1 Introduction

The main results in the inverse spectral problems for classical Sturm–Liouville operators can be found in the monographs [1, 2]. Some of the main methods in the inverse problem theory for classical Sturm–Liouville operators turned out to be unsuitable for operators with delays. In this paper, we use the method of Fourier coefficients. This method is based on the determination of direct relations between Fourier coefficients of the potential or functions containing the potential, and Fourier coefficients of some known function. Some of the results of the inverse spectral problem for Sturm–Liouville operators with a delay can be found in [3,4,5,6,7,8,9,10]. Studying of the spectral problems for Differential Operators with two or more constant delays is of recent origin and some of the results can be found in [11,12,13,14,15,16]. One of the interesting features of the case with two (and more) delays is the requirement of specifying the spectra for two (and accordingly more) different differential equation. Negative answer to the question whether one can find an appropriate inverse problem statement involving only one equation can be found in [17]. The paper [18] is devoted to the studying of direct problems for operators with N constant delays. In what follows, we always take \( i=0,1\) and \(k=1,2\).

We consider the boundary value problems \(D_{i,k}\)

$$\begin{aligned}&-y''(x)+q_1 (x)y(x-\tau _1)+(-1)^{i}q_2 (x) y(x-\tau _2)=\lambda y(x), \quad x \in [0,\pi ], \end{aligned}$$
(1.1)
$$\begin{aligned}&y'(0)-hy(0)=0, \end{aligned}$$
(1.2)
$$\begin{aligned}&y'(\pi )+H_k y(\pi )=0, \end{aligned}$$
(1.3)

where \(\dfrac{2\pi }{5} \le \tau _2 \le \tau _1 < \dfrac{\pi }{2}\), \(h,H_k\in \mathbb {R}\), \(H_1 \ne H_2\), and \(\lambda \) is a spectral parameter. We assume that \(q_1\), \(q_2\) are complex valued potential functions from \(L_2 [0,\pi ]\) such that \(q_1(x)=0\) as \(x \in [0,\tau _1)\) and \(q_2(x)=0\) as \(x \in [0,\tau _2)\).

We study the inverse spectral problem of recovering operators from the spectra of \(D_{i,k}\) and generalize the results from the paper [3] which deals with operators with one constant delay from the interval \(\left[ \dfrac{2\pi }{5}, \dfrac{\pi }{2}\right] \) to the operators with two constant delays from the same interval.

Let \((\lambda _{n,i,k})^\infty _{n=0}\) be the eigenvalues of the boundary value problems \(D_{i,k}\). The inverse problem is formulated as follows.

Inverse problem 1: Given \((\lambda _{n,i,k})^\infty _{n=0}\) find delays \(\tau _k\), parameters \(h,H_k\), and potential functions \(q_k\).

The organization of the paper is the following. In Sect. 2, we study the spectral properties of the boundary value problems \(D_{i,k}\). In Sect. 3, we prove that delays and parameters are uniquely determined by the spectra. Then we prove that the potentials are uniquely determined by the system of two Volterra linear integral equations.

2 Spectral properties

It can be easily shown that the differential Eq. (1.1) under the initial condition (1.2) along with the normalizing condition \(y(0)=1\) and conditions \(q_k (x)=0\) as \(x \in [0,\tau _k)\) is equivalent to the integral equation

$$\begin{aligned} y_i (x,z)=&\cos xz+\dfrac{h}{z} \sin zx+\dfrac{1}{z} \int ^x_{\tau _1} q_1 (t) \sin z(x-t) y(t-\tau _1,z) \,\mathrm{d}t\nonumber \\&+\dfrac{(-1)^{i}}{z} \int ^x_{\tau _2} q_2 (t) \sin z(x-t) y(t-\tau _2, z)\,\mathrm{d}t. \end{aligned}$$
(2.1)

Here and in the sequel, we take \(\lambda =z^2\). By the method of steps, it can be easily verified that the solution of the integral equation (2.1) on the interval \((2 \tau _1, \pi ]\) is

$$\begin{aligned} y_i (x,z)&=\cos zx+ \dfrac{h}{z} \sin zx+\frac{1}{z} \bigg (b_{sc}^{(1)}(x,z)+(-1)^{i} b_{sc}^{(2)} (x,z)\bigg )\nonumber \\&\quad + \dfrac{h}{z^2} \bigg ( b_{s^2}^{(1)}(x,z)+(-1)^{i} b^{(2)}_{s^{2}}(x,z)\bigg )\nonumber \\&\quad + \dfrac{1}{z^2} \bigg (b_{s^2 c}^{(1)} (x,z)+b_{s^2 c}^{(2)} (x,z)+(-1)^{i}b_{s^2 c}^{(1,2)}(x,z)+(-1)^{i}b_{s^2 c}^{(2,1)}(x,z)\bigg )\nonumber \\&\quad + \dfrac{h}{z^3}\bigg (b_{s^3}^{(1)}(x,z)+b_{s^3}^{(2)}(x,z)+(-1)^{i}b_{s^3}^{(1,2)}(x,z)+(-1)^{i}b_{s^3}^{(2,1)}(x,z)\bigg ) \end{aligned}$$
(2.2)

where the integral terms in this equation are given by

$$\begin{aligned} b_{sc}^{(k)}(x,z)&=\int ^x_{\tau _k} q_k (t) \sin z(x-t) \cos z(t-\tau _k)\,\mathrm{d}t,\\ b_{s^2}^{(k)}(x,z)&=\int ^x_{\tau _k} q_k (t) \sin z (x-t) \sin z(t-\tau _k) \,\mathrm{d}t,\\ b_{s^2 c}^{(k)}(x,z)&=\int ^x_{2 \tau _k}q_k (t) \sin z(x-t) b_{sc}^{(2)} (t-\tau _k, z)\,\mathrm{d}t,\\ b_{s^2 c}^{(k,l)}(x,z)&=\int ^x_{\tau _1+\tau _2} q_k (t) \sin z(x-t) b_{sc}^{(l)} (t-\tau _k, z)\,\mathrm{d}t, \quad (l=1,2, \ k\ne l),\\ b_{s^3}^{(k)}(x,z)&=\int _{2 \tau _k}^x q_k (t) \sin z(x-t) b_{s^2}^{(2)}(t-\tau _k,z)\,\mathrm{d}t,\\ b_{s^{3}}^{(k,l)}(x,z)&=\int ^x_{\tau _1 +\tau _2} q_k (t) \sin z(x-t)b_{s^2}^{(l)} (t-\tau _k, z)\,\mathrm{d}t, \quad (l=1,2, \ k\ne l). \end{aligned}$$

Denote

$$\begin{aligned} \Delta _{i,k}=F_{i,k}(z)=y'_i (\pi , z)+H_k y(\pi ,z) \end{aligned}$$

From (2.2), we obtain

$$\begin{aligned} F_{i,k}(z)&= \bigg (-z+\dfrac{hH_k}{z} \bigg )\sin \pi z+ (h+H_k) \cos \pi z+ b_{c^2}^{(1)} (z)+(-1)^{i} b_{c^2}^{(2)}(z) \\&\quad + \dfrac{h}{z}\bigg (b_{cs}^{(1)}(z)+(-1)^{i}b_{cs}^{(2)}(z)\bigg )+\dfrac{H_k}{z} \bigg (b_{sc}^{(1)}(z)+(-1)^{i}b_{sc}^{(2)}(z)\bigg ) \\&\quad + \dfrac{H_k h}{z^2} \bigg (b_{s^2}^{(1)}(z)+(-1)^{i}b_{s^2}^{(2)}(z)\bigg ) \\& \quad +\dfrac{1}{z} \bigg (b_{csc}^{(1)}(z)+b_{csc}^{(2)}(z)+(-1)^{i}b_{csc}^{(1,2)}(z)+(-1)^{i}b_{csc}^{(2,1)}(z)\bigg )\\&\quad +\dfrac{h}{z^2}\bigg (b_{cs^{2}}^{(1)}(z)+b_{cs^2}^{(2)}(z)+(-1)^{i}b_{cs^2}^{(1,2)}(z)+(-1)^i b_{cs^2}^{(2,1)}(z)\bigg )\\&\quad +\dfrac{H_k}{z^2}\bigg (b_{s^2 c}^{(1)}(z)+b_{s^2 c}^{(2)}(z)+(-1)^{i}b_{s^2 c}^{(1,2)}(z)+(-1)^{i}b_{s^2 c}^{(2,1)}(z)\bigg )\\& \quad +\dfrac{Hh}{z^3}\bigg (b_{s^3}^{(1)}(z)+b_{s^3}^{(2)}(z)+(-1)^{i}b_{s^3}^{(1,2)}(z)+(-1)^{(i)}b_{s^3}^{(2,1)}(z)\bigg ) \end{aligned}$$

where

$$\begin{aligned} b_{cs}^{(k)}(z)&=\int ^\pi _{\tau _k} q_k (t) \cos z(\pi -t)\sin z(t-\tau _k)\,\mathrm{d}t,\\ b_{c^2}^{(k)}(z)&=\int _{\tau _k}^{\pi } q_k(t) \cos z(\pi -t) \cos z(t-\tau _k)\,\mathrm{d}t,\\ b_{csc}^{(k)}(z)&=\int ^\pi _{2 \tau _k} q_k (t) \cos z (\pi -t)b_{sc}^{(k)}(t-\tau _k,z)\,\mathrm{d}t,\\ b_{csc}^{(k,l)}(z)&=\int ^x_{\tau _1+\tau _2}q_k (t) \cos z(\pi -t)b_{sc}^{(l)}(t-\tau _k,z)\,\mathrm{d}t, \quad (k\ne l),\\ b_{cs^2}^{(k)}(z)&=\int ^\pi _{2 \tau _2}q_k (t) \cos z (\pi -t)b_{s^2}^{(k)}(t-\tau _k,z)\,\mathrm{d}t,\\ b_{cs^2}^{(k,l)}(z)&= \int ^x_{\tau _1+\tau _2} q_k (t) \cos z(x-t)b_{s^2}^{(l)}(t-\tau _k,z)\,\mathrm{d}t, \quad (k\ne l). \end{aligned}$$

Here, for the sake of simplifying the above-given equations for \(F_{i,k}(z)\), we write (z) as the argument of the functions instead of \((\pi ,z)\) .

To simplify further consideration, we define the so-called transitional function \(\tilde{q}_i\) as follows:

$$\begin{aligned} \tilde{q}_i (t)=\left\{ \begin{array}{ll} q_1 \bigg (t+\dfrac{\tau _1}{2}\bigg )+(-1)^{i}q_2 \bigg (t+\dfrac{\tau _2}{2}\bigg ), &{} t \in \bigg [\dfrac{\tau _1}{2}, \pi -\dfrac{\tau _1}{2}\bigg ], \\ (-1)^{i}q_2 \bigg (t+\dfrac{\tau _2}{2}\bigg ), &{} t \in \bigg [\dfrac{\tau _2}{2},\tau _2\bigg ) \cup \bigg (\pi -\dfrac{\tau _1}{2}, \pi -\dfrac{\tau _2}{2}\bigg ],\\ 0, &{} t \in \bigg [0,\dfrac{\tau _2}{2}\bigg )\cup \bigg (\pi -\dfrac{\tau _2}{2},\pi \bigg ]. \end{array} \right. \end{aligned}$$

Let us also define the function

$$\begin{aligned} K_i (t)=K^{(1)}(t)+K^{(2)}(t)+(-1)^{i}K^{(1,2)}(t)+(-1)^{i}K^{(2,1)}(t), \end{aligned}$$

where

$$\begin{aligned}&K^{(k)}(t)=q_k (t+\tau _k)\int ^t_{\tau _k}q_k (s)\,\mathrm{d}s-q_k (t)\int ^\pi _{t+\tau _k}q_k (s)\,\mathrm{d}s\\&\quad -\int ^\pi _{t+\tau _k}q_k (s-t)q_k (s)\,\mathrm{d}s, \quad t \in [\tau _k, \pi -\tau _k],\\&K^{(k)}(t)=0, \quad t\in [0,\tau _k)\cup (\pi -\tau _k,\pi ], \\&K^{(k,l)}(t)=q_k\bigg (t+\dfrac{\tau _1+\tau _2}{2}\bigg )\int ^t_{\tau _l}q_l (s)\,\mathrm{d}s-q_l \bigg (t-\frac{\tau _k}{2}+\dfrac{\tau _l}{2}\bigg )\int _{t+\frac{\tau _1+\tau _2}{2}}^{\pi } q_k (s)\,\mathrm{d}s\\&\quad -\int _{t+\frac{\tau _1+\tau _2}{2}}^{\pi }q_l \bigg (s-t-\dfrac{\tau _k}{2}+\dfrac{\tau _l}{2}\bigg )q_k (s)\,\mathrm{d}s, \quad t \in \bigg [\dfrac{\tau _1+\tau _2}{2}, \pi -\dfrac{\tau _1+\tau _2}{2}\bigg ], \end{aligned}$$

and

$$\begin{aligned} K^{(k,l)}(t)=0, \quad t\in \bigg [0, \dfrac{\tau _1+\tau _2}{2}\bigg )\cup \bigg (\pi -\dfrac{\tau _1+\tau _2}{2}, \pi \bigg ]. \end{aligned}$$

Moreover, if we introduce the notations

$$\begin{aligned} J_1^{(k)}&=\int ^\pi _{\tau _k} q_k (t)\,\mathrm{d}t, \quad J_2^{(k)}=\int ^\pi _{2 \tau _k}q_k (t)\bigg (\int ^{t-\tau _k}_{\tau _k} q_k (s)\,\mathrm{d}s\bigg )\,\mathrm{d}t,\\ J_2^{(k,l)}&=\int ^\pi _{\tau _1+\tau _2} q_k (t)\bigg (\int ^{t-\tau _k}_{\tau _l}q_l (s)\,\mathrm{d}s\bigg )\,\mathrm{d}t\end{aligned}$$

and the functions

$$\begin{aligned} \tilde{a}_{i,c}(z)&=\int ^\pi _0 \tilde{q}_i (t) \cos z(\pi -2t)\,\mathrm{d}t, \quad \tilde{a}_{i,s} (z)= \int ^\pi _0 \tilde{q}_i (t)\sin z(\pi -2t)\,\mathrm{d}t,\\ k_{i,s}(z)&=\int ^{\pi }_0 K_i (t) \sin z(\pi -2t)\,\mathrm{d}t, \quad k_c(z)=\int ^\pi _0 K_i (t) \cos z (\pi -2t)\,\mathrm{d}t,\\ u_{i,s}(z)&=\int ^\pi _0 U_i (t) \sin z (\pi -2t)\,\mathrm{d}t, \quad u_c (z)= \int ^\pi _0 U_i (t) \cos z (\pi -2t)\,\mathrm{d}t\end{aligned}$$

where the functions \(U_i\) in the last two equations are defined as

$$\begin{aligned} U(t)=U^{(1)}(t)+U^{(2)}(t)+(-1)^{i}U^{(1,2)}(t)+(-1)^{i} U^{(2,1)}(t) \end{aligned}$$

we can easily show that the following relations hold:

$$\begin{aligned} \left\{ \begin{array}{ll} \int ^{\pi -\tau _k}_{\tau _k} K^{(k)}(t)\,\mathrm{d}t=-J_2^{(k)}, \quad \int ^{\pi -\tau _k}_{\tau _k} U^{(2)}(t)\,\mathrm{d}t=J_2^{(2)}, \\ \int ^{\pi -\frac{\tau _1+\tau _2}{2}}_{\frac{\tau _1+\tau _2}{2}} K^{(k,l)}(t)\,\mathrm{d}t=-J_2^{(k,l)}, \quad \int _{\frac{\tau _1+\tau _2}{2}}^{\pi -\frac{\tau _1+\tau _2}{2}} U^{(k,l)}(t)\,\mathrm{d}t=J_2^{(k,l)}. \end{array} \right. \end{aligned}$$
(2.3)

Here we note that the functions \(U^{(k,l)}(t)\) differ from functions \(K^{(k,l)}(t)\) only with the sign in front of the third integral in (2.3).

Using the aforementioned notations and relations given with (2.3), we can rewrite the characteristic functions \(F_{i,k}(z)\) as follows

$$\begin{aligned} F_{i,k}(z)=&\bigg (-z+\dfrac{hH_k}{z}\bigg )\sin \pi z+(h+H_k) \cos \pi z \nonumber \\&+\dfrac{1}{2}\big (\tilde{a}_{i,c}(z)+J_{i,c}(z)\big )+\dfrac{h}{2z}\big (-\tilde{a}_{i,s}(z)+J_{i,s}(z)\big )\nonumber \\&+\dfrac{H_k}{2z}\big (\tilde{a}_{i,s}(z)+J_{i,s}(z)\big )+\dfrac{h H_k}{2z^2}\big (\tilde{a}_{i,c}(z)-J_{i,c}(z)\big )\nonumber \\&+\dfrac{1}{4z}\big (J_{2,i,s}(z)-u_{i,s}(z)\big )-\dfrac{h}{4z^2}\big (J_{2,i,c}(z)+k_{i,c}(z)\big )\nonumber \\&-\dfrac{H_k}{4z^2}\big (J_{2,i,c}(z)-u_{i,c}(z)\big )-\dfrac{h H_k}{4z^3}\big (J_{2,i,s}(z)+k_{i,s}(z)\big ) \end{aligned}$$
(2.4)

where

$$\begin{aligned} J_{i,c}(z)&=J_1^{(1)}\cos z(\pi -\tau _1)+(-1)^{i} J_1^{(2)} \cos z(\pi -\tau _2), \\ J_{i,s}(z)&=J_1^{(1)} \sin z(\pi -\tau _1)+(-1)^{i}J_1^{(2)}\sin z(\pi -\tau _2),\\ J_{2,i,c}(z)=&J_2^{(1)} \cos z(\pi -2\tau _1)+J_2^{(2)} \cos z(\pi -2\tau _2)\\&+(-1)^{i} \big (J_2^{(1,2)}+J_2^{(2,1)}\big )\cos z(\pi -\tau _1-\tau _2),\\ J_{2,i,s}(z)=&J_2^{(1)} \sin z(\pi -2\tau _1)+J_2^{(2)} \sin z(\pi -2\tau _2)\\&+(-1)^{i} \big (J_2^{(1,2)}+J_2^{(2,1)}\big )\sin z(\pi -\tau _1-\tau _2). \end{aligned}$$

The functions \(F_{i,k}(z)\) have one singular point \(z=0\). It can be easily proved that \(z=0\) is an apparent singularity. Also, the functions \(F_{i,k}(z)\) are entire in \(\lambda \) of order 1/2. Indeed, it is well known that the functions \(\sin z\pi \) and \(\cos z\pi \) are entire of order 1/2. Since any entire function has the form of \(f(z)=\sum _{n=0}^\infty a_n z^n\) and we can determine its order as \(\underset{n\rightarrow \infty }{\limsup } \dfrac{n \ln n}{-\ln |a_n|}\), we conclude that every given function in (2.4) has the same order as the functions \(\sin z \pi \) and \(\cos z \pi \). Further, using (2.4) and a well-known method (see [1]), we obtain the asymptotic formulas for the eigenvalues \((\lambda _{n,i,k})^\infty _{n=0}\) of the boundary value problems \(D_{i,k}\) as

$$\begin{aligned} \lambda _{n,i,k}=n^2+\dfrac{2(h+H_k)}{\pi }+\dfrac{J_1^{(1)}}{\pi } \cos n\tau _1+\dfrac{(-1)^{i}J_1^{(2)}}{\pi } \cos n \tau _2+o(1), \ n \rightarrow \infty . \end{aligned}$$
(2.5)

Now, by Hadamard’s factorization theorem, from the spectra of \(D_{i,k}\), we can construct the characteristic functions \(F_{i,k}(z)\). The next lemma holds.

Lemma 1

The specification of the spectrum \((\lambda _{n,i,k})^\infty _{n=0}\) of the boundary value problems \(D_{i,k}\) uniquely determines the characteristic functions \(F_{i,k}(z)\) by the formulas

$$\begin{aligned} F_{i,k}(z)=\pi \big (\lambda _{0,i,k}-z^2\big )\prod _{n=1}^\infty \dfrac{\lambda _{n,i,k}-z^2}{n^2}. \end{aligned}$$
(2.6)

3 Main results

   In this section, we prove that the delays and the parameters are uniquely determined by the spectra.

Lemma 2

The delays \(\tau _k\), the integrals \(J_1^{(k)}\), and the sums \(h+H_k\) are uniquely determined by the eigenvalues \((\lambda _{n,i,k})^\infty _{n=0}\).

Proof

Let us consider the sequences \(\rho _{n,k}= \dfrac{1}{2} \big (\lambda _{n,0,k}+\lambda _{n,1,k}\big )\) and \(\sigma _n=\dfrac{1}{2} \big (\lambda _{n,0,1}-\lambda _{n,1,1}\big )\). From (2.5), we have

$$\begin{aligned} \rho _{n,k}=n^2+\dfrac{2}{\pi }(h+H_k)+\dfrac{J_1^{(1)}}{\pi } \cos n \tau _1+o(1) \end{aligned}$$

and

$$\begin{aligned} \sigma _{n,i}=\dfrac{J_1^{(2)}}{\pi } \cos n \tau _2 +o(1) \end{aligned}$$

as \(n \rightarrow \infty \). Obviously, the delays \(\tau _1\), \(\tau _2\) and integrals \(J_1^{(1)}\), \(J_1^{(2)}\) can be determined from sequences \(\big (\rho _{n,k}\big )_{n=0}^\infty \) and \(\big (\sigma _{n}\big )_{n=0}^\infty \) in the same way as for the operators with one delay (see [9]). Lemma is proved.

Lemma 3

Parameters h and \(H_k\) are uniquely determined by the eigenvalues \((\lambda _{n,0,k})^\infty _{n=0}\).

Proof

Functions \(J_{1,c}^0(z)\) and \(J_{1,s}^0(z)\) are known by virtue of Lemma 2. Since the characteristic functions are uniquely determined by the spectra, by writing \(\lambda =\big (\dfrac{4m+1}{2}\big )^2\) in (2.6), we can define the functions

$$\begin{aligned} F^*_{0,k}(m)=F_{0,k} \big (\dfrac{4m+1}{2}\big )+\dfrac{4m+1}{2}-\dfrac{1}{2} J_{0,c} \big (\dfrac{4m+1}{2}\big )-\dfrac{H_k+h}{4m+1} J_{0,s} \big (\dfrac{4m+1}{2}\big ). \end{aligned}$$

Then, using the form of the characteristic functions \(F_{0,k}\) given in (2.4), we get

$$\begin{aligned} h=\dfrac{1}{2} \underset{m \rightarrow \infty }{\lim } \dfrac{4m+1}{H_2 -H_1} \big (F^*_{0,2}(m)-F*_{0,1}(m)\big ). \end{aligned}$$

Thus, we determine \(H_k\) from \(h+H_k\) and prove the lemma. \(\square \)

To recover the potential functions from the spectra, we should transform the characteristic functions (2.4). For this purpose, we use the method of integration by parts in (2.4) once in the integrals denoted by \(\tilde{a}_{i,s} (z)\), \(\tilde{a}_{i,c}(z)\), \(u_s (z)\), and \(u_c (z)\), and twice in the integrals denoted by \(k_c (z)\) and \(k_s (z)\). This is how the following function appears

$$\begin{aligned} K_i^*(t)=K^{(1)*}(t)+K^{(2)*}(t)+(-1)^{i}K^{(1,2)*}(t)+(-1)^{(i)}K^{(2,1)*}(t), \end{aligned}$$

where

$$\begin{aligned} K^{(k)*}(t)=\left\{ \begin{array}{ll} \int ^t_{\tau _k} K^{(k)}(u) \,\mathrm{d}u, &{} t \in [\tau _k, \pi -\tau _k], \\ 0, &{} t\in [0,\tau _k)\cup (\pi -\tau _k, \pi ], \end{array} \right. \end{aligned}$$

and

$$\begin{aligned} K^{(k,l)*}(t)=\left\{ \begin{array}{ll} \int ^{t}_{\frac{\tau _1+\tau _2}{2}} K^{(k,l)}(u)\,\mathrm{d}u, &{} t \in \big [\frac{\tau _1+\tau _2}{2},\pi -\frac{\tau _1+\tau _2}{2}\big ],\\ 0, &{} t \in \big [0,\frac{\tau _1+\tau _2}{2}\big )\cup \big (\pi -\frac{\tau _1+\tau _2}{2},\pi \big ]. \end{array} \right. \end{aligned}$$

One can show that the following relations hold

$$\begin{aligned}&\int ^{\pi -\tau _k}_{\tau _k} \big (\int ^t_{\tau _k}K^{(k)}(u)\,\mathrm{d}u\big )\,\mathrm{d}t=-(\pi -2\tau _k)J_2^{(k)},\\&\int ^{\pi -\frac{\tau _1+\tau _2}{2}}_{\frac{\tau _1+\tau _2}{2}}\big (\int ^t_{\pi -\frac{\tau _1+\tau _2}{2}}(K^{(1,2)}(u)+K^{(2,1)}(u))\,\mathrm{d}u\big )\,\mathrm{d}t\\&\quad =-(\pi -\tau _1-\tau _2)\big (J_2^{(1,2)}+J_2^{(2,1)}\big ). \end{aligned}$$

Then we obtain the characteristic functions in the form

$$\begin{aligned} F_{i,k}(z)=&\big (-z+\dfrac{H_k h}{z}\big )\sin \pi z+(h+H_k) \cos \pi z \nonumber \\&+\dfrac{1}{2}\big (\tilde{a}_{i,c}(z)+\dfrac{H_k}{z}\tilde{a}_{i,s} (z)\big )-h \bigg (\tilde{q}_{i,c}^{(1)}(z)+\dfrac{H_k}{z}\tilde{q}_{i,s}^{(1)}(z)\bigg )\nonumber \\&-\dfrac{1}{2}\bigg (u^{*}_{i,c}(z)+\dfrac{H_k}{z}u^{*}_{i,s} (z)\bigg )+ h\bigg (k^{**}_{i,c}(z)+\dfrac{H_k}{z}k^{**}_{i,s}(z)\bigg )\nonumber \\&+ \dfrac{J_{i,c}(z)}{2}+\dfrac{2h+H_k}{2z}J_{i,s}(z)\nonumber \\&+\dfrac{1}{2z}\bigg (1+(\pi -2\tau )h-\dfrac{H_k h}{z^2}\bigg )J_{2,i,s}(z)\nonumber \\&+\dfrac{H_k h (\pi -2\tau )}{2z^2}J_{2,i,c}(z) \end{aligned}$$
(3.1)

where

$$\begin{aligned} \tilde{q}_{i,c}^{(1)}(z)&=\int ^{\pi -\frac{\tau _2}{2}}_{\frac{\tau _2}{2}} \bigg (\int ^t_{\frac{\tau _2}{2}}\tilde{q}_i (s)ds\bigg ) \cos z(\pi -2t)\,\mathrm{d}t, \\ \tilde{q}_{i,s}^{(1)}(z)&=\int ^{\pi -\frac{\tau _2}{2}}_{\frac{\tau _2}{2}} \bigg (\int ^t_{\frac{\tau _2}{2}}\tilde{q}_i (s)ds\bigg )\sin z(\pi -2t)\,\mathrm{d}t,\\ u_{i,c}^* (z)&=\int ^{\pi -\tau _2}_{\tau _2}\bigg (\int _{\tau _2}^t U_i (s)ds\bigg )\cos z (\pi -2t)\,\mathrm{d}t,\\ u_{i,s}^* (z)&=\int ^{\pi -\tau _2}_{\tau _2}\bigg (\int _{\tau _2}^t U_i (s)ds\bigg )\sin z (\pi -2t)\,\mathrm{d}t,\\ k_{i,c}^{**}(z)&=\int ^{\pi -\tau _2}_{\tau _2} \bigg (\int ^t_{\tau _2} K_i^* (s)ds\bigg )\cos z(\pi -2t)\,\mathrm{d}t,\\ k_{i,s}^{**}(z)&=\int _{\tau _2}^{\pi -\tau _2} \bigg (\int ^t_{\tau _2}K_i^* (s)ds\bigg )\sin z(\pi -2t)\,\mathrm{d}t\end{aligned}$$

and

$$\begin{aligned} (\pi -2 \tau )J_{2,i,s} (z)=&\sum _{k=1}^2 (\pi -2 \tau _k) J_2^{(k)} \sin z(\pi -2 \tau _k)\\&+(-1)^{i}(\pi -\tau _1-\tau _2) \big (J_2^{(1,2)}+J_2^{(2,1)}\big ) \sin z (\pi -\tau _1-\tau _2),\\ (\pi -2\tau )J_{2,i,c} (z)=&-\sum ^2_{k=1} (\pi -2 \tau _k)J_2 ^{(k)} \cos z(\pi -2\tau _k)\\&- (-1)^{i}(\pi -\tau _1-\tau _2) \big (J_2^{(1,2)}+J_2^{(2,1)}\big ) \cos z (\pi -\tau _1-\tau _2). \end{aligned}$$

Since we have the transformed version of the characteristic functions (2.4), we are now ready to recover the potential functions from the spectra. To do this, first we need to define the functions

$$\begin{aligned} A_i (z)=&\dfrac{2}{H_2-H_1}\big (H_2 F_{i,1}(z)-H_1 F_{i,2} (z)\big )+2 z\sin \pi z-2h \cos \pi z\\&-J_{i,c}(z)-\dfrac{2h}{z}J_{i,s}(z) \end{aligned}$$

and

$$\begin{aligned} B_i (z)=\dfrac{2z}{H_2-H_1}\big ( F_{i,2}(z)- F_{i,1} (z)\big )-2h \sin \pi z-2z \cos \pi z-J_{i,s}(z). \end{aligned}$$

From (3.1), we obtain

$$\begin{aligned}&A_i(z)=\tilde{a}_{i,c}(z)-2h\tilde{q}_{i,c}^{(1)}(z)-u^{*}_{i,c}(z)+2hk_{i,c}^{**}(z)+\alpha _i (z), \end{aligned}$$
(3.2)
$$\begin{aligned}&B_i (z)=\tilde{a}_{i,s} (z)-2h \tilde{q}_{i,s}^{(1)}(z)-u^{*}_{i,s}(z)+2hk_{i,s}^{**} (z)+\beta _i (z) \end{aligned}$$
(3.3)

where

$$\begin{aligned} \alpha _{i}(z)=\dfrac{1+(\pi -2\tau )h}{2z}J_{2,i,s}(z) \end{aligned}$$
(3.4)

and

$$\begin{aligned} \beta _i (z)=\dfrac{h(\pi -2\tau )}{z} J_{2,i,c}(z)-\dfrac{h}{z^2}J_{2,i,s}(z). \end{aligned}$$
(3.5)

The two below-given equations hold:

$$\begin{aligned} \beta _{0,i}=&\underset{z \rightarrow 0}{\lim } \beta _i (z)=0,\\ \alpha _{0,i}=&\underset{z \rightarrow 0}{\lim } \alpha _i (z)=\sum _{k=1}^2 \big (h(\pi -2\tau _k)^2+(\pi -2\tau _k)\big )J_2^{(k)}\sin z(\pi -2 \tau _k)\\&+ (-1)^{i} \big (h (\pi -\tau _1-\tau _2)^2+(\pi -\tau _1-\tau _2)\big )\bigg (J_2 ^{(1,2)}+J_2^{(2,1)}\bigg ) \\&\times \sin z (\pi -\tau _1-\tau _2). \end{aligned}$$

If we put \(z=m\), \(m \in \mathbb {N}\) into (3.2) and (3.3) and denote

$$\begin{aligned} A_{2m,i}&=\dfrac{2m}{\pi }(-1)^m A_i (m), \quad B_{2m,i}=\dfrac{2}{\pi }(-1)^{m+1}B_i (m),\\ \alpha _{2m,i}&=\dfrac{2}{\pi }(-1)^m \alpha _i (m), \quad \beta _{2m,i}=\dfrac{2}{\pi }(-1)^{m+1} \beta _i (m) \end{aligned}$$

, we obtain

$$\begin{aligned}&A_{2m,i}=\dfrac{2}{\pi }\tilde{a}_{2m,i}-\dfrac{4}{\pi } h \tilde{q}_{2m,i,c}^{(1)}-\dfrac{2}{\pi }u^{*}_{2m,i,c}+\dfrac{4}{\pi }hk^{**}_{2m,i,c}+\alpha _{2m,i} \end{aligned}$$
(3.6)
$$\begin{aligned}&B_{2m}=\dfrac{2}{\pi }\tilde{b}_{2m,i}-\dfrac{4}{\pi }h \tilde{q}^{(1)}_{2m,i,s}-\dfrac{2}{m}u^{*}_{2m,i,s}+\dfrac{4}{\pi } hk^{**}_{2m,i,s}+\beta _{2m,i} \end{aligned}$$
(3.7)

where

$$\begin{aligned} \tilde{a}_{2m,i}&=\int ^\pi _0 \tilde{q}_i(t)\cos 2mt \,\mathrm{d}t,\\ \tilde{b}_{2m,i}&= \int ^\pi _0 \tilde{q}_i (t) \sin 2mt \,\mathrm{d}t,\\ u^*_{2m,i,s}&=\int ^{\pi -\frac{\tau _2}{2}}_{\frac{\tau _2}{2}}\bigg (\int ^t_{\frac{\tau _2}{2}}U_i (t_2) dt_2\bigg )\sin 2mt \,\mathrm{d}t, \\ u^*_{2m,i,c}&=\int ^{\pi -\frac{\tau _2}{2}}_{\frac{\tau _2}{2}}\bigg (\int ^t_{\frac{\tau _2}{2}}U_i (t_2) dt_2\bigg )\cos 2mt \,\mathrm{d}t,\\ k_{2m,i,c}^{**}&= \int ^{\pi -\tau _2}_{\tau _2} \bigg (\int ^t_{\tau _2}K_i^* (t_2)dt_2\bigg )\cos 2mt \,\mathrm{d}t,\\ k_{2m,i,s}^{**}&= \int ^{\pi -\tau _2}_{\tau _2} \bigg (\int ^t_{\tau _2}K_i^* (t_2)dt_2\bigg )\sin 2mt \,\mathrm{d}t. \end{aligned}$$

Further, we have

$$\begin{aligned} A_{0,i}=\dfrac{2}{\pi }\underset{m \rightarrow 0}{\lim } A_i (m)= \dfrac{2}{\pi }\tilde{a}_{0,i}-\dfrac{4}{\pi } h\tilde{q}^{(1)}_{0,i,c}-\dfrac{2}{\pi }u^*_{0,i,c}+\dfrac{4}{\pi }h k_{0,i,c}^{**}+\alpha _{0,i}. \end{aligned}$$
(3.8)

Since sequences \(\{\alpha _{2m,i}\}\), \(\{\beta _{2m,i}\}\), \(\{A_{2m,i}\}\), and \(\{B_{2m,i}\}\) belong to the space \(l_2\), by virtue of Riesz–Fischer theorem, there exist functions \(f_i\) and \(\varphi _i\) from \(L_2 [0,\pi ]\) such that

$$\begin{aligned} f_i (t)=\dfrac{A_{0,i}}{2} +\sum ^\infty _{m=1} A_{2m,i} \cos 2mt + B_{2m,i} \sin 2mt, \quad t \in [0,\pi ] \end{aligned}$$

and

$$\begin{aligned} \varphi _{i} (t)=\dfrac{\alpha _{0,i}}{2}+\sum ^\infty _{m=1} \alpha _{2m,i} \cos 2mt +\beta _{2m,i} \sin 2mt, \quad t \in [0,\pi ] \end{aligned}$$

hold. Now multiplying (3.8) with \(\frac{1}{2}\), (3.6) with \(\cos 2mt\), (3.7) with \(\sin 2mt\), and then summing up from \(m=1\) to \(m=\infty \), we get the system of integral equations

$$\begin{aligned} \tilde{q}_i (t)-2h \int ^t_{\frac{\tau _i}{2}} \tilde{q}_i (s)\,\mathrm{d}s-\int ^{t}_{\tau _2} U(s)\,\mathrm{d}s+2h \int ^{t}_{\tau _2} K^*(s)\,\mathrm{d}s+\varphi _{i}(t)=f_i (t). \end{aligned}$$
(3.9)

Substituting functions U and \(K^*\) into (3.9), we obtain

$$\begin{aligned}&\tilde{q}_i (t)-2h\int ^t_{\frac{\tau _2}{2}} \tilde{q}_i (s) \,\mathrm{d}s- \int ^{t}_{\tau _2} U^{(2)}(s) \,\mathrm{d}s-\int ^t_{\tau _1} U^{(1)}(s)\,\mathrm{d}s\nonumber \\&\quad -(-1)^{i} \int ^t_{\frac{\tau _1+\tau _2}{2}}\big (U^{(1,2)}(s)+U^{(2,1)}(s)\big )\,\mathrm{d}s\nonumber \\&\quad +2h\int ^{t}_{\tau _2} K^{(2)*}(s)\,\mathrm{d}s+2h\int ^t_{\tau _1}K^{(1)*}(s)\,\mathrm{d}s\nonumber \\&\quad +(-1)^{i} \int ^t_{\frac{\tau _1+\tau _2}{2}}\big (K^{(1,2)*}(s)+K^{(2,1)*}(s)\big )\,\mathrm{d}s\nonumber \\&\quad + \varphi _i (t)=f_i (t). \end{aligned}$$
(3.10)

From (3.4) and (3.5), we have

$$\begin{aligned} \varphi _i (t)=J^{(1)}_2 S^{(1)}(t)+J_2^{(2)}S^{(2)}(t)+(-1)^{i}\big (J_2^{(1,2)}+J_2^{(2,1)}\big )S^{(1,2)}(t) \end{aligned}$$

where

$$\begin{aligned} S^{(k)}&=\dfrac{\pi -2 \tau _k}{\pi }\big (h(\pi -2\tau _k)+1\big )-\dfrac{2\big (h(\pi -2\tau _k)+1\big )}{\pi } \sum ^{\infty }_{m=1}\dfrac{\sin 2m \tau _i}{m} \cos 2mt\\&-\dfrac{2h(\pi -2\tau _k)}{\pi }\sum ^\infty _{m=1} \dfrac{\cos 2m \tau _i}{m} \sin 2mt-\dfrac{2h}{\pi }\sum ^\infty _{m=1} \dfrac{\sin 2m \tau _i}{m^2} \sin 2mt \end{aligned}$$

and

$$\begin{aligned} S^{(1,2)}(t)&= \dfrac{\pi -\tau _2-\tau _1}{\pi }\big (h(\pi -\tau _2-\tau _1)+1\big )\\&-\dfrac{2\big (h(\pi -\tau _2-\tau _1)+1\big )}{\pi } \sum ^{\infty }_{m=1}\dfrac{\sin m(\tau _2+\tau _1)}{m} \cos 2mt\\&- \dfrac{2h(\pi -\tau _2-\tau _1)}{\pi } \sum ^\infty _{m=1} \dfrac{\cos m(\tau _2+\tau _1)}{m}\sin 2mt\\&-\dfrac{2h}{\pi }\sum _{m=1}^\infty \dfrac{\sin m(\tau _2+\tau _1)}{m^2} \sin 2mt. \end{aligned}$$

Further, we have

$$\begin{aligned}&\sum ^\infty _{m=1} \dfrac{\sin 2ma}{m}\cos 2mt= \left\{ \begin{array}{ll} -a, &{} \quad t\in (a,\pi -a),\\ \pi /2-a, &{} \quad t \in (0,a)\cup (\pi -a,\pi ),\\ \pi /4-a, &{} \quad t=a, \ t=\pi -a, \end{array} \right. \\&\sum ^\infty _{m=1} \dfrac{\cos 2ma}{m}\sin 2mt= \left\{ \begin{array}{ll} -t, &{} \quad t\in (0,a),\\ \pi /2-t, &{} \quad t \in (a, \pi -a),\\ \pi -t, &{} \quad t\in (\pi -a,\pi ),\\ \pi /4-a, &{} \quad t=a, \\ -\pi /4+a, &{}\quad t=\pi -a, \end{array} \right. \end{aligned}$$

and

$$\begin{aligned} \sum ^\infty _{m=1} \dfrac{\sin 2ma}{m^2}\sin 2mt= \left\{ \begin{array}{ll} (\pi -2a)t, &{} \quad t\in (0,a),\\ a(\pi -2t), &{} \quad t \in (a, \pi -a),\\ (\pi -2a)(t-\pi ), &{} \quad t\in (\pi -a,\pi ),\\ (\pi -2a)a, &{} \quad t=a, \\ -(\pi -2a)a, &{}\quad t=\pi -a. \end{array} \right. \end{aligned}$$

Then we get

$$\begin{aligned} S^{(k)}(t)= \left\{ \begin{array}{ll} 0, &{} \quad t\in (0,\tau _k)\cup (\pi -\tau _k,\pi ),\\ 1+2h (t-\tau _k), &{} \quad t \in (\tau _k, \pi -\tau _k),\\ 1/2, &{} \quad t=\tau _k,\\ 1/2+h(\pi -2\tau _k), &{} \quad t=\pi -\tau _k, \end{array} \right. \end{aligned}$$
(3.11)

and

$$\begin{aligned} S^{(1,2)}(t)= \left\{ \begin{array}{ll} 0, &{} \quad t\in \bigg (0,\dfrac{\tau _1+\tau _2}{2}\bigg )\cup \bigg (\pi -\dfrac{\tau _1+\tau _2}{2},\pi \bigg ),\\ 1+2h \big (t-(\tau _1+\tau _2)/2\big ), &{} \quad t \in \bigg (\dfrac{\tau _1+\tau _2}{2}, \pi -\dfrac{\tau _1+\tau _2}{2}\bigg ),\\ 1/2, &{} \quad t=(\tau _1+\tau _2)/2,\\ 1/2+h(\pi -\tau _1-\tau _2), &{} \quad t=\pi -(\tau _1+\tau _2)/2. \end{array} \right. \end{aligned}$$
(3.12)

Now, after summing and subtracting integral equations (3.10) and then introducing substitution of variables, we get the system of integral equations

$$\begin{aligned} q_1 (x)&-2h \int ^x_{\tau _1} q_1 (u)\,\mathrm{d}u-\int ^x_{\tau _2+\frac{\tau _1}{2}} U^{(2)}\big (u-\frac{\tau _1}{2}\big )\,\mathrm{d}u- \int ^x_{\frac{3 \tau _1}{2}} U^{(1)}\big (u-\frac{\tau _1}{2}\big )\,\mathrm{d}u\nonumber \\&+2h \int ^x_{\tau _2+\frac{\tau _1}{2}}K^{(2)*}\big (u-\frac{\tau _1}{2}\big )\,\mathrm{d}u+ 2h\int ^x_{\frac{3\tau _1}{2}}K^{(1)*}\big (u-\frac{\tau _1}{2}\big )\,\mathrm{d}u\nonumber \\&+J_2^{(1)}S^{(1)}\big (x-\frac{\tau _1}{2}\big )+J_2^{(2)}S^{(2)}\big (x-\frac{\tau _1}{2}\big )\nonumber \\&=\frac{1}{2} \big (f_0 \big (x-\frac{\tau _1}{2}\big )+f_1 \big (x-\frac{\tau _1}{2}\big )\big ) \end{aligned}$$
(3.13)

and

$$\begin{aligned} q_2(x)&-2h\int ^x_{\tau _2} q_2(u)\,\mathrm{d}u-\int ^x_{\tau _2+\frac{\tau _1}{2}}\bigg (U^{(1,2)}\big (u-\frac{\tau _2}{2}\big )+U^{(2,1)}\big (u-\frac{\tau _2}{2}\big )\bigg )\,\mathrm{d}u\nonumber \\&+ 2h \int ^x_{\tau _2+\frac{\tau _1}{2}} \bigg (K^{(1,2)*}\big (u-\frac{\tau _2}{2}\big )+K^{(2,1)*}\big (u-\frac{\tau _2}{2}\big )\bigg )\,\mathrm{d}u\nonumber \\&+ \big (J_2^{(1,2)}+J_2^{(2,1)}\big )S^{(1,2)}\big (x-\frac{\tau _2}{2}\big )\nonumber \\&= \frac{1}{2} \bigg (f_0 \big (x-\frac{\tau _2}{2}\big )-f_1 \big (x-\frac{\tau _2}{2}\big )\bigg ). \end{aligned}$$
(3.14)

Notice that each integral Eqs. (3.13) and (3.14) contains both potential functions \(q_1\) and \(q_2\). To prove uniqueness of the solutions of integral equations, we reduce integral Eqs. (3.13) and (3.14) to the linear integral equations of Volterra type. For that purpose, we will prove that all functions containing \(q_2\) in integral equation (3.13) are known, as well as all functions containing \(q_1\) in integral equation (3.14).

However, it turns out not to be true in general and depends on the relations between delays. In integral Eqs. (3.13) and (3.14) there are unknown integrals \(J_2^{(1)}\), \(J_2^{(2)}\), \(J_2^{(2,1)}\), \(J_2^{(1,2)}\). Although they are not known, we show that they do not exist in the integral equations on certain sub-intervals created at the beginning and at the end of \([\tau _k, \pi ]\). After solving integral equations on these intervals, depending on the case we consider, we will reveal that all integrals or some of them are known. Then, after solving the integral equations on certain intervals, we prove that the remaining integrals are known.

It is well known that the point \(\frac{2\pi }{5}\) is a point that separates linear and nonlinear cases of the boundary value problem. The boundary value problem becomes linear on the right side of this point and becomes non-linear on the left. Furthermore, the point \(\frac{2\pi }{5}\) is a point of separation in terms of the uniqueness of the solution of the boundary value problem. On the right side of \(\frac{2\pi }{5}\), the boundary value problem has a unique solution, while this is not the case on the left (see [19,20,21]).

This is the motivation behind the following process. First, we show that the integral equations (3.13) and (3.14) have unique solutions on certain sub-intervals of \([\tau _k, \pi ].\)

Theorem 1

Let \(q_k \in L_2 [\tau _k,\pi ]\), \(q_k (x)=0\) as \(x \in [0,\tau _k)\). If \(\frac{2}{5}\pi \le \tau _2< \tau _1 < \frac{\pi }{2}\), then integral equation (3.13) has a unique solution \(q_1\) on intervals \(\bigg (\tau _1, \tau _2 +\frac{\tau _1}{2}\bigg )\) and \(\bigg (\pi +\frac{\tau _1}{2}-\tau _2,\pi \bigg )\), and integral equation (3.14) has a unique solution \(q_2\) on intervals \(\bigg [\tau _2, \tau _2+\frac{\tau _1}{2}\bigg ]\) and \(\bigg (\pi -\frac{\tau _1}{2},\pi \bigg ]\).

Proof

From (3.11) and (3.12), we have

$$\begin{aligned}&S^{(1)}\bigg (x-\frac{\tau _1}{2}\bigg )=\left\{ \begin{array}{ll} 0, &{} t \in \big (\frac{\tau _1}{2}, \frac{3\tau _1}{2}\big )\cup \big (\pi -\frac{\tau _1}{2}, \pi +\frac{\tau _1}{2}\big ),\\ 1+2h\big (x-\frac{3\tau _1}{2}\big ), &{} t \in \big (\frac{3\tau _1}{2}, \pi -\frac{\tau _1}{2}\big ), \\ 1/2, &{} t=\frac{3\tau _1}{2},\\ 1/2+h\big (\pi -2\tau _1\big ), &{} t=\pi -\frac{3\tau _1}{2}, \end{array} \right. \\&S^{(2)}\bigg (x-\frac{\tau _1}{2}\bigg )=\left\{ \begin{array}{ll} 0, &{} t \in \big (\frac{\tau _1}{2}, \tau _2+\frac{\tau _1}{2}\big )\cup \big (\pi -\tau _2+\frac{\tau _1}{2}, \pi +\frac{\tau _1}{2}\big ),\\ 1+2h\big (x-\tau _2-\frac{\tau _1}{2}\big ), &{} t \in \big (\tau _2+\frac{\tau _1}{2}, \pi -\tau _2+\frac{\tau _1}{2}\big ),\\ 1/2, &{} t=\tau _2+\frac{\tau _1}{2},\\ 1/2+h(\pi -2\tau _2), &{} t=\pi -\tau _2+\frac{\tau _1}{2}, \end{array} \right. \end{aligned}$$

and

$$\begin{aligned}&S^{(1,2)}\bigg (x-\dfrac{\tau _2}{2}\bigg )=\left\{ \begin{array}{ll} 0, &{} x \in \big (\frac{\tau _2}{2}, \tau _2+\frac{\tau _1}{2}\big )\cup \big (\pi -\frac{\tau _1}{2}, \pi +\frac{\tau _2}{2}\big ),\\ 1+2h\big (x-\tau _2-\frac{ \tau _2}{2}\big ), &{} x \in \big (\tau _2+\frac{\tau _1}{2}, \pi -\frac{\tau _1}{2}\big ),\\ \frac{1}{2}, &{} x=\tau _2+\frac{\tau _1}{2},\\ \frac{1}{2}+h(\pi -\tau _1-\tau _2), &{} x=\pi -\frac{\tau _1}{2}. \end{array} \right. \end{aligned}$$

Now we consider the following cases:

Case 1. If \(x \in \big [\tau _2, \tau _2+\frac{\tau _1}{2}\big ]\) we have \(S^{(1,2)}\big (x-\frac{\tau _2}{2}\big )=0\) and from (3.14) we obviously get the Volterra linear integral equation

$$\begin{aligned} q_2(x)=\frac{1}{2}\big (f_0 \big (x-\frac{\tau _2}{2}\big )-f_1\big (x-\frac{\tau _2}{2}\big )\big )+2h \int ^x_{\tau _2} q_2 (u)\,\mathrm{d}u\end{aligned}$$

which has a unique solution on \(\big [\tau _2, \tau _2+\frac{\tau _1}{2}\big ]\).

Case 2. Let \(x \in \big (\pi -\frac{\tau _1}{2},\pi \big ]\) in (3.14). Then we both have \(S^{(1,2)}\big (x-\frac{\tau _2}{2}\big )=0\) and

$$\begin{aligned} \int ^x_{\tau _2+\frac{\tau _1}{2}} U^{(k,l)}\big (u-\frac{\tau _2}{2}\big )\,\mathrm{d}t=\int ^x_{\tau _2+\frac{\tau _1}{2}} K^{(k,l)*} \big (u-\frac{\tau _2}{2}\big )\,\mathrm{d}t, \quad x\in \big (\pi -\frac{\tau _1}{2},\pi \big ]. \end{aligned}$$

Since,

$$\begin{aligned} \int ^x_{\tau _2} q_2 (u)\,\mathrm{d}u= \int ^\pi _{\tau _2} q_2 (u)\,\mathrm{d}u-\int ^\pi _{x} q_2 (u)\,\mathrm{d}u=J_1^{(2)}-\int ^\pi _{x}q_2(u)\,\mathrm{d}u\end{aligned}$$

we obtain integral equation

$$\begin{aligned} q_2(x)=g_1 (x)-2h \int ^\pi _x q_2 (u)\,\mathrm{d}u\end{aligned}$$
(3.15)

where

$$\begin{aligned} g_1(x)=\frac{1}{2}\bigg (f_0\big (x-\frac{\tau _2}{2}\big )-f_1\big (x-\frac{\tau _2}{2}\big )\bigg )+2hJ_1^{(2)}. \end{aligned}$$

Integral equation (3.15) has a unique solution on \(\big (\pi -\frac{\tau _1}{2},\pi \big ]\).

Case 3. On the interval \(\big (\tau _1,\tau _2+\frac{\tau _1}{2}\big )\), we have \(S^{(1)}\big (x-\frac{\tau _1}{2}\big )=S^{(2)}\big (x-\frac{\tau _1}{2}\big )=0\). Then integral equation (3.13) becomes a linear integral equation of Volterra type with a kernel equal to one which has a unique solution \(q_1\) on \(\big (\tau _1, \tau _2+\frac{\tau _1}{2}\big )\).

Case 4. In the same way as in Case 2 using \(S^{(1)}\big (x-\frac{\tau _1}{2}\big )=S^{(2)}\big (x-\frac{\tau _1}{2}\big )=0\) on \(\big (\pi +\frac{\tau _1}{2}-\tau _2, \pi \big ]\), we obtain the unique solution \(q_1\) of integral equation (3.13) on interval \(\big (\pi +\frac{\tau _1}{2}-\tau _2,\pi \big ]\). The theorem is proved.

Now, let us show that the integrals \(J_2^{(1)}\), \(J_2^{(2)}\), \(J_2^{(2,1)}\), and \(J_2^{(1,2)}\) are known. The next lemma holds.

Lemma 4

Integrals \(J_2^{(1)}\), \(J_2^{(2)}\), \(J_2^{(2,1)}\) and \(J_2^{(1,2)}\) are determined by potentials \(q_2\) on \(\big (\tau _2, \tau _2+\frac{\tau _1}{2}\big )\cup \big (\pi -\frac{\tau _1}{2},\pi \big ]\) and \(q_1\) on \(\big (\tau _1,\tau _2+\frac{\tau _1}{2}\big )\cup \big (\pi +\frac{\tau _1}{2}-\tau _2,\pi \big ]\).

Proof

The arguments of \(J_2^{(2)}\) belong to the intervals \(\big (2\tau _2,\pi \big ) \subset \big (\pi -\frac{\tau _1}{2},\pi \big )\) and \(\big (\tau _2, \pi -\tau _2\big ) \subset \big (\tau _2,\tau +\frac{\tau _1}{2}\big )\), so \(J_2^{(2)}\) is known. In the same way, we get that arguments of \(J_2^{(1)}\) belong to the intervals \((2\tau _1, \pi ) \subset \big (\pi +\frac{\tau _1}{2}-\tau _2,\pi \big ]\) and \((\tau _1, \pi -\tau _1)\subset \big (\tau _1, \tau +\frac{\tau _1}{2}\big )\) and integral \(J_2^{(1)}\) is known, too. Argument of \(q_2\) in the integral \(J_2^{(2,1)}\) belongs to the interval \((\tau _1+\tau _2,\pi )\subset \big (\pi -\frac{\tau _1}{2},\pi \big ]\). Consequently, integral \(J_2^{(2,1)}\) is known. Argument of \(q_1\) in the integral \(J_2^{(1,2)}\) belongs to the interval \((\tau _1+\tau _2, \pi ) \subset \big (\pi +\frac{\tau _1}{2}-\tau _2,\pi \big ]\) and argument of \(q_1\) to the interval \((\tau _2,\pi -\tau _1) \subset \big (\tau _2,\tau _2+\frac{\tau _1}{2}\big )\) and then integral \(J_2^{(1,2)}\) is known. Then, in further considerations, we take that summand \(J^{(1)}_2 S^{(1)}\big (x-\frac{\tau _1}{2}\big )+J_2^{(2)}S^{(2)}\big (x-\frac{\tau _1}{2}\big )\) in integral equation (3.13) and summand \(\big (J^{(1,2)}_2+J_2^{(2,1)}\big ) S^{(1,2)}\big (x-\frac{\tau _2}{2}\big )\) in integral equation (3.14) are known.

Now we come to our main result and prove that Inverse problem 1 has a unique solution.

Theorem 2

Let \(q_k \in L_2 [\tau _k,\pi ]\), \(q_k (x)=0\) for \(x \in [0,\tau _k)\) and \(\frac{2\pi }{5}\le \tau _2 <\tau _1 \le \frac{\pi }{2}\). The four spectra of the boundary value problem \(D_{i,k}\) uniquely determine delays \(\tau _k\), parameters \(h,H_k\), and potential functions \(q_k\).

Proof

Taking Lemma 1, Lemma 2, and Theorem 1 into account, it remains to show that integral equations (3.13) and (3.14) have unique solutions \(q_1\) on the interval \(\big ( \tau _2 +\dfrac{\tau _1}{2}, \pi +\dfrac{\tau _1}{2}-\tau _2\big )\) and \(q_2\) on the interval \(\big (\tau _2+\frac{\tau _1}{2},\pi -\frac{\tau _1}{2}\big )\), respectively.

Case 1. Let \(x \in \big (\tau _2+\frac{\tau _1}{2},\frac{3 \tau _1}{2}\big ]\). From (3.13), we get the integral equation

$$\begin{aligned} q_1 (x)&-2h\int ^{\tau _2+\frac{\tau _1}{2}}_{\tau _1} q_1 (u)\,\mathrm{d}u-2h\int ^x_{\tau _2+\frac{\tau _1}{2}}q_1(u)\,\mathrm{d}u\\&-\int ^{x}_{\tau _2+\frac{\tau _1}{2}}U^{(2)}\big (u-\frac{\tau _1}{2}\big )\,\mathrm{d}u+2h\int ^x_{\tau _2+\frac{\tau _1}{2}}K^{(2)*} \big (u-\frac{\tau _1}{2}\big )\,\mathrm{d}u\\&=\frac{1}{2}\bigg (f_0 \big (x-\frac{\tau _1}{2}\big )+f_1\big (x-\frac{\tau _1}{2}\big )\bigg )-J_2^{(1)}S^{(1)} \big (x-\frac{\tau _1}{2}\big )-J_2^{(2)}S^{(2)}\big (x-\frac{\tau _1}{2}\big ). \end{aligned}$$

One can easily show that arguments of the potential \(q_2\) appearing in the function

$$\begin{aligned} \int ^x_{\tau _2+\frac{\tau _1}{2}}U^{(2)}\big (u-\frac{\tau _1}{2}\big )\,\mathrm{d}u\end{aligned}$$

belong to the intervals \([2\tau _2,\tau _1+\tau _2] \subset \big [\pi -\frac{\tau _1}{2},\pi \big ]\), \([\tau _2, \tau _1] \subset \big [\tau _2, \tau _2+\frac{\tau _1}{2}\big ]\), \([2\tau _2, \pi ] \subset \big [\pi -\frac{\pi }{2},\pi \big ]\), and \([\tau _2, \pi -\tau _2] \subset \big [\tau _2, \tau _2+\frac{\tau _1}{2}\big ]\). We have the same for the function \(\int ^x_{\tau _2+\frac{\tau _1}{2}}K^{(2)*}\big (u-\frac{\tau _1}{2}\big )\,\mathrm{d}u.\) Then we get the integral equation

$$\begin{aligned} q_1(x)=g_2 (x)+2h\int ^x_{\tau _2 +\frac{\tau _1}{2}} q_1(u)\,\mathrm{d}u\end{aligned}$$
(3.16)

where

$$\begin{aligned} g_2 (x)&=\frac{1}{2} \bigg (f_0 \big (x-\frac{\tau _1}{2}\big )+f_1\big (x-\frac{\tau _1}{2}\big )\bigg )+\int ^x_{\tau _2+\frac{\tau _1}{2}}U^{(2)}\big (u-\frac{\tau _2}{2}\big )\,\mathrm{d}u\\&-2h \int ^x_{\tau _2+\frac{\tau _1}{2}}K^{(2)*}\big (u-\frac{\tau _2}{2}\big )\,\mathrm{d}u+2h \int ^{\tau _2+\frac{\tau _1}{2}}_{\tau _1} q_1 (u)\,\mathrm{d}u\\&-J_2^{(1)}S^{(1)}\big (x-\frac{\tau _1}{2}\big )+J_2^{(2)}S^{(2)}\big (x-\frac{\tau _1}{2}\big ) \end{aligned}$$

is a known function. Integral equation (3.16) has a unique solution \(q_1\) on \(\big (\tau _2+\frac{\tau _1}{2}, \frac{3\tau _1}{2}\big ]\).

Case 2. For \(x \in \big (\frac{3 \tau _1}{2},\pi -\frac{\tau _1}{2}\big ]\), from (3.14) we get

$$\begin{aligned} q_1(x)&-2h\int ^{\frac{3\tau _1}{2}}_{\tau _1} q_1 (u)\,\mathrm{d}u-2h \int ^x_{\frac{3\tau _1}{2}}q_1 (u)\,\mathrm{d}u\\&-\int ^{\frac{3\tau _1}{2}}_{\tau _2+\frac{\tau _1}{2}}U^{(2)}\big (u-\frac{\tau _1}{2}\big )\,\mathrm{d}u-\int ^x_{\frac{3\tau _1}{2}}U^{(2)}\big (u-\frac{\tau _1}{2}\big )\,\mathrm{d}u\\&-\int ^x_{\frac{3\tau _1}{2}}U^{(1)}\big (u-\frac{\tau _1}{2}\big )+2h \int ^{\frac{3\tau _1}{2}}_{\tau _2+\frac{\tau _1}{2}}K^{(2)*}\big (u-\frac{\tau _1}{2}\big )\,\mathrm{d}u\\&+2h\int ^x_{\frac{3\tau _1}{2}}K^{(2)*}\big (u-\frac{\tau _1}{2}\big )\,\mathrm{d}u+2h\int ^x_{\frac{3\tau _1}{2}}K^{(1)*}\big (u-\frac{\tau _1}{2}\big )\,\mathrm{d}u\\&=\frac{1}{2}\bigg (f_0\big (x-\frac{\tau _1}{2}\big )+f_1\big (x-\frac{\tau _1}{2}\big )\bigg )-J_2^{(1)}S^{(1)}\big (x-\frac{\tau _1}{2}\big )\\&-J_2^{(2)}S^{(2)}\big (x-\frac{\tau _1}{2}\big ). \end{aligned}$$

In the same way as in Case 1 one can show that functions

$$\begin{aligned} \int ^x_{\frac{3\tau _1}{2}}U^{(2)}\big (u-\frac{\tau _1}{2}\big )\,\mathrm{d}u\quad \texttt {and} \quad \int ^x_{\frac{3\tau _1}{2}}K^{(2)*}\big (u-\frac{\tau _1}{2}\big )\,\mathrm{d}u\end{aligned}$$

are known as well as the functions

$$\begin{aligned} \int ^x_{\frac{3\tau _1}{2}}U^{(1)}\big (u-\frac{\tau _1}{2}\big )\,\mathrm{d}u\quad \texttt {and} \quad \int ^{x}_{\frac{3\tau _1}{2}}K^{(1)*} \big (u-\frac{\tau _1}{2}\big )\,\mathrm{d}u. \end{aligned}$$

Then, we get the integral equation

$$\begin{aligned} q_1 (x)=g_3 (x)+2h \int ^x_{\frac{3\tau _1}{2}}q_1 (u)\,\mathrm{d}u\end{aligned}$$
(3.17)

where

$$\begin{aligned} g_3 (x)&=\frac{1}{2}\bigg (f_0\big (x-\frac{\tau _1}{2}\big )+f_1\big (x-\frac{\tau _1}{2}\big )\bigg )+\int ^x_{\tau _2+\frac{\tau _1}{2}}U^{(2)}\big (u-\frac{\tau _2}{2}\big )\,\mathrm{d}u\\&-2h\int ^{x}_{\tau _2+\frac{\tau _1}{2}}K^{(2)*}\big (u-\frac{\tau _2}{2}\big )\,\mathrm{d}u+2h\int ^{\frac{3\tau _1}{2}}_{\tau _1}q_1 (u)\,\mathrm{d}u\\&-J_2^{(1)}S^{(1)}\big (x-\frac{\tau _1}{2}\big )-J_2^{(2)}S^{(2)}\big (x-\frac{\tau _1}{2}\big ) \end{aligned}$$

is a known function. Integral equation (3.17) has a unique solution \(q_1\) on \(\big (\frac{3\tau _1}{2},\pi -\frac{\tau _1}{2}\big ]\).

Case 3. In the same way we show that integral equation (3.13) has a unique solution on \(\big (\pi -\frac{\tau _1}{2},\pi +\frac{\tau _1}{2}-\tau _2\big ]\).

Case 4. Finally, we prove that the integral equation (3.14) has a unique solution \(q_2\) on \(\big (\tau _2+\frac{\tau _1}{2}, \pi -\frac{\tau _1}{2}\big ]\). Notice that the potential \(q_1\) is known. From (3.14) we get

$$\begin{aligned} q_2 (x)&-2h \int ^{\tau _2+\frac{\tau _1}{2}}_{\tau _2} q_1 (u) \,\mathrm{d}u-2h \int ^x_{\tau _2+\frac{\tau _1}{2}}q_1 (u)\,\mathrm{d}u\\&-\int ^x_{\tau _1+\frac{\tau _1}{2}}\bigg (U^{(1,2)}\big (u-\frac{\tau _2}{2}\big )+U^{(2,1)}\big (u-\frac{\tau _2}{2}\big )\bigg )\,\mathrm{d}u\\&+2h \int ^x_{\tau _2+\frac{\tau _1}{2}}\bigg (K^{(1,2)*}\big (u-\frac{\tau _2}{2}\big )+K^{(2,1)*}\big (u-\frac{\tau _2}{2}\big )\bigg )\,\mathrm{d}u\\&=\frac{1}{2}\bigg (f_0 \big (x-\frac{\tau _2}{2}\big )-f_1 \big (x-\frac{\tau _2}{2}\big )\bigg )\\&-\big (J_2^{(1,2)}+J_2^{(2,1)}\big )S^{(1,2)}\big (x-\frac{\tau _2}{2}\big ). \end{aligned}$$

It can be easily shown that the arguments of the potential \(q_2\) in the function

$$\begin{aligned} \int ^x_{\tau _2+\frac{\tau _1}{2}}U^{(2,1)}\big (u-\frac{\tau _2}{2}\big ) \end{aligned}$$

belong to the interval \([\tau _2,\pi -\tau _1]\subset \big [\tau _2, \tau _2+\frac{\tau _1}{2}\big ]\) and arguments of the potential \(q_2\) in the function

$$\begin{aligned} \int ^x_{\tau _2+\frac{\tau _1}{2}} U^{(1,2)}\big (u-\frac{\tau _2}{2}\big )\,\mathrm{d}u\end{aligned}$$

belong to the interval \([\tau _1+\tau _2,\pi ]\subset \bigg [\pi -\frac{\tau _1}{2},\pi \bigg ].\) So, we get the integral equation

$$\begin{aligned} q_2 (x)=g_4 (x)x+2h\int ^x_{\tau _2+\frac{\tau _1}{2}} q_2 (u)\,\mathrm{d}u\end{aligned}$$
(3.18)

where

$$\begin{aligned} g_4(x)&=\frac{1}{2}\bigg (f_0\big (x-\frac{\tau _2}{2}\big )-f_1 \big (x-\frac{\tau _2}{2}\big )\bigg )+2h\int ^{\tau _2+\frac{\tau _1}{2}}_{\tau _2} q_1 (u)\,\mathrm{d}u\\&+\int ^x_{\tau _2+\frac{\tau _1}{2}}\bigg (U^{(1,2)}\big (u-\frac{\tau _2}{2}\big )+U^{(2,1)}\big (u-\frac{\tau _2}{2}\big )\bigg )\,\mathrm{d}u\\&-\big (J_2^{(1,2)}+J_2^{(2,1)}\big )S^{(1,2)}\big (x-\frac{\tau _2}{2}\big )\\&-2h\int ^x_{\tau _2+\frac{\tau _1}{2}}\bigg (K^{(1,2)*}\big (u-\frac{\tau _2}{2}\big )+K^{(2,1)*}\big (u-\frac{\tau _2}{2}\big )\bigg )\,\mathrm{d}u\end{aligned}$$

is a known function. Integral equation (3.18) has a unique solution on \(\big (\tau _2+\frac{\tau _1}{2},\pi -\frac{\tau _1}{2}\big ]\). Thus, the theorem is proved.