[firedrake] .dx without operand

[Christian Jacobs]

As Martin pointed out in a previous thread, .dx(*((0,)*k)) would do the job
for expressing d^ku/dx^k, but I can't hard code something like this in as I
don't know what the individual terms would look like in advance for
arbitrary k.

On 29 June 2015 at 19:58, Andrew McRae <a.mcrae12 at imperial.ac.uk> wrote:

> I don't *think* so (I might be wrong -- Martin?), but what about
> .dx(0).dx(0)?
>
> On 29 June 2015 at 19:54, Jacobs, Christian T <c.jacobs10 at imperial.ac.uk>
> wrote:
>
>>  I'm trying to express the following in UFL:
>> http://amcg.ese.ic.ac.uk/~ctj10/images/k.png where A and B are matrices
>> and u is a vector of solution variables.
>>
>> For the case of k=1 I can write e.g.
>>
>> -1*([1*u[0].dx(0) + 2*u[1].dx(0) + 5*u[0].dx(1) + 6*u[1].dx(1),
>> 3*u[0].dx(0) + 4*u[1].dx(0) + 7*u[0].dx(1) + 8*u[0].dx(0)])
>>
>> But how can I write this for k>1? Having e.g. (1*u[0].dx(0) +
>> 2*u[0].dx(1) + ...)**2 will result in terms that represent (du/dx)**2 not
>> (d**2)u/(dx**2). Is there a way of expressing .dx() (of arbitrary order)
>> without an operand in UFL, so we can just have d^k/dx^k ?
>>
>
>
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