Lyapunov Functions, Energy, and Invariance, differential equations lesson

A marble rolling in a bowl with friction ends up at the bottom, and we know this without solving any equation of motion, because its energy can only go down. Can the same reasoning prove that an equilibrium of a nonlinear system is stable, even where linearization says nothing or the question is about distant starts? This lesson develops that idea into Lyapunov’s method, then shows how LaSalle’s invariance principle handles the common case in which the energy stops decreasing for an instant while the motion goes on.

Key terms

Positive definite function

Definition Throughout this lesson $t$ is the independent variable, primes denote derivatives with respect to $t$ , $X' = F(X)$ is an autonomous system with $F$ continuously differentiable on an open set $D \subseteq \mathbb{R}^n$ , and the equilibrium under study is the origin, so $F(0) = 0$ . Any other equilibrium $X_*$ can be moved there with the shift $X = X_* + h$ . A function $V$ defined on a neighborhood $U \subseteq D$ of $0$ is positive definite on $U$ if $V(0) = 0$ and $V(X) > 0$ for every $X \in U$ with $X \ne 0$ . It is positive semidefinite on $U$ if $V(0) = 0$ and only $V(X) \ge 0$ holds there, and it is negative definite or negative semidefinite if $-V$ is positive definite or positive semidefinite. A function $V$ on all of $\mathbb{R}^n$ is radially unbounded if $V(X) \to \infty$ as $\lVert X \rVert \to \infty$ , where $\lVert \cdot \rVert$ is the Euclidean length.

In plain words A positive definite function measures how far the state is from the equilibrium in some chosen way. It is zero at the equilibrium and positive at every other nearby state, so near the equilibrium its graph has its lowest point there, like the bottom of a bowl.

Example The function $x^2 + xy + y^2 = (x + \tfrac{1}{2}y)^2 + \tfrac{3}{4}y^2$ is positive definite on the plane, although it is not the squared distance. The function $x^2$ is only positive semidefinite on the plane, since it vanishes on the whole $y$ -axis. The pendulum energy $1 - \cos x + \tfrac{1}{2}y^2$ is positive definite on the strip $\lvert x \rvert < 2\pi$ but not on the plane, because it is also zero at $(2\pi, 0)$ . The energy

V = \tfrac{1}{2}y^2 + \tfrac{1}{2}x^2 + \tfrac{1}{4}x^4

is positive definite on all of $\mathbb{R}^2$ , since its terms are nonnegative and all of them vanish only at the origin. It is radially unbounded, because $V \ge \tfrac{1}{2}(x^2 + y^2)$ .

Orbital derivative

Definition For a continuously differentiable function $V$ on $U$ , the orbital derivative of $V$ along the system $X' = F(X)$ is the function

\begin{aligned} \dot V(X) &= \nabla V(X) \cdot F(X) \\[0.4em] &= \sum_{i=1}^{n} \frac{\partial V}{\partial x_i}\,F_i, \end{aligned}

where the dot marks a rate of change with respect to $t$ . By the chain rule, every solution $X(t)$ that stays in $U$ satisfies

\frac{d}{dt}V(X(t)) = \dot V(X(t)).

So $\dot V$ gives, at each state, the rate at which $V$ changes along the trajectory through that state, and it is computed from $F$ alone, without solving the system. It is a function of the state, not of time, and it is zero at every equilibrium.

In plain words Stand at a state, look at the arrow of the field there, and ask how fast $V$ changes when you move along that arrow. A negative answer means the solution through that state is moving downhill in $V$ at that moment.

Example For a damped nonlinear oscillator with displacement $x$ and velocity $y$ ,

\begin{gathered} x' = y, \\[0.4em] y' = -x - x^3 - y, \end{gathered}

and the energy $V$ above, the gradient is $\nabla V = (x + x^3, y)^{T}$ , so

\begin{aligned} \dot V &= (x + x^3)\,y \\ &\quad + y(-x - x^3 - y) \\[0.4em] &= -y^2. \end{aligned}

The spring terms cancel and only the damping term remains. At the state $(1, 1)$ , for instance, $\nabla V = (2, 1)^{T}$ and $F(1, 1) = (1, -3)^{T}$ , so $\dot V(1, 1) = 2 - 3 = -1$ , which agrees with $-y^2$ .

Lyapunov function and Lyapunov stability

Definition A Lyapunov function for the equilibrium $0$ is a continuously differentiable function $V$ that is positive definite on a neighborhood $U$ of $0$ and whose orbital derivative satisfies $\dot V(X) \le 0$ for every $X \in U$ . Lyapunov’s stability theorem says that if such a function exists, then $0$ is Lyapunov stable. That is, for every $\varepsilon > 0$ there is a $\delta > 0$ such that every solution with $\lVert X(0) \rVert < \delta$ exists for all $t \ge 0$ and satisfies $\lVert X(t) \rVert < \varepsilon$ for all $t \ge 0$ . If in addition $\dot V(X) < 0$ for every $X \in U$ with $X \ne 0$ , then $0$ is locally asymptotically stable, and $V$ is called a strict Lyapunov function. The proof of stability traps the solution inside a level set of $V$ . Shrink $\varepsilon$ so that the closed ball of radius $\varepsilon$ lies in $U$ , let $m > 0$ be the smallest value of $V$ on the sphere $\lVert X \rVert = \varepsilon$ , and choose $\delta \le \varepsilon$ with $V < m$ on the ball of radius $\delta$ . Since $V$ cannot increase along a solution, a solution that starts in the small ball never reaches the sphere, and staying inside a closed ball keeps it defined for all $t \ge 0$ .

In plain words If a quantity that measures distance from the equilibrium can never increase, a state that starts on a low enough contour is trapped inside it. The function does not have to be the physical energy or the squared distance. Any function with these two properties will do.

Example For the damped oscillator, $V = \tfrac{1}{2}y^2 + \tfrac{1}{2}x^2 + \tfrac{1}{4}x^4$ is the kinetic energy plus the potential energy of the spring, since $\tfrac{1}{2}x^2 + \tfrac{1}{4}x^4$ is the integral of the restoring force $x + x^3$ . It is positive definite with $\dot V = -y^2 \le 0$ , so it is a Lyapunov function and the origin is stable. The bound can be made explicit. Since $\lVert X \rVert^2 \le 2V(X)$ , and since $x^4 \le x^2$ when $\lVert X \rVert \le 1$ ,

\begin{aligned} \lVert X(t) \rVert^2 &\le 2V(X(0)) \\[0.4em] &\le \tfrac{3}{2}\lVert X(0) \rVert^2 \end{aligned}

for all $t \ge 0$ whenever $\lVert X(0) \rVert \le 1$ . So $\delta = \min(1, \varepsilon\sqrt{2/3})$ works. The derivative $-y^2$ is zero on the whole $x$ -axis, not only at the origin, so $V$ is not a strict Lyapunov function and the theorem by itself does not give asymptotic stability.

Negative semidefinite derivative

Definition The orbital derivative of $V$ is negative semidefinite on $U$ if $\dot V(X) \le 0$ for all $X \in U$ , and negative definite if in addition $\dot V(X) < 0$ for all $X \in U$ with $X \ne 0$ . A semidefinite derivative that also vanishes at states other than the origin still gives Lyapunov stability when $V$ is positive definite, but it does not by itself give asymptotic stability. A solution may pass through the zero set $\{X \in U : \dot V(X) = 0\}$ and keep losing $V$ afterward, or it may stay in that set forever with $V$ constant. Which of these happens is decided by the invariant sets inside the zero set.

In plain words A derivative that is only less than or equal to zero allows pauses in the decrease. A pause that lasts one instant is harmless. A pause that lasts forever can keep the state away from the equilibrium for good.

Example For the damped oscillator, $\dot V = -y^2$ vanishes on the line $y = 0$ . These states are the turning points, where the mass has zero velocity for an instant, and the energy loss stops only for that instant. For the undamped oscillator $x' = y$ , $y' = -x$ with $V = \tfrac{1}{2}(x^2 + y^2)$ , the derivative is $\dot V = xy + y(-x) = 0$ everywhere. The solution $x = \cos t$ , $y = -\sin t$ starts at $(1, 0)$ and keeps $V = \tfrac{1}{2}$ forever, so it never approaches the origin. There the origin is stable, with $\delta = \varepsilon$ , but not asymptotically stable. Both derivatives are negative semidefinite, and the conclusions differ.

Invariant set

Definition A set $M \subseteq D$ is invariant for $X' = F(X)$ if every solution that starts in $M$ stays in $M$ for all $t$ in its interval of existence, forward and backward in time. It is positively invariant if every solution that starts in $M$ stays in $M$ for all $t \ge 0$ in its interval of existence. Equilibria, whole trajectories and unions of trajectories are invariant. If $V$ is continuously differentiable on $D$ with $\dot V \le 0$ there, every sublevel set $\{X \in D : V(X) \le c\}$ is positively invariant, because $V$ cannot increase along a solution. The largest invariant subset of a set $E$ is the union of all invariant sets contained in $E$ . It consists of the states whose entire trajectory lies in $E$ .

In plain words An invariant set is made of whole trajectories, so a solution in it has always been in it and always will be. To find the largest invariant part of a set, ask which solutions can stay in the set forever, not which ones merely touch it.

Example For the damped oscillator the line $E = \{y = 0\}$ is not invariant. At $(1, 0)$ the field is $F(1, 0) = (0, -2)^{T}$ , which points off the line. A solution that stays in $E$ must have $y(t) = 0$ for all $t$ , hence $y'(t) = 0$ , and the second equation then forces $x + x^3 = x(1 + x^2) = 0$ . For real $x$ the factor $1 + x^2$ is positive, so $x = 0$ . The largest invariant subset of $E$ is therefore the origin alone. In the predator-prey model $x' = x(2 - y)$ , $y' = y(x - 1)$ , the open first quadrant and each positive half axis are invariant sets.

LaSalle’s invariance principle

Definition Let $\Omega \subseteq D$ be compact and positively invariant, and let $V$ be continuously differentiable on $D$ with $\dot V(X) \le 0$ for every $X \in \Omega$ . Let $E = \{X \in \Omega : \dot V(X) = 0\}$ , and let $M$ be the largest invariant subset of $E$ . Then every solution that starts in $\Omega$ exists for all $t \ge 0$ , and its distance to $M$ tends to $0$ as $t \to \infty$ . Here $V$ need not be positive definite. Two consequences are used most. If $V$ is a Lyapunov function on $U$ and no solution other than $X(t) = 0$ stays in $\{X \in U : \dot V(X) = 0\}$ for all $t \ge 0$ , then $0$ is locally asymptotically stable. If in addition $U = D = \mathbb{R}^n$ and $V$ is radially unbounded, then $0$ is globally asymptotically stable.

In plain words A trapped solution must end up near motions that keep $V$ constant forever. If the only such motion is resting at the equilibrium, every trapped solution goes there, even though $V$ may stop decreasing for an instant along the way.

Example For the damped oscillator take $\Omega = \{V \le c\}$ for any $c > 0$ . This set is closed, it is bounded because $V \ge \tfrac{1}{2}\lVert X \rVert^2$ , and it is positively invariant because $\dot V = -y^2 \le 0$ . Its zero set is $E = \Omega \cap \{y = 0\}$ , and as shown above the largest invariant subset of $E$ is $M = \{0\}$ . So every solution that starts in $\Omega$ tends to the origin. Every state lies in such a set for a large enough $c$ , and the origin is stable by the Lyapunov function, so the origin is globally asymptotically stable. Linearization gives only the local part of this result, through the eigenvalues $(-1 \pm i\sqrt{3})/2$ of the Jacobian at the origin. For the undamped oscillator the zero set is all of $\Omega$ , which is invariant, so the principle gives no information, in agreement with the circles traced there.

Check yourself

Decide whether each of the functions $2x^2 - 2xy + y^2$ , $(x + y)^2$ and $x^2 + y^2 - y^3$ is positive definite on some neighborhood of the origin in the plane, and whether it is positive definite on the whole plane.

Show the answer

The first function equals $x^2 + (x - y)^2$ , which is zero only when $x = 0$ and $x = y$ , so it is positive definite on the whole plane. The second is zero on the entire line $y = -x$ , so it is positive semidefinite but not positive definite on any neighborhood of the origin. The third can be written $x^2 + y^2(1 - y)$ . When $y < 1$ both terms are nonnegative, and they vanish together only at the origin, so it is positive definite on the half plane $y < 1$ and in particular on the open disk of radius $1$ . It is not positive definite on the whole plane, since its value at $(0, 2)$ is $4 - 8 = -4$ .

Use $V = x^2 + y^2$ to show that the origin of $x' = -y - x^3$ , $y' = x - y^3$ is asymptotically stable. Why can linearization not settle this?

Show the answer

The orbital derivative is

\begin{aligned} \dot V &= 2x(-y - x^3) \\ &\quad + 2y(x - y^3) \\[0.4em] &= -2x^4 - 2y^4, \end{aligned}

which is negative at every state other than the origin. So $V$ is a strict Lyapunov function, and the origin is locally asymptotically stable. Because $V$ is also radially unbounded and $\dot V < 0$ off the origin on the whole plane, every disk $\{V \le c\}$ is compact and positively invariant with zero set $\{0\}$ , and LaSalle’s principle shows that the origin is globally asymptotically stable. The Jacobian at the origin has rows $(0, -1)$ and $(1, 0)$ and eigenvalues $\pm i$ , so the equilibrium is nonhyperbolic and its linearization, a center, cannot decide stability.

For $x' = y$ , $y' = -x - y^3$ , the energy $V = \tfrac{1}{2}(x^2 + y^2)$ has $\dot V = -y^4$ . A classmate says that the origin is only stable, since $\dot V$ is zero on the whole $x$ -axis. Decide whether the origin is asymptotically stable.

Show the answer

The derivative is $\dot V = xy + y(-x - y^3) = -y^4 \le 0$ , so $V$ is a Lyapunov function and the origin is stable. That is where the classmate stopped. Now apply LaSalle’s principle on a disk $\Omega = \{V \le c\}$ , which is compact and positively invariant. Its zero set is $E = \Omega \cap \{y = 0\}$ . A solution that stays in $E$ has $y(t) = 0$ for all $t$ , so $y'(t) = 0$ , and the second equation gives $-x = 0$ . So the largest invariant subset of $E$ is $\{0\}$ , and every solution that starts in $\Omega$ tends to the origin. Since every state lies in such a disk, the origin is globally asymptotically stable. Solutions only cross the $x$ -axis, at turning points, so the zeros of $\dot V$ there do not stop the convergence. Linearization would not have helped either, because the Jacobian at the origin has eigenvalues $\pm i$ .

Lyapunov Functions, Energy, and Invariance

Key terms

Positive definite function

Orbital derivative

Lyapunov function and Lyapunov stability

Negative semidefinite derivative

Invariant set

LaSalle’s invariance principle

Check yourself

How this lesson connects

Builds on

Used later in