Open Access

The existence of eigenvalue problems for the waveguide theory

Boundary Value Problems20122012:133

DOI: 10.1186/1687-2770-2012-133

Received: 5 April 2012

Accepted: 1 November 2012

Published: 13 November 2012

Abstract

In this paper, the existence of the eigenvalue problem for the waveguide theory is investigated. We used the Fourier transformation method for the solution of this problem. Also, we applied this problem to a dielectric waveguide. In this study, four theorems and two lemmas are obtained.

MSC: 35A22, 35P10.

Keywords

partial differential equations eigenvalue problems Fourier transformation method

1 Basic preliminaries

A dielectric waveguide is a composite of its own index of refraction for each layer. If Ω j https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-133/MediaObjects/13661_2012_Article_234_IEq1_HTML.gif is a layer, where the index of refraction is k j https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-133/MediaObjects/13661_2012_Article_234_IEq2_HTML.gif and μ is a spectral parameter, then the waveguide process can be written in the following form:
u j ( x ) = ( k j + μ ) u j ( x ) ; x Ω j , https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-133/MediaObjects/13661_2012_Article_234_Equ1_HTML.gif
(1)
where
= ( x 1 ) 2 + + ( x n ) 2 . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-133/MediaObjects/13661_2012_Article_234_Equa_HTML.gif
In order to obtain u j ( x ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-133/MediaObjects/13661_2012_Article_234_IEq3_HTML.gif and u r ( x ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-133/MediaObjects/13661_2012_Article_234_IEq4_HTML.gif, the process in all the waveguide for the common boundary of domains Ω j https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-133/MediaObjects/13661_2012_Article_234_IEq1_HTML.gif and Ω r https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-133/MediaObjects/13661_2012_Article_234_IEq5_HTML.gif is evaluated. u j ( x ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-133/MediaObjects/13661_2012_Article_234_IEq3_HTML.gif and u r ( x ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-133/MediaObjects/13661_2012_Article_234_IEq4_HTML.gif must be joined in the way that the obtained known functions u j , r ( x ) = u j ( x ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-133/MediaObjects/13661_2012_Article_234_IEq6_HTML.gif for x Ω j https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-133/MediaObjects/13661_2012_Article_234_IEq7_HTML.gif and u j , r ( x ) = u r ( x ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-133/MediaObjects/13661_2012_Article_234_IEq8_HTML.gif for x Ω r https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-133/MediaObjects/13661_2012_Article_234_IEq9_HTML.gif will be the generalized solution of the equation
v ( x ) = ( g ( x ) + μ ) v ( x ) ; x Ω j Ω r , https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-133/MediaObjects/13661_2012_Article_234_Equ2_HTML.gif
(2)

in which g ( x ) = k j https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-133/MediaObjects/13661_2012_Article_234_IEq10_HTML.gif for all x Ω j https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-133/MediaObjects/13661_2012_Article_234_IEq7_HTML.gif and g ( x ) = k r https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-133/MediaObjects/13661_2012_Article_234_IEq11_HTML.gif for all x Ω r https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-133/MediaObjects/13661_2012_Article_234_IEq9_HTML.gif. If the boundary Γ j , r https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-133/MediaObjects/13661_2012_Article_234_IEq12_HTML.gif is sufficiently smooth, the condition of this junction may be put down in a natural form. Indeed, the contraction of Γ j , r https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-133/MediaObjects/13661_2012_Article_234_IEq12_HTML.gif is noninfinitely smooth in Ω j https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-133/MediaObjects/13661_2012_Article_234_IEq1_HTML.gif and Ω r https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-133/MediaObjects/13661_2012_Article_234_IEq5_HTML.gif, the functions which deteriorate their smoothness where the conditions themselves could be impossible to write. That is how the solution of this problem was progressing.

If the boundaries of domains are bad and there are several of them, it is not clear what the condition of the junction looks like. In this situation (connection), we need another approach to the solution of the set problem.

Since results of the junction must preserve the property of solution (being a generalized solution), we propose a new circuit system to solve the set problem. In general case, it is not solved.

The existence of eigenvalue is proved in [1] for the special case n = 2 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-133/MediaObjects/13661_2012_Article_234_IEq13_HTML.gif, N = 2 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-133/MediaObjects/13661_2012_Article_234_IEq14_HTML.gif, Ω 1 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-133/MediaObjects/13661_2012_Article_234_IEq15_HTML.gif - the circle. For more details, see [25] and [6].

Consider the problem
u ( x ) = ( g ( x ) + μ ) u ( x ) ; x R n , u L 2 ( R n ) , https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-133/MediaObjects/13661_2012_Article_234_Equ3_HTML.gif
(3)
in which g ( x ) = k j https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-133/MediaObjects/13661_2012_Article_234_IEq10_HTML.gif for all x Ω j https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-133/MediaObjects/13661_2012_Article_234_IEq16_HTML.gif, and
R n = j = 1 N ( Ω j Γ j ) . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-133/MediaObjects/13661_2012_Article_234_Equ4_HTML.gif
(4)

It is obvious that if we prove the existence of the eigenvalue (3), we obtain the following solution of the problem (1) u j ( x ) = u ( x ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-133/MediaObjects/13661_2012_Article_234_IEq17_HTML.gif; x Ω j https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-133/MediaObjects/13661_2012_Article_234_IEq7_HTML.gif, j = 1 , , N https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-133/MediaObjects/13661_2012_Article_234_IEq18_HTML.gif, where they are found automatically joined by a required form.

2 Formulation of the problem

We consider the eigenvalue problem (3) where R n = j = 1 N ( Ω j Γ j ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-133/MediaObjects/13661_2012_Article_234_IEq19_HTML.gif and Ω j https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-133/MediaObjects/13661_2012_Article_234_IEq1_HTML.gif, j = 1 , , N https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-133/MediaObjects/13661_2012_Article_234_IEq20_HTML.gif are mutually exclusive (disjoint) measurable sets with a positive measure. If we introduce a new spectral parameter λ = μ + k j https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-133/MediaObjects/13661_2012_Article_234_IEq21_HTML.gif, then the problem (1) takes the form
u ( x ) = ( c ( x ) + λ ) u ( x ) ; x R n , u L 2 ( R n ) , https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-133/MediaObjects/13661_2012_Article_234_Equ5_HTML.gif
(5)

in which c ( x ) = 0 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-133/MediaObjects/13661_2012_Article_234_IEq22_HTML.gif, if x Ω j https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-133/MediaObjects/13661_2012_Article_234_IEq7_HTML.gif.

The problem (5) is self-adjoint. This can be easily seen if we use the Fourier transformation. However, it does not influence the eigenvalue existence. Some examples of the problem (5) are known (with concrete k j https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-133/MediaObjects/13661_2012_Article_234_IEq2_HTML.gif, N and Ω j https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-133/MediaObjects/13661_2012_Article_234_IEq1_HTML.gif) both with and without eigenvalues.

To use the Fourier transformation ( F u ( x ) ) ( z ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-133/MediaObjects/13661_2012_Article_234_IEq23_HTML.gif of the distribution (generalized) function u ( x ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-133/MediaObjects/13661_2012_Article_234_IEq24_HTML.gif of slow growth, we must be aware of the following well-known Parseval equality:
( F u ( x ) ) ( z ) ( F v ( x ) ) ( z ) d z = u ( x ) v ( z x ) d x , https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-133/MediaObjects/13661_2012_Article_234_Equb_HTML.gif
and Plancherel’s theorem: ( F v ( x ) ) ( z ) L 2 ( R n ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-133/MediaObjects/13661_2012_Article_234_IEq25_HTML.gif if and only if v L 2 ( R n ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-133/MediaObjects/13661_2012_Article_234_IEq26_HTML.gif, where
v ( x ) = ( F v ( x ) ) ( z ) , https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-133/MediaObjects/13661_2012_Article_234_Equc_HTML.gif

for all u and v L 2 ( R n ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-133/MediaObjects/13661_2012_Article_234_IEq26_HTML.gif.

From now on, if it is not specifically indicated, the notation https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-133/MediaObjects/13661_2012_Article_234_IEq27_HTML.gif is the norm in the space L 2 ( R n ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-133/MediaObjects/13661_2012_Article_234_IEq28_HTML.gif.

3 The existence of negative eigenvalues for the general case

Let us consider the problem:
P ( D ) u ( x ) = ( c ( x ) + λ ) u ; x R n , u ( x ) L 2 ( R n ) , https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-133/MediaObjects/13661_2012_Article_234_Equ6_HTML.gif
(6)
in which c ( x ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-133/MediaObjects/13661_2012_Article_234_IEq29_HTML.gif is a measurable function, c ( x ) > 0 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-133/MediaObjects/13661_2012_Article_234_IEq30_HTML.gif for all x Ω https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-133/MediaObjects/13661_2012_Article_234_IEq31_HTML.gif, | c ( x ) | d < + https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-133/MediaObjects/13661_2012_Article_234_IEq32_HTML.gif almost everywhere in Ω, c ( x ) = 0 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-133/MediaObjects/13661_2012_Article_234_IEq22_HTML.gif outside Ω, Ω is measurable and P ( D ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-133/MediaObjects/13661_2012_Article_234_IEq33_HTML.gif is a linear pseudo-differential operator with constant coefficients. Here P ( i z ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-133/MediaObjects/13661_2012_Article_234_IEq34_HTML.gif argument quasi-polynomial z R n https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-133/MediaObjects/13661_2012_Article_234_IEq35_HTML.gif, not depending on x and satisfying the following conditions for all z R n https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-133/MediaObjects/13661_2012_Article_234_IEq35_HTML.gif:
https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-133/MediaObjects/13661_2012_Article_234_Equ7_HTML.gif
(7)
https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-133/MediaObjects/13661_2012_Article_234_Equ8_HTML.gif
(8)
https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-133/MediaObjects/13661_2012_Article_234_Equ9_HTML.gif
(9)
We suppose that
| z | δ | P ( i z ) μ | 2 d z + , if  μ 0 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-133/MediaObjects/13661_2012_Article_234_Equ10_HTML.gif
(10)

for each sufficiently small δ > 0 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-133/MediaObjects/13661_2012_Article_234_IEq36_HTML.gif and μ 0 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-133/MediaObjects/13661_2012_Article_234_IEq37_HTML.gif.

Theorem 1 The problem (6) has at least one negative eigenvalue if Ω is bounded.

It is necessary to introduce several lemmas before proving this theorem.

In each case, we consider μ < 0 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-133/MediaObjects/13661_2012_Article_234_IEq38_HTML.gif. By virtue of (8), there is a function h ( x , μ ) L 2 ( R n ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-133/MediaObjects/13661_2012_Article_234_IEq39_HTML.gif of the Fourier transformation which coincides with [ P ( i z ) μ ] 1 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-133/MediaObjects/13661_2012_Article_234_IEq40_HTML.gif. Considering (7), the real and even function h ( x , μ ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-133/MediaObjects/13661_2012_Article_234_IEq41_HTML.gif could be obtained.

Lemma 1 Let μ < 0 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-133/MediaObjects/13661_2012_Article_234_IEq38_HTML.gif. The problem (6) has a nonzero solution if and only if the nonzero solution v ( x ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-133/MediaObjects/13661_2012_Article_234_IEq42_HTML.gif has the form
v ( x ) = Ω c ( t ) v ( t ) h ( x t , μ ) d t , v L 2 ( μ ) . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-133/MediaObjects/13661_2012_Article_234_Equ11_HTML.gif
(11)
Proof Applying the Fourier transformation for (6) yields
( P ( i z ) μ ) ( F u ( x ) ) ( z ) = [ F ( c ( x ) u ( x ) ) ] ( z ) L 2 ( R n ) . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-133/MediaObjects/13661_2012_Article_234_Equd_HTML.gif
Hence, in particular, the integral
( F ( u ( x ) ) ) ( z ) e i x z d z https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-133/MediaObjects/13661_2012_Article_234_Eque_HTML.gif
converges absolutely. From now on, i x z = ( i x 1 z 1 , , i x n z n ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-133/MediaObjects/13661_2012_Article_234_IEq43_HTML.gif. It follows from latter relations
u ( x ) = 1 2 π [ F ( u ( x ) ) ] ( z ) e i x z d z = [ F ( c ( t ) u ( t ) ) ] ( z ) [ F t h ( x + t , μ ) ] ( z ) d z , https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-133/MediaObjects/13661_2012_Article_234_Equf_HTML.gif
where F t https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-133/MediaObjects/13661_2012_Article_234_IEq44_HTML.gif means that the Fourier transformation has been determined under t. Hence, by virtue of Parseval’s equality, it follows that
u ( x ) = c ( t ) u ( t ) h ( x t , μ ) d t . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-133/MediaObjects/13661_2012_Article_234_Equg_HTML.gif

Since c ( t ) = 0 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-133/MediaObjects/13661_2012_Article_234_IEq45_HTML.gif outside Ω, then u ( x ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-133/MediaObjects/13661_2012_Article_234_IEq24_HTML.gif; x Ω https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-133/MediaObjects/13661_2012_Article_234_IEq31_HTML.gif is the solution of the problem (11). If u ( t ) = 0 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-133/MediaObjects/13661_2012_Article_234_IEq46_HTML.gif where in Ω we obtain [ c ( t ) u ( t ) ] = 0 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-133/MediaObjects/13661_2012_Article_234_IEq47_HTML.gif for R n https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-133/MediaObjects/13661_2012_Article_234_IEq48_HTML.gif, by virtue of the latter equality u ( x ) = 0 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-133/MediaObjects/13661_2012_Article_234_IEq49_HTML.gif. The necessity is proved.

Let us prove the sufficiency. Let v ( x ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-133/MediaObjects/13661_2012_Article_234_IEq42_HTML.gif be the nonzero solution of the problem (11). Consider the new problem
P ( D ) u = μ u + c ( x ) f ( x ) , x R n ; u L 2 ( R n ) , https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-133/MediaObjects/13661_2012_Article_234_Equ12_HTML.gif
(12)
in which f ( x ) = v ( x ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-133/MediaObjects/13661_2012_Article_234_IEq50_HTML.gif for all x Ω https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-133/MediaObjects/13661_2012_Article_234_IEq51_HTML.gif and f ( x ) = 0 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-133/MediaObjects/13661_2012_Article_234_IEq52_HTML.gif outside Ω. Since c ( x ) f ( x ) L 2 ( R n ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-133/MediaObjects/13661_2012_Article_234_IEq53_HTML.gif, applying the Fourier transformation for (12), we obtain
[ P ( i z μ ) ] [ F u ( x ) ] ( z ) = [ F ( c ( x ) f ( x ) ) ] z L 2 ( R n ) ; u ( x ) = 1 2 π [ F ( c ( x ) f ( x ) ) ] ( z ) [ F t h ( x + t , μ ) ] ( z ) d z = c ( t ) f ( t ) h ( x t , μ ) d t . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-133/MediaObjects/13661_2012_Article_234_Equh_HTML.gif
From Parseval’s equality, the solution of the problem (12) exists and it is unique. In particular, when x Ω https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-133/MediaObjects/13661_2012_Article_234_IEq31_HTML.gif, we have
u ( x ) = Ω c ( t ) v ( t ) h ( x t , μ ) d t = v ( x ) . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-133/MediaObjects/13661_2012_Article_234_Equi_HTML.gif

Considering this inequality and (12), we obtain c ( x ) f ( x ) = c ( x ) u ( x ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-133/MediaObjects/13661_2012_Article_234_IEq54_HTML.gif, i.e., u ( x ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-133/MediaObjects/13661_2012_Article_234_IEq24_HTML.gif is the solution of the problem (6). Thus, the lemma is proved. □

In the case when μ < 0 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-133/MediaObjects/13661_2012_Article_234_IEq38_HTML.gif, we consider A ( μ ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-133/MediaObjects/13661_2012_Article_234_IEq55_HTML.gif as an integral operator, where
A ( μ ) ω ( x ) = c ( x ) Ω c ( t ) h ( x t , μ ) ω ( t ) d t ; ω L b 2 ( Ω ) . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-133/MediaObjects/13661_2012_Article_234_Equj_HTML.gif
We remember that the operator A ( μ ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-133/MediaObjects/13661_2012_Article_234_IEq55_HTML.gif is defined only when μ < 0 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-133/MediaObjects/13661_2012_Article_234_IEq38_HTML.gif. Since P ( i z ) = P ( i z ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-133/MediaObjects/13661_2012_Article_234_IEq56_HTML.gif, thus the Fourier transformation for the functions h ( t , μ ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-133/MediaObjects/13661_2012_Article_234_IEq57_HTML.gif, h ( t , μ ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-133/MediaObjects/13661_2012_Article_234_IEq58_HTML.gif coincides. That is why h ( t , μ ) = h ( t , μ ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-133/MediaObjects/13661_2012_Article_234_IEq59_HTML.gif. If Ω is bounded, then the kernel ( c ( x ) c ( t ) h ( x t , μ ) ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-133/MediaObjects/13661_2012_Article_234_IEq60_HTML.gif of the integrated operator A ( μ ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-133/MediaObjects/13661_2012_Article_234_IEq55_HTML.gif belongs to L b 2 ( Ω × Ω ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-133/MediaObjects/13661_2012_Article_234_IEq61_HTML.gif. It follows that the operator A ( μ ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-133/MediaObjects/13661_2012_Article_234_IEq55_HTML.gif is completely continuous. Its self-adjointness and positiveness are obvious. This enables us to write down the eigenvalues of the operator A ( μ ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-133/MediaObjects/13661_2012_Article_234_IEq55_HTML.gif:
λ 1 ( μ ) > λ 2 ( μ ) > . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-133/MediaObjects/13661_2012_Article_234_Equ13_HTML.gif
(13)
It is well known that (see [7])
λ 1 ( μ ) = Sup < A ( μ ) f , https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-133/MediaObjects/13661_2012_Article_234_Equ14_HTML.gif
(14)

where Sup is determined for all the function f L b 2 ( Ω ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-133/MediaObjects/13661_2012_Article_234_IEq62_HTML.gif, for which f 1 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-133/MediaObjects/13661_2012_Article_234_IEq63_HTML.gif.

From the known results for self-adjoint and quite continuous operators (see [7]), it follows that λ k ( μ ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-133/MediaObjects/13661_2012_Article_234_IEq64_HTML.gif continuously depends on μ, where
| λ 1 ( μ ) | 2 = A ( μ ) 2 = Ω Ω c ( x ) c ( t ) ( h ( x t , μ ) ) 2 d x d t . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-133/MediaObjects/13661_2012_Article_234_Equ15_HTML.gif
(15)
Lemma 2 Let Ω be bounded when x Ω https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-133/MediaObjects/13661_2012_Article_234_IEq51_HTML.gif. Then
  1. (1)

    λ k ( μ ) 0 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-133/MediaObjects/13661_2012_Article_234_IEq65_HTML.gif at μ https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-133/MediaObjects/13661_2012_Article_234_IEq66_HTML.gif,

     
  2. (2)

    λ 1 ( μ ) + https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-133/MediaObjects/13661_2012_Article_234_IEq67_HTML.gif at μ 0 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-133/MediaObjects/13661_2012_Article_234_IEq37_HTML.gif.

     
Proof Since λ j ( μ ) f j ( x , μ ) = A ( μ ) f ( x , μ ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-133/MediaObjects/13661_2012_Article_234_IEq68_HTML.gif, f j ( x , μ ) Ω = 1 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-133/MediaObjects/13661_2012_Article_234_IEq69_HTML.gif, and | c ( x ) | a < + https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-133/MediaObjects/13661_2012_Article_234_IEq70_HTML.gif, we have
| λ j ( μ ) | A ( μ ) = [ Ω Ω c ( x ) c ( t ) ( h ( x t , μ ) ) 2 d x d t ] 1 2 a [ Ω c ( t ) ( h ( x t , μ ) 2 d x ) d t ] 1 2 = a h ( x , μ ) ( Ω c ( t ) d t ) 1 2 a h ( x , μ ) c ( t ) Ω = a 1 ( d z ( P ( i z ) μ ) 2 ) 1 2 . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-133/MediaObjects/13661_2012_Article_234_Equk_HTML.gif

Hence, the first statement follows from (9).

Let us prove the second statement. By virtue of (13), with c ( x ) = 0 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-133/MediaObjects/13661_2012_Article_234_IEq22_HTML.gif outside Ω and
( λ 1 ( μ ) ) 2 = [ c ( x ) h ( x t , μ ) ] [ c ( t ) h ( x t , μ ) ] d x d t , https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-133/MediaObjects/13661_2012_Article_234_Equl_HTML.gif
which is applied to the last integral in Parseval’s inequality, we obtain
( λ 1 ( μ ) ) 2 = ( F [ c ( x ) h ( x t , μ ) ] ) ( z , ξ ) ( F [ c ( t ) h ( x t , μ ) ] ) ( z , ξ ) d z d ξ . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-133/MediaObjects/13661_2012_Article_234_Equm_HTML.gif
The following equations are correct:
( F [ c ( x ) h ( x t , μ ) ] ) ( z , ξ ) = ( F x [ c ( x ) e i x ξ P ( i ξ ) μ ] ) ( z , ξ ) = 1 P ( i ξ ) μ ( F x [ c ( x ) e i x ξ ] ) ( z , ξ ) = 1 2 π 1 P ( i ξ ) μ Ω c ( x ) e i x ( ξ + z ) d x . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-133/MediaObjects/13661_2012_Article_234_Equn_HTML.gif
In a similar way, we obtain
( F [ c ( t ) h ( x t , μ ) ] ) ( z , ξ ) = 1 2 π 1 P ( i ξ ) μ Ω c ( t ) e i t ( ξ + z ) d t . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-133/MediaObjects/13661_2012_Article_234_Equo_HTML.gif
Thus, we have proved the following:
( λ 1 ( μ ) ) 2 = 1 2 π | Ω c ( x ) e i x ( ξ + z ) d x | 2 d z d ξ [ P ( i ξ μ ) ] 2 . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-133/MediaObjects/13661_2012_Article_234_Equp_HTML.gif
The following estimate is obvious:
( λ 1 ( μ ) ) 2 1 2 π | z | δ | ξ | δ | Ω c ( x ) e i x ( ξ + z ) d x | 2 d z d ξ [ P ( i ξ ) μ ] 2 , https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-133/MediaObjects/13661_2012_Article_234_Equ16_HTML.gif
(16)
where
| Ω c ( x ) e i x ( ξ + z ) d x | 2 = | Ω c ( x ) d x + Ω c ( x ) [ e i x ( ξ + z ) ] d x | 2 , https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-133/MediaObjects/13661_2012_Article_234_Equ17_HTML.gif
(17)

δ will be chosen in a way such that | e i x ( z + ξ ) 1 | 1 2 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-133/MediaObjects/13661_2012_Article_234_IEq71_HTML.gif for all x Ω https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-133/MediaObjects/13661_2012_Article_234_IEq31_HTML.gif and | z | δ https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-133/MediaObjects/13661_2012_Article_234_IEq72_HTML.gif, | ξ | δ https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-133/MediaObjects/13661_2012_Article_234_IEq73_HTML.gif. Since Ω is bounded, we may always obtain the latter.

Considering (16) and (17), we obtain
( λ 1 ( μ ) ) 2 1 4 2 π ( Ω c ( x ) d x ) 2 | z | δ d z | ξ | δ d ξ [ P ( i ξ μ ) ] 2 = b 1 ( δ ) | z | δ d ξ ( P ( i ξ ) μ ) 2 . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-133/MediaObjects/13661_2012_Article_234_Equq_HTML.gif

Hence, by virtue of (10), the lemma is proved. □

Proof of Theorem 1 At the first stage, we suppose that c ( x ) ν > 0 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-133/MediaObjects/13661_2012_Article_234_IEq74_HTML.gif for all x Ω https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-133/MediaObjects/13661_2012_Article_234_IEq31_HTML.gif. By virtue of Lemmas 1 and 2, where λ 1 ( μ 0 ( ν ) ) = 1 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-133/MediaObjects/13661_2012_Article_234_IEq75_HTML.gif for μ 0 ( ν ) < 0 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-133/MediaObjects/13661_2012_Article_234_IEq76_HTML.gif, if f 1 ( x ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-133/MediaObjects/13661_2012_Article_234_IEq77_HTML.gif is the eigenfunction corresponding to the eigenvalue λ 1 ( μ 0 ( ν ) ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-133/MediaObjects/13661_2012_Article_234_IEq78_HTML.gif, then
ϕ ( x ) = f 1 ( x ) / c ( x ) L 2 ( Ω ) . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-133/MediaObjects/13661_2012_Article_234_Equr_HTML.gif

When μ = μ 0 ( ν ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-133/MediaObjects/13661_2012_Article_234_IEq79_HTML.gif, we have the nonzero solution of the equation (11). It follows from Lemma 1 that μ 0 ( ν ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-133/MediaObjects/13661_2012_Article_234_IEq80_HTML.gif is the eigenvalue of the problem (6).

For the general case, we put c ( ν , x ) = c ( x ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-133/MediaObjects/13661_2012_Article_234_IEq81_HTML.gif if x Ω https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-133/MediaObjects/13661_2012_Article_234_IEq51_HTML.gif and c ( x ) ν https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-133/MediaObjects/13661_2012_Article_234_IEq82_HTML.gif; c ( ν , x ) = ν https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-133/MediaObjects/13661_2012_Article_234_IEq83_HTML.gif when x Ω https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-133/MediaObjects/13661_2012_Article_234_IEq31_HTML.gif and c ( x ) ( 0 , ν ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-133/MediaObjects/13661_2012_Article_234_IEq84_HTML.gif. The nonzero solutions of the equation
ϕ ( ν , x ) = Ω c ( ν , x ) h ( x t , μ 0 ( ν ) ) ϕ ( ν , t ) d t https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-133/MediaObjects/13661_2012_Article_234_Equ18_HTML.gif
(18)

are chosen in such a way that ϕ ( ν , x ) Ω https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-133/MediaObjects/13661_2012_Article_234_IEq85_HTML.gif.

The integral operators defined by the right-hand sides of (11) and (18) are defined in B ( μ ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-133/MediaObjects/13661_2012_Article_234_IEq86_HTML.gif, B ( ν , μ 0 ( ν ) ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-133/MediaObjects/13661_2012_Article_234_IEq87_HTML.gif respectively. Since Ω is bounded, then both { c ( ν , x ) } https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-133/MediaObjects/13661_2012_Article_234_IEq88_HTML.gif and { h ( x , μ 0 ( ν ) ) } https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-133/MediaObjects/13661_2012_Article_234_IEq89_HTML.gif uniformly converge by norm to c ( x ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-133/MediaObjects/13661_2012_Article_234_IEq29_HTML.gif and h ( x , μ 0 ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-133/MediaObjects/13661_2012_Article_234_IEq90_HTML.gif respectively. If μ 0 ( ν ) μ 0 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-133/MediaObjects/13661_2012_Article_234_IEq91_HTML.gif, then
B ( μ 0 ) B ( ν , μ 0 ( ν ) ) 0 when  μ 0 ( ν ) μ 0 . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-133/MediaObjects/13661_2012_Article_234_Equ19_HTML.gif
(19)

Considering the choice ϕ ( ν , x ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-133/MediaObjects/13661_2012_Article_234_IEq92_HTML.gif and the property h ( x , μ ) 0 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-133/MediaObjects/13661_2012_Article_234_IEq93_HTML.gif, if μ https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-133/MediaObjects/13661_2012_Article_234_IEq94_HTML.gif, we can easily prove the boundedness of μ 0 ( ν ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-133/MediaObjects/13661_2012_Article_234_IEq95_HTML.gif. Noting that when μ 0 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-133/MediaObjects/13661_2012_Article_234_IEq96_HTML.gif and ν 0 0 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-133/MediaObjects/13661_2012_Article_234_IEq97_HTML.gif for which μ 0 ( ν j ) μ 0 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-133/MediaObjects/13661_2012_Article_234_IEq98_HTML.gif, the operator B ( μ 0 ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-133/MediaObjects/13661_2012_Article_234_IEq99_HTML.gif is completely continuous. In this case, as we know, the set B ( μ 0 ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-133/MediaObjects/13661_2012_Article_234_IEq99_HTML.gif, ϕ ( ν , x ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-133/MediaObjects/13661_2012_Article_234_IEq92_HTML.gif contains the subsequence B ( μ 0 ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-133/MediaObjects/13661_2012_Article_234_IEq99_HTML.gif, ϕ ( ν j , x ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-133/MediaObjects/13661_2012_Article_234_IEq100_HTML.gif which converges by norm where μ μ 0 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-133/MediaObjects/13661_2012_Article_234_IEq101_HTML.gif.

From (18) and (19) it follows that { ϕ ( ν j , x ) } https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-133/MediaObjects/13661_2012_Article_234_IEq102_HTML.gif converges to ϕ ( x ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-133/MediaObjects/13661_2012_Article_234_IEq103_HTML.gif by norm where u ( x ) Ω = 1 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-133/MediaObjects/13661_2012_Article_234_IEq104_HTML.gif. Then { B ( ν , μ 0 ( ν j 1 ) ) ϕ ( ν j 1 , x ) } https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-133/MediaObjects/13661_2012_Article_234_IEq105_HTML.gif converges to B ( μ 0 ) ϕ ( x ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-133/MediaObjects/13661_2012_Article_234_IEq106_HTML.gif by norm and satisfies the equality u ( x ) = B ( μ 0 ) ϕ ( x ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-133/MediaObjects/13661_2012_Article_234_IEq107_HTML.gif, i.e., when μ = μ 0 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-133/MediaObjects/13661_2012_Article_234_IEq108_HTML.gif, the equation (11) has a nonzero solution. Hence, the theorem is proved. □

4 Application to the problem of a dielectric waveguide

In the case of
P ( i z ) = ( i z 1 ) 2 ( i z 2 ) 2 ( i z n ) 2 = | z 1 | 2 + | z 2 | 2 + + | z n | 2 = | z | 2 0 , https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-133/MediaObjects/13661_2012_Article_234_Equs_HTML.gif
where
P ( i z ) = P ( i z ) for all  z R n , https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-133/MediaObjects/13661_2012_Article_234_Equt_HTML.gif
the condition (7) takes the form
R n ( | z | 2 μ ) 2 d z < + when  μ < 0 . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-133/MediaObjects/13661_2012_Article_234_Equ20_HTML.gif
(20)

It is clear that in the case of n arbitrary, these requirements are not satisfied. However, it takes place in the case n 3 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-133/MediaObjects/13661_2012_Article_234_IEq109_HTML.gif important for the application. It can easily be proved when we use the spherical coordinates. Moreover, for the case when n 3 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-133/MediaObjects/13661_2012_Article_234_IEq109_HTML.gif, (9) also takes place. Let us make sure that (10) is satisfied when n 4 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-133/MediaObjects/13661_2012_Article_234_IEq110_HTML.gif.

Let
I = | z | δ d z ( | z | 2 μ 2 ) 2 . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-133/MediaObjects/13661_2012_Article_234_Equu_HTML.gif
Consider the spherical coordinates
z 1 = r cos θ 1 , z 2 = r sin θ 1 cos θ 2 , , z n 1 = r sin θ 1 sin θ n 2 cos θ n 1 , z n = r sin θ 1 sin θ 2 sin θ n 2 sin θ n 1 . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-133/MediaObjects/13661_2012_Article_234_Equ21_HTML.gif
(21)
The left-hand side of (20) takes the form
0 + r n 1 c ( θ ) ( r 2 μ ) 2 ( sin θ 1 ) n 2 ( sin θ 2 ) n 3 ( sin θ n 2 ) d r d θ 1 d θ 2 d θ n 1 = a 2 0 + r n 1 ( r 2 μ ) 2 d r , https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-133/MediaObjects/13661_2012_Article_234_Equv_HTML.gif
where
c ( θ ) = { θ 1 [ 0 , 2 π ] , θ j [ π 2 , π 2 ] , j = 2 , , n 1 } . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-133/MediaObjects/13661_2012_Article_234_Equw_HTML.gif
It follows that
I = 0 δ c ( θ ) r n 1 sin n 2 θ 1 sin n 3 θ 2 sin θ n 2 ( r 2 μ ) 2 d r d θ 1 d θ n , https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-133/MediaObjects/13661_2012_Article_234_Equx_HTML.gif
where
c ( θ ) = { 0 θ 1 2 π , π 2 θ j < π 2 , j = 2 , , n } . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-133/MediaObjects/13661_2012_Article_234_Equy_HTML.gif
Hence,
I = b 2 0 δ r n 1 ( r 2 μ ) 2 d r . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-133/MediaObjects/13661_2012_Article_234_Equz_HTML.gif
We can see that when n 4 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-133/MediaObjects/13661_2012_Article_234_IEq111_HTML.gif,
0 δ r n 1 ( r 2 μ ) 2 d r + ; μ 0 . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-133/MediaObjects/13661_2012_Article_234_Equaa_HTML.gif
Taking into account that (10) is satisfied and denoting index j m https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-133/MediaObjects/13661_2012_Article_234_IEq112_HTML.gif in which k j https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-133/MediaObjects/13661_2012_Article_234_IEq2_HTML.gif is the minimum, the problem (5) can be rewritten in the following form:
u ( x ) = [ ( g ( x ) k j m ) + ( μ + k j m ) ] u ( x ) , x R n , u ( x ) L 2 ( R n ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-133/MediaObjects/13661_2012_Article_234_Equ22_HTML.gif
(22)

when g ( x ) k j m = c ( x ) = 0 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-133/MediaObjects/13661_2012_Article_234_IEq113_HTML.gif for all x Ω j m https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-133/MediaObjects/13661_2012_Article_234_IEq114_HTML.gif; i.e., c ( x ) = 0 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-133/MediaObjects/13661_2012_Article_234_IEq22_HTML.gif, outside ( R n Ω j m ) = Ω https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-133/MediaObjects/13661_2012_Article_234_IEq115_HTML.gif, where c ( x ) > 0 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-133/MediaObjects/13661_2012_Article_234_IEq116_HTML.gif at x Ω https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-133/MediaObjects/13661_2012_Article_234_IEq51_HTML.gif.

The theorem may be applied to the problem (21). As a consequence of this theorem, we get the following:

Theorem 2 If Ω is bounded, the problem (3) has an eigenvalue μ for which μ + k j m < 0 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-133/MediaObjects/13661_2012_Article_234_IEq117_HTML.gif.

Let j M https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-133/MediaObjects/13661_2012_Article_234_IEq118_HTML.gif be the index at which k j https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-133/MediaObjects/13661_2012_Article_234_IEq2_HTML.gif is maximum. Then the problem (3) may take the form
u ( x ) = [ ( g ( x ) k j M ) + ( μ + k j M ) ] u ( x ) ; x R n , u ( x ) L 2 ( R n ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-133/MediaObjects/13661_2012_Article_234_Equ23_HTML.gif
(23)
when
g ( x ) k j k < 0 , if  x Ω j M https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-133/MediaObjects/13661_2012_Article_234_Equab_HTML.gif
and
g ( x ) k j k = 0 , if  x Ω j M . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-133/MediaObjects/13661_2012_Article_234_Equac_HTML.gif

Now, we formulate the following theorem.

Theorem 3 The problem (3) does not have an eigenvalue μ for which https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-133/MediaObjects/13661_2012_Article_234_IEq119_HTML.gif .

Proof Multiplying the equality (22) by u ( x ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-133/MediaObjects/13661_2012_Article_234_IEq24_HTML.gif and integrating it in R n https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-133/MediaObjects/13661_2012_Article_234_IEq48_HTML.gif, we have
( u ( x ) ) u ( x ) d x = [ ( x 1 ) 2 + + ( x n ) 2 ] u ( x ) d x = ( g ( x ) k j M ) ( u ( x ) ) 2 d x + ( μ + k j M ) ( u ( x ) ) 2 d x . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-133/MediaObjects/13661_2012_Article_234_Equad_HTML.gif

If g ( x ) k j M 0 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-133/MediaObjects/13661_2012_Article_234_IEq120_HTML.gif, μ + k j M 0 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-133/MediaObjects/13661_2012_Article_234_IEq121_HTML.gif, then by virtue of the condition u ( x ) = 0 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-133/MediaObjects/13661_2012_Article_234_IEq49_HTML.gif, the latter is not impossible. □

By virtue of Theorems 2 and 3, we have

Theorem 4 Let ( R n Ω j m ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-133/MediaObjects/13661_2012_Article_234_IEq122_HTML.gif be bounded. Then the problem (3) has an eigenvalue μ which satisfies the condition https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-133/MediaObjects/13661_2012_Article_234_IEq123_HTML.gif .

Remark If the condition that the bounded set ( R n Ω j m ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-133/MediaObjects/13661_2012_Article_234_IEq122_HTML.gif is not valid, then the problem may not have eigenvalues.

5 Conclusions

This paper deals with the existence of eigenvalue problems for the waveguide theory. These problems are very important in the study of the mathematical analysis and mathematical physics. In this paper, we introduced four theorems and two lemmas.

Declarations

Acknowledgements

We wish to thank the referees for their valuable comments which improved the original manuscript.

Authors’ Affiliations

(1)
Department of Mathematics, Faculty of Science, Assiut University
(2)
Department of Applied Mathematics, Kazan State University

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Copyright

© Maher and Karachevskii; licensee Springer 2012

This article is published under license to BioMed Central Ltd. This is an Open Access article distributed under the terms of the Creative Commons Attribution License (http://​creativecommons.​org/​licenses/​by/​2.​0), which permits unrestricted use, distribution, and reproduction in any medium, provided the original work is properly cited.