Corrigendum to Braverman, E.; Chatzarakis, G. E.; Stavroulakis, I. P. Iterative oscillation tests for difference equations with several non-monotone arguments. J. Difference Equ. Appl. 21 (2015), no. 9, 854–874

Abstract

In Section 3 of our paper, the confusion with indices resulted in the incorrect statement of Theorems 3.3, 3.4, 3.3${}^{\prime }$ and 3.4${}^{\prime }$ (pp. 864–867), as well as influenced Example 4.3 (pp. 872–873). We sincerely apologize and present the corrected version below.

This article refers to:

Iterative oscillation tests for difference equations with several non-monotone arguments

3. Bounded deviations of arguments

1.1. Retarded difference equations

We present a corrected sufficient oscillation condition for (E${}_{\text{R}}$), under the assumption that all the delays are bounded and (1.1${}^{\prime }$) holds.

Theorem 3.3:

Assume that $(p_i(n))$, $1\le i\le m$, are sequences of nonnegative real numbers, (1.1${}^{\prime }$) holds and for each $k=1,\cdots ,m$,(3.3) $$\begin{array}{c}\hfill \underset{n\to \infty }{lim\; inf}\sum _{i=1}^m\sum _{j={\mathit{\tau }}_k(n)}^{n-1}{p}_i(j)>{\left(\frac{M_k}{M_k+1}\right)}^{M_k+1}\text{,}\end{array}$$(3.3)

then all solutions of (E${}_{\text{R}}$) oscillate.

Proof Assume, for the sake of contradiction, that ${(x(n))}_{n\ge -w}$ is a nonoscillatory solution of (E${}_{\text{R}}$). Without loss of generality, we can assume that $x(n)>0$ for all $n\ge n_1$. Then, there exists $n_2\ge n_1$ such that $x({\mathit{\tau }}_i(n))>0$, $\forall n\ge n_2$, $1\le i\le m$. In view of this, Equation (E${}_{\text{R}}$) becomes$$\begin{array}{c}\hfill \mathrm{\Delta }x(n)=-\sum _{i=1}^m{p}_i(n)x({\mathit{\tau }}_i(n))\le 0,\forall n\ge n_2,\end{array}$$

which means that the sequence ${(x(n))}_{n\ge n_2}$ is non-increasing. The sequences$$\begin{array}{c}\hfill b_k(n)={\left(\frac{n-{\mathit{\tau }}_k(n)}{n-{\mathit{\tau }}_k(n)+1}\right)}^{n-{\mathit{\tau }}_k(n)+1}\end{array}$$

satisfy the inequality(3.4) $$\begin{array}{c}\hfill \frac{1}{4}\le b_k(n)\le {\left(\frac{M_k}{M_k+1}\right)}^{M_k+1},n\ge 1,1\le k\le m.\end{array}$$(3.4)

Due to (3.3), for each $k=1,\cdots ,m$, we can choose $n_0(k)\ge n_2$ and ${\mathit{\epsilon }}_k>0$ such that for $n\ge n_0(k)$,$$\sum _{i=1}^m\sum _{j={\mathit{\tau }}_k(n)}^{n-1}{p}_i(j)>{\left(\frac{M_k}{M_k+1}\right)}^{M_k+1}+{\mathit{\epsilon }}_k$$

and$$d_k:={\left(\frac{M_k}{M_k+1}\right)}^{-M_k-1}\left[{\left(\frac{M_k}{M_k+1}\right)}^{M_k+1}+{\mathit{\epsilon }}_k\right]>1.$$

Denoting$${\mathit{\epsilon }}_0:=\underset{1\le k\le m}{min}{\mathit{\epsilon }}_k>0,d:=\underset{1\le k\le m}{min}{d}_k>1,$$

we obtain$$\begin{array}{c}\hfill \sum _{i=1}^m\sum _{j={\mathit{\tau }}_k(n)}^{n-1}{p}_i(j)>{\left(\frac{M_k}{M_k+1}\right)}^{M_k+1}+{\mathit{\epsilon }}_0\end{array}$$

and$$\begin{array}{c}\hfill {\left(\frac{M_k}{M_k+1}\right)}^{-M_k-1}\left[{\left(\frac{M_k}{M_k+1}\right)}^{M_k+1}+{\mathit{\epsilon }}_0\right]\ge d>1,\end{array}$$

which together with (3.3) immediately implies for any $k=1,\cdots ,m$ $$\begin{array}{c}\hfill \sum _{i=1}^m\sum _{j={\mathit{\tau }}_k(n)}^{n-1}\frac{p_i(j)}{b_k(n)}\ge {\left(\frac{M_k}{M_k+1}\right)}^{-M_k-1}\sum _{i=1}^m\sum _{j={\mathit{\tau }}_k(n)}^{n-1}{p}_i(j)>d>1.\end{array}$$

Dividing (E${}_{\text{R}}$) by x(n) we have$$\begin{array}{c}\hfill \frac{x(n+1)}{x(n)}=1-\sum _{i=1}^m{p}_i(n)\frac{x({\mathit{\tau }}_i(n))}{x(n)},n\ge n_0.\end{array}$$

Multiplying and taking into account that $x({\mathit{\tau }}_i(n))/x(n)\ge 1$, we obtain the estimate$$\begin{array}{c}\hfill \frac{x(n)}{x({\mathit{\tau }}_k(n))}=\prod _{j={\mathit{\tau }}_k(n)}^{n-1}\frac{x(j+1)}{x(j)}\le \prod _{j={\mathit{\tau }}_k(n)}^{n-1}\left(1-\sum _{i=1}^m{p}_i(j)\right).\end{array}$$

Using this estimate and the relation between the arithmetic and the geometric means, we have(3.5) $$\begin{array}{c}\hfill \frac{x(n)}{x({\mathit{\tau }}_k(n))}\le {\left[1-\frac{1}{n-{\mathit{\tau }}_k(n)}\sum _{i=1}^m\sum _{j={\mathit{\tau }}_k(n)}^{n-1}{p}_i(j)\right]}^{n-{\mathit{\tau }}_k(n)}.\end{array}$$(3.5)

Observe that the function $f:\left(0,1\right)\to \mathbb{R}$ defined as$$\begin{array}{c}\hfill f(y):=y{(1-y)}^{\mathit{\rho }},\mathit{\rho }\in \mathbb{N},\end{array}$$

attains its maximum at $y=\frac{1}{1+\mathit{\rho }}$, which equals ${\displaystyle f_{max}=\frac{{\mathit{\rho }}^{\mathit{\rho }}}{{(1+\mathit{\rho })}^{1+\mathit{\rho }}}\text{.}}$ In the inequality$$\begin{array}{c}\hfill y{(1-y)}^{\mathit{\rho }}\le \frac{{\mathit{\rho }}^{\mathit{\rho }}}{{(1+\mathit{\rho })}^{1+\mathit{\rho }}},y\in (0,1),\mathit{\rho }\in \mathbb{N},\end{array}$$

assuming $\mathit{\rho }=n-{\mathit{\tau }}_k(n)$, $y=x/\mathit{\rho }$, where ${\displaystyle x={\sum }_{i=1}^m{\sum }_{j={\mathit{\tau }}_k(n)}^{n-1}{p}_i(j)}$, we obtain from (3.5)$$\begin{array}{c}\hfill \frac{x({\mathit{\tau }}_k(n))}{x(n)}\ge \sum _{i=1}^m\sum _{j={\mathit{\tau }}_k(n)}^{n-1}{p}_i(j){\left(\frac{n-{\mathit{\tau }}_k(n)+1}{n-{\mathit{\tau }}_k(n)}\right)}^{n-{\mathit{\tau }}_k(n)+1}=\sum _{i=1}^m\sum _{j={\mathit{\tau }}_k(n)}^{n-1}\frac{p_i(j)}{b_k(n)}>d\end{array}$$

for any $n\ge n_0(k)$. Denote $n_0=max\{n_0(1),\cdots ,n_0(m)\}$. If we continue this procedure assuming $n\ge n_0+M$, where $M={max}_{1\le i\le m}{M}_i$, and using the properties of the geometric and the algebraic mean, we have$$\begin{array}{c}\hfill \frac{x(n)}{x({\mathit{\tau }}_k(n))}=\prod _{j={\mathit{\tau }}_k(n)}^{n-1}\frac{x(j+1)}{x(j)}=\prod _{j={\mathit{\tau }}_k(n)}^{n-1}\left(1-\sum _{i=1}^m{p}_i(j)\frac{x({\mathit{\tau }}_i(j))}{x(j)}\right)\end{array}$$ $$\begin{array}{c}\hfill \le \prod _{j={\mathit{\tau }}_k(n)}^{n-1}\left(1-d\sum _{i=1}^m{p}_i(j)\right)\le {\left[1-\frac{d}{n-{\mathit{\tau }}_k(n)}\sum _{i=1}^m\sum _{j={\mathit{\tau }}_k(n)}^{n-1}{p}_i(j)\right]}^{n-{\mathit{\tau }}_k(n)}.\end{array}$$

Applying the same argument, we obtain$$\begin{array}{c}\hfill \frac{x({\mathit{\tau }}_k(n))}{x(n)}\ge d\sum _{i=1}^m\sum _{j={\mathit{\tau }}_k(n)}^{n-1}\frac{p_i(j)}{b_k(n)}>d^2,n\ge n_0+2M,k=1,\cdots ,m,\end{array}$$ $$\begin{array}{c}\hfill \frac{x({\mathit{\tau }}_k(n))}{x(n)}>d^r,n\ge n_0+rM,k=1,\cdots ,m.\end{array}$$

Due to (3.3), we observe that for any k $$\underset{n\to \infty }{lim\; sup}\sum _{i=1}^m{p}_i(n)\ge \frac{1}{M_k}{\left(\frac{M_k}{M_k+1}\right)}^{M_k+1}.$$

Since the function ${\displaystyle h(x):=\frac{1}{x}\frac{x^{x+1}}{{(x+1)}^{x+1}}=\frac{x^x}{{(x+1)}^{x+1}}}$, $x\ge 1$, is decreasing and $M_k\le M$, we conclude that(3.6) $$\begin{array}{c}\hfill \underset{n\to \infty }{lim\; sup}\sum _{i=1}^m{p}_i(n)\ge c:=\frac{1}{M}{\left(\frac{M}{M+1}\right)}^{M+1}\end{array}$$(3.6)

and choose a subsequence $(\mathit{\theta }(n))$ of $\mathbb{N}$ such that$$\begin{array}{c}\hfill \sum _{i=1}^m{p}_i(\mathit{\theta }(n))\ge c>0.\end{array}$$

Since$$\begin{array}{c}\hfill 0<\frac{x(n+1)}{x(n)}=1-\sum _{i=1}^m{p}_i(n)\frac{x({\mathit{\tau }}_i(n))}{x(n)},\end{array}$$

we have$$\begin{array}{c}\hfill \sum _{i=1}^m{p}_i(n)\frac{x({\mathit{\tau }}_i(n))}{x(n)}<1.\end{array}$$

In particular,$$\begin{array}{c}\hfill \underset{1\le k\le m}{min}\frac{x({\mathit{\tau }}_k(\mathit{\theta }(n)))}{x(\mathit{\theta }(n))}\sum _{i=1}^m{p}_i(\mathit{\theta }(n))<1.\end{array}$$

Choosing $r\in \mathbb{N}$ such that $d^r>\frac{1}{c}$, where c was defined in (3.6), $\mathit{\theta }(n)\ge n_0+rM$ and noticing that$$\begin{array}{c}\hfill d^r<\underset{1\le k\le m}{min}\frac{x({\mathit{\tau }}_k(\mathit{\theta }(n)))}{x(\mathit{\theta }(n))}\le {\left(\sum _{i=1}^m{p}_i(\mathit{\theta }(n))\right)}^{-1}\le \frac{1}{c},\end{array}$$

we obtain a contradiction, which concludes the proof.

The following result is valid as ${\displaystyle {\left(\frac{n}{n+1}\right)}^{n+1}<\frac{1}{e}}$. In (3.7) a non-strict inequality is also sufficient.

Theorem 3.4:

Assume that $(p_i(n))$, $1\le i\le m$, are sequences of nonnegative real numbers and (1.1) holds. If(3.7) $$\begin{array}{c}\hfill \underset{n\to \infty }{lim\; inf}\sum _{i=1}^m\sum _{j={\mathit{\tau }}_k(n)}^{n-1}{p}_i(j)>\frac{1}{e}\text{,}k=1,\cdots ,m,\end{array}$$(3.7)

then all solutions of (E${}_{\text{R}}$) oscillate.

1.2. Advanced difference equations

Similar oscillation theorems for the (dual) advanced difference equation (E${}_{\text{A}}$) can be derived easily. The proofs of these theorems are omitted, since they follow the schemes of Subsection 3.1.

Theorem 3.3:

Assume that $(p_i(n))$, $1\le i\le m$, are sequences of nonnegative real numbers, all the advances are bounded, ($1.2^{\prime }$) holds,(3.8) $$\begin{array}{c}\hfill \underset{n\to \infty }{lim\; inf}\sum _{i=1}^m\sum _{j=n+1}^{{\mathit{\sigma }}_k(n)}{p}_i(j)>{\left(\frac{{\mathit{\mu }}_k}{{\mathit{\mu }}_k+1}\right)}^{{\mathit{\mu }}_k+1}\text{,}k=1,\cdots ,m,\end{array}$$(3.8)

then all solutions of (E${}_{\text{A}}$) oscillate.

Theorem 3.4:

Assume that $(p_i(n))$, $1\le i\le m$, are sequences of nonnegative real numbers, all the advances are bounded and (1.2) holds. If(3.9) $$\begin{array}{c}\hfill \underset{n\to \infty }{lim\; inf}\sum _{i=1}^m\sum _{j=n+1}^{{\mathit{\sigma }}_k(n)}{p}_i(j)>\frac{1}{e}\text{,}k=1,\cdots ,m,\end{array}$$(3.9)

then all solutions of (E${}_{\text{A}}$) oscillate.

Example 4.3:

Consider the delay difference equation(4.3) $$\begin{array}{c}\hfill \mathrm{\Delta }x(n)+a(n)x\left({\mathit{\tau }}_1(n)\right)+\frac{1}{125}x\left({\mathit{\tau }}_2(n)\right)=0,n\ge 0\text{,}\end{array}$$(4.3)

with$$\begin{array}{c}\hfill {\mathit{\tau }}_1(n)=\left\{\begin{array}{cc}n-1\hfill & \text{if}n\text{is even,}\hfill \\ n-2\hfill & \text{if}n\text{is odd,}\hfill \end{array}\right.{\mathit{\tau }}_2(n)=\left\{\begin{array}{cc}n-2\hfill & \text{if}n\text{is even,}\hfill \\ n-3\hfill & \text{if}n\text{is odd,}\hfill \end{array}\right.\end{array}$$ $$\begin{array}{c}\hfill a(n)=\left\{\begin{array}{cc}\frac{3}{125}\hfill & \text{if}n\text{is even,}\hfill \\ \hfill \\ \frac{37}{125}\hfill & \text{if}n\text{is odd.}\hfill \end{array}\right.\end{array}$$

Evidently$$\begin{array}{c}\hfill n-{\mathit{\tau }}_1(n)\le 2=M_1\text{and}n-{\mathit{\tau }}_2(n)\le 3=M_2\end{array}$$

and therefore for $k=1$ $$\underset{n\to \infty }{lim\; inf}\sum _{i=1}^2\sum _{j={\mathit{\tau }}_1(n)}^{n-1}{p}_i(j)=\frac{37}{125}+\frac{1}{125}=\frac{38}{125}=0.304>{\left(\frac{M_1}{M_1+1}\right)}^{M_1+1}\approx 0.2962963,$$

while for $k=2$ $$\underset{n\to \infty }{lim\; inf}\sum _{i=1}^2\sum _{j={\mathit{\tau }}_2(n)}^{n-1}{p}_i(j)=\frac{37}{125}+\frac{3}{125}+\frac{2}{125}=0.336>{\left(\frac{M_2}{M_2+1}\right)}^{M_2+1}\approx 0.31640625\text{,}$$

that is, condition (3.3) is satisfied and, by Theorem 3.3, all solutions of Equation (4.3) oscillate.