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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 $(*)$

jcneek
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  • Presumably it is a Taylor expansion, $P(r)=P(r_-)+P'(r_-) r_1 +o(r_1)$. For linear stability you can drop $o(r_1)$ terms. I can't immediately decipher why $P(r_-)=0$, though. – Ian Aug 03 '21 at 16:04
  • OK, and we can exclude the $P(r_-)$ term as it is a constant? – jcneek Aug 03 '21 at 16:09
  • No, if $P(r_-)$ is not zero then there is something weird happening, because then for $r_1$ small you have $\ddot{r_1}$ not small (even for short time). The linearized equation for the evolution of the perturbation should be homogeneous. – Ian Aug 03 '21 at 16:13
  • I am a bit confused where the equilibrium came from in the first place actually. It seems to me that you can have an equilibrium in this equation when $h^2/r^3=gf_r$ (in which case the equation is satisfied with $\dot{r}=\ddot{r}=0$), but I think this does not lead to your $r_-$ expression. – Ian Aug 03 '21 at 16:18
  • I forgot to include that the value of $h^2$ is given, I have edited the question. Solving that equality gives possibilities for equilibrium $r^4=r_-^4$ and $\frac{3\alpha}{4\beta}$ – jcneek Aug 03 '21 at 16:21
  • OK, then that explains it, you have $P(r_-)=0$ so you linearize $P(r)$ around $P(r_-)$ for the linear stability analysis. The only catch now is that your perturbations can't be just any phase space perturbation in order to justify neglecting the $\dot{r}^2$ term. You have to perturb the initial velocity by a small enough amount relative to how much you perturb the initial radius. – Ian Aug 03 '21 at 16:29
  • Could the neglection be a simplification, since this is from a first year undergraduate paper? – jcneek Aug 03 '21 at 16:33
  • You mean neglecting the catch I just mentioned? That depends on the context. It's totally plausible that the perturbations of interest are just pure radius perturbations with no velocity perturbation, for instance. And then dropping that term is totally legitimate. – Ian Aug 03 '21 at 16:35
  • Ah OK, from what I've been taught on the course 'small perturbations' is always referring to coordinates, whether it is $r,\theta,x,y,$ or $z$. Thank you for your help with this! – jcneek Aug 03 '21 at 16:39

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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.

Ian
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