We now consider the Jost solutions to problem (3.3). Since this problem can be considered as two half-line problems, solutions to (3.3) can be given in terms of the classical Jost solutions {f}_{+}(x,\zeta ) and {f}_{-}(x,\zeta ), *i.e.*, when M=I.

The solutions {f}_{+,M}(x,\zeta ) and {f}_{-,M}(x,\zeta ), as per Definition 2.1, can be expressed as

{f}_{+,M}(x,\zeta ):=\{\begin{array}{ll}{f}_{+}(x,\zeta ),& x>0,\\ {h}_{1}(x,\zeta ),& x<0,\end{array}

(4.1)

{f}_{-,M}(x,\zeta ):=\{\begin{array}{ll}{f}_{-}(x,\zeta ),& x<0,\\ {h}_{2}(x,\zeta ),& x>0,\end{array}

(4.2)

where {h}_{1}(x,\zeta ) is a solution of (1.1) on (-\mathrm{\infty},0) obeying the condition

\left(\begin{array}{c}{h}_{1}({0}^{-},\zeta )\\ {h}_{1}^{\prime}({0}^{-},\zeta )\end{array}\right)={M}^{-1}\left(\begin{array}{c}{f}_{+}({0}^{+},\zeta )\\ {f}_{+}^{\prime}({0}^{+},\zeta )\end{array}\right),

and {h}_{2}(x,\zeta ) is a solution of (1.1) on (0,\mathrm{\infty}) obeying the condition

\left(\begin{array}{c}{h}_{2}({0}^{+},\zeta )\\ {h}_{2}^{\prime}({0}^{+},\zeta )\end{array}\right)=M\left(\begin{array}{c}{f}_{-}({0}^{-},\zeta )\\ {f}_{-}^{\prime}({0}^{-},\zeta )\end{array}\right).

Here, it should be noted that the existence of an extension of {f}_{+} from (0,\mathrm{\infty}) to the solution {f}_{+,M} on \mathbb{R}\setminus \{0\} relies on *M* being non-singular.

As in the classical case, for \zeta =\xi \in \mathbb{R}, we may find the conjugate Jost solution. In this case, for {\overline{f}}_{+,M}(x,\xi ), we have

{\overline{f}}_{+,M}(x,\xi )=\{\begin{array}{ll}{\overline{f}}_{+}(x,\xi )={f}_{+}(x,-\xi ),& x>0,\\ {\overline{h}}_{1}(x,\xi )={h}_{1}(x,-\xi ),& x<0,\end{array}

(4.3)

where the transfer condition holds at x=0.

Since {f}_{+,M}(x,\xi ) and {\overline{f}}_{+,M}(x,\xi ) are equal to {f}_{+}(x,\xi ) and {\overline{f}}_{+}(x,\xi ) on (0,\mathrm{\infty}), we see that {f}_{+,M} and {\overline{f}}_{+,M} are linearly independent on (0,\mathrm{\infty}) for all \xi \in \mathbb{R}\setminus \{0\}. Therefore {h}_{2} can be expressed as a linear combination of {f}_{+,M} and {\overline{f}}_{+,M} on (0,\mathrm{\infty}) with coefficients A(\xi ) and B(\xi ) giving that on (0,\mathrm{\infty})

{f}_{-,M}(x,\xi )=A(\xi ){\overline{f}}_{+,M}(x,\xi )+B(\xi ){f}_{+,M}(x,\xi ).

(4.4)

Note that (4.4) also holds on (-\mathrm{\infty},0) as detM=1.

For \mathrm{\Im}(\zeta )\ge 0, \zeta \ne 0, we extend A(\xi ) to {\mathbb{C}}^{+} by

A(\zeta )=\frac{-1}{2i\zeta}W[{f}_{+,M}(x,\zeta ),{f}_{-,M}(x,\zeta )],

(4.5)

and for \xi \in \mathbb{R}\setminus \{0\},

B(\xi )=\frac{1}{2i\xi}W[{f}_{+,M}(x,-\xi ),{f}_{-,M}(x,\xi )].

(4.6)

**Theorem 4.1** *For* \xi \in \mathbb{R}\setminus \{0\}, A(\xi ) *and* B(\xi ) *satisfy the following equality*:

{|A(\xi )|}^{2}-{|B(\xi )|}^{2}=1.

*Proof* We begin by obtaining an expression for the solution {f}_{+,M}(x,\xi ) in terms of the conjugate solutions {f}_{-,M}(x,\xi ) and {\overline{f}}_{-,M}(x,\xi ) for \xi \in \mathbb{R}\setminus \{0\}. In a similar manner to the classical case, we obtain

det\left[\begin{array}{cc}{\overline{f}}_{-,M}& {f}_{-,M}\\ {\overline{f}}_{-,M}^{\prime}& {f}_{-,M}^{\prime}\end{array}\right]=W[{\overline{f}}_{-,M},{f}_{-,M}]=-2i\xi .

Thus, {\overline{f}}_{-,M}(x,\xi ) and {f}_{-,M}(x,\xi ) are linearly independent for \xi \in \mathbb{R}\setminus \{0\} and consequently,

{f}_{+,M}(x,\xi )={G}_{1}(\xi ){\overline{f}}_{-,M}(x,\xi )+{G}_{2}(\xi ){f}_{-,M}(x,\xi ),

(4.7)

where {G}_{1}(\xi ), {G}_{2}(\xi ) are independent of *x*. Equation (4.7) and its *x*-derivative give the matrix equation

\left[\begin{array}{c}{f}_{+,M}\\ {f}_{+,M}^{\prime}\end{array}\right]=\left[\begin{array}{cc}{\overline{f}}_{-,M}& {f}_{-,M}\\ {\overline{f}}_{-,M}^{\prime}& {f}_{-,M}^{\prime}\end{array}\right]\left[\begin{array}{c}{G}_{1}\\ {G}_{2}\end{array}\right],

which has a solution

\begin{array}{rcl}\left[\begin{array}{c}{G}_{1}\\ {G}_{2}\end{array}\right]& =& {\left[\begin{array}{cc}{\overline{f}}_{-,M}& {f}_{-,M}\\ {\overline{f}}_{-,M}^{\prime}& {f}_{-,M}^{\prime}\end{array}\right]}^{-1}\left[\begin{array}{c}{f}_{+,M}\\ {f}_{+,M}^{\prime}\end{array}\right]\\ =& \frac{i}{2\xi}\left[\begin{array}{cc}{f}_{-,M}^{\prime}& -{f}_{-,M}\\ -{\overline{f}}_{-,M}^{\prime}& {\overline{f}}_{-,M}\end{array}\right]\left[\begin{array}{c}{f}_{+,M}\\ {f}_{+,M}^{\prime}\end{array}\right].\end{array}

Thus, from (4.5),

\begin{array}{rcl}{G}_{1}(\xi )& =& \frac{i}{2\xi}({f}_{+,M}{f}_{-,M}^{\prime}-{f}_{-,M}{f}_{+,M}^{\prime})\\ =& \frac{i}{2\xi}W[{f}_{+,M},{f}_{-,M}]\\ =& A(\xi ),\end{array}

and, by (4.6),

\begin{array}{rcl}{G}_{2}(\xi )& =& \frac{i}{2\xi}({\overline{f}}_{-,M}{f}_{+,M}^{\prime}-{\overline{f}}_{-,M}^{\prime}{f}_{+,M})\\ =& \frac{i}{2\xi}W[{\overline{f}}_{-,M},{f}_{+,M}]\\ =& -\overline{B}(\xi ).\end{array}

Combining the expressions for {G}_{1}(\xi ) and {G}_{2}(\xi ) with (4.7) gives

{f}_{+,M}(x,\xi )=A(\xi ){\overline{f}}_{-,M}(x,\xi )-\overline{B}(\xi ){f}_{-,M}(x,\xi ).

(4.8)

Substituting (4.8) into (4.4) gives

\begin{array}{rcl}{f}_{-,M}(x,\xi )& =& A(\xi ){\overline{f}}_{+,M}(x,\xi )+B(\xi ){f}_{+,M}(x,\xi )\\ =& A(\xi )[\overline{A}(\xi ){f}_{-,M}(x,\xi )-B(\xi ){\overline{f}}_{-,M}(x,\xi )]\\ +B(\xi )[A(\xi ){\overline{f}}_{-,M}(x,\xi )-\overline{B}(\xi ){f}_{-,M}(x,\xi )]\\ =& ({|A(\xi )|}^{2}-{|B(\xi )|}^{2}){f}_{-,M}(x,\xi ).\end{array}

Now, since {f}_{-,M}(x,\xi )\not\equiv 0 for \xi \in \mathbb{R}\setminus \{0\}, x\in \mathbb{R},

{|A(\xi )|}^{2}-{|B(\xi )|}^{2}=1.

□

The reflection coefficient can be defined for this case as

R(\xi )=\frac{B(\xi )}{A(\xi )},\phantom{\rule{1em}{0ex}}\xi \in \mathbb{R}.

(4.9)

By Theorem 3.2, all eigenvalues, {\zeta}^{2}, for the scattering problem (1.1)-(1.2) on the line are negative. So, for each eigenvalue, *ζ* is of the form

\zeta =i\eta \phantom{\rule{1em}{0ex}}\text{for some}\eta \in {\mathbb{R}}^{+}.

Let \zeta =i\eta, \eta \in {\mathbb{R}}^{+}. Then, for large x>0, {f}_{+,M}(x,i\eta )={f}_{+}(x,i\eta )={e}^{-\eta x}(1+o(1)) by (2.1). So there exists *a* such that |{f}_{+}(x,i\eta )|>0 for all x\ge a. Direct computation gives that

y(x)={\int}_{a}^{x}\frac{{f}_{+}(x)}{{f}_{+}^{2}(t)}\phantom{\rule{0.2em}{0ex}}dt=\frac{{e}^{\eta x}}{2\eta}(1+o(1))

is a solution of differential equation (1.1) on [a,\mathrm{\infty}) which in not in {L}^{2}[a,\mathrm{\infty}). Elementary existence theory for differential equations gives that *y* can be extended to a solution of differential equation (1.1) with transfer condition (1.2) on (-\mathrm{\infty},0)\cup (0,\mathrm{\infty}). Since the transfer matrix *M* is invertible and (1.1) is second-order, the solution space to (1.1) with transfer condition (1.2) on (-\mathrm{\infty},0)\cup (0,\mathrm{\infty}) is 2-dimensional. Combining these facts gives that the geometric multiplicity of an eigenvalue is 1. Thus, if {\zeta}^{2} is an eigenvalue, then for some k\in \mathbb{C}\setminus \{0\}, {f}_{-,M}(x,\zeta )=k{f}_{+,M}(x,\zeta ) for all x\in {\mathbb{R}}^{+}. Consequently, {\zeta}^{2} is an eigenvalue if and only if W[{f}_{-,M},{f}_{+,M}](x,\zeta )=0, *i.e.*, from (4.5), A(\zeta )=0. Hence, for \zeta =i\eta, \eta \in {\mathbb{R}}^{+} with {\zeta}^{2} an eigenvalue of (3.1)-(3.2), we have

Conversely, if A(\zeta )=0, then {f}_{-,M}(x,\zeta ) and {f}_{+,M}(x,\zeta ) are linearly dependent making {f}_{+,M},{f}_{-,M}\in {L}^{2}(-\mathrm{\infty},\mathrm{\infty}). Thus {\zeta}^{2} is an eigenvalue and \zeta =i\eta for some \eta \in {\mathbb{R}}^{+}.

**Theorem 4.2** *The function* A(\zeta ) *has a finite set of zeros*.

*Proof* We know, from the above reasoning, that if A(\zeta )=0 then \zeta =i\eta for some \eta \in {\mathbb{R}}^{+}. It should be noted that A(\zeta ) is analytic on the upper half-plane and continuous there and up to the boundary.

We now compute the asymptotic form of A(\zeta ). For |\zeta | large in the upper half-plane, taking x\to {0}^{+} in (4.5), we obtain

\begin{array}{rcl}A(\zeta )& =& \frac{-1}{2i\zeta}det\left[\begin{array}{cc}{f}_{+,M}({0}^{+},\zeta )& {f}_{-,M}({0}^{+},\zeta )\\ {f}_{+,M}^{\prime}({0}^{+},\zeta )& {f}_{-,M}^{\prime}({0}^{+},\zeta )\end{array}\right]\\ =& \frac{-1}{2i\zeta}det\left[\begin{array}{cc}\begin{array}{c}{f}_{+,M}({0}^{+},\zeta )\\ {f}_{+,M}^{\prime}({0}^{+},\zeta )\end{array}& M\left(\begin{array}{c}{f}_{-,M}({0}^{-},\zeta )\\ {f}_{-,M}^{\prime}({0}^{-},\zeta )\end{array}\right)\end{array}\right]\\ =& \frac{-1}{2i\zeta}det\left[\begin{array}{cc}{f}_{+}(0,\zeta )& {m}_{11}{f}_{-}(0,\zeta )+{m}_{12}{f}_{-}^{\prime}(0,\zeta )\\ {f}_{+}^{\prime}(0,\zeta )& {m}_{21}{f}_{-}(0,\zeta )+{m}_{22}{f}_{-}^{\prime}(0,\zeta )\end{array}\right]\\ =& \frac{-1}{2i\zeta}\{{f}_{+}(0,\zeta )[{m}_{21}{f}_{-}(0,\zeta )+{m}_{22}{f}_{-}^{\prime}(0,\zeta )]-{f}_{+}^{\prime}(0,\zeta )[{m}_{11}{f}_{-}(0,\zeta )+{m}_{12}{f}_{-}^{\prime}(0,\zeta )]\}.\end{array}

Substituting into the above the asymptotic approximations to the Jost solutions for large values of |\zeta |, we get, for \mathrm{\Im}(\zeta )\ge 0,

A(\zeta )=\frac{{m}_{12}}{2i}\zeta +\frac{{m}_{11}+{m}_{22}}{2}+\frac{{m}_{12}}{2}{\int}_{-\mathrm{\infty}}^{\mathrm{\infty}}cos(-\zeta \tau )q(\tau ){e}^{i\zeta |\tau |}\phantom{\rule{0.2em}{0ex}}d\tau +O\left(\frac{1}{1+|\zeta |}\right).

(4.11)

Here, we observe that |cos(-\zeta \tau ){e}^{i\zeta |\tau |}|\le 1 and q\in {L}^{1}(\mathbb{R}), thus the integral term in (4.11) is uniformly bounded in the upper half-plane. In addition, as detM=1 and *M* has real entries, not both of {m}_{12} and {m}_{11}+{m}_{22} can simultaneously be zero. Hence |A(\zeta )| is bounded away from zero for large *ζ* in the upper half-plane giving that A(\zeta ) does not have an unbounded sequence of zeros.

Using the same approach as in [6], by Lemma 4.1, \frac{1}{|A(\zeta )|} is bounded in a neighbourhood of the real axis, so the zeros of A(\zeta ) have no finite accumulation point and are thus a finite set. □

From Theorem 4.2 and the reasoning preceding Theorem 4.2, we obtain the following corollary.

**Corollary 4.3** *The scattering problem with a transfer condition has a finite number of eigenvalues*, *all of which are simple*.

We note for reference the following properties of B(\xi ).

From (4.6), we have

B(\xi )=\frac{1}{2i\xi}W[{\overline{f}}_{+,M}(x,\xi ){f}_{-,M}(x,\xi )],\phantom{\rule{1em}{0ex}}\xi \in \mathbb{R}\setminus \{0\}.

Taking x\to {0}^{+}, we obtain

\begin{array}{rcl}B(\xi )& =& \frac{-1}{2i\xi}det\left[\begin{array}{cc}{\overline{f}}_{+,M}({0}^{+},\xi )& {f}_{-,M}({0}^{+},\xi )\\ {\overline{f}}_{+,M}^{\prime}({0}^{+},\xi )& {f}_{-,M}^{\prime}({0}^{+},\xi )\end{array}\right]\\ =& \frac{-1}{2i\xi}det\left[\begin{array}{cc}\begin{array}{c}{\overline{f}}_{+}(0,\xi )\\ {\overline{f}}_{+}^{\prime}(0,\xi )\end{array}& M\left(\begin{array}{c}{f}_{-}(0,\xi )\\ {f}_{-}^{\prime}(0,\xi )\end{array}\right)\end{array}\right].\end{array}

For |\xi | large, we get

B(\xi )=-\frac{{m}_{12}}{2i}\xi +\frac{{m}_{22}-{m}_{11}}{2}-\frac{{m}_{12}}{2}{\int}_{-\mathrm{\infty}}^{\mathrm{\infty}}cos(\xi \tau )q(\tau ){e}^{-i\xi \tau}\phantom{\rule{0.2em}{0ex}}d\tau +O\left(\frac{1}{1+|\xi |}\right).

(4.12)