数理解析研究所講究録 1445 巻 2005 年 56-66

58
Asymptotic Stability for the Linear Integro-Differential
Equation $\dot{x}(t)=ax(t)$ $-b \int_{t-h}^{t}x(s)ds$
舟久保 稔 (Minoru Funakubo) *大阪府立大学大学院工学研究科

原惟行 (Tadayuki Hara) \dag er

Graduate School of Engineering, Osaka Prefecture University
大阪電気通信大学工学部

坂田 定久 (Sadahisa Sakata) \d ag er

Faculty of Engineering, Osaka Electro-Communication University

Introduction

1

In this paper} we will discuss the uniform asymptotic stability of the zero solution of
the linear integro-diflerential equation$\dot{x}(t)=ax(t)-b$ $f_{t-h}^{t}x(s)ds$

where

$a$

and

$b$

(E)

,

are real and $h>0$ . As a special case, for $a=0$ , (E) becomes(I)$i(t)=-b \int_{t-h}^{t}x(s)ds$

and in [4] it is shown that the zero solution of (I) is uniformly asymptotically stable if andonly if$0<bh^{2}< \frac{\pi^{2}}{2}$.

There are also some stability results for (I) with a generalized continuously distributed
delay which is expressed in Stieltjes integral [3]. In case $a<0$ , some sufficient stability
conditions for $a<0$ are obtained by using Liapunov functionals in [1].
But, there exist no results on the stability of (E) for the case $a>0$ as far as theauthors know. So, we will study (E) for $a>0$ and give results on the uniform asymptotic

stability of (E).

Main results

2

We obtain the following theorems on the uniform asymptotic stability of$\dot{x}(t)$*$=ax(t)-b \int_{t-h}^{t}x(s)ds$

Email address: funamino@ms.osakafu-u.ac.jp
Email address: hara@ms.osakafu-u.ac.jp
Email address:sakata@isc.osakac.ac.jp

,

(E)

57

where $a>0$ ,$b$is real and $h>0$ .

Theorem 2.1. Let $a^{2}<2b$ . Then, the zero solution of (E) is uniformly asymptoticallystable if and only if$\frac{a}{b}<h<\frac{1}{\sqrt{2b-a^{2}}}$(2.1)$\mathrm{C}\mathrm{o}\mathrm{s}^{-1}\frac{a^{2}-b}{b}$is satisfied.

Theorem 2.2. Let $a^{2}>2b$ . Then, the zero solution of (E) is not uniformly asymptotically stable for all h $>0$ .

To give the proofs of Theorems 2.1 and 2.2 with the root analysis, we need to consider
the characteristic equation of (E) which is expressed in the form
(C)$\lambda=a-b\int_{-h}^{0}e^{\lambda s}ds$

and we introduce the following results which are used without proofs.

Theorem A. [2] The zero solution of (E) is uniformly asymptotically stable if and only
if any root of (C) has a negative real partNow, let $\nu(h)$ be the number of roots of (C) including multiplicity whose real parts

are positive at . Then, the following property holds.
$h$Theorem B. [5] Let $(\mathrm{i}=0,1,2, \cdots)$ be constants at which (C) has a root on theis constant on each interval
imaginary axis of the complex plane. Then, the number$h_{i}$$\nu(h)$$h_{l}<h<h_{i+1}$.

Lemma C. [6]If $a- bh\geq 0$ , then (C) has a nonnegative real root.

implies that if $0<h\leq a/b$ then the zero solution of (E) is not uniformly
asymptotically stable by Theorem A. in case $b\leq 0$ , (C) has a nonnegative real root for$bh>0$ is satisfied. Thus, in case $b\leq 0$ , we also see
all $h>0$ because the conditionLemma$\mathrm{C}$$a$–easily that the zero solution of (E) is not uniformly asymptotically stable. Hereafter, we

assume b $>0$ .

3

The proof of main results

In this section, we will prove Theorems 2.1 and 2.2 regarding and are fixed constants
and is a variable. At first, we give a proof of Theorem 2.1.
$a$

$h$

$b$

58

Proof of Theorem 2,1. (Sufficiency.) Our purpose is to show $\nu(h)=0$ whensatisfies the condition (2.1). However, it is difficult to get $\nu(h)=0$ directly only by using
the condition (2.1). So at first we will discuss the case $h\in(0, a/b)$ .
Before proving sufficiency of Theorem 2.1 we give four lemmas. Now, we consider thecase where increases minutely from zero. Then, we have
$h$

$h$

Lemma 3.1. For a sufficiently small $h>0$ , (C) has no pairs of imaginar$ry$ roots A
such that $x>0$ , $y>0$ .

$=x\pm \mathrm{i}y$

Proof. Suppose that there exists a pair of imaginary roots $=x\pm iy(x>0, y>0)$ .
Here, we note that (C) has a root of complex conjugate. Thus, it is sufficient to discuss
for (C), we haveonly. Then, substitutingA$\lambda$$\lambda=x+\mathrm{i}y$$=x+\mathrm{i}y$$x=a-b \int_{-h}^{0}e^{xs}\cos ysds$

(3.1)

,

(3.2)

$y=-b\{\begin{array}{l}0\mathrm{e}^{xs}\mathrm{s}\mathrm{i}\mathrm{n}ysds-h\end{array}\mathrm{t}$Prom (3.2),$y \leq b|\int_{-h}^{0}e^{xs}\sin ysds|\leq b\int_{-h}^{0}e^{xs}|\sin ys|ds$.

Also, from $|\sin ys|\leq y|s|$ ,$y\leq by$ $f_{-h}^{0}e^{xs}|s|ds<bhy \int_{-h}^{0}e^{xs}ds=bhy\mathrm{x}$$\frac{1}{x}(1-e^{-xh})<\frac{bhy}{x}$.

Hence,$0<x<bh$ ,. However, from (3.1), we have $xarrow a-$ asasso thata contradiction. This completes the proof of Lemma 3.1.
$xarrow+\mathrm{O}$$0$$harrow+\mathrm{O}$$harrow+\mathrm{O}$, which is$\square$Lemma 3.2. Suppose $a^{2}<2b$ and $0<h<a/b$ . Then, (C) has a positive real root
Moreover, the positive real root is simple.
Proof.

Suppose$\lambda=x$

is a root of (C). Now, we difine the characteristic function of (E)expressed as$p(\lambda, a, b, h):=$A$-a+b \int_{-h}^{0}e^{\lambda s}ds$

.

(3.3)

Then from (3.3),$p(x, a, b, h)=x-a+b \int_{-h}^{0}e^{xs}ds=0$

.

(3.4)

59

(3.4) is reduced to$\frac{1}{x}\{x^{2}-ax+b-be^{-xh}\}=0$.

Here, we define the function $q(x)$ as$q(x):=x^{2}-ax+b-be$ $-xh$ .

(3.5)

Thus, it is sufficient to show that there exists a root $x>0$ which satisfies $q(x)=0$ . Prom

(3.5),(3.6)$q^{J}(x)=2x-a+bhe^{-xh}$ ,$q’(x)=2-bh^{2}e^{-xh}$

.

(3.7)

Prom (3.6) and (3.7), we obtain$\mathrm{q}(\mathrm{x})=0$,$q(a)=b(1-e^{-ah})>0$ .

$q’(0)=-a+bh<0$ ,Then, there exists the root $x^{*}\in(0, a)$ which satisfies $q(x^{*})=0$ . Moreover we note thatis monotone increasing for all $x\geq 0$ and$q^{tt}(x)$$q’(0)=2-bh^{2}>2-b( \frac{a}{b})^{2}=2-\frac{a^{2}}{b}>0$

so we have $q^{l}(x)>0$ on
is determined uniquely.

$[0, \infty)$,. Therefore, $q’(x)$ is monotone increasing on $[0, \infty)$ and$x^{*}$

which satisfies
Here, by Rolle’s theorem we also see that there exists the root
, we have $q^{t}(x^{*})>0$ , which implies
is monotone increasing on. Since$\tilde{x}\in(0, x^{*})$$q’(\overline{x})=0$$[0, \infty)$$q^{\mathit{1}}(x)$that $x=x^{*}>0$ is a simple root of $q(x)=0$ .

$\square$Now, we consider a case where (C) has a root on the imaginary axis of the complex
plane. At first, we assume that (C) has a pair of purely imaginary roots, we; constant) at the first time $h=h^{*}$ when is increases from zero. Then for$\lambda$$\lambda\neq 0$

$h$

$0$

$=\pm \mathrm{i}\omega(\omega>$

rewrite (C) as follows:$\lambda^{2}=a\lambda$$-b(1-e^{-\lambda h})$where $a>0$ , $b>0$ and $h>0$ . Substituting$-\omega^{2}=\mathrm{i}a\omega$$\lambda=$,$(\mathrm{C}^{*})$

iw and $h=h^{*}$ for$-b(1-e^{-i\omega h}.)$$(\mathrm{C}^{*})$,.

From the above,$-\omega^{2}=-b+b\cos\omega h^{*}$$a\omega$$=b$ $\sin\omega h’$

.

,

(3.8)

(3.9)

60

Rrom (3.8) and (3.9),$( \frac{b-\omega^{2}}{b})^{2}+(\frac{a\omega^{2}}{b})^{2}=1$

Because of$\omega$$>0$ ,.

we have $a^{2}<2b$ and$=\sqrt{2b-a^{2}}$$\omega$.

(3.10)

Substituting (3.10) for (3.8) and (3.9),$\cos\sqrt{2b-a^{2}}h^{*}=\frac{a^{2}-b}{b}$(3.11),(3.12).

$\sin\sqrt{2b-a^{2}}h^{*}=\frac{a\sqrt{2b-a^{2}}}{b}>0$

Prom (3.11) and (3.12),$h^{*}= \frac{1}{\sqrt{2b-a^{2}}}$

Cos -1$\frac{a^{2}-b}{b}$.

at $h=h^{*}$ . However,We see that (C) has a pair of purely imaginary roots. Next, we prove
we have not yet made reference to the case where (C) has the root$\lambda=\pm \mathrm{i}\omega$$\lambda=0$

the following lemma on the root

Lemma 3.3. (C) has the root
roots at $h=a/b$ .

$\lambda=0$$\lambda=0$.

at $h=a/b$ . Moreover, there exist no positive real

We will give a proof with the characteristic function $p(x, a, b, h)$ expressed by(3.3), where $x\in$ R. We have

Proof,$p$then (C) has the root

(0, a , , ) $=-a+b \{0-(-\frac{a}{b})\}=0$ ,$\lambda=0$$b$$\frac{a}{b}$at $h=a/b$ .

Now, we assume that (C) has a positive real root $x=x$ ’ at $h=a/b$ , that is,$p(x^{*}, a, b, a/b)=0$. Then,.

$\frac{\partial}{\partial x}p(x, a, b, \frac{a}{b})=\frac{1}{x^{2}}\{x^{2}+axe^{-\frac{a}{b}x}-b +be^{-\frac{a}{b}x}\}$Here, we define the function $f(x)$ as$f(x):=x^{2}+axe^{-\frac{a}{b}x}-b+$$be^{-\frac{a}{b}x}$

Then,$f’(x)=2x(1- \frac{a^{2}}{2b}e^{-\frac{a}{6}x})$

.

.

81

Because of $a^{2}<2b$ , $f’(x)>0$ holds for all $x>0$ . Hence, $f(x)$ is monotone increasing
for au $x>0$ . Moreover, from $f(\mathrm{O})=0$ , we have $f(x)>0$ for all $x>0$ . This implies
for all $x>0$ . Since $p(0, a, b, a/b)=0$ andis satisfied for all $x>0$ , so we obtain $p(x^{*}, a, b, a/b)>0$ , which contradicts the initialthat$\mathrm{a}\mathrm{e}\partial p(x, a,b, a/b)>0$$\frac{\partial}{\partial x}p(x, a, b, a/b)>0$

assumption. Therefore, (C) has no positive real root at $h=a/b$ . It is clear that (C) has$[]$
at $h=a/b$ . Thus, the proof of Lemma 3.3 is complete.
the root$\lambda=0$

Now we show that $h^{*}>a/b$ if $0<a^{2}<2b$ . Let$b$be fixed and difine the function $g(a)$as$g(a):= \sqrt{2b-a^{2}}(\frac{1}{\sqrt{2b-a^{2}}}\mathrm{C}\mathrm{o}\mathrm{s}^{-1}\frac{a^{2}-b}{b}-\frac{a}{b})$

.

Then,
.

$g(a)= \mathrm{C}\mathrm{o}\mathrm{s}^{-1}\frac{a^{2}-b}{b}-\frac{a\sqrt{2b-a^{2}}}{b}$

It is easily seen that $g(0)=\pi,$ $g(\sqrt{2b})=0$ and $g’(a)=-2\sqrt{2b-a^{2}}/b<0$ for all $a\in$. Therefore, we have that $g(a)>0$ for all $a\in(0, \sqrt{2b})$ , so $h^{*}>a/b$ holds.
and 3.2, (C) has a nonnegative and simple real root for $h\in(0, a/b]$ .
By Lemmas
We also see that (C) has no imaginary roots for a sufEiciently small $h>0$ by Lemma 3.1.
Moreover, (C) has the zero root but no positive real roots by Lemma 3.3. Here, notingthat (C) has no pair of purely imaginary roots during moves from zero to , we seethat a root of whose real part is nonnegative for $h\in(0, a/b]$ is unique and the real root$(0, \sqrt{2b})$$\mathrm{C}$$h^{*}$$h$determined by Lemma C. Thus, we have $\nu(h)=1$ for $h\in(0, a/b)$ .
Finally, we investigate $l/(h)$ for the case $h>a/b$ . Now, we will investigate the behavior
with $\lambda(a/b)=0$ . Then, we show the following lemma for the behavior ofof the root$\lambda(h)$$\lambda(h)$.

Lemma 3.4. For the root
Proof.

$\lambda(h)$with $\lambda(a/b)=0$ ,${\rm Re}(d\lambda/dh)|_{\lambda}$

We again use the characteristic function$h=a/b=\mathfrak{a},<0$$p(\lambda, a, b, h)$holds.

given by (3.3). Then, we

obtain$\frac{\partial}{\partial\lambda}p(\lambda, a, b, h)=1+b\int_{-h}^{0}se^{\lambda s}ds$,$\frac{\partial}{\partial h}p(\lambda, a, b, h)=be^{-\lambda h}$.

By the theorem on implicit function, we obtain$\frac{d\lambda}{dh}=-\frac{be^{-\lambda h}}{1+b\int_{-h}^{0}se^{\lambda s}ds}$.

(3.13)

Here, we investigate the behavior of the root A $(a/b)$ when$h=a/b$ . Then, from (3.13),${\rm Re} \frac{d\lambda}{dh}|_{\lambda=}h=\frac{0_{a}}{b}=-\frac{b}{1-\frac{a^{2}}{2b}}<0$.

$h$increases minutely from62

This completes the proof of Lemma 3.4.

$\mathrm{C}1$moves into the left half-plane of the complex plane
Lemma 3.4 shows that the rootwhen increases minutely from $a/b$ . Since $\nu(h)=1$ for $h\in(0, a/b)$ , we have $\nu(h)=0$
for $h\in(a/b, h^{*})$ . Thus, the zero solution of (E) is uniformly asymptotically stable on theinterval $a/b<h<h^{*}$ by Theorem A. From the above, the proof of sufficiency of Theorem$\lambda(h)$$h$2.1 is complete.
(Necessity.) We will show that the uniform asymptotic stability of (E) implies the condition (2.1). We consider the contraposition of this statement, that is,
Proposition 3.1. Suppose $a^{2}<2b$ . If $h\leq a/b$ oris not unifo rmly asymptotically stable.

$h\geq h^{*}$. Then the zero solution of (E)Proof. First, we consider a case of $h\leq a/b$ . Then by Lemma , (C) has a nonnegative
real root, which implies $\nu(h)>0$ . Hence, the zero solution of (E) is not uniformly
asymptotically stable.
Next, we consider a case of $h\geq h^{*}$ . We proved that the (C) had a pair of purely
at $h=h^{*}$ in the proof of sufficiency. Here, we investigate theimaginary roots A$\mathrm{C}$$=\pm \mathrm{i}\omega$when increases minutely fromwithbehavior of the root
, the characteristic function of (E) is defined as follows:Rom$h$$\lambda(h^{*})=\pm \mathrm{i}\omega$$\lambda(h)$$h^{*}$.

$(\mathrm{C}^{*})$$p^{*}(\lambda, a, b, h):=\lambda^{2}-a\lambda+b(1-e^{-\lambda h})$.

(3.14)

From (3.14),$\frac{\partial}{\partial\lambda}p^{*}(\lambda, a, b, h)=2\lambda-a+bhe^{-\lambda h}$,$\frac{\partial}{\partial h}p^{*}(\lambda, a, b, h)=b\lambda e^{-\lambda h}$

.

Then,
‘

(3.15)$\frac{d\lambda}{dh}|_{\lambda=iw}h=h^{*}=\frac{-\mathrm{i}b\omega e^{-i_{\mathfrak{l}d}h^{*}}}{2\mathrm{i}\omega-a+bh^{*}e^{-i\omega h^{*}}}=\frac{-\mathrm{i}b\omega e^{-uvh^{*}}(-2\mathrm{i}\omega-a+bh^{*}e^{ivh^{*}})}{|2\mathrm{i}\omega-a+bh^{*}e^{-i\omega h^{*}}|^{2}}$by the theorem on implicit function. Now, we define the function$h_{1}(\omega):=-\mathrm{i}b\omega e^{-i\omega h^{*}}(-2\mathrm{i}\omega-a+bh^{*}e^{i\omega h^{*}})$

Then,$h_{1}(\omega)$$h_{1}(\omega)$

as follows:.

is reduced to$h_{1}(\omega)=b\omega\{(-2\omega\cos\omega h^{*}+a\sin\omega h^{*})+\mathrm{i}(2\omega\sin\omega h^{*}+a\cos\omega h^{*}-bh^{*})\}$

Considering the real part of$h_{1}(\omega)$, we have${\rm Re} h_{1}(\omega)=b\omega$( $-2\omega\cos\omega h^{*}+$ a$\sin\omega h^{*}$$=b \omega(-2\omega\frac{b-\omega^{2}}{b}+a\frac{a\omega}{b})$$=\omega^{4}>0$

)

.

63

from (3.8)-(3.10). This implies that the real part of the numerator of (3.15) is positive.

We also define the function$h_{2}(\omega)$

as follows:$h_{2}(\omega):=2\mathrm{i}\omega$Then,$h_{2}(\omega)$$-a+bh^{*}e^{-i\omega h^{*}}$

becomesA2 (u)$=(-a+bh^{*}\cos\omega h^{*})+\mathrm{i}(2\omega-bh^{*}\sin\omega h^{*})$(3.16)

Here, we consider a case where both the real and imaginary part of #2 (u) are zero. Thenfrom (3.16),$-a+bh^{*}$

coswh’ $=0$ ,(3.17)$-bh^{*}$

sinuh’ $=0$ .

(3.18)$2\omega$

By (3.17)$\mathrm{x}$$\sin$

uh$’+(3.18)\rangle\langle\cos\omega h^{*}$, we obtain-asinuh’
Therefore, from (3.8)-(3.10), we haveshowed$\frac{\partial}{\partial\lambda}p^{*}(\mathrm{i}\omega,a, b, h^{*})\neq 0$$\omega$$+2\omega$

coswh” $=0$ .

$=0$ only,

which contradicts $\omega>0$ . Hence, we.

Thus, He $(d\lambda/dh)|\lambda=:\omega>0$ holds. This means that a pair of purely imaginary roots
move into the right half-plane of the complex plane when increases minutely from 1*.
$h=h^{*}$$h$$\lambda(h)$Therefore, since we have $\nu(h)>0$ , the zero solution of (E) is not uniformly asymptoticallystable by Theorem A. This completes the proof of Proposition 3.1.

$\square$FYom Proposition 3.1, we can show the necessity of Theorem 2.1. Thus, the proof of
Theorem 2.1 is finished completely.
$\square$

Proof of Theorem 2.2. Next, we give a proof of Theorem 2.2. we note that $\nu(h)=1$
for $h\in(0, a/b)$ and (C) has the root $\lambda=0$ at $h=a/b$ in the same way as Theorem 2.1.
with $\lambda(a/b)=0$ andThus, it is sufficient to investigate the behavior of the root
for all $h>a/b$ . Here, for all $h>a/b$ , the following lemmas holds.
$\lambda(h)$$\nu(h)$Lemma 3.5. Suppose $a^{2}\geq 26$ . Then, (C) has no roots on the imaginary axis for allh $>a/b$ .
is constant for all $h>a/b$ . Since (C) has no pairs ofLemma 3.5 implies that
purely imaginary roots and no the zero root for all $h>a/b$ by using the characteristic, it is easy to give a proof of Lemma 3.5. In this paPer, we omit thefunction
details of this proof.
with $\lambda(a/b)=0$ , which it is quiet
Now, we will investigate the behavior offor all $h>a/b$ . Then, we must prove the following property
important on$\nu(h)$$p(\lambda, a, b, h)$$\lambda(h)$$\nu(h)$64

Lemma 3.6. Suppose $a^{2}>26$ . Then, we have $\nu(h)>0$ for all h $>a/b$ .

Proof. In case $a^{2}>26$ , we consider a sign of${\rm Re}(d\lambda/dh)|$,$h=a/b\lambda=0$

in the same way as

Lemma 3.4. Then,.

${\rm Re} \frac{d\lambda}{dh}|_{\lambda=}h=\frac{0_{a}}{b}=-\frac{b}{1-\frac{a^{2}}{2b}}>0$

moves to the right half-plane when is increased from $a/b$ minutely
It implies thatand the root which exists in the right half-plane for $h\in(0, a/b)$ remains in the righthalf-plane. Hence, we have $\nu(h)$ $=2>0$ for all $h>a/b$ from Lemma 3.5.
$h$$\lambda(h)$$\square$Thus, by Lemmas 3.5 and 3.6 we show that the zero solution of (E) is not uniformly
asymptotically stable for all $h>0$ , so the proof of Theorem 2.2 is complete.
$\square$4

Critical case $a^{2}=2b$ and conjectures

In this section, we consider the case $a^{2}=26$ . Here, in the same way as case $a^{2}>2b$ ,at $h=a/b$ .
we can see easily that $\nu(h)=1$ for $h\in(0, a/b)$ and (C) has the root$\lambda=0$

However, if we introduce the characteristic function $p(\lambda, a, b, h)$ . Then,$p(0, a, b, \frac{a}{b})=0$,$\frac{\partial}{\partial\lambda}p(0, a, b, \frac{a}{b})=0$,.

$\frac{\partial^{2}}{\partial\lambda^{2}}p(0, a, b, \frac{a}{b})=\frac{a^{3}}{3b^{2}}\neq 0$

Therefore, we see that the root $\lambda=0$ is a double root of (C), and so we cannot analyze

the behavior of$\lambda(h)$with $\lambda(a/b)=0$ by using the derivative${\rm Re} \frac{d\lambda}{dh}|_{\lambda=}h=\frac{0_{a}}{b}=-\frac{b}{1-\frac{a^{2}}{2b}}$.

Thus, we need to discuss the case $a^{2}=2b$ by another method. But, to our negret wecannot find the new method now. By the numerical examples (Figures 1 through 3), weare convinced that the zero solution of (E) is not uniformly asymptotically stable in case$a^{2}=26$. Then, we have the following conjecture.

Conjecture 4.1. Let $a^{2}=26$ . Then, the zero solution of (E) is not uniformly asyrnptotically stable for all $h>0$ .
If we can prove Conjecture 4.1, then we can show immediately the following statement
by Theorems 2.1 and 2.2

65

Conjecture 4.2. The zero solution of (E) is unifo rmly asymptotically stable if and only

if conditions $a^{2}<2b$ and$\frac{a}{b}<h<\frac{1}{\sqrt{2b-a^{2}}}$$\mathrm{C}\mathrm{o}\mathrm{s}^{-1}\frac{a^{2}-b}{b}$are satisfied.
Finally, we will show the behavior of solutions numerically for the case $a^{2}=2b$ which

illustrate Conjecture 4.2. Then, we fix $a=4$ and $b=8$ and take the initial function as$\phi(t)=100$ $+20$. We put the parameter as follows and illustrate Conjecture 4.2 with$\sin t$$h$drawing the solution curves.

Figure 1: $h=0.45(0<h<a/b)$
Figure 2: $h=0.5(h=a/b)$
Figure 3: $h=0.55(h>a/b)$

Figures 1 through 3 suggest that the zero solution of (E) is not uniformly asymptoticallystable for all $h>0$ in case $a^{2}=2b$ .

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