Re: Functional differentiation on lattice

*To*: mathgroup at smc.vnet.net*Subject*: [mg39777] Re: Functional differentiation on lattice*From*: Jens-Peer Kuska <kuska at informatik.uni-leipzig.de>*Date*: Fri, 7 Mar 2003 03:30:53 -0500 (EST)*Organization*: Universitaet Leipzig*References*: <b46uaf$92m$1@smc.vnet.net>*Reply-to*: kuska at informatik.uni-leipzig.de*Sender*: owner-wri-mathgroup at wolfram.com

Hi, S[A_] := Sum[(A[x + 1] - A[x])^2, {x, -Infinity, Infinity}] Unprotect[D]; D[Sum[b_, {x_, -Infinity, Infinity}], a[q_]] /; ! FreeQ[b, a] := Module[{forms}, forms = Cases[b, a[_], Infinity] /. a[r_] :> r; forms = Solve[q == #, x][[1]] & /@ forms; Plus @@ (D[#, a[q]] & /@ (b /. forms)) ] Protect[D]; and D[S[a], a[y]] // Simplify gives -2 (a[-1 + y] - 2 a[y] + a[1 + y]) Regards Jens Norbert Nemec wrote: > > Hi there, > > I've just recently decided that the maths I have to do at the moment really > demand the use of a CAS. I'm absolutely new to Mathematica, but the problem > I have is probably a bit hard to get moving on. Perhaps someone can give me > a simple solution to start out on? > > What I need to do could probably be called a "Functional differentiation on > a lattice". To give one very simply example: > > I have the functional > > S[A] := sum_x (A(x+1)-A(x))^2 > > (where x is integer - in my case there are periodic boundaries, but that > does not matter at that point) > > now I want to calculate > > dS/dA(y) > > which should result in > > - 2(A(y+1)-A(y)) + 2(A(y)-A(y-1)) > > or simplified > > -2A(y+1) + 4A(y) - 2A(y-1) > > Lateron, the whole thing will get 4-dimensional and A will get indices that > will be summed over as well. > > Is there a simple way to do that in Mathematica? I would really appreciate a > piece of code to get me started on. > > Thanks, > Nobbi