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 and , i.e., when .
The solutions and , as per Definition 2.1, can be expressed as
(4.1)
(4.2)
where is a solution of (1.1) on obeying the condition
and is a solution of (1.1) on obeying the condition
Here, it should be noted that the existence of an extension of from to the solution on relies on M being non-singular.
As in the classical case, for , we may find the conjugate Jost solution. In this case, for , we have
(4.3)
where the transfer condition holds at .
Since and are equal to and on , we see that and are linearly independent on for all . Therefore can be expressed as a linear combination of and on with coefficients and giving that on
(4.4)
Note that (4.4) also holds on as .
For , , we extend to by
(4.5)
and for ,
(4.6)
Theorem 4.1 For , and satisfy the following equality:
Proof We begin by obtaining an expression for the solution in terms of the conjugate solutions and for . In a similar manner to the classical case, we obtain
Thus, and are linearly independent for and consequently,
(4.7)
where , are independent of x. Equation (4.7) and its x-derivative give the matrix equation
which has a solution
Thus, from (4.5),
and, by (4.6),
Combining the expressions for and with (4.7) gives
(4.8)
Substituting (4.8) into (4.4) gives
Now, since for , ,
□
The reflection coefficient can be defined for this case as
(4.9)
By Theorem 3.2, all eigenvalues, , for the scattering problem (1.1)-(1.2) on the line are negative. So, for each eigenvalue, ζ is of the form
Let , . Then, for large , by (2.1). So there exists a such that for all . Direct computation gives that
is a solution of differential equation (1.1) on which in not in . 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 . 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 is 2-dimensional. Combining these facts gives that the geometric multiplicity of an eigenvalue is 1. Thus, if is an eigenvalue, then for some , for all . Consequently, is an eigenvalue if and only if , i.e., from (4.5), . Hence, for , with an eigenvalue of (3.1)-(3.2), we have
Conversely, if , then and are linearly dependent making . Thus is an eigenvalue and for some .
Theorem 4.2 The function has a finite set of zeros.
Proof We know, from the above reasoning, that if then for some . It should be noted that is analytic on the upper half-plane and continuous there and up to the boundary.
We now compute the asymptotic form of . For large in the upper half-plane, taking in (4.5), we obtain
Substituting into the above the asymptotic approximations to the Jost solutions for large values of , we get, for ,
(4.11)
Here, we observe that and , thus the integral term in (4.11) is uniformly bounded in the upper half-plane. In addition, as and M has real entries, not both of and can simultaneously be zero. Hence is bounded away from zero for large ζ in the upper half-plane giving that does not have an unbounded sequence of zeros.
Using the same approach as in [6], by Lemma 4.1, is bounded in a neighbourhood of the real axis, so the zeros of 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 .
From (4.6), we have
Taking , we obtain
For large, we get
(4.12)