Back to the general case, consider *p* ∈ ℕ^{∗}, *a* ∈ *ℓ*(ℤ;*p*), *c* ∈ *ℓ*^{∗}(ℤ;*p*) and the associate self–adjoint operator *Δ*_{a,c}. Although this scenario seems to be far away from the easiest one analyzed in the previous section, we will show that in fact Chebyshev equations contain all the information needed to conclude the existence of bounded solutions for the difference equation *Δ*_{a,c}(*z*) = 0. This is true because the main result in [5] establishes that (irreductible) second order difference equations (not necessarily self-adjoint) with periodic coefficients are basically equivalent to some Chebyshev equation. For the setting concerning to this paper we have the following facts.

#### Lemma 4.1

([5, Theorem 3.3]). *Given* *p* ∈ ℕ^{∗}, *a* ∈ *ℓ*(ℤ;*p*) *and* *c* ∈ *ℓ*^{∗}(ℤ;*p*); *there exists q*(*a*, *c*;*p*) ∈ ℝ, *depending only on the coefficients a and c and on the period p*, *such that z* ∈ *ℓ*(ℤ) *is a solution of the equation* *Δ*_{a,c}(*z*) = 0 *iff for any* *m* ∈ ℤ, *z*_{p,m} *is a solution of the
Chebyshev equation with parameter q*(*a*, *c*;*p*); *that is*

$$\begin{array}{}{\displaystyle v(k+1)-2q(a,c;p)v(k)+v(k-1)=0,\phantom{\rule{.25cm}{0ex}}k\in \mathbb{Z}.}\end{array}$$

As the boundedness of *z* is equivalent to the boundedness of the sequences *z*_{p,m}, *m* = 0, …, *p* – 1, we can conclude that existence of bounded solutions for the equation *Δ*_{a,c}(*z*) = 0, depends only on the knowledge of the specific value *q*(*a*, *c*;*p*). Since in [5, Theorem 3.3] the existence of this parameter was proved by induction the above result is not useful in practice. For this reason, most of the above mentioned paper was devoted to the explicit computation of the so–called *Floquet function*; that is, the function assigning the value *q*(*a*, *c*;*p*) to any *a* ∈ *ℓ*(ℤ;*p*) and *c* ∈ *ℓ*^{∗}(ℤ;*p*). Notice that, in fact, the value *q*(*a*, *c*;*p*) only depends on *a*(*j*),*c*(*j*), *j* = 0, …,*p* – 1. Once this function was obtained, the characterization of the existence of periodic solutions for the equation *Δ*_{a,c}(*z*) = 0 appears as a simple by–product, since from Lemma 2.1, they are characterized as being constant the sequences *z*_{p,m}, 0 ≤ *m* ≤ *p* – 1, see [5, Corollary 4.8]. So, the main novelty of this paper is to derive the characterization of the existence of bounded solutions for the equation *Δ*_{a,c}(*z*) = 0, from the value *q*(*a*, *c*;*p*). To do this, we need to introduce some notations and concepts.

A *binary multi-index of order p* is a *p*-tuple *α* = (*α*_{0}, …, *α*_{p – 1}) ∈ {0,1}^{p} and its *length* is defined as
$\begin{array}{}{\displaystyle |\alpha |=\sum _{j=0}^{p-1}{\alpha}_{j}\le p.}\end{array}$
So |*α*| = *m* iff exactly *m* components of *α* are equal to 1 and exactly *p* – *m* components of *α* are equal to 0.

Given a binary multi-index of order *p*, *α* ∈ {0, 1}^{p} such that |*α*| = *m* ≥ 1, we consider 0 ≤ *i*_{1} < … < *i*_{m}≤ *p* – 1 such that *α*_{i1} = … = *α*_{im} = 1. Given *p* ∈ ℕ^{∗}, we define the following subsets of the set {0, 1}^{p} of binary multi-indexes of order *p*:

$\begin{array}{}{\displaystyle {\mathrm{\Lambda}}_{p}^{0}=\{(0,\dots ,0)\},}\end{array}$
for *p* ≥ 1.

$\begin{array}{}{\displaystyle {\mathrm{\Lambda}}_{p}^{1}=\{\alpha :|\alpha |=1\},}\end{array}$
for *p* ≥ 2.

$\begin{array}{}{\displaystyle {\mathrm{\Lambda}}_{p}^{m}=\{\alpha :|\alpha |=m,\phantom{\rule{.15cm}{0ex}}{i}_{j+1}-{i}_{j}\ge 2,\phantom{\rule{.15cm}{0ex}}1\le j\le m-1\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\text{and}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}{i}_{m}\le p-2\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\text{if}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}{i}_{1}=0\}}\end{array}$
for *p* ≥ 4, and
$\begin{array}{}{\displaystyle m=2,\dots ,\lfloor \frac{p}{2}\rfloor ,}\end{array}$
where 0 ≤ *i*_{1} < … < *i*_{m} ≤ *p* – 1 are the indexes such that *α*_{i1} = … = *α*_{im} = 1.

In addition, if *p* ≥ 2, *m* = 1,…,
$\begin{array}{}{\displaystyle \lfloor \frac{p}{2}\rfloor}\end{array}$ and
$\begin{array}{}{\displaystyle \alpha \in {\mathrm{\Lambda}}_{p}^{m},}\end{array}$ let 0 ≤ *i*_{1} < … < *i*_{m} ≤ *p* – 1 be the indexes such that *α*_{i1} = … = *α*_{im} = 1. Then, we define the binary multi-index *α* of order *p* as

$$\begin{array}{}{\displaystyle {\overline{\alpha}}_{{i}_{j}}={\overline{\alpha}}_{{i}_{j}+1}=0,\phantom{\rule{.15cm}{0ex}}j=1,\dots ,m,\phantom{\rule{.25cm}{0ex}}\text{and}\phantom{\rule{.25cm}{0ex}}{\overline{\alpha}}_{i}=1\phantom{\rule{.25cm}{0ex}}\text{otherwise},}\end{array}$$

where if *i*_{m} = *p* – 1, then *α*_{p – 1} = *α*_{0} = 0. Moreover, if *α* =(0,…,0); that is, if
$\begin{array}{}{\displaystyle \alpha \in {\mathrm{\Lambda}}_{p}^{0},}\end{array}$ then we define *α* = (1,…,1). It is clear that, in any case, |*α*| = *p* – 2*m*.

We are now ready to show the expression for the value of *q*(*a*, *c*; *p*). In the sequel, we always assume that 0^{0} = 1 and also the usual convention that empty sums and empty products are defined as 0 and 1, respectively.

#### Lemma 4.2

([5, Theorem 4.4]). *Given p* ∈ ℕ^{*}, *a* ∈ *ℓ*(ℤ; *p*) *and c* ∈ *ℓ*^{*} (ℤ; *p*), *then*

$$\begin{array}{}q(a,c;p)={\displaystyle \frac{1}{2}}(\prod _{i=0}^{p-1}c(i){)}^{-1}\sum _{j=0}^{\lfloor \frac{p}{2}\rfloor}(-1{)}^{j}\sum _{\alpha \in {\mathrm{\Lambda}}_{p}^{j}}\prod _{i=0}^{p-1}c(i{)}^{2{\alpha}_{i}}a(i{)}^{{\overline{\alpha}}_{i}}.\end{array}$$

Observe that when *p* = 1, the above identity becomes
$\begin{array}{}{\displaystyle q(a,c;1)=\frac{a}{2c};}\end{array}$ that is, the value corresponding to the case in which the coefficients *a* and *c* are constant; or equivalent both have period *p* = 1.

Our main result appears now as a consequence of the Proposition 3.1 together with Lemma 2.1 and also the above Lemma.

#### Theorem 4.3

*Given p* ∈ ℕ^{*}, *a* ∈ *ℓ*(ℤ; *p*) *and c* ∈ *ℓ*^{*}(ℤ; *p*), *then the equation*

$$\begin{array}{}{\displaystyle c(k)z(k+1)+c(k-1)z(k-1)-a(k)z(k)=0,\phantom{\rule{.25cm}{0ex}}k\in \mathbb{Z}}\end{array}$$

*has bounded solutions iff*

$$\begin{array}{}{\displaystyle |\sum _{j=0}^{\lfloor \frac{p}{2}\rfloor}(-1{)}^{j}\sum _{\alpha \in {\mathrm{\Lambda}}_{p}^{j}}\prod _{i=0}^{p-1}c(i{)}^{2{\alpha}_{i}}a(i{)}^{{\overline{\alpha}}_{i}}|\le 2\prod _{i=0}^{p-1}|c(i)|}\end{array}$$

*and when the inequality is strict*, *all the solutions are bounded*. *Moreover*, *if*

$$\begin{array}{}{\displaystyle \sum _{j=0}^{\lfloor \frac{p}{2}\rfloor}(-1{)}^{j}\sum _{\alpha \in {\mathrm{\Lambda}}_{p}^{j}}\prod _{i=0}^{p-1}c(i{)}^{2{\alpha}_{i}}a(i{)}^{{\overline{\alpha}}_{i}}=2\prod _{i=0}^{p-1}c(i)}\end{array}$$

*then the equation has periodic solutions with period p and these are the unique bounded solutions*.

#### Corollary 4.4

*Given E*, *λ* ∈ ℝ, *θ* ∈ [0, 2*π*) *and*
$\begin{array}{}{\displaystyle \omega =\frac{m}{p},}\end{array}$ *where p* ∈ ℕ^{*}, *m* ∈ ℤ *and* (*p*, *m*) = 1, *then the Mathieu equation*

$$\begin{array}{}{\displaystyle z(k+1)+z(k-1)+\lambda \phantom{\rule{thinmathspace}{0ex}}\mathrm{cos}(2\pi \omega k+\theta )z(k)=Ez(k),\phantom{\rule{.25cm}{0ex}}k\in \mathbb{Z}}\end{array}$$

*has bounded solutions iff*

$$\begin{array}{}{\displaystyle |\sum _{j=0}^{\lfloor \frac{p}{2}\rfloor}(-1{)}^{j}\sum _{\alpha \in {\mathrm{\Lambda}}_{p}^{j}}\prod _{i=0}^{p-1}(E-\lambda \phantom{\rule{thinmathspace}{0ex}}\mathrm{cos}(2\pi \omega i+\theta ){)}^{{\overline{\alpha}}_{i}}|\le 2}\end{array}$$

*and when the inequality is strict*, *all the solutions are bounded*. *Moreover*, *if*

$$\begin{array}{}{\displaystyle \sum _{j=0}^{\lfloor \frac{p}{2}\rfloor}(-1{)}^{j}\sum _{\alpha \in {\mathrm{\Lambda}}_{p}^{j}}\prod _{i=0}^{p-1}(E-\lambda \phantom{\rule{thinmathspace}{0ex}}\mathrm{cos}(2\pi \omega i+\theta ){)}^{{\overline{\alpha}}_{i}}=2}\end{array}$$

*then the Mathieu equation has periodic solutions with period p and these are the unique bounded solutions*.

Clearly, the main difficulty to apply the above characterizations is to obtain the binary multi–indexes involved in them. In general, this is a difficult task and, in fact, the number of multi–indexes in
$\begin{array}{}{\displaystyle {\mathrm{\Lambda}}_{p}^{j},0\le j\le \lfloor \frac{p}{2}\rfloor ,}\end{array}$ grows dramatically with *p*. Specifically, we have
$\begin{array}{}{\displaystyle |{\mathrm{\Lambda}}_{p}^{j}|=\frac{p}{p-j}\left(\genfrac{}{}{0ex}{}{p-j}{j}\right)\text{for any}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}0\le j\le \lfloor \frac{p}{2}\rfloor ,}\end{array}$ which for any *m* ∈ ℕ^{*} implies that
$\begin{array}{}{\displaystyle \sum _{j=0}^{m}|{\mathrm{\Lambda}}_{2m}^{j}|=2{T}_{m}(\frac{3}{2})\text{and that}\sum _{j=0}^{m}|{\mathrm{\Lambda}}_{2m+1}^{j}|={W}_{m}(\frac{3}{2}),}\end{array}$, see [5, Proposition 4.2].

We end this paper with some specific examples using the given characterization for the existence of bounded solutions for difference equations with periodic coefficients with period up to 4. Remember that the case *p* = 1, the easiest case, has been analyzed in the previous sections.

#### Corollary 4.5

(Period *p* = 2) *Given a* ∈ *ℓ*(ℤ; 2) *and c* ∈ *ℓ*^{*}(ℤ; 2), *then the equation*

$$\begin{array}{}{\displaystyle c(k)z(k+1)+c(k-1)z(k-1)-a(k)z(k)=0,\phantom{\rule{.25cm}{0ex}}k\in \mathbb{Z}}\end{array}$$

*has bounded solutions iff*

$$\begin{array}{}{\displaystyle (|c(0)|-|c(1)|{)}^{2}\le a(0)a(1)\le (|c(0)|+|c(1)|{)}^{2}}\end{array}$$

*and when both inequalities are strict*, *then all the solutions are bounded*. *In particular*, *given E*, *λ*∈ ℝ, *θ* ∈ [0, 2*π*), *then the Mathieu equation*

$$\begin{array}{}{\displaystyle z(k+1)+z(k-1)+\lambda \phantom{\rule{thinmathspace}{0ex}}\mathrm{cos}(\pi k+\theta )z(k)=Ez(k),\phantom{\rule{.25cm}{0ex}}k\in \mathbb{Z}}\end{array}$$

*has bounded solutions iff*

$$\begin{array}{}{\displaystyle 0\le {E}^{2}-{\lambda}^{2}{\mathrm{cos}}^{2}(\theta )\le 4.}\end{array}$$

#### Proof

In this case we have
$\begin{array}{}{\displaystyle {\mathrm{\Lambda}}_{2}^{0}}\end{array}$ = {(0, 0)},
$\begin{array}{}{\displaystyle {\mathrm{\Lambda}}_{2}^{1}}\end{array}$ = {(1, 0),(0, 1)},
$\begin{array}{}{\displaystyle {\overline{\mathrm{\Lambda}}}_{2}^{0}}\end{array}$ = {(1, 1)} and
$\begin{array}{}{\displaystyle {\overline{\mathrm{\Lambda}}}_{2}^{1}}\end{array}$ = {(0, 0),(0, 0)}, which implies that
$\begin{array}{}{\displaystyle q(a,c;2)=\frac{a(0)a(1)-c(0{)}^{2}-c(1{)}^{2}}{2c(0)c(1)}.}\end{array}$

In particular, for the Mathieu which coefficient has period 2, the frequency is
$\begin{array}{}{\displaystyle \omega =\frac{m}{2},}\end{array}$ where *m* ∈ ℤ is odd and hence
$\begin{array}{}{\displaystyle \omega =n+\frac{1}{2}}\end{array}$ with *n* ∈ ℤ. Therefore, the coefficient is *a*(*k*) = *E* – *λ* cos(*π* *k* + *θ*), which implies that *a*(0) = *E* – *λ* cos(*θ*), whereas *a*(1) = *E* – *λ* cos(*π* + *θ*) = *E* + *λ* cos(*θ*).

□

#### Corollary 4.6

(Period *p* = 3). *Given a* ∈ *ℓ*(ℤ; 3) *and c* ∈ *ℓ*^{*}(ℤ; 3), *then the equation*

$$\begin{array}{}{\displaystyle c(k)z(k+1)+c(k-1)z(k-1)-a(k)z(k)=0,\phantom{\rule{.25cm}{0ex}}k\in \mathbb{Z}}\end{array}$$

*has bounded solutions iff*

$$\begin{array}{}{\displaystyle |a(0)a(1)a(2)-c(0{)}^{2}a(2)-c(1{)}^{2}a(0)-c(2{)}^{2}a(1)|\le 2|c(0)c(1)c(2)|}\end{array}$$

*and when the inequality is strict*, *all the solutions are bounded*. *In particular*, *given E*, *λ* ∈ ℝ, *θ* ∈ [0, 2*π*) *and*
$\begin{array}{}{\displaystyle \omega =\frac{m}{3},}\end{array}$ *where m* ∈ ℤ *and* (3, *m*) = 1, *then the Mathieu equation*

$$\begin{array}{}{\displaystyle z(k+1)+z(k-1)+\lambda \phantom{\rule{thinmathspace}{0ex}}\mathrm{cos}(2\pi \omega k+\theta )z(k)=Ez(k),\phantom{\rule{.25cm}{0ex}}k\in \mathbb{Z}}\end{array}$$

*has bounded solutions iff*

$$\begin{array}{}{\displaystyle |(E-\lambda \phantom{\rule{thinmathspace}{0ex}}\mathrm{cos}(\theta ))({E}^{2}-\frac{3}{4}{\lambda}^{2}+{\lambda}^{2}{\mathrm{cos}}^{2}(\theta )+E\lambda \phantom{\rule{thinmathspace}{0ex}}\mathrm{cos}(\theta ))-3E|\le 2.}\end{array}$$

#### Proof

In this case we have
$\begin{array}{}{\displaystyle {\mathrm{\Lambda}}_{3}^{0}}\end{array}$ = {(0, 0, 0)} and
$\begin{array}{}{\displaystyle {\mathrm{\Lambda}}_{3}^{1}}\end{array}$ = {(1, 0, 0),(0, 1, 0),(0, 0, 1)}, which
implies that
$\begin{array}{}{\displaystyle {\overline{\mathrm{\Lambda}}}_{3}^{0}}\end{array}$ = {(1, 1, 1)} and
$\begin{array}{}{\displaystyle {\overline{\mathrm{\Lambda}}}_{3}^{1}}\end{array}$ = {(0, 0, 1),(1, 0, 0),(0, 1, 0)} and hence,

$$\begin{array}{}{\displaystyle q(a,c;3)=\frac{a(0)a(1)a(2)-c(0{)}^{2}a(2)-c(1{)}^{2}a(0)-c(2{)}^{2}a(1)}{2c(0)c(1)c(2)}.}\end{array}$$

In particular, for the Mathieu equation the condition for the existence of bounded solutions becomes

$$\begin{array}{}{\displaystyle |a(0)a(1)a(2)-a(0)-a(1)-a(2)|\le 2.}\end{array}$$

On the other hand, the frequency is
$\begin{array}{}{\displaystyle \omega =\frac{m}{3}}\end{array}$ where (*m*, 3) = 1 which implies that
$\begin{array}{}{\displaystyle \omega =n+\frac{r}{3},}\end{array}$ where *n* ∈ ℤ and *r* = 1, 2. Therefore, the coefficient is given by
$\begin{array}{}{\displaystyle {a}_{r}(k)=E-\lambda \phantom{\rule{thinmathspace}{0ex}}\mathrm{cos}(\pi \frac{2r}{3}\phantom{\rule{thinmathspace}{0ex}}k+\theta ),}\end{array}$, and hence

$$\begin{array}{}{\displaystyle {a}_{1}(0)={a}_{2}(0)=E-\lambda \phantom{\rule{thinmathspace}{0ex}}\mathrm{cos}(\theta ),}\\ {a}_{1}(1)={a}_{2}(2)=E-\lambda \phantom{\rule{thinmathspace}{0ex}}\mathrm{cos}(\pi \frac{2}{3}+\theta )=E+{\displaystyle \frac{\lambda}{2}}\phantom{\rule{thinmathspace}{0ex}}[\mathrm{cos}(\theta )+\sqrt{3}\phantom{\rule{thinmathspace}{0ex}}\mathrm{sin}(\theta )],.\\ {a}_{1}(2)={a}_{2}(1)=E-\lambda \phantom{\rule{thinmathspace}{0ex}}\mathrm{cos}(\pi \frac{4}{3}+\theta )=E+{\displaystyle \frac{\lambda}{2}}\phantom{\rule{thinmathspace}{0ex}}[\mathrm{cos}(\theta )-\sqrt{3}\phantom{\rule{thinmathspace}{0ex}}\mathrm{sin}(\theta )]\end{array}$$

□

#### Corollary 4.7

(Period *p* = 4). *Given a* ∈ *ℓ*(ℤ; 4) *and c* ∈ *ℓ*^{*}(ℤ; 4), *then the equation*

$$\begin{array}{}{\displaystyle c(k)z(k+1)+c(k-1)z(k-1)-a(k)z(k)=0,\phantom{\rule{.25cm}{0ex}}k\in \mathbb{Z}}\end{array}$$

*has bounded solutions iff*

$$\begin{array}{}{\displaystyle |a(0)a(1)a(2)a(3)-c(0{)}^{2}a(2)a(3)-c(1{)}^{2}a(0)a(3)-c(2{)}^{2}a(0)a(1)}\\ -c(3{)}^{2}a(1)a(2)+c(0{)}^{2}c(2{)}^{2}+c(1{)}^{2}c(3{)}^{2}|\le 2|c(0)c(1)c(2)c(3)|\end{array}$$

*and when the inequality is strict*, *all the solutions are bounded*. *In particular*, *given E*, *λ* ∈ ℝ, *θ* ∈ [0, 2*π*) *and*
$\begin{array}{}{\displaystyle \omega =\frac{m}{4},}\end{array}$ *where m* ∈ ℤ *and* (4, *m*) = 1, *then the Mathieu equation*

$$\begin{array}{}{\displaystyle z(k+1)+z(k-1)+\lambda \phantom{\rule{thinmathspace}{0ex}}\mathrm{cos}(2\pi \omega k+\theta )z(k)=Ez(k),\phantom{\rule{.25cm}{0ex}}k\in \mathbb{Z}}\end{array}$$

*has bounded solutions iff*

$$\begin{array}{}{\displaystyle 4({E}^{2}-1)\le ({E}^{2}-{\lambda}^{2}{\mathrm{cos}}^{2}(\theta ))({E}^{2}-{\lambda}^{2}{\mathrm{sin}}^{2}(\theta ))\le 4{E}^{2}.}\end{array}$$

#### Proof

In this case we have
$\begin{array}{}{\displaystyle {\mathrm{\Lambda}}_{4}^{0}}\end{array}$ = {(0, 0, 0, 0)},

$$\begin{array}{}{\displaystyle {\mathrm{\Lambda}}_{4}^{1}=\{(1,0,0,0),(0,1,0,0),(0,0,1,0),(0,0,0,1)\},\phantom{\rule{.15cm}{0ex}}{\mathrm{\Lambda}}_{4}^{2}=\{(1,0,1,0),(0,1,0,1)\}}\end{array}$$

which implies that
$\begin{array}{}{\displaystyle {\overline{\mathrm{\Lambda}}}_{4}^{0}}\end{array}$ = {(1, 1, 1, 1)} and

$$\begin{array}{}{\displaystyle {\overline{\mathrm{\Lambda}}}_{4}^{1}=\{(0,0,1,1),(1,0,0,1),(1,1,0,0),(0,1,1,0)\},\phantom{\rule{.15cm}{0ex}}{\overline{\mathrm{\Lambda}}}_{4}^{2}=\{(0,0,0,0),(0,0,0,0)\}}\end{array}$$

and hence,

$$\begin{array}{}{\displaystyle q(a,c;4)=}& {\displaystyle \frac{1}{2c(0)c(1)c(2)c(3)}}[a(0)a(1)a(2)a(3)-c(0{)}^{2}a(2)a(3)-c(1{)}^{2}a(0)a(3)\\ & -c(2{)}^{2}a(0)a(1)-c(3{)}^{2}a(1)a(2)+c(0{)}^{2}c(2{)}^{2}+c(1{)}^{2}c(3{)}^{2}]\end{array}$$

In particular, for the Mathieu equation the condition for the existence of bounded solutions becomes

$$\begin{array}{}{\displaystyle -4\le a(0)a(1)a(2)a(3)-a(2)a(3)-a(0)a(3)-a(0)a(1)-a(1)a(2)\le 0.}\end{array}$$

On the other hand, the frequency is
$\begin{array}{}{\displaystyle \omega =\frac{m}{4}}\end{array}$ where (*m*, 4) = 1 which implies that
$\begin{array}{}{\displaystyle \omega =n+\frac{r}{4},}\end{array}$ where *n* ∈ ℤ and *r* = 1, 3. Therefore, the coefficient is given by *a*_{r}(*k*) =
$\begin{array}{}{\displaystyle E-\lambda \phantom{\rule{thinmathspace}{0ex}}\mathrm{cos}(\pi \frac{r}{2}\phantom{\rule{thinmathspace}{0ex}}k+\theta ),}\end{array}$ and hence

$$\begin{array}{}{\displaystyle {a}_{1}(0)={a}_{3}(0)=E-\lambda \phantom{\rule{thinmathspace}{0ex}}\mathrm{cos}(\theta ),}\\ {a}_{1}(1)={a}_{3}(3)=E+\lambda \phantom{\rule{thinmathspace}{0ex}}\mathrm{sin}(\theta ),\\ {a}_{1}(2)={a}_{3}(2)=E+\lambda \phantom{\rule{thinmathspace}{0ex}}\mathrm{cos}(\theta ),\\ {a}_{1}(3)={a}_{3}(1)=E-\lambda \phantom{\rule{thinmathspace}{0ex}}\mathrm{sin}(\theta )\end{array}$$

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