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In atomic units the Time-independent Schrodinger equation isThe displayed equation is the Radial Schrodinger equation for the hydrogen atom. Its radial derivative terms come from the Laplacian in spherical coordinates, the Coulomb term is the potential $-1/r$, and separation into spherical harmonics uses the orbital angular momentum eigenvalue $l(l+1)$, producing the centrifugal term $-l(l+1)R/r^2$.

For $E=-1/(2n^2)$, substitute $R=r^{\alpha }{e}^{-r/n}$. Dividing the equation by $R$ and comparing powers of $r$ givesThe solution regular at the origin has $\alpha =l$, and when $l=n-1$ both equations agree. This is the circular Coulomb bound state. ThusThe radial probability measure is $|R(r){|}^2{r}^{2}dr$. With $a=2/n$ and the Gamma integral, the mean radius of a circular Coulomb bound state is

At fixed principal quantum number $n$, the allowed orbital angular momentum quantum numbers are $l=0,\dots ,n-1$, and each has magnetic quantum number $m=-l,\dots ,l$. Hence the orbital quantum degeneracy isIncluding electron spin would double this to $2n^2$, but spin is absent from the stated wavefunctions.

For a spherically symmetric function $\Phi (r)$, the Laplacian in spherical coordinates isIntegrating from zero to $r$ and using regularity at the origin gives

For the stated density, integration and the boundary condition $\Phi (a)=0$ giveThe two pieces and their first derivatives agree at $r=b$. If another solution existed, its difference from $\Phi$ would be harmonic with zero boundary data, so part (a) proves Uniqueness of the Dirichlet problem.

On the shell $U(b,a)$, normalize the outer radial part to obtain the harmonic functionBy the Dirichlet principle, this function minimizes the energy among all $w$ with the same boundary values. Sinceits energy isTherefore

In a static spherical star, hydrostatic equilibrium and the Poisson equation giveor equivalently $r^{-2}d(r^{2}g)/dr=4\pi G\rho$.

Let $\mathbf{\xi }$ be the fluid displacement field, so the velocity perturbation is $\delta \mathbf{u}={\partial }_t\mathbf{\xi }$. Linearizing the ideal-fluid momentum equation about the static state and cancelling the background hydrostatic terms givesConservation of mass says that the Lagrangian density perturbation is ${\Delta }_L\rho =-\rho \nabla \cdot \mathbf{\xi }$. The relation between Eulerian and Lagrangian fluid perturbations and adiabatic compression gives, with $\Delta =\nabla \cdot \mathbf{\xi }$,while linearized self-gravity gives $\boxed{{\nabla }^2\delta \Phi =4\pi G\delta \rho }$.

For the stated spherical harmonic displacement, the radial divergence is $r^{-2}d(r^2{\tilde{\xi }}_r)/dr$, while $\nabla {\tilde{\xi }}_h$ is radial and orthogonal to the angular gradient of $Y_l^m$. HenceEquating the radial and horizontal coefficients of $Y_l^m$ and $\nabla Y_l^m$, and applying the separated Laplacian in spherical coordinates to $\widetilde{\delta \Phi }(r)Y_l^m$, gives

Define the stellar buoyancy frequency byEliminating $\tilde{\Delta }$ between the density and pressure perturbations givesSubstitution in the radial equation, followed by use of $dp/dr=-\rho g$, yieldsRegular spherical profiles near the center have $\rho ={\rho }_c+O(r^2)$ and $p=p_c+O(r^2)$, while $m(r)=4\pi {\rho }_c{r}^3/3+O(r^5)$ and hence $g=O(r)$. Both logarithmic gradients in the definition of $N^2$ are $O(r)$, so $N^2=A{r}^2+O(r^4)$. A Sun-like radiative central stratification is stable, making $A>0$.