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Differentiate under the integral and use ${\nabla }_x|x-x^{\prime }{|}^{-1}=-{\nabla }_{x^{\prime }}|x-x^{\prime }{|}^{-1}$:The omitted boundary term vanishes if the current is localized sufficiently rapidly. A steady current obeys charge conservation ${\nabla }^{\prime }\cdot J=0$, hence

For $r=|x|$ much larger than the source size,Localization and ${\nabla }^{\prime }\cdot J=0$ imply $\int J{d}^3{x}^{\prime }=0$. They also implyby integrating ${\partial }_k^{\prime }(x_i^{\prime }{x}_j^{\prime }{J}_k)$. Thus the first nonzero moment is antisymmetric and can be written using the magnetic dipole momentConsequently the Coulomb-gauge vector potential of a localized steady current has far fieldThe dimensions are

Continuity at $x=\xi$ is automatic in the proposed expression. Integrating the differential equation through $x=\xi$ gives the derivative jumpThusIn terms of the Wronskianone hasThe boundary conditions hold because $y_1^{\prime }(0)=0$ and $y_2^{\prime }(1)=0$.

The Abel identity gives $W^{\prime }=-\alpha W$. If $\alpha =0$, the Wronskian and hence $c$ are constant. The two branches of the displayed formula are then interchanged by $x\leftrightarrow \xi$, provingThis is the symmetry of the Neumann Green function for a second-order ordinary differential equation in the self-adjoint case.

For $G^{\prime \prime }-G=\delta (x-\xi )$, chooseTheir Wronskian isWriting $x_{<}=\min (x,\xi )$ and $x_{>}=\max (x,\xi )$ givesThe solution of the inhomogeneous problem is $y(x)={\int }_0^{1}G(x,\xi )\xi d\xi$. Equivalently, solving directly givesThe two Neumann conditions yield $B=1$ andTherefore