Abstract

We consider all of the transmission eigenvalues for one-dimensional media. We give some conditions under which complex eigenvalues exist. In the case when the index of refraction is constant, it is shown that all the transmission eigenvalues are real if and only if the index of refraction is an odd number or reciprocal of an odd number.

1. 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 [2–4]. 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 one-dimensional case: We refer to the set of all for which (1) has nontrivial solutions as the transmission eigenvalues. The corresponding nontrivial solutions are called the transmission eigenfunctions. Throughout this paper, we assume that

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.

2. 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 .

Suppose that the fundamental solutions and satisfy the initial value problems: Then the solutions and which satisfy (1) can be written as for constants , , , and .

Lemma 1. The value is a transmission eigenvalue if and only if , where where and 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 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 The proof is now complete.

The determinant condition 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 , and , . 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 and , the solutions , to (3), and their -derivatives satisfy where .

So we have the following asymptotic expansions.

Lemma 3. Assume that satisfies (2). Then for all , as in ,  , , and their -derivatives satisfy where .

After substituting the asymptotic expansions of and into (5), we have as , where

According to the fundamental inequality for , we know and . If or , then and the term in (16) becomes which is a periodic function if is rational and almost periodic if is irrational (see [7]). The fact that means that, for large enough , has infinitely many real zeros if we assume . Otherwise, if , we have that So if only one of the values and is , then , and the values are not transmission eigenvalues in the case when .

Our aim here is to find conditions under which has an infinite number of complex zeros when , . The search for zeros of leads us to look for the zeros of the polynomial We assume that is a rational number, and , . Replacing with , we get It is obviously true that if ; that is, implies 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 and ; that is, . Multiplying (22) by 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 . The above polynomial is a self-inversive polynomial because Note that the zeros of a self-inversive polynomial either lie on or are symmetric with respect to the unit circle. We further have the following lemma (see [3]).

Lemma 4 (Cohn). Let be a self-inversive polynomial. Then all the zeros of lie on the unit circle if and only if all the zeros of lie in .

Then we have the following result.

Theorem 5. Assume that satisfies (2), , and is a rational number greater than . Then if either (28) or (33) is valid, the eigenvalue problem (1) has an infinite number of complex eigenvalues and all these eigenvalues lie in a strip parallel to the real axis.

Proof. Based on Cohn’s theorem, our first aim is to look for conditions of under which there are some zeros of lying outside . 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 , there is at least one zero lying outside the unite circle . Hence we give the first condition In other words, By Rouché’s theorem, we can derive another condition for (25) to have zeros outside the unite circle . Set Then we need to prove that has zeros in . On the unite circle , Therefore if Rouché’s theorem implies that has zeros inside the unit disc. So has zeros outside the unit disc from (29). Condition (32) can be stated as 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 ( has complex zeros in this case), then has zeros outside the unit circle, and the zeros inside the unite circle have a positive distance from the origin. Suppose those zeros are , , where and are two integers which are used to denote . Based on the substitution , we know that each corresponds to the complex zeros , , for . Then the corresponding zeros of are , where , and is a positive integer. So all these complex transmission eigenvalues stay inside the strip since . 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 . 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 ,  , 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.

Lemma 6 (Paley-Wiener). Let be an entire function such that for positive constants and and all values of , and Then there exists a function in such that

Theorem 7. Assume that satisfies (2) and , . Then if complex eigenvalues exist, all of them lie in a strip parallel to the real axis.

Proof. From the identities and , (19) implies that Hence if , , then . In this case, is an entire function of of exponential type . It follows from Lemma 2 that , ,, and are entire functions of of exponential type . From (5), we get that is an entire functions of of exponential type at most . Then (35) implies that is also an entire function of of exponential type at most .
Furthermore, at , ,; then is an function on the real axis. So meets the conditions in the Paley-Wiener theorem. Then there exists a function such that Using Schwarz inequality, we have that Hence which implies that goes to zero as tends to infinity. Furthermore, we have Hence which implies that goes to as goes to infinity.
Suppose that ,  , are transmission eigenvalues, a sequence of the zeros of such that as . Then from the above discussion, we have and as , but ; this leads to a contradiction. Hence if complex eigenvalues exist, all of them lie in a strip parallel to the real axis.

3. The Existence of Complex Transmission Eigenvalues for Constant

We consider the eigenvalue problem (1) again and make the assumption that the graph of is symmetric about ; that is, . Using the change of variables we can transform (1) into

In the symmetric domain, we can separate (47) into two problems.

Lemma 8. In the case when , is a transmission eigenvalue if and only if satisfies or , where and are defined in (49).

Proof. By the fact that for , we know satisfies the first equation in (47) if is a solution to that equation. Hence there exist an even function and an odd function such that So the solutions and for (47) can be written as where , , , and are constants. Basing on the properties of even and odd functions, we get that, for all , From the boundary conditions in (47), we get From (51), (52) and (53) yield that Using (51), (54), and (55), we obtain that If , then . Equations (56) and (58) imply that In the case when , then . From (57) and (59), we have that If is a transmission eigenvalue, and cannot be zero simultaneously. Hence, or . This completes the proof.

Remark 9. In the case when , the set of transmission eigenvalues is the union of the sets of -values such that and .

From now on, we further assume that is a positive constant not equal to on , by using to denote that constant value. In this case, Substituting and into (48), we have that

Our goal is to determine under what conditions there exist complex eigenvalues when is a constant. In this case, , , , and . From (28) and (33), we know that there is an infinite number of complex transmission eigenvalues when and . A question naturally arises: are there some complex transmission eigenvalues when ? The following results (see [3]) play central roles in what follows. We give them as lemmas.

Lemma 10 (Laguerre). Let be a real entire-valued function of order less than with all its zeros being real. Then the critical points of  i.e., the zeros of are also real and interlace those of .

Lemma 11. Let be a real-valued entire function of order less than . Then if all its zeros are real, it cannot have more than one critical point inside an interval where it does not change sign.

Lemma 12. Let be a real-valued entire function of order less than . Suppose that has infinitely many real zeros and only a finite number of complex ones. Then has a single critical point on each interval formed by two consecutive real zeros of when the interval is sufficiently far away from the origin.

In order to find the transmission eigenvalues, we study the roots of and . First of all, we give some illustrative examples.

Example 13. First, when , we have that hence has a simple zero at , an infinite set of real zeros of multiplicity at -values for , and an infinite set of simple complex zeros at -values that are given by Second, when , we get hence has a simple zero at and an infinite set of real zeros of multiplicity at -values for . Third, when , then So has a simple zero at , an infinite set of real zeros of multiplicity at -values for , an infinite set of simple complex zeros at -values and an infinite set of complex zeros at -values such that

As we can see from the above examples, this problem may only have real transmission eigenvalues under some conditions. The following theorem presents a sufficient and necessary condition for the nonexistence of complex transmission eigenvalues in the case when is constant.

Theorem 14. Let be a constant not equal to . Then all of the transmission eigenvalues are real when is an odd number or a reciprocal of an odd number. Otherwise, (1) has infinitely many real and complex transmission eigenvalues.

Proof. We see that if , then . Set , and we have So it suffices to consider the case only. The zeros of and are the critical points of the functions separately. Obviously, is a real-valued entire function of order less than with all its zeros being real when is an odd number, and if is an integer, is a real-valued entire function of order less than with all its zeros being real. It follows from Laguerre’s theorem that all roots of are real when is an odd number, and the roots of are real when is an integer. Hence, from Lemma 8, (47) only has real transmission eigenvalues when is an odd number.
For the second part of this theorem, we only need to show that has an infinite number of complex roots when is not an odd number. We note that which has zeros at According to Lemmas 11 and 12, our aim here is to argue that there are infinitely many intervals where does not change sign and has at least two consecutive critical points inside.
From this point on, take to be a fixed integer. If , there is an integer , such that ; that is, Then we have From (74), we have So and are two consecutive critical points of . At , At the critical point , We see that the signs of and are identical since and ; that is, cannot change sign in .
In the case when , there is an integer such that We have Then for some . In the same way as done above, we have that cannot change sign in .

Conflict of Interests

The authors declare that there is no conflict of interests regarding the publication of this paper.

Acknowledgment

This research was supported by the National Natural Science Foundation of China under Grant no. 61170019.