I am working through a problem of a bead moving without friction on a surface $z=f(r)$ in cylindrical polar coordinates $(r,\theta,z)$, under the influence of gravity. I have shown that $h=r^2\dot\theta$ is constant and that $$(*)\;\;\;\large (1+f_r^2)\ddot r+f_rf_{rr}\dot r^2-\frac{h^2}{r^3}+gf_r=0.$$ Then $f$ and $h^2$ are given as $$f(r)=\frac{1}{2}\alpha r^2-\frac{1}{6}\beta r^6,\;\;h^2=\frac{3\alpha^2g}{16\beta}$$ and I must now determine the linear stability of the equilibrium point $r_-=\large(\frac{\alpha}{4\beta})^\frac{1}{4}.$
Writing $r=r_-+r_1$ for small $r_1$, we see $\ddot r=\ddot r_1$, $f_r=\alpha r-\beta r^5$, and we can neglect the $\dot r^2=\dot r_1^2$ term as it is small.
In the solution though, they immediately go on to write $$(1+f_r^2)\ddot r_1+P'(r_-)r_1=0$$ where $P(r)=-\frac{h^2}{r^3}+gf_r$, determining the stability of the point by checking the sign of $P'(r_-).$
Could someone explain how they have deduced this linearised equation and why we can introduce the derivative of the $P(r)$ term? I cannot see how this follows from $(*)$
The equation itself may be written $(1+f'(r)^2) \ddot{r}+ f'(r) f''(r) \dot{r}^2 + P(r)=0$. The state $(r,\dot{r})=(r_-,0)$ is an equilibrium here, since $P(r_-)=0$.
Pedantically speaking, the system should be viewed as
$$\begin{bmatrix} \dot{v} \\ \dot{u} \end{bmatrix} = \begin{bmatrix} -\frac{f'(u) f''(u)}{1+f'(u)^2} v^2 - \frac{P(u)}{1+f'(u)^2} \\ v \end{bmatrix}.$$
Then the linearization around $(u,v)=(r_-,0)$ is
$$\begin{bmatrix} \dot{v} \\ \dot{u} \end{bmatrix} = \begin{bmatrix} 0 & -\frac{P'(r_-)}{1+f'(r_-)^2} \\ 1 & 0 \end{bmatrix} \begin{bmatrix} v \\ u \end{bmatrix}$$
Thus the linear stability is dictated by the eigenvalues of this matrix. The trace is $0$ and the determinant is $\frac{P'(r_-)}{1+f'(r_-)^2}$, so the eigenvalues are both pure imaginary if $P'(r_-)>0$ and one of them is real and positive if $P'(r_-)<0$. Therefore the system is unstable when $P'(r_-)<0$ since the linearization is unstable. When $P'(r_-) \geq 0$ the system may or may not be stable; whether it is stable depends on the behavior of the nonlinearity.