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. □