ufl team mailing list archive

[Kristian Oelgaard <k.b.oelgaard@xxxxxxxxx>]

On 6 July 2010 15:32, Patrick Riesen <priesen@xxxxxxxxxxxxxxx> wrote:
> Kristian Oelgaard wrote:
>>
>> On 6 July 2010 14:52, Patrick Riesen <priesen@xxxxxxxxxxxxxxx> wrote:
>>>
>>> Kristian Oelgaard wrote:
>>>>
>>>> On 6 July 2010 14:07, Patrick Riesen <priesen@xxxxxxxxxxxxxxx> wrote:
>>>>>
>>>>> Garth N. Wells wrote:
>>>>>>
>>>>>> On 06/07/10 13:25, Patrick Riesen wrote:
>>>>>>>
>>>>>>> hello,
>>>>>>>
>>>>>>> i want to take the gradient of a coefficient with respect to some
>>>>>>> other
>>>>>>> quantity than the spatial coordinates (i.e. some other coefficient).
>>>>>>>
>>>>>>> i tried to construct a custom operator derived from diff() and the
>>>>>>> SpatialCoordinate/Derivative classes in ufl but after some point
>>>>>>> beyond
>>>>>>> my knowledge it always gets evaluated to zero by FFC then stops
>>>>>>> compiling with FFC : division by zero!
>>>>>>>
>>>>>>> does somebody of you experts know an approach to this?
>>>>>>>
>>>>>> Did you try using 'variable'?
>>>>>>
>>>>>> � �. . .
>>>>>> � �v �= variable(v)
>>>>>> � �f �= v*v
>>>>>> � �df = diff(f, v)
>>>>>
>>>>> yes, but if f is a coefficient and not an expression of v, this won't
>>>>> work,
>>>>> i.e.,
>>>>>
>>>>> f = Coefficient(element)
>>>>> f = variable(f)
>>>>> v = SpatialCoordinates(cell)
>>>>> v = variable(v)
>>>>> df = diff(f, v)
>>>>>
>>>>> will end up as Type Zero, although this is equivalent to grad(f).
>>>>> that's
>>>>> the
>>>>> one case, and it v = Coefficient(element) as well, it will also be
>>>>> zero.
>>>>> if would be nice if where not zero :-)
>>>>
>>>> The first example I get although I don't know why it does not work at
>>>> the moment, but it could be a bug. The second case where v is a
>>>> coefficient I don't get.
>>>> If f(x) = x**2, then df/dy = 0, I think that makes sense.
>>>
>>> yes, but if f is a coefficient, i.e., it is a eulerian quantity f=f(x),
>>> and
>>> df/dx is the spatial gradient, ufl computes this from either grad(f) or
>>> f.dx(i) or Dx(f,i).
>>>
>>> but the vector of spatial coordinates x is
>>> hidden in ufl. i would like to do something like f.dx(v[i]), i.e. compute
>>> df_i/dv_j where v represents some other coordinates, possibly given by
>>> another coefficient?
>>
>> My bad for choosing 'x' in my example. The point I was trying to make
>> is simply that if UFL does not know that a given expression depend on
>> a variable, then taking derivatives w.r.t. this variable will always
>> turn out zero:
>>
>> f(v) = 5*v --> d_f/d_foo = 0.
>>
>> Did you take a look at the blueprint I linked to?
>
> yes, if it got it correctly  i would define
>
> dfdX = Function(element)
> f = Function(element, gradient=dfdX)
>
> and dfdX must come from another form, where i have other spatial coordinates
> (say X, ).

Yes, something like that. Note that it is a blueprint and AFAIK nobody
is about to implement it in the near future.

Kristian

> i don't know how x (the spatial coordinates) are represented in ufl, but  if
> they are somehow indexed and acessible, one could define a new Grad(f, X[i])
> / f.dx(X[i]) , where the partial derivative of f is computed with respect to
> component X[i] instead of spatial variable component i as in f.dx(i).
>
> patrick
>
>>
>> Kristian
>>
>>> Perhaps what
>>>>
>>>> you want (or is trying to do) is described in this blueprint?
>>>>
>>>> https://blueprints.launchpad.net/ufl/+spec/function-derivatoves
>>>>
>>>> Kristian
>>>>
>>>>> patrick
>>>>>
>>>>>> Garth
>>>>>>
>>>>>>> many thanks and best regards,
>>>>>>> patrick
>>>>>>>
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>
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