To study the spectrum of the operator \(L_{G}\) at first we calculate the kernel of the resolvent of the operator \((L_{G}-\lambda^{2} E)\). Note that every solution \(\Psi_{k} ( {x_{k} ,\lambda } ) \) of the problem on the edge \(N_{k}=[0,\infty )\), \(k=1,2,3\), is a linear combination of the functions \(y_{kk} (x_{k} ,\lambda )\), \(y_{jk} (x _{k} ,\lambda )\), \(j \ne k\), \(j,k=1,2,3\), and can be written in the form
$$\Psi_{k} ( {x_{k} ,\lambda } ) = C_{0j}^{ ( k ) }(x _{k}) y_{kk} ( {x_{k} ,\lambda } ) + C_{1j}^{ ( k ) }(x _{k}) y_{jk} ( {x_{k} ,\lambda } ) ,\quad j \ne k,j,k = 1,2,3, $$
where \(C_{0j}^{ ( k ) }(x_{k})\) and \(C_{1j}^{ ( k ) }(x _{k})\) are such that conditions (8)-(9) hold for \(\Psi_{k} ( {x_{k} ,\lambda } ) \).
We will construct the resolvent of the operator \(L_{G}\) for \(\operatorname{Im} \lambda >0\).
To this aim, we solve the problem
$$\begin{aligned} &{-} y_{k} ^{\prime \prime }(x_{k},\lambda ) +2 \lambda p_{k} (x_{k} )y _{k}(x_{k}, \lambda ) +q_{k} (x_{k} )y_{k}(x_{k}, \lambda )= \lambda ^{2} y_{k} (x_{k},\lambda )+ \varphi_{k}(x_{k}) , \\ &\quad k = 1,2,3, \end{aligned}$$
(19)
in the space \(L_{2}[0_{k},\infty )\). Here \(\varphi_{k}(x_{k})\) is an arbitrary function belonging to the space \(L_{2}[0_{k},\infty )\), \(k = 1,2,3\).
By taking into account the relation
$$W\bigl[y_{kk} ( {x_{k} ,\lambda } ) ,y_{jk} ( {x_{k} , \lambda } ) \bigr] = 2i\lambda T_{jk} (\lambda ) $$
to find \(C_{0j}^{ ( k ) }(x_{k})\) and \(C_{1j}^{ ( k ) }(x _{k})\) we have
$$\begin{aligned} &C_{0j}^{ ( k ) } (x_{k} ) = \frac{1}{{2i\lambda T_{jk} ( \lambda )}} \int_{x_{k} }^{\infty} {y_{jk} ( {t_{k} , \lambda } ) \varphi_{k} (t_{k} )} \,dt_{k} + C_{0j}^{ ( k ) }( \infty ), \\ &C_{1j}^{ ( k ) } (x_{k} ) = \frac{1}{{2i\lambda T_{jk} ( \lambda )}} \int_{0_{k} }^{x_{k} } {y_{kk} ( {t_{k} , \lambda } ) \varphi_{k} (t_{k} )} \,dt_{k} + C_{1j}^{ ( k ) }(o _{k}), \end{aligned}$$
where \(x_{k} \in [0_{k},\infty )=N_{k}\) and \(C_{0j}^{ ( k ) }( \infty )\), \(C_{1j}^{ ( k ) }(o_{k})\) are arbitrary numbers.
Then
$$\begin{aligned} \Psi_{k} ( {x_{k} ,\lambda } ) ={}& \frac{1}{{2i\lambda T_{jk} (\lambda )}} \int_{x_{k} }^{\infty} {y_{kk} ( {x_{k} , \lambda } ) y_{jk} ( {t_{k} ,\lambda } ) \varphi_{k} (t _{k} )} \,dt_{k} + C_{0j}^{ ( k ) }(\infty )y_{kk} ( {x_{k} , \lambda } ) \\ {}&+ \frac{1}{{2i\lambda T_{jk} (\lambda )}}\int_{0_{k} }^{x_{k} } {y_{jk} ( {x_{k} ,\lambda } ) y_{kk} ( {t_{k} ,\lambda } ) \varphi_{k} (t_{k} )} \,dt_{k} \\ &{}+ C_{1j}^{ ( k ) }(o_{k}) y_{jk} ( {x_{k} ,\lambda } ) , \quad j \ne k,j,k = 1,2,3. \end{aligned}$$
By virtue of the condition \(\Psi_{k} ( \bullet ,\lambda ) \in L_{2} [0_{k} ,\infty ), y_{kk} (x_{k} ,\lambda ) \notin L_{2} [o_{k} , \infty )\), \(y_{jk} (x_{k} ,\lambda ) \in L_{2} [o_{k} , \infty )\) we find that \(C_{0j}^{ ( k ) }(\infty ) = 0\).
Then
$$\begin{aligned} \Psi_{k} ( {x_{k} ,\lambda } ) = {}& \frac{1}{{2i\lambda T_{jk} (\lambda )}}\biggl[ \int_{x_{k} }^{\infty} {y_{kk} ( {x_{k} , \lambda } ) y_{jk} ( {t_{k} ,\lambda } ) \varphi_{k} (t _{k} )} \,dt_{k} \\ &{}+ \int_{o_{k} }^{x_{k} } {y_{jk} ( {x_{k} ,\lambda } ) y _{kk} ( {t_{k} ,\lambda } ) \varphi_{k} (t_{k} )} \,dt_{k}\biggr] \\ &{} + C_{1j}^{ ( k ) }(o_{k}) y_{jk} ( {x_{k} ,\lambda } ) , \quad j \ne k,j,k = 1,2,3, \end{aligned}$$
(20)
and
$$\begin{aligned} &\Psi_{k} ( {o_{k} ,\lambda } ) = \frac{1}{{2i\lambda T_{jk} (\lambda )}} \int_{0_{k} }^{\infty} {y_{kk} ( {o_{k} , \lambda } ) y_{jk} ( {t_{k} ,\lambda } ) \varphi_{k} (t _{k} )} \,dt_{k} + C_{1j}^{ ( k ) }(o_{k}) y_{jk} ( {o _{k} ,\lambda } ) , \\ &\quad j \ne k,j,k = 1,2,3. \end{aligned}$$
Let us denote
$$\begin{aligned} &I_{k}(\lambda )=\frac{1}{{2i\lambda T_{jk} (\lambda )}} \int _{0_{k} }^{\infty} {y_{jk} ( {t_{k} ,\lambda } ) \varphi_{k} (t_{k} )} \,dt_{k}, \\ &C_{1}^{ ( k ) }(\lambda )= C_{1j}^{ ( k ) }(o_{k})T _{jk}(\lambda ); \end{aligned}$$
then by taking into account (17) we have
$$\Psi_{k} ( {o_{k} ,\lambda } ) =y_{kk} ( {o_{k} , \lambda } ) I_{k}(\lambda )+ C_{1}^{ ( k ) }( \lambda )f _{k}^{+} ( {o_{k},\lambda } ) . $$
For finding the constants \(C_{1}^{ ( k ) }(\lambda )\), \(k=1,2,3\), we will use the boundary conditions (8)-(9) and obtain
$$\begin{aligned} &C_{1}^{1}(\lambda ) f_{1}^{+} (o_{1} ,\lambda ) - C_{1}^{2}(\lambda ) f_{2}^{+} (o_{2} ,\lambda ) = y_{22} (o_{2} ,\lambda )I_{2} ( \lambda ) - y_{11} (o_{1} ,\lambda )I_{1} ( \lambda ) , \\ &C_{1}^{1} (\lambda )f_{1}^{+} (o_{1} ,\lambda ) - C_{1}^{3}(\lambda ) f_{3}^{+} (o_{3} ,\lambda ) = y_{33} (o_{3} ,\lambda )I_{3} ( \lambda ) - y_{11} (o_{1} ,\lambda )I_{1} ( \lambda ) , \\ &C_{1}^{1}(\lambda ) f_{1}^{\prime +} (o_{1} ,\lambda ) + C_{1}^{2} (\lambda )f_{2}^{\prime +} (o_{2} ,\lambda ) + C_{1}^{3 }(\lambda )f_{3}^{\prime +} (o_{3} ,\lambda ) \\ &\quad = - y'_{11} (o_{1} ,\lambda )I_{1} ( \lambda ) - y'_{22} (o _{2} ,\lambda )I_{2} ( \lambda ) - y'_{33} (o_{3} ,\lambda )I_{3} ( \lambda ) . \end{aligned}$$
The system of equations can be written as
$$F(\lambda ) \cdot C(\lambda ) = Y(\lambda ) $$
where for \(f_{k}^{+}=f_{k}^{+}(o_{k},\lambda )\), \(y_{kk}= y_{kk} (o _{k} ,\lambda )\)
$$\begin{aligned} &F(\lambda ) = \begin{pmatrix} {f_{1}^{+} } & { - f_{2}^{+} } & 0 \\ {f_{1}^{+} } & 0 & { - f_{3}^{+} } \\ {f_{1}^{\prime +} } & {f_{2}^{\prime +} } & {f_{3}^{\prime +} } \end{pmatrix}, \\ &C (\lambda )= \begin{pmatrix} {C_{1}^{1} (\lambda )} \\ {C_{1}^{2} (\lambda )} \\ {C_{1}^{3}(\lambda ) } \end{pmatrix}, \end{aligned}$$
and
$$Y(\lambda ) = - \begin{pmatrix} {y_{22} I_{2} ( \lambda ) - y_{11} I_{1} ( \lambda ) } \\ {y_{33} I_{3} ( \lambda ) - y_{11} I_{1} ( \lambda ) } \\ {y'_{11} I_{1} ( \lambda ) + y'_{22} I_{2} ( \lambda ) + y'_{33} I_{3} ( \lambda ) } \end{pmatrix}. $$
From this we find that
$$C(\lambda ) = F^{ - 1} (\lambda ) \cdot Y(\lambda ) $$
with
$$F^{ - 1} (\lambda ) = \frac{ \left({\scriptsize\begin{matrix}{} {f_{2}^{\prime +} f_{3}^{ +} } & { - f_{2}^{+} f_{3}^{\prime +} } & {f_{2}^{ +} f _{3}^{+} } \cr {f_{1}^{\prime +} f_{3}^{+} - f_{1}^{+} f_{3}^{\prime +} } & {f_{1}^{+} f_{3}^{\prime +} } & {f_{1}^{+} f_{3}^{ +} } \cr {f_{1}^{ +} f_{2}^{+} } & {f_{1}^{\prime +} f_{2}^{+ } - f_{1}^{+} f_{2} ^{\prime +} } & { - f_{1}^{+} f_{2}^{+} } \end{matrix}}\right) }{\Delta } $$
and
$$\Delta =\operatorname{det} F (\lambda )=\bigl[f_{1}^{ + } \cdot f_{2}^{+} \cdot f_{3}^{+} \bigr]^{\prime }. $$
By using the last relation we can find the coefficients \(C_{1}^{ ( k ) }(\lambda )\), \(k=1,2,3\), as
$$C_{1}^{k} (\lambda ) = \beta_{1}^{k} ( \lambda ) I_{1} ( \lambda ) + \beta_{2}^{k} ( \lambda ) I_{2} ( \lambda ) + \beta_{3}^{k} ( \lambda ) I_{3} ( \lambda ) . $$
To be specific, suppose \(k=1\), then
$$\begin{aligned} &\beta_{1}^{1} (\lambda ) = \frac{{ - y_{11} f_{2}^{+ \prime } f_{3} ^{+} + y_{11} f_{2}^{+} f_{3}^{+\prime } - y_{11} ^{\prime} f_{2}^{+} f_{3}^{+} }}{{\Delta }}, \\ &\beta_{2}^{1} (\lambda ) = \frac{{y_{22} f_{2}^{+\prime } f_{3}^{+} - y_{22} ^{\prime} f_{2}^{+} f_{3}^{+} }}{{\Delta }}, \\ &\beta_{3}^{1} (\lambda ) = \frac{{y_{33} f_{2}^{+} f_{3}^{+ \prime } - y_{33} ^{\prime} f_{2}^{+} f_{3}^{+} }}{{\Delta }}. \end{aligned}$$
By taking into account (17) we can rewrite equation (20) as
$$\begin{aligned} \Psi_{k} ( {x_{k} ,\lambda } ) ={}& \frac{1}{{2i\lambda }}\biggl[ \int_{x_{k} }^{\infty} {y_{kk} ( {x_{k} ,\lambda } ) f ^{+}_{k} ( {t_{k} ,\lambda } ) \varphi_{k} (t_{k} )} \,dt_{k} + \int_{o_{k} }^{x_{k} } {f^{+}_{k} ( {x_{k} ,\lambda } ) y _{kk} ( {t_{k} ,\lambda } ) \varphi_{k} (t_{k} )} \,dt_{k}\biggr] \\ &{}+ \sum_{j = 1}^{3} {\frac{{\beta_{j}^{k} (\lambda ) }}{{2i\lambda }}} \int_{o_{j} }^{\infty} {f_{j}^{+} (t_{j} ,\lambda )} f_{k} ^{+} (x_{k} ,\lambda )\varphi ( {t_{j} } ) \,dt_{j}, \quad k = 1,2,3. \end{aligned}$$
It is readily seen that the function \(\Psi (x,\lambda ) = (\Psi_{1} (x _{1} ,\lambda ),\Psi_{2} (x_{2} ,\lambda ),\Psi_{3} (x_{3} ,\lambda ))\) where
$$\Psi_{k} (x_{k} ,\lambda ) = \sum _{j = 1}^{3} \int_{o _{j} }^{\infty} G_{kj} (x_{k} ,t_{j} ,\lambda )\varphi ( {t_{j} } ) \,dt_{j} , \quad k = 1,2,3, $$
with
$$ G_{kk} (x_{k} ,t_{k} ,\lambda ) = \frac{1}{{2i\lambda }} \textstyle\begin{cases} {[y_{kk} (x_{k} ,\lambda ) + \beta_{k}^{k} (\lambda ) f_{k}^{+} (x _{k} ,\lambda )]f_{k}^{+} (t_{k} ,\lambda )}, & {o_{k}< x_{k} < t_{k} < \infty ,} \\ {[y_{kk} (t_{k} ,\lambda ) + \beta_{k}^{k} (\lambda ) f_{k}^{+} (t _{k} ,\lambda )]f_{k}^{+} (x_{k} ,\lambda )} ,& {o_{k}< t_{k} < x_{k}< \infty , } \end{cases} $$
(21)
and
$$\begin{aligned} &G_{jk} (x_{k} ,t_{j} ,\lambda ) = \frac{{\beta_{j}^{k} (\lambda )}}{ {2i\lambda }}f_{k}^{+} (x_{k} ,\lambda )f_{j}^{+} (t_{j} ,\lambda ),\quad o_{k}< x_{k}< \infty ,o_{j}< t_{j}< \infty , j \ne k, j,k = 1,2,3, \end{aligned}$$
(22)
are sufficiently smooth and satisfy the boundary conditions (8) and (9), i.e. they are contained in the domain of the operator \(L_{G}\). Thus, the constructed ‘spectral’ Green’s function
$$G(x,t,\lambda ) = \textstyle\begin{cases} {G_{kk} (x_{k} ,t_{k} ,\lambda )}, \\ {G_{jk} (x_{k} ,t_{j} ,\lambda )}, \end{cases}\displaystyle \quad j \ne k, j,k = 1,2,3, $$
is the kernel of the resolvent \((L_{G} - \lambda^{2} E)^{ - 1} \), which is an integral operator. The poles of the resolvent (poles of the Green’s function) are eigenvalues of the operator \(L_{G}\) and can be found as zeros of the determinants of the matrices that participate in the construction of the Green’s function.
A point \(\lambda_{0} \in \sigma (L_{G})\) where \(\sigma (L_{G})\) is the set of spectrum of the operator \(L_{G}\) we call a spectral singularity of the operator \(L_{G}\), in the sense of Naimark [24], if it is not an isolated eigenvalue of \(L_{G}\), but \(G(x,t,\lambda ) \to \infty \) as \(\lambda \in \rho (\lambda )\) (\(\rho (\lambda )\) is the set of all regular points of the operator \(L_{G}\)) and \(\lambda \to \lambda_{0}\).
Note that self-adjoint operators have no spectral singularities and for non-self-adjoint operators the spectral singularities correspond to resonance states with vanishing spectral width [25].
Thus, the procedure described above makes it possible to obtain explicitly the resolvent and calculate its poles.
So, we proved the following theorem.
Theorem 1
Assume
\(F(\lambda )\)
is nonsingular i.e. \(F^{ - 1} (\lambda )\)
exists, then for any
$$\varphi = \{ \varphi_{1} ,\varphi_{2} , \varphi_{3} \} , \qquad \varphi_{k} \in L_{2}(N_{k}),\quad k=1,2,3, $$
the unique solution
\(\Psi (x,\lambda ) = (\Psi_{1} (x_{1} ,\lambda ), \Psi_{2} (x_{2} ,\lambda ),\Psi_{3} (x_{3} ,\lambda ))\)
of (7), (8)-(9) is given by
$$\Psi_{k} (x_{k} ,\lambda ) = \sum _{j = 1}^{3} { \int_{o _{j} }^{\infty} {G_{kj} (x_{k} ,t_{j} ,\lambda )\varphi ( {t_{j} } ) \,dt_{j} , \quad k = 1,2,3,} } $$
where
\(G_{jk} (x_{k} ,t_{j} ,\lambda ),j,k = 1,2,3\), are determined by (21)-(22).
Theorem 2
The operator
\(L_{G}\)
has no real eigenvalue.
Proof
We recall that equation (7) has fundamental solutions \(f_{k}^{\pm } ( {x_{k}},\lambda ) \) in the case \(\lambda \ne 0, \pm n/2\). Then for the case \(\lambda \ne 0, \pm n/2\) the solution of equation (7) on the edge \(N_{k}\), \(k=1,2,3\), can be written in the form
$$y_{k} (x_{k} ,\lambda ) = C_{1} f_{k}^{+} (x_{k} ,\lambda ) + C_{2} f _{k}^{-} (x_{k} ,\lambda ). $$
So, the solution of equation (7) belonging to \(L_{2}(G)= \bigoplus_{k = 1}^{3} L_{2} [o_{k} ,\infty )\) and satisfying the conditions (8)-(9) is necessarily has \(C_{1}=0\) and \(C_{2}=0\), \(y_{k}(x_{k},\lambda )=0\). That shows that equation (7) has only a trivial solution belonging to \(L_{2}(G)=\bigoplus _{k = 1}^{3} L_{2} [o_{k} ,\infty )\) for \(\lambda \in ( - \infty , + \infty )\), \(\lambda \ne 0, \pm n/2\).
If as linearly independent solutions (15) and (16) of (7) according to \(\lambda = \pm n/2\) or \(\lambda =0\) are taken instead of \(f_{k}^{ \pm } ( {x_{k}},\lambda ) \) then a similar result also will be valid. So we proved that \(L_{G}\) has no real eigenvalue. □
Theorem 3
The eigenvalues of operator
\(L_{G}\)
are finite and coincide with the zeros of the function
\(\Delta (\lambda )\).
Proof
From equation (10) it is easy to see that for \(j=1,2,3\)
$$\begin{aligned} \bigl\vert {f_{j}^{+} (0,\lambda )} \bigr\vert &= 1 + \sum_{n = 1}^{ \infty} {\bigl\vert { V_{n}^{( + j)} } \bigr\vert } + \sum _{n = 1}^{\infty} {\sum_{\alpha = n}^{\infty} {\biggl\vert {\frac{{V_{\alpha n}^{ ( { + j} ) } }}{{n + 2\lambda }}} \biggr\vert } } \\ &< 1 + \sum_{n = 1}^{\infty} {\bigl\vert { V_{n}^{( + j)} } \bigr\vert } + \sum _{n = 1}^{\infty} {\sum_{\alpha = n}^{\infty} {\frac{{\vert {V_{\alpha n}^{ ( { + j} ) } } \vert }}{{\sqrt{(n + 2\operatorname{Re} \lambda )^{2} + 4\operatorname{Im}^{2} \lambda } }}} } \\ &< 1 + \sum_{n = 1}^{\infty} {\bigl\vert { V_{n}^{( + j)}} \bigr\vert } + \frac{1}{{\vert {\operatorname{Im} \lambda } \vert }} \sum_{n = 1}^{\infty} {\sum _{\alpha = n}^{\infty} {\frac{\alpha }{n}} } \bigl\vert {V_{\alpha n}^{ ( { + j} ) } } \bigr\vert . \end{aligned}$$
Therefore, as \(\vert \lambda \vert \to \infty \) we obtain \(f_{j}^{+} (0,\lambda ) = C_{j} + o(1)\), \(j=1,2,3\). Then for \(\Delta = \det F(\lambda )\) we get the following asymptotic equalities:
$$\Delta (\lambda )=3i\lambda C+o(1), $$
where C is a constant.
This asymptotic equality shows that the eigenvalues of the operator \(L_{G}\) are finite and coincide with the zeros of the function \(\Delta (\lambda )\).
The theorem is proved. □
Theorem 4
The spectrum of the operator
\(L_{G}\)
consists of the continuum spectrum filling the axis
\({ - \infty < \lambda < + \infty }\)
on which there may exist spectral singularities coinciding with the numbers
\(\pm n/2 \), \(n \in N\).
Proof
In order for all numbers \(\lambda \in ( - \infty , + \infty )\) to belong to the continuous spectra it suffices to show that the operator has no real eigenvalue, the domain of \(R_{L_{G}-\lambda^{2}I}\) (the resolvent set) of the operator \(( L_{G}-\lambda^{2} I ) \) is dense in \(L_{2}(G)\), and the range of \(R_{L_{G}-\lambda^{2}I}\) is not equal to \(L_{2}(G)\).
The absence of real spectra of \(L_{G}\) was proved above in Theorem 2.
To show that the domain of \(R_{L_{G}-\lambda^{2}I}\) (the resolvent set) of the operator \(( L_{G}-\lambda^{2} I ) \) is dense in \(L_{2}(G)\) we must prove that the orthogonal complement of the set \(R_{L_{G}-\lambda^{2} E}\) consists of only the zero element.
It is well known that the orthogonal complement of the set \(R_{L_{G}- \lambda^{2} E}\) coincides with the space of the solutions of the equation \(L^{*}_{G}f=\lambda^{2}f\) where the operator \(L^{*}_{G}\) is adjoint to the operator \(L_{G}\).
Let \(\psi_{k} ( x_{k} ) \in L_{2} [ o_{k} ,+\infty ) \), \(\psi_{k} ( x_{k} ) \ne 0\) and
$$ \int_{0_{k} }^{+\infty } \bigl( L_{G}f_{k}- \lambda^{2} f_{k} \bigr) \overline{ \psi_{k} ( x_{k} ) } \,dx_{k}=0,\quad k=1,2,3, $$
(23)
be satisfied for any \(f_{k} ( x_{k} ) \in D ( L_{G} ) \).
From (23) it follows that \(\psi_{k} ( x_{k} ) \in D ( L ^{*}_{G} ) \) and \(\psi_{k} ( x_{k} ) \) are eigenfunctions of the operator \(L^{*}_{G} \) corresponding to the eigenvalues λ.
In fact \(\overline{\psi_{k} ( x_{k} ) }\) is the solution of the equation
$$ -z_{k}''+\bigl[i \lambda p_{k}(x_{k})+q_{k} ( x_{k} ) \bigr]z_{k}=\lambda ^{2}z_{k} $$
(24)
belonging to \(L_{2}(G)\). We found that \(\psi_{k} ( x_{k} ) =0\), since the operator generated by the expression standing at the left hand side of (24) is an operator of type \(L_{G}\).
This contradiction shows that the domain of \(R_{L_{G}-\lambda^{2}I}\) of the operator \(( L_{G}-\lambda^{2} I ) \) is everywhere dense in \(L_{2}( G)\).
Now let us prove that the range of \(R_{L_{G}-\lambda^{2}I}\) is not equal to \(L_{2}(G)\). For this purpose we have to show that there is a function \(f(x)\) from the space \(L_{2}(G)\) for which the equation
has no solution.
Indeed for the compact supported function \(f(x)={ ( f_{1}(x_{1}),f _{2}(x_{2}),f_{3}(x_{3}) ) }\) defined on \(L_{2}(G)\) by
$$f(x) = \textstyle\begin{cases} {\varphi (x)} & \mbox{if } 0 \le x \le a, \\ 0 & \mbox{if }x > a, \end{cases} $$
where
$$\varphi (x) = \bigl(\varphi_{1} (x_{1} ), \varphi_{2} (x_{2} ),\varphi_{3} (x _{3} ) \bigr) $$
is a solution of the following problem:
$$\begin{aligned} & L_{G} \varphi = 0, \\ &\varphi (0,\lambda ) = 1, \qquad \varphi '(0,\lambda ) = 0 \end{aligned}$$
on \(L_{2}(0,\infty )\) equation (25) has no solution. To prove this fact we assume the contrary. Let equation (25) have a solution belonging to \(L_{2}(G)\). Then from Theorem 2 it follows that for \(x>a\) the function \(y(x,\lambda )\) will be a solution of (25) only under the condition
$$y(x,\lambda ) = y'(x,\lambda ) = 0. $$
Then from (25) we obtain
$$\begin{aligned} (f,\bar{\varphi} ) ={}& (L_{G} y,\bar{\varphi} )\\ ={}& \int_{0}^{ + \infty } {L_{1} y_{1} (x_{1} )} \varphi_{1} (x_{1} ) \,dx_{1} + \int _{0}^{ + \infty } {L_{2} y_{2} (x_{2} )} \varphi_{2} (x_{2} )\,dx _{2} + \int_{0}^{ + \infty } {L_{3} y_{3} (x_{3} )} \varphi_{3} (x _{3} )\,dx_{3}\\ ={}& {\bigl[y'_{1} (x_{1} ) \varphi_{1} (x_{1} ) - y_{1} (x _{1} ) \varphi '_{1} (x_{1} )\bigr]} \vert _{x_{1} = 0}^{a} + {\bigl[y'_{2} (x_{2} ) \varphi_{2} (x_{2} ) - y_{2} (x_{2} ) \varphi '_{2} (x_{2} )\bigr]} \vert _{x_{2} = 0}^{a} \\ &{}+ {\bigl[y'_{3} (x_{3} ) \varphi_{3} (x_{3} ) - y_{3} (x_{3} )\varphi '_{3} (x_{3} )\bigr]} \vert _{x_{3} = 0}^{a} + \int_{0}^{a} {y_{1} (x_{1} )} L_{1} \varphi_{1} (x_{1} )\,dx _{1} \\ &{}+ \int_{0}^{a} {y_{2} (x_{2} )} L_{2} \varphi_{2} (x_{2} )\,dx _{2} + \int_{0}^{a} {y_{3} (x_{3} )} L_{3} \varphi_{3} (x_{3} )\,dx _{3} \\ ={}& \bigl[y_{1} (0)\varphi '_{1} (0) + y_{2} (0)\varphi '_{2} (0) + y_{3} (0) \varphi '_{3} (0)\bigr] - \bigl[y'_{1} (0)\varphi_{1} (0) + y'_{2} (0)\varphi _{2} (0) + y'_{3} (0)\varphi_{3} (0)\bigr] \\ ={}& \bigl[y_{1} (0) \cdot 0 + y_{2} (0) \cdot 0 + y_{3} (0) \cdot 0\bigr] - \bigl[y'_{1} (0) + y'_{2} (0) + y'_{3} (0)\bigr] = 0; \end{aligned}$$
on the other hand it is easy to see that
$$(f,\bar{\varphi} ) = \sum_{k = 1}^{3} { \int_{0}^{ + \infty } {\bar{\varphi} _{k} (x_{k} )} \varphi_{k} (x_{k} ) \,dx_{k} } = \sum_{k = 1}^{3} { \int_{0}^{a} {\bigl\vert {\varphi_{k} (x _{k} )} \bigr\vert ^{2} } \,dx_{k} } > 0. $$
This contradiction shows that equation (25) has no solution belonging to \(L_{2}(G)\). So it is proved that the range of \(R_{L_{G}-\lambda ^{2}I}\) is not equal to \(L_{2}(G)\).
From (14) we find that the functions
$$f_{k}^{\pm} (x_{k} ,\lambda ) = \frac{{f_{nk}^{\pm} (x_{k} )}}{{n \pm 2\lambda }} + \Phi_{k}^{\pm} (x_{k} ,\lambda ), $$
where \(\Phi_{k}^{\pm} (x_{k} ,\lambda )\), have no poles at the points \(\mp n/2\), \(n \in N\).
By using this fact it is easy to show that the Green’s function \(G(x,t,\lambda )\) has poles of first order at the points \(\lambda_{0}= \pm \frac{n}{2},n \in N\). Therefore \(\lambda =\pm \frac{n}{2}\), \(n \in N\) is a spectral singularity of the operator \(L_{G}\).
The theorem is proved. □