# Perturbation Determinant and levinson’s formula for Schrödinger operators
with generalized Point interaction

Muhammad Usman and Muhammad Danish Zia  Muhammad Usman, Department of
Mathematics, Lahore University of Management Sciences (LUMS), Lahore, Pakistan
<EMAIL_ADDRESS>Muhammad Danish Zia, Department of Basic Sciences, School of
Civil Engineering, National University of Sciences and Technology (NUST),
Islamabad, Pakistan<EMAIL_ADDRESS>
###### Abstract.

We consider the one dimensional Schrödinger operator with properly connecting
generalized point interaction at the origin. We derive a trace formula for
trace of difference of resolvents of perturbed and unperturbed Schrödinger
operators in terms of a Wronskian which results into an explicit expression
for perturbation determinant. Using the estimate for large time real argument
on the trace norm of the resolvent difference of the perturbed and unperturbed
Schrödinger operators we express the spectral shift function in terms of
perturbation determinant. Under certain integrability condition on the
potential function, we calculate low energy asymptotics for the perturbation
determinant and prove an analog of Levinson’s formula.

###### Key words and phrases:

Schrödinger operators; Generalized point interaction; Trace formula;
Perturbation determinant; Spectral shift function; Levinson’s formula

###### 2010 Mathematics Subject Classification:

Primary: 34L25, 34L40; Secondary: 35P25, 81Q10

##  1\. Introduction and main results

Let $H_{V}^{\mathcal{A}}$ be the Schrödinger operator

$H_{V}^{\mathcal{A}}=H^{\mathcal{A}}_{0}+V,\quad
H^{\mathcal{A}}_{0}=-\frac{d^{2}}{dx^{2}}$ (1.1)

on the real-line which is realized as a union of two positive semi-axis
$e_{j}=[0,\infty)$, $j=1,2$ coupled at $0$. The potential function $V$ is
assumed to be real-valued and satisfies

$\int_{e_{j}}|V_{j}(x)|dx<\infty,$ (1.2)

where, $V_{j}$ is the restriction of $V$ on $e_{j}$, $j=1,2$. The operator
(1.1) is self-adjoint for all functions from the space
$H^{2}([0,\infty))\oplus H^{2}([0,\infty))$ satisfying the conditions

$\displaystyle\begin{bmatrix}\psi_{1}(0)\\\
\psi_{1}^{\prime}(0)\end{bmatrix}=\mathcal{A}\begin{bmatrix}\psi_{2}(0)\\\
\psi_{2}^{\prime}(0)\end{bmatrix}$ (1.3)

where $\mathcal{A}=e^{i\phi}\begin{bmatrix}a&b\\\ c&d\end{bmatrix}$ and
$\psi=(\psi_{1},\psi_{2})^{T}\in L^{2}([0,\infty))\oplus L^{2}([0,\infty))$
and the real parameters $\phi\in[-\pi/2,\pi/2]$, $a,b,c,d\in\mathbb{R}$
satisfying $ad-bc=-1$.

The above conditions are commonly known as the generalized point interaction,
first introduced by Šeba [18]. There are several other equivalent formulations
for the generalized point interaction, see for example [9] and references
therein. The free Hamiltonian $H_{0}^{\mathcal{A}}$ with conditions (1.3)
describes certain second order differential operators with generalized
functions in coefficients. The Schrödinger operator with $\delta$ potential of
strength $\alpha$, for instance, corresponds to the unperturbed operator
$H_{0}^{\mathcal{A}}$ with $\phi=0,\,a=1,\,b=0,\,c=\alpha$ and $d=-1$.
Similarly, $\delta^{\prime}$ potential of strength $\beta$ corresponds to
$\phi=0,\,a=-1,\,b=\beta,\,c=0$ and $d=1$. Differential operators of this kind
are closely related to exactly solvable models in quantum mechanics, atomic
physics, and acoustics [1, 7].

The present article is devoted to deriving explicit expression for
perturbation determinant for the pair of operators $H_{V}^{\mathcal{A}}$ and
$H_{0}^{\mathcal{A}}$ and its relationship with spectral shift function and
Levinson’s theorem. These mathematical objects play an important role in the
study of direct and inverse scattering theory [11, 12, 13, 10] as well as in
solid state physics in connection with Friedel sum rule and excess charge
[14]. Krĕin [15] (see also [17]) introduced the concept of perturbation
determinant and it is an important tool in studying trace formulas of higher
order. For a detailed study on these topics for Schrödinger operators on the
half-line as well as on the whole real line we refer to the monograph [23]
(see also reference [6]) and for a quantum star graph we refer to [5]. Our aim
in this paper is to generalize these results for Schrödinger operators on the
real-line, which can also be seen as a two edge star graph, satisfying the
most general properly connecting self-adjoint matching conditions at zero.

Using Krĕin’s resolvent formula (see [1] and [8]), we first prove a trace
formula that expresses the trace of the difference between the perturbed and
corresponding unperturbed resolvents in terms of a Wronskian. This trace
formula allows us to derive an explicit expression for the perturbation
determinant. The perturbation determinant is an interesting object in the
spectral theory of Schrödinger operators, as the zeros of perturbation
determinant coincide with the eigenvalues of the Schrödinger operator and the
multiplicity of each eigenvalue is equal to the order of the corresponding
zero of the perturbation determinant [3]. Estimates on the trace norm of the
resolvent difference of the operators $H_{V}^{\mathcal{A}}$ and
$H_{0}^{\mathcal{A}}$ obtained in [21] allow one to express the spectral shift
function in terms of perturbation determinant. For a real argument,
perturbation determinant is used to define the phase shift function. The so
called Levinson’s formula is derived for the Schrödinger operator
$H_{V}^{\mathcal{A}}$ in terms of its phase shift function. This formula, also
called a zero order trace formula gives a relationship between the number of
negative eigenvalues and the scattering data via phase shift function.

To state our main results we first consider the half-line Schrödinger
operators $H^{D}_{e_{j}}=-\frac{d^{2}}{dx^{2}}+V_{j}$, $j=1,2$, with Dirichlet
boundary condition at $0$ and let $H^{D}_{V}=H^{D}_{e_{1}}\oplus
H^{D}_{e_{2}}$ denote the decoupled Schrödinger operator on the whole line.
The resolvent $(H^{D}_{V}-z)^{-1}$ is denoted by $R^{D}_{V}(z)$. Under the
condition (1.2) the differential equation

$-u_{j}^{\prime\prime}+V_{j}u=\zeta^{2}u,\quad j=1,2$

has two particular solutions, namely, the regular solution and the Jost
solution. The regular solution $\phi_{j}$ is characterized by the conditions

$\phi_{j}(0,\zeta)=0,\quad\phi^{\prime}_{j}(0,\zeta)=1$

and the Jost solution $\theta_{j}$ by the asymptotics $\theta_{j}(x,\zeta)\sim
e^{i\zeta x}$, as $x\rightarrow\infty$. We denote the resolvent
$(H_{V}^{\mathcal{A}}-z)^{-1}$ of the perturbed operator $H_{V}^{\mathcal{A}}$
by $R^{\mathcal{A}}_{V}(z)$ and the resolvent $(H^{\mathcal{A}}_{0}-z)^{-1}$
of the unperturbed operator $H^{\mathcal{A}}_{0}$ by $R^{\mathcal{A}}_{0}(z)$.
Moreover, the (modified) perturbation determinant is defined as

$D(z):=\mbox{det}(\mathbb{I}+\sqrt{V}R_{0}^{\mathcal{A}}(z)\sqrt{|V|}),\quad
z\in\rho(H_{0}^{\mathcal{A}}),$

where, $\sqrt{V}=\mbox{sgn}V\sqrt{|V|}$.

Our first main result is the following trace formula for the difference of two
resolvents in terms of Jost solutions $\theta_{j}$ and their derivatives

######  Theorem 1.1.

If the potential $V$ satisfies (1.2) then the following trace formula holds

$\displaystyle\operatorname{Tr}(R^{\mathcal{A}}_{V}(z)-R^{\mathcal{A}}_{0}(z))=-\dfrac{1}{2\zeta}\left(\dfrac{d}{d\zeta}\ln\left(\dfrac{w_{1}(\zeta)w_{2}(\zeta)L(\zeta)}{(a-d)\zeta+(b\zeta^{2}+c)i}\right)\right),\quad\zeta=z^{1/2},\quad\emph{Im}\,\zeta>0$
(1.4)

where

$L(\zeta)=a\frac{\theta_{1}^{\prime}(0,\zeta)}{\theta_{1}(0,\zeta)}-d\frac{\theta_{2}^{\prime}(0,\zeta)}{\theta_{2}(0,\zeta)}+b\frac{\theta_{1}^{\prime}(0,\zeta)\theta_{2}^{\prime}(0,\zeta)}{\theta_{1}(0,\zeta)\theta_{2}(0,\zeta)}-c$

and $w_{j}(\zeta)=\theta_{j}(0,\zeta)$, $j=1,2$. Moreover, for
$z\in\rho(H_{V})$ and $\emph{Im}\,z^{1/2}>0,$ the perturbation determinant
$D(\zeta)$ of the operator $H_{V}^{\mathcal{A}}$ with respect to
$H^{\mathcal{A}}_{0}$ is given by

$D(z)=\dfrac{L(\sqrt{z})w_{1}(\sqrt{z})w_{2}(\sqrt{z})}{(a-d)\sqrt{z}i-(bz+c)}.$
(1.5)

In Section 4 we study the behaviour of $D(z)$ as $|z|\rightarrow 0$ and prove
the zero order trace formula, commonly known as the Levinson’s formula, for
the operator $H_{V}^{\mathcal{A}}$. We will need the following constants:

$\displaystyle\alpha_{1}$
$\displaystyle=\left(b\left(\dfrac{\theta_{1}^{\prime}(0,0)}{\theta_{2}(0,0)}+\dfrac{\theta_{2}^{\prime}(0,0)}{\theta_{1}(0,0)}\right)-a\dfrac{\theta_{2}(0,0)}{\theta_{1}(0,0)}+d\dfrac{\theta_{1}(0,0)}{\theta_{2}(0,0)}\right)i,$
(1.6) $\displaystyle\alpha_{2}$
$\displaystyle=a\theta_{1}^{\prime}(0,0)\theta_{2}(0,0)+b\theta_{1}^{\prime}(0,0)\theta_{2}^{\prime}(0,0),$
$\displaystyle\alpha_{3}$
$\displaystyle=(c\theta_{2}(0,0)+d\theta_{2}^{\prime}(0,0))\dot{\theta}_{1}(0,0),$
$\displaystyle\alpha_{4}$
$\displaystyle=a\theta_{1}^{\prime}(0,0)\dot{\theta}_{2}(0,0)-d\dot{\theta}_{1}(0,0)\theta_{2}^{\prime}(0,0),$

where dot denotes the derivative with respect to $\zeta$.

######  Theorem 1.2.

Assume that

$\displaystyle\int\limits_{e_{j}}(1+x)|V_{j}(x)|\,dx<\infty,\hskip
8.5359ptj=1,2$ (1.7)

and let $N$ be the number of negative eigenvalues of the operator
$H_{V}^{\mathcal{A}}$. Then, the following formula holds

$\eta(\infty)-\eta(0)=\begin{cases}\pi\left(N-\dfrac{P-Q}{2}\right),&w_{j}(0)\neq
0$, $j=1,2,\\\ \pi\left(N-\dfrac{P-R}{2}\right),&w_{1}(0)=0,w_{2}(0)\neq 0,\\\
\pi\left(N-\dfrac{P-S}{2}\right),&w_{1}(0)\neq 0,w_{2}(0)=0,\\\
\pi\left(N-\dfrac{P-T}{2}\right),&w_{j}(0)=0$, $j=1,2.\end{cases}$

Here, $\eta$ is the so-called phase shift function, $P,Q,R,S$ and $T$ are
functions which are defined as

$\displaystyle P$ $\displaystyle=\begin{cases}0&c\neq 0\\\ 1&c=0,a-d\neq 0\\\
2&c=0,a-d=0,b\neq 0,\end{cases}\qquad Q=\begin{cases}0&L(0)\neq 0\\\
1&L(0)=0,\alpha_{1}\neq 0\\\ 2&L(0)=0,\alpha_{1}=0,b\neq 0,\end{cases}$
$\displaystyle R$ $\displaystyle=\begin{cases}0&\alpha_{2}\neq 0\\\
1&\alpha_{2}=0,\end{cases}\qquad S=\begin{cases}0&\alpha_{3}\neq 0\\\
1&\alpha_{3}=0,\end{cases}$ $\displaystyle T$
$\displaystyle=\begin{cases}0&b\neq 0\\\ 1&b=0,\alpha_{4}\neq 0\\\
2&b=0,\alpha_{4}=0,c\neq 0.\end{cases}$

##  2\. Perturbation determinant

Our main purpose in this section is to derive the trace formula (1.4) and the
explicit expression (1.5) for the perturbation determinant of operator
$H_{V}^{\mathcal{A}}$ with respect to $H_{0}$ in terms of the Jost solutions
$\theta_{j}$ and their derivatives $\theta_{j}^{\prime}$. For this purpose we
use Krĕin’s resolvent formula which allows us to describe the kernel of
$R_{V}^{\mathcal{A}}(z)$ in terms of the kernel of $R^{D}_{V}(z)$. More
precisely, for $\mbox{Im}\,\zeta\geq 0$, the Krein’s formula states

$\displaystyle
R^{\mathcal{A}}_{j,l}(x,y;z):=R_{j,l}^{D}(x,y;z)+\lambda_{j,l}\theta_{j}(x,\zeta)\theta_{l}(y,\zeta),\quad
j,l=1,2,\quad z=\zeta^{2}\in\rho(H^{\mathcal{A}}_{V})\cap\rho(H^{D}_{V}),$
(2.1)

where $R^{\mathcal{A}}_{j,l}$ and $R^{D}_{j,l}$ denote the kernel of $2\times
2$ matrix integral operators $R_{V}^{\mathcal{A}}$ and $R^{D}_{V}$,
respectively. The values of the coefficients $\lambda_{j,l}$ are given by the
following lemma

######  Lemma 2.1.

Let $\psi(x):=\int_{0}^{\infty}R^{\mathcal{A}}(x,y;z)f(y)\,dy$ and
$f=(f_{1}(x),f_{2}(x))^{T}\in L^{2}([0,\infty))\oplus L^{2}([0,\infty))$. If
$\psi(x)$ satisfies $H^{\mathcal{A}}_{V}\psi=\zeta^{2}\psi+f$ and conditions
(1.3) then $\lambda_{j,l}$ in (2.1) are given by

$\displaystyle\begin{bmatrix}\lambda_{1,1}&\lambda_{1,2}\\\
\lambda_{2,1}&\lambda_{2,2}\\\
\end{bmatrix}=\begin{bmatrix}-\dfrac{a+b\dfrac{\theta_{2}^{\prime}(0,\zeta)}{\theta_{2}(0,\zeta)}}{L(\zeta)\theta_{1}^{2}(0,\zeta)}&\dfrac{(ad-
bc)e^{i\phi}}{L(\zeta)\theta_{1}(0,\zeta)\theta_{2}(0,\zeta)}\\\
\dfrac{e^{-i\phi}}{L(\zeta)\theta_{1}(0,\zeta)\theta_{2}(0,\zeta)}&-\dfrac{b\dfrac{\theta_{1}^{\prime}(0,\zeta)}{\theta_{1}(0,\zeta)}-d}{L(\zeta)\theta_{2}^{2}(0,\zeta)}\\\
\end{bmatrix}$

where
$L(\zeta)=a\frac{\theta_{1}^{\prime}(0,\zeta)}{\theta_{1}(0,\zeta)}-d\frac{\theta_{2}^{\prime}(0,\zeta)}{\theta_{2}(0,\zeta)}+b\frac{\theta_{1}^{\prime}(0,\zeta)\theta_{2}^{\prime}(0,\zeta)}{\theta_{1}(0,\zeta)\theta_{2}(0,\zeta)}-c$.

###### Proof.

It is a well-known fact (see for example [5]), that the resolvent of the
operator $H_{e_{j}}^{D}$ is an integral operator with symmetric kernel

$\displaystyle
R_{e_{j}}^{D}(x,y;z)=\frac{\phi_{j}(x,\zeta)\theta_{j}(y,\zeta)}{w_{j}(\zeta)},\hskip
28.45274ptx\leq y,\hskip 28.45274pt\zeta=\sqrt{z}.$ (2.2)

Therefore the resolvent $R^{D}_{V}$ is a matrix integral operator with kernel

$\displaystyle R_{j,l}^{D}(x,y;z)=\delta_{j,l}R_{j}^{D}(x,y;z)\hskip
28.45274ptj,l=1,2.$ (2.3)

For simplicity, we use the following notation

$R^{\mathcal{A}}_{j,l}(x,y;z)=R^{\mathcal{A}}_{j,l},\quad\phi_{j}(x,\zeta)=\phi_{j}(x),\quad\theta_{j}(y,\zeta)=\theta_{j}(y)\quad\mbox{and}\quad
w_{j}(\zeta)=w_{j}.$

By substituting (2.2) and (2.3) into (2.1) and expressing it in matrix form we
get

$\displaystyle\begin{bmatrix}R^{\mathcal{A}}_{1,1}&R^{\mathcal{A}}_{1,2}\\\
R^{\mathcal{A}}_{2,1}&R^{\mathcal{A}}_{2,2}\end{bmatrix}=\begin{bmatrix}\frac{\phi_{1}(x)\theta_{1}(y)+w_{1}\lambda_{1,1}\theta_{1}(x)\theta_{1}(y)}{w_{1}}&\lambda_{1,2}\theta_{1}(x)\theta_{2}(y)\\\
\lambda_{2,1}\theta_{2}(x)\theta_{1}(y)&\frac{\phi_{2}(x)\theta_{2}(y)+w_{2}\lambda_{2,2}\theta_{2}(x)\theta_{2}(y)}{w_{2}}\end{bmatrix}.$

Multiplying $f$ from right and using the definition of function $\psi$ the
above expression becomes

$\displaystyle\begin{bmatrix}\psi_{1}(x)\\\
\psi_{2}(x)\end{bmatrix}=\begin{bmatrix}\int_{0}^{\infty}\\{\frac{\phi_{1}(x)\theta_{1}(y)}{w_{1}}f_{1}(y)+\sum_{i=1}^{2}\lambda_{1,i}\theta_{1}(x)\theta_{i}(y)f_{i}(y)\\}dy\\\
\int_{0}^{\infty}\\{\frac{\phi_{2}(x)\theta_{2}(y)}{w_{2}}f_{2}(y)+\sum_{i=1}^{2}\lambda_{2,i}\theta_{2}(x)\theta_{i}(y)f_{i}(y)\\}dy\end{bmatrix}$

which is equivalent to the following system of equations

$\displaystyle\psi_{j}(x)=\int\limits_{0}^{\infty}\left\\{\frac{\phi_{j}(x)\theta_{j}(y)}{w_{j}}f_{j}(y)+\sum_{i=1}^{n}\lambda_{j,i}\theta_{j}(x)\theta_{i}(y)f_{i}(y)\right\\}dy,\hskip
14.22636ptj=1,2.$

In order to determine $\lambda_{j,l}$ uniquely, we first express conditions
(1.3) as a system of two equations and first apply the sub-condition

$\psi_{1}(0)=e^{i\phi}(a\psi_{2}(0)+b\psi_{2}^{\prime}(0))$

on $\psi$ and Dirichlet condition on $\phi$. This yields the following
expression

$\displaystyle\int\limits_{0}^{\infty}\left\\{\sum_{i=1}^{2}\lambda_{1,i}\theta_{1}(0)\theta_{i}(y)f_{i}(y)\right\\}dy=$
(2.4) $\displaystyle
e^{i\phi}\left(a\int\limits_{0}^{\infty}\left\\{\sum_{i=1}^{2}\lambda_{2,i}\theta_{2}(0)\theta_{i}(y)f_{i}(y)\right\\}dy+b\int\limits_{0}^{\infty}\left\\{\frac{\theta_{2}(y)f_{2}(y)}{\theta_{2}(0)}+\sum_{i=1}^{2}\lambda_{2,i}\theta_{2}^{\prime}(0)\theta_{i}(y)f_{i}(y)\right\\}dy\right).$

Applying the second sub-condition

$\psi_{1}^{\prime}(0)=e^{i\phi}(c\psi_{2}(0)+d\psi_{2}^{\prime}(0))$

on $\psi$ and Dirichlet condition on $\phi$, we obtain

$\displaystyle\int\limits_{0}^{\infty}\left\\{\frac{\theta_{1}(y)f_{1}(y)}{\theta_{1}(0)}+\sum_{i=1}^{2}\lambda_{1,i}\theta_{1}^{\prime}(0)\theta_{i}(y)f_{i}(y)\right\\}dy=$
(2.5) $\displaystyle
e^{i\phi}\left(c\int\limits_{0}^{\infty}\left\\{\sum_{i=1}^{2}\lambda_{2,i}\theta_{2}(0)\theta_{i}(y)f_{i}(y)\right\\}dy+d\int\limits_{0}^{\infty}\left\\{\frac{\theta_{2}(y)f_{2}(y)}{\theta_{2}(0)}+\sum_{i=1}^{2}\lambda_{2,i}\theta_{2}^{\prime}(0)\theta_{i}(y)f_{i}(y)\right\\}dy\right).$

Comparing the coefficients of $f_{i}(y)\theta_{i}(y)$ in (2.4) and (2.5), we
get the following set of equations

$\displaystyle\theta_{1}(0)\lambda_{1,2}-e^{i\phi}\left(a\theta_{2}(0)\lambda_{2,2}-b\left(\frac{1}{\theta_{2}(0)}+\theta_{2}^{\prime}(0)\lambda_{2.2}\right)\right)$
$\displaystyle=0$
$\displaystyle\theta_{1}(0)\lambda_{1,1}-e^{i\phi}\left(a\theta_{2}(0)\lambda_{2,1}+b\theta_{2}^{\prime}(0)\lambda_{2,1}\right)$
$\displaystyle=0$
$\displaystyle\theta_{1}^{\prime}(0)\lambda_{1,2}-e^{i\phi}\left(c\theta_{2}(0)\lambda_{2,2}-d\left(\frac{1}{\theta_{2}(0)}+\theta_{2}^{\prime}(0)\lambda_{2.2}\right)\right)$
$\displaystyle=0$
$\displaystyle\frac{1}{\theta_{1}(0)}+\theta_{1}^{\prime}(0)\lambda_{1,1}-e^{i\phi}\left(c\theta_{2}(0)\lambda_{2,1}+d\theta_{2}^{\prime}(0)\lambda_{2,1}\right)$
$\displaystyle=0$

We obtain the values of $\lambda_{j,l}$ $j,l=1,2$ by solving the above set of
equations. ∎

Using (2.3) and Lemma (2.1) we can express the resolvent kernel of
$H_{V}^{\mathcal{A}}$ given by (2.1) as

$\displaystyle\begin{bmatrix}R^{\mathcal{A}}_{1,1}&R^{\mathcal{A}}_{1,2}\\\
R^{\mathcal{A}}_{2,1}&R^{\mathcal{A}}_{2,2}\\\
\end{bmatrix}=\begin{bmatrix}R_{e_{1}}^{D}-\dfrac{a+b\dfrac{\theta_{2}^{\prime}(0)}{\theta_{2}(0)}}{L(\zeta)\theta_{1}^{2}(0)}\theta_{1}^{2}(x)&\dfrac{(ad-
bc)e^{i\phi}}{L(\zeta)\theta_{1}(0)\theta_{2}(0)}\theta_{1}(x)\theta_{2}(y)\\\
\dfrac{e^{-i\phi}}{L(\zeta)\theta_{1}(0)\theta_{2}(0)}\theta_{2}(x)\theta_{1}(y)&R_{e_{2}}^{D}-\dfrac{b\dfrac{\theta_{1}^{\prime}(0)}{\theta_{1}(0)}-d}{L(\zeta)\theta_{2}^{2}(0)}\theta_{2}^{2}(x)\\\
\end{bmatrix}.$ (2.6)

Here, for simplicity, we used the notation
$R^{\mathcal{A}}_{i,j}(x,y;\zeta)=R^{\mathcal{A}}_{i,j}$,
$R_{e_{i}}^{D}(x,y;\zeta)=R_{e_{i}}^{D}$ and
$\theta_{i}(x,\zeta)=\theta_{i}(x)$.

For unperturbed operator i.e., when $V_{j}(x)\equiv 0$, we may let

$\displaystyle\theta_{j}(x,\zeta)=e^{ix_{j}\zeta}\Rightarrow
L(\zeta)=(a-d)i\zeta-\left(b\zeta^{2}+c\right).$

Hence

$\displaystyle\begin{bmatrix}R^{\mathcal{A}}_{0,(1,1)}&R^{\mathcal{A}}_{0,(1,2)}\\\
R^{\mathcal{A}}_{0,(2,1)}&R^{\mathcal{A}}_{0,(2,2)}\end{bmatrix}=\begin{bmatrix}R^{D}_{0,e_{1}}-\frac{a+i\zeta
b}{(a-d)i\zeta-\left(b\zeta^{2}+c\right)}e^{2i\zeta x}&\frac{(ad-
bc)}{(a-d)i\zeta-\left(b\zeta^{2}+c\right)}e^{i(x+y+\phi)\zeta}\\\
\frac{1}{(a-d)i\zeta-\left(b\zeta^{2}+c\right)}e^{i(x+y-\phi)\zeta}&R^{D}_{0,e_{2}}-\frac{i\zeta
b-d}{(a-d)i\zeta-\left(b\zeta^{2}+c\right)}e^{2i\zeta x}\end{bmatrix}.$ (2.7)

we now prove Theorem (1.1).

###### Proof of Theorem 1.1.

The operator difference $R^{D}_{V}(z)-R^{D}_{0}(z)$ is a trace class operator
and the second term on the right hand side of (2.1) is a finite rank
perturbation. Therefore, the difference
$R_{V}^{\mathcal{A}}(z)-R_{0}^{\mathcal{A}}$ is a trace class operator. The
operator $H_{V}^{\mathcal{A}}$ can be seen as a particular case of a more
general matrix valued Schrödinger operator considered in [21] on the half-line
when potential is a diagonal matrix. The fact that
$R_{V}^{\mathcal{A}}(z)-R_{0}^{\mathcal{A}}$ is trace class also follows from
Lemma 9.1 of [21]. Representation (2.1) with values of $\lambda_{j,l}$ given
by lemma (2.1) allows us to compute the trace of the resolvent difference of
operators $H_{V}^{\mathcal{A}}$ and $H_{0}^{\mathcal{A}}$.

Using (2.6) and (2.7) the trace of the difference of $R^{\mathcal{A}}_{V}$ and
$R^{\mathcal{A}}_{0}$ can be written as

$\displaystyle\operatorname{Tr}(R^{\mathcal{A}}_{V}(z)-R^{\mathcal{A}}_{0}(z))$
$\displaystyle=\sum\limits_{j=1}^{2}\int\limits_{0}^{\infty}\left(R^{D}_{e_{j}}(x,x,z)-R^{D}_{0,e_{j}}(x,x,z)\right)dx$
$\displaystyle-\int\limits_{0}^{\infty}\left(\dfrac{a+b\dfrac{\theta_{2}^{\prime}(0)}{\theta_{2}(0)}}{L(\zeta)\theta_{1}^{2}(0)}\theta_{1}^{2}(x)+\dfrac{b\dfrac{\theta_{1}^{\prime}(0)}{\theta_{1}(0)}-d}{L(\zeta)\theta_{2}^{2}(0)}\theta_{2}^{2}(x)\right)dx$
$\displaystyle+\int\limits_{0}^{\infty}\left(\dfrac{(a-d+2i\zeta
b)}{(a-d)i\zeta-(b\zeta^{2}+c)}e^{2ix\zeta}\right)dx.$ (2.8)

The first term on the right hand side of (2.8) is given by (c.f. Remark 1.2 of
[5])

$\displaystyle\sum\limits_{j=1}^{2}\int\limits_{0}^{\infty}\left(R^{D}_{e_{j}}(x,x,z)-R^{D}_{0,e_{j}}(x,x,z)\right)dx=-\frac{1}{2\zeta}\sum\limits_{j=1}^{2}\frac{\dot{w}_{j}(\zeta)}{w_{j}(\zeta)}.$
(2.9)

To compute the first term in the second integral on the right hand side of
(2.8) we use the following equation which is true for any arbitrary solutions
$f_{j}(x,\zeta)$ and $g_{j}(x,\zeta)$ of
$H^{D}_{e_{j}}\psi_{j}=\zeta^{2}\psi_{j}$

$\displaystyle
f_{j}(x,\zeta)g_{j}(x,\zeta)=\dfrac{(f^{\prime}_{j}(x,\zeta)\dot{g}_{j}(x,\zeta)-f_{j}(x,\zeta)\dot{g}^{\prime}_{j}(x,\zeta))^{\prime}}{2\zeta}.$
(2.10)

If we let $f_{j}=g_{j}=\theta_{j}$ we obtain

$\displaystyle\int\limits_{0}^{\infty}\theta_{j}^{2}(x,\zeta)dx=\dfrac{[\theta^{\prime}_{j}(x,\zeta)\dot{\theta}_{j}(x,\zeta)-\theta_{j}(x,\zeta)\dot{\theta}^{\prime}_{j}(x,\zeta)]_{0}^{\infty}}{2\zeta}.$

If $V_{j}$ is compactly supported potential then the Jost solution
$\theta_{j}(x,\zeta)=e^{i\zeta x}$ and hence

$\displaystyle\theta^{\prime}_{j}(x,\zeta)\dot{\theta}_{j}(x,\zeta)-\theta_{j}(x,\zeta)\dot{\theta}^{\prime}_{j}(x,\zeta)=ie^{2i\zeta
x}.$

For $\operatorname{Im}\zeta>0$, $ie^{2i\zeta x}\rightarrow 0$ as
$x\rightarrow\infty$. Therefore, we are left with

$\displaystyle\int\limits_{0}^{\infty}\theta_{j}^{2}(x,\zeta)dx=\dfrac{\theta_{j}(0,\zeta)\dot{\theta}^{\prime}_{j}(0,\zeta)-\theta^{\prime}_{j}(0,\zeta)\dot{\theta}_{j}(0,\zeta)}{2\zeta}$
$\displaystyle=\dfrac{\theta_{j}^{2}(0,\zeta)}{2\zeta}\dfrac{d}{d\zeta}\dfrac{\theta^{\prime}_{j}(0,\zeta)}{\theta_{j}(0,\zeta)}.$

This implies

$\displaystyle\int\limits_{0}^{\infty}\left(\dfrac{a+b\dfrac{\theta_{2}^{\prime}(0)}{\theta_{2}(0)}}{L(\zeta)\theta_{1}^{2}(0)}\theta_{1}^{2}(x)+\dfrac{b\dfrac{\theta_{1}^{\prime}(0)}{\theta_{1}(0)}-d}{L(\zeta)\theta_{2}^{2}(0)}\theta_{2}^{2}(x)\right)dx$
$\displaystyle=\dfrac{1}{2\zeta
L(\zeta)}\left(\left(a+b\dfrac{\theta_{2}^{\prime}(0)}{\theta_{2}(0)}\right)\dfrac{d}{d\zeta}\dfrac{\theta^{\prime}_{1}(0,\zeta)}{\theta_{1}(0,\zeta)}+\left(b\dfrac{\theta_{1}^{\prime}(0)}{\theta_{1}(0)}-d\right)\dfrac{d}{d\zeta}\dfrac{\theta^{\prime}_{2}(0,\zeta)}{\theta_{2}(0,\zeta)}\right)$
$\displaystyle=\dfrac{\dot{L}(\zeta)}{2\zeta L(\zeta)}.$ (2.11)

By a density argument the above result can be extended to all potentials
satisfying the condition $\int_{0}^{\infty}|V_{j}(x)|<\infty$.

Now it only remains to compute the last integral on the right hand side of
(2.8), which is

$\displaystyle\int\limits_{0}^{\infty}\left(\dfrac{(a-d+2i\zeta
b)}{(a-d)i\zeta-b\zeta^{2}+c}e^{2ix\zeta}\right)dx=\dfrac{(a-d+2i\zeta
b)}{2\zeta((a-d)\zeta+(b\zeta^{2}+c)i)}.$ (2.12)

Substituting values from (2.9), (2.11) and (2.12) in (2.8), we get

$\displaystyle\operatorname{Tr}(R^{\mathcal{A}}_{V}(z)-R^{\mathcal{A}}_{0}(z))$
$\displaystyle=-\frac{1}{2\zeta}\left(\sum\limits_{j=1}^{2}\frac{\dot{w}_{j}(\zeta)}{w_{j}(\zeta)}+\dfrac{\dot{L}(\zeta)}{L(\zeta)}-\dfrac{(a-d+2i\zeta
b)}{(a-d)\zeta+(b\zeta^{2}+c)i}\right)$
$\displaystyle=-\frac{1}{2\zeta}\left(\sum\limits_{j=1}^{2}\frac{d}{d\zeta}\ln({w_{j}(\zeta)})+\frac{d}{d\zeta}\ln({L(\zeta)})-\frac{d}{d\zeta}\ln({(a-d)\zeta+(b\zeta^{2}+c)i})\right)$
$\displaystyle=-\frac{1}{2\zeta}\left(\dfrac{d}{d\zeta}\ln\left(\dfrac{L(\zeta)}{(a-d)\zeta+(b\zeta^{2}+c)i}\prod\limits_{j=1}^{2}w_{j}(\zeta)\right)\right).$

The second term on the right hand side of (2.1) with values of $\lambda_{j,l}$
given by lemma (2.1) is a finite rank operator and if the potential function
$V$ satisfy condition (1.2) then the operator
$\sqrt{|V|}(H^{\mathcal{A}}_{0})^{-1/2}$ is Hilbert-Schmidt and therefore,
$\sqrt{V}R_{0}^{\mathcal{A}}\sqrt{|V|}$ is trace class. This implies that the
perturbation determinant $D(z)$ is well defined. The perturbation determinant
and trace of the resolvent difference
$R^{\mathcal{A}}_{V}(z)-R_{0}^{\mathcal{A}}(z)$ are related by the following
expression

$\operatorname{Tr}(R^{\mathcal{A}}_{V}(z)-R_{0}^{\mathcal{A}}(z))=\frac{\frac{d}{dz}D(z)}{D(z)},\quad
z\in\rho(H_{V}^{\mathcal{A}})\cap\rho(H_{0}^{\mathcal{A}}).$

In order to find an explicit expression for the perturbation determinant we
choose $\zeta=\sqrt{z}$ such that Im $(z)>0$. It follows from Theorem 1.1 that

$\displaystyle\dfrac{\dfrac{d}{dz}(D(z))}{D(z)}=\frac{\dfrac{d}{dz}\left(\dfrac{L(\sqrt{z})}{(a-d)\sqrt{z}+(bz+c)i}\prod\limits_{j=1}^{2}w_{j}(\sqrt{z})\right)}{\dfrac{L(\sqrt{z})}{(a-d)\sqrt{z}+(bz+c)i}\prod\limits_{j=1}^{2}w_{j}(\sqrt{z})}.$

From here we deduce that

$\displaystyle
D(z)=A\dfrac{L(\sqrt{z})}{(a-d)\sqrt{z}+(bz+c)i}\prod\limits_{j=1}^{2}w_{j}(\sqrt{z}),\hskip
14.22636ptA\in\mathbb{C}.$

To find the value of the complex coefficient $A$, we use the following
asymptotic of perturbation determinant

$\displaystyle\lim_{|\text{Im}(z)|\rightarrow\infty}D(z)=1.$

This asymptotic holds if $\sqrt{|V|R^{\mathcal{A}}_{0}(z)^{-1}}$ is Hilbert-
Schmidt operator (see [23]). For the asymptotic behaviour of $w_{j}(\sqrt{z})$
as $|\sqrt{z}|\rightarrow\infty$ , we refer to [6], which provides

$\displaystyle w_{j}(\sqrt{z})=\theta_{j}(0,\sqrt{z})=1+O(|\sqrt{z}|^{-1}).$

Further, it is easy to find that

$\displaystyle
L(\sqrt{z})=(a-d)\sqrt{z}i-(bz+c)+O(1),\quad|\sqrt{z}|\rightarrow\infty.$
(2.13)

This implies, $A=\dfrac{1}{i}$. ∎

###  2.1. Spectral shift function

In this section we briefly discuss the relationship between perturbation
determinant $D(z)$ and the spectral shift function
$\xi(\lambda;H_{V}^{\mathcal{A}},H_{0}^{\mathcal{A}})$. The spectral shift
function for the pair of operators $H_{V}^{\mathcal{A}}$ and
$H_{0}^{\mathcal{A}}$ can be defined by the trace formula

$\operatorname{Tr}(f(H_{V}^{\mathcal{A}}-H_{0}^{\mathcal{A}})):=\int_{-\infty}^{\infty}\xi(\lambda;H_{V}^{\mathcal{A}},H_{0}^{\mathcal{A}})f^{\prime}(\lambda)\,d\lambda$
(2.14)

for all $f\in C_{0}^{\infty}(\mathbb{R})$.

Theorem 9.2 of [21] explicitly characterize the spectral shift function in the
more general case of matrix valued Schrödinger operators with general self-
adjoint conditions at the origin. It was proved that the trace formula (2.14)
holds for a bigger class of functions satisfying
$f^{\prime}(\lambda)=O(\lambda^{-1/2-\varepsilon}),\,f^{\prime\prime}(\lambda)=O(\lambda^{-1-\varepsilon}),\,\varepsilon>0,\,\lambda\rightarrow\infty$
and

$\int_{-\infty}^{\infty}|\xi(\lambda;H^{\mathcal{A}}_{V},H^{\mathcal{A}}_{0})|(1+|\lambda|)^{-1/2-\varepsilon}\,d\lambda<\infty,\quad\varepsilon>0.$

One of the main ingredients in the proof of Theorem 9.2 of [21] is the
following estimate111The interested reader can consult Appendix (A) where
explicit derivation of this estimate is provided. (cf. inequality (9.2) of
[21], see also (5.11) of [23] )

$||R^{\mathcal{A}}_{V}(-t)-R^{\mathcal{A}}_{0}(-t)||_{1}\leq
C_{\varepsilon}t^{-\frac{3}{2}+\varepsilon},\quad\varepsilon>0,\quad
t\rightarrow\infty.$ (2.15)

The above estimate implies

$\int_{1}^{\infty}t^{-m}||R^{\mathcal{A}}_{V}(-t)-R^{\mathcal{A}}_{0}(-t)||_{1}<\infty$

which holds for all $m>-1/2$. Proposition (3.3) of [5] then implies

$\xi(\lambda;\,H^{\mathcal{A}}_{V},H^{\mathcal{A}}_{0})=\pi^{-1}\lim_{\varepsilon\rightarrow
0}\emph{arg}\,D(\lambda+i\varepsilon),$

where $\emph{arg}\,D(z)=\operatorname{Im}\ln D(z)$ is defined by the condition
$\ln D(z)\rightarrow 0$ as
$\operatorname{dist}\\{z,\sigma(H^{\mathcal{A}}_{0})\\}\rightarrow\infty$.
Moreover,

$\ln
D(z)=\int_{-\infty}^{\infty}\xi(\lambda;\,H^{\mathcal{A}}_{V},H^{\mathcal{A}}_{0})(\lambda-z)^{-1}\,d\lambda,\quad
z\in\rho(H^{\mathcal{A}}_{0})\cap\rho(H^{\mathcal{A}}_{V}).$

##  3\. Levinson’s formula

In this section we derive a zero order trace formula, commonly known as
Levinson’s formula, which describes the number of negative eigenvalues of an
operator in terms of phase shift function which is denoted by $\eta$ and it is
defined, for real $k$, as $\eta(k):=\arg D(k)$. The Levinson’s formula derived
in this section is closely related to the Levinson’s formulas obtained in
Theorem 9.3 of [3] and Theorem 9.3 of [21]. Levinson’s formula obtained in [3]
relates the number of negative eigenvalues with the determinant of the
scattering matrix and certain parameters that depend on the boundary
conditions. Formula obtained in [21] relates the number of negative
eigenvalues with the spectral shift function. The vertex conditions of this
paper can be expressed as a particular case of more general conditions
considered in references [3] and [21] and therefore the Levinson’s formulas
obtained there also hold for the operator $H_{V}^{\mathcal{A}}$. For higher
order trace formulas of integer and half-integer order we refer to [23, 22,
19].

We will need the following low-energy asymptotics of the perturbation
determinant

######  Lemma 3.1 (Low energy asymptotics).

Let the potential $V$ satisfies the condition

$\displaystyle\int\limits_{e_{j}}(1+x)|V_{j}(x)|dx<\infty,\hskip 8.5359pt1\leq
j\leq n.$ (3.1)

Then as $\zeta\rightarrow 0$ the perturbation determinant satisfies

$D(\zeta)=\begin{cases}\zeta^{(Q-P)}(1+o(1))&w_{j}(0)\neq 0$, $j=1,2,\\\
\zeta^{(R-P)}(1+o(1))&w_{1}(0)=0,w_{2}(0)\neq 0,\\\
\zeta^{(S-P)}(1+o(1))&w_{1}(0)\neq 0,w_{2}(0)=0,\\\
\zeta^{(T-P)}(1+o(1))&w_{j}(0)=0$, $j=1,2.\end{cases}$

Where the functions $P,Q,R,S$ and $T$ are defined in Theorem (1.2).

###### Proof.

Expression for perturbation determinant is

$\displaystyle
D(\zeta)=\dfrac{L(\zeta)\prod\limits_{j=1}^{2}w_{j}(\zeta)}{(a-d)i\zeta-(b\zeta^{2}+c)}.$

We consider the following five different cases

  1. (1)

$w_{j}(0)\neq 0$, $j=1,2$ and $L(0)\neq 0$.

  2. (2)

$w_{j}(0)\neq 0$, $j=1,2$ and $L(0)=0$.

  3. (3)

$w_{1}(0)=0$ and $w_{2}(0)\neq 0$.

  4. (4)

$w_{1}(0)\neq 0$ and $w_{2}(0)=0$.

  5. (5)

$w_{j}(0)=0$, $j=1,2$.

In the first case $D(\zeta)$ has no zeros and we can let

$\displaystyle L(0)\prod\limits_{j=1}^{2}w_{j}(0)=\tilde{c}\neq 0$
$\displaystyle\Rightarrow\lim\limits_{\zeta\rightarrow
0}D(\zeta)=\lim\limits_{\zeta\rightarrow
0}\dfrac{\tilde{c}}{(a-d)i\zeta-(b\zeta^{2}+c)}$
$\displaystyle\Rightarrow\lim\limits_{\zeta\rightarrow
0}D(\zeta)((a-d)i\zeta-(b\zeta^{2}+c))=\tilde{c}$
$\displaystyle\Rightarrow\lim\limits_{\zeta\rightarrow
0}\dfrac{D(\zeta)((a-d)i\zeta-(b\zeta^{2}+c))-\tilde{c}}{\tilde{c}}=0$
$\displaystyle\Rightarrow
D(\zeta)=\frac{\tilde{c}}{(a-d)i\zeta-(b\zeta^{2}+c)}(1+o(1)).$

In the second case, we make use of the following small energy asymptotics
(c.f. Theorem 2.3 (iii)of [2])

$\displaystyle\dfrac{\theta_{j}^{\prime}(0,\zeta)}{\theta_{j}(0,\zeta)}=\dfrac{\theta_{j}^{\prime}(0,0)}{\theta_{j}(0,0)}-\dfrac{i\zeta}{\theta_{j}^{2}(0,0)}+o(\zeta)$

which implies

$\displaystyle L(\zeta)=$ $\displaystyle
a\left(\frac{\theta_{1}^{\prime}(0,0)}{\theta_{1}(0,0)}-\frac{i\zeta}{\theta_{1}^{2}(0,0)}+o(\zeta)\right)-d\left(\frac{\theta_{2}^{\prime}(0,0)}{\theta_{2}(0,0)}-\frac{i\zeta}{\theta_{2}^{2}(0,0)}+o(\zeta)\right)$
$\displaystyle+b\left(\frac{\theta_{1}^{\prime}(0,0)}{\theta_{1}(0,0)}-\frac{i\zeta}{\theta_{1}^{2}(0,0)}+o(\zeta)\right)\left(\frac{\theta_{2}^{\prime}(0,0)}{\theta_{2}(0,0)}-\frac{i\zeta}{\theta_{2}^{2}(0,0)}+o(\zeta)\right)+c$
$\displaystyle=$ $\displaystyle
a\frac{\theta_{1}^{\prime}(0,0)}{\theta_{1}(0,0)}-d\frac{\theta_{2}^{\prime}(0,0)}{\theta_{2}(0,0)}+b\frac{\theta_{1}^{\prime}(0,0)}{\theta_{1}(0,0)}\frac{\theta_{2}^{\prime}(0,0)}{\theta_{2}(0,0)}-c+a\left(\frac{-i\zeta}{\theta_{1}^{2}(0,0)}+o(\zeta)\right)$
$\displaystyle-d\left(\frac{-i\zeta}{\theta_{2}^{2}(0,0)}+o(\zeta)\right)+b\left[\left(\frac{\theta_{1}^{\prime}(0,0)}{\theta_{1}(0,0)\theta_{2}^{2}(0,0)}+\frac{\theta_{2}^{\prime}(0,0)}{\theta_{1}^{2}(0,0)\theta_{2}(0,0)}\right)i\zeta\right]$
$\displaystyle+b\left(o(\zeta)+o(\zeta^{2})-\frac{\zeta^{2}}{\theta_{1}(0,0)\theta_{2}(0,0)}\right)$

This yields

$\displaystyle L(\zeta)=$ $\displaystyle
L(0)+a\left(\frac{-i\zeta}{\theta_{1}^{2}(0,0)}+o(\zeta)\right)-d\left(\frac{-i\zeta}{\theta_{2}^{2}(0,0)}+o(\zeta)\right)$
$\displaystyle+b\left[\left(\frac{\theta_{1}^{\prime}(0,0)}{\theta_{1}(0,0)\theta_{2}^{2}(0,0)}+\frac{\theta_{2}^{\prime}(0,0)}{\theta_{1}^{2}(0,0)\theta_{2}(0,0)}\right)i\zeta+o(\zeta)+o(\zeta^{2})-\frac{\zeta^{2}}{\theta_{1}(0,0)\theta_{2}(0,0)}\right].$

Therefore,

$\displaystyle
D(\zeta)=\dfrac{\alpha_{1}\zeta+o(\zeta)+b(\zeta^{2}+(o(\zeta^{2})))}{(a-d)i\zeta-(b\zeta^{2}+c)},$

where,
$\alpha_{1}=\left(b\left(\dfrac{\theta_{1}^{\prime}(0,0)}{\theta_{2}(0,0)}+\dfrac{\theta_{2}^{\prime}(0,0)}{\theta_{1}(0,0)}\right)-a\dfrac{\theta_{2}(0,0)}{\theta_{1}(0,0)}+d\dfrac{\theta_{1}(0,0)}{\theta_{2}(0,0)}\right)i$.
The above calculations show that the order of the numerator of $D(\zeta)$
depends on the value of the parameter $b$. If $b=0$, then it will be of order
$\zeta$ otherwise it will be of order $\zeta^{2}$.

The study of first two cases can be summarized as follows

  * •

If $w_{j}(0)\neq 0$, $j=1,2$ and $L(0)\neq 0$ then the numerator of $D(\zeta)$
is a non zero constant and it will have no zeros.

  * •

If $w_{j}(0)\neq 0$, $j=1,2$ and $L(0)=0$ then the numerator of $D(\zeta)$ for
small $\zeta$ will have one zero in case $\alpha_{1}\neq 0$ and it will have
two zeros in case $\alpha_{1}=0$, $b\neq 0$.

This gives us the following low energy asymptotics in the first two cases

$\displaystyle D(\zeta)=\zeta^{(Q-P)}(1+o(1)),$

where

$Q=Q(b,\alpha_{1},L(0))=\begin{cases}0&L(0)\neq 0,\\\ 1&L(0)=0,\alpha_{1}\neq
0\\\ 2&L(0)=0,\alpha_{1}=0,b\neq 0\end{cases}$

and

$P=P(a,b,c,d)=\begin{cases}0&c\neq 0,\\\ 1&c=0,a-d\neq 0\\\ 2&c=0,a-d=0,b\neq
0.\end{cases}$

The function $P$ gives the number of poles of $D(\zeta)$ in the given cases.

Let us now consider the third case, i.e., when $w_{1}(0)=0$ and $w_{2}(0)\neq
0$. We rewrite $D(\zeta)$ as

$\displaystyle D(\zeta)$
$\displaystyle=\dfrac{\left(a\dfrac{\theta_{1}^{\prime}(0,\zeta)}{\theta_{1}(0,\zeta)}-d\dfrac{\theta_{2}^{\prime}(0,\zeta)}{\theta_{2}(0,\zeta)}+b\prod\limits_{k=1}^{2}\dfrac{\theta_{k}^{\prime}(0,\zeta)}{\theta_{k}(0,\zeta)}-c\right)\prod\limits_{j=1}^{2}\theta_{j}(0,\zeta)}{(a-d)i\zeta-(b\zeta^{2}+c)}$
$\displaystyle=\dfrac{\left(a\theta_{1}^{\prime}(0,\zeta)\theta_{2}(0,\zeta)+b\theta_{1}^{\prime}(0,\zeta)\theta_{2}^{\prime}(0,\zeta)-d\theta_{1}(0,\zeta)\theta_{2}^{\prime}(0,\zeta)-c\prod\limits_{j=1}^{2}\theta_{j}(0,\zeta)\right)}{(a-d)i\zeta-(b\zeta^{2}+c)}$
(3.2)

and use the following asymptotic for $\theta_{1}(x,\zeta)$ (c.f. [4])

$\displaystyle\theta_{1}(0,\zeta)=\zeta\dot{\theta}_{1}(0,0)+o(\zeta).$

This implies

$\displaystyle
D(\zeta)=\dfrac{\left(\alpha_{2}-c_{3}\zeta+o(\zeta)\right)}{(a-d)i\zeta-(b\zeta^{2}+c)}$

Where
$\alpha_{2}=a\theta_{1}^{\prime}(0,0)\theta_{2}(0,0)+b\theta_{1}^{\prime}(0,0)\theta_{2}^{\prime}(0,0)$
and $c_{3}=(c\theta_{2}(0,0)+d\theta_{2}^{\prime}(0,0))\dot{\theta}_{1}(0,0)$.
The numerator of $D(\zeta)$ is equal to $\alpha_{2}-c_{3}\zeta+o(\zeta)$ as
$\zeta\rightarrow 0$. If $\alpha_{2}\neq 0$ then the numerator of $D(\zeta)$
has no zeros for small $\zeta$ and if $\alpha_{2}=0$, then the numerator of
$D(\zeta)$ has one zero for small $\zeta$. Hence,

$\displaystyle D(\zeta)=\zeta^{(R-P)}(1+o(1)),\quad\zeta\rightarrow 0,$

where,

$R=R(\alpha_{2})=\begin{cases}0&\alpha_{2}\neq 0,\\\
1&\alpha_{2}=0.\end{cases}$

For the fourth case we substitute

$\displaystyle\theta_{2}(0,\zeta)=\zeta\dot{\theta}_{2}(0,0)+o(\zeta),\zeta\rightarrow
0$

into (3.2) and obtain

$\displaystyle
D(\zeta)=\dfrac{\left(\alpha_{3}+c_{5}\zeta+o(\zeta)\right)}{(a-d)i\zeta-(b\zeta^{2}+c)}.$

Here,
$\alpha_{3}=b\theta_{1}^{\prime}(0,0)\theta_{2}^{\prime}(0,0)-d\theta_{1}(0,0)\theta_{2}^{\prime}(0,0)$
and $c_{5}=(a\theta_{1}^{\prime}(0,0)-c\theta_{1}(0,0))\dot{\theta}_{2}(0,0)$.
The numerator of $D(\zeta)$ for small $\zeta$ has no zeros if $\alpha_{3}\neq
0$ and it has one zero if $\alpha_{3}=0$. Therefore,

$\displaystyle D(\zeta)=\zeta^{(S-P)}(1+o(1)),\quad\zeta\rightarrow 0,$

where,

$S=S(\alpha_{3})=\begin{cases}0&\alpha_{3}\neq 0,\\\
1&\alpha_{3}=0.\end{cases}$

Finally, for the fifth case we substitute the asymptotics

$\displaystyle\theta_{1}(0,\zeta)=\zeta\dot{\theta}_{1}(0,0)+o(\zeta)\text{
and
}\theta_{2}(0,\zeta)=\zeta\dot{\theta}_{2}(0,0)+o(\zeta),\quad\zeta\rightarrow
0$

into (3.2) and obtain

$\displaystyle
D(\zeta)=\dfrac{c\left(c_{6}\zeta^{2}+o(\zeta^{2})\right)+\alpha_{4}\zeta+bc_{8}+o(\zeta)}{(a-d)i\zeta-(b\zeta^{2}+c)},\quad\zeta\rightarrow
0.$

Here, $c_{6}=\dot{\theta}_{1}(0,0)\dot{\theta}_{2}(0,0)\neq 0$,
$c_{8}=\theta_{1}^{\prime}(0,0)\theta_{2}^{\prime}(0,0)\neq 0$ (because
$\theta(0,0)=0\Rightarrow\dot{\theta}(0,0)\theta^{\prime}(0,0)=-i$, the proof
is given in [5, Lemma 4.6.(1)], hence $\dot{\theta}(0,0)\neq 0$ and
$\theta^{\prime}(0,0)\neq 0$)
$\alpha_{4}=a\theta_{1}^{\prime}(0,0)\dot{\theta}_{2}(0,0)-d\dot{\theta}_{1}(0,0)\theta_{2}^{\prime}(0,0)$.
This implies the numerator of $D(\zeta)$ for small $\zeta$ has no zeros if
$b\neq 0$, it has one zero if $b=0,\alpha_{4}\neq 0$ and it has two zeros if
$b=0$, $\alpha_{4}=0$, $c\neq 0$. This yields

$\displaystyle D(\zeta)=\zeta^{(T-P)}(1+o(1)),\quad\zeta\rightarrow 0,$

where

$T=T(b,c,\alpha_{4})=\begin{cases}0&b\neq 0,\\\ 1&b=0,\alpha_{4}\neq 0\\\
2&b=0,\alpha_{4}=0,c\neq 0.\end{cases}$

∎

Using the low energy asymptotics of $D(\zeta)$ we now prove the analogue of
Levinson’s formula for the negative eigenvalues of $H_{V}^{\mathcal{A}}$,
stated in Theorem (1.2).

###### Proof of Theorem 1.2.

The function $D(\zeta)$ has a zero in $\zeta$ of order $r$ if and only if
$\zeta^{2}$ is an eigenvalue of multiplicity $r$ of the operator
$H_{V}^{\mathcal{A}}$ [3]. As $H_{V}^{\mathcal{A}}$ is a self-adjoint
operator, so the zeros of $D(\zeta)$ lie on the positive imaginary axis and it
may have only real eigenvalues.

Let $\Gamma_{R,\varepsilon}$ denotes the contour (counterclockwise) consisting
of semicircles $C^{+}_{R}=\\{|\zeta|=R,\quad\mbox{Im }\zeta\geq 0\\}$ and
$C^{+}_{\varepsilon}=\\{|\zeta|=\varepsilon,\quad\mbox{Im }\zeta\geq 0\\}$ and
the intervals $(\varepsilon,R)$ and $(-R,-\varepsilon)$. $R$ and $\varepsilon$
are chosen such that all $N$ negative eigenvalues lie inside the contour
$\Gamma_{R,\varepsilon}$.

The function $w_{j}(\zeta)$ is analytic in the upper halfp-plane, therefore,
$D(\zeta)$ is analytic inside and on the contour $\Gamma_{R,\varepsilon}$.
Hence,

$\displaystyle\int\limits_{\Gamma_{R,\varepsilon}}\dfrac{\frac{d}{d\zeta}D(\zeta)}{D(\zeta)}d\zeta=2\pi
iN.$ (3.3)

Note that,

$\displaystyle\lim\limits_{\mbox{Im
}\zeta\rightarrow\infty}D(\zeta)=\lim\limits_{\mbox{Im
}\zeta\rightarrow\infty}\dfrac{L(\zeta)\prod\limits_{j=1}^{2}w_{j}(\zeta)}{(a-d)i\zeta-(b\zeta^{2}+c)}=1+O(|\zeta|^{-1}).$
(3.4)

The branch of the function $\ln D(\zeta)$ can be fixed by the condition $\ln
D(\zeta)\rightarrow 0$ as $|\zeta|\rightarrow\infty$. For $k\in\mathbb{R}$, we
set $D(k)=a(k)e^{i\eta(k)}$, where $a(k)=|D(k)|$ and $\eta(k)=\arg D(k)$. Now

$\displaystyle\mbox{var}_{\Gamma_{R,\varepsilon}}\arg
D(\zeta)=\eta(R)-\eta(\varepsilon)+\mbox{var}_{C_{R}^{+}}\arg
D(\zeta)+\\{\eta(-\varepsilon)-\eta(-R)\\}+\mbox{var}_{C_{\varepsilon}^{+}}\arg
D(\zeta).$

It follows from the representation of $D(k)$ that $\eta(-k)=-\eta(k)$. Hence,

$\displaystyle\mbox{var}_{\Gamma_{R,\varepsilon}}\arg
D(\zeta)=2(\eta(R)-\eta(\varepsilon))+\mbox{var}_{C_{R}^{+}}\arg
D(\zeta)+\mbox{var}_{C_{\varepsilon}^{+}}\arg D(\zeta)$

Equation (3.3) implies that

$\displaystyle\mbox{var}_{\Gamma_{R,\varepsilon}}\arg D(\zeta)=2\pi N.$

This implies

$\displaystyle 2\pi N=2(\eta(R)-\eta(\varepsilon))+\mbox{var}_{C_{R}^{+}}\arg
D(\zeta)+\mbox{var}_{C_{\varepsilon}^{+}}\arg D(\zeta).$

Using (3.4), one can see that
$\lim\limits_{R\rightarrow\infty}\mbox{var}_{C_{R}^{+}}\arg D(\zeta)=0$. We
are left with

$\displaystyle\eta(\infty)-\eta(\varepsilon)=\pi
N-\dfrac{1}{2}\mbox{var}_{C_{\varepsilon}^{+}}\arg D(\zeta).$

Now letting $\varepsilon\rightarrow 0$ and using Lemma (3.1), we arrive at

$\lim\limits_{\varepsilon\rightarrow 0}\mbox{var}_{C_{\varepsilon}^{+}}\arg
D(\zeta)=\begin{cases}-(Q-P)\pi&w_{j}(0)\neq 0$, $j=1,2,\\\
-(R-P)\pi&w_{1}(0)=0,w_{2}(0)\neq 0,\\\ -(S-P)\pi&w_{1}(0)\neq
0,w_{2}(0)=0,\\\ -(T-P)\pi&w_{j}(0)=0$, $j=1,2.\end{cases}$

Therefore,

$\eta(\infty)-\eta(0)=\begin{cases}\pi(N+\dfrac{Q-P}{2})&w_{j}(0)\neq 0$,
$j=1,2,\\\ \pi(N+\dfrac{R-P}{2})&w_{1}(0)=0,w_{2}(0)\neq 0,\\\
\pi(N+\dfrac{S-P}{2})&w_{1}(0)\neq 0,w_{2}(0)=0,\\\
\pi(N+\dfrac{T-P}{2})&w_{j}(0)=0$, $j=1,2.\end{cases}$

∎

##  4\. examples

In this section we illustrate the trace formula and perturbation determinant
obtained in theorem (1.1) with some examples.

######  Example 4.1.

For the $\delta$ interaction of strength $\alpha\in\mathbb{R}$, we can choose
$\phi=0,\,a=1,\,b=0,\,c=\alpha$ and $d=-1$. Then according to theorem (1.1)
the trace of the resolvent difference is given by

$\operatorname{Tr}(R_{V}^{\mathcal{A}}(\zeta)-R_{0}^{\mathcal{A}}(\zeta))=-\frac{1}{2\zeta}\ln\left(\frac{\theta_{1}^{\prime}(0,\zeta)\theta_{2}(0,\zeta)+\theta_{1}(0,\zeta)\theta_{2}^{\prime}(0,\zeta)-\alpha\theta_{1}(0,\zeta)\theta_{2}(0,\zeta)}{2\zeta+i\alpha}\right)$

and the expression for perturbation determinant is

$D(\zeta)=\frac{1}{2i\zeta-\alpha}\left(\theta_{1}^{\prime}(0,\zeta)\theta_{2}(0,\zeta)+\theta_{1}(0,\zeta)\theta_{2}^{\prime}(0,\zeta)-\alpha\theta_{1}(0,\zeta)\theta_{2}(0,\zeta)\right).$

For $\alpha=0$ the $\delta$ conditions are commonly known as Kirchhoff
conditions and in this case the above expressions coincide with the results
obtained in Theorem (1.1) and Corollary (3.1) of [5] with $n=2$ (see also
[20]).

######  Example 4.2.

The $\delta^{\prime}$ conditions are obtained by choosing
$\phi=0,\,a=-1,\,b=\beta,\,c=0$ and $d=1$ in (1.3). In this case theorem (1.1)
implies

$\operatorname{Tr}(R_{V}^{\mathcal{A}}(\zeta)-R_{0}^{\mathcal{A}}(\zeta))=-\frac{1}{2\zeta}\ln\left(\frac{\theta_{1}^{\prime}(0,\zeta)\theta_{2}(0,\zeta)+\theta_{1}(0,\zeta)\theta_{2}^{\prime}(0,\zeta)-\beta\theta^{\prime}_{1}(0,\zeta)\theta^{\prime}_{2}(0,\zeta)}{2\zeta-i\beta\zeta^{2}}\right)$

and perturbation determinant is given by

$D(\zeta)=\frac{1}{2i\zeta+\beta\zeta^{2}}\left(\theta_{1}^{\prime}(0,\zeta)\theta_{2}(0,\zeta)+\theta_{1}(0,\zeta)\theta_{2}^{\prime}(0,\zeta)-\beta\theta^{\prime}_{1}(0,\zeta)\theta^{\prime}_{2}(0,\zeta)\right).$

In the next two examples we use the notation $D_{x}\phi=\phi^{(1)}$ for the
generalized derivative in the sense of [16].

######  Example 4.3.

The Schrödinger operator with the singular density
$-D_{x}\left(1+\sigma\delta\right)D_{x}$ is equivalent to the operator
$H_{V}^{\mathcal{A}}$ with $\phi=0,\,a=1,\,b=-\sigma\in\mathbb{R},\,c=0$ and
$d=-1$ and the perturbation determinant is given by

$D(\zeta)=\frac{1}{2i\zeta+\sigma\zeta^{2}}\left(\theta_{1}^{\prime}(0,\zeta)\theta_{2}(0,\zeta)+\theta_{1}(0,\zeta)\theta^{\prime}_{2}(0,\zeta)-\sigma\theta^{\prime}_{1}(0,\zeta)\theta_{2}^{\prime}(0,\zeta)\right).$

######  Example 4.4.

For real $\sigma_{1}$ and $\sigma_{2}$ the two parameters family of
Schrödinger operators with singular potential
$-D_{x}^{2}+\sigma_{1}\delta+\sigma_{2}\delta^{(1)}+V$ is equivalent to the
operator $H_{V}^{\mathcal{A}}$ with
$\phi=0,\,a=\frac{2+\sigma_{2}}{2-\sigma_{2}},\,b=0,\,c=\frac{4\sigma_{1}}{4-\sigma_{2}^{2}},\,d=-\left(\frac{2-\sigma_{2}}{2+\sigma_{2}}\right)$
(c.f. Section 3.2.4 of [16]). Theorem (1.1) implies the following expression
for the perturbation determinant

$D(\zeta)=\dfrac{\left(\left(\frac{2+\sigma_{2}}{2-\sigma_{2}}\right)\theta_{1}^{\prime}(0,\zeta)\theta_{2}(0,\zeta)+\left(\frac{2-\sigma_{2}}{2+\sigma_{2}}\right)\theta_{1}(0,\zeta)\theta_{2}^{\prime}(0,\zeta)-\left(\frac{4\sigma_{1}}{4-\sigma_{2}^{2}}\right)\theta_{1}(0,\zeta)\theta_{2}(0,\zeta)\right).}{\left(\frac{8+2\sigma_{2}^{2}}{4-\sigma_{2}^{2}}\right)i\zeta-\left(\frac{4\sigma_{1}}{4-\sigma_{2}^{2}}\right)}$

##  5\. conclusions

The Schrödinger operators on the real-line with generalized point interaction
at zero can be imagined as a $2\times 2$ matrix-valued Schrödinger operator on
semi-axis with general self-adjoint conditions at zero. Scattering theory of
matrix-valued Schrödinger operators on semi-axis with general self-adjoint
conditions at zero has been studied in much detail in the last few years. On
the other hand, explicit derivation of perturbation determinants and its
connections with spectral shift function and Levinson’s formulas has received
relatively less attention. The present article is devoted to explicit
derivation of perturbation determinant via trace formula for Schrödinger
operators and expressing Levinson’s formula and spectral shift function in
terms of perturbation determinant. Our results complement the previous studies
in this regard. We used operator theoretic approach in our derivations similar
to Demirel’s work which deals with star graphs with Kirchhoff conditions at
the vertex. The expressions for perturbation determinants are further used to
derive LevinsonÂ’s formulas connecting the number of negative eigenvalues with
phase shift. Phase shift is defined in terms of perturbation determinant.
Looking forward, it would be interesting to extend and generalize the results
on perturbation determinant to matrix-valued Schrödinger operators with most
general self-adjoint conditions at zero.

##  Appendix A Estimate on trace norm of resolvent difference

Here we’ll provide and explicit derivation of the estimate

$\displaystyle||R_{V}^{\mathcal{A}}(-t)-R_{0}^{\mathcal{A}}(-t)||_{1}\leq
C_{\varepsilon}t^{-\frac{3}{2}+\varepsilon}$

for all $\varepsilon>0$ and large $t$. We will need the following lemma.

######  Lemma A.1 ([5], Lemma 3.4 ).

Let $\mathcal{H}$ be a Hilbert space and $f,g\in\mathcal{H}$. Assume that
$\mathcal{R}=(\cdot,\,f)f-(\cdot,\,g)g$ is an operator of rank two on
$\mathcal{H}$. Then, the trace norm of $\mathcal{R}$ is given by

$\displaystyle||\mathcal{R}||_{1}=\sqrt{(||f||^{2}+||g||^{2})^{2}-4|(f,g)|^{2}}.$
(A.1)

Moreover, if we let $g=f+h$ then,

$\displaystyle(||f||^{2}+||g||^{2})^{2}-4|(f,g)|^{2}\leq
6||h||^{2}||f||^{2}+3||h||^{4}.$ (A.2)

######  Lemma A.2.

Assume that the potential $V$ satisfies condition (1.2). Then the resolvents
$R_{V}^{\mathcal{A}}$ of $H_{V}^{\mathcal{A}}$ and $R_{0}^{\mathcal{A}}$ of
$H_{0}^{\mathcal{A}}$ satisfy, for all $\varepsilon>0$ and for large $t,$

$\displaystyle||R_{V}^{\mathcal{A}}(-t)-R_{0}^{\mathcal{A}}(-t)||_{1}\leq
C_{\varepsilon}t^{-\frac{3}{2}+\varepsilon}.$

###### Proof.

Let
$\mathcal{R}_{1}=R^{\mathcal{A}}_{V}(-t)-R^{D}_{V}(-t)+R^{D}_{0}(-t)-R^{\mathcal{A}}_{0}(-t)$
and $\mathcal{R}_{2}=R^{D}_{V}(-t)-R^{D}_{0}(-t)$ then,

$\displaystyle
R^{\mathcal{A}}_{V}(-t)-R^{\mathcal{A}}_{0}(-t)=\mathcal{R}_{1}+\mathcal{R}_{2}.$

Hence

$\displaystyle||R^{\mathcal{A}}_{V}(-t)-R^{\mathcal{A}}_{0}(-t)||_{1}\leq||\mathcal{R}_{1}||_{1}+||\mathcal{R}_{2}||_{1}.$
(A.3)

Trace norm on $\mathcal{R}_{2}$ can be estimated as
$||\mathcal{R}_{2}||_{1}\leq ct^{-\frac{3}{2}+\varepsilon}$ (c.f. Lemma 4.5.6.
[23]). We only need to find estimate on the trace norm of rank two operator
$\mathcal{R}_{1}$. By adding and subtracting the quantity

$\displaystyle\dfrac{\theta_{j}^{2}(x_{j},i\sqrt{t})}{L(i\sqrt{t})\theta_{j}^{2}(0,i\sqrt{t})}+\dfrac{e^{-2x_{j}\sqrt{t}}}{(d-a)\sqrt{t}+bt-c}\hskip
28.45274ptj=1,2$

in the diagonal of $\mathcal{R}_{1}$ we can re-write it as

$\displaystyle\mathcal{R}_{1}=\mathcal{S}_{1}+\mathcal{S}_{2},$

where

$\displaystyle\mathcal{S}_{1}=\begin{bmatrix}\frac{(a-\sqrt{t}b+1)e^{-2x_{1}\sqrt{t}}}{(d-a)\sqrt{t}+bt-c}-\frac{\left(a+b\frac{\theta_{2}^{\prime}(0,i\sqrt{t})}{\theta_{2}(0,i\sqrt{t})}+1\right)\theta_{1}^{2}(x_{1},i\sqrt{t})}{L(i\sqrt{t})\theta_{1}^{2}(0,i\sqrt{t})}&0\\\
0&\frac{(1-\sqrt{t}b-d)e^{-2x_{2}\sqrt{t}}}{(d-a)\sqrt{t}+bt-c}-\frac{\left(b\frac{\theta_{1}^{\prime}(0,i\sqrt{t})}{\theta_{1}(0,i\sqrt{t})}-d+1\right)\theta_{2}^{2}(x_{2},i\sqrt{t})}{L(i\sqrt{t})\theta_{2}^{2}(0,i\sqrt{t})}\end{bmatrix}$

and

$\displaystyle(\mathcal{S}_{2})_{j,k}=\begin{bmatrix}\frac{\theta_{j}(x_{j},i\sqrt{t})\theta_{k}(x_{k},i\sqrt{t})}{L(i\sqrt{t})\theta_{j}(0,i\sqrt{t})\theta_{k}(0,i\sqrt{t})}-\frac{e^{-(x_{j}+x_{k})\sqrt{t}}}{(d-a)\sqrt{t}+bt-c}\end{bmatrix}\hskip
28.45274ptj,k=1,2.$

Now, as $\mathcal{S}_{1}$ is a diagonal matrix and in order to apply
definition of trace norm, we can let $\mathcal{S}_{1}=I\mathcal{S}_{1}I$,
where $I$ is second order identity matrix. This implies,

$\displaystyle||\mathcal{S}_{1}||_{1}=|\sigma_{1}|+|\sigma_{2}|$

where

$\displaystyle|\sigma_{1}|=\int_{0}^{\infty}\left|\frac{(a-\sqrt{t}b+1)e^{-2x_{1}\sqrt{t}}}{(d-a)\sqrt{t}+bt-c}-\frac{\left(a+b\frac{\theta_{2}^{\prime}(0,i\sqrt{t})}{\theta_{2}(0,i\sqrt{t})}+1\right)\theta_{1}^{2}(x_{1},i\sqrt{t})}{L(i\sqrt{t})\theta_{1}^{2}(0,i\sqrt{t})}\right|dx_{1}$

and

$\displaystyle|\sigma_{2}|=\int_{0}^{\infty}\left|\frac{(1-\sqrt{t}b-d)e^{-2x_{2}\sqrt{t}}}{(d-a)\sqrt{t}+bt-c}-\frac{\left(b\frac{\theta_{1}^{\prime}(0,i\sqrt{t})}{\theta_{1}(0,i\sqrt{t})}-d+1\right)\theta_{2}^{2}(x_{2},i\sqrt{t})}{L(i\sqrt{t})\theta_{2}^{2}(0,i\sqrt{t})}\right|dx_{2}.$

For $t$ large enough, these expressions can be simplified to

$\displaystyle|\sigma_{1}|$
$\displaystyle\approx\left|\frac{(a-\sqrt{t}b+1)}{(d-a)\sqrt{t}+bt-c}\right|\int_{0}^{\infty}\left|e^{-2x_{1}\sqrt{t}}-\theta_{1}^{2}(x_{1},i\sqrt{t})\right|dx_{1}$
$\displaystyle=\left|\frac{(a-\sqrt{t}b+1)}{(d-a)\sqrt{t}+bt-c}\right|\int_{0}^{\infty}\left|\theta_{1}(x_{1},i\sqrt{t})-e^{-x_{1}\sqrt{t}}\right|\left|\theta_{1}(x_{1},i\sqrt{t})+e^{-x_{1}\sqrt{t}}\right|dx_{1}$

and

$\displaystyle|\sigma_{2}|\approx\left|\frac{(1-\sqrt{t}b-d)}{(d-a)\sqrt{t}+bt-c}\right|\int_{0}^{\infty}\left|\theta_{2}(x_{2},i\sqrt{t})-e^{-x_{2}\sqrt{t}}\right|\left|\theta_{2}(x_{2},i\sqrt{t})+e^{-x_{2}\sqrt{t}}\right|dx_{2}.$

To find estimates for $|\sigma_{1}|$ and $|\sigma_{2}|$ we use the following
estimates for large $t$

$\left|\theta_{k}(x_{k},i\sqrt{t})-e^{-\sqrt{t}x_{k}}\right|\leq\dfrac{m_{1}}{\sqrt{t}}e^{-\sqrt{t}x_{k}}$
(A.4)

and

$\left|\theta_{k}(x_{k},i\sqrt{t})+e^{-\sqrt{t}x_{k}})\right|\leq\left(\frac{m_{2}}{\sqrt{t}}+2\right)e^{-\sqrt{t}x_{k}}$
(A.5)

This yields the following bounds

$\displaystyle|\sigma_{1}|\leq\left|\frac{(a-\sqrt{t}b+1)}{(d-a)\sqrt{t}+bt-c}\right|\frac{m_{1}(m_{2}+2\sqrt{t})}{2t^{3/2}}=|-m_{1}t^{-3/2}+O(t^{-2})|,$
$\displaystyle|\sigma_{2}|\leq\left|\frac{(1-\sqrt{t}b-d)}{(d-a)\sqrt{t}+bt-c}\right|\frac{m_{1}(m_{2}+2\sqrt{t})}{2t^{3/2}}=|-m_{1}t^{-3/2}+O(t^{-2})|.$

Therefore

$\displaystyle|\sigma_{1}|=|\sigma_{2}|=O(t^{-3/2})$

and hence

$\displaystyle||\mathcal{S}_{1}||_{1}=O(t^{-3/2}).$

To find estimate on norm of $\mathcal{S}_{2}$, we use Lemma (A.1). Clearly,
$\mathcal{S}_{2}=(\cdot,f)f-(\cdot,g)g$ is an operator of rank two, where

$f_{k}(x_{k})=\dfrac{-e^{-x_{k}\sqrt{t}}}{\sqrt{(d-a)\sqrt{t}+bt-c}}\quad\text{
and }\quad
g_{k}(x_{k})=\dfrac{-\theta_{k}(x_{k},i\sqrt{t})}{\sqrt{L(i\sqrt{t})}\theta_{k}(0,i\sqrt{t})}.$

Let $h=g-f$, then

$\displaystyle h_{k}(x_{k})$
$\displaystyle=\dfrac{e^{-x_{k}\sqrt{t}}}{\sqrt{(d-a)\sqrt{t}+bt-c}}-\dfrac{\theta_{k}(x_{k},i\sqrt{t})}{\sqrt{L(i\sqrt{t})}\theta_{k}(0,i\sqrt{t})}$
$\displaystyle=-\left(\dfrac{\theta_{k}(x_{k},\sqrt{t}i)-e^{-\sqrt{t}x_{k}}}{\sqrt{L(i\sqrt{t})}\theta_{k}(0,i\sqrt{t})}+\left(\dfrac{1}{\sqrt{L(i\sqrt{t})}\theta_{k}(0,i\sqrt{t})}-\dfrac{1}{\sqrt{(d-a)\sqrt{t}+bt-c}}\right)e^{-\sqrt{t}x_{k}}\right).$

To compute estimate on
$||h||^{2}=\sum\limits_{k=1}^{2}\int\limits_{0}^{\infty}\left|h_{k}(x_{k})\right|^{2}dx_{k}$,
we use (2.13) and (A.4) and after some simplifications we obtain

$\displaystyle||h||^{2}\leq\dfrac{m_{1}^{2}}{\left((d-a)\sqrt{t}+bt-c\right){t}^{3/2}}.$

Similarly,

$\displaystyle||f||^{2}=\sum\limits_{k=1}^{2}\int\limits_{0}^{\infty}\left|\dfrac{-e^{-\sqrt{t}x_{k}}}{\sqrt{(d-a)\sqrt{t}+bt-c}}\right|^{2}dx_{k}=\dfrac{1}{\left((d-a)\sqrt{t}+bt-c\right)\sqrt{t}}.$

The estimate on the right hand side of inequality (A.2) is now given by

$\displaystyle 6||h||^{2}||f||^{2}+3||h||^{4}$
$\displaystyle\leq\frac{6m_{1}^{2}}{\left((d-a)\sqrt{t}+bt-c\right)^{2}t^{2}}+\frac{3m_{1}^{4}}{\left((d-a)\sqrt{t}+bt-c\right)^{2}t^{3}}$
$\displaystyle=\frac{3m_{1}^{4}+6m_{1}^{2}t}{\left((d-a)\sqrt{t}+bt-c\right)^{2}t^{3}}$
$\displaystyle=\frac{6m_{1}^{2}}{b^{2}t^{4}}+\frac{12m_{1}^{2}(a-d)}{b^{3}t^{\frac{9}{2}}}+O(t^{-5})$
$\displaystyle=O(t^{-4}).$

This implies

$\displaystyle||\mathcal{S}_{2}||_{1}=O(t^{-2})$

and hence

$\displaystyle||\mathcal{R}_{1}||_{1}=O(t^{-\frac{3}{2}}).$

∎

## Declarations

### Ethics approval and consent to participate

Not applicable.

### Consent for publication

Not applicable.

### Availability of data and material

Not applicable

### Competing interests

The authors declare no competing interests.

### Funding

The research leading to these results received no funding.

### Authors’ contributions

M.U. conceptualized the study and drafted the manuscript. MD. Z. carried out
the computations. All authors read and approved the final manuscript.

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