Statement

Let a(s,z),b(s,z),c(s,z)a(s,z),b(s,z),c(s,z) be on an open neighborhood of the relevant points with s0s\ge0. An axisymmetric field on r>0r>0 with

ur=ra(r2,z),uθ=rb(r2,z),uz=c(r2,z)u_r=r\,a(r^2,z),\qquad u_\theta=r\,b(r^2,z),\qquad u_z=c(r^2,z)

extends smoothly through the axis.

Cartesian verification

Putting s=x2+y2s=x^2+y^2, its Cartesian components are

ux=xa(s,z)yb(s,z),uy=ya(s,z)+xb(s,z),uz=c(s,z).u_x=x\,a(s,z)-y\,b(s,z),\qquad u_y=y\,a(s,z)+x\,b(s,z),\qquad u_z=c(s,z).

These are compositions and products of smooth functions. This proves the stated sufficient criterion directly, including parameter-dependent versions when a,b,ca,b,c are jointly smooth in those parameters.

Why radial smoothness alone is insufficient

The field er=(x/r,y/r,0)e_r=(x/r,y/r,0) has constant cylindrical radial component but no continuous extension at zero. Even a scalar f(r)=rf(r)=r, smooth on the radial half-line in its one-sided sense, becomes x2+y2\sqrt{x^2+y^2}, which is not differentiable at the axis.