That term is really the partial derivative of \phi with respect to the quantity
(x^2)? Strange. I don't think you have any choice but to do that piece with the
chain rule. After that, you'll get
(\frac{1}{2 x}\frac{\partial \phi}{\partial x})^2
but that's still not a form that FiPy can really do much to render implicitly.
You can rearrange a bit to get part of it as a DiffusionTerm, but you'll still
be left with some explicit pieces, so you should probably do that whole piece
explicitly, e.g.,
tmp = (1./(2*mesh.x)) * phi.grad
term = tmp.dot(tmp)
It's possible something more clever could be done, but I'm not seeing it.
On Nov 4, 2015, at 5:12 PM, Fausto Arinos de A. Barbuto
<[email protected]> wrote:
>
> Hi,
>
> How do I represent the 2nd term in the RHS of the following dimensionless
> PDE? I've tried a couple of approaches but none worked.
>
> \frac{\partial \phi}{\partial t} = (1 +b \phi) \frac{\partial^2
> \phi}{\partial x^2}
> + ( \frac{\partial \phi}{\partial x^2} )^2
>
> Thanks,
>
> Fausto
>
>
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