> On 15 May 2018, at 11:49, Onno Bokhove <O.Bokhove@leeds.ac.uk> wrote:
>
> HI David and Stephan,
>
> Thanks.
>
> Per the below:
> (i) The expression is a^1.5 but was written as a*sqrt(a); now that is corrected the nonlinear solver seems to work fine; apparently that rewrite matters for the Newton iteration.

The Jacobian is computed by automatic differentiation, so although mathematically a^1.5 and a*sqrt(a) are the same, they are differentiated differently to:

1.5 a^0.5

or

0.5 a / sqrt(a) + sqrt(a)

The latter has a numerical singularity at 0.  In floating point, if a is zero, you compute 0.0 / 0.0 => NaN

Lawrence
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