Complex Transmission Eigenvalues in One Dimension

and Applied Analysis 3 Lemma 3. Assume that n satisfies (2). Then for all x ∈ [0, 1], as |k| → ∞ inC, u 1 (x), u 2 (x), and their x-derivatives satisfy


Introduction
The transmission eigenvalue problem appears in the inverse scattering theory for acoustic and electromagnetic waves [1].It is a nonlinear boundary value problem for a coupled set of equations defined on the support of the scattering object.Since this eigenvalue problem is not self-adjoint, there exists the possibility of complex eigenvalues which has been proved for the spherically stratified media under some conditions in [2][3][4].But so far only a small part of the transmission eigenvalues (real eigenvalues or complex eigenvalues) has been considered; this is of great limitation in the inverse scattering problem.If we consider all of the transmission eigenvalues, this problem, even for the one-dimensional media, is not simple.For one-dimensional problem, Sylvester [5] has shown how to locate all the transmission eigenvalues in the complex plane for a constant index of refraction.In this work we contribute more to the discussion of the onedimensional case: +  2  ()  = 0,  ∈ [0, 1] , = V (1) ,   (0) = V  (0) ,   (1) = V  (1) . ( We refer to the set of all  ∈ C for which (1) has nontrivial solutions as the transmission eigenvalues.The corresponding nontrivial solutions (, V) are called the transmission eigenfunctions.Throughout this paper, we assume that  () > 0,  ∈ [0, 1] ;  ∈  1 (0, 1) ;   ∈  2 (0, 1) ;   (0) = 0. ( Since only real eigenvalues can be determined from the scattering data and the physical properties of the scattering object can be obtained from the transmission eigenvalues, it is of interest to find the existence conditions of the complex transmission eigenvalues and to research the conditions under which there are no complex eigenvalues at all.For the existence of the transmission eigenvalues, to the author's knowledge, only the sufficient conditions are given. The plan of our paper is as follows.In Section 2, motivated by [3] which gives the existence conditions of complex transmission eigenvalues for three-dimensional media, we turn our attention to the case when () is a variable and show that complex eigenvalues can exist.Then, in Section 3, using the methods in [3,5], we give the necessary and sufficient condition for the existence of complex transmission eigenvalues when the index of refraction is a constant.

The Existence of Complex Transmission
Eigenvalues for Variable () Our research methods rely on transforming the first Helmholtz equation in (1) into a Sturm-Liouville form that separates  and (). ( Then the solutions  and V which satisfy (1) can be written as for constants  1 , where  1 and  2 are the solutions of (3).
Proof.The boundary conditions in (1) imply that So in order for the value  to be a transmission eigenvalue there must exist a nontrivial pair ( 1 ,  2 ) satisfying For this to be true the determinant of the coefficient matrix must be zero: Using the Wronskian identity for the fundamental solutions of the Sturm-Liouville problems we have sin The proof is now complete.
The determinant condition () = 0 gives us an algebraic relation that must be satisfied by the transmission eigenvalues.This reduces the study of the transmission eigenvalue problem to a root finding problem.Next, we will give the expansions of  1 ,   1 and  2 ,   2 .We make the change of variables Since  satisfies (2), the problem (3) becomes where With the help of the Liouville transformation (11) and some basic estimates in [6] for the corresponding Schrödinger equations in (12), we get the following lemma.Lemma 2. Assume that  satisfies (2).Then there exists a positive constant  such that, for all  ∈ [0, 1] and  ∈ C, the solutions  1 (),  2 () to (3), and their -derivatives satisfy where () = ∫  0 √().
So we have the following asymptotic expansions.
Our aim here is to find conditions under which () has an infinite number of complex zeros when (0) ̸ = 1, (1) ̸ = 1.The search for zeros of () leads us to look for the zeros of the polynomial  () := −2 +  sin () sin  +  cos () cos . ( We assume that  is a rational number, and  = / > 1, ,  ∈ Z + .Replacing  with , we get It is obviously true that Based on this fact, a substitution of  =   into (20) is used.According to Euler's formula, we have So (20) becomes We assume that (0) ̸ = 1 and (1) ̸ = 1; that is,  ̸ = .Multiplying ( 22) by (4/( − )) + leads us to look for zeros of the polynomial In order to show the conditions under which () has complex zeros, we only need to research in what situations () cannot have all roots lying on the unit circle || = 1.The above polynomial () is a self-inversive polynomial because Note that the zeros of a self-inversive polynomial either lie on || = 1 or are symmetric with respect to the unit circle.We further have the following lemma (see [3]).

Lemma 4 (Cohn). Let 𝑝(𝑧) be a self-inversive polynomial. Then all the zeros of 𝑝(𝑧) lie on the unit circle if and only if
all the zeros of   () lie in || ≤ 1.
Then we have the following result.
Proof.Based on Cohn's theorem, our first aim is to look for conditions of  under which there are some zeros of   () lying outside || = 1.For () that was defined in (23), we have With the help of Vieta theorem, the product of all zeros of (25) equals If the absolute value of this product is greater than So far, we obtain two conditions (28) and (33) which guarantee the existence of complex zeros of ().
The following proof for the existence of complex zeros for () is the same as that stated in [3].To facilitate reading, we state it once again.If () has zeros not on the unite circle || = 1 (() has complex zeros in this case), then () has zeros outside the unit circle, and the zeros inside the unite circle || = 1 have a positive distance from the origin.Suppose those zeros are   =      ,  = 1, . . ., ℎ, ℎ ≤  + , where  and  are two integers which are used to denote .Based on the substitution  =   , we know that each   =      corresponds to the complex zeros   =   + 2 −  log   ,  = 1, 2, . .., for ().Then the corresponding zeros of () are   + 2 −  log   , where  = 1, 2, . .., and  is a positive integer.So all these complex transmission eigenvalues stay inside the strip since 0 <   < 1.Let   be a small circle surrounding   , lie inside the unit circle, and isolate   from the other zeros of ().Under the transformation  =   , the circle   corresponds to a periodic array of closed Jordan curves surrounding each of the corresponding zeros of (), and, on these curves, |()| >   for some constant   > 0. From (16), we have that for  being large enough.Using (34), we get that is valid for  being large and lying on some closed Jordan curves.It follows from Rouché's theorem that () has a complex zero inside each Jordan curve when  is large.
Next, inspired by the results and methods in [4], we will show that, when (0) ̸ = 1, (1) ̸ = 1, if transmission eigenvalues exist they must lie in a strip parallel to the real axis; that is, we remove some assumptions on () which were required in the above theorem.The major tool we use is the following result from [8] for an entire function.
Using Schwarz inequality, we have that Hence ( as  → ∞, but 0 = (  ) = (  ) + (  ); this leads to a contradiction.Hence if complex eigenvalues exist, all of them lie in a strip parallel to the real axis.