ufl team mailing list archive

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

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?

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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