Solution of the Rational Difference Equation xn+1=xn131+xn1xn3xn5xn7xn9xn11{x_{n + 1}} = {{{x_{n - 13}}} \over {1 + {x_{n - 1}}{x_{n - 3}}{x_{n - 5}}{x_{n - 7}}{x_{n - 9}}{x_{n - 11}}}}

Dagistan Simsek 1 , 2 , Burak Ogul 1  and Fahreddin Abdullayev 1 , 3
  • 1 Kyrgyz Turkish Manas University, Bishkek, Kyrgyzstan
  • 2 Konya Technical University, Konya, Turkey
  • 3 Mersin University, Mersin, Turkey
Dagistan Simsek
  • Kyrgyz Turkish Manas University, Bishkek, Kyrgyzstan
  • Konya Technical University, Konya, Turkey
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, Burak Ogul and Fahreddin Abdullayev
  • Kyrgyz Turkish Manas University, Bishkek, Kyrgyzstan
  • Mersin University, Mersin, Turkey
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Abstract

In this paper, solution of the following difference equation is examined

xn+1=xn131+xn1xn3xn5xn7xn9xn11,
where the initial conditions are positive real numbers.

1 Introduction

Difference equations appear naturally as discrete analogs and as numerical solutions of differential and delay differential equations, having applications in biology, ecology, physics.

Recently, a high attention to studying the periodic nature of nonlinear difference equations has been attracted. For some recent results concerning the periodic nature of scalar nonlinear difference equations, among other problems, see, for example, [1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,17,18,19,20,21,22,23,24,25,26,27,28,29,30,31,32,33].

Amleh [1], studied the global stability, the boundedness character, and the periodic nature of the positive solutions of the difference equation:

xn+1=α+xn1xn
where α ∈ [0, ∞) and where the initial conditions x−1 and x0 are arbitrary positive real numbers.

De Vault [8], studied the following problems

xn+1=Axn+1xn2,n=0,1,,
and showed every positive solution of the equation where A ∈ (0, ∞).

Elsayed [15], studied the global result, boundedness, and periodicity of solutions of the difference equation

xn+1=a+bxn1+cxnkdxn1+exnk,n=0,1,,
where the parameters a, b, c, d and e are positive real numbers and the initial conditions xt,xt+1,...,x0 are positive real numbers where t = max{l,k}, lk.

Ibrahim [18], studied the solutions of non-linear difference equation

xn+1=xnxn2xn1(a+bxnxn2),n=0,1,,
where the initial values x0,x−1 and x−2 non-negative real numbers with bx0x−2 ≠ −a and x−1 ≠ 0. He investigated some properties for this difference equation such as the local stability and the boundedness for the solutions.

Simsek et. al. [25,26,27] and [30], studied the following problems with positive initial values

xn+1=xn31+xn1,n=0,1,,xn+1=xn51+xn2,n=0,1,,xn+1=xn51+xn1xn3,n=0,1,,xn+1=xn31+xnxn1xn2,n=0,1,,
respectively.

In this work, the following non-linear difference equation is studied

xn+1=xn131+xn1xn3xn5xn7xn9xn11,n=0,1,,
where x−13,x−12,...,x−1,x0 ∈ (0, ∞) is investigated.

2 Main Result

Let x¯ be the unique positive equilibrium of the equation 1, then clearly,

x¯=x¯1+x¯.xxxxx¯x¯+x¯7=x¯x¯7=0x¯0,
so, x¯=0 can be obtained. For any k ≥ 0 and m > k, notation i=k,m¯ means i = k,k + 1,...,m.

Theorem 1

Consider the difference equation 1. Then the following statements are true:

  1. a)The sequences (x14n−13), (x14n−12),...(x14n−1), (x14n) are decreased and a1,..., a14 ≥ 0 is existed in such that:
    limnx14n13+k=a1+k,k=0,13¯.
  2. b)(a1,a2,a3,a4,a5,a6,a7,a8,a9,a10,a11,a12,a13,a14,...) is a solution of 1 having period fourteen.
  3. c)
    k=06limnx14nj+2k=0,j=0,1¯;
    or
    k=06a2k+i=0,i=0,1¯.
  4. d)If there exist n0 ∈ ℕ such that xn+1xn−11 for all nn0, then,
    limnxn=0.
  5. e)The following formulas below:
    x14n+k+1=x13+k(1x1+kx3+kx5+kx7+kx9+kx11+k1+x1+kx3+kx5+kx7+kx9+kx11+k×j=0ni=17j11+x2i11+kx2i9+kx2i7+kx2i5+kx2i3+kx2i1+k),
    x14n+k+3=x11+k(1x1+kx3+kx5+kx7+kx9+kx13+k1+x1+kx3+kx5+kx7+kx9+kx11+k×j=0ni=17j+111+x2i11+kx2i9+kx2i7+kx2i5+kx2i3+kx2i1+k),
    x14n+k+5=x9+k(1x1+kx3+kx5+kx7+kx11+kx13+k1+x1+kx3+kx5+kx7+kx9+kx11+k×j=0ni=17j+211+x2i11+kx2i9+kx2i7+kx2i5+kx2i3+kx2i1+k),
    x14n+k+7=x7+k(1x1+kx3+kx5+kx9+kx11+kx13+k1+x1+kx3+kx5+kx7+kx9+kx11+k×j=0ni=17j+311+x2i11+kx2i9+kx2i7+kx2i5+kx2i3+kx2i1+k),
    x14n+k+9=x5+k(1x1+kx3+kx7+kx9+kx11+kx13+k1+x1+kx3+kx5+kx7+kx9+kx11+k×j=0ni=17j+411+x2i11+kx2i9+kx2i7+kx2i5+kx2i3+kx2i1+k),
    x14n+k+11=x3+k(1x1+kx5+kx7+kx9+kx11+kx13+k1+x1+kx3+kx5+kx7+kx9+kx11+k×j=0ni=17j+511+x2i11+kx2i9+kx2i7+kx2i5+kx2i3+kx2i1+k),
    x14n+k+13=x1+k(1x3+kx5+kx7+kx9+kx11+kx13+k1+x1+kx3+kx5+kx7+kx9+kx11+k×j=0ni=17j+611+x2i11+kx2i9+kx2i7+kx2i5+kx2i3+kx2i1+k),
    k=0,1¯ , holds.
  6. f)If x14n+1+kak+1 ≠ 0, x14n+3+kak+3 ≠ 0, x14n+5+kak+5 ≠ 0, x14n+7+kak+7 ≠ 0, x14n+9+kak+9 ≠ 0, x14n+11+kak+11 ≠ 0 then x14n+13+k → 0 as n → ∞, k=0,1¯ .

Proof

  1. a)Firstly, from 1,
    xn+1(1+xn1xn3xn5xn7xn9xn11)=xn13,
    is obtained. If xn−1xn−3xn−5xn−7xn−9xn−11 ∈ (0, +∞), then 1 + xn−1xn−3xn−5xn−7xn−9xn−11 ∈ (1, +∞).
    Since
    xn+1<xn13,
    n ∈ ℕ
    limnx14n13+k=a1+k,fork=0,13¯
    existed formulas are obtained.
  2. b)(a1,a2,a3,a4,a5,a6,a7,a8,a9,a10,a11,a12,a13,a14,...) is a solution of 1 having period fourteen.
  3. c)In view of the 1,
    n=14nx14n+1=x14n131+k=05limnx14n11+2k,
    is obtained. If the limits are put on both sides of the above equality,
    k=06limnx14n13+2k=0ork=06a2k+1=0,
    is obtained. Similarly for n = 14n + 1,
    n=14n+1x14n+2=x14n121+k=05limnx14n10+2k,
    is obtained. If the limits are put on both sides of the above equality,
    k=06limnx14n12+2k=0ork=06a2k+2=0,
    is obtained.
  4. d)If there exist n0 ∈ ℕ such that xn+1xn−11 for all nn0, then, a1a3a5a7a9a11a13a1, a2a4a6a8a10a12a14a2. Using (c) we get
    k=06a2k+i=0,i=1,2¯.
    Then we see that,
    limnxn=0.
    Hence the proof of (d) completed.
  5. e)Subracting xn−13 from the left and right-hand sides in 1,
    xn+1xn13=11+xn1xn3xn5xn7xn9xn11(xn1xn15),
    is obtained and the following formula is produced below, for n ≥ 2
    x2n3x2n17=(x1x13)i=1n211+x2i1x2i3x2i5x2i7x2i9x2i11,x2n2x2n16=(x1x12)i=1n211+x2ix2i2x2i4x2i6x2i8x2i10,
    7 j inserted in 2 by replacing n, j = 0 to j = n is obtained by summing, for k=0,1¯ ,
    x14n+1+kx13+k=(x1+kx13+k)j=0ni=17j11+x2i1+kx2i3+kx2i5+kx2i7+kx2i9+kx2i11+k.
    Also, 7 j + 1 inserted in 2 by replacing n, j = 0 to j = n is obtained by summing, for k=0,1¯ ,
    x14n+3+kx11+k=(x3+kx11+k)j=0ni=17j+111+x2i1+kx2i3+kx2i5+kx2i7+kx2i9+kx2i11+k.
    Also, 7 j + 2 inserted in 2 by replacing n, j = 0 to j = n is obtained by summing, for k=0,1¯ ,
    x14n+5+kx9+k=(x5+kx9+k)j=0ni=17j+211+x2i1+kx2i3+kx2i5+kx2i7+kx2i9+kx2i11+k.
    Also, 7 j + 3 inserted in 2 by replacing n, j = 0 to j = n is obtained by summing, for k=0,1¯ ,
    x14n+7+kx7+k=(x7+kx7+k)j=0ni=17j+311+x2i1+kx2i3+kx2i5+kx2i7+kx2i9+kx2i11+k.
    Also, 7 j + 4 inserted in 2 by replacing n, j = 0 to j = n is obtained by summing, for k=0,1¯ ,
    x14n+9+kx5+k=(x9+kx5+k)j=0ni=17j+411+x2i1+kx2i3+kx2i5+kx2i7+kx2i9+kx2i11+k.
    Also, 7 j + 5 inserted in 2 by replacing n, j = 0 to j = n is obtained by summing, for k=0,1¯ ,
    x14n+11+kx3+k=(x11+kx3+k)j=0ni=17j+511+x2i1+kx2i3+kx2i5+kx2i7+kx2i9+kx2i11+k.
    Also, 7 j + 6 inserted in 2 by replacing n, j = 0 to j = n is obtained by summing, for k=0,1¯ ,
    x14n+13+kx1+k=(x13+kx1+k)j=0ni=17j+611+x2i1+kx2i3+kx2i5+kx2i7+kx2i9+kx2i11+k.
    Now we obtained of the above formulas:
    x14n+k+1=x13+k(1x1+kx3+kx5+kx7+kx9+kx11+k1+x1+kx3+kx5+kx7+kx9+kx11+k×j=0ni=17j11+x2i11+kx2i9+kx2i7+kx2i5+kx2i3+kx2i1+k),
    x14n+k+3=x11+k(1x1+kx3+kx5+kx7+kx9+kx13+k1+x1+kx3+kx5+kx7+kx9+kx11+k×j=0ni=17j+111+x2i11+kx2i9+kx2i7+kx2i5+kx2i3+kx2i1+k),
    x14n+k+5=x9+k(1x1+kx3+kx5+kx7+kx11+kx13+k1+x1+kx3+kx5+kx7+kx9+kx11+k×j=0ni=17j+211+x2i11+kx2i9+kx2i7+kx2i5+kx2i3+kx2i1+k),
    x14n+k+7=x7+k(1x1+kx3+kx5+kx9+kx11+kx13+k1+x1+kx3+kx5+kx7+kx9+kx11+k×j=0ni=17j+311+x2i11+kx2i9+kx2i7+kx2i5+kx2i3+kx2i1+k),
    x14n+k+9=x5+k(1x1+kx3+kx7+kx9+kx11+kx13+k1+x1+kx3+kx5+kx7+kx9+kx11+k×j=0ni=17j+411+x2i11+kx2i9+kx2i7+kx2i5+kx2i3+kx2i1+k),
    x14n+k+11=x3+k(1x1+kx5+kx7+kx9+kx11+kx13+k1+x1+kx3+kx5+kx7+kx9+kx11+k×j=0ni=17j+511+x2i11+kx2i9+kx2i7+kx2i5+kx2i3+kx2i1+k),
    x14n+k+13=x1+k(1x3+kx5+kx7+kx9+kx11+kx13+k1+x1+kx3+kx5+kx7+kx9+kx11+k×j=0ni=17j+611+x2i11+kx2i9+kx2i7+kx2i5+kx2i3+kx2i1+k).
  6. f)Suppose that a1 = a3 = a5 = a7 = a9 = a11 = a13 = 0. By (e), we have
    limnx14n+1=limnx13(1x1x3x5x7x9x111+x1x3x5x7x9x11j=0ni=17j11+x2i1x2i3x2i5x2i7x2i9x2i11),a1=x13(1x1x3x5x7x9x111+x1x3x5x7x9x11j=0ni=17j11+x2i1x2i3x2i5x2i7x2i9x2i11)a1=01+x1x3x5x7x9x11x1x3x5x7x9x11=j=0ni=17j11+x2i1x2i3x2i5x2i7x2i9x2i11.
    Similarly,
    a3=01+x1x3x5x7x9x11x1x3x5x7x9x13=j=0ni=17j+111+x2i1x2i3x2i5x2i7x2i9x2i11.
    Similarly,
    a5=01+x1x3x5x7x9x11x1x3x5x7x11x13=j=0ni=17j+211+x2i1x2i3x2i5x2i7x2i9x2i11.
    Similarly,
    a7=01+x1x3x5x7x9x11x1x3x5x9x11x13=j=0ni=17j+311+x2i1x2i3x2i5x2i7x2i9x2i11.
    Similarly,
    a9=01+x1x3x5x7x9x11x1x3x7x9x11x13=j=0ni=17j+411+x2i1x2i3x2i5x2i7x2i9x2i11.
    Similarly,
    a11=01+x1x3x5x7x9x11x1x5x7x9x11x13=j=0ni=17j+511+x2i1x2i3x2i5x2i7x2i9x2i11.
    Similarly,
    a13=01+x1x3x5x7x9x11x3x5x7x9x11x13=j=0ni=17j+611+x2i1x2i3x2i5x2i7x2i9x2i11.
    From 3 and 4,
    1+x1x3x5x7x9x11x1x3x5x7x9x11=j=0ni=17j11+x2i1x2i3x2i5x2i7x2i9x2i11>1+x1x3x5x7x9x11x1x3x5x7x9x13=j=0ni=17j+111+x2i1x2i3x2i5x2i7x2i9x2i11,
    thus, x−13 > x−11. From 4 and 5
    1+x1x3x5x7x9x11x1x3x5x7x9x13=j=0ni=17j+111+x2i1x2i3x2i5x2i7x2i9x2i11>1+x1x3x5x7x9x11x1x3x5x7x11x13=j=0ni=17j+211+x2i1x2i3x2i5x2i7x2i9x2i11,
    thus, x−11 > x−9. From 5 and 6,
    1+x1x3x5x7x9x11x1x3x5x7x11x13=j=0ni=17j+211+x2i1x2i3x2i5x2i7x2i9x2i11>1+x1x3x5x7x9x11x1x3x5x9x11x13=j=0ni=17j+311+x2i1x2i3x2i5x2i7x2i9x2i11,
    thus, x−9 > x−7. From 6 and 7,
    1+x1x3x5x7x9x11x1x3x5x9x11x13=j=0ni=17j+311+x2i1x2i3x2i5x2i7x2i9x2i11>1+x1x3x5x7x9x11x1x3x7x9x11x13=j=0ni=17j+411+x2i1x2i3x2i5x2i7x2i9x2i11
    thus, x−7 > x−5. From 7 and 8,
    1+x1x3x5x7x9x11x1x3x7x9x11x13=j=0ni=17j+411+x2i1x2i3x2i5x2i7x2i9x2i11>1+x1x3x5x7x9x11x1x5x7x9x11x13=j=0ni=17j+511+x2i1x2i3x2i5x2i7x2i9x2i11
    thus, x−5 > x−3. From 8 and 9,
    1+x1x3x5x7x9x11x1x5x7x9x11x13=j=0ni=17j+511+x2i1x2i3x2i5x2i7x2i9x2i11>1+x1x3x5x7x9x11x3x5x7x9x11x13=j=0ni=17j+611+x2i1x2i3x2i5x2i7x2i9x2i11,
    thus, x−3 > x−1. From here we obtain x−13 > x−11 > x−9 > x−7 > x−5 > x−3 > x−1. Similarly, we can obtain x−12 > x−10 > x−8 > x−6 > x−4 > x−2 > x0. We arrive at a contradiction which completes the proof of theorem.

Example 2

If the initial conditions are selected as follows:

x13=0.98,x12=0.97,x11=0.96,x10=0.95,x9=0.94,x8=0.93,x7=0.92,x6=0.91,x5=0.9,x4=0.89,x3=0.88,x2=0.87,x1=0.86,x0=0.85.

The graph of the solution is given below, xn = {0.62601, 0.634341, 0.701375, 0.701974, 0.737178, 0.734195, 0.753815, 0.748866, 0.759718, 0.753567, 0.759006, 0.752062, 0.753929, 0.746432, 0.535307, 0.545309, 0.621059, 0.622964, 0.664753, 0.66286, 0.687738, 0.683736, 0.698929, 0.693618, 0.702724, 0.696539, 0.701557, 0.694753, 0.487602, 0.498104, 0.576757, 0.579084, 0.623219, 0.621705, 0.64857, 0.644917, 0.661839, 0.656857, 0.667494,...}

Fig. 1
Fig. 1

xn graph of the solution.

Citation: Applied Mathematics and Nonlinear Sciences 5, 1; 10.2478/amns.2020.1.00047

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    Simsek D., Cinar C., Yalcinkaya I., (2006), On the recursive sequence x n + 1 = x n 3 1 + x n 1 {x_{n + 1}} = {{{x_{n - 3}}} \over {1 + {x_{n - 1}}}} , Internat. J. Contemp. 9(12), 475–480.

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    Simsek D., Cinar C., Yalcinkaya I., (2008), On the recursive sequence x n + 1 = x n ( 5 k + 9 ) 1 + x n 4 x n 9 x n ( 5 k + 4 ) {x_{n + 1}} = {{{x_{n - (5k + 9)}}} \over {1 + {x_{n - 4}}{x_{n - 9}}{x_{n - (5k + 4)}}}} Taiwanese Journal of Mathematics, 12(5), 1087–1098.

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    Simsek D., Ogul B., Abdullayev F., (2017), Solutions of the rational difference equations x n + 1 = x n 11 1 + x n 2 x n 5 x n 8 {x_{n + 1}} = {{{x_{n - 11}}} \over {1 + {x_{n - 2}}{x_{n - 5}}{x_{n - 8}}}} , AIP Conference Proceedings 1880(1) 040003.

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    Simsek D., Abdullayev F., (2017), On the Recursive Sequence x n + 1 = α x n 4 k + 3 1 + t = 0 2 x n ( k + 1 ) t k {x_{n + 1}} = {{\alpha {x_{n - 4k + 3}}} \over {1 + \prod\nolimits_{t = 0}^2 {x_{n - (k + 1)t - k}}}} , Journal of Mathematics Sciences, 222(6), 762–771.

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