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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.

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.

The spatial factor $r^l{Y}_l^m$ is a regular solid harmonic. Besides the Laplace equation ${\nabla }^{2}F=0$, homogeneity gives $\mathbf{r}\cdot \nabla F=lF$. The divergence of the fluid displacement isDefine the scalar dilation amplitude $D_r=r{U}^{\prime }+(l+3)U+(l/r)V^{\prime }$; this is the quantity denoted $\Delta$ in the question, not the notation ${\Delta }_L$ for a Lagrangian pressure perturbation. Then mass conservation gives $\hat{\rho }=-\rho D_r$.

The equilibrium pressure gradient and homogeneity identity giveConsequently the adiabatic equation of state yieldsFor the force equations, the product rule givesEquating the coefficients of $F{\mathbf{e}}_r$ and $\nabla F$ in the self-gravitating adiabatic displacement equations givesFinally, applying the Laplacian to the gravitational perturbation givesso the Poisson equation becomesThese are the required interior equations for uniform-density stellar oscillation.

There is a radial degeneracy at $l=0$: $F$ is spatially constant and $\nabla F=0$, so the displacement is independent of $V$. The second force equation then cannot be inferred by equating independent vectors. For nonzero $\omega$ it may be imposed as an auxiliary definition of $V$, but it is not an additional physical radial equation. At zero frequency the radial equations should be used directly. This distinction matters for interpreting the zero factor in (c).

For the even polynomial family of degree $n\ge 2$, the physical radial squared frequency is $[\gamma n(n+1)/2-4]{\omega }_d^2$. The fundamental $n=2$ mode is stable when $\gamma >4/3$ and marginal at $\gamma =4/3$. At angular degree zero the gradient of the regular solid harmonic vanishes, so the auxiliary displacement amplitude multiplying it is redundant.

A regular solid harmonic is $r^l{Y}_l^m(\theta ,\varphi )$: it is homogeneous of degree $l$ and solves the Laplace equation. Thus $\mathbf{r}\cdot \nabla F=lF$ and ${\nabla }^{2}F=0$. These identities turn angular spherical harmonic expansions of displacement, density and Newtonian gravitational potential into radial equations.

A constant-density gas sphere in hydrostatic equilibrium has $g={\omega }_d^{2}r$ and $p=\rho {\omega }_d^2(R^2-r^2)/2$. Its stellar oscillations can be expanded using regular solid harmonics and radial displacement amplitudes. Uniform density does not imply uniform specific entropy, so this equilibrium can be radially stable and still exhibit convective instability.