A Novel Integrable Fourth-Order Difference Equation Admitting Three Invariants
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3 A Novel Example
Consider the following differential-difference equation:
∂x n
∂t
= (x n + 1)
x n+2 x n (x n+1 + 1) 2
x n+1
−
x n−2 x n (x n−1 + 1) 2
x n−1
+ (1 + 2x n )(x n+1 − x n−1 )
.
(12)
Equation (12) has been found in [2] and it is related to the discrete Sawada–Kotera
equation found in [1, 39]. Equation (12) emerged again in [15] where the authors
classified the fourth-order Volterra-like equations (1) linear in x n±2 . Imposing
∂x n /∂t ≡ 0 in (12) we obtain its stationary reduction:
x n+2 x n (x n+1 + 1) 2
x n+1
−
x n−2 x n (x n−1 + 1) 2
x n−1
= −(1 + 2x n )(x n+1 − x n−1 ).
(13)
Equation (13) is not resummable, in the sense of [3], to a second-order difference
equation, nor it is deflatable, in the sense of [26], to a third-order difference
equation.
In [19] it was heuristically shown that Eq. (12) has quadratic growth [10].
Since (13) is a reduction of (12) we have that it can have at most quadratic growth.
Computing the growth of degrees of equation (13) we obtain
1, 5, 15, 35, 67, 113, 167, 229, 297, 375, 463, 561, 667,
785, 911, 1047, 1193, 1349, 1511, 1681, 1859, 2051,
2255, 2469, 2689, 2917, 3151, 3395, 3651, 3921, 4199 . . .
(14)
whose generating function is
g (z) =
⎡
⎣
5z
13
+ 5z
12
+ 15z
11
+ 17z
10
+ 29z
9
+ 25z
8
+32z
7
+ 26z
6
+ 27z
5
+ 19z
4
+ 13z
3
+ 7z
2
+ 3z + 1
⎤
⎦
(1 − z) 3 (z + 1)(z 2 + 1)(z 4 + 1)(z 2 − z + 1)(z 2 + z + 1)
.
(15)
Since all the roots of the denominator of (15) lie on the unit circle we have, as
expected, that the algebraic entropy of equation (13) vanishes. Moreover, due to the
presence of the factor (1 − z)
3 we obtain that its growth is asymptotically quadratic
[14]. That is, Eq. (13) is integrable in the sense of algebraic entropy and it is not
expected to be linearizable, since its growth is quadratic.