The Analysis of Contour Integrals

Recommended by Stephen L. Clark For any n, the contour integral y cosh x ∮ C cosh zs / sinh z − sinhx n 1 dz, s2 −λ, is associated with differential equation d2y x /dx2 λ n n 1 /coshx y x 0. Explicit solutions for n 1 are obtained. For n 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.


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 − d 2 y x dx 2 q x y x λy x 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 λ is the case q x 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: which can be solved explicitly in elementary terms when n 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 −π/2, π/2 , which leads, when ν n, 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 f is "bistable".That is, there exist three numbers where A and F are positive constants and F < A, so that Such equations arise in a number of different applications, for example, in dislocation theory where u n is the displacement of the nth atom in some material, or in neurobiology where u n is typically the electric potential of the nth nerve cell, and in both these applications the interest is in monotonic solutions u n with The basic question is whether there exist such solutions with the form of a travelling wave, u n − ct , c / 0, or of a steady solution or standing wave, where c 0, and there is an important distinction between these two cases.For a travelling wave, c / 0, u n is clearly a function of the continuous variable t, and indeed because of 1.4 , a differentiable function of t.This leads to the difference-differential equation If, however, c 0, 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 u x with values of u x δ for |δ| < 1.It is best therefore to think of the solution of 1.9 as a number of monotonic sequences u α n indexed by α, each satisfying The simplest case would be that there is just one such sequence modulo the translation n → n k, k 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 12 and in order to make sense of the scaling, in 3 the authors introduced a factor 2 in front of f.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 f that is bistable, there is just one possible wave-speed c, and this value of c is 0, that is, there is a steady solution if and only if The proof is a simple phase plane argument, and c 0 implies 1.15 follows by multiplying 1.14 by u 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 1.16 The case corresponding to 1.15 is F 0, but the authors, in 3 , have shown that for F sufficiently small, say |F| < F crit , there exist precisely two steady solutions of 1.16 , and F crit , which of course depends on , can be evaluated for small .Specifically, where the constant B 64π π 0 sin 2 s /s ds is given.For |F| > F crit , 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 If we linearize 1.18 about U, we obtain φ − cos U φ 0.

1.19
But multiplying 1.18 by U and integrating lead to 1/2 U 2 1 − cos U, so that since U 2sech x , we have cos U 1 − 2sech 2 x .The linearization 1.19 thus becomes φ −1 2sech 2 x φ 0, which is of course 1.2 with λ −1 and n 1.Thus the selfadjoint operator T given in L 2 −∞, ∞ by Tφ −φ − 2sech 2 x φ has an eigenvalue at −1, with eigenfunction U differentiation of 1.18 shows that U satisfies 1.19 .This fact, together with the additional fact that the spectrum of T is continuous above 0 since sech 2 x ∈ L 0, ∞ , 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 n using contour integrals different from what Kamke did is known to 7 .For more information on this problem one can see 7, 8 .

Preliminaries
We want to know expansion of an arbitrary function f x in terms of eigenfunctions.So one needs to know the following.Let θ x, λ and φ x, λ be the solutions of 1.1 such that where The spectrum is defined by means of the function which exists for all real λ and k λ is a nondecreasing function.The expansion of a function f x in terms of the spectral function depends on the following lemmas taken from 1 .

Main results
We are now dealing with 1.2 in the case where n is integral.Without loss of generality, we may suppose n ≥ 0, but since n 0 reduces 1.2 to the simple trigonometric case, we are in fact interested only in n > 0. We first prove that a solution is given by where the contour C is taken round the point z x and no other zero of sinh z − sinh x.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 n is not an integer, but the difficulty then is to choose a suitable contour, since the integrand has a branch point at z x.
Remark 3.1.We also remark that it is obvious that we can express the solution 3.1 equivalently ignoring some multiplicative constants as Proof.Proof is the same as the above theorem.So we omit it.
Remark 3.5.Furthermore, once the integrands have poles at z x, the solution can be evaluated by calculating the relevant residues.For example, in the trivial case n 0, when we should recover the trigonometric functions, the residues of cosh zs sinh z − sinh x 3.8 are cosh xs cosh x , 3.9 so that the solution 3.7 becomes multiples of cos x √ λ similarly sin x √ λ , as we expect.We can generalize Theorems 3.2 and 3.4 by defining the following operator: where f is a differentiable function as long as one can pick up residue.

The explicit solution given by residues for n 1
We now require the residues of cosh zs Proof.

furthermore, one and only one eigenvalue lies in the interval
for every integral k / 0.
Proof.One can see from Remark 4.3 that if y 0 0, then c 2 0. So that y b 0 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 b and set Notice that h x is an even and does not intersect x-axis, where −1/2 < x < 1/2.If x < −1/2, then h x < 0 and if x > 1/2, then h x > 0. So the monotonicity of h x implies that h x has only one zero belonging to the interval 5.2 for every integral k / 0.
The orthonormalized eigenfunctions denoted by Remark 5.4.An arbitrary function f x in terms of eigenfunctions follows: So it is obvious that the zeros eigenvalues belong to the interval 5.
It is enough to show that h x is monotonic: We see that h x > 0 everywhere.We therefore conclude that h x is monotonic.
The orthonormalized eigenfunctions denoted by Remark 6.4.Therefore, an arbitrary f x in terms of orthonormalized eigenfunctions is 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

7.14
If m λ has poles, then they are the eigenvalues.If so, the eigenfunctions are φ x, λ 1 and φ x, λ 2 .Finally, there is one thing to be proved in this case α π/4.To do this, we expand the zeros around α π/4 and at the end α → π/4:

7.16
That completes the proof.

Expansion over −∞, ∞
Now consider the interval −∞, ∞ instead of 0, ∞ .From 2.7 the following can be calculated: where c 1 is a constant.

Remark 5 . 2 . 4 Corollary 5 . 3 .
So y k is the eigenfunction in the form of 4.4 : y k λ k sin x λ k tanh x cos x λ k .5.One can orthonormalize the eigenfunctions.Proof.b 0

and eigenfunctions for n 1 when y 0 y b 0 Theorem 6 . 1 .implies λ tan b λ sech 2 b tan b λ λ tanh b 0
The eigenvalues with y 0 y b 0 are the zeros of 6.1 ; furthermore, there exists one and only one eigenvalue lying in the interval 5.2 for every integral k / 0.Proof.One can see from Remark 4.3 that if y 0 0, then either c 1 0 or λ −1.If λ −1, then the associated eigenfunction is zero.So this is useless.Hence, c 1 0. y b 0

Remark 6 . 2 . 4 Corollary 6 . 3 .
The associated eigenfunctions are y k λ k cos x λ k − tanh x sin x λ k .6.One can orthonormalize the eigenfunctions.Proof.b 0 We now point out that the factor cosh zs played little part in the argument.Certainly, the argument would have washed equally well if we had replaced cosh zs by sinh zs : Theorem 3.2.The contour integral 3.1 satisfies the differential equation 1.2 .Proof.We see that y n 1 tanh x y n 1 cosh n 2 x C cosh zs sinh z − sinh x n 2 dz, 3.3 y n n 1 sech 2 x y n 1 cosh n 1 x C cosh zs n 1 sinh 2 z sinh z sinh x n 2 sinh z − sinh x n 3 dz.
Remark 4.1.The solution can also be obtained from 3.2 .For we have already seen, from our brief discussion of the case n 0, that the integral in 3.2 is just a multiple of cos x √ λ / cosh x or of sin x √ λ / cosh x if we replace cosh zs by sinh zs hence in the first case, 3.The general solution is y x c 1 y 1 x c 2 y 2 x .
Lemma 4.4.y 1 x is an odd function but y 2 x is an even function.