Fractional-order optimal control model for the equipment management optimization problem with preventive maintenance

Abstract

The current quality status of most machinery and equipment is based on its accumulated historical status, but the influence of the past quality status on the current status of equipment is often overlooked in optimization management. This paper uses a Caputo-type fractional derivative to characterize this property. By refining the nature and characteristics of the equipment maintenance effect function and considering the memory characteristics of equipment quality, the existing model is improved, and a fractional-order optimal control model for equipment maintenance and replacement is constructed. Theoretical analyses verify the effectiveness of the fractional-order equipment maintenance management model. Furthermore, the results of numerical experiments also reflect this difference between integer-order and fractional-order equipment maintenance management models. The result shows that with an increase of the order $\alpha$, the optimal target value of the equipment maintenance management problem will also increase with the weakening of the memory effect.

Appendices

Appendix

Important properties of fractional calculus

Lemma 6

[43] Let there be a function $h(t) \in C(0, 1)\bigcap L(0, 1)$ with a $\nu (\nu >0)$-order Riemann–Liouville type fractional derivative, and let $h\in C(0, 1)\bigcap$ L(0, 1); then,

$$\begin{aligned} I^{\nu }_{0}D_t^{\nu }h(t)=h(t)+C_1t^{\nu -1}+C_2t^{\nu -2}+ \cdots +C_Nt^{\nu -N}, \end{aligned}$$

(39)

where $C_i \in {\mathbb {R}}, i=1, 2,\ldots ,N$ and N is the smallest positive integer that satisfies $N \geqslant \nu$.

Lemma 7

[40,41,42] If $\nu _1, \nu _2,\nu >0, t \in [0,1]$ and $h(t) \in L [0,1],$ we have

$$\begin{aligned} I^{\nu _1}I^{\nu _2}h(t)=I^{\nu _1+\nu _2}h(t),_{0}D_t^{\nu }I^{\nu }h(t)=h(t). \end{aligned}$$

(40)

Lemma 8

[40, 44] If $h(t) \in C [0,1]$ and $\nu >0$, we have

$$\begin{aligned} \left[ I^{\nu }h(t) \right] _{t=0} = 0, \text { or } \lim _{t\rightarrow 0}\frac{1}{{\varGamma }(\nu )}\int ^t_0(t-s)^{\nu -1}h(s) {\mathrm{d}} s=0. \end{aligned}$$

(41)

Lemma 9

[40] Let $\nu >0$; then, for the fractional differential equation

$$\begin{aligned} ^C_{0}D_t^{\nu }h(t)=0, \end{aligned}$$

(42)

there is solution of the following form:

$$\begin{aligned} h(t)&=c_0+c_1t+c_2t^2+\cdots +c_{n-1}t^{n-1}, c_i \in {\mathbb {R}},\\ i&=0,1,2,\cdots ,n-1, n=[\nu ]+1. \end{aligned}$$

Lemma 10

[40] Let $\nu >0$; then, we have

$$\begin{aligned} I^{\nu } {}^C_{0}D_t^{\nu }h(t)=h(t)+c_0+c_1t+c_2t^2+\cdots +c_{n-1}t^{n-1}, \end{aligned}$$

where $c_i \in {\mathbb {R}}, i=0,1, 2,\cdots ,n-1, n=[\nu ]+1.$

Lemma 11

[41, 42] If $\nu _1, \nu _2,\nu >0, t \in [0,1]$ and function $h(t) \in L [0,1]$, then we have

$$\begin{aligned} {}^C_{0}D_t^{\nu }I^{\nu }h(t)=h(t). \end{aligned}$$

(43)

Lemma 12

[41, 45] Let $h(t) \in L^1(0,+\infty )$, $\nu _1, \nu _2,\nu >0$; then, we have

$$\begin{aligned} I^{\nu _1}I^{\nu _2}h(t)=I^{\nu _1+\nu _2}h(t), ^C_{0}D_t^{\nu }I^{\nu }h(t)=h(t). \end{aligned}$$

(44)

Lemma 13

[45] Let $_{0}^C D_t^{\nu }h(t) \in L^1(0,+\infty ), \nu >0$; then, we have

$$\begin{aligned}&I^{\nu } {}_{0}^C D_t^{\nu }h(t)=h(t)+C_1t^{\nu -1}+C_2t^{\nu -2}+ \cdots +C_Nt^{\nu -N},\, t>0, \end{aligned}$$

(45)

where $C_i \in {\mathbb {R}}, i=1, 2,\ldots ,N,$ and N is the smallest positive integer greater than or equal to $\nu$.

Remark 2

If the value of the function h in the above definition and lemma is in a Banach space E, then the integral involved in the above definition and lemma refers to the integral in the Bochner sense. If the abstract function g is measurable and its norm is integrable in the Lebesgue sense, then it is Bochner integrable.

Important tools

Here, we mainly introduce the concept of compact sets and related theorems, some commonly used conclusions and theorems in functional analysis, several fixed point theorems, and the Gronwall inequality. These concepts are important tools for proving the main results of this article.

First, a series of concepts and theorems of compact sets are given. The relevant details can be found in [49, 51].

Definition 1

[49] Let X be a nonempty compact set, and let $\{A_{\alpha }\}$ be a family subset of X, $A\subset X$; if $A\subset \bigcup _{\alpha }A_{\alpha }$, there is a family set $\{A_{\alpha }\}$ covering A. If the intersection of any finite set in $\{A_{\alpha }\}$ is not empty, then $\{A_{\alpha }\}$ has the finite intersection property.

Definition 2

[49] Let X be a topological space, $A\subset X$; if in any open set family covering A, a finite number of open coverings A can always be taken, then A is said to be a compact set in X.

Remark 3

[49] Two small conclusions regarding compact sets are given as follows:

• 1. A closed subset of a compact set is a compact set.

• 2. A compact set in a Hausdorff space X must be a closed set.

Theorem 5

[49] Assuming f is a continuous mapping from topological space $X_1$ to topological space $X_2$, the compact set A in $X_1$ is similar to f(A) and is a compact set in $X_2$.

Definition 3

[49] Assume X is a distance space, $M\subset X$; if for any point in M, denoted as $\{x_n\}$, there is a convergent subcolumn $\{x_{n_k} in$X$\}$, then M is a column compact set in X. If $\forall \epsilon >0$, there is a finite subset A of M, making $M\subset \bigcup _{x\in A}O(x,\epsilon )$, then M is said to be completely bounded, and it is called A, as a limited $\epsilon -$ net of M.

Remark 4

[49] Three small conclusions about column compact sets are given as follows:

• 1. The closure of a column compact set is still a column compact set.

• 2. The sequence compact set in the distance space must be completely bounded; a completely bounded set must be bounded.

• 3. Assume X is a distance space, $M\subset X$; then, the necessary and sufficient condition for M to be a compact set is that M is a sequence tightly closed set.

The continuous functions on the compact space X are all C(X). For $f\in C(X)$, we have

$$\begin{aligned} \Vert f\Vert =\max _{x}|f(x)|. \end{aligned}$$

According to $\Vert \cdot \Vert$, C(X) is a Banach space.

Definition 4

[49] Let $M\subset C(X)$; for arbitrary $\epsilon >0$ and $\delta >0$, when $\rho (x,y)<\delta$, we have

$$\begin{aligned} |f(x)-f(y)|<\epsilon , \forall f\in M. \end{aligned}$$

Then, M is an equally continuous family of functions.

The Arzela–Ascoli theorem is given below.

Theorem 6

[49] Let E be a real Banach space, $J_0=[a,b]$, and let

$$\begin{aligned} C[J_0,E]=\{x|x:J_0\rightarrow E \text { is continuous}\}. \end{aligned}$$

Its norm is

$$\begin{aligned} \Vert x\Vert =\sup _{t\in [0,T]}\Vert x(t)\Vert . \end{aligned}$$

Then, $C[J_0,E]$ is a Banach space.

The necessary and sufficient condition for $H\subset C[J_0,E]$ to be a relatively compact set is that H is an equicontinuous function of the family and that for an arbitrary $t\in J_0$, $H(t)=\{x(t)|x\in H\}$ is a relatively compact set in E.

The generalized Arzela–Ascoli theorem is given below.

Theorem 7

[50] Let X be a compact distance space, $M\subset X$; the necessary and sufficient condition for M to be compact is that M is a bounded and equally continuous function family.

Definition 5

[51] Let $(X,\rho )$ be a distance space; therefore, $T:(X,\rho )\rightarrow (X,\rho )$ is a contraction map. If $0<\alpha <1$, make $\rho (Tx,Ty)\leqslant \alpha \rho (x,y)$ $(\forall x,y \in X)$.

A commonly used fixed point theorem is given below: the Banach fixed point theorem, which is the principle of contraction mapping.

Theorem 8

[51] Let $(X,\rho )$ be a complete distance space, and let T be a compressed mapping from $(X,\rho )$ to itself; then, T has the only immovable X point. That is, there is only one $x\in X$ such that $Tx=x$.

For details on weakly singular Gronwall inequalities, see [52,53,54].

Theorem 9

[52, 53] Let u(t) be a continuous function that is nonnegative on [0, T]. If

$$\begin{aligned} u(t)\leqslant \varphi (t)+M\int _0^t\frac{u(s)}{(t-s)^{\alpha }} {\mathrm{d}} s, 0\leqslant t\leqslant T, \end{aligned}$$

where $0\leqslant \alpha <1$, $\varphi (t)$ is a nonnegative monotonically increasing continuous function on [0, T], and M is a positive constant.

$$\begin{aligned} u(t)\leqslant \varphi (t)E_{1-\alpha } (M{\varGamma }(1-\alpha )t^{1-\alpha }), 0\leqslant t\leqslant T, \end{aligned}$$

where $E_{1-\alpha }(z)$ is a Mittag-Leffler function defined on $0\leqslant \alpha <1$.

$$\begin{aligned} E_{1-\alpha }(z):=\sum _{n=0}^{\infty }\frac{z^n}{{\varGamma } (n(1-\alpha )+1)}. \end{aligned}$$

Theorem 10

[54] Let $u(t)\in PC(J,{\mathbb {R}})$ satisfy the following inequality:

$$\begin{aligned} |u(t)|\leqslant c_1(t)&+c_2\int _0^t(t-s)^{q-1}|u(s)|{\mathrm{d}} s +\sum _{0<t_k<t}\theta _k|u(t_k^-)|, 0\leqslant t\leqslant T, \end{aligned}$$

where $0< q <1$, $c_1$ is a nonnegative monotonically increasing continuous function on [0, T], and $c_2, \theta _k(0<t_k<t)$ is a positive constant.

$$\begin{aligned}&|u(t)|\leqslant c_1(t)\left( 1+\theta E_{q} (c_2{\varGamma }(q)t^{q})\right) ^k E_q (c_2{\varGamma }(q)t^{q}),\, t_k< t\leqslant t_{k+1}, \end{aligned}$$

where

$$\begin{aligned} \theta =\max _{0<t_k<t}\theta _k. \end{aligned}$$

Proof of the partial theorem lemma

Proof of Lemma 1

“Necessity”. Let $y\in PC_m[0,T]$ be the solution of equation (18). By Lemma 10, when $t\in [0,t_1]$, we have

$$\begin{aligned} y(t)=c_0+\frac{1}{{\varGamma }{(\alpha )}}\int _{a_0}^t(t-s)^{\alpha -1}c(s) {\mathrm{d}} s, c_0\in {\mathbb {R}}^m. \end{aligned}$$

By $y(0)=y^0$, we have $c_1=y^0$; thus,

$$\begin{aligned} y(t)=y^0+\frac{1}{{\varGamma }{(\alpha )}}\int _{a_0}^t(t-s)^{\alpha -1}c(s) {\mathrm{d}} s, t\in [0,t_1]. \end{aligned}$$

When $t\in (t_1,t_2]$, we have

$$\begin{aligned} _{a_1}^CD_t^{\alpha }y(t)=c(t), {\varDelta } y(t_1)=y(t_1^+)-y(t_1^-)=\psi _1(y(t_1^-)). \end{aligned}$$

By Lemma 10, when $t\in (t_1,t_2]$, we have

$$\begin{aligned} y(t)=c_1+\frac{1}{{\varGamma }{(\alpha )}}\int _{a_1}^t(t-s)^{\alpha -1}c(s) {\mathrm{d}} s, c_1\in {\mathbb {R}}^m. \end{aligned}$$

By

$$\begin{aligned} y(t_1^+)-y(t_1^-)&=c_1+\frac{1}{{\varGamma }{(\alpha )}}\int _{a_1}^{t_1}(t_1-s)^{\alpha -1}c(s) {\mathrm{d}} s-y^0 \\&-\quad \,\frac{1}{{\varGamma }{(\alpha )}}\int _{a_0}^{t_1}(t_1-s)^{\alpha -1}c(s) {\mathrm{d}} s=c_1-y^0 \\&\quad -\frac{1}{{\varGamma }{(\alpha )}}\int _{a_0}^{a_1}(t_1-s)^{\alpha -1}c(s) {\mathrm{d}} s=\psi _1(y(t_1^-))=\psi _1(y(t_1)), \end{aligned}$$

we have

$$\begin{aligned} c_1=y^0+\psi _1(y(t_1))+\frac{1}{{\varGamma }{(\alpha )}}\int _{a_0}^{a_1}(t_1-s)^{\alpha -1}c(s) {\mathrm{d}} s. \end{aligned}$$

Thus, for $t\in (t_1,t_2]$, we have

$$\begin{aligned} y(t)&=y^0+\psi _1(y(t_1))+\frac{1}{{\varGamma }{(\alpha )}}\int _{a_0}^{a_1}(t_1-s)^{\alpha -1}c(s) {\mathrm{d}} s \\&\quad +\frac{1}{{\varGamma }{(\alpha )}}\int _0^t(t-s)^{\alpha -1}c(s) {\mathrm{d}} s. \end{aligned}$$

By analogy, we have $t\in (t_k,t_{k+1}]$, and thus,

$$\begin{aligned} y(t)&=y^0+\sum _{i=1}^{k}\psi _i(y(t_i))\\&\quad +\sum _{j=1}^{k}\frac{1}{{\varGamma }{(\alpha )}}\int _{a_{j-1}}^{a_{j}}(t_j-s)^{\alpha -1}c(s) {\mathrm{d}} s \\&\quad \,+\frac{1}{{\varGamma }{(\alpha )}}\int _{a_k}^t(t-s)^{\alpha -1}c(s) {\mathrm{d}} s. \end{aligned}$$

Therefore,

$$(t) = \left\{ {\begin{array}{*{20}l} {y^{0} + \frac{1}{{(\alpha )}}\int_{{a_{0} }}^{t} {(t - s)^{{\alpha - 1}} } c(s)ds,t \in [0,t_{1} ],} \hfill \\ {y^{0} + \psi _{1} (y(t_{1} )) + \frac{1}{{(\alpha )}}\int_{{a_{0} }}^{{a_{1} }} {(t_{1} - s)^{{\alpha - 1}} } c(s)ds} \hfill \\ {\quad {\mkern 1mu} + \frac{1}{{(\alpha )}}\int_{{a_{1} }}^{t} {(t - s)^{{\alpha - 1}} } c(s)ds,t \in (t_{1} ,t_{2} ],} \hfill \\ {\qquad \qquad \vdots ,} \hfill \\ {y^{0} + \sum\limits_{{i = 1}}^{k} {\psi _{i} (y(t_{i} ))} + \sum\limits_{{j = 1}}^{k} {\frac{1}{{(\alpha )}}} \int_{{a_{{j - 1}} }}^{{a_{j} }} {(t_{j} - s)^{{\alpha - 1}} } c(s)ds} \hfill \\ {\quad {\mkern 1mu} + \frac{1}{{(\alpha )}}\int_{{a_{k} }}^{t} {(t - s)^{{\alpha - 1}} } c(s)ds,t \in (t_{k} ,t_{{k + 1}} ],} \hfill \\ {\qquad \qquad \vdots ,} \hfill \\ {y^{0} + \sum\limits_{{i = 1}}^{p} {\psi _{i} (y(t_{i} ))} + \sum\limits_{{j = 1}}^{p} {\frac{1}{{(\alpha )}}} \int_{{a_{{j - 1}} }}^{{a_{j} }} {(t_{j} - s)^{{\alpha - 1}} } c(s)ds} \hfill \\ {\quad {\mkern 1mu} + \frac{1}{{(\alpha )}}\int_{{a_{p} }}^{t} {(t - s)^{{\alpha - 1}} } c(s)ds,t \in (t_{p} ,T].} \hfill \\ \end{array} } \right.$$

“Adequacy”. Assume y satisfies the function (19). By Lemma 11, $y\in PC_m[0,T]$ is the solution of function (18). The proof is complete. $\square$

Proof of Lemma 2

For the convenience of description and writing, let $a_0=0$. $\forall u\in U$, $\forall f \in Y_K$, by Lemma 1, $y\in PC_m[0,T]$ is the solution of function (20) if and only if $y\in PC_m[0,T]$ is the solution of the following integral equation:

$$(t) = \left\{ {\begin{array}{*{20}l} {y^{0} + \frac{1}{{(\alpha )}}\int_{{a_{0} }}^{t} {(t - s)^{{\alpha - 1}} } f(t,y(s),u(s))ds,t \in [0,t_{1} ],} \hfill \\ {y^{0} + \psi _{1} (y(t_{1} )) + \frac{1}{{(\alpha )}}\int_{{a_{1} }}^{t} {(t - s)^{{\alpha - 1}} } f(t,y(s),u(s))ds} \hfill \\ {\quad {\mkern 1mu} + \frac{1}{{(\alpha )}}\int_{{a_{0} }}^{{a_{1} }} {(t_{1} - s)^{{\alpha - 1}} } f(t,y(s),u(s))ds,t \in (t_{1} ,t_{2} ],} \hfill \\ { \vdots ,} \hfill \\ {y^{0} + \sum\limits_{{i = 1}}^{k} {\psi _{i} (y(t_{i} ))} + \sum\limits_{{j = 1}}^{k} {\frac{1}{{(\alpha )}}\int_{{a_{{j - 1}} }}^{{a_{j} }} {(t_{j} - s)^{{\alpha - 1}} } f(t,y(s),u(s))ds} } \hfill \\ {\quad {\mkern 1mu} + \frac{1}{{(\alpha )}}\int_{{a_{k} }}^{t} {(t - s)^{{\alpha - 1}} } f(t,y(s),u(s))ds,t \in (t_{k} ,t_{{k + 1}} ],} \hfill \\ { \vdots ,} \hfill \\ {y^{0} + \sum\limits_{{i = 1}}^{p} {\psi _{i} (y(t_{i} ))} + \sum\limits_{{j = 1}}^{p} {} \frac{1}{{(\alpha )}}\int_{{a_{{j - 1}} }}^{{a_{j} }} {(t_{j} - s)^{{\alpha - 1}} } f(t,y(s),u(s))ds} \hfill \\ {\quad {\mkern 1mu} + \frac{1}{{(\alpha )}}\int_{{a_{p} }}^{t} {(t - s)^{{\alpha - 1}} } f(t,y(s),u(s))ds,t \in (t_{p} ,T].} \hfill \\ \end{array} } \right.$$

For a given $u\in U$, $f \in Y_K$, consider the following operator:

$$(Ty)(t) = \left\{ {\begin{array}{*{20}l} {y^{0} + \frac{1}{{(\alpha )}}\int_{{a_{0} }}^{t} {(t - s)^{{\alpha - 1}} } f(t,y(s),u(s))ds,t \in [0,t_{1} ],} \hfill \\ {y^{0} + \psi _{1} (y(t_{1} )) + \frac{1}{{(\alpha )}}\int_{{a_{1} }}^{t} {(t - s)^{{\alpha - 1}} } f(t,y(s),u(s))ds} \hfill \\ {\quad {\mkern 1mu} + \frac{1}{{(\alpha )}}\int_{{a_{0} }}^{{a_{1} }} {(t_{1} - s)^{{\alpha - 1}} } f(t,y(s),u(s))ds,t \in (t_{1} ,t_{2} ],} \hfill \\ { \vdots ,} \hfill \\ {y^{0} + \sum\limits_{{i = 1}}^{k} {\psi _{i} (y(t_{i} ))} + \sum\limits_{{j = 1}}^{k} {\frac{1}{{(\alpha )}}\int_{{a_{{j - 1}} }}^{{a_{j} }} {(t_{j} - s)^{{\alpha - 1}} } f(t,y(s),u(s))ds} } \hfill \\ {\quad {\mkern 1mu} + \frac{1}{{(\alpha )}}\int_{{a_{k} }}^{t} {(t - s)^{{\alpha - 1}} } f(t,y(s),u(s))ds,t \in (t_{k} ,t_{{k + 1}} ],} \hfill \\ { \vdots ,} \hfill \\ {y^{0} + \sum\limits_{{i = 1}}^{p} {\psi _{i} (y(t_{i} ))} + \sum\limits_{{j = 1}}^{p} {\frac{1}{{(\alpha )}}\int_{{a_{{j - 1}} }}^{{a_{j} }} {(t_{j} - s)^{{\alpha - 1}} } f(t,y(s),u(s))ds} } \hfill \\ {\quad {\mkern 1mu} + \frac{1}{{(\alpha )}}\int_{{a_{p} }}^{t} {(t - s)^{{\alpha - 1}} } f(t,y(s),u(s))ds,t \in (t_{p} ,T].} \hfill \\ \end{array} } \right.$$

With the binding condition $(F_K)$, we have $T:PC_m[0,T]\rightarrow PC_m[0,T]$.

Clearly, $y\in PC_m[0,T]$ is the solution of (20) if and only if $y\in PC_m[0,T]$ is the fixed point of operator T on $PC_m[0,T]$.

Next, we will use the Banach fixed point theorem 8 to prove that the operator T has a unique fixed point in the Banach space $PC_m[0,T]$.

First, since the condition $(H_{\phi })$ is established, the following equivalent norm can be defined in the Banach space $PC_m[0,T]$:

$$\begin{aligned} \Vert x \Vert _*=\max _{0\leqslant t \leqslant T} {\mathrm{e}}^{-\chi _Kt} \Vert x \Vert , \end{aligned}$$

where $\chi _K>0$ and

$$\sum\limits_{{i = 1}}^{k} \Theta _{i} e^{{ - \chi _{K} (t_{k} - t_{i} )}} + \frac{{D_{K} }}{{\chi _{K}^{\alpha } [(\alpha )]^{2} }} + e^{{ - \chi _{K} t}} \sum\limits_{{j = 1}}^{k} {\frac{{D_{K} }}{{(\alpha + 1)}}} [(t_{j} - a_{{j - 1}} )^{\alpha } - (t_{j} - a_{j} )^{\alpha } < 1.$$

In fact,

$$\begin{aligned} {\mathrm{e}}^{-\chi _KT} \Vert x \Vert \leqslant \Vert x \Vert _*=\max _{0\leqslant t \leqslant T} {\mathrm{e}}^{-\chi _Kt} \Vert x \Vert \leqslant \Vert x \Vert . \end{aligned}$$

It can be seen that $\Vert \cdot \Vert _1$ and $\Vert \cdot \Vert$ are equivalent norms. Next, we use the norm $\Vert \cdot \Vert _1$ in related discussions.

For a given positive constant $K>0$, we know that there is a constant $D_K>0$ such that $\forall (t,y_1,u), (t,y_2,u)\in I_K$, we have

$$\begin{aligned} \Vert f(t,y_1,u)-f(t,y_2,u) \Vert \leqslant D_K \Vert y_1-y_2 \Vert , \end{aligned}$$

(46)

where

$$\begin{aligned} I_K=[0,T]\times {\mathbb {R}}^m \times B_K \end{aligned}$$

and

$$\begin{aligned} B_K= \{u\in {\mathbb {R}}^n:\Vert u\Vert \leqslant K \}. \end{aligned}$$

For arbitrary $\vartheta _1, \vartheta _2\in PC_m[0,T]$, let $\vartheta _1\ne \vartheta _2$, and let $\Vert \vartheta _1-\vartheta _2\Vert _*=\xi _0>0$; by the definition of $\Vert \cdot \Vert _*$, $\forall t\in [0,T]$, we have

$$\begin{aligned} {\mathrm{e}}^{-\chi _Kt} \Vert \vartheta _1(t)-\vartheta _2(t) \Vert&\leqslant {\mathrm{e}}^{-\chi _Kt} \Vert \vartheta _1-\vartheta _2 \Vert \leqslant \Vert \vartheta _1-\vartheta _2 \Vert _*=\xi _0, \end{aligned}$$

and thus, $\forall t\in [0,T]$,

$$\begin{aligned} \Vert \vartheta _1(t)-\vartheta _2(t) \Vert \leqslant {\mathrm{e}}^{\chi _Kt}\xi _0. \end{aligned}$$

When $t\in [0,t_1]$, we have

$$\begin{aligned}&\Vert {\mathrm{e}}^{-L_Kt} [(T\vartheta _1)(t)-(T\vartheta _2)(t)] \Vert \\&\quad ={\mathrm{e}}^{-L_Kt}\frac{1}{{\varGamma }{(\alpha )}}\left\| \int _0^t(t-s)^{\alpha -1}f(s,\vartheta _1(s),u(s)) {\mathrm{d}} s \right. \\&\qquad \,\left. -\int _0^t(t-s)^{\alpha -1}f(s,\vartheta _2(s),u(s)) {\mathrm{d}} s\right\| \\&\quad \leqslant {\mathrm{e}}^{-L_Kt}\frac{1}{{\varGamma }{(\alpha )}}\int _0^t\left\| (t-s)^{\alpha -1} (f(s,\vartheta _1(s),u(s)) \right. \\&\qquad \left. -f(s,\vartheta _2(s),u(s)))\right\| {\mathrm{d}} s\\&\quad \leqslant {\mathrm{e}}^{-L_Kt}\frac{D_K}{{\varGamma }{(\alpha )}}\int _0^t(t-s)^{\alpha -1} \Vert \vartheta _1(s)-\vartheta _2(s) \Vert {\mathrm{d}} s\\&\quad \leqslant {\mathrm{e}}^{-L_Kt}\frac{\xi _0 D_K}{{\varGamma }{(\alpha )}}\int _0^t(t-s)^{\alpha -1}{\mathrm{e}}^{L_Ks} {\mathrm{d}} s\\&\quad =\frac{\xi _0 D_K}{{\varGamma }{(\alpha )}}\int _0^t(t-s)^{\alpha -1}{\mathrm{e}}^{-\chi _K(t-s)} {\mathrm{d}} s =\frac{\xi _0 D_K}{{\varGamma }{(\alpha )}}\int _0^t\tau ^{\alpha -1}{\mathrm{e}}^{-\chi _K\tau } d\tau \\&\quad =\frac{\xi _0 D_K}{\chi _K^{\alpha }{\varGamma }{(\alpha )}}\int _0^t(\chi _K\tau )^{\alpha -1}{\mathrm{e}}^{-\chi _K\tau } {\mathrm{d}}(\chi _K\tau ) =\frac{\xi _0 D_K}{\chi _K^{\alpha }{\varGamma }{(\alpha )}}\int _0^{\chi _Kt}s^{\alpha -1}{\mathrm{e}}^{-s} {\mathrm{d}} s\\&\quad \leqslant \frac{\xi _0 D_K}{\chi _K^{\alpha }{\varGamma }{(\alpha )}}\int _0^{+\infty }s^{\alpha -1}{\mathrm{e}}^{-s} {\mathrm{d}} s = \frac{D_K}{\chi _K^{\alpha }[{\varGamma }{(\alpha )}]^2}\xi _0 \\&\quad \leqslant \left( \sum _{i=1}^{k} {\varTheta }_i{\mathrm{e}}^{-\chi _K(t_k-t_i)}+\frac{D_K}{\chi _K^{\alpha }[{\varGamma }{(\alpha )}]^2}\right) \xi _0, \end{aligned}$$

and thus, for $t\in [0,t_1]$,

$$\begin{aligned}&\Vert {\mathrm{e}}^{-L_Kt} [(T\vartheta _1)(t)-(T\vartheta _2)(t)] \Vert \\&\quad \leqslant \left( \sum _{i=1}^{k} {\varTheta }_i{\mathrm{e}}^{-\chi _K(t_k-t_i)}+\frac{D_K}{\chi _K^{\alpha }[{\varGamma }{(\alpha )}]^2}\right) \xi _0. \end{aligned}$$

Next, consider that for $t\in (t_k,t_{k+1}]$, $\forall y\in PC_m[0,T]$, we have

$$\begin{aligned} \begin{aligned} (Ty)(t)&= y^0+\sum _{i=1}^{k} \psi _i(y(t_i)) +\sum _{j=1}^{k}\frac{1}{{\varGamma }{(\alpha )}}\int _{a_{j-1}}^{a_{j}}(t_j-s)^{\alpha -1}f(t,y(s),u(s)) {\mathrm{d}} s\\&\quad \, +\frac{1}{{\varGamma }{(\alpha )}}\int _{a_k}^t(t-s)^{\alpha -1}f(t,y(s),u(s)) {\mathrm{d}} s. \end{aligned} \end{aligned}$$

and we have

$$\begin{aligned}&\Vert {\mathrm{e}}^{-\chi _Kt} [(T\vartheta _1)(t)-(T\vartheta _2)(t)] \Vert \leqslant {\mathrm{e}}^{-\chi _Kt}\left\| \sum _{i=1}^{k} \psi _i(\vartheta _1(t_i))-\sum _{i=1}^{k} \psi _i(\vartheta _2(t_i))\right\| \\&\qquad \, + {\mathrm{e}}^{-\chi _Kt}\frac{1}{{\varGamma }{(\alpha )}}\left\| \int _{a_k}^t(t-s)^{\alpha -1}f(s,\vartheta _1(s),u(s)) {\mathrm{d}} s \right. \\&\qquad \, \left. -\int _{a_k}^t(t-s)^{\alpha -1}f(s,\vartheta _2(s),u(s)) {\mathrm{d}} s\right\| \\&\qquad \, +{\mathrm{e}}^{-\chi _Kt}\sum _{j=1}^{k}\frac{1}{{\varGamma }{(\alpha )}}\left\| \int _{a_{j-1}}^{a_{j}}(t_j-s)^{\alpha -1}(f(s,\vartheta _1(s),u(s)) -f(s,\vartheta _2(s),u(s))) {\mathrm{d}} s\right\| \\&\quad \leqslant {\mathrm{e}}^{-\chi _Kt}\sum _{i=1}^{k} {\varTheta }_i\left\| \vartheta _1(t_i)-\vartheta _2(t_i)\right\| \\&\qquad \, + {\mathrm{e}}^{-\chi _Kt}\frac{1}{{\varGamma }{(\alpha )}}\int _{a_k}^t\left\| (t-s)^{\alpha -1} (f(s,\vartheta _1(s),u(s)) -f(s,\vartheta _2(s),u(s)))\right\| {\mathrm{d}} s\\&\qquad \, +{\mathrm{e}}^{-\chi _Kt}\sum _{j=1}^{k}\frac{1}{{\varGamma }{(\alpha )}}\int _{a_{j-1}}^{a_{j}}(t_j-s)^{\alpha -1} \left\| f(s,\vartheta _1(s),u(s))-f(s,\vartheta _2(s),u(s))\right\| {\mathrm{d}} s \\&\quad \leqslant {\mathrm{e}}^{-\chi _Kt}\sum _{i=1}^{k} {\varTheta }_i{\mathrm{e}}^{\chi _Kt_i}\xi _0 + {\mathrm{e}}^{-\chi _Kt}\frac{D_K}{{\varGamma }{(\alpha )}}\int _{a_k}^t(t-s)^{\alpha -1} \Vert \vartheta _1(s)-\vartheta _2(s) \Vert {\mathrm{d}} s\\&\qquad \, +{\mathrm{e}}^{-\chi _Kt}\sum _{j=1}^{k}\frac{D_K}{{\varGamma }{(\alpha )}}\int _{a_{j-1}}^{a_{j}}(t_j- s)^{\alpha -1} \Vert \vartheta _1(s) - \vartheta _2(s) \Vert {\mathrm{d}} s\\&\quad \leqslant \sum _{i=1}^{k} {\varTheta }_i{\mathrm{e}}^{-\chi _K(t-t_i)}\xi _0+{\mathrm{e}}^{-\chi _Kt}D_K\frac{\xi _0}{{\varGamma }{(\alpha )}} \int _{a_k}^t(t-s)^{\alpha -1}{\mathrm{e}}^{L_Ks} {\mathrm{d}} s\\&\qquad \, +\xi _0{\mathrm{e}}^{-\chi _Kt}\sum _{j=1}^{k}\frac{D_K}{{\varGamma }{(\alpha +1)}} [(t_j-a_{j-1})^{\alpha }-(t_j-a_{j})^{\alpha }]\\&\quad \leqslant \left( \sum _{i=1}^{k} {\varTheta }_i{\mathrm{e}}^{-\chi _K(t_k-t_i)}\right) \xi _0+D_K\frac{\xi _0}{{\varGamma }{(\alpha )}} \int _{a_k}^t(t-s)^{\alpha -1}{\mathrm{e}}^{-\chi _K(t-s)} {\mathrm{d}} s\\&\qquad \, +\xi _0{\mathrm{e}}^{-\chi _Kt}\sum _{j=1}^{k}\frac{D_K}{{\varGamma }{(\alpha +1)}} [(t_j-a_{j-1})^{\alpha }-(t_j-a_{j})^{\alpha }]\\&\quad =\left( \sum _{i=1}^{k} {\varTheta }_i{\mathrm{e}}^{-\chi _K(t_k-t_i)}\right) \xi _0+D_K\frac{\xi _0}{{\varGamma }{(\alpha )}} \int _{a_k}^t\tau ^{\alpha -1}{\mathrm{e}}^{-\chi _K\tau } d\tau \\&\qquad \, +\xi _0{\mathrm{e}}^{-\chi _Kt}\sum _{j=1}^{k}\frac{D_K}{{\varGamma }{(\alpha +1)}} [(t_j-a_{j-1})^{\alpha }-(t_j-a_{j})^{\alpha }]\\&\quad =\left( \sum _{i=1}^{k} {\varTheta }_i{\mathrm{e}}^{-\chi _K(t_k-t_i)}\right) \xi _0 +D_K\frac{\xi _0}{\chi _K^{\alpha }{\varGamma }{(\alpha )}}\int _0^t(\chi _K\tau )^{\alpha -1}{\mathrm{e}}^{-\chi _K\tau } {\mathrm{d}}(\chi _K\tau )\\&\qquad \, +\xi _0{\mathrm{e}}^{-\chi _Kt}\sum _{j=1}^{k}\frac{D_K}{{\varGamma }{(\alpha +1)}} [(t_j-a_{j-1})^{\alpha }-(t_j-a_{j})^{\alpha }]\\&\quad =\left( \sum _{i=1}^{k} {\varTheta }_i{\mathrm{e}}^{-\chi _K(t_k-t_i)}\right) \xi _0+D_K\frac{\xi _0}{\chi _K^{\alpha }{\varGamma }{(\alpha )}}\int _0^{\chi _Kt}s^{\alpha -1}{\mathrm{e}}^{-s} {\mathrm{d}} s\\&\qquad \, +\xi _0{\mathrm{e}}^{-\chi _Kt}\sum _{j=1}^{k}\frac{D_K}{{\varGamma }{(\alpha +1)}} [(t_j-a_{j-1})^{\alpha }-(t_j-a_{j})^{\alpha }]\\&\leqslant \left( \sum _{i=1}^{k} {\varTheta }_i{\mathrm{e}}^{-\chi _K(t_k-t_i)}\right) \xi _0+D_K\frac{\xi _0}{\chi _K^{\alpha }{\varGamma }{(\alpha )}}\int _0^{+\infty }s^{\alpha -1}{\mathrm{e}}^{-s} {\mathrm{d}} s\\&\qquad \, +\xi _0{\mathrm{e}}^{-\chi _Kt}\sum _{j=1}^{k}\frac{D_K}{{\varGamma }{(\alpha +1)}} [(t_j-a_{j-1})^{\alpha }-(t_j-a_{j})^{\alpha }]\\&= \left( \sum _{i=1}^{k} {\varTheta }_i{\mathrm{e}}^{-\chi _K(t_k-t_i)}\right) \xi _0+\frac{D_K}{\chi _K^{\alpha }[{\varGamma }{(\alpha )}]^2}\xi _0 \\&\qquad \, +\xi _0{\mathrm{e}}^{-\chi _Kt}\sum _{j=1}^{k}\frac{D_K [(t_j-a_{j-1})^{\alpha }-(t_j-a_{j})^{\alpha }]}{{\varGamma }{(\alpha +1)}}\\&\quad \leqslant \left( \sum _{i=1}^{k} {\varTheta }_i{\mathrm{e}}^{-\chi _K(t_k-t_i)}+\frac{D_K}{\chi _K^{\alpha }[{\varGamma }{(\alpha )}]^2}\right. \\&\qquad \, \left. +{\mathrm{e}}^{-\chi _Kt}\sum _{j=1}^{k}\frac{D_K}{{\varGamma }{(\alpha +1)}} [(t_j-a_{j-1})^{\alpha }-(t_j-a_{j})^{\alpha }]\right) \xi _0. \end{aligned}$$

In summary, we have

$$\begin{aligned}&\Vert T\vartheta _1-T\vartheta _2 \Vert _*\leqslant \left( \sum _{i=1}^{k} {\varTheta }_i{\mathrm{e}}^{-\chi _K(t_k-t_i)}+\frac{D_K}{\chi _K^{\alpha }[{\varGamma }{(\alpha )}]^2} \right. \\&\quad \left. +{\mathrm{e}}^{-\chi _Kt}\sum _{j=1}^{k}\frac{D_K}{{\varGamma }{(\alpha +1)}} [(t_j-a_{j-1})^{\alpha }-(t_j-a_{j})^{\alpha }]\right) \Vert \vartheta _1-\vartheta _2 \Vert _*. \end{aligned}$$

Because we have

$$\begin{aligned}&\sum _{i=1}^{k} {\varTheta }_i{\mathrm{e}}^{-\chi _K(t_k-t_i)}+\frac{D_K}{\chi _K^{\alpha }[{\varGamma }{(\alpha )}]^2}\\&\quad +{\mathrm{e}}^{-\chi _Kt}\sum _{j=1}^{k}\frac{D_K}{{\varGamma }{(\alpha +1)}} [(t_j-a_{j-1})^{\alpha }-(t_j-a_{j})^{\alpha }]<1, \end{aligned}$$

T is a contraction mapping on $PC_m[0,T]$; there is a principle of contraction mapping 8, and it can be seen that the operator T has a unique fixed point in $PC_m[0,T]$. Therefore, Equation (20) has a unique solution in $PC_m[0,T]$. The proof is complete. $\square$

Proof of theorem 4

According to the definition form of the performance index function J, we have

$$\begin{aligned} J_{f^j}(u^j)=k(y^j(0),y^j(T))+\int _0^T h(t,y^j(t),u^j(t)) {\mathrm{d}} t \end{aligned}$$

and

$$\begin{aligned} J_{f^*}(u^*)=k(y^*(0),y^*(T))+\int _0^T h(t,y^*(t),u^*(t)) {\mathrm{d}} t, \end{aligned}$$

where

$$y^{j} (t) = S_{{f^{j} }} (u^{j} )(t) = \left\{ {\begin{array}{*{20}l} {y^{0} + \frac{1}{{(\alpha )}}\int_{0}^{t} {(t - s)^{{\alpha - 1}} } f^{j} (s,y^{j} (s),u^{j} (s))ds,t \in [0,t_{1} ],} \hfill \\ {y^{0} + \psi _{1} (y(t_{1} )) + \frac{1}{{(\alpha )}}\int_{{a_{1} }}^{t} {(t - s)^{{\alpha - 1}} } f^{j} (s,y^{j} (s),u^{j} (s))ds} \hfill \\ {\quad {\mkern 1mu} + \frac{1}{{(\alpha )}}\int_{{a_{0} }}^{{a_{1} }} {(t_{1} - s)^{{\alpha - 1}} } f^{j} (s,y^{j} (s),u^{j} (s))ds,t \in (t_{1} ,t_{2} ],} \hfill \\ { \vdots ,} \hfill \\ {y^{0} + \sum\limits_{{i = 1}}^{k} {\psi _{i} (y(t_{i} ))} + \sum\limits_{{j = 1}}^{k} {\frac{1}{{(\alpha )}}\int_{{a_{{j - 1}} }}^{{a_{j} }} {(t_{j} - s)^{{\alpha - 1}} } f^{j} (s,y^{j} (s),u^{j} (s))ds} } \hfill \\ {\quad {\mkern 1mu} + \frac{1}{{(\alpha )}}\int_{{a_{k} }}^{t} {(t - s)^{{\alpha - 1}} } f^{j} (s,y^{j} (s),u^{j} (s))ds,t \in (t_{k} ,t_{{k + 1}} ],} \hfill \\ { \vdots ,} \hfill \\ {y^{0} + \sum\limits_{{i = 1}}^{p} {\psi _{i} (y(t_{i} ))} + \sum\limits_{{j = 1}}^{p} {\frac{1}{{(\alpha )}}\int_{{a_{{j - 1}} }}^{{a_{j} }} {(t_{j} - s)^{{\alpha - 1}} } f^{j} (s,y^{j} (s),u^{j} (s))ds} } \hfill \\ {\quad {\mkern 1mu} + \frac{1}{{(\alpha )}}\int_{{a_{p} }}^{t} {(t - s)^{{\alpha - 1}} } f^{j} (s,y^{j} (s),u^{j} (s))ds,t \in (t_{p} ,T]} \hfill \\ \end{array} } \right.$$

and

$$y^{*} (t) = S_{{f^{*} }} (u^{*} )(t) = \left\{ {\begin{array}{*{20}l} {y^{0} + \frac{1}{{(\alpha )}}\int_{0}^{t} {(t - s)^{{\alpha - 1}} } f^{*} (s,y^{*} (s),u^{*} (s))ds,t \in [0,t_{1} ],} \hfill \\ {y^{0} + \psi _{1} (y(t_{1} )) + \frac{1}{{(\alpha )}}\int_{{a_{1} }}^{t} {(t - s)^{{\alpha - 1}} } f^{j} (s,y^{j} (s),u^{j} (s))ds} \hfill \\ {\quad {\mkern 1mu} + \frac{1}{{(\alpha )}}\int_{{a_{0} }}^{{a_{1} }} {(t_{1} - s)^{{\alpha - 1}} } f^{*} (s,y^{*} (s),u^{*} (s))ds,t \in (t_{1} ,t_{2} ],} \hfill \\ {\qquad \qquad \vdots ,} \hfill \\ {y^{0} + \sum\limits_{{i = 1}}^{k} {\psi _{i} (y(t_{i} ))} + \sum\limits_{{j = 1}}^{k} {\frac{1}{{(\alpha )}}\int_{{a_{{j - 1}} }}^{{a_{j} }} {(t_{j} - s)^{{\alpha - 1}} } f^{*} (s,y^{*} (s),u^{*} (s))ds} } \hfill \\ {\quad {\mkern 1mu} + \frac{1}{{(\alpha )}}\int_{{a_{k} }}^{t} {(t - s)^{{\alpha - 1}} } f^{*} (s,y^{*} (s),u^{*} (s))ds,t \in (t_{k} ,t_{{k + 1}} ],} \hfill \\ {\qquad \qquad \vdots ,} \hfill \\ {y^{0} + \sum\limits_{{i = 1}}^{p} {\psi _{i} (y(t_{i} ))} + \sum\limits_{{j = 1}}^{p} {\frac{1}{{(\alpha )}}\int_{{a_{{j - 1}} }}^{{a_{j} }} {(t_{j} - s)^{{\alpha - 1}} } f^{*} (s,y^{*} (s),u^{*} (s))ds} } \hfill \\ {\quad {\mkern 1mu} + \frac{1}{{(\alpha )}}\int_{{a_{p} }}^{t} {(t - s)^{{\alpha - 1}} } f^{*} (s,y^{*} (s),u^{*} (s))ds,t \in (t_{p} ,T].} \hfill \\ \end{array} } \right.$$

From $f^j\rightarrow f^*(j\rightarrow +\infty )$ and $u^j\rightarrow u^*(j\rightarrow +\infty )$, by Lemma 4, we have $y^j\rightarrow y^*(j\rightarrow +\infty )$.

The functions k and h satisfy the conditions $(H_k)$ and $(H_h)$, respectively. k and h are continuous functions, and we have

$$\begin{aligned} k(y^j(0),y^j(T))\rightarrow k(y^*(0),y^*(T)), j\rightarrow +\infty \end{aligned}$$

and

$$\begin{aligned}&\int _0^T h(t,y^j(t),u^j(t)) {\mathrm{d}} t \rightarrow \int _0^T h(t,y^*(t),u^*(t)) {\mathrm{d}} t, j\rightarrow +\infty . \end{aligned}$$

Therefore,

$$\begin{aligned} \begin{aligned}&k(y^j(0),y^j(T))+\int _0^T h(t,y^j(t),u^j(t)) {\mathrm{d}} t\\&\rightarrow k(y^*(0),y^*(T))+\int _0^T h(t,y^*(t),u^*(t)) {\mathrm{d}} t, j\rightarrow +\infty , \end{aligned} \end{aligned}$$

Therefore, we have

$$\begin{aligned} J_{f^j}(u^j)\rightarrow J_{f^*}(u^*), j\rightarrow +\infty . \end{aligned}$$

The proof is complete. $\square$