Stability

From Scholarpedia

Philip Holmes and Eric T. Shea-Brown (2006), Scholarpedia, 1(10):1838.

revision #60388 [link to/cite this article]

(Redirected from Unstable)

Curator: Dr. Philip Holmes, Princeton University, NJ, USA
Curator: Dr. Eric T. Shea-Brown, Courant Institute, New York University

The stability of an orbit of a dynamical system characterizes whether nearby (i.e., perturbed) orbits will remain in a neighborhood of that orbit or be repelled away from it. Asymptotic stability additionally characterizes attraction of nearby orbits to this orbit in the long-time limit. The distinct concept of structural stability is treated elsewhere, and concerns changes in the family of all solutions due to perturbations to the functions defining the dynamical system.

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Setup

We mainly consider autonomous ordinary differential equations (ODEs), written in vector notation as:

(1)

$\frac{d}{dt} \mathbf{x} {=} \dot{\mathbf{x}} = \mathbf{f}(\mathbf{x}) ; \ \mathbf{x} \in {\mathbb R}^n ,$

where

$\mathbf{f}(\mathbf{x}) = \begin{pmatrix} f_1(x_1,\ldots,x_n) \\ \vdots \\ f_n(x_1,\ldots,x_n) \end{pmatrix} ,~ \mathbf{x} = \begin{pmatrix} x_1 \\ \vdots \\ x_n \end{pmatrix} .$

We denote a solution to (1) by $\mathbf{x}(t)$ , with initial conditions $\mathbf{x}(0)$ . Equilibria $\mathbf{x}^e$ (sometimes called equilibrium points or fixed points), are special constant solutions: $\mathbf{x}(t) \equiv \mathbf{x}^e$ where $\dot{\mathbf{x^e}} = \mathbf{f}(\mathbf{x}^e) = \mathbf{0}$ , which is equivalent to requiring $f_j(x_1^e, \ldots, x_n^e) = 0$ for all $1 \leq j \leq n$ .

Below, we treat the stability of equilibria in detail, and then mention extensions to the stability of more general solutions $\mathbf{x}(t)$ . We also give some analogous results for maps.

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Definitions: Stability of an Equilibrium

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Lyapunov stability

Figure 1: Stability.

$\mathbf{x}^e$ is a stable equilibrium if for every neighborhood $U$ of $\mathbf{x}^e$ there is a neighborhood $V \subseteq U$ of $\mathbf{x}^e$ such that every solution $\mathbf{x}(t)$ starting in $V$ $(\mathbf{x}(0) \in V)$ remains in $U$ for all $t \geq 0$ . Notice that $\mathbf{x}(t)$ need not approach $\mathbf{x}^e$ .

If $\mathbf{x}^e$ is not stable, it is unstable.

Figure 2: Asymptotic stability: a sink (see below).

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Asymptotic stability

An equilibrium $\mathbf{x}^e$ is asymptotically stable if it is Lyapunov stable and additionally $V$ can be chosen so that $|\mathbf{x}(t) - \mathbf{x}^e | \to 0$ as $t \to \infty$ for all $\mathbf{x}(0) \in V$ .

An equilibrium that is Lyapunov stable but not asymptotically stable is sometimes called neutrally stable. See Figs. 1 and 2 for illustrations.

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Exponential stability

An equilibrium $\mathbf{x}^e$ is exponentially stable if there is a neighborhood $V$ of $\mathbf{x}^e$ and a constant $a>0$ such that $|\mathbf{x}(t) - \mathbf{x}^e | < e^{-a t}$ as $t \to \infty$ for all $\mathbf{x}(0) \in V$ . Exponentially stable equilibria are also asymptotically stable, and hence Lyapunov stable.

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Linearization and Stability of Equilibria

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Linearization

Suppose that $\mathbf{x} = \mathbf{x}^e$ is an equilibrium, so that if $\mathbf{x}(0) = \mathbf{x}^e$ , then $\mathbf{x}(t) \equiv \mathbf{x}^e$ . Let $\mathbf{x}(t) = \mathbf{x}^e + \boldsymbol{\xi}(t)$ , where $\mathbf{\xi}(t) = (\xi_1, \ldots, \xi_n)$ is a small perturbation: $|\boldsymbol{\xi}(t)| << 1$ . Substitute into (1) and expand $\mathbf{f}$ in a multivariable, vector-valued Taylor series to obtain:

(2)

$\dot{\mathbf{x}}^e + \dot{\boldsymbol{\xi}} = \mathbf{f}(\mathbf{x}^e + \boldsymbol{\xi}) = \mathbf{f}(\mathbf{x}^e) + D\mathbf{f}(\mathbf{x}^e)\boldsymbol{\xi} + {\mathcal{O}}(|\boldsymbol{\xi}|^2)$ .

(We assume that $\mathbf{f}$ is sufficiently differentiable so that Taylor's Theorem with remainder applies to each component.) Here, $D\mathbf{f}(\mathbf{x}^e)$ denotes the $n \times n$ Jacobian matrix of partial derivatives $[\partial f_i / \partial x_j]$ , evaluated at the equilibrium $\mathbf{x}^e$ , and ${\mathcal{O}} (|\boldsymbol{\xi}|^2)$ denotes terms of quadratic and higher order in the components $\xi_1, \ldots, \xi_n$ . Specifically, if $\mathbf{g(\boldsymbol{\xi})} = {\mathcal{O}}(|\boldsymbol{\xi}|^p)$ then

$\lim_{|\boldsymbol{\xi}| \to 0} \frac{g(\boldsymbol{\xi})}{|\boldsymbol{\xi}|^p} \leq k < \infty$ .

Thus, for small enough $|\boldsymbol{\xi}|$ , the first order term $D\mathbf{f}(\mathbf{x}^e) \boldsymbol{\xi}$ dominates. Taking into account that $\dot{\mathbf{x}}^e$ and $\mathbf{f}(\mathbf{x}^e)$ vanish and ignoring the small term $O(|\boldsymbol{\xi}|^2)$ , we get the linear system:

(3)

$\dot{\boldsymbol{\xi}} = D\mathbf{f}(\mathbf{x}^e)\boldsymbol{\xi}$ .

This is called the linearization of (1). It can be solved by standard methods (Boyce and DiPrima, 1997).

The general solution $\boldsymbol{\xi}(t)$ of Eqn. (3) is determined by the eigenvalues and eigenvectors of the Jacobian matrix $D\mathbf{f}(\mathbf{x}^e)$ . Here we are concerned with qualitative properties rather than complete solutions. In particular, in studying stability we want to know whether the size of solutions grows, stays constant, or shrinks as $t \to \infty$ . This can usually be answered just by examining the eigenvalues.

Recall that, if $\lambda$ is a real eigenvalue with eigenvector $\mathbf{v}$ , then there is a solution to the linearized equation of the form:

$\boldsymbol{\xi}(t) = c\mathbf{v} e^{\lambda t} .$

If $\lambda = \alpha \pm i\beta$ is a complex conjugate pair with eigenvectors $\mathbf{v} = \mathbf{u} \pm i\mathbf{w}$ (where $\mathbf{u}, \mathbf{w}$ are real) then

$\boldsymbol{\xi}_1(t) = e^{\alpha t}(\mathbf{u} \cos \beta t - \mathbf{w} \sin \beta t)$

and

$\boldsymbol{\xi}_2(t) = e^{\alpha t}(\mathbf{u} \sin \beta t + \mathbf{w} \cos \beta t)$

are two linearly-independent solutions. In both cases the real part of $\lambda$ (almost) determines stability. Since any solution of the linearized equation can be written as a linear superposition of terms of these forms (except in the case of multiple eigenvalues), we can deduce that

If all eigenvalues of $D\mathbf{f}(\mathbf{x}^e)$ have strictly negative real parts, then $|\boldsymbol{\xi}(t)| \to 0$ as $t \to \infty$ for all solutions.

If at least one eigenvalue of $D\mathbf{f}(\mathbf{x}^e)$ has a positive real part, then there is a solution $\boldsymbol{\xi}(t)$ with $|\boldsymbol{\xi}(t)| \to +\infty$ as $t \to \infty$ .

If some pairs of complex-conjugate eigenvalues have zero real parts with distinct imaginary parts, then the corresponding solutions for $|\boldsymbol{\xi}(t)| \to +\infty$ oscillate and neither decay nor grow as $t \to \infty$ .

Note: The eigenvalues of the linearization are preserved under (smooth) changes of coordinates (Arnold, 1973).

Note: When multiple eigenvalues exist and there are not enough linearly-independent eigenvectors to span $\mathbb{R}^n$ , solutions behave like $|\boldsymbol{\xi}(t)| \sim t^k e^{\lambda t}$ , so that they still decay for sufficiently long times if $\lambda < 0$ and grow if $\lambda > 0$ .

Note: The form $t^k e^{\lambda t}$ implies that transient growth occurs over initial times even if $\lambda < 0$ . This can also occur in the case of distinct eigenvalues. See (Trefethen and Embree, 2005) for more on this, but consider the example

$\begin{matrix} \dot{\xi_1} &=& - 2 \xi_1 + \alpha \xi_2 \\ \dot{\xi_2} &=& - \xi_2 \end{matrix}$

for large $|\alpha|$ . This system has eigenvalues $-1$ and $-2$ . However, taking $\xi_1(0)=0,~\xi_2(0)=1$ , the first coordinate $\xi_1(t)= \alpha (e^{-t}-e^{-2t})$ initially grows from zero to a maximum value of $\alpha/4$ . For sufficiently large $\alpha$ , the growth of $\xi_1$ will initially overwhelm the decay of $\xi_2=e^{-t}$ so that the trajectory transiently moves farther from the fixed point before approaching it as $t \rightarrow \infty$ . This also illustrates the need for the two neighborhoods $U$ and $V$ in the definitions of stability.

This motivates the concept of:

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Hyperbolic equilibria

Definition: $\mathbf{x}^e$ is a hyperbolic or non-degenerate equilibrium if all the eigenvalues of $D\mathbf{f}(\mathbf{x}^e)$ have non-zero real parts.

Equipped with the linear analysis sketched above, and recognizing that the remainder terms ignored in passing from Eqn. (2) to (3) can be made as small as we wish by selecting a sufficiently small neighborhood of $\mathbf{x}^e$ , we can determine the stability of hyperbolic equilibria from their linearization:

Proposition: If $\mathbf{x}^e$ is an equilibrium of $\dot{\mathbf{x}} = \mathbf{f}(\mathbf{x})$ and all the eigenvalues of the Jacobian matrix $D\mathbf{f}(\mathbf{x}^e)$ have strictly negative real parts, then $\mathbf{x}^e$ is exponentially (and hence asymptotically) stable. If at least one eigenvalue has strictly positive real part, then $\mathbf{x}^e$ is unstable.

Moreover, the Hartman-Grobman Theorem says that the full nonlinear system (1) is topologically equivalent to the linearized system (3) in a small neighborhood of a hyperbolic equilibrium.

Borrowing from fluid mechanics, we say that if all nearby solutions approach an equilibrium (e.g. all eigenvalues have negative real parts), it is a sink; if all nearby solutions recede from it, it is a source, and if some approach and some recede, it is a saddle point. When the equilibrium is surrounded by nested closed orbits, we call it a center.

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Degenerate Equilibria

One might hope to claim that Lyapunov stability (per the definition above) holds even if (some) eigenvalues have zero real part, but the following counter examples demonstrates that this is not the case:

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Example 1

Consider

(4)

$\dot{x} = \alpha x^3, \quad \alpha \neq 0$ .

Here $x = 0$ is the equilibrium and the linearization at 0 is

(5)

$\dot{\xi} = 3\alpha x^2 \big|_{x = 0}\, \xi = 0,$

with solution $\xi(t) = \xi(0) = \textrm{const.}$ , so certainly $x = 0$ is Lyapunov stable for Eqn. (5), but not asymptotically stable.

The exact solution of the nonlinear ODE (4) may be found by separating variables:

$\begin{matrix} \int_{x(0)}^{x(t)}\frac{dx}{x^3} = \int \alpha dt \quad & \Rightarrow \quad -\left( \frac{1}{2x(t)^2} - \frac{1}{2x(0)^2} \right) = \alpha t \\ & \Rightarrow \quad x(t) = \frac{x(0)}{\sqrt{1 - 2\alpha \, x(0)^2 t}} . \end{matrix}$

We therefore deduce that

$\begin{matrix} & |x(t)| \to \infty & \textrm{as} & t \to \frac{1}{2\alpha \, x(0)^2} & \textrm{if} \quad \alpha > 0 & \quad \textrm{(blow up!} \quad \textrm{instability)} \\ \textrm{but} & |x(t)| \to 0 &\textrm{as}& t \to \infty & \textrm{if} \quad \alpha < 0 & \quad \textrm{(asymptotic \; stability)} \end{matrix}$

The linearized system (5) is degenerate and the nonlinear "remainder terms", ignored in our linearized analysis, determine the outcome in this case. Here it is obvious, at least in retrospect, that ignoring these terms is perilous, since while they are indeed ${\mathcal{O}}(\xi^2)$ (in fact, ${\mathcal{O}}(\xi^3)$ ), the linear ${\mathcal{O}}(\xi)$ term is identically zero!

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Example 2

Consider the two-dimensional system

$\begin{matrix} \dot{x} &=& y + \alpha (x^2 + y^2) x , \\ \dot{y} &=& -x + \alpha (x^2 + y^2) y . \end{matrix}$

Note that the linearization is simply a harmonic oscillator with eigenvalues $\pm i$ . Is the equilibrium $(x,y) = (0,0)$ of this system stable or unstable? To answer this, it is convenient to transform to polar coordinates $x = r \cos \theta, y = r \sin \theta$ , which gives the uncoupled system:

$\dot{r} = \alpha r^3 , \ \dot{\theta} = -1 .$

The first equation is as in the example above, so we conclude: $\alpha > 0 \Rightarrow$ unstable; $\alpha = 0 \Rightarrow$ stable; $\alpha < 0 \Rightarrow$ asymptotically stable. The linearization gives no information if $\alpha = 0$ .

How can we prove stability in such degenerate cases, in which one or more eigenvalues has zero real part? One method requires construction of a function, often called a Lyapunov function, which remains constant, or decreases, along solutions. For mechanical systems the total (kinetic plus potential) energy is often a good candidate. This allows one to prove stability and even asymptotic stability in certain cases, via describe Lyapunov's second method or direct method:

Theorem: Suppose that $d\mathbf{x}/dt = \dot{\mathbf{x}} = \mathbf{f}(\mathbf{x})$ has an isolated equilibrium at $\mathbf{x} = \mathbf{0}$ (without loss of generality one can move an equilibrium $\mathbf{x}^e$ to $\mathbf{0}$ by letting $\mathbf{y} = \mathbf{x} - \mathbf{x}^e$ ). If there exists a differentiable function $V(\mathbf{x})$ , which is positive definite in a neighborhood of $0$ (in the sense that $V(0)=0$ and $V(\mathbf{x})>0$ for $\mathbf{x} \neq 0$ ) and for which $dV/dt = \mathbf{\nabla}V \cdot \mathbf{f}$ is negative definite on some domain $D \ni \mathbf{0}$ , then $\mathbf{0}$ is asymptotically stable. If $dV/dt$ is negative semidefinite (i.e., $dV/dt = 0$ is allowed), then $\mathbf{0}$ is Lyapunov-stable.

For a proof, see, e.g., (Hirsch, Smale, and Devaney (2004)).

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Linearized stability for maps

Analogous results exist for stability of maps of the form

(6)

$\mathbf{x} \mapsto \mathbf{f}(\mathbf{x}) \ \ \mbox{or} \ \ \mathbf{x}_{n+1} = \mathbf{f}(\mathbf{x}_n)$ .

but here the magnitude rather than the sign of the eigenvalues is important. A solution $\mathbf{x}^e$ is called a fixed point if $\mathbf{x}^e = \mathbf{f}(\mathbf{x}^e)$ .

Definition: $\mathbf{x}^e$ is a hyperbolic or non-degenerate fixed point of the map $\mathbf{f}(\mathbf{x})$ if no eigenvalue of $D\mathbf{f}(\mathbf{x}^e)$ has magnitude 1.

Proposition: If $\mathbf{x}^e$ is a fixed point of the map $\mathbf{f}(\mathbf{x})$ and all the eigenvalues of $D\mathbf{f}(\mathbf{x}^e)$ have magnitudes strictly strictly less than 1, then $\mathbf{x}^e$ is asymptotically stable. If at least one eigenvalue has magnitude greater than 1, then $\mathbf{x}^e$ is unstable.

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Example: Logistic map

The following one-dimensional example illustrates this. Consider the logistic map

$x_{n+1} = \mu x_n (1 - x_n)$

which has two fixed points: $x^e = 0$ and $x^e = 1-1/\mu$ . The linearization at $x^e = 0$ is

$\xi_{n+1} = \mu \xi_n$

which may be solved to give

$\xi_n = \mu^n \xi_0$ .

Clearly, $|\xi_n|$ grows without bound if $|\mu| > 1$ , and decays to 0 if $|\mu| < 1$ .

Similarly, we linearize at $x^e = 1-1/\mu$ and obtain

$\xi_{n+1} = (2-\mu) \xi_n$

so that this fixed point is asymptotically stable if $1<\mu<3$ .

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Stability of General Orbits

Figure 3: Orbital stability.

The notions of stability may be generalized to non-constant orbits (periodic, quasiperiodic or non-periodic) of ODEs.

First, we give some definitions and notation. Let $\gamma_t(\mathbf{x}) = \mathbf{x}(t)$ , given the initial value $\mathbf{x}(0)=\mathbf{x}.$ Then, the (forward) orbit is the set of all values that this trajectory obtains: $\gamma(\mathbf{x}) = \left\{ \gamma_t(\mathbf{x}) | t \ge 0 \right\}$ . Next, we have

Definition: Two orbits $\gamma(\textbf{x})$ and $\gamma(\hat{\mathbf{{x}}})$ are $\epsilon$ -close if there is a reparameterization of time (a smooth, monotonic function) $\hat{t}(t)$ such that

$| \gamma_t(\mathbf{x}) - \gamma_{\hat{t}(t)}(\hat{\mathbf{{x}}}) | < \epsilon \mbox{ for all } t \ge 0$ .

We say that an orbit is orbitally stable if all orbits with nearby initial points remain close in this sense:

Definition: An orbit $\gamma(\textbf{x})$ is orbitally stable if, for any $\epsilon>0$ , there is a neighborhood $V$ of $\mathbf{x}$ so that, for all $\hat{\mathbf{x}}$ in $V$ , $\gamma(\mathbf{x})$ and $\gamma(\hat{\mathbf{x}})$ are $\epsilon$ -close.

Definition: If additionally $V$ may be chosen so that, for all $\hat{\mathbf{x}} \in V$ , there exists a constant $\tau(\hat{\mathbf{x}})$ so that

$| \gamma_t(\mathbf{x}) - \gamma_{t-\tau(\hat{\mathbf{x}})}(\hat{\mathbf{x}}) | \rightarrow 0 \mbox{ as } t \rightarrow \infty .$

then $\gamma_t(\mathbf{x})$ is asymptotically stable.

See Figures 3-4, which show (a segment of) the orbit ${\gamma}(\mathbf{x})$ as well as a neighboring orbit ${\gamma}(\hat{\mathbf{x}})$ . The black lines indicate the boundary of an $\epsilon$ -neighborhood of ${\gamma}(\mathbf{x})$ .

Figure 4: Asymptotic orbital stability (an $\epsilon$ -neighborhood of ${\gamma}(\mathbf{x})$ is shown in yellow).

We note that the linearization techniques discussed above for equilibria and fixed points can be extended to apply to asymptotic stability of Periodic Orbits, as described in the corresponding article.

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Example: The nonlinear pendulum

Consider the pendulum equations

$\begin{matrix} \dot{x} &=& y \\ \dot{y} &=& -\sin(x) \end{matrix}$

Orbits lie on the energy level sets shown in Fig 5. Neighboring orbits have different periods. However, the two orbits animated in the figure are $\epsilon$ -close, as the corresponding trajectories remain close under a reparameterization of time (under which their periods would become equal). As this is true for all orbits in a neighborhood of either of the animated trajectories, they are both orbitally stable. In fact, all orbits are orbitally stable for this system, except for the saddle points and their connections.

Figure 5: Orbital stability for the nonlinear pendulum.

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Example: Linear flows on the torus

The flow on the two-torus

$\begin{matrix} \dot{\theta_1}&=&0 \\ \dot{\theta_2}&=&\sin(\theta_1) \end{matrix}$

is similar to the pendulum example above: here, all orbits are orbitally stable, as their neighbors are $\epsilon$ -close under reparameterization of time.

However, upon adding a third coordinate with constant velocity

$\begin{matrix} \dot{\theta_1}&=&0 \\ \dot{\theta_2}&=&\sin(\theta_1) \\ \dot{\theta_3}&=&1 \end{matrix}$

the situation changes dramatically. Consider two neighboring orbits with sightly different initial values of $\theta_1$ . These two orbits are linear flows on invariant two-tori with different, fixed values of $\theta_1$ . Generically, the two flows are irrational, so that each orbit is dense on its two-torus. Therefore, the two orbits are close as sets. However, time cannot be reparameterized so that the orbits will be $\epsilon-$ close under the definition above, because the flows have different slopes: see Fig. 6.

Figure 6: Lack of orbital stability for the three-torus example.

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Example: The two-body problem

Consider the following equations, written in polar coordinates for $R^2 \times R^2$ , which describe a limiting case of the gravitational dynamics of two bodies:

$\begin{matrix} \dot{r_1} &=& 0 \\ \dot{r_2}&=&0 \\ \dot{\theta_1}&=&r_1^{-3/2} \\ \dot{\theta_2}&=&r_2^{-3/2} \end{matrix}$

As for the last example above, orbits with different initial points are linear flows on invariant two-tori which generally have different frequency ratios (i.e., different slopes). Therefore, no orbits (except for the equilibrium at the origin) are orbitally stable.

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References

We thank both referees for their careful reading and suggestions, and one in in particular for her/his correction of our definition of orbital stability and for providing one of the examples above.

W.E. Boyce and R.C. DiPrima (1997). Elementary Differential Equations and Boundary Value Problems. Wiley, New York.

M.W. Hirsch, S. Smale, and R.L. Devaney (2004). Differential Equations, Dynamical Systems and an Introduction to Chaos. Academic Press/Elsevier, San Diego.

V.I. Arnold (1973). Ordinary Differential Equations. MIT Press, Cambridge, MA.

L.N. Trefethen and M. Embree (2005). Spectra and Pseudospectra: The Behavior of Nonnormal Matrices and Operators. Princeton Univ. Press, Princeton, NJ.

Internal references

John W. Milnor (2006) Attractor. Scholarpedia, 1(11):1815.
Edward Ott (2006) Basin of attraction. Scholarpedia, 1(8):1701.
Jeff Moehlis, Kresimir Josic, Eric T. Shea-Brown (2006) Periodic orbit. Scholarpedia, 1(7):1358.

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External Links

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Stability

From Scholarpedia

Contents

Setup

Definitions: Stability of an Equilibrium

Lyapunov stability

Asymptotic stability

Exponential stability

Linearization and Stability of Equilibria

Linearization

Hyperbolic equilibria

Degenerate Equilibria

Example 1

Example 2

Linearized stability for maps

Example: Logistic map

Stability of General Orbits

Example: The nonlinear pendulum

Example: Linear flows on the torus

Example: The two-body problem

References

External Links

See also

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