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For the three-dimensional ball $|x|<a$, the displayed Dirichlet Green function has convention ${\Delta }_x{G}_D=+\delta (x-y)$. Its method of images uses the exterior inverse point $y^{\ast }=a^{2}y/|y{|}^2$. On $|x|=a$, the distance identity $|x-y^{\ast }|=(a/|y|)|x-y|$ makes the boundary value zero. The image lies outside the ball and therefore adds no source singularity to the domain. At a central source the weighted image term tends to $1/a$, giving $G_D(x,0)=-(4\pi |x|{)}^{-1}+(4\pi a{)}^{-1}$. Spherical inversion with the required radial weight is the Kelvin transform.

An irregular solid harmonic is harmonic on punctured three-dimensional Euclidean space and generally singular at the origin. The Kelvin transform maps a degree-$n$ regular solid harmonic to a constant multiple of its irregular counterpart.

The image charge for a grounded conducting sphere is obtained by the Kelvin transform:It lies inside the electrical conductor, outside the region where it represents the physical electric field. The electric potential isAt the physical point charge, the electric potential has its usual Coulomb's law singularity. There is no physical singularity at the image charge: the interior expression is identically zero.

For $|\mathbf{r}|=a$, direct expansion of squared distances givesConsequently the two terms cancel exactly, proving the zero Dirichlet boundary condition on the sphere. At infinity each term is $O(r^{-1})$, and more explicitlyThus the Dirichlet boundary condition at infinity also holds. The candidate solves the Poisson equation with the one physical point charge; uniqueness of Poisson equation, with these Dirichlet boundary conditions, identifies it as the required electric potential.

Put $\lambda =(r/a{)}^2$, so the inverted observation point is $\overline{\mathbf{r}}=\mathbf{r}/\lambda$ and $\overline{r}=r/\lambda$. Let $\mathbf{R}=\overline{\mathbf{r}}-{\mathbf{r}}_0$. The Kelvin transform correction in the first bracket is transformed usingThus its dipole contribution is exactlyFor the remaining correction use the dipole line-integral identity. Writing ${\mathbf{P}}_c=\mathbf{r}-c{\mathbf{r}}_0$ and $P_c=|{\mathbf{P}}_c|$, it isIt can be proved by differentiating the right side with respect to $c$ and checking its zero value at $c=0$. Alternatively resolve into directions parallel and perpendicular to ${\mathbf{r}}_0=b\mathbf{e}$: if $\mathbf{r}=\rho {\mathbf{e}}_{\rho }+z\mathbf{e}$, the two integrals are $(z/r-(z-bc)/P_c)/(b\rho )$ and $(1/P_c-1/r)/b$, respectively, which combine to the displayed expression. The limiting cases follow by continuity.

Set $c=\lambda$. Then ${\mathbf{P}}_c=\lambda \mathbf{R}$, $P_c=\lambda R$, $\mathbf{r}=\lambda \overline{\mathbf{r}}$ and $r=\lambda \overline{r}$. Substitution givesDotting with $\mathbf{Q}/(4\pi \sigma )$ reproduces the second correction. Including the primary dipole therefore proves the Spherical Neumann dipole images representation,The upper limit is $(r/a{)}^2$ in the PDF; extracting the small stacked fraction as text can incorrectly invert it. The extra terms are one scaled dipole and a continuous dipole line in this parametrized method of images. Their positions and strengths depend on the observation radius, so they should not be treated as a fixed set of independent physical sources. For an interior point $r<a$, the entire image segment satisfies $t|{\mathbf{r}}_0|\le (r/a{)}^2|{\mathbf{r}}_0|<r$, so it introduces no new interior singularity. The only physical interior dipole singularity remains at ${\mathbf{r}}_0$.

Let $\overline{\mathbf{r}}=a^2\mathbf{r}/r^2$ and $\hat{\mathbf{r}}=\mathbf{r}/r$. Its Jacobian isThe chain rule therefore yieldsTaking its cross product with $\mathbf{r}$ kills the radial term and proves that spherical inversion preserves angular derivatives:This is also the invariance of the differential expression underlying the angular momentum operator.

A scalar spherical harmonic $Y$ depends only on direction, so $Y(\hat{\mathbf{r}})=Y(\widehat{\overline{\mathbf{r}}})$ and its gradient is tangential. Thus $\hat{\mathbf{r}}\cdot {\nabla }_{\overline{\mathbf{r}}}Y=0$, andEach of the three normalized vector spherical harmonic families is consequently unchanged as an angular basis function:The normalization requires $n\ge 1$ for $\mathbf{B}$ and $\mathbf{C}$; at $n=0$ the gradient vanishes and only $\mathbf{P}$ remains.

This substitution into the basis functions is distinct from the pushforward of a vector field of a physical vector field by inversion. Under that Jacobian map the radial vector picks up $-a^2/r^2$ and the two tangential vectors pick up $+a^2/r^2$. Likewise, the weighted Kelvin transform of a scalar includes the extra factor $a/r$. Neither operation changes the angular differential identity just proved.

For a degree-$n$ homogeneous polynomial, the Euler homogeneous function theorem gives $\mathbf{r}\cdot \nabla H_n=n{H}_n$. On $r>0$ in three-dimensional Euclidean space,Apply the product rule for the Laplacian:For $s=-(2n+1)$ the second term vanishes. Hence homogeneous harmonic inversion givesThe nonzero factor $r^{-(2n+1)}$ proves both directions: the transformed function is harmonic exactly when $\Delta H_n=0$ away from the origin. Since $\Delta H_n$ is a polynomial, vanishing on the punctured space makes it identically zero everywhere.

Equivalently, homogeneity shows that the Kelvin transform isIt sends a regular solid harmonic to an irregular solid harmonic, exchanging the radial powers $r^n$ and $r^{-n-1}$. The origin is excluded from the harmonicity statement: for example, the $n=0$ transform $1/r$ is harmonic only away from zero and has a distributional Dirac delta function source there.

Put $r_0=|{\mathbf{x}}_0|$ and ${\mathbf{x}}_0^{\ast }=a^2{\mathbf{x}}_0/r_0^2$ for $0<r_0<a$. The Dirichlet Green function inside a sphere isOn $|\mathbf{x}|=a$, the distance identity $|\mathbf{x}-{\mathbf{x}}_0^{\ast }|=(a/r_0)|\mathbf{x}-{\mathbf{x}}_0|$ cancels the two terms. The inverse point lies outside the sphere, so it introduces no interior source. This weighted inversion is the Kelvin transform. For a central source, take the limit to get $G_B(\mathbf{x};0)=-(4\pi {)}^{-1}(1/|\mathbf{x}|-1/a)$.

Let $r_0=|x_0|$ with $0<r_0<a$. The Dirichlet Green function inside a sphere uses the exterior inverse point $x_{\ast }=a^2{x}_0/r_0^2$ and strength $s_{\ast }=-a/r_0$. ThusFor $|x|=a$, the distance identity $|x-x_{\ast }|=(a/r_0)|x-x_0|$ makes the bracket zero. The image is outside the sphere, so it adds no interior singularity. This is the radial weight associated with the Kelvin transform.

The printed unit-strength restriction must be relaxed in this part: the required single image has strength $-a/r_0$, whose magnitude is generally not one. A reflected unit charge does not give the spherical boundary cancellation. For a centered source, the limiting formula isIts correcting term is constant and corresponds to the weighted image moving to infinity, rather than a finite unit point image.

Suppose the incident axisymmetric flow is an incompressible flow and an irrotational flow, with Stokes streamfunction $\Psi$ regular inside $r<a$. The displayed transformation adds an impermeable sphere at $r=a$ without new finite exterior singularities. Writing $s=a^2/r$ and $G=(r/a)\Psi (s,\theta )$, direct differentiation showsHence the Stokes streamfunction equation for irrotational flow is preserved. At $r=a$, $\tilde{\Psi }=0$, making the radial velocity zero. Exterior points sample only the regular interior under inversion. If $\Psi$ vanishes at the origin and is $O(r)$ there with smooth regular derivatives, the induced velocity decays at infinity. This is an inversion construction for an axisymmetric streamfunction, distinct from a scalar Kelvin transform.

For a current dipole in a homogeneous insulating sphere of radius $a$, let $\lambda =r^2/a^2$ and let $\Psi$ be its primary electric potential. The displayed representation solves the interior Neumann boundary condition up to an additive constant. It rewrites the spherical Kelvin transform correction and a distributed dipole correction using the dipole line-integral identity. The image parameters depend on the observation radius; this is an exact parametrized representation rather than a fixed collection of image source positions.