Inverse problem for dirac system with singularities in interior points

Abstract

We study the non-selfadjoint Dirac system on a finite interval having non-integrable regular singularities in interior points with additional matching conditions at these points. Properties of spectral characteristics are established, and the inverse spectral problem is investigated. We provide a constructive procedure for the solution of the inverse problem, and prove its uniqueness. Moreover, necessary and sufficient conditions for the global solvability of this nonlinear inverse problem are obtained.

1 Introduction

Consider the boundary value problem $L=L(Q_{\omega }(x), Q(x), \alpha ,\beta )$ for the Dirac system on a finite interval with $N$ regular singularities inside the interval:

$$\begin{aligned}&BY'+\Big (Q_\omega (x)+Q(x)\Big )Y=\lambda Y,\quad 0<x<\pi ,\end{aligned}$$

(1)

$$\begin{aligned}&(\cos \alpha ,\sin \alpha )Y(0)=(\cos \beta ,\sin \beta )Y(\pi )=0, \end{aligned}$$

(2)

where

$$\begin{aligned} Y(x)=\left( \begin{array}{l} y_1(x) \\ y_2(x) \end{array}\right) ,\quad B=\left( \begin{array}{ll} 0&{}\quad 1\\ -1&{}\quad 0 \end{array}\right) ,\quad Q(x)=\left( \begin{array}{ll} q_1(x)&{}\quad q_2(x)\\ q_2(x)&{}\quad -q_1(x) \end{array}\right) , \end{aligned}$$

$Q_\omega (x)=Q^{\left\langle {k}\right\rangle }_\omega (x)=\displaystyle \frac{\mu _k}{x-\gamma _k}\left( \begin{array}{ll} \sin 2\eta _k&{}\cos 2\eta _k \\ \cos 2\eta _k&{}\sin 2\eta _k\end{array}\right)$ for $x\in \omega _{k+1/2}\cup \gamma _{k+1/2},\;k=\overline{1,N}$.

Here $0<\gamma _1<\gamma _2<\cdots <\gamma _N<\pi ,\; \omega _p=(\gamma _{p},\,\gamma _{p+1}), \; \gamma _{k+1/2}=(\gamma _{k+1}+\gamma _k)/2,\; k=\overline{1,\,N-1},\; \gamma _{1/2}=\gamma _0=0,\; \gamma _{N+1/2}=\gamma _{N+1}=\pi$, $q_j(x)$ are complex-valued functions, and $\mu _k$ are complex numbers. Let for definiteness, $\alpha ,\;\beta ,\;\eta _k\in [-\pi /2,\pi /2],$ $\hbox {Re}\,\mu _k>0, \;\mu _k+1/2\notin \mathbb {N}.$ Let $q_j(x)$ be absolutely continuous on $[0,\pi ]$ and $|q_j(x)|\prod \nolimits _{k=1}^N |x-\gamma _k|^{-2Re\mu _k}\in L(0,\pi )$. If $Q_{\omega }(x), Q(x), \alpha ,\beta$ satisfy these conditions, we will say that $L\in W.$

In this paper we establish properties of spectral characteristics and investigate the inverse spectral problem of recovering $L$ from the given spectral data. We provide a constructive procedure for the solution of the inverse problem, and prove its uniqueness. Moreover, necessary and sufficient conditions for the global solvability of this nonlinear inverse problem are obtained.

Differential equations with singularities inside the interval play an important role in various areas of mathematics as well as in applications. Moreover, a wide class of differential equations with turning points can be reduced to equations with singularities. For example, such problems appear in electronics for constructing parameters of heterogeneous electronic lines with desirable technical characteristics [1–3]. Boundary value problems with discontinuities in an interior point appear in geophysical models for oscillations of the Earth [4]. Differential equations with turning points arise in various physical and technical problems; see [5] where further references and links to applications can be found. We also note that in different problems of natural sciences we face different kind of matching conditions in singular points.

The case when a singular point lies at the endpoint of the interval was investigated fairly completely for various classes of differential equations in [6–10] and other works. The presence of singularity inside the interval produces essential qualitative modifications in the investigation (see [11]).

A few words on the structure of the paper. In Sect. 2 properties of spectral characteristics are studied. For this we use the results from [12] where special fundamental systems of solutions are constructed with prescribed analytic and asymptotic properties. In Sect. 3 we provide a constructive procedure for the solution of the inverse problem, and prove its uniqueness. Necessary and sufficient conditions for the global solvability of the inverse problem are presented in Sect. 4.

2 Properties of the spectrum

System (1) has non-integrable singularities at the points $\gamma _k$, hence it is necessary to require additional matching conditions for solutions on the intervals $\omega _{k-1}$ and $\omega _k$. We will do it as follows. It was shown in [12] that for $x\in \omega _{k-1}\cup \omega _k$ there exist a fundamental system of solutions $S^{\left\langle {k}\right\rangle }(x,\lambda )=(S_1^{\left\langle {k}\right\rangle }(x,\lambda ),\; S_2^{\left\langle {k}\right\rangle }(x,\lambda ))$ such that

$$\begin{aligned} S^{\left\langle {k}\right\rangle }_1(x,\lambda )\sim (x-\gamma _k)^{-\mu _k} \left( \begin{array}{c}0\\ c_{01}\end{array}\right) ,\ S^{\left\langle {k}\right\rangle }_2(x,\lambda )\sim (x-\gamma _k)^{\mu _k} \left( \begin{array}{c}c_{02} \\ 0\end{array}\right) \quad \hbox {for}\ x\rightarrow \gamma _k. \end{aligned}$$

where $c_{01}c_{02}=1.$ Let $Y(x,\lambda )=a_1(\lambda )S^{\left\langle {k}\right\rangle }_1(x,\lambda )+ a_2(\lambda )S^{\left\langle {k}\right\rangle }_2(x,\lambda )$ be a solution of system (1) for $x\in \omega _{k-1}$. Then we put by definition

$$\begin{aligned} Y(x,\lambda )=a_1(\lambda )S^{\left\langle {k}\right\rangle }_1(x,\lambda )A^{\left\langle {k}\right\rangle }(\lambda )+ a_2(\lambda )S^{\left\langle {k}\right\rangle }_2(x,\lambda )A^{\left\langle {k}\right\rangle }(\lambda ), \end{aligned}$$

for $x\in \omega _k$, where $A^{\left\langle {k}\right\rangle }(\lambda )$ is a fixed given transition matrix for $\gamma _k$. For example, if $A^{\left\langle {k}\right\rangle }(\lambda )=I$ ($I$ is the identity matrix) and $Q(x)$ is analytic at $\gamma _k$, then this continuation of the solution coincides with the analytic continuation through the upper half-plane $Im x>0.$ If $A^{\left\langle {k}\right\rangle }(\lambda )= {\left( \begin{array}{cc}e^{2i\pi \mu _k}&{}0 \\ 0&{}e^{-2i\pi \mu _k}\end{array}\right) }$, then it corresponds to the analytic continuation through the lower half-plane $Im x <0.$

Let $S(x,\lambda )=(S_1(x,\lambda ),\;S_2(x,\lambda ))$ be the fundamental matrix for system (1) with the initial condition $S(0,\lambda )=I$ and with the above mentioned matching conditions. For definiteness, everywhere below $A^{\left\langle {k}\right\rangle }(\lambda )=I,\,k=\overline{1,N}$. The construction of this fundamental matrix can be described as follows. If $x\in \omega _0\cup \omega _1$, then we put $S(x,\lambda )=S^{\left\langle {1}\right\rangle }(x,\lambda )\Big (S^{\left\langle {1}\right\rangle }(0,\lambda )\Big )^{-1}$; moreover, if $x\in \omega _1$, then $S(x,\lambda )=S^{\left\langle {2}\right\rangle }(x,\lambda )C^{\left\langle {1}\right\rangle }(\lambda ).$ Fix $x_1\in \omega _1$. Then $S^{\left\langle {1}\right\rangle }(x_1,\lambda )\Big (S^{\left\langle {1}\right\rangle }(0,\lambda )\Big )^{-1} =S^{\left\langle {2}\right\rangle }(x_1,\lambda )C^{\left\langle {1}\right\rangle }(\lambda )$, i.e.

$$\begin{aligned} S(x,\lambda )=S^{\left\langle {2}\right\rangle }(x,\lambda )\Big (S^{\left\langle {2}\right\rangle }(x_1,\lambda )\Big )^{-1} S^{\left\langle {1}\right\rangle }(x_1,\lambda )\Big (S^{\left\langle {1}\right\rangle }(0,\lambda )\Big )^{-1},\; x\in \omega _1. \end{aligned}$$

Analogously, one gets for $x\in \omega _k$:

$$\begin{aligned} S(x,\lambda )\!&= \!S^{\left\langle {k+1}\right\rangle }(x,\lambda ) \left( \prod \limits _{j=1}^{k}\Big (S^{\left\langle {j+1}\right\rangle }(x_j,\lambda )\Big )^{-1} S^{\left\langle {j}\right\rangle }(x_j,\lambda )\right) (S^{\left\langle {1}\right\rangle }(0,\lambda ))^{-1}, x_j\in \omega _j.\nonumber \\ \end{aligned}$$

(3)

Lemma 1

For $x\in \omega _k$ and $|\lambda (x-\gamma _k)|\ge 1,$

$$\begin{aligned} S(x,\lambda )&= \frac{1}{2i}\left( e^{i\lambda x}\left[ \begin{array}{ll} i&{}\quad -1\\ 1&{}\quad i \end{array}\right] _{\left\langle {k}\right\rangle } + e^{-i\lambda x} \left[ \begin{array}{ll} i&{}\quad 1\\ -1&{}\quad i \end{array}\right] _{\left\langle {k}\right\rangle }\right) \nonumber \\&+\sum \limits _{j=1}^k\sin \pi \mu _je^{-il\lambda (x-2\gamma _j)+2il\eta _j} \left[ \begin{array}{ll} -i&{}\quad l\\ l&{}\quad i \end{array}\right] _{\left\langle {k}\right\rangle }, \end{aligned}$$

$\left[ \Big (a_{ij}\Big )_{i,j=1}^{n,m}\right] _{\left\langle {k}\right\rangle } := \Bigg (a_{ij}+O\Big (|\lambda (x-\gamma _k)|^{-\nu }\Big )\Bigg )_{i,j=1}^{n,m},\,\, \nu =\min \{1,\;2\mathrm{Re}\mu _1,\;2\mathrm{Re}\mu _2,\ldots ,2\mathrm{Re}\mu _N\},\ l=\left\{ \begin{array}{ll} 1,&{}\quad \arg \lambda \in \Pi _{-1}\cup \Pi _1,\\ -1,&{}\quad \arg \lambda \in \Pi _0, \end{array}\right. ,\,\, \displaystyle \Pi _k=\left\{ \lambda \;\Big |\;\arg \lambda \in \Big (\pi \frac{5k-3}{6-2k},\right. \left. \pi \frac{5k+3}{6+2k}\Big ]\right\} ,\,\,k=0,\;\pm 1.$

We prove the lemma by induction. According to [12], the matrix $S^{\left\langle {k}\right\rangle }(x,\lambda )$ can be represented by $S^{\left\langle {k}\right\rangle }(x,\lambda )=E^{\left\langle {k}\right\rangle }(x,\lambda )\beta ^{\left\langle {k}\right\rangle }(\lambda ),$ where $E^{\left\langle {k}\right\rangle }(x,\lambda )$ is the Birkhoff-type fundamental matrix$,$ and $\beta ^{\left\langle {k}\right\rangle }(\lambda )$ are Stockes multipliers.

Let $x\in \omega _0$. Then $S(x,\lambda )=S^{\left\langle {1}\right\rangle }(x,\lambda )(S^{\left\langle {1}\right\rangle }(0,\lambda ))^{-1}$ and (see [12])

$$\begin{aligned} S(x,\lambda )=\frac{1}{[2i]_{\left\langle {1}\right\rangle }} \left( \begin{array}{ll}e^{i\lambda x)}[i]_{\left\langle {1}\right\rangle } + e^{-i\lambda x)}[i]_{\left\langle {1}\right\rangle }&{} \quad e^{i\lambda x)}[-1]_{\left\langle {1}\right\rangle } + e^{-i\lambda x)}[1]_{\left\langle {1}\right\rangle }\\ e^{i\lambda x)}[1]_{\left\langle {1}\right\rangle } + e^{-i\lambda x)}[-1]_{\left\langle {1}\right\rangle } &{}\quad e^{i\lambda x)}[i]_{\left\langle {1}\right\rangle } +e^{-i\lambda x)}[i]_{\left\langle {1}\right\rangle }\end{array}\right) . \end{aligned}$$

Suppose that the assertion of the lemma is true for $x\in \omega _{k-1}$. Let us prove it for $x\in \omega _k$. It follows from (3) that for $x\in \omega _k$,

$$\begin{aligned} S(x,\lambda )=S^{\left\langle {k}\right\rangle }(x,\lambda )(S^{\left\langle {k}\right\rangle }(x_{k-1},\lambda ))^{-1}S(x_{k-1},\lambda ). \end{aligned}$$

(4)

We find the asymptotics for $S^{\left\langle {k}\right\rangle }(x,\lambda )(S^{\left\langle {k}\right\rangle }(x_{k-1},\lambda ))^{-1},$ using the asymptotics from [12]. Denote $l^+=l^{\left\langle {k}\right\rangle },\;m^+=m^{\left\langle {k}\right\rangle }$ for $x>\gamma _k$, and $l^-=l^{\left\langle {k}\right\rangle },\;m^-=m^{\left\langle {k}\right\rangle }$ for $x<\gamma _k$. One has

$$\begin{aligned}&S^{\left\langle {k}\right\rangle }(x,\lambda )B\Big (S^{\left\langle {k}\right\rangle }(x_{k-1},\lambda )\Big )^TB^T\\&\quad =\left( e^{-i\lambda (x-\gamma _k)+i\eta _k}\left[ \begin{array}{ll} -i&{}\quad -i\\ 1&{}\quad 1 \end{array}\right] _{\left\langle {k}\right\rangle }+e^{i\lambda (x-\gamma _k)-i\eta _k}\left[ \begin{array}{ll} i&{}\quad -i\\ 1&{}\quad -1 \end{array}\right] _{\left\langle {k}\right\rangle }H(e^{i\pi \mu _kl^+})\right) \\&\qquad \times H(e^{2i\pi \mu _km^+})H(\lambda ^{-\mu _k})\beta ^{\left\langle {k}\right\rangle }B\beta ^{\left\langle {k}\right\rangle }H(\lambda ^{-\mu _k}) H(e^{2i\pi \mu _km^-})\\&\qquad \times \left( e^{-i \lambda (x_{k-1}-\gamma _k)+i\eta _k}\left[ \begin{array}{ll} - i &{}\quad 1 \\ - i &{}\quad 1 \end{array}\right] _{\left\langle {k}\right\rangle }\right. \\&\qquad +\left. e^{i\lambda (x_{k-1}-\gamma _k)-i\eta _k}H(e^{i\pi \mu _kl^{-}}) \left[ \begin{array}{ll}i &{}\quad 1 \\ - i &{}\quad -1\end{array}\right] _{\left\langle {k}\right\rangle }\right) B^T, \end{aligned}$$

where $H(z)=\left( \begin{array}{ll} z^{-1}&{}0\\ 0&{}z \end{array}\right)$, and $T$ is the sign for the transposition. Since $BH(z)B^T=H(z^{-1})$ and $\beta ^{\left\langle {k}\right\rangle }B\beta ^{\left\langle {k}\right\rangle }B^T =\beta ^{\left\langle {k}\right\rangle }_1\beta ^{\left\langle {k}\right\rangle }_2$, it follows that

$$\begin{aligned}&S^{\left\langle {k}\right\rangle }(x,\lambda )B\Big (S^{\left\langle {k}\right\rangle }(x_{k-1},\lambda )\Big )^TB^T\\&\quad =\beta ^{\left\langle {k}\right\rangle }_1\beta ^{\left\langle {k}\right\rangle }_2\left( e^{-i\lambda (x-\gamma _k)+i\eta _k}\left[ \begin{array}{ll} -i&{}\quad -i\\ 1&{}\quad 1 \end{array}\right] _{\left\langle {k}\right\rangle }H(e^{2i\pi \mu _k(m^+-m^-)})\right. \\&\qquad \left. +\,e^{2i\lambda (x-\gamma _k)-i\eta _k}\left[ \begin{array}{ll}i&{}\quad -i \\ 1&{}\quad -1 \end{array}\right] _{\left\langle {k}\right\rangle }H(e^{i\pi \mu _k(l^++2m^+-2m^-)})\right) \\&\qquad \times \left( e^{-i\lambda (x_{k-1}-\gamma _k)\!+\!i\eta _k}\left[ \begin{array}{ll}1 &{}\quad \!\! i \\ -1&{}\quad \!\! -i \end{array}\right] _{\left\langle {k}\right\rangle }\!+e^{i\lambda (x_{k-1}-\gamma _k)-i\eta _k}H(e^{-i\pi \mu _kl^-}) \left[ \!\begin{array}{ll} - 1 &{}\quad \!\! i \\ -1&{}\quad \!\! i\end{array}\!\right] _{\left\langle {k}\right\rangle }\right) \!. \end{aligned}$$

Taking the relation $\beta ^{\left\langle {k}\right\rangle }_1\beta ^{\left\langle {k}\right\rangle }_2=(4i\cos \pi \mu _k)^{-1}$ into account, we calculate

$$\begin{aligned}&S^{\left\langle {k}\right\rangle }(x,\lambda )\Big (S^{\left\langle {k}\right\rangle }(x_{k-1},\lambda )\Big )^{-1}\nonumber \\&\quad =\frac{1}{4i\cos \pi \mu _k}\; e^{-i\lambda (x+x_{k-1}-2\gamma _k)+2i\eta _k} \left[ 2i\sin \Big (2\pi \mu _k(m^--m^+)\Big ) \left( \begin{array}{ll} -i &{}\quad 1 \\ 1 &{}\quad i \end{array}\right) \right] _{\left\langle {k}\right\rangle }\\&\qquad +\frac{1}{4i\cos \pi \mu _k}\; e^{-i\lambda (x-x_{k-1})} \left[ 2\cos \Big (\pi \mu _k(l^-+2m^--2m^+)\Big ) \left( \begin{array}{ll} i &{}\quad 1 \\ - 1 &{}\quad i \end{array}\right) \right] _{\left\langle {k}\right\rangle }\\&\qquad +\frac{1}{4i\cos \pi \mu _k}\; e^{i\lambda (x-x_{k-1})} \left[ 2\cos \Big (\pi \mu _k(l^++2m^+-2m^-)\Big ) \left( \begin{array}{ll} i&{}\quad -1\\ 1&{}\quad i \end{array}\right) \right] _{\left\langle {k}\right\rangle }\\&\qquad +\frac{1}{4i\cos \pi \mu _k}\; e^{i\lambda (x+x_{k-1}-2\gamma _k)-2i\eta _k}\\&\qquad \times \left[ 2\sin \Big (\pi \mu _k(l^{-}-l^{+}+2m^{-}-2m^{+})\Big ) \left( \begin{array}{ll} -i&{}\quad -1\\ -1&{}\quad i \end{array}\right) \right] _{\left\langle {k}\right\rangle }. \end{aligned}$$

Consider three cases:

• 1. If $\lambda \in \Pi _1$, then $m^-=1,\,m^+=0,\,l^-=-1,\,l^+=1,$ and

  $$\begin{aligned}&S^{\left\langle {k}\right\rangle }(x,\lambda )\Big (S^{\left\langle {k}\right\rangle }(x_{k-1},\lambda )\Big )^{-1}\\&\quad = \sin (\pi \mu _k) e^{-i\lambda (x+x_{k-1}-2\gamma _k)+2i\eta _k} \left[ \begin{array}{ll} -i&{}\quad 1\\ 1&{}\quad i \end{array}\right] _{\left\langle {k}\right\rangle }\!+\! \frac{1}{2i}\; e^{-i\lambda (x-x_{k-1})} \left[ \begin{array}{ll} i&{}\quad 1\\ -1&{}\quad i \end{array} \right] _{\left\langle {k}\right\rangle }\\&\qquad +\frac{1}{2i}\; e^{i\lambda (x-x_{k-1})} \left[ \begin{array}{ll} i &{}\quad - 1 \\ 1 &{}\quad i \end{array}\right] _{\left\langle {k}\right\rangle }+ e^{i\lambda (x+x_{k-1}-2\gamma _k)-2i\eta _k} \left[ \begin{array}{ll} 0 &{}\quad 0 \\ 0 &{} \quad 0 \end{array}\right] _{\left\langle {k}\right\rangle }. \end{aligned}$$

• 2. If $\lambda \in \Pi _{-1}$, then $m^-=0,m^+=-1,l^-=-1,l^+=1,$ and

  $$\begin{aligned}&S^{\left\langle {k}\right\rangle }(x,\lambda )\Big (S^{\left\langle {k}\right\rangle }(x_{k-1},\lambda )\Big )^{-1}\\&\quad = \sin (\pi \mu _k) e^{-i\lambda (x+x_{k-1}-2\gamma _k)+2i\eta _k} \left[ \begin{array}{ll} - i &{}\quad 1 \\ 1 &{}\quad i \end{array}\right] _{\left\langle {k}\right\rangle }\!+\! \frac{1}{2i}\; e^{-i\lambda (x-x_{k-1})} \left[ \begin{array}{ll} i &{}\quad 1 \\ -1 &{}\quad i \end{array}\right] _{\left\langle {k}\right\rangle }\\&\qquad +\frac{1}{2i}\; e^{i\lambda (x-x_{k-1})} \left[ \begin{array}{ll} i &{}\quad -1 \\ 1 &{}\quad i \end{array}\right] _{\left\langle {k}\right\rangle }+ e^{i\lambda (x+x_{k-1}-2\gamma _k)-2i\eta _k} \left[ \begin{array}{ll} 0 &{}\quad 0 \\ 0 &{}\quad 0 \end{array}\right] _{\left\langle {k}\right\rangle }. \end{aligned}$$

• 3. If $\lambda \in \Pi _0$, then $m^-=0,m^+=0,l^-=1,l^+=-1,$ and

  $$\begin{aligned}&S^{\left\langle {k}\right\rangle }(x,\lambda )\Big (S^{\left\langle {k}\right\rangle }(x_{k-1},\lambda )\Big )^{-1} = e^{-i\lambda (x+x_{k-1}-2\gamma _k)+2i\eta _k} \left[ \begin{array}{ll} 0 &{}\quad 0 \\ 0 &{}\quad 0 \end{array}\right] _{\left\langle {k}\right\rangle }\nonumber \\&\quad + \frac{1}{2i}\; e^{-i\lambda (x-x_{k-1})} \left[ \begin{array}{ll} i &{}\quad 1 \\ - 1 &{}\quad i \end{array} \right] _{\left\langle {k}\right\rangle }\\&\quad +\frac{1}{2i}\; e^{i\lambda (x-x_{k-1})} \left[ \begin{array}{ll} i &{}\quad -1 \\ 1 &{}\quad i \end{array}\!\right] _{\left\langle {k}\right\rangle }\!+\! \sin (\pi \mu _k) e^{i\lambda (x+x_{k-1}-2\gamma _k)-2i\eta _k} \left[ \!\begin{array}{ll} -i&{}\quad -1\\ -1&{}\quad i \end{array}\!\right] _{\left\langle {k}\right\rangle }\!. \end{aligned}$$

Since $x_{k-1}<\gamma _k,\,x>\gamma _k$, it follows that $x+x_{k-1}-2\gamma _k= x-x_{k-1}+2(x_{k-1}-\gamma _k)<x-x_{k-1}$, $x+x_{k-1}-2\gamma _k= x_{k-1}-x+2(x-\gamma _k)>x_{k-1}-x$, and the exponentials $e^{\pm i\lambda (x+x_{k-1}-2\gamma _k)}$ grow not faster than $e^{\pm i\lambda (x-x_{k-1})}$. Thus,

$$\begin{aligned}&S^{\left\langle {k}\right\rangle }(x,\lambda )\Big (S^{\left\langle {k}\right\rangle }(x_{k-1},\lambda )\Big )^{-1}= \frac{1}{2i}\; e^{i\lambda (x-x_{k-1})} \left[ \begin{array}{ll} i&{}\quad -1 \\ 1&{}\quad i\end{array}\right] _{\left\langle {k}\right\rangle }\\&\qquad +\frac{1}{2i}\; e^{-i\lambda (x-x_{k-1})} \left[ \begin{array}{ll} i&{}\quad 1\\ -1&{}\quad i \end{array}\right] _{\left\langle {k}\right\rangle }+ \sin (\pi \mu _k) e^{-li\lambda (x+x_{k-1}-2\gamma _k)+2li\eta _k} \left[ \!\begin{array}{ll} -i&{}\quad l\\ l&{}\quad i \end{array}\!\right] _{\left\langle {k}\right\rangle }. \end{aligned}$$

Substituting this asymptotics into (4), we get

$$\begin{aligned} S(x,\lambda )&= \left( \frac{1}{2i}\; e^{i\lambda (x-x_{k-1})} \left[ \begin{array}{ll} i&{}\quad -1\\ 1&{}\quad i \end{array} \right] _{\left\langle {k}\right\rangle }+ \frac{1}{2i}\; e^{-i\lambda (x-x_{k-1})} \left[ \begin{array}{ll} i&{}\quad 1\\ -1&{}\quad i \end{array}\right] _{\left\langle {k}\right\rangle }\right. \\&\left. +\sin (\pi \mu _k) e^{-li\lambda (x+x_{k-1}-2\gamma _k)+2li\eta _k} \left[ \begin{array}{ll} - i &{}\quad l \\ l &{}\quad i \end{array}\right] _{\left\langle {k}\right\rangle } \right) \\&\times \left( \frac{1}{2i}\;e^{i\lambda x_{k-1}}\left[ \begin{array}{ll} i &{}\quad -1 \\ 1 &{}\quad i \end{array}\right] _{\left\langle {k-1}\right\rangle } +\frac{1}{2i}\;e^{-i\lambda x_{k-1}} \left[ \begin{array}{ll} i &{}\quad 1 \\ - 1 &{}\quad i \end{array}\right] _{\left\langle {k-1}\right\rangle }\right. \\&\left. + \sum \limits _{j=1}^{k-1}\sin \pi \mu _je^{-il\lambda (x_{k-1}-2\gamma _j)+2il\eta _j} \left[ \begin{array}{ll} - i &{}\quad l \\ l &{}\quad i \end{array}\right] _{\left\langle {k-1}\right\rangle }\right) . \end{aligned}$$

Since $0<x_{k-1}<x,$ it follows that $0<2x_{k-1}<2x,\; -x<2x_{k-1}-x<x,$ and $e^{\pm i\lambda (2x_{k-1}-x)}$ grow not faster than $e^{\pm i\lambda x}$. Therefore

$$\begin{aligned} S(x,\lambda )&= \frac{1}{2i}\;e^{i\lambda x}\left[ \begin{array}{ll} i &{}\quad -1 \\ 1 &{}\quad i \end{array}\right] _{\left\langle {k}\right\rangle }+ \frac{1}{2i}\;e^{-i\lambda x}\left[ \begin{array}{ll} i &{}\quad 1 \\ - 1 &{}\quad i \end{array}\right] _{\left\langle {k}\right\rangle }\\&+\frac{1}{2i}\sum \limits _{j=1}^{k-1}\sin (\pi \mu _j) e^{i\lambda (x-x_{k-1})-li\lambda (x_{k-1}-2\gamma _j)+2li\eta _j} \left[ (1-l)\left( \begin{array}{ll} 1 &{}\quad -i \\ - i &{}\quad -1 \end{array}\right) \right] _{\left\langle {k}\right\rangle }\\&+\frac{1}{2i}\sum \limits _{j=1}^{k-1}\sin (\pi \mu _j) e^{-i\lambda (x-x_{k-1})-li\lambda (x_{k-1}-2\gamma _j)+2li\eta _j} \left[ (1+l)\left( \begin{array}{ll} 1 &{}\quad i \\ i &{}\quad -1 \end{array}\right) \right] _{\left\langle {k}\right\rangle }\\&+\frac{1}{2i}\sin (\pi \mu _k) e^{i\lambda x_{k-1}-li\lambda (x+x_{k-1}-2\gamma _k)+2li\eta _k} \left[ (1+l)\left( \begin{array}{ll} 1 &{}\quad i \\ i &{}\quad -1 \end{array}\right) \right] _{\left\langle {k}\right\rangle }\\&+\frac{1}{2i}\sin (\pi \mu _k) e^{-i\lambda x_{k-1}-li\lambda (x+x_{k-1}-2\gamma _k)+2li\eta _k} \left[ (1-l)\left( \begin{array}{ll} 1 &{}\quad -i \\ - i &{}\quad -1 \end{array}\right) \right] _{\left\langle {k}\right\rangle }\\&+\sum \limits _{j=1}^{k-1}\sin (\pi \mu _k)\sin (\pi \mu _j) e^{-li\lambda (x+2x_{k-1}-2\gamma _k-2\gamma _j)+4li\eta _j} \left[ \begin{array}{ll} 0 &{}\quad 0 \\ 0 &{}\quad 0 \end{array}\right] _{\left\langle {k}\right\rangle }. \end{aligned}$$

Let $l=-1.$ Then

$$\begin{aligned} S(x,\lambda )&= \frac{1}{2i}\;e^{i\lambda x} \left[ \begin{array}{ll} i &{}\quad -1 \\ 1 &{}\quad i \end{array}\right] _{\left\langle {k}\right\rangle }+ \frac{1}{2i}\;e^{-i\lambda x} \left[ \begin{array}{ll} i &{}\quad 1 \\ - 1 &{}\quad i \end{array}\right] _{\left\langle {k}\right\rangle }\\&+\sum \limits _{j=1}^{k-1}\sin (\pi \mu _j) e^{i\lambda x-2i\lambda \gamma _j-2i\eta _j} \left[ \begin{array}{ll} -i &{}\quad -1 \\ - 1 &{}\quad i \end{array}\right] _{\left\langle {k}\right\rangle }\\&+\sum \limits _{j=1}^{k-1}\sin (\pi \mu _j) e^{-i\lambda x+2i\lambda x_{k-1}-2i\lambda \gamma _j-2i\eta _j} \left[ \begin{array}{ll} 0 &{}\quad 0 \\ 0 &{}\quad 0 \end{array}\right] _{\left\langle {k}\right\rangle }\\&+\, e^{2i\lambda x_{k-1}+i\lambda x-2i\lambda \gamma _k-2i\eta _k} \left[ \begin{array}{ll} 0 &{}\quad 0 \\ 0 &{}\quad 0 \end{array}\right] _{\left\langle {k}\right\rangle }\\&+ \sin (\pi \mu _k) e^{i\lambda x-2i\lambda \gamma _k-2i\eta _k} \left[ \begin{array}{ll} 1 &{}\quad -i \\ - i &{}\quad -1 \end{array}\right] _{\left\langle {k}\right\rangle }\\&+\sum \limits _{j=1}^{k-1}\sin (\pi \mu _k)\sin (\pi \mu _j) e^{i\lambda (x+2x_{k-1}-2\gamma _k-2\gamma _j)-4i\eta _j} \left[ \begin{array}{ll} 0 &{}\quad 0 \\ 0 &{}\quad 0 \end{array}\right] _{\left\langle {k}\right\rangle }. \end{aligned}$$

This yields

$$\begin{aligned} S(x,\lambda )&= \frac{1}{2i}\left( e^{i\lambda x}\left[ \begin{array}{ll} i &{}\quad -1 \\ 1 &{}\quad i \end{array}\right] _{\left\langle {k}\right\rangle } +e^{-i\lambda x}\left[ \begin{array}{ll} i &{}\quad 1 \\ - 1 &{}\quad i \end{array}\right] _{\left\langle {k}\right\rangle }\right) \\&+\sum \limits _{j=1}^k\sin (\pi \mu _j) e^{i\lambda x-2i\lambda \gamma _j-2i\eta _j} \left[ \begin{array}{ll} -i &{}\quad -1 \\ - 1 &{}\quad i \end{array}\right] _{\left\langle {k}\right\rangle }. \end{aligned}$$

The case $l=1$ is treated similarly. Lemma 1 is proved.

The following assertion is proved analogously.

Lemma 2

For $x\in \omega _k$ and $|\lambda (x-\gamma _k)|\ge 1$

$$\begin{aligned} \frac{\partial }{\partial \lambda } S(x,\lambda )&= \frac{x}{2i} \left( e^{i\lambda x}\left[ \begin{array}{ll} -1&{}\quad -i\\ i&{}\quad -1 \end{array}\right] _{\left\langle {k}\right\rangle } +e^{-i\lambda x}\left[ \begin{array}{ll} 1&{}\quad -i\\ i&{}\quad 1 \end{array}\right] _{\left\langle {k}\right\rangle }\right) \\&+\sum \limits _{j=1}^k(x-2\gamma _j)\sin \pi \mu _je^{-il\lambda (x-2\gamma _j)+2il\eta _j} \left[ \begin{array}{ll} -l&{}\quad -i\\ -i&{}\quad l \end{array}\right] _{\left\langle {k}\right\rangle }. \end{aligned}$$

Definition A function $Y(x,\lambda )$ is called the solution of system (1), if there exist constants $C_1(\lambda ),\,C_2(\lambda )$ such that $Y(x,\lambda )=C_1(\lambda )S_1(x,\lambda )+C_2(\lambda )S_2(x,\lambda ),$ $x\in (0,\pi ){\setminus }\bigcup \nolimits _{k=1}^N\{\gamma _k\}.$

We introduce the functions

$$\begin{aligned} \varphi (x,\lambda )&= \Big (\varphi _1(x,\lambda ),\;\varphi _2(x,\lambda )\Big )=S(x,\lambda )V(\alpha ),\nonumber \\ \quad V(\alpha )&= \Big (V_1(\alpha ),V_2(\alpha )\Big )=\left( \begin{array}{ll}\cos \alpha &{}\quad -\sin \alpha \\ \sin \alpha &{}\quad \cos \alpha \end{array}\right) ,\\ \Delta (\lambda )&= \left( \begin{array}{cc}\Delta _{11}(\lambda )&{}\quad \Delta _{12}(\lambda )\\ \Delta _{21}(\lambda )&{}\quad \Delta _{22}(\lambda )\end{array}\right) =V^T(\beta )S(\pi ,\lambda )V(\alpha ),\\ \psi (x,\lambda )&= \Big (\psi _1(x,\lambda ),\;\psi _2(x,\lambda )\Big )=S(x,\lambda )S^{-1}(\pi ,\lambda )V(\beta ). \end{aligned}$$

Clearly, $\varphi (x,\lambda ),\,\psi (x,\lambda )$ are fundamental matrices for system (1). Denote ${\left\langle {Y,Z}\right\rangle }:=Y^TBZ.$ If $Y(x,\lambda ), Z(x,\lambda )$ are solutions of system (1), then ${\left\langle {Y(x,\lambda ),Z(x,\lambda )}\right\rangle }:=\det \{Y(x,\lambda ),Z(x,\lambda )\}$ is their Wronskian. Obviously,

$$\begin{aligned} {\left\langle {\psi _2(x,\lambda ),\varphi _2(x,\lambda )}\right\rangle }=-\Delta _{12}(\lambda ). \end{aligned}$$

(5)

A number $\lambda _0$ is called an eigenvalue of problem (1)–(2), if there exist constants $A_1,A_2$ ($|A_1|+|A_2|>0$) such that the function $A_1S_1(x,\lambda _0)+A_2S_2(x,\lambda _0)$ satisfies the boundary conditions (2).

Lemma 3

Zeros of $\Delta _{12}(\lambda )$ coincide with the eigenvalues of the boundary value problem (1)–(2). If $\lambda _0$ is an eigenvalue$,$ then $\varphi (x,\lambda _0)$ and $\psi (x,\lambda _0)$ are eigenfunctions$,$ and $\psi (x,\lambda _0) = b_0\varphi (x,\lambda _0)$.

Proof

1. Let $\lambda _0$ be a zero of $\Delta _{12}(\lambda ),$ i.e. $V_1^T(\beta )S(\pi ,\lambda _0)V_2(\alpha )=0.$ Therefore, $\varphi _2(x,\lambda _0)=S(x,\lambda _0)V_2(\alpha )$ is an eigenfunction, and $\lambda _0$ is an eigenvalue. It follows from (5) that $\varphi _2(x,\lambda _0)$ and $\psi _2(x,\lambda _0)$ are linear dependent.

2. Let $\lambda _0$ be an eigenvalue, and let $Y_0(x)$ be the corresponding eigenfunction. Since $\varphi _1(x,\lambda ),\varphi _2(x,\lambda )$ form a fundamental system of solutions, it follows that $Y_0(x)=D_1\varphi _1(x,\lambda _0)+D_2\varphi _2(x,\lambda _0).$ Substituting this relation into the first boundary condition, we obtain $D_1V_1^T(\alpha )V_1(\alpha )+D_2V_1^T(\alpha )V_2(\alpha )=0,$ hence $D_1=0$. Using the second boundary condition, we find $D_2V_1^T(\beta )\varphi _2(\pi ,\lambda _0)=0.$ Since $Y_0(x)\not \equiv 0,$ one has $D_2\ne 0,$ i.e. $V_1^T(\beta )\varphi _2(\pi ,\lambda _0)=0.$ Lemma 3 is proved.

We note that the functions $\Delta _{jk}(\lambda ),\;j,k=1,2,$ are the characteristic functions for the boundary value problems $L_{jk}$ foe system (1) with boundary conditions. $V^T_{3-k}(\alpha )Y(0)=V^T_j(\beta )Y(\pi )=0.$ Denote

$$\begin{aligned} \Phi (x,\lambda )&= \Big (\Phi _1(x,\lambda ),\,\Phi _2(x,\lambda )\Big ),\quad \Phi _1(x,\lambda )=-\frac{1}{\Delta _{12}(\lambda )}\psi _2(x,\lambda ),\\ \Phi _2(x,\lambda )&= \varphi _2(x,\lambda ). \end{aligned}$$

It follows from (5) that $\det \Phi (x,\lambda )\equiv 1.$ The functions $\Phi _1(x,\lambda ),\,\Phi _2(x,\lambda )$ are called the Weyl solutions, and the matrix ${\mathfrak {M}}(\lambda ):=V^T_2(\lambda )\Phi _1(0,\lambda )$ is called the Weyl matrix for the problem (1)–(2).

Lemma 4

The following relations hold

$$\begin{aligned} \Phi (x,\lambda )=\varphi (x,\lambda )M(\lambda ),\quad \hbox {where } M(\lambda )=\left( \begin{array}{cc} 1 &{}\quad 0 \\ {\mathfrak {M}}(\lambda )&{}\quad 1\end{array}\right) ,\ {\mathfrak {M}}(\lambda )=-\frac{\Delta _{11}(\lambda )}{\Delta _{12}(\lambda )}. \end{aligned}$$

Only formula for $\Phi _1(x,\lambda )$ is needed to be proved. Let $\Phi _1(x,\lambda )=D_1(\lambda )\varphi _1(x,\lambda )+D_2(\lambda )\varphi _2(x,\lambda ).$ Then

$$\begin{aligned} -(\Delta _{12}(\lambda ))^{-1}\psi _2(0,\lambda )=D_1(\lambda )V_1(\alpha )+D_2(\lambda )V_2(\alpha ). \end{aligned}$$

(6)

Multiplying (6) by $V_1^T(\alpha ),$ we infer $-(\Delta _{12}(\lambda ))^{-1}V_1^T(\alpha )\psi _2(0,\lambda )=D_1(\lambda ).$ Since

$$\begin{aligned} V_1^T(\alpha )\psi _2(0,\lambda )=V_1^T(\alpha )B^TS^T(\pi ,\lambda )BV_2(\beta ), \end{aligned}$$

it follows that $V_1^T(\alpha )\psi _2(0,\lambda )=-V_1^T(\beta )S(\pi ,\lambda )V_2(\alpha ) =-\Delta _{12}(\lambda ),$ i.e. $D_1(\lambda )=1$. Multiplying (6) by $V_2^T(\alpha ),$ we find $D_2(\lambda )=V_2^T(\alpha )\Phi _1(0,\lambda )={\mathfrak {M}}(\lambda ).$ Taking the relation $V_2^T(\alpha )\psi _2(0,\lambda )=\Delta _{11}(\lambda )$ into account, we obtain the assertion of the lemma.

Thus, ${\mathfrak {M}}(\lambda )$ is a meromorphic function; its poles coincide with the eigenvalues of $L,$ and its zeros coincide with the eigenvalues of $L_{11}$.

Lemma 5

For $x\in \omega _k$ and $|\lambda (x-\gamma _k)|\ge 1,\,|\lambda (x-\gamma _{k+1})|\ge 1,$ one has

$$\begin{aligned} \varphi (x,\lambda )&= \frac{1}{2i}\left( e^{i\lambda x+i\alpha }\left[ \begin{array}{ll} i &{}\quad -1 \\ 1 &{}\quad i \end{array}\right] _{\left\langle {k}\right\rangle }+e^{-i\lambda x-i\alpha } \left[ \begin{array}{ll} i &{}\quad 1 \\ - 1 &{}\quad i \end{array}\right] _{\left\langle {k}\right\rangle }\right) \nonumber \\&+\sum _{j=1}^k \sin \pi \mu _je^{-il\lambda (x-2\gamma _j)+2il\eta _j-il\alpha } \left[ \begin{array}{ll} -i &{}\quad l \\ l &{}\quad i \end{array}\right] _{\left\langle {k}\right\rangle },\end{aligned}$$

(7)

$$\begin{aligned} \psi (x,\lambda )&= -\frac{1}{2i}\left( e^{i\lambda (\pi -x)-i\beta } \left[ \begin{array}{ll} i &{}\quad 1 \\ - 1 &{}\quad i \end{array}\right] _{\left\langle {k+1}\right\rangle } + e^{-i\lambda (\pi -x)+i\beta } \left[ \begin{array}{ll} i &{}\quad -1 \\ 1 &{}\quad i \end{array}\right] _{\left\langle {k+1}\right\rangle }\right) \nonumber \\&-\sum _{j=k+1}^N \sin \pi \mu _je^{-il\lambda (x+\pi -2\gamma _j)+2il\eta _j-il\beta } \left[ \begin{array}{ll} -i &{}\quad -l \\ - l &{}\quad i \end{array}\right] _{\left\langle {k+1}\right\rangle }. \end{aligned}$$

(8)

Indeed, since $\varphi (x,\lambda )=S(x,\lambda )V(\alpha )$, relation (7) follows from Lemma 1. To prove (8) we make the substitution $x\rightarrow \pi -x$ and repeat the arguments.

Taking the relation $\Delta (\lambda )=V(\beta )\varphi (\pi ,\lambda )$ into account we arrive at the following assertion.

Corollary 1

For the characteristic function $\Delta _{12}(\lambda ),$ the following asymptotics holds

$$\begin{aligned} \Delta _{12}(\lambda )&= \frac{1}{2i}e^{-i(\lambda \pi +\alpha -\beta )}[1]- \frac{1}{2i}e^{i(\lambda \pi +\alpha -\beta )}[1]\nonumber \\&+\sum _{j=1}^N \sin \pi \mu _j e^{-il\lambda (\pi -2\gamma _j)+il(2\eta _j-\alpha -\beta )}[l], \end{aligned}$$

(9)

where $\left[ \left( a_{kj}\right) _{k,j=1}^{n,m}\right] := \Big (a_{kj}+O\left( |\lambda |^{-\nu }\right) \Big )_{k,j=1}^{n,m}$ for $|\lambda |\rightarrow \infty$.

By the well-known method (see for example [13–15]) one obtains the following properties:

• 1. $\Delta _{12}(\lambda )=O(e^{\pi |Im\lambda |})$.

• 2. All eigenvalues $\lambda _{k},\;k\in \mathbb {Z}$ of the problem (1)–(2) lie in the strip $|Im\lambda |\le h.$

• 3. Let $N_a$ be a number of eigenvalues in the rectangle $\Big \{\lambda \,|\,Re\lambda \in [a,a+1),\;|Im\lambda |\le h\Big \}.$ Then $N_a$ is uniformly bounded.

• 4. Denote $G_\delta =\{\lambda \,:\,|\lambda -\lambda _{k}|\ge \delta \;\forall \, k\}$. Then $|\Delta _{12}(\lambda )|\ge C_\delta e^{\pi |Im\lambda |}$ for $\lambda \in G_{\delta }.$

• 5. For sufficiently small $\delta ,$ there exists a sequence $R_n\rightarrow \infty$ such that the circles $\Gamma _n=\Big \{\lambda \,:\,|\lambda | =R_n\Big \}$ lie in $G_\delta .$

• 6. Let $\left\{ \lambda ^0_k\right\} _{k=-\infty }^\infty$ be zeros of the function

  $$\begin{aligned} \Delta ^0_{12}(\lambda )&= \frac{1}{2i}e^{-i(\lambda \pi +\alpha -\beta )}- \frac{1}{2i}e^{i(\lambda \pi +\alpha -\beta )}\nonumber \\&+\, l\sum _{j=1}^N \sin \pi \mu _j e^{-il\lambda (\pi -2\gamma _j)+il(2\eta _j-\alpha -\beta )}. \end{aligned}$$

  (10)

  Then $\lambda _{k}=\lambda ^0_k+O(|\lambda ^0_k|^{-\nu }).$

For simplicity, we confine ourselves to the case when all eigenvalues of $L$ are simple, i.e. the function $\Delta _{12}(\lambda )$ has only simple zeros. In particular, it is always true for the self-adjoint case. Denote $a_k:=\mathop {\text {Res}}\limits \nolimits _{\lambda =\lambda _{k}}\,{\mathfrak {M}}(\lambda ).$ The data $\{a_{k},\;\lambda _{k}\}_{k=-\infty }^{+\infty }$ are called the spectral data for $L.$ The inverse problem is formulated as follows.

Inverse Problem 1

Given $\{a_k,\lambda _{k}\}_{k=-\infty }^{+\infty },$ construct $L,$ i.e. $Q(x),Q_\omega (x),\alpha ,\beta .$

In Sects. 3 and 4 we give an algorithm for the global solution of this nonlinear inverse problem and provide necessary and sufficient conditions for its solvability.

Lemma 6

Let ${\mathfrak {M}}^0(\lambda )$ be the Weyl function for the problem $L^0$ of the form (1)–(2) but with the zero potential $Q(x)\equiv 0.$ Then

$$\begin{aligned} {\mathfrak {M}}(\lambda )={\mathfrak {M}}^0(\lambda )+\sum _{k=-\infty }^{+\infty }\left( \frac{a_k}{\lambda -\lambda _{k}}- \frac{a^0_k}{\lambda -\lambda ^0_k}\right) ,\quad \displaystyle \sum _{k=-\infty }^{+\infty }= \displaystyle \lim _{n\rightarrow \infty }\displaystyle \sum _{|\lambda _{k}|<R_n,\,|\lambda ^0_k|<R_n}. \end{aligned}$$

Proof

Consider the integral $J_n(\lambda )=\frac{1}{2\pi i}\int _{\Gamma _n} \frac{{\mathfrak {M}}(\xi )-{\mathfrak {M}}^0(\xi )}{\xi -\lambda }d\xi$, $\lambda \in \hbox {int}\,\Gamma _n$. Using Lemmas 4–5 and Corollary 1, we obtain ${\mathfrak {M}}(\xi )-{\mathfrak {M}}^0(\xi )=O(|\xi |^{-\nu })$ for $\xi \in G_\delta ,$ and consequently, $J_n(\lambda )\rightarrow 0$ as $n\rightarrow \infty .$ On the other hand, by residue’s theorem,

$$\begin{aligned} J_n(\lambda )=\mathop {\text {Res}}\limits _{\xi =\lambda }\frac{{\mathfrak {M}}(\xi )-{\mathfrak {M}}^0(\xi )}{\xi -\lambda }+ \sum _{|\lambda |<R_n,\,|\lambda ^{0}_{k}|<R_n}\left( \mathop {\text {Res}}\limits _{\xi ={\lambda }_{k}} \frac{{\mathfrak {M}}(\xi )}{\xi -\lambda }- \mathop {\text {Res}}\limits _{\xi =\lambda ^{0}_{k}}\frac{{\mathfrak {M}}^0(\xi )}{\xi -\lambda }\right) , \end{aligned}$$

hence

$$\begin{aligned} J_n(\lambda )={\mathfrak {M}}(\lambda )-{\mathfrak {M}}^0(\lambda )+ \sum _{|\lambda |<R_n,\,|\lambda ^0_k|<R_n}\left( \frac{a_k}{\lambda _{k}-\lambda }- \frac{a^0_k}{\lambda ^0_k-\lambda }\right) . \end{aligned}$$

If $n\rightarrow \infty ,$ we arrive at the assertion of the lemma.

Together with $L$ we consider a boundary value problem $\widetilde{L}$ of the same form (1)–(2) but with different $\widetilde{Q}(x),\;\widetilde{Q}_\omega (x)$, $\widetilde{\alpha },\;\widetilde{\beta }$. We agree that if a certain symbol $v$ denotes an object related to $L,$ then $\widetilde{v}$ will denote an analogous object related to $\widetilde{L}.$

Lemma 7

If $\lambda _{k}=\widetilde{\lambda }_{k}$ for all $k,$ then $\Delta _{12}(\lambda )\equiv \widetilde{\Delta }_{12}(\lambda ).$

Proof

The functions $\Delta _{12}(\lambda )$ and $\widetilde{\Delta }_{12}(\lambda )$ are entire in $\lambda$ of exponential type. Using Hadamard’s factorization theorem, we get $\Delta _{12}(\lambda )=e^{a\lambda +b}\widetilde{\Delta }_{12}(\lambda ).$ Let us show that $a=0,\;b=0.$ In view of (9),

$$\begin{aligned}&\frac{1}{2i}e^{-i(\lambda \pi +\alpha -\beta )}[1]-\frac{1}{2i}e^{i(\lambda \pi +\alpha -\beta )}[1] +\sum _{j=1}^N \sin \pi \mu _j e^{-il\lambda (\pi -2\gamma _j)+il(2\eta _j-\alpha -\beta )}[l]\nonumber \\&\quad =\frac{1}{2i}e^{-i(\lambda \pi +\widetilde{\alpha }-\widetilde{\beta })+a\lambda +b}[1]- \frac{1}{2i}e^{i(\lambda \pi +\widetilde{\alpha }-\widetilde{\beta })+a\lambda +b}[1]\nonumber \\&\qquad +\sum _{j=1}^N \sin \pi \widetilde{\mu }_je^{-il\lambda (\pi -2\widetilde{\gamma }_j) +il(2\widetilde{\eta }_j-\widetilde{\alpha }-\widetilde{\beta })+a\lambda +b}[l]. \end{aligned}$$

(11)

Let $\lambda =\sigma +i\tau .$ If $\tau =0$ and $\sigma \rightarrow +\infty ,$ then the right-hand side in (11) is bounded; hence $Re\,a\le 0$; for $\tau =0$ and $\sigma \rightarrow -\infty ,$ we get $Re\,a\ge 0,$ i.e. $Re\,a=0.$ Furthermore, the right-hand side in (11) is $O(e^{-Im\lambda \tau }),$ but the left-hand side is $O(e^{-Im\lambda \tau -Im\,aIm\lambda })$ for $\tau \le 0.$ For $\tau \rightarrow -\infty$ we have $Im\,a\le 0.$ If $\tau \ge 0,$ then it follows from (11) that $O(e^{\pi \tau })=O(e^{\pi \tau -Im\,a\tau }).$ This means that $Im\,a\ge 0,$ i.e. $Im\,a=0.$ Thus, $a=0.$ Similarly, one gets that $b=0.$ Lemma is proved.

Corollary 2

If $\lambda _{k}=\widetilde{\lambda }_{k}$ for all $k,$ then $\Delta ^0_{12}(\lambda )\equiv \widetilde{\Delta }^0_{12}(\lambda ),$ i.e. $\alpha -\beta =\widetilde{\alpha }-\widetilde{\beta },\;\gamma _k=\widetilde{\gamma }_k,$ $\sin \pi \mu _ke^{il(2\eta _k-\alpha -\beta )}= \sin \pi \widetilde{\mu }_ke^{il(2\widetilde{\eta }_k-\widetilde{\alpha }- \widetilde{\beta })}$. Here $\Delta ^0_{12}(\lambda )$ is defined by (10).

Lemma 8

If $\alpha -\widetilde{\alpha }=\beta -\widetilde{\beta }=\widetilde{\eta }_k-\eta _k,\; \mu _k=\widetilde{\mu }_k,\;\gamma _k=\widetilde{\gamma }_k,\;k=\overline{1,N}$ and $Q(x)=\widetilde{Q}(x)V^2(\widetilde{\alpha }-\alpha ),$ then ${\mathfrak {M}}(\lambda )=\widetilde{\mathfrak {M}}(\lambda )$.

Proof

Denote $\delta :=\alpha -\widetilde{\alpha }=\beta -\widetilde{\beta }=\widetilde{\eta }_k-\eta _k$. Let us show that if $Y(x,\lambda )$ is a solution of (1), then $\widetilde{Y}(x,\lambda )= V(-\delta )Y(x,\lambda )$ is a solution of $\widetilde{(1)}$. Indeed, substituting $V(\delta )\widetilde{Y}(x,\lambda )$ into (1), we obtain

$$\begin{aligned} BV(\delta )\widetilde{Y}'(x,\lambda )+\Big (Q(x)+Q_\omega (x)\Big )V(\delta )\widetilde{Y}(x,\lambda )=\lambda V(\delta )\widetilde{Y}(x,\lambda ). \end{aligned}$$

Multiplying by $V^T(\delta )=V(-\delta )$ and taking the relation $V^T(\delta )Q(x)=Q(x)V(\delta )$ into account, we get

$$\begin{aligned} B\widetilde{Y}'(x,\lambda )+\Big (Q(x)+Q_\omega (x)\Big ) V^2(\delta )\widetilde{Y}(x,\lambda )=\lambda \widetilde{Y}(x,\lambda ). \end{aligned}$$

One has $V(-\delta )S(0,\lambda )V(\delta )=I.$ Since the Cauchy problem has the unique solution, we infer $\widetilde{S}(x,\lambda )=V(-\delta )S(x,\lambda )V(\delta ).$ Then $\widetilde{\Delta }(\lambda )\!=\!V^T(\widetilde{\beta })V(-\delta )S(\pi ,\lambda )V(\delta ) V(\widetilde{\alpha })$ or

$$\begin{aligned} \widetilde{\Delta }(\lambda )=V^T(\widetilde{\beta }+\delta )S(\pi ,\lambda )V(\delta +\widetilde{\alpha }). \end{aligned}$$

This yields $\widetilde{\Delta }_{jk}(\lambda )=\Delta _{jk}(\lambda ).$ Lemma is proved.

3 Solution of the inverse problem

Let us first prove the uniqueness theorem.

Theorem 1

If ${\mathfrak {M}}(\lambda )=\widetilde{\mathfrak {M}}(\lambda ),$ then $\alpha -\widetilde{\alpha }=\beta -\widetilde{\beta }=\widetilde{\eta }_k-\eta _k,\; \mu _k=\widetilde{\mu }_k, \;\gamma _k=\widetilde{\gamma }_k,\;k=\overline{1,N}$ and $Q(x)=\widetilde{Q}(x)V^2(\widetilde{\alpha }-\alpha ).$

Proof

By virtue of Lemma 8, it is sufficient to prove the theorem for the case $\widetilde{\beta }=\beta =0.$ Consider the function $P(x,\lambda )= \Phi (x,\lambda )\widetilde{\Phi }^{-1}(x,\lambda ).$

Since ${\mathfrak {M}}(\lambda )=\widetilde{\mathfrak {M}}(\lambda ),$ it follows that these functions have the same poles. In view of Lemma 7, one gets $\Delta _{12}(\lambda )= \widetilde{\Delta }_{12}(\lambda ).$ By Corollary 2, $\widetilde{\alpha }=\alpha ,\; \widetilde{\gamma }_k=\gamma _k,\; \sin \pi \mu _ke^{il(2\eta _k-\alpha )}= \sin \pi \widetilde{\mu }_ke^{il(2\widetilde{\eta }_k-\widetilde{\alpha })}.$ This yields

$$\begin{aligned} \varphi (x,\lambda )-\widetilde{\varphi }(x,\lambda )\!=\!O(e^{|Im\lambda |x}|\lambda |^{-\nu }),\quad \psi (x,\lambda )-\widetilde{\psi }(x,\lambda )=O(e^{|Im\lambda |(\pi -x)}|\lambda |^{-\nu }).\nonumber \\ \end{aligned}$$

(12)

Since $\Phi _1(x,\lambda )=-(\Delta _{12}(\lambda ))^{-1}\psi _2(x,\lambda ),$ it follows that

$$\begin{aligned} \Phi _1(x,\lambda )=O(e^{-x|Im\lambda |}),\quad \lambda \in G_\delta . \end{aligned}$$

(13)

Taking (12) into account, we infer

$$\begin{aligned} \Phi _1(x,\lambda )-\widetilde{\Phi }_1(x,\lambda )= O(e^{-x|Im\lambda |}|\lambda |^{-\nu }),\quad \lambda \in G_\delta . \end{aligned}$$

(14)

Obviously, $P(x,\lambda )-I= (\Phi (x,\lambda )-\widetilde{\Phi }(x,\lambda ))B\widetilde{\Phi }(x,\lambda )B^T.$ Using (12)–(14), we obtain for $\lambda \in G_\delta$:

$$\begin{aligned} P(x,\lambda )-I\!&= \!|\lambda |^{-\nu } \left( \begin{array}{ll}O(e^{-|Im\lambda |x})&{}\quad O(e^{|Im\lambda |x})\\ O(e^{-|Im\lambda |x})&{}\quad O(e^{|Im\lambda |x})\end{array}\right) \left( \begin{array}{ll}O(e^{|Im\lambda |x})&{}\quad O(e^{|Im\lambda |x})\\ O(e^{-|Im\lambda |x})&{}\quad O(e^{-|Im\lambda |x})\end{array}\right) \nonumber \\&= O(|\lambda |^{-\nu }). \end{aligned}$$

(15)

Since $\Phi (x,\lambda )=\varphi (x,\lambda )M(\lambda ),$ we get $P(x,\lambda )= \varphi (x,\lambda )M(\lambda )\widetilde{M}^{-1}(\lambda )\widetilde{\varphi }^{-1}(x,\lambda ),$ or $P(x,\lambda )=\varphi (x,\lambda )\widetilde{\varphi }^{-1}(x,\lambda ).$ Therefore, $P(x,\lambda )$ is entire in $\lambda .$ Using (15), maximum modulus principle and Liouville’s theorem, we conclude that $P(x,\lambda )=I,$ i.e. $\widetilde{\Phi }(x,\lambda )=\Phi (x,\lambda ).$ Then $Q(x)+Q_\omega (x)=\widetilde{Q}(x)+\widetilde{Q}_{\widetilde{\omega }}(x),$ and consequently, $Q(x)=\widetilde{Q}(x),$ $Q_\omega (x)=\widetilde{Q}_{\widetilde{\omega }}(x).$ Theorem is proved.

Corollary 3

If $a_k=\widetilde{a}_k,\;\lambda _{k}= \widetilde{\lambda }_{k}$ for all $k,$ then $L=\widetilde{L}.$

Corollary 4

If $\lambda ^{\left\langle {11}\right\rangle }_k=\widetilde{\lambda }^{\left\langle {11}\right\rangle }_k,\;\lambda _{k}=\widetilde{\lambda }_{k}$ for all $k,$ then $L=\widetilde{L}.$ Here $\{\lambda ^{\left\langle {11}\right\rangle }_k\}$ and $\{\widetilde{\lambda }^{\left\langle {11}\right\rangle }_k\}$ are eigenvalues of $L_{11}$ and $\widetilde{L}_{11}$, respectively.

Indeed, according to Lemma 7, $\Delta _{12}(\lambda )=\widetilde{\Delta }_{12}(\lambda ).$ Analogously, we obtain $\Delta _{11}(\lambda )=\widetilde{\Delta }_{11}(\lambda ).$ By Lemma 4, ${\mathfrak {M}}(\lambda )=\widetilde{\mathfrak {M}}(\lambda ).$

Let us now go on to constructing the solution of the nonlinear Inverse Problem 1. The central role here is played by the so-called main equation of the inverse problem, which is a linear equation in the corresponding Banach space. Let us derive the main equation.

Let the problem $L$ with a simple spectrum be given. We choose a model boundary value problem $\widetilde{L}$ with a simple spectrum such that $\omega =\tilde{\omega },$ $Q_\omega (x)=\widetilde{Q}_{\omega }(x)$ and

$$\begin{aligned} \Lambda := \sum _{k=-\infty }^{+\infty }|\widetilde{a}_k|\xi _k <\infty ,\quad \xi _k:=|\lambda _{k}-\widetilde{\lambda }_{k}|+ |\widetilde{a}^{-1}_ka_k-1|. \end{aligned}$$

(16)

For definiteness, we assume that $\alpha =\widetilde{\alpha }=0.$ Then $\beta =\widetilde{\beta }.$ Denote $\Omega _\varepsilon :=\{x\,:\,x\in (0,\pi ),\,|x-\gamma _k|\ge \varepsilon ,\; k=\overline{1,N}\}$, $\lambda _{k0}=\lambda _{k},\,\lambda _{k1} =\widetilde{\lambda }_{k},\;a_{k0}=a_k,\,a_{k1}=\widetilde{a}_k$,

$$\begin{aligned}&\widetilde{D}^{\left\langle {l}\right\rangle }(x,\lambda ,\theta ):=\frac{{\left\langle {\widetilde{\Phi }_l(x,\lambda ), \; \widetilde{\varphi }_2(x,\theta )}\right\rangle }}{\lambda -\theta },\quad l=1,2,\qquad \widetilde{D}^{\left\langle {2}\right\rangle }_{kj}(x,\lambda )=\widetilde{D}^{\left\langle {2}\right\rangle }(x,\lambda ,\lambda _{kj}),\\&\widetilde{P}_{ni,kj}(x)=\widetilde{D}^{\left\langle {2}\right\rangle }(x,\lambda _{ni},\lambda _{kj})a_{kj}, \quad \varphi _{2,kj}(x)=\varphi _2(x,\lambda _{kj}),\quad \widetilde{\varphi }_{2,kj}(x)=\widetilde{\varphi }_2(x,\lambda _{kj}), \end{aligned}$$

where ${\left\langle {Y,Z}\right\rangle }:=\det (Y,Z)=Y^TBZ.$ Analogously we define $D^{\left\langle {l}\right\rangle }(x,\lambda ,\theta ),$ $D^{\left\langle {2}\right\rangle }_{kj}(x,\lambda )$ and $P_{ni,kj}(x).$

Lemma 9

For $x\in \Omega _\varepsilon$ and $\lambda$ on compact sets$,$

$$\begin{aligned}&|\widetilde{\varphi }^{(m)}_{2,kj}(x)|\le C(1+|\lambda ^0_k|)^m,\quad |\widetilde{\varphi }^{(m)}_{2,k1}(x)-\widetilde{\varphi }^{(m)}_{2,k0}(x)| \le C\xi _k(1\!+\!|\lambda ^0_k|)^m,\quad m\!=\!0,1,\end{aligned}$$

(17)

$$\begin{aligned}&\left. \begin{array}{l} |\widetilde{D}^{\left\langle {2}\right\rangle }_{kj}(x,\lambda )| \le \displaystyle \frac{C}{1+|\lambda -\lambda ^0_k|},\quad |\widetilde{D}^{\left\langle {2}\right\rangle }_{k0}(x,\lambda )a_{k0}- \widetilde{D}^{\left\langle {2}\right\rangle }_{k1}(x,\lambda )a_{k1}| \le \displaystyle \frac{C|a_{k1}|\xi _k}{1\!+\!|\lambda \!-\!\lambda ^0_k|},\\ |(\widetilde{D}^{\left\langle {2}\right\rangle }_{kj}(x,\lambda ))'|\le C,\quad |(\widetilde{D}^{\left\langle {2}\right\rangle }_{k0}(x,\lambda )a_{k0}- \widetilde{D}^{\left\langle {2}\right\rangle }_{k1}(x,\lambda )a_{k1})'|\le C|a_{k1}|\xi _k. \end{array}\!\right\} \!.\nonumber \\ \end{aligned}$$

(18)

The same estimates are valid for $\varphi _{2,kj}(x),\;D^{\left\langle {2}\right\rangle }_{kj}(x,\lambda )$.

In order to prove the lemma, we need the following generalization of Schwarz’s lemma:

Let the function $f(z)$ be analytic inside the circle $|z-z_0|\le R$ and continuous in the whole circle. Moreover, $|f(z)|\le C$ on the boundary, and $f(z_0)=0.$ Then $|f(z)|\le C|z-z_0|/R$ in the circle $|z-z_0|\le R$.

• 1. It follows from (7) that

  $$\begin{aligned} |\widetilde{\varphi }_2(x,\lambda )|\le C e^{|Im\lambda |x},\quad x\in \Omega _\varepsilon . \end{aligned}$$

  (19)

  The eigenvalues lie in the strip $|Im\lambda |\le \max \{h,\,\widetilde{h}\}$; it follows from (19) that $|\widetilde{\varphi }^{(m)}_{2,kj}(x)|\le C(1+|\lambda _{kj}|)^m.$ Using (10), we obtain the first estimate in (17) for $m=0.$ Applying Schwarz’s lemma, we find $|\widetilde{\varphi }_2(x,\lambda )-\widetilde{\varphi }_{2,k1}(x)|\le Ce^{|Im\lambda |x}|\lambda -\lambda _{k1}|.$ Hence the second estimate in (17) holds (17) for $m=0.$ For $m=1,$ the arguments are similar.

• 2. Since $\widetilde{D}^{\left\langle {2}\right\rangle }(x,\lambda ,\theta )=(\lambda -\theta )^{-1} (\widetilde{\varphi }_2(x,\lambda ))^T B\widetilde{\varphi }_2(x,\theta ),$ it follows from (19) for $\lambda \ne \theta$, $|\lambda |\le R,\;|\theta |\le R$ that $|\widetilde{D}(x,\lambda ,\theta )|\le C|\lambda -\theta |^{-1}.$ If $\lambda =\theta ,$ then $\widetilde{D}(x,\lambda ,\lambda )=(\widetilde{\varphi }_2(x,\lambda ))^T B\dot{\widetilde{\varphi }}_2(x,\lambda ),$ where $\dot{\widetilde{\varphi }}_2(x,\lambda )= \displaystyle \frac{\partial }{\partial \lambda }\widetilde{\varphi }_2(x,\lambda ).$ Using Lemma 2, we obtain $|\dot{\widetilde{\varphi }}_2(x,\lambda )|\le Cxe^{|Im\lambda |x}$ for $x\in \Omega _\varepsilon .$ Then $|\widetilde{D}(x,\lambda ,\lambda )|\le C$ for $|\lambda |\le R.$ Thus,

  $$\begin{aligned} |\widetilde{D}(x,\lambda ,\theta )|\le \frac{C}{1+|\lambda -\theta |}, \quad x\in \Omega _\varepsilon ,\ |\lambda |\le R,\ |\theta |\le R. \end{aligned}$$

  (20)

Furthermore, ${\left\langle {\widetilde{\Phi }_j(x,\lambda ),\widetilde{\varphi }_2(x,\theta )}\right\rangle }'= (\widetilde{\Phi }^T_j(x,\lambda ))'B\widetilde{\varphi }_2(x,\theta )+ \widetilde{\Phi }^T_j(x,\lambda )B\widetilde{\varphi }\,'_2(x,\theta ).$ Then

$$\begin{aligned} {\left\langle {\widetilde{\Phi }_j(x,\lambda ),\widetilde{\varphi }_2(x,\theta )}\right\rangle }'= -(B\widetilde{\Phi }'_j(x,\lambda ))^T\widetilde{\varphi }_2(x,\theta )+ \widetilde{\Phi }^T_j(x,\lambda )B\widetilde{\varphi }\,'_2(x,\theta ). \end{aligned}$$

Since $\widetilde{\Phi }_j, \widetilde{\varphi }_2$ are solutions of the system, it follows that ${\left\langle {\widetilde{\Phi }_j(x,\lambda ),\widetilde{\varphi }_2(x,\theta )}\right\rangle }'= (\theta -\lambda ) \widetilde{\Phi }^T_j(x,\lambda )\widetilde{\varphi }_2(x,\theta ).$ This yields

$$\begin{aligned} (\widetilde{D}^{\left\langle {j}\right\rangle }(x,\lambda ,\theta ))' =-\widetilde{\Phi }^T_j(x,\lambda )\widetilde{\varphi }_2(x,\theta ). \end{aligned}$$

(21)

Taking (21) and (19) into account, we arrive at the third estimate in (18).

Using Schwarz’s lemma and (20), we infer $|\widetilde{D}^{\left\langle {2}\right\rangle }_{k0}(x,\lambda )-\widetilde{D}^{\left\langle {2}\right\rangle }_{k1}(x,\lambda )| \!\le \!\displaystyle \frac{C|\lambda _{k0}\!-\!\lambda _{k1}|}{1\!+\!|\lambda \!-\!\lambda ^0_k|}.$ Since

$$\begin{aligned} |\widetilde{D}^{\left\langle {2}\right\rangle }_{k0}(x,\lambda )a_{k0}-\widetilde{D}^{\left\langle {2}\right\rangle }_{k1}(x,\lambda )a_{k1}|&\le |\widetilde{D}^{\left\langle {2}\right\rangle }_{k0}(x,\lambda )(a_{k0}-a_{k1})|\\&+ |(\widetilde{D}^{\left\langle {2}\right\rangle }_{k0}(x,\lambda )-\widetilde{D}^{\left\langle {2}\right\rangle }_{k1}(x,\lambda ))a_{k1}|, \end{aligned}$$

one gets the second estimate in (18). Other estimates are obtained analogously. Lemma is proved.

Similarly one can prove the following assertion.

Lemma 10

For $x\in \Omega _\varepsilon$ and $\lambda$ on compact sets$,$

$$\begin{aligned}&|\widetilde{P}_{ni,kj}(x)|\le \frac{C|a_{k1}|}{1+|\lambda ^0_n-\lambda ^0_k|}, \quad |\widetilde{P}'_{ni,kj}(x)|\le C|a_{k1}|,\\&|\widetilde{P}_{ni,k1}(x)-\widetilde{P}_{ni,k0}(x)|\le \frac{C|a_{k1}|\xi _k}{1+|\lambda ^0_n-\lambda ^0_k|},\quad |\widetilde{P}'_{ni,k1}(x)-\widetilde{P}'_{ni,k0}(x)|\le C|a_{k1}|\xi _k,\\&|\widetilde{P}_{n1,kj}(x)-\widetilde{P}_{n0,kj}(x)|\le \frac{C|a_{k1}|\xi _n}{1+|\lambda ^0_n-\lambda ^0_k|},\quad |\widetilde{P}'_{n1,kj}(x)-\widetilde{P}'_{n0,kj}(x)|\le C|a_{k1}|\xi _n,\\&|\widetilde{P}_{n1,k1}(x)-\widetilde{P}_{n1,k0}(x)-\widetilde{P}_{n0,k1}(x) +\widetilde{P}_{n0,k0}(x)| \le \frac{C|a_{k1}|\xi _k\xi _n}{1+|\lambda ^0_n-\lambda ^0_k|},\\&|\widetilde{P}'_{n1,k1}(x)-\widetilde{P}'_{n1,k0}(x)-\widetilde{P}'_{n0,k1}(x) +\widetilde{P}'_{n0,k0}(x)| \le C|a_{k1}|\xi _k\xi _n. \end{aligned}$$

Moreover, if $\lambda \in G_\delta =\Big \{\lambda \,: \,|\lambda -\widetilde{\lambda }_{k}|\ge \delta ,\;k\in \mathbb {Z}\Big \},$ then

$$\begin{aligned}&|\widetilde{D}^{\left\langle {1}\right\rangle }_{kj}(x,\lambda )|\le \frac{C_\delta }{|\lambda -\lambda _{kj}|},\quad |(\widetilde{D}^{\left\langle {1}\right\rangle }_{kj}(x,\lambda ))'|\le C_\delta ,\\&|\widetilde{D}^{\left\langle {1}\right\rangle }_{k0}(x,\lambda )a_{k0}-\widetilde{D}^{\left\langle {1}\right\rangle }_{k1}(x,\lambda ) a_{k1}|\le C_\delta |a_{k1}|\xi _k\left( \frac{1}{|\lambda -\lambda _{k0}|}+ \frac{1}{|\lambda -\lambda _{k1}|}\right) ,\\&|(\widetilde{D}^{\left\langle {1}\right\rangle }_{k0}(x,\lambda )a_{k0}-\widetilde{D}^{\left\langle {1}\right\rangle }_{k1}(x,\lambda ) a_{k1})'|\le C_\delta |a_{k1}|\xi _k, \end{aligned}$$

where $C$ and $C_\delta$ depend on $\varepsilon .$ The same estimates are valid for $D^{\left\langle {l}\right\rangle }_{kj}(x,\lambda ),\;P_{ni,kj}(x)$.

Lemma 11

The following relations hold

$$\begin{aligned} \Phi _j(x,\lambda )&= \widetilde{\Phi }_j(x,\lambda )+\sum _{k=-\infty }^{+\infty }\Big ( \widetilde{D}^{\left\langle {j}\right\rangle }_{k0}(x,\lambda )a_{k0}\varphi _{2,k0}(x)- \widetilde{D}^{\left\langle {j}\right\rangle }_{k1}(x,\lambda )a_{k1}\varphi _{2,k1}(x)\Big ),\nonumber \\&\quad j=1,2, \end{aligned}$$

(22)

the series converge absolutely and uniformly for $x\in \Omega _\varepsilon$ and $\lambda$ on compact sets without the spectra of $L$ and $\widetilde{L}.$

Proof

Consider the function $P(x,\lambda )=\Phi (x,\lambda )\widetilde{\Phi }^{-1}(x,\lambda ).$ Denote

$$\begin{aligned} J_n(x,\lambda )=\displaystyle \frac{1}{2\pi i}\int _{\Gamma _n}\displaystyle \frac{1}{\xi -\lambda } \Big (P(x,\xi )-I\Big )d\xi ,\quad \Gamma _n:=\Big \{\lambda \,:\,|\lambda |=R_n\Big \}. \end{aligned}$$

The functions $\Phi (x,\lambda )$ and $\widetilde{\Phi }(x,\lambda )$ have the same main term in the asymptotics. Therefore, for a fixed $x\ne \gamma _k$, one has $P(x,\xi )-I=O(|\xi |^{-\nu }),$ and $J_n(x,\lambda )\rightarrow 0$ as $n\rightarrow \infty$ uniformly in $\lambda$ on the compact sets. Integration on $\Gamma _n$ is divided into integration on the contours $\Gamma ^{\left\langle {1}\right\rangle }_n=\Gamma ^{\left\langle {3}\right\rangle }_n\bigcup \Gamma ^{\left\langle {5}\right\rangle }_n$, $\Gamma ^{\left\langle {2}\right\rangle }_n=\Gamma ^{\left\langle {4}\right\rangle }_n\bigcup \Gamma ^{\left\langle {5}\right\rangle }_n$ (with counterclockwise circuit), where $\Gamma ^{\left\langle {3}\right\rangle }_n=\{\lambda \,:\,|Im\lambda |\le h\}\bigcap \Gamma _n$, $\Gamma ^{\left\langle {4}\right\rangle }_n=\Gamma _n{\setminus }\Gamma ^{\left\langle {3}\right\rangle }_n= \{\lambda \,:\,|Im\lambda |>h\}\bigcap \Gamma _n$, $\Gamma ^{\left\langle {5}\right\rangle }_n=\{\lambda \,:\,|Im\lambda |=h\}\bigcap \hbox {int}\,\Gamma _n$. Let $\lambda \in \hbox {int}\,\Gamma ^{\left\langle {2}\right\rangle }_n$. By the Cauchy integral formula,

$$\begin{aligned} \frac{1}{2\pi i}\int _{\Gamma ^{\left\langle {2}\right\rangle }_n}\frac{1}{\xi -\lambda } \Big (P(x,\xi )-I\Big )d\xi = P(x,\lambda )-I. \end{aligned}$$

Clearly,

$$\begin{aligned} \frac{1}{2\pi i}\int _{\Gamma ^{\left\langle {1}\right\rangle }_n}\frac{1}{\xi -\lambda }Id\xi = 0. \end{aligned}$$

Then

$$\begin{aligned} P(x,\lambda )=I+\frac{1}{2\pi i} \int _{\Gamma ^{\left\langle {1}\right\rangle }_n}\frac{1}{\lambda -\xi }P(x,\xi )d\xi -J_n(x,\lambda ). \end{aligned}$$

(23)

Since $\Phi (x,\lambda )=\varphi (x,\lambda )M(\lambda ),$ it follows that

$$\begin{aligned} P(x,\xi )&= \varphi (x,\xi )M(\xi )\widetilde{M}^{-1}(\xi )\widetilde{\varphi }^{-1}(x,\xi )\\&= \varphi (x,\xi )\widetilde{\varphi }^{-1}(x,\xi )- ({\mathfrak {M}}(\xi )-\widetilde{\mathfrak {M}}(\xi ))\varphi (x,\xi )B_{\left\langle {1}\right\rangle }\widetilde{\varphi }^{-1}(x,\xi ),\\ B_{\left\langle {1}\right\rangle }&= \left( \begin{array}{ll} 0 &{}\quad 0 \\ - 1 &{}\quad 0 \end{array}\right) . \end{aligned}$$

The function $\varphi (x,\xi )\widetilde{\varphi }^{-1}(x,\xi )$ is entire in $\xi .$ Therefore,

$$\begin{aligned} \frac{1}{2\pi i}\int _{\Gamma ^{\left\langle {1}\right\rangle }_n}\varphi (x,\xi )\widetilde{\varphi }^{-1}(x,\xi ) \frac{d\xi }{\lambda -\xi }=0, \end{aligned}$$

since $\lambda$ is outside $\Gamma ^{\left\langle {1}\right\rangle }_n$. Thus, it follows from (23) that

$$\begin{aligned} P(x,\lambda )=I-\frac{1}{2\pi i}\int _{\Gamma ^{\left\langle {1}\right\rangle }_n} ({\mathfrak {M}}(\xi )-\widetilde{\mathfrak {M}}(\xi ))\varphi (x,\xi )B_{\left\langle {1}\right\rangle }\widetilde{\varphi }^{-1}(x,\xi ) \frac{d\xi }{\lambda -\xi }-J_n(x,\lambda ). \end{aligned}$$

One has $\Phi (x,\lambda )=P(x,\lambda )\widetilde{\Phi }(x,\lambda ),$ hence $\Phi _j(x,\lambda )=P(x,\lambda )\widetilde{\Phi }_j(x,\lambda ).$ Then

$$\begin{aligned}&\Phi _j(x,\lambda )=\widetilde{\Phi }_j(x,\lambda )\\&\quad -\frac{1}{2\pi i}\int _{\Gamma ^{\left\langle {1}\right\rangle }_n} ({\mathfrak {M}}(\xi )-\widetilde{\mathfrak {M}}(\xi ))\varphi (x,\xi )B_{\left\langle {1}\right\rangle }\widetilde{\varphi }^{-1}(x,\xi ) \widetilde{\Phi }_j(x,\lambda )\frac{d\xi }{\lambda -\xi }+\varepsilon _n(x,\lambda ), \end{aligned}$$

and $\varepsilon _n(x,\lambda )\rightarrow 0$ as $n\rightarrow \infty$ uniformly for $x\in \Omega _\varepsilon$. Furthermore,

$$\begin{aligned} B_{\left\langle {1}\right\rangle }\widetilde{\varphi }^{-1}(x,\xi ) \widetilde{\Phi }_j(x,\lambda )= \Big (\widetilde{\varphi }_{12}(x,\xi )\widetilde{\Phi }_{2j}(x,\lambda )- \widetilde{\varphi }_{22}(x,\xi )\widetilde{\Phi }_{1j}(x,\lambda )\Big ) \left( \begin{array}{c} 0 \\ 1\end{array}\right) , \end{aligned}$$

and consequently,

$$\begin{aligned} \Phi _j(x,\lambda )\!&= \!\widetilde{\Phi }_j(x,\lambda )-\frac{1}{2\pi i}\int _{\Gamma ^{\left\langle {1}\right\rangle }_n} ({\mathfrak {M}}(\xi )-\widetilde{\mathfrak {M}}(\xi )){\left\langle {\widetilde{\Phi }_j(x,\lambda ), \widetilde{\varphi }_2(x,\xi )}\right\rangle }\varphi _2(x,\xi )\frac{d\xi }{\lambda -\xi }\\&+\varepsilon _n(x,\lambda ). \end{aligned}$$

Calculating the integral by residue’s theorem and taking $n\rightarrow \infty ,$ we arrive at (22). Lemma is proved.

Consider (22) for $j=2$ and $\lambda =\lambda _{ni}$:

$$\begin{aligned} \widetilde{\varphi }_{m2,ni}(x)&= \varphi _{m2,ni}(x)- \sum _{k=-\infty }^{+\infty }\Big (\widetilde{P}_{ni,k0}(x)\varphi _{m2,k0}(x)- \widetilde{P}_{ni,k1}(x)\varphi _{m2,k1}(x)\Big ),\nonumber \\ \quad m&= 1,2, \end{aligned}$$

(24)

where $\varphi _{2,kj}(x)= \left( \begin{array}{c}\varphi _{12,kj}(x) \\ \varphi _{22,kj}(x)\end{array}\right)$. The last relation is not convenient for our purpose, since the series converges only “with brackets”. We transform (24) as follows:

$$\begin{aligned}&\widetilde{\varphi }_{m2,n0}(x)-\widetilde{\varphi }_{m2,n1}(x)= \varphi _{m2,n0}(x)-\varphi _{m2,n1}(x)\nonumber \\&\quad -\sum _{k=-\infty }^{+\infty }\Big ((\widetilde{P}_{n0,k0}(x)- \widetilde{P}_{n1,k0}(x))(\varphi _{m2,k0}(x)-\varphi _{m2,k1}(x))\\&\quad +(\widetilde{P}_{n0,k0}(x)-\widetilde{P}_{n1,k0}(x)- \widetilde{P}_{n0,k1}(x)+\widetilde{P}_{n1,k1}(x))\varphi _{m2,k1}(x)\Big ),\\&\widetilde{\varphi }_{m2,n1}(x)=\varphi _{m2,n1}(x)-\sum _{k=-\infty }^{+\infty } \Big (\widetilde{P}_{n1,k0}(x)(\varphi _{m2,k0}(x)-\varphi _{m2,k1}(x))\nonumber \\&\qquad \qquad \qquad \!\!+(\widetilde{P}_{n1,k0}(x)-\widetilde{P}_{n1,k1}(x))\varphi _{m2,k1}(x)\Big ). \end{aligned}$$

Denote

$$\begin{aligned} \Psi ^{\left\langle {m}\right\rangle }_{n0}(x)&= \chi _n\Big (\varphi _{m2,n0}(x)-\varphi _{m2,n1}(x)\Big ),\\ \quad \chi _n&= \left\{ \begin{array}{ll}0,&{}\quad \xi _n=0,\\ \xi ^{-1}_n,&{}\quad \xi _n\ne 0, \end{array}\right. \; \Psi ^{\left\langle {m}\right\rangle }_{n1}(x)=\varphi _{m2,n1}(x),\\ \widetilde{H}_{n0,k0}(x)&= (\widetilde{P}_{n0,k0}(x)- \widetilde{P}_{n1,k0}(x))\chi _n\xi _k,\\ \widetilde{H}_{n0,k1}(x)&= (\widetilde{P}_{n0,k0}(x)-\widetilde{P}_{n1,k0}(x) -\widetilde{P}_{n0,k1}(x)+\widetilde{P}_{n1,k1}(x))\chi _n,\\ \widetilde{H}_{n1,k0}(x)&= \widetilde{P}_{n1,k0}(x)\xi _k,\; \widetilde{H}_{n1,k1}(x)=\widetilde{P}_{n1,k0}(x)-\widetilde{P}_{n1,k1}(x). \end{aligned}$$

Then

$$\begin{aligned} \widetilde{\Psi }^{\left\langle {m}\right\rangle }_{ni}(x)=\Psi ^{\left\langle {m}\right\rangle }_{ni}(x)-\sum _{k=-\infty }^{+\infty } \Big (\widetilde{H}_{ni,k0}(x)\Psi ^{\left\langle {m}\right\rangle }_{k0}(x)+ \widetilde{H}_{ni,k1}(x)\Psi ^{\left\langle {m}\right\rangle }_{k1}(x)\Big ). \end{aligned}$$

(25)

Using Lemmas 9 and 10, we obtain the estimates

$$\begin{aligned} \left. \begin{array}{l} |\widetilde{\Psi }^{\left\langle {m}\right\rangle }_{ni}(x)|\le C,\; |(\widetilde{\Psi }^{\left\langle {m}\right\rangle }_{ni}(x))'|\le C(1+|\lambda ^0_k|),\\ |\widetilde{H}_{ni,kj}(x)| \le \displaystyle \frac{C|a_{k1}|\xi _k}{1+|\lambda ^0_n-\lambda ^0_k|},\; |\widetilde{H}'_{ni,kj}(x)|\le C|a_{k1}|\xi _k. \end{array}\right\} \end{aligned}$$

(26)

The same estimates are valid for $\Psi ^{\left\langle {m}\right\rangle }_{ni}(x), \,\,H_{ni,kj}(x)$. Denote

$$\begin{aligned} \Psi ^{\left\langle {m}\right\rangle }(x)&= \left( \begin{array}{c}\Psi ^{\left\langle {m}\right\rangle }_{n0}(x)\\ \Psi ^{\left\langle {m}\right\rangle }_{n1}(x)\end{array}\right) _{n=-\infty }^{+\infty }\\&= \Big (\ldots ,\Psi ^{\left\langle {m}\right\rangle }_{-1,1}(x),\quad \Psi ^{\left\langle {m}\right\rangle }_{00}(x),\!\Psi ^{\left\langle {m}\right\rangle }_{01}(x),\quad \Psi ^{\left\langle {m}\right\rangle }_{10}(x),\ldots \Big )^T. \end{aligned}$$

Similarly we define the block-matrix

$$\begin{aligned} \widetilde{H}(x)=\left( \begin{array}{cc}\widetilde{H}_{n0,k0}(x)&{} \quad \widetilde{H}_{n0,k1}(x) \\ \widetilde{H}_{n1,k0}(x)&{}\quad \widetilde{H}_{n1,k1}(x)\end{array}\right) _{n,k=-\infty }^{+\infty }. \end{aligned}$$

Then we rewrite (25) as follows

$$\begin{aligned} \widetilde{\Psi }^{\left\langle {m}\right\rangle }(x)=(I-\widetilde{H}(x))\Psi ^{\left\langle {m}\right\rangle }(x),\quad m=1,2, \end{aligned}$$

(27)

where $I$ is the identity operator. It follows from (26) that $\Psi ^{\left\langle {m}\right\rangle }(x),\;\widetilde{\Psi }^{\left\langle {m}\right\rangle }(x)\in \mathbf{m}$ for each fixed $x\ne \gamma _k,\;k=\overline{1,N},$ where $\mathbf{m}$ is the Banach space of bounded sequences. The operator $\widetilde{H}(x),$ acting from $\mathbf{m}$ to $\mathbf{m},$ is a linear bounded operator, and

$$\begin{aligned} \Vert \widetilde{H}(x)\Vert _{m\rightarrow m}\le C\sup _n\sum _{k=-\infty }^{+\infty } \frac{|a_{k1}|\xi _k}{1+|\lambda _n^0-\lambda ^0_k|}\le C\sum _{k=-\infty }^{+\infty }|a_{k1}|\xi _k<\infty . \end{aligned}$$

For each fixed $x,$ relation (27) is a linear equation in $\mathbf{m}$ with respect to $\Psi ^{\left\langle {m}\right\rangle }(x).$ This equation is called the main equation of the inverse problem.

Lemma 12

The following relation holds

$$\begin{aligned} Q(x)=\widetilde{Q}(x)+B\mathrm{\ae }(x)-\mathrm{\ae }(x)B, \end{aligned}$$

(28)

where

$$\begin{aligned} \mathrm{\ae }(x)=\sum _{k=-\infty }^{+\infty }\Big (a_{k0}\widetilde{\varphi }_{2,k0}(x) \varphi ^T_{2,k0}(x)-a_{k1}\widetilde{\varphi }_{2,k1}(x)\varphi ^T_{2,k1}(x)\Big ), \end{aligned}$$

(29)

and the series converges uniformly for $x\in \Omega _\varepsilon .$

Proof

Differentiating (22), we calculate

$$\begin{aligned} \Phi '_j(x,\lambda )&= \widetilde{\Phi }'_j(x,\lambda )+\sum _{k=-\infty }^{+\infty }\Big ((\widetilde{D}^{\left\langle {j}\right\rangle }_{k0}(x,\lambda ))'a_{k0}\varphi _{2,k0}(x)+ \widetilde{D}^{\left\langle {j}\right\rangle }_{k0}(x,\lambda )a_{k0}\varphi \,'_{2,k0}(x)\\&-(\widetilde{D}^{\left\langle {j}\right\rangle }_{k1}(x,\lambda ))'a_{k1}\varphi _{2,k1}(x)- \widetilde{D}^{\left\langle {j}\right\rangle }_{k1}(x,\lambda )a_{k1}\varphi \,'_{2,k1}(x)\Big ). \end{aligned}$$

Multiplying this relation by $B$ and using (21), we obtain

$$\begin{aligned}&\Big (\lambda I-Q(x)-Q_\omega (x)\Big )\Phi _j(x,\lambda )= \Big (\lambda I-\widetilde{Q}(x)-\widetilde{Q}_\omega (x)\Big )\widetilde{\Phi }_j(x,\lambda )\\&\quad +\sum _{k=-\infty }^{+\infty } \Big (-\widetilde{\Phi }^T_j(x,\lambda )\widetilde{\varphi }_{2,k0}(x)a_{k0}B\varphi _{2,k0}(x)\\&\quad +\widetilde{D}^{\left\langle {j}\right\rangle }_{k0}(x,\lambda )a_{k0}\Big (\lambda _{k0}I\!-\!Q(x)\!-\!Q_\omega (x)\Big ) \varphi _{2,k0}(x) \!+\!\widetilde{\Phi }^T_j(x,\lambda )\widetilde{\varphi }_{2,k1}(x)a_{k1}B\varphi _{2,k1}(x)\\&\quad - \widetilde{D}^{\left\langle {j}\right\rangle }_{k1}(x,\lambda )a_{k1}\Big (\lambda _{k1}I-Q(x)-Q_\omega (x)\Big ) \varphi _{2,k1}(x)\Big ), \end{aligned}$$

and consequently,

$$\begin{aligned}&(Q(x)-\widetilde{Q}(x))\widetilde{\Phi }_j(x,\lambda )+\sum _{k=-\infty }^{+\infty } (-\widetilde{\Phi }^T_j(x,\lambda )\widetilde{\varphi }_{2,k0}(x)a_{k0}B\varphi _{2,k0}(x)\\&\quad +\widetilde{D}^{\left\langle {j}\right\rangle }_{k0}(x,\lambda )a_{k0}(\lambda _{k0}-\lambda )\varphi _{2,k0}(x)\\&\quad +\widetilde{\Phi }^T_j(x,\lambda )\widetilde{\varphi }_{2,k1}(x)a_{k1}B\varphi _{2,k1}(x) -\widetilde{D}^{\left\langle {j}\right\rangle }_{k1}(x,\lambda )a_{k1}(\lambda _{k1}-\lambda )\varphi _{2,k1}(x))=0. \end{aligned}$$

Since $\widetilde{D}^{\left\langle {j}\right\rangle }_{ni}(x,\lambda )= \displaystyle \frac{\widetilde{\Phi }^T_j(x,\lambda )B\widetilde{\varphi }_{2,ni}(x)}{\lambda -\lambda _{ni}}$, it follows that

$$\begin{aligned}&(Q(x)-\widetilde{Q}(x))\widetilde{\Phi }_j(x,\lambda )+\sum _{k=-\infty }^{+\infty } \Big (-\{\widetilde{\Phi }^T_j(x,\lambda )\widetilde{\varphi }_{2,k0}(x)a_{k0}\}B\varphi _{2,k0}(x)\\&\quad -\{\widetilde{\Phi }^T_j(x,\lambda )B\widetilde{\varphi }_{2,k0}(x)a_{k0}\}\varphi _{2,k0}(x) +\{\widetilde{\Phi }^T_j(x,\lambda )\widetilde{\varphi }_{2,k1}(x)a_{k1}\}B\varphi _{2,k1}(x)\\&\quad +\{\widetilde{\Phi }^T_j(x,\lambda )B\widetilde{\varphi }_{2,k1}(x)a_{k1}\}\varphi _{2,k1}(x)\Big )=0. \end{aligned}$$

The matrices $Q(x)$ and $\widetilde{Q}(x)$ are symmetrical. Then

$$\begin{aligned}&\widetilde{\Phi }^T(x,\lambda )\Big \{Q(x)-\widetilde{Q}(x)+\sum _{k=-\infty }^{+\infty } \Big (\Big (a_{k0}\widetilde{\varphi }_{2,k0}(x)\varphi ^T_{2,k0}(x)- a_{k1}\widetilde{\varphi }_{2,k1}(x)\varphi ^T_{2,k1}(x)\Big )B\\&\quad -B\Big (a_{k0}\widetilde{\varphi }_{2,k0}(x)\varphi ^T_{2,k0}(x)- a_{k1}\widetilde{\varphi }_{2,k1}(x)\varphi ^T_{2,k1}(x)\Big )\Big )\Big \}=0. \end{aligned}$$

Multiplying by $(\widetilde{\Phi }^T(x,\lambda ))^{-1},$ we arrive at (28). It follows from the estimate

$$\begin{aligned}&|a_{k0}\widetilde{\varphi }_{2,k0}(x)\varphi ^T_{2,k0}(x)- a_{k1}\widetilde{\varphi }_{2,k1}(x)\varphi ^T_{2,k1}(x)|\\&\quad \le |\widetilde{\varphi }_{2,k0}(x)\varphi ^T_{2,k0}(x)- \widetilde{\varphi }_{2,k1}(x)\varphi ^T_{2,k1}(x)|\cdot |a_{k1}|\\&\quad \quad +|\widetilde{\varphi }_{2,k1}(x)\varphi ^T_{2,k1}(x)| \cdot |a_{k0}-a_{k1}|\le C|a_{k1}|\xi _k \end{aligned}$$

that the series in (29) converges uniformly. Lemma is proved.

Let us now study the solvability of the main equation. For this purpose we need the following assertion.

Lemma 13

The following relation holds

$$\begin{aligned} D^{\left\langle {2}\right\rangle }(x,\lambda ,\theta )&= \widetilde{D}^{\left\langle {2}\right\rangle }(x,\lambda ,\theta )+\sum _{k=-\infty }^{+\infty } (\widetilde{D}^{\left\langle {2}\right\rangle }_{k0}(x,\lambda )D^{\left\langle {2}\right\rangle }_{k0}(x,\theta )a_{k0}\nonumber \\&-\widetilde{D}^{\left\langle {2}\right\rangle }_{k1}(x,\lambda ) D^{\left\langle {2}\right\rangle }_{k1}(x,\theta )a_{k1}), \end{aligned}$$

(30)

and the series converges uniformly for $x\in \Omega _\varepsilon$ and $\lambda$ on compact sets.

Proof

According to (23) we have for $\lambda ,\,\theta \in \Gamma ^{\left\langle {2}\right\rangle }_n$:

$$\begin{aligned} P(x,\lambda )-P(x,\theta )=\frac{1}{2\pi i}\int _{\Gamma ^{\left\langle {1}\right\rangle }_n} \left( \frac{1}{\lambda -\xi }-\frac{1}{\theta -\xi }\right) P(x,\xi )d\xi +J_n(x,\lambda ,\theta ), \end{aligned}$$

where $J_n(x,\lambda ,\theta )\rightarrow 0$ as $n\rightarrow \infty$ uniformly for $x\in \Omega _\varepsilon$ and $\lambda ,\;\theta$ on compact sets. Therefore,

$$\begin{aligned} \frac{1}{\lambda -\theta }\Big (P^T(x,\lambda )-P^T(x,\theta )\Big )&= \frac{1}{2\pi i}\int _{\Gamma ^{\left\langle {1}\right\rangle }_n} \frac{1}{(\lambda -\xi )(\xi -\theta )}P^T(x,\xi )d\xi \nonumber \\&+J^1_n(x,\lambda ,\theta ). \end{aligned}$$

(31)

Since $P(x,\xi )=\Phi (x,\xi )\widetilde{\Phi }^{-1}(x,\xi )= -\Phi (x,\lambda )B\widetilde{\Phi }^T(x,\xi )B,$ it follows that

$$\begin{aligned} \widetilde{\varphi }^T_2(x,\lambda )P^T(x,\xi )B\varphi _2(x,\theta )= -\widetilde{\varphi }^T_2(x,\lambda )B\widetilde{\Phi }(x,\xi ) B\Phi ^T(x,\xi )B\varphi _2(x,\theta ). \end{aligned}$$

One has ${\left\langle {y,z}\right\rangle }=y^TBz,$ and consequently,

$$\begin{aligned}&\widetilde{\varphi }^T_2(x,\lambda )P^T(x,\xi )B\varphi _2(x,\theta )= {\left\langle {\widetilde{\varphi }_2(x,\lambda ),\widetilde{\varphi }_2(x,\xi )}\right\rangle } {\Big \langle {\Phi _1(x,\xi ),\varphi _2(x,\theta )}\Big \rangle }\nonumber \\&\quad -{\left\langle {\widetilde{\varphi }_2(x,\lambda ),\widetilde{\Phi }_1(x,\xi )}\right\rangle } {\left\langle {\varphi _2(x,\xi ),\varphi _2(x,\theta )}\right\rangle }. \end{aligned}$$

(32)

Since ${\Big \langle {\Phi _1(x,\lambda ),\varphi _2(x,\lambda )}\Big \rangle }\equiv 1$, ${\Big \langle {\varphi _2(x,\lambda ),\varphi _2(x,\lambda )}\Big \rangle }\equiv 0,$ we infer

$$\begin{aligned}&\widetilde{\varphi }^T_2(x,\lambda )P^T(x,\lambda )B\varphi _2(x,\theta ) = {\Big \langle {\varphi _2(x,\lambda ),\varphi _2(x,\theta )}\Big \rangle },\\&\widetilde{\varphi }^T_2(x,\lambda )P^T(x,\theta )B\varphi _2(x,\theta ) = {\Big \langle {\widetilde{\varphi }_2(x,\lambda ),\widetilde{\varphi }_2(x,\theta )}\Big \rangle }. \end{aligned}$$

Multiplying (31) by $\widetilde{\varphi }^T_2(x,\lambda )$ from the left, and by $B\varphi _2(x,\theta )$ from the right, and using (32), we calculate

$$\begin{aligned}&\frac{{\Big \langle {\varphi _2(x,\lambda ),\varphi _2(x,\theta )}\Big \rangle }}{\lambda -\theta }- \frac{{\Big \langle {\widetilde{\varphi }_2(x,\lambda ),\widetilde{\varphi }_2(x,\theta )}\Big \rangle }}{\lambda -\theta }\\&\quad =\frac{1}{2\pi i}\int _{\Gamma ^{\left\langle {1}\right\rangle }_n}\Big ( \frac{{\left\langle {\widetilde{\varphi }_2(x,\lambda ),\widetilde{\varphi }_2(x,\xi )}\right\rangle } {\Big \langle {\Phi _1(x,\xi ),\varphi _2(x,\theta )}\Big \rangle }}{(\lambda -\xi )(\xi -\theta )}\\&\qquad -\frac{{\left\langle {\widetilde{\varphi }_2(x,\lambda ),\widetilde{\Phi }_1(x,\xi )}\right\rangle } {\left\langle {\varphi _2(x,\xi ),\varphi _2(x,\theta )}\right\rangle }}{(\lambda -\xi )(\xi -\theta )}\Big )d\xi +J^2_n(x,\lambda ,\theta ). \end{aligned}$$

By Lemma 4, $\Phi _1(x,\xi )=\varphi _1(x,\xi )+{\mathfrak {M}}(\xi )\varphi _2(x,\xi ).$ This yields

$$\begin{aligned}&\frac{{\Big \langle {\varphi _2(x,\lambda ),\varphi _2(x,\theta )}\Big \rangle }}{\lambda -\theta }- \frac{{\Big \langle {\widetilde{\varphi }_2(x,\lambda ),\widetilde{\varphi }_2(x,\theta )}\Big \rangle }}{\lambda -\theta }\\&\quad =\frac{1}{2\pi i}\int _{\Gamma ^{\left\langle {1}\right\rangle }_n} \frac{{\Big \langle {\widetilde{\varphi }_2(x,\lambda ),\widetilde{\varphi }_2(x,\xi )}\Big \rangle } {\Big \langle {\varphi _2(x,\xi ),\varphi _2(x,\theta )}\Big \rangle }}{(\lambda -\xi )(\xi -\theta )} \Big ({\mathfrak {M}}(\xi )-\widetilde{\mathfrak {M}}(\xi )\Big )d\xi \\&\qquad + J^2_n(x,\lambda ,\theta ), \end{aligned}$$

since the integrals from analytic functions are equal to zero. Calculating the integral by residue’s theorem and taking $n\rightarrow \infty ,$ we arrive at (30) firstly for $|\lambda |\ge h,$ and by analytic continuation for all $\lambda .$ Lemma is proved.

Taking $\lambda =\lambda _{ni},\;\;\theta =\lambda _{lj}$ in (30) and multiplying by $a_{lj}$, we obtain

$$\begin{aligned} P_{ni,lj}(x)-\widetilde{P}_{ni,lj}(x)-\sum _{k=-\infty }^{+\infty } \Big (\widetilde{P}_{ni,k0}(x)P_{k0,lj}(x)- \widetilde{P}_{ni,k1}(x)P_{k1,lj}(x)\Big )=0.\quad \end{aligned}$$

(33)

Symmetrically, one has

$$\begin{aligned} P_{lj,ni}(x)-\widetilde{P}_{lj,ni}(x)-\sum _{k=-\infty }^{+\infty } \Big (P_{lj,k0}(x)\widetilde{P}_{k0,ni}(x)- P_{lj,k1}(x)\widetilde{P}_{k1,ni}(x)\Big )=0.\quad \end{aligned}$$

(34)

It follows from (33)–(34) that

$$\begin{aligned}&H_{ni,lj}(x)-\widetilde{H}_{ni,lj}(x)\!-\!\sum _{k=-\infty }^{+\infty } \Big (\widetilde{H}_{ni,k0}(x)H_{k0,lj}(x)-\widetilde{H}_{ni,k1}(x)H_{k1,lj}(x)\Big )\!=\!0,\qquad \end{aligned}$$

(35)

$$\begin{aligned}&H_{ni,lj}(x)-\widetilde{H}_{ni,lj}(x)\!-\!\sum _{k=-\infty }^{+\infty } \Big (H_{ni,k0}(x)\widetilde{H}_{k0,lj}(x)- H_{ni,k1}(x)\widetilde{H}_{k1,lj}(x)\Big )\!=\!0.\qquad \end{aligned}$$

(36)

We rewrite relations (35) and (36) in the matrix form

$$\begin{aligned} H(x)-\widetilde{H}(x)-\widetilde{H}(x)H(x)=0,\quad H(x)-\widetilde{H}(x)-H(x)\widetilde{H}(x)=0 \end{aligned}$$

or $(I-\widetilde{H}(x))(I+H(x))=I,\; (I+H(x))(I-\widetilde{H}(x))=I.$ Thus, we have proved the following assertion.

Theorem 2

For each fixed $x$ $(x\ne \gamma _k,\; k=\overline{1,N}),$ the linear bounded operator $I-\widetilde{H}(x),$ acting from $\mathbf{m}$ to $\mathbf{m},$ has the unique inverse operator$,$ and the main equation (27) is uniquely solvable in $\mathbf{m}$.

The solution of Inverse Problem 1 can be constructed by the following algorithm.

Algorithm 1

Given the spectral data $\{\lambda _{k},\;a_k\}_{k=-\infty }^{+\infty }$ of the problem $L.$

• 1. Choose a model boundary value problem $\tilde{L},$ for example, with the zero potential.

• 2. Construct $\widetilde{\Psi }^{\left\langle {m}\right\rangle }(x)$ and $\widetilde{H}(x).$

• 3. Solving the linear main Eq. (27), find $\Psi ^{\left\langle {m}\right\rangle }(x),$ and then calculate $\varphi _{2,kj}(x).$

• 4. Construct $Q(x)$ by (28), and $\alpha =\widetilde{\alpha },\;\;\beta =\widetilde{\beta }.$

4 Necessary and sufficient conditions for the solvability of the inverse problem

Theorem 3

For numbers $\{\lambda _{k},\;a_k\}_{k=-\infty }^{+\infty },$ $a_k\ne 0,$ $\lambda _{k}\ne \lambda _n,$ $(k\ne n),$ to be the spectral data for a certain problem $L\in W,$ it is necessary and sufficient that the following conditions hold

• 1. $($Asymptotics$){:}$ There exists $\widetilde{L}\in W$ such that (16) holds$;$

• 2. $($Condition S$){:}$ For each fixed $x\ne \gamma _k,\;k=\overline{1,N},$ the linear bounded operator $I-\tilde{H}(x)$ has the unique inverse operator$;$

• 3. $(B\mathrm{\ae }(x)-\mathrm{\ae }(x)B)|x-\gamma _k|^{-2Re\mu _k}\in L(w_{k+1/2}),$ where $\mathrm{\ae }(x)$ is constructed by (29).

Under these conditions the potential $Q(x)$ is constructed by (28) and $\alpha =\widetilde{\alpha }$, $\beta =\widetilde{\beta }$.

The necessity part of the theorem was proved above. Let us prove the sufficiency. Let numbers $\{\lambda _{k},\;a_k\}_{k=-\infty }^{+\infty }$ be given such that $a_k\ne 0$ and $\lambda _{k}\ne \lambda _n,$ $(k\ne n).$ Let $\widetilde{L}=L(Q_\omega (x),\widetilde{Q}(x), 0, \beta )\in W$ be chosen such that (16) holds. Let $\{\Psi ^{\left\langle {m}\right\rangle }_{ni}(x)\}$ be the solution of the main equation (25). The following assertion is proved in [14].

Lemma 14

Consider the equations

$$\begin{aligned} (I+A_0)y_0=f_0,\quad (I+A)y=f, \end{aligned}$$

in a Banach space $\mathfrak {B},$ where $A_0, A$ are linear bounded operators, acting from $\mathfrak {B}$ to $\mathfrak {B},$ and $I$ is the identity operator. Suppose that there exists the linear bounded operator $R_0:=(I+A_0)^{-1}.$ If $\Vert A-A_0\Vert \le (2\Vert R_0\Vert )^{-1},$ then there exists the linear bounded operator $R=(I+A)^{-1},$ and $\Vert R\Vert \le 2\Vert R_0\Vert ,\Vert R-R_0\Vert \le 2\Vert R_0\Vert ^2\Vert A-A_0\Vert .$

Lemma 15

The following relations hold

$$\begin{aligned} \begin{aligned}&\Psi ^{\left\langle {m}\right\rangle }_{ni}(x)\in C(\Omega _\varepsilon ),\quad |\Psi ^{\left\langle {m}\right\rangle }_{ni}(x)|\le C_\varepsilon ,\\&|\Psi ^{\left\langle {m}\right\rangle }_{ni}(x)-\widetilde{\Psi }^{\left\langle {m}\right\rangle }_{ni}(x)|\le C_\varepsilon \Lambda \theta _n, \end{aligned} \end{aligned}$$

(37)

$$\begin{aligned}&\quad \theta _n=\left( \sum _{k=-\infty }^{+\infty }\frac{1}{(1+|\lambda _n^0-\lambda ^0_k|)^2 (1+|\lambda ^0_k|)^2}\right) ^{1/2},\ x\in \Omega _\varepsilon , \end{aligned}$$

(38)

$$\begin{aligned}&|(\Psi ^{\left\langle {m}\right\rangle }_{ni}(x))'|\le C_\varepsilon (1+|\lambda _{k}^0|)\quad x\in \Omega _\varepsilon ,\nonumber \\&\quad |(\Psi ^{\left\langle {m}\right\rangle }_{ni}(x))'-(\widetilde{\Psi }^{\left\langle {m}\right\rangle }_{ni}(x))'| \le C_\varepsilon \Lambda ,\quad x\in \Omega _\varepsilon . \end{aligned}$$

(39)

Proof

Using (26), we infer

$$\begin{aligned} |\widetilde{H}_{ni,kj}(x)\!-\!\widetilde{H}_{ni,kj}(x_0)|\le | \widetilde{H}'_{ni,kj}(\xi )|\,|x-x_0|\le C_\varepsilon |a_{k1}|\xi _k|x-x_0|,\quad x,x_0,\xi \in \Omega _\varepsilon , \end{aligned}$$

hence

$$\begin{aligned} \Vert \widetilde{H}(x)-\widetilde{H}(x_0)\Vert \le C_\varepsilon \sum _{k=-\infty }^{+\infty } |a_{k1}|\xi _k|x-x_0|\le C_\varepsilon \Lambda |x-x_0|. \end{aligned}$$

Choose $\delta _0>0$ such that $\Vert \widetilde{H}(x)-\widetilde{H}(x_0)\Vert \le (2\Vert \widetilde{R}(x_0)\Vert )^{-1}$ for $|x-x_0|\delta _0$, where $\widetilde{R}(x)=(I-\widetilde{H}(x))^{-1}.$ By Lemma 14, $\Vert \widetilde{R}(x)-\widetilde{R}(x_0)\Vert \le \ |\widetilde{R}(x_0)\Vert C_\varepsilon |x-x_0|,\;\;x\in \Omega _\varepsilon .$ Therefore, the function $f(x)=\Vert \widetilde{R}(x)\Vert$ is continuous and bounded on $\Omega _\varepsilon .$ Then

$$\begin{aligned} \Vert \widetilde{R}(x)\Vert \le C_\varepsilon ,\quad \Vert \widetilde{R}(x)-\widetilde{R}(x_0)\Vert \le C_\varepsilon |x-x_0|,\qquad x\in \Omega _\varepsilon , \end{aligned}$$

where the constant $C_\varepsilon$ does not depend on $x, x_0$. Since $\Psi ^{\left\langle {m}\right\rangle }(x)=\widetilde{R}(x)\widetilde{\Psi }^{\left\langle {m}\right\rangle }(x),$ it follows that $\Vert \Psi ^{\left\langle {m}\right\rangle }(x)\Vert \le \Vert \widetilde{R}(x)\Vert \,\Vert \widetilde{\Psi }^{\left\langle {m}\right\rangle }(x)\Vert \le C_\varepsilon .$ Thus, (37) is proved.

Taking (25)–(26) into account, we obtain

$$\begin{aligned} |\Psi ^{\left\langle {m}\right\rangle }_{ni}(x)-\widetilde{\Psi }^{\left\langle {m}\right\rangle }_{ni}(x)|\le \sum _{k=-\infty }^{+\infty }\sum _{j=0}^1|\widetilde{H}_{ni,kj}(x)|\,|\Psi ^{\left\langle {m}\right\rangle }_{kj}(x)| \le C_\varepsilon \sum _{k=-\infty }^{+\infty }\frac{|a_{k1}|\xi _k}{1+|\lambda ^0_k-\lambda ^0_n|}, \end{aligned}$$

and consequently,

$$\begin{aligned} |\Psi ^{\left\langle {m}\right\rangle }_{ni}(x)-\widetilde{\Psi }^{\left\langle {m}\right\rangle }_{ni}(x)|\le C_\varepsilon \theta _n\sum _{k=-\infty }^{+\infty }\Big (|a_{k1}|\xi _k(1+|\lambda ^0_k|)\Big )^2, \end{aligned}$$

i.e. (38) holds. Estimates (39) are proved similarly. Lemma is proved.

Construct the functions $\varphi _{2,ni}(x)=\Big (\varphi _{12,ni}(x),\; \varphi _{22,ni}(x)\Big )^T$ by the formula

$$\begin{aligned} \varphi _{m2,n0}(x)=\Psi ^{\left\langle {m}\right\rangle }_{n0}(x)\xi _n+\Psi ^{\left\langle {m}\right\rangle }_{n1}(x),\quad \varphi _{m2,n1}(x)=\Psi ^{\left\langle {m}\right\rangle }_{n1}(x),\qquad m=1,2.\quad \end{aligned}$$

(40)

It follows from (40) and Lemma 15 that

$$\begin{aligned}&|\varphi ^{(m)}_{2,ni}(x)|\le C_\varepsilon (1\!+\!|\lambda ^0_n|)^m,\quad |\varphi ^{(m)}_{2,n0}(x)\!-\!\varphi ^{(m)}_{2,n1}(x)| \le C_\varepsilon \xi _n(1+|\lambda ^0_n|)^m,\quad m=0,1,\end{aligned}$$

(41)

$$\begin{aligned}&|\widetilde{\varphi }_{2,ni}(x)-\varphi _{2,ni}(x)|\le C_\varepsilon \Lambda \theta _n,\quad |\widetilde{\varphi }\,'_{2,ni}(x)-\varphi \,'_{2,ni}(x)|\le C_\varepsilon \Lambda . \end{aligned}$$

(42)

Lemma 16

The function $Q(x),$ constructed by (28), is absolutely continuous on $[0,\pi ].$

Proof

In view of (41), the series in (29) converges uniformly on $\Omega _\varepsilon .$ According to Lemma 15, the functions $\varphi _{2,ni}(x)$ are continuous, and consequently, $\mathrm{\ae }(x)$ is continuous on $\Omega _\varepsilon .$ One has

$$\begin{aligned} \mathrm{\ae }(x)&= A_1(x)+A_2(x),\quad A_1(x)= \sum _{k=-\infty }^{+\infty }(a_{k0}-a_{k1})\widetilde{\varphi }_{2,k0}(x)\varphi ^T_{2,k0}(x),\\ A_2(x)&= \sum _{k=-\infty }^{+\infty }a_{k1}\Big (\widetilde{\varphi }_{2,k0}(x)\varphi ^T_{2,k0}(x) -\widetilde{\varphi }_{2,k1}(x)\varphi ^T_{2,k1}(x)\Big ). \end{aligned}$$

Taking (41) and (17) into account, we infer $|(a_{k0}-a_{k1})\widetilde{\varphi }_{2,k0}(x)\varphi ^T_{2,k0}(x)| \le |a_{k1}|\xi _kC_\varepsilon (1+|\lambda ^0_k|).$ This yields that the series for $A_1(x)$ converges uniformly on $\Omega _\varepsilon$ and $A'_1(x)\in L(0,\pi ).$ Since

$$\begin{aligned}&(\widetilde{\varphi }_{2,k0}(x)\varphi ^T_{2,k0}(x)- \widetilde{\varphi }_{2,k1}(x)\varphi ^T_{2,k1}(x))'= (\widetilde{\varphi }\,'_{2,k0}(x)-\widetilde{\varphi }\,'_{2,k1}(x))\varphi ^T_{2,k0}(x)\\&\quad +\,\widetilde{\varphi }\,'_{2,k1}(x)(\varphi _{2,k0}(x)-\varphi _{2,k1}(x))^T+ \widetilde{\varphi }_{2,k0}(x)(\varphi \,'_{2,k0}(x)-\varphi \,'_{2,k1}(x))^T\\&\quad +\,(\widetilde{\varphi }_{2,k0}(x)-\widetilde{\varphi }_{2,k1}(x)) (\varphi ^T_{2,k1}(x))', \end{aligned}$$

it follows from (41) and (17) that $|a_{k1}(\widetilde{\varphi }_{2,k0}(x)\varphi ^T_{2,k0}(x)- \widetilde{\varphi }_{2,k1}(x)\varphi ^T_{2,k1}(x))'| \le C_\varepsilon |a_{k1}|\xi _k(1+|\lambda ^0_k|).$ This yields $A'_2(x)\in L(0,\pi ).$ Thus, $\mathrm{\ae }(x)$ is absolutely continuous on $[0,\pi ]$ and $\mathrm{\ae }'(x)\in L(0,\pi ).$ Lemma is proved.

Let us now show that the given numbers $\{\lambda _{k}\}_{k=-\infty }^{+\infty }$ are eigenvalues of the constructed boundary value problem $L(Q_\omega (x),Q(x),0,\beta )$.

Lemma 17

The following relations hold

$$\begin{aligned}&\ell \varphi _{2,kj}(x)=\lambda _{kj}\varphi _{2,kj}(x), \quad \ell \Phi _j(x,\lambda )=\lambda \Phi _j(x,\lambda ),\end{aligned}$$

(43)

$$\begin{aligned}&\Phi _2(0,\lambda )=V_2(0),\quad V^T_1(0)\Phi _1(0,\lambda )=1,\quad V^T_1(\beta )\Phi _1(\pi ,\lambda )=0,\quad \Delta _{12}(\lambda _{k})\!=\!0.\nonumber \\ \end{aligned}$$

(44)

Proof

• 1. We construct $\Phi _j(x,\lambda )$ by (22). In view of (41)–(42), the series in (22) converges uniformly in $\Omega _\varepsilon$. Moreover, differentiating (22) and taking (21) into account, we obtain

  $$\begin{aligned}&\Phi '_j(x,\lambda )\!=\!\widetilde{\Phi }'_j(x,\lambda )\!+\!\sum _{k=-\infty }^{+\infty } \Big (\widetilde{D}^{\left\langle {j}\right\rangle }_{k0}(x,\lambda )a_{k0}\varphi \,'_{2,k0}(x)\!-\! \widetilde{D}^{\left\langle {j}\right\rangle }_{k1}(x,\lambda )a_{k1}\varphi \,'_{2,k1}(x)\Big )\\&\quad -\sum _{k=-\infty }^{+\infty }\Big (\widetilde{\Phi }^T_j(x,\lambda ) \widetilde{\varphi }_{2,k0}(x)a_{k0}\varphi _{2,k0}(x)- \widetilde{\Phi }^T_j(x,\lambda )\widetilde{\varphi }_{2,k1}(x)a_{k1}\varphi _{2,k1}(x)\Big ), \end{aligned}$$

  and consequently,

  $$\begin{aligned}&(\Phi '_j(x,\lambda ))^T=(\widetilde{\Phi }'_j(x,\lambda ))^T-\widetilde{\Phi }^T_j(x,\lambda )\mathrm{\ae }(x)\\&\quad +\sum _{k=-\infty }^{+\infty }\Big (\widetilde{D}^{\left\langle {j}\right\rangle }_{k0}(x,\lambda )a_{k0} (\varphi \,'_{2,k0}(x))^T-\widetilde{D}^{\left\langle {j}\right\rangle }_{k1}(x,\lambda )a_{k1}(\varphi \,'_{2,k1}(x))^T\Big ). \end{aligned}$$

  This yields

  $$\begin{aligned}&\Big (B\Phi '_j(x,\lambda )+Q(x)\Phi _j(x,\lambda )\Big )^T=(\widetilde{\Phi }'_j(x,\lambda ))^T +\widetilde{\Phi }^T_j(x,\lambda )\Big (Q(x)+\mathrm{\ae }(x)B\Big )\\&\quad +\sum _{k=-\infty }^{+\infty }\Big (\widetilde{D}^{\left\langle {j}\right\rangle }_{k0}(x,\lambda )a_{k0}\Big (B\varphi \,'_{2,k0}(x)+Q(x)\varphi _{2,k0}(x)\Big )^T\\&\quad -\widetilde{D}^{\left\langle {j}\right\rangle }_{k1}(x,\lambda )a_{k1}\Big (B\varphi \,'_{2,k1}(x)+Q(x)\varphi _{2,k1}(x)\Big )^T\Big ). \end{aligned}$$

  Since $Q(x)=\widetilde{Q}(x)+B\mathrm{\ae }(x)-\mathrm{\ae }(x)B$, $\widetilde{\Phi }^T_j(x,\lambda )B\widetilde{\varphi }_{2,ni}(x) =\widetilde{D}^{\left\langle {j}\right\rangle }_{ni}(x,\lambda )(\lambda -\lambda _{ni})$, it follows that

  $$\begin{aligned} \widetilde{\Phi }^T_j(x,\lambda )B\mathrm{\ae }(x)&= \sum _{k=-\infty }^{+\infty }\Big (\widetilde{D}^{\left\langle {j}\right\rangle }_{k0}(x,\lambda )(\lambda -\lambda _{k0})a_{k0}\varphi ^T_{2,k0}(x)\\&-\widetilde{D}^{\left\langle {j}\right\rangle }_{k1}(x,\lambda )(\lambda -\lambda _{k1})a_{k1}\varphi ^T_{2,k1}(x)\Big ), \end{aligned}$$

  hence

  $$\begin{aligned}&\Big (B\Phi '_j(x,\lambda )+Q(x)\Phi _j(x,\lambda )\Big )^T=\Big (\widetilde{\Phi }'_j(x,\lambda ) +\widetilde{Q}(x)\widetilde{\Phi }_j(x,\lambda )\Big )^T\\&\quad +\sum _{k=-\infty }^{+\infty }\Big (\widetilde{D}^{\left\langle {j}\right\rangle }_{k0}(x,\lambda )a_{k0} \Big (B\varphi \,'_{2,k0}(x)+Q(x)\varphi _{2,k0}(x)+(\lambda -\lambda _{k0})\varphi _{2,k0}(x))\Big )^T\\&\quad -\widetilde{D}^{\left\langle {j}\right\rangle }_{k1}(x,\lambda )a_{k1}\Big (B\varphi \,'_{2,k1}(x)+Q(x)\varphi _{2,k1}(x) +(\lambda -\lambda _{k1})\varphi _{2,k1}(x)\Big )^T\Big ). \end{aligned}$$

  Taking (22) into account, we calculate

  $$\begin{aligned}&\ell \Phi _j(x,\lambda )-\lambda \Phi _j(x,\lambda )=\sum _{k=-\infty }^{+\infty } \Bigg (\widetilde{D}^{\left\langle {j}\right\rangle }_{k0}(x,\lambda )a_{k0}\Big (\ell \varphi _{2,k0}(x)- \lambda _{k0}\varphi _{2,k0}(x)\Big )\nonumber \\&\quad -\widetilde{D}^{\left\langle {j}\right\rangle }_{k1}(x,\lambda )a_{k1}\Big (\ell \varphi _{2,k1}(x) -\lambda _{k1}\varphi _{2,k1}(x)\Big )\Bigg ). \end{aligned}$$

  (45)

  Consider (45) for $j=2$ and $\lambda =\lambda _{ni}$:

  $$\begin{aligned} z_{ni}(x)-\sum _{k=-\infty }^{+\infty }\Big (\widetilde{P}_{ni,k0}(x)z_{k0}(x) -\widetilde{P}_{ni,k1}(x)z_{k1}(x)\Big )=0, \end{aligned}$$

  where $z_{ni}(x)=\ell \varphi _{2,ni}(x)-\lambda _{ni}\varphi _{2,ni}(x),$ or

  $$\begin{aligned} Z^{\left\langle {m}\right\rangle }_{ni}(x)-\sum _{k,j}\widetilde{H}_{ni,kj}(x)Z^{\left\langle {m}\right\rangle }_{kj}(x)=0, \end{aligned}$$

  (46)

  where $Z^{\left\langle {m}\right\rangle }_{n0}=\Big (z_{m,n0}(x)-z_{m,n1}(x)\Big )\chi _n,\; Z^{\left\langle {m}\right\rangle }_{n1}(x)=z_{m,n1}(x)$, $z_{ni}(x)=(z_{1,ni}(x),\;z_{2,ni}(x))^T$, $m=1,2$. Taking (41) into account, we get $|Z^{\left\langle {m}\right\rangle }_{ni}(x)|\le C_\varepsilon (1+|\lambda ^0_n|)$. Using (46) and (26), we infer

  $$\begin{aligned} |Z^{\left\langle {m}\right\rangle }_{ni}(x)|\le C_\varepsilon \sum _{k=-\infty }^{+\infty } \frac{|a_{k1}|\xi _k(|\lambda ^0_k|+1)}{1+|\lambda ^0_n-\lambda ^0_k|}\le C_\varepsilon \Lambda , \end{aligned}$$

  and consequently, $\{Z^{\left\langle {m}\right\rangle }_{ni}(x)\}\in \mathfrak {m}$. Equation (46) has only trivial solution, i.e. $Z^{\left\langle {m}\right\rangle }_{ni}(x)=0,$ hence $\ell \varphi _{2,ni}(x)-\lambda _{ni}\varphi _{2,ni}(x)=0.$ The second relation (43) follows now from (45).

• 2. Since $\widetilde{D}^{\left\langle {2}\right\rangle }_{kj}(x,\lambda )=\displaystyle \frac{1}{\lambda -\lambda _{kj}} \det (\widetilde{\varphi }_2(x,\lambda ),\widetilde{\varphi }_{2,kj}(x)),$ it follows that

  $$\begin{aligned} \widetilde{D}^{\left\langle {2}\right\rangle }_{kj}(0,\lambda )=\displaystyle \frac{1}{\lambda -\lambda _{kj}}\det (V_2(0),V_2(0))=0. \end{aligned}$$

Using (22), we find $\Phi _2(0,\lambda )=\widetilde{\Phi }_2(0,\lambda )=V_2(0).$ Furthermore, taking $j=1$, $x=0$ in (22) and multiplying by $V^T_1(0)$, we calculate

$$\begin{aligned}&V^T_1(0)\Phi _1(0,\lambda )=V^T_1(0)\widetilde{\Phi }_1(0,\lambda )\nonumber \\&\quad +\sum _{k=-\infty }^{+\infty } \Big (\widetilde{D}^{\left\langle {1}\right\rangle }_{k0}(0,\lambda )a_{k0}V^T_1(0)\varphi _{2,k0}(0)- \widetilde{D}^{\left\langle {1}\right\rangle }_{k1}(0,\lambda )a_{k1}V^T_1(0)\varphi _{2,k1}(0)\Big ). \end{aligned}$$

Since $V^T_1(0)\varphi _{2,kj}(0)=V^T_1(0)V_2(0)=0,$ one gets $V^T_1(0)\Phi _1(0,\lambda )=V^T_1(0)\widetilde{\Phi }_1(0,\lambda )=1$. Thus, $\Phi _2(x,\lambda )=\varphi _2(x,\lambda )$ is a solution of (1) with the initial condition $\varphi _2(0,\lambda )=V_2(0).$ Then $\Delta _{12}(\lambda )=V^T_1(\beta )\varphi _2(\pi ,\lambda ).$ Let us show that $\Delta _{12}(\lambda _{n0})=0,$ i.e. $\{\lambda _n\}_{n=-\infty }^{+\infty }$ are eigenvalues of $L.$ For this purpose we take $j=2$, $x=\pi$ in (22) and multiply by $V^T_1(\beta ).$ This yields

$$\begin{aligned} \Delta _{12}(\lambda )=\widetilde{\Delta }_{12}(\lambda )+\sum _{k=-\infty }^{+\infty }\Big (\widetilde{D}^{\left\langle {2}\right\rangle }_{k0}(\pi ,\lambda )a_{k0}\Delta _{12}(\lambda _{k0})-\widetilde{D}^{\left\langle {2}\right\rangle }_{k1}(\pi ,\lambda )a_{k1}\Delta _{12}(\lambda _{k1})\Big ), \end{aligned}$$

and consequently,

$$\begin{aligned} \Delta _{12}(\lambda _{ni})\!=\!\widetilde{\Delta }_{12}(\lambda _{ni})+\sum _{k=-\infty }^{+\infty } \Big (\widetilde{P}_{ni,k0}(\pi )\Delta _{12}(\lambda _{k0}) -\widetilde{P}_{ni,k1}(\pi )\Delta _{12}(\lambda _{k1})\Big ). \end{aligned}$$

(47)

By definition $\Delta (\lambda )=V^T(\beta )\varphi (\pi ,\lambda )$, then $\det \Delta (\lambda )\equiv 1,$ or

$$\begin{aligned} \Delta _{11}(\lambda )\Delta _{22}(\lambda )-\Delta _{12}(\lambda )\Delta _{21}(\lambda )\equiv 1. \end{aligned}$$

(48)

Furthermore, ${\left\langle {\widetilde{\varphi }_2(\pi ,\lambda ),\,\widetilde{\varphi }_{2,kj}(\pi )}\right\rangle } =\widetilde{\varphi }^T_2(\pi ,\lambda )B\widetilde{\varphi }_{2,kj}(\pi ).$ Since $V(\beta )V^T(\beta )=I,$ it follows that ${\Big \langle {\widetilde{\varphi }_2(\pi ,\lambda ),\,\widetilde{\varphi }_{2,kj}(\pi )}\Big \rangle }= \Big (V^T(\beta )\varphi _2(\pi ,\lambda )\Big )^TB V^T(\beta )\varphi _2(\pi ,\lambda _{kj})$, and consequently,

$$\begin{aligned} {\Big \langle {\widetilde{\varphi }_2(\pi ,\lambda ),\,\widetilde{\varphi }_{2,kj}(\pi )}\Big \rangle }= \widetilde{\Delta }_{12}(\lambda )\widetilde{\Delta }_{22}(\lambda _{kj})- \widetilde{\Delta }_{12}(\lambda _{kj})\widetilde{\Delta }_{22}(\lambda ). \end{aligned}$$

Thus,

$$\begin{aligned} \widetilde{D}^{\left\langle {2}\right\rangle }_{kj}(\pi ,\lambda )=\frac{1}{\lambda -\lambda _{kj}} \Big (\widetilde{\Delta }_{12}(\lambda )\widetilde{\Delta }_{22}(\lambda _{kj})- \widetilde{\Delta }_{12}(\lambda _{kj})\widetilde{\Delta }_{22}(\lambda )\Big ). \end{aligned}$$

(49)

From (49) for $n\ne k$, we find

$$\begin{aligned} \widetilde{P}_{n1,k1}(\pi )=\displaystyle \frac{a_{k1}}{\lambda _{n1}-\lambda _{k1}} \Big (\widetilde{\Delta }_{12}(\lambda _{n1})\widetilde{\Delta }_{22}(\lambda _{k1})- \widetilde{\Delta }_{12}(\lambda _{k1})\widetilde{\Delta }_{22}(\lambda _{n1})\Big )=0. \end{aligned}$$

For $n=k,$ one has $\widetilde{P}_{n1,n1}(\pi )=a_{n1} \dot{\widetilde{\Delta }}_{12}(\lambda _{n1})\widetilde{\Delta }_{22}(\lambda _{n1})$, where $\dot{\widetilde{\Delta }}_{12}(\lambda ):=\frac{d}{d\lambda }{\widetilde{\Delta }}_{12}(\lambda ).$ Since $a_{n1}=\mathop {\text {Res}}\limits \nolimits _{\lambda =\lambda _{n1}}\widetilde{\mathfrak {M}}(\lambda )= -(\widetilde{\Delta }_{11}(\lambda _{n1}))(\dot{\widetilde{\Delta }}_{12}(\lambda _{n1}))^{-1}$, it follows that $\widetilde{P}_{n1,n1}(\pi )=-\widetilde{\Delta }_{11}(\lambda _{n1}) \widetilde{\Delta }_{22}(\lambda _{n1})$. From (48) for $\lambda =\lambda _{n1}$ we infer $\widetilde{P}_{n1,n1}(\pi )=-1.$ Thus,

$$\begin{aligned} \widetilde{P}_{n1,k1}(\pi )=-\delta _{nk}, \end{aligned}$$

(50)

where $\delta _{nk}$ is the Kronecker symbol. From (49) for $\lambda _{n0}\ne \lambda _{k1}$ one has

$$\begin{aligned} \widetilde{P}_{n0,k1}(\pi )&= \displaystyle \frac{a_{k1}}{\lambda _{n0}-\lambda _{k1}} \Big (\widetilde{\Delta }_{12}(\lambda _{n0})\widetilde{\Delta }_{22}(\lambda _{k1})- \widetilde{\Delta }_{12}(\lambda _{k1})\widetilde{\Delta }_{22}(\lambda _{n0})\Big )\\&= a_{k1}\frac{\widetilde{\Delta }_{12}(\lambda _{n0}) \widetilde{\Delta }_{22}(\lambda _{k1})}{\lambda _{n0}-\lambda _{k1}}. \end{aligned}$$

By virtue of (48), $\widetilde{\Delta }_{22}(\lambda _{k1})= (\widetilde{\Delta }_{11}(\lambda _{k1}))^{-1}$. Moreover, $\widetilde{P}_{n0,k1}(\pi )=-1$ for $\lambda _{n0}=\lambda _{k1}$. Thus,

$$\begin{aligned} \widetilde{P}_{n0,k1}(\pi )\!=\!-1\hbox { for }\lambda _{n0}\!=\!\lambda _{k1},\quad \widetilde{P}_{n0,k1}(\pi )\!=\!-\displaystyle \frac{\widetilde{\Delta }_{12}(\lambda _{n0})}{\dot{\widetilde{\Delta }}_{12}(\lambda _{k1})(\lambda _{n0}-\lambda _{k1})} \quad \hbox {for }\lambda _{n0}\!\ne \!\lambda _{k1}.\nonumber \\ \end{aligned}$$

(51)

Consider the function $Z(\lambda )=(\Delta _{12}(\lambda )-\widetilde{\Delta }_{12}(\lambda )) (\widetilde{\Delta }_{12}(\lambda ))^{-1}$. Since $Q_\omega (x)=\widetilde{Q}_\omega (x)$, $\alpha =\widetilde{\alpha }=0$, $\beta =\widetilde{\beta }=0,$ it follows that $\Delta _{12}(\lambda )-\widetilde{\Delta }_{12}(\lambda )=O\Big (e^{\pi |Im\lambda |}|\lambda |^{-\nu }\Big ),$ hence $|Z(\lambda )|\le C_\delta |\lambda |^{-\nu }$ for $\lambda \in \widetilde{G}_\delta .$ In particular, this yields $\int _{|\xi |=\widetilde{R}_n} \frac{Z(\xi )}{\xi -\lambda }d\xi \rightarrow 0$ as $n\rightarrow \infty .$ Calculating the integral by residue’s theorem, we obtain for $n\rightarrow \infty ,$

$$\begin{aligned} \Delta _{12}(\lambda )=\widetilde{\Delta }_{12}(\lambda )+\sum _{k=-\infty }^{+\infty } \frac{\widetilde{\Delta }_{12}(\lambda )}{(\lambda -\lambda _{k1}) \dot{\widetilde{\Delta }}_{12}(\lambda _{k1})} \Delta _{12}(\lambda _{k1}). \end{aligned}$$

Putting $\lambda =\lambda _{n0}$ and taking (51) into account, we infer

$$\begin{aligned} \Delta _{12}(\lambda _{n0})=\widetilde{\Delta }_{12}(\lambda _{n0})- \sum _{k=-\infty }^{+\infty }\widetilde{P}_{n0,k1}(\pi )\Delta _{12}(\lambda _{k1}). \end{aligned}$$

Together with (47) and (50) this yields

$$\begin{aligned} \sum _{k=-\infty }^{+\infty }\widetilde{P}_{ni,k0}(\pi )\Delta _{12}(\lambda _{k0})=0, \end{aligned}$$

and consequently, $\Delta _{12}(\lambda _{n0})=0.$ Now we take $j=1$, $x=\pi$ in (22) and multiply by $V^T_1(\beta ).$ Then

$$\begin{aligned} V^T_1(\beta )\Phi _1(\pi ,\lambda )&= V^T_1(\beta )\widetilde{\Phi }_1(\pi ,\lambda )\\&+\sum _{k=-\infty }^{+\infty }\Big (\widetilde{D}^{\left\langle {1}\right\rangle }_{k0}(\pi ,\lambda )a_{k0} \Delta _{12}(\lambda _0)-\widetilde{D}^{\left\langle {1}\right\rangle }_{k1}(\pi ,\lambda )a_{k1}\Delta _{12}(\lambda _1)\Big ). \end{aligned}$$

Furthermore, $\widetilde{D}^{\left\langle {1}\right\rangle }_{k1}(\pi ,\lambda )=\displaystyle \frac{1}{\lambda -\lambda _{k1}} (V^T(\beta )\widetilde{\Phi }_1(\pi ,\lambda ))^T BV^T(\beta )\widetilde{\varphi }_2(\pi ,\lambda _{k1})$ or

$$\begin{aligned} \widetilde{D}^{\left\langle {1}\right\rangle }_{k1}(\pi ,\lambda )=\frac{1}{\lambda -\lambda _{k1}} \Big (V^T_1(\beta )\widetilde{\Phi }_1(\pi ,\lambda )\widetilde{\Delta }_{22}(\lambda _{k1})- V^T_2(\beta )\widetilde{\Phi }_1(\pi ,\lambda )\widetilde{\Delta }_{12}(\lambda _{k1})\Big ), \end{aligned}$$

hence, $\widetilde{D}^{\left\langle {1}\right\rangle }_{k1}(\pi ,\lambda )=0.$ Thus, $V^T_1(\beta )\Phi _1(\pi ,\lambda )=V^T_1(\beta )\widetilde{\Phi }_1(\pi ,\lambda )=0$, and all relations (44) are valid. Lemma 17 is proved.

It remains to show that $a_k=\mathrm{Res}_{\lambda =\lambda _{k}}{\mathfrak {M}}(\lambda ),$ where ${\mathfrak {M}}(\lambda )=V^T_2(0)\Phi _1(0,\lambda ).$ Taking in (22) $j=1$, $x=0$ and multiplying by $V^T_2(0),$ we obtain

$$\begin{aligned} {\mathfrak {M}}(\lambda )&= \widetilde{\mathfrak {M}}(\lambda )+\sum _{k=-\infty }^{+\infty } \Big (\widetilde{D}^{\left\langle {1}\right\rangle }_{k0}(0,\lambda )a_{k0}V^T_2(0)\varphi _{2,k0}(0)\nonumber \\&- \widetilde{D}^{\left\langle {1}\right\rangle }_{k1}(0,\lambda )a_{k1}V^T_2(0)\varphi _{2,k1}(0)\Big ). \end{aligned}$$

(52)

Using Lemma 17, we calculate $V^T_2(0)\varphi _{2,kj}(0)=V^T_2(0)V_2(0)=1$. Since

$$\begin{aligned} \widetilde{D}^{\left\langle {1}\right\rangle }_{kj}(0,\lambda )=\displaystyle \frac{1}{\lambda -\lambda _{kj}} \widetilde{\Phi }^T_1(0,\lambda )B\widetilde{\varphi }_{2,kj}(0),\quad \widetilde{\varphi }_{2,kj}(0)=V_2(0), \end{aligned}$$

it follows that $\widetilde{D}^{\left\langle {1}\right\rangle }_{kj}(0,\lambda )=\displaystyle \frac{1}{\lambda -\lambda _{kj}}$. Thus, (52) takes the form

$$\begin{aligned} {\mathfrak {M}}(\lambda )=\widetilde{\mathfrak {M}}(\lambda )+\sum _{k=-\infty }^{+\infty }\left( \frac{a_{k0}}{\lambda -\lambda _{k0}} -\frac{a_{k1}}{\lambda -\lambda _{k1}}\right) . \end{aligned}$$

With the help of Lemma 6, this yields $a_k=\mathrm{Res}_{\lambda =\lambda _{k}}{\mathfrak {M}}(\lambda ).$ Theorem 3 is proved.