# Necessary and Sufficient Conditions for an Equilibrium to be Stable

In the 4th section The condition that convection be absent of the book Fluid Mechanics by Landau and Lifshitz, they give the following statement:

For the (mechanical) equilibrium to be stable, it is necessary (though not in general sufficient) that the resulting force on the element should tend to return it to its original position.

My question is, why this condition is not sufficient in general?

In the context, they consider a fluid element moving along $$z$$-axis, see the image below.

By Lyapunov's theorem, if the potential function $$U(z) \in C^1$$ has a strict minimum at an equilibrium $$z_0$$, then $$z_0$$ is stable. (For example, see What is the state of the equilibrium for a second derivative equal to zero?)

But, if assume $$U(z)$$ is $$C^{\infty}$$, then the strict minimality of $$U$$ at $$z_0$$ is equivalent to the attractivity of $$F$$ in some neighborhood of $$z_0$$: by Taylor's formula, we have $$$$U(z) = U(z_0) + a_k z^k + O(z^{k + 1}),$$$$ where $$a_k \neq 0$$, then $$$$F(z) = - \partial U / \partial z = - k a_k z^{k-1} + O(z^{k}).$$$$ So $$U(z_0)$$ is strictly minimal at $$z_0$$ $$\iff$$ $$a_k > 0$$ and $$k$$ is even $$\iff$$ $$F$$ is attractive in some neighborhood of $$z_0$$.

In other words, the local attractivity of $$F$$ should also be sufficient for stability. What am I missing here?

• I think that it is sufficient condition only in the 1 dimension case. – ziv Mar 18 at 16:39
• @ziv The context is a fluid element moving along $z$-axis. – Zephyr Mar 18 at 21:50

• The equilibrium may be not a single point, but a range, e.g., if we take potential $$U(x) = \begin{cases} (x+1)^2, x<-1,\\ 0, -1 < x < 1,\\ (x-1)^2, x>1 \end{cases}$$
• The potential might be non-analytical, e.g., $$U(x) = \frac{1}{x}$$
The hydrodynamic variables are fields, $$\rho(x,t)$$, $$\vec{v}(x,t)$$, $$T(x,t)$$. In order to study stability of a solution to the hydrostatic equations $$(\rho_0(x),\vec{v}=0,T(x))$$, we have to consider general fluctuations $$\rho(x,t)=\rho_0(x)+\delta\rho(x,t), \quad \vec{v}(x,t)=\delta\vec{v}(x,t), \quad T(x,t)=T_0(x)+\delta T(x,t).$$ The stability of a slab under displacements is just a special case of this, so not in general sufficient to establish stability. Indeed, there are many well-known cases where the unstable mode has a finite wave number.