 Solution

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 $A>0$.