Situation of Stable, Neutral and Unstable Equilibrium Recently, I was reading about stability of equilibrium.
I came across the definitions for different types of equilibrium.
Neutral Equilibrium: The kind of equilibrium of a body so placed that when moved slightly it neither tends to return to its former position not depart more widely from it.
Unstable Equilibrium: A kind of equilibrium in which, if the system is displaced an arbitrarily small distance from the equilibrium state, the forces of the system cause it to move even farther away.
Stable Equilibrium: A kind of equilibrium in which, if the system is displaced an arbitrarily small distance from the equilibrium state, the forces of the system cause it to move back to the original equilibrium position.
In all the books and resources which I came across, these equilibrium were
analyzed only for 1 dimension.
So I was thinking what will be the case in 3 Dimension if along all the three orthogonal directions we had different equilibrium.
When I asked my teacher he said in order for a  system to be in stable equilibrium it must be so in all the directions possible.
Hence my conclusion was even if in one direction the system was in unstable equilibrium the system in general is in unstable equilibrium.

Then I tried to think what if in all directions the system is in stable equilibrium except a particular direction in which the equilibrium is neutral.
For example, look at the shape above.
If we place this rigid structure on the floor and a ball in the middle of this structure we obtain a configuration. Now I am unable to figure out whether the configuration is Stable or Neutral Equilibrium(I'm pretty sure this is not Unstable).
Either way I tried to classify it and I find the classification either way is unsatisfactory and I find no book or website mention this kind of situation.
So my question is, Is this configuration Stable or Neutral?
 A: I suppose the simplest way to deal with this is in the Lagrangian formalism, assuming the Lagrangian is of the form
$$
L=\frac{1}{2}m\dot{x}^2 - V(x)
$$
The critical points of the Lagrangian are those for which 
$$
\left(\frac{\partial L}{\partial \dot{x}}, \frac{\partial L}{\partial x}\right)=0
=\left(m\dot{x},-\frac{\partial V}{\partial x}\right)\, ,
$$
which immediately imply the equilibrium positions are at $\dot{x}=0$ and at the critical points of $V$.  
Let $x_0$ be such that $V'(x_0)=0$ and expand $V(x)$ near $x_0$ by writing
$x=x_0+ \delta x$.  This yields $\dot{x}=\dot{(\delta x)}$
$$
V(x)\approx V(x_0)+(\delta x)^2 \frac{1}{2}V''(x_0) \tag{1}
$$
since by definition $V'(x_0)=0$.  The approximate Lagrangian is thus 
$$
L=\frac{1}{2}m(\dot{\delta x})^2 -  V(x_0)-(\delta x)^2 \frac{1}{2}V''(x_0)
$$
and the resulting EOM  is the so-called linearized equation of motion:
$$
m\ddot{(\delta x)}= -V''(x_0)\delta x\, .
$$
Thus, if $V''(x_0)>0$, the solution is a harmonic oscillation of the type
$$
\delta x(t)= A\cos(\omega t+\varphi) \tag{2}
$$
This is the stable case in the sense that, if the particle starts "close" to $x_0$, it will always remain "close" to $x_0$.  
If $V''(x_0)<0$, then the solution is exponential:
$$
\delta x(t) =Ae^{\lambda t} + Be^{-\lambda t}\, . \tag{3}
$$
This is unstable as a particle starting near $x_0$ will runaway exponentially fast never to return close to $x_0$.
If $V''(x_0)=0$, then we cannot decide: this is the neutral case and we must expand $V(x)$ beyond the second order term in $\delta$ in (1).
The same argument holds if $V=V(x,y)$.  If $(x_0,y_0)$ is a critical point of $V$, then expand $V$ to second order.  The resulting equations of motions for $\delta x$ and $\delta y$ are typically linearly coupled, i.e. they are of the form
$$
\left(\begin{array}{c}
\ddot{\delta x}\\
\ddot{\delta y}\end{array}\right)=
\left(\begin{array}{cc}
\alpha & \beta \\
\beta &\gamma \end{array}\right)\left(\begin{array}{c}
{\delta x}\\
{\delta y}\end{array}\right) = M\left(\begin{array}{c}
{\delta x}\\
{\delta y}\end{array}\right) \, . \tag{4}
$$
The numbers $\alpha,\beta,\gamma$ are related $V_{xx}(x_0,y_0), V_{xy}(x_0,y_0)$ and $V_{yy}(x_0,y_0)$.
To uncouple these linear equations, one finds the linear combinations of $\delta x$ and $\delta y$ that make the matrix $M$ diagonal.  This amounts to find eigenvectors $\delta X$ and $\delta Y$ in terms of $\delta x$ and $\delta y$.  If $D_M$ is the diagonal form of the matrix $M$, the uncoupled equations of motion are of the form
$$
\left(\begin{array}{c}
\ddot{\delta X}\\
\ddot{\delta Y}\end{array}\right)=
\left(\begin{array}{cc}
A& 0 \\
0 &G \end{array}\right)\left(\begin{array}{c}
{\delta X}\\
{\delta Y}\end{array}\right) = D_M\left(\begin{array}{c}
{\delta X}\\
{\delta Y}\end{array}\right) \, . \tag{5}
$$
where $A$ and $G$ are the eigenvalues of $M$.  The stability character of the motion for $\delta X$ and $\delta Y$ can be studied as per the 1d motion example.
Clearly, if the eigenvalues $A$ and $G$ are both negative, $\delta X$ and $\delta Y$ will have the form of (3) and the particle will remain always close to its initial position: the critical point is stable. 
If one or both of the eigenvalues is positive, then one or both of $\delta X$ and $\delta Y$ will have an exponential solution and the particle will NOT remain close to its initial position: it will run away exponentially in at least one direction.
If one or both eigenvalues are $0$, we can't decide since there is no guarantee that the particle will remain close to its initial position.
The case of $3$ or more degrees of freedom is treated by immediate generalization of the 2d case.
