advanced31 min studyLesson 355 in Mathematics

Lyapunov equations and stability margins

Build an energy certificate for a Hurwitz matrix and quantify a sufficient perturbation bound in that chosen metric.

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What you’ll understand

  • Solve a continuous-time Lyapunov equation and verify positive definiteness.
  • Derive a state-norm decay bound from a quadratic energy.
  • Compute a sufficient norm perturbation margin without confusing it with a gain or phase margin.

Stable eigenvalues establish eventual decay for a finite-dimensional linear system, but they do not show how a particular Euclidean norm behaves at every instant. A nonnormal matrix may briefly amplify states before they decay. A quadratic Lyapunov function changes the measuring geometry so that a chosen energy decreases monotonically. Solving a linear matrix equation constructs that geometry, and the same energy can quantify tolerance to an additive matrix perturbation. The resulting margin is a sufficient certificate in specified coordinates, not an exact universal robustness radius.

Integrate future energy to construct the metric

Let A be a real square matrix and Q a symmetric positive definite matrix. If A is Hurwitz, define P as the integral of e^{Aᵀt}Qe^{At} over all positive time. Exponential decay makes the integral finite. Every nonzero initial vector produces positive integrated Q-energy, so P is positive definite. Differentiating the integrand and using its vanishing limit at infinity gives AᵀP+PA=−Q. The solution is unique: the homogeneous equation would make e^{Aᵀt}Pe^{At} constant, yet its limit is zero.

Conversely, if some P>0 and Q>0 satisfy this equation, V(x)=xᵀPx has derivative −xᵀQx along ẋ=Ax, proving exponential stability. Thus a Hurwitz matrix is characterized by positive-definite Lyapunov solutions for every positive-definite Q. A semidefinite Q alone is not enough without additional observability-type information, since the derivative can vanish on a nondecaying mode. In discrete time the equation is AᵀPA−P=−Q, with A Schur stable; using the continuous-time equation there would analyze the wrong dynamics.

A⊤P+PA=−Q,P=∫0∞eA⊤tQeAt dt,V˙=−x⊤QxA^\top P+PA=-Q,\qquad P=\int_0^\infty e^{A^\top t}Qe^{At}\,dt,\qquad \dot V=-x^\top Qx
For real Hurwitz A and Q>0, the integral produces the unique P>0. The equation and stability region are continuous-time statements.↔ Scroll wide equations horizontally.

Translate energy decay into a norm estimate

The smallest and largest eigenvalues of P bound V between two multiples of the squared Euclidean norm. Similarly, xᵀQx is at least λ_min(Q) times the squared norm. Hence V̇≤−λ_min(Q)V/λ_max(P). Integrating this scalar differential inequality gives an exponential energy bound and, after taking a square root, a state-norm bound. The prefactor involves the square root of P’s condition number. This prefactor accommodates possible transient growth in Euclidean norm even though V decreases at every instant.

Consider A with diagonal entries −1 and an upper-right coupling 4. Both eigenvalues are −1, but e^{At}=exp(−t)[[1,4t],[0,1]]. Starting from (0,1), the norm at t=1 is √17/e, greater than its initial value one. There is no contradiction with asymptotic stability or a Lyapunov certificate. Different norms and metrics answer different questions. Quoting only the largest real eigenvalue as a bound on every transient state norm would be false for this nonnormal example.

∥x(t)∥2≤λmax⁡(P)λmin⁡(P)exp⁡ ⁣(−λmin⁡(Q)2λmax⁡(P)t)∥x(0)∥2\|x(t)\|_2\le\sqrt{\frac{\lambda_{\max}(P)}{\lambda_{\min}(P)}}\exp\!\left(-\frac{\lambda_{\min}(Q)}{2\lambda_{\max}(P)}t\right)\|x(0)\|_2
This certified bound may be conservative. Its prefactor reflects the mismatch between Euclidean distance and the Lyapunov energy geometry.↔ Scroll wide equations horizontally.

Reuse the certificate under a perturbation

Replace A by A+E, where E is a real additive matrix perturbation and the induced Euclidean norm is used throughout. The derivative gains xᵀ(EᵀP+PE)x, whose magnitude is at most 2||P||₂||E||₂||x||₂². Therefore the same P proves stability whenever 2||P||₂||E||₂<λ_min(Q). This is a uniform sufficient bound for all perturbation directions within that norm ball. It can be much smaller than the actual distance to instability because it insists on preserving one chosen energy function.

Changing state units changes the matrix norm and P, so a numerical margin is meaningful only with its coordinate scaling stated. The bound is also different from classical gain margin or phase margin, which describe structured loop changes rather than an arbitrary additive E. Time-varying perturbations also give uniform exponential decay if there is one fixed positive epsilon with 2||P||₂||E(t)||₂≤λ_min(Q)−epsilon for all t; a merely pointwise strict inequality can lose its decay margin over time, but delays, actuator saturation and neglected dynamics do not automatically fit a static additive matrix model. A certificate’s strength comes from stating exactly the uncertainty class it controls.

A nonnormal energy metric

Solve AᵀP+PA=−I for A=[[−1,4],[0,−1]], then obtain a sufficient additive perturbation bound.

  1. Write P=[[p,q],[q,r]]. Matching entries gives −2p=−1, 4p−2q=0 and 8q−2r=−1, hence p=1/2,q=1,r=9/2.

  2. The trace is 5 and determinant 5/4, so the eigenvalues are (5±20)/2=5/2±5(5\pm\sqrt{20})/2=5/2\pm\sqrt5, both positive. Thus P is a valid energy metric.

  3. With Q=I, the sufficient condition is ∥E∥2<1/(2λmax⁡(P))=1/(5+25)\|E\|_2<1/(2\lambda_{\max}(P))=1/(5+2\sqrt5), approximately 0.1056.

The result

The nominal energy decreases at rate V̇=−||x||² even though the Euclidean norm can initially grow. The perturbation bound certifies a norm ball and is not claimed to be the exact stability radius.

Explore the model

Interactive lab

Interactive lab · oscillator

Dissipation removes oscillatory energy

Pause and Reset, then set damping ratio ζ=0 and natural frequency ω₀=1. Predict constant energy E=(velocity²+x²)/2=0.5. Step and evaluate it from position and velocity. For a fresh comparison Reset and set ζ=0.5,ω₀=1; predict nonincreasing energy with derivative −velocity². The instrument’s physical oscillator energy is a particular Lyapunov function, not the nonnormal matrix example or a general matrix-equation solver.

Dissipation removes oscillatory energyUnit 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 suitable energy can have a simple monotonic law even while individual state coordinates oscillate.

Check your understanding

What does 2||P||₂||E||₂<λ_min(Q) establish for the given Lyapunov solution?

Try it yourself

For the scalar system ẋ=−3x and Q=2, solve for P and use the norm bound to certify real scalar perturbations ẋ=(−3+e)x. Compare with the exact stability condition.

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

Give me a hint

The scalar Lyapunov equation is −6P=−2. Keep the absolute value in the sufficient perturbation condition.

Compare with the solution

P=1/3 and V̇=−2x². The same-certificate bound is 2(1/3)|e|<2, namely |e|<3. The exact real-scalar condition is −3+e<0, or e<3. Thus the symmetric uncertainty ball |e|<3 is sharp as a symmetric radius, but it excludes arbitrarily negative e even though such individual perturbations remain stable. A worst-direction norm guarantee and a one-sided parameter condition are different descriptions.

Take it with you

  • A Hurwitz matrix has a positive-definite quadratic energy certificate for every Q>0.
  • Monotone energy does not require monotone Euclidean norm.
  • A reused Lyapunov metric yields a sufficient perturbation margin with explicit norm and coordinates.

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