I want to compute the Frechet derivative for a vector field, and that code works perfectly for functions with two variables x and t.
var ("t x")
v = function ("v")
u = function ("u")
w1 = function ("w1")
w2 = function ("w2")
eqsys = [diff(v(x,t), x) - u(x,t), diff(v(x,t), t) - diff(u(x,t), x)/(u(x,t)**2)]
def FrechetD (support, dependVar, independVar, testfunction):
frechet = []
eps = var ("eps")
v = independVar[:] + [eps]
for j in range (len(support)):
deriv = []
for i in range (len(support)):
r0 (x, t, eps) = dependVar[i](*independVar)+ testfunction[i](*independVar) * eps
s = support[j].substitute_function (dependVar[i], r0)
deriv.append (diff(s, eps).subs ({eps: 0}))
frechet.append (deriv)
return frechet
FrechetD (eqsys, [u,v], [x,t], [w1,w2])
[[-w1(x, t), diff(w2(x, t), x)],
[2*w1(x, t)*diff(u(x, t), x)/u(x, t)^3 - diff(w1(x, t), x)/u(x, t)^2,
diff(w2(x, t), t)]]
but my problem is the line
r0 (x, t, eps) = dependVar[i](*independVar)+ testfunction[i](*independVar) * eps
because this depends on the hardcoded x and t. Without it the following derivation for eps always is 0. Even when I add someting like
_r0 = function('ro')(*v)
doesn't work. What i want to do is to something like
r0 (*v) = dependVar[i](*independVar)+ testfunction[i](*independVar) * eps
to get rid of the hardcoded variables and/or the number of variables. Is that possible ?
