Solution of the Maximum of Difference Equation xn+1=max{Axn−1,ynxn};yn+1=max{Ayn−1,xnyn}\matrix{ {x_{n + 1} = max \left\{ {{A \over {x_{n - 1} }},{{y_n } \over {x_n }}} \right\};} & {y_{n + 1} = max \left\{ {{A \over {y_{n - 1} }},{{x_n } \over {y_n }}} \right\}}}

Let I be an interval of reel numbers and let f : I^s+1 → I be a continuously differentiable function where s is a non-negative integer. Consider the difference equation (2)xn+1=f(xn,xn−1,...,xn−s)forn=0,1,...,\matrix{ {x_{n + 1} = f\left( {x_n ,x_{n - 1} ,...,x_{n - s} } \right){\rm{for}}} & {n = 0,1,...,}} with the initial values x_−s,...,x₀ ∈ I. A point x¯\overline x called an equilibrium point of equation 2. if x¯=f(x¯,...,x¯)\overline x = f\left( {\overline x ,...,\overline x } \right) .

A positive semi sycle of a solution {xn}n∞=−s\left\{ {x_n } \right\}_n^\infty = - s of 2 consist of a string of terms {x_l,x_l+1,...,x_m} all greater than or equal to equilibrium x¯\overline x with l ≥ −s and m ≤ ∞ such that either l = −s or l > s and xl−1<x¯x_{l - 1} < \overline x and either m = ∞ or m < ∞ and xm+1<x¯x_{m + 1} < \overline x .

A negative semisycle of a solution {xn}n∞=−s\left\{ {x_n } \right\}_n^\infty = - s of 2 consist of a string of terms {x_l,x_l+1,...,x_m} all less than or equal to equilibrium x¯\overline x with l ≥ −s and m ≤ ∞ such that either l = −s or l > − s and xl−1≥x¯x_{l - 1} \ge \overline x and either m = ∞ or m ≤ ∞ and xm+1≥x¯x_{m + 1} \ge \overline x .

Assume that, A and x₀,x₋₁,y₀,y₋₁are positive integer sequence for 1

A > x₀ > x₋₁ > y₀ > y₋₁,A > x₀ > y₀ > x₋₁ > y₋₁,A > y₀ > x₀ > x₋₁ > y₋₁,

Then the following statements are true:

n ≥ 0 for x_n and n ≥ 1 for y_na)

Every positive semi-cycle consist two term.

A > x₀ > x₋₁ > y₀ > y₋₁,A > x₀ > y₀ > x₋₁ > y₋₁,A > y₀ > x₀ > x₋₁ > y₋₁ The solution x_n and y_n can be obtained as follows: x1=max{Ax−1,y0x0}=Ax−1<x¯; y1=max{Ay−1,x0y0}=Ay−1>y¯,x_1 = max \left\{ {{A \over {x_{ - 1} }},{{y_0 } \over {x_0 }}} \right\} = {A \over {x_{ - 1} }} < \overline x ;\quad y_1 = max \left\{ {{A \over {y_{ - 1} }},{{x_0 } \over {y_0 }}} \right\} = {A \over {y_{ - 1} }} > \overline y ,x2=max{Ax0,y1x1}=max{Ax0,x−1y−1}=x−1y−1<x¯; y2=max{Ay0,x1y1}=max{Ay0,y−1x−1}=Ay0<y¯,x_2 = max \left\{ {{A \over {x_0 }},{{y_1 } \over {x_1 }}} \right\} = max \left\{ {{A \over {x_0 }},{{x_{ - 1} } \over {y_{ - 1} }}} \right\} = {{x_{ - 1} } \over {y_{ - 1} }} < \overline x ;\quad y_2 = max \left\{ {{A \over {y_0 }},{{x_1 } \over {y_1 }}} \right\} = max \left\{ {{A \over {y_0 }},{{y_{ - 1} } \over {x_{ - 1} }}} \right\} = {A \over {y_0 }} < \overline y ,x3=max{Ax1,y2x2}=max{x−1,Ay−1x−1y0}=x−1>x¯; y3=max{Ay1,x2y2}=max{y−1,x−1y0Ay−1}=y−1<y¯,x_3 = max \left\{ {{A \over {x_1 }},{{y_2 } \over {x_2 }}} \right\} = max \left\{ {x_{ - 1} ,{{Ay_{ - 1} } \over {x_{ - 1} y_0 }}} \right\} = x_{ - 1} > \overline x ;\quad y_3 = max \left\{ {{A \over {y_1 }},{{x_2 } \over {y_2 }}} \right\} = max \left\{ {y_{ - 1} ,{{x_{ - 1} y_0 } \over {Ay_{ - 1} }}} \right\} = y_{ - 1} < \overline y ,x4=max{Ax2,y3x3}=max{Ay−1x−1,y−1x−1}=Ay−1x−1>x¯; y4=max{Ay2,x3y3}=max{y0,x−1y−1}=y0>y¯,x_4 = max \left\{ {{A \over {x_2 }},{{y_3 } \over {x_3 }}} \right\} = max \left\{ {{{Ay_{ - 1} } \over {x_{ - 1} }},{{y_{ - 1} } \over {x_{ - 1} }}} \right\} = {{Ay_{ - 1} } \over {x_{ - 1} }} > \overline x ;\quad y_4 = max \left\{ {{A \over {y_2 }},{{x_3 } \over {y_3 }}} \right\} = max \left\{ {y_0 ,{{x_{ - 1} } \over {y_{ - 1} }}} \right\} = y_0 > \overline y ,x5=max{Ax3,y4x4}=max{Ax−1,x−1y0Ay−1}=Ax−1<x¯; y5=max{Ay3,x4y4}=max{Ay−1,Ay−1x−1y0}=Ay−1>y¯,x_5 = max \left\{ {{A \over {x_3 }},{{y_4 } \over {x_4 }}} \right\} = max \left\{ {{A \over {x_{ - 1} }},{{x_{ - 1} y_0 } \over {Ay_{ - 1} }}} \right\} = {A \over {x_{ - 1} }} < \overline x ;\quad y_5 = max \left\{ {{A \over {y_3 }},{{x_4 } \over {y_4 }}} \right\} = max \left\{ {{A \over {y_{ - 1} }},{{Ay_{ - 1} } \over {x_{ - 1} y_0 }}} \right\} = {A \over {y_{ - 1} }} > \overline y ,x6=max{Ax4,y5x5}=max{x−1y−1,x−1y−1}=x−1y−1<x¯; y6=max{Ay4,x5y5}=max{Ay0,y−1x−1}=Ay0<y¯,x_6 = max \left\{ {{A \over {x_4 }},{{y_5 } \over {x_5 }}} \right\} = max \left\{ {{{x_{ - 1} } \over {y_{ - 1} }},{{x_{ - 1} } \over {y_{ - 1} }}} \right\} = {{x_{ - 1} } \over {y_{ - 1} }} < \overline x ;\quad y_6 = max \left\{ {{A \over {y_4 }},{{x_5 } \over {y_5 }}} \right\} = max \left\{ {{A \over {y_0 }},{{y_{ - 1} } \over {x_{ - 1} }}} \right\} = {A \over {y_0 }} < \overline y ,x7=max{Ax5,y6x6}=max{x−1,Ay−1y0x−1}=x−1>x¯; y7=max{Ay5,x6y6}=max{y−1,y0x−1Ay−1}=y−1<y¯,x_7 = max \left\{ {{A \over {x_5 }},{{y_6 } \over {x_6 }}} \right\} = max \left\{ {x_{ - 1} ,{{Ay_{ - 1} } \over {y_0 x_{ - 1} }}} \right\} = x_{ - 1} > \overline x ;\quad y_7 = max \left\{ {{A \over {y_5 }},{{x_6 } \over {y_6 }}} \right\} = max \left\{ {y_{ - 1} ,{{y_0 x_{ - 1} } \over {Ay_{ - 1} }}} \right\} = y_{ - 1} < \overline y ,x8=max{Ax6,y7x7}=max{Ay−1x−1,y−1x−1}=Ay−1x−1>x¯; y8=max{Ay6,x7y7}=max{y−1,x−1y−1}=y0>y¯,⋮\matrix{ {x_8 = max \left\{ {{A \over {x_6 }},{{y_7 } \over {x_7 }}} \right\} = max \left\{ {{{Ay_{ - 1} } \over {x_{ - 1} }},{{y_{ - 1} } \over {x_{ - 1} }}} \right\} = {{Ay_{ - 1} } \over {x_{ - 1} }} > \overline x ;\quad y_8 = max \left\{ {{A \over {y_6 }},{{x_7 } \over {y_7 }}} \right\} = max \left\{ {y_{ - 1} ,{{x_{ - 1} } \over {y_{ - 1} }}} \right\} = y_0 > \overline y ,} \cr \vdots}

Hence we obtained. x1<x¯,x2<x¯,x3>x¯,x4>x¯,x5<x¯,x6<x¯,x7>x¯,x8>x¯,...x_1 < \overline x ,x_2 < \overline x ,x_3 > \overline x ,x_4 > \overline x ,x_5 < \overline x ,x_6 < \overline x ,x_7 > \overline x ,x_8 > \overline x ,...

y1>y¯,y2<y¯,y3<y¯,y4>y¯,y5>y¯,y6<y¯,y7<y¯,y8>y¯,...y_1 > \overline y ,y_2 < \overline y ,y_3 < \overline y ,y_4 > \overline y ,y_5 > \overline y ,y_6 < \overline y ,y_7 < \overline y ,y_8 > \overline y ,...

Hence, the solution n ≥ 0 for x_n and n ≥ 1 for y_n, every positive semi-cycle consists of two terms, every negative semi-cycle consists of two terms

Assume that, A and x₀,x₋₁,y₀,y₋₁are positive integer sequence for 1

A > x₀ > y₀ > y₋₁ > x₋₁,A > y₀ > x₀ > y₋₁ > x₋₁,A > y₀ > y₋₁ > x₀ > x₋₁,

Then the following statements are true:

n ≥ 1 for x_n and n ≥ 0 for y_na)

Every positive semi-cycle consist two term.

Lemma 2 proof’s can be obtained similarly Lemma 1

Assume that, A and x₀,x₋₁,y₀,y₋₁are positive integer sequence for 1

A > x₋₁ > y₋₁ > x₀ > y₀,A > y₋₁ > x₋₁ > x₀ > y₀,A > y₋₁ > x₀ > x₋₁ > y₀,

Then the following statements are true:

n ≥ 0 for x_n and n ≥ 1 for y_na)

Every positive semi-cycle consist two term.

Lemma 3 proof’s can be obtained similarly Lemma 1

Let (x_n, y_n) be a solution of 1 for

A > x₀ > x₋₁ > y₀ > y₋₁,A > x₀ > y₀ > x₋₁ > y₋₁,A > y₀ > x₀ > x₋₁ > y₋₁.

Then for n = 0, 1,... we have, xn={Ax−1,x−1y−1,x−1,Ay−1x−1,...},yn={Ay−1,Ay0,y−1,y0,...}.\matrix{ {x_n = \left\{ {{A \over {x_{ - 1} }},{{x_{ - 1} } \over {y_{ - 1} }},x_{ - 1} ,{{Ay_{ - 1} } \over {x_{ - 1} }},...} \right\},} \cr {y_n = \left\{ {{A \over {y_{ - 1} }},{A \over {y_0 }},y_{ - 1} ,y_0 ,...} \right\}.}}

We obtain, x1=max{Ax−1,y0x0}=Ax−1; y1=max{Ay−1,x0y0}=Ay−1,x_1 = max \left\{ {{A \over {x_{ - 1} }},{{y_0 } \over {x_0 }}} \right\} = {A \over {x_{ - 1} }};\quad y_1 = max \left\{ {{A \over {y_{ - 1} }},{{x_0 } \over {y_0 }}} \right\} = {A \over {y_{ - 1} }}, x2=max{Ax0,y1x1}=max{Ax0,x−1y−1}=x−1y−1; y2=max{Ay0,x1y1}=max{Ay0,y−1x−1}=Ay0,x_2 = max \left\{ {{A \over {x_0 }},{{y_1 } \over {x_1 }}} \right\} = max \left\{ {{A \over {x_0 }},{{x_{ - 1} } \over {y_{ - 1} }}} \right\} = {{x_{ - 1} } \over {y_{ - 1} }};\quad y_2 = max \left\{ {{A \over {y_0 }},{{x_1 } \over {y_1 }}} \right\} = max \left\{ {{A \over {y_0 }},{{y_{ - 1} } \over {x_{ - 1} }}} \right\} = {A \over {y_0 }},x3=max{Ax1,y2x2}=max{x−1,Ay−1x−1y0}=x−1; y3=max{Ay1,x2y2}=max{y−1,x−1y0Ay−1}=y−1,x_3 = max \left\{ {{A \over {x_1 }},{{y_2 } \over {x_2 }}} \right\} = max \left\{ {x_{ - 1} ,{{Ay_{ - 1} } \over {x_{ - 1} y_0 }}} \right\} = x_{ - 1} ;\quad y_3 = max \left\{ {{A \over {y_1 }},{{x_2 } \over {y_2 }}} \right\} = max \left\{ {y_{ - 1} ,{{x_{ - 1} y_0 } \over {Ay_{ - 1} }}} \right\} = y_{ - 1} ,x4=max{Ax2,y3x3}=max{Ay−1x−1,y−1x−1}=Ay−1x−1; y4=max{Ay2,x3y3}=max{y0,x−1y−1}=y0,x_4 = max \left\{ {{A \over {x_2 }},{{y_3 } \over {x_3 }}} \right\} = max \left\{ {{{Ay_{ - 1} } \over {x_{ - 1} }},{{y_{ - 1} } \over {x_{ - 1} }}} \right\} = {{Ay_{ - 1} } \over {x_{ - 1} }};\quad y_4 = max \left\{ {{A \over {y_2 }},{{x_3 } \over {y_3 }}} \right\} = max \left\{ {y_0 ,{{x_{ - 1} } \over {y_{ - 1} }}} \right\} = y_0 ,x5=max{Ax3,y4x4}=max{Ax−1,x−1y0Ay−1}=Ax−1; y5=max{Ay3,x4y4}=max{Ay−1,Ay−1x−1y0}=Ay−1,x_5 = max \left\{ {{A \over {x_3 }},{{y_4 } \over {x_4 }}} \right\} = max \left\{ {{A \over {x_{ - 1} }},{{x_{ - 1} y_0 } \over {Ay_{ - 1} }}} \right\} = {A \over {x_{ - 1} }};\quad y_5 = max \left\{ {{A \over {y_3 }},{{x_4 } \over {y_4 }}} \right\} = max \left\{ {{A \over {y_{ - 1} }},{{Ay_{ - 1} } \over {x_{ - 1} y_0 }}} \right\} = {A \over {y_{ - 1} }},x6=max{Ax4,y5x5}=max{x−1y−1,x−1y−1}=x−1y−1; y6=max{Ay4,x5y5}=max{Ay0,y−1x−1}=Ay0,x_6 = max \left\{ {{A \over {x_4 }},{{y_5 } \over {x_5 }}} \right\} = max \left\{ {{{x_{ - 1} } \over {y_{ - 1} }},{{x_{ - 1} } \over {y_{ - 1} }}} \right\} = {{x_{ - 1} } \over {y_{ - 1} }};\quad y_6 = max \left\{ {{A \over {y_4 }},{{x_5 } \over {y_5 }}} \right\} = max \left\{ {{A \over {y_0 }},{{y_{ - 1} } \over {x_{ - 1} }}} \right\} = {A \over {y_0 }},x7=max{Ax5,y6x6}=max{x−1,Ay−1y0x−1}=x−1; y7=max{Ay5,x6y6}=max{y−1,y0x−1Ay−1}=y−1,x_7 = max \left\{ {{A \over {x_5 }},{{y_6 } \over {x_6 }}} \right\} = max \left\{ {x_{ - 1} ,{{Ay_{ - 1} } \over {y_0 x_{ - 1} }}} \right\} = x_{ - 1} ;\quad y_7 = max \left\{ {{A \over {y_5 }},{{x_6 } \over {y_6 }}} \right\} = max \left\{ {y_{ - 1} ,{{y_0 x_{ - 1} } \over {Ay_{ - 1} }}} \right\} = y_{ - 1} ,x8=max{Ax6,y7x7}=max{Ay−1x−1,y−1x−1}=Ay−1x−1; y8=max{Ay6,x7y7}=max{y−1,x−1y−1}=y0,⋮\matrix{{x_8 = \max \left\{ {{A \over {x_6 }},{{y_7 } \over {x_7 }}} \right\} = \max \left\{ {{{Ay_{ - 1} } \over {x_{ - 1} }},{{y_{ - 1} } \over {x_{ - 1} }}} \right\} = {{Ay_{ - 1} } \over {x_{ - 1} }};\quad y_8 = \max \left\{ {{A \over {y_6 }},{{x_7 } \over {y_7 }}} \right\} = \max \left\{ {y_{ - 1} ,{{x_{ - 1} } \over {y_{ - 1} }}} \right\} = y_0 ,} \cr \vdots}

Thus, xn={Ax−1,x−1y−1,x−1,Ay−1x−1,...},yn={Ay−1,Ay0,y−1,y0,...},\matrix{ {x_n = \left\{ {{A \over {x_{ - 1} }},{{x_{ - 1} } \over {y_{ - 1} }},x_{ - 1} ,{{Ay_{ - 1} } \over {x_{ - 1} }},...} \right\},} \cr {y_n = \left\{ {{A \over {y_{ - 1} }},{A \over {y_0 }},y_{ - 1} ,y_0 ,...} \right\},}} the solutions are shown to be 4-peirod

Let (x_n, y_n) be a solution of 1 for

A > x₀ > y₀ > y₋₁ > x₋₁,A > y₀ > x₀ > y₋₁ > x₋₁,A > y₀ > y₋₁ > x₀ > x₋₁

Then for n = 0, 1,... we have, xn={Ax−1,Ax0,x−1,x0,...},yn={Ay−1,y−1x−1,y−1,Ax−1y−1,...}.\matrix{ {x_n = \left\{ {{A \over {x_{ - 1} }},{A \over {x_0 }},x_{ - 1} ,x_0 ,...} \right\},} \cr {y_n = \left\{ {{A \over {y_{ - 1} }},{{y_{ - 1} } \over {x_{ - 1} }},y_{ - 1} ,{{Ax_{ - 1} } \over {y_{ - 1} }},...} \right\}. }}

Proof of the Theorem 5 can be obtain similar way to the Theorem 4

Let (x_n, y_n) be a solution of 1 for

A > x₋₁ > y₋₁ > x₀ > y₀,A > y₋₁ > x₋₁ > x₀ > y₀,A > y₋₁ > x₀ > x₋₁ > y₀

Then for n = 0, 1,... we have, xn={Ax−1,Ax0,x−1,x0,...},yn={x0y0,Ay0,Ay0x0,y0,...}.\matrix { {x_n = \left\{ {{A \over {x_{ - 1} }},{A \over {x_0 }},x_{ - 1} ,x_0 ,...} \right\},} \cr {y_n = \left\{ {{{x_0 } \over {y_0 }},{A \over {y_0 }},{{Ay_0 } \over {x_0 }},y_0 ,...} \right\}. }}

Proof of the Theorem 6 can be obtain similar way to the Theorem 4

If the initial conditions are selected follows for Lemma 1 A > x₀ > x₋₁ > y₀ > y₋₁: A = 36;x[−1] = 25;x[0] = 30;y[−1] = 15;y[0] = 20;

The graph of the solution is given below:

x_n = {1.44, 1.66667, 25., 21.6, 1.44, 1.66667, 25., 21.6, 1.44, 1.66667, 25., 21.6, 1.44, 1.66667, 25., 21.6, 1.44, 1.66667, 25., 21.6, 1.44, 1.66667, 25., 21.6, 1.44, 1.66667, 25., 21.6, 1.44, 1.66667, 25., 21.6, 1.44, 1.66667, 25., 21.6, 1.44, 1.66667, 25., 21.6, 1.44, 1.66667, 25., 21.6, 1.44, 1.66667, 25., 21.6, 1.44, 1.66667, 25., 21.6, 1.44, 1.66667, 25., 21.6, 1.44, 1.66667, 25., 21.6, 1.44,...}.