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**On monotone solutions of some classes of difference equations.**
*(English)*
Zbl 1109.39013

Assume that \(\alpha_{i} \geq 0\) \(i = 0,\ldots,k-1,\) \(\sum_{i=0}^{k-1}\alpha_{i} = 1.\) The main result of this paper is the proof that

1) if \(p > -1,\) the equation

\[ x_{n+1} = p + \frac{x_{n-k}}{\sum_{i=0}^{k-1}\alpha_{i}x_{n-i}} \quad (n = 0,1,\ldots) \]

has a positive solution which remains above the equilibrium \(\overline x_{1} = p+1\);

2) the equation \[ x_{n+1} = \frac{1+x_{n-k}}{\sum_{i=0}^{k-1}\alpha_{i}x_{n-i}} \quad (n = 0,1,\ldots) \] has a nontrivial positive solution which decreases to the equilibrium \(\overline x_{2} = (1 + \sqrt 5)/2;\)

3) if \(\alpha > 0,\) the equation \[ x_{n+1} =\frac{\alpha + x_{n-k}}{\sum_{i=0}^{k-1}\alpha_{i}x_{n-i}} \quad (n = 0,1,\ldots) \] has a nontrivial positive solution which decreases to the equilibrium \(\overline x_{3} = \sqrt \alpha\).

This solves positively some earlier conjectures.

1) if \(p > -1,\) the equation

\[ x_{n+1} = p + \frac{x_{n-k}}{\sum_{i=0}^{k-1}\alpha_{i}x_{n-i}} \quad (n = 0,1,\ldots) \]

has a positive solution which remains above the equilibrium \(\overline x_{1} = p+1\);

2) the equation \[ x_{n+1} = \frac{1+x_{n-k}}{\sum_{i=0}^{k-1}\alpha_{i}x_{n-i}} \quad (n = 0,1,\ldots) \] has a nontrivial positive solution which decreases to the equilibrium \(\overline x_{2} = (1 + \sqrt 5)/2;\)

3) if \(\alpha > 0,\) the equation \[ x_{n+1} =\frac{\alpha + x_{n-k}}{\sum_{i=0}^{k-1}\alpha_{i}x_{n-i}} \quad (n = 0,1,\ldots) \] has a nontrivial positive solution which decreases to the equilibrium \(\overline x_{3} = \sqrt \alpha\).

This solves positively some earlier conjectures.

Reviewer: Jean Mawhin (Louvain-La-Neuve)

### MSC:

39A11 | Stability of difference equations (MSC2000) |

39A20 | Multiplicative and other generalized difference equations |

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\textit{S. Stević}, Discrete Dyn. Nat. Soc. 2006, No. 2, Article ID 53890, 9 p. (2006; Zbl 1109.39013)

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