数理解析研究所講究録1083 巻 1999 年 105-112

105

The asymptotic stability of $x_{n+1}-X_{n}+Ax_{n-k}=0$(Hideaki Matsunaga)

松永秀章

大阪府立大学工学部

1. Introduction and main results
The asymptotic stability of delay difference equations has been investigated by manyauthers. In scalar case, many results can be found in several books and papers [1-5].
Levin et. $al[5]$ and Kuruklis [4] have shown the nice result as follows:
The delay difference equation$x_{n+1}-X_{n}+qx_{n-k}=0$ ,where is a real number andonly if$q$$k$$n=0,1,$$\cdots$,

is a nonnegative integer, is as ymptotically stable if and$0<q<2 \cos\frac{k\pi}{2k+1}$.

In this paper we give some new necessary and sufficient conditions for the asymptotic
stability of a 2-dimensional linear delay difference system$x_{n+1}-x_{n}+Ax_{n-k}=0$ ,$n=0,1,$$\cdots$,(1)

constant matrix.
where is a nonnegative integer and is a
By the transformation $x_{n}=Py_{n}$ with an appropriate regular matrix $P$ , we canrewrite (1) as$A$$k$$2\cross 2$$y_{n+1}-y_{n}+P^{-1}APy_{n-k}=0$ ,$n=0,1,$Thus, we only have to consider (1) where the matrix
matrices :(I) $A=qR(\theta)\equiv q$ ,whereandFor the case (I), we have$q,$$q_{1},$$q_{2},$$b$$\theta(|\theta|\leq\frac{\pi}{2})$are real numbers.

$A$(II)$\cdots$

is either of the folloing two$A=$ ,

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Theorem 1. The system (1) is asymptotically stable if and only if$0<q<2 \cos\frac{k\pi+|\theta|}{2k+1}$.

(2)

For the case (II), we have
Theorem 2. The system (1) is asymptotically stable if and only ifand$0<q_{1}<2 \cos\frac{k\pi}{2k+1}$$0<q_{2}<2 \cos\frac{k\pi}{2k+1}$.

(3)

In this paper we only give the proof of the Theorem 1 since both theorems are provedin very similar way.

2. Proof of Theorem 1
Theorem 1 is proved by using the fact that the system (1) is asymptotically stable
if and only if all the roots of its characteristic equation(4)$F(\lambda)\equiv\det(\lambda^{k+1}I-\lambda kI+qR(\theta))=0$

are inside the unit disk. Hence, we investigate the characteristic roots of (4) to prove
Theorem 1.
Let$f^{+}(\lambda)\equiv\lambda k+1-\lambda k+qei\theta$,$f^{-}(\lambda)\equiv\lambda^{k+1}-\lambda^{k}+qe^{-i}\theta$.

Then we have$F(\lambda)=\det(^{\lambda^{k+1}-\lambda q\mathrm{c}\mathrm{o}}q\sin\theta k+\mathrm{s}\theta$$\lambda^{k+1}-\lambda^{k}+q\cos-q\sin\theta\theta)$$=(\lambda^{k+1}-\lambda k\mathrm{c}+q\mathrm{o}\mathrm{s}\theta)^{2}+q^{2}\sin^{2}\theta$$=(\lambda^{k+1}-\lambda^{k}+q\cos\theta+iq\sin\theta)(\lambda k+1-\lambda^{k}+q\cos\theta-iq\sin\theta)$$=f^{+}(\lambda)f^{-}(\lambda)$.

Note that $f^{-}(\lambda)=0$ implies, where is the complex conjugate of any
complex . Also,, substituting
in $f^{+}(\lambda)=0$ and $f^{-}(\lambda)=0$
implies. Therefore, we only have to consider the case $f^{+}(\lambda)=0$ under thecondition
to investigate the characteristic roots of (4). We also notice that(4) has no real roots if $q\neq 0$ and.
When $k=0$ , it follows from (4) that $\lambda=1-q\cos\theta\pm iq\sin\theta$ . Then$f^{+}(\overline{\lambda})=0$$\overline{\lambda}$$\tilde{\theta}=-\theta$$\lambda$$\mathrm{w}\mathrm{h}\mathrm{e}\mathrm{n}-\frac{\pi}{2}\leq\theta\leq 0$$0 \leq\tilde{\theta}\leq\frac{\pi}{2}$$0 \leq\theta\leq\frac{\pi}{2}$$\theta\neq 0$$|\lambda|^{2}=(1-q\cos\theta)^{2}+(q\sin\theta)^{2}=q^{2}-2q\cos\theta+1$.

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if and only if $0<q<2\cos\theta$ , and so, Theorem 1 stands if
It is easy to see that
$k=0$ . Hereafter, letbe a positive integer.
As a beginning, we shall examine the existence region of the arguments of complex
roots of $f^{+}(\lambda)=0$ .
$|\lambda|<1$$k$

Lemma 1. Assume that $q>0$ andbe a complex root of $f^{+}(\lambda)=0$ . Then$0 \leq\theta\leq\frac{\pi}{2}$$\frac{\theta+2p\pi}{k}<\omega<\frac{\theta+(2p+1)\pi}{k+1},$$\frac{\theta-\pi}{k+1}<\omega<0,$$p=0,1,$. Let$\cdots,$$re^{i\omega}$, with $r>0$ and $0<|\omega|<\pi$ ,$[ \frac{k-1}{2}]$$\frac{\theta-(2p+1)\pi}{k+1}<\omega<\frac{\theta-2p\pi}{k},$, if $\omega>0$ ;$p=1,2,$$\cdots,$$[ \frac{k}{2}]$, if $\omega<0$ ,where $[a]$ represents th.e integer part of a.

Proof. From $f^{+}(re^{iv}‘)=0$ , we have)$-1i((k+1)(vrkie(k\omega-\theta)+q=0$$r^{k+}e-\theta$, namely$r^{k}\{\cos(k\omega-\theta)-r\cos((k+1)\omega-\theta)\}=q$,(5)$r^{k}\{\sin(k\omega-\theta)-r\sin((k+1)\omega-\theta)\}=0$.

(6)

andIt is obvious that $\sin((k+1)\omega-\theta)\neq 0$ , so (6) implies$r= \frac{\sin(k\omega-\theta)}{\sin((k+1)\omega-\theta)}$(7).

(5) and (7) yield$q=r^{k_{\frac{\sin((k+1)\omega-\theta)\cos(k\omega-\theta)-\sin(k\omega-\theta)\cos((k+1)\omega-\theta)}{\sin((k+1)\omega-\theta)}}}$$=r^{k_{\frac{\sin\omega}{\sin((k+1)\omega-\theta)}}}$(8).

, the proof is similar.) From (7)We consider the case $0<\omega<\pi$ . (Inand (8), we must have $\sin(k\omega-\theta)>0$ and $\sin((k+1)\omega-\theta)>0$ because of $q>0,r>0$
and $\sin\omega>0$ . Hence,$\mathrm{c}\mathrm{a}\mathrm{S}\mathrm{e}-\pi<\omega<0$$\frac{\theta+2m\pi}{k}<\omega<\frac{\theta+(2m+1)\pi}{k}$,$m=0,1,$$\cdots,$$[ \frac{k-1}{2}]$and$\frac{\theta+2n\pi}{k+1}<\omega<\frac{\theta+(2n+1)\pi}{k+1}$,$n=0,1,$$[ \frac{k}{2}]$$\cdots,$,

which imply$\frac{\theta+2p\pi}{k}<\omega<\frac{\theta+(2p+1)\pi}{k+1}$

The proof is complete.

,$p=0,1,$$\cdots,$$[ \frac{k-1}{2}]$.

,

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When $q<0$ , we have the following analogous result.

Lemma 2. Assume that $q<0$ andbe a complex root of $f^{+}(\lambda)=0$ . Then$0 \leq\theta\leq\frac{\pi}{2}$$0< \omega<\frac{\theta}{k+1},$. Let$\frac{\theta+(2p-1)\pi}{k}<\omega<\frac{\theta+2p\pi}{k+1},$$\frac{\theta-(2p+2)\pi}{k+1}<\omega<\frac{\theta-(2p+1)\pi}{k},$$p=0,1,$

The next lemma determines the value ofcircle.

$q$

Lemma 3. Assume that $q>0$ andof $f^{+}(\lambda)=0$ on the unit circle are given by$re^{\iota\omega}$, with $r>0$ and $0<|\omega|<\pi$ ,$p=1,2,$$\cdots,$$[ \frac{k-1}{2}]$$\cdots,$$[ \frac{k}{2}]$, if $\omega>0$ ;, if $\omega<0$ .

and the root’s argument$0 \leq\theta\leq\frac{\pi}{2}$. Then theorwhere$argument_{\mathit{8}}$$\omega_{p}^{+}$$\omega$on the unitof complex roots$\omega_{p^{f}}^{-}$$\omega_{p}^{+}\equiv\frac{2\theta+(4p+1)\pi}{2k+1}>0$,$p=0,1,$$\cdots,$$\omega_{p}^{-}\equiv\frac{2\theta-(4p+1)\pi}{2k+1}<0$,$p=0,1,$$\cdots,$$[ \frac{k-1}{2}]$$[ \frac{k}{2}]$,.

Moreover, the following relation stands:$q=\sqrt{2-2\cos\omega}$.

(9)

Proof. Substituting $r=1$ into (5) and (6), we get$\cos(k\omega-\theta)-\cos((k+1)\omega-\theta)=q$

,

(10)

$\sin(k\omega-\theta)-\sin((k+1)\omega-\theta)=0$

.

(11)

and(11) implies that 2$\cos\frac{(2k+1)(v-2\theta}{2}\sin\frac{\omega}{2}=0$. Since$\sin\frac{\omega}{2}\neq 0$$\cos\frac{(2k+1)\omega-2\theta}{2}=0$, we have.

(12)

We consider the case $0<\omega<\pi$ . (In case $-\pi<\omega<0$ , the proof is similar.) Then
(12) yields$\omega=\frac{2\theta+(2n+1)\pi}{2k+1}$

By Lemma 1, (13) is suitable when$n$,$n=0,1,$$\cdot$,. , $k-1$ .

(13)

is only even, and therefore, we obtain$\omega=\omega_{p}^{+}\equiv\frac{2\theta+(4p+1)\pi}{2k+1}$,$p=0,1,$$\cdots,$$[ \frac{k-1}{2}]$.

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Next, by squaring both sides of (10) and (11), and adding them together, we have$q^{2}=\{\cos(k\omega-\theta)-\cos((k+1)\omega-\theta)\}2+\{\sin(k\omega-\theta)-\sin((k+1)\omega-\theta)\}^{2}$$=2-2\{\cos((k+1)\omega-\theta)\cos(k\omega-\theta)+\sin((k+1)\omega-\theta)\sin(k\omega-\theta)\}$$=2-2\cos\omega$.

Hence, (9) stands. The proof is complete.

andRemark 1. In view of the definitions of. Taking account ofis minimum when$\omega_{p}^{+}$$\omega_{p}^{-}$, the value of$\sqrt{2-2\cos\omega}=2|\sin\frac{\omega}{2}|$$\omega=\omega_{0}^{-}$

given by (9), we obtain that$q(\omega)$$q( \omega_{0}^{-})=2\sin|\frac{2\theta-\pi}{2(2k+1)}|=2\sin(^{\frac{\pi}{2}}-\frac{k\pi+\theta}{2k+1})=2\cos\frac{k\pi+\theta}{2k+1}$

When $q<0$ , we have the$\mathrm{f}\mathrm{o}\dot{1}\mathrm{l}\mathrm{o}\mathrm{w}\mathrm{i}\mathrm{n}\mathrm{g}$.

result which is analogous to Lemma 3.

Lemma 4. Assume that $q<0$ andof $f^{+}(\lambda)=0$ on the unit circle are given by$0 \leq\theta\leq\frac{\pi}{2}$. Then the$\alpha_{p}^{+}$

or$\alpha_{\mathrm{p}}^{-}$$argument_{\mathit{8}}$of complex roots, where$\alpha_{p}^{+}\equiv\frac{2\theta+(4p-1)\pi}{2k+1}>0$,$p=1,2,$ $\cdots,$$\alpha_{p}^{-}\equiv\frac{2\theta-(4p+3)\pi}{2k+1}<0$,$p=0,1,$ $\cdots,$$[ \frac{k}{2}]$,$[ \frac{k-1}{2}]$.

Furthermore, we shall observe the crossing of the unit circle by the roots of $f^{+}(\lambda)=0$
when the value of varies.
$q$. Then the simple root $\lambda=1$ of $f^{+}(\lambda)=0$ with
Lemma 5. Assume that) as $qincrease\mathit{8}$ fromunit disk (resp. outside the unit
$q=0$ moves inside(resp. decreases from ).
$0 \leq\theta<\frac{\pi}{2}$$0$$di\mathit{8}k$$the\vee$$0$

Proof. It suffices to show that $(dr/dq)|_{r=1}q=0<0$ . If$\theta=0$, let$\lambda=r$

be a positive rootof $f^{+}(\lambda)=0$ , then$r^{k+1}-r+q=k0$ .

(14)Note that $r=1$ implies $q=0$ . Taking the derivative of

$r$

with

$(k+1)r^{k_{\frac{dr}{dq}-kr^{k}\frac{dr}{dq}}}-11+=0$

or$\frac{dr}{dq}=\frac{1}{kr^{k-1}-(k+1)r^{k}}$.

,$q$on (14), we have110
Hence, we arrive at$\frac{dr}{dq}|_{r=1}=\frac{1}{k-(k+1)}=-1<0$

If$0< \theta<\frac{\pi}{2}$.

, it follows from (8) that$\frac{dr}{dq}=\frac{dr}{d\omega}(\frac{dq}{h})^{-1}$$= \frac{dr}{d\omega}(kr^{k-1}\frac{dr}{d\omega}\frac{\sin\omega}{\sin((k+1)\omega-\theta)}+r^{k_{\frac{d}{d\omega}}}(\frac{\sin\omega}{\sin((k+1)\omega-\theta)}))^{-1}$(15)$= \frac{dr}{d\omega}(\frac{qk}{r}\frac{dr}{d\omega}+r^{k_{\frac{d}{dv}}}(\frac{\sin\omega}{\sin((k+1)\omega-\theta)}))^{-1}$Also, by (7), we have$\frac{dr}{d\omega}=\frac{k\cos(k\omega-\theta)\sin((k+1)\omega-\theta)-(k+1)\sin(k\omega-\theta)\cos((k+1)\omega-\theta)}{\backslash \sin^{2}((k+1)\omega-\theta)}$,$= \frac{k\sin\omega-\sin(k\omega-\theta)\cos((k+1)\omega-\theta)}{\sin^{2}((k+1)\omega-\theta)}$and$\frac{d}{d\omega}(\frac{\sin\omega}{\sin((k+1)\omega-\theta)})$$= \frac{\cos\omega\sin((k+1)\omega-\theta)-(k+1)\sin\omega\cos((k+1)\omega-\theta)}{\sin^{2}((k+1)\omega-\theta)}$$= \frac{\sin(k\omega-\theta)-k\sin\omega\cos((k+1)\omega-\theta)}{\sin^{2}((k+1)\omega-\theta)}$.

Denote,(16)$H(\omega)\equiv\sin(k\omega-\theta)-k\sin\omega\cos((k+1)\omega-\theta)$(17)$G(\omega)\equiv k\sin\omega-\sin(k\omega-\theta)\cos((k+1)\omega-\theta)$

andthen (15) yields$\frac{dr}{dq}=\frac{G(\omega)}{(qk/r)c(\omega)+rHk(\omega)}$

.

(18)

Noticing that $q=0$ is equivalent to $\omega=0$ , we obtain that$\frac{dr}{dq}|_{r=1}=q=0\frac{G(0)}{H(0)}=\frac{-\sin(-\theta)\cos(-\theta)}{\sin(-\theta)}=-\cos\theta<0$

The proof is complete.

.

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Lemma 6. Assume that $q>0$ andunit circle move outside as increases.

$0 \leq\theta\leq\frac{\pi}{2}$. Then all the roots of$f^{+}(\lambda)=0$

on the$q$

Proof. By Lemma 3, it suffices to show that$\frac{dr}{dq}|_{r1}=>0q=q(\omega^{+})\mathrm{p}$$p=0,1,$’$\cdots,$,$[ \frac{k-1}{2}]$

(19)

and
$\frac{dr}{dq}|_{r=1 ,q=q}(\omega_{\mathrm{p}}^{-})>0$,$p=0,1,$$[ \frac{k}{2}]$$\cdots,$.

(20)

From (18), we have$\frac{dr}{dq}|_{qq(\omega_{\mathrm{p}}^{+})}r=1==\frac{G(\omega_{p}^{+})}{qkG(\omega_{p\mathrm{P}}^{+})+H(\omega^{+})}$,are defined by (16) and (17) respectively. Note that
andwhereand $\sin(k\omega_{p}^{+}-\theta)>0$ because of Lemma 1. Also, since$H(\omega)$$G(\omega)$$\frac{2\theta+(4p+1)\pi}{2(k+1)}<\omega_{\mathrm{p}}^{+}<\frac{\theta+(2p+1)\pi}{k+1}$$\sin\omega_{p}^{+}>0$,

it is easy to see that $\cos((k+1)\omega_{p}^{+}-\theta)<0$ . Thus, we obtain that $G(\omega_{p}^{+})>0$ and, and so (19) holds. Similarly, we can show (20). The proof is complete.
$H(\omega_{\mathrm{p}}^{+})>0$

When $q<0$ , using Lemmas 2 and 4 instead of Lemmas 1 and 3, we have the following
result which is analogous to Lemma 6.
that $q<0$ andLemma 7.
increases.
unit circle move outside as$A\mathit{8}Sume$$0 \leq\theta\leq\frac{\pi}{2}$. Then all the roots of $f^{+}(\lambda)=0$ on the$|q|$

Now we are ready to prove Theorem 1.
Proof of Theorem 1. Let be a characteristic root of (4). Here, recalling theargument above, we only have to consider the value of satisfying $f^{+}(\lambda)=0$ with$\lambda$$\lambda$$0 \leq\theta\leq\frac{\pi}{2}$.

. Thus, by virtue of Lemma 5
From (2), we must haveand continuity of with respect to , we notice that, if $q>0$ is sufficient small, thenholds for any and the system (1) is asymptotically stable.
ofIf the increasing of leads the system (1) to instability, there exists a root$f^{+}(\lambda)=0$ such thatis the minimum. By Remark 1, we find that 2
holds for. This fact indicates that, if (2) is true, thenvalue of when
any and the system (1) is asymptotically stable.
(Sufficiency)$0 \leq\theta<\frac{\pi}{2}$$\lambda$$q$$|\lambda|<1$$\lambda$$\lambda^{*}$$q$$|\lambda^{*}|=1$$q$$\lambda$$|\lambda^{*}|=1$$\cos\frac{k\pi+\theta}{2k+1}$$|\lambda|<1$112(Necessity) Suppose that the system (1) is asymptotically stable, that is, for any ,$\lambda$$|\lambda|<1$.

(21)

For the sake of contradiction, (2) is false. Then we consider two cases.
of $f^{+}(\lambda)=0$ such that. By Lemma 6, there exists a root
Case 1:, which contradicts (21).
as increases from 2
Case 2: $q\leq 0$ . By Lemma 5, the simple root $\lambda=1$ with $q=0$ moves outside theunit disk asincreases from . Hence, in view of the fact above and Lemma 7 there, which also contradicts (21).
of $f^{+}(\lambda)=0$ such thatexists a root$\lambda^{*}$$q \geq 2\cos\frac{k\pi+\theta}{2k+1}$$|\lambda^{*}|\geq 1$$\cos\frac{k\pi+\theta}{2k+1}$

$q$

$|q|$

$\lambda^{*}$$0$$|\lambda^{*}|\geq 1$

References[1] F. Brauer, Continuous and discrete delayed-recruitment population models, Dynam. Contin. Discrete Impuls. Systems, 3 (1997), 245-252.
[2] S. N. Elaydi, “An Introduction to Difference Equations”, Springer-Verlag, NewYork, 1996.

[3] V. L. Kocic and G. Ladas, “Global Behavior of Nonlinear Difference Equations ofHigher Order with Applications”, Kluwer Academic Publishers, Dordrecht, 1993.
[4] S. A. Kuruklis, The asymptotic stability of $x_{n+1}-ax+nbxn-k=0$ , J. Math. Anal.
Appl., 188 (1994), 719-731.

[5] S. A. Levin and R. M. May, A note on difference-delay equations, Theor. Popol.
Biol., 9 (1976), 178-187.