Hi,
On Wednesday, November 4, 2015 8:49 PM, "Guyer, Jonathan E. Dr."
<[email protected]> wrote:
>That term is really the partial derivative of \phi with respect to the
>quantity >(x^2)? Strange.
No, that was a typo I introduced in the formula while editing it. Sorryfor
goofing it up. That term in fact the square of the 1st derivative ofphi,
[d(phi)/dx]². The equation should actually read as follows:
\frac{\partial \phi}{\partial t} = (1 +b \phi) \frac{\partial^2 \phi}{\partial
x^2} + ( \frac{\partial \phi}{\partial x} )^2
Hopefully it is correct now.
Given that, what would you suggest? I tried
T.grad()*T.grad()
but that obviously didn't work (got a rank mismatch).
Fausto
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