On a class of nonlinear max-type difference equations

Abstract

We investigate the solutions to the following system of nonlinear difference equations, $x_{n+1}=max\left\{y_{n-1}^2,\frac{A}{y_{n-1}}\right\},y_{n+1}=max\left\{x_{n-1}^2,\frac{A}{x_{n-1}}\right\}\text{for}n\in N_0$, where $x_{-1}=\mathit{\alpha }$, $y_{-1}=\mathit{\beta }$, $x_0=\mathit{\lambda }$, and $y_0=\mathit{\mu }$ are constants and $A>0$.

Keywords:

• nonlinear system
• max-type
• periodic solutions
• general solution

Public Interest Statement

Difference equations are pervasive in mathematics and recent advances give insight into other fields such as biology and engineering as well. Understanding the analysis of such equations is crucial in many applications in today’s society. In particular, many equations are discrete in nature and need a solid understanding of the theory of difference equations. We develop conditions for which a system of difference equations will be periodic, oscillatory, or nonperiodic. Also, the general solutions of such systems are given.

1. Introduction

Many authors have studied the solutions of max-type difference equations (see El-Dessoky, MR3400335; Elsayed, Iricanin, & Stevic, elsayed2010max; Kent, Kustesky, Nguyen, & Nguyen, kent2003eventually; Kent & Radin, kent2003boundedness; Stević, MR2577582, MR3037547; Yalçinkaya, Iricanin, & Cinar, yalccinkaya2008max).

Next are a few papers on systems of max-type difference equations:

Stević, Alghamdi, Alotaibi, and Naseer Shahzad (MR3262799) studied the boundedness characteristic of the following max-type difference equations$$\begin{array}{c}\hfill x_{n+1}=max\left\{A,\frac{y_n^p}{x_{n-1}^q}\right\},y_{n+1}=max\left\{A,\frac{x_n^p}{y_{n-1}^q}\right\}\text{for}n\in N_0.\end{array}$$

Stević (MR2916163) studied the solutions to the following max-type difference equation$$\begin{array}{c}\hfill x_{n+1}=max\left\{\frac{A}{x_n},\frac{y_n}{x_n}\right\},y_{n+1}=max\left\{\frac{A}{y_n},\frac{x_n}{y_n}\right\}\text{for}n\in N_0.\end{array}$$

We shall study the solutions to the following system of nonlinear difference equations,(1.1) $$\begin{array}{c}\hfill x_{n+1}=max\left\{y_{n-1}^2,\frac{A}{y_{n-1}}\right\},y_{n+1}=max\left\{x_{n-1}^2,\frac{A}{x_{n-1}}\right\}\text{for}n\in N_0,\end{array}$$(1.1)

where $x_{-1}=\mathit{\alpha }$, $y_{-1}=\mathit{\beta }$, $x_0=\mathit{\lambda }$, and $y_0=\mathit{\mu }$ are constants and $A>0$.

2. Main results

Lemma 2.1

Assume that $0<\mathit{\alpha },\mathit{\beta },\mathit{\lambda },\mathit{\mu }<A<1$. Then $\forall n\in N_0$ the following equalities hold:$$\begin{array}{cc}\hfill max\left\{{\left(\frac{A}{\mathit{\beta }}\right)}^{2^{2n}},A{\left(\frac{\mathit{\beta }}{A}\right)}^{2^{2n-1}}\right\}& ={\left(\frac{A}{\mathit{\beta }}\right)}^{2^{2n}}\hfill \\ \hfill max\left\{{\left(\frac{A}{\mathit{\alpha }}\right)}^{2^{2n}},A{\left(\frac{\mathit{\alpha }}{A}\right)}^{2^{2n-1}}\right\}& ={\left(\frac{A}{\mathit{\alpha }}\right)}^{2^{2n}}\hfill \\ \hfill max\left\{{\left(\frac{A}{\mathit{\mu }}\right)}^{2^{2n}},A{\left(\frac{\mathit{\mu }}{A}\right)}^{2^{2n-1}}\right\}& ={\left(\frac{A}{\mathit{\mu }}\right)}^{2^{2n}}\hfill \\ \hfill max\left\{{\left(\frac{A}{\mathit{\lambda }}\right)}^{2^{2n}},A{\left(\frac{\mathit{\lambda }}{A}\right)}^{2^{2n-1}}\right\}& ={\left(\frac{A}{\mathit{\lambda }}\right)}^{2^{2n}}.\hfill \end{array}$$

Proof

Show that$$\begin{array}{c}\hfill max\left\{{\left(\frac{A}{\mathit{\beta }}\right)}^{2^{2n}},A{\left(\frac{\mathit{\beta }}{A}\right)}^{2^{2n-1}}\right\}={\left(\frac{A}{\mathit{\beta }}\right)}^{2^{2n}}\forall n\in N_0.\end{array}$$

We shall proceed by induction on n. Let $n=0$, then ${\left(\frac{A}{\mathit{\beta }}\right)}^{2^0}=\frac{A}{\mathit{\beta }}>1$ and $A{\left(\frac{\mathit{\beta }}{A}\right)}^{\frac{1}{2}}=A^{\frac{1}{2}}{\mathit{\beta }}^{\frac{1}{2}}<1^{\frac{1}{2}}{1}^{\frac{1}{2}}=1.$ Therefore, $max\left\{\frac{A}{\mathit{\beta }},A{\left(\frac{\mathit{\beta }}{A}\right)}^{2^{-1}}\right\}=\frac{A}{\mathit{\beta }}.$ So the result holds for $n=0$. For $n=1,$${\left(\frac{\mathit{\beta }}{A}\right)}^{2^1}=\frac{{\mathit{\beta }}^2}{A^2}<1.$ Now suppose for some $k\in N$, we have$$\begin{array}{c}\hfill {\left(\frac{A}{\mathit{\beta }}\right)}^{2^{2k}}>1\end{array}$$

and$$\begin{array}{c}\hfill {\left(\frac{\mathit{\beta }}{A}\right)}^{2^{2k-1}}<1.\end{array}$$

Show that$$\begin{array}{c}\hfill max\left\{{\left(\frac{A}{\mathit{\beta }}\right)}^{2^{2k+2}},A{\left(\frac{\mathit{\beta }}{A}\right)}^{2^{2k+1}}\right\}={\left(\frac{A}{\mathit{\beta }}\right)}^{2^{2k+2}}.\end{array}$$

Observe,$$\begin{array}{cc}\hfill {\left(\frac{A}{\mathit{\beta }}\right)}^{2^{2k+2}}& ={\left(\frac{A}{\mathit{\beta }}\right)}^{2^{2k}·2^2}={\left(\frac{A}{\mathit{\beta }}\right)}^{2^{2k}·4}={\left(\frac{A}{\mathit{\beta }}\right)}^{2^{2k}+2^{2k}+2^{2k}+2^{2k}}\hfill \\ \hfill & ={\left(\frac{A}{\mathit{\beta }}\right)}^{2^{2k}}{\left(\frac{A}{\mathit{\beta }}\right)}^{2^{2k}}{\left(\frac{A}{\mathit{\beta }}\right)}^{2^{2k}}{\left(\frac{A}{\mathit{\beta }}\right)}^{2^{2k}}>1·1·1·1=1\hfill \end{array}$$

and$$\begin{array}{cc}\hfill A{\left(\frac{\mathit{\beta }}{A}\right)}^{2^{2k+1}}& =A{\left(\frac{\mathit{\beta }}{A}\right)}^{2^{2k-1}·2^2}=A{\left(\frac{\mathit{\beta }}{A}\right)}^{2^{2k-1}}{\left(\frac{\mathit{\beta }}{A}\right)}^{2^{2k-1}}{\left(\frac{\mathit{\beta }}{A}\right)}^{2^{2k-1}}{\left(\frac{\mathit{\beta }}{A}\right)}^{2^{2k-1}}\hfill \\ \hfill & <A·1·1·1·1=A<1.\hfill \end{array}$$

Hence,$$\begin{array}{c}\hfill max\left\{{\left(\frac{A}{\mathit{\beta }}\right)}^{2^{2k+2}},A{\left(\frac{\mathit{\beta }}{A}\right)}^{2^{2k+1}}\right\}={\left(\frac{A}{\mathit{\beta }}\right)}^{2^{2k+2}}.\end{array}$$

Therefore, the result is true $\forall n\in N_0.$ Similarly, the remaining equalities hold. $\square$

Theorem 2.1

Assume that $0<x_{-1},y_{-1},x_0,y_0<A<1$. Also, let $\left\{x_n,y_n\right\}$ be a solution of the system of Equation (1.1) with $x_{-1}=\mathit{\alpha }$, $y_{-1}=\mathit{\beta }$, $x_0=\mathit{\lambda }$, and $y_0=\mathit{\mu }$. Then all solutions of (1.1) are of the following:$$\begin{array}{cc}\hfill x_{4n-3}& ={\left(\frac{A}{\mathit{\beta }}\right)}^{2^{2n-2}},y_{4n-3}={\left(\frac{A}{\mathit{\alpha }}\right)}^{2^{2n-2}}\hfill \\ \hfill x_{4n-2}& ={\left(\frac{A}{\mathit{\mu }}\right)}^{2^{2n-2}},y_{4n-2}={\left(\frac{A}{\mathit{\lambda }}\right)}^{2^{2n-2}}\hfill \\ \hfill x_{4n-1}& ={\left(\frac{A}{\mathit{\alpha }}\right)}^{2^{2n-1}},y_{4n-1}={\left(\frac{A}{\mathit{\beta }}\right)}^{2^{2n-1}}\hfill \\ \hfill x_{4n}& ={\left(\frac{A}{\mathit{\lambda }}\right)}^{2^{2n-2}},y_{4n}={\left(\frac{A}{\mathit{\mu }}\right)}^{2^{2n-2}}.\hfill \end{array}$$

Proof

For $n=1$, we have$$\begin{array}{cc}\hfill x_1& =max\left\{{\mathit{\beta }}^2,\frac{A}{\mathit{\beta }}\right\}=\frac{A}{\mathit{\beta }},y_1=max\left\{{\mathit{\alpha }}^2,\frac{A}{\mathit{\alpha }}\right\}=\frac{A}{\mathit{\alpha }}\hfill \\ \hfill x_2& =max\left\{{\mathit{\mu }}^2,\frac{A}{\mathit{\mu }}\right\}=\frac{A}{\mathit{\mu }},y_2=max\left\{{\mathit{\lambda }}^2,\frac{A}{\mathit{\lambda }}\right\}=\frac{A}{\mathit{\lambda }}\hfill \\ \hfill x_3& =max\left\{\frac{A^2}{{\mathit{\alpha }}^2},\mathit{\alpha }\right\}=\frac{A}{\mathit{\alpha }},y_3=max\left\{\frac{A^2}{{\mathit{\beta }}^2},\mathit{\beta }\right\}=\frac{A}{\mathit{\beta }}\hfill \\ \hfill x_4& =max\left\{\frac{A^2}{{\mathit{\lambda }}^2},\mathit{\lambda }\right\}=\frac{A}{\mathit{\lambda }},y_4=max\left\{\frac{A^2}{{\mathit{\mu }}^2},\mathit{\mu }\right\}=\frac{A}{\mathit{\mu }}.\hfill \end{array}$$

So the result holds for $n=1$. Now suppose the result is true for some $k>0$, that is,$$\begin{array}{cc}\hfill x_{4k-3}& ={\left(\frac{A}{\mathit{\beta }}\right)}^{2^{2k-2}},y_{4k-3}={\left(\frac{A}{\mathit{\alpha }}\right)}^{2^{2k-2}}\hfill \\ \hfill x_{4k-2}& ={\left(\frac{A}{\mathit{\mu }}\right)}^{2^{2k-2}},y_{4k-2}={\left(\frac{A}{\mathit{\lambda }}\right)}^{2^{2k-2}}\hfill \\ \hfill x_{4k-1}& ={\left(\frac{A}{\mathit{\alpha }}\right)}^{2^{2k-1}},y_{4k-1}={\left(\frac{A}{\mathit{\beta }}\right)}^{2^{2k-1}}\hfill \\ \hfill x_{4k}& ={\left(\frac{A}{\mathit{\lambda }}\right)}^{2^{2k-2}},y_{4k}={\left(\frac{A}{\mathit{\mu }}\right)}^{2^{2k-2}}.\hfill \end{array}$$

Also, for $k+1$ we have the following:$$\begin{array}{cc}\hfill x_{4k+1}& =max\left\{y_{4k-1}^2,\frac{A}{y_{4k-1}}\right\}=max\left\{{\left(\frac{A}{\mathit{\beta }}\right)}^{2^{2k}},A{\left(\frac{\mathit{\beta }}{A}\right)}^{2^{2k-1}}\right\}={\left(\frac{A}{\mathit{\beta }}\right)}^{2^{2k}}\hfill \\ \hfill y_{4k+1}& =max\left\{x_{4k-1}^2,\frac{A}{x_{4k-1}}\right\}=max\left\{{\left(\frac{A}{\mathit{\alpha }}\right)}^{2^{2k}},A{\left(\frac{\mathit{\alpha }}{A}\right)}^{2^{2k-1}}\right\}={\left(\frac{A}{\mathit{\alpha }}\right)}^{2^{2k}}\hfill \\ \hfill x_{4k+2}& =max\left\{y_{4k}^2,\frac{A}{y_{4k}}\right\}=max\left\{{\left(\frac{A}{\mathit{\mu }}\right)}^{2^{2k}},A{\left(\frac{\mathit{\mu }}{A}\right)}^{2^{2k-1}}\right\}={\left(\frac{A}{\mathit{\mu }}\right)}^{2^{2k}}\hfill \\ \hfill y_{4k+2}& =max\left\{x_{4k}^2,\frac{A}{x_{4k}}\right\}=max\left\{{\left(\frac{A}{\mathit{\lambda }}\right)}^{2^{2k}},A{\left(\frac{\mathit{\lambda }}{A}\right)}^{2^{2k-1}}\right\}={\left(\frac{A}{\mathit{\lambda }}\right)}^{2^{2k}}\hfill \\ \hfill x_{4k+3}& =max\left\{y_{4k+1}^2,\frac{A}{y_{4k+1}}\right\}=max\left\{{\left(\frac{A}{\mathit{\alpha }}\right)}^{2^{2k+1}},A{\left(\frac{\mathit{\alpha }}{A}\right)}^{2^{2k}}\right\}={\left(\frac{A}{\mathit{\alpha }}\right)}^{2^{2k+1}}\hfill \\ \hfill y_{4k+3}& =max\left\{x_{4k+1}^2,\frac{A}{x_{4k+1}}\right\}=max\left\{{\left(\frac{A}{\mathit{\beta }}\right)}^{2^{2k+1}},A{\left(\frac{\mathit{\beta }}{A}\right)}^{2^{2k}}\right\}={\left(\frac{A}{\mathit{\beta }}\right)}^{2^{2k+1}}\hfill \\ \hfill x_{4k+4}& =max\left\{y_{4k+2}^2,\frac{A}{y_{4k+2}}\right\}=max\left\{{\left(\frac{A}{\mathit{\lambda }}\right)}^{2^{2k+1}},A{\left(\frac{\mathit{\lambda }}{A}\right)}^{2^{2k}}\right\}={\left(\frac{A}{\mathit{\lambda }}\right)}^{2^{2k+1}}\hfill \\ \hfill y_{4k+4}& =max\left\{x_{4k+2}^2,\frac{A}{x_{4k+2}}\right\}=max\left\{{\left(\frac{A}{\mathit{\mu }}\right)}^{2^{2k+1}},A{\left(\frac{\mathit{\mu }}{A}\right)}^{2^{2k}}\right\}={\left(\frac{A}{\mathit{\mu }}\right)}^{2^{2k+1}}.\hfill \end{array}$$

Therefore the result is true for every $k\in N$. This concludes the proof. $\square$

Lemma 2.2

Assume that $\mathit{\alpha },\mathit{\beta },\mathit{\lambda },\mathit{\mu }\le -1$ and $0<A<1$. Then $\forall n\in N_0$ the following equalities hold:$$\begin{array}{cc}\hfill max\left\{{\mathit{\beta }}^{2^{2n+1}},\frac{A}{{\mathit{\beta }}^{2^{2n}}}\right\}& ={\mathit{\beta }}^{2^{2n+1}}\hfill \\ \hfill max\left\{{\mathit{\alpha }}^{2^{2n+1}},\frac{A}{{\mathit{\alpha }}^{2^{2n}}}\right\}& ={\mathit{\alpha }}^{2^{2n+1}}\hfill \\ \hfill max\left\{{\mathit{\mu }}^{2^{2n+1}},\frac{A}{{\mathit{\mu }}^{2^{2n}}}\right\}& ={\mathit{\mu }}^{2^{2n+1}}\hfill \\ \hfill max\left\{{\mathit{\lambda }}^{2^{2n+1}},\frac{A}{{\mathit{\lambda }}^{2^{2n}}}\right\}& ={\mathit{\lambda }}^{2^{2n+1}}.\hfill \end{array}$$

Proof

Show that$$\begin{array}{c}\hfill max\left\{{\mathit{\beta }}^{2^{2n+1}},\frac{A}{{\mathit{\beta }}^{2^{2n}}}\right\}={\mathit{\beta }}^{2^{2n+1}}\forall n\in N_0.\end{array}$$

We shall proceed by induction on n. Let $n=0$, then ${\mathit{\beta }}^{2^1}={\mathit{\beta }}^2\ge 1>0$ and $\frac{A}{{\mathit{\beta }}^{2^0}}=\frac{A}{\mathit{\beta }}<0<1.$ Therefore, $max\left\{{\mathit{\beta }}^2,\frac{A}{\mathit{\beta }}\right\}={\mathit{\beta }}^2.$ So the result holds for $n=0$. For $n=1,\frac{1}{{\mathit{\beta }}^{2^2}}=\frac{1}{{\mathit{\beta }}^4}\le 1.$ Now suppose for some $k\in N$, we have$$\begin{array}{c}\hfill {\mathit{\beta }}^{2^{2k+1}}\ge 1\end{array}$$

and$$\begin{array}{c}\hfill \frac{1}{{\mathit{\beta }}^{2^{2k}}}\le 1.\end{array}$$

Show that$$\begin{array}{c}\hfill max\left\{{\mathit{\beta }}^{2^{2k+3}},\frac{A}{{\mathit{\beta }}^{2^{2k+2}}}\right\}={\mathit{\beta }}^{2^{2k+3}}.\end{array}$$

Observe,$$\begin{array}{cc}\hfill {\mathit{\beta }}^{2^{2k+3}}& ={\mathit{\beta }}^{2^{2k+1}·2^2}={\mathit{\beta }}^{2^{2k+1}·4}={\mathit{\beta }}^{2^{2k+1}+2^{2k+1}+2^{2k+1}+2^{2k+1}}\hfill \\ \hfill & ={\mathit{\beta }}^{2^{2k+1}}·{\mathit{\beta }}^{2^{2k+1}}·{\mathit{\beta }}^{2^{2k+1}}·{\mathit{\beta }}^{2^{2k+1}}\hfill \\ \hfill & \ge 1·1·1·1=1\hfill \end{array}$$

and$$\begin{array}{cc}\hfill \frac{A}{{\mathit{\beta }}^{2^{2k+2}}}& =A\left(\frac{1}{{\mathit{\beta }}^{2^{2k}·2^2}}\right)=A\left(\frac{1}{{\mathit{\beta }}^{2^{2k}·4}}\right)=A\left(\frac{1}{{\mathit{\beta }}^{2^{2k}+2^{2k}+2^{2k}+2^{2k}}}\right)\hfill \\ \hfill & =A\left(\frac{1}{{\mathit{\beta }}^{2^{2k}}}\right)\left(\frac{1}{{\mathit{\beta }}^{2^{2k}}}\right)\left(\frac{1}{{\mathit{\beta }}^{2^{2k}}}\right)\left(\frac{1}{{\mathit{\beta }}^{2^{2k}}}\right)<A<1.\hfill \end{array}$$

Hence,$$\begin{array}{c}\hfill max\left\{{\mathit{\beta }}^{2^{2k+3}},\frac{A}{{\mathit{\beta }}^{2^{2k+2}}}\right\}={\mathit{\beta }}^{2^{2k+3}}.\end{array}$$

Therefore, the result is true $\forall n\in N_0.$ Similarly, the remaining equalities hold. $\square$

Theorem 2.2

Assume that $x_{-1},y_{-1},x_0,y_0\le -1$ and $0<A<1$. Also, let $\left\{x_n,y_n\right\}$ be a solution of the system of Equations (1.1) with $x_{-1}=\mathit{\alpha }$, $y_{-1}=\mathit{\beta }$, $x_0=\mathit{\lambda }$, and $y_0=\mathit{\mu }$. Then all solutions of (1.1) are of the following:$$\begin{array}{cc}\hfill x_{4n-3}& ={\mathit{\beta }}^{2^{2n-1}},y_{4n-3}={\mathit{\alpha }}^{2^{2n-1}}\hfill \\ \hfill x_{4n-2}& ={\mathit{\mu }}^{2^{2n-1}},y_{4n-2}={\mathit{\lambda }}^{2^{2n-1}}\hfill \\ \hfill x_{4n-1}& ={\mathit{\alpha }}^{2^{2n}},y_{4n-1}={\mathit{\beta }}^{2^{2n}}\hfill \\ \hfill x_{4n}& ={\mathit{\lambda }}^{2^{2n}},y_{4n}={\mathit{\mu }}^{2^{2n}}.\hfill \end{array}$$

Proof

For $n=1$, we have$$\begin{array}{cc}\hfill x_1& =max\left\{{\mathit{\beta }}^2,\frac{A}{\mathit{\beta }}\right\}={\mathit{\beta }}^2,y_1=max\left\{{\mathit{\alpha }}^2,\frac{A}{\mathit{\alpha }}\right\}={\mathit{\alpha }}^2\hfill \\ \hfill x_2& =max\left\{{\mathit{\mu }}^2,\frac{A}{\mathit{\mu }}\right\}={\mathit{\mu }}^2,y_2=max\left\{{\mathit{\lambda }}^2,\frac{A}{\mathit{\lambda }}\right\}={\mathit{\lambda }}^2\hfill \\ \hfill x_3& =max\left\{\frac{A}{{\mathit{\alpha }}^2},{\mathit{\alpha }}^4\right\}={\mathit{\alpha }}^4,y_3=max\left\{\frac{A}{{\mathit{\beta }}^2},{\mathit{\beta }}^4\right\}={\mathit{\beta }}^4\hfill \\ \hfill x_4& =max\left\{\frac{A}{{\mathit{\lambda }}^4},\mathit{\lambda }\right\}={\mathit{\lambda }}^4,y_4=max\left\{\frac{A}{{\mathit{\mu }}^4},\mathit{\mu }\right\}={\mathit{\mu }}^4.\hfill \end{array}$$

So the result holds for $n=1$. Now suppose the result is true for some $k>0$, that is,$$\begin{array}{cc}\hfill x_{4k-3}& ={\mathit{\beta }}^{2^{2k-1}},y_{4k-3}={\mathit{\alpha }}^{2^{2k-1}}\hfill \\ \hfill x_{4k-2}& ={\mathit{\mu }}^{2^{2k-1}},y_{4k-2}={\mathit{\lambda }}^{2^{2k-1}}\hfill \\ \hfill x_{4k-1}& ={\mathit{\alpha }}^{2^{2k}},y_{4k-1}={\mathit{\beta }}^{2^{2k}}\hfill \\ \hfill x_{4k}& ={\mathit{\lambda }}^{2^{2k}},y_{4k}={\mathit{\mu }}^{2^{2n}}.\hfill \end{array}$$

Also, for $k+1$ we have the following:$$\begin{array}{cc}\hfill x_{4k+1}& =max\left\{y_{4k-1}^2,\frac{A}{y_{4k-1}}\right\}=max\left\{{\mathit{\beta }}^{2^{2k+1}},\frac{A}{{\mathit{\beta }}^{2^{2k}}}\right\}={\mathit{\beta }}^{2^{2k+1}}\hfill \\ \hfill y_{4k+1}& =max\left\{x_{4k-1}^2,\frac{A}{x_{4k-1}}\right\}=max\left\{{\mathit{\alpha }}^{2^{2k+1}},\frac{A}{{\mathit{\alpha }}^{2^{2k}}}\right\}={\mathit{\alpha }}^{2^{2k+1}}\hfill \\ \hfill x_{4k+2}& =max\left\{y_{4k}^2,\frac{A}{y_{4k}}\right\}=max\left\{{\mathit{\mu }}^{2^{2k+1}},\frac{A}{{\mathit{\mu }}^{2^{2k}}}\right\}={\mathit{\mu }}^{2^{2k+1}}\hfill \\ \hfill y_{4k+2}& =max\left\{x_{4k}^2,\frac{A}{x_{4k}}\right\}=max\left\{{\mathit{\lambda }}^{2^{2k+1}},\frac{A}{{\mathit{\lambda }}^{2^{2k}}}\right\}={\mathit{\lambda }}^{2^{2k+1}}\hfill \\ \hfill x_{4k+3}& =max\left\{y_{4k+1}^2,\frac{A}{y_{4k+1}}\right\}=max\left\{{\mathit{\alpha }}^{2^{2k+2}},\frac{A}{{\mathit{\alpha }}^{2^{2k+1}}}\right\}={\mathit{\alpha }}^{2^{2k+2}}\hfill \\ \hfill y_{4k+3}& =max\left\{x_{4k+1}^2,\frac{A}{x_{4k+1}}\right\}=max\left\{{\mathit{\beta }}^{2^{2k+2}},\frac{A}{{\mathit{\beta }}^{2^{2k+1}}}\right\}={\mathit{\beta }}^{2^{2k+2}}\hfill \\ \hfill x_{4k+4}& =max\left\{y_{4k+2}^2,\frac{A}{y_{4k+2}}\right\}=max\left\{{\mathit{\lambda }}^{2^{2k+2}},\frac{A}{{\mathit{\lambda }}^{2^{2k+1}}}\right\}={\mathit{\lambda }}^{2^{2k+2}}\hfill \\ \hfill y_{4k+4}& =max\left\{x_{4k+2}^2,\frac{A}{x_{4k+2}}\right\}=max\left\{{\mathit{\mu }}^{2^{2k+2}},\frac{A}{{\mathit{\mu }}^{2^{2k+1}}}\right\}={\mathit{\mu }}^{2^{2k+2}}.\hfill \end{array}$$

Therefore the result is true for every $k\in N$. This concludes the proof. $\square$

Theorem 1.3 Let $\left\{x_n,y_n\right\}$ be a solution of the system of Equations (1.1) with $x_{-1}=x_0=\mathit{\lambda }$, and $y_{-1}=y_0=\mathit{\mu }$. Assume that $A^2<{\mathit{\mu }}^3<A$ and $A^2<{\mathit{\lambda }}^3<A$ where $0<A<1.$ Then all solutions of (1.1) are periodic with period 4 and given by the following:$$\begin{array}{cc}\hfill x_{4n-3}& =\frac{A}{\mathit{\mu }},y_{4n-3}=\frac{A}{\mathit{\lambda }}\hfill \\ \hfill x_{4n-2}& =\frac{A}{\mathit{\mu }},y_{4n-2}=\frac{A}{\mathit{\lambda }}\hfill \\ \hfill x_{4n-1}& =\mathit{\lambda },y_{4n-1}=\mathit{\mu }\hfill \\ \hfill x_{4n}& =\mathit{\lambda },y_{4n}=\mathit{\mu }.\hfill \end{array}$$Proof For $n=1$, we have$$\begin{array}{cc}\hfill x_1& =max\left\{{\mathit{\mu }}^2,\frac{A}{\mathit{\mu }}\right\}=\frac{A}{\mathit{\mu }},y_1=max\left\{{\mathit{\lambda }}^2,\frac{A}{\mathit{\lambda }}\right\}=\frac{A}{\mathit{\lambda }}\hfill \\ \hfill x_2& =max\left\{{\mathit{\mu }}^2,\frac{A}{\mathit{\mu }}\right\}=\frac{A}{\mathit{\mu }},y_2=max\left\{{\mathit{\lambda }}^2,\frac{A}{\mathit{\lambda }}\right\}=\frac{A}{\mathit{\lambda }}\hfill \\ \hfill x_3& =max\left\{\frac{A^2}{{\mathit{\lambda }}^2},\mathit{\lambda }\right\}=\mathit{\lambda },y_3=max\left\{\frac{A^2}{{\mathit{\mu }}^2},\mathit{\mu }\right\}=\mathit{\mu }\hfill \\ \hfill x_4& =max\left\{\frac{A^2}{{\mathit{\lambda }}^2},\mathit{\lambda }\right\}=\mathit{\lambda },y_4=max\left\{\frac{A^2}{{\mathit{\mu }}^2},\mathit{\mu }\right\}=\mathit{\mu }.\hfill \end{array}$$

So the result holds for $n=1$. Now suppose the result is true for some $k>0$, that is,$$\begin{array}{cc}\hfill x_{4k-3}& =\frac{A}{\mathit{\mu }},y_{4k-3}=\frac{A}{\mathit{\lambda }}\hfill \\ \hfill x_{4k-2}& =\frac{A}{\mathit{\mu }},y_{4k-2}=\frac{A}{\mathit{\lambda }}\hfill \\ \hfill x_{4k-1}& =\mathit{\lambda },y_{4k-1}=\mathit{\mu }\hfill \\ \hfill x_{4k}& =\mathit{\lambda },y_{4k}=\mathit{\mu }.\hfill \end{array}$$

Also, for $k+1$ we have the following:$$\begin{array}{cc}\hfill x_{4k+1}& =max\left\{y_{4k-1}^2,\frac{A}{y_{4k-1}}\right\}=max\left\{{\mathit{\mu }}^2,\frac{A}{\mathit{\mu }}\right\}=\frac{A}{\mathit{\mu }}\hfill \\ \hfill y_{4k+1}& =max\left\{x_{4k-1}^2,\frac{A}{x_{4k-1}}\right\}=max\left\{{\mathit{\lambda }}^2,\frac{A}{\mathit{\lambda }}\right\}=\frac{A}{\mathit{\lambda }}\hfill \\ \hfill x_{4k+2}& =max\left\{y_{4k}^2,\frac{A}{y_{4k}}\right\}=max\left\{{\mathit{\mu }}^2,\frac{A}{\mathit{\mu }}\right\}=\frac{A}{\mathit{\mu }}\hfill \\ \hfill y_{4k+2}& =max\left\{x_{4k}^2,\frac{A}{x_{4k}}\right\}=max\left\{{\mathit{\lambda }}^2,\frac{A}{\mathit{\lambda }}\right\}=\frac{A}{\mathit{\lambda }}\hfill \\ \hfill x_{4k+3}& =max\left\{y_{4k+1}^2,\frac{A}{y_{4k+1}}\right\}=max\left\{\frac{A^2}{{\mathit{\lambda }}^2},\mathit{\lambda }\right\}=\mathit{\lambda }\hfill \\ \hfill y_{4k+3}& =max\left\{x_{4k+1}^2,\frac{A}{x_{4k+1}}\right\}=max\left\{\frac{A^2}{{\mathit{\mu }}^2},\mathit{\mu }\right\}=\mathit{\mu }\hfill \\ \hfill x_{4k+4}& =max\left\{y_{4k+2}^2,\frac{A}{y_{4k+2}}\right\}=max\left\{\frac{A^2}{{\mathit{\lambda }}^2},\mathit{\lambda }\right\}=\mathit{\lambda }\hfill \\ \hfill y_{4k+4}& =max\left\{x_{4k+2}^2,\frac{A}{x_{4k+2}}\right\}=max\left\{\frac{A^2}{{\mathit{\mu }}^2},\mathit{\mu }\right\}=\mathit{\mu }.\hfill \end{array}$$

Therefore the result is true for every $k\in N$. This concludes the proof. $\square$

To see the periodic behavior of $\left\{x_n,y_n\right\}$, observe the following three diagrams with $x_1=\frac{1}{4}$, $x_2=\frac{1}{4}$, $y_1=\frac{1}{3}$, $y_2=\frac{1}{3}$ and $A=\frac{1}{9}$:

Lemma 1.3 Assume that $-1\le \mathit{\alpha },\mathit{\beta },\mathit{\lambda },\mathit{\mu }<0$ and $A\ge 1$. Then $\forall n\in N_0$ the following equalities hold:$$\begin{array}{cc}\hfill max\left\{\frac{A^{2^{2n}}}{{\mathit{\alpha }}^{2^{2n+1}}},A\left(\frac{{\mathit{\alpha }}^{2^{2n}}}{A^{2^{2n-1}}}\right)\right\}& =\frac{A^{2^{2n}}}{{\mathit{\alpha }}^{2^{2n+1}}}\hfill \\ \hfill max\left\{\frac{A^{2^{2n}}}{{\mathit{\beta }}^{2^{2n+1}}},A\left(\frac{{\mathit{\beta }}^{2^{2n}}}{A^{2^{2n-1}}}\right)\right\}& =\frac{A^{2^{2n}}}{{\mathit{\beta }}^{2^{2n+1}}}\hfill \\ \hfill max\left\{\frac{A^{2^{2n}}}{{\mathit{\lambda }}^{2^{2n+1}}},A\left(\frac{{\mathit{\lambda }}^{2^{2n}}}{A^{2^{2n-1}}}\right)\right\}& =\frac{A^{2^{2n}}}{{\mathit{\lambda }}^{2^{2n+1}}}\hfill \\ \hfill max\left\{\frac{A^{2^{2n}}}{{\mathit{\mu }}^{2^{2n+1}}},A\left(\frac{{\mathit{\mu }}^{2^{2n}}}{A^{2^{2n-1}}}\right)\right\}& =\frac{A^{2^{2n}}}{{\mathit{\mu }}^{2^{2n+1}}}.\hfill \end{array}$$Proof First, note that ${\mathit{\alpha }}^2\le 1$ this implies that $1\le \frac{1}{{\mathit{\alpha }}^2}$. Multiplying by A and using the assumption $A\ge 1$ means that $\frac{A}{{\mathit{\alpha }}^2}\ge A\ge 1$. So, for $n=0$ we have $\frac{A^{2^0}}{{\mathit{\alpha }}^{2^1}}=\frac{A}{{\mathit{\alpha }}^2}\ge 1$ and $A\left(\frac{{\mathit{\alpha }}^{2^0}}{A^{2^{-1}}}\right)=A^{\frac{1}{2}}·\mathit{\alpha }<0<1.$ Therefore, $max\left\{\frac{A}{{\mathit{\alpha }}^2},A^{\frac{1}{2}}·\mathit{\alpha }\right\}=\frac{A}{{\mathit{\alpha }}^2}.$ So the result holds for $n=0$. Now suppose for some $k>0$, we have$$\begin{array}{c}\hfill \frac{A^{2^{2k}}}{{\mathit{\alpha }}^{2^{2k+1}}}\ge 1\end{array}$$

and$$\begin{array}{c}\hfill A\left(\frac{{\mathit{\alpha }}^{2^{2k}}}{A^{2^{2k-1}}}\right)<1,\end{array}$$

this means that $\frac{{\mathit{\alpha }}^{2^{2k}}}{A^{2^{2k-1}}}<\frac{1}{A}\le 1.$ Show that$$\begin{array}{c}\hfill max\left\{\frac{A^{2^{2k+2}}}{{\mathit{\alpha }}^{2^{2k+3}}},A\left(\frac{{\mathit{\alpha }}^{2^{2k+2}}}{A^{2^{2k+1}}}\right)\right\}=\frac{A^{2^{2k+2}}}{{\mathit{\alpha }}^{2^{2k+3}}}.\end{array}$$

Observe,$$\begin{array}{cc}\hfill \frac{A^{2^{2k+2}}}{{\mathit{\alpha }}^{2^{2k+3}}}& =\frac{A^{2^{2k}·2^2}}{{\mathit{\alpha }}^{2^{2k+1}·2^2}}=\frac{A^{2^{2k}·4}}{{\mathit{\alpha }}^{2^{2k+1}·4}}=\frac{A^{2^{2k}+2^{2k}+2^{2k}+2^{2k+1}}}{{\mathit{\alpha }}^{2^{2k+1}+2^{2k+1}+2^{2k+1}+2^{2k+1}}}\hfill \\ \hfill & =\frac{A^{2^{2k}}}{{\mathit{\alpha }}^{2^{2k+1}}}·\frac{A^{2^{2k}}}{{\mathit{\alpha }}^{2^{2k+1}}}·\frac{A^{2^{2k}}}{{\mathit{\alpha }}^{2^{2k+1}}}·\frac{A^{2^{2k}}}{{\mathit{\alpha }}^{2^{2k+1}}}\ge 1\hfill \end{array}$$

and$$\begin{array}{cc}\hfill A\left(\frac{{\mathit{\alpha }}^{2^{2k+2}}}{A^{2^{2k+1}}}\right)& =A\left(\frac{{\mathit{\alpha }}^{2^{2k}+2^{2k}+2^{2k}+2^{2k}}}{A^{2^{2k-1}+2^{2k-1}+2^{2k-1}+2^{2k-1}}}\right)\hfill \\ \hfill & =A\left(\frac{{\mathit{\alpha }}^{2^{2k}}}{A^{2^{2k-1}}}\right)\frac{{\mathit{\alpha }}^{2^{2k}}}{A^{2^{2k-1}}}\frac{{\mathit{\alpha }}^{2^{2k}}}{A^{2^{2k-1}}}\frac{{\mathit{\alpha }}^{2^{2k}}}{A^{2^{2k-1}}}<1.\hfill \end{array}$$

Hence,$$\begin{array}{c}\hfill max\left\{\frac{A^{2^{2k+2}}}{{\mathit{\alpha }}^{2^{2k+3}}},A\left(\frac{{\mathit{\alpha }}^{2^{2k+2}}}{A^{2^{2k+1}}}\right)\right\}=\frac{A^{2^{2k+2}}}{{\mathit{\alpha }}^{2^{2k+3}}}.\end{array}$$

Therefore, the result is true $\forall n\in N_0.$ Similarly, the remaining equalities hold.

$\square$

Theorem 1.4 Assume that $-1\le x_{-1},y_{-1},x_0,y_0<0$ and $A\ge 1$. Also, let $\left\{x_n,y_n\right\}$ be a solution of the system of Equations (1.1) with $x_{-1}=\mathit{\alpha }$, $y_{-1}=\mathit{\beta }$, $x_0=\mathit{\lambda }$, and $y_0=\mathit{\mu }$. Then all solutions of (1.1) are of the following:$$\begin{array}{cc}\hfill x_1& ={\mathit{\beta }}^2,y_1={\mathit{\alpha }}^2\hfill \\ \hfill x_2& ={\mathit{\mu }}^2,y_2={\mathit{\lambda }}^2\hfill \\ \hfill \end{array}$$

For $n\in N$,$$\begin{array}{cc}\hfill x_{4n-1}& =\frac{A^{2^{2n-2}}}{{\mathit{\alpha }}^{2^{2n-1}}},y_{4n-1}=\frac{A^{2^{2n-2}}}{{\mathit{\beta }}^{2^{2n-1}}}\hfill \\ \hfill x_{4n}& =\frac{A^{2^{2n-2}}}{{\mathit{\lambda }}^{2^{2n-1}}},y_{4n}=\frac{A^{2^{2n-2}}}{{\mathit{\mu }}^{2^{2n-1}}}\hfill \\ \hfill x_{4n+1}& =\frac{A^{2^{2n-1}}}{{\mathit{\beta }}^{2^{2n}}},y_{4n+1}=\frac{A^{2^{2n-1}}}{{\mathit{\alpha }}^{2^{2n}}}\hfill \\ \hfill x_{4n+2}& =\frac{A^{2^{2n-1}}}{{\mathit{\mu }}^{2^{2n}}},y_{4n+2}=\frac{A^{2^{2n-1}}}{{\mathit{\lambda }}^{2^{2n}}}.\hfill \end{array}$$Proof First,$$\begin{array}{cc}\hfill x_1& =max\left\{{\mathit{\beta }}^2,\frac{A}{\mathit{\beta }}\right\}={\mathit{\beta }}^2,y_1=max\left\{{\mathit{\alpha }}^2,\frac{A}{\mathit{\alpha }}\right\}={\mathit{\alpha }}^2\hfill \\ \hfill x_2& =max\left\{{\mathit{\mu }}^2,\frac{A}{\mathit{\mu }}\right\}={\mathit{\mu }}^2,y_2=max\left\{{\mathit{\lambda }}^2,\frac{A}{\mathit{\lambda }}\right\}={\mathit{\lambda }}^2.\hfill \end{array}$$

Since, $-1\le \mathit{\alpha },\mathit{\beta },\mathit{\lambda },\mathit{\mu }<0$. Next, we shall proceed by induction on n. For $n=1$, we have$$\begin{array}{cc}\hfill x_3& =max\left\{{\mathit{\alpha }}^4,\frac{A}{{\mathit{\alpha }}^2}\right\}=\frac{A}{{\mathit{\alpha }}^2},y_3=max\left\{{\mathit{\beta }}^4,\frac{A}{{\mathit{\beta }}^2}\right\}=\frac{A}{{\mathit{\beta }}^2}\hfill \\ \hfill x_4& =max\left\{{\mathit{\lambda }}^4,\frac{A}{{\mathit{\lambda }}^2}\right\}=\frac{A}{{\mathit{\lambda }}^2},y_4=max\left\{{\mathit{\mu }}^4,\frac{A}{{\mathit{\mu }}^2}\right\}=\frac{A}{{\mathit{\mu }}^2}\hfill \\ \hfill x_5& =max\left\{\frac{A^2}{{\mathit{\beta }}^4},{\mathit{\beta }}^2\right\}=\frac{A^2}{{\mathit{\beta }}^4},y_5=max\left\{\frac{A^2}{{\mathit{\alpha }}^4},{\mathit{\alpha }}^2\right\}=\frac{A^2}{{\mathit{\alpha }}^4}\hfill \\ \hfill x_6& =max\left\{\frac{A^2}{{\mathit{\mu }}^4},{\mathit{\mu }}^2\right\}=\frac{A^2}{{\mathit{\mu }}^4},y_6=max\left\{\frac{A^2}{{\mathit{\lambda }}^4},{\mathit{\lambda }}^2\right\}=\frac{A^2}{{\mathit{\lambda }}^4}.\hfill \end{array}$$

So the result holds for $n=1$. Now suppose the result is true for some $k>0$, that is:$$\begin{array}{cc}\hfill x_{4k-1}& =\frac{A^{2^{2k-2}}}{{\mathit{\alpha }}^{2^{2k-1}}},y_{4k-1}=\frac{A^{2^{2k-2}}}{{\mathit{\beta }}^{2^{2k-1}}}\hfill \\ \hfill x_{4k}& =\frac{A^{2^{2k-2}}}{{\mathit{\lambda }}^{2^{2k-1}}},y_{4k}=\frac{A^{2^{2k-2}}}{{\mathit{\mu }}^{2^{2k-1}}}\hfill \\ \hfill x_{4k+1}& =\frac{A^{2^{2k-1}}}{{\mathit{\beta }}^{2^{2k}}},y_{4k+1}=\frac{A^{2^{2k-1}}}{{\mathit{\alpha }}^{2^{2k}}}\hfill \\ \hfill x_{4k+2}& =\frac{A^{2^{2k-1}}}{{\mathit{\mu }}^{2^{2k}}},y_{4k+2}=\frac{A^{2^{2k-1}}}{{\mathit{\lambda }}^{2^{2k}}}.\hfill \end{array}$$

Also, for $k+1$ we have the following:$$\begin{array}{cc}\hfill x_{4k+3}& =max\left\{y_{4k+1}^2,\frac{A}{y_{4k+1}}\right\}=max\left\{\frac{A^{2^{2k}}}{{\mathit{\alpha }}^{2^{2k+1}}},A\left(\frac{{\mathit{\alpha }}^{2^{2k}}}{A^{2^{2k-1}}}\right)\right\}=\frac{A^{2^{2k}}}{{\mathit{\alpha }}^{2^{2k+1}}}\hfill \\ \hfill y_{4k+3}& =max\left\{x_{4k+1}^2,\frac{A}{x_{4k+1}}\right\}=max\left\{\frac{A^{2^{2k}}}{{\mathit{\beta }}^{2^{2k+1}}},A\left(\frac{{\mathit{\beta }}^{2^{2k}}}{A^{2^{2k-1}}}\right)\right\}=\frac{A^{2^{2k}}}{{\mathit{\beta }}^{2^{2k+1}}}\hfill \\ \hfill x_{4k+4}& =max\left\{y_{4k+2}^2,\frac{A}{y_{4k+2}}\right\}=max\left\{\frac{A^{2^{2k}}}{{\mathit{\lambda }}^{2^{2k+1}}},A\left(\frac{{\mathit{\lambda }}^{2^{2k}}}{A^{2^{2k-1}}}\right)\right\}=\frac{A^{2^{2k}}}{{\mathit{\lambda }}^{2^{2k+1}}}\hfill \\ \hfill y_{4k+4}& =max\left\{x_{4k+2}^2,\frac{A}{x_{4k+2}}\right\}=max\left\{\frac{A^{2^{2k}}}{{\mathit{\mu }}^{2^{2k+1}}},A\left(\frac{{\mathit{\mu }}^{2^{2k}}}{A^{2^{2k-1}}}\right)\right\}=\frac{A^{2^{2k}}}{{\mathit{\mu }}^{2^{2k+1}}}\hfill \\ \hfill x_{4k+5}& =max\left\{y_{4k+3}^2,\frac{A}{y_{4k+3}}\right\}=max\left\{\frac{A^{2^{2k+1}}}{{\mathit{\beta }}^{2^{2k+2}}},A\left(\frac{{\mathit{\beta }}^{2^{2k+1}}}{A^{2^{2k}}}\right)\right\}=\frac{A^{2^{2k+1}}}{{\mathit{\beta }}^{2^{2k+2}}}\hfill \\ \hfill y_{4k+5}& =max\left\{x_{4k+3}^2,\frac{A}{x_{4k+3}}\right\}=max\left\{\frac{A^{2^{2k+1}}}{{\mathit{\alpha }}^{2^{2k+2}}},A\left(\frac{{\mathit{\alpha }}^{2^{2k+1}}}{A^{2^{2k}}}\right)\right\}=\frac{A^{2^{2k+1}}}{{\mathit{\alpha }}^{2^{2k+2}}}\hfill \\ \hfill x_{4k+6}& =max\left\{y_{4k+4}^2,\frac{A}{y_{4k+4}}\right\}=max\left\{\frac{A^{2^{2k+1}}}{{\mathit{\mu }}^{2^{2k+2}}},A\left(\frac{{\mathit{\mu }}^{2^{2k+1}}}{A^{2^{2k}}}\right)\right\}=\frac{A^{2^{2k+1}}}{{\mathit{\mu }}^{2^{2k+2}}}\hfill \\ \hfill y_{4k+6}& =max\left\{x_{4k+4}^2,\frac{A}{x_{4k+4}}\right\}=max\left\{\frac{A^{2^{2k+1}}}{{\mathit{\lambda }}^{2^{2k+2}}},A\left(\frac{{\mathit{\lambda }}^{2^{2k+1}}}{A^{2^{2k}}}\right)\right\}=\frac{A^{2^{2k+1}}}{{\mathit{\lambda }}^{2^{2k+2}}}.\hfill \end{array}$$

Therefore the result is true for every $k\in N$. This concludes the proof. $\square$

Theorem 1.5 Let $\left\{x_n,y_n\right\}$ be a solution of the system of Equations (1.1) with $x_{-1}=x_0=\mathit{\lambda },$ and $y_{-1}=y_0=\mathit{\mu }$. Assume that ${\mathit{\mu }}^3<A^2<{\mathit{\lambda }}^3<A$ and $A^5<{\mathit{\mu }}^6<A^4$ where $0<A<1.$ Then all solutions of (1.1) are periodic with period 4 and given by the following:$$\begin{array}{cc}\hfill x_1& =\frac{A}{\mathit{\mu }},y_1=\frac{A}{\mathit{\lambda }}\hfill \\ \hfill x_2& =\frac{A}{\mathit{\mu }},y_2=\frac{A}{\mathit{\lambda }}\hfill \end{array}$$

For $n\in N$,$$\begin{array}{cc}\hfill x_{4n-1}& =x_{4n}=\mathit{\lambda },y_{4n-1}=y_{4n}=\frac{A^2}{{\mathit{\mu }}^2}\hfill \\ \hfill x_{4n+1}& =x_{4n+2}=\frac{{\mathit{\mu }}^2}{A},y_{4n+1}=y_{4n+2}=\frac{A}{\mathit{\lambda }}.\hfill \end{array}$$Proof The result follows by the principle of mathematical induction. $\square$

To see the periodic behavior of $\left\{x_n,y_n\right\}$, observe the following three diagrams with $x_1=\frac{1}{2}$, $x_2=\frac{1}{2}$, $y_1=\frac{1}{3}$,$y_2=\frac{1}{3}$, and $A=\frac{1}{4}$:

Correction

This article was originally published with errors. This version has been amended to remove erroneous text after the first equation on page two of the article.

Additional information

Funding

The author received no direct funding for this research.

Notes on contributors

J.L. Williams

J.L. Williams received his PhD in mathematics from Mississippi State University (MSU) August 2013 and his master’s in mathematical sciences from MSU May 2008. In particular, his research area is in partial differential equations (PDE). In PDE, he mainly studies the existence and nonexistence to elliptic equations. He is the first African-American to receive a PhD in Mathematics from MSU.

Williams received his bachelor’s degree in mathematics with a minor in Business from the University of Mississippi (Olemiss) in 2006, where he was a member of Pi Mu Epsilon (Honorary math society). Williams is an assistant professor of mathematics at Texas Southern University (TSU), where he started in fall 2013. He has published several papers in professional journals since being at TSU. Also, he received the College of Science, Engineering, & Technology Distinguished Teaching Award and the McCleary Teaching Excellent Award at the university level for teaching.