数理解析研究所講究録 1246 巻 2002 年 139-147

139
Eigenvalue Problem related to Euler-Bernoulli Equation with
Joint Boundary Condition
埼玉大学・理学部

辻岡

邦夫 (Kunio Tsujioka )Faculty of Science, Saitama University
崎玉大学・理工学研究科
内海 順一 (Junichi Uchiumi)
Graduate School of Science and Engineering,
Saitama University
Abstract

We consider the control system of two vibrating beams which are coupled at a
joint. The displacement of the beam is described by an Euler-Bernoulli equation
with control applied at acoupled point. Our purpose is to argue the contorollabilityof the system. To this purpose, we discuss the eigenvalue problem related to thissystem.

1Introduction
Let us consider the controllability problem for asystem coupled by Euler-Bernoulli beams.
For $m\in(0,1)$ , we put $x_{0}=0$ , $x_{1}=m$ and $x_{2}=1$ . The displacement of each beam attime is described by $y_{i}(x, t)$ on $I_{i}=(x_{i-1}, x_{i})$ , $i=1,2$ , and satisfies the Euler-Bernoulli
equation:$t$$\rho_{i}\dot{y}_{i}+T_{i}y_{i}^{(4)}=0$

where,
flexural rigidity respectively on$\dot{y}_{i}(x, t)=\partial y_{i}(x, t)/\partial t$

on $I_{i}\cross(0, T)$(1).
is mass density and. Let both ends be clamped:$y_{i}^{(k)}(x, t)=\partial^{k}y_{i}(x, t)/\partial x^{k}$$\rho_{i}$$I_{i}$$(B_{0}y)(t):=(y_{1}(0, t)$,$y_{1}^{(1)}(0, t)$, $y_{2}(1, t)$ , $y_{2}^{(1)}(1, t))=0$ .

$T_{i}$

is(2)

At the coupled point $x=m$ , we apply control $F=(f_{1}, f_{2}, f_{3}, f_{4})$ as follows:$(B_{1}y)(t):=y_{1}(m, t)-y_{2}(m, t)=f_{1}(t)$ ,$(B_{2}y)(t):=y_{1}^{(1)}(m, t)-y_{2}^{(1)}(m, t)=f_{2}(t)$,$(B_{4}y)(t).\cdot.=T_{1}y_{1}^{(3)}(m(B_{3}y)(t).=T_{1}y_{1}^{(2)}(m,’ t)-T_{2}y_{2}^{(3)}(m,t)=f_{4}(t)t)-T_{2}y_{2}^{(2)}(m,t)=f_{3}(t).’\}$(3)

Initial condition is given as follows$y_{i}(x, 0)=y_{i}^{0}(x)$,$\dot{y}_{i}(x, 0)=y_{i}^{1}(x)$,$x\in I_{i}$, $i=1,2$ .

(4)

We assume that controls
belong to $L^{2}(0, T)$ , $i=1,2,3,4$ . In this paper, we treat
controllability of the above system. Roughly speaking, the system (1) (2)(3)(4) is controllable if for any initial value
and final value, $i=1,2$ , there exists a
control $F=(f_{1}, f_{2}, f_{3}, f_{4})$ such that the corresponding solution of the system (1) (2) (3) (4)satisfies the final condition $(y_{i}(x, T),\dot{y}_{i}(x, T))=(z_{i}^{0}(x), z_{i}^{1}(x))$ , $i=1,2$ .
$f_{i}$$(y_{i}^{0}, y_{i}^{1})$$(z_{i}^{0}, z_{i}^{1})$140
2Eigenvalue Problem
Let us identify $v\in L^{2}(I)$ with$v|_{I}:’

i=1,2$$(\begin{array}{l}v_{1}v_{2}\end{array})$

where$\in H=L^{2}(I)=L^{2}(I_{1})\cross L^{2}(I_{2})$$v:=$, $I_{1}=(0, m)$ , $I_{2}=(m, 1)$ . Then $H$ becomes aHilbert space with innerproduct(v,$w)=\rho_{1}(v_{1}, w_{1})_{L^{2}(I_{1})}+\hslash(v_{2}, w_{2})_{L^{2}(I_{2})}$

for v

,w

$=(\begin{array}{l}v_{1}v_{2}\end{array})$$=(\begin{array}{l}w_{1}w_{2}\end{array})$$\in H$.

We define an operator A in H byfor

$Av=((T_{2}/\rho_{2})v_{2}^{(4)}(T_{1}/\rho_{1})v_{1}^{(4)})$$v=(\begin{array}{l}v_{1}v_{2}\end{array})$$\in D(A):=H^{4}(I_{1})\cross H^{4}(I_{2})$and an operator $A$ by restricting $A$ to$D(A):=$ { $v\in H^{4}(I_{1})\cross H^{4}$ (I2);$B_{0}v=0$ , $Bv:=(B_{1}v,$ $B_{2}v$ , $B_{3}v$ , $B_{4}v)=0$}.

For this operator $A$ , we have
Lemma 1The operator $A$ is a selfadjoint operator in $H$ with compact resolvent.
The proof of this lemma is easy to verify.
Let Abe an eigenvalue for A with corresponding eigenfunction . Then we have$\phi$$A\phi$$=\lambda\phi$,$B\phi$(1)with boundary conditions$B_{0}\phi=0$

We introduce functions$\mathrm{C}_{\pm}$,$\mathrm{S}_{\pm}$$I_{\dot{l}}$$\phi_{\dot{1}}$$=\phi|_{I}.$is given by,$\alpha:=(\rho:/T_{\dot{l}})^{\frac{1}{4}}$

.

(2)

by

$\mathrm{C}_{\pm}(\theta):=\frac{\cosh\theta\pm\cos\theta}{2}$

Let

$=0$

,

for

$\mathrm{S}_{\pm}(\theta):=\frac{\sinh\theta\pm\sin\theta}{2}$$\theta\in \mathrm{R}$.

, $i=1,2$ . Asystem of fundamental solutions to (1) in each
and we have$\{\mathrm{C}_{\pm}(\alpha:\omega(x-x_{i-1})), \mathrm{S}_{\pm}(\alpha:\omega(x-x:-1))\}$,$=(p_{1}^{\dot{l}}\mathrm{C}_{+}+p_{2}^{\dot{l}}\mathrm{S}_{+}+p_{3}^{\dot{1}}\mathrm{C}_{-}+p_{4}^{l}\mathrm{S}_{-})(\alpha:\omega(x-x:-1))$$\}$$\phi_{\dot{1}}(x)(\phi_{\dot{l}})^{(2)}(x)(\phi_{\dot{l}})^{(1)}(x)(\phi_{\dot{l}})^{(3)}(x)$$===\alpha_{\dot{1}}^{3}\omega^{3}(p_{1}^{1}\mathrm{S}_{+}+p_{2-+p_{3}^{\dot{1}}\mathrm{S}_{-}+p_{4}^{\dot{l}}\mathrm{C}_{+})(\alpha_{\dot{l}}\omega(x-x_{\dot{l}-1}))}^{l}\alpha_{\dot{l}}^{2}\omega^{2}(p_{!}^{\dot{l}}\mathrm{C}_{-}+p_{2}^{l}\mathrm{S}_{-}+p_{3}.\cdot \mathrm{C}_{+}+p_{4}^{\dot{l}}\mathrm{S}_{+})(\alpha_{\dot{l}}\omega(x-\dot{.}x_{\dot{|}-1})’)\alpha_{\dot{l}}\omega(p^{\dot{l}}1\mathrm{S}-+p_{2+}^{\dot{1}}\mathrm{C}+p_{3}^{1}\mathrm{S}_{+}+p_{4-}^{\dot{l}}\mathrm{C})(\alpha_{\dot{l}}\omega(x-x_{-1}))\mathrm{C}’$for$x\in I_{\dot{1}}$where$=\lambda^{\frac{1}{4}}$$\omega$. By (2), we have$p_{1}^{1}=p_{2}^{1}=0$

and therefore$=(p_{3}^{1}\mathrm{C}_{-}+p_{4}^{1}\mathrm{S}_{-})(\alpha_{1}\omega x)$

,

,
$\}$

$\phi_{1}(x)(\phi_{1})^{(2)}(x)(\phi_{1})^{(1)}(x)(\phi_{1})^{(3)}(x)$.’$===\alpha_{1}^{3}\omega^{3}(p_{3}^{1}\mathrm{S}_{-}+p_{4}^{1}\mathrm{C}_{+})(\alpha_{1}\omega x)\alpha_{1}^{2}\omega^{2}(p_{3}^{1}\mathrm{C}_{+}+p_{4}^{1}\mathrm{S}_{+})(\alpha_{1}\omega x)\alpha 1\omega(p_{3}^{1}\mathrm{S}_{+}+p_{4}^{1}\mathrm{C}_{-})(\alpha_{1}\omega x)$

(3)

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By (2), we have$\gamma_{2}\alpha_{2}p^{2}=\gamma_{2}\alpha_{2}(p_{4}^{2}p_{2}^{2}p_{3}^{2}p_{1}^{2})=(\gamma_{1}\alpha_{2}\mathrm{C}(\omega)\gamma_{2}\alpha_{2}\mathrm{C}_{\frac{1}{+1+1}}(\omega)\gamma_{1}\alpha_{1}\mathrm{S}_{-}^{1}(\omega)\gamma_{2}\alpha_{1}\mathrm{S}(\omega)$where$\gamma_{i}=T_{i}\alpha_{i}^{2}$, $i=1,2$ ,$\beta_{1}=\mathrm{a}\mathrm{i}\mathrm{m}$,(4)$\gamma_{1}\alpha_{1}\mathrm{C}(\omega)\gamma_{2}\alpha_{1}\mathrm{C}_{\frac{1}{+1+1}}(\omega)\gamma_{1}\alpha_{2}\mathrm{S}(\omega)\gamma_{2}\alpha 2\mathrm{S}_{\frac{1}{}}(\omega))(_{p_{4}^{1}}p_{3}^{1})$$\mathrm{S}_{\pm}^{1}(\omega)=\mathrm{S}_{\pm}(\beta_{1}\omega)$see,$\mathrm{C}_{\pm}^{1}(\omega)=\mathrm{C}_{\pm}(\beta_{1}\omega)$. By (2), we$p_{1}^{2}\backslash$($\mathrm{C}_{+}^{2}(\omega)\mathrm{S}_{-}^{2}(\omega)$$\mathrm{C}_{+}^{2}(\omega)\mathrm{S}_{+}^{2}(\omega)$$\mathrm{C}_{\frac{2}{2+}}(\omega)\mathrm{S}(\omega)$$\mathrm{C}(\omega)\mathrm{S}_{\frac{2}{2-}}(\omega)$)$p_{2}^{2}$

$=0$

$\{$

$p_{4}^{2}$

where $\beta_{2}=\alpha_{2}(1-m)$ ,$\mathrm{S}_{\pm}^{2}(\omega)=\mathrm{S}_{\pm}(\beta_{2}\omega)$

,

,

(5)

$p_{3}^{2}$$\mathrm{C}_{\pm}^{2}(\omega)=\mathrm{C}_{\pm}(\beta_{2}\omega)$

,

. Let

$D(\omega)=(\begin{array}{ll}D_{11}(\omega) D_{12}(\omega)D_{21}(\omega) D_{22}(\omega)\end{array})$$.=($$\mathrm{C}_{+}^{2}(\omega)\mathrm{S}_{-}^{2}(\omega)$$\mathrm{C}_{+}^{2}(\omega)\mathrm{S}_{+}^{2}(\omega)$$\mathrm{C}_{\frac{2}{2+}}(\omega)\mathrm{S}(\omega)$$\mathrm{C}(\omega)\mathrm{S}_{\frac{2}{2-}}(\omega)$)$(\gamma_{1}\alpha_{2}\mathrm{C}(\omega)\gamma_{2}\alpha_{2}\mathrm{C}_{\frac{1}{+1+1}}(\omega)\gamma_{1}\alpha_{1}\mathrm{S}_{-}^{1}(\omega)\gamma_{2}\alpha_{1}\mathrm{S}(\omega)$.

$\gamma_{1}\alpha_{1}\mathrm{C}(\omega)\gamma_{2}\alpha_{1}\mathrm{C}_{\frac{1}{+1+1}}(\omega)\gamma_{1}\alpha_{2}\mathrm{S}(\omega)\gamma_{22}\alpha \mathrm{S}_{\frac{1}{}}(\omega))$Then’$D_{12}(\omega)D_{11}(\omega)==(\gamma_{2}\alpha_{2}\mathrm{C}_{+}^{2}\cdot \mathrm{S}_{-}^{1}+\gamma_{2}\alpha_{1}\mathrm{S}_{+}^{2}\cdot \mathrm{C}_{-}^{1}+\gamma_{1}\alpha_{2}\mathrm{C}_{-}^{2}\cdot \mathrm{S}_{+}^{1}+\gamma_{1}\alpha_{1}\mathrm{S}_{-}^{2}\cdot \mathrm{C}_{+}^{1})(\omega)(\gamma_{2}\alpha_{2}\mathrm{C}_{+}^{2}\cdot \mathrm{C}_{-}^{1}+\gamma_{2}\alpha_{1}\mathrm{S}_{+}^{2}\cdot \mathrm{S}_{+}^{1}+\gamma_{1}\alpha_{2}\mathrm{C}_{-}^{2}\cdot \mathrm{C}_{+}^{1}+\gamma_{1}\alpha_{1}\mathrm{S}_{-}^{2}\cdot \mathrm{S}_{-}^{1})(\omega),$,$D_{21}(\omega)=(\gamma_{2}\alpha_{2}\mathrm{S}_{-}^{2}\cdot \mathrm{C}_{-}^{1}+\gamma_{2}\alpha_{1}\mathrm{C}_{+}^{2}\cdot \mathrm{S}_{+}^{1}+\gamma_{1}\alpha_{2}\mathrm{S}_{+}^{2}\cdot \mathrm{C}_{+}^{1}+\gamma_{1}\alpha_{1}\mathrm{C}_{-}^{2}\cdot \mathrm{S}_{-}^{1})(\omega)$.

$D_{22}(\omega)=(\gamma_{2}\alpha_{2}\mathrm{S}_{-}^{2}\cdot \mathrm{S}_{-}^{1}+\gamma_{2}\alpha_{1}\mathrm{C}_{+}^{2}\cdot \mathrm{C}_{-}^{1}+\gamma_{1}\alpha_{2}\mathrm{S}_{+}^{2}\cdot \mathrm{S}_{+}^{1}+\gamma_{1}\alpha_{1}\mathrm{C}_{-}^{2}\cdot \mathrm{C}_{+}^{1})(\omega)$

We put$d(\omega):=4\det D(\omega)$$=4\gamma_{2}^{2}\alpha_{2}\alpha_{1}(\mathrm{S}_{+}^{2}\cdot \mathrm{S}_{-}^{2}-\mathrm{C}_{+}^{2}\cdot \mathrm{C}_{+}^{2})\cdot$ $(\mathrm{S}_{+}^{1}\cdot \mathrm{S}_{-}^{1}-\mathrm{C}_{-}^{1}\cdot \mathrm{C}_{-}^{1})(\omega)$$+4\gamma_{2}\gamma_{1}\alpha_{2}^{2}$$(\mathrm{S}_{+}^{2}\cdot \mathrm{C}_{+}^{2}-\mathrm{S}_{-}^{2}\cdot \mathrm{C}_{-}^{2})\cdot$ $(\mathrm{S}_{+}^{1}\cdot \mathrm{C}_{-}^{1}-\mathrm{C}_{+}^{1}\cdot \mathrm{S}_{-}^{1})(\omega)$$+8\gamma_{2}\gamma_{1}\alpha_{2}\alpha_{1}(\mathrm{S}_{+}^{2}\cdot \mathrm{S}_{+}^{2}-\mathrm{C}_{+}^{2}\cdot \mathrm{C}_{-}^{2})\cdot$$+4\gamma_{2}\gamma_{1}\alpha_{1}^{2}(\mathrm{S}_{+}^{2}\cdot \mathrm{C}_{-}^{2}-\mathrm{C}_{+}^{2}\cdot \mathrm{S}_{-}^{2})\cdot$$(\mathrm{S}_{+}^{1}\cdot \mathrm{S}_{+}^{1}-\mathrm{C}_{+}^{1}\cdot \mathrm{C}_{-}^{1})(\omega)$$(\mathrm{S}_{+}^{1}\cdot \mathrm{C}_{+}^{1}-\mathrm{S}_{-}^{1}\cdot \mathrm{C}_{-}^{1})(\omega)$$+4\gamma_{1}^{2}\alpha_{1}\alpha_{2}(\mathrm{S}_{+}^{2}\cdot \mathrm{S}_{-}^{2}-\mathrm{C}_{-}^{2}\cdot \mathrm{C}_{-}^{2})\cdot$ $(\mathrm{S}_{+}^{1}\cdot \mathrm{S}_{-}^{1}-\mathrm{C}_{+}^{1}\cdot \mathrm{C}_{+}^{1})(\omega)$.

(6)

By (4), (5), we have$D(\omega)(_{p_{4}^{1}}p_{3}^{1})=0$.

Since is an eigenfunction if and only if, we see thatof $A$ if $d(\omega)=0$ , $\omega>0$ . Let ,, is the -th positive zero of$(p_{3}^{1}, p_{4}^{1})\neq 0$$\phi$$\omega_{n}$$n\in \mathrm{N}$$n$is an eigenvalue. Then,$\mathrm{A}=\omega^{4}$$d(\omega)$$\lambda_{n}:=\omega_{n}^{4}$

142
is asimple-th eigenvalue of A. We can verify thatin
H. In theAn,tonormalized
corresponding
be an eigenfunction
be afunction defined by$0<\mathrm{u}_{1}<\mathrm{u}_{2}<\cdots$

eigenvalue. Let
following, let$\langle 6^{n}$$\ovalbox{\tt\small REJECT}$, is the$\mathrm{A}_{n}$$\mathrm{r}\mathrm{z}$$\varphi(\omega)=A\cos\beta_{1}\omega\cos\beta_{2}\omega-B\sin\beta_{1}\omega\sin\beta_{2}\omega+C\sin(\beta_{1}-\beta_{2})\omega$where Adenote the -th positive zero of$=(\gamma_{1}\alpha_{1}+\gamma_{2}\alpha_{2})(\gamma_{1}\alpha_{2}+\gamma_{2}\alpha_{1})$$n$Lemma 2$d(\omega)$$\varphi$by, B$\mu_{n}$$=\gamma_{1}\gamma_{2}(\alpha_{1}+\alpha_{2})^{2}$. We

is written as$h(\omega)\in \mathrm{C}^{1}(\mathrm{R})$Proof By (6),.

$=\gamma_{1}\gamma_{2}(\alpha_{1}^{2}-\alpha_{2}^{2})$.

$d(\omega)=e^{(\beta_{1}+\beta_{2})\{d}(\varphi(\omega)-h(\omega))$

where

, C

(7)

and

$h(\omega)arrow 0$,$\omega\in \mathrm{R}$, $h’(\omega)arrow 0$ exponentially as$h(\omega)=\varphi(\omega)-e^{-(\beta_{1}+\beta_{2})\omega}d(\omega)$and$\omega$$arrow\infty$.

converge to 0exponenrially as$h’(\omega)$$\omegaarrow\infty$

To discuss controllability, we treat the moment problem on the system (1)(2)(3)(4).
According to Krabs [4] or Russell [12], to solve the moment problem, we need the following
conditions:$\lim_{narrow}\inf_{\infty}(\omega_{n+1}^{2}-\omega_{n}^{2})>\frac{2\pi}{T}$

(8)

,

(9)

$\lim\sup\lim_{xyarrow\inftyarrow}\sup_{\infty}\frac{d(x+y)-d(x)}{y}<\frac{T}{2\pi}$where$d(x)=\mathrm{n}\mathrm{u}\mathrm{m}\mathrm{b}\mathrm{e}\mathrm{r}$

of

with

$\omega_{j}$$\omega_{j}<x^{2}$. The aim of this paper is to prove the following

Theorem 1We have(1) There exist $M$ and $N$ such that $\omega_{M+n}-\mu_{N+n}arrow 0$ ,(2)

$0< \frac{1}{a}(\pi-\sin^{-1}k)\leq\lim_{narrow}\inf_{\infty}(\omega_{n+1}-\omega_{n})\leq\lim_{narrow}\sup_{\infty}(\omega_{n+1}-\omega_{n})<\infty$(3)$\lim_{narrow\infty}(\omega_{n+1}^{2}-\omega_{n}^{2})=\infty$,.

satisfies the codition (8). Moreover, the condition
By this theorem, it is clear that(8) verify the condition (9).
are given in the following
Some simple facts for$\{\omega_{n}\}_{n\in \mathrm{N}}$$\varphi(\omega)$

Lemma 3In the for mula (7), we have(1) $A\geq B>0$ ,(2) $A=B$ if and only if(3) $C=0$ if and only if$\rho_{1}T_{1}=\rho_{2}T_{2}$$\rho_{1}T_{2}=\mathrm{P}2\mathrm{T}1$(4) $A=B$ , $C=0$ if and only if.

.

$(\rho_{1}, T_{1})=(\rho_{2}, T_{2})$(5) $A>B$ or $C\neq 0$ if and only if.

$(\rho_{1}, T_{1})\neq(\rho_{2}, T_{2})$

.

143
(6)

$\varphi$

is written as$\varphi(\omega)=D(\cos a\omega+k\sin(b\omega+\tau))$

where $D=(A+B)/2$ ,$a=\beta_{1}+\beta_{2}$$\tau=\cos^{-1}(C/R)\in[0, \pi]$for,$R\neq 0$$b=\beta_{1}-\beta_{2}$

,

(10)

$0\leq k<1$

, $k=R/D$ , $R=\sqrt{((A-B)/2)^{2}+C^{2}}$ ,and $\tau=0$ for $R=0$ .

Proof We see (1) from $A=B+\alpha_{1}\alpha_{2}(\gamma_{1}-\gamma_{2})^{2}\geq B>0$ . The assertions (2), (3), (4)and (5) are clear. We have (6) since$\varphi(\omega)$$=$$\frac{A+B}{2}\cos(\beta_{1}+\beta_{2})\omega+\frac{A-B}{2}\cos(\beta_{1}-\beta_{2})\omega+C\sin(\beta_{1}-\beta_{2})\omega$$=$$D\cos a\omega+R\sin(b\omega+\tau)=D(\cos a\omega+k\sin(b\omega+\tau))$.

where $k\geq 0$ satisfies$k^{2}= \frac{R^{2}}{D^{2}}=\frac{(A+B)^{2}-4(AB-C^{2})}{(A+B)^{2}}=1-\frac{4(\gamma_{1}+\gamma_{2})^{2}\gamma_{1}\gamma_{2}(\alpha_{1}+\alpha_{2})^{2}\alpha_{1}\alpha_{2}}{(A+B)^{2}}<1$.

for simplicity. So we have $a>b\underline{>}0$ . We put. Since $f_{k}(\omega)=0$ implies $|\cos a\omega|=|k\sin(b\omega+\tau)|\leq k$ , all the positive
aw-l
zeros of,are in the set$s_{n}(k)=(2n-1)\pi/2a-\sin^{-1}k/a$$J_{n}=[(n-1)\pi/a, n\pi/a]$ , $n=1,2$ ,, $t_{n}(k)=$where$(2n-1)\pi/2a+\sin^{-1}k/a$ . We write, $s_{n}=s_{n}(k)$ , $t_{n}=t_{n}(k)$ and $f(\omega)=f_{k}(\omega)$ .
In Theorem 2below, we prove that there exists exactly one zero of in each.

We assume$\cos$$k$$\beta_{1}\geq\beta_{2}>0$$f_{k}(\omega)=$$\sin(b\omega+\tau)$$\mathrm{I}_{n}(k)--[s_{n}(k), t_{n}(k)]\subseteq$$\{\omega>0;|\cos a\omega|\leq k\}=\bigcup_{n=1}^{\infty}\mathrm{I}_{n}(k)$$\varphi(\omega)$$\cdots$$\mathrm{I}_{n}=\mathrm{I}_{n}(k)$$f$Theorem 2For each$n\in \mathrm{N}$, there exist$u_{n}$,$v_{n}\in \mathrm{I}_{n}$$\mathrm{I}_{n}\subset J_{n}$such that(1) $f(u_{n})=1-k=-f(v_{n})$ and$f(\omega)$

is monotone decreasing on$[u_{n}, v_{n}]$(2) $f(u_{n})=k-1=-f(v_{n})$ and$f(\omega)$

is monotone increasing on$[u_{n}, v_{n}]$(3) $|f(\omega)|\geq 1-k$ for(4) only zero of$f$$\omega$for odd ,for even ,
$n$

$n$

$\in J_{n}\backslash [u_{n}, v_{n}]$exists in$(u_{n}, v_{n})$for anyFirst, we show, for sufficiently small ,
$k$

$n$

, which implies that$\mu_{n}\in \mathrm{I}_{n}$for every$n\in \mathrm{N}$$\mu_{n}\in \mathrm{I}_{n}$for any.

Lemma 4Let $k\in[0,1/\sqrt{2}]$ . Then we have(1) for odd ,$f(s_{n})\geq 0\underline{>}f(t_{n})$

$n$

and

$f(\omega)$(2) for even , $f(t_{n})\geq 0\geq f(s_{n})$ and$n$Consequently, in $Jn$ ,$f(\omega)$$f(\omega)$

is monotone decreasing on$[s_{n}, t_{n}]$is monotone increasing on

has only one zero in$[s_{n}, t_{n}]$,$[s_{n}, t_{n}]$,.

Proof We only show (1). (2) is proved similarly.
$f(s_{n})$$f(t_{n})$In$[s_{n}, t_{n}]$$=$$\cos(\frac{(2n-1)\pi}{2}-\mathrm{s}.\mathrm{n}^{-1}k)-k\sin(bs_{n}+\tau)$$=$$(-1)^{n+1}k-k\sin(bs_{n}+\tau)=k-k\sin(bs_{n}+\tau)\geq 0$$=$$(-1)^{n}k-k\sin(bs_{n}+\tau)=-k-k\sin(bs_{n}+\tau)\leq-k+k=0$ .

, we have $\sin a\omega\geq 1/\sqrt{2}$ and.

$f’(\omega)=-a\sin a\omega+kb\cos(b\omega+\tau)\leq-a/\sqrt{2}+b/\sqrt{2}<0$$n\in \mathrm{N}$

.

144
, s.
In the following, we put$f(w)$and S { kC [0,1);$\ovalbox{\tt\small REJECT}_{\ovalbox{\tt\small REJECT}}kb^{2}/a^{2}$$\ovalbox{\tt\small REJECT} 7_{\mathrm{i}}^{\ovalbox{\tt\small REJECT}}(\ovalbox{\tt\small REJECT} \mathrm{u})$$\overline{k}\in \mathrm{S}$,$\ovalbox{\tt\small REJECT}\ovalbox{\tt\small REJECT}\ovalbox{\tt\small REJECT}$$\mathrm{p}.(k)\mathrm{E}\mathrm{L}\ovalbox{\tt\small REJECT}(\mathrm{A})$$\ovalbox{\tt\small REJECT}$

Lemma 5For$s_{n}(\mathrm{A})$$\ovalbox{\tt\small REJECT}$c$J\ovalbox{\tt\small REJECT}$$t_{n}(k)$,Q$\ovalbox{\tt\small REJECT}$$\mathrm{Q}(7\mathrm{C})$, and$I^{e_{n}\ovalbox{\tt\small REJECT} \mathrm{p}.(\mathrm{A})}$,for each nC N}.

, the conclusion of Theorem 2is valid.

Proof We have$f’(\omega)$$=-a^{2}\cos a\omega-kb^{2}\sin(!+\tau)$$=$$f_{k}’(Ci)$$=$$-a^{2}(\cos a\omega+\overline{k}\sin(!+\tau))=-a^{2}f_{\overline{k}}(\omega)=-a^{2}\overline{f}(\omega)$.

is even is also treated similarly. Then $\overline{f}((n-1)\pi/a)\geq$$1-k>0>k-1\geq\overline{f}(\pi/a)$ . Since
is the only zero of in $((n-1)\pi/a, n\pi/a)$ ,, $f’(\omega)>0$ for. Thuswe have $f’(\omega)<0$ for$(\mu_{n},n\pi/a)$$z_{n}\in
J_{n}$$((n-1)\pi/a,\mu_{n})$. Let ,
withand convex on
is concave on. Then, we find ,andwith $s_{n}\leq y_{n}<\mu_{n}<v_{n}\leq z_{n}\leq t_{n}$ such that $f(u_{n})=1-k$ and $f(v_{n})=k-1$ . Thus,.
is monotone decreasing on
Let

$n$

be odd. The case where$n$$\overline{f}$$\overline{\mu_{n}}$$\omega\in((n-1)\pi/a, \overline{\mu_{n}})$$\omega\in(\overline{\mu_{n}}, n\pi/a)$$f(\omega)$$y_{n}$$f(z_{n})= \min_{\omega\in \mathrm{I}_{n}}f(\omega)\leq k-1$

$f(y_{n})=\mathrm{e}\mathrm{v}\mathrm{e}\mathrm{n}f(\omega)\geq 1-k$$u_{n}$$v_{n}$$f$$[u_{n}, v_{n}]\subset[y_{n}, z_{n}]$Proof of Theorem 2There exists, $N$ . Then
and $k_{:}=k(b/a)^{2:}$ , $i=0,1,2$ ,, $N-1$ . In particular,, $i=1,2$ ,again, we can prove Theorem 2.

$N\in \mathrm{N}$$\ldots$$k_{i}\in \mathrm{S}$such that $0\leq(b/a)^{2N}\leq 1/2$ . Let $k\in[0,1)$
by Lemma 4. Therefore, by Lemma 5,. Thus, by using Lemma 5$k_{N}\in \mathrm{S}$$k_{1}=k(b/a)^{2}=\overline{k}\in \mathrm{S}$$\ldots$

Next, we want to show that, for sufficiently large n, there exists only one zero ofin each. More precisely, we have$d(\omega)$$J_{n}$

Theorem 3There exists $M$ ,$N\in \mathrm{N}$such that $\omega_{M+n}\in J_{N+n}$

for $n=0,1$ ,$\ldots$.

To prove the above thoerem, we prepare Lemma 6and lemma 7given below:

Lemma 6For any$n\in \mathrm{N}$, the following inequality holds:$|f’(\mu_{n})|\geq\delta=\sqrt{(1-k^{2})(a^{2}-b^{2})}$

Proof Since$\mu_{n}$,n$\in \mathrm{N}$

.

(11)

, are zeros of f, we have$f(\mu_{n})$$=\cos a\mu_{n}+k\sin(b\mu_{n}+\tau)=0$ ,$f’(\mu_{n})$$=$$-a\sin a\mu_{n}+kb\cos(b\mu_{n}+\tau)$

.

(12)
(13)

If $b=0$ , then $(f’(\mu_{n}))^{2}=a^{2}\sin^{2}a\mu_{n}=a^{2}(1-\cos^{2}\tau)\geq a^{2}(1-k^{2})$ . Hence we have (11).
If $b\neq 0$ , then$f’(\mu_{n})^{2}$$=$$\geq$

Thus we have (11).

$\frac{(a^{2}-b^{2})}{b^{2}}(\sin a\mu_{n}-\frac{af’(\mu_{n})}{a^{2}-b^{2}})^{2}+(1-k^{2})(a^{2}-b^{2})$$(1-k^{2})(a^{2}-b^{2})$.

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Lemma 7There exists an interval$|f(\omega)|\leq l$,$|f’(\omega)|\geq\delta/2$$|f(\omega)|\geq l$

Proof By uniform continuity of$1-k$ such thatand $l\in[0,1-k)$ such that$[a_{n}, b_{n}]\subset[u_{n}, v_{n}]$$f’(\omega)$

for

for

$\omega\in[a_{n}, b_{n}]$$\omega\in J_{n}\backslash [a_{n}, b_{n}]$

,

(14)
(15)

.

and Lemma 6, there exists $c>0$ with $l=\delta c/2<$

for $\omega\in[\mu_{n}-c, \mu_{n}+c]$ .

$|f’(\omega)|\geq\delta/2$(16)

Therefore, we have $|f(\omega)|\geq(\delta/2)|\omega-\mu_{n}|$ on $[\mu_{n}-c, \mu_{n}+c]$ . If is odd (resp. even),
we define an, with $\mu_{n}-c<a_{n}<b_{n}<\mu_{n}+c$ by $f(a_{n})=l$ (resp. $-l$ ) and $f(b_{n})=-l$(resp. ). Hence we have$n$$b_{n}$$l$$\{\omega\in[\mu_{n}-c, \mu_{n}+c];|f(\omega)|\leq l\}=[a_{n}, b_{n}]$

.

(17)

By (16) and (17), we see (14), and by Theorem 3, (15).

We put) $-h(\omega)$ . Since, $h’(\omega)arrow 0$ as, thereexistssuch that $|h(\omega)|<Dl$ and $|h’(\omega)|<D\delta/2$ for $\omega>(N-1)\pi/a$ .
Let be odd with $n>N$ . Then, by Lemma 7, we have $f(a_{n})=l$ , $f(bn)=-l$
and $f’(\omega)<-\delta/2$ for. Hence $/(an)=Df$ (an) $-h(an)=Dl-h(an)>$$Dl-Dl=0$ and $g(b_{n})=Df(b_{n})-h(b_{n})=Dl-h(b_{n})<-Dl+Dl=0$ . Thus, for, $g’(\omega)=Df’(\omega)-h’(\omega)\leq-D\delta/2+|h’(\omega)|<-D\delta/2+D\delta/2=0$ which
implies thathas aunique zero in. For, by (14), we have$|g(\omega)|\geq|Df(\omega)|-|h(\omega)|\geq Dl-Dl=0$ . Therefore,
has aunique zero in. The$n\geqN$case with evenis also similarly proved. Let
be azero ofin. Thus$\omega_{M+n}\in J_{N+n}$
for $n=0,1,2$ ,.
$g(\omega)=\varphi(\omega)-h(\omega)=\mathrm{D}\mathrm{f}\{\mathrm{u}$$h(\omega)$$\omegaarrow\infty$$N\in \mathrm{N}$$n$$\omega\in[a_{n}, b_{n}]$$\omega\in[a_{n}, b_{n}]$$g(\omega)$$(a_{n}, b_{n})$$\omega$$\in J_{n}\backslash [a_{n}, b_{n}]$$g(\omega)$$J_{n}$$g(\omega)$$\omega_{M}$$J_{N}$$\ldots$

Proof of Theorem 1Since $f(\mu_{N+n})=g(\omega_{M+n})=0$ , we have$h(\omega_{M+n})=Df(\omega_{M+n})-g(\omega_{M+n})=D(f(\omega_{h\mathrm{f}+n})-f(\mu_{N+n}))$

.

(18)

By Mean Value Theorem, there exists $\theta\in(0,1)$ such that $f(\omega_{M+n})-f(\mu_{N+n})=(\omega_{\mathrm{A}I+n}-$$\mu_{N+n})f’(\mu_{N+n}+\theta(\omega_{M+n}-\mu_{M+n}))$ . Thus, by (14) and (18),$| \omega_{N+n}-\mu_{N+n}|\leq\frac{|f(\omega_{M+n})-f(\mu_{N+n})|}{|f’(\mu_{N+n}+\theta(\omega_{M+n}-\mu_{M+n}))|}\leq\frac{2}{D\delta}|h(\omega_{M+n})|arrow 0$

as$narrow\infty$. Therefore,$\lim_{narrow}\inf_{\infty}(\omega_{n+1}-\omega_{n})=\lim_{narrow}\inf_{\infty}(\omega_{M+n+1}-\omega_{M+n})$$=$$\lim_{narrow}\inf_{\infty}(\omega_{M+n+1}-\mu_{N+n+1}+\mu_{N+n+1}-\mu_{N+n}+\mu_{N+n}-\omega_{M+n})$$=$$\lim_{narrow}\inf_{\infty}(\mu_{N+n+1}-\mu_{N+n})=\lim_{narrow}\inf_{\infty}(\mu_{n+1}-\mu_{n})$.

By $s_{n}<\mu_{n}<t_{n}<s_{n+1}<\mu_{n+1}<t_{n+1}$ , we have$\mu_{n+1}-\mu_{n}\geq s_{n+1}-t_{n}$$=$$( \frac{2n+1}{2a}\pi-\frac{\sin^{-1}k}{a})-(\frac{2n-1}{2a}\pi+\frac{\sin^{-1}k}{a})$$=$$\frac{\pi}{a}-\frac{2\sin^{-1}k}{a}=\frac{1}{a}(\pi-2\sin^{-1}k)$

The above theorem implies that$\lim_{narrow}\inf_{\infty}(\omega_{n+1}^{2}-\omega_{n}^{2})\geq\lim_{narrow}\inf_{\infty}(\omega_{n+1}+\omega_{n})\lim_{narrow}\inf_{\infty}(\omega_{n+1}-\omega_{n})=\infty$

.

(14)

146
3Concluding Remarks
This paper is only afirst step to the controllability theory for the Euler-Bernoulli equation
using the moment problem method [4], [12].

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