\documentclass[reqno]{amsart}

\AtBeginDocument{{\noindent\small
2004 Conference on Diff. Eqns. and Appl. in Math. Biology,  Nanaimo, BC, Canada.\newline
{\em Electronic Journal of Differential Equations},
Conference 12, 2005, pp. 57--64.\newline
ISSN: 1072-6691. URL: http://ejde.math.txstate.edu or
http://ejde.math.unt.edu
\newline ftp ejde.math.txstate.edu  (login: ftp)}
\thanks{\copyright 2005 Texas State University - San Marcos.}
\vspace{9mm}}
\setcounter{page}{57}

\begin{document}

\title[\hfilneg EJDE/Conf/12 \hfil On isospectral beams]
{On the isospectral beams}

\author[K. Ghanbari \hfil EJDE/Conf/12 \hfilneg]
{Kazem Ghanbari}

\address{Department of Mathematics,
Sahand University of Technology, Tabriz, Iran}
\email{kghanbari@sut.ac.ir}

\date{}
\thanks{Published April 20, 2005.}
\subjclass[2000]{34B05,34B10} 
\keywords{Isospectral, Euler-Bernoulli
equation for the vibrating beam, beam operator}

\begin{abstract}
 The free undamped infinitesimal transverse vibrations of a thin
 straight beam are modelled by a forth-order differential
 equation.  This paper investigates the families of
 fourth-order systems which have one spectrum in common,
 and correspond to four different sets of  end-conditions. The
 analysis is based on the transformation of the beam operator into
 a fourth-order self-adjoint linear differential operator. This
 operator is factorized as a product $L=H^{*}H$, where $H$ is a
 second-order differential operator of the form $H=D^2+rD+s$, and
 $H^{*}$ is its adjoint operator.
\end{abstract}

\maketitle
\numberwithin{equation}{section}
\newtheorem{theorem}{Theorem}[section]
\newtheorem{lemma}[theorem]{Lemma}
\newtheorem{remark}[theorem]{Remark}

\section{Beam Equation}

The free undamped infinitesimal transverse vibrations, of
frequency $\omega$, of a thin straight beam of length $\ell$ are
governed by the Euler-Bernoulli equation
\begin{equation} \label{e1}
(EI(x)u''(x))''=\omega^2\rho A(x)u(x),\quad 0\leq x\leq\ell.
\end{equation}
Here $E$ is the Young's modulus, $\rho$ is the density, both assumed
constant, $A(x)$ is the cross-sectional area at $x$, $I(x)$ is the
second moment of this area about the axis through the centroid at
right angle to the plane of vibration (the natural axis).
Euler-Bernoulli equation \eqref{e1} can be written even in simpler form
by the following substitutions. Put
\begin{gather*}
x=\ell s,\quad y(s)=u(x),\quad r(s)=\frac{I(x)}{I(x_0)},\\
a(s)=\frac{A(x)}{A(x_0)},\quad \lambda=
\frac{\rho \ell^4\omega^2A(x_0)}{EI(x_0)},
\end{gather*}
where $x_0$ is a fixed point in $[0,1]$. Then \eqref{e1} becomes
\begin{equation} \label{e2}
(r(s)y''(s))''=\lambda a (s)y(s),\quad 0\leq s\leq1,
\end{equation}
where $r(s),a(s)>0$ for all $s \in [0,1]$. For the beam equation
\eqref{e2} the most commonly used end-conditions are
\begin{itemize}
\item [(F)] $y''=0=(ry'')'$ (Free)
\item[(C)] $y=0=y'$ (Clamped)
\item[(P)] $y=0=y''$ (Pinned)
\item[(S)] $y'=0=(ry'')'$ (Sliding).
\end{itemize}
By using a transformation, we can obtain another form of equation
\eqref{e2} which is useful in practice. This transformation is as
follows:
\begin{equation} \label{e3}
z=\int_s^1\big[\frac{a(t)}{r(t)}\big]^{1/4}dt,\quad
b(z)=\big[\frac{a(s)}{r(s)}\big]^{1/4}, \quad
c^2(z)=[r^3(s)a(s)]^{1/4}.
\end{equation}
We have
$$
\frac{d}{ds}=\frac{d}{dz}\frac{dz}{ds}
=-\big[\frac{a(s)}{r(s)}\big]^{1/4}\frac{d}{dz}
=-b(z)\frac{d}{dz},
$$
and
$$r(s)\frac{d^2}{ds^2}=r(s)b(z)\frac{d}{dz}\big[b(z)\frac{d}{dz}\big]
=c^2(z)\frac{d}{dz}\big[b(z)\frac{d}{dz}\big].
$$
Therefore, the \eqref{e2} becomes
$$
b\frac{d}{dz}\big\{b\frac{d}{dz}\big[c^2\frac{d}{dz}\big(b\frac{dy}{dz}\big)
\big]\big\}=\lambda b^3c^2y,
$$
which can be written as
\begin{equation} \label{e4}
\big(b\big(c^2(by' (z))'\big)'\big)'= \lambda b^2c^2y(z),\quad 0\leq z\leq L,
\end{equation}
where $L=\int_0^1[\frac{a(t)}{r(t)}]^{1/4}dt$. For more details
see \cite{BARC} or \cite{GW2}. Equation \eqref{e2} can be written
in the form $v^{(4)}+(Av')'+Bv=\lambda v$ as follows. Put
$y=\frac{v}{bc}$, then
$$
by'=b\big[\frac{v'}{bc}-\frac{v}{b^2c^2}(bc'+b' c)\big].
$$
If we put $\frac{b'}{b}=\beta$ and
$\frac{c'}{c}=\gamma$, then after some algebraic
calculations we find
\begin{equation} \label{e5}
\begin{aligned}
by'&=\frac{1}{c}[v'-(\beta+\gamma)v]\\
(by')'&= \frac{1}{c}[v''-(\beta+\gamma)v'-(\beta'+\gamma
')v-\gamma v'+\gamma(\beta+\gamma)v]\\
(c^2(by')')'&= c\{v'''-(\beta+\gamma)v''
-[2\beta'+3\gamma'+\gamma^2]v'-(\beta''+\gamma'')v\\
&\quad + [\gamma(\beta+\gamma)'+\gamma'(\beta+\gamma)-\gamma(\beta+\gamma)'+
\gamma^2(\beta+\gamma)]v\}\\
\big(b\big(c^2(by')'\big)'\big)'&= bc\{v^{(4)}-
[(\beta+\gamma)'+2\beta'+3\gamma'+\gamma^2+(\beta+\gamma)^2]v'' \\
&\quad -[2\gamma\gamma'+\beta''+\gamma''-
\gamma'(\beta+\gamma)-\gamma^2(\beta+\gamma)]v'\\
&\quad -[2\beta''+3\gamma''+2\beta'(\beta+\gamma)+
3\gamma'(\beta+\gamma)+\gamma^2(\beta+\gamma)]v'\\
&\quad +[\gamma'(\beta+\gamma)'+\gamma''(\beta+\gamma)+[\gamma^2(\beta+
\gamma)]']v\\
&\quad -[\beta'''+\gamma'''+ (\beta+\gamma)(\beta''+\gamma'')]v\\
&\quad +[\gamma'(\beta+\gamma)^2+\gamma^2(\beta+\gamma)^2]v\}.
\end{aligned}
\end{equation}
It is clear that clamped is the only end-condition that remains
invariant under the transformation (3), i.e., $v=0=v'$. By
using (5) we can write the equation (4) in $v$ as follows:
\begin{equation} \label{e6}
v^{(4)}+(Av')'+Bv=\lambda v,
\end{equation}
where
\begin{gather} \label{e7}
A=-[3\beta'+4\gamma'+\gamma^2+(\beta+\gamma)^2], \\
\label{e8} \begin{aligned}
B&=-(\beta+\gamma)'''-\beta''(\beta+\gamma)+\gamma'
(\beta+\gamma)'+\gamma'(\beta+\gamma)^2\\
 &\quad +[\gamma^2(\beta+\gamma)]'+ \gamma^2(\beta+\gamma)^2.
\end{aligned}
\end{gather}
Put $\theta=\beta+\gamma$ and $\phi=\gamma'+\gamma^2$, then
$A$ and $B$ can be written as follows
\begin{equation} \label{e9}
\begin{aligned}
A&=-3\theta'-\theta^2-\phi\\
B&=-\theta'''-(\theta''-\gamma'')\theta+
\gamma'\theta'+\gamma'\theta^2+(\gamma^2\theta)'+\gamma^2\theta^2\\
&=-\theta'''-\theta''\theta+(\gamma'\theta)'+
\gamma'\theta^2+(\gamma^2\theta)'+\gamma^2\theta^2\\
&=-\theta'''-\theta''\theta+[(\gamma'+\gamma^2)\theta]'+(\gamma'+\gamma^2)\theta^2\\
&=-\theta'''-\theta''\theta+(\theta\phi)'+\theta^2\phi\\
&= (-\theta''+\theta\phi)'+(-\theta''+\theta\phi)\theta.
\end{aligned}
\end{equation}
If we substitute $\beta=\frac{b'}{b}$ and $\gamma=\frac{c'}{c}$
back in \eqref{e7} and \eqref{e8} then $A$ and $B$ can be written
in terms of $b,c$ and their derivatives as follows:
\begin{equation} \label{e10}
A=2{c'}^2/c^2-4c''/c-3b''/b+2{b'}^2/b^2-2b' c'/bc
\end{equation}
\begin{equation}\label{e11}
\begin{aligned}
B&=4{b'}^4/b^4-c^{(4)}/c+4{c'}^4/c^4+4{c''}^2/c^2-b^{(4)}/b\\
&\quad + 3{b''}^2/b^2-10{c'}^2c''/c^3+4c' c''' /c^2+3b'
b'''/b^2\\
&\quad -9b''{b'}^2/b^3-2b'{c'}^3/bc^3+3b''b'c' /b^2c-c' b'''/bc\\
&\quad -2{b'}^3c'/b^3c+4b' c' c''/bc^2+b'' c''/bc.
\end{aligned}
\end{equation}

\section{Isospectral rods}

The linear, free, undamped longitudinal vibrations of a thin
elastic rod with variable cross section $A=A(x)$ are governed by
the equation
\begin{equation} \label{e12}
(Au')'+\lambda Au=0.
\end{equation}
Gladwell and Morassi \cite{GW1} found isospectral rods,
rods with identical spectrum, by using the Darboux lemma. In this
section we apply a different method, namely factorizing the rod
operator, to find the the isospectral rods. Then we develop this
method to find isospectral beams in the following section. This
method first was proposed by P$\ddot{o}$chel J and Trubowitz E
\cite{TW} to Sturm-Liouville operator. The rod equation \eqref{e12} can
be written in the Sturm-Liouville form. The cross section function
$A$ is positive; write
$A=a^2$, $y=au$.
Then
$$
Au'=a^2u'=ay'-a' y
$$
so that \eqref{e12} reduces
to the Sturm-Liouville form
\begin{equation} \label{e13}
-y''+qy=\lambda y,\quad \text{where }q=\frac{a''}{a}.
\end{equation}
Define the rod operator
\begin{equation} \label{e14}
L=-D^2+q.
\end{equation}
We want to write the rod operator $L$ as a product of two first
order differential operators. Every linear differential operator
of the form
\begin{equation} \label{e15}
H(v)=\sum_{k=0}^{n}p_k(x)D^k(v)
\end{equation}
has the adjoint operator $H^*$ of the form
\begin{equation} \label{e16}
H^*(u)=\sum_{k=0}^{n}D^k\left[(-1)^kp_k(x)u\right].
\end{equation}
See for example Lanczos \cite{LAN}. Therefore the adjoint operator
of $H=D+\alpha$ is $H^*=-D+\alpha$, where $\alpha$ is a real
function. Now consider the rod operator $L$ given by \eqref{e14}. We want
to write $L$ as a product of the form $L=H^*H$, where
$H=D+\alpha$. Let
$$
L=-D^2+q=-D^2+\frac{a''}{a}=H^*H=(-D+\alpha)(D+\alpha).
$$
Thus
\begin{equation} \label{e17}
q=\alpha^2-\alpha'=\frac{a''}{a}.
\end{equation}
The operator $L=H^*H$ and $\hat{L}=HH^*$ are isospectral. Now let
$$
\hat{L}=HH^*=(D+\alpha)(-D+\alpha)=-D^2+\hat{q}=-D^2+\frac{\hat{a}''}
{\hat{a}}.
$$
Hence
\begin{equation} \label{e18}
\hat{q}=\alpha^2+\alpha'.
\end{equation}
Substituting the solution $\alpha=-\frac{a'}{a}$ in
\eqref{e18} yields
\begin{equation} \label{e19}
\hat{q}=2\frac{{a'}^2}{a^2}-\frac{a''}{a}.
\end{equation}
On the other hand
$$  \hat{q}=\frac{{\hat{a}}''}{\hat{a}}.
$$
In order to determine the isospectral rods we need the general
solution of the following differential equation
\begin{equation} \label{e20}
\frac{{\hat{a}}''}{\hat{a}}=2\frac{{a'}^2}{a^2}-
\frac{a''}{a}.
\end{equation}
The general solution of the equation \eqref{e20} is
$$
\hat{a}=\frac{1}{a}\{1+K\int_0^xa^2(s)ds\},\quad \text{where }K\in\mathbb{R}.
$$
This family of isospectral rods coincide with those obtained by
 Gladwell and Morassi \cite{GW1}.

\begin{remark} \label{rmk1} \rm
In general if we consider the Sturm-Liouville
operator $L$ given by \eqref{e14}, using a similar argument we can
factor $L-\mu$ as follows
\begin{equation} \label{e21}
L-\mu=H^*H=(-D-\frac{g'}{g})(D-\frac{g'}{g}),
\end{equation}
where $g$ is a nonzero solution of the Sturm-Liouville equation,
i.e.,
$$
-g''+(q-\mu)g=0.
$$
If we reverse the factors
in \eqref{e21} then we obtain the isospectral Sturm-Liouville operator
\begin{equation} \label{e22}
\hat{L}-\mu=HH^*=-D^2+(\hat{q}-\mu),
\end{equation}
where
$\hat{q}=q-2\frac{d^2}{dx^2}(\ln g)$.
For more details see \cite{TW}.
\end{remark}

\section{Isospectral beams}

According to the beam equation \eqref{e6} we define the beam operator as
follows
\begin{equation} \label{e24}
L=H^*H=D^4+D(AD)+B,
\end{equation}
where $A$ and $B$ are given by \eqref{e7} and \eqref{e8}.
Using \eqref{e15} and \eqref{e16} it
can be easily checked that the adjoint operator of a general
operator of the form $H=aD^2+bD+c$ is
$H^*=aD^2+(2a'-b)D+(a''-b'+c)$. The idea
is to find the isospectral beams by factorizing the beam operator
$L$ and then reversing the factors to find the isospectral beams.
We therefore suppose that $L$ can be factorized as follows
$$
L=H^*H=[D^2-rD+(s-r')][D^2+rD+s]
$$
that is equivalent to
the nonlinear system:
$$
\{2s+r'-r^2=A, ~s^2+s''-(rs)'=B\}.
$$
Define $\hat{L}=HH^*$. Then $L$ and $\hat{L}$ are isospectral and
we have
\begin{equation} \label{e25}
L=H^*H=D^4+D(AD)+B,~\hat{L}=HH^*=D^4+D(\hat{A}D)+\hat{B},
\end{equation}
where
\begin{equation} \label{e26}
\begin{gathered}
A=2s+r'-r^2\\ B=s^2+s''-(rs)'\,,
\end{gathered}
\end{equation}
and
\begin{equation} \label{e27}
\begin{gathered}
\hat{A}=2s-3r'-r^2 \\
\hat{B}=s^2+s''-r'''-rr''+rs'-sr'.
\end{gathered}
\end{equation}
Let us call the system \eqref{e26} the {\it principal} system because the
solutions of \eqref{e26} will produce isospectral beams. It is easy to
verify that $r=-\beta-2\gamma$, and
$s=\gamma^2+\beta\gamma-\beta'-\gamma'$ is a solution
to the nonlinear system \eqref{e26}. Comparing \eqref{e26} and \eqref{e27} we obtain
\begin{equation} \label{e28}
A-\hat{A}=4r',
B-\hat{B}=r'''+rr''-2rs'.
\end{equation}
Substituting $r=-\beta-2\gamma$, and
$s=\gamma^2+\beta\gamma-\beta'-\gamma'$ in \eqref{e28} we
find
\begin{equation} \label{e29}
\hat{A}=\beta'+4\gamma'-\beta^2-2\gamma^2-2\beta\gamma,
\end{equation}
and
\begin{equation}\label{e30}
\begin{aligned}
\hat{B}&= \gamma'''+\beta''\gamma+\gamma'(\beta+\gamma)^
\prime-4\beta\gamma\gamma'-\beta^2\gamma'-4\gamma'\gamma^2\cr
&\quad -3\beta'\gamma^2-2\beta\beta'\gamma+\gamma^2(\beta+\gamma)^2.
\end{aligned}
\end{equation}
It is clear that
\begin{gather*}
\hat{A}(\beta,\gamma)=A(\beta,-\beta-\gamma),\\
\hat{B}(\beta,\gamma)= B(\beta,-\beta-\gamma).
\end{gather*}
This implies that
$\hat{\beta}=\beta$ and $\hat{\gamma}=-\beta-\gamma$.
Therefore, $\hat{b}=b$ and $\hat{c}=\frac{1}{bc}$.


\begin{remark} \label{rmk2} \rm
 We can verify this isospectral family as
follows. Consider the original beam equation
\[
\big(b\big(c^2(by')'\big)'\big)'= \lambda b^2c^2y.
\]
Put $H_1=\frac{1}{b^2c^2}(DbD)$ and $H_2=c^2(DbD)$. Then
$$
\big(b\big(c^2(by')'\big)'\big)'=
\lambda b^2c^2y\Longleftrightarrow H_1H_2(y)=\lambda y.
$$
Therefore $H_2H_1(H_2y)=\lambda(H_2y)$, which can be written as
$H_2H_1(v)=\lambda v$, where $v=H_2y.$
\end{remark}


\begin{remark} \label{rmk3} \rm
 Similar to Remark \ref{rmk1}, if we factorize the
operator $L-\mu$ instead the beam operator $L$ given by \eqref{e24} then
the corresponding principal system will be
\begin{equation} \label{e31}
\begin{gathered}
2s+r'-r^2=A\\
s^2+s''-(rs)'=B-\mu=-\frac{g^{(4)}}{g}-\frac{(Ag')'}{g}
\end{gathered}
\end{equation}
where $g$ is a nonzero solution of the beam equation, i.e.,
$$
g^{(4)}+(Ag')'+(B-\mu)g=0.
$$
\end{remark}
In contrast to the principal system \eqref{e26}, the system \eqref{e31} does not
seem to be easy to solve. The operator $\hat{L}$ coincide with the
beam operator given by  Gottlieb \cite{GOT}. There he finds
seven classes of beams isospectral to the standard unit coefficient
beam equation
\begin{equation} \label{e32}
D^4v=\lambda v.
\end{equation}
Since $L$ and $\hat{L}$ are isospectral, by using the operator $L$
we can find another seven class of beams isospectral to the
standard unit-coefficient beam. Referring to Gottlieb's notations
we define
\begin{equation} \label{e33}
b=\eta(\delta z+\epsilon)^\mu,\quad c=\zeta(\delta z+\epsilon)^\nu,
\end{equation}
where $\delta,\epsilon,\eta,\zeta$ are constants, and $\mu,\nu$
are parameters to be determined from the expressions of $A,B$ in
terms of $b,c$ and their derivatives given by \eqref{e10} and \eqref{e11}. After
some algebraic calculations we find
\begin{gather} \label{e34}
A=-\delta^2(\delta z+\epsilon)^{-2}[\mu^2+2\mu
\nu+2\nu^2-4\nu-3\mu]\\
\begin{aligned}
 B&=\delta^4(\delta z+\epsilon)^{-4}[\nu^4-4\nu^3+\mu^2\nu^2+\nu^2]\\
&\quad \times
 \big[2\mu\nu^3-2\mu^2-\nu\mu^2-\mu\nu-5\mu\nu^2+6(\mu+\nu)\big].
\end{aligned}
\end{gather}
Using the transformation given by \eqref{e3}, we get
\begin{equation} \label{e36}
\begin{aligned}
1-s&=\int_0^z\frac{dz}{b(z)}\\
&=\int_0^z\frac{1}{\eta}(\delta z+\epsilon)^{-\mu}dz\\
&= \frac{1}{\eta(1-\mu)}(\delta z+\epsilon)^{1-\mu}-\frac{1}{\eta(1-\mu)}
\epsilon^{1-\mu}.
\end{aligned}
\end{equation}
Now put
\begin{equation} \label{e37}
Z=\delta
z+\epsilon,\quad\text{and}\quad \xi(s)=\epsilon^{1-\mu}+\eta(1-\mu)(1-s).
\end{equation}
Then
 %\label{e38}
$Z=[\xi(s)]^\chi$, where $\chi=\frac{1}{1-\mu}$,
and hence
\begin{equation} \label{e39}
z(s)=\delta^{-1}\left([\xi(s)]^\chi-\epsilon\right).
\end{equation}
Thus
\begin{gather} \label{e40}
b\left(z(s)\right)=\eta[\xi(s)]^\lambda,\quad \text{with }\lambda=\frac{\mu}{1-\mu}\\
c\left(z(s)\right)=\zeta[\xi(s)]^\kappa,\quad \text{with }\kappa=\frac{\nu}{1-\mu}.
\label{e41}
\end{gather}
Therefore, the seven class of beams isospectral to the standard
unit-coefficient beam is corresponding to $A=B=0$; so that we can
determine these classes by solving the algebraic equations
\begin{equation} \label{e42}
\begin{gathered}
\mu^2+2\mu\nu+2\nu^2-4\nu-3\mu=0\\
\nu^4-4\nu^3+\mu^2\nu^2+\nu^2+2\mu
\nu^3-2\mu^2-\nu\mu^2-\mu\nu-5\mu\nu^2+6(\mu+\nu)=0
\end{gathered}
\end{equation}
which gives the solutions cited in the following table:
\begin{center}
\begin{tabular}{|c|ccccc|} \hline
class &$\mu$ &$\nu$ &$\chi$ &$\lambda$ &$\kappa$ \\ \hline
(0)&0 &0 &1 &0 &0 \\
(1)&0 &2 &1 &0 &2 \\
(2)&$-1$ &2 &1/2 &$-1$/2&1 \\
(3)&$-1$ &1&1/2 &$-1/2$ &1/2 \\
(4)&2 &1 &$-1$ &$-2$ &$-1$ \\
(5)&2 &$-1$ &$-1$ &$-2$ &1 \\
(6)&3 &$-1$ &$-1/2$ &$-3/2$ &1/2 \\
(7)&3 &0 &$-1/2$ &$-3/2$ &0 \\ \hline
\end{tabular}
\end{center}

\subsection*{More isospectral beams}
For this purpose, we put $\theta=\beta+\gamma$ and
$\phi=\gamma'+\gamma^2$. Then \eqref{e9} implies
\begin{equation} \label{e43}
\begin{gathered}
A=A(\theta,\phi)=-3\theta'-\theta^2+\phi\cr
B=B(\theta,\phi)=(-\theta''+\theta\phi)'+
(-\theta''+\theta\phi)\theta.
\end{gathered}
\end{equation}
Let $\tilde{A}=A(\tilde{\theta},\tilde{\phi})$, and
$\tilde{B}=B(\tilde{\theta},\tilde{\phi})$. If $A=\tilde{A}$ and
$B=\tilde{B}$ then $L$ and $\tilde{L}$ are isospectral, where $L$
is given by \eqref{e24} and
\begin{equation} \label{e44}
\tilde{L}=D^4+D(\tilde{A}D)+\tilde{B}.
\end{equation}
Let $\theta=\tilde{\theta}$, $\phi=\tilde{\phi}$, then
$A=\tilde{A}$ and $B=\tilde{B}$. Thus we obtain isospectral class
as follows
\begin{equation} \label{e45}
\theta=\tilde{\theta}\Longleftrightarrow bc=\tilde{b}\tilde{c}.
\end{equation}
Now the equation $\phi=\tilde{\phi}$ implies
\begin{equation} \label{e47}
\frac{c''}{c}=\frac{{\tilde{c}}''}{\tilde{c}}\Longleftrightarrow
c''\tilde{c}-{\tilde{c}}''c=0\Longleftrightarrow
c' \tilde{c}-{\tilde{c}}' c=\text{const.}
\end{equation}
This equation implies that for arbitrary real constants $K_1,K_2$
we find
\[
\tilde{c}=c\int\frac{K_1}{c^2}ds+K_2c.
\]
Now with a similar way we can find another family isospectral to
$\hat{L}$ which is isospectral to $L$. Put
$\phi=\theta^2-\theta'$. After some algebraic computation we
can write $\hat{A}$ and $\hat{B}$ as follows
\begin{equation} \label{e48}
\begin{gathered}
\hat{A}=3\gamma'-\gamma^2-\phi\cr
\hat{B}=(\gamma''-\gamma\phi)'-\gamma(\gamma''-\gamma\phi).
\end{gathered}
\end{equation}
If $\gamma=\tilde{\gamma}$ and $\phi=\tilde{\phi}$ then $\hat{L}$
and $\tilde{{\hat{L}}}$ are isospectreal. The equation
$\gamma=\tilde{\gamma}$ implies  $c=\tilde{c}$. Now we find
$\tilde{b}$ by solving the equation $\phi=\tilde{\phi}$. We have
\[
\phi=\tilde{\phi}\Longleftrightarrow
(\tilde{\theta}-\theta)'={\tilde{\theta}}^2-\theta^2\Longleftrightarrow
\frac{(\tilde{\theta}-\theta)'}{\tilde{\theta}-\theta}=\tilde{\theta}+\theta.
\]
This equation  implies that
$\tilde{\theta}-\theta=bc\tilde{b}\tilde{c}=b\tilde{b}c^2$,
which is equivalent to
\[
\frac{{\tilde{b}}'}{\tilde{b}}-\frac{b'}{b}=b\tilde{b}c^2\Longleftrightarrow
{\tilde{b}}' b-b'
\tilde{b}=(b\tilde{b})^2c^2\Longleftrightarrow
\Big(\frac{\tilde{b}}{b}\Big)'={\tilde{b}}^2c^2.
\]
This  equation can be easily solved. In fact, putting
$u=\frac{\tilde{b}}{b}$, the above equation becomes
\[
u'=u^2b^2c^2.
\]
By solving this differential equation, we find
\[
\tilde{b}=\frac{b}{K_1-\int
b^2c^2ds},\quad \text{where}~K_1\in\mathbb{R}.
\]
Suppose
\[
b=c^2\Big(\int\frac{K_1}{c^2}ds+K_2\Big).
\]
Then we can define another isospectral class as follows: Define
$\tilde{b}=c^2$ and $\tilde{c}=\sqrt{b}$. Then
$bc^2=\tilde{b}{\tilde{c}}^2$. Hence $r=\tilde{r}$ where
$r=\beta+2\gamma$. It is clear that $s=\tilde{s}$, thus we have
$A=\tilde{A}$ and $B=\tilde{B}$ according to the principal system
\eqref{e26}.

\subsection*{Conclusion}
Factoring the beam operator
$L=D^4+D(AD)+B$ into the product of the form $L=HH^*$, where $H$
is a second order differential operator of the form $H=D^2+rD+s$
and $H^*$ is the adjoint operator of $H$, and reversing the order
of this product we managed to find some 12 classes of isospectral
beams. However, determining all classes is linked to nonlinear
system \eqref{e26} for
which we found particular solution.

\subsection*{Acknowledgement}
 The author would like to thank anonymous referee for careful reading of the
 manuscript an giving valuable comments and helpful references.

\begin{thebibliography}{0}
\bibitem{BARC} V. Barcilon; Inverse Problem for a vibrating beam
in the free-clamped configuration,
 {\it Phil. Trans. R. Soc. London A} \textbf{304}(1982), 211-251.

\bibitem{GW1} G. M. Gladwell and A. Morassi; On the isospectral rods,
horns and strings, {\it Inverse Problems}, \textbf{11} (1995)
533-554.

\bibitem{GW2} G. M. L. Gladwell; {\it Inverse Problems in Vibration}, Kluwer,
Dordrecht  1986.

\bibitem{GW3} G. M. L. Gladwell; Isospectral vibrating beams,
Proc. R. Soc. London A \textbf{458}(2002), 2691-2703.

\bibitem{GOT} H. P. W. Gottlieb; Isospectral Euler-Bernoulli beams
with continous density and rigidity functions, {\it Proceeding of
Royal Society of London A} \textbf{423} (1987) 235-250.

\bibitem{LAN} C. Lanczos, 1961 {\it Linear Differential Operators},
D. Van Nostrand Company, Ltd, London.

\bibitem{TW} P\"oschel J and Trubwitz E; {\it Inverse
Spectral Theory} (London: Academic), 1987.

\bibitem{SUB} G. Subramanian and A. Raman; Isopectral systems for
tapered beams, J. Sound and Vibration \textbf{198} (1996) 257-266.
\end{thebibliography}


\end{document}