1 Introduction

An intensive development of the spectral theory for various classes of differential and integral operators as well as for operators in abstract spaces took place in the second half of twentieth century and twenty-first century. Within this theory inverse spectral problems take a special place. Inverse spectral problems are problems studying operator from some of its spectral characteristics. The inverse spectral problem can be regarded from three aspects: existence, uniqueness and reconstruction of the operators with specific properties of eigenvalues and eigenfunctions. The Sturm–Liouville-type operators are generated by second-order differential expression and boundary conditions (see [7, 9] and references therein). In this paper, we study Sturm–Liouville operators with one constant delay under Dirichlet–Dirichlet and Dirichlet/Polynomial boundary conditions. In the papers [1,2,3, 6, 8, 13,14,15,16] authors study this differential expression under different types of boundary conditions. Also in many papers authors study boundary conditions with spectral parameter (see [10,11,12]). We will show uniqueness and we will recover the operators from two spectra.

In this paper we generalize the results in the paper [13] and [14].

In this paper, we study two boundary value problems \(L_{k},k = {0,1}\), \(L_{0}\) generated by (1.1),(1.2),(1.3) and \(L_{1}\) generated by (1.1),(1.2),(1.4)

$$\begin{aligned}&-y^{\prime \prime }(x)+q(x)y(x-\tau )=\lambda y(x), \, x\in [0, \pi ], \end{aligned}$$
(1.1)
$$\begin{aligned}&y(0) = 0, \end{aligned}$$
(1.2)
$$\begin{aligned}&y(\pi )=0, \end{aligned}$$
(1.3)
$$\begin{aligned}&y^{\prime }(\pi )+P(\lambda )y(\pi )=0, \end{aligned}$$
(1.4)

where \(\lambda \) is the spectral parameter, \(\tau \in \left[ \frac{2\pi }{5},\pi \right) \), Function q(x) is a complex-valued function which we call potential, such that \(q \in L^{2}(0,\pi )\), and \(q(x)\equiv 0\) for \(x \in [0,\tau ]\). Function \(P(\lambda )\) is normalized polynomial with degree \(s,s\in N\) and complex coefficients. We will separately study two cases, the first when \(\tau \in \left[ \frac{\pi }{2},\pi \right) \) and the second when \(\tau \in \left[ \frac{2\pi }{5},\frac{\pi }{2}\right) \).

The spectra of \(L_{0}\) and \(L_{1}\) are countable. We will prove that the potential q and polynomial P are uniquely determined from the spectra of \(L_{0}\) and \(L_{1}\). Let \((\lambda _{n})_{n=1}^{\infty }\) be the eigenvalues of \(L_{0}\) and \((\mu _{n})_{n=1}^{\infty }\) be the eigenvalues of \(L_{1}\).

The inverse problem is to prove that q(x) and \(P(\lambda )\) are uniquely determined from \((\lambda _{n})_{n=1}^{\infty }\) and \((\mu _{n})_{n=1}^{\infty }\), and determine q(x) and \(P(\lambda )\) from \((\lambda _{n})_{n=1}^{\infty }\) and \((\mu _{n})_{n=1}^{\infty }\).

2 Preliminaries

Let the function Y(x) be the solution of the differential equation (1.1) under initial condition \(Y(0)=0, Y'(0)=1\), then the function Y(x) satisfy an integral equation

$$\begin{aligned} Y(x)=\frac{1}{z}\int \limits _{0}^{x}q(t)Y(t-\tau ) \sin z(x-t){\text{ d }}t+\frac{\sin zx}{z}, \end{aligned}$$
(2.1)

where \(\lambda =z^{2}\). We will be solving equation (2.1) using \(q(x)\equiv 0\) for \(x \in [0,\tau ]\).

For \(x \in [0,\tau ]\), the solution is

$$\begin{aligned} Y(x)= \frac{1}{z} \int \limits _{0} ^{x}q(t)Y(t-\tau ) \sin z(x-t){\text{ d }}t + \frac{\sin zx}{z}= \frac{\sin zx}{z}. \end{aligned}$$

For \(x \in (\tau , 2\tau ]\), the solution is

$$\begin{aligned} Y(x)= \frac{1}{z^{2}} \int \limits _{\tau } ^{x}q(t) \sin z(t-\tau ) \sin z(x-t){\text{ d }}t + \frac{\sin zx}{z}. \end{aligned}$$
(2.2)

For \(x\in (2\tau ,3\tau ]\), the solution is

$$\begin{aligned} Y(x)= & {} \frac{\sin zx}{z}+\frac{1}{z^{2}}\int \limits _{\tau }^{x}q(t) \sin z(t-\tau ) \sin z(x-t){\text{ d }}t\nonumber \\&+\frac{1}{z^{3}}\int \limits _{2\tau }^{x}\int \limits _{\tau }^{t-\tau }q(t)q(t_{1}) \sin z(x-t) \sin z(t_{1}-\tau ) \sin z(t-\tau -t_{1}){\text{ d }}t_{1}{\text{ d }}t.\nonumber \\ \end{aligned}$$
(2.3)

3 Main Results

3.1 Linear Case, \(\tau \in \left[ \frac{\pi }{2},\pi \right) \)

In the case, when \(\tau \in \left[ \frac{\pi }{2},\pi \right) \) we have \(\pi \in (\tau ,2\tau ]\), and using (1.3) and (1.4) from Eq. (2.2), we get

$$\begin{aligned} \Delta _{0}(\lambda )=y(\pi )&= \frac{\sin z\pi }{z} + \frac{1}{z^{2}} \int \limits _{\tau } ^{\pi }q(t) \sin z(t-\tau ) \sin z(\pi -t){\text{ d }}t,\nonumber \\ \quad \Delta _{1}(\lambda )&= y^{\prime }(\pi )+P(\lambda )y(\pi )=\cos z\pi +\frac{1}{z}\int \limits _{\tau }^{\pi }q(t) \sin z(t-\tau ) \cos z(\pi -t){\text{ d }}t+\nonumber \\ \end{aligned}$$
(3.1)
$$\begin{aligned}&P(z^{2})\left( \frac{\sin z\pi }{z} + \frac{1}{z^{2}} \int \limits _{\tau } ^{\pi }q(t) \sin z(t-\tau ) \sin z(\pi -t){\text{ d }}t\right) . \end{aligned}$$
(3.2)

The functions \(\Delta _{0}(\lambda ), \Delta _{1}(\lambda )\) are entire in \(\lambda \) of order 1/2. It is clear that the set of zeros of functions \(\Delta _{0}(\lambda ), \Delta _{1}(\lambda )\) is equivalent to the spectrum of boundary spectral problems \(L_{0}, L_{1}\), respectively. For the spectrum \((\lambda _{n})_{n=1}^{\infty }\) of boundary spectral problems \(L_{0}\), we have the asymptotic formula (see [8]):

$$\begin{aligned} \lambda _{n}= n^{2} + \frac{\cos \tau n}{\pi } \int _{\tau }^{\pi } q(t){\text{ d }}t + o(1) , \; (n\rightarrow \infty ). \end{aligned}$$

For the spectrum \((\mu _{n})_{n=1}^{\infty }\) of boundary spectral problems \(L_{1}\), using well-known method (see [8]), based on Rouche‘s theorem, we have asymptotic formula which depends on the degree of the polynomial \(P(\lambda )\). When degree of the polynomial \(P(\lambda )\) is equal to 1, \(s=1\) we have asymptotic formula

$$\begin{aligned} \mu _{n}= n^{2} + \frac{\cos \tau n}{\pi } \int \limits _{\tau }^{\pi } q(t){\text{ d }}t-\frac{2}{\pi } + o(1) , \; (n\rightarrow \infty ). \end{aligned}$$

When degree of the polynomial \(P(\lambda )\) is different from 1, \(s>1\), we have asymptotic formula

$$\begin{aligned} \mu _{n}= n^{2} + \frac{\cos \tau n}{\pi } \int \limits _{\tau }^{\pi } q(t){\text{ d }}t + o(1) , \; (n\rightarrow \infty ). \end{aligned}$$

Using Hadamard’s factorization theorem we conclude that spectra uniquely determine functions \(\Delta _{0}(\lambda ), \Delta _{1}(\lambda )\). We introduce notation

$$\begin{aligned} F_{0}(z) = z \Delta _{0}(\lambda ), \, F_{1}(z) = z \Delta _{1}(\lambda ). \end{aligned}$$

The delay \(\tau \) and the integral \(\int _{\tau }^{\pi } q(t) {\text{ d }}t=I_{1}\) are uniquely determined from the spectrum \((\lambda _{n})_{n=1}^{\infty }\)(see [1]).

Lemma 3.1

The spectra \((\mu _{n})_{n=1}^{\infty }\) and \((\lambda _{n})_{n=1}^{\infty }\) of boundary value problem \(L_{1}\) and \(L_{0}\) uniquely determine polynomial \(P(\lambda )\).

Proof

Function \(F_{1}(z)\) is uniquely determined by spectrum \((\mu _{n})_{n=1}^{\infty }\) and function \(F_{0}(z)\) is uniquely determined by spectrum \((\lambda _{n})_{n=1}^{\infty }\).

Let \(P(\lambda )=\lambda ^{s}+p_{s-1}\lambda ^{s-1}+\cdots +p_{0},p_{i}\in C\,.\) We have

$$\begin{aligned} F_{1}(z)&=z \cos z\pi +\int \limits _{\tau }^{\pi }q(t) \sin z(t-\tau ) \cos z(\pi -t){\text{ d }}t \\&\quad + (z^{2s}+p_{s-1}z^{2s-2}+\cdots +p_{0}) F_{0}(z). \end{aligned}$$

Now, we put \(z=2m+\frac{1}{2},m\in N\) and we have

$$\begin{aligned}&F_{1}\left( 2m+\frac{1}{2}\right) - \left( 2m+\frac{1}{2} \right) ^{2s} F_{0}\left( 2m+\frac{1}{2}\right) \\&\quad =\int \limits _{\tau }^{\pi } q(t) \sin \left( \left( 2m+\frac{1}{2}\right) (t-\tau )\right) \cos \left( \left( 2m+\frac{1}{2}\right) (\pi -t)\right) {\text{ d }}t\\&\quad +\left( p_{s-1}\left( 2m+\frac{1}{2}\right) ^{2s-2}+\cdots +p_{0}\right) F_{0}\left( 2m+\frac{1}{2}\right) . \end{aligned}$$

Since

$$\begin{aligned}&\lim \limits _{m\rightarrow \infty } F_{0} \left( 2m+\frac{1}{2} \right) \\&\quad =\lim \limits _{m\rightarrow \infty } \left( 1+ \frac{1}{2m+ \frac{1}{2}} \int \limits _{\tau }^{\pi } q(t) \sin \left( \frac{(4m+1)(t-\tau )}{2}\right) \sin \left( \frac{(4m+1)(\pi -t)}{2}\right) {\text{ d }}t \right) = 1, \end{aligned}$$

we conclude \( \lim \limits _{m\rightarrow \infty } \frac{ \left( p_{s-2}\left( 2m+\frac{1}{2}\right) ^{2s-4}+\cdots +p_{0}\right) F_{0}\left( 2m+\frac{1}{2}\right) }{\left( 2m+\frac{1}{2}\right) ^{2s-2}} =0 \, ,\) and from this equation, we have

$$\begin{aligned} p_{s-1}=\lim \limits _{m\rightarrow \infty }\frac{F_{1}\left( 2m+\frac{1}{2}\right) -\left( 2m+\frac{1}{2}\right) ^{2s}F_{0}\left( 2m+\frac{1}{2}\right) }{\left( 2m+\frac{1}{2}\right) ^{2s-2}}. \end{aligned}$$

Now, when we showed that coefficient \(p_{s-1}\) is known from spectra \((\mu _{n})_{n=1}^{\infty }\) and \((\lambda _{n})_{n=1}^{\infty }\), we have

$$\begin{aligned} p_{s-2}= \lim \limits _{m\rightarrow \infty }\frac{F_{1}\left( 2m+\frac{1}{2}\right) -\left( \left( 2m+\frac{1}{2}\right) ^{2s}+ p_{s-1}\left( 2m+\frac{1}{2}\right) ^{2s-2} \right) F_{0}\left( 2m+\frac{1}{2}\right) }{\left( 2m+\frac{1}{2}\right) ^{2s-4}}. \end{aligned}$$

We repeat this procedure and we have all coefficients \(p_{s-1},...,p_{1}\) are determined by spectra \((\mu _{n})_{n=1}^{\infty }\) and \((\lambda _{n})_{n=1}^{\infty }\).

Now, we will show that coefficient \(p_{0}\) is determined by spectra \((\mu _{n})_{n=1}^{\infty }\) and \((\lambda _{n})_{n=1}^{\infty }\).

First we transform product trigonometric functions to sum and we have

$$\begin{aligned}&\int \limits _{\tau }^{\pi }q(t) \sin z(t-\tau ) \cos z(\pi -t){\text{ d }}t=\frac{\sin z(\pi -\tau )}{2}I_{1}\\&\quad +\frac{\cos z(\pi +\tau )}{2}\int \limits _{\tau }^{\pi }q(t) \sin 2zt{\text{ d }}t-\frac{\sin z(\pi +\tau )}{2}\int \limits _{\tau }^{\pi }q(t) \cos 2zt{\text{ d }}t, \end{aligned}$$

if we put \(z=2m+\frac{1}{2},m\in N\), from Riemann–Lebesgue lemma we have

$$\begin{aligned} \lim \limits _{m\rightarrow \infty }\left( \int \limits _{\tau }^{\pi }q(t) \sin \frac{(4m+1)(t-\tau )}{2} \cos \frac{(4m+1)(\pi -t)}{2}{\text{ d }}t-\frac{\sin \frac{(4m+1)(\pi -\tau )}{2}}{2}I_{1}\right) =0. \end{aligned}$$

Finally, we have

$$\begin{aligned} p_{0}= & {} \lim \limits _{m\rightarrow \infty } \frac{F_{1}\left( 2m+\frac{1}{2}\right) -\frac{\sin \frac{(4m+1)(\pi -\tau )}{2}}{2}I_{1}}{F_{0}\left( 2m+\frac{1}{2} \right) } -\\&\left( \left( 2m+\frac{1}{2}\right) ^{2s}+p_{s-1}\left( 2m+\frac{1}{2}\right) ^{2s-2}+\cdots +p_{1}\left( 2m+\frac{1}{2}\right) ^{2} \right) , \end{aligned}$$

and polynomial function \(P(\lambda )\) is ordered by spectra \((\mu _{n})_{n=1}^{\infty }\) and \((\lambda _{n})_{n=1}^{\infty }\). \(\square \)

Since \(q(x) = 0\) for \(x\in [0, \tau )\), we have \(a_{0}=\int \limits _{0}^{\pi } q(t) {\text{ d }}t=\int \limits _{\tau }^{\pi } q(t) {\text{ d }}t = I_{1}\) and coefficient \(a_{0}\) is ordered by spectrum \((\lambda _{n})_{n=1}^{\infty }\). Now we will prove that the other coefficients \(a_{m}=\int \limits _{0}^{\pi }q(t) \cos 2mt {\text{ d }}t \) and \(b_{m}=\int \limits _{0}^{\pi } q(t) \sin 2mt {\text{ d }}t\) of the potential q are also uniquely determined by spectra.

Theorem 3.2

Let \((\lambda _{n}) _{n=1}^{\infty }\) and \((\mu _{n}) _{n=1}^{\infty }\) be the spectra of boundary spectral problems \(L_{k}, k=0,1\), respectively, then potential q is uniquely determined by \((\lambda _{n}) _{n=1}^{\infty }\) and \((\mu _{n}) _{n=1}^{\infty }\) if \(\frac{\pi }{2} \leqslant \tau < \pi \).

Proof

From (3.1) and (3.2), we have

$$\begin{aligned} F_{0}(z)= & {} \sin z\pi +\frac{1}{z}\int \limits _{\tau }^{\pi }q(t) \sin z(t-\tau ) \sin z(\pi -t){\text{ d }}t\\ F_{1}(z)= & {} z \cos z\pi + P(z^{2}) \sin z\pi +\int \limits _{\tau }^{\pi } q(t) \sin z(t-\tau ) \cos z(\pi -t){\text{ d }}t\\&+\frac{P(z^{2})}{z}\int \limits _{\tau }^{\pi }q(t) \sin z(t-\tau ) \sin z(\pi -t){\text{ d }}t. \end{aligned}$$

Using transformations product of trigonometric functions to sum and addition formulas, we have

$$\begin{aligned} F_{0}(z)= & {} \sin z\pi -\frac{\cos z(\pi -\tau )}{2z}\int \limits _{\tau }^{\pi } q(t){\text{ d }}t +\frac{ \cos z(\pi +\tau )}{2z}\int \limits _{\tau }^{\pi }q(t) \cos 2zt{\text{ d }}t\\&+\frac{\sin z(\pi +\tau )}{2z}\int \limits _{\tau }^{\pi }q(t) \sin 2zt{\text{ d }}t, \end{aligned}$$

and

$$\begin{aligned} F_{1}(z)= & {} z \cos z\pi +P(z^{2}) \sin z\pi +\frac{\sin z(\pi -\tau )}{2}\int \limits _{\tau }^{\pi }q(t){\text{ d }}t-\frac{\sin z(\pi +\tau )}{2}\int \limits _{\tau }^{\pi }q(t) \cos 2zt{\text{ d }}t\\&+\frac{ \cos z(\pi +\tau )}{2}\int \limits _{\tau }^{\pi }q(t) \sin 2zt{\text{ d }}t- \frac{P(z^{2}) \cos z(\pi -\tau )}{2z} \int \limits _{\tau }^{\pi }q(t){\text{ d }}t\\&+\frac{P(z^{2}) \cos z(\pi +\tau )}{2z}\int \limits _{\tau }^{\pi }q(t) \cos 2zt{\text{ d }}t+ \frac{P(z^{2}) \sin z(\pi +\tau )}{2z}\int \limits _{\tau }^{\pi }q(t) \sin 2zt{\text{ d }}t. \end{aligned}$$

Now we put \(z = m, m \in N\) and using \(q(x)=0\) for \(x\in [0, \tau )\), we have

$$\begin{aligned}&F_{0}(m)+\frac{\cos m(\pi -\tau )}{2m}\int \limits _{\tau }^{\pi }q(t){\text{ d }}t= \frac{ \cos m(\pi +\tau )}{2m}a_{m}+\frac{\sin m(\pi +\tau )}{2m}b_{m},\\&\quad F_{1}(m)-m(-1)^{m}-\frac{\sin m(\pi -\tau )}{2}\int _{\tau }^{\pi }q(t){\text{ d }}t+\frac{P(m^{2})\cos m(\pi -\tau )}{2m}\int \limits _{\tau }^{\pi }q(t){\text{ d }}t\\&\quad =\left( \frac{P(m^{2}) \cos m(\pi +\tau )}{2m}-\frac{\sin m(\pi +\tau )}{2} \right) a_{m}\\&\quad +\left( \frac{P(m^{2}) \sin m(\pi +\tau )}{2m}+\frac{\cos m(\pi +\tau )}{2} \right) b_{m}. \end{aligned}$$

This is linear system of two variables \(a_{m}\) and \(b_{m}\) with determinant \(D=\frac{1}{4m}\ne 0\), and which has unique solution

$$\begin{aligned} a_{m}= & {} \left( (-1)^{m}F_{0}(m)+\frac{\cos m\tau }{2m} \int \limits _{\tau }^{\pi }q(t){\text{ d }}t\right) \left( 2m \cos m\tau +P(m^{2}) \sin m\tau \right) \\&-2\left( (-1)^{m}F_{1}(m)-m+\frac{\sin m\tau }{2} \int _{\tau }^{\pi }q(t){\text{ d }}t+ \frac{P(m^{2}) \cos m\tau }{2m}\int \limits _{\tau }^{\pi }q(t){\text{ d }}t\right) \sin m\tau \\ b_{m}= & {} 2 \cos m\tau \left( (-1)^{m}F_{1}(m)-m+\frac{ \sin m\tau }{2}\int \limits _{\tau }^{\pi }q(t){\text{ d }}t+ \frac{P(m^{2}) \cos m\tau }{2m}\int \limits _{\tau }^{\pi }q(t){\text{ d }}t\right) \\&-2\left( P(m^{2}) \cos m\tau -m \sin m\tau \right) \left( (-1)^{m}F_{0}(m)+\frac{ \cos m\tau }{2m}\int \limits _{\tau }^{\pi }q(t){\text{ d }}t\right) . \end{aligned}$$

Since \(\tau \), \(\int _{\tau }^{\pi }q(t){\text{ d }}t\), polynomial \(P(\lambda )\) and functions \(F_{0}, F_{1}\) are determined from spectra \((\lambda _{n})_{n=1}^{\infty }\) and \((\mu _{n})_{n=1}^{\infty }\), coefficients \(a_{m}\) and \(b_{m}\) are also determined from this spectra. Since \(q\in L^{2}[0, \pi ]\), we have

$$\begin{aligned} q(x)=\sum _{m=-\infty }^{m=+\infty } c_{m} {\text{ e }}^{2imx}, \end{aligned}$$

where is \(c_{m}= \frac{1}{\pi }a_{m} - \frac{i}{\pi }b_{m} \), and we finally conclude that potential q is uniquely determine from spectra \((\lambda _{n})_{n=1}^{\infty }\) and \((\mu _{n})_{n=1}^{\infty }\). \(\square \)

3.2 Nonlinear Case, \(\tau \in \left[ \frac{2\pi }{5},\frac{\pi }{2}\right) \)

In the case when \(\tau \in \left[ \frac{2\pi }{5},\frac{\pi }{2}\right) \) we have \(\pi \in (2\tau ,3\tau ]\), and using (1.3) and (1.4) from Eq. (2.3), we get

$$\begin{aligned} \Delta _{0}(\lambda )= & {} y(\pi )= \frac{\sin z\pi }{z} + \frac{1}{z^{2}} \int \limits _{\tau } ^{\pi }q(t) \sin z(t-\tau ) \sin z(\pi -t){\text{ d }}t\nonumber \\&+\frac{1}{z^{3}}\int \limits _{2\tau }^{\pi }\int \limits _{\tau }^{t-\tau }q(t)q(t_{1}) \sin z(\pi -t) \sin z(t_{1}-\tau ) \sin z(t-\tau -t_{1}) {\text{ d }}t_{1}{\text{ d }}t.\nonumber \\ \end{aligned}$$
(3.3)
$$\begin{aligned} \Delta _{1}(\lambda )= & {} y^{\prime }(\pi )+P(\lambda )y(\pi )= \cos z\pi +\frac{1}{z}\int \limits _{\tau }^{\pi }q(t) \cos z(\pi -t) \sin z(t-\tau ){\text{ d }}t\nonumber \\&+\frac{1}{z^{2}}\int \limits _{2\tau }^{\pi }\int \limits _{\tau }^{t-\tau }q(t)q(t_{1}) \cos z(\pi -t) \sin z(t_{1}-\tau ) \sin z(t-\tau -t_{1}) {\text{ d }}t_{1}{\text{ d }}t\nonumber \\&+P(z^{2})\frac{\sin z\pi }{z} + \frac{P(z^{2})}{z^{2}} \int \limits _{\tau } ^{\pi }q(t) \sin z(t-\tau ) \sin z(\pi -t){\text{ d }}t\nonumber \\&+\frac{P(z^{2})}{z^{3}}\int \limits _{2\tau }^{\pi }\int \limits _{\tau }^{t-\tau }q(t)q(t_{1}) \sin z(\pi -t) \sin z(t_{1}-\tau ) \sin z(t-\tau -t_{1}) {\text{ d }}t_{1}{\text{ d }}t.\nonumber \\ \end{aligned}$$
(3.4)

The functions \(\Delta _{0}(\lambda ), \Delta _{1}(\lambda )\) are entire in \(\lambda \) of order 1/2. It is clear that the set of zeros of functions \(\Delta _{0}(\lambda ), \Delta _{1}(\lambda )\) is equivalent to the spectrum of boundary spectral problems \(L_{0}, L_{1}\), respectively. Like in previous case specification of the spectra uniquely determines functions \(\Delta _{0}(\lambda ), \Delta _{1}(\lambda )\). We introduce notation

$$\begin{aligned} F_{0}(z) = z \Delta _{0}(\lambda ), \, F_{1}(z) = z \Delta _{1}(\lambda ) \end{aligned}$$

and

$$\begin{aligned} I_{1}=\int \limits _{\tau } ^{\pi }q(t){\text{ d }}t,\quad I_{2}=\int \limits _{2\tau } ^{\pi } q(t) \int \limits _{\tau } ^{t-\tau }q(t_{1}){\text{ d }}t_{1} {\text{ d }}t. \end{aligned}$$

Similar to the first case we have that delay \(\tau \) and the integral \(I_{1}\) are uniquely ordered from the spectrum \((\lambda _{n})_{n=1}^{\infty }\)(see [1]).

Lemma 3.3

The spectra \((\mu _{n})_{n=1}^{\infty }\) and \((\lambda _{n})_{n=1}^{\infty }\) of boundary value problem \(L_{1}\) and \(L_{0}\) uniquely determine polynomial \(P(\lambda )\).

Proof

Similar like in first case. \(\square \)

Now we transform the products of trigonometric functions into sums/differences and we have

$$\begin{aligned} F_{0}(z)= & {} \sin z\pi - \frac{\cos z(\pi -\tau )}{2z}I_{1} + \frac{1}{2z} \int \limits _{\tau }^{\pi } q(t)\cos z(\pi +\tau -2t) {\text{ d }}t - \frac{\sin z(\pi -2\tau )}{4z^{2}} I_{2}\\&+\frac{1}{4z^{2}}\int \limits _{2\tau }^{\pi }\int \limits _{\tau }^{t-\tau } q(t) q(u)\left( \sin z(\pi -2u)- \sin z(\pi -2t+2\tau ) +\sin z(\pi -2t+2u)\right) {\text{ d }}u{\text{ d }}t,\\ F_{1}(z)= & {} z \cos z\pi + \frac{\sin z(\pi -\tau )}{2}I_{1} -\frac{1}{2} \int \limits _{\tau }^{\pi } q(t)\sin z(\pi +\tau -2t) {\text{ d }}t\\&+\frac{1}{4z}\int \limits _{2\tau }^{\pi }\int \limits _{\tau }^{t-\tau } q(t)q(u)\left( \cos z(\pi -2u)- \cos z(\pi -2t+2\tau )+ \cos z(\pi -2t+2u)\right) {\text{ d }}u{\text{ d }}t\\&-\frac{\cos z(\pi -2\tau )}{4z} I_{2}+ P(z^{2}) \sin z\pi + \frac{P(z^{2})}{2z}\int \limits _{\tau }^{\pi }q(t) \cos z(\pi +\tau -2t){\text{ d }}t\\&-\frac{P(z^{2}) \cos z(\pi -\tau )}{2z}I_{1} -\frac{P(z^{2}) \sin z(\pi -2\tau )}{4z^{2}}I_{2}\\&+\frac{P(z^{2})}{4z^{2}}\int \limits _{2\tau } ^{\pi }\int \limits _{\tau }^{t-\tau } q(t)q(u)( \sin z(\pi -2u) - \sin z(\pi -2t+2\tau )\\&+ \sin z(\pi -2t+2u)) {\text{ d }}u{\text{ d }}t. \end{aligned}$$

We define functions \(K:[0,\pi ]\rightarrow R\), \(\widetilde{q}:[0,\pi ]\rightarrow R\)

$$\begin{aligned} K(t)= & {} \left\{ \begin{array}{rl} q(t+\tau )\int _{\tau }^{t}q(u){\text{ d }}u-q(t)\int _{t+\tau }^{\pi }q(u){\text{ d }}u -\int _{t+\tau }^{\pi }q(u-t)q(u){\text{ d }}u, &{} t\in [\tau , \pi -\tau ]\\ 0, &{} \mathrm{else} \end{array}\right. \\ \widetilde{q}(t)= & {} \left\{ \begin{array}{rl} q\left( t+\frac{\tau }{2}\right) , &{} t\in \left[ \frac{\tau }{2}, \pi -\frac{\tau }{2}\right] \\ 0, &{} \mathrm{else} \end{array}\right. \end{aligned}$$

It is obvious that \(K(t)\in L^{2}[0,\pi ]\) and \(\widetilde{q}(t)\in L^{2}[0,\pi ]\).

Using the following formulas as well as analogous formulas with cosine function function

$$\begin{aligned}&\int \limits _{2\tau }^{\pi }\int \limits _{\tau } ^{t-\tau }q(t) q(u)\sin z(\pi -2u){\text{ d }}u{\text{ d }}t = \int \limits _{\tau }^{\pi -\tau } \sin z(\pi -2t)q(t)\int \limits _{t+\tau }^{\pi } q(u){\text{ d }}u{\text{ d }}t,\\&\quad \int \limits _{2\tau }^{\pi }\int \limits _{\tau }^{t-\tau }q(t)q(u)\sin z(\pi -2t+2\tau ){\text{ d }}u{\text{ d }}t =\int \limits _{\tau }^{\pi -\tau } \sin z(\pi -2t)q(t+\tau )\int \limits _{\tau }^{t}q(u){\text{ d }}u{\text{ d }}t,\\&\quad \int \limits _{2\tau }^{\pi }\int \limits _{\tau }^{t-\tau }q(t)q(u) \sin z(\pi -2t+2u){\text{ d }}u{\text{ d }}t=\int \limits _{\tau }^{\pi -\tau } \sin z(\pi -2t)\int \limits _{t+\tau }^{\pi } q(u)q(u-t){\text{ d }}u{\text{ d }}t, \end{aligned}$$

and notation

$$\begin{aligned} \widetilde{a_{c}}(z)= & {} \int \limits _{\frac{\tau }{2}}^{\pi -\frac{\tau }{2}} \widetilde{q}(t) \cos z(\pi -2t) {\text{ d }}t,\quad \widetilde{a_{s}}(z) = \int \limits _{\frac{\tau }{2}}^{\pi -\frac{\tau }{2}} \widetilde{q}(t) \sin z(\pi -2t) {\text{ d }}t,\\ k_{c}(z)= & {} \int _{\tau }^{\pi -\tau } K(t) \cos z(\pi -2t) {\text{ d }}t,\quad k_{s}(z) = \int _{\tau }^{\pi -\tau } K(t) \sin z(\pi -2t) {\text{ d }}t, \end{aligned}$$

we have

$$\begin{aligned} F_{0}(z)= & {} \sin z\pi + \frac{1}{2z}\bigg ( \widetilde{a_{c}}(z) - I_{1}\cos z(\pi -\tau ) \bigg ) - \frac{1}{4z^{2}}\bigg ( k_{s}(z) + I_{2}\sin z(\pi -2\tau ) \bigg ), \nonumber \\ \end{aligned}$$
(3.5)
$$\begin{aligned} F_{1}(z)= & {} z\cos z\pi - \frac{1}{2}\bigg ( \widetilde{a_{s}}(z) - I_{1}\sin z(\pi -\tau ) \bigg ) - \frac{1}{4z}\bigg ( k_{c}(z) + I_{2}\cos z(\pi -2\tau ) \bigg )\nonumber \\&+P(z^{2})\sin z\pi +\frac{P(z^{2})}{2z}\bigg (\widetilde{a_{c}}(z) - I_{1} \cos z(\pi -\tau ) \bigg ) \nonumber \\&- \frac{P(z^{2})}{4z^{2}}\bigg ( k_{s}(z) + I_{2}\sin z(\pi -2\tau ) \bigg ). \end{aligned}$$
(3.6)

One can easily show that \(\int _{\tau }^{\pi -\tau }K(t){\text{ d }}t=-I_{2}\). Using this formula and integration by parts from (3.5) and (3.6), we have

$$\begin{aligned} F_{0}(z)= \sin z\pi + \frac{1}{2z}\bigg ( \widetilde{a_{c}}(z) -I_{1}\cos z(\pi -\tau ) \bigg ) - \frac{1}{2z} K_{c}^{\circ }(z) - \frac{1}{2z^{2}} I_{2} \sin z(\pi -2\tau ), \end{aligned}$$
(3.7)

and

$$\begin{aligned} F_{1}(z)= & {} z\cos z\pi - \frac{1}{2}\bigg ( \widetilde{a_{s}}(z) - I_{1}\sin z(\pi -\tau ) \bigg ) + \frac{1}{2} K_{s}^{\circ }(z) + P(z^{2})\sin z\pi \nonumber \\&+\frac{P(z^{2})}{2z}\bigg (\widetilde{a_{c}}(z) - I_{1}\cos z(\pi -\tau ) \bigg ) - \frac{P(z^{2})}{2z^{2}} I_{2} \sin z(\pi -2\tau ) - \frac{P(z^{2})}{2z} K_{c}^{\circ }(z), \end{aligned}$$
(3.8)

where is

$$\begin{aligned} K_{c}^{\circ }(z)= & {} \int \limits _{\tau }^{\pi -\tau } \int _{\tau }^{t} K(s) {\text{ d }}s \cos z(\pi -2t){\text{ d }}t,\\ K_{s}^{\circ }(z)= & {} \int \limits _{\tau }^{\pi -\tau } \int _{\tau }^{t} K(s) {\text{ d }}s \sin z(\pi -2t){\text{ d }}t. \end{aligned}$$

Using \(\int _{\tau }^{\pi -\tau }K(t){\text{ d }}t=-I_{2}\), we have

$$\begin{aligned} \int \limits _{\tau }^{t}K(s){\text{ d }}s=-I_{2}-\int \limits _{t}^{\pi -\tau }K(s){\text{ d }}s, \end{aligned}$$

and

$$\begin{aligned} K_{c}^{\circ }(z)= & {} \int \limits _{\tau }^{\pi -\tau } \int \limits _{\tau }^{t} K(s){\text{ d }}s \cos z(\pi -2t){\text{ d }}t=\int \limits _{\tau }^{\pi -\tau } \left( -I_{2}-\int \limits _{t}^{\pi -\tau }K(s)\right) {\text{ d }}s \cos z(\pi -2t){\text{ d }}t\\= & {} -I_{2}\int \limits _{\tau }^{\pi -\tau } \cos z(\pi -2t){\text{ d }}t-\int \limits _{\tau }^{\pi -\tau }\int \limits _{t}^{\pi -\tau }K(s){\text{ d }}s \cos z(\pi -2t){\text{ d }}t\\= & {} -\frac{I_{2} \sin z(\pi -2\tau )}{z}+K_{c}^{*}(z),\\ K_{s}^{\circ }(z)= & {} K_{s}^{*}(z), \end{aligned}$$

where is

$$\begin{aligned} K_{c}^{*}(z)= & {} -\int \limits _{\tau }^{\pi -\tau }\int \limits _{t}^{\pi -\tau }K(s){\text{ d }}s \cos z(\pi -2t){\text{ d }}t,\\ K_{s}^{*}(z)= & {} -\int \limits _{\tau }^{\pi -\tau }\int \limits _{t}^{\pi -\tau }K(s){\text{ d }}s \sin z(\pi -2t){\text{ d }}t. \end{aligned}$$

Now we transform (3.7) and (3.8) and we have

$$\begin{aligned} F_{0}(z)= \sin z\pi + \frac{1}{2z}\bigg ( \widetilde{a_{c}}(z) -I_{1}\cos z(\pi -\tau ) \bigg ) - \frac{1}{2z} K_{c}^{*}(z), \end{aligned}$$
(3.9)

and

$$\begin{aligned} F_{1}(z)= & {} z\cos z\pi - \frac{1}{2}\bigg ( \widetilde{a_{s}}(z) - I_{1}\sin z(\pi -\tau ) \bigg ) + \frac{1}{2} K_{s}^{*}(z) + P(z^{2})\sin z\pi \nonumber \\&+\frac{P(z^{2})}{2z}\bigg (\widetilde{a_{c}}(z) - I_{1}\cos z(\pi -\tau ) \bigg ) - \frac{P(z^{2})}{2z} K_{c}^{*}(z). \end{aligned}$$
(3.10)

We define function A(z)

$$\begin{aligned} A(z)= 2z F_{0}(z) - 2z \sin z\pi + I_{1} \cos z(\pi - \tau ). \end{aligned}$$

Function A(z) is determined by spectrum \((\lambda _{n})_{n=1}^{\infty }\) and from (3.10), we have

$$\begin{aligned} A(z)=\widetilde{a_{c}}(z) - K_{c}^{*}(z). \end{aligned}$$
(3.11)

Also, we define function \(B_{1}(z)\)

$$\begin{aligned} B_{1}(z)= -2 F_{1}(z)+ 2z \cos z\pi + I_{1} \sin z(\pi -\tau ) + 2P(z^{2})\sin z\pi - \frac{P(z^{2})}{z}I_{1}\cos z(\pi -\tau ), \end{aligned}$$

this function is ordered by spectrum \((\mu _{n})_{n=1}^{\infty }\) and we have

$$\begin{aligned} B_{1}(z)= \widetilde{a_{s}}(z)-K_{s}^{*}(z)-\frac{P(z^{2})}{z}A(z). \end{aligned}$$

Finally, we define function

$$\begin{aligned} B(z)=B_{1}(z)+\frac{P(z^{2})}{z}A(z), \end{aligned}$$

which is determined by spectra \((\lambda _{n})_{n=1}^{\infty }\) and \((\mu _{n})_{n=1}^{\infty }\), and

$$\begin{aligned} B(z)=\widetilde{a_{s}}(z) - K_{s}^{*}(z). \end{aligned}$$
(3.12)

We define function \(K^{*}:[0,\pi ]\rightarrow R\)

$$\begin{aligned} K^{*}(t) = \left\{ \begin{array}{rl} \int _{t}^{\pi -\tau }K(s){\text{ d }}s, &{} t\in (\tau ,\pi -\tau )\\ 0, &{} \mathrm{else} \end{array} \right. \end{aligned}$$

Put \(z = m, m \in N\) into (3.11) and (3.12), we have

$$\begin{aligned} (-1)^{m}A(m)= & {} \int \limits _{\frac{\tau }{2}}^{\pi -\frac{\tau }{2}}\widetilde{q}(t) \cos 2mt {\text{ d }}t - \int \limits _{\tau }^{\pi -\tau }\int \limits _{t}^{\pi -\tau }K(s){\text{ d }}s \cos 2mt{\text{ d }}t \end{aligned}$$
(3.13)
$$\begin{aligned} (-1)^{m+1}B(m)= & {} \int \limits _{\frac{\tau }{2}}^{\pi -\frac{\tau }{2} }\widetilde{q}(t) \sin 2mt {\text{ d }}t - \int \limits _{\tau }^{\pi -\tau } \int \limits _{t}^{\pi -\tau }K(s){\text{ d }}s \sin 2mt{\text{ d }}t. \end{aligned}$$
(3.14)

From (3.11), we have

$$\begin{aligned} \lim \limits _{z\rightarrow 0}A(z)=\int \limits _{\frac{\tau }{2}}^{\pi -\frac{\tau }{2}}\widetilde{q}(t){\text{ d }}t - \int \limits _{\tau }^{\pi -\tau } \int \limits _{t}^{\pi -\tau }K(s){\text{ d }}s{\text{ d }}t. \end{aligned}$$
(3.15)

We multiply Eq. (3.13) by \(\frac{1}{\pi }{\text{ e }}^{2imt} \), Eq. (3.14) by \(\frac{-i}{\pi }{\text{ e }}^{2imt} \) and Eq. (3.15) by \( \frac{1}{\pi } \) and then sum them, using definition of function \(\widetilde{q}(t)\) and \(K^{*}(t)\) we get the integral equation

$$\begin{aligned} \widetilde{q}(t) - K^{*}(t) = f(t),t\in [0,\pi ], \end{aligned}$$
(3.16)

where

$$\begin{aligned} f(t)=\frac{ 1 }{\pi }\lim _{z\rightarrow 0} A(z) + \frac{1}{\pi } \sum _{m\in Z\smallsetminus {0}}((-1)^{m}A(m) + i(-1)^{m}B(m)){\text{ e }}^{2imt}. \end{aligned}$$

Theorem 3.4

Let \((\lambda _{n}) _{n=1}^{\infty }\) and \((\mu _{n}) _{n=1}^{\infty }\) be the spectra of boundary spectral problems \(L_{k}, k=0,1\), respectively, then potential q is uniquely determined by \((\lambda _{n}) _{n=1}^{\infty }\) and \((\mu _{n}) _{n=1}^{\infty }\) if \(\frac{2\pi }{5} \leqslant \tau < \frac{\pi }{2}\).

Proof

The potential q satisfies integral equation (3.16), we will show uniqueness of solution of this equation.

  • For \(t\in \left( \pi -\tau , \pi -\frac{\tau }{2}]\right) \), since \(K^{*}(t)=0,t\in \left( \pi -\tau , \pi -\frac{\tau }{2}\right] \), integral equation (3.16) have a form:

    $$\begin{aligned} \widetilde{q}(t)=f(t) \end{aligned}$$

    Function f is determined by \((\lambda _{n}) _{n=1}^{\infty }\) and \((\mu _{n}) _{n=1}^{\infty }\), then potential q(x) is determined for \(x\in (\pi -\frac{\tau }{2}, \pi ]\).

  • For \(t\in \left( \frac{\tau }{2}, \tau \right] \), since \(K^{*}(t)=0,t\in \left( \frac{\tau }{2},\tau \right] \), integral equation (3.16) have a form:

    $$\begin{aligned} \widetilde{q}(t)=f(t). \end{aligned}$$

    Function f is determined by \((\lambda _{n}) _{n=1}^{\infty }\) and \((\mu _{n}) _{n=1}^{\infty }\), then potential q is determined for \(x\in [\tau , \frac{3\tau }{2}]\).

  • For \(t\in (\tau , \pi - \tau ]\), from (3.16), we have equation

    $$\begin{aligned} \widetilde{q}(t) - \int \limits _{t}^{\pi -\tau }K(s){\text{ d }}s=f(t). \end{aligned}$$

    One can easily show that arguments of the potential q appearing in the function

    $$\begin{aligned} \int \limits _{t}^{\pi -\tau }K(s){\text{ d }}s = \int \limits _{t}^{\pi -\tau }\left( q(s+\tau )\int \limits _{\tau }^{s}q(u){\text{ d }}u-q(s)\int \limits _{s+\tau }^{\pi }q(u){\text{ d }}u-\int \limits _{s+\tau }^{\pi }q(u-s)q(u){\text{ d }}u\right) {\text{ d }}s \end{aligned}$$

    belong to the intervals \([2\tau , \pi ] \subset [\pi -\frac{\tau }{2}, \pi ]\) and \( [\tau , \pi - \tau ] \subset [\tau , \frac{3\tau }{2}] \). Then the function \( \int \limits _{t}^{\pi -\tau }K(s){\text{ d }}s \) is known. Therefore from (3.16) for \(t\in (\tau , \pi - \tau ]\), we get

    $$\begin{aligned} \widetilde{q}(t) = \int \limits _{t}^{\pi -\tau }K(s){\text{ d }}s+f(t). \end{aligned}$$

    Function f is determined by \((\lambda _{n}) _{n=1}^{\infty }\) and \((\mu _{n}) _{n=1}^{\infty }\), then potential q(x) is determined for \(x\in \left( \frac{3\tau }{2}, \pi -\frac{\tau }{2}\right] \).

\(\square \)

When \(\frac{\pi }{3} \leqslant \tau < \frac{\pi }{2}\), we have same integral equation like 3.13, but in the case when \(\frac{\pi }{3} \leqslant \tau < \frac{2\pi }{5}\) not satisfied \([2\tau , \pi ] \subset [\pi -\frac{\tau }{2}, \pi ]\) and \( [\tau , \pi - \tau ] \subset [\tau , \frac{3\tau }{2}] \), and we cannot prove Theorem 3.4 on this way. Moreover, in the case when \(\frac{\pi }{3} \leqslant \tau < \frac{2\pi }{5}\) theorem of uniqueness not true. For this conclusion, the main arguments are the results published in the papers [4] and [5]. In the case when \(\frac{\pi }{3}\leqslant \tau < \frac{2\pi }{5}\) then critical interval \(\left( \frac{3\tau }{2},\pi -\tau \right) \) not equal to \(\varnothing \).