A partial inverse problem for the Dirac system with an integral delay

ABSTRACT

The Dirac system with an integral delay in the form of convolution is considered. We suppose that the convolution kernel, having four independent components, is known a priori on a part of the interval. A partial inverse problem is studied, which consists in recovering the kernel on the remaining part of the interval from some fractional parts of two spectra. Uniqueness theorems are proved, and a constructive algorithm is developed for solving this inverse problem.

KEYWORDS:

• Integro-differential Dirac system
• inverse spectral problem
• partial inverse problem
• nonlocal operator
• Riesz basis

AMS MATHEMATICS SUBJECT CLASSIFICATION (2010):

• 34A55
• 34B05
• 34B09
• 34B10
• 34L40
• 47E05

1. Introduction and main results

The paper concerns the inverse spectral theory for integro-differential operators. Operators of this kind have many applications in science and engineering, in particular, in biological population models, in the theory of nuclear reactors, in the elasticity theory (see [1,2]). Integro-differential operators are nonlocal, and therefore they are often more adequate for description of physical processes, than differential operators. However, the spectral theory for integro-differential operators has not been completely constructed yet, and contains only a few fragmentary results.

Inverse problems of spectral analysis consist in recovering operators from their spectral characteristics. For differential operators, there exist powerful methods, that reduce nonlinear inverse spectral problems to some linear equations (e.g. Gelfand-Levitan method, the method of spectral mappings, see [3–5]). For integro-differential operators, those methods do not work because of nonlocality. In recent years, the new method has been developed for solving inverse problems for integral operators [6] and, later on, for integro-differential operators [7–15]. This method reduces inverse spectral problems to uniquely solvable nonlinear integral equations, and allows to prove uniqueness theorems, obtain algorithms for solution together with necessary and sufficient conditions for solvability of inverse problems. In particular, this method yielded the results for the Sturm-Liouville (second order) integro-differential operators [7,8], Dirac integro-differential systems [9–11], first-order operators with discontinuities [12], integro-differential operators of arbitrary order [13] and fractional order [14,15]. Some other issues of inverse problem theory for integro-differential operators were studied in [16–21]. However, this topic still contains many open questions.

This paper is devoted to the Dirac system with the integral delay in the form of convolution: (1) $B{y}^{\prime }+{\int }_0^{x}M(x-t)y\left(t\right)\mathrm{d}t=\lambda y,x\in (0,\pi ),$(1) where $B=\left(\begin{array}{cc}0& 1\\ -1& 0\end{array}\right),M\left(x\right)=\left(\begin{array}{cc}{M}_1\left(x\right)& M_2\left(x\right)\\ M_3\left(x\right)& M_4\left(x\right)\end{array}\right),y\left(x\right)=\left(\begin{array}{c}{y}_1\left(x\right)\\ y_2\left(x\right)\end{array}\right),$ the functions $(\pi -x)M_k\left(x\right)$, $k=\overline{1,4}$, belong to $L_2(0,\pi )$ and are complex-valued. For j=1, 2, denote by $D_j=D_j\left(M\right)$ the boundary value problems for the system (1(1) $B{y}^{\prime }+{\int }_0^{x}M(x-t)y\left(t\right)\mathrm{d}t=\lambda y,x\in (0,\pi ),$(1) ) with the boundary conditions (2) $y_1\left(0\right)=y_j\left(\pi \right)=0.$(2) The spectra of these boundary value problems are described by the following lemma.

Lemma 1.1

[9,10]

The spectra of $D_j,$ $j=1,2,$ are countable sets of complex eigenvalues, which can be numbered as $\{{\lambda }_{nj}{\}}_{n\in \mathbb{Z}}$ $($counting with their multiplicities$),$ so that ${\lambda }_{n1}=n+{\varkappa }_{n1},{\lambda }_{n2}=n+\frac{1}{2}+{\varkappa }_{n2},\left\{{\varkappa }_{nj}\right\}\in l_2.$

In [10] the inverse problem has been studied, that consists in recovering the kernel $M\left(x\right)$ from two spectra $\{{\lambda }_{nj}{\}}_{n\in \mathbb{Z}}$, j=1, 2. In the earlier paper [9], the special symmetric case was considered, when $M_1\left(x\right)=M_4\left(x\right)$, $M_2\left(x\right)=-M_3\left(x\right)$. In that case, the only spectrum $\{{\lambda }_{n1}{\}}_{n\in \mathbb{Z}}$ is sufficient for reconstruction of $M\left(x\right)$.

In the present paper, we study the partial inverse problem, formulated below. Let a be a real number, $\pi /2\le a<\pi$, and let ${\mathcal{I}}_j$, j=1, 2, be some subsets of $\mathbb{Z}$.

Inverse Problem 1

Suppose that the matrix-function $M\left(x\right)$ for $x\in (0,a)$ is known a priori. Given the subspectra $\{{\lambda }_{nj}{\}}_{n\in {\mathcal{I}}_j}$, $j=1,2$, construct $M\left(x\right)$ on $(a,\pi )$.

Inverse Problem 1 is an analogue of the famous Hochstadt-Lieberman problem, which consists in recovering the potential of the differential Sturm-Liouville operator from the spectrum, while the potential is known a priori on a half of the interval. The Hochstadt-Lieberman problem and its generalizations for differential operators were investigated in [22–29,30]. In the paper [8], the first results has been obtained in this direction for integro-differential operators (namely, for the integro-differential Sturm-Liouville operator). The goals of the present paper are to prove uniqueness theorems and to develop a constructive algorithm for solving Inverse Problem 1 under some assumptions on the sets ${\mathcal{I}}_j$, j=1, 2.

We proceed to the formulation of the main results. Along with $D_j$, $j=1,2$, we consider the boundary value problems ${\tilde{D}}_j=D_j\left(\tilde{M}\right)$, j=1, 2, of the same form (1(1) $B{y}^{\prime }+{\int }_0^{x}M(x-t)y\left(t\right)\mathrm{d}t=\lambda y,x\in (0,\pi ),$(1) )–(2(2) $y_1\left(0\right)=y_j\left(\pi \right)=0.$(2) ), but with a different convolution kernel $\tilde{M}\left(x\right)$. We agree that if a certain symbol γ denotes an object related to $D_j$, then $\tilde{\gamma }$ with tilde denotes an analogous object related to ${\tilde{D}}_j$. Define $b:=\pi -a$. Denote by $\mathcal{H}$ the complex Hilbert space of vector functions $L_2(0,b)\oplus L_2(0,b)$ with the scalar product $\begin{array}{rl}(g,h{)}_{\mathcal{H}}& ={\int }_0^b\left(\overline{g_1\left(x\right)}{h}_1\right(x)+\overline{g_2\left(x\right)}{h}_2(x\left)\right)\mathrm{d}x,g=\left(\begin{array}{c}{g}_1\\ g_2\end{array}\right),h=\left(\begin{array}{c}{h}_1\\ h_2\end{array}\right),\\ & g_j,h_j\in L_2(0,b),j=1,2.\end{array}$ Define the sequences of vector functions (3) ${\mathcal{S}}_j:={\left\{\left(\begin{array}{c}\sin {\lambda }_{nj}x\\ \cos {\lambda }_{nj}x\end{array}\right)\right\}}_{n\in {\mathcal{I}}_j},j=1,2.$(3) The uniqueness theorem for Inverse Problem 1 is formulated as follows.

Theorem 1.1

Suppose that $M\left(x\right)=\tilde{M}\left(x\right)$ a.e. on $(0,a)$, ${\lambda }_{nj}={\tilde{\lambda }}_{nj}$, $n\in {\mathcal{I}}_j$, j=1, 2, and the sequences ${\mathcal{S}}_j$, j=1, 2, are complete in $\mathcal{H}$. Then $M\left(x\right)=\tilde{M}\left(x\right)$ a.e. on $(a,\pi )$.

Remark 1.1

The numeration of eigenvalues in the sequences $\{{\lambda }_{nj}{\}}_{n\in \mathbb{Z}}$ and $\{{\tilde{\lambda }}_{nj}{\}}_{n\in \mathbb{Z}}$, j=1, 2, is not uniquely fixed by Lemma 1.1. Therefore the formulation of Theorem 1.1 should be understood in the following sense: suppose that there exists such numeration of the eigenvalues, that the conditions of the theorem hold.

Remark 1.2

The completeness of the sequences ${\mathcal{S}}_j$ implies, that the eigenvalues in the subspectra $\{{\lambda }_{nj}{\}}_{n\in {\mathcal{I}}_j}$ are distinct for each fixed j=1, 2. However, our results can be generalized for the case of multiple eigenvalues, by using the approach of [11].

For the case when the sequences ${\mathcal{S}}_j$, j=1, 2, are Riesz bases in $\mathcal{H}$, we develop a constructive procedure for solving Inverse Problem 1 (see Algorithm 3.1). One can find more information about Riesz bases in [5, Section 1.8.5].

We also prove the following theorem for the special case.

Theorem 1.2

Let $a=\pi -\pi /k$, $k\in \mathbb{N}$, $k\ge 2$, and ${\mathcal{I}}_j=\{n=kl+s_j:l\in \mathbb{Z}\}$, $s_j\in \mathbb{R}$, $j=1,2$. Suppose that $M\left(x\right)=\tilde{M}\left(x\right)$ a.e. on $(0,a)$, ${\lambda }_{nj}={\tilde{\lambda }}_{nj}$, $n\in {\mathcal{I}}_j$, j=1, 2, and for j=1, 2, the subspectrum $\{{\lambda }_{nj}{\}}_{n\in {\mathcal{I}}_j}$ consists of distinct eigenvalues. Then $M\left(x\right)=\tilde{M}\left(x\right)$ a.e. on $(a,\pi )$. Thus, in this case the solution of Inverse Problem 1 is unique and can be found by using Algorithm 3.1.

Roughly speaking, if the convolution kernel $M\left(x\right)$ is unknown on the interval of length $\pi /k$, then we need a $1/k$th part of each spectrum to solve the inverse problem.

Our approach is based on the so-called main equation, derived in [10]. It is a nonlinear integral equation with respect to the unknown convolution kernels $M_{\nu }\left(x\right)$, $\nu =\overline{1,4}$. We show that, if the matrix-function $M\left(x\right)$ is known a priori on the interval $(0,a)$, $a\ge \pi /2$, the main equation is linear on the remaining interval $(a,\pi )$. Under the certain conditions on the subspectra $\{{\lambda }_{nj}{\}}_{n\in {\mathcal{I}}_j}$, j=1, 2, this equation can be used for solving the partial Inverse Problem 1.

The paper is organized as follows. In Section 2, we provide the nonlinear main equation and other preliminaries. In Section 3, we prove the uniqueness Theorems 1.1, 1.2 and develop an algorithm for solving the partial inverse problem. In Section 4, we illustrate our method by a numerical example.

2. Preliminaries

In this section, we formulate some results of [10], that will be important for investigation of the partial inverse problem.

Let $S(x,\lambda )=\left(S_1\right(x,\lambda ),S_2(x,\lambda ){)}^{\mathrm{T}}$ be the solution of the system (1(1) $B{y}^{\prime }+{\int }_0^{x}M(x-t)y\left(t\right)\mathrm{d}t=\lambda y,x\in (0,\pi ),$(1) ), satisfying the initial conditions $S_1(0,\lambda )=0$, $S_2(0,\lambda )=-1$. Here and below T is the transposition sign. Note that $S_j(x,\lambda )$ are entire functions of λ for each fixed $x\in [0,\pi ]$, j=1, 2. For j=1, 2, the eigenvalues of the boundary value problem $D_j$ coincide with the zeros of the characteristic function ${\mathrm{\Delta }}_j\left(\lambda \right):=S_j(\pi ,\lambda )$, counting with their multiplicities. The characteristic functions admit the representation (4) ${\mathrm{\Delta }}_j\left(\lambda \right)={\mathrm{\Delta }}_j^0\left(\lambda \right)+{\int }_0^{\pi }\left(v_{j1}\right(x)\sin \lambda x+v_{j2}(x)\cos \lambda x)\mathrm{d}x,j=1,2,$(4) where ${\mathrm{\Delta }}_1^0\left(\lambda \right):=\sin \lambda \pi$, ${\mathrm{\Delta }}_2^0\left(\lambda \right):=-\cos \lambda \pi$, and $v_{lm}$, l, m=1, 2, are functions from $L_2(0,\pi )$.

Introduce the notations $\begin{array}{rl}(f\ast g)\left(x\right)& :={\int }_0^{x}f(x-t)g\left(t\right)\mathrm{d}t,\\ g_n\left(x\right)& :=\frac{x^n}{n!},n\ge 0,f\ast g_{-1}:=f.\end{array}$ Denote by $Q_{nj}\left[M\right]$, $j=\overline{1,m_n}$, all possible convolution monomials of the form (5) $Q_{nj}\left[M\right]=M_{i_1}\ast M_{i_2}\ast \cdots \ast M_{i_n},1\le i_1\le i_2\le \cdots \le i_n\le 4.$(5)

In [10] the following nonlinear integral main equation with respect to the convolution kernels $M_{\nu }$, $\nu =\overline{1,4}$, has been derived: (6) $w_{\nu }\left(x\right)=(\pi -x)M_{\nu }\left(x\right)+\sum _{n=2}^{\mathrm{\infty }}\sum _{j=1}^{m_n}\sum _{k=1}^n{b}_{njk}^{\left(\nu \right)}{g}_k(\pi -x)\left(Q_{nj}\right[M]\ast g_{n-1-k})\left(x\right),\nu =\overline{1,4}.$(6) Here $b_{njk}^{\left(\nu \right)}$ are certain constant coefficients, independent of M and satisfying the estimate $\sum _{j=1}^{m_n}\sum _{k=1}^n|b_{njk}^{\left(\nu \right)}|\le Cn{4}^n,n\ge 2,\nu =\overline{1,4},$ where C is an absolute constant, and (7) $\left.\begin{array}{ll}{w}_1\left(x\right)& =-v_{21}(\pi -x)-(\pi -x)(v_{12}-v_{21}{)}^{\prime }(\pi -x),\\ w_2\left(x\right)& =-v_{11}(\pi -x)-(\pi -x)(v_{11}+v_{22}{)}^{\prime }(\pi -x),\\ w_3\left(x\right)& =v_{11}(\pi -x)-(\pi -x)(v_{11}+v_{22}{)}^{\prime }(\pi -x),\\ w_4\left(x\right)& =-v_{21}(\pi -x)+(\pi -x)(v_{12}-v_{21}{)}^{\prime }(\pi -x).\end{array}\right\}$(7)

It has been shown in [10], that, if $(\pi -x)M_{\nu }\left(x\right)\in L_2(0,\pi )$, $\nu =\overline{1,4}$, then the functions $(v_{12}-v_{21})$ and $(v_{11}+v_{22})$ are absolutely continuous on $(0,\pi ]$ and $w_{\nu }\in L_2(0,\pi )$, $\nu =\overline{1,4}$. Furthermore, one can calculate the functions $v_{lm}$, l, m=1, 2, by the following formulae: (8) $\left.\begin{array}{ll}{v}_{11}\left(x\right)& =-\frac{x}{2}(M_2-M_3)(\pi -x)+u_{11}\left(x\right),\\ v_{12}\left(x\right)& =-\frac{x}{2}(M_1+M_4)(\pi -x)+\frac{1}{2}{\int }_0^{\pi -x}(M_1-M_4)\left(t\right)\mathrm{d}t-u_{12}\left(x\right),\\ v_{21}\left(x\right)& =-\frac{x}{2}(M_1+M_4)(\pi -x)+u_{21}\left(x\right),\\ v_{22}\left(x\right)& =\frac{x}{2}(M_2-M_3)(\pi -x)+\frac{1}{2}{\int }_0^{\pi -x}(M_2+M_3)\left(t\right)\mathrm{d}t-u_{22}\left(x\right),\end{array}\right\}$(8) where $u_{lm}\left(x\right)=\sum _{n=2}^{\mathrm{\infty }}\sum _{j=1}^{m_n}\sum _{k=0}^n{a}_{njk}^{lm}{g}_k\left(x\right)\left(Q_{nj}\right[M]\ast g_{n-1-k})(\pi -x),$ the coefficients $a_{njk}^{lm}$ are certain constants and $\sum _{j=1}^{m_n}\sum _{k=0}^n|a_{njk}^{lm}|\le C{4}^n,n\in \mathbb{N},l,m=1,2.$

3. Proofs of the main results

Suppose that the functions $M_{\nu }\left(x\right)$, $\nu =\overline{1,4}$, are known on the subinterval $(0,a)$, $\pi /2\le a<\pi$. First we modify the main equation (6(6) $w_{\nu }\left(x\right)=(\pi -x)M_{\nu }\left(x\right)+\sum _{n=2}^{\mathrm{\infty }}\sum _{j=1}^{m_n}\sum _{k=1}^n{b}_{njk}^{\left(\nu \right)}{g}_k(\pi -x)\left(Q_{nj}\right[M]\ast g_{n-1-k})\left(x\right),\nu =\overline{1,4}.$(6) ) for this case. Represent the functions $M_{\nu }\left(x\right)$ in the following form: (9) $M_{\nu }\left(x\right)=M_{\nu ,1}\left(x\right)+M_{\nu ,2}\left(x\right),$(9) where $M_{\nu ,1}\left(x\right)\equiv 0$ on $(a,\pi )$ and $M_{\nu ,2}\left(x\right)\equiv 0$ on $(0,a)$, $\nu =\overline{1,4}$. Note that (10) $M_{i_1,2}\ast M_{i_2,2}\equiv 0,i_1,i_2\in \{1,2,3,4\}.$(10) Substituting (9(9) $M_{\nu }\left(x\right)=M_{\nu ,1}\left(x\right)+M_{\nu ,2}\left(x\right),$(9) ) into (5(5) $Q_{nj}\left[M\right]=M_{i_1}\ast M_{i_2}\ast \cdots \ast M_{i_n},1\le i_1\le i_2\le \cdots \le i_n\le 4.$(5) ) and using (10(10) $M_{i_1,2}\ast M_{i_2,2}\equiv 0,i_1,i_2\in \{1,2,3,4\}.$(10) ), we get the following representation for $Q_{nj}\left[M\right]$, $n\ge 2$, $j=\overline{1,m_n}$: $Q_{nj}\left[M\right]=Q_{nj}\left[M_{\left(1\right)}\right]+\sum _{\nu =1}^4{M}_{\nu ,2}\ast \sum _{s=1}^{m_{n-1}}{\beta }_{njs}^{\left(\nu \right)}{Q}_{n-1,s}\left[M_{\left(1\right)}\right],$ where ${\beta }_{njs}^{\left(\nu \right)}$ are certain integer coefficients from the interval $[0,n]$ and $Q_{nj}\left[M_{\left(1\right)}\right]=M_{i_1,1}\ast M_{i_2,1}\ast \cdots \ast M_{i_n,1},1\le i_1\le i_2\le \cdots \le i_n\le 4.$ Then the main equations (6(6) $w_{\nu }\left(x\right)=(\pi -x)M_{\nu }\left(x\right)+\sum _{n=2}^{\mathrm{\infty }}\sum _{j=1}^{m_n}\sum _{k=1}^n{b}_{njk}^{\left(\nu \right)}{g}_k(\pi -x)\left(Q_{nj}\right[M]\ast g_{n-1-k})\left(x\right),\nu =\overline{1,4}.$(6) ) for $x\in [a,\pi ]$ can be rewritten in the form $\begin{array}{rl}{w}_{\nu }\left(x\right)& =(\pi -x)M_{\nu ,2}\left(x\right)+\sum _{n=2}^{\mathrm{\infty }}\sum _{j=1}^{m_n}\sum _{k=1}^n{b}_{njk}^{\left(\nu \right)}{g}_k(\pi -x)\left(Q_{nj}\right[M_{\left(1\right)}]\ast g_{n-1-k})\left(x\right)\\ & +\sum _{n=2}^{\mathrm{\infty }}\sum _{j=1}^{m_n}\sum _{k=1}^n{b}_{njk}^{\left(\nu \right)}{g}_k(\pi -x)\sum _{\xi =1}^4{M}_{\xi ,2}\ast \sum _{s=1}^{m_{n-1}}{\beta }_{njs}^{\left(\xi \right)}\left(Q_{n-1,s}\right[M_{\left(1\right)}]\ast g_{n-1-k})\left(x\right),\\ & \nu =\overline{1,4}.\end{array}$ Thus, we obtain the Volterra integral equation of the second kind (11) $\mu \left(x\right)=z\left(x\right)+{\int }_a^{x}H(x,t)z\left(t\right)\mathrm{d}t,a<x<\pi ,$(11) where $\mu \left(x\right)=\left({\mu }_{\nu }\right(x){)}_{\nu =1}^4$ and $z\left(x\right)=\left(z_{\nu }\right(x){)}_{\nu =1}^4$ are column vector functions and $H(x,t)=\left(H_{\nu \xi }\right(x,t){)}_{\nu ,\xi =1}^4$ is the $4\times 4$ matrix function with the elements defined as follows: (12) $\begin{array}{rl}{z}_{\nu }\left(x\right)& =(\pi -x)M_{\nu ,2}\left(x\right),\\ {\mu }_{\nu }\left(x\right)& =w_{\nu }\left(x\right)-\sum _{n=2}^{\mathrm{\infty }}\sum _{j=1}^{m_n}\sum _{k=1}^n{b}_{njk}^{\left(\nu \right)}{g}_k(\pi -x)\left(Q_{nj}\right[M_{\left(1\right)}]\ast g_{n-1-k})\left(x\right),\end{array}$(12) (13) $\begin{array}{rl}{H}_{\nu \xi }(x,t)& =\frac{1}{\pi -t}\sum _{n=2}^{\mathrm{\infty }}\sum _{j=1}^{m_n}\sum _{k=1}^n{b}_{njk}^{\left(\nu \right)}{g}_k(\pi -x)\\ & \times \sum _{s=1}^{m_{n-1}}{\beta }_{njs}^{\left(\xi \right)}\left(Q_{n-1,s}\right[M_{\left(1\right)}]\ast g_{n-1-k})(x-t).\end{array}$(13) One can easily show that ${\mu }_{\nu }\in L_2(a,\pi )$ and $H_{\nu \xi }\in L_2\left(\right(a,\pi )\times (a,\pi \left)\right)$, where we suppose that $H_{\nu \xi }(x,t)\equiv 0$ for $t>\xi$, $\nu ,\xi =\overline{1,4}$. Consequently, Equation (11(11) $\mu \left(x\right)=z\left(x\right)+{\int }_a^{x}H(x,t)z\left(t\right)\mathrm{d}t,a<x<\pi ,$(11) ) has the unique solution z, such that $z_{\nu }\in L_2(a,\pi )$.

In order to solve Equation (11(11) $\mu \left(x\right)=z\left(x\right)+{\int }_a^{x}H(x,t)z\left(t\right)\mathrm{d}t,a<x<\pi ,$(11) ), we have to construct the functions $w_{\nu }\left(x\right)$, $\nu =\overline{1,4}$, for $x\in (a,\pi )$. Using the functions $M_{\nu }\left(x\right)$, $\nu =\overline{1,4}$, on $(0,a)$, we can find the functions $v_{lm}\left(x\right)$, l, m=1, 2, on $(b,\pi )$, $b=\pi -a$, by the formulae (8(8) $\left.\begin{array}{ll}{v}_{11}\left(x\right)& =-\frac{x}{2}(M_2-M_3)(\pi -x)+u_{11}\left(x\right),\\ v_{12}\left(x\right)& =-\frac{x}{2}(M_1+M_4)(\pi -x)+\frac{1}{2}{\int }_0^{\pi -x}(M_1-M_4)\left(t\right)\mathrm{d}t-u_{12}\left(x\right),\\ v_{21}\left(x\right)& =-\frac{x}{2}(M_1+M_4)(\pi -x)+u_{21}\left(x\right),\\ v_{22}\left(x\right)& =\frac{x}{2}(M_2-M_3)(\pi -x)+\frac{1}{2}{\int }_0^{\pi -x}(M_2+M_3)\left(t\right)\mathrm{d}t-u_{22}\left(x\right),\end{array}\right\}$(8) ). Note that $v_{lm}\in L_2(b,\pi )$, l, m=1, 2. Using (4(4) ${\mathrm{\Delta }}_j\left(\lambda \right)={\mathrm{\Delta }}_j^0\left(\lambda \right)+{\int }_0^{\pi }\left(v_{j1}\right(x)\sin \lambda x+v_{j2}(x)\cos \lambda x)\mathrm{d}x,j=1,2,$(4) ), we get the relation $\begin{array}{rl}{\mathrm{\Delta }}_j\left({\lambda }_{nj}\right)& ={\mathrm{\Delta }}_j^0\left({\lambda }_{nj}\right)+{\int }_0^b\left(v_{j1}\right(x)\sin {\lambda }_{nj}x+v_{j2}(x)\cos {\lambda }_{nj}x)\mathrm{d}x\\ & +{\int }_b^{\pi }\left(v_{j1}\right(x)\sin {\lambda }_{nj}x+v_{j2}(x)\cos {\lambda }_{nj}x)\mathrm{d}x=0,n\in {\mathcal{I}}_j,j=1,2.\end{array}$ Define (14) $E_{nj}:=-{\mathrm{\Delta }}_j^0\left({\lambda }_{nj}\right)-{\int }_b^{\pi }\left(v_{j1}\right(x)\sin {\lambda }_{nj}x+v_{j2}(x)\cos {\lambda }_{nj}x)\mathrm{d}x,n\in {\mathcal{I}}_j,j=1,2.$(14) Then (15) $E_{nj}={\int }_0^b\left(v_{j1}\right(x)\sin {\lambda }_{nj}x+v_{j2}(x)\cos {\lambda }_{nj}x)\mathrm{d}x,n\in {\mathcal{I}}_j,j=1,2.$(15) Therefore one can determine the numbers $E_{nj}$, $n\in {\mathcal{I}}_j$, $j=1,2$, by (14(14) $E_{nj}:=-{\mathrm{\Delta }}_j^0\left({\lambda }_{nj}\right)-{\int }_b^{\pi }\left(v_{j1}\right(x)\sin {\lambda }_{nj}x+v_{j2}(x)\cos {\lambda }_{nj}x)\mathrm{d}x,n\in {\mathcal{I}}_j,j=1,2.$(14) ), and then use them to find the functions $v_{lm}\left(x\right)$, l, m=1, 2, on $(0,b)$ from (15(15) $E_{nj}={\int }_0^b\left(v_{j1}\right(x)\sin {\lambda }_{nj}x+v_{j2}(x)\cos {\lambda }_{nj}x)\mathrm{d}x,n\in {\mathcal{I}}_j,j=1,2.$(15) ). Now we are ready to prove the uniqueness theorem for Inverse Problem 1.

Proof of Theorem 1.1.

Let boundary value problems L and $\tilde{L}$ satisfy the conditions of Theorem 1.1. In particular, $M_{\nu }\left(x\right)={\tilde{M}}_{\nu }\left(x\right)$ a.e. on $(0,a)$, $\nu =\overline{1,4}$. Then, in view of (8(8) $\left.\begin{array}{ll}{v}_{11}\left(x\right)& =-\frac{x}{2}(M_2-M_3)(\pi -x)+u_{11}\left(x\right),\\ v_{12}\left(x\right)& =-\frac{x}{2}(M_1+M_4)(\pi -x)+\frac{1}{2}{\int }_0^{\pi -x}(M_1-M_4)\left(t\right)\mathrm{d}t-u_{12}\left(x\right),\\ v_{21}\left(x\right)& =-\frac{x}{2}(M_1+M_4)(\pi -x)+u_{21}\left(x\right),\\ v_{22}\left(x\right)& =\frac{x}{2}(M_2-M_3)(\pi -x)+\frac{1}{2}{\int }_0^{\pi -x}(M_2+M_3)\left(t\right)\mathrm{d}t-u_{22}\left(x\right),\end{array}\right\}$(8) ), $v_{lm}\left(x\right)={\tilde{v}}_{lm}\left(x\right)$ a.e. on $(b,\pi )$ for l, m=1, 2. Since ${\lambda }_{nj}={\tilde{\lambda }}_{nj}$, $n\in {\mathcal{I}}_j$, j=1, 2, by virtue of (14(14) $E_{nj}:=-{\mathrm{\Delta }}_j^0\left({\lambda }_{nj}\right)-{\int }_b^{\pi }\left(v_{j1}\right(x)\sin {\lambda }_{nj}x+v_{j2}(x)\cos {\lambda }_{nj}x)\mathrm{d}x,n\in {\mathcal{I}}_j,j=1,2.$(14) ), we have $E_{nj}={\tilde{E}}_{nj}$, $n\in {\mathcal{I}}_j$, j=1, 2. Then it follows from (15(15) $E_{nj}={\int }_0^b\left(v_{j1}\right(x)\sin {\lambda }_{nj}x+v_{j2}(x)\cos {\lambda }_{nj}x)\mathrm{d}x,n\in {\mathcal{I}}_j,j=1,2.$(15) ), that ${\int }_0^b\left(\right(v_{j1}\left(x\right)-{\tilde{v}}_{j1}\left(x\right))\sin {\lambda }_{nj}x+(v_{j2}\left(x\right)-{\tilde{v}}_{j2}\left(x\right))\cos {\lambda }_{nj}x)\mathrm{d}x=0,n\in {\mathcal{I}}_j,j=1,2.$ Recall that the sequences ${\mathcal{S}}_j$, j=1, 2, defined by (3(3) ${\mathcal{S}}_j:={\left\{\left(\begin{array}{c}\sin {\lambda }_{nj}x\\ \cos {\lambda }_{nj}x\end{array}\right)\right\}}_{n\in {\mathcal{I}}_j},j=1,2.$(3) ), are complete in $\mathcal{H}$. Hence $v_{lm}\left(x\right)={\tilde{v}}_{lm}\left(x\right)$ a.e. on $(0,b)$, l, m=1, 2. Consequently, the relations (7(7) $\left.\begin{array}{ll}{w}_1\left(x\right)& =-v_{21}(\pi -x)-(\pi -x)(v_{12}-v_{21}{)}^{\prime }(\pi -x),\\ w_2\left(x\right)& =-v_{11}(\pi -x)-(\pi -x)(v_{11}+v_{22}{)}^{\prime }(\pi -x),\\ w_3\left(x\right)& =v_{11}(\pi -x)-(\pi -x)(v_{11}+v_{22}{)}^{\prime }(\pi -x),\\ w_4\left(x\right)& =-v_{21}(\pi -x)+(\pi -x)(v_{12}-v_{21}{)}^{\prime }(\pi -x).\end{array}\right\}$(7) ) imply $w_{\nu }\left(x\right)={\tilde{w}}_{\nu }\left(x\right)$ a.e. on $(a,\pi )$ for $\nu =\overline{1,4}$. Thus, Equation (11(11) $\mu \left(x\right)=z\left(x\right)+{\int }_a^{x}H(x,t)z\left(t\right)\mathrm{d}t,a<x<\pi ,$(11) ) for L coincides with the similar equation for $\tilde{L}$. By virtue of the unique solvability of (11(11) $\mu \left(x\right)=z\left(x\right)+{\int }_a^{x}H(x,t)z\left(t\right)\mathrm{d}t,a<x<\pi ,$(11) ), we conclude that $M_{\nu }\left(x\right)={\tilde{M}}_{\nu }\left(x\right)$ a.e. on $(a,\pi )$, $\nu =\overline{1,4}$.

Suppose that ${\mathcal{S}}_j$, j=1, 2, are Riesz bases in $\mathcal{H}$. Summarizing the above arguments, we arrive at the following algorithm for solving Inverse Problem 1.

Algorithm 3.1

Let the functions $M_{\nu }\left(x\right)$, $x\in (0,a)$, $\nu =\overline{1,4}$, and the eigenvalues $\{{\lambda }_{nj}{\}}_{n\in {\mathcal{I}}_j}$, j=1, 2, be given.

1. Construct the functions $v_{lm}\left(x\right)$, $x\in (b,\pi )$, $l,m=1,2$, by formulae (8(8) $\left.\begin{array}{ll}{v}_{11}\left(x\right)& =-\frac{x}{2}(M_2-M_3)(\pi -x)+u_{11}\left(x\right),\\ v_{12}\left(x\right)& =-\frac{x}{2}(M_1+M_4)(\pi -x)+\frac{1}{2}{\int }_0^{\pi -x}(M_1-M_4)\left(t\right)\mathrm{d}t-u_{12}\left(x\right),\\ v_{21}\left(x\right)& =-\frac{x}{2}(M_1+M_4)(\pi -x)+u_{21}\left(x\right),\\ v_{22}\left(x\right)& =\frac{x}{2}(M_2-M_3)(\pi -x)+\frac{1}{2}{\int }_0^{\pi -x}(M_2+M_3)\left(t\right)\mathrm{d}t-u_{22}\left(x\right),\end{array}\right\}$(8) ).

2. Find the numbers $\{E_{nj}{\}}_{n\in {\mathcal{I}}_j,j=1,2}$, using (14(14) $E_{nj}:=-{\mathrm{\Delta }}_j^0\left({\lambda }_{nj}\right)-{\int }_b^{\pi }\left(v_{j1}\right(x)\sin {\lambda }_{nj}x+v_{j2}(x)\cos {\lambda }_{nj}x)\mathrm{d}x,n\in {\mathcal{I}}_j,j=1,2.$(14) ).

3. Construct the functions $v_{lm}\left(x\right)$ on $(0,b)$ by the formula $\left(\begin{array}{c}{v}_{j1}\left(x\right)\\ v_{j2}\left(x\right)\end{array}\right)=\sum _{n\in {\mathcal{I}}_j}{E}_{nj}{s}_{nj}^{\ast }\left(x\right),j=1,2,$ where $\{s_{nj}^{\ast }{\}}_{n\in {\mathcal{I}}_j}$ are the biorthonormal bases to ${\mathcal{S}}_j$, j=1, 2.

4. Find the functions $w_{\nu }\left(x\right)$, $x\in (a,\pi )$, $\nu =\overline{1,4}$, using (7(7) $\left.\begin{array}{ll}{w}_1\left(x\right)& =-v_{21}(\pi -x)-(\pi -x)(v_{12}-v_{21}{)}^{\prime }(\pi -x),\\ w_2\left(x\right)& =-v_{11}(\pi -x)-(\pi -x)(v_{11}+v_{22}{)}^{\prime }(\pi -x),\\ w_3\left(x\right)& =v_{11}(\pi -x)-(\pi -x)(v_{11}+v_{22}{)}^{\prime }(\pi -x),\\ w_4\left(x\right)& =-v_{21}(\pi -x)+(\pi -x)(v_{12}-v_{21}{)}^{\prime }(\pi -x).\end{array}\right\}$(7) ).

5. Find $\mu \left(x\right)=\left({\mu }_{\nu }\right(x){)}_{\nu =1}^4$ and $H(x,t)=\left(H_{\nu \xi }\right(x,t){)}_{\nu ,\xi =1}^4$, using (12(12) $\begin{array}{rl}{z}_{\nu }\left(x\right)& =(\pi -x)M_{\nu ,2}\left(x\right),\\ {\mu }_{\nu }\left(x\right)& =w_{\nu }\left(x\right)-\sum _{n=2}^{\mathrm{\infty }}\sum _{j=1}^{m_n}\sum _{k=1}^n{b}_{njk}^{\left(\nu \right)}{g}_k(\pi -x)\left(Q_{nj}\right[M_{\left(1\right)}]\ast g_{n-1-k})\left(x\right),\end{array}$(12) ) and (13(13) $\begin{array}{rl}{H}_{\nu \xi }(x,t)& =\frac{1}{\pi -t}\sum _{n=2}^{\mathrm{\infty }}\sum _{j=1}^{m_n}\sum _{k=1}^n{b}_{njk}^{\left(\nu \right)}{g}_k(\pi -x)\\ & \times \sum _{s=1}^{m_{n-1}}{\beta }_{njs}^{\left(\xi \right)}\left(Q_{n-1,s}\right[M_{\left(1\right)}]\ast g_{n-1-k})(x-t).\end{array}$(13) ).

6. Solve the Volterra integral equation of the second kind (11(11) $\mu \left(x\right)=z\left(x\right)+{\int }_a^{x}H(x,t)z\left(t\right)\mathrm{d}t,a<x<\pi ,$(11) ), and find $M_{\nu }\left(x\right)=z_{\nu }\left(x\right)/(\pi -x)$, $x\in (a,\pi )$, $\nu =\overline{1,4}$.

Further in this section we assume $a=\pi /k$, $k\in \mathbb{N}$, $k\ge 2$, and ${\mathcal{I}}_j=\{n=kl+s_j:l\in \mathbb{Z}\}$, $s_j\in \mathbb{R}$, j=1, 2. The proof of Theorem 1.2 is based on the following lemma.

Lemma 3.1

Under the conditions of Theorem 1.2, the sequences ${\mathcal{S}}_j,$ $j=1,2,$ are Riesz bases in $\mathcal{H}$.

The assertion of Lemma 3.1 easily follows from Lemmas 3.2 and 3.3.

Lemma 3.2

Let $\{{\theta }_n{\}}_{n\in \mathbb{Z}}$ be a sequence of distinct complex numbers. Then the sequence of vector functions $\left\{\right(\sin {\theta }_{n}x,\cos {\theta }_{n}x{)}^{\mathrm{T}}{\}}_{n\in \mathbb{Z}}$ is a Riesz basis in $\mathcal{H}$ if and only if $\{\exp (i{\theta }_{n}x){\}}_{n\in \mathbb{Z}}$ is a Riesz basis in $L_2(-b,b)$.

Proof.

The sequence $\left\{\right(\sin {\theta }_{n}x,\cos {\theta }_{n}x{)}^{\mathrm{T}}{\}}_{n\in \mathbb{Z}}$ is transformed into $\{\exp (i{\theta }_{n}x){\}}_{n\in \mathbb{Z}}$ by the linear bounded transform $A:\mathcal{H}\to L_2(-b,b)$, acting as follows: $\left(Ah\right)\left(x\right)=\left\{\begin{array}{ll}{h}_2\left(x\right)+i{h}_1\left(x\right),& x\in (0,b),\\ h_2(-x)-i{h}_1(-x),& x\in (-b,0),\end{array}\right.h=\left(\begin{array}{c}{h}_1\\ h_2\end{array}\right)\in \mathcal{H}.$ The inverse transform is also bounded: $\left(A^{-1}e\right)\left(x\right)={\left(\frac{e\left(x\right)+e(-x)}{2},\frac{e\left(x\right)-e(-x)}{2i}\right)}^{\mathrm{T}},e\in L_2(-b,b),$ so it preserves the Riesz basis property.

Lemma 3.3

Let $k\in \mathbb{N},$ $s\in \mathbb{R},$ $\{{\varkappa }_n{\}}_{n\in \mathbb{Z}}\in l_2,$ $b=\pi /k$. Then the system of exponents $\{\exp (kn+s+{\varkappa }_n)t{\}}_{n\in \mathbb{Z}}$ is a Riesz basis in $L_2(-b,b)$.

Proof.

Note that the shift s does not influence the Riesz basis property, since the linear transform $\exp t\to \exp (t+s)$ is bounded and has a bounded inverse. Thus, it is sufficient to study the system $\{\exp (kn+{\varkappa }_n)t{\}}_{n\in \mathbb{Z}}$. This system is $L_2$-close to the orthonormal system $\{\exp knt{\}}_{n\in \mathbb{Z}}$. Using the Theorem by Levin and Ljubarskiĭ [31], one can easily show that the system $\{\exp (kn+{\varkappa }_n)t{\}}_{n\in \mathbb{Z}}$ in complete in $L_2(-b,b)$. Consequently, it is a Riesz basis.

Lemma 3.1 implies that the sequences ${\mathcal{S}}_j$, $j=1,2$, are complete in $\mathcal{H}$, so the uniqueness assertion of Theorem 1.2 immediately follows from Theorem 1.1. Since ${\mathcal{S}}_j$, j=1, 2, are Riesz bases, one can apply Algorithm 3.1 to solve Inverse Problem 1 for the considered case.

4. Example

In this section, we apply Algorithm 3.1 to the following example. Suppose that (16) $\begin{array}{rl}{M}_{\nu }\left(x\right)& =0,x\in \left(0,\frac{\pi }{2}\right),\nu =\overline{1,4},{\mathcal{I}}_j=\{2n:n\in \mathbb{Z}\},j=1,2,\\ {\lambda }_{2,1}& =\gamma ,{\lambda }_{-2,1}=-\gamma ,{\lambda }_{2n,1}=2n,n\ne \pm 1,{\lambda }_{2n,2}=2n+\frac{1}{2},n\in \mathbb{Z},\end{array}$(16) where $\gamma \in (0,2)$ is a given number.

Using formulae (8(8) $\left.\begin{array}{ll}{v}_{11}\left(x\right)& =-\frac{x}{2}(M_2-M_3)(\pi -x)+u_{11}\left(x\right),\\ v_{12}\left(x\right)& =-\frac{x}{2}(M_1+M_4)(\pi -x)+\frac{1}{2}{\int }_0^{\pi -x}(M_1-M_4)\left(t\right)\mathrm{d}t-u_{12}\left(x\right),\\ v_{21}\left(x\right)& =-\frac{x}{2}(M_1+M_4)(\pi -x)+u_{21}\left(x\right),\\ v_{22}\left(x\right)& =\frac{x}{2}(M_2-M_3)(\pi -x)+\frac{1}{2}{\int }_0^{\pi -x}(M_2+M_3)\left(t\right)\mathrm{d}t-u_{22}\left(x\right),\end{array}\right\}$(8) ), we get $v_{lm}\left(x\right)=0$ for $x\in (\pi /2,\pi )$, l, m=1, 2. Therefore the relations (14(14) $E_{nj}:=-{\mathrm{\Delta }}_j^0\left({\lambda }_{nj}\right)-{\int }_b^{\pi }\left(v_{j1}\right(x)\sin {\lambda }_{nj}x+v_{j2}(x)\cos {\lambda }_{nj}x)\mathrm{d}x,n\in {\mathcal{I}}_j,j=1,2.$(14) ) and (15(15) $E_{nj}={\int }_0^b\left(v_{j1}\right(x)\sin {\lambda }_{nj}x+v_{j2}(x)\cos {\lambda }_{nj}x)\mathrm{d}x,n\in {\mathcal{I}}_j,j=1,2.$(15) ) together with (16(16) $\begin{array}{rl}{M}_{\nu }\left(x\right)& =0,x\in \left(0,\frac{\pi }{2}\right),\nu =\overline{1,4},{\mathcal{I}}_j=\{2n:n\in \mathbb{Z}\},j=1,2,\\ {\lambda }_{2,1}& =\gamma ,{\lambda }_{-2,1}=-\gamma ,{\lambda }_{2n,1}=2n,n\ne \pm 1,{\lambda }_{2n,2}=2n+\frac{1}{2},n\in \mathbb{Z},\end{array}$(16) ) imply (17) $\begin{array}{rl}& {\int }_0^{\pi /2}\left(\right(-1{)}^j{v}_{11}\left(x\right)\sin \gamma x+v_{12}\left(x\right)\cos \gamma x)\mathrm{d}x=(-1{)}^{j+1}\sin \gamma \pi ,j=1,2,\end{array}$(17) (18) $\begin{array}{rl}& {\int }_0^{\pi /2}\left(v_{11}\right(x)\sin 2nx+v_{12}(x)\cos 2nx)\mathrm{d}x=-\sin 2n\pi =0,\end{array}$(18) (19) $\begin{array}{rl}& {\int }_0^{\pi /2}\left(v_{21}\left(x\right)\sin \left(2n+\frac{1}{2}\right)x+v_{22}\left(x\right)\cos \left(2n+\frac{1}{2}\right)x\right)\mathrm{d}x=\cos \left(2n+\frac{1}{2}\right)\pi =0.\end{array}$(19)

In view of (18(18) $\begin{array}{rl}& {\int }_0^{\pi /2}\left(v_{11}\right(x)\sin 2nx+v_{12}(x)\cos 2nx)\mathrm{d}x=-\sin 2n\pi =0,\end{array}$(18) ), the Fourier coefficients of the vector functions $(v_{11},v_{12}{)}^{\mathrm{T}}$ with respect to the orthogonal system $\left\{\right(\sin 2nx,\cos 2nx{)}^{\mathrm{T}}{\}}_{n\in \mathbb{Z}}$ in $\mathcal{H}=L_2(0,\frac{\pi }{2})\oplus L_2(0,\frac{\pi }{2})$ equal zeros for all $n\ne \pm 1$, so we have $\left(\begin{array}{c}{v}_{11}\left(x\right)\\ v_{12}\left(x\right)\end{array}\right)=a_1\left(\begin{array}{c}\sin 2x\\ \cos 2x\end{array}\right)+a_{-1}\left(\begin{array}{c}\sin (-2x)\\ \cos (-2x)\end{array}\right).$ Substituting the latter representation into (17(17) $\begin{array}{rl}& {\int }_0^{\pi /2}\left(\right(-1{)}^j{v}_{11}\left(x\right)\sin \gamma x+v_{12}\left(x\right)\cos \gamma x)\mathrm{d}x=(-1{)}^{j+1}\sin \gamma \pi ,j=1,2,\end{array}$(17) ) and solving the linear system, we get $a_1=a_{-1}=\theta$, $\theta :=-\frac{\sin \gamma \pi }{\frac{\sin (2-\gamma )\frac{\pi }{2}}{2-\gamma }-\frac{\sin (2+\gamma )\frac{\pi }{2}}{2+\gamma }},$ so $v_{11}\left(x\right)=2\theta \sin 2x$, $v_{12}\left(x\right)=0$, $x\in (0,\pi /2)$. Similarly we obtain $v_{2j}\left(x\right)=0$, $x\in (0,\pi /2)$, j=1, 2, from (19(19) $\begin{array}{rl}& {\int }_0^{\pi /2}\left(v_{21}\left(x\right)\sin \left(2n+\frac{1}{2}\right)x+v_{22}\left(x\right)\cos \left(2n+\frac{1}{2}\right)x\right)\mathrm{d}x=\cos \left(2n+\frac{1}{2}\right)\pi =0.\end{array}$(19) ). Then one can easily calculate $w_{\nu }\left(x\right)$, $x\in (\pi /2,\pi )$, $\nu =\overline{1,4}$, using (7(7) $\left.\begin{array}{ll}{w}_1\left(x\right)& =-v_{21}(\pi -x)-(\pi -x)(v_{12}-v_{21}{)}^{\prime }(\pi -x),\\ w_2\left(x\right)& =-v_{11}(\pi -x)-(\pi -x)(v_{11}+v_{22}{)}^{\prime }(\pi -x),\\ w_3\left(x\right)& =v_{11}(\pi -x)-(\pi -x)(v_{11}+v_{22}{)}^{\prime }(\pi -x),\\ w_4\left(x\right)& =-v_{21}(\pi -x)+(\pi -x)(v_{12}-v_{21}{)}^{\prime }(\pi -x).\end{array}\right\}$(7) ). Since $M_{\left(1\right)}\left(x\right)\equiv 0$, the main equation (11(11) $\mu \left(x\right)=z\left(x\right)+{\int }_a^{x}H(x,t)z\left(t\right)\mathrm{d}t,a<x<\pi ,$(11) ) takes the form $w_{\nu }\left(x\right)=(\pi -x)M_{\nu }\left(x\right)$, $x\in (\pi /2,\pi )$, $\nu =\overline{1,4}$. Hence we get (20) $\left.\begin{array}{ll}{M}_1\left(x\right)& =M_4\left(x\right)=0,\\ M_2\left(x\right)& =-\frac{2\theta \sin 2(\pi -x)}{\pi -x}-4\theta \cos 2(\pi -x),\\ M_3\left(x\right)& =\frac{2\theta \sin 2(\pi -x)}{\pi -x}-4\theta \cos 2(\pi -x),\end{array}\right\}$(20) for $x\in (\pi /2,\pi )$. Note that the functions $M_2$ and $M_3$ are continuous at $x=\pi$.

Now let us check for a particular value $\gamma =0.5$, that the boundary value problems $D_j$, j=1, 2, with $M\left(x\right)\equiv 0$, $x\in (0,\pi /2)$, and $M\left(x\right)$, given by (20(20) $\left.\begin{array}{ll}{M}_1\left(x\right)& =M_4\left(x\right)=0,\\ M_2\left(x\right)& =-\frac{2\theta \sin 2(\pi -x)}{\pi -x}-4\theta \cos 2(\pi -x),\\ M_3\left(x\right)& =\frac{2\theta \sin 2(\pi -x)}{\pi -x}-4\theta \cos 2(\pi -x),\end{array}\right\}$(20) ) for $x\in (\pi /2,\pi )$, have eigenvalues in the form (16(16) $\begin{array}{rl}{M}_{\nu }\left(x\right)& =0,x\in \left(0,\frac{\pi }{2}\right),\nu =\overline{1,4},{\mathcal{I}}_j=\{2n:n\in \mathbb{Z}\},j=1,2,\\ {\lambda }_{2,1}& =\gamma ,{\lambda }_{-2,1}=-\gamma ,{\lambda }_{2n,1}=2n,n\ne \pm 1,{\lambda }_{2n,2}=2n+\frac{1}{2},n\in \mathbb{Z},\end{array}$(16) ). For $x\in (0,\pi /2)$, we have $S_1(x,\lambda )=\sin \lambda x$, $S_2(x,\lambda )=-\cos \lambda x$. For $x\in (\pi /2,\pi )$, the Dirac system (1(1) $B{y}^{\prime }+{\int }_0^{x}M(x-t)y\left(t\right)\mathrm{d}t=\lambda y,x\in (0,\pi ),$(1) ) for $S(x,\lambda )$ takes the form $\begin{array}{rl}{S}_1^{\prime }(x,\lambda )& =-\lambda S_2(x,\lambda )+{\int }_0^{x-\pi /2}{M}_3(x-t)\sin \lambda t\mathrm{d}t,\\ S_2^{\prime }(x,\lambda )& =\lambda S_1+{\int }_0^{x-\pi /2}{M}_2(x-t)\cos \lambda t\mathrm{d}t.\end{array}$

Solving the latter system by the Runge-Kutta method, using Simpson's formula for integration, we find the following real eigenvalues in the interval $[-10,10]$:

Table

λn1	λn2
−9.022	−9.500
−8.000	−8.500
−6.962	−7.500
−6.000	−6.500
−5.078	−5.500
−4.000	−4.500
−0.500	−3.500
0.000	−2.500
0.500	−1.500
4.000	−0.500
5.077	0.500
6.000	1.500
6.962	2.500
8.000	3.500
9.022	4.500
10.000	5.500
	6.500
	7.500
	8.500
	9.500

Obviously, the spectra include subsequences in the form (16(16) $\begin{array}{rl}{M}_{\nu }\left(x\right)& =0,x\in \left(0,\frac{\pi }{2}\right),\nu =\overline{1,4},{\mathcal{I}}_j=\{2n:n\in \mathbb{Z}\},j=1,2,\\ {\lambda }_{2,1}& =\gamma ,{\lambda }_{-2,1}=-\gamma ,{\lambda }_{2n,1}=2n,n\ne \pm 1,{\lambda }_{2n,2}=2n+\frac{1}{2},n\in \mathbb{Z},\end{array}$(16) ).

Disclosure statement

No potential conflict of interest was reported by the author.

Additional information

Funding

This work was supported by Grant 17-11-01193 of the Russian Science Foundation.