Hindawi Publishing Corporation Abstract and Applied Analysis Volume 2014, Article ID 561349, 8 pages http://dx.doi.org/10.1155/2014/561349

Research Article

Complex Transmission Eigenvalues in One Dimension

Yalin Zhang,1 Yanling Wang,1 Guoliang Shi,1 and Shizhong Liao2

1 Department of Mathematics, Tianjin University, Tianjin 300072, China

2 School of Computer Science and Technology, Tianjin University, Tianjin 300072, China

Correspondence should be addressed to Yanling Wang; mathtju@163.com Received 15 May 2014; Accepted 18 June 2014; Published 6 July 2014 Academic Editor: Ali H. Bhrawy

Copyright © 2014 Yalin Zhang et al. This is an open access article distributed under the Creative Commons Attribution License, which permits unrestricted use, distribution, and reproduction in any medium, provided the original work is properly cited.

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:

w" + k2n (x)u = 0, x € [0,1].

v" + k2v =0, x € [0,1], u (0) = v (0), u (1) = v (1), u (0) = v (0), u (1) = v' (1).

nontrivial solutions (w, v) are called the transmission eigen-functions. Throughout this paper, we assume that

n(x) > 0, x € [0,1] ; n€CJ (0,1); n" €l2 (0,1); n (0) = 0.

We refer to the set of all k e C for which (1) has nontrivial solutions as the transmission eigenvalues. The corresponding

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 n(r) 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 n(x)

Our research methods rely on transforming the first Helmholtz equation in (1) into a Sturm-Liouville form that separates k and n(x).

Suppose that the fundamental solutions u1 and u2 satisfy the initial value problems:

u' + k2n (x) ut = 0, i = 1,2,

u1 (0) = u2 (0) = 0, uf (0) = U2 (0) = 1. Then the solutions u and v which satisfy (1) can be written as u (x) = a1u1 (x) + a2u2 (x), v (x) = b1 sin kx + b2 cos kx, for constants a1, a2, b1, and b2.

Lemma 1. The value k is a transmission eigenvalue if and only ifd(k) = 0, where

f sink

d(k) = -2+ u1 (1)ksink - u2 (1) —-—

+ u1 (1) cos k + u2 (1) cos k, where u1 and u2 are the solutions of (3). Proof. The boundary conditions in (1) imply that

a2 = b2,

a1 = b1k,

a1u1 (1) + a2u2 (1) = b1 sink + b2 cosk, a1u1 (1) + a2u2 (1) = b1kcosk - b2ksink.

So in order for the value k to be a transmission eigenvalue there must exist a nontrivial pair (a1, a2) satisfying

u1 (1) - k sin k u2 (1) - cos k\(a1 v u[ (1) - cosk u2 (1)+ksinkj^2

For this to be true the determinant of the coefficient matrix must be zero:

d(k):= det (u1(1)-k sink u2 (1)-cosk) = 0. (8) \ u[ (1)- cosk u2 (1) + ksink/

Using the Wronskian identity for the fundamental solutions of the Sturm-Liouville problems

W (u1,u2) (x) := u1 (x) u2 (x) - u[ (x) u2 (x) = -1 in [0,1],

: have

d(k) = -2+ u1 (1)k sin k - u2 (1)

' sin k

*2W k + u1 (1) cos k + u2 (1) cos k. The proof is now complete.

The determinant condition d( k) = 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 u1, u[ and u2, u2. We make the change of variables

\ Vn (s)ds, J0

y{ (t) = (n(x))l/iui (x), i = 1,2. Since n satisfies (2), the problem (3) becomes

y't' (t) + (k2 -p(t))y, (t) = 0 for te [0,8]; y1 (0) = 0, y; (0) = n(0)-1/4, y2 (0) = n(0)1/4, y; (0) = 0,

= 1 n (x) - 5 n;(x)2 P( ) 4 n(x)2 16 n(x)3 ,

8 = \ Vn (s)ds.

With the help of the Liouville transformation (11) and some basic estimates in [6] for the corresponding Schrodinger equations in (12), we get the following lemma.

Lemma 2. Assume that n satisfies (2). Then there exists a positive constant C such that, for all x e [0,1] and k e C, the solutions u1 (x), u2(x) to (3), and their x-derivatives satisfy

u1 (x) -

[n(0)n(x)]1/4k

sin (kt (x))

< ^ exp (\Imk\t(x)),

; fn(x)\1/4 u[ (x)-^nö)) cos(kt(x))

< — exp (\Imk\t(x)),

, ^ fn(0)\1/4 „ , „

u2 (x) - \n(x) ) cos (kt(x))

<— exp (\Imk\t(x)), \k\

\u2 (x) + k(n (0) n (x))1/4 sin (kt (x))| <C exp (\lmk\t(x)), where t(x) = J* ^n(s)ds.

So we have the following asymptotic expansions.

Lemma 3. Assume that n satisfies (2). Thenfor all x e [0,1], as \k\ ^ >x> in C, u1 (x), u2(x), and their x-derivatives satisfy

u1 (x) =

sin (kt (x))

[n(0)n(x)]1/4k

exp (|Imk| t (x))

ul (x) = (nx) cos(kt(x))

exp (|Imk| t (x))

U2 (x) = (nxj) cos (kt(x))

' exp (|Imk| t (x))

u2 (x) = -k(n (0) n(x)) sin (kt (x)) + O (exp (\Imk\t(x))), where t(x) = J0 ^n(s)ds.

After substituting the asymptotic expansions of u1 and u2 into (5), we have

d(k) = -2 + Bsin (kS) sink + C cos (kS) cosk

( exp (\Imk\(S+1))\ (16)

as \k\ ^ œ>, where

+ (n(0)n(1))

(n(0)n(1))1/4

c==(nJHr + (nM)

1/4 + (nM\1/4

\n(0)J (n(1)J '

I Vn (s)ds Jo

According to the fundamental inequality«2+b2 > 2ab for a,b > 0, we know B > 2 and C > l. If n(0) = 1 or n(1) = 1, then B = C and the term -2 + B sin(kS) sin k + C cos(kS) cos k in (16) becomes -2 + C cos(k(S - 1)) which is a periodic function if S is rational and almost periodic if S is irrational (see [7]). The fact that C > 2 means that, for large enough k, d(k) has infinitely many real zeros if we assume S= 1. Otherwise, if S = 1,we have that

d(k)-0

exp (|Imk| (5+1))

= -2 + B sin2k + C cos2k

„ B+C C-B = -2 +-+-cos (2k).

So if only one of the values n(0) and n(1) is 1, then B = C > 2, and the values k e R are not transmission eigenvalues in the case when \k\ ^ >x>.

Our aim here is to find conditions under which d( k) has an infinite number of complex zeros when n(0) = 1, n(1) = 1. The search for zeros of d(k) leads us to look for the zeros of the polynomial

T(k):=-2 + B sin (kS) sin k + C cos (kS) cos k. (19)

We assume that S is a rational number, and S = m/n > 1, m,n e Z+. Replacing k with nk, we get

T (nk) = -2 + B sin (mk) sin (nk) + C cos (mk) cos (nk).

It is obviously true that \eik\ = 1 if k e R; that is, \eik\ = 1 implies that k £ R. Based on this fact, a substitution of z = e'k into (20) is used. According to Euler's formula, we have

eimk - e-imk zm - z-

eimk + e-imk zm + z-

sin (mk) =

cos (mk) = So (20) becomes

-2-B (zm - z-m) (zn - z-n) + C.(zm + z-m) (zn + z-n).

We assume that n(0) = 1 and n(1) = 1; that is, B = C. Multiplying (22) by (4/(B - C))zm+n leads us to look for zeros of the polynomial

-8zm+n - B (z2m - 1) (z2n - 1) + C (z2m + 1) (z2n + 1)

2(m+n) ~ ■ 2m ~ rn+n ~ ■ 2n -,

= -zy ' +-z--z +-z - 1.

B-C B-C B-C

In order to show the conditions under which T( k) has complex zeros, we only need to research in what situations p(z) cannot have all roots lying on the unit circle \z\ = 1. The above polynomial p(z) is a self-inversive polynomial because

2(m+n)

p{D=p(z).

Note that the zeros of a self-inversive polynomial either lie on \ z\ = 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 p'(z) lie in \z\ < 1.

Then we have the following result.

Theorem 5. Assume thatn satisfies (2), n(0) = 1, n(1) = 1, and 8 = J y!n(s)ds is a rational number greater than 1. 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 n under which there are some zeros of p'(z) lying outside \z\ = 1. For p(z) that was defined in (23), we have

J / N T / \ 2(m+n) - B + C 2m 8 (m + n) m+n

zp (z) = -2(m + n)z + 2m-z--z

y B-C B-C

B + C 2 n

+ 2n-z .

With the help of Vieta theorem, the product of all zeros of (25) equals

m + n B - C

If the absolute value of this product is greater than 1, there is at least one zero lying outside the unite circle \z\ = 1. Hence we give the first condition

n B + C

m + n B - C

In other words,

>1 + 5.

By Rouché's theorem, we can derive another condition for (25) to have zeros outside the unite circle \z\ = 1. Set

q(z) :=

z2(m+n) 1 ,( 1

2 zP (z

B + C 2n

= -(m + n) + m-—z

4(m + n) m+n

Then we need to prove that q(z) has zeros in \z\ < 1. On the unite circle \ z\ = 1,

B + C 2n

a (z) - m-z

a( ) B-C

4 (m + n) m+n B + C

-(m + n)--z +n-;

( ) B-C B-C

4 (m + n) B + C <m + n +---+ n-

\B-C\ \B-C\

Therefore

B + C 2n

a (z) - m-z

a( ) B-C

B + C !-.

4 (m + n) B + C B + C m + n+---+ n- < m-

\B-C\ \B-C\

Rouché's theorem implies that q(z) has 2n zeros inside the unit disc. So p'(z) has 2n zeros outside the unit disc from (29). Condition (32) can be stated as

\B-C\+4 5-1 <

So far, we obtain two conditions (28) and (33) which guarantee the existence of complex zeros of T(k).

The following proof for the existence of complex zeros for d(k) is the same as that stated in [3]. To facilitate reading, we state it once again. If p(z) has zeros not on the unite circle \z\ = 1 (T(k) has complex zeros in this case), then p(z) has zeros outside the unit circle, and the zeros inside the unite circle\z\ = 1 haveapositivedistancefromtheorigin. Suppose those zeros are zj = rje' >, j = 1,...,h, h < m + n, where m and n are two integers which are used to denote 8. Based on the substitution z = e' , we know that each zj = rje' > corresponds to the complex zeros kj = dj + 2ln - i logrj, I = 1,2,..., for T(nk). Then the corresponding zeros of T(k) are ndj + 2nln - in log rj, where l = 1,2,..., and n is a positive integer. So all these complex transmission eigenvalues stay inside the strip

\Imfc\ < nmaxllogr.-

since 0 < rj <1. Let Cj be a small circle surrounding zj, lie inside the unit circle, and isolate zj from the other

zeros of p(z). Under the transformation z = e'k, the circle Cj corresponds to a periodic array of closed Jordan curves surrounding each of the corresponding zeros of T(k), and, on these curves, \ T(k)\ > dj for some constant dj > 0. From (16), we have that

d(k) = T(k) + 0

exp (\Imk\ (5+1))

for k being large enough. Using (34), we get that \d(k)-T(k)\ < \T(k)\,

is valid for k being large and lying on some closed Jordan curves. It follows from Rouche's theorem that d(k) has a complex zero inside each Jordan curve when k is large. □

Next, inspired by the results and methods in [4], we will show that, when n(0) = 1, n(1) = 1, if transmission eigenvalues exist they must lie in a strip parallel to the real axis; that is, we remove some assumptions on n(x) 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 f(z) be an entire function such that

\f(z)\<CeAlzl,

for positive constants A and C and all values of z, and

i™ 2

\f(z)\ dx < >x>.

Then there exists a function $ in L [-A, A] such that

f(z)=\ $(t)eiztdt.

Theorem 7. Assume that n satisfies (2) and n(0) = 1, n(1) = 1. Then if complex eigenvalues exist, all of them lie in a strip parallel to the real axis.

Proof. From the identities cos(a -b) = cos a cos b + sin a sin b and cos(a + b) = cos a cos b - sin a sin b, (19) implies that

T (k) = -2 + cos (k (8 - 1)) - cos (k (8 + 1)).

Hence if n(0) = 1, n(1) = 1, then B = C. In this case, T(k) is an entire function of k of exponential type 8 + 1. It follows from Lemma 2 that u1(1), u[(1), u2(1), and u2(1) are entire functions of k of exponential type 8. From (5), we get that d(k) is an entire functions of k of exponential type at most 8+1. Then (35) implies that E(k) := 0((exp(| Imk\(8 + 1)))/k) is also an entire function of k of exponential type at most 8+1.

Furthermore, at k = 0, d(k) = 0,\T(k)\ = \ - 2 + C\ < rn; then E(k) is an L2 function on the real axis. So E(k) meets the conditions in the Paley-Wiener theorem. Then there exists a function $ e L2[-(8 + 1), (8 + 1)] such that

$ (t) eiktdt.

-(5+1)

Using Schwarz inequality, we have that r5+1

\E(k)\2 < ||0||2

-(ä+1) \

||2 / 2| Imfc|(5+1)

-2| Im fc|(5+1)^

2\Imk\

e-2|ImW+1)\E{k)\2 <

-4| Im fc|(5+1)

2 \ Im k\

which implies that e 1 Imfcl(5+1)\E(k)\ goes to zero as ! Imk\ tends to infinity. Furthermore, we have

\T(k)\ <2 + B+C \cos (k(8-1))\

cos (k(8+1))\

B + C e~| ImfcK5-1) +e| ImfcK5-1)

22 \B-C\ e-| Im fc|(ä+1) +e| Im fc|(5+1)

e-|Imfc|(5+1) \T{k)\

| Im fc|(5+1)

+ B + C (e-2| Im fc|S + e-2| Im fc| )

-2| Im fc|(5+1)

which implies that e-1 Imfc|(5+1)\T(k)\goes to \B-C\/4 as \ Im k\ goes to infinity.

Suppose that kj, j = 1,2..., are transmission eigenvalues, a sequence of the zeros of d(k) such that \ Imk^ ^ >x> as j ^ >x>. Then from the above discussion, we have

,-| Imfc,|(S+1)

0 and e-Im'"'^^(k) ^ \B-C\/4

as j ^ <x>, but 0 = d(kj) = T(kj) + E(kj); 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 n(x)

We consider the eigenvalue problem (1) again and make the assumption that the graph of n(x) is symmetric about x = 1/2; that is, n(x) = n(1 - x). Using the change of variables

V = x--,

f(y) = V (x), we can transform (1) into

((y)=u (x),

m (y) = n(x- ^

( + k2m (y) ( = 0, ye

" i 2 ^ f + k f = 0, ye

-2)=V(--2

('(J2)=f' (4

1 1 2'2

f(-2)=f(~2 <( 1\ <(1

(pK-2) = f\.2

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

Lemma8. Inthecasewhen n(x) = n(1-x),k is a transmission eigenvalue if and only if k satisfies d1 (k) = 0 or d2(k) = 0, where

d1 (k) :=k91 (i) sin (k)+f'1 (i) cos

d2 (k) '-=k(P2 cos (^O-?2 (1) sin (k).

f1 and f2 are defined in (49).

Proof. By the fact that m(-y) = m(y) for y e [-1/2,1/2], we know f(-y) satisfies the first equation in (47) if cp(y) is a solution to that equation. Hence there exist an even function (p1(y) and an odd function cp2(y) such that

~2' 2 J

, i = 1,2,

<p'{ + k2m (y) = 0, ye

Vi (0) = 1, cp[ (0) = 0, <P2 (0) = 0, <p'2 (0) = 1. So the solutions (p and f for (47) can be written as

f(y) = cifi (y) + c2f2 (У),

y (y) = d1 cos (ky) + d2 sin (ky),

where q, c2, d1, and d2 are constants. Basing on the properties of even and odd functions, we get that, for all y e [-1/2,1/2],

fi (-y) = fi (y), <P2 (-y) = -f2 (y), v1 (-y) = -v1 (y), v2 (-y) = v2 (y) ■

From the boundary conditions in (47), we get

ciVi (-2) + c2lP2 \-2)=dl cos \2)-d2sin [2

2) + c2V2 (2) =ai cos [2)+a2 sinV 2

ci?1 i-2) + c292 (-2) = dik sin (2) + d2k cos ij)'

CiVi (1)+C2(p2 Q) = -d1k sin (fc)+d2ki

From (51), (52) and (53) yield that

ci<Pi (2)=aicos [2

C2?2 ( 2)=d2 sin V 2

Using (51), (54), and (55), we obtain that cif'i = diksin

= d2k cos

(54) )■

^2^2 (2 J z \2

If c1 = 0, then d1 = 0. Equations (56) and (58) imply that

di (k)=kf1 (2) sin (¡)+f[ (1) cos (fc) = 0. (60)

In the case when c2 = 0, then d2 = 0. From (57) and (59), we have that

d2 (k) = kf2( cos [l2

2 (1) sin [21 = °.

If k is a transmission eigenvalue, c1 and c2 cannot be zero simultaneously. Hence, d1(k) = 0 or d2(k) = 0. This completes the proof. □

Remark 9. In the case when n(x) = n(1 - x), the set of transmission eigenvalues is the union of the sets of fc-values such that d1(k) = 0 and d2 (k) = 0.

From now on, we further assume that m(y) = n(x - 1/2) is a positive constant not equal to 1 on [0,1], by using n to denote that constant value. In this case,

r \ t r- \ t \ sin (kjny) cp1 (y) = cos (kjny), cp2 (y) = --¡—p—. (62)

Substituting cp1 and (p2 into (48), we have that di (k) :=k cos (—j^) sin

. (k^i\ (k

■ kynsin ( - 1 cos ( -

[ 2 1 [2

, 1 . fkjE\ fk

i2 (k) := —¡= sin ( —:— 1 cos ( — 1 - cos

Our goal is to determine under what conditions there exist complex eigenvalues k when n=1 is a constant. In this case, B = jn + 1/ jn, C = 2, S = jn, and B > C. From (28) and (33), we know that there is an infinite number of complex transmission eigenvalues when 0 < jn < 1 and 1 < jn < 3. A question naturally arises: are there some complex transmission eigenvalues when jn> 3? The following results (see [3]) play central roles in what follows. We give them as lemmas.

Lemma 10 (Laguerre). Let f(z) be a real entire-valued function of order less than 2 with all its zeros being real. Then the critical points of f(z) (i.e., the zeros of f'(z)) are also real and interlace those off(z).

Lemma 11. Let f(z) be a real-valued entire function of order less than 2. 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 f(z) be a real-valued entire function of order less than 2. Suppose that f(z) has infinitely many real zeros and only a finite number of complex ones. Then f( z) has a single critical point on each interval (ai,ai+1) formed by two consecutive real zeros of f(z) when the interval is sufficiently far away from the origin.

In order to find the transmission eigenvalues, we study the roots of d1 = 0 and d2 = 0. First of all, we give some illustrative examples.

Example 13. First, when Vn = 2,we have that

d1 (k) = -k sin (^)(2 + cos (k)), d2 (k) = sin (Ç) (1 - cos (k)) ;

hence d(k) has a simple zero at k = 0, an infinite set of real zeros of multiplicity 3 at fc-values 2jn for j e N, and an infinite set of simple complex zeros at fc-values that are given

(2j-1)n+ilog(2±V3), je N.

Second, when Vn = 3,we get

d1 (k) = -2k sin k(1 + cos k), d2 (k) = 2 sin k(-1 + cos k) ;

hence d(k) has a simple zero at k = 0 and an infinite set of real zeros of multiplicity 3 at fc-values jn for j e N. Third, when Vn = 2/3, then

d1 (k) = ^ sin (2cos2

M fk, „

- ) + cos ( - ) + 2 3) (3

*2 (k) = -sin3 ( + 2 cos ( ^

So d(k) has a simple zero at k = 0, an infinite set of real zeros of multiplicity 4 at fc-values 6jn for j e N, an infinite set of simple complex zeros at fc-values

3(2j+1)n + i3 log , je N,

and an infinite set of complex zeros at fc-values such that

2cos2 ( - )+ cos ( - ) + 2 = 0.

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 n is constant.

Theorem 14. Let n be a constant not equal to 1. Then all ofthe transmission eigenvalues are real when Vn 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 0 < n < 1, then q := 1/n > 1. Set A = Vnk, and we have

A \ / aM

d1 (k) = x(v^cos (~2) sin

'A\ /V^

- sin ( — ) cos ( -

.2) ( 2

¿2 (k) = Vn sin ( 2) cos^ 2 'M. ( Vnx\

- cos ( — ) sin ( - ) .

.2) ( 2 )

So it suffices to consider the case n > 1 only. The zeros of d11(k) := d1(k)/k and d2(k) are the critical points of the functions

fl (k) :=

cos (k^n/2) cos(k/2)

/2 (k) :=

sin (kVn/2) sin (k/2) '

separately. Obviously, fi(k) is a real-valued entire function of order less than 2 with all its zeros being real when Vn is an odd number, and if Vn is an integer, f2(k) is a real-valued entire function of order less than 2 with all its zeros being real. It follows from Laguerre's theorem that all roots of d11(k) = 0 are real when Vn is an odd number, and the roots of d2(k) = 0 are real when Vn is an integer. Hence, from Lemma 8, (47) only has real transmission eigenvalues when Vn is an odd number.

For the second part of this theorem, we only need to show that d11(k) has an infinite number of complex roots when Vn is not an odd number. We note that

, 1-n k^n k

a,, (k) =-cos-cos —,

11 w 2 2 2

which has zeros at

{n,3n,...,(2N + 1)n,...},

(2j+ l)n

n 3n Vn Vn'

N, j e N.

According to Lemmas 11 and 12, our aim here is to argue that there are infinitely many intervals where d11(k) does not change sign and has at least two consecutive critical points inside.

From this point on, take N > 0 to be a fixed integer. If Vn(N+\/2)-l/2 i Z,thereisaninteger j,suchthat Vn(N+ 1/2) - 1/2 < j < Vn(N+ 1/2) + 1 /2; that is,

Vn^N + < (j + < Vn^N + 1)n + n. (74) Then we have

0 + 1i)n = Vü(n + 1)n + en for some e e (0,1).

From (74), we have

(2j+lW 2 n

V» yn

So (2N + l)n and (2j + 1)^/ V« are two consecutive critical points of dn (fc). At (2N + 1 )n,

d11 ((2N+ 1)rc) = (-1)N cos (Vn(N+

= (-1)N cos ((j+ i^-erc) = (-1)N+1 sin (jrc - erc)

= (-1)N+J sin (e^). At the critical point (2j + 1)^/ V«, (2j+

+1 i(j+1/2)tt = (-1) V« cos ( -■=-

= (-1)j+1Vwcos ((n + - +

= (-1); V« sin ( Nrc +

= (-1)-^ sin (VI)

We see that the signs of dn ((2N +1)^) and dn ((2j +1)^/ V«) are identical since e e (0,1) and V« > 1;thatis, dn(fc) cannot change sign in ((2j + 1)^/ V«, (2N + 1)^).

Inthecasewhen V«(^+1/2)-1/2 e Z,thereisaninteger j such that

We have

j < + i) - 1 < j + 1

rc + rc.

n + erc,

for some e e (0,1). In the same way as done above, we have

that dn(fc) cannot change sign in ((2N + 1)^, (2j + 1)^/ V«).

References

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[8] R. M. Young, An Introduction to Nonharmonic Fourier Series, Academic Press, San Diego, Calif, USA, 2001.

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.

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