The Resolvent of Impulsive Singular Hahn–Sturm–Liouville Operators

Bilender P. Allahverdiev; Hüseyin Tuna; Hamlet A. Isayev

doi:10.2478/amsil-2024-0001

1. Introduction

Impulsive differential equations are one of the interesting topics in the theory of differential equations. These equations serve as basic models to study the dynamics of processes that are subject to sudden changes in their states. These types of problems are especially encountered in heat and mass transfer problems ([17]). There are many studies on this subject in the literature [2, 6, 7, 8, 9, 12, 13, 14, 19, 23].

W. Hahn introduced the concept of the Hahn derivative to the literature in 1949 [10]. With this definition he made, he gathered two important operators under a single structure. These are the q-difference and forward difference operators. In 2018, Annaby et al. [4] using this definition instead of the classical derivative, investigated the fundamental properties of the Sturm–Liouville problems. In [5], the authors studied singular q-Sturm-Liouville equations. In [18], the author studied a q-analog of the singular Dirac problem. Recently in [21], the author proved a spectral expansion theorem by constructing the spectral function of the Hahn–Sturm–Liouville equation in the singular case under impulsive conditions.

In this paper, our aim is to consider Hahn–Sturm–Liouville problems under impulsive boundary conditions. The integral representation of the resolvent operator corresponding to this type of problem will be obtained using Weyl’s method [16, 22, 24].

2. Preliminaries

Now, we provide a concise overview of the Hahn calculus [3, 4, 10, 11]. Let q ∈ (0, 1), ω₀ := ω/ (1 − q) , ω > 0, and let Ψ : J ⊂ ℝ ω ℝ be a function such that ω₀ ∈ J.

Definition 2.1 ([10], [11]).

The Hahn derivative D_ω,q Ψ is defined by

D_{ω, q} Ψ (ζ) = \begin{array}{l} \frac{Ψ (ω + q ζ) - Ψ (ζ)}{ω + (q - 1) ζ}, & ζ \neq ω_{0}, \\ Ψ^{'} (ω_{0}), & ζ = ω_{0}, \end{array}

where the expression Ψ^′ (ω₀) shows the ordinary derivative of Ψ at ω₀.

Definition 2.2 ([3]).

Let a, b, ω₀ ∈ J. The Hahn integral (ω, q-integral) is defined by

\int_{a}^{b} Ψ (ζ) d_{ω, q} ζ : = \int_{ω_{0}}^{b} Ψ (ζ) d_{ω, q} ζ - \int_{ω_{0}}^{a} Ψ (ζ) d_{ω, q} ζ,

where

\int_{ω_{0}}^{ζ} Ψ (t) d_{ω, q} t : = ((1 - q) ζ - ω) \sum_{n = 0}^{\infty} q^{n} Ψ (ω \frac{1 - q^{n}}{1 - q} + ζ q^{n}), ζ \in J,

provided that the series converges at ζ = a and ζ = b.

Definition 2.3.

The ω, q-Wronskian of Ψ₁ and Ψ₂ is defined by

W_{ω, q} (Ψ_{1}, Ψ_{2}) : = Ψ_{1} D_{ω, q} Ψ_{2} - Ψ_{2} D_{ω, q} Ψ_{1} .

3. Main results

Let us consider the following impulsive boundary-value problem (BVP)

(3.1)

- \frac{1}{q} D_{- \frac{ω}{q}, \frac{1}{q}} D_{ω, q} y (ζ) + v (ζ) y (ζ) = λ y (ζ), ζ \in (ω_{0}, d) \cup (d, \frac{1}{q^{n}}),

(3.2)

y (ω_{0}, λ) \cos β + D_{- \frac{ω}{q}, \frac{1}{q}} y (ω_{0}, λ) \sin β = 0,

(3.3)

y (d -) = η y (d +),

(3.4)

D_{- \frac{ω}{q}, \frac{1}{q}} y (d -) = \frac{1}{η} D_{- \frac{ω}{q}, \frac{1}{q}} y (d +),

(3.5)

y (\frac{1}{q^{n}}, λ) \cos γ + D_{- \frac{ω}{q}, \frac{1}{q}} y (\frac{1}{q^{n}}, λ) \sin γ = 0,

where q ∈ (0, 1), ω₀ := ω/ (1 − q), ω > 0, γ, β ∈ ℝ,

\frac{1}{q^{n}} > d

, n ∈ ℕ := {1, 2, 3, . . .}, η > 0, λ ∈ ℂ, y(d±) := lim_ζd± y (ζ) , v is a real-valued continuous function on [ω₀, d) ∪ (d, ∞), and has finite limits v(d±).

A similar problem has been studied by the authors without impulsive boundary conditions ([1]).

$H_{n} = L_{ω, q}^{2} (ω_{0}, d) \dot{+} L_{ω, q}^{2} (d, \frac{1}{q^{n}})$ , $\frac{1}{q^{n}} > d$ , $n \in ℕ (H = L_{ω, q}^{2} (ω_{0}, d) \dot{+} L_{ω, q}^{2} (d, \infty))$ is a Hilbert space endowed with the following inner product

\begin{array}{l} {〈 y, z 〉}_{n} : = \int_{ω_{0}}^{d} y^{(1)} \bar{z^{(1)}} d_{ω, q} ζ + \int_{d}^{\frac{1}{q^{n}}} y^{(2)} \bar{z^{(2)}} d_{ω, q} ζ, \\ (〈 y, z 〉 : = \int_{ω_{0}}^{d} y^{(1)} \bar{z^{(1)}} d_{ω, q} ζ + \int_{d}^{\infty} y^{(2)} \bar{z^{(2)}} d_{ω, q} ζ) \end{array}

where

y (ζ) = \begin{array}{l} y^{(1)} (ζ), & ζ \in ω_{0}, d), \\ y^{(2)} (ζ), & ζ \in (d, \infty), \end{array}

and

z (ζ) = \begin{array}{l} z^{(1)} (ζ), & ζ \in ω_{0}, d), \\ z^{(2)} (ζ), & ζ \in (d, \infty) . \end{array}

Let

ψ (ζ, λ) = \begin{array}{l} ψ^{(1)} (ζ, λ), & ζ \in ω_{0}, d), \\ ψ^{(2)} (ζ, λ), & ζ \in (d, \infty), \end{array}

and

θ (ζ, λ) = \begin{array}{l} θ^{(1)} (ζ, λ), & ζ \in ω_{0}, d), \\ θ^{(2)} (ζ, λ), & ζ \in (d, \infty), \end{array}

be solutions of Eq. (3.1) satisfying the following conditions

\begin{array}{l} ψ^{(1)} (ω_{0}, λ) = \cos β, D_{- \frac{ω}{q}, \frac{1}{q}} ψ^{(1)} (ω_{0}, λ) = \sin β, \\ θ^{(1)} (ω_{0}, λ) = \sin β, D_{- \frac{ω}{q}, \frac{1}{q}} θ^{(1)} (ω_{0}, λ) = - \cos β, \end{array}

and

\begin{array}{l} θ (d -, λ) & = & η θ (d +, λ), \\ D_{- \frac{ω}{q}, \frac{1}{q}} θ (d -, λ) & = & \frac{1}{η} D_{- \frac{ω}{q}, \frac{1}{q}} θ (d +, λ), \\ ψ (d -, λ) & = & η ψ (d +, λ), \\ D_{- \frac{ω}{q}, \frac{1}{q}} ψ (d -, λ) & = & \frac{1}{η} D_{- \frac{ω}{q}, \frac{1}{q}} ψ (d +, λ) . \end{array}

Then the solution of Eq. (3.1) be represented

ψ (ζ, λ) + 𝓁 (λ, \frac{1}{q^{n}}) θ (ζ, λ)

which satisfies the boundary condition

\begin{array}{l} (D_{- \frac{ω}{q}, \frac{1}{q}} ψ^{(2)} (\frac{1}{q^{n}}, λ) + 𝓁 (λ, \frac{1}{q^{n}}) D_{- \frac{ω}{q}, \frac{1}{q}} θ^{(2)} (\frac{1}{q^{n}}, λ)) \sin γ \\ + (ψ^{(2)} (\frac{1}{q^{n}}, λ) + 𝓁 (λ, \frac{1}{q^{n}}) θ^{(2)} (\frac{1}{q^{n}}, λ)) \cos γ = 0. \end{array}

Hence

𝓁 (λ, \frac{1}{q^{n}}) = - \frac{ψ^{(2)} (\frac{1}{q^{n}}, λ) cot γ + D_{- \frac{ω}{q}, \frac{1}{q}} ψ^{(2)} (\frac{1}{q^{n}}, λ)}{θ (2) (\frac{1}{q^{n}}, λ) cot γ + D_{- \frac{ω}{q}, \frac{1}{q}} θ ψ^{(2)} (\frac{1}{q^{n}}, λ)} .

Lemma 3.1.

Let

Z_{\frac{1}{q^{n}}} (ζ, λ) = ψ (ζ, λ) + 𝓁 (λ, \frac{1}{q^{n}}) θ (ζ, λ),

where

Z_{\frac{1}{q^{n}}} \in H_{n}

and

\frac{1}{q^{n}} > d

, n ∈ ℕ. Then, for each nonreal λ, the following relations hold:

\begin{array}{l} Z_{\frac{1}{q^{n}}} (ζ, λ) \to Z (ζ, λ), n \to \infty, \\ \int_{ω_{0}}^{d} {Z_{\frac{1}{q^{n}}} (ζ, λ)|}^{2} d_{ω, q} ζ + \int_{d}^{\frac{1}{q^{n}}} {Z_{\frac{1}{q^{n}}} (ζ, λ)|}^{2} d_{ω, q} ζ \\ \to \int_{ω_{0}}^{d} {Z (ζ, λ)|}^{2} d_{ω, q} ζ + \int_{d}^{\infty} {(ζ, λ)|}^{2} d_{ω, q} ζ, n \to \infty . \end{array}

Proof

It is immediate that

Z_{\frac{1}{q^{n}}} (ζ, λ) = Z (ζ, λ) + 𝓁 (λ, \frac{1}{q^{n}}) - m (λ)] θ (ζ, λ)

where Z(·, λ) ∈ H and m(λ) is the Titchmarsh–Weyl function.

𝓁 (λ, \frac{1}{q^{n}})

varies on a circle with a finite radius

r_{\frac{1}{q^{n}}}

in the plane. In the limit-circle case,

𝓁 (λ, \frac{1}{q^{n}}) \to m (λ) (n \to \infty)

; therefore

Z_{\frac{1}{q^{n}}} (ζ, λ) \to Z (ζ, λ) (n \to \infty) .

Hence

\begin{array}{l} \int_{ω_{0}}^{d} {Z_{\frac{(1) 1}{q^{n}}} (ζ, λ)|}^{2} d_{ω, q} ζ + \int_{d}^{\frac{1}{q^{n}}} {Z_{\frac{(2) 1}{q^{n}}} (ζ, λ)|}^{2} d_{ω, q} ζ \\ \to \int_{ω_{0}}^{d} {Z^{(1)} (ζ, λ)|}^{2} d_{ω, q} ζ + \int_{d}^{\infty} {Z^{(2)} (ζ, λ)|}^{2} d_{ω, q} ζ (n \to \infty), \end{array}

due to Z(·, λ) ∈ H. In the limit-point case, we find

\begin{array}{l} 𝓁 (λ, \frac{1}{q^{n}}) - m (λ)| \leq r_{\frac{1}{q^{n}}} \\ = {(2 Im λ \int_{ω_{0}}^{d} {θ^{(1)} (ζ, λ)|}^{2} d_{ω, q} ζ + \int_{d}^{\frac{1}{q^{n}}} {θ^{(2)} (ζ, λ)|}^{2} d_{ω, q} ζ])}^{- 1}, \end{array}

where Im λ ≠ 0. As

r_{\frac{1}{q^{n}}} \to 0

Z_{\frac{1}{q^{n}}} (ζ, λ) \to Z (ζ, λ) (n \to \infty)

. Moreover, we have

\begin{array}{l} \int_{ω_{0}}^{d} {| {ℓ (λ, \frac{1}{q^{n}}) - m (λ)} θ^{(1)} (ζ, λ) |}^{2} d_{ω, q} ζ \\ + \int_{d}^{\frac{1}{q^{n}}} {| {𝓁 (λ, \frac{1}{q^{n}}) - m (λ)} θ^{(2)} (ζ, λ) |}^{2} d_{ω, q} ζ \\ = {| 𝓁 (λ, \frac{1}{q^{n}}) - m (λ) |}^{2} (\int_{ω_{0}}^{d} {| θ^{(1)} (ζ, λ) |}^{2} d_{ω, q} ζ + \int_{d}^{\frac{1}{q^{n}}} {| θ^{(2)} (ζ, λ) |}^{2} d_{ω, q} ζ) \\ \leq {(4 {(Im λ)}^{2} [\int_{ω_{0}}^{d} {| θ^{(1)} (ζ, λ) |}^{2} d_{ω, q} ζ + \int_{d}^{\frac{1}{q^{n}}} {| θ^{(2)} (ζ, λ) |}^{2} d_{ω, q} ζ])}^{- 1}, \end{array}

which implies that

\begin{array}{l} \int_{ω_{0}}^{d} {| Z_{\frac{(1) 1}{q^{n}}} (ζ, λ) |}^{2} d_{ω, q} ζ + \int_{d}^{\frac{1}{q^{n}}} {| Z_{\frac{(2) 1}{q^{n}}} (ζ, λ) |}^{2} d_{ω, q} ζ \\ \to \int_{ω_{0}}^{d} {| Z^{(1)} (ζ, λ) |}^{2} d_{ω, q} ζ + \int_{d}^{\infty} {| Z^{(2)} (ζ, λ) |}^{2} d_{ω, q} ζ . \end{array}

Let $f \in H_{n} (\frac{1}{q^{n}} > d, n \in ℕ)$ . Define

(3.6)

\begin{array}{l} G_{\frac{1}{q^{n}}} (ζ, ς, λ) = \begin{array}{l} Z_{\frac{1}{q^{n}}} (ζ, λ) θ (ζ, λ), & ς \leq ζ, \\ θ (ζ, λ) Z_{\frac{1}{q^{n}}} (ς, λ), & ς > ζ, \end{array} \\ \begin{array}{l} (R_{\frac{1}{q^{n}}} f) (ζ, λ) = \int_{ω_{0}}^{d} G_{\frac{1}{q^{n}}} (ζ, ς, λ) f^{(1)} (ς) d_{ω, q} ς \\ + \int_{d}^{\frac{1}{q^{n}}} G_{\frac{1}{q^{n}}} (ζ, ς, λ) f^{(2)} (ς) d_{ω, q} ς, λ \in ℂ . \end{array} \end{array}

Without loss of generality, we can assume that λ = 0 is not an eigenvalue of the BVP (3.1)–(3.5). Now let us prove that the resolvent operator is compact.

Theorem 3.2.

$G_{\frac{1}{q^{n}}} (ζ, ς) (λ = 0) (\frac{1}{q^{n}} > d, n \in ℕ)$ defined as (3.6) is a ω, q-Hilbert–Schmidt kernel, i.e.,

\begin{array}{l} \int_{ω_{0}}^{d} \int_{ω_{0}}^{d} | G_{\frac{1}{q^{n}}} (ζ, ς) |^{2} d_{ω, q} ζ d_{ω, q} ς < + \infty, \\ \int_{d}^{\frac{1}{q^{n}}} \int_{d}^{\frac{1}{q^{n}}} | G_{\frac{1}{q^{n}}} (ζ, ς) |^{2} d_{ω, q} ζ d_{ω, q} ς < + \infty . \end{array}

Proof

By (3.6), it is obvious that

\begin{array}{l} \int_{ω_{0}}^{d} d_{ω, q} ζ \int_{ω_{0}}^{d} | G_{\frac{1}{q^{n}}} (ζ, ς) |^{2} d_{ω, q} ς < + \infty, \\ \int_{d}^{\frac{1}{q^{n}}} d_{ω, q} ζ \int_{d}^{\frac{1}{q^{n}}} | G_{\frac{1}{q^{n}}} (ζ, ς) |^{2} d_{ω, q} ς < + \infty, \end{array}

due to

Z_{\frac{1}{q^{n}}}

, (·, λ), θ (·, λ) ∈ H_n

(\frac{1}{q^{n}} > d, n \in ℕ)

. Hence

(3.7)

\begin{array}{l} \int_{ω_{0}}^{d} \int_{ω_{0}}^{d} | G_{\frac{1}{q^{n}}} (ζ, ς) |^{2} d_{ω, q} ζ d_{ω, q} ς < + \infty, \\ \int_{d}^{\frac{1}{q^{n}}} \int_{d}^{\frac{1}{q^{n}}} | G_{\frac{1}{q^{n}}} (ζ, ς) |^{2} d_{ω, q} ζ d_{ω, q} ς < + \infty . \end{array}

Theorem 3.3 ([20]).

Let A {t_i} = {x_i}, i ∈ ℕ, where

(3.8)

x_{i} = \sum_{k = 1}^{\infty} η_{i k} t_{k}, i, k \in ℕ .

(3.9)

\sum_{i, k = 1}^{\infty} | η_{i k} |^{2} < + \infty,

then the operator A is compact in l².

Theorem 3.4.

Let 𝒯 be the ω, q-integral operator 𝒯: H_n → H_n ( $\frac{1}{q^{n}} > d, n \in ℕ$ ),

(T f) (ζ) = \begin{array}{l} \int_{ω_{0}}^{d} G_{\frac{1}{q^{n}}} (ζ, ς) f^{(1)} (ς) d_{ω, q} ς, ζ \in ω_{0}, d), \\ \int_{d}^{\frac{1}{q^{n}}} G_{\frac{1}{q^{n}}} (ζ, ς) f^{(2)} (ς) d_{ω, q} ς, ζ \in d, \frac{1}{q^{n}}], \end{array}

where

f (ζ) = \begin{array}{l} f^{(1)} (ζ), ζ \in ω_{0}, d), \\ f^{(2)} (ζ), ζ \in d, \frac{1}{q^{n}}] . \end{array}

Then 𝒯 is a compact self-adjoint operator in space H_n.

Proof

Let

φ_{i} : = φ_{i} (ζ) = \begin{array}{l} φ_{i}^{(1)} (ζ), & ζ \in ω_{0}, d), \\ φ_{i}^{(2)} (ζ), & ζ \in d, \frac{1}{q^{n}}], \end{array} (i, n \in ℕ, \frac{1}{q^{n}} > d)

be a complete, orthonormal basis of H_n. Let i, k, n ∈ ℕ,

\frac{1}{q^{n}} > d

. Write

\begin{array}{l} t_{i} = {〈 f, φ_{i} 〉}_{n} = \int_{ω_{0}}^{d} f^{(1)} (ζ) \bar{φ_{i}^{(1)} (ζ)} d_{ω, q} ζ \\ + \int_{d}^{\frac{1}{q^{n}}} f^{(2)} (ζ) \bar{φ_{i}^{(2)} (ζ)} d_{ω, q} ζ, \\ x_{i} = {〈 g, φ_{i} 〉}_{n} = \int_{ω_{0}}^{d} g^{(1)} (ζ) \bar{φ_{i}^{(1)} (ζ)} d_{ω, q} ζ \\ + \int_{d}^{\frac{1}{q^{n}}} g^{(2)} (ζ) \bar{φ_{i}^{(2)} (ζ)} d_{ω, q} ζ, \\ η_{i k} = \int_{ω_{0}}^{d} \int_{ω_{0}}^{d} G_{\frac{1}{q^{n}}} (ζ, ς) φ_{i}^{(1)} (ζ) \bar{φ_{k}^{(1)} (ς)} d_{ω, q} ζ d_{ω, q} ς \\ + \int_{d}^{\frac{1}{q^{n}}} \int_{d}^{\frac{1}{q^{n}}} G_{\frac{1}{q^{n}}} (ζ, ς) φ_{i}^{(2)} (ζ) \bar{φ_{k}^{(2)} (ς)} d_{ω, q} ζ d_{ω, q} ς . \end{array}

H_n is mapped isometrically on to l². By this mapping, 𝒯 transforms into the operator A defined by (3.8) in l² and (3.7) is translated into (3.9). By Theorems 3.2 and 3.3, we see that A and 𝒯 are compact operators.

Let h, g ∈ H_n and $\frac{1}{q^{n}} > d$ , n ∈ ℕ. Then we have

\begin{array}{l} {〈 𝒯 h, g 〉}_{n} & = \int_{d}^{ω_{0}} (𝒯 h^{(1)}) (ζ) \bar{g^{(1)} (ζ}) d_{ω, q} ζ + \int_{\frac{1}{q^{n}}}^{d} (𝒯 h^{(2)}) (ζ) \bar{g^{(2)} (ζ}) d_{ω, q} ζ \\ = \int_{d}^{ω_{0}} \int_{d}^{ω_{0}} G_{\frac{1}{q^{n}}} (ζ, ς) h^{(1)} (ς) d_{ω, q} ς \bar{g^{(1)} (ζ)} d_{ω, q} ζ \\ + \int_{\frac{1}{q^{n}}}^{d} \int_{\frac{1}{q^{n}}}^{d} G_{\frac{1}{q^{n}}} (ζ, ς) h^{(2)} (ς) d_{ω, q} ς \bar{g^{(2)} (ζ)} d_{ω, q} ζ \\ = \int_{d}^{ω_{0}} h^{(1)} (ς) (\int_{d}^{ω_{0}} G_{\frac{1}{q^{n}}} (ς, ζ) \bar{g^{(1)} (ζ)} d_{ω, q} ζ) d_{ω, q} ς \\ + \int_{\frac{1}{q^{n}}}^{d} h^{(2)} (ς) (\int_{\frac{1}{q^{n}}}^{d} G_{\frac{1}{q^{n}}} (ς, ζ) \bar{g^{(2)} (ζ)} d_{ω, q} ζ) d_{ω, q} ς \\ = {〈 h, T g 〉}_{n}, \end{array}

since

G_{\frac{1}{q^{n}}} (ζ, γ)

is a symmetric function.

From Theorem 3.4, we conclude that 𝒯 has a discrete spectrum. Let $λ_{m, \frac{1}{q^{n}}}$ and

θ_{m, \frac{1}{q^{n}}} (ζ) : = \begin{array}{l} θ_{m, \frac{1}{q^{n}}}^{(1)} (ζ, λ_{m, \frac{1}{q^{n}}}), ζ \in ω_{0}, d), \\ θ_{m, \frac{1}{q^{n}}}^{(2)} (ζ, λ_{m, \frac{1}{q^{n}}}), ζ \in d, \frac{1}{q^{n}}], \end{array} (m, n \in ℕ, \frac{1}{q^{n}} > d)

be the eigenvalues and eigenfunctions of the BVP (3.1)–(3.5) and

α_{m, \frac{1}{q^{n}}}^{2} = \int_{ω_{0}}^{d} θ_{m, \frac{1}{q^{n}}}^{(1) 2} (ζ) d_{ω, q} ς + \int_{d}^{\frac{1}{q^{n}}} θ_{m, \frac{1}{q^{n}}}^{(2) 2} (ζ) d_{ω, q} ς .

By Theorem 3.4 and the Hilbert–Schmidt theorem, we infer that

(3.10)

\begin{array}{l} \int_{ω_{0}}^{d} | f^{(1)} (ζ) |^{2} d_{ω, q} ζ + \int_{d}^{\frac{1}{q^{n}}} | f^{(2)} (ζ) |^{2} d_{ω, q} ζ \\ = \sum_{m = 1}^{\infty} \frac{1}{α_{m, \frac{1}{q^{n}}}^{2}} {\int_{ω_{0}}^{d} f^{(1)} (ζ) ϕ_{m, \frac{1}{q^{n}}}^{(1)} (ζ) d_{ω, q} ζ + \int_{d}^{\frac{1}{q^{n}}} f^{(2)} (ζ) ϕ_{m, \frac{1}{q^{n}}}^{(2)} (ζ) d_{ω, q} ζ|}^{2} . \end{array}

Define

ϱ_{\frac{1}{q^{n}}} (λ) = \begin{array}{l} - \sum_{λ < λ_{m, \frac{1}{n}} < 0} \frac{1}{α_{m, \frac{1}{q^{n}}}^{2}}, & for λ \leq 0, \\ \sum_{0 \leq λ_{m, \frac{1}{q^{n}}} < λ} \frac{1}{α_{m, \frac{1}{q^{n}}}^{2}}, & for λ > 0. \end{array}

Then, (3.10) can be written as

(3.11)

\int_{ω_{0}}^{d} {f^{(1)} (ζ)|}^{2} d_{ω, q} ζ + \int_{d}^{\frac{1}{q^{n}}} {f^{(2)} (ζ)|}^{2} d_{ω, q} ζ = \int_{- \infty}^{\infty} | F (λ) |^{2} d ϱ_{\frac{1}{q^{n}}} (λ),

where

F (λ) = \int_{ω_{0}}^{d} f^{(1)} (ζ) ϕ_{m, \frac{1}{q^{n}}}^{(1)} (ζ) d_{ω, q} ζ + \int_{d}^{\frac{1}{q^{n}}} f^{(2)} (ζ) ϕ_{m, \frac{1}{q^{n}}}^{(2)} (ζ) d_{ω, q} ζ .

Lemma 3.5.

For any positive S, there is a positive number B = B (S) not depending on n so that

- ∨_{- S}^{S} ϱ_{\frac{1}{q^{n}}} (λ) = \sum_{- S \leq λ_{m, \frac{1}{q^{n}}} < S} \frac{1}{α_{m, \frac{1}{q^{n}}}^{2}} = ϱ_{\frac{1}{q^{n}}} (S) - ϱ_{\frac{1}{q^{n}}} (- S) < B .

Proof

Let sin β ≠ 0. Since θ(ζ, λ) is continuous in domain −S ≤ λ ≤ S, $ω_{0}, d) \cup (d, \frac{1}{q^{n}}]$ , and the condition θ⁽¹⁾(ω₀, λ) = sin β, there exists a positive number h such that for |λ| < S,

(3.12)

\frac{1}{h^{2}} {(\int_{ω_{0}}^{ω_{0} + h} θ^{(1)} (ζ, λ) d_{ω, q} ζ)}^{2} > \frac{1}{2} {sin}^{2} β .

Let

f_{h} (ζ) = \begin{array}{l} \frac{1}{h}, & ω_{0} \leq ζ \leq ω_{0} + h, \\ 0, & ζ > ω_{0} + h . \end{array}

From (3.12), we find

\begin{array}{l} \int_{ω_{0}}^{ω_{0} + h} f_{h}^{2} (ζ) d_{ω, q} ζ & = \frac{1}{h} = \int_{- \infty}^{\infty} {(\frac{1}{h} \int_{ω_{0}}^{ω_{0} + h} θ^{(1)} (ζ, λ) d_{ω, q} ζ)}^{2} d ϱ_{\frac{1}{q^{n}}} (λ) \\ \geq \int_{- S}^{S} {(\frac{1}{h} \int_{ω_{0}}^{ω_{0} + h} θ^{(1)} (ζ, λ) d_{ω, q} ζ)}^{2} d ϱ_{α} (λ) \\ > \frac{1}{2} {sin}^{2} β ϱ_{\frac{1}{q^{n}}} (S) - ϱ_{\frac{1}{q^{n}}} (- S)\} . \end{array}

If sin β = 0, then we define f_h(ζ) as

\begin{array}{l} f_{h} (ζ) = \begin{array}{l} \frac{1}{h^{2}}, & ω_{0} \leq ζ \leq ω_{0} + h, \\ 0, & ζ > ω_{0} + h . \end{array} \\ \begin{array}{l}  \end{array} \end{array}

This proves the lemma.

Now, we will give an expansion into a Fourier series of resolvent. By ω, q-integration by parts, we obtain

\begin{array}{l} \int_{d}^{ω_{0}} [\frac{1}{q} D_{- \frac{ω}{q}, \frac{1}{q}} D_{ω, q} y^{(1)} (ζ, λ) - v (ζ) y^{(1)} (ζ, λ)] θ_{m, \frac{1}{q^{n}}}^{(1)} (ζ) d_{ω, q} ζ \\ + \int_{\frac{1}{q^{n}}}^{d} [\frac{1}{q} D_{- \frac{ω}{q}, \frac{1}{q}} D_{ω, q} y^{(2)} (ζ, λ) - v (ζ) y^{(2)} (ζ, λ)] θ_{m, \frac{1}{q^{n}}}^{(2)} (ζ) d_{ω, q} ζ \\ = \int_{d}^{ω_{0}} [\frac{1}{q} D_{- \frac{ω}{q}, \frac{1}{q}} D_{ω, q} ϕ_{m, \frac{1}{q^{n}}}^{(1)} (ζ) - v (ζ) θ_{m, \frac{1}{q^{n}}}^{(1)} (ζ)] y^{(1)} (ζ, λ) d_{ω, q} ζ \\ + \int_{\frac{1}{q^{n}}}^{d} [\frac{1}{q} D_{- \frac{ω}{q}, \frac{1}{q}} D_{ω, q} ϕ_{m, \frac{1}{q^{n}}}^{(2)} (ζ) - v (ζ) θ_{m, \frac{1}{q^{n}}}^{(2)} (ζ)] y^{(2)} (ζ, λ) d_{ω, q} ζ \\ = - λ_{m, \frac{1}{q^{n}}} \int_{d}^{ω_{0}} y^{(1)} (ζ, λ) θ_{m, \frac{1}{q^{n}}}^{(1)} (ζ) d_{ω, q} ζ - λ_{m, \frac{1}{q^{n}}} \int_{\frac{1}{q^{n}}}^{d} y^{(2)} (ζ, λ) θ_{m, \frac{1}{q^{n}}}^{(2)} (ζ) d_{ω, q} ζ \\ = - λ_{m, \frac{1}{q^{n}}} φ_{m} (λ), \end{array}

where m ∈ ℕ. Let

\begin{array}{l} y (ζ, λ) = \sum_{m = 1}^{\infty} φ_{m} (λ) ψ_{m, \frac{1}{q^{n}}} (ζ), \\ a_{m} = \int_{ω_{0}}^{d} f (ζ) ψ_{m, \frac{1}{q^{n}}}^{(1)} (ζ) d_{ω, q} ζ + \int_{d}^{\frac{1}{q^{n}}} f (ζ) ψ_{m, \frac{1}{q^{n}}}^{(2)} (ζ) d_{ω, q} ζ, \end{array}

where m ∈ ℕ. Since y(ζ, λ) satisfies the equation

- \frac{1}{q} D_{- \frac{ω}{q}, \frac{1}{q}} D_{ω, q} y (ζ, λ) + (v (ζ) - λ) y (ζ, λ) = f (ζ),

we find

\begin{array}{l} a_{m} = \int_{ω_{0}}^{d} - \frac{1}{q} D_{- \frac{ω}{q}, \frac{1}{q}} D_{ω, q} y^{(1)} (ζ, λ) + (v (ζ) - λ) y^{(1)} (ζ, λ)] θ_{m, \frac{1}{q^{n}}}^{(1)} (ζ) d_{ω, q} ζ \\ + \int_{d}^{\frac{1}{q^{n}}} - \frac{1}{q} D_{- \frac{ω}{q}, \frac{1}{q}} D_{ω, q} y^{(2)} (ζ, λ) + (v (ζ) - λ) y^{(2)} (ζ, λ)] θ_{m, \frac{1}{q^{n}}}^{(2)} (ζ) d_{ω, q} ζ \\ = λ_{m, \frac{1}{q^{n}}} φ_{m} (λ) - λ φ_{m} (λ), m, n \in ℕ, \frac{1}{q^{n}} > d . \end{array}

Thus, we get

φ_{m} (λ) = \frac{a_{m}}{λ_{m, \frac{1}{q^{n}}} - λ} (m, n \in ℕ, \frac{1}{q^{n}} > d),

and

y (ζ, λ) = {G_{\frac{1}{q^{n}}} (ζ, \cdot, λ), \bar{f (\cdot)}〉}_{n} = \sum_{m = 1}^{\infty} \frac{a_{m} θ_{m, \frac{1}{q^{n}}} (ζ)}{λ_{m, \frac{1}{q^{n}}} - λ} .

Hence

(3.13)

\begin{array}{l} (R_{\frac{1}{q^{n}}} f) (ζ, z) = \sum_{m = 1}^{\infty} \frac{θ_{m, \frac{1}{q^{n}}} (ζ)}{α_{m, \frac{1}{q^{n}}}^{2} (λ_{m, \frac{1}{q^{n}}} - z)} {f (\cdot), θ_{m, \frac{1}{q^{n}}} (\cdot)〉}_{n} \\ = \int_{- \infty}^{\infty} \frac{θ (ζ, λ)}{λ - z} {f (\cdot), θ_{m, \frac{1}{q^{n}}} (\cdot)〉}_{n}\} d ϱ_{\frac{1}{q^{n}}} (λ) . \end{array}

Lemma 3.6.

For each nonreal z and fixed ζ, the following relation holds

(3.14)

\int_{- \infty}^{\infty} {\frac{θ (ζ, λ)}{z - λ}|}^{2} d ϱ_{\frac{1}{q^{n}}} (λ) < S .

Proof

Writing

f (ς) = \frac{θ_{m, \frac{1}{q^{n}}} (ς)}{α_{m, \frac{1}{q^{n}}}}

yields

(3.15)

\frac{1}{α_{m, \frac{1}{q^{n}}}} {G_{\frac{1}{q^{n}}} (ζ, \cdot, λ), θ_{m, \frac{1}{q^{n}}} (\cdot)〉}_{n} = \frac{θ_{m, \frac{1}{q^{n}}} (ζ)}{α_{m, \frac{1}{q^{n}}} (λ_{m, \frac{1}{q^{n}}} - z)},

due to the eigenfunctions

θ_{m, \frac{1}{q_{n}}} (ζ)

are orthogonal. Combining (3.15) and (3.10), we see that

\begin{array}{l} \int_{ω_{0}}^{d} {G_{\frac{1}{q^{n}}} (ζ, ς, z)|}^{2} d_{ω, q} ς + \int_{d}^{\frac{1}{q^{n}}} {G_{\frac{1}{q^{n}}} (ζ, ς, z)|}^{2} d_{ω, q} ς \\ = \sum_{m = 1}^{\infty} \frac{{θ_{m, \frac{1}{q^{n}}} (ζ)|}^{2}}{α_{m, \frac{1}{q^{n}}}^{2} {λ_{m, \frac{1}{q^{n}}} - z|}^{2}} = \int_{- \infty}^{\infty} {\frac{θ (ζ, λ)}{λ - z}|}^{2} d ϱ_{\frac{1}{q^{n}}} (λ) . \end{array}

By Lemma 3.1, the integral on the left converges and the result is immediate.

It follows from Lemma 8 that the set $ϱ_{\frac{1}{q^{n}}} (λ)\}$ is bounded. Using Helly’s theorems ([15]), one can find a sequence {1/q^nk} such that $ϱ_{\frac{1}{q^{n} k}} (λ)$ converges to a monotone function ϱ(λ) (as n_k → ∞).

Lemma 3.7.

Let z be a nonreal number and ζ be a fixed number. Then we have

(3.16)

\int_{- \infty}^{\infty} {\frac{θ (ζ, λ)}{z - λ}|}^{2} d ϱ (λ) \leq S .

Proof

For arbitrary η > 0, it follows from (3.14) that

\int_{- η}^{η} {\frac{ϕ (ς, λ)}{z - λ}|}^{2} d ϱ_{\frac{1}{q^{n}}} (λ) < S .

Letting η → ∞ and n → ∞, we get the desired result.

Lemma 3.8.

For arbitrary η > 0, we have

\int_{- \infty}^{- η} \frac{d ϱ (λ)}{| z - λ |^{2}} < \infty, \int_{η}^{\infty} \frac{d ϱ (λ)}{| z - λ |^{2}} < \infty .

Proof

Let sin β ≠ 0. Writing ζ = 0 in (3.16), we obtain

\int_{- \infty}^{\infty} \frac{d ϱ (λ)}{| z - λ |^{2}} < \infty .

Let sin β = 0. Then

\frac{1}{α_{m, \frac{1}{q^{n}}}} {D_{q, ζ} G_{\frac{1}{q^{n}}} (ζ, \cdot, z), θ_{m, \frac{1}{q^{n}}} (\cdot)〉}_{n} = \frac{D_{q, ζ} θ_{m, \frac{1}{q^{n}}} (ζ)}{α_{m, \frac{1}{q^{n}}} (λ_{m, \frac{1}{q^{n}}} - z)} .

By (3.11), we find

\begin{array}{l} \int_{ω_{0}}^{d} {D_{q, ζ} G_{\frac{1}{q^{n}}} (ζ, ς, z)|}^{2} d_{ω, q} ζ + \int_{d}^{\frac{1}{q^{n}}} {D_{q, ζ} G_{\frac{1}{q^{n}}} (ζ, ς, z)|}^{2} d_{ω, q} ζ \\ = \int_{- \infty}^{\infty} {\frac{D_{q, ζ} θ (ζ, λ)}{z - λ}|}^{2} d ϱ_{\frac{1}{q^{n}}} (λ) . \end{array}

Lemma 3.9.

Let

(R f) (ζ, z) = \int_{ω_{0}}^{\infty} G (ζ, ς, z) f (ς) d_{ω, q} ς,

where f ∈ H, and

G (ζ, ς, z) = \begin{array}{l} Z (ζ, z) θ (ς, z), & ς \leq ζ, ζ \neq d, ς \neq d, \\ θ (ζ, z) Z (ς, z), & ς > ζ, ζ \neq d, ς \neq d . \end{array}

Then, we have

\begin{array}{l} \int_{ω_{0}}^{d} | (R f) (ζ, z) |^{2} d_{ω, q} ζ + \int_{d}^{\infty} | (R f) (ζ, z) |^{2} d_{ω, q} ζ \\ \leq \frac{1}{v^{2}} \int_{ω_{0}}^{d} {f^{(1)} (ζ)|}^{2} d_{ω, q} ζ + \int_{d}^{\infty} {f^{(2)} (ζ)|}^{2} d_{ω, q} ζ, \end{array}

where v = Im z.

Proof

Combining (3.13) and (3.10), for each $\frac{1}{q^{n}} > d$ , n ∈ ℕ, we obtain

\begin{array}{l} \int_{ω_{0}}^{d} {(R_{\frac{1}{q^{n}}} f) (ζ, z)|}^{2} d_{ω, q} ζ + \int_{d}^{\frac{1}{q^{n}}} {(R_{\frac{1}{q^{n}}} f) (ζ, z)|}^{2} d_{ω, q} ζ \\ = \sum_{m = 1}^{\infty} \frac{{{〈 f (\cdot), θ_{m, \frac{1}{q^{n}}} (\cdot, z) 〉}_{n}|}^{2}}{α_{m, \frac{1}{q^{n}}}^{2} | λ_{m, \frac{1}{q^{n}}} - z |^{2}} \\ = \frac{1}{v^{2}} \int_{ω_{0}}^{d} {f^{(1)} (ς)|}^{2} d_{ω, q} ς + \frac{1}{v^{2}} \int_{d}^{\frac{1}{q^{n}}} {f^{(2)} (ς)|}^{2} d_{ω, q} ς . \end{array}

Letting n → ∞, we get the desired result.

Theorem 3.10 (Integral Representation of the Resolvent).

For every non-real z and for each f ∈ H, we obtain

(R f) (ζ, z) = \int_{- \infty}^{\infty} \frac{θ (ζ, λ)}{λ - z} F (λ) d ϱ (λ),

where

F (λ) = \int_{ω_{0}}^{d} f^{(1)} (ζ) θ^{(1)} (ζ, λ) d_{ω, q} ζ + \lim_{σ \to \infty} \int_{d}^{σ} f^{(2)} (ζ) θ^{(2)} (ζ, λ) d_{ω, q} ζ .

Proof

Suppose that f(ζ) = f_σ(ζ) satisfies (3.2)–(3.4) and vanishes outside the set [ω₀, d) ∪ (d, σ], where $d < σ < \frac{1}{q^{n}}$ , n ∈ ℕ. Let

F_{σ} (λ) = \int_{ω_{0}}^{d} f_{σ}^{(1)} (ζ) θ^{(1)} (ζ, λ) d_{ω, q} ζ + \int_{d}^{σ} f_{σ}^{(2)} (ζ) θ^{(2)} (ζ, λ) d_{ω, q} ζ .

By (3.13), we see that

(3.17)

\begin{array}{l} \begin{array}{l} (R_{\frac{1}{q^{n}}} f_{σ}) (ζ, z) & = \int_{- \infty}^{\infty} \frac{θ (ζ, λ)}{λ - z} F_{σ} (λ) d ϱ_{\frac{1}{q^{n}}} (λ) \\ = \int_{- \infty}^{- a} \frac{θ (ζ, λ)}{λ - z} F_{σ} (λ) d ϱ_{\frac{1}{q^{n}}} (λ) + \int_{- a}^{a} \frac{θ (ζ, λ)}{λ - z} F_{σ} (λ) d ϱ_{\frac{1}{q^{n}}} (λ) \\ + \int_{a}^{\infty} \frac{θ (ζ, λ)}{λ - z} F_{σ} (λ) d ϱ_{\frac{1}{q^{n}}} (λ) = I_{1} + I_{2} + I_{3} . \end{array} \end{array}

Firstly, we will estimate I₁. From (3.13), we deduce that

\begin{array}{l} I_{1} & = \int_{- \infty}^{- a} \frac{θ (ζ, λ)}{z - λ} F_{σ} (λ) d ϱ_{\frac{1}{q^{n}}} (λ) \\ = \sum_{λ_{k, \frac{1}{q^{n}}} < - a} \frac{θ_{k, \frac{1}{q^{n}}} (ζ)}{α_{k, \frac{1}{q^{n}}}^{2} (z - λ_{k, \frac{1}{q^{n}}})} {\int_{ω_{0}}^{d} f_{σ}^{(1)} (ζ) θ_{k, \frac{}{q^{n}}}^{(1)} (ζ) d_{ω, q} ζ + \int_{d}^{σ} f_{σ}^{(2)} (ζ) θ_{k, \frac{}{q^{n}}}^{(2)} (ζ) d_{ω, q} ζ} \\ \leq {(\sum_{λ_{k, \frac{1}{q^{n}}} < - a} \frac{θ_{k, \frac{1}{q^{n}}}^{2} (ζ)}{α_{k, \frac{1}{q^{n}}}^{2} {| z - λ_{k, \frac{1}{q^{n}}} |}^{2}})}^{1 / 2} \\ \times {(\sum_{λ_{k, \frac{1}{q^{n}}} < - a} \frac{1}{α_{k, \frac{1}{q^{n}}}^{2}} {| \int_{ω_{0}}^{d} f_{σ}^{(1)} (ζ) θ_{k, \frac{}{q^{n}}}^{(1)} (ζ) d_{ω, q} ζ + \int_{d}^{σ} f_{σ}^{(2)} (ζ) θ_{k, \frac{}{q^{n}}}^{(2)} (ζ) d_{ω, q} ζ |}^{2})}^{1 / 2} . \end{array}

Integrating twice by parts, we find

\begin{array}{l} \int_{ω_{0}}^{d} f_{σ}^{(1)} (ζ) θ_{k, \frac{}{q^{n}}}^{(1)} (ζ) d_{ω, q} ζ + \int_{σ}^{d} f_{σ}^{(2)} (ζ) θ_{k, \frac{}{q^{n}}}^{(2)} (ζ) d_{ω, q} ζ \\ = - \frac{1}{λ_{k, \frac{1}{q^{n}}}} \int_{ω_{0}}^{d} f_{σ}^{(1)} (ζ) {\frac{1}{q} D_{- \frac{ω}{q}, \frac{1}{q}} D_{ω, q} θ_{k, \frac{}{q^{n}}}^{(1)} (ζ) - v (ζ) θ_{k, \frac{}{q^{n}}}^{(1)} (ζ)} d_{ω, q} ζ \\ - \frac{1}{λ_{k, \frac{1}{q^{n}}}} \int_{d}^{σ} f_{σ}^{(2)} (ζ) {\frac{1}{q} D_{- \frac{ω}{q}, \frac{1}{q}} D_{ω, q} θ_{k, \frac{}{q^{n}}}^{(2)} (ζ) - v (ζ) θ_{k, \frac{}{q^{n}}}^{(2)} (ζ)} d_{ω, q} ζ \\ = - \frac{1}{λ_{k, \frac{1}{q^{n}}}} \int_{ω_{0}}^{d} {\frac{1}{q} D_{- \frac{ω}{q}, \frac{1}{q}} D_{ω, q} f_{σ}^{(1)} (ζ) - v (ζ) f_{σ}^{(1)} (ζ)} θ_{k, \frac{}{q^{n}}}^{(1)} (ζ) d_{ω, q} ζ \\ - \frac{1}{λ_{k, \frac{1}{q^{n}}}} \int_{d}^{σ} {\frac{1}{q} D_{- \frac{ω}{q}, \frac{1}{q}} D_{ω, q} f_{σ}^{(2)} (ζ) - v (ζ) f_{σ}^{(2)} (ζ)} θ_{k, \frac{}{q^{n}}}^{(2)} (ζ) d_{ω, q} ζ . \end{array}

By Lemma 3.6, we get

\begin{array}{l} I_{1} & \leq \frac{K^{1 / 2}}{a} \\ \times \sum_{λ_{k, \frac{1}{q^{n}}} < - a} \frac{1}{α_{k, \frac{1}{q^{n}}}^{2}} \int_{ω_{0}}^{d} {\frac{1}{q} D_{- \frac{ω}{q}, \frac{1}{q}} D_{ω, q} f_{σ}^{(1)} (ζ) - v (ζ) f_{σ}^{(1)} (ζ)} θ_{k, \frac{1}{q^{n}}}^{(1)} (ζ) d_{ω, q} ζ \\ {{+ \int_{d}^{σ} \frac{1}{q} D_{- \frac{ω}{q}, \frac{1}{q}} D_{ω, q} f_{σ}^{(2)} (ζ) - v (ζ) f_{σ}^{(2)} (ζ)\} θ_{k, \frac{1}{q^{n}}}^{(2)} (ζ) d_{ω, q} ζ|}^{2})}^{1 / 2} . \end{array}

Using Bessel inequality, we see that

\begin{array}{l} I_{1} \leq \frac{K^{1 / 2}}{a} \int_{ω_{0}}^{σ} {\frac{1}{q} D_{- \frac{ω}{q}, \frac{1}{q}} D_{ω, q} f_{σ}^{(1)} (ζ) - v (ζ) f_{σ}^{(1)} (ζ)|}^{2} d_{ω, q} ζ \\ + {\int_{d}^{σ} {\frac{1}{q} D_{- \frac{ω}{q}, \frac{1}{q}} D_{ω, q} f_{σ}^{(2)} (ζ) - v (ζ) f_{σ}^{(2)} (ζ)|}^{2} d_{ω, q} ζ]}^{1 / 2} = \frac{C}{a} . \end{array}

It is proved similarly that $I_{3} \leq \frac{C}{a}$ . Then I₁ and I₃ tend to zero as a → ∞, uniformly in $\frac{1}{q_{n}}$ . It follows from the Helly selection theorem and (3.17) that

(3.18)

(R f_{σ}) (ζ, z) = \int_{- \infty}^{\infty} \frac{θ (ζ, λ)}{z - λ} F_{σ} (λ) d ϱ (λ) .

As is known, if f(·) ∈ H, then we find a sequence

{f_{σ} (ς)\}}_{σ = 1}^{\infty}

that satisfies the previous conditions and tends to f(ζ) as σ → ∞. From (3.10), the sequence of Fourier transform converges to the transform of f(ζ). Using Lemmas 3.7 and 3.9, we can pass to the limit σ → ∞ in (3.18). Thus, we get the desired result.

Remark 3.11.

Using Theorem 3.10, we infer that

\begin{array}{l} \int_{ω_{0}}^{\infty} (R f^{(1)}) (ς, z) g^{(1)} (ς) d_{ω, q} ς + \int_{d}^{\infty} (R f^{(2)}) (ς, z) g^{(2)} (ς) d_{ω, q} ς \\ = \int_{- \infty}^{\infty} \frac{F (λ) G (λ)}{z - λ} d ϱ (λ), \end{array}

where

F (λ) = \int_{ω_{0}}^{d} f^{(1)} (ζ) θ^{(1)} (ζ, λ) d_{ω, q} ζ + \lim_{σ \to \infty} \int_{d}^{σ} f^{(2)} (ζ) θ^{(2)} (ζ, λ) d_{ω, q} ζ,

and

G (λ) = \int_{ω_{0}}^{d} g^{(1)} (ζ) θ^{(1)} (ζ, λ) d_{ω, q} ζ + \lim_{σ \to \infty} \int_{ω_{0}}^{σ} g^{(2)} (ζ) θ^{(2)} (ζ, λ) d_{ω, q} ζ .

Notes

[1] Statements and Declarations.

This work does not have any conflict of interested result.

[2] Availability of data and materials.

Data sharing not applicable to this article as no datasets were generated or analysed during the current study.

The Resolvent of Impulsive Singular Hahn–Sturm–Liouville Operators

Full Article

1. Introduction

2. Preliminaries

Definition 2.1 ([10], [11]).

Definition 2.2 ([3]).

Definition 2.3.

3. Main results

(3.1)

(3.2)

(3.3)

(3.4)

(3.5)

Lemma 3.1.

Proof

(3.6)

Theorem 3.2.

Proof

(3.7)

Theorem 3.3 ([20]).

(3.8)

(3.9)

Theorem 3.4.

Proof

(3.10)

(3.11)

Lemma 3.5.

Proof

(3.12)

(3.13)

Lemma 3.6.

(3.14)

Proof

(3.15)

Lemma 3.7.

(3.16)

Proof

Lemma 3.8.

Proof

Lemma 3.9.

Proof

Theorem 3.10 (Integral Representation of the Resolvent).

Proof

(3.17)

(3.18)

Remark 3.11.

Notes