**2. Preliminaries**

In this paper, we denote <sup>R</sup><sup>+</sup> := [0, <sup>∞</sup>), *<sup>J</sup>*<sup>1</sup> := [*s*<sup>0</sup> <sup>−</sup> *<sup>ζ</sup>*,*s*<sup>0</sup> <sup>+</sup> *<sup>η</sup>*], *<sup>J</sup>*<sup>2</sup> := [*s*0,*s*<sup>0</sup> <sup>+</sup> *<sup>η</sup>*], *<sup>J</sup>*<sup>3</sup> :<sup>=</sup> [*s*<sup>0</sup> <sup>−</sup> *<sup>ζ</sup>*,*s*0]. Let *<sup>C</sup>*(*J*1, <sup>R</sup>) be real Banach space of all continuous functions with norm:

$$\|\|u\|\| = \sup\{|\mu(s)| \colon s \in J\_1\}.$$

**Definition 1.** *Equation (3) is Hyers–Ulam stable on J*<sup>1</sup> *if there exists C* > 0 *such that for θ* > 0 *and each solution v*(*s*) *of the inequality*

$$\left| \left\{ \left| v^{(n)}(s) - z \left( s, v^{(0)}, \dots, v^{(n-1)}(\lambda), \int\_{s\_0}^{s} g(\tau, v^{(0)}, \dots, v^{(n-1)}) d\tau \right) \right| \le \theta, \ s \in \mathbb{I}\_{2^\prime} \right. \tag{4} \right. \tag{4}$$

*there exists a solution u*(*s*) <sup>∈</sup> *<sup>C</sup>*(*J*1, <sup>R</sup>) *Cn*(*J*2, R) *of Equation (3) with*

$$|v(s) - \mathfrak{u}(s)| \le \mathcal{C} \cdot \theta, \ s \in J\_1.$$

**Definition 2.** *Equation (3) is Hyers–Ulam–Rassias stable with respect to σ*(*s*) *on J*<sup>1</sup> *if there exists Kz*,*g*,*<sup>σ</sup>* > 0 *such that for each solution v*(*s*) *of the inequality*

$$\left| \left\{ \left| v^{(n)}(s) - z(s, v^{(0)}, \dots, v^{(n-1)}(\lambda), \int\_{s\_0}^{s} g(\tau, v^{(0)}, \dots, v^{(n-1)}) d\tau \right| \right| \le \sigma(s), \ s \in I\_{2, \epsilon} \right. \tag{5}$$

*there exists a solution u*(*s*) <sup>∈</sup> *<sup>C</sup>*(*J*1, <sup>R</sup>) *Cn*(*J*2, R) *of Equation (3) with*

$$|v(s) - \mu(s)| \le K\_{z, \emptyset, \sigma} \cdot \sigma(s), \ s \in J\_1.$$

**Lemma 1** (see [28])**.** *Assume that f* <sup>∈</sup> *<sup>C</sup>*(R, *<sup>Q</sup>*)*, then n-th repeated integrable of f based at s*0*,*

$$f^{(-n)}(s) = \int\_{s\_0}^{s} \int\_{s\_0}^{s\_1} \int\_{s\_0}^{s\_2} \int\_{s\_0}^{s\_3} \dots \int\_{s\_0}^{s\_{n-1}} f(s\_n) ds\_n ds\_{n-1} \dots ds\_2 ds\_{14}$$

*is given by*

$$f^{(-n)}(s) = \frac{1}{(n-1)!} \int\_{s0}^{s} (s-\tau)^{n-1} f(\tau) d\tau.$$

**Theorem 1.** *A function <sup>u</sup>*(*s*)<sup>∈</sup> *<sup>C</sup>*(*J*1, <sup>R</sup>) *Cn*(*J*2, R) *is a solution of the delay integro-differential equation*

$$\begin{cases} u^{(n)}(s) = z\left(s, \mu^{(0)}, \dots, \mu^{(0)}(\lambda), \dots, \mu^{(n-1)}(\lambda), \int\_{s\_0}^{s} g(\tau, \mu^{(0)}, \dots, \mu^{(n-1)}) d\tau\right), \ s \in \text{I}, \\ u(s) = \chi(s), \ s \in \text{I}\mathfrak{z}, \\ u^{(j)}(s\_0) = \chi^{(j)}(s\_0), \ j = 1, \dots, n-1, \end{cases} \tag{6}$$

*if and only if u*(*s*) *is a solution of the integral equation*

$$u(\mathbf{s}) = \begin{cases} \sum\_{j=0}^{n-1} \frac{(s-s\_0)\_j^{\prime} \chi^{(j)}(\lambda s)}{j!} + \frac{1}{(n-1)!} \int\_{s\_0}^{s} (s-\tau)^{n-1} \\ \cdot \mathcal{Z}\left(\tau, \mu^{(0)}, \dots, \mu^{(0)} (\lambda), \dots, \mu^{(n-1)} (\lambda), \int\_{s\_0}^{\tau} \mathcal{g}(r, \mu^{(0)}, \dots, \mu^{(n-1)}) dr\right) d\tau, \ s \in \mathcal{I}\_2, \\ \chi(s), \ s \in \mathcal{I}\_3. \end{cases} \tag{7}$$

**Proof.** For *n* = 1, from (6), we have

$$\begin{aligned} u^{(1)}(s) &= z \left( s, u^{(0)}, u^{(0)}(\lambda), \int\_{s\_0}^s g(\tau, u^{(0)}) d\tau \right), \\ u(s\_0) &= \chi(s\_0), \end{aligned}$$

then by integral formula, we have

$$
\mu(\mathbf{s}) = \chi(\mathbf{s}\_0) + \int\_{\mathfrak{s}\_0}^{\mathfrak{s}} z \Big( \pi, \mu^{(0)}, \mu^{(0)}(\lambda), \int\_{\mathfrak{s}\_0}^{\pi} \lg(r, \mu^{(0)}) dr \Big) d\pi, \text{ s} \in f\_2.
$$

This means that (7) holds for *n* = 1. For *n* = *k*, from (6), we have

$$\begin{aligned} u^{(k)}(\mathbf{s}) &= \mathbf{z} \Big( \mathbf{s}, u^{(0)}, \dots, u^{(0)}(\boldsymbol{\lambda}), \dots, u^{(k-1)}(\boldsymbol{\lambda}), \int\_{\mathbf{s}\_0}^{\mathbf{s}} \mathbf{g}(\mathbf{r}, u^{(0)}, \dots, u^{(k-1)}) d\mathbf{r} \Big), \\ u^{(j)}(\mathbf{s}\_0) &= \chi^{(j)}(\mathbf{s}\_0), j = 1, \dots, k - 1. \end{aligned}$$

Assume for *n* = *k*, (7) holds; that is,

$$u(s) = \sum\_{j=0}^{k-1} \frac{(s-s\_0)^j \chi^{(j)}(s\_0)}{j!} + \frac{1}{(k-1)!} \int\_{s\_0}^s (s-\tau)^{k-1} \cdot z\left(\tau, u^{(0)}, \dots, u^{(0)}(\lambda), \dots, u\right)$$

$$u^{(k-1)}(\lambda), \int\_{s\_0}^\tau g(r, u^{(0)}, \dots, u^{(k-1)}) dr) dr, \text{ s} \in \mathcal{J}\_2.$$

Hence, for *n* = *k* + 1, from (6), we have

$$u^{(k+1)}(s) = z\left(s, u^{(0)}, \dots, u^{(0)}(\lambda), \dots, u^{(k)}(\lambda), \int\_{s\_0}^s g(\tau, u^{(0)}, \dots, u^{(k)})d\tau\right),$$

then by the integral formula, we have

$$u^{(k)}(s) = \chi^{(j)}(s\_0) + \int\_{s\_0}^s z\left(\tau, u^{(0)}, \dots, u^{(0)}(\lambda), \dots, u^{(k)}(\lambda), \int\_{s\_0}^\tau g(r, u^{(0)}, \dots, u^{(k)}) dr\right) d\tau.$$

From the inductive hypothesis and Lemma 1, we have

$$\begin{split} u(s) &= \sum\_{j=0}^{k-1} \frac{(s-s\_0)^j \chi^{(j)}(s\_0)}{j!} + \frac{1}{(k-1)!} \int\_{s\_0}^s (s-\tau)^{k-1} \cdot \chi^{(j)}(s\_0) d\tau \\ &\quad + \frac{1}{(k-1)!} \int\_{s\_0}^s (s-\tau)^{k-1} \int\_{s\_0}^\tau z \Big(r, \mu^{(0)}, \dots, \mu^{(0)}(\lambda), \dots, \mu^{(k)}(\lambda)\Big) d\tau \\ &\quad \int\_{s\_0}^\tau \mathcal{g}(t, \mu^{(0)}, \dots, \mu^{(k)}) dt \Big) d\tau d\tau \\ &= \sum\_{j=0}^k \frac{(s-s\_0)^j \chi^{(j)}(s\_0)}{j!} + \frac{1}{k!} \int\_{s\_0}^s (s-\tau)^k \cdot z \Big(\tau, \mu^{(0)}, \dots, \mu^{(0)}(\lambda), \tau, \dots, \mu^{(k)}(\lambda)\Big) d\tau. \end{split}$$
 
$$\mu^{(k)}(\lambda) \Big/ \int\_{s\_0}^\tau \mathcal{g}(r, \mu^{(0)}, \dots, \mu^{(k)}) dr \Big) d\tau.$$

Hence, by mathematical induction, the conclusion is estabilished.

**Lemma 2** (see [23])**.** *(abstract Gronwall lemma) Let* (*Y*, *d*) *be an ordered metric space and A*: *Y* −→ *Y be an increasing Picard operator (FA* = {*xA*∗}*). Then, for x* ∈ *Y, x* ≤ *A*(*x*) *implies x* ≤ *xA*<sup>∗</sup> *and x* ≥ *A*(*x*) *implies x* ≥ *xA*∗*.*

**Lemma 3** (see [29])**.** *(Gronwall lemma) Assume that <sup>u</sup>*(*s*), *<sup>b</sup>*(*s*) <sup>∈</sup> *<sup>C</sup>*([*a*, <sup>+</sup>∞), <sup>R</sup>+)*, <sup>T</sup>* <sup>&</sup>gt; <sup>0</sup> *is constant. If u*(*s*) <sup>∈</sup> *<sup>C</sup>*([*a*, <sup>+</sup>∞), <sup>R</sup>+) *satisfies*

$$
\mu(s) \le T + \int\_a^s b(\tau)\mu(\tau)d\tau,\ s \in [a\_\prime + \infty),
$$

*then*

$$u(s) \le T \exp\left(\int\_a^s b(\tau)d\tau\right), \ s \in [a, +\infty).$$

**Lemma 4** (see [30])**.** *Assume that <sup>w</sup>*(*s*), *<sup>d</sup>*(*s*), *<sup>l</sup>*(*s*), *<sup>m</sup>*(*s*), *<sup>n</sup>*(*s*) <sup>∈</sup> *<sup>C</sup>*(R+, <sup>R</sup>+) *and <sup>d</sup>*(*s*), *<sup>l</sup>*(*s*) *are nondecreasing functions on* R+*. If w*(*s*) *satisfies the delay integral inequality*

$$w^h(s) \le d(s) + l(s) \int\_0^s \left[ m(\tau)w^a(\sigma(\tau)) + n(\tau)w^b(\tau) + \int\_0^\tau z(r)w^c(r) dr \right] d\tau, \ s \in \mathbb{R}^+ \ne \infty$$

*with initial conditions*

$$w(s) = r(s), \ s \in [\mathfrak{a}, 0]; r(\sigma(s)) \le d(s)^{\frac{1}{\mathfrak{a}}}, \ s \in \mathbb{R}^+ and \ \sigma(s) \le 0,$$

*where <sup>h</sup>* <sup>=</sup> 0, *<sup>h</sup>* <sup>≥</sup> *<sup>a</sup>* <sup>≥</sup> 0, *<sup>h</sup>* <sup>≥</sup> *<sup>b</sup>* <sup>≥</sup> 0, *<sup>h</sup>* <sup>≥</sup> *<sup>c</sup>* <sup>≥</sup> <sup>0</sup> *and <sup>h</sup>*, *<sup>a</sup>*, *<sup>b</sup>*, *<sup>c</sup> are constants; <sup>σ</sup>*(*s*) <sup>∈</sup> *<sup>C</sup>*(R+, <sup>R</sup>) *and <sup>σ</sup>*(*s*) <sup>≤</sup> *s;* <sup>−</sup><sup>∞</sup> <sup>&</sup>lt; *<sup>α</sup>* <sup>=</sup> inf{*σ*(*s*),*<sup>s</sup>* <sup>∈</sup> <sup>R</sup>+} ≤ <sup>0</sup> *and r*(*s*) <sup>∈</sup> *<sup>C</sup>*([*α*, 0], <sup>R</sup>+)*, then*

$$w(s) \le \left[ d(s) + l(s) \mathcal{U}(s) \exp\left( \int\_0^s V(\tau) d\tau \right) \right]^{\frac{1}{k}},$$

*where, for any H* <sup>&</sup>gt; <sup>0</sup>*, s* <sup>∈</sup> <sup>R</sup>+*,*

$$\begin{split} \mathcal{U}(s) &= \int\_{0}^{s} m(\tau) \left[ \left( \frac{a}{h} H^{\frac{a-h}{h}} d(\tau) + \frac{h-a}{h} H^{\frac{a}{h}} \right) + \mathcal{g}(\tau, s) \left( \frac{b}{h} H^{\frac{b-h}{h}} d(\tau) + \frac{h-b}{h} H^{\frac{b}{h}} \right) \right] \\ &+ \int\_{0}^{\tau} z(\tau) \left( \frac{c}{h} H^{\frac{c-h}{h}} d(\tau) + \frac{h-c}{h} H^{\frac{c}{h}} \right) d\tau, \\ \mathcal{V}(s) &= \left( \frac{a}{h} H^{\frac{a-h}{h}} m(s) + \frac{b}{h} H^{\frac{b-h}{h}} n(s) \right) \mathcal{l}(s) + \int\_{0}^{s} z(\tau) \mathcal{l}(\tau) \frac{c}{h} H^{\frac{c-h}{h}} d\tau. \end{split}$$
