... given a point

*P*and a new ink drop of radius

*r*centered at

*C*, move the point radially from

*C*to:

C + (P − C) · sqrt | ( | 1 + | r^{2}|| P − C||^{2} | ) |

One correspondent complains that, because the divergence of this transform is not zero, it can't be incompressible.

Divergence is defined for a continuously differentiable vector field. But this transform is not continuously differentiable around *C*; thus its divergence isn't well-defined. The common definition of incompressible is in terms of the divergence of a vector field. For a vector-field where the divergence isn't well-defined, the definition is silent.

However, I can show that this transform preserves the area of all neighborhoods not containing *C*. Consider the annulus centered on *C* having inner radius sqrt(*a*/*π*) and outer radius sqrt((*a*+*b*)/*π*). Its area is *b*.

If ink is injected at *C* forming a new circular region centered on *C* having area *e*, the annulus having area *b* will expand to have an inner radius of sqrt((*a*+*e*)/*π*) and an outer radius of sqrt((*a*+*b*+*e*)/*π*). The area of the expanded annulus is still *b*. Note that the expanded annulus is thinner than the original.

Because this transform is radially symmetric, any annulus slice with sides forming angles with *C* of *θ* and *η* will map to the expanded annulus slice between the angles of *θ* and *η*. Because the annulus and slices can be made arbitrarily small, all neighborhoods not containing *C* map to regions having the same area.