In this section, we consider the boundary value problem (2.1), (2.2) with

$g=0$. We count all eigenvalues with their proper multiplicities and develop a formula for the asymptotic distribution of the eigenvalues, which is used to obtain the corresponding formula for general

*g*. Observe that for

$g=0$, the quasi-derivatives

${y}^{[j]}$ coincide with the standard derivatives

${y}^{(j)}$. We take the canonical fundamental system

${y}_{j}(\cdot ,\lambda )$,

$j=1,\dots ,4$, of (2.1) with

${y}_{j}^{(m)}(0)={\delta}_{j,m+1}$ if

$j\ge 2$ for

$m=0,\dots ,3$. It is well known that the functions

${y}_{j}(\cdot ,\lambda )$ are analytic on ℂ with respect to

*λ*. Putting

$M(\lambda )={({B}_{i}(\lambda ){y}_{j}(\cdot ,\lambda ))}_{i,j=1}^{4},$

the eigenvalues of the boundary value problem (2.1), (2.2) are the eigenvalues of the analytic matrix function *M*, where the corresponding geometric and algebraic multiplicities coincide, see [[15], Theorem 6.3.2].

Setting

$\lambda ={\mu}^{2}$ and

$y(x,\mu )=\frac{1}{2{\mu}^{3}}sinh(\mu x)-\frac{1}{2{\mu}^{3}}sin(\mu x)$

it is easy to see that

${y}_{j}(x,\lambda )={y}^{(4-j)}(x,\mu ),\phantom{\rule{1em}{0ex}}j=1,\dots ,4.$

The second row of $M(\lambda )$ has exactly two non-zero entries (for $\lambda \ne 0$), and these non-zero entries are:

In Cases 1 and 2, ${B}_{2}(\lambda ){y}_{2}(\cdot ,\lambda )=-i\alpha {\mu}^{2}$ and ${B}_{2}(\lambda ){y}_{3}(\cdot ,\lambda )=1$;

In Cases 3 and 4, ${B}_{2}(\lambda ){y}_{1}(\cdot ,\lambda )=i\alpha {\mu}^{2}$ and ${B}_{2}(\lambda ){y}_{4}(\cdot ,\lambda )=1$.

Since the first row of

$M(\lambda )$ has exactly one entry 1 and all other entries zero, an expansion of

$M(\lambda )$ with respect to the second row shows that

$detM(\lambda )=\pm \varphi (\mu )$, where

$\begin{array}{rcl}\varphi (\mu )& =& i\alpha {\mu}^{2}det\left(\begin{array}{cc}{B}_{3}({\mu}^{2}){y}_{\sigma (1)}(\cdot ,\mu )& {B}_{3}({\mu}^{2}){y}_{\sigma (2)}(\cdot ,\mu )\\ {B}_{4}({\mu}^{2}){y}_{\sigma (1)}(\cdot ,\mu )& {B}_{4}({\mu}^{2}){y}_{\sigma (2)}(\cdot ,\mu )\end{array}\right)\\ +det\left(\begin{array}{cc}{B}_{3}({\mu}^{2}){y}_{\sigma (3)}(\cdot ,\mu )& {B}_{3}({\mu}^{2}){y}_{\sigma (4)}(\cdot ,\mu )\\ {B}_{4}({\mu}^{2}){y}_{\sigma (3)}(\cdot ,\mu )& {B}_{4}({\mu}^{2}){y}_{\sigma (4)}(\cdot ,\mu )\end{array}\right),\end{array}$

with

$(\sigma (1),\sigma (2),\sigma (3),\sigma (4))=\{\begin{array}{cc}(1,3,1,2)\hfill & \text{in Case 1},\hfill \\ (3,4,2,4)\hfill & \text{in Case 2},\hfill \\ (3,4,1,3)\hfill & \text{in Case 3},\hfill \\ (2,4,1,2)\hfill & \text{in Case 4}.\hfill \end{array}$

In view of (2.7), (2.8) this gives

$\begin{array}{rcl}\varphi (\mu )& =& i\alpha {\mu}^{2}[({y}_{\sigma (1)}^{\u2033}(a,\mu )+i\alpha {\mu}^{2}{y}_{\sigma (1)}^{\prime}(a,\mu ))({y}_{\sigma (2)}^{(3)}(a,\mu )-i\alpha {\mu}^{2}{y}_{\sigma (2)}(a,\mu ))\\ -({y}_{\sigma (2)}^{\u2033}(a,\mu )+i\alpha {\mu}^{2}{y}_{\sigma (2)}^{\prime}(a,\mu ))({y}_{\sigma (1)}^{(3)}(a,\mu )-i\alpha {\mu}^{2}{y}_{\sigma (1)}(a,\mu ))]\\ +({y}_{\sigma (3)}^{\u2033}(a,\mu )+i\alpha {\mu}^{2}{y}_{\sigma (3)}^{\prime}(a,\mu ))({y}_{\sigma (4)}^{(3)}(a,\mu )-i\alpha {\mu}^{2}{y}_{\sigma (4)}(a,\mu ))\\ -({y}_{\sigma (4)}^{\u2033}(a,\mu )+i\alpha {\mu}^{2}{y}_{\sigma (4)}^{\prime}(a,\mu ))({y}_{\sigma (3)}^{(3)}(a,\mu )-i\alpha {\mu}^{2}{y}_{\sigma (3)}(a,\mu ))\\ =& i\alpha {\mu}^{2}\left[i\alpha {\mu}^{2}\right\{{y}_{\sigma (1)}^{\prime}(a,\mu ){y}_{\sigma (2)}^{(3)}(a,\mu )-{y}_{\sigma (2)}^{\prime}(a,\mu ){y}_{\sigma (1)}^{(3)}(a,\mu )\\ +{y}_{\sigma (2)}^{\u2033}(a,\mu ){y}_{\sigma (1)}(a,\mu )-{y}_{\sigma (1)}^{\u2033}(a,\mu ){y}_{\sigma (2)}(a,\mu )\}\\ +{\alpha}^{2}{\mu}^{4}\{{y}_{\sigma (1)}^{\prime}(a,\mu ){y}_{\sigma (2)}(a,\mu )-{y}_{\sigma (2)}^{\prime}(a,\mu ){y}_{\sigma (1)}(a,\mu )\}\\ +{y}_{\sigma (1)}^{\u2033}(a,\mu ){y}_{\sigma (2)}^{(3)}(a,\mu )-{y}_{\sigma (2)}^{\u2033}(a,\mu ){y}_{\sigma (1)}^{(3)}(a,\mu )]\\ +i\alpha {\mu}^{2}[{y}_{\sigma (3)}^{\prime}(a,\mu ){y}_{\sigma (4)}^{(3)}(a,\mu )-{y}_{\sigma (4)}^{\prime}(a,\mu ){y}_{\sigma (3)}^{(3)}(a,\mu )\\ +{y}_{\sigma (4)}^{\u2033}(a,\mu ){y}_{\sigma (3)}(a,\mu )-{y}_{\sigma (3)}^{\u2033}(a,\mu ){y}_{\sigma (4)}(a,\mu )]\\ +{\alpha}^{2}{\mu}^{4}({y}_{\sigma (3)}^{\prime}(a,\mu ){y}_{\sigma (4)}(a,\mu )-{y}_{\sigma (4)}^{\prime}(a,\mu ){y}_{\sigma (3)}(a,\mu ))\\ +{y}_{\sigma (3)}^{\u2033}(a,\mu ){y}_{\sigma (4)}^{(3)}(a,\mu )-{y}_{\sigma (4)}^{\u2033}(a,\mu ){y}_{\sigma (3)}^{(3)}(a,\mu ).\end{array}$

Each of the summands in

*ϕ* is a product of a power in

*μ* and a product of two sums of a trigonometric and a hyperbolic functions. The terms with the highest

*μ*-powers in

$\varphi (\mu )$ are non-zero constant multiples of

${\varphi}_{0}(\mu )=\{\begin{array}{cc}2{\mu}^{4}({y}_{\sigma (1)}^{\prime}(a,\mu ){y}_{\sigma (2)}^{(3)}(a,\mu )-{y}_{\sigma (2)}^{\prime}(a,\mu ){y}_{\sigma (1)}^{(3)}(a,\mu ))\hfill & \text{in Cases 1, 2},\hfill \\ 2{\mu}^{2}({y}_{\sigma (3)}^{\prime}(a,\mu ){y}_{\sigma (4)}^{(3)}(a,\mu )-{y}_{\sigma (4)}^{\prime}(a,\mu ){y}_{\sigma (3)}^{(3)}(a,\mu ))\hfill & \text{in Cases 3, 4}.\hfill \end{array}$

For the above four cases, we obtain:

We next give the asymptotic distributions of the zeros of ${\varphi}_{0}(\mu )$ with proper counting.

**Lemma 3.1** Case 1:

${\varphi}_{0}$ *has a zero of multiplicity* 8

*at* 0,

*simple zeros at* ${\tilde{\mu}}_{k}=(k-2)\frac{\pi}{a},\phantom{\rule{1em}{0ex}}k=3,4,\dots ,$

*simple zeros at* $-{\tilde{\mu}}_{k}$, ${\tilde{\mu}}_{-k}=i{\tilde{\mu}}_{k}$ *and* $-i{\tilde{\mu}}_{k}$ *for* $k=3,4,\dots $, *and no other zeros*.

Case 2:

${\varphi}_{0}$ *has a zero of multiplicity* 4

*at* 0,

*exactly one simple zero* ${\tilde{\mu}}_{k}$ *in each interval* $((k-\frac{1}{2})\frac{\pi}{a},(k+\frac{1}{2})\frac{\pi}{a})$ *for positive integers* *k* *with asymptotics* ${\tilde{\mu}}_{k}=(4k-5)\frac{\pi}{4a}+o(1),\phantom{\rule{1em}{0ex}}k=2,3,\dots ,$

*simple zeros at* $-{\tilde{\mu}}_{k}$, ${\tilde{\mu}}_{-k}=i{\tilde{\mu}}_{k}$ *and* $-i{\tilde{\mu}}_{k}$ *for* $k=2,3,\dots $ , *and no other zeros*.

Case 3:

${\varphi}_{0}$ *has a zero of multiplicity* 6

*at* 0,

*simple zeros at* ${\tilde{\mu}}_{k}=(k-1)\frac{\pi}{a},\phantom{\rule{1em}{0ex}}k=2,3,\dots ,$

*simple zeros at* $-{\tilde{\mu}}_{k}$, ${\tilde{\mu}}_{-k}=i{\tilde{\mu}}_{k}$ *and* $-i{\tilde{\mu}}_{k}$ *for* $k=2,3,\dots $, *and no other zeros*.

Case 4:

${\varphi}_{0}$ *has a zero of multiplicity* 6

*at* 0,

*exactly one simple zero* ${\tilde{\mu}}_{k}$ *in each interval* $((k-\frac{1}{2})\frac{\pi}{a},(k+\frac{1}{2})\frac{\pi}{a})$ *for positive integers* *k* *with asymptotics* ${\tilde{\mu}}_{k}=(4k-5)\frac{\pi}{4a}+o(1),\phantom{\rule{1em}{0ex}}k=2,3,\dots ,$

*simple zeros at* $-{\tilde{\mu}}_{k}$, ${\tilde{\mu}}_{-k}=i{\tilde{\mu}}_{k}$ *and* $-i{\tilde{\mu}}_{k}$ *for* $k=2,3,\dots $, *and no other zeros*.

*Proof* The result is obvious in Cases 1 and 3. Cases 2 and 4 only differ in the factor with the power of *μ*, and the multiplicity of the corresponding zero of ${\varphi}_{0}$ at 0 is easy to verify. The choice of the indexing for the non-zero zeros of ${\varphi}_{0}$ in each case will become apparent later.

It, therefore, remains to describe the behavior of the non-zero zeros of

${\varphi}_{0}$ in Case 2. First, we are going to find the zeros of

${\varphi}_{0}$ on the positive real axis. One can observe that for

$\mu \ne 0$,

${\varphi}_{0}(\mu )=0$ implies

$cosh(\mu a)\ne 0$ and

$cos(\mu a)\ne 0$, whence the positive zeros of

${\varphi}_{0}$ are those

$\mu >0$ for which

$tan(\mu a)+tanh(\mu a)=0$. Since

${tan}^{\prime}(\mu a)\ge 1$ and

${tanh}^{\prime}(\mu a)>0$ for all

$x\in \mathbb{R}$ where the functions are defined, the function

$\mu \mapsto tan(\mu a)+tanh(\mu a)$ is increasing with a positive derivative on each interval

$((k-\frac{1}{2})\frac{\pi}{a},(k+\frac{1}{2})\frac{\pi}{a})$,

$k\in \mathbb{Z}$. On each of these intervals, the function moves from −∞ to ∞, thus we have exactly one simple zero

${\tilde{\mu}}_{k}$ of

$tan(\mu a)+tanh(\mu a)$ in each interval

$((k-\frac{1}{2})\frac{\pi}{a},(k+\frac{1}{2})\frac{\pi}{a})$, where

*k* is a positive integer, and no zero in

$(0,\frac{\pi}{2a})$. Since

$tanh(\mu a)\to 1$ as

$\mu \to \mathrm{\infty}$, we have

${\tilde{\mu}}_{k}=(4k-5)\frac{\pi}{4a}+o(1),\phantom{\rule{1em}{0ex}}k=1,2,\dots .$

The location of the zeros on the other three half-axes follows by repeated application of ${\varphi}_{0}(i\mu )=-{\varphi}_{0}(\mu )$.

The proof will be complete if we show that all zeros of

${\varphi}_{0}$ lie on the real or the imaginary axis. To this end, we observe that the product-to-sum formula for trigonometric functions gives

$\begin{array}{rcl}{\varphi}_{0}(\mu )& =& {\mu}^{3}[cosh(\mu a)sin(\mu a)+sinh(\mu a)cos(\mu a)]\\ =& \frac{1}{2}{\mu}^{3}[sin((1+i)\mu a)+sin((1-i)\mu a)\\ -isin((1+i)\mu a)+isin((1-i)\mu a)]\\ =& \frac{1}{2}{\mu}^{3}[(1-i)sin((1+i)\mu a)+(1+i)sin((1-i)\mu a)].\end{array}$

(3.1)

Putting

$(1+i)\mu a=x+iy$,

$x,y\in \mathbb{R}$, it follows for

$\mu \ne 0$ that

Since

${cosh}^{2}x+{cos}^{2}x=\frac{1}{2}cosh(2x)+\frac{1}{2}cos(2x)+1$ has a positive derivative on

$(0,\mathrm{\infty})$, this function is strictly increasing, and

${\varphi}_{0}(\mu )=0$ therefore, implies by (3.2) that

$|y|=|x|$ and thus

$y=\pm x$. Then

$\mu =\frac{x+iy}{(1+i)a}=\frac{1\pm i}{1+i}\frac{x}{a}$

is either real or pure imaginary. □

**Proposition 3.2** *For* $g=0$,

*there exists a positive integer* ${k}_{0}$ *such that the eigenvalues* ${\stackrel{\u02c6}{\lambda}}_{k}$,

*counted with multiplicity*,

*of the problem* (2.1), (2.5)-(2.8),

*where* $k\in \mathbb{Z}\setminus \{0\}$ *in Cases* 1

*and * 2

*and* $k\in \mathbb{Z}$ *in Cases* 3

*and* 4,

*can be enumerated in such a way that the eigenvalues* ${\stackrel{\u02c6}{\lambda}}_{k}$ *are pure imaginary for* $|k|<{k}_{0}$,

*and* ${\stackrel{\u02c6}{\lambda}}_{-k}=-\overline{{\stackrel{\u02c6}{\lambda}}_{k}}$ *for* $k\ge {k}_{0}$.

*For* $k>0$,

*we can write* ${\stackrel{\u02c6}{\lambda}}_{k}={\stackrel{\u02c6}{\mu}}_{k}^{2}$,

*where the* ${\stackrel{\u02c6}{\mu}}_{k}$ *have the following asymptotic representation as* $k\to \mathrm{\infty}$:

*In particular*, *the number of pure imaginary eigenvalues is even in Cases* 1 *and* 2 *and odd in Cases* 3 *and* 4.

*Proof* Case 4: A straightforward calculation gives

$\begin{array}{rcl}\varphi (\mu )& =& -\frac{1}{2}i(2\alpha +{\alpha}^{3}){\mu}^{3}[cosh(\mu a)sin(\mu a)-sinh(\mu a)cos(\mu a)]\\ -\frac{1}{2}i\alpha {\mu}^{5}[sinh(\mu a)cos(\mu a)+cosh(\mu a)sin(\mu a)]\\ -\frac{1}{2}{\alpha}^{2}{\mu}^{4}[3cosh(\mu a)cos(\mu a)+1]-\frac{1}{2}{\mu}^{4}[cosh(\mu a)cos(\mu a)-1]\\ -{\alpha}^{2}{\mu}^{2}sin(\mu a)sinh(\mu a).\end{array}$

(3.3)

Up to the constant factor

$\frac{1}{2}i\alpha $, the second term equals

${\varphi}_{0}(\mu )$. It follows that for

*μ* outside the zeros of

${\varphi}_{0}$,

$cos(\cdot a)$ and

$cosh(\cdot a)$, we have

$\begin{array}{rcl}{\varphi}_{1}(\mu )=\frac{2\varphi (\mu )-i\alpha {\varphi}_{0}(\mu )}{i\alpha {\varphi}_{0}(\mu )}& =& \frac{{\alpha}^{2}-1}{i\alpha \mu}\frac{1}{cosh(\mu a)cos(\mu a)}\frac{1}{tan(\mu a)+tanh(\mu a)}\\ +\frac{3{\alpha}^{2}+1}{i\alpha \mu}\frac{1}{tan(\mu a)+tanh(\mu a)}+\frac{2\alpha}{i{\mu}^{3}}\frac{tan(\mu a)tanh(\mu a)}{tan(\mu a)+tanh(\mu a)}\\ +\frac{(2+{\alpha}^{2})}{{\mu}^{2}}[1-2\frac{tanh(\mu a)}{tan(\mu a)+tanh(\mu a)}].\end{array}$

(3.4)

Fix

$\epsilon \in (0,\frac{\pi}{4a})$ and for

$k=2,3,\dots $ let

${R}_{k,\epsilon}$ be the squares determined by the vertices

$(4k-5)\frac{\pi}{4a}\pm \epsilon \pm i\epsilon $,

$k\in \mathbb{Z}$. These squares do not intersect due to

$\epsilon <\frac{\pi}{2a}$. Since

$tanz=-1$ if and only if

$z=j\pi -\frac{\pi}{4}$ and

$j\in \mathbb{Z}$, it follows from the periodicity of tan that the number

${C}_{1}(\epsilon )=2min\{|tan(\mu a)+1|:\mu \in {R}_{k,\epsilon}\}$

is positive and independent of

*ε*. Since

$tanh(\mu a)\to 1$ uniformly in the strip

$\{\mu \in \mathbb{C}:Re\mu \ge 1,|Im\mu |\le \frac{\pi}{4a}\}$ as

$|\mu |\to \mathrm{\infty}$, there is an integer

${k}_{1}(\epsilon )$ such that

$|tan(\mu a)+tanh(\mu a)|\ge {C}_{1}(\epsilon )\phantom{\rule{1em}{0ex}}\text{for all}\mu \in {R}_{k,\epsilon}\text{with}k{k}_{1}(\epsilon ).$

By periodicity, there are numbers ${C}_{2}(\epsilon )>0$ and ${C}_{3}(\epsilon )>0$ such that $|tan(\mu a)|<{C}_{2}(\epsilon )$ and $|cos(\mu a)|>{C}_{3}(\epsilon )$ for all $\mu \in {R}_{k,\epsilon}$ and all *k*. Observing $|cosh(\mu a)|\ge |sinh((Re\mu )a)|$, it follows that there is ${k}_{2}(\epsilon )\ge {k}_{1}(\epsilon )$ such that for all *μ* on the squares ${R}_{k,\epsilon}$ with $k>{k}_{2}(\epsilon )$ the estimate $|{\varphi}_{1}(\mu )|<1$ holds. Further, we may assume by Lemma 3.1 that ${\tilde{\mu}}_{k}$ is inside ${R}_{k,\epsilon}$ for $k>{k}_{2}(\epsilon )$ and that no other zero of ${\varphi}_{0}$ has this property. Hence, it follows by Rouché’s theorem that there is exactly one (simple) zero ${\stackrel{\u02c6}{\mu}}_{k}$ of *ϕ* in each ${R}_{k}$ for $k\ge {k}_{2}(\epsilon )$. Replacing *μ* with *iμ* only changes the sign of the second term in (3.3) and thus the sign of ${\varphi}_{1}$. Hence, the same estimates apply to corresponding squares along the other three half-axes, and we therefore have that *ϕ* has zeros $\pm {\stackrel{\u02c6}{\mu}}_{k}$, $\pm {\stackrel{\u02c6}{\mu}}_{-k}$ for $k>{k}_{2}(\epsilon )$ with the same asymptotic behavior as the zeros $\pm {\tilde{\mu}}_{k}$, $\pm i{\tilde{\mu}}_{k}$ of ${\varphi}_{0}$ as discussed in Lemma 3.1.

Next, we are going to estimate

${\varphi}_{1}$ on the squares

${S}_{k}$,

$k\in \mathbb{N}$, whose vertices are

$\pm k\frac{\pi}{a}\pm ik\frac{\pi}{a}$. For

$k\in \mathbb{Z}$ and

$\gamma \in \mathbb{R}$,

$tan\left((\frac{k\pi}{a}+i\gamma )a\right)=tan(i\gamma a)=itanh(\gamma a)\in i\mathbb{R}.$

(3.5)

Therefore, we have for

$\mu =\frac{k\pi}{a}+i\gamma $, where

$k\in \mathbb{Z}$ and

$\gamma \in \mathbb{R}$, that

$|tan(\mu a)|<1\phantom{\rule{1em}{0ex}}\text{and}\phantom{\rule{1em}{0ex}}|tan(\mu a)\pm 1|\ge 1.$

(3.6)

For

$\mu =x+iy$,

$x,y\in \mathbb{R}$ and

$x\ne 0$, we have

$tanh(\mu a)=\frac{{e}^{(ax+iay)}-{e}^{-(ax+iay)}}{{e}^{(ax+iay)}+{e}^{-(ax+iay)}}\to \pm 1$

uniformly in

*y* as

$x\to \pm \mathrm{\infty}$. Hence, there exists

${\stackrel{\u02c6}{k}}_{0}\in \mathbb{N}$ such that for all

$k\in \mathbb{Z}$,

$|k|\ge {\stackrel{\u02c6}{k}}_{0}$ and

$\gamma \in \mathbb{R}$,

$|tanh\left((\frac{k\pi}{a}+i\gamma )a\right)-sgn(k)|<\frac{1}{2}.$

(3.7)

It follows from (3.6) and (3.7) for

$\mu =\frac{k\pi}{a}+i\gamma $,

$k\in \mathbb{Z}$,

$|k|\ge {\stackrel{\u02c6}{k}}_{0}$ and

$\gamma \in \mathbb{R}$ that

$|tan(\mu a)+tanh(\mu a)|\ge \frac{1}{2}.$

(3.8)

Furthermore, we are going to make use of the estimates

which hold for all $k\in \mathbb{Z}$ with $|k|\ge {\stackrel{\u02c6}{k}}_{0}$ and all $\gamma \in \mathbb{R}$. Therefore, it follows from (3.6), (3.8)-(3.10) and the corresponding estimates with *μ* replaced by *iμ* that there is ${\stackrel{\u02c6}{k}}_{1}$ such that $|{\varphi}_{1}(\mu )|<1$ for all $\mu \in {S}_{k}$ with $k>{\stackrel{\u02c6}{k}}_{1}$. Again from the definition of ${\varphi}_{1}$ in (3.4) and Rouché’s theorem, we conclude that the functions ${\varphi}_{0}$ and *ϕ* have the same number of zeros in the square ${S}_{k}$, for $k\in \mathbb{N}$ with $k\ge {\stackrel{\u02c6}{k}}_{1}$.

Since ${\varphi}_{0}$ has $4k+2$ zeros inside ${S}_{k}$ and thus $4k+2+4$ zeros inside ${S}_{k+1}$, it follows that *ϕ* has no large zeros other than the zeros $\pm {\stackrel{\u02c6}{\mu}}_{k}$ found above for $|k|$ sufficiently large, and that ${\stackrel{\u02c6}{\lambda}}_{k}={\stackrel{\u02c6}{\mu}}_{k}^{2}$ account for all eigenvalues of the problem (2.1)-(2.2) since each of these eigenvalues gives rise to two zeros of *ϕ*, counted with multiplicity. By Proposition 2.3, all eigenvalues with non-zero real part occur in pairs ${\stackrel{\u02c6}{\lambda}}_{k}$, $-\overline{{\stackrel{\u02c6}{\lambda}}_{k}}$, which shows that we can index all such eigenvalues as ${\stackrel{\u02c6}{\lambda}}_{-k}=-\overline{{\stackrel{\u02c6}{\lambda}}_{k}}$. Since there is an odd number of remaining indices, the number of pure imaginary eigenvalues must be odd.

Case 2: The function

*ϕ* in this case is

$\begin{array}{rcl}\varphi (\mu )& =& -\frac{1}{2}(2{\alpha}^{2}+1)\mu [cosh(\mu a)sin(\mu a)-sinh(\mu a)cos(\mu a)]\\ -\frac{1}{2}{\alpha}^{2}{\mu}^{3}[sinh(\mu a)cos(\mu a)+cosh(\mu a)sin(\mu a)]\\ +\frac{1}{2}i\alpha {\mu}^{2}[3cosh(\mu a)cos(\mu a)+1]+\frac{1}{2}i{a}^{3}{\mu}^{2}[cosh(\mu a)cos(\mu a)-1]\\ +i\alpha {\mu}^{2}sin(\mu a)sinh(\mu a).\end{array}$

Then all the estimates are as in Case 4, and the result in Case 2 immediately follows from that in Case 4 if we observe that each ${S}_{k}$ for *k* large enough contains two fewer zeros of *ϕ* than in Case 4.

Case 1: A straightforward calculation gives

$\begin{array}{rcl}\varphi (\mu )& =& {\alpha}^{2}{\mu}^{6}sin(\mu a)sinh(\mu a)-\frac{1}{2}(1+3{\alpha}^{2}){\mu}^{4}cos(\mu a)cosh(\mu a)\\ -\frac{1}{2}i(2\alpha +{\alpha}^{3}){\mu}^{5}(sin(\mu a)cosh(\mu a)+cos(\mu a)sinh(\mu a))\\ -\frac{1}{2}i\alpha {\mu}^{3}(sin(\mu a)cosh(\mu a)-cos(\mu a)sinh(\mu a))+\frac{1}{2}(1-{\alpha}^{2}){\mu}^{4}.\end{array}$

Then

$\begin{array}{rcl}{\varphi}_{1}(\mu )& =& \frac{2\varphi (\mu )+{\alpha}^{2}{\varphi}_{0}(\mu )}{{\varphi}_{0}(\mu )}\\ =& \frac{1+3{\alpha}^{2}}{2{\mu}^{2}}cot(\mu a)coth(\mu a)+\frac{(2\alpha +{\alpha}^{3})i}{2\mu}[coth(\mu a)+cot(\mu a)]\\ +\frac{i\alpha}{2{\mu}^{3}}[coth(\mu a)-cot(\mu a)]-\frac{1-{\alpha}^{2}}{2{\mu}^{2}}\frac{1}{sin(\mu a)sinh(\mu a)}.\end{array}$

The result follows with reasonings and estimates as in the proof of Case 4, replacing *μ* by $\mu \pm \frac{\pi}{2}$ and $\mu \pm i\frac{\pi}{2}$, respectively.

Case 3: The function

*ϕ* in this case is

$\begin{array}{rcl}\varphi (\mu )& =& -i\alpha {\mu}^{4}sin(\mu a)sinh(\mu a)+\frac{1}{2}i(3\alpha +{\alpha}^{3}){\mu}^{2}cos(\mu a)cosh(\mu a)\\ -\frac{1}{2}(2{\alpha}^{2}+1){\mu}^{3}(sin(\mu a)cosh(\mu a)+cos(\mu a)sinh(\mu a))\\ -\frac{1}{2}{\alpha}^{2}\mu (sin(\mu a)cosh(\mu a)-cos(\mu a)sinh(\mu a))+\frac{1}{2}i(\alpha -{\alpha}^{3}){\mu}^{2},\end{array}$

and a reasoning as in Case 1 completes the proof. □