dolfin team mailing list archive

[Anders Logg <logg@xxxxxxxxx>]

On Wed, Jun 11, 2008 at 08:11:56AM -0400, mspieg wrote:
> Hi all,
>   is there a simple way to implement a known constant vector (e.g. the unit
> vector jhat = [ 0 1]') into a FFC form?
> 
> my specific  issue is that the right hand side of my PDE looks like
> 
> -\Div f\jhat = - df/dy 
> 
> where f(x,y) is a user defined scalar function (and df/dy is not easily
> evaluated)
> 
> Given a test function v, the weak form (in pseudo-latex) is 
> 
> \int dv/dy f dx - \int v f \jhat\dot ds
> 
> The question is how to evaluate the 2nd term in the FFC form language (the
> volume integral works fine as  v.dx(1)*f*dx)
> 
> ideally something like
> 
> jhat=??
> L = v.dx(1)*f*dx) - v*f*jhat*ds
> 
> (or more generally for a known unit vector k  L=dot(grad(v),k)*f*dx - v*f*k*ds
> )
> 
> would be nice (in the same way as simple known constants can be built in
> directly)
> 
> But I'm flexible...all help greatly appreciated
> marc

It looks like you expect ds to be a vector. It's not. It's a scalar.

-- 
Anders

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