2 Main Results
In some cases of parameter A and initial conditions, the solution of the system of max type difference equation has been studied. Let
x ¯ \overline x
and
y ¯ \overline y
be the unique positive equilibrium of 1 , then clearly,
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x ¯ = max { A x ¯ , y ¯ x ¯ } ; y ¯ = max { A y ¯ , x ¯ y ¯ } . \overline x = max \left\{ {{A \over {\overline x }},{{\overline y } \over {\overline x }}} \right\};\overline y = max \left\{ {{A \over {\overline y }},{{\overline x } \over {\overline y }}} \right\}.
The parameter A is the greatest value in all initial conditions that we select, so
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x ¯ = A x ¯ ⇒ x ¯ 2 = A ⇒ x ¯ = ± A ; y ¯ = A y ¯ ⇒ y ¯ 2 = A ⇒ y ¯ = ± A , \overline x = {A \over {\overline x }} \Rightarrow \overline x ^2 = A \Rightarrow \overline x = \pm \sqrt A ;\quad \overline y = {A \over {\overline y }} \Rightarrow \overline y ^2 = A \Rightarrow \overline y = \pm \sqrt A ,
we can obtain
x ¯ = A \overline x = \sqrt A
and
y ¯ = A \overline y = \sqrt A
.
Lemma 1
Assume that, A and x 0 ,x −1 ,y 0 ,y −1 are positive integer sequence for 1
A > x 0 > x −1 > y 0 > y −1 ,A > x 0 > y 0 > x −1 > y −1 ,A > y 0 > x 0 > x −1 > y −1 ,
Then the following statements are true:
n ≥ 0 for xn and n ≥ 1 for yn a) Every positive semi-cycle consist two term.
b) Every negative semi-cycle consist two term.
c) Every positive semi-cycle of length two is followed by a negative semi-cycle of length two.
d) Every negative semi-cycle of length two is followed by a positive semi-cycle of length two.
Proof
A > x 0 > x −1 > y 0 > y −1 ,A > x 0 > y 0 > x −1 > y −1 ,A > y 0 > x 0 > x −1 > y −1 The solution xn and yn can be obtained as follows:
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x 1 = max { A x − 1 , y 0 x 0 } = A x − 1 < x ¯ ; y 1 = max { A y − 1 , x 0 y 0 } = A y − 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 , ![]()
x 2 = max { A x 0 , y 1 x 1 } = max { A x 0 , x − 1 y − 1 } = x − 1 y − 1 < x ¯ ; y 2 = max { A y 0 , x 1 y 1 } = max { A y 0 , y − 1 x − 1 } = A y 0 < 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 , ![]()
x 3 = max { A x 1 , y 2 x 2 } = max { x − 1 , Ay − 1 x − 1 y 0 } = x − 1 > x ¯ ; y 3 = max { A y 1 , x 2 y 2 } = max { y − 1 , x − 1 y 0 Ay − 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 , ![]()
x 4 = max { A x 2 , y 3 x 3 } = max { Ay − 1 x − 1 , y − 1 x − 1 } = Ay − 1 x − 1 > x ¯ ; y 4 = max { A y 2 , x 3 y 3 } = max { y 0 , x − 1 y − 1 } = y 0 > 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 , ![]()
x 5 = max { A x 3 , y 4 x 4 } = max { A x − 1 , x − 1 y 0 Ay − 1 } = A x − 1 < x ¯ ; y 5 = max { A y 3 , x 4 y 4 } = max { A y − 1 , Ay − 1 x − 1 y 0 } = A y − 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 , ![]()
x 6 = max { A x 4 , y 5 x 5 } = max { x − 1 y − 1 , x − 1 y − 1 } = x − 1 y − 1 < x ¯ ; y 6 = max { A y 4 , x 5 y 5 } = max { A y 0 , y − 1 x − 1 } = A y 0 < 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 , ![]()
x 7 = max { A x 5 , y 6 x 6 } = max { x − 1 , Ay − 1 y 0 x − 1 } = x − 1 > x ¯ ; y 7 = max { A y 5 , x 6 y 6 } = max { y − 1 , y 0 x − 1 Ay − 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 , ![]()
x 8 = max { A x 6 , y 7 x 7 } = max { Ay − 1 x − 1 , y − 1 x − 1 } = Ay − 1 x − 1 > x ¯ ; y 8 = max { A y 6 , x 7 y 7 } = max { y − 1 , x − 1 y − 1 } = y 0 > 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.
x 1 < x ¯ , x 2 < x ¯ , x 3 > x ¯ , x 4 > x ¯ , x 5 < x ¯ , x 6 < x ¯ , x 7 > x ¯ , x 8 > 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 ,...
y 1 > y ¯ , y 2 < y ¯ , y 3 < y ¯ , y 4 > y ¯ , y 5 > y ¯ , y 6 < y ¯ , y 7 < y ¯ , y 8 > 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 xn and n ≥ 1 for yn , every positive semi-cycle consists of two terms, every negative semi-cycle consists of two terms
Lemma 2
Assume that, A and x 0 ,x −1 ,y 0 ,y −1 are positive integer sequence for 1
A > x 0 > y 0 > y −1 > x −1 ,A > y 0 > x 0 > y −1 > x −1 ,A > y 0 > y −1 > x 0 > x −1 ,
Then the following statements are true:
n ≥ 1 for xn and n ≥ 0 for yn a) Every positive semi-cycle consist two term.
b) Every negative semi-cycle consist two term.
c) Every positive semi-cycle of length two is followed by a negative semi-cycle of length two.
d) Every negative semi-cycle of length two is followed by a positive semi-cycle of length two.
Proof
Lemma 2 proof’s can be obtained similarly Lemma 1
Lemma 3
Assume that, A and x 0 ,x −1 ,y 0 ,y −1 are positive integer sequence for 1
A > x −1 > y −1 > x 0 > y 0 ,A > y −1 > x −1 > x 0 > y 0 ,A > y −1 > x 0 > x −1 > y 0 ,
Then the following statements are true:
n ≥ 0 for xn and n ≥ 1 for yn a) Every positive semi-cycle consist two term.
b) Every negative semi-cycle consist two term.
c) Every positive semi-cycle of length two is followed by a negative semi-cycle of length two.
d) Every negative semi-cycle of length two is followed by a positive semi-cycle of length two.
Proof
Lemma 3 proof’s can be obtained similarly Lemma 1
Theorem 4
Let (xn , yn ) be a solution of 1 for
A > x 0 > x −1 > y 0 > y −1 ,A > x 0 > y 0 > x −1 > y −1 ,A > y 0 > x 0 > x −1 > y −1 .
Then for n = 0, 1,... we have ,
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x n = { A x − 1 , x − 1 y − 1 , x − 1 , Ay − 1 x − 1 , ... } , y n = { A y − 1 , A y 0 , y − 1 , y 0 , ... } . \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\}.}}
Proof
We obtain,
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x 1 = max { A x − 1 , y 0 x 0 } = A x − 1 ; y 1 = max { A y − 1 , x 0 y 0 } = A y − 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} }},
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x 2 = max { A x 0 , y 1 x 1 } = max { A x 0 , x − 1 y − 1 } = x − 1 y − 1 ; y 2 = max { A y 0 , x 1 y 1 } = max { A y 0 , y − 1 x − 1 } = A y 0 , 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 }}, ![]()
x 3 = max { A x 1 , y 2 x 2 } = max { x − 1 , Ay − 1 x − 1 y 0 } = x − 1 ; y 3 = max { A y 1 , x 2 y 2 } = max { y − 1 , x − 1 y 0 Ay − 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} , ![]()
x 4 = max { A x 2 , y 3 x 3 } = max { Ay − 1 x − 1 , y − 1 x − 1 } = Ay − 1 x − 1 ; y 4 = max { A y 2 , x 3 y 3 } = max { y 0 , x − 1 y − 1 } = y 0 , 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 , ![]()
x 5 = max { A x 3 , y 4 x 4 } = max { A x − 1 , x − 1 y 0 Ay − 1 } = A x − 1 ; y 5 = max { A y 3 , x 4 y 4 } = max { A y − 1 , Ay − 1 x − 1 y 0 } = A y − 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} }}, ![]()
x 6 = max { A x 4 , y 5 x 5 } = max { x − 1 y − 1 , x − 1 y − 1 } = x − 1 y − 1 ; y 6 = max { A y 4 , x 5 y 5 } = max { A y 0 , y − 1 x − 1 } = A y 0 , 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 }}, ![]()
x 7 = max { A x 5 , y 6 x 6 } = max { x − 1 , Ay − 1 y 0 x − 1 } = x − 1 ; y 7 = max { A y 5 , x 6 y 6 } = max { y − 1 , y 0 x − 1 Ay − 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} , ![]()
x 8 = max { A x 6 , y 7 x 7 } = max { Ay − 1 x − 1 , y − 1 x − 1 } = Ay − 1 x − 1 ; y 8 = max { A y 6 , x 7 y 7 } = max { y − 1 , x − 1 y − 1 } = y 0 , ⋮ \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,
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x n = { A x − 1 , x − 1 y − 1 , x − 1 , Ay − 1 x − 1 , ... } , y n = { A y − 1 , A y 0 , y − 1 , y 0 , ... } , \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
Theorem 5
Let (xn , yn ) be a solution of 1 for
A > x 0 > y 0 > y −1 > x −1 ,A > y 0 > x 0 > y −1 > x −1 ,A > y 0 > y −1 > x 0 > x −1
Then for n = 0, 1,... we have ,
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x n = { A x − 1 , A x 0 , x − 1 , x 0 , ... } , y n = { A y − 1 , y − 1 x − 1 , y − 1 , Ax − 1 y − 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
Proof of the Theorem 5 can be obtain similar way to the Theorem 4
Theorem 6
Let (xn , yn ) be a solution of 1 for
A > x −1 > y −1 > x 0 > y 0 ,A > y −1 > x −1 > x 0 > y 0 ,A > y −1 > x 0 > x −1 > y 0
Then for n = 0, 1,... we have ,
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x n = { A x − 1 , A x 0 , x − 1 , x 0 , ... } , y n = { x 0 y 0 , A y 0 , Ay 0 x 0 , y 0 , ... } . \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
Proof of the Theorem 6 can be obtain similar way to the Theorem 4
Example 7
If the initial conditions are selected follows for Lemma 1 A > x 0 > x −1 > y 0 > y −1 : A = 36;x [−1] = 25;x [0] = 30;y [−1] = 15;y [0] = 20;
The graph of the solution is given below: xn = {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,...}.
yn = {2.4, 1.8, 15., 20., 2.4, 1.8, 15., 20., 2.4, 1.8, 15., 20., 2.4, 1.8, 15., 20., 2.4, 1.8, 15., 20., 2.4, 1.8, 15., 20., 2.4, 1.8, 15., 20., 2.4, 1.8, 15., 20., 2.4, 1.8, 15., 20., 2.4, 1.8, 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, 15., 20., 2.4, 1.8, 15., 20., 2.4, 1.8, 15., 20., 2.4, 1.8, 15., 20., 2.4, 1.8, 15., 20., 2.4, 1.8, 15., 20., 2.4,...}.
Fig. 1 xn graph solution.
Fig. 2 yn graph solution.