[firedrake] .dx without operand

[Andrew McRae]

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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