Our first result shows that there is a closed-form formula for general solutions to equation (1) when \((q_{n})_{n\in \mathbb {N}_{0}}\) is a constant sequence. From the theoretical point of view, we know that for each inhomogeneous linear difference equation with constant coefficients, such a formula exists. On the other hand, we know that the polynomial equations of order greater than or equal to five need not be solved by radicals, which implies that there are linear difference equations with constant coefficients for which we cannot find a closed-form formula for their general solutions. The result shows that there is a class of linear equations of arbitrary order for which it is possible to find such a closed-form formula. Moreover, the result gives only one formula that includes all the solutions to the equation.
Lemma 1
Consider the difference equation
$$ x_{n+k}-qx_{n}=f_{n},\quad n\in \mathbb {N}_{0}, $$
(7)
where
\(k\in \mathbb {N}\), \(q\in \mathbb {C}\setminus\{0\}\), and
\((f_{n})_{n\in \mathbb {N}_{0}}\)
is a given sequence of complex numbers. Then, the general solution to the equation is
$$ x_{n}= \sum_{s=0}^{k-1}\bigl( \sqrt[k]{q} \varepsilon ^{s}\bigr)^{n} \Biggl(c_{s}+ \frac{1}{k}\sum_{j=0}^{n-1} \frac{\varepsilon ^{-sj}f_{j}}{(\sqrt[k]{q})^{j+k}} \Biggr),\quad n\in \mathbb {N}_{0}, $$
(8)
where
\(c_{s}\), \(s=\overline {0,k-1}\), are arbitrary numbers, and
\(\sqrt[k]{q}\)
is one of the
kth roots of
q.
Proof
Based on (3), we try to find the general solution to equation (7) in the form
$$ x_{n}=\sum_{s=0}^{k-1}c_{n}^{(s)} \bigl(\sqrt[k]{q} \varepsilon ^{s}\bigr)^{n},\quad n\in \mathbb {N}_{0}, $$
(9)
where \((c_{n}^{(s)})_{n\in \mathbb {N}_{0}}\), \(s=\overline {0,k-1}\), are some undetermined sequences.
To do this, we pose the following conditions:
$$ \begin{aligned} &x_{n+1}= \sum _{s=0}^{k-1}c_{n+1}^{(s)}\bigl( \sqrt[k]{q} \varepsilon ^{s}\bigr)^{n+1}=\sum _{s=0}^{k-1}c_{n}^{(s)}\bigl( \sqrt[k]{q} \varepsilon ^{s}\bigr)^{n+1}, \\ &x_{n+2}= \sum_{s=0}^{k-1}c_{n+1}^{(s)} \bigl(\sqrt[k]{q} \varepsilon ^{s}\bigr)^{n+2}=\sum _{s=0}^{k-1}c_{n}^{(s)}\bigl( \sqrt[k]{q} \varepsilon ^{s}\bigr)^{n+2}, \\ &\vdots \\ &x_{n+k-1}= \sum_{s=0}^{k-1}c_{n+1}^{(s)} \bigl(\sqrt[k]{q} \varepsilon ^{s}\bigr)^{n+k-1}=\sum _{s=0}^{k-1}c_{n}^{(s)}\bigl( \sqrt[k]{q} \varepsilon ^{s}\bigr)^{n+k-1} \end{aligned} $$
(10)
for \(n\in \mathbb {N}_{0}\), from which it follows that
$$ \begin{aligned} &\sum_{s=0}^{k-1} \bigl(c_{n+1}^{(s)}-c_{n}^{(s)}\bigr) \bigl( \sqrt[k]{q} \varepsilon ^{s}\bigr)^{n+1}=0, \\ &\sum_{s=0}^{k-1}\bigl(c_{n+1}^{(s)}-c_{n}^{(s)} \bigr) \bigl(\sqrt[k]{q} \varepsilon ^{s}\bigr)^{n+2}=0, \\ & \vdots \\ &\sum_{s=0}^{k-1}\bigl(c_{n+1}^{(s)}-c_{n}^{(s)} \bigr) \bigl(\sqrt[k]{q} \varepsilon ^{s}\bigr)^{n+k-1}=0 \end{aligned} $$
(11)
for \(n\in \mathbb {N}_{0}\).
From (7), (9), and the last equality in (10) with \(n\to n+1\), we easily get
$$ \sum_{s=0}^{k-1}\bigl(c_{n+1}^{(s)}-c_{n}^{(s)} \bigr) \bigl(\sqrt [k]{q} \varepsilon ^{s}\bigr)^{n+k}=f_{n} $$
(12)
for \(n\in \mathbb {N}_{0}\).
Since \(q\ne0\), for each fixed \(n\in \mathbb {N}_{0}\), equations (11) and (12) are equivalent to the following k-dimensional linear system:
$$ \begin{aligned} &\sum_{s=0}^{k-1} \bigl(c_{n+1}^{(s)}-c_{n}^{(s)}\bigr) \bigl( \varepsilon ^{s}\bigr)^{n+1}=0, \\ &\sum_{s=0}^{k-1}\bigl(c_{n+1}^{(s)}-c_{n}^{(s)} \bigr) \bigl(\varepsilon ^{s}\bigr)^{n+2}=0, \\ & \vdots \\ &\sum_{s=0}^{k-1}\bigl(c_{n+1}^{(s)}-c_{n}^{(s)} \bigr) \bigl(\varepsilon ^{s}\bigr)^{n+k-1}=0, \\ &\sum_{s=0}^{k-1}\bigl(c_{n+1}^{(s)}-c_{n}^{(s)} \bigr) \bigl(\varepsilon ^{s}\bigr)^{n+k}=\frac {f_{n}}{(\sqrt[k]{q})^{n+k}} \end{aligned} $$
(13)
in unknown variables \(c_{n+1}^{(s)}-c_{n}^{(s)}\), \(s=\overline {0,k-1}\).
The determinant of the system is
$$\begin{aligned} \Delta_{k}(n)&=\left \vert \begin{matrix} 1 & \varepsilon ^{n+1} & \varepsilon ^{2(n+1)} & \cdots& \varepsilon ^{(k-1)(n+1)}\\ 1 & \varepsilon ^{n+2} & \varepsilon ^{2(n+2)} & \cdots& \varepsilon ^{(k-1)(n+2)}\\ 1 & \varepsilon ^{n+3} & \varepsilon ^{2(n+3)} & \cdots& \varepsilon ^{(k-1)(n+3)}\\ \vdots& \vdots& \vdots& \ddots& \vdots\\ 1 & \varepsilon ^{n+k} & \varepsilon ^{2(n+k)} & \cdots&\varepsilon ^{(k-1)(n+k)} \end{matrix} \right \vert \\ &=\varepsilon ^{\frac{(n+1)(k-1)k}{2}}\left \vert \begin{matrix} 1 & 1 & 1 & \cdots& 1\\ 1 & \varepsilon & \varepsilon ^{2} & \cdots& \varepsilon ^{k-1}\\ 1 & \varepsilon ^{2} & \varepsilon ^{4} & \cdots& \varepsilon ^{2(k-1)}\\ \vdots& \vdots& \vdots& \ddots& \vdots\\ 1 & \varepsilon ^{k-1} & \varepsilon ^{2(k-1)} & \cdots&\varepsilon ^{(k-1)(k-1)} \end{matrix} \right \vert \\ &=\varepsilon ^{\frac{(n+1)(k-1)k}{2}}V_{k}\bigl(1,\varepsilon ,\ldots, \varepsilon ^{k-1} \bigr). \end{aligned}$$
(14)
From (13) and (14), by some calculation and using properties of determinants, we get
$$\begin{aligned} c_{n+1}^{(s)}-c_{n}^{(s)} =& \frac{\varepsilon ^{\frac{(n+1)(1-k)k}{2}}}{V_{k}(1,\varepsilon ,\ldots, \varepsilon ^{k-1})} \\ &{}\times \left \vert \begin{matrix} 1 & \varepsilon ^{n+1} & \cdots&\varepsilon ^{(s-1)(n+1)} & 0 & \varepsilon ^{(s+1)(n+1)}&\cdots & \varepsilon ^{(k-1)(n+1)}\\ 1 & \varepsilon ^{n+2} & \cdots&\varepsilon ^{(s-1)(n+2)} & 0 &\varepsilon ^{(s+1)(n+2)}&\cdots & \varepsilon ^{(k-1)(n+2)}\\ 1 & \varepsilon ^{n+3} & \cdots&\varepsilon ^{(s-1)(n+3)} & 0 &\varepsilon ^{(s+1)(n+3)}&\cdots & \varepsilon ^{(k-1)(n+3)}\\ \vdots& \vdots& &\vdots& \vdots& \vdots& &\vdots\\ 1 & \varepsilon ^{n+k-1} & \cdots&\varepsilon ^{(s-1)(n+k-1)} &0 &\varepsilon ^{(s+1)(n+k-1)}&\cdots&\varepsilon ^{(k-1)(n+k-1)}\\ 1 & \varepsilon ^{n+k} & \cdots&\varepsilon ^{(s-1)(n+k)} &\frac{f_{n}}{(\sqrt [k]{q})^{n+k}} &\varepsilon ^{(s+1)(n+k)}&\cdots&\varepsilon ^{(k-1)(n+k)} \end{matrix} \right \vert \\ =&\frac{\varepsilon ^{-s(n+1)}}{V_{k}(1,\varepsilon ,\ldots, \varepsilon ^{k-1})} \\ &{}\times \left \vert \begin{matrix} 1 & 1 & \cdots&1 & 0 & 1&\cdots& 1\\ 1 & \varepsilon & \cdots&\varepsilon ^{s-1} & 0 &\varepsilon ^{s+1}&\cdots& \varepsilon ^{k-1}\\ 1 & \varepsilon ^{2} & \cdots&(\varepsilon ^{(s-1)})^{2} & 0 &(\varepsilon ^{s+1})^{2}&\cdots& (\varepsilon ^{(k-1)})^{2}\\ \vdots& \vdots& &\vdots& \vdots& \vdots& &\vdots\\ 1 & \varepsilon ^{k-2} & \cdots&(\varepsilon ^{(s-1)})^{k-2} &0 &(\varepsilon ^{(s+1)})^{k-2}&\cdots&(\varepsilon ^{(k-1)})^{k-2}\\ 1 & \varepsilon ^{k-1} & \cdots&(\varepsilon ^{(s-1)})^{k-1} &\frac{f_{n}}{(\sqrt [k]{q})^{n+k}} &(\varepsilon ^{(s+1)})^{k-1}&\cdots&(\varepsilon ^{(k-1)})^{k-1} \end{matrix} \right \vert \\ =&\frac{(-1)^{k+s+1}\varepsilon ^{-s(n+1)}f_{n}}{(\sqrt[k]{q})^{n+k}V_{k}(1,\varepsilon ,\ldots, \varepsilon ^{k-1})} \\ &{}\times \left \vert \begin{matrix} 1 & 1 & \cdots&1 & 1&\cdots& 1\\ 1 & \varepsilon & \cdots&\varepsilon ^{s-1} & \varepsilon ^{s+1}&\cdots& \varepsilon ^{k-1}\\ 1 & \varepsilon ^{2} & \cdots&(\varepsilon ^{(s-1)})^{2} & (\varepsilon ^{s+1})^{2}&\cdots& (\varepsilon ^{(k-1)})^{2}\\ \vdots& \vdots& &\vdots& \vdots& &\vdots\\ 1 & \varepsilon ^{k-2} & \cdots&(\varepsilon ^{(s-1)})^{k-2} &(\varepsilon ^{(s+1)})^{k-2}&\cdots&(\varepsilon ^{(k-1)})^{k-2} \end{matrix} \right \vert \\ =&(-1)^{k+s+1}\varepsilon ^{-s(n+1)}f_{n}\frac{V_{k-1}(1,\varepsilon ,\ldots, \varepsilon ^{s-1},\varepsilon ^{s+1},\ldots, \varepsilon ^{k-1})}{(\sqrt[k]{q})^{n+k}V_{k}(1,\varepsilon ,\ldots, \varepsilon ^{k-1})}. \end{aligned}$$
(15)
By using formula (6) it follows that
$$\begin{aligned} &\frac{V_{k-1}(1,\varepsilon ,\ldots, \varepsilon ^{s-1},\varepsilon ^{s+1},\ldots, \varepsilon ^{k-1})}{V_{k}(1,\varepsilon ,\ldots, \varepsilon ^{k-1})} \\ &\quad =\frac{1}{(\varepsilon ^{s}-1)\cdots(\varepsilon ^{s}-\varepsilon ^{s-1})(\varepsilon ^{s+1}-\varepsilon ^{s})\cdots (\varepsilon ^{k-1}-\varepsilon ^{s})} \\ &\quad =\frac{(-1)^{k-1-s}}{(\varepsilon ^{s}-1)\cdots(\varepsilon ^{s}-\varepsilon ^{s-1})(\varepsilon ^{s}-\varepsilon ^{s+1})\cdots(\varepsilon ^{s}-\varepsilon ^{k-1})}. \end{aligned}$$
(16)
Now note that
$$ z^{k}-1=(z-1)\cdots\bigl(z-\varepsilon ^{s-1}\bigr) \bigl(z- \varepsilon ^{s}\bigr) \bigl(z-\varepsilon ^{s+1}\bigr)\cdots\bigl(z-\varepsilon ^{k-1}\bigr), $$
(17)
from which it follows that
$$\begin{aligned} &\bigl(\varepsilon ^{s}-1\bigr)\cdots\bigl(\varepsilon ^{s}-\varepsilon ^{s-1} \bigr) \bigl(\varepsilon ^{s}-\varepsilon ^{s+1}\bigr)\cdots \bigl( \varepsilon ^{s}-\varepsilon ^{k-1}\bigr) \\ &\quad =\lim_{z\to \varepsilon ^{s}}(z-1)\cdots\bigl(z-\varepsilon ^{s-1}\bigr) \bigl(z-\varepsilon ^{s+1}\bigr)\cdots \bigl(z-\varepsilon ^{k-1}\bigr) \\ &\quad =\lim_{z\to \varepsilon ^{s}}\frac{z^{k}-1}{z-\varepsilon ^{s}}=k\varepsilon ^{s(k-1)}. \end{aligned}$$
(18)
From (15), (16), and (18), since \(\varepsilon ^{k}=1\), it follows that
$$ c_{n+1}^{(s)}-c_{n}^{(s)}= \frac{\varepsilon ^{-sn}f_{n}}{k(\sqrt [k]{q})^{n+k}} $$
(19)
for \(n\in \mathbb {N}_{0}\) and \(s=\overline {0,k-1}\).
Summing up (19) from 0 to \(n-1\), we obtain
$$ c_{n}^{(s)}=c_{s}+\frac{1}{k}\sum _{j=0}^{n-1}\frac{\varepsilon ^{-sj}f_{j}}{(\sqrt [k]{q})^{j+k}} $$
(20)
for \(n\in \mathbb {N}_{0}\) and \(s=\overline {0,k-1}\), where \(c_{s}:=c_{0}^{(s)}\), \(s=\overline {0,k-1}\).
Employing (20) in (9), we get formula (8), as desired. □
Remark 1
It is interesting that the determinant \(V_{k}(1,\varepsilon ,\ldots, \varepsilon ^{k-1})\) can be calculated in closed form (see, for example, [26, p. 46], [27, p. 61]). Namely, we have the following formula:
$$ V_{k}\bigl(1,\varepsilon ,\ldots, \varepsilon ^{k-1}\bigr)=k^{\frac{k}{2}}e^{\frac {\pi i}{4}(k-1)(3k-2)}. $$
(21)
From (16), (18), and (21) we obtain
$$\begin{aligned} V_{k-1}^{(s)}&:=V_{k-1}\bigl(1,\varepsilon ,\ldots, \varepsilon ^{s-1},\varepsilon ^{s+1},\ldots, \varepsilon ^{k-1} \bigr) \\ &=(-1)^{k-1-s}k^{\frac{k-2}{2}}e^{\frac{\pi i}{4k}(k-1)(3k^{2}-2k-8s)} \end{aligned}$$
for \(s=\overline {0,k-1}\).
Our next result gives an application of Lemma 1 in the investigation of the existence of a bounded solution to equation (7) when \(|q|>1\) and \((f_{n})_{n\in \mathbb {N}_{0}}\) is a bounded sequence of complex numbers.
Theorem 1
Assume that
\(|q|>1\)
and
\(f:=(f_{n})_{n\in \mathbb {N}_{0}}\subset \mathbb {C}\)
is a given bounded sequence. Then, there is a unique bounded solution to equation (7).
Proof
Employing (8), it follows that
$$\begin{aligned} x_{km+l}&=\sum_{s=0}^{k-1}\bigl( \sqrt[k]{q} \varepsilon ^{s}\bigr)^{km+l} \Biggl(c_{s}+ \frac{1}{k}\sum_{j=0}^{km+l-1} \frac{\varepsilon ^{-sj}f_{j}}{(\sqrt [k]{q})^{j+k}} \Biggr) \\ &=\bigl(\sqrt[k]{q}\bigr)^{km+l} \Biggl(\sum _{s=0}^{k-1}c_{s}\varepsilon ^{sl}+ \frac{1}{k}\sum_{s=0}^{k-1}\sum _{j=0}^{km+l-1}\frac{\varepsilon ^{s(l-j)}f_{j}}{(\sqrt [k]{q})^{j+k}} \Biggr) \end{aligned}$$
(22)
for all \(m\in \mathbb {N}_{0}\) and \(l=\overline {0,k-1}\).
Since \(|q|>1\) and f is bounded, we have
$$ \Biggl\vert \sum_{s=0}^{k-1}\sum _{j=0}^{km+l-1}\frac{\varepsilon ^{s(l-j)}f_{j}}{(\sqrt[k]{q})^{j+k}} \Biggr\vert \le k\sum _{j=0}^{\infty}\frac {\| f\|_{\infty}}{|\sqrt[k]{q}|^{j+k}}= \frac{k\|f\|_{\infty}}{|\sqrt [k]{q}|^{k-1}(|\sqrt[k]{q}|-1)}< \infty $$
(23)
for each \(l=\overline {0,k-1}\).
From (22), (23), and the assumption \(|q|>1\), we see that, for a bounded solution \((x_{n})_{n\in \mathbb {N}_{0}}\) to (7), there must be
$$ \sum_{s=0}^{k-1}c_{s} \varepsilon ^{sl}=-\frac{1}{k}\sum_{j=0}^{\infty}\sum_{s=0}^{k-1}\frac{\varepsilon ^{s(l-j)}f_{j}}{(\sqrt[k]{q})^{j+k}}=:S_{l} $$
(24)
for \(l=\overline {0,k-1}\).
Equalities (24) are a k-dimensional linear system in variables \(c_{s}\), \(s=\overline {0,k-1}\), whose determinant is \(V_{k}(1,\varepsilon ,\ldots, \varepsilon ^{k-1})=:V_{k}\). By solving the system we have
$$\begin{aligned} c_{s-1}&=\frac{1}{V_{k}}\left \vert \begin{matrix} 1 & 1 & \cdots&1 & S_{0} & 1&\cdots& 1\\ 1 & \varepsilon & \cdots&\varepsilon ^{s-2} & S_{1} &\varepsilon ^{s}&\cdots& \varepsilon ^{k-1}\\ 1 & \varepsilon ^{2} & \cdots&(\varepsilon ^{(s-2)})^{2} & S_{2} &(\varepsilon ^{s})^{2}&\cdots& (\varepsilon ^{(k-1)})^{2}\\ \vdots& \vdots& &\vdots& \vdots& \vdots& &\vdots\\ 1 & \varepsilon ^{k-2} & \cdots&(\varepsilon ^{(s-2)})^{k-2} &S_{k-2} &(\varepsilon ^{(s)})^{k-2}&\cdots&(\varepsilon ^{(k-1)})^{k-2}\\ 1 & \varepsilon ^{k-1} & \cdots&(\varepsilon ^{(s-2)})^{k-1} & S_{k-1} &(\varepsilon ^{(s)})^{k-1}&\cdots&(\varepsilon ^{(k-1)})^{k-1} \end{matrix} \right \vert \\ &=\frac{\sum_{l=1}^{k}(-1)^{l+s}S_{l-1}W_{ls}}{V_{k}(1,\varepsilon ,\ldots, \varepsilon ^{k-1})} \end{aligned}$$
(25)
for \(s=\overline {1,k}\), where \(W_{ls}\), \(l,s=\overline {1,k}\), are \((k-1)\)-dimensional minors of the determinant in (25) corresponding to the element on the position \((l, s)\).
They can be obtained by the coefficients of the following polynomial of \((k-1)\)th order, which is defined by the Vandermonde determinant:
$$\begin{aligned} P_{k-1}(x) :=&\left \vert \begin{matrix} 1 & 1 & \cdots&1 & 1 & 1&\cdots& 1\\ 1 & \varepsilon & \cdots&\varepsilon ^{s-2} & x &\varepsilon ^{s}&\cdots& \varepsilon ^{k-1}\\ 1 & \varepsilon ^{2} & \cdots&(\varepsilon ^{(s-2)})^{2} & x^{2} &(\varepsilon ^{s})^{2}&\cdots& (\varepsilon ^{(k-1)})^{2}\\ \vdots& \vdots& &\vdots& \vdots& \vdots& &\vdots\\ 1 & \varepsilon ^{k-2} & \cdots&(\varepsilon ^{(s-2)})^{k-2} & x^{k-2} &(\varepsilon ^{(s)})^{k-2}&\cdots&(\varepsilon ^{(k-1)})^{k-2}\\ 1 & \varepsilon ^{k-1} & \cdots&(\varepsilon ^{(s-2)})^{k-1} & x^{k-1} &(\varepsilon ^{(s)})^{k-1}&\cdots&(\varepsilon ^{(k-1)})^{k-1} \end{matrix} \right \vert \\ =&(-1)^{k+s}x^{k-1}W_{ks}+(-1)^{k+s-1}x^{k-2}W_{k-1\,s}+ \cdots +(-1)^{s+1}W_{1s} \\ =&(-1)^{k-s}(x-1)\cdots\bigl(x-\varepsilon ^{s-2}\bigr) \bigl(x- \varepsilon ^{s}\bigr)\cdots\bigl(x-\varepsilon ^{k-1}\bigr) \\ &{}\times V_{k-1}\bigl(1,\varepsilon ,\ldots, \varepsilon ^{s-2}, \varepsilon ^{s},\ldots, \varepsilon ^{k-1}\bigr), \end{aligned}$$
(26)
where the second equality is obtained by expanding the determinant along the sth column, whereas the third one follows from (6).
First, note that from (26) it follows that
$$ W_{ks}=V_{k-1}\bigl(1,\varepsilon ,\ldots, \varepsilon ^{s-2}, \varepsilon ^{s},\ldots ,\varepsilon ^{k-1}\bigr). $$
(27)
Now note the following equality:
$$ \sum_{j=0}^{k-1}\varepsilon ^{sj}= \textstyle\begin{cases} 0, & s \ne km, \\ k, & s=km, \end{cases} $$
(28)
for \(m\in \mathbb {Z}\).
From (28) and (26) we have
$$\begin{aligned} \begin{aligned}[b] W_{k-1\,s}&=V_{k-1}\bigl(1,\varepsilon ,\ldots, \varepsilon ^{s-2}, \varepsilon ^{s},\ldots, \varepsilon ^{k-1}\bigr)\sum _{j=0, j\ne s-1}^{k-1}\varepsilon ^{j} \\ &=V_{k-1}\bigl(1,\varepsilon ,\ldots, \varepsilon ^{s-2},\varepsilon ^{s}, \ldots, \varepsilon ^{k-1}\bigr) \Biggl(\sum_{j=0}^{k-1} \varepsilon ^{j}-\varepsilon ^{s-1} \Biggr) \\ &=-\varepsilon ^{s-1}V_{k-1}\bigl(1,\varepsilon ,\ldots, \varepsilon ^{s-2}, \varepsilon ^{s},\ldots, \varepsilon ^{k-1}\bigr). \end{aligned} \end{aligned}$$
(29)
Now note that from (17) and the Viète formula, it follows that
$$ \sum_{0\le j_{1}< j_{2}< \cdots< j_{t}\le k-1} \varepsilon ^{j_{1}}\varepsilon ^{j_{2}} \cdots \varepsilon ^{j_{t}}=0 $$
(30)
for \(t=\overline {1,k-1}\).
From (26), (30) with \(t=2\), and the calculation in (29) we have
$$\begin{aligned} W_{k-2\,s}&=V_{k-1}\bigl(1,\varepsilon ,\ldots, \varepsilon ^{s-2}, \varepsilon ^{s},\ldots, \varepsilon ^{k-1}\bigr)\sum _{1\le j_{1}< j_{2}\le k-1, j_{1},j_{2}\ne s-1}\varepsilon ^{j_{1}}\varepsilon ^{j_{2}} \\ &=V_{k-1}\bigl(1,\varepsilon ,\ldots, \varepsilon ^{s-2},\varepsilon ^{s}, \ldots, \varepsilon ^{k-1}\bigr) \Biggl(\sum_{0\le j_{1}< j_{2}\le k-1} \varepsilon ^{j_{1}}\varepsilon ^{j_{2}}-\varepsilon ^{s-1}\sum _{j=0, j\ne s-1}^{k-1}\varepsilon ^{j} \Biggr) \\ &=\varepsilon ^{2(s-1)}V_{k-1}\bigl(1,\varepsilon ,\ldots, \varepsilon ^{s-2}, \varepsilon ^{s},\ldots, \varepsilon ^{k-1}\bigr). \end{aligned}$$
(31)
Assume that, for an \(m\in\{2,\ldots,k-2\}\), we have proved that
$$ W_{k-m\,s}=(-1)^{m}\varepsilon ^{m(s-1)}V_{k-1}\bigl(1, \varepsilon ,\ldots, \varepsilon ^{s-2},\varepsilon ^{s},\ldots ,\varepsilon ^{k-1} \bigr) $$
(32)
and
$$ \sum_{0\le j_{1}< \cdots< j_{m}\le k-1, j_{1},\ldots,j_{m}\ne s-1}\varepsilon ^{j_{1}}\cdots \varepsilon ^{j_{m}}=(-1)^{m}\varepsilon ^{m(s-1)}. $$
(33)
Then, from (26), (30) with \(t=m+1\), and (33) we have
$$\begin{aligned} W_{k-m-1\,s} =&V_{k-1}\bigl(1,\varepsilon ,\ldots, \varepsilon ^{s-2}, \varepsilon ^{s},\ldots, \varepsilon ^{k-1}\bigr)\sum _{0\le j_{1}< \cdots< j_{m+1}\le k-1, j_{1},\ldots,j_{m+1}\ne s-1}\varepsilon ^{j_{1}}\cdots \varepsilon ^{j_{m+1}} \\ =&V_{k-1}\bigl(1,\varepsilon ,\ldots, \varepsilon ^{s-2},\varepsilon ^{s}, \ldots, \varepsilon ^{k-1}\bigr) \\ &{}\times \biggl(\sum_{0\le j_{1}< \cdots< j_{m+1}\le k-1}\varepsilon ^{j_{1}} \cdots \varepsilon ^{j_{m+1}}-\varepsilon ^{s-1}\sum_{0\le\widehat{j_{1}}< \cdots< \widehat {j_{m}}\le k-1, \widehat{j_{1}},\ldots,\widehat{j_{m}}\ne s-1} \varepsilon ^{\widehat {j_{1}}}\cdots \varepsilon ^{\widehat{j_{m}}} \biggr) \\ =&(-1)^{m+1}\varepsilon ^{(m+1)(s-1)}V_{k-1}\bigl(1,\varepsilon ,\ldots, \varepsilon ^{s-2},\varepsilon ^{s},\ldots ,\varepsilon ^{k-1} \bigr). \end{aligned}$$
(34)
From (29), (34), and the method of induction we see that (32) holds.
Employing (18), (24), and (32) in (25), we get
$$\begin{aligned} \begin{aligned}[b] c_{s}&=-\frac{1}{k^{2}}\sum_{j=0}^{\infty}\frac{f_{j}}{(\sqrt [k]{q})^{j+k}}\sum_{l=0}^{k-1} \varepsilon ^{-ls}\sum_{t=0}^{k-1} \varepsilon ^{-t(j-l)} \\ &=-\frac{1}{k}\sum_{j=0}^{\infty}\frac{\varepsilon ^{-sj}f_{j}}{(\sqrt [k]{q})^{j+k}} \end{aligned} \end{aligned}$$
(35)
for \(s=\overline {0,k-1}\), where we have also used that
$$\sum_{l=0}^{k-1}\varepsilon ^{-ls}\sum _{t=0}^{k-1}\varepsilon ^{-t(j-l)}=\sum _{t=0}^{k-1}\varepsilon ^{-tj}\sum _{l=0}^{k-1}\varepsilon ^{(t-s)l}, $$
and, then applied (28) in the cases \(t=s\) and \(t\ne s\) (note also that \(|t-s|< k\)).
Using (35) in (8), we get
$$ x_{n}=-\frac{1}{k}\sum_{s=0}^{k-1} \bigl(\sqrt[k]{q} \varepsilon ^{s}\bigr)^{n}\sum _{j=n}^{\infty}\frac{\varepsilon ^{-sj}f_{j}}{(\sqrt[k]{q})^{j+k}} $$
(36)
for \(n\in \mathbb {N}_{0}\).
Using (28), by a direct calculation we verify that (36) presents a solution to equation (7). Also, we have that
$$|x_{n}|\le\frac{\|f\|_{\infty}}{|\sqrt[k]{q}|^{k-1}(|\sqrt [k]{q}|-1)}< \infty,\quad n\in \mathbb {N}_{0}, $$
showing the boundedness of the solution. The uniqueness of the bounded solution follows from the unique choice of constants \(c_{s}\), \(s=\overline {0,k-1}\), in (35). □
Remark 2
Note that by using (28) in (24) it follows that
$$S_{l}=-\sum_{m=0}^{\infty}\frac{f_{l+mk}}{(\sqrt[k]{q})^{l+mk+k}} $$
for \(l=\overline {0,k-1}\).
Now, motivated by (36) and some operator theory technique, we prove a result on the unique existence of bounded solutions to equation (1).
Theorem 2
Consider equation (1) where
$$ 1< a\le q_{n}\le b,\quad n\in \mathbb {N}_{0}, $$
(37)
or
$$ -b\le q_{n}\le-a< -1,\quad n\in \mathbb {N}_{0}, $$
(38)
for some positive numbers
a
and
b, and
\((f_{n})_{n\in \mathbb {N}_{0}}\)
is a bounded sequence of complex numbers. Then the equation has a unique bounded solution.
Proof
We may assume that (37) holds. The reasoning in the case (38) is similar. Choose a number q such that
$$ q\in \bigl(\max\bigl\{ a,(b+1)/2\bigr\} , b\bigr) $$
(39)
and write (1) as follows:
$$ x_{n+k}-qx_{n}=(q_{n}-q)x_{n}+f_{n}, \quad n\in \mathbb {N}_{0}. $$
(40)
Now we introduce the following operator:
$$ A(u)= \Biggl(-\bigl(\sqrt[k]{q}\bigr)^{n}\frac{1}{k}\sum _{j=n}^{\infty}\frac{\sum_{s=0}^{k-1}\varepsilon ^{s(n-j)}((q_{j}-q)u_{j}+f_{j})}{(\sqrt[k]{q})^{j+k}} \Biggr)_{n\in \mathbb {N}_{0}}. $$
(41)
Assume that \(u\in l^{\infty}(\mathbb {N}_{0})\). Then (41), together with some simple estimates, implies
$$\begin{aligned} \bigl\Vert A(u) \bigr\Vert _{\infty}&=\sup_{n\in \mathbb {N}_{0}} \Biggl\vert \bigl(\sqrt [k]{q}\bigr)^{n}\frac{1}{k}\sum _{j=n}^{\infty}\frac{\sum_{s=0}^{k-1}\varepsilon ^{s(n-j)}((q_{j}-q)u_{j}+f_{j})}{(\sqrt[k]{q})^{j+k}} \Biggr\vert \\ &\le\sup_{n\in \mathbb {N}_{0}}\frac{1}{k}\sum _{j=n}^{\infty}\frac {k((q_{k}+q)|u_{k}|+|f_{k}|)}{|\sqrt[k]{q}|^{j+k-n}} \\ &\le\frac{(b+q)\|u\|_{\infty}+\|f\|_{\infty}}{|\sqrt [k]{q}|^{k-1}(|\sqrt [k]{q}|-1)}< \infty. \end{aligned}$$
Hence, \(A(l^{\infty}(\mathbb {N}_{0}))\subseteq l^{\infty}(\mathbb {N}_{0})\).
Assume that \(u,v\in l^{\infty}(\mathbb {N}_{0})\). Then, using (28) and (41), we have
$$\begin{aligned} \bigl\Vert A(u)-A(v) \bigr\Vert _{\infty}&=\sup_{n\in \mathbb {N}_{0}} \Biggl\vert \bigl(\sqrt[k]{q}\bigr)^{n}\frac{1}{k}\sum _{j=n}^{\infty}\frac{\sum_{s=0}^{k-1}\varepsilon ^{s(n-j)}(q_{j}-q)(u_{j}-v_{j})}{(\sqrt[k]{q})^{j+k}} \Biggr\vert \\ &=\sup_{n\in \mathbb {N}_{0}} \Biggl\vert \bigl(\sqrt[k]{q} \bigr)^{n}\sum_{j=0}^{\infty}\frac {(q_{n+kj}-q)(u_{n+kj}-v_{n+kj})}{(\sqrt[k]{q})^{n+kj+k}} \Biggr\vert \\ &\le\sup_{n\in \mathbb {N}_{0}}\sum_{j=0}^{\infty}\frac {|q_{n+kj}-q||u_{n+kj}-v_{n+kj}|}{q^{j+1}} \\ &\le\frac{\max\{q-a,b-q\}}{q-1}\|u-v\|_{\infty}. \end{aligned}$$
(42)
By the choice of q it follows that
$$\hat{q}:=\frac{\max\{q-a,b-q\}}{q-1}\in(0,1), $$
so (42) can be written as
$$ \bigl\Vert A(u)-A(v) \bigr\Vert _{\infty}\le\hat{q} \Vert u-v \Vert _{\infty} $$
(43)
for \(u,v\in l^{\infty}(\mathbb {N}_{0})\), which means that \(A:l^{\infty}(\mathbb {N}_{0})\to l^{\infty}(\mathbb {N}_{0})\) is a contraction.
The Banach fixed point theorem says that the operator has a unique fixed point, say \(x^{*}=(x_{n}^{*})_{n\in \mathbb {N}_{0}}\in l^{\infty}(\mathbb {N}_{0})\), that is, \(A(x^{*})=x^{*}\), or equivalently
$$ x_{n}^{*}=-\bigl(\sqrt[k]{q}\bigr)^{n}\frac{1}{k}\sum _{j=n}^{\infty}\frac {\sum_{s=0}^{k-1}\varepsilon ^{s(n-j)}((q_{j}-q)x^{*}_{j}+f_{j})}{(\sqrt[k]{q})^{j+k}},\quad n\in \mathbb {N}_{0}. $$
(44)
It is not difficult to verify that (44) is a bounded solution to (1) for \(n\in \mathbb {N}_{0}\). □