Spectral analysis of discontinuous Sturm-Liouville operators with Herglotzs transmission

1.   Introduction

In the present work, we shall investigate the spectra of the Sturm-Liouville equation

with the boundary condition

The spectral parameters not only appear in boundary condition, but also depend on the Herglotzs function

Here, $p(x) = \frac{1}{p^{2}_{1}}, x\in[-a, 0)$ and $p(x) = \frac{1}{p^{2}_{2}}, x\in(0, b]$; $q(x)$ is real valued continuous function in $J$; $p_{i}$, $\alpha_{i}$, $\beta_{i}$, $\alpha'_{i}$ and $\beta'_{i}\ (i = 1, 2)$ are nonzero real numbers. In the Herglotzs function, all parameters satisfy the following conditions: $a_{j}, b_{i} > 0, c_{i} < c_{i+1}, d_{j} < d_{j+1}, i = \overline{1, N-1}, j = \overline{1, M-1}$; $\eta$, $\kappa > 0$, $\xi$, $\zeta\in \mathbb{R}$ and $N$, $M\in\mathbb{N}_{0}$. Let

Then according to the properties of Herglotzs function (see ^[1]), we know that

where $\sigma$, $\tau\in\mathbb{R}$ and $\varepsilon_{i}$, $\epsilon_{j} > 0$, for $i = \overline{1, N'-1}$, $j = \overline{1, M'-1}$; $\gamma_{i} < \gamma_{i+1}$, $\delta_{j} < \delta_{j+1}$. Therefore

In many mathematical and physical models, it is necessary to study the eigenvalue of Sturm-Liouville problem and its corresponding eigenfunction. When the spectral parameters appear not only in differential equations, but also in boundary conditions, excellent results have been obtained (see ^[2,3,4]). Binding firstly studied the eigenvalue problem of Sturm-Liouville operator with boundary conditions dependent on spectral parameters

In addition, similar problems for differential equations with continuous coefficient $(p(x)\equiv1)$ and boundary conditions with spectral parameter were investigated in ^[5,6,7] and other works.

It is noteworthy that the boundary condition or the coefficient in the equation in the above results are all continuous. Now, the question is when the coefficient in the equation and the boundary condition are both discontinuous, could we still obtain the spectral properties of the linear eigenvalue problems? We know that discontinuous boundary value problems can also be found in many physical problems, such as, diffraction problem (see ^[8]), heat and mass transfer problem (see ^[9]) and vibrating problem (see ^[10]). To deal with the discontinuities, some conditions are necessary, such as, point interactions, impulsive conditions, transmission conditions, jump conditions or interface conditions (see^[11,12,13]). For example, in ^[], the author considered transmission conditions at one point and found asymptotic formulas of eigenvalues and corresponding eigenfunctions. Moreover, for similar problems, the work of literature ^[15] focused on Sturm-Liouville operators with a finite number of transmission conditions and established the self-adjointness of linear operator in a suitable Hilbert space.

Aspired by the above results, we consider the eigenvalue problems (1.1)–(1.5), where the coefficient $p(x)$ is discontinuous. In details, we consider the Sturm-Liouville equation in which the first coefficient may have the discontinuity at one point. Moreover, we allow boundary conditions and transmission conditions (Nevanlinna-Herglotz functions) to depend on spectral parameters. In Section 2, linear operator formulation is established, and the problems (1.1)–(1.5) can be interpreted as the eigenvalue problem of linear operator. The fundamental solutions and characteristic determinant are given in Section 3. Based on the operator formulation in the Hilbert space, the resolvent operator and self-adjointness of linear operator are constructed in the last Section.

Finally, for the sake of reader's convenience, let us introduce the properties of the Nevanlinna-Herglotz function as follows.

(i) if $\lambda = c_{i}$ ($c_{i}$ is the pole of $\mu(\lambda)$), then transmission condition (1.4) and (1.5) degenerates into $y(0^{+}) = 0$ and $y'(0^{-})\nu(\lambda) = -y(0^{-})$;

(ii) if $\mu(\lambda) = 0$ ($\lambda\neq c_{i}$ is the zero of $\mu(\lambda)$), then $\Delta'y = 0$;

(iii) if $\lambda = d_{j}$ ($d_{j}$ is the pole of $\nu(\lambda)$), then transmission condition (1.4) and (1.5) degenerates into $y'(0^{-}) = 0$ and $y(0^{+})\mu(\lambda) = y'(0^{+})$;

(iv) if $\nu(\lambda) = 0$ ($\lambda\neq d_{j}$ is the zero of $\nu(\lambda)$), then $\Delta y = 0$.

2.   The operator eigenvalue problem

In this section, we define a special inner product in the Hilbert space

and a linear operator $A$ defined on $\mathcal{H}$. Moreover, the Sturm-Liouville problems (1.1)–(1.5) can be considered as the operator eigenvalue problem.

Define

For convenience's sake, we use the following notations:

For $\eta$, $\kappa > 0$, we define a new inner product in $\mathcal{H}$ by

for

where $\langle\cdot, \cdot\rangle_{1}$ denotes Euclidean inner product.

In the Hilbert space $\mathcal{H}$, we consider the operator $A$ which is defined by

with the domain

where $[\gamma_{i}]: = {\rm{diag}}(\gamma_{1}, \cdot\cdot\cdot, \gamma_{N'})$, $[\delta_{j}]: = {\rm{diag}}(\delta_{1}, \cdot\cdot\cdot, \delta_{M'})$, $\boldsymbol{\varepsilon}: = (\varepsilon_{i})$ and $\boldsymbol{\epsilon}: = (\epsilon_{j})$.

Lemma 2.1. The domain $D(A)$ is dense in $\mathcal{H}$.

Proof. Let $W = (w, f_{1}, f_{2}, \textbf{f}^{1}, \textbf{f}^{2})^{T}\in \mathcal{H}$, where $w\in C^{\infty}[-a, 0)\cup(0, b]$ satisfying

and the condition

Meanwhile,

Then, it is easy to verify that $W\in D(A)$. Next, as long as it is proved that the elements in $\mathcal{H}$ can be approximated by the elements in $D(A)$, the desired result can be obtained.

Since

and

there exists a sequence $\{m_{n}\}\in C_{0}^{\infty}(-a, 0)\oplus C_{0}^{\infty}(0, b)$ with $m_{n}\rightarrow f-w$ as $n\rightarrow \infty$, where $M_{n}: = (m_{n}, 0, 0, \textbf{0}, \textbf{0})^{T}\in D(A)$. Therefore, $W+M_{n}\rightarrow F$ as $n\rightarrow \infty$ giving that $\overline{D(A)}\supset\mathcal{H}$.

Theorem 2.1. The operator eigenvalue problem $AF = \lambda F$ and the considered Sturm-Liouville problems (1.1)–(1.5) are equivalent and the eigenfunction is the first components of the corresponding eigenelements of the operator $A$. Moreover, for $\eta$, $\kappa > 0$, we have following results:

(i) if $\lambda\neq\gamma_{i} \; \forall i = \overline{1, N'}$, then ${\bf{f}}^{1} = (\lambda I-[\gamma_{i}])^{-1}\boldsymbol{\varepsilon}\Delta'f;$ if $\lambda = \gamma_{I} \; \exists I\in\{\overline{1, N'}\}$, then ${\bf{f}}^{1} = \frac{-f(0^{+})}{\varepsilon_{I}}e^{I};$

(ii) if $\lambda\neq\delta_{j} \; \forall j = \overline{1, M'}$, then ${\bf{f}}^{2} = (\lambda I-[\delta_{i}])^{-1}\boldsymbol{\epsilon}\Delta f$; if $\lambda = \gamma_{J} \; \exists J\in\{\overline{1, M'}\}$, then ${\bf{f}}^{2} = \frac{f(0^{-})}{\epsilon_{J}}e^{J},$ where $\textbf{e}^{n}$ is the vector in $\mathbb{R}^{n}$ with and except the $n$-th element is 1, all other elements are 0.

Proof. We just need to show that the eigenelement $f$ of the operator $A$ obeys the boundary conditions (1.2) and (1.3) and transfer conditions (1.4) and (1.5). It is clear that $f$ satisfies (1.2) and (1.3). The definition of $A$ implies $\gamma_{i}f^{1}_{i}+\varepsilon_{i}\Delta'f = \lambda f^{1}_{i}$ for all $i$. Meanwhile, the domain of $A$ gives $-f(0^{+})+\sigma\Delta'f-\langle\textbf{f}^{1}, \boldsymbol{\varepsilon}\rangle_{1} = 0$. Thus, if $\lambda\neq\gamma_{i}$ for all $i$, then

If $\lambda = \gamma_{I}$ for some $I\in\{1, \cdot\cdot\cdot, N'\}$, then $-f(0^{+})-\langle f_{I}^{1}, \varepsilon_{I}\rangle_{1} = 0$. That is, $f^{1}_{I} = \frac{-f(0^{+})}{\varepsilon_{I}}$. Hence, $f$ satisfies (1.4).

Similarly, if $\lambda\neq\delta_{j}$ for all $j$, then $f^{2}_{j} = \frac{\epsilon_{j}}{\lambda-\delta_{j}}\Delta f$ and

While $\lambda = \delta_{J}$ for some $j\in\{1, \cdot\cdot\cdot, M'\}$, the domain condition forces $f^{2}_{J} = \frac{f'(0^{-})}{\epsilon_{J}}$ from which (1.5) follows.

Theorem 2.2. The linear operator $A$ is symmetric.

Proof. Let $F$, $G\in D(A)$. Then it follows from the problem (1.1) and the relation (2.1) that

Moreover, we have

Meanwhile, the vector components satisfy

and the domain condition $D(A)$ implies

Therefore,

It can be obtained by simple calculation

Thus, $\langle AF, G\rangle-\langle F, AG\rangle = 0$ and so $A$ is symmetric.

Corollary 2.1. All eigenvalues of the Sturm-Liouville problems (1.1)–(1.5) are real.

Corollary 2.2. Let $\lambda_{1}$ and $\lambda_{2}$ be two different eigenvalues of the Sturm-Liouville problems (1.1)–(1.5). Then the corresponding the eigenfunctions $u_{1}(x)$ and $u_{2}(x)$ are orthogonal, i.e.,

3.   Fundamental solutions and characteristic determinant

Lemma 3.1. (^[5]) All eigenvalues of Sturm-Liouville problems (1.1)–(1.5), not at poles of $\mu(\lambda)$ or $\nu(\lambda)$, are geometrically simple. In the case, Herglotz condition (1.4) and (1.5) can be transformed into

Lemma 3.2. (^[16]) Let $q(x)\in C[-a, b]$, and $f(\lambda)$ and $g(\lambda)$ be given entire functions. Then for $\forall \lambda\in\mathbb{C}$, the equation

has a unique solution $u = u(x, \lambda)$ satisfying the initial conditions

Moreover, for each fixed $x\in[-a, b]$, $u = u(x, \lambda)$ is an entire function of $\lambda$.

Lemma 3.3. Let $u_{-}(x, \lambda), x\in[-a, 0)$ be the solution of the Sturm-Liouville problem (1.1) satisfying conditions

and $v(x, \lambda), x\in (0, b]$ denote the solution of the Sturm-Liouville problem (1.1) satisfying the conditions

Then the Wronskian $W[u_{-}, v_{+}] = u_{-}v'_{+}-v_{+}u'_{-}$ is independent of $x$.

Proof. Direct computation, we have

It follows that the Wronskian $W[u_{-}(x, \lambda), v_{+}(x, \lambda)]$ is constant on $[-a, 0)\cup(0, b]$ and by virtue of Lemma 3.2, it is a function of $\lambda$.

We know that the problem (1.1) exists two fundamental solutions on whole $[-a, 0)\cup(0, b]$ satisfying the boundary conditions (1.2)–(1.5). First, we extend $u_{-}(x, \lambda)$ and $v_{+}(x, \lambda)$ by the zero function to $[-a, 0)\cup(0, b]$, i.e., we define

and

Now by virtue of Lemma 3.1, it's workable to extend $u_{-}(x, \lambda)$, $x\in[-a, 0)$ and $v_{+}(x, \lambda)$, $x\in(0, b]$ by nontrivial solution $u_{+}(x, \lambda)$, $x\in(0, b]$ and $v_{-}(x, \lambda)$, $x\in[-a, 0)$ satisfying the conditions

and

Moreover, we define two linearly independent solutions of the problem (1.1) on the whole $[-a, 0)\cup(0, b]$ as

It note that $u$ and $v$ must satisfy boundary conditions (1.2)–(1.5). Let

According to (1.4) and (1.5), we have

Moreover, we define

and any solution to problem (1.1) on $[-a, 0)\cup(0, b]$ satisfying the boundary conditions (1.2) and (1.3) must be of the form (3.5). Let

Therefore, $\omega(\lambda)$ will be referred to as the characteristic determinant of (1.1)–(1.5). It is shown in Theorem 3.1 below that $\omega(\lambda)$ has the properties expected of the characteristic determinant.

Theorem 3.1. The eigenvalue $\lambda$ of the Sturm-Liouville problems (1.1)–(1.5) consists of the zero of the characteristic determinant.

Proof. The relation (3.6) implies

That is, $\lambda$ is an eigenvalue of Sturm-Liouville problems (1.1)–(1.5) if and only if $U_{1}(y, \lambda) = 0$, $U_{2}(y, \lambda) = 0$. Moreover, these two equations exist nontrivial solution $\varphi(\lambda)$ and $\psi(\lambda)$ if and only if

Therefore, we proved that the eigenvalue of Sturm-Liouville problems (1.1)–(1.5) coincides with the zero of $\omega(\lambda)$.

Similar to Lemma 3.3, let $W[u(x, \lambda), v(x, \lambda)] = :\varpi(\lambda)$. In view of Theorem 3.1, we define

and the Green's function of the Sturm-Liouville problems (1.1)–(1.5) is given by

Theorem 3.2. Let

Then $g(x, \lambda)$ is the solution of operator equation $(\lambda-L)g = ph$ on J. Moreover, $g$ satisfies the boundary conditions (1.2)–(1.5).

Proof. The relation (3.8) implies

Furthermore, we have

and

Therefore, $(\lambda-L)g = ph$ holds.

It remains only to show that $g$ satisfies (1.2) and (1.3), (1.4) and (1.5). By (3.9) and (3.10), we obtain

Moreover, we know that $u$ satisfies (1.2). Then, $g$ satisfies (1.2). Similarly, $g$ satisfies (1.3). Moreover,

Obviously, (1.4) and (1.5) are obeyed.

4.   The resolvent operator of $A$

In this section, we study the resolvent operator in the Hilbert space $\mathcal{H}$. We first consider nonhomogeneous conditions

Meanwhile, the domain of the operator $A$ implies

If $\lambda\neq\gamma_{i}$ for all $i$, then from (4.1) we have

By inner product calculation, we get

Therefore, by (1.7), we have

If $\lambda = \gamma_{I}$ for some $I\in\{1, \cdot\cdot\cdot, N'\}$, then from (4.1) we have $\Delta'f = -\frac{ph^{1}_{I}}{\boldsymbol{\varepsilon}_{I}}$. For $i\in\{1, \cdot\cdot\cdot, N'\}\setminus I$, $f^{1}_{i} = \frac{h^{1}_{i}+\varepsilon_{i}\Delta'f}{\gamma_{I}-\gamma_{i}}$. Thus, from (4.3) we get

Similarly, if $\lambda\neq\delta_{j}$ for all $j$, then

If $\lambda = \delta_{J}$ for some $J\in\{1, \cdot\cdot\cdot, M'\}$, then

Therefore, the operator equation

is equivalent to the discontinuous nonhomogeneous BVP

together with inhomogeneous boundary condition

and transmission conditions (the case of $\lambda\neq\gamma_{i}$ and $\lambda\neq\delta_{j}$)

We consider the resolvent set $\rho(A) = \{\lambda\in \mathbb{C}|(\lambda I-A)^{-1}\in D(A)\}$. Then, we need to show $(\lambda I-A)^{-1}$ is the resolvent operator, just prove $(\lambda I-A)^{-1}\in D(A)$.

Theorem 4.1. Let $\lambda$ not be an eigenvalue of operator $A$. Then

Proof. Obviously, the resolvent operator $(\lambda I-A)^{-1}$ exists. It remains only to show $\tilde{G}h\in D(A)$. The definition of $T_{\lambda}h$ and Theorem 3.2 imply that $g\in AC[-a, b]$, $g'\in AC[-a, 0)\cup(0, b]$ and obeys the boundary conditions (1.2) and (1.3). Moreover, the equalities

and

hold. Meanwhile, the properties of the Nevanlinna-Herglotz function imply that if $\lambda = \gamma_{I}$ for some $I\in \{1, \cdot\cdot\cdot, N'\}$, then $(\Delta'T_{\lambda}h) = 0$ and ${\bf{g}}^{1} = \frac{-y(0^{+})}{\varepsilon_{I}}e^{I}.$ Therefore,

Similarly, we have

Thus, $\tilde{G}h\in D(A)$ and the desired result holds.

Theorem 4.2. Let $R(\lambda, A) = (\lambda I-A)^{-1}$. Then

holds, where $\forall\lambda\in\mathbb{C}$ satisfies ${\rm{Im}}\lambda\neq0$.

Proof. For $\forall\ H = (ph, ph_{1}, ph_{2}, p\textbf{h}_{3}, p\textbf{h}_{4})^{T}\in\mathcal{H}$ and $Y = R(\lambda, A)H$. Since $(\lambda I-A)Y = H$, we have

and

which imply $|{\rm{Im}}\lambda|\|Y\|^{2} = |{\rm{Im}}(H, Y)|$. On the other side, in view of Cauchy-Schwartz inequality, we get

Therefore, the inequality

holds.

Theorem 4.3. In Hilbert space $\mathcal{H}$, the operator $A$ is self-adjoint.

Proof. Obviously, $A$ is a dense symmetric operator. To show that $A$ is self-adjoint, it remains only to verify that $D(A^{\ast}) = D(A)$. Let $H\in D(A^{\ast})$. Then

It follows from (4.5) that

Note that $\lambda = -i$ is a regular point. Then, we have

Substituting (4.7) in (4.6), we have

Similarly, from $\lambda = i$ is a regular point, let $G = R(i, A)(H-Y)$. Then by (4.8), we have $H = Y$ and thus $H\in D(A)$.

Acknowledgments

This work was supported by National Natural Science Foundation of China [Grant No.11961060].

Conflict of interest

The authors declare there is no conflicts of interest.