1 Introduction

With the emergence of various applications in natural sciences, there appeared a growing interest in inverse problems for differential operator with constant delay. Regarding the inverse problems of these delay differential operator models, the main focus is on how to reconstruct the operator from given spectral data. Specifically, for \(i=0,1\), denote by \(\{\lambda _{n,i}\}^{\infty }_{n=1}\) the spectrum of the boundary value problem of the form

$$\begin{aligned} -y''(x)+&q(x)y(x-\tau )=\lambda y(x),\ \ x\in (0,\pi ), \end{aligned}$$
(1.1)
$$\begin{aligned}&y(0)=y^{(i)}(\pi )=0,\ i=0, 1 \end{aligned}$$
(1.2)

with the delay \(\tau \in (0,\pi )\) and complex-value potential \(q(x)\in L_{2}(\tau ,\pi )\). It is significant in some specific mathematical models, such as, long-term economic forecasting and automatic control theory. These mathematical models with the delay \(\tau >0\) can describe the real processes more comprehensively than the case of \(\tau =0\). We refer the reader to monographs [1,2,3] and the references therein.

For different delay value \(\tau\), the inverse problems of delay differential operator can be divided into linear inverse problems and nonlinear inverse problems. In details, for \(\tau \geqslant \pi /2\), the linear inverse problems mean that the characteristic functions depend linearly on the potential and the various aspects of linear inverse problem (1.1)–(1.2) were widely studied in [4,5,6] and other works.

If \(0<\tau <\pi /2\), then the inverse problem (1.1)–(1.2) is nonlinear. As well for sufficiently small constant delay \(\tau \rightarrow 0\), the inverse problem (1.1)–(1.2) is closer to the classical Sturm–Liouville inverse problem [7] and many scholars expect to obtain unique solvability and non-unique solvability of this kind of problems, see [8,9,10]. Therefore, it is much more complicated to give unique results for this kind of problems to some extent than linear ones. Recently, an affirmative answer has been given in [11] to the solvability of inverse problem (1.1)–(1.2) in nonlinear case. It is worth noting whether it is linear case or nonlinear case, the classical methods (transformation operator method, method of spectral mappings) of the inverse spectral problems do not give reliable results for various classes of nonlocal operators (see [12, 13]).

While there are a number of results about both direct and inverse problems for various difference or differential operators, see [14,15,16,17,18,19,20,21,22], there are just a few results related to differential operators with two or more delays, see [23, 24]. Inspired by the above mentioned literatures, the purpose of this paper is to study the nonlinear inverse problem for Sturm–Liouville operator with multiple delays. We set out to investigate the boundary value problems \(S{-}L_{i}(q_{j}), (i=0,1, j=1,2)\) of the form

$$\begin{aligned} -y''(x)+q_{1}(x)y(x&-\tau _{1})+q_{2}(x)y(x-\tau _{2})=\lambda y(x),\ \ x\in (0,\pi ), \end{aligned}$$
(1.3)
$$\begin{aligned}&y(0)=y^{(i)}(\pi )=0,\ i=0,1, \end{aligned}$$
(1.4)

where \(\lambda\) is the spectral parameter, \(2\pi /5\leqslant \tau _{1}<\tau _{2}<\pi /2\), \(q_{j}(x)\) (\(j=1,2\)) are complex-valued functions such that \(q_{j}(x)\in L_{2}[\tau _{j}, \pi ]\) and \(q_{j}(x)\equiv 0, x\in [0,\tau _{j})\). More precisely, the main goal is to recover Sturm–Liouville operator from given spectra. Let \(\{\lambda _{ni}\}^{\infty }_{n=1},i=0,1\) be the eigenvalues of \(S{-}L_{i}(q_{j}), (i=0,1, j=1,2)\) and the inverse problem is formulated as follows:

Inverse problem Given the spectra \(\{\lambda _{n0}\}^{\infty }_{n=1}\) and \(\{\lambda _{n1}\}^{\infty }_{n=1}\), determine the potential \(q_{j}(x)\), \(j=1,2\).

We try to develop a global program for the inverse problem of Sturm–Liouville operator with multiple delays in nonlinear case and establish a uniqueness theorem. During the whole process of algorithm construction, in order to overcome the obstacles caused by multiple delays and complex integral terms, we propose a new technique and method (see Algorithm 3.2, Step 7): zeroing the potential terms in the proper integrand interval, so as to remove some undesirable nonlinear integral terms. Moreover, we generalize the result of [25] and develop the concept of [11] for inverse Srurm–Liouville problem with constant delay. For this purpose, we study the relationship between given spectra and characteristic function of \(S{-}L_{i}(q_{j}),(i=0,1,j=1,2)\) and present some auxiliary assertions about eigenvalues and characteristic functions in Sect. 2. Compared with the results of [23] and [24], we obtain the uniqueness theorem and reconstruction algorithm for \(S{-}L_{i}(q_{j}), (i=0,1, j=1,2)\) in nonlinear case for a smaller range of delay \(\tau _{j},j=1,2\). The main result Theorem 3.1 and constructive Algorithm 3.2 are given in Sect. 3.

2 Preliminary

Let \(S(x,\lambda )\) be the solution of equation (1.1) satisfying the initial conditions

$$\begin{aligned} \begin{aligned} S(0,\lambda )=0,\ S'(0,\lambda )=1. \end{aligned} \end{aligned}$$

Then the function \(S(x,\lambda )\) satisfies the integral equation

$$\begin{aligned} \begin{aligned} S(x,\lambda )=\frac{\sin \sqrt{\lambda } x}{\sqrt{\lambda }} +\frac{1}{\sqrt{\lambda }}\sum ^{2}_{k=1}\int ^{x}_{\tau _{k}}q_{k}(t)\sin \sqrt{\lambda }(x-t)S(t-\tau _{k},\lambda )dt. \end{aligned} \end{aligned}$$
(2.1)

We solve the integral equation (2.1) by the method of successive approximations (see [25]) and following lemma holds.

Lemma 2.1

The Volterra-type integral equation (2.1) has a unique solution

$$\begin{aligned} \begin{aligned} S(x,\lambda )=S_{0}(x,\lambda )+S_{1}(x,\lambda )+S_{2}(x,\lambda ),\ \ x\geqslant 2\tau _{2}, \end{aligned} \end{aligned}$$
(2.2)

where

$$\begin{aligned} \begin{aligned}&S_{0}(x,\lambda )=\frac{\sin \sqrt{\lambda }x}{\sqrt{\lambda }},\\&S_{1}(x,\lambda )=\frac{1}{\lambda }\sum ^{2}_{k=1}\int ^{x}_{\tau _{k}}q_{k}(t) \sin \sqrt{\lambda }(x-t)\sin \sqrt{\lambda }(t-\tau _{k})dt,\\&S_{2}(x,\lambda )=\frac{1}{\sqrt{\lambda }}\sum ^{2}_{k=1}\int ^{x}_{2\tau _{k}}q_{k}(t) \sin \sqrt{\lambda }(x-t)S_{1}(t-\tau _{k},\lambda )dt. \end{aligned} \end{aligned}$$

Denote the \(\Delta _{i}(\lambda ):=S^{(i)}(\pi ,\lambda ), i=0,1\). Then from (2.2) we have

$$\begin{aligned} \begin{aligned} \Delta _{0}(\lambda )&=\frac{\sin \sqrt{\lambda }\pi }{\sqrt{\lambda }} +\frac{1}{\lambda }\sum ^{2}_{k=1}\int ^{\pi }_{\tau _{k}}q_{k}(t)\sin \sqrt{\lambda }(\pi -t)\sin \sqrt{\lambda }(t-\tau _{k})dt+S_{2}(\pi ,\lambda ),\\ \Delta _{1}(\lambda )&=\cos \sqrt{\lambda }\pi +\frac{1}{\sqrt{\lambda }} \sum ^{2}_{k=1}\int ^{\pi }_{\tau _{k}}q_{k}(t)\cos \sqrt{\lambda }(\pi -t) \sin \sqrt{\lambda }(t-\tau _{k})dt+S'_{2}(\pi ,\lambda ). \end{aligned} \end{aligned}$$
(2.3)

Lemma 2.2

The eigenvalues \(\{\lambda _{ni}\}^{\infty }_{n=1}\) of the boundary value problems \(S{-}L_{i}(q_{j}),(i=0,1,j=1,2)\) coincide with the zeros of characteristic function \(\Delta _{i}(\lambda ), i=0,1\).

Proof

From Lemma 2.1, it can be known that \(S(x,\lambda )\) is a solution of initial value problem. Let \(\lambda\) be an eigenvalue of boundary value problem (1.1)–(1.2), and \(S(x,\lambda )\) is the corresponding eigenfunction satisfying the boundary condition (1.2). In view of classical Sturm–Liouville theory, the eigenvalues of the boundary value problem \(S{-}L_{i}(q_{j}),(i=0,1,j=1,2)\) coincide with characteristic function \(\Delta _{i}(\lambda ), (i=0,1)\) and it is entire function of \(\lambda\). \(\square\)

Note that the characteristic functions \(\Delta _{i}(\lambda ), (i=0,1)\) depend nonlinearly on the potential \(q_{j}(x), (j=1,2)\). By the method mentioned in reference [26] (Ch.1), we obtain the asymptotical formulae for the eigenvalues \(\{\lambda _{ni}\}^{\infty }_{n=1}, (i=0,1)\) for \(n\rightarrow \infty\) given by following lemma:

Lemma 2.3

The eigenvalues \(\{\lambda _{ni}\}^{\infty }_{n=1}\), (\(i=0,1\)) of the boundary value problem \(S{-}L_{i}(q_{j}),(i=0,1,j=1,2)\) satisfy asymptotic formula:

$$\begin{aligned} \sqrt{\lambda _{n0}}&=n+\sum ^{2}_{k=1}\frac{\int ^{\pi }_{\tau _{k}}q_{k}(t)dt\cos n \tau _{k}}{2\pi n}+o\Big (\frac{1}{n}\Big ), \end{aligned}$$
(2.4)
$$\begin{aligned} \sqrt{\lambda _{n1}}&=\Big (n-\frac{1}{2}\Big )+\sum ^{2}_{k=1}\frac{\int ^{\pi }_{\tau _{k}}q_{k}(t)dt\cos (n-\frac{1}{2})\tau _{k}}{2\pi n}+o\Big (\frac{1}{n}\Big ). \end{aligned}$$
(2.5)

In view of Hadamard’s factorization theorem (see [1]), characteristic function is uniquely determined up to a multiplicative constant by its zeros. Here, the following lemma holds.

Lemma 2.4

(see [25]) The specification of the spectrum \(\{\lambda _{ni}\}^{\infty }_{n=1}\), (\(i=0,1\)) uniquely determines the characteristic function \(\Delta _{i}(\lambda )\) by the formulas

$$\begin{aligned} \begin{aligned} \Delta _{0}(\lambda )=\pi \prod ^{\infty }_{n=1}\frac{\lambda _{n0}-\lambda }{n^{2}},\ \Delta _{1}(\lambda )=\prod ^{\infty }_{n=1}\frac{\lambda _{n1}-\lambda }{(n-1/2)^{2}}. \end{aligned} \end{aligned}$$
(2.6)

Now, from the spectra of \(S{-}L_{i}(q_{j}), (i=0,1, j=1,2)\), combined with Lemmas 2.1–2.4, we construct the connection between the potential \(q_{j}(x), j=1,2\) and the characteristic functions \(\Delta _{i}(\lambda ), i=0,1\). Firstly, we denote

$$\begin{aligned} {\tilde{\Delta }}_{0}(\lambda ):= & {} \lambda \left( \Delta _{0}(\lambda )-\frac{\sin \sqrt{\lambda }\pi }{\sqrt{\lambda }}\right) ,\nonumber \\ {\tilde{\Delta }}_{1}(\lambda ):= & {} \sqrt{\lambda }(\Delta _{1}(\lambda )-\cos \sqrt{\lambda }\pi ). \end{aligned}$$
(2.7)

Then

$$\begin{aligned} \begin{aligned} {\tilde{\Delta }}_{0}(\lambda )&=\sum ^{2}_{k=1}\int ^{\pi }_{\tau _{k}}q_{k}(t)\sin \sqrt{\lambda }(\pi -t)\sin \sqrt{\lambda }(t-\tau _{k})dt+\lambda S_{2}(\pi ,\lambda ), \\ {\tilde{\Delta }}_{1}(\lambda )&=\sum ^{2}_{k=1}\int ^{\pi }_{\tau _{k}}q_{k}(t)\cos \sqrt{\lambda }(\pi -t)\sin \sqrt{\lambda }(t-\tau _{k})dt+\sqrt{\lambda } S'_{2}(\pi ,\lambda ). \end{aligned} \end{aligned}$$
(2.8)

Next, in view of Lemmas 2.1–2.4 and relation (2.7)–(2.8), we need to separate the constant term from the potential function integral term. In order to achieve this goal, we calculate

$$\begin{aligned} \begin{aligned}&\lambda S_{2}(\pi ,\lambda )\\&\quad =A_{1}\frac{\sin \sqrt{\lambda }(\pi -2\tau _{1})}{4}+A_{2}\frac{\sin \sqrt{\lambda }(\pi -\tau _{1}-\tau _{2})}{4}\\&\qquad+A_{3}\frac{\sin \sqrt{\lambda }(\pi -\tau _{1}-\tau _{2})}{4}+A_{4}\frac{\sin \sqrt{\lambda }(\pi -2\tau _{2})}{4}\\&\qquad +\int ^{\pi -2\tau _{1}}_{2\tau _{1}-\pi }(B_{1}(x_{1})+C_{1}(y_{1})+D_{1}(z_{1}))\frac{\sin \sqrt{\lambda }\xi }{8}d\xi \\&\qquad +\int ^{\pi -2\tau _{2}}_{2\tau _{2}-\pi }(B_{4}(x_{4})+C_{4}(y_{4})+D_{4}(z_{4}))\frac{\sin \sqrt{\lambda }\xi }{8}d\xi \\&\qquad +\int _{3\tau _{1}-\tau _{2}-\pi }^{\pi -\tau _{1}-\tau _{2}}(B_{2}(x_{2})+C_{22}(y_{2}))\frac{\sin \sqrt{\lambda }\xi }{8}d\xi \\&\qquad +\int _{3\tau _{2}-\tau _{1}-\pi }^{\pi -\tau _{1}-\tau _{2}}(B_{3}(x_{3})+C_{32}(y_{3}))\frac{\sin \sqrt{\lambda }\xi }{8}d\xi \\&\qquad +\int _{\tau _{1}+\tau _{2}-\pi }^{3\tau _{1}-\tau _{2}-\pi }(C_{21}(y_{2}))\frac{\sin \sqrt{\lambda }\xi }{8}d\xi +\int _{\tau _{1}+\tau _{2}-\pi }^{3\tau _{2}-\tau _{1}-\pi }(C_{31}(y_{3}))\frac{\sin \sqrt{\lambda }\xi }{8}d\xi \\&\qquad +\int ^{\pi -\tau _{1}-\tau _{2}}_{\pi -3\tau _{1}+\tau _{2}}D_{21}(z_{2})\frac{\sin \sqrt{\lambda }\xi }{8}d\xi +\int ^{\pi -\tau _{1}-\tau _{2}}_{\tau _{1}+\tau _{2}-\pi }D_{22}(z_{2})\frac{\sin \sqrt{\lambda }\xi }{8}d\xi \\&\qquad +\int ^{\pi -\tau _{1}-\tau _{2}}_{\pi -3\tau _{2}+\tau _{1}}D_{31}(z_{3})\frac{\sin \sqrt{\lambda }\xi }{8}d\xi +\int ^{\pi -3\tau _{2}+\tau _{1}}_{\tau _{1}+\tau _{2}-\pi }D_{32}(z_{3})\frac{\sin \sqrt{\lambda }\xi }{8}d\xi ,\\ \end{aligned} \end{aligned}$$
(2.9)
$$\begin{aligned}&\sqrt{\lambda }S'_{2}(\pi ,\lambda )\nonumber \\&\quad =A_{1}\frac{\cos \sqrt{\lambda }(\pi -2\tau _{1})}{4}+A_{2}\frac{\cos \sqrt{\lambda }(\pi -\tau _{1}-\tau _{2})}{4}\nonumber \\&\qquad +A_{3}\frac{\cos \sqrt{\lambda }(\pi -\tau _{1}-\tau _{2})}{4}+A_{4}\frac{\cos \sqrt{\lambda }(\pi -2\tau _{2})}{4}\nonumber \\&\qquad -\int ^{\pi -2\tau _{1}}_{2\tau _{1}-\pi }(B_{1}(x_{1})+C_{1}(y_{1})-D_{1}(z_{1}))\frac{\cos \sqrt{\lambda }\xi }{8}d\xi \nonumber \\&\qquad -\int ^{\pi -2\tau _{2}}_{2\tau _{2}-\pi }(B_{4}(x_{4})+C_{4}(y_{4})-D_{4}(z_{4}))\frac{\cos \sqrt{\lambda }\xi }{8}d\xi \nonumber \\&\qquad +\int _{3\tau _{1}-\tau _{2}-\pi }^{\pi -\tau _{1}-\tau _{2}}(-B_{2}(x_{2})-C_{22}(y_{2}))\frac{\cos \sqrt{\lambda }\xi }{8}d\xi \nonumber \\&\qquad +\int _{3\tau _{2}-\tau _{1}-\pi }^{\pi -\tau _{1}-\tau _{2}}(-B_{3}(x_{3})-C_{32}(y_{3}))\frac{\cos \sqrt{\lambda }\xi }{8}d\xi \nonumber \\&\qquad +\int _{\tau _{1}+\tau _{2}-\pi }^{3\tau _{1}-\tau _{2}-\pi }(-C_{21}(y_{2}))\frac{\cos \sqrt{\lambda }\xi }{8}d\xi +\int _{\tau _{1}+\tau _{2}-\pi }^{3\tau _{2}-\tau _{1}-\pi }(-C_{31}(y_{3}))\frac{\cos \sqrt{\lambda }\xi }{8}d\xi \nonumber \\&\qquad +\int ^{\pi -\tau _{1}-\tau _{2}}_{\pi -3\tau _{1}+\tau _{2}}D_{21}(z_{2})\frac{\cos \sqrt{\lambda }\xi }{8}d\xi +\int ^{\pi -\tau _{1}-\tau _{2}}_{\tau _{1}+\tau _{2}-\pi }D_{22}(z_{2})\frac{\cos \sqrt{\lambda }\xi }{8}d\xi \nonumber \\&\qquad +\int ^{\pi -\tau _{1}-\tau _{2}}_{\pi -3\tau _{2}+\tau _{1}}D_{31}(z_{3})\frac{\cos \sqrt{\lambda }\xi }{8}d\xi +\int ^{\pi -3\tau _{2}+\tau _{1}}_{\tau _{1}+\tau _{2}-\pi }D_{32}(z_{3})\frac{\cos \sqrt{\lambda }\xi }{8}d\xi . \end{aligned}$$
(2.10)

At this point, we have completed the separation of integral terms of potential. In (2.9) and (2.10), for detailed data on the integrand, see the Appendix.

We assume that \(q_{j}(x)\in AC[\tau _{j},\pi ], j=1,2\). Generally, some minor technical modifications are needed. Denote \(p_{1}(x)=q'_{1}(x)\) and \(p_{2}(x)=q'_{2}(x)\). Taking (2.8)–(2.10) into account and applying integration by part, we arrive at

$$\begin{aligned}&4\sqrt{\lambda }{\tilde{\Delta }}_{0}(\lambda )\nonumber \\&\quad =\sqrt{\lambda }\Bigg (\int _{\tau _{1}+\tau _{2}-\pi }^{3\tau _{1}-\tau _{2}-\pi }(C_{21}(y_{2}))\sin \sqrt{\lambda }\xi d\xi +\int ^{\pi -\tau _{1}-\tau _{2}}_{\pi -3\tau _{1}+\tau _{2}}\nonumber D_{21}(z_{2})\sin \sqrt{\lambda }\xi d\xi +\int ^{\pi -\tau _{1}-\tau _{2}}_{\tau _{1}+\tau _{2}-\pi }D_{22}(z_{2})\sin \sqrt{\lambda }\xi d\xi \Bigg )\nonumber \\&\qquad +\sqrt{\lambda }\Bigg (\int ^{\pi -2\tau _{1}}_{2\tau _{1}-\pi }(B_{1}(x_{1})+C_{1}(y_{1})+D_{1}(z_{1}))\sin \sqrt{\lambda }\xi d\xi \nonumber+\int _{3\tau _{1}-\tau _{2}-\pi }^{\pi -\tau _{1}-\tau _{2}}(B_{2}(x_{2})+C_{22}(y_{2}))\sin \sqrt{\lambda }\xi d\xi \Bigg )\nonumber \\&\qquad -4\sqrt{\lambda }L_{1}\cos \sqrt{\lambda }(\pi -\tau _{1})+N_{1}\sin \sqrt{\lambda }(\pi -\tau _{1})-\int _{\tau _{1}-\pi }^{\pi -\tau _{1}}p_{01}(\xi )\sin \sqrt{\lambda }\xi d\xi \nonumber \\&\qquad +2\sqrt{\lambda }\Big (A_{1}\sin \sqrt{\lambda }(\pi -2\tau _{1})+A_{2}\sin \sqrt{\lambda }(\pi -\tau _{1}-\tau _{2})\Big ), \end{aligned}$$
(2.11)

where \(N_{1}=2(q_{1}(\pi )+q_{1}(\tau _{1}))\), \(p_{01}(\xi )=p_{1}(\frac{\pi +\tau _{1}-\xi }{2})\), \(L_{1}=\int ^{\pi }_{\tau _{1}}q_{1}(t)dt\).

In relation (2.11), we do zero function extension for partial integrand, that is, it is always zero outside its corresponding interval (for example \(D_{21}(z_2)\equiv 0\) outside the interval \((\pi -\tau _{1}-\tau _{2},\pi -3\tau _{1}+\tau _{2})\)). Therefore, we have

$$\begin{aligned} 4\sqrt{\lambda }{\tilde{\Delta }}_{0}(\lambda )&=\sqrt{\lambda }\int _{3\tau _{1}-\tau _{2}-\pi }^{\pi -3\tau _{1}+\tau _{2}}\Big (-C_{21}(y_{2})-D_{21}(z_{2})+D_{22}(z_{2})+B_{2}(x_{2})+C_{22}(y_{2})\Big )\sin \sqrt{\lambda }\xi d\xi \\&\quad -4\sqrt{\lambda }L_{1}\cos \sqrt{\lambda }(\pi -\tau _{1})+N_{1}\sin \sqrt{\lambda }(\pi -\tau _{1})-\int _{\tau _{1}-\pi }^{\pi -\tau _{1}}p_{01}(\xi )\sin \sqrt{\lambda }\xi d\xi \\&\quad +\sqrt{\lambda }\int ^{\pi -2\tau _{1}}_{2\tau _{1}-\pi }\Big (B_{1}(x_{1})+C_{1}(y_{1})+D_{1}(z_{1})\Big )\sin \sqrt{\lambda }\xi d\xi \\&\quad +2\sqrt{\lambda }\Big (A_{1}\sin \sqrt{\lambda }(\pi -2\tau _{1})+A_{2}\sin \sqrt{\lambda }(\pi -\tau _{1}-\tau _{2})\Big ). \end{aligned}$$

In a similar way, we get

$$\begin{aligned} 4\sqrt{\lambda }{\tilde{\Delta }}_{0}(\lambda )&=\sqrt{\lambda }\int _{\tau _{1}+\tau _{2}-\pi }^{\pi -\tau _{1}-\tau _{2}} \Big (C_{31}(y_{3})+D_{31}(z_{3})+D_{32}(z_{3})+B_{3}(x_{3})+C_{32}(y_{3})\Big )\sin \sqrt{\lambda }\xi d\xi \nonumber \\&\quad -4\sqrt{\lambda }L_{1}\cos \sqrt{\lambda }(\pi -\tau _{1})+N_{1}\sin \sqrt{\lambda }(\pi -\tau _{1})-\int _{\tau _{2}-\pi }^{\pi -\tau _{2}}p_{02}(\xi )\sin \sqrt{\lambda }\xi d\xi \nonumber \\&\quad +\sqrt{\lambda }\int ^{\pi -2\tau _{2}}_{2\tau _{2}-\pi }\Big (B_{4}(x_{4})+C_{4}(y_{4})+D_{4}(z_{4})\Big )\sin \sqrt{\lambda }\xi d\xi \nonumber \\&+2\sqrt{\lambda }\Big (A_{3}\sin \sqrt{\lambda }(\pi -\tau _{1}-\tau _{2})+A_{4}\sin \sqrt{\lambda }(\pi -2\tau _{1})\Big ),\end{aligned}$$
(2.12)

where \(N_{2}=2(q_{2}(\pi )+q_{2}(\tau _{2}))\), \(p_{02}(\xi )=p_{2}(\frac{\pi +\tau _{2}-\xi }{2})\), \(L_{2}=\int ^{\pi }_{\tau _{2}}q_{2}(t)dt\).

Similar to the processing techniques in (2.11) and (2.12), we do further calculations for \({\tilde{\Delta }}_{1}(\lambda )\). Therefore, we have

$$ \begin{aligned} \begin{aligned} 4\sqrt{\lambda }{\tilde{\Delta }}_{1}(\lambda )&=\sqrt{\lambda }\int _{3\tau _{1}-\tau _{2}-\pi }^{\pi -3\tau _{1}+\tau _{2}}\Big (C_{21}(y_{2})+D_{21}(z_{2})+D_{22}(z_{2})- B_{2}(x_{2})-C_{22}(y_{2})\Big )\cos \sqrt{\lambda }\xi d\xi \\&\quad -4\sqrt{\lambda }L_{1}\cos \sqrt{\lambda }(\pi -\tau _{1})+N_{1}\sin \sqrt{\lambda }(\pi -\tau _{1})-\int _{\tau _{1}-\pi }^{\pi -\tau _{1}}p_{01}(\xi )\sin \sqrt{\lambda }\xi d\xi \\&\quad -\sqrt{\lambda }\Big (\int ^{\pi -2\tau _{1}}_{2\tau _{1}-\pi }(B_{1}(x_{1})+C_{1}(y_{1})-D_{1}(z_{1}))\cos \sqrt{\lambda }\xi d\xi \Big )\\&\quad +2\sqrt{\lambda }\Big (A_{1}\cos \sqrt{\lambda }(\pi -2\tau _{1})+A_{2}\cos \sqrt{\lambda }(\pi -\tau _{1}-\tau _{2})\Big ),\\ \end{aligned} \end{aligned}$$
(2.13)

where \(M_{1}=-2(q_{1}(\pi )-q_{1}(\tau _{1}))\), \(p_{01}(\xi )=p_{1}(\frac{\pi +\tau _{1}-\xi }{2})\), \(L_{1}=\int ^{\pi }_{\tau _{1}}q_{1}(t)dt\).

$$\begin{aligned} 4\sqrt{\lambda }{\tilde{\Delta }}_{1}(\lambda )&=\sqrt{\lambda }\int _{\tau _{1}+\tau _{2}-\pi }^{\pi -\tau _{1}-\tau _{2}}\Big (-C_{31}(y_{2})+D_{31}(z_{3})+D_{32}(z_{3})-B_{3}(x_{3})-C_{32}(y_{2})\Big )\cos \sqrt{\lambda }\xi d\xi \nonumber \\&\quad +4\sqrt{\lambda }L_{2}\sin \sqrt{\lambda }(\pi -\tau _{1})+M_{2}\cos \sqrt{\lambda }(\pi -\tau _{1})-\int _{\tau _{2}-\pi }^{\pi -\tau _{2}}p_{02}(\xi )\cos \sqrt{\lambda }\xi d\xi \nonumber \\&\quad -\sqrt{\lambda }\Big (\int ^{\pi -2\tau _{2}}_{2\tau _{2}-\pi }(B_{4}(x_{4})+C_{4}(y_{4})-D_{4}(z_{4}))\cos \sqrt{\lambda }\xi d\xi \Big )\nonumber \\&\quad +2\sqrt{\lambda }\Big (A_{3}\cos \sqrt{\lambda }(\pi -\tau _{1}-\tau _{2})+A_{4}\cos \sqrt{\lambda }(\pi -2\tau _{1})\Big ), \end{aligned}$$
(2.14)

where \(M_{2}=-2(q_{2}(\pi )-q_{2}(\tau _{2}))\), \(p_{02}(\xi )=p_{2}(\frac{\pi +\tau _{2}-\xi }{2})\), \(L_{2}=\int ^{\pi }_{\tau _{2}}q_{2}(t)dt\).

Next, we separate the constant term. Denote

$$\begin{aligned} \begin{aligned} b_{0}(\lambda )&:=4\sqrt{\lambda }{\tilde{\Delta }}_{0}(\lambda )+4\sqrt{\lambda }L_{1}\cos \sqrt{\lambda }(\pi -\tau _{1})-N_{1}\sin \sqrt{\lambda }(\pi -\tau _{1}) \\&\quad -2\sqrt{\lambda }\Big (A_{1}\sin \sqrt{\lambda }(\pi -2\tau _{1})+A_{2}\sin \sqrt{\lambda }(\pi -\tau _{1}-\tau _{2})\Big ),\\ \end{aligned} \end{aligned}$$
(2.15)
$$\begin{aligned} \begin{aligned} b_{1}(\lambda )&:=4\sqrt{\lambda }{\tilde{\Delta }}_{0}(\lambda )+4\sqrt{\lambda }L_{2}\cos \sqrt{\lambda }(\pi -\tau _{2})-N_{2}\sin \sqrt{\lambda }(\pi -\tau _{2}) \\&\quad -2\sqrt{\lambda }\Big (A_{3}\sin \sqrt{\lambda }(\pi -\tau _{1}-\tau _{2})+A_{4}\sin \sqrt{\lambda }(\pi -2\tau _{2})\Big );\\ \end{aligned} \end{aligned}$$
(2.16)
$$\begin{aligned} \begin{aligned} c_{0}(\lambda )&:=4\sqrt{\lambda }{\tilde{\Delta }}_{1}(\lambda )-4\sqrt{\lambda }L_{1}\sin \sqrt{\lambda }(\pi -\tau _{1})-M_{1}\cos \sqrt{\lambda }(\pi -\tau _{1}) \\&\quad -2\sqrt{\lambda }\Big (A_{1}\cos \sqrt{\lambda }(\pi -2\tau _{1})+A_{2}\cos \sqrt{\lambda }(\pi -\tau _{1}-\tau _{2})\Big ), \end{aligned} \end{aligned}$$
(2.17)
$$\begin{aligned} \begin{aligned} c_{1}(\lambda )&:=4\sqrt{\lambda }{\tilde{\Delta }}_{1}(\lambda )-4\sqrt{\lambda }L_{2}\sin \sqrt{\lambda }(\pi -\tau _{1})-M_{2}\cos \sqrt{\lambda }(\pi -\tau _{1}) \\&\quad -2\sqrt{\lambda }\Big (A_{3}\cos \sqrt{\lambda }(\pi -\tau _{1}-\tau _{2})+A_{4}\cos \sqrt{\lambda }(\pi -2\tau _{1})\Big ). \end{aligned} \end{aligned}$$
(2.18)

Let

$$\begin{aligned} M_{1}(\xi )= & {} \sqrt{\lambda }(B_{1}(x_{1})+C_{1}(y_{1})+D_{1}(z_{1})),\\ M_{2}(\xi )= & {} \sqrt{\lambda }(-C_{21}(y_{2})-D_{21}(z_{2})+D_{22}(z_{2})+B_{2}(x_{2})+C_{22}(y_{2})). \end{aligned}$$

Then

$$ \begin{aligned} b_{0}(\lambda )= & {} -\int _{\tau _{1}-\pi }^{\pi -\tau _{1}}p_{01}(\xi )\sin \sqrt{\lambda }\xi d\xi \nonumber+\int ^{\pi -2\tau _{1}}_{2\tau _{1}-\pi }M_{1}(\xi )\sin \sqrt{\lambda }\xi d\xi +\int ^{\pi -3\tau _{1}+\tau _{2}}_{3\tau _{1}-\tau _{2}-\pi }M_{2}(\xi )\sin \sqrt{\lambda }\xi d\xi \nonumber \\= & {} \int _{\tau _{1}-\pi }^{\pi -\tau _{1}}\Big (-p_{01}(\xi )+M_{1}(\xi )+M_{2}(\xi )\Big )\sin \sqrt{\lambda }\xi d\xi \nonumber \\= & {} \int _{\tau _{1}-\pi }^{\pi -\tau _{1}}M(\xi )\sin \sqrt{\lambda }\xi d\xi , \end{aligned}$$
(2.19)

where \(M(\xi )=p_{01}(\xi )+M_{1}(\xi )+M_{2}(\xi )\), \(M_{1}(\xi )\equiv 0\) outside the interval \((2\tau _{1}-\pi ,\pi -2\tau _{1})\) and \(M_{2}(\xi )\equiv 0\) outside the interval \((3\tau _{1}-\tau _{2}-\pi ,\pi -3\tau _{1}+\tau _{2})\). Then we get

$$\begin{aligned} \begin{aligned} p_{01}(\xi )=M_{1}(\xi )+M_{2}(\xi )-M(\xi ). \end{aligned} \end{aligned}$$
(2.20)

Let

$$\begin{aligned} R_{1}(\xi )= & {} \sqrt{\lambda }(B_{4}(x_{4})+C_{4}(y_{4})+D_{4}(z_{4})),\\ R_{2}(\xi )= & {} \sqrt{\lambda }(C_{31}(y_{3})+D_{31}(z_{3})+D_{32}(z_{3})+B_{3}(x_{3})+C_{32}(y_{3})). \end{aligned}$$

Then

$$\begin{aligned} \begin{aligned} b_{1}(\lambda )&=-\int _{\tau _{2}-\pi }^{\pi -\tau _{2}}p_{02}(\xi )\sin \sqrt{\lambda }\xi d\xi +\int ^{\pi -2\tau _{2}}_{2\tau _{2}-\pi }R_{1}(\xi )\sin \sqrt{\lambda }\xi d\xi+\int ^{\pi -\tau _{1}-\tau _{2}}_{\pi -\tau _{1}-\tau _{2}}R_{2}(\xi )\sin \sqrt{\lambda }\xi d\xi \\&=\int _{\tau _{2}-\pi }^{\pi -\tau _{2}}\Big (-p_{02}(\xi )+R_{1}(\xi )+R_{2}(\xi )\Big )\sin \sqrt{\lambda }\xi d\xi \\&=\int _{\tau _{2}-\pi }^{\pi -\tau _{2}}R(\xi )\sin \sqrt{\lambda }\xi d\xi , \end{aligned} \end{aligned}$$
(2.21)

where \(R(\xi )=-p_{02}(\xi )+R_{1}(\xi )+R_{2}(\xi )\), \(R_{1}(\xi )\equiv 0\) outside the interval \((2\tau _{2}-\pi ,\pi -2\tau _{2})\) and \(R_{2}(\xi )\equiv 0\) outside the interval \((\tau _{1}+\tau _{2}-\pi ,\pi -\tau _{1}-\tau _{2})\). Therefore, we have

$$\begin{aligned} \begin{aligned} p_{02}(\xi )=R_{1}(\xi )+R_{2}(\xi )-R(\xi ). \end{aligned} \end{aligned}$$
(2.22)

Let

$$\begin{aligned} O_{1}(\xi )= & {} \sqrt{\lambda }(-B_{1}(x_{1})-C_{1}(y_{1})+D_{1}(z_{1})),\\ O_{2}(\xi )= & {} \sqrt{\lambda }(C_{21}(y_{2})+D_{21}(z_{2})+D_{22}(z_{2})-B_{2}(x_{2})-C_{22}(y_{2})). \end{aligned}$$

Then

$$ \begin{aligned} \begin{aligned} c_{0}(\lambda ) =&-\int _{\tau _{1}-\pi }^{\pi -\tau _{1}}p_{01}(\xi )\cos \sqrt{\lambda }\xi d\xi+\int ^{\pi -2\tau _{1}}_{2\tau _{1}-\pi }O_{1}(\xi )\cos \sqrt{\lambda }\xi d\xi +\int ^{\pi -3\tau _{1}+\tau _{2}}_{3\tau _{1}-\tau _{2}-\pi }O_{2}(\xi )\cos \sqrt{\lambda }\xi d\xi \\ =&\int _{\tau _{1}-\pi }^{\pi -\tau _{1}}\Big (-p_{01}(\xi )+O_{1}(\xi )+O_{2}(\xi )\Big )\cos \sqrt{\lambda }\xi d\xi \\ =&\int _{\tau _{1}-\pi }^{\pi -\tau _{1}}O(\xi )\cos \sqrt{\lambda }\xi d\xi , \end{aligned} \end{aligned}$$
(2.23)

where \(O(\xi )=-p_{01}(\xi )+O_{1}(\xi )+O_{2}(\xi )\), \(O_{1}(\xi )\equiv 0\) outside the interval \((2\tau _{1}-\pi ,\pi -2\tau _{1})\) and \(O_{2}(\xi )\equiv 0\) outside the interval \((3\tau _{1}-\tau _{2}-\pi ,\pi -3\tau _{1}+\tau _{2})\). Then we get

$$\begin{aligned} \begin{aligned} p_{01}(\xi )=-O(\xi )+O_{1}(\xi )+O_{2}(\xi ). \end{aligned} \end{aligned}$$
(2.24)

Let

$$\begin{aligned} Q_{1}(\xi )= & {} \sqrt{\lambda }(-B_{4}(x_{4})-C_{4}(y_{4})+D_{4}(z_{4})),\\ Q_{2}(\xi )= & {} \sqrt{\lambda }(-C_{31}(y_{2})+D_{31}(z_{3})+D_{32}(z_{3})-B_{3}(x_{3})-C_{32}(y_{2})). \end{aligned}$$

Then

$$\begin{aligned} \begin{aligned} c_{1}(\lambda ) =&-\int _{\tau _{2}-\pi }^{\pi -\tau _{2}}p_{02}(\xi )\cos \sqrt{\lambda }\xi d\xi +\int ^{\pi -2\tau _{2}}_{2\tau _{2}-\pi }Q_{1}(\xi )\cos \sqrt{\lambda }\xi d\xi +\int ^{\pi -\tau _{1}-\tau _{2}}_{\tau _{1}+\tau _2-\pi }Q_{2}(\xi )\cos \sqrt{\lambda }\xi d\xi \\ =&\int _{\tau _{2}-\pi }^{\pi -\tau _{2}}\Big (-p_{02}(\xi )+Q_{1}(\xi )+Q_{2}(\xi )\Big )\cos \sqrt{\lambda }\xi d\xi \\ =&\int _{\tau _{2}-\pi }^{\pi -\tau _{2}}Q(\xi )\cos \sqrt{\lambda }\xi d\xi , \end{aligned} \end{aligned}$$
(2.25)

where \(Q(\xi )=-p_{02}(\xi )+Q_{1}(\xi )+Q_{2}(\xi )\), \(Q_{1}(\xi )\equiv 0\) outside the interval \((2\tau _{2}-\pi ,\pi -2\tau _{2})\) and \(Q_{2}(\xi )\equiv 0\) outside the interval \((\tau _{1}+\tau _{2}-\pi ,\pi -\tau _{1}-\tau _{2})\). Then we get

$$\begin{aligned} \begin{aligned} p_{02}(\xi )=-Q(\xi )+Q_{1}(\xi )+Q_{2}(\xi ). \end{aligned} \end{aligned}$$
(2.26)

Therefore, we obtain a linear system

$$\begin{aligned} \begin{aligned} {\left\{ \begin{array}{ll} p_{01}(\xi )=M_{1}(\xi )+M_{2}(\xi )-M(\xi ),\\ p_{02}(\xi )=R_{1}(\xi )+R_{2}(\xi )-R(\xi ),\\ p_{01}(\xi )=O_{1}(\xi )+O_{2}(\xi )-O(\xi ),\\ p_{02}(\xi )=Q_{1}(\xi )+Q_{2}(\xi )-Q(\xi ). \end{array}\right. } \end{aligned} \end{aligned}$$
(2.27)

Moreover,

$$\begin{aligned} \begin{aligned} p_{01}(\xi )&=\frac{-M(\xi )-O(\xi )+M_{1}(\xi )+O_{1}(\xi )+M_{2}(\xi )+O_{2}(\xi )}{2},\ \xi \in (2\tau _1-\pi ,\pi -2\tau _1), \\ p_{02}(\xi )&=\frac{-R(\xi )-Q(\xi )+R_{1}(\xi )+Q_{1}(\xi )+R_{2}(\xi )+Q_{2}(\xi )}{2},\ \xi \in (2\tau _2-\pi ,\pi -2\tau _2). \end{aligned} \end{aligned}$$
(2.28)

In particular, this yields

$$\begin{aligned} \begin{aligned} {\left\{ \begin{array}{ll}\!\! p_{1}(x)\!=\!\frac{M(\pi +\tau _{1}-2x)-O(\pi +\tau _{1}-2x)-M_{1}(\pi +\tau _{1}-2x)+O_{1}(\pi +\tau _{1}-2x)-M_{2}(\pi +\tau _{1}-2x)+O_{2}(\pi +\tau _{1}-2x)}{2}, x\in (\frac{3\tau _{1}}{2},\pi -\frac{\tau _{1}}{2}),\\ \!\!p_{2}(x)\!=\!\frac{R(\pi +\tau _{2}-2x)-Q(\pi +\tau _{2}-2x)-R_{1}(\pi +\tau _{2}-2x)+Q_{1}(\pi +\tau _{2}-2x)-R_{2}(\pi +\tau _{2}-2x)+Q_{2}(\pi +\tau _{2}-2x)}{2}, x\in (\frac{3\tau _{2}}{2},\pi -\frac{\tau _{2}}{2}). \end{array}\right. } \end{aligned} \end{aligned}$$
(2.29)

3 Inverse Problem

In this section, we present main results: Theorem 3.1 and constructive Algorithm 3.2.

Theorem 3.1

The two spectra \(\{\lambda _{ni}\}^{\infty }_{n=1}\) of \(S{-}L_{i}(q_{j}), (i=0,1, j=1,2)\) can uniquely determine the potential \(q_{1}(x)\) and \(q_{2}(x)\).

Through the analysis of Sect. 2, the proof of Theorem 3.1 comes from the Algorithm 3.2:

Algorithm 3.2
figure a

Solve Inverse problem. Require: Let the spectra \(\{\lambda_{ni}\}^{\infty}_{n=1} of S{-}L_{i}(q_{j}), (i=0,1, j=1,2)\) be given.

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Remark 3.2

In the whole construction of the algorithm, we solve the uniqueness of the inverse problem on some intervals by means of zero function extension. In details, as in Step 7 and Step 8, we reconstruct \(p_{1}(x)\) for \(x\in (\tau _{1},\frac{3\tau _{1}}{2})\cup (\pi -\frac{\tau _{1}}{2},\pi )\), \(p_{2}(x)\) for \(x\in (\tau _{2},\frac{3\tau _{2}}{2})\cup (\pi -\frac{\tau _{2}}{2},\pi )\). Besides, by linear system (2.27), we determine the uniqueness of the potential function on the residual interval. It says that the uniqueness result for nonlinear inverse problem for Sturm–Liouville operator with multiple delays from two spectra can be established on the whole interval.