Inverse problems for the impulsive Sturm–Liouville operator with jump conditions

ABSTRACT

The inverse problem for impulsive Sturm–Liouville operators with discontinuity conditions is considered. We have shown that all parameters used in the boundary conditions as well as $q\left(x\right)$ can be uniquely established by a set of values of eigenfunctions at the mid-point $x=\pi /2$ and one spectrum. Moreover, we discuss Gesztesy–Simon theorem and show that if the potential function $q\left(x\right)$ is prescribed on the interval $\left[\pi /2\right(1-\alpha ),\pi ]$ for some $\alpha \in (0,1)$, then parts of a finite number of spectra suffice to determine $q\left(x\right)$ on $[0,\pi ]$.

KEYWORDS:

• Inverse problem
• impulsive Sturm–Liouville operator
• discontinuity

2010 MSC:

• 34A36
• 34B24
• 65L09

1. Introduction

Inverse problem theory is a research field that the coefficients applied in the boundary value problems are determined by some information about the spectral data. These kinds of problems can be seen in various branches of sciences like geophysics, seismic tomography, optics and graph theory (see e.g. [1–4]). The more detailed studies on inverse problems for several classes of Sturm–Liouville operators can be found in some articles [5–8].

We denote the boundary value problem L generated by differential equation (1) $-y^{\mathrm{\prime \prime }}+q\left(x\right)y=\lambda r\left(x\right)y,x\in (0,\pi ),$(1) subject to the boundary conditions (2) $\begin{array}{rl}U\left(y\right)& :=y^{\mathrm{\prime }}\left(0\right)-hy\left(0\right)=0,\end{array}$(2) (3) $\begin{array}{rl}V\left(y\right)& :=y^{\mathrm{\prime }}\left(\pi \right)+Hy\left(\pi \right)=0,\end{array}$(3) and the jump conditions (4) $y\left(\frac{\pi }{2}+0,\rho \right)=a_{1}y\left(\frac{\pi }{2}-0,\rho \right),y^{\mathrm{\prime }}\left(\frac{\pi }{2}+0,\rho \right)=a_1^{-1}{y}^{\mathrm{\prime }}\left(\frac{\pi }{2}-0,\rho \right)+a_{2}y\left(\frac{\pi }{2}-0,\rho \right).$(4) Here $r\left(x\right)=\left\{\begin{array}{l}{\alpha }_1^2,x<\frac{\pi }{2},\\ {\alpha }_2^2,x>\frac{\pi }{2},\end{array}\right.$ in which ${\alpha }_1<{\alpha }_2$ has been assumed. The parameters $h,H,a_1,a_2$ are real, $\lambda ={\rho }^2$ is a spectral parameter and the real-valued function $q\left(x\right)\in L^2(0,\pi )$.

Inverse problems for discontinuous Sturm–Liouville equations were investigated by a number of scholars [9–14]. The methods of Gelfand-Levitan, spectral mappings, zeros of the eigenfunctions (nodal points) and half inverse are techniques that are used in the investigation of the inverse problem (see [6,7,15,16]). The inverse problem for interior spectral data is another method that many authors have been applied for studying the inverse problem in the next years [17,18]. We have to mention that two spectra of two different boundary value problems are applied to reconstruct the parameter used in the boundary value problem in the usual case in this method. The taking of one spectrum and some information of eigenfunctions at the midpoint of the interval was Mochizuki and Trooshin's initiative in solving of the inverse problem [17]. This method has been applied by some researchers in the works [7,14,19,20]. As far as we know interior inverse problems for the boundary value problem (1(1) $-y^{\mathrm{\prime \prime }}+q\left(x\right)y=\lambda r\left(x\right)y,x\in (0,\pi ),$(1) )–(4(4) $y\left(\frac{\pi }{2}+0,\rho \right)=a_{1}y\left(\frac{\pi }{2}-0,\rho \right),y^{\mathrm{\prime }}\left(\frac{\pi }{2}+0,\rho \right)=a_1^{-1}{y}^{\mathrm{\prime }}\left(\frac{\pi }{2}-0,\rho \right)+a_{2}y\left(\frac{\pi }{2}-0,\rho \right).$(4) ) have not been studied yet. We are going to apply this method for studying the discontinuous Sturm–Liouville operators with the weight function. We also prove that parts of one spectrum can present the function $q\left(x\right)$ on the whole interval provided that this function is given on $\left[\pi /2\right(1-\alpha ),\pi ]$ for some $\alpha \in (0,1)$ which is a generalization of Gesztesy–Simon type theorem. In [21] Gesztesy and Simon have showed that if $q\left(x\right)$ is given on $[0,\frac{1}{2}+\alpha /2]$ for some $\alpha \in [0,1)$ a priori, then parts of one spectrum can establish $q\left(x\right)$ on the whole interval.

In this article we investigate two inverse problems for L. The purpose of this paper is to study the inverse problem for L by Mochizuki–Trooshin and Gesztesy–Simon methods. The techniques applied in this paper are based on those expressed in [17,21]. We establish the coefficients of the boundary value problem L by one spectrum and a set of values of eigenfunctions in a interior point. Moreover, we investigate Gesztesy–Simon type theorem for the inverse problem of L.

2. Preliminaries

We consider the function $y(x,\rho )$ as the solution of the equation (1(1) $-y^{\mathrm{\prime \prime }}+q\left(x\right)y=\lambda r\left(x\right)y,x\in (0,\pi ),$(1) ) under the initial conditions $y(0,\rho )=1$ and $y^{\prime }(0,\rho )=h$ and the jump conditions (4(4) $y\left(\frac{\pi }{2}+0,\rho \right)=a_{1}y\left(\frac{\pi }{2}-0,\rho \right),y^{\mathrm{\prime }}\left(\frac{\pi }{2}+0,\rho \right)=a_1^{-1}{y}^{\mathrm{\prime }}\left(\frac{\pi }{2}-0,\rho \right)+a_{2}y\left(\frac{\pi }{2}-0,\rho \right).$(4) ). This solution satisfies the following integral equation: (5) $\begin{array}{rl}y(x,\rho )& =A\left(\rho \right)\cos \rho {\alpha }_{1}x+B\left(\rho \right)\sin \rho {\alpha }_{1}x+{\int }_0^x\frac{\sin \rho {\alpha }_1(x-t)}{{\alpha }_1\rho }q\left(t\right)y(t,\rho )\mathrm{d}t,x<\frac{\pi }{2},\end{array}$(5) (6) $\begin{array}{rl}y(x,\rho )& =A\left(\rho \right)\cos \rho \gamma \left(x\right)+B\left(\rho \right)\sin \rho \gamma \left(x\right)+{\int }_0^x\frac{K(x,t,\rho )}{\rho }q\left(t\right)y(t,\rho )\mathrm{d}t,x>\frac{\pi }{2},\end{array}$(6) where $K(x,t,\rho )=\left\{\begin{array}{l}\begin{array}{c}\frac{1}{\sqrt{r\left(t\right)}}\left(b_{+}\sin \rho \left(\gamma \right(x)-\gamma (t\left)\right)\phantom{}\right.\\ +\left.b_{-}\sin \rho \left(2\gamma \left(\frac{\pi }{2}\right)-\gamma \left(x\right)-\gamma \left(t\right)\right)\right),\end{array}0<t<\frac{\pi }{2},\\ \frac{1}{\sqrt{r\left(t\right)}}\sin \rho \left(\gamma \right(x)-\gamma (t\left)\right),\frac{\pi }{2}<t<x,\end{array}\right.$ and $\gamma \left(x\right)={\int }_0^x\sqrt{r\left(t\right)}\mathrm{d}t$. The following asymptotic relations as sufficiently large ρ, can be obtained for above mentioned integral equations (7) $\begin{array}{rl}y(x,\rho )& =\cos \rho {\alpha }_{1}x+\left(h+\frac{1}{2{\alpha }_1}{\int }_0^{x}q\left(t\right)\mathrm{d}t\right)\frac{\sin \rho {\alpha }_{1}x}{\rho }\\ & +o\left(\frac{1}{\rho }\exp \left(|\mathrm{Im}\rho |{\alpha }_{1}x\right)\right),x<\frac{\pi }{2},\end{array}$(7) (8) $\begin{array}{rl}y(x,\rho )& =b_{+}\cos \rho \gamma \left(x\right)+b_{-}\cos \rho \left(2\gamma \left(\frac{\pi }{2}\right)-\gamma \left(x\right)\right)\\ & +\left(b_{+}\left(h+\frac{1}{2}{\int }_0^x\frac{q\left(t\right)}{\sqrt{r\left(t\right)}}\mathrm{d}t\right)+\frac{a_2}{2{\alpha }_2}\right)\frac{\sin \rho \gamma \left(x\right)}{\rho }\\ & +\left(b_{-}\left(h-\frac{1}{2}{\int }_0^x\frac{q\left(t\right)}{\sqrt{r\left(t\right)}}\mathrm{d}t+\frac{1}{{\alpha }_1}{\int }_0^{\pi /2}q\left(t\right)\mathrm{d}t\right)-\frac{a_2}{2{\alpha }_2}\right)\\ & \times \frac{\sin \rho \left(2\gamma \left(\frac{\pi }{2}\right)-\gamma \left(x\right)\right)}{\rho }+o\left(\frac{1}{\rho }\exp \left(|\mathrm{Im}\rho |\gamma \right(x\left)\right)\right),x>\frac{\pi }{2}.\end{array}$(8) Here $b_{\pm }=\frac{1}{2}(a_1\pm {\alpha }_1/{\alpha }_2{a}_1)$.

We define (9) $\mathrm{\Delta }\left(\rho \right):=-V\left(y\right(x,\rho \left)\right).$(9) as the characteristic function of L where is entire in ρ. The zeros of this function coincide with the eigenvalues of L. It is a result of (3(3) $\begin{array}{rl}V\left(y\right)& :=y^{\mathrm{\prime }}\left(\pi \right)+Hy\left(\pi \right)=0,\end{array}$(3) ), (8(8) $\begin{array}{rl}y(x,\rho )& =b_{+}\cos \rho \gamma \left(x\right)+b_{-}\cos \rho \left(2\gamma \left(\frac{\pi }{2}\right)-\gamma \left(x\right)\right)\\ & +\left(b_{+}\left(h+\frac{1}{2}{\int }_0^x\frac{q\left(t\right)}{\sqrt{r\left(t\right)}}\mathrm{d}t\right)+\frac{a_2}{2{\alpha }_2}\right)\frac{\sin \rho \gamma \left(x\right)}{\rho }\\ & +\left(b_{-}\left(h-\frac{1}{2}{\int }_0^x\frac{q\left(t\right)}{\sqrt{r\left(t\right)}}\mathrm{d}t+\frac{1}{{\alpha }_1}{\int }_0^{\pi /2}q\left(t\right)\mathrm{d}t\right)-\frac{a_2}{2{\alpha }_2}\right)\\ & \times \frac{\sin \rho \left(2\gamma \left(\frac{\pi }{2}\right)-\gamma \left(x\right)\right)}{\rho }+o\left(\frac{1}{\rho }\exp \left(|\mathrm{Im}\rho |\gamma \right(x\left)\right)\right),x>\frac{\pi }{2}.\end{array}$(8) ) and (9(9) $\mathrm{\Delta }\left(\rho \right):=-V\left(y\right(x,\rho \left)\right).$(9) ) for $|\rho |\to \mathrm{\infty },$ (10) $\mathrm{\Delta }\left(\rho \right)=\rho {\alpha }_2\left(b_{+}\sin \rho ({\alpha }_1+{\alpha }_2)\frac{\pi }{2}+b_{-}\sin \rho ({\alpha }_1-{\alpha }_2)\frac{\pi }{2}\right)+O\left(\exp \left(|\mathrm{Im}\rho |\gamma \right(\pi \left)\right)\right),$(10) and the eigenvalues ${\rho }_n$ admit the classical asymptotic form (11) ${\rho }_n=\frac{2n}{{\alpha }_1+{\alpha }_2}+O\left(1\right).$(11)

Let be $\delta >0$ and fixed. Define $G_{\delta }:=\{\rho \in \mathbb{C};\mid \rho -{\rho }_n\mid \ge \delta ,\mathrm{\forall }n\}$. The following inequality can be deduced using the asymptotic formula for $\mathrm{\Delta }\left(\rho \right)$, (12) $\mathrm{\Delta }\left(\rho \right)\ge C_1|\rho |\exp \left(({\alpha }_1+{\alpha }_2)|\mathrm{Im}\rho |\frac{\pi }{2}\right),\rho \in G_{\delta },$(12) for some positive constant $C_1$.

In the next sections, we describe two uniqueness theorems that are the main results of this paper. For this purpose, we assume that together with $L=L(q,h,H),$ $\tilde{L}=L(\tilde{q},\tilde{h},\tilde{H})$ be a boundary value problem of the form (1(1) $-y^{\mathrm{\prime \prime }}+q\left(x\right)y=\lambda r\left(x\right)y,x\in (0,\pi ),$(1) )–(4(4) $y\left(\frac{\pi }{2}+0,\rho \right)=a_{1}y\left(\frac{\pi }{2}-0,\rho \right),y^{\mathrm{\prime }}\left(\frac{\pi }{2}+0,\rho \right)=a_1^{-1}{y}^{\mathrm{\prime }}\left(\frac{\pi }{2}-0,\rho \right)+a_{2}y\left(\frac{\pi }{2}-0,\rho \right).$(4) ) with different coefficients. The symbol with tilde denotes the object in $\tilde{L}$ that the same object is denoted with same symbol without tilde in L.

3. Mochizuki–Trooshin's method for BVP(L)

Now, we express the Mochizuki–Trooshin's theorem and demonstrate its proof. Firstly, we denote $<y,z>\left(x\right):=y\left(x\right)z^{\prime }\left(x\right)-y^{\prime }\left(x\right)z\left(x\right),$ as the Wronskian of two continuously differentiable functions $y\left(x\right)$ and $z\left(x\right)$ on the intervals $[0,\pi /2)$ and $(\pi /2,\pi ]$.

Theorem 3.1

If for any $n\in \mathbb{N},$ ${\lambda }_n={\tilde{\lambda }}_n,<y_n,{\tilde{y}}_n{>}_{x=\pi /2-0}=0,$ then $q\left(x\right)=\tilde{q}\left(x\right)$ a. e. on $[0,\pi ]$ and $h=\tilde{h},$ $H=\tilde{H}$.

Proof.

Our proof starts with the observation that the solution $y(x,\rho )$ satisfies the integral representation as follows (see [9,12,15,22]): (13) $y(x,\rho )=\cos \rho {\alpha }_{1}x+{\int }_0^{x}A(x,t)\cos \rho {\alpha }_{1}t\mathrm{d}t,x\in \left(0,\frac{\pi }{2}\right).$(13) Then we have (14) $y(x,\rho )\tilde{y}(x,\rho )=\frac{1}{2}\left(\cos 2\rho {\alpha }_{1}x+1\right)+{\int }_0^{x}B(x,t)\cos 2\rho {\alpha }_{1}t\mathrm{d}t,x\in \left(0,\frac{\pi }{2}\right),$(14) where $A(x,t)$ and $B(x,t)$ are two continuous functions which do not depend on ρ.

According to the considered assumptions for $y(x,\rho )$, we can get (15) $-y^{\mathrm{\prime \prime }}+q\left(x\right)y=\lambda r\left(x\right)y,y\left(0\right)=1,y^{\prime }\left(0\right)=h,$(15) and similarly we have (16) $-{\tilde{y}}^{\mathrm{\prime \prime }}+q\left(x\right)\tilde{y}=\lambda r\left(x\right)\tilde{y},y\left(0\right)=1,y^{\prime }\left(0\right)=\tilde{h}.$(16) After multiplying (15(15) $-y^{\mathrm{\prime \prime }}+q\left(x\right)y=\lambda r\left(x\right)y,y\left(0\right)=1,y^{\prime }\left(0\right)=h,$(15) ) by $\tilde{y}(x,\rho )$ and (16(16) $-{\tilde{y}}^{\mathrm{\prime \prime }}+q\left(x\right)\tilde{y}=\lambda r\left(x\right)\tilde{y},y\left(0\right)=1,y^{\prime }\left(0\right)=\tilde{h}.$(16) ) by $y(x,\rho )$, we subtract these equations from each other. Then by integrating on $[0,\pi /2)$, we will have (17) ${\int }_0^{\pi /2}(q\left(x\right)-\tilde{q}\left(x\right))y\left(x\right)\tilde{y}\left(x\right)\mathrm{d}x=\left(y^{\mathrm{\prime }}\right(x\left)\tilde{y}\right(x)-y(x\left){\tilde{y}}^{\mathrm{\prime }}\right(x\left)\right){|}_0^{\pi /2}.$(17) Therefore, (18) ${\int }_0^{\pi /2}(q\left(x\right)-\tilde{q}\left(x\right))y\left(x\right)\tilde{y}\left(x\right)\mathrm{d}x+h-\tilde{h}=y^{\mathrm{\prime }}\left(\frac{\pi }{2}\right)\tilde{y}\left(\frac{\pi }{2}\right)-y\left(\frac{\pi }{2}\right){\tilde{y}}^{\mathrm{\prime }}\left(\frac{\pi }{2}\right).$(18) With assumption (19) $H\left(\rho \right):={\int }_0^{\pi /2}(q\left(x\right)-\tilde{q}\left(x\right))y\left(x\right)\tilde{y}\left(x\right)\mathrm{d}x+h-\tilde{h},$(19) one can write (20) $H\left(\rho \right)=y^{\mathrm{\prime }}\left(\frac{\pi }{2}\right)\tilde{y}\left(\frac{\pi }{2}\right)-y\left(\frac{\pi }{2}\right){\tilde{y}}^{\mathrm{\prime }}\left(\frac{\pi }{2}\right).$(20)

Taking the assumptions of the theorem, we arrive at $H\left({\rho }_n\right)=0$. In the sequel, we have to prove that $H\left(\rho \right)=0$ for values of the non-eigenvalue ρ.

It comes from (14(14) $y(x,\rho )\tilde{y}(x,\rho )=\frac{1}{2}\left(\cos 2\rho {\alpha }_{1}x+1\right)+{\int }_0^{x}B(x,t)\cos 2\rho {\alpha }_{1}t\mathrm{d}t,x\in \left(0,\frac{\pi }{2}\right),$(14) ) that for $x\in (0,\pi /2)$ (21) $|y(x,\rho )\tilde{y}(x,\rho )|\le C_2\exp \left(|\mathrm{Im}\rho |2{\alpha }_{1}x\right),$(21) for some positive constant $C_2$. Thus (22) $|H\left(\rho \right)|\le C_2\exp \left(|\mathrm{Im}\rho |{\alpha }_1\pi \right).$(22)

Now, we define (23) $\phi \left(\rho \right)=\frac{H\left(\rho \right)}{\mathrm{\Delta }\left(\rho \right)}.$(23) Using (12(12) $\mathrm{\Delta }\left(\rho \right)\ge C_1|\rho |\exp \left(({\alpha }_1+{\alpha }_2)|\mathrm{Im}\rho |\frac{\pi }{2}\right),\rho \in G_{\delta },$(12) ), (22(22) $|H\left(\rho \right)|\le C_2\exp \left(|\mathrm{Im}\rho |{\alpha }_1\pi \right).$(22) ) and Liouville's theorem [23], we get that the entire function $\phi \left(\rho \right)$ is zero for all ρ and then $H\left(\rho \right)=0$ for all ρ.

Let us consider $Q\left(x\right)=q\left(x\right)-\tilde{q}\left(x\right)$. From (14(14) $y(x,\rho )\tilde{y}(x,\rho )=\frac{1}{2}\left(\cos 2\rho {\alpha }_{1}x+1\right)+{\int }_0^{x}B(x,t)\cos 2\rho {\alpha }_{1}t\mathrm{d}t,x\in \left(0,\frac{\pi }{2}\right),$(14) ) and (19(19) $H\left(\rho \right):={\int }_0^{\pi /2}(q\left(x\right)-\tilde{q}\left(x\right))y\left(x\right)\tilde{y}\left(x\right)\mathrm{d}x+h-\tilde{h},$(19) ), we can write $\begin{array}{rl}& {\int }_0^{\pi /2-0}Q\left(x\right)g[\cos 2\rho {\alpha }_{1}x+1g]\mathrm{d}x\\ & +{\int }_0^{\pi /2-0}Q\left(x\right)g[{\int }_0^{x}B(x,t)\cos 2\rho {\alpha }_{1}t\mathrm{d}tg]\mathrm{d}x+h-\tilde{h}=0.\end{array}$ Rewriting the above relation, we will have ${\int }_0^{\pi /2-0}Q\left(x\right)\mathrm{d}x+{\int }_0^{\pi /2-0}\cos 2\rho {\alpha }_{1}tg[Q\left(t\right)+{\int }_t^{\pi /2-0}Q\left(x\right)B(x,t)\mathrm{d}xg]\mathrm{d}t+h-\tilde{h}=0.$ For $|\rho |\to \mathrm{\infty },$ we can conclude the following result from the Riemann–Lebesgue lemma (24) $\begin{array}{rl}& {\int }_0^{\pi /2-0}\cos 2\rho {\alpha }_{1}tg[Q\left(t\right)+{\int }_t^{\pi /2-0}Q\left(x\right)B(x,t)\mathrm{d}xg]\mathrm{d}t=0,\end{array}$(24) (25) $\begin{array}{rl}& {\int }_0^{\pi /2-0}Q\left(x\right)\mathrm{d}x+h-\tilde{h}=0.\end{array}$(25) Since the function ‘cos’ is complete [15], we see that (26) $Q\left(t\right)+{\int }_t^{\pi /2-0}Q\left(x\right)B(x,t)\mathrm{d}x=0.$(26) Because of the zero solution is only the solution of the above homogeneous Volterra integral equation, $Q\left(x\right)=0$ on $[0,\pi /2]$. This means that $q\left(x\right)=\tilde{q}\left(x\right)$ almost everywhere on $[0,\pi /2]$. Also from (25(25) $\begin{array}{rl}& {\int }_0^{\pi /2-0}Q\left(x\right)\mathrm{d}x+h-\tilde{h}=0.\end{array}$(25) ), we can result that $h=\tilde{h}$.

To prove the problem on the interval $(\pi /2,\pi )$, it is enough to consider the supplementary problem $\hat{L}$ for $x\in (0,\pi )$ (27) $\begin{array}{rl}& -y^{\mathrm{\prime \prime }}+q_1\left(x\right)y=\lambda r_1\left(x\right)y,q_1\left(x\right)=q(\pi -x),r_1\left(x\right)=r(\pi -x),\end{array}$(27) (28) $\begin{array}{rl}& U\left(y\right):=y^{\mathrm{\prime }}\left(0\right)-Hy\left(0\right)=0,\end{array}$(28) (29) $\begin{array}{rl}& V\left(y\right):=y^{\mathrm{\prime }}\left(\pi \right)+hy\left(\pi \right)=0,\end{array}$(29) (30) $\begin{array}{rl}& y\left(\frac{\pi }{2}+0,\rho \right)=a_1^{-1}y\left(\frac{\pi }{2}-0,\rho \right),\\ & y^{\mathrm{\prime }}\left(\frac{\pi }{2}+0,\rho \right)=a_1{y}^{\mathrm{\prime }}\left(\frac{\pi }{2}-0,\rho \right)+a_1{a}_{2}y\left(\frac{\pi }{2}-0,\rho \right).\end{array}$(30) Considering $\hat{y}\left(x\right):=y(\pi -x)$ as the solution of the supplementary problem $\hat{L}$, we can write ${\hat{y}}_n(\pi /2-0):=y_n(\pi /2+0)$ and the assumptions of the theorem are true. Therefore, in the previous same manner we can give $Q_1\left(x\right)=Q(\pi -x)=0$ on $(0,\pi /2)$ and so $q\left(x\right)=\tilde{q}\left(x\right)$ a. e. on $[\pi /2,\pi ]$ and $H=\tilde{H}$. This completes the proof.

Remark 3.2

Let the functions $y\left(x\right)$ and $z\left(x\right)$ be the solutions of L. Since $<y,z{>}_{{|}_{\pi /2-0}}=<y,z{>}_{{|}_{\pi /2+0}}$, we can replace $<y,z{>}_{{|}_{\pi /2-0}}=0$ by $<y,z{>}_{{|}_{\pi /2+0}}=0$ in Theorem 3.1. In other words, we can give Theorem 3.1 by $<y,z{>}_{{|}_{\pi /2+0}}=0$ instead of $<y,z{>}_{{|}_{\pi /2-0}}=0$.

4. Gesztesy–Simon's method for BVP(L)

Finally, we state the Gesztesy–Simon's theorem and bring its proof.

Theorem 4.1

Let $\sigma \left(L\right)$ be the spectrum of L. If for some $\alpha \in (0,1),$ $q\left(x\right)=\tilde{q}\left(x\right),x\in \left[\pi /2(1-\alpha ),\pi \right],H=\tilde{H},$ and $S\subseteq \sigma \left(L\right)\cap \sigma \left(\tilde{L}\right)$ satisfying $\mathrm{♯}\{\lambda \in S;\lambda \le {\lambda }_0\}\ge (1-\alpha )\mathrm{♯}\{\lambda \in \sigma (L);\lambda \le {\lambda }_0\}+\frac{\alpha }{2},$ for all sufficiently large ${\lambda }_0\in \mathbb{R},$ then $q\left(x\right)=\tilde{q}\left(x\right),\mathrm{a}.\mathrm{e}.x\in [0,\pi ],h=\tilde{h}.$

Proof.

If we multiply (1(1) $-y^{\mathrm{\prime \prime }}+q\left(x\right)y=\lambda r\left(x\right)y,x\in (0,\pi ),$(1) ) by $\tilde{y}(x,\rho )$, and Equation (1(1) $-y^{\mathrm{\prime \prime }}+q\left(x\right)y=\lambda r\left(x\right)y,x\in (0,\pi ),$(1) ) corresponding with tilde by $y(x,\rho )$ and subtract, then integrate on $[0,\pi ]$, we can infer (31) $\begin{array}{rl}{\int }_0^{\pi }(q\left(x\right)-\tilde{q}\left(x\right))y\left(x\right)\tilde{y}\left(x\right)\mathrm{d}x& =\left(y^{\mathrm{\prime }}\right(x\left)\tilde{y}\right(x)-y(x\left){\tilde{y}}^{\mathrm{\prime }}\right(x\left)\right){|}_0^{\pi /2-0}\\ & +\left(y^{\mathrm{\prime }}\right(x\left)\tilde{y}\right(x)-y(x\left){\tilde{y}}^{\mathrm{\prime }}\right(x\left)\right){|}_{\pi /2+0}^{\pi }.\end{array}$(31) Because $q\left(x\right)=\tilde{q}\left(x\right),x\in \left[\pi /2\right(1-\alpha ),\pi ]$, we can write (32) ${\int }_0^{\pi /2(1-\alpha )}(q\left(x\right)-\tilde{q}\left(x\right))y\left(x\right)\tilde{y}\left(x\right)\mathrm{d}x=\tilde{h}-h+y^{\mathrm{\prime }}\left(\pi \right)\tilde{y}\left(\pi \right)-y\left(\pi \right){\tilde{y}}^{\mathrm{\prime }}\left(\pi \right).$(32) Assuming (33) $H_1\left(\rho \right):={\int }_0^{\pi /2(1-\alpha )}(q\left(x\right)-\tilde{q}\left(x\right))y\left(x\right)\tilde{y}\left(x\right)\mathrm{d}x+h-\tilde{h},$(33) we will have (34) $H_1\left(\rho \right)=y^{\mathrm{\prime }}\left(\pi \right)\tilde{y}\left(\pi \right)-y\left(\pi \right){\tilde{y}}^{\mathrm{\prime }}\left(\pi \right).$(34)

According to the conditions of the theorem, we obtain $H_1\left({\rho }_n\right)=0$ for ${\rho }_n\in S$. Now we should prove that $H_1\left(\rho \right)=0$ for all ρ.

From (14(14) $y(x,\rho )\tilde{y}(x,\rho )=\frac{1}{2}\left(\cos 2\rho {\alpha }_{1}x+1\right)+{\int }_0^{x}B(x,t)\cos 2\rho {\alpha }_{1}t\mathrm{d}t,x\in \left(0,\frac{\pi }{2}\right),$(14) ) and (33(33) $H_1\left(\rho \right):={\int }_0^{\pi /2(1-\alpha )}(q\left(x\right)-\tilde{q}\left(x\right))y\left(x\right)\tilde{y}\left(x\right)\mathrm{d}x+h-\tilde{h},$(33) ), we can give that (35) $H_1\left(\rho \right)\le C_3\exp \left(|\mathrm{Im}\rho |{\alpha }_1\right(1-\alpha \left)\pi \right),$(35) for some positive constant $C_3$. Moreover if $\lambda =ib,$ (36) $H_1\left(ib\right)\le C_3\exp \left(\mathrm{Im}\sqrt{i}|\sqrt{b}|{\alpha }_1\right(1-\alpha \left)\pi \right).$(36)

Define (37) ${\phi }_1\left(\rho \right)=\frac{H_1\left(\rho \right)}{{\mathrm{\Delta }}_1\left(\rho \right)},$(37) where (38) ${\mathrm{\Delta }}_1\left(\rho \right)=\prod _{{\lambda }_n\in S}\left(1-\frac{\lambda }{{\lambda }_n}\right).$(38) This function is an entire function in ρ. For convenience, we denote (39) $N_{{\mathrm{\Delta }}_1}\left({\rho }_0\right)=\mathrm{♯}\{\lambda \in S;\lambda \le {\lambda }_0\},N_{\mathrm{\Delta }}\left({\rho }_0\right)=\mathrm{♯}\{\lambda \in \sigma (L);\lambda \le {\lambda }_0\}.$(39) By the assumption of the theorem, we will have (40) $N_{{\mathrm{\Delta }}_1}\left({\rho }_0\right)\ge (1-\alpha )N_{\mathrm{\Delta }}\left({\rho }_0\right)+\frac{\alpha }{2}.$(40) Since the characteristic function $\mathrm{\Delta }\left(\rho \right)$ is an entire function of order $\frac{1}{2},$ we infer (41) $N_{{\mathrm{\Delta }}_1}\left({\rho }_0\right)\le N_{\mathrm{\Delta }}\left({\rho }_0\right)\le C\sqrt{\lambda },$(41) for some positive constant C. By standard calculations (see [21,24]), it is concluded that (42) $\ln |{\mathrm{\Delta }}_1\left(ib\right)|=(1-\alpha )\ln |\mathrm{\Delta }\left(ib\right)|+\frac{\alpha }{4}\ln (1+b^2).$(42) Since $\sigma \left(L\right)$ is a complete set of eigenvalues for L, we can write for sufficiently large b (43) $\mathrm{\Delta }\left(ib\right)\ge C\sqrt{|b|}\exp \left(\mathrm{Im}\sqrt{i}|\sqrt{b}|({\alpha }_1+{\alpha }_2)\frac{\pi }{2}\right).$(43) Therefore, by (42(42) $\ln |{\mathrm{\Delta }}_1\left(ib\right)|=(1-\alpha )\ln |\mathrm{\Delta }\left(ib\right)|+\frac{\alpha }{4}\ln (1+b^2).$(42) ) and (43(43) $\mathrm{\Delta }\left(ib\right)\ge C\sqrt{|b|}\exp \left(\mathrm{Im}\sqrt{i}|\sqrt{b}|({\alpha }_1+{\alpha }_2)\frac{\pi }{2}\right).$(43) ), we have (44) ${\mathrm{\Delta }}_1\left(ib\right)\ge C\sqrt{|b|}\exp \left(\mathrm{Im}\sqrt{i}|\sqrt{b}|({\alpha }_1+{\alpha }_2)(1-\alpha )\frac{\pi }{2}\right).$(44) Taking (36(36) $H_1\left(ib\right)\le C_3\exp \left(\mathrm{Im}\sqrt{i}|\sqrt{b}|{\alpha }_1\right(1-\alpha \left)\pi \right).$(36) ), (37(37) ${\phi }_1\left(\rho \right)=\frac{H_1\left(\rho \right)}{{\mathrm{\Delta }}_1\left(\rho \right)},$(37) ) and (44(44) ${\mathrm{\Delta }}_1\left(ib\right)\ge C\sqrt{|b|}\exp \left(\mathrm{Im}\sqrt{i}|\sqrt{b}|({\alpha }_1+{\alpha }_2)(1-\alpha )\frac{\pi }{2}\right).$(44) ), we get (45) ${\phi }_1\left(ib\right)\le C\frac{1}{\sqrt{|b|}}\exp \left(\mathrm{Im}\sqrt{i}|\sqrt{b}|({\alpha }_1-{\alpha }_2)(1-\alpha )\frac{\pi }{2}\right).$(45) By Phragmén–Lindelöf theorem, we have ${\phi }_1\left(\rho \right)=0$ for all ρ and then $H_1\left(\rho \right)=0$ for all ρ.

We now apply the argument again such as Theorem 3.1, to obtain $q\left(x\right)=\tilde{q}\left(x\right)$ a. e. on $[0,\pi /2(1-\alpha \left)\right]$ and $h=\tilde{h}$. The proof is completed.

5. Examples

Example 5.1

Define $r\left(x\right)={\alpha }_1^2$ for $x<\pi /2$ and $r\left(x\right)={\alpha }_2^2$ for $x>\pi /2,$ and then consider the boundary value problem $L_0,$ $\begin{array}{rl}-y^{\mathrm{\prime \prime }}+q\left(x\right)y& =\lambda r\left(x\right)y,x\in (0,\pi ),\\ U\left(y\right)& :=y^{\mathrm{\prime }}\left(0\right)-hy\left(0\right)=0,\\ V\left(y\right)& :=y^{\mathrm{\prime }}\left(\pi \right)+Hy\left(\pi \right)=0,\end{array}$ with the jump conditions $y\left(\frac{\pi }{2}+0,\rho \right)=a_{1}y\left(\frac{\pi }{2}-0,\rho \right),y^{\mathrm{\prime }}\left(\frac{\pi }{2}+0,\rho \right)=a_1^{-1}{y}^{\mathrm{\prime }}\left(\frac{\pi }{2}-0,\rho \right)+a_{2}y\left(\frac{\pi }{2}-0,\rho \right).$ Also consider the boundary value problem $L_1,$ $\begin{array}{rl}-y^{\mathrm{\prime \prime }}& =\lambda r\left(x\right)y,x\in (0,\pi ),\\ U\left(y\right)& :=y^{\mathrm{\prime }}\left(0\right)=0,\\ V\left(y\right)& :=y^{\mathrm{\prime }}\left(\pi \right)=0,\end{array}$ with the same jump conditions. Let ${\lambda }_{n0}$ and ${\lambda }_{n1}$ be the eigenvalues of BVP($L_0$) and BVP($L_1$), respectively. Based on Theorem 3.1, if ${\lambda }_{n0}={\lambda }_{n1}$ and the Wronskian of the functions $y_{n0}(x,\rho )$ and $y_{n1}(x,\rho )$ at the point $x=\pi /2$ be equal to zero, then $q\left(x\right)=0$ and h=H=0.

Example 5.2

We consider the boundary value problem $L_0$ in Example 5.1. Also consider the boundary value problem $L_2,$ $\begin{array}{rl}-y^{\mathrm{\prime \prime }}+q\left(x\right)y& =\lambda r\left(x\right)y,x\in (0,\pi ),\\ U\left(y\right)& :=y^{\mathrm{\prime }}\left(0\right)=0,\\ V\left(y\right)& :=y^{\mathrm{\prime }}\left(\pi \right)+y\left(\pi \right)=0,\end{array}$ with the jump conditions $y\left(\frac{\pi }{2}+0,\rho \right)=a_{1}y\left(\frac{\pi }{2}-0,\rho \right),y^{\mathrm{\prime }}\left(\frac{\pi }{2}+0,\rho \right)=a_1^{-1}{y}^{\mathrm{\prime }}\left(\frac{\pi }{2}-0,\rho \right)+a_{2}y\left(\frac{\pi }{2}-0,\rho \right).$ Here $q\left(x\right)=\left\{\begin{array}{cc}{\mathbf{q}}_{\mathbf{1}}\left(x\right),& x<\pi /2(1-\alpha ),\\ 0,& x>\pi /2(1-\alpha ),\end{array}\right.$ for some $\alpha \in (0,1)$. Considering $S\subseteq \left\{{\lambda }_{n0}\right\}\cap \left\{{\lambda }_{n2}\right\}$, if $\mathrm{♯}\{\lambda \in S;\lambda \le {\lambda }_0\}\ge (1-\alpha )\mathrm{♯}\{\lambda \in \{{\lambda }_{n0}\};\lambda \le {\lambda }_0\}+\frac{\alpha }{2},$ for all sufficiently large ${\lambda }_0\in \mathbb{R},$ by regarding to Theorem 4.1, then ${\mathbf{q}}_{\mathbf{1}}\left(x\right)=0$ and h=0.

Disclosure statement

No potential conflict of interest was reported by the authors.