Asymptotical stability of the exact solutions and the numerical solutions for impulsive neutral differential equations

Abstract

In this paper, we not only study asymptotical stability of a class of linear impulsive neutral delay differential equations(INDDEs), but also study stability and asymptotical stability of nonlinear INDDEs. Asymptotical stability of zero solution of linear INDDEs is studied by the properties of simple autonomous linear neutral delay differential equations(NDDEs) without impulsive perturbations. Base on this idea, numerical methods of INDDEs are constructed. The constructed numerical methods preserve asymptotical stability of linear INDDEs if corresponding methods are A-stable. Moreover, some stability and asymptotical stability criteria are established for nonlinear INDDEs, respectively. The constructed numerical methods which can preserve stability and asymptotical stability of the exact solutions under these criteria are obtained. Some numerical examples are given to confirm the theoretical results.

1 Introduction

Impulsive differential equations arise widely in the study of medicine, biology, economics, engineering, and so forth. In recent years, INDDEs are attracting more and more attention. In papers Anguraj and Karthikeyan (2009) and Cuevasa et al. (2009), existence uniqueness and continuous dependence of INDDEs are investigated. In papers Li and Rogovchenko (2015, 2016) and Li and Deng (2017), oscillation of first-order, second-order, even-order of INDDEs are studied, respectively. In papers Li and Deng (2017) and Li and Rogovchenko (2020), stability of first-order, third-order INDDEs are studied, respectively. Mean square exponential input-to state stability of stochastic Markovian reaction diffusion systems with impulsive perturbations has also been studied in paper Xue et al. (2023).

On the other hand, stability and asymptotical stability of numerical methods for NDDEs without impulsive perturbations have been widely studied (see Liu 1999; Enright and Hayashi 1998; Engelborghs et al. 2001; Wang and Li 2008 etc). Wang and Li (2008) studied stability and asymptotic stability of $\theta$-methods for nonlinear NDDEs with constant delay and proportional delay. Enright and Hayashi (1998) estabilished sufficient conditions for order of convergence results about continuous Runge–Kutta methods for NDDEs with state dependent delays. But to the best of our knowledge, up to now, there are few articles referring to stability of numerical methods for INDDEs.

The aim of this paper is to provide asymptotical stability criteria for the exact solutions and the numerical solutions of INDDEs. Applying asymptotical stability of NDDEs without impulsive perturbations, asymptotical stability criteria are obtained for the exact solutions of linear INDDEs and nonlinear INDDEs, respectively. Numerical schemes for INDDEs are constructed based on the relationship between INDDEs and NDDEs. Moreover, we proved that some numerical methods can preserve asymptotical stability of linear and nonlinear INDDEs, respectively. The rest of this paper is organized as follows. In Sect. 2, asymptotical stability of zero solution of linear INDDEs is studied by the properties of simple linear NDDEs with constant coefficients. Base on this idea, numerical methods of INDDEs are constructed. The constructed numerical methods furnished A-stable Runge–Kutta methods can preserve asymptotical stability of linear INDDEs. In Sect. 3, some stability and asymptotical stability criteria are established for nonlinear INDDEs by the properties of nonlinear NDDEs. Under these stability and asymptotical stability criteria, the constructed numerical methods furnished by implicit Euler method or 2-stage Lobatto IIIC method are stability and asymptotical stability. In Sect. 4, we provide some numerical examples to confirm our theoretical results.

2 Linear INDDEs

In this section, we consider the following scalar linear INDDE:

$$\begin{aligned} {\left\{ \begin{array}{ll} x'(t) = ax(t)+bx(t-\tau )+cx'(t-\tau ), &{} t\ge 0, t \ne k\tau ,k\in \mathbb {N},\\ x(t)=\lambda x(t^-), &{}t=k\tau ,k\in \mathbb {N},\\ x(t)=\phi (t), &{} t\in [-\tau ,0), \end{array}\right. } \end{aligned}$$

(2.1)

where $\tau$ is a positive constant, a, b, c and $\lambda$ are complex constants, the initial function $\phi (t)$ is continuous differentiable on $[-\tau ,0)$, $x'(t)$ denotes the right-hand derivative of x(t) and $\mathbb {N}=\{0,1,2,\cdots \}$.

In the following of this paper, always assume $\lambda \ne 1$ and $\lambda \ne 0$. When $\lambda =1$, INDDE (2.1) is changed into NDDE without impulsive perturbations. When $\lambda =0$, the solution x(t) of INDDE (2.1) satisfies $x(k\tau )=0$ for all $k\in \mathbb {N}$.

2.1 Relations between INDDEs and NDDEs

Theorem 2.1

Assume that x(t) is the solution of (2.1) and $y(t)=\lambda ^{\{t/\tau \}}x(t)$, $t\in [-\tau ,\infty )$, then y(t) is the solution of the following equation without impulsive perturbations

$$\begin{aligned} {\left\{ \begin{array}{ll} y'(t) = \alpha y(t)+\beta y(t-\tau )+cy'(t-\tau ), &{} t\ge 0,\\ y(t)=\Phi (t), &{} t\in [-\tau ,0], \end{array}\right. } \end{aligned}$$

(2.2)

where $\alpha =a+\frac{\ln \lambda }{\tau }$, $\beta =b-\frac{c\ln \lambda }{\tau }$, $\{\frac{t}{\tau }\}=\frac{t}{\tau }-\lfloor \frac{t}{\tau }\rfloor$, $\lfloor \cdot \rfloor$ denotes the greatest integer function and

$$\begin{aligned} \Phi (t)= {\left\{ \begin{array}{ll} \lambda ^{(\frac{t}{\tau }+1)}\phi (t), &{} t\in [-\tau ,0),\\ \lambda \phi (0^-), &{} t=0. \end{array}\right. } \end{aligned}$$

On the other hand, if y(t) is the solution of (2.2) and $x(t)=\lambda ^{-\{t/\tau \}}y(t)$, $t\in [-\tau ,\infty )$, then x(t) is the solution of (2.1).

Proof

We will prove that y(t) is continuous if x(t) is the solution of (2.1). Obviously,

$$\begin{aligned} y(t)=\lambda ^{\{t/\tau \}}x(t)=\lambda ^{t/\tau -k}x(t) \end{aligned}$$

(2.3)

is continuous on $[k\tau ,(k+1)\tau )$, $k=-1,0, 1,2,\ldots$. Therefore,

$$\begin{aligned}{} & {} y(0^-)=\lim _{t\rightarrow 0^-}\lambda ^{t/\tau +1}x(t)=\lambda x(0^-)=\lambda \phi (0^-),\\{} & {} y(0)=x(0)=\lambda x(0^-)=\lambda \phi (0^-). \end{aligned}$$

Hence $y(0)=y(0^-)$, which implies that y(t) is continuous at $t=0$. It follows from

$$\begin{aligned}{} & {} y(k\tau )=\lambda ^{\{\frac{k\tau }{\tau }\}}x(k\tau )=x(k\tau )=\lambda x(k\tau ^-),\\{} & {} y(k\tau ^-)=\lim _{t\rightarrow k\tau ^-}\lambda ^{\{\frac{t}{\tau }\}}x(t)=\lambda x(k\tau ^-), \end{aligned}$$

that y(t) is continuous at $t=k\tau$, $k=1,2,\ldots$. Consequently, y(t) is continuous on $[-\tau ,\infty )$.

Next, we will prove that y(t) is the solution of (2.2) if x(t) is the solution of (2.1). For $t\in [k\tau ,(k+1)\tau )$, $k=0,1,2,\ldots$, we can obtain that

$$\begin{aligned} \begin{aligned} y'(t)&=\lambda ^{t/\tau -k}x(t)\frac{\ln \lambda }{\tau }+\lambda ^{t/\tau -k}x'(t)\\&=\frac{y(t)\ln \lambda }{\tau }+\lambda ^{t/\tau -k}x'(t), \end{aligned} \end{aligned}$$

and

$$\begin{aligned} \begin{aligned} y'(t-\tau )&=\lambda ^{(t-\tau )/\tau -(k-1)}x(t-\tau )\frac{\ln \lambda }{\tau }+\lambda ^{(t-\tau )/\tau -(k-1)}x'(t-\tau )\\&=\frac{y(t-\tau )\ln \lambda }{\tau }+\lambda ^{t/\tau -k}x'(t-\tau ), \end{aligned} \end{aligned}$$

which implies that

$$\begin{aligned} \begin{aligned}&~~~~y'(t) \\&=\frac{y(t)\ln \lambda }{\tau }+\lambda ^{t/\tau -k}\left[ ax(t)+bx(t-\tau )+cx'(t-\tau )\right] \\&=\frac{y(t)\ln \lambda }{\tau }+ay(t)+by(t-\tau )+c\lambda ^{t/\tau -k}x'(t-\tau )\\&=\left( a+\frac{\ln \lambda }{\tau }\right) y(t)+by(t-\tau )+c\left[ y'(t-\tau )-\frac{y(t-\tau )\ln \lambda }{\tau }\right] \\&=\left( a+\frac{\ln \lambda }{\tau }\right) y(t)+\left( b-\frac{c\ln \lambda }{\tau }\right) y(t-\tau )+cy'(t-\tau )\\&=\alpha y(t)+\beta y(t-\tau )+cy'(t-\tau ). \end{aligned} \end{aligned}$$

Finally, we will prove that x(t) is the solution of (2.1) if y(t) is the solution of (2.2). For $t\in [k\tau ,(k+1)\tau )$, $k=0,1,2,\cdots$, we can obtain that

$$\begin{aligned} \begin{aligned} x'(t)&=-\lambda ^{-(t/\tau -k)}y(t)\frac{\ln \lambda }{\tau }+\lambda ^{-(t/\tau -k)}y'(t)\\&=-\frac{x(t)\ln \lambda }{\tau }+\lambda ^{-(t/\tau -k)}y'(t), \end{aligned} \end{aligned}$$

and

$$\begin{aligned} \begin{aligned} x'(t-\tau )&=-\lambda ^{-[(t-\tau )/\tau -(k-1)]}y(t-\tau )\frac{\ln \lambda }{\tau }+\lambda ^{-[(t-\tau )/\tau -(k-1)]}y'(t-\tau )\\&=-\frac{x(t-\tau )\ln \lambda }{\tau }+\lambda ^{-(t/\tau -k)}y'(t-\tau ), \end{aligned} \end{aligned}$$

which implies that

$$\begin{aligned} \begin{aligned}&~~~~x'(t) \\&=-\frac{x(t)\ln \lambda }{\tau }+\lambda ^{-(t/\tau -k)}\left[ \left( a+\frac{\ln \lambda }{\tau }\right) y(t)+\left( b-\frac{c\ln \lambda }{\tau }\right) y(t-\tau )+cy'(t-\tau )\right] \\&=-\frac{x(t)\ln \lambda }{\tau }+\left( a+\frac{\ln \lambda }{\tau }\right) x(t)+\left( b-\frac{c\ln \lambda }{\tau }\right) x(t-\tau )+c\lambda ^{-(t/\tau -k)}y'(t-\tau )\\&=ax(t)+\left( b-\frac{c\ln \lambda }{\tau }\right) x(t-\tau )+c\left[ x'(t-\tau )+\frac{x(t-\tau )\ln \lambda }{\tau }\right] \\&=ax(t)+bx(t-\tau )+cx'(t-\tau ). \end{aligned} \end{aligned}$$

Obviously, it follows from

$$\begin{aligned}{} & {} x(k\tau )=\lambda ^{-\{\frac{k\tau }{\tau }\}}y(k\tau )=y(k\tau ),\\{} & {} x(k\tau ^-)=\lim _{t\rightarrow k\tau ^-}\lambda ^{\{\frac{x}{\tau }\}}x(t)=\lambda ^{-1} y(k\tau ^-), \end{aligned}$$

that $x(k\tau )=\lambda x(k\tau ^-)$, $k=0,1,2,\ldots$. Hence x(t) is the solution of (2.1). $\square$

2.2 Asmpotical stability of linear INDDEs

By (Bellen et al. 1988, Theorem 2.1) and Theorem 2.1 of present paper, we can obtain the following results.

Theorem 2.2

Assume that

$$\begin{aligned} |\alpha \bar{c}-\bar{\beta }|+|\alpha c+\beta |<-2\Re (\alpha ). \end{aligned}$$

Then the solution y(t) of NDDE (2.2) tends to zero as $t\rightarrow \infty$. Hence if

$$\begin{aligned} \left| \left( a+\frac{\ln \lambda }{\tau }\right) \bar{c}-\overline{\left( b-\frac{c\ln \lambda }{\tau }\right) }\right| +\left| \left( a+\frac{\ln \lambda }{\tau }\right) c+\left( b-\frac{c\ln \lambda }{\tau }\right) \right| <-2\Re \left( a+\frac{\ln \lambda }{\tau }\right) , \end{aligned}$$

then the solution x(t) of INDDE (2.1) tends to zero as $t\rightarrow \infty$.

Maybe some readers are interested in linear INDDE as the following form.

$$\begin{aligned} {\left\{ \begin{array}{ll} x'(t) = ax(t)+bx(t-\tau )+cx'(t-\tau ), &{} t\ge 0, t \ne k\tau ,k\in \mathbb {Z}^+,\\ x(t)=\lambda x(t^-), &{}t=k\tau ,k\in \mathbb {Z}^+,\\ x(t)=\phi (t), &{} t\in [-\tau ,0], \end{array}\right. } \end{aligned}$$

(2.4)

where $\mathbb {Z}^+=\{1,2,\ldots \}$. In fact, on interval $[-\tau ,\tau )$, the Eq. (2.4) is a delay differential equations without impulsive perturbations. If the solution x(t), $t\in [0,\tau )$ is seen as initial function, the Eq. (2.4) can be seen the same as (2.1) for $t\ge \tau$. So the asymptotically stable results of Theorem 2.2 can be extended as follows.

Corollary 2.3

If

$$\begin{aligned} \left| \left( a+\frac{\ln \lambda }{\tau }\right) \bar{c}-\overline{\left( b-\frac{c\ln \lambda }{\tau }\right) }\right| +\left| \left( a+\frac{\ln \lambda }{\tau }\right) c+\left( b-\frac{c\ln \lambda }{\tau }\right) \right| <-2\Re \left( a+\frac{\ln \lambda }{\tau }\right) , \end{aligned}$$

then the solution x(t) of INDDE (2.4) tends to zero as $t\rightarrow \infty$.

In fact, all the stable and asymptotically stable results of INDDE in this paper can be extended similarly, we do not introduce in detail for concise.

2.3 Stability analysis of numerical methods for linear INDDEs

Based on the relations between INDDE (2.1) and NDDE (2.2), the numerical method for linear INDDE (2.1) can be constructed as follows

$$\begin{aligned} {\left\{ \begin{array}{ll} &{}y_{n+1}=y_n+\sum _{i=1}^s w_i z_{n+1}^{(i)},\;\;n\in \mathbb {N}\\ &{}z_{n+1}^{(i)}=\alpha \left( y_n+h\sum _{j=1}^sa_{ij}z_{n+1}^{(j)}\right) +\beta \left( y_{n-m}+h\sum _{j=1}^sb_{ij}z_{n-m+1}^{(j)}\right) \\ &{}~~~~~~~~~\quad +c\sum _{j=1}^sc_{ij}z_{n-m+1}^{(j)}\\ &{}x_n=\lambda ^{-\{\frac{t_n}{\tau }\}}y_n, \end{array}\right. } \end{aligned}$$

(2.5)

where $t_n=nh$, $n\in \mathbb {N}$, $h=\frac{\tau }{m}$ and m is a positive integer. Here, the vector $w=[w_1,w_2,\ldots ,w_s]^T$ and the matrix $A=[a_{ij}]_{i,j=1}^s$ define a Runge–Kutta method for ODEs. For $\forall n \in \mathbb {N}$, $y_n$ is an approximation to $y(t_n)$ of (2.2), $x_n$ is an approximation to $x(t_n)$ of (2.1), $y_n+h\sum _{j=1}^sa_{ij}z_{n+1}^{(j)}$ is an approximation to $y(t_n+c_ih)$, $y_{n-m}+h\sum _{j=1}^sb_{ij}z_{n-m+1}^{(j)}$ is an approximation to $y(t_{n-m}+c_ih)$, $\sum _{j=1}^sc_{ij}z_{n-m+1}^{(j)}$ is an approximation to $y'(t_{n-m}+c_ih)$, where $c_i=\sum _{j=1}^s a_{ij}$, $i=1,2,\ldots ,s$.

Usually, $b_{ij}=b_j(c_i)$ and $c_{ij}=b_j'(c_i)$, where $b_j(\theta )$, $j=1,2,\ldots ,s$, are polynomials which define the natural continuous extension of the Runge–Kutta method, i.e. polynomials such that the approximate solution $y_h$ defined on the whole interval of integration is given by

$$\begin{aligned} y_h(t_n+\theta h)=y_h(t_n)+h\sum _{j=1}^sb_j(\theta )z_{n+1}^{(j)},\;n\in \mathbb {N},\theta \in [0,1]. \end{aligned}$$

Put $B=[b_{ij}]_{i,j=1}^s$ and $C=[c_{ij}]_{i,j=1}^s$. Let $\tilde{\alpha }=h\alpha$, $\tilde{\beta }=h\beta$ and denote by $\{y_n(m;\alpha ,\beta ,c)\}_{n=0}^{\infty }$ can be described by quadrupple $\{w,A,B,C\}$. So the numerical method (2.5) for linear INDDE (2.1) can be rewritten as follows

$$\begin{aligned} {\left\{ \begin{array}{ll} &{}y_{n+1}=y_n+hw^T z_{n+1},\;\;n\in \mathbb {N}\\ &{}z_{n+1}=\alpha (y_ne+hAz_{n+1})+\beta (y_{n-m}e+hBz_{n-m+1})+cCz_{n-m+1}\\ &{}x_n=\lambda ^{-\{\frac{t_n}{\tau }\}}y_n, \end{array}\right. } \end{aligned}$$

(2.6)

where $z_n$ stands for $[z_n^{(1)},z_n^{(2)},\ldots ,z_n^{(s)}]^T$ and $e=[1,1,\ldots ,1]^T$.

A-stable Runge–Kutta methods for ODEs can be extended to asympotically stable numerical method for INDDEs. By (Bellen et al. 1988, Theorem 3.4), we can obtain the following theorem.

Theorem 2.4

Assume that the Runge–Kutta method $\{w,A\}$ for ODE is A-stable. Then the corresponding method (2.6) for INDDE (2.1) with $B=A$ and $C=I$, which satisfies

$$\begin{aligned} \left| \left( a+\frac{\ln \lambda }{\tau }\right) \bar{c}-\overline{\left( b-\frac{c\ln \lambda }{\tau }\right) }\right| +\left| \left( a+\frac{\ln \lambda }{\tau }\right) c+\left( b-\frac{c\ln \lambda }{\tau }\right) \right| <-2\Re \left( a+\frac{\ln \lambda }{\tau }\right) . \end{aligned}$$

Then the numerical solution $x_n$ of (2.6) tends to zero as $n\rightarrow \infty$.

3 Nonlinear INDDEs

In this section, we will consider the following nonlinear INDDE:

$$\begin{aligned} {\left\{ \begin{array}{ll} x'(t)=f(t,x(t),x(t-\tau ),x'(t-\tau )), &{} {t\ge 0,\; t\ne k\tau ,\;k\in \mathbb {N}}\\ x(t)=\lambda x(t^-), &{} {t= k\tau ,\;k\in \mathbb {N}}\\ x(t)=\phi (t),&{} {t\in [-\tau , 0),} \end{array}\right. } \end{aligned}$$

(3.1)

and the same equation with another initial function:

$$\begin{aligned} {\left\{ \begin{array}{ll} \tilde{x}'(t)=f(t,\tilde{x}(t),\tilde{x}(t-\tau ),\tilde{x}'(t-\tau )), &{} {t\ge 0,\; t\ne k\tau ,\;k\in \mathbb {N}}\\ \tilde{x}(t)=\lambda \tilde{x}(t^-), &{} {t= k\tau ,\;k\in \mathbb {N}}\\ \tilde{x}(t)=\tilde{\phi } (t),&{} {t\in [-\tau , 0),} \end{array}\right. } \end{aligned}$$

(3.2)

where $\lambda \ne 1$, $\lambda \ne 0$, $\tau >0$, $\phi$ and $\tilde{\phi }$ are continuous functions on $[-\tau , 0)$, $\lim \limits _{t\rightarrow 0^-}\phi (t)$ and $\lim \limits _{t\rightarrow 0^-}\tilde{\phi }(t)$ exist, $x'(t)$ denotes the right-hand derivative of x(t). Let $\langle \cdot ,\cdot \rangle$ be a given inner product on $\mathbb {C}^{d}$ and $\Vert \cdot \Vert$ the corresponding norm. Assume that the function $f: [0, \infty )\times \mathbb {C}^d\times \mathbb {C}^d\times \mathbb {C}^d\rightarrow \mathbb {C}^d$ is continuous in t and satisfies the following conditions: for arbitrary $x, u, v, x_1, x_2, u_1, u_2, v_1, v_2 \in \mathbb {C}^d$ and $\forall t \in [0,+\infty )$,

$$\begin{aligned}{} & {} \Re (\langle x_1-x_2, f(t,x_1,u,v)-f(t,x_2,u,v)\rangle )\le R(t) \Vert x_1-x_2\Vert ^2 \end{aligned}$$

(3.3)

$$\begin{aligned}{} & {} \Vert f(t,x,u_1,v)-f(t,x,u_2,v)\Vert \le \beta (t) \Vert u_1-u_2\Vert \end{aligned}$$

(3.4)

$$\begin{aligned}{} & {} \Vert f(t,x,u,v_1)-f(t,x,u,v_2)\Vert \le \gamma (t) \Vert v_1-v_2\Vert \end{aligned}$$

(3.5)

$$\begin{aligned}{} & {} \Vert U(t,x,u_1,v,w)-U(t,x,u_2,v,w)\Vert \le \sigma (t) \Vert u_1-u_2\Vert \end{aligned}$$

(3.6)

where R(t), $\beta (t)$, $\gamma (t)$ and $\sigma (t)$ are continuous real functions and

$$\begin{aligned} U(t,x,u,v,w)=f(t,x,u,f(t-\tau ,u,v,w)). \end{aligned}$$

3.1 Relations between INDDEs and NDDEs

Assume that the scalar function $\alpha :[-\tau ,\infty )\rightarrow \mathbb {C}$ satisfies the following conditions:

• (1) for any $t \in [0, \infty )$, $\alpha (t)=\alpha (t-\tau )$;

• (2) $\alpha (t)$ is infinite smooth on $[0, \tau )$;

• (3) $\alpha (0)=1$ and $\alpha (0^-)=\lambda$;

• (4) $\inf _{t \in [0, \tau )} |\alpha (t)|\ge m>0$.

Theorem 3.1

Assume that x(t) is the solution of (2.1) and $y(t)=\alpha (t)x(t),\;t\in [-\tau ,+\infty )$. Then y(t) is the solution of

$$\begin{aligned} {\left\{ \begin{array}{ll} y'(t)=F(t,y(t),y(t-\tau ),y'(t-\tau )), &{} t\ge 0,\\ y(t)=\Phi (t), &{} t\in [-\tau ,0], \end{array}\right. } \end{aligned}$$

(3.7)

where

$$\begin{aligned} F(t,y,u,v)=\frac{\alpha '(t)y}{\alpha (t)}+\alpha (t)f(t,\frac{y}{\alpha (t)},\frac{u}{\alpha (t)},\frac{v}{\alpha (t)}-\frac{\alpha '(t)u}{\alpha ^2(t)}) \end{aligned}$$

and

$$\begin{aligned} \Phi (t)= \left\{ \begin{array}{l} \alpha (t)\phi (t), ~~~~t\in [-\tau ,0),\\ \alpha (0^-)\phi (0^-),~~~~~t=0. \end{array} \right. \end{aligned}$$

On the other hand, assume that y(t) is the solution of (3.7) and $x(t)= \frac{y(t)}{\alpha (t)},\;t\in [-\tau ,+\infty )$. Then x(t) is the solution of (2.1).

Proof

(i) Because $\alpha (t)$ and x(t) are continuous on $[k\tau , (k+1)\tau )$, y(t) is continuous on $[k\tau , (k+1)\tau )$, where $k=0,1,\ldots$. Obviously, we have

$$\begin{aligned} y(k\tau )= & {} y(k\tau ^+) = \alpha (k\tau ^+)x(k\tau ^+)\\= & {} \alpha (k\tau )\lambda x(k\tau ^-)= \alpha (0)\lambda x(k\tau ^-)\\= & {} \lambda x(k\tau ^-) \end{aligned}$$

and

$$\begin{aligned} y(k\tau ^-)= \alpha (k\tau ^-)x(k\tau ^-)=\lambda x(k\tau ^-). \end{aligned}$$

So $y(k\tau )= y(k\tau ^+)=y(k\tau ^-),\;k=0,1,2,\ldots$. Hence y(t) is continuous on $[-\tau , \infty )$.

Obviously, for $t\in [k\tau , (k+1)\tau ), k=0,1,\ldots$, we can obtain that

$$\begin{aligned} \begin{aligned}&~~~~x'(t)= (\frac{y(t)}{\alpha (t)})'=\frac{y'(t)}{\alpha (t)}-\frac{\alpha '(t)y(t)}{\alpha ^2(t)} \\&\Rightarrow x'(t-\tau )=\frac{y'(t-\tau )}{\alpha (t)}-\frac{\alpha '(t)y(t-\tau )}{\alpha ^2(t)}, \end{aligned} \end{aligned}$$

which implies

$$\begin{aligned} \begin{aligned}&~~~~y'(t)=\alpha '(t)x(t)+\alpha (t)x'(t) \\&=\alpha '(t)x(t)+\alpha (t)f(t,x(t),x(t-\tau ),x'(t-\tau ))\\&=\frac{\alpha '(t)y(t)}{\alpha (t)}+\alpha (t)f\left( t,\frac{y(t)}{\alpha (t)},\frac{y(t-\tau )}{\alpha (t-\tau )},\frac{y'(t-\tau )}{\alpha (t-\tau )}-\frac{\alpha '(t-\tau )y(t-\tau )}{\alpha ^2(t-\tau )}\right) \\&=\frac{\alpha '(t)y(t)}{\alpha (t)}+\alpha (t)f\left( t,\frac{y(t)}{\alpha (t)},\frac{y(t-\tau )}{\alpha (t)},\frac{y'(t-\tau )}{\alpha (t)}-\frac{\alpha '(t)y(t-\tau )}{\alpha ^2(t)}\right) \\&=F(t,y(t),y(t-\tau ),y'(t-\tau )). \end{aligned} \end{aligned}$$

(ii) Assume that y(t) is the solution of (3.7). For $t\in [k\tau , (k+1)\tau ), k=0,1,\ldots$,

$$\begin{aligned} \begin{aligned}&~~~~x'(t)= (\frac{y(t)}{\alpha (t)})'=\frac{y'(t)}{\alpha (t)}-\frac{\alpha '(t)y(t)}{\alpha ^2(t)}\\&=\frac{1}{\alpha (t)}\left( \frac{\alpha '(t)y(t)}{\alpha (t)}+\alpha (t)f\left( t,\frac{y(t)}{\alpha (t)},\frac{y(t-\tau )}{\alpha (t)},\frac{y'(t-\tau )}{\alpha (t)}-\frac{\alpha '(t)y(t-\tau )}{\alpha ^2(t)}\right) \right) -\frac{\alpha '(t)y(t)}{\alpha ^2(t)}\\&=f\left( t,\frac{y(t)}{\alpha (t)},\frac{y(t-\tau )}{\alpha (t)},\frac{y'(t-\tau )}{\alpha (t)}-\frac{\alpha '(t)y(t-\tau )}{\alpha ^2(t)}\right) \\&=f(t,x(t),x(t-\tau ),x'(t-\tau )). \end{aligned} \end{aligned}$$

Obviously,

$$\begin{aligned} x(k\tau )=\frac{y(k\tau )}{\alpha (k\tau )}=\frac{y(k\tau )}{\alpha (0)}=y(k\tau ) \end{aligned}$$

and

$$\begin{aligned} x(k\tau ^-)=\lim _{t\rightarrow k\tau ^-}\frac{y(t)}{\alpha (t)}=\frac{y(k\tau )}{\alpha (k\tau ^-)}=\frac{y(k\tau )}{\alpha (\tau ^-)}=\frac{y(k\tau )}{\lambda }. \end{aligned}$$

So $x(k\tau )=\lambda x(k\tau ^-), k=0,1,\cdots$. Obviously, we have $x(t)=\frac{y(t)}{\alpha (t)}=\phi (t)$, $t\in [-\tau , 0)$. Hence x(t) is the solution of (3.1). $\square$

3.2 Asymptotical stability of INDDEs

Theorem 3.2

Assume that IDDEs (3.1) and (3.2) fulfill the inequalities (3.3)-(3.6), and $R(t)\le 0$ for $\forall t\ge 0$,

$$\begin{aligned} \Re \left( \frac{\alpha '(t)}{\alpha (t)}\right) + \frac{R(t)}{\omega ^2}<0,\;\;\frac{\omega \sigma (t)-\omega \gamma (t)\left[ \Re \left( \frac{\alpha '(t)}{\alpha (t)}\right) +\frac{R(t-\tau )}{\omega ^2}\right] }{-\left[ \Re \left( \frac{\alpha '(t)}{\alpha (t)}\right) +\frac{R(t)}{\omega ^2}\right] }\le 1, \end{aligned}$$

(3.8)

then

$$\begin{aligned} \Vert x(t)-\tilde{x}(t)\Vert \le \ \max \{\frac{\lambda \Vert \phi (0^-)-\tilde{\phi }(0^-)\Vert }{m},\kappa \}, \end{aligned}$$

(3.9)

where $\omega =\frac{\sup _{t\in [-\tau , 0)}|\alpha (t)|}{\inf _{t\in [-\tau , 0)}|\alpha (t)|}$, $\zeta =\sup \nolimits _{t\in [-\tau , 0)}|\frac{\alpha '(t)}{\alpha (t)}|$, $\delta \phi (s)=\Vert \phi (s-\tau )-\tilde{\phi }(s-\tau )\Vert$, $\delta \dot{\phi }(s)=\Vert \phi '(s-\tau )-\tilde{\phi }'(s-\tau )\Vert$ and

$$\begin{aligned} \kappa =\sup _{0\le s < \tau }\frac{\omega ^2(\beta (s)+2\zeta \gamma (s))\delta \phi (s)+\omega ^2 \gamma (s)\delta \dot{\phi }(s)}{-\left( {\Re \left( \frac{\alpha '(t)}{\alpha (t)}\right) +\frac{R(t)}{\omega ^2}}\right) }. \end{aligned}$$

Moreover, assume that

$$\begin{aligned} \Re \left( \frac{\alpha '(t)}{\alpha (t)}\right) + \frac{R(t)}{\omega ^2} \le R_0<0,\;\forall t\ge 0, \end{aligned}$$

(3.10)

and

$$\begin{aligned}{} & {} \frac{\omega \gamma (t)\left( \Re \left( \frac{\alpha '(t)}{\alpha (t)}\right) + \frac{R(t-\tau )}{\omega ^2}\right) }{\Re \left( \frac{\alpha '(t)}{\alpha (t)}\right) +\frac{R(t)}{\omega ^2} }=r(t)\le \bar{\xi }<1,\;\forall t\ge 0, \end{aligned}$$

(3.11)

$$\begin{aligned}{} & {} \frac{\omega \sigma (t)}{-\left( \Re \left( \frac{\alpha '(t)}{\alpha (t)}\right) + \frac{R(t)}{\omega ^2}\right) }\le k(1-r(t)),\;k<1,\;\forall t\ge 0, \end{aligned}$$

(3.12)

then we have

$$\begin{aligned} \lim \limits _{t\rightarrow \infty } \Vert x(t)-\tilde{x}(t) \Vert =0. \end{aligned}$$

Proof

We will apply inequalities (3.3)-(3.6) to prove that the function $F: [0, \infty )\times \mathbb {C}^d\times \mathbb {C}^d\times \mathbb {C}^d\rightarrow \mathbb {C}^d$ satisfies the following conditions respectively: for arbitrary $y, u, v, y_1, y_2, u_1, u_2, v_1, v_2 \in \mathbb {C}^d$ and $\forall t \in [0,+\infty )$,

$$\begin{aligned}{} & {} \Re (\langle y_1-y_2, F(t,y_1,u,v)-F(t,y_2,u,v)\rangle )\le \left( \Re \left( \frac{\alpha '(t)}{\alpha (t)}\right) + \frac{R(t)}{\omega ^2} \right) \Vert y_1-y_2\Vert ^2 \end{aligned}$$

(3.13)

$$\begin{aligned}{} & {} \Vert F(t,y,u_1,v)-F(t,y,u_2,v)\Vert \le \omega \left( \beta (t) +\zeta \gamma (t)\right) \Vert u_1-u_2\Vert \end{aligned}$$

(3.14)

$$\begin{aligned}{} & {} \Vert F(t,y,u,v_1)-F(t,y,u,v_2)\Vert \le \omega \gamma (t) \Vert v_1-v_2\Vert \end{aligned}$$

(3.15)

$$\begin{aligned}{} & {} \Vert V(t,y,u_1,v,w)-V(t,y,u_2,v,w)\Vert \le \omega \sigma (t) \Vert u_1-u_2\Vert \end{aligned}$$

(3.16)

where $V(t,y,u,v,w)=F(t,y,u,F(t-\tau ,u,v,w)).$

First of all, we will prove inequality (3.13) as follows

$$\begin{aligned} \begin{aligned}&~~~~\Re \langle y_1-y_2, F(t,y_1,u,v)-F(t,y_2,u,v)\rangle \\&=\Re \langle y_1-y_2, \frac{\alpha '(t)y_1}{\alpha (t)}+\alpha (t)f\left( t,\frac{y_1}{\alpha (t)},\frac{u}{\alpha (t)},\frac{v}{\alpha (t)}-\frac{\alpha '(t)u}{\alpha ^2(t)}\right) -\frac{\alpha '(t)y_2}{\alpha (t)}\\&~~~~~-\alpha (t)f\left( t,\frac{y_2}{\alpha (t)},\frac{u}{\alpha (t)},\frac{v}{\alpha (t)}-\frac{\alpha '(t)u}{\alpha ^2(t)}\right) \rangle \\&=\Re \langle y_1-y_2, \frac{\alpha '(t)}{\alpha (t)}(y_1-y_2)\rangle +|\alpha (t)|^2\Re \langle \frac{y_1}{\alpha (t)}-\frac{y_2}{\alpha (t)}, f\left( t,\frac{y_1}{\alpha (t)},\frac{u}{\alpha (t)},\frac{v}{\alpha (t)}-\frac{\alpha '(t)u}{\alpha ^2(t)}\right) \\&~~~~~-f\left( t,\frac{y_2}{\alpha (t)},\frac{u}{\alpha (t)},\frac{v}{\alpha (t)}-\frac{\alpha '(t)u}{\alpha ^2(t)}\right) \rangle \\&\le \Re \left( \frac{\alpha '(t)}{\alpha (t)}\right) \Vert y_1-y_2\Vert ^2+|\alpha (t)|^2 R(t)\Vert \frac{y_1}{\alpha (t)}-\frac{y_2}{\alpha (t)}\Vert ^2 \end{aligned} \end{aligned}$$

which implies, if $R(t)\le 0$ for $\forall t$,

$$\begin{aligned} \Re \langle y_1-y_2, F(t,y_1,u,v)-F(t,y_2,u,v)\rangle \le [\Re \left( \frac{\alpha '(t)}{\alpha (t)}\right) +\frac{R(t)}{\omega ^2}]\Vert y_1-y_2\Vert ^2, \end{aligned}$$

(3.17)

and if $R(t)\le \hat{R}$ and $\hat{R}>0$ for $\forall t$,

$$\begin{aligned} \Re \langle y_1-y_2, F(t,y_1,u,v)-F(t,y_2,u,v)\rangle \le [\Re \left( \frac{\alpha '(t)}{\alpha (t)}\right) +\omega ^2\hat{R}]\Vert y_1-y_2\Vert ^2. \end{aligned}$$

(3.18)

Second, the inequality (3.14) can be proved as follows

$$\begin{aligned} \begin{aligned}&~~~~\Vert F(t,y,u_1,v)-F(t,y,u_2,v)\Vert \\&=\Vert \frac{\alpha '(t)y}{\alpha (t)}+\alpha (t)f\left( t,\frac{y}{\alpha (t)},\frac{u_1}{\alpha (t)},\frac{v}{\alpha (t)}-\frac{\alpha '(t)u_1}{\alpha ^2(t)}\right) \\&~~~~-\left[ \frac{\alpha '(t)y}{\alpha (t)}+\alpha (t)f\left( t,\frac{y}{\alpha (t)},\frac{u_2}{\alpha (t)},\frac{v}{\alpha (t)}-\frac{\alpha '(t)u_2}{\alpha ^2(t)}\right) \right] \Vert \\&\le (\sup _{-\tau \le t<0}|\alpha (t)|)\Vert f\left( t,\frac{y}{\alpha (t)},\frac{u_1}{\alpha (t)},\frac{v}{\alpha (t)}-\frac{\alpha '(t)u_1}{\alpha ^2(t)}\right) -f\left( t,\frac{y}{\alpha (t)},\frac{u_2}{\alpha (t)},\frac{v}{\alpha (t)}-\frac{\alpha '(t)u_2}{\alpha ^2(t)}\right) \Vert \\&\le (\sup _{-\tau \le t<0}|\alpha (t)|)\Vert f\left( t,\frac{y}{\alpha (t)},\frac{u_1}{\alpha (t)},\frac{v}{\alpha (t)}-\frac{\alpha '(t)u_1}{\alpha ^2(t)}\right) -f\left( t,\frac{y}{\alpha (t)},\frac{u_2}{\alpha (t)},\frac{v}{\alpha (t)}-\frac{\alpha '(t)u_1}{\alpha ^2(t)}\right) \Vert \\&~~~+(\sup _{-\tau \le t<0}|\alpha (t)|)\Vert f\left( t,\frac{y}{\alpha (t)},\frac{u_2}{\alpha (t)},\frac{v}{\alpha (t)}-\frac{\alpha '(t)u_1}{\alpha ^2(t)}\right) -f\left( t,\frac{y}{\alpha (t)},\frac{u_2}{\alpha (t)},\frac{v}{\alpha (t)}-\frac{\alpha '(t)u_2}{\alpha ^2(t)}\right) \Vert \\&\le \left( \sup _{-\tau \le t<0}|\alpha (t)|\right) \left[ \beta (t)\Vert \frac{u_1}{\alpha (t)}-\frac{u_2}{\alpha (t)}\Vert +\gamma (t)\Vert \frac{v}{\alpha (t)}-\frac{\alpha '(t)u_1}{\alpha ^2(t)}-\left( \frac{v}{\alpha (t)}-\frac{\alpha '(t)u_2}{\alpha ^2(t)}\right) \Vert \right] \\&\le \omega (\beta (t)+\gamma (t)\zeta )\Vert u_1-u_2\Vert \end{aligned} \end{aligned}$$

Third, the inequality (3.15) can be proved as follows

$$\begin{aligned} \begin{aligned}&~~~~\Vert F(t,y,u,v_1)-F(t,y,u,v_2)\Vert \\&=\Vert \frac{\alpha '(t)y}{\alpha (t)}+\alpha (t)f\left( t,\frac{y}{\alpha (t)},\frac{u}{\alpha (t)},\frac{v_1}{\alpha (t)}-\frac{\alpha '(t)u}{\alpha ^2(t)}\right) \\&~~~~~-\left[ \frac{\alpha '(t)y}{\alpha (t)}+\alpha (t)f\left( t,\frac{y}{\alpha (t)},\frac{u}{\alpha (t)},\frac{v_2}{\alpha (t)}-\frac{\alpha '(t)u}{\alpha ^2(t)}\right) \right] \Vert \\&\le \left( \sup _{-\tau \le t<0}|\alpha (t)|\right) \gamma (t)\Vert \frac{v_1}{\alpha (t)}-\frac{\alpha '(t)u}{\alpha ^2(t)}-\left( \frac{v_2}{\alpha (t)}-\frac{\alpha '(t)u}{\alpha ^2(t)}\right) \Vert \\&\le \omega \gamma (t)\Vert v_1-v_2\Vert \end{aligned} \end{aligned}$$

Finally, the inequality (3.16) can be proved as follows

$$\begin{aligned} \begin{aligned}&~~~~\Vert V(t,y,u_1,v,w)-V(t,y,u_2,v,w)\Vert \\&=\Vert F(t,y,u_1,F(t-\tau ,u_1,v,w))-F(t,y,u_2,F(t-\tau ,u_2,v,w))\Vert \\&=\left( \sup _{-\tau \le t<0}|\alpha (t)|\right) \Vert f\left( t,\frac{y}{\alpha (t)},\frac{u_1}{\alpha (t)},f\left( t-\tau ,\frac{u_1}{\alpha (t)},\frac{v}{\alpha (t)},\frac{w}{\alpha (t)}-\frac{\alpha '(t)v}{\alpha ^2(t)}\right) \right) \\&~~~~~~~~~~~~~~~~~~~~~~~-f\left( t,\frac{y}{\alpha (t)},\frac{u_2}{\alpha (t)},f\left( t-\tau ,\frac{u_2}{\alpha (t)},\frac{v}{\alpha (t)},\frac{w}{\alpha (t)}-\frac{\alpha '(t)v}{\alpha ^2(t)}\right) \right) \Vert \\ {}&=\left( \sup _{-\tau \le t<0}|\alpha (t)|\right) \Vert U\left( t,\frac{y}{\alpha (t)},\frac{u_1}{\alpha (t)},\frac{v}{\alpha (t)},\frac{w}{\alpha (t)}-\frac{\alpha '(t)v}{\alpha ^2(t)}\right) \\&~~~~~~~~~~~~~~~~~~~~~~-U\left( t,\frac{y}{\alpha (t)},\frac{u_2}{\alpha (t)},\frac{v}{\alpha (t)},\frac{w}{\alpha (t)}-\frac{\alpha '(t)v}{\alpha ^2(t)}\right) \Vert \\&\le \left( \sup _{-\tau \le t<0}|\alpha (t)|\right) \sigma (t) \Vert \frac{u_1}{\alpha (t)}-\frac{u_2}{\alpha (t)}\Vert \\&\le \omega \sigma (t) \Vert u_1-u_2\Vert \end{aligned} \end{aligned}$$

By the condition (3.8) and (Wang et al. 2009, Theorem 3.1), (Wang and Li 2004, Theorem 1), we can obtain that

$$\begin{aligned} \Vert y(t)-\tilde{y}(t)\Vert \le \max \{\Vert \Phi (0)-\tilde{\Phi }(0)\Vert ,\hat{\kappa }\}, \end{aligned}$$

which implies (3.9) holds, where

$$\begin{aligned} \hat{\kappa }=\sup _{0\le s \le \tau }\frac{(\omega \beta (s)+\omega \zeta \gamma (t))\Vert \Phi (s-\tau )-\tilde{\Phi }(s-\tau )\Vert +\omega \gamma (s)\Vert \Phi '(s-\tau )-\tilde{\Phi }'(s-\tau )\Vert }{-\left[ \Re \left( \frac{\alpha '(t)}{\alpha (t)}\right) + \frac{R(t)}{\omega ^2} \right] }. \end{aligned}$$

Moreover, by the condition (3.10)-(3.12) and (Wang et al. 2009, Theorem 3.2), (Wang and Li 2004, Theorem 4), we can obtain that

$$\begin{aligned} \lim \limits _{t\rightarrow \infty } \Vert y(t)-\tilde{y}(t) \Vert =0, \end{aligned}$$

which implies

$$\begin{aligned} \lim \limits _{t\rightarrow \infty } \Vert x(t)-\tilde{x}(t) \Vert =0. \end{aligned}$$

$\square$

Due to the difference in (3.17) and (3.18), similar to Theorem 3.2, by (Wang et al. 2009, Theorem 3.2) or (Wang and Li 2004, Theorem 4), we can obtain the following theorem.

Theorem 3.3

Assume that IDDEs (3.1) and (3.2) fulfill the inequalities (3.3)-(3.6), $R(t)\le \hat{R}$, $\hat{R}>0$, and for $\forall t\ge 0$,

$$\begin{aligned} \Re \left( \frac{\alpha '(t)}{\alpha (t)}\right) +\omega ^2 \hat{R}<0,\;\;\frac{\omega \sigma (t)-\omega \gamma (t)\left[ \Re \left( \frac{\alpha '(t)}{\alpha (t)}\right) +\omega ^2\hat{R}\right] }{-\left[ \Re \left( \frac{\alpha '(t)}{\alpha (t)}\right) +\omega ^2\hat{R}\right] }\le 1, \end{aligned}$$

(3.19)

then

$$\begin{aligned} \Vert x(t)-\tilde{x}(t)\Vert \le \ \max \{\frac{\lambda \Vert \phi (0^-)-\tilde{\phi }(0^-)\Vert }{m},\bar{\kappa }\}, \end{aligned}$$

(3.20)

where

$$\begin{aligned} \bar{\kappa }=\sup _{0\le s < \tau }\frac{\omega ^2(\beta (s)+2\zeta \gamma (s))\delta \phi (s)+\omega ^2\gamma (s)\delta \dot{\phi }(s)}{-\left[ \Re \left( \frac{\alpha '(t)}{\alpha (t)}\right) +\omega ^2 \hat{R}\right] }. \end{aligned}$$

Moreover, assume that

$$\begin{aligned}{} & {} \Re \left( \frac{\alpha '(t)}{\alpha (t)}\right) +\omega ^2 \hat{R}\le \bar{R}<0,\; \omega \gamma (t)\le \bar{\xi }<1,\;\forall t\ge 0, \end{aligned}$$

(3.21)

$$\begin{aligned}{} & {} \frac{\omega \sigma (t)}{-\left[ \Re \left( \frac{\alpha '(t)}{\alpha (t)}\right) +\omega ^2 \hat{R}\right] }\le \bar{k}(1-\omega \gamma (t)),\;\bar{k}<1,\;\forall t\ge 0, \end{aligned}$$

(3.22)

then we have

$$\begin{aligned} \lim \limits _{t\rightarrow \infty } \Vert x(t)-\tilde{x}(t) \Vert =0. \end{aligned}$$

3.3 Special cases

In this subsection, we only introduce three special cases and some sufficient conditions for stability and asymptotical stability of (3.1) and (3.2) are obtained from them. Many different functions can be chosen as $\alpha (t)$ (see Zhang et al. 2015), which implies many special cases can be obtained.

Special case I. Obviously, $\alpha _1(t)=\lambda ^{\{\frac{t}{\tau }\}}$, $t\in [-\tau , \infty )$. By Theorem 3.2, when $\alpha (t)=\alpha _1(t)=\lambda ^{\{\frac{t}{\tau }\}}$, we can obtain the following theorem.

Theorem 3.4

Assume that x(t) is the solution of (2.1) and $y_1(t)=\lambda ^{\{\frac{t}{\tau }\}}x(t),\;t\in [-\tau ,+\infty )$. Then $y_1(t)$ is the solution of

$$\begin{aligned} {\left\{ \begin{array}{ll} y_1'(t)=F_1(t,y_1(t),y_1(t-\tau ),y_1'(t-\tau )), &{} t\ge 0,\\ y_1(t)=\Phi _1(t), &{} t\in [-\tau ,0], \end{array}\right. } \end{aligned}$$

(3.23)

where

$$\begin{aligned} F_1(t,y,u,v)=\left( \frac{\ln \lambda }{\tau }\right) y+\lambda ^{\{\frac{t}{\tau }\}}f\left( t,\lambda ^{-\{\frac{t}{\tau }\}}y,\lambda ^{-\{\frac{t}{\tau }\}}u,\lambda ^{-\{\frac{t}{\tau }\}}\left( v-\frac{u\ln \lambda }{\tau }\right) \right) \end{aligned}$$

and

$$\begin{aligned} \Phi _1(t)= \left\{ \begin{array}{l} \lambda ^{\frac{t}{\tau }+1}\phi (t), ~~~~t\in [-\tau ,0),\\ \lambda \phi (0^-),~~~~~~t=0. \end{array} \right. \end{aligned}$$

On the other hand, assume that y(t) is the solution of (3.23) and $x(t)=\lambda ^{-\{\frac{t}{\tau }\}}y(t),\;t\in [-\tau ,+\infty )$. Then x(t) is the solution of (2.1).

Theorem 3.5

Assume that IDDEs (3.1) and (3.2) fulfill the inequalities (3.3)-(3.6), and $R(t)\le 0$ for $\forall t\ge 0$,

$$\begin{aligned} \frac{\ln |\lambda |}{\tau }+ \frac{R(t)}{\omega _1^2}<0,\;\;\frac{\omega _1\sigma (t)-\omega _1\gamma (t)\left( \frac{\ln |\lambda |}{\tau }+\frac{R(t-\tau )}{\omega _1^2}\right) }{-\left( \frac{\ln |\lambda |}{\tau }+\frac{R(t)}{\omega _1^2}\right) }\le 1, \end{aligned}$$

then

$$\begin{aligned} \Vert x(t)-\tilde{x}(t)\Vert \le \ \max \{\frac{\lambda \Vert \phi (0^-)-\tilde{\phi }(0^-)\Vert }{m_1},\kappa _1\}, \end{aligned}$$

where $\omega _1=\max \{|\lambda |,\frac{1}{|\lambda |}\}$, $m_1=\min \{1,|\lambda |\}$, $\zeta _1=|\frac{\ln \lambda }{\tau }|$ and

$$\begin{aligned} \kappa _1=\sup _{0\le s < \tau }\frac{\omega _1^2(\beta (s)+2\zeta \gamma (s))\delta \phi (s)+\omega _1^2 \gamma (s)\delta \dot{\phi }(s)}{-\left( {\frac{\ln |\lambda |}{\tau }+\frac{R(t)}{\omega _1^2}}\right) }. \end{aligned}$$

Moreover, assume that

$$\begin{aligned} \frac{\ln |\lambda |}{\tau }+\frac{R(t)}{\omega _1^2} \le R_1<0,\;\forall t\ge 0, \end{aligned}$$

and

$$\begin{aligned}{} & {} \frac{\omega _1\gamma (t)\left( \frac{\ln |\lambda |}{\tau }+ \frac{R(t-\tau )}{\omega _1^2}\right) }{\frac{\ln |\lambda |}{\tau }+\frac{R(t)}{\omega _1^2} }=r_1(t)\le \bar{\xi }_1<1,\;\forall t\ge 0,\\{} & {} \frac{\omega _1\sigma (t)}{-\left( \frac{\ln |\lambda |}{\tau }+ \frac{R(t)}{\omega _1^2}\right) }\le k_1(1-r_1(t)),\;k_1<1,\;\forall t\ge 0, \end{aligned}$$

then we have

$$\begin{aligned} \lim \limits _{t\rightarrow \infty } \Vert x(t)-\tilde{x}(t) \Vert =0. \end{aligned}$$

Theorem 3.6

Assume that IDDEs (3.1) and (3.2) fulfill the inequalities (3.3)–(3.6), $R(t)\le \hat{R}$, $\hat{R}>0$, and for $\forall t\ge 0$,

$$\begin{aligned} \frac{\ln |\lambda |}{\tau }+\omega _1^2 \hat{R}<0,\;\;\frac{\omega _1\sigma (t)-\omega _1\gamma (t)\left( \frac{\ln |\lambda |}{\tau }+\omega _1^2 \hat{R}\right) }{-\left( \frac{\ln |\lambda |}{\tau }+\omega _1^2 \hat{R}\right) }\le 1, \end{aligned}$$

then

$$\begin{aligned} \Vert x(t)-\tilde{x}(t)\Vert \le \max \{\frac{\lambda \Vert \phi (0^-)-\tilde{\phi }(0^-)\Vert }{m_1},\check{\kappa }_1\}, \end{aligned}$$

where

$$\begin{aligned} \check{\kappa }_1=\sup _{0\le s < \tau }\frac{\omega _1^2(\beta (s)+2\zeta _1\gamma (s))\delta \phi (s)+\omega _1^2\gamma (s)\delta \dot{\phi }(s)}{-\left( \frac{\ln |\lambda |}{\tau }+\omega _1^2 \hat{R}\right) }. \end{aligned}$$

Moreover, assume that

$$\begin{aligned} \frac{\ln |\lambda |}{\tau }+\omega _1^2 \hat{R}<0,\;\omega _1\gamma (t)\le \bar{\xi }_1<1,\;\forall t\ge 0, \end{aligned}$$

and

$$\begin{aligned} \frac{\omega _1\sigma (t)}{-\left( \frac{\ln |\lambda |}{\tau }+\omega _1^2 \hat{R}\right) }\le \bar{k}_1(1- \omega _1\gamma (t)),\;\bar{k}_1<1,\;\forall t\ge 0, \end{aligned}$$

then we have

$$\begin{aligned} \lim \limits _{t\rightarrow \infty } \Vert x(t)-\tilde{x}(t) \Vert =0. \end{aligned}$$

Special Case II: When $\lambda \in \mathbb {R}$ and $\lambda >0$, $\alpha _2(t)=1+(\lambda -1)\{\frac{t}{\tau }\}$, $t\in [-\tau , \infty )$. By Theorem 3.2, when $\alpha (t)=\alpha _2(t)$, we can obtain the following theorem.

Theorem 3.7

Assume that x(t) is the solution of (2.1) and $y_2(t)=\alpha _2(t)x(t),\;t\in [-\tau ,+\infty )$. Then $y_2(t)$ is the solution of

$$\begin{aligned} {\left\{ \begin{array}{ll} y_2'(t)=F_2(t,y_2(t),y_2(t-\tau ),y_2'(t-\tau )), &{} t\ge 0,\\ y_2(t)=\Phi _2(t), &{} t\in [-\tau ,0], \end{array}\right. } \end{aligned}$$

(3.24)

where

$$\begin{aligned} F_2(t,y,u,v)=\frac{(\lambda -1)y}{\tau \alpha _2(t)}+\alpha _2(t)f\left( t,\frac{y}{\alpha _2(t)},\frac{u}{\alpha _2(t)},\frac{v}{\alpha _2(t)}-\frac{(\lambda -1)u}{\tau \alpha _2^2(t)}\right) \end{aligned}$$

and

$$\begin{aligned} \Phi _2(t)= \left\{ \begin{array}{l} [1+(\lambda -1)(\frac{t}{\tau }+1)]\phi (t), ~~~~t\in [-\tau ,0),\\ \lambda \phi (0^-),~~~~~t=0. \end{array} \right. \end{aligned}$$

On the other hand, assume that y(t) is the solution of (3.24) and $x(t)= \frac{y_2(t)}{1+(\lambda -1)\{\frac{t}{\tau }\}},\;t\in [-\tau ,+\infty )$. Then x(t) is the solution of (2.1).

Theorem 3.8

When $\lambda \in \mathbb {R}$, $\lambda >0$ and $\lambda \ne 1$, assume that IDDEs (3.1) and (3.2) fulfill the inequalities (3.3)–(3.6), $R(t)\le 0$, for $\forall t\ge 0$,

$$\begin{aligned} \frac{\lambda -1}{\tau +(\lambda -1)\tau \{\frac{t}{\tau }\}}+ \frac{R(t)}{\omega _2^2}<0,\;\;\frac{\omega _2\sigma (t)-\omega _2\gamma (t)\left( \frac{\lambda -1}{\tau +(\lambda -1)\tau \{\frac{t}{\tau }\}}+\frac{R(t-\tau )}{\omega _2^2}\right) }{-\left( \frac{\lambda -1}{\tau +(\lambda -1)\tau \{\frac{t}{\tau }\}}+\frac{R(t)}{\omega _2^2}\right) }\le 1, \end{aligned}$$

then

$$\begin{aligned} \Vert x(t)-\tilde{x}(t)\Vert \le \ \max \{\frac{\lambda \Vert \phi (0^-)-\tilde{\phi }(0^-)\Vert }{m_2},\kappa _2\}, \end{aligned}$$

where $\omega _2=\max \{\lambda ,\frac{1}{\lambda }\}$, $m_2=\min \{1,\lambda \}$, $\zeta _2=\max \{|\frac{\lambda -1}{\tau }|,|\frac{\lambda -1}{\lambda \tau }|\}$ and

$$\begin{aligned} \kappa _2=\sup _{0\le s < \tau }\frac{\omega _2^2[\beta (s)+2\zeta \gamma (s)]\delta \phi (s)+\omega _2^2 \gamma (s)\delta \dot{\phi }(s)}{-\left( {\frac{\lambda -1}{\tau +(\lambda -1)\tau \{\frac{t}{\tau }\}}+\frac{R(t)}{\omega _2^2}}\right) }. \end{aligned}$$

Moreover, assume that

$$\begin{aligned} \frac{\lambda -1}{\tau +(\lambda -1)\tau \{\frac{t}{\tau }\}}+ \frac{R(t)}{\omega _2^2} \le R_2<0,\;\forall t\ge 0, \end{aligned}$$

and

$$\begin{aligned} \frac{\omega _1\gamma (t)\left( \frac{\lambda -1}{\tau +(\lambda -1)\tau \{\frac{t}{\tau }\}}+ \frac{R(t-\tau )}{\omega _2^2}\right) }{\frac{\lambda -1}{\tau +(\lambda -1)\tau \{\frac{t}{\tau }\}}+ \frac{R(t)}{\omega _2^2}}=r_2(t)\le \xi _2<1,\;\forall t\ge 0, \end{aligned}$$

$$\begin{aligned} \frac{\omega _1\sigma (t)}{-\left( \frac{\lambda -1}{\tau +(\lambda -1)\tau \{\frac{t}{\tau }\}}+ \frac{R(t)}{\omega _2^2}\right) }\le k_2(1-r_2(t)),\;k_2<1,\;\forall t\ge 0, \end{aligned}$$

then we have

$$\begin{aligned} \lim \limits _{t\rightarrow \infty } \Vert x(t)-\tilde{x}(t) \Vert =0. \end{aligned}$$

Theorem 3.9

When $\lambda \in \mathbb {R}$ and $\lambda >0$, assume that IDDEs (3.1) and (3.2) fulfill the inequalities (3.3)–(3.6), $R(t)\le \hat{R}$, $\hat{R}>0$, and for $\forall t\ge 0$,

$$\begin{aligned} \frac{\lambda -1}{\tau +(\lambda -1)\tau \{\frac{t}{\tau }\}}+\omega _2^2 \hat{R}<0, \end{aligned}$$

and

$$\begin{aligned} \frac{\omega _2\sigma (t)-\omega _2\gamma (t)\left( \frac{\lambda -1}{\tau +(\lambda -1)\tau \{\frac{t}{\tau }\}}+\omega _2^2\hat{R}\right) }{-\left( \frac{\lambda -1}{\tau +(\lambda -1)\tau \{\frac{t}{\tau }\}}+\omega _2^2\hat{R}\right) }\le 1, \end{aligned}$$

then

$$\begin{aligned} \Vert x(t)-\tilde{x}(t)\Vert \le \max \{\frac{\lambda \Vert \phi (0^-)-\tilde{\phi }(0^-)\Vert }{m_2},\check{\kappa }_2\}, \end{aligned}$$

where

$$\begin{aligned} \check{\kappa }_2=\sup _{0\le s < \tau }\frac{\omega _2^2\left( \beta (s)+2\zeta _2\gamma (t)\right) \delta \phi (s)+\omega _2^2\gamma (s)\delta \dot{\phi }(s)}{-\left( \frac{\lambda -1}{\tau +(\lambda -1)\tau \{\frac{t}{\tau }\}}+\omega _2^2\hat{R}\right) }. \end{aligned}$$

Moreover, assume that

$$\begin{aligned}{} & {} \frac{\lambda -1}{\tau +(\lambda -1)\tau \{\frac{t}{\tau }\}}+\omega _2^2 \hat{R}\le \bar{R}_2<0,\;\omega _2\gamma (t)\le \bar{\xi }_2<1,\;\forall t\ge 0,\\{} & {} \frac{\omega _2\sigma (t)}{-\left( \frac{\lambda -1}{\tau +(\lambda -1)\tau \{\frac{t}{\tau }\}}+\omega _2^2 \hat{R}\right) }\le \bar{k}_2(1-\omega _2\gamma (t)),\;\bar{k}_2<1,\;\forall t\ge 0, \end{aligned}$$

then we have

$$\begin{aligned} \lim \limits _{t\rightarrow \infty } \Vert x(t)-\tilde{x}(t) \Vert =0. \end{aligned}$$

Because $\frac{\lambda -1}{\tau +(\lambda -1)\tau \{\frac{t}{\tau }\}}\le \frac{\lambda -1}{\tau }$ for all $\lambda >0$, $\forall t\in \mathbb {R}$, by Theorem 3.9, we can obtain the following corollary.

Corollary 3.10

When $\lambda \in \mathbb {R}$, $\lambda >0$ and $\lambda \ne 1$, assume that IDDEs (3.1) and (3.2) fulfill the inequalities (3.3)–(3.6), $R(t)\le 0$, for $\forall t\ge 0$,

$$\begin{aligned} \frac{\lambda -1}{\tau }+ \frac{R(t)}{\omega _2^2}<0,\;\;\frac{\omega _2\sigma (t)-\omega _2\gamma (t)\left( \frac{\lambda -1}{\tau }+\frac{R(t-\tau )}{\omega _2^2}\right) }{-\left( \frac{\lambda -1}{\tau }+\frac{R(t)}{\omega _2^2}\right) }\le 1, \end{aligned}$$

then

$$\begin{aligned} \Vert x(t)-\tilde{x}(t)\Vert \le \ \max \{\frac{\lambda \Vert \phi (0^-)-\tilde{\phi }(0^-)\Vert }{m_2},\kappa _2\}. \end{aligned}$$

Moreover, assume that

$$\begin{aligned} \frac{\lambda -1}{\tau }+ \frac{R(t)}{\omega _2^2} \le \check{R}_2<0,\;\forall t\ge 0, \end{aligned}$$

and

$$\begin{aligned}{} & {} \frac{\omega _1\gamma (t)\left( \frac{\lambda -1}{\tau }+ \frac{R(t-\tau )}{\omega _2^2}\right) }{\frac{\lambda -1}{\tau }+ \frac{R(t)}{\omega _2^2}}=\check{r}_2(t)\le \check{\xi }_2<1,\;\forall t\ge 0,\\{} & {} \frac{\omega _1\sigma (t)}{-\left( \frac{\lambda -1}{\tau }+ \frac{R(t)}{\omega _2^2}\right) }\le \check{k}_2(1-\check{r}_2(t)),\;\check{k}_2<1,\;\forall t\ge 0, \end{aligned}$$

then we have

$$\begin{aligned} \lim \limits _{t\rightarrow \infty } \Vert x(t)-\tilde{x}(t) \Vert =0. \end{aligned}$$

Corollary 3.11

When $\lambda \in \mathbb {R}$ and $\lambda >0$, assume that IDDEs (3.1) and (3.2) fulfill the inequalities (3.3)–(3.6), $R(t)\le \hat{R}$, $\hat{R}>0$, and for $\forall t\ge 0$,

$$\begin{aligned} \frac{\lambda -1}{\tau }+\omega _2^2 \hat{R}<0, \end{aligned}$$

and

$$\begin{aligned} \frac{\omega _2\sigma (t)-\omega _2\gamma (t)\left( \frac{\lambda -1}{\tau }+\omega _2^2\hat{R}\right) }{-\left( \frac{\lambda -1}{\tau }+\omega _2^2\hat{R}\right) }\le 1, \end{aligned}$$

then

$$\begin{aligned} \Vert x(t)-\tilde{x}(t)\Vert \le \max \{\frac{\lambda \Vert \phi (0^-)-\tilde{\phi }(0^-)\Vert }{m_2},\kappa _2\}. \end{aligned}$$

Moreover, assume that

$$\begin{aligned}{} & {} \frac{\lambda -1}{\tau }+\omega _2^2\hat{R}<0,\; \omega _2\gamma (t)\le \tilde{\xi }_2<1,\;\;\forall t\ge 0,\\{} & {} \frac{\omega _2\sigma (t)}{-\left( \frac{\lambda -1}{\tau }+\omega _2^2 \hat{R}\right) }\le \tilde{k}_2(1- \omega _2\gamma (t)),\;\tilde{k}_2<1,\;\forall t\ge 0, \end{aligned}$$

then we have

$$\begin{aligned} \lim \limits _{t\rightarrow \infty } \Vert x(t)-\tilde{x}(t) \Vert =0. \end{aligned}$$

Special Case III: When $\lambda \in \mathbb {R}$ and $\lambda >0$, $\alpha (t)=\alpha _3(t)=-\{\frac{t}{\tau }\}^2+\lambda \{\frac{t}{\tau }\}+1$, $t\in [-\tau , \infty )$. By Theorem 3.2, we can obtain the following theorem.

Theorem 3.12

Assume that x(t) is the solution of (2.1) and $y_3(t)=(-\{\frac{t}{\tau }\}^2+\lambda \{\frac{t}{\tau }\}+1)x(t),\;t\in [-\tau ,+\infty )$. Then $y_3(t)$ is the solution of

$$\begin{aligned} {\left\{ \begin{array}{ll} y_3'(t)=F_3(t,y_3(t),y_3(t-\tau ),y_3'(t-\tau )), &{} t\ge 0,\\ y_3(t)=\Phi _3(t), &{} t\in [-\tau ,0], \end{array}\right. } \end{aligned}$$

(3.25)

where

$$\begin{aligned} F_3(t,y,u,v)=\frac{[-2\{\frac{t}{\tau }\}+\lambda ]y}{\tau \alpha _3(t)}+\alpha _3(t)f(t,\frac{y}{\alpha _3(t)},\frac{u}{\alpha _3(t)},\frac{v}{\alpha _3(t)}-\frac{(\lambda -1)u}{\tau \alpha _3^2(t)}) \end{aligned}$$

and

$$\begin{aligned} \Phi _3(t)= \left\{ \begin{array}{l} [-(\frac{t}{\tau }+1)^2+\lambda (\frac{t}{\tau }+1)+1]\phi (t), ~~~~t\in [-\tau ,0),\\ \lambda \phi (0^-),~~~~~t=0. \end{array} \right. \end{aligned}$$

On the other hand, assume that $y_3(t)$ is the solution of (3.25) and $x(t)= \frac{y_3(t)}{-\{\frac{t}{\tau }\}^2+\lambda \{\frac{t}{\tau }\}+1},\;t\in [-\tau ,+\infty )$. Then x(t) is the solution of (2.1).

Theorem 3.13

When $\lambda \in \mathbb {R}$ and $\lambda >0$, assume that IDDEs (3.1) and (3.2) fulfill the inequalities (3.3)–(3.6), $R(t)\le 0$, for $\forall t\ge 0$,

$$\begin{aligned} \frac{-2\{\frac{t}{\tau }\}+\lambda }{-\{\frac{t}{\tau }\}^2\tau +\lambda \tau \{\frac{t}{\tau }\}+\tau }+\frac{R(t)}{\omega _3^2}<0, \end{aligned}$$

and

$$\begin{aligned} \frac{\omega _3\sigma (t)-\omega _3\gamma (t)\left( \frac{-2\{\frac{t}{\tau }\}+\lambda }{-\{\frac{t}{\tau }\}^2\tau +\lambda \tau \{\frac{t}{\tau }\}+\tau }+\frac{R(t-\tau )}{\omega _3^2}\right) }{-\left( \frac{-2\{\frac{t}{\tau }\}+\lambda }{-\{\frac{t}{\tau }\}^2\tau +\lambda \tau \{\frac{t}{\tau }\}+\tau }+\frac{R(t)}{\omega _3^2}\right) }\le 1, \end{aligned}$$

then

$$\begin{aligned} \Vert x(t)-\tilde{x}(t)\Vert \le \max \{\frac{\lambda \Vert \phi (0^-)-\tilde{\phi }(0^-)\Vert }{m_3},\kappa _3\}, \end{aligned}$$

where $m_3=\min \{1,\lambda \}$, $\zeta _3=\sup \nolimits _{t\in [-\tau , 0)}|\frac{\alpha _3'(t)}{\alpha _3(t)}|=\max \{|\frac{\lambda -2}{\lambda \tau }|,\frac{\lambda }{\tau }\}$,

$$\begin{aligned} \omega _3= \left\{ \begin{array}{l} \frac{\lambda }{4}+\frac{1}{\lambda },~~~~0<\lambda \le 1,\\ \frac{\lambda ^2}{4}+1,~~~~1<\lambda \le 2, \\ \lambda ,~~~~~\lambda >2, \end{array} \right. \end{aligned}$$

and

$$\begin{aligned} \kappa _3=\sup _{0\le s < \tau }\frac{\omega _3^2(\beta (s)+2\zeta _3\gamma (s))\delta \phi (s)+\omega _3^2\gamma (s)\delta \dot{\phi }(s)}{-\left[ \frac{-2\{\frac{t}{\tau }\}+\lambda }{-\{\frac{t}{\tau }\}^2\tau +\lambda \tau \{\frac{t}{\tau }\}+\tau }+\frac{R(s)}{\omega _3^2}\right] }. \end{aligned}$$

Moreover, assume that

$$\begin{aligned} \frac{-2\{\frac{t}{\tau }\}+\lambda }{-\{\frac{t}{\tau }\}^2\tau +\lambda \tau \{\frac{t}{\tau }\}+\tau }+\frac{R(t)}{\omega _3^2} \le R_3<0,\;\forall t\ge 0, \end{aligned}$$

and

$$\begin{aligned}{} & {} \frac{\omega _3\gamma (t)\left( \frac{-2\{\frac{t}{\tau }\}+\lambda }{-\{\frac{t}{\tau }\}^2\tau +\lambda \tau \{\frac{t}{\tau }\}+\tau }+\frac{R(t-\tau )}{\omega _3^2}\right) }{\frac{-2\{\frac{t}{\tau }\}+\lambda }{-\{\frac{t}{\tau }\}^2\tau +\lambda \tau \{\frac{t}{\tau }\}+\tau }+\frac{R(t)}{\omega _3^2}}\le \bar{\xi }_3<1,\;\forall t\ge 0,\\{} & {} \frac{\omega _3\sigma (t)}{-\left( \frac{-2\{\frac{t}{\tau }\}+\lambda }{-\{\frac{t}{\tau }\}^2\tau +\lambda \tau \{\frac{t}{\tau }\}+\tau }+\frac{R(t)}{\omega _3^2}\right) }\le k_3(1-\omega _3\gamma (t)),\;k_3<1,\;\forall t\ge 0, \end{aligned}$$

then we have

$$\begin{aligned} \lim \limits _{t\rightarrow \infty } \Vert x(t)-\tilde{x}(t) \Vert =0. \end{aligned}$$

Because $\frac{-2\{\frac{t}{\tau }\}+\lambda }{-\{\frac{t}{\tau }\}^2\tau +\lambda \tau \{\frac{t}{\tau }\}+\tau }\le \frac{\lambda }{\tau }$ for all $\lambda >0$, $\forall t\in \mathbb {R}$, by Theorem 3.13, we can obtain the following corollary.

Corollary 3.14

When $\lambda \in \mathbb {R}$ and $\lambda >0$, assume that IDDEs (3.1) and (3.2) fulfill the inequalities (3.3)–(3.6), $R(t)\le 0$, for $\forall t\ge 0$,

$$\begin{aligned} \frac{\lambda }{\tau }+\frac{R(t)}{\omega _3^2}<0, \end{aligned}$$

and

$$\begin{aligned} \frac{\omega _3\sigma (t)-\omega _3\gamma (t)\left( \frac{\lambda }{\tau }+\frac{R(t-\tau )}{\omega _3^2}\right) }{-\left( \frac{\lambda }{\tau }+\frac{R(t)}{\omega _3^2}\right) }\le 1, \end{aligned}$$

then

$$\begin{aligned} \Vert x(t)-\tilde{x}(t)\Vert \le \max \{\frac{\lambda \Vert \phi (0^-)-\tilde{\phi }(0^-)\Vert }{m_3},\kappa _3\}. \end{aligned}$$

Moreover, assume that

$$\begin{aligned} \frac{\lambda }{\tau }+\frac{R(t)}{\omega _3^2} \le R_3<0,\;\forall t\ge 0, \end{aligned}$$

and

$$\begin{aligned}{} & {} \frac{\omega _3\gamma (t)\left( \frac{\lambda }{\tau }+\frac{R(t-\tau )}{\omega _3^2}\right) }{\frac{\lambda }{\tau }+\frac{R(t)}{\omega _3^2}}\le \bar{\xi }_3<1,\;\forall t\ge 0,\\{} & {} \frac{\omega _3\sigma (t)}{-\left( \frac{\lambda }{\tau }+\frac{R(t)}{\omega _3^2}\right) }\le k_3(1-\omega _3\gamma (t)),\;k_3<1,\;\forall t\ge 0, \end{aligned}$$

then we have

$$\begin{aligned} \lim \limits _{t\rightarrow \infty } \Vert x(t)-\tilde{x}(t) \Vert =0. \end{aligned}$$

3.4 Numerical methods for nonlinear INDDEs (3.1) and (3.2)

The numerical method for nonlinear INDDE (3.1) can be constructed as follows

$$\begin{aligned} {\left\{ \begin{array}{ll} Y_{n+1}^i=\bar{y}(t_n)+h\sum _{j=1}^sa_{ij}F\left( t_{n+1}^{j-},Y_{n+1}^j,Y_{n-m+1}^j,z'\left( t^{j-}_{n-m+1}\right) \right) , &{}i=1,2,\ldots ,s,\\ \bar{y}(t_n+\theta h)=\bar{y}(t_n)+h\sum _{i=1}^sb_{i}(\theta )F\left( t_{n+1}^{i-},Y_{n+1}^i,Y_{n-m+1}^i,z'\left( t^{i-}_{n-m+1}\right) \right) , &{} n\in \mathbb {N}, \end{array}\right. } \end{aligned}$$

(3.26)

and

$$\begin{aligned} \bar{x}_n=\frac{\bar{y}(t_n)}{\alpha (t_n)},\;\;n\in \mathbb {N}, \end{aligned}$$

(3.27)

where $Y_{n-m+1}^i=\Phi (t_{n-m+1}^i)$ if $t_{n-m+1}^i\le 0$, $i=1,2,\cdots ,s$,

$$\begin{aligned}z'(t^i_{n-m+1})= {\left\{ \begin{array}{ll} \Phi '(t^i_{n-m+1}), &{}t^i_{n-m+1}\le 0,\\ F(t_{n-m+1}^i,Y_{n-m+1}^i,\Phi _{n-2m+1}^i,\Phi '(t^i_{n-2m+1})), &{}0\le t^i_{n-m+1}\le \tau ,\\ F(t_{n-m+1}^i,Y_{n-m+1}^i,Y_{n-2m+1}^i,z'(t^i_{n-2m+1})), &{}\tau \le t^i_{n-m+1}, \end{array}\right. } \end{aligned}$$

$t_n=nh$, $n\in \mathbb {N}$, $h=\frac{\tau }{m}$ and m is a positive integer.

Similarly, the numerical method for (3.2) can be constructed as follows

$$\begin{aligned} {\left\{ \begin{array}{ll} \bar{Y}_{n+1}^i=\bar{\tilde{y}}(t_n)+h\sum _{j=1}^sa_{ij}F(t_{n+1}^{j-},\bar{Y}_{n+1}^j,\bar{Y}_{n-m+1}^j,\tilde{z}'(t^{j-}_{n-m+1})), &{}i=1,2,\cdots ,s,\\ \bar{\tilde{y}}(t_n+\theta h)=\bar{\tilde{y}}(t_n)+h\sum _{i=1}^sb_{i}(\theta )F(t_{n+1}^{i-},\bar{Y}_{n+1}^i,\bar{Y}_{n-m+1}^i,\tilde{z}'(t^{i-}_{n-m+1})), &{} n\in \mathbb {N}, \end{array}\right. }\nonumber \\ \end{aligned}$$

(3.28)

and

$$\begin{aligned} \bar{\tilde{x}}_n=\frac{\bar{\tilde{y}}(t_n)}{\alpha (t_n)},\;\;n\in \mathbb {N}, \end{aligned}$$

(3.29)

where

$$\begin{aligned}\tilde{z}'(t^i_{n-m+1})= {\left\{ \begin{array}{ll} \tilde{\Phi }'(t^i_{n-m+1}), &{}t^i_{n-m+1}\le 0,\\ F(t_{n-m+1}^i,Y_{n-m+1}^i,\tilde{\Phi }_{n-2m+1}^i,\tilde{\Phi }'(t^i_{n-2m+1})), &{}0\le t^i_{n-m+1}\le \tau ,\\ F(t_{n-m+1}^i,Y_{n-m+1}^i,Y_{n-2m+1}^i,\tilde{z}'(t^i_{n-2m+1})), &{}\tau \le t^i_{n-m+1}. \end{array}\right. } \end{aligned}$$

Theorem 3.15

Under the condition of (3.8), the constructed numerical methods (3.26)–(3.27) and (3.28)–(3.29) furnished by backward Euler method with linear interpolation (2-stage Lobatto IIIC method with linear interpolation) are stable, in the following sense,

$$\begin{aligned} \Vert \bar{x}_n-\bar{\tilde{x}}_n\Vert \le \max \{\frac{\lambda \Vert \phi (0^-)-\tilde{\phi }(0^-)\Vert }{m},\kappa \}. \end{aligned}$$

Moreover, under the condition of Theorem 3.2, the constructed numerical method (3.26)–(3.27) and (3.28)–(3.29) furnished by backward Euler method with linear interpolation (or 2-stage Lobatto IIIC method with linear interpolation) are also asymptotically stable, that is,

$$\begin{aligned} \lim _{n\rightarrow +\infty } \Vert \bar{x}_n-\bar{\tilde{x}}_n\Vert =0. \end{aligned}$$

Proof

By (Wang et al. 2009, Theorem 4.8), RVCRK formulae (3.26) and (3.28) furnished by backward Euler method with linear interpolation (or 2-stage Lobatto IIIC method with linear interpolation) are RN-stable, that is

$$\begin{aligned} \Vert \bar{y}(t_n)-\bar{\tilde{y}}(t_n)\Vert \le \max \{\Vert \Phi (0)-\tilde{\Phi }(0)\Vert ,\kappa \}. \end{aligned}$$

Moreover, (Wang et al. 2009, Theorem 4.9), under the condition of Theorem 3.2, the RVCRK formulae (3.26) and (3.28) furnished by backward Euler method with linear interpolation (or 2-stage Lobatto IIIC method with linear interpolation) are also asymptotically stable, that is,

$$\begin{aligned} \lim _{n\rightarrow +\infty } \Vert \bar{y}(t_n)-\bar{\tilde{y}}(t_n)\Vert =0. \end{aligned}$$

Because the relation between the numerical solutions INDDE and NDDE without impulsive perturbations, that is (3.27) and (3.29), the theorem holds. $\square$

Similar to Theorem 3.15, we can obtain the following result.

Theorem 3.16

Under the condition of (3.19), the constructed numerical methods (3.26)–(3.27) and (3.28)–(3.29) furnished by backward Euler method with linear interpolation (or 2-stage Lobatto IIIC method with linear interpolation) are stable, in the following sense

$$\begin{aligned} \Vert \bar{x}_n-\bar{\tilde{x}}_n\Vert \le \frac{\max \{\lambda \Vert \phi (0^-)-\tilde{\phi }(0^-)\Vert ,\bar{\kappa }\}}{m}. \end{aligned}$$

Moreover, under the condition of Theorem 3.3, the constructed numerical method (3.26)–(3.27) and (3.28)–(3.29) furnished by backward Euler method with linear interpolation (or 2-stage Lobatto IIIC method with linear interpolation) are also asymptotically stable, that is

$$\begin{aligned} \lim _{n\rightarrow +\infty } \Vert \bar{x}_n-\bar{\tilde{x}}_n\Vert =0. \end{aligned}$$

The following results are provided to analysis the difference between the linear equations and nonlinear equations.

Remark 3.17

• 1. The meanings of asymptotic stability of the exact solutions of linear INDDE (2.1) and nonlinear INDDE (3.1) are different. The defintions of asymptotic stability of the exact solutions of linear INDDE (2.1) and nonlinear INDDE (3.1) are provided as follows.

  • (1) The exact solution x(t) of linear INDDE (2.1) is said to be asymptotically stable if

    $$\begin{aligned} \lim _{t\rightarrow \infty }x(t)=0. \end{aligned}$$

  • (2) For $\forall \epsilon >0$, if there exists a constant $\delta >0$ such that $\Vert \Phi -\tilde{\Phi }\Vert \le \delta$ and $\Vert \Phi '-\tilde{\Phi }'\Vert \le \delta$ imply that

    $$\begin{aligned} \Vert x(t)-\tilde{x}(t)\Vert <\epsilon , \end{aligned}$$

    then we call the exact solution x(t) of nonlinear INDDE (3.1) and $\tilde{x}(t)$ of (3.2) are stable.

  • (3) The exact solution x(t) of nonlinear INDDE (3.1) and $\tilde{x}(t)$ of (3.2) are said to be asymptotically stable, if they are stable and fulfil

    $$\begin{aligned} \lim _{t\rightarrow \infty }\Vert x(t)-\tilde{x}(t)\Vert =0. \end{aligned}$$

• 2. The meanings of asymptotic stability of numerical method (2.5) for linear equation (2.1) and numerical method (3.26)–(3.27) for nonlinear equation (3.1) are also different. The asymptotic stability of numerical method (2.5) and numerical method (3.26)–(3.27) are given as follows.

  • (1) Numerical method (2.5) for linear INDDE (2.1) is said to be asymptotically stable if $x_n$ obtained from (2.5) satisfies

    $$\begin{aligned} \lim _{n\rightarrow \infty }x_n=0. \end{aligned}$$

  • (2) For $\forall \epsilon >0$, if there exists a constant $\delta >0$ such that $\Vert \Phi -\tilde{\Phi }\Vert \le \delta$ and $\Vert \Phi '-\tilde{\Phi }'\Vert \le \delta$ imply that $\bar{x}_n$ obtained from (3.26)–(3.27) and $\bar{\tilde{x}}_n$ obtained from (3.28)–(3.29) satisfy

    $$\begin{aligned} \Vert \bar{x}_n-\bar{\tilde{x}}_n\Vert <\epsilon , \end{aligned}$$

    then we call numerical methods (3.26)–(3.27) for nonlinear INDDE (3.1) and (3.28)-(3.29) for (3.2) are stable.

  • (3) Numerical methods (3.26)–(3.27) for nonlinear INDDE (3.1) and (3.28)–(3.29) for (3.2) are said to be asymptotically stable, if they are stable and fulfil

    $$\begin{aligned} \lim _{n\rightarrow \infty }\Vert \bar{x}_n-\bar{\tilde{x}}_n\Vert =0. \end{aligned}$$

• 3. For linear equations, the constructed numerical methods furnished by A-stable Runge–Kutta methods can preserve asymptotic stability of the exact solutions. But, for nonlinear equations, the constructed numerical methods furnished by implicit Euler method (or 2-stage Lobatto IIIC method) can preserve asymptotic stability of the exact solutions.

4 Numerical experiments

Example 4.1

Consider the following scalar linear INDDE:

$$\begin{aligned} {\left\{ \begin{array}{ll} x'(t)=-x(t)+2x(t-1)-\frac{1}{4} x'(t-1),&{}t\ge 0, t\ne k,k\in \mathbb {N},\\ x(k)=\frac{x(k^-)}{{\textrm{e}}},&{}\\ x(t)=\phi (t), &{} t\in [-1,0). \end{array}\right. } \end{aligned}$$

(4.1)

By Theorem 2.1, assume that $y(t)= {\textrm{e}}^{-\{t\}}x(t)$, $t\ge -1$, then x(t) is the solution of INDDE (4.1) if and only if y(t) is the solotion of the following equation:

$$\begin{aligned} {\left\{ \begin{array}{ll} y'(t)=-2y(t)+\frac{7}{4} y(t-1)-\frac{1}{4} y'(t-1),&{}t\ge 0, t\ne k,k\in \mathbb {N},\\ x(t)=\Phi (t), &{} t\in [-1,0], \end{array}\right. } \end{aligned}$$

(4.2)

where

$$\begin{aligned} \Phi (t)= {\left\{ \begin{array}{ll} \textrm{e}^{-(t+1)}\phi (t), &{} t\in [-1,0),\\ \frac{\phi (0^-)}{{\textrm{e}}}, &{} t=0. \end{array}\right. } \end{aligned}$$

By Theorem 2.2, both the solution x(t) of (4.1) and the solution y(t) of (4.2) tend to zeros as $t\rightarrow \infty$. By Theorem 2.4, the numerical methods (2.5) for (4.1) furnished by A-stable Runge-Kutta methods are stable and asymptotically stable (see Figs. 1 and 2).

Table 1 The errors between the numerical solutions obtained from (2.5) and the exact solution of (4.1) at $t=10$

Fig. 1

The numerical methods (2.5) for (4.1) furnished by implicit Euler method with the stepsize $h=\frac{1}{10}$

Fig. 2

The numerical methods (2.5) for (4.1) furnished by Lobatto IIIC method with the stepsize $h=\frac{1}{10}$

Example 4.2

Consider the following scalar nonlinear INDDE:

$$\begin{aligned} {\left\{ \begin{array}{ll} x'(t)=ax(t)+b\cos (x'(t-1))\sin ^2(x(t-1)),&{}t\ge 0, t\ne k,k\in \mathbb {N},\\ x(k)=\lambda x(k^-),&{}\\ x(t)=\phi (t), &{}{ t\in [-1,0)}, \end{array}\right. } \end{aligned}$$

(4.3)

where a, b and $\lambda$ are real constants. It is easy to prove that the first equation of (4.3) satisfies inequalities (3.3)–(3.6) with $R(t)=\Re (a)$, $\beta (t)=2|b|$, $\gamma (t)=|b|$, $\sigma (t)=(|a|+2)|b|$.

• (i) When $a=1/10$, $b=1/30$, $\lambda =\frac{1}{{\textrm{e}}}$, by Theorem 3.6, the solution of (4.3) is stable and asymptotically stable. By Theorem 3.16, the numerical methods (3.26)–(3.27) for (4.3) furnished by backward Euler method (2-stage Lobatto IIIC method) are stable and asymptotically stable (See Fig. 3).

• (ii) When $a=1/50$, $b=1/50$, $\lambda =\frac{1}{3}$, by Theorem 3.6, Theorem 3.9 or Corollary 3.11 the solution of (4.3) is stable and asymptotically stable. By Theorem 3.16, the numerical methods (3.26)–(3.27) for (4.3) furnished by backward Euler method (2-stage Lobatto IIIC method) are stable and asymptotically stable.

• (iii) When $a=-50$, $b=1/50$, $\lambda =\textrm{e}$, by Theorem 3.5, Theorem 3.8, Corollary 3.10, Theorem 3.13 or Corollary 3.14, the solution of (4.3) is bound and asymptotically stable. By Theorem 3.15, the numerical methods (3.26)–(3.27) for (4.3) furnished by backward Euler method (2-stage Lobatto IIIC method) are stable and asymptotically stable.

Fig. 3

The numerical methods (2.5) for (4.3) furnished by implicit Euler method and Lobatto IIIC method, respectively, with the stepsize $h=\frac{1}{10}$

Table 2 The errors between the numerical solutions obtained from (3.26)–(3.27) and the exact solution of (4.3) when $a=1/10$, $b=1/30$, $\lambda =\frac{1}{{\textrm{e}}}$, $t = 6$

Tables 1 and 2 roughly illuminate that the constructed method furnished by backward Euler method is convergent of order 1 and by 2-stage Lobatto IIIC method is convergent of order 2.

5 Future work

The special case of (2.1) and (3.1) (when $\lambda =0$) and the general case of (2.1) and (3.1) (when the impulsive interval does not equal to the delay $\tau$) will be studied in the future.