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The existence of eigenvalue problems for the waveguide theory
Boundary Value Problems volume 2012, Article number: 133 (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.
1 Basic preliminaries
A dielectric waveguide is a composite of its own index of refraction for each layer. If {\mathrm{\Omega}}_{j} is a layer, where the index of refraction is {k}_{j} and μ is a spectral parameter, then the waveguide process can be written in the following form:
where
In order to obtain {u}_{j}(x) and {u}_{r}(x), the process in all the waveguide for the common boundary of domains {\mathrm{\Omega}}_{j} and {\mathrm{\Omega}}_{r} is evaluated. {u}_{j}(x) and {u}_{r}(x) must be joined in the way that the obtained known functions {u}_{j,r}(x)={u}_{j}(x) for x\in {\mathrm{\Omega}}_{j} and {u}_{j,r}(x)={u}_{r}(x) for x\in {\mathrm{\Omega}}_{r} will be the generalized solution of the equation
in which g(x)={k}_{j} for all x\in {\mathrm{\Omega}}_{j} and g(x)={k}_{r} for all x\in {\mathrm{\Omega}}_{r}. If the boundary {\mathrm{\Gamma}}_{j,r} is sufficiently smooth, the condition of this junction may be put down in a natural form. Indeed, the contraction of {\mathrm{\Gamma}}_{j,r} is noninfinitely smooth in {\mathrm{\Omega}}_{j} and {\mathrm{\Omega}}_{r}, 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, N=2, {\mathrm{\Omega}}_{1}  the circle. For more details, see [2–5] and [6].
Consider the problem
in which g(x)={k}_{j} for all x\in {\mathrm{\Omega}}_{j}, and
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); x\in {\mathrm{\Omega}}_{j}, j=1,\dots ,N, where they are found automatically joined by a required form.
2 Formulation of the problem
We consider the eigenvalue problem (3) where {R}^{n}={\bigcup}_{j=1}^{N}({\mathrm{\Omega}}_{j}\cup {\mathrm{\Gamma}}_{j}) and {\mathrm{\Omega}}_{j}, j=1,\dots ,N are mutually exclusive (disjoint) measurable sets with a positive measure. If we introduce a new spectral parameter \lambda =\mu +{k}_{j}, then the problem (1) takes the form
in which c(x)=0, if x\in {\mathrm{\Omega}}_{j}.
The problem (5) is selfadjoint. 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}, N and {\mathrm{\Omega}}_{j}) both with and without eigenvalues.
To use the Fourier transformation (Fu(x))(z) of the distribution (generalized) function u(x) of slow growth, we must be aware of the following wellknown Parseval equality:
and Plancherel’s theorem: (Fv(x))(z)\in {L}^{2}({R}^{n}) if and only if v\in {L}^{2}({R}^{n}), where
for all u and v\in {L}^{2}({R}^{n}).
From now on, if it is not specifically indicated, the notation \parallel \cdot \parallel is the norm in the space {L}^{2}({R}^{n}).
3 The existence of negative eigenvalues for the general case
Let us consider the problem:
in which c(x) is a measurable function, c(x)>0 for all x\in \mathrm{\Omega}, c(x)\le d<+\mathrm{\infty} almost everywhere in Ω, c(x)=0 outside Ω, Ω is measurable and P(D) is a linear pseudodifferential operator with constant coefficients. Here P(iz)\ge argument quasipolynomial z\in {R}^{n}, not depending on x and satisfying the following conditions for all z\in {R}^{n}:
We suppose that
for each sufficiently small \delta >0 and \mu \to {0}^{}.
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 \mu <0. By virtue of (8), there is a function h(x,\mu )\in {L}^{2}({R}^{n}) of the Fourier transformation which coincides with {[P(iz)\mu ]}^{1}. Considering (7), the real and even function h(x,\mu ) could be obtained.
Lemma 1 Let \mu <0. The problem (6) has a nonzero solution if and only if the nonzero solution v(x) has the form
Proof Applying the Fourier transformation for (6) yields
Hence, in particular, the integral
converges absolutely. From now on, ixz=(i{x}_{1}{z}_{1},\dots ,i{x}_{n}{z}_{n}). It follows from latter relations
where {F}_{t} means that the Fourier transformation has been determined under t. Hence, by virtue of Parseval’s equality, it follows that
Since c(t)=0 outside Ω, then u(x); x\in \mathrm{\Omega} is the solution of the problem (11). If u(t)=0 where in Ω we obtain [c(t)u(t)]=0 for {R}^{n}, by virtue of the latter equality u(x)=0. The necessity is proved.
Let us prove the sufficiency. Let v(x) be the nonzero solution of the problem (11). Consider the new problem
in which f(x)=v(x) for all x\in \mathrm{\Omega} and f(x)=0 outside Ω. Since c(x)f(x)\in {L}^{2}({R}^{n}), applying the Fourier transformation for (12), we obtain
From Parseval’s equality, the solution of the problem (12) exists and it is unique. In particular, when x\in \mathrm{\Omega}, we have
Considering this inequality and (12), we obtain c(x)f(x)=c(x)u(x), i.e., u(x) is the solution of the problem (6). Thus, the lemma is proved. □
In the case when \mu <0, we consider A(\mu ) as an integral operator, where
We remember that the operator A(\mu ) is defined only when \mu <0. Since P(iz)=P(iz), thus the Fourier transformation for the functions h(t,\mu ), h(t,\mu ) coincides. That is why h(t,\mu )=h(t,\mu ). If Ω is bounded, then the kernel (\sqrt{c(x)}\sqrt{c(t)}h(xt,\mu )) of the integrated operator A(\mu ) belongs to {L}_{b}^{2}(\mathrm{\Omega}\times \mathrm{\Omega}). It follows that the operator A(\mu ) is completely continuous. Its selfadjointness and positiveness are obvious. This enables us to write down the eigenvalues of the operator A(\mu ):
It is well known that (see [7])
where Sup is determined for all the function f\in {L}_{b}^{2}(\mathrm{\Omega}), for which \parallel f\parallel \le 1.
From the known results for selfadjoint and quite continuous operators (see [7]), it follows that {\lambda}_{k}(\mu ) continuously depends on μ, where
Lemma 2 Let Ω be bounded when x\in \mathrm{\Omega}. Then

(1)
{\lambda}_{k}(\mu )\to 0 at \mu \to \mathrm{\infty},

(2)
{\lambda}_{1}(\mu )\to +\mathrm{\infty} at \mu \to {0}^{}.
Proof Since {\lambda}_{j}(\mu ){f}_{j}(x,\mu )=A(\mu )f(x,\mu ), {\parallel {f}_{j}(x,\mu )\parallel}_{\mathrm{\Omega}}=1, and c(x)\le a<+\mathrm{\infty}, we have
Hence, the first statement follows from (9).
Let us prove the second statement. By virtue of (13), with c(x)=0 outside Ω and
which is applied to the last integral in Parseval’s inequality, we obtain
The following equations are correct:
In a similar way, we obtain
Thus, we have proved the following:
The following estimate is obvious:
where
δ will be chosen in a way such that {e}^{ix(z+\xi )}1\le \frac{1}{2} for all x\in \mathrm{\Omega} and z\le \delta, \xi \le \delta. Since Ω is bounded, we may always obtain the latter.
Considering (16) and (17), we obtain
Hence, by virtue of (10), the lemma is proved. □
Proof of Theorem 1 At the first stage, we suppose that c(x)\ge \nu >0 for all x\in \mathrm{\Omega}. By virtue of Lemmas 1 and 2, where {\lambda}_{1}({\mu}_{0}(\nu ))=1 for {\mu}_{0}(\nu )<0, if {f}_{1}(x) is the eigenfunction corresponding to the eigenvalue {\lambda}_{1}({\mu}_{0}(\nu )), then
When \mu ={\mu}_{0}(\nu ), we have the nonzero solution of the equation (11). It follows from Lemma 1 that {\mu}_{0}(\nu ) is the eigenvalue of the problem (6).
For the general case, we put c(\nu ,x)=c(x) if x\in \mathrm{\Omega} and c(x)\ge \nu; c(\nu ,x)=\nu when x\in \mathrm{\Omega} and c(x)\in (0,\nu ). The nonzero solutions of the equation
are chosen in such a way that {\parallel \varphi (\nu ,x)\parallel}_{\mathrm{\Omega}}.
The integral operators defined by the righthand sides of (11) and (18) are defined in B(\mu ), B(\nu ,{\mu}_{0}(\nu )) respectively. Since Ω is bounded, then both \{c(\nu ,x)\} and \{h(x,{\mu}_{0}(\nu ))\} uniformly converge by norm to c(x) and h(x,{\mu}_{0}) respectively. If {\mu}_{0}(\nu )\to {\mu}_{0}, then
Considering the choice \varphi (\nu ,x) and the property \parallel h(x,\mu )\parallel \to 0, if \mu \to \mathrm{\infty}, we can easily prove the boundedness of {\mu}_{0}(\nu ). Noting that when {\mu}_{0} and {\nu}_{0}\to 0 for which {\mu}_{0}({\nu}_{j})\to {\mu}_{0}, the operator B({\mu}_{0}) is completely continuous. In this case, as we know, the set B({\mu}_{0}), \varphi (\nu ,x) contains the subsequence B({\mu}_{0}), \varphi ({\nu}_{j},x) which converges by norm where \mu \to {\mu}_{0}.
From (18) and (19) it follows that \{\varphi ({\nu}_{j},x)\} converges to \varphi (x) by norm where {\parallel u(x)\parallel}_{\mathrm{\Omega}}=1. Then \{B(\nu ,{\mu}_{0}({\nu}_{{j}_{1}}))\varphi ({\nu}_{{j}_{1}},x)\} converges to B({\mu}_{0})\varphi (x) by norm and satisfies the equality u(x)=B({\mu}_{0})\varphi (x), i.e., when \mu ={\mu}_{0}, 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
where
the condition (7) takes the form
It is clear that in the case of n arbitrary, these requirements are not satisfied. However, it takes place in the case n\le 3 important for the application. It can easily be proved when we use the spherical coordinates. Moreover, for the case when n\le 3, (9) also takes place. Let us make sure that (10) is satisfied when n\le 4.
Let
Consider the spherical coordinates
The lefthand side of (20) takes the form
where
It follows that
where
Hence,
We can see that when n\le 4,
Taking into account that (10) is satisfied and denoting index {j}_{m} in which {k}_{j} is the minimum, the problem (5) can be rewritten in the following form:
when g(x){k}_{{j}_{m}}=c(x)=0 for all x\in {\mathrm{\Omega}}_{{j}_{m}}; i.e., c(x)=0, outside ({R}^{n}{\mathrm{\Omega}}_{{j}_{m}})=\mathrm{\Omega}, where c(x)>0 at x\in \mathrm{\Omega}.
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 \mu +{k}_{{j}_{m}}<0.
Let {j}_{M} be the index at which {k}_{j} is maximum. Then the problem (3) may take the form
when
and
Now, we formulate the following theorem.
Theorem 3 The problem (3) does not have an eigenvalue μ for which .
Proof Multiplying the equality (22) by u(x) and integrating it in {R}^{n}, we have
If g(x){k}_{{j}_{M}}\le 0, \mu +{k}_{{j}_{M}}\le 0, then by virtue of the condition u(x)=0, the latter is not impossible. □
By virtue of Theorems 2 and 3, we have
Theorem 4 Let ({R}^{n}{\mathrm{\Omega}}_{{j}_{m}}) be bounded. Then the problem (3) has an eigenvalue μ which satisfies the condition .
Remark If the condition that the bounded set ({R}^{n}{\mathrm{\Omega}}_{{j}_{m}}) 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.
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Acknowledgements
We wish to thank the referees for their valuable comments which improved the original manuscript.
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The idea of this paper was introduced by the first author. The second author shared the first author in calculations.
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Maher, A., Karachevskii, E. The existence of eigenvalue problems for the waveguide theory. Bound Value Probl 2012, 133 (2012). https://doi.org/10.1186/168727702012133
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DOI: https://doi.org/10.1186/168727702012133