数理解析研究所講究録
978 巻 1997 年 183-189

183

A FUZZY RELATIONAL EQUATIONIN
DYNAMIC FFUZZY SYSTEMS
M. KU(Chiba University), M. YASUDA(
J. NAKAGAMI(Chiba University) &Y. YOSHIDA($1\mathrm{R}\mathrm{A}\mathrm{N}\mathrm{O}$$\mathrm{c}\mathrm{h}\mathrm{i}\mathrm{b}\mathrm{a}$University),
University)$\mathrm{K}\mathrm{i}\mathrm{t}\mathrm{a}\mathrm{k}\mathrm{y}\mathrm{u}\mathrm{s}\mathrm{h}\mathrm{u}$Abstract : For a dynamic
system, the fundamental method is to analysis its recursive relation of
the fuzzy states. It is similar as the Bellman equation is the important tool in the dynamic programming.
Here we will consider the existence and the umiqueness of solution of the fuzzy
al equation. Two
examples which satisfies our conditions, are given to illustrate the results.
$\mathrm{F}\backslash \mathrm{l}\mathrm{z}\mathrm{z}\mathrm{y}$$\mathrm{r}\mathrm{e}\mathrm{l}\mathrm{a}\mathrm{t}\mathrm{i}_{0\dot{\mathrm{n}}}$1

Introduction and notations

We use the notations in [4]. Let $X$ be a compact metric space. We denote by
the collection of all
subsets of $X$ , and denote by $C(X)$ the collection of all closed subsets of $X$ . Let be the Hausdorff metricon. Then it is well-known ([3]) that $(C(X), \rho)$ is a compact metric space. Let $F(X)$ be the set of all
fuzzy sets : $Xarrow[0,1]$ which are upper semi-continuous and satisfy $\sup_{x\in X}s(\sim X)=1$ . Let : $X\cross Xarrow$.
$[0,1]$ be a continuous fuzzy relation on $X$ .
in the following fuzzy
In this paper, we consider the existence and uniqueness of solution
relational equation (1.1) for given a continuous fuzzy relation on $X$ (see [4]) :$2^{X}$$\rho$$2^{X}$$\tilde{s}$$\tilde{q}$$\tilde{p}\in \mathcal{F}(X)$$\acute{q}$$\tilde{p}(y)=\sup_{x\in X}$

where$a\wedge b=\mathrm{m}\mathrm{i}\mathfrak{n}\{a, b\}$for real numbers{$\tilde{q}_{\alpha}(D):=\{$$a$$y|\tilde{q}(x,$ $y)\geq\alpha$$\mathrm{c}1${$y|\tilde{q}(x,${$\tilde{p}(x)$A$\tilde{q}(x,$$y\grave{)}$},$y\in X$and . We define a map$b$for some $x\in D$ }$\tilde{q}_{\alpha}$.

$r$: $2^{X}arrow 2^{X}(\alpha\in[0,1])$ by$D\in 2^{X},$forfor $\alpha=0,$ $D\in 2^{X},$,for $0\leq\alpha\leq 1,$$\alpha\neq 0,$$y)>0$ for some $x\in D$ }(1.1)

$D\neq\emptyset$$D\neq\emptyset$

,
,

(1.2)

$D=\emptyset$$X$

where cl denotes the closure of a set. Especially, we put$\tilde{q}_{\alpha}(x):=\tilde{q}_{\alpha}(\{x\})$for $x\in X$ . We note that$\tilde{q}_{\alpha}$:$C(X)arrow C(X)$ .

Lemma 1.1 ([4, Lenuma 2]). For each $\alpha\in[0,1]$ , the map
to .

$\tilde{q}_{\alpha}$: $C(X)arrow C(X)$ is contin uous with respect$\rho$

For$\tilde{s}\in \mathcal{F}(X)$, the$\alpha$-cut$s_{\alpha}\sim,$$\alpha\in[0,1]$

is defined by$s_{\alpha}\sim:=\{x\in X|s(\sim X)\geq\alpha\}(\alpha\neq 0)$

and$\tilde{s}_{0}:=\mathrm{c}1\{x\in X|\tilde{s}(x)>0\}$.

Lemma 1.2.
(i) For$\tilde{s}\in \mathcal{F}(X),\tilde{s}$satisfies (1.1) if and only if$\tilde{q}_{\alpha}(/\tilde{s}_{C\backslash })=\tilde{s}_{\alpha}$(ii) We suppose th at a family of(b) and (c):$s\mathrm{u}$,$\alpha\in[0,1]$

.

(1.3)

\’osets $\{D_{-\alpha}|\alpha\in[0, l]\}(\subseteq C,(X))$ satisfies the following conditions (a),

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(a)

$D_{\alpha}\subset D_{\alpha’}$(b)$\lim_{\alpha’\uparrow\alpha}D\alpha’=D_{\alpha}$(c)for $0\leq\alpha’<\alpha\leq 1$ ;for$\alpha\neq 0$;

for $\alpha\in[0,1]$ .

$\tilde{q}_{\alpha}(\tilde{S}_{\alpha})=\tilde{S}_{\alpha}$Then}, $x\in X$ , sa tisfies{ A1
characteristic function of a set $D\in 2^{X}$ .
$\tilde{s}(x):=\sup_{\alpha\in[0,1]}$$D_{\alpha}(x)$$\alpha$Proof. (i) is trivial. (ii) is from (i) and [4, Lemma 3].

2$\tilde{s}\in \mathcal{F}(X)$and (1.1), where$1_{D}$

denotes the$\square$

The existence of solutions

For $\alpha\in[0,1]$ and $x\in X$ , a sequence$\tilde{q}_{\alpha}^{0}(x):=\{x\}$Then, let,$\{\tilde{q}_{\alpha}^{k}(X)\}_{k=}1,2,\cdots$$\tilde{q}_{\alpha}^{1}(x):=\tilde{q}_{\alpha}(x)$$G_{a}(x):= \bigcup_{k=1}^{\infty}\tilde{q}_{\alpha}^{k}(x)$andis defined iteratively by$\tilde{q}_{\alpha}^{k+1}(X):=\tilde{q}_{\alpha}(\tilde{q}_{\alpha}^{k}(x))$

for $k=1,2,$$\cdots$.

and$F_{\alpha}(x):= \bigcup_{k=0}\tilde{q}_{\alpha}^{k}(X)=\infty\{x\}\cup G_{\alpha}(X)$

.

(2.1)

We now consider a class of invariant points for this iteration procedure, that is, $x\in G_{\alpha}(x)$ . So put$R_{\alpha}:=\{x\in X|x\in G_{\alpha}(x)\}$Each state ofis called as anand (ii) holds clearly:$R_{\alpha}$(i)$\tilde{q}_{a}(F_{\alpha}(x))=G_{\alpha}(x)$(ii)$R_{\alpha}\subset R_{\alpha’}$for$‘(\alpha$

for $\alpha\in[0,1]$ .

(2.2)-recurrent” state and it is studied by [7]. The following properties (i)$\alpha\neq 0$

and $x\in X$ ;for $0\leq\alpha’<\alpha\leq 1$ .

Lemma 2.1. If $z\in R_{1}$ , the following (i) and (ii) hold:(i)$\tilde{q}_{\alpha}(F_{\alpha}(z))=F_{\alpha}(z)$(ii)$F_{\alpha}(z)\subseteq F_{\alpha’}(Z)$Proof. Sincefor$\alpha\in[0,1]$

;

(2.3)

for $0\leq\alpha’<\alpha\leq 1$ .

$z\in R_{1}\subset R_{\alpha)}$

we have$\overline{q}_{\alpha}(F_{\alpha}(z))=^{c_{\alpha}(_{Z)F(}}=\alpha z)$

So, we obtain (i). (ii) is trivial.

.

$\square$For $z\in R_{1}$ , we define$\hat{F}_{\alpha}^{1}(z):=,\bigcap_{a<\alpha}\mathrm{c}1\{Fa(\prime Z)\}(\alpha\neq 0)$

where$\mathrm{c}1\{F_{\alpha}(Z)\}$anddenotes the closure of $F_{a}(z)$ .

Lemma 2.2. If $z\in R_{1}$ , the following (i), (ii) and (iii) hold:
(i)
(ii)
(iii)

$\tilde{q}_{\alpha}(\hat{F}_{\alpha}(\mathcal{Z}))=\hat{F}_{\alpha}(z)$$\hat{F}_{\alpha}(z)\subset\hat{F}_{\alpha’}(z)$for $\alpha\in[0,1]$ ;for $0\leq\alpha’<\alpha\leq 1$ ;$\hat{F}_{a}(z)=\lim_{a’\uparrow\alpha}\hat{F}\alpha’(z)$for$\alpha\neq 0$.

$\hat{F}_{0}(z):=\mathrm{c}1\{F_{0(}Z)\}$

,

(2.4)

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Proof. (ii) is trivial from Lemma 2.1 and (iii) is also trivial from the definition. To prove (i), let. From Lemma 2.1 (i), we have
. By the continuity of , we can checkin similar way to the proof of [4, Lemma 1]. Therefore,.
Let $\alpha>0$ and. By, we have$0$$\tilde{q}_{\alpha}(F_{\mathrm{O}}(\mathcal{Z}))=F_{0}(z)$$\alpha=$$\tilde{q}_{\alpha}(\mathrm{c}1\{F0(Z)\})=$$\tilde{q}$$\tilde{q}0(\hat{F}_{\alpha}(Z))=\hat{F}_{0}(z)$$\mathrm{c}1\{F_{0}(z)\}$$y\in\tilde{q}_{\alpha}(\hat{F}_{\alpha}(Z))$$\mathrm{I}_{\mathit{1}}\mathrm{e}\mathrm{m}\mathrm{m}\mathrm{a}1.1$.

$y \in,\cap\tilde{q}\alpha(\mathrm{C}1\{F_{\alpha}’(\alpha<\alpha Z)\})=\bigcap_{n=1}^{\infty}\tilde{q}_{\alpha}(\mathrm{C}1\{F_{(/n}-1)\alpha \mathrm{v}\mathrm{o}(z)\})$

From the continuity of , for $n\geq 1$ , there exists $x_{n}\in F_{(\alpha-1/n}$)Lemma 2.1 (i),$\tilde{q}$$\mathrm{v}\mathrm{o}(Z)$such that $\tilde{q}(x_{n}, y)\geq\alpha-1/n$ . By
for all $n\geq 1$ .

$y\in\tilde{q}_{(\alpha-}1/n)\mathrm{v}0(F_{(}\alpha-1/n)\mathrm{v}\mathrm{o}(Z))=F_{(\alpha-1/}n)0(Z)\subset \mathrm{c}1\{F_{(1}\alpha-/n)\mathrm{v}\mathrm{o}(\mathcal{Z})\}$

So,$y\in\hat{F}_{\alpha}(z)$. Therefore, we obtain$\tilde{q}_{\alpha}(\hat{F}_{\alpha}(z))\subset\hat{F}\alpha(z)$.

While, we havefor$\mathrm{c}1\{F_{\alpha^{\prime()\}}}Z\subset\tilde{q}_{\alpha’}(\mathrm{c}1\{F\alpha\prime\prime(z)\})$$\alpha’’<\alpha’<\alpha$

.

Then
for

$\hat{F}_{\alpha}(z)=,\bigcap_{a<\alpha}\mathrm{c}1\{F_{\alpha}’(Z)\}\subset,\bigcap_{\alpha<\alpha}\tilde{q}_{\alpha^{\prime(\mathrm{c}}}1\{F_{\alpha^{\prime!}}(z)\})=\tilde{q}_{\alpha}(\mathrm{c}1\{F_{\alpha}\prime\prime(Z)\})$$\alpha’’<\alpha$.

So, we get.

$\hat{F}_{\alpha}(z)\subset,\bigcap_{\alpha’<\alpha}\tilde{q}\alpha(\mathrm{d}\{F_{\alpha^{l}}l(_{Z})\})=\tilde{q}_{\alpha}(,\bigcap_{\alpha’<\alpha}\mathrm{c}1\{F_{\alpha^{l}}’(Z)\})=\tilde{q}_{\alpha}(\hat{F}_{\alpha}(z))$

Therefore, we can obtain (i).

Let $z\in R_{1}$ . Since
following theorem.

$\square$$\{\hat{F}_{\alpha}(z)|\alpha\in[0,1]\}$

satisfies the conditions$(\mathrm{a})-(\mathrm{c})$of Lemma 2.1 (ii), we obtain the

Theorem 2.1.
(i) If$R_{1}\neq\emptyset$, then there exists a$sol\mathrm{u}$tion of (1.1).

(ii) Define a fuzzy state$\tilde{s}^{z}(x):=\sup_{\alpha\in[0,1]}\{\alpha\wedge 1_{\hat{F}_{\alpha}(z)}(x)\}$Then$s^{z}\sim\in \mathcal{F}(X)$

Assume that

. We introduce an equivalent$z_{1}\sim z_{2}$Then we could identify the states ofmeans that
$R_{\alpha}$$\mathrm{r}\mathrm{e}\mathrm{l}\mathrm{a}\mathrm{t}\mathrm{i}_{\mathrm{o}\mathrm{n}}\sim \mathrm{o}\mathrm{n}R_{\alpha}$$z_{1}\in F_{\alpha}(z_{2})$$z_{2}\in R_{1}$

(2.5)

and

$z_{2}\in F_{\alpha}(z_{1})$$\mathrm{t}\mathrm{o}\sim$$z_{1},$$z_{2}\in R_{\alpha}$,.

, and so put.

,$z_{1}\sim z_{2}$if and only if$F_{\alpha}(z_{1})=F_{a}(z_{2})$. Then, we have
Proof. Let, we obtain. Since we haveof
The reverse proof is trivial.
$z_{1}\sim z_{2}$.

as follows: For

which is equivalent with respect$R_{\alpha}^{\sim}:=R_{\alpha}/\sim$$z_{1},$$x\in X$sa tisfies (1.1).

$R_{1}\neq\emptyset$

Lemma 2.3. For,$z_{1}\in F_{1}(z_{2})\subset F_{\alpha}(z_{2})$$F_{a}(z_{1})$$F_{\alpha}(z_{1})\subset F_{\alpha}(Z_{2})\square$for all $\alpha\in[0,1]$ .

for any$\alpha\in[0,1]$$F_{a}(z_{2})\subset F_{\alpha}(z_{1})$. From the definition (2.1)

similarly,$F_{\alpha}(z_{1})=F_{\alpha}(z_{2})$holds.

Ihom Theorem 2.1 and Lemma 2.3, the number of solutions of (1.1) is greater than or equals to the numis a solution of (1.1)}.
ber of -recurrent” sets. To consider the class of solution (1.1), let $P:=$ {Then $P$ has the following property:$‘(1$

Theorem 2.2. Let $\tilde{p}^{k}\in P(k=1,2, \cdots, l)$ . Then:$\tilde{p}\in \mathcal{F}(X)|\tilde{p}$186(i) Put$\tilde{p}(x):=k1,2,\cdot,\iota\max_{=}..\tilde{p}(kX),$

Then$\tilde{p}\in P$$x\in X$.

.

(ii) Let $\{\alpha^{k}\in[0,1]|k=1,2, \cdots, l\}s\mathrm{a}tisfi\gamma\max_{k1,2}=,\cdots,\mathrm{t}^{\alpha^{k}}=1$ . Put$\tilde{p}(x):=k1,2\max_{=,\cdot\iota}..,\{\alpha^{k}\wedge\tilde{p}^{k}(x)\}$Then$\tilde{p}\in P$

for $x\in X$ .

.

Proof. (ii) Taking the -cut of$\alpha$$\tilde{p}\in \mathcal{F}(X)$, we have$\tilde{p}_{\alpha}:=\bigcup_{\geq k\cdot\alpha^{k}\alpha}\tilde{p}_{\alpha}^{k}$

.

Then,

.

$\tilde{q}_{\alpha}(\tilde{p}_{\alpha})=\tilde{q}_{\alpha}(_{k\alpha^{k}\geq\alpha}\cup\tilde{p}\alpha k)=\cup\tilde{q}\alpha(\tilde{p}^{k}\alpha)ka^{k}\geq\alpha=\bigcup_{k:\alpha^{k}\geq\alpha}\tilde{p}_{\alpha}=\tilde{p}\alpha k$

Therefore, we obtain (ii) from Lernma3$1.2(\mathrm{i})$.

$(\mathrm{i})$is proved similarly.

$\square$

The uniqueness of solutions

In this section, we discuss the uniqueness of a solution of the equation (1.1) under convexity and compactness. Let
be a convex subset of
and $C_{c}(B)$ the class of all closed and convex subsets of .
Throughout this section, we assume that the state space $X$ is a convex and compact subset of. The
fuzzy setis called convex if its -cut
is convex fro each $\alpha\in[0,1]$ . Let.
.
Let, applying Kakutani’s fixed point theorem ([2]), we have the following.
$B$

$B$

$\mathcal{R}^{n}$$\mathcal{R}^{n}$$\tilde{s}\in \mathcal{F}(X)$$\mathcal{F}_{c}(X):=\{\overline{s}\in \mathcal{F}$$\tilde{s}_{\alpha}$$\alpha$$\tilde{s}\mathrm{i}_{\mathrm{S}\mathrm{C}\mathrm{o}11\mathrm{v}\mathrm{e}}\mathrm{X}\}$

Lemma 3.1.
Let $\alpha\in[0,1]$ and, there exists an $x\in X$ such that$\tilde{q}(x)$$\tilde{q}_{\alpha}(A)$is convex for eacl] $x\in X.$ Then, for any $A\in C_{c}(X)$ with $A=$$\tilde{q}(x, x)\geq\alpha$.

Proof. The map: $Aarrow C_{c}(A)$ withfor all $x\in A$ is continuous from Lenuna 1.1, so
Kakutani’s fixed point theorem gurantees the existence of an element $x\in A$ such that, whichimplies. This completes the proof.
$\tilde{q}_{\alpha}(x)\in C_{\mathrm{C}}^{1}(A)$$\tilde{q}_{\alpha}$$x\in\tilde{q}_{\alpha}(x)$$\tilde{q}(x, x)\geq\alpha$As a consequence, we have a property of the convex solution of (1.1).
Proposition 3.1. Letwith $q(x, x)\geq\alpha$ .
$\mathrm{P}\mathrm{r}\mathrm{o}\mathrm{o}\mathrm{f}\square$. By Lemma 1.2,

$p\in \mathcal{F}_{c}(X)$be a soluti on of (1.1). Then, for each $\alpha\in[0,1]$ , there exists an$\tilde{p}_{\alpha}=\tilde{q}_{\alpha}(\overline{p}_{\alpha})$$x\in p_{\alpha}$

for each $\alpha\in[0,1]$ . Thus, Lemma 3.1 clearly proves the desired result.

Now, we give sufficient conditions for the uniqueness of a convex solution of (1.1). Let $U_{\alpha}:=\{x\in$.

$X|q(_{X}, X)\geq\alpha\}$Assumption A. The following AI –A3 holds.

Al. The set

A2. [$U_{\alpha},$$\alpha=1$$I_{\alpha}\subset F_{\alpha}(u)$A3. If$A=\tilde{q}_{\alpha}(A)$is one point set, say . That is, $U_{1}=\{u\}$ .

for each $\alpha\in[0,1]$ , where

$u$

$u$

is in the above Al andfor any $A\in F_{c}(X))0\leq\alpha\leq 1$ , then
$A= \bigcup_{x\underline{\in}U_{\alpha^{\cap A}}}F\alpha(A)$$F_{\alpha}(u)$is defined by (2.1).

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Theorem 3.1. Under Assumption $A$ , the $eq$ uation (1.1) has a unique solu tion in

Proof. Letandpointset,
now show that
. Therefore,$\tilde{p},\tilde{p}’\in F_{\mathrm{c}}(X)$$u\in\overline{p}_{1}$be solution of (1.1). By Lemma 3.1,
. Thus, by A2 and A3,
and$0<\alpha\leq 1$ . Sincefor$\tilde{p}_{1}\cap U_{1}\neq$$\tilde{p}_{1}=F_{1}(u)$$u\in\tilde{p}_{1}’$$\tilde{p}_{\alpha}=\hat{F}_{\alpha}(u)$

.ach

and

$u\in\tilde{p}_{\alpha’}=\tilde{q}_{\alpha’}(\tilde{p}_{\alpha’})$$\mathrm{e}$and$\tilde{p}_{1}’=F_{1}(u)$$\mathcal{F}_{c}(X)$.

. Since, which implies
is closed, it holds$\tilde{p}_{1}’\cap U_{1}\neq$$U_{1}$

is one. We$\tilde{p}_{1}=\tilde{p}_{1}’$$\tilde{p}_{\alpha}$$F_{\alpha};(u)\subset$$\tilde{p}_{\alpha’}$.

$\hat{F}_{\alpha}(u)=\alpha’\bigcup_{<\alpha}\mathrm{c}1\{F\alpha’(u)\}\subset,\bigcup_{\alpha<\alpha}\tilde{p}\alpha’(u)=\tilde{p}\alpha$On the other hand, we have$\tilde{p}_{\alpha}$,$= \bigcup_{x\in U_{\alpha}\cap\overline{p}_{\alpha}}\mathrm{c}1\{F_{\alpha}(X)\}$,$\subset\bigcup_{x\in F_{\alpha^{\cap}}}\overline{p}\alpha \mathrm{c}1\{F\alpha\langle X)\}$$\subset\bigcup_{x\in}\hat{F}_{\alpha}\mathrm{C}1\{F\alpha(X)\}$

From that$x\in\hat{F}_{\alpha}$means$\hat{F}_{\alpha}(x)\subset\hat{F}_{\alpha}(u)$from A3
from A2.

, it holds that$\tilde{p}_{\alpha}\subseteq \mathrm{c}1\{F_{\alpha}(u)\}\subset\hat{F}\alpha(u)$The above shows

4$\tilde{p}_{\alpha}=\hat{F}(u)$. Similarly$\tilde{p}_{\alpha}’=\hat{F}_{\alpha}(u)$. Thus,.
. This completes the proof.

$\tilde{p}_{\alpha}=\tilde{p}_{\alpha}’$$\square$Numerical example

Here two numerical examples are given to comprehend computational aspect of this paper.
Example 1. Let $X=[0,1]$ . For any $g;[0,1]arrow[0,1]$ , let$\tilde{q}(x, y):=(1-|y-g(x)|)\vee 0$ .

is strictly increasing and there exists a unique $x_{0}\in[0,1]$ with $x_{0}=g(x_{0})$ . Under the
We assume thatabove condition, $R_{1}=\{x_{0}\}$ and for each $\alpha\in[0,1)$ ,$g(\cdot)$$U_{\alpha}=[\underline{x}_{\alpha},\overline{x}_{\alpha}]$,is a unique solution of $x=g(x)-(1-\alpha),x=g(x)+(1-\alpha)$ respectively and $arrow x=0,$,
when
if the solution does not exist in $[0,1]$ .
in $C_{c}([0,1])$ , so that Assumption A in
Clearly,equals a unique solution of the equation
Section 3 holds in this case. Thus, by Theorem 3.1, we have$arrow x$$\overline{x}_{\alpha}=1$$\overline{x}_{\alpha}$$A=\tilde{p}_{\alpha}(A)$$U_{\alpha}$(4.1)$\tilde{s}(x)=\sup\{\alpha\wedge I_{U\alpha}(_{X}\alpha\in[0,1])\}$

is a unique convex solution of (1.1). For a concrete example such as $g(x)=(2x^{2}+1)/4$, then it is seen that
$R_{1}=\{(2-\sqrt{2})/2\}$ and$1-\sqrt{5/2-2\alpha})\vee 0$ ,$3/4<\alpha$1,$\overline{x}_{\alpha}=\underline{x}_{\alpha}=\{$,$1-\sqrt{2\alpha-3}/2$$3/4\leq\alpha\leq 1$.

By (4.1), the unique solution is as follows (Fig.1):$\tilde{s}(x)=\{$$-x^{2}/2+x+3/4$ ,$0\leq x\leq 1-\sqrt{2}/2$$x^{2}/2-x+5/4$ ,$1-\sqrt{2}/2<x\leq 1$

Fig.1 The unique solution$\tilde{s}$

.

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Example 2. This example has two peaks for the fuzzy relation. Let $X=[0,1]$ and$\tilde{q}(x, y)=(1-|y-(x^{2}+1)/4|)\vee(1-|y-(x^{2}+2)/4|)$ .

Then, $R_{1}=\{a, b\}$ , where $a=2-\sqrt{3},$ $b=2-\sqrt{2}$. By simple calculation, we getand$\hat{F}_{\alpha}(a)=[\underline{x}_{\alpha)\alpha}^{a}\overline{x}^{a}]$$\hat{F}_{\alpha}(b)=[\underline{x}_{\alpha}^{b},\overline{X}_{\alpha}^{b}1$

for $\alpha\in[0,1]$ , where$0$,$0\leq\alpha\leq 3/4$$\underline{x}_{\alpha}^{a}=\{$$2-\sqrt{7-4\alpha}$,

1,$0\leq\alpha\leq 7/8$$\underline{x}_{\alpha}^{a}=\{$$2-\sqrt{4\alpha-1}$$0$$3/4<\alpha\leq 1$,,$7/8<\alpha\leq 1$$0\leq\alpha\leq 7/8$$\underline{x}_{\alpha}^{b}=\{$$2-\sqrt{6-4\alpha}$,1,$7/8<\alpha\leq 1$$0\leq\alpha\leq 3/4$$\underline{x}_{\alpha}^{b}=\{$$2-\sqrt{4\alpha-2}$,$3/4<\alpha\leq 1$

By Theorem 2.1, the solution of (1.1) are given as follows(Fig2.):’$\frac{-x^{2}+2_{X}+3}{4}$$x^{2}-4_{X+}5$$\tilde{s}^{a}(x)=$,.

$2-\sqrt{3}\leq x\leq 2-\sqrt{10}/2$$\overline{4}’$. 7/8,$2-\sqrt{10}/2\leq x\leq 1$

7/8,$0\leq x\leq 2-\sqrt{10}/2$$\frac{-x^{2}+4_{X}+2}{4}$$\tilde{s}^{b}(_{X)=}\{$$\frac{x^{2}-4_{X}+6}{4}$$0\leq x\leq 2-\sqrt{3}$,,$2-\sqrt{10}/2\leq x\leq 2-\sqrt{2}$$2-\sqrt{2}\leq x\leq 1$

Fig.2 The unique solutions$\tilde{s}^{a}$and$\tilde{s}^{b}$

.

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References[1] R.E.Bellman and L.A.Zadeh, Decision-making in a fuzzy environment, Management Sci. Ser B. 17(1970) 141-164.
[2] N.Dunford and J.T.Schwartz, Linear Operators, Part 1: General Theory (Interscience Publishers,
New York, 1958).
[3] K.Kuratowski, Topology I (Academic Press, New York, 1966).

[4] M.Kurano, M.Yasuda, J.Nakagami and Y.Yoshida, A limit theorem in some dynamic fuzzy systems,
Fuzzy Sets and Systems 51 (1992) 83-88.
[5] V.Nov\’ak, Fuzzy Sets and Their Applications (Adam Hilder, Bristol-Boston, 1989).
[6] Y.Yoshida, M.Yasuda, J.Nakagami and M.Kurano, A limit theorem in some dynamic fuzzy systems
with a monotone property, accepted in Fuzzy Sets and Systems.

[7] Y.Yoshida, The recurrence of dynamic fuzzy systems, RIFIS Technical Report 80 (1994), ResearchInstitute of Fundarnental Information Science, Kyushu University 33 Fukuoka 812, Japan.