I'm not sure if there is a general algorithm to solve, the "real" tensor
equation, e.g., like
(g([i,n],[])+S([])*levi_civita([i,m,n],[])*k([],[m]))*u([],[n])=rhs([i],[])
to solve about vector u, where g is metric S - arbitrary scalar and k -
arbitrary vector. If g is an Euclidean like metric (Kronneker symbol) then the
inverse operator will be
(g([i,n],[])-S([])*levi_civita([i,m,n],[])*k([],[m]))/(1+S([])^2)
Situation become already more complicated if we multiple the orginal
operator look like this
(S([])*delta([i,n],[])+O([],[j])*k([j],[])*levi_civita([i,m,n],[])*k([],[m]))
The inverse operator will look similar as before but with complicatier
denominator. Such things appears in analytical work on the
magnetohydrodinamical turbulence.
If somebody on the list aware about the algorithms to solve tensorial
equation, it would be intersting to hear about
Thanks
Valery
> Although solve() is not really designed to deal with itensor expressions,
> it can be used, so long as you avoid reusing the same dummy index in your
> equation. I.e., try
>
> solve(X([],[i])*F([i,j],[])*X([],[j]) + a*
> X([],[u])*F([u,v],[])*Y([],[v]),a );
>
> (Note also that I corrected the position of the index in the second
> occurrence of X in the numerator and in Y in the denominator; as you
> originally wrote it, the same index occurred twice in a covariant position,
> which is not correct.)
>
>
> Viktor
>
>
>
>
> -----Original Message-----
> From: maxima-bounces at math.utexas.edu
> [mailto:maxima-bounces at math.utexas.edu] On Behalf Of Niitsuma Hirotaka
> Sent: Thursday, September 02, 2010 12:50 AM
> To: maxima at math.utexas.edu
> Subject: How to solve equation using itensor
>
> solve(x.F.x+a*x.F.y,a)
> /*=> a=x.F.x/x.F.y*/
>
> can write as the following using itensor
>
> solve(X([],[i])*F([i,j],[])*X([j],[]) + a*
> X([],[i])*F([i,j],[])*Y([j],[]),a )
> /*=> a=X([j], [])/ Y([j], [])*/
>
> But, result is not same.
>
> How to solve equations which use itensor?
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