Inverse nodal problems for Sturm–Liouville equation with eigenparameter-dependent boundary and jump conditions

Abstract

In this work, the Sturm–Liouville problem with boundary and jump conditions dependent on the spectral parameter linearly is studied. We show that all coefficients of the problem can be uniquely determined by nodal points. Moreover, we give an algorithm for reconstruction of the potential function and the coefficient in the jump conditions.

Keywords:

• Sturm–Liouville equation
• inverse nodal problem
• parameter dependent boundary condition
• discontinuity condition

AMS Subject Classifications:

• 34A55
• 34B07
• 34B24
• 34B37

1 Introduction

We consider the boundary value problem $L=L\left(q,,,H_0,,,H_1,,,H_2,,,\mathit{\alpha }\right)$ generated by the Sturm–Liouville equation(1) $$\begin{array}{c}\hfill \ell y:=-y^{″}+q(x)y=\mathit{\lambda }y,x\in \mathrm{\Omega }=(0,1)\end{array}$$(1) subject to the boundary conditions(2) $$\begin{array}{cc}& U,(,y,),:,=,y,(,0,),=,0\hfill \end{array}$$(2) (3) $$\begin{array}{cc}& V,(y),:,=,\mathit{\lambda },(y^{\prime }(1)+H_{0}y(1)),-,H_1,y^{\prime },(1),-,H_2,y,(1),=,0\hfill \end{array}$$(3) and the jump conditions(4) $$\begin{array}{c}\hfill \left\{\begin{array}{c}y,\left(\frac{1}{2},+,0\right),=,\mathit{\alpha },y,\left(\frac{1}{2},-,0\right)\hfill \\ y^{\prime },\left(\frac{1}{2},+,0\right),=,{\mathit{\alpha }}^{-1},y^{\prime },(\frac{1}{2}-0),-,\left(\mathit{\lambda },+,\mathit{\beta }\right),y,\left(\frac{1}{2},-,0\right),\hfill \end{array}\right.\end{array}$$(4) where $q(x)$ is a real-valued function in $L_2(0,1);$$\mathit{\alpha }$, $\mathit{\beta }$ and $H_i$, $i=0,1,2,$ are real numbers; $\mathit{\alpha }>0,$$H_0{H}_1-H_2>0$ and $\mathit{\lambda }$ is the spectral parameter. We assume that ${\int }_0^{1}q(x)dx=0.$ Otherwise, the term $q(x)-{\int }_0^{1}q(x)dx$ is determined uniquely, instead of $q(x).$

The inverse nodal Sturm–Liouville problem was solved firstly by McLaughlin [1] in 1988. She showed that the potential of the Sturm–Liouville problem can be determined by a given dense subset of nodal points of the eigenfunctions. In 1989, Hald and McLaughlin generalized this result to more general boundary conditions and gave some numerical schemes for the reconstruction of the potential [2]. Inverse nodal problems for Sturm–Liouville operators with the classical boundary conditions have been studied in the several papers.[2–10] The first result on inverse nodal problems for the Sturm–Liouville operators with a discontinuity condition was obtained in [11]. Inverse nodal problem for Sturm–Liouville operator with boundary conditions linearly dependent on the spectral parameter was investigated firstly by Browne and Sleeman [12]. C-F. Yang generalized their result to a Sturm–Liouville problem with nonlinear boundary conditions.[13] Additionally, the studies [14, 15] include inverse nodal problem for differential pencils.

Inverse problems according to the classical spectral data for various differential equations with the eigenparameter-dependent jump conditions, like (4(4) $$\begin{array}{c}\hfill \left\{\begin{array}{c}y,\left(\frac{1}{2},+,0\right),=,\mathit{\alpha },y,\left(\frac{1}{2},-,0\right)\hfill \\ y^{\prime },\left(\frac{1}{2},+,0\right),=,{\mathit{\alpha }}^{-1},y^{\prime },(\frac{1}{2}-0),-,\left(\mathit{\lambda },+,\mathit{\beta }\right),y,\left(\frac{1}{2},-,0\right),\hfill \end{array}\right.\end{array}$$(4) ), were studied in [16–18].

In the present paper, we consider the boundary value problem (1(1) $$\begin{array}{c}\hfill \ell y:=-y^{″}+q(x)y=\mathit{\lambda }y,x\in \mathrm{\Omega }=(0,1)\end{array}$$(1) )–(4(4) $$\begin{array}{c}\hfill \left\{\begin{array}{c}y,\left(\frac{1}{2},+,0\right),=,\mathit{\alpha },y,\left(\frac{1}{2},-,0\right)\hfill \\ y^{\prime },\left(\frac{1}{2},+,0\right),=,{\mathit{\alpha }}^{-1},y^{\prime },(\frac{1}{2}-0),-,\left(\mathit{\lambda },+,\mathit{\beta }\right),y,\left(\frac{1}{2},-,0\right),\hfill \end{array}\right.\end{array}$$(4) ) and solve inverse nodal problem to reconstruct the coefficients.

2 Preliminaries

Let a function $\mathit{\phi }(x,\mathit{\lambda })$ be the solution of (1(1) $$\begin{array}{c}\hfill \ell y:=-y^{″}+q(x)y=\mathit{\lambda }y,x\in \mathrm{\Omega }=(0,1)\end{array}$$(1) ) under the initial conditions(5) $$\begin{array}{c}\hfill \mathit{\phi }(0,\mathit{\lambda })=0,{\mathit{\phi }}^{\prime }(0,\mathit{\lambda })=1\end{array}$$(5) and the jump conditions (4(4) $$\begin{array}{c}\hfill \left\{\begin{array}{c}y,\left(\frac{1}{2},+,0\right),=,\mathit{\alpha },y,\left(\frac{1}{2},-,0\right)\hfill \\ y^{\prime },\left(\frac{1}{2},+,0\right),=,{\mathit{\alpha }}^{-1},y^{\prime },(\frac{1}{2}-0),-,\left(\mathit{\lambda },+,\mathit{\beta }\right),y,\left(\frac{1}{2},-,0\right),\hfill \end{array}\right.\end{array}$$(4) ). It can be calculated that $\mathit{\phi }(x,\mathit{\lambda })$ satisfies the following integral equations:

for $x<\frac{1}{2}$(6) $$\begin{array}{c}\hfill \mathit{\phi }(x,\mathit{\lambda })=\frac{sin\sqrt{\mathit{\lambda }}x}{\sqrt{\mathit{\lambda }}}+\frac{1}{\sqrt{\mathit{\lambda }}}{\int }_0^{x}sin\sqrt{\mathit{\lambda }}(x-t)q(t)\mathit{\phi }(t,\mathit{\lambda })dt,\end{array}$$(6) for $x>\frac{1}{2}$(7) $$\begin{array}{cc}\hfill \mathit{\phi }(x,\mathit{\lambda })& =\frac{\mathit{\lambda }+\mathit{\beta }}{2\mathit{\lambda }}\left(cos,\sqrt{\mathit{\lambda }},x,-,cos,\sqrt{\mathit{\lambda }},(1-x)\right)+{\mathit{\alpha }}^{+}\frac{sin\sqrt{\mathit{\lambda }}x}{\sqrt{\mathit{\lambda }}}+{\mathit{\alpha }}^{-}\frac{sin\sqrt{\mathit{\lambda }}(1-x)}{\sqrt{\mathit{\lambda }}}\hfill \\ \hfill & +\frac{\mathit{\lambda }+\mathit{\beta }}{2\mathit{\lambda }}{\int }_0^{1/2}\left(cos,\sqrt{\mathit{\lambda }},\left(x,-,t\right),-,cos,\sqrt{\mathit{\lambda }},(1-x-t)\right)q(t)\mathit{\phi }(t,\mathit{\lambda })dt\hfill \\ \hfill & +\frac{1}{\sqrt{\mathit{\lambda }}}{\int }_0^{1/2}\left({\mathit{\alpha }}^{+},sin,\sqrt{\mathit{\lambda }},\left(x,-,t\right),+,{\mathit{\alpha }}^{-},sin,\sqrt{\mathit{\lambda }},(1-x-t)\right)q(t)\mathit{\phi }(t,\mathit{\lambda })dt\hfill \\ \hfill & +\frac{1}{\sqrt{\mathit{\lambda }}}{\int }_{1/2}^{x}sin\sqrt{\mathit{\lambda }}\left(x,-,t\right)q(t)\mathit{\phi }(t,\mathit{\lambda })dt\hfill \end{array}$$(7) where ${\mathit{\alpha }}^{\pm }={\displaystyle \frac{1}{2}}\left(\mathit{\alpha },\pm ,{\displaystyle \frac{1}{\mathit{\alpha }}}\right)$. Using above integral equations, we can obtain the following asymptotic relations for $\left|\mathit{\lambda }\right|\to \infty .$

For $x<\frac{1}{2}$(8) $$\begin{array}{c}\hfill \mathit{\phi }(x,\mathit{\lambda })=\frac{sin\sqrt{\mathit{\lambda }}x}{\sqrt{\mathit{\lambda }}}-\frac{cos\sqrt{\mathit{\lambda }}x}{2\mathit{\lambda }}{\int }_0^{x}q(t)dt+o\left({\displaystyle \frac{1}{\mathit{\lambda }}},exp,\mathit{\tau },x\right),\end{array}$$(8) for $x>\frac{1}{2}$(9) $$\begin{array}{cc}\hfill \mathit{\phi }(x,\mathit{\lambda })& ={\displaystyle \frac{1}{2}}\left(cos,\sqrt{\mathit{\lambda }},x,-,cos,\sqrt{\mathit{\lambda }},(1-x)\right)\hfill \\ \hfill & +I_1(x)\frac{sin\sqrt{\mathit{\lambda }}x}{\sqrt{\mathit{\lambda }}}+I_2(x)\frac{sin\sqrt{\mathit{\lambda }}\left(1,-,x\right)}{\sqrt{\mathit{\lambda }}}+o\left({\displaystyle \frac{1}{\sqrt{\mathit{\lambda }}}},exp,\mathit{\tau },x\right)\hfill \end{array}$$(9) where$$\begin{array}{cc}\hfill I_1(x)& ={\mathit{\alpha }}^{+}+{\displaystyle \frac{1}{4}}{\int }_0^{x}q(t)dt,\hfill \\ \hfill I_2(x)& ={\mathit{\alpha }}^{-}-{\displaystyle \frac{1}{2}}{\int }_0^{1/2}q(t)dt+{\displaystyle \frac{1}{4}}{\int }_0^{x}q(t)dt\hfill \end{array}$$and $\mathit{\tau }=\left|I,m,\sqrt{\mathit{\lambda }}\right|$.

Let ${\left\{{\mathit{\lambda }}_n\right\}}_{n\ge 0}$ be the set of eigenvalues of (1(1) $$\begin{array}{c}\hfill \ell y:=-y^{″}+q(x)y=\mathit{\lambda }y,x\in \mathrm{\Omega }=(0,1)\end{array}$$(1) )–(4(4) $$\begin{array}{c}\hfill \left\{\begin{array}{c}y,\left(\frac{1}{2},+,0\right),=,\mathit{\alpha },y,\left(\frac{1}{2},-,0\right)\hfill \\ y^{\prime },\left(\frac{1}{2},+,0\right),=,{\mathit{\alpha }}^{-1},y^{\prime },(\frac{1}{2}-0),-,\left(\mathit{\lambda },+,\mathit{\beta }\right),y,\left(\frac{1}{2},-,0\right),\hfill \end{array}\right.\end{array}$$(4) ) and $\mathit{\phi }(x,{\mathit{\lambda }}_n)$ be the eigenfunction corresponding to the eigenvalue ${\mathit{\lambda }}_n.$ It is given in [16] that the numbers ${\mathit{\lambda }}_n$ are real and simple. Moreover, it can be proven using classical methods that the sequence ${\left\{{\mathit{\lambda }}_n\right\}}_{n\ge 0}$ satisfies the following asymptotic relation for $n\to \infty$:(10) $$\begin{array}{cc}\hfill \sqrt{{\mathit{\lambda }}_n}& =(n-1)\mathit{\pi }+\frac{A}{(n-1)\mathit{\pi }}+o\left({\displaystyle \frac{1}{n}}\right)\hfill \end{array}$$(10) (11) $$\begin{array}{cc}\hfill \frac{1}{\sqrt{{\mathit{\lambda }}_n}}& =\frac{1}{(n-1)\mathit{\pi }}-\frac{A}{{(n-1)}^2{\mathit{\pi }}^2}\frac{1}{(n-1)\mathit{\pi }}+o\left({\displaystyle \frac{1}{n^3}}\right)\hfill \end{array}$$(11) where $A=(w_1+{(-1)}^{n-1}{w}_2)$ and $w_1=2I_1(1)+H_0$, $w_2=2I_2(1)+H_0.$

3 Main results

Let $X=\left\{x_n^j,:,n,=,2,m,,,m,\in ,\mathbb{N}\right\}$ be the set of nodal points with even indexes of the eigenfunctions and $X_0$ be a dense subset of $X$ in $\left(0,,,1\right)$. In this section, we prove that the coefficients of the problem (1(1) $$\begin{array}{c}\hfill \ell y:=-y^{″}+q(x)y=\mathit{\lambda }y,x\in \mathrm{\Omega }=(0,1)\end{array}$$(1) )–(4(4) $$\begin{array}{c}\hfill \left\{\begin{array}{c}y,\left(\frac{1}{2},+,0\right),=,\mathit{\alpha },y,\left(\frac{1}{2},-,0\right)\hfill \\ y^{\prime },\left(\frac{1}{2},+,0\right),=,{\mathit{\alpha }}^{-1},y^{\prime },(\frac{1}{2}-0),-,\left(\mathit{\lambda },+,\mathit{\beta }\right),y,\left(\frac{1}{2},-,0\right),\hfill \end{array}\right.\end{array}$$(4) ) can be uniquely determined by $X_0.$

Lemma 1

The elements of $X_0$ satisfy the following asymptotic formulae for sufficiently large $n$, for $x_n^j\in \left(0,,,{\displaystyle \frac{1}{2}}\right)$(12) $$\begin{array}{c}\hfill x_n^j=\frac{j}{n-1}-\frac{w_1-w_2}{{\left(n,-,1\right)}^2{\mathit{\pi }}^2}\frac{j}{n-1}+\frac{{\int }_0^{x_n^j}q(t)dt}{2{\left(n,-,1\right)}^2{\mathit{\pi }}^2}+o\left(\frac{1}{n^2}\right),\end{array}$$(12) and for $x_n^j\in \left({\displaystyle \frac{1}{2}},,,1\right)$(13) $$\begin{array}{c}\hfill x_n^j=\frac{\left(j,+,1,/,2\right)}{n-1}-\frac{w_1-w_2}{{\left(n,-,1\right)}^2{\mathit{\pi }}^2}\frac{\left(j,+,1,/,2\right)}{n-1}+\frac{w_1-w_2+2I_1(x_n^j)+2I_2(x_n^j)}{2{\left(n,-,1\right)}^2{\mathit{\pi }}^2}+o\left(\frac{1}{n^2}\right).\end{array}$$(13)

Proof

Let $x_n^j\in \left({\displaystyle \frac{1}{2}},,,1\right).$ Use the asymptotic formula (9) to get$$\begin{array}{cc}\hfill 0& =\mathit{\phi }(x_n^j,{\mathit{\lambda }}_n)={\displaystyle \frac{1}{2}}\left(cos,\sqrt{{\mathit{\lambda }}_n},x_n^j,-,cos,\sqrt{{\mathit{\lambda }}_n},\left(1,-,x_n^j\right)\right)\hfill \\ \hfill & +I_1(x_n^j)\frac{sin\sqrt{{\mathit{\lambda }}_n}{x}_n^j}{\sqrt{{\mathit{\lambda }}_n}}+I_2(x_n^j)\frac{sin\sqrt{{\mathit{\lambda }}_n}\left(1,-,x_n^j\right)}{\sqrt{{\mathit{\lambda }}_n}}+o\left({\displaystyle \frac{1}{\sqrt{{\mathit{\lambda }}_n}}},exp,\mathit{\tau },x_n^j\right)\hfill \end{array}$$for $x_n^j>\frac{1}{2}.$ Therefore, it is calculated that$$\begin{array}{c}\hfill tan\left(\frac{\mathit{\pi }}{2},-,\sqrt{{\mathit{\lambda }}_n},x_n^j\right)={\displaystyle \frac{{(-1)}^{n-1}(w_1-w_2)-\frac{2I_1(x_n^j)}{\left(n,-,1\right)\mathit{\pi }}-\frac{2I_2(x_n^j)}{\left(n,-,1\right)\mathit{\pi }}+o\left({\displaystyle \frac{1}{n}}\right)}{2-\frac{2I_2(x_n^j)}{\left(n,-,1\right)\mathit{\pi }}(w_1-w_2)+o\left({\displaystyle \frac{1}{n}}\right)}}+o\left({\displaystyle \frac{1}{n}}\right)\end{array}$$This yields$$\begin{array}{cc}\hfill x_n^j& =\frac{\left(j,+,1,/,2\right)}{n-1}-\frac{w_1-w_2}{{\left(n,-,1\right)}^2{\mathit{\pi }}^2}\frac{\left(j,+,1,/,2\right)}{n-1}\hfill \\ \hfill & +{\displaystyle \frac{(w_1-w_2)+2I_1(x_n^j)+2I_2(x_n^j)}{2{\left(n,-,1\right)}^2{\mathit{\pi }}^2}}+o\left(\frac{1}{n^2}\right)\hfill \end{array}$$We complete the proof of (13(13) $$\begin{array}{c}\hfill x_n^j=\frac{\left(j,+,1,/,2\right)}{n-1}-\frac{w_1-w_2}{{\left(n,-,1\right)}^2{\mathit{\pi }}^2}\frac{\left(j,+,1,/,2\right)}{n-1}+\frac{w_1-w_2+2I_1(x_n^j)+2I_2(x_n^j)}{2{\left(n,-,1\right)}^2{\mathit{\pi }}^2}+o\left(\frac{1}{n^2}\right).\end{array}$$(13) ). The proof of (12(12) $$\begin{array}{c}\hfill x_n^j=\frac{j}{n-1}-\frac{w_1-w_2}{{\left(n,-,1\right)}^2{\mathit{\pi }}^2}\frac{j}{n-1}+\frac{{\int }_0^{x_n^j}q(t)dt}{2{\left(n,-,1\right)}^2{\mathit{\pi }}^2}+o\left(\frac{1}{n^2}\right),\end{array}$$(12) ) is similar.$\square$

For each fixed $x$ in $\left(0,,,1\right)$, there exists a sequence $\left(x_n^{j(n)}\right)$ which converges to $x.$ Therefore, we can show from Lemma 1 that the following limits exist and are finite:(14) $$\begin{array}{cc}& \underset{n\to \infty }{lim},{\left(n,-,1\right)}^2,{\mathit{\pi }}^2,\left(x_n^{j(n)},-,\frac{j(n)}{n-1}\right)\hfill \\ \hfill & =,-,(w_1-w_2),x,+,{\displaystyle \frac{1}{2}},{\int }_0^x,q,(t),d,t,,,x,\in ,\left(0,,,{\displaystyle \frac{1}{2}}\right)\hfill \end{array}$$(14) (15) $$\begin{array}{cc}& \underset{n\to \infty }{lim},{\left(n,-,1\right)}^2,{\mathit{\pi }}^2,\left(x_n^{j(n)},-,\frac{\left(j,(n),+,\frac{1}{2}\right)}{n-1}\right)\hfill \\ \hfill & =,-,(w_1-w_2),x,+,{\displaystyle \frac{(w_1-w_2)}{2}},+,I_1,(x),+,I_2,(x)\hfill \\ \hfill & =,-,(w_1-w_2),x,+,{\displaystyle \frac{(w_1-w_2)}{2}},+,{\mathit{\alpha }}^{+},+,{\mathit{\alpha }}^{-},+,{\displaystyle \frac{1}{2}},{\int }_{\frac{1}{2}}^x,q,(t),d,t,,,x,\in ,\left({\displaystyle \frac{1}{2}},,,1\right)\hfill \end{array}$$(15) Denote,(16) $$\begin{array}{c}\hfill F(x)=\left\{\begin{array}{c}{f}^{-}(x),x<{\displaystyle \frac{1}{2}}\\ f^{+}(x),x>{\displaystyle \frac{1}{2}}\end{array}\right.\end{array}$$(16) where $f^{-}(x)=-(w_1-w_2)x+{\displaystyle \frac{1}{2}}{\int }_0^{x}q(t)dt,$$f^{+}(x)=-(w_1-w_2)x+{\displaystyle \frac{(w_1-w_2)}{2}}+{\mathit{\alpha }}^{+}+{\mathit{\alpha }}^{-}+{\displaystyle \frac{1}{2}}{\int }_{\frac{1}{2}}^{x}q(t)dt.$

Consider the problem $\tilde{L}=L\left(\tilde{q},,,{\tilde{H}}_0,,,{\tilde{H}}_1,,,{\tilde{H}}_2,,,\tilde{\mathit{\alpha }}\right)$ under the same assumptions with $L.$ It is assumed in what follows that if a certain symbol $s$ denotes an object related to the problem $L$, then $\tilde{s}$ denotes the corresponding object related to the problem $\tilde{L}$.

Theorem 1

If $X_0={\tilde{X}}_0$ then $q(x)=\tilde{q}(x)$ a.e. in $\left(0,,,1\right)$, $H_0={\tilde{H}}_0,$$H_1={\tilde{H}}_1,$$H_2={\tilde{H}}_2$ and $\mathit{\alpha }=\tilde{\mathit{\alpha }}.$ Thus, the potential $q(x),$ a.e. in $\left(0,,,1\right)$, the coefficients $H_0,$$H_1,$$H_2$ and $\mathit{\alpha }$ are uniquely determined by the subset $X_0$. Moreover, $q(x)$ and $\mathit{\alpha }$ can be reconstructed by the following formulae:$$\begin{array}{cc}\hfill q(x)& =2\left[F^{\prime },(x),+,F,\left(\frac{1}{2},+,0\right),-,F,\left(\frac{1}{2},-,0\right),-,F,(1)\right]\hfill \\ \hfill \mathit{\alpha }& =F\left(\frac{1}{2},+,0\right).\hfill \end{array}$$

Proof

Direct calculations in (14(14) $$\begin{array}{cc}& \underset{n\to \infty }{lim},{\left(n,-,1\right)}^2,{\mathit{\pi }}^2,\left(x_n^{j(n)},-,\frac{j(n)}{n-1}\right)\hfill \\ \hfill & =,-,(w_1-w_2),x,+,{\displaystyle \frac{1}{2}},{\int }_0^x,q,(t),d,t,,,x,\in ,\left(0,,,{\displaystyle \frac{1}{2}}\right)\hfill \end{array}$$(14) )–(16(16) $$\begin{array}{c}\hfill F(x)=\left\{\begin{array}{c}{f}^{-}(x),x<{\displaystyle \frac{1}{2}}\\ f^{+}(x),x>{\displaystyle \frac{1}{2}}\end{array}\right.\end{array}$$(16) ) yield$$\begin{array}{cc}\hfill \mathit{\alpha }& =F\left(\frac{1}{2},+,0\right),\hfill \\ \hfill A& =w_1-w_2=F\left(\frac{1}{2},+,0\right)-F\left(\frac{1}{2},-,0\right)-F(1)\hfill \\ \hfill q(x)& =2A+2F^{\prime }(x).\hfill \end{array}$$Since $X_0={\tilde{X}}_0$, then $F(x)=\tilde{F}(x),$$A=\tilde{A}$ and so, $q(x)=\tilde{q}(x)$, $\mathit{\alpha }=\tilde{\mathit{\alpha }}.$

Let the function $\mathit{\psi }(x,\mathit{\lambda })$ be the solution of (1(1) $$\begin{array}{c}\hfill \ell y:=-y^{″}+q(x)y=\mathit{\lambda }y,x\in \mathrm{\Omega }=(0,1)\end{array}$$(1) ) under the initial conditions$$\begin{array}{c}\hfill \mathit{\psi }(1,\mathit{\lambda })=H_1-\mathit{\lambda },{\mathit{\psi }}^{\prime }(1,\mathit{\lambda })=\mathit{\lambda }{H}_0-H_2\end{array}$$and the jump conditions (4(4) $$\begin{array}{c}\hfill \left\{\begin{array}{c}y,\left(\frac{1}{2},+,0\right),=,\mathit{\alpha },y,\left(\frac{1}{2},-,0\right)\hfill \\ y^{\prime },\left(\frac{1}{2},+,0\right),=,{\mathit{\alpha }}^{-1},y^{\prime },(\frac{1}{2}-0),-,\left(\mathit{\lambda },+,\mathit{\beta }\right),y,\left(\frac{1}{2},-,0\right),\hfill \end{array}\right.\end{array}$$(4) ). It is clear that $\mathit{\psi }(x,{\mathit{\lambda }}_n)={\mathit{\chi }}_n\mathit{\phi }(x,{\mathit{\lambda }}_n),$ where ${\mathit{\chi }}_n={\mathit{\psi }}^{\prime }(0,{\mathit{\lambda }}_n).$

To complete the proof, consider a sequence $\left\{x_n^j\right\}\subset X_0$ that converges to $1$ and write Equation (1(1) $$\begin{array}{c}\hfill \ell y:=-y^{″}+q(x)y=\mathit{\lambda }y,x\in \mathrm{\Omega }=(0,1)\end{array}$$(1) ) for $\mathit{\psi }(x,{\mathit{\lambda }}_n)$ and $\tilde{\mathit{\psi }}(x,{\tilde{\mathit{\lambda }}}_n)$ as follows$$\begin{array}{cc}\hfill -{\tilde{\mathit{\psi }}}^{″}\left(x,,,{\tilde{\mathit{\lambda }}}_n\right)+q(x)\tilde{\mathit{\psi }}\left(x,,,{\tilde{\mathit{\lambda }}}_n\right)& ={\tilde{\mathit{\lambda }}}_n\tilde{\mathit{\psi }}\left(x,,,{\tilde{\mathit{\lambda }}}_n\right),\hfill \\ \hfill -{\mathit{\psi }}^{″}\left(x,,,{\mathit{\lambda }}_n\right)+q(x)\mathit{\psi }\left(x,,,{\mathit{\lambda }}_n\right)& ={\mathit{\lambda }}_n\mathit{\psi }\left(x,,,{\mathit{\lambda }}_n\right).\hfill \end{array}$$If these equations are (i): multiplied by $\mathit{\psi }\left(x,,,{\mathit{\lambda }}_n\right)$ and $\tilde{\mathit{\psi }}\left(x,,,{\tilde{\mathit{\lambda }}}_n\right)$, respectively; (ii): subtracted from each other and (iii): integrated over the interval $\left(x_n^j,,,1\right)$, the equality$$\begin{array}{c}\hfill {\mathit{\psi }}^{\prime }\left(1,,,{\mathit{\lambda }}_n\right)\tilde{\mathit{\psi }}\left(1,,,{\tilde{\mathit{\lambda }}}_n\right)-{\tilde{\mathit{\psi }}}^{\prime }\left(1,,,{\tilde{\mathit{\lambda }}}_n\right)\mathit{\psi }\left(1,,,{\mathit{\lambda }}_n\right)=\left({\tilde{\mathit{\lambda }}}_n,-,{\mathit{\lambda }}_n\right){\int }_{x_n^j}^1\tilde{\mathit{\psi }}\left(x,,,{\tilde{\mathit{\lambda }}}_n\right)\mathit{\psi }\left(x,,,{\mathit{\lambda }}_n\right)dx\end{array}$$is obtained. Using (10(10) $$\begin{array}{cc}\hfill \sqrt{{\mathit{\lambda }}_n}& =(n-1)\mathit{\pi }+\frac{A}{(n-1)\mathit{\pi }}+o\left({\displaystyle \frac{1}{n}}\right)\hfill \end{array}$$(10) ), we get the following estimate for sufficiently large $n$$$\begin{array}{cc}& {\left(n,-,1\right)}^4,{\mathit{\pi }}^4,\left({\tilde{H}}_0,-,H_0\right),\left[1,+,o,\left(\frac{1}{n}\right)\right]\hfill \\ \hfill & +,{\left(n,-,1\right)}^2,{\mathit{\pi }}^2,\left(H_0,{\tilde{H}}_1,-,{\tilde{H}}_2,-,{\tilde{H}}_0,H_1,+,H_2\right),\left[1,+,o,\left(\frac{1}{n^2}\right)\right]\hfill \\ \hfill & +,H_1,{\tilde{H}}_2,-,{\tilde{H}}_1,H_2,=,o,(1),.\hfill \end{array}$$The last equality yields $H_0={\tilde{H}}_0,$$\left({\tilde{H}}_0,{\tilde{H}}_1,-,{\tilde{H}}_2\right)\left(H_2,-,{\tilde{H}}_2\right)=0$and $H_2{\tilde{H}}_1={\tilde{H}}_2{H}_1.$ Since ${\tilde{H}}_0{\tilde{H}}_1-{\tilde{H}}_2>0$, $H_2={\tilde{H}}_2$ and so $H_1={\tilde{H}}_1.$ This completes the proof.$\square$

Example 1

Consider the BVP$$\begin{array}{c}\hfill L:\left\{\begin{array}{c}\ell ,y,:,=,-,y^{″},+,q,(x),y,=,\mathit{\lambda },y,,,x,\in ,\left(0,,,\frac{1}{2}\right),\cup ,\left(\frac{1}{2},,,1\right)\\ y(0)=0\\ \mathit{\lambda }(y^{\prime }(\mathit{\pi })+H_{0}y(\mathit{\pi }))-H_1{y}^{\prime }(\mathit{\pi })-H_{2}y(\mathit{\pi })=0\\ y,\left(\frac{1}{2},+,0\right),=,\mathit{\alpha },y,\left(\frac{1}{2},-,0\right)\\ y^{\prime },\left(\frac{1}{2},+,0\right),=,{\mathit{\alpha }}^{-1},y^{\prime },\left(\frac{1}{2},-,0\right),-,\left(\mathit{\lambda },+,\mathit{\beta }\right),y,\left(\frac{1}{2},-,0\right)\end{array}\right.\end{array}$$where $q(x)\in L_2(0,1)$ and $\mathit{\alpha }$ are unknown coefficients that confirm to the assumptions of the problem (1(1) $$\begin{array}{c}\hfill \ell y:=-y^{″}+q(x)y=\mathit{\lambda }y,x\in \mathrm{\Omega }=(0,1)\end{array}$$(1) )–(4(4) $$\begin{array}{c}\hfill \left\{\begin{array}{c}y,\left(\frac{1}{2},+,0\right),=,\mathit{\alpha },y,\left(\frac{1}{2},-,0\right)\hfill \\ y^{\prime },\left(\frac{1}{2},+,0\right),=,{\mathit{\alpha }}^{-1},y^{\prime },(\frac{1}{2}-0),-,\left(\mathit{\lambda },+,\mathit{\beta }\right),y,\left(\frac{1}{2},-,0\right),\hfill \end{array}\right.\end{array}$$(4) ). Let $X_0=\left\{x_n^j,:,n,=,2,m\right\}$ be the given subset of nodal points which satisfy the following asymptotics$$x_n^j=\left\{\begin{array}{c}{\displaystyle \frac{j}{n-1}-\frac{4}{{\left(n,-,1\right)}^2{\mathit{\pi }}^2}\frac{j}{n-1}+\frac{1}{2{\left(n,-,1\right)}^2{\mathit{\pi }}^2}\frac{j}{n-1}\left(1,-,\frac{j}{n-1}\right)}\hfill \\ +o\left(\frac{1}{n^2}\right),x<{\displaystyle \frac{1}{2}}\hfill \\ {\displaystyle \frac{j+1/2}{n-1}-\frac{4}{{\left(n,-,1\right)}^2{\mathit{\pi }}^2}\frac{j+1/2}{n-1}+\frac{\frac{31}{4}}{2{\left(n,-,1\right)}^2{\mathit{\pi }}^2}\frac{j+1/2}{n-1}\left(1,-,\frac{j+1/2}{n-1}\right)}\hfill \\ +o\left(\frac{1}{n^2}\right),x>{\displaystyle \frac{1}{2}}.\hfill \end{array}\right.$$One can calculate that$$\begin{array}{cc}\hfill F(x)& =\left\{\begin{array}{c}{f}^{-}(x),x<{\displaystyle \frac{1}{2}}\\ f^{+}(x),x>{\displaystyle \frac{1}{2}}\end{array}\right.\hfill \\ \hfill f^{-}(x)& =-4x+\frac{1}{2}x(1-x),\hfill \\ \hfill f^{+}(x)& =-4x+\frac{31}{8}+\frac{1}{2}x(1-x)\hfill \end{array}$$According to Theorem 1, we find$$\begin{array}{cc}& \mathit{\alpha },=,F,\left(\frac{1}{2},+,0\right),=,2,,\hfill \\ \hfill & A,=,F,\left(\frac{1}{2},+,0\right),-,F,\left(\frac{1}{2},-,0\right),-,F,(1),=,4\hfill \\ \hfill & F^{\prime },(x),=,-,A,+,\frac{1}{2},q,(x),=,-,4,+,\frac{1}{2},q,(x)\hfill \\ \hfill & q,(x),=,2,F^{\prime },(x),+,F,\left(\frac{1}{2},+,0\right),-,F,\left(\frac{1}{2},-,0\right),-,F,(1),=,1,-,2,x,.\hfill \end{array}$$

Acknowledgements

The authors would like to thank the referees for their valuable comments which helped to improve the manuscript.