Let and . Their cross partial derivatives agree: . On the simply connected plane, this makes the differential form exact. Integrating gives ; matching then requires , so one potential isThe ordinary differential equation states along a solution. Its initial condition fixes the constant to . Hence the solution is specified implicitly byAt the initial point , so the implicit function theorem guarantees a unique local branch through that point.
For , multiply by to obtain the first-order linear ordinary differential equation . Its integrating factor is . The product rule then givesIntegration yields , soThe constant of integration is arbitrary on each connected interval of the original equation's domain. The formula extends smoothly through zero, but the printed equation itself contains and is not defined there.
The real equation has domain . Its integrating factor isThus the linear ordinary differential equation becomes . After integration, . The initial condition gives , hence
Write the scalar field as , where and . Its radial contribution to the Laplacian in spherical coordinates isFor the angular contribution, and , soCombining the radial and angular terms provesThe calculation uses to avoid coordinate singularities. In Cartesian coordinates, , which extends smoothly across the polar axis for . It therefore satisfies Laplace's equation there as well. The origin is a genuine singularity and is excluded.
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