Show stable node or spiral cannot occur If I have the equation:
$$\ddot{x} + f(\dot{x}) + g(x) = 0$$
where $f$ is even and $f$ and $g$ are both smooth, how do I show that the equilibrium points cannot be stable nodes or spirals?
What I've done so far is:
$$\dot{x} = y$$
$$\dot{y} = F(x,y) = -f(y) - g(x)$$
I can try to take the Jacobian at the fixed points
$$J = \begin{pmatrix}
0 & 1 \\
\frac{\partial F}{\partial x} & \frac{\partial F}{\partial y}
\end{pmatrix}$$
but without knowing the values of $\frac{\partial F}{\partial x}$ or $\frac{\partial F}{\partial y}$ I can't tell what type of fixed points can occur.
My best guess is that, based on the information in the question, there must be some restriction on the values $\frac{\partial F}{\partial x}$ or $\frac{\partial F}{\partial y}$ can take, which would then rule out the possibility of stable nodes or spirals. However, I can't seem to find any such restrictions. Any help would be appreciated!
 A: Given the equation
$\ddot x + f(\dot x) + g(x) \tag{1}$
with $f$, $g$ smooth and $f$ even, we can follow the lead of put OP AlasPoorYorick ( which, by the way, is a great user name!) and set
$y = \dot x, \tag{2}$
so that (1) becomes
$\dot y = - f(y) - g(x), \tag{3}$
and the single second order equation (1) is replaced by the first order system comprised of (2) and (3).  We first observe that any equilibrium point of this system satisfies
$y = \dot x = 0, \tag{4}$
although the $x$-coordinates of the equilibria, satisfying as they do the equation
$g(x) = -f(0), \tag{5}$
cannot be determined without further knowledge of $g(x)$.  (We know (5) holds at equilibrium points by virtue of the fact that $y = \dot x = 0$ and $\dot y = 0$ at such points; substituting these values into (3) yields (5).)  We secondly observe that, $f(y)$ being even, we have $f'(0) = 0$, since $f'(0) \ne 0$ implies $f(y)$ is either increasing or decreasing at $y = 0$, which further implies the existence of an interval $(-\delta, \delta)$ (and here $\delta > 0$) with $f(y)< f(0)$ for $y \in (-\delta, 0)$ and $f(y) > f(0)$ for $y \in (0, \delta)$, or vice-versa; in either event we cannot have $f(-y) = f(y)$ for $0 \ne y \in (-\delta, \delta)$, contradicting the evenness of $f$; so $f'(0)= 0$.
We examine the Jacobean $J(x, y)$ of the system at possible equilibria.  To begin, we observe that the Jacobean matrix at any point $(x, y)$ is given by
$J(x, y) = \begin{bmatrix} \dfrac{\partial \dot x}{\partial x} & \dfrac{\partial \dot x}{\partial y} \\ \dfrac{\partial \dot y}{\partial x} & \dfrac{\partial \dot y}{\partial x} \end{bmatrix} = \begin{bmatrix} 0 & 1 \\ -g'(x) & -f'(y) \end{bmatrix}; \tag{6}$
at an equilibrium, as we have seen we must have $y = 0$, whence as we have further seen $f'(y) = f'(0) = 0$ so the Jacobean becomes
$J(x, 0) = \begin{bmatrix} 0 & 1 \\ -g'(x) & 0 \end{bmatrix}. \tag{7}$
The eigenvalues of $J(x, 0)$ are easily had from (7); indeed the characteristic poloynomial $p_x(\lambda)$ of $J(x, 0)$ is
$p_x(\lambda) = \det J(x, 0) - \lambda I) = \det(\begin{bmatrix} -\lambda & 1 \\ -g'(x) & -\lambda \end{bmatrix} = \lambda^2 + g'(x); \tag{8}$
we see from (8) that for $g'(x) < 0$ we have
$\lambda = \pm \sqrt{-g'(x)}; \tag{9}$
the values of $\lambda$ are real and of opposite signs; $(x, 0)$ is a saddle.  When $g'(x) > 0$, then
$\lambda = \pm i \sqrt{g'(x)}; \tag{10}$
in this case, the eigenvalues form a conjugate pair of non-zero, purely imaginary numbers; $(x, 0)$ is an example of an elliptic point.  Since the eigenvalues have zero real part, such an equilibrium is neither a node nor a spiral point.  In the event that $g'(x) = 0$, then both values of $\lambda$ vanish and so $(x, 0)$ is once again neither a node or a spiral point.
To synopsize:  one shows an equilibrium to be a spiral or node by establishing that the eigenvalues of the Jacobean at the point have non-vanishing real parts of the same sign (note this includes the case of complex conjugate eigenvalues).
Evaluating the Jacobean in the present case is facilitated by the fact that $f(y)$ is an even function of $y$, making $J(x, y)$ particularly simple to evaluate when $y = 0$.
Hope this helps.  Cheers,
and as ever,
Fiat Lux!!!
A: Equilibrium points are simultaneous solutions to $y = 0$ and $-f(y)-g(x)=0$.
We have $tr(J) =  \frac{\partial F}{\partial y} = \frac{\partial}{\partial y}(-f(y)-g(x))$
You can see that at any equilibrium point we will have $tr(J)=0$. Ruling out the possibility of nodes or spirals (regardless of their stability).
