# Closed form solution and transmutation operators for Schrödinger equations
with finitely many $\delta$-interactions

Vladislav V. Kravchenko1, Víctor A. Vicente-Benítez2
1Department of Mathematics, Cinvestav, Campus Querétaro
Libramiento Norponiente # 2000,
Fracc. Real de Juriquilla, Querétaro, Qro. C.P. 76230 México
2Department of Physics, Universidad de Guanajuato, Campus León
Lomas del Bosque 103, Lomas del Campestre, León, Guanajuato C.P. 37150 México
<EMAIL_ADDRESS><EMAIL_ADDRESS>
###### Abstract

A closed form solution for the one-dimensional Schrödinger equation with a
finite number of $\delta$-interactions

$\mathbf{L}_{q,\mathfrak{I}_{N}}y:=-y^{\prime\prime}+\left(q(x)+\sum_{k=1}^{N}\alpha_{k}\delta(x-x_{k})\right)y=\lambda
y,\quad 0<x<b,\;\lambda\in\mathbb{C}$

is presented in terms of the solution of the unperturbed equation

$\mathbf{L}_{q}y:=-y^{\prime\prime}+q(x)y=\lambda y,\quad
0<x<b,\;\lambda\in\mathbb{C}$

and a corresponding transmutation operator $\mathbf{T}_{\mathfrak{I}_{N}}^{f}$
is obtained in the form of a Volterra integral operator. With the aid of the
spectral parameter power series method, a practical construction of the image
of the transmutation operator on a dense set is presented, and it is proved
that the operator $\mathbf{T}_{\mathfrak{I}_{N}}^{f}$ transmutes the second
derivative into the Schrödinger operator $\mathbf{L}_{q,\mathfrak{I}_{N}}$ on
a Sobolev space $H^{2}$. A Fourier-Legendre series representation for the
integral transmutation kernel is developed, from which a new representation
for the solutions and their derivatives, in the form of a Neumann series of
Bessel functions, is derived.

Keywords:  One-dimensional Schrödinger equation, point interactions,
transmutation operator, Fourier-Legendre series, Neumann series of Bessel
functions.

MSC Classification: 34A25; 34A45; 46F10; 47G10; 81Q05.

##  1 Introduction

We consider the one-dimensional Schrödinger equation with a finite number of
$\delta$-interactions

$-y^{\prime\prime}+\left(q(x)+\sum_{k=1}^{N}\alpha_{k}\delta(x-x_{k})\right)y=\lambda
y,\quad 0<x<b,\;\lambda\in\mathbb{C},$ (1)

where $q\in L_{2}(0,b)$ is a complex valued function, $\delta(x)$ is the Dirac
delta distribution, $0<x_{1}<x_{2}<\dots<x_{N}<b$ and
$\alpha_{1},\dots,\alpha_{N}\in\mathbb{C}\setminus\\{0\\}$. Schrödinger
equations with distributional coefficients supported on a set of measure zero
naturally appear in various problems of mathematical physics [3, 4, 5, 6, 16,
44] and have been studied in a considerable number of publications and from
different perspectives. In general terms, Eq. (1) can be interpreted as a
regular equation, i.e., with the regular potential $q\in L_{2}(0,b)$, whose
solutions are continuous and such that their first derivatives satisfy the
jump condition $y^{\prime}(x_{k}+)-y^{\prime}(x_{k}-)=\alpha_{k}y(x_{k})$ at
special points [25, 26]. Another approach consists in considering the interval
$[0,b]$ as a quantum graph whose edges are the segments $[x_{k},x_{k+1}]$,
$k=0,\dots,N$, (setting $x_{0}=0$, $x_{N+1}=b$), and the Schrödinger operator
with the regular potential $q$ as an unbounded operator on the direct sum
$\bigoplus_{k=0}^{N}H^{2}(x_{k},x_{k+1})$, with the domain given by the
families $(y_{k})_{k=0}^{N}$ that satisfy the condition of continuity
$y_{k}(x_{k}-)=y_{k+1}(x_{k}+)$ and the jump condition for the derivative
$y_{k+1}^{\prime}(x_{k}+)-y_{k}^{\prime}(x_{k}-)=\alpha_{k}y_{k}(x_{k})$ for
$k=1,\dots N$ (see, e.g., [18, 34, 35]). This condition for the derivative is
known in the bibliography of quantum graphs as the $\delta$-type condition
[9]. Yet another approach implies a regularization of the Schrodinger operator
with point interactions, that is, finding a subdomain of the Hilbert space
$L_{2}(0,b)$, where the operator defines a function in $L_{2}(0,b)$. For this,
note that the potential $q(x)+\sum_{k=1}^{N}\alpha_{k}\delta(x-x_{k})$ defines
a functional that belongs to the Sobolev space $H^{-1}(0,b)$. In [11, 20, 23,
42] these forms of regularization have been studied, rewriting the operator by
means of a factorization that involves a primitive $\sigma$ of the potential.

Theory of transmutation operators, also called transformation operators, is a
widely used tool in studying differential equations and spectral problems
(see, e.g., [8, 29, 36, 39, 43]), and it is especially well developed for
Schrödinger equations with regular potentials. It is known that under certain
general conditions on the potential $q$ the transmutation operator transmuting
the second derivative into the Schrödinger operator can be realized in the
form of a Volterra integral operator of the second kind, whose kernel can be
obtained by solving a Goursat problem for the Klein-Gordon equation with a
variable coefficient [14, 36, 39]. Furthermore, functional series
representations of the transmutation kernel have been constructed and used for
solving direct and inverse Sturm-Liouville problems [29, 30]. For Schrödinger
equations with $\delta$-point interactions, there exist results about
equations with a single point interaction and discontinuous conditions
$y(x_{1}+)=ay(x_{1}-)$,
$y^{\prime}(x_{1}+)=\frac{1}{a}y^{\prime}(x_{1}-)+dy(x_{1}-)$, $a,b>0$ (see
[22, 46]), and for equations in which the spectral parameter is also present
in the jump condition (see [1, 37, 38]). Transmutation operators have also
been studied for equations with distributional coefficients belonging to the
$H^{-1}$-Sobolev space in [11, 23, 42]. In [14], the possibility of extending
the action of the transmutation operator for an $L_{1}$-potential to the space
of generalized functions $\mathscr{D}^{\prime}$, was studied.

The aim of this work is to present a construction of a transmutation operator
for the Schrödinger equation with a finite number of point interactions. The
transmutation operator appears in the form of a Volterra integral operator,
and with its aid we derive analytical series representations for solutions of
(1). For this purpose, we obtain a closed form of the general solution of (1).
From it, the construction of the transmutation operator is deduced, where the
transmutation kernel is ensembled from the convolutions of the kernels of
certain solutions of the regular equation (with the potential $q$), in a
finite number of steps. Next, the spectral parameter power series (SPPS)
method is developed for Eq. (1). The SPPS method was developed for continuous
([27, 31]) and $L_{1}$-potentials ([10]), and it has been used in a piecewise
manner for solving spectral problems for equations with a finite number of
point interactions in [6, 7, 41]. Following [15], we use the SPPS method to
obtain an explicit construction of the image of the transmutation operator
acting on polynomials. Similarly to the case of a regular potential [30], we
obtain a representation of the transmutation kernel as a Fourier series in
terms of Legendre polynomials and as a corollary, a representation for the
solutions of equation (1) in terms of a Neumann series of Bessel functions.
Similar representations are obtained for the derivatives of the solutions. It
is worth mentioning that the methods based on Fourier-Legendre representations
and Neumann series of Bessel functions have shown to be an effective tool in
solving direct and inverse spectral problems for equations with regular
potentials, see, e.g., [29, 30, 33].

In Section 2, basic properties of the solutions of (1) are compiled, studying
the equation as a distributionional sense in $\mathscr{D}^{\prime}(0,b)$ and
deducing properties of its regular solutions. Section 3 presents the
construction of the closed form solution of (1). In Section 4, the
construction of the transmutation operator and the main properties of the
transmutation kernel are developed. In Section 5, the SPPS method is
presented, with the mapping and transmutation properties of the transmutation
operator. Section 6 presents the Fourier-Legendre series representations for
the transmutation kernels and the Neumann series of Bessel functions
representations for solutions of (1), and a recursive integral relation for
the Fourier-Legendre coefficients is obtained. Finally, in Section 7, integral
and Neumann series of Bessel functions representations for the derivatives of
the solutions are presented.

##  2 Problem setting and properties of the solutions

We use the standard notation $W^{k,p}(0,b)$ ($b>0$) for the Sobolev space of
functions in $L_{p}(0,b)$ that have their first $k$ weak derivatives in
$L_{p}(0,b)$, $1\leqslant p\leqslant\infty$ and $k\in\mathbb{N}$. When $p=2$,
we denote $W^{k,2}(0,b)=H^{k}(0,b)$. We have that $W^{1,1}(0,b)=AC[0,b]$, and
$W^{1,\infty}(0,b)$ is precisely the class of Lipschitz continuous functions
in $[0,b]$ (see [12, Ch. 8]). The class of smooth functions with compact
support in $(0,b)$ is denoted by $C_{0}^{\infty}(0,b)$, then we define
$W_{0}^{1,p}(0,b)=\overline{C_{0}^{\infty}(0,b)}^{W^{1,p}}$ and
$H_{0}^{1}(0,b)=W_{0}^{1,2}(0,b)$. Denote the dual space of $H^{1}_{0}(0,b)$
by $H^{-1}(0,b)$. By $L_{2,loc}(0,b)$ we denote the class of measurable
functions $f:(0,b)\rightarrow\mathbb{C}$ such that
$\int_{\alpha}^{\beta}|f(x)|^{2}dx<\infty$ for all subintervals
$[\alpha,\beta]\subset(0,b)$.

The characteristic function of an interval $[A,B]\subset\mathbb{R}$ is denoted
by $\chi_{[A,B]}(t)$. In order to simplify the notation, for the case of a
symmetric interval $[-A,A]$, we simply write $\chi_{A}$. The Heaviside
function is given by $H(t)=\chi_{(0,\infty)}(t)$. The lateral limits of the
function $f$ at the point $\xi$ are denoted by
$f(\xi\pm)=\lim_{x\rightarrow\xi{\pm}}f(x)$. We use the notation
$\mathbb{N}_{0}=\mathbb{N}\cup\\{0\\}$. The space of distributions
(generalized functions) over $C_{0}^{\infty}(0,b)$ is denoted by
$\mathscr{D}^{\prime}(0,b)$, and the value of a distribution
$f\in\mathscr{D}^{\prime}(0,b)$ at $\phi\in C_{0}^{\infty}(0,b)$ is denoted by
$(f,\phi)_{C_{0}^{\infty}(0,b)}$.

Let $N\in\mathbb{N}$ and consider a partition $0<x_{1}<\dots<x_{N}<b$ and the
numbers $\alpha_{1},\dots,\alpha_{N}\in\mathbb{C}\setminus\\{0\\}$. The set
$\mathfrak{I}_{N}=\\{(x_{j},\alpha_{j})\\}_{j=1}^{N}$ contains the information
about the point interactions of Eq. (1). Denote

$q_{\delta,\mathfrak{I}_{N}}(x):=\sum_{k=1}^{N}\alpha_{k}\delta(x-x_{k}),\quad\mathbf{L}_{q}:=-\frac{d^{2}}{dx^{2}}+q(x),\quad\mathbf{L}_{q,\mathfrak{I}_{N}}:=\mathbf{L}_{q}+q_{\delta,\mathfrak{I}_{N}}(x).$

For $u\in L_{2,loc}(0,b)$, $\mathbf{L}_{q,\mathfrak{I}_{N}}u$ defines a
distribution in $\mathscr{D}^{\prime}(0,b)$ as follows

$(\mathbf{L}_{q,\mathfrak{I}_{N}}u,\phi)_{C_{0}^{\infty}(0,b)}:=\int_{0}^{b}u(x)\mathbf{L}_{q}\phi(x)dx+\sum_{k=1}^{N}\alpha_{k}u(x_{k})\phi(x_{k})\quad\mbox{for
}\,\phi\in C_{0}^{\infty}(0,b).$

Note that the function $u$ must be well defined at the points $x_{k}$,
$k=1,\dots,N$. Actually, for a function $u\in H^{1}(0,b)$, the distribution
$\mathbf{L}_{q,\mathfrak{I}_{N}}u$ can be extended to a functional in
$H^{-1}(0,b)$ as follows

$(\mathbf{L}_{q,\mathfrak{I}_{N}}u,v)_{H_{0}^{1}(0,b)}:=\int_{0}^{b}\\{u^{\prime}(x)v^{\prime}(x)+q(x)u(x)v(x)\\}dx+\sum_{k=1}^{N}\alpha_{k}u(x_{k})v(x_{k})\quad\mbox{for
}\,v\in H_{0}^{1}(0,b).$

We say that a distribution $F\in\mathscr{D}^{\prime}(0,b)$ is $L_{2}$-regular,
if there exists a function $g\in L_{2}(0,b)$ such that
$(F,\phi)_{C_{0}^{\infty}(0,b)}=(g,\phi)_{C_{0}^{\infty}(0,b)}:=\int_{0}^{b}g(x)\phi(x)dx$
for all $\phi\in{C_{0}^{\infty}(0,b)}$.

Denote $x_{0}=0$, $x_{N+1}=b$. We recall the following characterization of
functions $u\in L_{2,loc}(0,b)$ for which $\mathbf{L}_{q,\mathfrak{I}_{N}}u$
is $L_{2}$-regular.

###### Proposition 1

If $u\in L_{2,loc}(0,b)$, then the distribution
$\mathbf{L}_{q,\mathfrak{I}_{N}}u$ is $L_{2}$-regular iff the following
conditions hold.

  1. 1.

For each $k=0,\dots,N$, $u|_{(x_{k},x_{k+1})}\in H^{2}(x_{k},x_{k+1})$.

  2. 2.

$u\in AC[0,b]$.

  3. 3.

The discontinuities of the derivative $u^{\prime}$ are located at the points
$x_{k}$, $k=1,\dots,N$, and the jumps are given by

$u^{\prime}(x_{k}+)-u^{\prime}(x_{k}-)=\alpha_{j}u(x_{k})\quad\mbox{for
}k=1,\cdots,N.$ (2)

In such case,

$(\mathbf{L}_{q,\mathfrak{I}_{N}}u,\phi)_{C_{0}^{\infty}(0,b)}=(\mathbf{L}_{q}u,\phi)_{C_{0}^{\infty}(0,b)}\quad\mbox{for
all }\phi\in{C_{0}^{\infty}(0,b)}.$ (3)

Proof. Suppose that $\mathbf{L}_{q,\mathfrak{I}_{N}}u$ is $L_{2}$-regular.
Then there exists $g\in L_{2}(0,b)$ such that

$(\mathbf{L}_{q,\mathfrak{I}_{N}}u,\phi)_{C_{0}^{\infty}(0,b)}=(g,\phi)_{C_{0}^{\infty}(0,b)}\quad\mbox{for
all }\phi\in{C_{0}^{\infty}(0,b)}.$ (4)

  1. 1.

Fix $k\in\\{1,\dots,N-1\\}$. Take a test function $\phi\in
C_{0}^{\infty}(0,b)$ with $\operatorname{Supp}(\phi)\subset(x_{k},x_{k+1})$.
Hence

$\int_{x_{k}}^{x_{k+1}}g(x)\phi(x)dx=(\mathbf{L}_{q,\mathfrak{I}_{N}}u,\phi)_{C_{0}^{\infty}(0,b)}=\int_{x_{k}}^{x_{k+1}}u(x)\mathbf{L}_{q}\phi(x)dx,$
(5)

because $\phi(x_{j})=0$ for $j=1,\dots,N$. From (5) we obtain

$\int_{x_{k}}^{x_{k+1}}u(x)\phi^{\prime\prime}(x)dx=\int_{x_{k}}^{x_{k+1}}\\{q(x)u(x)-g(x)\\}\phi(x)dx.$

Set $v(x)=\int_{0}^{x}\int_{0}^{t}\\{q(s)u(s)-g(s)\\}dsdt$. Hence $v\in
W^{2,1}(x_{j},x_{j+1})$, $v^{\prime\prime}(x)=q(x)u(x)-g(x)$ a.e.
$x\in(x_{j},x_{j+1})$, and we get the equality

$\int_{x_{k}}^{x_{k+1}}(u(x)-v(x))\phi^{\prime\prime}(x)dx=0\quad\forall\phi\in
C_{0}^{\infty}(x_{k},x_{k+1}).$ (6)

Equality (6) implies that $u(x)=v(x)+Ax+B$ a.e. $x\in(x_{k},x_{k+1})$ for some
constants $A$ and $B$ ([45, pp. 85]). In consequence $u\in
W^{2,1}(x_{k},x_{k+1})$ and

$-u^{\prime\prime}(x)+q(x)u(x)=g(x)\quad\mbox{a.e. }x\in(x_{k},x_{k+1}).$ (7)

Furthermore, $u\in C[x_{k},x_{k+1}]$, hence $qu\in L_{2}(x_{k},x_{k+1})$ and
then $u^{\prime\prime}=qu-g\in L_{2}(x_{k},x_{k+1})$. In this way
$u|_{(x_{k},x_{k+1})}\in H^{2}(x_{k},x_{k+1})$.

Now take $\varepsilon>0$ and an arbitrary $\phi\in
C_{0}^{\infty}(\varepsilon,x_{1})$. We have that

$(\mathbf{L}_{q,\mathfrak{I}_{N}}u,\phi)_{C_{0}^{\infty}(0,b)}=\int_{\varepsilon}^{x_{1}}\\{-u(x)\phi^{\prime\prime}(x)+q(x)u(x)\phi(x)\\}dx=\int_{\varepsilon}^{x_{1}}g(x)\phi(x)dx.$

Applying the same procedure as in the previous case we obtain that $u\in
H^{2}(\varepsilon,x_{1})$ and satisfies Eq. (7) in the interval
$(\varepsilon,x_{1})$. Since $\varepsilon$ is arbitrary, we conclude that $u$
satisfies (7) for a.e. $x\in(0,x_{1})$. Since $q,g\in L_{2}(0,x_{1})$, then
$u|_{(0,x_{1})}\in H^{2}(0,x_{1})$ (see [47, Th. 3.4]). The proof for the
interval $(x_{N},b)$ is analogous.

Since $u\in C^{1}[x_{k},x_{k+1}]$, $k=0,\dots,N$, the following equality is
valid (see formula (6) from [24, pp. 100])

$\displaystyle\int_{0}^{b}u(x)\phi^{\prime\prime}(x)dx$
$\displaystyle=\sum_{k=1}^{N}\left\\{u^{\prime}(x_{k}+)-u^{\prime}(x_{k}-)\right\\}\phi(x_{k})$
(8)
$\displaystyle\quad-\sum_{k=1}^{N}\left\\{u(x_{k}+)-u(x_{k}-)\right\\}\phi^{\prime}(x_{k})+\int_{0}^{b}u^{\prime\prime}(x)\phi(x)dx,\qquad\forall\phi\in
C_{0}^{\infty}(0,b).$

Fix $k\in\\{1,\cdots,N\\}$ arbitrary and take $\varepsilon>0$ small enough
such that $(x_{k}-\varepsilon,x_{k}+\varepsilon)\subset(x_{k-1},x_{k+1})$.
Choose a cut-off function $\psi\in
C_{0}^{\infty}(x_{k}-\varepsilon,x_{k}+\varepsilon)$ satisfying
$0\leqslant\psi\leqslant 1$ on $(x_{k}-\varepsilon,x_{k}+\varepsilon)$ and
$\psi(x)=1$ for
$x\in(x_{k}-\frac{\varepsilon}{3},x_{k}+\frac{\varepsilon}{3})$.

  2. 2.

By statement 1, it is enough to show that $u(x_{k}+)=u(x_{k}-)$. Set
$\phi(x)=(x-x_{k})\psi(x)$, in such a way that $\phi(x_{k})=0$ and
$\phi^{\prime}(x_{k})=1$. Hence

$(\mathbf{L}_{q,\mathfrak{I}_{N}}u,\phi)_{C_{0}^{\infty}(0,b)}=\int_{x_{k}-\varepsilon}^{x_{k}+\varepsilon}u(x)\mathbf{L}_{q}\phi(x)dx.$

By (8) we have

$\int_{x_{k}-\varepsilon}^{x_{k}+\varepsilon}u(x)\phi^{\prime\prime}(x)dx=u(x_{k}-)-u(x_{k}+)+\int_{x_{k}-\varepsilon}^{x_{k}+\varepsilon}u^{\prime\prime}(x)\phi(x)dx,$

because $\phi(x_{k})=0$ and $\phi^{\prime}(x_{k})=1$. Since $u$ satisfies (4),
we have

$\int_{x_{k}-\varepsilon}^{x_{k}+\varepsilon}(\mathbf{L}_{q}u(x)-g(x))\phi(x)dx+u(x_{k}+)-u(x_{k}-)=0.$

By statement 1, $\mathbf{L}_{q}u=g$ on both intervals $(x_{k-1},x_{k})$,
$(x_{k},x_{k+1})$. Then we obtain that $u(x_{k}+)-u(x_{k}-)$=0.

  3. 3.

Now take $\psi$ as the test function. Hence

$(\mathbf{L}_{q,\mathfrak{I}_{N}}u,\psi)_{C_{0}^{\infty}(0,b)}=\int_{x_{k}-\varepsilon}^{x_{k}+\varepsilon}u(x)\mathbf{L}_{q}\psi(x)dx+\alpha_{k}u(x_{k}),$

because
$\operatorname{Supp}(\psi)\subset(x_{k}-\varepsilon,x_{k}+\varepsilon)$ and
$\psi\equiv 1$ on $(x_{k}-\frac{\varepsilon}{3},x_{k}+\frac{\varepsilon}{3})$.
On the other hand, by (8) we obtain

$\int_{x_{k}-\varepsilon}^{x_{k}+\varepsilon}u(x)\psi^{\prime\prime}(x)dx=u^{\prime}(x_{k}+)-u^{\prime}(x_{k}-)+\int_{x_{k}-\varepsilon}^{x_{k}+\varepsilon}u^{\prime\prime}(x)\psi(x)dx,$

because $\psi^{\prime}(x_{k})=0$. Thus, by (4) we have

$\int_{x_{k}-\varepsilon}^{x_{k}+\varepsilon}(\mathbf{L}_{q}u(x)-g(x))\psi(x)dx+u^{\prime}(x_{k}-)-u^{\prime}(x_{k}+)+\alpha_{k}u(x_{k})=0.$

Again, by statement 1, we obtain (2).

Reciprocally, if $u$ satisfies conditions 1,2 and 3, equality (8) implies (3).
By condition 1, $\mathbf{L}_{q,\mathfrak{I}_{N}}u$ is $L_{2}$-regular.

###### Definition 2

The $L_{2}$-regularization domain of $\mathbf{L}_{q,\mathfrak{I}_{N}}$,
denoted by $\mathcal{D}_{2}\left(\mathbf{L}_{q,\mathfrak{I}_{N}}\right)$, is
the set of all functions $u\in L_{2,loc}(0,b)$ satisfying conditions 1,2 and 3
of Proposition 1.

If $u\in L_{2,loc}(0,b)$ is a solution of (1), then
$\mathbf{L}_{q-\lambda,\mathfrak{I}_{N}}u$ equals the regular distribution
zero. Then we have the next characterization.

###### Corollary 3

A function $u\in L_{2,loc}(0,b)$ is a solution of Eq. (1) iff
$u\in\mathcal{D}_{2}\left(\mathbf{L}_{q,\mathfrak{I}_{N}}\right)$ and for each
$k=0,\dots,N$, the restriction $u|_{(x_{k},x_{k+1})}$ is a solution of the
regular Schrödinger equation

$-y^{\prime\prime}(x)+q(x)y(x)=\lambda y(x)\quad\mbox{for }x_{k}<x<x_{k+1}.$
(9)

###### Remark 4

Let $f\in\mathcal{D}_{2}\left(\mathbf{L}_{q,\mathfrak{I}_{N}}\right)$. Given
$g\in C^{1}[0,b]$, we have

$\displaystyle(fg)^{\prime}(x_{k}+)-(fg)^{\prime}(x_{k}-)$
$\displaystyle=f^{\prime}(x_{k}+)g(x_{k})+f(x_{k})g^{\prime}(x_{k}+)-f^{\prime}(x_{k}-)g(x_{k})-f(x_{k})g^{\prime}(x_{k}-)$
$\displaystyle=\left[f^{\prime}(x_{k}+)-f^{\prime}(x_{k}-)\right]g(x_{k})=\alpha_{k}f(x_{k})g(x_{k})$

for $k=1,\dots,N$. In particular,
$fg\in\mathcal{D}_{2}\left(\mathbf{L}_{q,\mathfrak{I}_{N}}\right)$ for $g\in
H^{2}(0,b)$.

###### Remark 5

Let $u_{0},u_{1}\in\mathbb{C}$. Consider the Cauchy problem

$\begin{cases}\mathbf{L}_{q,\mathfrak{I}_{N}}u(x)=\lambda u(x),\quad 0<x<b,\\\
u(0)=u_{0},\;u^{\prime}(0)=u_{1}.\end{cases}$ (10)

If the solution of the problem exists, it must be unique. It is enough to show
the assertion for $u_{0}=u_{1}=0$. Indeed, if $w$ is a solution of such
problem, by Corollary 3, $w$ is a solution of (9) on $(0,x_{1})$ satisfying
$w(0)=w^{\prime}(0)=0$. Hence $w\equiv 0$ on $[0,x_{1}]$. By the continuity of
$w$ and condition (2), we have $w(x_{1})=w^{\prime}(x_{1}-)=0$. Hence $w$ is a
solution of (9) satisfying these homogeneous conditions. Thus, $w\equiv 0$ on
$[x_{1},x_{2}]$. By continuing the process until the points $x_{k}$ are
exhausted, we arrive at the solution $w\equiv 0$ on the whole segment $[0,b]$.

The uniqueness of the Cauchy problem with conditions $u(b)=u_{0}$,
$u^{\prime}(b)=u_{1}$ is proved in a similar way.

###### Remark 6

Suppose that $u_{0}=u_{0}(\lambda)$ and $u_{1}=u_{1}(\lambda)$ are entire
functions of $\lambda$ and denote by $u(\lambda,x)$ the corresponding unique
solution of (10). Since $u$ is the solution of the Cauchy problem
$\mathbf{L}_{q}u=\lambda u$ on $(0,x_{1})$ with the initial conditions
$u(\lambda,0)=u_{1}(\lambda)$, $u^{\prime}(\lambda,0)=u_{1}(\lambda)$, both
$u(\lambda,x)$ and $u^{\prime}(\lambda,x+)$ are entire functions for any
$x\in[0,x_{1}]$ (this is a consequence of [47, Th. 3.9] and [10, Th. 7]).
Hence
$u^{\prime}(\lambda,x_{1}-)=u^{\prime}(\lambda,x_{1}+)-\alpha_{1}u(\lambda,x_{1})$
is entire in $\lambda$. Since $u$ is the solution of the Cauchy problem
$\mathbf{L}_{q}u=\lambda u$ on $(x_{1},x_{2})$ with initial conditions
$u(\lambda,x_{1})$ and $u^{\prime}(\lambda,x_{1}+)$, we have that
$u(\lambda,x)$ and $u^{\prime}(\lambda,x+)$ are entire functions for
$x\in[x_{1},x_{2}]$. By continuing the process we prove this assertion for all
$x\in[0,b]$.

##  3 Closed form solution

In what follows, denote the square root of $\lambda$ by $\rho$, so
$\lambda=\rho^{2}$, $\rho\in\mathbb{C}$. For each $k\in\\{1,\cdots,N\\}$ let
$\widehat{s}_{k}(\rho,x)$ be the unique solution of the Cauchy problem

$\begin{cases}-\widehat{s}_{k}^{\prime\prime}(\rho,x)+q(x+x_{k})\widehat{s}_{k}(\rho,x)=\rho^{2}\widehat{s}_{k}(\rho,x)\quad\mbox{
for }0<x<b-x_{k},\\\
\widehat{s}_{k}(\rho,0)=0,\;\widehat{s}_{k}^{\prime}(\rho,0)=1.\end{cases}$
(11)

In this way, $\widehat{s}_{k}(\rho,x-x_{k})$ is a solution of
$\mathbf{L}_{q}u=\rho^{2}u$ on $(x_{k},b)$ with initial conditions
$u(x_{k})=0$, $u^{\prime}(x_{k})=1$. According to [45, Ch. 3, Sec. 6.3],
$(\mathbf{L}_{q}-\rho^{2})\left(H(x-x_{k})\widehat{s}_{k}(\rho,x-x_{k})\right)=-\delta(x-x_{k})$
for $x_{k}<x<b$.

We denote by $\mathcal{J}_{N}$ the set of finite sequences
$J=(j_{1},\dots,j_{l})$ with $1<l\leqslant N$,
$\\{j_{1},\dots,j_{l}\\}\subset\\{1,\dots,N\\}$ and $j_{1}<\cdots<j_{l}$.
Given $J\in\mathcal{J}_{N}$, the length of $J$ is denoted by $|J|$ and we
define $\alpha_{J}:=\alpha_{j_{1}}\cdots\alpha_{j_{|J|}}$.

###### Theorem 7

Given $u_{0},u_{1}\in\mathbb{C}$, the unique solution $\displaystyle
u_{\mathfrak{I}_{N}}\in\mathcal{D}_{2}\left(\mathbf{L}_{q,\mathfrak{I}_{N}}\right)$
of the Cauchy problem (10) has the form

$\displaystyle u_{\mathfrak{I}_{N}}(\rho,x)$
$\displaystyle=\widetilde{u}(\rho,x)+\sum_{k=1}^{N}\alpha_{k}\widetilde{u}(\rho,x_{k})H(x-x_{k})\widehat{s}_{k}(\rho,x-x_{k})$
$\displaystyle\;\;\;\;+\sum_{J\in\mathcal{J}_{N}}\alpha_{J}H(x-x_{j_{|J|}})\widetilde{u}(\rho,x_{j_{1}})\left(\prod_{l=1}^{|J|-1}\widehat{s}_{j_{l}}(\rho,x_{j_{l+1}}-x_{j_{l}})\right)\widehat{s}_{j_{|J|}}(\rho,x-x_{j_{|J|}}),$
(12)

where $\widetilde{u}(\rho,x)$ is the unique solution of the regular
Schrödinger equation

$\mathbf{L}_{q}\widetilde{u}(\rho,x)=\rho^{2}\widetilde{u}(\rho,x),\quad
0<x<b,$ (13)

satisfying the initial conditions
$\widetilde{u}(\rho,0)=u_{1},\;\widetilde{u}^{\prime}(\rho,0)=u_{1}$.

Proof. The proof is by induction on $N$. For $N=1$, the proposed solution has
the form

$u_{\mathfrak{I}_{1}}(\rho,x)=\widetilde{u}(\rho,x)+\alpha_{1}H(x-x_{1})\widetilde{u}(\rho,x_{1})\widehat{s}_{1}(\rho,x-x_{1}).$

Note that $u_{\mathfrak{I}_{1}}(\rho,x)$ is continuous, and
$u_{\mathfrak{I}_{1}}(\rho,x_{1})=\widetilde{u}(\rho,x_{1})$. Hence

$(\mathbf{L}_{q}-\rho^{2})u_{\mathfrak{I}_{1}}(\rho,x)=\alpha_{1}\widetilde{u}(\rho,x_{1})(\mathbf{L}_{q}-\rho^{2})\left(H(x-x_{1})\widehat{s}_{1}(\rho,x-x_{1})\right)=-\alpha_{1}\widetilde{u}(\rho,x_{1})\delta(x-x_{1}),$

that is, $u_{\mathfrak{I}_{1}}(\rho,x)$ is a solution of (1) with $N=1$.
Suppose the result is valid for $N$. Let $u_{\mathfrak{I}_{N+1}}(\rho,x)$ be
the proposed solution given by formula (12). It is clear that
$u_{\mathfrak{I}_{N+1}}(\rho,\cdot)|_{(x_{k},x_{k+1})}\in
H^{2}(x_{k},x_{k+1})$, $k=0,\cdots,N$, $u_{\mathfrak{I}_{N+1}}(\rho,x)$ is a
solution of (9) on each interval $(x_{k},x_{k+1})$, $k=0,\dots,N+1$, and
$u_{\mathfrak{I}_{N+1}}^{(j)}(\rho,0)=\widetilde{u}^{(j)}(\rho,0)=u_{j}$,
$j=0,1$. Furthermore, we can write

$u_{\mathfrak{I}_{N+1}}(\rho,x)=u_{\mathfrak{I}_{N}}(\rho,x)+H(x-x_{N+1})f_{N}(\rho,x),$

where
$\mathfrak{I}_{N}=\mathfrak{I}_{N+1}\setminus\\{(x_{N+1},\alpha_{N+1})\\}$,
$u_{\mathfrak{J}_{N}}(\rho,x)$ is the proposed solution for the interactions
$\mathfrak{I}_{N}$, and the function $f_{N}(\rho,x)$ is given by

$\displaystyle f_{N}(\rho,x)$
$\displaystyle=\alpha_{N+1}\widetilde{u}(\rho,x_{N+1})\widehat{s}_{N+1}(x-x_{N+1})$
$\displaystyle\quad+\sum_{\overset{J\in\mathcal{J}_{N+1}}{j_{|J|}=N+1}}\alpha_{J}\widetilde{u}(\rho,x_{j_{1}})\left(\prod_{l=1}^{|J|-1}\widehat{s}_{j_{l}}(\rho,x_{j_{l+1}}-x_{j_{l}})\right)\widehat{s}_{N+1}(\rho,x-x_{N+1}),$

where the sum is taken over all the sequences
$J=(j_{1},\dots,j_{|J|})\in\mathcal{J}_{N}$ with $j_{|J|}=N+1$. From this
representation we obtain
$u_{\mathfrak{I}_{N+1}}(\rho,x_{N+1}\pm)=u_{\mathfrak{I}_{N}}(\rho,x_{N+1})$
and hence $u_{\mathfrak{I}_{N+1}}\in AC[0,b]$. By the induction hypothesis,
$u_{\mathfrak{I}_{N}}(\rho,x)$ is the solution of (1) for $N$, then in order
to show that $u_{\mathfrak{I}_{N+1}}(\rho,x)$ is the solution for $N+1$ it is
enough to show that
$(\mathbf{L}_{q}-\rho^{2})\hat{f}_{N}(\rho,x)=-\alpha_{N}u_{N}(x_{N+1})\delta(x-x_{N+1})$,
where $\hat{f}_{N}(\rho,x)=H(x-x_{N+1})f_{N}(\rho,x)$. Indeed, we have

$\displaystyle(\rho^{2}-\mathbf{L}_{q})\hat{f}_{N}(\rho,x)$
$\displaystyle=\alpha_{N+1}\widetilde{u}(\rho,x_{N+1})\delta(x-x_{N+1})+$
$\displaystyle\quad\;+\sum_{\overset{J\in\mathcal{J}_{N+1}}{j_{|J|}=N+1}}\alpha_{J}\widetilde{u}(\rho,x_{j_{1}})\left(\prod_{l=1}^{|J|-1}\widehat{s}_{j_{l}}(\rho,x_{j_{l+1}}-x_{j_{l}})\right)\delta(x-x_{N+1})$
$\displaystyle=\alpha_{N+1}\delta(x-x_{N+1})\Bigg{[}\widetilde{u}(\rho,x_{N+1})+\sum_{k=1}^{N}\alpha_{k}\widetilde{u}(\rho,x_{N+1})\widehat{s}_{k}(\rho,x_{N+1}-x_{k})$
$\displaystyle\quad\;\;\,+\sum_{J\in\mathcal{J}_{N}}\alpha_{J}\widetilde{u}(\rho,x_{j_{1}})\left(\prod_{l=1}^{|J|-1}\widehat{s}_{j_{l}}(\rho,x_{j_{l+1}}-x_{j_{l}})\right)\widehat{s}_{j_{|J|}}(\rho,x_{N+1}-x_{j_{|J|}})\Bigg{]}$
$\displaystyle=\alpha_{N+1}u_{\mathfrak{I}_{N}}(\rho,x_{N+1})\delta(x-x_{N+1})=\alpha_{N+1}u_{\mathfrak{I}_{N+1}}(\rho,x_{N+1})\delta(x-x_{N+1}),$

where the second equality is due to the fact that

$\\{J\in\mathcal{J}_{N+1}\,|\,j_{|J|}=N+1\\}=\\{(J^{\prime},N+1)\,|\,J^{\prime}\in\mathcal{J}_{N}\\}\cup\\{(j,N+1)\\}_{j=1}^{N}.$

Hence $u_{\mathfrak{I}_{N+1}}(\rho,x)$ is the solution of the Cauchy problem.

###### Example 8

Consider the case $q\equiv 0$. Denote by $e_{\mathfrak{I}_{N}}^{0}(\rho,x)$
the unique solution of

$-y^{\prime\prime}+\left(\sum_{k=1}^{N}\alpha_{k}\delta(x-x_{k})\right)y=\rho^{2}y,\quad
0<x<b,$ (14)

satisfying $e_{\mathfrak{I}_{N}}^{0}(\rho,0)=1$,
$e_{\mathfrak{I}_{N}}^{0}(\rho,0)=i\rho$. In this case we have
$\widehat{s}_{k}(\rho,x)=\frac{\sin(\rho x)}{\rho}$ for $k=1,\dots,N$. Hence,
according to Theorem 7, the solution $e_{\mathfrak{I}_{N}}^{0}(\rho,x)$ has
the form

$\displaystyle e_{\mathfrak{I}_{N}}^{0}(\rho,x)$ $\displaystyle=e^{i\rho
x}+\sum_{k=1}^{N}\alpha_{k}e^{i\rho
x_{k}}H(x-x_{k})\frac{\sin(\rho(x-x_{k}))}{\rho}$
$\displaystyle\quad+\sum_{J\in\mathcal{J}_{N}}\alpha_{J}H(x-x_{j_{|J|}})e^{i\rho
x_{j_{1}}}\left(\prod_{l=1}^{|J|-1}\frac{\sin(\rho(x_{j_{l+1}}-x_{j_{l}}))}{\rho}\right)\frac{\sin(\rho(x-x_{j_{|J|}}))}{\rho}.$
(15)

##  4 Transmutation operators

###  4.1 Construction of the integral transmutation kernel

Let $h\in\mathbb{C}$. Denote by $\widetilde{e}_{h}(\rho,x)$ the unique
solution of Eq. (13) satisfying $\widetilde{e}_{h}(\rho,0)=1$,
$\widetilde{e}^{\prime}_{h}(\rho,0)=i\rho+h$. Hence the unique solution
$e_{\mathfrak{I}_{N}}^{h}(\rho,x)$ of Eq. (1) satisfying
$e_{\mathfrak{I}_{N}}^{h}(\rho,0)=1$,
$(e_{\mathfrak{I}_{N}}^{h})^{\prime}(\rho,0)=i\rho+h$ is given by

$\displaystyle e_{\mathfrak{I}_{N}}^{h}(\rho,x)$
$\displaystyle=\widetilde{e}_{h}(\rho,x)+\sum_{k=1}^{N}\alpha_{k}\widetilde{e}_{h}(\rho,x_{k})H(x-x_{k})\widehat{s}_{k}(\rho,x-x_{k})$
(16)
$\displaystyle\quad+\sum_{J\in\mathcal{J}_{N}}\alpha_{J}H(x-x_{j_{|J|}})\widetilde{e}_{h}(\rho,x_{j_{1}})\left(\prod_{l=1}^{|J|-1}\widehat{s}_{j_{l}}(\rho,x_{j_{l+1}}-x_{j_{l}})\right)\widehat{s}_{j_{|J|}}(\rho,x-x_{j_{|J|}}).$

It is known that there exists a kernel $\widetilde{K}^{h}\in
C(\overline{\Omega})\cap H^{1}(\Omega)$, where
$\Omega=\\{(x,t)\in\mathbb{R}^{2}\,|\,0<x<b,|t|<x\\}$, such that
$\widetilde{K}^{h}(x,x)=\frac{h}{2}+\frac{1}{2}\int_{0}^{x}q(s)ds$,
$\widetilde{K}^{h}(x,-x)=\frac{h}{2}$ and

$\widetilde{e}_{h}(\rho,x)=e^{i\rho
x}+\int_{-x}^{x}\widetilde{K}^{h}(x,t)e^{i\rho t}dt$ (17)

(see, e.g., [36, 39]). Actually, $\widetilde{K}^{h}(x,\cdot)\in L_{2}(-x,x)$
and it can be extended (as a function of $t$) to a function in
$L_{2}(\mathbb{R})$ with a support in $[-x,x]$. For each $k\in\\{1,\dots,N\\}$
there exists a kernel $\widehat{H}_{k}\in C(\overline{\Omega_{k}})\cap
H^{1}(\Omega_{k})$ with
$\Omega_{k}=\\{(x,t)\in\mathbb{R}^{2}\,|\,0<x<b-x_{k},\;|t|\leqslant x\\}$,
and $\widehat{H}_{k}(x,x)=\frac{1}{2}\int_{x_{k}}^{x+x_{k}}q(s)ds$,
$\widehat{H}_{k}(x,-x)=0$, such that

$\widehat{s}_{k}(\rho,x)=\frac{\sin(\rho
x)}{\rho}+\int_{0}^{x}\widehat{H}_{k}(x,t)\frac{\sin(\rho t)}{\rho}dt$ (18)

(see [19, Ch. 1]). From this we obtain the representation

$\widehat{s}_{k}(\rho,x-x_{k})=\frac{\sin(\rho(x-x_{k}))}{\rho}+\int_{0}^{x-x_{k}}\widehat{H}_{k}(x-x_{k},t)\frac{\sin(\rho
t)}{\rho}dt=\int_{-(x-x_{k})}^{x-x_{k}}\widetilde{K}_{k}(x,t)e^{i\rho t}dt,$
(19)

where

$\widetilde{K}_{k}(x,t)=\frac{1}{2}\chi_{x-x_{k}}(t)+\displaystyle\frac{1}{2}\int_{|t|}^{x-x_{k}}\widehat{H}_{k}(x-x_{k},s)ds.$
(20)

We denote the Fourier transform of a function $f\in L_{1}(\mathbb{R})$ by
$\mathcal{F}f(\rho)=\int_{\mathbb{R}}f(t)e^{i\rho t}dt$ and the convolution of
$f$ with a function $g\in L_{1}(\mathbb{R})$ by $f\ast
g(t)=\int_{\mathbb{R}}f(t-s)g(s)ds$. We recall that $\mathcal{F}(f\ast
g)(\rho)=\mathcal{F}f(\rho)\cdot\mathcal{F}g(\rho)$. Given
$f_{1},\dots,f_{M}\in L_{2}(\mathbb{R})$ with compact support, we denote their
convolution product by
$\left(\prod_{l=1}^{M}\right)^{\ast}f_{l}(t):=(f_{1}\ast\cdots\ast f_{M})(t)$.
For the kernels $\widetilde{K}^{h}(x,t),\widetilde{K}_{k}(x,t)$, the
operations $\mathcal{F}$ and $\ast$ will be applied with respect to the
variable $t$.

###### Lemma 9

Let $A,B>0$. If $f\in C[-A,A]$ and $g\in C[-B,B]$, then
$(\chi_{A}f)\ast(\chi_{B}g)\in C(\mathbb{R})$ with
$\operatorname{Supp}\left((\chi_{A}f)\ast(\chi_{B}g)\right)\subset[-(A+B),A+B]$.

Proof. The assertion
$\operatorname{Supp}\left((\chi_{A}f)\ast(\chi_{B}g)\right)\subset[-(A+B),A+B]$
is due to [12, Prop. 4.18]. Since $(\chi_{A}f)\in L_{1}(\mathbb{R})$ and
$(\chi_{B}g)\in L_{\infty}(\mathbb{R})$, it follows from [17, Prop. 8.8] that
$(\chi_{A}f)\ast(\chi_{B}g)\in C(\mathbb{R})$.

###### Theorem 10

There exists a kernel $K_{\mathfrak{I}_{N}}^{h}(x,t)$ defined on $\Omega$ such
that

$e_{\mathfrak{I}_{N}}^{h}(\rho,x)=e^{i\rho
x}+\int_{-x}^{x}K_{\mathfrak{I}_{N}}^{h}(x,t)e^{i\rho t}dt.$ (21)

For any $0<x\leqslant b$, $K_{\mathfrak{J}_{N}}^{h}(x,t)$ is piecewise
absolutely continuous with respect to the variable $t\in[-x,x]$ and satisfies
$K_{\mathfrak{I}_{N}}^{h}(x,\cdot)\in L_{2}(-x,x)$. Furthermore,
$K_{\mathfrak{I}_{N}}^{h}\in L_{\infty}(\Omega)$.

Proof. Susbtitution of formulas (17) and (19) in (16) leads to the equality

$\displaystyle e_{\mathfrak{I}_{N}}^{h}(\rho,x)=e^{i\rho
x}+\int_{-x}^{x}\widetilde{K}^{h}(x,t)e^{i\rho t}dt+$
$\displaystyle+\sum_{k=1}^{N}\alpha_{k}H(x-x_{k})\left(e^{i\rho
x_{k}}+\int\limits_{-x_{k}}^{x_{k}}\widetilde{K}^{h}(x_{k},t)e^{i\rho
t}dt\right)\left(\int\limits_{-(x-x_{k})}^{x-x_{k}}\widetilde{K}_{k}(x,t)e^{i\rho
t}dt\right)$
$\displaystyle+\sum_{J\in\mathcal{J}_{N}}\alpha_{J}H(x-x_{j_{|J|}})\Bigg{[}\left(e^{i\rho
x_{j_{1}}}+\int\limits_{-x_{j_{1}}}^{x_{j_{1}}}\widetilde{K}^{h}(x_{j_{1}},t)e^{i\rho
t}dt\right)\left(\prod_{l=1}^{|J|-1}\int\limits_{-(x_{j_{l+1}}-x_{j_{l}})}^{x_{j_{l+1}}-x_{j_{l}}}\widetilde{K}_{k}(x_{j_{l+1}},t)e^{i\rho
t}dt\right)$
$\displaystyle\qquad\qquad\qquad\qquad\cdot\int\limits_{-(x-x_{j_{|J|}})}^{x-x_{j_{|J|}}}\widetilde{K}_{k}(x,t)e^{i\rho
t}dt\Bigg{]}$

Note that

$\displaystyle\prod_{l=1}^{|J|-1}\int\limits_{-(x_{j_{l+1}}-x_{j_{l}})}^{x_{j_{l+1}}-x_{j_{l}}}\widetilde{K}_{k}(x_{j_{l+1}},t)e^{i\rho
t}dt$
$\displaystyle=\mathcal{F}\left\\{\left(\prod_{l=1}^{|J|-1}\right)^{\ast}\left(\chi_{x_{j_{l+1}}-x_{j_{l}}}(t)\widetilde{K}_{k}(x_{j_{l+1}},t)\right)\right\\}.$

In a similar way, if we denote $I_{A,B}=\left(e^{i\rho
A}+\int\limits_{-A}^{A}\widetilde{K}^{h}(A,t)e^{i\rho
t}dt\right)\left(\int\limits_{-B}^{B}\widetilde{K}_{k}(B,t)e^{i\rho
t}dt\right)$ with $A,B\in(0,b)$, then

$\displaystyle I_{A,B}=$ $\displaystyle e^{i\rho
A}\int\limits_{-B}^{B}\widetilde{K}_{k}(B,t)e^{i\rho
t}dt+\mathcal{F}\left(\chi_{A}(t)\widetilde{K}^{h}(A,t)\ast\chi_{B}(t)\widetilde{K}_{k}(B,t)\right)$
$\displaystyle=$
$\displaystyle\mathcal{F}\left(\chi_{[A-B,B+A]}(t)\widetilde{K}_{k}(B,t-A)+\chi_{A}(t)\widetilde{K}^{h}(A,t)\ast\chi_{B}(t)\widetilde{K}_{k}(B,t)\right).$

Set $R_{N}(\rho,x)=e_{N}(\rho,x)-e^{i\rho x}$. Thus,

$\displaystyle R_{N}(\rho,x)=$
$\displaystyle\mathcal{F}\Bigg{[}\chi_{x}(t)\widetilde{K}^{h}(x,t)$
$\displaystyle+\sum_{k=1}^{N}\alpha_{k}H(x-x_{k})\left(\chi_{[2x_{k}-x,x]}(t)\widetilde{K}_{k}(x,t-x_{k})+\chi_{x_{k}}(t)\widetilde{K}^{h}(x_{k},t)\ast\chi_{x-x_{k}}(t)\widetilde{K}_{k}(x,t)\right)$
$\displaystyle+\sum_{J\in\mathcal{J}_{N}}\alpha_{J}H(x-x_{j_{|J|}})\left(\prod_{l=1}^{|J|-1}\right)^{\ast}\left(\chi_{x_{j_{l+1}}-x_{j_{l}}}(t)\widetilde{K}_{k}(x_{j_{l+1}},t)\right)$
$\displaystyle\ast\Big{(}\chi_{[x_{j_{|J|}}+x_{j_{1}}-x,x-(x_{j_{|J|}}-x_{j_{1}})]}(t)\widetilde{K}_{j_{|J|}}(x,t-x_{j_{1}})$
$\displaystyle\qquad\;+\chi_{x_{j_{1}}}(t)\widetilde{K}^{h}(x_{j_{1}},t)\ast\chi_{x-x_{j_{|J|}}}(t)\widetilde{K}_{j_{|J|}}(x,t)\Big{)}\Bigg{]}$

According to Lemma 9, the support of
$\left(\prod_{l=1}^{|J|-1}\right)^{\ast}\left(\chi_{x_{j_{l+1}}-x_{j_{l}}}(t)\widetilde{K}_{k}(x_{j_{l+1}},t)\right)$
lies in
$[x_{j_{1}}-x_{j_{|J|}},x_{j_{|J|}}-x_{j_{1}}]$ and
$\chi_{x-(x_{j_{|J|}}-x_{j_{1}})}(t)\widetilde{K}_{j_{|J|}}(x,t-x_{j_{1}})+\chi_{x_{j_{1}}}(t)\widetilde{K}^{h}(x_{j_{1}},t)\ast\chi_{x-x_{j_{|J|}}}(t)\widetilde{K}_{j_{|J|}}(x,t)$
has its support in $[x_{j_{|J|}}+x_{j_{1}}-x,x-(x_{j_{|J|}}-x_{j_{1}})]$.
Hence the convolution in the second sum of $R_{N}(\rho,x)$ has its support in
$[-x,x]$. On the other hand,
$\chi_{x_{k}}(t)\widetilde{K}^{h}(x_{k},t)\ast\chi_{x-x_{k}}(t)\widetilde{K}_{k}(x,t)$
has its support in $[-x,x]$, and since $[2x_{k}-x,x]\subset[-x,x]$, we
conclude that
$\operatorname{Supp}\left(\mathcal{F}^{-1}R_{N}(\rho,x)\right)\subset[-x,x]$.

Thus, we obtain (21) with

$\displaystyle K_{\mathfrak{I}_{N}}^{h}(x,t)=$
$\displaystyle\chi_{x}(t)\widetilde{K}^{h}(x,t)$
$\displaystyle+\sum_{k=1}^{n}\alpha_{k}H(x-x_{k})\left(\chi_{[2x_{k}-x,x]}(t)\widetilde{K}_{k}(x,t-x_{k})+\chi_{x_{k}}(t)\widetilde{K}^{h}(x_{k},t)\ast\chi_{x-x_{k}}(t)\widetilde{K}_{k}(x,t)\right)$
$\displaystyle+\sum_{J\in\mathcal{J}_{N}}\alpha_{J}H(x-x_{j_{|J|}})\left(\prod_{l=1}^{|J|-1}\right)^{\ast}\left(\chi_{x_{j_{l+1}}-x_{j_{l}}}(t)\widetilde{K}_{j_{l}}(x_{j_{l+1}},t)\right)$
(22)
$\displaystyle\qquad\ast\Big{(}\chi_{x-(x_{j_{|J|}}-x_{j_{1}})}(t)\widetilde{K}_{j_{|J|}}(x,t-x_{j_{1}})+\chi_{x_{j_{1}}}(t)\widetilde{K}^{h}(x_{j_{1}},t)\ast\chi_{x-x_{j_{|J|}}}(t)\widetilde{K}_{j_{|J|}}(x,t)\Big{)},$

and $K_{\mathfrak{I}_{N}}(x,\cdot)\in L_{2}(x,-x)$. By formula (22) and the
definitions of $\widehat{K}^{h}(x,t)$ and $\widetilde{K}_{k}(x,t)$,
$K_{\mathfrak{I}_{N}}(x,t)$ is piecewise absolutely continuous for
$t\in[-x,x]$. Since $\widehat{K}^{h},\widetilde{K}_{k}\in L_{\infty}(\Omega)$,
is clear that $K_{\mathfrak{I}_{N}}^{f}\in L_{\infty}(\Omega)$.

As a consequence of (21), $e_{\mathfrak{I}_{N}}^{h}(\rho,x)$ is an entire
function of exponential type $x$ on the spectral parameter $\rho$.

###### Example 11

Consider (15) with $N=1$. In this case the solution
$e_{\mathfrak{I}_{1}}^{0}(\rho,x)$ is given by

$e_{\mathfrak{I}_{1}}^{0}(\rho,x)=e^{i\rho x}+\alpha_{1}e^{i\rho
x_{1}}H(x-x_{1})\frac{\sin(\rho(x-x_{1}))}{\rho}.$

We have

$e^{i\rho
x_{1}}\frac{\sin(\rho(x-x_{1}))}{\rho}=\frac{1}{2}\int_{x_{1}-x}^{x-x_{1}}e^{i\rho(t+x_{1})}dt=\frac{1}{2}\int_{2x_{1}-x}^{x}e^{i\rho
t}dt.$

Hence

$e_{\mathfrak{I}_{1}}^{0}(\rho,x)=e^{i\rho
x}+\int_{-x}^{x}K_{\mathfrak{I}_{1}}^{0}(x,t)e^{i\rho t}dt\quad\mbox{with
}\,K_{\mathfrak{I}_{1}}^{0}(x,t)=\frac{\alpha_{1}}{2}H(x-x_{1})\chi_{[2x_{1}-x,x]}(t).$

###### Example 12

Consider again Eq. (15) but now with $N=2$. In this case the solution
$e_{\mathfrak{I}_{2}}^{0}(\rho,x)$ is given by

$\displaystyle e_{\mathfrak{I}_{2}}^{0}(\rho,x)=$ $\displaystyle e^{i\rho
x}+\alpha_{1}e^{i\rho
x_{1}}H(x-x_{1})\frac{\sin(\rho(x-x_{1}))}{\rho}+\alpha_{2}e^{i\rho
x_{2}}H(x-x_{2})\frac{\sin(\rho(x-x_{2}))}{\rho}$
$\displaystyle+\alpha_{1}\alpha_{2}e^{i\rho
x_{1}}H(x-x_{2})\frac{\sin(\rho(x_{2}-x_{1}))}{\rho}\frac{\sin(\rho(x-x_{2}))}{\rho},$

and the transmutation kernel $K_{\mathfrak{I}_{2}}^{0}(x,t)$ has the form

$\displaystyle K_{\mathfrak{I}_{2}}^{0}(x,t)$
$\displaystyle=\frac{\alpha_{1}H(x-x_{1})}{2}\chi_{[2x_{1}-x,x]}(t)+\frac{\alpha_{2}H(x-x_{2})}{2}\chi_{[2x_{1}-x,x]}(t)$
$\displaystyle\qquad+\frac{\alpha_{1}\alpha_{2}H(x-x_{2})}{4}\left(\chi_{x_{2}-x_{1}}\ast\chi_{x-x_{2}}\right)(t-x_{1}).$

Direct computation shows that

$\chi_{x_{2}-x_{1}}\ast\chi_{x-x_{2}}(t-x_{1})=\\\
\begin{cases}0,&t\not\in[2x_{1}-x,x],\\\
t+x-2x_{1},&2x_{1}-x<t<-|2x_{2}-x-x_{1}|+x_{1},\\\
x-x_{1}-|2x_{2}-x-x_{1}|,&-|2x_{2}-x-x_{1}|+x_{1}<t<|2x_{2}-x-x_{1}|+x_{1}\\\
x-t,&|2x_{2}-x-x_{1}|+x_{1}<t<x.\end{cases}$

In Figure 1, we can see some level curves of the kernel
$K_{\mathfrak{I}_{2}}^{0}(x,t)$ (as a function of $t$),
$\mathfrak{I_{2}}=\\{(0.25,1),(0.75,2)\\}$, for some values of $x$.

Figure 1: The graphs of $K_{\mathfrak{I}_{2}}^{0}(x,t)$, as a function of
$t\in[-1,1]$, for some points $x\in(0,1)$ and
$\mathfrak{I_{2}}=\\{(0.25,1),(0.75,2)\\}$.

For the general case we have the following representation for the kernel.

###### Proposition 13

The transmutation kernel $K_{\mathfrak{I}_{N}}^{0}(\rho,x)$ for the solution
$e_{\mathfrak{I}_{N}}^{0}(\rho,x)$ of (15) is given by

$\displaystyle K_{\mathfrak{I}_{N}}^{0}(x,t)$
$\displaystyle=\sum_{k=0}^{N}\frac{\alpha_{k}H(x-x_{k})}{2}\chi_{[2x_{k}-x,x]}(t)$
$\displaystyle+\sum_{J\in\mathcal{J}_{N}}\frac{\alpha_{J}H(x-x_{j_{|J|}})}{2^{|J|}}\left(\left(\prod_{l=1}^{|J|-1}\right)^{\ast}\chi_{x_{j_{l+1}}-x_{j_{l}}}(t)\right)\ast\chi_{x-x_{j_{|J|}}}(t-x_{j_{1}})$
(23)

Proof. In this case $\widetilde{e}_{0}(\rho,x)=e^{i\rho x}$,
$\widehat{s}_{k}(\rho,x-x_{k})=\frac{\sin(\rho(x-x_{k}))}{\rho}$, hence
$\widetilde{K}^{0}(x,t)\equiv 0$,
$\widetilde{K}_{k}(x,t)=\frac{1}{2}\chi_{x-x_{k}}(t)$. Substituting these
expressions into (22) and taking into account that
$\chi_{x_{j_{|J|}}+x_{j_{1}}-x,x-(x_{j_{|J|}}-x_{j_{1}})}(t)=\chi_{x-x_{j_{|J|}}}(t-x_{j_{1}})$
we obtain (23)

Let

$\mathbf{T}_{\mathfrak{I}_{N}}^{h}u(x):=u(x)+\int_{-x}^{x}K_{\mathfrak{I}_{N}}^{h}(x,t)u(t)dt.$
(24)

By Theorem 10,
$\mathbf{T}_{\mathfrak{I}_{N}}^{f}\in\mathcal{B}\left(L_{2}(-b,b)\right)$ and

$e_{\mathfrak{I}_{N}}^{h}(\rho,x)=\mathbf{T}_{\mathfrak{I}_{N}}^{h}\left[e^{i\rho
x}\right].$ (25)

###  4.2 Goursat conditions

Let us define the function

$\sigma_{\mathfrak{I}_{N}}(x):=\sum_{k=1}^{N}\alpha_{k}H(x-x_{k}).$ (26)

Hence $\sigma_{\mathfrak{I}_{N}}^{\prime}(x)=q_{\delta,\mathfrak{I}_{n}}(x)$
in the distributional sense (
$(\sigma_{\mathfrak{I}_{N}},\phi)_{C_{0}^{\infty}(0,b)}=-(q_{\delta,\mathfrak{I}_{N}},\phi^{\prime})_{C_{0}^{\infty}(0,b)}$
for all $\phi\in C_{0}^{\infty}(0,b)$). Note that in Examples 11 and 12 we
have

$K_{\mathfrak{J}_{N}}^{0}(x,x)=\frac{1}{2}\left(\int_{0}^{x}q(s)ds+\sigma_{\mathfrak{I}_{N}}(x)\right)\;\,\mbox{
and }\;\,K_{\mathfrak{I}_{N}}^{0}(x,-x)=0\;\,\mbox{ for }N=1,2.$

More generally, the following statement is true.

###### Proposition 14

The integral transmutation kernel $K_{\mathfrak{I}_{N}}^{h}$ satisfies the
following Goursat conditions for $x\in[0,b]$

$K_{\mathfrak{J}_{N}}^{h}(x,x)=\frac{1}{2}\left(h+\int_{0}^{x}q(s)ds+\sigma_{\mathfrak{I}_{N}}(x)\right)\qquad\mbox{
and }\qquad K_{\mathfrak{I}_{N}}^{h}(x,-x)=\frac{h}{2}.$ (27)

Proof. Fix $x\in[0,b]$ and take $\xi\in\\{-x,x\\}$. By formula (22) we can
write

$K_{\mathfrak{I}_{N}}^{h}(x,\xi)=\widetilde{K}^{h}(x,\xi)+\sum_{k=1}^{N}\alpha_{k}H(x-x_{k})\chi_{[2x_{k}-x,x]}(\xi)\widetilde{K}_{k}(x,\xi-
x_{k})+F(x,\xi),$

where

$\displaystyle F(x,t)=$
$\displaystyle\sum_{k=1}^{n}\alpha_{k}H(x-x_{k})\chi_{x_{k}}(t)\widetilde{K}^{h}(x_{k},t)\ast\chi_{x-x_{k}}(t)\widetilde{K}_{k}(x,t)$
$\displaystyle+\sum_{J\in\mathcal{J}_{N}}\alpha_{J}H(x-x_{j_{|J|}})\left(\prod_{l=1}^{|J|-1}\right)^{\ast}\left(\chi_{x_{j_{l+1}}-x_{j_{l}}}(t)\widetilde{K}_{j_{l}}(x_{j_{l+1}},t)\right)$
$\displaystyle\qquad\ast\Big{(}\chi_{x-(x_{j_{|J|}}-x_{j_{1}})}(t)\widetilde{K}_{j_{|J|}}(x,t-x_{j_{1}})+\chi_{x_{j_{1}}}(t)\widetilde{K}^{h}(x_{j_{1}},t)\ast\chi_{x-x_{j_{|J|}}}(t)\widetilde{K}_{j_{|J|}}(x,t)\Big{)}.$

In the proof of Theorem 10 we obtain that
$\operatorname{Supp}(F(x,t))\subset[-x,x]$. Since $\widetilde{K}^{h}(x_{j},t)$
and $\widetilde{K}_{k}(x_{j},t)$ are continuous with respect to $t$ in the
intervals $[-x_{j},x_{j}]$ and $[x_{k}-x_{j},x_{j}-x_{k}]$ respectively for
$j=1,\dots,N$, $k\leqslant j$, by Lemma 9 the function $F(x,t)$ is continuous
for all $t\in\mathbb{R}$. Thus $F(x,\xi)=0$. For the case $\xi=x$, we have
that $\widetilde{K}^{h}(x,x)=\frac{h}{2}+\frac{1}{2}\int_{0}^{x}q(s)ds$,
$\chi_{[2x_{k}-x,x]}(x)=1$ and

$\widetilde{K}_{k}(x,x-x_{k})=\frac{1}{2}\chi_{x-x_{k}}(x-x_{k})+\displaystyle\frac{1}{2}\int_{|x-x_{k}|}^{x-x_{k}}\widehat{H}_{k}(x-x_{k},s)ds=\frac{1}{2}$

(we assume that $x\geqslant x_{k}$ in order to have $H(x-x_{k})=1$). Thus
$K_{\mathfrak{I}_{N}}^{h}(x,x)=\frac{1}{2}\left(h+\int_{0}^{x}q(s)ds+\sigma_{\mathfrak{I}_{N}}(x)\right)$.
For the case $\xi=-x$, $\widetilde{K}^{h}(x,-x)=\frac{h}{2}$ and
$\chi_{[2x_{k}-x,x]}(-x)=0$. Hence
$K_{\mathfrak{I}_{N}}^{h}(x,x)=\frac{h}{2}$.

###### Remark 15

According to Proposition 14, $2K_{\mathfrak{I}_{N}}^{h}(x,x)$ is a
(distributional) antiderivative of the potential
$q(x)+q_{\delta,\mathfrak{I}_{N}}(x)$.

###  4.3 The transmuted Cosine and Sine solutions

Let $c_{\mathfrak{I}_{N}}^{h}(\rho,x)$ and $s_{\mathfrak{I}_{N}}(\rho,x)$ be
the solutions of Eq. (1) satisfying the initial conditions

$\displaystyle c_{\mathfrak{I}_{N}}^{h}(\rho,0)=1,$
$\displaystyle(c_{\mathfrak{I}_{N}}^{h})^{\prime}(\rho,0)=h,$ (28)
$\displaystyle s_{\mathfrak{I}_{N}}(\rho,0)=0,$ $\displaystyle
s_{\mathfrak{I}_{N}}^{\prime}(\rho,0)=1.$ (29)

Note that
$c_{\mathfrak{I}_{N}}^{h}(\rho,x)=\frac{e_{\mathfrak{I}_{N}}^{h}(\rho,x)+e_{\mathfrak{I}_{N}}^{h}(-\rho,x)}{2}$
and
$s_{\mathfrak{I}_{N}}(\rho,x)=\frac{e_{\mathfrak{I}_{N}}^{h}(\rho,x)-e_{\mathfrak{I}_{N}}^{h}(-\rho,x)}{2i\rho}$.

###### Remark 16

By Corollary 3,
$c_{\mathfrak{I}_{N}}^{h}(\rho,\cdot),s_{\mathfrak{I}_{N}}(\rho,\cdot)\in
AC[0,b]$ and both functions are solutions of Eq. (9) on $[0,x_{1}]$, hence
their Wronskian is constant for $x\in[0,x_{1}]$ and

$\displaystyle 1$
$\displaystyle=W\left[c_{\mathfrak{I}_{N}}^{h}(\rho,x),s_{\mathfrak{I}_{N}}(\rho,x)\right](0)=W\left[c_{\mathfrak{I}_{N}}^{h}(\rho,x),s_{\mathfrak{I}_{N}}(\rho,x)\right](x_{1}-)=\begin{vmatrix}c_{\mathfrak{I}_{N}}^{h}(\rho,x_{1})&s_{\mathfrak{I}_{N}}(\rho,x_{1})\\\
(c_{\mathfrak{I}_{N}}^{h})^{\prime}(\rho,x_{1}-)&s^{\prime}_{\mathfrak{I}_{N}}(\rho,x_{1}-)\end{vmatrix}$
$\displaystyle=\begin{vmatrix}c_{\mathfrak{I}_{N}}^{h}(\rho,x_{1})&s_{\mathfrak{I}_{N}}(\rho,x_{1})\\\
(c_{\mathfrak{I}_{N}}^{h})^{\prime}(\rho,x_{1}+)-\alpha_{1}c_{\mathfrak{I}_{N}}^{h}(\rho,x_{1})&s^{\prime}_{\mathfrak{I}_{N}}(\rho,x_{1}+)-\alpha_{1}s_{\mathfrak{I}_{N}}(\rho,x_{1})\end{vmatrix}$
$\displaystyle=\begin{vmatrix}c_{\mathfrak{I}_{N}}^{h}(\rho,x_{1})&s_{\mathfrak{I}_{N}}(\rho,x_{1})\\\
(c_{\mathfrak{I}_{N}}^{h})^{\prime}(\rho,x_{1}+)&s^{\prime}_{\mathfrak{I}_{N}}(\rho,x_{1}+)\end{vmatrix}=W\left[c_{\mathfrak{I}_{N}}^{h}(\rho,x),s_{\mathfrak{I}_{N}}(\rho,x)\right](x_{1}+)$

(the equality in the second line is due to (2)). Since
$c_{\mathfrak{I}_{N}}^{h}(\rho,x),s_{\mathfrak{I}_{N}}(\rho,x)$ are solutions
of (9) on $[x_{1},x_{2}]$, then
$W\left[C_{\mathfrak{I}_{N}}^{h}(\rho,x),s_{\mathfrak{I}_{N}}(\rho,x)\right]$
is constant for $x\in[x_{1},x_{2}]$. Thus,
$W\left[c_{\mathfrak{I}_{N}}^{h}(\rho,x),s_{\mathfrak{I}_{N}}(\rho,x)\right](x)=1$
for all $x\in[0,x_{2}]$. Continuing the process we obtain that the Wronskian
equals one in the whole segment $[0,b]$. Thus,
$c_{\mathfrak{I}_{N}}^{h}(\rho,x),s_{\mathfrak{I}_{N}}(\rho,x)$ are linearly
independent. Finally, if $u$ is a solution of (1), by Remark 5, $u$ can be
written as
$u(x)=u(0)c_{\mathfrak{I}_{N}}^{h}(\rho,x)+u^{\prime}(0)s_{\mathfrak{I}_{N}}(\rho,x)$.
In this way,
$\left\\{c_{\mathfrak{I}_{N}}^{h}(\rho,x),s_{\mathfrak{I}_{N}}(\rho,x)\right\\}$
is a fundamental set of solutions for (1).

Similarly to the case of the regular Eq. (13) (see [39, Ch. 1]), from (21) we
obtain the following representations.

###### Proposition 17

The solutions $c_{\mathfrak{I}_{N}}^{h}(\rho,x)$ and
$s_{\mathfrak{I}_{N}}(\rho,x)$ admit the following integral representations

$\displaystyle c_{\mathfrak{I}_{N}}^{h}(\rho,x)$ $\displaystyle=\cos(\rho
x)+\int_{0}^{x}G_{\mathfrak{I}_{N}}^{h}(x,t)\cos(\rho t)dt,$ (30)
$\displaystyle s_{\mathfrak{I}_{N}}(\rho,x)$ $\displaystyle=\frac{\sin(\rho
x)}{\rho}+\int_{0}^{x}S_{\mathfrak{I}_{N}}(x,t)\frac{\sin(\rho t)}{\rho}dt,$
(31)

where

$\displaystyle G_{\mathfrak{I}_{N}}^{h}(x,t)$
$\displaystyle=K_{\mathfrak{I}_{N}}^{h}(x,t)+K_{\mathfrak{I}_{N}}^{h}(x,-t),$
(32) $\displaystyle S_{\mathfrak{I}_{N}}(x,t)$
$\displaystyle=K_{\mathfrak{I}_{N}}^{h}(x,t)-K_{\mathfrak{I}_{N}}^{h}(x,-t).$
(33)

###### Remark 18

By Proposition 14, the cosine and sine integral transmutation kernels satisfy
the conditions

$G_{\mathfrak{I}_{N}}^{h}(x,x)=h+\frac{1}{2}\left(\int_{0}^{x}q(s)ds+\sigma_{\mathfrak{I}_{N}}(x)\right),$
(34)

$S_{\mathfrak{I}_{N}}(x,x)=\frac{1}{2}\left(\int_{0}^{x}q(s)ds+\sigma_{\mathfrak{I}_{N}}(x)\right)\quad\mbox{and
}\;\;S_{\mathfrak{I}_{N}}(x,0)=0.$ (35)

Introducing the cosine and sine transmutation operators

$\mathbf{T}_{\mathfrak{I}_{N},h}^{C}u(x)=u(x)+\int_{0}^{x}G_{\mathfrak{I}_{N}}^{h}(x,t)u(t)dt,\quad\mathbf{T}_{\mathfrak{I}_{N}}^{S}u(x)=u(x)+\int_{0}^{x}S_{\mathfrak{I}_{N}}(x,t)u(t)dt$
(36)

we obtain

$c_{\mathfrak{I}_{N}}^{h}(\rho,x)=\mathbf{T}_{\mathfrak{I}_{N},h}^{C}\left[\cos(\rho
x)\right],\quad
s_{\mathfrak{I}_{N}}(\rho,x)=\mathbf{T}_{\mathfrak{I}_{N}}^{S}\left[\frac{\sin(\rho
x)}{\rho}\right].$ (37)

###### Remark 19

According to Remark 16, the space of solutions of (1) has dimension 2, and
given $f,g\in\mathcal{D}_{2}\left(\mathbf{L}_{q,\mathfrak{I}_{N}}\right)$
solutions of (1), repeating the same procedure of Remark 16, $W[f,g]$ is
constant in the whole segment $[0,b]$. The solutions $f,g$ are a fundamental
set of solutions iff $W[f,g]\neq 0$.

##  5 The SPPS method and the mapping property

###  5.1 Spectral parameter powers series

As in the case of the regular Schrödinger equation [10, 31], we obtain a
representation for the solutions of (1) as a power series in the spectral
parameter (SPPS series). Assume that there exists a solution
$f\in\mathcal{D}_{2}\left(\mathbf{L}_{q,\mathfrak{I}_{N}}\right)$ that does
not vanish in the whole segment $[0,b]$.

###### Remark 20

Given $g\in L_{2}(0,b)$, a solution
$u\in\mathcal{D}_{2}\left(\mathbf{L}_{q,\mathfrak{I}_{N}}\right)$ of the non-
homogeneous Cauchy problem

$\begin{cases}\mathbf{L}_{q,\mathfrak{I}_{N}}u(x)=g(x),\quad 0<x<b\\\
u(0)=u_{0},\;u^{\prime}(0)=u_{1}\end{cases}$ (38)

can be obtained by solving the regular equation $\mathbf{L}_{q}u(x)=g(x)$ a.e.
$x\in(0,b)$ as follows. Consider the Polya factorization
$\mathbf{L}_{q}u=-\frac{1}{f}Df^{2}D\frac{u}{f}$, where $D=\frac{d}{dx}$. A
direct computation shows that $u$ given by

$u(x)=-f(x)\int_{0}^{x}\frac{1}{f^{2}(t)}\int_{0}^{t}f(s)g(s)ds+\frac{u_{0}}{f(0)}f(x)+(f(0)u_{1}-f^{\prime}(0)u_{0})f(x)\int_{0}^{x}\frac{dt}{f^{2}(t)}$
(39)

satisfies (38) (actually, $f(x)\int_{0}^{x}\frac{1}{f^{2}(t)}dt$ is the second
linearly independent solution of $\mathbf{L}_{q}u=0$ obtained from $f$ by
Abel’s formula). By Remark 4,
$u\in\mathcal{D}_{2}\left(\mathbf{L}_{q,\mathfrak{J}_{N}}\right)$ and by
Proposition 1 and Remark 5, formula (39) provides the unique solution of (38).
Actually, if we denote $\mathcal{I}u(x):=\int_{0}^{x}u(t)dt$ and define
$\mathbf{R}_{\mathfrak{I}_{N}}^{f}:=-f\mathcal{I}f^{2}\mathcal{I}$, then
$\mathbf{R}_{\mathfrak{I}_{N}}^{f}\in\mathcal{B}\left(L_{2}(0,b)\right)$,
$\mathbf{R}_{\mathfrak{I}_{N}}^{f}\left(L_{2}(0,b)\right)\subset\mathcal{D}_{2}\left(\mathbf{L}_{q,\mathfrak{I}_{N}}\right)$
and is a right-inverse for $\mathbf{L}_{q,\mathfrak{I}_{N}}$, i.e.,
$\mathbf{L}_{q,\mathfrak{I}_{N}}\mathbf{R}_{\mathfrak{I}_{N}}^{f}g=g$ for all
$g\in L_{2}(0,b)$.

Following [31] we define the following recursive integrals:
$\widetilde{X}^{(0)}\equiv X^{(0)}\equiv 1$, and for $k\in\mathbb{N}$

$\displaystyle\widetilde{X}^{(k)}(x)$
$\displaystyle:=k\int_{0}^{x}\widetilde{X}^{(k-1)}(s)\left(f^{2}(s)\right)^{(-1)^{k-1}}ds,$
(40) $\displaystyle X^{(k)}(x)$
$\displaystyle:=k\int_{0}^{x}X^{(k-1)}(s)\left(f^{2}(s)\right)^{(-1)^{k}}ds.$
(41)

The functions $\\{\varphi_{f}^{(k)}(x)\\}_{k=0}^{\infty}$ defined by

$\varphi_{f}^{(k)}(x):=\begin{cases}f(x)\widetilde{X}^{(k)}(x),\quad\mbox{if
}k\mbox{ even},\\\ f(x)X^{(k)}(x),\quad\mbox{if }k\mbox{ odd}.\end{cases}$
(42)

for $k\in\mathbb{N}_{0}$, are called the formal powers associated to $f$.
Additionally, we introduce the following auxiliary formal powers
$\\{\psi_{f}^{(k)}(x)\\}_{k=0}^{\infty}$ given by

$\psi_{f}^{(k)}(x):=\begin{cases}\frac{\widetilde{X}^{(k)}(x)}{f(x)},\quad\mbox{if
}k\mbox{ odd},\\\ \frac{X^{(k)}(x)}{f(x)},\quad\mbox{if }k\mbox{
even}.\end{cases}$ (43)

###### Remark 21

For each $k\in\mathbb{N}_{0}$,
$\varphi_{f}^{(k)}\in\mathcal{D}_{2}\left(\mathbf{L}_{q,\mathfrak{I}_{N}}\right)$.
Indeed, direct computations show that the following relations hold for all
$k\in\mathbb{N}_{0}$:

$\displaystyle D\varphi_{f}^{(k)}$
$\displaystyle=\frac{f^{\prime}}{f}\varphi_{f}^{(k)}+k\psi_{f}^{(k-1)}$ (44)
$\displaystyle D^{2}\varphi_{f}^{(k)}$
$\displaystyle=\frac{f^{\prime\prime}}{f}\varphi_{f}^{(k)}+k(k-1)\varphi_{f}^{(k-2)}$
(45)

Since $\varphi_{f}^{(k)},\psi_{f}^{(k)}\in C[0,b]$, using the procedure from
Remark 4 and (44) we obtain
$\varphi_{f}^{(k)}\in\mathcal{D}_{2}\left(\mathbf{L}_{q,\mathfrak{I}_{N}}\right)$.

######  Theorem 22 (SPPS method)

Suppose that $f\in\mathcal{D}_{2}\left(\mathbf{L}_{q,\mathfrak{I}_{N}}\right)$
is a solution of (1) that does not vanish in the whole segment $[0,b]$. Then
the functions

$u_{0}(\rho,x)=\sum_{k=0}^{\infty}\frac{(-1)^{k}\rho^{2k}\varphi_{f}^{(2k)}(x)}{(2k)!},\quad
u_{1}(\rho,x)=\sum_{k=0}^{\infty}\frac{(-1)^{k}\rho^{2k}\varphi_{f}^{(2k+1)}(x)}{(2k+1)!}$
(46)

belong to $\mathcal{D}_{2}\left(\mathbf{L}_{q,\mathfrak{I}_{N}}\right)$, and
$\\{u_{0}(\rho,x),u_{1}(\rho,x)\\}$ is a fundamental set of solutions for (1)
satisfying the initial conditions

$\displaystyle u_{0}(\rho,0)=f(0),$ $\displaystyle
u^{\prime}_{0}(\rho,0)=f^{\prime}(0),$ (47) $\displaystyle u_{1}(\rho,0)=0,$
$\displaystyle u^{\prime}_{1}(\rho,0)=\frac{1}{f(0)},$ (48)

The series in (46) converge absolutely and uniformly on $x\in[0,b]$, the
series of the derivatives converge in $L_{2}(0,b)$ and the series of the
second derivatives converge in $L_{2}(x_{j},x_{j+1})$, $j=0,\cdots,N$. With
respect to $\rho$ the series converge absolutely and uniformly on any compact
subset of the complex $\rho$-plane.

Proof. Since $f\in C[0,b]$, the following estimates for the recursive
integrals $\\{\widetilde{X}^{(k)}(x)\\}_{k=0}^{\infty}$ and
$\\{X^{(k)}(x)\\}_{k=0}^{\infty}$ are known:

$|\widetilde{X}^{(n)}(x)|\leqslant M_{1}^{n}b^{n},\;|X^{(n)}(x)|\leqslant
M_{1}^{n}b^{n}\quad\mbox{for all }x\in[0,b],$ (49)

where $M_{1}=\|f^{2}\|_{C[0,b]}\cdot\left\|\frac{1}{f^{2}}\right\|_{C[0,b]}$
(see the proof of Theorem 1 of [31]). Thus, by the Weierstrass $M$-tests, the
series in (46) converge absolutely and uniformly on $x\in[0,b]$, and for
$\rho$ on any compact subset of the complex $\rho$-plane. We prove that
$u_{0}(\rho,x)\in\mathcal{D}_{2}\left(\mathbf{L}_{q,\mathfrak{I}_{N}}\right)$
and is a solution of (1) (the proof for $u_{1}(\rho,x)$ is analogous). By
Remark 21, the series of the derivatives of $u_{0}(\rho,x)$ is given by
$\frac{f^{\prime}}{f}\sum_{k=0}^{\infty}\frac{(-1)^{k}\rho^{2k}\varphi_{f}^{(2k)}}{(2k)!}+\sum_{k=1}^{\infty}\frac{(-1)^{k}\rho^{2k}\psi_{f}^{(2k-1)}}{(2k-1)!}$.
By (49), the series involving the formal powers $\varphi_{f}^{(k)}$ and
$\psi_{f}^{(k)}$ converge absolutely and uniformly on $x\in[0,b]$. Hence,
$\sum_{k=0}^{\infty}\frac{(-1)^{k}\rho^{k}D\varphi_{f}^{(2k)}(x)}{(2k)!}$
converges in $L_{2}(0,b)$. Due to [10, Prop. 3], $u_{0}(\rho,\cdot)\in
AC[0,b]$ and
$u_{0}^{\prime}(\rho,x)=\frac{f^{\prime}(x)}{f(x)}\sum_{k=0}^{\infty}\frac{(-1)^{k}\rho^{2k}\varphi_{f}^{(2k)}}{(2k)!}+\sum_{k=1}^{\infty}\frac{(-1)^{k}\rho^{2k}\psi_{f}^{(2k-1)}}{(2k-1)!}$
in $L_{2}(0,b)$. Since the series involving the formal powers defines
continuous functions, then $u_{0}(\rho,x)$ satisfies the jump condition (2).
Applying the same reasoning it is shown that
$u_{0}^{\prime\prime}(\rho,x)=\sum_{k=0}^{\infty}\frac{(-1)^{k}\rho^{2k}D^{2}\varphi_{f}^{(2k)}}{(2k)!}$,
the series converges in $L_{2}(x_{j},x_{j+1})$ and
$u_{0}(\rho,\cdot)|_{(x_{j},x_{j+1})}\in H^{2}(x_{j},x_{j+1})$, $j=0,\dots,N$.

Since $\widetilde{X}^{(n)}(0)=0$ for $n\geqslant 1$, we have (47). Finally, by
(45)

$\displaystyle\mathbf{L}_{q}u_{0}(\rho,x)$
$\displaystyle=\sum_{k=0}^{\infty}\frac{(-1)^{k}\rho^{2k}\mathbf{L}_{q}\varphi_{f}^{(2k)}(x)}{(2k)!}=\sum_{k=2}^{\infty}\frac{(-1)^{k+1}\rho^{2k}\varphi_{f}^{(2k-2)}(x)}{(2k-2)!}$
$\displaystyle=\rho^{2}\sum_{k=0}^{\infty}\frac{(-1)^{k}\rho^{2k}\varphi_{f}^{(2k)}(x)}{(2k)!}=\rho^{2}u(\rho,x),$

this for a.e. $x\in(x_{j},x_{j+1})$, $j=0,\dots,N$.

Using (47) and (48) we obtain $W[u_{0}(\rho,x),u_{1}(\rho,x)](0)=1$. Since the
Wronskian is constant (Remark 19), $\\{u_{0}(\rho,x),u_{1}(\rho,x)\\}$ is a
fundamental set of solutions.

###  5.2 Existence and construction of the non-vanishing solution

The existence of a non-vanishing solution is well known for the case of a
regular Schrödinger equation with continuous potential (see [31, Remark 5] and
[13, Cor. 2.3]). The following proof adapts the one presented in [21, Prop.
2.9] for the Dirac system.

######  Proposition 23 (Existence of non-vanishing solutions)

Let $\\{u,v\\}\in\mathcal{D}_{2}\left(\mathbf{L}_{q,\mathfrak{I}_{N}}\right)$
be a fundamental set of solutions for (1). Then there exist constants
$c_{1},c_{2}\in\mathbb{C}$ such that the solution $f=c_{1}u+c_{2}v$ does not
vanish in the whole segment $[0,b]$.

Proof. Let
$\\{u,v\\}\in\mathcal{D}_{2}\left(\mathbf{L}_{q,\mathfrak{I}_{N}}\right)$ be a
fundamental set of solutions for (1). Then $u$ and $v$ cannot have common
zeros in $[0,b]$. Indeed, if $u(\xi)=v(\xi)=0$ for some $\xi\in[0,b]$, then
$W[u,v](\xi+)=u(\xi)v^{\prime}(\xi+)-v(\xi)u^{\prime}(\xi+)=0$. Since $W[u,v]$
is constant in $[0,b]$, this contradicts that $\\{u,v\\}$ is a fundamental
system.

This implies that in each interval $[x_{j},x_{j+1}]$, $j=0,\cdots,N$, the map
$F_{j}:[x_{j},x_{j+1}]\rightarrow\mathbb{CP}^{1}$,
$F_{j}(x):=\left[u|_{[x_{j},x_{j+1}]}(x):v|_{[x_{j},x_{j+1}]}(x)\right]$
(where $\mathbb{CP}^{1}$ is the complex projective line, i.e., the quotient of
$\mathbb{C}^{2}\setminus\\{(0,0)\\}$ under the action of $\mathbb{C}^{*}$, and
$[a:b]$ denotes the equivalent class of the pair $(a,b)$) is well defined and
differentiable. In [13, Prop. 2.2] it was established that a differentiable
function $f:I\rightarrow\mathbb{CP}^{1}$, where $I\subset\mathbb{R}$ is an
interval, is never surjective, using that Sard’s theorem implies that $f(I)$
has measure zero.

Suppose that $(\alpha,\beta)\in\mathbb{C}^{2}\setminus\\{(0,0)\\}$ is such
that $\alpha u(\xi)-\beta v(\xi)=0$ for some $\xi\in[0,b]$. Hence
$\begin{vmatrix}u(\xi)&\beta\\\ v(\xi)&\alpha\end{vmatrix}=0$, that is,
$(u(\xi),v(\xi))$ and $(\alpha,\beta)$ are proportional. Since
$\xi\in[x_{j},x_{j+1}]$ for some $j\in\\{0,\cdots,N\\}$, hence
$[\alpha:-\beta]\in F_{j}\left([x_{j},x_{j+1}]\right)$.

Thus, the set
$C:=\left\\{[\alpha:\beta]\in\mathbb{CP}^{1}\,|\,\exists\xi\in[0,b]\,:\,\alpha
u(\xi)+\beta v(\xi)=0\right\\}$ is contained in
$\cup_{j=0}^{N}F_{j}\left([x_{j},x_{j+1}]\right)$, and then $C$ has measure
zero. Hence we can obtain a pair of constants
$(c_{1},c_{2})\in\mathbb{C}^{2}\setminus\\{(0,0)\\}$ with
$[c_{1}:-c_{2}]\in\mathbb{CP}^{1}\setminus C$ and $f=c_{1}u+c_{2}v$ does not
vanish in the whole segment $[0,b]$.

###### Remark 24

If $q$ is real valued and
$\alpha_{1},\cdots,\alpha_{N}\in\mathbb{R}\setminus\\{0\\}$, taking a real-
valued fundamental system of solutions for the regular equation
$\mathbf{L}_{q}y=0$ and using formula (12), we can obtain a real-valued
fundamental set of solutions $\\{u,v\\}$ for
$\mathbf{L}_{q,\mathfrak{I}_{N}}y=0$. In the proof of Proposition 23 we obtain
that $u$ and $v$ have no common zeros. Hence $f=u+iv$ is a non vanishing
solution.

For the complex case, we can choose randomly a pair of constants
$(c_{1},c_{2})\in\mathbb{C}^{2}\setminus\\{(0,0)\\}$ and verify if the linear
combination $c_{1}u+c_{2}v$ has no zero. If there is a zero, we repeat the
process until we find the non vanishing solution. Since the set $C$ (from the
proof of Proposition 23) has measure zero, is almost sure to find the
coefficients $c_{1},c_{2}$ in the first few tries.

By Proposition 23, there exists a pair of constants
$(c_{1},c_{2})\in\mathbb{C}^{2}\setminus\\{(0,0)\\}$ such that

$\displaystyle y_{0}(x)$
$\displaystyle=c_{1}+c_{2}x+\sum_{k=1}^{N}\alpha_{k}(c_{1}+c_{2}x_{k})H(x-x_{k})(x-x_{k})$
$\displaystyle\quad+\sum_{J\in\mathcal{J}_{N}}\alpha_{J}(c_{1}+c_{2}x_{j_{1}})H(x-x_{j_{|J|}})\left(\prod_{l=1}^{|J|-1}(x_{j_{l+1}}-x_{j_{1}})\right)(x-x_{j_{|J|}})$
(50)

is a non-vanishing solution of (1) for $\rho=0$ (if
$\alpha_{1},\dots,\alpha_{k}\in(0,\infty)$, it is enough with take $c_{1}=1$,
$c_{2}=0$). Below we give a procedure based on the SPPS method ([10, 31]) to
obtain the non-vanishing solution $f$ from $y_{0}$.

###### Theorem 25

Define the recursive integrals $\\{Y^{(k)}\\}_{k=0}^{\infty}$ and
$\\{\tilde{Y}^{(k)}\\}_{k=0}^{\infty}$ as follows:
$Y^{(0)}\equiv\tilde{Y}^{(0)}\equiv 1$, and for $k\geqslant 1$

$\displaystyle Y^{(k)}(x)$
$\displaystyle=\begin{cases}\int_{0}^{x}Y^{(k)}(s)q(s)y_{0}^{2}(s)ds,&\mbox{
if }k\mbox{ is even},\\\ \int_{0}^{x}\frac{Y^{(k)}(s)}{y_{0}^{2}(s)}ds,&\mbox{
if }k\mbox{ is odd},\end{cases}$ (51) $\displaystyle\tilde{Y}^{(k)}(x)$
$\displaystyle=\begin{cases}\int_{0}^{x}\tilde{Y}^{(k)}(s)q(s)y_{0}^{2}(s)ds,&\mbox{
if }k\mbox{ is odd},\\\
\int_{0}^{x}\frac{\tilde{Y}^{(k)}(s)}{y_{0}^{2}(s)}ds,&\mbox{ if }k\mbox{ is
even}.\end{cases}$ (52)

Define

$f_{0}(x)=y_{0}(x)\sum_{k=0}^{\infty}\tilde{Y}^{(2k)}(x),\qquad
f_{1}(x)=y_{0}(x)\sum_{k=0}^{\infty}Y^{(2k+1)}(x).$ (53)

Then
$\\{f_{0},f_{1}\\}\subset\mathcal{D}_{2}\left(\mathbf{L}_{q,\mathfrak{I}_{N}}\right)$
is a fundamental set of solution for $\mathbf{L}_{q,\mathfrak{I}_{N}}u=0$
satisfying the initial conditions $f_{0}(0)=c_{1}$, $f^{\prime}_{0}(0)=c_{2}$,
$f_{1}(0)=0$, $f^{\prime}_{1}(0)=1$. Both series converge uniformly and
absolutely on $x\in[0,b]$. The series of the derivatives converge in
$L_{2}(0,b)$, and on each interval $[x_{j},x_{j+1}]$, $j=0,\dots,N$, the
series of the second derivatives converge in $L_{2}(x_{j},x_{j+1})$. Hence
there exist constants $C_{1},C_{2}\in\mathbb{C}$ such that
$f=C_{1}f_{0}+C_{2}f_{1}$ is a non-vanishing solution of
$\mathbf{L}_{q,\mathfrak{I}_{N}}u=0$ in $[0,b]$.

Proof. Using the estimates

$|\tilde{Y}^{(2k-j)}(x)|\leqslant\frac{M_{1}^{(n-j)}M_{2}^{n}}{(n-j)!n!},\quad|Y^{(2k-j)}(x)|\leqslant\frac{M_{1}^{n}M_{2}^{(n-j)}}{n!(n-j)!},\quad
x\in[0,b],\;j=0,1,\;k\in\mathbb{N},$

where $M_{1}=\left\|\frac{1}{y_{0}^{2}}\right\|_{L_{1}(0,b)}$ and
$M_{2}=\|qy_{0}^{2}\|_{L_{1}(0,b)}$, from [10, Prop. 5], the series in (53)
converge absolutely and uniformly on $[0,b]$. The proof of the convergence of
the derivatives and that
$\\{f_{0},f_{1}\\}\in\mathcal{D}_{2}\left(\mathbf{L}_{q,\mathfrak{I}_{N}}\right)$
is a fundamental set of solutions is analogous to that of Theorem 22 (see also
[31, Th. 1]) and [10, Th. 7] for the proof in the regular case).

###  5.3 The mapping property

Take a non vanishing solution
$f\in\mathcal{D}_{2}\left(\mathbf{L}_{q,\mathfrak{I}_{N}}\right)$ normalized
at zero, i.e., $f(0)=1$, and set $h=f^{\prime}(0)$. Then the corresponding
transmutation operator and kernel $\mathbf{T}^{h}_{\mathfrak{I}_{N}}$ and
$K_{\mathfrak{I}_{N}}^{h}(x,t)$ will be denoted by
$\mathbf{T}^{f}_{\mathfrak{I}_{N}}$ and $K_{\mathfrak{I}_{N}}^{f}(x,t)$ and
called the canonical transmutation operator and kernel associated to $f$,
respectively (same notations are used for the cosine and sine transmutations).

###### Theorem 26

The canonical transmutation operator $\mathbf{T}^{f}_{\mathfrak{I}_{N}}$
satisfies the following relations

$\mathbf{T}_{\mathfrak{I}_{N}}^{f}\left[x^{k}\right]=\varphi_{f}^{(k)}(x)\qquad\forall
k\in\mathbb{N}_{0}.$ (54)

The canonical cosine and sine transmutation operators satisfy the relations

$\displaystyle\mathbf{T}_{\mathfrak{I}_{N},f}^{C}\left[x^{2k}\right]$
$\displaystyle=\varphi_{f}^{(2k)}(x)\qquad\forall k\in\mathbb{N}_{0}.$ (55)
$\displaystyle\mathbf{T}_{\mathfrak{I}_{N}}^{S}\left[x^{2k+1}\right]$
$\displaystyle=\varphi_{f}^{(2k+1)}(x)\qquad\forall k\in\mathbb{N}_{0}.$ (56)

Proof. Consider the solution $e_{\mathfrak{I}_{N}}^{h}(\rho,x)$ with
$h=f^{\prime}(0)$. By the conditions (47) and (48), solution
$e_{\mathfrak{I}_{N}}^{h}(\rho,x)$ can be written in the form

$\displaystyle e_{\mathfrak{I}_{N}}^{h}(\rho,x)$
$\displaystyle=u_{0}(\rho,x)+i\rho u_{1}(\rho,x)$
$\displaystyle=\sum_{k=0}^{\infty}\frac{(-1)^{k}\rho^{2k}\varphi_{f}^{(2k)}(x)}{(2k)!}+\sum_{k=0}^{\infty}\frac{i(-1)^{k}\rho^{2k+1}\varphi_{f}^{(2k+1)}(x)}{(2k+1)!}$
$\displaystyle=\sum_{k=0}^{\infty}\frac{(i\rho)^{2k}\varphi_{f}^{(2k)}(x)}{(2k)!}+\sum_{k=0}^{\infty}\frac{(i\rho)^{2k+1}\varphi_{f}^{(2k+1)}(x)}{(2k+1)!}$
$\displaystyle=\sum_{k=0}^{\infty}\frac{(i\rho)^{k}\varphi_{f}^{(k)}(x)}{k!}$
(57)

(The rearrangement of the series is due to absolute and uniform convergence,
Theorem 22). On the other hand

$e_{\mathfrak{I}_{N}}^{h}(\rho,x)=\mathbf{T}_{\mathfrak{I}_{N}}^{f}\left[e^{i\rho
x}\right]=\mathbf{T}_{\mathfrak{I}_{N}}^{f}\left[\sum_{k=0}^{\infty}\frac{(i\rho)^{k}x^{k}}{k!}\right]$

Note that
$\displaystyle\int_{-x}^{x}K_{\mathfrak{I}_{N}}^{f}(x,t)\left(\sum_{k=0}^{\infty}\frac{(i\rho)^{k}t^{k}}{k!}\right)dt=\sum_{k=0}^{\infty}\frac{(i\rho)^{k}}{k!}\int_{-x}^{x}K_{\mathfrak{I}_{N}}^{f}(x,t)t^{k}dt$,
due to the uniform convergence of the exponential series in the variable
$t\in[-x,x]$. Thus,

$e_{\mathfrak{I}_{N}}^{h}(\rho,x)=\sum_{k=0}^{\infty}\frac{(i\rho)^{k}\mathbf{T}_{\mathfrak{I}_{N}}^{f}\left[x^{k}\right]}{k!}.$
(58)

Comparing (58) and (57) as Taylor series in the complex variable $\rho$ we
obtain (54). Relations (55) and (56) follows from (54), (32), (33) and the
fact that $G_{\mathfrak{I}_{N}}^{f}(x,t)$ and $S_{\mathfrak{I}_{N}}(x,t)$ are
even and odd in the variable $t$, respectively.

###### Remark 27

The formal powers $\\{\varphi_{f}^{(k)}(x)\\}_{k=0}^{\infty}$ satisfy the
asymptotic relation
$\varphi_{f}^{(k)}(x)=x^{k}(1+o(1))$, $x\rightarrow 0^{+}$, $\forall
k\in\mathbb{N}$.

Indeed, by Theorem 26 and the Cauchy-Bunyakovsky-Schwarz inequality we have

$\displaystyle|\varphi_{f}^{(k)}(x)-x^{k}|$
$\displaystyle=\left|\int_{-x}^{x}K_{\mathfrak{I}_{N}}^{f}(x,t)t^{k}dt\right|\leqslant\left(\int_{-x}^{x}\left|K_{\mathfrak{I}_{N}}^{f}(x,t)\right|^{2}dt\right)^{\frac{1}{2}}\left(\int_{-x}^{x}|t|^{2k}dt\right)^{\frac{1}{2}}$
$\displaystyle\leqslant\sqrt{2b}\left\|K_{\mathfrak{I}_{N}^{f}}\right\|_{L_{\infty}(\Omega)}\sqrt{\frac{2}{2k+1}}x^{k+\frac{1}{2}}$

(because $K_{\mathfrak{I}_{N}}^{f}\in L_{\infty}(\Omega)$ by Theorem 10).
Hence

$\left|\frac{\varphi_{f}^{(k)}(x)}{x^{k}}-1\right|\leqslant\sqrt{2b}\left\|K_{\mathfrak{I}_{N}^{f}}\right\|_{L_{\infty}(\Omega)}\sqrt{\frac{2}{2k+}}x^{\frac{1}{2}}\rightarrow
0,\qquad x\rightarrow 0^{+}.$

###### Remark 28

Denote $\mathcal{P}(\mathbb{R})=\mbox{Span}\\{x^{k}\\}_{k=0}^{\infty}$.
According to Remark 21 and Proposition 1 we have that
$\mathbf{T}_{\mathfrak{I}_{N}}^{f}\left(\mathcal{P}(\mathbb{R})\right)=\mbox{Span}\left\\{\varphi_{f}^{(k)}(x)\right\\}_{k=0}^{\infty}$,
and by (45) we have the relation

$\mathbf{L}_{q,\mathfrak{I}_{N}}\mathbf{T}_{\mathfrak{I}_{N}}^{f}p=-\mathbf{T}_{\mathfrak{I}_{N}}^{f}D^{2}p\qquad\forall
p\in\mathcal{P}(\mathbb{R}).$ (59)

According to [14], $\mathbf{T}_{q,\mathfrak{I}_{N}}^{f}$ is a transmutation
operator for the pair $\mathbf{L}_{q,\mathfrak{I}_{N}}$, $-D^{2}$ in the
subspace $\mathcal{P}(\mathbb{R})$, and
$\\{\varphi_{f}^{(k)}(x)\\}_{k=0}^{\infty}$ is an
$\mathbf{L}_{q,\mathfrak{I}_{N}}$-basis. Since
$\varphi_{f}^{(K)}(0)=D\varphi_{f}^{(k)}(0)=0$ for $k\geqslant 2$,
$\\{\varphi_{f}^{(k)}(x)\\}_{k=0}^{\infty}$ is called a standard
$\mathbf{L}_{q,\mathfrak{J}_{N}}$-basis, and
$\mathbf{T}_{\mathfrak{I}_{N}}^{f}$ a standard transmutation operator. By
Remark 20 we can recover $\varphi_{f}^{(k)}$ for $k\geqslant 2$ from
$\varphi_{f}^{(0)}$ and $\varphi_{f}^{(0)}$ by the formula

$\varphi_{f}^{(k)}(x)=-k(k-1)\mathbf{R}_{\mathfrak{I}_{N}}^{f}\varphi_{f}^{(k)}(x)=k(k-1)f(x)\int_{0}^{x}\frac{1}{f^{2}(t)}\int_{0}^{t}f(s)\varphi_{f}^{(k-2)}(s)ds$
(60)

(compare this formula with [14, Formula (8), Remark 9]).

The following result adapts Theorem 10 from [14], proved for the case of an
$L_{1}$-regular potential.

###### Theorem 29

The operator $\mathbf{T}_{\mathfrak{I}_{N}}^{f}$ is a transmutation operator
for the pair $\mathbf{L}_{q,\mathfrak{I}_{N}}$, $-D^{2}$ in $H^{2}(-b,b)$,
that is,
$\mathbf{T}_{\mathfrak{I}_{N}}^{f}\left(H^{2}(-b,b)\right)\subset\mathcal{D}_{2}\left(\mathbf{L}_{q,\mathfrak{I}_{N}}\right)$
and

$\mathbf{L}_{q,\mathfrak{I}_{N}}\mathbf{T}_{\mathfrak{I}_{N}}u=-\mathbf{T}_{\mathfrak{I}_{N}}D^{2}u\qquad\forall
u\in H^{2}(-b,b)$ (61)

Proof. We show that

$\mathbf{T}_{\mathfrak{I}_{N}}u(x)=u(0)\varphi_{f}^{(0)}(x)+u^{\prime}(0)\varphi_{f}^{(1)}(x)-\mathbf{R}_{\mathfrak{I}_{N}}^{f}\mathbf{T}_{\mathfrak{I}_{N}}^{f}u^{\prime\prime
2}(-b,b).$ (62)

Let us first see that (62) is valid for $p\in\mathcal{P}(\mathbb{R})$. Indeed,
set $p(x)=\sum_{k=0}^{M}c_{k}x^{k}$. By the linearity of
$\mathbf{T}_{\mathfrak{I}_{N}}^{f}$, Theorem 26 and (60) we have

$\displaystyle\mathbf{T}_{\mathfrak{I}_{N}}^{f}p(x)$
$\displaystyle=c_{0}\varphi_{f}^{(0)}+c_{1}\varphi_{f}^{(1)}(x)+\sum_{k=2}^{M}c_{k}\varphi_{f}^{(k)}(x)$
$\displaystyle=p(0)\varphi_{f}^{(0)}+p^{\prime}(0)\varphi_{f}^{(1)}(x)-\sum_{k=2}^{M}c_{k}k(k-1)\mathbf{R}_{\mathfrak{I}_{N}}^{f}\varphi_{f}^{(k-2)}(x)$
$\displaystyle=p(0)\varphi_{f}^{(0)}+p^{\prime}(0)\varphi_{f}^{(1)}(x)-\sum_{k=2}^{M}c_{k}k(k-1)\mathbf{R}_{\mathfrak{I}_{N}}^{f}\mathbf{T}_{\mathfrak{I}_{N}}^{f}\left[x^{k-2}\right]$
$\displaystyle=p(0)\varphi_{f}^{(0)}+p^{\prime}(0)\varphi_{f}^{(1)}(x)-\mathbf{R}_{\mathfrak{I}_{N}}^{f}\mathbf{T}_{\mathfrak{I}_{N}}^{f}p^{\prime\prime}(x)$

This establishes (62) for $p\in\mathcal{P}(\mathbb{R})$. Take $u\in
H^{2}(-b,b)$ arbitrary. There exists a sequence
$\\{p_{n}\\}\subset\mathcal{P}(\mathbb{R})$ such that
$p_{n}^{(j)}\overset{[-b,b]}{\rightrightarrows}u^{(j)}$, $j=0,1$, and
$p^{\prime\prime}_{n}\rightarrow u$ in $L_{2}(-b,b)$, when
$n\rightarrow\infty$ (see [14, Prop. 4]). Since
$\mathbf{R}_{\mathfrak{I}_{N}}^{f}\mathbf{T}_{\mathfrak{I}_{N}}^{f}\in\mathcal{B}\left(L_{2}(-b,b),L_{2}(0,b)\right)$
we have

$\displaystyle\mathbf{T}_{\mathfrak{I}_{N}}^{f}u(x)$
$\displaystyle=\lim_{n\rightarrow\infty}\mathbf{T}_{\mathfrak{I}_{N}}^{f}p_{n}(x)=\lim_{n\rightarrow\infty}\left[p_{n}(0)\varphi_{f}^{(0)}+p^{\prime}_{n}(0)\varphi_{f}^{(1)}(x)-\mathbf{R}_{\mathfrak{I}_{N}}^{f}\mathbf{T}_{\mathfrak{I}_{N}}^{f}p^{\prime\prime}_{n}(x)\right]$
$\displaystyle=u(0)\varphi_{f}^{(0)}(x)+u^{\prime}(0)\varphi_{f}^{(1)}(x)-\mathbf{R}_{\mathfrak{I}_{N}}^{f}\mathbf{T}_{\mathfrak{I}_{N}}^{f}u^{\prime\prime}(x)$

and we obtain (62). Hence, by Remark 20,
$\mathbf{T}_{\mathfrak{I}_{N}}^{f}\left(H^{2}(-b,b)\right)\subset\mathcal{D}_{2}\left(\mathbf{L}_{q,\mathfrak{I}_{N}}\right)$,
and since $\mathbf{L}_{q,\mathfrak{I}_{N}}\varphi_{f}^{(k)}=0$ for $k=0,1$,
applying $\mathbf{L}_{q,\mathfrak{I}_{N}}$ in both sides of (62) we have (61).

##  6 Fourier-Legendre and Neumann series of Bessel functions expansions

###  6.1 Fourier-Legendre series expansion of the transmutation kernel

Fix $x\in(0,b]$. Theorem 10 establishes that
$K_{\mathfrak{I}_{N}}^{h}(x,\cdot)\in L_{2}(-x,x)$, then
$K_{\mathfrak{I}_{N}}^{h}(x,t)$ admits a Fourier series in terms of an
orthogonal basis of $L_{2}(-x,x)$. Following [30], we choose the orthogonal
basis of $L_{2}(-1,1)$ given by the Legendre polynomials
$\\{P_{n}(z)\\}_{n=0}^{\infty}$. Thus,

$K_{\mathfrak{I}_{N}}^{h}(x,t)=\sum_{n=0}^{\infty}\frac{a_{n}(x)}{x}P_{n}\left(\frac{t}{x}\right)$
(63)

where

$a_{n}(x)=\left(n+\frac{1}{2}\right)\int_{-x}^{x}K_{\mathfrak{I}_{N}}^{h}(x,t)P_{n}\left(\frac{t}{x}\right)dt\qquad\forall
n\in\mathbb{N}_{0}.$ (64)

The series (63) converges with respect to $t$ in the norm of $L_{2}(-x,x)$.
Formula (64) is obtained multiplying (63) by $P_{n}\left(\frac{t}{x}\right)$,
using the general Parseval’s identity [2, pp. 16] and taking into account that
$\|P_{n}\|_{L_{2}(-1,1)}^{2}=\frac{2}{2n+1}$, $n\in\mathbb{N}_{0}$.

###### Example 30

Consider the kernel
$K_{\mathfrak{I}_{1}}^{0}(x,t)=\frac{\alpha_{1}}{2}H(x-x_{1})\chi_{[2x_{1}-x,x]}$
from Example 11. In this case, the Fourier-Legendre coefficients has the form

$a_{n}(x)=\frac{\alpha_{1}}{2}\left(n+\frac{1}{2}\right)H(x-x_{1})\int_{2x_{1}-x}^{x}P_{n}(t)dt=\frac{\alpha_{1}}{2}\left(n+\frac{1}{2}\right)xH(x-x_{1})\int_{2\frac{x_{1}}{x}-1}^{1}P_{n}(t)dt.$

From this we obtain $a_{0}(x)=\frac{\alpha_{1}}{2}H(x-x_{1})(x-x_{1})$. Using
formula
$P_{n}(t)=\frac{1}{2n+1}\frac{d}{dt}\left(P_{n+1}(t)-P_{n-1}(t)\right)$ for
$n\in\mathbb{N}$, and that $P_{n}(1)=0$ for all $n\in\mathbb{N}$, we have

$a_{n}(x)=\frac{\alpha_{1}}{4}xH(x-x_{1})\left[P_{n-1}\left(\frac{2x_{1}}{x}-1\right)-P_{n+1}\left(\frac{2x_{1}}{x}-1\right)\right]$

###### Remark 31

From (64) we obtain that the first coefficient $a_{0}(x)$ is given by

$\displaystyle a_{0}(x)$
$\displaystyle=\frac{1}{2}\int_{-x}^{x}K_{\mathfrak{I}_{N}}^{h}(x,t)P_{0}\left(\frac{t}{x}\right)dt=\frac{1}{2}\int_{-x}^{x}K_{\mathfrak{I}_{N}}^{h}(x,t)dt$
$\displaystyle=\frac{1}{2}\mathbf{T}_{\mathfrak{I}_{N}}^{h}[1]-\frac{1}{2}=\frac{1}{2}(e_{\mathfrak{I}_{N}}^{h}(0,x)-1).$

Thus, we obtain the relations

$a_{0}(x)=\frac{1}{2}(e_{\mathfrak{I}_{N}}^{h}(0,x)-1),\qquad
e_{\mathfrak{I}_{N}}^{h}(0,x)=2a_{n}(x)+1.$ (65)

For the kernels $G_{\mathfrak{I}_{N}}^{h}(x,t)$ and
$S_{\mathfrak{I}_{N}}(x,t)$ we obtain the series representations in terms of
the even and odd Legendre polynomials, respectively,

$\displaystyle G_{\mathfrak{I}_{N}}^{h}(x,t)$
$\displaystyle=\sum_{n=0}^{\infty}\frac{g_{n}(x)}{x}P_{2n}\left(\frac{t}{x}\right),$
(66) $\displaystyle S_{\mathfrak{I}_{N}}(x,t)$
$\displaystyle=\sum_{n=0}^{\infty}\frac{s_{n}(x)}{x}P_{2n+1}\left(\frac{t}{x}\right),$
(67)

where the coefficients are given by

$\displaystyle g_{n}(x)$
$\displaystyle=2a_{2n}(x)=(4n+1)\int_{0}^{x}G_{\mathfrak{I}_{N}}^{h}(x,t)P_{2n}\left(\frac{t}{x}\right)dt,$
(68) $\displaystyle s_{n}(x)$
$\displaystyle=2a_{2n+1}(4n+3)\int_{0}^{x}S_{\mathfrak{I}_{N}}(x,t)P_{2n+1}\left(\frac{t}{x}\right)dt.$
(69)

The proof of these facts is analogous to that in the case of Eq. (9), see [30]
or [29, Ch. 9].

###### Remark 32

Since $g_{0}(x)=2a_{0}(x)$, then $g_{0}(x)=e_{\mathfrak{I}_{N}}^{h}(0,x)-1$.
Since $e_{\mathfrak{I}_{N}}^{h}(0,x)$ is the solution of (1) with $\rho=0$
satisfying $e_{\mathfrak{I}_{N}}^{h}(0,0)=1$,
$(e_{\mathfrak{I}_{N}}^{h})^{\prime}(0,0)=h$, hence by Remark 5,
$e_{\mathfrak{I}_{N}}^{h}(0,x)=c_{\mathfrak{I}_{N}}^{h}(0,x)$ and

$g_{0}(x)=c_{\mathfrak{I}_{N}}^{h}(0,x)-1.$ (70)

On the other hand, for the coefficient $s_{0}(x)$ we have the relation

$s_{0}(x)=3\int_{0}^{x}H_{\mathfrak{I}_{N}}(x,t)P_{1}\left(\frac{t}{x}\right)dt=\frac{3}{x}\int_{0}^{x}H_{\mathfrak{I}_{N}}(x,t)tdt.$

Since $\frac{\sin(\rho x)}{\rho}\big{|}_{x=0}=x$, from (31) we have

$s_{0}(x)=3\left(\frac{s_{\mathfrak{I}_{N}}(0,x)}{x}-1\right).$ (71)

For every $n\in\mathbb{N}_{0}$ we write the Legendre polynomial $P_{n}(z)$ in
the form $P_{n}(z)=\sum_{k=0}^{n}l_{k,n}z^{k}$. Note that if $n$ is even,
$l_{k,n}=0$ for odd $k$, and $P_{2n}(z)=\sum_{k=0}^{n}\tilde{l}_{k,n}z^{2k}$
with $\tilde{l}_{k,n}=l_{2k,2n}$. Similarly
$P_{2n+1}(z)=\sum_{k=0}^{n}\hat{l}_{k,n}z^{2k+1}$ with
$\hat{l}_{k,n}=l_{2k+1,2n+1}$. With this notation we write an explicit formula
for the coefficients (64) of the canonical transmutation kernel
$K_{\mathfrak{J}_{N}}^{f}(x,t)$.

###### Proposition 33

The coefficients $\\{a_{n}(x)\\}_{n=0}^{\infty}$ of the Fourier-Legendre
expansion (63) of the canonical transmutation kernel
$K_{\mathfrak{I}_{N}}^{f}(x,t)$ are given by

$a_{n}(x)=\left(n+\frac{1}{2}\right)\left(\sum_{k=0}^{n}l_{k,n}\frac{\varphi_{f}^{(k)}(x)}{x^{k}}-1\right).$
(72)

The coefficients of the canonical cosine and sine kernels satisfy the
following relations for all $n\in\mathbb{N}_{0}$

$\displaystyle g_{n}(x)$
$\displaystyle=(4n+1)\left(\sum_{k=0}^{n}\tilde{l}_{k,n}\frac{\varphi_{f}^{(2k)}(x)}{x^{2k}}-1\right),$
(73) $\displaystyle s_{n}(x)$
$\displaystyle=(4n+3)\left(\sum_{k=0}^{n}\hat{l}_{k,n}\frac{\varphi_{f}^{(2k+1)}(x)}{x^{2k+1}}-1\right),$
(74)

Proof. From (64) we have

$\displaystyle a_{n}(x)$
$\displaystyle=\left(n+\frac{1}{2}\right)\int_{-x}^{x}K_{\mathfrak{I}_{N}}^{f}(x,t)\left(\sum_{k=0}^{n}l_{k,n}\left(\frac{t}{x}\right)^{k}\right)dt$
$\displaystyle=\left(n+\frac{1}{2}\right)\sum_{k=0}^{n}\frac{l_{k,n}}{x^{k}}\int_{0}^{x}K_{\mathfrak{I}_{N}}^{f}(x,t)t^{k}dt$
$\displaystyle=\left(n+\frac{1}{2}\right)\sum_{k=0}^{n}\frac{l_{k,n}}{x^{k}}\left(\mathbf{T}_{\mathfrak{I}_{N}}^{f}\left[x^{k}\right]-x^{k}\right).$

Hence (72) follows from Theorem 26 and that $P_{n}(z)=1$. Since
$g_{n}(x)=2a_{2n}(x)$, $s_{n}(x)=2a_{2n+1}(x)$, $l_{2k+1,2n}=0$,
$l_{2k,2n+1}=0$ and $l_{2k,2n}=\tilde{l}_{k,n}$,$l_{2k+1,2n+1}=\hat{l}_{k,n}$,
we obtain (73) and (74).

###### Remark 34

By Remark 27, formula (72) is well defined at $x=0$. Note that $x^{n}a_{n}(x)$
belongs to $\mathcal{D}_{2}\left(\mathbf{L}_{q,\mathfrak{I}_{N}}\right)$ for
all $n\in\mathbb{N}_{0}$.

###  6.2 Representation of the solutions as Neumann series of Bessel functions

Similarly to the case of the regular Eq. (13) [30], we obtain a representation
for the solutions in terms of Neumann series of Bessel functions (NSBF). For
$M\in\mathbb{N}$ we define

$K_{\mathfrak{I}_{N},M}^{h}(x,t):=\sum_{n=0}^{M}\frac{a_{n}(x)}{x}P_{n}\left(\frac{t}{x}\right),$

that is, the $M$-partial sum of (63).

###### Theorem 35

The solutions $c_{\mathfrak{I}_{N}}^{h}(\rho,x)$ and
$s_{\mathfrak{I}_{N}}(\rho,x)$ admit the following NSBF representations

$\displaystyle c_{\mathfrak{I}_{N}}^{h}(\rho,x)=\cos(\rho
x)+\sum_{n=0}^{\infty}(-1)^{n}g_{n}(x)j_{2n}(\rho x),$ (75) $\displaystyle
s_{\mathfrak{I}_{N}}(\rho,x)=\frac{\sin(\rho
x)}{\rho}+\frac{1}{\rho}\sum_{n=0}^{\infty}(-1)^{n}s_{n}(x)j_{2n+1}(\rho x),$
(76)

where $j_{\nu}$ stands for the spherical Bessel function
$j_{\nu}(z)=\sqrt{\frac{\pi}{2z}}J_{\nu+\frac{1}{2}}(z)$ (and $J_{\nu}$ stands
for the Bessel function of order $\nu$). The series converge pointwise with
respect to $x$ in $(0,b]$ and uniformly with respect to $\rho$ on any compact
subset of the complex $\rho$-plane. Moreover, for $M\in\mathbb{N}$ the
functions

$\displaystyle c_{\mathfrak{I}_{N},M}^{h}(\rho,x)=\cos(\rho
x)+\sum_{n=0}^{M}(-1)^{n}g_{n}(x)j_{2n}(\rho x),$ (77) $\displaystyle
s_{\mathfrak{I}_{N},M}(\rho,x)=\frac{\sin(\rho
x)}{\rho}+\frac{1}{\rho}\sum_{n=0}^{M}(-1)^{n}s_{n}(x)j_{2n+1}(\rho x),$ (78)

obey the estimates

$\displaystyle|c_{\mathfrak{I}_{N}}^{h}(\rho,x)-c_{\mathfrak{I}_{N},M}^{h}(\rho,x)|$
$\displaystyle\leqslant 2\epsilon_{2M}(x)\sqrt{\frac{\sinh(2bC)}{C}},$ (79)
$\displaystyle|\rho s_{\mathfrak{I}_{N}}(\rho,x)-\rho
s_{\mathfrak{I}_{N},M}(\rho,x)|$ $\displaystyle\leqslant
2\epsilon_{2M+1}(x)\sqrt{\frac{\sinh(2bC)}{C}},$ (80)

for any $\rho\in\mathbb{C}$ belonging to the strip
$|\operatorname{Im}\rho|\leqslant C$, $C>0$, and where
$\epsilon_{M}(x)=\|K_{\mathfrak{I}_{N}}^{h}(x,\cdot)-K_{\mathfrak{I}_{N},2M}^{h}(x,\cdot)\|_{L_{2}(-x,x)}$.

Proof. We show the results for the solution $c_{\mathfrak{I}_{N}}^{h}(\rho,x)$
(the proof for $s_{\mathfrak{I}_{N}}(\rho,x)$ is similar). Substitution of the
Fourier-Legendre series (66) in (30) leads us to

$\displaystyle c_{\mathfrak{I}_{N}}^{h}(\rho,x)$ $\displaystyle=\cos(\rho
x)+\int_{0}^{x}\left(\sum_{n=0}^{\infty}\frac{g_{n}(x)}{x}P_{2n}\left(\frac{t}{x}\right)\right)\cos(\rho
t)dt$ $\displaystyle=\cos(\rho
x)+\sum_{n=0}^{\infty}\frac{g_{n}(x)}{x}\int_{0}^{x}P_{2n}\left(\frac{t}{x}\right)\cos(\rho
t)dt$

(the exchange of the integral with the summation is due to the fact that the
integral is nothing but the inner product of the series with the function
$\overline{\cos(\rho t)}$ and the series converges in $L_{2}(0,x)$). Using
formula 2.17.7 in [40, pp. 433]

$\int_{0}^{a}\left\\{\begin{matrix}P_{2n+1}\left(\frac{y}{a}\right)\cdot\sin(by)\\\
P_{2n}\left(\frac{y}{a}\right)\cdot\cos(by)\\\
\end{matrix}\right\\}dy=(-1)^{n}\sqrt{\frac{\pi
a}{2b}}J_{2n+\delta+\frac{1}{2}}(ab),\quad\delta=\left\\{\begin{matrix}1\\\
0\end{matrix}\right\\},\;a>0,$

we obtain the representation (75). Take $C>0$ and $\rho\in\mathbb{C}$ with
$|\operatorname{Im}\rho|\leqslant C$. For $M\in\mathbb{N}$ define
$G_{\mathfrak{I}_{N},M}^{h}(x,t):=K_{\mathfrak{I}_{N},2M}^{h}(x,t)-K_{\mathfrak{I}_{N},2M}^{h}(x,-t)=\sum_{n=0}^{M}\frac{g_{n}(x)}{x}P_{2n}\left(\frac{t}{x}\right)$,
the $M$-th partial sum of (66). Then

$c_{\mathfrak{I}_{N},M}^{h}(\rho,x)=\cos(\rho
x)+\int_{0}^{x}G_{\mathfrak{I}_{N},M}^{h}(x,t)\cos(\rho t)dt.$

Using the Cauchy-Bunyakovsky-Schwarz inequality we obtain

$\displaystyle|c_{\mathfrak{I}_{N}}^{h}(\rho,x)-C_{\mathfrak{I}_{N},M}^{h}(\rho,x)|$
$\displaystyle=\left|\int_{0}^{x}\left(G_{\mathfrak{I}_{N}}^{h}(x,t)-G_{\mathfrak{I}_{N},M}^{h}(x,t)\right)\cos(\rho
t)dt\right|$
$\displaystyle=\left|\left\langle\overline{G_{\mathfrak{I}_{N}}^{h}(x,t)-G_{\mathfrak{I}_{N},M}^{h}(x,t)},\cos(\rho
t)\right\rangle_{L_{2}(0,x)}\right|$
$\displaystyle\leqslant\|G_{\mathfrak{I}_{N}}^{h}(x,\cdot)-G_{\mathfrak{I}_{N},M}^{h}(x,\cdot)\|_{L_{2}(0,x)}\|\cos(\rho
t)\|_{L_{2}(0,x)}.$

Since
$\|K_{\mathfrak{I}_{N}}^{h}(x,\cdot)-K_{\mathfrak{I}_{N},2M}^{h}(x,\cdot)\|_{L_{2}(-x,x)}=\frac{1}{2}\|G_{\mathfrak{I}_{N}}^{h}(x,\cdot)-G_{M,n}^{h}(x,\cdot)\|_{L_{2}(0,x)}$,

$\displaystyle\int_{0}^{x}|\cos(\rho t)|^{2}dt$
$\displaystyle\leqslant\frac{1}{4}\int_{0}^{x}\left(|e^{i\rho t}|+|e^{-i\rho
t}|\right)^{2}dt\leqslant\frac{1}{2}\int_{0}^{x}\left(e^{-2t\operatorname{Im\rho}}+e^{2t\operatorname{Im\rho}}\right)dt$
$\displaystyle=\int_{-x}^{x}e^{-2\operatorname{Im}\rho
t}dt=\frac{\sinh(2x\operatorname{Im}\rho)}{\operatorname{Im}\rho}$

and the function $\frac{\sinh(\xi x)}{\xi}$ is monotonically increasing in
both variables when $\xi,x\geqslant 0$, we obtain (79).

Given $H\in\mathbb{C}$, we look for a pair of solutions
$\psi_{\mathfrak{I}_{N}}^{H}(\rho,x)$ and
$\vartheta_{\mathfrak{I}_{N}}(\rho,x)$ of (1) satisfying the conditions

$\displaystyle\psi_{\mathfrak{I}_{N}}^{H}(\rho,b)=1,$
$\displaystyle(\psi_{\mathfrak{I}_{N}}^{H})^{\prime}(\rho,b)=-H,$ (81)
$\displaystyle\vartheta_{\mathfrak{I}_{N}}(\rho,b)=0,$
$\displaystyle\vartheta_{\mathfrak{I}_{N}}^{\prime}(\rho,b)=1.$ (82)

###### Theorem 36

The solutions $\psi_{\mathfrak{I}_{N}}^{H}(\rho,x)$ and
$\vartheta_{\mathfrak{I}_{N}}(\rho,x)$ admit the integral representations

$\displaystyle\psi_{\mathfrak{I}_{N}}^{H}(\rho,x)$
$\displaystyle=\cos(\rho(b-x))+\int_{x}^{b}\widetilde{G}_{\mathfrak{I}_{N}}^{H}(x,t)\cos(\rho(b-t))dt,$
(83) $\displaystyle\vartheta_{\mathfrak{I}_{N}}(\rho,x)$
$\displaystyle=\frac{\sin(\rho(b-x))}{\rho}+\int_{x}^{b}\widetilde{S}_{\mathfrak{I}_{N}}^{H}(x,t)\frac{\sin(\rho(b-t))}{\rho}dt,$
(84)

where the kernels $\widetilde{G}_{\mathfrak{I}_{N}}^{H}(x,t)$ and
$\widetilde{S}_{\mathfrak{I}_{N}}(x,t)$ are defined in $\Omega$ and satisfy
$\widetilde{G}_{\mathfrak{I}_{N}}^{H}(x,\cdot),\widetilde{S}_{\mathfrak{I}_{N}}(x,\cdot)\in
L_{2}(0,x)$ for all $x\in(0,b]$. In consequence, the solutions
$\psi_{\mathfrak{I}_{N}}^{H}(\rho,x)$ and
$\vartheta_{\mathfrak{I}_{N}}(\rho,x)$ can be written as NSBF

$\psi_{\mathfrak{I}_{N}}^{H}(\rho,x)=\cos(\rho(b-x))+\sum_{n=0}^{\infty}(-1)^{n}\tau_{n}(x)j_{2n}(\rho(b-x)),$
(85)
$\vartheta_{\mathfrak{I}_{N}}(\rho,x)=\frac{\sin(\rho(b-x))}{\rho}+\sum_{n=0}^{\infty}(-1)^{n}\zeta_{n}(x)j_{2n}(\rho(b-x)),$
(86)

with some coefficients $\\{\tau_{n}(x)\\}_{n=0}^{\infty}$ and
$\\{\zeta_{n}(x)\\}_{n=0}^{\infty}$.

Proof. We prove the results for $\psi_{\mathfrak{I}_{N}}^{H}(\rho,x)$ (the
proof for $\vartheta_{\mathfrak{I}_{N}}(\rho,x)$ is similar). Set
$y(\rho,x)=\psi_{\mathfrak{I}_{N}}^{H}(\rho,b-x)$. Note that $y(\rho,0)=1$,
$y^{\prime}(\rho,0)=H$ and for $\phi\in C_{0}^{\infty}(0,b)$ we have

$\displaystyle(y^{\prime\prime 2}y(x),\phi(x))_{C_{0}^{\infty}(0,b)}$
$\displaystyle=(\psi_{\mathfrak{I}_{N}}^{H}(\rho,x),\phi^{\prime\prime
2}\phi(b-x))_{C_{0}^{\infty}(0,b)}$
$\displaystyle=(q(x)\psi_{\mathfrak{I}_{N}}^{H}(\rho,x),\phi(b-x))_{C_{0}^{\infty}(0,b)}+\sum_{k=0}^{N}\alpha_{k}\psi_{\mathfrak{I}_{N}}^{H}(\rho,x_{k})\phi(b-x_{k})$
$\displaystyle=(q(b-x)y(x),\phi(x))_{C_{0}^{\infty}(0,b)}+\sum_{k=0}^{N}\alpha_{k}y(b-x_{k})\phi(b-x_{k}),$

that is, $\psi_{\mathfrak{I}_{N}}^{H}(\rho,x)$ is a solution of (1) iff
$y(x)=\psi_{\mathfrak{I}_{N}}^{H}(\rho,b-x)$ is a solution of

$-y^{\prime\prime}(x)+\left(q(b-x)+\sum_{k=0}^{N}\alpha_{k}\delta(x-(b-x_{k}))\right)y(x)=\rho^{2}y(x).$
(87)

Since $0<b-x_{N}<\cdots<b-x_{0}<b$, hence (87) is of the type (1) with the
point interactions
$\mathfrak{I}_{N}^{*}=\\{(b-x_{N-j},\alpha_{N-j})\\}_{j=0}^{N}$ and
$\psi_{\mathfrak{I}_{N}}^{H}(\rho,b-x)$ is the corresponding solution
$c_{\mathfrak{I}_{N}^{*}}^{H}(\rho,x)$ for (87). Hence

$\psi_{\mathfrak{I}_{N}}^{H}(\rho,b-x)=\cos(\rho
x)+\int_{0}^{x}G_{\mathfrak{I}_{N}^{*}}^{H}(x,t)\cos(\rho t)dt$ (88)

for some kernel $G_{\mathfrak{I}_{N}^{*}}^{H}(x,t)$ defined on $\Omega$ with
$\widetilde{G}_{\mathfrak{I}_{N}}^{H}(x,\cdot)\in L_{2}(0,x)$ for $x\in(0,b]$.
Thus,

$\displaystyle\psi_{\mathfrak{I}_{N}}(\rho,x)$
$\displaystyle=\cos(\rho(b-x))+\int_{0}^{b-x}G_{\mathfrak{I}_{N}^{*}}^{H}(b-x,t)\cos(\rho
t)dt$
$\displaystyle=\psi_{\mathfrak{I}_{N}}(\rho,x)=\cos(\rho(b-x))+\int_{x}^{b}G_{\mathfrak{I}_{N}^{*}}^{H}(b-x,b-t)\cos(\rho(b-t))dt,$

where the change of variables $x\mapsto b-x$ was used. Hence we obtain (83)
with
$\widetilde{G}_{\mathfrak{I}_{N}^{*}}^{H}(x,t)=G_{\mathfrak{I}_{N}^{*}}^{H}(b-x,b-t)$
In consequence, by Theorem 35 we obtain (85).

###### Remark 37

As in Remark 32

$\tau_{0}(x)=\psi_{\mathfrak{I}_{N}}^{H}(0,x)-1\quad\mbox{and
}\;\zeta_{0}(x)=3\left(\frac{\vartheta_{\mathfrak{I}_{N}}(0,x)}{b-x}-1\right).$
(89)

###### Remark 38

Let $\lambda\in\mathbb{C}$ and $\lambda=\rho^{2}$.

  * (i)

The functions $\widehat{s}_{k}(\rho,x-x_{k})$ are entire with respect to
$\rho$. Then from (12) $c_{\mathfrak{I}_{N}}^{h}(\rho,x)$,
$s_{\mathfrak{I}_{N}}(\rho,x)$ and $\psi_{\mathfrak{I}_{N}}^{H}(\rho,x)$ are
entire as well.

  * (ii)

Suppose that $q$ is real valued and
$\alpha_{0},\dots,\alpha_{N},u_{0},u_{1}\in\mathbb{R}$. If $u(\lambda,x)$ is a
solution of $u^{(k)}(\lambda,0)=u_{k}$, $k=0,1$, then by the uniqueness of the
Cauchy problem $\overline{u(\lambda,x)}=u(\overline{\lambda},x)$. In
particular, for $\rho,h,H\in\mathbb{R}$, the solutions
$c_{\mathfrak{I}_{N}}^{h}(\rho,x)$, $s_{\mathfrak{I}_{N}}(\rho,x)$ and
$\psi_{\mathfrak{I}_{N}}^{H}(\rho,x)$ are real valued.

###  6.3 A recursive integration procedure for the coefficients
$\\{a_{n}(x)\\}_{n=0}^{\infty}$

Similarly to the case of the regular Schrödinger equation [29, 30, 32], we
derive formally a recursive integration procedure for computing the Fourier-
Legendre coefficients $\\{a_{n}(x)\\}_{n=0}^{\infty}$ of the canonical
transmutation kernel $K_{\mathfrak{J}_{N}}^{f}(x,t)$. Consider the sequence of
functions $\sigma_{n}(x):=x^{n}a_{n}(x)$ for $n\in\mathbb{N}_{0}$. According
to Remark 34,
$\\{\sigma_{n}(x)\\}_{n=0}^{\infty}\subset\mathcal{D}_{2}\left(\mathbf{L}_{q,\mathfrak{J}_{N}}\right)$.

###### Remark 39

  * (i)

By Remark 32,

$\sigma_{0}(x)=\frac{f(x)-1}{2}.$ (90)

  * (ii)

By (72), $a_{1}(x)=\frac{3}{2}\left(\frac{\varphi_{f}^{(1)}(x)}{x}-1\right)$.
Thus, from (42) and (43) we have

$\sigma_{1}(x)=\frac{3}{2}\left(f(x)\int_{0}^{x}\frac{dt}{f^{2}(t)}-x\right).$
(91)

  * (iii)

For $n\geqslant 2$, $\sigma_{n}(0)=0$, and by (72) we obtain

$\displaystyle D\sigma_{n}(x)$
$\displaystyle=\left(n+\frac{1}{2}\right)\sum_{k=0}^{n}l_{k,n}D\left(x^{n-k}\varphi_{f}^{(k)}(x)\right)$
$\displaystyle=\left(n+\frac{1}{2}\right)\left(\sum_{k=0}^{n-1}l_{k,n}(n-k)x^{n-k-1}\varphi_{f}^{(k)}(x)+\sum_{k=0}^{n}l_{k,n}x^{n-k}D\varphi_{f}^{(k)}(x)\right).$

By (44) and (43), $D\varphi_{f}^{(k)}(0)=0$ for $k\geqslant 1$. Hence,
$\sigma_{n}^{\prime}(0)=0$.

Denote by $c_{\mathfrak{J}_{N}}^{f}(\rho,x)$ the solution of (1) satisfying
(28) with $h=f^{\prime}(0)$. On each interval $[x_{k},x_{k+1}]$,
$k=0,\cdots,N$, $c_{\mathfrak{J}_{N}}^{f}(\rho,x)$ is a solution of the
regular equation (9). In [30, Sec. 6] by substituting the Neumann series (75)
of $c_{\mathfrak{J}_{N}}^{f}(\rho,x)$ into Eq. (9) it was proved that the
functions $\\{\sigma_{2n}(x)\\}_{n=0}^{\infty}$ must satisfy, at least
formally, the recursive relations

$\mathbf{L}_{q}\sigma_{2n}(x)=\frac{4n+1}{4n-3}x^{4n-1}\mathbf{L}_{q}\left[\frac{\sigma_{2n-2}(x)}{x^{4n-3}}\right],\quad
x_{k}<x<x_{k}$ (92)

for $k=0,\cdots,N$. Similarly, substitution of the Neumann series (76) of
$s_{\mathfrak{J}_{N}}(\rho,x)$ into (9) leads to the equalities

$\mathbf{L}_{q}\sigma_{2n+1}(x)=\frac{4n+3}{4n+1}x^{4n+3}\mathbf{L}_{q}\left[\frac{\sigma_{2n+1}(x)}{x^{4n+1}}\right],\quad
x_{k}<x<x_{k}.$ (93)

Taking into account that
$\sigma_{n}\in\mathcal{D}_{2}\left(\mathbf{L}_{q,\mathfrak{J}_{N}}\right)$ and
combining (92), by Remark 39(iii) and (93) we obtain that the functions
$\sigma_{n}(x)$, $n\geqslant 2$, must satisfy (at least formally) the
following Cauchy problems

$\begin{cases}\displaystyle\mathbf{L}_{q,\mathfrak{J}_{N}}\sigma_{n}(x)=\frac{2n+1}{2n-3}x^{2n-1}\mathbf{L}_{q}\left[\frac{\sigma_{n-2}(x)}{x^{2n-3}}\right],\quad
0<x<b,\\\ \sigma_{n}(0)=\sigma^{\prime}_{n}(0)=0.\end{cases}$ (94)

###### Remark 40

If $g\in\mathcal{D}_{2}\left(L_{q,\mathfrak{J}_{N}}\right)$, then
$\frac{g}{f}\in H^{2}(0,b)$.

Indeed, $\frac{g}{f}\in C[0,b]$, and the jump of the derivative at $x_{k}$ is
given by

$\displaystyle\left(\frac{g}{f}\right)^{\prime}(x_{k}+)-\left(\frac{g}{f}\right)^{\prime}(x_{k}-)$
$\displaystyle=\frac{g^{\prime}(x_{k}+)f(x_{k})-f^{\prime}(x_{k}+)g(x_{k})}{f^{2}(x_{k})}-\frac{g^{\prime}(x_{k}-)f(x_{k})-f^{\prime}(x_{k}-)g(x_{k})}{f^{2}(x_{k})}$
$\displaystyle=\frac{1}{f^{2}(x_{k})}\left[\left(g^{\prime}(x_{k}+)-g^{\prime}(x_{k}-)\right)f(x_{k})-g(x_{k})\left(f^{\prime}(x_{k}+)-f^{\prime}(x_{k}-)\right)\right]$
$\displaystyle=\frac{1}{f^{2}(x_{k})}\left[\alpha_{k}g(x_{k})f(x_{k})-\alpha_{k}g(x_{k})f(x_{k})\right]=0.$

Hence $\frac{g}{f}\in AC[0,b]$, and then $\frac{g}{f}\in H^{2}(0,b)$.

###### Proposition 41

The sequence $\\{\sigma_{n}(x)\\}_{n=0}^{\infty}$ satisfying the recurrence
relation (94) for $n\geqslant 2$, with $\sigma_{0}(x)=\frac{f(x)-1}{2}$ and
$\sigma_{1}(x)=\frac{3}{2}\left(f(x)\int_{0}^{x}\frac{dt}{f^{2}(t)}-x\right)$,
is given by

$\sigma_{n}(x)=\frac{2n+1}{2n-3}\left(x^{2}\sigma_{n-2}(x)+2(2n-1)\theta_{n}(x)\right),\quad
n\geqslant 2,$ (95)

where

$\theta_{n}(x):=\int_{0}^{x}\left(\eta_{n}(t)-tf(t)\sigma_{n-2}(t)\right)\frac{dt}{f^{2}(t)},\quad
n\geqslant 2,$ (96)

and

$\eta_{n}(x):=\int_{0}^{x}\left((n-1)f(t)+tf^{\prime}(t)\right)\sigma_{n-2}(t)dt,\quad
n\geqslant 2.$ (97)

Proof. Set $g\in\mathcal{D}_{2}\left(\mathbf{L}_{q,\mathfrak{J}_{N}}\right)$
and $n\geqslant 2$. Consider the Cauchy problem

$\begin{cases}\displaystyle\mathbf{L}_{q,\mathfrak{J}_{N}}u_{n}(x)=\frac{2n+1}{2n-3}x^{2n-1}\mathbf{L}_{q}\left[\frac{g(x)}{x^{2n-3}}\right],\quad
0<x<b,\\\ u_{n}(0)=u^{\prime}_{n}(0)=0.\end{cases}$ (98)

By formula (39) and the Polya factorization
$\mathbf{L}_{q}=-\frac{1}{f}Df^{2}D\frac{1}{f}$ we obtain that the unique
solution of the Cauchy problem (98) is given by

$u_{n}(x)=\frac{2n+1}{2n-3}f(x)\int_{0}^{x}\frac{1}{f^{2}(t)}\left(\int_{0}^{t}s^{2n-1}Df^{2}(s)D\left[\frac{g(s)}{s^{2n-3}f(s)}\right]ds\right)dt.$

Consider an antiderivative $\int
s^{2n-1}Df^{2}(s)D\left[\frac{g(s)}{s^{2n-3}f(s)}\right]ds$. Integration by
parts gives

$\displaystyle\int s^{2n-1}Df^{2}(s)D\left[\frac{g(s)}{s^{2n-3}f(s)}\right]ds$
$\displaystyle=s^{2n-1}f^{2}(s)D\left(\frac{g(s)}{s^{2n-3}f(s)}\right)-(2n-1)sf(s)g(s)$
$\displaystyle\quad+\int\left((2n-1)(2n-2)f(s)+2(2n-1)sf^{\prime}(s)\right)g(s)ds.$

Note that

$\displaystyle s^{2n-1}f^{2}(s)D\left(\frac{g(s)}{s^{2n-3}f(s)}\right)$
$\displaystyle=s^{2n-1}f^{2}(s)\frac{D\left(\frac{g(s)}{f(s)}\right)}{s^{2n-3}}-s^{2n-1}f^{2}(s)\frac{\frac{g(s)}{f(s)}}{s^{4n-6}}(2n-3)s^{2n-4}$
$\displaystyle=s^{2}f^{2}(s)D\left(\frac{g(s)}{f(s)}\right)-(2n-3)sf(s)g(s).$

Since $g\in\mathcal{D}_{2}\left(\mathbf{L}_{q,\mathfrak{J}_{N}}\right)$, by
Remark 40, $D\left(\frac{g(s)}{f(s)}\right)$ is continuous in $[0,b]$. Thus,

$\displaystyle\int s^{2n-1}Df^{2}(s)D\left[\frac{g(s)}{s^{2n-3}f(s)}\right]ds$
$\displaystyle=s^{2}f^{2}(s)D\left(\frac{g(s)}{f(s)}\right)-(4n-4)sf(s)g(s)$
$\displaystyle\quad+2(2n-1)\int\left((n-1)f(s)+sf^{\prime}(s)\right)ds$

is well defined at $s=0$ and is continuous in $[0,b]$. Then we obtain that

$\displaystyle\Phi(t)$
$\displaystyle:=\int_{0}^{t}s^{2n-1}Df^{2}(s)D\left[\frac{g(s)}{s^{2n-3}f(s)}\right]ds$
$\displaystyle=t^{2}f^{2}(t)D\left(\frac{g(t)}{f(t)}\right)-(4n-4)tf(t)g(t)+2(2n-1)\Theta_{n}[g](t),$

with
$H_{n}[g](t):=\displaystyle\int_{0}^{t}\left((n-1)f(s)+sf^{\prime}(s)\right)g(s)ds$,
is a continuous function in $[0,b]$. Now,

$\displaystyle\int_{0}^{x}\Phi(t)\frac{dt}{f^{2}(t)}$
$\displaystyle=\int_{0}^{x}t^{2}D\left[\frac{g(t)}{f(t)}\right]dt-(4n-4)\int_{0}^{t}t\frac{g(t)}{f(t)}dt+2(2n-1)H_{n}[g](t)$
$\displaystyle=x^{2}\frac{g(x)}{f(x)}-2(2n-1)\int_{0}^{x}\left[H_{n}[g](t)-tf(t)g(t)\right]dt.$

Hence

$u_{n}(x)=\frac{2n+1}{2n-3}\left(x^{2}g(x)-2(2n-1)\Theta_{n}[g](x)\right),$
(99)

with
$\Theta_{n}[g](x):=\displaystyle\int_{0}^{x}\left[H_{n}[g](t)-tf(t)g(t)\right]dt$.

Finally, since
$\sigma_{0},\sigma_{1}\in\mathcal{D}_{2}\left(\mathbf{L}_{q,\mathfrak{J}_{N}}\right)$,
formula (95) is obtained for all $n\geqslant 2$ by induction, taking
$g=\sigma_{2n-2}$ in (98) and $\eta_{n}(x)=H_{n}[\sigma_{n-2}](x)$,
$\theta_{n}(x)=\Theta_{n}[\sigma_{n-2}](x)$ in (99).

Integral relations of type (95) are effective for the numerical computation of
the partial sums (77) and (78), as seen in [30, 32].

##  7 Integral representation for the derivative

Since $e_{\mathfrak{I}_{N}}^{h}(\rho,\cdot)\in AC[0,b]$, it is worthwhile
looking for an integral representation of the derivative of
$e_{\mathfrak{I}_{N}}^{h}(\rho,x)$. Differentiating (16) we obtain

$\displaystyle(e_{\mathfrak{I}_{N}}^{h})^{\prime}(\rho,x)$
$\displaystyle=\widetilde{e}_{h}^{\prime}(\rho,x)+\sum_{k=0}^{N}\alpha_{k}\widetilde{e}_{h}(\rho,x_{k})H(x-x_{k})\widehat{s}^{\prime}_{k}(\rho,x-x_{k})$
$\displaystyle\quad+\sum_{J\in\mathcal{I}_{N}}\alpha_{J}H(x-x_{j_{|J|}})\widetilde{e}_{h}(\rho,x_{j_{1}})\left(\prod_{l=1}^{|J|-1}\widehat{s}_{j_{l}}(\rho,x_{j_{l+1}}-x_{j_{l}})\right)\widehat{s}_{j_{|J|}}^{\prime}(\rho,x-x_{j_{|J|}}).$

Differentiating (18) and using that
$\widehat{H}_{k}(x,x)=\frac{1}{2}\int_{0}^{x}q(t+x_{k})dt$, we obtain

$\displaystyle\widehat{s}_{k}^{\prime}(\rho,x)$ $\displaystyle=\cos(\rho
x)+\frac{1}{2}\frac{\sin(\rho
x)}{\rho}\int_{0}^{x}q(t+x_{k})dt+\int_{0}^{x}\partial_{x}\widehat{H}_{k}(x,t)\frac{\sin(\rho
t)}{\rho}dt.$

Denote

$w(y,x):=\frac{1}{2}\int_{y}^{x}q(s)ds\quad\mbox{ for }\;x,y\in[0,b].$ (100)

Hence, the derivative $\widehat{s}_{k}^{\prime}(\rho,x-x_{k})$ can be written
as

$\widehat{s}_{k}^{\prime}(\rho,x-x_{k})=\cos(\rho(x-x_{k}))+\int_{-(x-x_{k})}^{x-x_{k}}\widetilde{K}_{k}^{1}(x,t)e^{i\rho
t}dt,$ (101)

where
$\displaystyle\widetilde{K}_{k}^{1}(x,t)=w(x_{k},x)+\frac{1}{2}\int_{|t|}^{x-x_{k}}\partial_{x}\widehat{H}_{k}(x,t)dt$.

On the other hand, differentiation of (17) and the Goursat conditions for
$\widetilde{K}^{h}(x,t)$ lead to the equality

$\tilde{e}^{\prime}_{h}(\rho,x)=(i\rho+w(0,x))e^{i\rho x}+h\cos(\rho
x)+\int_{-x}^{x}\partial_{x}\widetilde{K}^{h}(x,t)e^{i\rho t}dt.$ (102)

Using the fact that

$\cos(\rho A)\int_{-B}^{B}f(t)e^{i\rho
t}dt=\int_{-(B+A)}^{B+A}\frac{1}{2}\left(\chi_{[-(B+A),B-A]}(t)f(t-A)+\chi_{[A-B,B+A]}(t)f(t+A)\right)e^{i\rho
t}dt$

for $A,B>0$ and $f\in L_{2}(\mathbb{R})$ with
$\operatorname{Supp}(f)\subset[-B,B]$, we obtain

$\displaystyle\tilde{e}_{h}(\rho,x_{j})\widehat{s}_{k}^{\prime}(\rho,x-x_{k})$
$\displaystyle=e^{i\rho
x_{j}}\cos(\rho(x-x_{k}))+\mathcal{F}\left[\widehat{K}_{x_{j},x_{k}}(x,t)\right],$

where

$\displaystyle\widehat{K}_{x_{j},x_{k}}(x,t)$
$\displaystyle=\chi_{[x_{k}-x-x_{j},x-x_{k}-x_{j}]}(t)\widetilde{K}_{k}^{1}(x,t-x_{j})+\chi_{x_{j}}(t)\widetilde{K}^{h}(x_{j},t)\ast\chi_{x-x_{k}}(t)\widehat{K}_{k}^{1}(x,t)$
$\displaystyle\qquad+\frac{1}{2}\chi_{[x_{k}-x_{j}-x,x_{j}-x+x_{k}]}(t)\widehat{K}^{h}(x_{j},t-x+x_{k})$
$\displaystyle\qquad+\frac{1}{2}\chi_{[x-x_{k}-x_{j},x-x_{k}+x_{j}]}(t)\widehat{K}^{h}(x_{j},t+x-x_{k})\Big{)}.$

By Lemma 9 the support of $\widehat{K}_{x_{j},x_{k}}(x,t)$ belongs to
$[x_{k}-x-x_{j},x-x_{k}+x_{j}]$. Using the equality

$\prod_{l=1}^{|J|-1}\widehat{s}_{j_{l}}(\rho,x_{j_{l+1}}-x_{j_{l}})=\mathcal{F}\left\\{\left(\prod_{l=1}^{|J|-1}\right)^{\ast}\left(\chi_{x_{j_{l+1}}-x_{j_{l}}}(t)\widetilde{K}_{k}(x_{j_{l+1}},t)\right)\right\\}$

we have

$(e_{\mathfrak{I}_{N}}^{h})^{\prime i\rho x}+h\cos(\rho
x)+\sum_{k=0}^{N}\alpha_{k}H(x-x_{k})e^{i\rho
x_{k}}\cos(\rho(x-x_{k}))+\mathcal{F}\left\\{E_{\mathfrak{I}_{N}}^{h}(x,t)\right\\}$

where

$\displaystyle E_{\mathfrak{I}_{N}}^{h}(x,t)$
$\displaystyle=\chi_{x}(t)\partial_{x}\widetilde{K}^{h}(x,t)+\sum_{k=0}^{N}\alpha_{k}H(x-x_{k})\widehat{K}_{x_{k},x_{k}}(x,t)$
$\displaystyle\quad+\sum_{J\in\mathcal{I}_{N}}\alpha_{J}H(x-x_{j_{|J|}})\widehat{K}_{x_{j_{1}},x_{j_{|J|}}}(x,t)\ast\left(\prod_{l=1}^{|J|-1}\right)^{\ast}\left(\chi_{x_{j_{l+1}}-x_{j_{l}}}(t)\widetilde{K}_{k}(x_{j_{l+1}},t)\right).$

Again, by Lemma 9 the support of $E_{\mathfrak{I}_{N}}^{h}(x,t)$ belongs to
$[-x,x]$. Since $e^{i\rho x_{k}}\cos(\rho(x-x_{k}))=\frac{1}{2}e^{i\rho
x}\left(1+e^{-2i\rho(x-x_{k})}\right)$, we obtain the following
representation.

###### Theorem 42

The derivative $(e_{\mathfrak{I}_{N}}^{h})^{\prime}(\rho,x)$ admits the
integral representation

$\displaystyle(e_{\mathfrak{I}_{N}}^{h})^{\prime}(\rho,x)$
$\displaystyle=\left(i\rho+w(0,x)+\frac{1}{2}\sigma_{\mathfrak{I}_{N}}(x)\right)e^{i\rho
x}+h\cos(\rho x)$
$\displaystyle\quad+\sum_{k=0}^{N}\frac{\alpha_{k}}{2}H(x-x_{k})e^{-2i\rho(x-x_{k})}+\int_{-x}^{x}E_{\mathfrak{I}_{N}}^{h}(x,t)e^{i\rho
t}dt,$ (103)

where $E_{\mathfrak{I}_{N}}^{h}(x,\cdot)\in L_{2}(-x,x)$ for all $x\in(0,b]$.

###### Corollary 43

The derivatives of the solutions $c_{\mathfrak{I}_{N}}^{h}(\rho,x)$ and
$s_{\mathfrak{I}_{N}}(\rho,x)$ admit the integral representations

$\displaystyle(c_{\mathfrak{I}_{N}}^{h})^{\prime}(\rho,x)$
$\displaystyle=-\rho\sin(\rho
x)+\left(h+w(0,x)+\frac{1}{2}\sigma_{\mathfrak{I}_{N}}(x)\right)\cos(\rho x)$
$\displaystyle\quad+\sum_{k=0}^{N}\frac{\alpha_{k}}{2}H(x-x_{k})\cos(2\rho(x-x_{k}))+\int_{0}^{x}M_{\mathfrak{I}_{N}}^{h}(x,t)\cos(\rho
t)dt,$ (104) $\displaystyle s^{\prime}_{\mathfrak{I}_{N}}(\rho,x)$
$\displaystyle=\cos(\rho
x)+\left(w(0,x)+\frac{1}{2}\sigma_{\mathfrak{I}_{N}}(x)\right)\frac{\sin(\rho
x)}{\rho}$
$\displaystyle\quad-\sum_{k=0}^{N}\alpha_{k}H(x-x_{k})\frac{\sin(2\rho(x-x_{k}))}{2\rho}+\int_{0}^{x}R_{\mathfrak{I}_{N}}(x,t)\frac{\sin(\rho
t)}{\rho}dt,$ (105)

where

$\displaystyle N_{\mathfrak{I}_{N}}^{h}(x,t)$
$\displaystyle=E_{\mathfrak{I}_{N}}^{h}(x,t)+E_{\mathfrak{I}_{N}}^{h}(x,-t)\quad$
(106) $\displaystyle R_{\mathfrak{I}_{N}}^{h}(x,t)$
$\displaystyle=E_{\mathfrak{I}_{N}}^{h}(x,t)-E_{\mathfrak{I}_{N}}^{h}(x,-t),$
(107)

defined for $x\in[0,b]$ and $|t|\leqslant x$.

###### Corollary 44

The derivatives of the solutions $c_{\mathfrak{I}_{N}}^{h}(\rho,x)$ and
$s_{\mathfrak{I}_{N}}(\rho,x)$ admit the NSBF representations

$\displaystyle(c_{\mathfrak{I}_{N}}^{h})^{\prime}(\rho,x)$
$\displaystyle=-\rho\sin(\rho
x)+\left(h+w(0,x)+\frac{1}{2}\sigma_{\mathfrak{I}_{N}}(x)\right)\cos(\rho x)$
$\displaystyle\quad+\sum_{k=0}^{N}\frac{\alpha_{k}}{2}H(x-x_{k})\cos(2\rho(x-x_{k}))+\sum_{n=0}^{\infty}(-1)^{n}l_{n}(x)j_{2n}(\rho
x),$ (108) $\displaystyle s^{\prime}_{\mathfrak{I}_{N}}(\rho,x)$
$\displaystyle=\cos(\rho
x)+\left(w(0,x)+\frac{1}{2}\sigma_{\mathfrak{I}_{N}}(x)\right)\frac{\sin(\rho
x)}{\rho}$
$\displaystyle\quad-\sum_{k=0}^{N}\alpha_{k}H(x-x_{k})\frac{\sin(2\rho(x-x_{k}))}{2\rho}+\sum_{n=0}^{\infty}(-1)^{n}r_{n}(x)j_{2n+1}(\rho
x),$ (109)

where $\\{l_{n}(x)\\}_{n=0}^{\infty}$ and $\\{r_{n}(x)\\}_{n=0}^{\infty}$ are
the coefficients of the Fourier-Legendre expansion of
$M_{\mathfrak{I}_{N}}^{h}(x,t)$ and $R_{\mathfrak{J}_{N}}(x,t)$ in terms of
the even and odd Legendre polynomials, respectively.

##  8 Conclusions

The construction of a transmutation operator that transmute the solutions of
equation $v^{\prime\prime}+\lambda v=0$ into solutions of (1) is presented.
The transmutation operator is obtained from the closed form of the general
solution of equation (1). It was shown how to construct the image of the
transmutation operator on the set of polynomials, this with the aid of the
SPPS method. A Fourier-Legendre series representation for the integral
transmutation kernel is obtained, together with a representation for the
solutions $c_{\mathfrak{I}_{N}}^{h}(\rho,x)$, $s_{\mathfrak{I}_{N}}(\rho,x)$
and their derivatives as Neumann series of Bessel functions, together with
integral recursive relations for the construction of the Fourier-Legendre
coefficients. The series (75), (76), (108), (109) are useful for solving
direct and inverse spectral problems for (1), as shown for the regular case
[28, 29, 30, 32].

## Acknowledgments

Research was supported by CONACYT, Mexico via the project 284470.

## References

  * [1] O. Akcay, The representation of the solution of Sturm-Liouville equation with discontinuity conditions, Acta Math. Scientia. 38B(4) (2018), 1195-1213.
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