We are interested in the numerical stability of the Euler-Maclaurin method for the following differential equation with piecewise constant arguments (EPCA):

$ \begin{equation} \label{eq:1} \left\{ \begin{aligned} &x'(t)=ax(t)+bx\left(3\left[\frac{t+1}{3}\right]\right), \\ &x(0)=x_0, \end{aligned} \right. \end{equation} $

(1.1)

where $t>0$, $a\neq 0$, $b$ and $x_0$ are real constants and $[\cdot]$ denotes the greatest integer function.

EPCA belongs to one special kind of delay differential equations [1-3]. They described hybrid dynamical systems and combine properties of both differential and difference equations. So EPCA had many applications in science and engineering. In the past twenty years, many researchers investigated the properties of the exact solution of EPCA (see [4-6] and the references therein). In particularly, stability of solutions of EPCA received much attention (see [7-9] and the extensive bibliography therein). For more information on this type of equations, the interested readers can refer Wiener's book [10]. Recently, special interest was shown to the properties of numerical solution of EPCA, such as stability [11, 12], dissipativity [13] and oscillation [14]. In this paper, we will study the stability of the numerical solution in the Euler-Maclaurin method for (1.1). Whether the numerical method preserves stability of the exact solution is considered. Two numerical examples for demonstrating the theoretical results are also provided.

The following results give the definition and stability of exact solution for (1.1).

Definition 1.1(see [10])  A solution of (1.1) on $[0, \infty)$ is a function $x(t)$ which satisfies the conditions

(ⅰ) $x(t)$ is continuous on $[0, \infty)$;

(ⅱ) the derivative $x'(t)$ exists at each point $t\in [0, \infty)$, with the possible exception of the points $t=3n-1$ for $n\in N$, where one-sided derivatives exist;

(ⅲ) (1.1) is satisfied on each interval $[3n-1, 3n+2)$ for $n\in N$.

Theorem 1.2(see [10])  Assume that $a$, $b$ and $x_0\in R$, then (1.1) has on $[0, \infty)$ a unique solution $x(t)$ given by

$ x(t)=\lambda (\Omega(t))\left(\frac{\lambda_1}{\lambda_{-1}}\right)^{\left[\frac{t+1}{3}\right]}x_0, $

where

$ \lambda(t)=e^{at}+\frac{b}{a}(e^{at}-1), \quad \Omega(t)=t-3\left[\frac{t+1}{3}\right], \quad \lambda_1=\lambda(2), \quad \lambda_{-1}=\lambda(-1). $

Theorem 1.3(see [10])  The solution $x(t)=0$ of (1.1) is asymptotically stable ( $x(t)\to 0$ as $t\to \infty$) if and only if any one of the following conditions is satisfied

$ \begin{equation} \label{eq:2} \begin{aligned} &-\frac{a(e^{3a}+1)}{\phi(a)}<b<-a, \quad a>\bar{a}, \\ &b>-\frac{a(e^{3a}+1)}{\phi(a)} \quad \mbox{or} \quad b<-a, \quad a<\bar{a}, \\ &b<-a, \quad a=\bar{a}, \end{aligned} \end{equation} $

(1.2)

where $\bar{a}$ is the nonzero solution of equation $\phi(x)=e^{3x}-2e^x+1=0.$

Let $h$ be a given stepsize, $m\geq 1$ be a given integer and satisfies $h=1/m$. The gridpoints $t_i$ be defined by $t_i=ih$ $(i=0, 1, 2, \cdots)$. Applying the Euler-Maclaurin formula to (1.1), we have

$ \begin{equation} \label{eq:3} x_{i+1}=x_{i}+\frac{ha}{2}(x_{i+1}+x_{i})-\sum\limits_{j=1}^{n}\frac{B_{2j}(ha)^{2j}}{(2j)!}(x_{i+1}-x_{i})+hbx_{i}^{(n)}, \end{equation} $

(2.1)

where $B_{2j}$ denotes the $2j$-th Bernoulli number, $x_i$ and $x_{i+1}$ are approximations to $x(t)$ at $t_n$ and $t_{n+1}$, respectively, $x_i^{(n)}$ is an approximation to $x(3[(t+1)/3])$ at $t_n$. Let us denote $i=3km+l$, $l=-m, -m+1, \cdots, 2m-1$ for $k\geq 1$ and $l=0, 1, \cdots, 2m-1$ for $k=0$. Then $x_i^{(n)}$ can be defined as $x_{3km}$ according to Definition 1.1. So we have

$ \begin{equation} \label{eq:4} \left[1-\frac{ha}{2}+\sum\limits_{j=1}^{n}\frac{B_{2j}(ha)^{2j}}{(2j)!}\right]x_{i+1}=\left[1+\frac{ha}{2}+\sum\limits_{j=1}^{n}\frac{B_{2j}(ha)^{2j}}{(2j)!}\right]x_{i}+hbx_{3km}, \end{equation} $

(2.2)

which is equivalent to

$ \begin{equation} \label{eq:5} x_{3km+l+1}=R(z)x_{3km+l}+\frac{b}{a}(R(z)-1)x_{3km}, \end{equation} $

(2.3)

where

$ \mathit{z}{\rm{ = }}\mathit{ha}{\rm{, }}\quad \mathit{R}{\rm{(}}\mathit{z}{\rm{) = 1 + }}\frac{\mathit{z}}{{{\rm{\Phi (}}\mathit{z}{\rm{)}}}}{\rm{, }}\quad {\rm{\Phi (}}\mathit{z}{\rm{) = 1 - }}\frac{\mathit{z}}{{\rm{2}}}{\rm{ + }}\sum\limits_{\mathit{j}{\rm{ = 1}}}^\mathit{n} {\frac{{{\mathit{B}_{{\rm{2}}\mathit{j}}}{\mathit{z}^{{\rm{2}}\mathit{j}}}}}{{{\rm{(2}}\mathit{j}{\rm{)!}}}}} {\rm{.}} $

Thus

$ \begin{equation} \label{eq:6} \begin{aligned} &x_{3km+l}=\left[R(z)^l+\frac{b}{a}(R(z)^l-1)\right]x_{3km}, \\ &x_{3(k+1)m}=\frac{R(z)^{2m}+\frac{b}{a}(R(z)^{2m}-1)}{R(z)^{-m}+\frac{b}{a}(R(z)^{-m}-1)}x_{3km}. \end{aligned} \end{equation} $

(2.4)

Similar to Theorem 2.2 in [14], we have the following result for convergence.

Theorem 2.1  For any given $n\in N$, the Euler-Maclaurin method is of order $2n+2$.

Definition 2.2  The Euler-Maclaurin method is called asymptotically stable at $(a, b)$ if there exists a constant $M_0$ such that $x_n$ defined by (2.3) tends to zero as $n\to \infty$ for all $h=1/m$ and any given $x_0$.

Lemma 2.3(see [15])  If $|z|<1$, then $\Phi(z)\geq 1/2$ for $z>0$ and $\Phi(z)\geq 1$ for $z<0$.

Lemma 2.4(see [15])  If $|z|<1$, then

$ \Phi(z)\leq \frac{z}{e^z-1} $

for $n$ is even and

$ \Phi(z)\geq \frac{z}{e^z-1} $

for $n$ is odd.

Theorem 2.5  The Euler-Maclaurin method is asymptotically stable if any one of the following conditions is satisfied

$ \begin{equation} \label{eq:7} \begin{aligned} &-\frac{a(R(z)^{3m}+1)}{\bar{\phi}(z)}<b<-a, \quad a>a^0, \\ &b>-\frac{a(R(z)^{3m}+1)}{\bar{\phi}(z)} \quad \mbox{or} \quad b<-a, \quad a<a^0, \\ &b<-a, \quad a=a^0, \end{aligned} \end{equation} $

(2.5)

where $z=ha$, $\bar{\phi}(z)=R(z)^{3m}-2R(z)^m+1$, $a^0$ is the nonzero solution of equation $\bar{\phi}(z)=0.$

Proof  Let

$ \overline{\lambda_1}=R(z)^{2m}+\frac{b}{a}(R(z)^{2m}-1) $

and

$ \overline{\lambda_{-1}}=R(z)^{-m}+\frac{b}{a}(R(z)^{-m}-1), $

so we need to verify

$ \begin{equation} \label{eq:8} \left|\frac{\overline{\lambda_1}}{\overline{\lambda_{-1}}}\right|<1. \end{equation} $

(2.6)

Assume $\overline{\lambda_{-1}}>0$, i.e.,

$ b<\frac{a}{R(z)^m-1}. $

Then (2.6) is equivalent to

$ \begin{aligned} &-\frac{a(R(z)^{3m}+1)}{\bar{\phi}(z)}<b<-a, \quad a>a^0, \\ &b<-a, \quad a<a^0, \\ &b<-a, \quad a=a^0. \end{aligned} $

Assume $\overline{\lambda_{-1}}<0$, i.e.,

$ b>\frac{a}{R(z)^m-1}. $

Then (2.6) is equivalent to

$ b>-\frac{a(R(z)^{3m}+1)}{\bar{\phi}(z)}, \quad a<a^0. $

The proof is completed.

The following two lemmas are given naturally.

Lemma 2.6  Let $f(r)=r^3-2r+1$, $r>0$, then

(a) the function $f(r)$ has a minimum at $r_1=\sqrt{2}/\sqrt{3}$, and $f(r)$ is decreasing in $[0, r_1)$ and increasing in $[r_1, +\infty)$;

(b) the function $f(r)$ has a unique solution $1>r_0\neq 1$;

(c) $f(r)<0$ if $r\in [r_0, 1)$ and $f(r)>0$ if $r \in [0, r_0)$ or $r\in [1, +\infty)$.

Lemma 2.7  Let

$ g(\omega)=\frac{\omega^3+1}{\omega^3-2\omega+1}, $

then

(a) the function $g(\omega)$ has extremum at $\omega_1=1/\sqrt[3]{2}$;

(b) $g(\omega)$ is increasing in $(0, r_0)$ and $(r_0, \omega_1)$;

(c) $g(\omega)$ is decreasing in $(\omega_1, 1)$ and $(1, +\infty)$.

By Lemmas 2.6 and 2.7, we obtain

Corollary 2.8  Assume that $r_0\neq 1$ is a unique solution of the function $f(r)=r^3-2r+1$, then $r_0<\omega_1<r_1<1$.

So we have the following result.

Theorem 2.9  Assume that (1.1) is asymptotically stable, then the Euler-Maclaurin method is asymptotically stable if one of the following conditions is satisfied

(a) $R(z)^m \leq e^a$ $(a\leq \ln \omega_1)$;

(b) $R(z)^m \geq e^a$ $(\ln \omega_1<a<0)$;

(c) $R(z)^m \leq e^a$ $(a\geq 0)$.

Proof  In view of Theorems 1.2 and 2.5, we will prove that condition (2.5) is satisfied under condition (1.2).

If (a) holds, then we know from Lemmas 2.3 and 2.4 that $f(r)$ is decreasing and $g(\omega)$ is increasing. Hence $\bar{a}<a^0$ and

$ \begin{equation} \label{eq:9} -\frac{a(R(z)^{3m}+1)}{\bar{\phi}(z)}\leq -\frac{a(e^{3a}+1)}{\phi(a)}. \end{equation} $

(2.7)

If $a>\bar{a}$, then the first inequality of (1.2) holds. Then by (2.7) we get the first inequality of (2.5). If $a<\bar{a}$, then the second inequality of (1.2) holds. Then by (2.7) we obtain the second inequality of (2.5). If $a=\bar{a}$, then the third inequality of (1.2) holds which implies the first inequality of (2.5). The other cases can be proved in the same way. The proof is completed.

From Lemmas 2.3, 2.4 and Theorem 2.9, we have the following main result in this paper.

Theorem 2.10  The Euler-Maclaurin method preserves the stability of (1.1) if one of the following conditions is satisfied

(a) $n$ is odd if $e^a>\omega_1$,

(b) $n$ is even if $e^a \leq \omega_1$.

Consider the following two problems

$ \begin{equation} \label{eq:10} \left\{ \begin{aligned} &x'(t)=x(t)-1.2x\left(3\left[\frac{t+1}{3}\right]\right), \\ &x(0)=1 \end{aligned} \right. \end{equation} $

(3.1)

and

$ \begin{equation} \label{eq:11} \left\{ \begin{aligned} &x'(t)=-x(t)+4x\left(3\left[\frac{t+1}{3}\right]\right), \\ &x(0)=1. \end{aligned} \right. \end{equation} $

(3.2)

In Figures 1 and 2, we plot the exact solution and the numerical solution for (3.1), respectively. Moreover, for (3.2), we also plot the exact solution and the numerical solution in Figures 3 and 4, respectively. We can see from these figures that the Euler-Maclaurin method preserves the stability of (3.1) and (3.2), which is coincide with Theorem 2.10.