Equilibrium point

An equilibrium point, in mathematics, is a constant solution to a differential equation. Given a generic ODE $\dot{\mathbf{x}}(t)=f(\mathbf{x}(t))$, an equilibrium point is a solution $\mathbf{x}(t)=\mathbf{c}$. They are called equilibrium points because their "trajectories" in phase space are actually just a single point, unlike the usual Curve. Unless perturbed, the system never moves out of the equilibrium point.

Properties#

• Equilibrium points constitute the set of all zeros of the Vector field $f$: if $\mathbf{c}$ is an equilibrium point, then $f(\mathbf{c})=0$ for all $t$; the converse also applies.
• An equilibrium point $\mathbf{c}$ is stable if for every neighborhood $U$ of $\mathbf{c}$, there exists another neighborhood $V$ of $\mathbf{c}$ such that any motion $\mathbf{x}(t;\mathbf{x}_{0})$ that starts in $V$ ($\mathbf{x}_{0}\in V$) remains in $U$ for all $t$. An equilibrium point that doesn't meet this condition is unstable.
• A less binding property is being stable in the future and in the past. These have the same definition, but motion only remains in $U$ for all $t>0$ (future) or $t<0$ (past).

Search#

In practice, given an ODE, finding its equilibrium points is just a matter of finding the zeros of $f(\mathbf{x}(t))$. More complicated is determining if a given equilibrium point is stable or not. This can be done through Ljapunov's theorem.

Linearization near equilibrium points#

It is possible to make statements about the behavior of differential equations near equilibrium points. Consider a first order autonomous system $\dot{\mathbf{x}}=f(\mathbf{x})$, where $f:\mathbb{R}^{N}\to \mathbb{R}^{N}$, and an equilibrium point $\mathbf{c}$. As usual, $f(\mathbf{c})=0$. We want to analyze the behavior of the system near $\mathbf{c}$, that is, for $\mathbf{x}$ in a neighborhood $\lVert \mathbf{x}-\mathbf{c} \rVert\ll 1$. We assume it stays this way for the entirety of motion, so that $\lVert \mathbf{x}(t)-\mathbf{c} \rVert\ll1$ for all $t$ (if it doesn't, this still applies to the intervals of $t$ where it does).

To do so, we start with a Taylor series of each $i$ component of $f$ centered in $\mathbf{c}$, truncating to first order:

This assumes that the derivatives of $f$ in the equilibrium point $\mathbf{c}$ are nonzero. If they are, then this approximation would claim that $f_{i}(\mathbf{x})=0$ everywhere, which is manifestly false and in the entire approximation fails. In such a case, we wouldn't be able to stop at the first term and we'd require the quadratic terms too (the second derivatives) which, in other words, means that we'd be stuck with nonlinear analysis. When working with linearization, we need to assume (or ideally, prove) that the derivatives are nonzero in $\mathbf{c}$.

In order to approximate the equation to a practical state, we claim that the ODE can be locally¹ approximated as a linear ODE $\dot{\mathbf{x}}=\mathrm{A}\mathbf{x}$, where $\mathrm{A}$ is an $N\times N$ matrix. Specifically, our matrix is going to be made of all of the first derivatives that are left in the equation above and we will evaluate it in $\mathbf{x}-\mathbf{c}$:

But this matrix is precisely the Jacobian $\mathrm{J}$ of the function $f$ when evaluated in $\mathbf{c}$, and so $\mathrm{A}\equiv \mathrm{J}(\mathbf{c})$. If we set $\boldsymbol{\xi}(t)=\mathbf{x}(t)-\mathbf{c}$, its derivative is $\dot{\boldsymbol{\xi}}=\dot{\mathbf{x}}=f(\mathbf{x})\simeq \mathrm{J}(\mathbf{c})(\mathbf{x}-\mathbf{c})=\mathrm{J}(\mathbf{c})\boldsymbol{\xi}$. As such, within a small enough neighborhood of an equilibrium point, any mechanical system with nonzero derivatives in $\mathbf{c}$ can be linearized to the form

In the simplest scenario of a one-dimensional system $\dot{x}=f(x)$, with $\xi(t)=x(t)-c$ , the Jacobian simplifies down to the only derivative of $f$, $\frac{df}{dx}(x)\equiv f'(x)$. In such a case, then

This is a one-dimensional, linear ODE in $\xi$. It's solution is easy: it's an exponential. Moreover, $f'(c)$ provides useful information on the behavior around the point. If the sign is negative, and thus the slope of $f(c)$ is downwards, then the point is stable. Else, it isn't. Either way, the magnitude of $f'(c)$ tells us how stable the point is, and its reciprocal $1/\lvert f'(c) \rvert$ is the characteristic time scale of the system, which gives a ballpark number of how long the system takes to vary significantly in the neighborhood of $c$.

Now, it would be great if there were a universal solution to this approximation. That way, we'd have a good-enough universal solution to all mechanical systems near equilibrium. Turns out, there actually is one. To find it, we make the following ansatz:

Basically, we claim that $\boldsymbol{\xi}(t)$ is a constant vector $\mathbf{u}$ scaled by a Scalar field $\rho(t)$. Plugging this into the linearized equation, knowing that $\dot{\boldsymbol{\xi}}(t)=\dot{\rho}(t)\mathbf{u}$, we get

For this to hold, $\mathbf{u}$ must be an eigenvector of $\mathrm{A}$, of some eigenvalue $\alpha$, so that

We can now match the coefficients of $\mathbf{u}$:

Thus, the partial solution of $\boldsymbol{\xi}(t)$ must in general be

for some constants $C$ and an eigenvalue $\alpha$ satisfying $\mathrm{A}\mathbf{u}=\alpha \mathbf{u}$. There is one of these for each eigenvector $\mathbf{u}$. The general solution will then be the Linear combination of all of these:

This approach works as long as $\mathrm{A}\equiv \mathrm{J}(\mathbf{c})$ is diagonalizable or at the very least admits a Basis of eigenvectors $\mathbf{u}_{1},\ldots,\mathbf{u}_{N}$. The behavior of $\boldsymbol{\xi}(t)$ near $\mathbf{c}$ can then be discerned by the real parts of $\alpha_{k}$ (which may very well be complex):

• if all $\alpha_{k}$ have negative real parts, $\mathbf{c}$ is stable;
• if all $\alpha_{k}$ have positive real parts, $\mathbf{c}$ is unstable;
• if any $\alpha_{k}$ has zero real part but nonzero imaginary part, then that signals oscillatory behavior around the equilibrium.

The idea here is that the spectrum of the Jacobian of $f$ determines how the system behaves around equilibrium points.

Examples#

Thus, $\mathbf{c}=\begin{pmatrix}c \\ 0\end{pmatrix}$ with $c$ such that $f(c,0)=0$.