Consider the following difference operator
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.
PS:
There seems to be a recurrence relation solver RSolve[]
, but it does not seem to be useful in the above problem.