\documentclass[reqno]{amsart}
\usepackage{hyperref}
\usepackage{graphicx}

\AtBeginDocument{{\noindent\small
\emph{Electronic Journal of Differential Equations},
Vol. 2017 (2017), No. 33, pp. 1--10.\newline
ISSN: 1072-6691. URL: http://ejde.math.txstate.edu or http://ejde.math.unt.edu}
\thanks{\copyright 2017 Texas State University.}
\vspace{8mm}}

\begin{document}
\title[\hfilneg EJDE-2017/33\hfil 
Inverse boundary problems for intermediate springs]
{Inverse boundary problems for intermediate springs on a
rod with geometrical symmetry}

\author[D. B. Nurakhmetov, S. A. Jumabayev, A. A. Aniyarov \hfil EJDE-2017/33\hfilneg]
{Daulet B. Nurakhmetov, Serik A. Jumabayev, Almir A. Aniyarov}

\address{Daulet B. Nurakhmetov \newline
Department of Information Systems,
S. Seifullin Kazakh agrotechnical university,
Astana 010011, Al-Farabi Kazakh national
university, Almaty 050040, Kazakhstan}
\email{dauletkaznu@gmail.com}

\address{Serik A. Jumabayev \newline
Department of Economic Management,
Academy of administration under the President
of the Republic of Kazakhstan, Astana 010011, Kazakhstan}
\email{ser\_jum@inbox.ru} 

\address{Almir A. Aniyarov \newline
Department of Information Systems,
S. Seifullin Kazakh agrotechnical university,
Astana 010011, Kazakhstan}
\email{aniyarov.a@gmail.com}

\dedicatory{Communicated by Ira Herbst}

\thanks{Submitted November 24, 2016. Published January 30, 2017.}
\subjclass[2010]{35R30, 47A75}
\keywords{Natural frequencies; rod; localized masses}

\begin{abstract}
 In this article we try to solve the inverse problem of determining
 the coefficients of stiffness of the intermediates on the springs on
 the rod from the two known natural frequencies. We find sufficient
 conditions for the existence of a unique solution to the
 inverse problem of determining the stiffness on the intermediates of
 the springs of non-terminal points of the rod from the two known natural
 frequencies. It was shown that for the determination of the coefficients
 of stiffness of the springs played an essential role in the geometric
 symmetry of the arrangement of the spring relative to the rod center.
\end{abstract}

\maketitle
\numberwithin{equation}{section}
\newtheorem{theorem}{Theorem}[section]
\newtheorem{lemma}[theorem]{Lemma}
\newtheorem{example}[theorem]{Example}
\allowdisplaybreaks


\section{Introduction}\label{1}

In various problems of mechanics and physics, many structural elements
can be represented by a set of different rods. Therefore, in some areas
of modern technology the problem of oscillations of rod systems,
including lumped masses or intermediate springs have to be solved.
As an example, one can specify the machine-building equipment,
aerospace engineering \cite{a1,b2,c1,g2,m1,m2,s1,s2}.
(See. Figures  \ref{fig1}, \ref{fig2}).

Since the 1930's, one of the most important practical problems was finding
the upper and lower bounds of eigenvalues of systems with a finite number
of degrees of freedom \cite{b2}. The motion of system with a finite number
of degrees of freedom is based on the methods of Rayleigh-Ritz and Weinstein,
which consist of the successive approximation of an elastic medium by
finite systems.

\begin{figure}[ht]
\begin{center}
 \includegraphics[width=0.7\textwidth]{fig1}
\end{center}
\caption{The rod with intermediate springs}\label{fig1}
\end{figure}

\begin{figure}[ht]
\begin{center}
 \includegraphics[width=0.7\textwidth]{fig2}
\end{center}
\caption{Beam with three lumped mass}\label{fig2}
\end{figure}

In recent years, methods for the analysis of direct and inverse problems
for differential operators of lumped mass and elastic connections were
actively developing \cite{a1,e1,g1,k1,m2,y1,y2}.
These methods are very important, as they provide an opportunity
to develop the technology and to ensure the safety of people.
For the solution of inverse problems, it is very important to have
a unique solution. These issues are also addressed in terms of boundary
controls \cite{n2}. Relation \cite{b1} studied the spectral and oscillation properties
of boundary value problems which describe the movement of partially
enshrined rod the concentrated on the free end of the additional weight.

\section{Problem of transverse vibrations of a rod with intermediate springs}

This paper considers a rod with two intermediate springs with stiffness
coefficients ${c_1}$ and ${c_2}$ ($\text{N}\,{\text{m}^{ - 1}}$).
Units of measurement and reduction of all physical parameters considered
in the paper as standard. In contrast to \cite{a1,k1,m2},
in this paper, we find sufficient conditions for the existence of 
a unique solution to the
inverse problem of determining the stiffness of the springs in the intermediate
non-terminal points of the rod from the known two natural frequencies.
The idea of the proof of sufficient conditions an ideological close to \cite{k1}.
In contrast to \cite{k1} in this paper it is proved 
(Theorem \ref{thm3.1}) that for an
unambiguous determination of the stiffness coefficients of the springs
plays an important role in the geometric symmetry of the arrangement of the
spring relative to the rod center.

Suppose the first spring is located on the left end of the rod at a distance
${a}$, second spring, respectively, at a distance  ${b}$   (fig.3).
As a result, the rod is divided into 3 sections: $ - l/2 < x < a - l/2$,
$a - l/2 < x < b - l/2$, $b - l/2 < x < l/2$.

\begin{figure}[ht]
\begin{center}
 \includegraphics[width=0.7\textwidth]{fig3}
\end{center}
\caption{The rod with intermediate springs}\label{fig3}
\end{figure}

The equation of the free transverse vibrations of a rod of length ${l}$ of
the following
$$
{\rho } A\frac{{{\partial ^2}{ w}(x,t)}}{{\partial {t^2}}}
+ EJ\frac{{{\partial ^4}{ w}(x,t)}}{{\partial {x^4}}} = 0,
$$
where ${ w}(x,t)$ is the transverse movement, (m) when
$ - l/2 < x < l/2$, $t > 0$; ${\rho}$  is the density of material,
($\text{kg} \,{\text{m}^{-3}}$);  $A$ is the cross-sectional area,
($\text{m}^2$);  $E$ is Young's modulus,($\text{N}\,{\text{m}^{-2}}$);
$J$ is the moment of inertia of the cross-sectional area of relative to
the neutral axis, ($\text{m}^2$).

This method is applicable to rods with different types of fixings.
To be specific, we consider only the hinged rod.

The problem of transverse vibrations of a rod of length ${l}$ with the
replacement of ${ w}(x,t) = y(x)\sin(\omega t)$ reduced to the following
 spectral problem:
\begin{gather}
\label{2-1}
EJy^{IV}(x) = {\omega ^2}\rho   Ay(x),\quad
 - l/2 < x < l/2, \; x \ne a - l/2, \; x \ne b - l/2, \\
 \label{2-2}
  [ EJy'''(x)]_{x = a - l /2}   = c_1 y(a - l/2), \\
\label{2-3}
[ y(x)]_{x = a - l/2} = 0, \quad [Ey'(x)]_{x = a - l/2} = 0,\quad
[EJy''(x)]_{x = a - l/2} = 0, \\
  \label{2-4}
[ EJy'''(x)]_{x = b - l/2}  = c_2y(b - l/  2) , \\
  \label{2-5}
[ {y(x)}]_{x = b - l/2} = 0,    \quad [ Ey'(x) ]_{x = b - l/2}= 0,  \quad
 [ EJy''(x) ]_{x = b - l/ 2 } = 0, \\
  \label{2-6}
y(x)|_{x =  -l / 2 } = 0,\quad
EJy''(x)|_{x =  - l/ 2 } = 0, \\
 \label{2-7}
y(x)|_{x = l/ 2}  = 0, \quad EJy''(x) |_{x = l/ 2} = 0,
  \end{gather}
where $[f(x)]_{x = c} = \lim_{\varepsilon \to  + 0}
[ f(c - \varepsilon) - f(c + \varepsilon)]$  it means the jump
of the function at the point $x = c$.

Let  $p^4 = \omega ^2\rho  A / EJ$, where $\omega $ is the frequency parameter,
Hz. Then the equation \eqref{2-1} has the form
\begin{equation}  \label{2-8}
 y^{IV}(x) = p^4y(x),\quad  - l/ 2 < x < l/ 2,\quad x \ne a - l/ 2,
\quad x \ne b - l/ 2.
\end{equation}
 We will need the following lemma.

\begin{lemma} \label{lem2.1}
 The frequencies of  problem \eqref{2-1}--\eqref{2-7} are defined
by the  equation
 \begin{equation}
  \label{2-9}
  \Delta(a,b,l,p,{c_1},{c_2}) = \alpha(a,b,l,p)\varphi \psi
 + \beta(a,l,p)\varphi  + \gamma(b,l,p)\psi + {\Delta _0}(l,p) = 0,
 \end{equation}
where
\begin{gather*}
\begin{aligned}
\alpha(a,b,l,p)
&= \cos (pl)\{ ch(p(2a - l)) + ch(p(2b - l)) - 2ch(pl) \} \\
&\quad +ch(pl)\{\cos (p(2a - l)) + \cos (p(2b - l)) \} \\
&\quad + \sin (pl)\{  - sh(p(2a - l)) + sh(p(2b - l)) + sh(p(2a - 2b + l)) \} \\
&\quad + sh(pl)\{\sin (p(2a - l)) - \sin (p(2a - 2b + l)) - \sin (p(a + b - l)) \} \\
&\quad + 2\cos (p(a + b - l))\{ ch(p(a + b - l)) - ch(p(a - b + l)) \} \\
&\quad + 2\cos (p(a - b + l))\{ ch(p(a - b + l)) - ch(p(a + b - l)) \} \\
&\quad - \cos (p(2b - l))ch(p(2a - l)) - \cos (p(2a - l))ch(p(2b - l)),
\end{aligned}\\
\beta(a,l,p) =  - 4{p^3}\{ sh(pl)( \cos(p(2a - l)) - \cos(pl))
+ \sin(pl)( ch(p(2a - l)) - ch(pl) ) \},
\\
\gamma(b,l,p) =  - 4{p^3}\{ sh(pl)( \cos(p(2b - l)) - \cos(pl))
+ \sin(pl)( ch(p(2b - l)) - ch(pl) ) \},
\\
\Delta _0(l,p) =  - 16{p^6}\sin(pl)sh(pl),\quad
 \varphi  = c_1 /EJ, \quad
\psi = c_2 /EJ.
\end{gather*}
\end{lemma}

 \begin{proof}
 To find the characteristic determinant of \eqref{2-1}--\eqref{2-7},
we must take into account that each of the intervals
$ - l/ 2  < x < a - l/ 2$, $a - l/ 2  < x < b - l/ 2 $,
$b - l/ 2 < x < l/ 2$ it has a differential equation of  fourth order
 with constant coefficients. The clearly written out the general solution
for each equation which has four free coefficients in considered three
intervals from twelve independent variables. From the boundary conditions
\eqref{2-6} and \eqref{2-7} are directly determined four coefficients.
Conditions ``conjugation'' at points  $a - l/ 2$,  $b - l/ 2$   give
a system of eight linear homogeneous equations with eight unknowns.
The determinant of this system is the characteristic determinant of
 problem \eqref{2-1}--\eqref{2-7}.

 Detailing our reasoning. We are looking for the solution of equation \eqref{2-8}
in the  form
\begin{align*}
y(x) &= {B_1}\cos \big(  p(x+\frac{l}{2}) \big)
 + {B_2}\sin \big(  p(x+\frac{l}{2}) \big)  \\
&\quad + {B_3}ch\big(  p(x+\frac{l}{2}) \big)
+ B_4 sh\big(  p(x+\frac{l}{2}) \big)
\end{align*}
by   $ -l/2<x<a-l/2$.
From conditions \eqref{2-6} it follows that ${B_1} = {B_3} = 0$.
We are looking for the solution of  \eqref{2-8} in the  form
\begin{align*}
y(x) &= D_1\cos \big( p(x-\frac{l}{2}) \big) + {D_2}\sin \big( p(x-\frac{l}{2}) \big) \\
&\quad + {D_3}ch\big( p(x-\frac{l}{2}) \big) + {D_4}sh\big( p(x-\frac{l}{2}) \big)
\end{align*}
by  $ b-l/2<x<l/2$.

From  conditions \eqref{2-7} it follows that ${D_1} = {D_3} = 0$.
We are looking for the solution of  \eqref{2-8} in the  form
\begin{align*}
y(x) &= {K_1}\cos \big( p(x-a+\frac{l}{2}) \big)
 + {K_2}\sin \big( p(x-a+\frac{l}{2}) \big)\\
&\quad + {K_3}ch\big( p(x-a+\frac{l}{2}) \big)
 + {K_4}sh\big( p(x-a+\frac{l}{2}) \big)
\end{align*}
by  $ a-l/2<x<b-l/2$.

From  conditions \eqref{2-2}, \eqref{2-3} we have
\begin{gather}
  \label{2-10}
  y(a-\frac{l}{2}-0) - y(a - \frac{l}{2} + 0)
 = B_2\sin (pa) + {B_4}sh(pa) - K_1 - K_3 = 0, \\
  \label{2-11}
y'( {a - \frac{l}{2} - 0} ) - y'( a - \frac{l}{2} + 0 )
= p ( B_2\cos (pa) + B_4ch(pa) - K_2 - K_4 ) = 0, \\
 \label{2-12}
\begin{aligned}
&y''( a - \frac{l}{2} - 0 ) - y''( a - \frac{l}{2} + 0 )\\
&= p^2 ( - B_2\sin (pa) + B_4 sh(pa) + K_1 - K_3 ) = 0,
\end{aligned} \\
 \label{2-13}
\begin{aligned}
& y'''( a - \frac{l}{2} - 0 ) - y'''( {a - \frac{l}{2} + 0} )
- \frac{c_1}{EJ} y( a - \frac{l}{2} - 0 )  \\
&=  - B_2(p^3\cos (pa) + \frac{c_1}{EJ}\sin (pa) ) + B_4(p^3ch(pa)
- \frac{c_1}{EJ}sh(pa) ) \\
 &\quad  + p^3(K_2 - K_4 ) = 0.
\end{aligned}
\end{gather}
From  conditions \eqref{2-4}, \eqref{2-5} we have
\begin{gather}
  \label{2-14}
\begin{aligned}
&y( b - \frac{l}{2} - 0 ) - y( b - \frac{l}{2} + 0 ) \\
&= K_1\cos (p(b - a)) + K_2\sin (p(b - a))    + K_3ch(p(b - a))  \\
&\quad + K_2sh(p(b - a)) - D_2\sin (p(b - l))
- {D_4}sh(p(b - l)) = 0,
\end{aligned}  \\ 
 \label{2-15}
\begin{aligned}
&y'( b - \frac{l}{2} - 0 ) - y'( b - \frac{l}{2} + 0 ) \\
&= p\big( K_1\sin (p(b - a)) + K_2\cos (p(b - a)) \big)     \\
&\quad + p\big(K_3sh(p(b - a)) 
 + K_2ch(p(b - a)) - D_2\cos (p(b - l)) \\
&\quad - D_4ch(p(b - l)) \big) = 0,
\end{aligned} \\ 
\label{2-16}
\begin{aligned}
 &y''( b - \frac{l}{2} - 0 ) - y''( b - \frac{l}{2} + 0 ) \\
&= p^2(  - K_1\cos (p(b - a)) - K_2\sin (p(b - a)) )\\
&\quad   + p^2\big( K_3ch(p(b - a)) 
 + {K_4}sh(p(b - a)) + D_2\sin (p(b - l)) \\
&\quad - D_4sh(p(b - l)) \big) = 0,
\end{aligned}  \\
\label{2-17}
\begin{aligned}
&y'''( b - \frac{l}{2} - 0 ) - y'''( b - \frac{l}{2} + 0 )
 - \frac{c_2}{EJ}y( b-\frac{l}{2}-0 ) \\
&= K_1( p^3\sin (p(b - a)) - \frac{c_2}{EJ}\cos (p(b - a)) )  \\ 
&\quad - K_2(p^3\cos (p(b - a)) + \frac{c_2}{EJ}\sin (p(b - a)) )  \\
&\quad + K_3(p^3sh(p(b - a)) - \frac{c_2}{EJ}ch(p(b - a)) )  \\
&\quad + K_4(p^3ch(p(b - a)) - \frac{c_2}{EJ}sh(p(b - a)) )  \\
&\quad  + p^3( D_2\cos (p(b - l)) - D_4ch(p(b - l)) ) = 0.
\end{aligned}
\end{gather}
Thus, we have obtained a system of eight linear homogeneous equations
 \eqref{2-10}-\eqref{2-17} with 8 unknowns
$B_2,B_4,K_i,D_2,D_4$, $i = \overline {1,4} $.
As is well known from the theory of linear operators that the system
\eqref{2-10}-\eqref{2-17} has nontrivial solutions if and only if
$\Delta(a,b,l,p,c_1,c_2)$, the characteristic determinant,
is equal to zero \cite{n1}. Where
\begin{gather*}
\Delta(a,b,l,p,c_1,c_2 ) = \begin{vmatrix}
B&K_{\rm up} &O\\
O&K_{\rm down} &D
\end{vmatrix},\\
B = \begin{pmatrix}
  \sin (pa) &sh(pa) \\
  \cos (pa) &ch(pa) \\
   - \sin (pa) &sh(pa) \\
   - p^3\cos (pa) - \frac{c_1}{EJ}\sin (pa)
 &p^3ch(pa) - \frac{c_1}{EJ} sh(pa)
\end{pmatrix},\\
O =\begin{pmatrix}
  0&0 \\
  0&0 \\
  0&0 \\
  0&0
\end{pmatrix},\quad 
K_{\rm up} = \begin{pmatrix}
  - 1&0& - 1&0 \\
  0& - 1&0& - 1 \\
  1&0& - 1&0 \\
  0&p^3&0& - p^3
\end{pmatrix}, \\
D = \begin{pmatrix}
 - \sin (p(b - l))& - sh(p(b - l)) \\
 - \cos (p(b - l))& - ch(p(b - l)) \\
 \sin (p(b - l))& - sh(p(b - l)) \\
  p^3\cos (p(b - l))& - p^3ch(p(b - l))
\end{pmatrix}, \\
K_{\rm down} = \begin{pmatrix}
  \cos (p(b - a)) &\sin (p(b - a)) &ch(p(b - a)) & sh(p(b - a)) \\
   - \sin (p(b - a)) &\cos (p(b - a)) &sh(p(b - a)) &ch(p(b - a)) \\
   - \cos (p(b - a)) & - \sin (p(b - a)) &ch(p(b - a)) &sh(p(b - a)) \\
  \frac{1}{p}{{k'}_{d1}}(p(b - a)) &k_{d1}(p(b - a)) &\frac{1}{p} k'_{d2}(p(b - a))
 &k_{d2}(p(b - a))
\end{pmatrix},\\
k_{d1}(p(x - a)) =  - p^3\cos (p(x - a)) - \frac{c_2}{EJ}\sin (p(x - a)),\\
k_{d2}(p(x - a)) = p^3 ch(p(x - a)) - \frac{c_2}{EJ}sh(p(x - a)).
\end{gather*}
We calculate the characteristic determinant 
$\Delta(a,b,l,p,c_1,c_2 )$  and obtain the formula \eqref{2-9}.
The proof is complete.
\end{proof}

In this setting ${\Delta _0}( {l,p} )$ is the characteristic determinant in the 
absence of springs,  
${\Delta _{{c_2}}}( {b,l,p} ) = \gamma( {b,l,p} )\psi + {\Delta _0}( {l,p} )$
is the characteristic determinant in the case of only the second spring, 
${\Delta _{{c_1}}}( {a,l,p} ) = \beta( {a,l,p} )\varphi  + {\Delta _0}( {l,p} )$  
is the characteristic determinant in the case of only the first spring. 
Thus,  to find frequency of oscillation  of the relation \eqref{2-9} 
is first determined value  and then to find 
$\omega  = \sqrt {EJ /(\rho A p^2)}$.

\section{Determination of the magnitude of the stiffness coefficient 
of the springs on the prescribed frequency of oscillation}

Now consider the inverse problem of determining the stiffness coefficients 
of the springs. Knowing all the physical parameters, the location of 
the intermediate springs, as well as the first two natural frequencies 
of transverse vibrations of a rod, it is then required to determine the value 
of intermediate stiffness coefficients of the springs.

Let $p_1,p_2$ be zeros of $\Delta (p): = \Delta( {a,b,l,p,{c_1},{c_2}} )$. 
Then the following two equations hold
\begin{equation}  \label{3-1}
\begin{gathered}
\alpha _1 \psi\varphi  + \beta _1\varphi  + \gamma _1\psi 
 + \Delta _{01} = 0,\\
\alpha _2 \psi\varphi  + \beta _2\varphi  + \gamma _2\psi 
+ \Delta _{02} = 0.
\end{gathered}
\end{equation}
There  $\alpha _i = \alpha (a,b,l,p_i )$, $i = 1,2$.
 Similar designations is valid for $\beta ,\gamma ,\Delta _0$.
The following Pl\"ucker relations
\begin{equation}  \label{3-2}
 N_1N_4 - N_2N_5 + N_3N_6 = 0
\end{equation}
are true for the matrix 
\[
A =\begin{pmatrix}
  \alpha _1 &\beta _1 &\gamma _1 &\Delta _{01} \\
  \alpha _2 &\beta _2 &\gamma _2 &\Delta _{02}
\end{pmatrix},
\]
where 
\begin{gather*}
N_1 = \begin{vmatrix}
\alpha_1 & \beta_1 \\
\alpha_2 & \beta_2
\end{vmatrix}, \quad
N_2 = \begin{vmatrix}
\alpha _1 &\gamma _1\\
\alpha _2 &\gamma _2
\end{vmatrix},\quad
N_3 = \begin{vmatrix}
\alpha _1 &\Delta _{01}\\
\alpha _2 &\Delta _{02}
\end{vmatrix}, \\
N_4 = \begin{vmatrix}
\gamma _1 &\Delta _{01}\\
\gamma _2 &\Delta _{02}
\end{vmatrix}, \quad
N_5 = \begin{vmatrix}
\beta _1 &\Delta _{01}\\
\beta _2 &\Delta _{02}
\end{vmatrix}, \quad
N_6 = \begin{vmatrix}
\beta _1 &\gamma _1\\
\beta _2 &\gamma _2
\end{vmatrix}.
\end{gather*}
The system of nonlinear equations \eqref{3-1} is reduced to a system of
 quadratic equations
\begin{equation}  \label{3-3}
\begin{gathered}
N_1\varphi ^2 + (N_3 - N_6)\varphi  + N_4 = 0,\\
N_2\psi ^2 + (N_3 + N_6)\psi  + N_5 = 0.
\end{gathered} 
\end{equation}
Note that between $\varphi$ and $\psi $ the linear relationship also holds
\begin{equation}   \label{3-4}
N_1\varphi  + N_2\psi  + N_3 = 0.
\end{equation}
The equality $D_\varphi - D_\psi = 0$ is right for discriminants 
$D_\varphi ,D_\psi $ of system of quadratic equations because 
$D_\varphi = (N_3 - N_6)^2- 4N_1N_4$, 
$D_\psi = (N_3 + N_6)^2 - 4N_2N_5$. Let $D_\varphi = D_\psi \ge 0$. 
Then the first equation in \eqref{3-3} has real roots. 
In order for quadratic equation has only one positive root is sufficient 
that 
\begin{equation}   \label{3-5}
\operatorname{sgn}({N_1})\operatorname{sgn}({N_4}) < 0.
\end{equation}
$\psi$ can be found from \eqref{3-4} selecting positive $\varphi$.  
In this case, there is a positive value $\psi$ from the following inequality
 \begin{equation}  \label{3-6}
 - 2N_2N_3 + N_2(N_3 - N_6) - N_2\sqrt {(N_3 - N_6)^2  - 4N_1N_4}  \ge 0.
  \end{equation}
The result is the solution of the inverse problem of determining the
 pair $\varphi$ and $\psi$ two natural frequencies 
- only, when ${D_\varphi } = {D_\psi } \ge 0$ and the conditions 
\eqref{3-5} and \eqref{3-6} met.

For examples \ref{examp3.1} and \ref{examp3.2} experimental model consisting of a steel 
rod having a radius of 0.01 m and a length of 6 m which is hinged 
- simply supported at the ends. Then  
$EJ = 1649.34\,\rm{N\,m}^2$,  ${\rho}  = 7800\,\text{kg\,m}^{- 3}$,
 $A = 3.14 \cdot {10^{- 4}}\,\text{m}^2$.


\begin{table}[ht]
\caption{Determination of the magnitude of the stiffness coefficient of the springs}
\label{table1}
\begin{tabular}{|c|c|c|c|c|c|} \hline
 $\omega _1$ & $\omega _2$ & $a$ & $b$ & $c _1$ & $c _2$\\ \hline
7.17 & 28.48 & 1 & 2 & 10.000 & 5.000\\ \hline
7.17 & 28.48 & 1 & 2 & -10528.221 & 32489.777\\ \hline
\end{tabular}
\end{table}

 \begin{example} \label{examp3.1} \rm
Table \ref{table1} shows that magnitude of the coefficient stiffness of 
the first spring is equal $10\,\text{N}\text{m}^{-1}$, and the 
coefficient stiffness of the second spring is equal 
$5\,\text{N}\text{m}^{-1}$ and they are defined uniquely by the following 
two values $\omega _1 = 7.17\,{\text{Hz}}$ and 
${\omega _2} = 28.48\,{\text{Hz}}$  (a negative value does not satisfy 
the physical meaning of the problem). Notice that to calculate the natural 
frequencies of the problem \eqref{2-1} - \eqref{2-7} take the accuracy 
of $\varepsilon  = {10^{ - 6}}$. For  $\varepsilon $, it is means that 
 $| {\Delta( {a,b,l,{\omega _i},{c_1},{c_2}} )} | < \varepsilon$, $i = 1,\,2$, 
for fixed values $a,b,l,c_1,c_2$.
\end{example}

In example \ref{examp3.1}, $N_1 =  - 0.00004$, $N_2 =  - 0.00001$, 
$N_3 = 0.0005$, ${N_4} = 4.54142$, $N_5 =  - 2.27322$, 
${N_6} = 0.45421$. Then
$$
N_1N_4 - N_2N_5 + N_3N_6 = 0,\quad \operatorname{sgn}({N_1})
\operatorname{sgn}({N_4}) < 0.
$$
Also
$$ 
- 2N_1N_3 + N_1(N_3 + N_6) - N_1\sqrt {(N_3 + N_6)^2 - 4N_2N_5}
  = 0.00001> 0.
$$
Consequently, the value of the stiffness coefficients of intermediate 
springs are uniquely determined.


\begin{table}[ht]
\caption{Determination of the magnitude of the stiffness coefficient of the springs}
\label{table2}
\begin{tabular}{|c|c|c|c|c|c|c|} \hline
 $\omega _1$ & $\omega _2$ & $a$ & $b$ & $c _1$ & $c _2$\\ \hline
7.29 & 28.49 & 2 & 4 & 14.997 & 10.002\\ \hline
7.29 & 28.49 & 2 & 4 & 10.002 & 14.997\\ \hline
\end{tabular}
\end{table}

\begin{example} \label{examp3.2}\rm
Table \ref{table2} shows that the value of the stiffness coefficients of 
the springs are not uniquely defined.
\end{example}

In example \ref{examp3.2}, $N_1 =  - 0.00079$, $N_2 =  - 0.00079$, $N_3 = 0.01988$, 
$N_4 =  - 0.11929$, $N_5 = - 0.11929$, $N_6 = 4 \cdot {10^{- 10}}$.
Then
$$
N_1N_4 - N_2N_5 + N_3N_6 = 0,\quad \operatorname{sgn}({N_1})
\operatorname{sgn}({N_4}) > 0,\quad
\operatorname{sgn}({N_2})\operatorname{sgn}({N_5}) > 0.
$$
Consequently, the value of the stiffness coefficients of the springs are not 
uniquely defined. But here it should be noted another very important moment, 
which gives us the opportunity to consider this situation from the other side.  
This is due to the locations of intermediate springs. 
In this case, we note that the intermediate springs located geometrically 
symmetrically about the center rod, i.e. with the condition
\begin{equation}  \label{3-7}
 a + b = l.
\end{equation}


In Example \ref{examp3.2}, $a = 2$, $b = 4$, $l = 6$ i.e. with the condition of 
geometric symmetry of fixing springs relative to the rod center \eqref{3-7}.
Note that analytical frequency in Tables 1, 2 calculated using the Maple. 
The following theorem is the main result of the paper.

\begin{theorem}\label{thm3.1}
If intermediate springs located geometrically symmetrically about the center 
rod then the quadratic equations of \eqref{3-3} have a pair of identical roots 
accurate to their permutations.
\end{theorem}

\begin{proof}
In view of  condition \eqref{3-7} the geometric symmetry of fixing springs 
relative to the rod center, we have
$$
\beta ( {l - b,l,p} ) = \gamma ( {b,l,p} ).
$$
Then  $N_6 = 0$. It remains to show that $N_1 = N_2$ and $N_4 = N_5$. 
Since the first columns $N_1$ and $N_2$ is equal and 
$\beta ( l - b,l,p_i ) = \gamma ( b,l,{p_i} )$, $i = 1,2$, it follows that
$N_1 = N_2$. Since the second columns ${N_4}$ and $N_5$ is equal and
 $\beta (l - b,l,p_i ) = \gamma ( b,l,{p_i} )$, $i = 1,2$ then 
$N_4 = N_5$. The proof is complete.
\end{proof}

In practice, intervals changes are already known for values of the 
stiffness of the first and second springs for a particular design. 
The physical meaning is that the problem of the stiffness coefficients 
can be only decrease. And this fact makes it possible to logically uniquely 
define the values of coefficients of stiffness of the first and second spring, 
in the case even where they are located symmetrically relative to the center rod.
Such problems arise when using the methods of vibration diagnostics relating 
to the methods of non-destructive testing.

\subsection*{Conclusions}
This article solved the inverse problem of determining the stiffness coefficients
 of the springs on the intermediate rod from the known the two natural frequencies. 
It is shown that in the general case this inverse problem has two solutions. 
We found sufficient conditions the inverse problem for the existence
 of a unique solution of determining the stiffness of the springs in the 
intermediate non-terminal points of the rod from the known the two eigen frequencies. 
It is proved that for the unique determination of the stiffness coefficients 
of springs geometric symmetry of the location of the spring relative to the
 center rod played an important role.

\subsection*{Acknowledgments}
The authors would like to thank Professor B. E. Kanguzhin for proposing the problem
 and for his valuable comments.
This work was financially supported by the Ministry of Education and 
Science of the Republic of Kazakhstan (project 2217 / GF3 MES of  RK).


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\end{document}