Forward scattering on the line with a transfer condition

We consider scattering on the line with a transfer condition at the origin. Under suitable growth conditions on the potential, the spectrum consists of a finite number of eigenvalues which are negative real numbers, while the remainder is continuous spectrum which is comprised of the positive real axis. Asymptotics are provided for the Jost solutions. Conditions which characterize transfer conditions resulting in self-adjoint problems are found. Properties are given of the scattering coefficient linking it to the spectrum.

MSC:34L25, 47N50, 34B10, 34B40.

The mathematical analysis of scattering theory has been a major area of interest in mathematics and physics research since the latter half of the twentieth century. In this work we investigate forward scattering for the differential equation

in L^2(-\mathrm{\infty },0)\oplus L^2(0,\mathrm{\infty })=L^2(\mathbb{R}) with the point transfer condition

Here, the entries of M are taken to be real, q\in L^2(\mathbb{R}) is assumed to be real-valued and obey the growth condition

Note that (1.3) gives that q\in L^1. Denote f(0^{+}):={lim}_{t\downarrow 0}f(t) and f(0^{-}):={lim}_{t\uparrow 0}f(t). The operator L in L^2(\mathbb{R}) is defined by

on \mathbb{R}\setminus \{0\} for y in the domain D(L) of L, where

Only point transfer matrices at the origin will be considered, and henceforth we will refer to them as transfer matrices. In [1] conditions for self-adjointness and limit point/limit circle criteria are considered for a more general problem.

In the physical context, the transfer matrix represents a change of medium which affects the incident wave as represented by components of the matrix. Livšic in [[2], p.7] gives a concrete physical example of a scattering problem of the form given above. He considers a uniform, infinite string, attached at the point x=0 to a transverse spring. The behaviour at the point of attachment is described by what we have called the transfer matrix M.

Our transfer matrices will be real constant transfer matrices, i.e., all components will be constants. The theory could be extended to eigenparameter-dependent transfer matrices, this will be considered in future studies. Gordon and Pearson, in [3], characterized the self-adjoint constant point transfer matrices as well as eigenparameter-dependent ones.

The Jost solutions of (1.1) represent oscillations moving left or right on ℝ. The scattering data is defined in terms of these solutions. The classical Jost solutions correspond to the case where the transfer matrix M is the identity. In particular, asymptotic approximations to the classical Jost solutions and their derivatives, as well as some basic relations regarding them, are given. The Jost solutions for the scattering problem with a transfer condition (M\ne I) are then expressed in terms of the classical Jost solutions. Consequently, we consider functional analytic aspects of the operator L such as location of eigenvalues and continuous spectrum, multiplicity of eigenvalues, properties of the scattering coefficients/scattering matrix and the reflection coefficient. We show that the operator L has finitely many eigenvalues, they are negative and simple, and that the positive real axis [0,\mathrm{\infty }) is the continuous spectrum of L.

This paper is structured as follows. The Jost solutions of the scattering problem (1.1) and (1.2) are defined in Section 2. In addition, some basic properties and asymptotic approximations of the classical Jost solutions and their derivatives are recalled. In Section 3, the scattering problem is reformulated as a system spectral problem, and we prove that L is self-adjoint if and only if detM=1. Moreover, it is shown that all the eigenvalues are negative and that the entire non-negative real half-line is the continuous spectrum of L. In Section 4, since the scattering problem (1.1) and (1.2) can be considered as two half-line problems, the Jost solutions are expressed in terms of the classical Jost solutions. Asymptotics for the scattering coefficients are determined, and we prove that the problem has finitely many eigenvalues all of which are simple. In the final section, Section 5, we give a relationship between the norming constants and the derivative of the scattering coefficient.

The results obtained in this paper provide the foundation needed in order to consider the associated inverse problem, this will be the topic of a subsequent paper.

In this section, asymptotic solutions y(x,\zeta ),(x,\zeta )\in \mathbb{R}\times {\overline{\mathbb{C}}}^{+}, will be developed for equation (1.1) for large values of |x|+|\zeta |, where {\overline{\mathbb{C}}}^{+}=\{\xi +i\eta \mid \xi ,\eta \in \mathbb{R},\eta \ge 0\}. The focus of this section will be on the so-called Jost solutions of (1.1) and their derivatives. The Jost solutions f_{+,M} and f_{-,M} of (1.1) and (1.2) are the solutions of (1.1) and (1.2) defined by their asymptotic behaviour at ±∞ as follows.

Definition 2.1 [[4], p.297]

The Jost solutions f_{+,M}(x,\zeta ) and f_{-,M}(x,\zeta ) are the solutions of (1.1) and (1.2) with

When M=I, the subscript M will be dropped. In this case, the existence and asymptotic behaviour of the Jost solutions are well known, see [5, 6]. In particular,

and

as |x|+|\zeta |\to \mathrm{\infty }, where \eta =\mathrm{\Im }(\zeta ). Here, C(x) is a non-negative, non-increasing function of x and

Let ζ be real and denote \zeta =\xi \in \mathbb{R}. Then as \overline{\xi }=\xi, {\overline{f}}_{+}(x,\xi ) obeys equations (1.1) and (1.2) with M=I. Taking the conjugate of the integral equation

which defines f_{+}, gives

From the above two expressions and the uniqueness of the solution of (2.5), we have

The asymptotic results given earlier in this section obviously hold for the conjugate solution {\overline{f}}_{+}(x,\xi ) but with the simplification that \eta =0 in this case. In particular,

where C(x) is non-increasing and ρ and σ are as defined earlier.

Note that the Wronskian, W[f_{+},{\overline{f}}_{+}], of f_{+} and {\overline{f}}_{+} for \xi \in \mathbb{R} and x\in \mathbb{R} is given by W[f_{+},{\overline{f}}_{+}](x,\xi )=-2i\xi, see [7]. Thus f_{+}(x,\xi ) and {\overline{f}}_{+}(x,\xi )=f_{+}(x,-\xi ) for \xi =\zeta \in \mathbb{R}\setminus \{0\} are independent.

Here we consider the scattering problem on the line with transfer condition (1.2) at x=0. In order for (1.1) and (1.2) to be self-adjoint in L^2((-\mathrm{\infty },0)\cup (0,\mathrm{\infty }))=L^2(\mathbb{R}), the transfer matrix M will have to be of a certain form. Here, we restrict our attention to the most interesting case of M invertible, the case of M non-invertible will be considered elsewhere. The scattering problem can then be treated as two classical half-line problems joined at the origin by matrix condition (1.2).

The operator eigenvalue problem associated with L, of (1.4), can be reformulated as a system eigenvalue problem as follows. Let y_1(t)=y(t), y_2(t)=y(-t) and Y(t)=\left(\begin{array}{c}{y}_1(t)\\ y_2(t)\end{array}\right) and consider the differential operator in L^2(0,\mathrm{\infty })\oplus L^2(0,\mathrm{\infty }) given by

where Q(t)=\left(\begin{array}{cc}q(t)& 0\\ 0& q(-t)\end{array}\right). The domain of T is given by

where U=\left(\begin{array}{cc}1& -m_{11}\\ 0& m_{21}\end{array}\right) and V=\left(\begin{array}{cc}0& -m_{12}\\ 1& m_{22}\end{array}\right). Here, m_{ij}, for i,j=1,2, are the entries of the transfer matrix M. As the norm on L^2(0,\mathrm{\infty })\oplus L^2(0,\mathrm{\infty }), we take

It should be noted here that L and T are unitarily equivalent by the map \mathrm{\Psi }:L^2(\mathbb{R})\to {(L^2(0,\mathrm{\infty }))}^2 given by \mathrm{\Psi }y=\left(\begin{array}{c}{y}_1\\ y_2\end{array}\right), where

It is easily verified that Ψ is a bijective isometry between L^2(\mathbb{R}) and {(L^2(0,\mathrm{\infty }))}^2 with the additional properties that \mathrm{\Psi }(D(L))=D(T) and {\mathrm{\Psi }}^{-1}T\mathrm{\Psi }=L.

The transfer matrix scattering problem can now be posed as

Let F,G\in D(T). Define

where

We now obtain conditions on the transfer matrix which ensure that T is a self-adjoint operator. We begin by defining the minimal and maximal operators associated with T.

The minimal operator associated with T is T_0 which is given by

with

The maximal operator associated with T is T_{\max }=T_0^{\ast }, where

Note that T_0 has deficiency indices (2,2).

Theorem 3.1 If detM\ne 0, then the operator L is a self-adjoint operator if and only if detM=1.

Proof We will prove the result for the operator T and, consequently, it will be true for the operator L.

Let F=\left(\begin{array}{c}{f}_1\\ f_2\end{array}\right),G=\left(\begin{array}{c}{g}_1\\ g_2\end{array}\right)\in {(C^2(0,\mathrm{\infty }))}^2 such that F(x)=G(x)=0 for all x\ge K. Then, since q\in L^2(\mathbb{R}), F,G\in D(T_{\max }). Also,

As the deficiency indices of T_0 are (2,2), we need to restrict the domain of T_{\max } by two boundary conditions to ensure self-adjointness, see Weidmann [[8], p.53]. For the operator to be self-adjoint with two linear domain conditions, they must ensure that (f_1{\overline{g}}_1^{\prime }-f_1^{\prime }{\overline{g}}_1+f_2{\overline{g}}_2^{\prime }-f_2^{\prime }{\overline{g}}_2)(0)=0.

We now show what the above condition implies in terms of the transfer matrix M,

Denote J=\left(\begin{array}{cc}0& I\\ -I& 0\end{array}\right), then we have that J^T=-J and J^2=-I.

Let U and V be as in (3.2), then if F,G\in D(T) we have

and

So

where \left[\begin{array}{c}detM\\ m_{22}\\ 0\\ m_{21}\end{array}\right] and \left[\begin{array}{c}0\\ -m_{12}\\ detM\\ -m_{11}\end{array}\right] are linearly independent. Thus we require

Thus {(detM)}^2=-m_{12}{m}_{21}+m_{11}{m}_{22}=detM giving that detM=1 since M is invertible. □

It should be noted that in [1] a condition on the transfer matrix M is given for self-adjointness; however, for the case of detM\ne 0, where the entries of M are real, the proof presented above is substantially simpler.

Throughout the remainder of the paper, we will now assume that detM=1.

Theorem 3.2 Let detM=1. All eigenvalues (if any) of L as defined in (1.4), (1.5), and consequently of T, are negative.

Proof For \lambda ={\zeta }^2\in \mathbb{R}\setminus \{0\}, f_{+} and {\overline{f}}_{+} constitute an independent set of solutions to equation (1.1). If L has a positive eigenvalue \lambda ={\xi }^2>0, where \xi >0, then L has an eigenfunction of the form

for x>0. From (2.1) and (2.7)

for |\xi |+x large, \xi \in \mathbb{R}, x>0. Hence

if |c_1|+|c_2|\ne 0 and L has no λ positive eigenvalues.

From the definition of f_{+}(x,0), we have that f_{+}(x,0)\to 1 as x\to \mathrm{\infty }, so f_{+}(x,0)\notin L^2(0,\mathrm{\infty }). Observe that

is a solution of (1.1) which is asymptotic to x for x\to \mathrm{\infty }, [[7], p.173], and therefore linearly independent of f_{+}(x,0). But a{f}_{+}(x,0)+bg(x) is asymptotic to a+bx as x\to \mathrm{\infty }, and thus not in L^2(0,\mathrm{\infty }) unless |a|+|b|=0. Hence {\xi }^2=0 is not an eigenvalue of L. □

We now study the continuous spectrum.

Theorem 3.3 The continuous spectrum of T (and thus of L) is [0,\mathrm{\infty }).

Proof From [[9], p.66] or [[10], p.251], as T is self-adjoint, the spectrum of T, \sigma (T), is comprised of continuous and point spectrum only, i.e., T has no residual spectrum. In addition, the continuous spectrum of T is real. By [6] the essential spectrum of the minimal operator generated by T is [0,\mathrm{\infty }) and this is the same as the essential spectrum of T, see [10]. Thus, as T has no residual spectrum, we have that the continuous spectrum of T is [0,\mathrm{\infty }). □

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

where h_1(x,\zeta ) is a solution of (1.1) on (-\mathrm{\infty },0) obeying the condition

and h_2(x,\zeta ) is a solution of (1.1) on (0,\mathrm{\infty }) obeying the condition

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

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 })

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

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

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

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

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

where G_1(\xi ), G_2(\xi ) 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 G_1(\xi ) and G_2(\xi ) with (4.7) gives

Substituting (4.8) into (4.4) gives

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

 □

The reflection coefficient can be defined for this case as

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

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

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

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

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

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

For |\xi | large, we get

Denote the eigenvalues {\zeta }^2, where \zeta =i\eta, \eta \in (0,\mathrm{\infty }), by . We define the norming constants c_k by

Theorem 5.1 The zeros of A(\zeta ) are simple. In addition, if {\zeta }^2=-{\eta }_k^2 is an eigenvalue of the scattering problem with a transfer condition, then

where f_{-,M}(x,i{\eta }_k)=d_k{f}_{+,M}(x,i{\eta }_k) for all x\in \mathbb{R}\setminus \{0\} and d_k\ne 0.

Proof Differentiating (4.5), we get

where the right-hand side is independent of x. So,

The functions f_{+,M} and f_{-,M} obey -f^{″}+qf={\zeta }^{2}f so, differentiating this equation with respect to ζ, gives

and similarly for f_{-,M}. Thus

and

Let y>x>0, integrating (5.2) over the interval [x,y] gives

where we should keep in mind that f_{-,M}=h_2 on the positive semi-axis, see (4.2). Similarly for , integrating (5.3) over the interval [-y,x], we get

Let Γ be a square (with side length equal to ε) contour oriented anticlockwise around i{\eta }_k (see Figure 1), where \epsilon >0 is sufficiently small that {\eta }_k>\epsilon and for all i=1,\dots ,N-1.

The square contour with centre \mathit{i}{\mathit{\eta }}_{\mathit{k}} .

Full size image

By Cauchy’s integral representation for analytic functions and their derivatives, we have

Since \frac{1}{|z-i{\eta }_k|}\le \frac{4}{{\epsilon }^2}, for z\in \mathrm{\Gamma }, we have

From (2.1), for x>0, we obtain

which is bounded on each interval of the form [y,\mathrm{\infty }). Similarly, as f_{+,M}(x,\zeta ) is analytic in the upper half-plane and twice differentiable with respect to x on ℝ, we have

Proceeding as above, we obtain

which by (2.2) is bounded on each interval of the form [y,\mathrm{\infty }). In exactly the same way, it can be shown that \frac{\partial f_{-,M}}{\partial \zeta }(x,i{\eta }_k) and \frac{\partial f_{-,M}^{\prime }}{\partial \zeta }(x,i{\eta }_k) are bounded functions on each interval of the form (-\mathrm{\infty },-y]. Now, for {\zeta }^2 an eigenvalue, we have, by the reasoning prior to Lemma 4.2, that \zeta =i{\eta }_k for {\eta }_k\in {\mathbb{R}}^{+} and f_{-,M}(x,i{\eta }_k)=d_k{f}_{+,M}(x,i{\eta }_k) for some d_k\in \mathbb{C}\setminus \{0\}.

Thus

and

as y\to \mathrm{\infty }. Letting y\to \mathrm{\infty } in (5.4) and (5.5), from the above equations, we get

and