On a uniqueness theorem of Sturm–Liouville equations with boundary conditions polynomially dependent on the spectral parameter

Abstract

Inverse nodal problems for Sturm–Liouville equations associated with boundary conditions polynomially dependent on the spectral parameter are studied. The authors show that a twin-dense subset $W_{B}([a,b])$ can uniquely determine the operator up to a constant translation of eigenparameter and potential, where $[a,b]$ is an arbitrary interval which contains the middle point of the domain of the operator and B is a subset of $\mathbb {N}$ which satisfies some condition (see Theorem 4.2).

1 Introduction

The inverse problems of the differential operator $L:=L(q,U_{0},U_{1})$:

are considered, where λ is called the spectral parameter, q is a real-valued $L^{2}$-function on $(0,1)$ and

$$ R_{\xi k}(\lambda)=\sum_{l=0}^{ r_{\xi k}}a_{\xi kl} \lambda^{ r_{\xi k}-l},\quad r_{\xi1}=r_{\xi0}=r_{\xi}\geq0, a_{\xi10}=1, \xi,k=0,1, $$

are arbitrary polynomials of degree $r_{\xi}$ with real coefficients such that $R_{\xi1}(\lambda)$ and $R_{\xi 0}(\lambda)$ have no common zeros for $\xi=0,1$. The inverse spectral problem for the Sturm–Liouville equation with boundary conditions dependent on the spectral parameter was studied in [1–7] respectively. In particular, Freiling and Yurko [4] studied three inverse spectral problems for L and showed that this operator L can be recovered either from the Weyl function, or from discrete spectral data, or from two spectra. Recently, the inverse spectral problem for L was studied with mixed spectral data in [7–9]. For the case $R_{00}(\lambda)=1$, $R_{01}(\lambda)=-h$ in (1.2) and $R_{10}(\lambda)=1$, ${R_{11}(\lambda)}=H$ in (1.3), the operator $U(q,U_{0},U_{1})$ turns to a classical Sturm–Liouville problem $L(q,h,H)$. Inverse spectral problems and inverse nodal problems of $L(q,h,H)$ have been well studied, the readers can refer to [2, 10–21] and the references therein.

The aim of this article is to investigate the inverse spectral and nodal problems for the BVP L. We show that the result on the Weyl m-function for L also holds by an alternative approach, which is a generalization of the result for the classical Sturm–Liouville operator in [16]. Moreover, the authors show that the operator $L(q,U_{0},U_{1})$ can be uniquely determined up to constant translation by the twin-dense subset in the interior interval.

2 Preliminaries

Let $u_{-}(x,\lambda)$ and $u_{+}(x,\lambda)$ be solutions of equation (1.1) with initial conditions

$$\begin{aligned} &u_{-}(0,\lambda)=R_{01}(\lambda),\qquad u_{-}'(0, \lambda)=-R_{00}(\lambda ), \\ &u_{+}(1,\lambda)=R_{11}(\lambda),\qquad u_{+}'(1, \lambda)=-R_{10}(\lambda). \end{aligned}$$

Denote $\lambda=\rho^{2}$, $\tau=|\operatorname{Im} \rho|$, for sufficiently large $|\lambda|$, we have

$$\begin{aligned} &u_{-} (x,\lambda)=\lambda^{r_{0}} \biggl(\cos\rho x+O \biggl( \frac{e^{\tau x}}{\rho} \biggr) \biggr), \end{aligned}$$

(2.1)

$$\begin{aligned} &u_{-}'(x,\lambda)=\lambda^{r_{0}}\bigl(-\rho\sin\rho x+O \bigl(e^{\tau x}\bigr)\bigr), \end{aligned}$$

(2.2)

$$\begin{aligned} &u_{+}(x,\lambda)=\lambda^{r_{1}} \biggl(\cos\rho (1-x)+O \biggl( \frac{e^{\tau(1-x)}}{\rho} \biggr) \biggr), \end{aligned}$$

(2.3)

$$\begin{aligned} &u_{+}'(x,\lambda)=\lambda^{r_{1}}\bigl(\rho\sin\rho (1-x)+O \bigl(e^{\tau(1-x)}\bigr)\bigr). \end{aligned}$$

(2.4)

Denote

$$ \Delta(\lambda):=[u_{+},u_{-}](x,\lambda), $$

where $[y,z](x):=y(x)z'(x)-y'(x)z(x)$ is the Wronskian of y and z. Then

$$ \Delta(\lambda)=-R_{01}(\lambda)u'_{+}(0, \lambda) -R_{00}(\lambda)u_{+}(0, \lambda) =-U_{0}(u_{+})=U_{1}(u_{-}), $$

(2.5)

which is called the characteristic function of L (see [4]). By virtue of (2.1), (2.2), and (2.5), we have

$$\begin{aligned} \Delta(\lambda)=\lambda^{r_{0}+r_{1}}\bigl(-\rho\sin\rho+\omega \cos\rho +o\bigl(e^{\tau}\bigr)\bigr). \end{aligned}$$

(2.6)

Define the Weyl m-function $m_{\pm}(x,\lambda)$ by

$$ m_{-}(x,\lambda)=-\frac{u_{-}'(x,\lambda)}{u_{-}(x,\lambda)},\qquad m_{+}(x,\lambda)=\frac{u_{+}'(x,\lambda)}{u_{+}(x,\lambda)}, $$

then

$$\begin{aligned} m_{-}(x,\lambda)& =i\rho+o(1)\quad\bigl(\mbox{resp. }m_{+}(x,\lambda)=i \rho+o(1)\bigr), \end{aligned}$$

(2.7)

$$\begin{aligned} \frac{1}{m_{-}(x,\lambda)}&=-\frac{i}{\rho}+o \biggl(\frac{1}{\rho^{2}} \biggr)\quad \biggl(\mbox{resp. }\frac{1}{m_{+}(x,\lambda)}=-\frac{i}{\rho}+o \biggl(\frac{1}{\rho^{2}} \biggr) \biggr) \end{aligned}$$

(2.8)

uniformly in $x\in[\delta,1]$ (resp. $x\in[0,1-\delta]$) for $|\lambda |\rightarrow\infty$ in any sector $\varepsilon<\arg(\lambda)<\pi-\varepsilon$ for $\varepsilon>0$, where $\delta\in(0,1)$.

Denote the spectrum $\sigma(L):=\{\lambda_{n}\}_{n=0}^{\infty}$ of L, $\sigma(L)$ consisting of the zeros (counting with multiplicities) of the entire function $\Delta(\lambda)$. For n sufficiently large, $\lambda_{n}$ are real and simple and satisfy the asymptotic formulae (see [4])

$$\begin{aligned} \rho_{n}:=\sqrt{\lambda_{n}}=(n-r_{0}-r_{1}) \pi+\frac{\omega}{n\pi}+\frac {\kappa_{n}}{n}, \quad \{\kappa_{n}\}\in l^{2}, \end{aligned}$$

(2.9)

where

$$ \omega=\frac{1}{2} \int_{0}^{1}{q(t)}\, {d}t-a_{000}+a_{100}. $$

(2.10)

3 Inverse spectral problems

For convenience, let $\widetilde{L}=L(\widetilde{q},\widetilde {U}_{0},\widetilde{U}_{1})$, where $L(\widetilde{q},\widetilde {U}_{0},\widetilde{U}_{1})$ is the operator of the same form as L. If a certain symbol γ denotes an object related to L, then the corresponding symbol γ̃ denotes the analogous object related to L̃ and $\widehat{\gamma}=\gamma-\widetilde {\gamma}$.

The following two theorems on the Weyl m-function of the BVP L are derived from [4], and they are generalizations of the analogical results for the classical Sturm–Liouville in [16].

Theorem 3.1

Let $m_{-}(a_{0},\lambda)$ be the Weyl m-function of the BVP L. Then $m_{-}(a_{0},\lambda)$ can uniquely determine functions $R_{0k}(\lambda)$ for $k=0 \textit{ and } 1$ as well as q (a.e.) on the interval $[0,a_{0}]$, $0< a_{0}\leq1$.

Proof

Denote by $L_{D}$ the boundary value problem (1.1), (1.2) together with $\Delta_{D}(\lambda):=u_{-}(a_{0},\lambda)=0$ and $\{\mu_{a_{0},n}\}_{n=1}^{\infty}$, the zeros (counting with multiplicities) of the entire function $\Delta_{D}(\lambda)$ (see [10]). Then $\mu_{a_{0},n}$ is real and simple for sufficiently large n and

$$\begin{aligned} u_{-}(a_{0},\lambda)=\lambda^{r_{0}} \biggl( \cos(a_{0}\rho)+\omega_{1}\frac{\sin (a_{0}\rho)}{\rho}+o \biggl( \frac{\mathrm{e}^{a_{0}\tau}}{\rho} \biggr) \biggr) \quad\text{for } \vert \rho \vert \gg1, \end{aligned}$$

(3.1)

where $\omega_{1}=\frac{1}{2}\int_{0}^{a_{0}}q(x)\,dx-a_{000}$. Thus we have

$$\begin{aligned} \sqrt{\mu_{a_{0},n}}=\frac{\pi}{a_{0}} \biggl(n-r_{0}+ \frac{1}{2}+\frac{\omega _{1}}{n\pi}+o \biggl(\frac{1}{n} \biggr) \biggr) \quad \text{for } n\gg1. \end{aligned}$$

By virtue of Hadamard’s factorization theorem,

$$\begin{aligned} u_{-}(a_{0},\lambda)=C_{a_{0},0}\lambda^{m_{0}} \prod_{\mu_{a_{0},n}\neq0} \biggl(1-\frac{\lambda}{\mu_{a_{0},n}} \biggr), \end{aligned}$$

(3.2)

where $C_{a_{0},0}$ is a constant and $m_{0}\geq0$. Let $G_{\delta_{0}}:=\{ \lambda:|\rho-\frac{\pi}{a_{0}} (k-r_{0}+\frac{1}{2} )|>\delta_{0}, k\in\mathbb{Z}\}$, where $\delta_{0}$ is sufficiently small, then there exists a constant $C_{a_{0},\delta_{0}}$ (see [10, 11]) such that

$$\begin{aligned} \bigl\vert \Delta_{D}(\lambda) \bigr\vert \geq C_{a_{0},\delta_{0}} \vert \rho \vert ^{2r_{0}}\mathrm{e}^{a_{0}\tau}, \quad\forall \lambda\in G_{\delta_{0}} \text{ and } \vert \lambda \vert \gg1. \end{aligned}$$

(3.3)

Similarly, denote by $L_{N}$ the boundary value problem (1.1), (1.2) together with $\Delta_{N}(\lambda):=u_{-}'(a_{0},\lambda)=0$ and $\lambda_{a_{0},n}$, the zeros (counting with multiplicities) of the entire function $\Delta_{N}(\lambda)$. Then $\{\mu_{a_{0},n}\}_{n=1}^{\infty}$ are real and simple for sufficiently large n and

$$\begin{aligned} u_{-}'(a_{0},\lambda)=\lambda^{r_{0}} \bigl(-\rho\sin(a_{0}\rho)+c_{0}\cos(a_{0}\rho )+o \bigl(\mathrm{e}^{a_{0}\tau}\bigr)\bigr), \end{aligned}$$

(3.4)

where $c_{0}=2r_{0}+\omega_{1}$. Therefore we have

$$\begin{aligned} \sqrt{\lambda_{a_{0},n}}=\frac{\pi}{a_{0}} \biggl(n-r_{0}+ \frac{\omega_{1}}{n\pi }+o \biggl(\frac{1}{n} \biggr) \biggr). \end{aligned}$$

Let $G_{\delta_{1}}:=\{\lambda:|\rho-\frac{(k-r_{0})\pi}{a_{0}}|>\delta_{1}, k\in\mathbb{Z}\}$, where $\delta_{1}$ is sufficiently small, then there exists a constant $C_{a_{0},\delta_{1}}$ such that, for sufficiently large $|\lambda|$,

$$\begin{aligned} \bigl\vert \Delta_{N}(\lambda) \bigr\vert \geq C_{a_{0},\delta_{1}} \vert \rho \vert ^{2r_{0}+1}\mathrm{e}^{a_{0}\tau}, \quad\forall \lambda\in G_{\delta_{1}}. \end{aligned}$$

(3.5)

Thus we have

$$\begin{aligned} u_{-}'(a_{0},\lambda)=C_{a_{0},1} \lambda^{m_{1}}\prod_{\lambda_{a_{0},n}\neq 0} \biggl(1- \frac{\lambda}{\lambda_{a_{0},n}} \biggr), \end{aligned}$$

(3.6)

where $m_{1}\geq0$, $C_{a_{0},1}$ is a constant. Under the assumption $m_{-}(a_{0},\lambda)=\widetilde{m}_{-}(a_{0},\lambda)$, we obtain

$$\begin{aligned} \frac{u_{-}'(a_{0},\lambda)}{u_{-}(a_{0},\lambda)}=\frac{\widetilde {u}_{-}'(a_{0},\lambda)}{\widetilde{u}_{-}(a_{0},\lambda)}. \end{aligned}$$

(3.7)

Since $u_{-}(a_{0},\lambda)$ and $u_{-}'(a_{0},\lambda)$ (resp. $\widetilde {u}_{-}(a_{0},\lambda)$ and $\widetilde{u}_{-}'(a_{0},\lambda)$) have no common zeros, $\frac{\widetilde{u}_{-}(a_{0},\lambda)}{u_{-}(a_{0},\lambda)}$ and $\frac {\widetilde{u}_{-}'(a_{0},\lambda)}{u_{-}'(a_{0},\lambda)}$ are two entire functions in λ and

$$\begin{aligned} \frac{\widetilde{u}_{-}(a_{0},\lambda)}{u_{-}(a_{0},\lambda)}=\frac{\widetilde {u}_{-}'(a_{0},\lambda)}{u_{-}'(a_{0},\lambda)}. \end{aligned}$$

(3.8)

By virtue of (3.8) together with (2.1), (2.10), (2.2), and (3.4), this yields

$$\begin{aligned} &\biggl\vert \frac{\widetilde{u}_{-}(a_{0},\lambda)}{u_{-}(a_{0},\lambda)} \biggr\vert \leq O(1),\quad\forall\lambda\in G_{\delta_{0}}, \\ &\biggl\vert \frac{\widetilde{u}_{-}'(a_{0},\lambda)}{u_{-}'(a_{0},\lambda)} \biggr\vert \leq O(1),\quad\forall\lambda\in G_{\delta_{1}}. \end{aligned}$$

Using the maximum modulus principle and Liouville’s theorem, we have

$$ \textstyle\begin{cases} \frac{\widetilde{u}_{-}(a_{0},\lambda)}{u_{-}(a_{0},\lambda)}\equiv c, & \forall\lambda\in\mathbb {C},\\ \frac{\widetilde{u}_{-}'(a_{0},\lambda)}{u_{-}'(a_{0},\lambda)}\equiv c, & \forall\lambda\in\mathbb {C}. \end{cases} $$

(3.9)

Letting $\rho=iy\rightarrow\infty$, then either the first formula in (3.9) together with (2.3) or the second formula in (3.9) together with (2.4) implies

$$\begin{aligned} c=\pm1. \end{aligned}$$

(3.10)

By virtue of (2.1), (2.2), and (3.10), we have

$$c=1. $$

Therefore,

$$\begin{aligned} u_{-}(a_{0},\lambda)=\widetilde{u}_{-}(a_{0},\lambda),\quad \text{and}\quad u_{-}'(a_{0},\lambda)=\widetilde{u}_{-}'(a_{0}, \lambda)\quad\forall\lambda\in \mathbb {C}. \end{aligned}$$

This implies

$$\begin{aligned} u_{-}(x,\lambda)=\widetilde{u}_{-}(x,\lambda),\quad\forall x \in[0,a_{0}]. \end{aligned}$$

(3.11)

Thus (3.11) shows

$$q(x)\overset{a.e.}{=}\widetilde{q}(x)\quad\text{on } [0,a_{0}]\quad\text{and}\quad R_{0k}(\lambda)=\widetilde{R}_{0k}(\lambda),\quad k=0,1. $$

Therefore the proof of Theorem 3.1 is completed. □

Analogously, we prove the following theorem on the Weyl m-function $m_{+}(a_{0},\lambda)$.

Theorem 3.2

Let $m_{+}(b_{0},\lambda)$ be the Weyl m-function of the BVP L. Then $m_{+}(b_{0},\lambda)$ can uniquely determine functions $R_{1k}(\lambda)$ for $k=0,1$ as well as q (a.e.) on the interval $[b_{0},1]$, $0\leq b_{0}<1$.

4 Inverse nodal problems

By virtue of Lemma 3.1 in [22], we see that, for $n\gg1$, the eigenfunction $u_{-}(x,\lambda_{n})$ has exactly $n-r_{0}-r_{1}$ zeros $0< x_{n}^{1}< x_{n}^{2}<\cdots<x_{n}^{j}<\cdots<x_{n}^{n-r_{0}-r_{1}}<1$ inside the interval $(0,1)$ and satisfy the following asymptotic formula:

$$\begin{aligned} x_{n}^{j}&=\frac{j-\frac{1}{2}}{n-r_{0}-r_{1}}- \frac{ (j-\frac{1}{2} )\omega}{n(n-r_{0}-r_{1})^{2}\pi^{2}}+\frac{\frac{1}{2}\int _{0}^{x_{n}^{j}}{q(t)}\,{d}t-a_{000}}{(n-r_{0}-r_{1})^{2}\pi^{2}} +O \biggl(\frac{1}{n^{3}} \biggr) \end{aligned}$$

(4.1)

for $0< j< n-r_{0}-r_{1}$, where w is as that in (2.10). Denote $x_{n}^{0}=0$, $x_{n}^{n-r_{0}-r_{1}+1}=1$. Note that $\sigma(L)$ might contain non-real eigenvalues, hence we write

$$\sigma(L)=\sigma_{R}(L)\cup\sigma_{C}(L), $$

where $\sigma_{R}(L)$ consists of real eigenvalues of L. Denote by X the collection of all zeros of all eigenfunctions of L. Let $B=\{n_{k}\}_{k=1}^{\infty}$ be a strictly increasing sequence in $\mathbb {N}$, where $\lambda_{n_{k}}\in{\sigma_{R}(L)}$. For $0\leq a< b\leq 1$, we call the subset $W_{B}([a,b])$ of $X\cap[a,b]$ an interior twin-dense nodal subset on the interval $[a,b]$ if the following conditions hold:

• (1) For all $n_{k}\in B$, there exists some $j_{k}$ such that $x_{n_{k}}^{j_{k}}\in W_{B}([a,b])$.

• (2) The nodal subset $W_{B}([a,b])$ is twin on the interval $[a,b]$, i.e., if $x_{n_{k}}^{j_{k}}\in W_{B}([a,b])$, then $x_{n_{k}}^{j_{k}+1}\in W_{B}([a,b])$ or $x_{n_{k}}^{j_{k}-1}\in W_{B}([a,b])$.

• (3) The nodal subset $W_{B}([a,b])$ is dense on the set $[a,b]$, i.e., $\overline{W}_{B}([a,b])=[a,b]$, where $\overline{W}_{B}([a,b])$ denotes the closure of $\overline{W}_{B}([a,b])$.

The following Lemma 4.1 is necessary for us to prove our main results.

Lemma 4.1

(Theorem 3.2 [8])

If $W_{B}([a,b])=\widetilde{W}_{\widetilde{B}}([a,b])$, then

Let $S_{B}=\{\lambda_{n_{k}}:n_{k}\in B\}$. For any sequence $S=\{x_{n}\} _{n=0}^{\infty}$ of positive real numbers, we define

$$N_{S}(t)=\#\bigl\{ n\in\mathbb {N}\cup\{0\}:x_{n}< t\bigr\} . $$

The following theorem is our main result which concerns the unique determination of the operator from a twin-dense nodal subset and a partial spectrum.

Theorem 4.2

$0< a<\frac{1}{2}<b<1$. Suppose $W_{B}([a,b])=\widetilde{W}_{\widetilde{B}}([a,b])$ and

$$ N_{S_{B}}(t)\ge2a_{1} N_{\sigma_{R}(L)}(t)+2(1-a_{1}) \biggl(r_{0}+r_{1}+\frac {1}{2}\biggr)+2a_{1}k_{0}-1, \quad t\gg1 $$

(4.6)

for $a_{1}=a$ and $1-b$, where $k_{0}$ is the number of elements in $\sigma _{c}(L)$. Then

$$ \widetilde{q}(x)\overset{a.e.}{=}{q}(x)-c\quad\textit{on } [0,1] \quad\textit{and}\quad \widetilde{R}_{\xi k}(\lambda)={R}_{\xi k}(\lambda +c)\quad\textit{for }\xi,k=0,1, $$

and some constant c.

Proof

From Lemma 4.1, we have $r_{0}+r_{1}=\widetilde {r}_{0}+\widetilde{r}_{1}$, and $r_{0}+r_{1}$ can be reconstructed from (4.1). By virtue of (3.2), one can reconstruct ω by

$$\begin{aligned} &\lim_{k\rightarrow\infty}{n_{k} \bigl(\sqrt{\lambda _{n_{k}}}-(n_{k}-r_{0}-r_{1})\pi \bigr)}= \lim_{k\rightarrow\infty}(\omega+\kappa _{n_{k}})=\omega, \end{aligned}$$

(4.7)

$$\begin{aligned} &\lim_{k\rightarrow\infty}{n_{k} \bigl(\sqrt{\widetilde{\lambda}_{n_{k}}}-(n_{k}-r_{0}-r_{1})\pi \bigr)}= \lim_{k\rightarrow\infty}(\omega+\kappa _{n_{k}})=\widetilde{\omega}, \end{aligned}$$

(4.8)

and

Denote

$$ F(x,u_{-},\widetilde{u}_{-},\lambda)=u_{-}(x,\lambda)\widetilde {u}_{-}'(x, \lambda-2\widehat{w})-u_{-}'(x,\lambda)\widetilde{u}_{-}(x,\lambda -2 \widehat{w}). $$

Then

$$\begin{aligned} F(a,u_{-},\widetilde{u}_{-},\lambda)&=u_{-}(a,\lambda)\widetilde{u}_{-}'(a, \lambda)-u_{-}'(a,\lambda)\widetilde{ u}_{-}(a,\lambda) \\ &=u_{-}'(a,\lambda)\widetilde{u}_{-}'(a,\lambda-2\widehat{w}) \bigl(m_{-}^{-1}(a,\lambda)-\widetilde{m}_{-}^{-1}(a,\lambda-2 \widehat{w})\bigr). \end{aligned}$$

(4.11)

From (2.2), (2.8), and (4.11), we obtain

$$ \bigl\vert F(a,u_{-},\widetilde{u}_{-},ix) \bigr\vert =o\bigl( \vert y \vert ^{r_{0}+\tilde{r}_{0}}\mathrm{e}^{ \vert 2\operatorname{Im}{\sqrt {ix}} \vert a}\bigr) \quad \text{for } \vert x \vert\gg1, x \in\mathbb {R}. $$

(4.12)

Moreover, we can choose $\{x_{n_{k}}^{j_{n_{k}}}\}\in W_{B}([a,b])$ and apply Green’s formula to obtain

$$\begin{aligned} & \bigl(\widetilde{u}(x,{\widetilde{\lambda}_{n_{k}}})u'(x, \lambda _{n_{k}})-\widetilde{u}'(x,{\widetilde{\lambda}_{n_{k}}})u(x,\lambda _{n_{k}}) \bigr) \big\vert _{a}^{x_{n_{k}}^{j_{n_{k}}}} \\ &\quad =- \int_{a}^{x_{n_{k}}^{j_{n_{k}}}}\bigl[\lambda_{n_{k}}-\widetilde{\lambda}_{n_{k}}-\bigl(q(x)-\widetilde{q}(x)\bigr)\bigr]u(x, \lambda_{n_{k}})\widetilde{u}(x,\widetilde{\lambda}_{n_{k}}) \,dx=0, \end{aligned}$$

i.e.,

$$\begin{aligned} F(a,u_{-},\widetilde{u}_{-},\lambda_{n_{k}})=0 \quad \text{for all } n_{k}\in B. \end{aligned}$$

(4.13)

Define the functions

$$\begin{aligned}& G_{B}(\lambda)=\prod_{n_{k}\in B}{ \biggl(1-\frac{\lambda}{\lambda_{n_{k}}}\biggr)}, \end{aligned}$$

(4.14)

$$\begin{aligned}& \Delta_{R}(\lambda) = \prod _{\lambda_{n}\in\sigma_{R}(L)}{\biggl(1-\frac{\lambda }{\lambda_{n}}\biggr)}, \end{aligned}$$

(4.15)

and

$$ \Delta_{C}(\lambda) = \prod _{\lambda_{n}\in\sigma_{C}(L)}{\biggl(1-\frac{\lambda }{\lambda_{n}}\biggr)}. $$

(4.16)

Then we know $\Delta_{C}(\lambda)$ is a polynomial of degree $k_{0}$ and

$$\Delta(\lambda)=K\Delta_{C}(\lambda)\Delta_{R}(\lambda) $$

for some constant K.

Next, we shall use the technique in Appendix B of [15] to get an estimate of $|G_{S_{B}}(ix)|$. Without loss of generality, we may assume $\lambda>1$ for $\lambda\in\sigma_{R}(L)$ (it can be done by a shift of the parameter λ in L). Then $N_{S_{B}(t)}=N_{\sigma_{R}(L)}(t)=0$ for $t\le1$, and

$$\begin{aligned} \begin{aligned}[b] \ln \bigl\vert G(ix) \bigr\vert &=\sum_{k=1}^{\infty}\frac{1}{2}\ln \biggl(1+\frac{x^{2}}{\lambda_{n_{k}}^{2}}\biggr)=\frac{1}{2} \int_{1}^{\infty}\ln\biggl(1+\frac {x^{2}}{t^{2}}\biggr) \,dN_{S_{B}}(t) \\ &=\frac{1}{2} \biggl(\ln\biggl(1+\frac{x^{2}}{t^{2}}\biggr)N_{S_{B}}(t) \bigg\vert _{0}^{\infty}- \int _{1}^{\infty}N_{S_{B}}(t)\frac{d}{dt} \biggl(\ln\biggl(1+\frac{x^{2}}{t^{2}}\biggr)\biggr) \,dt \biggr) \\ &=-\frac{1}{2} \int_{1}^{\infty}N_{S_{B}}(t)\frac{d}{dt}\biggl( \ln\biggl(1+\frac {x^{2}}{t^{2}}\biggr) \biggr)\,dt= \int_{1}^{\infty}\biggl(\frac{x^{2}}{t^{3}+tx^{2}} \biggr)N_{S_{B}}(t) \,dt. \end{aligned} \end{aligned}$$

(4.17)

Hence

$$\begin{aligned} \ln{\frac{ \vert G(ix) \vert }{ \vert \Delta_{R}(ix) \vert ^{2a}}}= \int_{1}^{\infty}\biggl(\frac{x^{2}}{t^{3}+tx^{2}} \biggr) \bigl(N_{S_{B}}(t)-2aN_{\sigma_{R}(L)}\bigr) \,dt. \end{aligned}$$

(4.18)

By (4.6), we know that there exist a $t_{0}$ and a positive number K so that

$$ \textstyle\begin{cases}N_{S_{B}}(t)-2a_{1}N_{\sigma_{R}(L)}(t) \ge-K& \text{for } 1\leq t\leq t_{0}, \\ N_{S_{B}}(t)-2a_{1}N_{\sigma_{R}(L)}(t) \ge2(1-a_{1})(r_{0}+r_{1}+\frac {1}{2})+2a_{1}k_{0}-1&\text{for } t> t_{0}, \end{cases} $$

for $a_{1}=a\text{ or } 1-b$. This leads to

$$\begin{aligned} \ln\frac{ \vert G(ix) \vert }{ \vert \Delta_{R}(ix) \vert ^{2a_{1}}}& \ge \int_{1}^{t_{0}} \bigl[(K+L)/2\bigr] \frac {d}{dt}\biggl( \ln\biggl(1+\frac{x^{2}}{t^{2}}\biggr)\biggr) \,dt \\ &\quad{}+L \int_{t_{0}}^{\infty}(-1/2)\frac{d}{dt}\biggl(\ln\biggl(1+ \frac{x^{2}}{t^{2}}\biggr)\biggr) \,dt, \end{aligned}$$

where $L=[2(1-a_{1})(r_{0}+r_{1}+\frac{1}{2})+2a_{1}k_{0}-1]$. Hence

$$\begin{aligned} \bigl\vert G(ix) \bigr\vert &=O\bigl( \bigl\vert \Delta_{R}(ix) \bigr\vert ^{2a_{1}} \bigl\vert \Delta_{C}(ix) \bigr\vert ^{2a_{1}} \bigl\vert \Delta _{C}(ix) \bigr\vert ^{-2a_{1}} \vert x \vert ^{2(1-a_{1}))(r_{0}+r_{1}+\frac{1}{2})+2a_{1}k_{0}-1}\bigr) \\ &=O\bigl( \vert x \vert ^{(2a_{1})(r_{0}+r1+1/2)}e^{2a_{1} \vert \operatorname{Im}\sqrt{ix} \vert } \vert x \vert ^{-2a_{1}k_{0}} \vert x \vert ^{2(1-a_{1})(r_{0}+r_{1}+\frac{1}{2})+2a_{1}k_{0}-1}\bigr) \\ &=O\bigl( \vert x \vert ^{2(r_{0}+r_{1})}e^{2a_{1} \vert \operatorname {Im}(\sqrt{ix}) \vert }\bigr), \end{aligned}$$

i.e.,

$$ \bigl\vert G_{S_{B}}(ix) \bigr\vert \geq c \vert x \vert ^{2(r_{0}+r_{1})}e^{2a_{1}\operatorname{Im}{\sqrt{i}} \vert x \vert ^{\frac{1}{2}} } \quad\text{for } \vert x \vert\gg1 $$

(4.19)

and $a_{1}=a$ and $1-b$, where c is a constant. Therefore

$$ K_{1}(\lambda):=\frac{F(a,u_{,}\tilde{u}_{,}\lambda)}{G(\lambda)} $$

(4.20)

is an entire function and

$$ K_{1}(ix)\to0 \quad\text{as } \vert x \vert \to \infty. $$

(4.21)

In addition, we easily prove that the following formula

$$\begin{aligned} \inf_{ \vert \lambda \vert =R_{k}}{ \bigl\vert G_{S_{B}}(\lambda ) \bigr\vert }\geq c_{0}\exp^{-C_{0}R_{k}^{\frac{1+\varepsilon_{0}}{2}}} \end{aligned}$$

holds for sufficiently large $R_{k}$, some $0<\varepsilon_{0}<1$, $c_{0}$ and $C_{0}$ are two positive constants. Thus, we have

$$\begin{aligned} \sup_{ \vert \lambda \vert =R_{k}}{ \bigl\vert K_{1}( \lambda) \bigr\vert }\leq c_{0}\exp^{C_{0}R_{k}^{\frac{1+\varepsilon_{0}}{2}}} \end{aligned}$$

(4.22)

for sufficiently large $R_{k}\rightarrow\infty$ as $k\rightarrow+\infty$. By virtue of (4.21) and (4.22) (see [11], Proposition B.6), we obtain

$$ K_{1}(\lambda)=0,\quad\forall\lambda\in\mathbb {C}. $$

(4.23)

Therefore, we get

$$ F(a,u_{-},\widetilde{u}_{-},\lambda)=0,\quad\forall\lambda\in\mathbb {C}. $$

This implies

$$ m_{-}(a,\lambda)=\widetilde{m}_{-}(a,\lambda-2\widehat{w}). $$

(4.24)

By Theorem 3.1 together with (4.24), we get

$$ \widetilde{q}(x)\overset{a.e.}{=}{q}(x)-2\widehat{w}\quad\text{on } [0,a], \quad\text{and}\quad\widetilde{R}_{0k}(\lambda)={R}_{0k}( \lambda +2\widehat{w}),\quad k=0,1. $$

(4.25)

Similarly, we can define

$$H(x,u_{+},\tilde{u}_{+},\lambda)=u_{+}(x,\lambda)\widetilde{u}_{+}'(x, \lambda -2\widehat{w})-u_{+}'(x,\lambda)\widetilde{u}_{+}(x,\lambda-2 \widehat{w}). $$

Then one can repeat the same arguments as above on $H(b,u_{+},\tilde{u}_{+},\lambda)$ to show

$$\frac{ H(b,u_{+},\tilde{u}_{+},\lambda)}{G_{B}(\lambda)}=0. $$

This leads to $H(b,u_{+},\tilde{u}_{+},\lambda)=0$ and

$$ m_{+}(b,\lambda)=\widetilde{m}_{+}(b,\lambda-2\widehat{w}). $$

(4.26)

Hence

$$ \widetilde{q}(x)\overset{a.e.}{=}{q}(x)-2\widehat{w} \quad\text{on } [b,1], \quad\text{and}\quad\widetilde{R}_{1k}(\lambda)={R}_{1k}( \lambda +2\widehat{w}),\quad k=0,1. $$

(4.27)

(4.9), (4.25), and (4.27) imply

$$ \widetilde{q}(x)\overset{a.e.}{=}{q}(x)-2\widehat{w}\quad\text{on } [0,1]\quad\text{and}\quad \widetilde{R}_{\xi k}(\lambda)={R}_{\xi k}(\lambda+2\widehat{w}),\quad \xi,k=0,1. $$

This completes the proof of Theorem 4.2. □

Corollary 4.3

Under the assumptions of Theorem 4.2, if $\lambda_{n_{k}}=\widetilde{\lambda}_{n_{k}}$ for $n_{k}\gg1$, then $q(x)=\widetilde{q}(x)$ and $R_{ij}(\lambda)=\widetilde{R}_{i,j}(\lambda )$ for $i,j=0,1$.

Remark

The readers might be interested in the inverse nodal problem for a more general equation

$$ \textstyle\begin{cases} u''+A(\lambda)u=0,\quad x\in(0,1),\\ R_{01}(\lambda)u'(0)+R_{00} u(0)=0,\\ R_{11}(\lambda)u'(1)+R_{10}u(1)=0, \end{cases} $$

(4.28)

where $\frac{d^{2}}{dx^{2}}+A(\lambda)$ is an operator on $H^{2}(a,b)$. Some of such problems arise from PDE (please refer to [23, 24] for details). Same arguments for Theorem 4.2 seem to work for (4.28) if $A(\lambda)$ is an appropriate operator.

5 Conclusion

In this paper, the authors show that a twin-dense subset $W_{B}([a,b])$, $0< a< 1/2<b <1$, can uniquely determine (up to a constant translation on both boundary conditions and potential) the Sturm–Liouville operator associated with boundary conditions polynomially dependent on the spectral parameter. The theorem leads to the same conclusion for classical Sturm–Liouville equation when the coefficient polynomials $R_{ij}(\lambda)$ are all of degree 0 (refer to [25]), but the translation effect on boundary conditions only appears when one of $R_{ij}(\lambda)$ is a non-trivial polynomial.