© Wuhan University 2025

0 Introduction

Dirac operators are important models in quantum mechanics. As a result, some conclusions about these operators have been obtained (see Refs. [1-4]). As a typical problem in the spectral theory of differential operators, Dirac operators are closely related to Sturm-Liouville operators, and they have many similarities in the properties and research methods of eigenvalues, see Refs. [5-9]. In particular, in Ref. [10], Li et al studied the continuous dependence of eigenvalue of self-adjoint Dirac system consisting of the symmetric differential operator

$\begin{array}{l}\mathit{B}{\mathcal{U}}^{\mathit{\text{'}}}\left(x\right)+\mathit{Q}\left(x\right)\mathcal{U}\left(x\right)=\lambda \mathcal{U}\left(x\right),\\ x\in J\equiv \left({\alpha }^{\text{'}},{\beta }^{\text{'}}\right),-\mathrm{\infty }\le {\alpha }^{\text{'}}<{\beta }^{\text{'}}\le +\mathrm{\infty },\end{array}$(1)

$\mathit{M}\mathcal{U}\left(\alpha \right)+\mathit{N}\mathcal{U}\left(\beta \right)=\mathrm{0},\mathcal{U}\left(x\right)={\left(u_{\mathrm{1}}\left(x\right),u_{\mathrm{2}}\left(x\right)\right)}^{\mathrm{T}},$(2)

where $\mathit{B}=\left(\begin{array}{cc}\mathrm{0}& \mathrm{1}\\ -\mathrm{1}& \mathrm{0}\end{array}\right),\mathit{Q}\left(x\right)=\left(\begin{array}{cc}\rho \left(x\right)& \sigma \left(x\right)\\ \sigma \left(x\right)& \varrho \left(x\right)\end{array}\right)\in L_{\mathrm{1}}\left(J\right)$ and $\mathit{Q}\left(x\right)$ is real-valued Lebesgue measurable function on $J;\mathit{M}$ and $\mathit{N}$ are $\mathrm{2}\times \mathrm{2}$ complex matrices such that rank$\left(\mathit{M},\mathit{N}\right)=\mathrm{2}$ and satisfy $\mathit{M}\mathit{E}{\mathit{M}}^{*}=\mathit{N}\mathit{E}{\mathit{N}}^{*}$, where ${\mathit{M}}^{\mathit{*}}$ denotes the complex conjugate transpose of $\mathit{M},\mathit{E}$ is the second order symplectic matrix. Besides, we are also interested in the spectral analysis of Dirac operator with the spectral parameter in boundary condition, which has a discrete spectrum consisting of an increasing infinite sequence of (real, simple) eigenvalues ${\lambda }_{\pm n}$ such that ${\lambda }_n\to \pm \mathrm{\infty }$ as $n\to \mathrm{\infty }$, see Ref. [11], and some remarkable works have been reached, see Refs. [12-13].

From the above literature, we notice that the research on problems (1)-(2) requires that the boundary conditions satisfy $\mathit{M}\mathit{E}{\mathit{M}}^{*}=\mathit{N}\mathit{E}{\mathit{N}}^{*}.$ Typically, the eigenparameter is only present in differential equations. Nonetheless, in numerous practical applications, including mechanics and acoustic scattering theory, it is necessary for the spectral parameter to be featured not only within the differential equations but also within the boundary conditions, see Ref. [14]. Recently, inverse spectral problems for Sturm-Liouville operators with non-self-adjoint eigenparameter-dependent boundary conditions have been studied via matrix representations and inverse matrix eigenvalue problems, see Ref. [15]. For the above model the condition $\mathit{M}\mathit{E}{\mathit{M}}^{*}=\mathit{N}\mathit{E}{\mathit{N}}^{*}$ may not be satisfied. Therefore, we hope to get the correlation spectrum properties of this kind of problems. For example, considering the self-adjointness of operators and the continuous dependence of eigenvalues, it is necessary to discuss the eigenvalue problem of a linear operator in a suitable Hilbert space. To address the limitation that these boundary conditions lack self-adjointness, we construct a suitable Banach space and an embedded mapping, and prove the continuity of the embedded mapping. Moreover, based on the definition of Fréchet derivative, we obtain the differentiability of eigenvalue bifurcation.

Our research plan is structured as follows. In Section 1, we introduce a self-adjoint operator eigenvalue problem and elucidate its critical spectral properties from various perspectives. In Section 2 and 3, leveraging the established continuity of the embedded mapping, we demonstrate the differentiability of the eigenvalue branch and present the corresponding derivative formulas.

1 Some Properties

This paper concerns eigenvalue problem of the form

$\begin{array}{l}\mathit{ℒ}\mathcal{U}:=\mathit{B}{\mathcal{U}}^{\mathit{\text{'}}}\left(x\right)+\mathit{Q}\left(x\right)\mathcal{U}\left(x\right)=\lambda \mathcal{U}\left(x\right),\\ -\mathrm{\infty }<a<b<+\mathrm{\infty },\end{array}$(3)

$\mathit{M}\mathcal{U}\left(a\right)+\mathit{N}\mathcal{U}\left(b\right)=\mathrm{0},\mathcal{U}\left(x\right)={\left(u_{\mathrm{1}}\left(x\right),u_{\mathrm{2}}(x)\right)}^{\mathrm{T}},$(4)

where $\mathit{Q}\left(x\right)=[p(x),q(x);q(x),p(x)]$ and every element in $\mathit{Q}(x)$ belongs to $L_{\mathrm{l}\mathrm{o}\mathrm{c}}^{\mathrm{1}}(a,b),$ $\lambda$ is the spectral parameter. Here, we require $\mathit{M},\mathit{N}$ to satisfy $\mathit{M}=[a_{\mathrm{0}}+a_{\mathrm{1}}\lambda ,-(b_{\mathrm{0}}+b_{\mathrm{1}}\lambda );\mathrm{0,0}],\mathit{N}=[\mathrm{0,0};c_o+c_{\mathrm{1}}\lambda ,-(d_{\mathrm{0}}+d_{\mathrm{1}}\lambda )]$ and $a_i,b_i,c_i,d_i,$ $i=\mathrm{0,1},$ are real numbers satisfying ${\sigma }_{\mathrm{0}}=a_{\mathrm{0}}{b}_{\mathrm{1}}-b_{\mathrm{0}}{a}_{\mathrm{1}}>\mathrm{0}$ and ${\sigma }_{\mathrm{1}}=c_{\mathrm{0}}{d}_{\mathrm{1}}-d_{\mathrm{0}}{c}_{\mathrm{1}}<\mathrm{0}.$

We first consider a linear operator eigenvalue problem derived from spectral problems (3)-(4). The inner product in the Hilbert space $H=L_{\mathrm{2}}(a,b)\oplus L_{\mathrm{2}}(a,b)⨁{\mathbb{C}}^{\mathrm{2}}$ associated with

$〈\mathit{U},\mathit{Z}〉={\int }_a^b(u_{\mathrm{1}}(x){\overline{z}}_{\mathrm{1}}(x)+u_{\mathrm{2}}(x){\overline{z}}_{\mathrm{2}}(x))\mathrm{d}x-\frac{\mathrm{1}}{{\sigma }_{\mathrm{0}}}{u}_{\mathrm{3}}{\overline{z}}_{\mathrm{3}}+\frac{\mathrm{1}}{{\sigma }_{\mathrm{1}}}{u}_{\mathrm{4}}{\overline{z}}_{\mathrm{4}}$

for $\mathit{U}=(u_{\mathrm{1}}(x),u_{\mathrm{2}}(x),u_{\mathrm{3}},u_{\mathrm{4}}{)}^{\mathrm{T}},\mathit{Z}=(z_{\mathrm{1}}(x),z_{\mathrm{2}}(x),z_{\mathrm{3}},z_{\mathrm{4}}{)}^{\mathrm{T}}\in H.$ Define the operator $\mathit{ℒ}$ acting in $H$ so that

$\mathit{ℒ}\mathit{U}=\left(\begin{array}{c}\mathit{B}{\mathcal{U}}^{\mathit{\text{'}}}(x)+\mathit{Q}(x)\mathcal{U}(x)\\ b_{\mathrm{0}}{u}_{\mathrm{2}}(a)-a_{\mathrm{0}}{u}_{\mathrm{1}}(a)\\ d_{\mathrm{0}}{u}_{\mathrm{2}}(b)-c_{\mathrm{0}}{u}_{\mathrm{1}}(b)\end{array}\right)$

with the domain

$D(\mathrm{ℒ})=\left\{\mathit{U}\in H:u_{\mathrm{1}},u_{\mathrm{2}}\in \mathrm{A}\mathrm{C}[a,b]\mathrm{a}\mathrm{n}\mathrm{d}\mathrm{s}\mathrm{a}\mathrm{t}\mathrm{i}\mathrm{s}\mathrm{f}\mathrm{y}\mathrm{b}\mathrm{o}\mathrm{u}\mathrm{n}\mathrm{d}\mathrm{a}\mathrm{r}\mathrm{y}\mathrm{c}\mathrm{o}\mathrm{n}\mathrm{d}\mathrm{i}\mathrm{t}\mathrm{i}\mathrm{o}\mathrm{n}(\mathrm{4}),\mathit{ℒ}\mathcal{U}\in L_{\mathrm{2}}(a,b)\oplus L_{\mathrm{2}}(a,b),u_{\mathrm{3}}=a_{\mathrm{1}}{u}_{\mathrm{1}}(a)-b_{\mathrm{1}}{u}_{\mathrm{2}}(a)\in \mathbb{C},u_{\mathrm{4}}=c_{\mathrm{1}}{u}_{\mathrm{1}}(b)-d_{\mathrm{1}}{u}_{\mathrm{2}}(b)\in \mathbb{C}\right\}$

Here $\mathrm{A}\mathrm{C}[a,b]$ denotes the set of absolutely continuous and complex-valued functions on $[a,b]$.

By immediate verification, we conclude that the problems (3)-(4) are equivalent to linear operator eigenvalue problem $\mathrm{ℒ}\mathit{U}=\lambda \mathit{U}$. We now focus on discussing the properties of linear operator $\mathrm{ℒ}$ as follows.

Theorem 1   $\mathrm{ℒ}$ is a self-adjoint operator in $H$.

Proof   In order to prove that the operator $\mathrm{ℒ}$ is self-adjoint, we first need to prove that the domain of operator $\mathit{ℒ}$ is dense in $H$. Suppose $\mathit{U}=(u_{\mathrm{1}}(x),u_{\mathrm{2}}(x),u_{\mathrm{3}},u_{\mathrm{4}}{)}^{\mathrm{T}}\in H$ and $\mathit{U}$ is orthogonal to $\mathit{Y}=(y_{\mathrm{1}}(x),y_{\mathrm{2}}(x),y_{\mathrm{3}},y_{\mathrm{4}}{)}^{\mathrm{T}}\in D(\mathrm{ℒ})$. We will prove $\mathit{U}={(\mathrm{0,0},\mathrm{0,0})}^{\mathrm{T}}$. Since $C_{\mathrm{0}}^{\mathrm{\infty }}\oplus C_{\mathrm{0}}^{\mathrm{\infty }}⨁\left\{\mathrm{0}\right\}⨁\left\{\mathrm{0}\right\}\subset D(\mathrm{ℒ})$, for arbitrary $\mathit{V}=(v_{\mathrm{1}}(x),v_{\mathrm{2}}{(x),\mathrm{0,0})}^{\mathrm{T}}\in C_{\mathrm{0}}^{\mathrm{\infty }}\oplus C_{\mathrm{0}}^{\mathrm{\infty }}⨁\left\{\mathrm{0}\right\}⨁\left\{\mathrm{0}\right\}$, we have $〈\mathit{U},\mathit{V}〉={\int }_a^b(u_{\mathrm{1}}(x){\overline{v}}_{\mathrm{1}}(x)+u_{\mathrm{2}}(x){\overline{v}}_{\mathrm{2}}(x))dx=\mathrm{0}.$ Since $C_{\mathrm{0}}^{\mathrm{\infty }}$ is dense in $L_{\mathrm{2}}(a,b)$, we have $u_{\mathrm{1}}(x)=u_{\mathrm{2}}(x)=\mathrm{0}$, so $\mathit{U}=(\mathrm{0},\mathrm{0},u_{\mathrm{3}},u_{\mathrm{4}}{)}^{\mathrm{T}}$. For any $\mathit{Y}$, through the inner product in $H$, we get $〈\mathit{U},\mathit{Y}〉=\frac{\mathrm{1}}{{\sigma }_{\mathrm{1}}}{u}_{\mathrm{4}}{\overline{y}}_{\mathrm{4}}-\frac{\mathrm{1}}{{\sigma }_{\mathrm{0}}}{u}_{\mathrm{3}}{\overline{y}}_{\mathrm{3}}=\mathrm{0}.$ Since $\mathit{Y}$ is arbitrary, we have $u_{\mathrm{3}}=u_{\mathrm{4}}=\mathrm{0}.$ Hence $\mathit{U}={(\mathrm{0,0},\mathrm{0,0})}^{\mathrm{T}}.$

Secondly, we need to show that the operator $\mathrm{ℒ}$ is symmetric. Let $\mathit{U}=(u_{\mathrm{1}}(x),u_{\mathrm{2}}(x),u_{\mathrm{3}},u_{\mathrm{4}}{)}^{\mathrm{T}}$. For any $\mathit{V}=(v_{\mathrm{1}}(x),v_{\mathrm{2}}(x),v_{\mathrm{3}},v_{\mathrm{4}}{)}^{\mathrm{T}}\in D(\mathrm{ℒ})$, then

$〈\mathrm{ℒ}\mathit{U},\mathit{V}〉-〈\mathit{U},\mathrm{ℒ}\mathit{V}〉={\overline{v}}_{\mathrm{1}}(b)u_{\mathrm{2}}(b)-{\overline{v}}_{\mathrm{2}}(b)u_{\mathrm{1}}(b)+u_{\mathrm{1}}(a){\overline{v}}_{\mathrm{2}}(a)-u_{\mathrm{2}}(a){\overline{v}}_{\mathrm{1}}(a)-\frac{\mathrm{1}}{{\sigma }_{\mathrm{0}}}(b_{\mathrm{0}}{u}_{\mathrm{2}}(a)-a_{\mathrm{0}}{u}_{\mathrm{1}}(a)){\overline{v}}_{\mathrm{3}}+\frac{\mathrm{1}}{{\sigma }_{\mathrm{1}}}(d_{\mathrm{0}}{u}_{\mathrm{2}}(b)-c_{\mathrm{0}}{u}_{\mathrm{1}}(b)){\overline{v}}_{\mathrm{4}}+\frac{\mathrm{1}}{{\sigma }_{\mathrm{0}}}(b_{\mathrm{0}}{\overline{v}}_{\mathrm{2}}(a)-a_{\mathrm{0}}{\overline{v}}_{\mathrm{1}}(a))u_{\mathrm{3}}-\frac{\mathrm{1}}{{\sigma }_{\mathrm{1}}}(d_{\mathrm{0}}{\overline{v}}_{\mathrm{2}}(b)-c_{\mathrm{0}}{\overline{v}}_{\mathrm{1}}(b))u_{\mathrm{4}}$(5)

By the boundary condition (4), we have

$\begin{array}{l}\frac{\mathrm{1}}{{\sigma }_{\mathrm{0}}}(b_{\mathrm{0}}{\overline{v}}_{\mathrm{2}}(a)-a_{\mathrm{0}}{\overline{v}}_{\mathrm{1}}(a))u_{\mathrm{3}}-\frac{\mathrm{1}}{{\sigma }_{\mathrm{0}}}(b_{\mathrm{0}}{u}_{\mathrm{2}}(a)-a_{\mathrm{0}}{u}_{\mathrm{1}}(a)){\overline{v}}_{\mathrm{3}}\\ =u_{\mathrm{2}}(a){\overline{v}}_{\mathrm{1}}(a)-{\overline{v}}_{\mathrm{2}}(a)u_{\mathrm{1}}(a)\end{array}$(6)

$\begin{array}{l}\frac{\mathrm{1}}{{\sigma }_{\mathrm{1}}}(d_{\mathrm{0}}{u}_{\mathrm{2}}(b)-c_{\mathrm{0}}{u}_{\mathrm{1}}(b)){\overline{v}}_{\mathrm{4}}-\frac{\mathrm{1}}{{\sigma }_{\mathrm{1}}}(d_{\mathrm{0}}{\overline{v}}_{\mathrm{2}}(b)-c_{\mathrm{0}}{\overline{v}}_{\mathrm{1}}(b))u_{\mathrm{4}}\\ =u_{\mathrm{1}}(b){\overline{v}}_{\mathrm{2}}(b)-{\overline{v}}_{\mathrm{1}}(b)u_{\mathrm{2}}(b)\end{array}$(7)

Consequently, combining (5)-(7), we obtain

$〈\mathrm{ℒ}\mathit{U},\mathit{V}〉=〈\mathit{U},\mathrm{ℒ}\mathit{V}〉,$

which implies that operator $\mathrm{ℒ}$ is symmetric.

Since $\mathrm{ℒ}$ is symmetric, it suffices to prove that for any $\mathit{Y}$ and some $\mathit{Z}\in D({\mathrm{ℒ}}^{*}),\mathit{U}\in H$ satisfying $〈\mathrm{ℒ}\mathit{Y},\mathit{Z}〉=〈\mathit{Y},\mathit{U}〉$, then $\mathit{Z}\in D(\mathrm{ℒ})$ and $\mathrm{ℒ}\mathit{Z}=\mathit{U}$, where $\mathit{Z}={\left(z_{\mathrm{1}}(x),z_{\mathrm{2}}(x),z_{\mathrm{3}},z_{\mathrm{4}}\right)}^{\mathrm{T}},\mathit{U}={\left(u_{\mathrm{1}}(x),u_{\mathrm{2}}(x),u_{\mathrm{3}},u_{\mathrm{4}}\right)}^{\mathrm{T}}$. It means that $\mathit{Z}$ satisfies following conditions:

(ⅰ) $z_i(x)\in \mathrm{A}\mathrm{C}[a,b],i=\mathrm{1,2};$

(ⅱ) $z_{\mathrm{3}}=a_{\mathrm{1}}{z}_{\mathrm{1}}(a)-b_{\mathrm{1}}{z}_{\mathrm{2}}(a),z_{\mathrm{4}}=c_{\mathrm{1}}{z}_{\mathrm{1}}(b)-d_{\mathrm{1}}{z}_{\mathrm{2}}(b);$

(ⅲ) $u_{\mathrm{3}}=b_{\mathrm{0}}{z}_{\mathrm{2}}(a)-a_{\mathrm{0}}{z}_{\mathrm{1}}(a),u_{\mathrm{4}}=d_{\mathrm{0}}{z}_{\mathrm{2}}(b)-c_{\mathrm{0}}{z}_{\mathrm{1}}(b);$

(ⅳ) $\mathrm{ℒ}(z_{\mathrm{1}},z_{\mathrm{2}}{)}^{\mathrm{T}}=(u_{\mathrm{1}},u_{\mathrm{2}}{)}^{\mathrm{T}}$.

Indeed, the above conditions imply that $D({\mathrm{ℒ}}^{*})\subset D(\mathrm{ℒ}).$

Step 1   For arbitrary $\mathit{V}=(v_{\mathrm{1}}(x),v_{\mathrm{2}}{(x),\mathrm{0,0})}^{\mathrm{T}}\in C_{\mathrm{0}}^{\mathrm{\infty }}\oplus$ $C_{\mathrm{0}}^{\mathrm{\infty }}⨁\left\{\mathrm{0}\right\}⨁\left\{\mathrm{0}\right\}$, there is $〈\mathrm{ℒ}\mathit{V},\mathit{Z}〉=〈\mathit{V},\mathit{U}〉$. Moreover, we arrive at

$\begin{array}{l}{\int }_a^b(v_{\mathrm{1}}(x){\overline{z}}_{\mathrm{1}}(x)+v_{\mathrm{2}}(x){\overline{z}}_{\mathrm{2}}(x))\mathrm{d}x\\ ={\int }_a^b(v_{\mathrm{1}}(x){\overline{u}}_{\mathrm{1}}(x)+v_{\mathrm{2}}(x){\overline{u}}_{\mathrm{2}}(x))dx.\end{array}$

Namely, $〈\mathrm{ℒ}\mathit{V},\mathit{Z}〉=〈\mathit{V},\mathit{U}〉$. Since $\mathrm{ℒ}$ is symmetric, combined with $\mathit{V}={\left(v_{\mathrm{1}}(x),v_{\mathrm{2}}(x),\mathrm{0,0}\right)}^{\mathrm{T}}\in C_{\mathrm{0}}^{\mathrm{\infty }}\oplus C_{\mathrm{0}}^{\mathrm{\infty }}⨁\left\{\mathrm{0}\right\}⨁\left\{\mathrm{0}\right\}$, we can obtain $〈\mathrm{ℒ}\mathit{V},\mathit{Z}〉=〈\mathit{V},\mathrm{ℒ}\mathit{Z}〉=〈\mathit{V},\mathit{U}〉$. In view of the classical theory of differential operators, (ⅰ) and (ⅳ) hold.

Step 2   According (ⅳ) and the relation $〈\mathrm{ℒ}\mathit{Y},\mathit{Z}〉=$ $〈\mathit{Y},\mathit{U}〉$, we have

$\begin{array}{l}〈\mathit{B}{\mathcal{U}}^{\mathit{\text{'}}}+\mathit{Q}(x)\mathcal{U},\mathcal{Z}〉-\frac{\mathrm{1}}{{\sigma }_{\mathrm{0}}}(b_{\mathrm{0}}{y}_{\mathrm{2}}(a)-a_{\mathrm{0}}{y}_{\mathrm{1}}(a)){\overline{z}}_{\mathrm{3}}\\ +\frac{\mathrm{1}}{{\sigma }_{\mathrm{1}}}(d_{\mathrm{0}}{y}_{\mathrm{2}}(b)-c_{\mathrm{0}}{y}_{\mathrm{1}}(b)){\overline{z}}_{\mathrm{4}}\end{array}$

$=〈\mathcal{y},\mathit{B}{\mathcal{Z}}^{\mathit{\text{'}}}+\mathit{Q}(x)\mathcal{Z}〉-\frac{\mathrm{1}}{{\sigma }_{\mathrm{0}}}{y}_{\mathrm{3}}{\overline{z}}_{\mathrm{3}}+\frac{\mathrm{1}}{{\sigma }_{\mathrm{1}}}{y}_{\mathrm{4}}{\overline{z}}_{\mathrm{4}}.$

In light of

$\begin{array}{l}〈\mathit{B}{\mathcal{U}}^{\mathit{\text{'}}}+\mathit{Q}(x)\mathcal{U},\mathcal{Z}〉-〈\mathcal{y},\mathit{B}{\mathcal{Z}}^{\mathit{\text{'}}}+\mathit{Q}(x)\mathcal{Z}〉\\ ={\overline{z}}_{\mathrm{1}}(b)y_{\mathrm{2}}(b)-{\overline{z}}_{\mathrm{2}}(b)y_{\mathrm{1}}(b)+y_{\mathrm{1}}(a){\overline{z}}_{\mathrm{2}}(a)-y_{\mathrm{2}}(a){\overline{z}}_{\mathrm{1}}(a)\end{array}$

Furthermore, one has

$\begin{array}{l}\frac{\mathrm{1}}{{\sigma }_{\mathrm{1}}}{y}_{\mathrm{4}}{\overline{z}}_{\mathrm{4}}-\frac{\mathrm{1}}{{\sigma }_{\mathrm{0}}}{y}_{\mathrm{3}}{\overline{z}}_{\mathrm{3}}+\frac{\mathrm{1}}{{\sigma }_{\mathrm{0}}}(b_{\mathrm{0}}{y}_{\mathrm{2}}(a)-a_{\mathrm{0}}{y}_{\mathrm{1}}(a)){\overline{z}}_{\mathrm{3}}\\ -\frac{\mathrm{1}}{{\sigma }_{\mathrm{1}}}(d_{\mathrm{0}}{y}_{\mathrm{2}}(b)-c_{\mathrm{0}}{y}_{\mathrm{1}}(b)){\overline{z}}_{\mathrm{4}}\end{array}$

$={\overline{z}}_{\mathrm{1}}(b)y_{\mathrm{2}}(b)-{\overline{z}}_{\mathrm{2}}(b)y_{\mathrm{1}}(b)+y_{\mathrm{1}}(a){\overline{z}}_{\mathrm{2}}(a)-y_{\mathrm{2}}(a){\overline{z}}_{\mathrm{1}}(a).$(8)

Using Naimark’s patching lemma^[16], there exists a $\mathit{Y}={\left(y_{\mathrm{1}}(x),y_{\mathrm{2}}(x),y_{\mathrm{3}},y_{\mathrm{4}}\right)}^{\mathrm{T}}\in D(\mathrm{ℒ})$ such that

$y_{\mathrm{1}}(b)=y_{\mathrm{2}}(b)=\mathrm{0},y_{\mathrm{2}}(a)=-a_{\mathrm{1}},y_{\mathrm{1}}(a)=-b_{\mathrm{1}}.$

Then by (8), we have $z_{\mathrm{3}}=a_{\mathrm{1}}{z}_{\mathrm{1}}(a)-b_{\mathrm{1}}{z}_{\mathrm{2}}(a)$. Similarly, we get

$y_{\mathrm{2}}(a)=y_{\mathrm{1}}(a)=\mathrm{0},y_{\mathrm{2}}(b)=c_{\mathrm{1}},y_{\mathrm{1}}(b)=d_{\mathrm{1}}.$

Moreover, $z_{\mathrm{4}}=c_{\mathrm{1}}{z}_{\mathrm{1}}(b)-d_{\mathrm{1}}{z}_{\mathrm{2}}(b)$. Therefore, (ⅱ) holds. We can prove (ⅲ) by using the similar method, hence we omit the details. The proof is completed.

Corollary 1   The following assertions are true:

1) The two vector eigenfunctions corresponding to different eigenvalues of ${\lambda }_{\mathrm{1}}$ and ${\lambda }_{\mathrm{2}}$ are orthogonal in the following sense

${\int }_a^b(u_{\mathrm{1}}(x){\overline{z}}_{\mathrm{1}}(x)+u_{\mathrm{2}}(x){\overline{z}}_{\mathrm{2}}(x))dx-\frac{\mathrm{1}}{{\sigma }_{\mathrm{0}}}{u}_{\mathrm{3}}{\overline{z}}_{\mathrm{3}}+\frac{\mathrm{1}}{{\sigma }_{\mathrm{1}}}{u}_{\mathrm{4}}{\overline{z}}_{\mathrm{4}}=\mathrm{0}$

2) All eigenvalues of the operator $\mathrm{ℒ}$ are real, and all vector-eigenfunctions are real-valued.

Theorem 2^[11] There exists an unboundedly decreasing sequence ${\left\{{\lambda }_{-n}\right\}}_{n=\mathrm{1}}^{\mathrm{\infty }}$ of negative eigenvalues and an unboundedly increasing sequence ${\left\{{\lambda }_n\right\}}_{n=\mathrm{1}}^{\mathrm{\infty }}$ of nonnegative eigenvalues of the boundary value problems (3)-(4):

$-\mathrm{\infty }\leftarrow \cdots <{\lambda }_{-n}\cdots <{\lambda }_{-\mathrm{2}}<{\lambda }_{-\mathrm{1}}<{\lambda }_{\mathrm{1}}<{\lambda }_{\mathrm{2}}<\cdots <{\lambda }_n<\cdots <\to +\mathrm{\infty }.$

Let $\mathit{\Phi }(x,\lambda )=({\varphi }_{\mathrm{1}}(x,\lambda ),{\varphi }_{\mathrm{2}}{(x,\lambda ))}^{\mathrm{T}}$ and $\mathit{\Psi }(x,\lambda )=({\psi }_{\mathrm{1}}(x,\lambda ),{\psi }_{\mathrm{2}}{(x,\lambda ))}^{\mathrm{T}}$ be the fundamental solutions of (3), which satisfy the initial condition

${\varphi }_{\mathrm{1}}(a,\lambda )={\psi }_{\mathrm{2}}(a,\lambda )=\mathrm{1},{\varphi }_{\mathrm{2}}(a,\lambda )={\psi }_{\mathrm{1}}(a,\lambda )=\mathrm{0}.$

Then, $\mathit{\Phi }(x,\lambda )$ and $\mathit{\Psi }(x,\lambda )$ are linearly independent and entire functions of $\lambda$.

Denote

${\mathit{\Lambda }}_{\lambda }(x,\lambda ):=\left(\begin{array}{cc}{\varphi }_{\mathrm{1}}(x,\lambda )& {\psi }_{\mathrm{1}}(x,\lambda )\\ {\varphi }_{\mathrm{2}}(x,\lambda )& {\psi }_{\mathrm{2}}(x,\lambda )\end{array}\right).$

Lemma 1   $\lambda \in \mathbb{C}$ is an eigenvalue of (3)-(4) if and only if $\mathrm{\Delta }(\lambda )=\mathrm{d}\mathrm{e}\mathrm{t}\left[{\mathit{M}}_{\lambda }+{\mathit{N}}_{\lambda }{\mathit{\Lambda }}_{\lambda }(b)\right]=\mathrm{0}$.

Proof   By using similar methods in Ref. [10], we can obtain the assertion holds.

2 Continuous Dependence of Eigenvalues and Eigenfunction

Denote $\mathit{R}=\left[a_{\mathrm{0}},a_{\mathrm{1}};b_{\mathrm{0}},b_{\mathrm{1}}\right],\mathit{G}=\left[c_{\mathrm{0}},c_{\mathrm{1}};d_{\mathrm{0}},d_{\mathrm{1}}\right]$. We introduce Banach space

$\mathrm{ℬ}:=L_{\mathrm{1}}(a,b)\oplus L_{\mathrm{1}}(a,b)⨁M_{\mathrm{2}\times \mathrm{2}}(\mathbb{C})⨁M_{\mathrm{2}\times \mathrm{2}}(\mathbb{C})$

equipped with the norm

$\Vert K\Vert :={\int }_a^b(|p|+|q|)\mathrm{d}x+{\Vert \mathit{R}\Vert }_{\mathbb{M}}+{\Vert \mathit{G}\Vert }_{\mathbb{M}}$

For any $K=(p(x),q(x),\mathit{R},\mathit{G})\in \mathrm{ℬ}$, where ${\Vert \cdot \Vert }_{\mathbb{M}}$denotes the matrix normal. Let’s construct a boundary condition space $\mathrm{\Omega }$={$K\in \mathrm{ℬ}$: the coefficient in (3)-(4), and ${\sigma }_{\mathrm{0}}>\mathrm{0},{\sigma }_{\mathrm{1}}<\mathrm{0}$ hold}. Then $\mathrm{\Omega }$ is a closed subset of $\mathrm{ℬ}$ and inherits its topology $\mathrm{ℬ}$.

Theorem 3   Let $\tilde{K}=(\tilde{p},\tilde{q},\tilde{\mathit{R}},\tilde{\mathit{G}})\in \mathrm{\Omega }$, and $\lambda (\tilde{K})$ be an eigenvalue of spectral problems (3)-(4). Then for any sufficiently small $\epsilon >\mathrm{0}$, there exists $\delta >\mathrm{0}$ such that if $\Vert \tilde{K}-K\Vert <\delta$, the spectral problems (3)-(4) have exactly one eigenvalue satisfying $|\lambda (\tilde{K})-\lambda (K)|<\epsilon .$

Proof   It is well known that $\lambda (K)$ is eigenvalue of (3)-(4) if and only if Lemma 1 holds. It is obvious that $\mathrm{\Delta }(\lambda (K))$ is not constant with regard to λ since λ is isolated eigenvalue. Furthermore, for any $K\in \mathrm{\Omega },\mathrm{\Delta }(\lambda (K))$ is an entire function of $\lambda$. Hence, there exists $\epsilon >\mathrm{0}$ such that $\mathrm{\Delta }(\lambda )\ne \mathrm{0}$ for $\lambda \in \left\{\lambda \in \mathbb{C}:|\lambda -\lambda (\tilde{K})|=\epsilon \right\}$. By the well known theorem on continuity of the roots of an equation as a function of parameters, see Ref. [17], the assertion holds.

The normalized eigenfunction $\mathit{U}={\left(u_{\mathrm{1}}(x),u_{\mathrm{2}}(x),u_{\mathrm{3}},u_{\mathrm{4}}\right)}^{\mathrm{T}}$ of spectral problems (3)-(4) is defined as follows:

$〈\mathit{U},\mathit{U}〉={\int }_a^b(u_{\mathrm{1}}(x){\overline{u}}_{\mathrm{1}}(x)+u_{\mathrm{2}}(x){\overline{u}}_{\mathrm{2}}(x))dx-\frac{\mathrm{1}}{{\sigma }_{\mathrm{0}}}{u}_{\mathrm{3}}{\overline{u}}_{\mathrm{3}}+\frac{\mathrm{1}}{{\sigma }_{\mathrm{1}}}{u}_{\mathrm{4}}{\overline{u}}_{\mathrm{4}}=\mathrm{1}.$

Theorem 4   Let $\lambda (K)$ is an eigenvalue of operator $\mathrm{ℒ}$ with $K\in \mathrm{\Omega }$ and ${\left(w_{\mathrm{1}}(x),w_{\mathrm{2}}(x),w_{\mathrm{3}},w_{\mathrm{4}}\right)}^{\mathrm{T}}\in H$ is a normalized eigenvector for $\lambda (K)$. Then there exists a normalized eigenvector ${\left(v_{\mathrm{1}}(x),v_{\mathrm{2}}(x),v_{\mathrm{3}},v_{\mathrm{4}}\right)}^{\mathrm{T}}\in H$ for $\lambda (\tilde{K})$ with $\tilde{K}\in \mathrm{\Omega }$ such that

$(v_{\mathrm{1}}(x),v_{\mathrm{2}}(x))\to (w_{\mathrm{1}}(x),w_{\mathrm{2}}(x)),v_{\mathrm{3}}\to w_{\mathrm{3}},v_{\mathrm{4}}\to w_{\mathrm{4}}$

as $\tilde{K}\to K$ both uniformly on $\left[a,b\right]$.

Proof   We know that $\lambda (K)$ is simple. Let ${\left(u_{\mathrm{1}}(x,K),u_{\mathrm{2}}(x,K),u_{\mathrm{3}},u_{\mathrm{4}}\right)}^{\mathrm{T}}$ be a normalized eigenvector of operator $\mathrm{ℒ}$. Then in view of inner product in the Hilbert space H, we have

$\Vert (u_{\mathrm{1}}(x),u_{\mathrm{2}}(x),u_{\mathrm{3}},u_{\mathrm{4}}{)}^T\Vert ={\int }_a^b(u_{\mathrm{1}}{\overline{u}}_{\mathrm{1}}+u_{\mathrm{2}}{\overline{u}}_{\mathrm{2}})dx-\frac{\mathrm{1}}{{\sigma }_{\mathrm{0}}}{u}_{\mathrm{3}}{\overline{u}}_{\mathrm{3}}+\frac{\mathrm{1}}{{\sigma }_{\mathrm{1}}}{u}_{\mathrm{4}}{\overline{u}}_{\mathrm{4}}=\mathrm{1}$

and $\lambda (K)$ is the corresponding eigenvalue. Theorem 3 means that there exists $\lambda (\tilde{K})$ such that $\lambda (\tilde{K})\to \lambda (K)$ as $\tilde{K}\to K$.

Denote the boundary conditions matrix as follows

$({\mathit{M}}_{\lambda },{\mathit{N}}_{\lambda })(K)=\left(\begin{array}{cc}{a}_{\mathrm{1}}\lambda (K)+a_{\mathrm{0}}& -(b_{\mathrm{1}}\lambda (K)+b_{\mathrm{0}})\\ \mathrm{0}& \mathrm{0}\end{array}\begin{array}{cc}\mathrm{0}& \mathrm{0}\\ c_{\mathrm{1}}\lambda (K)+c_{\mathrm{0}}& -(d_{\mathrm{1}}\lambda (K)+d_{\mathrm{0}})\end{array}\right).$

Then $({\mathit{M}}_{\lambda },{\mathit{N}}_{\lambda })(\tilde{K})\to ({\mathit{M}}_{\lambda },{\mathit{N}}_{\lambda })(K)$ as $\tilde{K}\to K.$ Besides, there exists ${\left(u_{\mathrm{1}}(x,\tilde{K}),u_{\mathrm{2}}(x,\tilde{K}),u_{\mathrm{3}}(\tilde{K}),u_{\mathrm{4}}(\tilde{K})\right)}^{\mathrm{T}}$satisfying the $({\mathit{M}}_{\lambda },{\mathit{N}}_{\lambda })(\tilde{K})$ for $\lambda (\tilde{K})$ and $\Vert {\left(u_{\mathrm{1}}(x,\tilde{K}),u_{\mathrm{2}}(x,\tilde{K}),u_{\mathrm{3}}(\tilde{K}),u_{\mathrm{4}}(\tilde{K})\right)}^T\Vert =\mathrm{1}$. Therefore, we obtain

$(u_{\mathrm{1}}(x,\tilde{K}),u_{\mathrm{2}}(x,\tilde{K}))\to (u_{\mathrm{1}}(x,K),u_{\mathrm{2}}(x,K)),u_{\mathrm{3}}(\tilde{K})\to u_{\mathrm{3}}(K),u_{\mathrm{4}}(\tilde{K})\to u_{\mathrm{4}}(K)$

as $\tilde{K}\to K$ both uniformly on $\left[a,b\right]$. Let

$(v_{\mathrm{1}}(x),v_{\mathrm{2}}(x),v_{\mathrm{3}},v_{\mathrm{4}}{)}^{\mathrm{T}}=\frac{(u_{\mathrm{1}}(x,\tilde{K}),u_{\mathrm{2}}(x,\tilde{K}),u_{\mathrm{3}}(\tilde{K}),u_{\mathrm{4}}{(\tilde{K}))}^{\mathrm{T}}}{\Vert {\left(u_{\mathrm{1}}(x,\tilde{K}),u_{\mathrm{2}}(x,\tilde{K}),u_{\mathrm{3}}(\tilde{K}),u_{\mathrm{4}}(\tilde{K})\right)}^T\Vert },$

$(w_{\mathrm{1}}(x),w_{\mathrm{2}}(x),w_{\mathrm{3}},w_{\mathrm{4}}{)}^{\mathrm{T}}=\frac{(u_{\mathrm{1}}(x,K),u_{\mathrm{2}}(x,K),u_{\mathrm{3}}(K),u_{\mathrm{4}}{(K))}^{\mathrm{T}}}{\Vert (u_{\mathrm{1}}(x,K),u_{\mathrm{2}}(x,K),u_{\mathrm{3}}(K),u_{\mathrm{4}}{(K))}^T\Vert }.$

Then the desired assertion holds. The proof is completed.

3 Differentiability of Eigenvalue

In this section, we will prove that the simple eigenvalue branch is differentiable for all parameters and obtain the differential expression for all parameter in the sense of Fréchet derivative.

Definition 1^[17] A map $\mathcal{T}$ from an open set $D$ of the Banach space $E_{\mathrm{1}}$ into the Banach space $E_{\mathrm{2}}$ is Fréchet differentiable at a point $x_{\mathrm{0}}\in D$ if there exists a bounded linear operator $\mathrm{d}{\mathcal{T}}_{x_{\mathrm{0}}}(s):E_{\mathrm{1}}\to E_{\mathrm{2}}$such that in some neighborhood of the $x_{\mathrm{0}}$ ,

$\Vert \mathcal{T}(x_{\mathrm{0}}+s)-\mathcal{T}(x_{\mathrm{0}})-\mathrm{d}{\mathcal{T}}_{x_{\mathrm{0}}}(s)\Vert =o(\Vert s\Vert )\mathrm{a}\mathrm{s}s\to \mathrm{0}.$

Theorem 5   Let $\lambda (K)$ be an eigenvalue for the operator $\mathrm{ℒ}$ for $K\in \mathrm{\Omega }$ and $(u_{\mathrm{1}}(x),u_{\mathrm{2}}(x),u_{\mathrm{3}},u_{\mathrm{4}}{)}^{\mathrm{T}}$ be a normalized eigenvector of $\lambda (K)$. Then $\lambda$ is Fréchet differentiable with respect to all parameters in $K$. Specifically, the derivative formulas of $\lambda$ are given as follows:

Result 1 Fix all components of $K$ except the boundary matrix $\mathit{R}$ and let $\lambda (\mathit{R})=\lambda (K)$ denote the eigenvalue. Then

$\mathrm{d}{\lambda }_{\mathit{R}}(\mathit{T})=(-u_{\mathrm{1}}(a),u_{\mathrm{2}}(a))\left[\mathit{I}-\mathit{R}{(\mathit{R}+\mathit{T})}^{-\mathrm{1}}\right]\left(\begin{array}{c}{\overline{u}}_{\mathrm{2}}(a)\\ {\overline{u}}_{\mathrm{1}}(a)\end{array}\right)$

for all $\mathit{T}$, where $\mathrm{d}\mathrm{e}\mathrm{t}(\mathit{R}+\mathit{T})=\mathrm{d}\mathrm{e}\mathrm{t}\mathit{R}={\sigma }_{\mathrm{0}}$.

Result 2 Fix all components of $K$ except the boundary matrix $\mathit{G}$ and let $\lambda (\mathit{G})=\lambda (K)$ denote the eigenvalue. Then $\mathrm{d}{\lambda }_{\mathit{G}}(\mathit{T})=(u_{\mathrm{1}}(b),-u_{\mathrm{2}}(b))\left[\mathit{I}-\mathit{G}{(\mathit{G}+\mathit{T})}^{-\mathrm{1}}\right]\left(\begin{array}{c}{\overline{u}}_{\mathrm{2}}(b)\\ {\overline{u}}_{\mathrm{1}}(b)\end{array}\right)$, for all $\mathit{T}$, where $\mathrm{d}\mathrm{e}\mathrm{t}(\mathit{G}+\mathit{T})=\mathrm{d}\mathrm{e}\mathrm{t}\mathit{G}={\sigma }_{\mathrm{1}}$.

Result 3 Fix all components of $K$ except $p$ or $q$ and let $\lambda (p)=\lambda (K)\mathrm{o}\mathrm{r}\lambda (q)=\lambda (K)$ denote the eigenvalue. Then

$\begin{array}{l}\mathrm{d}{\lambda }_p(s)={\int }_a^b\mathrm{2}\mathrm{R}\mathrm{e}(u_{\mathrm{1}}{\overline{u}}_{\mathrm{2}})s\mathrm{d}x,\mathrm{o}\mathrm{r}\\ \mathrm{d}{\lambda }_q(s)={\int }_a^b\mathrm{2}\mathrm{R}\mathrm{e}(u_{\mathrm{1}}{\overline{u}}_{\mathrm{2}})s\mathrm{d}x,s\in L_{\mathrm{1}}(a,b).\end{array}$

Result 4 Fix all components of $K$ except $\mathit{Q}$, and let $\lambda (\mathit{Q})=\lambda (K)$ denote the eigenvalue. Then

$\mathrm{d}{\lambda }_{\mathit{Q}}(\mathit{P})={\int }_a^b{\mathcal{U}}^{\mathit{*}}\mathit{P}\mathcal{U}\mathrm{d}x,\mathrm{w}\mathrm{h}\mathrm{e}\mathrm{r}\mathrm{e}\mathit{P}\in M_{\mathrm{2}\times \mathrm{2}}(\mathbb{R})$

such that $\mathit{Q}+\mathit{P}$ is symmetric and every element in $\mathit{Q}+\mathit{P}$ belong to $L_{\mathrm{1}}(a,b)$.

Proof   1) Let $\lambda (\mathit{R}+\mathit{T}),(v_{\mathrm{1}}(x),v_{\mathrm{2}}(x),v_{\mathrm{3}},v_{\mathrm{4}}{)}^{\mathrm{T}}$ denote the eigenvalue and corresponding normalize eigenvector. Direct computation yields

$\lambda (\mathit{R}+\mathit{T}){\mathcal{v}}^{\mathit{*}}=\mathit{B}({\mathcal{v}}^{\mathit{*}}{)}^{\text{'}}+\mathit{Q}{\mathcal{v}}^{\mathit{*}},$(9)

$\lambda (\mathit{R})\mathcal{U}=\mathit{B}{\mathcal{U}}^{\text{'}}+\mathit{Q}\mathcal{U}.$(10)

By (9)-(10) and integration by parts, we obtain

$\left[\lambda (\mathit{R}+\mathit{T})-\lambda (\mathit{R})\right]{\int }_a^b{\mathcal{v}}^{\mathit{*}}\mathcal{U}dx=-{\overline{v}}_{\mathrm{1}}(b)u_{\mathrm{2}}(b)+{\overline{v}}_{\mathrm{2}}(b)u_{\mathrm{1}}(b)+u_{\mathrm{2}}(a){\overline{v}}_{\mathrm{1}}(a)-u_{\mathrm{1}}(a){\overline{v}}_{\mathrm{2}}(a)$(11)

Let

$\mathit{R}+\mathit{T}=\left(\begin{array}{cc}{\tilde{a}}_{\mathrm{0}}& {\tilde{a}}_{\mathrm{1}}\\ {\tilde{b}}_{\mathrm{0}}& {\tilde{b}}_{\mathrm{1}}\end{array}\right).$

By (4), we have

$\left[\lambda (\mathit{R}+\mathit{T})-\lambda (\mathit{R})\right]\frac{\mathrm{1}}{{\sigma }_{\mathrm{0}}}{u}_{\mathrm{3}}{\overline{v}}_{\mathrm{3}}=(u_{\mathrm{1}}(a),-u_{\mathrm{2}}(a))\frac{\mathrm{1}}{{\sigma }_{\mathrm{0}}}\left(\begin{array}{cc}{a}_{\mathrm{1}}{\tilde{b}}_{\mathrm{0}}-a_{\mathrm{0}}{\tilde{b}}_{\mathrm{1}}& a_{\mathrm{0}}{\tilde{a}}_{\mathrm{1}}-a_{\mathrm{1}}{\tilde{a}}_{\mathrm{0}}\\ b_{\mathrm{1}}{\tilde{b}}_{\mathrm{0}}-b_{\mathrm{0}}{\tilde{b}}_{\mathrm{1}}& b_{\mathrm{0}}{\tilde{a}}_{\mathrm{1}}-b_{\mathrm{1}}{\tilde{a}}_{\mathrm{0}}\end{array}\right)\left(\begin{array}{c}{\overline{v}}_{\mathrm{2}}(a)\\ {\overline{v}}_{\mathrm{1}}(a)\end{array}\right)=(u_{\mathrm{1}}(a),-u_{\mathrm{2}}(a))\left[-\mathit{R}{(\mathit{R}+\mathit{T})}^{-\mathrm{1}}\right]\left(\begin{array}{c}{\overline{v}}_{\mathrm{2}}(a)\\ {\overline{v}}_{\mathrm{1}}(a)\end{array}\right).$(12)