Is the dynamical system stable at an equilibrium point?

[Chris L T521]

Here's this week's problem.

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Problem: Consider a constant mass $m$ moving under the influence of a conservative force field $-\mathrm{grad}\,\Phi(x)$ defined by a potential function $\Phi:W_0\rightarrow\Bbb{R}$ on an open set $W_0\subset\Bbb{R}^3$. The corresponding dynamical system on the state space $W=W_0\times\Bbb{R}^3 \subset \Bbb{R}^3\times\Bbb{R}^3$ for $(x,v)\in W_0\times\Bbb{R}^3$ is given by

\[\left\{\begin{aligned}\frac{dx}{dt} &= v\\ \frac{dv}{dt} &= -\mathrm{grad}\,\Phi(x) \end{aligned}\right.\]

Let $(\overline{x},\overline{v})\in W_0\times\Bbb{R}^3$ be an equilibrium point of the above dynamical system. Determine when the dynamical system is stable at $(\overline{x},\overline{v})$.

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[Chris L T521]

No one answered this week's question. You can find the solution below.

[sp]If $(\overline{x},\overline{v})\in W_0 \times \Bbb{R}^3$ is an equilibrium point, then $\overline{v}=0$ and $\mathrm{grad}\,\Phi(\overline{x})=0$. To investigate stability at $(\overline{x},0)$, let us use the total energy
\[E(x,v) = \frac{1}{2}mv^2+\Phi(x)\]
to construct a Lyapunov function. Since a Lyapunov function must vanish at $(\overline{x},0)$, we subtract from $E(x,v)$ the energy of the state, which is $\Phi(\overline{x})$, and define $V: W_0\times\Bbb{R}^3\rightarrow\Bbb{R}$ by
\[\begin{aligned}V(x,v) &= E(x,v) - E(\overline{x},0)\\ &= \frac{1}{2}mv^2+\Phi(x) - \Phi(\overline{x}).\end{aligned}\]
By conservation of energy, $\dot{V}=0$. Since $\frac{1}{2}mv^2\geq 0$, we assume $\Phi(x)>\Phi(\overline{x})$ for $x$ near $\overline{x}$, $x\neq \overline{x}$, in order to make $V$ a Lyapunov function.

With this, we proved a well-known theorem of Lagrange: an equilibrium $(\overline{x},0)$ of a conservative force field is stable if the potential energy has a local absolute minimum at $\overline{x}$.[/sp]