Divergence of a Euclidean vector field

For a $C^1$ vector field $u:U\to\mathbb R^n$ , its divergence is

\operatorname{div}u=\nabla\cdot u=\sum_{i=1}^n\partial_{x_i}u_i.

It is a scalar field. The condition $\operatorname{div}u=0$ defines a divergence-free field.

Product identity

For a scalar $f$ , the product rule gives $\operatorname{div}(fu)=\nabla f\cdot u+f\operatorname{div}u$ .

Cylindrical expression

Writing $u=u_r e_r+u_\theta e_\theta+u_z e_z$ in the cylindrical frame gives, for $r>0$ ,

\operatorname{div}u=\frac1r\partial_r(ru_r)+\frac1r\partial_\theta u_\theta+\partial_z u_z.

The $u_r/r$ term arises from differentiating the moving basis. Coordinate singularities at $r=0$ require a separate Cartesian regularity check.

References