}{(t-s)^{1-\alpha }}\bigg ]ds \end{aligned}$$

(3.14)

We get

$$\begin{aligned} \vert Y(t)\vert&\le \frac{1}{\Gamma (\alpha )}\int _{0}^{t}\frac{\varphi (s)}{(t-s)^{1-\alpha }}ds+\frac{1}{\Gamma (\alpha )}\int _{0}^{t}\frac{L(s)}{(t-s)^{1-\alpha }}\vert Y(s)\vert ds\\&\quad +\frac{1}{\Gamma (\alpha )}\int _{0}^{t}\frac{L(s)}{(t-s)^{1-\alpha }}\vert Y(s-h)\vert ds, \quad t\in [0,T]. \end{aligned}$$

By applying the Gronwall’s inequality (Lemma 2.1), mention $q>\frac{1}{\alpha }$ and $L(\cdot )\in L^{\frac{1}{\alpha }+}(0,T)$,

$$\begin{aligned} \vert Y(t)\vert&\le \frac{1}{\Gamma (\alpha )}\int _{0}^{t}\frac{\varphi (s)}{(t-s)^{1-\alpha }}ds+\frac{K}{(\Gamma (\alpha ))^{2}}\int _{0}^{t}\frac{L(s)}{(t-s)^{1-\alpha }}\int _{0}^{s}\frac{\varphi (\tau )}{(s-\tau )^{1-\alpha }}d\tau ds\\&\quad +\frac{K}{(\Gamma (\alpha ))^{2}}\int _{0}^{t}\frac{L(s)}{(t-s)^{1-\alpha }}\int _{0}^{s-h}\frac{\varphi (\tau )}{(s-h-\tau )^{1-\alpha }}d\tau ds\\&\quad +\frac{K}{(\Gamma (\alpha ))^{2}}\int _{0}^{t-h}\frac{L(s)}{(t-h-s)^{1-\alpha }}\int _{0}^{s}\frac{\varphi (\tau )}{(s-\tau )^{1-\alpha }}d\tau ds\\&\quad +\frac{K}{(\Gamma (\alpha ))^{2}}\int _{0}^{t-h}\frac{L(s)}{(t-h-s)^{1-\alpha }}\int _{0}^{s-h}\frac{\varphi (\tau )}{(s-h-\tau )^{1-\alpha }}d\tau ds\\&\quad +\cdots +\frac{K}{(\Gamma (\alpha ))^{2}}\int _{0}^{t-(n-1)h}\frac{L(s)}{(t-(n-1)h-s)^{1-\alpha }}\int _{0}^{s}\frac{\varphi (\tau )}{(s-\tau )^{1-\alpha }}d\tau ds\\&\quad +\frac{K}{(\Gamma (\alpha ))^{2}}\int _{0}^{t-(n-1)h}\frac{L(s)}{(t-(n-1)h-s)^{1-\alpha }}\int _{0}^{s-(n-1)h}\frac{\varphi (\tau )}{(s-h-\tau )^{1-\alpha }}d\tau ds\\&\le K^{\prime }\Vert \varphi (\cdot )\Vert _{q}, \quad t\in [0,T]. \end{aligned}$$

Therefore,

$$\begin{aligned} Y^{\delta }(t)-Y(t)&=\frac{1}{\Gamma (\alpha )}\int _{0}^{t}\bigg [ \frac{f^{\delta }_{y}(s)}{(t-s)^{1-\alpha }}Y^{\delta }(s)\\&\quad -\frac{f_{y}(s,y^{*}(s),y^{*}(s-h),u^{*}(s))}{(t-s)^{1-\alpha }}Y(s)\bigg ]ds\\&\quad +\frac{1}{\Gamma (\alpha )}\int _{0}^{t}\bigg [\frac{f^{\delta }_{y_{h}}(s)}{(t-s)^{1-\alpha }}Y^{\delta }(s-h)\\&\quad -\frac{f_{y_{h}}(s,y^{*}(s),y^{*}(s-h),u^{*}(s))}{(t-s)^{1-\alpha }}Y(s-h)\bigg ]ds\\&\quad +\frac{1}{\Gamma (\alpha )}\int _{0}^{t}\bigg (\frac{1_{E_{\delta }}(s)}{\delta }-1\bigg )\frac{{\widehat{f}}(s)}{(t-s)^{1-\alpha }}ds\\&=\frac{1}{\Gamma (\alpha )}\int _{0}^{t}\frac{f^{\delta }_{y}(s)}{(t-s)^{1-\alpha }}[Y^{\delta }(s)-Y(s)]ds\\&\quad +\int _{0}^{t}\frac{f^{\delta }_{y_{h}}(s)}{(t-s)^{1-\alpha }}[Y^{\delta }(s-h)-Y(s-h)]ds\\&\quad +\frac{1}{\Gamma (\alpha )}\int _{0}^{t}\frac{f^{\delta }_{y}(s)-f_{y}(s,y^{*}(s),y^{*}(s-h),u^{*}(s))}{(t-s)^{1-\alpha }}Y(s)ds\\&\quad +\frac{1}{\Gamma (\alpha )}\int _{0}^{t}\frac{f^{\delta }_{y_{h}}(s)-f_{y_{h}}(s,y^{*}(s),y^{*}(s-h),u^{*}(s))}{(t-s)^{1-\alpha }}Y(s-h)ds\\&\quad +\int _{0}^{t}\bigg (\frac{1_{E_{\delta }}(s)}{\delta }-1\bigg )\frac{{\widehat{f}}(s)}{(t-s)^{1-\alpha }}ds\\&\equiv \frac{1}{\Gamma (\alpha )}\int _{0}^{t}\frac{f^{\delta }_{y}(s)}{(t-s)^{1-\alpha }}[Y^{\delta }(s)-Y(s)]ds\\&\quad +\frac{1}{\Gamma (\alpha )}\int _{0}^{t}\frac{f^{\delta }_{y_{h}}(s)}{(t-s)^{1-\alpha }}[Y^{\delta }(s-h)-Y(s-h)]ds\\&\quad +a^{\delta }_{1}(t)+a^{\delta }_{2}(t)+a^{\delta }_{3}(t), \quad t\in [0,T]. \end{aligned}$$

Since

$$\begin{aligned}&\Big \vert \frac{f^{\delta }_{y}(s)-f_{y}(s,y^{*}(s),y^{*}(s-h),u^{*}(s))}{(t-s)^{1-\alpha }}Y(s)\Big \vert \\&\quad \le \frac{2L(s)}{(t-s)^{1-\alpha }}\vert Y(s)\vert , \quad a.e.\quad s\in [0,t),\\&\Big \vert \frac{f^{\delta }_{y_{h}}(s)-f_{y_{h}}(s,y^{*}(s),y^{*}(s-h),u^{*}(s))}{(t-s)^{1-\alpha }}Y(s-h)\Big \vert \\&\quad \le \frac{2L(s)}{(t-s)^{1-\alpha }}\vert Y(s-h)\vert , \quad a.e.\quad s\in [0,t). \end{aligned}$$

with

$$\begin{aligned} \int _{0}^{t}\frac{2L(s)}{(t-s)^{1-\alpha }}\vert Y(s)\vert ds<\infty ,\\ \int _{0}^{t}\frac{2L(s)}{(t-s)^{1-\alpha }}\vert Y(s-h)\vert ds<\infty . \end{aligned}$$

By using the dominated convergence theorem, we get

$$\begin{aligned} \lim _{\delta \rightarrow 0}a^{\delta }_{1}(t)=0, \quad \lim _{\delta \rightarrow 0}a^{\delta }_{2}(t)=0, \quad t\in [0,T]. \end{aligned}$$

In addition, by (3.12),

$$\begin{aligned} a^{\delta }_{3}(t)\equiv \Big \vert \frac{1}{\Gamma (\alpha )}\int _{0}^{t}\bigg (\frac{1_{E_{\delta }}(s)}{\delta }-1\bigg )\frac{{\widehat{f}}(s)}{(t-s)^{1-\alpha }}ds \Big \vert \le \delta , \quad t\in [0,T]. \end{aligned}$$

Therefore,

$$\begin{aligned} \vert Y^{\delta }(t)-Y(t) \vert&\le \frac{1}{\Gamma (\alpha )}\int _{0}^{t}\frac{L(s)}{(t-s)^{1-\alpha }}\vert Y^{\delta }(s)-Y(s)\vert ds\\&\quad + \frac{1}{\Gamma (\alpha )}\int _{0}^{t}\frac{L(s)}{(t-s)^{1-\alpha }}\vert Y^{\delta }(s-h)-Y(s-h)\vert ds\\&\quad +\vert a^{\delta }_{1}(t)\vert +\vert a^{\delta }_{2}(t)\vert +\vert a^{\delta }_{3}(t)\vert , \quad t\in [0,T]. \end{aligned}$$

Applying the Gronwall’s inequality,

$$\begin{aligned}&\vert Y^{\delta }(t)-Y(t) \vert \le \vert a^{\delta }_{1}(t)\vert +\vert a^{\delta }_{2}(t)\vert +\vert a^{\delta }_{3}(t)\vert \\&\quad +\frac{K}{\Gamma (\alpha )}\int _{0}^{t}\frac{L(s)}{(t-s)^{1-\alpha }}(\vert a^{\delta }_{1}(s)\vert +\vert a^{\delta }_{2}(s)\vert +\vert a^{\delta }_{3}(s)\vert ) ds\\&\quad + \frac{K}{\Gamma (\alpha )} \int _{0}^{t}\frac{L(s)}{(t-s)^{1-\alpha }}(\vert a^{\delta }_{1}(s-h)\vert +\vert a^{\delta }_{2}(s-h)\vert +\vert a^{\delta }_{3}(s-h)\vert ) ds\\&\quad + \frac{K}{\Gamma (\alpha )}\int _{0}^{t-h}\frac{L(s)}{(t-h-s)^{1-\alpha }}(\vert a^{\delta }_{1}(s)\vert +\vert a^{\delta }_{2}(s)\vert +\vert a^{\delta }_{3}(s)\vert ) ds\\&\quad +\frac{K}{\Gamma (\alpha )} \int _{0}^{t-h}\frac{L(s)}{(t-h-s)^{1-\alpha }}(\vert a^{\delta }_{1}(s-h)\vert +\vert a^{\delta }_{2}(s-h)\vert +\vert a^{\delta }_{3}(s-h)\vert ) ds\\&\quad +\cdots +\frac{K}{\Gamma (\alpha )}\int _{0}^{t-(n-1)h}\frac{L(s)}{(t-(n-1)h-s)^{1-\alpha }}(\vert a^{\delta }_{1}(s)\vert +\vert a^{\delta }_{2}(s)\vert +\vert a^{\delta }_{3}(s)\vert ) ds\\&\quad +\frac{K}{\Gamma (\alpha )} \int _{0}^{t-(n-1)h}\frac{L(s)}{(t-(n-1)h-s)^{1-\alpha }}(\vert a^{\delta }_{1}(s-h)\vert \\&\quad +\vert a^{\delta }_{2}(s-h)\vert +\vert a^{\delta }_{3}(s-h)\vert ) ds, \quad t\in [0,T]. \end{aligned}$$

As a result, using the dominated convergence theorem, we get

$$\begin{aligned} \lim _{\delta \rightarrow 0} \vert Y^{\delta }(t)-Y(t) \vert =0, \quad t\in [0,T]. \end{aligned}$$

With optimality of $(y^{*}(\cdot ),u^{*}(\cdot ))$, we have

$$\begin{aligned} 0\le&\frac{J(u^{\delta }(\cdot )-J(u^{*}(\cdot )))}{\delta }=\int _{0}^{T}\left[ g^{\delta }_{y}(t)Y^{\delta }(t)+g^{\delta }_{y_{h}}(t)Y^{\delta }(t-h)+\frac{1}{\delta }1_{E_{\delta }}(t){\widehat{g}}(t)\right] dt \end{aligned}$$

where ${\widehat{g}}(\cdot )$ is denoted in (3.11), and

$$\begin{aligned} g^{\delta }_{y}(t)&=\int _{0}^{1} g_{y}(t,y^{*}(t)+\tau (y^{\delta }(t)-y^{*}(t),y^{*}(t-h)\\&\quad +\tau (y^{\delta }(t-h)-y^{*}(t-h)),u^{\delta }(t))d\tau ,\\ g^{\delta }_{y_{h}}(t)&= \int _{0}^{1} g_{y_{h}}(t,y^{*}(t)+\tau (y^{\delta }(t)-y^{*}(t),y^{*}(t-h)\\&\quad +\tau (y^{\delta }(t-h)-y^{*}(t-h)),u^{\delta }(t))d\tau , \quad t\in [0,T]. \end{aligned}$$

Therefore, by utilizing (3.12) and the convergence $y^{\delta }(t)\rightarrow y^{*}(t)$,    $t\in [0,T]$, it follows

$$\begin{aligned} 0&\le \int _{0}^{T}\Big ( g_{y}(t,y^{*}(t),y^{*}(t-h),u^{*}(t))Y(t)+g_{y_{h}}(t,y^{*}(t),y^{*}(t-h),u^{*}(t))Y(t-h)\\&\quad +g(t,y^{*}(t),y^{*}(t-h),u(t))-g(t,y^{*}(t),y^{*}(t-h),u^{*}(t))\Big )dt. \end{aligned}$$

Step 2. Duality. Suppose $\psi (\cdot )$ be the solution of the adjoint equation (3.9). Than we get

$$\begin{aligned} 0&\le \int _{0}^{T}\Big (g_{y}(s,y^{*}(s),y^{*}(s-h),u^{*}(s))Y(s)+g_{y_{h}}(s,y^{*}(s),y^{*}(s-h),u^{*}(s))Y(s-h)\\&\quad +g(s,y^{*}(s),y^{*}(s-h),u(s))-g(s,y^{*}(s),y^{*}(s-h),u^{*}(s))\Big )ds\\&=\int _{0}^{T}\Big (g_{y}(s,y^{*}(s),y^{*}(s-h),u^{*}(s))\\&\quad + \chi _{(0,T-h)}(s) g_{y_{h}}(s+h,y^{*}(s+h),y^{*}(s),u^{*}(s+h))\Big )Y(s)ds\\&\quad +\int _{0}^{T}g(s,y^{*}(s),y^{*}(s-h),u(s))-g(s,y^{*}(s),y^{*}(s-h),u^{*}(s))ds\\&=\int _{0}^{T}\Big (-{^{C}}D^{\alpha }_{T}\psi (s)+f_{y}(s,y^{*}(s),y^{*}(s-h),u^{*}(s))\psi (s)\\&\quad +\chi _{(0,T-h)}(s)f_{y_{h}}(s+h,y^{*}(s+h),y^{*}(s),u^{*}(s+h))\psi (s+h)\Big )Y(s)ds\\&\quad +\int _{0}^{T}g(s,y^{*}(s),y^{*}(s-h),u(s))-g(s,y^{*}(s),y^{*}(s-h),u^{*}(s))ds\\&=\int _{0}^{T}\Big (-{^{C}}D^{\alpha }_{T}\psi (s)\Big )Y(s)ds+\int _{0}^{T}\Big (f_{y}(s,y^{*}(s),y^{*}(s-h),u^{*}(s))\psi (s)\\&\quad +\chi _{(0,T-h)}(s)f_{y_{h}}(s+h,y^{*}(s+h),y^{*}(s),u^{*}(s+h))\psi (s+h)\Big )Y(s)ds\\&\quad +\int _{0}^{T}g(s,y^{*}(s),y^{*}(s-h),u(s))-g(s,y^{*}(s),y^{*}(s-h),u^{*}(s))ds\\&=\int _{0}^{T}\psi (s)\Big (-{^{C}}D^{\alpha }_{0}Y(s)\Big )ds+\int _{0}^{T}f_{y}(s,y^{*}(s),y^{*}(s-h),u^{*}(s))\psi (s)Y(s)ds\\&\quad +\int _{0}^{T}\chi _{(0,T-h)}(s)f_{y_{h}}(s+h,y^{*}(s+h),y^{*}(s),u^{*}(s+h))\psi (s+h)\Big )Y(s)ds\\&\quad +\int _{0}^{T}g(s,y^{*}(s),y^{*}(s-h),u(s))-g(s,y^{*}(s),y^{*}(s-h),u^{*}(s))ds\\&=\int _{0}^{T}\psi (s)\bigg [ -{^{C}}D^{\alpha }_{0}Y(s)+f_{y}(s,y^{*}(s),y^{*}(s-h),u^{*}(s))Y(s)\bigg ]ds\\&\quad +\int _{h}^{T+h}\chi _{(h,T)}(s-h)f_{y_{h}}(s,y^{*}(s),y^{*}(s-h),u^{*}(s))\psi (s)Y(s-h) ds\\&\quad +\int _{0}^{T}g(s,y^{*}(s),y^{*}(s-h),u(s))-g(s,y^{*}(s),y^{*}(s-h),u^{*}(s))ds\\&=\int _{0}^{T}\psi (s)\bigg [ -{^{C}}D^{\alpha }_{0}Y(s)+f_{y}(s,y^{*}(s),y^{*}(s-h),u^{*}(s))Y(s)\\&\quad +\chi _{(h,T)}(s-h)f_{y_{h}}(s,y^{*}(s),y^{*}(s-h),u^{*}(s))Y(s-h) \bigg ]ds\\&\quad +\int _{0}^{T}g(s,y^{*}(s),y^{*}(s-h),u(s))-g(s,y^{*}(s),y^{*}(s-h),u^{*}(s))ds\\&=\int _{0}^{T}-\psi ^{\top }(s){\widehat{f}}(s)ds+\int _{0}^{T}{\widehat{g}}(s)ds=\int _{0}^{T}[{\widehat{g}}(s)-\psi ^{\top }(s){\widehat{f}}(s)]ds\\&\equiv \int _{0}^{T}[\Psi (s,u^{*}(s))-\Psi (s,u(s))]ds \end{aligned}$$

where

$$\begin{aligned}&\Psi (t,u(t))=\psi ^{\top }(t)f(t,y^{*}(t),y^{*}(t-h),u(t))-g(t,y^{*}(t),y^{*}(t-h),u(t)) \end{aligned}$$

and $Y(\cdot )$ is the solution of the following differential equation:

$$\begin{aligned} {\left\{ \begin{array}{ll} &{}{^{C}}D^{\alpha }_{0}Y(s)=f_{y}(t,y^{*}(t),y^{*}(t-h),u^{*}(t))Y(t) \\ &{}+\chi _{(h,T)}(t-h)f_{y_{h}}(t,y^{*}(t),y^{*}(t-h),u^{*}(t))Y(t-h)+{\widehat{f}}(t)\\ &{}Y(t)=0, \quad -h\le t\le 0, \quad h>0. \end{array}\right. } \end{aligned}$$

Let $u\in U$ be fixed and let $s_{0}\in (0,T)$ be a Lebesgue point of $\Psi (s,u^{*}(s))-\Psi (s,u(s))$. Then for any $\delta >0$, let

$$\begin{aligned} u(s)= {\left\{ \begin{array}{ll} u^{*}(s),\quad s\in [0,T]\setminus (s_{0}-\delta , s_{0}+\delta ),\\ u,\quad t\in (s_{0}-\delta , s_{0}+\delta ). \end{array}\right. } \end{aligned}$$

Therefore, from the above, we get

$$\begin{aligned} 0&\le \frac{1}{2\delta }=\int _{s_{0}-\delta }^{s_{0}+\delta }[\Psi (s,u^{*}(s))-\Psi (s,u(s))]ds\rightarrow \Psi (s_{0},u^{*}(s_{0}))\\&\quad -\Psi (s_{0},u(s_{0})),\quad \delta \rightarrow 0. \end{aligned}$$

Consequently, by the Lebesgue point theorem, we get the maximum condition (3.10). $\square$

3.2 Maximum principle for the delayed Volterra equation with the weak singular kernel

Let us research the following issue

$$\begin{aligned} {\left\{ \begin{array}{ll} y(t)=\eta (t)+\int _{0}^{t}\frac{f(t,s,y(s),y(s-h),u(s))}{(t-s)^{1-\alpha }}ds, \quad a.e.\quad t\in [0,T],\\ y(t)=0,\quad -h\le t\le 0,\quad h>0. \end{array}\right. } \end{aligned}$$

(3.15)

where $\eta (\cdot )\in L^{p}(0,T;R)$.

$$\begin{aligned} J(u(\cdot ))=\int _{0}^{T} g(t,y(t),y(t-h),u(t))dt. \end{aligned}$$

(3.16)

Now, we discuss the optimal control problem for (3.15) with payoff functional (3.16). We introduce certain assumptions that are considered to be more than enough for our requirements. Therefore, we will make use of these conditions without any loss of generality.

($H1^{\prime }$): Let $f: \Delta \times R^{n}\times R^{n}\times U\rightarrow R^{n}$ be a transformation with $(t,s)\mapsto f(t,s,y,y_{h},u)$ being measurable, $(y,y_{h})\mapsto f(t,s,y,y_{h},u)$ being continuously differentiable, $(y,y_{h},u)\mapsto f(t,s,y,y_{h},u)$, $(y,y_{h},u)\mapsto f_{y}(t,s,y,y_{h},u)$ and $(y,y_{h},u)\mapsto f_{y_{h}}(t,s,y,y_{h},u)$being continuous. Furthermore, there are non-negative functions involved in these conditions. $L_{0}(\cdot ), L(\cdot )$ with

$$\begin{aligned}&L_{0}(\cdot )\in L^{\frac{1}{\alpha }{+}}(0,T),\quad L(\cdot )\in L^{\frac{p}{p\alpha -1}{+}}(0,T) \end{aligned}$$

(3.17)

for some $p>\frac{1}{\alpha }$ and $\alpha \in (0,1), u_{0}\in U$.

$$\begin{aligned}&\vert f(t,s,0,0,u_{0})\vert \le L_{0}(s), \quad (t,s)\in \Delta , \end{aligned}$$

(3.18)

$$\begin{aligned}&\vert f(t,s,y,y_{h},u)-f(t,s,y^{\prime },y^{\prime }_{h},u^{\prime })\vert \le L(s)[\vert y-y^{\prime }\vert +\vert y_{h}-y^{\prime }_{h}\vert +\rho (u,u^{\prime })],\nonumber \\&\quad (t,s)\in \Delta , \quad y,y^{\prime }, y_{h},y^{\prime }_{h}\in R^{n}, \quad u,u^{\prime }\in U. \end{aligned}$$

(3.19)

We point out (3.18)–(3.19) declare

$$\begin{aligned} \vert f(t,s,y,y_{h},u)\vert&\le L_{0}(s)+L(s)[\vert y\vert +\vert y_{h}\vert +\rho (u,u_{0})],\nonumber \\&\quad (t,s,y,y_{h},u)\in \Delta \times R^{n}\times R^{n} \times U. \end{aligned}$$

(3.20)

$$\begin{aligned} \vert f(t,s,y,y_{h},u)-f(t^{\prime },y,y_{h},u)\vert&\le K\omega (\vert t-t^{\prime }\vert )(1+\vert y\vert +\vert y_{h}\vert ), \nonumber \\&\quad (t,s),(t^{\prime },s)\in \Delta ,\quad y,y_{h}\in R^{n}, \quad u\in U, \end{aligned}$$

(3.21)

for some modulus of continuity $\omega (\cdot )$. Moreover, it is evident that L is included in a smaller space, compared to the space to which $L_{0}$ belongs.

($H2^{\prime }$): Let $g: [0,T]\times R^{n}\times R^{n}\times U\rightarrow R$ be a transformation with $t\mapsto g(t,y,y_{h},u)$ being measurable, $(y,y_{h})\mapsto g(t,y,y_{h},u)$ being continuously differentiable, and $(y,y_{h},u)\mapsto (g(t,y,y_{h},u),g_{y}(t,y,y_{h},u),g_{y_{h}}(t,y,y_{h},u))$ being continuous. There is a constant $L>0$ such that

$$\begin{aligned}&\vert g(t,0,0,u)\vert + \vert g_{y}(t,0,0,u)\vert + \vert g_{y_{h}}(t,0,0,u)\vert \\&\quad \le L, \quad (t,y,y_{h},u)\in [0,T]\times R^{n}\times R^{n}\times U. \end{aligned}$$

Given that conditions ($H1^{\prime }$) and ($H2^{\prime }$) hold, we may proceed to analyze the optimal control problem (OCP) defined by the payoff functional (3.16).

Problem ($P^{\prime }$) Find a $u^{*}(\cdot )\in U^{p}[0,T]$ such that

$$\begin{aligned} J(u^{*}(\cdot ))=\inf _{u(\cdot )\in U^{p}[0,T]}J(u(\cdot )). \end{aligned}$$

(3.22)

An optimal control for Problem ($P^{\prime }$) is any arbitrary $u^{*}(\cdot )$ that satisfies Eq. (3.22). The corresponding state $y^{*}(\cdot )$ is known as an optimal state, and when combined with the optimal control $u^{*}$, they form the optimal pair $(y^{*},u^{*})$.

Now, Pontryagin’s maximum principle can be applied to Problem ($P^{\prime }$) and can be restated as follows.

Lemma 3.1

Let ($H1^{\prime }$) and ($H2^{\prime }$) satisfy. Then the following result holds:

$$\begin{aligned} \sup _{t\in [0,T]}\vert y^{\delta }(t)-y^{*}-\delta Y(t)\vert =o(\delta ), \end{aligned}$$

(3.23)

where Y is the solution to the first variational equation related to the optimal pair $(y^{*},u^{*}(\cdot ))\in R^n\times U^p[0,T]$ given by

$$\begin{aligned} Y(t)=&\int _{0}^{t}\bigg [ \frac{f_{y}(t,s,y^{*}(s),y^{*}(s-h),u^{*}(s))}{(t-s)^{1-\alpha }}Y(s)\nonumber \\&+\frac{f_{y_{h}}(t,s,y^{*}(s),y^{*}(s-h),u^{*}(s))}{(t-s)^{1-\alpha }}Y(s-h) +\frac{{\widehat{f}}(t,s)}{(t-s)^{1-\alpha }}\bigg ]ds \end{aligned}$$

(3.24)

with for any $u(\cdot )\in U^p[0,T],$

$$\begin{aligned} {\widehat{f}}(t,s)=&\,f(t,s,y^{*}(s),y^{*}(s-h),u(s))\nonumber \\&\quad -f(t,s,y^{*}(s),y^{*}(s-h),u^{*}(s)), \quad (t,s) \in \Delta . \end{aligned}$$

(3.25)

Proof

It is obvious from step 1 in the proof of Theorem 3.1. $\square$

Theorem 3.2

Let ($H1^{\prime }$) and ($H2^{\prime }$) satisfy. Assume $(y^{*}(\cdot ),u^{*}(\cdot ))$ is an optimal pair of Problem($P^{\prime }$). Then there is a solution $\varphi \in L^{\frac{p}{p-1}}(0,T;R^{n})$ of the following adjoint equation

$$\begin{aligned} \psi (t)&=-g_{y}(t,y^{*}(t),y^{*}(t-h),u^{*}(t))^{\top }-g_{y_{h}}(t,y^{*}(t),y^{*}(t-h),u^{*}(t))^{\top }\nonumber \\&\quad +\int _{s}^{T}\frac{f_{y}(t,s,y^{*}(s),y^{*}(s-h),u^{*}(s))^{\top }}{(t-s)^{1-\alpha }}\psi (t)dt \nonumber \\&\quad +\int _{s+h}^{T}\frac{f_{y_{h}}(t,s+h,y^{*}(s+h),y^{*}(s),u^{*}(s+h))^{\top }}{(t-h-s)^{1-\alpha }}\psi (t)dt \quad a.e.\quad t\in [0,T],\nonumber \\ \psi (t)&=0, \quad T\le t \le T+h, \quad h>0. \end{aligned}$$

(3.26)

such that the following maximum condition satisfies:

$$\begin{aligned}&\int _{s}^{T}\psi (t)^{\top } \frac{f(t,s,y^{*}(s),y^{*}(s-h),u^{*}(s))}{(t-s)^{1-\alpha }}dt -g(s,y^{*}(s),y^{*}(s-h),u^{*}(s))\nonumber \\&\quad =\max _{u\in U}\Bigg [\int _{s}^{T}\psi (t)^{\top } \frac{f(t,s,y^{*}(s),y^{*}(s-h),u(s))}{(t-s)^{1-\alpha }}dt \nonumber \\&\qquad -g(s,y^{*}(s),y^{*}(s-h),u(s))\Bigg ], \quad \forall u\in U,\quad a.e. \quad t\in [0,T]. \end{aligned}$$

(3.27)

Proof

Based on Theorem 3.1, and Lemma 3.1, we have

$$\begin{aligned} 0&\le \int _{0}^{T}\Big ( g_{y}(t,y^{*}(t),y^{*}(t-h),u^{*}(t))Y(t)+g_{y_{h}}(t,y^{*}(t),y^{*}(t-h),u^{*}(t))Y(t-h)\\&\quad + g(t,y^{*}(t),y^{*}(t-h),u(t))-g(t,y^{*}(t),y^{*}(t-h),u^{*}(t))\Big )dt. \end{aligned}$$

Similarly, assume $\psi (\cdot )$ be the solution of the adjoint equation (3.26). Than we get

$$\begin{aligned} 0&\le \int _{0}^{T}\Big ( g_{y}(s,y^{*}(s),y^{*}(s-h),u^{*}(s))Y(s)+g_{y_{h}}(s,y^{*}(s),y^{*}(s-h),u^{*}(s))Y(s-h)\\&\quad +g(s,y^{*}(s),y^{*}(s-h),u(s))-g(s,y^{*}(s),y^{*}(s-h),u^{*}(s))\Big )ds\\&=\int _{0}^{T}\big ( -\psi (s)+\int _{s}^{T}\frac{f_{y}(t,s,y^{*}(s),y^{*}(s-h),u^{*}(s))}{(t-s)^{1-\alpha }}\psi (t)dt\\&\quad +\int _{s+h}^{T}\frac{f_{y_{h}}(t,s+h,y^{*}(s+h),y^{*}(s),u^{*}(s+h))}{(t-h-s)^{1-\alpha }}\psi (t)dt\big )^{\top }Y(s)ds\\&\quad +\int _{0}^{T}g(s,y^{*}(s),y^{*}(s-h),u(s))-g(s,y^{*}(s),y^{*}(s-h),u^{*}(s))\Big )ds\\&=\int _{0}^{T}\psi (t)^{\top }\big (-Y(t)+\int _{0}^{t}\frac{f_{y}(t,s,y^{*}(s),y^{*}(s-h),u^{*}(s))}{(t-s)^{1-\alpha }}Y(s)ds\\&\quad +\int _{0}^{t}\frac{f_{y_{h}}(t,s,y^{*}(s),y^{*}(s-h),u^{*}(s))}{(t-s)^{1-\alpha }}Y(s-h)ds\big )dt\\&\quad +\int _{0}^{T}g(s,y^{*}(s),y^{*}(s-h),u(s))-g(s,y^{*}(s),y^{*}(s-h),u^{*}(s))\Big )ds\\&=\int _{0}^{T}\bigg [-\psi (t)^{\top } \int _{0}^{t} \frac{f(t,s,y^{*}(s),y^{*}(s-h),u(s))-f(t,s,y^{*}(s),y^{*}(s-h),u^{*}(s))}{(t-s)^{1-\alpha }}ds\bigg ]dt\\&\quad +\int _{0}^{T}g(s,y^{*}(s),y^{*}(s-h),u(s))-g(s,y^{*}(s),y^{*}(s-h),u^{*}(s))\Big )ds\\&= \int _{0}^{T}\bigg [ \int _{s}^{T}-\psi (t)^{\top }\bigg ( \frac{f(t,s,y^{*}(s),y^{*}(s-h),u(s))-f(t,s,y^{*}(s),y^{*}(s-h),u^{*}(s))}{(t-s)^{1-\alpha }}\bigg )dt\\&\quad +g(s,y^{*}(s),y^{*}(s-h),u(s))-g(s,y^{*}(s),y^{*}(s-h),u^{*}(s))\bigg ]ds\\&\equiv \int _{0}^{T}[\Psi (s,u^{*}(s))-\Psi (s,u(s))]ds. \end{aligned}$$

Here

$$\begin{aligned} \Psi (s,u(s))=&\int _{s}^{T}\psi (t)^{\top } \frac{f(t,s,y^{*}(s),y^{*}(s-h),u(s))}{(t-s)^{1-\alpha }}dt -g(s,y^{*}(s),y^{*}(s-h),u(s)) \end{aligned}$$

Let $u\in U$ be fixed and let $s_{0}\in (0,T)$ be a Lebesgue point of $\Psi (s,u^{*}(s))-\Psi (s,u(s))$. Then for any $\delta >0$, let

$$\begin{aligned} u(s)= {\left\{ \begin{array}{ll} u^{*}(s),\quad s\in [0,T]\setminus (s_{0}-\delta , s_{0}+\delta ),\\ u,\quad t\in (s_{0}-\delta , s_{0}+\delta ). \end{array}\right. } \end{aligned}$$

Therefore, from the above, we get

$$\begin{aligned} 0\le \frac{1}{2\delta }=\int _{s_{0}-\delta }^{s_{0}+\delta }[\Psi (s,u^{*}(s))-\Psi (s,u(s))]ds\rightarrow \Psi (s_{0},u^{*}(s_{0}))-\Psi (s_{0},u(s_{0})),\quad \delta \rightarrow 0. \end{aligned}$$

Consequently, by the Lebesgue point theorem, we get the maximum condition (3.27). $\square$

4 Example

Let us consider the following fractional delay optimal control problem (FDOCP).

$$\begin{aligned}&(D^{\alpha }_{0+}y)(t)=Ay(t-h)+Bu(t),\quad t \in [0,2] \quad a.e. \end{aligned}$$

(4.1)

$$\begin{aligned}&\quad y(t)=0,\quad -h\le t\le 0,\quad h>0, \end{aligned}$$

(4.2)

$$\begin{aligned}&\quad u(t)\in [0,1],\quad t\in [0,2], \end{aligned}$$

(4.3)

$$\begin{aligned}&\quad J(y,u)=\int _{0}^{2}(y_{1}(t-h)-y_{2}(t-h)+u(t))dt\rightarrow \min . \end{aligned}$$

(4.4)

$where \quad y\in R^{2},\quad A=\begin{bmatrix} 0 &{}\quad 1 \\ 0 &{}\quad 0 \\ \end{bmatrix}, \quad B=\begin{bmatrix} -1 \\ -1 \\ \end{bmatrix}, \quad \alpha =\frac{1}{2},\quad h=\frac{1}{2}.$

In this case

$$\begin{aligned} g(t,y(t),y(t-h),u(t))=Ay(t-h)+Bu(t),\\ g_{0}(t,y(t),y(t-h),u(t))=<(1,-1),y(t-h)>+u. \end{aligned}$$

Clearly,

$$\begin{aligned}&A^{k}=(A^{T})^{k}=0,\quad k\ge 2,\\&g_{y_{h}}(t,y(t),y(t-h),u)=A,\quad (g_{0})_{y_{h}}(t,y(t),y(t-h),u)=[1,-1]. \end{aligned}$$

Consequently, if $(y^{*},u^{*})$ is a locally optimal solution to problem (4.1)–(4.4), hence there is $\lambda$ such that

$$\begin{aligned}&\left( ^{C}D^{1/2}_{0+}\lambda \right) (t)=A^{T}\lambda \left( t-\frac{1}{2}\right) +\begin{bmatrix} -1 \\ -1 \\ \end{bmatrix},\end{aligned}$$

(4.5)

$$\begin{aligned}&\quad \lambda (t)=0,\quad t\in \left[ -\frac{1}{2}, 0\right] . \end{aligned}$$

(4.6)

Moreover,

$$\begin{aligned} u^{*}(t)-\lambda (t)Bu^{*}(t)=\min _{u\in [0,1]}\lbrace u-\lambda (t) Bu\rbrace \end{aligned}$$

(4.7)

for $t\in [0,2]$ a.e. Now, we must prove the following theorem, which will be used later on.

Theorem 4.1

[11] Let $\alpha \in (0,1)$. Assume

$$\begin{aligned} {\left\{ \begin{array}{ll} ^{C}D^{\alpha }_{0+} x(t)=Ax(t)+B x(t-h)+Cu(t),\quad t\in [0,T],\\ x(t)=\phi (t),\quad -h < t\le 0. \end{array}\right. } \end{aligned}$$

(4.8)

where $x\in R^{n},\quad u\in R^{m}$, A and B are $n\times n$ matrices, C is an $n\times m$ matrix with $n >m$. The solution of (4.8) is given as

$$\begin{aligned} x(t)&=X_{\alpha }(t)\phi (0)+B\int _{-h}^{0}(t-s-h)^{\alpha -1}X_{\alpha ,\alpha }(t-s-h)\phi (s)ds\\&\quad +\int _{0}^{t}(t-s)^{\alpha -1}X_{\alpha ,\alpha }(t-s)Cu(s)ds \end{aligned}$$

where

$$\begin{aligned} X_{\alpha }(t)&={\mathscr {L}}^{-1}\lbrace s^{\alpha -1}(s^{\alpha }I-A-Be^{-sh})^{-1}\rbrace (t),\quad and\\ X_{\alpha ,\alpha }(t)&=t^{1-\alpha }\int _{0}^{t}\frac{(t-s)^{\alpha -2}}{\Gamma (\alpha -1)}X_{\alpha }(s)ds. \end{aligned}$$

Theorem 4.2

Let $\alpha \in (0,1)$. If $A \in R^{n\times n}$ and $W\in R^{n}$.

$$\begin{aligned} {\left\{ \begin{array}{ll} ^{C}D^{\alpha }_{0+} x(t)=A x(t-h)+W,\quad t\in [0,T],\\ x(t)=0,\quad -h \le t\le 0. \end{array}\right. } \end{aligned}$$

has a unique solution x given the by formula

$$\begin{aligned} x(t)=\sum _{k=0}^{\infty }\frac{A^{k}(t-kh)^{\alpha (k+1)}}{\Gamma (\alpha (k+1)+1)}W,\quad t\in [0,T]. \end{aligned}$$

Proof

Using Laplace transform, we obtain

$$\begin{aligned} {\mathscr {L}}\lbrace ^{C}D^{\alpha }_{0+} x(t)\rbrace (s)=&{\mathscr {L}}\bigg \lbrace {^{RL}}I^{1-\alpha }_{0+} \bigg (\frac{dx(t)}{dt}\bigg )\bigg \rbrace (s)=s^{\alpha -1}{\mathscr {L}}\bigg \lbrace \frac{dx(t)}{dt}\bigg \rbrace (s)\\ =&s^{\alpha -1}\bigg (s{\widehat{x}}(s)-x(0)\bigg )=s^{\alpha }{\widehat{x}}(s), \end{aligned}$$

where ${\widehat{x}}(s)={\mathscr {L}}\lbrace x\rbrace (s)$.

$$\begin{aligned} {\mathscr {L}}\lbrace x(t-h)\rbrace (s)&=\int _{0}^{\infty }e^{-st}x(t-h)dt\\&\quad (If \quad we\quad make\quad the\quad substitution\quad t-h=\theta )\\&\quad =e^{-sh}\int _{-h}^{\infty }e^{-s\theta }x(\theta )d\theta =e^{-sh}\int _{0}^{\infty }e^{-s\theta }x(\theta )d\theta =e^{-sh}{\widehat{x}}(s). \\&\quad {\mathscr {L}}\lbrace W\rbrace (s)=\int _{0}^{\infty }e^{-st}Wdt=s^{-1}W \\&\quad s^{\alpha }{\widehat{x}}(s)=Ae^{-sh}{\widehat{x}}(s)+s^{-1}W\\&\quad {\widehat{x}}(s)=\frac{W}{s^{\alpha +1}-Ae^{-sh}s} \end{aligned}$$

by using inverse Laplace transform

$$\begin{aligned}&x(t)={\mathscr {L}}^{-1} \bigg \lbrace \frac{W}{s^{\alpha +1}-Ae^{-sh}s}\bigg \rbrace (t)=W {\mathscr {L}}^{-1} \bigg \lbrace \frac{1}{s^{\alpha +1}-Ae^{-sh}s}\bigg \rbrace (t) \\&\quad \frac{1}{s^{\alpha +1}-Ae^{-sh}s}=\frac{1}{s^{\alpha +1}}\frac{1}{1-\frac{Ae^{-sh}}{s^{\alpha }}}=\frac{1}{s^{\alpha +1}}\sum _{k=0}^{\infty }\frac{A^{k}e^{-skh}}{s^{\alpha k}} =\sum _{k=0}^{\infty }\frac{A^{k}e^{-skh}}{s^{\alpha (k+1)+1}} \\&\quad {\mathscr {L}}^{-1} \bigg \lbrace \frac{1}{s^{\alpha +1}-Ae^{-sh}s}\bigg \rbrace (t)=\sum _{k=0}^{\infty } A^{k}\frac{(t-kh)^{\alpha (k+1)}}{\Gamma (\alpha (k+1)+1)} \end{aligned}$$

then

$$\begin{aligned} x(t)=\sum _{k=0}^{\infty } A^{k}\frac{(t-kh)^{\alpha (k+1)}}{\Gamma (\alpha (k+1)+1)}W. \end{aligned}$$

$\square$

From Theorem 4.2, it follows that a solution of problem (4.5)–(4.6) is given by

$$\begin{aligned} \lambda (t)=&\begin{bmatrix} \lambda _{1}(t) \\ \lambda _{2}(t) \\ \end{bmatrix}=\frac{t^{1/2}}{\Gamma (3/2)} \begin{bmatrix} -1 \\ 1 \\ \end{bmatrix}+\begin{bmatrix} 0 &{} 1 \\ 0 &{} 0 \\ \end{bmatrix}\frac{(t-1/2)}{\Gamma (2)}\begin{bmatrix} -1 \\ 1 \\ \end{bmatrix}\\ =&\begin{bmatrix} \lambda _{1}(t) \\ \lambda _{2}(t) \\ \end{bmatrix}=\frac{t^{1/2}}{\Gamma (3/2)} \begin{bmatrix} -1 \\ 1 \\ \end{bmatrix}+\frac{(t-1/2)}{\Gamma (2)}\begin{bmatrix} 1 \\ 0 \\ \end{bmatrix}\\ =&\begin{bmatrix} \frac{(t-1/2)}{\Gamma (2)}-\frac{t^{1/2}}{\Gamma (3/2)} \\ \frac{t^{1/2}}{\Gamma (3/2)} \\ \end{bmatrix},\quad t\in [0,2]. \end{aligned}$$

thus,

$$\begin{aligned} \begin{bmatrix} \lambda _{1}(t) \\ \lambda _{2}(t) \\ \end{bmatrix}=&\begin{bmatrix} \frac{(t-1/2)}{\Gamma (2)}-\frac{t^{1/2}}{\Gamma (3/2)} \\ \frac{t^{1/2}}{\Gamma (3/2)} \\ \end{bmatrix},\quad t\in [0,2]. \end{aligned}$$

Consequently, condition (4.7) is equivalent to the following one

$$\begin{aligned} u^{*}(t)-(t-1/2)u^{*}(t)=\min _{u\in [0,1]}\lbrace u-(t-1/2)u\rbrace ={\left\{ \begin{array}{ll} \frac{3}{2}-t, \quad t\in [0,1],\\ 0,\quad \quad t\in [1,2]. \end{array}\right. } \end{aligned}$$

for $t\in [0,2]$ a.e. Thus

$$\begin{aligned} u^{*}(t)={\left\{ \begin{array}{ll} 1, \quad t\in [0,1] \quad a.e.\\ 0, \quad t\in [1,2] \quad a.e. \end{array}\right. } \end{aligned}$$

From Theorem 4.1, it follows that a solution of (4.1)–(4.2), corresponding to $u^{*}(\cdot )$ is given by

$$\begin{aligned} y^{*}(t)=&X_{\alpha }(t)y(0)+A\int _{-h}^{0}(t-s)^{\alpha -1}X_{\alpha ,\alpha }(t-s)y(s)ds\\&+\int _{0}^{t}(t-s)^{\alpha -1}X_{\alpha ,\alpha }(t-s)Bu^{*}(s)ds\\ =&{\left\{ \begin{array}{ll} \int _{0}^{t}(t-s)^{\alpha -1}X_{\alpha ,\alpha }(t-s)Bds,\quad t\in [0,1]\quad a.e.\\ \int _{0}^{1}(t-s)^{\alpha -1}X_{\alpha ,\alpha }(t-s)Bds,\quad t\in [1,2]\quad a.e. \end{array}\right. } \end{aligned}$$

For our problem, it is easy to check that, $X_{\alpha }(t)$ and $X_{\alpha ,\alpha }(t-s)$ are given the following form

$$\begin{aligned} X_{\alpha }(t)=\sum _{k=0}^{\infty }A^{k}\frac{(t-kh)^{\alpha k}}{\Gamma (\alpha k+1)},\quad and\quad X_{\alpha ,\alpha }(t-s)=\sum _{k=0}^{\infty }A^{k}\frac{(t-s-kh)^{\alpha ( k+1)}}{\Gamma (\alpha (k+1)+1)} \end{aligned}$$

then, we get

$$\begin{aligned} y^{*}(t)=&{\left\{ \begin{array}{ll}\begin{bmatrix} \frac{t^{1/2}}{\Gamma (2)}+\frac{t}{\Gamma (3/2)}-\frac{2t^{3/2}}{3} \\ \frac{t^{1/2}}{\Gamma (2)}-\frac{t}{\Gamma (3/2)}-\frac{2t^{3/2}}{3} \\ \end{bmatrix} ,\quad \quad \quad \quad \quad \quad \quad \quad t\in [0,1]\quad a.e.\\ \begin{bmatrix} \frac{(t-1)^{1/2}-t^{1/2}}{\Gamma (2)}-\frac{1}{\Gamma (3/2)}+\frac{2((t-1)^{3/2}-t^{3/2})}{3} \\ \frac{2((t-1)^{3/2}-t^{3/2})}{3}-\frac{(t-1)^{1/2}-t^{1/2}}{\Gamma (2)}-\frac{1}{\Gamma (3/2)} \\ \end{bmatrix},\quad t\in [1,2]\quad a.e. \end{array}\right. } \end{aligned}$$

It means that the pair $(y^{*}(t),u^{*}(t))$ is the only pair, which can be a locally optimal solution of problem (4.1)–(4.4).

5 Conclusion

This article addressed two separate issues. To begin with, we analyzed Pontryagin’s maximum principle concerning fractional differential equations with delay. In addition, we investigated the most efficient method for solving the control problem associated with Eq. (1.1) and its corresponding payoff function (1.2). Subsequently, we investigated the Pontryagin maximum principle in the framework of Volterra integral equations with delay (1.3). We have improved the results of our investigation by presenting illustrative examples at the conclusion of the article.

Further research in this area could involve several avenues:

• Investigate controllability and optimal control strategies for nonlinear systems, which can provide insights into real-world applications where linear approximations may not suffice.

• Develop efficient algorithms for solving the optimal control problem associated with first-order systems, considering computational complexity and scalability issues.