advanced34 min studyLesson 334 in Mathematics

Lyapunov functions and LaSalle invariance

Prove stability without explicit trajectories, use compact invariant sublevels, and identify the largest invariant subset of the zero-dissipation set.

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What you’ll understand

  • Compute the derivative of a candidate energy along trajectories.
  • Distinguish nonincreasing energy from strict asymptotic decay.
  • Apply LaSalle using a compact positively invariant set and its largest invariant zero-derivative subset.

An energy-like scalar can prove what thousands of trajectory plots cannot: every permitted nearby initial condition remains controlled. Lyapunov's method compares the value of that scalar along motion, without solving the equation. Sometimes its derivative vanishes at many non-equilibrium states, leaving strict decrease unavailable. LaSalle's principle then asks a sharper question: which of those zero-derivative states can support an entire invariant motion? Compactness supplies the long-time accumulation needed to turn that geometric question into convergence.

A scalar barrier can prove stability

Translate the equilibrium to zero and assume a locally Lipschitz autonomous vector field, with a continuously differentiable candidate V on a neighborhood. A positive-definite V is zero at the origin and positive at every other nearby point. Its derivative along motion is the gradient of V dotted with f. If that derivative is nonpositive in the neighborhood, sufficiently small sublevel sets trap solutions and prove Lyapunov stability: a small initial displacement stays small for all future time. Stability does not itself say the displacement vanishes.

If the derivative is strictly negative away from zero, the usual local Lyapunov theorem gives asymptotic stability. For a global conclusion, the hypotheses must hold globally and boundedness cannot be left implicit. A positive-definite, radially unbounded V has compact sublevel sets in finite dimensions; nonincrease then prevents escape, supplies forward existence for a smooth field, and supports global attraction under the appropriate strict or invariance argument. A candidate defined only inside one region cannot automatically prove a claim about every initial state.

V˙(x)=∇V(x)⋅f(x),V(x)>0 (x≠0),V˙(x)≤0\dot V(x)=\nabla V(x)\cdot f(x),\qquad V(x)>0\ (x\ne0),\qquad \dot V(x)\le0
Positive definiteness and nonpositive orbital derivative are local stability conditions. Asymptotic attraction requires further information.↔ Scroll wide equations horizontally.

Zero instantaneous dissipation is not invariant motion

Let K be a compact positively invariant subset of the domain, let V be continuously differentiable on a neighborhood of K, and suppose its orbital derivative is nonpositive on K. LaSalle's invariance principle states that every trajectory starting in K approaches the largest invariant subset M of the zero-derivative set E within K. Approaches means distance to M tends to zero; it does not generally promise convergence to a single point when M contains many states or recurrent motions.

The key step is invariance. A point can have zero derivative for an instant and immediately leave E, so it cannot belong to an entire invariant trajectory contained in E. For a damped oscillator, all states with zero velocity have zero instantaneous energy loss. Most have nonzero acceleration, so their motion immediately acquires velocity and dissipates. Keeping only the states that remain in the zero-loss set often reduces it to the equilibrium. Positive definiteness then supplies stability, and the reduced invariant set supplies attraction.

E={x∈K:V˙(x)=0},M=largest invariant subset of E,dist⁡(ϕt(x),M)→0E=\{x\in K:\dot V(x)=0\},\qquad M=\text{largest invariant subset of }E,\qquad \operatorname{dist}(\phi_t(x),M)\to0
Compact positive invariance, regularity, and nonincrease are essential assumptions. The conclusion concerns an invariant set, which need not be a single equilibrium.↔ Scroll wide equations horizontally.

Check every hypothesis in a nonlinear oscillator

For position x and velocity v, consider x dot = v and v dot = −x−x cubed−v. Choose V as half v squared plus half x squared plus one quarter x to the fourth. The chain rule cancels the conservative terms and leaves minus v squared. The function is positive definite and radially unbounded, so every closed sublevel is compact and positively invariant. This gives forward completeness and the compact setting LaSalle needs, rather than assuming a bounded phase plot proves bounded motion.

Here E is the line v=0 intersected with the chosen sublevel. A trajectory staying in that line must also have v dot = −x−x cubed equal to zero. The only real possibility is x=0, so the largest invariant subset is the origin. Every initial state belongs to some compact sublevel, yielding global asymptotic stability. The proof gives no uniform exponential rate by itself. Linearization may provide a local rate, while other systems with zero Jacobian can converge only algebraically despite possessing a strict Lyapunov function.

A turning point still loses energy later

For x dot = v, v dot = −x−x³−v, analyze the initial state (1,0) using V=v²/2+x²/2+x⁴/4. Explain why its zero initial energy derivative does not obstruct convergence.

  1. The initial value is V=0+1/2+1/4=3/4. The solution remains in this compact sublevel because direct differentiation gives V dot = −v².

  2. At (1,0), V dot is indeed zero, but v dot = −1−1=−2. The vector field immediately moves away from the zero-velocity line, so this point is not part of an invariant motion contained in E.

  3. Within E, remaining at zero velocity requires x+x³=0, whose only real root is zero. LaSalle therefore gives distance to the origin tending to zero. Positive-definite energy supplies stability as well as this attraction.

The result

The state converges to the origin despite zero instantaneous dissipation at its turning point. Energy need not decrease strictly at every non-equilibrium instant for LaSalle to apply.

Explore the model

Interactive lab

Interactive lab · oscillator

Distinguish a turning point from a conserved orbit

Pause and reset, then set both controls to damping ζ = 0.5 and frequency ω₀ = 1. The initial state is x = 1, v = 0; predict zero instantaneous loss but decreasing later values of E = (x²+v²)/2, and use Step to compute E from the readouts. For a fresh comparison, pause and reset, then set ζ = 0 and ω₀ = 1. Predict constant E = 1/2 and sustained oscillation. This instrument solves the linear damped oscillator, not the cubic oscillator above; it illustrates the same zero-velocity versus invariant-set distinction.

Distinguish a turning point from a conserved orbitUnit mass, x(0) = 1 and v(0) = 0. The analytic solution includes underdamped, critically damped and overdamped cases; time is measured in seconds. Current readings: Displacement x(t): 1; Velocity v(t): 0; Regime: underdamped. All parameters are adjustable with the labeled controls below.02.557.510−1.15−0.5700.571.15Time · sDisplacementunderdamped
Displacement x(t)Current time
Displacement x(t)
1
Velocity v(t)
0
Regime
underdamped
t = 0 s

Unit mass, x(0) = 1 and v(0) = 0. The analytic solution includes underdamped, critically damped and overdamped cases; time is measured in seconds. Parameter changes pause playback and recompute the model at the current elapsed time. Reset restores both parameters and returns time to zero. Playback starts paused. Adjust either parameter, then choose Play or Step.

Look forA zero derivative at one instant differs from an entire motion on which the derivative stays zero.

Check your understanding

What does LaSalle require you to identify inside the set where V dot equals zero?

Try it yourself

For x dot = −x³ and y dot = −2y³, use V=(x²+y²)/2 to prove global asymptotic stability. Starting from (1,0), compute the decay and decide whether it is exponential.

Take a moment to make a prediction or work it out on paper. Open a hint when you need one.

Give me a hint

Check radial unboundedness and V dot first. Separate the scalar x equation for the specified initial condition.

Compare with the solution

V is positive definite and radially unbounded, and V dot = −x⁴−2y⁴ is strictly negative away from zero. Its compact sublevels trap all forward solutions, yielding global asymptotic stability by Lyapunov's theorem or LaSalle with E equal to the origin. Starting from (1,0), y stays zero and x(t)=1/sqrt(1+2t). This tends to zero algebraically. It cannot satisfy an exponential upper bound Ce^(−at) for fixed positive a and finite C for all large t. Global asymptotic stability therefore does not imply a global exponential rate.

Take it with you

  • A positive-definite nonincreasing scalar proves stability, while attraction needs strict decay or an invariance argument.
  • LaSalle requires compact positive invariance and approaches the largest invariant zero-loss subset.
  • Radial unboundedness can supply global boundedness; convergence rates still need their own proof.

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