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Abstract and Applied Analysis
Volumeย 2008ย (), Article IDย 765920, 12 pages
Research Article

The Analysis of Contour Integrals

1Department of Mathematics, Harran University, Osmanbey Campus, Sanlurfa 63100, Turkey
2Department of Mathematics, University of Pittsburgh, 301 Thackeray Hall, Pittsburgh, PA 15260, USA

Received 10 November 2007; Accepted 19 January 2008

Academic Editor: Stephen L.ย Clark

Copyright ยฉ 2008 Tanfer Tanriverdi and JohnBryce Mcleod. 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.


For any ๐‘›, the contour integral ๐‘ฆ=cosh๐‘›+1๐‘ฅโˆฎ๐ถ(cosh(๐‘ง๐‘ )/(sinh๐‘งโˆ’sinh๐‘ฅ)๐‘›+1๐‘‘๐‘ง,๐‘ 2=โˆ’๐œ†, is associated with differential equation ๐‘‘2๐‘ฆ(๐‘ฅ)/๐‘‘๐‘ฅ2+(๐œ†+๐‘›(๐‘›+1)/cosh2๐‘ฅ)๐‘ฆ(๐‘ฅ)=0. Explicit solutions for ๐‘›=1 are obtained. For ๐‘›=1, eigenvalues, eigenfunctions, spectral function, and eigenfunction expansions are explored. This differential equation which does have solution in terms of the trigonometric functions does not seem to have been explored and it is also one of the purposes of this paper to put it on record.

1. Introduction

When one considers eigenfunction expansions associated with second-order ordinary differential equations, as Titchmarsh does in his book [1], one is concerned with solutions of the equation โˆ’๐‘‘2๐‘ฆ(๐‘ฅ)๐‘‘๐‘ฅ2+๐‘ž(๐‘ฅ)๐‘ฆ(๐‘ฅ)=๐œ†๐‘ฆ(๐‘ฅ)(1.1)along with certain boundary conditions, and one tends to say that the only case in which one can solve this equation explicitly in elementary terms for all ๐œ† is the case ๐‘ž(๐‘ฅ)=0, when the solutions are of course trigonometric functions.

Now in fact this is not true, and there is in particular one problem which does not seem to have been explored, and it is the purpose of this paper to put it on record. Here is the problem: ๐‘‘2๐‘ฆ(๐‘ฅ)๐‘‘๐‘ฅ2+๎€ท๐œ†+๐‘›(๐‘›+1)sech2๐‘ฅ๎€ธ๐‘ฆ(๐‘ฅ)=0,(1.2) which can be solved explicitly in elementary terms when ๐‘› is integral. The explicit solution was known to Kamke [2], but Kamke does not anyway explore the consequences for eigenfunction expansions nor does Titchmarsh discuss this problem, although he does discuss problems close to it, for example, ๐‘‘2๐‘ฆ(๐‘ฅ)๐‘‘๐‘ฅ2+๎‚€๎‚€๐œˆ๐œ†+2โˆ’14๎‚sec2๐‘ฅ๎‚๐‘ฆ(๐‘ฅ)=0,(1.3) on (โˆ’๐œ‹/2,๐œ‹/2), which leads, when ๐œˆ=๐‘›, to an expansion in series involving associated Legendre functions.

It is perhaps worth remarking how our interest in this problem arises. In [3] there is the question of travelling waves and steady solutions for a discrete reaction-diffusion equation of the type ๐‘ข๎…ž๐‘›=๐‘ข๐‘›+1โˆ’2๐‘ข๐‘›+๐‘ข๐‘›โˆ’1+๐‘“(๐‘ข๐‘›),(1.4) where the function ๐‘“ is โ€œbistableโ€. That is, there exist three numbers ๐‘ˆ1, ๐‘ˆ2, ๐‘ˆ3, ๐‘ˆ1<๐‘ˆ2<๐‘ˆ3, such that ๐‘“(๐‘ˆ1)=๐‘“(๐‘ˆ2)=๐‘“(๐‘ˆ3)=0(1.5) with ๐‘“<0 in (๐‘ˆ1,๐‘ˆ2) and ๐‘“>0 in (๐‘ˆ2,๐‘ˆ3). A prototypical ๐‘“ would be ๐‘“(๐‘ข)=โˆ’๐ดsin(๐‘ข)+๐น,(1.6) where ๐ด and ๐น are positive constants and ๐น<๐ด, so that ๐‘ˆ1=sinโˆ’1๎‚€๐น๐ด๎‚,๐‘ˆ2=๐œ‹+sinโˆ’1๎‚€๐น๐ด๎‚,๐‘ˆ3=2๐œ‹+sinโˆ’1๎‚€๐น๐ด๎‚.(1.7) Such equations arise in a number of different applications, for example, in dislocation theory where ๐‘ข๐‘› is the displacement of the ๐‘›th atom in some material, or in neurobiology where ๐‘ข๐‘› is typically the electric potential of the ๐‘›th nerve cell, and in both these applications the interest is in monotonic solutions ๐‘ข๐‘› with ๐‘ข๐‘›โ†’๐‘ˆ1 as ๐‘›โ†’โˆ’โˆž, ๐‘ข๐‘›โ†’๐‘ˆ3 as ๐‘›โ†’โˆž [3โ€“6].

The basic question is whether there exist such solutions with the form of a travelling wave, ๐‘ข(๐‘›โˆ’๐‘๐‘ก), ๐‘โ‰ 0, or of a steady solution or standing wave, where ๐‘=0, and there is an important distinction between these two cases. For a travelling wave, ๐‘โ‰ 0, ๐‘ข๐‘› is clearly a function of the continuous variable ๐‘ก, and indeed because of (1.4), a differentiable function of ๐‘ก. This leads to the difference-differential equation ๐‘ข๐‘ก๎€ท๎€ธ(๐‘ฅ,๐‘ก)=๐‘ข(๐‘ฅ+1,๐‘ก)โˆ’2๐‘ข(๐‘ฅ,๐‘ก)+๐‘ข(๐‘ฅโˆ’1,๐‘ก)+๐‘“๐‘ข(๐‘ฅ,๐‘ก).(1.8) If, however, ๐‘=0, then, as in [3], we have to study the purely difference equation ๎€ท๎€ธ๐‘ข(๐‘ฅ+1,๐‘ก)โˆ’2๐‘ข(๐‘ฅ,๐‘ก)+๐‘ข(๐‘ฅโˆ’1,๐‘ก)+๐‘“๐‘ข(๐‘ฅ,๐‘ก)=0,(1.9) and the solutions may be discontinuous since there is nothing that now connects values of ๐‘ข(๐‘ฅ) with values of ๐‘ข(๐‘ฅ+๐›ฟ) for |๐›ฟ|<1. It is best therefore to think of the solution of (1.9) as a number of (monotonic) sequences ๐‘ข๐›ผ๐‘› indexed by ๐›ผ, each satisfying ๐‘ข๐‘›+1โˆ’2๐‘ข๐‘›+๐‘ข๐‘›โˆ’1+๐‘“(๐‘ข๐‘›)=0.(1.10) The simplest case would be that there is just one such sequence (modulo the translation ๐‘›โ†’๐‘›+๐‘˜, ๐‘˜ integral), but it is possible that there may be a finite number, or even a partial or total continuum.

In view of applications, where the distance between atoms or nerve cells is small, it is more natural to think of (1.8) in the form ๐‘ข๐‘ก๎€ท๎€ธ(๐‘ฅ,๐‘ก)=๐‘ข(๐‘ฅ+๐œ–,๐‘ก)โˆ’2๐‘ข(๐‘ฅ,๐‘ก)+๐‘ข(๐‘ฅโˆ’๐œ–,๐‘ก)+๐‘“๐‘ข(๐‘ฅ,๐‘ก),(1.11) where ๐œ– is small and represents the distance between atoms or nerve cells. A tempting approximation is then ๐‘ข๐‘ก(๐‘ฅ,๐‘ก)โˆผ๐œ–2๐‘ข๐‘ฅ๐‘ฅ๎€ท๎€ธ,(๐‘ฅ,๐‘ก)+๐‘“๐‘ข(๐‘ฅ,๐‘ก)(1.12) and in order to make sense of the scaling, in [3] the authors introduced a factor ๐œ–2 in front of ๐‘“. This therefore leads to a comparison between the solutions of ๐‘ข๎…ž๐‘›=๐‘ข๐‘›+1โˆ’2๐‘ข๐‘›+๐‘ข๐‘›โˆ’1+๐œ–2๐‘“(๐‘ข๐‘›๐‘ข)=0,(1.13)๐‘ก(๐‘ฅ,๐‘ก)=๐‘ข๐‘ฅ๐‘ฅ(๐‘ฅ,๐‘ก)+๐‘“(๐‘ข).(1.14) For the continuous diffusion problem, the answer is both simple and well known [4, 5].

Given a function ๐‘“ that is bistable, there is just one possible wave-speed ๐‘, and this value of ๐‘ is 0, that is, there is a steady solution if and only if ๎€œ๐‘ˆ3๐‘ˆ1๐‘“(๐‘ข)๐‘‘๐‘ข=0.(1.15) (The proof is a simple phase plane argument, and ๐‘=0 implies (1.15) follows by multiplying (1.14) by ๐‘ข๎…ž and integrating.)

The solution in the discrete case is however different, as discussed in [6]. There may continue to be steady solutions where (1.15) no longer holds. Consider specifically the case (1.6), so that ๐‘ข๎…ž๐‘›=๐‘ข๐‘›+1โˆ’2๐‘ข๐‘›+๐‘ข๐‘›โˆ’1โˆ’๐œ–2sin๐‘ข๐‘›+๐น.(1.16) The case corresponding to (1.15) is ๐น=0, but the authors, in [3], have shown that for ๐น sufficiently small, say |๐น|<๐นcrit, there exist precisely two steady solutions of (1.16), and ๐นcrit, which of course depends on ๐œ–, can be evaluated for small ๐œ–. Specifically, ๐นcritโˆผ๐ต๐‘’โˆ’๐œ‹2/๐œ–,(1.17) where the constant โˆซ๐ต=64๐œ‹๐œ‹0(sin2(๐‘ )/๐‘ )๐‘‘๐‘  is given. For |๐น|>๐นcrit, the solutions move and the equation has travelling wave solutions instead of steady solutions.

In order to prove results such as (1.17), one has to regard (1.16) as a singular perturbation of the steady continuous-diffusion equation ๐‘ข๐‘ฅ๐‘ฅโˆ’sin๐‘ข=0,(1.18) for which the solution (satisfying ๐‘ข(โˆ’โˆž)=0, ๐‘ข(โˆž)=2๐œ‹) is ๐‘ˆ=4tanโˆ’1๐‘’๐‘ฅ. If we linearize (1.18) about ๐‘ˆ, we obtain ๐œ™๎…ž๎…žโˆ’cos(๐‘ˆ)๐œ™=0.(1.19) But multiplying (1.18) by ๐‘ˆ๎…ž and integrating lead to (1/2)(๐‘ˆ๎…ž)2=1โˆ’cos๐‘ˆ, so that since ๐‘ˆ๎…ž=2sech(๐‘ฅ), we have cos๐‘ˆ=1โˆ’2sech2(๐‘ฅ). The linearization (1.19) thus becomes ๐œ™๎…ž๎…ž+(โˆ’1+2sech2๐‘ฅ)๐œ™=0, which is of course (1.2) with ๐œ†=โˆ’1 and ๐‘›=1. Thus the selfadjoint operator ๐‘‡ given (in ๐ฟ2(โˆ’โˆž,โˆž)) by ๐‘‡๐œ™=โˆ’๐œ™๎…ž๎…žโˆ’2sech2(๐‘ฅ)๐œ™ has an eigenvalue at โˆ’1, with eigenfunction ๐‘ˆ๎…ž (differentiation of (1.18) shows that ๐‘ˆ๎…ž satisfies (1.19)). This fact, together with the additional fact that the spectrum of ๐‘‡ is continuous above 0 (since sech2(๐‘ฅ)โˆˆ๐ฟ(0,โˆž)), is highly relevant to the work in [3] and led to our interest more generally in the spectral problem (1.2).

The explicit solution for any ๐‘› using contour integrals different from what Kamke did is known to [7]. For more information on this problem one can see [7, 8].

2. Preliminaries

We want to know expansion of an arbitrary function ๐‘“(๐‘ฅ) in terms of eigenfunctions. So one needs to know the following. Let ๐œƒ(๐‘ฅ,๐œ†) and ๐œ™(๐‘ฅ,๐œ†) be the solutions of (1.1) such that ๐œ™(0,๐œ†)=sin๐›ผ,๐œ™๎…ž(0,๐œ†)=โˆ’cos๐›ผ,๐œƒ(0,๐œ†)=cos๐›ผ,๐œƒ๎…ž(0,๐œ†)=sin๐›ผ,(2.1) where ๐›ผ is real. ๐‘Š๐‘ฅ(๐œ™,๐œƒ)=๐‘Š0(๐œ™,๐œƒ)=1. The general solution of (1.1) is of the form ๐œ“(๐‘ฅ,๐œ†)=๐œƒ(๐‘ฅ,๐œ†)+๐‘š(๐œ†)๐œ™(๐‘ฅ,๐œ†)โˆˆ๐ฟ2(0,โˆž).(2.2) The spectrum is defined by means of the function ๐‘˜(๐œ†)=lim๐›ฟโ†’0โˆซ๐œ†0โˆ’โ„‘{๐‘š(๐‘ข+๐‘–๐›ฟ)}๐‘‘๐‘ข,(2.3) which exists for all real ๐œ† and ๐‘˜(๐œ†) is a nondecreasing function. The expansion of a function ๐‘“(๐‘ฅ) in terms of the spectral function depends on the following lemmas taken from [1].

Lemma 2.1. Without detailing, let the interval be (0,โˆž): 1๐‘“(๐‘ฅ)=๐œ‹โˆซโˆž0โˆซ๐œ™(๐‘ฅ,๐œ†)๐‘‘๐‘˜(๐œ†)โˆž0๐œ™(๐‘ก,๐œ†)๐‘“(๐‘ก)๐‘‘๐‘ก.(2.4) If ๐‘š(๐œ†) has poles, then ๐‘“(๐‘ฅ)=โˆžโˆ‘๐‘=0๐œ™๐‘โˆซ(๐‘ก)โˆž0๐œ™๐‘(๐‘ก,๐œ†)๐‘“(๐‘ก)๐‘‘๐‘ก.(2.5)

Lemma 2.2. Without detailing, let the interval be (โˆ’โˆž,โˆž). If ๐‘ž(๐‘ฅ) is an even function, then ๐‘š1(๐œ†)=โˆ’๐‘š2(๐œ†). So the expansion formula is 1๐‘“(๐‘ฅ)=๐œ‹โˆซโˆžโˆ’โˆžโˆซ๐œƒ(๐‘ฅ,๐œ†)๐‘‘๐œ‰โˆžโˆ’โˆž1๐œƒ(๐‘ฆ,๐œ†)๐‘“(๐‘ฆ)๐‘‘๐‘ฆ+๐œ‹โˆซโˆžโˆ’โˆžโˆซ๐œ™(๐‘ฅ,๐œ†)๐‘‘๐œโˆžโˆ’โˆž๐œ™(๐‘ฆ,๐œ†)๐‘“(๐‘ฆ)๐‘‘๐‘ฆ,(2.6) where ๐œ‰(๐œ†)=lim๐›ฟโ†’0๎€œ๐œ†0๎‚†1โˆ’โ„‘๐‘š1(๐‘ข+๐‘–๐›ฟ)โˆ’๐‘š2๎‚‡(๐‘ข+๐‘–๐›ฟ)๐‘‘๐‘ข,๐œ‰๎…ž๎‚†1(๐œ†)=โ„‘2๐‘š2๎‚‡,(๐œ†)๐œ(๐œ†)=lim๐›ฟโ†’0๎€œ๐œ†0๎‚†๐‘šโˆ’โ„‘1(๐‘ข+๐‘–๐›ฟ)๐‘š2(๐‘ข+๐‘–๐›ฟ)๐‘š1(๐‘ข+๐‘–๐›ฟ)โˆ’๐‘š2๎‚‡(๐‘ข+๐‘–๐›ฟ)๐‘‘๐‘ข,๐œ๎…ž1(๐œ†)=โˆ’2โ„‘๎€ฝ๐‘š2๎€พ.(๐œ†)(2.7)

3. Main Results

We are now dealing with (1.2) in the case where ๐‘› is integral. Without loss of generality, we may suppose ๐‘›โ‰ฅ0, but since ๐‘›=0 reduces (1.2) to the simple trigonometric case, we are in fact interested only in ๐‘›>0. We first prove that a solution is given by ๐‘ฆ=cosh๐‘›+1๐‘ฅ๎€Ÿ๐ถcosh(๐‘ง๐‘ )(sinh๐‘งโˆ’sinh๐‘ฅ)๐‘›+1๐‘‘๐‘ง,๐‘ 2=โˆ’๐œ†,(3.1) where the contour ๐ถ is taken round the point ๐‘ง=๐‘ฅ and no other zero of sinh๐‘งโˆ’sinh๐‘ฅ. This is slight variant of a form which Titchmarsh uses in his discussion of (1.3). The proof below will show (3.1), being continuous at least formally, to be a solution of (1.2) where ๐‘› is not an integer, but the difficulty then is to choose a suitable contour, since the integrand has a branch point at ๐‘ง=๐‘ฅ.

Remark 3.1. We also remark that it is obvious that we can express the solution (3.1) equivalently ignoring some multiplicative constants as ๐‘ฆ(๐‘ฅ)=cosh๐‘›+1๐‘ฅ๐‘‘๐‘›(cosh๐‘ฅ๐‘‘๐‘ฅ)๐‘›๎€Ÿ๐ถcosh(๐‘ง๐‘ )sinh๐‘งโˆ’sinh๐‘ฅ๐‘‘๐‘ง.(3.2)

Theorem 3.2. The contour integral (3.1) satisfies the differential equation (1.2).

Proof. We see that ๐‘ฆ๎…ž=(๐‘›+1)tanh(๐‘ฅ)๐‘ฆ+(๐‘›+1)cosh๐‘›+2๐‘ฅ๎€Ÿ๐ถcosh(๐‘ง๐‘ )(sinh๐‘งโˆ’sinh๐‘ฅ)๐‘›+2๐‘ฆ๐‘‘๐‘ง,(3.3)๎…ž๎…ž+๐‘›(๐‘›+1)sech2(๐‘ฅ)๐‘ฆ=(๐‘›+1)cosh๐‘›+1๐‘ฅ๎€Ÿ๐ถ๎€ฝcosh(๐‘ง๐‘ )(๐‘›+1)sinh2๎€พ๐‘ง+sinh๐‘งsinh๐‘ฅ+๐‘›+2(sinh๐‘งโˆ’sinh๐‘ฅ)๐‘›+3๐‘‘๐‘ง.(3.4) Integrating (3.1) by parts, ๐‘ฆ(๐‘ฅ)=๐‘›+1๐‘ 2๎€Ÿ๐ถ๎€ฝcosh(๐‘ง๐‘ )(๐‘›+1)sinh2๎€พ๐‘ง+sinh๐‘งsinh๐‘ฅ+๐‘›+2(sinh๐‘งโˆ’sinh๐‘ฅ)๐‘›+3๐‘‘๐‘ง,(3.5) so that ๐œ†๐‘ฆ=โˆ’(๐‘›+1)cosh๐‘›+1๐‘ฅ๎€Ÿ๐ถ๎€ฝcosh(๐‘ง๐‘ )(๐‘›+1)sinh2๎€พ๐‘ง+sinh๐‘งsinh๐‘ฅ+๐‘›+2(sinh๐‘งโˆ’sinh๐‘ฅ)๐‘›+3๐‘‘๐‘ง.(3.6) Comparing (3.4) and (3.6), we see that ๐‘ฆ๎…ž๎…ž+๐‘›(๐‘›+1)sech2(๐‘ฅ)๐‘ฆ=โˆ’๐œ†๐‘ฆ, so that ๐‘ฆ(๐‘ฅ) satisfies (1.2), as required.

Remark 3.3. We now point out that the factor cosh(๐‘ง๐‘ ) played little part in the argument. Certainly, the argument would have washed equally well if we had replaced cosh(๐‘ง๐‘ ) by sinh(๐‘ง๐‘ ): ๐‘ฆ2(๐‘ฅ)=cosh๐‘›+1๐‘ฅ๎€Ÿ๐ถsinh(๐‘ง๐‘ )(sinh๐‘งโˆ’sinh๐‘ฅ)๐‘›+1๐‘‘๐‘ง.(3.7)

Theorem 3.4. The contour integral (3.7) satisfies the differential equation (1.2).

Proof. Proof is the same as the above theorem. So we omit it.

Remark 3.5. Furthermore, once the integrands have poles at ๐‘ง=๐‘ฅ, the solution can be evaluated by calculating the relevant residues. For example, in the trivial case ๐‘›=0, when we should recover the trigonometric functions, the residues of cosh(๐‘ง๐‘ )sinh๐‘งโˆ’sinh๐‘ฅ(3.8) are cosh(๐‘ฅ๐‘ ),cosh๐‘ฅ(3.9) so that the solution (3.7) becomes multiples of โˆšcos(๐‘ฅ๐œ†) (similarly โˆšsin(๐‘ฅ๐œ†)), as we expect.
We can generalize Theorems (3.2) and (3.4) by defining the following operator:๐‘‡๐‘“(๐‘ฅ)=cosh๐‘›+1๐‘ฅ๎€Ÿ๐ถ๐‘“(๐‘ง)(sinh๐‘งโˆ’sinh๐‘ฅ)๐‘›+1๐‘‘๐‘ง,(3.10)where ๐‘“ is a differentiable function as long as one can pick up residue.

Corollary 3.6. If ๐‘“(๐‘ง)=cosh(๐‘ง๐‘ )(sinh(๐‘ง๐‘ )), then we obtain Theorems (3.2) and (3.4). The operator ๐‘‡ is also linear.

4. The Explicit Solution Given by Residues for ๐‘›=1

We now require the residues ofcosh(๐‘ง๐‘ )(sinh๐‘งโˆ’sinh๐‘ฅ)2,sinh(๐‘ง๐‘ )(sinh๐‘งโˆ’sinh๐‘ฅ)2.(4.1) Since cosh(๐‘ง๐‘ )(sinh๐‘งโˆ’sinh๐‘ฅ)2={cosh(๐‘ฅ๐‘ )+(๐‘งโˆ’๐‘ฅ)๐‘ sinh(๐‘ฅ๐‘ )+โ‹ฏ}{1โˆ’(๐‘งโˆ’๐‘ฅ)tanh(๐‘ฅ)+โ‹ฏ}(๐‘งโˆ’๐‘ฅ)2cosh2,(๐‘ฅ)(4.2) we see that the residue at ๐‘ง=๐‘ฅ is๐‘ sinh(๐‘ฅ๐‘ )โˆ’tanh(๐‘ฅ)cosh(๐‘ฅ๐‘ )cosh2,(๐‘ฅ)(4.3) so that one solution is๐‘ฆ1=โˆšโˆš๐œ†sin(๐‘ฅโˆš๐œ†)+tanh(๐‘ฅ)cos(๐‘ฅ๐œ†).(4.4) By examining the residue of the second equation of (4.1), we see that a second solution is๐‘ฆ2=โˆšโˆš๐œ†cos(๐‘ฅโˆš๐œ†)โˆ’tanh(๐‘ฅ)sin(๐‘ฅ๐œ†).(4.5)

Remark 4.1. The solution can also be obtained from (3.2). For we have already seen, from our brief discussion of the case ๐‘›=0, that the integral in (3.2) is just a multiple of โˆšcos(๐‘ฅ๐œ†)/cosh๐‘ฅ (or of โˆšsin(๐‘ฅ๐œ†)/cosh๐‘ฅ if we replace cosh(๐‘ง๐‘ ) by sinh(๐‘ง๐‘ )) hence in the first case, (3.2) gives a multiple of ๐‘‘cosh(๐‘ฅ)๎‚€โˆš๐‘‘๐‘ฅcos(๐‘ฅ๐œ†)๎‚โˆšcosh(๐‘ฅ)=โˆ’โˆš๐œ†sin(๐‘ฅ๐œ†)โˆ’tanh(๐‘ฅ)cos(๐‘ฅ๐œ†),(4.6)in accordance with (4.4).

Remark 4.2. Wronskian ๐‘Š(๐‘ฆ1(๐‘ฅ),๐‘ฆ2โˆš(๐‘ฅ))=โˆ’๐œ†(๐œ†+1). We now have two linearly independent solutions.

Remark 4.3. The general solution is ๐‘ฆ(๐‘ฅ)=๐‘1๐‘ฆ1(๐‘ฅ)+๐‘2๐‘ฆ2(๐‘ฅ).

Lemma 4.4. ๐‘ฆ1(๐‘ฅ) is an odd function but ๐‘ฆ2(๐‘ฅ) is an even function.

Proof. ๐‘ฆ1(โˆ’๐‘ฅ)=โˆ’๐‘ฆ1(๐‘ฅ),๐‘ฆ2(โˆ’๐‘ฅ)=๐‘ฆ2(๐‘ฅ).(4.7)

5. Eigenvalues and Eigenfunctions for ๐‘›=1 When ๐‘ฆ(0)=๐‘ฆ(๐‘)=0

Theorem 5.1. Eigenvalues with associated boundary conditions ๐‘ฆ(0)=๐‘ฆ(๐‘)=0 are the zeros of โˆšโˆš๐œ†tan(๐‘๐œ†)+tanh(๐‘)=0;(5.1) furthermore, one and only one eigenvalue lies in the interval ๎‚€1๐‘˜โˆ’2๎‚โˆš๐œ‹<๐‘๎‚€1๐œ†<๐‘˜+2๎‚๐œ‹(5.2) for every integral ๐‘˜โ‰ 0.

Proof. One can see from Remark 4.3 that if ๐‘ฆ(0)=0, then ๐‘2=0. So that ๐‘ฆ(๐‘)=0 implies โˆšโˆš๐œ†sin(๐‘โˆš๐œ†)+tanh(๐‘)cos(๐‘โˆš๐œ†)=0,โˆš๐œ†tan(๐‘๐œ†)=โˆ’tanh(๐‘).(5.3) One can see immediately that the eigenvalues belong to the interval (5.2).
To prove the second part we use the following strategy. Multiply (5.1) by ๐‘ and set โˆš๐‘ฅ=๐‘๐œ†, then denote โ„Ž(๐‘ฅ)=๐‘ฅtan(๐‘ฅ)+๐‘tanh(๐‘). So โ„Ž๎…ž(๐‘ฅ)=sec2(๐‘ฅ){(1/2)sin(2๐‘ฅ)+๐‘ฅ}. Notice that โ„Ž(๐‘ฅ) is an even and does not intersect ๐‘ฅ-axis, where โˆ’1/2<๐‘ฅ<1/2. If ๐‘ฅ<โˆ’1/2, then โ„Ž๎…ž(๐‘ฅ)<0 and if ๐‘ฅ>1/2, then โ„Ž๎…ž(๐‘ฅ)>0. So the monotonicity of โ„Ž(๐‘ฅ) implies that โ„Ž(๐‘ฅ) has only one zero belonging to the interval (5.2) for every integral ๐‘˜โ‰ 0.

Remark 5.2. So ๐‘ฆ๐‘˜ is the eigenfunction in the form of (4.4): ๐‘ฆ๐‘˜=โˆš๐œ†๐‘˜โˆšsin(๐‘ฅ๐œ†๐‘˜โˆš)+tanh(๐‘ฅ)cos(๐‘ฅ๐œ†๐‘˜).(5.4)

Corollary 5.3. One can orthonormalize the eigenfunctions.

Proof. โˆซ๐‘0๐‘ฆ2๐‘˜โˆš๐‘‘๐‘ฅ=โˆ’4๐œ†๐‘˜tanh(๐‘)cos2โˆš(๐‘๐œ†๐‘˜โˆš)+2๐œ†๐‘˜(๐œ†๐‘˜+1)๐‘โˆ’(๐œ†๐‘˜โˆšโˆ’1)sin(2๐‘๐œ†๐‘˜)4โˆš๐œ†๐‘˜.(5.5) The orthonormalized eigenfunctions denoted by ฮฆ(๐‘ฅ,๐œ†๐‘˜), ฮฆ(๐‘ฅ,๐œ†๐‘˜โˆš)=(๐œ†๐‘˜โˆšsin(๐‘ฅ๐œ†๐‘˜โˆš)+tanh(๐‘ฅ)cos(๐‘ฅ๐œ†๐‘˜๎”))/โˆซ๐‘0๐‘ฆ2๐‘˜๐‘‘๐‘ฅ.

Remark 5.4. An arbitrary function ๐‘“(๐‘ฅ) in terms of eigenfunctions follows: ๐‘“(๐‘ฅ)=โˆž๎“๐‘˜=0๐ถ๐‘˜ฮฆ(๐‘ฅ,๐œ†๐‘˜),where๐ถ๐‘˜=๎€œ๐‘0๐‘“(๐‘ฅ)ฮฆ(๐‘ฅ,๐œ†๐‘˜)๐‘‘๐‘ฅ.(5.6)

6. Eigenvalues and Eigenfunctions for ๐‘›=1 When ๐‘ฆ๎…ž(0)=๐‘ฆ๎…ž(๐‘)=0

Theorem 6.1. The eigenvalues with ๐‘ฆ๎…ž(0)=๐‘ฆ๎…ž(๐‘)=0 are the zeros of (6.1); furthermore, there exists one and only one eigenvalue lying in the interval (5.2) for every integral ๐‘˜โ‰ 0.

Proof. One can see from Remark 4.3 that if ๐‘ฆ๎…ž(0)=0, then either ๐‘1=0 or ๐œ†=โˆ’1. If ๐œ†=โˆ’1, then the associated eigenfunction is zero. So this is useless. Hence, ๐‘1=0. ๐‘ฆ๎…ž(๐‘)=0 implies โˆš๐œ†tan(๐‘๐œ†)+sech2โˆš(๐‘)tan(๐‘โˆš๐œ†)+๐œ†tanh(๐‘)=0.(6.1)So it is obvious that the zeros (eigenvalues) belong to the interval (5.2).
Set โˆš๐‘ฅ=๐‘๐œ†. (๐‘2๐œ†=๐‘ฅ2๐œ†=๐‘ฅ2/๐‘2). Equation (6.1) is denoted by โ„Ž(๐‘ฅ): ๎‚€๐‘ฅโ„Ž(๐‘ฅ)=2๐‘2+sech2๎‚๐‘ฅ(๐‘)tan(๐‘ฅ)+๐‘tanh(๐‘).(6.2) It is enough to show that โ„Ž(๐‘ฅ) is monotonic: โ„Ž๎…ž๐‘ฅ(๐‘ฅ)=2+๐‘2sech2(๐‘)+๐‘ฅsin(2๐‘ฅ)๐‘2cos2+1(๐‘ฅ)๐‘tanh(๐‘).(6.3) We see that โ„Ž๎…ž(๐‘ฅ)>0 everywhere. We therefore conclude that โ„Ž(๐‘ฅ) is monotonic.

Remark 6.2. The associated eigenfunctions are ๐‘ฆ๐‘˜=โˆš๐œ†๐‘˜โˆšcos(๐‘ฅ๐œ†๐‘˜โˆš)โˆ’tanh(๐‘ฅ)sin(๐‘ฅ๐œ†๐‘˜).(6.4)

Corollary 6.3. One can orthonormalize the eigenfunctions.

Proof. ๎€œ๐‘0๐‘ฆ2๐‘˜โˆš๐‘‘๐‘ฅ=โˆ’4๐œ†๐‘˜tanh(๐‘)sin2โˆš(๐‘๐œ†๐‘˜โˆš)+2๐œ†๐‘˜(๐œ†๐‘˜+1)๐‘+(๐œ†๐‘˜โˆšโˆ’1)sin(2๐‘๐œ†๐‘˜)4โˆš๐œ†๐‘˜.(6.5) The orthonormalized eigenfunctions denoted by ฮจ(๐‘ฅ,๐œ†๐‘˜), ฮจ(๐‘ฅ,๐œ†๐‘˜โˆš)=(๐œ†๐‘˜โˆšcos(๐‘ฅ๐œ†๐‘˜โˆš)โˆ’tanh(๐‘ฅ)sin(๐‘ฅ๐œ†๐‘˜๎”))/โˆซ๐‘0๐‘ฆ2๐‘˜๐‘‘๐‘ฅ.

Remark 6.4. Therefore, an arbitrary ๐‘“(๐‘ฅ) in terms of orthonormalized eigenfunctions is ๐‘“(๐‘ฅ)=โˆž๎“๐‘˜=0๐ถ๐‘˜ฮจ(๐‘ฅ,๐œ†๐‘˜),๐ถ๐‘˜=๎€œ๐‘0๐‘“(๐‘ฅ)ฮจ(๐‘ฅ,๐œ†๐‘˜)๐‘‘๐‘ฅ.(6.6)

7. Spectral Function ๐‘š(๐œ†) over (0,โˆž) and Expansion

Now let ๐œƒ(๐‘ฅ,๐œ†) and ๐œ™(๐‘ฅ,๐œ†) be the solutions of ๐‘‘2๐‘ฆ๐‘‘๐‘ฅ2+๎€ฝ๐œ†+2sech2๎€พ(๐‘ฅ)๐‘ฆ=0,(7.1) which satisfy (2.1); so that โˆš๐œ™(๐‘ฅ,๐œ†)=โˆ’cos๐›ผ{โˆš๐œ†sin(๐‘ฅโˆš๐œ†)+tanh๐‘ฅcos(๐‘ฅ๐œ†)}+โˆš๐œ†+1sin๐›ผ{โˆš๐œ†cos(๐‘ฅโˆš๐œ†)โˆ’tanh๐‘ฅsin(๐‘ฅ๐œ†)}โˆš๐œ†โˆš,(7.2)๐œƒ(๐‘ฅ,๐œ†)=sin๐›ผ{โˆš๐œ†sin(๐‘ฅโˆš๐œ†)+tanh๐‘ฅcos(๐‘ฅ๐œ†)}+โˆš๐œ†+1cos๐›ผ{โˆš๐œ†cos(๐‘ฅโˆš๐œ†)โˆ’tanh๐‘ฅsin(๐‘ฅ๐œ†)}โˆš๐œ†.(7.3) Now we need to find ๐‘š(๐œ†). This suggests that ๐œ“(๐‘ฅ,๐œ†)=๐œƒ(๐‘ฅ,๐œ†)+๐‘š(๐œ†)๐œ™(๐‘ฅ,๐œ†)โˆˆ๐ฟ2(0,โˆž).(7.4) To get demanded result, one needs to find the asymptotics of ๐œƒ(๐‘ฅ,๐œ†) and ๐œ™(๐‘ฅ,๐œ†) as ๐‘ฅโ†’โˆž and โ„‘โˆš๐œ†>0: โˆš๐œ™(๐‘ฅ,๐œ†)โˆผ{โˆ’โˆš๐œ†cos๐›ผ+๐œ†(๐œ†+1)sin๐›ผ+๐‘–(โˆ’๐œ†cos๐›ผโˆ’(๐œ†+1)sin๐›ผ)}๐‘’โˆšโˆ’๐‘–๐‘ฅ๐œ†2โˆš๐œ†(๐œ†+1)=๐‘€1(๐œ†)๐‘’โˆšโˆ’๐‘–๐‘ฅ๐œ†,{โˆš๐œƒ(๐‘ฅ,๐œ†)โˆผโˆš๐œ†sin๐›ผ+๐œ†(๐œ†+1)cos๐›ผ+๐‘–(๐œ†sin๐›ผโˆ’(๐œ†+1)cos๐›ผ)}๐‘’โˆšโˆ’๐‘–๐‘ฅ๐œ†2โˆš๐œ†(๐œ†+1)=๐‘€(๐œ†)๐‘’โˆšโˆ’๐‘–๐‘ฅ๐œ†.(7.5) Finally, we must arrange the linear combination so that the terms ๐‘’โˆšโˆ’๐‘–๐‘ฅ๐œ† cancel. That is, ๐‘€(๐œ†)+๐‘š(๐œ†)๐‘€1(๐œ†)=0. Hence, โˆš๐‘š(๐œ†)=โˆ’๐‘–๐œ†(๐œ†+1)โˆ’(๐œ†2+๐œ†+1)sin๐›ผcos๐›ผ๐œ†cos2๐›ผ+(๐œ†+1)2sin2๐›ผ,โˆ’โˆšโ„‘๐‘š(๐œ†)=๐œ†(๐œ†+1)๐œ†cos2๐›ผ+(๐œ†+1)2sin2๐›ผwhen๐œ†>0,0when๐œ†<0.(7.6) We see that the spectrum is continuous for ๐œ†>0. But we have a point spectrum for ๐œ†<0. So that an arbitrary ๐‘“(๐‘ฅ) is a linear combination of integrand and series. The spectral function is calculated from (2.3). Hence, โŽงโŽชโŽจโŽชโŽฉโˆš๐‘‘๐‘˜(๐œ†)=๐œ†(๐œ†+1)๐œ†cos2๐›ผ+(๐œ†+1)2sin2๐›ผwhen๐œ†>0,0when๐œ†<0.(7.7)(1) In particular, if ๐›ผ=0, thenโŽงโŽชโŽจโŽชโŽฉ๐‘‘๐‘˜(๐œ†)=๐œ†+1โˆš๐œ†when๐œ†>0,0when๐œ†<0.(7.8) So that in this case there is no eigenvalue for ๐œ†<0. From Lemma 2.1, one can see the expansion of1๐‘“(๐‘ฅ)=๐œ‹๎€œโˆž0โˆšโˆš๐œ†sin(๐‘ฅโˆš๐œ†)+tanh๐‘ฅcos(๐‘ฅ๐œ†)โˆš๐œ†๎€œ๐‘‘๐œ†โˆž0โˆšโˆš๐œ†sin(๐‘ฆโˆš๐œ†)+tanh๐‘ฆcos(๐‘ฆ๐œ†)๐œ†+1๐‘“(๐‘ฆ)๐‘‘๐‘ฆ.(7.9) Similarly, if ๐›ผ=๐œ‹/2, then โŽงโŽชโŽจโŽชโŽฉโˆ’โˆšโ„‘๐‘š(๐œ†)=๐œ†๐œ†+1when๐œ†>0,0when๐œ†<0..(7.10)(2) Hence, there exits only one eigenvalue at ๐œ†=โˆ’1 and the corresponding eigenfunction ๐œ™(๐‘ฅ,โˆ’1) where ๐œ™(๐‘ฅ,๐œ†) is (7.2). So the spectrum calculated by (2.3) isโŽงโŽชโŽจโŽชโŽฉโˆš๐‘‘๐‘˜(๐œ†)=๐œ†๐œ†+1when๐œ†>0,0when๐œ†<0..(7.11) Therefore, from Lemma 2.1, an expansion of function ๐‘“(๐‘ฅ) in terms of eigenfunctions and spectral function follows:1๐‘“(๐‘ฅ)=๐œ‹๎€œโˆž0โˆšโˆš๐œ†cos(๐‘ฅโˆš๐œ†)โˆ’tanh๐‘ฅsin(๐‘ฅ๐œ†)ร—๎€œ๐œ†+1๐‘‘๐œ†โˆž0โˆšโˆš๐œ†cos(๐‘ฆโˆš๐œ†)โˆ’tanh๐‘ฆsin(๐‘ฆ๐œ†)โˆš๐œ†๐‘“(๐‘ฆ)๐‘‘๐‘ฆ+๐‘1sech(๐‘ฅ),(7.12) where ๐‘1 is a constant. So we have proved the following theorem regarding the nature of ๐‘š(๐œ†).

Theorem 7.1. If ๐›ผ=0, there is no eigenvalue where ๐œ†<0. If ๐›ผ=๐œ‹/2, then there exists only one eigenvalue at ๐œ†=โˆ’1 and the corresponding eigenfunction is ๐œ™(๐‘ฅ,โˆ’1)=๐‘ ๐‘’๐‘โ„Ž(๐‘ฅ).

Finally, one can ask what is the range of ๐›ผ in the case of ๐œ†<0 and ๐‘›=1.

Theorem 7.2. If 0<๐›ผ<๐œ‹/2, then there are precisely two eigenvalues except at ๐›ผ=๐œ‹/4. Hence there are two eigenfunctions, namely, ๐œ™(๐‘ฅ,๐œ†1) and ๐œ™(๐‘ฅ,๐œ†2).

Proof. We check the zeros of both the numerator and denominator of (7.6). After working out the algebra, we see that the zeros of the numerator and the denominator are ๐œ‡1=โˆ’(2+tan2โˆš๐›ผ)+tan๐›ผ4+tan2๐›ผ+12,๐œ‡2=โˆ’(2+tan2โˆš๐›ผ)โˆ’tan๐›ผ4+tan2๐›ผ+12,๐œ†1=โˆ’(2tan2โˆš๐›ผ+1)+4tan2๐›ผ+12tan2๐›ผ,๐œ†2=โˆ’(2tan2โˆš๐›ผ+1)โˆ’4tan2๐›ผ+12tan2๐›ผ.(7.13) If ๐‘š(๐œ†) has poles, then they are the eigenvalues. If so, the eigenfunctions are ๐œ™(๐‘ฅ,๐œ†1) and ๐œ™(๐‘ฅ,๐œ†2). Finally, there is one thing to be proved in this case ๐›ผ=๐œ‹/4. To do this, we expand the zeros around ๐›ผ=๐œ‹/4 and at the end ๐›ผโ†’๐œ‹/4: ๐œ†1=โˆšโˆ’3+52+โˆš20โˆ’125๎‚€๐œ‹10๐›ผโˆ’4๎‚+โˆšโˆ’400+3525๎‚€๐œ‹100๐›ผโˆ’4๎‚2๐œ‹+๐‘‚๎‚†๎‚€๐›ผโˆ’4๎‚3๎‚‡,๐œ†2=โˆšโˆ’3โˆ’52+โˆš20+125๎‚€๐œ‹10๐›ผโˆ’4๎‚+โˆšโˆ’400โˆ’3525๎‚€๐œ‹100๐›ผโˆ’4๎‚2๐œ‹+๐‘‚๎‚†๎‚€๐›ผโˆ’4๎‚3๎‚‡.(7.14) Similarly, ๐œ‡1=โˆšโˆ’3+62+7โˆš6โˆ’126๎‚€๐œ‹๐›ผโˆ’4๎‚+โˆšโˆ’288+1186๎‚€๐œ‹36๐›ผโˆ’4๎‚2๐œ‹+๐‘‚๎‚†๎‚€๐›ผโˆ’4๎‚3๎‚‡,๐œ‡2=โˆšโˆ’3โˆ’62+โˆšโˆ’76โˆ’126๎‚€๐œ‹๐›ผโˆ’4๎‚+โˆšโˆ’288โˆ’1186๎‚€๐œ‹36๐›ผโˆ’4๎‚2๐œ‹+๐‘‚๎‚†๎‚€๐›ผโˆ’4๎‚3๎‚‡.(7.15) That completes the proof.

8. Expansion over (โˆ’โˆž,โˆž)

Now consider the interval (โˆ’โˆž,โˆž) instead of (0,โˆž). From (2.7) the following can be calculated: ๐œ‰๎…žโŽงโŽชโŽจโŽชโŽฉโˆš(๐œ†)=๐œ†(๐œ†+1)2๐œ†sin2๐›ผ+2(๐œ†+1)2cos2๐›ผ๐œwhen๐œ†>0,0when๐œ†<0,๎…žโŽงโŽชโŽจโŽชโŽฉโˆš(๐œ†)=(๐œ†+1)๐œ†2๐œ†cos2๐›ผ+2(๐œ†+1)2sin2๐›ผwhen๐œ†>0,0when๐œ†<0.(8.1) Now one can use (7.2), (7.3), and (8.1) gets the following expansion from Lemma (2.2): 1๐‘“(๐‘ฅ)=๐œ‹๎‚†โˆซโˆž0โˆšโˆš๐œ†cos(๐‘ฅโˆš๐œ†)โˆ’tanh๐‘ฅsin(๐‘ฅ๐œ†)โˆซ2(๐œ†+1)๐‘‘๐œ†โˆžโˆ’โˆžโˆšโˆš๐œ†cos(๐‘ฆโˆš๐œ†)โˆ’tanh๐‘ฆsin(๐‘ฆ๐œ†)โˆš๐œ†+โˆซ๐‘“(๐‘ฆ)๐‘‘๐‘ฆโˆž0โˆšโˆš๐œ†sin(๐‘ฅโˆš๐œ†)+tanh๐‘ฅcos(๐‘ฅ๐œ†)2โˆš๐œ†โˆซ๐‘‘๐œ†โˆžโˆ’โˆžโˆšโˆš๐œ†sin(๐‘ฆโˆš๐œ†)+tanh๐‘ฆcos(๐‘ฆ๐œ†)๎‚‡๐œ†+1๐‘“(๐‘ฆ)๐‘‘๐‘ฆ+๐‘1sech(๐‘ฅ),(8.2) where ๐‘1 is a constant.


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