Consider the following difference operator

enter image description here

For n=2 this operator action is given by:

n = 2;
tup = Tuples[{1, -1}, n];
LHS = Sum[ Product[((1 - Subscript[u, 1] Subscript[x, i]^tup[[qq, i]]) (1 - Subscript[u, 2] Subscript[x, i]^tup[[qq, i]]))/(1 - Subscript[ x, i]^(2 tup[[qq, i]])), {i, 1, n}] Product[(1 - t Subscript[x, i]^tup[[qq, i]] Subscript[x, j]^tup[[qq, j]])/(1 - Subscript[x, i]^tup[[qq, i]] Subscript[x, j]^tup[[qq, j]]), {j, 2, n}, {i, 1, j - 1}] f[Subscript[x, 1] q^(tup[[qq, 1]]/2), Subscript[x, 2] q^(tup[[qq, 2]]/2)], {qq, 1, Length[tup]}]


Now I would like to use Mathematica to solve the difference equation

LHS==EE f[Subscript[x,1],Subscript[x,2]]

to find eigenfunctions f and eigenvalues EE. Is there a command in Mathematica that can do this? I looked online but could not find it.


There seems to be a recurrence relation solver RSolve[], but it does not seem to be useful in the above problem.


1 Answer 1


One simple solution can be obtained by assuming q == 1.

Simplify[LHS /. {q -> 1}]
(* f[Subscript[x, 1], Subscript[x, 2]] 
   (-1 + Subscript[u, 1] Subscript[u, 2]) (-1 + t Subscript[u, 1] Subscript[u, 2]) *)

So, the second line of this answer is EE in this case. This immediately suggests a more general result, namely the eigenfunction f a constant and

EE = (-1 + Subscript[u, 1] Subscript[u, 2]) (-1 + t Subscript[u, 1] Subscript[u, 2])

as may be verified by evaluating

Simplify[LHS /. f[_, _] -> c]

It is not obvious (to me, at least) whether other solutions exist. If they do, I am unaware of any single Mathematica function to find them. One might be able to make some progress, however, by expanding LHS as a power series in q about q == 1.

  • $\begingroup$ Yes, that is my plan as well (expanding in q->1 and effectively work with differential operators instead). I hoped there was a better way to automate the calculation for the full difference equation, but if it is not the case this will have to suffice. $\endgroup$
    – Kagaratsch
    Apr 6, 2016 at 13:20

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