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The interior Helmholtz equation for a polytrope of index one also admits a positive solution vanishing on the faces of a cube:Its central value is ${\rho }_c$ and its average is $\bar{\rho }=8{\rho }_c/{\pi }^3$. Together with $\Phi =C-2K\rho$, it solves the interior hydrostatic equilibrium and Poisson equation for Newtonian gravity. A local interior solution need not match the external field of its own mass.

Applying the Poisson equation for Newtonian gravity to a flattened power-law gravitational potential yieldsFor $0<\beta <1$, nonnegative mass density everywhere outside the origin requires and is ensured by $\beta \le {\lambda }^2\le 2/(\beta +1)$. The central density singularity is locally integrable and carries no point mass, although the total mass diverges at large radius in this scale-free model.

Set $A={\Phi }_0{R}_0^{\beta }<0$, $q={\lambda }^2$ and $s=R^2+q{z}^2$. The flattened power-law gravitational potential is $\Phi =A{s}^{-\beta /2}$. In axisymmetric cylindrical coordinates, the Poisson equation for Newtonian gravity givesThe first derivatives are ${\partial }_R\Phi =-\beta AR{s}^{-(\beta +2)/2}$ and ${\partial }_z\Phi =-\beta Aqz{s}^{-(\beta +2)/2}$. Differentiating again and collecting powers givesTherefore the required mass density isThis is a formal density for arbitrary $\lambda$, but a physical dark matter distribution must be nonnegative. The density positivity for a flattened power-law potential condition isNecessity follows by evaluating on the midplane and symmetry axis; sufficiency follows because both numerator coefficients are then nonnegative. In this range the origin is a locally integrable density cusp: the mass enclosed near radius $r$ scales as $r^{1-\beta }$ and tends to zero, so no point mass needs adding. The scale-free distribution has infinite total mass at large radius and represents an idealized background, not a finite isolated halo.

For the cold, non-self-gravitating disk, radial balance of a circular orbit is $R{\Omega }^2={\partial }_R\Phi (R,0)$. ThusThe displayed positive root chooses the rotation orientation; the opposite orientation has the negative of this angular frequency.

The same polytrope of index one interior equation is ${\nabla }^2\rho +(2\pi G/K)\rho =0$. A positive separated solution with zero density on all six cube faces is the cubic polytropic interiorIts Laplacian is $-3{\pi }^2\rho /L^2$, so the required side length isWith $\Phi =C-2K\rho$ and $P=K{\rho }^2$, it satisfies the interior hydrostatic equilibrium and Poisson equation for Newtonian gravity. Integrating each sine factor yieldsThis formal interior solution is not an isolated physical cubic star. The cubic polytrope fails isolated gravitational matching: at a vertex, the product of sines has $\nabla \rho =0$, so the interior potential predicts zero gravitational acceleration. At the vertex $(0,0,0)$, however, the field generated by its own positive mass isand each component is strictly positive. There can be no continuous matching to the isolated external field without additional forces or mass sources. Thus solving the interior density equation and imposing zero face values is insufficient. Fluid stars also have no rigid structure to maintain sharp cubic faces, and observed stellar shapes are approximately spherical or rotationally flattened rather than cubic.

For $P=K{\rho }^2$, hydrostatic equilibrium gives $2K\nabla \rho =-\nabla \Phi$. The Poisson equation for Newtonian gravity therefore gives the Helmholtz equationFor spherical symmetry, the regular polytrope of index one solution isThe singular $\cos (kr)/r$ solution is excluded by finite central mass density. Taking the first zero as the stellar surface yieldsThe mass integral isDividing by the volume $4\pi R^3/3$ gives

A stellar polytrope with polytropic index one has $\nabla \Phi =-2K\nabla \rho$ in hydrostatic equilibrium. Combining this with the Poisson equation for Newtonian gravity gives the Helmholtz equationThe spherical solution regular at the origin is $\rho (r)={\rho }_c\sin (kr)/(kr)$. Its first zero gives $R=\pi /k=\sqrt{\pi K/(2G)}$, independent of central mass density, and $\bar{\rho }/{\rho }_c=3/{\pi }^2$.