Abstract and Applied Analysis

Volumeย 2008ย (), Article IDย 765920, 12 pages

http://dx.doi.org/10.1155/2008/765920

## The Analysis of Contour Integrals

^{1}Department of Mathematics, Harran University, Osmanbey Campus, Sanlurfa 63100, Turkey^{2}Department 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.

#### Abstract

For any , the contour integral , is associated with differential equation . Explicit solutions for are obtained. For , 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 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 is the case , 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: which can be solved explicitly in elementary terms when 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, on , which leads, when , 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 where the function is โbistableโ. That is, there exist three numbers , , , , such that with in and in . A prototypical would be where and are positive constants and , so that Such equations arise in a number of different applications, for example, in dislocation theory where is the displacement of the atom in some material, or in neurobiology where is typically the electric potential of the nerve cell, and in both these applications the interest is in monotonic solutions with as , as [3โ6].

The basic question is whether there exist such solutions with the form of a travelling wave, , , or of a steady solution or standing wave, where , and there is an important distinction between these two cases. For a travelling wave, , is clearly a function of the continuous variable , and indeed because of (1.4), a differentiable function of . This leads to the difference-differential equation If, however, , then, as in [3], we have to study the purely difference equation and the solutions may be discontinuous since there is nothing that now connects values of with values of for . It is best therefore to think of the solution of (1.9) as a number of (monotonic) sequences indexed by , each satisfying The simplest case would be that there is just one such sequence (modulo the translation , 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 where is small and represents the distance between atoms or nerve cells. A tempting approximation is then and in order to make sense of the scaling, in [3] the authors introduced a factor in front of . This therefore leads to a comparison between the solutions of For the continuous diffusion problem, the answer is both simple and well known [4, 5].

Given a function that is bistable, there is just one possible wave-speed , and this value of is , that is, there is a steady solution if and only if (The proof is a simple phase plane argument, and implies (1.15) follows by multiplying (1.14) by 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 The case corresponding to (1.15) is , but the authors, in [3], have shown that for sufficiently small, say , there exist precisely two steady solutions of (1.16), and , which of course depends on , can be evaluated for small . Specifically, where the constant is given. For , 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 for which the solution (satisfying , ) is . If we linearize (1.18) about , we obtain But multiplying (1.18) by and integrating lead to , so that since , we have . The linearization (1.19) thus becomes , which is of course (1.2) with and . Thus the selfadjoint operator given (in ) by has an eigenvalue at , with eigenfunction (differentiation of (1.18) shows that satisfies (1.19)). This fact, together with the additional fact that the spectrum of is continuous above (since ), 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 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 in terms of eigenfunctions. So one needs to know the following. Let and be the solutions of (1.1) such that where is real. . The general solution of (1.1) is of the form The spectrum is defined by means of the function which exists for all real and is a nondecreasing function. The expansion of a function in terms of the spectral function depends on the following lemmas taken from [1].

Lemma 2.1. *Without detailing, let the interval be :
** If has poles, then
*

Lemma 2.2. *Without detailing, let the interval be . If
is an even
function, then . So the expansion formula is
**
where
*

#### 3. Main Results

We are now dealing with (1.2) in the case where is integral. Without loss of generality, we may suppose , but since reduces (1.2) to the simple trigonometric case, we are in fact interested only in . We first prove that a solution is given by where the contour is taken round the point and no other zero of . 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 is not an integer, but the difficulty then is to choose a suitable contour, since the integrand has a branch point at .

*Remark 3.1. *We also
remark that it is obvious that we can express the solution (3.1)
equivalently
ignoring some multiplicative constants as

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

*Proof. *We
see that
Integrating (3.1) by parts,
so that
Comparing (3.4) and (3.6), we see that , so that satisfies (1.2), as required.

*Remark 3.3. *We now point
out that the factor
played little
part in the argument. Certainly, the argument would have washed equally well if
we had replaced by :

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 , the solution can be evaluated by calculating the
relevant residues. For example, in the trivial case , when we should recover the trigonometric functions,
the residues of
are
so that the
solution (3.7) becomes
multiples of (similarly ), as we
expect.

We can
generalize Theorems (3.2) and (3.4) by defining the
following operator:where is a
differentiable function as long as one can pick up residue.

Corollary 3.6. *If then we obtain
Theorems (3.2) and (3.4). The operator
is also linear. *

#### 4. The Explicit Solution Given by Residues for

We now require the residues of Since we see that the residue at is so that one solution is By examining the residue of the second equation of (4.1), we see that a second solution is

*Remark 4.1. *The solution
can also be obtained from (3.2). For we have already seen, from our brief discussion
of the case , that the integral in (3.2)
is just a
multiple of (or of if we replace by ) hence in the
first case, (3.2)
gives a
multiple of
in accordance
with (4.4).

*Remark 4.2. *Wronskian . We now have two linearly independent solutions.

*Remark 4.3. *The general solution is .

Lemma 4.4. *
is an odd
function but
is an even
function. *

*Proof. *

#### 5. Eigenvalues and Eigenfunctions for When

Theorem 5.1. *
Eigenvalues with associated boundary conditions
are the zeros
of
**
furthermore,
one and only one eigenvalue lies in the interval
**
for every
integral .*

*Proof. *One
can see from Remark 4.3 that if then . So that implies
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 and set , then denote . So Notice that is an even and
does not intersect -axis, where . If , then and if , then . So the monotonicity of implies that has only one
zero belonging to the interval (5.2) for every
integral .

*Remark 5.2. *So is the
eigenfunction in the form of (4.4):

Corollary 5.3. *One can
orthonormalize the eigenfunctions. *

*Proof. * The
orthonormalized eigenfunctions denoted by , .

*Remark 5.4. *An arbitrary
function
in terms of
eigenfunctions follows:

#### 6. Eigenvalues and Eigenfunctions for When

Theorem 6.1. *The
eigenvalues with
are the zeros
of (6.1); furthermore, there exists one and only one
eigenvalue lying in the interval (5.2)
for every
integral . *

*Proof. *One
can see from Remark 4.3 that if then either or . If then the
associated eigenfunction is zero. So this is useless. Hence, . implies
So it is
obvious that the zeros (eigenvalues) belong to the interval (5.2).

Set . (). Equation (6.1) is denoted by :
It is enough to
show that is monotonic:
We see that everywhere. We
therefore conclude that is monotonic.

*Remark 6.2. * The
associated eigenfunctions are

Corollary 6.3. *One can orthonormalize the eigenfunctions. *

*Proof. * The
orthonormalized eigenfunctions denoted by , .

*Remark 6.4. *Therefore, an arbitrary
in terms of
orthonormalized eigenfunctions is

#### 7. Spectral Function over and Expansion

Now let and be the solutions of which satisfy (2.1); so that Now we need to find . This suggests that To get demanded result, one needs to find the asymptotics of and as and : Finally, we must arrange the linear combination so that the terms cancel. That is, . Hence, We see that the spectrum is continuous for . But we have a point spectrum for . So that an arbitrary is a linear combination of integrand and series. The spectral function is calculated from (2.3). Hence, In particular, if , then So that in this case there is no eigenvalue for . From Lemma 2.1, one can see the expansion of Similarly, if , then Hence, there exits only one eigenvalue at and the corresponding eigenfunction where is (7.2). So the spectrum calculated by (2.3) is Therefore, from Lemma 2.1, an expansion of function in terms of eigenfunctions and spectral function follows: where is a constant. So we have proved the following theorem regarding the nature of .

Theorem 7.1. *If there is no
eigenvalue where . If , then there exists only one eigenvalue at and the
corresponding eigenfunction is . *

Finally, one can ask what is the range of in the case of and .

Theorem 7.2. *If , then there are precisely two eigenvalues except at . Hence there are two eigenfunctions, namely, and . *

*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
If has poles, then
they are the eigenvalues. If so, the eigenfunctions are and . Finally, there is one thing to be proved in this
case . To do this, we expand the zeros around and at the end :
Similarly, That completes
the proof.

#### 8. Expansion over

Now consider the interval instead of . From (2.7) the following can be calculated: Now one can use (7.2), (7.3), and (8.1) gets the following expansion from Lemma (2.2): where is a constant.

#### References

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