constant $a$, solves (\[Hamil\]) for $R
\geq a$, and imposes $${\partial \phi \over \partial R} +{1 \over 2R} \phi =0 
\;\; \mbox{for} \;\; R= a,   \label{robin}$$ and $$\phi > 0 
\;\; , \;\; \lim_{R \rightarrow \infty} \phi = 1.   \label{bounds}$$

Equation (\[robin\]) implies that the inner boundary is an extremal surface, as was shown in reference [@BoYo] and leads to a well posed elliptic problem when used with equations (\[Hamil\],\[bounds\]). One can also use the boundary condition (\[robin\]) to generate initial data that obeys an isometry condition, provided one chooses an extrinsic curvature —different from the one we are choosing here— that satisfies the isometry condition. This was explored in great detail by Cook [@Cook] in his numerical work constructing solutions for the conformal factor for multiple black holes. The symmetrization procedure yields non-trivial results even for a single black hole as we are considering here [@CoYo]. We have chosen, for simplicity, not to symmetrize the extrinsic curvature. Experience has shown that symmetrization does not change significantly the amount of radiated energy in head-on black hole collisions [@GlNiPrPuboost2]. Therefore our choice should not crucially influence the central conclusions we are attempting to obtain about the conformally flat black hole solutions.

