数理解析研究所講究録984 巻 1997 年 75-85

75

Characteristic equation and asymptotic behavior of
2-dimensional delay-differential equations

大阪府立大学工学部

宮崎

倫子 (Rinko Miyazaki)

ABSTRACT. Consider 2-dimensional delay-differential equations$\dot{x}(t)=A\int_{-\Gamma}^{0}x(t+s)d\eta(S)$

where

$A$

is a

$2\cross 2$

constant matrix,$r$,

is a positive constant, and$\eta:[-r, \mathrm{O}]arrow \mathrm{R}$is monotone

on $[-r, 0]$ and continuous to the left on $(-r, 0)$ . The purpose of this work is to show thata necessary and sufficient condition under which the zero solution of
asymptotically stable can be obtained, if we impose a restriction on$\eta(s)+\eta(-r-S)=\eta(0)+\eta(-r)$

for$\mathrm{a}.\mathrm{e}$.

$\eta$$(\mathrm{A}\mathrm{L})$is uniformly

as follows:$s\in[-r, 0]$.

The proof will be give by using the characteristic equation.

1. Main Results
Consider 2-dimensional equations$\dot{x}(t)=A\int_{-r}^{0_{X(}}t+s)d\eta(s)$

where

$A$

is a

$2\cross 2$

constant matrix,$r$,$(\mathrm{A}\mathrm{L})$is a positive constant, and$\eta:[-r, \mathrm{O}]arrow \mathrm{R}$is monotone

on $[-r, 0]$ and continuous to the left on $(-r, 0)$ . Moreover, we assume that$\eta(s)+\eta(-r-S)=\eta(.0)+\eta(-r)$

In [5] we have discussed 1-dimensional case:for$\mathrm{a}.\mathrm{e}$.

$s\in[-r, 0]$

.

(H1)

76
Theorem A. Suppose (H1) holds. The zero solution of a scalar equation(1.1)$\dot{x}(t)=-\int_{-r}^{0_{X(}}t+s)d\eta(S)$is uniforndy asymptoticaly stable if and ondy if $\eta(0)>\eta(-r)$ and$\int_{-r}^{0}\sin(-\frac{s}{r}\pi)d\eta(S)<\frac{\pi}{r}$.

In [1] we also have shown the following:
Theorem B. The zero solution of 2-dimensional equations$\dot{x}(t)=-a\sum_{k=1}^{N}x(t-\tau k)$ ,where$\tau_{k}$$k=1,2,$is an arithmetric sequence, that is, $\tau_{k}=\tau+(k-1)l$ with$\ldots,$$N,$ $\tau_{N}>0$

and$| \theta|<\frac{\pi}{2}$$\tau\geq 0$

and $l>0$ foris uniformly asymptoticaly stable if and$on\mathit{4}y$

if

$a>0$

and

$\frac{a(_{\mathcal{T}_{1}+}\mathcal{T}_{N})}{2}\frac{\sin(\frac{Nl}{\tau_{1}+\tau_{N}}(\frac{\pi}{2}-|\theta|)_{\mathrm{I}}}{\sin(\frac{l}{\tau_{1}+\tau_{N}}(\frac{\pi}{2}-|\theta|)_{\mathrm{I}}}<\frac{\pi}{2}-|\theta|$.

By using the ideas of the proofs of Theorems A and , we will give more extended results$\mathrm{B}$(Theorems 1.1 and 1.2.)

By the transfomation $x(t)=Py(t)$ with an appropriate regular matrix$(\mathrm{A}\mathrm{L})$$P$, we can rewriteas$\dot{y}(t)=P^{-1}AP\int_{-r}^{0}y(t+s)d\eta(S)$Consequently, we only have to consider the equationsof the following two matrices:$(\mathrm{A}\mathrm{L})$.
where the matrix$A$is either

77(I) in the case matrix $A$ has real eigenvalues,

$A=-$
where

$a_{1},$$a_{2}$

andare real numbers;$b$(II) in the case matrix$A$

has complex eigenvalues,$A=-R(\theta)=-$where$| \theta|\leq\frac{\pi}{2}$.

For the case (I), we have

Theorem 1.1. Suppose (H1) holds. The zero solution ofis unifomdy asymptoti-$(\mathrm{A}\mathrm{L})$cally stable if and only if $\eta(0)>\eta(-r)$ ,and$a_{i}>0$$i=1,2$ .

$a_{i} \int_{-r}^{0}\sin(-\frac{s}{r}\pi)d\eta(S)<\frac{\pi}{r}$

For the case (II), we have

Theorem 1.2. Suppose (H1) holds. The zero solution of$(\mathrm{A}\mathrm{L})$is unifomdy asymptoti-cally stable if and only if $\eta(0)>\eta(-r)$ and$\int_{-r}^{0}\cos\{\frac{r+2s}{r}(\frac{\pi}{2}-|\theta|)\}d\eta(_{S})<\frac{\pi-2|\theta|}{r}$Remark 1.1. If$\theta=0$

section that Theorem

.

(1.2)

, Theorem 1.2 coincides with Theorem A. We state in the following$\mathrm{B}$is included by Theorem 1.2.

Remark 1.2. In the case

$A$

is an

$\mathrm{n}\cross \mathrm{n}$matrix, we can obtain the necessary and sufficientcondition for the uniform as ymptotic stability of the zero solution of$(\mathrm{A}\mathrm{L})$by applying

Theorems 1.1, 1.2 and theory of Jordan canonical form. (cf. [2, Theorem 3.4]).

78
The proofs of Theorems 1.1 and 1.2 are very similar, so that we give only the proof of
Theorem 1.2.

We prepare a lemma to prove the theorem.
Lemma 1.1. For any integer , if$n$$| \theta|\leq\frac{\pi}{2}$and$|\alpha|\leq 1$, then the following inequality

holds:$| \cos(\frac{(2n+1)\pi+2\theta}{2}\alpha)|<|2n+1|\cos\{(\frac{\pi}{2}-|\theta|)|\alpha|\}$Proof of Lemma 1.1. First of all we note that$\phi\in$$|\sin k\phi|\leq|k||\sin\theta|$

for any integer

$k$

and

R. Then we have$| \cos\{(2k+1)\phi\}|=|\sin\{(2k+1)(\frac{\pi}{2}-\phi)\}|$$\leq|2k+1||\sin(\frac{\pi}{2}-\phi)|$$=|2k+1||\cos\phi|$.

Using these relations, we have$| \cos(\frac{(2n+1)\pi+2\theta}{2}\alpha)|$$=| \cos\{(2n+1)\frac{\pi}{2}\alpha\}\cos(\theta\alpha)-\sin\{(2n+1)\frac{\pi}{2}\alpha\}\sin(\theta\alpha)|$$\leq|2n+1||\cos(\frac{\pi}{2}\alpha)||\cos(\theta\alpha)|+|2n+1||\sin(\frac{\pi}{2}\alpha)||\sin(\theta\alpha)|$$=|2n+1| \{\cos(\frac{\pi}{2}|\alpha|)\cos(|\theta||\alpha|)+\sin(\frac{\pi}{2}|\alpha|)\sin(|\theta||\alpha||)\}$$=|2n+1| \cos\{(\frac{\pi}{2}-|\theta|)|\alpha|\}$This completes the proof.

.

$\square$Lemma 1.2. Suppose (H1) holds. If$f:[-r, \mathrm{O}]arrow \mathrm{R}$is continuous, then$\int_{-r}^{0}f(S)d\eta(s)=-\int_{-r}^{0}f(s)d\eta(-r-s)$

Proof of Lemma 1.2. Define$\tilde{\eta}:[-r, \mathrm{O}]arrow \mathrm{R}$.

by $\tilde{\eta}(s)=\eta(-r-s)$ . Then$[-r, 0]$ and $f$ is Riemann-Stieltjes integrable with respect to$\tilde{\eta}$$\tilde{\eta}$

is monotone on. By the assumption (H1),

79
for any positive integer$n$there exist $t_{k}\in(-r, 0)$ such

and $\eta(t_{k})=-\tilde{\eta}(t_{k})+\eta(0)+\eta(-r)$ for $k=1,2,$$\cdots,$ $n$$\mathrm{t}\mathrm{h}\mathrm{a}\mathrm{t}-r(1-\frac{k-1}{n})<t_{k}<-r(1-\frac{k}{n})$. For a partition$D_{n}$: $-r=t_{0}<t_{1}<$$<t_{n}<t_{n+1}=0$ of $[-r, 0]$ and any choice of $\xi_{k}\in[t_{k-1}, t_{k}]$ , we consider the Riemann

sum $S(f;\eta;D;\xi)n$ . Then we have$S(f; \eta;D_{n};\xi)=\sum_{k=1}^{n}f(+1\xi_{k})[\eta(tk)-\eta(t_{k1}-)]$$=- \sum_{k=1}^{n}f(\xi_{k})[+1\tilde{\eta}(t_{k})-\tilde{\eta}(t_{k}-1)]=-S(f;\tilde{\eta};D;\xi)n$.

Noticing that $d(D_{n})= \max_{1\leq k\leq}n+1(t_{k}-tk-1)<\frac{2r}{n}$ , we have$\int_{-r}^{0}f(S)d\eta(s)=-\int_{-r}^{0}f(S)d\tilde{\eta}(S)$as$narrow\infty$. This completes the proof.

$\square$

Proof of Theorem 1.2. The characteristic equation of$(\mathrm{A}\mathrm{L})$is the following form;$\triangle(\lambda)=\det[\lambda I+R(\theta)\int_{-r}^{0}e^{\lambda_{S}}d\eta(s)]=0$

where

$I$

is the$2\cross 2$,

identity matrix. We use the fact that the zero solution of(1.3)$(\mathrm{A}\mathrm{L})$isuniformly asymptotically stable if and only if all the roots of the characteristic equation(1.3) lie in the left half of the complex plane, that is, the real part of every characteristicroot$\lambda$of (1.3) is negative.

Let$p^{\pm}( \lambda;\mu)=\lambda+\rho e\pm i\theta\int_{-r}^{0}e^{\mu\lambda}ds\eta(S)$for a parameter $\mu\in[0,1]$ . Then the characteristic equation of (1.3) can be expressed in
the following form:$\triangle(\lambda)=\det[_{\sin}^{\lambda+\mathrm{c}_{\theta}}\mathrm{o}\mathrm{s}\int\theta\int_{S) ,-r}-r(e^{\lambda s_{d\eta}}S)0e0\lambda_{S}d\eta( -\sin\cos\theta\int_{\lambda+\theta\int_{-r}d\eta}-r_{0})e^{\lambda_{S}}d\eta(s]0e^{\lambda_{S}}(_{S)}$$=( \lambda+\cos\theta\int^{0}-red\lambda S(\eta s))2+(\sin\theta\int_{-}^{0}r(ed\lambda_{S}s\eta))^{2}$$=( \lambda+e^{i\theta}\int_{-r}^{0}e^{\lambda}dS(\eta s))(\lambda+e^{-i\theta}\int_{-r}^{0}e^{\lambda s_{d(s}}\eta))$80$=p^{+}(\lambda;1)p^{-}(\lambda;1)=0$.

Now we consider the distribution of the zeros of$p^{+}(\lambda;\mu)$and$p^{-}(\lambda;\mu)$in the complex

plane.
(Sufficiency) If $\mu=0$ , then$| \theta|<\frac{\pi}{2}$$\lambda$. Then we have$\lambda=-e^{\pm i\theta}\{\eta(\mathrm{o})-\eta(-r)\}$. By (1.2) and $\eta(0)>\eta(-r)$ ,. Noticing that each branch of${\rm Re}\lambda=-\cos(\pm\theta)\{\eta(\mathrm{o})-\eta(-r)\}<0$

is continuous in the parameter , we only show that there is no zeros on the imaginary$\mu$axis for any $\mu\in(0,1]$ . If
When$\overline{\lambda}$$\triangle(i\omega)=0$

for an$\omega\in \mathrm{R}$, we have $p^{+}(i\omega;\mu)=0$ or $p^{-}(i\omega;\mu)=0$ .

is the complex conjugate of any complex , the relation$\lambda$(1.4)$p^{+}(\lambda;\mu)=\overline{p-(\overline{\lambda}\cdot,\mu)}$stands and $p^{-}(i\omega;\mu)=0$ implies $p^{+}(-i\omega;\mu)=0$ . Thus, we only have to consider the case$p^{+}(i\omega;\mu)=0$. Calculating$p^{+}(i\omega :\mu)$, we have$p^{+}(i \omega;\mu)=i\omega+e^{i\theta}\int_{-r}^{0}e^{i}\eta(\mu\omega s_{d})s$$= \int_{-r}^{0}\cos(\theta+\mu\omega S)d\eta(s)+i(\omega+\int_{-r}^{0}\sin(\theta+\mu\omega S)d\eta(s))$.

Therefore we have$\int_{-r}^{0}\cos(\theta+\mu\omega s)d\eta(S)=0$

and$\omega=-\int_{-r}^{0}\sin(\theta+\mu\omega s)d\eta(S)$.

Using the assumption (H1) and Lemma 1.2, we have$\int_{-r}^{0}\cos(\theta+\mu\omega s)d\eta(_{S)\frac{1}{2}}=[\int_{-r}0\theta\cos(+\mu\omega S)d\eta(_{S})\cdot+\int_{0}-r(\cos\{\theta+\mu\omega(-r-s)\}d\eta-r-S)]$$= \int_{-r}^{0}\frac{\cos(\theta+\mu\omega s)+\cos\{\theta+\mu\omega(-r-S)\}}{2}d\eta(s)$$= \cos\frac{2\theta-\mu\omega r}{2}\int_{-r}^{0}\cos\frac{\mu\omega(r+2S)}{2}d\eta(_{S)}$and$- \int_{-r}^{0}\sin(\theta+\mu\omega S)d\eta(s)=-\int_{-r}^{0}\frac{\sin(\theta+\mu\omega s)+\sin\{\theta+\mu\omega(-r-S)\}}{2}d\eta(s)$.

$=- \sin\frac{2\theta-\mu\omega r}{2}\int_{-r}^{0}\cos\frac{\mu\omega(r+2S)}{2}d\eta(_{S)}$81
Therefore we have
(1.5)$\cos\frac{2\theta-\mu\omega r}{2}\int_{-r}^{0}\cos\frac{\mu\omega(r+2S)}{2}d\eta(s)=0$

and.

$\omega=-\sin\frac{2\theta-\mu\omega r}{2}\int_{-r}^{0}\cos\frac{\mu\omega(r+2S)}{2}d\eta(_{S)}$By (1.5) and$| \theta|<\frac{\pi}{2}$$\eta(0)>\eta(-r)$

obtain. Then$\cos\frac{2\theta-\mu\omega r}{2}=0$(1.6), we get $\int_{-r}^{0}d\eta(s)=0$ for $\omega=0$ . This contradicts the inequality$\omega\neq 0$, and hence$\int_{-r}^{0}\cos\frac{\mu\langle_{\mathrm{A}}J(r+2s)}{2}d\eta(S)\neq 0$by (1.6). Therefore wefrom (1.5), that is,for some integer .

$\omega=\frac{(2n+1)\pi+2\theta}{\mu r}$

Substituting the above$\omega$$n$in (1.6), we have$\omega=\frac{(2n+1)\pi+2\theta}{\mu r}=(-1)n\int^{0}-r\cos\{\frac{(2n+1)\pi+2\theta}{2}\frac{r+2s}{r}\}d\eta(S)$

.

(1.7)

From Lemma 1.1 and $0<\mu\leq 1$ , we obtain$| \frac{(2n+1)\pi+2\theta}{r}|\leq\mu\int_{-}^{0}r|\cos\{\frac{(2n+1)\pi+2\theta}{2}\frac{r+2s}{r}\}|d\eta(s)$$\leq|2n+1|\int_{-r}^{0}\cos\{(\frac{\pi}{2}-|\theta|)\frac{r+2s}{r}\}d\eta(S)$$<|2n+1| \frac{\pi-2|\theta|}{r}<\frac{|2n+1|\pi-2|\theta|}{r}$,where we used (1.2). This is a contradiction. Therefore there is no characteristic root on
the imaginary axis when $\mu\in(0,1]$ .
(Necessity) If $\eta(0)=\eta(-r)$ , then$x(t0, \phi)$of$(\mathrm{A}\mathrm{L}),$$\eta(0)\neq\eta(-r)$$x(t_{0,\phi})(t)\equiv\phi(0)$$\eta(s)\equiv 0$

for all

for $s\in[-r, 0]$ . In this case, for any solution$t\geq t_{0}$. This is a contradiction and we obtain.

Suppose that $\eta(0)<\eta(-r)$ or$\int_{-r}^{0}\cos\{\frac{r+2s}{r}(\frac{\pi}{2}-|\theta|)\}d\eta(S)\geq\frac{\pi-2|\theta|}{r}$Claim 1. There exist a $\mu_{0}\in[0,1]$ and a$\lambda_{0}\in \mathrm{C}$such that.

$p^{+}(\lambda_{0};\mu 0)=0$

(1.8)

and

${\rm Re}\lambda_{0}\geq 0$.

82
Proof of Claim 1. If$| \theta|=\frac{\pi}{2}$, then $\mu_{0}=0$ and $\lambda_{0}=-i(\sin\theta)\{\eta(0)-\eta(-r)\}$ . If

and $\eta(0)<\eta(-r)$ , then $\mu_{0}=0$ and$\lambda_{0}=-e^{i\theta}\{\eta(\mathrm{o})-\eta(-r)\}$. If$| \theta|<\frac{\pi}{2}$$| \theta|<\frac{\pi}{2}$and $\eta(0)>\eta(-r)$ ,then there exists a $\mu_{0}\in(0,1]$ such that$\mu_{0}\int_{-r}^{0}\cos\{\frac{r+2s}{r}(\frac{\pi}{2}-|\theta|)\}d\eta(s)=\frac{\pi-2|\theta|}{r}$by (1.8). This yeilds that$\lambda_{0}=i\frac{\pi-2|\theta|}{\mu \mathrm{o}r}$.

Claim 2. Assume that there exist an
Consider the zero ofthrough$\lambda=i\omega$at$p^{+}(\lambda;\mu)$$\mu=\tilde{\mu}$$\omega\in \mathrm{R}$and awith parameter $\mu\in[0,1]$ and let. Then${\rm Re}( \frac{\partial\lambda}{\partial\mu})>0$

when$\lambda=i\omega$

Proof of Claim 2. Taking the partial derivative of$\lambda$such that$\tilde{\mu}\in[0,1]$

and

with

$\mu$

.

be a branch of the zero$\lambda$$\mu=\tilde{\mu}$on$p^{+}(i\omega;\tilde{\mu})=0$.

$p^{+}(\lambda :\mu)=0$, we obtain$- \lambda e^{i\theta}\int_{-r}^{0_{Se^{\mu}}}\lambda s_{d\eta}(s)$$\frac{\partial\lambda}{\partial\mu}=\overline{1+\mu e^{i\theta}\int_{-}^{0}r\eta se^{\mu}\lambda s_{d}(s)}$If$\tilde{\mu}=0$

and$| \theta|<\frac{\pi}{2}$, there is no$\omega=-\sin\theta\{\eta(\mathrm{o})-\eta(-r)\}$$\omega\in \mathrm{R}$such that $p^{+}(i\omega;0)--0$ . If$\tilde{\mu}=0$

and$| \theta|=\frac{\pi}{2}$, thenand${\rm Re}( \frac{\partial\lambda}{\partial\mu})|_{\lambda=i\omega}={\rm Re}\{-i\omega e^{i}\theta\int-rse^{\mu}d\lambda s\eta 0(s)\}$$= \omega\sin\theta\int_{-r}0_{Sd\eta(s)=}\frac{r\omega^{2}}{2}>0$

If$\tilde{\mu}\in(0,1]$, we have$\int_{-r^{S\mathrm{c}}}^{0}\mathrm{o}\mathrm{s}(\theta+\tilde{\mu}\omega s)d\eta(S)$$s)\}=\sin(\theta+\tilde{\mu}\omega s)$$\omega=\frac{(2n+1)\pi+2\theta}{\overline{\mu}r}$andfor some integer$L_{s}= \int_{-r^{S\mathrm{s}}}^{0}\mathrm{i}\mathrm{n}(\theta+\tilde{\mu}\omega s)d\eta(S)$$n$from (1.7).

. Noticing that

Letting $L_{c}=$$\sin\{\theta+\tilde{\mu}\omega(-r-$and the as sumption (H1), we have$L_{s}= \int_{-r}^{0}s\sin(\theta+\tilde{\mu}\omega S)d\eta(S)+\int_{0}^{-r}(-r-s)\sin\{\theta+\tilde{\mu}\omega(-r-S)\}d\eta(-r-S)$$=- \frac{r}{2}\int_{-r}^{0}\sin(\theta+\tilde{\mu}\omega S)d\eta(s)=\frac{r\omega}{2}$where we used the equality,$\omega=-\int_{-r}^{0}\sin(\theta+\tilde{\mu}\omega s)d\eta(S)$Therefore we obtain${\rm Re}( \frac{\partial\lambda}{\partial\mu})|_{\lambda=\dot{u}v}={\rm Re}\{\frac{-i\omega(L_{c}+iL_{s})}{1+\tilde{\mu}(L_{C}+iL_{s})}\}$

obtained from${\rm Im} p^{+}(i\omega;\tilde{\mu})=0$.

83$={\rm Re}[ \frac{-i\omega\{L_{c}(1+\tilde{\mu}L_{c})+\tilde{\mu}L^{2}\}s\omega+L_{s}}{(1+\tilde{\mu}L_{C})^{2}+(\tilde{\mu}L_{S})^{2}}]$$= \frac{\frac{\omega^{2}r}{2}}{(1+\tilde{\mu}L_{c})2+(\tilde{\mu}L_{s})2}>0$

Claims 1 and 2 yield that the branch of the zero of$\mu=\mu_{0}$.

$p^{+}(\lambda;\mu)$through the point$\lambda=\lambda_{0}$atcontinues to lie in the right half of the complex plane for $\mu\in(\mu_{0},1]$ . So the zero

solution of$(\mathrm{A}\mathrm{L})$is not uniformly asymptotically stable. This is a contradiction and theproof is completed.

$\square$

2. Applications

We will give some applications of Theorems 1.1 and 1.2.
Example 2.1. Consider 2-dimensional delaly differential equations with $N$ delays$\dot{x}(t)=-R(\theta)\sum a_{k^{X}}(t-\tau_{k})kN=1$

where $R(\theta)=$ andfor $k=1,2,$$\cdots$

for $i,j=1,2,$, $N$ . Here$\cdots,$$N$$\tau\geq 0$$| \theta|\leq\frac{\pi}{2}$

(2.1)

’

. Suppose $\tau_{k}=\tau+(k-1)l$ and $a_{N-k+1}=a_{k}$and $l>0$ are constants. We also suppose that$a_{i}a_{j}\geq 0$

and $\tau_{N}>0$ . Let $r=\tau_{N}+\tau_{1}$ and$0$$\eta(s)=k\sum_{=1}eNk(S)$,,$s\in[-r, -\mathcal{T}_{k}]$$e_{k}(s)=\{$$a_{k}$$s\in(-\mathcal{T}k, \mathrm{o}]$.

Then the assumption (H1) holds, and Theorem 1.2 is applicable. Let us compute theformula in the condition (1.2).
$\int_{-r}^{0}\cos\{\frac{r+2s}{r}(\frac{\pi}{2}-|\theta|)\}d\eta(_{S})=\sum_{k=1}a_{k}\cos N\{$

Then we have the following:$\frac{r-2\tau_{k}}{r}(\frac{\pi}{2}-|\theta|)\}$

84
Corollary 2.1. The zero solution of (2.1) is$unif_{\mathit{0}}mry$asymptotically stable if and onlyザ$\sum_{k=1}^{N}a_{k}>0$

Remark 2.1. Thorem

and$\mathrm{B}$$\sum_{k=1}^{N}a_{k}\cos\{\frac{\tau_{N-k+1}-\tau k}{\tau_{1}+\tau_{N}}(\frac{\pi}{2}-|\theta|)\}<\frac{\pi-2|\theta|}{\tau_{1}+\tau_{N}}$.

is given by putting $a_{k}=a$ in this corollary. St\’ep\’an [6, p. 65]

and Kuang [4, p. 87] proved that the zero solution of the scalar delay differential equation
with two delays$x’(t)=-ax(t-\mathcal{T}_{1})-ax(t-\mathcal{T}_{2})$where $a>0,$$\tau_{1},$$\tau_{2}\geq 0,$$\tau_{1}+\tau_{2}>0$,, is uniformly asymptotically stable if and only if$2a( \tau_{1}+\tau_{2})\cos(\frac{\tau_{1}-\tau_{2}}{\tau_{1}+\tau_{2}}\frac{\pi}{2})<\pi$

.

(2.2)

If $\theta=0,$ $a_{k}=a$ and $N=2$ , the condition in Corollary 2.1 coincides with (2.2). The proofof Theorem 1.2 is given by generalizing the proof given by Kuang.
Example 2.2. Consider a scalar integro-differential equation$\dot{x}(t)=-\int_{t-r}^{t}c(t-S)X(s)ds$

(2.3)

,

where $c:[0, r]arrow[0, \infty)$ is continuous satisfying $c(s)=c(r-S)$ and$r$

is a positive constant.

If we choose $\eta(s)=\int_{0}^{s}c(-\xi)d\xi$ for $s\in[-r, 0]$ , then the assumption (H1) holds. Applying
Theorem 1.2 for $\theta=0$ , we have
Corollary 2.2. The zero solution of (2.3) is uniforrnly $asymptoti_{Cal}l,y$ stable if and only
ザ$0< \int_{0}^{r}c(s)\sin(\frac{s}{r}\pi)dS<\frac{\pi}{r}$,$\mathrm{t}$

.

(2.4)

85
Remark 2.2. Krisztin [3] gives the following excellent sufficient condition as far as theauthor knows: If(2.5)$0< \int_{-r}^{0}|_{S}|d\eta(s)<\frac{\pi}{2’}$then the zero solution of (1.1) is asymptotically stable. In fact, it becomes a necessary andsufficient condition in case $N=1$ and $\theta=0$ in Example 2.1. However, let $c(s)=1$ for$s\in[0, r]$in (2.3), then conditions (2.4) and (2.5) are reduced to $0<r<\pi/\sqrt{2}=2.221\ldots$

and $0<r<\sqrt{\pi}=$ 1.772. . ., respectively. This gap suggests us it should be difficult

to obtain an explicit necessary and sufficient condition ensuring the uniform asymptotic

stability of the zero solution of$(\mathrm{A}\mathrm{L})$without some restriction on .
$\eta$

References[1] Hara, T., Miyazaki, R. and Morii, T., Asymptotic stability condition for linear differential-difference

equations with n delays, to appear.
[2] Hara, T. and Sugie, J., Stability region for systems of differential-difference equations, hnkcial. Ekwac.,
39(1996), 69-89.

[3] Krisztin, T., Stability for functional differential equations and some variational problems, T\^ohoku

Math. J., 42(1990), 407-417.
[4] Kuang, Y., Delay Differential Equations with Applications in Population Dynamics, Academic Press,

San Diego, 1993.
[5] Miyazaki, R., Characteristic equation and asymptotic behavior of delay-differential equation, to ap-

pear.
[6] St\’ep\’an, G., Retarde Dynamical Systems: Stability and Characteristic Functions, Longman Scientificand Technical, U.K., 1989.